[
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{2}+\\frac{y^{2}}{3}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2/3 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 43]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{m^{2}}=1$ $(m>0)$ is $5 x-2 y=0$, then $m=$?",
    "fact_expressions": "G: Hyperbola;m: Number;m>0;Expression(G) = (x^2/4 - y^2/m^2 = 1);Expression(OneOf(Asymptote(G))) = (5*x - 2*y = 0)",
    "query_expressions": "m",
    "answer_expressions": "5",
    "fact_spans": "[[[2, 49]], [[71, 74]], [[5, 49]], [[2, 49]], [[2, 69]]]",
    "query_spans": "[[[71, 76]]]",
    "process": "The asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{m^{2}}=1$ $(m>0)$ are given by $y=\\pm\\frac{m}{2}x$. The equation of the line $5x-2y=0$ can be rewritten as $y=\\frac{5}{2}x$, so $m=5$."
  },
  {
    "text": "Given that the distance from the focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ to one of its asymptotes equals the length of the real axis, what is the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Distance(Focus(G), OneOf(Asymptote(G))) = Length(RealAxis(G))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 59], [80, 83]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[2, 59]], [[2, 76]]]",
    "query_spans": "[[[80, 90]]]",
    "process": ""
  },
  {
    "text": "Given that point $F$ is a focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, a circle centered at point $F$ is tangent to the asymptotes of $C$ and intersects $C$ at points $A$ and $B$. If $AF \\perp x$-axis, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;A: Point;B:Point;F: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);OneOf(Focus(C)) = F;Center(G)=F;IsTangent(Asymptote(C),G);Intersection(G,C)={A,B};IsPerpendicular(LineSegmentOf(A,F),xAxis)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[7, 68], [85, 88], [97, 100], [129, 132]], [[15, 68]], [[15, 68]], [[83, 84]], [[102, 105]], [[106, 109]], [[75, 79], [2, 6]], [[15, 68]], [[15, 68]], [[7, 68]], [[2, 73]], [[74, 84]], [[83, 94]], [[83, 111]], [[113, 127]]]",
    "query_spans": "[[[129, 138]]]",
    "process": "The distance from the focus $ F $ of the hyperbola to the asymptote is $ b $. Since $ AF \\perp x $-axis, we have $ \\frac{b^{2}}{a} = b' a = b $, so $ c = \\sqrt{a^{2} + b^{2}} = \\sqrt{2} a $, and $ e = \\frac{c}{a} = \\sqrt{2} $."
  },
  {
    "text": "The parabola $x^{2}=a y$ passes through the point $A(1, \\frac{1}{4})$, then the distance from point $A$ to the focus of this parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = a*y);a: Number;A: Point;Coordinate(A) = (1, 1/4);PointOnCurve(A, G)",
    "query_expressions": "Distance(A, Focus(G))",
    "answer_expressions": "5/4",
    "fact_spans": "[[[0, 14], [43, 46]], [[0, 14]], [[3, 14]], [[15, 35], [37, 41]], [[15, 35]], [[0, 35]]]",
    "query_spans": "[[[37, 54]]]",
    "process": "\\because the parabola x^{2}=ay passes through point A(1,\\frac{1}{4}), \\therefore 1^{2}=a\\times\\frac{1}{4}, solving gives a=4. Therefore, the equation of the parabola is x^{2}=4y, yielding its focus at F(0,1) and directrix equation y=-1. \\because the distance from any point on the parabola to the focus equals the distance from that point to the directrix of the parabola, \\therefore the distance from point A to the focus of this parabola is y_{4}-(-1)=\\frac{1}{4}+1=\\frac{5}{4}"
  },
  {
    "text": "What is the length of the imaginary axis of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1)",
    "query_expressions": "Length(ImageinaryAxis(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 34]]]",
    "process": "From the given condition, we have $a^{2}=3$, $b^{2}=1$, so the length of the imaginary axis is $2b=2$."
  },
  {
    "text": "If the foci of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{4}=1$ lie on the $x$-axis and the focal distance is $2$, then the value of the real number $m$ is?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (y^2/4 + x^2/m = 1);PointOnCurve(Focus(G), xAxis);FocalLength(G) = 2",
    "query_expressions": "m",
    "answer_expressions": "5",
    "fact_spans": "[[[1, 38]], [[56, 61]], [[1, 38]], [[1, 47]], [[1, 54]]]",
    "query_spans": "[[[56, 65]]]",
    "process": "Since the foci of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{4}=1$ lie on the $x$-axis and the focal distance is $2$, it follows that $m>4$ and $m=4+(\\frac{2}{2})^{2}=5$. Therefore, the value of the real number $m$ is $5$."
  },
  {
    "text": "Let a focus of the hyperbola $m{x}^{2}+n y^{2}=1$ coincide with the focus of the parabola $y=\\frac{1}{8} x^{2}$, and let the eccentricity be $2$. Then the asymptotes of this hyperbola have the equation?",
    "fact_expressions": "G: Hyperbola;m: Number;n:Number;H: Parabola;Expression(G) = (m*x^2+n*n^2=1);Expression(H) = (y = x^2/8);OneOf(Focus(G))=Focus(H);Eccentricity(G)=2",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(3)/3)*x",
    "fact_spans": "[[[1, 24], [70, 73]], [[4, 24]], [[4, 24]], [[30, 54]], [[1, 24]], [[30, 54]], [[1, 59]], [[1, 67]]]",
    "query_spans": "[[[70, 81]]]",
    "process": ""
  },
  {
    "text": "If the foci of the hyperbola $\\frac{x^{2}}{7}-\\frac{y^{2}}{t}=1$ are the same as the foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, then the real number $t$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/7 - y^2/t = 1);t: Real;H: Ellipse;Expression(H) = (x^2/25 + y^2/9 = 1);Focus(G) = Focus(H)",
    "query_expressions": "t",
    "answer_expressions": "9",
    "fact_spans": "[[[1, 39]], [[1, 39]], [[88, 93]], [[43, 81]], [[43, 81]], [[1, 86]]]",
    "query_spans": "[[[88, 95]]]",
    "process": "From the ellipse equation, the foci of the ellipse are: $(\\pm4,0)$, $\\therefore 7+t=16$, solving gives: $t=9$."
  },
  {
    "text": "The coordinates of the two foci of the ellipse $\\frac{y^{2}}{m+2}+\\frac{x^{2}}{m-2}=1$ are?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/(m - 2) + y^2/(m + 2) = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, pm*2)",
    "fact_spans": "[[[0, 41]], [[2, 41]], [[0, 41]]]",
    "query_spans": "[[[0, 52]]]",
    "process": ""
  },
  {
    "text": "Let $A$ and $B$ be points on the parabola $y^{2}=4x$ and the circle $C$: $(x-4)^{2}+y^{2}=1$, respectively. If there exists a real number $\\lambda$ such that $\\overrightarrow{AB}=\\lambda \\overrightarrow{BC}$, then the minimum value of $|\\lambda|$ is?",
    "fact_expressions": "G: Parabola;C: Circle;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Expression(C) = (y^2 + (x - 4)^2 = 1);PointOnCurve(A,G);PointOnCurve(B,C);lambda:Real;VectorOf(A,B)=lambda*VectorOf(B,C)",
    "query_expressions": "Min(Abs(lambda))",
    "answer_expressions": "2*sqrt(3)-1",
    "fact_spans": "[[[11, 25]], [[26, 50]], [[1, 4]], [[5, 8]], [[11, 25]], [[26, 50]], [[1, 53]], [[1, 53]], [[57, 68], [123, 134]], [[70, 121]]]",
    "query_spans": "[[[123, 140]]]",
    "process": "Let $ A\\left(\\frac{1}{4}m^{2}, m\\right) $, $ C: (x-4)^{2} + y^{2} = 1 $ with center coordinate $ C(4, 0) $, $ \\therefore |AC|^{2} = \\left(\\frac{1}{4}m^{2}-4\\right)^{2} + m^{2} = \\frac{1}{16}(m^{2}-8)^{2} + 12 \\geqslant 12 $, $ \\therefore |AC| \\geqslant 2\\sqrt{3} $. Since $ B $ is any point on the circle $ (x-4)^{2} + y^{2} = 1 $, $ \\therefore $ the minimum value of $ |AB| $ is $ 2\\sqrt{3} - 1 $. From $ \\overrightarrow{AB} = \\lambda \\overrightarrow{BC} $, we get $ |\\overrightarrow{AB}| = |\\lambda||\\overrightarrow{BC}| = |\\lambda| $, $ \\therefore $ the minimum value of $ |\\lambda| $ is the minimum value of $ |AB| $, which is $ 2\\sqrt{3} - 1 $."
  },
  {
    "text": "Given that $P$ is a moving point on the parabola $y^{2}=4x$ and $F$ is the focus of the parabola, then the equation of the locus of the midpoint of segment $PF$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G);F: Point;Focus(G) = F",
    "query_expressions": "LocusEquation(MidPoint(LineSegmentOf(P, F)))",
    "answer_expressions": "y^2=2*x-1",
    "fact_spans": "[[[6, 20], [29, 32]], [[6, 20]], [[2, 5]], [[2, 24]], [[25, 28]], [[25, 35]]]",
    "query_spans": "[[[37, 53]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ lies on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola respectively, and $|F_{1} F_{2}|=\\frac{b^{2}}{a}$. Let $I$ be the incenter of $\\Delta P F_{1} F_{2}$. If $S_{\\Delta I P F_{1}}=S_{\\Delta I P F_{2}}+\\lambda S_{\\Delta I F_{1} F_{2}}$ holds, then what is the value of $\\lambda$?",
    "fact_expressions": "P: Point;PointOnCurve(P, RightPart(G)) = True;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(F1, F2)) = b^2/a;I: Point;Incenter(TriangleOf(P,F1,F2)) = I;Area(TriangleOf(I,P,F1))=Area(TriangleOf(I,P,F2))+lambda*Area(TriangleOf(I,F1,F2));lambda: Number",
    "query_expressions": "lambda",
    "answer_expressions": "sqrt(2) - 1",
    "fact_spans": "[[[2, 6]], [[2, 68]], [[7, 63], [87, 90]], [[7, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[69, 76]], [[77, 84]], [[69, 96]], [[69, 96]], [[98, 129]], [[131, 134]], [[131, 160]], [[162, 238]], [[242, 251]]]",
    "query_spans": "[[[242, 255]]]",
    "process": "Let the inradius of triangle $ APF_{1}F_{2} $ be $ r $. By the definition of the hyperbola, $ |PF_{1}| - |PF_{2}| = 2a $, $ |F_{1}F_{2}| = 2c $. $ \\frac{S_{AHPF_{1}}}{2}|PF_{1}|\\cdot r = \\frac{1}{2}|PF_{2}|\\cdot r + \\lambda cr $, so $ \\lambda = \\frac{|PF_{1}|-|PF_{2}^{1}|}{2c} = \\frac{a}{c} $. Since $ |F_{1}F_{2}| = \\frac{b^{2}}{a} $, so $ \\frac{-a^{2}}{a} $, thus $ \\left(\\frac{a}{c}\\right)^{2} + \\frac{2a}{c} - 1 = 0 $, so $ \\frac{a}{c} = \\sqrt{2} - 1 $, i.e., $ \\lambda = \\sqrt{2} - 1 $. Application of simple geometric properties. [Method Insight] This problem mainly examines the definition, standard equation, and application of simple geometric properties of hyperbolas, as well as the calculation of triangle area and properties of incircles. The key to solving the problem is using the properties of the incircle of a triangle to express the areas of $ AIF_{1}F_{2} $, $ AIPF_{1} $, $ AIPF_{2} $, and then using the relation to find the expression for $ \\lambda $. It also emphasizes students' abilities in analyzing and solving problems, classified as a medium-difficulty question."
  },
  {
    "text": "The left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a, b>0)$ are $F_{1}$ and $F_{2}$, respectively. The chord $AB$ passes through the focus $F_{2}$ and is perpendicular to the $x$-axis. If $\\angle A F_{1} B=90^{\\circ}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;A: Point;B: Point;F1: Point;F2: Point;b>0;a>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;IsPerpendicular(LineSegmentOf(A,B),xAxis);PointOnCurve(F2,LineSegmentOf(A,B));AngleOf(A, F1, B) = ApplyUnit(90, degree);IsChordOf(LineSegmentOf(A,B),G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)+1",
    "fact_spans": "[[[0, 54], [139, 142]], [[3, 54]], [[3, 54]], [[102, 106]], [[102, 106]], [[63, 70]], [[84, 91], [72, 80]], [[3, 54]], [[3, 54]], [[0, 54]], [[0, 80]], [[0, 80]], [[92, 106]], [[81, 106]], [[108, 137]], [[0, 106]]]",
    "query_spans": "[[[139, 148]]]",
    "process": "From the given conditions, the latus rectum of the hyperbola is: \\frac{2b^{2}}{a}. Since the chord AB passing through the focus F_{2} and perpendicular to the x-axis satisfies \\angle AF_{1}B = 90^{\\circ}, it follows that 2c = \\frac{b^{2}}{a}. Therefore, 2ca = c^{2} - a^{2}. Since e = \\frac{c}{a}, we have e^{2} - 2e - 1 = 0. Solving gives e = 1 \\pm \\sqrt{2}. Because e > 1, we have e = 1 + \\sqrt{2}."
  },
  {
    "text": "If the hyperbola equation is $x^{2}-y^{2}=1$, then the coordinates of the foci of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(pm*sqrt(2),0)",
    "fact_spans": "[[[1, 4], [24, 27]], [[1, 22]]]",
    "query_spans": "[[[24, 34]]]",
    "process": ""
  },
  {
    "text": "Given that the focus of the parabola $y=\\frac{1}{4} x^{2}$ is $F$, the directrix is $l$, and if $l$ intersects the two asymptotes of the hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0, b>0)$ at points $A$ and $B$ respectively, and $|A B|=4|O F|$ ($O$ is the origin), then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;A: Point;B: Point;O: Origin;F: Point;l: Line;a>0;b>0;Expression(G) = (-x^2/b^2 + y^2/a^2 = 1);Expression(H) = (y = x^2/4);Focus(H) = F;Directrix(H) = l;L1:Line;L2:Line;Asymptote(G)={L1,L2};Intersection(l, L1) = A;Intersection(l,L2)=B;Abs(LineSegmentOf(A, B)) = 4*Abs(LineSegmentOf(O, F))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[46, 102], [147, 150]], [[49, 102]], [[49, 102]], [[2, 26]], [[112, 116]], [[117, 121]], [[138, 141]], [[30, 33]], [[37, 40], [42, 45]], [[49, 102]], [[49, 102]], [[46, 102]], [[2, 26]], [[2, 33]], [[2, 40]], [], [], [[46, 108]], [[42, 121]], [[42, 121]], [[123, 137]]]",
    "query_spans": "[[[147, 156]]]",
    "process": "The parabola $ y = \\frac{1}{4}x^{2} $, that is, $ x^{2} = 4y $, hence its directrix $ l $ has the equation $ y = -1 $, $ F(0,1) $. The asymptotes of the hyperbola are $ y = \\pm\\frac{a}{b}x $. Then we have $ A(-\\frac{b}{a},-1) $, $ B(\\frac{b}{a},-1) $, $ \\therefore |AB| = \\frac{2b}{a} = 4 $, $ \\frac{b}{a} = 2^{n} $, $ \\therefore e = \\frac{c}{a} = \\sqrt{1+(\\frac{b}{a})^{2}} = \\sqrt{5} $"
  },
  {
    "text": "Given that the eccentricity of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{4}=1$ is $\\frac{\\sqrt {2}}{2}$, then what is the length of the major axis of this ellipse?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (y^2/4 + x^2/m = 1);Eccentricity(G) = sqrt(2)/2",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "{4, 4*sqrt(2)}",
    "fact_spans": "[[[2, 39], [68, 70]], [[4, 39]], [[2, 39]], [[2, 65]]]",
    "query_spans": "[[[68, 76]]]",
    "process": ""
  },
  {
    "text": "The distance from a point $P(-1,4)$ on the parabola $y^{2}=a x  (a \\neq 0)$ to its focus $F$ is $|P F|=$?",
    "fact_expressions": "G: Parabola;a: Number;P: Point;F: Point;Expression(G) = (y^2 = a*x);Negation(a=0);Coordinate(P) = (-1, 4);PointOnCurve(P, G);Focus(G) = F;Distance(P, F) = Abs(LineSegmentOf(P, F))",
    "query_expressions": "Abs(LineSegmentOf(P,F))",
    "answer_expressions": "5",
    "fact_spans": "[[[0, 26], [40, 41]], [[3, 26]], [[30, 39]], [[43, 46]], [[0, 26]], [[3, 26]], [[30, 39]], [[0, 39]], [[40, 47]], [[30, 56]]]",
    "query_spans": "[[[49, 58]]]",
    "process": "P(-1,4) is a point on the parabola C: y^2 = ax, thus 4^{2} = -a, a = -16, the equation of the parabola is y^{2} = -16x, the focus is (-4,0), hence |PF| = \\sqrt{(-1+4)^{2}+4^{2}} = 5"
  },
  {
    "text": "A point $M$ on the parabola $y^{2}=4 x$ is at a distance of $2$ from the focus. What is the horizontal coordinate of point $M$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);M: Point;PointOnCurve(M, G);Distance(M, Focus(G)) = 2",
    "query_expressions": "XCoordinate(M)",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[17, 20], [32, 36]], [[0, 20]], [[0, 30]]]",
    "query_spans": "[[[32, 42]]]",
    "process": "Let the horizontal coordinate of point M be $ x $, then $ |MF| = x + \\frac{p}{2} = x + 1 = 2 $. Solving gives $ x = 1 $, so the horizontal coordinate of point M is $ 1 $."
  },
  {
    "text": "What is the standard equation of a parabola with focus $(3,0)$?",
    "fact_expressions": "G: Parabola;H: Point;Coordinate(H) = (3, 0);Focus(G) = H",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = 12*x",
    "fact_spans": "[[[11, 14]], [[3, 10]], [[3, 10]], [[0, 14]]]",
    "query_spans": "[[[11, 21]]]",
    "process": "Since the focus of the parabola is (3,0), we have \\frac{p}{2}=3, p=6, 2p=12, so the standard equation of the parabola with focus (3,0) is y^2=12x."
  },
  {
    "text": "The standard equation of the hyperbola that shares the same asymptotes as $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ and passes through the point $A(6,8 \\sqrt{2})$ is?",
    "fact_expressions": "G: Hyperbola;A: Point;C:Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);Coordinate(A) = (6, 8*sqrt(2));Asymptote(G)=Asymptote(C);PointOnCurve(A, C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/64 - y^2/36 = 1",
    "fact_spans": "[[[1, 40]], [[51, 69]], [[70, 73]], [[1, 40]], [[51, 69]], [[0, 73]], [[50, 73]]]",
    "query_spans": "[[[70, 80]]]",
    "process": ""
  },
  {
    "text": "Given that the center of the ellipse is at the origin, the foci are on the coordinate axes, the major axis has length $12$, and the eccentricity is $\\frac{1}{3}$, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;PointOnCurve(Focus(G),axis);Length(MajorAxis(G)) = 12;Eccentricity(G)=1/3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/36+y^2/32=1,x^2/32+y^2/36=1}",
    "fact_spans": "[[[2, 4], [48, 50]], [[8, 10]], [[2, 10]], [[2, 18]], [[2, 28]], [[2, 46]]]",
    "query_spans": "[[[48, 55]]]",
    "process": ""
  },
  {
    "text": "The point $P(8,1)$ bisects a chord of the hyperbola $x^{2}-4 y^{2}=4$. Then, the slope of the line containing this chord is?",
    "fact_expressions": "P: Point;Coordinate(P) = (8, 1);G: Hyperbola;Expression(G) = (x^2 - 4*y^2 = 4);H: LineSegment;IsChordOf(H, G);MidPoint(H) = P",
    "query_expressions": "Slope(OverlappingLine(H))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 9]], [[0, 9]], [[11, 31]], [[11, 31]], [], [[11, 35]], [[0, 35]]]",
    "query_spans": "[[[11, 49]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola equation $x^{2}- \\frac{y^{2}}{2}=1$, a line passing through the fixed point $P(2,1)$ intersects the hyperbola at two points $P_{1}$ and $P_{2}$, such that $P$ is the midpoint of $P_{1} P_{2}$. Then the equation of this line is?",
    "fact_expressions": "G: Hyperbola;H: Line;P1: Point;P2: Point;P: Point;Coordinate(P) = (2, 1);Expression(G) = (x^2 - y^2/2 = 1);PointOnCurve(P, H);Intersection(H, G) = {P1, P2};MidPoint(LineSegmentOf(P1, P2)) = P",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y = 4*x - 7",
    "fact_spans": "[[[2, 5], [50, 53]], [[47, 49], [97, 99]], [[54, 61]], [[62, 69]], [[38, 46], [74, 77]], [[38, 46]], [[2, 34]], [[35, 49]], [[47, 71]], [[74, 94]]]",
    "query_spans": "[[[97, 103]]]",
    "process": "Let $ P_{1}(x_{1},y_{1}) $, $ P_{2}(x_{2},y_{2}) $, we get $ 2x_{2}^{2-y_{2}} = \\frac{2}{2} = 2^{x} $. Subtracting the two equations and simplifying gives the slope of the line, thus obtaining the equation of the line. From the problem, $ 2x^{2} - y^{2} = 2 $. Let $ P_{1}(x_{1},y_{1}) $, $ P_{2}(x_{2},y_{2}) $, so \n$$\n\\begin{cases}\n2x_{1}^{2} - y_{1}^{2} = 2 \\\\\n2x_{2}^{2} - y_{2}^{2} = 2\n\\end{cases}\n$$\nSubtracting the two equations gives $ 2(x_{1}+x_{2})(x_{1}-x_{2}) - (y_{1}+y_{2})(y_{1}-y_{2}) = 0 $. From the problem, $ x_{1}+x_{2} = 4 $, $ y_{1}+y_{2} = 2 $, so $ 8(x_{1}-x_{2}) - 2(y_{1}-y_{2}) = 0 $. Since $ x_{1} \\neq x_{2} $, we have $ \\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = k = 4 $. Therefore, the equation of the line is $ y - 1 = 4(x - 2) $, that is, $ y = 4x - 7 $."
  },
  {
    "text": "Given that the vertex of the parabola $C$ is at the origin and its focus lies on the $x$-axis, the line $y = x$ intersects the parabola $C$ at points $A$ and $B$. If $P(2,\\ 2)$ is the midpoint of $AB$, then the equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;O: Origin;Vertex(C) = O;PointOnCurve(Focus(C), xAxis) = True;G: Line;Expression(G) = (y = x);Intersection(G, C) = {A, B};A: Point;B: Point;P: Point;Coordinate(P) = (2, 2);MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[2, 8], [34, 40], [74, 80]], [[14, 16]], [[2, 16]], [[2, 25]], [[26, 33]], [[26, 33]], [[26, 51]], [[42, 45]], [[46, 49]], [[53, 63]], [[53, 63]], [[53, 72]]]",
    "query_spans": "[[[74, 85]]]",
    "process": ""
  },
  {
    "text": "Let a point $P$ on the parabola $y^{2}=8x$ be at a distance of $4$ from the $y$-axis. Then, the distance from point $P$ to the focus of the parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);P: Point;PointOnCurve(P, G) = True;Distance(P, yAxis) = 4",
    "query_expressions": "Distance(P, Focus(G))",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 15], [41, 44]], [[1, 15]], [[18, 21], [35, 39]], [[1, 21]], [[18, 33]]]",
    "query_spans": "[[[35, 51]]]",
    "process": "The focus of the parabola $ y^{2}=8x $ is $ F(2,0) $, the equation of the directrix is $ x=-2 $. As shown in the figure, $ |PA|=4 $, $ |AB|=2 $, $ |F|=|PB|=|PA|+|AB|=6 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with right focus $F$, a line passing through the origin intersects the hyperbola $C$ at points $A$ and $B$. The segments $AF$ and $BF$ are connected. If $|AF|=6$, $|BF|=8$, $\\angle AFB=\\frac{\\pi}{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;A: Point;F: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;PointOnCurve(O, G);Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, F)) = 6;Abs(LineSegmentOf(B, F)) = 8;AngleOf(A, F, B) = pi/2;O: Origin",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "5",
    "fact_spans": "[[[2, 63], [79, 85], [168, 171]], [[10, 63]], [[10, 63]], [[76, 78]], [[88, 91]], [[68, 71]], [[92, 95]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 71]], [[72, 78]], [[76, 97]], [[114, 123]], [[125, 134]], [[136, 165]], [[73, 75]]]",
    "query_spans": "[[[168, 177]]]",
    "process": "In right triangle AABF, by the Pythagorean theorem we have |AB|^{2}=|AF|^{2}+|BF|^{2}=6^{2}+8^{2}=100, solving gives |AB|=10. Let the other focus of the hyperbola be F, connect BF and AF respectively. According to the symmetry of the hyperbola, quadrilateral AFBF is a rectangle. Using the properties of a rectangle, we obtain 2c=10, so c=5. By the definition of a hyperbola, 2a=|AF|-|AF|=8-6=2, solving gives a=1. Therefore, the eccentricity of the hyperbola is e=\\frac{c}{a}=5."
  },
  {
    "text": "Let $P$ be an arbitrary point on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, excluding the vertices, and let $F_{1}$, $F_{2}$ be the left and right foci respectively. The incircle of $\\Delta F_{1} PF_{2}$ intersects the real axis at point $M$. Then the value of $|F_{1} M| \\cdot| MF_{2} |$ is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F1: Point;P: Point;F2: Point;M: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);Negation(P=Vertex(G));LeftFocus(G) = F1;RightFocus(G) = F2;Intersection(InscribedCircle(TriangleOf(F1, P, F2)), RealAxis(G)) = M",
    "query_expressions": "Abs(LineSegmentOf(F1, M))*Abs(LineSegmentOf(M, F2))",
    "answer_expressions": "b^2",
    "fact_spans": "[[[5, 62]], [[8, 62]], [[8, 62]], [[73, 80]], [[1, 4]], [[81, 88]], [[125, 129]], [[8, 62]], [[8, 62]], [[5, 62]], [[1, 72]], [[1, 72]], [[5, 95]], [[5, 95]], [[5, 129]]]",
    "query_spans": "[[[131, 161]]]",
    "process": ""
  },
  {
    "text": "$P$ is a point on the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$, and $F_{1}$, $F_{2}$ are its two foci. If $\\angle F_{1} P F_{2}=30^{\\circ}$, then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/5 + y^2/4 = 1);Focus(G)={F1,F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(30, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "8-4*sqrt(3)",
    "fact_spans": "[[[4, 41], [61, 62]], [[45, 52]], [[0, 3]], [[53, 60]], [[4, 41]], [[45, 66]], [[0, 44]], [[68, 101]]]",
    "query_spans": "[[[103, 130]]]",
    "process": "a^{2}=5, b^{2}=4, c^{2}=1, let |PF_{1}|=m_{1}, |PF_{2}|=n, then by the definition of the ellipse we have m+n=2a=2\\sqrt{5}, so m^{2}+n^{2}=20-2mn. Since \\angle F_{1}PF_{2}=30^{\\circ}, by the law of cosines we get |F_{1}F_{2}|^{2}=m^{2}+n^{2}-2mn\\cos\\angle F_{1}PF_{2}, then mn=\\frac{16}{2+\\sqrt{3}}, so S_{\\Delta F_{1}PF_{2}}=\\frac{1}{2}mn\\sin 30^{\\circ}=\\frac{1}{2}\\times\\frac{16}{2+\\sqrt{3}}\\times\\frac{1}{2}=8-4\\sqrt{3}"
  },
  {
    "text": "What is the distance from the focus to the directrix of the parabola $y^{2}=4 x$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 25]]]",
    "process": "Focus F(1,0), directrix equation x=-1, ∴ the distance from the focus to the directrix is 2."
  },
  {
    "text": "The parabola $y^{2}=4 x$, the line $l$ passes through the focus $F$ of the parabola and intersects the parabola at points $A$ and $B$ ($A$ is in the first quadrant), and $\\overrightarrow{B A}=4 \\overrightarrow{B F}$. Then the area of triangle $A O B$ ($O$ is the origin) is?",
    "fact_expressions": "l: Line;G: Parabola;B: Point;A: Point;F: Point;O: Origin;Expression(G) = (y^2 = 4*x);Focus(G)=F;PointOnCurve(F,l);Intersection(l,G)={A,B};Quadrant(A)=1;VectorOf(B, A) = 4*VectorOf(B, F)",
    "query_expressions": "Area(TriangleOf(A,O,B))",
    "answer_expressions": "4*sqrt(3)/3",
    "fact_spans": "[[[16, 21]], [[0, 14], [34, 37], [24, 27]], [[43, 46]], [[39, 42], [50, 54]], [[30, 33]], [[122, 125]], [[0, 14]], [[24, 33]], [[16, 33]], [[16, 48]], [[50, 59]], [[63, 108]]]",
    "query_spans": "[[[111, 137]]]",
    "process": "The focus of the parabola $ y^{2} = 4x $ is $ (1,0) $. Let the line $ l $ be $ x = my + 1 $. Substituting into the parabola equation yields $ y^{2} - 4my - 4 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ y_{1} + y_{2} = 4m $, $ y_{1}y_{2} = -4 $. From $ \\overrightarrow{BA} = 4\\overrightarrow{BF} $, we get $ y_{1} = -3y_{2} $. By substitution, we obtain $ m^{2} = \\frac{1}{3} $. Also, the area of $ \\triangle AOB $ is $ S = \\frac{1}{2}|OF| \\cdot |y_{1} - y_{2}| = \\frac{1}{2}\\sqrt{(y_{1} + y_{2})^{2} - 4y_{1}y_{2}} = \\frac{1}{2}\\sqrt{16m^{2} + 16} = \\frac{4\\sqrt{3}}{3} $."
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$. A line $l$ passing through $F_{2}$ intersects the right branch of the hyperbola at points $M$, $N$. When $|F_{1} M|+|F_{1} N|$ attains its minimum value, what is the perimeter of $\\Delta F_{1} M N$?",
    "fact_expressions": "l: Line;G: Hyperbola;F1: Point;M: Point;N: Point;F2: Point;Expression(G) = (x^2/2 - y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, l);Intersection(l, RightPart(G)) = {M, N};WhenMin(Abs(LineSegmentOf(F1,M))+Abs(LineSegmentOf(F1,N)))",
    "query_expressions": "Perimeter(TriangleOf(F1,M,N))",
    "answer_expressions": "6*sqrt(2)",
    "fact_spans": "[[[60, 65]], [[16, 44], [66, 69]], [[0, 7]], [[74, 77]], [[78, 81]], [[8, 15], [52, 59]], [[16, 44]], [[0, 50]], [[0, 50]], [[51, 65]], [[60, 81]], [[83, 110]]]",
    "query_spans": "[[[111, 134]]]",
    "process": "From the given conditions, 2a=2\\sqrt{2}, |F_{1}M|=2a+|F_{2}M|, |F_{1}N|=2a+|F_{2}N| \\therefore |F_{1}M|+|F_{1}N|=4a+|F_{2}M|+|F_{2}N|=4\\sqrt{2}+|MN|. When |MN| is minimized, |F_{1}M|+|F_{1}N| reaches its minimum value. From the conditions, the latus rectum is the shortest, i.e., when line l \\bot x-axis, |MN| is minimized, at which point |MN|=\\frac{2b^{2}}{a}=\\frac{2}{\\sqrt{2}}=\\sqrt{2}, so the minimum value of |F_{M}|+|F_{N}| is 5\\sqrt{2}, and the perimeter of \\triangle F_{1}MN at this time is |F_{1}M|+|F_{N}|+|MN|=6\\sqrt{2}. The answer is: 6.5"
  },
  {
    "text": "The right focus of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ is $F$, and $O$ is the coordinate origin. A circle centered at $F$ with radius $FO$ intersects the two asymptotes of this hyperbola at points $A$ and $B$ (distinct from $O$). Then $|AB|=$?",
    "fact_expressions": "G: Hyperbola;H: Circle;F: Point;O: Origin;A: Point;B: Point;Expression(G) = (x^2 - y^2/3 = 1);RightFocus(G) = F;Center(H)=F;Radius(H)=LineSegmentOf(F,O);L1:Line;L2:Line;Asymptote(G)={L1,L2};Intersection(L1,H)=A;Intersection(L2,H)=B;Negation(A=O);Negation(B=O)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[0, 28], [67, 70]], [[64, 65]], [[48, 51], [33, 36]], [[37, 40], [92, 96]], [[80, 84]], [[85, 88]], [[0, 28]], [[0, 36]], [[47, 65]], [[55, 65]], [], [], [[67, 76]], [[64, 88]], [[64, 88]], [[80, 97]], [[80, 97]]]",
    "query_spans": "[[[99, 107]]]",
    "process": ""
  },
  {
    "text": "Given that the right focus of the hyperbola is $(5,0)$ and one asymptote has the equation $2x - y = 0$, then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;F: Point;Coordinate(F) = (5, 0);RightFocus(G) = F;Expression(OneOf(Asymptote(G))) = (2*x - y = 0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5 - y^2/20 = 1",
    "fact_spans": "[[[2, 5], [38, 41]], [[10, 17]], [[10, 17]], [[2, 17]], [[2, 35]]]",
    "query_spans": "[[[38, 48]]]",
    "process": ""
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ has two foci $F_{1}(-c, 0)$, $F_{2}(c, 0)$, and $M$ is a point on the ellipse satisfying $\\overrightarrow{F_{1} M} \\cdot \\overrightarrow{F_{2} M}=0$. Then the range of eccentricity $e$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;c: Number;F1: Point;F2: Point;Focus(G) = {F1, F2};Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);M: Point;PointOnCurve(M, G);DotProduct(VectorOf(F1, M), VectorOf(F2, M)) = 0;e: Number;Eccentricity(G) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "[\\sqrt{2}/2, 1)",
    "fact_spans": "[[[0, 52], [93, 95]], [[0, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[58, 72]], [[58, 72]], [[74, 87]], [[0, 87]], [[58, 72]], [[74, 87]], [[89, 92]], [[89, 98]], [[101, 160]], [[165, 168]], [[93, 168]]]",
    "query_spans": "[[[165, 175]]]",
    "process": "Let $ M(x,y) $, then $ \\overrightarrow{F_{1}M}=(x+c,y) $, $ \\overrightarrow{F_{2}M}=(x-c,y) $, from $ \\overrightarrow{F_{1}M}\\cdot\\overrightarrow{F_{2}M}=0 \\Rightarrow x^{2}+y^{2}=c^{2} \\Rightarrow y^{2}=c^{2}-x^{2} $, and since $ M $ is on the ellipse, $ \\therefore y^{2}=b^{2}-\\frac{b^{2}}{a^{2}}x^{2} $, $ \\therefore c^{2}-x^{2}=b^{2}-\\frac{b^{2}}{a^{2}}x^{2} \\Rightarrow x^{2}=a^{2}-\\frac{a^{2}b^{2}}{c^{2}} $. Also $ 0\\leqslant x^{2}\\leqslant a^{2} $, $ \\therefore 0<2-\\frac{1}{e^{2}}\\leqslant1 \\Rightarrow \\frac{\\sqrt{2}}{2}\\leqslant e\\leqslant1 $. Since $ 0<e<1 $, $ \\therefore \\frac{\\sqrt{2}}{2}\\leqslant e<1 $."
  },
  {
    "text": "There is a point $P$ on the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ such that the distance from $P$ to the left directrix is $\\frac{16}{5}$. Then, what is the distance from $P$ to the right focus?",
    "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2/9 - y^2/16 = 1);PointOnCurve(P, G);Distance(P, LeftDirectrix(G)) = 16/5",
    "query_expressions": "Distance(P, RightFocus(G))",
    "answer_expressions": "34/3",
    "fact_spans": "[[[0, 39]], [[43, 46], [71, 74]], [[0, 39]], [[0, 46]], [[0, 68]]]",
    "query_spans": "[[[0, 83]]]",
    "process": ""
  },
  {
    "text": "Let the line $x - 3y + m = 0$ ($m \\neq 0$) intersect the two asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ ($a > b > 0$) at points $A$ and $B$, respectively. If the point $P(m, 0)$ satisfies $|PA| = |PB|$, then what is the eccentricity of the hyperbola?",
    "fact_expressions": "H: Line;Expression(H) = (m + x - 3*y = 0);m: Number;Negation(m=0);G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;A: Point;B: Point;L1: Line;L2: Line;Asymptote(G) = {L1, L2};Intersection(H, L1) = A;Intersection(H, L2) = B;P: Point;Coordinate(P) = (m, 0);Abs(LineSegmentOf(P, A)) = Abs(LineSegmentOf(P, B))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[1, 24]], [[1, 24]], [[3, 24]], [[3, 24]], [[25, 80], [128, 131]], [[25, 80]], [[28, 80]], [[28, 80]], [[28, 80]], [[28, 80]], [[89, 93]], [[94, 97]], [], [], [[25, 85]], [[1, 97]], [[1, 97]], [[99, 109]], [[99, 109]], [[111, 124]]]",
    "query_spans": "[[[128, 137]]]",
    "process": "From the equation of the hyperbola, its asymptotes are $ y = \\frac{b}{a}x' $ and $ y = -\\frac{b}{a}x^{,} $. Intersecting each with $ x - 3y + m = 0 $, solving the systems gives $ A\\left(\\frac{-am}{a-3b}, \\frac{-bm}{a-3b}\\right) $, $ B\\left(\\frac{-am}{a+3b}, \\frac{bm}{a+3b}\\right) $. From $ |PA| = |PB| $, let $ Q $ be the midpoint of $ AB $, then $ Q\\left(\\frac{-am}{2} + \\frac{-am}{2}, \\frac{-bm}{2} + \\frac{bm}{a+3b}\\right) $. Since $ PQ $ is perpendicular to the given line, we have $ \\frac{\\frac{-bm}{a-3b} + \\frac{bm}{a+3b}}{\\frac{-3am}{a-3b} + \\frac{am}{a+3b}} = -3 $. Solving yields $ 2a^{2} = 8b^{2} = 8(c^{2} - a^{2}) $, i.e., $ \\frac{c^{2}}{a^{2}} = \\frac{5}{4} $, $ \\frac{c}{a} = \\frac{\\sqrt{5}}{2} $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. The line $l$ passing through $F_{2}$ with an inclination angle of $\\frac{\\pi}{4}$ intersects the ellipse $C$ at points $A$ and $B$. Then, the area of $\\Delta F_{1} A B$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;l: Line;PointOnCurve(F2, l);Inclination(l) = pi/4;A: Point;B: Point;Intersection(l, C) = {A, B}",
    "query_expressions": "Area(TriangleOf(F1, A, B))",
    "answer_expressions": "12*sqrt(2)/7",
    "fact_spans": "[[[2, 44], [104, 109]], [[2, 44]], [[53, 60]], [[61, 68], [70, 77]], [[2, 68]], [[2, 68]], [[98, 103]], [[69, 103]], [[78, 103]], [[110, 113]], [[114, 117]], [[98, 119]]]",
    "query_spans": "[[[121, 144]]]",
    "process": "First, find the equation of line $ l $, then solve it simultaneously with the ellipse equation, eliminate $ x $, and find $ |y_{1}-y_{2}| $. Using $ S_{\\triangle F_{1}AB} = \\frac{1}{2}|F_{1}F_{2}||y_{1}-y_{2}| $, the area of $ \\triangle F_{1}AB $ can be found. Solution: From the given conditions: line $ l: y = x - 1 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then:\n$$\n\\begin{cases}\ny = x - 1 \\\\\n3x^{2} + 4y^{2} = 12\n\\end{cases}\n$$\nEliminating $ x $ gives: $ 7y^{2} + 6y - 9 = 0 $,\n$$\n= \\frac{1}{2} \\times 2 \\times \\sqrt{(y_{1} + y_{2})^{2} - 4y_{1}y_{2}} = \\frac{1}{2} \\times 2 \\times \\frac{12\\sqrt{2}}{7} = \\frac{12\\sqrt{2}}{7}\n$$\nThus, the area of $ \\triangle F_{1}AB $ is"
  },
  {
    "text": "The sum of the distances from a moving point $P$ in the plane to fixed points $F_{1}$ and $F_{2}$ is equal to $|F_{1} F_{2}|$. What is the trajectory of point $P$?",
    "fact_expressions": "F1: Point;F2: Point;P: Point;Distance(P,F1) + Distance(P,F2) = Abs(LineSegmentOf(F1,F2))",
    "query_expressions": "Locus(P)",
    "answer_expressions": "LineSegmentOf(F1,F2)",
    "fact_spans": "[[[12, 19]], [[20, 27]], [[6, 9], [50, 54]], [[6, 48]]]",
    "query_spans": "[[[50, 59]]]",
    "process": "The sum of the distances from a moving point $P$ on the plane to fixed points $F_{1}$ and $F_{2}$ equals $|F_{1}F_{2}|$, which does not satisfy the definition of an ellipse; the trajectory is the line segment $F_{1}F_{2}$."
  },
  {
    "text": "If the vertex of a parabola is at the origin and its focus lies on the $y$-axis, and it passes through the point $(1,4)$, then what is the equation of the parabola?",
    "fact_expressions": "G: Parabola;O:Origin;H: Point;Coordinate(H) = (1, 4);Vertex(G)=O;PointOnCurve(Focus(G),yAxis);PointOnCurve(H,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2=(1/4)*y",
    "fact_spans": "[[[1, 4], [1, 4]], [[8, 12]], [[25, 33]], [[25, 33]], [[1, 12]], [[1, 21]], [[1, 33]]]",
    "query_spans": "[[[36, 44]]]",
    "process": "Let the equation of the parabola be $x^{2}=2py$ according to the given conditions. Since the parabola passes through the point $(1,4)$, we have $1^{2}=2\\times4\\times p$, so $p=\\frac{1}{8}$. Thus, the equation of the parabola is $x^{2}=\\frac{1}{4}y$."
  },
  {
    "text": "Given point $M(4 , 2)$, $F$ is the focus of the parabola $y^{2}=2 x$, and point $P$ moves on the parabola. When $|PM|+|PF|$ is minimized, what are the coordinates of point $P$?",
    "fact_expressions": "G: Parabola;M: Point;P: Point;F: Point;Expression(G) = (y^2 = 2*x);Coordinate(M) = (4, 2);Focus(G) = F;PointOnCurve(P, G);WhenMin(Abs(LineSegmentOf(P,M))+Abs(LineSegmentOf(P,F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2,2)",
    "fact_spans": "[[[19, 33], [42, 45]], [[2, 13]], [[37, 41], [65, 69]], [[15, 18]], [[19, 33]], [[2, 13]], [[15, 36]], [[37, 46]], [[49, 64]]]",
    "query_spans": "[[[65, 74]]]",
    "process": "As shown in the figure, let line $ l $ be the directrix of the parabola, with equation: $ x = -\\frac{1}{2} $. Draw $ PA \\perp l $ from point $ P $, with foot at point $ A $, then $ |PA| = |PF| $. Therefore, when points $ A $, $ P $, and $ M $ are collinear, $ |PM| + |PF| $ reaches the minimum value $ |AM| $, where $ |AM| = 4 - (-\\frac{1}{2}) = \\frac{9}{2} $. Substituting $ y = 2 $ into the parabola equation gives: $ 2^2 = 2x $, solving yields $ x = 2 $. Therefore, $ P(2, 2) $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left vertex is denoted as $A$. From point $A$, perpendiculars are drawn to the two asymptotes of the hyperbola, with feet of the perpendiculars being $M$ and $N$, respectively, and $|M N|=\\frac{4}{5}|O A|$ ($O$ is the origin). Then, the eccentricity of this hyperbola is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;A: Point;LeftVertex(C) = A;L1: Line;L2: Line;PointOnCurve(A, L1) = True;PointOnCurve(b, L2) = True;J1: Line;J2: Line;Asymptote(C) = {J1, J2};IsPerpendicular(L1, J1) = True;IsPerpendicular(L2, J2) = True;FootPoint(L1, J1) = M;FootPoint(L2, J2) = N;M: Point;N: Point;O: Origin;Abs(LineSegmentOf(M, N)) = (4/5)*Abs(LineSegmentOf(O, A))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 60], [74, 77], [138, 141]], [[2, 60]], [[10, 60]], [[10, 60]], [[10, 60]], [[10, 60]], [[65, 68], [70, 73]], [[2, 68]], [], [], [[69, 85]], [[69, 85]], [], [], [[74, 82]], [[69, 85]], [[69, 85]], [[69, 98]], [[69, 98]], [[91, 94]], [[95, 98]], [[126, 129]], [[100, 125]]]",
    "query_spans": "[[[138, 147]]]",
    "process": "By the given condition, A(-a,0), the hyperbola C: \\frac{x^2}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>b>0) has asymptotes given by: y=\\pm\\frac{b}{a}x. Without loss of generality, assume AM is perpendicular to the line y=\\frac{b}{a}x and AN is perpendicular to the line y=-\\frac{b}{a}x, then k_{AM}=-\\frac{a}{b}, k_{AN}=\\frac{a}{b}. Thus, the equation of line AM is: y=-\\frac{a}{b}(x+a); the equation of line AN is: y=\\frac{a}{b}(x+a). Solving \\begin{cases} y=-\\frac{a}{b}(x+a) \\\\ y=\\frac{b}{a}x \\end{cases} yields \\begin{cases} x=-\\frac{a^3}{c^{2}} \\\\ y=-\\frac{ba^{2}}{c^{2}} \\end{cases} (where c^{2}=a^{2}+b^{2}), so M(-\\frac{a^3}{c^{2}}, -\\frac{ba^2}{c^{2}}). Solving \\begin{cases} y=\\frac{a}{b}(x+a) \\\\ y=-\\frac{b}{a}x \\end{cases} yields \\begin{cases} x=-\\frac{a^3}{c^{2}} \\\\ y=\\frac{ba^{2}}{c^{2}} \\end{cases}, thus N(-\\frac{a^{3}}{c^{2}}, \\frac{ba^2}{c^{2}}). Therefore, |MN|=\\frac{2ba^{2}}{c^2}. Since |MN|=\\frac{4}{5}|OA|, we have \\frac{2ba^{2}}{c^{2}}=\\frac{4}{5}a, i.e., 2c^{2}=5ab, or 2a^{2}-5ab+2b^{2}=0. Solving gives a=2b or a=\\frac{b}{2} (which does not satisfy a>b). Hence, the eccentricity of this hyperbola is e=\\frac{c}{a}=\\sqrt{\\frac{a^{2}+b^{2}}{a^{2}}}=\\sqrt{\\frac{5b^{2}}{4b^{2}}}=\\frac{\\sqrt{5}}{2}."
  },
  {
    "text": "The length of the major axis of an ellipse is $2$ times the length of the minor axis, and one of its foci is $(\\sqrt{3}, 0)$. Then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;Length(MajorAxis(G)) =2*Length(MinorAxis(G));Coordinate(OneOf(Focus(G)))=(sqrt(3),0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4+y^2=1",
    "fact_spans": "[[[0, 2], [16, 17], [40, 42]], [[0, 15]], [16, 37]]",
    "query_spans": "[[[40, 49]]]",
    "process": "According to the problem, 2a=4b, c=\\sqrt{3}, and a^{2}=b^{2}+c^{2}, \\therefore a=2, b=1"
  },
  {
    "text": "If the line $y=x+m$ and the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ have two common points, then the range of real values for $m$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;m: Real;Expression(G) = (x^2/4 - y^2 = 1);Expression(H) = (y = m + x);NumIntersection(H, G)=2",
    "query_expressions": "Range(m)",
    "answer_expressions": "{(sqrt(3),+oo),(-oo,-sqrt(3))}",
    "fact_spans": "[[[11, 39]], [[1, 10]], [[47, 52]], [[11, 39]], [[1, 10]], [[1, 45]]]",
    "query_spans": "[[[47, 59]]]",
    "process": "Solving the system \\begin{cases}\\frac{x^{2}}{4}-y^{2}=1,\\\\y=x+m,\\end{cases} by eliminating $ y $, we obtain $ 3x^{2}+8mx+4m^{2}+4=0 $. According to the problem, we have $ \\Delta=(8m)^{2}-4\\times3\\times(4m^{2}+4)>0 $, that is, $ m^{2}>3 $. Solving gives $ m>\\sqrt{3} $ or $ m<-\\sqrt{3} $."
  },
  {
    "text": "The line $l$ passing through the focus of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. Tangents $l_{1}$ and $l_{2}$ to the parabola are drawn at points $A$ and $B$, respectively. Then, the horizontal coordinate of the intersection point of $l_{1}$ and $l_{2}$ is?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;l1:Line;l2:Line;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), l);Intersection(l, G) = {A, B};TangentOfPoint(A,G)=l1;TangentOfPoint(B,G)=l2",
    "query_expressions": "XCoordinate(Intersection(l1,l2))",
    "answer_expressions": "-1",
    "fact_spans": "[[[19, 24]], [[1, 15], [25, 28], [51, 54]], [[29, 32], [42, 45]], [[33, 36], [46, 50]], [[57, 64], [75, 82]], [[66, 73], [83, 90]], [[1, 15]], [[0, 24]], [[19, 38]], [[39, 73]], [[39, 73]]]",
    "query_spans": "[[[75, 99]]]",
    "process": ""
  },
  {
    "text": "Draw a line $l$ through the focus $F$ of the parabola $y^{2}=2 p x(p>0)$, intersecting the parabola at points $A$ and $B$. If $|F A|=3|F B|$, then the inclination angle of line $l$ is?",
    "fact_expressions": "l: Line;G: Parabola;p: Number;F: Point;A: Point;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(F, l);Intersection(l,G)={A,B};Abs(LineSegmentOf(F, A)) = 3*Abs(LineSegmentOf(F, B))",
    "query_expressions": "Inclination(l)",
    "answer_expressions": "{ApplyUnit(60,degree),ApplyUnit(120,degree)}",
    "fact_spans": "[[[29, 34], [67, 72]], [[1, 22], [36, 39]], [[4, 22]], [[25, 28]], [[40, 43]], [[44, 47]], [[4, 22]], [[1, 22]], [[1, 28]], [[0, 34]], [[29, 49]], [[51, 65]]]",
    "query_spans": "[[[67, 78]]]",
    "process": "As shown in the figure. For the parabola $ y^{2}=2px $ ($ p>0 $), the focus is $ F $, the directrix is $ x=-\\frac{p}{2} $. Draw perpendiculars from $ A $, $ B $ to the directrix, with feet at $ A $, $ B $ respectively. Line $ l $ intersects the directrix at $ C $. As shown in the figure, $ |AA|=|AF| $, $ |BB|=|BF| $, $ |FA|=3|FB| $, so $ |AM|=2|BF| $, $ |AB|=4|BF| $, thus $ \\angle ABM=30^{\\circ} $, i.e., the inclination angle of line $ l $ equals $ \\angle AFx=60^{\\circ} $. Similarly, when the inclination angle of line $ l $ is obtuse, it is $ 120^{\\circ} $."
  },
  {
    "text": "Given points $M(-2,0)$, $N(2,0)$, a moving point $P$ satisfies the condition $|P M|-|P N|=2 \\sqrt{2}$. What is the trajectory equation of the moving point $P$?",
    "fact_expressions": "M: Point;N: Point;P: Point;Coordinate(M) = (-2, 0);Coordinate(N) = (2, 0);Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)) = 2*sqrt(2)",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(x^2/2-y^2/2=1)&(x>0)",
    "fact_spans": "[[[2, 12]], [[13, 21]], [[24, 27]], [[2, 12]], [[13, 21]], [[31, 55]]]",
    "query_spans": "[[[58, 70]]]",
    "process": "Since |PM| - |PN| = 2\\sqrt{2} < |MN| = 4, by the definition of a hyperbola, the locus of point P is the right branch of a hyperbola with foci M and N. For this hyperbola, 2a = 2\\sqrt{2}, 2c = 4, \\therefore a = \\sqrt{2}, c = 2, \\therefore b^{2} = 2, and the equation of the trajectory is \\frac{x^{2}}{2} - \\frac{y^{2}}{2} = 1 (x > 0)."
  },
  {
    "text": "Given that the circle $x^{2}+y^{2}-6 x-7=0$ is tangent to the directrix of the parabola $y^{2}=2 p x$ ($p>0$), then the value of $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;H: Circle;p>0;Expression(G) = (y^2 = 2*p*x);Expression(H) = (-6*x + x^2 + y^2 - 7 = 0);IsTangent(H,Directrix(G))",
    "query_expressions": "p",
    "answer_expressions": "1/16",
    "fact_spans": "[[[25, 46]], [[53, 56]], [[2, 24]], [[28, 46]], [[25, 46]], [[2, 24]], [[2, 51]]]",
    "query_spans": "[[[53, 60]]]",
    "process": "The center of the circle $x^{2}+y^{2}-6x-7=0$ is $(3,0)$, and the radius $r=4$. The directrix of the parabola $y^{2}=2px$ ($p>0$) is $x=-\\frac{p}{2}$. According to the problem, $\\left|3-\\left(-\\frac{p}{2}\\right)\\right|=4$, therefore $p=2$ or $p=-14$ (discarded)."
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{45}+\\frac{y^{2}}{20}=1$ are $F_{1}$ and $F_{2}$ respectively. A line passing through the origin $O$ intersects the ellipse at points $A$ and $B$. If the area of $\\triangle ABF_{2}$ is $20$, then the equation of line $AB$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/45 + y^2/20 = 1);F1: Point;F2: Point;Focus(G) = {F1,F2};O: Origin;H: Line;PointOnCurve(O,H) = True;Intersection(H,G) = {A,B};A: Point;B: Point;Area(TriangleOf(A,B,F2)) = 20",
    "query_expressions": "Expression(LineOf(A,B))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 39], [71, 73]], [[0, 39]], [[45, 52]], [[53, 60]], [[0, 60]], [[62, 67]], [[68, 70]], [[61, 70]], [[68, 85]], [[76, 79]], [[80, 83]], [[87, 114]]]",
    "query_spans": "[[[116, 128]]]",
    "process": ""
  },
  {
    "text": "The equation of an ellipse that shares a common focus with the hyperbola $\\frac{x^{2}}{25}-\\frac{y^{2}}{24}=1$ and whose directrix is at a distance of $8$ from the center is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/25 - y^2/24 = 1);H: Ellipse;Focus(G) = Focus(H);Distance(Directrix(H),Center(H)) = 8",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2/56+y^2/7=1",
    "fact_spans": "[[[1, 41]], [[1, 41]], [[59, 61]], [[0, 61]], [[47, 61]]]",
    "query_spans": "[[[59, 65]]]",
    "process": ""
  },
  {
    "text": "Write the standard equation of an ellipse whose eccentricity is the reciprocal of the eccentricity of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$?",
    "fact_expressions": "C: Hyperbola;G: Ellipse;Expression(C) = (x^2 - y^2/3 = 1);InterReciprocal(Eccentricity(C),Eccentricity(G))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4+y^2/3=1",
    "fact_spans": "[[[8, 41]], [[50, 52]], [[8, 41]], [[4, 52]]]",
    "query_spans": "[[[50, 58]]]",
    "process": "The hyperbola $ C: x^{2} - \\frac{y^{2}}{3} = 1 $ has eccentricity $ e = \\frac{\\sqrt{1+3}}{1} = 2 $, then the eccentricity of the ellipse is $ \\frac{1}{2} $, so the standard equation of the ellipse can be $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, $P$ is a point on the directrix, and $P F_{1} \\perp P F_{2}$, $P F_{1} \\cdot P F_{2}=4 a b$. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, Directrix(G));IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));LineSegmentOf(P, F1)*LineSegmentOf(P, F2) = 4*(a*b)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 58], [149, 152]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[67, 74]], [[75, 82]], [[2, 82]], [[2, 82]], [[83, 86]], [[2, 92]], [[94, 117]], [[118, 147]]]",
    "query_spans": "[[[149, 158]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse, respectively, $B$ is the upper vertex of the ellipse, and $\\angle F_{1} B F_{2}=120^{\\circ}$, then the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;B: Point;UpperVertex(G) = B;AngleOf(F1, B, F2) = ApplyUnit(120, degree);e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[20, 22], [33, 35], [77, 79]], [[2, 9]], [[10, 17]], [[2, 28]], [[2, 28]], [[29, 32]], [[29, 39]], [[41, 75]], [[82, 85]], [[77, 85]]]",
    "query_spans": "[[[82, 87]]]",
    "process": "Since $\\angle F_{1}BF_{2}=120^{\\circ}$, it follows that $\\angle BF_{1}F_{2}=30^{\\circ}$, so $e=\\frac{c}{a}=\\cos30^{\\circ}=\\frac{\\sqrt{3}}{2}$"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $P$ is a point on the ellipse $C$, and $\\overrightarrow{P F_{1}} \\perp \\overrightarrow{P F_{2}}$. If the area of $\\Delta P F_{1} F_{2}$ is $4$, then $b=$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Focus(C) = {F1, F2};PointOnCurve(P, C);IsPerpendicular(VectorOf(P,F1),VectorOf(P,F2));Area(TriangleOf(P, F1, F2)) = 4",
    "query_expressions": "b",
    "answer_expressions": "2",
    "fact_spans": "[[[18, 75], [85, 90]], [[18, 75]], [[186, 189]], [[25, 75]], [[81, 84]], [[2, 9]], [[10, 17]], [[25, 75]], [[25, 75]], [[2, 80]], [[81, 94]], [[96, 153]], [[155, 184]]]",
    "query_spans": "[[[186, 191]]]",
    "process": "According to the area formula of the focal triangle in an ellipse, substitute the values to obtain: since $\\overrightarrow{PF}_{1}\\bot\\overrightarrow{PF}_{2}$, then $\\angle F_{1}PF_{2}=90^{\\circ}$. By the area formula of the focal triangle in an ellipse, we have $S=b^{2}\\tan\\frac{\\angle F_{1}PF_{2}}{2}$, that is, $4=b^{2}\\times\\tan45^{\\circ}=b^{2}$, solving gives $b=2$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left focus is $F_{1}$. A line passing through $F_{1}$ intersects the asymptotes of the hyperbola at points $A$ and $B$. The circle with diameter $AB$ passes through the origin. Then, the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;H: Line;A: Point;B: Point;F1: Point;O:Origin;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;PointOnCurve(F1, H);Intersection(H, Asymptote(C)) = {A, B};IsDiameter(LineSegmentOf(A,B),G);PointOnCurve(O,G)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 62], [88, 91], [125, 128]], [[9, 62]], [[9, 62]], [[117, 118]], [[85, 87]], [[97, 100]], [[101, 104]], [[67, 74], [77, 84]], [[119, 123]], [[9, 62]], [[9, 62]], [[2, 62]], [[2, 74]], [[76, 87]], [[85, 106]], [[107, 118]], [[117, 123]]]",
    "query_spans": "[[[125, 134]]]",
    "process": "According to the problem, the circle with AB as diameter passes through the origin, so $\\angle AOB = 90^{\\circ}$. By the symmetry of the asymptotes, the equations of the asymptotes are $y = \\pm x$, thus $\\frac{b^{2}}{a^{2}} = \\frac{c^{2}-a^{2}}{a^{2}} = e^{2}-1 = 1$. Hence $e = \\sqrt{2}$."
  },
  {
    "text": "Given that the line $AB$ passing through the point $M(1, 0)$ intersects the parabola $y^2 = 2x$ at points $A$ and $B$, $O$ is the origin, and the sum of the slopes of $OA$ and $OB$ is $1$, then the equation of the line $AB$ is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;O: Origin;M: Point;Expression(G) = (y^2 = 2*x);Coordinate(M) = (1, 0);PointOnCurve(M,LineOf(A,B));Intersection(LineOf(A,B), G) = {A, B};Slope(LineSegmentOf(O, A))+Slope(LineSegmentOf(O, B)) = 1",
    "query_expressions": "Expression(LineOf(A,B))",
    "answer_expressions": "2*x+y-2=0",
    "fact_spans": "[[[23, 37]], [[39, 42]], [[43, 46]], [[49, 52]], [[3, 14]], [[23, 37]], [[3, 14]], [[2, 22]], [[15, 48]], [[59, 81]]]",
    "query_spans": "[[[83, 94]]]",
    "process": "Let the equation of line AB be $ x = ty + 1 $. Substituting into $ y^2 = 2x $, we get $ y^2 - 2ty - 2 = 0 $. Let $ A(x_1, y_1) $, $ B(x_2, y_2) $, then $ y_1y_2 = -2 $, $ y_1 + y_2 = 2t $. Therefore, $ k_{OA} + k_{OB} = \\frac{y_1}{x_1} + \\frac{y_2}{x_2} = \\frac{2}{y_1} + \\frac{2}{y_2} = \\frac{2(y_1 + y_2)}{y_1y_2} = \\frac{4t}{-2} = -2t $. Hence, $ -2t = 1 $, solving gives $ t = -\\frac{1}{2} $. Therefore, the equation of line AB is $ x = -\\frac{1}{2}y + 1 $, or $ 2x + y - 2 = 0 $."
  },
  {
    "text": "It is known that the line $l$ passes through the point $(1,2)$ and is perpendicular to the $x$-axis. If the segment of $l$ intercepted by the parabola $y^{2}=4 a x$ has length $4$, then the coordinates of the focus of the parabola are?",
    "fact_expressions": "l: Line;G: Parabola;a: Number;H: Point;Expression(G) = (y^2 = 4*(a*x));Coordinate(H) = (1, 2);PointOnCurve(H,l);IsPerpendicular(l,xAxis);Length(InterceptChord(l,G))=4",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(1,0)",
    "fact_spans": "[[[2, 7], [26, 29]], [[30, 46], [57, 60]], [[33, 46]], [[8, 16]], [[30, 46]], [[8, 16]], [[2, 16]], [[2, 24]], [[26, 55]]]",
    "query_spans": "[[[57, 67]]]",
    "process": "In the equation of the parabola $ y^{2} = 4ax $, let $ x = 1 $, then $ y = \\pm 2\\sqrt{a} $, hence $ 4\\sqrt{a} = 4 $, so $ a = 1 $. Therefore, the equation of the parabola is $ y^{2} = 4x $, and its focus coordinates are $ (1, 0) $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and $A$, $B$ are the two intersection points of a line passing through $F_{1}$ with the ellipse, then what is the perimeter of $\\Delta A F_{1} F_{2}$? What is the perimeter of $\\triangle ABF_{2}$?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(G) = {F1, F2};PointOnCurve(F1,H);Intersection(H,G)={A,B}",
    "query_expressions": "Perimeter(TriangleOf(A,F1,F2));Perimeter(TriangleOf(A,B,F2))",
    "answer_expressions": "18\n20",
    "fact_spans": "[[[18, 56], [82, 84]], [[79, 81]], [[62, 65]], [[66, 69]], [[10, 17]], [[2, 9], [71, 78]], [[18, 56]], [[2, 61]], [[70, 81]], [[62, 89]]]",
    "query_spans": "[[[91, 119]], [[119, 142]]]",
    "process": ""
  },
  {
    "text": "Given that hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ and hyperbola $C_{2}$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{16}=1$ have the same asymptotes, and the right focus of $C_{1}$ is $F(\\sqrt{5}, 0)$, then $a=?$ $b=?$",
    "fact_expressions": "C1: Hyperbola;C2:Hyperbola;b: Number;a: Number;F: Point;Expression(C1) = (-y^2/b^2 + x^2/a^2 = 1);Expression(C2) = (-y^2/16 + x^2/4 = 1);Coordinate(F) = (sqrt(5), 0);Asymptote(C1)=Asymptote(C2);RightFocus(C1) = F",
    "query_expressions": "a;b",
    "answer_expressions": "1\n2",
    "fact_spans": "[[[2, 57], [115, 122]], [[58, 106]], [[151, 154]], [[146, 149]], [[127, 143]], [[2, 57]], [[58, 106]], [[127, 143]], [[2, 113]], [[115, 143]]]",
    "query_spans": "[[[146, 151]], [[151, 156]]]",
    "process": ""
  },
  {
    "text": "The parabola $y^{2}=2 p x(p>0)$ passes through the center of the circle $x^{2}+y^{2}-4 x+8 y+19=0$, then the equation of the directrix of this parabola is?",
    "fact_expressions": "G: Parabola;p: Number;H: Circle;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (8*y - 4*x + x^2 + y^2 + 19 = 0);PointOnCurve(Center(H), G)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x = -2",
    "fact_spans": "[[[0, 21], [55, 58]], [[3, 21]], [[22, 49]], [[3, 21]], [[0, 21]], [[22, 49]], [[0, 52]]]",
    "query_spans": "[[[55, 65]]]",
    "process": "The standard equation of the circle is (x-2)^{2}+(y+4)^{2}=1, and the center coordinates are (2,-4). Substituting the center coordinates into the parabola equation gives 2p\\times2=(-4)^{2}, solving for p yields p=4. Therefore, the directrix equation of this parabola is x=-2."
  },
  {
    "text": "If $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and a line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$, then the perimeter of $\\triangle A B F_{2}$ is?",
    "fact_expressions": "F1: Point;F2: Point;Focus(G) = {F1, F2};G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);H: Line;A: Point;B: Point;PointOnCurve(F1, H);Intersection(H, G) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "20",
    "fact_spans": "[[[1, 8], [62, 69]], [[9, 16]], [[1, 60]], [[17, 55], [73, 75]], [[17, 55]], [[70, 72]], [[77, 80]], [[83, 86]], [[61, 72]], [[70, 88]]]",
    "query_spans": "[[[89, 115]]]",
    "process": ""
  },
  {
    "text": "It is known that the foci of hyperbola $C$ lie on the $y$-axis, the length of the imaginary axis is $4$, and it has the same asymptotes as the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$. Then the equation of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Hyperbola;Expression(G) = (x^2/4-y^2/3 = 1);PointOnCurve(Focus(C), yAxis);Length(ImageinaryAxis(C)) = 4;Asymptote(C) = Asymptote(G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/3 - x^2/4 = 1",
    "fact_spans": "[[[2, 8], [74, 80]], [[28, 66]], [[28, 66]], [[2, 17]], [[2, 25]], [[2, 72]]]",
    "query_spans": "[[[74, 85]]]",
    "process": "According to the problem, let the desired hyperbola be $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$ and $\\lambda\\neq0$, that is, $\\frac{x^{2}}{4\\lambda}-\\frac{y^{2}}{3\\lambda}=1$. Since the foci lie on the $y$-axis and the length of the imaginary axis is 4, we have $\\begin{cases}\\lambda<0\\\\2\\sqrt{-42}=4\\end{cases}$, solving gives $\\lambda=-1$, thus the equation of hyperbola $C$ is $\\frac{y^{2}}{3}-\\frac{x^{2}}{4}=1$."
  },
  {
    "text": "Given that the focus of the parabola $C$ is $F(0,1)$, then the standard equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;F: Point;Coordinate(F) = (0, 1);Focus(C) = F",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2=4*y",
    "fact_spans": "[[[2, 8], [22, 28]], [[12, 20]], [[12, 20]], [[2, 20]]]",
    "query_spans": "[[[22, 35]]]",
    "process": "Analysis: Given that the focus of parabola C is F(0,1), we have \\frac{p}{2}=1, 2p=4, thus obtaining the standard equation of parabola C. Since the focus of the parabola lies on the positive y-axis, the standard equation is x^{2}=2py. Given the focus F(0,1), we get \\frac{p}{2}=1, 2p=4, \\therefore x^{2}=4y."
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{5}=1$ ($a$ is a constant and $a>\\sqrt{5}$) has its left focus at $F$. The line $x=m$ intersects the ellipse at points $A$, $B$. The maximum perimeter of $\\Delta F A B$ is $12$. Then, what is the eccentricity of this ellipse?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/5 + x^2/a^2 = 1);a: Number;a>sqrt(5);F: Point;LeftFocus(G) = F;H: Line;Expression(H) = (x = m);m: Number;A: Point;B: Point;Intersection(G,H) = {A,B}\t;Max(Perimeter(TriangleOf(F,A,B))) = 12",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2/3",
    "fact_spans": "[[[0, 41], [81, 83], [127, 129]], [[0, 41]], [[42, 45]], [[50, 62]], [[69, 72]], [[0, 72]], [[73, 80]], [[73, 80]], [[75, 80]], [[86, 90]], [[92, 95]], [[73, 95]], [[98, 124]]]",
    "query_spans": "[[[127, 135]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(4/3)*x",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 47]]]",
    "process": "Directly solve from the hyperbola equation [Detailed explanation] Since the hyperbola equation is \\frac{x^2}{9}-\\frac{y^{2}}{16}=1, the asymptotes of the hyperbola are given by \\frac{x^2}{9}-\\frac{y^{2}}{16}=0, that is, y=\\pm\\frac{4}{3}x."
  },
  {
    "text": "Given that points $A$ and $D$ are the left vertex and upper vertex of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively, and the left and right foci of the ellipse are $F_{1}$ and $F_{2}$, respectively. Point $P$ is a moving point on the segment $AD$. If the maximum value of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ is $2$ and the minimum value is $-\\frac{2}{3}$, then what is the standard equation of the ellipse $C$?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;D: Point;A: Point;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(C)=A;UpperVertex(C)=D;LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P,LineSegmentOf(A,D));Max(DotProduct(VectorOf(P, F1),VectorOf(P, F2)))=2;Min(DotProduct(VectorOf(P,F1),VectorOf(P,F2)))=-2/3",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4 + y^2/2 = 1",
    "fact_spans": "[[[13, 69], [79, 81], [216, 221]], [[19, 69]], [[19, 69]], [[7, 10]], [[2, 6]], [[106, 110]], [[89, 96]], [[97, 104]], [[19, 69]], [[19, 69]], [[13, 69]], [[2, 77]], [[2, 77]], [[79, 104]], [[79, 104]], [[106, 122]], [[126, 190]], [[126, 210]]]",
    "query_spans": "[[[216, 229]]]",
    "process": "As shown in the figure; $\\therefore$ the equation of line $AD$ is $\\frac{x}{-a}+\\frac{y}{b}=1$, $x\\in[-a,0]$; $\\therefore \\overrightarrow{PF_{1}}=(-c-x,-y)$, $\\overrightarrow{PF_{2}}=(c-x,-y)$, $\\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=x^{2}-c^{2}+y^{2}=x^{2}+y^{2}-c^{2}$; let $t=x^{2}+y^{2}$, then $\\sqrt{t}$ represents the distance from point $P$ to the origin $O$, $\\therefore$ when $P$ is at point $A$, $\\sqrt{t}$ is maximum, at this time $\\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=(-a)^{2}-0^{2}-c^{2}=b^{2}=2$; when $P$ is at the point where the distance from $O$ to line $AD$ is minimum, $\\sqrt{t}$ is minimum, $\\therefore \\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=\\frac{2a^{2}}{a^{2}+2}-(a^{2}-2)=-\\frac{2}{3}$, rearranging yields $3a^{4}-8a^{2}-16=0$, solving gives $a^{2}=4$, or $a^{2}=-\\frac{4}{3}$ (discarded). In conclusion, $a^{2}=4$, $b^{2}=2$, the equation of the ellipse is $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$."
  },
  {
    "text": "If the line $y=kx+1$ ($k>0$) intersects the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$ at exactly one point, then the value of $k$ is?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*x + 1);k: Number;k > 0;G: Hyperbola;Expression(G) = (x^2 - y^2/2 = 1);NumIntersection(H, G) = 1",
    "query_expressions": "k",
    "answer_expressions": "{sqrt(2),sqrt(3)}",
    "fact_spans": "[[[1, 17]], [[1, 17]], [[56, 59]], [[3, 17]], [[18, 46]], [[18, 46]], [[1, 54]]]",
    "query_spans": "[[[56, 63]]]",
    "process": "\\Square Solving the system of the linear equation and the hyperbola equation gives (2-k^{2})x^{2}-2kx-3=0. According to the problem, the equation has only one solution; discuss in two cases: 2-k^{2}=0 and 2-k^{2}\\neq0. Substituting the equation y=kx+1 into x^{2}-\\frac{y^{2}}{2}=1, we obtain (2-k^{2})x^{2}-2kx-3=0. Since the line y=kx+1 (k>0) intersects the hyperbola x^{2}-\\frac{y^{2}}{2}=1 at exactly one point, when k=-\\sqrt{2}, we get x=\\frac{3\\sqrt{2}}{4}, which holds; when k=\\sqrt{2}, we get x=-\\frac{3\\sqrt{2}}{1}, which holds; when k\\neq\\pm\\sqrt{2}, A=4k^{2}+12(2-k^{2})=0, solving gives k=\\pm\\sqrt{3}. Since k>0, we have k=\\sqrt{2} or k=\\sqrt{3}."
  },
  {
    "text": "The center of the hyperbola is at the origin, the eccentricity is $2$, and the coordinates of one focus are $(2, 0)$. Then the equations of the asymptotes of this hyperbola are?",
    "fact_expressions": "G: Hyperbola;O: Origin;Center(G) = O;Eccentricity(G) = 2;Coordinate(OneOf(Focus(G))) = (2, 0)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*sqrt(3)*x",
    "fact_spans": "[[[0, 3], [43, 46]], [[7, 11]], [[0, 11]], [[0, 21]], [[0, 40]]]",
    "query_spans": "[[[43, 54]]]",
    "process": ""
  },
  {
    "text": "The eccentricity $e \\in (1,2)$ of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{k}=1$, then the range of real number $k$ is?",
    "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (x^2/4 - y^2/k = 1);Eccentricity(G) =e;e:Number;In(e,(1,2))",
    "query_expressions": "Range(k)",
    "answer_expressions": "(0,12)",
    "fact_spans": "[[[0, 38]], [[56, 61]], [[0, 38]], [[0, 54]], [[42, 54]], [[42, 54]]]",
    "query_spans": "[[[56, 68]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. A line passing through point $F_{1}$ intersects ellipse $E$ at points $A$ and $B$, with $|A F_{1}|=3|F_{1} B|$, $|A B|=4$, and the perimeter of $\\Delta A B F_{2}$ being $16$. Then $|A F_{2}|=$?",
    "fact_expressions": "E: Ellipse;b: Number;a: Number;G: Line;A: Point;B: Point;F2: Point;F1: Point;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(E) = F1;RightFocus(E) = F2;PointOnCurve(F1, G);Intersection(G, E) = {A, B};Abs(LineSegmentOf(A, F1)) = 3*Abs(LineSegmentOf(F1, B));Abs(LineSegmentOf(A, B)) = 4;Perimeter(TriangleOf(A, B, F2)) = 16",
    "query_expressions": "Abs(LineSegmentOf(A, F2))",
    "answer_expressions": "5",
    "fact_spans": "[[[19, 76], [96, 101]], [[26, 76]], [[26, 76]], [[93, 95]], [[102, 105]], [[106, 109]], [[9, 16]], [[1, 8], [84, 92]], [[26, 76]], [[26, 76]], [[19, 76]], [[1, 82]], [[1, 82]], [[83, 95]], [[93, 111]], [[112, 134]], [[136, 145]], [[147, 173]]]",
    "query_spans": "[[[175, 188]]]",
    "process": "From |AF₁| = 3|F₁B|, |AB| = 4, we obtain |AF₁| = 3, |F₁B| = 1. Since the perimeter of △ABF₂ is 16, it follows that |AB| + |AF₂| + |BF₂| = 4a = 16, solving for a gives a = 4. Also, |AF₁| + |AF₂| = 2a = 8, so |AF₂| = 5."
  },
  {
    "text": "It is known that the focus of the parabola $y^{2}=8x$ coincides with the right focus of $\\frac{x^{2}}{a^{2}}+y^{2}=1$ $(a>0)$, then $a$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);C:Curve;Expression(C)=(x^2/a^2+y^2=1);a>0;a:Number;Focus(G)=RightFocus(C)",
    "query_expressions": "a",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[20, 54]], [[20, 54]], [[20, 54]], [[62, 65]], [[2, 60]]]",
    "query_spans": "[[[62, 67]]]",
    "process": "From the given conditions, the focus of the parabola has coordinates (2,0). It is known from the conditions that $\\frac{x^{2}}{a^{2}}+y^{2}=1$ represents an ellipse with foci on the x-axis. For the ellipse: $b^{2}=1$, $c^{2}=a^{2}-1=4$, so $a^{2}=5$. Since $a>0$, it follows that $a=\\sqrt{5}$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$, a line $l$ is drawn through the focus $F$, intersecting the parabola $C$ at points $P$ and $Q$. Then the range of $|PQ|$ is?",
    "fact_expressions": "l: Line;C: Parabola;P: Point;Q: Point;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {P, Q}",
    "query_expressions": "Range(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "[4, +oo)",
    "fact_spans": "[[[29, 34]], [[2, 21], [35, 41]], [[43, 46]], [[47, 50]], [[2, 21]], [[25, 28]], [[2, 28]], [[22, 34]], [[29, 52]]]",
    "query_spans": "[[[54, 68]]]",
    "process": ""
  },
  {
    "text": "Given an ellipse and a hyperbola with the same foci $F_{1}$, $F_{2}$ intersecting at point $P$, $|F_{1} F_{2}|=2|P O|$, the eccentricities of the ellipse and hyperbola are $e_{1}$, $e_{2}$ respectively, then $\\frac{1}{e_{1}^{2}}+\\frac{1}{e_{2}^{2}}$ = (point $O$ is the coordinate origin) ?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;F1: Point;F2: Point;P: Point;O: Origin;Focus(G) = Focus(H);Focus(G) = {F1, F2};Focus(H) = {F1, F2};Intersection(H, G) = P;Abs(LineSegmentOf(F1, F2)) = 2*Abs(LineSegmentOf(P, O));e1: Number;e2: Number;Eccentricity(H) = e1;Eccentricity(G) = e2",
    "query_expressions": "1/e1^2 + 1/e2^2",
    "answer_expressions": "2",
    "fact_spans": "[[[26, 29], [63, 66]], [[23, 25], [60, 62]], [[7, 14]], [[15, 22]], [[31, 35]], [[137, 141]], [[2, 29]], [[2, 29]], [[2, 29]], [[23, 36]], [[37, 59]], [[73, 80]], [[83, 90]], [[60, 90]], [[60, 90]]]",
    "query_spans": "[[[93, 149]]]",
    "process": "Let the semi-major axis of the ellipse be $ a_{1} $, the semi-transverse axis of the hyperbola be $ a_{2} $, their semi-focal distance be $ c $, and let $ |PF_{1}| = m $, $ |PF_{2}| = n $, with $ m > n $. According to the definitions of the ellipse and hyperbola, we obtain $ m + n = 2a_{1} $, $ m - n = 2a_{2} $, solving gives $ m = a_{1} + a_{2} $, $ n = a_{1} - a_{2} $. $ \\because |F_{1}F_{2}| = 2|PO| $, $ \\therefore PF_{1} \\perp PF_{2} $, by the Pythagorean theorem we get $ |PF_{1}|^{2} + |PF_{2}|^{2} = |F_{1}F_{2}|^{2} $, $ \\therefore (a_{1} + a_{2})^{2} + (a_{1} - a_{2})^{2} = (2c)^{2} $, simplifying yields $ a_{1}^{2} + a_{2}^{2} = 2c^{2} $, $ \\frac{\\sqrt{2}}{2} \\frac{1}{e_{2}^{2}} = 2 $"
  },
  {
    "text": "Given the hyperbola $C$ with equation $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1$ $(a>0)$, a line $l$ passing through the origin $O$ intersects the hyperbola $C$ at points $A$ and $B$, point $F$ is the left focus of the hyperbola $C$, and $A F \\perp B F$. Then the area of $\\triangle A B F$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;A: Point;F: Point;B: Point;O:Origin;Expression(C) = (-y^2/9 + x^2/a^2 = 1);PointOnCurve(O, l);Intersection(l, C) = {A, B};LeftFocus(C) = F;IsPerpendicular(LineSegmentOf(A, F), LineSegmentOf(B, F));a:Number;a>0",
    "query_expressions": "Area(TriangleOf(A, B, F))",
    "answer_expressions": "9",
    "fact_spans": "[[[64, 69]], [[2, 8], [70, 76], [94, 100]], [[79, 82]], [[89, 93]], [[83, 86]], [[58, 63]], [[2, 56]], [[57, 69]], [[64, 88]], [[89, 104]], [[106, 121]], [[2, 56]], [[2, 56]]]",
    "query_spans": "[[[123, 145]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the right directrix intersects the $x$-axis at point $A$, and point $B$ has coordinates $(0, a)$. If a point $M$ on the ellipse satisfies $\\overrightarrow {A B}=3 \\overrightarrow {A M}$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;A: Point;B: Point;M: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(B) = (0, a);Intersection(RightDirectrix(C), xAxis) = A;VectorOf(A,B)=3*VectorOf(A,M);PointOnCurve(M, C)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[2, 60], [94, 96], [153, 158]], [[8, 60]], [[8, 60]], [[71, 75]], [[76, 80]], [[98, 102]], [[8, 60]], [[8, 60]], [[2, 60]], [[76, 92]], [[2, 75]], [[104, 151]], [[94, 102]]]",
    "query_spans": "[[[153, 165]]]",
    "process": "From $\\overline{A}\\overline{B}=3\\overline{A}\\overline{M}$, we obtain $M(\\frac{2a^{2}}{3c},\\frac{1}{3}a)$. Substituting the coordinates of $M$ into $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, and combining with $b^{2}=a^{2}-c^{2}$, it simplifies to $\\frac{4a^{2}}{9c^{2}}+\\frac{a^{2}}{9a^{2}-9c^{2}}=1$, leading to $\\frac{4}{9e^{2}}+\\frac{1}{9-9e^{2}}=1$. Solving gives $e^{2}=\\frac{2}{3}$, $e=\\frac{\\sqrt{6}}{3}$. [Key Ideas] This problem primarily examines the simple geometric properties of ellipses and the property of vector collinearity, making it a difficult question. First, from $\\overline{AB}=3\\overline{AM}$, using the property of collinear vectors and the given coordinates $A(\\frac{a^{2}}{c},0)$, $B(0,a)$, we find the coordinates of point $M$ as $M(\\frac{2a^{2}}{3c},\\frac{1}{3}a)$. Since this point lies on the ellipse, substituting into the equation $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{3^{2}}=1$, we transform it into an equation in terms of $a$ and $c$, from which the eccentricity of the ellipse is calculated. Key points: 1. Collinear vectors; 2. Eccentricity of ellipse"
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$ is $\\frac{\\sqrt{10}}{5}$, then the value of the real number $m$ is?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (x^2/5 + y^2/m = 1);Eccentricity(G) = sqrt(10)/5",
    "query_expressions": "m",
    "answer_expressions": "3, 25/3",
    "fact_spans": "[[[0, 37]], [[65, 70]], [[0, 37]], [[0, 63]]]",
    "query_spans": "[[[65, 74]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ is a point on the ellipse such that $\\angle F_{1} P F_{2}=120^{\\circ}$, and the area of $\\Delta F_{1} P F_{2}$ is $\\sqrt{3}$. Then the length of the minor axis of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;P: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(120, degree);Area(TriangleOf(F1, P, F2)) = sqrt(3)",
    "query_expressions": "MinorAxis(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 54], [83, 85], [83, 85]], [[4, 54]], [[4, 54]], [[62, 69]], [[78, 82]], [[70, 77]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 77]], [[2, 77]], [[78, 88]], [[89, 123]], [[125, 161]]]",
    "query_spans": "[[[163, 170]]]",
    "process": "Since the area of $AF_{1}PF_{2}$ is $\\sqrt{3}$, we have $b^{2}\\tan\\frac{120^{\\circ}}{2}=\\sqrt{3}$, hence $b=1$, and the length of the minor axis is $2b=2$."
  },
  {
    "text": "The tangents to the parabola $y=x^{2}$ at points $A(1,1)$ and $B(-2,4)$ intersect at point $M$. Then, the area of $\\Delta M A B$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2);A: Point;B: Point;Coordinate(A) = (1, 1);Coordinate(B) = (-2, 4);M: Point;Intersection(TangentOnPoint(A,G),TangentOnPoint(B,G))=M;PointOnCurve(A,G);PointOnCurve(B,G)",
    "query_expressions": "Area(TriangleOf(M, A, B))",
    "answer_expressions": "27/4",
    "fact_spans": "[[[0, 12]], [[0, 12]], [[15, 23]], [[25, 34]], [[15, 23]], [[25, 34]], [[40, 44]], [[0, 44]], [0, 32], [0, 32]]",
    "query_spans": "[[[46, 65]]]",
    "process": ""
  },
  {
    "text": "Given an ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$ with foci on the $x$-axis and eccentricity $e=\\frac{\\sqrt{10}}{5}$, what is the value of $m$?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/5 + y^2/m = 1);PointOnCurve(Focus(G), xAxis);e:Number;Eccentricity(G)=e;e = sqrt(10)/5",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[11, 48]], [[77, 80]], [[11, 48]], [[2, 48]], [[52, 75]], [[11, 75]], [[52, 75]]]",
    "query_spans": "[[[77, 84]]]",
    "process": "Since the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$ has its foci on the $x$-axis, it follows that $a^{2}=5$, $b^{2}=m$ $(0<m<5)$, so $c^{2}=a^{2}-b^{2}=5-m$. Given the eccentricity $e=\\frac{\\sqrt{10}}{5}$, we have $e=\\frac{c}{a}=\\frac{\\sqrt{5-m}}{\\sqrt{5}}=\\frac{\\sqrt{10}}{5}$. Solving the equation gives: $m=3$. Hence, the answer is: $3$."
  },
  {
    "text": "Point $P$ moves on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and points $Q$, $R$ move on the circles $(x+1)^{2}+y^{2}=1$ and $(x-1)^{2}+y^{2}=1$ respectively. Then the maximum value of $|PQ|+|PR|$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P, G);G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);H1: Circle;Expression(H1) = (y^2 + (x + 1)^2 = 1);PointOnCurve(Q, H1) = True;Q: Point;R: Point;H2:Circle;Expression(H2) = (y^2 + (x - 1)^2 = 1);PointOnCurve(R, H2) = True",
    "query_expressions": "Max(Abs(LineSegmentOf(P, Q)) + Abs(LineSegmentOf(P, R)))",
    "answer_expressions": "6",
    "fact_spans": "[[[0, 4]], [[0, 45]], [[5, 42]], [[5, 42]], [[57, 77]], [[57, 77]], [[46, 100]], [[46, 49]], [[50, 53]], [[78, 97]], [[78, 97]], [[46, 100]]]",
    "query_spans": "[[[102, 121]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with left and right foci $F_{1}(-c, 0)$ and $F_{2}(c, 0)$ respectively, if there exists a point $P$ on the ellipse such that $\\frac{a}{\\sin \\angle P F_{1} F_{2}}=\\frac{c}{\\sin \\angle P F_{2} F_{1}}$, then the range of values for the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;F2: Point;c: Number;P: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);RightFocus(G) = F2;LeftFocus(G) = F1;PointOnCurve(P, G);a/Sin(AngleOf(P, F1, F2)) = c/Sin(AngleOf(P, F2, F1))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(\\sqrt{2} - 1, 1)",
    "fact_spans": "[[[2, 54], [95, 97], [182, 184]], [[4, 54]], [[4, 54]], [[63, 78]], [[80, 93]], [[80, 93]], [[102, 105]], [[4, 54]], [[4, 54]], [[2, 54]], [[63, 78]], [[80, 93]], [[2, 93]], [[2, 93]], [[95, 105]], [[106, 179]]]",
    "query_spans": "[[[182, 195]]]",
    "process": ""
  },
  {
    "text": "If the line $y = x + t$ intersects the parabola $y^2 = 4x$ at two distinct points $A$, $B$, and the horizontal coordinate of the midpoint of chord $AB$ is $3$, then $t = $?",
    "fact_expressions": "G: Parabola;H: Line;t: Number;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (y = t + x);Intersection(H, G) = {A, B};IsChordOf(LineSegmentOf(A,B),G);XCoordinate(MidPoint(LineSegmentOf(A, B))) = 3;Negation(A=B)",
    "query_expressions": "t",
    "answer_expressions": "-1",
    "fact_spans": "[[[11, 25]], [[1, 10]], [[60, 63]], [[32, 36]], [[37, 40]], [[11, 25]], [[1, 10]], [[1, 40]], [[11, 48]], [[43, 58]], [27, 39]]",
    "query_spans": "[[[60, 65]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and $A$ is its left vertex. If there exists a point $P$ on the hyperbola such that $3 \\overrightarrow{P A}=2 \\overrightarrow{P F_{1}}+\\overrightarrow{P F_{2}}$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;A:Point;LeftVertex(G)=A;PointOnCurve(P,G);3*VectorOf(P,A)=2*VectorOf(P,F1)+VectorOf(P,F2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3",
    "fact_spans": "[[[18, 74], [85, 86], [92, 95], [183, 186]], [[21, 74]], [[21, 74]], [[98, 102]], [[2, 9]], [[10, 17]], [[21, 74]], [[21, 74]], [[18, 74]], [[2, 80]], [[2, 80]], [[81, 84]], [[81, 89]], [[92, 102]], [[104, 180]]]",
    "query_spans": "[[[183, 192]]]",
    "process": "Let P(x,y), and A(-a,0), F_{1}(-c,0), F_{2}(c,0), then 3(-a-x,-y)=2(-c-x,-y)+(c-x,-y), (-3a-3x,-3y)=(-c-3x,-3y), hence -3a-3x=-c-3x \\therefore e=\\frac{c}{a}=3."
  },
  {
    "text": "Given that the point $P(1 , 2)$ is the midpoint of the segment cut from the line $l$ by the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{8}=1$, then the equation of the line $l$ is?",
    "fact_expressions": "P: Point;Coordinate(P) = (1, 2);l: Line;G: Ellipse;Expression(G) = (x^2/4 + y^2/8 = 1);MidPoint(InterceptChord(l, G)) = P",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x+y-3=0",
    "fact_spans": "[[[2, 13]], [[2, 13]], [[14, 19], [68, 73]], [[20, 57]], [[20, 57]], [[2, 66]]]",
    "query_spans": "[[[68, 78]]]",
    "process": "Let the line $ l $ intersect the ellipse at points $ A $ and $ B $, where $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Therefore,  \n\\[\n\\begin{cases}\n\\frac{x^{2}}{4}+\\frac{y^{2}}{8}=1 \\\\\n\\frac{x^{2}}{2}+\\frac{y^{2}}{8}=1\n\\end{cases}\n\\]\nso  \n\\[\n\\frac{x^{2}}{4}-\\frac{x^{2}}{4}=-(\\frac{y_{1}^{2}}{8}-\\frac{y^{2}}{8})\n\\]\nthus  \n\\[\n-2\\cdot\\frac{x_{1}^{8}+x_{2}}{y_{1}+y_{0}}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}},\n\\]\nand  \n\\[\nx_{1}+x_{2}=2x_{P}=2, \\quad y_{1}+y_{2}=2y_{P}=4\n\\]\nso  \n\\[\nk_{1}=\\frac{y_{1}-y_{2}^{2}}{x_{1}-x_{2}}=-2\\cdot\\frac{2}{4}=-1,\n\\]\ntherefore  \n\\[\nl: y-2=-(x-1), \\quad \\text{i.e.,} \\quad x+y-3=0\n\\]"
  },
  {
    "text": "Given $\\frac{1}{m}+\\frac{2}{n}=1$ $(m>0, n>0)$, when $mn$ attains its minimum value, the number of intersection points between the line $y=-\\sqrt{2} x+2$ and the curve $\\frac{x|x|}{m}+\\frac{y|y|}{n}=1$ is?",
    "fact_expressions": "G: Line;Expression(G) = (y = 2 - sqrt(2)*x);H: Curve;Expression(H) = ((y*Abs(y))/n + (x*Abs(x))/m = 1);m: Number;n: Number;m>0;n>0;2/n + 1/m = 1;WhenMin(m*n)",
    "query_expressions": "NumIntersection(G, H)",
    "answer_expressions": "2",
    "fact_spans": "[[[52, 71]], [[52, 71]], [[72, 107]], [[72, 107]], [[2, 39]], [[2, 39]], [[2, 39]], [[2, 39]], [[2, 39]], [[40, 51]]]",
    "query_spans": "[[[52, 114]]]",
    "process": ""
  },
  {
    "text": "The two foci of the ellipse $x^{2}+k y^{2}=1$ lie on the circle $x^{2}+y^{2}=4$. Then the real number $k$=?",
    "fact_expressions": "G: Ellipse;k: Real;H: Circle;Expression(G) = (k*y^2 + x^2 = 1);Expression(H) = (x^2 + y^2 = 4);PointOnCurve(Focus(G), H)",
    "query_expressions": "k",
    "answer_expressions": "1/5",
    "fact_spans": "[[[0, 19]], [[44, 49]], [[25, 41]], [[0, 19]], [[25, 41]], [[0, 42]]]",
    "query_spans": "[[[44, 51]]]",
    "process": "Since the two foci of the ellipse $x^{2}+ky^{2}=1$ lie on the circle $x^{2}+y^{2}=4$, we have $c=2$. Because $x^{2}+\\frac{y^{2}}{\\frac{1}{k}}=1$, when $0<k<1$, $a^{2}=\\frac{1}{k}$, $b^{2}=1$, and from $a^{2}=b^{2}+c^{2}$ we obtain $k=\\frac{1}{5}$. When $k>1$, $b^{2}=\\frac{1}{k}$, $a^{2}=1$, which does not hold. Therefore, $k=\\frac{1}{5}$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, with left and right foci $F_{1}$, $F_{2}$ respectively, right vertex at $A(\\sqrt{5}, 0)$, and eccentricity $\\frac{\\sqrt{5}}{5}$, find the length of the minor axis?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;A: Point;F1: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (sqrt(5), 0);LeftFocus(C) = F1;RightFocus(C) = F2;RightVertex(C) = A;Eccentricity(C) = sqrt(5)/5",
    "query_expressions": "Length(MinorAxis(C))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 59]], [[9, 59]], [[9, 59]], [[88, 104]], [[68, 75]], [[76, 83]], [[9, 59]], [[9, 59]], [[2, 59]], [[88, 104]], [[2, 83]], [[2, 83]], [[2, 104]], [[2, 130]]]",
    "query_spans": "[[[2, 137]]]",
    "process": "By the given condition, the right vertex of the ellipse $ C: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $) is $ A(\\sqrt{5}, 0) $, so $ a = \\sqrt{5} $. Also, the eccentricity of the ellipse is $ \\frac{\\sqrt{5}}{5} $, that is, $ \\frac{c}{a} = \\frac{\\sqrt{5}}{5} $, which gives $ c = 1 $. Therefore, $ 2b = 4 $, that is, the minor axis length of the ellipse is 4."
  },
  {
    "text": "Draw a line $l$ perpendicular to the $x$-axis through the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$. The line $l$ intersects the hyperbola at points $A$ and $B$, and intersects the asymptotes of the hyperbola at points $C$ and $D$. If $3|AB|=2|CD|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "l: Line;G: Hyperbola;b: Number;a: Number;A: Point;B: Point;C: Point;D: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(RightFocus(G),l);IsPerpendicular(l,xAxis);Intersection(l, G) = {A, B};Intersection(l,Asymptote(G))={C,D};3*Abs(LineSegmentOf(A, B)) = 2*Abs(LineSegmentOf(C, D))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3*sqrt(5)/5",
    "fact_spans": "[[[70, 75], [76, 81]], [[1, 57], [82, 85], [98, 101], [135, 138]], [[4, 57]], [[4, 57]], [[87, 90]], [[91, 94]], [[107, 110]], [[111, 114]], [[4, 57]], [[4, 57]], [[1, 57]], [[0, 75]], [[62, 75]], [[76, 96]], [[76, 116]], [[118, 133]]]",
    "query_spans": "[[[135, 144]]]",
    "process": "From the given conditions, we have $3b=2c$, $\\therefore 9a^{2}=5c^{2}$, then the eccentricity of the hyperbola is $e=\\frac{3\\sqrt{5}}{5}$. Therefore, fill in $\\frac{3\\sqrt{5}}{5}$."
  },
  {
    "text": "Given that point $P$ moves on the parabola $y^{2}=4x$, $F$ is the focus of the parabola, and the coordinates of point $M$ are $(3,2)$. When $PM+PF$ takes the minimum value, what are the coordinates of point $P$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G);F: Point;Focus(G) = F;M: Point;Coordinate(M) = (3, 2);WhenMin(LineSegmentOf(P, M) + LineSegmentOf(P, F))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1, 2)",
    "fact_spans": "[[[7, 21], [29, 32]], [[7, 21]], [[2, 6], [65, 69]], [[2, 24]], [[25, 28]], [[25, 35]], [[36, 40]], [[36, 51]], [[52, 65]]]",
    "query_spans": "[[[65, 74]]]",
    "process": "From the definition of a parabola, we know that PF equals the distance from point P to the directrix of the parabola $ x = -1 $, denoted as $ d $. Therefore, $ PM + PF = PM + d $. By the triangle inequality that the sum of any two sides is greater than the third side, $ PM + PF $ reaches its minimum when point P lies at the intersection of the parabola and the perpendicular line drawn from M to the directrix of the parabola. At this time, the point P is found to be $ (1, 2) $."
  },
  {
    "text": "Given a fixed point $A(-1 , 0)$ on the parabola $x^{2}=y+1$ and two moving points $P$, $Q$. When $PA \\perp PQ$, what is the range of the abscissa of point $Q$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = y + 1);A: Point;Coordinate(A) = (-1, 0);P: Point;Q: Point;IsPerpendicular(LineSegmentOf(P, A), LineSegmentOf(P, Q));PointOnCurve(A, G)",
    "query_expressions": "Range(XCoordinate(Q))",
    "answer_expressions": "(-oo, -3]+[1, +oo)",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[20, 31]], [[20, 31]], [[35, 38]], [[39, 42], [58, 62]], [[43, 56]], [2, 30]]",
    "query_spans": "[[[58, 73]]]",
    "process": ""
  },
  {
    "text": "Given that one focus of the ellipse $k x^{2}+4 k y^{2}=1$ is $(-3,0)$, what is the value of $k$?",
    "fact_expressions": "G: Ellipse;k: Number;H: Point;Expression(G) = (k*x^2 + 4*(k*y^2) = 1);Coordinate(H) = (-3, 0);OneOf(Focus(G)) = H",
    "query_expressions": "k",
    "answer_expressions": "1/12",
    "fact_spans": "[[[2, 25]], [[41, 44]], [[31, 39]], [[2, 25]], [[31, 39]], [[2, 39]]]",
    "query_spans": "[[[41, 48]]]",
    "process": "Transform the ellipse equation into: $\\frac{x^{2}}{1}+\\frac{y^{2}}{\\frac{1}{4k}}=1$, therefore $\\frac{1}{k}-\\frac{1}{4k}=9$, solving gives: $k=\\frac{1}{12}$"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$, a line passing through the focus $F$ with an inclination angle of $\\frac{\\pi}{3}$ intersects $C$ at points $A$ and $B$. Then, the distance from the midpoint of chord $AB$ to the directrix is?",
    "fact_expressions": "C: Parabola;G: Line;A: Point;B: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C)=F;Inclination(G)=pi/3;PointOnCurve(F,G);Intersection(G, C) = {A, B};IsChordOf(LineSegmentOf(A, B), C)",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A,B)),Directrix(C))",
    "answer_expressions": "8/3",
    "fact_spans": "[[[2, 21], [52, 55]], [[49, 51]], [[56, 59]], [[60, 63]], [[25, 28]], [[2, 21]], [[2, 28]], [[29, 51]], [[22, 51]], [[49, 65]], [[52, 73]]]",
    "query_spans": "[[[52, 84]]]",
    "process": "By the given condition, the parabola $ C: y^{2} = 4x $ has focus $ F(1,0) $ and directrix equation $ x = -1 $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, and the equation of line $ AB $ be $ \\sqrt{3}x - y - \\sqrt{3} = 0 $. Solving the system of equations \n\\[\n\\begin{cases}\n\\sqrt{3}x - y - \\sqrt{3} = 0 \\\\\ny^{2} = 4x\n\\end{cases}\n\\]\nwe obtain $ 3x^{2} - 10x + 3 = 0 $. Then $ x_{1} + x_{2} = \\frac{10}{3} $, so the horizontal coordinate of the midpoint of chord $ AB $ is $ \\frac{5}{3} $, and the distance from the midpoint of chord $ AB $ to the directrix is $ \\frac{5}{3} + 1 = \\frac{8}{3} $."
  },
  {
    "text": "Given the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1$ has eccentricity $\\frac{\\sqrt{5}}{2}$, the left focus is $F_{1}$. When point $P$ moves on the right branch of the hyperbola and point $Q$ moves on the circle $x^{2}+(y-1)^{2}=1$, the minimum value of $|P Q|+|P F_{1}|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/b^2 = 1);b: Number;Eccentricity(G) = sqrt(5)/2;F1: Point;LeftFocus(G) = F1;P: Point;PointOnCurve(P, RightPart(G));H: Circle;Expression(H) = (x^2 + (y - 1)^2 = 1);Q: Point;PointOnCurve(Q, H)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "5/2",
    "fact_spans": "[[[2, 34], [78, 81]], [[2, 34]], [[5, 34]], [[2, 59]], [[64, 71]], [[2, 71]], [[73, 77]], [[73, 86]], [[92, 112]], [[92, 112]], [[87, 91]], [[87, 115]]]",
    "query_spans": "[[[117, 140]]]",
    "process": "According to the problem, $ a=1 $, $ b=\\frac{1}{2} $, let $ B(0,1) $, from $ |PF_{1}| - |PF_{2}| = 2 $ we obtain $ |PQ| + |PF_{1}| = |PQ| + |PF_{2}| + 2 \\geqslant |QF_{2}| + 2 $. The problem is transformed into finding the minimum value from point $ F_{2} $ to a point on circle $ B $, that is, $ |QF_{2}|_{\\min} = |BF_{2}| - 1 = \\frac{3}{2} - 1 = \\frac{1}{2} $, hence $ (|PQ| + |PF_{1}|)_{\\min} = \\frac{1}{2} + 2 = \\frac{5}{2} $."
  },
  {
    "text": "What is the length of the real axis of the hyperbola $\\frac{x^{2}}{9}-y^{2}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2 = 1)",
    "query_expressions": "Length(RealAxis(G))",
    "answer_expressions": "6",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 34]]]",
    "process": "From \\frac{x^{2}}{9}-y^{2}=1, we know a^{2}=9, so a=3, therefore the length of the real axis is 2a=6."
  },
  {
    "text": "The standard equation of an ellipse with eccentricity $e=\\frac{\\sqrt{5}}{3}$ and a directrix $x=3$ is?",
    "fact_expressions": "G: Ellipse;e: Number;Eccentricity(G) = e;e = sqrt(5)/3;Expression(OneOf(Directrix(G))) = (x = 3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5 + 9*y^2/20 = 1",
    "fact_spans": "[[[37, 39]], [[3, 25]], [[0, 39]], [[3, 25]], [[26, 39]]]",
    "query_spans": "[[[37, 46]]]",
    "process": ""
  },
  {
    "text": "Let the two foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{20}=1$ be $F_{1}$ and $F_{2}$, and let point $P$ lie on the hyperbola. If $|P F_{1}|=9$, then $|P F_{2}|=$?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/16 - y^2/20 = 1);Focus(G) = {F1,F2};PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 9",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "17",
    "fact_spans": "[[[1, 41], [68, 71]], [[63, 67]], [[47, 54]], [[55, 62]], [[1, 41]], [[1, 62]], [[63, 72]], [[74, 87]]]",
    "query_spans": "[[[89, 102]]]",
    "process": "From the hyperbola $\\frac{x^2}{16}-\\frac{y^{2}}{20}=1$, we have $a=4$. By the definition of a hyperbola, $||PF_{1}|-|PF_{2}||=2a=8$, so $|PF_{1}|-|PF_{2}|=8$ or $|PF_{1}|-|PF_{2}|=-8$. Since $|PF_{1}|=9$, it follows that $|PF_{2}|=17$ or $1$. Moreover, since $|PF_{2}|\\geqslant c-a=2$, $|PF_{2}|=1$ is discarded. Therefore, $|PF_{2}|=17$."
  },
  {
    "text": "Given that $O$ is the coordinate origin, the ellipse $T$: $\\frac{x^2}{a^2}+\\frac{y^2}{b^2} = 1$ has eccentricity $\\frac{\\sqrt{2}}{2}$, and a vertex $B(0,1)$. Two lines $PA$, $PC$ passing through a point $P$ on the ellipse intersect the ellipse at $A$, $C$ respectively. Let $D$, $E$ be the midpoints of $PA$, $PC$ respectively, and let the slopes of $PA$, $PC$ be $k_1$, $k_2$ $(k_1,k_2<0)$. If the sum of the slopes of lines $OD$ and $OE$ is $2$, then the maximum value of $4 k_{1}+k_{2}$ is?",
    "fact_expressions": "O: Origin;T: Ellipse;Expression(T) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;Eccentricity(T) = sqrt(2)/2;B: Point;Coordinate(B)=(0,1);OneOf(Vertex(T)) = B;P: Point;PointOnCurve(P, T);PointOnCurve(P, LineOf(P, A));PointOnCurve(P, LineOf(P, C));A: Point;Intersection(LineOf(P, A), T) = A;C: Point;Intersection(LineOf(P, C), T) = C;D: Point;MidPoint(LineSegmentOf(P, A)) = D;E: Point;MidPoint(LineSegmentOf(P, C)) = E;k1: Number;k2: Number;Slope(LineOf(P, A)) = k1;Slope(LineOf(P, C)) = k2;k1<0;k2<0;Slope(LineOf(O, D))+Slope(LineOf(O, E)) = 2",
    "query_expressions": "Max(4*k1 + k2)",
    "answer_expressions": "-9/4",
    "fact_spans": "[[[2, 5]], [[11, 54], [93, 95], [118, 120]], [[11, 54]], [[17, 54]], [[17, 54]], [[11, 78]], [[83, 91]], [[83, 91]], [[11, 91]], [[98, 101]], [[93, 101]], [[92, 115]], [[92, 115]], [[122, 125]], [[106, 129]], [[126, 129]], [[106, 129]], [[146, 149]], [[131, 153]], [[150, 153]], [[131, 153]], [[171, 176]], [[177, 182]], [[154, 182]], [[154, 182]], [[182, 195]], [[182, 195]], [[197, 216]]]",
    "query_spans": "[[[218, 239]]]",
    "process": ""
  },
  {
    "text": "Given point $M(0,2)$, a line $AB$ passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. If $\\overrightarrow{A M} \\cdot \\overrightarrow{F M}=0$, then the ordinate of point $B$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (0, 2);G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;PointOnCurve(F, LineOf(A,B)) = True;Intersection(LineOf(A,B),G) = {A,B};A: Point;B: Point;DotProduct(VectorOf(A, M), VectorOf(F, M)) = 0",
    "query_expressions": "YCoordinate(B)",
    "answer_expressions": "-1",
    "fact_spans": "[[[2, 11]], [[2, 11]], [[13, 27], [42, 45]], [[13, 27]], [[30, 33]], [[13, 33]], [[12, 41]], [[34, 55]], [[46, 49]], [[50, 53], [110, 114]], [[57, 108]]]",
    "query_spans": "[[[110, 120]]]",
    "process": "Since point M(0,2) and the focus F(1,0) of the parabola y^{2}=4x, we have k_{MF}=\\frac{2-0}{0-1}=-2. From \\overrightarrow{AM}\\cdot\\overrightarrow{FM}=0, it follows that AM\\bot FM, so the slope of line AM is k_{AM}=\\frac{1}{2}. Thus, the equation of line AM is y-2=\\frac{1}{2}x, or y=\\frac{1}{2}x+2. Solving \\begin{cases}y=\\frac{1}{2}x+2\\\\y^{2}=4x\\end{cases}, simplifying gives x^{2}-8x+16=0, solving yields x=4, thus point A(4,4). Therefore, the slope of line AF is k_{AF}=\\frac{4}{4-1}=\\frac{4}{3}, so the equation of line AF is y=\\frac{4}{3}(x-1). Solving the system \\begin{cases}y^{2}=4x\\\\y=\\frac{4}{3}(x-1)\\end{cases}, eliminating x gives y^{2}-3y-4=0, solving yields y=-1 or y=4, so the y-coordinate of point R is -1."
  },
  {
    "text": "$AB$ is a chord passing through the focus of $C$: $y^{2}=4x$, and $|AB|=10$, then the horizontal coordinate of the midpoint of $AB$ is?",
    "fact_expressions": "A: Point;B: Point;Abs(LineSegmentOf(A, B)) = 10;C:Curve;Expression(C)=(y^2=4*x);IsChordOf(LineSegmentOf(A,B),C);PointOnCurve(Focus(C),LineSegmentOf(A,B))",
    "query_expressions": "XCoordinate(MidPoint(LineSegmentOf(A, B)))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 5]], [[0, 5]], [[29, 39]], [[7, 23]], [[7, 23]], [[0, 27]], [[0, 27]]]",
    "query_spans": "[[[41, 54]]]",
    "process": ""
  },
  {
    "text": "Given that the upper vertex of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ is $M$, and the line $l$ intersects the ellipse at points $P$ and $Q$, such that the point $(1,0)$ is exactly the orthocenter of $\\triangle P Q M$, then the equation of line $l$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);M: Point;UpperVertex(G) = M;l: Line;P: Point;Q: Point;Intersection(l, G) = {P, Q};H: Point;Coordinate(H) = (1, 0);Orthocenter(TriangleOf(P, Q, M)) = H",
    "query_expressions": "Expression(l)",
    "answer_expressions": "y=x-4/3",
    "fact_spans": "[[[2, 29], [45, 47]], [[2, 29]], [[34, 37]], [[2, 37]], [[38, 43], [92, 97]], [[49, 52]], [[53, 56]], [[38, 58]], [[60, 68]], [[60, 68]], [[60, 90]]]",
    "query_spans": "[[[92, 102]]]",
    "process": "The upper vertex is $ M(0,1) $, the right focus $ F $ is the orthocenter $ (1,0) $. Since $ k_{FM} = -1 $ and $ FM \\perp l $, we have $ k_1 = 1 $. Let the line $ PQ $ be $ y = x + m $, and let $ P(x_1, y_1) $, $ Q(x_2, y_2) $. From\n\\[\n\\begin{cases}\ny = x + m \\\\\n\\frac{x^2}{2} + y^2 = 1\n\\end{cases}\n\\]\neliminating $ y $, we get $ 3x^2 + 4mx + 2m^2 \\cdot 2 = 0 $. $ \\Delta = 16m^2 \\cdot 12(2m^2 \\cdot 2) > 0 $, so $ m^2 < 3 $, $ x_1 + x_2 = \\frac{4m}{3} $, $ x_1 x_2 = \\frac{2n}{}\\frac{12-2}{3} $. $ y_1 y_2 = (x_1 + m)(x_2 + m) = x_1 x_2 + m(x_1 + x_2)_{+m^{2}}2\\frac{m^{2}-2}{3}-\\frac{4m^{2}}{3}+m^{2}=\\frac{m^{2}-2}{3} $. Also, since $ F $ is the orthocenter of $ \\triangle MPQ $, $ \\therefore PF \\perp MQ $, $ \\therefore \\overrightarrow{PF} \\cdot \\overrightarrow{MO} = 0 $. Again $ \\frac{}{PF}\\sqrt{2}-y_{1} $, $ \\overrightarrow{MQ} = (x_2, y_2 - 1) $. $ \\therefore \\overrightarrow{PF} \\cdot \\overrightarrow{MQ} = x_2 + y_1 - x_1 x_2 - y_1 y_2 = x_2 + x_1 + m - x_1 x_2 - y_1 y_2 = -\\frac{4}{3}m + m - \\frac{2m^{2}-2}{3} - \\frac{m^{2}-2}{3} = 0 $. $ \\therefore -\\frac{m}{3} - m^2 + \\frac{4}{3} = 0 $, $ \\therefore 3m^2 + m - 4 = 0 $, $ m = -\\frac{4}{2} $, $ m = 1 $. Upon verification, $ m^2 < 3 $ is satisfied. $ \\therefore $ there exist lines $ l $ satisfying the condition: $ x - y + 1 = 0 $, $ 3x - 3y - 4 = 0 $. Since $ x \\cdot y + 1 = 0 $ passes through point $ M $, i.e., $ MP $ coincides, which does not form a triangle, $ \\therefore 3x \\cdot 3v \\cdot 4 = 0 $ satisfies the requirement."
  },
  {
    "text": "$P$ is a moving point on the right branch of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, and $F$ is the right focus of the hyperbola. Given $A(3, 1)$, find the minimum value of $|P A|+|P F|$.",
    "fact_expressions": "G: Hyperbola;A: Point;P: Point;F: Point;Expression(G) = (x^2/3 - y^2 = 1);Coordinate(A) = (3, 1);PointOnCurve(P, RightPart(G));RightFocus(G) = F",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "sqrt(26)-2*sqrt(3)",
    "fact_spans": "[[[4, 32], [44, 47]], [[54, 64]], [[0, 3]], [[40, 43]], [[4, 32]], [[54, 64]], [[0, 39]], [[40, 51]]]",
    "query_spans": "[[[66, 85]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $A$ lies on the hyperbola, point $M$ has coordinates $(\\frac{2}{3}, 0)$, and the distances from $M$ to the lines $A F_{1}$ and $A F_{2}$ are equal. Then $|A F_{1}|$=?",
    "fact_expressions": "C: Hyperbola;F1: Point;A: Point;M: Point;F2: Point;Expression(C) = (x^2 - y^2/3 = 1);Coordinate(M) = (2/3, 0);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(A, C);Distance(M, LineOf(A, F1)) = Distance(M, LineOf(A, F2))",
    "query_expressions": "Abs(LineSegmentOf(A, F1))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 35], [64, 67]], [[43, 50]], [[59, 63]], [[69, 73], [97, 100]], [[51, 58]], [[2, 35]], [[69, 95]], [[2, 58]], [[2, 58]], [[59, 68]], [[97, 129]]]",
    "query_spans": "[[[131, 144]]]",
    "process": "From the given conditions, F(-2,0), F_{2}(2,0), and point A lies on the right branch of the hyperbola. The coordinates of point M are (\\frac{2}{3},0). Therefore, |F_{1}M| = 2 + \\frac{2}{3} = \\frac{8}{3}, |MF_{2}| = 2 - \\frac{2}{3} = \\frac{4}{3}. The figure is drawn as shown, with MP \\perp AF_{1}, MQ \\perp AF_{2}, and the feet of perpendiculars being P and Q respectively. AM is the angle bisector of \\angle F_{1}AF_{2}, so \\frac{|AF_{1}|}{|AF_{2}|} = \\frac{|F_{1}M|}{|MF_{2}|} = 2, that is, |AF_{1}| = 2|AF_{2}|. Also, |AF_{1}| - |AF_{2}| = 2, therefore |AF_{1}| = 4, |AF_{2}| = 2."
  },
  {
    "text": "The circle with the right focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ as its center and radius $a$ is exactly tangent to the two asymptotes of the hyperbola. Then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(G) = F;H: Circle;Center(H) = F;Radius(H) = a;IsTangent(H,Asymptote(G)) = True",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 57], [93, 96], [79, 82]], [[1, 57]], [[4, 57]], [[68, 71]], [[4, 57]], [[4, 57]], [[61, 64]], [[1, 64]], [[75, 76]], [[0, 67]], [[68, 76]], [[75, 90]]]",
    "query_spans": "[[[93, 102]]]",
    "process": "From the given condition, we have a = b, and c^{2} = a^{2} + b^{2} = 2a^{2}, then c = \\sqrt{2}a, so the eccentricity is e = \\frac{c}{a} = \\sqrt{2},"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively, a line passing through $F_{1}$ with an inclination angle of $30^{\\circ}$ intersects the right branch of the hyperbola at point $P$. If $P F_{2} \\perp F_{1} F_{2}$, then the equation of the asymptotes of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Line;PointOnCurve(F1, H);Inclination(H) = ApplyUnit(30, degree);P: Point;Intersection(H, RightPart(G)) = P;IsPerpendicular(LineSegmentOf(P, F2), LineSegmentOf(F1, F2))",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(2)*x",
    "fact_spans": "[[[20, 76], [111, 114], [153, 156]], [[20, 76]], [[23, 76]], [[23, 76]], [[23, 76]], [[23, 76]], [[2, 9], [83, 90]], [[10, 17]], [[2, 81]], [[2, 81]], [[108, 110]], [[82, 110]], [[91, 110]], [[118, 121]], [[108, 121]], [[123, 150]]]",
    "query_spans": "[[[153, 164]]]",
    "process": "Let |PF_{1}| = m, |PF_{2}| = n, |F_{1}F_{2}| = 2c. By the definition of the hyperbola and properties in a right triangle, the relationship between m and n can be obtained. From the relationship among a, b, c, we can find b, and then obtain the required asymptotes of the hyperbola. [Detailed explanation] Let |PF_{1}| = m, |PF_{2}| = n, |F_{1}F_{2}| = 2c. In the right triangle \\triangle PF_{1}F_{2}, \\angle PF_{1}F_{2} = 30^{\\circ}, we get m = 2n, so m - n = 2a = n, thus a = \\frac{1}{2}n, 2c = \\sqrt{3}n, so c = \\frac{\\sqrt{3}}{2}n, b = \\sqrt{c^{2} - a^{2}} = \\frac{\\sqrt{2}}{2}n. The asymptotes of the hyperbola are given by y = \\pm\\frac{b}{a}x, which yields y = \\pm\\sqrt{2}x."
  },
  {
    "text": "Given that the line $AB$ passes through the focus of the parabola $y^{2}=4x$, intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $x_{1}+x_{2}=5$, then $|AB|$ equals?",
    "fact_expressions": "PointOnCurve(Focus(G),LineOf(A,B)) = True;G: Parabola;Expression(G) = (y^2 = 4*x);Intersection(LineOf(A,B),G) = {A,B};A: Point;x1: Number;y1: Number;B: Point;x2: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);x1 + x2 = 5",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "7",
    "fact_spans": "[[[2, 27]], [[10, 24], [29, 32]], [[10, 24]], [[2, 72]], [[33, 50]], [[33, 50]], [[33, 50]], [[53, 70]], [[53, 70]], [[53, 70]], [[33, 50]], [[53, 70]], [[74, 89]]]",
    "query_spans": "[[[91, 101]]]",
    "process": "From the given information, |AB| = x_{1} + x_{2} + p = 5 + 2 = 7"
  },
  {
    "text": "Given the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$ with foci $F_{1}$ and $F_{2}$, point $M$ lies on the hyperbola such that $M F_{1} \\cdot M F_{2}=0$. Find the distance from point $M$ to the $x$-axis.",
    "fact_expressions": "G: Hyperbola;M: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2/2 = 1);Focus(G)={F1,F2};PointOnCurve(M, G);LineSegmentOf(M, F1)*LineSegmentOf(M, F2) = 0",
    "query_expressions": "Distance(M, xAxis)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 30], [55, 58]], [[50, 54], [88, 92]], [[34, 41]], [[42, 49]], [[2, 30]], [[2, 49]], [[50, 59]], [[61, 86]]]",
    "query_spans": "[[[88, 102]]]",
    "process": ""
  },
  {
    "text": "Given that the line $l$ intersects the ellipse $4 x^{2} + 9 y^{2}=36$ at points $A$ and $B$, and the midpoint of chord $AB$ has coordinates $(1,1)$, then the equation of line $l$ is?",
    "fact_expressions": "l: Line;G: Ellipse;A: Point;B: Point;Expression(G) = (4*x^2 + 9*y^2 = 36);Intersection(l, G) = {A, B};Coordinate(MidPoint(LineSegmentOf(A,B)))=(1,1);IsChordOf(LineSegmentOf(A,B),G)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "4*x+9*y-13=0",
    "fact_spans": "[[[2, 7], [68, 73]], [[8, 32]], [[35, 38]], [[39, 42]], [[8, 32]], [[2, 44]], [[47, 65]], [[8, 52]]]",
    "query_spans": "[[[68, 78]]]",
    "process": ""
  },
  {
    "text": "Given the line $l$: $y = m x - 4 m$ intersects the parabola $y^{2} = 2 p x$ ($p > 0$) at points $A$, $B$, and the circle with diameter $AB$ passes through the origin, then the equation of the parabola is?",
    "fact_expressions": "l: Line;Expression(l) = (y = m*x - 4*m);m: Number;G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;A: Point;B: Point;Intersection(l, G) = {A, B};H: Circle;IsDiameter(LineSegmentOf(A, B), H);O: Origin;PointOnCurve(O, H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[2, 20]], [[2, 20]], [[9, 20]], [[21, 42], [71, 74]], [[21, 42]], [[24, 42]], [[24, 42]], [[44, 48]], [[49, 52]], [[2, 52]], [[64, 65]], [[54, 65]], [[67, 69]], [[64, 69]]]",
    "query_spans": "[[[71, 79]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since the circle with AB as diameter passes through the origin, OA\\bot OB, i.e., x_{1}x_{2}+y_{1}y_{2}=0. From \\begin{cases}y+4m=mx\\\\y^{2}=2px\\end{cases}, eliminating x gives my^{2}-2py-8mp=0, \\therefore y_{1}y_{2}=-8p, (y_{1}y_{2})^{2}=4p^{2}x_{1}x_{2}|x|=16. \\therefore 16-8p=0, solving yields p=2, \\therefore the equation of the parabola is y^{2}=4x."
  },
  {
    "text": "Given that the line $y=(a+1) x-1$ and the parabola $y^{2}=a x$ ($a \\neq 0$) have exactly one common point, then $a=$?",
    "fact_expressions": "H: Line;Expression(H) = (y = x*(a + 1) - 1);G: Parabola;Expression(G) = (y^2 = a*x);a: Number;Negation(a=0);NumIntersection(H, G)=1",
    "query_expressions": "a",
    "answer_expressions": "{-4/5,-1}",
    "fact_spans": "[[[2, 17]], [[2, 17]], [[18, 42]], [[18, 42]], [[51, 54]], [[21, 42]], [[2, 49]]]",
    "query_spans": "[[[51, 56]]]",
    "process": "When $ a+1=0 $, that is, when $ a=-1 $, the equation of the line is $ y=-1 $, and the equation of the parabola is $ y^{2}=-x $. Solving the system of equations  \n\\[\n\\begin{cases}\ny=-1 \\\\\ny^{2}=x\n\\end{cases}\n\\]\nyields  \n\\[\n\\begin{cases}\nx=1 \\\\\ny=-1\n\\end{cases}\n\\]\nAt this time, the line and the parabola have only one intersection point. When $ a+1 \\neq 0 $ and $ a \\neq 0 $, that is, when $ a \\neq -1 $ and $ a \\neq 0 $, solving the system  \n\\[\n\\begin{cases}\ny=(a+1)x-1 \\\\\ny^{2}=x\n\\end{cases}\n\\]\ngives $ (a+1)y^{2}-ay-a=0 $, then $ \\Delta = a^{2}+4a(a+1) = a(5a+4) = 0 $, solving yields $ a=-\\frac{4}{5} $. Therefore, $ a=-\\frac{4}{5} $ or $ -1 $."
  },
  {
    "text": "Given that the distance from point $P$ to the $y$-axis is $1$ less than its distance to the point $(1,0)$, what equation does point $P$ satisfy?",
    "fact_expressions": "G: Point;Coordinate(G) = (1, 0);P: Point;Distance(P, yAxis) = Distance(P, G) - 1",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "{(y^2=4*x),(y=0)&(x<0)}",
    "fact_spans": "[[[17, 25]], [[17, 25]], [[2, 6], [15, 16], [34, 38]], [[2, 32]]]",
    "query_spans": "[[[34, 45]]]",
    "process": "Let P(x,y), then |x|+1=\\sqrt{(}. If x\\geqslant0, then x+1=\\sqrt{(}, square both sides and simplify to get y^{2}=4x. If x<0, then 1-x=\\sqrt{(x-1)^{2}+y^{2}}, square both sides and simplify to get y=0. \\therefore The trajectory equation of point P is y=0 (x<0) or y^{2}=4x;"
  },
  {
    "text": "Let the focus of the parabola $y^{2}=8x$ be $F$, the directrix be $l$, and $P$ be a point on the parabola. Let $PA \\perp l$, where $A$ is the foot of the perpendicular. If the slope of the line $AF$ is $-\\sqrt{3}$, then the coordinates of point $P$ are?",
    "fact_expressions": "G: Parabola;A: Point;F: Point;P: Point;l: Line;Expression(G) = (y^2 = 8*x);Focus(G) = F;Directrix(G)=l;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P,A),l);FootPoint(LineSegmentOf(P,A),l)=A;Slope(LineOf(A,F)) = -sqrt(3)",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(6,4*sqrt(3))",
    "fact_spans": "[[[1, 15], [34, 37]], [[56, 59]], [[19, 22]], [[30, 33], [87, 91]], [[26, 29]], [[1, 15]], [[1, 22]], [[1, 29]], [[30, 40]], [[41, 54]], [[41, 62]], [[64, 85]]]",
    "query_spans": "[[[87, 96]]]",
    "process": "From the problem, we know: the focus of the parabola $ y^{2} = 8x $ is $ F(2,0) $, and the directrix of the parabola $ y^{2} = 8x $ is $ x = -2 $. Let $ P(m,n) $, then $ A(-2,n) $, $ 8m = n^{2} $. The slope of line $ AF $ is $ -\\sqrt{3} $, so $ \\angle AFx = \\frac{2\\pi}{3} $, thus $ |AB| = |BF|\\tan60^{\\circ} = 4\\sqrt{3} $, therefore $ n = 4\\sqrt{3} $, giving $ 8m = 48 $, solving yields $ m = 6 $."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ with focus $F$, let $l$ be the tangent line to the parabola at point $P(2, y_{0})$, and let $m$ be the line passing through $P$ parallel to the $x$-axis. A line passing through $F$ and parallel to $l$ intersects $m$ at $M$. If $|P M|=5$, then the value of $p$ is?",
    "fact_expressions": "m: Line;G: Parabola;p: Number;l: Line;P: Point;M: Point;F: Point;p>0;y0:Number;Expression(G) = (y^2 = 2*(p*x));Coordinate(P) = (2, y0);Focus(G) = F;TangentOfPoint(P,G)=l;PointOnCurve(P,m);IsParallel(m,xAxis);L:Line;PointOnCurve(F,L);IsParallel(L,l);Intersection(L,m)=M;Abs(LineSegmentOf(P, M)) = 5",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[72, 77], [93, 96]], [[2, 23], [32, 35]], [[113, 116]], [[54, 57], [86, 89]], [[36, 50], [59, 63]], [[97, 100]], [[27, 30], [79, 82]], [[5, 23]], [[37, 50]], [[2, 23]], [[36, 50]], [[2, 30]], [[31, 57]], [[58, 77]], [[64, 77]], [[90, 92]], [[78, 92]], [[83, 92]], [[90, 100]], [[102, 111]]]",
    "query_spans": "[[[113, 120]]]",
    "process": "Let P(2,2\\sqrt{p}). From y=\\sqrt{2px}, we get y'=\\sqrt{2p}\\frac{1}{2\\sqrt{x}}. Then when x=2, y'=\\frac{\\sqrt{p}}{2}. Thus, the equation of the line passing through F and parallel to l is y=\\frac{\\sqrt{p}}{2}(x-\\frac{p}{2}). Substituting M(7,2\\sqrt{p}), we obtain 7-\\frac{p}{2}=4. Solving gives p=6."
  },
  {
    "text": "The ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has eccentricity $\\frac{1}{2}$ and focal length $2$. Then, what is the length of the minor axis of the ellipse?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Eccentricity(C) = 1/2;FocalLength(C) = 2",
    "query_expressions": "Length(MinorAxis(C))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[0, 56], [83, 85]], [[6, 56]], [[6, 56]], [[6, 56]], [[6, 56]], [[0, 56]], [[0, 74]], [[0, 81]]]",
    "query_spans": "[[[83, 91]]]",
    "process": "Since the eccentricity of the ellipse is $\\frac{1}{2}$, the focal distance is $2$, $c=1$, $e=\\frac{c}{a}=\\frac{1}{a}=\\frac{1}{2}$, so $a=2$, from $b^{2}=a^{2}-c^{2}=4-1=3$, we get $b=\\sqrt{3}$, then the minor axis length of the ellipse is $2b=2\\sqrt{3}$,"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, let $F_{1}$ and $F_{2}$ be its left and right foci respectively, and $P$ be a point on the hyperbola such that $|P F_{1}|=7$. Then the value of $|P F_{1}|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P,G) = True;Abs(LineSegmentOf(P, F1)) = 7",
    "query_expressions": "Abs(LineSegmentOf(P, F1))",
    "answer_expressions": "13",
    "fact_spans": "[[[2, 41], [63, 64], [75, 78]], [[2, 41]], [[44, 51]], [[53, 60]], [[44, 70]], [[44, 70]], [[71, 74]], [[71, 81]], [[83, 96]]]",
    "query_spans": "[[[98, 113]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the hyperbola $x^{2}-4 y^{2}=1$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - 4*y^2 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[0, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 26]]]",
    "process": ""
  },
  {
    "text": "A line passing through the left focus $F$ of an ellipse, with an inclination angle of $\\frac{\\pi}{3}$, intersects the ellipse at points $A$ and $B$. If $|F A|=2|F B|$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;H: Line;F: Point;A: Point;B: Point;LeftFocus(G) = F;Inclination(H)=pi/3;Intersection(H, G) = {A, B};Abs(LineSegmentOf(F, A)) = 2*Abs(LineSegmentOf(F, B));PointOnCurve(F,H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2/3",
    "fact_spans": "[[[1, 3], [33, 35], [63, 65]], [[30, 32]], [[6, 9]], [[36, 39]], [[40, 43]], [[1, 9]], [[10, 32]], [[30, 45]], [[47, 61]], [[0, 32]]]",
    "query_spans": "[[[63, 71]]]",
    "process": ""
  },
  {
    "text": "A point $P$ on the hyperbola $y^{2}-4 x^{2}=64$ is at a distance of $1$ from one of its foci; then the distance from $P$ to the other focus is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (-4*x^2 + y^2 = 64);PointOnCurve(P, G);Distance(P,F1) = 1;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2)",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "17",
    "fact_spans": "[[[0, 21], [28, 29], [48, 49]], [[24, 27], [44, 47]], [], [], [[0, 21]], [[0, 27]], [[24, 42]], [[28, 34]], [[48, 55]], [[28, 55]]]",
    "query_spans": "[[[44, 62]]]",
    "process": ""
  },
  {
    "text": "The equation of the locus of the center of a moving circle that is externally tangent to the circle $(x-2)^{2}+y^{2}=1$ and tangent to the line $x+1=0$ is?",
    "fact_expressions": "G: Circle;H: Line;C:Circle;Expression(G) = (y^2 + (x - 2)^2 = 1);Expression(H) = (x + 1 = 0);IsOutTangent(G,C);IsTangent(H,C)",
    "query_expressions": "LocusEquation(Center(C))",
    "answer_expressions": "y^2=8*x",
    "fact_spans": "[[[1, 21]], [[26, 35]], [[38, 40]], [[1, 21]], [[26, 35]], [[0, 40]], [[25, 40]]]",
    "query_spans": "[[[38, 49]]]",
    "process": "From the circle $(x-2)^{2}+y^{2}=1$, we obtain the center $F(2,0)$ and radius $r=1$. Let $P(x,y)$ be the center of the desired moving circle. Draw $PM\\bot$ line $l: x+1=0$, where $M$ is the foot of the perpendicular. Then $|PF|-r=|PM|$, so $|PF|=|PM|+1$. Therefore, the point $P$ is equidistant from the fixed point $F(2,0)$ and the line $l: x=-2$, which allows us to solve the problem. [Detailed solution] From the circle $(x-2)^{2}+y^{2}=1$, we obtain the center $F(2,0)$ and radius $r=1$. Let $P(x,y)$ be the center of the desired moving circle. Draw $PM\\bot$ line $l: x+1=0$, where $M$ is the foot of the perpendicular. Then $|PF|-r=|PM|$, so $|PF|=|PM|+1$. Therefore, the locus of point $P$ consists of points equidistant from the fixed point $F(2,0)$ and the line $l: x=-2$. By the definition of a parabola, the locus of point $P$ is a parabola with focus $F(2,0)$ and directrix $l: x=-2$. Thus, the equation of the parabola is $y^{2}=8x$."
  },
  {
    "text": "Given that the center of the ellipse is at the origin, one focus is $F(-2 \\sqrt{3}, 0)$, and the major axis is twice the length of the minor axis, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;F: Point;Coordinate(F) = (-2*sqrt(3), 0);OneOf(Focus(G)) = F;Length(MajorAxis(G)) = 2*Length(MinorAxis(G))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16+y^2/4=1",
    "fact_spans": "[[[2, 4], [50, 52]], [[7, 9]], [[2, 9]], [[15, 34]], [[15, 34]], [[2, 34]], [[2, 47]]]",
    "query_spans": "[[[50, 59]]]",
    "process": ""
  },
  {
    "text": "The standard equation of a circle centered at the right focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ and tangent to its asymptotes is?",
    "fact_expressions": "G: Hyperbola;C: Circle;Expression(G) = (x^2/9 - y^2/16 = 1);Center(C) = RightFocus(G);IsTangent(Asymptote(G), C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "(x-5)^2+y^2=16",
    "fact_spans": "[[[51, 52], [1, 40]], [[58, 59]], [[1, 40]], [[0, 59]], [[50, 59]]]",
    "query_spans": "[[[58, 66]]]",
    "process": "In the hyperbola, a=3, b=4, c=5, the right focus is F(5,0), therefore r=d=4, therefore the equation of the circle is (x-5)^{2}+y^{2}=16"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and $P Q$ is a chord of the hyperbola passing through $F_{1}$ and perpendicular to the $x$-axis. If $\\angle P F_{2} Q=90^{\\circ}$, then the eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "F2: Point;F1: Point;Focus(G) = {F1, F2};G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;P: Point;Q: Point;IsChordOf(LineSegmentOf(P, Q), G);PointOnCurve(F1, LineSegmentOf(P, Q));IsPerpendicular(LineSegmentOf(P, Q), xAxis);AngleOf(P, F2, Q) = ApplyUnit(90, degree);e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "1+sqrt(2)",
    "fact_spans": "[[[10, 17]], [[2, 9], [88, 95]], [[2, 79]], [[18, 74], [104, 107], [143, 146]], [[18, 74]], [[21, 74]], [[21, 74]], [[21, 74]], [[21, 74]], [[80, 85]], [[80, 85]], [[80, 109]], [[80, 95]], [[80, 103]], [[112, 141]], [[149, 152]], [[143, 152]]]",
    "query_spans": "[[[149, 156]]]",
    "process": "\\because PQ is a chord of the hyperbola passing through F_{1} and perpendicular to the x-axis, \\angle PF_{2}Q = 90^{\\circ}, \\therefore |PF_{1}| = |F_{1}F_{2}|, \\frac{b^{2}}{a} = 2c', \\therefore e^{2} - 2e - 1 = 0, \\therefore e = 1 \\pm \\sqrt{2}, \\because e > 1, \\therefore e = 1 + \\sqrt{2}"
  },
  {
    "text": "Given the parabola $y^{2}=4x$, a line is drawn through the focus $F$ intersecting the parabola at points $A$ and $B$. If $|AF|=4$, then the coordinates of point $A$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);A: Point;B: Point;Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, F)) = 4",
    "query_expressions": "Coordinate(A)",
    "answer_expressions": "(3, pm*2*sqrt(3))",
    "fact_spans": "[[[2, 16], [27, 30]], [[2, 16]], [[20, 23]], [[2, 23]], [[24, 26]], [[17, 26]], [[32, 36], [55, 59]], [[37, 40]], [[24, 42]], [[44, 53]]]",
    "query_spans": "[[[55, 64]]]",
    "process": "As shown in the figure, $F(1,0)$, since $|AF|=4$, $\\therefore x_{A}+1=4$, solving gives $x_{A}=3$. Substituting into the parabolic equation yields $y_{A}=2\\sqrt{3}$ or $y_{A}=-2\\sqrt{3}$. Hence, the coordinates of point $A$ are $(3,2\\sqrt{3})$ or $(3,-2\\sqrt{3})$."
  },
  {
    "text": "If the focal distance of the ellipse $\\frac{x^{2}}{10-m}+\\frac{y^{2}}{m+4}=1$ is $4$, then what is the value of $m$?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/(10 - m) + y^2/(m + 4) = 1);FocalLength(G) = 4",
    "query_expressions": "m",
    "answer_expressions": "{1,5}",
    "fact_spans": "[[[2, 44]], [[53, 56]], [[2, 44]], [[2, 51]]]",
    "query_spans": "[[[53, 60]]]",
    "process": "According to the problem, if the foci are on the x-axis, then $10 - m - (m + 4) = 4$, solving gives $m = 1$; if the foci are on the y-axis, then $m + 4 - (10 - m) = 4$, solving gives $m = 5$; in conclusion, $m = 1$ or $5$."
  },
  {
    "text": "If $x^{2}+k y^{2}=2$ represents an ellipse with foci on the $y$-axis, then the range of real values for $k$ is?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (k*y^2 + x^2 = 2);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(0,1)",
    "fact_spans": "[[[30, 32]], [[34, 39]], [[2, 32]], [[21, 32]]]",
    "query_spans": "[[[34, 46]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x$ ($p>0$) with focus $F$, and let $A$ be the intersection point of the directrix and the $x$-axis. Let point $B$ be one of the intersection points of the circle centered at $F$ with radius $AF$ and the parabola $C$, and let $O$ be the origin. Denote $\\angle A B O = \\theta$. Then $\\tan \\theta =$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;Intersection(Directrix(C), xAxis) = A;A: Point;G: Circle;Center(G) = F;Radius(G) = LineSegmentOf(A,F);OneOf(Intersection(G,C)) = B;B: Point;O: Origin;AngleOf(A,B,O) = theta;theta: Number",
    "query_expressions": "Tan(theta)",
    "answer_expressions": "1/3",
    "fact_spans": "[[[2, 28], [75, 81]], [[2, 28]], [[10, 28]], [[10, 28]], [[32, 35], [57, 60]], [[2, 35]], [[2, 50]], [[47, 50]], [[73, 74]], [[56, 74]], [[64, 74]], [[51, 86]], [[51, 55]], [[87, 90]], [[97, 118]], [[120, 133]]]",
    "query_spans": "[[[120, 135]]]",
    "process": "The parabola $ C: y^{2} = 2px $ ($ p > 0 $) has focus $ F\\left(\\frac{p}{2}, 0\\right) $. Given $ |AF| = p $, the circle centered at $ F $ is $ \\left(x - \\frac{p}{2}\\right)^{2} + y^{2} = p^{2} $. Solving the system \n\\[\n\\begin{cases}\n\\left(x - \\frac{p}{2}\\right)^{2} + y^{2} = p^{2} \\\\\ny^{2} = 2px\n\\end{cases}\n\\]\nyields $ 4x^{2} + 4px - 3p^{2} = 0 $. Since $ x_{B} > 0 $, it follows that $ x_{B} = \\frac{p}{2} = x_{F'} $. Hence, $ BF \\perp AF $, and $ BF = y_{B} = p $, so $ \\angle BAF = 45^{\\circ} $, $ \\tan \\angle BOF = \\frac{BF}{OF} = \\frac{p}{\\frac{p}{2}} = 2 $. Then $ \\angle ABO = \\theta = \\angle BOF - \\angle BAF = \\angle BOF - 45^{\\circ} $, and $ \\angle ABO = \\theta = \\angle BOF - \\angle BAF = \\angle BOF - 45^{\\circ} $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ with focus $F$, a line $l$ with slope $k$ passes through point $F$ and intersects $C$ at points $A$ and $B$. If $|A B|=6|O F|$ ($O$ is the coordinate origin), then $k^{2}=$?",
    "fact_expressions": "l: Line;C: Parabola;p: Number;A: Point;B: Point;O: Origin;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;k:Number;Slope(l)=k;PointOnCurve(F,l);Intersection(l,C)={A,B};Abs(LineSegmentOf(A, B)) = 6*Abs(LineSegmentOf(O, F))",
    "query_expressions": "k^2",
    "answer_expressions": "2",
    "fact_spans": "[[[43, 48]], [[2, 28], [56, 59]], [[10, 28]], [[61, 64]], [[65, 68]], [[88, 91]], [[32, 35], [49, 53]], [[10, 28]], [[2, 28]], [[2, 35]], [[39, 42]], [[36, 48]], [[43, 53]], [[43, 70]], [[72, 87]]]",
    "query_spans": "[[[99, 108]]]",
    "process": "Let $ l: y = k\\left(x - \\frac{p}{2}\\right) $ ($ k \\neq 0 $). By combining the equation of the line and the parabola, using the definition of the parabola and the relationship between roots and coefficients, and combining with the given conditions to set up equations, solve for $ k^{2} $. [Detailed solution] From the problem, let $ l: y = k\\left(x - \\frac{p}{2}\\right) $ ($ k \\neq 0 $), $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. From $ |AB| = 6|OF| $, we get $ x_{1} + x_{2} + p = 6 \\cdot \\frac{p}{2} $, that is, $ x_{1} + x_{2} = 2p $. Substitute $ y = k\\left(x - \\frac{p}{2}\\right) $ into $ y^{2} = 2px $, rearranging gives $ k^{2}x^{2} - (pk^{2} + 2p)x + \\frac{k^{2}p^{2}}{4} = 0 $, then $ x_{1} + x_{2} = \\frac{pk^{2} + 2p}{k^{2}} = 2p $, solving yields $ k^{2} = 2 $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2px$ ($p>0$) with focus $F$, a line $l$ passing through $F$ intersects the parabola $C$ at points $A$ and $B$. If $\\overrightarrow{AF}=2\\overrightarrow{FB}$, and the ordinate of the midpoint of chord $AB$ is $\\frac{\\sqrt{2}}{2}$, then the equation of the parabola $C$ is?",
    "fact_expressions": "l: Line;C: Parabola;p: Number;A: Point;B: Point;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};VectorOf(A, F) = 2*VectorOf(F, B);YCoordinate(MidPoint(LineSegmentOf(A, B))) = sqrt(2)/2;IsChordOf(LineSegmentOf(A, B),C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[41, 46]], [[2, 28], [47, 53], [150, 156]], [[9, 28]], [[55, 58]], [[59, 62]], [[32, 35], [37, 40]], [[9, 28]], [[2, 28]], [[2, 35]], [[36, 46]], [[41, 64]], [[66, 113]], [[116, 148]], [[47, 121]]]",
    "query_spans": "[[[150, 161]]]",
    "process": "Let the line $ l $ be $ x = my + \\frac{p}{2} $, and combine it with the equation of the parabola $ C $. Eliminate $ x $, and using $ \\overrightarrow{AF} = 2\\overrightarrow{FB} $ and Vieta's formulas, derive the relationship between $ y_A $ and $ y_B $, then solve for $ p $. Let the line $ l $ passing through $ F $ be $ x = my + \\frac{p}{2} $, combined with the parabola $ C: y^2 = 2px $, we get:\n\\[\n\\begin{cases}\nx = my + \\frac{p}{2} \\\\\ny^2 = 2px\n\\end{cases}\n\\]\nEliminating $ x $, we obtain $ y^2 = 2p(my + \\frac{p}{2}) $, simplifying gives $ y^2 - 2pmy - p^2 = 0 $, thus $ y_A y_B = -p^2 $. Since $ \\overrightarrow{AF} = 2\\overrightarrow{FB} $, we have $ y_A = -2y_B $; $ y_A y_B = -p^2 = -2y_B^2 $. Given that the vertical coordinate of the midpoint of chord $ AB $ is $ \\frac{\\sqrt{2}}{2} > 0 $, therefore $ A $ lies above the $ x $-axis, $ B $ below the $ x $-axis, $ y_B = -\\frac{\\sqrt{2}}{2}p $, $ y_A = \\sqrt{2}p $. $ y_A + y_B = \\frac{\\sqrt{2}}{2}p = \\sqrt{2} $, thus $ p = 2 $, hence the equation of the parabola $ C $ is $ y^2 = 4x $."
  },
  {
    "text": "In the rectangular coordinate system $xOy$, the focus of the parabola $y^{2}=2 x$ is $F$. Let $M$ be a moving point on the parabola. Then the maximum value of $\\frac{M O}{M F}$ is?",
    "fact_expressions": "E: Parabola;F: Point;M: Point;Expression(E) = (y^2 = 2*x);Focus(E) = F;PointOnCurve(M, E);O: Origin",
    "query_expressions": "Max(LineSegmentOf(M,O)/LineSegmentOf(F,M))",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[15, 29]], [[33, 36]], [[38, 41]], [[15, 29]], [[15, 36]], [[38, 49]], [[51, 68]]]",
    "query_spans": "[[[51, 74]]]",
    "process": ""
  },
  {
    "text": "Let the upper and lower foci of the ellipse $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1(a>b>0)$ be $F_{1}$ and $F_{2}$, respectively, and the right vertex be $B$. If $|B F_{2}|=|F_{1} F_{2}|=2$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;B: Point;F2: Point;F1: Point;a > b;b > 0;Expression(G) = (x^2/b^2 + y^2/a^2 = 1);UpperFocus(G) = F1;LowerFocus(G) = F2;RightVertex(G) = B;Abs(LineSegmentOf(B,F2)) = Abs(LineSegmentOf(F1, F2));Abs(LineSegmentOf(F1, F2)) = 2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/4 + x^2/3 = 1",
    "fact_spans": "[[[1, 53], [118, 120]], [[3, 53]], [[3, 53]], [[82, 85]], [[70, 77]], [[62, 69]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 77]], [[1, 77]], [[1, 85]], [[88, 115]], [[88, 115]]]",
    "query_spans": "[[[118, 127]]]",
    "process": "From |BF₂| = |F₁F₂| = 2, we know a = 2c = 2, that is, a = 2, c = 1, b² = a² - c² = 3. Hence, the standard equation of the ellipse is \\frac{y^{2}}{4}+\\frac{x^{2}}{3}=1. Therefore, fill in \\frac{y^{2}}{4}+\\frac{x^{2}}{3}=1. [Note] This problem examines the standard equation of an ellipse. Mastering the values of each segment in an ellipse is key to solving this problem, and it is a basic question."
  },
  {
    "text": "Let $P$ be a point on the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{27}=1$, and let $F_{1}$, $F_{2}$ be the left and right foci, respectively. If $|P F_{1}|=7$, then $|P F_{2}|=$?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 - y^2/27 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(P, F1)) = 7",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "13",
    "fact_spans": "[[[5, 44]], [[1, 4]], [[48, 55]], [[56, 63]], [[5, 44]], [[1, 47]], [[5, 70]], [[5, 70]], [[72, 85]]]",
    "query_spans": "[[[88, 101]]]",
    "process": "According to the problem, the hyperbola is $\\frac{x^2}{9}-\\frac{y^{2}}{27}=1$, where $a=3$, $c=6$. Since $P$ is a point on the hyperbola, we have $||PF_{1}|-|PF_{2}||=2a=6$. Given $|PF_{1}|=7$, then $|PF_{2}|=1<c-a=3$ (discarded) or $13$."
  },
  {
    "text": "If the ellipse $m x^{2}+n y^{2}=1$ intersects the line $x+y=1$ at points $A$ and $B$, point $M$ is the midpoint of $AB$, the slope of line $OM$ is $2$ (where $O$ is the origin), and $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}=\\frac{1}{3}$, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (m*x^2 + n*y^2 = 1);n: Number;m: Number;H: Line;Expression(H) = (x + y = 1);A: Point;B: Point;Intersection(G, H) = {A, B};M: Point;MidPoint(LineSegmentOf(A,B)) = M;Slope(LineOf(O,M)) = 2;O: Origin;DotProduct(VectorOf(O, A), VectorOf(O, B)) = 1/3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "2*x^2 + y^2 = 1",
    "fact_spans": "[[[1, 22], [147, 149]], [[1, 22]], [[3, 22]], [[3, 22]], [[23, 32]], [[23, 32]], [[34, 37]], [[38, 41]], [[1, 43]], [[44, 48]], [[44, 57]], [[58, 72]], [[75, 78]], [[83, 144]]]",
    "query_spans": "[[[147, 153]]]",
    "process": "From \\begin{cases}x+y=1\\\\mx^{2}+ny^{2}=1\\end{cases}, eliminating $y$ and simplifying yields $(m+n)x^{2}-2nx+n-1=0$. Let the coordinates of points $A$ and $B$ be $(x_{1},y_{1})$, $(x_{2},y_{2})$, and the coordinates of point $M$ be $(x_{0},y_{0})$. Then, $x_{1}+x_{2}=\\frac{2n}{m+n}$, $x_{1}x_{2}=\\frac{n-1}{m+n}$. Thus, $x_{0}=\\frac{x_{1}+x_{2}}{2}=\\frac{n}{m+n}$, $y_{0}=1-x_{0}=1-\\frac{n}{m+n}=\\frac{m}{m+n}$. Therefore, the coordinates of point $M$ are $\\left(\\frac{n}{m+n},\\frac{m}{m+n}\\right)$. Hence, $k_{OM}=\\frac{y_{0}}{x_{0}}=\\frac{m}{n}=2^{n}$, so $m=2n$.\\textcircled{1} Also, $\\overrightarrow{OA}\\cdot\\overrightarrow{OB}=x_{1}x_{2}+y_{1}y_{2}=x_{1}x_{2}+(1-x_{1})(1-x_{2})=2x_{1}x_{2}-(x_{1}+x_{2})+1=\\frac{1}{3}$. Therefore, $2\\times\\frac{n-1}{m+n}-\\frac{2n}{m+n}=-\\frac{2}{3}$. Simplifying gives $m+n=3$.\\textcircled{2} From \\textcircled{1}\\textcircled{2}, solving yields $m=2$, $n=1$. Thus, the ellipse equation is $2x^{2}+y^{2}=1$. Answer: $2x^{2}+y^{2}=1$"
  },
  {
    "text": "Given that the line $y=2x-3$ passes through the focus of the parabola $C$: $y^{2}=2px$ $(p>0)$, then $p=$?",
    "fact_expressions": "G: Line;Expression(G) = (y = 2*x - 3);C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;PointOnCurve(Focus(C), G)",
    "query_expressions": "p",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 13]], [[2, 13]], [[14, 40]], [[14, 40]], [[45, 48]], [[21, 40]], [[2, 43]]]",
    "query_spans": "[[[45, 50]]]",
    "process": "Since the line $y=2x-3$ intersects the $x$-axis at the point $(\\frac{3}{2},0)$, and $y=2x-3$ passes through the focus of the parabola $C$, then $(\\frac{3}{2},0)$ is the focus of the parabola. Therefore, $\\frac{p}{2}=\\frac{3}{2}$, hence $p=3$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$, draw a line $l$ through its focus $F$ with a slope greater than $0$. The line $l$ intersects the parabola at points $M$ and $N$, and $|MF|=3|NF|$. Find the slope of the line $l$.",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l: Line;Slope(l) > 0;Intersection(l, C) = {M, N};M: Point;N: Point;Abs(LineSegmentOf(M, F)) = 3*Abs(LineSegmentOf(N, F));PointOnCurve(F, l)",
    "query_expressions": "Slope(l)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 21], [23, 24], [52, 55]], [[2, 21]], [[26, 29]], [[23, 29]], [[40, 45], [48, 51], [84, 89]], [[32, 45]], [[48, 66]], [[57, 60]], [[61, 64]], [[68, 82]], [[22, 45]]]",
    "query_spans": "[[[84, 94]]]",
    "process": "As shown in the figure: draw perpendiculars from points M and N to the directrix, with feet at P and Q respectively; draw a perpendicular from point N to PM, with foot at S. Let |NF| = k, then |MF| = 3k. By the definition of the parabola, we obtain |MP| = 3k, |NQ| = k, thus |MS| = 2k, |MN| = k + 3k = 4k, \\sin\\angleMNS=\\frac{1}{2}, hence \\angleMNS=30^{\\circ}, therefore the inclination angle of line l is 60^{\\circ}, and the slope of line l is \\sqrt{3}."
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{k-5}+\\frac{y^{2}}{3-k}=-1$ represents an ellipse, what is the range of values for $k$?",
    "fact_expressions": "G: Ellipse;Expression(G)= (x^2/(k - 5) + y^2/(3 - k) = -1) ;k:Number",
    "query_expressions": "Range(k)",
    "answer_expressions": "(3,4)+(4,5)",
    "fact_spans": "[[[46, 48]], [[2, 48]], [[50, 53]]]",
    "query_spans": "[[[50, 60]]]",
    "process": "\\frac{x^2}{k-5}+\\frac{y^{2}}{3-k}=-1 \\therefore \\frac{x^{2}}{5-k}+\\frac{y^{2}}{k-3}=1 \\therefore \\begin{cases} 5-k>0 \\\\ k-3>0 \\\\ 5-k \\neq k-3 \\end{cases} Solving the inequalities gives the range of values (3,4)\\cup(4,5)"
  },
  {
    "text": "What is the standard equation of a parabola passing through the point $(1,1)$?",
    "fact_expressions": "G: Parabola;H: Point;Coordinate(H) = (1, 1);PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=x,x^2=y}",
    "fact_spans": "[[[10, 13]], [[1, 9]], [[1, 9]], [[0, 13]]]",
    "query_spans": "[[[10, 20]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $y^{2}=8 x$ coincides with the right focus of the hyperbola $\\frac{x^{2}}{m}-y^{2}=1$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;m: Number;H: Parabola;Expression(G) = (-y^2 + x^2/m = 1);Expression(H) = (y^2 = 8*x);Focus(H) = RightFocus(G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[19, 47], [55, 58]], [[22, 47]], [[1, 15]], [[19, 47]], [[1, 15]], [[1, 53]]]",
    "query_spans": "[[[55, 64]]]",
    "process": ""
  },
  {
    "text": "The directrix of the parabola $y^{2}=-2 p x(p>0)$ passes through the right focus of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, then $p$=?",
    "fact_expressions": "G: Parabola;p: Number;H: Ellipse;p>0;Expression(G) = (y^2 = -2*p*x);Expression(H) = (x^2/9 + y^2/5 = 1);PointOnCurve(RightFocus(H), Directrix(G))",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 22]], [[70, 73]], [[27, 64]], [[3, 22]], [[0, 22]], [[27, 64]], [[0, 68]]]",
    "query_spans": "[[[70, 75]]]",
    "process": "From the ellipse equation, its right focus is (2,0). Since the parabola's directrix passes through the right focus of the ellipse, $\\frac{p}{2}=2$, solving gives $p=4$."
  },
  {
    "text": "The standard equation of an ellipse passing through the point $(2, -3)$ and sharing the same foci as the ellipse $9x^{2}+4y^{2}=36$ is?",
    "fact_expressions": "G: Ellipse;H: Point;C:Ellipse;Expression(G) = (9*x^2 + 4*y^2 = 36);Coordinate(H) = (2,-3);PointOnCurve(H, C);Focus(G)=Focus(C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/15 + x^2/10 = 1",
    "fact_spans": "[[[13, 35]], [[1, 11]], [[42, 44]], [[13, 35]], [[1, 11]], [[0, 44]], [[12, 44]]]",
    "query_spans": "[[[42, 51]]]",
    "process": ""
  },
  {
    "text": "Given that the vertex of parabola $C_{1}$ is at the origin and its directrix is $x=-3$, and circle $C_{2}$: $(x-3)^{2}+y^{2}=1$. A line $l$ passing through the center of circle $C_{2}$ intersects parabola $C_{1}$ at points $A$, $B$, and intersects circle $C_{2}$ at points $M$, $N$, with $|A M|<|A N|$. Then the minimum value of $|A M|+\\frac{1}{4}| B M |$ is?",
    "fact_expressions": "l: Line;C1: Parabola;C2: Circle;A: Point;M: Point;N: Point;B: Point;O: Origin;Expression(Directrix(C1))=(x=-3);Expression(C2)=((x-3)^2+y^2=1);Vertex(C1) = O;Center(C2)=C;C:Point;PointOnCurve(C,l);Intersection(l,C1)={A,B};Intersection(l,C2) = {M,N};Abs(LineSegmentOf(A, M)) < Abs(LineSegmentOf(A, N))",
    "query_expressions": "Min(Abs(LineSegmentOf(A, M)) + Abs(LineSegmentOf(B, M))/4)",
    "answer_expressions": "6",
    "fact_spans": "[[[71, 76], [102, 105]], [[2, 12], [77, 87]], [[31, 59], [106, 114]], [[89, 93]], [[116, 120]], [[121, 124]], [[96, 99]], [[16, 20]], [[2, 30]], [[31, 59]], [[2, 20]], [[31, 70]], [[63, 70]], [[60, 76]], [[71, 100]], [[102, 124]], [[126, 139]]]",
    "query_spans": "[[[141, 173]]]",
    "process": "Let the equation of the parabola be: $ y^{2} = 2px $ ($ p > 0 $). From the directrix equation $ x = -3 $, we get $ \\frac{p}{2} = 3 $, so $ p = 6 $. The standard equation of the parabola is $ y^{2} = 12x $, and the focus has coordinates $ F(3,0) $. The circle $ C_{2}: (x-3)^{2} + y^{2} = 1 $ has center $ (3,0) $ and radius 1. Since line $ AB $ passes through the focus of the parabola, using polar coordinates, we can set $ A(\\rho_{1},\\theta) $, $ B(\\rho_{2},\\pi+\\theta) $. From $ \\rho = \\frac{p}{1 - \\cos\\theta} $, we obtain $ \\frac{1}{|AF|} + \\frac{1}{|BF|} = \\frac{1 - \\cos\\theta}{p} + \\frac{1 + \\cos\\theta}{p} = \\frac{2}{p} = \\frac{1}{3} $. Then $ |AM| + \\frac{1}{4}|BM| = |AF| \\cdot 1 + \\frac{1}{4}(|BF| + 1) = |AF| + \\frac{1}{4}|BF| - \\frac{3}{4} = 3\\left( \\frac{1}{|AF|} + \\frac{1}{|BF|} \\right)\\left( |AF| + \\frac{1}{4}|BF| \\right) - \\frac{3}{4} = 3\\left( \\frac{5}{4} + 2\\sqrt{\\frac{1}{4}} \\right) - \\frac{3}{4} = 6 $, with equality if and only when $ |BF| = 2|AF| = 9 $. Therefore, the minimum value of $ |AM| + \\frac{1}{4}|BM| $ is 6."
  },
  {
    "text": "Given that the focal distance of the hyperbola $\\frac{y^{2}}{m}-\\frac{x^{2}}{9}=1$ is $4 \\sqrt{3}$, find the value of the real number $m$.",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/9 + y^2/m = 1);m: Real;FocalLength(G) = 4*sqrt(3)",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 40]], [[2, 40]], [[58, 63]], [[2, 56]]]",
    "query_spans": "[[[58, 67]]]",
    "process": "Since $ m > 0 $ by the given condition, we have $ c = \\sqrt{m + 9} = \\frac{4\\sqrt{3}}{3} $, solving which gives $ m = 3 $."
  },
  {
    "text": "Given a point $P$ on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the distances from $P$ to the two asymptotes are $d_{1}$ and $d_{2}$ respectively. If $d_{1} d_{2}=\\frac{2}{5} a b$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;L1: Line;L2: Line;d1: Number;d2: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);Asymptote(G) = {L1, L2};Distance(P, L1) = d1;Distance(P, L2) = d2;d1*d2 = (2/5)*a*b",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{sqrt(5), sqrt(5)/2}",
    "fact_spans": "[[[2, 58], [124, 127]], [[5, 58]], [[5, 58]], [[61, 64]], [], [], [[75, 82]], [[84, 91]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 64]], [[2, 69]], [[2, 91]], [[2, 91]], [[93, 122]]]",
    "query_spans": "[[[124, 133]]]",
    "process": "The equations of the two asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ are $bx-ay=0$ or $bx+ay=0$. The product of the distances from point $P(x_{0},y_{0})$ to the two asymptotes is $\\frac{|bx_{0}-ay_{0}|}{\\sqrt{a^{2}+b^{2}}} \\cdot \\frac{|bx_{0}+ay_{0}|}{\\sqrt{a^{2}+b^{2}}} = \\frac{2}{5}ab$, that is, $\\frac{|b^{2}x_{0}^{2}-a^{2}y_{0}^{2}|}{a^{2}+b^{2}} = \\frac{2}{5}ab$. Since point $P(x_{0},y_{0})$ satisfies the equation of the hyperbola, $\\therefore b^{2}x_{0}^{2}-a^{2}y_{0}^{2}=a^{2}b^{2}$, $\\therefore \\frac{b^{2}a^{2}}{a^{2}+b^{2}} = \\frac{2}{5}ab$, i.e., $2a^{2}+2b^{2}=5ab$, $\\therefore b=2a$ or $b=\\frac{1}{2}a$, then $c=\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{5}$ or $\\frac{\\sqrt{5}}{2}$. Hence fill in $\\sqrt{5}$ or $\\frac{\\sqrt{5}}{2}$."
  },
  {
    "text": "Given that the foci of ellipse $C$ are $F_{1}(-1,0)$, $F_{2}(1,0)$, a line passing through $F_{2}$ intersects $C$ at points $A$ and $B$. If $|A F_{2}|=2|F_{2} B|$ and $|A B|=|B F_{1}|$, then the equation of $C$ is?",
    "fact_expressions": "C: Ellipse;F1: Point;F2: Point;Coordinate(F1) = (-1, 0);Coordinate(F2) = (1, 0);Focus(C) = {F1, F2};G: Line;PointOnCurve(F2, G);A: Point;B: Point;Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, F2)) = 2*Abs(LineSegmentOf(F2, B));Abs(LineSegmentOf(A, B)) = Abs(LineSegmentOf(B, F1))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/3+y^2/2=1",
    "fact_spans": "[[[2, 7], [51, 54], [107, 110]], [[11, 24]], [[26, 38], [40, 47]], [[11, 24]], [[26, 38]], [[2, 38]], [[48, 50]], [[39, 50]], [[55, 58]], [[59, 62]], [[48, 62]], [[64, 86]], [[88, 105]]]",
    "query_spans": "[[[107, 115]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the points on the parabola $x^{2}=4 y$ whose distance to the focus is equal to $6$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);P: Point;PointOnCurve(P, G);Distance(P, Focus(G)) = 6",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(pm*2*sqrt(5), 5)",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[27, 28]], [[0, 28]], [[0, 28]]]",
    "query_spans": "[[[27, 33]]]",
    "process": "The parabola has focus at (0,1) and directrix y = -1. When the distance from the point to the focus is 6, we have y + 1 = 6, so y = 5, and solving gives x = \\pm2\\sqrt{5}."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. If a point $P$ on the ellipse satisfies $|P F_{2}|=|F_{1} F_{2}|$, and the circle centered at the origin $O$ with radius $b$ has a common point with the line $P F_{1}$, then the range of the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Circle;F1: Point;P: Point;F2: Point;O:Origin;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P,G);Abs(LineSegmentOf(P,F2))=Abs(LineSegmentOf(F1,F2));Center(H)=O;Radius(H)=b;IsIntersect(LineOf(P,F1),H);Eccentricity(G)=e;e:Number",
    "query_expressions": "Range(e)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[17, 69], [77, 79], [152, 154]], [[125, 128]], [[19, 69]], [[132, 133]], [[1, 8]], [[82, 85]], [[9, 16]], [[115, 120]], [[19, 69]], [[19, 69]], [[17, 69]], [[1, 74]], [[1, 74]], [[77, 85]], [[87, 113]], [[114, 133]], [[124, 133]], [[132, 149]], [[152, 160]], [[157, 160]]]",
    "query_spans": "[[[157, 167]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ has left and right foci $F_{1}$, $F_{2}$ respectively, and $P$ is a point on the right branch such that $|\\overrightarrow{P F_{1}}|=6$, $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$. Then the asymptotes of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/b^2 = 1);b: Number;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, RightPart(G));Abs(VectorOf(P, F1)) = 6;DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(6)/2)*x",
    "fact_spans": "[[[0, 42], [170, 173]], [[0, 42]], [[3, 42]], [[50, 57]], [[58, 65]], [[0, 65]], [[0, 65]], [[66, 69]], [[0, 75]], [[77, 107]], [[109, 168]]]",
    "query_spans": "[[[170, 178]]]",
    "process": "From the given condition, |\\overrightarrow{PF_{2}}|=2, and by the focal triangle area formula, S=b^{2}\\cot45^{\\circ}=\\frac{1}{2}\\times PF_{1}\\times PF_{2}=6, so b^{2}=6, and the asymptotes are y=\\pm\\frac{\\sqrt{6}}{2}x"
  },
  {
    "text": "From a point $A$ on the graph of the parabola $y^{2}=4x$, draw a perpendicular to the directrix of the parabola, with foot at $B$, and $|AB|=5$. Let $F$ be the focus of the parabola. Then the area of $\\triangle ABF$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;PointOnCurve(A, G);Z: Line;PointOnCurve(A, Z);IsPerpendicular(Z, Directrix(G));B: Point;FootPoint(Z, Directrix(G)) = B;Abs(LineSegmentOf(A, B)) = 5;F: Point;Focus(G) = F",
    "query_expressions": "Area(TriangleOf(A, B, F))",
    "answer_expressions": "10",
    "fact_spans": "[[[1, 15], [24, 27], [56, 59]], [[1, 15]], [[20, 23]], [[1, 23]], [], [[0, 32]], [[0, 32]], [[36, 39]], [[0, 39]], [[41, 50]], [[52, 55]], [[52, 62]]]",
    "query_spans": "[[[64, 86]]]",
    "process": ""
  },
  {
    "text": "If the distance from point $P$ to the line $y=-1$ is $2$ less than its distance to the point $(0,3)$, then the equation of the trajectory of point $P$ is?",
    "fact_expressions": "G: Line;H: Point;P: Point;Expression(G) = (y = -1);Coordinate(H) = (0, 3);Distance(P, G) = Distance(P, H) - 2",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2 = 12*y",
    "fact_spans": "[[[6, 14]], [[20, 28]], [[1, 5], [18, 19], [37, 41]], [[6, 14]], [[20, 28]], [[1, 35]]]",
    "query_spans": "[[[37, 48]]]",
    "process": "\\because the distance from point P to the line y=-1 is 2 less than its distance to the point (0,3). \\therefore the distance from point P to the line y=-3 is equal to its distance to the point (0,3). \\therefore the locus of point P is a parabola with focus (0,3) and directrix l: y=-3. Therefore, let the equation of the locus of P be x^{2}=2py, (p>0), we obtain \\frac{1}{2}p=3, solving gives p=6, 2p=12, \\therefore the equation of the locus of moving point P is x^{2}=12y."
  },
  {
    "text": "It is known that a hyperbola passes through the point $(2 \\sqrt{2}, 1)$, and one of its asymptotes has the equation $y=\\frac{1}{2} x$. Then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (2*sqrt(2), 1);PointOnCurve(H, G);Expression(OneOf(Asymptote(G))) = (y = x/2)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4-y^2=1",
    "fact_spans": "[[[2, 5], [26, 27], [55, 58]], [[7, 25]], [[7, 25]], [[2, 25]], [[26, 52]]]",
    "query_spans": "[[[55, 65]]]",
    "process": "Let the hyperbola equation be: $mx^{2}+ny^{2}=1$ ($mn<0$). From the given conditions, we have:\n\\[\n\\begin{cases}\n8m+n=1 \\\\\n\\sqrt{-\\frac{m}{n}}=\\frac{1}{2}\n\\end{cases}\n\\]\nSolving gives:\n\\[\n\\begin{cases}\nm=\\frac{1}{4} \\\\\nn=1\n\\end{cases}\n\\]\nThe equation of the hyperbola is: $x^{2}-y^{2}=1$."
  },
  {
    "text": "What is the focal distance of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/5 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "6",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 43]]]",
    "process": "In the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, $a=2$, $b=\\sqrt{5}$, then $c=\\sqrt{a^{2}+b^{2}}=3$. Therefore, the focal distance of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ is $2c=6$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{5}-\\frac{y^{2}}{m}=1$ has eccentricity $e=\\frac{\\sqrt {10}}{5}$, then the value of $m$ is?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/5 - y^2/m = 1);Eccentricity(G) = e;e=sqrt(10)/5;e:Number",
    "query_expressions": "m",
    "answer_expressions": "{-3,-25/3}",
    "fact_spans": "[[[2, 39]], [[69, 72]], [[2, 39]], [[2, 67]], [[43, 67]], [[43, 67]]]",
    "query_spans": "[[[69, 76]]]",
    "process": ""
  },
  {
    "text": "The distance from the focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1$ to its asymptote is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/4 = 1)",
    "query_expressions": "Distance(Focus(G), Asymptote(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 50]]]",
    "process": "According to the problem, the right focus of the hyperbola is $ F(\\sqrt{13},0) $, and the equation of one of its asymptotes is $ y = \\frac{2}{3}x \\Rightarrow 2x - 3y = 0 $. Therefore, the distance from the focus to the asymptote is $ d = \\frac{2\\sqrt{13}}{\\sqrt{2^{2} + 3^{2}}} = 2 $."
  },
  {
    "text": "Given that on the circle $C$: $x^{2}+y^{2}-10 y+16=0$, there are exactly three points whose distance to an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $1$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "C: Circle;Expression(C) = (-10*y + x^2 + y^2 + 16 = 0);A: Point;B: Point;D: Point;PointOnCurve(A, C);PointOnCurve(B, C);PointOnCurve(D, C);Distance(A, OneOf(Asymptote(G))) = 1;Distance(B, OneOf(Asymptote(G))) = 1;Distance(D, OneOf(Asymptote(G))) = 1;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/2",
    "fact_spans": "[[[2, 31]], [[2, 31]], [], [], [], [[2, 39]], [[2, 39]], [[2, 39]], [[2, 99]], [[2, 99]], [[2, 99]], [[40, 86], [102, 105]], [[40, 86]], [[43, 86]], [[43, 86]]]",
    "query_spans": "[[[102, 111]]]",
    "process": "The equation of circle $ C $ can be rewritten as $ x^{2}+(y-5)^{2}=3^{2} $, so the center is $ (0,5) $ and the radius $ r=3 $. Since there are exactly three points on circle $ C $ whose distance to one asymptote of the hyperbola $ \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 $ is 1, the distance from the center to the asymptote is 2. Without loss of generality, let one asymptote of the hyperbola be $ y=\\frac{b}{a}x $, i.e., $ bx-ay=0 $. By the point-to-line distance formula, we have $ \\frac{|-5a|}{\\sqrt{a^{2}+b^{2}}}=\\frac{5a}{c}=2 $, $ e=\\frac{c}{a}=\\frac{5}{2} $"
  },
  {
    "text": "Given that point $P$ is a moving point on the parabola $y^{2}=4x$, and point $Q$ is a moving point on the circle $x^{2}+(y-4)^{2}=1$. When the sum of the distance from point $P$ to point $Q$ and the distance from point $P$ to the directrix of the parabola is minimized, what is the horizontal coordinate of point $P$?",
    "fact_expressions": "G: Parabola;H: Circle;P: Point;Q: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (x^2 + (y - 4)^2 = 1);PointOnCurve(P, G);PointOnCurve(Q, H);WhenMin(Distance(P, Q) + Distance(P, Directrix(G)))",
    "query_expressions": "XCoordinate(P)",
    "answer_expressions": "(9 - sqrt(17))/8",
    "fact_spans": "[[[7, 21], [76, 79]], [[31, 51]], [[2, 6], [58, 62], [71, 75], [91, 95]], [[27, 30], [63, 67]], [[7, 21]], [[31, 51]], [[2, 26]], [[27, 56]], [[57, 90]]]",
    "query_spans": "[[[91, 101]]]",
    "process": "According to the definition of a parabola, the distance from point P to the directrix of the parabola $ y^{2} = 4x $ equals its distance to the focus $ F(1,0) $. Therefore, the sum of the distance from point P to point Q and the distance from point P to the directrix of the parabola equals the sum of $ |PF| $ and the distance from P to the center $ C(0,4) $ of the circle $ x^{2} + (y - 4)^{2} = 1 $, minus the radius 1. The equation of line FC is $ y = -4x + 4 $. From the system of equations \n\\[\n\\begin{cases}\ny = -4 \\\\\ny = 4x\n\\end{cases}\n\\]\nwe obtain $ x = \\frac{9 - \\sqrt{17}}{8} $."
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(3/4)*x",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 47]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the left focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, $A(1,4)$, and $P$ is a moving point on the right branch of the hyperbola, then the minimum value of $|PF|+|PA|$ is?",
    "fact_expressions": "G: Hyperbola;A: Point;P: Point;F: Point;Expression(G) = (x^2/4 - y^2/12 = 1);Coordinate(A) = (1, 4);LeftFocus(G) = F;PointOnCurve(P, RightPart(G))",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "9",
    "fact_spans": "[[[6, 45], [64, 67]], [[50, 58]], [[60, 63]], [[2, 5]], [[6, 45]], [[50, 58]], [[2, 49]], [[60, 73]]]",
    "query_spans": "[[[75, 92]]]",
    "process": ""
  },
  {
    "text": "Given $A(3,0)$, if point $P$ is any point on the parabola $y^{2}=8x$, and point $Q$ is any point on the circle $(x-2)^{2}+y^{2}=1$, then the minimum value of $\\frac{|PA|^{2}}{|PQ|}$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (3, 0);P: Point;PointOnCurve(P, G) = True;G: Parabola;Expression(G) = (y^2 = 8*x);H: Circle;Expression(H) = (y^2 + (x - 2)^2 = 1);Q: Point;PointOnCurve(Q, H) = True",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A))^2/Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "4*sqrt(3)-4",
    "fact_spans": "[[[2, 10]], [[2, 10]], [[12, 16]], [[12, 37]], [[17, 31]], [[17, 31]], [[43, 63]], [[43, 63]], [[38, 42]], [[38, 68]]]",
    "query_spans": "[[[70, 100]]]",
    "process": "By the given conditions, the parabola $ y^{2} = 8x $ has focus $ F(2,0) $ and directrix $ x = -2 $. Since point $ P $ is a point on the parabola and point $ Q $ is any point on the circle $ (x-2)^{2} + y^{2} = 1 $, it follows that $ |PQ|_{\\max} = |PF| + 1 $. Therefore, $ \\frac{|PA|^{2}}{|PQ|} \\geqslant \\frac{|PA|^{2}}{|PF|+1} $. Let $ t = |PF| + 1 $, and let the coordinates of point $ P $ be $ (x_{P}, y_{P}) $. Then $ x_{P} = |PF| - 2 = t - 3 $, so $ |PA|^{2} = (x_{P}-3)^{2} + y_{P}^{2} = (x_{P}-3)^{2} + 8x_{P} = (t-3-3)^{2} + 8(t-3) = t^{2} - 4t + 12 $. Thus, $ \\frac{|PA|^{2}}{|PQ|} = \\frac{t^{2} - 4t + 12}{t} = t + \\frac{12}{t} - 4 \\geqslant 2\\sqrt{t \\cdot \\frac{12}{t}} - 4 = 4\\sqrt{3} - 4 $, with equality holding if and only if $ t = \\frac{12}{t} $, i.e., when $ t = 2\\sqrt{3} $. The minimum value of $ \\frac{|PA|^{2}}{|PQ|} $ is $ 4\\sqrt{3} - 4 $."
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ is $3x+2y=0$, then $b=$?",
    "fact_expressions": "G: Hyperbola;b: Number;b>0;Expression(G) = (x^2/4 - y^2/b^2 = 1);Expression(OneOf(Asymptote(G))) = (3*x + 2*y = 0)",
    "query_expressions": "b",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 49]], [[71, 74]], [[5, 49]], [[2, 49]], [[2, 69]]]",
    "query_spans": "[[[71, 76]]]",
    "process": "Analysis: First write the asymptote equations of the hyperbola, then solve for the value of b. From the problem, the asymptote equations of the hyperbola are $ y = -\\frac{b}{2}x $, $\\therefore -\\frac{b}{2} = -\\frac{3}{2}$, $\\therefore b = 3$"
  },
  {
    "text": "Given point $A(4,0)$ and the focus $F$ of the parabola $y^{2}=4x$, if a point $P$ on the parabola satisfies $|P A|=2|P F|$, then $|P F|=$?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (4, 0);PointOnCurve(P, G);Focus(G)=F;Abs(LineSegmentOf(P,A))=2*Abs(LineSegmentOf(P,F))",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "2*sqrt(2-1)",
    "fact_spans": "[[[12, 26], [34, 37]], [[2, 11]], [[39, 43]], [[29, 32]], [[12, 26]], [[2, 11]], [[34, 43]], [[12, 32]], [[45, 59]]]",
    "query_spans": "[[[61, 70]]]",
    "process": "Let $ P(x_{0},y_{0}) $, then $ y_{0}^{2}=4x_{0} $, $ |PF|=x_{0}+1 $. Since the focus of the parabola $ y^{2}=4x $ is $ F $, point $ A(4,0) $, and $ |PA|=2|PF| $, $ \\sqrt{(x_{0}-4)^{2}+y_{0}^{2}}=2(x_{0}+1) $, that is, $ 4(x_{0}+1)^{2}=(x_{0}-4)^{2}+4x_{0} $, $ x_{0}>0 $, $ \\therefore x_{0}=-2+2\\sqrt{2} $, $ \\therefore |PF|=2\\sqrt{2}-2+1=2\\sqrt{2} $."
  },
  {
    "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{m}=1$ are given by $y=\\pm \\frac{\\sqrt{5}}{3} x$, then the distance from the focus $F$ of the hyperbola to an asymptote is?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/9 - y^2/m = 1);Expression(Asymptote(G)) = (y = pm*x*(sqrt(5)/3));F: Point;Focus(G) = F",
    "query_expressions": "Distance(F, Asymptote(G))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 39], [76, 79]], [[4, 39]], [[1, 39]], [[1, 74]], [[81, 84]], [[76, 84]]]",
    "query_spans": "[[[76, 93]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ is an arbitrary point on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and points $F_{1}$, $F_{2}$ are the left and right foci of the ellipse respectively, then the maximum value of $\\sin \\angle P F_{1} F_{2}+\\sin \\angle P F_{2} F_{1}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2",
    "query_expressions": "Max(Sin(AngleOf(P, F1, F2)) + Sin(AngleOf(P, F2, F1)))",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[7, 34], [61, 63]], [[2, 6]], [[41, 49]], [[50, 57]], [[7, 34]], [[2, 40]], [[41, 69]], [[41, 69]]]",
    "query_spans": "[[[71, 130]]]",
    "process": "Using the Law of Sines, express $\\sin\\angle PF_{2}F_{1} + \\sin\\angle PF_{1}F_{2} = \\frac{4}{t}$, then find $t$, and then use $t = \\frac{|F_{1}F_{2}|}{\\sin\\angle F_{1}PF_{2}}$ to find the maximum value of $\\sin\\angle F_{1}PF_{2}$. In $\\triangle PF_{1}F_{2}$, by the Law of Sines, we have $\\frac{|PF_{2}|}{\\sin\\angle PF_{1}F_{2}} = \\frac{|PF_{1}|}{\\sin\\angle PF_{2}F_{1}} = \\frac{|F_{1}F_{2}|}{\\sin\\angle F_{1}PF_{2}} = t$, so $\\frac{|PF_{1}|}{t} = \\sin\\angle PF_{2}F_{1}$, $\\frac{|PF_{2}|}{t} = \\sin\\angle PF_{1}F_{2}$, that is, find $t$'s minimum value. And $\\frac{|F_{1}F_{2}|}{\\sin\\angle F_{1}PF_{2}} = \\frac{2\\sqrt{3}}{\\sin\\angle F_{1}PF_{2}} = t$, that is, when $\\sin\\angle F_{1}PF_{2}$ is maximum. By the property of the ellipse, $\\angle F_{1}PF_{2}$ is maximum when $P$ is at the top vertex of the ellipse. At this time, $a=2$, $b=1$, $c=\\sqrt{3}$, so $\\angle F_{1}PF_{2} = 120^{\\circ}$, thus the maximum value of $\\sin\\angle F_{1}PF_{2}$ is $1$, $t = \\frac{2\\sqrt{3}}{1} = 2\\sqrt{3}$, so $\\sin\\angle PF_{2}F_{1} + \\sin\\angle PF_{1}F_{2} = \\frac{4}{2\\sqrt{3}} = \\frac{2\\sqrt{3}}{3}$."
  },
  {
    "text": "A point $M$ on the curve $y^{2}=4 \\sqrt{2} x$ is at a distance of $4 \\sqrt{2}$ from its focus $F$, and $O$ is the coordinate origin. Then the area of $\\triangle M F O$ is?",
    "fact_expressions": "G: Curve;M: Point;F: Point;O: Origin;Expression(G) = (y^2 = 4*(sqrt(2)*x));PointOnCurve(M, G);Focus(G) = F;Distance(M, F) = 4*sqrt(2)",
    "query_expressions": "Area(TriangleOf(M, F, O))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[0, 22], [29, 30]], [[25, 28]], [[33, 36]], [[54, 58]], [[0, 22]], [[0, 28]], [[29, 36]], [[25, 52]]]",
    "query_spans": "[[[65, 87]]]",
    "process": "Analysis: Find the focus coordinates of the parabola, and use the definition of the parabola to transform and solve. The focus coordinates of $ y^{2}=4\\sqrt{2}x $ are $ (\\sqrt{2},0) $. For a point $ M $ on the curve $ y^{2}=4\\sqrt{2}x $ whose distance to its focus $ F $ is 4, the horizontal coordinate of $ M $ is $ 3\\sqrt{2} $, and the vertical coordinate is $ \\pm2\\sqrt{6} $. Let $ O $ be the origin; then the area of $ \\triangle MFO $ is $ \\frac{1}{2}\\times\\sqrt{2}\\times2\\sqrt{6}=2\\sqrt{3} $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with eccentricity $\\frac{\\sqrt{3}}{2}$, intersecting with a line passing through the right focus $F$ and having slope $k$ $(k>0)$ at points $A$ and $B$. If $\\overrightarrow{A F}=3 \\overrightarrow{F B}$, then $k=$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Eccentricity(C) = sqrt(3)/2;F: Point;RightFocus(C) = F;G: Line;PointOnCurve(F,G) = True;Slope(G) = k;k: Number;k>0;Intersection(G,C) = {A,B};A: Point;B: Point;VectorOf(A, F) = 3*VectorOf(F, B)",
    "query_expressions": "k",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 59]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[2, 84]], [[90, 93]], [[2, 93]], [[106, 108]], [[86, 108]], [[94, 108]], [[172, 175], [97, 105]], [[97, 105]], [[2, 122]], [[111, 114]], [[117, 120]], [[125, 170]]]",
    "query_spans": "[[[172, 177]]]",
    "process": ""
  },
  {
    "text": "Given that $x$, $y$ satisfy $2 x^{2}+3 y^{2}=6$, then the maximum value of $2 x+3 y$ is?",
    "fact_expressions": "x_: Number;y_: Number;2*x_^2 + 3*y_^2 = 6",
    "query_expressions": "Max(2*x_ + 3*y_)",
    "answer_expressions": "sqrt(30)",
    "fact_spans": "[[[2, 6]], [[8, 11]], [[13, 32]]]",
    "query_spans": "[[[34, 49]]]",
    "process": "Let t = 2x + 3y. Combining the system \\begin{cases} t = 2x + 3y \\\\ 2x^{2} + 3y^{2} = 6 \\end{cases}, eliminate y to obtain a quadratic equation in x, use \\Delta \\geqslant 0 to find the range of t, thus obtaining the result. Let t = 2x + 3y. Combining the system \\begin{cases} t = 2x + 3y \\\\ 2x^{2} + 3y^{2} = 6 \\end{cases}, eliminating y yields 10x^{2} - 4tx + t^{2} - 18 = 0. Then \\Delta = 16t^{2} - 4 \\times 10 \\times (t^{2} - 18) = 24 \\times (30 - t^{2}) \\geqslant 0, solving gives -\\sqrt{30} \\leqslant t \\leqslant \\sqrt{30}. Therefore, the maximum value of 2x + 3y is \\sqrt{30}."
  },
  {
    "text": "Given the parabola equation $x=2 y^{2}$, what are the coordinates of its focus?",
    "fact_expressions": "G: Parabola;Expression(G) = (x = 2*y^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(1/8,0)",
    "fact_spans": "[[[2, 5], [19, 20]], [[2, 18]]]",
    "query_spans": "[[[19, 26]]]",
    "process": "From $x=2y^2$, we obtain the standard equation of the parabola as $y^{2}=\\frac{1}{2}x$, from which we can see that its focus coordinates are $(\\frac{1}{8},0)$."
  },
  {
    "text": "Given a point $P$ on the parabola $C$: $x^{2}=2 p y(p>0)$, the distance from $P$ to the $x$-axis is 3 less than the distance from $P$ to the focus. Find $p=?$",
    "fact_expressions": "C: Parabola;p: Number;p>0;P:Point;Expression(C) = (x^2 = 2*(p*y));PointOnCurve(P,C);Distance(P, xAxis) = Distance(P,Focus(C)) -3",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 28]], [[56, 59]], [[9, 28]], [[31, 34], [43, 44]], [[2, 28]], [[2, 34]], [[2, 54]]]",
    "query_spans": "[[[56, 61]]]",
    "process": "According to the definition of a parabola, the distance from point P to the focus is equal to the distance from point P to the directrix. Setting up the equation based on this definition, we know that the distance from point P to the focus equals the distance to the directrix $ y = -\\frac{p}{2} $. Therefore, $ \\frac{p}{2} = 3 $, yielding $ p = 6 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}(-c, 0)$ and $F_{2}(c, 0)$ respectively, and the point $N(-c, \\frac{3 b^{2}}{2 a})$. If for any point $M$ on the left branch of the hyperbola $C$, the inequality $|M F_{2}|+|M N|>4 b$ holds, then the range of the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;LeftFocus(C) = F1;RightFocus(C) = F2;F1: Point;F2: Point;Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);c: Number;N: Point;Coordinate(N) = (-c, (3*b^2)/(2*a));M: Point;PointOnCurve(M, LeftPart(C)) = True;Abs(LineSegmentOf(M, F2)) + Abs(LineSegmentOf(M, N)) > 4*b",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(1, sqrt(13)/3) + (sqrt(5), +oo)",
    "fact_spans": "[[[2, 63], [135, 141], [178, 184]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 102]], [[2, 102]], [[72, 86]], [[89, 102]], [[72, 86]], [[89, 102]], [[89, 102]], [[104, 133]], [[104, 133]], [[149, 152]], [[135, 152]], [[155, 176]]]",
    "query_spans": "[[[178, 195]]]",
    "process": ""
  },
  {
    "text": "If $AB$ is a moving chord of the parabola $y^{2}= 2 p x$ $(p>0)$, and $|AB|=a$ $(a>2 p)$, then the shortest distance from the midpoint $M$ of $AB$ to the $y$-axis is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;A: Point;B: Point;M: Point;a:Number;a>2*p;IsChordOf(LineSegmentOf(A,B),G);Abs(LineSegmentOf(A, B)) = a;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Min(Distance(M, yAxis))",
    "answer_expressions": "a/2 - p/2",
    "fact_spans": "[[[7, 29]], [[7, 29]], [[10, 29]], [[10, 29]], [[1, 6]], [[1, 6]], [[60, 63]], [[34, 50]], [[34, 50]], [[1, 32]], [[34, 50]], [[52, 63]]]",
    "query_spans": "[[[60, 75]]]",
    "process": ""
  },
  {
    "text": "The line $3x + 5y - 1 = 0$ intersects the ellipse $C$: $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1$ $(a > b > 0)$ at points $M$ and $N$. Let $P$ be the midpoint of $MN$, and suppose the slope of the line $OP$ is $\\frac{3}{5}$, where $O$ is the origin. What is the eccentricity of the ellipse $C$?",
    "fact_expressions": "C: Ellipse;a: Number;b: Number;G: Line;O: Origin;P: Point;M: Point;N: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (3*x + 5*y - 1 = 0);Intersection(G, C) = {M, N};MidPoint(LineSegmentOf(M, N)) = P;Slope(LineOf(O, P)) = 3/5",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "4/5",
    "fact_spans": "[[[16, 73], [135, 140]], [[23, 73]], [[23, 73]], [[0, 15]], [[125, 128]], [[93, 96]], [[74, 77]], [[78, 81]], [[23, 73]], [[23, 73]], [[16, 73]], [[0, 15]], [[0, 83]], [[84, 96]], [[97, 122]]]",
    "query_spans": "[[[135, 145]]]",
    "process": "From \\begin{cases}3x+5y-1=0\\\\\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1\\end{cases}, we obtain (25b^{2}+9a^{2})x^{2}+6a^{2}x+a^{2}-25a^{2}b^{2}=0. By Vieta's formulas, we have: x_{1}+x_{2}= . Thus, the midpoint of MN is P(-\\frac{3a^{2}}{25b^{2}+9a^{2}}, ). Since the slope of line OP equals \\frac{3}{5}, we have k_{OP}=\\frac{5b^{2}}{3a^{2}}=\\frac{3}{5}. Solving gives \\frac{b^{2}}{a^{2}}=\\frac{9}{25}. Therefore, e=\\frac{c}{a}=\\sqrt{1-\\frac{b^{2}}{a^{2}}}=\\frac{4}{5}."
  },
  {
    "text": "The number of intersection points between the line $y=x+3$ and the curve $\\frac{y^{2}}{9}-\\frac{x|x|}{4}=1$?",
    "fact_expressions": "G: Line;H: Curve;Expression(G) = (y = x + 3);Expression(H) = (-x*Abs(x)/4 + y^2/9 = 1)",
    "query_expressions": "NumIntersection(G, H)",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 9]], [[10, 46]], [[0, 9]], [[10, 46]]]",
    "query_spans": "[[[0, 53]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of a hyperbola are $y=\\pm \\frac{4}{3} x$ and the focal distance is $20$, then the standard equation of the hyperbola is?",
    "fact_expressions": "E: Hyperbola;Expression(Asymptote(E)) = (y = pm*4*1/3*x);FocalLength(E) = 20",
    "query_expressions": "Expression(E)",
    "answer_expressions": "{x**2/36-y**2/64=1, y**2/64-x**2/36=1}",
    "fact_spans": "[[[2, 5], [49, 52]], [[2, 34]], [[2, 46]]]",
    "query_spans": "[[[49, 59]]]",
    "process": ""
  },
  {
    "text": "The focal distance of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ is? The coordinates of the foci are?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/9 = 1)",
    "query_expressions": "FocalLength(G);Coordinate(Focus(G))",
    "answer_expressions": "2*sqrt(7)\n(pm*2*sqrt(7),0)",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 43]], [[0, 49]]]",
    "process": ""
  },
  {
    "text": "Given the equation of the hyperbola is $\\frac{x^{2}}{20}-\\frac{y^{2}}{5}=1$, what is its focal distance?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/20 - y^2/5 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "10",
    "fact_spans": "[[[2, 5], [48, 49]], [[2, 45]]]",
    "query_spans": "[[[48, 54]]]",
    "process": "Since the equation of the hyperbola is \\frac{x^{2}}{20}-\\frac{y^{2}}{5}=1, we have a^{2}=20, b^{2}=5, c^{2}=a^{2}+b^{2}=25 \\Rightarrow 2c=10, so the focal distance of the hyperbola is 2c=10."
  },
  {
    "text": "What is the equation of the directrix of the parabola $x^{2}+12 y=0$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 + 12*y = 0)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=3",
    "fact_spans": "[[[0, 17]], [[0, 17]]]",
    "query_spans": "[[[0, 24]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{81}=1  (a>0)$ is $y=3x$, then $a$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/81 + x^2/a^2 = 1);a: Number;a>0;Expression(OneOf(Asymptote(G))) = (y = 3*x)",
    "query_expressions": "a",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 52]], [[2, 52]], [[70, 73]], [[5, 52]], [[2, 68]]]",
    "query_spans": "[[[70, 75]]]",
    "process": "Since the asymptotes of the hyperbola are $ y = 3x $, we know $ \\frac{b}{a} = 3 $. Also, from the given information, $ b = 9 $. Therefore, solving gives $ a = 3 $. Answer: $ 3 $"
  },
  {
    "text": "Given that $AB$ is a chord passing through the left focus $F_1$ of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and $|A F_{2}|+|B F_{2}|=12$, where $F_2$ is the right focus of the ellipse, then the length of chord $AB$ is?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2/25 + y^2/16 = 1);LeftFocus(G) = F1;PointOnCurve(F1, LineSegmentOf(A, B));Abs(LineSegmentOf(A, F2)) + Abs(LineSegmentOf(B, F2)) = 12;RightFocus(G) = F2;IsChordOf(LineSegmentOf(A,B),G)",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[9, 48], [97, 99]], [[2, 7]], [[2, 7]], [[89, 96]], [[51, 58]], [[9, 48]], [[9, 58]], [[2, 60]], [[62, 86]], [[89, 103]], [[2, 60]]]",
    "query_spans": "[[[106, 115]]]",
    "process": "In the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, the semi-major axis length is $a=5$. $AB$ is a chord passing through the left focus $F_{1}$, and $|AF_{2}|+|BF_{2}|=12$. According to the definition of an ellipse, $|AF_{2}|+|BF_{2}|+|AB|=4a=20$. Then the length of chord $AB$ is $20-12=8$. The answer is: $8$."
  },
  {
    "text": "What are the coordinates of the focus of the parabola $x=a y^{2}(a>0)$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x = a*y^2);a: Number;a>0",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(1/(4*a), 0)",
    "fact_spans": "[[[0, 19]], [[0, 19]], [[3, 19]], [[3, 19]]]",
    "query_spans": "[[[0, 26]]]",
    "process": ""
  },
  {
    "text": "The center is at the origin, and one asymptote of the hyperbola with foci on the $x$-axis is $y = \\frac{3}{4}x$. The distance from a focus to the asymptote is $3$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;O: Origin;Center(G) = O;PointOnCurve(Focus(G), xAxis);Expression(OneOf(Asymptote(G))) = (y = (3/4)*x);Distance(Focus(G), Asymptote(G)) = 3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16 - y^2/9 = 1",
    "fact_spans": "[[[15, 18], [59, 62]], [[3, 5]], [[0, 18]], [[6, 18]], [[15, 42]], [[15, 56]]]",
    "query_spans": "[[[59, 67]]]",
    "process": "A hyperbola has an asymptote $ y = \\frac{3}{4}x = \\frac{b}{a}x $, then $ 3a = 4b $, $ 3x - 4y = 0 $. The distance from the focus $ F(c,0) $ to the asymptote $ 3x - 4y = 0 $ is 3, then $ d = \\frac{|3c|}{5} = 3 $, $ c = 5 $. From \n\\[\n\\begin{cases}\n3a = 4b \\\\\nb^{2} = c^{2} - a^{2} = 25 - a^{2}\n\\end{cases}\n\\]\nwe obtain\n\\[\n\\begin{cases}\na^{2} = 16 \\\\\nb^{2} = 9\n\\end{cases}\n\\]\nthen the equation of the hyperbola is $ \\frac{x^{2}}{16} - \\frac{y^{2}}{9} = 1 $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>b>0)$ has its right focus at $F(1,0)$, and $A$, $B$ are the left and right vertices of the ellipse $C$, with $|A F|=3|F B|$, then the equation of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;F: Point;A: Point;B: Point;a > b;b > 0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (1, 0);RightFocus(C) = F;LeftVertex(C)=A;RightVertex(C)=B;Abs(LineSegmentOf(A, F)) = 3*Abs(LineSegmentOf(F, B))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4+y^2/3=1",
    "fact_spans": "[[[2, 61], [87, 92], [115, 120]], [[9, 61]], [[9, 61]], [[66, 74]], [[77, 81]], [[83, 86]], [[9, 61]], [[9, 61]], [[2, 61]], [[66, 74]], [[2, 75]], [[77, 97]], [[77, 97]], [[99, 113]]]",
    "query_spans": "[[[115, 125]]]",
    "process": ""
  },
  {
    "text": "The minimum value of the sum of distances from a moving point on the parabola $x^{2}=-4 y$ to the points $F(0,-1)$ and $E(1,-3)$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = -4*y);K: Point;PointOnCurve(K, G);F: Point;Coordinate(F) = (0, -1);E: Point;Coordinate(E) = (1, -3)",
    "query_expressions": "Min(Distance(K, F)+Distance(K, E))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 15]], [[0, 15]], [[17, 19]], [[0, 19]], [[20, 30]], [[20, 30]], [[32, 41]], [[32, 41]]]",
    "query_spans": "[[[17, 52]]]",
    "process": "From the standard equation of the parabola, it is known that point $ F $ is the focus of the parabola. Draw the directrix $ l $ as shown in the figure. Take any point $ P $ on the parabola, and draw $ PP' \\perp l $ at point $ P' $, then $ PF = PP' $. Hence: $ PE + PF = PE + PP' $. Combining geometric relationships, it is known that the sum of distances reaches the minimum value when points $ E $, $ P $, $ P' $ are collinear: $ 3 + 1 = 4 $."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x$ ($p>0$) with focus $F$. If there exists a point $A$ on the parabola such that the horizontal coordinate of the midpoint of segment $A F$ is $1$, then $|A F|$=?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(A, G);XCoordinate(MidPoint(LineSegmentOf(A,F))) = 1",
    "query_expressions": "Abs(LineSegmentOf(A, F))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 23], [33, 36]], [[5, 23]], [[39, 43]], [[27, 30]], [[5, 23]], [[2, 23]], [[2, 30]], [[33, 43]], [[47, 65]]]",
    "query_spans": "[[[67, 76]]]",
    "process": "By the given condition, $ F\\left(\\frac{p}{2},0\\right) $, let $ A(x,y) $. Since the x-coordinate of the midpoint of $ AF $ is 1, we have $ \\frac{p}{2} + x = 2 $, thus $ x = 2 - \\frac{p}{2} $. Because the distance from any point on the parabola to the focus equals the distance to the directrix, $ |AF| = x + \\frac{p}{2} = 2 - \\frac{p}{2} + \\frac{p}{2} = 2 $. Hence, the answer is $ 2 $. [This question mainly examines the definition and simple properties of a parabola, typical problem type]"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, respectively, and a line $l$ with inclination angle $\\frac{\\pi}{3}$ passes through $F_{1}$ and intersects the ellipse at points $A$ and $B$, then the perimeter of $\\triangle F_{2} A B$ is?",
    "fact_expressions": "l: Line;G: Ellipse;F2: Point;A: Point;B: Point;F1: Point;Expression(G) = (x^2/25 + y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;Inclination(l) = pi/3;PointOnCurve(F1, l);Intersection(l, G) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(F2, A, B))",
    "answer_expressions": "20",
    "fact_spans": "[[[84, 89]], [[20, 58], [101, 103]], [[10, 17]], [[105, 108]], [[109, 112]], [[2, 9], [91, 98]], [[20, 58]], [[2, 63]], [[2, 63]], [[64, 89]], [[84, 98]], [[84, 114]]]",
    "query_spans": "[[[116, 142]]]",
    "process": "From the ellipse equation, we know: a=5, and |AF_{1}|+|AF_{2}|=|BF_{1}|+|BF_{2}|=2a. Moreover, the perimeter of \\triangle ABF_{2} is (|AF_{1}|+|AF_{2}|)+(|BF_{1}|+|BF_{2}|)=4a=20."
  },
  {
    "text": "Let point $M(m, 0)$ lie on the major axis of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$, and let point $P$ be any point on the ellipse. When $|M P|$ is minimized, point $P$ exactly falls on the right vertex of the ellipse. Then the range of real number $m$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (m, 0);m: Real;PointOnCurve(M, MajorAxis(G)) = True;G: Ellipse;Expression(G) = (x^2/16 + y^2/12 = 1);P: Point;PointOnCurve(P, G) = True;WhenMin(Abs(LineSegmentOf(M, P)));RightVertex(G) = P",
    "query_expressions": "Range(m)",
    "answer_expressions": "[1,4]",
    "fact_spans": "[[[1, 11]], [[1, 11]], [[97, 102]], [[1, 55]], [[12, 51], [61, 63], [89, 91]], [[12, 51]], [[56, 60], [81, 85]], [[56, 68]], [[69, 80]], [[81, 95]]]",
    "query_spans": "[[[97, 109]]]",
    "process": "Let $ P(x,y) $ be a moving point on the ellipse. Since the equation of the ellipse is $ \\frac{x^{2}}{16} + \\frac{y^{2}}{12} = 1 $, it follows that $ -4 \\leqslant x \\leqslant 4 $.  \n$ |MP|^{2} = (x - m)^{2} + y^{2} = (x - m)^{2} + 12\\left(1 - \\frac{x^{2}}{16}\\right) = \\frac{1}{4}(x - 4m)^{2} + 12 - 3m^{2} $.  \nSince $ |MP| $ is minimized when point $ P $ coincides with the right vertex of the ellipse, i.e., $ |MP|^{2} $ attains its minimum value when $ x = 4 $, and since $ x \\in [-4,4] $, we have $ 4m \\geqslant 4 $, which gives $ m \\geqslant 1 $.  \nAlso, point $ M $ lies on the major axis of the ellipse, so $ -4 \\leqslant m \\leqslant 4 $.  \nTherefore, the range of real number $ m $ is $ [1,4] $."
  },
  {
    "text": "The hyperbola $C$ has the same foci as the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$, and the line $y=\\sqrt{3} x$ is an asymptote of $C$. Then the equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Ellipse;H: Line;Expression(G) = (x^2/8 + y^2/4 = 1);Expression(H) = (y = sqrt(3)*x);Focus(C) = Focus(G);OneOf(Asymptote(C))=H",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[0, 6], [68, 71], [79, 85]], [[7, 44]], [[51, 67]], [[7, 44]], [[51, 67]], [[0, 50]], [[51, 77]]]",
    "query_spans": "[[[79, 90]]]",
    "process": ""
  },
  {
    "text": "If the focus $F$ of the parabola $y^{2}=8x$ coincides with one focus of the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{n}=1$, then the value of $n$ is?",
    "fact_expressions": "H: Parabola;G: Hyperbola;n: Number;F: Point;Expression(H) = (y^2 = 8*x);Expression(G) = (x^2/3 - y^2/n = 1);Focus(H) = F;OneOf(Focus(G))=F",
    "query_expressions": "n",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 15]], [[22, 60]], [[69, 72]], [[18, 21]], [[1, 15]], [[22, 60]], [[1, 21]], [18, 66]]",
    "query_spans": "[[[69, 76]]]",
    "process": "Given the parabola $ y^{2} = 8x $, its focus $ F $ has coordinates $ (2, 0) $. The right focus of the hyperbola $ \\frac{x^{2}}{3} - \\frac{y^{2}}{n} = 1 $ is $ (\\sqrt{3+n}, 0) $. Therefore, $ \\sqrt{3+n} = 2 $, solving gives $ n = 1 $."
  },
  {
    "text": "The equation of an ellipse centered at the origin, with foci on the coordinate axes, eccentricity $\\frac{\\sqrt{3}}{2}$, and passing through the point $(2,0)$ is?",
    "fact_expressions": "G: Ellipse;H: Point;O: Origin;Coordinate(H) = (2, 0);Center(G) = O;PointOnCurve(Focus(G),axis);PointOnCurve(H, G);Eccentricity(G)=sqrt(3)/2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/4+y^2=1,x^2/4+y^2/16=1}",
    "fact_spans": "[[[50, 52]], [[41, 49]], [[3, 5]], [[41, 49]], [[0, 52]], [[6, 52]], [[40, 52]], [[14, 52]]]",
    "query_spans": "[[[50, 57]]]",
    "process": ""
  },
  {
    "text": "The major axis of the ellipse is $A_{1} A_{2}$, and $B$ is one endpoint of the minor axis. If $\\angle A_{1} B A_{2}=120^{\\circ}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;MajorAxis(G) = LineSegmentOf(A1, A2);A1: Point;A2: Point;OneOf(Endpoint(MinorAxis(G))) = B;B: Point;AngleOf(A1, B, A2) = ApplyUnit(120, degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[0, 2], [68, 70]], [[0, 19]], [[6, 19]], [[6, 19]], [[0, 30]], [[20, 24]], [[32, 66]]]",
    "query_spans": "[[[68, 76]]]",
    "process": "By the given condition, the major axis of the ellipse is $A_{1}A_{2}$, and $B$ is one endpoint of the minor axis. If $\\angle A_{1}BA_{2}=120^{\\circ}$, then $\\angle A_{1}A_{2}B=30^{\\circ}$, hence $\\angle A_{2}BO=60^{\\circ}$. From this, it follows that $\\frac{b}{a}=\\frac{\\sqrt{3}}{3}$, i.e., $\\frac{b^{2}}{a^{2}}=\\frac{1}{3}$, that is, $\\frac{a^{2}-c^{2}}{a^{2}}=\\frac{1}{3}$. Simplifying gives $1-e^{2}=\\frac{1}{3}$, solving yields $e=\\frac{\\sqrt{6}}{3}$."
  },
  {
    "text": "The coordinates of the focus of the parabola $y^{2}=-2 x$ are?",
    "fact_expressions": "E: Parabola;Expression(E) = (y^2 = -2*x)",
    "query_expressions": "Coordinate(Focus(E))",
    "answer_expressions": "(-1/2,0)",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 22]]]",
    "process": ""
  },
  {
    "text": "If $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively, and a circle with diameter $F_{1}F_{2}$ intersects the right branch of the hyperbola at point $P$, and if $\\angle P F_{1} F_{2}=\\alpha$, then what is the eccentricity of the hyperbola? (Express the result in terms of $\\alpha$)",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;F2: Point;F1: Point;P: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;IsDiameter(LineSegmentOf(F1,F2),H);Intersection(H,RightPart(G))=P;AngleOf(P, F1, F2) = alpha;alpha:Number",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/(Cos(alpha)-Sin(alpha))",
    "fact_spans": "[[[19, 75], [104, 107], [147, 150]], [[22, 75]], [[22, 75]], [[102, 103]], [[9, 16]], [[1, 8]], [[110, 114]], [[22, 75]], [[22, 75]], [[19, 75]], [[1, 81]], [[1, 81]], [[82, 103]], [[102, 114]], [[116, 145]], [[116, 145]]]",
    "query_spans": "[[[147, 157]]]",
    "process": "By the given condition, $ PF_{1} \\perp PF_{2} $, hence $ |PF_{1}| = 2c \\cdot \\cos\\alpha $, $ |PF_{2}| = 2c \\cdot \\sin\\alpha $, therefore $ e = \\frac{2c}{2a} = \\frac{2c}{|PF_{1}| - |PF_{2}|} = \\frac{1}{\\cos\\alpha - \\sin\\alpha} $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ with left focus $F$, point $P$ lies on the ellipse in the second quadrant, $M$ is the midpoint of segment $PF$, and $|OF|=|OM|$. Then the slope of line $PF$ is?",
    "fact_expressions": "G: Ellipse;F: Point;P: Point;O: Origin;M: Point;Expression(G) = (x^2/9 + y^2/5 = 1);LeftFocus(G) = F;PointOnCurve(P,G);Quadrant(P)=2;MidPoint(LineSegmentOf(P,F))=M;Abs(LineSegmentOf(O,F))=Abs(LineSegmentOf(O,M))",
    "query_expressions": "Slope(LineOf(P,F))",
    "answer_expressions": "sqrt(15)",
    "fact_spans": "[[[2, 39], [53, 55]], [[44, 47]], [[48, 52]], [[77, 90]], [[73, 76]], [[2, 39]], [[2, 47]], [[48, 56]], [[48, 62]], [[63, 76]], [[77, 90]]]",
    "query_spans": "[[[92, 104]]]",
    "process": "Since point M is the midpoint of segment PF, and the origin O of the ellipse is the midpoint of the two foci, a midline appears, and using the properties of the ellipse, we can derive an equality relationship. Let the right focus be F, connect PF. $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1\\Rightarrow a=3,b=\\sqrt{5},c=2$. Since O and M are the midpoints of FF' and PF respectively, $PF'=2OM=2OF=2c=4$, and $PF+PF'=2a=6 \\Rightarrow PF=2$, $FF'=4$. By the law of cosines: $\\cos\\angle PFF' = \\frac{PF^{2} + PF'^{2} - FF'^{2}}{2 \\cdot PF \\cdot PF'} = \\frac{4 + 16 - 16}{2 \\cdot 2 \\cdot 4} = \\frac{1}{4}$. Therefore, $k = \\tan\\angle PFF' = \\sqrt{15}$. So fill in $\\sqrt{15}$."
  },
  {
    "text": "Given the hyperbola $ C $: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, let the left vertex be $ A $, the right focus be $ F $, and the eccentricity be $ e $. If a moving point $ B $ lies on the right branch of the hyperbola $ C $ and does not coincide with the right vertex, such that $\\frac {\\angle B F A} {\\angle B A F}=e$ always holds, then what is the equation of the asymptotes of the hyperbola $ C $?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;B: Point;F: Point;A: Point;LeftVertex(C)=A;RightFocus(C)=F;e:Number;Eccentricity(C)=e;PointOnCurve(B,RightPart(C));Negation(B=RightVertex(C));AngleOf(B,F,A)/AngleOf(B,A,F)=e",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[2, 63], [96, 102], [161, 167]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[92, 95]], [[76, 79]], [[68, 71]], [[2, 71]], [[2, 79]], [[84, 87]], [[2, 87]], [[90, 106]], [[90, 114]], [[117, 156]]]",
    "query_spans": "[[[161, 176]]]",
    "process": "Take the special position where $ BF \\perp x $-axis, at this time $ \\angle BFA = 90^{\\circ} $, $ \\angle BAF = \\left(\\frac{90}{e}\\right)^{\\circ} $, $ AF = a + c $, substitute $ x = c $ into the hyperbola to get $ y = \\pm \\frac{b^{2}}{a} $, so $ B\\left(c, \\frac{b^{2}}{a}\\right) $, $ \\tan \\angle BAF = \\frac{BF}{AF} = \\frac{b^{2}}{a+c} = e - 1 $, we obtain $ \\tan\\left(\\frac{90}{e}\\right)^{\\circ} = e - 1 $. Discuss separately $ e > 2 $, $ e = 2 $, $ 1 < e < 2 $, we get $ e = 2 $, then the asymptotes equation can be found as $ y = \\pm \\frac{b}{a}x $. As shown in figure: since $ \\frac{\\angle BFA}{\\angle BAF} = e $ always holds, take the special position $ BF \\perp x $-axis, at this time $ \\angle BFA = 90^{\\circ} $, so $ \\angle BAF = \\left(\\frac{90}{e}\\right)^{\\circ} $. In right triangle $ \\triangle ABF $, $ \\tan \\angle BAF = \\frac{BF}{AF} $. For hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0 $, $ b > 0 $), $ AF = a + c $. Substitute $ x = c $ into the hyperbola equation to get $ \\frac{c^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $, simplifying yields: $ y = \\pm \\frac{b^{2}}{a} $. Take point $ B\\left(c, \\frac{b^{2}}{a}\\right) $ located in the first quadrant, so $ BF = \\frac{b^{2}}{a} $, then $ \\tan \\angle BAF = \\frac{BF}{AF} = \\frac{\\frac{b^{2}}{a}}{a + c} = \\frac{b^{2}}{a(a + c)} = \\frac{c^{2} - a^{2}}{a(a + c)} = \\frac{c - a}{a} = e - 1 $. So $ \\tan \\angle BAF = \\tan\\left(\\frac{90}{e}\\right)^{\\circ} = e - 1 $. When $ e > 2 $, $ e - 1 > 1 $, $ \\tan\\left(\\frac{90}{e}\\right)^{\\circ} < \\tan 45^{\\circ} = 1 $, which does not satisfy the condition, thus invalid. When $ 1 < e < 2 $, $ e - 1 < 1 $, $ \\tan\\left(\\frac{90}{e}\\right)^{\\circ} > \\tan 45^{\\circ} = 1 $, which does not satisfy the condition, thus invalid. When $ e = 2 $, $ \\tan\\left(\\frac{90}{e}\\right)^{\\circ} = \\tan 45^{\\circ} = 1 = e - 1 $, so $ e = 2 $, i.e., $ \\frac{c}{a} = 2 $, we get $ \\frac{c^{2}}{a^{2}} = \\frac{a^{2} + b^{2}}{a^{2}} = 4 $, so $ b^{2} = 3a^{2} $. Therefore $ \\frac{b^{2}}{a^{2}} = 3 $, $ \\frac{b}{a} = \\pm \\sqrt{3} $. Thus the asymptotes of the hyperbola are $ y = \\pm \\sqrt{3}x $."
  },
  {
    "text": "Given that the distance from the right focus $F_{2}$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ to one of its asymptotes $y=\\frac{b}{a} x$ is equal to $2 a$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G)=F2;L:Line;OneOf(Asymptote(G))=L;Expression(L)=(y = x*(b/a));Distance(F2, L) = 2*a",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 58], [106, 109]], [[5, 58]], [[5, 58]], [[62, 69]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 69]], [[77, 94]], [[2, 94]], [[77, 94]], [[62, 104]]]",
    "query_spans": "[[[106, 115]]]",
    "process": ""
  },
  {
    "text": "What is the distance from the focus to the directrix of the parabola $x^{2}=-2 y$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = -2*y)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 26]]]",
    "process": "The focus of the parabola $x^{2}=-2y$ is $(0,-\\frac{1}{2})$, and the equation of the directrix is $y=\\frac{1}{2}$. Therefore, the distance from the focus to the directrix of this parabola is $1$."
  },
  {
    "text": "Given that $F_{1}$, $F_{2}$ are the foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, and $P Q$ is a chord passing through focus $F_{1}$ with an inclination angle of $60^{\\circ}$, then the value of $|P F_{2}|+|Q F_{2}|-|P Q|$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;Q: Point;F1:Point;F2: Point;Expression(G) = (x^2/16 - y^2/9 = 1);Focus(G)={F1,F2};PointOnCurve(F1, LineSegmentOf(P, Q));IsChordOf(LineSegmentOf(P, Q), G);Inclination(LineSegmentOf(P, Q)) = ApplyUnit(60, degree)",
    "query_expressions": "Abs(LineSegmentOf(P, F2)) + Abs(LineSegmentOf(Q, F2)) - Abs(LineSegmentOf(P, Q))",
    "answer_expressions": "16",
    "fact_spans": "[[[18, 57]], [[61, 66]], [[61, 66]], [[2, 9], [70, 77]], [[10, 17]], [[18, 57]], [[2, 60]], [[61, 79]], [[18, 79]], [[81, 103]]]",
    "query_spans": "[[[106, 137]]]",
    "process": "Since |PF_{2}| + |QF_{2}|PQ|(PF_{2}| - |PF_{1}D + (QF_{2}| - |QF_{1}D = 4a = 16"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, $O$ is the coordinate origin. A line passing through the right focus $F$ intersects the two asymptotes of hyperbola $C$ at points $P$ and $Q$, respectively. If $O P \\perp P Q$ and $\\overrightarrow{Q F}=3 \\overrightarrow{F P}$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;O: Origin;F: Point;RightFocus(C) = F;G: Line;PointOnCurve(F, G);Z1: Line;Z2: Line;Asymptote(C) = {Z1, Z2};P: Point;Intersection(G, Z1) = P;Q: Point;Intersection(G, Z2) = Q;IsPerpendicular(LineSegmentOf(O, P), LineSegmentOf(P, Q));VectorOf(Q, F) = 3*VectorOf(F, P)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[2, 63], [86, 92], [175, 181]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[65, 69]], [[79, 82]], [[2, 82]], [[83, 85]], [[75, 85]], [], [], [[86, 98]], [[102, 105]], [[83, 109]], [[106, 109]], [[83, 109]], [[111, 126]], [[128, 173]]]",
    "query_spans": "[[[175, 187]]]",
    "process": ""
  },
  {
    "text": "Given that a point $P$ on an asymptote of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$ is at a distance of $3$ from the center of the hyperbola, what is the distance from point $P$ to the $y$-axis?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/2 = 1);P: Point;PointOnCurve(P, OneOf(Asymptote(G)));Distance(P, Center(G)) = 3",
    "query_expressions": "Distance(P, yAxis)",
    "answer_expressions": "sqrt(6)",
    "fact_spans": "[[[2, 40], [53, 56]], [[2, 40]], [[49, 52], [67, 71]], [[2, 52]], [[49, 65]]]",
    "query_spans": "[[[67, 81]]]",
    "process": "Let the coordinates of point P be P(x, y). According to the problem, the asymptotes of the hyperbola are given by: y = \\pm\\frac{\\sqrt{2}}{2}x. Since the distance from P to the center of the hyperbola is 3, we have x^{2} + y^{2} = 9, that is, x^{2} + y^{2} = \\frac{3}{2}x^{2} = 9. Solving this yields: |x| = \\sqrt{6}, so the distance from point P to the y-axis is \\sqrt{6}."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x  (p>0)$, the focus is $F$, the directrix is $l$, $M$ is a point on $l$, $N$ is the intersection point of segment $M F$ and $C$, if $\\overrightarrow{M N}=2 \\overrightarrow{N F}$, $O$ is the origin, and the area $S$ of $\\Delta O F N$ is $\\sqrt{3}$, then the value of $p$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;l: Line;Directrix(C) = l;M: Point;PointOnCurve(M, l);N: Point;Intersection(LineSegmentOf(M, F), C) = N;VectorOf(M, N) = 2*VectorOf(N, F);O: Origin;S: Number;Area(TriangleOf(O, F, N)) = S;S = sqrt(3)",
    "query_expressions": "p",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[2, 30], [69, 72]], [[2, 30]], [[166, 169]], [[10, 30]], [[34, 37]], [[2, 37]], [[41, 44], [50, 53]], [[2, 44]], [[46, 49]], [[46, 56]], [[57, 60]], [[57, 75]], [[77, 122]], [[123, 126]], [[150, 153]], [[133, 153]], [[150, 164]]]",
    "query_spans": "[[[166, 173]]]",
    "process": "From AMPF∼ANQF, and $\\overrightarrow{MN}=2\\overrightarrow{NF}$, we get $x=\\frac{1}{6}p$\\textcircled{1}; from $S_{AOFN}=\\frac{1}{2}|OF||NQ|=\\sqrt{3}$, we get $x=\\frac{24}{p^{3}}$\\textcircled{2}; combining \\textcircled{1}\\textcircled{2}, we obtain the answer. Assume point M is in the upper half of the directrix, the intersection point of the directrix and the x-axis is P, draw NQ perpendicular to the x-axis with foot Q, let point N be $(x,y)$. It is easy to see that AMPF∼4NQF, and $\\overrightarrow{MN}=2\\overrightarrow{NF}$, so $|QF|=\\frac{1}{3}|PF|=\\frac{1}{3}p$, then $x=\\frac{1}{6}p$\\textcircled{1}; also $S_{AOFN}=\\frac{1}{2}|OF||NQ|=\\frac{1}{2}\\cdot\\frac{p}{2}\\cdot y=\\sqrt{3}$, we get $y=\\frac{4\\sqrt{3}}{D}$; substituting into the parabola equation $y^{2}=2px$ ($p>0$), we get $x=\\frac{24}{n3}$\\textcircled{2}; solving \\textcircled{1}\\textcircled{2} together gives $p=2\\sqrt{3}$."
  },
  {
    "text": "Parabola $C$: $y^{2}=2 p x  (p>0)$, the intersection point of the directrix and the $x$-axis is $M$. From point $M$, draw two tangents to $C$, with points of tangency $P$ and $Q$, respectively. Then $\\angle P M Q$=?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;M: Point;Intersection(Directrix(C), xAxis) = M;Z1: Line;Z2: Line;TangentOfPoint(M, C) = {Z1, Z2};P: Point;Q: Point;TangentPoint(Z1, C) = P;TangentPoint(Z2, C) = Q",
    "query_expressions": "AngleOf(P, M, Q)",
    "answer_expressions": "pi/2",
    "fact_spans": "[[[0, 28], [50, 53]], [[0, 28]], [[8, 28]], [[8, 28]], [[40, 43], [45, 49]], [[0, 43]], [], [], [[44, 58]], [[64, 67]], [[68, 71]], [[44, 71]], [[44, 71]]]",
    "query_spans": "[[[73, 89]]]",
    "process": "From the given conditions, we have M(-\\frac{p}{2},0). Let the tangent line passing through point M be x=my-\\frac{p}{2}. Substituting into y^{2}=2px gives y^{2}-2pmy+p^{2}=0. \\therefore 4=4p^{2}m^{2}-4p^{2}=0 \\therefore m=\\pm1, i.e., k=\\pm1, MQ\\bot MP, therefore \\angle PMQ=\\frac{\\pi}{2}"
  },
  {
    "text": "A point $P$ on the hyperbola $4 x^{2}-y^{2}+64=0$ is at a distance of $1$ from one focus. What is the distance from point $P$ to the other focus?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (4*x^2 - y^2 + 64 = 0);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Distance(P, F1) = 1;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2)",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "17",
    "fact_spans": "[[[0, 23]], [[0, 23]], [[27, 30], [46, 50]], [[0, 30]], [], [], [[0, 43]], [0, 34], [0, 55], [0, 55]]",
    "query_spans": "[[[0, 61]]]",
    "process": "First, convert the given hyperbola equation into the standard form $\\frac{y^{2}}{64}-\\frac{x^{2}}{16}=1$. Then, according to the definition of a hyperbola, the absolute value of the difference in distances from a point $P$ on the hyperbola to the two foci is 16. Thus, the distance from point $P$ to the other focus can be found to be 17."
  },
  {
    "text": "A line passing through the focus of the parabola $y^{2}=2 x$ intersects the parabola at points $A$ and $B$. If the distance from the midpoint $M$ of $AB$ to the directrix of the parabola is $5$, then the length of segment $AB$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);H: Line;PointOnCurve(Focus(G), H);A: Point;B: Point;Intersection(H, G) = {A, B};M: Point;MidPoint(LineSegmentOf(A, B)) = M;Distance(M, Directrix(G)) = 5",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "10",
    "fact_spans": "[[[1, 15], [21, 24], [50, 53]], [[1, 15]], [[18, 20]], [[0, 20]], [[25, 28]], [[29, 32]], [[18, 34]], [[45, 48]], [[37, 48]], [[45, 62]]]",
    "query_spans": "[[[65, 77]]]",
    "process": "As shown in the figure, let $ l $ be the directrix of the parabola, draw $ BE \\perp l $ at point $ E $, $ MF \\perp l $ at point $ F $, $ AD \\perp l $ at point $ D $, then $ AD + BE = 2MF = 10 $. By the definition of the parabola, we have: $ AB = AD + BE = 10 $."
  },
  {
    "text": "What are the asymptote equations of the hyperbola $x^{2}-y^{2}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "x + pm*y = 0",
    "fact_spans": "[[[0, 18]], [[0, 18]]]",
    "query_spans": "[[[0, 26]]]",
    "process": "Since $x^{2}-y^{2}=1$, the asymptotic line equation is $x^{2}-y^{2}=0$, which is $x\\pm y=0$."
  },
  {
    "text": "If the distance from the focus of the parabola $x^{2}=4 y$ to the asymptotes of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is equal to $\\frac{\\sqrt{5}}{5}$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Distance(Focus(G), Asymptote(C)) = sqrt(5)/5",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[1, 15]], [[1, 15]], [[19, 80], [110, 116]], [[19, 80]], [[26, 80]], [[26, 80]], [[26, 80]], [[26, 80]], [[1, 108]]]",
    "query_spans": "[[[110, 122]]]",
    "process": "Analysis: Find the focus coordinates of the parabola, the asymptote equations of the hyperbola, and using the point-to-line distance formula, obtain the relationship between $a$ and $b$. Then, using the eccentricity formula, compute to get the result. The focus of the parabola $x^{2}=4y$ is $(0,1)$. One asymptote of the hyperbola $C_{2}: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) is $bx+ay=0$. Then the distance from the focus to the asymptote is $d=\\frac{a}{\\sqrt{a^{2}+b^{2}}}=\\frac{\\sqrt{5}}{5}$, thus $b^{2}=4a^{2}$, so $c^{2}=5a^{2}$. Hence, the eccentricity of the hyperbola is: $\\sqrt{5}$."
  },
  {
    "text": "The equation of an ellipse that has the same foci as the hyperbola $\\frac{x^{2}}{5}-\\frac{y^{2}}{4}=-1$ and has eccentricity $\\frac{3}{5}$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(G) = (x^2/5 - y^2/4 = -1);Eccentricity(H)=3/5;Focus(H) = Focus(G)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2/16+y^2/25=1",
    "fact_spans": "[[[1, 40]], [[65, 67]], [[1, 40]], [[47, 67]], [[0, 67]]]",
    "query_spans": "[[[65, 71]]]",
    "process": "The hyperbola $\\frac{x^{2}}{5}-\\frac{y^{2}}{4}=-1$ is rewritten as $\\frac{y^{2}}{4}-\\frac{x^{2}}{5}=1$, $\\therefore c^{2}=a^{2}+b^{2}=9$, $\\therefore c=3$, foci $(0,\\pm3)$, in the ellipse $\\frac{c}{a}=\\frac{3}{5}$, $\\therefore a=5$, the equation of the ellipse is $\\frac{x^{2}}{16}+\\frac{y^{2}}{25}=1$."
  },
  {
    "text": "There is a chord $PQ$ passing through the right focus $F_{2}$ of the hyperbola $x^{2}-y^{2}=8$, with $PQ=7$, and $F_{1}$ is the left focus. Then, what is the perimeter of $\\triangle F_{1} PQ$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 8);F2: Point;RightFocus(G) = F2;P: Point;Q: Point;PointOnCurve(F2, LineSegmentOf(P, Q)) ;IsChordOf(LineSegmentOf(P, Q), G) ;LineSegmentOf(P, Q) = 7;F1: Point;LeftFocus(G) = F1",
    "query_expressions": "Perimeter(TriangleOf(F1, P, Q))",
    "answer_expressions": "14 + 8*sqrt(2)",
    "fact_spans": "[[[1, 19]], [[1, 19]], [[23, 30]], [[1, 30]], [[34, 38]], [[34, 38]], [[0, 38]], [[1, 38]], [[41, 47]], [[50, 57]], [[1, 61]]]",
    "query_spans": "[[[64, 89]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through the right focus $F_{2}$ intersects the right branch of the hyperbola at points $A$ and $B$. If $|A B|=5$, then the perimeter of $\\triangle A B F_{1}$ is?",
    "fact_expressions": "l: Line;G: Hyperbola;A: Point;B: Point;F1: Point;F2: Point;Expression(G) = (x^2/16 - y^2/9 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(F2, l);Intersection(l,RightPart(G))={A, B};Abs(LineSegmentOf(A, B))=5",
    "query_expressions": "Perimeter(TriangleOf(A, B, F1))",
    "answer_expressions": "26",
    "fact_spans": "[[[79, 84]], [[2, 41], [85, 88]], [[93, 96]], [[97, 100]], [[50, 57]], [[58, 65], [71, 78]], [[2, 41]], [[2, 65]], [[2, 65]], [[67, 84]], [[79, 102]], [[104, 113]]]",
    "query_spans": "[[[115, 141]]]",
    "process": ""
  },
  {
    "text": "Given that the vertex of a parabola is at the origin, its axis of symmetry is a coordinate axis, and it passes through the point $(6,6)$, then the standard equation of the parabola is?",
    "fact_expressions": "G: Parabola;O: Origin;Vertex(G) = O;SymmetryAxis(G) = axis;H: Point;Coordinate(H) = (6, 6);PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=6*x, x^2=6*y}",
    "fact_spans": "[[[3, 6], [37, 40]], [[10, 14]], [[3, 14]], [[3, 22]], [[26, 34]], [[26, 34]], [[3, 34]]]",
    "query_spans": "[[[37, 47]]]",
    "process": "Since the point (6,6) lies in the first quadrant, the parabola opens to the right or upward. Therefore, we may assume the standard equation of the parabola is $ y^{2} = 2px $ ($ p > 0 $) or $ x^{2} = 2py $ ($ p > 0 $). Substituting the point (6,6) into each equation yields $ p = 3 $. Thus, the standard equation of the parabola is $ y^{2} = 6x $ or $ x^{2} = 6y $."
  },
  {
    "text": "Given that $P$ is a moving point on the parabola $y^{2}=4x$, then the minimum distance from point $P$ to the line $y=x+2$ is?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (y = x + 2);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, H))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[6, 20]], [[31, 40]], [[2, 5], [26, 30]], [[6, 20]], [[31, 40]], [[2, 24]]]",
    "query_spans": "[[[26, 49]]]",
    "process": "Find the equation of the line parallel to the line $ y = x + 2 $ and tangent to the parabola, then the distance between the two parallel lines is the required value. Let the equation of the line parallel to $ y = x + 2 $ and tangent to the parabola be $ y = x + m $. Combining with the parabola equation $ y^2 = 4x $, we obtain $ x^2 + (2m - 4)x + m^2 = 0 $. From $ \\Delta = (2m - 4)^2 - 4m^2 = 0 $, we get $ m = 1 $. Thus, the equation of the line is $ y = x + 1 $. Therefore, the distance between the two parallel lines is $ d = \\frac{|1 - 2|}{\\sqrt{2}} = \\frac{\\sqrt{2}}{2} $, which is the required value."
  },
  {
    "text": "It is known that circle $C$ passes through the focus of the parabola $y^{2}=4x$, and its center lies on the directrix of this parabola. If the center of circle $C$ does not lie on the $x$-axis and the circle is tangent to the line $x+\\sqrt{3}y-3=0$, then the radius of circle $C$ is?",
    "fact_expressions": "C: Circle;G: Parabola;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), C);PointOnCurve(Center(C), Directrix(G));Negation(PointOnCurve(Center(C), xAxis));H: Line;Expression(H) = (x + sqrt(3)*y - 3 = 0);IsTangent(H, C)",
    "query_expressions": "Radius(C)",
    "answer_expressions": "14",
    "fact_spans": "[[[2, 6], [39, 43], [80, 84]], [[7, 21], [30, 33]], [[7, 21]], [[2, 24]], [[2, 37]], [[39, 53]], [[56, 76]], [[56, 76]], [[39, 78]]]",
    "query_spans": "[[[80, 89]]]",
    "process": "Since the directrix of the parabola is $ x = -1 $ and the focus has coordinates $ F(1,0) $, let the center of the circle have coordinates $ C(-1,t) $ ($ t \\neq 0 $). According to the problem, the radius of the circle is $ r = \\sqrt{4 + t^{2}} = \\frac{|-4 + \\sqrt{3}t|}{2} $. Solving this gives $ t = -8\\sqrt{3} $. Therefore, the radius of the circle is $ r = \\sqrt{4 + t^{2}} = \\sqrt{196} = 14 $. The answer to be filled in is $ 14 $."
  },
  {
    "text": "Given that $F$ is the right focus of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, point $P$ lies on $C$, $O$ is the coordinate origin, $|O P|=2 b$, and $\\angle P O F=\\frac{\\pi}{3}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;O: Origin;P: Point;F: Point;b>0;Expression(C) = (x^2 - y^2/b^2 = 1);RightFocus(C) = F;PointOnCurve(P, C);Abs(LineSegmentOf(O, P)) = 2*b;AngleOf(P, O, F) = pi/3",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[6, 48], [58, 61], [117, 120]], [[14, 48]], [[63, 66]], [[53, 57]], [[2, 5]], [[14, 48]], [[6, 48]], [[2, 52]], [[53, 62]], [[73, 84]], [[87, 115]]]",
    "query_spans": "[[[117, 126]]]",
    "process": "Let $ P(x_{0},y_{0}) $. From the given conditions, we have $ x_{0} > 0 $, assume $ y_{0} > 0 $, $ \\overrightarrow{OP} = (x_{0},y_{0}) $. By the condition $ |OP| = 2b $, we get $ x_{0}^{2} + y_{0}^{2} = 4b^{2} $. $ \\overrightarrow{OF} = (c,0) $. Since $ \\angle POF = \\frac{\\pi}{3} $, it follows that $ \\cos\\angle POF = \\frac{1}{2} = \\frac{\\overrightarrow{OP}}{|\\overrightarrow{OP}|} \\cdot |\\overrightarrow{OF}| = \\frac{x_{0} \\cdot c}{2b \\cdot c} $, which gives $ x_{0} = b $, $ y_{0}^{2} = 3b^{2} $, $ y_{0} > 0 $. Substituting the coordinates of point $ P $ into the hyperbola equation yields: $ \\frac{b^{2}}{a^{2}} - 3 = 1 $, so $ b^{2} = 4a^{2} $. Therefore, the eccentricity of the hyperbola is $ e = \\sqrt{\\frac{c^{2}}{a^{2}}} = \\sqrt{\\frac{a^{2} + b^{2}}{a^{2}}} = \\sqrt{5} $."
  },
  {
    "text": "If point $P(4,4)$ lies on the parabola $y^{2}=2 p x$, then the coordinates of the focus of the parabola are? The distance from point $P$ to the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;p: Number;P: Point;Expression(G) = (y^2 = 2*p*x);Coordinate(P) = (4, 4);PointOnCurve(P, G)",
    "query_expressions": "Coordinate(Focus(G));Distance(P,Directrix(G))",
    "answer_expressions": "(1, 0)\n5",
    "fact_spans": "[[[11, 27], [32, 35], [46, 49]], [[14, 27]], [[1, 10], [41, 45]], [[11, 27]], [[1, 10]], [[1, 30]]]",
    "query_spans": "[[[32, 41]], [[41, 57]]]",
    "process": "Substituting the point $ P(4,4) $ into the parabola equation yields $ 4^{2} = 2p \\Rightarrow p = 2 $, then the focus coordinates of the parabola are $ (1,0) $. The distance from point $ P $ to the directrix of the parabola is $ 5 $."
  },
  {
    "text": "The parabola $y^{2}=m x$ ($m$ is a constant) passes through the point $(-1,1)$, then the coordinates of the focus of the parabola are?",
    "fact_expressions": "G: Parabola;m: Number;H: Point;Expression(G) = (y^2 = m*x);Coordinate(H) = (-1, 1);PointOnCurve(H,G)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(-1/4, 0)",
    "fact_spans": "[[[0, 14], [35, 38]], [[15, 18]], [[24, 33]], [[0, 14]], [[24, 33]], [[0, 33]]]",
    "query_spans": "[[[35, 45]]]",
    "process": "Substitute the given point into the parabola equation to find the value of m, then convert the original equation into the standard form of a parabola to determine the value of p, identify the focus location, and finally obtain the coordinates of the focus. Substituting the point (-1,1) into the parabola y^{2}=mx yields m=-1, \\therefore the parabola equation is y^{2}=-x=-2\\times\\frac{1}{2}x, \\therefore p=\\frac{1}{2}, the focus lies on the negative x-axis, \\therefore the focus coordinates of the parabola are (-\\frac{1}{4},0)."
  },
  {
    "text": "Let the focus of the parabola $x^{2}=2 p y(p>0)$ be $F$, and the directrix be $l$. From a point $A$ on the parabola, draw a perpendicular to $l$, with foot $B$. Let $C(0, \\frac{9}{2} p)$. If $AF$ and $BC$ intersect at point $E$, $|CF|=2|AF|$, and the area of $\\triangle ACE$ is $\\sqrt{3}$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;F: Point;B: Point;C: Point;E: Point;l: Line;L: Line;p>0;Expression(G) = (x^2 = 2*(p*y));Coordinate(C) = (0, (9/2)*p);Focus(G) = F;Directrix(G) = l;PointOnCurve(A,L);IsPerpendicular(l,L);FootPoint(l,L)=B;Intersection(LineSegmentOf(A, F), LineSegmentOf(B, C)) = E;Abs(LineSegmentOf(C, F)) = 2*Abs(LineSegmentOf(A, F));Area(TriangleOf(A, C, E)) = sqrt(3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2=sqrt(6)*y",
    "fact_spans": "[[[1, 22], [38, 41], [154, 157]], [[4, 22]], [[44, 47]], [[26, 29]], [[58, 61]], [[63, 84]], [[100, 104]], [[33, 36]], [], [[4, 22]], [[1, 22]], [[63, 84]], [[1, 29]], [[1, 36]], [[37, 54]], [[37, 54]], [[37, 61]], [[86, 104]], [[105, 119]], [[121, 152]]]",
    "query_spans": "[[[154, 162]]]",
    "process": "From the given conditions, |AF| = 2p. Using the definition of the parabola, find the x-coordinate of point A. Based on similarity, S_{\\triangle ACF} = 3\\sqrt{3}. The result can be obtained using the triangle area formula. [Detailed solution] Let A(x_{A}, y_{A}), F(0, \\frac{p}{2}), |CF| = \\frac{9}{2}p - \\frac{p}{2} = 4p. Also, |CF| = 2|AF|, so |AF| = 2p. By the definition of the parabola, |AB| = 2p, thus y_{A} = \\frac{3}{2}p, and |x_{A}| = \\sqrt{3}p. Since CF // AB, \\frac{EF}{EA} = \\frac{CF}{AB}, i.e., \\frac{EF}{EA} = \\frac{A}{A}\\frac{F^{2}}{|F} = 2. Therefore, S_{\\triangle CEF} = 2S_{\\triangle CEA} = 2\\sqrt{3}, and S_{\\triangle ACF} = S_{\\triangle AEC} + S_{\\triangle CFE} = 3\\sqrt{3}. Hence, \\frac{1}{2} \\times 4p \\times \\sqrt{3}p = 3\\sqrt{3}, solving gives: p = \\frac{\\sqrt{6}}{2}"
  },
  {
    "text": "If on the parabola $x^{2}=-2 p y(p>0)$ the point with ordinate $-4$ has a distance of $5$ to the focus, then what is the distance from the focus to the directrix?",
    "fact_expressions": "G: Parabola;p: Number;p>0;Expression(G) = (x^2 = -2*p*y);P:Point;YCoordinate(P)=-4;PointOnCurve(P, G);Distance(P, Focus(G)) = 5",
    "query_expressions": "Distance(Focus(G),Directrix(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 23]], [[4, 23]], [[4, 23]], [[1, 23]], [[33, 34]], [[1, 34]], [[1, 34]], [[1, 44]]]",
    "query_spans": "[[[1, 56]]]",
    "process": "Since the point on the parabola $x^{2}=-2py$ $(p>0)$ with ordinate $-4$ has a distance of 5 to the focus, by using the definition of the parabola, we have $\\frac{p}{2}+4=5$, $\\therefore p=2$. The distance from the focus to the directrix is $p$, so fill in 2."
  },
  {
    "text": "It is known that the eccentricity of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{m}=1$ with foci on the $x$-axis is $\\frac{1}{2}$. Then what is the value of $m$?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/16 + y^2/m = 1);PointOnCurve(Focus(G), xAxis);Eccentricity(G) = 1/2",
    "query_expressions": "m",
    "answer_expressions": "12",
    "fact_spans": "[[[11, 49]], [[69, 72]], [[11, 49]], [[2, 49]], [[11, 67]]]",
    "query_spans": "[[[69, 75]]]",
    "process": "From the equation, we have $a^{2}=16$, $b^{2}=m$, $\\therefore c^{2}=16-m$, $\\therefore e^{2}=\\frac{16-m}{16}=\\frac{1}{4}$, $\\therefore m=12$."
  },
  {
    "text": "Given that the point $P(x, y)$ lies on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, then the maximum value of $x^{2}+2 x-y^{2}$ is?",
    "fact_expressions": "P:Point;x1:Number;y1:Number;G:Ellipse;Expression(G)=(x^2/4+y^2=1);Coordinate(P)=(x1,y1);PointOnCurve(P,G)",
    "query_expressions": "Max(-y1^2 + x1^2 + 2*x1)",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 12]], [[3, 12]], [[3, 12]], [[13, 40]], [[13, 40]], [[2, 12]], [[2, 41]]]",
    "query_spans": "[[[43, 66]]]",
    "process": ""
  },
  {
    "text": "Given that point $P(2,1)$ bisects a chord of the parabola $y^{2}=4x$, then the equation of the line containing this chord is?",
    "fact_expressions": "G: Parabola;H: LineSegment;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(P) = (2, 1);IsChordOf(H, G);MidPoint(H)=P",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "y=2*x-3",
    "fact_spans": "[[[13, 27]], [[30, 31], [35, 36]], [[2, 11]], [[16, 27]], [[3, 11]], [[13, 31]], [[2, 31]]]",
    "query_spans": "[[[33, 45]]]",
    "process": ""
  },
  {
    "text": "It is known that an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ passes through the point $(1,2)$. What is the length of the chord formed by the intersection of this asymptote with the circle $(x+1)^{2}+(y-2)^{2}=4$?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;P: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = ((x + 1)^2 + (y - 2)^2 = 4);Coordinate(P) = (1, 2);L:Line;OneOf(Asymptote(G))=L;PointOnCurve(P,L)",
    "query_expressions": "Length(InterceptChord(L,H))",
    "answer_expressions": "4*sqrt(5)/5",
    "fact_spans": "[[[2, 58]], [[5, 58]], [[5, 58]], [[81, 105]], [[66, 74]], [[5, 58]], [[5, 58]], [[2, 58]], [[81, 105]], [[66, 74]], [], [[2, 64]], [[2, 74]]]",
    "query_spans": "[[[2, 114]]]",
    "process": "Since the line $ bx - ay = 0 $ passes through the point $ (1, 2) $, we have $ b - 2a = 0 $. The asymptote is $ 2x - y = 0 $. The distance from the center of the circle to this line is $ d = \\frac{4}{\\sqrt{5}} $. Hence, the chord length is $ 2\\sqrt{4 - \\frac{16}{5}} = \\frac{4\\sqrt{5}}{5} $. The answer to fill in is $ \\frac{4\\sqrt{5}}{5} $."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=4 x$ be $F$, $P$ a point on it, and $O$ the origin. If $|O P|=|P F|$, then the area of $\\triangle O P F$ is?",
    "fact_expressions": "G: Parabola;O: Origin;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G)=F;PointOnCurve(P,G);Abs(LineSegmentOf(O, P)) = Abs(LineSegmentOf(P, F))",
    "query_expressions": "Area(TriangleOf(O, P, F))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[1, 15], [27, 28]], [[33, 36]], [[23, 26]], [[19, 22]], [[1, 15]], [[1, 22]], [[23, 32]], [[43, 56]]]",
    "query_spans": "[[[58, 80]]]",
    "process": "The focus of the parabola $ y^{2}=4x $ is $ F(1,0) $. Since $ P $ is a point on it, $ O $ is the origin, and $ |OP|=|PF| $, the x-coordinate of point $ P $ is $ \\frac{1}{2} $. Thus, when $ x=\\frac{1}{2} $, $ y^{2}=4\\times\\frac{1}{2}=2 $, yielding $ y=\\pm\\sqrt{2} $. Therefore, the area of $ \\triangle OPF $ is $ \\frac{1}{2}\\times1\\times\\sqrt{2}=\\frac{\\sqrt{2}}{2} $."
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $y=\\sqrt{3} x$, and one of its foci coincides with the focus of the parabola $y^{2}=32 x$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x);H: Parabola;Expression(H) = (y^2 = 32*x);OneOf(Focus(G)) = Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16 - y^2/48 = 1",
    "fact_spans": "[[[2, 58], [83, 84], [113, 116]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 81]], [[90, 105]], [[90, 105]], [[83, 110]]]",
    "query_spans": "[[[113, 121]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the left focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, point $A(1,4)$ is a fixed point, and point $P$ is a moving point on the right branch of the hyperbola, then the minimum value of $|P F|+|P A|$ is?",
    "fact_expressions": "F: Point;LeftFocus(G) = F;G: Hyperbola;Expression(G) = (x^2/4 - y^2/12 = 1);Coordinate(A) = (1, 4);A: Point;P: Point;PointOnCurve(P, RightPart(G)) = True",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "9",
    "fact_spans": "[[[2, 5]], [[2, 49]], [[6, 45], [66, 69]], [[6, 45]], [[52, 60]], [[52, 60]], [[61, 65]], [[61, 75]]]",
    "query_spans": "[[[77, 96]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the upper and lower foci of the hyperbola $C$: $\\frac{y^{2}}{4}-\\frac{x^{2}}{3}=1$, respectively, a line perpendicular to the $y$-axis is drawn through point $F_{2}$ intersecting the hyperbola $C$ at points $P$ and $Q$. Then, the area of $\\triangle P F_{1} Q$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-x^2/3 + y^2/4 = 1);F1: Point;F2: Point;UpperFocus(C) = F1;LowerFocus(C) = F2;Z: Line;PointOnCurve(F2, Z);IsPerpendicular(Z, yAxis);P: Point;Q: Point;Intersection(Z, C) = {P, Q}",
    "query_expressions": "Area(TriangleOf(P, F1, Q))",
    "answer_expressions": "3*sqrt(7)",
    "fact_spans": "[[[20, 63], [88, 94]], [[20, 63]], [[2, 9]], [[10, 17], [71, 79]], [[2, 69]], [[2, 69]], [], [[70, 87]], [[70, 87]], [[95, 98]], [[99, 102]], [[70, 104]]]",
    "query_spans": "[[[106, 132]]]",
    "process": "The hyperbola $ C: \\frac{y^{2}}{4} - \\frac{x^{2}}{3} = 1 $ has upper and lower foci $ F_{1}(0,\\sqrt{7}) $, $ F_{2}(0,-\\sqrt{7}) $. Letting $ y = -\\sqrt{7} $ be substituted into $ \\frac{y^{2}}{4} - \\frac{x^{2}}{3} = 1 $, we solve to get: $ x = \\pm \\frac{3}{2} $, so $ P(-\\frac{3}{2}, -\\sqrt{7}) $, $ Q(\\frac{3}{2}, -\\sqrt{7}) $. Thus $ |PQ| = 3 $, and the area of $ \\triangle PF_{1}Q $ is $ \\frac{1}{2} \\times 3 \\times 2\\sqrt{7} = 3\\sqrt{7} $."
  },
  {
    "text": "If the focal distance of the hyperbola $x^{2}+m y^{2}=1$ is equal to 3 times the length of the imaginary axis, then the value of $m$ is?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (m*y^2 + x^2 = 1);Length(ImageinaryAxis(G))*3 = FocalLength(G)",
    "query_expressions": "m",
    "answer_expressions": "-8",
    "fact_spans": "[[[1, 21]], [[36, 39]], [[1, 21]], [[1, 34]]]",
    "query_spans": "[[[36, 43]]]",
    "process": "First, convert the hyperbola into standard form, thereby obtaining $a^{2}=1$, $b^{2}=-\\frac{1}{m}$, $c^{2}=1-\\frac{1}{m}$, then set up an equation according to the given conditions and solve for $m$.  \nDetailed solution: Convert $x^{2}+my^{2}=1$ into standard form: $x^{2}-\\frac{y^{2}}{-\\frac{1}{m}}=1$, then $a^{2}=1$, $b^{2}=-\\frac{1}{m}$, hence $c^{2}=1-\\frac{1}{m}$. Thus we obtain: $2\\sqrt{1-\\frac{1}{m}}=6\\sqrt{-\\frac{1}{m}}$, solving gives: $m=-8$."
  },
  {
    "text": "The ellipse passes through the points $(2,-\\sqrt{6})$ and $(2 \\sqrt{2}, \\sqrt{3})$. Then the standard equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;H: Point;Coordinate(H) = (2, -sqrt(6));I: Point;Coordinate(I) = (2*sqrt(2), sqrt(3));PointOnCurve(H, G);PointOnCurve(I, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/12+y^2/9=1",
    "fact_spans": "[[[0, 2], [48, 50]], [[4, 20]], [[4, 20]], [[21, 45]], [[21, 45]], [[0, 20]], [[0, 45]]]",
    "query_spans": "[[[48, 57]]]",
    "process": "From the given conditions: the standard equation of the ellipse is $mx^{2}+ny^{2}=1$ $(m>0,n>0,m\\neq n)$. Since the ellipse passes through the points $(2,-\\sqrt{6})$ and $(2\\sqrt{2},\\sqrt{3})$, substituting $(2,-\\sqrt{6})$ and $(2\\sqrt{2},\\sqrt{3})$ into $mx^{2}+ny^{2}=1$ gives  \n$$\n\\begin{cases}\n4m+6n=1 \\\\\n8m+3n=1\n\\end{cases}\n\\Rightarrow\n\\begin{cases}\nn=\\frac{1}{9} \\\\\nm=\\frac{1}{12}\n\\end{cases}\n$$\nTherefore, the standard equation of the ellipse is $\\frac{x^{2}}{12}+\\frac{y^{2}}{9}=1$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus is $F$, the right vertex is $A$. If the upper vertex $B$ lies on the ellipse $C$ and $B F \\perp B A$, then what is the eccentricity $e$ of the ellipse $C$?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;B: Point;F: Point;A: Point;e: Number;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;RightVertex(C)=A;UpperVertex(C)=B;IsPerpendicular(LineSegmentOf(B, F), LineSegmentOf(B, A));Eccentricity(C) = e",
    "query_expressions": "e",
    "answer_expressions": "(\\sqrt{5} - 1)/2",
    "fact_spans": "[[[2, 59], [77, 82], [107, 112]], [[9, 59]], [[9, 59]], [[85, 88]], [[64, 67]], [[72, 75]], [[116, 119]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 67]], [[2, 75]], [[77, 88]], [[90, 105]], [[107, 119]]]",
    "query_spans": "[[[116, 123]]]",
    "process": "First write the coordinates of A, B, F, then use BF\\botBA and the dot product of vectors equal to zero to obtain b^{2}=ac, then convert it into the homogeneous equation e^{2}+e-1=0; solve the equation to get the solution. Solution: According to the problem, A(a,0), B(0,b), F(-c,0); \\overrightarrow{BA}=(a,-b), \\overrightarrow{BF}=(-c,-b). \\because BF\\bot BA, \\cdot \\overrightarrow{BA}\\cdot\\overrightarrow{BF}=0, i.e., (a,-b)\\cdot(-c,-b)=0, \\therefore b^{2}=ac. Also \\because c^{2}=a^{2}-b^{2}, \\therefore c^{2}-a^{2}+ac=0; dividing both sides by a^{2} yields (\\frac{c}{a})^{2}+\\frac{c}{a}-1=0, i.e., e^{2}+e-1=0. Solving the equation gives e=\\frac{\\sqrt{5}+1}{2} (discarded) or e=\\frac{\\sqrt{5}-1}{2}."
  },
  {
    "text": "The distance from the vertex of the hyperbola $3 x^{2}-y^{2}=3$ to its asymptote is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (3*x^2 - y^2 = 3)",
    "query_expressions": "Distance(Vertex(G), Asymptote(G))",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[0, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 32]]]",
    "process": "The standard equation of the hyperbola is $x^{2}-\\frac{y^{2}}{3}=1$, so the vertices of the hyperbola are $(\\pm1,0)$, and the asymptotes are $y=\\pm\\sqrt{3}x$. The distance from the point $(1,0)$ to the line $\\sqrt{3}x-y=0$ is $\\frac{\\sqrt{3}}{2}$. Hence, fill in $\\frac{\\sqrt{3}}{2}$."
  },
  {
    "text": "Given $B(-1 , 0)$, $C(1 , 0)$, $|A B|+|A C|=10$, then the trajectory equation of point $A$ is?",
    "fact_expressions": "B: Point;C: Point;A: Point;Coordinate(B) = (-1, 0);Coordinate(C) = (1, 0);Abs(LineSegmentOf(A, B)) + Abs(LineSegmentOf(A, C)) = 10",
    "query_expressions": "LocusEquation(A)",
    "answer_expressions": "x^2/25+y^2/24=1",
    "fact_spans": "[[[2, 14]], [[16, 27]], [[46, 50]], [[2, 14]], [[16, 27]], [[28, 44]]]",
    "query_spans": "[[[46, 57]]]",
    "process": ""
  },
  {
    "text": "Given that point $P(x_{0}, y_{0})$ lies on the parabola $C$: $x^{2}=4y$, and the distance from point $P$ to the focus of parabola $C$ is $3$, then $|x_{0}|=$?",
    "fact_expressions": "C: Parabola;P: Point;x0:Number;y0:Number;Expression(C) = (x^2 = 4*y);Coordinate(P) = (x0, y0);PointOnCurve(P, C);Distance(P, Focus(C)) = 3",
    "query_expressions": "Abs(x0)",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[21, 40], [50, 56]], [[2, 20], [45, 49]], [[67, 76]], [[3, 20]], [[21, 40]], [[2, 20]], [[2, 43]], [[45, 65]]]",
    "query_spans": "[[[67, 78]]]",
    "process": "According to the focal radius formula of the parabola, solve for $ y_{0} = 2 $. Then, since the point lies on the parabola, the coordinates of the point satisfy the equation of the parabola, yielding $ |x_{0}| = 2\\sqrt{2} $. [Solution] Let the parabola $ C: x^{2} = 4y $ have focus $ F $, then $ F(0,1) $. According to the focal radius formula of the parabola, we have: $ PF = y_{0} + 1 = 3 $, so $ y_{0} = 2 $. Substituting into the parabola equation gives: $ x_{0}^{2} = 4y_{0} = 8 $. Hence, $ |x_{0}| = 2\\sqrt{2} $."
  },
  {
    "text": "Given the ellipse $C_{1}: \\frac{x^{2}}{4} + \\frac{y^{2}}{b^{2}} = 1$ $(0 < b < 2)$ has eccentricity $\\frac{1}{2}$, $F_{1}$ and $F_{2}$ are the left and right foci of $C_{1}$, $M$ is a moving point on $C_{1}$, point $N$ lies on the extension of segment $F_{1}M$, $|MN| = |MF_{2}|$, and $P$ is the midpoint of segment $F_{2}N$. Then the maximum value of $|F_{1}P|$ is?",
    "fact_expressions": "C1: Ellipse;b:Number;0<b;b<2;F1: Point;M: Point;F2: Point;N: Point;P: Point;Expression(C1) = (x^2/4 + y^2/b^2 = 1);Eccentricity(C1)=1/2;LeftFocus(C1) = F1;RightFocus(C1) = F2;PointOnCurve(M, C1);PointOnCurve(N, OverlappingLine(LineSegmentOf(F1,M)));Abs(LineSegmentOf(M, N)) = Abs(LineSegmentOf(M, F2));MidPoint(LineSegmentOf(F2,N)) = P",
    "query_expressions": "Max(Abs(LineSegmentOf(F1, P)))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 58], [94, 101], [111, 118]], [[12, 58]], [[12, 58]], [[12, 58]], [[78, 85]], [[107, 110]], [[86, 93]], [[123, 127]], [[178, 181]], [[2, 58]], [[2, 76]], [[78, 106]], [[78, 106]], [[107, 122]], [[123, 144]], [[145, 162]], [[163, 181]]]",
    "query_spans": "[[[183, 200]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is any point on the parabola $y^{2}=4x$, $F$ is the focus of the parabola, and $M(2,1)$ is a fixed point in the plane, find the minimum value of $|PF|+|PM|$.",
    "fact_expressions": "G: Parabola;M: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(M) = (2, 1);PointOnCurve(P, G);Focus(G) = F",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F)) + Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "3",
    "fact_spans": "[[[6, 20], [30, 33]], [[37, 45]], [[2, 5]], [[26, 29]], [[6, 20]], [[37, 45]], [[2, 25]], [[26, 36]]]",
    "query_spans": "[[[54, 73]]]",
    "process": "Using the definition of a parabola |PF| = |PD|, and combining with the graph, we can solve it. [Detailed solution] From the problem, the directrix of the parabola $ y^{2} = 4x $ is $ x = -1 $. Let point $ P $'s projection on the directrix be $ D $. Then, according to the definition of the parabola, $ |PF| = |PD| $. Therefore, to minimize $ |PF| + |PM| $, we need to minimize $ |PM| + |PD| $. The minimum occurs when points $ D $, $ P $, and $ M $ are collinear, and the minimum value is $ 2 - (-1) = 3 $."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{8-k}+\\frac{y^{2}}{k-6}=1$ represents an ellipse, then the range of real values for $k$ is?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G)=(x^2/(8-k)+y^2/(k-6)=1)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(6,7)+(7,8)",
    "fact_spans": "[[[44, 46]], [[48, 53]], [[1, 46]]]",
    "query_spans": "[[[48, 60]]]",
    "process": "Since the equation $\\frac{x^{2}}{8-k}+\\frac{y^{2}}{k-6}=1$ represents an ellipse, we have $\\begin{cases}8-k>0\\\\k-6>0\\\\8-k\\neq k-6\\end{cases}$, solving which yields $6<k<8$ and $k\\neq7$. Therefore, the range of real values for $k$ is $(6,7)\\cup(7,8)$."
  },
  {
    "text": "A line passing through the focus of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. If $|AB|=10$, then the distance from the midpoint $P$ of $AB$ to the $y$-axis is equal to?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Line;PointOnCurve(Focus(G), H);A: Point;B: Point;Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 10;P: Point;MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Distance(P, yAxis)",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 15], [21, 24]], [[1, 15]], [[18, 20]], [[0, 20]], [[25, 28]], [[29, 32]], [[18, 34]], [[36, 46]], [[56, 59]], [[48, 59]]]",
    "query_spans": "[[[56, 70]]]",
    "process": "The parabola $ y^{2} = 4x $ has focus $ E(1,0) $ and directrix $ x = -1 $. Since the midpoint of $ AB $ is $ P $, draw perpendiculars from $ A $, $ P $, $ B $ to the directrix, with feet at $ C $, $ F $, $ D $ respectively. $ PF $ intersects the vertical axis at point $ H $, as shown in the figure. By the definition of a parabola, $ AC = AE $, $ BD = BE $. Since $ PF $ is the median of the right trapezoid, $ PF = \\frac{AC + BD}{2} = \\frac{AE + EB}{2} = \\frac{AB}{2} = 5 $, $ \\therefore PH = PF - FH = 5 - 1 = 4 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$, the left and right foci are $F_{1}(-1,0)$ and $F_{2}(1,0)$ respectively. $P$ is any point on the hyperbola. If the range of the eccentricity of the hyperbola is $[2,4]$, then the range of the minimum value of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;F1: Point;P: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-1, 0);Coordinate(F2)=(1, 0);LeftFocus(C) =F1;RightFocus(C)=F2;PointOnCurve(P, C);Range(Eccentricity(C))=[2,4]",
    "query_expressions": "Range(Min(DotProduct(VectorOf(P, F1), VectorOf(P, F2))))",
    "answer_expressions": "[-15/16, -3/4]",
    "fact_spans": "[[[2, 64], [108, 111], [117, 120]], [[10, 64]], [[10, 64]], [[73, 86]], [[104, 107]], [[89, 102]], [[10, 64]], [[10, 64]], [[2, 64]], [[73, 86]], [[89, 102]], [[2, 101]], [[2, 101]], [[103, 115]], [[117, 137]]]",
    "query_spans": "[[[139, 207]]]",
    "process": "Let P(m,n), then \\frac{m^{2}}{a^{2}}-\\frac{n^{2}}{b^{2}}=1, i.e., m^{2}=a^{2}(1+\\frac{n^{2}}{b^{2}}). Also F_{1}(-1,0), F_{2}(1,0), then =(-1-m,-n), =(1-m,-n), \\cdot=n^{2}+m^{2}-1=n^{2}+a^{2}(1+\\frac{n^{2}}{b^{2}})-1=n^{2}(1+\\frac{a^{2}}{b^{2}})+a^{2}-1\\geqslant a^{2}-1, with equality if and only if n=0. Thus the minimum value is a^{2}-1. From 2\\leqslant\\frac{1}{a}\\leqslant4, we get \\frac{1}{4}\\leqslanta\\leqslant\\frac{1}{2}. Hence -\\frac{15}{16}\\leqslanta^{2}-1\\leqslant-\\frac{3}{4}, i.e., the range of minimum values of \\overrightarrow{PF}\\cdot\\overrightarrow{PF_{2}} is \\left[-\\frac{15}{16}, -\\frac{3}{4}\\right]."
  },
  {
    "text": "What is the focal distance of the ellipse $x^{2}+4 y^{2}=16$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 4*y^2 = 16)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[0, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 25]]]",
    "process": "Transform the equation into the standard form of an ellipse, and then find c; the result can be obtained from the focal distance 2c. [Detailed solution] From x^{2}+4y^{2}=16, we get: \\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1. \\therefore c^{2}=16-4=12, solving yields: c=2\\sqrt{3}. \\therefore The focal distance is: 2c=4\\sqrt{3}"
  },
  {
    "text": "The standard equation of a hyperbola with focal length $4$ passing through the point $(\\sqrt{3}, 0)$ is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (sqrt(3), 0);FocalLength(G) = 4;PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/3 - y^2/1 = 1",
    "fact_spans": "[[[26, 29]], [[8, 25]], [[8, 25]], [[0, 29]], [[7, 29]]]",
    "query_spans": "[[[26, 36]]]",
    "process": ""
  },
  {
    "text": "The two foci of an ellipse are given as $F_{1}(0, -8)$ and $F_{2}(0, 8)$, and the sum of the distances from any point on the ellipse to the two foci is $20$. What is the equation of this ellipse?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Coordinate(F1) = (0,-8);Coordinate(F2) = (0, 8);Focus(G) = {F1,F2};P: Point;PointOnCurve(P,G) = True;Distance(P,F1) + Distance(P,F2) = 20",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/100+y^2/36=1",
    "fact_spans": "[[[2, 4], [45, 47], [68, 70]], [[12, 26]], [[29, 43]], [[12, 26]], [[29, 43]], [[2, 43]], [], [[45, 50]], [[45, 65]]]",
    "query_spans": "[[[68, 75]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the common foci of an ellipse and a hyperbola, $P$ is one of their common points, and $\\angle F_{1} P F_{2}=\\frac{\\pi}{4}$, then the minimum value of the product of the eccentricities of the ellipse and the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;F1: Point;P: Point;F2: Point;Focus(G)={F1,F2};Focus(H)={F1,F2};OneOf(Intersection(G,H))=P;AngleOf(F1, P, F2) = pi/4",
    "query_expressions": "Min(Eccentricity(G)*Eccentricity(H))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[21, 24], [85, 88]], [[18, 20], [82, 84]], [[2, 9]], [[30, 33]], [[10, 17]], [[2, 29]], [[2, 29]], [[30, 42]], [[44, 80]]]",
    "query_spans": "[[[82, 100]]]",
    "process": "As shown in the figure, let the semi-major axis of the ellipse be $ a_{1} $, and the semi-transverse axis of the hyperbola be $ a_{2} $. Then according to the definitions of the ellipse and hyperbola: $ |PF_{1}| + |PF_{2}| = 2a_{1} $, $ |PF_{1}| - |PF_{2}| = 2a_{2} $. Therefore, $ |PF_{1}| = a_{1} + a_{2} $, $ |PF_{2}| = a_{1} - a_{2} $. Let $ |F_{1}F_{2}| = 2c $, $ \\angle F_{1}PF_{2} = \\frac{\\pi}{4} $, then: in $ \\triangle PF_{1}F_{2} $, by the cosine law we get, $ 4c^{2} = (a_{1}+a_{2})^{2} + (a_{1}-a_{2})^{2} - 2(a_{1}+a_{2})(a_{1}-a_{2})\\cos\\frac{\\pi}{4} $. Simplifying yields: $ (2-\\sqrt{2})a^{2} + (2+\\sqrt{2})a_{2}^{2} = 4c^{2} $, i.e., $ \\frac{2-\\sqrt{2}}{e} + \\frac{2+\\sqrt{2}}{e} = 4'2\\cdot\\frac{2-\\sqrt{2}}{e_{1}^{2}} + \\frac{2+\\sqrt{2}}{e_{2}^{2}} \\geqslant 2\\sqrt{2}\\cdot\\frac{1}{e_{1}e}000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 $ and the minimum value of the product of the eccentricities of the hyperbola is $ \\frac{\\sqrt{2}}{2} $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and $P$ is a point on the ellipse such that $\\angle F_{1} P F_{2}=90^{\\circ}$, then what is the area of $\\triangle P F_{1} F_{2}$?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/25 + y^2/9 = 1);Focus(C) = {F1, F2};PointOnCurve(P, C);AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "9",
    "fact_spans": "[[[18, 61], [71, 73]], [[67, 70]], [[2, 9]], [[10, 17]], [[18, 61]], [[2, 66]], [[67, 76]], [[78, 111]]]",
    "query_spans": "[[[113, 142]]]",
    "process": "According to the equation of the ellipse, obtain c, and get |F_{1}F_{2}|. Let |PF_{1}|=t_{1}, |PF_{2}|=t_{2}. Using the Pythagorean theorem and the definition of the ellipse, the value of t_{1}t_{2} can be found, thus the area of the triangle can be determined. Detailed solution: \\because a=5, b=3; \\therefore c=4. Let |PF_{1}|=t_{1}, |PF_{2}|=t_{2}, then t_{1}+t_{2}=10\\textcircled{1}, t_{1}^{2}+t_{2}^{2}=8^{2}\\textcircled{2}. From \\textcircled{1}^{2}-\\textcircled{2}, we get t_{1}t_{2}=18. \\therefore S_{\\Delta F_{1}PF_{2}}=\\frac{1}{2}t_{1}t_{2}=\\frac{1}{2}\\times 18=9."
  },
  {
    "text": "If an asymptote of a hyperbola passes through the point $(8,-6)$, then its eccentricity is equal to?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (8,-6);PointOnCurve(H,OneOf(Asymptote(G)))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{5/4, 5/3}",
    "fact_spans": "[[[1, 4], [23, 24]], [[12, 21]], [[12, 21]], [[1, 21]]]",
    "query_spans": "[[[23, 30]]]",
    "process": "Let an asymptote equation be $ y = kx $. From the given condition, $ -6 = 8k $, we get $ k = -\\frac{3}{4} $. Therefore, the asymptote equation is $ y = -\\frac{3}{4}x $. If the focus lies on the $ x $-axis, then $ \\frac{b}{a} = \\frac{3}{4} $, so the eccentricity $ e = \\frac{c}{a} = \\sqrt{1+}\\frac{2}{12} $. If the focus lies on the $ y $-axis, then $ \\frac{a}{b} = \\frac{3}{4} $, so the eccentricity $ e = \\frac{c}{a} = \\sqrt{1+(\\frac{b}{a})^{2}} = \\frac{5}{3} $."
  },
  {
    "text": "A line passing through the left focus $F$ of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$ intersects the left branch of the hyperbola at points $M$ and $N$, and $F_{2}$ is its right focus. Then the value of $|M F_{2}|+|NF_{2}|-|M N|$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;M: Point;F2: Point;N: Point;F: Point;Expression(G) = (x^2/4 - y^2/3 = 1);LeftFocus(G) = F;PointOnCurve(F, H);Intersection(H, LeftPart(G)) = {M, N};RightFocus(G) = F2",
    "query_expressions": "Abs(LineSegmentOf(M, F2)) + Abs(LineSegmentOf(N, F2)) - Abs(LineSegmentOf(M, N))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 39], [74, 75], [49, 52]], [[46, 48]], [[56, 59]], [[66, 73]], [[60, 63]], [[42, 45]], [[1, 39]], [[1, 45]], [[0, 48]], [[46, 65]], [[66, 78]]]",
    "query_spans": "[[[80, 110]]]",
    "process": "According to the first definition of hyperbola, |MF_{2}|\\cdot|MF|=2a, |NF_{2}|\\cdot|NF|=2a, adding these two equations gives |MF_{2}|+|NF_{2}|\\cdot|MN|. According to the definition of hyperbola, |MF_{2}|\\cdot|MF|=2a, |NF_{2}|\\cdot|NF|=2a, adding these two equations gives |MF_{2}|+|NF_{2}|\\cdot|MN|=4a=8. Answer: 8."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ and the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$ have the same left and right foci $F_{1}$, $F_{2}$, and point $P$ is the intersection point of the ellipse and the hyperbola in the first quadrant, then the value of $\\frac{|P F_{1}|}{|P F_{2}|}$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/2 - y^2 = 1);Expression(H) = (x^2/4 + y^2 = 1);LeftFocus(G) = F1;LeftFocus(H) = F1;RightFocus(H) = F2;RightFocus(G) = F2;Quadrant(P) = 1;Intersection(H, G) = P",
    "query_expressions": "Abs(LineSegmentOf(P, F1))/Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "3 + 2*sqrt(2)",
    "fact_spans": "[[[30, 58], [91, 94]], [[2, 29], [88, 90]], [[83, 87]], [[67, 74]], [[75, 82]], [[30, 58]], [[2, 29]], [[2, 82]], [[2, 82]], [[2, 82]], [[2, 82]], [[83, 102]], [[83, 102]]]",
    "query_spans": "[[[104, 137]]]",
    "process": "Let |PF_{1}| = m, |PF_{2}| = n. According to the definition of the ellipse, we have m + n = 4; according to the definition of the hyperbola, we have m - n = 2\\sqrt{2}. Solving gives m = 2 + \\sqrt{2}, n = 2 - \\sqrt{2}, \\frac{m}{n} = \\frac{2 + \\sqrt{2}}{2 - \\sqrt{2}} = 3 + 2\\sqrt{2}"
  },
  {
    "text": "The distance from the point $M(-1,0)$ to the asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{9}=1$ is?",
    "fact_expressions": "G: Hyperbola;M: Point;Expression(G) = (x^2 - y^2/9 = 1);Coordinate(M) = (-1, 0)",
    "query_expressions": "Distance(M,Asymptote(G))",
    "answer_expressions": "3*sqrt(10)/10",
    "fact_spans": "[[[11, 39]], [[0, 10]], [[11, 39]], [[0, 10]]]",
    "query_spans": "[[[0, 48]]]",
    "process": "x^2 - \\frac{y^{2}}{9} = 1 has an asymptote equation: y = 3x, that is, 3x - y = 0. By symmetry, the distances are equal. Using the point-to-line distance formula, we get: d = \\frac{|-3|}{\\sqrt{10}} = \\frac{3\\sqrt{10}}{10}"
  },
  {
    "text": "Given a fixed point $A(2 , 0)$, what is the trajectory equation of the midpoint $M$ of the line segment connecting $A$ and a moving point $P$ on the parabola $y^{2}=x$?",
    "fact_expressions": "G: Parabola;A: Point;Expression(G) = (y^2 = x);Coordinate(A) = (2, 0);P: Point;PointOnCurve(P, G);M: Point;MidPoint(LineSegmentOf(A, P)) = M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "2*y^2 = x - 1",
    "fact_spans": "[[[17, 29]], [[4, 14], [15, 16]], [[17, 29]], [[4, 14]], [[33, 36]], [[17, 36]], [[41, 44]], [[15, 44]]]",
    "query_spans": "[[[41, 51]]]",
    "process": ""
  },
  {
    "text": "If the line $l$ passes through the focus of the parabola $y = a x^{2}$ $(a > 0)$ and is perpendicular to the $y$-axis, and the segment of $l$ intercepted by the parabola has length $4$, then $a = $?",
    "fact_expressions": "l: Line;G: Parabola;Expression(G) = (y = a*x^2);a: Number;a>0;PointOnCurve(Focus(G), l);IsPerpendicular(l, yAxis);Length(InterceptChord(l, G)) = 4",
    "query_expressions": "a",
    "answer_expressions": "1/4",
    "fact_spans": "[[[1, 6], [42, 45]], [[7, 27], [46, 49]], [[7, 27]], [[61, 64]], [[10, 27]], [[1, 30]], [[1, 40]], [[42, 59]]]",
    "query_spans": "[[[61, 66]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{10}+\\frac{y^{2}}{6}=1$, a line with an inclination angle of $60^{\\circ}$ intersects the ellipse at points $A$ and $B$ such that $|A B|=\\frac{8 \\sqrt{30}}{9}$. Point $C$ is a point on the ellipse distinct from $A$ and $B$. Then, the maximum area of $\\triangle A B C$ is?",
    "fact_expressions": "G:Ellipse;Expression(G)=(x^2/10+y^2/6=1);l:Line;Inclination(l)=ApplyUnit(60,degree);A:Point;B:Point;Intersection(l,G)={A,B};Abs(LineSegmentOf(A,B))=8*sqrt(30)/9;PointOnCurve(C,G);C:Point;Negation(C=A);Negation(C=B)",
    "query_expressions": "Max(Area(TriangleOf(A,B,C)))",
    "answer_expressions": "16*sqrt(30)/9",
    "fact_spans": "[[[2, 40], [61, 63], [112, 114]], [[2, 40]], [[58, 60]], [[41, 60]], [[67, 70], [118, 121]], [[71, 74], [122, 125]], [[58, 76]], [[77, 106]], [[107, 127]], [[107, 111]], [[107, 127]], [[107, 127]]]",
    "query_spans": "[[[129, 154]]]",
    "process": "According to the problem, let the equation of line AB be $ y = \\sqrt{3}x + m $, and let point $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Solving the system of equations\n\\[\n\\begin{cases}\ny = \\sqrt{3}x + m \\\\\n\\frac{x^{2}}{10} + \\frac{y^{2}}{6} = 1\n\\end{cases}\n\\]\nwe obtain $ 18x^{2} + 10\\sqrt{3}mx + 5m^{2} - 30 = 0 $. Thus,\n$ x_{1} + x_{2} = \\frac{-5\\sqrt{3}m}{9} $, $ x_{1}x_{2} = \\frac{5m^{2} - 30}{18} $.  \nSince $ |AB| = \\frac{8\\sqrt{30}}{9} $, that is,\n$ \\sqrt{(1+3)\\left[(x_{1} + x_{2})^{2} - 4x_{1}x_{2}\\right]} = \\frac{8\\sqrt{30}}{9} $,  \nsubstituting and simplifying gives $ m^{2} = 4 $, so $ m = \\pm 2 $. Without loss of generality, take $ m = 2 $, then the equation of line AB is $ y = \\sqrt{3}x + 2 $.  \nLet the line parallel to AB and tangent to the ellipse be $ y = \\sqrt{3}x + t $. Solving the system\n\\[\n\\begin{cases}\ny = \\sqrt{3}x + t \\\\\n\\frac{x^{2}}{10} + \\frac{y^{2}}{6} = 1\n\\end{cases}\n\\]\nwe get $ 18x^{2} + 10\\sqrt{3}tx + 5t^{2} - 30 = 0 $. From $ \\Delta = 300t^{2} - 72(5t^{2} - 30) = 0 $, we solve $ t = \\pm 6 $. Taking $ t = -6 $, the distance $ d $ between the line $ y = \\sqrt{3}x + t $ (tangent and parallel to AB) and line AB is $ d = \\frac{8}{\\sqrt{1 + (\\sqrt{3})^{2}}} $. Therefore, the maximum area of $ \\triangle ABC $ is\n$ S = \\frac{1}{2}d|AB| = \\frac{1}{2} \\times 4 \\times \\frac{8\\sqrt{30}}{9} = \\frac{16\\sqrt{30}}{9} $."
  },
  {
    "text": "If point $P$ is a point on the parabola $y^{2}=10 x$, then the minimum distance from point $P$ to the line $x+y+5=0$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 10*x);P: Point;PointOnCurve(P, G);H: Line;Expression(H) = (x + y + 5 = 0)",
    "query_expressions": "Min(Distance(P, H))",
    "answer_expressions": "5*sqrt(2)/4",
    "fact_spans": "[[[6, 21]], [[6, 21]], [[1, 5], [26, 30]], [[1, 24]], [[31, 42]], [[31, 42]]]",
    "query_spans": "[[[26, 50]]]",
    "process": ""
  },
  {
    "text": "The vertex of the parabola is at the origin, the axis of symmetry is a coordinate axis, and the focus lies on the line $x - y + 4 = 0$. Then the equation of this parabola is?",
    "fact_expressions": "G: Parabola;H: Line;O: Origin;Expression(H) = (x - y + 4 = 0);Vertex(G) = O;SymmetryAxis(G)=axis;PointOnCurve(Focus(G), H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=-16*x,x^2=16*y}",
    "fact_spans": "[[[0, 3], [0, 3]], [[22, 33]], [[7, 9]], [[22, 33]], [[0, 9]], [[0, 17]], [[0, 34]]]",
    "query_spans": "[[[37, 44]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, and the line $y=k(x-1)$ intersects the parabola $C$ at points $P$ and $Q$, then $\\frac{1}{|F P|}+\\frac{1}{|F Q|}$=?",
    "fact_expressions": "C: Parabola;G: Line;F: Point;P: Point;Q: Point;k:Number;Expression(C) = (y^2 = 4*x);Expression(G) = (y = k*(x - 1));Focus(C) = F;Intersection(G, C) = {P, Q}",
    "query_expressions": "1/Abs(LineSegmentOf(F, Q)) + 1/Abs(LineSegmentOf(F, P))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 21], [42, 48]], [[29, 41]], [[25, 28]], [[51, 54]], [[55, 58]], [[31, 41]], [[2, 21]], [[29, 41]], [[2, 28]], [[29, 60]]]",
    "query_spans": "[[[62, 97]]]",
    "process": "Let P(x_{1},y_{1}), Q(x_{2},y_{2}) be given by \\begin{cases} y = k(x - \\\\ y2 = 4x \\end{cases} k(x - 1) = \\frac{1}{4x} yields ky^{2} - 4y - 4k = 0, so y_{1} + y_{2} = \\frac{4}{k}, y_{1}y_{2} = -4 \\frac{1}{2} \\frac{4}{k^{2}} + 2, x_{1}x_{2} = \\frac{1}{16}(y_{2}y_{2})^{2} = 1,"
  },
  {
    "text": "Given $|AB|=4$, $O$ is the midpoint of segment $AB$, and point $P$ moves in the plane containing $A$ and $B$ such that $|PA|+|PB|=6$. Then the maximum and minimum values of $|PO|$ are respectively?",
    "fact_expressions": "A: Point;B: Point;Abs(LineSegmentOf(A, B)) = 4;O: Origin;MidPoint(LineSegmentOf(A, B)) = O;P: Point;Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B)) = 6",
    "query_expressions": "Max(Abs(LineSegmentOf(P, O)));Min(Abs(LineSegmentOf(P, O)))",
    "answer_expressions": "3\nsqrt(5)",
    "fact_spans": "[[[31, 34]], [[35, 38]], [[2, 10]], [[12, 15]], [[12, 25]], [[26, 30]], [[49, 62]]]",
    "query_spans": "[[[64, 82]], [[64, 82]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with right focus $F$, a line passing through the origin intersects the ellipse at points $A$ and $B$, $|AB|=4$, $|BF|=2\\sqrt{3}$, $\\angle ABF=30^{\\circ}$. Find the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;A: Point;B: Point;F: Point;a > b;b > 0;L:Line;O:Origin;RightFocus(G) = F;PointOnCurve(O,L);Intersection(L,G) = {A, B};Abs(LineSegmentOf(A, B)) = 4;Abs(LineSegmentOf(B, F)) = 2*sqrt(3);AngleOf(A, B, F) = ApplyUnit(30, degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[2, 54], [70, 72], [142, 144]], [[2, 54]], [[4, 54]], [[4, 54]], [[73, 76]], [[77, 80]], [[59, 62]], [[4, 54]], [[4, 54]], [[67, 69]], [[64, 66]], [[2, 62]], [[63, 69]], [[67, 82]], [[83, 92]], [[94, 112]], [[115, 140]]]",
    "query_spans": "[[[142, 150]]]",
    "process": "Let the left focus of the ellipse be $F_{1}$. According to the given conditions and by symmetry, we obtain $|AF_{1}|=|BF|=2\\sqrt{3}$, $|BF_{1}|=|AF|$. Combining with the definition of the ellipse, we get $|AF|$. Using the law of cosines in triangle $ABF$, set up an equation to solve for $a$; in triangle $OBF$, use the law of cosines to solve for $c$, and then find the eccentricity. Denote the left focus of the ellipse as $F_{1}$. Since a line passing through the origin intersects the ellipse at points $A$ and $B$, $|AB|=4$, $|BF|=2\\sqrt{3}$. By symmetry, we have $|AF_{1}|=|BF|=2\\sqrt{3}$, $|BF_{1}|=|AF|$. From the definition of the ellipse, $|AF|=2a-|AF_{1}|=2a-|BF|=2a-2\\sqrt{3}$. In triangle $ABF$, $\\angle ABF=30^{\\circ}$. Thus, by the law of cosines:  \n$$\n|AF|^{2}=|AB|^{2}+|BF|^{2}-2|AB|\\cdot|BF|\\cos\\angle ABF\n$$  \nHence,\n$$\n(2a-2\\sqrt{3})^{2}=4^{2}+(2\\sqrt{3})^{2}-2\\times4\\times2\\sqrt{3}\\cos30^{\\circ}\n$$  \nSolving gives $a=\\sqrt{3}+1$. In triangle $OBF$, $\\angle ABF=30^{\\circ}$, $|OB|=2$. By the law of cosines:\n$$\n|OF|^{2}=|OB|^{2}+|BF|^{2}-2|OB|\\cdot|BF|\\cos\\angle ABF=4+12-2\\times2\\times2\\sqrt{3}\\times\\frac{\\sqrt{3}}{2}=4\n$$  \nSo $c=|OF|=2$, therefore $e=\\frac{c}{a}=\\frac{2}{\\sqrt{3}+1}=\\sqrt{3}-1$"
  },
  {
    "text": "Given the parabola $E$: $y^{2}=2px$ ($p>0$) with focus $F$, a line $l$ passing through point $F$ intersects the parabola $E$ at points $A$ and $B$, and intersects the circle $x^{2}-px+y^{2}-\\frac{3}{4}p^{2}=0$ at points $C$ and $D$. If $|AB|=2|CD|$, then the slope of line $l$ is?",
    "fact_expressions": "l: Line;E: Parabola;p: Number;G: Circle;A: Point;B: Point;C: Point;D: Point;F: Point;p>0;Expression(E) = (y^2 = 2*p*x);Expression(G) = (x^2-p*x+y^2-(3/4)*p^2=0);Focus(E) = F;PointOnCurve(F, l);Intersection(l, E) = {A, B};Intersection(l,G)={C,D};Abs(LineSegmentOf(A, B)) = 2*Abs(LineSegmentOf(C, D))",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*1",
    "fact_spans": "[[[41, 46], [66, 71], [138, 143]], [[2, 27], [47, 53]], [[10, 27]], [[72, 110]], [[55, 58]], [[59, 62]], [[112, 115]], [[116, 119]], [[31, 34], [36, 40]], [[10, 27]], [[2, 27]], [[72, 110]], [[2, 34]], [[35, 46]], [[41, 64]], [[66, 121]], [[122, 136]]]",
    "query_spans": "[[[138, 148]]]",
    "process": "From the given conditions, the equation of the circle is $(x-\\frac{p}{2})^{2}+y^{2}=p^{2}$, so the center of the circle is $(\\frac{p}{2},0)$, which is the focus of the parabola. Therefore, $|CD|=2p$, and thus $|AB|=4p$. Let the line $l: x=ty+\\frac{p}{2}$; substituting into $y^{2}=2px$ $(p>0)$ gives $y^{2}-2pty-p^{2}=0$. Let the intersection points of line $l$ and parabola $E$ be $A(x_{1},y_{1})$, $B(x_{2},y_{2})$. Then $y_{1}+y_{2}=2pt$, $y_{1}y_{2}=-p^{2}$. Then $|AB|=\\sqrt{(1+t^{2})(4p^{2}t^{2}+4p^{2})}=2p(1+t^{2})=4p$, so $1+t^{2}=2$, solving yields $t=\\pm1$."
  },
  {
    "text": "Given that point $P(2, -3)$ lies on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and the distance between the two foci of the hyperbola is $4$, what is the equation of this hyperbola?",
    "fact_expressions": "P: Point;Coordinate(P) = (2, -3);G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;PointOnCurve(P, G) = True;F1: Point;F2: Point;Focus(G) = {F1, F2} ;Distance(F1, F2) = 4",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/1 - y^2/3 = 1",
    "fact_spans": "[[[2, 13]], [[2, 13]], [[14, 70], [74, 77], [93, 96]], [[14, 70]], [[17, 70]], [[17, 70]], [[17, 70]], [[17, 70]], [[2, 73]], [], [], [[74, 81]], [[74, 90]]]",
    "query_spans": "[[[93, 100]]]",
    "process": ""
  },
  {
    "text": "Let the circle $x^{2}+y^{2}-2 x-4 y+4=0$ be tangent to the asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1$, then the real number $b$=?",
    "fact_expressions": "H: Circle;Expression(H) = (-4*y - 2*x + x^2 + y^2 + 4 = 0);G: Hyperbola;Expression(G) = (x^2 - y^2/b^2 = 1);b: Real;IsTangent(H,Asymptote(G)) = True",
    "query_expressions": "b",
    "answer_expressions": "{-3/4, 3/4}",
    "fact_spans": "[[[1, 27]], [[1, 27]], [[28, 60]], [[28, 60]], [[68, 73]], [[1, 66]]]",
    "query_spans": "[[[68, 75]]]",
    "process": "According to the problem, the center of the circle has coordinates (1,2) and radius r=1. The asymptotes of the hyperbola are given by: bx+y=0 or -bx+y=0.  \n\\frac{|b+2|}{\\sqrt{1+b^{2}}}=1 or \\frac{|-b+2|}{\\sqrt{1+b^{2}}}=1. Solving yields: b=-\\frac{3}{4} or b=\\frac{3}{4}."
  },
  {
    "text": "The equation of an ellipse with foci on the $x$-axis, the distance between the two directrices being $\\frac{18 \\sqrt{5}}{5}$, and focal length $2 \\sqrt{5}$ is?",
    "fact_expressions": "G: Ellipse;PointOnCurve(Focus(G), xAxis);l1: Line;l2: Line;Directrix(G) = {l1,l2};Distance(l1,l2) = 18*sqrt(5)/5;FocalLength(G) = 2*sqrt(5)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 + y^2/4 = 1",
    "fact_spans": "[[[56, 58]], [[0, 58]], [], [], [[8, 58]], [[8, 58]], [[40, 58]]]",
    "query_spans": "[[[56, 62]]]",
    "process": "Set up a system of equations based on the distance from the directrix and the focal length, solve the system to find a and b, and thus obtain the ellipse equation. Let the ellipse equation be \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0). According to the given conditions:\n\n\\begin{cases}2\\times\\frac{a^{2}}{c}=\\frac{18\\sqrt{5}}{5}\\\\2c=2\\sqrt{5}\\\\a^{2}=b^{2}+c^{2}\\end{cases}\n\nSolving gives a=3, b=2, c=\\sqrt{5}. Therefore, the ellipse equation is \\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1"
  },
  {
    "text": "Given two points $A$ and $B$ on the parabola $y^{2}=4x$ with focus $F$ such that $\\overrightarrow{A F}=2 \\overrightarrow{F B}$, what is the distance from the midpoint of chord $AB$ to the directrix?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(A,G);PointOnCurve(B,G);VectorOf(A, F) = 2*VectorOf(F, B);IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A,B)),Directrix(G))",
    "answer_expressions": "9/4",
    "fact_spans": "[[[10, 24]], [[28, 31]], [[32, 35]], [[3, 6]], [[10, 24]], [[2, 24]], [[10, 35]], [[10, 35]], [[37, 82]], [[10, 89]]]",
    "query_spans": "[[[10, 99]]]",
    "process": "Let $ A(x_{1},y_{1}), B(x_{2},y_{2}) $, with midpoint $ D(x_{0},y_{0}) $, $ F(1,0) $. From $ \\overrightarrow{AF}=2\\overrightarrow{FB} $ we get $ f(x)=\\frac{1}{2}(-y_{1}=2y_{2} \\therefore \\begin{cases} x_{1}+2x_{2}=3 \\\\ y_{1}+2y_{2}=0 \\end{cases} \\frac{3-x_{2}}{2} $, since $ \\begin{cases} y_{1}=4x_{1} \\\\ y_{2}^{2}=4x_{2} \\end{cases} $, subtracting the two equations gives $ (y_{1}-y_{2})(y_{1}+y_{2})=4(x_{1}-x_{2}) $. Hence $ k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{4}{y_{1}+y_{2}} $, $ \\therefore y_{2}(y_{1}+y_{2})=4(x_{2}-1) $, that is $ -y_{2}^{2}=4(x_{2}-1) $, and $ -y_{2}^{2}=-4x_{2} $, $ \\therefore 4(x_{2}-1)=-4x_{2} $, we get $ x_{2}=\\frac{1}{2} $, $ \\therefore x_{0}=\\frac{3-x_{2}}{2}=\\frac{5}{4} $, the distance from the midpoint of $ AB $ to the directrix of the parabola is $ d=x_{0}+1=\\frac{9}{4} $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$, and the line $m$ passing through the point $(-1,0)$ with slope $1$, what is the length of the chord intercepted on the ellipse $C$ by the line $m$?",
    "fact_expressions": "m: Line;C: Ellipse;G: Point;Expression(C) = (x^2/2 + y^2 = 1);Coordinate(G) = (-1, 0);PointOnCurve(G, m);Slope(m) = 1",
    "query_expressions": "Length(InterceptChord(C,m))",
    "answer_expressions": "4*sqrt(2)/3",
    "fact_spans": "[[[35, 40], [65, 70]], [[2, 34], [59, 64]], [[41, 50]], [[2, 34]], [[41, 50]], [[35, 50]], [[35, 57]]]",
    "query_spans": "[[[59, 77]]]",
    "process": "From the problem, the equation of the line is: y = x + 1. Combining with the ellipse equation: \\frac{x^{2}}{2} + y^{2} = 1, eliminating y and simplifying gives a quadratic equation in x: 3x^{2} + 4x = 0. Let the intersection points be A and B. It is easy to solve for A(0,1), B(-\\frac{4}{3}, -\\frac{1}{3}). Using the distance formula between two points, the chord length _{AB} = \\sqrt{(\\frac{4}{3})^{2} + (\\frac{4}{3})^{2}} = \\frac{4\\sqrt{2}}{3}"
  },
  {
    "text": "In $\\triangle A B C$, the coordinates of $B$ and $C$ are known to be $(-3 , 0)$ and $(3 , 0)$, respectively, and the perimeter of $\\triangle A B C$ is equal to $16$. Then the trajectory equation of vertex $A$ is?",
    "fact_expressions": "A: Point;B: Point;C: Point;Coordinate(B) = (-3, 0);Coordinate(C) = (3, 0);Perimeter(TriangleOf(A, B, C)) = 16",
    "query_expressions": "LocusEquation(A)",
    "answer_expressions": "(x^2/25 + y^2/16 = 1)&(Negation(y = 0))",
    "fact_spans": "[[[86, 89]], [[21, 24]], [[25, 28]], [[21, 54]], [[21, 54]], [[56, 82]]]",
    "query_spans": "[[[86, 96]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, and $P$ is a point on the hyperbola, then the range of $P F_{1}-P F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G)",
    "query_expressions": "Range(LineSegmentOf(P,F1)-LineSegmentOf(P,F2))",
    "answer_expressions": "[-b^2,+\\infty)",
    "fact_spans": "[[[18, 75], [85, 88]], [[21, 75]], [[21, 75]], [[81, 84]], [[2, 9]], [[10, 17]], [[21, 75]], [[21, 75]], [[18, 75]], [[2, 80]], [[81, 92]]]",
    "query_spans": "[[[94, 118]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the focus of the parabola $y=9 x^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 9*x^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": ""
  },
  {
    "text": "The range of the slope of a line passing through the origin and intersecting the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=-1$ at two points is?",
    "fact_expressions": "G: Hyperbola;H: Line;Expression(G) = (x^2/4 - y^2/3 = -1);O: Origin;PointOnCurve(O, H);NumIntersection(H, G) = 2",
    "query_expressions": "Range(Slope(H))",
    "answer_expressions": "{(-oo, -sqrt(3)/2), (sqrt(3)/2, +oo)}",
    "fact_spans": "[[[4, 43]], [[48, 50]], [[4, 43]], [[1, 3]], [[0, 50]], [[3, 50]]]",
    "query_spans": "[[[48, 60]]]",
    "process": ""
  },
  {
    "text": "When $\\alpha \\in(0, \\frac{\\pi}{2})$, the equation $x^{2} \\sin \\alpha+y^{2} \\cos \\alpha=1$ represents an ellipse with foci on the $x$-axis. Then the range of values for $\\alpha$ is?",
    "fact_expressions": "G: Ellipse;alpha:Number;Expression(G) = (x^2*Sin(alpha) + y^2*Cos(alpha) = 1);PointOnCurve(Focus(G), xAxis);In(alpha,(0,pi/2))",
    "query_expressions": "Range(alpha)",
    "answer_expressions": "(0, \\pi/4)",
    "fact_spans": "[[[85, 87]], [[89, 97]], [[33, 87]], [[76, 87]], [[1, 31]]]",
    "query_spans": "[[[89, 104]]]",
    "process": "x^{2}\\sin\\alpha+y2\\cos\\alpha=1, i.e. \\frac{x^{2}}{\\sin\\alpha}+\\frac{y^{2}}{\\cos\\alpha}=1,\\ \\alpha\\in(0,\\frac{\\pi}{2}), hence \\frac{1}{\\sin\\alpha}>0,\\ \\frac{1}{\\cos\\alpha}>0. The equation x^{2}\\sin\\alpha+y2\\cos\\alpha=1 represents an ellipse with foci on the x-axis, hence \\frac{1}{\\sin\\alpha}>\\frac{1}{\\cos\\alpha}, i.e. \\cos\\alpha>\\sin\\alpha, thus \\alpha\\in(0,\\frac{\\pi}{4})"
  },
  {
    "text": "Given that one asymptote of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ is $y=2x$, then the distance from the focus to this asymptote is?",
    "fact_expressions": "G: Hyperbola;b: Number;b>0;Expression(G) = (x^2 - y^2/b^2 = 1);OneOf(Asymptote(G)) = Z;Expression(Z) = (y = 2*x);Z: Line",
    "query_expressions": "Distance(Focus(G), Z)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 39]], [[5, 39]], [[5, 39]], [[2, 39]], [[2, 53]], [[46, 53]], [[46, 53]]]",
    "query_spans": "[[[2, 69]]]",
    "process": "\\because the hyperbola x^{2}-\\frac{y^{2}}{b^{2}}=1 (b>0) has an asymptote y=2x, \\therefore \\frac{b}{1}=2, solving gives: b=2, \\therefore c=\\sqrt{1+2^{2}}=\\sqrt{5}, \\therefore the right focus of the hyperbola is (\\sqrt{5},0), \\therefore the distance from the focus to this asymptote is: \\frac{2\\sqrt{5}}{\\sqrt{1+2^{2}}}=2. The correct answer to this question: 2["
  },
  {
    "text": "The equation of the locus of points equidistant from the $x$-axis and the line $4x - 3y = 0$ is?",
    "fact_expressions": "G: Line;Expression(G) = (4*x - 3*y = 0);K: Point;Distance(K, xAxis) = Distance(K, G)",
    "query_expressions": "LocusEquation(K)",
    "answer_expressions": "{x-2*y=0, 2*x+y=0}",
    "fact_spans": "[[[6, 19]], [[6, 19]], [[25, 26]], [[0, 26]]]",
    "query_spans": "[[[25, 33]]]",
    "process": "Let the coordinates of the moving point be (x, y). Since the distance to the line 4x - 3y = 0 is equal to the distance to the x-axis, we have |y| = \\frac{|4x - 3y|}{\\sqrt{4^{2} + (-3)^{2}}}. Therefore, |4x - 3y| = 5|y|, which implies 4x - 3y = \\pm5y, that is, x - 2y = 0 or 2x + y = 0"
  },
  {
    "text": "Given that line $l$ passes through the focus of the parabola $C$: $y^{2}=x$, and intersects parabola $C$ at points $A$ and $B$, with $|A B|=2$, then the horizontal coordinate of the midpoint $G$ of chord $A B$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = x);l: Line;PointOnCurve(Focus(C), l);A: Point;B: Point;Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, B)) = 2;IsChordOf(LineSegmentOf(A, B), C);G: Point;MidPoint(LineSegmentOf(A, B)) = G",
    "query_expressions": "XCoordinate(G)",
    "answer_expressions": "3/4",
    "fact_spans": "[[[8, 25], [31, 37]], [[8, 25]], [[2, 7]], [[2, 28]], [[38, 41]], [[42, 45]], [[2, 47]], [[48, 57]], [[31, 65]], [[67, 70]], [[60, 70]]]",
    "query_spans": "[[[67, 76]]]",
    "process": "From the given conditions, the equation of the directrix $ m $ of the parabola is $ x = -\\frac{1}{4} $. As shown in the figure: draw $ GD \\perp m $, $ AA' \\perp m $, $ BB' \\perp m $ from points $ G $, $ A $, $ B $ respectively. By the definition of the parabola, we have: $ |AA'| + |BB'| = |AB| = 2 $. Since $ G $ is the midpoint of $ AB $, it follows that $ |GD| = \\frac{|AA'| + |BB'|}{2} = \\frac{1}{2}|AB| = 1 $. Therefore, the horizontal coordinate of point $ G $ is: $ 1 - \\frac{1}{4} = \\frac{3}{4} $."
  },
  {
    "text": "The line $l$ intersects the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) at points $A$ and $B$, and $M$ is the midpoint of segment $AB$. If the product of the slope of $l$ and the slope of $OM$ ($O$ is the origin) equals $1$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "l: Line;C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Intersection(l, C) = {A, B};A: Point;B: Point;M: Point;MidPoint(LineSegmentOf(A, B)) = M;Slope(l)*Slope(LineSegmentOf(O, M)) = 1;O: Origin",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 5], [95, 98]], [[6, 67], [127, 130]], [[6, 67]], [[13, 67]], [[13, 67]], [[13, 67]], [[13, 67]], [[0, 78]], [[69, 72]], [[73, 76]], [[79, 82]], [[79, 93]], [[95, 124]], [[105, 108]]]",
    "query_spans": "[[[127, 136]]]",
    "process": ""
  },
  {
    "text": "The line $y=\\frac{b}{2 a} x$ intersects the left and right branches of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ at points $A$ and $B$ respectively, and $F$ is the right focus. If $A B \\perp B F$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "H: Line;Expression(H) = (y = x*(b/(2*a)));G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;A: Point;B: Point;Intersection(H, LeftPart(G)) = A;Intersection(H, RightPart(G)) = B;F: Point;RightFocus(G) = F;IsPerpendicular(LineSegmentOf(A, B), LineSegmentOf(B, F))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[0, 21]], [[0, 21]], [[22, 78], [125, 128]], [[22, 78]], [[25, 78]], [[25, 78]], [[25, 78]], [[25, 78]], [[88, 91]], [[92, 95]], [[0, 97]], [[0, 97]], [[98, 101]], [[22, 105]], [[107, 122]]]",
    "query_spans": "[[[125, 134]]]",
    "process": "Let the right focus of the hyperbola be $ F(c,0) $. Solving $ y = \\frac{b}{2a}x $ and $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ simultaneously, we get $ \\frac{3}{4} \\times \\frac{x^{2}}{a^{2}} = 1 $, so $ x^{2} = \\frac{4a^{2}}{3} $, thus $ B\\left( \\frac{2a}{\\sqrt{3}}, \\frac{b}{\\sqrt{3}} \\right) $, then $ k_{BF} = \\frac{\\frac{b}{\\sqrt{3}}}{\\frac{2a}{\\sqrt{3}} - c} = \\frac{b}{2a - \\sqrt{3}c} $. Since $ AB \\perp BF $, we have $ \\frac{b}{2a - \\sqrt{3}c} \\times \\frac{b}{2a} = \\frac{c^{2} - a^{2}}{4a^{2} - 2\\sqrt{3}ac} = -1 $. Rearranging gives $ c^{2} - 2\\sqrt{3}ac + 3a^{2} = 0 $, i.e., $ e^{2} - 2\\sqrt{3}e + 3 = 0 $, so $ e = \\sqrt{3} $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{3}-\\frac{y^{2}}{3}=1$, $F$ is the right focus of the hyperbola $C$, point $A$ lies on the left branch of the hyperbola $C$, and point $B$ lies on the circle $D$: $x^{2}+(y+3 \\sqrt{2})^{2}=3$. Then the minimum value of $|A B|+|A F|$ is?",
    "fact_expressions": "C: Hyperbola;D: Circle;A: Point;B: Point;F: Point;Expression(C) = (x^2/3 - y^2/3 = 1);Expression(D) = (x^2 + (y + 3*sqrt(2))^2 = 3);RightFocus(C) = F;PointOnCurve(A, LeftPart(C));PointOnCurve(B, D)",
    "query_expressions": "Min(Abs(LineSegmentOf(A, B)) + Abs(LineSegmentOf(A, F)))",
    "answer_expressions": "2*sqrt(6) + sqrt(3)",
    "fact_spans": "[[[2, 45], [52, 58], [68, 74]], [[87, 121]], [[63, 67]], [[82, 86]], [[48, 51]], [[2, 45]], [[87, 121]], [[47, 62]], [[63, 81]], [[82, 124]]]",
    "query_spans": "[[[126, 145]]]",
    "process": "Hyperbola $ C: \\frac{x^{2}}{3} - \\frac{y^{2}}{3} = 1 $, $ a = b = \\sqrt{3} $, $ c = \\sqrt{6} $, let $ F_{1}(-\\sqrt{6}, 0) $ be the left focus of the hyperbola, the circle $ x^{2} + (y + 3\\sqrt{2})^{2} = 3 $ has center $ D(0, -3\\sqrt{2}) $ and radius $ r = \\sqrt{3} $. According to the definition of the hyperbola, $ |AB| + |AF| = |AB| + |AF_{1}| + 2a = |AB| + |AF_{1}| + 2\\sqrt{3} $. Since $ B $ is a point on the circle $ x^{2} + (y + 3\\sqrt{2})^{2} = 3 $, and $ F_{1}(-\\sqrt{6}, 0) $ is a fixed point, $ |AB| + |AF_{1}| $ is minimized when $ F_{1} $, $ A $, $ B $, $ D $ are collinear. Thus, the minimum value of $ |AB| + |AF_{1}| $ is $ |DF_{1}| - r = \\sqrt{6 + 18} - \\sqrt{3} = 2\\sqrt{6} - \\sqrt{3} $. Therefore, the minimum value of $ |AB| + |AF| $ is $ 2\\sqrt{6} - \\sqrt{3} + 2\\sqrt{3} = 2\\sqrt{6} + \\sqrt{3} $."
  },
  {
    "text": "The hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ and the hyperbola $C_{2}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=-1$ have eccentricities $e_{1}$ and $e_{2}$, respectively. Then $\\frac{1}{e_{1}^{2}}+\\frac{1}{e_{2}^{2}}$=?",
    "fact_expressions": "C1: Hyperbola;Expression(C1) = (x^2/a^2 - y^2/b^2 = 1);C2: Hyperbola;Expression(C2) = (x^2/a^2 - y^2/b^2 = -1);a: Number;b: Number;e1: Number;e2:Number;Eccentricity(C1) = e1;Eccentricity(C2) = e2",
    "query_expressions": "1/e1^2 + 1/e2^2",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 55]], [[0, 55]], [[56, 112]], [[56, 112]], [[12, 55]], [[12, 55]], [[119, 126]], [[127, 134]], [[0, 134]], [[0, 134]]]",
    "query_spans": "[[[136, 179]]]",
    "process": "From the given conditions: $ C_{2}:\\frac{y^{2}}{b^{2}}-\\frac{x^{2}}{a^{2}}=1 $, $ e_{1}=\\sqrt{\\frac{a^{2}+b^{2}}{a^{2}}} $, $ e_{2}=\\sqrt{\\frac{a^{2}+b^{2}}{b^{2}}} $, so $ \\frac{1}{e_{1}^{2}}+\\frac{1}{e_{2}^{2}}=\\frac{a^{2}}{a^{2}+b^{2}}+\\frac{b^{2}}{a^{2}+b^{2}}=1 $."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, respectively. If there exists a point $P$ on the hyperbola such that $|P F_{1}|+|P F_{2}|=3 b$ and $|P F_{1}| \\cdot|P F_{2}|=\\frac{9}{4} a b$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;P: Point;PointOnCurve(P, G) = True;Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 3*b;Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = (9/4)*a*b",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[1, 8]], [[11, 18]], [[1, 86]], [[1, 86]], [[21, 80], [88, 91], [174, 177]], [[21, 80]], [[24, 80]], [[24, 80]], [[24, 80]], [[24, 80]], [[96, 99]], [[88, 99]], [[102, 127]], [[129, 171]]]",
    "query_spans": "[[[174, 183]]]",
    "process": "Since point $ P $ lies on the hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $), we have $ ||PF_{1}| - |PF_{2}|| = 2a $, that is, $ (|PF_{1}| - |PF_{2}|)^{2} = 4a^{2} $. Because $ (|PF_{1}| - |PF_{2}|)^{2} = (|PF_{1}| + |PF_{2}|)^{2} - 4|PF_{1}| \\cdot |PF_{2}| $, it follows that $ 9b^{2} - 9ab = 4a^{2} $. Solving this gives $ b = \\frac{4}{3}a $ or $ b = -\\frac{1}{3}a $ (discarded). Therefore, the eccentricity of the hyperbola is $ e = \\frac{c}{a} = \\sqrt{\\frac{c^{2}}{a^{2}}} = \\sqrt{\\frac{a^{2} + b^{2}}{a^{2}}} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{1 + \\frac{16}{9}} = \\frac{5}{3} $. Hence, the answer should be $ \\frac{5}{3} $."
  },
  {
    "text": "The line $l$ intersects the parabola $y^{2}=2x$ at points $M(x_{1}, y_{1})$, $N(x_{2}, y_{2})$, and passes through the focus. Then the value of $y_{1} y_{2}$ is?",
    "fact_expressions": "l: Line;G: Parabola;Expression(G) = (y^2 = 2*x);M: Point;N: Point;x1:Number;x2:Number;y1:Number;y2:Number;Coordinate(M) = (x1, y1);Coordinate(N) = (x2, y2);Intersection(l, G) = {M, N};PointOnCurve(Focus(G), l)",
    "query_expressions": "y1*y2",
    "answer_expressions": "-1",
    "fact_spans": "[[[0, 5]], [[6, 20]], [[6, 20]], [[21, 39]], [[41, 59]], [[21, 39]], [[41, 59]], [[21, 39]], [[41, 59]], [[21, 39]], [[41, 59]], [[0, 59]], [[0, 64]]]",
    "query_spans": "[[[66, 83]]]",
    "process": ""
  },
  {
    "text": "The right focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $F$, a line passing through the origin $O$ intersects the ellipse at points $A$ and $P$, and $PF$ is perpendicular to the $x$-axis. The line $AF$ intersects the ellipse at point $B$, and $PB \\perp PA$. Then the eccentricity $e$ of the ellipse is $?$.",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;RightFocus(G) = F;O: Origin;H: Line;PointOnCurve(O, H);A: Point;P: Point;Intersection(H, G) = {A, P};IsPerpendicular(LineSegmentOf(P, F), xAxis) ;Intersection(LineOf(A, F), G) = B;B: Point;IsPerpendicular(LineSegmentOf(P, B), LineSegmentOf(P, A));e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[0, 52], [71, 73], [105, 107], [132, 134]], [[0, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[57, 60]], [[0, 60]], [[62, 67]], [[68, 70]], [[61, 70]], [[74, 78]], [[79, 82]], [[68, 82]], [[84, 96]], [[97, 112]], [[108, 112]], [[113, 129]], [[138, 141]], [[132, 141]]]",
    "query_spans": "[[[138, 143]]]",
    "process": "This problem examines the properties of ellipses and knowledge about linear equations, using the conclusion: if the equation of the ellipse is \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0), i.e., the foci lie on the x-axis, and if a line l intersects the ellipse, with the chord intercepted by the ellipse being AB and its midpoint denoted as P, then the product of the slope of the line and the slope of the line connecting the midpoint of the chord to the origin is a constant, i.e., k_{l}\\cdot k_{PO}=-\\frac{b^{2}}{a^{2}}; this makes solving simpler. From the given information, P(c,\\frac{b^{2}}{a}), A(-c,-\\frac{b^{2}}{a}), F(c,0), taking M as the midpoint of AB, it follows that OM//PB \\Rightarrow k_{OM}=k_{PB}. Also, since PB\\bot PA, we have k_{PB}=-\\frac{1}{k_{PA}}=-\\frac{ac}{b^{2}} \\Rightarrow k_{OM}=-\\frac{ac}{b^{2}}. Moreover, k_{AB}=\\frac{0-(-\\frac{b^{2}}{a})}{c-(-c)}=\\frac{b^{2}}{2ac}. From k_{OM} \\cdot k_{AB}=-\\frac{b^{2}}{a^{2}} \\Rightarrow \\frac{b^{2}}{a^{2}}=\\frac{1}{2} \\Rightarrow e=\\sqrt{1-\\frac{b^{2}}{a^{2}}}=\\frac{\\sqrt{2}}{2}"
  },
  {
    "text": "The equation of the hyperbola passing through the point $A(2,-2)$ and having common asymptotes with the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$ is?",
    "fact_expressions": "G: Hyperbola;A: Point;H:Hyperbola;Expression(G) = (x^2/2 - y^2 = 1);Coordinate(A) = (2, -2);PointOnCurve(A, H);Asymptote(G)=Asymptote(H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2/2 - x^2/4 = 1",
    "fact_spans": "[[[14, 42]], [[2, 12]], [[49, 52]], [[14, 42]], [[2, 12]], [[0, 52]], [[13, 52]]]",
    "query_spans": "[[[49, 56]]]",
    "process": ""
  },
  {
    "text": "What is the minimum value of the focal distance of the ellipse $\\frac{x^{2}}{m^{2}+6}+\\frac{y^{2}}{4 m+1}=1$?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/(m^2 + 6) + y^2/(4*m + 1) = 1)",
    "query_expressions": "Min(FocalLength(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 47]], [[2, 47]], [[0, 47]]]",
    "query_spans": "[[[0, 56]]]",
    "process": "From the given condition, we have: $m^{2}+6-(4m+1)=(m-2)^{2}+1\\geqslant1,\\therefore$ focal length $=2\\sqrt{(m-2)^{2}+1}\\geqslant2.$"
  },
  {
    "text": "The line $m$ intersects the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ at points $P_{1}$ and $P_{2}$, with the midpoint of segment $P_{1} P_{2}$ being $P$. Let the slope of line $m$ be $k_{1}$ $(k_{1} \\neq 0)$, and the slope of line $O P$ be $k_{2}$. Then $k_{1} k_{2} = $?",
    "fact_expressions": "m: Line;G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);P1: Point;P2: Point;Intersection(m, G) = {P1, P2};P: Point;MidPoint(LineSegmentOf(P1, P2)) = P;k1: Number;k2: Number;Slope(m) = k1;Negation(k1=0);O: Origin;Slope(LineOf(O, P)) = k2",
    "query_expressions": "k1*k2",
    "answer_expressions": "-1/4",
    "fact_spans": "[[[0, 5], [75, 80]], [[6, 33]], [[6, 33]], [[35, 42]], [[43, 50]], [[0, 50]], [[70, 73]], [[51, 73]], [[84, 105]], [[117, 124]], [[75, 105]], [[84, 105]], [[108, 113]], [[106, 124]]]",
    "query_spans": "[[[126, 141]]]",
    "process": "Let $P_{1}(x_{1},y_{1})$, $P_{2}(x_{2},y_{2})$, and the midpoint be $P(x_{0},y_{0})$. Then $k_{1}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$, $k_{2}=\\frac{y_{0}}{x_{0}}=\\frac{y_{1}+y_{2}}{x_{1}+x_{2}}$. Rearranging gives $\\frac{x_{1}}{4}+y_{2}=1$, $\\frac{x_{2}}{4}+y_{2}=1$, and rearranging further yields $\\frac{y^{2}-y_{2}^{2}}{x_{1}^{2}-x_{2}^{2}}$, so that $k_{1}k_{2}=-\\frac{1}{4}$."
  },
  {
    "text": "Let the focus of the parabola $x^{2}=2 y$ be $F$, and let a line $l$ passing through point $P(1,3)$ intersect the parabola at points $A$ and $B$, such that point $P$ is exactly the midpoint of $AB$. Then $|\\overrightarrow{A F}|+|\\overrightarrow{B F}|$=?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;P: Point;F: Point;Expression(G) = (x^2 = 2*y);Coordinate(P) = (1, 3);Focus(G) = F;PointOnCurve(P, l);Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Abs(VectorOf(A, F)) + Abs(VectorOf(B, F))",
    "answer_expressions": "7",
    "fact_spans": "[[[35, 40]], [[1, 15], [41, 44]], [[47, 50]], [[51, 54]], [[25, 34], [58, 62]], [[19, 22]], [[1, 15]], [[25, 34]], [[1, 22]], [[23, 40]], [[35, 56]], [[58, 72]]]",
    "query_spans": "[[[74, 123]]]",
    "process": "|\\overrightarrow{AF}|+|\\overrightarrow{BF}|=y_{A}+\\frac{1}{2}+y_{B}+\\frac{1}{2}=2y_{P}+1=2\\times3+1=7."
  },
  {
    "text": "$P$ is a moving point on the right branch of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, $F$ is the right focus of the hyperbola, given $A(3,1)$, then the minimum value of $|P A|+|P F|$ is?",
    "fact_expressions": "G: Hyperbola;A: Point;P: Point;F: Point;Expression(G) = (x^2/3 - y^2 = 1);Coordinate(A) = (3, 1);PointOnCurve(P, RightPart(G));RightFocus(G)=F",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "sqrt(26)-2*sqrt(3)",
    "fact_spans": "[[[4, 32], [44, 47]], [[55, 63]], [[0, 3]], [[40, 43]], [[4, 32]], [[55, 63]], [[0, 39]], [[40, 51]]]",
    "query_spans": "[[[66, 85]]]",
    "process": "Let the left focus of the hyperbola be $ F_{2} $, then $ |PA| + |PF| = |PF_{2}| - 2a + |PA| $. The minimum value occurs when points $ P $, $ F_{2} $, and $ A $ are collinear. At this time, $ F_{2}(-2,0) $, $ A(3,1) $, so $ |PF_{2}| + |PA| = |AF_{2}| = \\sqrt{26} $. For this hyperbola, $ 2a = 2\\sqrt{3} $, so the minimum value is $ \\sqrt{26} - 2\\sqrt{3} $."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $y^{2}=12x$, point $M(-1,0)$, and points $A$, $B$ on the parabola satisfy $\\overrightarrow{M A}=\\lambda \\overrightarrow{M B}(\\lambda>0)$. When the sum of the areas of $\\triangle A B O$ and $\\triangle A F O$ is minimized (where $O$ is the origin), $\\lambda=$?",
    "fact_expressions": "G: Parabola;M: Point;A: Point;B: Point;O: Origin;F: Point;Expression(G) = (y^2 = 12*x);Coordinate(M) = (-1, 0);Focus(G) = F;PointOnCurve(A,G);PointOnCurve(B,G);VectorOf(M,A)=lambda*VectorOf(M,B);lambda:Number;lambda>0;WhenMin(Area(TriangleOf(A,B,O))+Area(TriangleOf(A,F,O)))",
    "query_expressions": "lambda",
    "answer_expressions": "1/2",
    "fact_spans": "[[[6, 21], [36, 39]], [[25, 35]], [[42, 45]], [[46, 49]], [[161, 164]], [[2, 5]], [[6, 21]], [[25, 35]], [[2, 24]], [[36, 49]], [[36, 49]], [[51, 113]], [[171, 180]], [[51, 113]], [[114, 157]]]",
    "query_spans": "[[[171, 182]]]",
    "process": "By the given condition, point $ F $ is the focus of the parabola $ y^{2} = 12x $, so $ F(3,0) $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ y_{1} = \\lambda y_{2} $. Let the equation of line $ AB $ be $ my = x + 1 $ (without loss of generality, assume $ m > 0 $). Solving the system of equations\n\\[\n\\begin{cases}\nmy = x + 1 \\\\\ny^{2} = 12x\n\\end{cases}\n\\]\nwe obtain $ y^{2} - 12my + 12 = 0 $, so $ y_{1}y_{2} = 12 $. Then\n\\[\nS_{\\Delta ABO} = |S_{\\Delta MBO} - S_{\\Delta MAO}| = \\frac{1}{2}|OM||y_{2} - y_{1}| = \\frac{1}{2}|y_{2} - y_{1}|.\n\\]\nSince the sum of the areas of $ \\triangle ABO $ and $ \\triangle AFO $ is minimized, we must have $ |y_{2}| > |y_{1}| $. By symmetry, without loss of generality, consider the case $ y_{2} > y_{1} > 0 $. At this time,\n\\[\nS_{\\triangle ABO} + S_{\\triangle AFO} = \\frac{1}{2}y_{2} + y_{1} \\geqslant 2\\sqrt{\\frac{y_{1}y_{2}}{2}} = 2\\sqrt{6},\n\\]\nwith equality if and only if $ \\frac{1}{2}y_{2} = y_{1} = \\sqrt{6} $, i.e., $ \\lambda = \\frac{1}{2} $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$, and a point $P$ on $C$ such that $|PF|=8$, then the distance from point $P$ to the $x$-axis is?",
    "fact_expressions": "C: Parabola;P: Point;F: Point;Expression(C) = (y^2 = 8*x);Focus(C) = F;PointOnCurve(P, C);Abs(LineSegmentOf(P, F)) = 8",
    "query_expressions": "Distance(P, xAxis)",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[2, 21], [30, 33]], [[37, 40], [53, 57]], [[25, 28]], [[2, 21]], [[2, 28]], [[29, 40]], [[42, 51]]]",
    "query_spans": "[[[53, 67]]]",
    "process": "From the definition of the parabola, we have: |PF| = x_{p} + 2 = 8, so x_{p} = 6. Substituting into y^{2} = 8x, we get y_{p}^{2} = 48, so |y_{p}| = 4\\sqrt{3}. Hence, the distance from point P to the c-axis is 4\\sqrt{3}."
  },
  {
    "text": "What is the focal distance of the hyperbola $\\frac{x^{2}}{25-k}+\\frac{y^{2}}{9-k}=1$?",
    "fact_expressions": "G: Hyperbola;k: Number;Expression(G) = (x^2/(25 - k) + y^2/(9 - k) = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "8",
    "fact_spans": "[[[0, 43]], [[3, 43]], [[0, 43]]]",
    "query_spans": "[[[0, 48]]]",
    "process": "The hyperbola $\\frac{x^2}{25-k}+\\frac{y^2}{9-k}=1$, according to the given condition $(25-k)(9-k)<0$, therefore $9<k<25$, thus $c^{2}=25-k+k-9=16$, therefore $c=4$, therefore $2c=8$,"
  },
  {
    "text": "The equation of the hyperbola with asymptotes $y=\\pm \\frac{1}{2} x$ and passing through the point $(2 , 2)$ is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (2, 2);Expression(Asymptote(G)) = (y=pm*(x/2));PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/3 - x^2/12 = 1",
    "fact_spans": "[[[39, 42]], [[28, 38]], [[28, 38]], [[0, 42]], [[27, 42]]]",
    "query_spans": "[[[39, 46]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the right focus is $F$, the line $x=\\frac{a}{2}$ intersects $C$ at points $A$ and $B$. If $\\angle A F B=120^{\\circ}$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;A: Point;F: Point;B: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x = a/2);RightFocus(C) = F;Intersection(G, C) = {A, B};AngleOf(A, F, B) = ApplyUnit(120, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "4/5",
    "fact_spans": "[[[2, 59], [86, 89], [130, 135]], [[9, 59]], [[9, 59]], [[68, 85]], [[91, 94]], [[64, 67]], [[95, 98]], [[9, 59]], [[9, 59]], [[2, 59]], [[68, 85]], [[2, 67]], [[68, 100]], [[102, 128]]]",
    "query_spans": "[[[130, 141]]]",
    "process": "According to the problem, substitute $ x = \\frac{a}{2} $ into $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $, we get $ y = \\pm\\frac{\\sqrt{3}b}{2} $. Without loss of generality, assume $ A(\\frac{a}{2}, \\frac{\\sqrt{3}b}{2}) $, and $ F(c,0) $. Then the distance from $ F $ to the line $ x = \\frac{a}{2} $ is $ |\\frac{a}{2} - c| $. From $ \\angle AFB = 120^{\\circ} $, we obtain $ \\tan\\frac{120^{\\circ}}{2} = \\frac{\\frac{\\sqrt{3}b}{2}}{|\\frac{a}{2} - c|} = \\sqrt{3} $, so $ b = |a - 2c| $. Squaring both sides and calculating yields $ \\frac{c}{a} = \\frac{4}{5} $."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ has focus $F$, directrix $l$, a line passing through point $F$ with slope $\\sqrt{3}$ intersects the parabola at point $M$ ($M$ in the first quadrant), $M N \\perp l$, with foot of perpendicular at $N$, line $N F$ intersects the $y$-axis at point $D$, if $|M D|=\\sqrt{6}$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;F: Point;N: Point;M: Point;D: Point;l: Line;p>0;Expression(G) = (y^2 = 2*p*x);Focus(G) = F;Directrix(G) = l;PointOnCurve(F, H);Slope(H) = sqrt(3);Intersection(H, G)=M;Quadrant(M)=1;IsPerpendicular(LineSegmentOf(M,N),l);FootPoint(LineSegmentOf(M,N),l)=N;Intersection(LineOf(N,F),yAxis)=D;Abs(LineSegmentOf(M, D)) = sqrt(6)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=2*sqrt(2)*x",
    "fact_spans": "[[[2, 23], [61, 64], [61, 64]], [[5, 23]], [[58, 60]], [[27, 30], [39, 43]], [[97, 100]], [[65, 69], [70, 73]], [[114, 118]], [[34, 37]], [[5, 23]], [[2, 23]], [[2, 30]], [[2, 37]], [[38, 60]], [[44, 60]], [[58, 69]], [[70, 78]], [[80, 93]], [[80, 100]], [[101, 118]], [[120, 136]]]",
    "query_spans": "[[[138, 146]]]",
    "process": "Draw the graph. Based on the shape of triangle MNF, we obtain |MN|, thus obtaining the equation of the parabola. As shown in the figure, by the definition of the parabola, |MN| = |MF|. Since the slope of MF is \\sqrt{3} and MN // OF, it follows that \\angle NMF = 60^{\\circ}, so triangle MNF is equilateral. In triangle ANF, it is clear that D is the midpoint of NF. Given |MD| = \\sqrt{6}, we have |MN| = |MF| = 2\\sqrt{2}, so M(\\sqrt{2} + \\frac{p}{2}, \\sqrt{6}). From 6 = 2p(\\sqrt{2} + \\frac{p}{2}), we get p = \\sqrt{2}."
  },
  {
    "text": "The equation of the right directrix of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1)",
    "query_expressions": "Expression(RightDirectrix(G))",
    "answer_expressions": "x=3/2",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 36]]]",
    "process": ""
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{4-k}+\\frac{y^{2}}{6+k}=1$ represents an ellipse, what is the range of real values for $k$?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G)=(x^2/(4-k)+y^2/(6+k)=1)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-6,-1)+(-1,4)",
    "fact_spans": "[[[45, 47]], [[49, 54]], [[2, 47]]]",
    "query_spans": "[[[49, 61]]]",
    "process": "\\because the equation \\frac{x^{2}}{4-k}+\\frac{y^{2}}{6+k}=1 represents an ellipse, then \\begin{cases}4-k>0\\\\6+k>0\\end{cases}, solving gives -6<k<4 and k\\neq-1, 4-k\\neq6+k"
  },
  {
    "text": "If one focus of an ellipse has coordinates $(2,0)$, and the length of the chord passing through this focus and perpendicular to the $x$-axis is equal to $\\frac{10}{3}$, then the standard equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;H:LineSegment;Coordinate(F)=(2,0);OneOf(Focus(G))=F;IsChordOf(H,G);PointOnCurve(F,H);IsPerpendicular(H,xAxis);Length(H)=10/3;F:Point",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9+y^2/5=1",
    "fact_spans": "[[[2, 4], [56, 58]], [], [[12, 19]], [[2, 19]], [[2, 34]], [[2, 34]], [[2, 34]], [[2, 52]], []]",
    "query_spans": "[[[56, 65]]]",
    "process": "Let the standard equation of the ellipse be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. From the given conditions:  \n\\begin{cases}  \nc=2 \\\\  \n\\frac{2b^{2}}{a}=\\frac{10}{3} \\\\  \na^{2}=b^{2}+c^{2}  \n\\end{cases}  \n\\Rightarrow  \n\\begin{cases}  \na=3 \\\\  \nb=\\sqrt{5} \\\\  \nc=2  \n\\end{cases}  \nThe required standard equation of the ellipse is $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$."
  },
  {
    "text": "It is known that $F$ is the left focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$. Let the moving point $P$ lie on the ellipse. If the slope of the line $FP$ is greater than $\\sqrt{3}$, then what is the range of the slope of the line $OP$ ($O$ being the origin)?",
    "fact_expressions": "F: Point;G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);LeftFocus(G) = F;P: Point;PointOnCurve(P, G) = True;Slope(LineOf(F, P)) > sqrt(3);O: Origin",
    "query_expressions": "Range(Slope(LineOf(O, P)))",
    "answer_expressions": "(-oo,-3/2)+(3*sqrt(3)/8,3/2)",
    "fact_spans": "[[[2, 5]], [[6, 43], [55, 57]], [[6, 43]], [[2, 47]], [[51, 54]], [[51, 58]], [[60, 82]], [[92, 95]]]",
    "query_spans": "[[[84, 110]]]",
    "process": "From the given conditions, F(-1,0). First, find the intersection points of the ellipse with the line passing through point F(-1,0) with slope \\sqrt{3} and the line with undefined slope. Then, based on how the slope changes as the line rotates around the fixed point, determine the position of point P that satisfies the conditions, and thus find the range of the slope of line OP. From the ellipse equation \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1, we know F(-1,0). When the line passes through point F(-1,0) with slope \\sqrt{3}, the corresponding line is y=\\sqrt{3}(x+1). Solving \\begin{cases}y=\\sqrt{3}(x+1)\\\\\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1\\end{cases} yields \\begin{cases}x=0\\\\y=\\sqrt{3}\\end{cases} or \\begin{cases}x=-\\frac{8}{5}\\\\y=-\\frac{3\\sqrt{3}}{5}\\end{cases}. Thus, the intersection points of the line y=\\sqrt{3}(x+1) and the ellipse \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1 are P_{1}(0,\\sqrt{3}), P_{2}(-\\frac{8}{5},-\\frac{3\\sqrt{3}}{5}). Since the vertical line through F intersecting the x-axis meets the ellipse at A_{1}(0,\\frac{3}{2}), A_{2}(0,-\\frac{3}{2}), when point P lies on arcs P_{1}A_{1} and P_{2}A_{2}, the conditions are satisfied. \\because k_{0A_{1}}=-\\frac{3}{2}, k_{0A_{2}}=\\frac{3}{2}, k_{0P_{2}}=\\frac{3\\sqrt{3}}{8}, \\therefore the range of the slope of OP is (-\\infty,-\\frac{3}{2})\\cup(\\frac{3\\sqrt{3}}{8},\\frac{3}{2})."
  },
  {
    "text": "In $\\triangle A B C$, $B(-2,0)$, $C(2,0)$, and the perimeter of $\\triangle A B C$ is $10$. Then the trajectory equation of point $A$ is?",
    "fact_expressions": "B: Point;C: Point;A: Point;Coordinate(B) = (-2, 0);Coordinate(C) = (2, 0);Perimeter(TriangleOf(A, B, C)) = 10",
    "query_expressions": "LocusEquation(A)",
    "answer_expressions": "(x^2/9+y^2/5=1)&Negation(y=0)",
    "fact_spans": "[[[20, 29]], [[30, 40]], [[69, 73]], [[20, 29]], [[30, 40]], [[42, 67]]]",
    "query_spans": "[[[69, 80]]]",
    "process": "\\because the perimeter of \\triangle ABC is 10, \\therefore |AB| + |AC| + |BC| = 10, where |BC| = 4, thus |AB| + |AC| = 6 > 4, \\therefore the locus of point A is an ellipse, excluding the two vertices on the major axis. \\because 2a = 6, 2c = 4, \\therefore a = 3, c = 2, b^{2} = 5. \\therefore the equation of the locus of point A is \\frac{x^{2}}{9} + \\frac{y^{2}}{5} = 1 (y \\neq 0). Therefore, the answer to this question is \\frac{x^{2}}{9} + \\frac{y^{2}}{5} = 1 (y \\neq 0). [Note] This question mainly examines the standard equation of an ellipse; the key is to properly use the definition and properties of the ellipse."
  },
  {
    "text": "The equation of the locus of points whose distance from the $y$-axis is equal to $2$ is?",
    "fact_expressions": "P: Point;Distance(P, yAxis) = 2",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "{(x = 2), (x = -2)}",
    "fact_spans": "[[[13, 14]], [[0, 14]]]",
    "query_spans": "[[[13, 21]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, $|F_{1} F_{2}|=2 \\sqrt{3}$. When point $P$ moves on $C$, the minimum value of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ is $-2$. Then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;F1: Point;F2: Point;P: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C)=F1;RightFocus(C)=F2;Abs(LineSegmentOf(F1, F2)) = 2*sqrt(3);PointOnCurve(P, C);Min(DotProduct(VectorOf(P, F1), VectorOf(P, F2))) = -2",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 63], [124, 127], [200, 206]], [[9, 63]], [[9, 63]], [[72, 80]], [[82, 89]], [[119, 123]], [[9, 63]], [[9, 63]], [[2, 63]], [[2, 89]], [[2, 89]], [[91, 117]], [[119, 130]], [[132, 198]]]",
    "query_spans": "[[[200, 212]]]",
    "process": "Let $ P(x_{0},y_{0}) $, then $ \\frac{x_{0}^{2}}{a^{2}} - \\frac{y_{0}^{2}}{b^{2}} = 1 $, $ \\therefore x_{0}^{2} = a^{2} + \\frac{a^{2}}{b^{2}} y_{0}^{2} $. $ \\therefore F_{1}(-\\sqrt{3},0) $, $ F_{2}(\\sqrt{3},0) $, $ c^{2} = 3 = a^{2} + b^{2} $. $ \\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}} = (x_{0} + \\sqrt{3})(x_{0} - \\sqrt{3}) + y_{0}^{2} = x_{0}^{2} + y_{0}^{2} - 3 = \\frac{c^{2}}{b^{2}} y_{0}^{2} + a^{2} - 3 \\geqslant a^{2} - 3 $, equality holds when $ y_{0} = 0 $. $ \\because \\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}} $'s minimum value is $ -2 $, $ \\therefore a^{2} - 3 = -2 $, solving gives $ a = 1 $, also $ c = \\sqrt{3} $, $ \\therefore e = \\frac{c}{a} = \\sqrt{3} $."
  },
  {
    "text": "Given that $A$, $B$, and $F$ are the upper vertex, lower vertex, and right focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. The line $AF$ intersects the right directrix of the ellipse at point $M$. If the line $MB$ is parallel to the $x$-axis, then the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "A: Point;F: Point;G: Ellipse;a: Number;b: Number;B: Point;M: Point;e: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);UpperVertex(G)=A;LowerVertex(G)=B;RightFocus(G)=F;Intersection(LineOf(A, F),RightDirectrix(G)) = M;IsParallel(LineOf(M,B),xAxis);Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 5]], [[10, 13]], [[16, 68], [86, 88], [118, 120]], [[18, 68]], [[18, 68]], [[6, 9]], [[94, 98]], [[124, 127]], [[18, 68]], [[18, 68]], [[16, 68]], [[2, 78]], [[2, 78]], [[2, 78]], [[79, 98]], [[100, 115]], [[118, 127]]]",
    "query_spans": "[[[124, 129]]]",
    "process": ""
  },
  {
    "text": "It is known that point $F$ is the left focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and the line $y=k x$ $(k>0)$ intersects $C$ at points $M$ and $N$ (where $M$ lies in the first quadrant). If $|M N|=2 \\sqrt{a^{2}-b^{2}}$ and $|F M| \\leq \\sqrt{3}|F N|$, then the maximum value of the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;k: Number;M: Point;N: Point;F: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);k>0;Expression(G) = (y = k*x);LeftFocus(C) = F;Intersection(G,C)={M,N};Quadrant(M)=1;Abs(LineSegmentOf(M, N)) = 2*sqrt(a^2 - b^2);Abs(LineSegmentOf(F, M)) <= sqrt(3)*Abs(LineSegmentOf(F, N))",
    "query_expressions": "Max(Eccentricity(C))",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[7, 64], [84, 87], [171, 174]], [[14, 64]], [[14, 64]], [[69, 83]], [[71, 83]], [[90, 93], [103, 106]], [[94, 97]], [[2, 6]], [[14, 64]], [[14, 64]], [[7, 64]], [[71, 83]], [[69, 83]], [[2, 68]], [[69, 100]], [[103, 111]], [[114, 142]], [[143, 169]]]",
    "query_spans": "[[[171, 184]]]",
    "process": "Let the right focus be F', connect MF' and NF'. By the symmetry of the ellipse, quadrilateral FMF'N is a parallelogram. Since |MN| = 2\\sqrt{a^{2}-b^{2}} = 2c = FF', FMF'N is a rectangle. |FM| \\leqslant \\sqrt{3}|FN| = \\sqrt{3}|FM|, |FM| + |FM| = 2a, i.e., 2a - |F'M| \\leqslant \\sqrt{3}|F'M|, \\therefore |F'M| \\geqslant \\frac{2a}{\\sqrt{3}} + 1. Also, (2a - |F'M|)^{2} + |F'M|^{2} = 4c^{2}, hence 0 < e \\leqslant \\sqrt{3} - 1"
  },
  {
    "text": "The standard equation of the ellipse passing through the points $P(\\sqrt{3},-2)$, $Q(-2 \\sqrt{3}, 1)$ is?",
    "fact_expressions": "G: Ellipse;P: Point;Q:Point;Coordinate(P) = (sqrt(3), -2);Coordinate(Q)=(-2*sqrt(3),1);PointOnCurve(P, G);PointOnCurve(Q, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/15 + y^2/5 = 1",
    "fact_spans": "[[[42, 44]], [[1, 18]], [[20, 39]], [[1, 18]], [[20, 39]], [[0, 44]], [[0, 44]]]",
    "query_spans": "[[[42, 50]]]",
    "process": "Let the standard equation of the ellipse be $mx^{2}+ny^{2}=1$ $(m,n>0,m\\neq n)$, then $\\begin{cases}3m+4n=1\\\\12m+n=1\\end{cases}$, solving yields $\\begin{cases}m=\\frac{1}{15}\\\\n=\\frac{1}{5}\\end{cases}$, thus the equation of the ellipse is $\\frac{x^{2}}{15}+\\frac{y^{2}}{5}=1$, fill in $\\frac{x^{2}}{15}+\\frac{y^{2}}{5}=1$. This question examines the method of finding the standard equation of an ellipse, and belongs to basic problems."
  },
  {
    "text": "The standard equation of a circle centered at the right focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{16}=1$ and intercepted by one of its asymptotes to form a chord of length $6$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/16 = 1);RightFocus(G) = Center(H);H:Circle;Length(InterceptChord(OneOf(Asymptote(G)),H))=6",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x-2*\\sqrt{5})^2+y^2=25",
    "fact_spans": "[[[1, 40]], [[1, 40]], [[0, 68]], [[67, 68]], [[0, 68]]]",
    "query_spans": "[[[67, 75]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $x^{2}=4 y$, with focus $F(0,1)$, let $A$ be a moving point on the parabola. The circle with diameter $A F$ is tangent to a fixed line. Then the equation of the line is?",
    "fact_expressions": "G: Parabola;H: Circle;I: Line;A: Point;F: Point;Expression(G) = (x^2 = 4*y);Coordinate(F) = (0, 1);Focus(G) = F;PointOnCurve(A, G);IsDiameter(LineSegmentOf(A, F), H);IsTangent(H, I)",
    "query_expressions": "Expression(I)",
    "answer_expressions": "y = 0",
    "fact_spans": "[[[2, 16], [35, 38]], [[53, 54]], [[56, 58], [62, 64]], [[30, 34]], [[20, 28]], [[2, 16]], [[20, 28]], [[2, 28]], [[30, 42]], [[43, 54]], [[53, 60]]]",
    "query_spans": "[[[62, 69]]]",
    "process": "It is easy to see that $ F(0,1) $ is the focus of the parabola $ x^{2} = 4y $. Let point $ A(x_{0}, y_{0}) $. By the definition of the parabola, we have $ |AF| = y_{0} + 1 $. Therefore, the distance from the center $ M $ to the $ x $-axis is always equal to the radius $ \\frac{|AF|}{2} $. Hence, the equation of the fixed line is $ y = 0 $."
  },
  {
    "text": "The parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$, directrix $l$, and $O$ is the origin. Point $P$ lies on the parabola such that $|O P|=|P F|$. Line segment $F P$ is extended to intersect the directrix $l$ at point $Q$. If the area of $\\triangle O F Q$ is $8 \\sqrt{2}$, then the equation of parabola $C$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;l: Line;Directrix(C) = l;O: Origin;P: Point;PointOnCurve(P, C);Abs(LineSegmentOf(O, P)) = Abs(LineSegmentOf(P, F));Q: Point;Intersection(OverlappingLine(LineSegmentOf(F, P)), l) = Q;Area(TriangleOf(O, F, Q)) = 8*sqrt(2)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=8*x",
    "fact_spans": "[[[0, 26], [55, 58], [133, 139]], [[0, 26]], [[8, 26]], [[8, 26]], [[30, 33]], [[0, 33]], [[37, 40], [88, 91]], [[0, 40]], [[42, 45]], [[51, 54]], [[51, 59]], [[61, 74]], [[92, 96]], [[75, 96]], [[98, 131]]]",
    "query_spans": "[[[133, 144]]]",
    "process": "From the given information, the equation of the directrix $ l $ of the parabola is $ x = -\\frac{p}{2} $, then the distance from the focus $ F\\left(\\frac{p}{2}, 0\\right) $ to the directrix is $ p $. It is known that $ |OP| = |PF| $, so point $ P $ lies on the perpendicular bisector of segment $ OF $. As shown in the figure, $ y_{P} = \\frac{\\sqrt{2}}{2}p $. We have $ \\tan\\angle OFP = \\frac{|y_{P}|}{\\frac{1}{4}p} = \\frac{|y_{Q}|}{p} $, that is, $ \\frac{\\frac{\\sqrt{2}}{2}p}{\\frac{1}{4}p} = \\frac{|y_{Q}|}{p} $. Therefore, $ S_{\\triangle OFQ} = \\frac{1}{2} \\cdot |OF| \\cdot |y_{Q}| = \\frac{1}{2} \\cdot \\frac{p}{2} \\cdot 2\\sqrt{2}p = 8\\sqrt{2} $. Solving gives $ p = 4 $. Hence, the equation of the parabola $ C $ is $ y^2 = 8x $."
  },
  {
    "text": "Given that the foci of the hyperbola $2 x^{2}-y^{2}=m$ lie on the $x$-axis, and one focus is $(\\sqrt{3}, 0)$, then the value of $m$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (2*x^2 - y^2 = m);m: Number;PointOnCurve(Focus(G),xAxis) = True;Coordinate(OneOf(Focus(G))) = (sqrt(3),0)",
    "query_expressions": "m",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 22]], [[2, 22]], [[54, 57]], [[2, 30]], [[2, 52]]]",
    "query_spans": "[[[54, 61]]]",
    "process": ""
  },
  {
    "text": "A point $A$ on the parabola $y^{2}=4x$ is such that the sum of the distances from $A$ to the point $B(3, 2)$ and to the focus is minimized. Then the coordinates of point $A$ are?",
    "fact_expressions": "G: Parabola;B: Point;A: Point;Expression(G) = (y^2 = 4*x);Coordinate(B) = (3, 2);PointOnCurve(A, G);WhenMin(Distance(A, B) + Distance(A, Focus(G)))",
    "query_expressions": "Coordinate(A)",
    "answer_expressions": "(1, 2)",
    "fact_spans": "[[[0, 14]], [[21, 32]], [[17, 20], [44, 48]], [[0, 14]], [[21, 32]], [[0, 20]], [[0, 42]]]",
    "query_spans": "[[[44, 53]]]",
    "process": ""
  },
  {
    "text": "Let the hyperbola $\\frac{x^{2}}{m}+\\frac{y^{2}}{n}=1$ have eccentricity $2$, and one focus coincides with the focus of the parabola $x^{2}=8 y$. Then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;m: Number;n: Number;H: Parabola;Expression(G) = (y^2/n + x^2/m = 1);Expression(H) = (x^2 = 8*y);Eccentricity(G) = 2;OneOf(Focus(G))=Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 - x^2/3 = 1",
    "fact_spans": "[[[1, 39], [76, 79]], [[4, 39]], [[4, 39]], [[54, 68]], [[1, 39]], [[54, 68]], [[1, 47]], [[1, 73]]]",
    "query_spans": "[[[75, 84]]]",
    "process": "The focus of the parabola has coordinates (0,2), so the foci of the hyperbola lie on the y-axis and c=2, thus the foci of the hyperbola are (0,2), (0,-2). Also, e=\\frac{c}{a}=2, so a=1, that is, n=1. Moreover, b^{2}=c^{2}-a^{2}=3, so m=-b^{2}=-3. Hence, the equation of the hyperbola is y^{2}-\\frac{x^{2}}{3}=1"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$, a line $l$ passes through point $F$ and intersects $C$ at points $A$ and $B$, and intersects the directrix of $C$ at point $P$. If $\\overrightarrow{AP}=3\\overrightarrow{BP}$, then the slope of $l$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 8*x);F: Point;Focus(C) = F;l: Line;PointOnCurve(F, l) ;Intersection(l, C) = {A, B};A: Point;B: Point;Intersection(l, Directrix(C)) = P;P: Point;VectorOf(A, P) = 3*VectorOf(B, P)",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*sqrt(3)",
    "fact_spans": "[[[2, 21], [40, 43], [56, 59]], [[2, 21]], [[25, 28], [35, 39]], [[2, 28]], [[29, 34], [117, 120]], [[29, 39]], [[29, 54]], [[45, 48]], [[49, 52]], [[29, 68]], [[64, 68]], [[70, 115]]]",
    "query_spans": "[[[117, 125]]]",
    "process": ""
  },
  {
    "text": "Given a point $M(1, m)$ on the parabola $y^{2}=2 p x(p>0)$ such that the distance from $M$ to its focus $F$ is $5$, and the distance from the vertex of the parabola to the line $M F$ is $d$, then the value of $d$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;M: Point;m: Number;Coordinate(M) = (1, m);PointOnCurve(M, G);F: Point;Focus(G) = F;Distance(M, F) = 5;d: Number;Distance(Vertex(G), LineOf(M, F)) = d",
    "query_expressions": "d",
    "answer_expressions": "16/5",
    "fact_spans": "[[[2, 23], [36, 37], [51, 54]], [[2, 23]], [[5, 23]], [[5, 23]], [[26, 35]], [[26, 35]], [[26, 35]], [[2, 35]], [[39, 42]], [[36, 42]], [[26, 49]], [[69, 72], [74, 77]], [[51, 72]]]",
    "query_spans": "[[[74, 81]]]",
    "process": "The distance from point M(1, m) on the parabola y^{2}=2px (p>0) to its focus F is 5, so 1+\\frac{p}{2}=5. Solving gives p=8, thus the parabola equation is y^{2}=16x, and the focus is F(4,0). Substituting point M(1,m) into y^{2}=16x yields: m=\\pm4. Without loss of generality, take m=4, then M(1,4). The equation of line MF is: y-0=-\\frac{4}{3}(x-4), or 4x+3y-16=0. The distance from the vertex of the parabola to line MF is d=\\frac{16}{\\sqrt{4^{2}+3^{2}}}=\\frac{16}{5}. By symmetry of the parabola, when m=-4, the distance from the vertex to line MF is also \\frac{16}{5}. Hence, the answer is: \\frac{16}{5}. Using geometric properties of the parabola to find the curve equation, and using the formula to find the distance from a point to a line."
  },
  {
    "text": "From a point $P$ on the parabola $y^{2}=4x$, draw a perpendicular to its directrix, with foot of perpendicular at $M$. Let the focus of the parabola be $F$, and $PF=5$. Then the area of $\\Delta{MPF}$ is?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;M: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(P,G);l1:Line;PointOnCurve(P,l1);IsPerpendicular(Directrix(G),l1);FootPoint(Directrix(G),l1)=M;Focus(G)=F;LineSegmentOf(P, F) = 5",
    "query_expressions": "Area(TriangleOf(M,P,F))",
    "answer_expressions": "10",
    "fact_spans": "[[[1, 15], [22, 23], [37, 40]], [[18, 21]], [[44, 47]], [[32, 35]], [[1, 15]], [[1, 21]], [], [[0, 28]], [[0, 28]], [[0, 35]], [[37, 47]], [[49, 56]]]",
    "query_spans": "[[[58, 78]]]",
    "process": ""
  },
  {
    "text": "Draw a line $l$ through the left focus $F_{1}$ of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, intersecting the ellipse at points $A$ and $B$. Then, what is the perimeter of $\\triangle A B F_{2}$ (where $F_{2}$ is the right focus of the ellipse)?",
    "fact_expressions": "l: Line;G: Ellipse;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2/4 + y^2/3 = 1);LeftFocus(G) = F1;RightFocus(G)=F2;PointOnCurve(F1, l);Intersection(l, G) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A,B,F2))",
    "answer_expressions": "8",
    "fact_spans": "[[[50, 55]], [[1, 38], [56, 58], [102, 104]], [[59, 62]], [[63, 66]], [[94, 101]], [[42, 49]], [[1, 38]], [[1, 49]], [[94, 108]], [[0, 55]], [[50, 68]]]",
    "query_spans": "[[[70, 115]]]",
    "process": "From the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, we get $a=2$; by the definition of an ellipse: $|AF_{1}|+|AF_{2}|=|BF_{1}|+|BF_{2}|=2a=4$. Therefore, the perimeter of $\\triangle ABF_{2}$ is $|AB|+|AF_{2}|+|BF_{2}|=|AF_{1}|+|BF_{1}|+|AF_{2}|+|BF_{2}|=8$."
  },
  {
    "text": "If a point $M$ on the parabola $y^{2}=4 x$ is at a distance of $\\sqrt{5}$ from the origin $O$, then what is the distance from point $M$ to the focus of the parabola?",
    "fact_expressions": "G: Parabola;M: Point;O: Origin;Expression(G) = (y^2 = 4*x);PointOnCurve(M, G);Distance(M, O) = sqrt(5)",
    "query_expressions": "Distance(M, Focus(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 15], [52, 55]], [[19, 22], [46, 50]], [[23, 30]], [[1, 15]], [[1, 22]], [[19, 44]]]",
    "query_spans": "[[[46, 62]]]",
    "process": "Let point $ M\\left(\\frac{y^{2}}{4}, y\\right) $. Since $ |MO| = \\sqrt{5} $, we have $ \\left(\\frac{y^{2}}{4} - 0\\right)^{2} + (y - 0)^{2} = 5 $, so $ y^2 = 4 $ or $ y^2 = -16 $ (discarded). Thus, $ x = \\frac{y^{2}}{4} = 1 $. The distance $ d $ from $ M $ to the directrix $ x = -1 $ of the parabola $ y^{2} = 4x $ is $ 1 - (-1) = 2 $. Since the distance from point $ M $ to the focus of this parabola equals the distance from point $ M $ to the directrix of the parabola $ y^2 = 4x $, the distance from point $ M $ to the focus of this parabola is: $ 2 $. Answer: $ 2 $"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with foci $F_{1}$, $F_{2}$ and eccentricity $\\frac{\\sqrt{6}}{2}$, if a point $P$ on $C$ satisfies $|P F_{1}|-|PF_{2}|=2 \\sqrt{6}$, then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Focus(C) = {F1, F2};Eccentricity(C) = sqrt(6)/2;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)) = 2*sqrt(6)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/6 - y^2/3 = 1",
    "fact_spans": "[[[2, 63], [109, 112], [153, 156]], [[10, 63]], [[10, 63]], [[115, 118]], [[67, 74]], [[75, 82]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 82]], [[2, 107]], [[109, 118]], [[120, 151]]]",
    "query_spans": "[[[153, 161]]]",
    "process": "\\because|PF_{1}|-|PF_{2}|=2\\sqrt{6}, by the definition of hyperbola we know a=\\sqrt{6}; from e=\\frac{c}{a}=\\frac{\\sqrt{6}}{2}, we get c=3, then b^{2}=c^{2}-a^{2}=3, so the equation of hyperbola C is \\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1"
  },
  {
    "text": "The distance from a point $M$ on the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{16}=1$ to the left focus $F_{1}$ is $2$. $N$ is the midpoint of the segment $MF_{1}$ ($O$ is the coordinate origin), then $|ON|=$?",
    "fact_expressions": "G: Ellipse;F1: Point;M: Point;O: Origin;N: Point;Expression(G) = (x^2/36 + y^2/16 = 1);PointOnCurve(M, G);LeftFocus(G) = F1;Distance(M, F1) = 2;MidPoint(LineSegmentOf(M, F1)) = N",
    "query_expressions": "Abs(LineSegmentOf(O, N))",
    "answer_expressions": "5",
    "fact_spans": "[[[0, 39]], [[49, 56]], [[42, 45]], [[85, 88]], [[66, 69]], [[0, 39]], [[0, 45]], [[0, 56]], [[42, 64]], [[66, 83]]]",
    "query_spans": "[[[96, 104]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the focus of the parabola $y^{2}=2 p x(p>0)$ are $(\\frac{1}{8}, 0)$, then $p$=?",
    "fact_expressions": "G: Parabola;p: Number;p>0;Expression(G) = (y^2 = 2*p*x);Coordinate(Focus(G))=(1/8,0)",
    "query_expressions": "p",
    "answer_expressions": "1/4",
    "fact_spans": "[[[0, 21]], [[48, 51]], [[3, 21]], [[0, 21]], [[0, 45]]]",
    "query_spans": "[[[48, 53]]]",
    "process": "According to the definition of a parabola, the answer can be obtained. \\because y^{2}=2px (p>0), the focus coordinate is (\\frac{1}{8},0) \\therefore \\frac{p}{2}=\\frac{1}{8}, solving gives: p=\\frac{1}{4}."
  },
  {
    "text": "Given that the focal distance of the ellipse $\\frac{x^{2}}{t^{2}}+\\frac{y^{2}}{5 t}=1$ is $2 \\sqrt{6}$, find the real number $t=?$",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/t^2 + y^2/(5*t) = 1);t: Real;FocalLength(G) = 2*sqrt(6)",
    "query_expressions": "t",
    "answer_expressions": "{2,3,6}",
    "fact_spans": "[[[2, 45]], [[2, 45]], [[63, 68]], [[2, 61]]]",
    "query_spans": "[[[63, 70]]]",
    "process": "If the ellipse's foci are on the x-axis, then \\begin{cases}t^{2}-5t=(\\frac{2\\sqrt{6}}{2})^{2}\\\\2-6,0\\end{cases}, solving gives: t=6. If the ellipse's foci are on the y-axis, then \\begin{cases}t^{2}>5t>0\\\\5t-t^{2}=(\\frac{2\\sqrt{6}}{2})\\\\5t>t^{2}>0\\end{cases}, solving gives: t=2 or 3."
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $\\frac{\\sqrt{6}}{2}$, what is the eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Ellipse;a > b;b > 0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = sqrt(6)/2",
    "query_expressions": "Eccentricity(H)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 55]], [[5, 55]], [[5, 55]], [[82, 127]], [[5, 55]], [[5, 55]], [[2, 55]], [[82, 127]], [[2, 80]]]",
    "query_spans": "[[[82, 133]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, a line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. If $|F_{2} A|+|F_{2} B|=12$, then $|A B|$=?",
    "fact_expressions": "G: Ellipse;H: Line;F2: Point;A: Point;B: Point;F1: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(G) = {F1, F2};PointOnCurve(F1, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(F2, A)) + Abs(LineSegmentOf(F2, B)) = 12",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[18, 56], [74, 76]], [[71, 73]], [[10, 17]], [[77, 80]], [[81, 84]], [[2, 9], [63, 70]], [[18, 56]], [[2, 61]], [[62, 73]], [[71, 86]], [[88, 112]]]",
    "query_spans": "[[[114, 123]]]",
    "process": ""
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{5-m}+\\frac{y^{2}}{m+3}=1$ represents an ellipse, what is the range of values for $m$?",
    "fact_expressions": "G: Ellipse;Expression(G)=(x^2/(5 - m) + y^2/(m + 3) = 1) ;m:Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-3,1)+(1,5)",
    "fact_spans": "[[[45, 47]], [[2, 47]], [[49, 52]]]",
    "query_spans": "[[[49, 59]]]",
    "process": "According to the characteristics of the standard equation of an ellipse, list the inequalities and solve them to obtain the result. Since the equation $\\frac{x^2}{5-m}+\\frac{y^{2}}{m+3}=1$ represents an ellipse, we have $\\begin{cases} m+3>0 \\\\ 5-m>0 \\end{cases}$, solving yields $-3<m<1$ or $1<m<5$, that is, the range of $m$ is $(-3,1)\\cup(1,5)$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, let point $F$ be the left focus and point $P$ be the lower vertex. A line $l$ parallel to $FP$ intersects the ellipse at points $A$ and $B$, and the midpoint of $A$ and $B$ is $M(1, \\frac{1}{2})$. Then the eccentricity of the ellipse is?",
    "fact_expressions": "l: Line;G: Ellipse;a: Number;b: Number;F: Point;P: Point;M: Point;A: Point;B: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(M) = (1, 1/2);LowerVertex(G)=P;LeftFocus(G) = F;IsParallel(LineSegmentOf(F, P),l);MidPoint(LineSegmentOf(A,B))=M;Intersection(l, G) = {A, B}",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[82, 87]], [[2, 54], [88, 90], [134, 136]], [[4, 54]], [[4, 54]], [[55, 59]], [[64, 68]], [[113, 132]], [[91, 94], [102, 105]], [[95, 98], [106, 109]], [[4, 54]], [[4, 54]], [[2, 54]], [[113, 132]], [[2, 72]], [[2, 63]], [[73, 87]], [[102, 132]], [[82, 100]]]",
    "query_spans": "[[[134, 142]]]",
    "process": "First, find the slope of line FP as $\\frac{-b-0}{0-(-c)}=-\\frac{b}{c}$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then use the point difference method to find the slope of line AB: $k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{2b^{2}}{a^{2}}$. Using the equality of slopes, obtain the relationship among $a$, $b$, and $c$, and combine with $a^{2}=b^{2}+c^{2}$ to find the eccentricity. From the problem, $F(-c,0)$, $P(0,-b)$, so the slope of line FP is $\\frac{-b-0}{0-(-c)}=-\\frac{b}{c}$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $\\frac{x_{1}^{2}}{a^{2}}+\\frac{y_{1}^{2}}{b^{2}}=1$①, $\\frac{x_{2}^{2}}{a^{2}}+\\frac{y_{2}^{2}}{b^{2}}=1$②. ①$-$② gives: $\\frac{x_{1}^{2}-x_{2}^{2}}{a^{2}}=-\\frac{y_{1}^{2}-y_{2}^{2}}{b^{2}}$, i.e., $\\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{a^{2}}=-\\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{b^{2}}$. Since $M(1,\\frac{1}{2})$ is the midpoint of $A,B$, $x_{1}+x_{2}=2$, $y_{1}+y_{2}=1$, so $\\frac{2(x_{1}-x_{2})}{a^{2}}=-\\frac{(y_{1}-y_{2})}{b^{2}}$, thus $k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{2b^{2}}{a^{2}}$. Since $AB \\parallel FP$, $-\\frac{b}{c}=-\\frac{2b^{2}}{a^{2}}$, i.e., $a^{2}=2bc$, so $b^{2}+c^{2}=2bc$, hence $b=c$, so $a^{2}=b^{2}+c^{2}=2c^{2}$, thus $e=\\frac{c}{a}=\\frac{\\sqrt{2}}{2}$."
  },
  {
    "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ are given by $y=\\pm \\frac{1}{2} x$, then what is the value of $b$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/b^2 = 1);b: Number;b>0;Expression(Asymptote(G)) = (y = pm*1/2*x)",
    "query_expressions": "b",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 48]], [[1, 48]], [[79, 82]], [[4, 48]], [[1, 77]]]",
    "query_spans": "[[[79, 85]]]",
    "process": ""
  },
  {
    "text": "$P$ is a point on the right branch of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, and $F_{1}$, $F_{2}$ are the left and right foci, respectively. If $|P F_{1}|+|P F_{2}|=10$, then $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$=?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2/3 = 1);PointOnCurve(P, RightPart(G));LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 10",
    "query_expressions": "DotProduct(VectorOf(P, F1), VectorOf(P, F2))",
    "answer_expressions": "18",
    "fact_spans": "[[[4, 32]], [[0, 3]], [[38, 45]], [[46, 53]], [[4, 32]], [[0, 37]], [[4, 59]], [[4, 59]], [[61, 85]]]",
    "query_spans": "[[[87, 146]]]",
    "process": "Let P(x,y), with F_{1} and F_{2} being the left and right foci, respectively, i.e., F_{1}(-2,0), F_{2}(2,0). \\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=(-2-x,y)(2-x,y)=-(4-x^{2})+y^{2}=-4+x^{2}+y^{2}. Since P lies on the hyperbola x^{2}-\\frac{y^{2}}{3}=1, i.e., 3x^{2}-y^{2}=3, \\therefore \\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=.\\frac{4+x^{2}+y^{2}=4x^{2}-7,}{|\\overrightarrow{PF}|}=\\sqrt{(-2-x)^{2}+y^{2}}, ||\\overrightarrow{PF_{2}}|}=\\sqrt{(2-x)^{2}+y^{2}}, then |\\overrightarrow{PF_{1}}|+|\\overrightarrow{PF_{2}}|=\\sqrt{(2+x^{2}+y_{2}+=\\sqrt{(2-x^{2}+y)^{2}}=10. Substituting 3x^{2}-y^{2}=3 into the above equation yields x=\\frac{5}{2}, hence \\overrightarrow{PF}\\cdot\\overrightarrow{PF_{2}}=-4+x^{2}+y^{2}=4x^{2}-7=25-7=18,"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the right focus is $F$, the left vertex is $A$. A circle centered at $F$ with radius $|F A|$ intersects the right branch of $C$ at points $M$ and $N$, and the perpendicular bisector of segment $A M$ passes through point $N$. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;M: Point;A: Point;F: Point;N: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;LeftVertex(C)=A;Center(G)=F;Radius(G)=Abs(LineSegmentOf(F,A));Intersection(G,RightPart(C))={M, N};PointOnCurve(N,PerpendicularBisector(LineSegmentOf(A,M)))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "4/3",
    "fact_spans": "[[[2, 63], [101, 104], [101, 104]], [[10, 63]], [[10, 63]], [[99, 100]], [[108, 111]], [[76, 79]], [[81, 84], [81, 84]], [[112, 115], [134, 138]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 71]], [[2, 79]], [[80, 100]], [[88, 100]], [[99, 117]], [[119, 138]]]",
    "query_spans": "[[[140, 149]]]",
    "process": "First prove that 4AMN is an equilateral triangle. In AMFF, by the law of cosines and combining with the definition of the hyperbola, we obtain |FF|^{2}+|FM|^{2}-2|FF||FM|\\cos120^{\\circ}=|FM|^{2}=(|FM|+2a)^{2}, which simplifies to 3c^{2}-ac-4a^{2}=0, thus obtaining the result. According to the problem, A(-a,0), F(c,0), and the other focus F(-c,0). By symmetry, AM=AN. Since the perpendicular bisector of segment AM passes through point N, it follows that AN=MN, so AAMN is an equilateral triangle, as shown in the figure. Connect MF, then AF=MF=a+c. By the symmetry of the figure, \\angleMAF=\\angleNAF=\\frac{1}{2}\\angleMAN=30^{\\circ}. Also, since 4AMF is an isosceles triangle, \\angleAFM=120^{\\circ}. In AMFF, by the law of cosines: |FF|^{2}+|FM|^{2}-2|FF||FM|\\cos120^{\\circ}=|FM|^{2}=(|FM|+2a)^{2}. The above equation can be rewritten as 4c^{2}+(a+c)^{2}-2\\times2c(a+c)(-\\frac{1}{2})=(3a+c)^{2}. Simplifying yields: 3c^{2}-ac-4a^{2}=0, i.e., (c+a)(3c-4a)=0. Since a>0, c>0, we have 3c-4a=0, c=\\frac{4}{3}a. Hence e=\\frac{c}{a}=\\frac{4}{3}."
  },
  {
    "text": "Given the asymptotes of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{n}=1$ are $5 x \\pm 3 y=0$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/n + x^2/m = 1);m: Number;n: Number;Expression(Asymptote(G)) = (pm*3*y + 5*x = 0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{sqrt(34)/3, sqrt(34)/5}",
    "fact_spans": "[[[2, 40], [65, 68]], [[2, 40]], [[5, 40]], [[5, 40]], [[2, 62]]]",
    "query_spans": "[[[65, 74]]]",
    "process": "According to the problem, the asymptotes of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{n}=1$ are given by $5x\\pm3y=0$, that is, $y=\\pm\\frac{5}{3}x$. When $m>0, n>0$, the foci of the hyperbola lie on the $x$-axis, so $\\frac{\\sqrt{n}}{\\sqrt{m}}=\\frac{5}{3}$, i.e., $\\frac{n}{m}=\\frac{25}{9}$. Thus, the eccentricity of the hyperbola is $e=\\frac{\\sqrt{m+n}}{\\sqrt{m}}=\\sqrt{1+\\frac{n}{m}}=\\sqrt{\\frac{34}{9}}=\\frac{\\sqrt{34}}{3}$. When $m<0, n<0$, the foci of the hyperbola lie on the $y$-axis, so $\\frac{\\sqrt{-n}}{\\sqrt{-m}}=\\frac{5}{3}$, i.e., $\\frac{n}{m}=\\frac{25}{9}$. Thus, the eccentricity of the hyperbola is $e=\\frac{\\sqrt{m+n}}{\\sqrt{n}}=\\sqrt{1+\\frac{m}{n}}=\\sqrt{\\frac{34}{25}}=\\frac{\\sqrt{34}}{5}$. Therefore, the eccentricity of the hyperbola is $\\frac{\\sqrt{34}}{3}$ or $\\frac{\\sqrt{34}}{5}$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, $F_{1}$, $F_{2}$ are the left and right foci of $C$, and $P$ is a moving point on the ellipse. Then the maximum value of the inradius of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/25 + y^2/16 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C)",
    "query_expressions": "Max(Radius(InscribedCircle(TriangleOf(F1,P,F2))))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[2, 46], [66, 69], [80, 82]], [[2, 46]], [[48, 56]], [[58, 65]], [[48, 75]], [[48, 75]], [[76, 79]], [[76, 86]]]",
    "query_spans": "[[[88, 121]]]",
    "process": ""
  },
  {
    "text": "Given point $M(-1,2)$ and parabola $C$: $y^{2}=4x$, a line passing through the focus of $C$ with slope $k$ intersects $C$ at points $A$ and $B$. If $\\angle AMB=90^{\\circ}$, then $k=?$",
    "fact_expressions": "M: Point;Coordinate(M) = (-1, 2);C: Parabola;Expression(C) = (y^2 = 4*x);G: Line;PointOnCurve(Focus(C), G);k: Number;Slope(G) = k;A: Point;B: Point;Intersection(G, C) = {A, B};AngleOf(A, M, B) = ApplyUnit(90, degree)",
    "query_expressions": "k",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 12]], [[2, 12]], [[13, 32], [34, 37], [51, 54]], [[13, 32]], [[48, 50]], [[33, 50]], [[94, 97], [44, 47]], [[41, 50]], [[56, 59]], [[60, 63]], [[48, 65]], [[67, 92]]]",
    "query_spans": "[[[94, 99]]]",
    "process": "By the given condition, the focus of $ C: y^{2} = 4x $ is $ (1,0) $, so we can set the line as $ y = k(x - 1) $. Combining with the parabola equation, $ \\therefore k^{2}x^{2} - 2(k^{2} + 2)x + k^{2} = 0 $. If $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, $ \\therefore x_{1} + x_{2} = 2 + \\frac{4}{k^{2}} $, $ x_{1}x_{2} = 1 $, then $ y_{1} + y_{2} = k(x_{1} + x_{2} - 2) = \\frac{4}{k} $, $ y_{1}y_{2} = k^{2}[x_{1}x_{2} - (x_{1} + x_{2}) + 1] = -4 $. Since $ \\angle AMB = 90^{\\circ} $, i.e., $ \\overrightarrow{MA} \\cdot \\overrightarrow{MB} = (x_{1} + 1, y_{1} - 2) \\cdot (x_{2} + 1, y_{2} - 2) = 0 $, $ \\therefore x_{1}x_{2} + x_{1} + x_{2} + 1 + y_{1}y_{2} - 2(y_{1} + y_{2}) + 4 = 0 $, then $ \\frac{1}{k^{2}} - \\frac{2}{k} + 1 = 0 $, $ \\therefore k = 1 $."
  },
  {
    "text": "The eccentricity of an ellipse is $\\frac{1}{2}$, one focus is $F(3 , 0)$, and the corresponding directrix is $x-1=0$. Then the equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;Eccentricity(G) = 1/2;F: Point;OneOf(Focus(G)) = F;Coordinate(F) = (3, 0);Expression(Directrix(G)) = (x-1 = 0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "3*x^2+4*y^2-22*x+35=0",
    "fact_spans": "[[[0, 2], [54, 56]], [[0, 20]], [[28, 38]], [[0, 38]], [[28, 38]], [[0, 50]]]",
    "query_spans": "[[[54, 60]]]",
    "process": ""
  },
  {
    "text": "Given two fixed points $A(0,-2)$, $B(0,2)$, and a point $P$ on the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{16}=1$ such that $|\\overrightarrow{A P}|-|\\overrightarrow{B P}|=2$, then $\\overrightarrow{A P} \\cdot \\overrightarrow{B P}$=?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;P: Point;Expression(G) = (x^2/12 + y^2/16 = 1);Coordinate(A) = (0,-2);Coordinate(B) = (0, 2);PointOnCurve(P, G);Abs(VectorOf(A, P)) - Abs(VectorOf(B, P)) = 2",
    "query_expressions": "DotProduct(VectorOf(A, P), VectorOf(B, P))",
    "answer_expressions": "9",
    "fact_spans": "[[[30, 69]], [[5, 14]], [[16, 24]], [[25, 29]], [[30, 69]], [[5, 14]], [[16, 24]], [[25, 70]], [[74, 123]]]",
    "query_spans": "[[[125, 176]]]",
    "process": ""
  },
  {
    "text": "Given that point $P(1,-2)$ lies on the line $y=k x+2$, then the eccentricity of the conic section $C$: $k x^{2}+3 y^{2}=1$ is?",
    "fact_expressions": "H: Line;k: Number;C:ConicSection;P: Point;Expression(H) = (y = k*x + 2);Expression(C) = (k*x^2 + 3*y^2 = 1);Coordinate(P) = (1, -2);PointOnCurve(P, H)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(7)/2",
    "fact_spans": "[[[13, 24]], [[36, 55]], [[27, 55]], [[2, 12]], [[13, 24]], [[27, 55]], [[2, 12]], [[2, 25]]]",
    "query_spans": "[[[27, 61]]]",
    "process": "Analysis: Since point $ P(1,-2) $ lies on the line $ y = kx + 2 $, we can find the value of $ k $, thereby determining the values of $ a $ and $ c $, and then obtain the result. Details: $ \\because P(1,-2) $ lies on $ y = kx + 2 $, $ \\therefore -2 = k + 2 $, $ k = -4 $, $ -4x^{2} + 3y^{2} = 1 $, rewritten as $ \\frac{y^{2}}{3} - \\frac{x^{2}}{\\frac{1}{4}} = 1 $, $ a = \\frac{\\sqrt{3}}{3} $, $ \\therefore c = \\sqrt{\\frac{1}{4} + \\frac{1}{3}} = \\frac{\\sqrt{7}}{2\\sqrt{3}} $, $ \\therefore e = \\frac{c}{a} = \\frac{\\sqrt{7}}{2} $,"
  },
  {
    "text": "A point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ has a distance of $\\frac{5}{2}$ to the left directrix; then the distance from point $P$ to the left focus is?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/25 + y^2/9 = 1);PointOnCurve(P, G);Distance(P, LeftDirectrix(G)) = 5/2",
    "query_expressions": "Distance(P,LeftFocus(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 38]], [[41, 44], [67, 71]], [[0, 38]], [[0, 44]], [[0, 65]]]",
    "query_spans": "[[[0, 80]]]",
    "process": ""
  },
  {
    "text": "Given that points $F$ and $B$ are the focus and the endpoint of the imaginary axis of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively, if the midpoint of segment $FB$ lies on the hyperbola $C$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;F: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);OneOf(Focus(C))=F;OneOf(Endpoint(ImageinaryAxis(C)))=B;PointOnCurve(MidPoint(LineSegmentOf(F,B)), C)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[13, 74], [95, 101], [104, 110]], [[20, 74]], [[20, 74]], [[2, 6]], [[7, 10]], [[20, 74]], [[20, 74]], [[13, 74]], [[2, 82]], [[2, 82]], [[84, 102]]]",
    "query_spans": "[[[104, 116]]]",
    "process": ""
  },
  {
    "text": "The right focus of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{6}=1$ is $F$, and chord $A B$ is perpendicular to the major axis. When the perimeter of $\\triangle A B F$ is maximized, what is the area of the triangle's circumcircle?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;F: Point;Expression(G) = (x^2/9 + y^2/6 = 1);RightFocus(G) = F;IsChordOf(LineSegmentOf(A, B), G);IsPerpendicular(LineSegmentOf(A, B), MajorAxis(G));WhenMax(Perimeter(TriangleOf(A, B, F)))",
    "query_expressions": "Area(CircumCircle(TriangleOf(A, B, F)))",
    "answer_expressions": "16*pi/3",
    "fact_spans": "[[[0, 37]], [[46, 51]], [[46, 51]], [[41, 44]], [[0, 37]], [[0, 44]], [[0, 51]], [[0, 55]], [[56, 80]]]",
    "query_spans": "[[[57, 91]]]",
    "process": "Let $ A(x_{0},y_{0}) $, and point $ A $ lies above the x-axis, $ x_{0}=3\\cos\\alpha $, $ y_{0}=\\sqrt{6}\\sin\\alpha $. From the problem, $ a=3 $, $ b=\\sqrt{6} $, $ c=\\sqrt{3} $. It follows that $ |AF|=|BF|=\\sqrt{(3\\cos\\alpha-\\sqrt{3})^{2}+(\\sqrt{6}\\sin\\alpha)^{2}}=2(3-\\sqrt{3}\\cos\\alpha) $. Then the perimeter of $ \\triangle ABF $ is $ l=|AF|+|BF|+|AB|=6+2\\sqrt{6}\\sin\\alpha-2\\sqrt{3}\\cos\\alpha=6+6\\sin(\\alpha+\\theta) $. The perimeter of $ \\triangle ABF $ reaches its maximum if and only if $ 2\\sqrt{6}\\sin\\alpha-2\\sqrt{3}\\cos\\alpha=6 $. Solving the system:\n$$\n\\begin{cases}\n2\\sqrt{6}\\sin\\alpha-2\\sqrt{3}\\cos\\alpha=6 \\\\\n\\sin^{2}\\alpha+\\cos^{2}\\alpha=1\n\\end{cases}\n$$\nwe obtain $ \\cos\\alpha=-\\frac{\\sqrt{3}}{3} $. Therefore, $ A(-\\sqrt{3},2) $, $ B(-\\sqrt{3},-2) $. At this moment, $ AB $ passes through the left focus. In $ \\triangle ABF $, $ AF=\\sqrt{2^{2}+(2\\sqrt{3})^{2}}=4 $, so $ \\sin A=\\frac{2\\sqrt{3}}{4}=\\frac{\\sqrt{3}}{2} $. By the law of sines, $ 2R=\\frac{BF}{\\sin A}=\\frac{4}{\\frac{\\sqrt{3}}{2}}=\\frac{8}{\\sqrt{3}} $. Thus, the area of the circumcircle of the triangle is $ S=\\pi R^{2} $. This problem mainly examines properties of ellipses and the calculation of the area of a triangle's circumcircle, aiming to assess students' understanding and mastery of these concepts."
  },
  {
    "text": "The line $l$: $x+2 y-4=0$ intersects the ellipse $C$: $x^{2}+4 y^{2}=16$ at points $A$ and $B$. Then the chord length $|A B|=$?",
    "fact_expressions": "l: Line;C: Ellipse;A: Point;B: Point;Expression(C) = (x^2 + 4*y^2 = 16);Expression(l) = (x + 2*y - 4 = 0);Intersection(l, C) = {A, B};IsChordOf(LineSegmentOf(A, B), C)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[0, 18]], [[19, 44]], [[46, 49]], [[50, 53]], [[19, 44]], [[0, 18]], [[0, 55]], [[19, 66]]]",
    "query_spans": "[[[59, 68]]]",
    "process": "Solve the system of equations of the line and the ellipse, use Vieta's formulas to determine the relationship between roots and coefficients, then apply the chord length formula |AB| = \\sqrt{(1+k^{2})[(x_{1}+x_{2})^{2}-4x_{1}x_{2}}] to find the chord length. Given the line l: x + 2y - 4 = 0 and the ellipse C: x^{2} + 4y^{2} = 16 intersecting at points A and B, let A(x_{1}, y_{1}), B(x_{2}, y_{2}), -4x_{1}x_{2}] = \\sqrt{[1+()]^{2}_{1\\times4^{2}} = 2\\sqrt{5} 2\\sqrt{5} Qing】When solving comprehensive problems involving lines and ellipses, note: (1) carefully examine every condition provided in the problem, clearly identifying the conditions that define the line and the ellipse; (2) strengthen computational skills after combining the equations of the line and ellipse into a quadratic equation, emphasizing the relationships between roots and coefficients, chord length, slope, and triangle area problems."
  },
  {
    "text": "$F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively. A line is drawn through point $F_{2}$ perpendicular to one asymptote of this hyperbola, with foot of perpendicular at $M$, satisfying $|\\overrightarrow{M F_{1}}|=3|\\overrightarrow{M F_{2}}|$. Then the equation of the asymptotes of this hyperbola is?",
    "fact_expressions": "F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Z: Line;PointOnCurve(F2, Z);IsPerpendicular(Z, OneOf(Asymptote(G)));M: Point;FootPoint(Z, OneOf(Asymptote(G))) = M;Abs(VectorOf(M, F1)) = 3*Abs(VectorOf(M, F2))",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(2)/2)*x",
    "fact_spans": "[[[0, 7]], [[8, 15], [82, 90]], [[0, 80]], [[0, 80]], [[18, 74], [92, 95], [172, 175]], [[18, 74]], [[21, 74]], [[21, 74]], [[21, 74]], [[21, 74]], [], [[81, 103]], [[81, 103]], [[107, 110]], [[81, 110]], [[113, 169]]]",
    "query_spans": "[[[172, 183]]]",
    "process": "Let the distance from point $F_{2}(c,0)$ to the asymptote $y=\\frac{b}{a}x$ be $d=\\frac{|bc|}{\\sqrt{1+(\\frac{b}{a})^{2}}}=b$, that is, $|\\overrightarrow{MF_{2}}|=b$, and $|\\overrightarrow{MF_{1}}|=3|\\overrightarrow{MF_{2}}|$. Thus, $|\\overrightarrow{MF_{1}}|=3b$. In $\\triangle MF_{1}O$, $|\\overrightarrow{OM}|=a$, $|\\overrightarrow{OF}|=c$, $\\cos\\angle F_{1}OM=-\\frac{a}{c}$, by the law of cosines: $\\frac{a^{2}+c^{2}-9b^{2}}{2ac}=-\\frac{a}{c}$. Since $c^{2}=a^{2}+b^{2}$, we obtain $a^{2}=2b^{2}$, that is, $\\frac{b}{a}=\\frac{\\sqrt{2}}{2}$. Therefore, the equations of the asymptotes are $y=\\pm\\frac{\\sqrt{2}}{2}x$."
  },
  {
    "text": "The two asymptotes of the hyperbola $\\frac{x^{2}}{144}-\\frac{y^{2}}{b^{2}}=1$ are perpendicular to each other, then its eccentricity is?",
    "fact_expressions": "G: Hyperbola;b: Number;l1: Line;l2: Line;Expression(G) = (x^2/144 - y^2/b^2 = 1);Asymptote(G) = {l1, l2};IsPerpendicular(l1, l2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 44], [57, 58]], [[3, 44]], [], [], [[0, 44]], [[0, 50]], [[0, 54]]]",
    "query_spans": "[[[57, 64]]]",
    "process": "\\because two asymptotes are perpendicular to each other, \\therefore \\frac{b}{12} \\times (-\\frac{b}{12}) = -1, \\therefore b^{2} = 144, \\therefore c^{2} = 288, \\therefore e = \\sqrt{2}."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, if points $A$ and $B$ lie on this parabola such that $\\angle AFB = \\frac{\\pi}{2}$, and the projection of the midpoint $M$ of segment $AB$ onto the directrix of the parabola is $N$, then the maximum value of $\\frac{|MN|}{|AB|}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;B: Point;F: Point;Focus(G) = F;M: Point;N: Point;PointOnCurve(A, G);PointOnCurve(B, G);AngleOf(A, F, B) = pi/2;M = MidPoint(LineSegmentOf(A, B));Projection(M, Directrix(G)) = N",
    "query_expressions": "Max(Abs(LineSegmentOf(M, N))/Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 16], [85, 88], [35, 38]], [[2, 16]], [[25, 29]], [[30, 33]], [[20, 23]], [[2, 23]], [[81, 84]], [[96, 99]], [[25, 41]], [[25, 41]], [[42, 70]], [[71, 84]], [[81, 99]]]",
    "query_spans": "[[[101, 128]]]",
    "process": ""
  },
  {
    "text": "Let a focus of the hyperbola be $F$, and an endpoint of the imaginary axis be $B$. If the line $FB$ is perpendicular to an asymptote of the hyperbola, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;B: Point;F: Point;OneOf(Focus(G)) = F;OneOf(Endpoint(ImageinaryAxis(G))) = B;IsPerpendicular(LineOf(F, B), OneOf(Asymptote(G)))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(sqrt(5) + 1)/2",
    "fact_spans": "[[[1, 4], [38, 41], [53, 56]], [[22, 25]], [[10, 13]], [[1, 13]], [[1, 25]], [[29, 49]]]",
    "query_spans": "[[[53, 62]]]",
    "process": "Let the hyperbola equation be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$, its asymptotes are given by $y=\\pm\\frac{b}{a}x$, and the focus is $F(c,0)$. Point $B(0,b)$ is an endpoint of the imaginary axis. $\\therefore$ the slope of line $FB$ is $k_{FB}=\\frac{0-b}{c-0}=-\\frac{b}{c}$; $\\because$ line $FB$ is perpendicular to the line $y=\\frac{b}{a}x$, $\\therefore -\\frac{b}{c}\\times\\frac{b}{a}=-1$, yielding $b^{2}=ac$; $\\because b^{2}=c^{2}-a^{2}$, $\\therefore c^{2}-a^{2}=ac$. Dividing both sides by $a^{2}$, we obtain $e^{2}-e-1=0$, solving gives $e=\\frac{\\sqrt{5}+1}{2}$ $(e>1)$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, a perpendicular is drawn from point $F_{2}$ to one asymptote of the hyperbola, with foot at point $A$, intersecting the other asymptote at point $B$, and $\\overrightarrow{A F_{2}}=-\\frac{1}{2} \\overrightarrow{F_{2} B}$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;A: Point;F2: Point;B: Point;F1: Point;l1: Line;l2: Line;a>0;b>0;L:Line;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;OneOf(Asymptote(G))=l1;OneOf(Asymptote(G))=l2;Negation(l1=l2);PointOnCurve(F2,L);IsPerpendicular(L,l1);FootPoint(L,l1)=A;Intersection(L,l2)=B;VectorOf(A, F2) = -VectorOf(F2, B)/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[18, 74], [90, 93], [192, 195]], [[21, 74]], [[21, 74]], [[106, 110]], [[10, 17], [81, 89]], [[119, 123]], [[2, 9]], [], [], [[21, 74]], [[21, 74]], [], [[18, 74]], [[2, 79]], [[2, 79]], [[90, 99]], [[90, 118]], [[90, 118]], [[80, 102]], [[80, 102]], [[80, 110]], [[80, 123]], [[125, 189]]]",
    "query_spans": "[[[192, 201]]]",
    "process": "Notice that in $\\triangle F_{2}BF_{1}$, $OA$ is the midline, so $\\triangle OBF_{1}$ is an equilateral triangle, thus $2a = c$, $e = 2$. [Detailed solution] From the given conditions, in $\\triangle BF_{1}F_{2}$, $A$ is the midpoint of side $BF_{2}$, $BF_{1} \\perp BF_{2}$, $|OA| = a$, $|BF| = 2a$, $|OB| = c$. Since $\\triangle OBF$ is an isosceles triangle, and $\\angle AOF_{2} = \\angle BOF_{1} = \\angle BF_{1}O$, therefore $\\triangle OBF$ is an equilateral triangle. Hence $2a = c$, $e = 2$."
  },
  {
    "text": "The coordinates of the foci of the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1$ are?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/3 + y^2/2 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(pm*1,0)",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 44]]]",
    "process": "From the ellipse equation $\\frac{x^2}{3}+\\frac{y^{2}}{2}=1$, we obtain $a^{2}=3$, $b^{2}=2$, so $c^{2}=a^{2}-b^{2}=1$, thus $c=1$, therefore the foci coordinates are $(-1,0)$, $(1,0)$."
  },
  {
    "text": "Given a point $M$ on the hyperbola $\\frac{x^{2}}{25}-\\frac{y^{2}}{9}=1$ such that the distance from $M$ to the left focus $F_{1}$ is $18$, then the distance from point $M$ to the right focus $F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;M: Point;F1: Point;F2:Point;Expression(G) = (x^2/25 - y^2/9 = 1);PointOnCurve(M, G);LeftFocus(G) = F1;Distance(M, F1) = 18;RightFocus(G)=F2",
    "query_expressions": "Distance(M, F2)",
    "answer_expressions": "{8,28}",
    "fact_spans": "[[[2, 41]], [[44, 47], [68, 72]], [[51, 58]], [[76, 83]], [[2, 41]], [[2, 47]], [[2, 58]], [[44, 66]], [[2, 83]]]",
    "query_spans": "[[[68, 88]]]",
    "process": "From the given conditions, we have a=5, b=3, c=\\sqrt{25+9}=\\sqrt{34}, and |MF_{1}|=18. By the definition of a hyperbola, ||MF_{1}|-|MF_{2}||=2a, that is, |18-|MF_{2}||=10. Solving this gives |MF_{2}|=8 or 28, both of which satisfy |MF_{2}|\\geqslant c-a."
  },
  {
    "text": "Let $0 \\leq \\alpha < 2\\pi$. If the equation $x^{2} \\sin \\alpha - y^{2} \\cos \\alpha = 1$ represents an ellipse with foci on the $y$-axis, then the range of values for $\\alpha$ is?",
    "fact_expressions": "G: Ellipse;alpha:Number;0<=alpha;alpha<2*pi;Expression(G) = (x^2*Sin(alpha) - y^2*Cos(alpha) = 1);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(alpha)",
    "answer_expressions": "(\\pi/2,3\\pi/4)+(3\\pi/2,7\\pi/4)",
    "fact_spans": "[[[76, 78]], [[80, 88]], [[1, 22]], [[1, 22]], [[24, 78]], [[67, 78]]]",
    "query_spans": "[[[80, 95]]]",
    "process": ""
  },
  {
    "text": "Point $P$ is a point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, and $F_{1}$, $F_{2}$ are its foci. If $\\angle F_{1} PF_{2}=90^{\\circ}$, what is the area of $\\Delta F_{1} PF_{2}$?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/16 + y^2/9 = 1);PointOnCurve(P, G);Focus(G) = {F1,F2};AngleOf(F1,P,F2)=ApplyUnit(90,degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "9",
    "fact_spans": "[[[5, 43], [63, 64]], [[47, 54]], [[0, 4]], [[55, 62]], [[5, 43]], [[0, 46]], [[47, 66]], [[68, 101]]]",
    "query_spans": "[[[103, 128]]]",
    "process": ""
  },
  {
    "text": "Given that one focus of the ellipse $x^{2}+k y^{2}=3 k(k>0)$ coincides with the focus of the parabola $x=\\frac{1}{12} y^{2}$, what is the eccentricity of the ellipse?",
    "fact_expressions": "H: Ellipse;Expression(H) = (k*y^2 + x^2 = 3*k);k: Number;k>0;G: Parabola;Expression(G) = (x = y^2/12);OneOf(Focus(H)) = Focus(G)",
    "query_expressions": "Eccentricity(H)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[2, 28], [67, 69]], [[2, 28]], [[4, 28]], [[4, 28]], [[34, 59]], [[34, 59]], [[2, 64]]]",
    "query_spans": "[[[67, 75]]]",
    "process": "From $ x = \\frac{1}{12}y^2 $, we have: $ y^{2} = 12x $, $ F(3,0) $, the ellipse equation is $ \\frac{x^{2}}{3k} + \\frac{y^{2}}{3} = 1 $ $ (k > 0) $, then $ a^{2} = 3k $, $ b^{2} = 3 $, $ c^{2} = 9 $, from $ a^{2} = b^{2} + c^{2} $, $ k = 4 $, $ e = \\frac{c}{a} = \\frac{\\sqrt{3}}{2} $"
  },
  {
    "text": "The focal distance of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{4}=1$ is $2$, then $m$=?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (y^2/4 + x^2/m = 1);FocalLength(G) = 2",
    "query_expressions": "m",
    "answer_expressions": "{3,5}",
    "fact_spans": "[[[0, 37]], [[46, 49]], [[0, 37]], [[0, 44]]]",
    "query_spans": "[[[46, 51]]]",
    "process": "Since the focal distance of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{4}=1$ is 2, we have $c=1$. If the foci lie on the $x$-axis, then $m=4+c^{2}$, solving gives $m=5$; if the foci lie on the $y$-axis, then $4=m+c^{2}$, solving gives $m=3$; in conclusion, $m=3$ or $5$."
  },
  {
    "text": "If the absolute difference of the distances from any point on hyperbola $C$ to its two foci is $6$, and the length of the imaginary axis of $C$ is $6 \\sqrt{3}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;P: Point;PointOnCurve(P, C);F1: Point;F2: Point;Focus(C) = {F1, F2};Abs(Distance(P, F1) - Distance(P, F2)) = 6;Length(ImageinaryAxis(C)) = 6*sqrt(3)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 7], [31, 34], [53, 56]], [], [[1, 12]], [], [], [[1, 16]], [[1, 29]], [[31, 51]]]",
    "query_spans": "[[[53, 62]]]",
    "process": "According to the problem, we have $2a=6$, $2b=6\\sqrt{3}$, so $a=3$, $b=3\\sqrt{3}$, thus $e=\\frac{c}{a}=\\frac{\\sqrt{a^{2}+b^{2}}}{a}=2$."
  },
  {
    "text": "The distance from a point $P$ on the parabola $x^{2}=m y(m>0)$ to the focus $F$ of the parabola and to the point $M(0,7)$ are both equal to $5$. Then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = m*y);m: Number;m>0;P: Point;PointOnCurve(P, G);F: Point;Focus(G) = F;M: Point;Coordinate(M) = (0, 7);Distance(P, F) = 5;Distance(P, M) = 5",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2=4*y",
    "fact_spans": "[[[0, 19], [26, 29], [60, 63]], [[0, 19]], [[3, 19]], [[3, 19]], [[21, 25]], [[0, 25]], [[32, 35]], [[26, 35]], [[40, 49]], [[40, 49]], [[21, 58]], [[21, 58]]]",
    "query_spans": "[[[60, 68]]]",
    "process": "The focus of the parabola $ x^{2} = my $ ($ m > 0 $) is $ (0, \\frac{m}{4}) $, then the distance from point $ P $ to the $ x $-axis is $ 5 - \\frac{m}{4} $, the midpoint coordinates of segment $ MF $ are $ (0, \\frac{7 + \\frac{m}{4}}{2}) $, from $ |PF| = |PM| $ we get $ 7 + \\frac{m}{4} = 5 - \\frac{m}{4} $, $ \\therefore m = 4 $, the equation of the parabola is $ x^{2} = 4y $, so the answer is $ x^{2} = 4y $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1(0<m<4)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. The line $l$ passes through point $F_{2}$ and intersects the ellipse $C$ at points $A$ and $B$. Then, what is the perimeter of $\\triangle A B F_{1}$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/m = 1);m: Number;0 < m;m < 4;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;l: Line;PointOnCurve(F2, l);A: Point;B: Point;Intersection(l, C) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, B, F1))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 51], [90, 95]], [[2, 51]], [[9, 51]], [[9, 51]], [[9, 51]], [[59, 66]], [[67, 74], [81, 89]], [[2, 74]], [[2, 74]], [[75, 80]], [[75, 89]], [[96, 99]], [[101, 105]], [[75, 107]]]",
    "query_spans": "[[[110, 136]]]",
    "process": "The solution process is omitted"
  },
  {
    "text": "Given that points $A(1, e)$ and $B(e, \\frac{\\sqrt{3}}{2})$ both lie on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, where $e$ is the eccentricity of the ellipse, then $e$=?",
    "fact_expressions": "A: Point;Coordinate(A) = (1, e);B: Point;Coordinate(B) = (e, sqrt(3)/2);G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;PointOnCurve(A, G) = True;PointOnCurve(B, G) = True;e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 12]], [[2, 12]], [[13, 40]], [[13, 40]], [[42, 94], [102, 104]], [[42, 94]], [[44, 94]], [[44, 94]], [[44, 94]], [[44, 94]], [[2, 95]], [[2, 95]], [[98, 101], [110, 113]], [[98, 108]]]",
    "query_spans": "[[[110, 115]]]",
    "process": ""
  },
  {
    "text": "Given the line $ l $: $ y = 1 $ intersects the $ y $-axis at point $ M $, and $ Q $ is a moving point on line $ l $ distinct from $ M $. Denote the horizontal coordinate of point $ Q $ as $ x_0 $ ($ x_0 \\neq 0 $). If there exists a point $ N $ on the ellipse $ \\frac{x^2}{2} + y^2 = 1 $ such that $ \\angle MQN = 45^\\circ $, then the range of values for $ x_0 $ is?",
    "fact_expressions": "l: Line;G: Ellipse;M: Point;Q: Point;N: Point;x0:Number;Expression(G) = (x^2/2 + y^2 = 1);Intersection(l,yAxis)=M;PointOnCurve(Q,l);Negation(Q=M);XCoordinate(Q) = x0;PointOnCurve(N, G);AngleOf(M, Q, N) = ApplyUnit(45, degree);Expression(l)=(y=1);Negation(x0=0)",
    "query_expressions": "Range(x0)",
    "answer_expressions": "{[-1-sqrt(3),0) ∪ (0,1+sqrt(3)]}",
    "fact_spans": "[[[2, 14], [30, 35]], [[79, 106]], [[21, 25], [38, 41]], [[26, 29], [46, 50]], [[109, 113]], [[143, 150], [55, 76]], [[79, 106]], [[2, 25]], [[26, 44]], [[26, 44]], [[46, 76]], [[79, 113]], [[116, 141]], [[2, 14]], [[55, 76]]]",
    "query_spans": "[[[143, 157]]]",
    "process": "Let $ Q(x_{0},1) $. If the inclination angle of $ QN $ is $ 45^{\\circ} $, then the equation of the line is $ y-1 = x - x_{0} $, that is, $ y = x - x_{0} + 1 $. From \n\\[\n\\begin{cases}\ny = x - x_{0} + 1 \\\\\n\\frac{x^{2}}{2} + y^{2} = 1\n\\end{cases},\n\\]\neliminating $ y $ gives $ 3x^{2} + 4(1 - x_{0})x + 2x_{0}^{2} - 4x_{0} = 0 $, so $ \\Delta = 16(1 - x_{0})^{2} - 12(2x_{0}^{2} - 4x_{0}) \\geqslant 0 $, solving yields $ 1 - \\sqrt{3} \\leqslant x_{0} \\leqslant 1 + \\sqrt{3} $. If the inclination angle of $ QN $ is $ 135^{\\circ} $, then the equation of the line is $ y - 1 = -(x - x_{0}) $, that is, $ y = -x + x_{0} + 1 $. From \n\\[\n\\begin{cases}\ny = -x + x_{0} + 1 \\\\\n\\frac{x^{2}}{2} + y^{2} = 1\n\\end{cases},\n\\]\neliminating $ y $ gives $ 3x^{2} - 4(1 + x_{0})x + 2x_{0}^{2} + 4x_{0} = 0 $, so $ \\Delta = 16(1 + x_{0})^{2} - 12(2x_{0}^{2} + 4x_{0}) \\geqslant 0 $, solving yields $ -1 - \\sqrt{3} \\leqslant x_{0} \\leqslant -1 + \\sqrt{3} $. When $ x_{0} = 0 $, $ Q $ coincides with $ M $, which does not satisfy the condition. In conclusion, the range of $ x_{0} $ is $ [-1 - \\sqrt{3}, 0) \\cup (0, 1 + \\sqrt{3}] $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}}=1$ $(a > b>0)$, the left and right foci are $F_{1}(-c, 0)$ and $F_{2}(c, 0)$ respectively. If there exists a point $P$ on the ellipse such that $\\frac{a}{\\sin \\angle P F_{1} F_{2}}=\\frac{c}{\\sin \\angle P F_{2} F_{1}}$, then the range of values for the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);c: Number;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);a/Sin(AngleOf(P, F1, F2)) = c/Sin(AngleOf(P, F2, F1))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(\\sqrt{2}-1, 1)",
    "fact_spans": "[[[2, 59], [98, 100], [185, 187]], [[2, 59]], [[4, 59]], [[4, 59]], [[4, 59]], [[4, 59]], [[68, 82]], [[83, 96]], [[68, 82]], [[83, 96]], [[109, 182]], [[2, 96]], [[2, 96]], [[105, 108]], [[98, 108]], [[109, 182]]]",
    "query_spans": "[[[185, 198]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola $y^{2}=2 p x$ is $F$, points $A(1,4)$ and points $B$, $C$ lie on the parabola, and $\\overrightarrow{F A}+\\overrightarrow{F B}+\\overrightarrow{F C}=\\overrightarrow{0}$. Then the equation of the line passing through points $B$ and $C$ is?",
    "fact_expressions": "G: Parabola;p: Number;L:Line;A: Point;F: Point;B: Point;C: Point;Expression(G) = (y^2 = 2*p*x);Coordinate(A) = (1, 4);Focus(G) = F;PointOnCurve(B,G);PointOnCurve(C,G);VectorOf(F, A) + VectorOf(F, B) + VectorOf(F, C) = 0;PointOnCurve(B,L);PointOnCurve(C,L);PointOnCurve(A,G)",
    "query_expressions": "Expression(L)",
    "answer_expressions": "4*x+y-20=0",
    "fact_spans": "[[[0, 16], [43, 46]], [[3, 16]], [[144, 146]], [[24, 33]], [[20, 23]], [[34, 38], [135, 139]], [[39, 42], [140, 143]], [[0, 16]], [[24, 33]], [[0, 23]], [[34, 47]], [[39, 47]], [[49, 132]], [[134, 146]], [[134, 146]], [[24, 47]]]",
    "query_spans": "[[[144, 150]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ has an eccentricity of $\\sqrt{3}$. Then the asymptotes of this hyperbola have the equations?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(C) = sqrt(3)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y = pm*sqrt(2)*x",
    "fact_spans": "[[[0, 63], [82, 85]], [[7, 63]], [[7, 63]], [[7, 63]], [[7, 63]], [[0, 63]], [[0, 79]]]",
    "query_spans": "[[[82, 93]]]",
    "process": ""
  },
  {
    "text": "The right focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $F(c, 0)$, and the upper and lower vertices are $A$ and $B$, respectively. The line $AF$ intersects the ellipse at another point $P$. If the slope of $PB$ is $\\frac{\\sqrt{3}}{4}$, then the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "A: Point;F: Point;G: Ellipse;a: Number;b: Number;P: Point;B: Point;e: Number;c: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (c, 0);RightFocus(G) = F;UpperVertex(G) = A;LowerVertex(G) = B;Intersection(LineOf(A, F), G) = P;Slope(LineSegmentOf(P, B)) = sqrt(3)/4;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "{1/2, sqrt(3)/2}",
    "fact_spans": "[[[74, 77]], [[57, 66]], [[0, 52], [91, 93], [133, 135]], [[2, 52]], [[2, 52]], [[97, 100]], [[80, 83]], [[139, 142]], [[57, 66]], [[2, 52]], [[2, 52]], [[0, 52]], [[57, 66]], [[0, 66]], [[0, 83]], [[0, 83]], [[84, 100]], [[102, 130]], [[133, 142]]]",
    "query_spans": "[[[139, 144]]]",
    "process": "The line AP: $ y = \\frac{b}{-c}(x - c) $, that is, $ y = -\\frac{b}{c}x + b $, together with the ellipse equation, yields: $ x_{p} = \\frac{2a^{2}c}{a^{2} + c^{2}} $. Thus, $ P\\left( \\frac{2a^{2}c}{a^{2} + c^{2}}, \\frac{-b^{3}}{a^{2} + c^{2}} \\right) $, so $ k_{PB} = \\frac{bc}{a^{2}} = \\frac{\\sqrt{3}}{4} $, solving gives the eccentricity $ \\frac{c}{a} = \\frac{1}{2} $ or $ \\frac{\\sqrt{3}}{2} $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F_{1}$, $F_{2}$ are the foci of the ellipse. If there exists a point $Q$ on $C$ such that $\\angle F_{1} Q F_{2}=120^{\\circ}$, then what is the range of values for the eccentricity $e$?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;F1: Point;Q: Point;F2: Point;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Focus(C)={F1,F2};PointOnCurve(Q, C);AngleOf(F1, Q, F2) = ApplyUnit(120, degree);Eccentricity(C)=e;e:Number;a > b;b > 0",
    "query_expressions": "Range(e)",
    "answer_expressions": "[\\sqrt{3}/2,1)",
    "fact_spans": "[[[2, 60], [80, 82], [89, 92]], [[8, 59]], [[8, 59]], [[61, 69]], [[95, 99]], [[71, 79]], [[2, 60]], [[62, 86]], [[89, 99]], [[101, 135]], [[89, 144]], [[141, 144]], [[9, 59]], [[9, 59]]]",
    "query_spans": "[[[141, 151]]]",
    "process": "When Q is at the upper and lower vertices of the ellipse, $\\angle F_{1}QF_{2}$ is maximized, so $120^{\\circ} \\leqslant \\angle F_{1}QF_{2} < 180^{\\circ}$, then $60^{\\circ} \\leqslant \\angle F_{1}QO < 90^{\\circ}$, thus $\\sin 60^{\\circ} \\leqslant \\sin \\angle F_{1}QO < \\sin 90^{\\circ}$. Since $|F_{1}Q| = a$, $|F_{1}O| = c$, we have $\\frac{\\sqrt{3}}{2} \\leqslant \\frac{c}{a} < 1$, that is, the range of the ellipse's eccentricity is $(\\frac{\\sqrt{3}}{2}, 1)$. Hence, choose: D."
  },
  {
    "text": "The parabola $y=p x^{2}$ passes through the point $(1,4)$, then the distance from the focus to the directrix of the parabola is equal to?",
    "fact_expressions": "G: Parabola;p: Number;H: Point;Expression(G) = (y = p*x^2);Coordinate(H) = (1, 4);PointOnCurve(H, G)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "1/8",
    "fact_spans": "[[[0, 14], [26, 29]], [[3, 14]], [[16, 24]], [[0, 14]], [[16, 24]], [[0, 24]]]",
    "query_spans": "[[[26, 41]]]",
    "process": "Since the point lies on the parabola, the equation of the parabola is $x^{2}=\\frac{1}{4}y$. Combining with the properties of the parabola, the equation of the directrix and the coordinates of the focus can be obtained, thus yielding the solution. [Solution] Since the point $(1,4)$ lies on the parabola $y=px^{2}$, we obtain $p=4$. Therefore, the equation of the parabola is $x^{2}=\\frac{1}{4}y$. Hence, the focus of the parabola is $(0,\\frac{1}{16})$, and the equation of the directrix is $y=-\\frac{1}{16}$. Thus, the distance from the focus to the directrix of the parabola equals $\\frac{1}{16}+\\frac{1}{16}=\\frac{1}{8}$."
  },
  {
    "text": "Given the ellipse $ T $: $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ $ (a > b > 0) $, the length of the major axis is twice the length of the minor axis. A line passing through the left focus $ F $ with an inclination angle of $ 45^{\\circ} $ intersects $ T $ at points $ A $ and $ B $. If $ |AB| = \\frac{8 \\sqrt{2}}{5} $, then the equation of the ellipse $ T $ is?",
    "fact_expressions": "T: Ellipse;Expression(T) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Length(MajorAxis(T)) = 2 * Length(MinorAxis(T));F: Point;LeftFocus(T) = F;PointOnCurve(F, G) = True;Inclination(G) = ApplyUnit(45, degree);G: Line;Intersection(G, T) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(A, B)) = (8*sqrt(2))/5",
    "query_expressions": "Expression(T)",
    "answer_expressions": "x^2/8+y^2/2=1",
    "fact_spans": "[[[2, 59], [101, 104], [146, 151]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[2, 72]], [[77, 80]], [[2, 80]], [[73, 100]], [[81, 100]], [[98, 100]], [[98, 114]], [[105, 108]], [[109, 112]], [[116, 144]]]",
    "query_spans": "[[[146, 156]]]",
    "process": "\\because a=2b, then c=\\sqrt{3}b, \\therefore ellipse T: \\frac{x^{2}}{4b^{2}}+\\frac{y^{2}}{b^{2}}=1, left focus F(-\\sqrt{3}b,0). Let the line: y=x+\\sqrt{3}b, A(x_{1},y_{1}), B(x_{2},y_{2}). Form the system of equations: \\begin{cases} y=x+\\sqrt{3}b \\\\ \\frac{x^{2}}{4a^{2}}+\\frac{y^{2}}{12}=1 \\end{cases}. Eliminating gives: 5x^{2}+8\\sqrt{3}bx+8b^{2}=0, x_{1}+x_{2}=-\\frac{8\\sqrt{3}}{5}b, x_{1}x_{2}=\\frac{8b^{2}}{5}"
  },
  {
    "text": "Given point $P$ is a point on the parabola $C$: $y^{2}=4x$. Let $d_{1}$ denote the distance from $P$ to the directrix $l$ of this parabola, and let $d_{2}$ denote the distance from $P$ to the points on the circle $(x+2)^{2}+(y+4)^{2}=4$. Find the minimum value of $d_{1}+d_{2}$.",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);P: Point;PointOnCurve(P, C);l: Line;Directrix(C) = l;d1: Number;Distance(P, l) = d1;G: Circle;Expression(G) = ((x + 2)^2 + (y + 4)^4 = 4);d2: Number;F: Point;PointOnCurve(F, G);Distance(P, F) = d2",
    "query_expressions": "Min(d1 + d2)",
    "answer_expressions": "3",
    "fact_spans": "[[[7, 26], [36, 39]], [[7, 26]], [[31, 34], [2, 6], [56, 60]], [[2, 29]], [[41, 44]], [[36, 44]], [[48, 55]], [[31, 55]], [[61, 85]], [[61, 85]], [[92, 99]], [[87, 88]], [[61, 88]], [[56, 99]]]",
    "query_spans": "[[[101, 120]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{m^{2}}-\\frac{y^{2}}{n^{2}}=1$ is $\\frac{4}{3}$, then the eccentricity of the hyperbola $-\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{n^{2}}=1$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/n^2 + x^2/m^2 = 1);m: Number;n: Number;Eccentricity(G) = 4/3;H: Hyperbola;Expression(H) = (y^2/n^2 - x^2/m^2 = 1)",
    "query_expressions": "Eccentricity(H)",
    "answer_expressions": "4*sqrt(7)/7",
    "fact_spans": "[[[2, 48]], [[2, 48]], [[5, 48]], [[5, 48]], [[2, 66]], [[68, 115]], [[68, 115]]]",
    "query_spans": "[[[68, 121]]]",
    "process": "Let $\\frac{x^{2}}{m^{2}}-\\frac{y^{2}}{n^{2}}=1$ have eccentricity $e_{1}$, and $-\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{n^{2}}=1$ have eccentricity $e_{2}$. Since they are conjugate hyperbolas, $\\frac{1}{e_{1}^{2}}+\\frac{1}{e_{2}^{2}}=1 \\Rightarrow \\frac{1}{(4)^{2}}+\\frac{1}{e_{2}^{2}}=1 \\Rightarrow \\frac{1}{e_{2}^{2}}=1-\\frac{9}{16}=\\frac{7}{16} \\Rightarrow e_{2}^{2}=\\frac{16}{7} \\Rightarrow e_{2}=\\sqrt{\\frac{16}{7}}=\\frac{4\\sqrt{7}}{7}$"
  },
  {
    "text": "If an ellipse passes through the point $(2 , 3)$, and has foci $F_{1}(-2,0)$ , $F_{2}(2,0)$, then the eccentricity of this ellipse is equal to?",
    "fact_expressions": "G: Ellipse;H: Point;Coordinate(H) = (2, 3);PointOnCurve(H, G) = True;F1: Point;F2: Point;Coordinate(F1) = (-2, 0);Coordinate(F2) = (2, 0);Focus(G) = {F1, F2}",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[1, 3], [52, 54]], [[5, 15]], [[5, 15]], [[1, 15]], [[20, 33]], [[36, 48]], [[20, 33]], [[36, 48]], [[1, 48]]]",
    "query_spans": "[[[52, 61]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{m^{2}-4}=1$ $(m>2)$, and point $P$ lies on the ellipse. If $|P F_{1}| \\cdot|P F_{2}|=2 \\sqrt{3} m$, then the range of values for the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;m: Number;P: Point;F1: Point;F2: Point;m>2;Expression(G) = (y^2/(m^2 - 4) + x^2/m^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 2*(sqrt(3)*m)",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(0, \\sqrt{3}/3]",
    "fact_spans": "[[[18, 70], [82, 84], [129, 131]], [[20, 70]], [[77, 81]], [[2, 9]], [[10, 17]], [[20, 70]], [[18, 70]], [[2, 76]], [[2, 76]], [[77, 85]], [[87, 126]]]",
    "query_spans": "[[[129, 141]]]",
    "process": "From the given conditions and the definition of an ellipse, c=2, |PF_{1}|+|PF_{2}|=2m, |PF_{1}|\\cdot|PF_{2}|\\leqslant\\left(\\frac{|PF_{1}|+|PF_{2}|}{2}\\right)^{2} \\Rightarrow 2\\sqrt{3}m\\leqslant m^{2} \\Rightarrow m\\geqslant 2\\sqrt{3}. Also, e=\\frac{c}{a}=\\frac{c}{m}, hence e\\in\\left(0,\\frac{\\sqrt{3}}{3}\\right]"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the right branch of the hyperbola, and $|P F_{1}|=4|P F_{2}|$. Then, the maximum value of the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F1)) = 4*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Max(Eccentricity(G))",
    "answer_expressions": "5/3",
    "fact_spans": "[[[2, 59], [89, 92], [123, 126]], [[2, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[68, 75]], [[76, 83]], [[2, 83]], [[2, 83]], [[84, 88]], [[84, 96]], [[98, 120]]]",
    "query_spans": "[[[123, 136]]]",
    "process": ""
  },
  {
    "text": "Draw a line $l$ through the focus of the parabola $y^{2}=-12x$. The line $l$ intersects the parabola at points $A$ and $B$. If the x-coordinate of the midpoint of segment $AB$ is $-9$, then $|AB|=$?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;Expression(G) = (y^2 = -12*x);PointOnCurve(Focus(G),l);Intersection(l, G) = {A, B};XCoordinate(MidPoint(LineSegmentOf(A,B))) = -9",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "24",
    "fact_spans": "[[[21, 26], [27, 32]], [[1, 17], [33, 36]], [[37, 40]], [[41, 44]], [[1, 17]], [[0, 26]], [[27, 46]], [[48, 66]]]",
    "query_spans": "[[[68, 77]]]",
    "process": "Given that $ p = 6 $, let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Since the horizontal coordinate of the midpoint of segment $ AB $ is $-9$, we have $ x_{1} + x_{2} = -18 $, and $ |AB| = p - (x_{1} + x_{2}) = 6 + 18 = 24 $."
  },
  {
    "text": "It is known that for the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with left and right foci $F_{1}$ and $F_{2}$ respectively, one of its asymptotes is perpendicular to the line $l$: $x-2y=0$. Point $P$ lies on the hyperbola $C$, and $|P F_{1}| - |P F_{2}| = 3$. Then, the focal distance of the hyperbola $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;a: Number;b: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, C);Expression(l)=(x-2*y=0);RightFocus(C)=F2;LeftFocus(C)=F1;IsPerpendicular(OneOf(Asymptote(C)),l);Abs(LineSegmentOf(P,F1))-Abs(LineSegmentOf(P,F2))=3",
    "query_expressions": "FocalLength(C)",
    "answer_expressions": "3*sqrt(5)",
    "fact_spans": "[[[95, 111]], [[26, 88], [121, 127], [157, 163]], [[34, 88]], [[34, 88]], [[116, 120]], [[10, 17]], [[18, 25]], [[34, 88]], [[34, 88]], [[26, 88]], [[116, 128]], [[95, 111]], [[2, 32]], [[2, 32]], [[26, 115]], [[130, 155]]]",
    "query_spans": "[[[157, 168]]]",
    "process": "Find the asymptotes of the hyperbola. Using the condition for two lines to be perpendicular: the product of their slopes is -1, we obtain b=2a. From the definition of the hyperbola, we can find a and b, then using the relationship among a, b, and c, we can find c, and thus obtain the focal distance. [Detailed solution] The hyperbola C: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0) has asymptotes y=\\pm\\frac{b}{a}x. One asymptote is perpendicular to the line 1: x\\cdot2y=0, which gives \\frac{b}{a}=2, so b=2a. From the definition of the hyperbola, we have 2a=|PF_{1}|-|PF_{2}|=3, yielding a=\\frac{3}{2}, b=3. Thus, c=\\sqrt{a^{2}+b^{2}}=\\sqrt{\\frac{9}{4}+9}=\\frac{3\\sqrt{5}}{2}, so the focal distance is 2c=3\\sqrt{5}."
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $25 x^{2}-16 y^{2}=400$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (25*x^2 - 16*y^2 = 400)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "pm*4*y+5*x=0",
    "fact_spans": "[[[0, 26]], [[0, 26]]]",
    "query_spans": "[[[0, 34]]]",
    "process": "Since the hyperbola is $25x^{2}-16y^{2}=400$, the standard form of the hyperbola equation is $\\frac{x^{2}}{16}-\\frac{y^{2}}{25}=1$. Therefore, the asymptotes of the hyperbola $25x^{2}-16y^{2}=400$ are given by $5x\\pm4y=0$."
  },
  {
    "text": "The curve of the equation $\\frac{x^{2}}{m}+\\frac{y^{2}}{4-m}=1$ is a hyperbola; then the range of values for $m$ is?",
    "fact_expressions": "G: Hyperbola;m:Number;H:Curve;Expression(H)=(x^2/m+y^2/(4-m)=1);H=G",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-oo,0)+(4,+oo)",
    "fact_spans": "[[[43, 46]], [[48, 51]], [[40, 42]], [[0, 42]], [[40, 46]]]",
    "query_spans": "[[[48, 58]]]",
    "process": "If the curve of the equation $\\frac{x^{2}}{m}+\\frac{y^{2}}{4-m}=1$ is a hyperbola with foci on the $x$-axis, then $\\begin{cases}m>0\\\\4-m<0\\end{cases}$, solving gives $m>4$. If the curve of the equation $\\frac{x^{2}}{m}+\\frac{y^{2}}{4-m}=1$ is a hyperbola with foci on the $y$-axis, then $\\begin{cases}m<0\\\\4-m>0\\end{cases}$, solving gives $m<0$. Therefore, the range of real number $m$ is $(-\\infty,0)\\cup(4,+\\infty)$."
  },
  {
    "text": "The length of the real axis of the hyperbola $9 x^{2}-16 y^{2}=-144$ is equal to? Its asymptotes are tangent to the circle $x^{2}+y^{2}-2 x+m=0$, then $m$=?",
    "fact_expressions": "G: Hyperbola;H: Circle;m: Number;Expression(G) = (9*x^2 - 16*y^2 = -144);Expression(H) = (m - 2*x + x^2 + y^2 = 0);IsTangent(Asymptote(G),H)",
    "query_expressions": "Length(RealAxis(G));m",
    "answer_expressions": "6\n16/25",
    "fact_spans": "[[[0, 26], [33, 34]], [[38, 60]], [[64, 67]], [[0, 26]], [[38, 60]], [[33, 62]]]",
    "query_spans": "[[[0, 33]], [[64, 69]]]",
    "process": "Convert the hyperbola equation to standard form: $\\frac{y2}{9}-\\frac{x^{2}}{16}=1,\\therefore$ the length of the real axis is $2\\cdot3=6$, the asymptote equations are $3x\\pm4y=0$. Convert the circle equation to standard form: $(x-1)^{2}+y^{2}=1-m,\\therefore\\frac{3}{\\sqrt{3^{2}+4^{2}}}=\\sqrt{1-m}\\Rightarrow m=\\frac{16}{25}$, thus fill in: $6,\\frac{16}{25}$"
  },
  {
    "text": "Given points $A(-2,0)$, $B(3,0)$, a moving point $P(x_1, y_1)$ satisfies $\\overrightarrow{P A} \\cdot \\overrightarrow{P B}=x^{2}-6$, then the trajectory equation of point $P$ is?",
    "fact_expressions": "A: Point;B: Point;Coordinate(A) = (-2,0);Coordinate(B) = (3,0);P: Point;x1: Number;y1: Number;Coordinate(P) = (x1,y1);DotProduct(VectorOf(P,A),VectorOf(P,B))=x1^2-6",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "y^2=x",
    "fact_spans": "[[[2, 12]], [[15, 23]], [[2, 12]], [[15, 23]], [[26, 37], [100, 103]], [[26, 37]], [[26, 37]], [[26, 37]], [[39, 96]]]",
    "query_spans": "[[[100, 110]]]",
    "process": "According to $\\overrightarrow{PA}\\cdot\\overrightarrow{PB}=x^{2}-6$, set up the equation and simplify to obtain the result. Since $A(-2,0)$, $B(3,0)$, $P(x,y)$, then $\\overrightarrow{PA}=(-2-x,-y)$, $\\overrightarrow{PB}=(3-x,-y)$. Given $\\overrightarrow{PA}\\cdot\\overrightarrow{PB}=x^{2}-6$, so $(-2-x)(3-x)+y^{2}=x^{2}-6$, simplifying yields $y^{2}=x$."
  },
  {
    "text": "Pass a line through the focus $F$ of the parabola $y^{2}=4x$ with an inclination angle of $\\frac{\\pi}{4}$, intersecting the parabola at points $A$ and $B$. Then $|AB|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;PointOnCurve(F, H) ;Inclination(H) = pi/4;H: Line;Intersection(G, H) = {A, B};A: Point;B: Point",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 15], [46, 49]], [[1, 15]], [[18, 21]], [[1, 21]], [[0, 44]], [[22, 44]], [[42, 44]], [[42, 59]], [[50, 53]], [[54, 57]]]",
    "query_spans": "[[[61, 70]]]",
    "process": "The focus of the parabola $ y^{2} = 4x $ is $ F(1,0) $. Since the line $ AB $ passes through the focus $ F(1,0) $ and has an inclination angle of $ \\frac{\\pi}{4} $, the equation of line $ AB $ can be found as: $ y = x - 1 $. Solving the system of equations, eliminating $ y $, we get: $ x^2 - 6x + 1 = 0 $, then: $ x_{1} + x_{2} = 6 $, therefore $ |AB| = x_{1} + x_{2} + p = 6 + 2 = 8 $."
  },
  {
    "text": "Given that the ellipse has coordinate axes as its axes of symmetry, with foci at $(0,-2)$ and $(0,2)$, and passes through the point $\\left(-\\frac{3}{2}, \\frac{5}{2}\\right)$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;H:Point;Coordinate(F1) = (0, -2);Coordinate(F2) = (0,2);Coordinate(H) = (-3/2,5/2);SymmetryAxis(G)=axis;Focus(G)={F1,F2};PointOnCurve(H,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/10 + x^2/6 = 1",
    "fact_spans": "[[[2, 4], [73, 75]], [[22, 30]], [[31, 38]], [[41, 71]], [[22, 30]], [[31, 38]], [[41, 71]], [[2, 12]], [[2, 17]], [[2, 71]]]",
    "query_spans": "[[[73, 82]]]",
    "process": ""
  },
  {
    "text": "Given a point $P$ on the ellipse $x^{2}+\\frac{y^{2}}{4}=1$, draw $PD \\perp x$-axis at point $D$, and let $E$ be the midpoint of segment $PD$. Then the equation of the trajectory of point $E$ is?",
    "fact_expressions": "G: Ellipse;P: Point;D: Point;E:Point;Expression(G) = (x^2 + y^2/4 = 1);PointOnCurve(P, G);PointOnCurve(P,LineSegmentOf(P,D));IsPerpendicular(LineSegmentOf(P,D),xAxis);FootPoint(LineSegmentOf(P,D),xAxis)=D;MidPoint(LineSegmentOf(P,D))=E",
    "query_expressions": "LocusEquation(E)",
    "answer_expressions": "x^2+y^2=1",
    "fact_spans": "[[[2, 29]], [[32, 35], [37, 41]], [[57, 61]], [[62, 65], [78, 82]], [[2, 29]], [[2, 35]], [[36, 56]], [[42, 56]], [[42, 61]], [[62, 76]]]",
    "query_spans": "[[[78, 89]]]",
    "process": "Let the coordinates of point E be (x, y), the coordinates of point P be (x_{0}, y_{0}), and the coordinates of point D be (x_{0}, 0). Then \\begin{cases}x=x_{0}\\\\y=\\frac{y_{0}}{2}\\end{cases}, that is, \\begin{cases}x_{0}=x\\\\y_{0}=2y'\\end{cases}. Since point P(x_{0}, y_{0}) lies on the ellipse x^{2}+\\frac{y^{2}}{4}=1, we have x^{2}+\\frac{(2y)^{2}}{4}=1, which simplifies to x^{2}+y^{2}=1."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$ and directrix $l$, let $P$ be a point on $C$, $PQ$ perpendicular to $l$ at point $Q$, and $M$, $N$ the midpoints of $PQ$ and $PF$ respectively. $MN$ intersects the $x$-axis at point $R$. If $\\angle NRF=60^{\\circ}$, then $FR$ equals?",
    "fact_expressions": "C: Parabola;P: Point;Q: Point;F: Point;M: Point;N: Point;R: Point;l: Line;Expression(C) = (y^2 = 4*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P,Q),l);FootPoint(LineSegmentOf(P,Q),l)=Q;MidPoint(LineSegmentOf(P,Q))=M;MidPoint(LineSegmentOf(P,F))=N;Intersection(LineSegmentOf(M,N),xAxis)=R;AngleOf(N, R, F) = ApplyUnit(60, degree)",
    "query_expressions": "LineSegmentOf(F, R)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 21], [42, 45]], [[38, 41]], [[60, 64]], [[25, 28]], [[65, 68]], [[69, 72]], [[105, 109]], [[32, 35], [56, 59]], [[2, 21]], [[2, 28]], [[2, 36]], [[38, 48]], [[49, 59]], [[49, 64]], [[65, 91]], [[65, 91]], [[92, 109]], [[111, 136]]]",
    "query_spans": "[[[138, 146]]]",
    "process": "By the given conditions: |FH| = 2, |PF| = |PQ|, MN/QF, PQ//OR. From ∠NRF = 60°, it follows that △PQF is an equilateral triangle, MF⊥PQ, and thus F is the midpoint of HR, so we need to find |FR|. Without loss of generality, assume point P lies in the first quadrant, as shown in the figure; connect MF, QF. ∵ the parabola C: y² = 4x has focus F and directrix l, and P is a point on C, ∴ |FH| = 2, |PF| = |PQ|. ∵ M, N are the midpoints of PQ and PF respectively, ∴ MN//OF. ∵ PQ is perpendicular to the point Q, PQ//OR, |PF| = |PQ|, ∠NRF = 60°, ∴ △PQF is an equilateral triangle, ∴ MF⊥PQ. It is clear that quadrilaterals MOHF and MQFR are both parallelograms, ∴ F is the midpoint of HR, ∴ |FR| = |FH| = 2."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, one of its asymptotes is perpendicular to the line $l$: $x+\\sqrt{3} y=0$, and the distance from a focus of $C$ to $l$ is $1$. Then, the length of the real axis of the curve $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(l) = (x + sqrt(3)*y = 0);IsPerpendicular(OneOf(Asymptote(C)), l);Distance(OneOf(Focus(C)), l) = 1",
    "query_expressions": "Length(RealAxis(C))",
    "answer_expressions": "2",
    "fact_spans": "[[[70, 93], [105, 108]], [[2, 63], [96, 99], [117, 122]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 63]], [[70, 93]], [[2, 95]], [[96, 115]]]",
    "query_spans": "[[[117, 128]]]",
    "process": "The slope of line $ l $ is $ -\\frac{\\sqrt{3}}{3} $. One asymptote is perpendicular to line $ l $, then $ \\frac{b}{a} = \\sqrt{3} $, $ b = \\sqrt{3}a $. Let the focus be $ F(c,0) $, then $ \\frac{c}{\\sqrt{1^{2}+(\\sqrt{3}})}\\frac{c}{F}^{\\frac{1}{2}} = 1 $, $ c = 2 $. $ \\therefore a^{2} + b^{2} = a^{2} + 3a^{2} = c^{2} = 4 $, solving gives $ a = 1 $, $ \\therefore $ the length of the real axis is $ 2a = 2 $."
  },
  {
    "text": "The focus of the parabola $E$: $y^{2}=8x$ is $F$. A line $l$ passing through point $F$ intersects the parabola $E$ at points $A$ and $B$. If $|AF|=6$, then what are the coordinates of point $A$?",
    "fact_expressions": "l: Line;E: Parabola;A: Point;F: Point;B: Point;Expression(E) = (y^2 = 8*x);Focus(E) = F;PointOnCurve(F, l);Intersection(l, E) = {A, B};Abs(LineSegmentOf(A, F)) = 6",
    "query_expressions": "Coordinate(A)",
    "answer_expressions": "(4, \\pm 4\\sqrt{2})",
    "fact_spans": "[[[33, 38]], [[0, 19], [39, 45]], [[47, 50], [69, 73]], [[23, 26], [28, 32]], [[51, 54]], [[0, 19]], [[0, 26]], [[27, 38]], [[33, 56]], [[58, 67]]]",
    "query_spans": "[[[69, 78]]]",
    "process": "Analysis: |AF| = x_{A} + 2 = 6 \\Rightarrow x_{A} = 4, substitute into y^{2} = 8x \\Rightarrow y_{A} = \\pm 4\\sqrt{2}, thus A(4, \\pm 4\\sqrt{2})"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has foci $F_{1}$, $F_{2}$, and point $B$ is one endpoint of the minor axis of the ellipse, with $\\angle F_{1} B F_{2}=90^{\\circ}$, then the eccentricity of the ellipse equals?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;B: Point;F2: Point;F1: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Focus(G)={F1,F2};OneOf(Endpoint(MinorAxis(G)))=B;AngleOf(F1,B,F2)=ApplyUnit(90,degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 56], [81, 83], [127, 129]], [[4, 56]], [[4, 56]], [[76, 80]], [[68, 75]], [[60, 67]], [[4, 56]], [[4, 56]], [[2, 56]], [[2, 75]], [[76, 90]], [[92, 125]]]",
    "query_spans": "[[[127, 136]]]",
    "process": ""
  },
  {
    "text": "Given a line passing through the point $P(2,1)$ intersects the parabola $y^{2}=2x$ at points $A$ and $B$, and $P$ is the midpoint of chord $AB$, then the equation of line $AB$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;P: Point;Expression(G) = (y^2 = 2*x);Coordinate(P) = (2, 1);PointOnCurve(P, H);Intersection(H, G) = {A, B};IsChordOf(LineSegmentOf(A, B), G);MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Expression(LineOf(A, B))",
    "answer_expressions": "y = x - 1",
    "fact_spans": "[[[18, 32]], [[15, 17]], [[34, 37]], [[38, 41]], [[44, 47], [5, 14]], [[18, 32]], [[5, 14]], [[4, 17]], [[15, 43]], [[18, 54]], [[44, 57]]]",
    "query_spans": "[[[59, 71]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. According to the given conditions, we have $ x_{1}+x_{2}=4 $, $ y_{1}+y_{2}=2 $. Also, since $ y_{1}^{2}=2x_{1} $, $ y_{2}^{2}=2x_{2} $, subtracting these two equations and simplifying yields $ \\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = \\frac{2}{y_{1}+y_{2}} = 1 $, which means the slope of line $ AB $ is 1. Therefore, the equation of line $ AB $ is $ y = x - 1 $."
  },
  {
    "text": "On the parabola $y^{2}=x$, the distance from a point $M$ to the focus is $1$. Then the horizontal coordinate of point $M$ is?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = x);PointOnCurve(M, G);Distance(M, Focus(G)) = 1",
    "query_expressions": "XCoordinate(M)",
    "answer_expressions": "3/4",
    "fact_spans": "[[[0, 12]], [[15, 18], [31, 35]], [[0, 12]], [[0, 18]], [[0, 28]]]",
    "query_spans": "[[[31, 41]]]",
    "process": "\\because the parabola equation is y^{2}=x, \\therefore the focus of the parabola is F(\\frac{1}{4},0)-----let point M(x_{0},y_{0}), then |MF|=\\sqrt{(x_{0}-\\frac{1}{4})^{2}+y_{0}^{2}}= substituting y_{0}^{2}=x_{0}, we get (x_{0}-\\frac{1}{4})^{2}+x_{0}=1, solving yields x_{0}=\\frac{3}{4} or x_{0}=-\\frac{5}{4} (discarded)"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, its two asymptotes intersect the directrix of the parabola $y^{2}=4x$ at points $A$ and $B$, respectively, and $O$ is the origin. If the area of $\\triangle AOB$ is $\\sqrt{3}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;A: Point;O: Origin;B: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 4*x);l1: Line;l2: Line;Asymptote(G) = {l1, l2};Intersection(l1, Directrix(H)) = A;Intersection(l2, Directrix(H)) = B;Area(TriangleOf(A, O, B)) = sqrt(3)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 58], [141, 144]], [[5, 58]], [[5, 58]], [[65, 79]], [[86, 89]], [[96, 99]], [[90, 93]], [[5, 58]], [[5, 58]], [[2, 58]], [[65, 79]], [], [], [[2, 64]], [[2, 95]], [[2, 95]], [[107, 138]]]",
    "query_spans": "[[[141, 150]]]",
    "process": ""
  },
  {
    "text": "It is known that an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ passes through the point $(2 , \\sqrt{3})$, and one focus of the hyperbola lies on the directrix of the parabola $y^{2}=4 \\sqrt{7} x$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;I: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 4*(sqrt(7)*x));Coordinate(I) = (2, sqrt(3));PointOnCurve(I,OneOf(Asymptote(G)));PointOnCurve(OneOf(Focus(G)), Directrix(H))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/3 = 1",
    "fact_spans": "[[[2, 59], [85, 88], [123, 126]], [[5, 59]], [[5, 59]], [[94, 117]], [[66, 83]], [[5, 59]], [[5, 59]], [[2, 59]], [[94, 117]], [[66, 83]], [[2, 83]], [[85, 121]]]",
    "query_spans": "[[[123, 131]]]",
    "process": "By the given condition, $\\frac{b}{a}=\\frac{\\sqrt{3}}{2}$; since the directrix of the parabola $y^{2}=4\\sqrt{7}x$ is $x=-\\sqrt{7}$, and one focus of the hyperbola lies on the directrix of the parabola $y^{2}=4\\sqrt{7}x$, we have $c=\\sqrt{7}$; thus $a^{2}+b^{2}=c^{2}=7$, so $a=2$, $b=\\sqrt{3}$; therefore, the equation of the hyperbola is $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=$"
  },
  {
    "text": "The ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1(b>0)$ has its right focus at $F(c, 0)$, and $O$ is the origin. The line $7 x-\\sqrt{14} y=0$ intersects $C$ at points $A$ and $B$. If $|\\overrightarrow{A O}|=|\\overrightarrow{A F}|$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;G: Line;F: Point;A: Point;O: Origin;B: Point;b>0;c:Number;Expression(C) = (x^2/4 + y^2/b^2 = 1);Expression(G) = (7*x - sqrt(14)*y = 0);Coordinate(F) = (c, 0);RightFocus(C) = F;Intersection(G, C) = {A, B};Abs(VectorOf(A, O)) = Abs(VectorOf(A, F))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[0, 51], [99, 102], [166, 171]], [[7, 51]], [[77, 98]], [[56, 65]], [[105, 108]], [[68, 71]], [[109, 112]], [[7, 51]], [[56, 65]], [[0, 51]], [[77, 98]], [[56, 65]], [[0, 65]], [[77, 114]], [[117, 164]]]",
    "query_spans": "[[[166, 177]]]",
    "process": "\\because\\because|\\overrightarrow{AO}|=|\\overrightarrow{AF}|,\\therefore point A is in the first quadrant, and the x-coordinate of point A is \\frac{c}{2}, substituting into 7x-\\sqrt{14}y=0 gives y=\\frac{\\sqrt{14}}{4}c, i.e., the coordinates of point A are (\\frac{c}{2},\\frac{\\sqrt{14}}{4}c). Substituting the coordinates of A into the ellipse equation gives \\frac{4}{16}+\\frac{14c^{2}}{16b^{2}}=1, obtaining c^{2}=\\frac{1}{b^{2}+14}, also c^{2}=4-b^{2}, so (4-b^{2})(b^{2}+14)=16b^{2}, simplifying gives b^{4}+26b^{2}-56=0, solving b^{2}=2 or b^{2}=-28 (rejected), so c^{2}=2,\\therefore the eccentricity of ellipse C is e=\\frac{\\sqrt{2}}{2}"
  },
  {
    "text": "Let $A$ and $B$ be two points on the ellipse $3 x^{2}+y^{2}=36$, and point $N(1,3)$ is the midpoint of segment $A B$. What is the equation of line $A B$?",
    "fact_expressions": "G: Ellipse;B: Point;A: Point;N: Point;Expression(G) = (3*x^2 + y^2 = 36);Coordinate(N) = (1, 3);PointOnCurve(A, G);PointOnCurve(B, G);MidPoint(LineSegmentOf(A,B))=N",
    "query_expressions": "Expression(LineOf(A,B))",
    "answer_expressions": "x+y-4=0",
    "fact_spans": "[[[9, 29]], [[5, 8]], [[1, 4]], [[34, 43]], [[9, 29]], [[34, 43]], [[1, 33]], [[1, 33]], [[34, 54]]]",
    "query_spans": "[[[55, 68]]]",
    "process": "Let the coordinates of points A and B be given. Since both points lie on the ellipse, substitute them into the ellipse equation, take the difference, and use the midpoint coordinate formula to simplify and find the slope of line AB. Then, using the slope and the fixed point on the line, write the point-slope form equation. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then  \n$$\n\\begin{cases}\n3x_{1}^{2}+y_{1}^{2}=36 \\\\\n3x_{2}^{2}+y_{2}^{2}=36\n\\end{cases}\n.\n$$\n$$\n3(x_{1}-x_{2})(x_{1}+x_{2})+(y_{1}-y_{2})(y_{1}+y_{2})=0\n$$\nBy the given condition, $ x_{1} \\ne x_{2} $,  \n$$\n\\therefore k_{AB}=-\\frac{3(x_{1}+x_{2})}{y_{1}+y_{2}}.\n$$\nSince $ N(1,3) $ is the midpoint of AB,  \n$$\n\\therefore x_{1}+x_{2}=2, y_{1}+y_{2}=6\n$$\nThus, $ k_{AB}=-1 $. Therefore, the equation of line AB is $ y-3=-(x-1) $, that is, $ x+y-4=0 $."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $x^{2}=4 y$, a line $l$ passing through point $F$ intersects the parabola $C$ at two distinct points $A$ and $B$. The tangents to the parabola $C$ at points $A$ and $B$ are $l_{1}$ and $l_{2}$, respectively, and $l_{1}$, $l_{2}$ intersect at point $P$. Let $|A B|=m$, then what is the value of $|P F|$? (Express the result in terms of $m$)?",
    "fact_expressions": "l1:Line;l2:Line;l:Line;C: Parabola;A: Point;B: Point;P: Point;F: Point;m:Number;Expression(C) = (x^2 = 4*y);Focus(C) = F;PointOnCurve(F,l);Intersection(l,C)={A,B};Negation(A=B);TangentOnPoint(A,C)=l1;TangentOnPoint(B,C)=l2;Intersection(l1,l2)=P;Abs(LineSegmentOf(A, B)) = m",
    "query_expressions": "Abs(LineSegmentOf(P,F))",
    "answer_expressions": "sqrt(m)",
    "fact_spans": "[[[86, 93], [103, 110]], [[94, 101], [111, 118]], [[35, 40]], [[6, 25], [63, 69], [41, 47]], [[70, 73], [55, 58]], [[74, 77], [59, 62]], [[121, 125]], [[2, 5], [30, 34]], [[128, 137]], [[6, 25]], [[2, 28]], [[29, 40]], [[35, 62]], [[50, 62]], [[63, 101]], [[63, 101]], [[103, 125]], [[128, 137]]]",
    "query_spans": "[[[139, 150]]]",
    "process": ""
  },
  {
    "text": "A point $M$ on the parabola $x^{2}=-4 y$ is at a distance of $7$ from the focus. What is the distance from point $M$ to the directrix?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (x^2 = -4*y);PointOnCurve(M, G);Distance(M, Focus(G)) = 7",
    "query_expressions": "Distance(M, Directrix(G))",
    "answer_expressions": "7",
    "fact_spans": "[[[0, 15]], [[18, 21], [33, 37]], [[0, 15]], [[0, 21]], [[0, 31]]]",
    "query_spans": "[[[0, 45]]]",
    "process": "According to the definition of a parabola, the distance from point M to the directrix can be determined. By the definition of a parabola: the distance from any point on the parabola to the focus is equal to the distance from that point to the directrix, so the distance from point M to the directrix is $7$."
  },
  {
    "text": "Given the standard equation of the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{9}=1$, what are the coordinates of the foci of the ellipse? What is the eccentricity?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/8 + y^2/9 = 1)",
    "query_expressions": "Coordinate(Focus(G));Eccentricity(G)",
    "answer_expressions": "(0, pm*1)\n1/3",
    "fact_spans": "[[[2, 4], [46, 48]], [[2, 44]]]",
    "query_spans": "[[[46, 55]], [[46, 60]]]",
    "process": ""
  },
  {
    "text": "Given point $P$ is a moving point on the hyperbola $C$: $\\frac{x^{2}}{36}-\\frac{y^{2}}{64}=1$, point $A(-10,0)$, point $B(10,0)$. If $|P A|=16$, then $|P B|$=?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/36 - y^2/64 = 1);P: Point;PointOnCurve(P, C);A: Point;Coordinate(A) = (-10, 0);B: Point;Coordinate(B) = (10, 0);Abs(LineSegmentOf(P, A)) = 16",
    "query_expressions": "Abs(LineSegmentOf(P, B))",
    "answer_expressions": "{28, 4}",
    "fact_spans": "[[[7, 52]], [[7, 52]], [[2, 6]], [[2, 56]], [[57, 68]], [[57, 68]], [[69, 79]], [[69, 79]], [[82, 92]]]",
    "query_spans": "[[[94, 103]]]",
    "process": "Obtain |PB| by using the definition of the hyperbola. According to the given conditions, for the hyperbola, a=6, b=8, c=\\sqrt{36+64}=10, so A and B are the left and right foci of the hyperbola. By the definition of the hyperbola, ||PA|-|PB||=|16-|PB||=2a=12, thus |PB|=28 or |PB|=4. Since c-a=10-6=4, it follows that |PB|\\geqslant4."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{2}=1$, its right focus coincides with the focus of the parabola $y^{2}=4 \\sqrt{3} x$. Then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;Expression(G) = (-y^2/2 + x^2/a^2 = 1);Expression(H) = (y^2 = 4*(sqrt(3)*x));RightFocus(G) = Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/2 = 1",
    "fact_spans": "[[[2, 44], [45, 46], [81, 84]], [[5, 44]], [[50, 73]], [[2, 44]], [[50, 73]], [[45, 78]]]",
    "query_spans": "[[[81, 88]]]",
    "process": "The focus of the parabola $ y^{2} = 4\\sqrt{3}x $ is at $ (\\sqrt{3}, 0) $. Therefore, the coordinates of the right focus of the hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{2} = 1 $ are $ (\\sqrt{3}, 0) $, so $ a^{2} + 2 = 3 $, yielding $ a^{2} = 1 $. Thus, the equation of this hyperbola is $ x^{2} - \\frac{y^{2}}{2} = 1 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the right focus is $F$, the right vertex is $A$, and $O$ is the origin. If $A$ is the midpoint of segment $O F$, then the asymptotes of $C$ have equations?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;O: Origin;F: Point;A: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;RightVertex(C) = A;MidPoint(LineSegmentOf(O,F)) = A",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[2, 62], [104, 107]], [[9, 62]], [[9, 62]], [[79, 82]], [[67, 70]], [[75, 78], [88, 91]], [[9, 62]], [[9, 62]], [[2, 62]], [[2, 70]], [[2, 78]], [[88, 102]]]",
    "query_spans": "[[[104, 115]]]",
    "process": "From the given conditions, F(c,0), A(a,0). Since A is the midpoint of segment OF, ∴ c=2a, and b=\\sqrt{c^{2}-a^{2}}=\\sqrt{4a^{2}-a^{2}}=\\sqrt{3}a, ∴ \\frac{b}{a}=\\sqrt{3}, ∴ the asymptotes of C are y=\\pm\\frac{b}{a}x=\\pm\\sqrt{3}x"
  },
  {
    "text": "The parabola $C$: $x^{2}=3 y$ with focus $F$ has its directrix intersecting the $y$-axis at point $A$. Point $M$ lies on the parabola $C$. The range of $\\frac{|M F|}{|M A|}$ is?",
    "fact_expressions": "C: Parabola;M: Point;F: Point;A: Point;Expression(C) = (x^2 = 3*y);Focus(C)=F;Intersection(Directrix(C),yAxis)=A;PointOnCurve(M, C)",
    "query_expressions": "Range(Abs(LineSegmentOf(M, F))/Abs(LineSegmentOf(M, A)))",
    "answer_expressions": "[sqrt(2)/2,1]",
    "fact_spans": "[[[7, 26], [46, 52]], [[41, 45]], [[3, 6]], [[36, 40]], [[7, 26]], [[0, 26]], [[7, 40]], [[41, 53]]]",
    "query_spans": "[[[55, 83]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus of $C$ is $F$, and $C$ intersects a line passing through the origin at points $A$ and $B$. Connect $A F$, $B F$. If $|A B|=10$, $|A F|=6$, $\\cos \\angle A B F=\\frac{4}{5}$, then the eccentricity $e$ of $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;A: Point;F: Point;B: Point;e: Number;a > b;b > 0;O: Origin;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;PointOnCurve(O, G);Intersection(C, G) = {A, B};Abs(LineSegmentOf(A, B)) = 10;Abs(LineSegmentOf(A, F)) = 6;Cos(AngleOf(A, B, F)) = 4/5;Eccentricity(C) = e",
    "query_expressions": "e",
    "answer_expressions": "5/7",
    "fact_spans": "[[[2, 59], [68, 71], [164, 167]], [[9, 59]], [[9, 59]], [[76, 78]], [[81, 84]], [[64, 67]], [[85, 88]], [[171, 174]], [[9, 59]], [[9, 59]], [[73, 75]], [[2, 59]], [[2, 67]], [[72, 78]], [[68, 90]], [[108, 118]], [[119, 128]], [[130, 161]], [[164, 174]]]",
    "query_spans": "[[[171, 176]]]",
    "process": "In triangle AFB, by the law of cosines, |AF|^{2} = |AB|^{2} + |BF|^{2} - 2|AB||BF|\\cos\\angle ABF. Substituting the values gives 36 = |BF|^{2} + 100 - 2 \\times 10 \\times |BF| \\times \\frac{4}{5}, solving yields |BF| = 8. Thus, triangle ABF is a right triangle, OF = 5, so c = 5. Since the ellipse is centrally symmetric, when the right focus is F_{2}, \\triangle AFB \\cong \\triangle ABF_{2}A, 2a = AF + AF_{2} = 14, a = 7, e = \\frac{5}{7}. [Topic Location] This problem examines the definition of an ellipse, knowledge related to solving triangles, and the geometric properties of an ellipse."
  },
  {
    "text": "The vertices of a hyperbola are the endpoints of the major axis of the ellipse $x^{2}+\\frac{y^{2}}{2}=1$, and the product of the eccentricity of the hyperbola and the eccentricity of the ellipse equals $1$. Then, what is the equation of the hyperbola?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2 + y^2/2 = 1);Endpoint(MajorAxis(H)) = Vertex(G);Eccentricity(G) * Eccentricity(H) = 1",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 - x^2 = 2",
    "fact_spans": "[[[2, 5], [43, 46], [67, 70]], [[9, 36], [51, 53]], [[9, 36]], [[2, 41]], [[43, 65]]]",
    "query_spans": "[[[67, 75]]]",
    "process": "(Analysis) According to the ellipse equation, obtain the coordinates of the endpoints of its major axis and its eccentricity; then derive the vertex and eccentricity of the hyperbola, find the lengths of the real semi-axis and imaginary semi-axis of the hyperbola, and thus obtain the equation of the hyperbola. From the given conditions, assume the hyperbola equation is \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1, with eccentricity e. The endpoints of the major axis of the ellipse x_{2}+\\frac{y^{2}}{2}=1 are (0,\\sqrt{2}), \\therefore a=\\sqrt{2}. \\because the eccentricity of the ellipse x^{2}+\\frac{y^{2}}{2}=1 is \\frac{1}{\\sqrt{2}}, \\therefore the eccentricity of the hyperbola is e=\\sqrt{2} \\Rightarrow c=2, \\therefore b=\\sqrt{2}, then the equation of the hyperbola is y^{2}-x^{2}=2."
  },
  {
    "text": "In the first quadrant, $P$ is a point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$. If the difference of the distances from point $P$ to the two foci $F_{1}$ and $F_{2}$ is $2$, then the coordinates of point $P$ are?",
    "fact_expressions": "P: Point;Quadrant(P) = 1;G: Ellipse;Expression(G) = (x^2/16 + y^2/12 = 1);PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};Distance(P, F1) - Distance(P, F2) = 2",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2, 3)",
    "fact_spans": "[[[6, 9], [54, 58], [88, 92]], [[0, 52]], [[10, 49]], [[10, 49]], [[6, 52]], [[62, 69]], [[70, 77]], [[10, 77]], [[54, 86]]]",
    "query_spans": "[[[88, 96]]]",
    "process": "In the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$, $a=4$, $b=2\\sqrt{3}$, $c=2$. If the sum of distances from point $P$ to the two foci $F_{1}$, $F_{2}$ is $|PF_{1}|+|PF_{2}|=8$, and $|PF_{1}|-|PF_{2}|=2$, then $|PF_{1}|=5$, $|PF_{2}|=3$, $|PF_{1}|^{2}=|PF_{2}|^{2}+|F_{1}F_{2}|^{2}$, $PF_{2}\\bot F_{1}F_{2}$, so the horizontal coordinate of point $P$ is $2$. Then $\\frac{2^{2}}{16}+\\frac{y^{2}}{12}=1$, so $y=3$, giving the coordinates of point $P$ as $(2,3)$."
  },
  {
    "text": "The asymptotes of the hyperbola $\\frac{y^{2}}{12}-\\frac{x^{2}}{24}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/24 + y^2/12 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(2)/2)*x",
    "fact_spans": "[[[0, 40]], [[0, 40]]]",
    "query_spans": "[[[0, 48]]]",
    "process": "From the hyperbola equation, we know: $ a=2\\sqrt{3}, b=2\\sqrt{6} $, so the asymptote equations are $ y=\\pm\\frac{a}{b}x=\\pm\\frac{\\sqrt{2}}{2}x $"
  },
  {
    "text": "It is known that the hyperbola $C$ passes through the point $(1,1)$, and one of its asymptotes has the equation $y=\\sqrt{3} x$. Then, what is the standard equation of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;G: Point;Coordinate(G) = (1, 1);PointOnCurve(G, C);Expression(OneOf(Asymptote(C))) = (y = sqrt(3)*x)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/(2/3)-y^2/2=1",
    "fact_spans": "[[[2, 8], [20, 21], [47, 53]], [[10, 18]], [[10, 18]], [[2, 18]], [[20, 44]]]",
    "query_spans": "[[[47, 60]]]",
    "process": "A hyperbola has an asymptote with equation $ y = \\sqrt{3}x $. Let the equation of the hyperbola be: $ x^{2} - \\frac{y^{2}}{3} = \\lambda $. Substituting the point $ (1,1) $ gives $ \\lambda = \\frac{2}{3} $, so $ x^{2} - \\frac{y^{2}}{3} = \\frac{2}{3} $. The standard equation of hyperbola $ C $ is: $ \\frac{x^{2}}{2} - \\frac{y^{2}}{2} = 1 $."
  },
  {
    "text": "The distance from a point $P$ on the parabola ${y}^{2}=8x$ to the focus is $6$, then the horizontal coordinate of point $P$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);P: Point;PointOnCurve(P, G);Distance(P, Focus(G)) = 6",
    "query_expressions": "XCoordinate(P)",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 16]], [[0, 16]], [[19, 22], [34, 38]], [[0, 22]], [[0, 32]]]",
    "query_spans": "[[[34, 44]]]",
    "process": ""
  },
  {
    "text": "Let $P$ be a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and let $M$, $N$ be points on the two circles $(x+4)^{2}+y^{2}=1$ and $(x-4)^{2}+y^{2}=1$, respectively. Then the minimum and maximum values of $P M+P N$ are?",
    "fact_expressions": "G: Ellipse;H: Circle;C:Circle;P: Point;M: Point;N: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Expression(H) = (y^2 + (x + 4)^2 = 1);Expression(C) = ((x-4)^2+y^2=1);PointOnCurve(P, G);PointOnCurve(M, H);PointOnCurve(N, C)",
    "query_expressions": "Min(LineSegmentOf(P, M) + LineSegmentOf(P, N));Max(LineSegmentOf(P, M) + LineSegmentOf(P, N))",
    "answer_expressions": "8\n12",
    "fact_spans": "[[[5, 43]], [[60, 79]], [[80, 99]], [[1, 4]], [[47, 50]], [[51, 54]], [[5, 43]], [[60, 79]], [[80, 99]], [[1, 46]], [[47, 102]], [[47, 102]]]",
    "query_spans": "[[[104, 125]], [[104, 125]]]",
    "process": "\\because the centers of the two circles $F_{1}(-4,0)$, $F_{2}(4,0)$ are exactly the foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, $\\therefore |PF_{1}|+|PF_{2}|=10$, the two circles have equal radii, both being 1, i.e., $r=1$, $\\therefore (|PM|+|PN|)\\min=|PF_{1}|+|PF_{2}|-2r=10-2=8$. $(|PM|+|PN|)\\max=|PF_{1}|+|PF_{2}|+2r=10+2=12$"
  },
  {
    "text": "Given that the left focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1$ $(a>0)$ with eccentricity $\\frac{3}{5}$ coincides with the focus of the parabola $y^{2}=2 m x$, find the real number $m=?$",
    "fact_expressions": "C: Hyperbola;a: Number;G: Parabola;m: Real;a>0;Expression(C) = (-y^2/4 + x^2/a^2 = 1);Expression(G) = (y^2 = 2*(m*x));Eccentricity(C) = 3/5;LeftFocus(C) = Focus(G)",
    "query_expressions": "m",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[20, 72]], [[28, 72]], [[77, 92]], [[99, 104]], [[28, 72]], [[20, 72]], [[77, 92]], [[2, 72]], [[20, 97]]]",
    "query_spans": "[[[99, 106]]]",
    "process": ""
  },
  {
    "text": "The foci of the conic section $x^{2}+m y^{2}=1$ lie on the $x$-axis, and the eccentricity is $\\frac{1}{2}$. Then the value of the real number $m$ is?",
    "fact_expressions": "G: ConicSection;m: Real;Expression(G) = (m*y^2 + x^2 = 1);PointOnCurve(Focus(G), xAxis);Eccentricity(G) = 1/2",
    "query_expressions": "m",
    "answer_expressions": "4/3",
    "fact_spans": "[[[0, 21]], [[50, 55]], [[0, 21]], [[0, 30]], [[0, 48]]]",
    "query_spans": "[[[50, 59]]]",
    "process": "Since the conic section $x^{2}+my^{2}=1$ has its foci on the x-axis and eccentricity $\\frac{1}{2}$, the curve is an ellipse with $a^{2}=1$, $b^{2}=\\frac{1}{m}$. Thus, $e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{a^{2}-b^{2}}{a^{2}}=1-\\frac{1}{m}=\\frac{1}{4}$. Solving gives $m=\\frac{4}{2}$, b female peace as .4"
  },
  {
    "text": "The ellipse $C_{1}$ has the equation $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, and the hyperbola $C_{2}$ has the equation $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$. The product of the eccentricities of $C_{1}$ and $C_{2}$ is $\\frac{\\sqrt{3}}{2}$. Then, what is the equation of the asymptotes of $C_{2}$?",
    "fact_expressions": "C1: Ellipse;Expression(C1) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;C2: Hyperbola;Expression(C2)=(-y^2/b^2+x^2/a^2=1);Eccentricity(C1)*Eccentricity(C2) = sqrt(3)/2",
    "query_expressions": "Expression(Asymptote(C2))",
    "answer_expressions": "y=pm*sqrt(2)*x/2",
    "fact_spans": "[[[0, 9], [116, 123]], [[0, 55]], [[12, 55]], [[12, 55]], [[56, 66], [126, 133], [162, 169]], [[56, 114]], [[116, 160]]]",
    "query_spans": "[[[162, 177]]]",
    "process": "The eccentricities of the two curves are $e_{1}=\\frac{\\sqrt{a^{2}-b^{2}}}{a}$, $e_{2}=\\frac{\\sqrt{a^{2}+b^{2}}}{a}$, and their product is $\\frac{\\sqrt{a^{4}-b^{4}}}{a^{2}}=\\frac{\\sqrt{3}}{2}$. Solving gives $\\frac{b^{4}}{a^{4}}=\\frac{1}{4}$, $\\frac{b}{a}=\\frac{\\sqrt{2}}{2}$, so the asymptotes of the hyperbola are $y=\\pm\\frac{b}{a}x=\\pm\\frac{\\sqrt{2}}{2}x$."
  },
  {
    "text": "It is known that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ passes through the point $(4,4 \\sqrt{3})$. What is the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;PointOnCurve(H, OneOf(Asymptote(G)));H: Point;Coordinate(H) = (4, 4*sqrt(3))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 58], [87, 90]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 83]], [[66, 83]], [[66, 83]]]",
    "query_spans": "[[[87, 96]]]",
    "process": ""
  },
  {
    "text": "Point $P$ is a point on the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$. The area of the triangle with vertices at point $P$ and the foci $F_{1}$, $F_{2}$ is equal to $1$. Then the coordinates of point $P$ are?",
    "fact_expressions": "P: Point;G: Ellipse;Expression(G) = (x^2/5 + y^2/4 = 1);PointOnCurve(P, G) = True;Focus(G) = {F1, F2};F1: Point;F2: Point;Area(TriangleOf(P, F1, F2)) = 1",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "{(sqrt(15)/2,1),(-sqrt(15)/2,1),(sqrt(15)/2,-1),(-sqrt(15)/2,-1)}",
    "fact_spans": "[[[0, 4], [47, 51], [87, 91]], [[5, 42]], [[5, 42]], [[0, 45]], [[5, 70]], [[55, 62]], [[63, 70]], [[46, 85]]]",
    "query_spans": "[[[87, 96]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively. The point $M(3, \\sqrt{2})$ lies on this hyperbola, and the distance from point $F_{2}$ to the line $M F_{1}$ is $\\frac{4 \\sqrt{6}}{9}$. Then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "F1: Point;F2: Point;C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;LeftFocus(C) = F1;RightFocus(C) = F2;M: Point;Coordinate(M) = (3, sqrt(2));PointOnCurve(M, C) = True;Distance(F2, LineOf(M, F1)) = 4*sqrt(6)/9",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[1, 8]], [[9, 16], [111, 119]], [[19, 80], [106, 109], [159, 165]], [[19, 80]], [[27, 80]], [[27, 80]], [[27, 80]], [[27, 80]], [[1, 86]], [[1, 86]], [[87, 104]], [[87, 104]], [[87, 110]], [[111, 157]]]",
    "query_spans": "[[[159, 171]]]",
    "process": "According to the problem, $F_{1}(-c,0)$, $F_{2}(c,0)$, $MF_{1}: y = \\frac{\\sqrt{2}}{3+c}(x+c)$, i.e., $\\sqrt{2}x - (3+c)y + \\sqrt{2}c = 0$. The distance from $F_{2}$ to $MF_{1}$ is $\\frac{2\\sqrt{2}c}{\\sqrt{2+(3+c)^{2}}} = \\frac{4\\sqrt{6}}{9}$, $\\therefore c=2$. Also $\\frac{9}{a^{2}} - \\frac{2}{b^{2}} = 1$, $b^{2} = c^{2} - a \\cdot 2$, $a = \\sqrt{3}$, $\\therefore e = \\frac{2}{\\sqrt{3}} = \\frac{2\\sqrt{3}}{3}$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has a focal distance of $2 c$, the right vertex is $A$, the focus of the parabola $x^{2}=2 p y(p>0)$ is $F$. If the segment length intercepted by the hyperbola on the directrix of the parabola is $2 c$, and $|F A|=c$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;c: Number;FocalLength(G) = 2*c;A: Point;RightVertex(G) = A;H: Parabola;Expression(H) = (x^2 = 2*(p*y));p: Number;p>0;F: Point;Focus(H) = F;Length(InterceptChord(Directrix(H), G)) = 2*c;Abs(LineSegmentOf(F, A)) = c",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 58], [106, 109], [140, 143]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[62, 67]], [[2, 67]], [[72, 75]], [[2, 75]], [[76, 97], [110, 113]], [[76, 97]], [[79, 97]], [[79, 97]], [[101, 104]], [[76, 104]], [[106, 127]], [[129, 138]]]",
    "query_spans": "[[[140, 149]]]",
    "process": "The focus of the parabola has coordinates $F(0,\\frac{p}{2})$, and the equation of the directrix is $y=-\\frac{p}{2}$. Since the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) intersects the directrix $y=-\\frac{p}{2}$ of the parabola $x^{2}=2py$ ($p>0$) to form a segment of length $2c$, the point $(c,-\\frac{p}{2})$ lies on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$. Therefore, $\\frac{c^{2}}{a^{2}}-\\frac{p^{2}}{4b^{2}}=1$. Since $|FA|=c$, we have $a^{2}+\\frac{p^{2}}{4}=c^{2}$, so $b^{2}=\\frac{p^{2}}{4}$. Substituting $b^{2}=\\frac{p^{2}}{4}$ into $\\frac{c^{2}}{a^{2}}-\\frac{p^{2}}{4b^{2}}=1$, we obtain $\\frac{c^{2}}{a^{2}}=2$, which means the eccentricity of the hyperbola is $\\sqrt{2}$. Thus, fill in $\\sqrt{2}$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ and the point $M(-2,2)$, a line passing through the focus of $C$ with slope $k$ intersects $C$ at points $A$ and $B$. If $\\overrightarrow{M A} \\cdot \\overrightarrow{M B}=0$, then $k=?$",
    "fact_expressions": "C: Parabola;G: Line;M: Point;A: Point;B: Point;Expression(C) = (y^2 = 8*x);Coordinate(M) = (-2, 2);k:Number;PointOnCurve(Focus(C), G);Slope(G) = k;Intersection(G, C) = {A, B};DotProduct(VectorOf(M, A), VectorOf(M, B)) = 0",
    "query_expressions": "k",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 21], [34, 37], [51, 54]], [[48, 50]], [[22, 32]], [[56, 59]], [[60, 63]], [[2, 21]], [[22, 32]], [[44, 47], [120, 123]], [[33, 50]], [[41, 50]], [[48, 65]], [[67, 118]]]",
    "query_spans": "[[[120, 125]]]",
    "process": "From the focus $ F(2,0) $ of the parabola $ C: y^{2} = 8x $, let the equation of line $ AB $ be $ y = k(x - 2) $.  \n$$\n\\begin{cases}\ny = k(x -\n\\end{cases}\n-2)\n$$\nEliminating $ x $ and simplifying yields $ ky^{2} - 8y - 16k = 0 $.  \nLet $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $,  \n$ \\therefore y_{1} + y_{2} = \\frac{8}{k} $, $ y_{1}y_{2} = -16 $,  \n$ \\overrightarrow{MA} = (x_{1} + 2, y_{1} - -2) $, $ \\overrightarrow{MB} = (x_{2} $  \n$ \\therefore \\overrightarrow{MA} \\cdot \\overrightarrow{MB} = (x_{1} + 2)(x_{2} + 2) + (y_{1} - -2(y_{2} $  \n$ = x_{1}x_{2} + 2(x_{1} + x_{2}) + 4 + y_{1}y_{2} - 2(y_{1} + y_{2}) + 4\\frac{2 + y_{2}^{2} + y_{2} - 2(y_{1} + y)^{2} + 8}{8_{2y}^{y_{2}}} + y_{y} - 2(y_{1} + y_{2} + 8 $  \nSimplifying yields $ k^{2} - 4k + 4 = 0 $, solving gives $ k = 2 $."
  },
  {
    "text": "The ellipse $x^{2}+n y^{2}=1$ intersects the line $y=1-x$ at points $M$ and $N$. The slope of the line passing through the origin and the midpoint of segment $MN$ is $\\frac{\\sqrt{2}}{2}$. Then the value of $n$ is?",
    "fact_expressions": "G: Ellipse;n: Number;H: Line;M: Point;N: Point;O:Origin;L:Line;Expression(G) = (n*y^2 + x^2 = 1);Expression(H) = (y = 1 - x);Intersection(G, H) = {M, N};PointOnCurve(O,L);PointOnCurve(MidPoint(LineSegmentOf(M,N)),L);Slope(L)=sqrt(2)/2",
    "query_expressions": "n",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 19]], [[83, 86]], [[20, 29]], [[31, 34]], [[35, 38]], [[42, 44]], [[56, 58]], [[0, 19]], [[20, 29]], [[0, 40]], [[41, 58]], [[41, 58]], [[56, 82]]]",
    "query_spans": "[[[83, 90]]]",
    "process": "Let M(x_{1},y_{1}), N(x_{2},y_{2}), and the midpoint be (x,y). The ellipse x^{2}+ny^{2}=1 intersects the line y=1-x at points M and N. Simplifying \\begin{cases}x^{2}+ny^{2}=1\\\\y=1-x\\end{cases} yields: (1+n)x^{2}-2nx-n-1=0, so x_{1}+x_{2}=\\frac{2n}{n+1}, hence x=\\frac{n}{n+1}, y=\\frac{1}{n+1}. Since the slope of the line passing through the origin and the midpoint of segment MN is \\frac{\\sqrt{2}}{2}, it follows that \\frac{1}{n}=\\frac{\\sqrt{2}}{2}, thus n=\\sqrt{2}."
  },
  {
    "text": "Given that the center of the hyperbola is at the origin, the foci lie on the $x$-axis, one of its asymptotes has the equation $x - \\sqrt{3} y = 0$, and it passes through the point $(2 \\sqrt{3}, \\sqrt{3})$, then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (2*sqrt(3), sqrt(3));O:Origin;Center(G)=O;PointOnCurve(Focus(G),xAxis);Expression(OneOf(Asymptote(G)))=(x-sqrt(3)*y=0);PointOnCurve(H,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/3-y^2=1",
    "fact_spans": "[[[2, 5], [23, 24], [80, 83]], [[52, 77]], [[52, 77]], [[9, 13]], [[2, 13]], [[2, 22]], [[23, 49]], [[23, 77]]]",
    "query_spans": "[[[80, 88]]]",
    "process": "Let the hyperbola equation be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, its asymptotes are given by $y=\\pm\\frac{b}{a}x$. Since one of them is $y=\\frac{\\sqrt{3}}{3}x$, we have $\\frac{b}{a}=\\frac{\\sqrt{3}}{3}$. Also, the point $(2\\sqrt{3},\\sqrt{3})$ lies on the hyperbola; substituting into the hyperbola equation gives $\\frac{12}{a^{2}}-\\frac{3}{b^{2}}=1$. Solving these equations simultaneously yields $a=\\sqrt{3}$, $b=1$, so the equation of the hyperbola is $\\frac{x^{2}}{3}-y^{2}=1$."
  },
  {
    "text": "Let point $P$ lie on the curve $y=x^{2}+2$, and point $Q$ lie on the curve $y=\\sqrt{x-2}$. Then the minimum value of $|P Q|$ is equal to?",
    "fact_expressions": "G: Curve;H:Curve;P: Point;Q: Point;Expression(G) = (y = x^2 + 2);Expression(H) = (y = sqrt(x - 2));PointOnCurve(P, G);PointOnCurve(Q, H)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "7*sqrt(2)/4",
    "fact_spans": "[[[6, 19]], [[26, 42]], [[1, 5]], [[21, 25]], [[6, 19]], [[26, 42]], [[1, 20]], [[21, 43]]]",
    "query_spans": "[[[45, 60]]]",
    "process": ""
  },
  {
    "text": "Let $A_{1}$ and $A_{2}$ be the left and right vertices of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. If there exists a point $P$ on the ellipse, distinct from $A_{1}$ and $A_{2}$, such that $\\overrightarrow{P O} \\cdot \\overrightarrow{P A_{1}}=0$, where $O$ is the origin, then the range of the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;O: Origin;A1: Point;A2: Point;e: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(G) = A1;RightVertex(G) = A2;PointOnCurve(P,G);Negation(A1=P);Negation(A2=P);DotProduct(VectorOf(P, O), VectorOf(P, A1)) = 0;Eccentricity(G) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "(\\sqrt{2}/2,1)",
    "fact_spans": "[[[17, 69], [78, 80], [176, 178]], [[19, 69]], [[19, 69]], [[101, 105]], [[166, 169]], [[1, 8], [85, 92]], [[9, 16], [93, 100]], [[182, 185]], [[19, 69]], [[19, 69]], [[17, 69]], [[1, 75]], [[1, 75]], [[78, 105]], [[81, 105]], [[81, 105]], [[108, 163]], [[176, 185]]]",
    "query_spans": "[[[182, 192]]]",
    "process": "Let point $ P(x,y) $. From $ \\overrightarrow{PO} \\cdot \\overrightarrow{PA_{1}} = 0 $, we obtain $ e^{2}x^{2} + ax + b^{2} = 0 $. Find the zero of function $ f(x) $ in the interval $ (-a,0) $ as $ -\\frac{ab^{2}}{c^{2}} $, simplify to get $ 0 < \\frac{b^{2}}{c^{2}} < 1 $, then solve for the range of $ e $. Let point $ P(x,y) $, then $ y^{2} = b^{2} - \\frac{b^{2}}{a^{2}}x^{2} $, it is known that point $ A_{1}(-a,0) $, $ \\overrightarrow{PO} = (-x,-y) $, $ \\overrightarrow{PA_{1}} = (-a-x,-y) $, $ \\overrightarrow{PO} \\cdot \\overrightarrow{PA_{1}} = -x(-a-x) + (-y)^{2} = x^{2} + y^{2} + ax = x^{2} + b^{2} - \\frac{b^{2}}{a^{2}}x^{2} + ax = \\frac{c^{2}}{a^{2}}x^{2} + ax + b^{2} = 0 $. Let $ f(x) = e^{2}x^{2} + ax + b^{2} $, then the function $ f(x) $ has a zero in the interval $ (-a,0) $. Since $ f(-a) = c^{2} - a^{2} + b^{2} = 0 $, then $ -a $ is a root of the equation $ e^{2}x^{2} + ax + b^{2} = 0 $. Let the zero of function $ f(x) $ in the interval $ (-a,0) $ be $ x_{1} $, by Vieta's formula we get $ -a x_{1} = \\frac{b^{2}}{e^{2}} = \\frac{a^{2}b^{2}}{c^{2}} $, so $ -a < -\\frac{ab^{2}}{c^{2}} < 0 $, i.e., $ 0 < \\frac{b^{2}}{c^{2}} < 1 $, rearranging gives $ a^{2} - c^{2} = b^{2} < c^{2} $. $ \\therefore a^{2} < 2c^{2} $, i.e., $ 2e^{2} > 1 $, since $ 0 < e < 1 $, solve to get $ \\frac{\\sqrt{2}}{2} < e < 1 $. Therefore, the range of eccentricity $ e $ of the ellipse is $ \\left( \\frac{\\sqrt{2}}{2}, 1 \\right) $."
  },
  {
    "text": "Given point $Q(2 \\sqrt{2}, 0)$ and a moving point $P(x, y)$ on the parabola $y=\\frac{x^{2}}{4}$, find the minimum value of $y+|P Q|$.",
    "fact_expressions": "Q: Point;Coordinate(Q) = (2*sqrt(2), 0);G: Parabola;Expression(G) = (y = x^2/4);P: Point;x1: Number;y1: Number;Coordinate(P) = (x1, y1);PointOnCurve(P, G)",
    "query_expressions": "Min(y1+ Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 21]], [[2, 21]], [[22, 44]], [[22, 44]], [[48, 57]], [[48, 57]], [[48, 57]], [[48, 57]], [[22, 57]]]",
    "query_spans": "[[[59, 74]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1$ $(0<b<2)$, with left and right foci $F_{1}$ and $F_{2}$ respectively, a line $l$ passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. If the maximum value of $|BF_{2}|+|AF_{2}|$ is $5$, then the value of $b$ is?",
    "fact_expressions": "l: Line;G: Ellipse;b: Number;B: Point;F2: Point;A: Point;F1: Point;0 < b;b < 2;Expression(G) = (x^2/4 + y^2/b^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, l);Intersection(l, G) = {A, B};Max(Abs(LineSegmentOf(A, F2)) + Abs(LineSegmentOf(B, F2))) = 5",
    "query_expressions": "b",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[84, 89]], [[2, 51], [90, 92]], [[134, 137]], [[97, 100]], [[67, 74]], [[93, 96]], [[59, 66], [76, 83]], [[4, 51]], [[4, 51]], [[2, 51]], [[2, 74]], [[2, 74]], [[75, 89]], [[84, 102]], [[104, 132]]]",
    "query_spans": "[[[134, 141]]]",
    "process": "From the definition of the ellipse, we have |BF_{2}| + |AF_{2}| = 2a - |BF_{1}| + 2a - |AF_{1}| = 4a - |AB| = 8 - |AB|. From the given condition, |BF_{2}| + |AF_{2}| = 8 - |AB|. Therefore, when |AB| takes the minimum value 3, |BF_{2}| + |AF_{2}| reaches the maximum value of 5. At this time, AB \\bot x-axis, and point A is (-c, \\frac{3}{2}). Substituting the coordinates of point A into the ellipse equation yields b = \\sqrt{3}."
  },
  {
    "text": "The area of the triangle formed by the two foci and one endpoint of the minor axis of the ellipse $C$: $x^{2}+2 y^{2}=4$ is equal to?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2 + 2*y^2 = 4);F1:Point;F2:Point;P:Point;Focus(C)={F1,F2};OneOf(Endpoint(MinorAxis(C)))=P",
    "query_expressions": "Area(TriangleOf(F1,F2,P))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 24]], [[0, 24]], [], [], [], [[0, 29]], [[0, 37]]]",
    "query_spans": "[[[0, 49]]]",
    "process": "The equation of the ellipse can be written as \\frac{x^2}{4}+\\frac{y^{2}}{2}=1. \\therefore a^{2}=4, b^{2}=2, thus c^{2}=2, c=\\sqrt{2}. Therefore, the two foci are (-\\sqrt{2},0), (\\sqrt{2},0), and the endpoints of the minor axis are (0,\\sqrt{2}). \\therefore The area of the triangle formed is \\frac{1}{2}\\times2\\sqrt{2}\\times\\sqrt{2}=2."
  },
  {
    "text": "Given that the right focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $F(3,0)$, and the point $(-3, \\frac{3 \\sqrt{2}}{2})$ lies on the ellipse $C$, then the standard equation of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Point;F: Point;a>0;b>0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(G) = (-3, (3*sqrt(2))/2);Coordinate(F) = (3, 0);RightFocus(C) = F;PointOnCurve(G, C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/18+y^2/9^2=1",
    "fact_spans": "[[[115, 120], [107, 112], [2, 62]], [[8, 62]], [[8, 62]], [[77, 106]], [[67, 75]], [[8, 62]], [[8, 62]], [[2, 62]], [[77, 106]], [[67, 75]], [[2, 75]], [[77, 113]]]",
    "query_spans": "[[[115, 127]]]",
    "process": ""
  },
  {
    "text": "Given the circle $C$: $x^{2}+y^{2}=25$, draw a line $l$ through the point $M(-2, 3)$ intersecting the circle $C$ at points $A$ and $B$. Tangents to the circle are drawn at points $A$ and $B$, respectively. When the two tangents intersect at point $Q$, what is the trajectory equation of point $Q$?",
    "fact_expressions": "l: Line;C: Circle;M: Point;A: Point;B: Point;Q:Point;Expression(C) = (x^2 + y^2 = 25);Coordinate(M) = (-2, 3);PointOnCurve(M, l);Intersection(l, C) = {A, B};L1:Line;L2:Line;TangentOfPoint(A,C)=L1;TangentOfPoint(B,C)=L2;Intersection(L1,L2)=Q",
    "query_expressions": "LocusEquation(Q)",
    "answer_expressions": "2*x-3*y+25=0",
    "fact_spans": "[[[39, 44]], [[2, 24], [45, 49], [73, 74]], [[26, 38]], [[50, 53], [63, 66]], [[54, 57], [67, 70]], [[86, 90], [92, 96]], [[2, 24]], [[26, 38]], [[25, 44]], [[39, 59]], [], [], [[60, 77]], [[60, 77]], [[60, 91]]]",
    "query_spans": "[[[92, 103]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, $ Q(x_{0},y_{0}) $. Since $ AQ $ is tangent to circle $ C $, then $ AQ \\perp CA $, i.e., $ x_{1}^{2} - x_{0}x_{1} + y_{1}^{2} - y_{0}y_{1} = 0 $. Furthermore, since point $ A(x_{1},y_{1}) $ lies on the circle, $ x_{1}^{2} + y_{1}^{2} = 25 $. Substituting into the previous equation and simplifying yields $ x_{0}x_{1} + y_{0}y_{1} = 25 $. Similarly, $ x_{0}x_{2} + y_{0}y_{2} = 25 $. Thus, the line passing through $ A $ and $ B $ has the equation $ x_{0}x + y_{0}y = 25 $. Combining this with the condition that line $ AB $ passes through point $ M(-2,3) $, we solve as follows: As shown in the figure, for circle $ C: x^{2} + y^{2} = 25 $, the center $ C $ is at $ (0,0) $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, $ Q(x_{0},y_{0}) $. Since $ AQ $ is tangent to circle $ C $, $ AQ \\perp CA $, so $ (x_{1}-x_{0})(x_{1}-0) + (y_{1}-y_{0})(y_{1}-0) = 0 $, i.e., $ x_{1}^{2} - x_{0}x_{1} + y_{1}^{2} - y_{0}y_{1} = 0 $. Since $ x_{1}^{2} + y_{1}^{2} = 25 $, it follows that $ x_{0}x_{1} + y_{0}y_{1} = 25 $. Similarly, $ x_{0}x_{2} + y_{0}y_{2} = 25 $. Therefore, the equation of the line passing through points $ A $ and $ B $ is $ x_{0}x + y_{0}y = 25 $. Since line $ AB $ passes through point $ M(-2,3) $, we have $ -2x_{0} + 3y_{0} = 25 $. Hence, the trajectory equation of point $ Q $ is $ 2x - 3y + 25 = 0 $."
  },
  {
    "text": "Given the line $l$: $x - y + m = 0$ intersects the hyperbola $x^{2} - \\frac{y^{2}}{2} = 1$ at two distinct points $A$, $B$. If the midpoint of segment $AB$ lies on the circle $x^{2} + y^{2} = 5$, then the value of $m$ is?",
    "fact_expressions": "l: Line;G: Hyperbola;H: Circle;A: Point;B: Point;m: Number;Expression(G) = (x^2 - y^2/2 = 1);Expression(H) = (x^2 + y^2 = 5);Expression(l) = (m + x - y = 0);Intersection(l, G) = {A, B};Negation(A=B);PointOnCurve(MidPoint(LineSegmentOf(A,B)), H)",
    "query_expressions": "m",
    "answer_expressions": "pm*1",
    "fact_spans": "[[[2, 18]], [[19, 47]], [[74, 90]], [[54, 57]], [[58, 61]], [[93, 96]], [[19, 47]], [[74, 90]], [[2, 18]], [[2, 61]], [[49, 61]], [[63, 91]]]",
    "query_spans": "[[[93, 100]]]",
    "process": "Let points A(x_{1},y_{1}), B(x_{2},y_{2}), and the midpoint of segment AB be M(x_{0},y_{0}). From \\begin{cases}x-y+m=0\\\\x^{2}-\\frac{y^{2}}{2}=1\\end{cases}, we obtain x^{2}-2mx-m^{2}-2=0 (discriminant \\triangle>0). Since x_{1}+x_{2}=2m, then x_{0}=\\frac{x_{1}+x_{2}}{2}=m, y_{0}=x_{0}+m=2m. Since point M(x_{0},y_{0}) lies on the circle x^{2}+y^{2}=5, it follows that m^{2}+(2m)^{2}=5, hence m=\\pm1."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, the distance from the left focus $F$ of the hyperbola $C$ to one of its asymptotes is $4$. Point $M$ lies on the right branch of the hyperbola $C$, and the coordinates of point $P$ are $(0, b)$. Then the minimum perimeter of $\\Delta P F M$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;P: Point;F: Point;M: Point;b>0;Expression(C) = (x^2/4 - y^2/b^2 = 1);Coordinate(P) = (0, b);LeftFocus(C)=F;Distance(F,OneOf(Asymptote(C)))=4;PointOnCurve(M, RightPart(C))",
    "query_expressions": "Min(Perimeter(TriangleOf(P, F, M)))",
    "answer_expressions": "16",
    "fact_spans": "[[[2, 54], [62, 68], [89, 95]], [[108, 116]], [[100, 104]], [[58, 61]], [[84, 88]], [[10, 54]], [[2, 54]], [[100, 116]], [[2, 61]], [[58, 83]], [[84, 99]]]",
    "query_spans": "[[[118, 141]]]",
    "process": "The equations of the asymptotes are $ bx \\pm 4y = 0 $, so the distance from $ F $ to the asymptote is $ \\frac{|bc|}{\\sqrt{b^{2}+4}} = b = 4 $. Thus, $ c = \\sqrt{4+16} = 2\\sqrt{5} $, and $ F(-2\\sqrt{5},0) $, $ F_{2}(2\\sqrt{5},0) $. Let the right focus of the hyperbola be $ F_{2} $. The perimeter $ l $ of $ APFM $ is $ |FP| + |MF| + |MP| = 6 + |MF| + |MP| = 10 + |MF_{2}| + |MP| $. Since $ |MF_{2}| + |MP| \\geqslant |PF_{2}| = 6 $, with equality if and only if points $ M $, $ F_{2} $, and $ P $ are collinear, the minimum value of $ l $ is 16."
  },
  {
    "text": "Given that point $P$ is a point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, and the area of the triangle with vertices at $P$ and the two foci is $2 \\sqrt{5}$, find the coordinates of point $P$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};Area(TriangleOf(P, F1, F2)) = 2*sqrt(5)",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(0, pm*2)",
    "fact_spans": "[[[7, 44]], [[7, 44]], [[2, 6], [84, 88], [51, 54]], [[2, 48]], [], [], [[7, 58]], [[7, 82]]]",
    "query_spans": "[[[84, 92]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$, point $F$ is its focus. For any line passing through point $A(a, 0)$ and intersecting the parabola $C$ at two distinct points $M$, $N$, it holds that $\\overrightarrow{F M} \\cdot \\overrightarrow{F N}<0$. Then the range of real values for $a$ is?",
    "fact_expressions": "C: Parabola;A: Point;F: Point;M: Point;N: Point;a: Real;l:Line;Expression(C) = (y^2 = 8*x);Focus(C)=F;Coordinate(A) = (a, 0);PointOnCurve(A,l);Intersection(l,C)={M,N};Negation(M=N);DotProduct(VectorOf(F,M),VectorOf(F,N))<0",
    "query_expressions": "Range(a)",
    "answer_expressions": "(6-4*sqrt(2),6+4*sqrt(2))",
    "fact_spans": "[[[2, 20], [26, 27], [46, 52]], [[34, 44]], [[21, 25]], [[60, 63]], [[64, 67]], [[127, 132]], [[70, 72]], [[2, 20]], [[21, 30]], [[34, 44]], [[33, 72]], [[45, 72]], [[53, 67]], [[74, 125]]]",
    "query_spans": "[[[127, 139]]]",
    "process": "Let the equation of the line be $ x = ty + a $, then  \n$$\n\\begin{cases}\ny^{2} = 8x \\\\\nx = ty + a\n\\end{cases}\n\\Rightarrow y^2 - 8ty - 8a = 0.\n$$\nLet $ M(x_{1}, y_{1}) $, $ N(x_{2}, y_{2}) $, $ F(2, 0) $. Then  \n$ y_{1} + y_{2} = 8t $, $ y_{1}y_{2} = -8a $,  \n$ x_{1} + x_{2} = t(y_{1} + y_{2}) + 2a = 8t^{2} + 2a $,  \n$ x_{1}x_{2} = \\frac{(y_{1}y_{2})^{2}}{64} = a^{2} $.  \nThen from $ \\overrightarrow{FM} \\cdot \\overrightarrow{FN} < 0 $ we get  \n$ (x_{1} - 2, y_{1}) \\cdot (x_{2} - 2, y_{2}) < 0 \\Rightarrow x_{1}x_{2} - 2(x_{1} + x_{2}) + 4 + y_{1}y_{2} < 0 $.  \nSubstituting Vieta's formulas gives  \n$ a^{2} - 2(8t^{2} + 2a) + 4 - 8a < 0 \\Rightarrow a^{2} - 12a + 4 < 16t^{2} $, which holds constantly.  \nHence $ a^{2} - 12a + 4 < 0 \\Rightarrow 6 - 4\\sqrt{2} < a < 6 + 4\\sqrt{2} $."
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ is $2$, then the value of $m$ is?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (-y^2 + x^2/m = 1);Eccentricity(G) = 2",
    "query_expressions": "m",
    "answer_expressions": "1/3",
    "fact_spans": "[[[2, 30]], [[40, 43]], [[2, 30]], [[2, 38]]]",
    "query_spans": "[[[40, 47]]]",
    "process": "From the given condition, the eccentricity of the hyperbola is $ e = \\frac{\\sqrt{m+1}}{\\sqrt{m}} = 2 $, solving yields $ m = \\frac{1}{3} $."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ passes through the point $M(2,2 \\sqrt{2})$, and the focus is $F$, then the slope of the line $M F$ is?",
    "fact_expressions": "G: Parabola;p: Number;M: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(M) = (2, 2*sqrt(2));PointOnCurve(M, G);Focus(G) = F",
    "query_expressions": "Slope(LineOf(M, F))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 23]], [[5, 23]], [[25, 43]], [[48, 51]], [[5, 23]], [[2, 23]], [[25, 43]], [[2, 43]], [[2, 51]]]",
    "query_spans": "[[[53, 65]]]",
    "process": "Substituting the coordinates of the given point into the parabola equation to find the value of p, the coordinates of the focus can be obtained, thus allowing the calculation of the slope of the line. According to the problem, (2\\sqrt{2})^{2}=2p\\times2, solving gives p=2, \\therefore the focus is F(1,0) \\therefore MF=\\frac{2\\sqrt{2}-0}{2-1}=2\\sqrt{2}"
  },
  {
    "text": "The vertex of the parabola is at the origin, and its focus is a focus of the ellipse $4 x^{2}+y^{2}=1$. Then, the distance from the focus of this parabola to its directrix is?",
    "fact_expressions": "G: Parabola;H: Ellipse;O: Origin;Expression(H) = (4*x^2 + y^2 = 1);Vertex(G) = O;Focus(G) = OneOf(Focus(H))",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[0, 3], [42, 45], [49, 50]], [[15, 34]], [[7, 11]], [[15, 34]], [[0, 11]], [[0, 39]]]",
    "query_spans": "[[[42, 57]]]",
    "process": "The ellipse $4x^{2}+y^{2}=1$ can be written as $y^{2}+\\frac{x^{2}}{\\frac{1}{4}}=1$, where $a=1$, $b=\\frac{1}{2}$, $c=\\sqrt{a^{2}-b^{2}}=\\frac{\\sqrt{3}}{2}$. Thus, one focus of the ellipse is $(0,\\frac{\\sqrt{3}}{2})$. Since the focus of the parabola is $(0,\\frac{\\sqrt{3}}{2})$, it follows that $\\frac{p}{2}=\\frac{\\sqrt{3}}{2} \\Rightarrow p=\\sqrt{3}$. Therefore, the distance from the focus to the directrix of the parabola is $\\sqrt{3}$."
  },
  {
    "text": "Given two points $A$ and $B$ on the parabola $y^{2}=4x$ with focus $F$ such that $\\overrightarrow{AF}=3\\overrightarrow{FB}$, then the distance from the midpoint of chord $AB$ to the directrix is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(A, G);PointOnCurve(B, G);VectorOf(A, F) = 3*VectorOf(F, B);IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A, B)), Directrix(G))",
    "answer_expressions": "8/3",
    "fact_spans": "[[[10, 24]], [[28, 31]], [[34, 37]], [[3, 6]], [[10, 24]], [[2, 24]], [[10, 37]], [[10, 37]], [[39, 84]], [[87, 92]]]",
    "query_spans": "[[[88, 103]]]",
    "process": ""
  },
  {
    "text": "The eccentricities of a confocal ellipse and hyperbola are $e_{1}$, $e_{2}$ respectively. If the minor axis length of the ellipse is twice the imaginary axis length of the hyperbola, then the minimum value of $e_{1} e_{2}$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;e1:Number;e2:Number;Eccentricity(H)=e1;Eccentricity(G)=e2;Focus(G)=Focus(H);Length(MinorAxis(H))=2*Length(ImageinaryAxis(G))",
    "query_expressions": "Min(e1*e2)",
    "answer_expressions": "4/5",
    "fact_spans": "[[[7, 10], [41, 44]], [[4, 6], [34, 36]], [[17, 24]], [[25, 32]], [[4, 32]], [[4, 32]], [[0, 10]], [[34, 53]]]",
    "query_spans": "[[[55, 74]]]",
    "process": "Let the semi-transverse axis, semi-conjugate axis, and semi-focal distance of the hyperbola be a, b, c, respectively. The eccentricities of the ellipse and hyperbola with common foci are e_{1}, e_{2}, respectively. If the minor axis length of the ellipse is twice the conjugate axis length of the hyperbola, then the semi-major axis, semi-minor axis, and semi-focal distance of the ellipse are \\sqrt{4b^{2}+c^{2}}, 2b, c, \\therefore e_{1}e_{2}=\\frac{c}{\\sqrt{4b^{2}+c^{2}}}\\times-4(\\frac{b}{c})^{4}+3(\\frac{b}{c})^{2}+1=\\frac{1}{\\sqrt{25-[(b)^{2}-3}^{2}}}\\frac{1}{2}\\geqslant\\frac{1}{4}=\\frac{4}{5}, that is, the minimum value of e_{1}e_{2} is \\frac{4}{5}."
  },
  {
    "text": "What is the distance from the point on the parabola $y^{2}=4 x$ with horizontal coordinate $2$ to its focus?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G);XCoordinate(P) = 2",
    "query_expressions": "Distance(P, Focus(G))",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 14], [25, 26]], [[0, 14]], [[23, 24]], [[0, 24]], [[15, 24]]]",
    "query_spans": "[[[23, 33]]]",
    "process": "According to the problem, $2p=4$, $p=2$, the required distance is $d=2+\\frac{p}{2}=2+\\frac{2}{2}=3$,"
  },
  {
    "text": "It is known that a hyperbola passes through the point $(1, 2\\sqrt{2})$, and one of its asymptotes has the equation $y = 2x$. Then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (1, 2*sqrt(2));PointOnCurve(H, G);Expression(OneOf(Asymptote(G))) = (y = 2*x)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/4-x^2=1",
    "fact_spans": "[[[2, 5], [25, 26], [44, 47]], [[7, 24]], [[7, 24]], [[2, 24]], [[25, 41]]]",
    "query_spans": "[[[44, 54]]]",
    "process": "Let the hyperbola equation be $x^{2}-\\frac{y^{2}}{4}=\\lambda(\\lambda\\neq0)$ based on the asymptote equations of the hyperbola. Since the point $(1,2\\sqrt{2})$ lies on the hyperbola, we have $1-\\frac{(2\\sqrt{2})^{2}}{4}=\\lambda$, so $\\lambda=-1$. Thus, the hyperbola equation is $\\frac{y^{2}}{4}-x^{2}=1$."
  },
  {
    "text": "Given the parabola $y = a x^{2}$ ($a > 0$) with focus $F(0, 2)$, a line $l$ passing through $F$ intersects the parabola at points $A$ and $B$. Then the minimum value of $|AB|$ is?",
    "fact_expressions": "l: Line;G: Parabola;a: Number;A: Point;B: Point;F:Point;a>0;Expression(G) = (y = a*x^2);Coordinate(F)=(0, 2) ;Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B}",
    "query_expressions": "Min(Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "8",
    "fact_spans": "[[[40, 45]], [[2, 21], [46, 49]], [[5, 21]], [[50, 53]], [[54, 57]], [[25, 34], [36, 39]], [[5, 21]], [[2, 21]], [[25, 34]], [[2, 34]], [[35, 45]], [[40, 59]]]",
    "query_spans": "[[[61, 74]]]",
    "process": "Convert the parabola into standard form, derive the equation of the parabola from the focus, and then find the latus rectum to obtain the result. $ y = ax^{2} (a > 0) \\Rightarrow x^{2} = \\frac{1}{a}y $. Since the focus of the parabola is $ F(0,2) $, we have $ \\frac{1}{4a} = 2 $, solving gives $ a = \\frac{1}{8} $. Thus, $ x^{2} = 8y $. When $ y = 2 $, then $ x^{2} = 16 $, solving gives $ x = \\pm 4 $. Draw a line $ l $ through $ F $ intersecting the parabola at points $ A $ and $ B $. The latus rectum is the shortest such chord, so the minimum value of $ |AB| $ is 8."
  },
  {
    "text": "A directrix of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ is given by the equation $x=\\frac{3}{2}$, then its eccentricity is?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(OneOf(Directrix(G))) = (x = 3/2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 37], [62, 63]], [[3, 37]], [[3, 37]], [[0, 37]], [[0, 60]]]",
    "query_spans": "[[[62, 68]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $C$: $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$, respectively, and let $O$ be the origin. Point $P$ lies on $C$ such that $P F_{1} \\perp P F_{2}$. Then the area of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/16 + y^2/7 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;O: Origin;P: Point;PointOnCurve(P, C) = True;IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2)) = True",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "7",
    "fact_spans": "[[[19, 62], [83, 86]], [[19, 62]], [[1, 8]], [[9, 16]], [[1, 68]], [[1, 68]], [[69, 72]], [[78, 82]], [[78, 87]], [[89, 112]]]",
    "query_spans": "[[[115, 145]]]",
    "process": "By the Pythagorean theorem, |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}, and combining with the definition of an ellipse, |PF_{1}||PF_{2}| can be found, thus obtaining the product. From the given, F_{1}(-3,0), F_{2}(3,0), a=4, c=3, and \\triangle F_{1}F_{2}P is a right triangle with the right angle at P, so |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}, that is, |PF_{1}|^{2}+|PF_{2}|^{2}=36. Also, |PF_{1}|+|PF_{2}|=2a=8, \\therefore 64=(|PF_{1}|+|PF_{2}|)^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}+2|PF_{1}||PF_{2}|=36+2|PF_{1}||PF_{2}|, solving gives |PF_{1}||PF_{2}|=14, \\therefore S_{\\Delta PF_{1}F_{2}}=\\frac{1}{2}|PF_{1}||PF_{2}|=7."
  },
  {
    "text": "If the right focus and right vertex of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a, b>0)$ coincide with the right vertex and right focus of the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{4}=1$, respectively, then $b$=?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Ellipse;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2/6 + y^2/4 = 1);RightFocus(G)=RightVertex(H);RightFocus(H)=RightVertex(G)",
    "query_expressions": "b",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 55]], [[4, 55]], [[115, 118]], [[66, 103]], [[4, 55]], [[4, 55]], [[1, 55]], [[66, 103]], [[1, 113]], [[1, 113]]]",
    "query_spans": "[[[115, 120]]]",
    "process": "From the problem, we know: the right focus and right vertex of the hyperbola are (c,0) and (a,0), respectively; the right vertex and right focus of the ellipse are (\\sqrt{6},0) and (\\sqrt{2},0), respectively. According to the problem, c=\\sqrt{6}, a=\\sqrt{2}. For the hyperbola, b=\\sqrt{c^{2}-a^{2}}=2"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{12}-\\frac{y^{2}}{4}=1$ has its right focus at point $F$. If a line passing through point $F$ intersects the right branch of the hyperbola at exactly one point, then what is the range of possible values for the slope of this line?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/12 - y^2/4 = 1);H: Line;F: Point;RightFocus(C) = F;PointOnCurve(F,H);NumIntersection(H,RightPart(C))=1",
    "query_expressions": "Range(Slope(H))",
    "answer_expressions": "[-sqrt(3)/3,sqrt(3)/3]",
    "fact_spans": "[[[2, 41], [60, 63]], [[2, 41]], [[57, 59], [57, 59]], [[46, 49], [52, 56]], [[2, 49]], [[51, 59]], [[57, 74]]]",
    "query_spans": "[[[77, 89]]]",
    "process": ""
  },
  {
    "text": "The line $l$ passing through the point $P(1,1)$ intersects the hyperbola $\\frac{y^{2}}{2}-x^{2}=1$ at points $M$ and $N$, and the point $P$ is exactly the midpoint of the segment $MN$. Then the equation of the line $l$ is?",
    "fact_expressions": "P: Point;Coordinate(P) = (1, 1);l: Line;PointOnCurve(P, l);G: Hyperbola;Expression(G) = (-x^2 + y^2/2 = 1);M: Point;N: Point;Intersection(l, G) = {M, N};MidPoint(LineSegmentOf(M, N)) = P",
    "query_expressions": "Expression(l)",
    "answer_expressions": "2*x-y-1=0",
    "fact_spans": "[[[1, 10], [58, 62]], [[1, 10]], [[11, 16], [77, 82]], [[0, 16]], [[17, 45]], [[17, 45]], [[47, 50]], [[51, 54]], [[11, 56]], [[58, 75]]]",
    "query_spans": "[[[77, 87]]]",
    "process": "Let $ M(x_{1},y_{1}) $, $ N(x_{2},y_{2}) $, substitute into the hyperbola equation, subtract the two equations, and simplify to obtain $ (x_{1}-x_{2})(x_{1}+x_{2}) = \\frac{1}{2}(y_{1}+y_{2})(y_{1}-y_{2}) $. Using the midpoint coordinate formula, find the slope of line $ MN $, then use the point-slope form to find the equation of the line. The line $ l $ passing through point $ P(1,1) $ intersects the hyperbola at points $ M $ and $ N $. Let $ M(x_{1},y_{1}) $, $ N(x_{2},y_{2}) $, $ \\therefore \\begin{cases} \\frac{y_{1}^{2}}{2} - x_{1}^{2} = 1 \\\\ \\frac{y_{2}^{2}}{2} - x_{2}^{2} = 1 \\end{cases} $. Subtracting the two equations gives: $ (x_{1}-x_{2})(x_{1}+x_{2}) = \\frac{1}{2}(y_{1}+y_{2})(y_{1}-y_{2}) $. Since $ P $ is the midpoint of $ MN $, $ \\therefore x_{1}+x_{2}=2 $, $ y_{1}+y_{2}=2 $, $ \\therefore 2(x_{1}-x_{2}) = y_{1}-y_{2} $. Then $ k_{MN} = \\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = 2 $, so the equation of line $ l $ is $ y-1 = 2(x-1) $, i.e., $ 2x - y - 1 = 0 $."
  },
  {
    "text": "The distance from point $M$ on the parabola ${y}^{2}=4x$ to its focus $F$ is $4$. What is the horizontal coordinate of point $M$?",
    "fact_expressions": "G: Parabola;M: Point;F: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(M, G);Focus(G) = F;Distance(M, F) = 4",
    "query_expressions": "XCoordinate(M)",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 16], [23, 24]], [[18, 22], [38, 42]], [[26, 29]], [[0, 16]], [[0, 22]], [[23, 29]], [[18, 36]]]",
    "query_spans": "[[[38, 48]]]",
    "process": ""
  },
  {
    "text": "Point $P$ is an arbitrary point on the ellipse, $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse respectively, the maximum value of $\\angle F_{1} P F_{2}$ is $60^{\\circ}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;P: Point;PointOnCurve(P, G);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;Max(AngleOf(F1, P, F2)) = ApplyUnit(60, degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[5, 7], [31, 33], [80, 82]], [[0, 4]], [[0, 12]], [[13, 20]], [[21, 28]], [[13, 38]], [[13, 38]], [[39, 78]]]",
    "query_spans": "[[[80, 90]]]",
    "process": "According to the problem, when P is at the upper or lower vertex of the ellipse, $\\angle F_{1}PF_{2}$ reaches its maximum value of $60^{\\circ}$, that is, $\\frac{c}{a}=\\sin30^{\\circ}=\\frac{1}{2}$"
  },
  {
    "text": "Given that $A$ and $B$ are two points on the circle $C$: $x^{2}+y^{2}-8 x-2 y+16=0$, and point $P$ lies on the parabola $x^{2}=2 y$. When $\\angle A P B$ reaches its maximum value, $|A B|=$?",
    "fact_expressions": "C: Circle;Expression(C) = (-2*y - 8*x + x^2 + y^2 + 16 = 0);A: Point;B: Point;PointOnCurve(A, C) = True;PointOnCurve(B, C) = True;P: Point;G: Parabola;Expression(G) = (x^2 = 2*y);PointOnCurve(P, G) = True;WhenMax(AngleOf(A, P, B)) = True",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "4*sqrt(5)/5",
    "fact_spans": "[[[10, 42]], [[10, 42]], [[2, 5]], [[6, 9]], [[2, 45]], [[2, 45]], [[46, 50]], [[51, 65]], [[51, 65]], [[46, 66]], [[67, 88]]]",
    "query_spans": "[[[89, 98]]]",
    "process": "According to the problem, when PA and PB are tangent to circle C, ∠APB reaches its maximum value, where A and B are the points of tangency. Let ∠APB = 2α, P(x₀, x₀²/2). Since circle C: x² + y² − 8x − 2y + 16 = 0, the center is C(4,1) and the radius is 1. Thus, sinα = |CB| / |PC| = 1 / |PC|. Since |PC|² = (4 − x₀)² + (1 − x₀²/2) = x₀⁴/4 − 8x₀ + 17, let f(x) = x⁴/4 − 8x + 17, then f′(x) = x³ − 8. When x < 2, f′(x) < 0, so function f(x) is decreasing on (−∞, 2); when x > 2, f′(x) > 0, so function f(x) is increasing on (2, +∞). Therefore, f(x)ₘᵢₙ = f(2) = 5, i.e., |PC|ₘᵢₙ = √5, (sinα)ₘₐₓ = √5/5, at which point ∠APB is maximized. Hence, cosα = 2√5/5 = 1/(2/|AB|), so |AB| = 4√5/5."
  },
  {
    "text": "If the distance from point $M$ on the parabola $y^{2}=4x$ to the focus is $11$, then what is the distance from $M$ to the $y$-axis?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(M, G);Distance(M, Focus(G)) = 11",
    "query_expressions": "Distance(M,yAxis)",
    "answer_expressions": "10",
    "fact_spans": "[[[1, 15]], [[17, 21], [34, 37]], [[1, 15]], [[1, 21]], [[1, 32]]]",
    "query_spans": "[[[34, 47]]]",
    "process": "The directrix of the parabola is $x = -1$. Since the distance from point $M$ to the focus is 11, the distance from point $M$ to the directrix $x = -1$ is 11, therefore the distance from point $M$ to the $y$-axis is 10."
  },
  {
    "text": "The equation of the hyperbola that shares a common focus with the parabola $y^{2}=-8 \\sqrt{3} x$ and has an asymptote given by $x+\\sqrt {3} y=0$ is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(H) = (y^2 = -8*sqrt(3)*x);Focus(G) = Focus(H);Expression(OneOf(Asymptote(G))) = (x + sqrt(3)*y = 0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/3 = 1",
    "fact_spans": "[[[58, 61]], [[1, 25]], [[1, 25]], [[0, 61]], [[32, 61]]]",
    "query_spans": "[[[58, 66]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $x^{2}=4 y$, a line $l$ with slope $\\frac{1}{2}$ passes through the focus of the parabola and intersects the parabola at points $A$ and $B$. If the circle with diameter $AB$ is tangent to the directrix of the parabola at point $P$, then the distance from point $P$ to the line $AB$ is?",
    "fact_expressions": "l: Line;G: Parabola;H: Circle;A: Point;B: Point;P: Point;Expression(G) = (x^2 = 4*y);Slope(l) = 1/2;PointOnCurve(Focus(G), l);Intersection(l, G) = {A, B};IsDiameter(LineSegmentOf(A, B), H);TangentPoint(H, Directrix(G)) = P",
    "query_expressions": "Distance(P, LineOf(A, B))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[34, 39]], [[2, 16], [40, 43], [49, 52], [80, 83]], [[78, 79]], [[55, 58]], [[59, 62]], [[89, 93], [95, 99]], [[2, 16]], [[17, 39]], [[34, 46]], [[34, 64]], [[66, 79]], [[78, 93]]]",
    "query_spans": "[[[95, 112]]]",
    "process": "Let the equation of line AB be $ y = -\\frac{1}{2}x + b $. Substituting into the parabola $ x^2 = 4y $, we obtain $ y^2 - (2b+1)y + b^2 = 0 $. Let $ A(x_1, y_1) $, $ B(x_2, y_2) $, and let $ M(x_0, y_0) $ be the midpoint of AB. Then $ y_1 + y_2 = 2b+1 $, $ y_1 y_2 = b^2 $, so $ y_0 = \\frac{y_1 + y_2}{2} = \\frac{2b+1}{2} $. Since the circle with diameter AB is tangent to the directrix of the parabola at point P, we have $ 2\\left( \\frac{2b+1}{2} + 1 \\right) = \\sqrt{5} \\sqrt{4b+1} $. Solving gives $ b = 1 $, so the equation of line AB is $ y = -\\frac{1}{2}x + 1 $, or $ x + 2y - 2 = 0 $. Thus, the coordinates of point P are $ (-1, -1) $, and the distance from point P to line AB is $ d = \\frac{|-1 - 2 - 2|}{\\sqrt{5}} = \\sqrt{5} $. This problem mainly examines the standard equation of a parabola and its simple geometric properties, as well as the application of the relationship between a line and a parabola. The key to the solution is setting up the line equation, substituting into the parabola's equation, using the relationship between roots and coefficients along with the given conditions to form an equation and solve for $ b $, thereby obtaining the line equation. It also examines the ability to analyze and solve problems, and is a medium-difficulty question."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$, with focus point $F$. Point $P$ is a moving point on the parabola $C$. A perpendicular is drawn from point $F$ to the line $(m+1)x + y - 4m - 6 = 0$, with foot of the perpendicular at point $Q$. Then the minimum value of $|PQ| + |PF|$ is?",
    "fact_expressions": "C: Parabola;G: Line;m: Number;P: Point;Q: Point;F: Point;l: Line;Expression(C) = (y^2 = 8*x);Expression(G) = (-4*m + x*(m + 1) + y - 6 = 0);Focus(C) = F;PointOnCurve(P, C);PointOnCurve(F, l);IsPerpendicular(l,G);FootPoint(l,G)=Q",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F)) + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "5-sqrt(2)",
    "fact_spans": "[[[2, 21], [22, 23], [36, 42]], [[53, 74]], [[55, 74]], [[31, 35]], [[81, 84]], [[26, 30], [48, 52]], [], [[2, 21]], [[53, 74]], [[22, 30]], [[31, 46]], [[47, 74]], [[47, 77]], [[47, 84]]]",
    "query_spans": "[[[86, 105]]]",
    "process": "Rewrite the given line $(m+1)x-4m+y-6=0$ as $m(x-4)+x+y-6=0$. When $x=4$, $y=2$, so the line passes through the fixed point $(4,2)$, denoted as point $M$. $\\because$ a perpendicular is drawn from point $F$ to the line $(m+1)x-4m+y-6=0$, with foot of perpendicular at $Q$, $\\therefore FQ\\bot$ the line $(m+1)x-4m+y-6=0$, i.e., $FQ\\bot MQ$, $\\angle FQM=90^{\\circ}$. Thus, the locus of point $Q$ is a circle with diameter $FM$, radius $r=\\sqrt{2}$, and center at the midpoint of $FM$, denoted as point $H$, $\\therefore H(3,1)$. $\\because P$ lies on the parabola $C: y^{2}=8x$, whose directrix is $x=-2$, $\\therefore |PF|$ equals the distance from $P$ to the directrix. Draw a perpendicular from $P$ to the directrix, with foot at $R$. To minimize $|PF|+|PQ|$, i.e., minimize $|PR|+|PQ|$, points $R$, $P$, $Q$ must be collinear, and the line $RQ$ passes through the circle's center $H$. As shown in the figure, at this time $(|PR|+|PQ|)=HR-r=5-\\sqrt{2}$."
  },
  {
    "text": "The line $l$: $x - y + 1 = 0$ intersects the parabola $y = x^{2}$ at points $A$ and $B$. If point $M(1,2)$, then the value of $|M A| \\cdot |M B|$ is?",
    "fact_expressions": "l: Line;G: Parabola;M: Point;A: Point;B: Point;Expression(G) = (y = x^2);Coordinate(M) = (1, 2);Expression(l)=(x-y+1=0);Intersection(l, G) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(M, A))*Abs(LineSegmentOf(M, B))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 16]], [[17, 29]], [[42, 51]], [[31, 34]], [[35, 38]], [[17, 29]], [[42, 51]], [[0, 16]], [[0, 40]]]",
    "query_spans": "[[[53, 75]]]",
    "process": "The parametric equations of the line are \\begin{cases}x=1+\\frac{\\sqrt{2}}{2}t\\\\y=2+\\frac{\\sqrt{2}}{2}t\\end{cases}. Substituting into the parabola and simplifying yields $t^{2}+\\sqrt{2}t-2=0$. Therefore, $t_{1}t_{2}=-2$, $|MA|\\cdot|MB|=|t_{1}|_{2}|=2$"
  },
  {
    "text": "Given the parabola $y^{2}=2 px(p>0)$ has focus $F$, and the intersection point of the directrix and the $x$-axis is $B$. A point $A(x_{0} , 2)$ on the parabola satisfies $|AB|=\\sqrt {2}|AF|$, then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;F: Point;x0:Number;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(A) = (x0, 2);Focus(G) = F;Intersection(Directrix(G), xAxis) = B;PointOnCurve(A,G);Abs(LineSegmentOf(A,B))=sqrt(2)*Abs(LineSegmentOf(A,F))",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 22], [44, 47]], [[88, 91]], [[50, 64]], [[40, 43]], [[26, 29]], [[50, 64]], [[5, 22]], [[2, 22]], [[50, 64]], [[2, 29]], [[2, 43]], [[44, 64]], [[66, 86]]]",
    "query_spans": "[[[88, 93]]]",
    "process": ""
  },
  {
    "text": "Given that $F(2,0)$ is a focus of the hyperbola $C$: $\\frac{x^{2}}{2 m}-\\frac{y^{2}}{2}=1$, then the equations of the asymptotes of $C$ are?",
    "fact_expressions": "C: Hyperbola;m: Number;F: Point;Expression(C) = (-y^2/2 + x^2/(2*m) = 1);Coordinate(F) = (2, 0);OneOf(Focus(C)) = F",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y = pm*x",
    "fact_spans": "[[[11, 56], [63, 66]], [[18, 56]], [[2, 10]], [[11, 56]], [[2, 10]], [[2, 61]]]",
    "query_spans": "[[[63, 74]]]",
    "process": "a^{2}=2m, b^{2}=2, c^{2}=4=a^{2}+b^{2}=2m+2, solving gives m=1, so the hyperbola equation is \\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1, thus the asymptotes are y=\\pm x"
  },
  {
    "text": "It is known that an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ is perpendicular to the line $2x+y-3=0$. Then, the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Line;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(H) = (2*x + y - 3 = 0);IsPerpendicular(OneOf(Asymptote(G)), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 39], [64, 67]], [[5, 39]], [[46, 59]], [[5, 39]], [[2, 39]], [[46, 59]], [[2, 61]]]",
    "query_spans": "[[[64, 73]]]",
    "process": "From the given, the equations of the asymptotes of the hyperbola are $ y = \\pm\\frac{x}{a} $. Since one asymptote of the hyperbola is perpendicular to the line $ 2x + y - 3 = 0 $, $ \\therefore \\frac{1}{a} = \\frac{1}{2} $, $ \\therefore a = 2 $, $ c = \\sqrt{5} $, $ \\therefore $ the eccentricity $ e = \\frac{c}{a} = \\frac{\\sqrt{5}}{2} $."
  },
  {
    "text": "Given that the directrix of the parabola ${y}^{2}=2 p x(p>0)$ is tangent to the circle $x^{2}+y^{2}-6 x-7=0$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;H: Circle;Expression(H) = (-6*x + x^2 + y^2 - 7 = 0);IsTangent(Directrix(G),H) = True",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 25]], [[2, 25]], [[55, 58]], [[5, 25]], [[29, 51]], [[29, 51]], [[2, 53]]]",
    "query_spans": "[[[55, 62]]]",
    "process": ""
  },
  {
    "text": "Given that $B_{1}$, $B_{2}$ are the two endpoints of the minor axis of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and $P$ is any point on the ellipse such that $|P B_{1}| \\leq|B_{1} B_{2}|$, then the range of values for the eccentricity of this ellipse is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;B1: Point;B2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Endpoint(MinorAxis(C))={B1,B2};PointOnCurve(P, C);Abs(LineSegmentOf(P, B1)) <= Abs(LineSegmentOf(B1, B2))",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(0,\\sqrt{2}/2]",
    "fact_spans": "[[[18, 74], [86, 88], [126, 128]], [[24, 74]], [[24, 74]], [[82, 85]], [[2, 9]], [[10, 17]], [[24, 74]], [[24, 74]], [[18, 74]], [[2, 81]], [[82, 93]], [[94, 123]]]",
    "query_spans": "[[[126, 138]]]",
    "process": "P is an arbitrary point on the ellipse, let P(a\\cos\\theta,b\\sin\\theta), \\theta\\in[0,2\\pi), from |PB_{1}|\\leqslant|B_{1}B_{2}|, using the distance formula between two points, we have \\sqrt{(a\\cos\\theta)^{2}+(b\\sin\\theta-b)^{2}}\\leqslant2b, simplifying yields (a^{2}-b^{2})\\sin^{2}\\theta-2b^{2}\\sin\\theta+3b^{2}-a^{2}\\geqslant0, i.e., the quadratic function in terms of \\sin\\theta is non-negative on the interval [-1,1]. Discussing the axis of symmetry of the function, the condition holds when the minimum value of the function is greater than or equal to 0; then using the eccentricity formula gives the result. Solution: According to the problem, set P(a\\cos\\theta,b\\sin\\theta), \\theta\\in[0,2\\pi), from |PB_{1}|\\leqslant|B_{1}B_{2}|, we obtain \\sqrt{(a\\cos\\theta)^{2}+(b\\sin\\theta-b)^{2}}\\leqslant2b, i.e., (a\\cos\\theta)^{2}+(b\\sin\\theta-b)^{2}\\leqslant4b^{2}, simplifying yields: (a^{2}-b^{2})\\sin^{2}\\theta-2b^{2}\\sin\\theta+3b^{2}-a^{2}\\geqslant0, which is a quadratic function in terms of \\sin\\theta on the interval [-1,1]. Since the quadratic function in terms of \\sin\\theta opens upwards, discuss whether the axis of symmetry lies within (-1,1). When 0<\\frac{b^{2}}{a^{2}-b^{2}}<1, the axis of symmetry lies within (-1,1), and when \\sin\\theta=1, (a^{2}-b^{2})\\sin^{2}\\theta-2b^{2}\\sin\\theta+3b^{2}-a^{2}=0, thus the minimum value of this quadratic function is less than 0. When \\frac{b^{2}}{a^{2}-b^{2}}\\geqslant1, the axis of symmetry lies outside (-1,1), and the minimum value of the quadratic function occurs at the endpoints. When \\sin\\theta=1, (a^{2}-b^{2})\\sin^{2}\\theta-2b^{2}\\sin\\theta+3b^{2}-a^{2}=0; when \\sin\\theta=-1, (a^{2}-b^{2})\\sin^{2}\\theta-2b^{2}\\sin\\theta+3b^{2}-a^{2}=4b^{2}>0. At this time, (a^{2}-b^{2})\\sin^{2}\\theta-2b^{2}\\sin\\theta+3b^{2}-a^{2}\\geqslant0 always holds, hence \\frac{b^{2}}{a^{2}-b^{2}}\\geqslant1, i.e., \\frac{a^{2}-c^{2}}{c^{2}}\\geqslant1, so a^{2}\\geqslant2c^{2}, (\\frac{c}{a})^{2}=\\frac{c^{2}}{a^{2}}\\leqslant\\frac{1}{2}, then e=\\frac{c}{a}\\leqslant\\frac{\\sqrt{2}}{2}, i.e., e\\in(0,\\frac{\\sqrt{2}}{2}]"
  },
  {
    "text": "The coordinates of the right focus of the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$ are?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/8 + y^2/4 = 1)",
    "query_expressions": "Coordinate(RightFocus(G))",
    "answer_expressions": "(2,0)",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 45]]]",
    "process": "Using the ellipse equation to find a, b, and then solving for c, the coordinates of the foci can be obtained. For the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$, we get $a=2\\sqrt{2}$, $b=2$, $c=2$. The coordinates of the right focus of the ellipse are: $(2,0)$"
  },
  {
    "text": "Given the hyperbola $C$ with equation $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, its left and right foci are $F_{1}$ and $F_{2}$ respectively. The point $M$ has coordinates $(2,1)$. A point $P(x_{0}, y_{0})$ on hyperbola $C$ ($x_{0}>0, y_{0}>0$) satisfies $\\frac{\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{M F_{1}}}{|\\overrightarrow{P F_{2}}|} = \\frac{\\overrightarrow{F_{2} F_{1}} \\cdot \\overrightarrow{M F_{1}}} {|\\overrightarrow{F_{2} F_{1}}|}$, then $S_{\\Delta P M F_{1}}-S_{\\Delta P M F_{2}}$=?",
    "fact_expressions": "C: Hyperbola;P: Point;F1: Point;M: Point;x0:Number;y0:Number;x0>0;y0>0;F2: Point;Expression(C) = (x^2/4 - y^2/5 = 1);Coordinate(M) = (2, 1);Coordinate(P) = (x0, y0);LeftFocus(C) =F1;RightFocus(C)=F2;DotProduct(VectorOf(P, F1), VectorOf(M, F1))/Abs(VectorOf(P, F2)) = DotProduct(VectorOf(F2, F1), VectorOf(M, F1))/Abs(VectorOf(F2, F1))",
    "query_expressions": "Area(TriangleOf(P, M, F1)) - Area(TriangleOf(P, M, F2))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 8], [48, 49], [89, 95]], [[96, 132]], [[57, 64]], [[75, 79]], [[97, 132]], [[97, 132]], [[97, 132]], [[97, 132]], [[65, 72]], [[2, 47]], [[75, 88]], [[96, 132]], [[48, 72]], [[48, 72]], [[134, 328]]]",
    "query_spans": "[[[330, 375]]]",
    "process": "From the given conditions, we have $F_{1}(-3,0)$, $F_{2}(3,0)$. Also, $\\frac{\\overrightarrow{PF}\\cdot\\overrightarrow{MF}}{|PF_{2}|}=\\frac{\\overrightarrow{F_{2}F}\\cdot\\overrightarrow{MF}}{|\\overrightarrow{F_{2}F_{1}}|}$. Using coordinate operations of vectors, we obtain $\\frac{15+5x_{0}+y_{0}}{\\sqrt{x_{0}^{2}+6x+9+y_{0}^{2}}}=5$, which implies $y_{0}=\\frac{5}{12}x_{0}+\\frac{15}{12}$. Since $P$ lies on the hyperbola, substitute $y_{0}=\\frac{5}{12}x_{0}+\\frac{15}{12}$ into the hyperbola equation $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, solving gives $x_{0}=3$ or $x_{0}=-\\frac{63}{31}$ (discarded). Thus, $(3,\\frac{5}{2})$, so the equation of line $PF_{1}$ is $5x-12y+15=0$. Therefore, the distance from point $M$ to line $PF$ is $d=\\frac{|5\\times2-12+15|}{\\sqrt{5^{2}+(-12)^{2}}}=1$. It is easy to see that the distances from point $M$ to the $x$-axis and line $PF_{2}$ are both 1. Hence, point $M$ is the incenter of $\\triangle PF_{1}F_{2}$. Therefore, $S_{APMF_{1}}-S_{APMF_{2}}=\\frac{1}{2}(|\\overrightarrow{PF_{1}}|-|\\overrightarrow{PF_{2}}|)\\times1=\\frac{1}{2}\\times4\\times1=2$. Thus, the answer should be $2$."
  },
  {
    "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ intersect the circle $x^{2}+y^{2}-4x+2=0$, then the range of the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (-4*x + x^2 + y^2 + 2 = 0);IsIntersect(Asymptote(G),H)",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1,\\sqrt{2}]",
    "fact_spans": "[[[2, 58], [92, 95]], [[5, 58]], [[5, 58]], [[63, 85]], [[5, 58]], [[5, 58]], [[2, 58]], [[63, 85]], [[2, 89]]]",
    "query_spans": "[[[92, 106]]]",
    "process": ""
  },
  {
    "text": "The distance from point $M$ to $F(0,-2)$ is $1$ less than its distance to the line $l$: $y-3=0$. The equation of the trajectory of point $M$ is?",
    "fact_expressions": "l: Line;F: Point;M: Point;Expression(l) = (y - 3 = 0);Coordinate(F) = (0, -2);Distance(M, l) - Distance(M, F) = 1",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x**2=-8*y",
    "fact_spans": "[[[20, 34]], [[5, 14]], [[0, 4], [18, 19], [42, 46]], [[20, 34]], [[5, 14]], [[0, 41]]]",
    "query_spans": "[[[42, 53]]]",
    "process": "Since the distance from point M to F(0,-2) is 1 less than its distance to the line l: y-3=0, the distance from point M to F(0,-2) equals the distance to the line y=2. By the definition of a parabola, \\frac{p}{2}=2, so p=4. Therefore, the trajectory equation of point M is the parabola x^{2}=-8y. Hence, the answer is x^{2}=-8y."
  },
  {
    "text": "An ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ has a focus $F_{1}$, $M$ is a point on the ellipse such that $|M F_{1}|=2$, and $N$ is the midpoint of segment $M F_{1}$. Then the length of $O N$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);F1: Point;OneOf(Focus(G))=F1;M: Point;PointOnCurve(M,G);Abs(LineSegmentOf(M, F1)) = 2;O: Origin;N: Point;MidPoint(LineSegmentOf(M,F1)) = N",
    "query_expressions": "Length(LineSegmentOf(O,N))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 38], [56, 58]], [[0, 38]], [[44, 51]], [[0, 51]], [[52, 55]], [[52, 61]], [[63, 77]], [[99, 104]], [[79, 82]], [[79, 97]]]",
    "query_spans": "[[[99, 108]]]",
    "process": "Draw the figure. According to the geometric relationship in the figure, ON is the midline of \\triangle F_{1}MF_{2}, so |ON|=\\frac{1}{2}|MF_{2}|, which can be used to solve the problem. Draw the figure as described. Since N is the midpoint of MF_{1}, ON is the midline of side MF_{2} in \\triangle F_{1}MF_{2}, that is, |ON|=\\frac{1}{2}|MF_{2}|. From the standard equation of the ellipse, we know a^{2}=25, a=5, |MF|+|MF_{2}|=2a=10, |MF|=2, |MF_{2}|=8, so ON=4. The answer is 4."
  },
  {
    "text": "Given that $F$ is the left focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $P$ is a point on the ellipse, $PF \\perp x$-axis, and $OP \\| AB$ ($O$ is the origin, $A$ is the right vertex, $B$ is the upper vertex), then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F: Point;A:Point;B:Point;a>0;b>0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P,F),xAxis);IsParallel(LineSegmentOf(O,P),LineSegmentOf(A,B));RightVertex(G)=A;UpperVertex(G)=B;O:Origin",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[6, 61], [70, 72], [128, 130]], [[8, 61]], [[8, 61]], [[66, 69]], [[2, 5]], [[109, 112]], [[117, 120]], [[8, 61]], [[8, 61]], [[6, 61]], [[2, 65]], [[66, 76]], [[77, 90]], [[91, 101]], [[70, 116]], [[70, 124]], [[102, 105]]]",
    "query_spans": "[[[128, 136]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{m}=1$ is $\\frac{1}{2}$, then $m=$?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/12 + y^2/m = 1);Eccentricity(G) = 1/2",
    "query_expressions": "m",
    "answer_expressions": "9, 16",
    "fact_spans": "[[[0, 38]], [[58, 61]], [[0, 38]], [[0, 56]]]",
    "query_spans": "[[[58, 63]]]",
    "process": "When the ellipse foci are on the x-axis, \\begin{cases}a^{2}=12\\\\b^{2}=m\\\\\\frac{c}{a}=\\frac{1}{2}\\end{cases}, solving gives when the ellipse foci are on the y-axis, \\begin{cases}a^{2}=b^{2}+c^{2}\\\\b^{2}=12\\\\c=\\frac{1}{2}\\end{cases}, solving gives m=16."
  },
  {
    "text": "Given that the distance from a vertex of the hyperbola $9 y^{2}-m^{2} x^{2}=1$ $(m>0)$ to one of its asymptotes is $\\frac{1}{5}$, find $m=?$",
    "fact_expressions": "G: Hyperbola;m: Number;m>0;Expression(G) = (-m^2*x^2 + 9*y^2 = 1);Distance(OneOf(Vertex(G)),OneOf(Asymptote(G)))=1/5",
    "query_expressions": "m",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 33], [39, 40]], [[65, 68]], [[5, 33]], [[2, 33]], [[2, 63]]]",
    "query_spans": "[[[65, 70]]]",
    "process": "First, obtain $a$ and $b$ from the hyperbola equation, then derive the asymptote equations and the coordinates of the fixed point. Given that the distance from the fixed point to the asymptote equals $\\frac{1}{5}$, solve for $m$. Detailed solution: From the hyperbola equation, we have $a = \\frac{1}{3}$, $b = \\frac{1}{m}$, so the asymptotes are $y = \\pm\\frac{m}{3}x$. Take the positive one: $\\frac{m}{3}x - y = 0$, and the vertex is $(0, \\frac{1}{3})$. Then the distance is $\\frac{|0 - \\frac{1}{3}|}{\\sqrt{(\\frac{m}{3})^{2} + 1}} = \\frac{1}{5}$. Solving gives $m^{2} = 16$, $\\therefore m = 4$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, the upper and lower vertices are $B_{1}$ and $B_{2}$ respectively. Point $C$ is the midpoint of $B_{1} F_{2}$. If $\\overrightarrow{B_{1} F_{1}} \\cdot \\overrightarrow{B_{1} F_{2}}=2$ and $C F_{1} \\perp B_{1} F_{2}$, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;B1: Point;B2: Point;UpperVertex(G) = B1;LowerVertex(G) = B2;C: Point;MidPoint(LineSegmentOf(B1, F2)) = C;DotProduct(VectorOf(B1, F1), VectorOf(B1, F2)) = 2;IsPerpendicular(LineSegmentOf(C, F1), LineSegmentOf(B1, F2))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4+y^2/3=1",
    "fact_spans": "[[[2, 54], [224, 226]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[63, 70]], [[71, 78]], [[2, 78]], [[2, 78]], [[87, 94]], [[95, 102]], [[2, 102]], [[2, 102]], [[103, 107]], [[103, 124]], [[126, 193]], [[195, 222]]]",
    "query_spans": "[[[224, 231]]]",
    "process": "From the given conditions, we have $F_{1}(-c,0)$, $F_{2}(c,0)$, $B_{1}(0,b)$, $B_{2}(0,-b)$, $C\\left(\\frac{c}{2},\\frac{b}{2}\\right)$, $\\overrightarrow{B_{1}F_{1}}\\cdot\\overrightarrow{B_{1}F_{2}}=(-c,-b)\\cdot(c,-b)=-c^{2}+b^{2}=2$①, $\\overrightarrow{CF_{1}}\\bot\\overrightarrow{B_{1}F_{2}}$, so $\\overrightarrow{CF_{1}}\\cdot\\overrightarrow{B_{1}F_{2}}=0$, which gives $\\left(-\\frac{3c}{2},-\\frac{b}{2}\\right)\\cdot(c,-b)=-\\frac{3}{2}c^{2}+\\frac{b^{2}}{2}=0$②. Solving these equations yields $c=1$, $b=\\sqrt{3}$, $a=\\sqrt{b^{2}+c^{2}}=2$, thus the equation of the ellipse is $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$."
  },
  {
    "text": "Let point $P(0, \\frac{9}{2})$, and moving points $A$, $B$ lie on the ellipse $\\frac{x^{2}}{18}+\\frac{y^{2}}{9}=1$ and satisfy $\\overrightarrow{P A}=\\lambda \\overrightarrow{P B}$. Find the range of $\\lambda$?",
    "fact_expressions": "P: Point;Coordinate(P) = (0, 9/2);G: Ellipse;Expression(G) = (x^2/18 + y^2/9 = 1);A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);lambda: Number;VectorOf(P, A) = lambda*VectorOf(P, B)",
    "query_expressions": "Range(lambda)",
    "answer_expressions": "[1/5, 5]",
    "fact_spans": "[[[1, 21]], [[1, 21]], [[32, 70]], [[32, 70]], [[24, 27]], [[28, 31]], [[24, 71]], [[28, 71]], [[127, 136]], [[74, 125]]]",
    "query_spans": "[[[127, 140]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Discuss in two cases depending on whether the slope of line $ AB $ exists. Combine the line equation $ y = kx + \\frac{9}{2} $ with the ellipse equation. Using the relationship between roots and coefficients, we obtain:  \n$ (1+\\lambda)x_{2} = \\frac{-18k}{2k^{2}+1} $,  \n$ x_{1}x_{2} = \\lambda x_{2}^{2} = \\frac{45}{2(2k^{2}+1)} $.  \nSimplifying yields $ \\frac{(1+\\lambda)^{2}}{\\lambda} = \\frac{72k^{2}}{5(2k^{2}+1)} $. Establish an inequality for $ \\lambda $ based on the monotonicity of the function on the right-hand side; solving it gives the range of real values for $ \\lambda $.  \n\nLet $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, $ P(0,\\frac{9}{2}) $. Since $ \\overrightarrow{PA} = \\lambda \\overrightarrow{PB} $, we have  \n$ (x_{1}, y_{1} - \\frac{9}{2}) = \\lambda (x_{2}, y_{2} - \\frac{9}{2}) $,  \nthus $ x_{1} = \\lambda x_{2} $.  \n\nWhen the slope of line $ AB $ does not exist, the equation of line $ AB $ is $ x = 0 $. Then $ A(0,3) $, $ B(0,-3) $ or $ A(0,-3) $, $ B(0,3) $.  \nIf $ A(0,3) $, $ B(0,-3) $, then $ \\lambda = \\frac{1}{5} $;  \nif $ A(0,-3) $, $ B(0,3) $, then $ \\lambda = 5 $.  \n\nWhen the slope of line $ AB $ exists, let the equation of line $ AB $ be $ y = kx + \\frac{9}{2} $. Substitute into the ellipse $ \\frac{x^{2}}{18} + \\frac{y^{2}}{9} = 1 $ and eliminate $ y $, obtaining  \n$ (2k^{2}+1)x^{2} + 18kx + \\frac{45}{2} = 0 $,  \n$ x_{1} + x_{2} = (1+\\lambda)x_{2} = \\frac{-18k}{2k^{2}+1} $, \\quad (1);  \n$ x_{1}x_{2} = \\lambda x_{2}^{2} = \\frac{45}{2(2k^{2}+1)} $. \\quad (2)  \n\nEliminate $ x_{2} $ from (1) and (2), yielding  \n$ \\frac{(1+\\lambda)^{2}}{\\lambda} = \\frac{72k^{2}}{5(2k^{2}+1)} $.  \n\nSince the equation $ (2k^{2}+1)x^{2} + 18kx + \\frac{45}{2} = 0 $ has real roots,  \n$ \\Delta = (18k)^{2} - 4(2k^{2}+1) \\times \\frac{45}{2} \\geqslant 0 $,  \nsimplifying to $ k^{2} \\geqslant \\frac{5}{8} $.  \n\nBecause $ F(k^{2}) = \\frac{72k^{2}}{5(2k^{2}+1)} = \\frac{36}{5} - \\frac{36}{5(2k^{2}+1)} $ is an increasing function of $ k^{2} $,  \nwe have $ 4 \\leqslant F(k^{2}) < \\frac{36}{5} $, thus  \n$ 4 \\leqslant \\frac{(1+\\lambda)^{2}}{\\lambda} < \\frac{36}{5} $. Solving gives $ \\frac{1}{5} < \\lambda < 5 $.  \n\nCombining all cases, the range of real values for $ \\lambda $ is $ [\\frac{1}{5}, 5] $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{3}+y^{2}=1$, then $|F_{1} F_{2}|$=?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Focus(G) = {F1, F2};Expression(G) = (x^2/3 + y^2 = 1)",
    "query_expressions": "Abs(LineSegmentOf(F1, F2))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[18, 45]], [[2, 9]], [[10, 17]], [[2, 50]], [[18, 45]]]",
    "query_spans": "[[[52, 69]]]",
    "process": "The ellipse $\\frac{x^2}{3}+y^{2}=1$ has $a=\\sqrt{3}, b=1, \\therefore c=\\sqrt{a^{2}-b^{2}}=\\sqrt{2}$. Thus, $|F_{1}F_{2}|=2\\sqrt{2}$."
  },
  {
    "text": "The line $l$ passing through the point $M\\left(\\frac{1}{2}, 0\\right)$ intersects the parabola $y^{2} = 2x$ at points $A$ and $B$, and $C(2,0)$. Then, the minimum area of $\\triangle ABC$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (1/2, 0);l: Line;PointOnCurve(M, l);G: Parabola;Expression(G) = (y^2 = 2*x);A: Point;B: Point;Intersection(l, G) = {A, B};C: Point;Coordinate(C) = (2, 0)",
    "query_expressions": "Min(Area(TriangleOf(A, B, C)))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[1, 21]], [[1, 21]], [[22, 27]], [[0, 27]], [[28, 42]], [[28, 42]], [[44, 47]], [[48, 51]], [[22, 53]], [[54, 62]], [[54, 62]]]",
    "query_spans": "[[[64, 89]]]",
    "process": "Let the equation of line $ l $ be: $ x = ty + \\frac{1}{2} $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Solving the system of equations  \n\\[\n\\begin{cases}\nx = ty + \\frac{1}{2} \\\\\ny^2 = 2x\n\\end{cases}\n\\]\nby eliminating $ x $, we get: $ y^2 - 2ty - 1 = 0 $,  \n$ \\therefore y_{1} + y_{2} = 2t $, $ y_{1}y_{2} = -1 $.  \n$ \\therefore S_{\\Delta ABC} = \\frac{1}{2} \\times (2 - \\frac{1}{2}) \\times |y_{1} - y_{2}| = \\frac{3}{4} \\times \\sqrt{(y_{1} + y_{2})^{2} - 4y_{1}y_{2}} = \\frac{3}{4} \\times \\sqrt{4t^{2} + 4} $.  \nWhen $ t = 0 $, $ S_{\\triangle ABC} $ reaches its minimum value, which is $ \\frac{3}{2} $."
  },
  {
    "text": "A point $P$ on the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ is at a distance of $9$ from one of its foci. What is the distance from point $P$ to the other focus?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 - y^2/16 = 1);PointOnCurve(P, G);OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2);Distance(P, F1) = 9",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "{3, 15}",
    "fact_spans": "[[[0, 39], [46, 47]], [[42, 45], [63, 67]], [], [], [[0, 39]], [[0, 45]], [[46, 52]], [[46, 73]], [[46, 73]], [[42, 60]]]",
    "query_spans": "[[[46, 79]]]",
    "process": "\\because the standard equation of the hyperbola is \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1, \\therefore a=3, c=5. Let the distance from point P to the other focus be x. \\because the distance from a point P on the hyperbola to one of its foci is equal to 9, \\therefore by the definition of a hyperbola: |x-9|=6, solving gives x=15, or x=3. \\because 3>c-a=2, \\therefore 3>c-a=2, the distance from point P to the other focus is 15 or 3"
  },
  {
    "text": "Given that the center of the ellipse is at the origin, one focus coincides with the focus of the parabola $y^{2}=8x$, and the coordinates of one vertex are $(0,2)$, then the equation of this ellipse is?",
    "fact_expressions": "G: Parabola;H: Ellipse;O: Origin;Expression(G) = (y^2 = 8*x);Center(H) = O;OneOf(Focus(H)) = Focus(G);Coordinate(OneOf(Vertex(H)))=(0,2)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2/8+y^2/4=1",
    "fact_spans": "[[[16, 30]], [[2, 4], [54, 56]], [[8, 10]], [[16, 30]], [[16, 30]], [[2, 35]], [[2, 51]]]",
    "query_spans": "[[[54, 60]]]",
    "process": "The equation of this ellipse is the standard form. The focus of the parabola is (2,0), indicating that the foci of the ellipse lie on the x-axis and c=2. The vertex coordinates are (0,2), indicating b=2, thus a^{2}=b^{2}+c^{2}=8. Therefore, the equation of the ellipse is \\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1"
  },
  {
    "text": "Given that the hyperbola passes through the point $(2,3)$ and has asymptotes $y=\\pm \\sqrt{3} x$, then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (2, 3);PointOnCurve(H, G);Expression(Asymptote(G)) = (y = pm*(sqrt(3)*x))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[2, 5], [42, 45]], [[6, 14]], [[6, 14]], [[2, 14]], [[2, 39]]]",
    "query_spans": "[[[42, 52]]]",
    "process": "Since the asymptotes are given by $ y = \\pm\\sqrt{3}x $, assume the hyperbola equation is $ 3x^{2} - y^{2} = \\lambda $. Since the hyperbola passes through the point $ (2,3) $, we have $ \\lambda = 3\\times4 - 9 = 3 $. Therefore, the standard equation of the hyperbola is $ x^{2} - \\frac{y^{2}}{3} = 1 $."
  },
  {
    "text": "Given that the foci of the hyperbola lie on the $x$-axis and its asymptotes are given by $\\sqrt{3} x \\pm y=0$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), xAxis);Expression(Asymptote(G)) = (sqrt(3)*x + pm*y = 0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 5], [45, 48], [15, 16]], [[2, 14]], [[15, 42]]]",
    "query_spans": "[[[45, 54]]]",
    "process": "\\because the foci are on the x-axis of the hyperbola, \\therefore let the standard equation of the hyperbola be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1, a>0, b>0. Since the asymptotes of the hyperbola are given by y=\\pm\\sqrt{3}x, \\therefore \\frac{b}{a}=\\sqrt{3}, \\therefore b^{2}=3a^{2}, \\therefore c^{2}-a^{2}=3a^{2}, e^{2}=(\\frac{c}{a})^{2}=4, so the eccentricity of this hyperbola is e=2"
  },
  {
    "text": "Given the ellipse $C$: $x^{2}+\\frac{y^{2}}{b^{2}}=1$ $(0<b<1)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through $F_{2}$ intersects the ellipse $C$ at points $A$ and $B$. Then, the perimeter of $\\triangle A B F_{1}$ is?",
    "fact_expressions": "l: Line;C: Ellipse;b: Number;A: Point;B: Point;F1: Point;F2: Point;0 < b;b < 1;Expression(C) = (x^2 + y^2/b^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F2, l);Intersection(l, C) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, B, F1))",
    "answer_expressions": "4",
    "fact_spans": "[[[79, 84]], [[2, 45], [85, 90]], [[9, 45]], [[91, 94]], [[95, 98]], [[54, 61]], [[62, 69], [71, 78]], [[9, 45]], [[9, 45]], [[2, 45]], [[2, 69]], [[2, 69]], [[70, 84]], [[79, 100]]]",
    "query_spans": "[[[102, 128]]]",
    "process": "From the ellipse $ C: x^{2} + \\frac{y^{2}}{b^{2}} = 1 $, we know $ a^{2} = 1 $, that is, $ a = 1 $. Since the line passes through $ F_{2} $ and intersects the ellipse $ C $ at points $ A $ and $ B $, we have $ |AF_{2}| + |AF| = 2a = 2 $, $ |BF_{2}| + |BF_{1}| = 2a = 2 $. Therefore, the perimeter of $ \\triangle ABF_{1} $ is $ |AB| + |BF_{1}| + |AF_{1}| = 4 $. (Note: This problem mainly examines the equation of an ellipse and the definition of an ellipse, and belongs to a medium-level question."
  },
  {
    "text": "Given the parabola $C$: $x^{2}=2 p y(p>0)$, point $A(x_{0}, \\frac{p}{2})$ lies on $C$, point $B$ has coordinates $(0,-\\frac{p}{2})$, and $|A B|=5 \\sqrt{2}$. Then the focus coordinates of $C$ are?",
    "fact_expressions": "C: Parabola;p: Number;A: Point;B: Point;p>0;x0:Number;Expression(C) = (x^2 = 2*p*y);Coordinate(B) = (0, -p/2);Coordinate(A) = (x0, p/2);PointOnCurve(A, C);Abs(LineSegmentOf(A, B)) = 5*sqrt(2)",
    "query_expressions": "Coordinate(Focus(C))",
    "answer_expressions": "(0,5/2)",
    "fact_spans": "[[[2, 28], [54, 57], [107, 110]], [[10, 28]], [[29, 53]], [[59, 63]], [[10, 28]], [[30, 53]], [[2, 28]], [[59, 85]], [[29, 53]], [[29, 58]], [[87, 105]]]",
    "query_spans": "[[[107, 117]]]",
    "process": "Since $ A(x_{0},\\frac{p}{2}) $ lies on $ C $, we have $ x_{0}^{2}=2p\\cdot\\frac{p}{2} \\Rightarrow p^{2}=x_{0}^{2} $. Also, since $ |AB|=5\\sqrt{2} $, it follows that $ \\sqrt{(x_{0}-0)^{2}+(\\frac{p}{2}+\\frac{p}{2})^{2}}=5\\sqrt{2} \\Rightarrow p^{2}=25 $. Since $ p>0 $, we have $ p=5 $. Therefore, the equation of the parabola is $ x^{2}=10y $, and thus the focus of $ C $ has coordinates $ (0,\\frac{5}{2}) $."
  },
  {
    "text": "The coordinates of the focus of the parabola $y^{2}=\\frac{1}{4} x$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = x/4)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(1/16, 0)",
    "fact_spans": "[[[0, 24]], [[0, 24]]]",
    "query_spans": "[[[0, 31]]]",
    "process": "Since the focus of the parabola \\( y^{2} = 2px \\) is \\( \\left( \\frac{p}{2}, 0 \\right) \\), the focus of the parabola \\( y^{2} = \\frac{1}{4}x \\) is \\( \\left( \\frac{1}{16}, \\right. \\)"
  },
  {
    "text": "Given a point $A(m, 4)$ on the parabola $C$: $x^{2}=2 p y(p>0)$ whose distance to the focus is $5$, then $m=$?",
    "fact_expressions": "C: Parabola;p: Number;A: Point;p>0;Expression(C) = (x^2 = 2*p*y);Coordinate(A) = (m, 4);PointOnCurve(A, C);Distance(A, Focus(C)) = 5;m:Number",
    "query_expressions": "m",
    "answer_expressions": "pm*4",
    "fact_spans": "[[[2, 28], [41, 42]], [[10, 28]], [[31, 40]], [[10, 28]], [[2, 28]], [[31, 40]], [[2, 40]], [[31, 51]], [[53, 56]]]",
    "query_spans": "[[[53, 58]]]",
    "process": "The vertex of the parabola is at the origin, and the focus lies on the y-axis. A point on the parabola is (m,4). Therefore, let the equation of the parabola be: $x^{2}=2py$ $(p>0)$. Thus, its directrix equation is: $y=-\\frac{p}{2}$. Since the distance from a point $P(m,4)$ on the parabola to the focus $F$ is equal to 5, by the definition of a parabola, we have: $|PF|=\\frac{p}{2}+4=5$. Hence, $p=2$. Therefore, the required equation of the parabola is $x^{2}=4y$. Substituting $x=m$ into the equation gives $m=\\pm4$."
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, respectively, $P$ is a point on the right branch of the hyperbola, $I$ is the incenter of $\\Delta P F_{1} F_{2}$, and $S_{\\triangle I P F_{1}}=S_{\\triangle I P F_{2}}-\\lambda S_{\\Delta I  F_{1} F_{2}}$. Then $\\lambda=?$",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2:Point;I: Point;lambda:Number;Expression(G) = (x^2/16 - y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));Incenter(TriangleOf(P,F1,F2)) = I;Area(TriangleOf(I,P,F1)) =Area(TriangleOf(I,P,F2))- lambda*Area(TriangleOf(I,F1,F2))",
    "query_expressions": "lambda",
    "answer_expressions": "-4/5",
    "fact_spans": "[[[19, 58], [69, 72]], [[65, 68]], [[0, 7]], [[8, 16]], [[78, 81]], [[194, 203]], [[19, 58]], [[0, 64]], [[0, 64]], [[65, 77]], [[78, 107]], [[109, 192]]]",
    "query_spans": "[[[194, 205]]]",
    "process": ""
  },
  {
    "text": "Given fixed circles $C_{1}$: $(x+5)^{2}+y^{2}=1$ and $C_{2}$: $(x-5)^{2}+y^{2}=225$, a moving circle $C_0$ is externally tangent to $C_{1}$ and internally tangent to $C_{2}$. Find the trajectory equation of the center $C$ of the moving circle.",
    "fact_expressions": "C0: Circle;C1: Circle;C2: Circle;C: Point;Expression(C1) = ((x + 5)^2 + y^2 = 1);Expression(C2) = ((x - 5)^2 + y^2 =225);IsOutTangent(C0, C1);IsInTangent(C0, C2);Center(C0) = C",
    "query_expressions": "LocusEquation(C)",
    "answer_expressions": "x^2/64 + y^2/39 = 1",
    "fact_spans": "[[[63, 69], [94, 96]], [[2, 31], [72, 79]], [[33, 62], [83, 90]], [[98, 101]], [[2, 31]], [[33, 62]], [[63, 81]], [[63, 92]], [[94, 101]]]",
    "query_spans": "[[[98, 108]]]",
    "process": "From circle $ C_{1}: B $, we obtain center $ B $, radius $ B $. From circle $ C_{2}: $ we obtain center $ \\otimes $, radius $ \\otimes $. Let the radius of circle $ C $ be $ r $. Since moving circle $ C $ is externally tangent to circle $ C_{1} $ and internally tangent to circle $ C_{2} $, then $ \\text{so}, \\text{so} $, hence $ B $. Therefore, the trajectory of point $ C $ is an ellipse with foci at $ $ and $ $, and $ \\text{region} $. Thus, $ a = 8 $, $ c = 5 $, $ B $. Hence, the trajectory equation of the center $ C $ of the moving circle is: $ \\otimes $"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with asymptotes given by $y=\\pm \\frac{1}{2} x$, and $F_{1}(-\\sqrt{5}, 0)$, $F_{2}(\\sqrt{5}, 0)$ being the left and right foci of $C$ respectively. If a moving point $P$ lies on the right branch of $C$, then the minimum value of $\\frac{|P F_{1}|^{2}}{|P F_{2}|}$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;F1: Point;F2: Point;P: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-sqrt(5), 0);Coordinate(F2) = (sqrt(5), 0);Expression(Asymptote(C)) = (y = pm*(x/2));LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, RightPart(C))",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F1))^2/Abs(LineSegmentOf(P, F2)))",
    "answer_expressions": "16",
    "fact_spans": "[[[2, 63], [141, 144], [158, 161]], [[10, 63]], [[10, 63]], [[94, 115]], [[118, 138]], [[154, 157]], [[10, 63]], [[10, 63]], [[2, 63]], [[94, 115]], [[118, 138]], [[2, 92]], [[94, 150]], [[94, 150]], [[154, 165]]]",
    "query_spans": "[[[167, 206]]]",
    "process": "Since the asymptotes of the hyperbola are given by $ y = \\pm\\frac{1}{2}x $, and the coordinates of the foci are $ F_{1}(-\\sqrt{5},0) $, $ F_{2}(\\sqrt{5},0) $, it follows that $ \\frac{b}{a} = \\frac{1}{2} $, $ c = \\sqrt{5} $, so $ a = 2 $, $ b = 1 $. Thus, the equation of the hyperbola is $ \\frac{x^{2}}{4} - y^{2} = 1 $. Let $ |PF_{2}| = m $, then $ |PF_{1}| = 4 + m $, and $ m \\geqslant \\sqrt{5} - 2 $. $ \\frac{F_{1}^{2}}{F_{1}} = \\frac{(4+m)^{2}}{m} = \\frac{16}{m} + m + 8 \\geqslant 2\\sqrt{\\frac{16}{m} \\cdot m} + 8 = 16 $, with equality holding if and only if $ \\frac{16}{m} = m $, i.e., $ m = 4 $. Therefore, the minimum value of $ \\frac{|PF_{1}|^{2}}{|PF_{2}|} $ is $ 16 $."
  },
  {
    "text": "Draw a line $l$ with slope $-\\frac{1}{3}$ passing through the point $M(1,1)$. The line $l$ intersects the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ at points $A$ and $B$. If $\\overrightarrow{A M}=\\overrightarrow{M B}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "l: Line;G: Ellipse;a: Number;b: Number;M: Point;A: Point;B: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(M) = (1, 1);PointOnCurve(M, l);Slope(l) = -1/3;Intersection(l, G) = {A, B};VectorOf(A, M) = VectorOf(M, B)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[29, 34], [35, 39]], [[40, 92], [151, 153]], [[42, 92]], [[42, 92]], [[1, 10]], [[95, 98]], [[99, 102]], [[42, 92]], [[42, 92]], [[40, 92]], [[1, 10]], [[0, 34]], [[11, 34]], [[36, 104]], [[106, 149]]]",
    "query_spans": "[[[151, 159]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Using the point difference method, we obtain  \n\\begin{cases}\\frac{x_{1}^{2}}{a^{2}}+\\frac{y_{1}^{2}}{b^{2}}=1\\\\\\frac{x_{2}^{2}}{a^{2}}+\\frac{y_{2}^{2}}{b^{2}}=1\\end{cases}  \n\\therefore \\frac{(x_{1}-x_{2})(x_{1}+x_{2})}{a^{2}} + \\frac{(y_{1}-y_{2})(y_{1}+y_{2})}{b^{2}} = 0.  \nSince \\overrightarrow{AM}=\\overrightarrow{MB}, M is the midpoint of AB,  \n\\begin{cases}x_{1}+x_{2}=2\\\\y_{1}+y_{2}=2\\end{cases}.  \nThe slope of line l is \\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = -\\frac{1}{3},  \nso \\frac{2}{a^{2}} - \\frac{2}{3b^{2}} = 0  \n\\therefore \\frac{b^{2}}{a^{2}} = \\frac{1}{3}  \n\\therefore e = \\sqrt{1 - \\frac{b^{2}}{a^{2}}} = \\frac{\\sqrt{6}}{3}"
  },
  {
    "text": "On the hyperbola $4 x^{2}-y^{2}+64=0$, the distance from a point $P$ to one of its foci is equal to $1$. Then, the distance from point $P$ to the other focus is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (4*x^2 - y^2 + 64 = 0);P: Point;PointOnCurve(P, G);OneOf(Focus(G)) = D1;OneOf(Focus(G)) = D2;Negation(D1 = D2);D1: Point;D2: Point;Distance(P, D1) = 1",
    "query_expressions": "Distance(P, D2)",
    "answer_expressions": "17",
    "fact_spans": "[[[0, 23], [30, 31]], [[0, 23]], [[26, 29], [46, 50]], [[0, 29]], [[30, 36]], [[30, 56]], [[30, 56]], [], [], [[26, 44]]]",
    "query_spans": "[[[30, 62]]]",
    "process": ""
  },
  {
    "text": "Given that the right focus of a hyperbola centered at the origin has coordinates $(\\sqrt{2}, 0)$, and its two asymptotes are perpendicular to each other, then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;O: Origin;Center(G) = O;Coordinate(RightFocus(G)) = (sqrt(2), 0);Z1: Line;Z2: Line;Asymptote(G) = {Z1, Z2};IsPerpendicular(Z1, Z2)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 = 1",
    "fact_spans": "[[[8, 11], [49, 52]], [[5, 7]], [[2, 11]], [[8, 33]], [], [], [[8, 41]], [[8, 45]]]",
    "query_spans": "[[[49, 59]]]",
    "process": "From the two asymptotes being perpendicular to each other, derive the equations of the asymptotes, that is, find the value of \\frac{b}{a}, and solve using the focus coordinates. According to the problem, the hyperbola's foci lie on the x-axis; let the hyperbola equation be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1, (a>0,b>0). Since the two asymptotes are perpendicular, -\\frac{b}{a}\\cdot\\frac{b}{a}=-1, thus a=b. Also, since the right focus is at (\\sqrt{2},0), it follows that a^{2}+b^{2}=2. Solving gives a=b=1. Therefore, the standard equation of the hyperbola is: x^{2}-y^{2}=1"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1$, what are the coordinates of the right focus of $C$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/6 - y^2/3 = 1)",
    "query_expressions": "Coordinate(RightFocus(C))",
    "answer_expressions": "(3,0)",
    "fact_spans": "[[[2, 45], [48, 51]], [[2, 45]]]",
    "query_spans": "[[[48, 60]]]",
    "process": "According to the hyperbola equation, directly find the coordinates of the foci. From the hyperbola equation, we know $a^{2}=6$, $b^{2}=3$, then $c^{2}=a^{2}+b^{2}=9$, so $c=3$, and the foci lie on the x-axis. The coordinates of the right focus of the hyperbola are $(3,0)$."
  },
  {
    "text": "If the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4 a-2}=1$ is $\\sqrt{3}$, then the value of the real number $a$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/(4*a - 2) + x^2/a^2 = 1);a: Real;Eccentricity(G) = sqrt(3)",
    "query_expressions": "a",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 47]], [[1, 47]], [[64, 69]], [[1, 62]]]",
    "query_spans": "[[[64, 73]]]",
    "process": "Since $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4a-2}=1$ represents a hyperbola, $4a-2>0$, and $b^{2}=4a-2$, $c=\\sqrt{a^{2}+4a-2}$, so $e=\\frac{c}{a}=\\frac{\\sqrt{a^{2}+4a-2}}{a}=\\sqrt{3}$. Solving gives $a=1$."
  },
  {
    "text": "If a point $P$ on the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ is at a distance of $8$ from the right focus, then what is the distance from $P$ to the left directrix?",
    "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2/4 - y^2/5 = 1);PointOnCurve(P, G) ;Distance(P, RightFocus(G)) = 8",
    "query_expressions": "Distance(P, LeftDirectrix(G))",
    "answer_expressions": "{8/3, 8}",
    "fact_spans": "[[[1, 39]], [[42, 45], [58, 61]], [[1, 39]], [[1, 45]], [[1, 56]]]",
    "query_spans": "[[[1, 70]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the tangent line to the parabola $y=x^{2}+x+1$ at the point $(0 , 1)$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2 + x + 1);H: Point;Coordinate(H) = (0, 1)",
    "query_expressions": "Expression(TangentOnPoint(H, G))",
    "answer_expressions": "{(x-y+1=0),(y=x+1)}",
    "fact_spans": "[[[0, 16]], [[0, 16]], [[17, 27]], [[17, 27]]]",
    "query_spans": "[[[0, 35]]]",
    "process": ""
  },
  {
    "text": "Given line $l_{1}$: $4x - 3y + 12 = 0$ and line $l_{2}$: $x = -1$, what is the minimum value of the sum of distances from a moving point $P$ on the parabola $y^{2} = 4x$ to line $l_{1}$ and line $l_{2}$?",
    "fact_expressions": "l1: Line;Expression(l1) = (4*x - 3*y + 12 = 0);l2: Line;Expression(l2) = (x = -1);G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G) = True",
    "query_expressions": "Min(Distance(P, l1) + Distance(P, l2))",
    "answer_expressions": "16/5",
    "fact_spans": "[[[2, 27], [69, 78]], [[2, 27]], [[28, 45], [79, 88]], [[28, 45]], [[47, 61]], [[47, 61]], [[65, 68]], [[47, 68]]]",
    "query_spans": "[[[65, 98]]]",
    "process": "Draw the graph, and according to the definition and properties of the parabola, transform the sum of distances into the minimum value of the sum of the distance from moving point P to the line $ l_{1} $ and the focus. By combining numerical and geometric reasoning, the minimum distance is the distance from the focus F to the line $ l_{1} $. The graph is shown below: the sum of distances to line $ l_{2} $ is $ PA + PB = PB + PF $, which reaches the minimum when point P is at the position P in the figure; at this time, the minimum value is the distance from focus F to the line $ l_{1} $. Using the distance formula: $ d = \\frac{|4 + 12|}{5} = \\frac{16}{5} $"
  },
  {
    "text": "The coordinates of the foci of the hyperbola $\\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/5 - y^2/4 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(pm*3, 0)",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 45]]]",
    "process": ""
  },
  {
    "text": "Write the equation of a hyperbola with eccentricity $2$?",
    "fact_expressions": "G: Hyperbola;Eccentricity(G) = 2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[12, 15]], [[4, 15]]]",
    "query_spans": "[[[12, 18]]]",
    "process": ""
  },
  {
    "text": "The ellipse passes through the point $(0 , 3)$, and the major axis is $3$ times the minor axis. Then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;H: Point;Coordinate(H) = (0, 3);PointOnCurve(H, G);MajorAxis(G) = 3*MinorAxis(G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2/9+x^2=1, x^2/81+y^9=1}",
    "fact_spans": "[[[0, 2], [28, 30]], [[4, 14]], [[4, 14]], [[0, 14]], [[0, 26]]]",
    "query_spans": "[[[28, 37]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has an eccentricity of $\\frac{5}{3}$, then the asymptotes of the hyperbola $C$ are?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(C) = 5/3",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*(4/3)*x",
    "fact_spans": "[[[2, 63], [83, 89]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[2, 63]], [[2, 81]]]",
    "query_spans": "[[[83, 97]]]",
    "process": "Since the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) has eccentricity $ \\frac{5}{3} $, it follows that $ e = \\frac{c}{a} = \\sqrt{1 + \\left( \\frac{b}{a} \\right)^{2}} = \\frac{5}{3} $. Solving gives $ \\frac{b}{a} = \\frac{4}{3} $, so the asymptotes of hyperbola $ C $ are $ y = \\pm \\frac{4}{3}x $."
  },
  {
    "text": "Given that the line $y=x$ and the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ have no common points, what is the range of values for the eccentricity of the hyperbola?",
    "fact_expressions": "H: Line;Expression(H) = (y = x);G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;NumIntersection(H, G) = 0",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1, \\sqrt{2}]",
    "fact_spans": "[[[2, 9]], [[2, 9]], [[10, 66], [72, 75]], [[10, 66]], [[13, 66]], [[13, 66]], [[13, 66]], [[13, 66]], [[2, 70]]]",
    "query_spans": "[[[72, 85]]]",
    "process": "Solving the line and hyperbola simultaneously gives $\\frac{x^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$, simplifying to $(b^{2}-a^{2})x^{2}=a^{2}b^{2}$. Clearly, $a^{2}b^{2}>0$, and since the equation has no solution, we obtain $b^{2}-a^{2}\\leqslant0$, that is, $b\\leqslant a$. Then $c^{2}=a^{2}+b^{2}\\leqslant2a^{2}$, $\\frac{c^{2}}{a^{2}}\\leqslant2$. Since the eccentricity is greater than 1, the range of eccentricity is $(1,\\sqrt{2}]$."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ has focus $F$, directrix $l$, a line passing through point $F$ intersects the parabola at points $A$ and $B$ (point $B$ is in the first quadrant), and intersects the directrix $l$ at point $P$. If $\\overrightarrow{A F}=\\frac{2}{3} \\overrightarrow{F B}$, $\\overrightarrow{A P}=\\lambda \\overrightarrow{A F}$, then $\\lambda=$?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;A: Point;F: Point;B: Point;P: Point;l: Line;lambda:Number;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;Directrix(G) = l;PointOnCurve(F, H);Intersection(H, G) = {A,B};Quadrant(B)=1;Intersection(H, l) = P;VectorOf(A, F) = (2/3)*VectorOf(F, B);VectorOf(A, P) = lambda*VectorOf(A, F)",
    "query_expressions": "lambda",
    "answer_expressions": "-5",
    "fact_spans": "[[[2, 23], [47, 50]], [[5, 23]], [[44, 46]], [[52, 55]], [[27, 30], [39, 43]], [[62, 66], [56, 59]], [[81, 85]], [[34, 37], [76, 79]], [[197, 206]], [[5, 23]], [[2, 23]], [[2, 30]], [[2, 37]], [[38, 46]], [[44, 61]], [[62, 71]], [[44, 85]], [[88, 143]], [[144, 195]]]",
    "query_spans": "[[[197, 208]]]",
    "process": "Through point A, draw AA\\bot l, with foot at A; through point B, draw BB\\bot l, with foot at B'. From the definition of a parabola, |AA|=|AF| and |BB|=|BF|. Assume |AF|=2x. Since \\overrightarrow{AF}=\\frac{2}{3}\\overrightarrow{FB}, it follows that |FB|=3x. Because \\triangle PAA \\sim \\triangle PBB', we have \\frac{|PA|}{|PB|}=\\frac{|AA|}{|BB|}=\\frac{|AF|}{|BF|}=\\frac{2}{3}, so \\frac{|PA|}{|PA|+|AB|}=\\frac{|PA|}{|PA|+5x}=\\frac{2}{3}. Thus, |PA|=10x, and therefore \\frac{|PA|}{|AF|}=\\frac{10x}{2x}=5. Since \\overrightarrow{AP} and \\overrightarrow{AF} are in opposite directions, \\lambda=-5."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1$ $(0<b<2)$ intersects the $y$-axis at points $A$ and $B$, and point $F$ is a focus of the ellipse, then the maximum area of $\\triangle A B F$ is?",
    "fact_expressions": "G: Ellipse;b: Number;A: Point;B: Point;F: Point;0 < b;b < 2;Expression(G) = (x^2/4 + y^2/b^2 = 1);Intersection(G, yAxis) = {A, B};OneOf(Focus(G)) = F",
    "query_expressions": "Max(Area(TriangleOf(A, B, F)))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 50], [73, 75]], [[4, 50]], [[57, 60]], [[61, 64]], [[67, 71]], [[4, 50]], [[4, 50]], [[2, 50]], [[2, 66]], [[67, 80]]]",
    "query_spans": "[[[82, 107]]]",
    "process": "Let the coordinates of point F be (\\sqrt{4-b^{2}},0), and |AB|=2b,\\leqslant\\frac{b^{2}+(4-b^{2})}{2}=2 (equality holds if and only if b^{2}=4-b^{2}, i.e., b^{2}=2). Hence, the maximum area of \\triangle ABF is 2."
  },
  {
    "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ have its right focus at $F(c, 0)$. A circle centered at $F(c, 0)$ with radius $a$ intersects an asymptote of $C$ at points $A$ and $B$. If $|AB|=\\frac{2}{3}c$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;c: Number;Coordinate(F) = (c, 0);RightFocus(C) = F;G: Circle;Center(G) = F;Radius(G) = a;A: Point;B: Point;Intersection(G, OneOf(Asymptote(C))) = {A, B};Abs(LineSegmentOf(A, B)) = (2/3)*c",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "3*sqrt(5)/5",
    "fact_spans": "[[[1, 62], [88, 91], [133, 139]], [[1, 62]], [[8, 62]], [[79, 82]], [[8, 62]], [[8, 62]], [[66, 75]], [[66, 75]], [[66, 75]], [[1, 75]], [[86, 87]], [[0, 87]], [[79, 87]], [[99, 102]], [[103, 106]], [[86, 108]], [[110, 131]]]",
    "query_spans": "[[[133, 145]]]",
    "process": "\\because the focus of the hyperbola is F(c,0), and one asymptote of the hyperbola is y=\\frac{b}{a}x, i.e., bx-ay=0, \\therefore the distance from the focus to the asymptote is d=\\frac{|bc|}{\\sqrt{a^{2}+b^{2}}}=\\frac{bc}{c}=b, \\because |AF|=|BF|=a, \\therefore |AD|=\\sqrt{AF^{2}-DF^{2}}=\\sqrt{a^{2}-b^{2}}, then |AB|=2|AD|=2\\sqrt{a^{2}-b^{2}}=\\frac{2}{3}c, squaring gives 4(a^{2}-b^{2})=\\frac{4}{9}c^{2}, i.e., a^{2}-c^{2}+a^{2}=\\frac{1}{9}c^{2}, then 2a^{2}=\\frac{10}{9}c^{2}, then c^{2}=\\frac{9}{5}a^{2}, then c=\\frac{3\\sqrt{5}}{5}a, thus the eccentricity e=\\frac{3\\sqrt{5}}{5},"
  },
  {
    "text": "A line $ l $ with slope $ \\sqrt{3} $ passes through the focus $ F $ of the parabola $ y^2 = 2px $ ($ p > 0 $) and intersects the parabola at points $ A $ and $ B $ (where point $ A $ is in the first quadrant). Then $ \\frac{|AF|}{|BF|} = $?",
    "fact_expressions": "l: Line;G: Parabola;p: Number;A: Point;F: Point;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Slope(l)=sqrt(3);Focus(G)=F;PointOnCurve(F,l);Intersection(l,G)={A,B};Quadrant(A)=1",
    "query_expressions": "Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(B, F))",
    "answer_expressions": "3",
    "fact_spans": "[[[14, 19]], [[21, 42], [50, 53]], [[24, 42]], [[56, 59], [68, 72]], [[45, 48]], [[60, 63]], [[24, 42]], [[21, 42]], [[0, 19]], [[21, 48]], [[14, 48]], [[14, 65]], [[68, 77]]]",
    "query_spans": "[[[80, 103]]]",
    "process": "Let the directrix of the parabola $ y^{2} = 2px $ ($ p > 0 $) be $ l: x = \\frac{p}{2} $. As shown in the figure, draw $ AM \\perp l $ and $ BN \\perp l $ from points $ A $ and $ B $, with feet of perpendiculars at $ M $ and $ N $, respectively. Draw $ BC \\perp AM $ intersecting at point $ C $. Then $ |AM| = |AF| $, $ |BN| = |BF| $. Since $ AM \\parallel x $-axis, $ \\angle BAC = \\angle AFx = 60^{\\circ} $. In right triangle $ \\triangle ABC $, $ |AC| = \\frac{1}{2}|AB| $. Also, $ |AM| - |BN| = |AC| $, hence $ |AF| - |BF| = \\frac{1}{2}(|AF| + |BF|) $, which simplifies to $ \\frac{|AF|}{|BF|} = 3 $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ and the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ have common foci $F_{1}$, $F_{2}$, and point $P$ is an intersection point of the two curves,  \nthen $|PF_{1}| \\cdot |PF_{2}|$=?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 - y^2 = 1);Expression(H) = (x^2/9 + y^2/4 = 1);Focus(H)={F1,F2};Focus(G)={F1,F2};OneOf(Intersection(H, G)) =P",
    "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "5",
    "fact_spans": "[[[40, 68]], [[2, 39]], [[89, 93]], [[73, 80]], [[81, 88]], [[40, 68]], [[2, 39]], [[2, 88]], [[2, 88]], [[89, 102]]]",
    "query_spans": "[[[105, 131]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $x^{2}=8 y$ coincides with one focus of the hyperbola $\\frac{y^{2}}{m}-x^{2}=1$, then $m=$?",
    "fact_expressions": "G: Hyperbola;m: Number;H: Parabola;Expression(G) = (-x^2 + y^2/m = 1);Expression(H) = (x^2 = 8*y);Focus(H) = OneOf(Focus(G))",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[19, 47]], [[56, 59]], [[1, 15]], [[19, 47]], [[1, 15]], [[1, 54]]]",
    "query_spans": "[[[56, 61]]]",
    "process": "\\because the focus coordinates of the parabola \\(x^{2}=8y\\) are \\((0,2)\\), \\(\\therefore m+1=2^{2} \\Rightarrow m=3\\)."
  },
  {
    "text": "The equation of the hyperbola that shares the same asymptotes as the hyperbola $x^{2}-4 y^{2}=4$ and passes through the point $(2, \\sqrt{5})$ is?",
    "fact_expressions": "G: Hyperbola;H: Point;C:Hyperbola;Expression(G) = (x^2 - 4*y^2 = 4);Coordinate(H) = (2, sqrt(5));Asymptote(C) = Asymptote(G);PointOnCurve(H,C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/4 - x^2/16 = 1",
    "fact_spans": "[[[1, 21]], [[32, 48]], [[49, 52]], [[1, 21]], [[32, 48]], [[0, 52]], [[30, 52]]]",
    "query_spans": "[[[49, 56]]]",
    "process": ""
  },
  {
    "text": "Draw a line through the focus of the parabola $y = a x^{2}  (a \\neq 0)$ parallel to the $x$-axis, intersecting the parabola at points $A$ and $B$. Let $O$ be the origin. If the area of $\\Delta O A B$ is $\\frac{1}{2}$, then $a = $?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = a*x^2);a: Number;Negation(a = 0);PointOnCurve(Focus(G), H);IsParallel(H, xAxis);H: Line;Intersection(H, G) = {A, B};A: Point;B: Point;O: Origin;Area(TriangleOf(O, A, B)) = 1/2",
    "query_expressions": "a",
    "answer_expressions": "pm*1/2",
    "fact_spans": "[[[1, 27], [42, 45]], [[1, 27]], [[99, 102]], [[4, 27]], [[0, 41]], [[31, 41]], [[39, 41]], [[39, 57]], [[48, 51]], [[52, 55]], [[58, 61]], [[67, 97]]]",
    "query_spans": "[[[99, 104]]]",
    "process": "Convert the parabolic equation into standard form to find the coordinates of the focus. Then obtain the equation of line AB and the length |AB|. The value of a can be found from the area. The parabola y=ax^{2}(a\\neq0) converted into standard form gives x^{2}=\\frac{1}{a}y, so the focus coordinates are (0,\\frac{1}{4a}). Thus, the equation of line AB is y=\\frac{1}{4a}. Substituting into the parabolic equation yields x^{2}=\\frac{1}{4a^{2}}, so x=\\pm\\frac{1}{2a}, then |AB|=\\frac{1}{a}. According to the problem, S_{AOAB}=\\frac{1}{2}\\times|AB|\\times\\frac{1}{4a}=\\frac{1}{2}. Substituting gives S_{AOAB}=\\frac{1}{2}\\times\\frac{1}{a}\\times\\frac{1}{4a}=\\frac{1}{2}. Solving yields a=\\pm\\frac{1}{2}"
  },
  {
    "text": "The asymptotes of the hyperbola are given by $3 x \\pm 2 y=0$, and the foci lie on the $y$-axis. Then the eccentricity of this hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (3*x+pm*2*y = 0);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(13)/3",
    "fact_spans": "[[[0, 3], [37, 40]], [[0, 25]], [[0, 34]]]",
    "query_spans": "[[[37, 47]]]",
    "process": "From the given condition, a:b = 3:2 ∴ a:c = 3:√13 ∴ e = √13 / 3"
  },
  {
    "text": "Given that point $P$ lies on the parabola $y^{2}=4x$, and point $Q$ lies on the circle $(x-5)^{2}+y^{2}=1$, then the minimum value of the length $PQ$ is?",
    "fact_expressions": "G: Parabola;H: Circle;P: Point;Q: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (y^2 + (x - 5)^2 = 1);PointOnCurve(P, G);PointOnCurve(Q, H)",
    "query_expressions": "Min(Length(LineSegmentOf(P, Q)))",
    "answer_expressions": "3",
    "fact_spans": "[[[7, 21]], [[28, 48]], [[2, 6]], [[23, 27]], [[7, 21]], [[28, 48]], [[2, 22]], [[23, 49]]]",
    "query_spans": "[[[51, 64]]]",
    "process": "Since both the parabola and the circle are symmetric about the horizontal axis, we may assume $ P(m, 2\\sqrt{m}) $ ($ m \\geqslant 0 $). Let the center of the circle $ (x-5)^{2} + y^{2} = 1 $ be $ A(5, 0) $ with radius 1. Therefore, $ PA = \\sqrt{(m-5)^{2} + (2\\sqrt{m})^{2}} = \\sqrt{(m-3)^{2} + 16} $. When $ m = 3 $, $ PA_{\\min} = \\sqrt{16} = 4 $. Hence, the minimum length of $ PQ $ is $ 4 - 1 = 3 $."
  },
  {
    "text": "A line $l$ passing through the point $M(-2,0)$ intersects the ellipse $x^{2}+2 y^{2}=2$ at points $P_{1}$ and $P_{2}$. The midpoint of segment $P_{1} P_{2}$ is $P$. Let the slope of line $l$ be $k_{1}$ ($k_{1} \\neq 0$), and the slope of line $O P$ be $k_{2}$ ($O$ is the origin). Then the value of $k_{1} \\cdot k_{2}$ is?",
    "fact_expressions": "l: Line;G: Ellipse;O: Origin;P: Point;P1: Point;P2: Point;M: Point;k1:Number;k2:Number;Negation(k1 = 0);Expression(G) = (x^2 + 2*y^2 = 2);Coordinate(M) = (-2, 0);PointOnCurve(M, l);Intersection(l, G) = {P1, P2};MidPoint(LineSegmentOf(P1, P2)) = P;Slope(l) = k1;Slope(LineOf(O, P)) = k2",
    "query_expressions": "k1*k2",
    "answer_expressions": "-1/2",
    "fact_spans": "[[[12, 17], [80, 85]], [[18, 37]], [[130, 133]], [[75, 78]], [[39, 46]], [[47, 54]], [[1, 11]], [[89, 110]], [[122, 129]], [[89, 110]], [[18, 37]], [[1, 11]], [[0, 17]], [[12, 56]], [[57, 78]], [[80, 110]], [[111, 129]]]",
    "query_spans": "[[[139, 162]]]",
    "process": "Let the equation of line $ l $ be: $ y = k_{1}(x + 2) $. From \n\\[\n\\begin{cases}\ny = k_{1}(x + 2) \\\\\nx^{2} + 2y^{2} = 1\n\\end{cases}\n\\]\n, simplifying yields: $ (1 + 2k_{1}^{2})x^{2} + 8k_{1}^{2}x + 8k_{1}^{2} - 1 = 0 $, so $ x_{1} + x_{2} = \\frac{-8k_{1}^{2}}{1 + 2k_{1}^{2}} $, $ x_{1}x_{2} = \\frac{8k_{1}^{2} - 1}{1 + 2k_{1}^{2}} $. Thus, $ y_{1} + y_{2} = k_{1}(x_{1} + 2) + k_{1}(x_{2} + 2) = k_{1}(x_{1} + x_{2} + 4) = \\frac{4k_{1}}{1 + 2k_{1}^{2}} $. Therefore, $ P\\left( \\frac{4k_{1}^{2}}{1 + 2k_{1}^{2}}, \\frac{-2k_{1}}{1 + 2k_{1}^{2}} \\right) $, $ k_{2} = \\frac{\\frac{-2k_{1}}{1 + 2k_{1}^{2}}}{\\frac{-4k_{1}^{2}}{1 + 2k_{1}^{2}}} = -\\frac{1}{2k_{1}} $, so $ k_{1}k_{2} = -\\frac{1}{2} $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, point $P$ lies on the hyperbola $C$, $P F_{2} \\perp x$-axis, $\\sin \\angle P F_{1} F_{2}=\\frac{1}{3}$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F2: Point;F1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P,F2),xAxis);Sin(AngleOf(P,F1,F2))=1/3",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[18, 79], [91, 97], [159, 165]], [[26, 79]], [[26, 79]], [[86, 90]], [[10, 17]], [[2, 9]], [[26, 79]], [[26, 79]], [[18, 79]], [[2, 85]], [[2, 85]], [[86, 98]], [[99, 117]], [[118, 157]]]",
    "query_spans": "[[[159, 171]]]",
    "process": "Point P lies on the hyperbola C, PF_{2} \\bot x-axis, let P(c,\\frac{b^{2}}{a}), and in right triangle PF_{1}F_{2}, \\sin\\angle PF_{1}F_{2} = \\frac{1}{3}, |F_{1}F_{2}| = 2c, then |PF_{2}| = \\frac{\\sqrt{2}}{2}c, i.e. \\frac{b^{2}}{a} = \\frac{\\sqrt{2}}{2}c \\Leftrightarrow c^{2} - a^{2} = \\frac{\\sqrt{2}}{2}ac. Dividing both sides of the equation by \\frac{1}{2}ac gives \\sqrt{2}e^{2} - e - \\sqrt{2} = 0, solving yields e = \\sqrt{2} or -\\frac{\\sqrt{2}}{2} (discarded), hence fill in \\sqrt{2}."
  },
  {
    "text": "The minimum distance from a moving point $Q$ on the parabola $y^{2}=2 p x  (p>0)$ to the focus is $1$, then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;Q: Point;p>0;Expression(G) = (y^2 = 2*(p*x));PointOnCurve(Q, G);Min(Distance(Q, Focus(G))) = 1",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 23]], [[46, 49]], [[27, 30]], [[3, 23]], [[0, 23]], [[0, 30]], [[0, 44]]]",
    "query_spans": "[[[46, 51]]]",
    "process": "Since the distance from a moving point on the parabola to the focus is equal to the distance from the moving point to the directrix, the shortest distance from a moving point on the parabola to the focus is the distance from the vertex to the directrix, which is \\frac{p}{2}=1, p=2."
  },
  {
    "text": "A point $M$ on the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$ has a distance of $3$ to its right focus. What is the horizontal coordinate of point $M$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/12 = 1);M: Point;PointOnCurve(M, G);Distance(M, RightFocus(G)) = 3",
    "query_expressions": "XCoordinate(M)",
    "answer_expressions": "5/2",
    "fact_spans": "[[[0, 39], [46, 47]], [[0, 39]], [[42, 45], [60, 64]], [[0, 45]], [[42, 58]]]",
    "query_spans": "[[[60, 70]]]",
    "process": "Analysis: Let the x-coordinate of point M be m. From the standard equation of the hyperbola, find the values of a, b, and c, then determine the equation of the right directrix and the eccentricity of the hyperbola. Next, according to the second definition of the hyperbola, the value of m can be obtained, thus finding the x-coordinate of point M. Let the x-coordinate of point M be m. ∵ The standard equation of the hyperbola is \\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1, ∴ a=2, b=2\\sqrt{3}. ∵ c^{2}=a^{2}+b^{2}, ∴ c=4. ∴ The equation of the right directrix of the hyperbola is x=\\frac{a^{2}}{c}=1, and the eccentricity of the hyperbola is e=\\frac{c}{a}=2. ∵ The distance from point M to the right focus of the hyperbola is 3, ∴ By the second definition of the hyperbola, we have \\frac{3}{m-1}=e=2, ∴ m=\\frac{5}{2}"
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the two foci of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, and point $P$ lies on the hyperbola such that $\\angle F_{1} P F_{2}=90^{\\circ}$. Then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[17, 45], [56, 59]], [[17, 45]], [[1, 8]], [[9, 16]], [[1, 50]], [[51, 55]], [[51, 60]], [[61, 94]]]",
    "query_spans": "[[[96, 123]]]",
    "process": ""
  },
  {
    "text": "The equation of the hyperbola that shares the same asymptotes as $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ and passes through the point $(-3, 2 \\sqrt{3})$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);Z: Hyperbola;Asymptote(Z) = Asymptote(G);H: Point;Coordinate(H) = (-3,2*sqrt(3));PointOnCurve(H, Z)",
    "query_expressions": "Expression(Z)",
    "answer_expressions": "x^2/(9/4)-y^2/4=1",
    "fact_spans": "[[[1, 40]], [[1, 40]], [[71, 74]], [[0, 74]], [[52, 70]], [[52, 70]], [[50, 74]]]",
    "query_spans": "[[[71, 78]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the focus of the parabola $y=4 x^{2}$, the point $P(x_{0}, y_{0})$ lies on the parabola, and $|P F|=2$, then $y_{0}$=?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;Expression(G) = (y = 4*x^2);Coordinate(P) = (x0, y0);Focus(G) = F;PointOnCurve(P, G);Abs(LineSegmentOf(P, F)) = 2;x0:Number;y0:Number",
    "query_expressions": "y0",
    "answer_expressions": "31/16",
    "fact_spans": "[[[6, 20], [43, 46]], [[24, 42]], [[2, 5]], [[6, 20]], [[24, 42]], [[2, 23]], [[24, 47]], [[49, 58]], [[25, 42]], [[60, 67]]]",
    "query_spans": "[[[60, 69]]]",
    "process": "The parabola y = 4x^2, i.e., x^{2} = \\frac{1}{4}y, has focus F(0, \\frac{1}{16}). Since point P(x_{0}, y_{0}) lies on the parabola and |PF| = 2, combining with the definition of a parabola, we easily obtain y_{0} = 2 - \\frac{1}{16} = \\frac{31}{16}."
  },
  {
    "text": "The line $l$: $4x - 5y = 20$ passes through one focus and one endpoint of the imaginary axis of the hyperbola $C$: $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ $(a > 0, b > 0)$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(l) = (4*x-5*y = 20);PointOnCurve(OneOf(Focus(C)),l);PointOnCurve(OneOf(Endpoint(ImageinaryAxis(C))),l)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[0, 19]], [[21, 82], [97, 103]], [[28, 82]], [[28, 82]], [[28, 82]], [[28, 82]], [[21, 82]], [[0, 19]], [[0, 95]], [[0, 95]]]",
    "query_spans": "[[[97, 109]]]",
    "process": "Since the line $ l: 4x - 5y = 20 $ intersects the $ x $-axis and $ y $-axis at points $ (5, 0) $ and $ (0, -4) $ respectively, and the line $ l: 4x - 5y = 20 $ passes through one focus and one endpoint of the imaginary axis of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0 $, $ b > 0 $), therefore $ (5, 0) $ is the focus of the hyperbola and $ (0, -4) $ is an endpoint of the imaginary axis of the hyperbola. Thus, $ c = 5 $, $ b = 4 $, so $ a = \\sqrt{c^{2} - b^{2}} = 3 $, and the eccentricity is $ e = \\frac{c}{a} = \\frac{5}{3} $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, with left and right foci $F_{1}$ and $F_{2}$ respectively. A perpendicular is drawn from $F_{2}$ to an asymptote of the hyperbola $C$, with foot of the perpendicular at $H$. If the area of $\\Delta F_{1} H F_{2}$ is $b^{2}$, then what is the equation of the asymptotes of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;F1: Point;H: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;L:Line;PointOnCurve(F2,L);IsPerpendicular(OneOf(Asymptote(C)),L);FootPoint(L,OneOf(Asymptote(C)))=H;Area(TriangleOf(F1, H, F2)) = b^2",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*x",
    "fact_spans": "[[[2, 63], [97, 103], [156, 162]], [[10, 63]], [[10, 63]], [[72, 79]], [[116, 119]], [[80, 87], [89, 96]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 87]], [[2, 87]], [], [[88, 112]], [[88, 112]], [[88, 119]], [[121, 154]]]",
    "query_spans": "[[[156, 170]]]",
    "process": "Since $F_{2}$ draws a perpendicular to an asymptote of hyperbola $C$, with foot of perpendicular at $H$, we obtain that $\\triangle OHF_{2}$ is a right triangle, and $|OH|=a$, $|HF_{2}|=b$, $|OF_{2}|=c$. Thus, the area of the triangle is $\\frac{1}{2}ab$. Since the area of quadrilateral $AF_{1}HF_{2}$ is $b^{2}$, it follows that the area of $\\triangle OHF_{2}$ is $\\frac{1}{2}b^{2}$. Therefore, we have $\\frac{1}{2}ab = \\frac{1}{2}b^{2}$, solving gives $a=b$. Hence, the equations of the asymptotes are $y=\\pm x$."
  },
  {
    "text": "It is known that the center of ellipse $C$ is at the origin, $F(3,0)$ is one of its foci, a line $l$ passing through $F$ intersects $C$ at points $A$ and $B$, and the midpoint of $AB$ is $N(2,-1)$. Then the equation of $C$ is?",
    "fact_expressions": "C: Ellipse;O: Origin;Center(C) = O;F: Point;Coordinate(F) = (3, 0);OneOf(Focus(C)) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, C) = {A, B};N: Point;Coordinate(N) = (2, -1);MidPoint(LineSegmentOf(A, B)) = N",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/18+y^2/9=1",
    "fact_spans": "[[[2, 7], [23, 24], [41, 44], [77, 80]], [[11, 13]], [[2, 13]], [[14, 22], [31, 34]], [[14, 22]], [[14, 29]], [[35, 40]], [[30, 40]], [[46, 49]], [[50, 53]], [[35, 55]], [[66, 75]], [[66, 75]], [[57, 75]]]",
    "query_spans": "[[[77, 85]]]",
    "process": "By combining the point difference method and simplifying, we can obtain the solution. [Detailed Solution] Let the equation of ellipse $ C $ be $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $), $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Then $ x_{1} + x_{2} = 4 $, $ y_{1} + y_{2} = -2k_{1} = k_{NF} = 1 $. Subtracting $ \\frac{x_{1}^{2}}{a^{2}} + \\frac{y_{1}^{2}}{b^{2}} = 1 $ and $ \\frac{x_{2}^{2}}{a^{2}} + \\frac{y_{2}^{2}}{b^{2}} = 1 $, we get $ \\frac{(x_{1} + x_{2})(x_{1} - x_{2})}{a^{2}} + \\frac{(y_{1} + y_{2})(y_{1} - y_{2})}{b^{2}} = 0 $, that is, $ \\frac{x_{1} + x_{2}}{a^{2}} + \\frac{y_{1} + y_{2}}{b^{2}} \\cdot k_{1} = 0 $, $ \\frac{4}{a^{2}} + \\frac{-2}{b^{2}} = 0 \\Rightarrow a = \\sqrt{2}b $. Also, $ c = 3 $, so $ a^{2} - b^{2} = 9 $, solving gives $ a^{2} = 18 $, $ b^{2} = 9 $. Hence, the equation of ellipse $ C $ is $ \\frac{x^{2}}{18} + \\frac{y^{2}}{9} = 1 $."
  },
  {
    "text": "A line $ l $ with slope $ 1 $ passes through the focus $ F $ of the parabola $ y^{2} = 4x $, and intersects the parabola at points $ A $ and $ B $. Then $ |AB| = $?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Slope(l)=1;PointOnCurve(F,l);Focus(G)=F;F: Point;Intersection(l,G)={A,B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[7, 12]], [[14, 28], [38, 41]], [[44, 47]], [[48, 51]], [[14, 28]], [[0, 12]], [[7, 34]], [[14, 34]], [[31, 34]], [[7, 53]]]",
    "query_spans": "[[[55, 64]]]",
    "process": "The focus coordinates of the parabola $ y^{2} = 4x $ are $ F(1,0) $. The equation of line $ l $ is $ y = x - 1 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Then, by the focal chord length formula of the parabola, we have: $ |AB| = x_{1} + x_{2} + p = x_{1} + x_{2} + 2 $. Since $ A $ and $ B $ are intersection points of the parabola and the line, solving the system \n\\[\n\\begin{cases}\ny^{2} = 4x \\\\\ny = x - 1\n\\end{cases}\n\\]\ngives $ x^{2} - 6x + 1 = 0 $, so $ x_{1} + x_{2} = 6 $. Therefore, $ |AB| = 8 $."
  },
  {
    "text": "If the curve $C$: $mx^{2}+(2-m) y^{2}=1$ is a hyperbola with foci on the $x$-axis, then the range of values for $m$ is?",
    "fact_expressions": "G: Hyperbola;C: Curve;m: Number;Expression(C) = (m*x^2 + y^2*(2 - m) = 1);PointOnCurve(Focus(G), xAxis);C = G",
    "query_expressions": "Range(m)",
    "answer_expressions": "(2, +oo)",
    "fact_spans": "[[[40, 43]], [[1, 30]], [[46, 49]], [[1, 30]], [[31, 43]], [[1, 43]]]",
    "query_spans": "[[[46, 56]]]",
    "process": "The curve $ C: mx^{2} + (2 - m)y^{2} = 1 $ is a hyperbola with foci on the x-axis, which can be written as $ \\frac{x^{2}}{m} - \\frac{y^{2}}{m - 2} = 1 $. Thus, we have $ m > 0 $, and $ 2 - m < 0 $, solving gives $ m > 2 $."
  },
  {
    "text": "If the asymptotes of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ are tangent to the circle $(x-2)^{2}+y^{2}=2$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y^2 + (x - 2)^2 = 2);IsTangent(Asymptote(C),G)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 62], [91, 97]], [[9, 62]], [[9, 62]], [[67, 87]], [[9, 62]], [[9, 62]], [[1, 62]], [[67, 87]], [[1, 89]]]",
    "query_spans": "[[[91, 103]]]",
    "process": "The hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) has asymptotes $ y = \\pm \\frac{b}{a}x $, i.e., $ bx \\pm ay = 0 $. The asymptotes are tangent to the circle $ (x - 2)^{2} + y^{2} = 2 $, so we obtain $ \\frac{2b}{\\sqrt{a^{2} + b^{2}}} = \\sqrt{2} $, solving which gives $ a = b $. Therefore, the eccentricity of the hyperbola is $ e = \\frac{c}{a} = \\frac{\\sqrt{2}a}{a} = \\sqrt{2} $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and point $P$ is a point on the ellipse $C$ such that $F_{1} P \\perp F_{2} P$, then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "C: Ellipse;F1: Point;P: Point;F2: Point;Expression(C) = (x^2/25 + y^2/9 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(F1, P), LineSegmentOf(F2, P))",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "9",
    "fact_spans": "[[[18, 61], [73, 78]], [[2, 9]], [[68, 72]], [[10, 17]], [[18, 61]], [[2, 67]], [[2, 67]], [[68, 81]], [[83, 106]]]",
    "query_spans": "[[[108, 135]]]",
    "process": "By the definition of an ellipse, we have $ PF_{1} + PF_{2} = 2a = 10 \\textcircled{1} $. By the Pythagorean theorem, we have $ PF_{1}^{2} + PF_{2}^{2} = 4c^{2} = 4 \\times (25 - 9) = 64 \\textcircled{2} $. Solving equations $ \\textcircled{1} $ and $ \\textcircled{2} $ simultaneously, we get $ PF_{1} \\cdot PF_{2} = 18 $. Therefore, the area of $ \\triangle F_{1}PF_{2} $ is $ S = \\frac{1}{2} \\times PF_{1} \\cdot PF_{2} = 9 $. Hence, fill in 9."
  },
  {
    "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{2}+\\frac{y^{2}}{m}=1$ with foci on the $x$-axis is $\\frac{1}{2}$, then what is the value of $m$?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/2 + y^2/m = 1);PointOnCurve(Focus(G), xAxis);Eccentricity(G) = 1/2",
    "query_expressions": "m",
    "answer_expressions": "3/2",
    "fact_spans": "[[[10, 47]], [[68, 71]], [[10, 47]], [[1, 47]], [[10, 65]]]",
    "query_spans": "[[[68, 74]]]",
    "process": "Combining $ e = \\frac{c}{a} = \\frac{\\sqrt{2 - m}}{\\sqrt{2}} = \\frac{1}{2} $, solving gives $ m = \\frac{3}{2} $."
  },
  {
    "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{n}-\\frac{y^{2}}{3-n}=1$ are given by $y=\\pm 2 x$. Then $n=$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/(3 - n) + x^2/n = 1);n: Number;Expression(Asymptote(G)) = (y=pm*2*x)",
    "query_expressions": "n",
    "answer_expressions": "3/5",
    "fact_spans": "[[[0, 40]], [[0, 40]], [[59, 62]], [[0, 58]]]",
    "query_spans": "[[[59, 64]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the left focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, $A(1,4)$, and $P$ is a moving point on the right branch of the hyperbola, then the minimum value of $|P F|+|P A|$ is?",
    "fact_expressions": "G: Hyperbola;A: Point;P: Point;F: Point;Expression(G) = (x^2/4 - y^2/12 = 1);Coordinate(A) = (1, 4);LeftFocus(G) = F;PointOnCurve(P, RightPart(G))",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "9",
    "fact_spans": "[[[6, 45], [64, 67]], [[50, 58]], [[60, 63]], [[2, 5]], [[6, 45]], [[50, 58]], [[2, 49]], [[59, 73]]]",
    "query_spans": "[[[75, 94]]]",
    "process": ""
  },
  {
    "text": "The equation $\\frac{x^{2}}{m}-\\frac{y^{2}}{2 m-6}=1$ represents an ellipse; then the range of values for $m$ is?",
    "fact_expressions": "G: Ellipse;Expression(G)=(-y^2/(2*m - 6) + x^2/m = 1);m:Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(0,3)&Negation(m=2)",
    "fact_spans": "[[[43, 45]], [[0, 45]], [[47, 50]]]",
    "query_spans": "[[[47, 57]]]",
    "process": ""
  },
  {
    "text": "If the parabola $y^{2}=2 p x(p>0)$ passes through the point $(1,2)$, then $p$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;H: Point;Coordinate(H) = (1, 2);PointOnCurve(H, G)",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 22]], [[1, 22]], [[34, 37]], [[4, 22]], [[24, 32]], [[24, 32]], [[1, 32]]]",
    "query_spans": "[[[34, 39]]]",
    "process": "Since the parabola \\( y^{2} = 2px \\) (\\( p > 0 \\)) passes through the point \\( (1, 2) \\), we have \\( 4 = 2p \\), so \\( p = 2 \\)."
  },
  {
    "text": "Given that $O$ is the coordinate origin, the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has left focus $F$, and $A$ is a point on $C$ such that $AF$ is perpendicular to the $x$-axis. If the area of $\\Delta FAO$ is $\\frac{b^{2}}{4}$, then the eccentricity of $C$ is?",
    "fact_expressions": "O: Origin;C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;LeftFocus(C) = F;A: Point;PointOnCurve(A, C)  ;IsPerpendicular(LineSegmentOf(A, F), xAxis) ;Area(TriangleOf(F, A, O)) = b^2/4",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 5]], [[11, 68], [81, 84], [140, 143]], [[11, 68]], [[18, 68]], [[18, 68]], [[18, 68]], [[18, 68]], [[73, 76]], [[11, 76]], [[77, 80]], [[77, 87]], [[88, 100]], [[103, 138]]]",
    "query_spans": "[[[140, 149]]]",
    "process": ""
  },
  {
    "text": "The right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $F$, and the line $y=\\frac{4}{3} x$ intersects the hyperbola at points $A$ and $B$. If $A F \\perp B F$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;A: Point;F: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y = (4/3)*x);RightFocus(C) = F;Intersection(G, C) = {A, B};IsPerpendicular(LineSegmentOf(A, F), LineSegmentOf(B, F))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[0, 61], [90, 93], [124, 130]], [[7, 61]], [[7, 61]], [[70, 89]], [[96, 99]], [[66, 69]], [[100, 103]], [[7, 61]], [[7, 61]], [[0, 61]], [[70, 89]], [[0, 69]], [[70, 105]], [[107, 122]]]",
    "query_spans": "[[[124, 136]]]",
    "process": "From the given conditions: the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ has its foci on the $x$-axis, and the right focus is $F(c,0)$. Then  \n$$\n\\begin{cases}\ny=\\frac{4}{3}x \\\\\n\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1\n\\end{cases}\n$$\nSimplifying yields: $(9b^{2}-16a^{2})x^{2}=9a^{2}b^{2}$, i.e., $x^{2}=\\frac{9a^{2}b^{2}}{9b^{2}-16a^{2}}$.  \n$\\therefore A$ and $B$ are symmetric about the origin. Let $A(x,\\frac{4}{3}x)$, $B(-x,-\\frac{4}{3}x)$,  \n$\\overrightarrow{FA}=(x-c,\\frac{4}{3}x)$, $\\overrightarrow{FB}=(-x-c,-\\frac{4}{3}x)$.  \nSince $AF \\perp BF$, $\\therefore \\overrightarrow{FA} \\cdot \\overrightarrow{FB} = 0$,  \ni.e., $(x-c)(-x-c)+\\frac{4}{3}x\\times(-\\frac{4}{3}x)=0$. Simplifying yields: $c^{2}=\\frac{25}{9}x^{2}$.  \n$\\therefore a^{2}+b^{2}=\\frac{25}{9}\\times\\frac{9a^{2}b^{2}}{9b^{2}-16a^{2}}$, i.e., $9b^{4}-32a^{2}b^{2}-16a^{4}=0$.  \n$\\therefore (b^{2}-4a^{2})(9b^{2}+4a^{2})=0$. Since $a>0$, $b>0$, $\\therefore 9b^{2}+4a^{2} \\neq 0$,  \n$\\therefore b^{2}-4a^{2}=0$, i.e., $c^{2}-a^{2}-4a^{2}=0 \\Rightarrow c^{2}=5a^{2} \\Rightarrow e=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{5}$.  \nHence, the answer is: $\\sqrt{5}$."
  },
  {
    "text": "It is known that the center of the hyperbola is at the origin of the coordinate system, the foci lie on the $x$-axis, and one asymptote is the line $\\sqrt{3} x + y = 0$. Then the eccentricity of this hyperbola is equal to?",
    "fact_expressions": "O: Origin;Center(G) = O;PointOnCurve(Focus(G), xAxis);G: Hyperbola;H: Line;Expression(H)=(sqrt(3)*x+y=0);OneOf(Asymptote(G))=H",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[9, 13]], [[2, 13]], [[2, 22]], [[2, 5], [51, 54]], [[30, 48]], [30, 47], [2, 47]]",
    "query_spans": "[[[51, 61]]]",
    "process": ""
  },
  {
    "text": "A line passing through point $P(-2,0)$ intersects the parabola $C$: $y^{2}=4x$ at points $A$ and $B$, and $|PA|=\\frac{1}{2}|AB|$. Then, what is the distance from point $A$ to the origin?",
    "fact_expressions": "C: Parabola;G: Line;P: Point;A: Point;B: Point;Expression(C) = (y^2 = 4*x);Coordinate(P) = (-2, 0);PointOnCurve(P, G);Intersection(G, C) = {A, B};Abs(LineSegmentOf(P, A)) = Abs(LineSegmentOf(A, B))/2;O:Origin",
    "query_expressions": "Distance(A,O)",
    "answer_expressions": "2*sqrt(7)/3",
    "fact_spans": "[[[15, 34]], [[12, 14]], [[1, 11]], [[37, 40], [74, 78]], [[41, 44]], [[15, 34]], [[1, 11]], [[0, 14]], [[12, 46]], [[49, 71]], [[79, 81]]]",
    "query_spans": "[[[74, 86]]]",
    "process": "Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Draw perpendiculars from points $ A $ and $ B $ to the line $ x = -2 $, with feet of perpendiculars $ D $ and $ E $, respectively. Since $ |PA| = \\frac{1}{2}|AB| $, it follows that $ |BE| = 3|AD| $, i.e.,  \n$$\n\\begin{cases}\n3(x_{1} + 2) = x_{2} + 2 \\\\\n3y_{1} = y_{2}\n\\end{cases}\n$$\n$ y_{2}^{2} = 4x_{2} $. Solving yields $ x_{1} = \\frac{2}{3} $, so  \n$ |OA| = \\sqrt{x_{1}^{2} + y_{1}^{2}} = \\sqrt{x_{1}^{2} + 4x_{1}} = \\sqrt{\\frac{4}{9} + \\frac{8}{3}} = \\frac{2\\sqrt{7}}{3} $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, a line passing through $F_{2}$ intersects the ellipse at points $A$ and $B$. Then, the minimum length of segment $A B$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};H: Line;PointOnCurve(F2, H);A: Point;B: Point;Intersection(H, G) = {A, B}",
    "query_expressions": "Min(Length(LineSegmentOf(A, B)))",
    "answer_expressions": "18/5",
    "fact_spans": "[[[18, 56], [74, 76]], [[18, 56]], [[2, 9]], [[10, 17], [63, 70]], [[2, 61]], [[71, 73]], [[62, 73]], [[77, 80]], [[81, 84]], [[71, 86]]]",
    "query_spans": "[[[88, 103]]]",
    "process": "From the ellipse equation, we know: a=5, b=3. \\because the shortest chord passing through the focus is the latus rectum, \\therefore|AB|_{\\min}=\\frac{2b^{2}}{a}=\\frac{18}{5}"
  },
  {
    "text": "If the line $y=k x+1$ intersects the curve $x = \\sqrt{1-4 y^{2}}$ at two distinct points, then what is the range of values for $k$?",
    "fact_expressions": "G: Line;k: Number;H: Curve;Expression(G) = (y = k*x + 1);Expression(H) = (x = sqrt(1 - 4*y^2));NumIntersection(G, H) = 2",
    "query_expressions": "Range(k)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 12]], [[47, 50]], [[13, 37]], [[1, 12]], [[13, 37]], [[1, 45]]]",
    "query_spans": "[[[47, 57]]]",
    "process": ""
  },
  {
    "text": "Given the line $l$: $y = k(x - 2)$ intersects the parabola $C$: $y^2 = 8x$ at points $A$ and $B$, and $F$ is the focus of the parabola $C$. If $|AF| = 3|BF|$, then the inclination angle of line $l$ is?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;F: Point;B: Point;Expression(C) = (y^2 = 8*x);Expression(l) = (y = k*(x - 2));Intersection(l, C) = {A, B};Focus(C) = F;Abs(LineSegmentOf(A, F)) = 3*Abs(LineSegmentOf(B, F));k:Number",
    "query_expressions": "Inclination(l)",
    "answer_expressions": "{pi/3,(2*pi)/3}",
    "fact_spans": "[[[2, 18], [80, 85]], [[19, 37], [53, 59]], [[39, 42]], [[49, 52]], [[43, 46]], [[19, 37]], [[2, 18]], [[2, 48]], [[49, 62]], [[64, 78]], [[8, 18]]]",
    "query_spans": "[[[80, 91]]]",
    "process": "Let the intersection points be A(x_{1},y_{1}), B(x_{2},y_{2}). Since the line l: y = k(x - 2) passes through the focus F(2,0), substituting y = k(x - 2) into y^{2} = 8x yields k^{2}x^{2} - (4k^{2} + 8)x + 4k^{2} = 0. Then x_{1} + x_{2} = 4 + \\frac{8}{k^{2}}, x_{1}x_{2} = 4. From the definition of the parabola, |AF| = x_{1} + 2, |BF| = x_{2} + 2. Thus, from the given condition, x_{1} = 3x_{2} + 4. Substituting into x_{1}x_{2} = 4 gives 3x_{2}^{2} + 4x_{2} - 4 = 0. Solving yields x_{2} = \\frac{2}{3} or x_{2} = -2 (discarded). Hence, when x_{2} = \\frac{2}{3}, x_{1} = 6. Substituting into x_{1} + x_{2} = 4 + \\frac{8}{k^{2}} gives k^{2} = 3 \\Rightarrow k = \\tan\\alpha = \\pm\\sqrt{3}. Therefore, the inclination angle of the line is \\alpha = \\frac{\\pi}{3} or \\alpha = \\frac{2\\pi}{3}. The answer should be \\frac{\\pi}{3} or \\frac{2\\pi}{3}."
  },
  {
    "text": "A line with slope $\\frac{1}{2}$ passes through the focus of the parabola $x^{2}=8 y$ and intersects the parabola at points $A$ and $B$. Then $|A B|$=?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;Expression(G) = (x^2 = 8*y);Slope(H)=1/2;PointOnCurve(Focus(G),H);Intersection(H,G)={A,B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "10",
    "fact_spans": "[[[41, 44], [21, 35]], [[17, 19]], [[47, 50]], [[51, 54]], [[21, 35]], [[0, 19]], [[17, 38]], [[17, 56]]]",
    "query_spans": "[[[58, 67]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). For the parabola x^{2}=8y, the focal chord length |AB| = p + (y_{1}+y_{2}). Since the focus of the parabola is at (0,2) and k_{AB} = \\frac{1}{2}, the equation of line AB is y = \\frac{1}{2}x + 2, or equivalently x = 2y - 4. Substituting x = 2y - 4 into the parabola equation gives y^{2} - 6y + 4 = 0, thus y_{1} + y_{2} = 6, so |AB| = 4 + 6 = 10."
  },
  {
    "text": "If the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{a^{2}}=1$ and the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{2}=1$ have the same foci, then $a$=?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Ellipse;Expression(G) = (-y^2/2 + x^2/a = 1);Expression(H) = (x^2/4 + y^2/a^2 = 1);Focus(G) = Focus(H)",
    "query_expressions": "a",
    "answer_expressions": "1",
    "fact_spans": "[[[44, 82]], [[91, 94]], [[2, 43]], [[44, 82]], [[2, 43]], [[2, 87]]]",
    "query_spans": "[[[91, 96]]]",
    "process": ""
  },
  {
    "text": "Given that the foci of a hyperbola lie on the coordinate axes, the center is at the origin, one asymptote has the equation $y=2x$, and the hyperbola passes through the point $(1,3)$, then the distance from the focus of the hyperbola to the asymptote is equal to?",
    "fact_expressions": "G: Hyperbola;P:Point;O:Origin;Coordinate(P) = (1, 3);Expression(OneOf(Asymptote(G))) = (y = 2*x);PointOnCurve(Focus(G),axis);Center(G)=O;PointOnCurve(P,G)",
    "query_expressions": "Distance(Focus(G), Asymptote(G))",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[16, 19], [49, 52]], [[39, 47]], [[13, 15]], [[40, 47]], [[16, 35]], [[2, 19]], [[10, 19]], [[16, 47]]]",
    "query_spans": "[[[49, 65]]]",
    "process": "The focus lies on the coordinate axis. When the focus is on the x-axis, the calculation does not satisfy the given conditions and is therefore discarded. When the focus is on the y-axis, according to the problem we have \\begin{cases}\\frac{a}{b}=2\\\\\\frac{9}{a^{2}}-\\frac{1}{b^{2}}=1\\end{cases}, solving yields \\begin{cases}a^{2}=5\\\\b^{2}=\\frac{5}{4}\\end{cases}. Thus, the coordinates of the foci are (0,\\pm\\frac{5}{2}). The distance from the focus to the asymptote is d=\\frac{\\frac{5}{2}}{\\sqrt{4+1}}=\\frac{\\sqrt{5}}{2}."
  },
  {
    "text": "Let the line $l$ passing through the focus $F$ of the parabola $y^{2}=4x$ intersect the parabola at points $A$ and $B$. If the sum of the horizontal coordinates of points $A$ and $B$ is $\\frac{10}{3}$, then $|AB|=$?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};XCoordinate(A) + XCoordinate(B) = 10/3",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[22, 27]], [[1, 15], [28, 31]], [[33, 36], [44, 47]], [[37, 40], [48, 51]], [[18, 21]], [[1, 15]], [[1, 21]], [[0, 27]], [[22, 42]], [[44, 74]]]",
    "query_spans": "[[[76, 85]]]",
    "process": "According to the property of a parabola, the distance from a point on the curve to the focus is equal to the distance from that point to the directrix, we obtain |AB| = x_{A} + x_{B} + \\frac{10}{3} = \\frac{16}{3}"
  },
  {
    "text": "Draw a vertical line from the left focus $F_{1}$ of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$ perpendicular to the $x$-axis, intersecting the ellipse at point $P$. Let $F_{2}$ be the right focus. If $\\angle F_{1} P F_{2}=60^{\\circ}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;P: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G)=F1;L:Line;PointOnCurve(F1, L);IsPerpendicular(L,xAxis);Intersection(L, G) = P;RightFocus(G)=F2;AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[1, 55], [75, 77], [130, 132]], [[3, 55]], [[3, 55]], [[59, 66]], [[78, 81]], [[82, 89]], [[3, 55]], [[3, 55]], [[1, 55]], [[1, 66]], [], [[0, 74]], [[0, 74]], [[0, 81]], [[75, 93]], [[95, 128]]]",
    "query_spans": "[[[130, 138]]]",
    "process": "Analysis: Substitute $x = -c$ into the ellipse equation to obtain the coordinates of point $P$, and then, based on $\\angle F_{1}PF_{2} = 60^\\circ$, deduce that $\\frac{2c}{\\frac{b^{2}}{a}} = \\sqrt{3}$. Rearranging yields $\\sqrt{3}e^{2} + 2e - \\sqrt{3} = 0$, from which the eccentricity $e$ of the ellipse is obtained. Specifically, from the given conditions, the coordinates of point $P$ are $(-c, \\frac{b^{2}}{a})$ or $(-c, -\\frac{b^{2}}{a})$. Since $\\angle F_{1}PF_{2} = 60^{\\circ}$, it follows that $\\frac{2c}{\\frac{b^{2}}{a}} = \\sqrt{3}$, i.e., $2ac = \\sqrt{3}b^{2} = \\sqrt{3}(a^{2} - c^{2})$. Thus, $\\sqrt{3}e^{2} + 2e - \\sqrt{3} = 0$, so $e = \\frac{\\sqrt{3}}{3}$ or $e = -\\sqrt{3}$ (discarded). Therefore, the answer is $\\frac{\\sqrt{3}}{3}$."
  },
  {
    "text": "Given an ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with a point $M\\left(\\frac{\\sqrt{2}}{2} a, \\frac{\\sqrt{2}}{2} b\\right)$, $F$ is the right focus, $B$ is the upper vertex, $O$ is the origin, and $S_{\\triangle B F O}=\\sqrt{2} S_{\\triangle B F M}$. Then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;M: Point;Coordinate(M) = (a*(sqrt(2)/2), b*(sqrt(2)/2));PointOnCurve(M, C);F: Point;RightFocus(C) = F;B: Point;UpperVertex(C) = B;O: Origin;Area(TriangleOf(B, F, O)) = sqrt(2)*Area(TriangleOf(B, F, M))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 59], [191, 196]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[63, 111]], [[63, 111]], [[2, 110]], [[113, 116]], [[2, 120]], [[121, 124]], [[2, 128]], [[129, 132]], [[139, 189]]]",
    "query_spans": "[[[191, 202]]]",
    "process": "From the given conditions, the equation of line BF is: $\\frac{x}{c}+\\frac{y}{b}=1$, that is, $bx+cy-cb=0$. Therefore, the distance from point M to line BF is $d=\\frac{\\frac{\\sqrt{2}}{2}ab+\\frac{\\sqrt{2}}{2}bc-bc}{\\sqrt{b^{2}+c^{2}}}=\\frac{1}{2}\\cdot\\frac{\\sqrt{2}b|a-(\\sqrt{2}-1)c|}{a}$. Since $|BF|=\\sqrt{b^{2}+c^{2}}=a$, we have $S_{\\triangle BFM}=\\frac{1}{2}|BF|\\cdot d=\\frac{\\sqrt{2}}{4}b|a-(\\sqrt{2}-1)c|$. While $S_{\\triangle BFO}=\\frac{1}{2}bc$, and since $S_{\\triangle BFO}=\\sqrt{2}S_{\\triangle BFM}$, it follows that $\\frac{1}{2}bc=\\sqrt{2}\\cdot\\frac{\\sqrt{2}}{4}b|a-(\\sqrt{2}-1)c|$. Simplifying yields $c=|a-(\\sqrt{2}-1)c|$, further simplification gives $a=\\sqrt{2}c$, solving yields $e=\\frac{\\sqrt{2}}{2}$."
  },
  {
    "text": "Draw a line $l$ through the left focus $F_{1}$ of the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, intersecting the left branch of hyperbola $C$ at points $M$ and $N$. When $l \\perp x$-axis, $|M N|=3$, then what is the distance from the right focus $F_{2}$ to the asymptote of hyperbola $C$?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;F1: Point;M: Point;N: Point;b>0;Expression(C) = (x^2/4 - y^2/b^2 = 1);LeftFocus(C)=F1;PointOnCurve(F1, l);Intersection(l,LeftPart(C)) = {M, N};IsPerpendicular(l,xAxis);Abs(LineSegmentOf(M,N))=3;F2:Point;RightFocus(C)=F2",
    "query_expressions": "Distance(F2,Asymptote(C))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[65, 70]], [[1, 53], [71, 77], [71, 77]], [[9, 53]], [[57, 64]], [[82, 85]], [[86, 89]], [[9, 53]], [[1, 53]], [[1, 64]], [[0, 70]], [[65, 91]], [[94, 106]], [[108, 117]], [[123, 130]], [[71, 130]]]",
    "query_spans": "[[[123, 146]]]",
    "process": "By the given condition, let the left focus of hyperbola $ C $ be $ F(-c,0) $ ($ c>0 $), then $ c^{2}=b^{2}+4 $. When $ l \\perp x $-axis, substitute the equation of line $ l $, $ x=-c $, into the hyperbola equation, simplifying yields $ y^2=\\frac{b^{4}}{4} $, i.e., $ y=\\pm\\frac{b^{2}}{2} $. Then from $ |MN|=b^{2}=3 $, we obtain $ c=\\sqrt{7} $. Thus, the distance from the right focus $ F_{2}(\\sqrt{7},0) $ to the asymptote $ \\sqrt{3}x\\pm2y=0 $ of hyperbola $ C $ is $ d=\\frac{|\\sqrt{3}\\times\\sqrt{7}|}{\\sqrt{3+4}}=\\sqrt{3} $."
  },
  {
    "text": "The length of the minor axis of an ellipse is $2$, and the major axis is $2$ times the minor axis. Then, what is the distance from the center of the ellipse to its directrix?",
    "fact_expressions": "G: Ellipse;Length(MinorAxis(G)) = 2;Length(MajorAxis(G)) = 2*Length(MinorAxis(G))",
    "query_expressions": "Distance(Center(G), Directrix(G))",
    "answer_expressions": "(4*sqrt(3))/3",
    "fact_spans": "[[[0, 2], [24, 26], [29, 30]], [[0, 10]], [[0, 22]]]",
    "query_spans": "[[[24, 37]]]",
    "process": ""
  },
  {
    "text": "If the left focus of the hyperbola $\\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1$ lies on the directrix of the parabola $y^{2}=2 p x(p>0)$, then $p=$?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;p>0;Expression(G) = (x^2/5 - y^2/4 = 1);Expression(H) = (y^2 = 2*p*x);PointOnCurve(LeftFocus(G),Directrix(H))",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 39]], [[44, 65]], [[47, 65]], [[47, 65]], [[1, 39]], [[44, 65]], [[1, 69]]]",
    "query_spans": "[[[71, 76]]]",
    "process": "By the given condition, for the hyperbola $\\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1$, we have $a^{2}=5$, $b^{2}=4$, so $c=\\sqrt{a^{2}+b^{2}}=3$, thus the left focus of the hyperbola is $F_{1}(-3,0)$. Also, for the parabola $y^{2}=2px$ ($p>0$), the equation of the directrix is $x=-\\frac{p}{2}$. According to the condition, we get $-\\frac{p}{2}=-3$, so $p=6$."
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $4 x-3 y=0$, find the eccentricity $e$ of the hyperbola.",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;e: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G)))=(4*x-3*y=0);Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "5/3",
    "fact_spans": "[[[2, 60], [83, 86]], [[5, 60]], [[5, 60]], [[90, 93]], [[5, 60]], [[5, 60]], [[2, 60]], [[2, 80]], [[83, 93]]]",
    "query_spans": "[[[90, 95]]]",
    "process": ""
  },
  {
    "text": "If the equation $\\frac{x^{2}}{10-k}+\\frac{y^{2}}{k-5}=1$ represents an ellipse with foci on the $y$-axis, then the range of values for $k$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(10 - k) + y^2/(k - 5) = 1);PointOnCurve(Focus(G), yAxis);k: Number",
    "query_expressions": "Range(k)",
    "answer_expressions": "(15/2, 10)",
    "fact_spans": "[[[54, 56]], [[1, 56]], [[45, 56]], [[58, 61]]]",
    "query_spans": "[[[58, 68]]]",
    "process": ""
  },
  {
    "text": "If the distance from the focus of the parabola $y^{2}=4 x$ to the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $\\frac{1}{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 4*x);G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Distance(Focus(H),Asymptote(G)) = 1/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[1, 15]], [[1, 15]], [[19, 75], [98, 101]], [[19, 75]], [[22, 75]], [[22, 75]], [[22, 75]], [[22, 75]], [[1, 96]]]",
    "query_spans": "[[[98, 107]]]",
    "process": "The focus of the parabola $ y^{2}=4x $ is $ (1,0) $, the asymptotes of the hyperbola $ \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 $ $ (a>0,b>0) $ are given by $ bx\\pm ay=0 $, so $ \\frac{b}{c}=\\frac{1}{2} $, $ a=\\frac{\\sqrt{3}}{2}c' $, $ e=\\frac{2\\sqrt{3}}{3} $."
  },
  {
    "text": "Given that one focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{4}=1  (a>0)$ is $(-\\sqrt{5}, 0)$, what is the length of the major axis of the ellipse?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/4 + x^2/a^2 = 1);a: Number;a>0;F: Point;OneOf(Focus(G)) = F;Coordinate(F) = (-sqrt(5), 0)",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 50], [75, 77]], [[2, 50]], [[4, 50]], [[4, 50]], [[56, 73]], [[2, 73]], [[56, 73]]]",
    "query_spans": "[[[75, 83]]]",
    "process": "Since one focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{4}=1$ $(a>0)$ is $(-\\sqrt{5},0)$, it follows that $a^{2}-4=5$, so $a=3$. Therefore, the major axis length of the ellipse is $2a=6$."
  },
  {
    "text": "$P$ is a point on the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$, and $F_{1}$, $F_{2}$ are its two foci. If $\\angle F_{1} P F_{2}=30^{\\circ}$, then the area of $\\triangle F_{1} P F_{2}$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P, G);G: Ellipse;Expression(G) = (x^2/5 + y^2/4 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};AngleOf(F1, P, F2) = ApplyUnit(30, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 3]], [[0, 44]], [[4, 41], [61, 62]], [[4, 41]], [[45, 52]], [[53, 60]], [[45, 66]], [[68, 100]]]",
    "query_spans": "[[[102, 132]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ is the left focus of the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$, and the line $l$: $y=x-1$ intersects the ellipse $C$ at points $A$ and $B$, then the value of $|F_{1} A|+|F_{1} B|$ is?",
    "fact_expressions": "l: Line;C: Ellipse;F1: Point;A: Point;B: Point;Expression(C) = (x^2/2 + y^2 = 1);Expression(l) = (y = x - 1);LeftFocus(C) = F1;Intersection(l, C) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(F1, A)) + Abs(LineSegmentOf(F1, B))",
    "answer_expressions": "8*sqrt(2)/3",
    "fact_spans": "[[[46, 59]], [[10, 41], [60, 65]], [[2, 9]], [[67, 70]], [[71, 74]], [[10, 41]], [[46, 59]], [[2, 45]], [[46, 76]]]",
    "query_spans": "[[[79, 104]]]",
    "process": ""
  },
  {
    "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a, b>0)$ are $y=\\pm x$, and one focus is $(2 \\sqrt{2}, 0)$. Then $a$=?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(H) = (2*sqrt(2), 0);OneOf(Focus(G))=H;Expression(Asymptote(G)) = (y = pm*x)",
    "query_expressions": "a",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 54]], [[3, 54]], [[95, 98]], [[76, 93]], [[3, 54]], [[3, 54]], [[0, 54]], [[76, 93]], [[0, 93]], [[0, 69]]]",
    "query_spans": "[[[95, 100]]]",
    "process": "From the given conditions, we have:  \n\\begin{cases}\\frac{b}{a}=1\\\\c=2\\sqrt{2}\\end{cases}, and since c^{2}=a^{2}+b^{2}, \\therefore a=2"
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{9}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/9 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*3*x",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 36]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the right focus is $F$, and $A$ is the right vertex of the hyperbola $C$. A line perpendicular to the $x$-axis is drawn through point $F$, intersecting the hyperbola $C$ at point $P$. If the slope of the line $AP$ is $\\sqrt{3}$ times the slope of one asymptote of the hyperbola $C$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(C) = F;A: Point;RightVertex(C) = A;Z: Line;PointOnCurve(F, Z);IsPerpendicular(Z, xAxis);P: Point;Intersection(Z, C) = P;Slope(LineOf(A, P)) = sqrt(3)*Slope(OneOf(Asymptote(C)))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 63], [76, 82], [102, 108], [126, 132], [154, 160]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 71], [88, 92]], [[2, 71]], [[72, 75]], [[72, 86]], [], [[87, 100]], [[87, 100]], [[110, 113]], [[87, 113]], [[115, 152]]]",
    "query_spans": "[[[154, 166]]]",
    "process": "Let the coordinates of focus F be (c,0), and the eccentricity of hyperbola C be e. Without loss of generality, assume point P lies in the first quadrant. From \\begin{cases}y=\\frac{b}{a}x\\end{cases}, the coordinates of point P are obtained as (c,\\frac{b^{2}}{a}). The coordinates of point A are (a,0). The slope of line AP is \\frac{b^{2}}{c-a} = \\frac{c^{2}-a^{2}}{a(c-a)} = \\frac{c+a}{a} = e+1. Also, from \\frac{b}{a} = \\sqrt{\\frac{c^{2}-a^{2}}{a^{2}}} = \\sqrt{e^{2}-1}, we have e+1 = \\sqrt{3(e^{2}-1)}. Rearranging gives e^{2}-e-2=0, solving yields e=2 or e=-1 (discarded)."
  },
  {
    "text": "Given that the ellipse $C_{1}$: $m^{2} x^{2}+y^{2}=1$ $(0<m<1)$ and the hyperbola $C_{2}$: $n^{2} x^{2}-y^{2}=1$ $(n>0)$ have coincident foci, and $e_{1}$, $e_{2}$ are the eccentricities of $C_{1}$, $C_{2}$ respectively, then the range of $e_{1} \\cdot e_{2}$ is?",
    "fact_expressions": "C2: Hyperbola;C1: Ellipse;Expression(C1) = (m^2*x^2 + y^2 = 1);m: Number;m>0;m<1;Expression(C2) = (n^2*x^2 - y^2 = 1);n: Number;n>0;Focus(C1) = Focus(C2);e1: Number;e2: Number;Eccentricity(C1) = e1;Eccentricity(C2) = e2",
    "query_expressions": "Range(e1*e2)",
    "answer_expressions": "(1, +\\infty)",
    "fact_spans": "[[[42, 80], [112, 119]], [[2, 41], [104, 111]], [[2, 41]], [[13, 41]], [[13, 41]], [[13, 41]], [[42, 80]], [[54, 80]], [[54, 80]], [[2, 85]], [[86, 93]], [[94, 101]], [[86, 123]], [[86, 123]]]",
    "query_spans": "[[[125, 151]]]",
    "process": "Since the foci of the two curves coincide, we obtain the relationship between m and n, then compute (e_{1}e_{2})^{2}=(1-m^{2})(1+n^{2}). Using the previously obtained relation to eliminate variables yields (e_{1}e_{2})^{2}=\\frac{(m^{2}-1)(m^{2}-1)}{1-2m^{2}}. Let t=1-2m^{2}, after substitution, use the monotonicity of the function to find the range. Note that the variable's range must be considered, otherwise errors will occur. [Detailed solution] Since the ellipse C_{1}: m^{2}x^{2}+y^{2}=1 (0<m<1) and the hyperbola C_{2}: n^{2}x^{2}-y^{2}=1 (n>0) have standard equations \\frac{x^{2}}{m^{2}}+y^{2}=1 and \\frac{x^{2}}{n^{2}}-y^{2}=1 respectively, and their foci coincide, then \\frac{1}{m^{2}}-1=\\frac{1}{n^{2}}+1, so \\frac{1}{m^{2}}-\\frac{1}{n^{2}}=2, \\therefore \\frac{1}{m^{2}}>2, 0<m^{2}<\\frac{1}{2}. On the other hand, (e_{1}e_{2})^{2}=(1-m^{2})(1+n^{2})=\\frac{(m^{2}-1)(m^{2}-1)}{1-2m^{2}}. Let 1-2m^{2}=t, then 0<t<1, (e_{1}e_{2})^{2}=\\frac{t^{2}+2t+1}{4t}=\\frac{1}{4}(t+2+\\frac{1}{t}), t\\in(0,1). Thus (e_{1}e_{2})^{2}>1, so e_{1}e_{2}>"
  },
  {
    "text": "Given $A(4,0)$, $B(1,0)$, and a moving point $P$ satisfying $PA = 2PB$. Let $d$ be the distance from point $P$ to point $C(-3,0)$. Then the range of values for $d$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (4, 0);B: Point;Coordinate(B) = (1, 0);P: Point;LineSegmentOf(P, A) = 2*LineSegmentOf(P, B);C: Point;Coordinate(C) = (-3, 0);Distance(P, C) = d;d: Number",
    "query_expressions": "Range(d)",
    "answer_expressions": "[1,5]",
    "fact_spans": "[[[2, 10]], [[2, 10]], [[13, 21]], [[13, 21]], [[24, 27], [43, 47]], [[29, 40]], [[48, 58]], [[48, 58]], [[43, 65]], [[62, 65], [67, 70]]]",
    "query_spans": "[[[67, 77]]]",
    "process": "Let P(x,y). According to the given condition, \\sqrt{(x-4)^{2}+y^{2}}=2\\sqrt{(x-1)^{2}+y^{2}}. Simplifying gives x^{2}+y^{2}=4. Since the distance from the center O(0,0) to the point C(-3,0) is 3, it follows that 3-2\\leqslant d\\leqslant3+2, i.e., 1\\leqslant d\\leqslant5."
  },
  {
    "text": "If the directrix of the parabola $y^{2}=2 p x(p>0)$ passes through the left vertex of the hyperbola $x^{2}-y^{2}=1$, then $p$=?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;Expression(G) = (x^2 - y^2 = 1);p>0;Expression(H) = (y^2 = 2*p*x);PointOnCurve(LeftVertex(G),Directrix(H))",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[27, 45]], [[1, 22]], [[51, 54]], [[27, 45]], [[4, 22]], [[1, 22]], [[1, 49]]]",
    "query_spans": "[[[51, 56]]]",
    "process": "The left vertex of the hyperbola $x^{2}-y^{2}=1$ is $(-1,0)$. The directrix of the parabola $y^{2}=2px$ ($p>0$) is $x=-\\frac{p}{2}$. Hence, $-\\frac{p}{2}=-1$, so $p=2$."
  },
  {
    "text": "Given that a point $P$ on the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ is at a distance of $10$ from one focus, then the distance from point $P$ to the other focus is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/16 - y^2/9 = 1);PointOnCurve(P, G);Distance(P, F1) = 10;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2)",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "{18, 2}",
    "fact_spans": "[[[2, 41]], [[44, 47], [62, 66]], [], [], [[2, 41]], [[2, 47]], [[2, 60]], [[2, 52]], [[2, 72]], [[2, 72]]]",
    "query_spans": "[[[2, 77]]]",
    "process": "Let the distance from point P to the other focus be $ d $, then according to the definition of the hyperbola, $ |d - 10| = 8 $, solving yields $ d = 2 $ or $ 18 $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{25}=1$, and a line passing through $F_{1}$ intersects the ellipse at points $M$ and $N$, then the perimeter of $\\Delta M N F_{2}$ is?",
    "fact_expressions": "F1: Point;F2: Point;Focus(G) = {F1, F2};G: Ellipse;Expression(G) = (x^2/16 + y^2/25 = 1);L: Line;PointOnCurve(F1, L);M: Point;N: Point;Intersection(L, G) = {M, N}",
    "query_expressions": "Perimeter(TriangleOf(M, N, F2))",
    "answer_expressions": "20",
    "fact_spans": "[[[2, 9], [64, 71]], [[10, 17]], [[2, 62]], [[18, 57], [75, 77]], [[18, 57]], [[72, 74]], [[63, 74]], [[79, 82]], [[83, 86]], [[72, 88]]]",
    "query_spans": "[[[90, 113]]]",
    "process": ""
  },
  {
    "text": "Given that the line $l$ has a slope of $1$ and is tangent to the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$ in the first quadrant at point $A$, then the coordinates of point $A$ are?",
    "fact_expressions": "l: Line;G: Hyperbola;A: Point;Expression(G) = (x^2/2 - y^2 = 1);Slope(l) = 1;TangentPoint(l, G) = A;Quadrant(A) = 1",
    "query_expressions": "Coordinate(A)",
    "answer_expressions": "(2, 1)",
    "fact_spans": "[[[2, 7]], [[17, 45]], [[53, 57], [59, 63]], [[17, 45]], [[2, 14]], [[2, 57]], [[48, 57]]]",
    "query_spans": "[[[59, 68]]]",
    "process": "Since the slope of line $ l $ is 1, let $ l: y = x + m $. Substituting into the hyperbola $ \\frac{x^{2}}{2} - y^{2} = 1 $ gives $ x^{2} + 4mx + 2m^{2} + 2 = 0 $. Since the line is tangent to the hyperbola, $ A = 0 $, that is, $ 16m^{2} - 4(2m^{2} + 2) = 0 $. Solving yields $ m = \\pm 1 $. When $ m = 1 $,  \n\\[\n\\begin{cases}\ny = x + 1 \\\\\n\\frac{x^{2}}{2} - y^{2} =\n\\end{cases}\n\\]\nsolving gives  \n\\[\n\\begin{cases}\nx = -2 \\\\\ny = -1\n\\end{cases}\n\\]\nWhen $ m = -1 $,  \n\\[\n\\begin{cases}\ny = x - 1 \\\\\n\\frac{x^{2}}{2} - y^{2} = 1\n\\end{cases}\n\\]\nsolving gives  \n\\[\n\\begin{cases}\nx = 2 \\\\\ny = 1\n\\end{cases}\n\\]\nSince the tangent point $ A $ is in the first quadrant, point $ A(2,1) $."
  },
  {
    "text": "The standard equation of a parabola with focus $F(-1,0)$ is?",
    "fact_expressions": "G: Parabola;F: Point;Coordinate(F) = (-1, 0);Focus(G)=F",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=-4*x",
    "fact_spans": "[[[13, 16]], [[3, 12]], [[3, 12]], [[0, 16]]]",
    "query_spans": "[[[13, 23]]]",
    "process": "According to the problem, let the standard equation of the parabola be $ y^{2} = -2px $, then $ -\\frac{p}{2} = -1 $, yielding $ p = 2 $. Therefore, the standard equation of the parabola is $ y^{2} = -4x $."
  },
  {
    "text": "Given that $M$ and $N$ are two moving points on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ such that $OM \\perp ON$ ($O$ being the origin), what is the minimum area of triangle $\\triangle OMN$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);M: Point;N: Point;PointOnCurve(M, G);PointOnCurve(N, G);IsPerpendicular(LineSegmentOf(O, M), LineSegmentOf(O, N));O: Origin",
    "query_expressions": "Min(Area(TriangleOf(O, M, N)))",
    "answer_expressions": "4/5",
    "fact_spans": "[[[10, 37]], [[10, 37]], [[2, 5]], [[6, 9]], [[2, 43]], [[2, 43]], [[44, 59]], [[60, 63]]]",
    "query_spans": "[[[71, 100]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, its directrix $l$ intersects the $x$-axis at point $K$. Point $P(x, y)$ $(y>0)$ is a point on $C$. When $\\frac{|PK|}{|PF|}$ is maximized, what is the slope of line $KP$?",
    "fact_expressions": "C: Parabola;l: Line;K: Point;P: Point;F: Point;x1:Number;y1:Number;y1>0;Expression(C) = (y^2 = 4*x);Coordinate(P) = (x1, y1);Focus(C) = F;Directrix(C)=l;Intersection(l, xAxis) = K;PointOnCurve(P, C);WhenMax(Abs(LineSegmentOf(P,K))/Abs(LineSegmentOf(P,F)))",
    "query_expressions": "Slope(LineOf(K,P))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 21], [29, 30], [64, 67]], [[32, 35]], [[44, 47]], [[48, 63]], [[25, 28]], [[49, 63]], [[49, 63]], [[49, 63]], [[2, 21]], [[48, 63]], [[2, 28]], [[29, 35]], [[32, 47]], [[48, 70]], [[71, 96]]]",
    "query_spans": "[[[97, 109]]]",
    "process": "According to the problem, find the coordinates of the focus and the equation of the directrix. Using the property of the parabola, the distance to the focus is converted into the distance to the directrix. When $\\frac{|PK|}{|PF|}$ is maximized, it is equivalent to finding the maximum value of $\\tan\\angle PKF$. Use the AM-GM inequality to find the coordinates of point $P$ when the maximum occurs. From the problem, the focus is $F(1,0)$, and the directrix equation is $x=$. Draw $PM$ perpendicular to the directrix from point $P$, with foot $M$. Then $|PF|=|PM|$, and $PM \\parallel KF$. Therefore, $\\frac{|PK|}{|PF|}=\\frac{|PK|}{|PM|}=\\frac{}{\\cos}$. Since $PM \\parallel KF$, $\\angle MPK = \\angle PKF$, so $\\frac{|PK|}{|PF|}=\\frac{1}{\\cos\\angle PKF}$ $(0\\leqslant\\angle PKF<\\frac{\\pi}{2})$. Finding the maximum value of $\\frac{|PK|}{|PF|}$ is equivalent to finding the minimum value of $\\cos\\angle PKF$, which is equivalent to finding the maximum value of $\\tan\\angle PKF$. Let $P(\\frac{a^{2}}{4},a)$, $a>0$, then $\\tan\\angle PKF=\\frac{a}{\\frac{a^{2}}{4}+1}=\\frac{1}{\\frac{a}{4}+\\frac{1}{a}}\\leqslant\\frac{1}{2\\sqrt{\\frac{a}{4}\\cdot\\frac{1}{a}}}=$. The equality holds if and only if $\\frac{a}{4}=\\frac{1}{a}$, that is, when $a=2$, meaning the slope of line $KP$ is $1$."
  },
  {
    "text": "Given that point $B$ is the upper vertex of the ellipse $C$: $\\frac{x^{2}}{4}+y^{2}=1$, a tangent line $l$ is drawn from $B$ to the circle $O$: $x^{2}+y^{2}=r^{2}$ $(0<r<\\frac{6}{7})$. The line $l$ intersects the ellipse $C$ at another point $Q$. If $O Q=\\frac{\\sqrt{13}}{2}$, then $r=$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2 = 1);B: Point;UpperVertex(C) = B;O1: Circle;r: Number;Expression(O1) = (x^2 + y^2 = r^2);0 < r;r < 6/7;l: Line;TangentOfPoint(B, O1) = l;Q: Point;OneOf(Intersection(l, C)) = Q;O: Origin;LineSegmentOf(O, Q) = sqrt(13)/2",
    "query_expressions": "r",
    "answer_expressions": "(2*sqrt(7))/7",
    "fact_spans": "[[[7, 39], [103, 108]], [[7, 39]], [[2, 6], [45, 48]], [[2, 43]], [[49, 91]], [[146, 149]], [[49, 91]], [[55, 91]], [[55, 91]], [[94, 97], [99, 102]], [[44, 97]], [[114, 117]], [[99, 117]], [[119, 144]], [[119, 144]]]",
    "query_spans": "[[[146, 151]]]",
    "process": "B(0,1), let Q(x_{0},y_{0}), then \\frac{x_{0}^{2}}{4}+y_{0}^{2}=1, OQ=\\sqrt{x_{0}^{2}+y_{0}^{2}}=\\sqrt{x_{0}^{2}+1-\\frac{x_{0}^{2}}{4}}=\\sqrt{\\frac{3}{4}x_{0}^{2}+1}=\\frac{\\sqrt{13}}{2} \\therefore x_{0}=\\pm\\sqrt{3}, without loss of generality, let x_{0}=\\sqrt{3}, then y_{0}=\\pm\\frac{1}{2}, when y_{0}=-\\frac{1}{2}, Q(\\sqrt{3},-\\frac{1}{2}), k_{BQ}=\\frac{1+\\frac{1}{2}}{0-\\sqrt{3}}=-\\frac{\\sqrt{3}}{2}, BQ: y=-\\frac{\\sqrt{3}}{2}x+1, i.e. \\frac{\\sqrt{3}}{2}x+y-1=0, the distance from O to BQ is d=\\frac{1}{\\sqrt{\\frac{7}{4}}}=\\frac{2}{\\sqrt{7}}=\\frac{2\\sqrt{7}}{7}<\\frac{6}{7}, in this case, \\frac{2\\sqrt{7}}{7}, when y_{0}=\\frac{1}{2}, Q(\\sqrt{3},\\frac{1}{2}), k_{BQ}=\\frac{1-\\frac{1}{2}}{0-\\sqrt{3}}=-\\frac{\\sqrt{3}}{6}, BQ: \\frac{\\sqrt{3}}{6}x+y-1=0, the distance from O to BQ is d=\\frac{1}{\\sqrt{\\frac{13}{36}+1}}=\\frac{1}{\\sqrt{\\frac{13}{36}+\\frac{36}{36}}}=\\frac{1}{\\sqrt{\\frac{49}{36}}}=\\frac{6}{7}, but corrected calculation gives d=\\frac{1}{\\sqrt{\\frac{13}{36}+1}}=\\frac{1}{\\sqrt{\\frac{49}{36}}}=\\frac{6}{7}, however original text states d=\\frac{1}{\\sqrt{\\frac{13}{2}}}=\\frac{\\sqrt{156}}{13}>\\frac{6}{7}, discarded, \\therefore r=\\frac{2\\sqrt{7}}{7}, the answer is: 2\\sqrt{7}"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, with left and right foci $F_{1}$ and $F_{2}$ respectively, $M$ is an arbitrary point on the ellipse $C$, and $N$ is an arbitrary point on the circle $E$: $(x-3)^{2}+(y-2)^{2}=1$. Then the minimum value of $|M N|-|M F_{1}|$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;M: Point;PointOnCurve(M,C) = True;E: Circle;Expression(E) = ((x - 3)^2 + (y - 2)^2 = 1);N: Point;PointOnCurve(N,E) = True",
    "query_expressions": "Min(-Abs(LineSegmentOf(M, F1)) + Abs(LineSegmentOf(M, N)))",
    "answer_expressions": "2*sqrt(2)-5",
    "fact_spans": "[[[2, 44], [73, 78]], [[2, 44]], [[53, 60]], [[61, 68]], [[2, 68]], [[2, 68]], [[69, 72]], [[69, 83]], [[88, 116]], [[88, 116]], [[84, 87]], [[84, 121]]]",
    "query_spans": "[[[123, 146]]]",
    "process": "As shown in the figure, since M is an arbitrary point on the ellipse C, then |MF_{1}| + |MF_{2}|. Also, N is an arbitrary point on the circle E: (x-3)^{2} + (y-2)^{2} = 1, then |MN| \\geqslant |ME| - 1 (equality holds if and only if M, N, E are collinear). |MN| - |MF_{1}| = |MN| - (4 - |MF_{2}|) = |MN| + |MF_{2}| - 4 \\geqslant |ME| + |MF_{2}| - 5 \\geqslant |EF_{2}| - 5, with equality if and only if M, N, E, F_{2} are collinear. \\because F_{2}(1,0), E(3,2), then |EF_{2}| = \\sqrt{(3-1)^{2} + (2-0)^{2}} = 2\\sqrt{2}. \\therefore The minimum value of |MN| - |MF_{1}| is 2\\sqrt{2} - 5."
  },
  {
    "text": "If the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are $F_{1}$ and $F_{2}$ respectively, and point $P$ lies on the right branch of the hyperbola such that $|P F_{1}|=4|P F_{2}|$, then the maximum value of the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;a>0;b>0;P: Point;F1: Point;F2: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P,F1))=4*Abs(LineSegmentOf(P,F2))",
    "query_expressions": "Max(Eccentricity(G))",
    "answer_expressions": "5/3",
    "fact_spans": "[[[2, 58], [123, 126], [89, 92]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[84, 88]], [[68, 75]], [[76, 83]], [[2, 58]], [[2, 83]], [[2, 83]], [[84, 96]], [[98, 120]]]",
    "query_spans": "[[[123, 136]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with a focus at $F(c, 0)$, the distance from focus $F$ to an asymptote is $d$, and $O$ is the origin. If $|O F|=2 d$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;c: Number;F: Point;O: Origin;d: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (c, 0);OneOf(Focus(C)) = F;Distance(F, Asymptote(C)) = d;Abs(LineSegmentOf(O, F)) = 2*d",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 63], [121, 127]], [[9, 63]], [[9, 63]], [[69, 78]], [[69, 78], [81, 84]], [[97, 101]], [[92, 95]], [[9, 63]], [[9, 63]], [[2, 63]], [[69, 78]], [[2, 78]], [[2, 95]], [[108, 119]]]",
    "query_spans": "[[[121, 133]]]",
    "process": "The equations of the asymptotes are: $bx\\pm ay=0$, the distance from $F$ to the asymptote is $d=\\frac{bc}{\\sqrt{a^{2}+b^{2}}}=b$, so $c=2b$, thus $c^{2}=4b^{2}=4(c^{2}-a^{2})$, i.e., $\\frac{c^{2}}{a^{2}}=\\frac{4}{3}$, the eccentricity $e=\\frac{c}{a}=\\frac{2\\sqrt{3}}{3}$"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has an eccentricity of $2$, and the line $x+y+2=0$ passes through the focus of the hyperbola $C$, then the equations of the asymptotes of the hyperbola $C$ are?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(C) = 2;G: Line;Expression(G) = (x + y + 2 = 0);PointOnCurve(Focus(C), G)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[2, 63], [85, 91], [96, 102]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 71]], [[72, 83]], [[72, 83]], [[72, 94]]]",
    "query_spans": "[[[96, 110]]]",
    "process": "The hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) has eccentricity 2, $ \\frac{c}{a} = 2 $. The line $ x + y + 2 = 0 $ passes through the focus of hyperbola $ C $, yielding $ c = 2 $. Thus, $ a = 1 $. From $ b^{2} = c^{2} - a^{2} = 3 $, we get $ b = \\sqrt{3} $. Since the foci of the hyperbola lie on the x-axis, the asymptotes of hyperbola $ C $ are given by: $ y = \\pm\\sqrt{3}x $."
  },
  {
    "text": "The distance from the focus of the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{1}=1$ to its asymptote is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1)",
    "query_expressions": "Distance(Focus(G), Asymptote(G))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 50]]]",
    "process": ""
  },
  {
    "text": "If the left and right foci of the hyperbola $E$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$ are $F_{1}$ and $F_{2}$ respectively, and point $P$ is a point on the hyperbola such that $|P F_{1}|=5$, then $|P F_{2}|=$?",
    "fact_expressions": "E: Hyperbola;F1: Point;F2: Point;P: Point;LeftFocus(E) = F1;RightFocus(E) = F2;PointOnCurve(P, E);Expression(E) = (x^2/4 - y^2/9 = 1);Abs(LineSegmentOf(P, F1)) = 5",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "9",
    "fact_spans": "[[[1, 44], [77, 80]], [[53, 60]], [[63, 70]], [[72, 76]], [[1, 70]], [[1, 70]], [[72, 84]], [[1, 44]], [[86, 99]]]",
    "query_spans": "[[[101, 114]]]",
    "process": ""
  },
  {
    "text": "Given that points $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively, points $A$ and $B$ lie on the right branch of $C$, and point $F_{2}$ is exactly the circumcenter of $\\Delta F_{1} A B$. If $(\\overrightarrow{B F_{1}}+\\overrightarrow{B A}) \\cdot \\overrightarrow{A F_{1}}=0$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;A: Point;B: Point;LeftFocus(C) = F1;RightFocus(C) = F2;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(A, RightPart(C));PointOnCurve(B, RightPart(C));Circumcenter(TriangleOf(F1, A, B)) = F2;DotProduct((VectorOf(B, A) + VectorOf(B, F1)), VectorOf(A, F1)) = 0",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(\\sqrt{3} + 1)/2",
    "fact_spans": "[[[21, 82], [98, 101], [225, 228]], [[29, 82]], [[29, 82]], [[29, 82]], [[29, 82]], [[2, 10]], [[11, 18], [107, 115]], [[89, 93]], [[94, 97]], [[2, 88]], [[2, 88]], [[21, 82]], [[89, 105]], [[89, 105]], [[107, 139]], [[141, 223]]]",
    "query_spans": "[[[225, 234]]]",
    "process": "Let $ C $ be the midpoint of $ AF_{1} $, connect $ BC $, $ AF_{2} $, $ BF_{2} $, as shown in the figure. Since $ (\\overrightarrow{BF}_{1} + \\overrightarrow{BA}) \\cdot \\overrightarrow{AF_{1}} = \\frac{1}{2} \\overrightarrow{BC} \\cdot \\overrightarrow{AF_{1}} = 0 $, it follows that $ BC \\perp AF_{1} $. Since $ C $ is the midpoint of $ AF_{1} $, $ \\triangle ABF_{1} $ is an isosceles triangle and $ BF_{1} = BA $. Since point $ F_{2} $ is exactly the circumcenter of $ \\triangle F_{1}AB $, point $ F_{2} $ lies on line $ BC $, and $ AF_{2} = BF_{2} = F_{1}F_{2} = 2c $. By the definition of a hyperbola, $ AF_{1} - AF_{2} = BF_{1} - BF_{2} = 2a $, so $ AF_{1} = BF_{1} = 2a + 2c $. Therefore, $ \\triangle ABF $ is an equilateral triangle, so $ BC = \\frac{3}{2} BF_{2} = 3c $. In $ \\triangle CBF_{1} $, $ CB^{2} + CF_{1}^{2} = BF_{1}^{2} $, that is, $ 9c^{2} + (a + c)^{2} = (2a + 2c)^{2} $. Simplifying gives $ 3a^{2} + 6ac - 6c^{2} = 0 $. Dividing both sides by $ a^{2} $ yields $ 2e^{2} - 2e - 1 = 0 $, solving gives $ e = \\frac{1 + \\sqrt{3}}{2} $ or $ \\frac{1 - \\sqrt{3}}{2} $ (discarded)."
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$ are $F_{1}$ and $F_{2}$, respectively. Chord $A B$ passes through point $F_{1}$. If the circumference of the incircle of triangle $A B F_{2}$ is $\\pi$, and the coordinates of points $A$ and $B$ are $(x_{1}, y_{1})$, $(x_{2}, y_{2})$, then $|y_{1}-y_{2}|=$?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;F2: Point;F1: Point;x1: Number;y1: Number;x2: Number;y2: Number;Expression(G) = (x^2/16 + y^2/7 = 1);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);LeftFocus(G) = F1;RightFocus(G) = F2;IsChordOf(LineSegmentOf(A, B),G);PointOnCurve(F1, LineSegmentOf(A, B));Perimeter(InscribedCircle(TriangleOf(A, B, F2))) = pi",
    "query_expressions": "Abs(y1 - y2)",
    "answer_expressions": "4/3",
    "fact_spans": "[[[0, 38]], [[114, 117]], [[118, 121]], [[55, 62]], [[47, 54], [70, 78]], [[129, 145]], [[129, 145]], [[146, 162]], [[146, 162]], [[0, 38]], [[114, 162]], [[114, 162]], [[0, 62]], [[0, 62]], [[0, 69]], [[64, 78]], [[80, 113]]]",
    "query_spans": "[[[164, 181]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and point $P$ in the first quadrant lies on the asymptote such that $\\angle F_{1} P F_{2}=\\frac{\\pi}{2}$. The line $P F_{1}$ intersects the left branch of the hyperbola at point $Q$. If point $Q$ is the midpoint of segment $P F_{1}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F1: Point;P: Point;F2: Point;Q: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;Quadrant(P)=1;PointOnCurve(P,Asymptote(G));AngleOf(F1, P, F2) = pi/2;Intersection(LineOf(P,F1), LeftPart(G)) = Q;MidPoint(LineSegmentOf(P,F1)) = Q",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)+1",
    "fact_spans": "[[[20, 76], [149, 152], [183, 186]], [[23, 76]], [[23, 76]], [[2, 9]], [[88, 92]], [[10, 17]], [[155, 159], [161, 165]], [[23, 76]], [[23, 76]], [[20, 76]], [[2, 82]], [[2, 82]], [[83, 92]], [[20, 97]], [[100, 136]], [[137, 159]], [[161, 180]]]",
    "query_spans": "[[[183, 192]]]",
    "process": ""
  },
  {
    "text": "If the focal distance of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{2 m-4}=1$ is $4 \\sqrt{2}$, then the real number $m$=?",
    "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (-y^2/(2*m - 4) + x^2/m = 1);FocalLength(G) = 4*sqrt(2)",
    "query_expressions": "m",
    "answer_expressions": "{4,-4/3}",
    "fact_spans": "[[[1, 43]], [[61, 66]], [[1, 43]], [[1, 59]]]",
    "query_spans": "[[[61, 68]]]",
    "process": "Classify according to whether the hyperbola's foci lie on the x-axis or y-axis, and thereby find the value of $ m $. [Solution] When the foci are on the x-axis, we have  \n$$\n\\begin{cases}\nm>0, \\\\\n2m-4>0, \\\\\n2m+2m-4=4\\sqrt{2},\n\\end{cases}\n$$\nyielding $ m=4 $; thus, $ m=4 $ or $ -\\frac{4}{3} $."
  },
  {
    "text": "Given the parabola $E$: $x^{2}=4 y$, and the line $y=x+1$ intersects the parabola $E$ at points $P$ and $Q$. Then the length of chord $|P Q|$ is?",
    "fact_expressions": "E: Parabola;Expression(E) = (x^2 = 4*y);G: Line;Expression(G) = (y = x + 1);P: Point;Q: Point;Intersection(G, E) = {P, Q};IsChordOf(LineSegmentOf(P, Q), E)",
    "query_expressions": "Abs(LineSegmentOf(P, Q))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 21], [32, 38]], [[2, 21]], [[22, 31]], [[22, 31]], [[41, 44]], [[45, 48]], [[22, 50]], [[32, 60]]]",
    "query_spans": "[[[53, 62]]]",
    "process": "From \\begin{cases}x^{2}=4y\\\\y=x+1\\end{cases}, we obtain x^{2}-4x-4=0, A=16+16=32. Let P(x_{1},y_{1}), Q(x_{2},y_{2}), then x_{1}+x_{2}=4, \\therefore|PQ|=\\sqrt{2}\\times\\sqrt{4^{2}-4\\times(-4)}=8"
  },
  {
    "text": "Let the right focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ be $F_{1}$, and the right directrix be $l_{1}$. If the length of the chord passing through $F_{1}$ and perpendicular to the $x$-axis is equal to the distance from point $F_{1}$ to $l_{1}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;RightFocus(G) = F1;l1: Line;RightDirectrix(G) = l1;L: LineSegment;PointOnCurve(F1, L);IsPerpendicular(L, xAxis);IsChordOf(L, G);Length(L) = Distance(F1, l1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[1, 53], [121, 123]], [[1, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[58, 65], [80, 87], [100, 108]], [[1, 65]], [[70, 77], [109, 116]], [[1, 77]], [], [[79, 97]], [[79, 97]], [[79, 97]], [[79, 119]]]",
    "query_spans": "[[[121, 129]]]",
    "process": ""
  },
  {
    "text": "The equation of the hyperbola that shares the same asymptotes as $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1$ and passes through the point $(-3 , 4)$?",
    "fact_expressions": "G: Hyperbola;C:Hyperbola;H: Point;Expression(G) = (x^2/9 - y^2/4 = 1);Coordinate(H) = (-3, 4);Asymptote(C) = Asymptote(G);PointOnCurve(H,C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/12 - x^2/27 = 1",
    "fact_spans": "[[[1, 39]], [[62, 65]], [[50, 61]], [[1, 39]], [[50, 61]], [[0, 65]], [[48, 65]]]",
    "query_spans": "[[[62, 68]]]",
    "process": ""
  },
  {
    "text": "Given that one focus of the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1$ coincides exactly with the focus $F$ of the parabola $x^{2}=2 p y$ ($p>0$), let the directrix $l$ of the parabola intersect the $y$-axis at point $M$. A line passing through $F$ intersects the parabola at points $A$ and $B$. If the circle with diameter $BM$ passes through point $A$, then $|A B|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 2*(p*y));p: Number;p>0;H: Ellipse;Expression(H) = (x^2/3 + y^2/4 = 1);C: Circle;M: Point;B: Point;A: Point;F: Point;l:Line;L:Line;Focus(G)=F;OneOf(Focus(H))=F;Directrix(G)=l;Intersection(l,yAxis)=M;PointOnCurve(F,L);Intersection(L, G) = {A, B};IsDiameter(LineSegmentOf(B,M),C);PointOnCurve(A, C)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "2+2*sqrt(5)",
    "fact_spans": "[[[46, 67], [75, 78], [105, 108]], [[46, 67]], [[49, 67]], [[49, 67]], [[2, 39]], [[2, 39]], [[133, 134]], [[93, 96]], [[114, 117]], [[110, 113], [135, 139]], [[70, 73], [98, 101]], [[81, 84]], [[102, 104]], [[46, 73]], [[2, 73]], [[75, 84]], [[81, 96]], [[97, 104]], [[102, 119]], [[121, 134]], [[133, 139]]]",
    "query_spans": "[[[141, 150]]]",
    "process": "From the ellipse equation, it is easy to know the focus coordinates are $ F(0,1) $, and the parabola equation is $ x^{2} = 4y $. Obviously, the slope of line $ AB $ exists and is not zero. Let the slope of line $ AB $ be $ k $, then the equation of $ AB $ is $ x = m(y - 1) $, where $ m = \\frac{1}{k} $. Combining the line equation with the parabola equation yields $ m^{2}y^{2} - (2m^{2} + 4)y + m^{2} = 0 $, solving gives: $ y = \\frac{m^{2} + 2 \\pm 2\\sqrt{m^{2} + 1}}{m^{2}} $, then $ y_{1} + y_{2} = \\frac{4}{m^{2}} + 2 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, the circle with diameter $ BM $ passes through point $ A $, then $ k_{AB} \\cdot k_{AM} = -1 $, i.e.: $ \\frac{1}{m} \\cdot \\frac{y_{1} + 1}{x_{1}} = -1 $. Combining with $ x_{1} = m(y_{1} - 1) $, we get $ y_{1} = \\frac{m^{2} - 1}{m^{2} + 1} $. Accordingly, $ \\frac{m^{2} - 1}{m^{2} + 1} = \\frac{m^{2} + 2 - 2\\sqrt{m^{2} + 1}}{m^{2}} $. Rearranging yields: $ m $. Using the chord length formula, we obtain: $ |AB| = y_{1} + y_{2} + p = \\frac{2}{m^{2}} + 4 - m^{2} - 1 = 0 $, solving gives: $ m^{2} = \\frac{1 + \\sqrt{5}}{2} $ (negative root discarded)."
  },
  {
    "text": "The standard equation of a circle centered at the focus of the parabola $x^{2}=4 y$ and tangent to the asymptotes of $\\frac{y^{2}}{4}-x^{2}=1$ is?",
    "fact_expressions": "G: Parabola;H: Circle;Expression(G) = (x^2 = 4*y);Focus(G) = Center(H);C:Curve;Expression(C)=(y^2/4-x^2=1);IsTangent(H,Asymptote(C))",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2+(y-1)^2=1/5",
    "fact_spans": "[[[1, 15]], [[56, 57]], [[1, 15]], [[0, 57]], [[24, 49]], [[24, 49]], [[23, 57]]]",
    "query_spans": "[[[56, 64]]]",
    "process": ""
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ are $F_{1}$ and $F_{2}$, and $P$ is a point on the ellipse such that $P F_{1} \\perp P F_{2}$. Find the area of $\\triangle F_{1} P F_{2}$.",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "9",
    "fact_spans": "[[[0, 38], [63, 65]], [[59, 62]], [[41, 48]], [[49, 57]], [[0, 38]], [[0, 57]], [[59, 69]], [[72, 95]]]",
    "query_spans": "[[[97, 127]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are tangent to the circle $x^{2}+y^{2}-4 x+2=0$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (-4*x + x^2 + y^2 + 2 = 0);IsTangent(Asymptote(G),H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 58], [90, 93]], [[5, 58]], [[5, 58]], [[63, 85]], [[5, 58]], [[5, 58]], [[2, 58]], [[63, 85]], [[2, 87]]]",
    "query_spans": "[[[90, 99]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of a hyperbola are $y=\\pm 2 x$, and it passes through the point $(\\sqrt{2}, 2)$, then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (sqrt(2), 2);Expression(Asymptote(G))=(y=pm*2*x);PointOnCurve(H,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/4 = 1",
    "fact_spans": "[[[2, 5], [43, 46]], [[24, 40]], [[24, 40]], [[2, 21]], [[2, 40]]]",
    "query_spans": "[[[43, 53]]]",
    "process": "Given the asymptotes of the hyperbola are $ y = \\pm 2x $, the equation of the hyperbola can be written as: $ 4x^{2} - y^{2} = \\lambda $. Since the hyperbola passes through the point $ (\\sqrt{2}, 2) $, we have $ \\therefore \\lambda = 4 \\times 2 - 4 = 4 $. Therefore, the standard equation of the hyperbola is: $ x^{2} - \\frac{y^{2}}{4} = 1 $. The correct answer to this problem is: $ x^{2} - \\frac{y^{2}}{4} = 1 $. This question examines the method of finding the equation of a hyperbola given its asymptotes and a point on the hyperbola, and is a basic-level problem."
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $\\frac{y^{2}}{5}-\\frac{x^{2}}{4}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/4 + y^2/5 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(5)/2)*x",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 46]]]",
    "process": "Since the hyperbola equation is \\frac{y^{2}}{5}-\\frac{x^{2}}{4}=1, let \\frac{y^{2}}{5}-\\frac{x^{2}}{4}=0, that is, \\frac{y^{2}}{5}=\\frac{x^{2}}{4}. Taking square roots on both sides gives y=\\pm\\frac{\\sqrt{5}}{2}x, so the asymptotes of the hyperbola are y=\\pm\\frac{\\sqrt{5}}{2}x"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through the origin $O$ with slope $\\sqrt{3}$ intersects the right branch of $C$ at point $A$. If $\\angle F_{1} A F_{2}=\\frac{2 \\pi}{3}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;F1: Point;A: Point;F2: Point;O: Origin;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(O, G);Slope(G)=sqrt(3);Intersection(G, RightPart(C)) = A;AngleOf(F1, A, F2) = (2*pi)/3",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(3*sqrt(2)+sqrt(10))/2",
    "fact_spans": "[[[2, 63], [111, 114], [164, 167]], [[9, 63]], [[9, 63]], [[108, 110]], [[72, 79]], [[118, 122]], [[80, 87]], [[89, 94]], [[9, 63]], [[9, 63]], [[2, 63]], [[2, 87]], [[2, 87]], [[88, 110]], [[95, 110]], [[108, 122]], [[124, 162]]]",
    "query_spans": "[[[164, 173]]]",
    "process": "By the given condition, $\\angle F_{1}OA = \\frac{2\\pi}{3}$; combining with the known conditions, it can be proved that $\\triangle F_{1}OA \\sim \\triangle F_{1}AF_{2}$; using $\\frac{F_{1}O}{F_{1}A} = \\frac{F_{1}A}{F_{1}F_{2}}$, we can calculate $F_{1}A = \\sqrt{2}c$; in $\\triangle F_{1}AF_{2}$, using the cosine law, we can compute $AF_{2} = \\frac{\\sqrt{10}-\\sqrt{2}}{2}c'$; from $e = \\frac{2c}{2a} = \\frac{F_{1}F_{2}}{AF_{1}-AF_{2}}$, the eccentricity can thus be obtained. By the given condition, the slope of line $OA$ is $\\sqrt{3}$, so $\\angle AOF_{2} = \\frac{\\pi}{3}$. Therefore, $\\angle F_{1}OA = \\frac{2\\pi}{3}$; also because $\\angle F_{1}AF_{2} = \\frac{2\\pi}{3}$, $\\angle Alr_{1}O = \\angle F_{2}F_{1}A_{1}$, so $\\triangle F_{1}OA \\sim \\triangle F_{1}AF_{2}$, hence $\\frac{F_{1}O}{F_{A}} = \\frac{F_{1}A}{F_{1}F_{2}}$, that is, $\\frac{c}{F_{1}} = \\frac{F_{1}A}{2c}$, yielding $F_{1}A = \\sqrt{2}c$; in $\\triangle F_{1}AF_{2}$, by the cosine law we obtain $F_{1}F_{2}^{2} = AF_{1}^{2} + AF_{2}^{2} - 2AF_{1} \\cdot AF_{2} \\cos \\frac{2\\pi}{3}$; solving gives: $AF_{2} = \\frac{\\sqrt{10}-\\sqrt{2}}{2}c'$; thus the hyperbola's eccentricity is $e = \\frac{2c}{2a} = \\frac{F_{1}F_{2}}{AF_{1}-AF_{2}} =$"
  },
  {
    "text": "Given that the center of the hyperbola is at the origin, the foci lie on the $x$-axis, the focal distance $2c = 4$, and it passes through the point $(2,3)$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;O: Origin;Coordinate(H) = (2, 3);Center(G) = O;PointOnCurve(Focus(G), xAxis);FocalLength(G) = 2*c;2*c = 4;c: Number;PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/3 = 1",
    "fact_spans": "[[[2, 5], [42, 45]], [[32, 40]], [[9, 11]], [[32, 40]], [[2, 11]], [[2, 20]], [[2, 30]], [[23, 30]], [[23, 30]], [[2, 40]]]",
    "query_spans": "[[[42, 52]]]",
    "process": ""
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2 x$ be $F$, and let a line passing through $F$ intersect the parabola at points $A$ and $B$. Then the minimum value of $|A F|+4|B F|$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 2*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B}",
    "query_expressions": "Min(Abs(LineSegmentOf(A, F)) + 4*Abs(LineSegmentOf(B, F)))",
    "answer_expressions": "9/2",
    "fact_spans": "[[[1, 15], [32, 35]], [[28, 30]], [[36, 39]], [[19, 22], [24, 27]], [[40, 43]], [[1, 15]], [[1, 22]], [[23, 30]], [[28, 43]]]",
    "query_spans": "[[[45, 65]]]",
    "process": "Let the focus of the parabola $ y^{2} = 2x $ be $ F\\left(\\frac{1}{2}, 0\\right) $. When $ AB \\perp x $-axis, $ |AF| + 4|BF| = 1 + 4 = 5 $. When the line $ AB $ has a slope, the equation of line $ AB $ can be written as $ y = k\\left(x - \\frac{1}{2}\\right) $. Substituting into the parabola equation gives $ 4k^{2}x^{2} - (4k^{2} + 2)x + k^{2} = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = \\frac{4k^{2} + 2}{4k^{2}} $, $ x_{1}x_{2} = \\frac{1}{4} $. $ |AF| + 4|BF| = x_{1} + \\frac{1}{2} + 4\\left(x_{2} + \\frac{1}{2}\\right) = x_{1} + 4x_{2} + \\frac{5}{2} \\geqslant 2\\sqrt{4x_{1}x_{2}} + \\frac{5}{2} = \\frac{9}{2} $, with equality if and only if $ x_{1} = 4x_{2} = 1 $, i.e., $ x_{1} = 1 $, $ x_{2} = \\frac{1}{4} $, so $ |AF| + 4|BF| $ attains the minimum value $ \\frac{9}{2} $. Therefore, fill in $ \\frac{9}{2} $."
  },
  {
    "text": "The minor axis of an ellipse is $6$, and the focal distance is $8$. Then its major axis length equals?",
    "fact_expressions": "G: Ellipse;Length(MinorAxis(G))= 6;FocalLength(G) = 8",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "10",
    "fact_spans": "[[[0, 2], [19, 20]], [[0, 10]], [[0, 17]]]",
    "query_spans": "[[[19, 27]]]",
    "process": "2b=6,b=3,2c=8,c=4,a^{2}=b^{2}+c^{2}=3^{2}+4^{2}=25,a=5,2a=10"
  },
  {
    "text": "Let $F$ be the focus of the parabola $C$: $y^{2}=8x$. A line passing through $F$ with an inclination angle of $60^{\\circ}$ intersects $C$ at points $A$ and $B$, and $O$ is the origin. Then the area of $\\triangle OAB$ is?",
    "fact_expressions": "C: Parabola;G: Line;O: Origin;A: Point;B: Point;F: Point;Expression(C) = (y^2 = 8*x);Focus(C) = F;PointOnCurve(F, G);Inclination(G) = ApplyUnit(60, degree);Intersection(G, C) = {A, B}",
    "query_expressions": "Area(TriangleOf(O, A, B))",
    "answer_expressions": "16*sqrt(3)/3",
    "fact_spans": "[[[5, 24], [53, 56]], [[50, 52]], [[67, 70]], [[57, 60]], [[61, 64]], [[1, 4], [29, 32]], [[5, 24]], [[1, 27]], [[28, 52]], [[33, 52]], [[50, 66]]]",
    "query_spans": "[[[77, 99]]]",
    "process": "First, from the parabola equation, obtain F(2,0), derive the equation of line AB, use the focal chord formula of the parabola to find the chord length |AB|, then use the point-to-line distance formula to obtain the result. Since F is the focus of the parabola C: y^{2}=8x, we have F(2,0). The line AB passes through point F and has an inclination angle of 60^{\\circ}. Thus, the equation of line AB is: y=\\sqrt{3}(x-2), or \\sqrt{3}x-y-2\\sqrt{3}=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}). From \\begin{cases}y=\\sqrt{3}(x-2)\\\\y^{2}=8x\\end{cases}, eliminating y gives 3(x-2)^{2}=8x, which simplifies to 3x^{2}-20x+12=0. Therefore, x_{1}+x_{2}=\\frac{20}{3}. Hence, |AB|=x_{1}+x_{2}+4=\\frac{20}{3}+4=\\frac{32}{3}. The distance from point O to the line \\sqrt{3}x-y-2\\sqrt{3}=0 is \\frac{|-2\\sqrt{3}|}{\\sqrt{3+1}}=\\sqrt{3}. Thus, the area of \\triangle OAB is S_{\\triangle OAB}=\\frac{1}{2}|AB|\\cdot d=\\frac{16\\sqrt{3}}{3}."
  },
  {
    "text": "Let $M(-5 , 0)$, $N(5 , 0)$, and the perimeter of $\\triangle M N P$ is $36$. Then the trajectory equation of vertex $P$ of $\\triangle M N P$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (-5, 0);N: Point;Coordinate(N) = (5, 0);Perimeter(TriangleOf(M, N, P)) = 36;P: Point",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(x^2/169+y^2/144=1)&Negation(y=0)",
    "fact_spans": "[[[1, 12]], [[1, 12]], [[14, 24]], [[14, 24]], [[26, 51]], [[73, 76]]]",
    "query_spans": "[[[73, 83]]]",
    "process": ""
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{a^{2}}=1$ and the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{2}=1$ have the same foci; then the real number $a$=?",
    "fact_expressions": "G: Hyperbola;a: Real;H: Ellipse;Expression(G) = (-y^2/2 + x^2/a = 1);Expression(H) = (x^2/4 + y^2/a^2 = 1);Focus(H) = Focus(G)",
    "query_expressions": "a",
    "answer_expressions": "1",
    "fact_spans": "[[[42, 80]], [[88, 93]], [[0, 41]], [[42, 80]], [[0, 41]], [[0, 86]]]",
    "query_spans": "[[[88, 95]]]",
    "process": "From the ellipse: $c^{2}=4-a^{2}$, from the hyperbola: $c^{2}=2+a$. Since the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{a^{2}}=1$ and the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{2}=1$ have the same foci, it follows that $a>0$, $4-a^{2}=2+a$, so $a=1$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$, respectively, a tangent line drawn from point $F_{2}$ to the circle $x^{2}+y^{2}=a^{2}$ intersects the left branch of the hyperbola at point $M$, and $\\angle F_{1} M F_{2}=60^{\\circ}$, then the asymptotes of this hyperbola have equations?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Circle;Expression(H) = (x^2 + y^2 = a^2);L: Line;TangentOfPoint(F2, H) = L;M: Point;Intersection(L, LeftPart(G)) = M;AngleOf(F1, M, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*((1+(sqrt(3)/3))*x)",
    "fact_spans": "[[[20, 79], [120, 123], [168, 171]], [[20, 79]], [[23, 79]], [[23, 79]], [[23, 79]], [[23, 79]], [[2, 9]], [[10, 17], [87, 95]], [[2, 85]], [[2, 85]], [[96, 116]], [[96, 116]], [], [[86, 119]], [[126, 130]], [[86, 130]], [[132, 165]]]",
    "query_spans": "[[[168, 179]]]",
    "process": "Let the point of tangency be $ A $, and draw $ F_{1}B \\perp MF_{2} $ from $ F_{1} $, with foot $ B $. From the given conditions, $ |OA| = a $, $ |OF_{2}| = c $, $ |AF_{2}| = \\sqrt{c^{2} - a^{2}} = b $. Since $ OA $ is the midline of $ \\triangle BF_{1}F_{2} $, we have $ |BF_{1}| = 2a $, $ |BF_{2}| = 2b $. Given $ \\angle F_{1}MF_{2} = 60^{\\circ} $, we obtain $ |MF_{1}| = \\frac{|BF_{1}|}{\\sin 60^{\\circ}} = \\frac{4a}{\\sqrt{3}} $, $ |MB| = \\frac{2a}{\\sqrt{3}} $, $ |MF_{2}| = |MB| + |BF_{2}| = \\frac{2a}{\\sqrt{3}} + 2b $. Also, $ |MF_{2}| - |MF_{1}| = \\frac{2a}{\\sqrt{3}} + 2b - \\frac{4a}{\\sqrt{3}} = 2a $, so $ b = \\left(1 + \\frac{\\sqrt{3}}{3}\\right)a $. Therefore, the asymptotes of the hyperbola are $ y = \\pm\\left(1 + \\frac{\\sqrt{3}}{3}\\right)x $."
  },
  {
    "text": "Ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, $F_{1}$, $F_{2}$ are the two foci of ellipse $C$, $P$ is a point on ellipse $C$, connect $P F_{1}$ and extend it to intersect the ellipse at another point $Q$, then the perimeter of $\\triangle P Q F_{2}$?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;Q: Point;F2: Point;Expression(C) = (x^2/4 + y^2/3 = 1);Focus(C) = {F1, F2};PointOnCurve(P, C);Intersection(OverlappingLine(LineSegmentOf(P, F1)), C) = Q",
    "query_expressions": "Perimeter(TriangleOf(P, Q, F2))",
    "answer_expressions": "8",
    "fact_spans": "[[[0, 43], [62, 67], [76, 81], [100, 102]], [[72, 75]], [[45, 52]], [[107, 110]], [[53, 61]], [[0, 43]], [[45, 71]], [[72, 84]], [[85, 110]]]",
    "query_spans": "[[[112, 137]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=16 x$, the line $l$ passing through $F$ is perpendicular to the line $x+\\sqrt{3} y-1=0$, and the line $l$ intersects the parabola $C$ at points $A$ and $B$, then $|A B|=$?",
    "fact_expressions": "l: Line;C: Parabola;G: Line;A: Point;B: Point;F: Point;Expression(C) = (y^2 = 16*x);Expression(G) = (x + sqrt(3)*y - 1 = 0);Focus(C) = F;PointOnCurve(F,l);IsPerpendicular(l,G);Intersection(l, C) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "64/3",
    "fact_spans": "[[[35, 40], [65, 70]], [[6, 26], [71, 77]], [[41, 61]], [[79, 82]], [[83, 86]], [[2, 5], [31, 34]], [[6, 26]], [[41, 61]], [[2, 29]], [[30, 40]], [[35, 63]], [[65, 88]]]",
    "query_spans": "[[[90, 99]]]",
    "process": "F is the focus of the parabola C: y^{2}=16x, \\therefore F(4,0). Since the line l passing through F is perpendicular to the line x+\\sqrt{3}y-1=0, the equation of line l is: y=\\sqrt{3}(x-4). Substituting into the parabola C: y^{2}=16x, we easily obtain: 3x^{2}-40x+48=0. Let A=(x_{1},y_{1}), B=(x_{2},y_{2}), then x_{1}+x_{2}=\\frac{40}{3}, x_{1}x_{2}=16. |AB|=\\sqrt{1+3}\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\frac{64}{2}"
  },
  {
    "text": "The line $l$ intersects the parabola $C$: $y^{2}=4x$ at points $M$ and $N$. If the product of the slopes of $OM$ and $ON$ is $-\\frac{1}{2}$, then the minimum value of $|MN|$ is?",
    "fact_expressions": "l: Line;C: Parabola;O: Origin;M: Point;N: Point;Expression(C) = (y^2 = 4*x);Intersection(l, C) = {M, N};Slope(LineSegmentOf(O, M))*Slope(LineSegmentOf(O, N)) = -1/2",
    "query_expressions": "Min(Abs(LineSegmentOf(M, N)))",
    "answer_expressions": "8*sqrt(2)",
    "fact_spans": "[[[0, 5]], [[6, 25]], [[38, 43]], [[27, 30]], [[31, 34]], [[6, 25]], [[0, 36]], [[38, 71]]]",
    "query_spans": "[[[73, 86]]]",
    "process": "Let $ M\\left(\\frac{y_1^2}{4}, y_1\\right), N\\left(\\frac{y_2^2}{4}, y_2\\right) $, then $ \\frac{y_{\\frac{y_2}{2y_2}}} = -\\frac{1}{2} $, hence $ y_1 y_2 = -32 $. By the AM-GM inequality, $ y_1^2 + y_2^2 \\geqslant 2|y_1 y_2| = 64 $, $ \\frac{y_1^2}{4} + \\frac{y_2^2}{4} = 32 $, equality holds if and only if $ y_1 = -y_2 $, i.e., $ y_1 = 4\\sqrt{2}, y_2 = -4\\sqrt{2} $ or $ y_1 = -4\\sqrt{2}, y_2 = 4\\sqrt{2} $. Thus, $ |MN|_{\\min} = 8\\sqrt{2} $."
  },
  {
    "text": "Given that the right focus of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$ is $F$, and $P$ is a point on the ellipse, point $A(0,3 \\sqrt{3})$. When the perimeter of $\\triangle A P F$ is maximized, the area of $\\triangle A P F$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/7 = 1);F: Point;RightFocus(G) = F;P: Point;PointOnCurve(P, G);A: Point;Coordinate(A) = (0, 3*sqrt(3));WhenMax(Perimeter(TriangleOf(A, P, F)))",
    "query_expressions": "Area(TriangleOf(A, P, F))",
    "answer_expressions": "120*sqrt(3)/11",
    "fact_spans": "[[[2, 40], [53, 55]], [[2, 40]], [[45, 48]], [[2, 48]], [[49, 52]], [[49, 58]], [[59, 77]], [[59, 77]], [[78, 102]]]",
    "query_spans": "[[[103, 125]]]",
    "process": "In the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$, $a=4$, $b=\\sqrt{7}$, $\\therefore c=3$. According to the problem, let $F'$ be the left focus, then the perimeter of $\\triangle APF$ is $|AF|+|AP|+|PF|=|AF|+|AP|+2a-|PF'|=4+6+|PA|-|PF'|\\leqslant10+|AF'|$ (equality holds when points $A$, $P$, $F'$ are collinear and $P$ lies on the extension of $AF$), at this time $k_{AP}=\\sqrt{3}$, $\\therefore \\angle AFF=\\frac{\\pi}{3}$, $\\therefore \\angle FFP=\\frac{2\\pi}{3}$. Let $|PF|=x$, then $|PF|=8-x$. By the law of cosines, $(8-x)^{2}=x^{2}+36-2\\times6x\\cdot\\cos\\frac{2\\pi}{3}$, $\\therefore x=\\frac{14}{11}$, so the area $S$ of $\\triangle APF$ is $S=S_{\\Delta AFF}+S_{\\Delta PFF}=\\frac{1}{2}\\times6\\times(3\\sqrt{3}+\\frac{14}{11}\\times\\frac{\\sqrt{3}}{2})=\\frac{120\\sqrt{3}}{11}$"
  },
  {
    "text": "Given the equation of the ellipse is $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, $F_{1}$, $F_{2}$ are the left and right foci of the ellipse respectively, the coordinates of point $A$ are $(2,1)$, and $P$ is a point on the ellipse. Then the maximum value of $|P A|+|P F_{2}|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;A: Point;F1:Point;F2: Point;Coordinate(A) = (2, 1);Expression(G) = (x^2/25 + y^2/16 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P,G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F2)))",
    "answer_expressions": "10+sqrt(26)",
    "fact_spans": "[[[2, 4], [2, 4], [2, 4]], [[95, 98]], [[77, 81]], [[48, 55]], [[58, 65]], [[77, 92]], [[2, 45]], [[48, 76]], [[48, 76]], [[95, 104]]]",
    "query_spans": "[[[106, 129]]]",
    "process": "\\because the equation of the ellipse is \\frac{x^2}{25}+\\frac{y^{2}}{16}=1, \\therefore a^{2}=25, c^{2}=a^{2}-b^{2}=9, then 2a=10, c=3, \\because A(2,1), F_{1}(-3,0), \\therefore |AF_{1}|=\\sqrt{(2+3)^{2}+(1-0)^{2}}=\\sqrt{26}, \\because |PF_{1}|+|PF_{2}|=2a=10, \\therefore |PA|+|PF_{2}|=10+|PA|-|PF_{1}|\\leqslant10+|AF_{1}|=10+\\sqrt{26}"
  },
  {
    "text": "Given that a line $L$ with slope $k$ intersects the ellipse $C$: $\\frac{x^{2}}{6}+\\frac{y^{2}}{3}=1$ at points $A$ and $B$, if the midpoint of segment $AB$ is $M(-1,1)$, then the value of $k$ is?",
    "fact_expressions": "C: Ellipse;L: Line;B: Point;A: Point;M: Point;Expression(C) = (x^2/6 + y^2/3 = 1);Coordinate(M) = (-1, 1);Slope(L)=k;k:Number;Intersection(L, C) = {A, B};MidPoint(LineSegmentOf(A,B)) = M",
    "query_expressions": "k",
    "answer_expressions": "1/2",
    "fact_spans": "[[[15, 57]], [[9, 14]], [[64, 67]], [[60, 63]], [[82, 91]], [[15, 57]], [[82, 91]], [[2, 14]], [[5, 8], [93, 96]], [[9, 69]], [[71, 91]]]",
    "query_spans": "[[[93, 100]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, substituting into the ellipse equation gives:\n\\[\n\\begin{cases}\n\\frac{x^{2}}{6}+\\frac{y^{2}}{3}=1 \\\\\n\\frac{x^{2}}{6}+\\frac{y^{2}}{3}=1\n\\end{cases}\n\\]\nSubtracting the two equations yields:\n\\[\n\\frac{x_{1}^{2}-x_{2}^{2}}{6}+y_{1}^{2}-y_{2}^{2}\n\\]\nThat is:\n\\[\n\\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{6}+\\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{3}=0\n\\Rightarrow k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{1}{2}\\cdot\\frac{x_{1}+x_{2}}{y_{1}+y_{2}}\n\\]\nSince the midpoint of segment $ AB $ is $ M(-1,1) $, we have:\n\\[\nx_{1}+x_{2}=-2, \\quad y_{1}+y_{2}=2\n\\Rightarrow k_{AB}=\\frac{1}{2}\n\\]\nThe correct result for this problem: $ \\frac{1}{2} $"
  },
  {
    "text": "Given that the focus of parabola $C$ is $F$, points $A$ and $B$ lie on $C$ such that $\\overrightarrow{A F}+\\overrightarrow{B F}=\\overrightarrow{0}$, and $\\overrightarrow{A F} \\cdot \\overrightarrow{B F}=-16$. Point $P$ is an arbitrary point on the directrix of the parabola. Then the area of $\\Delta P A B$ is?",
    "fact_expressions": "C: Parabola;F: Point;Focus(C) = F;A: Point;B: Point;PointOnCurve(A, C);PointOnCurve(B, C);VectorOf(A,F)+VectorOf(B,F) = 0;DotProduct(VectorOf(A, F),VectorOf(B, F)) = -16;P: Point;PointOnCurve(P, Directrix(C))",
    "query_expressions": "Area(TriangleOf(P, A, B))",
    "answer_expressions": "16",
    "fact_spans": "[[[2, 8], [155, 158], [25, 28]], [[12, 15]], [[2, 15]], [[16, 20]], [[21, 24]], [[16, 29]], [[21, 29]], [[32, 94]], [[96, 149]], [[150, 154]], [[150, 166]]]",
    "query_spans": "[[[168, 187]]]",
    "process": "Let the parabola $ C: y^2 = 2px $ ($ p > 0 $). Since $ \\overrightarrow{AF} + \\overrightarrow{BF} = \\overrightarrow{0} $, it follows that $ \\overrightarrow{AF} = \\overrightarrow{FB} $, so $ F $ is the midpoint of segment $ AB $, and thus $ AB $ is perpendicular to the $ x $-axis. Then $ \\overrightarrow{AF} \\cdot \\overrightarrow{BF} = -p^2 = -16 $, so $ p = 4 $, $ |AB| = 2p = 8 $, and the distance from point $ P $ to $ AB $ is $ p = 4 $. Therefore, $ S_{\\triangle PAB} = \\frac{1}{2} \\times 8 \\times 4 = 16 $."
  },
  {
    "text": "The line $\\sqrt{3} x-y-\\sqrt{3}=0$ intersects the parabola $y^{2}=4 x$ at points $A$ and $B$, and intersects the $x$-axis at point $F$. If $\\overrightarrow{O F}=\\lambda \\overrightarrow{O A}+\\mu \\overrightarrow{O B}$, then $9 \\mu^{2}-\\lambda^{2}$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Line;Expression(H) = (sqrt(3)*x - y - sqrt(3) = 0);A: Point;B: Point;Intersection(H, G) = {A, B};O: Origin;F: Point;Intersection(H,xAxis) = F;lambda: Number;mu: Number;VectorOf(O, F) = lambda*VectorOf(O, A) + mu*VectorOf(O, B)",
    "query_expressions": "-lambda^2 + 9*mu^2",
    "answer_expressions": "5",
    "fact_spans": "[[[28, 42]], [[28, 42]], [[0, 27]], [[0, 27]], [[44, 47]], [[48, 51]], [[0, 53]], [[67, 143]], [[61, 65]], [[0, 65]], [[67, 143]], [[67, 143]], [[67, 143]]]",
    "query_spans": "[[[145, 170]]]",
    "process": "From \\begin{cases}\\sqrt{3}x-y-\\sqrt{3}=0\\\\y^{2}=4x\\end{cases} we obtain: 3x^{2}-10x+3=0, i.e., (3x-1)(x-3)=0. Solving gives: x=3 or x=\\frac{1}{3}. When x=3, y=2\\sqrt{3}; when x=\\frac{1}{3}, y=-\\frac{2\\sqrt{3}}{3}. Thus, A(3,2\\sqrt{3}), B(\\frac{1}{3},-\\frac{2\\sqrt{3}}{3}). In the line \\sqrt{3}x-y-\\sqrt{3}=0, setting y=0 yields: x=1, so F(1,0). Since \\overrightarrow{OF}=\\lambda\\overrightarrow{OA}+\\mu\\overrightarrow{OB}, we have (1,0)=(3\\lambda,2\\sqrt{3}\\lambda)+(\\frac{1}{3}\\mu,-\\frac{2\\sqrt{3}}{3}\\mu). Therefore, \\begin{cases}3\\lambda+\\frac{1}{3}\\mu=1\\\\2\\sqrt{3}\\lambda-\\frac{2\\sqrt{3}}{3}\\mu=0\\end{cases}, solving gives \\begin{cases}\\lambda=\\frac{1}{4}\\\\\\mu=\\frac{3}{4}\\end{cases}. Hence, 9\\mu^{2}-\\lambda^{2}=9\\times\\frac{9}{16}-\\frac{1}{16}=5."
  },
  {
    "text": "There are three points on the plane: $A(-2 , y)$, $B(0, \\frac{y}{2})$, $C(x , y)$. If $\\overrightarrow{A B} \\perp \\overrightarrow{B C}$, then what is the trajectory equation of the moving point $C$?",
    "fact_expressions": "A: Point;B: Point;C: Point;x:Number;y:Number;Coordinate(A) = (-2,y) ;Coordinate(B) =(0, y/2);Coordinate(C) = (x, y);IsPerpendicular(VectorOf(B, C),VectorOf(A, B))",
    "query_expressions": "LocusEquation(C)",
    "answer_expressions": "(y^2 = 8*x)&(Negation(x = 0))",
    "fact_spans": "[[[7, 19]], [[21, 40]], [[43, 53], [108, 111]], [[43, 53]], [[43, 53]], [[7, 19]], [[21, 40]], [[43, 53]], [[55, 104]]]",
    "query_spans": "[[[108, 117]]]",
    "process": "From the given conditions, $\\overrightarrow{AB}=(2,-\\frac{y}{2})$, $\\overrightarrow{BC}=(x,\\frac{y}{2})$. Since $\\overrightarrow{AB}\\bot\\overrightarrow{BC}$, it follows that $\\overrightarrow{AB}\\cdot\\overrightarrow{BC}=0$, i.e., $(2,-\\frac{y}{2})\\cdot(x,\\frac{y}{2})=0$. Simplifying yields $y^{2}=-8x$ ($x\\neq0$)."
  },
  {
    "text": "From a point $P$ on the parabola $y^{2}=4 x$, draw a perpendicular to the directrix of the parabola, with foot of perpendicular at $M$, and $| PM |=5$. Let the focus of the parabola be $F$. Then the area of $\\triangle MPF$ is?",
    "fact_expressions": "G: Parabola;M: Point;P: Point;F: Point;l: Line;Expression(G) = (y^2 = 4*x);PointOnCurve(P, G);PointOnCurve(P, l);IsPerpendicular(l, Directrix(G));FootPoint(l, Directrix(G)) = M;Abs(LineSegmentOf(P, M)) = 5;Focus(G) = F",
    "query_expressions": "Area(TriangleOf(M, P, F))",
    "answer_expressions": "10",
    "fact_spans": "[[[1, 15], [22, 25], [52, 55]], [[34, 37]], [[18, 21]], [[59, 62]], [], [[1, 15]], [[1, 21]], [[0, 30]], [[22, 30]], [[22, 37]], [[39, 49]], [[52, 62]]]",
    "query_spans": "[[[64, 84]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ with focus $F$, a line passing through the focus $F$ intersects the parabola at points $P$ and $Q$, and $\\frac{1}{|P F|}+\\frac{1}{|F Q|}=8$. Then the equation of the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);Intersection(H, G) = {P, Q};P: Point;Q: Point;1/Abs(LineSegmentOf(P, F)) + 1/Abs(LineSegmentOf(F, Q)) = 8",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=-1/8",
    "fact_spans": "[[[2, 23], [41, 44], [93, 96]], [[2, 23]], [[5, 23]], [[5, 23]], [[27, 30], [34, 37]], [[2, 30]], [[38, 40]], [[31, 40]], [[38, 54]], [[45, 48]], [[49, 52]], [[56, 91]]]",
    "query_spans": "[[[93, 103]]]",
    "process": "Let the angle between the line PQ and the x-axis be $\\alpha$ $(0<\\alpha\\leqslant\\frac{\\pi}{2})$. According to the symmetry of the parabola, assume $|PF|\\geqslant|QF|$, as shown in the figure. Let E be the intersection point of the directrix of the parabola and the x-axis. Draw perpendiculars from point P to the directrix and the x-axis, with feet of perpendiculars denoted as P' and H, respectively. Draw perpendiculars from point Q to the directrix and the x-axis, with feet of perpendiculars denoted as Q' and G, respectively. By the definition of the parabola, $|PF|=|PP'|=|EH|=|EF|+|FH|=p+|PF|\\cos\\alpha \\Rightarrow |PF|=\\frac{p}{1-\\cos\\alpha}$. Similarly: $|QF|=|QQ'|=|EG|=|EF|-|GF|=p-|QF|\\cos\\alpha \\Rightarrow |QF|=\\frac{p}{1+\\cos\\alpha}$. Thus, $\\frac{1}{|PF|}+\\frac{1}{|QF|}=\\frac{1-\\cos\\alpha}{p}+\\frac{1+\\cos\\alpha}{p}=\\frac{2}{p}=8 \\Rightarrow p=\\frac{1}{4}$, then the equation of the directrix of the parabola is: $x=-\\frac{1}{8}$."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ with focus $F$, a line $l$ passing through $F$ intersects the parabola at points $A$ and $B$, and intersects the directrix at point $E$. Point $F$ is the midpoint of $AE$, and $|B F|=2$. Then $|B E|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, G) = {A, B};E: Point;Intersection(l, Directrix(G)) = E;MidPoint(LineSegmentOf(A, E)) = F;Abs(LineSegmentOf(B, F)) = 2",
    "query_expressions": "Abs(LineSegmentOf(B, E))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 23], [43, 46]], [[2, 23]], [[5, 23]], [[5, 23]], [[27, 30], [32, 36], [72, 76]], [[2, 30]], [[37, 42], [58, 63]], [[31, 42]], [[48, 51]], [[52, 55]], [[37, 57]], [[67, 71]], [[43, 71]], [[72, 85]], [[87, 96]]]",
    "query_spans": "[[[98, 107]]]",
    "process": "From the given conditions, the graph is shown as follows: Draw BB and AA perpendicular to the directrix, with feet at B and A respectively. According to the definition of a parabola: |BF| = |BB| = 2, |AA| = |AF|. Since F is the midpoint of AE, we have: |AA| = 2|FF| = 2p. Therefore, |AF| = |EF| = |AA| = 2p. From BB // AA, it follows that: \\frac{|BE|}{|AE|} = \\frac{|BB|}{|AA|}, i.e., \\frac{2p-2}{4p} = \\frac{2}{2p} \\Rightarrow p = 3. Therefore, |BE| = 2p - 2 = 4. The correct answer is: 4"
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{4-m}-\\frac{y^{2}}{2+m}=1$ represents a hyperbola, what is the range of real values for $m$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(4 - m) - y^2/(m + 2) = 1);m: Real",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-2, 4)",
    "fact_spans": "[[[45, 48]], [[2, 48]], [[50, 55]]]",
    "query_spans": "[[[50, 62]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, with vertices $M$ and $N$, and a point $P$ on $C$ such that the product of the slopes of lines $PM$ and $PN$ is $3$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;M: Point;N: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Vertex(C)={M,N};PointOnCurve(P,C);Slope(LineOf(P, N))*Slope(LineOf(P,M)) = 3",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 63], [81, 84], [113, 119]], [[10, 63]], [[10, 63]], [[77, 80]], [[69, 72]], [[73, 76]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 76]], [[77, 87]], [[88, 111]]]",
    "query_spans": "[[[113, 125]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$ and directrix $l$, let $P$ be a point on $l$, and the line $PF$ intersects the curve $C$ at points $M$ and $N$. If $\\overrightarrow{P M}=3 \\overrightarrow{M F}$, then $|M N|=$?",
    "fact_expressions": "C: Parabola;F: Point;P: Point;M: Point;N: Point;l: Line;Expression(C) = (y^2=8*x);Focus(C) = F;Directrix(C)=l;PointOnCurve(P, l);Intersection(LineOf(P,F),C)={M,N};VectorOf(P,M)=3*VectorOf(M,F)",
    "query_expressions": "Abs(LineSegmentOf(M, N))",
    "answer_expressions": "9",
    "fact_spans": "[[[2, 21], [57, 62]], [[25, 28]], [[38, 41]], [[65, 68]], [[69, 72]], [[32, 35], [42, 45]], [[2, 21]], [[2, 28]], [[2, 35]], [[37, 48]], [[49, 74]], [[76, 121]]]",
    "query_spans": "[[[123, 132]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}(-c, 0)$, $F_{2}(c, 0)$ are the foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, if there exists a point $P$ on the ellipse $C$ such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=2 c^{2}$, then the range of eccentricity of the ellipse $C$ is?",
    "fact_expressions": "F1: Point;Coordinate(F1) = (-c, 0);c: Number;F2: Point;Coordinate(F2) = (c, 0);C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C) = True;DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 2*c^2",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "[1/2, sqrt(3)/3]",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[19, 32]], [[19, 32]], [[19, 32]], [[33, 83], [88, 93], [169, 174]], [[33, 83]], [[39, 83]], [[39, 83]], [[2, 86]], [[96, 100]], [[88, 100]], [[102, 167]]]",
    "query_spans": "[[[169, 185]]]",
    "process": "Let P(x,y), from the coordinate representation of the dot product we obtain $\\overrightarrow{PF}_{1}\\cdot\\overrightarrow{PF_{2}}=x^{2}-c^{2}+y^{2}=2c^{2}$, then from point P lying on the ellipse we obtain $y^{2}=b^{2}-\\frac{b^{2}}{a^{2}}x^{2}$, solving the two equations together yields $x^{2}=\\frac{(4c^{2}-a^{2})a^{2}}{c^{2}}$, then from $x^{2}\\in[0,a^{2}]$ simplifying gives $3c^{2}\\leqslant a^{2}\\leqslant 4c^{2}$, combining with the eccentricity formula allows solution. Let P(x,y), then $\\overrightarrow{PF}_{1}\\cdot\\overrightarrow{PF_{2}}=(-c-x,-y)\\cdot(c-x,-y)=x^{2}-c^{2}+y^{2}=2c^{2\\textcircled{1}}$, substitute $y^{2}=b^{2}-\\frac{b^{2}}{\\frac{2}{2}}x^{2}$ into equation $\\textcircled{1}$ and solve to get $\\frac{(c^{2}-b^{2})a^{2}}{c^{2}}=\\frac{(4c^{2}-a^{2})a^{2}}{c^{2}}\\in[0,a^{2}]$, i.e., $0\\leqslant\\frac{(c^{2}-a^{2})a^{2}}{c^{2}}\\leqslant a^{2}$, $\\therefore 3c^{2}\\leqslant a^{2}\\leqslant 4c^{2}$, $\\therefore e=\\frac{c}{a}\\in[\\frac{1}{2},\\frac{\\sqrt{3}}{3}]$."
  },
  {
    "text": "The minimum value of $\\frac{|A B|}{3}-\\frac{12}{|C D|}$, where two perpendicular chords $A B$ and $C D$ are drawn through the focus of the parabola $y^{2}=4 x$, is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;C: Point;D: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G),LineSegmentOf(A,B));PointOnCurve(Focus(G),LineSegmentOf(C,D));IsChordOf(LineSegmentOf(A,B),G);IsChordOf(LineSegmentOf(C,D),G);IsPerpendicular(LineSegmentOf(A,B),LineSegmentOf(C,D))",
    "query_expressions": "Min(Abs(LineSegmentOf(A, B))/3 - 12/Abs(LineSegmentOf(C, D)))",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 15]], [[25, 30]], [[25, 30]], [[32, 37]], [[32, 37]], [[1, 15]], [[0, 37]], [[0, 37]], [[1, 37]], [[1, 37]], [[19, 37]]]",
    "query_spans": "[[[39, 79]]]",
    "process": "From the parabola $ y^{2} = 4x $, we know $ 2p = 4 $. Let the inclination angle of line $ AB $ be $ \\theta $, then the inclination angle of line $ CD $ is $ \\frac{\\pi}{2} + \\theta $. Through the focus, $ \\frac{2p}{2(\\frac{\\pi}{2}+\\theta)}\\frac{2p}{os2\\theta}\\frac{s^{2}\\theta}{n} = \\frac{1}{4} $, $ (\\frac{1}{4})^{2} - \\frac{1}{|AB|} ) = \\frac{|AB|}{3} + \\frac{12}{|AB|} - 3 \\geqslant 2\\sqrt{4} - 3 = 1 $, with equality if and only if $ |AB| = 6 $, achieving the minimum value."
  },
  {
    "text": "Given that point $P$ is a moving point on the parabola $x^{2}=8 y$, find the minimum value of the sum of the distance from point $P$ to point $A(2,0)$ and the distance from point $P$ to the directrix of the parabola.",
    "fact_expressions": "G: Parabola;A: Point;P: Point;Expression(G) = (x^2 = 8*y);Coordinate(A) = (2, 0);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, A) + Distance(P, Directrix(G)))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[7, 21], [48, 51]], [[34, 43]], [[2, 6], [29, 33]], [[7, 21]], [[34, 43]], [[2, 27]]]",
    "query_spans": "[[[29, 65]]]",
    "process": "Let the projection of point P on the directrix of the parabola be point P', and let the focus of the parabola be F, then F(0,2). By the definition of the parabola, the distance from point P to the directrix of the parabola is |PP'| = |PF|. Then the sum of the distance from point P to point A(2,0) and the distance to the directrix of the parabola is d = |PA| + |PF| \\geqslant |AF| = \\sqrt{2^{2}+2^{2}} = 2\\sqrt{2}"
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{2}-y^{2}=1$. A line $l$ passing through point $F_{1}$ intersects the left and right branches of $C$ at points $A$, $B$ respectively. If $l \\perp F_{2} B$, then $\\overrightarrow{F_{2} A} \\cdot \\overrightarrow{F_{2} B}$=?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/2 - y^2 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;l: Line;PointOnCurve(F1, l);A: Point;Intersection(l, LeftPart(C)) = A;B: Point;Intersection(l, RightPart(C)) = B;IsPerpendicular(l, LineSegmentOf(F2, B))",
    "query_expressions": "DotProduct(VectorOf(F2, A), VectorOf(F2, B))",
    "answer_expressions": "6-2*sqrt(5)",
    "fact_spans": "[[[16, 49], [72, 75]], [[16, 49]], [[0, 7], [57, 65]], [[8, 15]], [[0, 55]], [[0, 55]], [[66, 71]], [[56, 71]], [[87, 90]], [[66, 96]], [[91, 94]], [[66, 96]], [[98, 115]]]",
    "query_spans": "[[[117, 176]]]",
    "process": "As shown in the figure: in hyperbola $ C $, $ a=1 $, $ b=\\sqrt{2} $, $ c=\\sqrt{3} $, then $ F_{1}(-\\sqrt{3},0) $, $ F_{2}(\\sqrt{3},0) $. Since line $ l $ passes through point $ F_{1} $, and according to the problem, the slope of line $ l $ exists and is not zero, $ \\because l\\bot F_{2}B $, so $ \\triangle F_{1}BF_{2} $ is a right triangle, therefore, $ |BF_{1}|^{2}+|BF_{2}|^{2}=(2\\sqrt{3})^{2}=12 $. Also, since $ |BF_{1}|-|BF_{2}|=2 $, we obtain $ (|BF_{2}|+2)^{2}+|BF_{2}|^{2}=12 $, simplifying gives $ |BF_{2}|^{2}+2|BF_{2}|-4=0 $. $ \\because |BF_{2}|>0 $, solving yields $ |BF_{2}|=\\sqrt{5}-1 $, $ \\therefore |BF_{1}|=\\sqrt{5}+1 $. Therefore, $ \\overrightarrow{F_{2}A}\\cdot\\overrightarrow{F_{2}B}=(\\overrightarrow{F_{2}B}+\\overrightarrow{BA})\\cdot\\overrightarrow{F_{2}B}=\\overrightarrow{F_{2}B}^{2}=(\\sqrt{5}-1)^{2}=6-2\\sqrt{5} $"
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-\\frac{2 y^{2}}{\\sqrt{5}+1}=1$, if a line $l$ passes through the center of $C$ and intersects $C$ at points $M$ and $N$, and $P$ is an arbitrary point on the curve $C$, if the slopes of lines $P M$ and $P N$ exist and are denoted by $k_{P M}$ and $k_{P N}$ respectively, then $k_{P M} \\cdot k_{P N}$=?",
    "fact_expressions": "l: Line;C: Hyperbola;P: Point;M: Point;N: Point;Expression(C) = (x^2 - 2*y^2/(1 + sqrt(5)) = 1);PointOnCurve(Center(C),l);PointOnCurve(P, C);Intersection(l,C)={M,N};Slope(LineOf(P,M))=k1;Slope(LineOf(P,N))=k2;k1:Number;k2:Number",
    "query_expressions": "k1*k2",
    "answer_expressions": "(\\sqrt{5}+1)/2",
    "fact_spans": "[[[48, 53]], [[2, 46], [54, 57], [63, 66], [82, 87]], [[78, 81]], [[68, 71]], [[72, 75]], [[2, 46]], [[48, 60]], [[78, 92]], [[48, 77]], [[94, 139]], [[94, 139]], [[119, 128]], [[130, 139]]]",
    "query_spans": "[[[141, 166]]]",
    "process": "Let the equation of line $ l $ be $ y = kx $, $ M(x_{1}, y_{1}) $, $ N(x_{2}, y_{2}) $, $ P(x_{0}, y_{0}) $, then $ k_{PM} = \\frac{y_{0} - y_{1}}{x_{0} - x_{1}} $, $ k_{PN} = \\frac{y_{0} - y_{2}}{x_{0} - x_{2}} $. From \n\\[\n\\begin{cases}\nx^{2} - \\frac{2y^{2}}{\\sqrt{5} + 1} = 1 \\\\\ny = kx\n\\end{cases}\n\\]\nwe obtain $ (\\sqrt{5} - 2k^{2})x^{2} - (\\sqrt{5} + 1) = 0 $, so we have $ x_{1} + x_{2} = 0 $, $ x_{1}x_{2} = \\frac{\\sqrt{5} + 1}{\\sqrt{5} - 2k^{2}} $, \n\\[\nk_{PM} \\cdot k_{PN} = \\frac{y_{0} - y_{1}}{x_{0} - x_{1}} \\times \\frac{y_{0} - y_{2}}{x_{0} - x_{2}} = \\frac{y_{0}^{2} - y_{0}(y_{1} + y_{2}) + y_{1}y_{2}}{x_{0}^{2} - x_{0}(x_{1} + x_{2}) + x_{1}x_{2}} = \\frac{y_{0}^{2} - ky_{0}(x_{1} + x_{2}) + k^{2}x_{1}x_{2}}{x_{0}^{2} + \\frac{\\sqrt{5} + 1}{\\sqrt{5} - 2k^{2}}} = \\frac{\\sqrt{5} + 1}{2}\n\\]\nHence, the answer is $ \\frac{\\sqrt{5} + 1}{2} $."
  },
  {
    "text": "If $x$, $y$ satisfy $2 x + y - 2 \\leq 0$, and $y^{2} - 2 x \\leq 0$, then the minimum value of $z = x + y$ is?",
    "fact_expressions": "x:Number;y:Number;2*x + y - 2 <= 0;-2*x + y^2 <= 0;z=x+y;z:Number",
    "query_expressions": "Min(z)",
    "answer_expressions": "-12",
    "fact_spans": "[[[1, 4]], [[6, 9]], [[11, 27]], [[30, 48]], [[50, 57]], [[50, 57]]]",
    "query_spans": "[[[50, 63]]]",
    "process": "First, draw the feasible region according to the constraints, as shown in the shaded part. $ z = x + y $, i.e., $ y = -x + z $. When the line $ z = x + y $ is tangent to the parabola, the intercept of the line $ z = x + y $ on the $ y $-axis is minimized. From \n\\[\n\\begin{cases}\ny^{2} = 2x \\\\\nz = x + y\n\\end{cases}\n\\]\neliminating $ x $, we get: $ y^2 + 2y - 2z = 0 $. From $ \\Delta = 0 $, we obtain: $ z = -\\frac{1}{2} $"
  },
  {
    "text": "Given that the distance between the two foci of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ is $4$, what is the equation of the asymptotes of the hyperbola?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;F1:Point;F2:Point;Focus(G)={F1,F2};Expression(G) = (-y^2 + x^2/a^2 = 1);Distance(F1,F2)=4",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(3)/3)*x",
    "fact_spans": "[[[2, 39], [53, 56]], [[5, 39]], [[5, 39]], [], [], [[2, 42]], [[2, 39]], [[2, 51]]]",
    "query_spans": "[[[53, 64]]]",
    "process": "Since the distance between the two foci of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ ($a>0$) is 4, we have $2c=4$, solving gives $c=2$, so $c^{2}=a^{2}+1=4$, $a=\\sqrt{3}$, the asymptotes of the hyperbola are $y=\\pm\\frac{\\sqrt{3}}{3}x$."
  },
  {
    "text": "The length of the minor axis of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1)",
    "query_expressions": "Length(MinorAxis(G))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 43]]]",
    "process": "Since 3<4, the foci of the ellipse lie on the x-axis, so b^{2}=3, then the minor axis length of the ellipse \\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1 is 2b=2\\sqrt{3}. The answer is: 2\\sqrt{3}"
  },
  {
    "text": "The distance from the point $(0,2)$ to one of the asymptotes of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ is?",
    "fact_expressions": "H: Point;Coordinate(H) = (0, 2);G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1)",
    "query_expressions": "Distance(H, OneOf(Asymptote(G)))",
    "answer_expressions": "8/5",
    "fact_spans": "[[[0, 8]], [[0, 8]], [[9, 48]], [[9, 48]]]",
    "query_spans": "[[[0, 59]]]",
    "process": "From the given conditions, the asymptotes of the hyperbola are: $ y = \\pm\\frac{3}{4}x $, that is, $ 3x \\pm 4y = 0 $. Considering symmetry, without loss of generality, consider the distance from the point $ (0,2) $ to the line $ 3x + 4y = 0 $: $ d = \\frac{0 + 8}{\\sqrt{9 + 16}} = \\frac{8}{5} $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the common foci of an ellipse and a hyperbola, $P$ is one of their common points, and $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$. Let the eccentricities of the ellipse and the hyperbola be $e_{1}$ and $e_{2}$ respectively, then the maximum value of $\\frac{1}{e_{1}}+\\frac{1}{e_{2}}$ is?",
    "fact_expressions": "F1: Point;F2: Point;Focus(E) = {F1, F2};Focus(G) = {F1, F2};E: Ellipse;G: Hyperbola;P: Point;OneOf(Intersection(E, G)) = P;AngleOf(F1, P, F2) = pi/3;e1: Number;e2: Number;Eccentricity(E) = e1;Eccentricity(G) = e2",
    "query_expressions": "Max(1/e1 + 1/e2)",
    "answer_expressions": "(4*sqrt(3))/3",
    "fact_spans": "[[[2, 9]], [[10, 17]], [[2, 29]], [[2, 29]], [[18, 20], [82, 84]], [[21, 24], [85, 88]], [[30, 33]], [[30, 42]], [[44, 80]], [[95, 102]], [[104, 111]], [[82, 111]], [[82, 111]]]",
    "query_spans": "[[[113, 152]]]",
    "process": ""
  },
  {
    "text": "Given that the left and right foci of ellipse $C$ are $F_{1}(-3,0)$ and $F_{2}(3,0)$, respectively, and its graph passes through the fixed point $M(0,4)$, then the eccentricity $e$ of $C$ is?",
    "fact_expressions": "C: Ellipse;F1: Point;F2: Point;Coordinate(F1) = (-3, 0);Coordinate(F2) = (3, 0);LeftFocus(C) = F1;RightFocus(C) = F2;M: Point;Coordinate(M) = (0, 4);PointOnCurve(M, C);e: Number;Eccentricity(C) = e",
    "query_expressions": "e",
    "answer_expressions": "3/5",
    "fact_spans": "[[[2, 7], [60, 63], [44, 45]], [[16, 29]], [[30, 42]], [[16, 29]], [[30, 42]], [[2, 42]], [[2, 42]], [[50, 58]], [[50, 58]], [[44, 58]], [[67, 70]], [[60, 70]]]",
    "query_spans": "[[[67, 72]]]",
    "process": "From the given conditions, c=3, b=4 ∴ a=5, e=\\frac{3}{5}"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, what is the minimum distance from a moving point $P$ on the ellipse to the left focus?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/4 + y^2/3 = 1);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, LeftFocus(G)))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 39], [40, 42]], [[45, 48]], [[2, 39]], [[40, 48]]]",
    "query_spans": "[[[40, 61]]]",
    "process": "Let P(x,y), and let the left focus be F. Using the distance formula between two points, |PF| can be found. Then, using the fact that P lies on the ellipse, |PF| can be simplified to obtain the relationship between |PF| and x. Since x ∈ [-2,2], the minimum value of |PF| can be determined. Let P(x,y), and let the left focus be F; then F(-1,0), so |PF| = \\sqrt{(x+1)^{2}+(y-0)^{2}} = \\sqrt{x^{2}+2x+1+y^{2}} = \\sqrt{x^{2}+2x+1+3(1-\\frac{x^{2}}{4})} = \\sqrt{\\frac{1}{4}x^{2}+2x+4} = |\\frac{1}{2}x+2|. Since x ∈ [-2,2], the minimum value of |PF| is 1."
  },
  {
    "text": "Given the parabola equation $y^{2} = -4x$, and the line $l$ with equation $2x + y - 4 = 0$, there is a moving point $A$ on the parabola. The distance from point $A$ to the $y$-axis is $m$, and the distance from point $A$ to line $l$ is $n$. Then, the minimum value of $m + n$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -4*x);l: Line;Expression(l) = (2*x + y - 4 = 0);A: Point;PointOnCurve(A, G);Distance(A, yAxis) = m;Distance(A, l) = n;m: Number;n: Number",
    "query_expressions": "Min(m + n)",
    "answer_expressions": "(6*sqrt(5)/5)-1",
    "fact_spans": "[[[2, 5], [43, 46]], [[2, 20]], [[21, 26], [77, 82]], [[21, 41]], [[51, 54], [55, 59], [72, 76]], [[42, 54]], [[55, 71]], [[72, 89]], [[68, 71]], [[86, 89]]]",
    "query_spans": "[[[91, 102]]]",
    "process": "As shown in the figure, draw AH\\bot l and AN perpendicular to the directrix of the parabola, then |AH|+|AN|=m+n+1. Connect AF, then |AF|+|AH|=m+n+1. By plane geometry, when points A, F, H are collinear, |AF|+|AH|=m+n+1 achieves the minimum value |FH|=\\frac{6}{\\sqrt{5}}=\\frac{6\\sqrt{5}}{5}, so the minimum value of m+n is \\frac{6\\sqrt{5}}{5}-1; thus fill in \\frac{6\\sqrt{5}}{5}-1"
  },
  {
    "text": "The foci of the ellipse are $F_{1}$, $F_{2}$. A line passing through point $F_{1}$ intersects the ellipse, and the shortest segment $M N$ intercepted by the ellipse has length $\\frac{32}{5}$. The perimeter of $\\Delta M F_{2} N$ is $20$. Then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;F2: Point;F1: Point;Focus(G) = {F1, F2};L: Line;PointOnCurve(F1,L) = True;IsIntersect(L,G) = True;M: Point;N: Point;Min(InterceptChord(L,G)) = LineSegmentOf(M,N);Length(LineSegmentOf(M,N)) = 32/5;Perimeter(TriangleOf(M,F2,N)) = 20",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3/5",
    "fact_spans": "[[[0, 2], [35, 37], [41, 43], [103, 105]], [[14, 21]], [[6, 13], [23, 31]], [[0, 21]], [[32, 34]], [[22, 34]], [[32, 39]], [[51, 56]], [[51, 56]], [[32, 56]], [[49, 72]], [[75, 101]]]",
    "query_spans": "[[[103, 111]]]",
    "process": "Assume the standard equation of the ellipse is \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1. Since the perimeter of \\triangle MF_{2}N is 20, we have 4a=20, so a=5. Since the shortest line segment cut by the ellipse through point F_{1} has length \\frac{32}{5}, it follows that \\frac{2b^{2}}{a}=\\frac{32}{5}. Solving gives b=4, c=3. Therefore, the eccentricity of the ellipse is e=\\frac{c}{a}=\\frac{3}{5}."
  },
  {
    "text": "Through the focus $F$ of the parabola $x^{2}=\\frac{1}{2} y$, a chord $AB$ is drawn, and point $M(x_{0}, y_{0})$ is the midpoint of chord $AB$. If $y_{0}=\\frac{7}{8}$, then the length of chord $AB$ is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;M: Point;F: Point;Expression(G) = (x^2 = y/2);Coordinate(M) = (x0, y0);x0:Number;y0:Number;PointOnCurve(F,LineSegmentOf(A, B));IsChordOf(LineSegmentOf(A, B),G);Focus(G) = F;MidPoint(LineSegmentOf(A, B)) = M;y0 = 7/8",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 25]], [[33, 38]], [[33, 38]], [[39, 58]], [[28, 31]], [[1, 25]], [[39, 58]], [[40, 58]], [[40, 58]], [[0, 38]], [[1, 38]], [[1, 31]], [[39, 68]], [[70, 89]]]",
    "query_spans": "[[[92, 101]]]",
    "process": "From the condition, we have $2p = \\frac{1}{2} \\Rightarrow p = \\frac{1}{4}$. Let $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$, $y_{1} + y_{2} = 2y_{0} = \\frac{7}{4}$. From the focal chord length formula, we have $AB = y_{1} + y_{2} + p = 2$."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a, b>0)$, respectively, and let $P$ be a point on the hyperbola such that $|P F_{1}|=2 a$ and $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$. Then the eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);e: Number;Abs(LineSegmentOf(P, F1)) = 2*a;AngleOf(F1, P, F2) = pi/3;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[19, 73], [84, 87], [148, 151]], [[19, 73]], [[22, 73]], [[22, 73]], [[22, 73]], [[22, 73]], [[1, 8]], [[9, 16]], [[1, 79]], [[1, 79]], [[80, 83]], [[80, 90]], [[155, 158]], [[92, 107]], [[109, 145]], [[148, 158]]]",
    "query_spans": "[[[155, 162]]]",
    "process": ""
  },
  {
    "text": "Through a point $P$ on the ellipse $\\frac{x^{2}}{a^{2}}+y^{2}=1$ ($a>1$) and the origin $O$, draw a line $l$ intersecting the circle $x^{2}+y^{2}=a^{2}+1$ at points $A$ and $B$. If there exists a point $P$ satisfying $a^{2}=|PA||PB|+1$, then the range of real values for $a$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2 + x^2/a^2 = 1);a: Real;a>1;P: Point;PointOnCurve(P, G);O: Origin;l: Line;PointOnCurve(P, l);PointOnCurve(O, l);H: Circle;Expression(H) = (x^2 + y^2 = a^2 + 1);A: Point;B: Point;Intersection(l, H) = {A, B};a^2 = Abs(LineSegmentOf(P, A))*Abs(LineSegmentOf(P, B)) + 1",
    "query_expressions": "Range(a)",
    "answer_expressions": "[\\sqrt{2}, +\\infty)",
    "fact_spans": "[[[1, 37]], [[1, 37]], [[125, 130]], [[3, 37]], [[40, 43], [98, 101]], [[1, 43]], [[44, 51]], [[52, 57]], [[0, 57]], [[0, 57]], [[58, 80]], [[58, 80]], [[82, 85]], [[86, 89]], [[52, 91]], [[103, 123]]]",
    "query_spans": "[[[125, 137]]]",
    "process": "As shown in the figure: $|PA||PB|=(|OA|-|OP|)(|OA|+|OP|)=a^2+1-|OP|^{2}$. Since $|OP|\\in[1,a^{2}]$, it follows that $1\\leqslant|PA|\\cdot|PB|\\leqslanta^{2}$. If there exists a point $P$ such that $a^{2}=|PA||PB|+1$, i.e., $1\\leqslanta^{2}-1\\leqslanta^{2}$, solving yields $a\\geqslant\\sqrt{2}$."
  },
  {
    "text": "Given a point $M(1, t)$ $(t>0)$ on the parabola $y^{2}=2 px$ $(p>0)$ such that the distance from $M$ to the focus is $5$, and the left vertex of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1$ $(a>0)$ is $A$. If one asymptote of the hyperbola is parallel to the line $AM$, then the value of the real number $a$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (1, t);t: Number;t>0;H: Parabola;Expression(H) = (y^2 = 2*p*x);p: Number;p>0;PointOnCurve(M, H);Distance(M, Focus(H)) = 5;G: Hyperbola;Expression(G) = (-y^2/9 + x^2/a^2 = 1);a: Real;a>0;A: Point;LeftVertex(G) = A;IsParallel(OneOf(Asymptote(G)), LineOf(A, M)) = True",
    "query_expressions": "a",
    "answer_expressions": "3",
    "fact_spans": "[[[26, 40]], [[26, 40]], [[26, 40]], [[26, 40]], [[2, 22]], [[2, 22]], [[5, 22]], [[5, 22]], [[2, 40]], [[2, 50]], [[51, 98], [108, 111]], [[51, 98]], [[128, 133]], [[54, 98]], [[103, 106]], [[51, 106]], [[108, 126]]]",
    "query_spans": "[[[128, 137]]]",
    "process": "From the given condition, $1 + \\frac{p}{2} = 5$, so $p = 8$. Without loss of generality, assume point $M$ is above the $x$-axis, then $M(1, 4)$. Since the left vertex of the hyperbola is $A(-a, 0)$, and $AM$ is parallel to an asymptote, it follows that $\\frac{4}{1 + a} = \\frac{3}{a}$, then $a = 3$."
  },
  {
    "text": "If a line passing through the upper focus $F_{1}$ of the ellipse $\\frac{y^{2}}{16}+\\frac{x^{2}}{12}=1$ intersects the ellipse at points $A$, $B$, and $F_{2}$ is the lower focus of the ellipse, then the perimeter of triangle $F_{2}AB$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/12 + y^2/16 = 1);F1: Point;UpperFocus(G) = F1;L: Line;PointOnCurve(F1, L) = True;Intersection(L, G) = {A, B};A: Point;B: Point;LowerFocus(G) = F2;F2: Point",
    "query_expressions": "Perimeter(TriangleOf(F2, A, B))",
    "answer_expressions": "16",
    "fact_spans": "[[[2, 41], [55, 57], [75, 77]], [[2, 41]], [[44, 51]], [[2, 51]], [[52, 54]], [[1, 54]], [[52, 66]], [[58, 62]], [[63, 66]], [[67, 80]], [[67, 74]]]",
    "query_spans": "[[[82, 101]]]",
    "process": "By the definition of the ellipse, $AF_{1}+AF_{2}=2a$, $BF_{1}+BF_{2}=2a$. According to the equation of the ellipse, the answer is found in the ellipse $\\frac{y^{2}}{16}+\\frac{x^{2}}{12}=1$, where $a=4$. By the definition of the ellipse, $AF_{1}+AF_{2}=2a$, $BF_{1}+BF_{2}=2a$. Therefore, $AF_{1}+AF_{2}+BF_{1}+BF_{2}=4a$, that is, $AF_{2}+BF_{2}+AB=4a=16$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, the left and right foci are denoted as $F_{1}$ and $F_{2}$, respectively. If a point $P$ on the hyperbola satisfies $\\angle F_{1} PF_{2}=90^{\\circ}$, then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/9 - y^2/16 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;AngleOf(F1,P,F2)=ApplyUnit(90,degree);PointOnCurve(P,G)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "16",
    "fact_spans": "[[[2, 41], [67, 70]], [[50, 57]], [[71, 75]], [[58, 65]], [[2, 41]], [[2, 65]], [[2, 65]], [[76, 108]], [[67, 75]]]",
    "query_spans": "[[[110, 137]]]",
    "process": "From the equation, we obtain a=3, b=4. Combining the definition of the hyperbola, we find c=5. Using the Pythagorean theorem, we can find mn=32, and then determine the area of triangle. From the hyperbola equation, we know a=3, b=4, so c=\\sqrt{a^{2}+b^{2}}=5. Let |PF_{1}|=m, |PF_{2}|=n. By the definition of the hyperbola, we have |m-n|=2a=6. \\because\\angle F_{1}PF_{2}=90^{\\circ}, \\therefore (2c)^{2}=10^{2}=m^{2}+n^{2}=(m-n)^{2}+2mn=6^{2}+2mn, \\therefore mn=32. \\therefore S_{\\Delta F_{1}PF_{2}}=\\frac{1}{2}mn=\\frac{1}{2}\\times32=16."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{7}=1$, $F$ is its left focus, a line $l$ passing through the origin $O$ intersects the ellipse at points $A$ and $B$, point $A$ lies in the second quadrant, and $\\angle F A B = \\angle B F O$. Then the slope of line $l$ is?",
    "fact_expressions": "l: Line;C: Ellipse;F: Point;A: Point;B: Point;O: Origin;Expression(C) = (x^2/9 + y^2/7 = 1);LeftFocus(C) = F;PointOnCurve(O, l);Intersection(l, C) = {A, B};Quadrant(A)=2;AngleOf(F, A, B) = AngleOf(B, F, O)",
    "query_expressions": "Slope(l)",
    "answer_expressions": "-sqrt(7)/3",
    "fact_spans": "[[[61, 66], [120, 125]], [[2, 44], [49, 50], [67, 69]], [[45, 48]], [[70, 73], [80, 84]], [[74, 77]], [[55, 60]], [[2, 44]], [[45, 53]], [[54, 66]], [[61, 79]], [[80, 89]], [[91, 118]]]",
    "query_spans": "[[[120, 130]]]",
    "process": "Let $ A(x_{0},y_{0}) $, then $ B(-x_{0},-y_{0}) $, $ x_{0}<0 $, $ y_{0}>0 $ and $ \\frac{x_{0}^{2}}{9}+\\frac{y_{0}^{2}}{7}=1 $. Since $ F $ is its left focus, $ \\therefore F(-\\sqrt{2},0) $, $ \\tan\\angle BFO = \\frac{y_{0}}{-x_{0}+\\sqrt{2}} $, the slope of line $ AB $ is $ k_{1} = \\frac{y_{0}}{x_{0}} $. After analysis, the slope of line $ AF $ must exist, denoted as $ k_{2} = \\frac{y_{0}}{x_{0}+\\sqrt{2}} $, then $ \\tan\\angle FAB = \\frac{k_{1}-k_{2}}{1+k_{1}k_{2}} = \\frac{\\sqrt{2}}{x_{0}^{2}+\\sqrt{2}x_{0}} $. Also $ \\angle FAB = \\angle BFO $, $ \\therefore \\_ $. $ \\therefore x_{0}^{2}+2\\sqrt{2}x_{0}+y_{0}^{2}= $. Solving yields: $ x_{0} = -\\frac{3\\sqrt{2}}{2} $, $ y_{0}^{2} $. $ \\therefore $ the slope of line $ l $ is $ \\frac{y_{0}}{x_{0}} = -\\frac{\\sqrt{7}}{3} $."
  },
  {
    "text": "The line $l$ passes through the focus of the parabola $x^{2}=2 p y(p>0)$ and intersects the parabola at points $A$ and $B$. If the length of segment $AB$ is $6$, and the distance from the midpoint of $AB$ to the $x$-axis is $1$, then the equation of this parabola is?",
    "fact_expressions": "l: Line;G: Parabola;Expression(G) = (x^2 = 2*p*y);p: Number;p>0;PointOnCurve(Focus(G), l) ;Intersection(l, G) = {A, B};B: Point;A: Point;Length(LineSegmentOf(A, B)) = 6;Distance(MidPoint(LineSegmentOf(A, B)), xAxis) = 1",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2=8*y",
    "fact_spans": "[[[0, 5]], [[6, 27], [33, 36], [88, 91]], [[6, 27]], [[9, 27]], [[9, 27]], [[0, 30]], [[0, 47]], [[42, 45]], [[38, 41]], [[49, 62]], [[65, 85]]]",
    "query_spans": "[[[88, 95]]]",
    "process": "From the equation of the parabola, the focus is $ F(0,\\frac{p}{2}) $, and the directrix is $ y=-\\frac{p}{2} $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, so the vertical coordinate of the midpoint of $ AB $ is $ \\frac{y_{1}+y_{2}}{2} $. According to the problem, the distance from the midpoint of $ AB $ to the x-axis is 1, thus $ \\frac{y_{1}+y_{2}}{2}=1 $, so $ y_{1}+y_{2}=2 $. By the property of the parabola, $ |AB|=y_{1}+y_{2}+p=6 $, hence $ p=4 $. Therefore, the equation of the parabola is: $ x^{2}=8y $."
  },
  {
    "text": "Given that the asymptotes of a hyperbola with foci on the $x$-axis are $y = \\pm 2x$, and the semi-focal length is $5$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), xAxis) = True;Expression(Asymptote(G)) = (y = pm*(2*x));HalfFocalLength(G) = 5",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5 - y^2/20 = 1",
    "fact_spans": "[[[10, 13], [39, 42]], [[2, 13]], [[10, 29]], [[10, 37]]]",
    "query_spans": "[[[39, 49]]]",
    "process": "According to the given conditions, we have  \n\\begin{cases}\\frac{b}{a}=2\\\\c=5\\\\c^2=a^2+b^{2}\\end{cases} \\Rightarrow a=\\sqrt{5}, b=2\\sqrt{5}.  \nTherefore, the standard equation of the hyperbola is $\\frac{x^{2}}{5}-\\frac{y^{2}}{20}=1$."
  },
  {
    "text": "The length of the major axis of the ellipse $x^{2}+m y^{2}=1$ is twice the length of the minor axis, then the value of $m$ is?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (m*y^2 + x^2 = 1);Length(MajorAxis(G))=2*Length(MinorAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "{4,1/4}",
    "fact_spans": "[[[0, 19]], [[32, 35]], [[0, 19]], [[0, 30]]]",
    "query_spans": "[[[32, 39]]]",
    "process": "Convert the ellipse equation into standard form, and discuss separately according to whether the foci lie on the x-axis or y-axis, thereby solving for the value of $ m $. [Detailed solution] Convert $ x^{2} + my^{2} = 1 $ into $ x^{2} + \\frac{y^{2}}{\\frac{1}{m}} = 1 $. When the foci are on the x-axis, the major axis length is 2, and the minor axis length is $ 2\\sqrt{\\frac{1}{m}} = 1 $, then $ m = 4 $. When the foci are on the y-axis, the minor axis length is 2, and the major axis length is $ 2\\sqrt{\\frac{1}{m}} = 4 $, then $ m = \\frac{1}{4} $. In conclusion, fill in $ 4 $ or $ \\frac{1}{4} $."
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{27}=1$, respectively, $A$ is a point on the ellipse, and $\\overrightarrow{O B}=\\frac{1}{2}(\\overrightarrow{O A}+\\overrightarrow{O F_{1}})$, $\\overrightarrow{O C}=\\frac{1}{2}(\\overrightarrow{O A}+\\overrightarrow{O F_{2}})$, then $|\\overrightarrow{O B}|+|\\overrightarrow{O C}|$=?",
    "fact_expressions": "F1: Point;F2: Point;G: Ellipse;Expression(G) = (x^2/36 + y^2/27 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;A: Point;PointOnCurve(A, G) = True;O: Origin;B: Point;VectorOf(O, B) = (VectorOf(O, A) + VectorOf(O, F1))/2;C: Point;VectorOf(O, C) = (VectorOf(O, A) + VectorOf(O, F2))/2",
    "query_expressions": "Abs(VectorOf(O, B)) + Abs(VectorOf(O, C))",
    "answer_expressions": "6",
    "fact_spans": "[[[0, 7]], [[8, 15]], [[18, 57], [68, 70]], [[18, 57]], [[0, 63]], [[0, 63]], [[64, 67]], [[64, 73]], [[75, 156]], [[75, 156]], [[75, 156]], [[158, 239]], [[158, 239]]]",
    "query_spans": "[[[241, 290]]]",
    "process": "According to the problem, |OB| // \\frac{1}{2}|AF_{2}|, |OC| // \\frac{1}{2}|AF_{1}|, hence |OB| + |OC| = a = 6"
  },
  {
    "text": "The focus of the parabola $y^{2}=4 x$ is $F$. A line passing through $F$ with an inclination angle of $\\frac{\\pi}{2}$ intersects the parabola at points $A$ and $B$. Then $|A B|=$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, H);Inclination(H) = pi/2;Intersection(H, G) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 14], [50, 53]], [[47, 49]], [[55, 58]], [[59, 62]], [[18, 21], [23, 26]], [[0, 14]], [[0, 21]], [[22, 49]], [[27, 49]], [[47, 64]]]",
    "query_spans": "[[[66, 75]]]",
    "process": "According to the problem: The coordinates of point F are (1,0). The equation of the line passing through F with an inclination angle of $\\frac{\\pi}{2}$ is $x=1$, so $A(1,2)$, $B(1,-2)$, therefore $|AB|=4$."
  },
  {
    "text": "If the distance from point $A(m, 1)$ on the parabola $x^{2}=2 p y(p>0)$ to the focus is $4$, then $|m|=$?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;p>0;Expression(G) = (x^2 = 2*p*y);Coordinate(A) = (m, 1);PointOnCurve(A, G);Distance(A, Focus(G)) = 4;m:Number",
    "query_expressions": "Abs(m)",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[1, 22]], [[4, 22]], [[24, 34]], [[4, 22]], [[1, 22]], [[24, 34]], [[1, 34]], [[1, 44]], [[46, 51]]]",
    "query_spans": "[[[46, 53]]]",
    "process": "Since the distance from point A(m,1) on the parabola $x^{2}=2py$ ($p>0$) to the focus is 4, we have $1+\\frac{p}{2}=4$, that is: $p=6$, $x^{2}=12y$, so $m^{2}=12$, $|m|=2\\sqrt{3}$"
  },
  {
    "text": "Let the length of the imaginary axis of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ be $2$, and the focal distance be $2 \\sqrt{3}$. Then the equations of the asymptotes of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;Length(ImageinaryAxis(G)) = 2;FocalLength(G) = 2*sqrt(3)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*((sqrt(2)/2)*x)",
    "fact_spans": "[[[1, 58], [84, 87]], [[1, 58]], [[4, 58]], [[4, 58]], [[4, 58]], [[4, 58]], [[1, 66]], [[1, 82]]]",
    "query_spans": "[[[84, 95]]]",
    "process": ""
  },
  {
    "text": "Given the equation of the hyperbola is $x^{2}-\\frac{y^{2}}{3}=1$, then its eccentricity is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 5], [36, 37]], [[2, 34]]]",
    "query_spans": "[[[36, 43]]]",
    "process": ""
  },
  {
    "text": "Draw a line with slope $2$ passing through the right focus of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$, intersecting the ellipse at points $A$ and $B$, and let $O$ be the origin. Then the area of $\\triangle OAB$ is?",
    "fact_expressions": "G: Ellipse;H: Line;O: Origin;A: Point;B: Point;Expression(G) = (x^2/5 + y^2/4 = 1);PointOnCurve(RightFocus(G),H);Slope(H)=2;Intersection(H,G)={A,B}",
    "query_expressions": "Area(TriangleOf(O, A, B))",
    "answer_expressions": "5/3",
    "fact_spans": "[[[1, 38], [55, 57]], [[52, 54]], [[71, 74]], [[59, 62]], [[65, 68]], [[1, 38]], [[0, 54]], [[45, 54]], [[52, 70]]]",
    "query_spans": "[[[81, 101]]]",
    "process": ""
  },
  {
    "text": "What is the distance from the point $(2,-1)$ on the parabola $x^{2}=-4 y$ to the directrix?",
    "fact_expressions": "G: Parabola;P: Point;Expression(G) = (x^2 = -4*y);Coordinate(P) = (2, -1);PointOnCurve(P,G)",
    "query_expressions": "Distance(P,Directrix(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 15]], [[17, 26]], [[0, 15]], [[17, 26]], [[0, 26]]]",
    "query_spans": "[[[0, 34]]]",
    "process": "From the given condition, the equation of the directrix of the parabola is $ y = 1 $, so the distance from the point $ (2, -1) $ to the directrix is $ 1 - (-1) = 2 $."
  },
  {
    "text": "If the distance from the right focus $F(2,0)$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ to one of its asymptotes is $\\sqrt{3}$, then its eccentricity is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (2, 0);RightFocus(G) = F;Distance(F, OneOf(Asymptote(G))) = sqrt(3)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 57], [91, 92]], [[4, 57]], [[4, 57]], [[61, 69]], [[4, 57]], [[4, 57]], [[1, 57]], [[61, 69]], [[1, 69]], [[1, 89]]]",
    "query_spans": "[[[91, 97]]]",
    "process": "Take an asymptote of the hyperbola. Given that the distance from the right focus $ F(2,0) $ to one asymptote is $ \\sqrt{3} $, we can find $ a $ and $ b $, and then determine the eccentricity of the hyperbola. [Detailed Solution] Without loss of generality, take the asymptote $ y = \\frac{b}{a}x $, i.e., $ bx - ay = 0 $, of the hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $). Then the distance from the right focus $ F(2,0) $ to the asymptote $ y = \\frac{b}{a}x $ is $ \\frac{|2b|}{\\sqrt{a^{2}+b^{2}}} = \\frac{2b}{c} = \\frac{2b}{2} = \\sqrt{3} $. Therefore, $ b = \\sqrt{3} $, and $ a = \\sqrt{c^{2}-b^{2}} = 1 $. Thus, the eccentricity of the hyperbola is $ e = \\frac{c}{a} = 2 $."
  },
  {
    "text": "The foci of the ellipse $x^{2} + m y^{2}=1$ lie on the $y$-axis, and the length of the major axis is twice the length of the minor axis. Then the value of $m$ is?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (m*y^2 + x^2 = 1);PointOnCurve(Focus(G), yAxis);Length(MajorAxis(G))=2*Length(MinorAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "1/4",
    "fact_spans": "[[[0, 21]], [[43, 46]], [[0, 21]], [[0, 30]], [[0, 41]]]",
    "query_spans": "[[[43, 50]]]",
    "process": ""
  },
  {
    "text": "Given the two foci of a hyperbola $F_{1}(-\\sqrt{10}, 0)$, $F_{2}(\\sqrt{10}, 0)$, and asymptotes $y=\\pm \\frac{1}{2} x$, what is the standard equation of the hyperbola?",
    "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;Coordinate(F1) = (-sqrt(10), 0);Coordinate(F2) = (sqrt(10), 0);Focus(G) = {F1, F2};Expression(Asymptote(G)) = (y = pm*(x/2))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 5], [85, 88]], [[11, 33]], [[36, 57]], [[11, 33]], [[36, 57]], [[2, 57]], [[2, 83]]]",
    "query_spans": "[[[85, 95]]]",
    "process": ""
  },
  {
    "text": "If the distance from the focus of the hyperbola $x^{2}-\\frac{y^{2}}{k}=1$ to its asymptote is $2 \\sqrt{2}$, then the value of the real number $k$ is?",
    "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (x^2 - y^2/k = 1);Distance(Focus(G),Asymptote(G))=2*sqrt(2)",
    "query_expressions": "k",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 29]], [[54, 59]], [[1, 29]], [[1, 52]]]",
    "query_spans": "[[[54, 63]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of ellipse $C$, and $P$ is any point on the ellipse. If the maximum value of $\\frac{|P F_{1}|}{|P F_{2}|}$ is $3$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "F1: Point;F2: Point;C: Ellipse;Focus(C) = {F1, F2};P: Point;PointOnCurve(P,C) = True;Max(Abs(LineSegmentOf(P,F1)) / Abs(LineSegmentOf(P,F2))) = 3",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 9]], [[10, 17]], [[20, 25], [35, 37], [83, 88]], [[2, 30]], [[31, 34]], [[31, 42]], [[44, 81]]]",
    "query_spans": "[[[83, 94]]]",
    "process": "According to the geometric meaning of the ellipse, the maximum distance from point P to the foci of ellipse C is $a+c$, and the minimum distance is $a-c$, thereby obtaining the maximum value of $\\frac{|PF_{1}|}{|PF_{2}|}$. Solution: The maximum distance from point P to the foci of ellipse C is $a+c$, and the minimum distance is $a-c$. Given that the maximum value of $\\frac{|PF_{1}|}{|PF_{2}|}$ is 3, $\\therefore \\frac{a+c}{a-c}=3$, $\\therefore e=\\frac{1}{2}$"
  },
  {
    "text": "If there exists a point $M$ on the ellipse such that its distance to the left focus is twice its distance to the right directrix, then the minimum value of the ellipse's eccentricity is?",
    "fact_expressions": "G: Ellipse;M: Point;PointOnCurve(M, G);Distance(M,LeftFocus(G)) = 2*Distance(M,RightDirectrix(G))",
    "query_expressions": "Min(Eccentricity(G))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 3], [35, 37]], [[8, 11], [21, 22], [12, 13]], [[1, 11]], [[1, 33]]]",
    "query_spans": "[[[35, 46]]]",
    "process": ""
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. Point $O$ is the origin. If $|AF|=3$, then the area of $\\triangle AOF$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;L: Line;PointOnCurve(F, L) = True;Intersection(L, G) = {A, B};A: Point;B: Point;O: Origin;Abs(LineSegmentOf(A, F)) = 3",
    "query_expressions": "Area(TriangleOf(A, O, F))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 15], [25, 28]], [[1, 15]], [[18, 21]], [[1, 21]], [[22, 24]], [[0, 24]], [[22, 38]], [[29, 32]], [[33, 36]], [[39, 43]], [[50, 59]]]",
    "query_spans": "[[[62, 84]]]",
    "process": "The directrix of the parabola \\( y^{2}=4x \\) is \\( x=-1 \\). Since \\( |AF|=3 \\), the distance from point \\( A \\) to the directrix \\( x=-1 \\) is 3, so \\( 1+x_{A}=3 \\), then \\( x_{A}=2 \\). Substituting into \\( y^{2}=4x \\) gives \\( y_{A}=\\pm2\\sqrt{2} \\). Therefore, \\( S_{\\Delta AOF}=\\frac{1}{2}\\times1\\times2\\sqrt{2}=\\sqrt{2} \\)."
  },
  {
    "text": "If the focal distance of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{m}=1$ is $10$, then what is the equation of the asymptotes of the hyperbola?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/16 - y^2/m = 1);FocalLength(G) = 10",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(3/4)*x",
    "fact_spans": "[[[1, 40], [50, 53]], [[4, 40]], [[1, 40]], [[1, 48]]]",
    "query_spans": "[[[50, 61]]]",
    "process": ""
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=2x$, draw two mutually perpendicular lines $l_{1}$ and $l_{2}$. If $l_{1}$ and $l_{2}$ intersect the parabola at points $A$, $B$ and $C$, $D$ respectively, then the minimum value of $|FA| \\cdot |FB| + |FC| \\cdot |FD|$ is?",
    "fact_expressions": "G: Parabola;D: Point;l1:Line;l2:Line;F: Point;A: Point;B: Point;C: Point;Expression(G) = (y^2 = 2*x);Focus(G)=F;PointOnCurve(F,l1);PointOnCurve(F,l2);IsPerpendicular(l1,l2);Intersection(l1,G)={A,B};Intersection(l2,G)={C,D}",
    "query_expressions": "Min(Abs(LineSegmentOf(F, A))*Abs(LineSegmentOf(F, B)) + Abs(LineSegmentOf(F, C))*Abs(LineSegmentOf(F, D)))",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 15], [69, 72]], [[85, 88]], [[29, 38], [50, 57]], [[40, 48], [58, 65]], [[18, 21]], [[73, 76]], [[77, 80]], [[81, 84]], [[1, 15]], [[1, 21]], [[0, 48]], [[0, 48]], [[22, 48]], [[50, 90]], [[50, 90]]]",
    "query_spans": "[[[92, 133]]]",
    "process": "The focus of the parabola $ y^{2}=2x $ is $ F\\left(\\frac{1}{2},0\\right) $. If one of the lines $ l_{1},l_{2} $ is parallel to the $ y $-axis, then the other line coincides with the $ x $-axis, but the $ x $-axis intersects the parabola $ y^{2}=2x $ at only one point, which does not satisfy the condition. Therefore, the slopes of lines $ l_{1},l_{2} $ both exist and are non-zero. Let the equation of line $ l_{1} $ be $ x=my+\\frac{1}{2} $ ($ m\\neq0 $), then the equation of line $ l_{2} $ is $ x=-\\frac{1}{m}y+\\frac{1}{2} $. Let points $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Solving the system \n\\[\n\\begin{cases}\ny^{2}=2x \\\\\nx=my+\\frac{1}{2}\n\\end{cases}\n\\]\nwe obtain $ y^{2}-2my-1=0 $, $ \\triangle=4m^{2}+4>0 $. By Vieta's formulas, we have $ y_{1}+y_{2}=2m $, $ y_{1}y_{2}=-1 $. Thus,\n\\[\n|FA|\\cdot|FB|=\\left(x_{1}+\\frac{1}{2}\\right)\\left(x_{2}+\\frac{1}{2}\\right)=(my_{1}+1)(my_{2}+1)=m^{2}y_{1}y_{2}+m(y_{1}+y_{2})+1=-m^{2}+2m^{2}+1=m^{2}+1\n\\]\nSimilarly, we get $ |FC|\\cdot|FD|=\\frac{1}{m^{2}}+1 $. Therefore,\n\\[\n|FA|\\cdot|FB|+|FC|\\cdot|FD|=m^{2}+\\frac{1}{m^{2}}+2\\geqslant2\\sqrt{m^{2}\\cdot\\frac{1}{m^{2}}}+2=4\n\\]\nEquality holds if and only if $ m=\\pm1 $. Hence, the minimum value of $ |FA|\\cdot|FB|+|FC|\\cdot|FD| $ is 4. The answer is: 4"
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and $Q$, $R$ are points on the circles $(x+4)^{2}+y^{2}=\\frac{1}{4}$ and $(x-4)^{2}+y^{2}=\\frac{1}{4}$ respectively, then the minimum value of $|P Q|+|P R|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;Z1: Circle;Z2: Circle;Expression(G) = (x^2/25 + y^2/9 = 1);Expression(Z1) = (y^2 + (x + 4)^2 = 1/4);Expression(Z2) = (y^2 + (x - 4)^2 = 1/4);Q: Point;R: Point;PointOnCurve(P, G);PointOnCurve(Q, Z1);PointOnCurve(R, Z2)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, Q)) + Abs(LineSegmentOf(P, R)))",
    "answer_expressions": "9",
    "fact_spans": "[[[6, 44]], [[2, 5]], [[59, 89]], [[90, 119]], [[6, 44]], [[59, 89]], [[90, 119]], [[49, 52]], [[53, 56]], [[2, 48]], [[49, 121]], [[49, 121]]]",
    "query_spans": "[[[123, 142]]]",
    "process": "From the ellipse equation, the foci of the ellipse are $ F_{1}(-4,0) $, $ F_{2}(4,0) $, and $ |PF_{1}| + |PF_{2}| = 10 $.  \nThe circle $ (x+4)^{2} + y^{2} = \\frac{1}{4} $ has center $ F_{1}(-4,0) $ and radius $ r_{1} = \\frac{1}{2} $.  \nThe circle $ (x-4)^{2} + y^{2} = \\frac{1}{4} $ has center $ F_{2}(4,0) $ and radius $ r_{2} = \\frac{1}{2} $.  \nSince $ |PQ| \\geqslant |PF_{1}| - r_{1} $ (equality holds if and only if points $ P, Q, F_{1} $ are collinear),  \n$ |PR| \\geqslant |PF_{2}| - r_{2} $ (equality holds if and only if points $ P, R, F_{2} $ are collinear),  \nthus $ |PQ| + |PR| \\geqslant |PF_{1}| + |PF_{2}| - (r_{1} + r_{2}) $.  \nTherefore, $ (|PQ| + |PR|)_{\\min} = 10 - 1 = 9 $."
  },
  {
    "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ have an asymptote given by $y=\\frac{1}{2} x$. Then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(C))) = (y = x/2)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[1, 62], [90, 96]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[1, 62]], [[1, 88]]]",
    "query_spans": "[[[90, 102]]]",
    "process": ""
  },
  {
    "text": "Let the focus of the parabola $x^{2}=12 y$ be $F$. A line $l$ passing through point $P(2,1)$ intersects the parabola at points $A$ and $B$. It is known that point $P$ is exactly the midpoint of $AB$. Then $|A F|+|B F|$=?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;P: Point;F: Point;Expression(G)=(x^2=12*y);Coordinate(P)=(2,1);Focus(G)=F;PointOnCurve(P,l);Intersection(l,G)={A,B};MidPoint(LineSegmentOf(A,B)) = P",
    "query_expressions": "Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F))",
    "answer_expressions": "8",
    "fact_spans": "[[[36, 41]], [[1, 16], [42, 45]], [[48, 51]], [[52, 55]], [[26, 35], [60, 64]], [[20, 23]], [[1, 16]], [[26, 35]], [[1, 23]], [[24, 41]], [[36, 57]], [[60, 74]]]",
    "query_spans": "[[[76, 91]]]",
    "process": "The focus of the parabola $ x^{2} = 12y $ is $ F(0,3) $, and the equation of the directrix is $ y = -3 $. As shown in the figure, let $ AM $, $ BN $, and $ PQ $ be perpendicular to the directrix at points $ M $, $ N $, and $ Q $, respectively. According to the definition of a parabola, $ |AF| = |AM| $, $ |BF| = |BN| $. Since the equation of the directrix of the parabola is $ y = -3 $, it follows that $ |PQ| = 4 $. By the property of the midline of a trapezoid, we obtain $ |AF| + |BF| = |AM| + |BN| = 2|PQ| = 8 $."
  },
  {
    "text": "The distance from a vertex of the minor axis of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ to one of its foci is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1)",
    "query_expressions": "Distance(OneOf(Endpoint(MinorAxis(G))), OneOf(Focus(G)))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 27]], [[0, 27]]]",
    "query_spans": "[[[0, 44]]]",
    "process": "By the given condition, $ a = \\sqrt{2} $, which is the distance from one endpoint of the minor axis to a focus."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$, the left focus is $F$, and a moving point $M$ lies on the ellipse. Then the range of $|M F|$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/16 + y^2/12 = 1);F: Point;LeftFocus(C) = F;M: Point;PointOnCurve(M, C) = True",
    "query_expressions": "Range(Abs(LineSegmentOf(M, F)))",
    "answer_expressions": "[2,6]",
    "fact_spans": "[[[2, 46], [61, 63]], [[2, 46]], [[51, 54]], [[2, 54]], [[57, 60]], [[57, 64]]]",
    "query_spans": "[[[66, 80]]]",
    "process": "Solution 1:  \n\\because the standard form of the ellipse: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1,  \n\\therefore C: \\frac{x^{2}}{16} + \\frac{y^{2}}{12} = 1 can be simplified to \\frac{x^{2}}{4^{2}} + \\frac{y^{2}}{(2\\sqrt{3})^{2}} = 1,  \nwe obtain: a = 4, b = 2\\sqrt{3}.  \n\\because a^{2} = b^{2} + c^{2}, \\therefore c = 2.  \nLet M(x_{0}, y_{0}), hence -a \\leqslant x_{0} \\leqslant a, i.e., -4 \\leqslant x_{0} \\leqslant 4.  \nSince point M lies on the ellipse C: \\frac{x_{0}^{2}}{16} + \\frac{y_{0}^{2}}{12} = 1, we get: y_{0}^{2} = 12(1 - \\frac{x_{0}^{2}}{16}).  \nThus, |MF| = \\sqrt{(x_{0} + 2)^{2} + (y_{0} - 0)^{2}} = \\sqrt{(x_{0} + 2)^{2} + 12(1 - \\frac{x_{0}^{2}}{16})} = \\frac{1}{2}x_{0} + 4.  \n\\therefore 2 \\leqslant \\frac{1}{2}x_{0} + 4 \\leqslant 6, i.e., 2 \\leqslant |MF| \\leqslant 6.  \n\nSolution 2:  \nBy the property of an ellipse, the distance from a point on the ellipse to the left focus is ex + a (this result can be proven using ellipse properties).  \n|MF| = ex_{0} + a.  \n\\because -a \\leqslant x_{0} \\leqslant a, i.e., -4 \\leqslant x_{0} \\leqslant 4,  \nwe obtain: a = 4, b = 2\\sqrt{3}, c = 2, \\therefore e = \\frac{1}{2}.  \nThen |MF| = ex_{0} + a = \\frac{1}{2}x_{0} + 4.  \n\\therefore 2 \\leqslant \\frac{1}{2}x_{0} + 4 \\leqslant 6, i.e., 2 \\leqslant |MF| \\leqslant 6."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, respectively, $P$ a point on the ellipse, $M$ the midpoint of $F_{1} P$, and $|OM|=3$. Then the distance from point $P$ to the left focus of the ellipse is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;O: Origin;M: Point;F2: Point;Expression(G) = (x^2/25 + y^2/16 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);MidPoint(LineSegmentOf(F1, P)) = M;Abs(LineSegmentOf(O, M)) = 3",
    "query_expressions": "Distance(P, LeftFocus(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[19, 58], [69, 71], [69, 71]], [[1, 8]], [[65, 68], [102, 106]], [[92, 100]], [[75, 78]], [[9, 16]], [[19, 58]], [[1, 64]], [[1, 64]], [[65, 74]], [[75, 91]], [[92, 100]]]",
    "query_spans": "[[[102, 117]]]",
    "process": ""
  },
  {
    "text": "Given an ellipse with foci on the $x$-axis, $\\frac{x^{2}}{2 m^{2}}+\\frac{y^{2}}{m+1}=1$, one of its foci lies on the line $\\sqrt{2} x-y+2=0$. Then, the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;m: Number;H: Line;Expression(G) = (y^2/(m + 1) + x^2/(2*m^2) = 1);Expression(H) = (sqrt(2)*x - y + 2 = 0);PointOnCurve(Focus(G), xAxis);PointOnCurve(OneOf(Focus(G)), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2/3",
    "fact_spans": "[[[11, 56], [85, 87]], [[13, 56]], [[62, 82]], [[11, 56]], [[62, 82]], [[2, 56]], [[11, 83]]]",
    "query_spans": "[[[85, 93]]]",
    "process": "Substituting $ y=0 $ into the line equation gives $ x=-\\sqrt{2} $, so the coordinates of one focus of the ellipse are $ (-\\sqrt{2},0) $. Thus, the semi-focal distance is $ c=\\sqrt{2} $. Also, since $ 2m^{2}-(m+1)=c^{2}=2 $, we have $ 2m^{2}-m-3=0 $. Solving this equation yields $ m=\\frac{3}{2} $ or $ m=-1 $ (discarded). Therefore, the length of the semi-major axis is $ a=\\sqrt{2m^{2}}=\\sqrt{2\\times\\frac{9}{4}}=\\frac{3\\sqrt{2}}{2} $. Hence, the eccentricity of the ellipse is $ e=\\frac{c}{a}=\\frac{\\sqrt{2}}{\\frac{3\\sqrt{2}}{2}}=\\frac{2}{3} $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ with foci $F_{1}$, $F_{2}$, and a point $P$ on the ellipse such that $\\angle P F_{1} F_{2}=90^{\\circ}$, then the area of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G) = True;AngleOf(P, F1, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "9",
    "fact_spans": "[[[2, 40], [64, 66]], [[2, 40]], [[44, 51]], [[52, 59]], [[2, 59]], [[60, 63]], [[60, 69]], [[69, 102]]]",
    "query_spans": "[[[104, 135]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the focus of the parabola $x^{2}-4 y-3=0$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 - 4*y - 3 = 0)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,1/4)",
    "fact_spans": "[[[0, 18]], [[0, 18]]]",
    "query_spans": "[[[0, 25]]]",
    "process": ""
  },
  {
    "text": "Given that one focus of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{3}=1$ has coordinates $(-1,0)$, then the value of $m$ is?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (y^2/3 + x^2/m = 1);Coordinate(OneOf(Focus(G))) = (-1, 0)",
    "query_expressions": "m",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 39]], [[57, 60]], [[2, 39]], [[2, 55]]]",
    "query_spans": "[[[57, 64]]]",
    "process": "Since the focus is known to be on the x-axis, we have m - 3 = 1^{2}, so m = 4"
  },
  {
    "text": "Given that $F_{1}$ is the left focus of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, $P$ is a moving point on the ellipse, and $A(1 , 1)$ is a fixed point, then the maximum value of $PA+PF_{1}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);F1: Point;LeftFocus(G) = F1;P: Point;PointOnCurve(P, G);A: Point;Coordinate(A) = (1, 1)",
    "query_expressions": "Max(LineSegmentOf(P, A) + LineSegmentOf(P, F1))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[10, 48], [57, 59]], [[10, 48]], [[2, 9]], [[2, 52]], [[53, 56]], [[53, 63]], [[64, 74]], [[64, 74]]]",
    "query_spans": "[[[80, 97]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>1 , b>0)$ has focal distance $2c$ and eccentricity $e$. If the sum of the distances from the points $(-1 , 0)$ and $(1 , 0)$ to the line $\\frac{x}{a}-\\frac{y}{b}=1$ satisfies $s \\geq \\frac{4}{5} c$, then the range of values for $e$ is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;P:Point;D:Point;e:Number;c:Number;a>1;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (-y/b + x/a = 1);Coordinate(P) = (-1, 0);Coordinate(D) = (1, 0);FocalLength(G)=2*c;Eccentricity(G)=e;Distance(P,H)+Distance(D,H)=s;s:Number;s>=(4/5)*c",
    "query_expressions": "Range(e)",
    "answer_expressions": "[sqrt(5)/2,sqrt(5)]",
    "fact_spans": "[[[2, 59]], [[5, 59]], [[5, 59]], [[100, 129]], [[78, 89]], [[90, 99]], [[158, 161], [158, 161]], [[63, 68]], [[5, 59]], [[5, 59]], [[2, 59]], [[100, 129]], [[78, 89]], [[90, 99]], [[2, 68]], [[2, 76]], [[78, 156]], [[134, 156]], [[134, 156]]]",
    "query_spans": "[[[158, 168]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the focus of the parabola $y^{2}=x$, and $A$, $B$ are two points on this parabola, if the distance from the midpoint of segment $AB$ to the $y$-axis is $\\frac{5}{4}$, then $|AF|+|BF|=$?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;F: Point;Expression(G) = (y^2 = x);Focus(G) = F;PointOnCurve(A, G);PointOnCurve(B, G);Distance(MidPoint(LineSegmentOf(A, B)), yAxis) = 5/4",
    "query_expressions": "Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F))",
    "answer_expressions": "3",
    "fact_spans": "[[[6, 18], [31, 34]], [[22, 25]], [[26, 29]], [[2, 5]], [[6, 18]], [[2, 21]], [[22, 38]], [[22, 38]], [[40, 72]]]",
    "query_spans": "[[[74, 89]]]",
    "process": "From the parabola equation $ y^{2} = x $, we know the focus is $ F\\left(\\frac{1}{4}, 0\\right) $ and the directrix is $ x = -\\frac{1}{4} $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. From the given condition, we have $ \\frac{x_{1} + x_{2}}{2} = \\frac{5}{4} $, so $ x_{1} + x_{2} = \\frac{5}{2} $. By the definition of a parabola, $ |AF| + |BF| = \\left(x_{1} + \\frac{1}{4}\\right) + \\left(x_{2} + \\frac{1}{4}\\right) = (x_{1} + x_{2}) + \\frac{1}{2} = 3 $."
  },
  {
    "text": "Let the coordinates of point $A$ be $(1, \\sqrt{15})$, and let point $P$ move along the parabola $y^{2}=8x$. The distance from $P$ to the line $x=-2$ is $d$. Then the minimum value of $d+|PA|$ is?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;A: Point;d:Number;Expression(G) = (y^2 = 8*x);Expression(H) = (x = -2);Coordinate(A) = (1, sqrt(15));PointOnCurve(P, G);Distance(P, H) = d",
    "query_expressions": "Min(d + Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "4",
    "fact_spans": "[[[31, 45]], [[53, 61]], [[49, 52], [26, 30]], [[1, 5]], [[65, 68]], [[31, 45]], [[53, 61]], [[1, 25]], [[26, 48]], [[49, 68]]]",
    "query_spans": "[[[70, 85]]]",
    "process": "According to the definition of a parabola, when points A, P, and F are collinear, $ d + |PA| $ reaches its minimum value. The focus of the parabola is $ (2, 0) $. By the definition of a parabola, $ PF = d $, so when points A, P, and F are collinear, $ d + |PA| $ attains its minimum value, which is $ |AF| = \\sqrt{1 + 15} = 4 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right vertices are $A$ and $B$ respectively. Point $P$ is an arbitrary point on the hyperbola distinct from $A$ and $B$. If the product of the slopes of lines $PA$ and $PB$ equals $2$, then the eccentricity of the hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;A: Point;P: Point;B: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftVertex(G) = A;RightVertex(G) = B;PointOnCurve(P, G);Negation(P = A);Negation(P = B);Slope(LineOf(P,A))*Slope(LineOf(P,B))=2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 58], [79, 82], [125, 128]], [[5, 58]], [[5, 58]], [[66, 69], [85, 89]], [[74, 78]], [[70, 73], [90, 93]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 73]], [[2, 73]], [[74, 98]], [[74, 98]], [[74, 98]], [[100, 122]]]",
    "query_spans": "[[[125, 135]]]",
    "process": ""
  },
  {
    "text": "Given a hyperbola centered at the origin with foci on the $x$-axis, one of its asymptotes is given by $\\sqrt{3} x - y = 0$. What is the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;O: Origin;Center(G) = O;PointOnCurve(Focus(G), xAxis);Expression(OneOf(Asymptote(G))) = (sqrt(3)*x - y = 0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[17, 20], [48, 51]], [[5, 7]], [[2, 20]], [[8, 20]], [[17, 45]]]",
    "query_spans": "[[[48, 57]]]",
    "process": "According to the given conditions, for a hyperbola centered at the origin with foci on the x-axis, one of its asymptotes is given by $\\sqrt{3}x - y = 0$, so $\\frac{b}{a} = \\sqrt{3}$. Then $e = \\frac{c}{a} = \\sqrt{1 + \\left(\\frac{b}{a}\\right)^2} = \\sqrt{1 + (\\sqrt{3})^2} = 2$, that is, the eccentricity of the hyperbola is $2$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, and points $M(x_{1}, y_{1})$, $N(x_{2}, y_{2})$ moving on the parabola $C$. If $x_{1}+x_{2}+2=2MN$, then the maximum value of $\\angle MFN$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;M: Point;N: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(M) = (x1, y1);Coordinate(N) = (x2, y2);PointOnCurve(M, C);PointOnCurve(N, C);x1 + x2 + 2 = 2*LineSegmentOf(M, N)",
    "query_expressions": "Max(AngleOf(M, F, N))",
    "answer_expressions": "{pi/3, ApplyUnit(60, degree)}",
    "fact_spans": "[[[2, 21], [67, 73]], [[2, 21]], [[25, 28]], [[2, 28]], [[30, 47]], [[49, 66]], [[30, 47]], [[49, 66]], [[30, 47]], [[49, 66]], [[30, 47]], [[49, 66]], [[30, 79]], [[30, 79]], [[81, 102]]]",
    "query_spans": "[[[104, 124]]]",
    "process": "As shown in the figure, by the definition of a parabola, we have |MF| = x_{1} + 1, |NF| = x_{2} + 1, x_{1} + x_{2} + 2 = 2|MN| \\Leftrightarrow |MF| + |NF| = 2|MN|. By the cosine law, \\cos\\angle MFN = \\frac{MF^{2} + NF^{2} - MN^{2}}{2MF \\cdot NF} = \\frac{MF^{2} + NF^{2} - \\frac{1}{4}(MF + NF)^{2}}{2MF \\cdot NF} = \\left(\\frac{3}{4}MF^{2} + \\frac{3}{4}NF^{2} - \\frac{1}{2}MF \\cdot NF\\right) \\div 2MF \\cdot NF \\geqslant \\frac{\\cdot NF - \\frac{1}{2}MF \\cdot NF}{2MF \\cdot NF} = \\frac{1}{2}, 0 < \\angle MFN \\leqslant \\frac{\\pi}{3}."
  },
  {
    "text": "A line passing through the focus of the parabola $C$: $y^{2}=4x$ with slope $1$ intersects $C$ at points $A$ and $B$. Let $M(-1, m)$ be such that $\\angle AMB=90^{\\circ}$. Then $m=?$",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);G: Line;PointOnCurve(Focus(C), G);Slope(G) = 1;Intersection(G, C) = {A, B};A: Point;B: Point;M: Point;Coordinate(M) = (-1, m);m: Number;AngleOf(A, M, B) = ApplyUnit(90, degree)",
    "query_expressions": "m",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 20], [34, 37]], [[1, 20]], [[31, 33]], [[0, 33]], [[24, 33]], [[31, 48]], [[39, 42]], [[43, 46]], [[50, 60]], [[50, 60]], [[89, 92]], [[62, 87]]]",
    "query_spans": "[[[89, 94]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). The focus of the parabola C: y^{2}=4x is (1,0), and the slope of line AB is 1, so the equation of line AB is y=x-\\frac{1}{1}. Rearranging gives y_{1}y_{2}=-4, y_{1}+y_{2}=4, x_{1}x_{2}=\\frac{y_{1}^{2}y_{2}^{2}}{16}=1, y_{1}+y_{2}+2=6. \\angle AMB=90^{\\circ}, \\therefore \\overrightarrow{MA} \\cdot \\overrightarrow{MB}=0, i.e., (x_{1}+1,y_{1}-m) \\cdot (x_{2}+1,y_{2}-m)=0, \\therefore x_{1}x_{2}+x_{1}+x_{2}+1+y_{1}y_{2}-m(y_{1}+y_{2})+m^{2}=0, m^{2}-4m+4=0, solving gives m=2"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$, if the line $y=x-1$ intersects the parabola $C$ at points $M$ and $N$, then $|MN|$=?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);L: Line;Expression(L) = (y = x - 1);Intersection(L, C) = {M, N};M: Point;N: Point",
    "query_expressions": "Abs(LineSegmentOf(M, N))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 21], [33, 39]], [[2, 21]], [[23, 32]], [[23, 32]], [[23, 51]], [[42, 45]], [[46, 49]]]",
    "query_spans": "[[[53, 62]]]",
    "process": "Let M(x_{1},y_{1}), N(x_{2},y_{2}). From \\begin{cases}y=x-1\\\\y^{2}=4x\\end{cases} we obtain x^{2}-6x+1=0, so x_{1}+x_{2}=6, x_{1}x_{2}=1, |MN|=\\sqrt{1+1^{2}}|x_{1}-x_{2}|=\\sqrt{2}\\cdot\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\sqrt{2}\\cdot\\sqrt{6^{2}-4}=8"
  },
  {
    "text": "The focal distance of the ellipse $4 x^{2}+m y^{2}=4 m$ is $2$, then $m$=?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (m*y^2 + 4*x^2 = 4*m);FocalLength(G) = 2",
    "query_expressions": "m",
    "answer_expressions": "{3,5}",
    "fact_spans": "[[[0, 23]], [[32, 35]], [[0, 23]], [[0, 30]]]",
    "query_spans": "[[[32, 37]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the points on the parabola $y^{2}=-8 x$ that are at a distance of $6$ from the focus are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -8*x);P:Point;Distance(P,Focus(G))=6;PointOnCurve(P,G)",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(-4, \\pm 4\\sqrt{2})",
    "fact_spans": "[[[0, 15]], [[0, 15]], [[27, 28]], [[0, 28]], [[0, 28]]]",
    "query_spans": "[[[27, 33]]]",
    "process": "\\because the parabola equation is y^{2}=8x, we have 2p=8, \\frac{p}{2}=2. \\therefore the focus of the parabola is F(-2,0), and the directrix is x=2. Let point P(m,n) on the parabola be such that its distance to focus F is equal to 6. According to the definition of a parabola, the distance from point P to F equals the distance from P to the directrix, i.e., |PF| = -m+2=6, solving gives m=-4, \\therefore n^{2}=8m=32, thus n=\\pm4\\sqrt{2}. Therefore, the coordinates of point P are (-4,\\pm4\\sqrt{2})."
  },
  {
    "text": "Let the ellipse $C$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, $F$ is the right focus, $l$ is a line passing through point $F$ (not parallel to the $y$-axis), intersecting the ellipse at points $A$ and $B$, $l^{\\prime}$ is the perpendicular bisector of $AB$, intersecting the major axis of the ellipse at a point $D$, then the value of $\\frac{D F}{A B}$ is?",
    "fact_expressions": "C: Ellipse;A: Point;B: Point;D: Point;F: Point;l: Line;Expression(C) = (x^2/25 + y^2/9 = 1);RightFocus(C) = F;PointOnCurve(F,l);IsParallel(l, yAxis) = False ;Intersection(l,C)={A,B};PerpendicularBisector(LineSegmentOf(A,B))=l1;Intersection(l1,MajorAxis(C))=D;l1:Line",
    "query_expressions": "LineSegmentOf(D, F)/LineSegmentOf(A, B)",
    "answer_expressions": "2/5",
    "fact_spans": "[[[1, 43], [119, 121], [119, 121]], [[84, 87]], [[90, 93]], [[127, 130]], [[45, 49], [59, 63]], [[54, 57]], [[1, 43]], [[1, 53]], [[54, 68]], [[54, 79]], [[54, 95]], [[96, 117]], [[96, 130]], [[96, 108]]]",
    "query_spans": "[[[132, 153]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>b>0)$ has its right focus at $F$. A line passing through $F$ with slope $\\sqrt{3}$ intersects $C$ at points $A$ and $B$, and $\\overrightarrow{A F}=4 \\overrightarrow{F B}$. Find the eccentricity of $C$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;RightFocus(C) = F;G: Line;Slope(G) = sqrt(3);PointOnCurve(F, G);A: Point;B: Point;Intersection(G, C) = {A, B};VectorOf(A, F) = 4*VectorOf(F, B)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "6/5",
    "fact_spans": "[[[2, 60], [91, 94], [154, 157]], [[2, 60]], [[10, 60]], [[10, 60]], [[10, 60]], [[10, 60]], [[65, 68], [70, 73]], [[2, 68]], [[88, 90]], [[74, 90]], [[69, 90]], [[95, 98]], [[99, 102]], [[88, 104]], [[106, 151]]]",
    "query_spans": "[[[154, 163]]]",
    "process": "Let $ A(x_{1},y_{1}), B(x_{2},y_{2}) $. Solving the equation of the line and the equation of the hyperbola simultaneously, when $ y_{1}+y_{2}=\\frac{2\\sqrt{3}b^{2}c}{3a^{2}-b^{2}} $, $ y_{1}y_{2}=\\frac{-3b^{4}}{3a^{2}-b^{2}} $, from $ \\overrightarrow{AF}=4\\overrightarrow{FB} $ we obtain $ y_{1}=-4y_{2} $. Combining these equations with $ b^{2}=c^{2}-a^{2} $ and simplifying yields $ c=\\frac{6}{5}a $. Since the line $ AB $ passes through the point $ F(c,0) $ and has slope $ \\sqrt{3} $, the equation of line $ AB $ is: $ y=\\sqrt{3}(x-c) $. Substituting into the hyperbola $ \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 $ and eliminating $ x $, we get $ (\\frac{1}{3}b^{2}-a^{2})y^{2}+\\frac{2\\sqrt{3}}{3}b^{2}cy+b^{4}=0 $. Let $ A(x_{1},y_{1}), B(x_{2},y_{2}) $. Since $ \\overrightarrow{AF}=4\\overrightarrow{FB} $, we get $ y_{1}=-4y_{2} $. Thus $ y_{1}+y_{2}=\\frac{2\\sqrt{3}b^{2}c}{3a^{2}-b^{2}} $, $ y_{1}y_{2}=\\frac{-3b^{4}}{3a^{2}-b^{2}} $. Substituting into the above gives $ -3y_{2}=\\frac{2\\sqrt{3}b^{2}c}{3a^{2}-b^{2}} $, $ -4y_{2}^{2}=\\frac{-3b^{4}}{3a^{2}-b^{2}} $. Eliminating $ y_{2} $ and simplifying yields: $ \\frac{4}{3}c^{2}=\\frac{3}{4}(3a^{2}-b^{2}) $. Substituting $ b^{2}=c^{2}-a^{2} $ and simplifying gives $ c^{2}=\\frac{36}{25}a^{2} $. Solving yields $ c=\\frac{6}{5}a $. Therefore, the eccentricity of the hyperbola is $ e=\\frac{c}{a}=\\frac{6}{5} $."
  },
  {
    "text": "Given a line passing through point $A(-2,0)$ intersecting $x=2$ at point $C$, and a line passing through point $B(2,0)$ intersecting $x=-2$ at point $D$, if line $CD$ is tangent to the circle $x^{2}+y^{2}=4$, then the trajectory equation of the intersection point $M$ of lines $AC$ and $BD$ is?",
    "fact_expressions": "G: Circle;C1:Line;C2:Line;C3:Line;C4:Line;D: Point;C: Point;A: Point;B: Point;M:Point;Expression(G) = (x^2 + y^2 = 4);PointOnCurve(A,C1);Expression(C2)=(x=2);Intersection(C1,C2)=C;PointOnCurve(B,C3);Expression(C4)=(x=-2);Intersection(C3,C4)=D;Coordinate(A) = (-2, 0);Coordinate(B) = (2, 0);IsTangent(LineOf(C,D),G);Intersection(LineOf(A,C),LineOf(B,D))=M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "(x^2/4+y^2=1)&Negation(y=0)",
    "fact_spans": "[[[67, 83]], [[14, 16]], [[17, 22]], [[41, 43]], [[44, 50]], [[53, 57]], [[25, 29]], [[3, 13]], [[31, 40]], [[103, 106]], [[67, 83]], [[2, 16]], [[17, 22]], [[14, 29]], [[30, 43]], [[44, 50]], [[41, 57]], [[3, 13]], [[31, 40]], [[59, 85]], [[88, 106]]]",
    "query_spans": "[[[103, 113]]]",
    "process": "Let the slopes of lines AC and BD be $k_{1}$ and $k_{2}$, respectively. Then the equations of lines AC and BD are: $y = k_{1}(x + 2)$, $y = k_{2}(x - 2)$. Thus we obtain: $C(2, 4k_{1})$, $D(-2, -4k_{2})$. Then:  \n$k_{CD} = \\frac{4k_{1} + 4k_{2}}{2 - (-2)} = k_{1} + k_{2}$.  \nThe equation of line CD is: $y - 4k_{1} = (k_{1} + k_{2})(x - 2)$.  \nRearranging gives: $(k_{1} + k_{2})x - y + 2(k_{1} - k_{2}) = 0$.  \nSince the line is tangent to the circle, we have:  \n$\\frac{|2(k_{1} - k_{2})|}{\\sqrt{(k_{1} + k_{2})^{2} + 1}} = 2$.  \nFrom this we obtain: $k_{1}k_{2} = -\\frac{1}{4}$.  \nSince: $y = k_{1}(x + 2)$, $y = k_{2}(x - 2)$,  \nmultiplying these two equations gives: $y^{2} = k_{1}k_{2}(x^{2} - 4) = -\\frac{1}{4}x^{2} + 1$,  \nthus the trajectory equation of the intersection point M of lines AC and BD is $\\frac{x^{2}}{4} + y^{2} = 1$ ($y \\neq 0$)."
  },
  {
    "text": "A moving point $P$ lies on the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left vertex of $C$ is $A$, and the right focus is $F$. When $P F \\perp A F$ and $2|P F|=|A F|$, then the asymptotes of the hyperbola $C$ are given by?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;P: Point;PointOnCurve(P, C);A: Point;LeftVertex(C) = A;F: Point;RightFocus(C) = F;IsPerpendicular(LineSegmentOf(P, F), LineSegmentOf(A, F));2*Abs(LineSegmentOf(P, F)) = Abs(LineSegmentOf(A, F))",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*(sqrt(5)/2)*x",
    "fact_spans": "[[[6, 67], [69, 72], [124, 130]], [[6, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[2, 5]], [[2, 68]], [[77, 80]], [[69, 80]], [[85, 88]], [[69, 88]], [[90, 105]], [[108, 122]]]",
    "query_spans": "[[[124, 138]]]",
    "process": "According to the problem, the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ gives $A(-a,0)$, $F(c,0)$. Without loss of generality, assume point $P$ lies in the first quadrant. Since $PF\\bot AF$, we obtain $P(c,\\frac{b^{2}}{a})$. From $2|PF|=|AF|$, we have $2\\times\\frac{b^{2}}{a}=a+c$, which yields $2b^{2}=a^{2}+ac$. Since $b^{2}=c^{2}-a^{2}$, it follows that $2c^{2}-2a^{2}=a^{2}+ac$. Rearranging gives $2c^{2}-ac-3a^{2}=0$. Solving yields $\\frac{c}{a}=\\frac{3}{2}$ or $\\frac{c}{a}=-1$ (discarded). From $\\frac{c^{2}}{a^{2}}=1+\\frac{b^{2}}{a^{2}}=\\frac{9}{4}$, solving gives $\\frac{b}{a}=\\frac{\\sqrt{5}}{2}$. Therefore, the asymptotes of the hyperbola are $y=\\pm\\frac{\\sqrt{5}}{2}x$."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, respectively, $P$ be any point on the ellipse, and the coordinates of point $M$ be $(6,4)$. Then the minimum value of $|P M|-|P F_{1}|$ is?",
    "fact_expressions": "F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);P: Point;PointOnCurve(P, G);M: Point;Coordinate(M) = (6, 4)",
    "query_expressions": "Min(-Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "-5",
    "fact_spans": "[[[1, 8]], [[9, 16]], [[1, 64]], [[1, 64]], [[19, 58], [69, 71]], [[19, 58]], [[65, 68]], [[65, 75]], [[76, 80]], [[76, 91]]]",
    "query_spans": "[[[93, 117]]]",
    "process": "From the ellipse equation, we have $a^2=25$, $b^{2}=16$, $\\therefore c^{2}=9$, $\\therefore a=5$, $c=3$, the coordinates of the two foci are $(\\pm3,0)$. By the definition of an ellipse, $|PM|-|PF|=|PM|-(2a-|PF_{2}|)=|PM|+|PF_{2}|-10$. Combining the triangle inequality, we know $|PM|+|PF_{2}|>|MF_{2}|=5$, so $|PM|+|PF_{2}|-10\\geqslant-5$, the maximum value is $-5$. Key point: Application of the ellipse equation and its definition"
  },
  {
    "text": "Given point $P$ lies on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola, such that $(\\overrightarrow{O P}+\\overrightarrow{O F_{2}}) \\cdot \\overrightarrow{F_{2} P}=0$ ($O$ is the coordinate origin) and $|P F_{1}|=\\sqrt{3}|P F_{2}|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;O: Origin;P: Point;F2: Point;F1: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, RightPart(G));LeftFocus(G) = F1;RightFocus(G) = F2;DotProduct((VectorOf(O,P)+VectorOf(O,F2)),VectorOf(F2,P))=0;Abs(LineSegmentOf(P,F1))=sqrt(3)*Abs(LineSegmentOf(P,F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)+1",
    "fact_spans": "[[[7, 63], [86, 89], [223, 226]], [[10, 63]], [[10, 63]], [[181, 184]], [[2, 6]], [[78, 85]], [[70, 77]], [[10, 63]], [[10, 63]], [[7, 63]], [[2, 69]], [[70, 95]], [[70, 95]], [[98, 180]], [[192, 221]]]",
    "query_spans": "[[[223, 231]]]",
    "process": "Since $(\\overrightarrow{OP}+\\overrightarrow{OF}_{2})\\cdot\\overrightarrow{F_{2}P}=0 \\Rightarrow (\\overrightarrow{OP}+\\overrightarrow{OF}_{2})\\cdot(\\overrightarrow{OP}-\\overrightarrow{OF_{2}})=0$, it follows that $\\overrightarrow{OP}^{2}-\\overrightarrow{OF_{2}}^{2}=0$, i.e., $|OP|=|OF_{2}|=c=|OF_{1}|$, so $PF_{1}\\bot PF_{2}$. In right triangle $APF_{1}F_{2}$, since $|PF_{1}|=\\sqrt{3}|PF_{2}|$, we have $\\angle PF_{1}F_{2}=30^{\\circ}$. By the definition of hyperbola, $|PF_{1}|-|PF_{2}|=2a \\Rightarrow |PF_{2}|=\\frac{2a}{\\sqrt{3}-1}$. Thus, $\\sin 30^{\\circ}=\\frac{|PF_{2}|}{|F_{1}F_{2}|}=\\frac{\\frac{2a}{\\sqrt{3}-1}}{2c}=\\frac{a}{c(\\sqrt{3}-1)}$. Therefore, $2a=c(\\sqrt{3}-1)$, so $e=\\frac{c}{a}=\\sqrt{3}+1$."
  },
  {
    "text": "The vertex of the parabola is at the origin, and the focus is a focus of the ellipse $x^{2}+4 y^{2}=1$. Then the equation of this parabola?",
    "fact_expressions": "G: Parabola;H: Ellipse;Expression(H) = (x^2 + 4*y^2 = 1);Vertex(G)=O;O:Origin;Focus(G)=OneOf(Focus(H))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=pm*2*sqrt(3)*x",
    "fact_spans": "[[[0, 3], [41, 44]], [[14, 33]], [[14, 33]], [[0, 10]], [[6, 10]], [[0, 38]]]",
    "query_spans": "[[[41, 47]]]",
    "process": "The ellipse $x^{2}+4y^{2}=1 \\therefore x^{2}+\\frac{y^{2}}{4}=1 \\therefore a^{2}=1, b^{2}=\\frac{1}{4} \\therefore c^{2}=\\frac{3}{4}$, the foci are $\\pm\\frac{\\sqrt{3}}{2}, 0$, so the parabola equation is $y^{2}=\\pm2\\sqrt{3}x$"
  },
  {
    "text": "Given that $P$ is a moving point on the parabola $y^{2}=4x$, then the minimum value of the sum of the distances from $P$ to the lines $l_{1}: 4x-3y+11=0$ and $l_{2}: x+1=0$ is?",
    "fact_expressions": "G: Parabola;l1:Line;l2:Line;P: Point;Expression(G) = (y^2 = 4*x);Expression(l1)=(4*x-3*y+11=0);Expression(l2)=(x+1=0);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P,l1)+Distance(P,l2))",
    "answer_expressions": "3",
    "fact_spans": "[[[6, 20]], [[32, 56]], [[57, 73]], [[2, 5], [28, 31]], [[6, 20]], [[32, 56]], [[57, 73]], [[2, 26]]]",
    "query_spans": "[[[28, 84]]]",
    "process": "First determine the positional relationship between the line $l_{1}$ and the parabola. Draw $PM\\bot l_{1}$ from point $P$ meeting at point $M$, $PN\\bot l_{2}$ meeting at point $N$, and connect $PF$. According to the definition of the parabola, we obtain $|PN|=|PF|$, thus deducing $|PN|+|PM|=|PF|+|PM|$. Combining with the figure, when $M$, $P$, $F$ are collinear, $|PF|+|PM|$ is minimized, thereby obtaining the result. From $\\begin{cases}y^2=4x\\\\4x-3y+11=0\\end{cases}$, eliminating $x$ yields $y^{2}-3y+11=0$. Since $\\triangle=(-3)^{2}-4\\times11<0$, the equation $y^{2}-3y+11=0$ has no solution, meaning the line $l_{1}:4x-3y+11=0$ does not intersect the parabola. Draw $PM\\bot l_{1}$ from point $P$ meeting at $M$, $PN\\bot l_{2}$ meeting at $N$. Let $F(1,0)$ be the focus of the parabola $y^{2}=4x$, and connect $PF$. Since $l_{2}:x+1=0$ is the directrix of the parabola $y^{2}=4x$, by the definition of a parabola, we have $|PN|=|PF|$. Then the sum of distances from $P$ to lines $l_{1}$ and $l_{2}$ is $|PN|+|PM|=|PF|+|PM|$. If points $M$, $P$, $F$ are not collinear, then $|PF|+|PM|>|FM|$. When $M$, $P$, $F$ are collinear and $P$ lies between $M$ and $F$, $|PF|+|PM|=|FM|$, so $|PF|+|PM|\\geqslant|FM|$. Also, $|FM|=\\frac{|4-0+11|}{\\sqrt{4^{2}+(-3)^{2}}}=3$, hence $|PN|+|PM|=|PF|+|PM|\\geqslant3$, meaning the minimum value of the required sum of distances is $3$."
  },
  {
    "text": "The coordinates of the focus of the parabola $x=a y^{2}  (a \\neq 0)$ are?",
    "fact_expressions": "G: Parabola;a: Number;Negation(a=0);Expression(G) = (x = a*y^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(1/(4*a),0)",
    "fact_spans": "[[[0, 26]], [[3, 26]], [[3, 26]], [[0, 26]]]",
    "query_spans": "[[[0, 33]]]",
    "process": "From the parabola $ x = ay^{2} $ ($ a \\neq 0 $), we get $ y^{2} = \\frac{1}{a}x $, then $ p = \\frac{1}{2a} $, so its focus coordinates are $ \\left( \\frac{1}{4a}, 0 \\right) $."
  },
  {
    "text": "Let $P$ be a point on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1$, where one asymptote of the hyperbola is $3 x+4 y=0$, and let $F_{1}$, $F_{2}$ be the left and right foci of the hyperbola, respectively. If $|P F_{1}|=10$, then $|P  F_{2}|$ equals?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/9 + x^2/a^2 = 1);a: Number;P: Point;PointOnCurve(P, G);Expression(OneOf(Asymptote(G))) = (3*x + 4*y = 0);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(P, F1)) = 10",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "{18, 2}",
    "fact_spans": "[[[5, 47], [52, 55], [96, 99]], [[5, 47]], [[8, 47]], [[1, 4]], [[1, 50]], [[52, 75]], [[76, 84]], [[86, 93]], [[77, 105]], [[77, 105]], [[107, 121]]]",
    "query_spans": "[[[123, 138]]]",
    "process": "According to the asymptote equations of the hyperbola, the relationship between a and b can be found, and then a can be determined. Based on the definition of hyperbola, \\because|PF_{1}|\\cdot|PF_{2}|=2a or |PF_{2}|\\cdot|PF_{1}|=2a, thus the answer can be obtained. \\because one asymptote equation of the hyperbola is 3x+4y=0, b=3\\cdot\\frac{3}{a}\\frac{3}{4}, \\therefore a=4, \\therefore|PF_{1}|\\cdot|PF_{2}|=2a=8 or |PF_{2}|\\cdot|PF_{1}|=2a=8, \\therefore|PF_{2}|=2 or 18,"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the right focus is $F$, the upper vertex is $B$, the line $BF$ intersects the ellipse $C$ at another point $A$, and $\\frac{AF}{BF}=\\frac{4}{5}$. Find the eccentricity of the ellipse $C$.",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;B: Point;F: Point;A: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;UpperVertex(C) = B;OneOf(Intersection(LineOf(B,F),C)) = A;LineSegmentOf(A, F)/LineSegmentOf(B, F) = 4/5",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "1/3",
    "fact_spans": "[[[2, 59], [84, 89], [132, 137]], [[9, 59]], [[9, 59]], [[72, 75]], [[64, 67]], [[96, 99]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 67]], [[2, 75]], [[76, 99]], [[101, 130]]]",
    "query_spans": "[[[132, 143]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $\\frac{2 \\sqrt{6}}{3}$, and its vertices coincide with the foci of the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{5}=1$, what are the coordinates of the foci of the hyperbola? What are the equations of the asymptotes?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Ellipse;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2/8 + y^2/5 = 1);Eccentricity(G) = (2*sqrt(6))/3;Vertex(G)=Focus(H)",
    "query_expressions": "Coordinate(Focus(G));Expression(Asymptote(G))",
    "answer_expressions": "(pm*2*sqrt(2),0)\ny=pm*(sqrt(15)/3)*x",
    "fact_spans": "[[[2, 48], [124, 127]], [[5, 48]], [[5, 48]], [[79, 116]], [[2, 48]], [[79, 116]], [[2, 75]], [[2, 121]]]",
    "query_spans": "[[[124, 134]], [[124, 141]]]",
    "process": ""
  },
  {
    "text": "Given an ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with two foci $F_{1}$ and $F_{2}$. If there exists a point $P$ on the ellipse such that $|P F_{1}+P F_{2}|=| F_{1} F_{2} |$, then what is the range of values for the eccentricity?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;F1: Point;F2: Point;P: Point;a > b;b > 0;Focus(G)={F1,F2};PointOnCurve(P,G);Abs(LineSegmentOf(P,F1)+LineSegmentOf(P,F2))=Abs(LineSegmentOf(F1,F2))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[\\sqrt{2}/2,1)",
    "fact_spans": "[[[2, 54], [79, 81]], [[2, 54]], [[4, 54]], [[4, 54]], [[62, 69]], [[70, 77]], [[84, 88]], [[4, 54]], [[4, 54]], [[2, 77]], [[79, 88]], [[90, 126]]]",
    "query_spans": "[[[79, 140]]]",
    "process": ""
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $C$: $y^{2}=2px$ ($p>0$) intersects the parabola at points $A$ and $B$. If $|AF|=5|BF|$, and $O$ is the origin, then $\\frac{|AF|}{|OF|}$=?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;A: Point;F: Point;B: Point;O: Origin;p>0;Expression(C) = (y^2 = 2*p*x);Focus(C) = F;PointOnCurve(F, G);Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, F)) = 5*Abs(LineSegmentOf(B, F))",
    "query_expressions": "Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(O, F))",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 27], [38, 41]], [[9, 27]], [[34, 36]], [[42, 45]], [[30, 33]], [[46, 49]], [[69, 73]], [[9, 27]], [[1, 27]], [[1, 33]], [[0, 36]], [[34, 51]], [[53, 67]]]",
    "query_spans": "[[[80, 103]]]",
    "process": "From the given conditions, $ y^{2} = 2px $, then $ F(0, \\frac{p}{2}) $, so $ |OF| = \\frac{p}{2} $. Let the equation of line $ AB $ be $ y = k(x - \\frac{p}{2}) $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, and $ x_{1} > x_{2} $. Since $ |AF| = 5|BF| $, we have $ \\overrightarrow{AF} = -5\\overrightarrow{BF} $, then $ y_{1} = -5y_{2} $, $\\textcircled{1}$ From $ \\begin{cases} y = k(x - \\frac{p}{2}) \\\\ y^{2} = 2px \\end{cases} $, simplifying gives $ y^{2} - \\frac{2p}{k}y - p^{2} = 0 $, so $ y_{1} + y_{2} = \\frac{2p}{k} $, $ y_{1}y_{2} = -p^{2} $, $\\textcircled{2}$ Solving $\\textcircled{1}$ and $\\textcircled{2}$ together yields $ k = \\frac{\\sqrt{5}}{2} $, thus the equation of line $ AB $ is $ y = \\frac{\\sqrt{5}}{2}(x - \\frac{p}{2}) $. Again, $ \\begin{cases} y = \\frac{\\sqrt{5}}{2}(x - \\frac{p}{2}) \\\\ y^{2} = 2px \\end{cases} $, simplifying gives $ 20x^{2} - 52px + 5p^{2} = 0 $, solving yields $ x = \\frac{5}{2}p $ or $ x = \\frac{1}{10}p $. Since $ y^{2} = 2px $, according to the definition of parabola, $ |AF| = \\frac{5}{2}p + \\frac{1}{2}p = 3p $, so $ \\frac{|AF|}{|OF|} = 6 $."
  },
  {
    "text": "If the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ has no intersection points with the line $y=3 x$, then the range of eccentricity $e$ is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = 3*x);NumIntersection(G, H)=0;Eccentricity(G)=e;e:Number",
    "query_expressions": "Range(e)",
    "answer_expressions": "(1,\\sqrt{10}]",
    "fact_spans": "[[[1, 58]], [[4, 58]], [[4, 58]], [[59, 68]], [[4, 58]], [[4, 58]], [[1, 58]], [[59, 68]], [[1, 71]], [[1, 79]], [[76, 79]]]",
    "query_spans": "[[[76, 86]]]",
    "process": "According to the positional relationship between a line and a hyperbola, obtain the relationship between $a$ and $b$, combined with the eccentricity formula, the range of eccentricity can be easily found. $\\because$ if the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0,b>0)$ has no common point with the line $y=3x$, $\\therefore$ it is equivalent to the slope $\\frac{b}{a}$ of the asymptote $y=\\frac{b}{a}x$ of the hyperbola satisfying $\\frac{b}{a}\\leqslant3$, i.e., $b\\leqslant3a$, which implies $b^{2}\\leqslant9a^{2}$, i.e., $c^{2}-a^{2}\\leqslant9a^{2}$, so $c^{2}\\leqslant10a^{2}$, then $c\\leqslant\\sqrt{10}a$, thus $e\\leqslant\\sqrt{10}$. $\\because e>1$, $\\therefore$ the eccentricity satisfies $1<e\\leqslant\\sqrt{10}$, i.e., the range of the eccentricity of the hyperbola is $(1,\\sqrt{10}]$"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the right branch of the hyperbola such that $|P F_{2}|=|F_{1} F_{2}|$. If point $Q$ is the midpoint of segment $P F_{1}$, then the range of $\\angle P F_{2} Q$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F2)) = Abs(LineSegmentOf(F1, F2));Q: Point;MidPoint(LineSegmentOf(P, F1)) = Q",
    "query_expressions": "Range(AngleOf(P, F2, Q))",
    "answer_expressions": "(\\pi/6, \\pi/2)",
    "fact_spans": "[[[2, 58], [88, 91]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[67, 74]], [[75, 82]], [[2, 82]], [[2, 82]], [[83, 87]], [[83, 95]], [[97, 122]], [[124, 128]], [[124, 143]]]",
    "query_spans": "[[[145, 170]]]",
    "process": "From the properties of an isosceles triangle, we have $ F_{2}Q\\bot PF_{1} $, and $ \\sin\\angle PF_{2}Q = \\frac{|PQ|}{|PF_{2}|} $. According to the definition of the hyperbola, these two segments can be expressed in terms of $ a $ and $ c $, thus allowing us to find the range of values for the ratio, and then the range of the angle. [Detailed explanation] Since $ |PF_{2}| = |F_{1}F_{2}| $, and point $ Q $ is the midpoint of segment $ PF_{1} $, therefore $ QF_{2}\\bot PF_{1} $, $ |PF_{2}| = 2c $, $ |PQ| = \\frac{1}{2}|PF_{1}| = \\frac{1}{2}(|PF_{2}| + 2a) = a + c $. Hence, $ \\sin\\angle PF_{2}Q = \\frac{|PQ|}{|PF_{2}|} = \\frac{a + c}{2c} = \\frac{1}{2} + \\frac{1}{2e} \\in \\left( \\frac{1}{2}, 1 \\right) $, therefore $ \\angle PF_{2}Q \\in \\left( \\frac{\\pi}{6}, \\frac{\\pi}{2} \\right) $."
  },
  {
    "text": "The eccentricity of the hyperbola $x^{2}-y^{2}=4$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 4)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 18]], [[0, 18]]]",
    "query_spans": "[[[0, 24]]]",
    "process": "Since the hyperbola is a rectangular hyperbola, the answer can be obtained. [Detailed solution] From $x^{2}-y^{2}=4$, we get $\\frac{x^{2}}{4}-\\frac{y^{2}}{4}=1$, so the hyperbola is a rectangular hyperbola, hence the eccentricity is $\\sqrt{2}$."
  },
  {
    "text": "The eccentricity of a hyperbola with asymptotes given by $x \\pm y=0$ is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (x+(pm*y) = 0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[18, 21]], [[0, 21]]]",
    "query_spans": "[[[18, 27]]]",
    "process": "Since the asymptotes of the hyperbola are given by $x \\pm y = 0$, whether the foci lie on the $x$-axis or the $y$-axis, we have $a = b$. Thus, $c = \\sqrt{2}a$, and according to the definition of eccentricity of a hyperbola, the result follows. [Solution] For a hyperbola with asymptotes $x \\pm y = 0$, we obtain $a = b$, so $c = \\sqrt{a^{2} + b^{2}} = \\sqrt{a^{2} + a^{2}} = \\sqrt{2}a$. Therefore, the eccentricity of the hyperbola is $e = \\frac{c}{a} = \\frac{\\sqrt{2}a}{a} = \\sqrt{2}$."
  },
  {
    "text": "Given points $A(3,0)$ and $B(-3,0)$, a moving point $P$ satisfies $|P B|-|P A|=4$, then the equation of the trajectory of $P$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (3, 0);B: Point;Coordinate(B) = (-3, 0);P: Point;Abs(LineSegmentOf(P, B))- Abs(LineSegmentOf(P, A)) = 4",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(x^2/4-y^2/5=1)&(x>0)",
    "fact_spans": "[[[2, 11]], [[2, 11]], [[12, 21]], [[12, 21]], [[24, 27], [46, 49]], [[29, 44]]]",
    "query_spans": "[[[46, 56]]]",
    "process": "Since A and B are fixed points, and |PB| - |PA| = 4 < |AB|, the trajectory of point P is the right branch of a hyperbola. Given 2a = 4, c = 3, we obtain b^{2} = c^{2} - a^{2} = 9 - 4 = 5, so the equation of the trajectory of point P is \\frac{x^{2}}{4} - \\frac{y^{2}}{5} = 1 (x > 0)."
  },
  {
    "text": "It is known that a line passing through the point $(1,0)$ intersects the parabola $x^{2}=y$ at points $A$ and $B$, and the perpendicular bisector of segment $AB$ passes through the point $(0,2)$. $F$ is the focus of the parabola. Then $|A F|+|B F|$=?",
    "fact_expressions": "G: Parabola;H: Line;B: Point;A: Point;P:Point;D:Point;F: Point;Expression(G) = (x^2 = y);Coordinate(P) = (1, 0);Coordinate(D) = (0, 2);PointOnCurve(P,H);Intersection(H, G) = {A, B};Focus(G)=F;PointOnCurve(D,PerpendicularBisector(LineSegmentOf(A,B)))",
    "query_expressions": "Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F))",
    "answer_expressions": "7/2",
    "fact_spans": "[[[15, 27], [67, 70]], [[12, 14]], [[33, 36]], [[29, 32]], [[3, 11]], [[54, 62]], [[63, 66]], [[15, 27]], [[3, 11]], [[54, 62]], [[2, 14]], [[12, 38]], [[63, 73]], [[39, 62]]]",
    "query_spans": "[[[75, 90]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ x_{1}^{2} = y_{1} $, $ x_{2}^{2} = y_{2} $. Subtracting these two equations gives: $ (x_{1}-x_{2})(x_{1}+x_{2}) = y_{1}-y_{2} $. Therefore, $ \\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = x_{1}+x_{2} $, i.e., the slope of $ AB $ is $ x_{1}+x_{2} $. Let $ |AF| + |BF| = m $, then $ y_{1} + y_{2} + \\frac{1}{2} = m $, so $ y_{1} + y_{2} = m - \\frac{1}{2} $, thus the midpoint of $ AB $ has coordinates $ \\left( \\frac{x_{1}+x_{2}}{2}, \\frac{m}{2} - \\frac{1}{4} \\right) $. The slope of the perpendicular bisector of $ AB $ is $ -\\frac{1}{x_{1}+x_{2}} $. The equation of the perpendicular bisector of $ AB $ is $ y - \\left( \\frac{m}{2} - \\frac{1}{4} \\right) = -\\frac{1}{x_{1}+x_{2}} \\left( x - \\frac{x_{1}+x_{2}}{2} \\right) $. Since the perpendicular bisector of segment $ AB $ passes through the point $ (0,2) $, solving gives $ m = \\frac{7}{3} $. The value of $ |AF| + |BF| $ is $ \\frac{7}{2} $."
  },
  {
    "text": "Given a moving circle $C$ internally tangent to circle $A$: $(x+3)^{2}+y^{2}=100$, and passing through the fixed point $B(3,0)$, then the trajectory equation of the center $C$ of the moving circle is?",
    "fact_expressions": "C1: Circle;C: Point;A: Circle;B: Point;Coordinate(B) = (3, 0);Expression(A) = ((x + 3)^2 + y^2 = 100);IsInTangent(C1, A);PointOnCurve(B, C1);Center(C1) = C",
    "query_expressions": "LocusEquation(C)",
    "answer_expressions": "x^2/25 + y^2/16 = 1",
    "fact_spans": "[[[5, 8]], [[56, 59]], [[11, 37]], [[42, 50]], [[42, 50]], [[11, 37]], [[5, 37]], [[5, 50]], [[52, 59]]]",
    "query_spans": "[[[52, 66]]]",
    "process": "Circle A: $(x+3)^{2}+y^{2}=100$ has center $A(-3,0)$. From the problem, we know: $|CA|+|CB|=10>6$, so the locus of center $C$ is an ellipse with foci on the x-axis. Let the ellipse equation be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. Thus, $2a=10$, $2c=6$, so $a=5$, $c=3$, and therefore $b^{2}=a^{2}-c^{2}=16$. The equation of the ellipse is: $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$."
  },
  {
    "text": "Given that the line $x + y - 2 = 0$ intersects the parabola $y^{2} = 8x$ at points $A$ and $B$, then $|AB| = $?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;Expression(G) = (y^2 = 8*x);Expression(H) = (x + y - 2 = 0);Intersection(H, G) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16",
    "fact_spans": "[[[14, 28]], [[2, 13]], [[30, 33]], [[34, 37]], [[14, 28]], [[2, 13]], [[2, 39]]]",
    "query_spans": "[[[41, 50]]]",
    "process": "Solving the system \\begin{cases}x+y-2=0\\\\y^2=8x\\end{cases} gives: (2-x)^{2}=8x, i.e., x^{2}-12x+4=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), then x_{1}+x_{2}=12, x_{1}x_{2}=4. Therefore, AB=\\sqrt{2}|x_{1}-x_{2}|=\\sqrt{2}\\times\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\sqrt{2}\\times\\sqrt{144-16}=16."
  },
  {
    "text": "If the curve of the equation $\\frac{x^{2}}{4-k}+\\frac{y^{2}}{6+k}=1$ is an ellipse with foci on the $x$-axis, then the range of real values for $k$ is?",
    "fact_expressions": "H: Curve;Expression(H) = (x^2/(4 - k) + y^2/(6 + k) = 1);k: Real;G: Ellipse;PointOnCurve(Focus(G), xAxis);H = G",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-6, -1)",
    "fact_spans": "[[[43, 45]], [[1, 45]], [[59, 64]], [[55, 57]], [[46, 57]], [[43, 57]]]",
    "query_spans": "[[[59, 71]]]",
    "process": "The curve of the equation $\\frac{x^{2}}{4-k}+\\frac{y^{2}}{6+k}=1$ is an ellipse with foci on the x-axis, then $4-k>6+k>0$, solving yields $-6<k<-1$, that is, the range of real number $k$ is $-6<k<-1$."
  },
  {
    "text": "The hyperbola $C$ shares common asymptotes with the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$, and $C$ passes through the point $(2,0)$. Then the standard equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;C1: Hyperbola;Expression(C1) = (x^2-y^2/4=1);Asymptote(C) = Asymptote(C1);G: Point;Coordinate(G) = (2, 0);PointOnCurve(G, C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4 - y^2/16 = 1",
    "fact_spans": "[[[0, 6], [44, 47], [58, 61]], [[7, 35]], [[7, 35]], [[0, 42]], [[48, 56]], [[48, 56]], [[44, 56]]]",
    "query_spans": "[[[58, 68]]]",
    "process": "Let the standard equation of C be $x^2 - \\frac{y^{2}}{4} = \\lambda$ ($\\lambda \\neq 0$). Since the point $(2, 0)$ lies on the hyperbola, $\\lambda = 2^{2} = 4$. Therefore, the standard equation of C is $x^{2} - \\frac{y^{2}}{4} = 4$, which is $\\frac{x^{2}}{4} - \\frac{y^{2}}{16} = 1$. Answer: $\\frac{x^{2}}{4} - \\frac{y^{2}}{16} = 1$"
  },
  {
    "text": "Given a fixed line segment $AB$ of length $4$ in the plane, a moving point $P$ satisfies $|PA| - |PB| = 3$, and $O$ is the midpoint of $AB$. What is the minimum value of $|OP|$?",
    "fact_expressions": "B: Point;A: Point;P: Point;O: Origin;Length(LineSegmentOf(A,B))=4;Abs(LineSegmentOf(P, A)) - Abs(LineSegmentOf(P, B)) = 3;MidPoint(LineSegmentOf(A, B)) = O",
    "query_expressions": "Min(Abs(LineSegmentOf(O, P)))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[17, 22]], [[17, 22]], [[25, 28]], [[46, 49]], [[7, 22]], [[30, 43]], [[45, 56]]]",
    "query_spans": "[[[58, 69]]]",
    "process": ""
  },
  {
    "text": "Given that the intersection point of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the circle $x^{2}+y^{2}=c^{2}$ in the second quadrant is point $P$, $F_{1}(-c, 0)$ is the left focus of the ellipse, $O$ is the origin, and the distance from $O$ to the line $P F_{1}$ is $\\frac{\\sqrt{3}}{2} c$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;H: Circle;Expression(H) = (x^2 + y^2 = c^2);c: Number;P: Point;Quadrant(P) = 2;Intersection(G, H) = P;F1: Point;Coordinate(F1) = (-c, 0);LeftFocus(G) = F1;O: Origin;Distance(O, LineOf(P, F1)) = sqrt(3)/2*c",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[2, 54], [104, 106], [163, 165]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[55, 75]], [[55, 75]], [[56, 75]], [[84, 88]], [[75, 88]], [[2, 88]], [[89, 103]], [[89, 103]], [[89, 110]], [[111, 114], [120, 123]], [[120, 161]]]",
    "query_spans": "[[[163, 171]]]",
    "process": "Draw the graph, and the expressions for |PF_{1}| and |PF_{2}| can be found separately. Combining |PF_{1}| + |PF_{2}| = 2a, the eccentricity can be determined. From the given condition, PF_{1} \\perp PF_{2}. Draw ON // PF_{2} from O, intersecting PF at point N. Since O is the midpoint of F_{1}F_{2}, ON is the midline of \\triangle PF_{1}F_{2}. Given that the distance from O to line PF is \\frac{\\sqrt{3}}{2}, it follows that |PF_{2}| = 2|ON| = \\sqrt{3}c. Thus, |PF| = \\sqrt{|F_{1}F_{2}|^{2} - |PF_{2}|^{2}} = \\sqrt{4c^{2} - 3c^{2}} = c. According to the definition of an ellipse, |PF| + |PF_{2}| = c + \\sqrt{3}c = 2a, so \\frac{c}{a} = \\frac{2}{\\sqrt{3}+1} = \\sqrt{3}-1."
  },
  {
    "text": "The distance from a point $P$ on the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ to the point $(5 , 0)$ is $8.5$. Then, what is the distance from point $P$ to the left directrix?",
    "fact_expressions": "G: Hyperbola;H: Point;P: Point;Expression(G) = (x^2/16 - y^2/9 = 1);Coordinate(H) = (5, 0);PointOnCurve(P, G);Distance(P, H) = 8.5",
    "query_expressions": "Distance(P,LeftDirectrix(G))",
    "answer_expressions": "66/5",
    "fact_spans": "[[[0, 39]], [[46, 56]], [[41, 45], [67, 71]], [[0, 39]], [[46, 56]], [[0, 45]], [[41, 65]]]",
    "query_spans": "[[[0, 80]]]",
    "process": "Given $ c = \\sqrt{16 + 9} = 5 $, so the point $ (5, 0) $ is the right focus of the hyperbola. It is easy to see that the minimum distance from a point on the left branch of the hyperbola to the right focus is $ 4 + 5 = 9 $. Therefore, $ P $ lies on the right branch, $ e = \\frac{5}{4} $, then the distance from $ P $ to the right directrix is $ d = \\frac{8.5}{\\frac{5}{4}} = 6.8 $. The equation of the right directrix is $ x = \\frac{16}{5} $, so the distance from $ P $ to the left directrix is $ 6.8 + \\frac{16}{5} \\times 2 = \\frac{66}{5} $."
  },
  {
    "text": "If the line $l$: $y = kx - 1$ and the hyperbola $C$: $\\frac{x^2}{4} - y^2 = 1$ have two common points, then the range of real values for $k$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/4 - y^2 = 1);l: Line;Expression(l) = (y = k*x - 1);k: Real;NumIntersection(l, C) = 2",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-sqrt(2)/2, -1/2) + (-1/2, 1/2) + (1/2,sqrt(2)/2)",
    "fact_spans": "[[[18, 51]], [[18, 51]], [[1, 17]], [[1, 17]], [[59, 64]], [[1, 57]]]",
    "query_spans": "[[[59, 71]]]",
    "process": "Solve the system of equations \\begin{cases} y = kx - 1 \\\\ \\frac{x^2}{4} - y^2 = 1 \\end{cases}, simplifying gives (1-4k^2)x^2 + 8kx - 8 = 0. Since the line $ l: y = kx - 1 $ intersects the hyperbola $ C: \\frac{x^2}{4} - y^2 = 1 $ at two distinct points, we have $ \\Delta = 64k^2 + 32(1 - 4k^2) > 0 $. Solving this inequality yields $ -\\frac{\\sqrt{2}}{2} < k < \\frac{\\sqrt{2}}{2} $, and $ k \\neq \\pm \\frac{1}{2} $. Therefore, the range of real values for $ k $ is $ \\left(-\\frac{\\sqrt{2}}{2}, -\\frac{1}{2}\\right) \\cup \\left(-\\frac{1}{2}, \\frac{1}{2}\\right) \\cup \\left(\\frac{1}{2}, \\frac{\\sqrt{2}}{2}\\right) $."
  },
  {
    "text": "Given that point $P$ is on the ellipse $2 x^{2} + 3 y^{2} = 6$, then the shortest distance from point $P$ to one of the foci of the ellipse is?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (2*x^2 + 3*y^2 = 6);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, OneOf(Focus(G))))",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[7, 28], [38, 40]], [[2, 6], [33, 37]], [[7, 28]], [[2, 31]]]",
    "query_spans": "[[[33, 52]]]",
    "process": "From $2x^{2}+3y^{2}=6$, we get $\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1$, so $a^{2}=3$, $b^{2}=2$, thus $a=\\sqrt{3}$, $b=\\sqrt{2}$, $c=1$, so the shortest distance from point $P$ to one focus of the ellipse is $a-c=\\sqrt{3}-1$."
  },
  {
    "text": "The coordinates of the right focus of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1)",
    "query_expressions": "Coordinate(RightFocus(G))",
    "answer_expressions": "(5, 0)",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 48]]]",
    "process": "From the standard equation of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, we get $a=4$, $b=3$. $\\therefore c=5$. So the right focus is $(5,0)$."
  },
  {
    "text": "Through the left focus $F(-c, 0)$ $(c>0)$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a tangent to the circle $x^{2}+y^{2}=\\frac{a^{2}}{4}$, with point of tangency $E$. Extend $FE$ to intersect the right branch of the hyperbola at point $P$. If $\\overrightarrow{O P}=2 \\overrightarrow{O E}-\\overrightarrow{O F}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;F: Point;E: Point;O: Origin;P: Point;c:Number;c>0;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + y^2 = a^2/4);Coordinate(F) = (-c, 0);LeftFocus(G) = F;l:Line;TangentOfPoint(F,H)=l;TangentPoint(l,H)=E;Intersection(OverlappingLine(LineSegmentOf(F,E)),RightPart(G))=P;VectorOf(O, P) = 2*VectorOf(O, E) - VectorOf(O, F)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)/2",
    "fact_spans": "[[[1, 57], [127, 130], [207, 210]], [[4, 57]], [[4, 57]], [[78, 108]], [[61, 76]], [[115, 118]], [[139, 205]], [[133, 137]], [[61, 76]], [[61, 76]], [[4, 57]], [[4, 57]], [[1, 57]], [[78, 108]], [[61, 76]], [[1, 76]], [], [[77, 111]], [[77, 118]], [[119, 137]], [[139, 205]]]",
    "query_spans": "[[[207, 216]]]",
    "process": "Let the right focus be $ F $. $ \\overrightarrow{OP} = 2\\overrightarrow{OE} - \\overrightarrow{OF} $, therefore $ \\overrightarrow{OP} + \\overrightarrow{OF} = 2\\overrightarrow{OE} $, since $ E $ is the midpoint of $ PF $, and $ O $ is the midpoint of $ FF $, thus $ |PF| = 2|OE| = a $ and $ PF' | OE $, by the definition of hyperbola we know $ |PF| = 3a $, because $ OE \\perp PF $, so $ PF \\perp PF $, hence $ (3a)^{2} + a^{2} = 4c^{2} $, i.e., $ e^{2} = \\frac{c^{2}}{a^{2}} = \\frac{10}{4} $, therefore $ e = \\frac{\\sqrt{10}}{2} $. The numerical answer is: $ \\sqrt{10} $"
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x(p>0)$ coincides with the right focus of the hyperbola $\\frac{x^{2}}{12}-\\frac{y^{2}}{4}=1$, then $p$=?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;Expression(G) = (x^2/12 - y^2/4 = 1);p>0;Expression(H) = (y^2 = 2*p*x);Focus(H) = RightFocus(G)",
    "query_expressions": "p",
    "answer_expressions": "8",
    "fact_spans": "[[[26, 65]], [[1, 22]], [[73, 76]], [[26, 65]], [[4, 22]], [[1, 22]], [[1, 71]]]",
    "query_spans": "[[[73, 78]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of hyperbola $C$ are $2x \\pm 3y = 0$, write a standard equation of hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;Expression(Asymptote(C)) = (2*x+pm*3*y = 0)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/9 - y^2/4 = 1",
    "fact_spans": "[[[2, 8], [33, 39]], [[2, 30]]]",
    "query_spans": "[[[33, 47]]]",
    "process": "The asymptotes of hyperbola C are given by $2x\\pm3y=0$, so the equation of the hyperbola is: $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=\\lambda$, $\\lambda\\in\\mathbb{R}$, and $\\lambda\\neq0$. The required hyperbola equation is: $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1$, the answer is not unique."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, one of its asymptotes passes through the point $(1,-3)$. What is the eccentricity of $C$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;PointOnCurve(G,OneOf(Asymptote(C)))  =True;G: Point;Coordinate(G) = (1, -3)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[2, 63], [81, 84]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 79]], [[70, 79]], [[70, 79]]]",
    "query_spans": "[[[81, 90]]]",
    "process": "The hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) has an asymptote passing through the point $ (1, -3) $. One asymptote equation of the hyperbola is $ bx + ay = 0 $, $ \\therefore b = 3a $. $ \\cdot c = \\sqrt{a^{2} + b^{2}} = \\sqrt{10}a $, $ \\therefore e = \\frac{c}{a} = \\sqrt{10} $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(b>a>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ with slope $1$ passing through $F_{1}$ intersects the two asymptotes of the hyperbola at points $A$ and $B$ respectively. If $|A F_{1}|=|A B|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;b > a;a > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;l: Line;PointOnCurve(F1, l);Slope(l) = 1;A: Point;B: Point;L1: Line;L2: Line;Asymptote(G) = {L1, L2};Intersection(l, L1) = A;Intersection(l, L2) = B;Abs(LineSegmentOf(A, F1)) = Abs(LineSegmentOf(A, B))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[2, 55], [102, 105], [145, 148]], [[2, 55]], [[5, 55]], [[5, 55]], [[5, 55]], [[5, 55]], [[64, 71], [81, 88]], [[72, 79]], [[2, 79]], [[2, 79]], [[96, 101]], [[80, 101]], [[89, 101]], [[115, 118]], [[119, 122]], [], [], [[102, 111]], [[96, 124]], [[96, 124]], [[126, 143]]]",
    "query_spans": "[[[145, 154]]]",
    "process": "Analysis: First, according to the problem, set the equation of the line and solve it simultaneously with the asymptotes of the hyperbola to find the coordinates of points A and B respectively. Then, based on the condition |AF_{1}| = |AB|, deduce that A is the midpoint of F_{1}B. Using the midpoint coordinate formula, obtain the relationship between their coordinates. With the relationship among a, b, c in the hyperbola, derive the eccentricity of the hyperbola. Specifically, let the equation of line l be y = x + c, and the equations of the two asymptotes be y = -\\frac{b}{a}x and y = \\frac{b}{a}x respectively. Solving the systems of equations yields A(-\\frac{ac}{a+b}, \\frac{bc}{a+b}), B(\\frac{ac}{b-a}, \\frac{bc}{b-a}). From |AF_{1}| = |AB|, it follows that A is the midpoint of F_{1}B, so -c + \\frac{ac}{b-a} = -\\frac{2ac}{a+b}. Simplifying gives b = 3a. Combining with the relationship among a, b, c in the hyperbola, we obtain e = \\frac{c}{a} = \\sqrt{\\frac{a^{2}+b^{2}}{a^{2}}} = \\sqrt{10}. Therefore, the answer is \\sqrt{10}."
  },
  {
    "text": "Draw a line $l$ through the focus $F$ of the parabola $C$: $x^{2}=4 y$ with an inclination angle of $\\frac{\\pi}{4}$. The line $l$ intersects the parabola $C$ at two distinct points $A$ and $B$. Then $|A B|=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 4*y);F: Point;Focus(C) = F;l: Line;PointOnCurve(F, l);Inclination(l) = pi/4;A: Point;B: Point;Intersection(l, C) = {A, B};Negation(A=B)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 20], [58, 64]], [[1, 20]], [[23, 26]], [[1, 26]], [[47, 52], [54, 57]], [[0, 52]], [[27, 52]], [[71, 75]], [[76, 79]], [[54, 79]], [[66, 79]]]",
    "query_spans": "[[[81, 90]]]",
    "process": "Obtain the equation of line $ l $ as $ y = x + 1 $. By solving the system of equations and using the relationship between roots and coefficients, find $ x_{1} + x_{2} $, then obtain the value of $ y_{1} + y_{2} $. Combining this with the property of the focal chord of a parabola, the solution is obtained. According to the problem, the parabola $ C: x^{2} = 4y $ has focus $ F(0,1) $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Since the inclination angle of line $ l $ is $ \\frac{\\pi}{4} $, the slope is $ k = 1 $, so the equation of the line is $ y = x + 1 $. Solving the system of equations\n$$\n\\begin{cases}\ny = x + 1 \\\\\nx^{2} = 4y\n\\end{cases}\n$$\nwe obtain $ x^{2} - 4x - 4 = 0 $, from which $ x_{1} + x_{2} = 4 $. Then $ y_{1} + y_{2} = x_{1} + x_{2} + 2 = 6 $. Using the property of the focal chord of the parabola, we get $ |AB| = y_{1} + y_{2} + p = 6 + 2 = 8 $."
  },
  {
    "text": "If the left focus of the hyperbola $\\frac{x^{2}}{3}-\\frac{16 y^{2}}{p^{2}}=1  (p>0)$ lies on the directrix of the parabola $y^{2}=2 p x$, then $p$=?",
    "fact_expressions": "G: Hyperbola;p: Number;H: Parabola;p>0;Expression(G) = (x^2/3 - 16*y^2/p^2 = 1);Expression(H) = (y^2 = 2*p*x);PointOnCurve(LeftFocus(G),Directrix(H))",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 53]], [[80, 83]], [[58, 74]], [[4, 53]], [[1, 53]], [[58, 74]], [[1, 78]]]",
    "query_spans": "[[[80, 85]]]",
    "process": "The hyperbola $\\frac{x^2}{3}-\\frac{16y^{2}}{p^{2}}=1$ $(p>0)$ has the left focus at $(-\\sqrt{3+\\frac{p^{2}}{16}},0)$. The left focus of the hyperbola $\\frac{x^{2}}{3}-\\frac{16y^{2}}{p^{2}}=1$ $(p>0)$ lies on the directrix of the parabola $y^{2}=2px$, so we obtain $\\sqrt{3+\\frac{p^{2}}{16}}=\\frac{p}{2}$. Solving gives $p=4$."
  },
  {
    "text": "Given that a line $l$ with inclination angle $60^{\\circ}$ passes through the focus $F$ of the parabola $y^{2}=4x$, and intersects the parabola at points $A$ and $B$, then the chord length $|AB|=$?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;F: Point;Inclination(l) = ApplyUnit(60, degree);Focus(G) = F;PointOnCurve(F, l);Expression(G) = (y^2 = 4*x);Intersection(l, G) = {A, B};IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[19, 24]], [[25, 39], [48, 51]], [[53, 56]], [[57, 60]], [[42, 45]], [[2, 24]], [[25, 45]], [[19, 45]], [[25, 39]], [[19, 62]], [[48, 73]]]",
    "query_spans": "[[[66, 75]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, and let the distances from $ A $ and $ B $ to the directrix be $ d_{A} $, $ d_{B} $. By the definition of a parabola, we have $ |AF| = d_{A} = x_{1} + 1 $, $ |BF| = d_{B} = x_{2} + 1 $. Thus, $ |AB| = |AF| + |BF| = x_{1} + x_{2} + 2 $. It is given that the focus of the parabola is $ F(1,0) $, and the slope $ k = \\tan 60^{\\circ} = \\sqrt{3} $. Therefore, the equation of line $ AB $ is $ y = \\sqrt{3}(x - 1) $. Substituting $ y = \\sqrt{3}(x - 1) $ into the equation $ y^{2} = 4x $, we obtain after simplification $ 3x^{2} - 10x + 3 = 0 $. Using the quadratic formula, we find $ x_{1} + x_{2} = \\frac{10}{3} $, so $ |AB| = |AF| + |BF| = x_{1} + x_{2} + 2 = \\frac{10}{3} + 2 = \\frac{16}{3} $."
  },
  {
    "text": "Given a point $P$ on the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{20}=1$ such that the distance from $P$ to the point $(6,0)$ is $9$, then the distance from point $P$ to the point $(-6,0)$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/20 = 1);P: Point;PointOnCurve(P, G) = True;H: Point;Coordinate(H) = (6, 0);Distance(P,H) = 9;I: Point;Coordinate(I) = (-6,0)",
    "query_expressions": "Distance(P,I)",
    "answer_expressions": "17",
    "fact_spans": "[[[2, 42]], [[2, 42]], [[44, 48], [66, 70]], [[2, 48]], [[49, 57]], [[49, 57]], [[44, 64]], [[71, 80]], [[71, 80]]]",
    "query_spans": "[[[66, 85]]]",
    "process": "It is easy to see that the point (6,0) is the right focus of the hyperbola, and (-6,0) is the left focus of the hyperbola. Given $ a=4 $, $ c=6 $, and the distance from point $ P $ to the point (6,0) is 9, where $ 9 < 6+4 $, thus $ P $ lies on the right branch. Therefore, the distance from point $ P $ to the point (-6,0) is $ 9+8=17 $."
  },
  {
    "text": "The equation of the hyperbola that has the same foci as the ellipse $x^{2}+4 y^{2}=16$ and one asymptote given by $x+\\sqrt{3} y=0$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2 + 4*y^2 = 16);Focus(G) = Focus(H);Expression(OneOf(Asymptote(G))) = (x + sqrt(3)*y = 0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/3 = 1",
    "fact_spans": "[[[52, 55]], [[1, 21]], [[1, 21]], [[0, 55]], [[29, 55]]]",
    "query_spans": "[[[52, 60]]]",
    "process": ""
  },
  {
    "text": "The line $AB$ passes through the focus $F$ of the parabola $y^2 = 4x$, and intersects the parabola at points $A$ and $B$. The abscissa of the midpoint of segment $AB$ is $3$. What is the slope of line $AB$?",
    "fact_expressions": "G: Parabola;B: Point;A: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, LineOf(A, B));Intersection(LineOf(A, B), G) = {A, B};XCoordinate(MidPoint(LineSegmentOf(A, B))) = 3",
    "query_expressions": "Slope(LineOf(A, B))",
    "answer_expressions": "{1, -1}",
    "fact_spans": "[[[8, 22], [31, 34]], [[40, 43]], [[36, 39]], [[25, 28]], [[8, 22]], [[8, 28]], [[0, 28]], [[0, 45]], [[47, 65]]]",
    "query_spans": "[[[67, 79]]]",
    "process": "According to the parabolic equation, we obtain F(1,0). Let the line equation be x = my + 1, and combine it with the parabolic equation to get \\underline{1}y^{2}-4my-4=0. Then, based on the fact that the horizontal coordinate of the midpoint of segment AB is 3, so x_{1}+x_{2}=6, solve for m, and thus obtain the slope of the line. [Detailed solution] Since line AB passes through the focus F(1,0) of the parabola y^{2}=4x and intersects the parabola at points A and B, the slope is non-zero. Let the equation of line AB be x = my + 1. Combining with the parabolic equation yields: y^{2}-4my-4=0. By Vieta's formulas: y_{1}+y_{2}=4m, y_{1}\\cdot y_{2}=-4. Therefore, x_{1}+x_{2}=4m(y_{1}+y_{2})+2=4m^{2}+2=2\\times3. Solving gives m=\\pm1. Thus, the equation of the line is x=\\pm y+1, so k_{AB}=\\pm1. The answer is 1 or -1."
  },
  {
    "text": "Given that $P$ is a moving point on the parabola $y^{2}=4x$, and $Q$ is a moving point on the circle $(x+2)^{2}+(y-4)^{2}=1$, what is the minimum value of the sum of the distance from point $P$ to point $Q$ and the distance from point $P$ to the directrix of the parabola?",
    "fact_expressions": "G: Parabola;H: Circle;P: Point;Expression(G) = (y^2 = 4*x);Expression(H) = ((x + 2)^2 + (y - 4)^2 = 1);PointOnCurve(P, G);PointOnCurve(Q, H);Q:Point",
    "query_expressions": "Min(Distance(P,Q)+Distance(P,Directrix(G)))",
    "answer_expressions": "4",
    "fact_spans": "[[[6, 20], [80, 83]], [[30, 54]], [[2, 5], [62, 66], [75, 79]], [[6, 20]], [[30, 54]], [[2, 25]], [[26, 59]], [[26, 29], [67, 71]]]",
    "query_spans": "[[[62, 96]]]",
    "process": "Connect PF. According to the definition of a parabola, the distance from point P to the directrix of the parabola is equal to the distance from point P to the focus F(1,0). Connect the center of the circle A(-2,4) with the focus F(1,0), intersecting the circle at point Q_{1} and the parabola at point P_{1}, as shown in the figure. At this moment, the sum of the distance from point P to point Q and the distance from point P to the directrix of the parabola is minimized. Here, AF=\\sqrt{9+16}=5, so Q_{1}F=5-1=4,"
  },
  {
    "text": "The focal distance of the hyperbola $\\frac{x^{2}}{\\lambda-2}+\\frac{y^{2}}{\\lambda}=1$ is?",
    "fact_expressions": "G: Hyperbola;lambda: Number;Expression(G) = (x^2/(lambda - 2) + y^2/lambda = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[0, 52]], [[3, 52]], [[0, 52]]]",
    "query_spans": "[[[0, 57]]]",
    "process": "Since $\\lambda > \\lambda - 2$, we have $a^{2} = \\lambda > 0$, $b^{2} = 2 - \\lambda > 0$, so $c^{2} = a^{2} + b^{2} = \\lambda + 2 - \\lambda = 2$, solving gives $c = \\sqrt{2}$, thus the focal distance of the hyperbola is $2c = 2\\sqrt{2}$. Hence the answer is: $2.5$"
  },
  {
    "text": "Given the parabola $y^{2}=4 x$, a line passing through the point $Q(4,0)$ intersects the parabola at two points $A(x_{1} , y_{1})$ and $B(x_{2} , y_{2})$. Then the minimum value of $y_{1}^{2}+y_{2}^{2}$ is?",
    "fact_expressions": "G: Parabola;H: Line;Q: Point;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Coordinate(Q) = (4, 0);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);x1:Number;y1:Number;x2:Number;y2:Number;PointOnCurve(Q, H);Intersection(H, G) = {A,B}",
    "query_expressions": "Min(y1^2 + y2^2)",
    "answer_expressions": "32",
    "fact_spans": "[[[2, 16], [31, 34]], [[28, 30]], [[18, 27]], [[37, 55]], [[58, 76]], [[2, 16]], [[18, 27]], [[37, 55]], [[58, 76]], [[37, 55]], [[37, 55]], [[58, 76]], [[58, 76]], [[17, 30]], [[28, 78]]]",
    "query_spans": "[[[80, 107]]]",
    "process": "Solve the system of equations of the line and the parabola, use Vieta's formulas to express $(y_{1}^{2}+y_{2}^{2})$ as a function, and then find the minimum value of the function.  \n【Detailed Solution】Let the equation of the line passing through the point $(4,0)$ be: $x=ay+4$.  \nFrom  \n$$\n\\begin{cases}\nx=ay+4 \\\\\ny^{2}=4x\n\\end{cases}\n$$  \nwe obtain: $y^{2}-4ay-16=0$, so $y_{1}y_{2}=-16$, $y_{1}+y_{2}=4a$.  \nThus,  \n$y_{1}^{2}+y_{2}^{2}=(y_{1}+y_{2})^{2}-2y_{1}y_{2}=16a^{2}+32 \\geqslant 32$,  \nand when $a=0$, $(y_{1}^{2}+y_{2}^{2})_{\\min}=32$."
  },
  {
    "text": "Let $P$ be a point on the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{6}=1$, and let $F_{1}$, $F_{2}$ be the left and right foci of this hyperbola. If $(\\overrightarrow{F_{1} P}+\\overrightarrow{F_{2} P}) \\cdot(\\overrightarrow{F_{1} P}-\\overrightarrow{F_{2} P})=72$, then $|\\overrightarrow{F_{1} P}|$=?",
    "fact_expressions": "G: Hyperbola;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/3 - y^2/6 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;DotProduct((VectorOf(F1, P) - VectorOf(F2, P)),(VectorOf(F1, P) + VectorOf(F2, P))) = 72",
    "query_expressions": "Abs(VectorOf(F1, P))",
    "answer_expressions": "7*sqrt(3)",
    "fact_spans": "[[[5, 43], [65, 68]], [[48, 55]], [[1, 4]], [[56, 63]], [[5, 43]], [[1, 47]], [[48, 74]], [[48, 74]], [[76, 189]]]",
    "query_spans": "[[[191, 221]]]",
    "process": ""
  },
  {
    "text": "Let the hyperbola $\\frac{x^{2}}{m}+\\frac{y^{2}}{n}=1$ have eccentricity $2$, and one focus coincides with the focus of the parabola $x^{2}=16 y$. Then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;m: Number;n: Number;H: Parabola;Expression(G) = (y^2/n + x^2/m = 1);Expression(H) = (x^2 = 16*y);Eccentricity(G) = 2;OneOf(Focus(G)) = Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/4 - x^2/12 = 1",
    "fact_spans": "[[[1, 39], [77, 80]], [[4, 39]], [[4, 39]], [[54, 69]], [[1, 39]], [[54, 69]], [[1, 47]], [[1, 74]]]",
    "query_spans": "[[[77, 85]]]",
    "process": "The focus of the parabola $x^{2}=16y$ is $(0,4)$ on the $y$-axis, so for the hyperbola $c=4$, and $\\frac{c}{a}=2\\Rightarrow a=2$, thus $b^{2}=c^{2}-a^{2}=12$. Hence, the equation of the hyperbola is $\\frac{y^{2}}{4}-\\frac{x^{2}}{12}=1$."
  },
  {
    "text": "Given a point $M(1, m)$ on the parabola $y^{2}=2 p x(p>0)$ such that the distance from $M$ to the focus is $5$, and the left vertex of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $A$ with eccentricity $\\frac{\\sqrt{5}}{2}$. If one asymptote of the hyperbola is perpendicular to the line $A M$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;p: Number;M: Point;A: Point;m:Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);p>0;Expression(H) = (y^2 = 2*p*x);Coordinate(M) = (1, m);PointOnCurve(M, H);Distance(M, Focus(H)) = 5;LeftVertex(G)=A;Eccentricity(G)=sqrt(5)/2;IsPerpendicular(OneOf(Asymptote(G)),LineOf(A,M))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2-4*y^2=1",
    "fact_spans": "[[[47, 103], [138, 141], [159, 162]], [[50, 103]], [[50, 103]], [[2, 23], [36, 37]], [[5, 23]], [[26, 35]], [[108, 111]], [[26, 35]], [[50, 103]], [[50, 103]], [[47, 103]], [[5, 23]], [[2, 23]], [[26, 35]], [[2, 35]], [[26, 46]], [[47, 111]], [[47, 136]], [[138, 157]]]",
    "query_spans": "[[[159, 167]]]",
    "process": "Let the focus of the parabola be $ F $. By the definition of a parabola, $ |MF| = 1 + \\frac{p}{2} = 5 $, solving gives $ p = 8 $, $ \\therefore $ the equation of the parabola is $ y^{2} = 16x $. Without loss of generality, take $ M $ in the first quadrant, then its coordinates are $ (1, 4) $. From the problem, $ A(-a, 0) $, the asymptotes of the hyperbola are $ y = \\pm\\frac{b}{a}x $. Since the eccentricity of the hyperbola is $ \\frac{\\sqrt{5}}{2} $, $ \\therefore \\frac{c}{a} = \\frac{\\sqrt{5}}{2} $, $ b = \\sqrt{c^{2} - a^{2}} = \\frac{1}{2}a $ ①. Since one asymptote of the hyperbola is perpendicular to the line $ AM $, $ \\therefore -\\frac{b}{a} \\cdot \\frac{4}{1+a} = -1 $ ②. From ① and ②, solving gives $ a = 1 $, $ b = \\frac{1}{2} $. $ \\therefore $ the equation of the hyperbola is $ x^{2} - \\frac{y^{2}}{1} = 1 $, i.e., $ x^{2} - 4y^{2} = 1 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, its two asymptotes intersect the directrix of the parabola $y^{2}=2 p x$ $(p>0)$ at points $A$ and $B$, respectively, and $O$ is the origin. If the eccentricity of the hyperbola is $2$, and the area of $\\triangle A O B$ is $\\sqrt{3}$, then $p=$?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Parabola;p: Number;A: Point;O: Origin;B: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);p>0;Expression(H) = (y^2 = 2*(p*x));l1:Line;l2:Line;Asymptote(G)={l1,l2};Intersection(l1, Directrix(H)) = A;Intersection(l2, Directrix(H)) = B;Eccentricity(G) = 2;Area(TriangleOf(A, O, B)) = sqrt(3)",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 59], [115, 118]], [[5, 59]], [[5, 59]], [[66, 87]], [[162, 165]], [[94, 97]], [[104, 107]], [[98, 101]], [[5, 59]], [[5, 59]], [[2, 59]], [[69, 87]], [[66, 87]], [], [], [[2, 65]], [[2, 103]], [[2, 103]], [[115, 126]], [[129, 160]]]",
    "query_spans": "[[[162, 167]]]",
    "process": ""
  },
  {
    "text": "Point $P$ lies on the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$, point $Q$ lies on the curve $x^{2}+(y-3)^{2}=4$, the midpoint of segment $PQ$ is $M$, and $O$ is the origin. Then the minimum length of segment $OM$ is?",
    "fact_expressions": "G: Hyperbola;H: Curve;P: Point;Q: Point;M: Point;O: Origin;Expression(G) = (x^2/2 - y^2 = 1);Expression(H) = (x^2 + (y - 3)^2 = 4);PointOnCurve(P, G);PointOnCurve(Q, H);MidPoint(LineSegmentOf(P, Q)) = M",
    "query_expressions": "Min(Length(LineSegmentOf(O, M)))",
    "answer_expressions": "sqrt(2) - 1",
    "fact_spans": "[[[5, 33]], [[40, 61]], [[0, 4]], [[35, 39]], [[74, 77]], [[78, 81]], [[5, 33]], [[40, 61]], [[0, 34]], [[35, 62]], [[63, 77]]]",
    "query_spans": "[[[88, 102]]]",
    "process": "Let $ P(x_{1},y_{1}) $, $ Q(x_{2},y_{2}) $, take the symmetric point of $ P $ about the origin $ P_{1}(-x_{1},-y_{1}) $, then $ |OM| = \\frac{1}{2}|P_{1}Q| $. According to $ |P_{1}Q| \\geqslant |P_{1}N| - 2 $, find the minimum value of $ |P_{1}N| $ using the distance formula to obtain the answer. Let $ P(x_{1},y_{1}) $, $ Q(x_{2},y_{2}) $, where $ -x_{1}, -y_{1} $, so $ \\frac{(-x_{1})^{2}}{2} - (-y_{1})^{2} = 1 $, i.e., $ x_{1}^{2} = 2y_{1}^{2} + 2 $, then $ |OM| = \\frac{1}{2}|P_{1}Q| $. Since $ |P_{1}Q| \\geqslant |P_{1}N| - 2 = \\sqrt{(-x_{1}-0)^{2} + (-y_{1}-3)^{2}} - 2 = \\sqrt{x_{1}^{2} + y_{1}^{2} + 6y_{1} + 9} - 2 = \\sqrt{x_{1}^{2} + y_{1}^{2} + 6y_{1} + 9} - 2 = \\sqrt{3y_{1}^{2} + 6y_{1} + 11} - 2 $, so $ |OM| \\geqslant \\sqrt{2} - 1 $, i.e., the minimum value of $ |OM| $ is $ \\sqrt{2} - 1 $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, with left and right foci $F_{1}(-c, 0)$, $F_{2}(c, 0)$ respectively. If there exists a point $P$ on the ellipse such that $\\frac{a}{\\sin \\angle P F_{1} F_{2}}=\\frac{c}{\\sin \\angle P F_{2} F_{1}}$, then the range of values for the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;P: Point;F2: Point;a > b;b > 0;c:Number;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(P,G);a/Sin(AngleOf(P,F1,F2))=c/Sin(AngleOf(P,F2,F1))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(\\sqrt{2}-1,1)",
    "fact_spans": "[[[2, 54], [95, 97], [181, 183]], [[4, 54]], [[4, 54]], [[63, 77]], [[100, 104]], [[79, 92]], [[4, 54]], [[4, 54]], [[63, 77]], [[2, 54]], [[63, 77]], [[79, 92]], [[2, 92]], [[2, 92]], [[95, 104]], [[105, 178]]]",
    "query_spans": "[[[181, 194]]]",
    "process": "From $\\frac{a}{\\sin\\anglePF_{1}F_{2}}=\\frac{c}{\\sin\\anglePF_{2}F_{1}}$ and the law of sines, we obtain $\\frac{c}{a}=\\frac{\\sin\\anglePF_{1}F_{2}}{\\sin\\anglePF_{2}F_{1}}=\\frac{|PF_{1}|}{|PF_{2}|}$, that is, $|PF_{1}|=\\frac{c}{a}|PF_{2}|$. Since $|PF_{1}|+|PF_{2}|=2a$, it follows that $|PF_{2}|=\\frac{2a^{2}}{a+c}$. Because $|PF_{2}|$ is a side of $\\triangle PF_{1}F_{2}$, then $a-c<\\frac{2a^{2}}{a+c}<a+c$, which implies $c^{2}+2ac-a^{2}>0 \\Rightarrow e^{2}+2e-1>0$ $(0<e<0)$, solving yields $\\sqrt{2}-1<e<1$."
  },
  {
    "text": "The minimum distance from a point on an ellipse to a directrix is exactly equal to the semi-focal distance of the ellipse. Then, the eccentricity of this ellipse is?",
    "fact_expressions": "G: Ellipse;P: Point;PointOnCurve(P, G) = True;Min(Distance(P, OneOf(Directrix(G)))) = HalfFocalLength(G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(\\sqrt{5}-1)/2",
    "fact_spans": "[[[0, 2], [21, 23], [29, 31]], [[4, 5]], [[0, 5]], [[0, 26]]]",
    "query_spans": "[[[29, 37]]]",
    "process": "Let the length of the major axis of the ellipse be 2a (a>0), and the focal distance be 2c (c>0). Then the eccentricity of the ellipse is e=\\frac{c}{a} with 0<e<1. The minimum distance from a point on the ellipse to the directrix is \\frac{a^2}{c}-a. According to the problem, we have \\frac{a^{2}}{c}-a=c, that is, a^{2}-ac=c^{2}. Dividing both sides by a^2 gives e^{2}=1-e, or e^{2}+e-1=0. Since 0<e<1, solving yields e=\\frac{\\sqrt{5}-1}{2}. Therefore, the eccentricity of the ellipse is \\frac{\\sqrt{5}-1}{2}."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus is $F_{1}$ and the left directrix is $l$. If the length of the chord passing through $F_{1}$ and perpendicular to the $x$-axis is equal to the distance from point $F_{1}$ to $l$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;l: Line;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;LeftDirectrix(G) = l;H:LineSegment;PointOnCurve(F1,H);IsPerpendicular(H,xAxis);Length(H)=Distance(F1,l);IsChordOf(H,G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 54], [113, 115]], [[4, 54]], [[4, 54]], [[58, 65], [76, 83], [96, 104]], [[70, 73], [105, 108]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 65]], [[2, 73]], [], [[2, 93]], [[2, 93]], [[2, 111]], [[2, 94]]]",
    "query_spans": "[[[113, 121]]]",
    "process": "The length of the chord passing through $F_{1}$ and perpendicular to the x-axis is equal to $\\underline{2b^{2}}$, the distance from point $F_{1}$ to $l$ is $\\frac{a^{2}}{c}-c$. Since the length of the chord passing through $F$ and perpendicular to the x-axis equals the distance from point $F$ to $1$, we have $\\frac{2b^{2}}{a}=\\frac{a^{2}}{c}-c'$, that is, $\\frac{2}{a}=\\frac{1}{c}$, $\\therefore\\frac{c}{a}=\\frac{1}{2}$,"
  },
  {
    "text": "Given that a line passing through the left focus $F$ of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with slope $\\frac{\\sqrt{3}}{3}$ intersects the ellipse $C$ at points $A$ and $B$, and if $\\overrightarrow{A F}=3 \\overrightarrow{F B}$, then the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;LeftFocus(C) = F;G: Line;PointOnCurve(F,G) = True;Slope(G) = sqrt(3)/3;A: Point;B: Point;Intersection(G, C) = {A, B};VectorOf(A, F) = 3*VectorOf(F, B)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[3, 60], [95, 100], [161, 166]], [[3, 60]], [[10, 60]], [[10, 60]], [[10, 60]], [[10, 60]], [[64, 67]], [[3, 67]], [[92, 94]], [[2, 94]], [[68, 94]], [[103, 106]], [[107, 110]], [[92, 112]], [[114, 159]]]",
    "query_spans": "[[[161, 172]]]",
    "process": "Since the line AB passes through the point F(-c,0) and has a slope of \\frac{\\sqrt{3}}{3}, the equation of line AB is: y=\\frac{\\sqrt{3}}{3}(x+c). Combining this with the ellipse C: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0), eliminating x gives (3b^{2}+a^{2})y^{2}-2\\sqrt{3}b^{2}cy-b^{4}=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}). Then y_{1}+y_{2}=\\frac{2\\sqrt{3}b^{2}c}{3b^{2}+a^{2}}, y_{1}y_{2}=\\frac{-b^{4}}{3b^{2}+a^{2}}. Since \\overrightarrow{AF}=3\\overrightarrow{FB}, we obtain y_{1}=-3y_{2}. Substituting into the above equations gives -2y_{2}=\\frac{2\\sqrt{3}b^{2}c}{3b^{2}+a^{2}}, -3y_{2}^{2}=\\frac{-b^{4}}{3b^{2}+a^{2}}. Eliminating y_{2} and simplifying yields: 9c^{2}=3b^{2}+a^{2}. Substituting b^{2}=c^{2}-a^{2} and simplifying gives: c^{2}=\\frac{1}{3}a^{2}. Solving gives c=\\frac{\\sqrt{3}}{3}a. Therefore, the eccentricity of the hyperbola is e=\\frac{c}{a}=\\frac{\\sqrt{3}}{3}."
  },
  {
    "text": "Let the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ have its right focus at $F$, and let a moving line $l$ passing through the origin $O$ intersect the ellipse $C$ at points $A$ and $B$. Then the range of the perimeter of $\\triangle A B F$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);F: Point;RightFocus(C) = F;O: Origin;PointOnCurve(O, l);l: Line;Intersection(l, C) = {A, B};A: Point;B: Point",
    "query_expressions": "Range(Perimeter(TriangleOf(A, B, F)))",
    "answer_expressions": "[4+2*sqrt(3),8)",
    "fact_spans": "[[[1, 43], [66, 71]], [[1, 43]], [[48, 51]], [[1, 51]], [[53, 58]], [[52, 65]], [[62, 65]], [[62, 82]], [[73, 76]], [[77, 80]]]",
    "query_spans": "[[[85, 112]]]",
    "process": "Let the left focus be $ F_{1} $, then it is easy to see that quadrilateral $ AFBF_{1} $ is a parallelogram, so the perimeter of $ \\triangle ABF $ is $ |AB| + |AF| + |BF| = |AB| + |AF| + |AF_{1}| = |AB| + 4 $. From the problem, the slope of the line is not zero, so we can set the line as $ x = my $. Solving the system of equations\n$$\n\\begin{cases}\nx = my \\\\\n\\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1\n\\end{cases}\n$$\nwe get $ y^{2} = \\frac{12}{3m^{2} + 4} $. Let $ A(x_{0}, y_{0}) $, then\n$$\n|AB| = 2\\sqrt{x_{0}^{2} + y_{0}^{2}} = 2\\sqrt{(m^{2} + 1)y_{0}^{2}} = 2\\sqrt{\\frac{12(m^{2} + 1)}{3m^{2} + 4}} = 4\\sqrt{1 - \\frac{1}{3m^{2} + 4}}.\n$$\nSince $ m^{2} \\geqslant 0 $, we have $ 3m^{2} + 4 \\geqslant 4 $, thus\n$$\n2\\sqrt{3} \\leqslant 4\\sqrt{1 - \\frac{1}{3m^{2} + 4}} < 4,\n$$\nso $ 4 + 2\\sqrt{3} \\leqslant |AB| + 4 < 8 $. Therefore, the range of the perimeter of $ \\triangle ABF $ is $ [4 + 2\\sqrt{3}, 8) $."
  },
  {
    "text": "A point $P(2, m)$ on the parabola $y^{2}=2 p x  (p>0)$ has a distance of $4$ to its focus $F$. Then $p=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p>0;p: Number;P: Point;Coordinate(P) = (2, m);m: Number;PointOnCurve(P, G) = True;F: Point;Focus(G) = F;Distance(P, F) = 4",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 23], [37, 38]], [[0, 23]], [[3, 23]], [[52, 55]], [[26, 36]], [[26, 36]], [[26, 36]], [[0, 36]], [[40, 43]], [[37, 43]], [[26, 50]]]",
    "query_spans": "[[[52, 57]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a , b>0)$ is equal to $2$, and the distance from its focus to the asymptote is equal to $1$, then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = 2;Distance(Focus(G),Asymptote(G))=1",
    "query_expressions": "Expression(G)",
    "answer_expressions": "3*x^2-y^2=1",
    "fact_spans": "[[[2, 57], [86, 89], [67, 68]], [[5, 57]], [[5, 57]], [[5, 57]], [[5, 57]], [[2, 57]], [[2, 66]], [[67, 83]]]",
    "query_spans": "[[[86, 94]]]",
    "process": "From the given conditions, we have $ e = \\frac{c}{a} = 2 $, and the distance from the focus to the asymptote is $ b = 1 $. Combining with $ c^{2} = a^{2} + b^{2} $, we can solve for $ a^{2} $, $ b^{2} $, and obtain the equation of the hyperbola. [Detailed solution] From the given conditions, $ e = \\frac{c}{a} = 2 $, so $ c = 2a $. Let one focus be $ F(c, 0) $, and the asymptotes be $ bx \\pm ay = 0 $. Then $ d = \\frac{bc}{\\sqrt{b^{2} + a^{2}}} = \\frac{bc}{c} = b = 1 $, and $ c^{2} = 4a^{2} = a^{2} + b^{2} $, solving gives $ a^{2} = \\frac{1}{3} $. Thus, the required equation of the hyperbola is $ 3x^{2} - y^{2} = $"
  },
  {
    "text": "Given the parabola $y^{2}=2 p x$ with focus $F$, let $A$ be a point on the parabola distinct from the vertex. Draw $AB$ perpendicular to the directrix from point $A$, with foot of perpendicular at $B$. If $|B F|=|A F|$ and the area of $\\triangle A B F$ is $12 \\sqrt{3}$, then the equation of this parabola is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 2*(p*x));Focus(G)=F;Negation(A=Vertex(G));PointOnCurve(A,LineSegmentOf(A,B));IsPerpendicular(LineSegmentOf(A,B),Directrix(G));FootPoint(LineSegmentOf(A,B),Directrix(G))=B;Abs(LineSegmentOf(B, F)) = Abs(LineSegmentOf(A, F));Area(TriangleOf(A, B, F)) = 12*sqrt(3);PointOnCurve(A,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=4*sqrt(3)*x",
    "fact_spans": "[[[2, 18], [30, 33], [119, 122]], [[5, 18]], [[26, 29], [43, 47]], [[62, 65]], [[22, 25]], [[2, 18]], [[2, 25]], [[26, 41]], [[42, 53]], [[30, 58]], [[30, 65]], [[67, 80]], [[82, 116]], [[26, 41]]]",
    "query_spans": "[[[119, 127]]]",
    "process": "First prove that triangle AABF is equilateral, then find |BF|, then find the value of p, thus obtaining the equation of this parabola. Therefore, $\\angle BFO = 60^{\\circ}$. Since the area of triangle AABF is $12\\sqrt{3} = \\frac{\\sqrt{3}}{4}|BF|^{2}$, therefore $|BF| = 4\\sqrt{3}$, then the distance from focus F to the directrix is $|BF|\\sin30^{\\circ} = 2\\sqrt{3}$. Therefore, $p = 2\\sqrt{3}$, and the equation of this parabola is $y^2 = 4\\sqrt{3}x$."
  },
  {
    "text": "Given that $A$ and $B$ are two points on the parabola $C$: $x^{2}=4 y$, and $M(-1,2)$, if $\\overrightarrow{A M}=\\overrightarrow{M B}$, then the equation of line $A B$ is?",
    "fact_expressions": "C: Parabola;B: Point;A: Point;M: Point;Expression(C) = (x^2 = 4*y);Coordinate(M) = (-1, 2);PointOnCurve(A, C);PointOnCurve(B, C);VectorOf(A, M) = VectorOf(M, B)",
    "query_expressions": "Expression(LineOf(A,B))",
    "answer_expressions": "x+2*y-3=0",
    "fact_spans": "[[[10, 29]], [[6, 9]], [[2, 5]], [[34, 43]], [[10, 29]], [[34, 43]], [[2, 33]], [[2, 33]], [[45, 88]]]",
    "query_spans": "[[[90, 102]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}) and M(-1,2). Since \\overrightarrow{AM}=\\overrightarrow{MB}, we have \\begin{cases}x_{1}+x_{2}=-2\\\\y_{1}+y_{2}=4\\end{cases}. Also, x_{1}^{2}=4y_{1}, x_{2}^{2}=4y_{2}, then x_{1}^{2}-x_{2}^{2}=4y_{1}-4y_{2}, yielding x_{1}+x_{2}=\\frac{4y_{1}-4y_{2}}{x_{1}-x_{2}}=-2. Thus, the slope of line AB is k=-\\frac{1}{2}, so the equation of line AB is y-2=-\\frac{1}{2}(x+1), which simplifies to x+2y-3=0. Solving the system \\begin{cases}x^{2}=4y\\\\x+2y-3=0\\end{cases} gives x^{2}+2x-6=0, \\Delta=28>0, indicating the line intersects the parabola at two points, which is valid."
  },
  {
    "text": "Given an ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ with a chord $AB$, the coordinates of point $P$ are $(0,1)$, and $2 \\overrightarrow{O A}+\\overrightarrow{O B}-3 \\overrightarrow{O P}=0$. Then, the distance from the midpoint of chord $AB$ to the line $y=-\\frac{1}{4}$ is?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;O:Origin;P: Point;H:Line;Expression(G) = (x^2/4 + y^2/3 = 1);IsChordOf(LineSegmentOf(A,B),G);Expression(H) = (y = -1/4);Coordinate(P) = (0, 1);2*VectorOf(O, A) + VectorOf(O, B) - 3*VectorOf(O, P) = 0",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A,B)), H)",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 39]], [[44, 49]], [[44, 49]], [[67, 137]], [[50, 54]], [[149, 167]], [[2, 39]], [[2, 49]], [[149, 167]], [[50, 65]], [[67, 137]]]",
    "query_spans": "[[[140, 172]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since 2\\overrightarrow{OA}+\\overrightarrow{OB}-3\\overrightarrow{OP}=0, it follows that 2x_{1}+x_{2}=0, 2y_{1}+y_{2}=3. Because A and B lie on the ellipse, \\frac{x_{1}^2}{4}+\\frac{y_{1}}{3}=1, \\frac{x_{2}}{4}+\\frac{y_{2}}{3}=1. Thus, \\frac{4x_{1}^2}{4}+\\frac{4y_{1}^2}{3}=4, \\frac{(-2x_{1})^{2}}{4}+\\frac{(3-2y)^{2}}{3}=1. Subtracting gives y_{1}=\\frac{3}{2}, \\therefore y_{2}=0. Therefore, the vertical coordinate of the midpoint of chord AB is \\underline{\\frac{y_{1}+y_{2}}{2}}=\\frac{3}{4}, and its distance to the line y=-\\frac{1}{4} is \\frac{3}{4}-(-\\frac{1}{4})=1."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has a real axis length of $16$, the left focus is $F$, and $M$ is a point on one of the asymptotes of hyperbola $C$ such that $OM \\perp MF$, where $O$ is the origin. If $S_{\\Delta OMF}=16$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Length(RealAxis(C)) = 16;F: Point;LeftFocus(C) = F;M: Point;PointOnCurve(M, OneOf(Asymptote(C)));O: Origin;IsPerpendicular(LineSegmentOf(O, M), LineSegmentOf(M, F));Area(TriangleOf(O, M, F)) = 16",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 63], [85, 91], [151, 157]], [[2, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[2, 72]], [[77, 80]], [[2, 80]], [[81, 84]], [[81, 100]], [[118, 121]], [[102, 117]], [[128, 149]]]",
    "query_spans": "[[[151, 163]]]",
    "process": "Analysis: Obtain the equation of an asymptote of hyperbola C as $ y = \\frac{b}{a}x $. Using the point-to-line distance formula, we get $ |MF| = b $. Combining the Pythagorean theorem and the area formula of a triangle, simplifying and solving the equation yields $ b = 4 $, thereby obtaining the eccentricity of the hyperbola. Given that the length of the real axis of hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) is 16, we have $ 2a = 16 $, so $ a = 8 $. Let $ F(-c, 0) $, and an asymptote of hyperbola C is $ y = \\frac{b}{a}x $. Then $ |MF| = \\frac{bc}{\\sqrt{a^{2} + b^{2}}} = b $, which implies $ |OM| = \\sqrt{c^{2} - b^{2}} = a $. From $ S_{\\triangle OMF} = 16 $, we obtain $ \\frac{1}{2}ab = 16 $, so $ b = 4 $. Also, $ c = \\sqrt{a^{2} + b^{2}} = \\sqrt{64 + 16} = 4\\sqrt{5} $. Solving gives $ a = 8 $, $ b = 4 $, $ c = 4\\sqrt{5} $, thus the eccentricity is: $ \\frac{c}{a} = \\frac{\\sqrt{5}}{2} $."
  },
  {
    "text": "The right focus of ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $F(c, 0)$. The line $x-2 \\sqrt{2} y=0$ intersects $C$ at points $A$ and $B$. If $\\overrightarrow{A F} \\cdot \\overrightarrow{B F}=0$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;c: Number;Coordinate(F) = (c, 0);RightFocus(C) = F;G: Line;Expression(G) = (x - 2*sqrt(2)*y = 0);A: Point;B: Point;Intersection(G, C) = {A, B};DotProduct(VectorOf(A, F), VectorOf(B, F)) = 0",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[0, 57], [93, 96], [163, 168]], [[0, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[62, 71]], [[62, 71]], [[62, 71]], [[0, 71]], [[72, 92]], [[72, 92]], [[99, 102]], [[103, 106]], [[72, 108]], [[110, 161]]]",
    "query_spans": "[[[163, 174]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through point $F_{1}$ intersects the hyperbola $E$ at points $A$ and $B$, satisfying $|A F_{2}|=|F_{1} F_{2}|$ and $\\overrightarrow{B F_{1}}=2 \\overrightarrow{F_{1} A}$. Then the eccentricity $e$ of the hyperbola $E$ is?",
    "fact_expressions": "E: Hyperbola;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F2: Point;F1: Point;LeftFocus(E) = F1;RightFocus(E) = F2;G: Line;PointOnCurve(F1, G);A: Point;B: Point;Intersection(G, E) = {A, B};Abs(LineSegmentOf(A, F2)) = Abs(LineSegmentOf(F1, F2));e: Number;VectorOf(B, F1) = 2*VectorOf(F1, A);Eccentricity(E) = e",
    "query_expressions": "e",
    "answer_expressions": "5/3",
    "fact_spans": "[[[2, 62], [100, 106], [202, 208]], [[2, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[79, 86]], [[71, 78], [88, 96]], [[2, 86]], [[2, 86]], [[97, 99]], [[87, 99]], [[108, 111]], [[112, 115]], [[97, 117]], [[120, 145]], [[212, 215]], [[147, 200]], [[202, 215]]]",
    "query_spans": "[[[212, 217]]]",
    "process": "As shown in the figure, by the definition of the hyperbola, we have |AF₂| - |AF₁| = 2a, |BF₂| - |BF₁| = 2a. Since |AF₂| = |F₁F₂| and \\overrightarrow{BF} = 2\\overrightarrow{F₁A}, we obtain |\\overrightarrow{BF}| = 3|\\overrightarrow{F₁A}|. Let |AF₁| = m, then |AF₂| = |F₁F₂| = 2a + m, |BF₁| = 2m, |BF₂| = 2a + 2m. In \\triangle AF₁F₂, we get \\cos\\angle AF₁F₂ = \\frac{|AF₁|^{2} + |F₁F₂|^{2} - |AF₂|^{2}}{2|AF₁||F₁F₂|} = \\frac{m^{2} + (2a + m)^{2} - (2a + m)^{2}}{2m(2a + m)} = \\frac{m}{2(2a + m)}. In \\triangle BF₁F₂, we get \\cos\\angle BF₁F₂ = \\frac{|BF₁|^{2} + |F₁F₂|^{2} - |BF₂|^{2}}{2|BF₁||F₁F₂|}. Moreover, since \\angle AF₁F₂ + \\angle BF₁F₂ = \\pi, it follows that \\frac{m}{2(2a + m)} = \\frac{4a - m}{4(2a + m)}, solving this gives m = \\frac{4}{3}a. Thus, |F₁F₂| = 2a + m = 2a + \\frac{4}{3}a = \\frac{10}{3}a, so 2c = \\frac{10}{3}a, i.e., e = \\frac{c}{a} = \\frac{5}{3}. Therefore, the eccentricity of hyperbola E is \\frac{5}{3}."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has eccentricity $2$, and the distance from one of its foci to an asymptote is $\\sqrt{3}$, then the standard equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;Distance(OneOf(Focus(C)), OneOf(Asymptote(C))) = sqrt(3);Eccentricity(C) = 2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[2, 53], [63, 64], [91, 97]], [[2, 53]], [[10, 53]], [[10, 53]], [[63, 89]], [[2, 61]]]",
    "query_spans": "[[[91, 104]]]",
    "process": "From the given, we have $\\frac{c}{a}=2$, one asymptote equation is $bx-ay=0$, and according to the distance from the focus to the asymptote $d=\\frac{|bc-0|}{\\sqrt{a^{2}+b^{2}}}=b$, then $b=\\sqrt{3}$, $a=1$, $c=2$. Hence, the standard equation of hyperbola $C$ is $x^{2}-\\frac{y^{2}}{3}=1$."
  },
  {
    "text": "If the directrix of the parabola $x^{2}=p y$ is tangent to the circle $x^{2}+y^{2}-4 x=0$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;H: Circle;Expression(G) = (x^2 = p*y);Expression(H) = (-4*x + x^2 + y^2 = 0);IsTangent(Directrix(G),H)",
    "query_expressions": "p",
    "answer_expressions": "pm*8",
    "fact_spans": "[[[1, 15]], [[45, 48]], [[21, 41]], [[1, 15]], [[21, 41]], [[1, 43]]]",
    "query_spans": "[[[45, 52]]]",
    "process": "The directrix equation of the parabola $x^{2}=py$ is $y=-\\frac{p}{4}$. The standard equation of the circle is $(x-2)^{2}+y^{2}=4$, with center coordinates $(2,0)$ and radius $2$. According to the problem, $-\\frac{p}{4}=2$, solving gives $p=\\pm8$."
  },
  {
    "text": "If a point $M$ on the parabola $y=x^{2}$ is at a distance of $1$ from the focus of the parabola, then what is the horizontal coordinate of point $M$?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y = x^2);PointOnCurve(M, G);Distance(M, Focus(G)) = 1",
    "query_expressions": "XCoordinate(M)",
    "answer_expressions": "pm*sqrt(3)/2",
    "fact_spans": "[[[1, 13], [20, 23]], [[16, 19], [34, 38]], [[1, 13]], [[1, 19]], [[16, 32]]]",
    "query_spans": "[[[34, 44]]]",
    "process": "The standard equation of the parabola is $x^{2}=y$, $2p=1$, that is, $p=\\frac{1}{2}$. Let $M(x_{0},y_{0})$, then $|MF|=y_{0}+\\frac{1}{4}=1$, $y_{0}=\\frac{3}{4}$, from $x_{0}^{2}=\\frac{3}{4}$ we get $x_{0}=\\pm\\frac{\\sqrt{3}}{2}$"
  },
  {
    "text": "If one of the asymptotes of the hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0, b>0)$ has an inclination angle of $\\frac{\\pi}{6}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-x^2/b^2 + y^2/a^2 = 1);Inclination(OneOf(Asymptote(G)))=pi/6",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 57], [85, 88]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 83]]]",
    "query_spans": "[[[85, 94]]]",
    "process": "The hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0,b>0)$ has an asymptote with an inclination angle of $\\frac{\\pi}{6}$, so $\\frac{a}{b}=\\tan\\frac{\\pi}{6}=\\frac{\\sqrt{3}}{3}$, thus $b=\\sqrt{3}a$, therefore the eccentricity of the hyperbola is $\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=2$."
  },
  {
    "text": "Given that $F(1,0)$ is the right focus of the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$, the line passing through the lower vertex $B$ of $E$ and $F$ intersects $E$ at another point $A$. If $4 \\overrightarrow{B F}=5 \\overrightarrow{F A}$, then $a$=?",
    "fact_expressions": "F: Point;Coordinate(F) = (1, 0);E: Ellipse;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;RightFocus(E) = F;G: Line;B: Point;LowerVertex(E) = B;PointOnCurve(B, G) ;PointOnCurve(F, G) ;A: Point;Intersection(G, E) = {B, A};4*VectorOf(B, F) = 5*VectorOf(F, A)",
    "query_expressions": "a",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 10], [87, 90]], [[2, 10]], [[11, 70], [76, 79], [94, 97]], [[11, 70]], [[157, 160]], [[17, 70]], [[17, 70]], [[17, 70]], [[2, 74]], [[91, 93]], [[83, 86]], [[76, 86]], [[75, 93]], [[75, 93]], [[103, 106]], [[83, 106]], [[108, 155]]]",
    "query_spans": "[[[157, 162]]]",
    "process": "From the ellipse equation, we get B(0,-b), F(1,0), so k_{BF} = \\frac{0-(-b)}{1-0} = b, thus the line BF: y = b(x-1). Solving the system \\begin{cases} y = b(x-1) \\\\ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{12} = 1 \\end{cases}, simplifying yields (1+a^{2})x^{2} - 2a^{2}x = 0, solving gives x = 0 or x = \\frac{2a^{2}}{1+a^{2}}, so x_{A} = \\frac{2a^{2}}{1+a^{2}}, thus y_{A} = \\frac{b(a^{2}-1)}{1+a^{2}}. Since 4\\overrightarrow{BF} = 5\\overrightarrow{FA}, i.e., 4(1,b) = 5(\\frac{2a^{2}}{1+a^{2}}-1, \\frac{b(a^{2}-1)}{1+a^{2}}),"
  },
  {
    "text": "Given fixed point $A(3,4)$, point $P$ is a moving point on the parabola $y^{2}=4x$, and the distance from point $P$ to the line $x=-1$ is $d$. Then the minimum value of $|PA|+d$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;P: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (x = -1);Coordinate(A) = (3, 4);PointOnCurve(P, G);Distance(P, H) = d;d:Number",
    "query_expressions": "Min(d + Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[18, 32]], [[42, 50]], [[4, 12]], [[13, 17], [37, 41]], [[18, 32]], [[42, 50]], [[4, 12]], [[13, 36]], [[37, 57]], [[54, 57]]]",
    "query_spans": "[[[59, 74]]]",
    "process": "Point A is a point outside the parabola $ y^{2} = 4x $, so $ |PA| + d = |PF| + |PA| \\geqslant |AF| = \\sqrt{(3-1)^{2} + 4^{2}} = 2\\sqrt{5} $. The equality holds if and only if point P is the intersection point of segment AF and the parabola. Thus, the minimum value of $ |PA| + d $ is $ 2\\sqrt{5} $."
  },
  {
    "text": "Given that planar vectors $\\vec{a}$, $\\vec{b}$ satisfy $|\\vec{a}|=1$, $4-\\vec{a} \\cdot \\vec{b}=2|\\vec{a}-\\vec{b}|$, then the range of values for $|\\vec{a}+\\vec{b}|$ is?",
    "fact_expressions": "a: Vector;b: Vector;Abs(a) = 1;4 - DotProduct(a, b) = 2*Abs(a-b)",
    "query_expressions": "Range(Abs(a + b))",
    "answer_expressions": "[1, 3]",
    "fact_spans": "[[[6, 15]], [[17, 27]], [[29, 42]], [[45, 89]]]",
    "query_spans": "[[[91, 117]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ with focus $F$, points $A$ and $B$ are two moving points on the parabola such that $\\angle A F B=90^{\\circ}$. From the midpoint $M$ of chord $A B$, a perpendicular $M N$ is drawn to the directrix $l$ of the parabola, with foot $N$. Then the maximum value of $\\frac{|\\overrightarrow{M N}|}{|\\overrightarrow{A B}|}$ is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;M: Point;N: Point;F: Point;l:Line;p>0;Expression(G) = (y^2 = 2*p*x);Directrix(G)=l;Focus(G) = F;PointOnCurve(A,G);PointOnCurve(B,G);AngleOf(A, F, B) = ApplyUnit(90, degree);IsChordOf(LineSegmentOf(A,B),G);MidPoint(LineSegmentOf(A,B))=M;PointOnCurve(M,LineSegmentOf(M,N));IsPerpendicular(LineSegmentOf(M,N),l);FootPoint(LineSegmentOf(M,N),l)=N",
    "query_expressions": "Max(Abs(VectorOf(M, N))/Abs(VectorOf(A, B)))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 23], [40, 43], [40, 43]], [[5, 23]], [[31, 35]], [[36, 39]], [[88, 91]], [[112, 115]], [[27, 30]], [[97, 100]], [[5, 23]], [[2, 23]], [[92, 100]], [[2, 30]], [[31, 49]], [[31, 49]], [[53, 78]], [[40, 85]], [[80, 91]], [[78, 108]], [[92, 108]], [[92, 115]]]",
    "query_spans": "[[[117, 178]]]",
    "process": "Let |AF| = a, |BF| = b. By the definition of the parabola, 2|\\overrightarrow{MN}| = a + b. By the cosine law:  \n\\begin{matrix}2\\\\\\end{matrix} = a^{2} + b^{2} = (a + b)^{2} - 2ab \\geqslant \\frac{1}{2}(a + b)^{2},  \nthe solution follows. Let |AF| = a, |BF| = b. By the definition of the parabola, |AF| = |AQ|, |BF| = |BP|. In trapezoid ABPQ,  \n\\therefore 2|\\overrightarrow{MN}| = |AQ| + |BP| = a + b.  \nBy the cosine law:  \n|\\overrightarrow{AB}|^{2} = a^{2} + b^{2} = (a + b)^{2} - 2ab \\geqslant \\frac{1}{2}(a + b)^{2}  \n\\therefore |\\overrightarrow{AB}| \\geqslant \\frac{\\sqrt{2}}{2}(a + b)  \n\\frac{|\\overrightarrow{MN}|}{|\\overrightarrow{AB}|} \\leqslant \\frac{\\sqrt{2}}{2}"
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ are $F_{1}$ and $F_{2}$, point $P$ lies on the ellipse, and the midpoint of segment $P F_{1}$ lies exactly on the $y$-axis, $|P F_{1}|=\\lambda|PF_{2}|$, then $\\lambda=$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;lambda:Number;Expression(G) = (x^2/4 + y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);PointOnCurve(MidPoint(LineSegmentOf(P,F1)), yAxis);Abs(LineSegmentOf(P, F1)) = lambda*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "lambda",
    "answer_expressions": "7",
    "fact_spans": "[[[0, 27], [52, 54]], [[47, 51]], [[31, 38]], [[39, 46]], [[109, 118]], [[0, 27]], [[0, 46]], [[47, 55]], [[57, 79]], [[80, 107]]]",
    "query_spans": "[[[109, 120]]]",
    "process": ""
  },
  {
    "text": "Given point $Q(-2,0)$ and the parabola $y^{2}=2 p x(p>0)$, a line passing through the focus of the parabola intersects the parabola at points $A$ and $B$, and intersects the $y$-axis at point $P$. If $\\overrightarrow{A B}=3 \\overrightarrow{B P}$, and the slope of line $Q A$ is $1$, then $p=$?",
    "fact_expressions": "Q: Point;Coordinate(Q) = (-2, 0);G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;H: Line;PointOnCurve(Focus(G), H);A: Point;B: Point;Intersection(H, G) = {A, B};P: Point;Intersection(H, yAxis) = P;VectorOf(A, B) = 3*VectorOf(B, P);Slope(LineOf(Q, A)) = 1",
    "query_expressions": "p",
    "answer_expressions": "2*(sqrt(2)+1)",
    "fact_spans": "[[[2, 12]], [[2, 12]], [[13, 34], [36, 39], [45, 48]], [[13, 34]], [[136, 139]], [[16, 34]], [[42, 44]], [[35, 44]], [[50, 53]], [[54, 57]], [[42, 59]], [[67, 71]], [[42, 71]], [[73, 118]], [[120, 134]]]",
    "query_spans": "[[[136, 141]]]",
    "process": "From the given conditions, A is in the first quadrant and B is in the fourth quadrant. Let A(x_{1},y_{1}), B(x_{2},y_{2}), P(0,y_{p}). From \\overrightarrow{AB}=3\\overrightarrow{BP}, we have (x_{2}-x_{1},y_{2}-y_{1})=3(0-x_{2},y_{p}-y_{2}), so x_{1}=4x_{2}, thus y_{1}=-2y_{2}. Since A, B, F are collinear, then \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{y_{2}}{x_{2}-\\frac{P}{2}}, that is, \\frac{-3y_{2}}{3x_{2}}=\\frac{y_{2}}{x_{2}-\\frac{p}{2}}, solving gives x_{2}=\\frac{p}{4}, so x_{1}=p. From y_{1}^{2}=2px_{1}, we get y_{1}=\\sqrt{2}p. Since the slope of line QA is 1, \\frac{y_{1}}{x_{1}+2}=1, i.e., \\frac{\\sqrt{2}p}{D+2}=1, solving gives p=2(\\sqrt{2}+1)"
  },
  {
    "text": "Let $F$ be the focus of the parabola $C$: $y^{2}=6 x$. A line passing through $F$ intersects the parabola $C$ at points $A$ and $B$, and $O$ is the origin. Then the minimum area of $\\triangle A O B$ is?",
    "fact_expressions": "C: Parabola;G: Line;A: Point;O: Origin;B: Point;F: Point;Expression(C) = (y^2 = 6*x);Focus(C) = F;PointOnCurve(F, G);Intersection(G, C) = {A, B}",
    "query_expressions": "Min(Area(TriangleOf(A, O, B)))",
    "answer_expressions": "9/2",
    "fact_spans": "[[[5, 24], [36, 42]], [[33, 35]], [[43, 46]], [[53, 56]], [[47, 50]], [[1, 4], [29, 32]], [[5, 24]], [[1, 27]], [[28, 35]], [[33, 52]]]",
    "query_spans": "[[[63, 88]]]",
    "process": "According to the problem, the focus of the parabola $ y^{2} = 6x $ is $ F\\left(\\frac{3}{2}, 0\\right) $. From the given conditions, the slope of line $ AB $ is not zero, so we can set the equation of line $ AB $ as $ x = my + \\frac{3}{2} $. Substituting into the system \n\\[\n\\begin{cases}\nx = my + \\frac{3}{2} \\\\\ny^{2} = 6x\n\\end{cases}\n\\]\nand eliminating $ x $, we obtain $ y^{2} - 6my - 9 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. By the relationship between roots and coefficients, we have $ y_{1} + y_{2} = 6m $, $ y_{1}y_{2} = -9 $. Therefore,\n\\[\n|AB| = \\sqrt{1 + m^{2}} \\cdot \\sqrt{(y_{1} + y_{2})^{2} - 4y_{1}y_{2}} = 6\\sqrt{1 + m^{2}}.\n\\]\nThe distance from $ O $ to line $ AB $ is $ d = \\frac{\\frac{3}{2}}{\\sqrt{1 + m^{2}}} $. Then the area of $ \\triangle OAB $ is\n\\[\nS = \\frac{1}{2}|AB| \\cdot d = \\frac{1}{2} \\times 6\\sqrt{1 + m^{2}} \\cdot \\frac{\\frac{3}{2}}{\\sqrt{1 + m^{2}}} = \\frac{6}{2} \\cdot \\frac{3}{2} = \\frac{9}{2}.\n\\]\nTherefore, when $ m = 0 $, $ S $ is minimized to $ \\frac{9}{2} $."
  },
  {
    "text": "The standard equation of a parabola with vertex at the origin and passing through the point $(-2 , 4)$ is?",
    "fact_expressions": "G: Parabola;H: Point;O: Origin;Coordinate(H) = (-2, 4);Vertex(G)=O;PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2=y,y^2=-8*x}",
    "fact_spans": "[[[20, 23]], [[8, 19]], [[3, 5]], [[8, 19]], [[0, 23]], [[7, 23]]]",
    "query_spans": "[[[20, 30]]]",
    "process": ""
  },
  {
    "text": "Given that points $A$ and $B$ are moving points on the rays $l_{1}: y = x$ $(x \\geq 0)$ and $l_{2}: y = -x$ $(x \\leq 0)$, respectively, and $O$ is the origin. The area of $\\triangle AOB$ is a constant $4$. Then the equation of the trajectory of the midpoint $M$ of segment $AB$ is?",
    "fact_expressions": "l1: Ray;l2: Ray;B: Point;A: Point;O: Origin;Expression(l1) = And((y = x), (x > 0));Expression(l2) = And((y = -x), (x <= 0));PointOnCurve(A, l1);PointOnCurve(B, l2);Area(TriangleOf(A, O, B)) = 4;MidPoint(LineSegmentOf(A, B)) = M;M: Point",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "(y^2/4 - x^2/4 = 1)&(y > 0)",
    "fact_spans": "[[[13, 38]], [[41, 66]], [[7, 10]], [[2, 6]], [[71, 74]], [[13, 38]], [[41, 66]], [[2, 70]], [[2, 70]], [[81, 107]], [[109, 121]], [[118, 121]]]",
    "query_spans": "[[[118, 128]]]",
    "process": ""
  },
  {
    "text": "$P$ is a point on $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$, $F_{1}$ and $F_{2}$ are the foci, $\\angle F_{1} P F_{2}=30^{\\circ}$, then the area of $\\Delta F_{1} P F_{2}$ is equal to?",
    "fact_expressions": "P: Point;Q: Curve;Expression(Q) = (x^2/5 + y^2/4 = 1);PointOnCurve(P, Q) = True;F1: Point;F2: Point;Focus(Q) = {F1, F2};AngleOf(F1, P, F2) = ApplyUnit(30, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "8-4*sqrt(3)",
    "fact_spans": "[[[0, 3]], [[4, 39]], [[4, 39]], [[0, 42]], [[43, 50]], [[51, 58]], [[4, 61]], [[62, 95]]]",
    "query_spans": "[[[97, 124]]]",
    "process": "From the given conditions, we have c=1, |PF|+|PF_{2}|=2\\sqrt{5}, |F_{1}F_{2}|=2. In \\Delta F_{1}PF_{2}, there is |PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}||PF_{2}|\\cdot\\cos30^{\\circ}=|F_{1}F_{2}|. \\therefore (|PF_{1}|+|PF_{2}|)^{2}-(2+\\sqrt{3})|PF_{1}|\\cdot|PF_{2}|=4. \\therefore |PF_{1}|\\cdot|PF_{2}|=16(2-\\sqrt{3}). \\therefore The area of \\triangle F_{1}PF_{2} equals \\frac{1}{2}|PF|\\cdot|PF_{2}|\\sin30^{\\circ}=4(2-\\sqrt{3})=8-4\\sqrt{3}."
  },
  {
    "text": "Given that the center of circle $C$ lies in the first quadrant and on the line $y=2x$, and that circle $C$ is tangent to both the directrix of the parabola $y^{2}=4x$ and the $x$-axis, then the equation of circle $C$ is?",
    "fact_expressions": "C: Circle;Quadrant(Center(C)) = 1;H: Line;Expression(H) = (y = 2*x);PointOnCurve(Center(C), H);G: Parabola;Expression(G) = (y^2 = 4*x);IsTangent(C, Directrix(G));IsTangent(C, xAxis)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "(x-1)^2 + (y-2)^2 = 4",
    "fact_spans": "[[[2, 6], [28, 32], [60, 64]], [[2, 14]], [[17, 26]], [[17, 26]], [[2, 27]], [[33, 47]], [[33, 47]], [[28, 58]], [[28, 58]]]",
    "query_spans": "[[[60, 69]]]",
    "process": "\\because the center of circle C lies in the first quadrant and on the line y=2x, we can let the center be C(a,2a), a>0. \\because circle C is tangent to both the directrix x=-1 of the parabola y^{2}=4x and the x-axis, the radius satisfies |a+1|=|2a|. Solving gives: a=1 or a=-\\frac{1}{3} (discarded). Thus, the center of the circle is (1,2) and the radius is 2. Therefore, the equation of circle C is: (x-1)^{2}+(y-2)^{2}=4"
  },
  {
    "text": "Given that the distance from point $A(x_0, 2)$ on the parabola $y = 4 a x^2$ $(a > 0)$ to the focus is equal to $3$, then $a = $?",
    "fact_expressions": "G: Parabola;a: Number;A: Point;a>0;x0: Number;Expression(G) = (y = 4*a*(x^2));Coordinate(A) = (x0, 2);PointOnCurve(A, G);Distance(A, Focus(G)) = 3",
    "query_expressions": "a",
    "answer_expressions": "1/16",
    "fact_spans": "[[[2, 23]], [[53, 56]], [[25, 40]], [[5, 23]], [[26, 40]], [[2, 23]], [[25, 40]], [[2, 40]], [[2, 51]]]",
    "query_spans": "[[[53, 58]]]",
    "process": ""
  },
  {
    "text": "An ellipse centered at the origin has eccentricity $\\frac{1}{2}$, with two foci $F_{1}$ and $F_{2}$ on the $y$-axis. $P$ is an arbitrary point on the ellipse. If the perimeter of $\\Delta P F_{1} F_{2}$ is $12$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;O:Origin;Center(G)=O;Eccentricity(G)=1/2;PointOnCurve(P, G);Focus(G)={F1,F2};PointOnCurve(F1,yAxis);PointOnCurve(F2,yAxis);Perimeter(TriangleOf(P, F1, F2)) = 12",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/16 + x^2/12 = 1",
    "fact_spans": "[[[8, 10], [61, 63], [11, 12], [103, 105]], [[56, 59]], [[34, 41]], [[42, 49]], [[3, 7]], [[0, 10]], [[11, 29]], [[56, 69]], [[11, 49]], [[34, 55]], [[42, 55]], [[71, 101]]]",
    "query_spans": "[[[103, 112]]]",
    "process": "According to the problem: $ e = \\frac{c}{a} = \\frac{1}{2} $, the perimeter of $ \\triangle PF_{1}F_{2} $ is $ 2a + 2c = 12 $, solving gives $ a = 4 $, $ c = 2 $. Thus $ b = \\sqrt{a^{2} - c^{2}} = 2\\sqrt{3} $, hence the equation of the ellipse is: $ \\frac{y^{2}}{16} + \\frac{x^{2}}{12} = 1 $"
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-y^{2}=1$ with left and right vertices $A_{1}$ and $A_{2}$ respectively, and a point $P$ on the hyperbola $C$. If the slope of the line $A_{1} P$ is $\\frac{1}{2}$, then what is the slope of the line $P A_{2}$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2 = 1);A1: Point;A2: Point;LeftVertex(C) = A1;RightVertex(C) = A2;P: Point;PointOnCurve(P, C);Slope(LineOf(A1, P)) = 1/2",
    "query_expressions": "Slope(LineOf(P, A2))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 25], [55, 61]], [[2, 25]], [[34, 41]], [[42, 49]], [[2, 49]], [[2, 49]], [[50, 54]], [[50, 62]], [[64, 92]]]",
    "query_spans": "[[[94, 110]]]",
    "process": "Let point P(x, y). Since point P lies on the hyperbola, we have x^{2}-y^{2}=1; it is easy to see that A_{1}(-1,0), B(1,0). Thus, we obtain k_{A_{1}P}\\cdotk_{PA_{2}}=\\frac{y}{x+1}\\times\\frac{y}{x-1}=\\frac{y^{2}}{x^{2}-1}. Since y^{2}=x^{2}-1, substituting into the above expression yields k_{A_{1}P}\\cdotk_{PA_{2}}=1. Given that k_{A_{1}P}=\\frac{1}{2}, it follows that k_{PA_{2}}=2."
  },
  {
    "text": "If point $P$ lies on the ellipse $x^{2}+2 y^{2}=2$, and $F_{1}$, $F_{2}$ are the two foci of the ellipse respectively, with $\\angle F_{1} P F_{2}=90^{\\circ}$, then the area of $\\triangle F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2 + 2*y^2 = 2);PointOnCurve(P, G);Focus(G)={F1,F2};AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[6, 25], [45, 47]], [[27, 34]], [[1, 5]], [[35, 42]], [[6, 25]], [[1, 26]], [[27, 51]], [[53, 86]]]",
    "query_spans": "[[[88, 118]]]",
    "process": ""
  },
  {
    "text": "Given that line $l$, which passes through a focus of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{8}=1$ and is perpendicular to the real axis, intersects the hyperbola at points $A$ and $B$, find the length of segment $AB$.",
    "fact_expressions": "l: Line;G: Hyperbola;A: Point;B: Point;Expression(G) = (x^2/16 - y^2/8 = 1);PointOnCurve(OneOf(Focus(G)),l);IsPerpendicular(l,RealAxis(G));Intersection(l, G) = {A, B}",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "4",
    "fact_spans": "[[[56, 61]], [[4, 43], [62, 65]], [[67, 70]], [[71, 74]], [[4, 43]], [[2, 61]], [[4, 61]], [[56, 76]]]",
    "query_spans": "[[[78, 89]]]",
    "process": "The hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{8}=1$ has $a=4$, $b=2\\sqrt{2}$, $c=\\sqrt{a^{2}+b^{2}}=2\\sqrt{6}$, yielding a focus at $(2\\sqrt{6},0)$. Substituting the line $l: x=2\\sqrt{6}$ into the hyperbola equation gives $\\frac{24}{16}-\\frac{y^{2}}{8}=1$, solving which yields $y=\\pm2$, so $AB=4$. Hence, the answer is $4+m+x=0$, $m(1)$ equations and properties, examining equation thinking and computational ability, classified as a basic problem."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has an eccentricity of $\\frac{\\sqrt{3}}{2}$. The line $l$ intersects the ellipse $C$ at points $A$ and $B$, and the midpoint of segment $AB$ is $M(-2,1)$. Then, what is the slope of line $l$?",
    "fact_expressions": "l: Line;C: Ellipse;b: Number;a: Number;A: Point;B: Point;M: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(M) = (-2, 1);Eccentricity(C) = sqrt(3)/2;Intersection(l, C) = {A, B};MidPoint(LineSegmentOf(A,B)) = M",
    "query_expressions": "Slope(l)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[85, 90], [131, 136]], [[2, 59], [91, 96]], [[9, 59]], [[9, 59]], [[98, 101]], [[102, 105]], [[120, 129]], [[9, 59]], [[9, 59]], [[2, 59]], [[120, 129]], [[2, 84]], [[85, 107]], [[109, 129]]]",
    "query_spans": "[[[131, 141]]]",
    "process": "From the eccentricity of the ellipse and the relationship among $a$, $b$, $c$, we obtain the relationship between $a$ and $b$. Then, using the point difference method, midpoint coordinate formula, and the slope formula for two points, the required value can be found. From the given condition, $e = \\frac{c}{a} = \\sqrt{1 - \\frac{b^{2}}{a^{2}}} = \\frac{\\sqrt{3}}{2}$, simplifying yields $a = 2b$. Let $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$, then $\\frac{x_{1}^{2}}{a^{2}} + \\frac{y_{1}^{2}}{b^{2}} = 1$, $\\frac{x_{2}^{2}}{a^{2}} + \\frac{y_{2}^{2}}{b^{2}} = 1$. Subtracting these two equations gives $\\frac{(x_{1} - x_{2})(x_{1} + x_{2})}{a^{2}} + \\frac{(y_{1} - y_{2})(y_{1} + y_{2})}{b^{2}} = 0$. Since the midpoint of $AB$ is $M(-2, 1)$, we have $x_{1} + x_{2} = -4$, $y_{1} + y_{2} = 2$, then the slope of the line $k = \\frac{y_{1} - y_{2}}{x_{1} - x_{2}} = -\\frac{b^{2}}{a^{2}} \\cdot \\frac{x_{1} + x_{2}}{y_{1} + y_{2}} = -\\frac{1}{4} \\times (-2) = \\frac{1}{2}$."
  },
  {
    "text": "Through the focus of the parabola $y^{2}=8x$, draw a line intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $|AB|=16$, then $x_{1}+x_{2}$=?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;x1: Number;x2: Number;y2: Number;Expression(G) = (y^2 = 8*x);Coordinate(A) = (x1, x2);Coordinate(B) = (x2, y2);PointOnCurve(Focus(G), H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 16",
    "query_expressions": "x1 + x2",
    "answer_expressions": "12",
    "fact_spans": "[[[1, 15], [22, 25]], [[19, 21]], [[26, 43]], [[46, 63]], [[26, 43]], [[26, 43]], [[46, 63]], [[1, 15]], [[26, 43]], [[46, 63]], [[0, 21]], [[19, 65]], [[67, 77]]]",
    "query_spans": "[[[79, 94]]]",
    "process": "According to the problem, p=4, the focal chord |AB|=x_{1}+x_{2}+p=16, then x_{1}+x_{2}=12."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, points $A$ and $F$ are the left vertex and left focus of ellipse $C$, respectively. The line $AP$ is tangent to the circle centered at the origin $O$ with radius $b$ at point $P$, and $PF \\perp x$-axis. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Circle;P: Point;A: Point;F: Point;O: Origin;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(C) = A;LeftFocus(C) = F;TangentPoint(LineOf(A, P), G) = P;Center(G) = O;Radius(G) = b;IsPerpendicular(LineSegmentOf(P, F), xAxis)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(\\sqrt{5} - 1)/2",
    "fact_spans": "[[[2, 61], [73, 78], [141, 144]], [[107, 110]], [[9, 61]], [[114, 115]], [[118, 122]], [[62, 66]], [[67, 70]], [[96, 103]], [[9, 61]], [[9, 61]], [[2, 61]], [[62, 86]], [[62, 86]], [[87, 122]], [[95, 115]], [[107, 115]], [[124, 138]]]",
    "query_spans": "[[[141, 150]]]",
    "process": "Let the semi-focal distance of the ellipse be $ c $. As shown in the figure, since $ AP $ is tangent to circle $ O $ at $ P $, we have $ PA \\perp OP $. Since $ |OF| = c $, $ |OA| = a $, and $ PF \\perp x $-axis, it follows that $ |PF| = \\sqrt{b^{2} - c^{2}} $, and $ |PF|^{2} = |OF| \\cdot |AF| = c(a - c) $. Therefore, $ b^{2} - c^{2} = c(a - c) $. Since $ a^{2} = b^{2} + c^{2} $, we have $ a^{2} - ac - c^{2} = 0 $. Because $ e = \\frac{c}{a} $, it follows that $ e^{2} + e - 1 = 0 $. Since $ e \\in (0,1) $, we obtain $ e = \\frac{\\sqrt{5} - 1}{2} $."
  },
  {
    "text": "The line $l$ passes through the focus $F$ of the parabola ${y}^{2}=8x$ and intersects the parabola at points $A$ and $B$. If the distance from the midpoint of segment $AB$ to the $y$-axis is $2$, then $|AB|=$?",
    "fact_expressions": "l: Line;Focus(G) = F;F: Point;PointOnCurve(F,l) = True;G: Parabola;Expression(G) = (y^2 = 8*x);Intersection(l,G) = {A,B};A: Point;B: Point;Distance(MidPoint(LineSegmentOf(A,B)),yAxis) = 2",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[0, 5]], [[6, 28]], [[25, 28]], [[0, 28]], [[6, 22], [31, 34]], [[6, 22]], [[0, 45]], [[36, 39]], [[40, 43]], [[47, 68]]]",
    "query_spans": "[[[70, 78]]]",
    "process": ""
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) intersects the parabola at points $A$ and $B$, and intersects its directrix at point $C$ (with $B$ between $F$ and $C$), and $|BC|=2|BF|$, $|AF|=12$. Find the value of $p$.",
    "fact_expressions": "G: Parabola;p: Number;H: Line;B: Point;C: Point;F: Point;A: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G)=F;PointOnCurve(F,H);Intersection(H,G) = {B, A};Intersection(H,Directrix(G))=C;Between(B,F,C)=True;Abs(LineSegmentOf(B, C)) = 2*Abs(LineSegmentOf(B, F));Abs(LineSegmentOf(A, F)) = 12",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 22], [32, 35], [48, 49]], [[99, 102]], [[29, 31]], [[43, 46], [57, 60]], [[52, 56], [65, 68]], [[25, 28], [61, 65]], [[36, 40]], [[4, 22]], [[1, 22]], [[1, 28]], [[0, 31]], [[29, 46]], [[29, 56]], [[57, 71]], [[74, 86]], [[88, 97]]]",
    "query_spans": "[[[99, 106]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, points $A$ and $B$ are the two endpoints of the major axis. If there exists a point $P$ on the ellipse such that $\\angle A P B=120^{\\circ}$, then the minimum value of the eccentricity of this ellipse is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;A: Point;P: Point;B: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Endpoint(MajorAxis(C))={A,B};PointOnCurve(P, C);AngleOf(A, P, B) = ApplyUnit(120, degree)",
    "query_expressions": "Min(Eccentricity(C))",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[2, 59], [78, 80], [119, 121]], [[8, 59]], [[8, 59]], [[60, 64]], [[83, 87]], [[65, 68]], [[8, 59]], [[8, 59]], [[2, 59]], [[2, 76]], [[78, 87]], [[90, 116]]]",
    "query_spans": "[[[119, 131]]]",
    "process": "If there exists a point P on the ellipse such that \\angle APB = 120^{\\circ}, then the maximum value of \\angle APB is greater than or equal to 120^{\\circ}, that is, when P is at the endpoint of the minor axis, \\angle APO \\geqslant 60^{\\circ} suffices. Points A and B are the two endpoints of the major axis. If there exists a point P on the ellipse such that \\angle APB = 120^{\\circ}, then the maximum value of \\angle APB being greater than or equal to 120^{\\circ} suffices, that is, when P is at the endpoint of the minor axis, \\angle APO > 60^{\\circ} suffices (as shown in the figure), \\tan\\angle APO = \\frac{a}{b} > \\tan 60^{\\circ} = \\sqrt{3}, \\therefore e = \\sqrt{1 - \\left(b^{2}\\right)} > \\frac{\\sqrt{6}}{3}"
  },
  {
    "text": "The focus of the parabola $y^{2}=8 x$ is $F$, point $A(5,4)$, and $P$ is a point on the parabola such that $P$ does not lie on the line $A F$. Then the minimum perimeter of $\\triangle P A F$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;A: Point;Coordinate(A) = (5, 4);P: Point;PointOnCurve(P, G);Negation(PointOnCurve(P, LineOf(A, F)))",
    "query_expressions": "Min(Perimeter(TriangleOf(P, A, F)))",
    "answer_expressions": "12",
    "fact_spans": "[[[0, 14], [38, 41]], [[0, 14]], [[18, 21]], [[0, 21]], [[22, 32]], [[22, 32]], [[34, 37], [46, 49]], [[34, 44]], [[46, 59]]]",
    "query_spans": "[[[61, 86]]]",
    "process": ""
  },
  {
    "text": "If the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1(a>0)$ coincides with the center of the circle $x^{2}+y^{2}-8 x=0$, then $a$=?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Circle;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(H) = (-8*x + x^2 + y^2 = 0);RightFocus(G)=Center(H)",
    "query_expressions": "a",
    "answer_expressions": "sqrt(15)",
    "fact_spans": "[[[1, 38]], [[70, 73]], [[43, 63]], [[4, 38]], [[1, 38]], [[43, 63]], [[1, 68]]]",
    "query_spans": "[[[70, 75]]]",
    "process": "Since the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ ($a>0$) coincides with the center of the circle $x^{2}+y^{2}-8x=0$, from the center $(4,0)$ of the circle $x^{2}+y^{2}-8x=0$, we obtain $c=4$ in the hyperbola. Therefore, $a=\\sqrt{4^{2}-1}=\\sqrt{15}$."
  },
  {
    "text": "The standard equation of the hyperbola that shares common foci with $\\frac{x^{2}}{16}-\\frac{y^{2}}{4}=1$ and passes through the point $(3 \\sqrt{2}, 2)$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/4 = 1);H: Point;Coordinate(H) = (3*sqrt(2), 2);Z: Hyperbola;Focus(G) = Focus(Z);PointOnCurve(H, Z)",
    "query_expressions": "Expression(Z)",
    "answer_expressions": "x^2/12 - y^2/8 = 1",
    "fact_spans": "[[[1, 40]], [[1, 40]], [[48, 66]], [[48, 66]], [[67, 70]], [[0, 70]], [[47, 70]]]",
    "query_spans": "[[[67, 77]]]",
    "process": "From the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{4}=1$, the coordinates of the foci are $(\\pm2\\sqrt{5},0)$. Let the equation of the desired hyperbola be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, $(a>b>0)$. According to the conditions, we have:  \n\\[\n\\begin{cases}\n\\frac{18}{a^{2}}-\\frac{4}{b^{2}}=1 \\\\\na^{2}+b^{2}=20\n\\end{cases}\n\\],\nsolving gives  \n\\[\n\\begin{cases}\na^{2}=12 \\\\\nb^{2}=8\n\\end{cases}\n\\]\nThus, the standard equation of the hyperbola is: $\\frac{x^{2}}{12}-\\frac{y^{2}}{8}=1$,"
  },
  {
    "text": "Let the line $y = x + 2$ intersect the ellipse $\\frac{x^{2}}{m} + \\frac{y^{2}}{3} = 1$ at two points. Then the range of values for $m$ is?",
    "fact_expressions": "G: Ellipse;m: Number;H: Line;Expression(G) = (y^2/3 + x^2/m = 1);Expression(H) = (y = x + 2);NumIntersection(H, G) = 2",
    "query_expressions": "Range(m)",
    "answer_expressions": "(1,3)+(3,+oo)",
    "fact_spans": "[[[11, 48]], [[56, 59]], [[1, 10]], [[11, 48]], [[1, 10]], [[1, 54]]]",
    "query_spans": "[[[56, 66]]]",
    "process": "\\because\\frac{x^{2}}{m}+\\frac{y^{2}}{3}=1 represents an ellipse \\therefore m>0 and m\\neq3. From \\begin{cases}y=x+2\\\\\\frac{x^{2}}{m}+\\frac{y^{2}}{3}=1\\end{cases}, we obtain: (m+3)x^{2}+4mx+m=0. \\therefore A=16m^{2}-4m(m+3)>0. \\therefore m>1 and m\\neq3. \\therefore the range of m is (1,3)\\cup(3,+\\infty)"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, let the left focus be $F$. If a line passing through point $F$ with an inclination angle of $\\frac{2\\pi}{3}$ intersects the left branch of the hyperbola at exactly one point, then the range of the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F;PointOnCurve(F, H);Inclination(H) = (2*pi)/3;NumIntersection(H, LeftPart(G)) = 1",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[2, +oo)",
    "fact_spans": "[[[2, 58], [99, 102], [116, 119]], [[5, 58]], [[5, 58]], [[96, 98]], [[63, 66], [69, 73]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 66]], [[68, 98]], [[74, 98]], [[96, 113]]]",
    "query_spans": "[[[116, 129]]]",
    "process": "Since the line passing through point F with an inclination angle of $\\frac{2\\pi}{3}$ intersects the left branch of the hyperbola at exactly one point, $-\\frac{b}{a}\\leqslant\\tan\\frac{2\\pi}{3}$, $\\therefore\\frac{b}{a}\\geqslant\\sqrt{3}$, $\\therefore b^{2}\\geqslant3a^{2}$, $c^{2}-a^{2}\\geqslant3a^{2}$, $e^{2}\\geqslant4$, $e\\geqslant2$."
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{2}=1$ are $F_{1}$, $F_{2}$, and point $P$ lies on the ellipse. If $|P F_{1}|=4$, then $|P F_{2}|$= (fill in with a number)?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/2 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 4",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 37], [62, 64]], [[0, 37]], [[41, 48]], [[49, 56]], [[0, 56]], [[57, 61]], [[57, 65]], [[68, 81]]]",
    "query_spans": "[[[83, 103]]]",
    "process": "The foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{2}=1$ are $F_{1}, F_{2}$, and point $P$ lies on the ellipse. By the definition of an ellipse, $|PF_{1}|+|PF_{2}|=2a=6$. Given $|PF_{1}|=4$, it follows that $|PF_{2}|=2$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(b>a>0)$, the line $l$ passes through points $A(a, 0)$ and $B(0, b)$, and the distance from the origin to line $l$ is $\\frac{\\sqrt{3}}{4}c$ ($c$ is the semi-focal length). Find the eccentricity of the hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;l: Line;A: Point;B: Point;O:Origin;c:Number;b > a;a > 0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (a, 0);Coordinate(B) = (0, b);HalfFocalLength(G)=c;Distance(O,l)=(sqrt(3)/4)*c;PointOnCurve(A,l);PointOnCurve(B,l)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 55], [133, 136]], [[5, 55]], [[5, 55]], [[56, 61], [90, 95]], [[62, 74]], [[75, 85]], [[87, 89]], [[123, 126]], [[5, 55]], [[5, 55]], [[2, 55]], [[62, 74]], [[75, 85]], [[2, 131]], [[87, 122]], [[56, 74]], [[56, 85]]]",
    "query_spans": "[[[133, 142]]]",
    "process": ""
  },
  {
    "text": "Given the line $l$: $2x - y + 1 = 0$ intersects the parabola $y^2 = 16x$ at points $A$ and $B$. Perpendiculars to $l$ are drawn through $A$ and $B$, intersecting the $x$-axis at points $C$ and $D$, respectively. Then $|CD|$ = ?",
    "fact_expressions": "l: Line;G: Parabola;C: Point;D: Point;A: Point;B: Point;Expression(G) = (y^2 = 16*x);Expression(l) = (2*x - y + 1 = 0);Intersection(l, G) = {A, B};L1:Line;L2:Line;PointOnCurve(A,L1);PointOnCurve(B,L2);IsPerpendicular(l,L1);IsPerpendicular(l,L2);Intersection(L1, xAxis) = C ;Intersection(L2, xAxis) = D",
    "query_expressions": "Abs(LineSegmentOf(C, D))",
    "answer_expressions": "10*sqrt(2)",
    "fact_spans": "[[[2, 20], [59, 62]], [[21, 36]], [[72, 75]], [[76, 79]], [[38, 41], [49, 52]], [[42, 45], [53, 56]], [[21, 36]], [[2, 20]], [[2, 47]], [], [], [[48, 65]], [[48, 65]], [[48, 65]], [[48, 65]], [[48, 81]], [[48, 81]]]",
    "query_spans": "[[[83, 92]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Solving the system of equations:  \n\\begin{cases}2x-y+1=0\\\\y^{2}=16x\\end{cases}  \nwe obtain: 4x^{2}-12x+1=0, so x_{1}+x_{2}=3, x_{1}x_{2}=\\frac{1}{4}.  \nSince \\tan\\theta=2, it follows that \\cos\\theta=\\frac{\\sqrt{5}}{5}.  \nTherefore, |CD|=\\frac{|AB|}{\\cos\\theta}=10\\sqrt{2}."
  },
  {
    "text": "If the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{m}=1$ and the line $x+2 y-2=0$ have two distinct intersection points, then what is the range of values for $m$?",
    "fact_expressions": "G: Ellipse;m: Number;H: Line;Expression(G) = (x^2/3 + y^2/m = 1);Expression(H) = (x + 2*y - 2 = 0);NumIntersection(G, H) = 2",
    "query_expressions": "Range(m)",
    "answer_expressions": "(1/4,+\\infty)",
    "fact_spans": "[[[1, 38]], [[62, 65]], [[39, 52]], [[1, 38]], [[39, 52]], [[1, 60]]]",
    "query_spans": "[[[62, 72]]]",
    "process": "By combining the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{m}=1$ and the line $x+2y-2=0$, we obtain $(3+4m)y^{2}-8my+m=0$. According to the condition that the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{m}=1$ and the line $x+2y-2=0$ have two distinct intersection points, we get  \n$$\n\\begin{cases}\nm\\neq3 \\\\\nm>0 \\\\\n\\end{cases}\n$$\nSolving gives $m>\\frac{1}{4}$."
  },
  {
    "text": "What is the distance from the focus to the directrix of the parabola $y=2 x^{2}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 2*x^2)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "1/4",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 25]]]",
    "process": ""
  },
  {
    "text": "A vertex $P(7,12)$ of $\\triangle P F_{1} F_{2}$ lies on the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1$, and the other two vertices $F_{1}$, $F_{2}$ are the left and right foci of this hyperbola, respectively. Then the coordinates of the incenter of $\\triangle P F_{1} F_{2}$ are?",
    "fact_expressions": "G: Hyperbola;b: Number;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2/b^2 = 1);Coordinate(P) = (7, 12);PointOnCurve(P,G);LeftFocus(G)=F1;RightFocus(G)=F2",
    "query_expressions": "Coordinate(Incenter(TriangleOf(P, F1, F2)))",
    "answer_expressions": "(1,3/2)",
    "fact_spans": "[[[40, 72], [98, 101]], [[43, 72]], [[30, 39]], [[79, 86]], [[89, 96]], [[40, 72]], [[30, 39]], [[30, 73]], [[79, 107]], [[79, 107]]]",
    "query_spans": "[[[109, 141]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a+8}+\\frac{y^{2}}{9}=1$ has eccentricity $e=\\frac{1}{2}$, then the value of $a$ is equal to?",
    "fact_expressions": "G: Ellipse;a: Number;e:Number;Expression(G) = (x^2/(a + 8) + y^2/9 = 1);Eccentricity(G) = e;e = 1/2",
    "query_expressions": "a",
    "answer_expressions": "{4,-5/4}",
    "fact_spans": "[[[2, 41]], [[62, 65]], [[45, 60]], [[2, 41]], [[2, 60]], [[45, 60]]]",
    "query_spans": "[[[62, 70]]]",
    "process": ""
  },
  {
    "text": "Let $O$ be the origin, $P$ a point on the parabola $x^{2}=4 y$, $F$ the focus, $|P F|=5$, then $|O P|$=?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;O: Origin;Expression(G) = (x^2 = 4*y);PointOnCurve(P, G);Focus(G) = F;Abs(LineSegmentOf(P, F)) = 5",
    "query_expressions": "Abs(LineSegmentOf(O, P))",
    "answer_expressions": "4*sqrt(2)",
    "fact_spans": "[[[12, 26]], [[8, 11]], [[30, 33]], [[1, 4]], [[12, 26]], [[8, 29]], [[12, 36]], [[37, 46]]]",
    "query_spans": "[[[48, 57]]]",
    "process": "Let P(m, n) according to the problem. Since |PF| = 5, the distance from point P to the directrix of the parabola is 5. Given that the equation of the directrix is y = -1, it follows that n = 4, thus m^{2} = 16. Hence, |OP| = \\sqrt{16 + 16} = 4\\sqrt{2}."
  },
  {
    "text": "Given that $A$ and $B$ are two moving points on the parabola $y^{2}=2 x$, $O$ is the origin and satisfies $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}=0$, the line $AB$ intersects the $x$-axis at point $M$. When $AM=2 BM$, what is the slope of line $AB$?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;O: Origin;M: Point;Expression(G) = (y^2=2*x);PointOnCurve(A,G);PointOnCurve(B,G);DotProduct(VectorOf(O,A),VectorOf(O,B))=0;Intersection(LineOf(A,B),xAxis)=M;LineSegmentOf(A,M)=2*LineSegmentOf(B,M)",
    "query_expressions": "Slope(LineOf(A,B))",
    "answer_expressions": "pm*sqrt(2)",
    "fact_spans": "[[[10, 24]], [[2, 5]], [[6, 9]], [[31, 34]], [[108, 112]], [[10, 24]], [[2, 30]], [[2, 30]], [[42, 93]], [[94, 112]], [[114, 125]]]",
    "query_spans": "[[[127, 138]]]",
    "process": "Let the equation of line AB be $x = my + t$, and by solving it simultaneously with the parabola equation, we obtain a quadratic equation in $y$. Let $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$. From $\\overrightarrow{OA} \\cdot \\overrightarrow{OB} = 0$, we get $x_{1}x_{2} + y_{1}y_{2} = 0$. Combining with Vieta's formulas, we can find $y_{1}y_{2}$. Given $AM = 2BM$, we have $y_{1} = -2y_{2}$, from which we can determine the values of $m$ and $t$. Since $k_{AB} = \\frac{1}{m}$, we can find the slope of line AB. Solution: According to the problem, let the equation of line AB be $x = my + t$. Solving simultaneously\n$$\n\\begin{cases}\ny^{2} = 2x \\\\\nx = my + t\n\\end{cases}\n$$\nwe obtain $y^{2} - 2my - 2t = 0$. Let $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$, then\n$$\n\\begin{cases}\ny_{1} + y_{2} = 2m \\\\\ny_{1}y_{2} = -2t\n\\end{cases}\n$$\nSince $\\overrightarrow{OA} \\cdot \\overrightarrow{OB} = 0$, we have $x_{1}x_{2} + y_{1}y_{2} = 0$, i.e., $\\frac{y_{1}^{2}y_{2}^{2}}{4} + y_{1}y_{2} = 0$, i.e., $y_{1}y_{2}\\left(\\frac{y_{1}y_{2}}{4} + 1\\right) = 0$. Since $y_{1}y_{2} \\neq 0$, we get $y_{1}y_{2} = -4$, so $t = 2$. From $AM = 2BM$, we have $y_{1} = -2y_{2}$. Thus $-2y_{2} + y_{2} = 2m$, solving gives $y_{2}^{2} = 2$, $m^{2} = \\frac{1}{2}$, so $m = \\pm\\frac{\\sqrt{2}}{2}$, hence $k_{AB} = \\frac{1}{m} = \\pm\\sqrt{2}$."
  },
  {
    "text": "The trajectory equation of the center of a moving circle passing through the point $F(1 , 0)$ and tangent to the line $l$: $x=-1$ is?",
    "fact_expressions": "l: Line;G: Circle;F: Point;Coordinate(F) = (1, 0);PointOnCurve(F, G);IsTangent(G, l);Expression(l) = (x = -1)",
    "query_expressions": "LocusEquation(Center(G))",
    "answer_expressions": "y^2 = 4*x",
    "fact_spans": "[[[14, 26]], [[29, 31]], [[1, 12]], [[1, 12]], [[0, 31]], [[13, 31]], [[14, 26]]]",
    "query_spans": "[[[29, 40]]]",
    "process": ""
  },
  {
    "text": "A circle centered at the vertex of parabola $C$ intersects $C$ at points $A$, $B$ and intersects the directrix of $C$ at points $D$, $E$. Given that $|A B|=4 \\sqrt{2}$, $|D E|=2 \\sqrt{5}$. Then the distance from the focus of $C$ to its directrix is?",
    "fact_expressions": "C: Parabola;Vertex(C) = Center(G);G: Circle;Intersection(G, C) = {A, B};A: Point;B: Point;Intersection(G, Directrix(C)) = {D, E};D: Point;E: Point;Abs(LineSegmentOf(A, B)) = 4*sqrt(2);Abs(LineSegmentOf(D, E)) = 2*sqrt(5)",
    "query_expressions": "Distance(Focus(C), Directrix(C))",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 7], [16, 19], [31, 34], [90, 93]], [[0, 15]], [[14, 15]], [[14, 29]], [[20, 23]], [[24, 27]], [[14, 47]], [[38, 41]], [[42, 45]], [[50, 68]], [[69, 87]]]",
    "query_spans": "[[[90, 104]]]",
    "process": "Let the equation of the parabola be $ y^{2} = 2px $. The graph is shown in the figure below. From $ |AB| = 4\\sqrt{2} $, $ |DE| = 2\\sqrt{5} $, and the properties of the circle, we obtain $ |AM| = 2\\sqrt{2} $, $ |DN| = \\sqrt{5} $. Let the coordinates of point $ A $ be $ (x, y) $, then $ y^{2} = 8 = 2px $, $ \\therefore x = \\frac{4}{p} $, i.e., $ |OM| = \\frac{4}{p} $. Also, $ |ON| = \\frac{P}{2} $, $ |OD| = |OA| $, $ |ON|^{2} + |DN|^{2} = |OM|^{2} + |AM|^{2} $, $ \\left(\\frac{P}{2}\\right)^{2} + (\\sqrt{5})^{2} = \\left(\\frac{4}{p}\\right)^{2} + (2\\sqrt{2})^{2} $. Solving gives $ p = 4 $, $ \\therefore $ the distance from the focus of parabola $ C $ to its directrix is $ |MN| = p = 4 $."
  },
  {
    "text": "Given that a hyperbola with symmetry axes as coordinate axes has an asymptote $2x - y = 0$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;SymmetryAxis(G)=axis;Expression(OneOf(Asymptote(G)))=(2*x-y=0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{sqrt(5),sqrt(5)/2}",
    "fact_spans": "[[[10, 13], [32, 35]], [[2, 13]], [[10, 29]]]",
    "query_spans": "[[[32, 41]]]",
    "process": ""
  },
  {
    "text": "Given point $A(4,4)$ and two points $B$, $C$ on the parabola $y^{2}=4x$ such that $AB \\perp BC$, then the range of the ordinate of point $C$ is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;C: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (4, 4);PointOnCurve(B,G);PointOnCurve(C,G);IsPerpendicular(LineSegmentOf(A, B), LineSegmentOf(B, C))",
    "query_expressions": "Range(YCoordinate(C))",
    "answer_expressions": "(-\\infty,-4] \\cup [12,+\\infty)",
    "fact_spans": "[[[12, 26]], [[2, 11]], [[29, 32]], [[33, 36], [56, 60]], [[12, 26]], [[2, 11]], [[12, 36]], [[12, 36]], [[39, 54]]]",
    "query_spans": "[[[56, 71]]]",
    "process": "Let $ B\\left(\\frac{y_{1}^{2}}{4}, y_{1}\\right), C\\left(\\frac{2}{4}, y_{2}\\right) $, then $ k_{AB} = \\frac{y_{1}-4}{\\frac{y_{1}^{2}}{1}-4} = \\frac{4}{y_{1}+4} $, similarly $ k_{BC} = \\frac{4}{y_{1}+y_{2}} $. From $ k_{AB} \\cdot k_{BC} = -1 $, we get $ (y_{1}+4)(y_{1}+y_{2}) = -16y_{2}+16=0 $ and $ y \\neq 4 $. According to the problem, this equation has real roots, so $ A = (y_{2}+4)^{2} - 4(4y_{2}+16) = y_{2}^{2} - 8y_{2} - 48 \\geqslant 0 $, solving gives $ y_{2} \\geqslant 12 $ or $ y_{2} \\leqslant -4 $. Checking, when $ y_{2} = -4 $, $ y_{1} = 0 $; when $ y_{2} = 12 $, $ y_{1} = -8 $, both satisfy the conditions. Therefore, the range of the vertical coordinate of point $ C $ is $ (-\\infty, -4] \\cup [12, +\\infty) $."
  },
  {
    "text": "There is a point $P$ on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ with foci $F_{1}$ and $F_{2}$, and $\\angle P F_{1} F_{2}=120^{\\circ}$. The area of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "F1: Point;F2: Point;Focus(G) = {F1, F2};G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);P: Point;PointOnCurve(P, G);AngleOf(P, F1, F2) = ApplyUnit(120, degree)",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "3*sqrt(3)/5",
    "fact_spans": "[[[5, 12]], [[13, 20]], [[2, 58]], [[21, 58]], [[21, 58]], [[62, 65]], [[21, 65]], [[67, 101]]]",
    "query_spans": "[[[103, 130]]]",
    "process": "From the ellipse equation $\\frac{x^2}{4} + \\frac{y^{2}}{3} = 1$, we have $a=2$, $b=\\sqrt{3}$, $c=1$, so the foci are $F_{1}(-1,0)$, $F_{2}(1,0)$. Let $|PF_{1}| = m$, $|PF_{2}| = n$. By the definition of an ellipse, $m+n=4$. Given $\\angle PF_{1}F_{2} = 120^{\\circ}$, using the law of cosines we get $n^{2} = m^{2} + (2c)^{2} - 2 \\cdot m \\cdot 2c \\cdot \\cos 120^{\\circ}$, which simplifies to $n^{2} = m^{2} + 4 + 2m$. Solving the system  \n\\[\n\\begin{cases}\nn^{2} = m^{2} + 4 + 2m \\\\\nm + n = 4\n\\end{cases}\n\\]\nyields $m = \\frac{6}{5}$, $n = \\frac{14}{5}$. Therefore,  \n\\[\nS_{\\triangle PF_{1}F_{2}} = \\frac{1}{2} |PF_{1}| |F_{1}F_{2}| \\sin \\angle PF_{1}F_{2} = \\frac{1}{2} \\times \\frac{6}{5} \\times 2 \\times \\frac{\\sqrt{3}}{2} = \\frac{3\\sqrt{3}}{5}\n\\]"
  },
  {
    "text": "Given two fixed points $A(-1,0)$ and $B(1,0)$, a moving point $P(x, y)$ moves along the line $l$: $y=x+2$. The ellipse $C$ has foci at $A$ and $B$ and passes through point $P$. Then, the maximum eccentricity of ellipse $C$ is?",
    "fact_expressions": "l: Line;C: Ellipse;A: Point;B: Point;P: Point;x1: Number;y1: Number;Coordinate(A) = (-1, 0);Coordinate(B) = (1, 0);Coordinate(P) = (x1, y1);Expression(l) = (y = x + 2);PointOnCurve(P, l);Focus(C) = {A, B};PointOnCurve(P, C)",
    "query_expressions": "Max(Eccentricity(C))",
    "answer_expressions": "sqrt(10)/5",
    "fact_spans": "[[[36, 50]], [[54, 59], [79, 84]], [[5, 14], [60, 63]], [[15, 23], [64, 67]], [[26, 35], [73, 77]], [[26, 35]], [[26, 35]], [[5, 14]], [[15, 23]], [[26, 35]], [[36, 50]], [[26, 51]], [[54, 70]], [[54, 77]]]",
    "query_spans": "[[[79, 94]]]",
    "process": "From the given conditions, we know that $ c = 1 $, so the eccentricity is $ e = \\frac{c}{a} = \\frac{1}{a} $. Since point $ P(x, y) $ moves on the line $ l: y = x + 2 $, we have $ 2a = |PA| + |PB| $. Draw the symmetric point $ C $ of point $ A $ with respect to the line $ l: y = x + 2 $. Then $ 2a = |PA| + |PB| \\leqslant |CD| + |DB| = |BC| $. At this time, $ a $ attains its minimum value. Using the midpoint coordinate formula, we obtain $ C(-2, 1) $. By the distance formula between two points, $ |BC| = \\sqrt{(-2 - 1)^2 + 1^2} = \\sqrt{10} $, so $ a = \\frac{\\sqrt{10}}{2} $, and thus $ e = \\frac{c}{a} = \\frac{1}{a} = \\frac{\\sqrt{10}}{5} $."
  },
  {
    "text": "The equation of the parabola with focus at point $(2 , 1)$ and the $y$-axis as directrix is?",
    "fact_expressions": "G: Parabola;H: Point;Coordinate(H) = (2, 1);Focus(G)=H;Directrix(G)=yAxis",
    "query_expressions": "Expression(G)",
    "answer_expressions": "(y-1)^2=4*(x-1)",
    "fact_spans": "[[[23, 26]], [[1, 11]], [[1, 11]], [[0, 26]], [[15, 26]]]",
    "query_spans": "[[[23, 30]]]",
    "process": ""
  },
  {
    "text": "If points $O$ and $F$ are the center and left focus of the ellipse $3 x^{2}+4 y^{2}=12$, respectively, and point $P$ is any point on the ellipse, then the maximum value of $\\overrightarrow{O P} \\cdot \\overrightarrow{F P}$ is?",
    "fact_expressions": "G: Ellipse;O: Point;P: Point;F: Point;Expression(G) = (3*x^2 + 4*y^2 = 12);PointOnCurve(P, G);Center(G)=O;LeftFocus(G)=F",
    "query_expressions": "Max(DotProduct(VectorOf(O, P), VectorOf(F, P)))",
    "answer_expressions": "6",
    "fact_spans": "[[[13, 35], [48, 50]], [[1, 5]], [[43, 47]], [[6, 10]], [[13, 35]], [[43, 55]], [[1, 42]], [[1, 42]]]",
    "query_spans": "[[[57, 111]]]",
    "process": "Transforming the ellipse equation $3x^{2}+4y^{2}=12$ yields $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, so $F(-1,0)$. Let $P(x_{0},y_{0})$, $(-2\\leqslant x_{0}\\leqslant 2)$, then $3x_{0}^{2}+4y_{0}^{2}=12$, $\\therefore y_{0}^{2}=3-\\frac{3}{4}x_{0}^{2}$. $\\overrightarrow{OP}=(x_{0},y_{0})$, $\\overrightarrow{FP}=(x_{0}+1,y_{0})$, $\\therefore \\overrightarrow{OP}\\cdot\\overrightarrow{FP}=x_{0}(x_{0}+1)+y_{0}^{2}=x_{0}^{2}+x_{0}+(3-\\frac{3}{4}x_{0}^{2})=\\frac{1}{4}x_{0}^{2}+x_{0}+3=\\frac{1}{4}(x_{0}+2)^{2}+2$. Since $-2\\leqslant x_{0}\\leqslant 2$, $2\\leqslant \\frac{1}{4}(x_{0}+2)^{2}+2\\leqslant 6$, that is, $\\overrightarrow{OP}\\cdot\\overrightarrow{FP}\\in[2,6]$. Therefore, the maximum value of $\\overrightarrow{OP}\\cdot\\overrightarrow{FP}$ is $6$."
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ are $F_{1}$ and $F_{2}$ respectively, the chord $AB$ passes through $F_{1}$, if the circumference of the incircle of $\\triangle ABF_{2}$ is $2\\pi$, and the coordinates of points $A$, $B$ are $(x_{1}, y_{1})$, $(x_{2}, y_{2})$, then $|y_{2}-y_{1}|=$?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;F2: Point;F1: Point;x1:Number;x2:Number;y1:Number;y2:Number;Expression(G) = (x^2/25 + y^2/16 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;IsChordOf(LineSegmentOf(A,B),G);Perimeter(InscribedCircle(TriangleOf(A,B,F2)))=2*pi;Coordinate(A)=(x1,y1);Coordinate(B)=(x2,y2);PointOnCurve(F1, LineSegmentOf(A, B))",
    "query_expressions": "Abs(y2-y1)",
    "answer_expressions": "10/3",
    "fact_spans": "[[[0, 39]], [[118, 122]], [[124, 127]], [[56, 63]], [[48, 55], [71, 78]], [[135, 151]], [[153, 169]], [[135, 151]], [[153, 169]], [[0, 39]], [[0, 63]], [[0, 63]], [[0, 70]], [[80, 116]], [[119, 169]], [[119, 169]], [[65, 78]]]",
    "query_spans": "[[[170, 187]]]",
    "process": "In the ellipse $\\frac{x^2}{25}+\\frac{y^2}{16}=1$, $a=5$, $b=4$, $c=3$. $\\because$ the circumference of the incircle of $\\triangle ABF_{2}$ is $2\\pi$, $\\therefore$ the inradius of $\\triangle ABF_{2}$ is $r=1$. By the definition of the ellipse, the perimeter of $\\triangle ABF_{2}$ is $|AB|+|BF_{1}|+|AF_{1}|=(|AF_{1}|+|AF_{2}|)+(|BF_{1}|+|BF_{2}|)=4a=20$. Also, $S_{\\Delta ABF_2}=\\frac{1}{2}r(|AB|+|BF_{1}|+|AF_{1}|)=\\frac{1}{2}\\times1\\times20=10$ and $S_{\\Delta ABF_2}=\\frac{1}{2}|F_{1}F_{2}||y_{1}-y_{2}|=3|y_{1}-y_{2}|$. $\\therefore 3|y_{1}-y_{2}|=10$, solving gives $|y_{1}-y_{2}|=\\frac{10}{3}$. Answer: $\\frac{10}{3}$"
  },
  {
    "text": "A line passing through the focus of the parabola $x^{2}=8 y$ with slope $2$ intersects the parabola at points $A$ and $B$. The length of the chord $|A B|$ is equal to?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 8*y);H: Line;PointOnCurve(Focus(G), H);Slope(H) = 2;A: Point;B: Point;Intersection(H, G) = {A, B};IsChordOf(LineSegmentOf(A,B),G)",
    "query_expressions": "Abs(LineSegmentOf(A,B))",
    "answer_expressions": "40",
    "fact_spans": "[[[1, 15], [29, 32]], [[1, 15]], [[26, 28]], [[0, 28]], [[19, 28]], [[34, 37]], [[38, 41]], [[26, 43]], [[29, 53]]]",
    "query_spans": "[[[46, 56]]]",
    "process": "The equation of the line is y=2x+2. By solving the line and the parabola simultaneously and using the definition of the parabola, the solution can be obtained. \\because the line passes through the focus (0,2) of the parabola x^{2}=8y and has a slope of 2, \\therefore the equation of the line is y=2x+2. Let A(x_{1},y), B(x_{2},y_{2}), and let F be the focus of the parabola. \\therefore according to the definition of the parabola, |AB|=|AF|+|BF|=y_{1}+2+y_{2}+2. Solving the system of equations \\begin{cases}x^{2}=8y\\\\y=2x+2\\end{cases}, simplifying yields y^{2}-36y+4=0=36. \\therefore|AB|=v,+2+v+2=40"
  },
  {
    "text": "Let $P$ be a moving point on the parabola $C$: $y^{2}=4x$, $F$ be the focus of $C$, and $A(3, m)$ be a point in the plane. If the minimum value of $|PF| + |PA|$ is $4$, then the range of real values of $m$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);P: Point;PointOnCurve(P, C);F: Point;Focus(C) = F;A: Point;m: Real;Coordinate(A) = (3, m);Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F))) = 4",
    "query_expressions": "Range(m)",
    "answer_expressions": "[-2*sqrt(3), 2*sqrt(3)]",
    "fact_spans": "[[[4, 23], [32, 35]], [[4, 23]], [[0, 3]], [[0, 27]], [[28, 31]], [[28, 38]], [[44, 53]], [[78, 83]], [[44, 53]], [[55, 76]]]",
    "query_spans": "[[[78, 90]]]",
    "process": "The directrix of the parabola $ C: y^{2} = 4x $ is $ l: x = -1 $. Let $ PB \\perp l $, with foot of perpendicular at $ B $. Let the coordinates of point $ P $ be $ \\left( \\frac{y^{2}}{4}, y \\right) $. According to the definition of a parabola, $ |PF| + |PA| = |PB| + |PA| $. When $ P $ lies on segment $ AB $, $ |PF| + |PA| $ attains its minimum value, which is $ 4 $, consistent with the given condition. At this time, $ 0 \\leqslant \\frac{y^{2}}{4} \\leqslant 3 $, $ y = m \\Rightarrow m \\in [-2\\sqrt{3}, 2\\sqrt{3}] $."
  },
  {
    "text": "Given that point $F$ is the right focus of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, $M$ is a point on this ellipse, and $A(2,2)$ is a fixed point, then the minimum value of $|M A|+\\frac{5}{4}|M F|$ is?",
    "fact_expressions": "G: Ellipse;A: Point;M: Point;F: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Coordinate(A) = (2, 2);RightFocus(G) = F;PointOnCurve(M, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(M, A)) + (5/4)*Abs(LineSegmentOf(M, F)))",
    "answer_expressions": "17/4",
    "fact_spans": "[[[7, 45], [56, 58]], [[62, 70]], [[50, 53]], [[2, 6]], [[7, 45]], [[62, 70]], [[2, 49]], [[50, 61]]]",
    "query_spans": "[[[77, 107]]]",
    "process": ""
  },
  {
    "text": "Given that $A$ and $B$ are distinct points on the parabola $x^{2}=4 y$, the midpoint of segment $AB$ is $P$. If $|AB|=6$, then the minimum distance from point $P$ to the $x$-axis is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;P: Point;Expression(G) = (x^2 = 4*y);PointOnCurve(A, G);PointOnCurve(B, G);Negation(A=B);MidPoint(LineSegmentOf(A,B)) = P;Abs(LineSegmentOf(A, B)) = 6",
    "query_expressions": "Min(Distance(P, xAxis))",
    "answer_expressions": "2",
    "fact_spans": "[[[10, 24]], [[2, 5]], [[6, 9]], [[42, 45], [58, 62]], [[10, 24]], [[2, 31]], [[2, 31]], [[2, 31]], [[32, 45]], [[47, 56]]]",
    "query_spans": "[[[58, 75]]]",
    "process": "F is the focus of the parabola. Using |AF| + |BF| \\geqslant |AB|, we obtain the minimum value of |AF| + |BF|. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), then we get the minimum value of y_{1} + y_{2}, and \\frac{y_{1} + y_{2}}{2} is the required value. [Detailed Solution] For the parabola x^{2} = 4y, p = 2. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), then the midpoint of AB is P\\left(\\frac{x_{1} + x_{2}}{2}, \\frac{y_{1} + y_{2}}{2}\\right). The distance from P to the x-axis is d = \\frac{y_{1} + y_{2}}{2}. Let the focus of the parabola be F(0, 1), then |AF| = y_{1} + 1, |BF| = y_{2} + 1, \\therefore |AF| + |BF| = y_{1} + y_{2} + 2 \\geqslant |AB| = 6, y_{1} + y_{2} \\geqslant 4, with equality holding if and only if the chord AB passes through the focus F. \\therefore the minimum value of y_{1} + y_{2} is 4, \\therefore the minimum distance from P to the x-axis is 2."
  },
  {
    "text": "Given a point $M$ on the parabola $y^{2}=2 p x(p>0)$, the distance from $M$ to the $x$-axis is $4$, and the distance from $M$ to the focus is $5$. Then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;M: Point;p>0;Expression(G) = (y^2 = 2*(p*x));PointOnCurve(M, G);Distance(M, xAxis) = 4;Distance(M,Focus(G))=5",
    "query_expressions": "p",
    "answer_expressions": "2,8",
    "fact_spans": "[[[2, 23]], [[56, 59]], [[26, 29]], [[5, 23]], [[2, 23]], [[2, 29]], [[26, 41]], [[2, 53]]]",
    "query_spans": "[[[56, 61]]]",
    "process": "Let M(x_{0},y_{0}), then |y_{0}|=4, \\therefore 16=2px_{0}.\\textcircled{1} Also, the distance from point M to the focus is 5, \\therefore x_{0}+\\frac{p}{2}=5.\\textcircled{2} Eliminating x_{0} from \\textcircled{1}\\textcircled{2} and simplifying yields p^{2}-10p+16=0, solving gives p=2 or p=8."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ has eccentricity $\\sqrt{3}$, then the distance from the point $(3,0)$ to the asymptotes of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(C) = sqrt(3);G: Point;Coordinate(G) = (3, 0)",
    "query_expressions": "Distance(G, Asymptote(C))",
    "answer_expressions": "sqrt(6)",
    "fact_spans": "[[[2, 64], [90, 96]], [[2, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[2, 79]], [[81, 89]], [[81, 89]]]",
    "query_spans": "[[[81, 104]]]",
    "process": "Since the eccentricity of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) is $ \\sqrt{3} $, that is, $ e = \\frac{c}{a} = \\sqrt{3} $, we have $ \\frac{c^{2}}{a^{2}} = 3 $, i.e., $ \\frac{a^{2} + b^{2}}{a^{2}} = 3 $, so $ \\frac{b^{2}}{a^{2}} = 2 $. Thus, the asymptotes of the hyperbola are $ y = \\pm \\sqrt{2}x $. Then the distance from the point $ (3, 0) $ to the asymptote is $ d = \\frac{|3\\sqrt{2}|}{\\sqrt{(\\sqrt{2})^{2} + (-1)^{2}}} = \\sqrt{6} $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right vertices are denoted as $A_{1}$ and $A_{2}$ respectively. From $A_{2}$, a perpendicular line $A_{2} P$ is drawn to an asymptote, with foot of perpendicular at $P$, such that $\\frac{|A_{1} P|}{|A_{2} P|} \\geq 2$. Then, the range of eccentricity of this hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;A2: Point;P: Point;A1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftVertex(C) = A1;RightVertex(C) = A2;PointOnCurve(A2, LineSegmentOf(A2, P));IsPerpendicular(Asymptote(C), LineSegmentOf(A2, P));FootPoint(Asymptote(C), LineSegmentOf(A2, P)) = P;Abs(LineSegmentOf(A1, P))/Abs(LineSegmentOf(A2, P)) >= 2",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(1, \\sqrt{21}/3]",
    "fact_spans": "[[[2, 63], [159, 162]], [[10, 63]], [[10, 63]], [[79, 86], [88, 95]], [[115, 118]], [[71, 78]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 86]], [[2, 86]], [[87, 111]], [[2, 111]], [[2, 118]], [[120, 156]]]",
    "query_spans": "[[[159, 172]]]",
    "process": "Take the asymptote as $ y = \\frac{b}{a}x $. Let the line $ A_{2}P $ be $ y = -\\frac{a}{b}(x - a) $. Solving together with the asymptote equation gives the coordinates of point $ P $ as $ \\left( \\frac{a^{3}}{c^{2}}, \\frac{a^{2}b}{c^{2}} \\right) $. Thus, we obtain $ A_{1}P^{2} = \\left( \\frac{a^{3}}{c^{2}} + a \\right)^{2} + \\left( \\frac{a^{2}b}{c^{2}} \\right)^{2} $, $ A_{2}P^{2} = \\left( \\frac{a^{3}}{c^{2}} - a \\right)^{2} + \\left( \\frac{a^{2}b}{c^{2}} \\right)^{2} $. Since $ \\frac{|A_{1}P|}{|A_{2}P|} \\geqslant 2 $, it follows that $ A_{1}P^{2} \\geqslant 4A_{2}P^{2} $. Rearranging yields $ 3a^{4} + 3c^{4} - 10a^{2}c^{2} + 3a^{2}b^{2} \\leqslant 0 $, i.e., $ 3c^{2} - 7a^{2} \\leqslant 0 $, then $ \\frac{c^{2}}{a^{2}} \\leqslant \\frac{7}{3} $, $ \\frac{c}{a} \\leqslant \\frac{\\sqrt{21}}{3} $. Hence, the range of the hyperbola's eccentricity is $ \\left( 1, \\frac{\\sqrt{21}}{3} \\right] $."
  },
  {
    "text": "If the coordinates of point $A$ are $(-3 , 2)$, $F$ is the focus of the parabola $y^{2}=-4 x$, and point $P$ is a moving point on the parabola, then when $|PA|+|PF|$ takes the minimum value, what are the coordinates of point $P$?",
    "fact_expressions": "A: Point;Coordinate(A) = (-3, 2);G: Parabola;Expression(G) = (y^2 = -4*x);F: Point;Focus(G) = F;P: Point;PointOnCurve(P, G);WhenMin(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(-1, 2)",
    "fact_spans": "[[[1, 5]], [[1, 19]], [[25, 40], [49, 52]], [[25, 40]], [[21, 24]], [[21, 43]], [[44, 48], [75, 78]], [[44, 56]], [[57, 74]]]",
    "query_spans": "[[[75, 83]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $\\sqrt{6}$, then the equations of the asymptotes of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(G) = sqrt(6)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(5)*x",
    "fact_spans": "[[[0, 56], [73, 76]], [[0, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[0, 71]]]",
    "query_spans": "[[[73, 84]]]",
    "process": "Using $\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}$, find $\\frac{b}{a}$, thus obtaining the asymptote equations. From $\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{6}$, we get $\\frac{b^{2}}{a^{2}}=5$, $\\therefore\\frac{b}{a}=\\sqrt{5}$, $\\therefore$ the asymptotes of the hyperbola are $y=\\pm\\sqrt{5}x$."
  },
  {
    "text": "Given that the tangent line $l$ at point $P(x_{0}, y_{0})$ on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has the equation $\\frac{x_{0} x}{a^{2}}+\\frac{y_{0} y}{b^{2}}=1$, if $l$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively, then when $|A B|$ is minimized, $|O P|=$ ($O$ is the origin)?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;A: Point;B: Point;O: Origin;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (x0, y0);x0:Number;y0:Number;l:Line;TangentOfPoint(P,G)=l;Expression(l)=(x0*x/a^2+y0*y/b^2=1);Intersection(l,xAxis)=A;Intersection(l,yAxis)=B;WhenMin(Abs(LineSegmentOf(A,B)));PointOnCurve(P,G)",
    "query_expressions": "Abs(LineSegmentOf(O, P))",
    "answer_expressions": "sqrt(a^2+b^2-a*b)",
    "fact_spans": "[[[3, 55]], [[5, 55]], [[5, 55]], [[58, 75]], [[151, 154]], [[155, 158]], [[183, 186]], [[5, 55]], [[5, 55]], [[3, 55]], [[58, 75]], [[58, 75]], [[58, 75]], [[81, 133], [135, 138]], [[2, 133]], [[81, 133]], [[135, 160]], [[135, 160]], [[162, 173]], [3, 72]]",
    "query_spans": "[[[174, 193]]]",
    "process": "Because the tangent line at point $ P(x_{0},y_{0}) $ is $ l:\\frac{x_{0}x}{a^{2}}+\\frac{y_{0}y}{b^{2}}=1 $, and if $ l $ intersects the $ x $-axis and $ y $-axis at points $ A $ and $ B $ respectively, then $ A(\\frac{a^{2}}{x_{0}},0) $, $ B(0,\\frac{b^{2}}{y_{0}}) $, $ \\therefore|AB|^{2}=|OA|^{2}+|OB|^{2}=\\frac{a^{4}}{x_{0}^{2}}+\\frac{b^{4}}{y_{0}^{2}} $. Also, since point $ P(x_{0},y_{0}) $ lies on the ellipse $ \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 $ ($ a>b>0 $), we have $ \\frac{x_{0}^{2}}{a^{2}}+\\frac{y_{0}^{2}}{b^{2}}=1 $. Therefore, $ |AB|^{2}=\\frac{a^{4}}{x_{0}^{2}}+\\frac{b^{4}}{y_{0}^{2}}=(\\frac{a^{4}}{x_{0}^{2}}+\\frac{b^{4}}{y_{0}^{2}})(\\frac{x_{0}^{2}}{a^{2}}+\\frac{y_{0}^{2}}{b^{2}})=a^{2}+b^{2}+\\frac{b^{4}x_{0}^{2}}{y_{0}^{2}a^{2}}+\\frac{a^{4}y_{0}^{2}}{x_{0}^{2}b^{2}}\\geqslant(a^{2}+b^{2}+2ab)=(a+b)^{2} $, with equality if and only if $ \\frac{b^{4}x_{0}^{2}}{y_{0}^{2}a^{2}}=\\frac{a^{4}y_{0}^{2}}{x_{0}^{2}b^{2}} $. Thus, $ \\begin{cases}\\frac{b^{4}x_{0}^{2}}{y_{0}^{2}a^{2}}=\\frac{a^{4}y_{0}^{2}}{x_{0}^{2}b^{2}}\\\\\\frac{a^{4}}{x_{0}^{2}}+\\frac{b^{4}}{y_{0}^{2}}=a+b\\end{cases} $. Solving gives $ x_{0}^{2}=\\frac{a^{3}}{a+b} $, $ y_{0}^{2}=\\frac{b^{3}}{a+b} $, $ \\therefore x_{0}^{2}+y_{0}^{2}=\\frac{a^{3}}{a+b}+\\frac{b^{3}}{a+b}=\\frac{a^{3}+b^{3}}{a+b}=a^{2}+b^{2}-ab $. $ \\therefore|OP|=\\sqrt{x_{0}^{2}+y_{0}^{2}}=\\sqrt{a^{2}+b^{2}-ab} $"
  },
  {
    "text": "Given point $P$ lies on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, and points $F_{1}$, $F_{2}$ are the left and right foci of the hyperbola, respectively. The slope of one asymptote of the hyperbola is $\\sqrt{7}$. If $M$ is the incenter of $\\Delta P F_{1} F_{2}$, and $S_{\\Delta P M F_{1}}=S_{\\Delta P M F_{2}}+\\lambda S_{\\Delta M F_{1} F_{2}}$, then the value of $\\lambda$ is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;P: Point;F1: Point;F2: Point;M: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, RightPart(G));LeftFocus(G) = F1;RightFocus(G) = F2;Slope(OneOf(Asymptote(G))) = sqrt(7);Incenter(TriangleOf(P, F1, F2)) = M;lambda: Number;Area(TriangleOf(P, M, F1)) = Area(TriangleOf(P, M, F2)) + lambda*Area(TriangleOf(M, F1, F2))",
    "query_expressions": "lambda",
    "answer_expressions": "sqrt(2)/4",
    "fact_spans": "[[[7, 63], [89, 92], [99, 102]], [[10, 63]], [[10, 63]], [[2, 6]], [[70, 78]], [[79, 86]], [[124, 127]], [[10, 63]], [[10, 63]], [[7, 63]], [[2, 69]], [[70, 98]], [[70, 98]], [[99, 122]], [[124, 153]], [[233, 242]], [[155, 231]]]",
    "query_spans": "[[[233, 246]]]",
    "process": "Let the inradius be R. From the given condition, $ S_{APMF_{1}} - S_{APMF_{2}} = \\lambda S_{\\triangle MF_{1}F_{2}} $, that is, $ \\frac{1}{2} \\cdot |PF_{1}| \\cdot R - \\frac{1}{2} \\cdot |PF_{2}| \\cdot R = \\frac{1}{2} \\cdot \\lambda |F_{1}F_{2}| \\cdot R $, which simplifies to $ \\frac{1}{2} \\cdot 2a \\cdot R = \\frac{1}{2} \\cdot 2c \\cdot R $, so $ e = \\frac{c}{a} = \\frac{1}{\\lambda} $. Since $ e^{2} = 1 + \\left( \\frac{b}{a} \\right)^{2} $, it follows that $ \\frac{1}{2^{2}} = 1 + 7 = 8 $, $ \\lambda = \\frac{\\sqrt{2}}{4} $."
  },
  {
    "text": "The vertex of parabola $C$ is at the origin, and its focus lies on the $x$-axis. The line $y = x$ intersects parabola $C$ at points $A$ and $B$. If $F(2,2)$ is the midpoint of $AB$, then the equation of parabola $C$ is?",
    "fact_expressions": "C: Parabola;G: Line;A: Point;B: Point;F: Point;O: Origin;Expression(G) = (y = x);Coordinate(F) = (2, 2);PointOnCurve(Focus(C), xAxis);Intersection(G, C) = {A,B};MidPoint(LineSegmentOf(A, B)) = F;Vertex(C) = O",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2 = 4*x",
    "fact_spans": "[[[0, 6], [32, 38], [72, 78]], [[24, 31]], [[40, 43]], [[45, 49]], [[53, 61]], [[12, 14]], [[24, 31]], [[53, 61]], [[0, 23]], [[24, 51]], [[53, 70]], [[0, 14]]]",
    "query_spans": "[[[72, 83]]]",
    "process": ""
  },
  {
    "text": "Given the circle $C$: $(x-1)^{2}+(y-1)^{2}=9$, drawing any chord of circle $C$ through point $A(2, 3)$, the equation of the trajectory of the midpoints $P$ of these chords is?",
    "fact_expressions": "C: Circle;Expression(C) = ((x - 1)^2 + (y - 1)^2 = 9);A: Point;Coordinate(A) = (2, 3);L: LineSegment;PointOnCurve(A, L) = True;IsChordOf(L, C) = True;MidPoint(L) = P;P: Point",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(x-3/2)^2+(y-2)^2=5/4",
    "fact_spans": "[[[2, 30], [44, 48]], [[2, 30]], [[32, 43]], [[32, 43]], [], [[31, 52]], [[31, 52]], [[31, 63]], [[60, 63]]]",
    "query_spans": "[[[60, 70]]]",
    "process": "From the circle $ C: (x-1)^{2}+(y-1)^{2}=9 $, we know the center of the circle is $ C(1,1) $. From the property of the circle, we have $ CP \\perp PA $. Let the midpoint of $ AC $ be $ B\\left(\\frac{3}{2},2\\right) $, then $ |BP| = \\frac{1}{2}|AC| = \\frac{1}{2} \\times \\sqrt{()} $. Therefore, the locus of the moving point $ P $ is a circle with center $ B $ and radius $ \\frac{\\sqrt{5}}{2} $. Hence, the trajectory equation of the midpoints $ P $ of these chords is $ \\left(x-\\frac{3}{2}\\right)^{2}+(y-2)^{2}=\\frac{5}{4} $."
  },
  {
    "text": "If the focal distance of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{8}=1$ is $2$, then the value of $m$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/8 + x^2/m = 1);m: Number;FocalLength(G) = 2",
    "query_expressions": "m",
    "answer_expressions": "{9, 7}",
    "fact_spans": "[[[1, 38]], [[1, 38]], [[47, 50]], [[1, 45]]]",
    "query_spans": "[[[47, 54]]]",
    "process": "Standard equation of the ellipse: \\frac{x^{2}}{m}+\\frac{y^{2}}{8}=1. Given the focal distance 2c=2, we have c=1. Since a^{2}-b^{2}=c^{2}, when the foci of the ellipse are on the x-axis, m-8=1, so m=9; when the foci are on the y-axis, 8-m=1, so m=7. Therefore, m=9 or m=7."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left vertex is $M$. An arbitrary line parallel to the $x$-axis intersects the hyperbola $C$ at points $A$ and $B$. If $\\overrightarrow{M A} \\cdot \\overrightarrow{M B}=0$ always holds, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;M: Point;LeftVertex(C) = M;G: Line;IsParallel(G, xAxis);A: Point;B: Point;Intersection(G, C) = {A, B};DotProduct(VectorOf(M, A), VectorOf(M, B)) = 0",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 63], [87, 93], [160, 166]], [[2, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[68, 71]], [[2, 71]], [[84, 86]], [[76, 86]], [[94, 97]], [[98, 101]], [[84, 103]], [[107, 158]]]",
    "query_spans": "[[[160, 172]]]",
    "process": "Let A(x,y), then B(-x,y), and y^{2}=b^{2}(\\frac{x^{2}}{a^{2}}-1). Also, M(-a,0), so \\overrightarrow{MA}=(x+a,y), \\overrightarrow{MB}=(-x+a,y), yielding \\overrightarrow{MA}\\cdot\\overrightarrow{MB}=-x^{2}+a^{2}+y^{2}=0, i.e., (a^{2}-b^{2})(\\frac{x^{2}}{a^{2}}-1)=0 holds for x<-a or x>a. Hence a^{2}=b^{2}, i.e., a=b, so the eccentricity of hyperbola C is e=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{2}."
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ has eccentricity $\\sqrt{2}$, then the equations of the asymptotes are?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = sqrt(2)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*x",
    "fact_spans": "[[[0, 59]], [[3, 59]], [[3, 59]], [[3, 59]], [[3, 59]], [[0, 59]], [[0, 74]]]",
    "query_spans": "[[[0, 83]]]",
    "process": ""
  },
  {
    "text": "Given that the two foci of the ellipse are $F_{1}(-2,0)$, $F_{2}(2,0)$, and the point $B(0,2)$ lies on the ellipse, what is the standard equation of this ellipse?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Coordinate(F1) = (-2, 0);Coordinate(F2) = (2, 0);Focus(G) = {F1, F2};B: Point;Coordinate(B) = (0, 2);PointOnCurve(B, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/8+y^2/4=1",
    "fact_spans": "[[[2, 4], [49, 51], [57, 59]], [[10, 23]], [[25, 37]], [[10, 23]], [[25, 37]], [[2, 37]], [[39, 48]], [[39, 48]], [[39, 52]]]",
    "query_spans": "[[[57, 66]]]",
    "process": "Let the equation of the ellipse be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. Since the two foci of the ellipse are $F_1(-2,0)$, $F_2(2,0)$, $\\therefore c=2$. Also, since point $B(0,2)$ lies on the ellipse, $\\therefore b=2$. $\\therefore a^{2}=b^{2}+c^{2}=8$. $\\therefore$ The standard equation of this ellipse is $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$."
  },
  {
    "text": "Given the ellipse $\\Gamma$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, a line $l$ passing through the point $P(1,1)$ intersects the ellipse $\\Gamma$ at points $A$ and $B$. If the chord $AB$ has point $P$ as its midpoint, then what is the equation of line $l$? (Write in general form)",
    "fact_expressions": "Gamma: Ellipse;Expression(Gamma) = (x^2/4 + y^2/3 = 1);P: Point;Coordinate(P) = (1, 1);l: Line;PointOnCurve(P, l);A: Point;B: Point;Intersection(l, Gamma) = {A, B};IsChordOf(LineSegmentOf(A, B), Gamma);MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Expression(l)",
    "answer_expressions": "3*x+4*y-7=0",
    "fact_spans": "[[[2, 49], [67, 77]], [[2, 49]], [[51, 60], [100, 104]], [[51, 60]], [[61, 66], [109, 114]], [[50, 66]], [[80, 83]], [[84, 87]], [[61, 89]], [[67, 97]], [[92, 107]]]",
    "query_spans": "[[[109, 119]]]",
    "process": "By the chord midpoint formula: the slope of the line is $ k = -\\frac{b^{2}x_{0}}{a^{2}y_{0}} = -\\frac{3\\times1}{4\\times1} = -\\frac{3}{4} $, so the equation of the line is: $ y - 1 = -\\frac{3}{4}(x - 1) $, simplifying: $ 3x + 4y - 7 = 0 $."
  },
  {
    "text": "It is known that the length of the chord intercepted by a circle $x^{2}+y^{2}-4 x+2=0$ on an asymptote of a hyperbola is $2$. Then, the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Circle;Expression(H) = (-4*x + x^2 + y^2 + 2 = 0);Length(InterceptChord(OneOf(Asymptote(G)), H)) = 2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{2*sqrt(3)/3, 2}",
    "fact_spans": "[[[2, 5], [46, 49]], [[12, 34]], [[12, 34]], [[2, 44]]]",
    "query_spans": "[[[46, 55]]]",
    "process": "By the given condition, the standard equation of the circle is $(x-2)^{2}+y^{2}=2$, that is, the center is $(2,0)$ and the radius is $r=\\sqrt{2}$. If the hyperbola is $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, then the asymptotes are $y=\\pm\\frac{b}{a}x$ with $a,b>0$, so the distance from the center to the asymptote of the hyperbola is $d=\\frac{\\frac{2b}{a}}{\\sqrt{1+(\\frac{b}{a})^{2}}}=\\frac{2b}{\\sqrt{a^{2}+b^{2}}}$. From the relation among chord length, distance from center to chord, and radius, we have: $d^{2}+1=r^{2}=2$, hence $\\frac{4b^{2}}{a^{2}+b^{2}}=1$, yielding $a^{2}=3b^{2}$. Since $a^{2}+b^{2}=c^{2}$, it follows that $\\frac{4a^{2}}{3}=c^{2}$, thus $e=\\frac{2\\sqrt{3}}{3}>1$. If the hyperbola is $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$, then the asymptotes are $y=\\pm\\frac{a}{b}x$ with $a,b>0$, so the distance from the center to the asymptote of the hyperbola is $d=\\frac{\\frac{2a}{b}}{\\sqrt{1+(\\frac{a}{b})^{2}}}=\\frac{2a}{\\sqrt{a^{2}+b^{2}}}$. From the relation among chord length, distance from center to chord, and radius, we have: $d^{2}+1=r^{2}=2$, hence $\\frac{4a^{2}}{a^{2}+b^{2}}=1$, yielding $b^{2}=3a^{2}$. Since $a^{2}+b^{2}=c^{2}$, it follows that $4a^{2}=c^{2}$, thus $e=2>1$. In conclusion, the eccentricity of the hyperbola is $\\frac{2\\sqrt{3}}{3}$ or $2$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{7}=1$, $F_{1}$, $F_{2}$ are its left and right foci. Circle $E$: $x^{2}+y^{2}-4 y+3=0$, point $P$ is a moving point on the right branch of hyperbola $C$, point $Q$ is a moving point on circle $E$. Then the minimum value of $|P Q|+|P F_{1}|$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/9 - y^2/7 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;E: Circle;Expression(E) = (-4*y + x^2 + y^2 + 3 = 0);P: Point;PointOnCurve(P, RightPart(C)) = True;Q: Point;PointOnCurve(Q, E) = True",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "5+2*sqrt(5)",
    "fact_spans": "[[[2, 45], [66, 67], [105, 111]], [[2, 45]], [[48, 55]], [[58, 65]], [[48, 71]], [[48, 71]], [[72, 99], [123, 127]], [[72, 99]], [[100, 104]], [[100, 117]], [[118, 122]], [[118, 131]]]",
    "query_spans": "[[[133, 156]]]",
    "process": "From the given conditions, F_{1}(-4,0), F_{2}(4,0), E(0,2), the radius of circle E is r=1. Since point P lies on the right branch of hyperbola C, |PF_{1}| = |PF_{2}| + 6. Therefore, |PF_{1}| + |PQ| = |PF_{2}| + |PQ| + 6. Hence, (|PF_{1}| + |PQ|)_{\\min} = (|PF_{2}| + |PQ|)_{\\min} + 6 = |F_{2}E| - r + 6 = 2\\sqrt{5} - 1 + 6 = 5 + 2\\sqrt{5}."
  },
  {
    "text": "If $\\frac{x^{2}}{|k|-2}+\\frac{y^{2}}{1-k}=1$ represents a hyperbola with foci on the $y$-axis, then the range of its semi-focal length $c$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(Abs(k) - 2) + y^2/(1 - k) = 1);PointOnCurve(Focus(G), yAxis) = True;c: Number;HalfFocalLength(G) = c",
    "query_expressions": "Range(c)",
    "answer_expressions": "(1,+\\infty)",
    "fact_spans": "[[[53, 56], [58, 59]], [[1, 56]], [[44, 56]], [[63, 66]], [[58, 66]]]",
    "query_spans": "[[[63, 73]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ lies on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and point $A$ has coordinates $(\\frac{1}{4}, 0)$, then the range of $|P A|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;A: Point;Coordinate(A) = (1/4, 0);Expression(G)= (x^2/4 + y^2/3 = 1);PointOnCurve(P,G)",
    "query_expressions": "Range(Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "[(3/4)*sqrt(5),9/4]",
    "fact_spans": "[[[7, 9]], [[2, 6]], [[48, 52]], [[48, 73]], [[7, 46]], [[2, 47]]]",
    "query_spans": "[[[75, 89]]]",
    "process": "By the given condition, let $ P(m,n) $, then $ \\frac{m^{2}}{4} + \\frac{n^{2}}{3} = 1 $ $ (-2 \\leqslant m \\leqslant 2) $, that is, $ n^{2} = 3(1 - \\frac{1}{4}m^{2}) $. Therefore,  \n$ |PA|^{2} = (m - \\frac{1}{4})^{2} + n^{2} = (m - \\frac{1}{4})^{2} + 3(1 - \\frac{1}{4}m^{2}) = \\frac{1}{4}(m - 1)^{2} + \\frac{45}{16} $.  \nWhen $ m = 1 $, $ |PA|^{2} $ reaches the minimum value $ \\frac{45}{16} $; when $ m = -2 $, $ |PA| $ reaches the maximum value $ \\frac{81}{16} $.  \nThus, $ |PA|^{2} \\in [\\frac{45}{16}, \\frac{81}{16}] $, that is, $ |PA| \\in [\\frac{3\\sqrt{5}}{4}, \\frac{9}{4}] $. Obviously, $ a^{n} 3.\\sqrt{5} $"
  },
  {
    "text": "The equation of a parabola passing through the intersection points of the line $y=x$ and the circle $x^{2}+y^{2}+6 x=0$, and having a coordinate axis as its axis of symmetry, is?",
    "fact_expressions": "L: Line;Expression(L) = (y=x);C: Circle;Expression(C) = (x^2+y^2+6*x=0);G: Parabola;PointOnCurve(Intersection(L,C),G) = True;SymmetryAxis(G) = axis",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{(y^2=-3*x^2),(x^2=-3*x^2)}",
    "fact_spans": "[[[2, 9]], [[2, 9]], [[10, 30]], [[10, 30]], [[43, 46]], [[0, 46]], [[35, 46]]]",
    "query_spans": "[[[43, 50]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the right focus is $F_{2}$, the lower vertex is $B_{2}$, and point $C$ lies on the ellipse such that $\\overrightarrow{F_{2} B_{2}}=2 \\overrightarrow{C F_{2}}$. Then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F2: Point;B2: Point;C: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F2;LowerVertex(G) = B2;PointOnCurve(C, G);VectorOf(F2, B2) = 2*VectorOf(C, F2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[2, 54], [88, 90], [152, 154]], [[4, 54]], [[4, 54]], [[61, 68]], [[75, 82]], [[83, 87]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 68]], [[2, 82]], [[83, 91]], [[93, 150]]]",
    "query_spans": "[[[152, 160]]]",
    "process": "From the given conditions, we have $F_{2}(c,0)$, $B_{2}(0,-b)$, let $C(x_{0},y_{0})$. Since $\\overrightarrow{F_{2}B_{2}}=2\\overrightarrow{CF_{2}}$, then $(-c,-b)=2(c-x_{0},-y_{0})$, so we obtain:  \n$$\n\\begin{cases}\nx_{0}=\\frac{3}{2}c \\\\\ny_{0}=\\frac{b}{2}\n\\end{cases}\n$$\nthat is, $C(\\frac{3}{2}c,\\frac{b}{2})$. Since $C$ lies on the ellipse,  \n$$\n\\frac{9c^{2}}{4a^{2}}+\\frac{b^{2}}{\\frac{4}{k^{2}}}=1,\n$$\nthus $\\frac{c^{2}}{a^{2}}=\\frac{1}{3}$, so the eccentricity $e=\\frac{c}{a}=\\frac{\\sqrt{3}}{3}$."
  },
  {
    "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ are tangent to the circle $(x-2)^{2}+y^{2}=1$, find the eccentricity of this hyperbola.",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;H: Circle;Expression(H) = (y^2 + (x - 2)^2 = 1);IsTangent(Asymptote(G), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(2*sqrt(3))/3",
    "fact_spans": "[[[2, 58], [88, 91]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[63, 83]], [[63, 83]], [[2, 85]]]",
    "query_spans": "[[[88, 97]]]",
    "process": "From the given condition, \\frac{|2b|}{\\sqrt{a^{2}+b^{2}}}=1, simplifying this equation gives the eccentricity of the hyperbola. Without loss of generality, assume the asymptotes of the hyperbola are given by y=\\frac{b}{a}x, \\therefore bx-ay=0. From the given condition, \\frac{|2b|}{\\sqrt{a^{2}+b^{2}}}=c, \\therefore 4b^{2}=c^{2}, \\therefore 4c^{2}-4a^{2}=c^{2}, \\therefore 3c^{2}=4a^{2}, \\therefore e=\\frac{2}{3}\\sqrt{3}"
  },
  {
    "text": "The line $l$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$, $O$ is the origin, and the product of the slopes of lines $OA$ and $OB$ is $-1$. A circle with radius $\\sqrt{2}$ centered at the midpoint of segment $AB$ intersects line $l$ at points $P$ and $Q$. Given $M(6,0)$, find the minimum value of $|MP|^{2}+|MQ|^{2}$.",
    "fact_expressions": "l: Line;G: Parabola;A: Point;O: Origin;B: Point;M: Point;P: Point;Q: Point;H: Circle;Expression(G) = (y^2 = 4*x);Coordinate(M) = (6, 0);Intersection(l, G) = {A, B};Slope(LineOf(O, A))*Slope(LineOf(O, B)) = -1;Center(H) = MidPoint(LineSegmentOf(A, B));Radius(H) = sqrt(2);Intersection(H, l) = {P, Q}",
    "query_expressions": "Min(Abs(LineSegmentOf(M, P))^2 + Abs(LineSegmentOf(M, Q))^2)",
    "answer_expressions": "10",
    "fact_spans": "[[[0, 5], [98, 103]], [[6, 20]], [[22, 25]], [[32, 35]], [[26, 29]], [[115, 123]], [[105, 108]], [[109, 112]], [[96, 97]], [[6, 20]], [[115, 123]], [[0, 31]], [[41, 66]], [[67, 97]], [[82, 97]], [[96, 114]]]",
    "query_spans": "[[[125, 152]]]",
    "process": "Let the equation of line AB be $x = my + t$, with points $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$.  \nFrom  \n\\[\n\\begin{cases}\ny^2 = 4x \\\\\nx = my + t\n\\end{cases}\n\\]  \nwe obtain $y^{2} - 4my - 4t = 0$, so $\\Delta = (4m)^{2} - 4(-4t) = 16(t + m^{2}) > 0$.  \nThus, $y_{1} + y_{2} = 4m$, $y_{1}y_{2} = -4t$,  \nso $x_{1} + x_{2} = m(y_{1} + y_{2}) + 2t = 4m^{2} + 2t$,  \n$x_{1} \\cdot x_{2} = \\frac{y_{1}^{2} y_{2}^{2}}{16} = t^{2}$.  \nSince the product of the slopes of lines OA and OB is $-1$, we have $OA \\perp OB$, i.e., $\\overrightarrow{OA} \\cdot \\overrightarrow{OB} = 0$,  \nso $x_{1}x_{2} + y_{1}y_{2} = t^{2} - 4t = 0$, hence $t = 4$.  \nThus, the equation of line AB is $x = my + 4$, and $x_{1} + x_{2} = 4m^{2} + 8$,  \nso the center of the circle is $O'(2m^{2} + 4, 2m)$.  \nBy the property that the sum of squares of the four sides of a parallelogram equals the sum of squares of the diagonals (easily proven using vectors),  \nwe get  \n$2(|MP|^{2} + |MQ|^{2}) = 4|MO|^{2} + |PQ|^{2} = 4|MO|^{2} + (2\\sqrt{2})^{2} = 4[4(m^{2} - 1)^{2} + 4m^{2}] + 8 = 16(m^{4} - m^{2} + 1) + 8 = 16(m^{2} - \\frac{1}{2})^{2} + 20$,  \nso $|MP|^{2} + |MQ|^{2} = 8(m^{2} - \\frac{1}{2})^{2} + 10$.  \nTherefore, when $m = \\pm \\frac{\\sqrt{2}}{2}$, the minimum value of $|MP|^{2} + |MQ|^{2}$ is $10$."
  },
  {
    "text": "$P$ is a point on the ellipse $\\frac{x^{2}}{27}+\\frac{y^{2}}{16}=1$, then the minimum distance from $P$ to the line $l$: $4x+3y-25=0$ is?",
    "fact_expressions": "G: Ellipse;l: Line;P: Point;Expression(G) = (x^2/27 + y^2/16 = 1);Expression(l) = (4*x + 3*y - 25 = 0);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, l))",
    "answer_expressions": "1/5",
    "fact_spans": "[[[4, 43]], [[52, 72]], [[48, 51], [0, 3]], [[4, 43]], [[52, 72]], [[0, 46]]]",
    "query_spans": "[[[48, 81]]]",
    "process": ""
  },
  {
    "text": "The chord length intercepted by the directrix of the parabola $y^{2}=2 p x(p>0)$ on the circle $x^{2}+y^{2}-2 y-1=0$ is $2$, then $p=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;H: Circle;Expression(H) = (-2*y + x^2 + y^2 - 1 = 0);Length(InterceptChord(Directrix(G), H)) = 2",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 21]], [[0, 21]], [[57, 60]], [[3, 21]], [[25, 47]], [[25, 47]], [[0, 55]]]",
    "query_spans": "[[[57, 62]]]",
    "process": "The directrix of the parabola $ y^{2} = 2px $ ($ p > 0 $) is $ x = -\\frac{p}{2} $, and the circle converted to standard form is $ x^{2} + (y - 1)^{2} = 2 $, with center $ M(0,1) $ and $ r = \\sqrt{2} $. The distance from the center to the directrix is $ \\frac{p}{2} $, so $ \\left( \\frac{p}{2} \\right)^{2} + \\left( \\frac{2}{2} \\right)^{2} = (\\sqrt{2})^{2} $, which gives $ p = 2 $."
  },
  {
    "text": "The trajectory equation of the center $M$ of a moving circle passing through the fixed point $F(1,0)$ and tangent to the line $x=-1$ is?",
    "fact_expressions": "G: Circle;M:Point;H: Line;F: Point;Expression(H) = (x = -1);Coordinate(F) = (1, 0);PointOnCurve(F,G);IsTangent(H,G);Center(G)=M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[24, 26]], [[28, 31]], [[13, 21]], [[3, 11]], [[13, 21]], [[3, 11]], [[0, 26]], [[12, 26]], [[24, 31]]]",
    "query_spans": "[[[28, 38]]]",
    "process": "Let the center of the moving circle be $ M(x,y) $. Since circle $ M $ passes through the point $ F(1,0) $ and is tangent to the line $ l: x=-1 $, the distance from point $ M $ to $ F $ is equal to the distance from point $ M $ to the line. By the definition of a parabola, the trajectory of $ M $ is a parabola with focus $ F $ and directrix $ l $. Let the equation be $ y^{2}=2px $ ($ p>0 $), then $ \\frac{p}{2}=1 $, $ 2p=4 $. Therefore, the trajectory equation of $ M $ is $ y^{2}=4x $."
  },
  {
    "text": "An ellipse with center at the origin and foci on the $x$-axis passes through the point $P(3,4)$. The two foci of the ellipse are $F_{1}$ and $F_{2}$. If $P F_{1} \\perp P F_{2}$, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;PointOnCurve(Focus(G), xAxis);P: Point;Coordinate(P) = (3, 4);PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/45 + y^2/20 = 1",
    "fact_spans": "[[[17, 19], [31, 33], [83, 85]], [[3, 7]], [[0, 19]], [[8, 19]], [[21, 30]], [[21, 30]], [[17, 30]], [[41, 48]], [[49, 56]], [[31, 56]], [[58, 81]]]",
    "query_spans": "[[[83, 90]]]",
    "process": "Let $ F_{1}(-c,0) $, $ F_{2}(c,0) $. From $ PF_{1} \\perp PF_{2} $, we obtain $ (3+c)^{2} + 4^{2} + (3-c)^{2} + 4^{2} = (2c)^{2} $, so $ c=5 $. Let the ellipse equation be $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $. Then  \n\\[\n\\begin{cases}\n\\frac{9}{a^{2}} + \\frac{14}{b^{2}} = 1 \\\\\na^{2} - b^{2} = 25\n\\end{cases}\n\\]\nSolving gives  \n\\[\n\\begin{cases}\na^{2} = 45 \\\\\nb^{2} = 20\n\\end{cases}\n\\]\nThus, the ellipse equation is $ \\frac{x^{2}}{45} + \\frac{y^{2}}{20} = 1 $."
  },
  {
    "text": "Given that $F$ is the right focus of the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, from point $F$ a perpendicular is drawn to one asymptote of the hyperbola $E$, with foot of the perpendicular at $A$, and it intersects the other asymptote at point $B$. If $|O F|=|F B|$, then the eccentricity of the hyperbola $E$ is?",
    "fact_expressions": "F: Point;E: Hyperbola;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;RightFocus(E) = F;L: Line;PointOnCurve(F, L);L1: Line;L2: Line;OneOf(Asymptote(E))=L1;OneOf(Asymptote(E))=L2;Negation(L1=L2);IsPerpendicular(L, L1);FootPoint(L, L1) = A;A: Point;Intersection(L, L2) = B;B: Point;Abs(LineSegmentOf(O, F)) = Abs(LineSegmentOf(F, B));O: Origin",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 5], [73, 77]], [[6, 67], [78, 84], [131, 137]], [[6, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[2, 71]], [], [[72, 93]], [], [], [[78, 90]], [[78, 109]], [[78, 109]], [[72, 93]], [[72, 100]], [[97, 100]], [[72, 114]], [[110, 114]], [[116, 129]], [[116, 129]]]",
    "query_spans": "[[[131, 143]]]",
    "process": "Find the asymptotes of the hyperbola, and using the properties of a right triangle and the symmetry of the asymptotes, obtain the relationship between $a$ and $b$, and then find the eccentricity. The hyperbola $E: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ has asymptotes given by $y = \\pm \\frac{b}{a}x$. If $|OF| = |FB|$, then in the right triangle $OAB$, since $\\angle AOF = \\angle BOF = \\angle ABO = 30^{\\circ}$, it follows that $\\frac{b}{a} = \\tan 30^{\\circ} = \\frac{\\sqrt{3}}{3}$."
  },
  {
    "text": "Given the equation of the ellipse is $\\frac{x^{2}}{16}+\\frac{y^{2}}{m^{2}}=1$ $(m>0)$. If a point of intersection $M$ between the line $y=\\frac{\\sqrt{2}}{2} x$ and the ellipse has its projection on the $x$-axis exactly at the right focus $F$ of the ellipse, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;H: Line;F:Point;m:Number;m>0;M:Point;Expression(H) = (y = x*(sqrt(2)/2));Expression(G) = (x^2/16 + y^2/m^2 = 1);OneOf(Intersection(H,G))=M;Projection(M,xAxis)=F;RightFocus(G)=F",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 4], [84, 86], [105, 107], [116, 118]], [[57, 83]], [[111, 114]], [[8, 53]], [[8, 53]], [[91, 94]], [[57, 83]], [[2, 53]], [[57, 94]], [[91, 114]], [[105, 114]]]",
    "query_spans": "[[[116, 124]]]",
    "process": "From the ellipse equation, the coordinates of the right focus are (\\sqrt{16-m^{2}}). Since the projection of the intersection point M between the line and the ellipse onto the x-axis is exactly the right focus F of the ellipse, we have MF\\botx-axis. Therefore, the x-coordinate of M is \\sqrt{16-m^{2}}. Substituting into the line equation gives the y-coordinate of M as \\sqrt{\\frac{16-m^{2}}{2}}, so M(\\sqrt{16-m^{2}},\\sqrt{\\frac{16-m^{2}}{2}}). Substituting the coordinates of M into the ellipse equation yields: \\frac{16-m^{2}}{16}+\\frac{16-m^{2}}{2m^{2}}=1. Simplifying gives: (m^{2})^{2}+8m^{2}-128=0, i.e., (m^{2}-8)(m^{2}+16)=0. Solving yields m^{2}=8, m^{2}=-16 (discarded). Since m>0, we have m=2\\sqrt{2}. Thus, the ellipse equation is \\frac{x^{2}}{16}+\\frac{y^{2}}{8}=1, so a^{2}=16, b^{2}=8, then c^{2}=a^{2}-b^{2}=8. Hence, e=\\frac{c}{a}=\\frac{2\\sqrt{2}}{4}=\\frac{\\sqrt{2}}{2}. Therefore, the answer is \\sqrt{2}."
  },
  {
    "text": "Given a point $P$ on the parabola $y^{2}=4x$ such that the distance from $P$ to the focus $F$ is $5$, then the area of $\\triangle PFO$ is?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;O: Origin;Expression(G) = (y^2 = 4*x);PointOnCurve(P, G);Focus(G) = F;Distance(P, F) = 5",
    "query_expressions": "Area(TriangleOf(P, F, O))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 16]], [[19, 22]], [[25, 28]], [[37, 54]], [[2, 16]], [[2, 22]], [[2, 28]], [[2, 35]]]",
    "query_spans": "[[[37, 59]]]",
    "process": "Since the directrix of the parabola is $x=1$, the horizontal coordinate of point $P$ is $4$, the vertical coordinate is $4$, and $OF=1$, so $S_{\\Delta PFO} = \\frac{1}{2} \\times 1 \\times 4 = 2$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has eccentricity $\\frac{2 \\sqrt{3}}{3}$, focal length $2 c$, and $2 a^{2}=3 c$. A point $P$ on the hyperbola satisfies $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=2$ ($F_{1}$, $F_{2}$ being the left and right foci), then $|\\overrightarrow{P F_{1}}| \\cdot|\\overrightarrow{P F_{2}}|$=?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;c:Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;Eccentricity(G) = (2*sqrt(3))/3;FocalLength(G) = 2*c;2*a^2 = 3*c;PointOnCurve(P,G);DotProduct(VectorOf(P,F1),VectorOf(P,F2))=2",
    "query_expressions": "DotProduct(Abs(VectorOf(P,F1)),Abs(VectorOf(P,F2)))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 58], [111, 114]], [[5, 58]], [[5, 58]], [[117, 120]], [[182, 189]], [[190, 197]], [[5, 58]], [[5, 58]], [[89, 94]], [[2, 58]], [[111, 204]], [[111, 204]], [[2, 85]], [[2, 94]], [[96, 109]], [[111, 120]], [[122, 181]]]",
    "query_spans": "[[[206, 268]]]",
    "process": "According to the given conditions, we have  \n\\[\n\\begin{cases}\n2a^{2}=3c \\\\\ne=\\frac{c}{a}=\\frac{2\\sqrt{3}}{3}\n\\end{cases},\n\\]  \nsolving yields  \n\\[\n\\begin{cases}\na=\\sqrt{3} \\\\\nc=2\n\\end{cases}\n\\quad \\text{or} \\quad \n\\begin{cases}\na=0 \\\\\nc=0\n\\end{cases}\n\\quad \\text{(discarded)},\n\\]  \nthen the equation of the hyperbola is  \n\\[\n\\frac{x^{2}}{3}-y^{2}=1.\n\\]  \nFrom the geometric properties of the hyperbola, we obtain  \n\\[\n|PF_1|-|PF_2|=2a=2\\sqrt{3}, \\quad |F_1F_2|=2c=4.\n\\]  \nSince  \n\\[\n\\overrightarrow{PF_1} \\cdot \\overrightarrow{PF_2} = 2,\n\\]  \nwe have  \n\\[\n\\cos\\angle F_1PF_2 = \\frac{\\overrightarrow{PF_1} \\cdot \\overrightarrow{PF_2}}{|\\overrightarrow{PF_1}| \\cdot |\\overrightarrow{PF_2}|} = \\frac{2}{|PF_1| \\cdot |PF_2|}.\n\\]  \nOn the other hand, by the law of cosines,  \n\\[\n\\cos\\angle F_1PF_2 = \\frac{|PF_1|^{2} + |PF_2|^{2} - |F_1F_2|^{2}}{2|PF_1| \\cdot |PF_2|},\n\\]  \nthus  \n\\[\n\\frac{|PF_1|^{2} + |PF_2|^{2} - 16}{2|PF_1| \\cdot |PF_2|} = \\frac{2}{|PF_1| \\cdot |PF_2|},\n\\]  \nwhich implies  \n\\[\n|PF_1|^{2} + |PF_2|^{2} - 16 = 4,\n\\]  \nso  \n\\[\n|PF_1|^{2} + |PF_2|^{2} = 20.\n\\]  \nSince  \n\\[\n|PF_1| - |PF_2| = 2\\sqrt{3},\n\\]  \nlet  \n\\[\n|PF_1| = x, \\quad |PF_2| = x - 2\\sqrt{3},\n\\]  \nthen  \n\\[\nx^2 + (x - 2\\sqrt{3})^2 = 20,\n\\]  \nexpanding gives  \n\\[\nx^2 + x^2 - 4\\sqrt{3}x + 12 = 20,\n\\]  \n\\[\n2x^2 - 4\\sqrt{3}x - 8 = 0,\n\\]  \n\\[\nx^2 - 2\\sqrt{3}x - 4 = 0,\n\\]  \nsolving yields  \n\\[\nx = \\sqrt{3} \\pm \\sqrt{7},\n\\]  \nbut only the positive root is valid. However, returning to the earlier step: from  \n\\[\n(|PF_1| - |PF_2|)^2 = (2\\sqrt{3})^2 = 12,\n\\]  \nand  \n\\[\n|PF_1|^2 + |PF_2|^2 = 20,\n\\]  \nwe get  \n\\[\n(|PF_1| - |PF_2|)^2 = |PF_1|^2 - 2|PF_1||PF_2| + |PF_2|^2 = 12,\n\\]  \nso  \n\\[\n20 - 2|PF_1||PF_2| = 12,\n\\]  \nthus  \n\\[\n2|PF_1||PF_2| = 8,\n\\]  \n\\[\n|PF_1||PF_2| = 4,\n\\]  \ntherefore  \n\\[\n\\overrightarrow{PF_1} \\cdot \\overrightarrow{PF_2} = 4.\n\\]"
  },
  {
    "text": "The parabola $y^{2}=2 p x$ passes through the point $M(2,2)$, then the distance from point $M$ to the focus of the parabola is?",
    "fact_expressions": "G: Parabola;p: Number;M: Point;Expression(G) = (y^2 = 2*(p*x));Coordinate(M) = (2, 2);PointOnCurve(M, G)",
    "query_expressions": "Distance(M, Focus(G))",
    "answer_expressions": "5/2",
    "fact_spans": "[[[0, 16], [33, 36]], [[3, 16]], [[17, 26], [28, 32]], [[0, 16]], [[17, 26]], [[0, 26]]]",
    "query_spans": "[[[28, 43]]]",
    "process": ""
  },
  {
    "text": "Let the line $l$ passing through the focus $F$ of the parabola $y^{2}=8x$ intersect the parabola at points $A$ and $B$. If the distance from the midpoint $M$ of $AB$ to the directrix of the parabola is $8$, then what is the slope of $l$?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;F: Point;M: Point;Expression(G) = (y^2 = 8*x);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = M;Distance(M, Directrix(G)) = 8",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*1",
    "fact_spans": "[[[23, 28], [70, 73]], [[3, 17], [29, 32], [56, 59]], [[34, 37]], [[38, 41]], [[19, 22]], [[52, 55]], [[3, 17]], [[3, 22]], [[1, 28]], [[23, 43]], [[45, 55]], [[52, 68]]]",
    "query_spans": "[[[70, 78]]]",
    "process": "Let the slope of the line be $ k $. The parabola $ y^{2} = 8x $ has focus $ F(2,0) $ and directrix equation $ x = -2 $. Thus, the line $ l $ has equation $ y = k(x - 2) $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, $ M(x_{0}, y_{0}) $, then $ x_{0} = \\frac{x_{1} + x_{2}}{2} $, $ y_{0} = \\frac{y_{1} + y_{2}}{2} $. Since the distance from the midpoint $ M $ of $ AB $ to the directrix of the parabola is 8, we have $ x_{0} + 2 = 8 \\Rightarrow \\frac{x_{1} + x_{2}}{2} + 2 = 8 $, so $ x_{1} + x_{2} = 12 $. Solving the system of equations \n\\[\n\\begin{cases}\ny = k(x - 2) \\\\\ny^{2} = 8x\n\\end{cases}\n\\]\nwe obtain $ k^{2}x^{2} - 4(k^{2} + 2)x + 4k^{2} = 0 $. By Vieta's formulas, $ x_{1} + x_{2} = \\frac{4(k^{2} + 2)}{k^{2}} $, thus $ \\frac{4(k^{2} + 2)}{k^{2}} = 12 $. Solving gives $ k^{2} = 1 $, so $ k = \\pm 1 $."
  },
  {
    "text": "If the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$ with foci on the $y$-axis has eccentricity $e=\\frac{\\sqrt{3}}{2}$, then $m=$?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/5 + y^2/m = 1);PointOnCurve(Focus(G), yAxis);Eccentricity(G)=e;e:Number;e=sqrt(3)/2",
    "query_expressions": "m",
    "answer_expressions": "20",
    "fact_spans": "[[[10, 47]], [[75, 78]], [[10, 47]], [[1, 47]], [[10, 73]], [[51, 73]], [[51, 73]]]",
    "query_spans": "[[[75, 80]]]",
    "process": "Since the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$ has its foci on the $y$-axis, $\\therefore a^{2}=m$, $b^{2}=5$, $5<m$, $c^{2}=m-5$, $\\therefore e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{m-5}{m}=(\\frac{\\sqrt{3}}{2})^{2}=\\frac{3}{4}$, solving gives $m=20$."
  },
  {
    "text": "Given the ellipse $C_{1}$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ and the circle $C_{2}$: $x^{2}+y^{2}=b^{2}$, if there exists a point $P$ on the ellipse $C_{1}$ such that from $P$ two tangents $PA$ and $PB$ are drawn to the circle, with points of tangency $A$ and $B$, satisfying $\\angle BPA = \\frac{\\pi}{3}$, then the range of eccentricity of the ellipse $C_{1}$ is?",
    "fact_expressions": "C1: Ellipse;Expression(C1) = (x^2/a^2 + y^2/b^2 = 1);b: Number;a: Number;a > b;b > 0;C2: Circle;Expression(C2) = (x^2 + y^2 = b^2);P: Point;PointOnCurve(P, C1);A: Point;B: Point;TangentOfPoint(P, C2) = {LineOf(P, A), LineOf(P, B)};TangentPoint(LineOf(P, A), C2) = A;TangentPoint(LineOf(P, B), C2) = B;AngleOf(B, P, A) = pi/3",
    "query_expressions": "Range(Eccentricity(C1))",
    "answer_expressions": "[\\sqrt{3}/2, 1)",
    "fact_spans": "[[[2, 63], [97, 106], [180, 189]], [[2, 63]], [[13, 63]], [[13, 63]], [[13, 63]], [[13, 63]], [[64, 93], [119, 120]], [[64, 93]], [[109, 113], [115, 118]], [[96, 113]], [[140, 143]], [[144, 147]], [[114, 136]], [[119, 147]], [[119, 147]], [[149, 177]]]",
    "query_spans": "[[[180, 200]]]",
    "process": "Since $\\angle BPA = \\frac{\\pi}{3}$, it follows that $\\angle BPO = \\frac{\\pi}{6}$ ($O$ is the coordinate origin), so $|OP| = 2|OB| = 2b$. Because $b < |OP| \\leqslant a$, we have $2b \\leqslant a$, so $a^{2} - 4b^{2} \\geqslant 0$. Also, $b^{2} = a^{2} - c^{2}$. Therefore, $a^{2} - 4a^{2} + 3c^{2} \\geqslant 0$, i.e., $3a^{2} \\leqslant 4c^{2}$, so $e = \\frac{c}{a} \\geqslant \\frac{\\sqrt{3}}{2}$. Since $0 < e < 1$, we have $\\frac{\\sqrt{3}}{2} \\leqslant e < 1$."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=12 x$, $M$ is a point on $C$, and the extension of $F M$ intersects the $y$-axis at point $N$. If $\\overrightarrow{F M}=2 \\overrightarrow{M N}$, then $|F N|$=?",
    "fact_expressions": "C: Parabola;F: Point;M: Point;N: Point;Expression(C) = (y^2 = 12*x);Focus(C) = F;PointOnCurve(M, C);Intersection(OverlappingLine(LineSegmentOf(F, M)), yAxis) = N;VectorOf(F, M) = 2*VectorOf(M, N)",
    "query_expressions": "Abs(LineSegmentOf(F, N))",
    "answer_expressions": "6",
    "fact_spans": "[[[6, 26], [34, 37]], [[2, 5]], [[30, 33]], [[56, 60]], [[6, 26]], [[2, 29]], [[30, 40]], [[41, 60]], [[63, 108]]]",
    "query_spans": "[[[110, 119]]]",
    "process": "According to the vector relation and the fact that point M lies on the parabola, the coordinates of point N are obtained using the proportional relationship. Using the distance formula between two points, we can find that from $\\overrightarrow{FM}=2\\overrightarrow{MN}$, it follows that M is the trisection point of FN closer to N. Given F(3,0), solving gives $x_{M}=1$, substituting into the parabola yields $y=\\pm2\\sqrt{3}$, hence $N(0,\\pm3\\sqrt{3})$. Given F(3,0), thus $|FN|=\\sqrt{9+27}=6$."
  },
  {
    "text": "Given that the asymptotes of a hyperbola are $3x \\pm 4y = 0$, then the eccentricity of the hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (3*x+pm*4*y = 0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{5/4, 5/3}",
    "fact_spans": "[[[2, 5], [29, 32]], [[2, 27]]]",
    "query_spans": "[[[29, 39]]]",
    "process": "Let the hyperbola equation be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0). Since the asymptotes are 3x\\pm4y=0, we have \\frac{b}{a}=\\frac{3}{4}, so \\sqrt{e^{2}-1}=\\frac{3}{4}, and e=\\frac{5}{4}. Let the hyperbola equation be \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1 (a>0,b>0). Then \\frac{a}{b}=\\frac{3}{4}, \\frac{b}{a}=\\frac{4}{3}, so e=\\frac{5}{3}. Therefore, the eccentricity is e=\\frac{5}{4} or e=\\frac{5}{3}. Hence, the answer is \\frac{5}{4} or \\frac{5}{3}."
  },
  {
    "text": "Points $A$ and $B$ are the left and right vertices of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, respectively. The line $x=m y+\\frac{6}{5}$ intersects the ellipse at points $P$ and $Q$. Denote the slopes of lines $AP$ and $BQ$ as $k_{1}$ and $k_{2}$, respectively. Then the minimum value of $k_{1}^{2}+\\frac{1}{k_{2}^{2}}$ is?",
    "fact_expressions": "G: Ellipse;H: Line;m: Number;A: Point;P: Point;B: Point;Q: Point;k1: Number;k2: Number;Expression(G) = (x^2/4 + y^2 = 1);Expression(H) = (x = m*y + 6/5);LeftVertex(G) = A;RightVertex(G) = B;Intersection(H, G) = {P, Q};Slope(LineOf(A, P)) = k1;Slope(LineOf(B, Q)) = k2",
    "query_expressions": "Min(k1^2 + 1/(k2^2))",
    "answer_expressions": "1/2",
    "fact_spans": "[[[11, 38], [68, 70]], [[46, 67]], [[48, 67]], [[0, 4]], [[73, 76]], [[5, 8]], [[77, 80]], [[105, 112]], [[115, 122]], [[11, 38]], [[46, 67]], [[0, 45]], [[0, 45]], [[46, 82]], [[84, 122]], [[84, 122]]]",
    "query_spans": "[[[124, 161]]]",
    "process": "Let P(x_{1},y_{1}), Q(x_{2},y_{2}), solve the system \\begin{cases}x=my+\\frac{6}{5}\\\\x^{2}+4y=4\\end{cases}, by Vieta's formulas we obtain y_{1}+y_{2}, y_{1}y_{2}, let the slope of line AQ be the answer. [Solution] Let P(x_{1},y_{1}), Q(x_{2},y_{2}) solve the system \\begin{cases}x=my+\\frac{6}{5}\\\\x^{2}+4y^{2}=4\\end{cases}, eliminate x and simplify to get (m^{2}+4)y^{2}+\\frac{12}{5}my-\\frac{64}{25}=0, by Vieta's formulas we obtain y_{1}+y_{2}=-\\frac{-}{5}\\frac{12}{(m^{2}+4})', y_{1}y_{2}=-\\frac{64}{25(m^{2}+4)}. Let the slope of line AQ be k, then k=\\frac{y_{2}}{x_{2}+2}, so, k\\cdot k_{2}=\\frac{y_{2}}{x_{2}+2}\\cdot\\frac{y_{2}}{x_{2}-2}=\\frac{y_{2}^{2}}{x_{2}^{2}-}, while k_{1}\\cdot k=\\frac{y_{1}}{y_{2}}.-2-\\frac{1}{116}, thus, k_{1}^{2}(m^{2}+4)^{-\\frac{192m^{2}}{25(m^{2}+4)}+\\frac{256}{25}}. Equality holds if and only if k=\\pm\\frac{1}{8}, therefore, the minimum value of k_{1}^{2}+\\frac{1}{k^{2}} is \\frac{1}{2}."
  },
  {
    "text": "If the eccentricity of the ellipse $x^{2}+m y^{2}=1(0<m<1)$ is $\\frac{\\sqrt {3}}{2}$, then what is its major axis length?",
    "fact_expressions": "G: Ellipse;m: Number;0 < m;m < 1;Expression(G) = (m*y^2 + x^2 = 1);Eccentricity(G) = sqrt(3)/2",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "sqrt(7)",
    "fact_spans": "[[[1, 27], [55, 56]], [[3, 27]], [[3, 27]], [[3, 27]], [[1, 27]], [[1, 53]]]",
    "query_spans": "[[[55, 62]]]",
    "process": ""
  },
  {
    "text": "A point in the first quadrant on the parabola $y^{2}=4 x$ has a distance of $3$ to the focus; then the coordinates of this point are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P:Point;Quadrant(P)=1;PointOnCurve(P,G);Distance(P,Focus(G))=3",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2,2*sqrt(2))",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[38, 39]], [[0, 25]], [[0, 25]], [[0, 35]]]",
    "query_spans": "[[[38, 43]]]",
    "process": "Analysis: Let the coordinates of the point be $(x_{0},y_{0})$. According to the definition of the parabola, we have $x_{0}-(-1)=3$, thus obtaining the values of $x_{0}$ and $y_{0}$, leading to the solution. Specifically, let the coordinates of the point be $(x_{0},y_{0})$. According to the definition of the parabola, we get $x_{0}-(-1)=3$, so $x_{0}=2$. Hence, $y^{2}=8$, $\\therefore y=\\pm2\\sqrt{2}$. Since the point lies in the first quadrant, the coordinates of the point are $(2,2\\sqrt{2})$."
  },
  {
    "text": "The equation of the parabola with the center of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ as its vertex and the right focus as its focus is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (x^2/16 - y^2/9 = 1);Vertex(H) = Center(G);Focus(H) = RightFocus(G)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2=20*x",
    "fact_spans": "[[[1, 40]], [[54, 57]], [[1, 40]], [[0, 57]], [[0, 57]]]",
    "query_spans": "[[[54, 61]]]",
    "process": "Find the center and the coordinates of the right focus of the hyperbola, and finally write the equation of the parabola. From the hyperbola equation $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, it is known that the center of the hyperbola is at the origin, $a=4$, $b=3$, and the right focus is at $(5,0)$. Since the center is at the origin, the parabola equation is assumed to be: $y^{2}=2px$ ($p>0$). Thus, $\\frac{p}{2}=5 \\Rightarrow p=10$, so the equation of the parabola is: $y^{2}=20x$."
  },
  {
    "text": "Given that one asymptote of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1$ ($b>0$) has the equation $y=x$, then $b$=?",
    "fact_expressions": "G: Hyperbola;b: Number;b>0;Expression(G) = (x^2 - y^2/b^2 = 1);Expression(OneOf(Asymptote(G))) = (y = x)",
    "query_expressions": "b",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 39]], [[56, 59]], [[5, 39]], [[2, 39]], [[2, 54]]]",
    "query_spans": "[[[56, 61]]]",
    "process": "Since the asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ are given by $x^{2}-\\frac{y^{2}}{b^{2}}=0$, that is, $y=\\pm bx$, therefore $b=1$."
  },
  {
    "text": "A line passing through the right focus $F_{2}$ of the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{6}=1$ with an inclination angle of $30^{\\circ}$ intersects the hyperbola at points $A$ and $B$. Let $O$ be the origin and $F_{1}$ the left focus. Then $|A B |$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2/6 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Line;PointOnCurve(F2, H);Inclination(H) = ApplyUnit(30, degree);A: Point;B: Point;Intersection(H, G) = {A, B};O: Origin",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16*sqrt(3)/5",
    "fact_spans": "[[[1, 39], [71, 74]], [[1, 39]], [[94, 101]], [[43, 50]], [[71, 105]], [[1, 50]], [[68, 70]], [[0, 70]], [[51, 70]], [[75, 78]], [[79, 82]], [[68, 84]], [[85, 88]]]",
    "query_spans": "[[[107, 117]]]",
    "process": "Analysis: From the hyperbola equation, we obtain $ a = \\sqrt{3} $, $ b = \\sqrt{6} $, so $ c = \\sqrt{a^{2} + b^{2}} = 3 $, $ F_{1}(-3,0) $, $ F_{2}(3,0) $. The equation of line $ AB $ is $ y = \\frac{\\sqrt{3}}{3}(x - 3) $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. From $ \\begin{cases} y = \\frac{\\sqrt{3}}{3}(x - 3), \\\\ \\frac{x^{2}}{3} - \\frac{y^{2}}{6} = 1 \\end{cases} $, we get $ 5x^{2} + 6x - 27 = 0 $. Therefore, $ x_{1} + x_{2} = -\\frac{6}{5} $, $ x_{1}x_{2} = -\\frac{27}{5} $, so $ AB = \\sqrt{1 + k^{2}}|x_{1} - x_{2}| = \\sqrt{1 + \\left(\\frac{\\sqrt{3}}{3}\\right)^{2}} \\sqrt{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}} = \\sqrt{\\frac{4}{3}} \\sqrt{\\frac{36}{25} + \\frac{108}{5}} = \\frac{16\\sqrt{3}}{5} $."
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and let $P$ be a point on the ellipse $C$ such that $P F_{1} \\perp P F_{2}$. If the area of $\\Delta P F_{1} F_{2}$ is $9$ and the perimeter is $18$, then the equation of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Focus(C) = {F1, F2};PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));Area(TriangleOf(P, F1, F2)) = 9;Perimeter(TriangleOf(P, F1, F2))=18",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/25+y^2/9=1",
    "fact_spans": "[[[17, 74], [84, 89], [160, 165]], [[23, 74]], [[23, 74]], [[80, 83]], [[1, 8]], [[9, 16]], [[23, 74]], [[23, 74]], [[17, 74]], [[1, 79]], [[80, 94]], [[96, 119]], [[121, 150]], [[121, 158]]]",
    "query_spans": "[[[160, 170]]]",
    "process": "\\because PF_{1}\\bot PF_{2}, \\therefore \\triangle PF_{1}F_{2} is a right triangle. Given that the area of \\triangle PF_{1}F_{2} is 9, \\therefore \\frac{1}{2}|PF_{1}||PF_{2}|=9, thus |PF_{1}||PF_{2}|=18. In right triangle \\triangle PF_{1}F_{2}, by the Pythagorean theorem, |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}. From the definition of the ellipse, |PF_{1}|+|PF_{2}|=2a, \\therefore (|PF_{1}|+|PF_{2}|)^{2}-2|PF_{1}||PF_{2}|=|F_{1}F_{2}|^{2}, i.e., 4a^{2}-36=4c^{2}, \\therefore a^{2}-c^{2}=9, i.e., b^{2}=9. Given b>0, \\therefore b=3. \\because the perimeter of \\triangle PF_{1}F_{2} is 18, \\therefore 2a+2c=18, i.e., a+c=9, \\textcircled{1}. Also given a^{2}-c^{2}=9, \\therefore a-c=1, \\textcircled{2}. From \\textcircled{1} and \\textcircled{2}, a=5, c=4, \\therefore the required ellipse equation is \\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1."
  },
  {
    "text": "If one focus of the ellipse $\\frac{x^{2}}{p}+\\frac{y^{2}}{4}=1$ is $(0,-1)$, then the value of $p$ is?",
    "fact_expressions": "G: Ellipse;p: Number;H: Point;Expression(G) = (y^2/4 + x^2/p = 1);Coordinate(H) = (0, -1);OneOf(Focus(G)) = H",
    "query_expressions": "p",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 38]], [[54, 57]], [[44, 52]], [[1, 38]], [[44, 52]], [[1, 52]]]",
    "query_spans": "[[[54, 61]]]",
    "process": "Since the focus is at (0,-1), the focus lies on the y-axis, so a^{2}=4, b^{2}=p, c=1, p=a^{2}-c^{2}=4-1=3"
  },
  {
    "text": "The equation of the hyperbola is $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, then its eccentricity $e$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1);e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "5/4",
    "fact_spans": "[[[0, 3], [45, 46]], [[0, 43]], [[49, 52]], [[45, 52]]]",
    "query_spans": "[[[49, 54]]]",
    "process": "From the given conditions, we have $ a=4 $, $ b=3 $, $ c=\\sqrt{a^{2}+b^{2}}=5 $. Hence, the eccentricity is $ e=\\frac{c}{a}=\\frac{5}{4} $."
  },
  {
    "text": "The circle with the minor axis of ellipse $C$ as its diameter passes through the foci of this ellipse. What is the eccentricity of ellipse $C$?",
    "fact_expressions": "C: Ellipse;G: Circle;IsDiameter(MinorAxis(C),G);PointOnCurve(Focus(C),G)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[1, 6], [17, 19], [24, 29]], [[13, 14]], [[1, 14]], [[13, 22]]]",
    "query_spans": "[[[24, 35]]]",
    "process": ""
  },
  {
    "text": "Given that the parabola $y^{2}=2 px$ passes through the point $M(2 , 2)$, what is the distance from point $M$ to the focus of the parabola?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;M: Point;Coordinate(M) = (2, 2);PointOnCurve(M, G)",
    "query_expressions": "Distance(M, Focus(G))",
    "answer_expressions": "5/2",
    "fact_spans": "[[[2, 17], [36, 39]], [[2, 17]], [[5, 17]], [[18, 29], [31, 35]], [[18, 29]], [[2, 29]]]",
    "query_spans": "[[[31, 46]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$, the right focus is $F$. A perpendicular is drawn from point $F$ to an asymptote, with foot of the perpendicular at $P$. The area of triangle $\\triangle O P F$ is $2$. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2 + x^2/a^2 = 1);a: Number;a>0;F: Point;RightFocus(G) = F;L: Line;PointOnCurve(F, L) = True;IsPerpendicular(L, OneOf(Asymptote(G))) = True;FootPoint(L, OneOf(Asymptote(G))) = P;P: Point;O: Origin;Area(TriangleOf(O, P, F)) = 2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(17)/4",
    "fact_spans": "[[[2, 39], [99, 102]], [[2, 39]], [[5, 39]], [[5, 39]], [[44, 47], [49, 53]], [[2, 47]], [], [[2, 62]], [[2, 62]], [[2, 69]], [[66, 70]], [[72, 89]], [[71, 96]]]",
    "query_spans": "[[[99, 108]]]",
    "process": "From the given conditions, b=1. Since a perpendicular is drawn from the focus F to an asymptote, with foot of perpendicular at P, we have OP=a, PF=b=1. Given that the area of quadrilateral AOPF is 2, then \\frac{1}{2}ab=2, so a=4. From c^{2}=a^{2}+b^{2}, we get c=\\sqrt{17}, hence e=\\frac{c}{a}=\\frac{\\sqrt{17}}{4}"
  },
  {
    "text": "Given that $F$ is the focus of the parabola $y^{2}=x$, and $A$, $B$ are two points on the parabola such that $|A F|+|B F|=3$, then the distance from the midpoint of segment $AB$ to the $y$-axis is?",
    "fact_expressions": "A: Point;B: Point;G: Parabola;F: Point;Expression(G) = (y^2 = x);Focus(G) = F;PointOnCurve(A, G);PointOnCurve(B, G);Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F)) = 3",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A, B)), yAxis)",
    "answer_expressions": "5/4",
    "fact_spans": "[[[22, 25]], [[26, 29]], [[6, 18], [31, 34]], [[2, 5]], [[6, 18]], [[2, 21]], [[22, 38]], [[22, 38]], [[39, 54]]]",
    "query_spans": "[[[56, 75]]]",
    "process": ""
  },
  {
    "text": "A point $A$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ has distance $4$ to the left focus $F$. Let $B$ be the midpoint of $AF$, and $O$ be the origin. Find the value of $|OB|$.",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);A: Point;PointOnCurve(A, G);F: Point;LeftFocus(G) = F;Distance(A, F) = 4;B: Point;MidPoint(LineSegmentOf(A, F)) = B;O: Origin",
    "query_expressions": "Abs(LineSegmentOf(O, B))",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 38]], [[0, 38]], [[41, 44]], [[0, 44]], [[48, 51]], [[0, 51]], [[41, 58]], [[59, 62]], [[59, 71]], [[72, 75]]]",
    "query_spans": "[[[82, 92]]]",
    "process": "According to the definition of the ellipse, |AF| = 10 - 4 = 6, and OB is the midline of $\\triangle AFF$, from which the value of |OB| can be found. Since the ellipse is $\\frac{x^{2}}{25} + \\frac{y^{2}}{9} = 1$, we have $a = 5$. Let the other focus of the ellipse be F, then |AF| = 10 - 4 = 6, and since OB is the midline of $\\triangle AFF$, it follows that $OB = \\frac{|AF|}{2} = 3$."
  },
  {
    "text": "The eccentricity of the hyperbola $x^{2}-y^{2}=a^{2}(a>0)$ is?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (x^2 - y^2 = a^2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 27]], [[3, 27]], [[3, 27]], [[0, 27]]]",
    "query_spans": "[[[0, 33]]]",
    "process": "From the given: x2 - y2 = a2 (a > 0), so the hyperbola equation is \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{a^{2}}=1, (a>0), c^{2}=2a^{2}, (c>0), thus e=\\frac{c}{a}=\\sqrt{2}"
  },
  {
    "text": "If the curve of the equation $x^{2}+k^{2} y^{2}-3 x-k y-4=0$ passes through the point $P(2,1)$, then $k=$?",
    "fact_expressions": "G: Curve;P: Point;Expression(G)=(x^2+k^2*y^2-3*x-k*y-4=0);Coordinate(P) = (2, 1);PointOnCurve(P,G);k:Number",
    "query_expressions": "k",
    "answer_expressions": "-2, 3",
    "fact_spans": "[[[35, 37]], [[38, 47]], [[1, 37]], [[38, 47]], [[35, 47]], [[49, 52]]]",
    "query_spans": "[[[49, 54]]]",
    "process": "From the given condition we have: $2^{2}+k^{2}\\times1^{2}-3\\times2-k\\times1-4=0$, that is: $k^{2}-k-6=0$. Solving this equation for the real number $k$ gives $k=-2$ or $k=3$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, respectively, and $P$ is a point on the right branch of the hyperbola such that $|P F_{2}|=|F_{1} F_{2}|$, and the line $P F_{1}$ intersects the circle $x^{2}+y^{2}=a^{2}$, then the maximum value of the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F2)) = Abs(LineSegmentOf(F1, F2));H: Circle;Expression(H) = (x^2 + y^2 = a^2);IsIntersect(LineOf(P, F1), H)",
    "query_expressions": "Max(Eccentricity(G))",
    "answer_expressions": "5/3",
    "fact_spans": "[[[20, 66], [77, 80], [152, 155]], [[20, 66]], [[23, 66]], [[23, 66]], [[2, 9]], [[10, 17]], [[2, 72]], [[2, 72]], [[73, 76]], [[73, 85]], [[88, 113]], [[126, 146]], [[126, 146]], [[114, 150]]]",
    "query_spans": "[[[152, 165]]]",
    "process": "Combining the properties of plane geometry, we obtain |F_{1}M|=\\frac{a+c}{2}, and then using the Pythagorean theorem, we find |OM|^{2}=c^{2}-(\\frac{a+c}{2})^{2}. Then, since the line PF_{1} intersects the circle x^{2}+y^{2}=a^{2} at a common point, we have |OM|^{2}\\leqslant a^{2}, thereby obtaining a homogeneous expression in a and c; solving the inequality yields the result. Let PF_{1} intersect at N; because |PF_{2}|=|F_{1}F_{2}|, it follows that |PN|=|F_{1}N|. Also, since |OF_{2}|=|F_{1}O|, we have |MN|=|F_{1}M|, hence |F_{1}M|=\\frac{1}{4}|F_{1}P|. Moreover, since |PF_{1}|-|PF_{2}|=2a and |PF_{2}|=|F_{1}F_{2}|=2c, it follows that |PF_{1}|=2a+2c, therefore |F_{1}M|=\\frac{a+c}{2}, so |OM|^{2}=c^{2}-(\\frac{a+c}{2})^{2}. Since the line PF_{1} has a common point with the circle x^{2}+y^{2}=a^{2}, we have |OM|^{2}\\leqslant a^{2}, thus c^{2}-(\\frac{a+c}{2})^{2}\\leqslant a^{2}, i.e., 3c^{2}-2ac-5a^{2}\\leqslant 0, which implies 3e^{2}-2e-5\\leqslant 0, so -1\\leqslant e\\leqslant \\frac{5}{3}. Since the eccentricity of a hyperbola satisfies e>1, we have 1<e\\leqslant \\frac{5}{3}; hence, the maximum eccentricity is \\frac{5}{3}."
  },
  {
    "text": "Given that the hyperbola passes through the point $(1,2)$ and has asymptotes with equations $y=\\pm \\sqrt{2} x$, then the focal length of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (1, 2);PointOnCurve(H, G);Expression(Asymptote(G)) = (y = pm*(sqrt(2)*x))",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[2, 5], [43, 46]], [[6, 14]], [[6, 14]], [[2, 14]], [[2, 40]]]",
    "query_spans": "[[[43, 51]]]",
    "process": ""
  },
  {
    "text": "The equation of the parabola with the center of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ as its vertex and the left directrix as its directrix is?",
    "fact_expressions": "H: Ellipse;Expression(H) = (x^2/9 + y^2/5 = 1);G: Parabola;Center(H) = Vertex(G);LeftDirectrix(H) = Directrix(G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=18*x",
    "fact_spans": "[[[1, 38]], [[1, 38]], [[52, 55]], [[0, 55]], [[0, 55]]]",
    "query_spans": "[[[52, 59]]]",
    "process": ""
  },
  {
    "text": "When the eccentricity of the hyperbola $M$: $\\frac{x^{2}}{m}-\\frac{y^{2}}{m^{2}+4}=1$ reaches its minimum value, what is the equation of the asymptotes of the hyperbola $M$?",
    "fact_expressions": "M: Hyperbola;m: Number;Expression(M) = (-y^2/(m^2 + 4) + x^2/m = 1);WhenMin(Eccentricity(M))",
    "query_expressions": "Expression(Asymptote(M))",
    "answer_expressions": "y=pm*2*x",
    "fact_spans": "[[[1, 49], [60, 66]], [[8, 49]], [[1, 49]], [[0, 59]]]",
    "query_spans": "[[[60, 74]]]",
    "process": "Hyperbola M: \\frac{x^{2}}{m}-\\frac{y^{2}}{m+4}=1, obviously m>0, and equality holds if and only if m=2; at this time, hyperbola M: \\frac{x^{2}}{2}-\\frac{y^{2}}{8}=1, then the asymptote equations are: y=\\pm 2x"
  },
  {
    "text": "If the foci of the ellipse $\\frac{x^{2}}{3-k}+\\frac{y^{2}}{1+k}=1$ lie on the $x$-axis, then the range of values for $k$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(3 - k) + y^2/(k + 1) = 1);k: Number;PointOnCurve(Focus(G), xAxis) = True",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-1,1)",
    "fact_spans": "[[[1, 42]], [[1, 42]], [[53, 56]], [[1, 51]]]",
    "query_spans": "[[[53, 63]]]",
    "process": "From the given condition: $3 - k > 1 + k > 0 \\Rightarrow -1 < k < 1$"
  },
  {
    "text": "For any point $P$ on the parabola $y^{2}=4 x$, draw tangent lines to the circle $(x-4)^{2}+y^{2}=2$, with points of tangency denoted by $A$. Then the minimum value of $|P A|$ equals?",
    "fact_expressions": "G: Parabola;H: Circle;P: Point;A: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (y^2 + (x - 4)^2 = 2);PointOnCurve(P, G);l: Line;TangentOfPoint(P, H) = l;TangentPoint(l, H) = A",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[1, 15]], [[24, 44]], [[20, 23]], [[51, 54]], [[1, 15]], [[24, 44]], [[1, 23]], [], [[0, 47]], [[0, 54]]]",
    "query_spans": "[[[56, 70]]]",
    "process": "Let $ P\\left(\\frac{y^{2}}{4}, y\\right) $, $ (x-4)^{2}+y^{2}=2 $, circle center $ C(4,0) $, radius $ r=2 $. $ |PA|^{2} = |PC|^{2} - r^{2} = \\left(\\frac{y^{2}}{4}-4\\right)^{2} + y^{2} - 2 = \\frac{1}{16}(y^{2}-8)^{2} + 10 \\geqslant 10 $, equality holds if and only if $ y = \\pm 2\\sqrt{2} $, that is, when point $ P(2, \\pm 2\\sqrt{2}) $ is taken. Hence, the minimum value of $ |PA| $ equals $ \\sqrt{10} $."
  },
  {
    "text": "Given that the length of the imaginary axis of the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{b^{2}}=1$ is $2$, its eccentricity is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2/b^2 = 1);b: Number;Length(ImageinaryAxis(G)) = 2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 44], [53, 54]], [[2, 44]], [[5, 44]], [[2, 52]]]",
    "query_spans": "[[[54, 59]]]",
    "process": "The hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{b^{2}}=1$ has an imaginary axis length of 2, so $b=1$, $a=\\sqrt{3}$, and thus $c=2$. Therefore, the eccentricity of the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{1}=1$ is: $e=\\frac{c}{a}=\\frac{2}{\\sqrt{3}}=\\frac{2\\sqrt{3}}{3}$"
  },
  {
    "text": "Given that point $F$ is the focus of the parabola $y = \\frac{1}{4}x^{2}$, point $A$ is an arbitrary point on the parabola other than the origin, line $AF$ intersects the parabola at point $B$, tangents to the parabola are drawn at points $A$ and $B$ respectively, and the two tangents intersect at point $P$. A circle $\\odot M$ is drawn with $AB$ as diameter, where $M$ is the center of the circle. Then the minimum length of segment $PM$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = (1/4)*x^2);F: Point;Focus(G) = F;A: Point;PointOnCurve(A, G);O: Origin;Negation(A=O);B: Point;Intersection(LineOf(A, F), G) = B;l1: Line;l2: Line;P: Point;TangentOnPoint(A, G) = l1;TangentOfPoint(B, G) = l2;Intersection(l1, l2) = P;M1: Circle;IsDiameter(LineSegmentOf(A, B), M1);M: Point;Center(M1) = M",
    "query_expressions": "Min(Length(LineSegmentOf(P, M)))",
    "answer_expressions": "2",
    "fact_spans": "[[[7, 31], [40, 43], [62, 65], [84, 87]], [[7, 31]], [[2, 6]], [[2, 34]], [[35, 39], [75, 79]], [[35, 53]], [[46, 48]], [[35, 53]], [[66, 70], [80, 83]], [[54, 70]], [], [], [[97, 101]], [[74, 90]], [[74, 90]], [[74, 101]], [[113, 122]], [[103, 122]], [[124, 127]], [[113, 130]]]",
    "query_spans": "[[[132, 147]]]",
    "process": ""
  },
  {
    "text": "If the line $x - m y + m = 0$ passes through the focus of the parabola $x^{2} = 2 p y$ ($p > 0$), then $p =$?",
    "fact_expressions": "H: Line;Expression(H) = (m - m*y + x = 0);m: Number;G: Parabola;Expression(G) = (x^2 = 2*(p*y));p: Number;p>0;PointOnCurve(Focus(G), H)",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 14]], [[1, 14]], [[3, 14]], [[16, 37]], [[16, 37]], [[42, 45]], [[19, 37]], [[1, 40]]]",
    "query_spans": "[[[42, 47]]]",
    "process": "\\because the line x-my+m=0 can be rewritten as x-m(y-1)=0, so the line x-my+m=0 passes through the point (0,1), which is the focus F of the parabola x^{2}=2py(p>0). \\therefore \\frac{p}{2}=1, then p=2,"
  },
  {
    "text": "Given that the distance from point $M$ to point $F(4 , 0)$ is $1$ less than its distance to the line $l$: $x+5=0$, and the coordinates of point $A$ are $(2 , 3)$, then the minimum value of $MA + MF$ is?",
    "fact_expressions": "l: Line;F: Point;M: Point;A: Point;Expression(l)=(x+5=0);Coordinate(A) = (2, 3);Coordinate(F) = (4, 0);Distance(M,F)+1=Distance(M,l)",
    "query_expressions": "Min(LineSegmentOf(M, A) + LineSegmentOf(M, F))",
    "answer_expressions": "6",
    "fact_spans": "[[[24, 38]], [[7, 18]], [[2, 6], [22, 23]], [[46, 50]], [[24, 38]], [[46, 63]], [[7, 18]], [[2, 45]]]",
    "query_spans": "[[[65, 80]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ is a moving point on the parabola $y^{2}=2x$, the distance from point $P$ to the directrix is $d$, and the projection of point $P$ on the $y$-axis is $M$. Point $A\\left(\\frac{7}{2}, 4\\right)$. Then the minimum value of $|PA|+|PM|$ is?",
    "fact_expressions": "G: Parabola;P: Point;A: Point;M: Point;d: Number;Expression(G) = (y^2 = 2*x);Coordinate(A) = (7/2, 4);PointOnCurve(P, G);Distance(P, Directrix(G)) = d;Projection(P, yAxis) = M",
    "query_expressions": "Min(Abs(LineSegmentOf(P,A))+Abs(LineSegmentOf(P,M)))",
    "answer_expressions": "7/2",
    "fact_spans": "[[[7, 21]], [[2, 6], [26, 30], [42, 46]], [[60, 81]], [[56, 59]], [[37, 40]], [[7, 21]], [[60, 81]], [[2, 25]], [[7, 40]], [[42, 59]]]",
    "query_spans": "[[[83, 100]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $M$: $x^{2}-\\frac{y^{2}}{b^{2}}=1$, draw a line $l$ with slope $1$ passing through the left vertex $A$ of the hyperbola. If $l$ intersects the two asymptotes of hyperbola $M$ at points $B$ and $C$, and $|AB|=|BC|$, then the eccentricity of hyperbola $M$ is?",
    "fact_expressions": "M: Hyperbola;Expression(M) = (x^2 - y^2/b^2 = 1);b: Number;A: Point;LeftVertex(M) = A;l: Line;Slope(l) = 1;PointOnCurve(A, l);B: Point;C: Point;Abs(LineSegmentOf(A, B)) = Abs(LineSegmentOf(B, C));L1:Line;L2:Line;Asymptote(M)={L1,L2};Intersection(l,L1)=B;Intersection(l,L2)=C",
    "query_expressions": "Eccentricity(M)",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[1, 38], [65, 71], [106, 112]], [[1, 38]], [[8, 38]], [[42, 45]], [[1, 45]], [[53, 58], [61, 64]], [[46, 58]], [[0, 58]], [[80, 83]], [[84, 87]], [[90, 103]], [-1, -1], [-1, -1], [64, 76], [[61, 87]], [[61, 87]]]",
    "query_spans": "[[[106, 118]]]",
    "process": ""
  },
  {
    "text": "The length of the segment cut from the line $y=x+\\frac{3}{2}$ by the curve $y=\\frac{1}{2} x^{2}$ is?",
    "fact_expressions": "G: Line;Expression(G) = (y = x + 3/2);H: Curve;Expression(H) = (y = (1/2)*x^2)",
    "query_expressions": "Length(InterceptChord(G, H))",
    "answer_expressions": "4*sqrt(2)",
    "fact_spans": "[[[0, 19]], [[0, 19]], [[20, 43]], [[20, 43]]]",
    "query_spans": "[[[0, 51]]]",
    "process": "Let the intersection points be A(x_{1},y_{1}), B(x_{2},y_{2}). Solving the system of equations gives \\begin{cases}y=x+\\frac{3}{2}\\\\y=\\frac{1}{2}x^{2}\\end{cases}\\Rightarrow x^{2}-2x-3=0, yielding A(-1,\\frac{1}{2}), B(3,\\frac{9}{2}), so |AB|=\\sqrt{4^{2}+4^{2}}=4\\sqrt{2}"
  },
  {
    "text": "The equation of an ellipse with foci on the $x$-axis is $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. Connecting one endpoint of the minor axis with the two foci forms a triangle, whose incircle has radius $\\frac{b}{3}$. Then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;a:Number;b:Number;F1:Point;F2:Point;P:Point;a>b;b>0;PointOnCurve(Focus(G), xAxis);Expression(G) = (y^2/b^2 + x^2/a^2 = 1);OneOf(Endpoint(MinorAxis(G)))=P;Focus(G)={F1,F2};Radius(InscribedCircle(TriangleOf(P,F1,F2)))=b/3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[9, 11], [113, 115]], [[14, 64]], [[98, 111]], [], [], [], [[14, 64]], [[14, 64]], [[0, 11]], [[9, 64]], [[9, 72]], [[9, 77]], [[9, 111]]]",
    "query_spans": "[[[113, 121]]]",
    "process": "By the given conditions, as shown in the figure: from the properties of the ellipse, AB = 2c, AC = BC = a, OC = b, S_{\\triangle ABC} = \\frac{1}{2}AB \\cdot OC = \\frac{1}{2} \\cdot 2c \\cdot b = bc, so S_{\\Delta ABC} = \\frac{1}{2}(a + a + 2c) \\times \\frac{b}{3} = \\frac{b(a + c)}{3}, \\frac{b(a + c)}{3} = bc, a = 2c^{n}, hence the eccentricity of the ellipse e = \\frac{c}{a} = \\frac{1}{2}"
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. If $\\overrightarrow{A F}=2 \\overrightarrow{F B}$, then $|A F|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);A: Point;B: Point;Intersection(H, G) = {A, B};VectorOf(A, F) = 2*VectorOf(F, B)",
    "query_expressions": "Abs(LineSegmentOf(A, F))",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 15], [25, 28]], [[1, 15]], [[18, 21]], [[1, 21]], [[22, 24]], [[0, 24]], [[29, 32]], [[33, 36]], [[22, 38]], [[41, 86]]]",
    "query_spans": "[[[89, 98]]]",
    "process": "A line passing through the focus $ F $ of the parabola $ y^{2} = 4x $ intersects the parabola at points $ A $ and $ B $, and $ \\overrightarrow{AF} = 2\\overrightarrow{FB} $. The slope of the line exists; let the line $ AB $ be $ y = k(x - 1) $. Thus,\n$$\n\\begin{cases}\ny = k(x - 1) \\\\\ny^{2} = 4x\n\\end{cases}\n$$\nRearranging gives $ k^{2}x^{2} - (2k^{2} + 4)x + k^{2} = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1}x_{2} = 1 $ (1). From $ \\overrightarrow{AF} = 2\\overrightarrow{FB} $, we have $ x_{1} + 1 = 2(x_{2} + 1) $ (2). Solving (1) and (2) together yields $ x_{1} = 2 $ or $ x_{1} = -1 $ (discarded). Therefore, $ |AF| = x_{1} + \\frac{p}{2} = x_{1} + 1 = 3 $."
  },
  {
    "text": "The standard equation of the ellipse passing through points $P(-3 , 0)$, $Q(0,-2)$ is?",
    "fact_expressions": "P: Point;Coordinate(P) = (-3, 0);Q: Point;Coordinate(Q) = (0, -2);G: Ellipse;PointOnCurve(P, G);PointOnCurve(Q, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9+y^2/4=1",
    "fact_spans": "[[[2, 14]], [[2, 14]], [[17, 26]], [[17, 26]], [[27, 29]], [[1, 29]], [[1, 29]]]",
    "query_spans": "[[[27, 36]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively. A circle centered at $F_{2}$ with radius $|F_{1} F_{2}|$ intersects the right branch of hyperbola $C$ at points $A$ and $B$. If $|A B|=\\sqrt{3}|F_{1} F_{2}|$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;G: Circle;Center(G) = F2;Radius(G) = Abs(LineSegmentOf(F1, F2));A: Point;B: Point;Intersection(G, RightPart(C)) = {A, B};Abs(LineSegmentOf(A, B)) = sqrt(3)*Abs(LineSegmentOf(F1, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(\\sqrt{3}+1)/2",
    "fact_spans": "[[[20, 81], [121, 127], [173, 179]], [[20, 81]], [[28, 81]], [[28, 81]], [[28, 81]], [[28, 81]], [[2, 9]], [[10, 17], [89, 96]], [[2, 87]], [[2, 87]], [[119, 120]], [[88, 120]], [[100, 120]], [[131, 134]], [[135, 138]], [[119, 140]], [[142, 171]]]",
    "query_spans": "[[[173, 185]]]",
    "process": "Let AB intersect the x-axis at point H, then |AH| = \\sqrt{3}c, so \\angle AF_{2}H = 60^{\\circ}. Therefore, \\angle AF_{1}H = 30^{\\circ}, so |AF_{1}| = 2\\sqrt{3}c, so 2\\sqrt{3}c - 2c = 2a."
  },
  {
    "text": "The focus of the parabola $y^{2}=4 x$ is $F$. A line $l$ passing through $F$ intersects the parabola at points $A$ and $B$. The directrix intersects the $x$-axis at $K$. If $|A F|=2|F K|$, then $|A B|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, G) = {A, B};K: Point;Intersection(Directrix(G), xAxis) = K;Abs(LineSegmentOf(A, F)) = 2*Abs(LineSegmentOf(F, K))",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[0, 14], [33, 36]], [[0, 14]], [[18, 21], [23, 26]], [[0, 21]], [[27, 32]], [[22, 32]], [[38, 41]], [[42, 45]], [[27, 47]], [[56, 59]], [[33, 59]], [[61, 75]]]",
    "query_spans": "[[[77, 86]]]",
    "process": "Draw the graph, use the definition of the parabola to find the slope of line $ l $, then find the equation of line $ l $. Solve the system of equations of line $ l $ and the parabola, write Vieta's formulas, and use the chord length formula for the intersection of a line and a parabola to find $ |AB| $. By symmetry of the parabola, assume without loss of generality that the slope $ k > 0 $. Draw $ A_{1}A \\perp $ directrix at $ A_{1} $, and draw $ FE \\perp AA_{1} $ at $ E $. Then $ |A_{1}E| = |FK| = \\frac{1}{2}|AF| = \\frac{1}{2}|AA_{1}| = |AE| $. Since $ |AE| = \\frac{1}{2}|AF| $ and $ EF \\perp AA_{1} $, it follows that $ \\angle EAF = \\frac{\\pi}{3} $, $ \\angle AFx = \\frac{\\pi}{3} $, so the slope of the line is $ k = \\sqrt{3} $. The equation of the line passing through $ F $ is $ y = \\sqrt{3}(x - 1) $. From \n\\[\n\\begin{cases}\ny^2 = 4x \\\\\ny = \\sqrt{3}(x - 1)\n\\end{cases}\n\\]\nwe obtain $ 3x^{2} - 10x + 3 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = \\frac{10}{3} $, so $ |AB| = x_{1} + x_{2} + 2 = \\frac{16}{3} $."
  },
  {
    "text": "Given $A(1,2)$, $B(-1,2)$, a moving point $P$ satisfies $\\overrightarrow{A P} \\perp \\overrightarrow{B P}$. If the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ have no common points with the trajectory of the moving point $P$, then the range of the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;A: Point;B: Point;P: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (1, 2);Coordinate(B) = (-1, 2);IsPerpendicular(VectorOf(B, P),VectorOf(A, P));NumIntersection(Asymptote(G),Locus(P))=0",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1, 2)",
    "fact_spans": "[[[81, 137], [157, 160]], [[84, 137]], [[84, 137]], [[2, 10]], [[12, 21]], [[24, 27], [144, 147]], [[84, 137]], [[84, 137]], [[81, 137]], [[2, 10]], [[12, 21]], [[29, 78]], [[81, 155]]]",
    "query_spans": "[[[157, 170]]]",
    "process": "The trajectory equation of the moving point P is $x^{2}+(y-2)^{2}=1$, and one asymptote of the hyperbola is $bx-ay=0$. Since it is separate from the circle, we have $\\frac{|0\\times b-a\\times 2|}{\\sqrt{a^{2}+b^{2}}}>1$, that is, $\\frac{2a}{c}>1$, which implies $\\frac{c}{a}<2$. Therefore, the eccentricity of the hyperbola is $(1,2)$."
  },
  {
    "text": "Given that point $P$ is a point on the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$, and the area of the triangle with vertices at point $P$ and the foci $F_{1}$, $F_{2}$ is equal to $2$, then the coordinates of point $P$ are?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/5 + y^2/4 = 1);PointOnCurve(P, G);Area(TriangleOf(P,F1,F2))=2;F1:Point;F2:Point;Focus(G)={F1,F2}",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(0,pm*2)",
    "fact_spans": "[[[7, 44]], [[2, 6], [51, 55], [90, 94]], [[7, 44]], [[2, 48]], [[50, 88]], [[58, 65]], [[66, 73]], [[7, 73]]]",
    "query_spans": "[[[90, 99]]]",
    "process": "Given that the focal distance of the ellipse |F_{1}F_{2}| = 2. Since the area of \\triangle PF_{1}F_{2} equals 2, it follows that \\frac{1}{2} \\times |F_{1}F_{2}| \\times |y_{p}| = 2, solving gives |y_{p}| = 2. Substituting into the equation of the ellipse yields \\frac{x^{2}}{5} + 1 = 1, solving gives x = 0. Therefore, the coordinates of point P are (0, \\pm2)."
  },
  {
    "text": "Given that the point $(x , y)$ lies on the parabola $y^{2}=4 x$, then the minimum value of $z=x^{2}+\\frac{1}{2} y^{2}+3$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Point;x0: Number;y0: Number;Coordinate(H) = (x0, y0);PointOnCurve(H, G);z: Number;z=x0^2+(1/2)*y0^2+3",
    "query_expressions": "Min(z)",
    "answer_expressions": "3",
    "fact_spans": "[[[13, 27]], [[13, 27]], [[2, 12]], [[3, 12]], [[3, 12]], [[2, 12]], [[2, 28]], [[30, 59]], [[30, 59]]]",
    "query_spans": "[[[30, 65]]]",
    "process": "Substitute the parabola equation and complete the square using the properties of quadratic functions to find the extremum. Since the point (x, y) lies on the parabola y^{2} = 4x, we have x \\geqslant 0. Because z = x^{2} + \\frac{1}{2}y^{2} + 3 = x^{2} + 2x + 3 = (x+1)^{2} + 2, when x = 0, z is minimized and its value is 3."
  },
  {
    "text": "A line passing through a focus $F_1$ of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ intersects the ellipse at points $A$ and $B$. Then, points $A$, $B$, and the other focus $F_2$ of the ellipse form $\\triangle AB{F}_{2}$. What is the perimeter of $\\triangle A B F_{2}$?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2/16 + y^2/9 = 1);OneOf(Focus(G)) = F1;PointOnCurve(F1, H);Intersection(H, G) = {A, B};OneOf(Focus(G)) = F2;Negation(F1 = F2)",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "16",
    "fact_spans": "[[[1, 39], [55, 57], [78, 80]], [[52, 54]], [[59, 62], [70, 73]], [[63, 66], [74, 77]], [[87, 94]], [[44, 51]], [[1, 39]], [[1, 51]], [[0, 54]], [[52, 68]], [[78, 94]], [[0, 94]]]",
    "query_spans": "[[[119, 145]]]",
    "process": ""
  },
  {
    "text": "If the vertices of a hyperbola are the endpoints of the major axis of the ellipse $x^{2}+\\frac{y^{2}}{2}=1$, and the product of the eccentricity of the hyperbola and the eccentricity of the ellipse is $1$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2 + y^2/2 = 1);Vertex(G) = Endpoint(MajorAxis(H));Eccentricity(G)*Eccentricity(H) = 1",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/2 - x^2/2 = 1",
    "fact_spans": "[[[1, 4], [42, 45], [65, 68]], [[8, 35], [51, 53]], [[8, 35]], [[1, 40]], [[42, 63]]]",
    "query_spans": "[[[65, 73]]]",
    "process": "The endpoints of the major axis of the ellipse $x^{2}+\\frac{y^{2}}{2}=1$ are $(0,\\pm\\sqrt{2})$, so the vertices of the hyperbola are $(0,\\pm\\sqrt{2})$, $\\therefore a=\\sqrt{2}$, the eccentricity of the ellipse is $e=\\frac{1}{\\sqrt{2}}$, so the eccentricity of the hyperbola is $\\sqrt{2}$, therefore the equation of the hyperbola is $\\frac{y^{2}}{2}-\\frac{x^{2}}{2}=1$."
  },
  {
    "text": "It is known that the foci and vertices of hyperbola $C$ are exactly the endpoints of the major axis and the foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, respectively. Then the asymptotes of hyperbola $C$ have the equation?",
    "fact_expressions": "C: Hyperbola;G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);Focus(C)=Endpoint(MajorAxis(G));Vertex(C)=Focus(G)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "4*x+pm*3*y=0",
    "fact_spans": "[[[2, 8], [68, 74]], [[19, 58]], [[19, 58]], [[2, 66]], [[2, 66]]]",
    "query_spans": "[[[68, 82]]]",
    "process": "From the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, we know the endpoints of the major axis are $(-5,0)$ and $(5,0)$, and the foci are $(-3,0)$, $(3,0)$. Thus, the foci of the hyperbola are $(-5,0)$, $(5,0)$, and the vertices are $(-3,0)$, $(3,0)$. Therefore, the equation of the hyperbola is $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, so the asymptotes are $4x\\pm3y=0$."
  },
  {
    "text": "The parabola $y^{2}=4 x$, line $l$ passes through the focus $F$, intersects it at points $A$ and $B$, and $\\overrightarrow{BA}=4 \\overrightarrow{BF}$. Then the area of $\\triangle AOB$ ($O$ is the coordinate origin) is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);l: Line;F: Point;Focus(G) = F;PointOnCurve(F, l) = True;Intersection(l, G) = {A, B};A: Point;B: Point;VectorOf(B, A) = 4*VectorOf(B, F);O: Origin",
    "query_expressions": "Area(TriangleOf(A, O, B))",
    "answer_expressions": "4*sqrt(3)/3",
    "fact_spans": "[[[0, 14], [28, 29]], [[0, 14]], [[15, 20]], [[23, 26]], [[0, 26]], [[15, 26]], [[15, 41]], [[31, 34]], [[36, 39]], [[43, 86]], [[104, 107]]]",
    "query_spans": "[[[88, 118]]]",
    "process": "The focus of the parabola $ y^{2} = 4x $ is $ (1,0) $. Let the line $ l $ be $ x = my + 1 $. Substituting into the parabola equation yields $ y^{2} - 4my - 4 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ y_{1} + y_{2} = 4m $, $ y_{1}y_{2} = -4 $. From $ \\overrightarrow{BA} = 4\\overrightarrow{BF} $, we get $ y_{1} = -3y_{2} $. By substitution, we obtain $ m^{2} = \\frac{1}{3} $. Therefore, the area of $ \\triangle AOB $ is $ S = \\frac{1}{2} |OF| |y_{1} - y_{2}| = \\frac{1}{2} \\sqrt{(y_{1} + y_{2})^{2} - 4y_{1}y_{2}} = \\frac{1}{2} \\sqrt{16m^{2} + 16} = \\frac{4\\sqrt{3}}{3} $."
  },
  {
    "text": "Given a point $P$ on the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{25}=1$ such that the distance from $P$ to one focus of the ellipse is $3$, then the distance from point $P$ to the other focus is?",
    "fact_expressions": "G: Ellipse;P: Point;F1:Point;F2:Point;Expression(G) = (x^2/36 + y^2/25 = 1);PointOnCurve(P, G);Distance(P,F1) = 3;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2)",
    "query_expressions": "Distance(P,F2)",
    "answer_expressions": "9",
    "fact_spans": "[[[2, 41], [48, 50]], [[44, 47], [63, 67]], [], [], [[2, 41]], [[2, 47]], [[2, 61]], [[48, 54]], [[48, 73]], [[48, 73]]]",
    "query_spans": "[[[48, 78]]]",
    "process": ""
  },
  {
    "text": "Given that a line $l$ passing through the origin $O$ intersects the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ at two distinct points $A$, $B$, and $F$ is the left focus of the hyperbola $C$, such that $|A F|=\\frac{5}{3}|B F|$ and $|O A|=b$, then the eccentricity of $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;A: Point;F: Point;B: Point;O: Origin;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(O, l);Intersection(l, C) = {A, B};Negation(A = B);LeftFocus(C) = F;Abs(LineSegmentOf(A, F)) = (5/3)*Abs(LineSegmentOf(B, F));Abs(LineSegmentOf(O, A)) = b",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "3",
    "fact_spans": "[[[9, 14]], [[15, 76], [95, 101], [145, 148]], [[23, 76]], [[23, 76]], [[83, 86]], [[91, 94]], [[87, 90]], [[3, 8]], [[23, 76]], [[23, 76]], [[15, 76]], [[2, 14]], [[9, 90]], [[78, 90]], [[91, 105]], [[109, 133]], [[134, 143]]]",
    "query_spans": "[[[145, 154]]]",
    "process": "Let the right focus of the hyperbola be $ F_{1} $. Connect $ AF_{1} $, $ BF_{1} $ as shown in the figure. Since both the line $ y = kx $ and the hyperbola are symmetric about the origin, the quadrilateral $ AFBF_{1} $ is a parallelogram. Thus, $ |BF| = |AF_{1}| $. By the definition of the hyperbola, we have: $ |AF| - |AF_{1}| = |AF| - |BF| = \\frac{2}{3}|BF| = 2a $, so $ |BF| = 3a $, $ |AF| = 5a $. In $ \\triangle BFA $, $ \\cos\\angle BFA = \\frac{25a^{2} + 9a^{2} - 4b^{2}}{2 \\times 5a \\times 3a} $. In $ \\triangle FAF_{1} $, $ \\cos\\angle FAF_{1} = \\frac{25a^{2} + 9a^{2} - 4c^{2}}{2 \\times 5a \\times 3a} $. Since $ -\\frac{4c^{2}}{3a} = -\\cos\\angle BFA $, simplifying and rearranging gives: $ 17a^{2} = b^{2} + c^{2} $. Using $ b^{2} = c^{2} - a^{2} $, we obtain $ c^{2} = 9a^{2} $, hence $ e = \\sqrt{\\frac{c^{2}}{a^{2}}} = 3 $."
  },
  {
    "text": "Given points $A(1,2)$, $B(x, y)(-6<y<-2)$ both lie on the parabola $C$: $y^{2}=2 p x(p>0)$, then the range of the slope of line $AB$ is?",
    "fact_expressions": "C: Parabola;p: Number;A: Point;B: Point;x1:Number;y1:Number;y1>-6;y1<-2;p>0;Expression(C) = (y^2 = 2*p*x);Coordinate(A) = (1, 2);Coordinate(B) = (x1, y1);PointOnCurve(A,C);PointOnCurve(B, C)",
    "query_expressions": "Range(Slope(LineOf(A,B)))",
    "answer_expressions": "(-\\infty,-1)",
    "fact_spans": "[[[34, 59]], [[41, 59]], [[2, 11]], [[13, 32]], [[13, 32]], [[13, 32]], [[13, 32]], [[13, 32]], [[41, 59]], [[34, 59]], [[2, 11]], [[14, 32]], [[2, 60]], [[2, 60]]]",
    "query_spans": "[[[62, 79]]]",
    "process": "Since point A(1,2) lies on the parabola, we have 4 = 2p, so the equation of the parabola is y^{2} = 4x. Thus, k_{AB} = \\frac{y-2}{x-1} = \\frac{y-2}{\\frac{y^{2}}{4}-1} = \\frac{4}{y+2}. Since -6 < y < -2, it follows that -4 < y+2 < 0, so k_{AB} = \\frac{4}{y+2} \\in (-\\infty, -1). The answer is: (-\\infty, -1)"
  },
  {
    "text": "A line $l$ passing through the focus of the parabola $y^{2}=8x$ intersects the parabola at points $A$ and $B$. If the radius of the circle with $AB$ as diameter is $8$, then what is the inclination angle of line $l$?",
    "fact_expressions": "l: Line;G: Parabola;H: Circle;A: Point;B: Point;Expression(G) = (y^2 = 8*x);PointOnCurve(Focus(G), l);Intersection(l, G) = {A, B};IsDiameter(LineSegmentOf(A,B),H);Radius(H)=8",
    "query_expressions": "Inclination(l)",
    "answer_expressions": "{ApplyUnit(45,degree),ApplyUnit(135,degree)}",
    "fact_spans": "[[[21, 26], [62, 67]], [[1, 15], [27, 30]], [[52, 53]], [[31, 34]], [[35, 38]], [[1, 15]], [[0, 26]], [[21, 40]], [[42, 53]], [[52, 60]]]",
    "query_spans": "[[[62, 73]]]",
    "process": "Classify and discuss the case when the slope of the line does not exist. When the radius of the circle is 8, find the slope of the line, thereby obtaining the inclination angle of the line. [Detailed solution] The focus coordinates of the parabola $ y^{2}=8x $ are $ (2,0) $. Therefore, when the slope of line $ l $ does not exist, the radius of the circle with $ AB $ as diameter is 4, which does not meet the condition. When the slope of line $ l $ exists, let the equation of line $ l $ be $ y=k(x-2) $. Substituting into $ y^{2}=8x $ yields $ k^{2}x^{2}-(4k^{2}+8)x+4k^{2}=0 $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ |AB|=x_{1}+x_{2}+p=\\frac{4k^{2}+8}{k^{2}}+4=16 $. Solving gives $ k=\\pm1 $, so the inclination angle of line $ l $ is $ 45^{\\circ} $ or $ 135^{\\circ} $."
  },
  {
    "text": "A point $P$ on the ellipse $\\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1$ is such that the lines connecting $P$ to the two foci $F_{1}$ and $F_{2}$ of the ellipse are perpendicular to each other. Then, the area of $\\Delta PF_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2:Point;Expression(G) = (x^2/49 + y^2/24 = 1);PointOnCurve(P,G);Focus(G)={F1,F2};IsPerpendicular(LineSegmentOf(P,F1),LineSegmentOf(P,F2))",
    "query_expressions": "Area(TriangleOf(P,F1,F2))",
    "answer_expressions": "24",
    "fact_spans": "[[[0, 39], [46, 48]], [[42, 45]], [[53, 60]], [[61, 68]], [[0, 39]], [[0, 45]], [[46, 68]], [[42, 75]]]",
    "query_spans": "[[[77, 103]]]",
    "process": ""
  },
  {
    "text": "It is known that the directrix of the parabola $y^{2}=8x$ passes through a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and the eccentricity of the hyperbola is $2$. Then the equation of the hyperbola is?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 8*x);G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;PointOnCurve(OneOf(Focus(G)), Directrix(H)) = True;Eccentricity(G) = 2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[20, 76], [84, 87], [98, 101]], [[20, 76]], [[23, 76]], [[23, 76]], [[23, 76]], [[23, 76]], [[2, 81]], [[84, 95]]]",
    "query_spans": "[[[98, 106]]]",
    "process": ""
  },
  {
    "text": "It is known that one focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ coincides with the focus of the parabola $x=\\frac{1}{4} y^{2}$, and the eccentricity of the hyperbola is $\\sqrt{5}$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x = y^2/4);OneOf(Focus(G))=Focus(H);Eccentricity(G) = sqrt(5)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "5*x^2-(5/4)*y^2=1",
    "fact_spans": "[[[2, 48], [85, 88], [107, 110]], [[5, 48]], [[5, 48]], [[54, 78]], [[2, 48]], [[54, 78]], [[2, 83]], [[85, 104]]]",
    "query_spans": "[[[107, 115]]]",
    "process": "From the parabola $ x = \\frac{1}{4}y^{2} $, the focus has coordinates $ F(1,0) $, so $ c = 1 $, and thus $ a^{2} + b^{2} = 1 $. Given that the eccentricity of the hyperbola is $ \\sqrt{5} $, then $ \\frac{c}{a} = \\sqrt{5} $, which implies $ a^{2} + b^{2} = 5a^{2} $. Solving gives $ a^{2} = \\frac{1}{5} $, $ b^{2} = \\frac{4}{5} $. Therefore, the equation of the hyperbola is $ 5x^{2} - \\frac{5}{4}y^{2} = 1 $."
  },
  {
    "text": "If the hyperbola $C$: $\\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1$, then what is the distance from the right focus of the hyperbola $C$ to its asymptote?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/6 - y^2/3 = 1)",
    "query_expressions": "Distance(RightFocus(C), Asymptote(C))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[1, 44], [46, 52], [57, 58]], [[1, 44]]]",
    "query_spans": "[[[46, 66]]]",
    "process": "The right focus of the hyperbola $ C: \\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1 $ is $ (3,0) $, and its asymptotes have equations $ y=\\pm\\frac{\\sqrt{2}}{2}x $. The distance from the right focus of hyperbola $ C $ to its asymptote is $ \\frac{\\sqrt{2}\\times3}{\\sqrt{1+\\frac{1}{2}}}=\\sqrt{3} $."
  },
  {
    "text": "The equation of the hyperbola passing through the point $(2 \\sqrt{3}, \\sqrt{3})$ and having the same asymptotes as the hyperbola $C$: $y^{2}-\\frac{x^{2}}{2}=1$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-x^2/2 + y^2 = 1);Z: Hyperbola;H: Point;Coordinate(H) = (2*sqrt(3), sqrt(3));PointOnCurve(H, Z);Asymptote(C) = Asymptote(Z)",
    "query_expressions": "Expression(Z)",
    "answer_expressions": "x^2/6 - y^2/3 = 1",
    "fact_spans": "[[[31, 64]], [[31, 64]], [[71, 74]], [[1, 26]], [[1, 26]], [[0, 74]], [[27, 74]]]",
    "query_spans": "[[[71, 78]]]",
    "process": "From the given conditions, suppose the equation of the hyperbola with the same asymptotes as the hyperbola $ C: y^{2} - \\frac{x^{2}}{2} = 1 $ is $ y^{2} - \\frac{x^{2}}{2} = \\lambda $ ($ \\lambda \\neq 0 $). Substituting the point $ (2\\sqrt{3}, \\sqrt{3}) $ into this equation allows us to find $ \\lambda $, and thus obtain the result. [Detailed solution] According to the problem, the asymptote equations of the hyperbola $ C: y^{2} - \\frac{x^{2}}{2} = 1 $ are $ y = \\pm \\frac{\\sqrt{2}}{2}x $. Therefore, the required hyperbola equation is $ y^{2} - \\frac{x^{2}}{2} = \\lambda $ ($ \\lambda \\neq 0 $). Since it passes through the point $ (2\\sqrt{3}, \\sqrt{3}) $, substituting into the equation gives $ \\lambda = -3 $. Hence, the hyperbola equation is $ \\frac{x^{2}}{6} - \\frac{y^{2}}{3} = 1 $."
  },
  {
    "text": "Given a hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, a line $l$ passing through the left focus $F_{1}$ intersects the asymptote of the hyperbola in the first quadrant at point $P$, and is tangent to the circle $x^{2}+y^{2}=a^{2}$ at point $Q$. If $|P Q|=2|F_{1} Q|$, then the eccentricity $e$ is?",
    "fact_expressions": "l: Line;E: Hyperbola;b: Number;a: Number;G: Circle;P: Point;Q: Point;F1: Point;a>0;b>0;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = a^2);LeftFocus(E) = F1;PointOnCurve(F1, l);Intersection(l, Asymptote(E)) = P;Quadrant(P) = 1;TangentPoint(l, G) = Q;Abs(LineSegmentOf(P, Q)) = 2*Abs(LineSegmentOf(F1, Q));e: Number;Eccentricity(E) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[74, 79]], [[1, 62], [86, 89]], [[9, 62]], [[9, 62]], [[100, 120]], [[94, 98]], [[123, 127]], [[66, 73]], [[9, 62]], [[9, 62]], [[1, 62]], [[100, 120]], [[1, 73]], [[0, 79]], [[74, 98]], [[81, 98]], [[74, 127]], [[130, 148]], [[153, 156]], [[86, 156]]]",
    "query_spans": "[[[153, 160]]]",
    "process": "Let the right focus of the hyperbola be $F_{2}$, let $\\angle POF_{2} = \\theta$, $\\angle PF_{1}O = \\alpha$, $\\angle F_{1}PO = \\beta$, then $\\theta = \\alpha + \\beta$. According to the problem, $\\tan\\beta = \\frac{|OQ|}{|PQ|} = \\frac{a}{2b}$, $\\tan\\alpha = \\frac{|OQ|}{|F_{1}Q|} = \\frac{a}{b}$, thus we can obtain $\\tan\\theta = \\tan(\\alpha + \\beta) = \\frac{3ab}{2b^{2} - a^{2}}$, so $\\frac{3ab}{2b^{2} - a^{2}} = \\frac{b}{a}$, thereby yielding $b^{2} = 2a^{2}$, and further we can find the eccentricity. Solution: Let the right focus of the hyperbola be $F_{2}$, in $\\triangle PF_{1}O$, $\\angle POF_{2}$ is an exterior angle of $\\triangle PF_{1}O$. Let $\\angle POF_{2} = \\theta$, $\\angle PF_{1}O = \\alpha$, $\\angle F_{1}PO = \\beta$, then $\\theta = \\alpha + \\beta$. Since line $PF_{1}$ is tangent to the circle $x^{2} + y^{2} = a^{2}$ at point $Q$, we have $OQ \\perp PF_{1}$. In right triangle $\\triangle OQF_{1}$, $|OQ| = a$, $|OF_{1}| = c$, so $|F_{1}Q| = \\sqrt{OF_{1}^{2} - OQ^{2}} = \\sqrt{c^{2} - a^{2}} = b$. Since $|PQ| = 2|F_{1}Q|$, we have $|PQ| = 2b$. Therefore, in right triangle $\\triangle POQ$, $\\tan\\beta = \\frac{|OQ|}{|PQ|} = \\frac{a}{2b}$. In right triangle $\\triangle OQF_{1}$, $\\tan\\alpha = \\frac{|OQ|}{|F_{1}Q|} = \\frac{a}{b}$. Since $\\theta = \\alpha + \\beta$, we have $\\tan\\theta = \\tan(\\alpha + \\beta) = \\frac{\\tan\\alpha + \\tan\\beta}{1 - \\tan\\alpha\\tan\\beta} = \\frac{3ab}{2b^{2} - a^{2}}$. Since $\\theta$ is the inclination angle of line $OP$, and line $OP$ is an asymptote of the hyperbola, we have $\\frac{3ab}{2b^{2} - a^{2}} = \\frac{b}{a}$, so $b^{2} = 2a^{2}$, hence $c^{2} = a^{2} + b^{2} = 3a^{2}$, so $c = \\sqrt{3}a$. Thus, the eccentricity is $e = \\frac{c}{a} = \\sqrt{3}$."
  },
  {
    "text": "What is the equation of the directrix of the parabola $y=a x^{2}(a \\neq 0)$?",
    "fact_expressions": "G: Parabola;a: Number;Negation(a=0);Expression(G) = (y = a*x^2)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=-1/(4*a)",
    "fact_spans": "[[[0, 24]], [[3, 24]], [[3, 24]], [[0, 24]]]",
    "query_spans": "[[[0, 31]]]",
    "process": ""
  },
  {
    "text": "If the foci of the ellipse $\\frac{x^{2}}{k-1}+\\frac{y^{2}}{3-k}=1$ lie on the $y$-axis, then the range of real values for $k$ is?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (x^2/(k - 1) + y^2/(3 - k) = 1);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(1,2)",
    "fact_spans": "[[[1, 42]], [[53, 58]], [[1, 42]], [[1, 51]]]",
    "query_spans": "[[[53, 65]]]",
    "process": "Because the foci of the ellipse $\\frac{x2}{k>k}+\\frac{y^{2}}{3-k}=1$ are on the y-axis,"
  },
  {
    "text": "The equation $x^{2} \\sin \\alpha - y^{2} \\cos \\alpha = 1$, $0 < \\alpha < \\pi$, represents an ellipse with foci on the $y$-axis. Then the range of values for $\\alpha$ is?",
    "fact_expressions": "G: Ellipse;Expression(G)=(x^2*Sin(alpha)-y^2*Cos(alpha)=1);alpha:Number;alpha>0;alpha<pi;PointOnCurve(Focus(G),yAxis)",
    "query_expressions": "Range(alpha)",
    "answer_expressions": "(\\pi/2,(3/4)*\\pi)",
    "fact_spans": "[[[69, 71]], [[0, 71]], [[73, 81]], [[42, 56]], [[42, 56]], [[60, 71]]]",
    "query_spans": "[[[73, 88]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the focus of the parabola $x=8 y^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (x = 8*y^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(1/32, 0)",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": ""
  },
  {
    "text": "If the line $y=k(x-2)+b$ and the curve $x^{2}-y^{2}=1$ always have common points regardless of the value of $k$, then what is the range of values for $b$?",
    "fact_expressions": "G: Line;Expression(G) = (y = b + k*(x - 2));k: Number;b: Number;H: Curve;Expression(H) = (x^2 - y^2 = 1);IsIntersect(G, H)",
    "query_expressions": "Range(b)",
    "answer_expressions": "[-sqrt(3), sqrt(3)]",
    "fact_spans": "[[[10, 24]], [[10, 24]], [[3, 6]], [[49, 52]], [[25, 42]], [[25, 42]], [[10, 47]]]",
    "query_spans": "[[[49, 59]]]",
    "process": ""
  },
  {
    "text": "Point $P$ is a moving point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, $F_{1}$ is the left focus of the ellipse, and fixed point $M(6,4)$. Then the maximum value of $P M+P F_{1}$ is?",
    "fact_expressions": "G: Ellipse;M: Point;P: Point;F1: Point;Expression(G) = (x^2/25 + y^2/16 = 1);Coordinate(M) = (6, 4);PointOnCurve(P, G);LeftFocus(G) = F1",
    "query_expressions": "Max(LineSegmentOf(P, F1) + LineSegmentOf(P, M))",
    "answer_expressions": "15",
    "fact_spans": "[[[5, 44], [57, 59]], [[66, 74]], [[0, 4]], [[49, 56]], [[5, 44]], [[66, 74]], [[0, 48]], [[49, 63]]]",
    "query_spans": "[[[76, 95]]]",
    "process": "From the ellipse equation, we have $ a^{2}=25 $, $ b^{2}=16 $, $ \\therefore c^{2}=25-16=9 $, $ \\therefore c=3 $. The left focus is $ F_{1}(-3,0) $, the right focus is $ F_{2}(3,0) $. From the definition of the ellipse, we know $ PF_{1}+PF_{2}=2a=10 $, $ \\therefore PF_{1}=10-PF_{2} $. $ \\therefore PM+PF_{1}=PM-PF_{2}+10 $. Analysis shows that point $ M $ lies outside the ellipse, so $ PM-PF_{2}\\leqslant MF_{2}=\\sqrt{(6-3)^{2}+4^{2}}=5 $, $ \\therefore PM+PF_{1}=PM-PF_{2}+10\\leqslant 5+10=15 $. Thus, the maximum value of $ PM+PF $ is 15."
  },
  {
    "text": "It is known that a hyperbola shares the same foci as the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{3}=1$, and one of its asymptotes has the equation $y=\\frac{3}{2} x$. What is the standard equation of this hyperbola?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/16 + y^2/3 = 1);Focus(G) = Focus(H);Expression(OneOf(Asymptote(G))) = (y = (3/2)*x)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/9 = 1",
    "fact_spans": "[[[2, 5], [82, 85]], [[6, 44]], [[6, 44]], [[2, 50]], [[2, 79]]]",
    "query_spans": "[[[82, 92]]]",
    "process": "The foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{3}=1$ are $(\\pm\\sqrt{13},0)$, so in the hyperbola $c=\\sqrt{13}$, $\\therefore a^{2}+b^{2}=13$ $\\because \\frac{b}{a}=\\frac{3}{2}$ $\\therefore a^{2}=4$, $b^{2}=9$, the hyperbola equation is $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=$"
  },
  {
    "text": "The left focus of the ellipse $\\frac{x^{2}}{4 a^{2}}+\\frac{y^{2}}{3 a^{2}}=1(a>0)$ is $F$, and the line $x=m$ intersects the ellipse at points $A$, $B$. When the perimeter of $\\triangle FAB$ is maximized, what is the area of $\\triangle FAB$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/((4*a^2)) + y^2/((3*a^2)) = 1);a: Number;a>0;F: Point;LeftFocus(G) = F;H: Line;Expression(H) = (x = m);m: Number;Intersection(H, G) = {A, B};A: Point;B: Point;WhenMax(Perimeter(TriangleOf(F,A,B))) = True",
    "query_expressions": "Area(TriangleOf(F, A, B))",
    "answer_expressions": "3*a^2",
    "fact_spans": "[[[0, 54], [71, 73]], [[0, 54]], [[2, 54]], [[2, 54]], [[59, 62]], [[0, 62]], [[63, 70]], [[63, 70]], [[65, 70]], [[63, 86]], [[76, 80]], [[83, 86]], [[87, 109]]]",
    "query_spans": "[[[110, 130]]]",
    "process": ""
  },
  {
    "text": "The focal distance of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/5 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 42]]]",
    "process": "From the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, we know $a^{2}=9$, $b^{2}=5$, so $c^{2}=a^{2}-b^{2}=4$, solving gives $c=2$, so the focal distance is $2c=4$. The answer is A."
  },
  {
    "text": "The line $4 k x - 4 y - k = 0$ intersects the parabola $y^2 = x$ at points $A$ and $B$. If $|AB| = 4$, then the distance from the midpoint of chord $AB$ to the line $x + \\frac{1}{2} = 0$ is equal to?",
    "fact_expressions": "H: Line;Expression(H) = (-k + 4*(k*x) - 4*y = 0);k: Number;G: Parabola;Expression(G) = (y^2 = x);A: Point;B: Point;Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 4;L: Line;Expression(L) = (x + 1/2 = 0);IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A, B)), L)",
    "answer_expressions": "9/4",
    "fact_spans": "[[[0, 17]], [[0, 17]], [[2, 17]], [[18, 30]], [[18, 30]], [[32, 35]], [[36, 39]], [[0, 41]], [[43, 52]], [[64, 83]], [[64, 83]], [[18, 60]]]",
    "query_spans": "[[[55, 89]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $x^{2}=4 y$ with focus $F$, a line passing through point $F$ with slope $1$ intersects the parabola at points $M$ and $N$. Let line $l$ be a tangent to the parabola $C$ such that $l \\| M N$, and let $P$ be a point on $l$. Then the minimum value of $\\overrightarrow{P M} \\cdot \\overrightarrow{P N}$ is?",
    "fact_expressions": "l: Line;C: Parabola;G: Line;M: Point;N: Point;P: Point;F: Point;Expression(C) = (x^2 = 4*y);Focus(C) = F;PointOnCurve(F, G);Slope(G) = 1;Intersection(G, C) = {M, N};IsTangent(l, C);IsParallel(l,LineSegmentOf(M,N));PointOnCurve(P, l)",
    "query_expressions": "Min(DotProduct(VectorOf(P, M), VectorOf(P, N)))",
    "answer_expressions": "-14",
    "fact_spans": "[[[62, 67], [94, 97]], [[2, 20], [44, 47], [68, 74]], [[41, 43]], [[50, 53]], [[54, 57]], [[90, 93]], [[24, 27], [29, 33]], [[2, 20]], [[2, 27]], [[28, 43]], [[34, 43]], [[41, 59]], [[62, 77]], [[79, 89]], [[90, 100]]]",
    "query_spans": "[[[102, 157]]]",
    "process": "Let the line passing through the focus $ F(0,1) $ with slope 1 be $ y = x + 1 $, intersecting the parabola $ x^{2} = 4y $ at points $ M(x_{1}, y_{1}) $, $ N(x_{2}, y_{2}) $. Solving the system  \n\\[\n\\begin{cases}\ny = x + 1 \\\\\nx^{2} = 4y\n\\end{cases}\n\\frac{+1}{4y},\n\\]  \nwe obtain $ x^{2} - 4x - 4 = 0 $, then $ x_{1} + x_{2} = 4 $, $ x_{1}x_{2} = -4 $. Let line $ l $ be tangent to the parabola $ y = \\frac{x^{2}}{4} $ at point $ Q(x_{0}, y_{0}) $, then $ y = \\frac{x}{2} $. Since $ l \\parallel MN $, we have $ \\frac{x_{0}}{2} = 1 $, so $ Q(2,1) $. The equation of line $ l $ is $ y - 1 = x - 2 $, i.e., $ y = x - 1 $. Let point $ P(a, a - 1) $, then  \n\\[\n\\overrightarrow{PM} \\cdot \\overrightarrow{PN} = (x_{1} - a)(x_{2} - a) + (y_{1} - a + 1)(y_{2} - a - 1) = 2a^{2} - 12a + 4 = 2(a - 3)^{2} - 14 > -14.\n\\]  \nThe answer is: $ -14 $."
  },
  {
    "text": "Given that the line $y=\\sqrt{3} x$ intersects the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ at two points, what is the range of the eccentricity of hyperbola $C$?",
    "fact_expressions": "G: Line;Expression(G) = (y = sqrt(3)*x);C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;NumIntersection(G, C) = 2",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(2,+oo)",
    "fact_spans": "[[[2, 18]], [[2, 18]], [[19, 80], [87, 93]], [[19, 80]], [[27, 80]], [[27, 80]], [[27, 80]], [[27, 80]], [[2, 85]]]",
    "query_spans": "[[[87, 104]]]",
    "process": "If the line y=\\sqrt{3}x intersects the hyperbola C:\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0) at two points, then the slope of the line y=\\sqrt{3}x must be less than the slope of the asymptote y=\\frac{b}{a}x. By establishing this inequality, the solution can be obtained. [Detailed Solution] The asymptotes of the hyperbola C:\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0) are given by y=\\pm\\frac{b}{a}x. If the line y=\\sqrt{3}x intersects the hyperbola at two points, then \\frac{b}{a}>\\sqrt{3}, that is, b^{2}>3a^{2}, or equivalently c^{2}-a^{2}>3a^{2}. Therefore, c^{2}>4a^{2}, e^{2}>4, which implies e>2."
  },
  {
    "text": "The slope of an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ is $-2$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Slope(OneOf(Asymptote(G))) = -2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[0, 59], [75, 78]], [[0, 59]], [[3, 59]], [[3, 59]], [[3, 59]], [[3, 59]], [[0, 73]]]",
    "query_spans": "[[[75, 84]]]",
    "process": ""
  },
  {
    "text": "Let a circle pass through a vertex and a focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, with its center lying on this hyperbola. Then, the distance from the center of the circle to the center of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Q:Circle;Expression(G) = (x^2/9 - y^2/16 = 1);PointOnCurve(Center(Q), G);PointOnCurve(OneOf(Vertex(G)),Q);PointOnCurve(OneOf(Focus(G)),Q)",
    "query_expressions": "Distance(Center(Q), Center(G))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[3, 42], [57, 60], [66, 69]], [[1, 2]], [[3, 42]], [[1, 61]], [[1, 47]], [[1, 52]]]",
    "query_spans": "[[[1, 76]]]",
    "process": "Let a vertex and a focus of the hyperbola be A(3,0) and F(5,0), respectively. Then the perpendicular bisector of BF is x=4. Let the coordinates of the circle center be (a,b), then a=4, b^{2}=\\frac{112}{9}, hence the distance from the circle center to the center of the hyperbola is \\sqrt{16+\\frac{112}{9}}=\\frac{16}{3}."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{12}=1$ has eccentricity $e=2$, then the focal distance of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;Expression(G) = (-y^2/12 + x^2/a^2 = 1);Eccentricity(G) = e;e:Number;e=2",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 45], [56, 59]], [[5, 45]], [[2, 45]], [[2, 54]], [[49, 54]], [[49, 54]]]",
    "query_spans": "[[[56, 64]]]",
    "process": "\\because e=2,\\therefore\\frac{12}{a^{2}}=e^{2}-1=3,\\therefore a^{2}=4,a=2,\\therefore c=4,\\therefore 2c=8"
  },
  {
    "text": "One focus of the ellipse $5 x^{2}-ky^{2}=5$ is $(0 , 2)$, then what is $k$=?",
    "fact_expressions": "G: Ellipse;k: Number;H: Point;Expression(G) = (-k*y^2 + 5*x^2 = 5);Coordinate(H) = (0, 2);OneOf(Focus(G)) = H",
    "query_expressions": "k",
    "answer_expressions": "-1",
    "fact_spans": "[[[0, 20]], [[38, 41]], [[26, 35]], [[0, 20]], [[26, 35]], [[0, 35]]]",
    "query_spans": "[[[38, 43]]]",
    "process": ""
  },
  {
    "text": "If the midpoint of the chord $P Q$ of the parabola $y^{2}=2 p x(p>0)$ is $M(x_{0}, y_{0})(y_{0} \\neq 0)$, then the slope of the line $P Q$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;P: Point;Q: Point;IsChordOf(LineSegmentOf(P, Q), G);x0: Number;y0: Number;Negation(y0=0);M: Point;Coordinate(M) = (x0, y0);MidPoint(LineSegmentOf(P, Q)) = M",
    "query_expressions": "Slope(LineOf(P, Q))",
    "answer_expressions": "p/y0",
    "fact_spans": "[[[1, 22]], [[1, 22]], [[4, 22]], [[4, 22]], [[24, 29]], [[24, 29]], [[1, 29]], [[33, 64]], [[33, 64]], [[33, 64]], [[33, 64]], [[33, 64]], [[24, 64]]]",
    "query_spans": "[[[66, 78]]]",
    "process": "Let the coordinates of points P and Q be: P(x_{1},y_{1}), Q(x_{2},y_{2}), then y_{1}^{2}=2px_{1}, y_{2}^{2}=2px_{2}. Subtracting these two equations gives: y_{1}^{2}-y_{2}^{2}=2p(x_{1}-x_{2}), that is: \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{2p}{y_{1}+y_{2}}. Also, since the midpoint of PQ is M(x_{0},y_{0}) (y_{0}\\neq0)"
  },
  {
    "text": "The point $p(1, m)$ lies on a parabola with vertex at the origin and focus on the $x$-axis, and its distance to the focus of the parabola is $2$. Then the value of $m$ is?",
    "fact_expressions": "G: Parabola;O:Origin;p:Point;m:Number;Vertex(G)=O;PointOnCurve(Focus(G),xAxis);PointOnCurve(p,G);Coordinate(p)=(1,m);Distance(p,Focus(G))=2",
    "query_expressions": "m",
    "answer_expressions": "{2,-2}",
    "fact_spans": "[[[26, 29], [26, 29]], [[14, 16]], [[0, 10], [33, 34]], [[50, 53]], [[11, 29]], [[17, 29]], [[0, 32]], [[0, 10]], [[33, 48]]]",
    "query_spans": "[[[50, 57]]]",
    "process": "Given a parabola with vertex at the origin and focus on the x-axis, and point P(1, m) having a positive x-coordinate, the parabola opens to the right. Thus, we can set y^{2}=2px (p>0), with directrix equation x=-\\frac{p}{2}. Then 1+\\frac{p}{2}=2, \\therefore p=2. The equation of the parabola is y^{2}=4x. Substituting point P(1, m) gives m=\\pm2."
  },
  {
    "text": "The standard equation of the ellipse passing through the points $(-5,0)$ and $(0, \\sqrt{7})$ is?",
    "fact_expressions": "G: Ellipse;H: Point;I: Point;Coordinate(H) = (-5, 0);Coordinate(I) = (0, sqrt(7));PointOnCurve(H,G);PointOnCurve(I,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/25+y^2/7=1",
    "fact_spans": "[[[28, 30]], [[2, 11]], [[12, 27]], [[2, 11]], [[12, 27]], [[0, 30]], [[0, 30]]]",
    "query_spans": "[[[28, 37]]]",
    "process": "Let the equation of the ellipse be $mx^{2}+ny^{2}=1$ $(m>0, n>0)$. Substituting the points $(-5,0)$ and $(0,\\sqrt{7})$, we get:  \n$$\n\\begin{cases}\n25m = 1 \\\\\n7n = 1\n\\end{cases}\n$$\nThus, $m = \\frac{1}{25}$, $n = \\frac{1}{7}$, so the standard equation of the ellipse is $\\frac{x^{2}}{25} + \\frac{y^{2}}{7} = 1$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, a circle centered at $F_{1}$ with radius equal to the semi-focal distance $c$ of the hyperbola intersects the hyperbola at points $P$ and $Q$. If $P F_{2}$ is tangent to the circle $F_{1}$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;Center(G) = F1;HalfFocalLength(C) = c;c: Number;Radius(G) = c;G: Circle;Intersection(G, C) = {P, Q};P: Point;Q: Point;IsTangent(LineSegmentOf(P, F2), G) = True",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)+1",
    "fact_spans": "[[[18, 79], [97, 100], [113, 116], [151, 157]], [[18, 79]], [[26, 79]], [[26, 79]], [[26, 79]], [[26, 79]], [[2, 9], [86, 93]], [[10, 17]], [[2, 84]], [[2, 84]], [[85, 112]], [[97, 107]], [[104, 107]], [[104, 112]], [[111, 112], [140, 147]], [[111, 127]], [[118, 121]], [[122, 125]], [[129, 149]]]",
    "query_spans": "[[[151, 163]]]",
    "process": "As shown in the figure, according to the given conditions, we have $ PF_{1} = c $, $ F_{1}F_{2} = 2c $. Connecting $ PF_{1} $, we obtain $ PF_{1} \\perp PF_{2} $, so $ PF_{1}^{2} + PF_{2}^{2} = F_{1}F_{2}^{2} $, solving gives $ PF_{2} = \\sqrt{3}c $. Since $ PF_{2} - PF_{1} = 2a $, it follows that $ \\sqrt{3}c - c = 2a $, thus $ e = \\frac{c}{a} = \\sqrt{3} + 1 $."
  },
  {
    "text": "It is known that the foci of ellipse $C$ lie on the $x$-axis, and the eccentricity is $\\frac{1}{3}$. Then, what could be the standard equation of ellipse $C$?",
    "fact_expressions": "C: Ellipse;PointOnCurve(Focus(C),xAxis);Eccentricity(C)=1/3",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/9+y^2/8=1",
    "fact_spans": "[[[2, 7], [2, 7]], [[2, 16]], [[2, 35]]]",
    "query_spans": "[[[38, 52]]]",
    "process": "Let the equation of the ellipse be \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1. From the eccentricity being \\frac{1}{3}, we get \\frac{c}{a}=\\frac{1}{3}. From a^{2}=b^{2}+c^{2}, we obtain \\frac{b^{2}}{a^{2}}=\\frac{8}{9}. We can take a^{2}=9, b^{2}=8, then the standard equation of ellipse C can be \\frac{x^{2}}{9}+\\frac{y^{2}}{8}=1."
  },
  {
    "text": "The equation of a hyperbola with an asymptote $y = \\frac{1}{2}x$ passing through the point $(4,1)$ is?",
    "fact_expressions": "G: Hyperbola;Expression(OneOf(Asymptote(G)))=(y=x/2);P:Point;Coordinate(P)=(4,1);PointOnCurve(P,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/12 - y^2/3 = 1",
    "fact_spans": "[[[36, 39]], [[0, 39]], [[27, 35]], [[27, 35]], [[26, 39]]]",
    "query_spans": "[[[36, 44]]]",
    "process": "Given the asymptote equation $ y = \\frac{1}{2}x $, the hyperbola equation can be assumed as: $ \\frac{x^{2}}{4} - \\frac{y^{2}}{1} = \\lambda $ ($ \\lambda \\neq 0 $). Substituting the point $ (4,1) $, we solve to get: $ \\lambda = 3 $. Simplifying, the hyperbola equation becomes: $ \\frac{x^{2}}{12} - \\frac{y^{2}}{3} = 1 $."
  },
  {
    "text": "Given that the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ passes through the points $P_{1}(\\sqrt{6}, 1)$ and $P_{2}(-\\sqrt{3},-\\sqrt{2})$, find the standard equation of this ellipse?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P1: Point;P2: Point;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(P1) = (sqrt(6), 1);Coordinate(P2) = (-sqrt(3), -sqrt(2));PointOnCurve(P1,G);PointOnCurve(P2,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9+y^2/3=1",
    "fact_spans": "[[[2, 47], [104, 106]], [[4, 47]], [[4, 47]], [[49, 69]], [[71, 99]], [[2, 47]], [[49, 69]], [[71, 99]], [[2, 69]], [[2, 99]]]",
    "query_spans": "[[[104, 112]]]",
    "process": "Let $\\frac{1}{a^{2}}=m$, $\\frac{1}{b^{2}}=n$, then the ellipse equation is $mx^{2}+ny^{2}=1$ ($m>0$, $n>0$, and $m\\neq n$). Since the ellipse passes through points $P_{1}$ and $P_{2}$, we have\n$$\n\\begin{cases}\n6m+n=1 \\\\\n3m+2n=1\n\\end{cases}\n$$\nSolving gives:\n$$\n\\begin{cases}\nm=\\frac{1}{9} \\\\\nn=\\frac{1}{3}\n\\end{cases}\n$$\nthat is,\n$$\n\\begin{cases}\na^{2}=9 \\\\\nb^{2}=3\n\\end{cases}\n$$\nHence, the required ellipse equation is $\\frac{x^{2}}{9}+\\frac{y^{2}}{3}=1$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{10}+\\frac{y^{2}}{5}=1$, what is the equation of the line containing the chord $MN$ with midpoint $D(-2,1)$?",
    "fact_expressions": "C: Ellipse;M: Point;N: Point;D: Point;Expression(C) = (x^2/10 + y^2/5 = 1);Coordinate(D) = (-2, 1);IsChordOf(LineSegmentOf(M,N),C);MidPoint(LineSegmentOf(M,N))=D",
    "query_expressions": "Expression(OverlappingLine(LineSegmentOf(M,N)))",
    "answer_expressions": "y=x+3",
    "fact_spans": "[[[2, 45]], [[63, 68]], [[63, 68]], [[48, 58]], [[2, 45]], [[48, 58]], [[2, 68]], [[47, 68]]]",
    "query_spans": "[[[63, 77]]]",
    "process": "D is inside the ellipse. Let M(x_{1},y_{1}), N(x_{2},y_{2}), then x_{1}+x_{2}=-4, y_{1}+y_{2}=2. Since M and N lie on the ellipse, we have \\frac{x_{1}^{2}}{10}+\\frac{y_{1}^{2}}{5}=1, \\frac{x_{2}^{2}}{10}+\\frac{y_{2}^{2}}{5}=1. Subtracting these two equations gives \\frac{(x_{1}-x_{2})(x_{1}+x_{2})}{10}+\\frac{(y_{1}-y_{2})(y_{1}+y_{2})}{5}=0. Therefore, k_{MN}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{5(x_{1}+x_{2})}{10(y_{1}+y_{2})}=-\\frac{5\\times(-4)}{10\\times2}=1. Hence, the equation of line MN is y-1=1\\cdot(x+2), i.e., y=x+3."
  },
  {
    "text": "If the distance from point $A(2, m)$ on the parabola $y^{2}=2 p x(p>0)$ to the focus is $6$, then $p=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;A: Point;m: Number;Coordinate(A) = (2, m);PointOnCurve(A, G);Distance(A, Focus(G)) = 6",
    "query_expressions": "p",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 22]], [[1, 22]], [[46, 49]], [[4, 22]], [[24, 34]], [[24, 34]], [[24, 34]], [[1, 34]], [[1, 44]]]",
    "query_spans": "[[[46, 51]]]",
    "process": "From the parabola equation, the directrix equation is $x = -\\frac{p}{2}$. Since the distance from point $A(2, m)$ on the parabola $y^2 = 2px$ ($p > 0$) to the focus is 6, according to the definition of a parabola, the distance from the point to the directrix is also 6. Thus, $2 + \\frac{p}{2} = 6$, $\\therefore p = 8$."
  },
  {
    "text": "The equation of a hyperbola centered at the origin, with one focus at $(3 , 0)$ and one asymptote given by $2 x-3 y=0$, is?",
    "fact_expressions": "E: Hyperbola;O: Origin;Center(E) = O;Coordinate(OneOf(Focus(E))) = (3, 0);Expression(OneOf(Asymptote(E))) = (2*x-3*y=0)",
    "query_expressions": "Expression(E)",
    "answer_expressions": "13*x^2/81 - 13*y^2/36 = 1",
    "fact_spans": "[[[41, 44]], [[3, 5]], [[0, 44]], [[6, 44]], [[21, 44]]]",
    "query_spans": "[[[41, 48]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, with left and right foci $F_{1}$ and $F_{2}$ respectively, the focus of the parabola $y^{2}=2 p x$ coincides with $F_{2}$. If point $P$ is a common point of the ellipse and the parabola, and $\\cos \\angle P F_{1} F_{2}=\\frac{5}{7}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;H: Parabola;Expression(H) = (y^2 = 2*(p*x));p: Number;Focus(H) = F2;P: Point;OneOf(Intersection(C, H)) = P;Cos(AngleOf(P, F1, F2)) = 5/7",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "{1/2, 1/3}",
    "fact_spans": "[[[2, 59], [120, 122], [174, 176]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[68, 75]], [[76, 83], [104, 111]], [[2, 83]], [[2, 83]], [[84, 100], [123, 126]], [[84, 100]], [[87, 100]], [[84, 113]], [[115, 119]], [[115, 132]], [[133, 172]]]",
    "query_spans": "[[[174, 182]]]",
    "process": "Since point P lies on the parabola, we have: $\\cos\\angle PF_{1}F_{2} = \\frac{|PF_{2}|}{|PF_{1}|} = \\frac{5}{7}$. Also, $|PF_{1}| + |PF_{2}| = 2a$, solving gives $|PF_{1}| = \\frac{7}{6}a$, $|PF_{2}| = \\frac{5}{6}a$. In triangle $PF_{1}F_{2}$, using the cosine law we obtain: $\\cos\\angle PF_{1}F_{2} = \\frac{\\frac{49}{36}a^{2} + 4c^{2} - \\frac{25}{36}a^{2}}{2 \\times \\frac{7}{6}a \\times 2c} = \\frac{5}{7}$. Simplifying yields: $a^{2} + 6c^{2} = 5ac$. Thus, $6e^{2} - 5e + 1 = 0$, solving gives $e = \\frac{1}{2}$ or $e = \\frac{1}{3}$, hence fill in $\\frac{1}{2}$ or $\\frac{1}{3}$."
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ are $F_{1}$ and $F_{2}$ respectively. The line $l$ passes through $F_{1}$ and intersects the ellipse at points $A$ and $B$. Then, the range of the radius $r$ of the incircle of $\\triangle A B F_{2}$ is?",
    "fact_expressions": "l: Line;G: Ellipse;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2/4 + y^2/3 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, l);Intersection(l, G) = {A, B};r:Number;Radius(InscribedCircle(TriangleOf(A,B,F2)))=r",
    "query_expressions": "Range(r)",
    "answer_expressions": "(0,3/4]",
    "fact_spans": "[[[62, 67]], [[0, 37], [76, 78]], [[79, 82]], [[83, 86]], [[54, 61]], [[46, 53], [68, 75]], [[0, 37]], [[0, 61]], [[0, 61]], [[62, 75]], [[62, 88]], [[118, 121]], [[90, 121]]]",
    "query_spans": "[[[118, 126]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). According to the definition of the ellipse, we have: a=2, b=\\sqrt{3}, c=1. |AF_{1}|+|AF_{2}|=2a=4, |BF_{1}|+|BF_{2}|=2a=4. Since |AF_{1}|+|BF|=|AB|, it follows that |AB|+|BF_{2}|+|AF_{2}|=8. Because S_{\\triangle ABF_{2}}=\\frac{1}{2}(|AB|+|BF_{2}|+|AF_{2}|)\\cdot r, so r=\\frac{|y_{1}-y_{2}|\\cdot|F_{1}F_{2}|}{|AB|+|BF_{2}|+|AF_{2}|}\\cdot|F_{1}F_{2}|. Let the equation of line l be x=my-1. From \\begin{cases} \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1 \\\\ x=my-1 \\end{cases}, eliminating x and simplifying gives: (3m^{2}+4)y^{2}-6my-9=0. \\therefore y_{1}+y_{2}=\\frac{6m}{3m^{2}+4}, y_{1}y_{2}=\\frac{-9}{3m^{2}+4}. \\Delta=36m^{2}+36(3m^{2}+4)>0. \\therefore |y_{1}-y_{2}|=\\sqrt{(y_{1}+y_{2})^{2}-4y_{1}y_{2}}=\\sqrt{\\left(\\frac{6m}{3m^{2}+4}\\right)^{2}-4\\cdot\\frac{-9}{3m^{2}+4}}=\\sqrt{\\frac{144(m^{2}+1)}{9m^{4}+24m^{2}+16}}. Let t=m^{2}+1 (t\\geqslant1), then |y_{1}-y_{2}|=\\sqrt{\\frac{144t}{9t^{2}+6t+1}}=\\sqrt{\\frac{144}{9t+\\frac{1}{t}+6}}. Since t\\geqslant1, 9t+\\frac{1}{t} is monotonically increasing, \\therefore when t=1, |y_{1}-y_{2}|_{\\max}=\\sqrt{\\frac{144}{9+1+6}}=3. \\therefore |y_{1}-y_{2}|\\in(0,3], therefore r\\in(0,\\frac{3}{4}]."
  },
  {
    "text": "Given that line $l$ intersects the ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ at points $A$ and $B$, and $F_{1}$ is the left focus of the ellipse. When line $l$ passes through the right focus, the perimeter of $\\triangle A B F_{1}$ is?",
    "fact_expressions": "l: Line;C: Ellipse;A: Point;B: Point;F1: Point;Expression(C) = (x^2/9 + y^2/5 = 1);Intersection(l, C) = {A, B};LeftFocus(C) = F1;PointOnCurve(RightFocus(C),l)",
    "query_expressions": "Perimeter(TriangleOf(A,B,F1))",
    "answer_expressions": "12",
    "fact_spans": "[[[2, 7], [77, 82]], [[8, 50], [69, 71]], [[51, 54]], [[55, 58]], [[61, 68]], [[8, 50]], [[2, 60]], [[61, 75]], [[69, 87]]]",
    "query_spans": "[[[89, 114]]]",
    "process": "Let the right focus of the ellipse be $ F_{2} $. The perimeter of $ \\triangle ABF_{1} $ is $ |AF_{1}| + |F_{1}B| + |AB| = (|AF| + |AF_{2}|) + (|BF_{1}| + |BF_{2}|) = 2a + 2a = 4a = 12 $. Answer: 12"
  },
  {
    "text": "It is known that the foci of ellipse $C$ are the same as those of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, and the eccentricity is $\\frac{1}{2}$. Then, what is the standard equation of ellipse $C$?",
    "fact_expressions": "G: Hyperbola;C: Ellipse;Expression(G) = (x^2 - y^2/3 = 1);Focus(G)=Focus(C);Eccentricity(C)=1/2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/16+y^2/12=1",
    "fact_spans": "[[[11, 39]], [[2, 7], [65, 70]], [[11, 39]], [[2, 44]], [[2, 63]]]",
    "query_spans": "[[[65, 77]]]",
    "process": ""
  },
  {
    "text": "The standard equation of a circle whose center lies on the parabola $x^{2}=4 y$ and which is tangent to both the directrix of the parabola and the $y$-axis is?",
    "fact_expressions": "G: Parabola;H: Circle;Expression(G) = (x^2 = 4*y);PointOnCurve(Center(H), G);IsTangent(Directrix(G),H);IsTangent(yAxis,H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x+pm*2)^2+(y-1)^2=4",
    "fact_spans": "[[[3, 17], [22, 25]], [[37, 38]], [[3, 17]], [[0, 38]], [[21, 38]], [[21, 38]]]",
    "query_spans": "[[[37, 45]]]",
    "process": "The directrix of the parabola given in the problem is $ y = -1 $. Let the coordinates of the circle's center be $ (x_{0}, y_{0}) $, then $ |x_{0}| = y_{0} + 1 $, and $ x_{0}^{2} = 4y_{0} $. Solving the system of equations  \n\\[\n\\begin{cases}\n|x_{0}| = y_{0} + 1 \\\\\nx_{0}^{2} = 4y_{0}\n\\end{cases}\n\\]\nyields  \n\\[\n\\begin{cases}\nx_{0} = \\pm 2, \\\\\ny_{0} = 1\n\\end{cases}.\n\\]\nTherefore, the equation of the required circle is $ (x \\pm 2)^{2} + (y - 1)^{2} = 2^{2} $, that is, $ (x \\pm 2)^{2} + (y - 1)^{2} = 4 $."
  },
  {
    "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ are $y=\\pm \\sqrt{3} x$, then its eccentricity is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(Asymptote(G))=(y=pm*(sqrt(3)*x))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 59], [86, 87]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[2, 59]], [[2, 84]]]",
    "query_spans": "[[[86, 93]]]",
    "process": "From the given condition, we have $ e = \\frac{c}{a} = \\sqrt{1 + \\left( \\frac{b}{a} \\right)^2} = \\sqrt{1 + 3} = 2 $."
  },
  {
    "text": "Given point $A(1,3)$, and point $B$ moves on the line $2x + 3y - 6 = 0$, then the trajectory equation of the midpoint $P$ of $AB$ is?",
    "fact_expressions": "G: Line;A: Point;B: Point;P: Point;Expression(G) = (2*x + 3*y - 6 = 0);Coordinate(A) = (1, 3);PointOnCurve(B, G);MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "4*x+6*y-17=0",
    "fact_spans": "[[[17, 32]], [[2, 11]], [[12, 16]], [[44, 47]], [[17, 32]], [[2, 11]], [[12, 35]], [[37, 47]]]",
    "query_spans": "[[[44, 54]]]",
    "process": "Let the midpoint of AB be P(x,y), and B(x',y'). Since A(1,3), we have x'+1=2x, y'+3=2y, thus x'=2x-1, y'=2y-3. Since point B moves on the line 2x+3y-6=0, we have 2x'+3y'-6=0, that is, 2(2x-1)+3(2y-3)-6=0. Simplifying yields 4x+6y-17=0, which is the trajectory equation of point P."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $A$ lies on the hyperbola $C$, and $A F_{2} \\perp x$-axis. If point $B(3 c, 0)$ is such that $\\angle F_{1} A B=90^{\\circ}$, where $c$ is the semi-focal length of the hyperbola $C$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;c:Number;B: Point;A: Point;F2: Point;F1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(B) = (3*c, 0);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(A, C);IsPerpendicular(LineSegmentOf(A,F2),xAxis);AngleOf(F1,A,B)=ApplyUnit(90,degree);HalfFocalLength(C)=c",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "1+sqrt(2)",
    "fact_spans": "[[[2, 62], [92, 98], [172, 178], [184, 190]], [[9, 62]], [[9, 62]], [[168, 171]], [[121, 133]], [[87, 91]], [[79, 86]], [[71, 78]], [[9, 62]], [[9, 62]], [[2, 62]], [[121, 133]], [[2, 86]], [[2, 86]], [[87, 99]], [[101, 119]], [[135, 164]], [[168, 182]]]",
    "query_spans": "[[[184, 196]]]",
    "process": "From the given conditions, |F_{1}F_{2}| = |F_{2}B|, \\therefore \\triangle AF_{1}F_{2} \\cong \\triangle ABF_{2}, \\therefore \\angle F_{1}AF_{2} = \\angle BAF_{2}. \\because \\angle F_{1}AB = 90^{\\circ}, \\therefore \\angle F_{1}AF_{2} = 45^{\\circ}, \\therefore |AF_{2}| = |F_{1}F_{2}|. \\because point A lies on the hyperbola C, and AF_{2} \\bot x-axis. \\therefore \\frac{c^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1, \\therefore y = \\pm \\frac{b^{2}}{a}, \\therefore \\frac{b^{2}}{a} = 2c, \\therefore b^{2} = 2ac, \\therefore c^{2} - a^{2} = 2ac, \\therefore e^{2} - 2e - 1 = 0. Solving gives e = 1 + \\sqrt{2} or e = 1 - \\sqrt{2} (discarded), \\therefore the eccentricity of C is 1 + \\sqrt{2}."
  },
  {
    "text": "If an ellipse passes through the point $(2,3)$ and has foci $F_{1}(-2,0)$, $F_{2}(2,0)$, then the eccentricity of this ellipse is equal to?",
    "fact_expressions": "G: Ellipse;H: Point;F1: Point;F2: Point;Coordinate(H) = (2, 3);Coordinate(F1) = (-2, 0);Coordinate(F2) = (2, 0);PointOnCurve(H, G);Focus(G)={F1,F2}",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[1, 3], [48, 50]], [[5, 13]], [[18, 31]], [[32, 44]], [[5, 13]], [[18, 31]], [[32, 44]], [[1, 13]], [[1, 44]]]",
    "query_spans": "[[[48, 57]]]",
    "process": "|PF_{1}|+|PF_{2}|=8=2a, so a=4, c=2, eccentricity e=\\frac{c}{a}=\\frac{1}{2}."
  },
  {
    "text": "Let the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ be $F_{1}$, $F_{2}$ respectively, and the upper and lower vertices be $A$, $B$ respectively. The line $A F_{2}$ intersects the ellipse at points $A$ and $M$. If $\\angle F_{1} A F_{2}=90^{\\circ}$, then what is the slope of the line $B M$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;A: Point;B: Point;UpperVertex(G) = A;LowerVertex(G) = B;M: Point;Intersection(LineOf(A, F2), G) = {A, M};AngleOf(F1, A, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Slope(LineOf(B, M))",
    "answer_expressions": "1/2",
    "fact_spans": "[[[1, 53], [105, 107]], [[1, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[61, 68]], [[69, 76]], [[1, 76]], [[1, 76]], [[84, 87], [109, 112]], [[88, 91]], [[1, 91]], [[1, 91]], [[113, 116]], [[92, 118]], [[120, 153]]]",
    "query_spans": "[[[155, 167]]]",
    "process": "\\because\\angle F_{1}AF_{2}=90^{\\circ},\\therefore a=\\sqrt{2}b, i.e., the ellipse equation is: \\frac{x^{2}}{2b^{2}}+\\frac{y^{2}}{b^{2}}=1. Let M(m,n),A(0,b),B(0,-b), and \\frac{m^{2}}{2b^{2}}+\\frac{n^{2}}{b^{2}}=1, i.e., n^{2}-b^{2}=-\\frac{m^{2}}{2}. k_{AM}\\cdot k_{BM}=\\frac{n-b}{m}\\cdot\\frac{n+b}{m}=\\frac{n^{2}-b^{2}}{m^{2}}=\\frac{-\\frac{m^{2}}{2}}{m^{2}}=-\\frac{1}{2}, Also k_{AM}=-1, \\therefore k_{BM}=\\frac{1}{2}."
  },
  {
    "text": "The line $m$: $y = kx + 1$ intersects the left branch of the hyperbola $x^2 - y^2 = 1$ at two distinct points. What is the range of values for $k$?",
    "fact_expressions": "m: Line;Expression(m) = (y = k*x + 1);k: Number;G: Hyperbola;Expression(G) = (x^2 - y^2 = 1);NumIntersection(m,LeftPart(G)) = 2",
    "query_expressions": "Range(k)",
    "answer_expressions": "(1,\\sqrt{2})",
    "fact_spans": "[[[0, 16]], [[0, 16]], [[46, 49]], [[17, 35]], [[17, 35]], [[0, 44]]]",
    "query_spans": "[[[46, 56]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1$, respectively, and $A$ is a point on the hyperbola in the first quadrant. If $|A F_{2} |=2$ and $\\angle F_{1} A F_{2}=45^{\\circ}$, extend $A F_{2}$ to intersect the right branch of the hyperbola at point $B$. Then the area of $\\Delta F_{1} A B$ equals?",
    "fact_expressions": "G: Hyperbola;b: Number;A: Point;F2: Point;F1: Point;B: Point;Expression(G) = (x^2 - y^2/b^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;Quadrant(A) = 1;PointOnCurve(A, G);Abs(LineSegmentOf(A, F2)) = 2;AngleOf(F1, A, F2) = ApplyUnit(45, degree);Intersection(OverlappingLine(LineSegmentOf(A,F2)),RightPart(G))=B",
    "query_expressions": "Area(TriangleOf(F1, A, B))",
    "answer_expressions": "4",
    "fact_spans": "[[[20, 52], [63, 66], [139, 142]], [[23, 52]], [[59, 62]], [[10, 17]], [[2, 9]], [[145, 149]], [[20, 52]], [[2, 58]], [[2, 58]], [[59, 75]], [[59, 75]], [[77, 91]], [[92, 125]], [[127, 149]]]",
    "query_spans": "[[[151, 175]]]",
    "process": ""
  },
  {
    "text": "The line $l$: $y = kx + 1$ has only one common point with the hyperbola $C$: $x^2 - y^2 = 1$, then $k = $?",
    "fact_expressions": "l: Line;Expression(l) = (y = k*x + 1);k: Number;C: Hyperbola;Expression(C) = (x^2 - y^2 = 1);NumIntersection(l, C) = 1",
    "query_expressions": "k",
    "answer_expressions": "{pm*1, pm*sqrt(2)}",
    "fact_spans": "[[[0, 16]], [[0, 16]], [[49, 52]], [[17, 40]], [[17, 40]], [[0, 47]]]",
    "query_spans": "[[[49, 54]]]",
    "process": "From \\begin{cases}y=kx+1\\\\x^{2}-y^{2}=1\\end{cases} we obtain: (1-k^{2})x^{2}-2kx-2=0. Then when: 1-k^{2}=0, i.e., k=\\pm1, the equation (1-k^{2})x^{2}-2kx-2=0 has one root, and the two curves have one common point. When: 1-k^{2}\\neq0, there is one solution, then \\triangle=4k^{2}+8(1-k^{2})=0, k=\\pm\\sqrt{2}. In summary, k=\\pm1 or k=\\pm\\sqrt{2}, at which time the line and the hyperbola have only one intersection point."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2 p x(p>0)$ be $F$, points $A$ and $B$ lie on the parabola, and $\\angle A F B=120^{\\circ}$. The projection of the midpoint $M$ of chord $A B$ onto the directrix $l$ is $M_{1}$. Then the maximum value of $\\frac{|M M_{1}|}{|A B|}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(G) = F;A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);AngleOf(A, F, B) = ApplyUnit(120, degree);IsChordOf(LineSegmentOf(A, B), G);M: Point;MidPoint(LineSegmentOf(A, B)) = M;l: Line;Directrix(G) = l;M1: Point;Projection(M, l) = M1",
    "query_expressions": "Max(Abs(LineSegmentOf(M, M1))/Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[1, 22], [39, 42]], [[1, 22]], [[4, 22]], [[4, 22]], [[26, 29]], [[1, 29]], [[30, 34]], [[35, 38]], [[30, 43]], [[35, 43]], [[45, 71]], [[39, 78]], [[80, 83]], [[73, 83]], [[86, 89]], [[39, 89]], [[94, 101]], [[80, 101]]]",
    "query_spans": "[[[103, 134]]]",
    "process": "Let |AF| = a, |BF| = b. By the definition of a parabola, 2|MM| = a + b. By the law of cosines, |AB|^{2} = a^{2} + b^{2} - 2ab\\cos\\frac{2\\pi}{3} = (a + b)^{2} - ab. Furthermore, since a + b \\geqslant 2\\sqrt{ab}, it follows that |AB| \\geqslant \\frac{\\sqrt{3}}{2}(a + b). Therefore, the maximum value of \\frac{|MM|}{|AB|} is \\frac{\\sqrt{3}}{3}. Hence, the answer should be: \\frac{\\sqrt{3}}{3}"
  },
  {
    "text": "Write the standard equation of an ellipse with foci at $(0, \\pm 1)$.",
    "fact_expressions": "G: Ellipse;Coordinate(Focus(G)) = (0, pm*1)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/2 + x^2 = 1",
    "fact_spans": "[[[23, 25]], [[3, 25]]]",
    "query_spans": "[[[23, 31]]]",
    "process": "From the given condition, the foci of the ellipse lie on the y-axis, c=1, so the standard equation satisfying the condition is \\frac{y^{2}}{2}+x^{2}=1 (the answer is not unique)"
  },
  {
    "text": "It is known that the eccentricity of an ellipse with foci on the $x$-axis is $\\frac{\\sqrt{2}}{2}$, and its major axis length equals the radius of the circle $O$: $x^{2}+y^{2}-4 x-12=0$. Then, what is the length of the minor axis of the ellipse?",
    "fact_expressions": "G: Ellipse;O: Circle;Expression(O) = (-4*x + x^2 + y^2 - 12 = 0);PointOnCurve(Focus(G), xAxis);Eccentricity(G) = (sqrt(2)/2);Length(MajorAxis(G))=Radius(O)",
    "query_expressions": "Length(MinorAxis(G))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[11, 13], [80, 82], [40, 41]], [[47, 75]], [[47, 75]], [[2, 13]], [[11, 38]], [[40, 78]]]",
    "query_spans": "[[[80, 88]]]",
    "process": "The equation of circle C can be written as (x-2)^{2}+y^{2}=16, with radius 4, \\therefore the major axis length of the ellipse is 2a=4, \\therefore a=2. Also, the eccentricity e=\\frac{c}{a}=\\frac{\\sqrt{2}}{2}, \\therefore c=\\sqrt{2}, b=\\sqrt{a^{2}-c^{2}}=\\sqrt{2}, \\therefore the minor axis length of the ellipse is 2\\sqrt{2}."
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=4x$, draw a line intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. Then $x_{1} x_{2}=$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;F:Point;x1:Number;x2:Number;y1:Number;y2:Number;Expression(G) = (y^2 = 4*x);Coordinate(A) = (x1,y1);Coordinate(B) = (x2,y2);PointOnCurve(F,H);Intersection(H,G) = {A,B};Focus(G)=F",
    "query_expressions": "x1*x2",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 15], [25, 28]], [[22, 24]], [[29, 47]], [[50, 67]], [[18, 21]], [[69, 82]], [[69, 82]], [[29, 47]], [[50, 67]], [[1, 15]], [[29, 47]], [[50, 67]], [[0, 24]], [[22, 67]], [[1, 21]]]",
    "query_spans": "[[[69, 84]]]",
    "process": ""
  },
  {
    "text": "Given that the line $y = kx$ ($k \\neq 0$) intersects the hyperbola $C$: $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ ($a > 0$, $b > 0$) at points $A$ and $B$, and the circle with diameter $AB$ passes exactly through the right focus $F$ of the hyperbola. If the area of $\\Delta ABF$ is $4a^{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;H:Line;k: Number;A: Point;B: Point;F: Point;a>0;b>0;Negation(k=0);Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = k*x);Intersection(H, C) = {A, B};IsDiameter(LineSegmentOf(A,B),G);RightFocus(C)=F;PointOnCurve(F,G);Area(TriangleOf(A, B, F)) = 4*a^2",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[22, 83], [110, 113], [151, 154]], [[30, 83]], [[30, 83]], [[105, 106]], [[2, 21]], [[4, 21]], [[85, 88]], [[89, 92]], [[117, 120]], [[30, 83]], [[30, 83]], [[4, 21]], [[22, 83]], [[2, 21]], [[2, 94]], [[95, 106]], [[110, 120]], [[105, 120]], [[122, 149]]]",
    "query_spans": "[[[151, 160]]]",
    "process": "First, according to the problem, find the equation of the circle with diameter AB. Using the symmetry of the graph of a direct proportional function and the hyperbola, and based on the definition of the hyperbola, the triangle area formula, and the Pythagorean theorem, the eccentricity of the hyperbola can be found. Since the circle with diameter AB passes exactly through the right focus $ F $ of the hyperbola, the equation of the circle with diameter AB is $ x^{2}+y^{2}=c^{2} $. Let $ |AF|=m $, $ |BF|=n $, then $ m-n=2a $. The area of $ \\triangle ABF $ is $ S_{\\triangle ABF}=\\frac{1}{2}m\\cdot n=4a^{2} $, and $ m^{2}+n^{2}=|AB|^{2}=4c^{2} $. Combining the three equations:  \n$$\n\\begin{cases}\nm-n=2a \\\\\nmn=8a^{2} \\\\\nm^{2}+n^{2}=4c^{2}\n\\end{cases}\n$$\nwe obtain  \n$$\n\\begin{cases}\nm=4a \\\\\nn=2a\n\\end{cases}\n$$\nHence, $ 20a^{2}=4c^{2} \\Rightarrow e=\\sqrt{5} $"
  },
  {
    "text": "Given that the distance from the point $(-2,3)$ to the focus of the parabola $y^{2}=2 p x(p>0)$ is $5$, then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;H: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(H) = (-2, 3);Distance(H, Focus(G)) = 5",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[12, 33]], [[46, 49]], [[2, 11]], [[15, 33]], [[12, 33]], [[2, 11]], [[2, 43]]]",
    "query_spans": "[[[46, 51]]]",
    "process": "From the parabola equation, its focus is $(\\frac{p}{2},0)$ $(p>0)$, $\\therefore \\sqrt{(-2-\\frac{p}{2})^{2}+(3-0)^{2}}=5$, solving gives: $p=-12$ (discarded) or $p=4$."
  },
  {
    "text": "Given the line $l_{1}$: $x-2 y-3=0$, and the parabola $C$: $y^{2}=4 x$. If the line $l_{2}$ passing through the point $(0,1)$ and perpendicular to the line $l_{1}$ intersects the parabola $C$ at points $M$ and $N$, then $|M N|$=?",
    "fact_expressions": "C: Parabola;H:Point;l1:Line;l2:Line;M: Point;N: Point;Expression(C) = (y^2 = 4*x);Coordinate(H) = (0, 1);Expression(l1) = (x - 2*y - 3 = 0);PointOnCurve(H,l2);IsPerpendicular(l1,l2);Intersection(l2, C) = {M, N}",
    "query_expressions": "Abs(LineSegmentOf(M, N))",
    "answer_expressions": "sqrt(15)",
    "fact_spans": "[[[25, 44], [78, 84]], [[47, 55]], [[2, 24], [56, 65]], [[68, 77]], [[86, 89]], [[90, 93]], [[25, 44]], [[47, 55]], [[2, 24]], [[46, 77]], [[55, 77]], [[68, 95]]]",
    "query_spans": "[[[97, 106]]]",
    "process": "According to the problem, let the equation of line $ l_{2} $ be $ 2x + y + m = 0 $. Substituting the point $ (0,1) $, we solve to get $ m = -1 $. Thus, the line $ l_{2}: 2x + y - 1 = 0 $. Solving simultaneously \n\\[\n\\begin{cases}\n2x + y - 1 = 0 \\\\\ny^2 = 4x\n\\end{cases}\n\\]\nwe obtain $ y^{2} + 2y - 2 = 0 $. Therefore, \n\\[\n|MN| = \\sqrt{1 + \\left( \\frac{1}{k} \\right)^{2}} |y_{1} - y_{2}| = \\sqrt{1 + \\left( \\frac{1}{k} \\right)^{2}} \\cdot \\sqrt{(y_{1} + y_{2})^{2}} \\frac{1}{4y_{1}y_{2}} = \\sqrt{1 + \\left( -\\frac{1}{2} \\right)^{2}} \\sqrt{2^{2} + 8} = \\frac{\\sqrt{5}}{2} \\times 2\\sqrt{3} = \\sqrt{15}\n\\]"
  },
  {
    "text": "Given that point $P$ lies on the left branch of the hyperbola $C$: $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, and $F_{1}$, $F_{2}$ are its left and right foci, respectively. If $\\angle F_{1} P F_{2}=60^{\\circ}$, then $\\frac{1}{|P F_{1}|}-\\frac{1}{|P F_{2}|}$=?",
    "fact_expressions": "P: Point;C: Hyperbola;Expression(C) = (x^2/16 - y^2/9 = 1);PointOnCurve(P, LeftPart(C)) = True;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "-1/Abs(LineSegmentOf(P, F2)) + 1/Abs(LineSegmentOf(P, F1))",
    "answer_expressions": "2/9",
    "fact_spans": "[[[2, 6]], [[7, 51], [71, 72]], [[7, 51]], [[2, 54]], [[55, 62]], [[63, 70]], [[55, 77]], [[55, 77]], [[79, 113]]]",
    "query_spans": "[[[115, 158]]]",
    "process": "Let $ r_{1} = |PF_{1}| $, $ r_{2} = |PF_{2}| $. By the definition of a hyperbola, we have $ r_{2} - r_{1} = 8 $. In $ \\triangle PF_{1}F_{2} $, by the cosine law, we obtain $ r_{2}^{2} + r_{1}^{2} - r_{1}r_{2} = 100 $. Solving gives $ r_{1}r_{2} = 36 $, so $ \\frac{1}{r_{1}} - \\frac{1}{r_{2}} = \\frac{2}{9} $."
  },
  {
    "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ have its right focus at $F(c, 0)$, and let the line $l$: $y=\\sqrt{2}(x-c)$ intersect the hyperbola $C$ at points $A$ and $B$. If $\\overrightarrow{A F}=t \\overrightarrow{F B}$ $(t>0)$, then what is the range of real values for $t$?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;F: Point;A: Point;B: Point;t: Real;c:Number;a>0;b>0;t>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(l)=(y=sqrt(2)*(x-c));Coordinate(F) = (c, 0);RightFocus(C) = F;Intersection(l,C)={A,B};VectorOf(A, F) = t*VectorOf(F, B)",
    "query_expressions": "Range(t)",
    "answer_expressions": "(0,2-sqrt(3))+(2+sqrt(3),+oo)",
    "fact_spans": "[[[77, 101]], [[1, 62], [102, 108]], [[9, 62]], [[9, 62]], [[67, 76]], [[110, 113]], [[114, 117]], [[172, 177]], [[67, 76]], [[9, 62]], [[9, 62]], [[120, 170]], [[1, 62]], [[77, 101]], [[67, 76]], [[1, 76]], [[77, 119]], [[120, 170]]]",
    "query_spans": "[[[172, 184]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. By combining the equation of line $ l $ and the hyperbola, eliminating $ x $, we obtain a quadratic equation. Using Vieta's formulas, we get $ y_{1}+y_{2}=-\\frac{2\\sqrt{2}b^{2}c}{b^{2}-2a^{2}} $\\textcircled{1}, $ y_{1}y_{2}=\\frac{2b^{4}}{b^{2}-2a^{2}} $\\textcircled{2}. Also, from $ \\overrightarrow{AF}=t\\overrightarrow{FB} $ ($ t>0 $), we obtain $ -y_{1}=ty_{2} $\\textcircled{3}. Substituting \\textcircled{3} into \\textcircled{1} and \\textcircled{2} and simplifying yields $ e^{2}=\\frac{3(1-t)^{2}}{(1+t)^{2}} $. Since $ e^{2}>1 $, the range of $ t $ can be solved (detailed explanation). Let $ A(x_{1},y_{1}) $, solve the system:  \n$$\n\\begin{cases}\ny=\\sqrt{2}(x-c) \\\\\n\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1\n\\end{cases}\n$$\nEliminating $ x $ and simplifying gives $ (b^{2}-2a^{2})y^{2}+2\\sqrt{2}b^{2}cy+2b^{4}=0 $. $ \\Delta>0 $, so $ y_{1}+y_{2}=-\\frac{2\\sqrt{2}b^{2}c}{b^{2}-2a^{2}} $\\textcircled{1}, $ y_{1}y_{2}=\\frac{2b^{4}}{b^{2}-2a^{2}} $\\textcircled{2}. Since $ \\overrightarrow{AF}=t\\overrightarrow{FB} $ ($ t>0 $), we have $ -y_{1}=ty_{2} $\\textcircled{3}. Substituting \\textcircled{3} into \\textcircled{1} and \\textcircled{2} and simplifying gives $ -\\left[\\frac{2\\sqrt{2}b^{2}c}{(b^{2}-2a^{2})(1-t)}\\right]=\\frac{2b^{4}}{b^{2}-2a^{2}} $, then $ e^{2}=\\frac{3(1-t)^{2}}{2} $. Also, the eccentricity of the hyperbola satisfies $ e\\in(1,+\\infty) $, so $ e^{2}=\\frac{3(1-t)^{2}}{(1+t)^{2}}\\in(1,+\\infty) $. Solving gives $ t\\in(0,2-\\sqrt{3})\\cup(2+\\sqrt{3},+\\infty) $."
  },
  {
    "text": "It is known that the vertex of the parabola $C$ is at the origin, and the focus lies on the $x$-axis. The line $y = x$ intersects the parabola $C$ at points $A$ and $B$. If $P(2,2)$ is the midpoint of $AB$, then the equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;O: Origin;Vertex(C) = O;PointOnCurve(Focus(C), xAxis);G: Line;Expression(G) = (y = x);A: Point;B: Point;Intersection(G, C) = {A, B};P: Point;Coordinate(P) = (2, 2);MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[2, 8], [34, 40], [72, 78]], [[12, 16]], [[2, 16]], [[2, 25]], [[26, 33]], [[26, 33]], [[43, 46]], [[47, 50]], [[26, 52]], [[54, 62]], [[54, 62]], [[54, 70]]]",
    "query_spans": "[[[72, 82]]]",
    "process": ""
  },
  {
    "text": "Let $C$: $y^{2}=2 p x(p>0)$ have focus $F$. A line $l$ passing through $F$ with inclination angle $45^{\\circ}$ intersects $C$ at points $A$ and $B$, and $l$ is tangent to the circle $(x-5)^{2}+y^{2}=r^{2}(r>0)$ at the midpoint of $AB$. Then the value of $r$ is?",
    "fact_expressions": "l: Line;C:Curve;G: Circle;r: Number;A: Point;B: Point;F: Point;r>0;Expression(G) = (y^2 + (x - 5)^2 = r^2);Expression(C)=(y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(F, l);Inclination(l) = ApplyUnit(45, degree);Intersection(l, C) = {A, B};TangentPoint(l,G)=MidPoint(LineSegmentOf(A,B));p: Number;p>0",
    "query_expressions": "r",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[54, 59], [75, 78]], [[1, 24], [60, 63]], [[80, 109]], [[122, 125]], [[64, 67]], [[68, 71]], [[28, 31], [33, 36]], [[81, 109]], [[80, 109]], [[1, 24]], [[1, 31]], [[32, 59]], [[37, 59]], [[54, 73]], [[75, 120]], [6, 23], [6, 23]]",
    "query_spans": "[[[122, 129]]]",
    "process": "C: y^{2}=2px(p>0) has focus F(\\frac{p}{2},0). Since the inclination angle of line l is 45^{\\circ}, the equation of line l is y=x-\\frac{p}{2}. Solving the system \\begin{cases}y=x-\\frac{p}{2}\\\\y^{2}=2px\\end{cases}, we obtain x^{2}-3px+\\frac{p^{2}}{4}=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), and the midpoint of AB be (x_{0},y_{0}). Then x_{1}+x_{2}=3p, x_{0}=\\frac{x_{1}+x_{2}}{2}=\\frac{3p}{2}, y_{0}=\\frac{3p}{2}-\\frac{p}{2}=p. Since l is tangent to the circle (x-5)^{2}+y^{2}=r^{2}(r>0) at the midpoint of AB, \\frac{3p}{2}-5=\\frac{3p}{2}, solving gives p=2, r=2\\sqrt{2}, p^{2}=r^{2}"
  },
  {
    "text": "Let the two foci of hyperbola $C$ be $(-\\sqrt{2}, 0)$, $(\\sqrt{2}, 0)$, and one vertex be $(1,0)$. Then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;F1: Point;F2: Point;Coordinate(F1) = (-sqrt(2), 0);Coordinate(F2) = (sqrt(2), 0);Focus(C) = {F1, F2};I: Point;OneOf(Vertex(C)) = I;Coordinate(I) = (1, 0)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2 = 1",
    "fact_spans": "[[[1, 7], [61, 64]], [[13, 29]], [[31, 46]], [[13, 29]], [[31, 46]], [[1, 46]], [[52, 59]], [[1, 59]], [[52, 59]]]",
    "query_spans": "[[[61, 69]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{36}-\\frac{y^{2}}{9}=1$, what are its asymptotes?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/36 - y^2/9 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(3/4)*x",
    "fact_spans": "[[[2, 41], [43, 44]], [[2, 41]]]",
    "query_spans": "[[[43, 52]]]",
    "process": "According to the problem: $a^{2}=36$, $b^{2}=9$, and the foci are on the x-axis, so the asymptotes are $y=\\pm\\frac{b}{a}x$, therefore the answer is $y=\\pm\\frac{3}{4}x$."
  },
  {
    "text": "What is the equation of the directrix of the parabola $y=-4 x^{2}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = -4*x^2)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y = 1/16",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 22]]]",
    "process": "From $ y = -4x^{2} \\Leftrightarrow x^{2} = \\frac{-1}{4}y $, therefore the focus lies on the $ y $-axis, and the equation of the directrix is $ y = \\frac{1}{16} $."
  },
  {
    "text": "A point $M(3, t)$ on the parabola $y^{2}=2 p x(p>0)$ has distance $|M F|=p$ from the focus $F$. Then, what is the distance from $M$ to the coordinate origin?",
    "fact_expressions": "G: Parabola;p: Number;M: Point;F: Point;O:Origin;p>0;t:Number;Expression(G) = (y^2 = 2*(p*x));Coordinate(M) = (3, t);PointOnCurve(M, G);Focus(G) = F;Distance(M, F) = Abs(LineSegmentOf(M, F)) ;Abs(LineSegmentOf(M, F)) = p",
    "query_expressions": "Distance(M, O)",
    "answer_expressions": "3*sqrt(5)",
    "fact_spans": "[[[0, 21]], [[3, 21]], [[24, 33], [53, 56]], [[36, 39]], [[57, 61]], [[3, 21]], [[24, 33]], [[0, 21]], [[24, 33]], [[0, 33]], [[0, 39]], [[24, 51]], [[42, 51]]]",
    "query_spans": "[[[53, 66]]]",
    "process": "The directrix of the parabola $ y^{2}=2px $ is: $ x=-\\frac{p}{2} $. By the definition of the parabola, we have: $ 3-(-\\frac{p}{2})=p $, solving gives $ p=6 $. The equation of the parabola is $ y^{2}=12x $. Since point $ M(3,t) $ lies on the parabola, then $ t^{2}=36 $. Let the origin be $ O $, so $ |MO|=\\sqrt{3^{2}+t^{2}}=3\\sqrt{5} $. Therefore, the distance from $ M $ to the coordinate origin is $ 3\\sqrt{5} $."
  },
  {
    "text": "What is the distance from the focus to the directrix of the parabola $y=\\frac{1}{4} x^{2}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = (1/4)*x^2)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 24]], [[0, 24]]]",
    "query_spans": "[[[0, 35]]]",
    "process": "Convert the parabola $ y = \\frac{1}{4}x^2 $ into standard form: $ x^2 = 4y $. Therefore, the parabola opens upward, satisfying $ 2p = 4 $. Since $ \\frac{p}{2} = 1 $, the focus is $ (0, \\frac{p}{2}) $. Hence, the focus of the parabola is $ (0, 1) $. Also, since the directrix of the parabola is $ y = -\\frac{p}{2} $, that is, $ y = -1 $, the distance between the focus and the directrix is $ d = 1 - (-1) = 2 $."
  },
  {
    "text": "The distance from the focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$ to its asymptote is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/3 = 1)",
    "query_expressions": "Distance(Focus(G), Asymptote(G))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[0, 38], [42, 43]], [[0, 38]]]",
    "query_spans": "[[[0, 51]]]",
    "process": "The distance from the focus of a hyperbola to its asymptote is $ b $, so the distance is $ b = \\sqrt{3} $."
  },
  {
    "text": "Given that point $P$ lies on the parabola $y^{2}=2x$, find the minimum value of the sum of the distance from point $P$ to point $Q(2,1)$ and the distance from point $P$ to the focus of the parabola.",
    "fact_expressions": "P: Point;G: Parabola;Expression(G) = (y^2 = 2*x);PointOnCurve(P, G) = True;Q: Point;Coordinate(Q) = (2, 1)",
    "query_expressions": "Min(Distance(P, Q) + Distance(P, Focus(G)))",
    "answer_expressions": "5/2",
    "fact_spans": "[[[2, 6], [24, 28], [42, 46]], [[7, 21], [47, 50]], [[7, 21]], [[2, 22]], [[29, 38]], [[29, 38]]]",
    "query_spans": "[[[24, 63]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and the focal distance is $2c$. If a line $y=-\\sqrt{3}(x-c)$ intersects the ellipse at a point $M$ such that $\\angle M F_{2} F_{1}=2 \\angle M F_{1} F_{2}$, then the eccentricity of the ellipse is equal to?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;c: Number;FocalLength(G) = 2*c;H: Line;Expression(H) = (y = -sqrt(3)*(-c + x));M: Point;OneOf(Intersection(H, G)) = M;AngleOf(M, F2, F1) = 2*AngleOf(M, F1, F2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[2, 54], [109, 111], [171, 173]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[62, 69]], [[70, 77]], [[2, 77]], [[2, 77]], [[81, 86]], [[2, 86]], [[88, 108]], [[88, 108]], [[116, 119]], [[88, 119]], [[121, 167]]]",
    "query_spans": "[[[171, 180]]]",
    "process": "By the given condition, $\\angle F_{1}MF_{2}=90^{\\circ}$, using the side-angle relationship in a right triangle, $|MF_{2}|$ and $|MF_{1}|$ can be obtained, then using the definition of the ellipse and the formula for eccentricity, the result follows. Let the inclination angle of the line $y=-\\sqrt{3}(x-c)$ be $\\alpha$, then $\\tan\\alpha=-\\sqrt{3}$. Since $0^{\\circ}\\leqslant\\alpha<180^{\\circ}$, $\\alpha=120^{\\circ}$. Therefore, $\\angle MF_{2}F_{1}=60^{\\circ}$. Since $\\angle MF_{2}F_{1}=2\\angle MF_{1}F_{2}$, $\\angle MF_{1}F_{2}=30^{\\circ}$. Hence, $\\angle F_{1}MF_{2}=90^{\\circ}$. In the right triangle $F_{1}MF_{2}$, let $c=1$, then $|MF_{2}|=1$, $|MF_{1}|=\\sqrt{2^{2}-1^{2}}=\\sqrt{3}$. By the definition of the ellipse, $2a=|MF_{1}|+|MF_{2}|=\\sqrt{3}+1$. Therefore, the eccentricity of the ellipse is $e=\\frac{2c}{2a}=\\frac{2}{\\sqrt{3}+1}=\\sqrt{3}-1$."
  },
  {
    "text": "If the vertex of the parabola is at the origin and the focus lies on the $y$-axis, and it passes through the point $(1,-4)$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;O: Origin;Vertex(G) = O;PointOnCurve(Focus(G),yAxis) = True;H: Point;Coordinate(H) = (1, -4);PointOnCurve(H,G) = True",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 = -y/4",
    "fact_spans": "[[[1, 4], [36, 39]], [[8, 12]], [[1, 12]], [[1, 21]], [[25, 34]], [[25, 34]], [[1, 34]]]",
    "query_spans": "[[[36, 44]]]",
    "process": "By the given conditions, the equation of the parabola can be written as $x^{2} = -2py$ ($p > 0$). Since the parabola passes through the point $(1, -4)$, we have $1 = 8p$, so $p = \\frac{1}{8}$. Therefore, the equation of the parabola is $x^{2} = -\\frac{1}{4}y$."
  },
  {
    "text": "Given that point $P$ moves on the parabola $x^{2}=4 y$, $F$ is the focus of the parabola, and point $A$ has coordinates $(2, 3)$, find the minimum value of $PA+PF$?",
    "fact_expressions": "G: Parabola;P: Point;A: Point;F: Point;Expression(G) = (x^2 = 4*y);Coordinate(A) = (2, 3);PointOnCurve(P, G);Focus(G) = F",
    "query_expressions": "Min(LineSegmentOf(P, A) + LineSegmentOf(P, F))",
    "answer_expressions": "4",
    "fact_spans": "[[[7, 21], [29, 32]], [[2, 6]], [[36, 40]], [[25, 28]], [[7, 21]], [[36, 52]], [[2, 24]], [[25, 35]]]",
    "query_spans": "[[[54, 67]]]",
    "process": ""
  },
  {
    "text": "It is known that the center of the ellipse is at the origin, the foci are on the $x$-axis, and the sum of the distances from point $P(3 \\sqrt{2}, 4)$ on the ellipse to the two foci is $12$. Then, what is the standard equation of the ellipse?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;PointOnCurve(Focus(G), xAxis);P: Point;Coordinate(P) = (3*sqrt(2), 4);PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};Distance(P, F1) + Distance(P, F2) = 12",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/36+y^2/32=1",
    "fact_spans": "[[[2, 4], [22, 24], [61, 63]], [[8, 12]], [[2, 12]], [[2, 21]], [[26, 45]], [[26, 45]], [[22, 45]], [], [], [[22, 49]], [[22, 59]]]",
    "query_spans": "[[[61, 70]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ is $\\frac{\\sqrt{2}}{2}$, then the real number $m$=?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (x^2/4 + y^2/m = 1);Eccentricity(G) = sqrt(2)/2",
    "query_expressions": "m",
    "answer_expressions": "{2,8}",
    "fact_spans": "[[[2, 39]], [[66, 71]], [[2, 39]], [[2, 64]]]",
    "query_spans": "[[[66, 73]]]",
    "process": "\\textcircled{1} If the foci are on the x-axis, then m<4, so a^{2}=4, b^{2}=m \\therefore c^{2}=a^{2}-b^{2}=4-m \\therefore \\frac{c^{2}}{a^{2}}=\\frac{4-m}{4}=\\frac{1}{2}, i.e., m=2. \\textcircled{2} If the foci are on the y-axis, then m>4, so a^{2}=m, b^{2}=4 \\therefore c^{2}=a^{2}-b^{2}=m-4 \\therefore \\frac{c^{2}}{a^{2}}=\\frac{m-4}{m}=\\frac{1}{2}, i.e., m=8."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=2 x$ is $F$, point $P$ is a moving point on the parabola, and there is a point $A(3,2)$.  \nThen the minimum value of $|P A|+|P F|$ is?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 2*x);Coordinate(A) = (3, 2);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "7/2",
    "fact_spans": "[[[2, 16], [29, 32]], [[39, 48]], [[24, 28]], [[20, 23]], [[2, 16]], [[39, 48]], [[2, 23]], [[24, 36]]]",
    "query_spans": "[[[51, 70]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, let the left focus be $F$. If a line passing through point $F$ with an inclination angle of $45^{\\circ}$ intersects the right branch of the hyperbola at exactly one point, then the range of values for the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F;PointOnCurve(F, H);Inclination(H) = ApplyUnit(45, degree);NumIntersection(H, RightPart(G)) = 1",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(\\sqrt{2},+\\infty)",
    "fact_spans": "[[[2, 58], [94, 97], [111, 114]], [[5, 58]], [[5, 58]], [[91, 93]], [[63, 66], [69, 73]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 66]], [[68, 93]], [[74, 93]], [[91, 108]]]",
    "query_spans": "[[[111, 125]]]",
    "process": "Let the equation of the line be y = x + c. Substituting into the hyperbola equation gives (b^{2}-a^{2})x^{2}-2a^{2}cx-a^{2}c^{2}-a^{2}b^{2}=0. According to the problem, the equation should have one positive root and one negative root, so the product of the roots is less than zero, i.e., \\frac{a^{2}c^{2}+a^{2}}{b^{2}-a^{2}}>0,\\therefore b^{2}-a^{2}>0,\\frac{b^{2}}{a^{2}}>1. Hence e=\\sqrt{1+\\frac{b^{2}}{a^{2}}}>\\sqrt{2}."
  },
  {
    "text": "The equation of a parabola with vertex at the origin, coordinate axes as axes of symmetry, and passing through the point $(2,-1)$ is?",
    "fact_expressions": "G: Parabola;H: Point;O: Origin;Vertex(G)=O;Coordinate(H) = (2, -1);SymmetryAxis(G)=axis;PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=(1/2)*x,x^2=-4*y}",
    "fact_spans": "[[[26, 29]], [[16, 25]], [[3, 5]], [[0, 29]], [[16, 25]], [[6, 29]], [[15, 29]]]",
    "query_spans": "[[[26, 33]]]",
    "process": "First, assume the parabola equation, then determine the parameters based on the coordinates of the point to obtain the result. Since the vertex is at the origin and the coordinate axis is the axis of symmetry, the parabola equation can be set as $ y^{2} = mx $ or $ x^{2} = ny $. Because the parabola passes through the point $ (2, -1) $, we have $ (-1)^{2} = 2m $ or $ 2^{2} = n(-1) $, that is, $ m = \\frac{1}{2} $, $ n = -4 $. Thus, the parabola equations are $ y^{2} = \\frac{1}{2}x $ or $ x^{2} = -4y $."
  },
  {
    "text": "Given that point $A$ lies on the parabola $y^{2}=3x$, the tangent to the parabola at point $A$ intersects the $x$-axis at point $B$, and the focus of the parabola is $F$. If $\\angle BAF=30^{\\circ}$, then the coordinates of $A$ are?",
    "fact_expressions": "G: Parabola;B: Point;A: Point;F: Point;Expression(G) = (y^2 = 3*x);PointOnCurve(A, G);Focus(G) = F;H:Line;TangentOnPoint(A,G)=H;Intersection(H,xAxis)=B;AngleOf(B,A,F)=ApplyUnit(30,degree)",
    "query_expressions": "Coordinate(A)",
    "answer_expressions": "(9/4, pm*(3*sqrt(3)/2))",
    "fact_spans": "[[[7, 21], [29, 32], [47, 50]], [[42, 46]], [[2, 6], [2, 6], [85, 88]], [[54, 57]], [[7, 21]], [[2, 22]], [[47, 57]], [], [[23, 35]], [[23, 46]], [[59, 84]]]",
    "query_spans": "[[[85, 93]]]",
    "process": "F(\\frac{3}{4},0) Let A(\\frac{y_{0}^{2}}{3},y_{0}), y_{0}\\neq0. According to the problem, the slope of the tangent line passing through point A exists and is not zero, denoted as k. Then the tangent line equation is y-y_{0}=k(x-\\frac{y_{0}^{2}}{3}), i.e., y=kx+y_{0}-\\frac{k}{x}\\frac{ky_{0}^{2}}{0}\\frac{ky_{0}^{2}}{0}. From \\begin{cases}y=kx+y_{0}-\\frac{k}{}\\\\y^{2}=3x\\end{cases}, simplifying yields k\\cdot y^{2}-3y+3y_{0}-ky_{0}^{2}=0, \\Delta=9-4k(3y_{0}-ky_{0}^{2})=0, 4k^{2}y_{0}^{2}-12ky_{0}+9=0, (2ky_{0}-3)^{2}=0, k=\\frac{3}{2y_{0}}. Thus, the tangent line equation is y-y_{0}=\\frac{3}{2y_{0}}(x-\\frac{y_{0}^{2}}{3}). Letting y=0 gives x=-\\frac{y_{0}^{2}}{3}, hence B(-\\frac{y_{0}^{2}}{3},0). \\overrightarrow{AB}=(\\frac{-2y_{0}^{2}}{3},-y_{0}), |\\overrightarrow{AB}|, \\overrightarrow{AF}=(\\frac{9-4y_{0}^{2}}{12},-y_{0}). According to the problem, \\cos\\angle BAF=\\frac{\\overrightarrow{AB}\\cdot\\overrightarrow{AF}}{|\\overrightarrow{AB}|\\cdot|\\overrightarrow{AF}|}=\\frac{\\sqrt{3}}{2}\\frac{y_{0}^{2}}{2}. Since y_{0}\\neq0, it follows that y_{0}^{2}=\\frac{27}{4}, y_{0}=\\pm\\frac{3\\sqrt{3}}{2}, at this time x_{0}=\\frac{y_{0}^{2}}{3}=\\frac{27}{4}\\times\\frac{1}{3}=\\frac{9}{4}. Therefore, the coordinates of point A are (\\frac{9}{4},\\pm\\frac{3\\sqrt{3}}{2})."
  },
  {
    "text": "A line with slope $2$ passes through the focus of the parabola $y^{2}=4x$ and intersects the parabola at points $A$ and $B$. Then $|AB|=$?",
    "fact_expressions": "H: Line;Slope(H) = 2;G: Parabola;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), H);A: Point;B: Point;Intersection(H, G) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "5",
    "fact_spans": "[[[7, 9]], [[0, 9]], [[11, 25], [30, 33]], [[11, 25]], [[7, 28]], [[35, 38]], [[41, 44]], [[7, 46]]]",
    "query_spans": "[[[48, 57]]]",
    "process": ""
  },
  {
    "text": "If the distance from the focus to the directrix of the parabola $y^{2}=2 p x(p>0)$ is $1$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;p: Number;p>0;Expression(G) = (y^2 = 2*p*x);Distance(Focus(G),Directrix(G))=1",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=2*x",
    "fact_spans": "[[[1, 22], [26, 27], [39, 42]], [[4, 22]], [[4, 22]], [[1, 22]], [[1, 36]]]",
    "query_spans": "[[[39, 47]]]",
    "process": "From the geometric meaning of p (p is the distance from the focus to the directrix), we obtain p=1, so the equation of the parabola is y^{2}=2x"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $P$ is an arbitrary point on the ellipse. The line $F_{2} M$ is perpendicular to $O P$ and intersects the segment $F_{1} P$ at point $M$. If $|F_{1} M|=2|M P|$, then the range of values for the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;a: Number;b: Number;M: Point;F2: Point;P: Point;F1: Point;O: Origin;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(G)=F2;LeftFocus(G)=F1;PointOnCurve(P,G);IsPerpendicular(LineOf(F2,M),LineOf(O,P));Intersection(LineOf(F2,M),LineSegmentOf(F1,P))=M;Abs(LineSegmentOf(F1, M)) = 2*Abs(LineSegmentOf(M, P))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1/2,1)",
    "fact_spans": "[[[2, 54], [83, 85], [151, 153]], [[4, 54]], [[4, 54]], [[124, 128]], [[71, 78]], [[79, 82]], [[63, 70]], [[105, 110]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 78]], [[2, 78]], [[79, 90]], [[91, 110]], [[91, 128]], [[130, 148]]]",
    "query_spans": "[[[150, 164]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ with focus $F$ and directrix $l$, a line passing through $F$ intersects the parabola at $M$ and the directrix at $N$, and $\\overrightarrow{M F}=\\frac{1}{3} \\overrightarrow{M N}$. Then the sine of the inclination angle of the line $M N$ is?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;M: Point;N: Point;F: Point;l: Line;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;Directrix(G) = l;PointOnCurve(F, H);Intersection(H, G) = M;Intersection(H, l) = N;VectorOf(M, F) = VectorOf(M, N)/3",
    "query_expressions": "Sin(Inclination(LineOf(M, N)))",
    "answer_expressions": "2*sqrt(2)/3",
    "fact_spans": "[[[2, 23], [46, 49]], [[5, 23]], [[43, 45]], [[50, 53]], [[58, 61]], [[27, 30], [39, 42]], [[34, 37]], [[5, 23]], [[2, 23]], [[2, 30]], [[2, 37]], [[38, 45]], [[43, 53]], [[43, 61]], [[63, 118]]]",
    "query_spans": "[[[120, 137]]]",
    "process": "As shown in the figure: draw $MD\\bot$ directrix $l$ from point $M$, then $|MF|=|MD|$, and $\\overrightarrow{MF}=\\frac{1}{3}\\overrightarrow{MN}$, so $|MF|=\\frac{1}{3}|MN|$, i.e., $|MD|=\\frac{1}{3}|MN|$, thus $\\frac{|MD|}{|MN|}=\\frac{1}{3}$. Let the angle between the line and the directrix be $\\theta$, and the inclination angle of the line be $\\alpha$, then $\\sin\\theta=\\frac{1}{3}$, and $\\sin\\theta=\\sin(\\frac{\\pi}{2}-\\alpha)=\\cos\\alpha$, $\\alpha\\in[0,\\pi)$, so $\\sin\\alpha=\\frac{2\\sqrt{2}}{3}$."
  },
  {
    "text": "Given that the center of the hyperbola is at the origin and its eccentricity is $\\sqrt{3}$, if one of its directrices coincides with the directrix of the parabola $y^{2}=4x$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Center(G) = O;O: Origin;Eccentricity(G) = sqrt(3);H: Parabola;Expression(H) = (y^2 = 4*x);OneOf(Directrix(G)) = Directrix(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/3 - y^2/6 = 1",
    "fact_spans": "[[[2, 5], [29, 30], [58, 61]], [[2, 11]], [[9, 11]], [[2, 26]], [[36, 50]], [[36, 50]], [[29, 55]]]",
    "query_spans": "[[[58, 66]]]",
    "process": ""
  },
  {
    "text": "The ellipse $E$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ has left and right foci $F_{1}$, $F_{2}$ respectively. A line $l$ passing through the origin intersects $E$ at points $A$, $B$. $A F_{1}$, $B F_{2}$ are both perpendicular to the $x$-axis. Then $|A B|=$?",
    "fact_expressions": "l: Line;E: Ellipse;A: Point;F1: Point;B: Point;F2: Point;O:Origin;Expression(E) = (x^2/4 + y^2/3 = 1);LeftFocus(E) = F1;RightFocus(E) = F2;PointOnCurve(O,l);Intersection(l, E) = {A, B};IsPerpendicular(LineSegmentOf(A,F1),xAxis);IsPerpendicular(LineSegmentOf(B,F2),xAxis)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "sqrt(13)",
    "fact_spans": "[[[71, 76]], [[0, 42], [77, 80]], [[82, 85]], [[51, 58]], [[86, 89]], [[59, 66]], [[68, 70]], [[0, 42]], [[0, 66]], [[0, 66]], [[67, 76]], [[71, 91]], [[92, 121]], [[92, 121]]]",
    "query_spans": "[[[123, 132]]]",
    "process": "In the given ellipse, \\( c^{2} = a^{2} - b^{2} = 4 - 3 = 1 \\). Since line \\( l \\) passes through the origin and intersects the ellipse at points \\( A \\) and \\( B \\), \\( A \\) and \\( B \\) are symmetric with respect to the origin. Also, \\( AF_{1} \\) and \\( BF_{2} \\) are both perpendicular to the x-axis. Let \\( A(-1, y_{1}) \\), \\( B(1, -y_{1}) \\), then \\( |AB| = \\sqrt{4 + (-y_{1} - y_{1})^{2}} = \\sqrt{4 + 4y_{1}^{2}} \\). Since \\( A \\) lies on the ellipse, \\( \\frac{1}{4} + \\frac{y_{1}^{2}}{3} = 1 \\), we obtain \\( y_{1}^{2} = \\frac{9}{4} \\)."
  },
  {
    "text": "The coordinates of the focus of the parabola $y=\\frac{1}{m} x^{2}(m \\neq 0)$ are?",
    "fact_expressions": "G: Parabola;m: Number;Expression(G) = (y = x^2/m);Negation(m = 0)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, m/4)",
    "fact_spans": "[[[0, 34]], [[3, 34]], [[0, 34]], [[3, 34]]]",
    "query_spans": "[[[0, 41]]]",
    "process": "According to the given conditions, the standard equation of the parabola is $x^{2}=my$, so the coordinates of the focus are $(0,\\frac{m}{4})$"
  },
  {
    "text": "Draw a line through the focus of the parabola $y^{2}=4x$ intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $x_{1}+x_{2}=6$, then $|AB|=$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;x1:Number;x2:Number;y1:Number;y2:Number;Expression(G)=(y^2 = 4*x);Coordinate(A)=(x1,y1);Coordinate(B)=(x2,y2);PointOnCurve(Focus(G),H);Intersection(H, G) ={A,B};x1+x2=6",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 15], [22, 25]], [[19, 21]], [[26, 43]], [[44, 62]], [[66, 81]], [[66, 81]], [[26, 43]], [[44, 62]], [[1, 15]], [[26, 43]], [[44, 62]], [[0, 21]], [[19, 64]], [[66, 81]]]",
    "query_spans": "[[[83, 91]]]",
    "process": ""
  },
  {
    "text": "Draw two lines $l_1$ and $l_2$ through the focus $F$ of the parabola $C$: $x^{2}=4 y$, with slopes $k_{1}$ and $k_{2}$ respectively, where $l_1$ intersects the parabola $C$ at points $A$, $B$, and $l_2$ intersects the parabola $C$ at points $D$, $E$. If $k_{1} k_{2}=-2$, then the minimum value of $|A B|+|D E|$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 4*y);F: Point;Focus(C) = F;l1: Line;l2: Line;k1: Number;k2: Number;Slope(l1) = k1;Slope(l2) = k2;PointOnCurve(F, l1);PointOnCurve(F, l2);A: Point;B: Point;Intersection(l1, C) = {A, B};D: Point;E: Point;Intersection(l2, C) = {D, E};k1*k2 = -2",
    "query_expressions": "Min(Abs(LineSegmentOf(A, B)) + Abs(LineSegmentOf(D, E)))",
    "answer_expressions": "24",
    "fact_spans": "[[[1, 20], [78, 84], [103, 109]], [[1, 20]], [[22, 25]], [[1, 25]], [[49, 58], [70, 77]], [[59, 67], [95, 102]], [[31, 38]], [[39, 46]], [[26, 67]], [[26, 67]], [[0, 67]], [[0, 67]], [[85, 88]], [[89, 92]], [[70, 94]], [[110, 113]], [[114, 117]], [[95, 119]], [[121, 137]]]",
    "query_spans": "[[[139, 158]]]",
    "process": "From the given conditions: $ F(0,1) $, then line $ l_{1} $ is: $ y-1 = k_{1}x $, intersecting with $ C: x^{2} = 4y $ gives: $ x^{2} - 4k_{1}x - 4 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = 4k_{1} $, $ x_{1}x_{2} = -4 $, $ y_{1} + y_{2} = k_{1}(x_{1} + x_{2}) + 2 = 4k_{1}^{2} + 2 $. Then $ |AB| = y_{1} + y_{2} + 2 = 4k_{1}^{2} + 4 $. Similarly, $ |DE| = 4k_{2}^{2} + 4 $. Therefore, $ |AB| + |DE| = 4(k_{1}^{2} + k_{2}^{2}) + 8 \\geqslant 8|k_{1}k_{2}| + 8 = 24 $, with equality holding if and only if $ |k_{1}| = |k_{2}| = \\sqrt{2} $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has a focal length of $2c$, and $A$ is the right vertex of $C$. On one of the asymptotes of $C$, there exist points $M$ and $N$ such that $|A M|=|A N|=c$ and $\\angle M A N=120^{\\circ}$. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;c: Number;FocalLength(C) = 2*c;A: Point;RightVertex(C) = A;M: Point;N: Point;PointOnCurve(M,l) = True;PointOnCurve(N,l) = True;Abs(LineSegmentOf(A, M)) = Abs(LineSegmentOf(A, N));AngleOf(M, A, N) = ApplyUnit(120, degree);Abs(LineSegmentOf(A, N)) = c;l: Line;OneOf(Asymptote(C)) = l",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 63], [79, 82], [88, 91], [157, 160]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[67, 72]], [[2, 72]], [[75, 78]], [[75, 86]], [[100, 103]], [[104, 107]], [[87, 109]], [[87, 109]], [[112, 127]], [[129, 155]], [[112, 127]], [], [[88, 97]]]",
    "query_spans": "[[[157, 166]]]",
    "process": "Let the asymptote equation be $ y = \\frac{b}{a}x $, then the distance from point $ A $ to the asymptote is $ d = \\frac{ab}{c} $. Given $ \\angle MAN = 120^{\\circ} $, $ |AM| = |AN| = c $, then $ \\frac{ab}{c} = \\frac{c}{2} $, so $ 2ab = c^{2} = a^{2} + b^{2} $, hence $ a = b $, $ e = \\sqrt{2} $."
  },
  {
    "text": "Given that point $P$ is a moving point on the parabola $y^{2}=2x$, then the minimum distance from point $P$ to the line $y=x+2$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);P: Point;PointOnCurve(P, G);H: Line;Expression(H) = (y = x + 2)",
    "query_expressions": "Min(Distance(P, H))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[7, 21]], [[7, 21]], [[2, 6], [27, 31]], [[2, 25]], [[32, 41]], [[32, 41]]]",
    "query_spans": "[[[27, 50]]]",
    "process": ""
  },
  {
    "text": "What is the length of the chord passing through the focus of the parabola $y=-2 x^{2}$ and perpendicular to the axis of symmetry?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = -2*x^2);L: LineSegment;IsChordOf(L, G);PointOnCurve(Focus(G), L);IsPerpendicular(L, SymmetryAxis(G))",
    "query_expressions": "Length(L)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[1, 16]], [[1, 16]], [], [[1, 28]], [[0, 28]], [[1, 28]]]",
    "query_spans": "[[[1, 31]]]",
    "process": "The standard equation of the parabola is $x^{2}=-\\frac{1}{2}y$, the coordinates of the focus are $(0,-\\frac{1}{8})$, the axis of symmetry is the $y$-axis, and the line passing through the focus and perpendicular to the axis of symmetry is $y=-\\frac{1}{8}$. Substituting into the equation $x^{2}=-\\frac{1}{2}y$ gives $x=\\pm\\frac{1}{4}$, so the length of the chord passing through the focus of $y=-2x^{2}$ and perpendicular to the axis of symmetry is $|\\frac{1}{4}-(-\\frac{1}{4})|=\\frac{1}{2}$ uniformly safe for.1"
  },
  {
    "text": "Given that $F$ is the right focus of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, $P$ is a moving point on the ellipse, and $A(0, \\frac{1}{2})$, then the maximum perimeter of $\\triangle A P F$ is?",
    "fact_expressions": "G: Ellipse;A: Point;P: Point;F: Point;Expression(G) = (x^2/2 + y^2 = 1);Coordinate(A) = (0, 1/2);RightFocus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Max(Perimeter(TriangleOf(A, P, F)))",
    "answer_expressions": "sqrt(5)+2*sqrt(2)",
    "fact_spans": "[[[6, 33], [42, 44]], [[49, 68]], [[38, 41]], [[2, 5]], [[6, 33]], [[49, 68]], [[2, 37]], [[38, 48]]]",
    "query_spans": "[[[70, 95]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$, draw a line through point $P(1,0)$ with slope $k$ ($k>0$) intersecting the parabola $C$ at points $A$ and $B$. If the chord lengths intercepted on the $x$-axis and $y$-axis by the circle with diameter $AB$ are equal, then $k$?",
    "fact_expressions": "C: Parabola;G: Circle;H: Line;A: Point;B: Point;P: Point;k:Number;k>0;Expression(C) = (y^2 = 8*x);Coordinate(P) = (1, 0);PointOnCurve(P, H);Slope(H) = k;Intersection(H, C) = {A, B};IsDiameter(LineSegmentOf(A,B),G);Length(InterceptChord(xAxis,G))=Length(InterceptChord(yAxis,G))",
    "query_expressions": "k",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 21], [49, 55]], [[77, 78]], [[46, 48]], [[56, 59]], [[60, 63]], [[24, 33]], [[97, 100]], [[37, 45]], [[2, 21]], [[24, 33]], [[23, 48]], [[34, 48]], [[46, 65]], [[67, 78]], [[77, 95]]]",
    "query_spans": "[[[97, 101]]]",
    "process": "The line passing through point P(1,0) with slope k (k>0) has the equation y = k(x-1). Combining this with the parabola C: y^{2} = 8x yields k^{2}x^{2} - (2k^{2}+8)x + k^{2} = 0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), and M(x_{0},y_{0}) be the midpoint of AB. Then x_{1}+x_{2} = \\frac{2k^{2}+8}{k^{2}}, so x_{0} = \\frac{k^{2}+4}{k^{2}}, y_{0} = k \\times \\left(\\frac{k^{2}+4}{k^{2}} - 1\\right) = \\frac{4}{k}. Since the chord lengths intercepted on the x-axis and y-axis by the circle with AB as diameter are equal, the distances from the center M(x_{0},y_{0}) to the x-axis and y-axis are equal. Therefore, \\frac{k^{2}+4}{12} = \\left|\\frac{4}{k}\\right|, solving gives k = 2 (discarding -2). The answer is: 2."
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ has two foci $F_{1}$, $F_{2}$. A line passes through $F_{1}$ and intersects the ellipse at points $P$, $Q$. Then the perimeter of $\\triangle P Q F_{2}$ is?",
    "fact_expressions": "G: Ellipse;H: Line;P: Point;Q: Point;F2: Point;F1: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(G) ={F1,F2};PointOnCurve(F1, H);Intersection(H, G) = {P, Q}",
    "query_expressions": "Perimeter(TriangleOf(P, Q, F2))",
    "answer_expressions": "20",
    "fact_spans": "[[[0, 38], [72, 74]], [[61, 63]], [[75, 78]], [[79, 82]], [[53, 60]], [[43, 51], [64, 71]], [[0, 38]], [[0, 60]], [[61, 71]], [[61, 82]]]",
    "query_spans": "[[[84, 110]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=4x$, a line $l$ with negative slope intersects the parabola at points $A(1,2)$ and $B$. Point $Q$ is the midpoint of segment $AB$. A line $l_{1}$ passing through point $Q$ and perpendicular to the $y$-axis intersects the parabola at point $C$. If point $P$ satisfies $2 \\overrightarrow{QC} = \\overrightarrow{CP}$, then the maximum value of the slope of line $OP$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);l: Line;Slope(l)<0;A: Point;Coordinate(A) = (1, 2);B: Point;Intersection(l, G) = {A, B};Q: Point;MidPoint(LineSegmentOf(A, B)) = Q;l1: Line;PointOnCurve(Q, l1);IsPerpendicular(l1, yAxis);C: Point;Intersection(l1, G) = C;P: Point;2*VectorOf(Q, C) = VectorOf(C, P);O: Origin",
    "query_expressions": "Max(Slope(LineOf(O, P)))",
    "answer_expressions": "(2-sqrt(2))/2",
    "fact_spans": "[[[2, 16], [31, 34], [93, 96]], [[2, 16]], [[25, 30]], [[17, 30]], [[35, 43]], [[35, 43]], [[46, 49]], [[25, 51]], [[52, 56], [69, 73]], [[52, 67]], [[82, 91]], [[68, 91]], [[74, 91]], [[97, 101]], [[82, 101]], [[103, 107]], [[109, 154]], [[158, 163]]]",
    "query_spans": "[[[156, 172]]]",
    "process": "Let $ l: kx - y + 2 - k = 0 $ ($ k < 0 $), substitute into $ y^{2} = 4x $ to get $ ky^{2} - 4y + 8 - 4k = 0 $. By Vieta's formulas, we have: $ y_{B} = \\frac{4 - 2k}{k} $, $ x_{B} = \\frac{k^{2} - 4k + 4}{k^{2}} $, $ y_{Q} = \\frac{2}{k} $. From $ \\frac{}{2}\\overrightarrow{QC} = \\overrightarrow{CP} $, we know $ y_{P} = y_{Q} = y_{C} = \\frac{2}{k} $, $ x_{C} = \\frac{1}{k^{2}} $, $ x_{Q} = \\frac{k^{2} - 2k + :}{k^{2}} - 2k^{2} + 4k - \\frac{\\frac{2}{k}}{2k^{2} + 4k - 1} = \\frac{}{-2k}\\underline{2} = \\frac{2 - \\sqrt{2}}{} \\frac{\\sqrt{2\\sqrt{2}}{2} + 4}{} $. The equality holds if and only if ``$ 2k = \\frac{1}{k} $'', i.e., $ k = -\\frac{\\sqrt{2}}{2} $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and $P$ is a point on the hyperbola $C$ such that $\\overrightarrow{P F_{1}} \\perp \\overrightarrow{P F_{2}}$. If the area of $\\Delta P F_{1} F_{2}$ is $9$, then $b=$?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Focus(C) = {F1, F2};PointOnCurve(P, C);IsPerpendicular(VectorOf(P, F1), VectorOf(P, F2));Area(TriangleOf(P, F1, F2)) = 9",
    "query_expressions": "b",
    "answer_expressions": "3",
    "fact_spans": "[[[18, 79], [89, 95]], [[192, 195]], [[26, 79]], [[85, 88]], [[2, 9]], [[10, 17]], [[26, 79]], [[26, 79]], [[18, 79]], [[2, 84]], [[85, 99]], [[101, 158]], [[161, 190]]]",
    "query_spans": "[[[192, 197]]]",
    "process": "From the given condition, we have ||PF_{1}|-|PF_{2}||=2a. \\therefore(|PF_{1}|-|PF_{2}|)^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}|\\cdot|PF_{2}|=4a^{2}. \\because PF_{1}\\bot PF_{2}, \\therefore|PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}=4c^{2}, \\therefore4c^{2}-2|PF_{1}|\\cdot|PF_{2}|=4a^{2}. \\because the area of \\triangle PF_{1}F_{2} is 9 and PF_{1}\\bot PF_{2}, \\therefore|PF_{1}|\\cdot|PF_{2}|=2\\times9=18, \\therefore4c^{2}-2\\times18=4a^{2}, \\therefore b^{2}=c^{2}-a^{2}=9, \\therefore b=3."
  },
  {
    "text": "If the horizontal coordinate of point $P$ on the parabola $y^{2}=8x$ is $3$, then what is the distance from point $P$ to the focus?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);P: Point;PointOnCurve(P, G);XCoordinate(P) = 3",
    "query_expressions": "Distance(P, Focus(G))",
    "answer_expressions": "5",
    "fact_spans": "[[[1, 15]], [[1, 15]], [[17, 21], [31, 35]], [[1, 21]], [[17, 29]]]",
    "query_spans": "[[[1, 43]]]",
    "process": "The focus of the parabola $ y^{2}=8x $ is $ F(2,0) $, and the directrix is $ x=-2 $. Since the x-coordinate of point $ P $ is 3, the distance from $ P $ to the directrix is 5. Therefore, the distance from point $ P $ to the focus is 5."
  },
  {
    "text": "Point $P$ moves on the circle $C$: $x^{2}+(y-2)^{2}=\\frac{1}{9}$, and point $Q$ moves on the ellipse $x^{2}+4 y^{2}=4$. Then the maximum value of $|P Q|$ is?",
    "fact_expressions": "G: Ellipse;C: Circle;P: Point;Q: Point;Expression(G) = (x^2 + 4*y^2 = 4);Expression(C) = (x^2 + (y - 2)^2 = 1/9);PointOnCurve(P, C);PointOnCurve(Q, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "1/3 + 2*sqrt(21)/3",
    "fact_spans": "[[[49, 68]], [[5, 40]], [[0, 4]], [[44, 48]], [[49, 68]], [[5, 40]], [[0, 41]], [[44, 69]]]",
    "query_spans": "[[[73, 86]]]",
    "process": "Let the coordinates of an arbitrary point Q on the ellipse \\( x^{2} + 4y^{2} = 4 \\) be \\( (x, y) \\), then \\( x^{2} + 4y^{2} = 4 \\). Hence, when \\( y = -\\frac{2}{3} \\), \\( d \\) attains its maximum value \\( \\sqrt{\\frac{28}{3}} = \\frac{2\\sqrt{21}}{3} \\), so the maximum value of \\( |PQ| \\) is \\( \\frac{1}{3} + \\frac{2\\sqrt{21}}{3} \\)."
  },
  {
    "text": "Given that the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1$ ($a>0$) with eccentricity $\\frac{2 \\sqrt{5}}{5}$ has its left focus coinciding with the focus of the parabola $y^{2}=m x$, then the real number $m$=?",
    "fact_expressions": "C: Hyperbola;a: Number;G: Parabola;m: Real;a>0;Expression(C) = (-y^2/4 + x^2/a^2 = 1);Expression(G) = (y^2 = m*x);Eccentricity(C)=2*sqrt(5)/5;LeftFocus(C)=Focus(G)",
    "query_expressions": "m",
    "answer_expressions": "-12",
    "fact_spans": "[[[29, 81]], [[37, 81]], [[86, 100]], [[107, 112]], [[37, 81]], [[29, 81]], [[86, 100]], [[2, 81]], [[29, 105]]]",
    "query_spans": "[[[107, 114]]]",
    "process": "First, find the value of $ a $ from the eccentricity of the hyperbola, thereby obtaining the left focus of the hyperbola; then find the focus coordinates of the parabola $ y^{2}=2mx $, and by equating them, solve for the real number $ m $. Since the eccentricity of the hyperbola $ \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1 $ ($ a>0 $) is $ \\frac{3\\sqrt{5}}{5} $, we have $ \\frac{\\sqrt{a^{2}+4}}{a}=\\frac{3\\sqrt{5}}{5} \\Rightarrow a^{2}=5 $. The left focus of hyperbola $ C $ is $ (-3,0) $, and the focus of the parabola $ y^{2}=2mx $ is $ \\left( \\frac{m}{4}, 0 \\right) $. Therefore, $ \\frac{m}{4}=-3 \\Rightarrow m=-12 $."
  },
  {
    "text": "If the focal distance of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{m+3}=1$ is $2$, then what is the value of the real number $m$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/(m + 3) = 1);m: Real;FocalLength(G) = 2",
    "query_expressions": "m",
    "answer_expressions": "{5, 7}",
    "fact_spans": "[[[1, 40]], [[1, 40]], [[49, 54]], [[1, 47]]]",
    "query_spans": "[[[49, 58]]]",
    "process": "When the foci of the ellipse are on the x-axis, $a^{2}=9$, $b^{2}=m+3$, $c^{2}=a^{2}-b^{2}=6-m$. Since the focal distance $2c=2$, we have $6-m=1$, solving gives: $m=5$; when the foci of the ellipse are on the y-axis, $a^{2}=m+3$, $b^{2}=9$, $c^{2}=a^{2}-b^{2}=m-6$. Since the focal distance $2c=2$, we have $m-6=1$, solving gives: $m=7$;"
  },
  {
    "text": "A point $P$ on the parabola $x^{2}=4 y$ is at a distance of $3$ from the focus. Then, the distance from point $P$ to the $y$-axis is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);P: Point;PointOnCurve(P, G);Distance(P, Focus(G)) = 3",
    "query_expressions": "Distance(P, yAxis)",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[17, 20], [32, 36]], [[0, 20]], [[17, 30]]]",
    "query_spans": "[[[32, 46]]]",
    "process": "The directrix of the parabola is given by $ y = -1 $. The distance from point $ P $ to the directrix is 3, so $ y_{P} - (-1) = 3 $, $ \\therefore y_{p} = 2 $. Thus, $ x^{2} = 8 $, so $ x = \\pm 2\\sqrt{2} $. Therefore, the distance from point $ P $ to the $ y $-axis is $ 2\\sqrt{2} $."
  },
  {
    "text": "The line $L$ intersects the parabola $x^{2}=8 y$ at points $A$ and $B$, and the midpoint of $AB$ is $M(1 , 1)$. Then the equation of $L$ is?",
    "fact_expressions": "G: Parabola;L: Line;A: Point;B: Point;M: Point;Expression(G) = (x^2 = 8*y);Coordinate(M) = (1, 1);Intersection(L,G)={A,B};MidPoint(LineSegmentOf(A,B))=M",
    "query_expressions": "Expression(L)",
    "answer_expressions": "x-4*y+3=0",
    "fact_spans": "[[[6, 20]], [[0, 5], [54, 57]], [[23, 26]], [[27, 30]], [[42, 52]], [[6, 20]], [[42, 52]], [[0, 32]], [[33, 52]]]",
    "query_spans": "[[[54, 62]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), then \\begin{cases}x_{1}^{2}=8y_{1}\\\\x_{2}^{2}=8y_{2}\\end{cases}. Subtracting gives: (x_{1}+x_{2})(x_{1}-x_{2})=8(y_{1}-y_{2}), so \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{x_{1}+x_{2}}{8}. Since the midpoint of AB is M(1,1), we have \\frac{x_{1}+x_{2}}{8}=\\frac{2}{8}=\\frac{1}{4}. Therefore, the equation of L is: y-1=\\frac{1}{4}(x-1), i.e., x-4y+3=0."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $C$: $x^{2}=4 y$ intersects the parabola $C$ at points $A$ and $B$. If the distance from the midpoint of chord $AB$ to the $x$-axis is $5$, then $|AB|=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 4*y);F: Point;Focus(C) = F;PointOnCurve(F, G);G: Line;Intersection(G, C) = {A, B};A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), C) ;Distance(MidPoint(LineSegmentOf(A, B)), xAxis) = 5",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "12",
    "fact_spans": "[[[1, 20], [30, 36]], [[1, 20]], [[23, 26]], [[1, 26]], [[0, 29]], [[27, 29]], [[27, 47]], [[38, 41]], [[42, 45]], [[30, 55]], [[50, 69]]]",
    "query_spans": "[[[72, 81]]]",
    "process": "Analysis: First, use the given conditions to find the distance from the midpoint of the chord to the directrix of the parabola. Then, draw perpendiculars from the endpoints of the chord to the directrix of the parabola. Using the property that the median line of a trapezoid equals half the sum of the upper and lower bases, and combining with the definition of the parabola, obtain the length of the focal chord. Specifically, according to the problem, the directrix of the parabola $ x^{2}=4y $ is $ y=-1 $, thus the distance from the midpoint of the chord to the directrix of the parabola is $ 5-(-1)=6 $. Now, draw perpendiculars from points $ A $ and $ B $ to the directrix, with feet at $ A' $ and $ B' $. According to the property of the median line of a trapezoid, we have $ |AA'|+|BB'|=2\\times12 $. By the definition of the parabola, $ |AB|=|AF|+|BF|=|AA'|+|BB'|=12 $. Hence, the answer is 12."
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{2-a}+\\frac{y^{2}}{a-1}=1$ represents an ellipse, what is the range of values for $a$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(2 - a) + y^2/(a - 1) = 1);a: Number",
    "query_expressions": "Range(a)",
    "answer_expressions": "(1, 3/2)+(3/2, 2)",
    "fact_spans": "[[[45, 47]], [[2, 47]], [[49, 52]]]",
    "query_spans": "[[[49, 59]]]",
    "process": "According to the characteristics of the standard equation of an ellipse, list the inequalities and solve them to obtain the result. Since the equation $\\frac{x^2}{2-a}+\\frac{y^{2}}{a-1}=1$ represents an ellipse, we have $\\begin{cases}2-a>0\\\\a-1>0\\\\2-a\\neqa-1\\end{cases}$, solving yields $1<a<2$ or $a\\neq\\frac{3}{2}$, that is, the range of $a$ is $(1,\\frac{3}{2})\\cup(\\frac{3}{2},2)$."
  },
  {
    "text": "It is known that a hyperbola passes through the point $(2,3)$, and one of its asymptotes has the equation $y=\\sqrt{3} x$. Then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (2, 3);PointOnCurve(H, G);Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[2, 5], [42, 45], [16, 17]], [[7, 15]], [[7, 15]], [[2, 15]], [[16, 39]]]",
    "query_spans": "[[[41, 52]]]",
    "process": "According to the problem, one asymptote of the hyperbola is given by $ y = \\sqrt{3}x $, so we can assume the hyperbola has the form $ x^{2} - \\frac{y^{2}}{3} = m $ ($ m \\neq 0 $). Since the hyperbola passes through the point $ (2, 3) $, we have $ 2^{2} - \\frac{3^{2}}{3} = m $, so $ m = 1 $. The standard equation of the hyperbola is $ x^{2} - \\frac{y^{2}}{3} = 1 $."
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, and let $P$ lie on the hyperbola such that $\\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_2}=0$. Then the value of $|\\overrightarrow{P F_{1}}| \\cdot |\\overrightarrow{P F_{2}}|$ is equal to?",
    "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;P: Point;Expression(G) = (x^2/4 - y^2 = 1);Focus(G)={F1,F2};PointOnCurve(P, G);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0",
    "query_expressions": "Abs(VectorOf(P, F1))*Abs(VectorOf(P, F2))",
    "answer_expressions": "2",
    "fact_spans": "[[[17, 45], [55, 58]], [[1, 8]], [[9, 16]], [[51, 54]], [[17, 45]], [[1, 50]], [[51, 59]], [[60, 115]]]",
    "query_spans": "[[[117, 183]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the foci of the ellipse $\\frac{x^{2}}{4+k}-\\frac{y^{2}}{3+k}=1(k<-\\frac{7}{2})$ are?",
    "fact_expressions": "G: Ellipse;k: Number;k < -7/2;Expression(G) = (x^2/(k + 4) - y^2/(k + 3) = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, pm*1)",
    "fact_spans": "[[[0, 57]], [[2, 57]], [[2, 57]], [[0, 57]]]",
    "query_spans": "[[[0, 64]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, the inclination angle of one asymptote $l$ is $\\frac{\\pi}{3}$, and the distance from a focus of $C$ to $l$ is $\\sqrt{3}$. Then the equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;l: Line;OneOf(Asymptote(C)) = l;Inclination(l) = pi/3;Distance(OneOf(Focus(C)), l) = sqrt(3)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2/3 = 1",
    "fact_spans": "[[[2, 53], [84, 87], [112, 118]], [[2, 53]], [[9, 53]], [[9, 53]], [[59, 62], [93, 96]], [[2, 62]], [[59, 82]], [[84, 110]]]",
    "query_spans": "[[[112, 123]]]",
    "process": "Since one asymptote $ l $ of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ has the equation $ y = \\frac{b}{a}x $, it follows that $ \\frac{b}{a} = \\tan\\frac{\\pi}{3} = \\sqrt{3} $, so $ b = \\sqrt{3}a $. Since one focus of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ is $ F(c,0) $ and the asymptote is $ l: y = \\frac{b}{a}x $, the distance from point $ F $ to the line $ l: y = \\frac{b}{a}x $ is $ \\frac{|bc|}{\\sqrt{a^{2}+b^{2}}} = b = \\sqrt{3} $. Solving gives $ a = 1 $, so the required equation of the hyperbola $ C $ is $ x^{2} - \\frac{y^{2}}{3} = 1 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, with foci $F_{1}$ and $F_{2}$. A line perpendicular to the $x$-axis is drawn through $F_{2}$, intersecting the hyperbola at points $A$ and $B$, and the inradius of triangle $\\triangle A B F_{1}$ is $a$. Then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;A: Point;B: Point;F1: Point;F2: Point;a>0;b>0;Focus(G) = {F1, F2};L:Line;IsPerpendicular(L,xAxis);Intersection(L,G)={A,B};PointOnCurve(F2, L);Radius(InscribedCircle(TriangleOf(A,B,F1)))=a",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(1+sqrt(5))/2",
    "fact_spans": "[[[2, 58], [96, 99], [145, 148]], [[2, 58]], [[5, 58]], [[139, 142]], [[100, 103]], [[104, 107]], [[63, 70]], [[71, 78], [80, 87]], [[5, 58]], [[5, 58]], [[2, 78]], [], [[79, 95]], [[79, 109]], [[79, 95]], [[111, 142]]]",
    "query_spans": "[[[145, 154]]]",
    "process": ""
  },
  {
    "text": "A line $l$ passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. If the distance from the midpoint $M$ of $AB$ to the directrix of the parabola is $6$, then what is the length of segment $AB$?",
    "fact_expressions": "G:Parabola;Expression(G) = (y^2 = 4*x);F:Point;Focus(G)=F;l:Line;A: Point;B: Point;Intersection(l,G)={A,B};M: Point;PointOnCurve(F, l);MidPoint(LineSegmentOf(A,B))=M;Distance(M,Directrix(G))=6",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "12",
    "fact_spans": "[[[1, 15], [28, 31], [55, 58]], [[1, 15]], [[18, 21]], [[1, 21]], [[22, 27]], [[32, 35]], [[36, 39]], [[22, 41]], [[51, 54]], [[0, 27]], [[44, 54]], [[51, 67]]]",
    "query_spans": "[[[69, 80]]]",
    "process": "From the definition of a parabola, the length of segment AB is $2 \\times 6 = 12$."
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{2}=1$ are $F_{1}$, $F_{2}$, and point $P$ lies on the ellipse. If $|P F_{1}|=4$, then what is $|P F_{2}|$? What is $\\angle F_{1} P F_{2}$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 + y^2/2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 4",
    "query_expressions": "Abs(LineSegmentOf(P, F2));AngleOf(F1, P, F2)",
    "answer_expressions": "2\n(2/3)*pi",
    "fact_spans": "[[[1, 38], [63, 65]], [[58, 62]], [[42, 49]], [[50, 57]], [[1, 38]], [[1, 57]], [[58, 66]], [[68, 81]]]",
    "query_spans": "[[[83, 96]], [[96, 120]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+y^{2}=1$ $(a>0)$ is $\\frac{\\sqrt{3}}{3}$, then the real number $a$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2 + x^2/a^2 = 1);a: Real;a>0;Eccentricity(G) = sqrt(3)/3",
    "query_expressions": "a",
    "answer_expressions": "{sqrt(6)/2,sqrt(6)/3}",
    "fact_spans": "[[[2, 38]], [[2, 38]], [[65, 70]], [[4, 38]], [[2, 63]]]",
    "query_spans": "[[[65, 72]]]",
    "process": "When $ a > 1 $, the eccentricity $ e = \\sqrt{\\frac{a^{2}-1}{a^{2}}} = \\frac{\\sqrt{3}}{3} $, solving gives: $ a = \\frac{\\sqrt{6}}{2} $. When $ 0 < a < 1 $, the eccentricity $ e = \\sqrt{\\frac{1-a^{2}}{1}} = \\frac{\\sqrt{3}}{3} $, solving gives: $ a = \\frac{\\sqrt{6}}{3} $. The correct result for this problem: $ \\frac{\\sqrt{6}}{2} $ or $ \\frac{\\sqrt{6}}{3} $."
  },
  {
    "text": "Let the parabola $C$: $y^{2}=2x$ have focus $F$, and let line $l$ pass through $F$ intersecting $C$ at points $A$ and $B$. If $|AF|=3|BF|$, then the equation of $l$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*x);F: Point;Focus(C)= F;l: Line;PointOnCurve(F, l) = True;Intersection(l, C) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(A, F)) = 3*Abs(LineSegmentOf(B, F))",
    "query_expressions": "Expression(l)",
    "answer_expressions": "y=pm*sqrt(3)*(x-1/2)",
    "fact_spans": "[[[1, 20], [38, 41]], [[1, 20]], [[24, 27], [34, 37]], [[1, 27]], [[28, 33], [70, 73]], [[28, 37]], [[28, 52]], [[43, 46]], [[47, 50]], [[54, 68]]]",
    "query_spans": "[[[70, 78]]]",
    "process": "According to the problem, the parabola $ y^{2} = 2x $ has focus $ F\\left(\\frac{1}{2}, 0\\right) $. Let line $ l: y = k\\left(x - \\frac{1}{2}\\right) $, and points $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. From $ |AF| = 3|BF| $, we obtain $ x_{1} + \\frac{1}{2} = 3\\left(x_{2} + \\frac{1}{2}\\right) $, which implies $ x_{1} = 3x_{2} + 1 $. Solving the system\n\\[\n\\begin{cases}\ny^{2} = 2x \\\\\ny = k\\left(x - \\frac{1}{2}\\right)\n\\end{cases}\n\\]\ngives $ k^{2}x^{2} - (k^{2} + 2)x + \\frac{1}{4}k^{2} = 0 $. Then $ x_{1}x_{2} = x_{2}(3x_{2} + 1) = \\frac{1}{4} $, solving yields $ x_{2} = \\frac{1}{6} $. Also, $ x_{1} + x_{2} = 4x_{2} + 1 = 1 + \\frac{2}{k^{2}} $, thus $ k^{2} = 3 $, $ k = \\pm\\sqrt{3} $. Therefore, the equation of line $ l $ is $ y = \\pm\\sqrt{3}\\left(x - \\frac{1}{2}\\right) $."
  },
  {
    "text": "Given the parabola $y^{2}=4x$, a line passing through the point $P(4,0)$ intersects the parabola at two points $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$. What is the minimum value of $y_{1}^{2}+y_{2}^{2}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Line;P: Point;Coordinate(P) = (4, 0);PointOnCurve(P, H);A: Point;x1: Number;y1: Number;Coordinate(A) = (x1, y1);B: Point;x2: Number;y2: Number;Coordinate(B) = (x2, y2);Intersection(H, G) = {A, B}",
    "query_expressions": "Min(y1^2 + y2^2)",
    "answer_expressions": "32",
    "fact_spans": "[[[2, 16], [32, 35]], [[2, 16]], [[29, 31]], [[19, 28]], [[19, 28]], [[18, 31]], [[38, 55]], [[38, 55]], [[38, 55]], [[38, 55]], [[57, 74]], [[57, 74]], [[57, 74]], [[57, 74]], [[29, 76]]]",
    "query_spans": "[[[79, 106]]]",
    "process": ""
  },
  {
    "text": "Given a point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ such that the distance from $P$ to one focus is $2$, then the distance from point $P$ to the other focus is?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/25 + y^2/9 = 1);PointOnCurve(P, G);F1:Point;F2:Point;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2);Distance(P, F1) = 2",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 40]], [[43, 46], [60, 64]], [[2, 40]], [[2, 46]], [], [], [[2, 51]], [[2, 70]], [[2, 70]], [[2, 58]]]",
    "query_spans": "[[[2, 75]]]",
    "process": "Since it is known that the distance from a point P on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ to one focus is 2, let the distance from point P to the other focus be $m$. According to the definition of an ellipse, $m+2=10$, solving gives: $m=8$. The answer is: $8$"
  },
  {
    "text": "If the line $x - y = 2$ intersects the parabola $y^{2} = 4x$ at points $A$ and $B$, then the coordinates of the midpoint of segment $AB$ are?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (x - y = 2);Intersection(H, G) = {A, B}",
    "query_expressions": "Coordinate(MidPoint(LineSegmentOf(A, B)))",
    "answer_expressions": "(4, 2)",
    "fact_spans": "[[[11, 25]], [[1, 10]], [[27, 30]], [[31, 34]], [[11, 25]], [[1, 10]], [[1, 36]]]",
    "query_spans": "[[[38, 52]]]",
    "process": ""
  },
  {
    "text": "The moving point $M$ is 1 unit farther from the fixed point $F(3, 0)$ than from the fixed line $l$: $x = -2$. Then, the trajectory equation of the moving point $M$ is?",
    "fact_expressions": "M: Point;F: Point;Coordinate(F) = (3, 0);l: Line;Expression(l) = (x = -2);Distance(M, F) = Distance(M, l) + 1",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "y^2=12*x",
    "fact_spans": "[[[2, 5], [44, 47]], [[8, 18]], [[8, 18]], [[23, 33]], [[23, 33]], [[2, 40]]]",
    "query_spans": "[[[44, 54]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^2}=1$ $(a>b>0)$ has its right focus at $F$, and $A$, $B$ are the upper vertex and right vertex of the ellipse, respectively. If $AB + BF = 2a$, then what is the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;B: Point;F: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2);RightFocus(G)=F;UpperVertex(G)=A;RightVertex(G)=B;LineSegmentOf(A, B) + LineSegmentOf(B, F) = 2*a",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(\\sqrt{3}-1)/2",
    "fact_spans": "[[[2, 52], [71, 73], [98, 100]], [[4, 52]], [[4, 52]], [[61, 64]], [[65, 68]], [[57, 60]], [[4, 52]], [[4, 52]], [[2, 52]], [[2, 60]], [[61, 80]], [[61, 80]], [[82, 95]]]",
    "query_spans": "[[[98, 105]]]",
    "process": ""
  },
  {
    "text": "Given that point $F$ is the focus of the parabola $y^{2}=2 p x$ $(p>0)$, a line passing through point $F$ with an inclination angle of $\\frac{\\pi}{4}$ intersects the parabola at points $A$ and $B$. The perpendicular bisector of segment $AB$ intersects the $x$-axis at point $M$. Then the value of $\\frac{4 p}{|F M|}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;H: Line;Inclination(H) = pi/4;PointOnCurve(F, H);A: Point;B: Point;Intersection(H, G) = {A, B};M: Point;Intersection(PerpendicularBisector(LineSegmentOf(A, B)), xAxis) = M",
    "query_expressions": "(4*p)/Abs(LineSegmentOf(F, M))",
    "answer_expressions": "2",
    "fact_spans": "[[[7, 28], [62, 65]], [[7, 28]], [[10, 28]], [[10, 28]], [[2, 6], [34, 38]], [[2, 31]], [[59, 61]], [[39, 61]], [[32, 61]], [[68, 71]], [[72, 76]], [[59, 76]], [[98, 102]], [[77, 102]]]",
    "query_spans": "[[[105, 128]]]",
    "process": "The focus of the parabola $ y^{2} = 2px $ ($ p > 0 $) is $ F\\left(\\frac{p}{2}, 0\\right) $, so the line passing through point $ F $ with an inclination angle of $ \\frac{\\pi}{4} $ is $ y = x - \\frac{p}{2} $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, and the midpoint of segment $ AB $ be $ N(x_{0}, y_{0}) $. From \n\\[\n\\begin{cases}\ny = x - \\frac{p}{2} \\\\\ny^{2} = 2px\n\\end{cases}\n\\]\nwe obtain $ x^{2} - 3px + \\frac{p^{2}}{4} = 0 $. Thus, $ x_{0} = \\frac{x_{1} + x_{2}}{2} = \\frac{3p}{2} $, $ y_{0} = p $. Therefore, the equation of the perpendicular bisector of segment $ AB $ is $ y - p = -\\left(x - \\frac{3p}{2}\\right) $. Letting $ y = 0 $, we get $ x = \\frac{5p}{2} $, so $ M\\left(\\frac{5p}{2}, 0\\right) $. Hence, $ |FM| = \\frac{5p}{2} - \\frac{p}{2} = 2p $, and $ \\frac{4p}{|FM|} = \\frac{4p}{2p} = 2 $."
  },
  {
    "text": "Given the ellipse $ C: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 $ $(a>b>0)$, $ A(0,2b) $, if for any point $ P $ on $ C $ it satisfies $ |PA| \\leq 3b $, then the range of the eccentricity of $ C $ is?",
    "fact_expressions": "C: Ellipse;a: Number;b: Number;A: Point;P: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (0, 2*b);PointOnCurve(P,C);Abs(LineSegmentOf(P,A))<=3*b",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(0,sqrt(6)/3]",
    "fact_spans": "[[[2, 60], [74, 77], [106, 109]], [[8, 60]], [[8, 60]], [[62, 72]], [[82, 85]], [[8, 59]], [[8, 60]], [[2, 60]], [[62, 72]], [[74, 85]], [[88, 104]]]",
    "query_spans": "[[[106, 120]]]",
    "process": "Let $ P(x,y) $, then $ |PA| = \\sqrt{x^{2} + (y - 2b)^{2}} = \\sqrt{ -\\frac{c^{2}}{b^{2}}y^{2} - 4by + a^{2} + 4b^{2} } = \\sqrt{ -\\frac{c^{2}}{b^{2}}\\left(y + \\frac{2b^{3}}{c^{2}}\\right)^{2} + a^{2} + 4b^{2} + \\frac{4b^{4}}{c^{2}} } \\leqslant 3b $. Since $ y \\in [-b, b] $, when $ -\\frac{2b^{3}}{c^{2}} > -b $, i.e., $ 2a^{2} < 3c^{2} $, $ |PA|_{\\max} = \\sqrt{a^{2} + 4b^{2} + \\frac{4b^{4}}{c^{2}}} \\leqslant 3b $, so $ a^{2} + 4b^{2} + \\frac{4b^{4}}{c^{2}} \\leqslant 9b^{2} $, $ a^{2} + \\frac{4b^{2}(c^{2} + b^{2})}{c^{2}} \\leqslant 9(a^{2} - c^{2}) $, thus $ a^{2}c^{2} + 4(a^{2} - c^{2})a^{2} \\leqslant 9a^{2}c^{2} - 9c^{4} $, i.e., $ 4a^{4} - 12a^{2}c^{2} + 9c^{4} \\leqslant 0 \\Rightarrow (2a^{2} - 3c^{2})^{2} \\leqslant 0 $, clearly this inequality does not hold. When $ -\\frac{2b^{3}}{c^{2}} \\leqslant -b $, i.e., $ 2a^{2} \\geqslant 3c^{2} $, $ |PA|_{\\max} = \\sqrt{9a^{2} - 9c^{2}} \\leqslant 3b $, which always holds. From $ 2a^{2} \\geqslant 3c^{2} $, we get $ \\frac{c^{2}}{a^{2}} \\leqslant \\frac{2}{3} $, so $ 0 < e \\leqslant \\frac{\\sqrt{6}}{3} $. In conclusion, the range of eccentricity is $ \\left(0, \\frac{\\sqrt{6}}{3}\\right] $."
  },
  {
    "text": "The line $l$ with slope $1$ passes through the right focus of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ and intersects the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ at points $A$ and $B$. Then the chord length $|AB|=$?",
    "fact_expressions": "l: Line;G: Ellipse;A: Point;B: Point;Expression(G) = (x^2/4 + y^2 = 1);PointOnCurve(RightFocus(G), l);Slope(l)=1;Intersection(l, G) = {A, B};IsChordOf(LineSegmentOf(A,B),G)",
    "query_expressions": "Length(Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "8/5",
    "fact_spans": "[[[41, 46]], [[1, 28], [47, 74]], [[77, 80]], [[81, 84]], [[47, 74]], [[0, 46]], [[34, 46]], [[41, 86]], [[47, 96]]]",
    "query_spans": "[[[88, 98]]]",
    "process": ""
  },
  {
    "text": "It is known that an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ is perpendicular to the line $x+2 y-1=0$, then the eccentricity of the hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x + 2*y - 1 = 0);IsPerpendicular(OneOf(Asymptote(G)), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 59], [83, 86]], [[5, 59]], [[5, 59]], [[66, 79]], [[5, 59]], [[5, 59]], [[2, 59]], [[66, 79]], [[2, 81]]]",
    "query_spans": "[[[83, 93]]]",
    "process": ""
  },
  {
    "text": "A point $M(x_{0}, y_{0})$ on the parabola $y^{2}=8x$ is at a distance of $6$ from its focus. Then, what is the distance from point $M$ to the $y$-axis?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = 8*x);Coordinate(M) = (x0, y0);PointOnCurve(M, G);Distance(M, Focus(G)) = 6;x0:Number;y0:Number",
    "query_expressions": "Distance(M, yAxis)",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 14], [35, 36]], [[17, 34], [47, 51]], [[0, 14]], [[17, 34]], [[0, 34]], [[17, 45]], [[17, 34]], [[17, 34]]]",
    "query_spans": "[[[47, 61]]]",
    "process": "According to the parabolic equation, first find the directrix equation; combining with the definition of a parabola, the distance from point M to the y-axis can be obtained. The parabola is $ y^{2} = 8x $, so the directrix equation is $ x = -2 $. According to the definition of a parabola, the distance from point $ M(x_{0}, y_{0}) $ to its focus is 6, then the distance from point M to its directrix is also 6, that is, $ x_{0} - (-2) = 6 $, which gives $ x_{0} = 4 $. Therefore, the distance from point M to the y-axis is 4."
  },
  {
    "text": "What is the distance from the focus $F$ of the parabola $y=4 x^{2}$ to the directrix $l$?",
    "fact_expressions": "G: Parabola;F: Point;l: Line;Expression(G) = (y = 4*x^2);Focus(G) = F;Directrix(G) = l",
    "query_expressions": "Distance(F, l)",
    "answer_expressions": "1/8",
    "fact_spans": "[[[0, 14]], [[17, 20]], [[23, 26]], [[0, 14]], [[0, 20]], [[0, 26]]]",
    "query_spans": "[[[17, 31]]]",
    "process": "From the given conditions, since the parabola is $ y = 4x^{2} $, that is, $ x^{2} = \\frac{1}{4}y $, $ \\therefore p = \\frac{1}{8} $, so the distance from the focus $ F $ to the directrix $ l $ is $ \\frac{1}{8} $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$ with left and right foci $F_{1}$ and $F_{2}$ respectively, and $A B$ being a chord of the ellipse passing through focus $F_{1}$, then the perimeter of $\\triangle A B F_{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/12 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);PointOnCurve(F1, LineSegmentOf(A, B))",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "16",
    "fact_spans": "[[[2, 41], [74, 76]], [[2, 41]], [[50, 57], [79, 86]], [[59, 66]], [[2, 66]], [[2, 66]], [[68, 73]], [[68, 73]], [[68, 88]], [[68, 86]]]",
    "query_spans": "[[[90, 116]]]",
    "process": "Solve according to the definition of an ellipse. From the definition of an ellipse, we have \\begin{cases}|BF_{1}|+|BF_{2}|=2a,\\\\|AF|+|AF_{2}|=2a,\\end{cases} therefore |AB|+|AF_{2}|+|BF_{2}|=4a=16"
  },
  {
    "text": "The length of the imaginary axis of the hyperbola $x^{2}-\\frac{y^{2}}{m}=1$ is twice the length of the real axis, then $m$=?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2 - y^2/m = 1);Length(ImageinaryAxis(G)) = 2*Length(RealAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 28]], [[43, 46]], [[0, 28]], [[0, 41]]]",
    "query_spans": "[[[43, 48]]]",
    "process": "According to the geometric properties of the hyperbola, the lengths of the real axis and imaginary axis are obtained, the equation is set up, and then solved. From the given conditions, the real axis length is 2, and the imaginary axis length is $2\\sqrt{m}$. Therefore, $2\\sqrt{m}=2\\times2$, we get $m=4$."
  },
  {
    "text": "Given that the foci of the ellipse lie on the $x$-axis, the length of the major axis $2a$ is $10$, and the eccentricity is $\\frac{3}{5}$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;PointOnCurve(Focus(G), xAxis);a: Number;Length(MajorAxis(G)) = 2*a;2*a = 10;Eccentricity(G) = 3/5",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/25+y^2/16=1",
    "fact_spans": "[[[2, 4], [48, 50]], [[2, 13]], [[17, 22]], [[2, 22]], [[17, 27]], [[2, 45]]]",
    "query_spans": "[[[48, 57]]]",
    "process": ""
  },
  {
    "text": "Let $O$ be the origin, and let $F_{1}$, $F_{2}$ be the foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$. If there exists a point $M$ on the hyperbola such that $\\angle F_{1} M F_{2}=120^{\\circ}$, $|O M|=\\sqrt{3} a$, and $S_{\\Delta F_{1} M F_{2}}=2 \\sqrt{3}$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F1: Point;M: Point;F2: Point;O: Origin;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Focus(G)={F1,F2};PointOnCurve(M, G);AngleOf(F1, M, F2) = ApplyUnit(120, degree);Abs(LineSegmentOf(O,M))=sqrt(3)*a;Area(TriangleOf(F1,M,F2))=2*sqrt(3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/6 = 1",
    "fact_spans": "[[[26, 82], [88, 91], [196, 199]], [[29, 82]], [[29, 82]], [[10, 17]], [[94, 98]], [[18, 25]], [[1, 4]], [[29, 82]], [[29, 82]], [[26, 82]], [[10, 85]], [[88, 98]], [[101, 135]], [[137, 155]], [[156, 193]]]",
    "query_spans": "[[[196, 204]]]",
    "process": "From the area formula of the hyperbolic focal triangle, $ b^{2} $ can be obtained. Using $ \\cos\\angle MOF_{1} = -\\cos\\angle MOF_{2} $, an equation can be constructed to yield $ |MF_{1}|^{2} + |MF_{2}|^{2} = 6a^{2} + 2c^{2} $. Combining with the definition of the hyperbola, $ a^{2} = 1 $ is obtained, thus giving the hyperbola equation. Since $ \\angle F_{1}MF_{2} = 120^{\\circ} $, $ S_{\\Delta F_{1}MF_{2}} = \\frac{b^{2}}{\\tan 60^{\\circ}} = 2\\sqrt{3} $, so $ b^{2} = 6 $. Since $ \\cos\\angle MOF_{1} = \\frac{|MO|^{2} + |OF_{1}|^{2}}{2|MO|\\cdot|OF_{1}|} $, $ \\cos\\angle MOF_{2} = \\frac{|MO|^{2} + |OF_{2}|^{2} - |MF_{2}|^{2}}{2|MO|\\cdot|OF_{2}|} = \\frac{3a^{2} + c^{2} - |MF_{2}|^{2}}{2\\sqrt{3}ac} $. Also $ \\cos\\angle MOF_{1} = -\\cos\\angle MOF_{2} $, so $ 3a^{2} + c^{2} - |MF_{1}|^{2} = -3a^{2} - c^{2} + |MF_{2}|^{2} $, hence $ |MF_{1}|^{2} + |MF_{2}|^{2} = 6a^{2} + 2c^{2} $, therefore $ (|MF_{1}| - |MF_{2}|)^{2} + 2|MF_{1}|\\cdot|MF_{2}| = (|MF_{1}| - |MF_{2}|)^{2} + 16 = 4a^{2} + 16 = 6a^{2} + 2c^{2} $. Rearranging gives: $ a^{2} + c^{2} = 8 $, so $ a^{2} + (a^{2} + b^{2}) = 2a^{2} + 6 = 8 $, solving yields $ a^{2} = 1 $, therefore the equation of the hyperbola is $ x^{2} - \\frac{y^{2}}{6} = 1 $."
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and point $P$ moves on the ellipse. Then the maximum value of $\\overrightarrow{P F_{1}}\\cdot \\overrightarrow{P F_{2}}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G)",
    "query_expressions": "Max(DotProduct(VectorOf(P, F1),VectorOf(P, F2)))",
    "answer_expressions": "1",
    "fact_spans": "[[[17, 44], [55, 57]], [[50, 54]], [[0, 7]], [[9, 16]], [[17, 44]], [[0, 49]], [[0, 49]], [[50, 60]]]",
    "query_spans": "[[[63, 125]]]",
    "process": ""
  },
  {
    "text": "Given that $O$ is the coordinate origin, $F$ is the focus of the parabola $C$: $y^{2}=2 x$, the line $l$: $y=m(2 x-1)$ intersects the parabola $C$ at two distinct points $A$ and $B$, with point $A$ in the first quadrant. If $|A F|=2|B F|$, then the length of $A B$ is?",
    "fact_expressions": "O: Origin;F: Point;Focus(C) = F;C: Parabola;Expression(C) = (y^2 = 2*x);l: Line;Expression(l) = (y = m*(2*x - 1));A: Point;B: Point;Intersection(l, C) = {A, B};Quadrant(A) = 1;Abs(LineSegmentOf(A, F)) = 2*Abs(LineSegmentOf(B, F));m: Number",
    "query_expressions": "Length(LineSegmentOf(A,B))",
    "answer_expressions": "9/4",
    "fact_spans": "[[[2, 5]], [[11, 14]], [[11, 37]], [[15, 34], [58, 64]], [[15, 34]], [[38, 57]], [[38, 57]], [[69, 72], [79, 83]], [[73, 76]], [[38, 78]], [[79, 88]], [[90, 104]], [[45, 57]]]",
    "query_spans": "[[[106, 115]]]",
    "process": "The focus of $ y^{2}=2x $ is $ F\\left(\\frac{1}{2},0\\right) $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, $ (x_{1}>0,y_{1}>0) $. The line $ l: y=m(2x-1) $ intersects the parabola $ y^{2}=2x $, yielding $ 4m^{2}x^{2}-(2+4m^{2})x+m^{2}=0 $, so that $ x_{1}x_{2}=\\frac{1}{4} \\textcircled{1} $, $ x_{1}+x_{2}=1+\\frac{1}{2m^{2}} \\textcircled{2} $. From the given condition $ \\overrightarrow{AF}=2\\overrightarrow{FB} $, $ \\left(\\frac{1}{2}-x_{1}, -y_{1}\\right)=2\\left(x_{2}-\\frac{1}{2}, y_{2}\\right) $, we obtain $ \\frac{1}{2}-x_{1}=2\\left(x_{2}-\\frac{1}{2}\\right) $, $ x_{1}+2x_{2}=\\frac{3}{2} \\textcircled{3} $. From $ \\textcircled{1} \\textcircled{3} $, we get $ x_{1}=1 $, $ x_{2}=\\frac{1}{4} $ (discarding $ x_{1}=x_{2}=\\frac{1}{2} $). Then the length $ |AB| $ is $ |AB|=x_{1}+x_{2}+p=1+\\frac{1}{4}+1=\\frac{9}{4} $."
  },
  {
    "text": "Through the focus of the parabola $y^{2}=2 p x(p>0)$, draw a line perpendicular to the axis of symmetry of the parabola, intersecting the parabola at points $A$ and $B$, and $|A B|=4$. Then $p=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;H: Line;PointOnCurve(Focus(G), H);IsPerpendicular(H, SymmetryAxis(G));A: Point;B: Point;Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 4",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 22], [27, 30], [39, 42]], [[1, 22]], [[65, 68]], [[4, 22]], [[36, 38]], [[0, 38]], [[26, 38]], [[43, 46]], [[47, 50]], [[36, 52]], [[54, 63]]]",
    "query_spans": "[[[65, 70]]]",
    "process": "According to the focal radius formula of the parabola, express |AB|; then, since |AB| = 4, solve directly for the value of p. Let the focus coordinates of the parabola be F(\\frac{p}{2},0). From the given condition, we have x_{A} = x_{B} = x_{F} = \\frac{p}{2}. Therefore, |AB| = |AF| + |BF| = x_{A} + \\frac{p}{2} + x_{B} + \\frac{p}{2} = 2p. Since |AB| = 4, it follows that p = 2."
  },
  {
    "text": "Given that the coordinates of point $A$ are $(4,3)$, $F$ is the focus of the parabola $y^{2}=4x$, and point $P$ moves along the parabola. When $|PA|+|PF|$ takes the minimum value, what are the coordinates of point $P$?",
    "fact_expressions": "G: Parabola;P: Point;A: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (4, 3);Focus(G) = F;PointOnCurve(P, G);WhenMin(Abs(LineSegmentOf(P,A))+Abs(LineSegmentOf(P,F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(9/4,3)",
    "fact_spans": "[[[22, 36], [46, 49]], [[41, 45], [73, 77]], [[2, 6]], [[18, 21]], [[22, 36]], [[2, 17]], [[18, 39]], [[41, 52]], [[54, 73]]]",
    "query_spans": "[[[73, 82]]]",
    "process": "Let the projection of point P on the directrix be D. By the definition of a parabola, |PF| = |PD|. To minimize |PA| + |PF|, we need to minimize |PA| + |PD|. The sum |PA| + |PD| is minimized only when points D, P, and A are collinear. Setting y = 3, we get x = \\frac{9}{4}. Therefore, the coordinates of point P at the minimum value are \\left(\\frac{9}{4}, 3\\right)."
  },
  {
    "text": "Given that the line $y=a$ intersects an asymptote of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ at point $P$, and the left and right vertices of hyperbola $C$ are $A_{1}$ and $A_{2}$ respectively, if $|P A_{2}|=\\frac{\\sqrt{5}}{2}|A_{1} A_{2}|$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "G: Line;Expression(G) = (y = a);C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;P: Point;Intersection(G, OneOf(Asymptote(C))) = P;A2: Point;A1: Point;LeftVertex(C) = A1;RightVertex(C) = A2;Abs(LineSegmentOf(P, A2)) = (sqrt(5)/2)*Abs(LineSegmentOf(A1, A2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "{sqrt(2), sqrt(10)/3}",
    "fact_spans": "[[[2, 9]], [[2, 9]], [[10, 71], [84, 90], [162, 168]], [[10, 71]], [[18, 71]], [[18, 71]], [[18, 71]], [[18, 71]], [[79, 83]], [[2, 83]], [[107, 114]], [[99, 106]], [[84, 114]], [[84, 114]], [[116, 159]]]",
    "query_spans": "[[[162, 174]]]",
    "process": "If the equation of the asymptote is $ y = \\frac{b}{a}x $, then the coordinates of point $ P $ are $ \\left( \\frac{a2}{b}, a \\right) $. Since $ |PA_{2}| = \\frac{\\sqrt{5}}{2}|A_{1}A_{2}| $, it follows that $ \\left( \\frac{a2}{b} - a \\right)^{2} + a^{2} = 5a2 $, then $ \\left( \\frac{a}{b} - 1 \\right)^{2} = 4 $, so $ \\frac{a}{b} = 3 $. If the equation of the asymptote is $ y = -\\frac{b}{a}x $, then the coordinates of point $ P $ are $ \\left( -\\frac{a^{2}}{b}, a \\right) $, and similarly we obtain $ e = \\sqrt{2} $. This problem examines the eccentricity of a hyperbola and tests operational solving ability and classification discussion in mathematics."
  },
  {
    "text": "The minimum distance from points on the parabola $x^{2}=-y$ to the line $4 x+3 y-8=0$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = -y);PointOnCurve(P, G) = True;P: Point;H: Line;Expression(H) = (4*x + 3*y - 8 = 0)",
    "query_expressions": "Min(Distance(P, H))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[0, 13]], [[0, 13]], [[0, 16]], [[15, 16]], [[17, 32]], [[17, 32]]]",
    "query_spans": "[[[15, 41]]]",
    "process": "First, find the line parallel to the line $4x+3y-8=0$ and tangent to the parabola, then use the distance between parallel lines to solve. Let the line $4x+3y+c=0$ be tangent to the parabola. From $\\begin{cases}4x+3y+c=0\\end{cases}$, we obtain $3x^{2}-4x-c=0$. From $\\Delta=16+12c=0$, we get $c=-\\frac{4}{3}$. Therefore, the distance between the two parallel lines is $\\frac{|-8+\\frac{4}{3}|}{\\sqrt{16+9}}=\\frac{4}{3}$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $P$ is a point on the ellipse, and $\\angle F_{1} PF_{2}=90^{\\circ}$, then the range of values for the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;P: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[\\sqrt{2}/2, 1)",
    "fact_spans": "[[[18, 70], [78, 80], [119, 121]], [[20, 70]], [[20, 70]], [[2, 9]], [[74, 77]], [[10, 17]], [[20, 70]], [[20, 70]], [[18, 70]], [[2, 73]], [[74, 83]], [[85, 117]]]",
    "query_spans": "[[[119, 132]]]",
    "process": ""
  },
  {
    "text": "The equation $\\frac{x^{2}}{3-m}+\\frac{y^{2}}{m-1}=1$ represents a hyperbola; what is the range of values for $m$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(3 - m) + y^2/(m - 1) = 1);m:Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-oo,1)+(3,+oo)",
    "fact_spans": "[[[43, 46]], [[0, 46]], [[48, 51]]]",
    "query_spans": "[[[48, 57]]]",
    "process": "From the given condition, (3-m)(m-1)<0, solving yields: m>3 or m<1. Hence, the range of values for m is (-\\infty,1)\\cup(3,+\\infty)"
  },
  {
    "text": "Given the parabola $x^{2}=4 y$ with focus $F$ and point $A(-1,8)$, $P$ is a point on the parabola. Then the minimum value of $|P A|+|P F|$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);F: Point;Focus(G) = F;A: Point;Coordinate(A) = (-1, 8);P: Point;PointOnCurve(P, G) = True",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "9",
    "fact_spans": "[[[2, 16], [40, 43]], [[2, 16]], [[19, 22]], [[2, 22]], [[23, 33]], [[23, 33]], [[36, 39]], [[36, 46]]]",
    "query_spans": "[[[48, 67]]]",
    "process": "According to the problem, draw a perpendicular line from point P to the directrix of the parabola, with foot of perpendicular at P'. By the definition of the parabola, |PF| = |PP'|, so the minimum value of |PA| + |PF| is equal to the minimum value of |PA| + |PP'|, which is the minimum sum of the distance from a moving point P on the parabola to a fixed point A(-1,8) and the distance to the fixed line y = -1. Obviously, the minimum value is the distance from point A to the line y = -1, which is 8 - (-1) = 9."
  },
  {
    "text": "If one focus of an ellipse has coordinates $(\\sqrt{2}, 0)$, and the length of the major axis is $\\sqrt{3}$ times the length of the minor axis, then the standard equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;Coordinate(OneOf(Focus(G))) = (sqrt(2), 0);Length(MajorAxis(G))=sqrt(3)*Length(MinorAxis(G))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/3+y^2=1",
    "fact_spans": "[[[2, 4], [51, 53]], [[2, 27]], [[2, 48]]]",
    "query_spans": "[[[51, 60]]]",
    "process": "According to the problem, since one focus of the ellipse has coordinates $(\\sqrt{2},0)$, we have $c=\\sqrt{2}$, and the foci lie on the x-axis. Also, the major axis is $\\sqrt{3}$ times the minor axis, that is, $2a = \\sqrt{3}(2b)$, so $a = \\sqrt{3}b$. Then $a^{2} - b^{2} = 2b^{2} = c^{2} = 2$, solving gives $a^{2} = 3$, $b^{2} = 1$. Thus, the standard equation of the ellipse is $\\frac{x^{2}}{3} + \\frac{y^{2}}{1} = 1$."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, the intersection point of the directrix and the $x$-axis is $M$, and $N$ is a point on the parabola satisfying $|MN|=2|NF|$, then $\\angle NMF$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;M: Point;Intersection(Directrix(G), xAxis) = M;N: Point;PointOnCurve(N, G);Abs(LineSegmentOf(M, N)) = 2*Abs(LineSegmentOf(N, F))",
    "query_expressions": "AngleOf(N, M, F)",
    "answer_expressions": "pi/3",
    "fact_spans": "[[[2, 16], [42, 45]], [[2, 16]], [[20, 23]], [[2, 23]], [[34, 37]], [[2, 37]], [[38, 41]], [[38, 49]], [[53, 67]]]",
    "query_spans": "[[[69, 85]]]",
    "process": "Draw NP perpendicular to the directrix from point N, intersecting the directrix at P. By the definition of a parabola, |NP| = |NF|. Therefore, in right triangle MPN, ∠MPN = 90°, |MN| = 2|PN|, ∴ ∠PMN = 30°, ∴ ∠NMF = \\frac{\\pi}{3}"
  },
  {
    "text": "The line passing through the right focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with slope $1$ intersects the hyperbola at exactly one point. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F: Point;H: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;PointOnCurve(F,H);Slope(H) = 1;NumIntersection(H,G) = 1",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 57], [75, 78], [88, 91]], [[4, 57]], [[4, 57]], [[61, 64]], [[72, 74]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 64]], [[0, 74]], [[65, 74]], [[72, 86]]]",
    "query_spans": "[[[88, 97]]]",
    "process": "\\because the line intersects the hyperbola at exactly one point, \\therefore the line is parallel to the asymptote of the hyperbola, \\therefore \\frac{b}{a}=1, i.e., a=b. \\therefore a^{2}=b^{2}=c^{2}-a^{2}, i.e., c^{2}=2a^{2}, \\therefore e=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{2}"
  },
  {
    "text": "A line $ l $ with slope $ k $ ($ k<0 $) passes through the point $ F(0,1) $, and intersects the curve $ y=\\frac{1}{4} x^{2} $ ($ x \\geq 0 $) and the line $ y=-1 $ at points $ A $ and $ B $, respectively. If $ |F B|=5|F A| $, then $ k $?",
    "fact_expressions": "Slope(l) = k;k: Number;k<0;l: Line;F: Point;Coordinate(F) = (0, 1);PointOnCurve(F, l) ;H: Curve;Expression(H) = (y = x^2/4)&(x>=0);G: Line;Expression(G) = (y = -1);Intersection(l, H) = A;Intersection(l, G) = B;B: Point;A: Point;Abs(LineSegmentOf(F, B)) = 5*Abs(LineSegmentOf(F, A))",
    "query_expressions": "k",
    "answer_expressions": "-sqrt(15)/sqrt(15)",
    "fact_spans": "[[[0, 17]], [[3, 11], [103, 106]], [[3, 11]], [[12, 17]], [[18, 27]], [[18, 27]], [[12, 27]], [[30, 63]], [[30, 63]], [[64, 72]], [[64, 72]], [[12, 85]], [[12, 85]], [[80, 83]], [[76, 79]], [[87, 101]]]",
    "query_spans": "[[[103, 108]]]",
    "process": ""
  },
  {
    "text": "Draw a line through point $A(1,0)$ with an inclination angle of $\\frac{\\pi}{4}$, intersecting the parabola $y^{2}=2x$ at points $M$ and $N$. Then the value of $|MN|$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;M: Point;N: Point;Expression(G) = (y^2 = 2*x);Coordinate(A) = (1, 0);PointOnCurve(A, H);Inclination(H)=pi/4;Intersection(H,G)={M,N}",
    "query_expressions": "Abs(LineSegmentOf(M, N))",
    "answer_expressions": "2*sqrt(6)",
    "fact_spans": "[[[35, 49]], [[31, 33]], [[1, 10]], [[51, 54]], [[55, 58]], [[35, 49]], [[1, 10]], [[0, 33]], [[11, 33]], [[31, 60]]]",
    "query_spans": "[[[62, 74]]]",
    "process": "Method 1: From the given conditions, the equation of the line is $ y = x - 1 $. Let $ M(x_{1}, y_{1}) $, $ N(x_{2}, y_{2}) $. From the system of equations \n\\[\n\\begin{cases}\ny = x - 1 \\\\\ny^{2} = 2x\n\\end{cases}\n\\]\nwe obtain $ x^{2} - 4x + 1 = 0 $, then $ x_{1} + x_{2} = 4 $, $ x_{1}x_{2} = 1 $, $ y_{1} - y_{2} = x_{1} - x_{2} $. Method 2: From Method 1, $ x^{2} - 4x + 1 = 0 $, we can find $ M(2 + \\sqrt{3}, 1 + \\sqrt{3}) $, $ N(2 - \\sqrt{3}, 1 - \\sqrt{3}) $, then $ 2\\sqrt{3}^{2} + (2\\sqrt{3})^{2} = 2\\sqrt{6} $."
  },
  {
    "text": "Given that the directrix of the parabola $y^{2}=-8x$ passes through the right focus of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{3}=1$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = -8*x);G: Hyperbola;Expression(G) = (-y^2/3 + x^2/m = 1);m: Number;PointOnCurve(RightFocus(G), Directrix(H))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 17]], [[2, 17]], [[21, 59], [65, 68]], [[21, 59]], [[24, 59]], [[2, 63]]]",
    "query_spans": "[[[65, 74]]]",
    "process": "According to the problem, since the directrix of the parabola $ y^{2} = -8x $ is $ x = 2 $, and it passes through the right focus $ (2, 0) $ of the hyperbola $ \\frac{x^{2}}{m} - \\frac{y^{2}}{3} = 1 $, we have $ m + 3 = 4 $, so $ m = 1 $. Thus, $ a = 1 $, $ c = 2 $, and therefore the eccentricity is $ 2 $. The answer is $ 2 $."
  },
  {
    "text": "The left and right foci of ellipse $C$ are $F_{1}$ and $F_{2}$, respectively. A line passing through point $F_{2}$ intersects ellipse $C$ at points $A$ and $B$. If $|A F_{1}|=|A B|=5$, $|F_{1} B|=6$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;G: Line;A: Point;F1: Point;B: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F2, G);Intersection(G, C) = {A, B};Abs(LineSegmentOf(A,F1))=Abs(LineSegmentOf(A,B));Abs(LineSegmentOf(A,B))=5;Abs(LineSegmentOf(F1, B)) = 6",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(10)/5",
    "fact_spans": "[[[2, 7], [44, 49], [97, 102]], [[41, 43]], [[51, 55]], [[15, 22]], [[56, 59]], [[23, 30], [32, 40]], [[2, 30]], [[2, 30]], [[31, 43]], [[41, 59]], [[61, 80]], [[61, 80]], [[82, 95]]]",
    "query_spans": "[[[97, 108]]]",
    "process": "Let the major axis length of the ellipse be 2a, then 4a = |AF₁| + |AF₂| + |BF₁| + |BF₂| = 16. Thus a = 4, |AF₂| = 3, |BF₂| = 2. In △ABF₁, by the cosine law we obtain cos∠F₁BF₂. In △BF₁F₂, by the cosine law we obtain 2c, which allows solving. [Detailed explanation] As shown: Let the major axis length of the ellipse be 2a, |AF₁| + |AF₂| = |BF₁| + |BF₂| = 2a. Since |AF₁| = |AB| = 5, |F₁B| = 6, ∴ 4a = |AF₁| + |AF₂| + |BF₁| + |BF₂| = 16, thus a = 4, |AF₂| = 3, |BF₂| = 2. In △ABF₁, by the cosine law we get cos∠F₁BF₂ = (36 + 25 - 25) / (2 × 6 × 5) = 3/5. ∴ e = c/a = √10 / 5"
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=4x$, two mutually perpendicular lines $l_{1}$ and $l_{2}$ are drawn through $F$. The line $l_{1}$ intersects $C$ at points $A$ and $B$, and the line $l_{2}$ intersects $C$ at points $D$ and $E$. Then the minimum value of $|AB|+|DE|$ is?",
    "fact_expressions": "C: Parabola;l1:Line;l2:Line;A: Point;B: Point;D: Point;E: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F,l1);PointOnCurve(F,l2);IsPerpendicular(l1,l2);Intersection(l1,C)={A,B};Intersection(l2,C)={D,E}",
    "query_expressions": "Min(Abs(LineSegmentOf(A, B)) + Abs(LineSegmentOf(D, E)))",
    "answer_expressions": "16",
    "fact_spans": "[[[6, 25], [70, 73], [94, 97]], [[41, 50], [60, 69]], [[52, 59], [84, 93]], [[74, 77]], [[78, 81]], [[99, 102]], [[103, 106]], [[2, 5], [30, 33]], [[6, 25]], [[2, 28]], [[29, 59]], [[29, 59]], [[34, 59]], [[60, 83]], [[84, 108]]]",
    "query_spans": "[[[110, 130]]]",
    "process": "Let the equation of line $ l_{1} $ be $ y = k(x - 1) $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Substitute the line equation into the parabola equation, simplify and apply Vieta's formulas, use the chord length formula to find the length $ |AB| $, replace $ k $ with $ -\\frac{1}{k} $ to obtain the chord length $ |DE| $, compute $ |AB| + |DE| $, and use the basic inequality to find the minimum value. Solution: According to the problem, the focus of the parabola is $ F(1, 0) $. Clearly, the slopes of lines $ l_{1} $ and $ l_{2} $ both exist and are non-zero. Let the equation of line $ l_{1} $ be $ y = k(x - 1) $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. From\n$$\n\\begin{cases}\ny^2 = 4x \\\\\ny = k(x - 1)\n\\end{cases}\n$$\nwe obtain $ k^{2}x^{2} - 2(k^{2} + 2)x + k^{2} = 0 $, so $ x_{1} + x_{2} = \\frac{2(k^{2} + 2)}{k^{2}} $, $ x_{1}x_{2} = 1 $. Similarly, we get $ |DE| = \\frac{4[1 + (-\\frac{1}{k})^{2}]}{(-\\frac{1}{k})^{2}} = 4(1 + k^{2}) $. Therefore, $ |AB| + |DE| = 4(1 + \\frac{1}{k^{2}}) + 4(1 + k^{2}) = 8 + 4(\\frac{1}{k^{2}} + k^{2}) \\geqslant 8 + 4 \\times 2\\sqrt{\\frac{1}{k^{2}} \\times k^{2}} = 16 $, with equality if and only if $ k = \\pm 1 $."
  },
  {
    "text": "Given that from a moving point $P$ on the ellipse $E$: $\\frac{x^{2}}{m}+\\frac{y^{2}}{m-1}=1$ ($m>5$), two tangents are drawn to the circle $C$ (with center $C$): $x^{2}-2x+y^{2}=0$, with points of tangency $A$ and $B$ respectively. If the minimum value of $\\angle ACB$ is $\\frac{2\\pi}{3}$, then the eccentricity of the ellipse $E$ is?",
    "fact_expressions": "E: Ellipse;Expression(E) = (y^2/(m - 1) + x^2/m = 1);m: Number;m>5;P: Point;PointOnCurve(P, E);C1: Circle;Expression(C1) = (x^2 - 2*x + y^2 = 0);C: Point;Center(C1) = C;L1: Line;L2: Line;TangentOfPoint(P, C1) = {L1, L2};A: Point;B: Point;TangentPoint(L1, C1) = A;TangentPoint(L2, C1) = B;Min(AngleOf(A, C, B)) = (2*pi)/3",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "1/3",
    "fact_spans": "[[[3, 52], [151, 156]], [[3, 52]], [[10, 52]], [[10, 52]], [[56, 59]], [[3, 59]], [[60, 93]], [[60, 93]], [[65, 68]], [[60, 72]], [], [], [[2, 98]], [[104, 107]], [[108, 111]], [[2, 111]], [[2, 111]], [[113, 149]]]",
    "query_spans": "[[[151, 162]]]",
    "process": "From the equation of ellipse E, its right focus is at (1,0); from the equation of circle C, the center is at C(1,0) and the radius is 1; when $\\angle ACB$ is minimized, $\\angle ACP$ is minimized, i.e., $\\angle ACP = \\frac{\\pi}{3}$, at which point $|PC|$ is minimized; $|PC|_{\\min} = 2$; $\\because$ when P is the right vertex of the ellipse, $|PC| = \\sqrt{m} - 1 = 2$, solving gives: $m = 9$. $\\therefore$ the eccentricity of ellipse E is $e = \\sqrt{\\frac{1}{m}} = \\frac{1}{3}$."
  },
  {
    "text": "Given that the coordinates of the right focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{a}=1$ are $(\\sqrt{13}, 0)$, then the equations of the asymptotes of this hyperbola are?",
    "fact_expressions": "G: Hyperbola;a: Number;Expression(G) = (x^2/9 - y^2/a = 1);Coordinate(RightFocus(G)) = (sqrt(13), 0)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "2*x+pm*3*y=0",
    "fact_spans": "[[[2, 40], [68, 71]], [[5, 40]], [[2, 40]], [[2, 64]]]",
    "query_spans": "[[[68, 79]]]",
    "process": ""
  },
  {
    "text": "The line $l$ passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. If the sum of the horizontal coordinates of the two points is $5$, then $|AB|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;l: Line;PointOnCurve(F, l) = True;Intersection(l, G) = {A, B};A: Point;B: Point;XCoordinate(A) + XCoordinate(B) = 5",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "7",
    "fact_spans": "[[[1, 15], [27, 30]], [[1, 15]], [[17, 20]], [[1, 20]], [[21, 26]], [[0, 26]], [[21, 40]], [[31, 34]], [[35, 38]], [[31, 54]]]",
    "query_spans": "[[[56, 65]]]",
    "process": "From the parabola equation, we obtain p=2, then by the definition of the parabola, |AB|=x_{A}+x_{B}+p=5+2=7."
  },
  {
    "text": "Given a point $P(3, y_{0})$ on the parabola $C$: $y^{2}=2 p x(p>0)$ is at a distance of $8$ from its directrix, then $p=?$",
    "fact_expressions": "C: Parabola;p: Number;P: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(P) = (3, y0);y0: Number;PointOnCurve(P, C);Distance(P, Directrix(C)) = 8",
    "query_expressions": "p",
    "answer_expressions": "10",
    "fact_spans": "[[[2, 28], [45, 46]], [[57, 60]], [[31, 44]], [[10, 28]], [[2, 28]], [[31, 44]], [[31, 44]], [[2, 44]], [[31, 55]]]",
    "query_spans": "[[[57, 62]]]",
    "process": "From the given condition, we have $3 + \\frac{p}{2} = 8$, so $p = 10$."
  },
  {
    "text": "Draw a vertical line through the left focus $F_{1}$ of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ intersecting the ellipse at point $P$, and let $F_{2}$ be the right focus. If $\\angle F_{1} P F_{2}=60^{\\circ}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;P: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;L:Line;PointOnCurve(F1, L);IsPerpendicular(L,xAxis);Intersection(L, G) = P;AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[1, 53], [73, 75], [130, 132]], [[3, 53]], [[3, 53]], [[57, 64]], [[76, 80]], [[81, 88]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 64]], [[73, 92]], [], [[0, 72]], [[0, 72]], [[0, 80]], [[94, 127]]]",
    "query_spans": "[[[130, 138]]]",
    "process": ""
  },
  {
    "text": "Given points $A(2,0)$, $B(0,1)$, and $O$ as the origin. A moving point $M$ satisfies $\\overrightarrow{O M}=\\lambda \\overrightarrow{O A}+\\mu \\overrightarrow{O B}$ $(\\lambda, \\mu) \\in \\mathbb{R}$, when $\\lambda+\\mu=1$, what is the trajectory equation of point $M$?",
    "fact_expressions": "O: Origin;M: Point;A: Point;B: Point;lambda:Real;mu:Real;Coordinate(A)=(2,0);Coordinate(B)=(0,1);VectorOf(O, M) = lambda*VectorOf(O, A) + mu*VectorOf(O, B);lambda+mu=1",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x+2*y-2=0",
    "fact_spans": "[[[22, 25]], [[33, 36], [154, 158]], [[2, 11]], [[13, 21]], [[38, 134]], [[38, 134]], [[2, 11]], [[13, 21]], [[38, 135]], [[137, 152]]]",
    "query_spans": "[[[154, 165]]]",
    "process": "Let M(x, y), then \\overrightarrow{OM}=(x,y), \\overrightarrow{OA}=(2,0), \\overrightarrow{OB}=(0,1). Since \\overrightarrow{OM}=\\lambda\\overrightarrow{OA}+\\mu\\overrightarrow{OB}, it follows that (x,y)=(2\\lambda,\\mu), i.e., x=2\\lambda, y=\\mu. When \\lambda+\\mu=1, i.e., \\frac{x}{2}+y=1, or x+2y-2=0."
  },
  {
    "text": "Given that point $M$ lies on the parabola $y^{2}=4x$, if a circle centered at point $M$ is tangent to both the $x$-axis and its directrix $l$, then the distance from point $M$ to its vertex $O$ is?",
    "fact_expressions": "G: Parabola;H: Circle;M: Point;l:Line;Expression(G) = (y^2 = 4*x);PointOnCurve(M, G);Center(H)=M;IsTangent(xAxis,H);Directrix(G)=l;IsTangent(l,H);O:Point;Vertex(G)=O",
    "query_expressions": "Distance(M, O)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[7, 21], [40, 41], [56, 57]], [[33, 34]], [[2, 6], [51, 55], [25, 29]], [[43, 46]], [[7, 21]], [[2, 22]], [[24, 34]], [[33, 49]], [[40, 46]], [[33, 49]], [[59, 62]], [[56, 62]]]",
    "query_spans": "[[[51, 67]]]",
    "process": "Use the given conditions to find the coordinates of point M, then solve for the distance from point M to its vertex O. Point M lies on the parabola $ y^{2} = 4x $. If a circle centered at point M is tangent to both the x-axis and its directrix, let $ M(x, x+1) $, we obtain $ (x+1)^{2} = 4x $, solving gives $ x = 1 $, so $ M(1, 2) $. The distance from point M to its vertex O is: $ \\sqrt{2^{2} + 1^{2}} = \\sqrt{5} $. The answer is: $ \\sqrt{5} $. P.tress.to.ter-----Simple geometric properties of a parabola, examines computational ability, classified as a medium-difficulty problem."
  },
  {
    "text": "Given that line $l$ passes through point $M(0,3)$, and $l$ intersects the parabola $y=x^{2}$ at points $E$ and $F$. When $l$ is not perpendicular to the $y$-axis, there exists a point $P(0, t)$ on the $y$-axis such that the incenter of $\\Delta P E F$ lies on the $y$-axis. Then the real number $t$=?",
    "fact_expressions": "l: Line;M: Point;Coordinate(M) = (0, 3);PointOnCurve(M, l);G: Parabola;Expression(G) = (y = x^2);E: Point;F: Point;Intersection(l, G) = {E, F};Negation(IsPerpendicular(l, yAxis));P: Point;t: Real;Coordinate(P) = (0, t);PointOnCurve(P, yAxis);PointOnCurve(Incenter(TriangleOf(P, E, F)), yAxis)",
    "query_expressions": "t",
    "answer_expressions": "-3",
    "fact_spans": "[[[2, 7], [20, 23], [49, 52]], [[8, 18]], [[8, 18]], [[2, 18]], [[24, 36]], [[24, 36]], [[38, 41]], [[42, 45]], [[20, 47]], [[49, 60]], [[72, 81]], [[109, 114]], [[72, 81]], [[63, 81]], [[84, 107]]]",
    "query_spans": "[[[109, 116]]]",
    "process": "Let the line $ l: y = kx + 3 $ ($ k \\neq 0 $) be substituted into $ y = x^{2} $, yielding: $ x^{2} - kx - 3 = 0 $. Let $ E(x_{1}, y_{1}) $, $ F(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = k \\neq 0 $, $ x_{1}x_{2} = -3 $. Let the incenter be $ I(0, m) $, $ k_{PE} = \\frac{y_{1}t}{x_{1}} = \\frac{x_{1}^{2} - t}{x_{1}} $, the line $ PE: y = \\frac{x_{1}^{2} - t}{x_{1}}x + t' $. The distance from the incenter to the line $ PE $ is similarly obtained. The distance from the incenter to the line $ PF $ is $ \\frac{|-m + t|}{\\sqrt{|x^{2} - t^{2}| + 1}} $. According to the problem, $ d_{1} = d_{2} $, i.e., $ \\frac{|-m + t|}{\\sqrt{|x^{2} - t^{2}| + 1}} = \\frac{|-m + t|}{\\sqrt{x_{1}^{2} - t^{2}} + 1} $. Therefore, $ \\begin{cases} x^{2} - 2^{2} = (x^{2} + 2)^{2} \\\\ -1 = 1 \\end{cases} $. Hence, $ \\frac{x_{1}^{2} - t}{x_{1}} = -\\frac{x^{2} - t}{x_{2}} $, $ x_{1} - \\frac{t}{x_{1}} + x_{2} - \\frac{t}{x_{2}} = 0 $, $ (x_{1} + x_{2}) - \\frac{t(x_{1} + x_{2})}{x_{1}x_{2}} = 0 $, since $ x_{1} + x_{2} = k \\neq 0 $, therefore $ 1\\underline{t} - 0t = 0 $, $ t = -3 $."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=4x$ be $F$. A line passing through point $F$ intersects the parabola at points $A$ and $B$. From the midpoint $M$ of $AB$, a perpendicular is drawn to the directrix, intersecting the parabola at point $P$. If $|PF|=\\frac{3}{2}$, then the chord length $AB$ equals?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};MidPoint(LineSegmentOf(A,B))=M;L:Line;PointOnCurve(M,L);IsPerpendicular(L,Directrix(G));Intersection(L,G)=P;IsChordOf(LineSegmentOf(A,B),G);Abs(LineSegmentOf(P, F)) = 3/2;M:Point",
    "query_expressions": "LineSegmentOf(A, B)",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 15], [32, 35], [65, 68]], [[29, 31]], [[37, 40]], [[41, 44]], [[70, 74]], [[19, 22], [24, 28]], [[1, 15]], [[1, 22]], [[23, 31]], [[29, 46]], [[48, 58]], [], [[32, 64]], [[32, 64]], [[32, 74]], [[65, 104]], [[76, 95]], [[55, 58]]]",
    "query_spans": "[[[99, 107]]]",
    "process": ""
  },
  {
    "text": "If point $P$ lies on an ellipse with foci $F_{1}$ and $F_{2}$, $P F_{2} \\perp F_{1} F_{2} $, and $\\tan \\angle P F_{1} F_{2}=\\frac{3}{4}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;P: Point;F2: Point;F1: Point;Focus(G) = {F1, F2};PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F2), LineSegmentOf(F1, F2));Tan(AngleOf(P, F1, F2)) = 3/4",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[26, 28], [101, 103]], [[1, 5]], [[15, 22]], [[7, 14]], [[6, 28]], [[1, 29]], [[30, 58]], [[60, 99]]]",
    "query_spans": "[[[101, 109]]]",
    "process": "Since $ PF_{2} \\perp F_{1}F_{2} $, in right triangle $ \\triangle PF_{1}F_{2} $, because $ |F_{1}F_{2}| = 2c $, $ \\tan\\angle PF_{1}F_{2} = \\frac{3}{4} $, so $ |PF_{2}| = \\frac{3}{4}|F_{1}F_{2}| = \\frac{3}{2}c $. Since point $ P $ lies on the ellipse, $ |PF_{1}| = 2a - |PF_{2}| = 2a - \\frac{3}{2}c $. From $ |PF_{1}|^{2} = |PF_{2}|^{2} + |F_{1}F_{2}|^{2} $, we obtain $ (2a - \\frac{3}{2}c)^{2} = (\\frac{3}{2}c)^{2} + (2c)^{2} $, simplifying yields $ 2c^{2} + 3ac - 2a^{2} = 0 $, solving gives $ c = \\frac{1}{2}a $ or $ c = -2a $ (discarded), thus $ e = \\frac{c}{a} = \\frac{1}{2} $."
  },
  {
    "text": "Given that $A B$ is a chord of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, and the chord $A B$ is perpendicular to the line $l$: $x+2 y-3=0$, $P$ is the midpoint of $A B$, and $O$ is the center of the ellipse, then the slope of the line $O P$ is?",
    "fact_expressions": "A: Point;B: Point;G: Ellipse;Expression(G) = (x^2/9+y^2/4 = 1);IsChordOf(LineSegmentOf(A, B), G);l: Line;Expression(l) = (x+2*y-3 = 0);IsPerpendicular(LineSegmentOf(A, B), l);P: Point;MidPoint(LineSegmentOf(A, B)) = P;O: Point;Center(G) = O",
    "query_expressions": "Slope(LineOf(O, P))",
    "answer_expressions": "-2/9",
    "fact_spans": "[[[2, 7]], [[2, 7]], [[8, 45], [95, 97]], [[8, 45]], [[2, 48]], [[57, 75]], [[57, 75]], [[51, 77]], [[78, 81]], [[78, 90]], [[91, 94]], [[91, 100]]]",
    "query_spans": "[[[102, 113]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), P(x_{0},y_{0}), then x_{1}+x_{2}=2x_{0}, y_{1}+y_{2}=2y_{0}  \n\\begin{cases}\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1\\\\\\frac{x^{2}}{2}+\\frac{y^{2}}{4}=1\\\\\\frac{y^{2}}{1}-x_{2}=-\\frac{4}{9}.\\frac{x+2}{y+y},=-\\frac{4}{9}.\\frac{2x}{2y},\\end{cases}  \nAlso, chord AB is perpendicular to the line l: x+2y-3=0, so \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=2, thus \\frac{x_{0}}{y_{0}}\\times(-\\frac{4}{9})=2, hence k_{OP}=\\frac{y_{0}}{x_{0}}=-\\frac{2}{9}"
  },
  {
    "text": "The left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a, b>0)$ are $F_1$ and $F_2$, respectively. The line passing through focus $F_2$ and perpendicular to the $x$-axis intersects the hyperbola at points $A$ and $B$. If $\\overrightarrow{F_{1} A} \\cdot \\overrightarrow{F_{1} B}=0$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, H) = True;IsPerpendicular(H, xAxis) = True;H: Line;Intersection(H, G) = {A, B};B: Point;A: Point;DotProduct(VectorOf(F1, A), VectorOf(F1, B)) = 0",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2) +1",
    "fact_spans": "[[[0, 54], [97, 100], [176, 179]], [[0, 54]], [[3, 54]], [[3, 54]], [[3, 54]], [[3, 54]], [[63, 68]], [[71, 76], [80, 85]], [[0, 76]], [[0, 76]], [[77, 96]], [[86, 96]], [[94, 96]], [[94, 113]], [[108, 111]], [[103, 106]], [[115, 174]]]",
    "query_spans": "[[[176, 185]]]",
    "process": ""
  },
  {
    "text": "The difference of the distances from point $P$ to point $A(-m, 0)$ and to point $B(m, 0)$ $(m>0)$ is $2$. If $P$ lies on the line $y=x$, then the range of real values for $m$ is?",
    "fact_expressions": "G: Line;A: Point;B: Point;P: Point;m: Real;m>0;Expression(G) = (y = x);Coordinate(A) = (-m, 0);Coordinate(B) = (m, 0);PointOnCurve(P, G);Distance(P,A)-Distance(P,B)=2",
    "query_expressions": "Range(m)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[50, 57]], [[5, 16]], [[18, 35]], [[46, 49], [0, 4]], [[60, 65]], [[19, 35]], [[50, 57]], [[5, 16]], [[18, 35]], [[46, 58]], [[0, 44]]]",
    "query_spans": "[[[60, 72]]]",
    "process": ""
  },
  {
    "text": "Given point $P(\\sqrt{3}, 1)$ and circle $O$: $x^{2}+y^{2}=16$, a moving line passing through point $P$ intersects circle $O$ at points $M$ and $N$. What is the trajectory equation of the midpoint $Q$ of chord $MN$?",
    "fact_expressions": "O: Circle;G: Line;M: Point;N: Point;P: Point;Expression(O) = (x^2 + y^2 = 16);Coordinate(P) = (sqrt(3), 1);PointOnCurve(P, G);Intersection(G, O) = {M, N};IsChordOf(LineSegmentOf(M, N), O);Q: Point;MidPoint(LineSegmentOf(M, N)) = Q",
    "query_expressions": "LocusEquation(Q)",
    "answer_expressions": "(x - \\sqrt{3}/2)^2 + (y - 1/2)^2 = 1",
    "fact_spans": "[[[20, 42], [53, 57]], [[49, 52]], [[59, 62]], [[63, 66]], [[2, 19], [44, 48]], [[20, 42]], [[2, 19]], [[43, 52]], [[49, 66]], [[53, 74]], [[77, 80]], [[69, 80]]]",
    "query_spans": "[[[77, 86]]]",
    "process": "Let the midpoint of M and N be C, connect OC and OP, both being fixed points. Then triangle OPC is a right triangle with OP as the diameter. Point C lies on the circle, so the center of the circle is the midpoint of OP $\\left(\\frac{\\sqrt{3}}{2},\\frac{1}{2}\\right)$, and the radius is half of OP. Thus, the standard equation of the circle can be obtained."
  },
  {
    "text": "Given that the asymptotes of hyperbola $C$ are $y = \\pm \\frac{5}{2}x$, and it passes through the point $(3,8)$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Point;Coordinate(G) = (3, 8);Expression(Asymptote(C)) = (y = pm*(5/2)*x);PointOnCurve(G, C)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(29)/5",
    "fact_spans": "[[[2, 8], [49, 55]], [[39, 47]], [[39, 47]], [[2, 36]], [[2, 47]]]",
    "query_spans": "[[[49, 61]]]",
    "process": "From the asymptotes of the hyperbola $ y = \\pm\\frac{5}{2}x $, assume the equation of the hyperbola is $ 25x^{2} - 4y^{2} = \\lambda $ ($ \\lambda \\neq 0 $). Substituting the point $ (3,8) $, it is found that the foci of the hyperbola lie on the $ y $-axis, and $ \\frac{a}{b} = \\frac{5}{2} $, from which the eccentricity can be calculated. Let the equation of hyperbola $ C $ be $ 25x^{2} - 4y^{2} = \\lambda $ ($ \\lambda \\neq 0 $). Substituting $ (3,8) $ into $ C $, we get $ \\lambda = 25 \\times 9 - 4 \\times 64 = -31 $. Thus, the foci of hyperbola $ C $ lie on the $ y $-axis, and $ \\frac{a}{b} = \\frac{5}{2} $. Therefore, the eccentricity of hyperbola $ C $ is $ e = \\sqrt{1 + \\frac{4}{25}} = \\frac{\\sqrt{29}}{5} $."
  },
  {
    "text": "Draw a line $l$ through the focus $F$ of the parabola $C$: $y^{2}=8x$, intersecting the parabola $C$ at points $A$ and $B$. If the distance from $A$ to the directrix of the parabola is $6$, then $|AB|=$?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;F: Point;Expression(C) = (y^2 = 8*x);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Distance(A, Directrix(C)) = 6",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "9",
    "fact_spans": "[[[27, 32]], [[1, 20], [33, 39], [55, 58]], [[40, 43], [51, 54]], [[44, 47]], [[23, 26]], [[1, 20]], [[1, 26]], [[0, 32]], [[27, 49]], [[51, 68]]]",
    "query_spans": "[[[70, 79]]]",
    "process": "The coordinates of the focus are (2,0). The distance from point A to the directrix is 6, so the x-coordinate of A is 4. Substituting into the parabolic equation, the y-coordinate is found to be \\pm4\\sqrt{2}. Without loss of generality, assume A(4,4\\sqrt{2}). Thus, the slope of line AB is \\frac{4\\sqrt{2}}{4-2}=2\\sqrt{2}, and the equation is y=2\\sqrt{2}(x-2). Substituting into the parabolic equation and simplifying yields x^{2}-5x+4=0, with x_{1}+x_{2}=5. Therefore, |AB|=x_{1}+x_{2}+4=9."
  },
  {
    "text": "The distance from point $M$ on the parabola $y^{2}=4x$ to the focus is $3$. What is the horizontal coordinate of point $M$?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(M, G);Distance(M, Focus(G)) = 3",
    "query_expressions": "XCoordinate(M)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 14]], [[15, 19], [30, 34]], [[0, 14]], [[0, 19]], [[0, 28]]]",
    "query_spans": "[[[30, 40]]]",
    "process": "Let the coordinates of point M be (x, y), then the distance from point M to the focus is x + 1 = 3, solving gives x = 2,"
  },
  {
    "text": "If the foci of the ellipse $\\frac{x^{2}}{a+1}+\\frac{y^{2}}{3-a}=1$ lie on the $x$-axis, then the range of real values for $a$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(a + 1) + y^2/(3 - a) = 1);a: Real;PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Range(a)",
    "answer_expressions": "(1,3)",
    "fact_spans": "[[[1, 42]], [[1, 42]], [[53, 58]], [[1, 51]]]",
    "query_spans": "[[[53, 65]]]",
    "process": ""
  },
  {
    "text": "The two asymptotes of the hyperbola $\\frac{x^{2}}{b^{2}}-\\frac{y^{2}}{a^{2}}=1$ are perpendicular to each other. Then, what is the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/b^2 - y^2/a^2 = 1);b: Number;a: Number;l1: Line;l2: Line;Asymptote(G) = {l1, l2};IsPerpendicular(l1, l2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 46], [60, 63]], [[0, 46]], [[3, 46]], [[3, 46]], [], [], [[0, 52]], [[0, 56]]]",
    "query_spans": "[[[60, 69]]]",
    "process": ""
  },
  {
    "text": "If the line $l$: $y = a x + 2$ passes through the point $P(-1,6)$, then the eccentricity of the conic section $C$: $\\frac{x^{2}}{a} + \\frac{y^{2}}{16} = 1$ is?",
    "fact_expressions": "l: Line;C: ConicSection;a: Number;P: Point;Expression(C) = (y^2/16 + x^2/a = 1);Coordinate(P) = (-1, 6);Expression(l)=(y=a*x+2);PointOnCurve(P,l)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[1, 17]], [[31, 76]], [[7, 17]], [[19, 29]], [[31, 76]], [[19, 29]], [[1, 17]], [[1, 29]]]",
    "query_spans": "[[[31, 82]]]",
    "process": "From the condition, 6 = -a + 2, a = -4, so C: \\frac{y^{2}}{16} - \\frac{x^{2}}{4} = 1 is a hyperbola with foci on the y-axis, hence the eccentricity e = \\frac{\\sqrt{16+4}}{4}"
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ are $F_{1}$, $F_{2}$, and $P$ is a point on the ellipse. If $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, then $|\\overrightarrow{P F_{1}}| \\cdot|\\overrightarrow{P F_{2}}|$=?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 + y^2/4 = 1);PointOnCurve(P,G);Focus(G)={F1, F2};DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0",
    "query_expressions": "Abs(VectorOf(P,F1))*Abs(VectorOf(P,F2))",
    "answer_expressions": "8",
    "fact_spans": "[[[0, 37], [61, 63]], [[57, 60]], [[41, 48]], [[49, 56]], [[0, 37]], [[57, 67]], [[0, 56]], [[68, 127]]]",
    "query_spans": "[[[129, 191]]]",
    "process": "Analysis: From the equation of the ellipse, we obtain $ c = \\sqrt{a^{2} - b^{2}} = \\sqrt{5} $. Since $ \\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}} = 0 $, it follows that $ \\triangle F_{1}PF_{2} $ is a right triangle. Applying the Pythagorean theorem in the right triangle $ \\triangle F_{1}PF_{2} $, we get $ |\\overrightarrow{PF_{1}}|^{2} + |\\overrightarrow{PF_{2}}|^{2} = 20 $. Furthermore, by the definition of the ellipse, $ |\\overrightarrow{PF_{1}}| + |\\overrightarrow{PF_{2}}| = 2a = 6 $. Solving these two equations simultaneously yields $ |\\overrightarrow{PF_{1}}| \\cdot |\\overrightarrow{PF_{2}}| = 8 $, from which the area of the right triangle $ \\triangle F_{1}PF_{2} $ is found to be $ S = 4 $.  \n\nSince the equation of the ellipse is $ \\frac{x^{2}}{9} + \\frac{y^{2}}{4} = 1 $, we have $ a^{2} = 9 $ and $ b^{2} = 4 $, so $ c = \\sqrt{a^{2} - b^{2}} = \\sqrt{5} $. Because $ \\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}} = 0 $, it follows that $ \\overrightarrow{PF_{1}} \\perp \\overrightarrow{PF_{2}} $, hence $ \\angle F_{1}PF_{2} = 90^{\\circ} $, and therefore $ |\\overrightarrow{PF_{1}}|^{2} + |\\overrightarrow{PF_{2}}|^{2} = 20 \\cdots \\textcircled{1} $. By the definition of the ellipse, $ |\\overrightarrow{PF_{1}}| + |\\overrightarrow{PF_{2}}| = 2a = 6 $, thus $ (|\\overrightarrow{PF_{1}}| + |\\overrightarrow{PF_{2}}|)^{2} = 36 \\cdots \\textcircled{2} $. Subtracting $ \\textcircled{1} $ from $ \\textcircled{2} $, we obtain $ 2|\\overrightarrow{PF_{1}}| \\cdot |\\overrightarrow{PF_{2}}| = 16 $, so $ |\\overrightarrow{PF_{1}}| \\cdot |\\overrightarrow{PF_{2}}| = 8 $. Thus, the answer is: $ 8 $."
  },
  {
    "text": "Given that $A$ and $B$ are the left and right vertices of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, respectively, $P$ is a point on the ellipse in the first quadrant, $|P A|=\\lambda|P B|$, and $\\angle P B A=2 \\angle P A B$, then $\\lambda=$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);A: Point;B: Point;LeftVertex(G) = A;RightVertex(G) = B;P: Point;PointOnCurve(P, G);Quadrant(P) = 1;lambda: Number;Abs(LineSegmentOf(P, A)) = lambda*Abs(LineSegmentOf(P, B));AngleOf(P, B, A) = 2*AngleOf(P, A, B)",
    "query_expressions": "lambda",
    "answer_expressions": "sqrt(30)/3",
    "fact_spans": "[[[12, 39], [50, 52]], [[12, 39]], [[2, 5]], [[6, 9]], [[2, 45]], [[2, 45]], [[46, 49]], [[46, 60]], [[46, 60]], [[115, 124]], [[61, 81]], [[84, 113]]]",
    "query_spans": "[[[115, 126]]]",
    "process": "From the given conditions, we have A(-\\sqrt{2},0), B(\\sqrt{2},0). Let P(x_{0},y_{0}), and let the slopes of lines PA and PB be k_{1}, k_{2} respectively. Then k_{1}k_{2} = \\frac{y_{0}}{x_{0}+\\sqrt{2}} \\cdot \\frac{y_{0}}{x_{0}-\\sqrt{2}} = \\frac{y_{0}^{2}}{x_{0}^{2}-2} = \\frac{1-\\frac{x_{0}^{2}}{2}}{x_{0}^{2}-2} = -\\frac{1}{2}. By the law of sines, \\lambda = \\frac{|PA|}{|PB|} = \\frac{\\sin\\angle PBA}{\\sin\\angle PAB} = 2\\cos\\angle PAB. From \\angle PBA = 2\\angle PAB, we get \\tan\\angle PBA = \\tan 2\\angle PAB = \\frac{2\\tan\\angle PAB}{1-\\tan^{2}\\angle PAB}. Therefore, -k_{2} = \\frac{2k_{1}}{1-k_{1}^{2}}, and since k_{1}k_{2} = -\\frac{1}{2}, it follows that k_{1}^{2} = \\frac{1}{5}, i.e., \\tan^{2}\\angle PAB = \\frac{1}{5}. Also, \\sin^{2}\\angle PAB + \\cos^{2}\\angle PAB = 1, solving gives: \\cos\\angle PAB = \\frac{\\sqrt{30}}{6}, so \\lambda = \\frac{\\sqrt{30}}{3}."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the foci of the ellipse $\\frac{x^{2}}{3}+y^{2}=1$, and let points $A$ and $B$ lie on the ellipse. If $\\overrightarrow{F_{1} A}=5 \\overrightarrow{F_{2} B}$, then the coordinates of point $A$ are?",
    "fact_expressions": "G: Ellipse;F1: Point;A: Point;F2: Point;B: Point;Expression(G) = (x^2/3 + y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(A, G);PointOnCurve(B, G);VectorOf(F1, A) = 5*VectorOf(F2, B)",
    "query_expressions": "Coordinate(A)",
    "answer_expressions": "(0,pm*1)",
    "fact_spans": "[[[20, 47], [60, 62]], [[2, 9]], [[120, 124], [51, 55]], [[10, 17]], [[56, 59]], [[20, 47]], [[2, 50]], [[51, 63]], [[56, 63]], [[65, 118]]]",
    "query_spans": "[[[120, 129]]]",
    "process": ""
  },
  {
    "text": "Let the parabola $C$: $y^{2}=2 p x(p>0)$ have focus $F$ and directrix $l$. Let point $A$ be a point on the parabola $C$. A circle centered at $F$ with radius $FA$ intersects $l$ at points $B$ and $D$. If $\\angle B F D=120^{\\circ}$ and the area of triangle $A B D$ is $\\sqrt{3}$, then $p=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(C) = F;l: Line;Directrix(C) = l;A: Point;PointOnCurve(A, C) = True;Center(G) = F;Radius(G) = LineSegmentOf(F, A);G: Circle;Intersection(G, l) = {B, D};B: Point;D: Point;AngleOf(B, F, D) = ApplyUnit(120, degree);Area(TriangleOf(A, B, D)) = sqrt(3)",
    "query_expressions": "p",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[1, 27], [47, 53]], [[1, 27]], [[144, 147]], [[9, 27]], [[31, 34], [58, 61]], [[1, 34]], [[38, 41], [76, 79]], [[1, 41]], [[42, 46]], [[42, 56]], [[57, 75]], [[65, 75]], [[74, 75]], [[74, 89]], [[80, 83]], [[84, 87]], [[91, 117]], [[118, 142]]]",
    "query_spans": "[[[144, 149]]]",
    "process": ""
  },
  {
    "text": "A point $M(4, m)$ on the parabola $y^{2}=2 p x$ is at a distance of $6$ from the directrix, then $p=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;M: Point;m: Number;Coordinate(M) = (4, m);PointOnCurve(M, G);Distance(M, Directrix(G)) = 6",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 16]], [[0, 16]], [[41, 44]], [[19, 29]], [[19, 29]], [[19, 29]], [[0, 29]], [[0, 39]]]",
    "query_spans": "[[[41, 46]]]",
    "process": ""
  },
  {
    "text": "Given the line $M N$: $y=\\frac{1}{3} x+2$ and the hyperbola $C$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1$ intersect at points $M$ and $N$, and $O$ is the origin, then the area of $\\triangle O M N$ is?",
    "fact_expressions": "C: Hyperbola;M: Point;N: Point;O: Origin;Expression(C) = (x^2/9-y^2/4 = 1);Expression(LineOf(M,N))=(y=x/3+2);Intersection(LineOf(M,N),C)={M,N}",
    "query_expressions": "Area(TriangleOf(O, M, N))",
    "answer_expressions": "4*sqrt(7)",
    "fact_spans": "[[[31, 74]], [[77, 80]], [[81, 84]], [[87, 90]], [[31, 74]], [[2, 30]], [[2, 86]]]",
    "query_spans": "[[[95, 116]]]",
    "process": "Solving the system of equations gives a quadratic equation in $ x $. By Vieta's formulas, substitute into the chord length formula to find the chord length $ |MN| $, then calculate the distance from point $ O $ to line $ MN $, and use $ S_{\\triangle OMN} = \\frac{1}{2}|MN|d $ to compute the area.  \nSolve the system  \n$$\n\\begin{cases}\n\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1 \\\\\ny=\\frac{1}{3}x+2\n\\end{cases}\n$$  \nto get $ x^{2}-4x-24=0 $. Let $ M(x_{1},y_{1}) $, $ N(x_{2},y_{2}) $, then $ x_{1}+x_{2}=4 $, $ x_{1}x_{2}=-24 $, so  \n$$\n|MN| = \\sqrt{1+\\left(\\frac{1}{3}\\right)^{2}} \\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}} = \\frac{4}{3}\\sqrt{70}.\n$$  \nAlso, the distance from point $ O $ to line $ MN $ is $ d = \\frac{6}{\\sqrt{10}} $, thus  \n$$\nS_{\\Delta OMN} = \\frac{1}{2}|MN|d = \\frac{1}{2} \\times \\frac{4}{3}\\sqrt{70} \\times \\frac{6}{\\sqrt{10}} = 4\\sqrt{7}.\n$$"
  },
  {
    "text": "Given that point $A$ is a moving point on the parabola $\\Gamma$: $y^{2}=4 x$ with focus $F$, and $A$ does not coincide with the origin $O$. The perpendicular bisector of segment $O A$ intersects the $x$-axis at point $B$. If $\\overrightarrow{A F}=2 \\overrightarrow{C F}$, then $|A B|-|A C|$=?",
    "fact_expressions": "A: Point;F: Point;Focus(Gamma) = F;Gamma: Parabola;Expression(Gamma) = (y^2 = 4*x);PointOnCurve(A, Gamma) = True;Negation(A = O);O: Origin;Intersection(PerpendicularBisector(LineSegmentOf(O, A)), xAxis) = B;B: Point;VectorOf(A, F) = 2*VectorOf(C, F)",
    "query_expressions": "Abs(LineSegmentOf(A, B)) - Abs(LineSegmentOf(A, C))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[2, 6]], [[10, 13]], [[7, 38]], [[14, 38]], [[14, 38]], [[2, 42]], [[2, 55]], [[46, 53]], [[56, 79]], [[75, 79]], [[82, 127]]]",
    "query_spans": "[[[129, 144]]]",
    "process": "Let $ A(x_{0},y_{0}) $, $ x_{0}>0 $, $ B(x_{B},0) $, we obtain the midpoint $ D(\\frac{x_{0}}{2},\\frac{y_{0}}{2}) $ of segment $ OA $, thus $ k_{OA} \\cdot k_{DB} = -1 $, we get $ |AB| = |OB| = \\frac{1}{2}x_{0} + 2 $, then from $ \\overrightarrow{AF} = 2\\overrightarrow{CF} $, we get $ |AC| = \\frac{1}{2}|AF| = \\frac{x_{0}+1}{2} $, thus we can solve. Solution: Method 1: From the problem, without loss of generality, let $ A(x_{0},y_{0}) $, $ x_{0}>0 $, $ B(x_{B},0) $, then $ y_{0}^{2} = 4x_{0} $, the midpoint of segment $ OA $ is $ D(\\frac{x_{0}}{2},\\frac{y_{0}}{2}) $, according to the condition we have $ k_{OA} \\cdot k_{DB} = -1 $, so $ \\frac{y_{0}}{x_{0}} \\cdot \\frac{\\frac{y_{0}}{2}}{\\frac{x_{0}}{2}-x_{B}} = -1 $, so $ \\frac{y^{2}}{x_{0}(x_{0}-2x_{B})} = -1 $, so $ \\frac{4x_{0}}{x_{0}(x_{0}-2x_{B})} = -1 $, we obtain $ x_{B} = 2 + \\frac{1}{2}x_{0} $, hence $ |AB| = |OB| = \\frac{1}{2}x_{0} + 2 $. Since $ \\overrightarrow{AF} = 2\\overrightarrow{CF} $, $ C $ is the midpoint of segment $ AF $, and $ F(1,0) $, so $ |AC| = \\frac{1}{2}|AF| = \\frac{x_{0}+1}{2} $, therefore $ |AB| - |AC| = \\frac{3}{2} $. Method 2: From the condition we have $ F(1,0) $, $ C $ is the midpoint of segment $ AF $, without loss of generality, let $ A(x_{0},y_{0}) $, $ x_{0}>0 $, then $ C(\\frac{x_{0}+1}{2},\\frac{y_{0}}{2}) $, the perpendicular bisector of segment $ OA $ is $ y = -\\frac{x_{0}}{y_{0}}(x - \\frac{x_{0}}{2}) + \\frac{y_{0}}{2} $, let $ y = 0 $, we get $ x_{B} = \\frac{y_{0}^{2}}{2x_{0}} + \\frac{x_{0}}{2} $, and since $ y_{0}^{2} = 4x_{0} $, we have $ x_{B} = 2 + \\frac{x_{0}}{2} $, then $ |AB| = |OB| = 2 + \\frac{x_{0}}{2} $, also $ |AC| = \\sqrt{(x_{0} - \\frac{x_{0}}{2})^{+1}}^{2} + (y_{0} - \\frac{y_{0}}{2})^{2} = \\frac{x_{0}+1}{2} $, so $ |AB| - |AC| = 2 + \\frac{x_{0}}{2} - \\frac{x_{0}+1}{2} = \\frac{3}{2} $"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=16 x$ with focus $F$, $P$ is a moving point on the parabola $C$, and point $B(-4,6)$. When $\\frac{|P B|}{|P F|}$ reaches its maximum value, the coordinates of point $P$ are?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 16*x);F: Point;Focus(C) = F;P: Point;PointOnCurve(P, C);B: Point;Coordinate(B) = (-4, 6);WhenMax(Abs(LineSegmentOf(P, B))/Abs(LineSegmentOf(P, F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1, -4)",
    "fact_spans": "[[[2, 22], [34, 40]], [[2, 22]], [[26, 29]], [[2, 29]], [[30, 33], [83, 87]], [[30, 43]], [[44, 54]], [[44, 54]], [[55, 82]]]",
    "query_spans": "[[[83, 92]]]",
    "process": "According to the definition of a parabola, $\\frac{|PB|}{|PF|}$ is transformed into $\\frac{|PB|}{|PQ|}$; by analyzing the graph, determine when $\\frac{|PB|}{|PF|}$ reaches its maximum value and then find the coordinates of point $P$. Solution: From the given conditions, the focus is $F(4,0)$, and $B(-4,6)$ lies on the directrix of the parabola. Let point $Q$ be the projection of point $P$ onto the directrix of the parabola. Then $|PF|=|PQ|$, hence $\\frac{|PB|}{|PF|}=\\frac{|PB|}{|PQ|}$. To maximize $\\frac{|PB|}{|PF|}$, it suffices to minimize $\\angle PBQ$, which occurs when line $PB$ is tangent to the parabola. Let line $PB$: $x+4=t(y-6)$, i.e., $x=ty-6t-4$. Solving simultaneously \n\\[\n\\begin{cases}\ny^2=16x \\\\\nx=ty-6t\n\\end{cases}\n\\]\ngives $y^{2}-16ty+96t+64=0$, $v-6t-4'$. Since line $PB$ is tangent to the parabola, we have $A=(16t)^{2}-4(96t+64)=0$, so $t=2$ or $t=-\\frac{1}{2}$. From the graph, when $t=-\\frac{1}{2}$, $\\angle PBQ$ is minimized. Hence $y^{2}+8v+16=0$, i.e., $v=-4$. Therefore, the coordinates of point $P$ are $(1,-4)$."
  },
  {
    "text": "Given that $M$ is a point on the parabola $y^{2}=8x$, $F$ is the focus of the parabola, $\\angle MFO=120^{\\circ}$, and $N(-2,0)$ ($O$ being the origin), then the area of $\\triangle MNF$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);M: Point;PointOnCurve(M, G);F: Point;Focus(G) = F;AngleOf(M, F, O) = ApplyUnit(120, degree);N: Point;Coordinate(N) = (-2, 0);O: Origin",
    "query_expressions": "Area(TriangleOf(M, N, F))",
    "answer_expressions": "8*sqrt(3)",
    "fact_spans": "[[[6, 20], [29, 32]], [[6, 20]], [[2, 5]], [[2, 24]], [[25, 28]], [[25, 35]], [[37, 63]], [[64, 73]], [[64, 73]], [[74, 77]]]",
    "query_spans": "[[[85, 107]]]",
    "process": "From the given conditions, we have $k_{MF} = \\tan 60^{\\circ} = \\sqrt{3}$, $y_{M} > 0$, and by the definition of a parabola, $MF = x_{M} + 2$. Therefore, $\\frac{x_{M} + 2}{2} + 2 = x_{M} \\Rightarrow x_{M} = 6$, $y_{M} = 4\\sqrt{3}$. Thus, $S_{AMNF} = \\frac{1}{2} \\times 4\\sqrt{3} \\times 4 = 8\\sqrt{3}$."
  },
  {
    "text": "If a point $P$ on the hyperbola $\\frac{x^{2}}{64}-\\frac{y^{2}}{36}=1$ is at a distance of $8$ from its left focus, then what is the distance from point $P$ to its right directrix?",
    "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2/64 - y^2/36 = 1);PointOnCurve(P, G);Distance(P,LeftFocus(G)) = 8",
    "query_expressions": "Distance(P,RightDirectrix(G))",
    "answer_expressions": "96/5",
    "fact_spans": "[[[2, 42], [49, 50], [69, 70]], [[45, 48], [64, 68]], [[2, 42]], [[2, 48]], [[45, 61]]]",
    "query_spans": "[[[64, 79]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ with two foci $F_{1}$, $F_{2}$, and $P$ a point on the ellipse $C$, then the perimeter of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/25 + y^2/16 = 1);Focus(C)={F1,F2};PointOnCurve(P,C)",
    "query_expressions": "Perimeter(TriangleOf(P, F1, F2))",
    "answer_expressions": "16",
    "fact_spans": "[[[2, 46], [72, 77]], [[68, 71]], [[52, 59]], [[60, 67]], [[2, 46]], [[2, 67]], [[68, 80]]]",
    "query_spans": "[[[82, 109]]]",
    "process": "By the definition of an ellipse, $ PF_{1} + PF_{2} = 2 \\times \\sqrt{25} = 10 $, hence the perimeter of $ \\triangle PF_{1}F_{2} $ is $ PF_{1} + PF_{2} + F_{1}F_{2} = 10 + 2\\sqrt{25 - 16} = 16 $."
  },
  {
    "text": "Find a point $P$ on the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$ such that its distance to the fixed point $Q(0,1)$ is maximized. Then the coordinates of $P$ are?",
    "fact_expressions": "G: Ellipse;Q: Point;P: Point;Expression(G) = (x^2/8 + y^2/4 = 1);Coordinate(Q) = (0, 1);PointOnCurve(P, G);WhenMax(Distance(P,Q))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(pm*sqrt(6),-1)",
    "fact_spans": "[[[1, 38]], [[52, 60]], [[42, 45], [48, 49], [66, 69]], [[1, 38]], [[52, 60]], [[1, 45]], [[48, 65]]]",
    "query_spans": "[[[66, 74]]]",
    "process": ""
  },
  {
    "text": "Let $AB$ be the real axis of hyperbola $\\Gamma$, point $C$ lies on $\\Gamma$, and $\\angle CAB = \\frac{\\pi}{4}$. If $AB = 4$, $BC = \\sqrt{26}$, then the focal length of the hyperbola is?",
    "fact_expressions": "Gamma: Hyperbola;A: Point;B: Point;RealAxis(Gamma) = LineSegmentOf(A, B);C: Point;PointOnCurve(C, Gamma);AngleOf(C, A, B) = pi/4;LineSegmentOf(A, B) = 4;LineSegmentOf(B, C) = sqrt(26)",
    "query_expressions": "FocalLength(Gamma)",
    "answer_expressions": "4*sqrt(6)",
    "fact_spans": "[[[7, 18], [27, 35], [94, 97]], [[1, 6]], [[1, 6]], [[1, 21]], [[22, 26]], [[22, 36]], [[38, 66]], [[68, 75]], [[77, 92]]]",
    "query_spans": "[[[94, 102]]]",
    "process": "As shown in the figure, in triangle ABC, by the law of cosines, we have $ CB^{2} = CA^{2} + BA^{2} - 2CA \\cdot BA \\cdot \\cos\\frac{\\pi}{4} $, solving gives $ CA = 5\\sqrt{2} $. Draw $ CM \\perp x $-axis at M, then $ CM = 5\\sqrt{2} \\times \\cos\\frac{\\pi}{4} = 5 $, $ AM = 5 $, so $ C(3,5) $. Substituting point C's coordinates into the hyperbola equation yields $ \\frac{9}{2^{2}} - \\frac{25}{b^{2}} = 1 $, solving gives $ b^{2} = 20 $, $ c = \\sqrt{a^{2} + b^{2}} = 2\\sqrt{6} $, the focal length of the hyperbola is $ 2c = 4\\sqrt{6} $. This problem examines the positional relationship between a line and a hyperbola; using plane geometry knowledge and the definition of conic sections is an effective method for solving such problems, belonging to medium-difficulty questions."
  },
  {
    "text": "One focus of the hyperbola $8 kx^{2}-ky^{2}=8$ is $(3 , 0)$, then $k=$?",
    "fact_expressions": "G: Hyperbola;k: Number;Expression(G) = (8*k*x^2 - k*y^2=8);Coordinate(OneOf(Focus(G)))=(3,0)",
    "query_expressions": "k",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 22]], [[39, 42]], [[0, 22]], [[0, 37]]]",
    "query_spans": "[[[39, 44]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, a line $l$ passing through point $F$ intersects the parabola at points $A$ and $B$, $O$ is the origin, and the lines $OA$, $OB$ intersect the directrix of the parabola at points $P$ and $Q$, respectively. Then the minimum value of $|PQ|$ is?",
    "fact_expressions": "l: Line;C: Parabola;Expression(C) = (y^2 = 4*x);A: Point;O: Origin;B: Point;PointOnCurve(F, l);Intersection(l, C) = {A, B};P: Point;Q: Point;F: Point;Focus(C) = F;Intersection(LineOf(O,A),Directrix(C))=P;Intersection(LineOf(O,B),Directrix(C))=Q",
    "query_expressions": "Min(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "4",
    "fact_spans": "[[[35, 40]], [[2, 21], [41, 44], [82, 85]], [[2, 21]], [[47, 50]], [[57, 60]], [[51, 54]], [[29, 40]], [[35, 56]], [[93, 97]], [[98, 101]], [[25, 28], [30, 34]], [[2, 28]], [[66, 101]], [[66, 101]]]",
    "query_spans": "[[[103, 116]]]",
    "process": "Let the equation of line $ l $ be $ my = x - 1 $. By solving the system of equations and using the relationship between roots and coefficients, we obtain $ y_{1} + y_{2} $, $ y_{1}y_{2} $, and $ x_{1}x_{2} = 1 $. Combining with the equations of lines $ OA $ and $ OB $, we derive the coordinates of points $ P $ and $ Q $, obtain $ |PD| \\cdot |QD| = 4 $, and then, together with the basic inequality, solve the problem. Let the coordinates of points $ A $ and $ B $ be $ (x_{1}, y_{1}) $, $ (x_{2}, y_{2}) $, respectively. The equation of line $ l $ is $ my = x - 1 $. Solving the system of equations\n$$\n\\begin{cases}\ny^2 = 4x \\\\\nmy = x - 1\n\\end{cases}\n$$\neliminating $ x $ and rearranging gives $ y^{2} - 4my - 4 = 0 $, from which we get $ y_{1} + y_{2} = 4m $, $ y_{1}y_{2} = -4 $, then $ x_{1}x_{2} = \\frac{y_{1}^{2}y_{2}^{2}}{16} = 1 $. The equation of line $ OA $ is $ y = \\frac{y_{1}}{x_{1}} $, so the coordinates of point $ P $ are $ (-1, \\frac{y_{1}}{x_{1}}) $. The equation of line $ OB $ is $ y = \\frac{y_{2}}{x_{2}} $, so the coordinates of point $ Q $ are $ (-1, \\frac{y_{2}}{x_{2}}) $. Let $ D $ be the intersection point of the parabola's directrix with the $ x $-axis. Then $ |PD| \\cdot |QD| = \\left| \\frac{y_{1}}{x_{1}} \\cdot \\frac{y_{2}}{x_{2}} \\right| = \\frac{4}{1} = 4 $. Therefore, the minimum value of $ |PQ| $ is $ 4 $."
  },
  {
    "text": "Let $\\lambda \\in{R}$, if $\\frac{x^{2}}{\\lambda-2}+\\frac{y^{2}}{\\lambda-3}=1$ represents a hyperbola, then the range of values for $\\lambda$ is?",
    "fact_expressions": "lambda: Real;G: Hyperbola;Expression(G) = (x^2/(lambda - 2) + y^2/(lambda - 3) = 1)",
    "query_expressions": "Range(lambda)",
    "answer_expressions": "(2,3)",
    "fact_spans": "[[[1, 17], [77, 86]], [[72, 75]], [[19, 70]]]",
    "query_spans": "[[[77, 93]]]",
    "process": "Because $\\frac{x^{2}}{\\lambda-2}+\\frac{y^{2}}{\\lambda-3}=1$, that is, $\\frac{x^{2}}{\\lambda-2}-\\frac{y^{2}}{3-\\lambda}=1$, according to the properties of hyperbola, we know $(\\lambda-2)(3-\\lambda)>0$, i.e., $(\\lambda-2)(\\lambda-3)<0$. Solving the inequality yields $2<\\lambda<3$, that is, the range of $\\lambda$ is $(2,3)$."
  },
  {
    "text": "$l$ is a line passing through the focus $F$ of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ and perpendicular to the real axis. $A$, $B$ are the two vertices of the hyperbola $C$. If there exists a point $P$ on $l$ such that $\\angle A P B = 60^{\\circ}$, then the maximum value of the eccentricity of the hyperbola is?",
    "fact_expressions": "l: Line;C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;OneOf(Focus(C)) = F;PointOnCurve(F, l) = True;IsPerpendicular(l, RealAxis(C)) = True;A: Point;B:Point;Vertex(C) = {A, B};P: Point;PointOnCurve(P, l) = True;AngleOf(A, P, B) = ApplyUnit(60, degree)",
    "query_expressions": "Max(Eccentricity(C))",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[0, 3], [105, 108]], [[6, 68], [91, 97], [145, 148]], [[6, 68]], [[14, 68]], [[14, 68]], [[14, 68]], [[14, 68]], [[70, 73]], [[6, 73]], [[0, 82]], [[0, 82]], [[83, 86]], [[87, 90]], [[83, 102]], [[113, 116]], [[105, 116]], [[118, 143]]]",
    "query_spans": "[[[145, 157]]]",
    "process": ""
  },
  {
    "text": "The left and right foci of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ are $F_{1}$ and $F_{2}$ respectively, point $P$ lies on the hyperbola, and $|P F_{1}||P F_{2}|=64$, then $S_{\\Delta F_{1} P F_{2}}$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 64",
    "query_expressions": "Area(TriangleOf(F1,P,F2))",
    "answer_expressions": "16*sqrt(3)",
    "fact_spans": "[[[0, 39], [69, 72]], [[0, 39]], [[48, 55]], [[56, 63]], [[0, 63]], [[0, 63]], [[64, 68]], [[64, 73]], [[75, 98]]]",
    "query_spans": "[[[100, 128]]]",
    "process": "By the given condition, $|PF_{1}|-|PF_{2}|=2\\sqrt{9}=6$. By the cosine law, $\\cos\\angle F_{1}PF_{2}=\\frac{|PF_{1}|^{2}+|PF_{2}|^{2}-|F_{1}F_{2}|^{2}}{2|PF_{1}||PF_{2}|}=\\frac{(|PF_{1}|-|PF_{2}|)^{2}+2|PF_{1}||PF_{2}|-|F_{1}F_{2}|^{2}}{2|PF_{1}||PF_{2}|}$. Also, $|PF_{1}|-|PF_{2}|=6$, $|PF_{1}||PF_{2}|=64$, $|F_{1}F_{2}|=2\\sqrt{9+16}=10$. Substituting these values yields $\\cos\\angle F_{1}PF_{2}=\\frac{36+128-100}{128}=\\frac{1}{2}$. Since $\\angle F_{1}PF_{2}\\in(0,\\pi)$, it follows that $\\angle F_{1}PF_{2}=\\frac{\\pi}{3}$. Hence, $S=\\frac{1}{2}|PF_{1}||PF_{2}|\\sin\\angle F_{1}PF_{2}=\\frac{1}{2}\\times64\\times\\frac{\\sqrt{3}}{2}=16\\sqrt{3}$."
  },
  {
    "text": "Given the focus $F$ of the parabola $y^{2}=4x$ and the point $A(1,1)$, when the point $P$ moves along the parabola, find the coordinates of point $P$ at which $|PA|+|PF|$ attains its minimum value.",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (1, 1);Focus(G) = F;PointOnCurve(P, G);WhenMin(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1/4,1)",
    "fact_spans": "[[[2, 16], [37, 40]], [[23, 31]], [[32, 36], [66, 70]], [[19, 22]], [[2, 16]], [[23, 31]], [[2, 22]], [[32, 44]], [[46, 65]]]",
    "query_spans": "[[[66, 75]]]",
    "process": "Draw PB perpendicular to the directrix from point P, and draw AH perpendicular to the directrix from point A. Then PA + PF = PA + PB \\leqslant AH, which is minimized at this moment. The y-coordinate of point P is the same as that of point A, so point P is (\\frac{1}{4},1). 5. This mainly examines simple geometric properties of parabolas and extreme values of parabolas, aiming to assess the mastery level of these knowledge points and the ability of combining numerical and graphical analysis and reasoning. (2) When solving conic section problems, seeing the focus and focal radius should remind one of the definition of the curve to improve problem-solving efficiency."
  },
  {
    "text": "Given a point $M(1, m)$ on the parabola $y^{2}=2 px(p>0)$ such that the distance from $M$ to its focus $F$ is $5$, and the distance from the vertex of the parabola to the line $MF$ is $d$, then the value of $d$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;M: Point;m: Number;Coordinate(M) = (1, m);PointOnCurve(M, G);F: Point;Focus(G) = F;Distance(M, F) = 5;d: Number;Distance(Vertex(G), LineOf(M, F)) = d",
    "query_expressions": "d",
    "answer_expressions": "(8*sqrt(34))/17",
    "fact_spans": "[[[2, 22], [50, 53], [35, 36]], [[2, 22]], [[5, 22]], [[5, 22]], [[25, 34]], [[25, 34]], [[25, 34]], [[2, 34]], [[38, 41]], [[35, 41]], [[25, 48]], [[67, 70], [72, 75]], [[50, 70]]]",
    "query_spans": "[[[72, 79]]]",
    "process": ""
  },
  {
    "text": "Let the parabola $C$: $y^{2}=4x$ have focus $F$, and let the directrix $l$ intersect the $x$-axis at point $M$. Let $P$ be a point on $C$ such that $|PF|=5$. Then $|PM|=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l: Line;Directrix(C) = l;M: Point;Intersection(l, xAxis) = M;P: Point;PointOnCurve(P, C);Abs(LineSegmentOf(P, F)) = 5",
    "query_expressions": "Abs(LineSegmentOf(P, M))",
    "answer_expressions": "sqrt(41)",
    "fact_spans": "[[[1, 20], [50, 53]], [[1, 20]], [[24, 27]], [[1, 27]], [[30, 33]], [[1, 33]], [[42, 45]], [[30, 45]], [[46, 49]], [[46, 56]], [[58, 67]]]",
    "query_spans": "[[[69, 78]]]",
    "process": "From the equation of the parabola $ C: y^2 = 4x $, we obtain the focus $ F(1,0) $ and the directrix $ l: x = -1 $. According to the problem, $ M(-1,0) $. Let $ P(x,y) $. By the property of the parabola, $ |PF| = x + 1 = 5 $, solving gives $ x = 4 $. Substituting into the equation of the parabola yields $ y^2 = 4 \\times 4 = 16 $. Therefore, $ |PM| = \\sqrt{(4+1)^2 + 16} = \\sqrt{41} $."
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ has two foci $F_{1}$, $F_{2}$. A line passes through $F_{1}$ and intersects the ellipse at points $P$, $Q$. Then, what is the perimeter of $\\Delta PQF_{2}$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};H: Line;PointOnCurve(F1, H);P: Point;Q: Point;Intersection(H, G) = {P, Q}",
    "query_expressions": "Perimeter(TriangleOf(P, Q, F2))",
    "answer_expressions": "4*a",
    "fact_spans": "[[[0, 38], [72, 74]], [[0, 38]], [[43, 50], [64, 71]], [[52, 59]], [[0, 59]], [[61, 63]], [[61, 71]], [[75, 78]], [[79, 82]], [[61, 82]]]",
    "query_spans": "[[[84, 105]]]",
    "process": "From the ellipse equation, we know $a^{2}=25$, $\\therefore a=5$. From the definition of the ellipse, the perimeter of $\\triangle PQF_{2}$ is $4a=20$."
  },
  {
    "text": "Given the ellipse $x^{2}+\\frac{y^{2}}{2}=1$, what is the length of the chord intercepted by the line $y=2x-1$ on the ellipse?",
    "fact_expressions": "G: Ellipse;H: Line;Expression(G) = (x^2 + y^2/2 = 1);Expression(H) = (y = 2*x - 1)",
    "query_expressions": "Length(InterceptChord(H,G))",
    "answer_expressions": "5*sqrt(2)/3",
    "fact_spans": "[[[2, 29], [31, 33]], [[34, 45]], [[2, 29]], [[34, 45]]]",
    "query_spans": "[[[31, 52]]]",
    "process": "Let the line $ y = 2x - 1 $ intersect the ellipse $ x^{2} + \\frac{y^{2}}{2} = 1 $ at points $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Solving the system \n\\[\n\\begin{cases}\ny = 2x - 1 \\\\\nx^{2} + \\frac{y^{2}}{2} = 1\n\\end{cases}\n\\]\nand eliminating $ y $, we obtain $ 6x^{2} - 4x - 1 = 0 $. $ A = 40 > 0 $. By Vieta's formulas, $ x_{1} + x_{2} = \\frac{2}{3} $, $ x_{1}x_{2} = -\\frac{1}{6} $. Using the chord length formula, \n\\[\n|AB| = \\sqrt{1 + 2^{2}} \\cdot |x_{1} - x_{2}| = \\sqrt{5} \\cdot \\sqrt{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}} = \\sqrt{5} \\times \\sqrt{\\left(\\frac{2}{3}\\right)^{2} - 4 \\left(-\\frac{1}{6}\\right)} = \\frac{5\\sqrt{2}}{3}\n\\]"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$ with foci $F_{1}$ and $F_{2}$, point $M$ lies on the ellipse such that $M F_{1} \\perp x$-axis. Find the distance from point $F_{1}$ to the line $F_{2} M$.",
    "fact_expressions": "G: Ellipse;F2: Point;M: Point;F1: Point;Expression(G) = (x^2/16 + y^2/12 = 1);Focus(G)={F1,F2};PointOnCurve(M, G);IsPerpendicular(LineSegmentOf(M,F1),xAxis)",
    "query_expressions": "Distance(F1,LineOf(F2,M))",
    "answer_expressions": "12/5",
    "fact_spans": "[[[2, 41], [66, 68]], [[53, 60]], [[61, 65]], [[45, 52], [91, 99]], [[2, 41]], [[2, 60]], [[61, 69]], [[71, 89]]]",
    "query_spans": "[[[91, 116]]]",
    "process": "From the ellipse equation, we have: $ a=4, b=2\\sqrt{3}, c=2 $. Since $ MF_{1} \\perp x $-axis, i.e., $ MF_{1} $ is the semi-latus rectum of the ellipse, $ |MF_{1}| = \\frac{b^{2}}{a} = 3 $. Therefore, $ |MF_{2}| = 2a - |MF_{1}| = 5 $. Let $ d $ be the distance from $ F_{1} $ to the line $ F_{2}M $, then $ S_{AF_{1}F_{2}M} = \\frac{1}{2}|F_{2}M| \\cdot d = \\frac{1}{2}|F_{1}F_{2}| \\cdot |MF_{1}| $, that is, $ \\frac{5}{2}d = 6 $, solving gives: $ d = \\frac{12}{5} $."
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{m+2}-\\frac{y^{2}}{2 m-1}=1$ represents a hyperbola, what is the range of real values for $m$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(m + 2) - y^2/(2*m - 1) = 1);m: Real",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-oo, -2) + (1/2, +oo)",
    "fact_spans": "[[[46, 49]], [[1, 49]], [[51, 56]]]",
    "query_spans": "[[[51, 63]]]",
    "process": "From the given condition, we have $(m+2)(2m-1)>0$, solving yields $m<-2$ or $m>\\frac{1}{2}$. Therefore, the range of $m$ is $(-\\infty,-2)\\cup(\\frac{1}{2},+\\infty)$."
  },
  {
    "text": "The distance from the focus of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1(a>0)$ to its asymptote is?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1)",
    "query_expressions": "Distance(Focus(G), Asymptote(G))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 37], [41, 42]], [[3, 37]], [[3, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 50]]]",
    "process": "By the symmetry of the hyperbola, without loss of generality, assume the focus is at $(c,0)$ $(c=\\sqrt{a^{2}+1})$, and one of its asymptotes is $x-ay=0$. Therefore, the required distance is $d=\\frac{|c|}{\\sqrt{1+a^{2}}}=\\frac{c}{c}=1$."
  },
  {
    "text": "Write the standard equation of an ellipse whose major axis length is equal to 8 times the eccentricity.",
    "fact_expressions": "G: Ellipse;Length(MajorAxis(G)) = 8*Eccentricity(G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 + y^2/3 = 1 (the answer is not unique)",
    "fact_spans": "[[[17, 19]], [[4, 19]]]",
    "query_spans": "[[[17, 25]]]",
    "process": "Assume the foci of the ellipse lie on the x-axis, and the standard equation of the ellipse is \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0). Since the length of the major axis equals 8 times the eccentricity, we have 2a=8\\frac{c}{a}, that is, a^{2}=4c. Let c=1, then a^{2}=4, b^{2}=3. Therefore, an ellipse satisfying the condition is given by \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1."
  },
  {
    "text": "The distance from the right focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ to the line $x+2 y-8=0$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;Expression(G) = (x^2/4 - y^2/5 = 1);Expression(H) = (x + 2*y - 8 = 0)",
    "query_expressions": "Distance(RightFocus(G), H)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[0, 38]], [[43, 56]], [[0, 38]], [[43, 56]]]",
    "query_spans": "[[[0, 61]]]",
    "process": "Given that $ c = \\sqrt{a^{2} + b^{2}} = \\sqrt{5 + 4} = 3 $, the right focus of the hyperbola is at $ (3, 0) $. Therefore, the distance from the right focus $ (3, 0) $ to the line $ x + 2y - 8 = 0 $ is $ \\frac{|3 + 2 \\times 0 - 8|}{\\sqrt{1^{2} + 2^{2}}} = \\frac{5}{\\sqrt{5}} = \\sqrt{5} $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ is a moving point on the ellipse. Then the range of values of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G)",
    "query_expressions": "Range(DotProduct(VectorOf(P, F1), VectorOf(P, F2)))",
    "answer_expressions": "[-1,4]",
    "fact_spans": "[[[2, 39], [71, 73]], [[2, 39]], [[48, 55]], [[57, 65]], [[2, 65]], [[2, 65]], [[66, 70]], [[66, 79]]]",
    "query_spans": "[[[81, 145]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, what is the equation of the circle centered at the right focus of $C$ and tangent to the asymptotes of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;E: Circle;Expression(C) = (x^2/16 - y^2/9 = 1);Center(E) = RightFocus(C);IsTangent(Asymptote(C), E)",
    "query_expressions": "Expression(E)",
    "answer_expressions": "(x - 5)^2 + y^2 = 9",
    "fact_spans": "[[[2, 46], [49, 52], [62, 68]], [[75, 76]], [[2, 46]], [[48, 76]], [[61, 76]]]",
    "query_spans": "[[[75, 81]]]",
    "process": "The hyperbola $ C: \\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1 $ has its right focus at $ (5,0) $, and the asymptotes are given by $ 3x+4y=0 $. Therefore, the center of the circle is $ (5,0) $, and the radius is $ r=\\frac{|3\\times5\\pm0|}{\\sqrt{16+9}}=3 $. Hence, the equation of the required circle is $ (x-5)^{2}+y^{2}=9 $."
  },
  {
    "text": "If the asymptotes of a hyperbola are given by $y = \\pm 2x$, and the foci lie on the $y$-axis, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*(2*x));PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[1, 4], [34, 37]], [[1, 22]], [[1, 32]]]",
    "query_spans": "[[[34, 43]]]",
    "process": "According to the relationship between asymptotes and eccentricity, the solution can be obtained. The foci of the hyperbola lie on the y-axis, and the equations of the asymptotes are $y = \\pm 2x$. The eccentricity of the hyperbola is $\\sqrt{1 + \\frac{1}{4}} = \\frac{\\sqrt{5}}{2}$."
  },
  {
    "text": "The focal distance of the ellipse $\\frac{x^{2}}{m^{2}}+y^{2}=1$ is $2$, then what is the value of $m$?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (y^2 + x^2/m^2 = 1);FocalLength(G) = 2",
    "query_expressions": "m",
    "answer_expressions": "pm*sqrt(2)",
    "fact_spans": "[[[0, 31]], [[40, 43]], [[0, 31]], [[0, 38]]]",
    "query_spans": "[[[40, 47]]]",
    "process": "Since the focal distance of the ellipse $\\frac{x^{2}}{m2}+y^{2}=1$ is 2, we have $c=1$, that is, $c^{2}=1$. Because $c^{2}=a^{2}-b^{2}$, it follows that $1=m^{2}-1$, solving gives $m=\\pm\\sqrt{2}$."
  },
  {
    "text": "The equation of the directrix of the parabola $y^{2}=-4 x$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -4*x)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=1",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 22]]]",
    "process": ""
  },
  {
    "text": "The standard equation of a parabola passing through the point $(2,4)$, with vertex at the origin and focus on the $y$-axis is?",
    "fact_expressions": "G: Parabola;H: Point;O: Origin;Coordinate(H) = (2, 4);PointOnCurve(H, G);Vertex(G) = O;PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 = y",
    "fact_spans": "[[[25, 28]], [[1, 9]], [[13, 15]], [[1, 9]], [[0, 28]], [[10, 28]], [[16, 28]]]",
    "query_spans": "[[[25, 35]]]",
    "process": "Let the standard equation of the parabola passing through (2,4), with vertex at the origin and focus on the y-axis, be x^{2}=2py, p>0. Substituting (2,4) gives: 4=8p, solving yields p=\\frac{1}{2}. Therefore, the standard equation of the parabola is x^{2}=v.b Answer: x2=v"
  },
  {
    "text": "The line $l$: $y=2(x-\\sqrt{5})$ passes through the right focus $F$ of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ and intersects the hyperbola $C$ at exactly one point. Then the eccentricity of $C$ is?",
    "fact_expressions": "l: Line;Expression(l) = (y=2*(x-sqrt(5)));C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;RightFocus(C) = F;F: Point;PointOnCurve(F,l) = True;NumIntersection(l,C) = 1",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[0, 24]], [[0, 24]], [[25, 86], [95, 101], [111, 114]], [[25, 86]], [[33, 86]], [[33, 86]], [[33, 86]], [[33, 86]], [[25, 93]], [[90, 93]], [[0, 93]], [[0, 108]]]",
    "query_spans": "[[[111, 120]]]",
    "process": "Combining the hyperbola property $\\frac{b}{a}=2, 0=2(c-\\sqrt{5})$, we can solve for $a$ and $c$. [Detailed solution] The asymptotes of the hyperbola $C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ $(a>0, b>0)$ are given by $y = \\frac{b}{a}x$. Since the line $y = 2(x - \\sqrt{5})$ passing through the right focus $F$ of the hyperbola $C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ $(a>0, b>0)$ intersects $C$ at exactly one point, it follows that $\\frac{b}{a} = 2$, $0 = 2(c - \\sqrt{5})$. Also, since $a^{2} + b^{2} = c^{2}$, solving gives $c = \\sqrt{5}, a = 1$, so $e = \\frac{c}{a} = \\sqrt{5}$."
  },
  {
    "text": "Given a hyperbola $C_{1}: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ $(a>0, b>0)$ with eccentricity $2$, its right focus $F$ coincides with the focus of the parabola $C_{2}$, the center of $C_{1}$ coincides with the vertex of $C_{2}$, and $M$ is a common point of $C_{1}$ and $C_{2}$. If $|MF| = 5$, then the standard equation of $C_{1}$ is?",
    "fact_expressions": "C1: Hyperbola;C2: Parabola;M: Point;F: Point;Expression(C1)=(x^2/a^2 - y^2/b^2 = 1);Eccentricity(C1)=2;RightFocus(C1)=F;Focus(C2)=F;Center(C1)=Vertex(C2);Intersection(C1,C2)=M;Abs(LineSegmentOf(M, F)) = 5;a:Number;b:Number;a>0;b>0",
    "query_expressions": "Expression(C1)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[10, 75], [99, 106], [127, 134], [159, 166]], [[83, 93], [110, 117], [135, 142]], [[123, 126]], [[79, 82]], [[10, 75]], [[2, 75]], [[10, 82]], [[79, 98]], [[99, 122]], [[123, 146]], [[148, 157]], [[10, 75]], [[10, 75]], [[10, 75]], [[10, 75]]]",
    "query_spans": "[[[159, 173]]]",
    "process": "e=\\frac{c}{a}=2\\Rightarrow c=2a, l=\\sqrt{3}a, F(2a,0) so the hyperbola equation is: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3a^{2}}=1 \\frac{p}{2}=2a\\Rightarrow 2p=8a, let the parabola equation be: y^{2}=8ax. Solving the system of equations: \\begin{cases} 3x^{2}-y^{2}= \\\\ y^{2}=8ax \\end{cases} -y^{2}=3a^{2}\\Rightarrow 3x^{2}-8ax-3a^{2}=0 solving gives x=\\frac{8a\\pm10a}{6}=3a or -\\frac{a}{3} (discarded) \\therefore MF=3a+\\frac{p}{2}=3a+2a=5a=5\\Rightarrow a=1 so the hyperbola equation is: x^{2}-\\frac{y^{2}}{3}=1"
  },
  {
    "text": "The equation $\\frac{x^{2}}{18-m}+\\frac{y^{2}}{6+m}=1$ represents an ellipse with foci on the $y$-axis. Then the range of values for $m$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(18 - m) + y^2/(m + 6) = 1);PointOnCurve(Focus(G), yAxis);m: Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(6,18)",
    "fact_spans": "[[[53, 55]], [[0, 55]], [[44, 55]], [[57, 60]]]",
    "query_spans": "[[[57, 67]]]",
    "process": ""
  },
  {
    "text": "The standard equation of an ellipse with foci on the coordinate axes, focal length $2 \\sqrt{6}$, and minor axis length $4$ is?",
    "fact_expressions": "G: Ellipse;PointOnCurve(Focus(G),axis);FocalLength(G) = 2*sqrt(6);Length(MinorAxis(G)) = 4",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/10+y^2/4=1,y^2/10+x^2/4=1}",
    "fact_spans": "[[[32, 34]], [[0, 34]], [[8, 34]], [[24, 34]]]",
    "query_spans": "[[[32, 41]]]",
    "process": "Let the focal distance of the ellipse be 2c, the minor axis length be 2b, and the major axis length be 2a, and 2c=2\\sqrt{6}, 2b=4, so a^{2}=b^{2}+c^{2}=10, b^{2}=4. When the foci are on the x-axis, the standard equation of the ellipse is: x^{2}. When the foci are on the y-axis, the standard equation of the ellipse is: \\frac{y^{2}}{4}+\\frac{x}{x}+\\frac{y^{2}}{4}=1"
  },
  {
    "text": "Given the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, with eccentricity $e$. If a point $P$ on the hyperbola satisfies $\\angle P F_{2} F_{1}=60^{\\circ}$, then the value of $\\overrightarrow{F_{2} P} \\cdot \\overrightarrow{F_{2} F_{1}}$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;F2: Point;F1: Point;e:Number;Expression(G) = (x^2 - y^2/3 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;Eccentricity(G)=e;PointOnCurve(P,G);AngleOf(P,F2,F1)=ApplyUnit(60,degree)",
    "query_expressions": "DotProduct(VectorOf(F2, P), VectorOf(F2, F1))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 30], [64, 67]], [[70, 73]], [[47, 54]], [[39, 46]], [[59, 62]], [[2, 30]], [[2, 54]], [[2, 54]], [[2, 62]], [[64, 73]], [[74, 107]]]",
    "query_spans": "[[[109, 174]]]",
    "process": "From the given, we have $F_{2}F_{1}=2c=4$. In $\\triangle PF_{1}F_{2}$, let $PF_{2}=x$, then $PF_{1}=x+2$ or $PF_{1}=x-2$. When $PF_{1}=x+2$, by the law of cosines, $(x+2)^{2}=x^{2}+4^{2}-2\\times4x\\times\\frac{1}{2}$, solving gives $x=\\frac{3}{2}$, so $\\overrightarrow{F_{2}P}\\cdot\\overrightarrow{F_{2}F_{1}}=\\frac{3}{2}\\times4\\times\\frac{1}{2}=3$. When $PF_{1}=x-2$, by the law of cosines, $(x-2)^{2}=x^{2}+4^{2}-2\\times4x\\times\\frac{1}{2}$, which has no solution. Hence, $\\overrightarrow{F_{2}P}\\cdot\\overrightarrow{F_{2}F_{1}}=3$."
  },
  {
    "text": "Given the focus of a parabola $F(a, 0)$ where $a < 0$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;F: Point;Coordinate(F) = (a, 0);Focus(G) = F;a: Number;a<0",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=4*a*x",
    "fact_spans": "[[[2, 5], [24, 27]], [[8, 22]], [[8, 22]], [[2, 22]], [[8, 22]], [[8, 22]]]",
    "query_spans": "[[[24, 32]]]",
    "process": "Given the focus of a parabola and the standard equation of a parabola, since the focus $ F(a,0) $ ($ a<0 $), it is known that the focus lies on the negative half of the x-axis, $ \\frac{p}{2} = -a $. Therefore, $ y^{2} = -2px = 4ax $."
  },
  {
    "text": "If the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has a major axis length of $4$ and a focal distance of $2$, then the standard equation of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Length(MajorAxis(C)) = 4;FocalLength(C) = 2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4+y^2/3=1",
    "fact_spans": "[[[1, 58], [75, 80]], [[1, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[1, 66]], [[1, 73]]]",
    "query_spans": "[[[75, 87]]]",
    "process": "Since the major axis of the ellipse is 4, then a=2, and the focal distance is 2, so c=1. From b^{2}=a^{2}-c^{2}=4-1=3, we get b=\\sqrt{3}. Thus, the standard equation of ellipse C is: \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1"
  },
  {
    "text": "Given that the product of the slopes of the lines joining a moving point $P(x, y)$ to two fixed points $M(-1,0)$ and $N(1,0)$ is equal to a constant $\\lambda$ ($\\lambda \\neq 0$), then the trajectory equation of the moving point $P$ is?",
    "fact_expressions": "P:Point;M:Point;N:Point;lambda:Number ;Coordinate(P)=(x1,y1);Coordinate(M) = (-1, 0);Coordinate(N) = (1, 0);Slope(LineOf(P,M))*Slope(LineOf(P,N))=lambda;Negation(lambda=0);x1: Number;y1: Number",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2 - y^2/lambda = 1, negation(lambda = 0), negation(x = pm*1)",
    "fact_spans": "[[[4, 13], [75, 78]], [[17, 26]], [[27, 35]], [[46, 71]], [[4, 13]], [[17, 26]], [[27, 35]], [[2, 71]], [[46, 71]], [[4, 13]], [[4, 13]]]",
    "query_spans": "[[[73, 85]]]",
    "process": "From the given conditions, the slopes of lines PM and PN exist and are both non-zero, so $k_{PM} \\cdot k_{PN} = \\frac{y}{x+1} \\cdot \\frac{y}{x-1} = \\lambda$. Rearranging gives $x^{2} - \\frac{y^{2}}{2} = 1$ $(\\lambda \\neq 0, x \\neq \\pm 1)$. Therefore, the trajectory equation of the moving point P is $x^{2} - \\frac{y^{2}}{2} = 1$ $(\\lambda \\neq 0, x \\neq \\pm 1)$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, respectively, and point $P$ is a point on the hyperbola such that $|P F_{1}|=3$, then the value of $|PF_{2}|$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 - y^2/12 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 3",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "7",
    "fact_spans": "[[[20, 59], [71, 74]], [[66, 70]], [[2, 9]], [[10, 17]], [[20, 59]], [[2, 65]], [[2, 65]], [[66, 77]], [[79, 92]]]",
    "query_spans": "[[[94, 109]]]",
    "process": ""
  },
  {
    "text": "Draw a line with slope $2$ passing through the right focus of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$, intersecting the ellipse at points $A$ and $B$, and let $O$ be the origin. Then the area of $\\triangle OAB$ is?",
    "fact_expressions": "G: Ellipse;H: Line;O: Origin;A: Point;B: Point;Expression(G) = (x^2/5 + y^2/4 = 1);PointOnCurve(RightFocus(G), H);Slope(H) = 2;Intersection(H, G) = {A, B}",
    "query_expressions": "Area(TriangleOf(O, A, B))",
    "answer_expressions": "5/3",
    "fact_spans": "[[[1, 38], [55, 57]], [[52, 54]], [[69, 72]], [[59, 62]], [[63, 66]], [[1, 38]], [[0, 54]], [[45, 54]], [[52, 68]]]",
    "query_spans": "[[[79, 99]]]",
    "process": ""
  },
  {
    "text": "If on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the point with abscissa $\\frac{a}{3}$ has its distance to the left focus greater than its distance to the right directrix, then the range of values for the ellipse's eccentricity is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;D: Point;XCoordinate(D) = a/3;PointOnCurve(D, G) = True;Distance(D, LeftFocus(G)) > Distance(D, RightDirectrix(G))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(\\sqrt{7}-2,1)",
    "fact_spans": "[[[1, 53], [92, 94]], [[1, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[72, 73], [82, 83]], [[54, 73]], [[1, 73]], [[1, 90]]]",
    "query_spans": "[[[92, 104]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $x^{2} + 2 y^{2}=4$, what is the length of the chord that has $(1 ,1 )$ as its midpoint?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 2*y^2 = 4);H:LineSegment;IsChordOf(H,G);Coordinate(MidPoint(H))=(1,1)",
    "query_expressions": "Length(H)",
    "answer_expressions": "sqrt(30)/3",
    "fact_spans": "[[[2, 23]], [[2, 23]], [], [[2, 42]], [[2, 42]]]",
    "query_spans": "[[[2, 47]]]",
    "process": ""
  },
  {
    "text": "If the curve represented by the equation $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{a^{2}-4}=1$ is a hyperbola, then the range of real values for $a$ is?",
    "fact_expressions": "G: Hyperbola;H: Curve;a: Real;Expression(H) = (y^2/(a^2 - 4) + x^2/a^2 = 1);G=H",
    "query_expressions": "Range(a)",
    "answer_expressions": "{(-2,0),(0,2)}",
    "fact_spans": "[[[54, 57]], [[51, 53]], [[59, 64]], [[1, 53]], [[51, 57]]]",
    "query_spans": "[[[59, 71]]]",
    "process": "From the given conditions, \\begin{cases}a2-4<0\\\\a\\neq0\\end{cases}, it follows that -2<a<0 or 0<a<2."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has an eccentricity of $\\frac{2}{3}$. A line $l$ with slope $k$ is drawn through the right focus $F$, intersecting the ellipse at points $A$ and $B$. If $\\overrightarrow{A F}=2 \\overrightarrow{F B}$, then $k=$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Eccentricity(G) = 2/3;F: Point;RightFocus(G) = F;l: Line;PointOnCurve(F,l) = True;Slope(l) = k;k: Number;Intersection(l,G) = {A,B};A: Point;B: Point;VectorOf(A, F) = 2*VectorOf(F, B)",
    "query_expressions": "k",
    "answer_expressions": "pm*sqrt(3)",
    "fact_spans": "[[[2, 54], [95, 97]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[2, 72]], [[78, 81]], [[2, 81]], [[89, 94]], [[74, 94]], [[82, 94]], [[85, 88], [158, 161]], [[89, 109]], [[100, 103]], [[104, 107]], [[111, 156]]]",
    "query_spans": "[[[158, 163]]]",
    "process": "Let $ l $ be the right directrix of the ellipse, and from points $ A $ and $ B $, draw $ AA_{1} $, $ BB_{1} $ perpendicular to $ l $, with $ A_{1} $, $ B_{1} $ as the feet of the perpendiculars. Draw $ BE \\perp AA_{1} $ meeting at $ E $. According to the second definition of the ellipse, we have $ |AA_{1}| = \\frac{|AF|}{e} $, $ |BB_{1}| = \\frac{|BF|}{e} $. Since $ \\overrightarrow{AF} = 2\\overrightarrow{FB} $, it follows that $ \\cos\\angle BAE = \\frac{|AE|}{|AB|} = \\frac{|BF|}{3|BF|} = \\frac{1}{3e} = \\frac{1}{2} $, therefore $ \\tan\\angle BAE = \\sqrt{3} $."
  },
  {
    "text": "Given circle $M$: $(x-3)^{2}+(y-4)^{2}=4$, $O$ is the origin. A tangent line is drawn from point $P$ to circle $M$, with tangent point $T$. If $PO = PT$, then the equation of the locus of point $P$ is?",
    "fact_expressions": "M: Circle;P: Point;O: Origin;T: Point;L:Line;Expression(M)=((x - 3)^2 + (y - 4)^2 = 4);PointOnCurve(P, L);TangentOfPoint(P, M)=L;TangentPoint(L,M)=T;LineSegmentOf(P, O) = LineSegmentOf(P, T)",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "6*x-8*y-21=0",
    "fact_spans": "[[[2, 31], [48, 52]], [[43, 47], [75, 79]], [[32, 36]], [[59, 62]], [], [[2, 31]], [[42, 55]], [[42, 55]], [[42, 62]], [[64, 73]]]",
    "query_spans": "[[[75, 86]]]",
    "process": "From $(x-3)^{2}+(y-4)^{2}=4$ we know: $M(3,4)$, radius is 2. Since $TP$ is tangent to circle $M$, $TP\\bot MT \\Rightarrow TP^{2}=PM^{2}-TM^{2}$. Let $P(x,y)$. Since $PO=PT$, $PO^{2}=PT^{2}$, i.e., $OP^{2}=PM^{2}-TM^{2}$. Therefore, $x^{2}+y^{2}=(x-3)^{2}+(y-4)^{2}-4 \\Rightarrow 6x+8y-21=0$,"
  },
  {
    "text": "Given that both asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) are tangent to the circle $C$: $x^{2}+y^{2}-6x+5=0$, then the eccentricity of this hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;C: Circle;a>0;b>0;l1: Line;l2: Line;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(C) = (-6*x + x^2 + y^2 + 5 = 0);Asymptote(G) = {l1, l2};IsTangent(l1, C);IsTangent(l2, C)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3*sqrt(5)/5",
    "fact_spans": "[[[2, 58], [98, 101]], [[5, 58]], [[5, 58]], [[66, 93]], [[5, 58]], [[5, 58]], [], [], [[2, 58]], [[66, 93]], [[2, 64]], [[2, 95]], [[2, 95]]]",
    "query_spans": "[[[98, 108]]]",
    "process": "First, convert the equation of the circle into standard form. Then, since both asymptotes of the hyperbola are tangent to the circle $ C: x^{2}+y^{2}-6x+5=0 $, use the fact that the distance from the center of the circle to a line equals the radius to establish a relationship between the geometric quantities, thereby determining the eccentricity of the hyperbola. [Detailed Solution] The two asymptotes of the hyperbola $ \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 $ ($ a>0,b>0 $) are given by $ y=\\pm\\frac{b}{a}x $, or equivalently $ bx\\pm ay=0 $. The circle $ C: x^{2}+y^{2}-6x+5=0 $ is converted into standard form as $ (x-3)^{2}+y^{2}=4 $. Therefore, $ C(3,0) $, with radius 2. Since both asymptotes of the hyperbola $ \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 $ ($ a>0,b>0 $) are tangent to the circle $ C: x^{2}+y^{2}-6x+5=0 $, we have $ \\frac{3b}{\\sqrt{a^{2}+b^{2}}}=2 $. Thus, $ 9b^{2}=4b^{2}+4a^{2} $, so $ 5b^{2}=4a^{2} $. Since $ b^{2}=c^{2}-a^{2} $, it follows that $ 5(c^{2}-a^{2})=4a^{2} $, hence $ 9a^{2}=5c^{2} $. Therefore, $ e=\\frac{c}{a}=\\frac{3\\sqrt{5}}{5} $. Thus, the eccentricity of the hyperbola equals $ \\frac{3\\sqrt{5}}{5} $."
  },
  {
    "text": "Let the hyperbola share the same foci as the ellipse $\\frac{x^{2}}{27}+\\frac{y^{2}}{36}=1$, and intersect the ellipse, with one intersection point at $(\\sqrt{15}, 4)$. Then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/27 + y^2/36 = 1);Focus(G) = Focus(H);IsIntersect(G,H) = True;Coordinate(OneOf(Intersection(G,H))) = (sqrt(15),4)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/4 - x^2/5 = 1",
    "fact_spans": "[[[1, 4], [84, 87]], [[5, 44], [53, 55]], [[5, 44]], [[1, 50]], [[1, 57]], [[1, 81]]]",
    "query_spans": "[[[84, 94]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of a hyperbola are $y=\\pm \\frac{4}{3} x$, and the focal distance is $20$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*(4/3)*x);FocalLength(G) = 20",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/36 - y^2/64 = 1, y^2/64 - x^2/36 = 1}",
    "fact_spans": "[[[2, 5], [47, 50]], [[2, 33]], [[2, 44]]]",
    "query_spans": "[[[47, 57]]]",
    "process": "Since the asymptotes of the hyperbola are given by $ y = \\pm\\frac{4}{3}x $, we assume the standard equation of the hyperbola is $ \\frac{x^{2}}{9} - \\frac{y^{2}}{16} = k $. When $ k > 0 $, we have $ \\frac{x^{2}}{9k} - \\frac{y^{2}}{16k} = 1 $. Given the focal distance is 20, it follows that $ 9k + 16k = \\left(\\frac{20}{2}\\right)^{2} = 100 \\Rightarrow k = 4 $. Thus, the standard equation of the hyperbola is $ \\frac{x^{2}}{36} - \\frac{y^{2}}{64} = 1 $. When $ k < 0 $, we have $ \\frac{y^{2}}{16k} - \\frac{x^{2}}{9k} = 1 $. Given the focal distance is 20, it follows that $ 9k + 16k = \\left(\\frac{20}{2}\\right)^{2} = 100 \\Rightarrow k = 4 $. Thus, the standard equation of the hyperbola is $ \\frac{y^{2}}{64} - \\frac{x^{2}}{36} = 1 $."
  },
  {
    "text": "Let $O$ be the origin, and the line $x=a$ intersects the two asymptotes of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) at points $D$ and $E$, respectively. If the focal length of $C$ is $4$, then the maximum area of $\\Delta O D E$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G:Line;O: Origin;D: Point;E: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x = a);l1:Line;l2:Line;Asymptote(C)={l1,l2};Intersection(l1,G)=D;Intersection(l2,G)=E;FocalLength(C) = 4",
    "query_expressions": "Max(Area(TriangleOf(O, D, E)))",
    "answer_expressions": "2",
    "fact_spans": "[[[18, 79], [100, 103]], [[26, 79]], [[26, 79]], [[10, 17]], [[1, 4]], [[89, 92]], [[93, 96]], [[26, 79]], [[26, 79]], [[18, 79]], [[10, 17]], [], [], [[18, 85]], [[10, 98]], [[10, 98]], [[100, 110]]]",
    "query_spans": "[[[112, 134]]]",
    "process": "Assume point D is in the first quadrant and point E is in the fourth quadrant. Solving the system of equations  \n\\begin{cases}x=a,\\\\y=\\frac{b}{a}x,\\end{cases}  \nwe obtain  \n\\begin{cases}x=a,\\\\y=b,\\end{cases}  \nthus D(a,b). Similarly, we get E(a,-b), so |ED|=2b. S_{\\triangle ODE}=\\frac{1}{2}a\\times2b=ab. Since the focal length of C is 4, we have c=2, and c^{2}=a^{2}+b^{2}\\geqslant2ab, solving gives ab\\leqslant2, with equality if and only if a=b=\\sqrt{2}. Therefore, the maximum value of S_{\\triangle ODE} is 2."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through point $F_{2}$ and perpendicular to the $x$-axis intersects the hyperbola at a point $P$, and $\\angle P F_{1} F_{2}=\\frac{\\pi}{6}$. Then the asymptotes of the hyperbola have equations?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;Z: Line;PointOnCurve(F2, Z);IsPerpendicular(Z, xAxis);P: Point;OneOf(Intersection(Z, G)) = P;AngleOf(P, F1, F2) = pi/6",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(2)*x",
    "fact_spans": "[[[2, 48], [93, 96], [146, 149]], [[2, 48]], [[5, 48]], [[5, 48]], [[56, 63]], [[64, 71], [73, 81]], [[2, 71]], [[2, 71]], [[90, 92]], [[72, 92]], [[82, 92]], [[101, 104]], [[90, 104]], [[107, 143]]]",
    "query_spans": "[[[146, 157]]]",
    "process": "Since $ PF_{2} = \\frac{b^{2}}{a} $, $ \\tan30^{\\circ} = \\frac{b^{2}}{2c} = \\frac{b^{2}}{2ac} = \\frac{\\sqrt{3}}{3} \\Rightarrow \\sqrt{3}b^{2} = 2ac \\Rightarrow \\sqrt{3}(c^{2}-a^{2}) = 2ac $, then $ e = \\sqrt{3} $, and $ \\frac{b^{2}}{a^{2}} = \\frac{c^{2}-a^{2}}{a^{2}} = e^{2}-1 = 2 \\Rightarrow \\frac{b}{a} = \\sqrt{2} $, then the asymptotes of the hyperbola are $ y = \\pm\\sqrt{2}x $."
  },
  {
    "text": "A line with an inclination angle of $60^{\\circ}$ intersects the parabola $x^{2}=2 py(p>0)$ at points $A$ and $B$, and the sum of the horizontal coordinates of points $A$ and $B$ is $3$. Then, the equation of the parabola is?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;A:Point;B:Point;p>0;Expression(G) = (x^2 = 2*p*y);Inclination(H)=ApplyUnit(60,degree);Intersection(H,G)={A,B};XCoordinate(A)+XCoordinate(B)=3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 = sqrt(3)*y",
    "fact_spans": "[[[20, 40], [76, 79]], [[23, 40]], [[17, 19]], [[42, 45], [53, 56]], [[48, 51], [59, 62]], [[23, 40]], [[20, 40]], [[0, 19]], [[17, 51]], [[53, 74]]]",
    "query_spans": "[[[76, 84]]]",
    "process": ""
  },
  {
    "text": "The equation of the hyperbola with the foci of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$ as its vertices and the vertices of the ellipse as its foci is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/4 + y^2/2 = 1);Focus(H) = Vertex(G);Vertex(H)=Focus(G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/2 - y^2/2 = 1",
    "fact_spans": "[[[41, 44], [55, 58], [64, 67]], [[1, 38], [49, 51]], [[1, 38]], [[0, 47]], [[48, 61]]]",
    "query_spans": "[[[64, 72]]]",
    "process": "From the standard equation of the ellipse given in the problem, we obtain the vertex coordinates of the hyperbola as (\\pm) on the x-axis. Then its focus coordinates are (\\pm2,0), meaning the foci of the hyperbola lie on the x-axis, and: a^{2}=2, c^{2}=4, b^{2}=c^{2}-a^{2}=2. Thus, the equation of this hyperbola is \\frac{x^{2}}{2} - \\frac{y^{2}}{2} = 1."
  },
  {
    "text": "It is known that one focus of the ellipse $x^{2}+k y^{2}=3 k(k>0)$ coincides with the focus of the parabola $y^{2}=12 x$. Then, the eccentricity of the ellipse is ?",
    "fact_expressions": "G: Parabola;H: Ellipse;k: Number;Expression(G) = (y^2 = 12*x);k>0;Expression(H) = (k*y^2 + x^2 = 3*k);OneOf(Focus(H)) = Focus(G)",
    "query_expressions": "Eccentricity(H)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[34, 49]], [[2, 28], [57, 59]], [[4, 28]], [[34, 49]], [[4, 28]], [[2, 28]], [[2, 54]]]",
    "query_spans": "[[[57, 66]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has a semi-focal length $c$, and satisfies $c^{2}-b^{2}+a c<0$, then the range of the eccentricity $e$ of this ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;c: Number;e: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);HalfFocalLength(G) = c;a*c - b^2 + c^2 < 0;Eccentricity(G) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "(0,1/2)",
    "fact_spans": "[[[2, 54], [88, 90]], [[4, 54]], [[4, 54]], [[59, 62]], [[94, 97]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 62]], [[66, 85]], [[88, 97]]]",
    "query_spans": "[[[94, 104]]]",
    "process": "From $ c^{2} - b^{2} + ac < 0 $, we get $ c^{2} - (a^{2} - c^{2}) + ac < 0 $, $ 2c^{2} + ac - a^{2} < 0 $. Dividing both sides by $ a^{2} $ gives $ 2e^{2} + e - 1 < 0 $, solving yields $ 0 < e < \\frac{1}{2} $."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, respectively, and let $P$ be a point on this ellipse that does not coincide with the left and right vertices. Then the perimeter of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/5 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);Negation(P=LeftVertex(G));Negation(P=RightVertex(G))",
    "query_expressions": "Perimeter(TriangleOf(F1, P, F2))",
    "answer_expressions": "10",
    "fact_spans": "[[[19, 56], [68, 70]], [[19, 56]], [[1, 8]], [[9, 16]], [[1, 62]], [[1, 62]], [[63, 66]], [[63, 73]], [[63, 84]], [[63, 84]]]",
    "query_spans": "[[[86, 113]]]",
    "process": "From the given ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, we know: $a=3$, $b=\\sqrt{5}$, $c=2$, the perimeter of $\\Delta PF_{1}F_{2}$ is $2a+2c=6+4=10$"
  },
  {
    "text": "The line passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) intersects the parabola at points $A$ and $B$. Perpendiculars are drawn from $A$ and $B$ to the directrix, with feet of perpendiculars denoted as $A^{\\prime}$ and $B^{\\prime}$, respectively. Then $\\angle A^{\\prime} F B^{\\prime}$=?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;F: Point;B: Point;A: Point;A1: Point;B1: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};L1: Line;L2: Line;PointOnCurve(A, L1);PointOnCurve(B, L2);IsPerpendicular(L1, Directrix(G));IsPerpendicular(L2, Directrix(G));FootPoint(L1, Directrix(G)) = A1;FootPoint(L2, Directrix(G)) = B1",
    "query_expressions": "AngleOf(A1, F, B1)",
    "answer_expressions": "ApplyUnit(90, degree)",
    "fact_spans": "[[[1, 22], [32, 35]], [[4, 22]], [[29, 31]], [[25, 28]], [[42, 45], [53, 56]], [[38, 41], [49, 52]], [[68, 80]], [[81, 93]], [[4, 22]], [[1, 22]], [[1, 28]], [[0, 31]], [[29, 47]], [], [], [[32, 62]], [[32, 62]], [[32, 62]], [[32, 62]], [[32, 93]], [[32, 93]]]",
    "query_spans": "[[[95, 129]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A perpendicular is drawn from $F_{1}$ to an asymptote, with foot of the perpendicular at $M$, and $O$ is the origin. If $\\tan \\angle M F_{1} O=\\frac{1}{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;L: Line;PointOnCurve(F1, L);IsPerpendicular(L, Asymptote(G));M: Point;FootPoint(L, Asymptote(G)) = M;O: Origin;Tan(AngleOf(M, F1, O)) = 1/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 58], [152, 155]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[66, 73], [83, 90]], [[74, 81]], [[2, 81]], [[2, 81]], [], [[82, 97]], [[82, 97]], [[101, 104]], [[82, 104]], [[105, 108]], [[115, 150]]]",
    "query_spans": "[[[152, 161]]]",
    "process": "From the given information, $ F_{2}(c,0) $, the asymptotes of the hyperbola are $ y = \\pm\\frac{b}{a}x $. As shown in the figure, the distance from the left focus to the asymptote is: $ MF_{1} = \\frac{|bc|}{\\sqrt{a^{2}+b^{2}}} = b $. Since $ OF_{1} = c $, in right triangle $ \\triangle PF_{2}O $, $ OP = a $, so $ \\tan\\angle MF_{1}O = \\frac{OM}{MF_{1}} = \\frac{a}{b} = \\frac{1}{2} $, thus $ b = 2a $. Therefore, the eccentricity is: $ e = \\sqrt{1+(\\frac{b}{a})^{2}} = \\sqrt{5} $."
  },
  {
    "text": "The focus of the parabola $y^{2}=2 p x(p>0)$ is $F$, and $P$, $Q$ are two moving points on the parabola. The midpoint of segment $P Q$ is $M$. A perpendicular is drawn from $M$ to the directrix of the parabola, with foot $N$. If $|M N|=|P Q|$, then the maximum value of $\\angle P F Q$ is?",
    "fact_expressions": "G: Parabola;p: Number;P: Point;Q: Point;M: Point;N: Point;F: Point;L: Line;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G)=F;PointOnCurve(P,G);PointOnCurve(Q,G);MidPoint(LineSegmentOf(P,Q)) = M;PointOnCurve(M, L);IsPerpendicular(L, Directrix(G));FootPoint(L, Directrix(G))=N;Abs(LineSegmentOf(M, N)) = Abs(LineSegmentOf(P, Q))",
    "query_expressions": "Max(AngleOf(P, F, Q))",
    "answer_expressions": "pi/3",
    "fact_spans": "[[[0, 21], [37, 40], [67, 70]], [[3, 21]], [[29, 32]], [[33, 36]], [[58, 61], [63, 66]], [[79, 82]], [[25, 28]], [], [[3, 21]], [[0, 21]], [[0, 28]], [[29, 46]], [[29, 46]], [[47, 61]], [[62, 75]], [[62, 75]], [[62, 82]], [[84, 97]]]",
    "query_spans": "[[[99, 119]]]",
    "process": "Analysis: Let |PF| = 2a, |QF| = 2b. By the definition of a parabola, we get |PQ| = a + b. Applying the cosine law gives (a + b)^{2} = 4a^{2} + 4b^{2} - 8ab\\cos\\theta. Then, using the basic inequality, we find the range of \\theta, thus obtaining the solution to this problem. Detailed solution: Let |PF| = 2a, |QF| = 2b. By the definition of the parabola, we have |PF| = |PA|, |QF| = |QB|. In trapezoid ABPQ, 2|MN| = |PA| + |QB| = 2a + 2b. Since |MN| = |PQ|, it follows that |PQ| = a + b. By the cosine law, let \\angle PFQ = \\theta, then (a + b)^{2} = 4a^{2} + 4b^{2} - 8ab\\cos\\theta, so a^{2} + b^{2} + 2ab = 4a^{2} + 4b^{2} - 8ab\\cos\\theta, thus \\cos\\theta = \\frac{3a^{2} + 3b^{2} - 2ab}{8ab} \\geqslant \\frac{6ab - 2ab}{8ab} = \\frac{1}{2}, with equality if and only if a = b. Therefore, \\theta \\leqslant \\frac{\\pi}{3}. (1) This problem mainly examines simple geometric properties of a parabola, the relationship between lines and parabolas, and the basic inequality, aiming to test students' mastery of these fundamental concepts and their analytical reasoning ability. (2) The key points in solving this problem are two: first, recalling the definition of a parabola to solve the problem, thereby elegantly finding MN and PQ; second, after obtaining \\cos\\theta = \\frac{3a^{2} + 3b^{2} - 2ab}{8ab}, being able to use the basic inequality to find the extremum."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the right vertex of the ellipse is $A$, and the line $x=\\frac{a}{2}$ intersects the ellipse $C$ at points $M$ and $N$. If $\\overrightarrow{A M} \\cdot \\overrightarrow{A N}=0$, then the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;A: Point;RightVertex(C) = A;G: Line;Expression(G) = (x = a/2);M: Point;N: Point;Intersection(G, C) = {M, N};DotProduct(VectorOf(A, M), VectorOf(A, N)) = 0",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[2, 59], [86, 91], [157, 162]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[64, 67]], [[2, 67]], [[68, 85]], [[68, 85]], [[93, 96]], [[97, 100]], [[68, 102]], [[104, 155]]]",
    "query_spans": "[[[157, 168]]]",
    "process": "Ellipse $ C: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $) has its right vertex at $ A $. The line $ x = \\frac{a}{2} $ intersects ellipse $ C $ at points $ M $ and $ N $. If $ \\overrightarrow{AM} \\cdot \\overrightarrow{AN} = 0 $, then $ A(a, 0) $. Without loss of generality, assume $ M $ lies in the first quadrant, so the y-coordinate of $ M $ is: we obtain: $ \\frac{\\sqrt{3}}{2}b = \\frac{1}{2}a $, that is, $ 3b^{2} = a^{2} $, we get $ 3a^{2} - 3c^{2} = a^{2} $, $ 2a^{2} = 3c^{2} $, so $ e = \\frac{c}{a} = \\sqrt{\\frac{c^{2}}{a^{2}}} = \\frac{\\sqrt{6}}{3} $"
  },
  {
    "text": "Given that point $M$ moves on the parabola $C$: $y^{2}=4x$, and circle $C$ passes through the points $(5,0)$, $(2, \\sqrt{3})$, $(3,-2)$. From point $M$, lines $l_{1}$ and $l_{2}$ are drawn tangent to circle $C$, with points of tangency $P$ and $Q$, respectively. Then the range of $|PQ|$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);M: Point;PointOnCurve(M, C);C1: Circle;A: Point;B: Point;D: Point;Coordinate(A) = (5,0);Coordinate(B) = (2,sqrt(3));Coordinate(D) = (3,-2);PointOnCurve(A,C1);PointOnCurve(B,C1);PointOnCurve(D,C1);l1: Line;l2: Line;PointOnCurve(M,l1);PointOnCurve(M,l2);IsTangent(l1,C1);IsTangent(l2,C1);TangentPoint(l1,C1) = P;TangentPoint(l2,C1) = Q;P: Point;Q: Point",
    "query_expressions": "Range(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "[2*sqrt(2), 4)",
    "fact_spans": "[[[7, 26]], [[7, 26]], [[2, 6], [72, 76]], [[2, 27]], [[30, 34], [97, 101]], [[35, 43]], [[45, 60]], [[62, 70]], [[35, 43]], [[45, 60]], [[62, 70]], [[30, 43]], [[30, 60]], [[30, 70]], [[77, 86]], [[89, 96]], [[71, 87]], [[71, 96]], [[77, 103]], [[88, 103]], [[77, 116]], [[77, 116]], [[109, 112]], [[113, 116]]]",
    "query_spans": "[[[118, 132]]]",
    "process": "Let the equation of circle $ C' $ be $ x^{2} + y^{2} + Dx + Ey + F = 0 $. Substituting $ (5,0) $, $ (2,\\sqrt{3}) $, $ (3,-2) $ respectively, we obtain  \n$ 13 + 3D - 2E + F = 0 $  \n$$\n\\begin{cases}\n25 + 5D + F = 0 \\\\\n7 + 2D + \\sqrt{3}E + F = 0, \\\\\n13 + 3D - 2F + F = 0\n\\end{cases}\n$$\nSolving gives  \n$$\n\\begin{cases}\nD = -6 \\\\\nE = 0 \\\\\nF = 5\n\\end{cases}\n$$\nThus, circle $ C' $: $ (x-3)^{2} + y^{2} = 4 $. As shown in the figure, connect $ MC' $, $ C'P $, $ C'Q $, $ PQ $. It is easy to see that $ C'P \\perp MP $, $ C'Q \\perp MQ $, $ MC' \\perp PQ $. Therefore, the area of quadrilateral $ MPC'Q $ is $ \\frac{1}{2}|MC'| \\cdot |PQ| $. Additionally, the area of quadrilateral $ MPC'Q $ is twice the area of $ \\triangle MPC' $, so  \n$$\n\\frac{1}{2}|MC'| \\cdot |PQ| = |MP| \\cdot |C'P|\n$$  \nHence,  \n$$\n|PQ| = \\frac{2|MP| \\cdot |C'P|}{|MC'|} = \\frac{4\\sqrt{|C'M|^{2} - 4}}{|C'M|} = 4\\sqrt{1 - \\frac{4}{|C'M|^{2}}}\n$$  \nTherefore, when $ |C'M| $ is minimized, $ |PQ| $ is minimized. Let $ M(x,y) $, then  \n$$\n|MC'| = \\sqrt{(x-3)^{2} + y^{2}} = \\sqrt{x^{2} - 2x + 9}\n$$  \nSo when $ x = 1 $, $ |MC'|_{\\min} = 2\\sqrt{2} $. As $ x \\to \\infty $, $ |PQ| $ approaches the diameter of the circle, which is 4. Hence, the range of $ |PQ| $ is $ [2\\sqrt{2}, 4) $."
  },
  {
    "text": "Let the equation of the hyperbola $C$ be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, and let the line passing through the focus of the parabola $y^{2}=4x$ and the point $(0, b)$ be $l$. If one asymptote of $C$ is parallel to $l$, and the other asymptote is perpendicular to $l$, then the equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Parabola;a:Number;b:Number;a>0;b>0;P: Point;l: Line;l1: Line;l2: Line;Expression(G) = (y^2 = 4*x);Coordinate(P) = (0, b);Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(Focus(G),l);PointOnCurve(P,l);IsParallel(l,l1);IsPerpendicular(l,l2);OneOf(Asymptote(C)) = l1;OneOf(Asymptote(C)) = l2;Negation(l1=l2)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2 = 1",
    "fact_spans": "[[[1, 7], [103, 106], [133, 139]], [[66, 80]], [[11, 64]], [[11, 64]], [[11, 64]], [[11, 64]], [[84, 93]], [[97, 100], [113, 116], [126, 129]], [], [], [[66, 80]], [[84, 93]], [[1, 64]], [[65, 100]], [[65, 100]], [[103, 118]], [[103, 131]], [101, 109], [101, 122], [101, 122]]",
    "query_spans": "[[[133, 144]]]",
    "process": "Since line $ l $ passes through the focus of the parabola and the point $ (0,b) $, the slope of the line is $ -b $. Given that one asymptote of the hyperbola is parallel to $ l $ and the other asymptote is perpendicular to $ l $, we can find $ a = b = 1 $, thus obtaining the hyperbola equation. The asymptotes of the hyperbola are $ y = \\pm\\frac{b}{a}x $. The focus of the parabola $ y^{2} = 4x $ is $ (1,0) $, so line $ l $ passes through points $ (1,0) $ and $ (0,b) $, with slope $ \\frac{b-0}{0-1} = -b $. Therefore, the equation of line $ l $ is $ y - 0 = -b(x - 1) $, or $ y = -bx + b $. Since line $ l $ is parallel to one asymptote and perpendicular to the other asymptote of the hyperbola, we have\n$$\n\\begin{cases}\n-b = -\\frac{b}{a} \\\\\n-b \\cdot \\frac{b}{a} = -1\n\\end{cases}\n$$\nSolving gives $ a = b = 1 $. Thus, the equation of the hyperbola is $ x^{2} - y^{2} = 1 $."
  },
  {
    "text": "If a focus of the hyperbola $\\frac{y^{2}}{m}-x^{2}=1$ is $(0,2)$, then what is $m$? What is the equation of the asymptotes of this hyperbola?",
    "fact_expressions": "G: Hyperbola;m: Number;H: Point;Expression(G) = (-x^2 + y^2/m = 1);Coordinate(H) = (0, 2);OneOf(Focus(G)) = H",
    "query_expressions": "m;Expression(Asymptote(G))",
    "answer_expressions": "3  \ny=pm*sqrt(3)*x",
    "fact_spans": "[[[1, 29], [50, 53]], [[44, 47]], [[35, 42]], [[1, 29]], [[35, 42]], [[1, 42]]]",
    "query_spans": "[[[44, 49]], [[50, 61]]]",
    "process": "From the given condition, $ m+1=2^{2} \\Rightarrow m=3 $, so the hyperbola equation is $ \\frac{y^{2}}{3}-x^{2}=1 $, and the asymptote equations are $ y=\\pm\\sqrt{3}x $."
  },
  {
    "text": "The distance from the vertex of the hyperbola $C$: $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ to its asymptote is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/16 - y^2/9 = 1)",
    "query_expressions": "Distance(Vertex(C), Asymptote(C))",
    "answer_expressions": "12/5",
    "fact_spans": "[[[0, 44], [48, 49]], [[0, 44]]]",
    "query_spans": "[[[0, 57]]]",
    "process": "Since the hyperbola $ C: \\frac{x^2}{16} - \\frac{y^{2}}{9} = 1 $ has vertices at $ (\\pm4,0) $, and asymptotes given by $ y = \\pm\\frac{b}{a}x = \\pm\\frac{3}{4}x $, i.e., $ 3x \\pm 4y = 0 $, the distance from a vertex to an asymptote is $ \\frac{|3\\times4|}{\\sqrt{3^{2}+4^{2}}} = \\frac{12}{5} $."
  },
  {
    "text": "Given that the directrix of the parabola $y^{2}=4x$ intersects an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ at point $P(x_{0}, -2)$, what is the eccentricity of the hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;P: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 4*x);Coordinate(P) = (x0, -2);Intersection(Directrix(H),OneOf(Asymptote(G)))=P;x0:Number",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[20, 76], [101, 104]], [[23, 76]], [[23, 76]], [[2, 16]], [[84, 99]], [[23, 76]], [[23, 76]], [[20, 76]], [[2, 16]], [[84, 99]], [[2, 99]], [[85, 99]]]",
    "query_spans": "[[[101, 110]]]",
    "process": "The directrix of the parabola $ y^{2}=4x $ is $ x=-1 $. The asymptotes of the hyperbola are $ y=\\pm\\frac{b}{a}x $. From the given condition, $ \\frac{b}{a}=2 $, so $ \\frac{b^{2}}{a^{2}}=\\frac{c^{2}-a^{2}}{a^{2}}=4 $, that is, $ \\frac{c^{2}}{a^{2}}=5 $, $ e=\\sqrt{5} $."
  },
  {
    "text": "It is known that the asymptotes of the hyperbola $C$ are given by $y = \\pm \\sqrt{3} x$, and the point $(2, 2\\sqrt{2})$ lies on the hyperbola $C$. Then, what is the standard equation of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;Expression(Asymptote(C)) = (y = pm*(sqrt(3)*x));G: Point;Coordinate(G) = (2, 2*sqrt(2));PointOnCurve(G, C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "3*x^2/4-y^2/4=1",
    "fact_spans": "[[[2, 8], [52, 58], [61, 67]], [[2, 33]], [[34, 51]], [[34, 51]], [[34, 59]]]",
    "query_spans": "[[[61, 74]]]",
    "process": "\\because the asymptotes of hyperbola C are given by y=\\pm\\sqrt{3}x, \\therefore we can assume the equation of hyperbola C as y^{2}-3x^{2}=\\lambda. \\because hyperbola C passes through the point (2,2\\sqrt{2}), \\therefore 8-12=\\lambda, \\therefore \\lambda=-4, \\therefore the equation of hyperbola C is y^{2}-3x^{2}=-4, which can be rewritten as \\frac{3x^{2}}{4}-\\frac{y^{2}}{4}=1."
  },
  {
    "text": "Given the parabola $C$: $x^{2}=2 p y$ with focus $F$, and a fixed point $M(2 \\sqrt{3}, 0)$. If the line $F M$ intersects the parabola $C$ at points $A$ and $B$ (with point $B$ between $F$ and $M$), and intersects the directrix of the parabola $C$ at point $N$, and if $|B N|=7|B F|$, then the length of $A F$ is?",
    "fact_expressions": "C: Parabola;p: Number;M: Point;F: Point;A: Point;B: Point;N: Point;Expression(C) = (x^2 = 2*p*y);Coordinate(M) = (2*sqrt(3), 0);Focus(C) = F;Intersection(LineOf(F,M),C)={A,B};Between(B,F,M);Intersection(LineOf(F,M),Directrix(C))=N;Abs(LineSegmentOf(B, N)) = 7*Abs(LineSegmentOf(B, F))",
    "query_expressions": "Length(LineSegmentOf(A, F))",
    "answer_expressions": "7/6",
    "fact_spans": "[[[2, 23], [61, 67], [98, 104]], [[10, 23]], [[33, 51], [89, 92]], [[27, 30], [85, 88]], [[70, 73]], [[74, 77], [80, 84]], [[109, 113]], [[2, 23]], [[33, 51]], [[2, 30]], [[53, 79]], [[80, 94]], [[53, 113]], [[115, 129]]]",
    "query_spans": "[[[131, 140]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $x^{2}=4 y$ and the circle $C$: $(x-1)^{2}+(y-2)^{2}=r^{2}$ $(r>0)$ have a common point $P$, and if the tangent to the parabola at point $P$ is also tangent to the circle $C$, then $r=$?",
    "fact_expressions": "G: Parabola;C: Circle;r: Number;P: Point;Expression(G) = (x^2 = 4*y);r>0;Expression(C) = ((x - 1)^2 + (y - 2)^2 = r^2);Intersection(G,C)=P;IsTangent(TangentOnPoint(P,G),C)",
    "query_expressions": "r",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 16], [63, 66]], [[17, 54], [76, 80]], [[85, 88]], [[58, 61], [67, 71]], [[2, 16]], [[22, 54]], [[17, 54]], [[2, 61]], [[63, 83]]]",
    "query_spans": "[[[85, 90]]]",
    "process": "Let the point of tangency be $(a,\\frac{a2}{4})$, the derivative is $y=\\frac{x}{2}$, so the slope of the tangent line is $\\frac{a}{2}$. Connecting the center of the circle and the point of tangency, the two lines are perpendicular, so the product of their slopes equals $-1$, that is, $\\frac{2-\\frac{a^{2}}{4}}{1-a} \\cdot \\frac{a}{2} = -1$. Solving gives $a=2$, radius $r=\\sqrt{(1-a)^{2}+(2-\\frac{a^{2}}{4})^{2}}=\\sqrt{2}$"
  },
  {
    "text": "Given that point $F$ is the focus of the parabola $y^{2}=4x$, and point $A$ on this parabola in the first quadrant is at a distance of $5$ from its directrix, then the slope of line $AF$ is?",
    "fact_expressions": "G: Parabola;A: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(A, G);Quadrant(A)=1;Distance(A, Directrix(G)) = 5",
    "query_expressions": "Slope(LineOf(A, F))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[7, 21], [26, 29], [42, 43]], [[37, 41]], [[2, 6]], [[7, 21]], [[2, 24]], [[26, 41]], [[30, 41]], [[37, 52]]]",
    "query_spans": "[[[54, 66]]]",
    "process": "From the definition of the parabola, we obtain: $x_{A}+1=5$, $x_{A}=4$. Since point $A$ lies in the first quadrant, $y_{A}=4$, thus $k_{AF}=\\frac{4-0}{4-1}=\\frac{4}{3}$."
  },
  {
    "text": "The point $M(x, y)$ moves in such a way that it always satisfies the equation $\\sqrt{(x+1)^{2}+y^{2}}+\\sqrt{(x-1)^{2}+y^{2}}=6$. What is the trajectory equation of point $M$?",
    "fact_expressions": "M: Point;Coordinate(M) = (x, y);x: Number;y: Number;sqrt((x + 1)^2 + y^2) + sqrt((x - 1)^2 + y^2) = 6",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2/9 + y^2/8 = 1",
    "fact_spans": "[[[0, 10], [73, 77]], [[0, 10]], [[1, 10]], [[1, 10]], [[23, 72]]]",
    "query_spans": "[[[73, 84]]]",
    "process": ""
  },
  {
    "text": "Given that point $P(2, y)$ lies on the parabola $y^{2}=4 x$, then the distance from point $P$ to the focus $F$ of the parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;y1: Number;Coordinate(P) = (2, y1);PointOnCurve(P, G);F: Point;Focus(G) = F",
    "query_expressions": "Distance(P, F)",
    "answer_expressions": "3",
    "fact_spans": "[[[13, 27], [35, 38]], [[13, 27]], [[2, 12], [30, 34]], [[3, 12]], [[2, 12]], [[2, 28]], [[40, 43]], [[35, 43]]]",
    "query_spans": "[[[30, 48]]]",
    "process": "\\because point P(2,y) lies on the parabola y^{2}=4x, \\therefore the distance from point P to the focus F is 2+1=3."
  },
  {
    "text": "Given that $O$ is the coordinate origin, $F$ is the focus of the parabola $y^{2}=2 p x$ $(p>0)$, and the parabola intersects the line $l$: $x-\\sqrt{3} y-\\frac{p}{2}=0$ in the first and fourth quadrants at points $A$ and $B$ respectively, find the value of $\\frac{(\\overrightarrow{O F}-\\overrightarrow{O A})^{2}}{(\\overrightarrow{O F}-\\overrightarrow{O B})^{2}}$.",
    "fact_expressions": "G: Parabola;p: Number;l: Line;O: Origin;F: Point;A: Point;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(l) = (-p/2 + x - sqrt(3)*y = 0);Focus(G) = F;Intersection(G, l) = {A, B};Quadrant(A)=1;Quadrant(B)=4",
    "query_expressions": "(-VectorOf(O, A) + VectorOf(O, F))^2/(-VectorOf(O, B) + VectorOf(O, F))^2",
    "answer_expressions": "97+56*sqrt(3)",
    "fact_spans": "[[[15, 36], [41, 44]], [[18, 36]], [[45, 80]], [[2, 5]], [[11, 14]], [[91, 94]], [[95, 98]], [[18, 36]], [[15, 36]], [[45, 80]], [[11, 39]], [[41, 100]], [[81, 100]], [[81, 100]]]",
    "query_spans": "[[[102, 211]]]",
    "process": ""
  },
  {
    "text": "Given point $M(\\sqrt{3}, 0)$, the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ and the line $y=k(x+\\sqrt{3})$ intersect at points $A$ and $B$, then the perimeter of $\\triangle A B M$ is?",
    "fact_expressions": "G: Ellipse;H: Line;M: Point;A: Point;B: Point;k:Number;Expression(G) = (x^2/4 + y^2 = 1);Expression(H) = (y = k*(x + sqrt(3)));Coordinate(M) = (sqrt(3), 0);Intersection(G, H) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, B, M))",
    "answer_expressions": "8",
    "fact_spans": "[[[21, 48]], [[49, 68]], [[2, 20]], [[70, 74]], [[75, 78]], [[51, 68]], [[21, 48]], [[49, 68]], [[2, 20]], [[21, 78]]]",
    "query_spans": "[[[80, 102]]]",
    "process": "The line $ y = k(x + \\sqrt{3}) $ passes through the fixed point $ N(\\sqrt{3}, 0) $. From the problem, we know that $ M $ and $ N $ are the foci of the ellipse. By the definition of an ellipse, $ AN + AM = 2a = 4 $, $ BM + BN = 2a = 4 $. The perimeter of $ \\triangle ABM $ is $ AB + BM + AM = (AN + BN) + BM + AM = (AN + AM) + (BN + BM) = 8 $."
  },
  {
    "text": "The distance between the two foci of an ellipse is $16$, and the distances from a certain point on the ellipse to the two foci are $9$ and $15$, respectively. Then, what is the standard equation of the ellipse?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;P: Point;Focus(G)={F1,F2};Distance(F1, F2) = 16;PointOnCurve(P, G);Distance(P,F1)=9;Distance(P,F2)=15",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/144+y^2/80=1,y^2/144+x^2/80=1}",
    "fact_spans": "[[[0, 2], [0, 2], [0, 2]], [], [], [], [[0, 5]], [[0, 14]], [[16, 22]], [[16, 41]], [[16, 41]]]",
    "query_spans": "[[[43, 52]]]",
    "process": "According to the problem, the standard equation of the ellipse can be set as \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1,(a>b>0) or \\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1,(a>b>0). From the given conditions, we have 2c=16, so c=8, and 2a=9+15, so a=12. Hence, b^{2}=a^{2}-c^{2}=80. Therefore, the standard equation of the ellipse is: \\frac{x^{2}}{144}+\\frac{y^{2}}{80}=1 or \\frac{y^{2}}{144}+\\frac{x^{2}}{80}=1."
  },
  {
    "text": "A focus of the hyperbola $5 x^{2}+k y^{2}=5$ is $(\\sqrt{6}, 0)$, then what is the value of the real number $k$?",
    "fact_expressions": "G: Hyperbola;k: Real;H: Point;Expression(G) = (k*y^2 + 5*x^2 = 5);Coordinate(H) = (sqrt(6), 0);OneOf(Focus(G)) = H",
    "query_expressions": "k",
    "answer_expressions": "-1",
    "fact_spans": "[[[0, 22]], [[46, 51]], [[28, 43]], [[0, 22]], [[28, 43]], [[0, 43]]]",
    "query_spans": "[[[46, 55]]]",
    "process": ""
  },
  {
    "text": "Given that the line $l$ passes through the point $(0,2)$ and intersects the parabola $y^{2}=4x$ at two points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$, then $\\frac{1}{y_{1}}+\\frac{1}{y_{2}}=$?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;H: Point;x1: Number;y1: Number;x2: Number;y2: Number;Expression(G) = (y^2 = 4*x);Coordinate(H) = (0, 2);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);PointOnCurve(H, l);Intersection(l, G) = {A, B}",
    "query_expressions": "1/y2 + 1/y1",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 7]], [[19, 33]], [[35, 51]], [[54, 72]], [[8, 16]], [[35, 51]], [[35, 51]], [[54, 72]], [[54, 72]], [[19, 33]], [[8, 16]], [[35, 51]], [[54, 72]], [[2, 16]], [[2, 74]]]",
    "query_spans": "[[[76, 111]]]",
    "process": ""
  },
  {
    "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ are given by $y=\\pm \\frac{1}{2}x$, then what is the value of $b$?",
    "fact_expressions": "G: Hyperbola;b: Number;b>0;Expression(G) = (x^2/4 - y^2/b^2 = 1);Expression(Asymptote(G)) = (y = pm*(x/2))",
    "query_expressions": "b",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 50]], [[81, 84]], [[4, 50]], [[1, 50]], [[1, 79]]]",
    "query_spans": "[[[81, 87]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{3}=1$, the line $l$: $y=k x-2$ intersects the ellipse $C$ at points $A$ and $B$, point $P(0,1)$, and $|P A|=|P B|$, then the equation of line $l$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/9 + y^2/3 = 1);l: Line;k:Number;Expression(l) = (y = k*x - 2);A: Point;B: Point;Intersection(l, C) = {A, B};P: Point;Coordinate(P) = (0, 1);Abs(LineSegmentOf(P, A)) = Abs(LineSegmentOf(P, B))",
    "query_expressions": "Expression(l)",
    "answer_expressions": "{x-y-2=0, x+y+2=0}",
    "fact_spans": "[[[2, 44], [62, 67]], [[2, 44]], [[45, 61], [105, 110]], [[52, 61]], [[45, 61]], [[69, 72]], [[73, 76]], [[45, 78]], [[79, 88]], [[79, 88]], [[90, 103]]]",
    "query_spans": "[[[105, 115]]]",
    "process": "Solving the system \\begin{cases}\\frac{x^2}{9}+\\frac{y^{2}}{3}=1\\\\y=kx-2\\end{cases} yields $(1+3k^{2})x^{2}-12kx+3=0$. Since the line intersects the ellipse at two distinct points, $\\triangle=144k^{2}-12(1+3k^{2})>0$, solving gives $k^{2}>\\frac{1}{9}$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $x_{1}+x_{2}=\\frac{12k}{1+3k^{2}}$, $x_{1}x_{2}=$, $y_{1}+y_{2}=k(x_{1}+x_{2})-4=k$. Thus, the midpoint coordinates of segment $AB$ are $E\\left| -3k^{2}-4 = -\\frac{+}{1+3k^{2}} \\right|$. Since $|PA|=|PB|$, we have $PE\\perp AB$, $\\therefore k_{PE}\\cdot k_{AB}=-\\cdots=-1$, $\\therefore k=\\pm1$. Upon verification, it satisfies the conditions. Therefore, the equation of line $l$ is $x-y-2=0$ or $x+y+2=0$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$ and directrix $l$, a line passing through point $F$ with an inclination angle of $60^{\\circ}$ intersects the parabola at points $A$ and $B$ (point $A$ is in the first quadrant). A perpendicular is drawn from point $A$ to the directrix $l$, with foot of perpendicular $M$. Then, the area of $\\triangle AFM$ is?",
    "fact_expressions": "C: Parabola;G: Line;A: Point;B:Point;F: Point;M: Point;l: Line;Expression(C) = (y^2 = 4*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(F, G);Inclination(G)=ApplyUnit(60,degree);Intersection(G, C) = {A,B};Quadrant(A)=1;L:Line;PointOnCurve(A,L);IsPerpendicular(l,L);FootPoint(l,L)=M",
    "query_expressions": "Area(TriangleOf(A, F, M))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[2, 21], [62, 65]], [[59, 61]], [[66, 69], [76, 80], [88, 92]], [[70, 73]], [[25, 28], [37, 41]], [[105, 108]], [[32, 35], [95, 98]], [[2, 21]], [[2, 28]], [[2, 35]], [[36, 61]], [[42, 61]], [[59, 75]], [[76, 85]], [], [[87, 101]], [[87, 101]], [[87, 108]]]",
    "query_spans": "[[[110, 132]]]",
    "process": "Given the problem, first derive the focal radius formula. As shown in the figure, let \\( y^{2} = 2px \\), \\( (p > 0) \\), with \\( |PF| = t \\), \\( \\angle PFx = \\theta \\). Draw a tangent from \\( P \\) to the directrix. Then, according to the parabolic focal radius formula, \\( FM = FA = \\frac{}{1}\\frac{2}{\\cos60^{\\circ}} = 4 \\). Also, \\( \\angle FAM = 60^{\\circ} \\), hence \\( S_{\\triangle AFM} = \\frac{1}{2} \\times 4 \\times 4 \\times \\sin60^{\\circ} = 4\\sqrt{3} \\)."
  },
  {
    "text": "The standard equation of an ellipse centered at the origin, with foci $F_{1}(-\\sqrt{2}, 0)$, $F_{2}(\\sqrt{2}, 0)$, and eccentricity $\\frac{1}{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;O: Origin;Coordinate(F1) = (-sqrt(2), 0);Coordinate(F2) = (sqrt(2), 0);Center(G) = O;Focus(G)={F1,F2};Eccentricity(G) = 1/2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/8+y^2/6=1",
    "fact_spans": "[[[75, 77]], [[13, 34]], [[35, 55]], [[3, 7]], [[13, 34]], [[35, 55]], [[0, 77]], [[8, 77]], [[57, 77]]]",
    "query_spans": "[[[75, 84]]]",
    "process": "According to the problem, the ellipse intersects the x-axis. Let the standard equation of the ellipse be \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0). From the given conditions, c=\\sqrt{2}, and the eccentricity e=\\frac{c}{a}=\\frac{1}{2}. Solving gives a=2\\sqrt{2}, so b^{2}=a^{2}-c^{2}=8-2=6. Therefore, the standard equation of the ellipse is \\frac{x^{2}}{8}+\\frac{y^{2}}{6}=1."
  },
  {
    "text": "Given that $M$ is a moving point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, $F$ is the left focus, and $A(-2,1)$ is a fixed point, when $|\\overrightarrow{A M}|+\\frac{3}{2}|\\overrightarrow{M F}|$ takes the minimum value, $\\tan \\angle A F M$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/5 = 1);M: Point;PointOnCurve(M, G);F: Point;LeftFocus(G) = F;A: Point;Coordinate(A) = (-2, 1);WhenMin(Abs(VectorOf(A, M)) + (3/2)*Abs(VectorOf(M, F)))",
    "query_expressions": "Tan(AngleOf(A, F, M))",
    "answer_expressions": "(6*sqrt(5)-10)/5",
    "fact_spans": "[[[6, 43]], [[6, 43]], [[2, 5]], [[2, 49]], [[50, 53]], [[6, 57]], [[58, 67]], [[58, 67]], [[72, 136]]]",
    "query_spans": "[[[137, 158]]]",
    "process": "Let MM be perpendicular to the left directrix of the ellipse, intersecting the left directrix at point M'. By the second definition of the ellipse, e = \\frac{|MF|}{|MM|} = \\frac{2}{3}, \\therefore |MM| = \\frac{3}{2}|MF|, \\therefore |\\overrightarrow{AM}| + \\frac{3}{2}|\\overrightarrow{MF}| = |\\overrightarrow{AM}| + |MM|. Therefore, when A, M, M' are collinear, |\\overrightarrow{AM}| + \\frac{3}{2}|\\overrightarrow{MF}| attains its minimum value. At this time, let M(x_{0}, 1), x_{0} < 0, \\therefore \\frac{x_{0}^2}{9} + \\frac{1}{5} = 1, we get x_{0} = -\\frac{6}{\\sqrt{5}}, \\therefore \\tan\\angle AFM = \\frac{\\sqrt{5}}{1} = \\frac{6\\sqrt{5}-10}{5}"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$ and directrix $l$, point $P$ lies on the parabola $C$, $PQ$ is perpendicular to $l$ at point $Q$, $QF$ intersects the $y$-axis at point $T$, $O$ is the origin, and $|OT|=2$, then $|PF|$?",
    "fact_expressions": "C: Parabola;P: Point;Q: Point;F: Point;O: Origin;T: Point;l: Line;Expression(C) = (y^2 = 4*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P, Q), l) ;FootPoint(LineSegmentOf(P, Q), l)= Q;Intersection(LineSegmentOf(Q, F), yAxis) = T;Abs(LineSegmentOf(O, T)) = 2",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "5",
    "fact_spans": "[[[2, 21], [41, 47]], [[36, 40]], [[60, 64]], [[25, 28]], [[84, 87]], [[79, 83]], [[32, 35], [56, 59]], [[2, 21]], [[2, 28]], [[2, 35]], [[36, 48]], [[49, 59]], [[49, 64]], [[67, 83]], [[94, 103]]]",
    "query_spans": "[[[105, 113]]]",
    "process": "According to the problem, we have $ F(1,0) $, $ l: x = -1 $. By the definition of a parabola, $ |PQ| = |PF| $. Let $ PQ $ intersect the $ y $-axis at point $ M $. Since $ |OT| = 2 $, and $ |OF| = |QM| $, it follows that $ \\triangle TMQ \\cong \\triangle TOF $. Therefore, $ T $ is the midpoint of $ OM $, so $ |OM| = 4 $, meaning the $ y $-coordinate of $ P $ is 4. In $ y^2 = 4x $, letting $ y = 4 $, we get $ x = 4 $. Thus, $ |PQ| = x + \\frac{p}{2} = 4 + 1 = 5 $, so $ |PF| = 5 $."
  },
  {
    "text": "If the focal distance of the hyperbola $\\frac{y^{2}}{5}-x^{2}=m(m>0)$ is $12$, then $m$=?",
    "fact_expressions": "G: Hyperbola;m: Number;m>0;Expression(G) = (-x^2 + y^2/5 = m);FocalLength(G) = 12",
    "query_expressions": "m",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 34]], [[44, 47]], [[4, 34]], [[1, 34]], [[1, 42]]]",
    "query_spans": "[[[44, 49]]]",
    "process": "Analysis: Convert the hyperbola equation into standard form, then use the fact that the hyperbola's focal distance is 12 to find $ m $.  \nSince the hyperbola is $ \\frac{y^{2}}{5} - x^{2} = m $ ($ m > 0 $),  \nthe standard form of the hyperbola is $ \\frac{y^{2}}{5m} - \\frac{x^{2}}{m} = 1 $.  \nSince the hyperbola's focal distance is 12,  \n$ 5m + m = \\left( \\frac{12}{2} \\right)^{2} = 36 $,  \nso $ m = 6 $.  \nHence, the answer is 6."
  },
  {
    "text": "Let the ellipse $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{4}=1$ pass through the point $(-2, \\sqrt{3})$, then the focal length is equal to?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/4 + x^2/m^2 = 1);m: Number;H: Point;Coordinate(H) = (-2, sqrt(3));PointOnCurve(H, G)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[1, 42]], [[1, 42]], [[3, 42]], [[43, 60]], [[43, 60]], [[1, 60]]]",
    "query_spans": "[[[1, 67]]]",
    "process": "Since the ellipse $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{4}=1$ passes through the point $(-2,\\sqrt{3})$, substituting it in gives $m^{2}=16$, so $c^{2}=16-4=12$, $c=2\\sqrt{3}$, hence the focal length $2c=4\\sqrt{3}$."
  },
  {
    "text": "If one asymptote of a hyperbola with foci on the $x$-axis is $y=\\frac{1}{2} x$, then its eccentricity $e=$?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), xAxis);Expression(OneOf(Asymptote(G))) = (y = x/2);e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[9, 12], [38, 39]], [[1, 12]], [[9, 36]], [[43, 46]], [[38, 46]]]",
    "query_spans": "[[[38, 48]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$, $F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, and point $P$ lies on the hyperbola such that $\\angle F_{1} P F_{2}=90^{\\circ}$, then $|P F_{1}|-|P F_{2}|$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "2",
    "fact_spans": "[[[18, 46], [57, 60]], [[18, 46]], [[2, 9]], [[10, 17]], [[2, 51]], [[52, 56]], [[52, 61]], [[65, 98]]]",
    "query_spans": "[[[100, 123]]]",
    "process": ""
  },
  {
    "text": "The length of the line segment cut from the line $y=5 x-3$ by the curve $y=2 x^{2}$ is?",
    "fact_expressions": "G: Line;Expression(G) = (y = 5*x - 3);H: Curve;Expression(H) = (y = 2*x^2)",
    "query_expressions": "Length(InterceptChord(G, H))",
    "answer_expressions": "sqrt(26)/2",
    "fact_spans": "[[[0, 11]], [[0, 11]], [[12, 25]], [[12, 25]]]",
    "query_spans": "[[[0, 33]]]",
    "process": "Solving the system of equations of the line and the curve gives  \n\\begin{cases}y=5x-3\\\\y=2x^{2}\\end{cases}  \nSolutions are  \n\\begin{cases}x=1\\\\y=2\\end{cases}, \\quad \\text{or} \\quad \\begin{cases}x=\\frac{3}{2}\\\\y=\\frac{9}{2}\\end{cases}  \n\\therefore The length of the line segment intercepted by the curve is  \n\\sqrt{\\left(\\frac{3}{2}-1\\right)^{2}+\\left(\\frac{9}{2}-2\\right)^{2}} = \\frac{\\sqrt{26}}{2}."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ has eccentricity $\\sqrt{2}$, and $C$ intersects the directrix of the parabola $y^{2}=8 x$ at points $A$ and $B$, with $|A B|=2$. Then the focal distance of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Parabola;A: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y^2 = 8*x);Eccentricity(C) = sqrt(2);Intersection(C,Directrix(G)) = {A, B};Abs(LineSegmentOf(A, B)) = 2",
    "query_expressions": "FocalLength(C)",
    "answer_expressions": "2*sqrt(6)",
    "fact_spans": "[[[2, 64], [81, 85], [126, 132]], [[10, 64]], [[10, 64]], [[86, 100]], [[105, 108]], [[109, 112]], [[10, 64]], [[10, 64]], [[2, 64]], [[86, 100]], [[2, 79]], [[82, 114]], [[115, 124]]]",
    "query_spans": "[[[126, 137]]]",
    "process": "According to the problem, since the parabola is given by $ y^{2} = 8x $, we have $ 2p = 8 $, solving gives $ p = 4 $, so $ \\frac{p}{2} = 2 $. Thus, the directrix of the parabola is $ x = -2 $. Let the two intersection points of hyperbola $ C $ and the directrix $ x = -2 $ of the parabola $ y^{2} = 8x $ be $ A(-2, y) $, $ B(-2, -y) $ ($ y > 0 $). Then $ |AB| = |y - (-y)| = 2y = 2 $, solving gives $ y = 1 $. Substituting $ x = -2 $, $ y = 1 $ into the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $, we get $ \\frac{4}{a^{2}} - \\frac{1}{b^{2}} = 1, \\cdots\\textcircled{1} $. Also, since the eccentricity of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) is $ \\sqrt{2} $, we have $ \\frac{c}{a} = \\sqrt{2} $, that is, $ \\frac{a^{2} + b^{2}}{a^{2}} = 2 $, so $ b^{2} = a^{2} \\cdots\\cdots\\textcircled{2} $. Solving equations $ \\textcircled{1} $ and $ \\textcircled{2} $ simultaneously, we obtain $ a^{2} = 3 $, $ b^{2} = 3 $. Since $ c = \\sqrt{a^{2} + b^{2}} = \\sqrt{6} $, then $ 2c = 2\\sqrt{6} $. Therefore, the focal length of hyperbola $ C $ is $ 2\\sqrt{6} $, which involves the application of simple geometric properties. The key to solving is being familiar with the geometric properties of parabolas and hyperbolas and performing accurate calculations. This problem mainly examines reasoning and computational ability."
  },
  {
    "text": "Draw a line with slope $-\\frac{1}{2}$ passing through the point $M(1,1)$, intersecting the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ at points $A$ and $B$. If $M$ is the midpoint of segment $AB$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (1, 1);G: Line;PointOnCurve(M, G) = True;Slope(G) = -1/2;C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Intersection(G, C) = {A, B};A: Point;B: Point;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[1, 10], [102, 105]], [[1, 10]], [[29, 31]], [[0, 31]], [[11, 31]], [[32, 89], [119, 124]], [[32, 89]], [[39, 89]], [[39, 89]], [[39, 89]], [[39, 89]], [[29, 100]], [[91, 94]], [[95, 98]], [[102, 117]]]",
    "query_spans": "[[[119, 130]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ \\frac{x_{1}^{2}}{a^{2}} + \\frac{y_{1}^{2}}{b^{2}} = 1 $, $ \\frac{x_{2}^{2}}{a^{2}} + \\frac{y_{2}^{2}}{b^{2}} = 1 $. Subtracting these two equations yields $ \\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = -\\frac{b^{2}}{a^{2}} \\times \\frac{x_{1}+x_{2}}{y_{1}+y_{2}} $. Drawing a line through point $ M(1,1) $ with slope $ -\\frac{1}{2} $ that intersects the ellipse $ C: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $) at points $ A $ and $ B $, where $ M $ is the midpoint of segment $ AB $, then $ \\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = -\\frac{b^{2}}{a^{2}} \\times \\frac{x_{1}+x_{2}}{y_{1}+y_{2}} = -\\frac{b^{2}}{a^{2}} = -\\frac{1}{2} $, so $ a = \\sqrt{2}b $, hence $ c = \\sqrt{a^{2}-b^{2}} = b $, then $ e = \\frac{c}{a} = \\frac{\\sqrt{2}}{2} $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ with asymptotes given by $y=\\pm \\sqrt{3} x$. If a moving point $P$ lies on the right branch of $C$, and $F_{1}$, $F_{2}$ are the left and right foci of $C$ respectively, and the minimum value of $\\overrightarrow{O P} \\cdot \\overrightarrow{O F_{2}}$ is $2 a$ (where $O$ is the origin), then the minimum value of $\\frac{|P F_{1}|^{2}}{|P F_{2}|}$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;O: Origin;P: Point;F2: Point;F1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(Asymptote(C)) = (y = pm*(sqrt(3)*x));PointOnCurve(P, RightPart(C));LeftFocus(C) = F1;RightFocus(C) = F2;Min(DotProduct(VectorOf(O,P),VectorOf(O,F2)))=2*a",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F1))^2/Abs(LineSegmentOf(P, F2)))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 64], [97, 100], [123, 126]], [[10, 64]], [[10, 64]], [[199, 202]], [[93, 96]], [[113, 120]], [[105, 112]], [[10, 64]], [[10, 64]], [[2, 64]], [[2, 89]], [[91, 104]], [[105, 132]], [[105, 132]], [[133, 196]]]",
    "query_spans": "[[[210, 249]]]",
    "process": "Let $ P(x,y) $, and $ x \\geqslant a $, $ F_{2}(c,0) $, then $ \\overrightarrow{OP} = (x,y) $, $ \\overrightarrow{OF_{2}} = (c,0) $, therefore $ \\overrightarrow{OP} \\cdot \\overrightarrow{OF_{2}} = cx $. When $ x = c $, $ \\overrightarrow{OP} \\cdot \\overrightarrow{OF_{2}} $ attains its minimum value, and the minimum value is $ ac = 2a $, so $ c = 2 $. Thus, \n\\[\n\\begin{cases}\n\\frac{b}{a} = \\sqrt{3} \\\\\nc = 2 \\\\\nc = 2\n\\end{cases}\n\\]\nSolving gives $ a = 1 $, $ b = \\sqrt{3} $. Let $ |PF_{2}| = t $ ($ t \\geqslant 1 $), then $ |PF_{1}| = t + 2 $. So \n\\[\n\\frac{|PF_{1}|^2}{|PF_{2}|} = \\frac{(t+2)^2}{t} = t + \\frac{4}{t} + 4 \\geqslant 2\\sqrt{t \\times \\frac{4}{t}} + 4 = 8,\n\\]\nequality holds when $ t = \\frac{4}{t} $, i.e., $ t = 2 $. Therefore, the minimum value of $ \\frac{|PF_{1}|^2}{|PF_{2}|} $ is 8."
  },
  {
    "text": "Given that point $M$ moves on the parabola $y^{2}=4 x$, and point $N$ moves on the line $2 x-y+4=0$, then the minimum distance $|M N|$ between the two points is?",
    "fact_expressions": "G: Parabola;H: Line;M: Point;N: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (2*x - y + 4 = 0);PointOnCurve(M, G);PointOnCurve(N, H)",
    "query_expressions": "Min(Abs(LineSegmentOf(M, N)))",
    "answer_expressions": "7*sqrt(5)/10",
    "fact_spans": "[[[8, 22]], [[30, 43]], [[4, 7]], [[26, 29]], [[8, 22]], [[30, 43]], [[4, 23]], [[26, 44]]]",
    "query_spans": "[[[50, 63]]]",
    "process": "First, according to the problem, let $ M\\left(\\frac{y_{0}^{2}}{4}, y_{0}\\right) $. Using the point-to-line distance formula, find the minimum distance from point $ M $ to the line $ 2x - y + 4 = 0 $. Based on the given conditions, the result can be obtained. [Solution] Since the moving point $ M $ lies on the parabola $ y^{2} = 4x $, set $ M\\left(\\frac{y_{0}^{2}}{4}, y_{0}\\right) $. Thus, the distance from point $ M $ to the line $ 2x - y + 4 = 0 $ is $ \\frac{7}{2} \\geqslant \\frac{7\\sqrt{5}}{10} $, so the minimum distance from point $ M $ to the line $ 2x - y + 4 = 0 $ is $ \\frac{7\\sqrt{5}}{10} $. Since the moving point $ N $ lies on the line $ 2x - y + 4 = 0 $, the minimum value of $ |MN| $ is exactly $ d_{\\min} = \\frac{7\\sqrt{5}}{10} $."
  },
  {
    "text": "Through the right focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$), draw a perpendicular line to an asymptote, with foot of perpendicular at $P$, and let the intersection point of this line with the $y$-axis be $Q$. If $|F P|<|O Q|$ ($O$ is the origin), then the range of the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(G) = F;L: Line;PointOnCurve(F, L);IsPerpendicular(L, Asymptote(G));P: Point;FootPoint(L, Asymptote(G)) = P;Q: Point;Intersection(L, yAxis) = Q;O: Origin;Abs(LineSegmentOf(F, P)) < Abs(LineSegmentOf(O, Q))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1, (1+sqrt(5))/2)",
    "fact_spans": "[[[1, 57], [122, 125]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[61, 64]], [[1, 64]], [[81, 83]], [[0, 71]], [[0, 71]], [[75, 78]], [[0, 78]], [[92, 95]], [[81, 95]], [[111, 114]], [[97, 110]]]",
    "query_spans": "[[[122, 136]]]",
    "process": "Without loss of generality, let the asymptote equation be $ y = \\frac{b}{a}x' $, and the right focus be $ F(c,0) $. Then the distance from point $ F(c,0) $ to the asymptote $ y = \\frac{b}{a}x $ is $ \\begin{matrix} |bc| \\\\ \\sqrt{a^{2}+b^{2}} \\end{matrix} = b $. Also, in the equation $ y = -\\frac{b}{a}(x-c) $, setting $ x = 0 $ gives $ y = \\frac{ac}{b} $, so $ Q(0,\\frac{ac}{b}) $. From $ |FP < OQ| $, we obtain $ |FP| < |OQ| $, which implies $ b < \\frac{ac}{b} \\Rightarrow b^{2} - ac < 0 \\Rightarrow c^{2} - a^{2} - ac < 0 $, leading to $ e^{2} - e - 1 < 0 \\Rightarrow e < \\frac{1+\\sqrt{5}}{2} $. Since $ e > 1 $, it follows that $ 1 < e < \\frac{1+\\sqrt{5}}{2} $."
  },
  {
    "text": "Given that the line $y = a x - 1$ intersects the curve $y^{2} = 2 x$ at only one point, find the real number $a = ?$",
    "fact_expressions": "G: Line;Expression(G) = (y = a*x - 1);a: Real;H: Curve;Expression(H) = (y^2 = 2*x);NumIntersection(G, H) = 1",
    "query_expressions": "a",
    "answer_expressions": "{0, -1/2}",
    "fact_spans": "[[[2, 13]], [[2, 13]], [[35, 40]], [[14, 27]], [[14, 27]], [[2, 33]]]",
    "query_spans": "[[[35, 42]]]",
    "process": ""
  },
  {
    "text": "Given that the equation of circle $C$ is $(x-1)^{2}+y^{2}=1$, and $P$ is a point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$. Two tangents are drawn from point $P$ to circle $C$, with points of tangency $A$ and $B$. Then the minimum value of $\\overrightarrow{P A} \\cdot \\overrightarrow{P B}$ is?",
    "fact_expressions": "G: Ellipse;C: Circle;P: Point;A: Point;B: Point;l1: Line;l2: Line;Expression(G) = (x^2/4 + y^2/3 = 1);Expression(C) = (y^2 + (x - 1)^2 = 1);PointOnCurve(P, G);TangentOfPoint(P, C) = {l1, l2};TangentPoint(l1, C) = A;TangentPoint(l2, C) = B",
    "query_expressions": "Min(DotProduct(VectorOf(P, A), VectorOf(P, B)))",
    "answer_expressions": "2*sqrt(2) - 3",
    "fact_spans": "[[[36, 73]], [[2, 6], [83, 87]], [[78, 82], [31, 35]], [[98, 101]], [[102, 105]], [], [], [[36, 73]], [[2, 29]], [[31, 76]], [[77, 92]], [[77, 105]], [[77, 105]]]",
    "query_spans": "[[[107, 162]]]",
    "process": "Let $\\angle APB = 2\\theta$, $\\overrightarrow{PA} \\cdot \\overrightarrow{PB} = |\\overrightarrow{PA}| \\cdot |\\overrightarrow{PB}| \\cos 2\\theta = |\\overrightarrow{PA}|^{2} (2\\cos^{2}\\theta - 1) = |\\overrightarrow{PA}|^{2} \\left(2\\frac{|\\overrightarrow{PA}|^{2}}{|\\overrightarrow{PC}|^{2}} - 1\\right)$. Let $|\\overrightarrow{PC}|^{2} = x$, $|\\overrightarrow{PA}|^{2} = x - 1$, then we obtain $\\overrightarrow{PA} \\cdot \\overrightarrow{PB} = x + \\frac{2}{x} - 3$, $x \\in (1, 9]$. Therefore, $\\overrightarrow{PA} \\cdot \\overrightarrow{PB} > 2\\sqrt{2} - 3$, with equality if and only if $x = \\sqrt{2}$."
  },
  {
    "text": "Given the hyperbola $x^{2}-y^{2}=1$, points $F_{1}$ and $F_{2}$ are its two foci, and point $P$ is a point on the hyperbola. If $PF_{1} \\perp PF_{2}$, then the value of $|PF_{1}|+|PF_{2}|$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2 = 1);Focus(G)={F1,F2};PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))",
    "query_expressions": "Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[2, 20], [41, 42], [52, 55]], [[47, 51]], [[22, 31]], [[33, 40]], [[2, 20]], [[22, 46]], [[47, 58]], [[59, 80]]]",
    "query_spans": "[[[83, 106]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the hyperbola such that $P F_{2} \\perp x$-axis. Find the distance from $F_{2}$ to the line $P F_{1}$.",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/5 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P,F2),xAxis) = True",
    "query_expressions": "Distance(F2,LineOf(P,F1))",
    "answer_expressions": "30/13",
    "fact_spans": "[[[2, 40], [70, 73]], [[2, 40]], [[49, 56]], [[57, 64], [97, 104]], [[2, 64]], [[2, 64]], [[65, 69]], [[65, 74]], [[76, 94]]]",
    "query_spans": "[[[97, 121]]]",
    "process": ""
  },
  {
    "text": "Suppose the vertex of the parabola is at the origin, and its focus $F$ lies on the $y$-axis. If the point $P(k, -2)$ on the parabola is at a distance of $4$ from the point $F$, then what is $k$?",
    "fact_expressions": "G: Parabola;P: Point;O: Origin;F: Point;k: Number;Coordinate(P) = (k, -2);Vertex(G) = O;PointOnCurve(F, yAxis);PointOnCurve(P, G);Focus(G) = F;Distance(P, F) = 4",
    "query_expressions": "k",
    "answer_expressions": "{4,-4}",
    "fact_spans": "[[[1, 4], [11, 12], [25, 28]], [[30, 41]], [[8, 10]], [[14, 17], [42, 46]], [[55, 58]], [[30, 41]], [[1, 10]], [[14, 23]], [[25, 41]], [[11, 17]], [[30, 53]]]",
    "query_spans": "[[[55, 61]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ and the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ have common foci $F_{1}$, $F_{2}$, and $P$ is one of their common points, then $S_{\\Delta F_{1} P F_{2}}$=?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Ellipse;F1: Point;F2:Point;P: Point;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(H) = (x^2/4 + y^2 = 1);Focus(G)={F1,F2};Focus(H)={F1,F2};OneOf(Intersection(G,H))=P",
    "query_expressions": "Area(TriangleOf(F1,P,F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[30, 62]], [[33, 62]], [[2, 29]], [[67, 74]], [[75, 82]], [[83, 86]], [[30, 62]], [[2, 29]], [[2, 82]], [[2, 82]], [[83, 95]]]",
    "query_spans": "[[[97, 125]]]",
    "process": "Given $ a^{2}=2 $, the point $ P $ satisfies \n\\[\n\\begin{cases}\n\\frac{x^{2}}{2}-y^{2}=1, \\\\\n\\frac{x^{2}}{4}+y^{2}=1,\n\\end{cases}\n\\]\nsolving yields $ y=\\pm\\frac{\\sqrt{3}}{3} $, then $ S_{\\triangle F_{1}PF_{2}}=c|y|=1 $."
  },
  {
    "text": "The directrix of the parabola $x=ay^{2}$ is $x=2$, then the value of $a$ is?",
    "fact_expressions": "G: Parabola;a: Number;Expression(G) = (x = a*y^2);Expression(Directrix(G)) = (x = 2)",
    "query_expressions": "a",
    "answer_expressions": "-1/8",
    "fact_spans": "[[[0, 13]], [[26, 29]], [[0, 13]], [[0, 24]]]",
    "query_spans": "[[[26, 33]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y=\\frac{1}{8} x^{2}$ has focus $F$, a line $l$ passing through $F$ intersects the parabola at points $A$ and $B$, the directrix of the parabola intersects the $y$-axis at point $M$. When $\\frac{|A M|}{|A F|}$ is maximized, what is the length of chord $A B$?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;M: Point;F: Point;Expression(G) = (y = x^2/8);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};Intersection(Directrix(G), yAxis) = M;WhenMax(Abs(LineSegmentOf(A,M))/Abs(LineSegmentOf(A,F)));IsChordOf(LineSegmentOf(A,B),G)",
    "query_expressions": "Length(LineSegmentOf(A,B))",
    "answer_expressions": "8",
    "fact_spans": "[[[39, 44]], [[2, 26], [45, 48], [60, 63]], [[50, 53]], [[54, 57]], [[73, 77]], [[30, 33], [35, 38]], [[2, 26]], [[2, 33]], [[34, 44]], [[39, 59]], [[60, 77]], [[78, 103]], [[60, 110]]]",
    "query_spans": "[[[105, 114]]]",
    "process": "Draw the graph, from point A draw AE perpendicular to the directrix of the parabola $ y = \\frac{1}{8}x^{2} $ at point E, it follows that $ \\frac{|AM|}{|AF|} = \\frac{1}{\\sin\\angle AME} $. When $ \\angle AME $ takes the minimum value, i.e., when line AM is tangent to the parabola, $ \\frac{|AM|}{|AF|} $ is maximized. The slope of line AM can be found, then the coordinates of point A can be determined. Using symmetry, the coordinates of point B can be obtained. Using the chord length formula for a parabola's focal chord, the length of chord AB can be calculated. Let point A be in the first quadrant. From point A, draw AE perpendicular to the directrix of the parabola $ y = \\frac{1}{8}x^{2} $ at point E, as shown in the figure below. By the definition of a parabola, $ |AE| = |AF| $, so $ \\frac{|AM|}{|AF|} = \\frac{|AM|}{|AE|} = \\frac{1}{\\sin\\angle AME} $. It follows that when $ \\angle AME $ takes the minimum value, i.e., when line AM is tangent to the parabola, $ \\frac{|AM|}{|AF|} $ is maximized. The focus of the parabola $ y = \\frac{1}{8}x^{2} $ is $ F(0,2) $, and point $ M(0,-2) $ is easily known. When line AM is tangent to the parabola $ y = \\frac{1}{8}x^{2} $, the slope of line AM exists. Let the equation of line AM be $ y = kx - 2 $. Solving the system:  \n$$\n\\begin{cases}\ny = kx - 2 \\\\\nx^{2} = 8y\n\\end{cases}\n$$\nEliminating $ y $ gives $ x^{2} - 8kx + 16 = 0 $. The discriminant $ \\Delta = 64k^{2} - 64 = 0 $. Since point A is in the first quadrant, $ k > 0 $, solving gives $ k = 1 $. The equation becomes $ x^{2} - 8x + 16 = 0 $, solving gives $ x = 4 $, and at this time $ y = \\frac{x^{2}}{8} = 2 $, so point A is $ (4,2) $. At this moment, $ AB \\perp y $-axis, by symmetry, point B is $ (-4,2) $. Therefore, $ AB = 4 + 4 = 8 $. Answer: $ 8 $. Method"
  },
  {
    "text": "The focal distance of the ellipse $\\frac{x^2}{m}+\\frac{y^2}{4}=1$ is $2$, then the value of $m$ equals?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (y^2/4 + x^2/m = 1);FocalLength(G) = 2",
    "query_expressions": "m",
    "answer_expressions": "{3,5}",
    "fact_spans": "[[[0, 33]], [[42, 45]], [[0, 33]], [[0, 40]]]",
    "query_spans": "[[[42, 50]]]",
    "process": ""
  },
  {
    "text": "Given real numbers $x$, $y$ satisfying the equation $x = \\sqrt{1 - 2 y^{2}}$, then the range of $\\frac{y+2}{x}$ is?",
    "fact_expressions": "x_: Real;y_: Real;x_ = sqrt(1 - 2*y_^2)",
    "query_expressions": "Range((y_ + 2)/x_)",
    "answer_expressions": "[sqrt(14)/2, +oo)",
    "fact_spans": "[[[2, 7]], [[9, 12]], [[16, 36]]]",
    "query_spans": "[[[38, 60]]]",
    "process": "x=\\sqrt{1-2y^{2}}, squaring gives x^{2}=1-2y^{2}, that is, x^{2}+\\frac{y^{2}}{2}=1 (x\\geqslant0). Let k=\\frac{y+2}{x}, then y=kx-2. From \\begin{cases}y=kx-2\\\\x^{2}+2y^{2}=1^{x}\\end{cases}, we obtain (2k^{2}+1)x^{2}-8kx+7=0 having non-negative roots, =64k^{2}-28(2k^{2}+1)\\geqslant0 \\begin{cases}A=64k'-\\\\\\frac{8k}{2k^{2}+1}\\geqslant0\\end{cases}, yielding k\\geqslant\\frac{\\sqrt{14}}{2}"
  },
  {
    "text": "Given that $F$ is a focus of the ellipse $C$, $B$ is an endpoint of the minor axis, the extension of the line segment $BF$ intersects $C$ at point $D$, and $\\overrightarrow{BF} = 2 \\overrightarrow{FD}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;F: Point;OneOf(Focus(C)) = F;B: Point;OneOf(Endpoint(MinorAxis(C))) = B;D: Point;Intersection(OverlappingLine(LineSegmentOf(B, F)), C) = D;VectorOf(B, F) = 2*VectorOf(F, D)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[6, 11], [41, 44], [98, 101]], [[2, 5]], [[2, 16]], [[17, 20]], [[6, 28]], [[45, 49]], [[29, 49]], [[51, 96]]]",
    "query_spans": "[[[98, 107]]]",
    "process": "Let the foci of ellipse C lie on the x-axis, as shown in the figure. Then B(0,b), F(c,0), D(x_{D},y_{D}), so \\overrightarrow{BF}=(c,-b), \\overrightarrow{FD}=(x_{D}-c,y_{D}). Since \\overrightarrow{BF}=2\\overrightarrow{FD}, it follows that \\begin{cases}c=2(x_{D}-c)\\\\-b=2y_{D}\\end{cases}"
  },
  {
    "text": "The minimum distance from a point on the parabola $y^{2}=8 x$ to its focus is equal to?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);P:Point;PointOnCurve(P,G)",
    "query_expressions": "Min(Distance(P,Focus(G)))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 14], [18, 19]], [[0, 14]], [[16, 17]], [[0, 17]]]",
    "query_spans": "[[[16, 32]]]",
    "process": ""
  },
  {
    "text": "The sum of the distances from two points $A$ and $B$ on the parabola $y^{2}=2x$ to the focus is $5$. Then, what is the distance from the midpoint of segment $AB$ to the $y$-axis?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;Expression(G) = (y^2 = 2*x);PointOnCurve(A, G);PointOnCurve(B, G);Distance(A, Focus(G))+Distance(B,Focus(G)) = 5",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A,B)), yAxis)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 14]], [[18, 21]], [[22, 25]], [[0, 14]], [[0, 25]], [[0, 25]], [[0, 37]]]",
    "query_spans": "[[[39, 59]]]",
    "process": "Find the equation of the directrix from the parabola's equation, use the definition of the parabola that the distance from any point on the parabola to the focus equals the distance to the directrix, set up an equation to solve for the x-coordinate of the midpoint of A and B, and find the distance from the midpoint of segment AB to the y-axis. Since F is the focus of the parabola $ y^{2}=2x $, we have $ F(\\frac{1}{2},0) $, and the directrix equation is $ x=-\\frac{1}{2} $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Therefore, $ |AF|+|BF|=x_{1}+\\frac{1}{2}+x_{2}+\\frac{1}{2}=5 $, solving gives $ x_{1}+x_{2}=4 $. Hence, the x-coordinate of the midpoint of segment AB is: 2. Thus, the distance from the midpoint of segment AB to the y-axis is 2. The answer is: 2"
  },
  {
    "text": "If point $P$ lies on an ellipse with foci $F_{1}$ and $F_{2}$, $PF_{2} \\perp F_{1} F_{2}$, $\\tan \\angle P F_{1} F_{2}=\\frac{3}{4}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;P: Point;F2: Point;F1: Point;PointOnCurve(P, G);Focus(G) = {F1, F2};IsPerpendicular(LineSegmentOf(P, F2), LineSegmentOf(F1, F2));Tan(AngleOf(P, F1, F2)) = 3/4",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[28, 30], [102, 104]], [[1, 5]], [[17, 24]], [[7, 15]], [[1, 31]], [[6, 30]], [[32, 58]], [[61, 100]]]",
    "query_spans": "[[[102, 110]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4+k}=1$ is $\\frac{4}{5}$, then the value of $k$ is?",
    "fact_expressions": "G: Ellipse;k: Number;Expression(G) = (x^2/9 + y^2/(k + 4) = 1);Eccentricity(G) = 4/5",
    "query_expressions": "k",
    "answer_expressions": "-19/25, 21",
    "fact_spans": "[[[0, 39]], [[59, 62]], [[0, 39]], [[0, 57]]]",
    "query_spans": "[[[59, 66]]]",
    "process": ""
  },
  {
    "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{2}=1$ is $\\frac{\\sqrt{2}}{2}$, then what is the length of the major axis of this ellipse?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (y^2/2 + x^2/m = 1);Eccentricity(G) = sqrt(2)/2",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "{4,2*sqrt(2)}",
    "fact_spans": "[[[1, 38], [66, 68]], [[3, 38]], [[1, 38]], [[1, 63]]]",
    "query_spans": "[[[66, 74]]]",
    "process": "From the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{2}=1$, the eccentricity is $\\frac{\\sqrt{2}}{2}$. When $m>2$, the ellipse's foci lie on the $x$-axis, $\\frac{c}{a}=\\frac{\\sqrt{2}}{2}=\\frac{\\sqrt{m-2}}{\\sqrt{m}}$, solving gives $m=4$, so the major axis length of the ellipse is $4$. When $0<m<2$, the ellipse's foci lie on the $y$-axis, $\\frac{c}{a}=\\frac{\\sqrt{2}}{2}=\\frac{\\sqrt{2-m}}{\\sqrt{2}}$, solving gives $m=1$, so the major axis length of the ellipse is $2\\sqrt{2}$."
  },
  {
    "text": "Let the ellipse $ C $: $ \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 $ ($ a>b>0 $) have two foci $ F_{1} $, $ F_{2} $, with $ |F_{1} F_{2}|=2 \\sqrt{2} $. Let $ P $ be a point on $ C $ such that $ |P F_{1}|-|P F_{2}|=a $ and $ P F_{2} \\perp F_{1} F_{2} $. Then the equation of the ellipse $ C $ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;Focus(C) = {F1, F2};Abs(LineSegmentOf(F1, F2)) = 2*sqrt(2);P: Point;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)) = a;IsPerpendicular(LineSegmentOf(P, F2), LineSegmentOf(F1, F2))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4 + y^2/2 = 1",
    "fact_spans": "[[[1, 58], [178, 183], [116, 119]], [[1, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[66, 73]], [[75, 82]], [[1, 82]], [[84, 110]], [[112, 115]], [[112, 122]], [[124, 147]], [[149, 176]]]",
    "query_spans": "[[[178, 188]]]",
    "process": "Using the given conditions combined with the definition of an ellipse, we can obtain |PF_{1}|=\\frac{3a}{2}, |PF_{2}|=\\frac{a}{2}; then using the Pythagorean theorem, we can find the value of a. Subsequently, the value of b can be determined, thus obtaining the equation of the ellipse C. From \\begin{cases}|PF_{1}|-|PF_{2}|=a\\\\|PF_{1}|+|PF_{2}|=2a\\end{cases}, solving gives \\begin{cases}|PF_{1}|=\\frac{3a}{2}\\\\|PF_{2}|=\\frac{a}{2}\\end{cases}. In \\triangle PF_{1}F_{2}, since PF_{2}\\bot F_{1}F_{2}, it follows that \\angle PF_{2}F_{1}=90^\\circ. By the Pythagorean theorem, we have |PF_{2}|^{2}+|F_{1}F_{2}|^{2}=|PF_{1}|^{2}, that is, \\frac{a^{2}}{4}+8=\\frac{9a^{2}}{4}, solving gives a=2. Therefore, the equation of the ellipse C is \\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1."
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{m^{2}}=1$ and the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{2}=1$ have the same foci; then the value of the real number $m$ is?",
    "fact_expressions": "G: Hyperbola;m: Real;H: Ellipse;Expression(G) = (-y^2/2 + x^2/m = 1);Expression(H) = (x^2/4 + y^2/m^2 = 1);Focus(H) = Focus(G)",
    "query_expressions": "m",
    "answer_expressions": "1",
    "fact_spans": "[[[42, 80]], [[88, 93]], [[0, 41]], [[42, 80]], [[0, 41]], [[0, 86]]]",
    "query_spans": "[[[88, 97]]]",
    "process": ""
  },
  {
    "text": "If the focus of a parabola with vertex at the origin coincides with the center of the circle $x^{2}+y^{2}-4 x=0$, then the equation of the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;H: Circle;Expression(H) = (-4*x + x^2 + y^2 = 0);O:Origin;Vertex(G)=O;Focus(G)=Center(H)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=-2",
    "fact_spans": "[[[7, 10], [42, 45]], [[14, 34]], [[14, 34]], [[4, 6]], [[1, 10]], [[7, 39]]]",
    "query_spans": "[[[42, 52]]]",
    "process": "\\because the focus of a parabola with vertex at the origin coincides with the center of the circle x^{2}+y^{2}-4x=0 \\therefore the focus of the parabola is F(2,0) \\therefore the directrix equation of this parabola is x=-:"
  },
  {
    "text": "Given that one asymptote of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ passes through the point $(1,3)$, what is the eccentricity of $C$?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;G: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(G) = (1, 3);PointOnCurve(G, OneOf(Asymptote(C)))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[2, 62], [79, 82]], [[9, 62]], [[9, 62]], [[69, 77]], [[9, 62]], [[9, 62]], [[2, 62]], [[69, 77]], [[2, 77]]]",
    "query_spans": "[[[79, 88]]]",
    "process": "Since one asymptote of the hyperbola $ C: \\frac{x^2}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) passes through the point $ (1,3) $, the equation of one asymptote is $ bx - ay = 0 $. Since this asymptote passes through the point $ (1,3) $, we have $ b = 3a $, then $ c = \\sqrt{a^{2} + b^{2}} = \\sqrt{10}a $. Therefore, $ e = \\frac{c}{a} = \\sqrt{10} $."
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=2 px(p>0)$, draw a line intersecting the parabola at points $A$ and $B$. If $|AF|=2|BF|=6$, then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;A: Point;F: Point;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));PointOnCurve(F,H);Intersection(H,G) = {B, A};Focus(G)=F;Abs(LineSegmentOf(A,F))=2*Abs(LineSegmentOf(B,F));2*Abs(LineSegmentOf(B,F))=6",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 21], [31, 34]], [[64, 67]], [[28, 30]], [[35, 39]], [[24, 27]], [[41, 44]], [[4, 21]], [[1, 21]], [[0, 30]], [[28, 46]], [[1, 27]], [[48, 62]], [[48, 62]]]",
    "query_spans": "[[[64, 69]]]",
    "process": ""
  },
  {
    "text": "If the chord of the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$ is bisected by the point $(4,2)$, then what is the equation of the line on which this chord lies?",
    "fact_expressions": "G: Ellipse;H: LineSegment;P: Point;Expression(G) = (x^2/36 + y^2/9 = 1);Coordinate(P) = (4, 2);MidPoint(H)=P;IsChordOf(H,G)",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "x+2*y-8=0",
    "fact_spans": "[[[2, 40]], [], [[43, 51]], [[2, 40]], [[43, 51]], [[2, 53]], [[2, 42]]]",
    "query_spans": "[[[2, 67]]]",
    "process": ""
  },
  {
    "text": "If one of the asymptotes of the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{9}=1$ has an inclination angle of $60^{\\circ}$, then the eccentricity of the hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;a: Number;Expression(G) = (-y^2/9 + x^2/a = 1);Inclination(OneOf(Asymptote(G))) = ApplyUnit(60, degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 39], [64, 67]], [[4, 39]], [[1, 39]], [[1, 62]]]",
    "query_spans": "[[[64, 74]]]",
    "process": ""
  },
  {
    "text": "Given that point $P(1, \\sqrt{3})$ lies on the asymptote of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $F$ is the right focus of $C$, $O$ is the origin, and $\\angle F P O = 90^{\\circ}$, then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F: Point;O: Origin;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (1, sqrt(3));PointOnCurve(P, Asymptote(C));RightFocus(C) = F;AngleOf(F, P, O) = ApplyUnit(90, degree)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4 - y^2/12 = 1",
    "fact_spans": "[[[20, 81], [91, 94], [134, 137]], [[28, 81]], [[28, 81]], [[2, 19]], [[87, 90]], [[99, 102]], [[28, 81]], [[28, 81]], [[20, 81]], [[2, 19]], [[2, 86]], [[87, 98]], [[107, 132]]]",
    "query_spans": "[[[134, 142]]]",
    "process": "Since the asymptotes of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) are given by $ y = \\pm \\frac{b}{a}x $, and point $ P(1, \\sqrt{3}) $ lies on the asymptote, it follows that $ \\frac{b}{a} = \\angle FOP = 60^{\\circ} $, so $ OF = c = 4 $, $ b = 2\\sqrt{3} $, $ a = 2 $. Therefore, the equation of $ C $ is $ \\frac{x^{2}}{4} - \\frac{y^{2}}{12} = 1 $."
  },
  {
    "text": "The point $P(x, y)$ is a moving point on the curve $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, then the range of $2 x+\\sqrt{3} y$ is?",
    "fact_expressions": "P: Point;x1: Number;y1: Number;Coordinate(P) = (x1, y1);C: Curve;Expression(C) = (x^2/4 + y^2/3 = 1);PointOnCurve(P, C)",
    "query_expressions": "Range(2*x1 + sqrt(3)*y1)",
    "answer_expressions": "[-5,5]",
    "fact_spans": "[[[0, 10]], [[1, 10]], [[1, 10]], [[0, 10]], [[11, 53]], [[11, 53]], [[0, 58]]]",
    "query_spans": "[[[60, 83]]]",
    "process": "Let point $ P(x, y) $ be a moving point on the curve $ C: \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $. We can set $ x = 2\\cos\\theta $, $ y = \\sqrt{3}\\sin\\theta $, $ \\theta \\in [0, 2\\pi) $. Then $ 2x + \\sqrt{3}y = 4\\cos\\theta + 3\\sin\\theta = 5\\sin(\\theta + \\alpha) $, where $ \\tan\\alpha = \\frac{4}{3} $. Since $ 5\\sin(\\theta + \\alpha) \\in [-5, 5] $, it follows that $ 2x + \\sqrt{3}y \\in [-5, 5] $."
  },
  {
    "text": "If the distance from the point $(x_{0}, 2)$ $(x_{0}>\\frac{p}{2})$ on the parabola $y^{2}=2 p x$ $(p>0)$ to its focus is $\\frac{5}{2}$, then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;H: Point;x0: Number;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(H) = (x0, 2);x0 > p/2;PointOnCurve(H, G);Distance(H, Focus(G)) = 5/2",
    "query_expressions": "p",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 22], [58, 59]], [[80, 83]], [[24, 57]], [[25, 57]], [[4, 22]], [[1, 22]], [[24, 57]], [[25, 57]], [[1, 57]], [[24, 78]]]",
    "query_spans": "[[[80, 85]]]",
    "process": "From the given conditions, $2px_{0}=4$ and $x_{0}+\\frac{p}{2}=\\frac{5}{2}$, eliminating $x_{0}$ yields $p^{2}-5p+4=0$, solving gives $p=1$ or $p=4$ (discarded)."
  },
  {
    "text": "If the distance from a focus of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{m-5}=1$ to the origin is $3$, then the value of $m$ is?",
    "fact_expressions": "G: Hyperbola;m: Number;O: Origin;Expression(G) = (-y^2/(m - 5) + x^2/m = 1);Distance(OneOf(Focus(G)), O) = 3",
    "query_expressions": "m",
    "answer_expressions": "{7, -2}",
    "fact_spans": "[[[1, 41]], [[60, 63]], [[47, 51]], [[1, 41]], [[1, 58]]]",
    "query_spans": "[[[60, 67]]]",
    "process": "According to the problem, c=3. When the foci of the hyperbola lie on the x-axis, m>5, c^{2}=m+m-5=9, so m=7. When the foci of the hyperbola lie on the y-axis, m<0, c^{2}=-m+5-m=9, so m=-2. In conclusion, m=7 or m=-2."
  },
  {
    "text": "If the eccentricity of the ellipse $\\frac{y^{2}}{k+8}+\\frac{x^{2}}{9}=1$ is $e=\\frac{1}{2}$, then what is the value of $k$?",
    "fact_expressions": "G: Ellipse;k: Number;Expression(G) = (x^2/9 + y^2/(k + 8) = 1);Eccentricity(G) = e ;e= 1/2;e:Number",
    "query_expressions": "k",
    "answer_expressions": "{4, -5/4}",
    "fact_spans": "[[[1, 40]], [[62, 65]], [[1, 40]], [[1, 60]], [[45, 60]], [[45, 60]]]",
    "query_spans": "[[[62, 70]]]",
    "process": "When the foci of the ellipse are on the x-axis, then $a^{2}=9$, $b^{2}=k+8$ ($k+8>0$), so $c^{2}=a^{2}-b^{2}=9-(k+8)=1-k$. Since the eccentricity of the ellipse is $e=\\frac{1}{2}$, it follows that $\\frac{c^{2}}{a^{2}}=\\frac{1-k}{9}=\\frac{1}{4}$, solving gives $k=-\\frac{5}{4}$. When the foci of the ellipse are on the y-axis, then $b^{2}=9$, $a^{2}=k+8$ ($k+8>9$), so $c^{2}=a^{2}-b^{2}=k+8-9=k-1$. Since the eccentricity of the ellipse is $e=\\frac{1}{2}$, it follows that $\\frac{c^{2}}{a^{2}}=\\frac{k-1}{k+8}=\\frac{1}{4}$, solving gives $k=4$. In conclusion, $k=-\\frac{5}{4}$ or $k=4$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with left and right foci $F_{1}(-2,0)$, $F_{2}(2,0)$, and point $P$ being any point on the hyperbola. If the minimum value of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ is $-2$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;Coordinate(F1) = (-2, 0);Coordinate(F2) = (2, 0);LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C);Min(DotProduct(VectorOf(P, F1), VectorOf(P, F2))) = -2",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 63], [102, 105], [180, 186]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[69, 82]], [[83, 96]], [[69, 82]], [[83, 96]], [[2, 96]], [[2, 96]], [[97, 101]], [[97, 110]], [[112, 178]]]",
    "query_spans": "[[[180, 192]]]",
    "process": "Let $ P(x_{0},y_{0}) $, then $ \\frac{x_{0}^{2}}{a^{2}} - \\frac{y_{0}^{2}}{b^{2}} = 1 \\Rightarrow x_{0}^{2} = a^{2} + \\frac{a^{2}}{b^{2}} y_{0}^{2} $. Since $ F_{1}(-2,0) $, $ F_{2}(2,0) $, $ c^{2} = 4 = a^{2} + b^{2} $, therefore $ \\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}} = (x_{0}+2)(x_{0}-2) + y_{0}^{2} = x_{0}^{2} + y_{0}^{2} - 4 = \\frac{c^{2}}{b^{2}} y_{0}^{2} + a^{2} - 4 \\geqslant a^{2} - 4 $, where equality holds when $ y_{0} = 0 $. Since the minimum value of $ \\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}} $ is $ -2 $, we have $ a^{2} - 4 = -2 $. Solving gives $ a = \\sqrt{2} $. Also $ c = 2 $, therefore $ e = \\frac{c}{a} = \\sqrt{2} $."
  },
  {
    "text": "Given the curve $x^{2}+2 y^{2}=4$, find the general form equation of the line containing the chord for which $(1,1)$ is the midpoint.",
    "fact_expressions": "G: LineSegment;H: Curve;Expression(H) = (x^2 + 2*y^2 = 4);Coordinate(MidPoint(G)) = (1, 1);IsChordOf(G, H)",
    "query_expressions": "Expression(OverlappingLine(G))",
    "answer_expressions": "x + 2*y - 3 = 0",
    "fact_spans": "[[], [[2, 21]], [[2, 21]], [[2, 36]], [[2, 36]]]",
    "query_spans": "[[[2, 48]]]",
    "process": "Let the endpoints of the chord of $x^{2}+2y^{2}=4$ be $A(x_{1},y_{1})$, $B(x_{2},y_{2})$. Therefore, \n$$\n\\begin{cases}\nx_{1}^{2}+2y_{1}^{2}=4 \\\\\nx_{2}^{2}+2y_{2}^{2}=4\n\\end{cases}\n$$\nSubtracting the two equations gives $(x_{1}+x_{2})(x_{1}-x_{2})+2(y_{1}+y_{2})(y_{1}-y_{2})=0$. Hence, $2\\times(x_{1}-x_{2})+2\\times2\\times(y_{1}-y_{2})=0$, so $2+4\\times\\frac{(y_{1}-y_{2})}{(x_{1}-x_{2})}=0$. $\\therefore k=-\\frac{1}{2}$. $\\therefore$ the equation of the line containing the chord with midpoint $P(1,1)$ is $y-1=-\\frac{1}{2}(x-1)$, i.e., $x+2y-3=0$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ and the point $P(3,-1)$, a line $l$ passes through point $P$ and intersects the hyperbola at points $A$ and $B$. When $P$ is exactly the midpoint of segment $AB$, what is the equation of line $l$?",
    "fact_expressions": "l: Line;G: Hyperbola;B: Point;A: Point;P: Point;Expression(G) = (x^2/4 - y^2 = 1);Coordinate(P) = (3, -1);PointOnCurve(P, l);Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A,B)) = P",
    "query_expressions": "Expression(l)",
    "answer_expressions": "3*x+4*y-5=0",
    "fact_spans": "[[[42, 47], [90, 93]], [[2, 30], [55, 58]], [[65, 68]], [[61, 64]], [[31, 41], [49, 53], [72, 75]], [[2, 30]], [[31, 41]], [[42, 53]], [[42, 70]], [[72, 88]]]",
    "query_spans": "[[[90, 98]]]",
    "process": "Let points $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Using the point difference method, the equation of line $ l $ can be obtained, and thus the line $ l $. Let points $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. If line $ l \\perp x $-axis, then points $ A $ and $ B $ are symmetric with respect to the $ x $-axis, so point $ P $ lies on the $ x $-axis, which contradicts the given condition. Since $ P(3,-1) $ is the midpoint of segment $ AB $, we have\n$$\n\\begin{cases}\n\\frac{x_{1}+x_{2}}{2}=3 \\\\\n\\frac{y_{1}+y_{2}}{2}=-1\n\\end{cases}\n$$\nwhich gives\n$$\n\\begin{cases}\nx_{1}+x_{2}=6 \\\\\ny_{1}+y_{2}=-2\n\\end{cases}\n$$\nSubstituting the coordinates of points $ A $ and $ B $ into the hyperbola's equation yields\n$$\n\\begin{cases}\n\\\\\n\\end{cases}\n\\frac{x_{1}^{2}}{4}-y_{1}^{2}=\n$$\nSubtracting the above two equations gives $ \\frac{x^{2}-x^{2}}{4}=y_{1}^{2}-y_{2}^{2} $, i.e., $ \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}.\\frac{y_{1}+y}{x_{1}+x} $. Therefore, $ \\frac{y_{1}-y_{2}}{x_{1}}-\\frac{1}{3}=\\frac{1}{4} $, so the slope of line $ l $ is $ \\frac{y_{1}}{x_{1}} $. Thus, the equation of line $ l $ is $ y+1=-\\frac{3}{4}(x-3) $, i.e., $ 3x+4y-5 $."
  },
  {
    "text": "Draw a line passing through the origin with an inclination angle $\\theta$, intersecting the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ at points $A$ and $B$. Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse. If $\\angle F_{1} A F_{2}=\\frac{\\pi}{2}$, and the eccentricity $e$ of the ellipse satisfies $e \\in\\left[\\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{6}}{3}\\right]$, then what is the range of values for $\\theta$?",
    "fact_expressions": "O: Origin;theta: Number;H: Line;PointOnCurve(O, H);Inclination(H) = theta;G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;A: Point;B: Point;Intersection(H, G) = {A, B};F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;AngleOf(F1, A, F2) = pi/2;e: Number;Eccentricity(G) = e;In(e, [sqrt(2)/2, sqrt(6)/3])",
    "query_expressions": "Range(theta)",
    "answer_expressions": "[\\pi/6, 5\\pi/6]",
    "fact_spans": "[[[1, 3]], [[10, 18], [206, 214]], [[19, 21]], [[0, 21]], [[6, 21]], [[22, 74], [102, 104], [151, 153]], [[22, 74]], [[24, 74]], [[24, 74]], [[24, 74]], [[24, 74]], [[76, 79]], [[80, 83]], [[19, 85]], [[86, 93]], [[94, 101]], [[86, 110]], [[86, 110]], [[112, 148]], [[157, 204]], [[151, 204]], [[157, 204]]]",
    "query_spans": "[[[206, 221]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, the left vertex is $A_{1}$, the right focus is $F_{2}$, and $P$ is a point on the right branch of the hyperbola. Then the minimum value of $\\overrightarrow{P A_{1}} \\cdot \\overrightarrow{P F_{2}}$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;A1: Point;F2: Point;Expression(G) = (x^2 - y^2/3 = 1);LeftVertex(G)=A1;RightFocus(G)=F2;PointOnCurve(P,RightPart(G))",
    "query_expressions": "Min(DotProduct(VectorOf(P, A1), VectorOf(P, F2)))",
    "answer_expressions": "-2",
    "fact_spans": "[[[2, 30], [59, 62]], [[55, 58]], [[35, 42]], [[47, 54]], [[2, 30]], [[2, 42]], [[2, 54]], [[55, 67]]]",
    "query_spans": "[[[69, 131]]]",
    "process": "A_{1}(-1,0), F_{2}(2,0), let P(x,y) (x \\geqslant 1), \\overrightarrow{PA_{1}} \\cdot \\overrightarrow{PF_{2}} = (-1-x,y) \\cdot (2-x,y) = x^{2}-x-2+y^{2}, and since x^{2}-\\frac{y^{2}}{3}=1, it follows that y^{2}=3(x^{2}-1), thus \\overrightarrow{PA_{1}} \\cdot \\overrightarrow{PF_{2}} = 4x^{2}-x-5 = 4(x-\\frac{1}{8})^{2}-5-\\frac{1}{16}, when x=1, the minimum value -2 is attained."
  },
  {
    "text": "If point $P$ is a moving point on the curve $C_{1}$: $y^{2}=16 x$, point $Q$ is a moving point on the curve $C_{2}$: $(x-4)^{2}+y^{2}=9$, and point $O$ is the origin, then the minimum value of $|\\frac{P Q}{O P}|$ is?",
    "fact_expressions": "P: Point;C1: Curve;Expression(C1) = (y^2 = 16*x);PointOnCurve(P, C1);Q: Point;C2: Curve;Expression(C2) = (y^2 + (x - 4)^2 = 9);PointOnCurve(Q, C2);O: Origin",
    "query_expressions": "Min(Abs(LineSegmentOf(P, Q)/LineSegmentOf(O, P)))",
    "answer_expressions": "sqrt(15)/8",
    "fact_spans": "[[[1, 5]], [[6, 29]], [[6, 29]], [[1, 33]], [[34, 38]], [[39, 68]], [[39, 68]], [[34, 72]], [[73, 77]]]",
    "query_spans": "[[[84, 109]]]",
    "process": "Curve $ C_{2}:(x-4)^{2}+y^{2}=9 $ has center $ C_{2}(4,0) $, which is the focus $ F $ of the parabola, and radius 3, so $ \\left|\\frac{PQ}{OP}\\right| \\geqslant \\frac{|PF|-3}{|OP|} $. This transforms into finding the minimum value of $ \\frac{|PF|-3}{|OP|} $. Let $ P(x,y) $, use the focal radius formula and the parabola equation to express $ \\frac{|PF|-3}{|OP|} $ as a function of $ x $, simplify and apply the extremum of a quadratic function to solve. [Solution] The parabola $ C_{1}: y^{2}=16x $ has focus $ F(4,0) $. Curve $ C_{2}:(x-4)^{2}+y^{2}=9 $ has center $ F(4,0) $ and radius 3. Therefore, $ \\left|\\frac{PQ}{OP}\\right| \\geqslant \\frac{|PF|-3}{|OP|} $, with equality when points $ P, Q, F $ are collinear. Let $ P(x,y) $, $ x>0 $, then $ \\frac{|PF|-3}{|OP|} = $ When $ t=\\frac{7}{15} $, i.e., $ x=\\frac{8}{7} $, $ \\frac{|PF|-3}{|OP|} $ attains its minimum value $ \\frac{\\sqrt{15}}{8} $. Thus, when $ x=\\frac{8}{7} $, $ \\left|\\frac{PQ}{OP}\\right| $ attains its minimum value $ \\frac{\\sqrt{15}}{8} $."
  },
  {
    "text": "Draw a line through the focus of the parabola $y^{2}=16x$ intersecting the parabola at points $A$ and $B$. If the horizontal coordinate of the midpoint of segment $AB$ is $10$, then $|AB|$ equals?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;Expression(G) = (y^2 = 16*x);PointOnCurve(Focus(G), H);Intersection(H, G) = {A, B};XCoordinate(MidPoint(LineSegmentOf(A, B))) = 10",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "28",
    "fact_spans": "[[[1, 16], [23, 26]], [[20, 22]], [[27, 30]], [[31, 34]], [[1, 16]], [[0, 22]], [[20, 36]], [[38, 57]]]",
    "query_spans": "[[[60, 70]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), and let the focus of the ellipse be F. By the definition of a parabola, we have: |AF| = x_{1} + 4, |BF| = x_{2} + 4. Then |AB| = |AF| + |BF| = x_{1} + x_{2} + 8. Given the x-coordinate of the midpoint \\frac{x_{1}+x_{2}}{2} = 10, we obtain |AB| = 8 + 20 = 28."
  },
  {
    "text": "The sum of the lengths of the real axis and the imaginary axis of a hyperbola equals $\\sqrt{2}$ times its focal distance, and one vertex has coordinates $(0, 2)$. Then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Coordinate(OneOf(Vertex(G))) = (0, 2);Length(RealAxis(G)) + Length(ImageinaryAxis(G))=FocalLength(G)*sqrt(2)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/4 - x^2/4 = 1",
    "fact_spans": "[[[0, 3], [15, 16], [52, 55]], [[15, 50]], [[0, 31]]]",
    "query_spans": "[[[52, 62]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the common foci of an ellipse and a hyperbola, $P$ is one of their common points, and $|P F_{1}| > |P F_{2}|$, the perpendicular bisector of segment $P F_{1}$ passes through $F_{2}$. If the eccentricity of the ellipse is $e_{1}$ and the eccentricity of the hyperbola is $e_{2}$, then the minimum value of $\\frac{3}{e_{1}} + \\frac{e_{2}}{4}$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;F1: Point;P: Point;F2: Point;e1:Number;e2:Number;Focus(G)={F1,F2};Focus(H)={F1,F2};OneOf(Intersection(G,H))=P;Abs(LineSegmentOf(P, F1)) > Abs(LineSegmentOf(P, F2));PointOnCurve(F2,PerpendicularBisector(LineSegmentOf(P,F1)));Eccentricity(H) = e1;Eccentricity(G) = e2",
    "query_expressions": "Min(e2/4 + 3/e1)",
    "answer_expressions": "6+sqrt(3)",
    "fact_spans": "[[[21, 24], [108, 111]], [[18, 20], [93, 95]], [[2, 9]], [[30, 33]], [[10, 17], [84, 91]], [[100, 107]], [[116, 123]], [[2, 29]], [[2, 29]], [[30, 42]], [[44, 65]], [[66, 91]], [[93, 107]], [[108, 123]]]",
    "query_spans": "[[[125, 164]]]",
    "process": "Since the perpendicular bisector of segment $ PF_{1} $ passes through $ F_{2} $, we have $ |F_{1}F_{2}| = |PF_{2}| $. Then, according to the definitions of hyperbola and ellipse, we find the expression for $ 2c $, and then use the basic inequality to find the minimum value. Let the parameters corresponding to the ellipse be $ a_{1}, b_{1}, c $, and those corresponding to the hyperbola be $ a_{2}, b_{2}, c $. Since the perpendicular bisector of segment $ PF_{1} $ passes through $ F_{2} $, we have $ |F_{1}F_{2}| = |PF_{2}| = 2c $. According to the definitions of hyperbola and ellipse, we have\n\\[\n\\begin{cases}\n|PF_{1}| + 2c = 2a_{1} \\\\\n|PF_{1}| - 2c = 2a_{2}\n\\end{cases}\n\\]\nSubtracting the two equations gives $ 4c = 2(a_{1} - a_{2}) $, that is, $ a_{1} - a_{2} = 2c \\Rightarrow a_{1} = 2c + a_{2} $. Therefore,\n\\[\n\\frac{3}{e_{1}} + \\frac{e_{2}}{4} = \\frac{3a_{1}}{c} + \\frac{c}{4a_{2}} = 6 + \\frac{3a_{2}}{c} + \\frac{c}{4a_{2}} \\geqslant 6 + 2\\sqrt{\\frac{3a_{2}}{c} \\cdot \\frac{c}{4a_{2}}} = 6 + \\sqrt{3}\n\\]\nEquality holds if and only if $ c = 2\\sqrt{3}a_{2} $, so the minimum value of $ \\frac{3}{e_{1}} + \\frac{e_{2}}{4} $ is $ 6 + \\sqrt{3} $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=6x$ with focus $F$, a line $l$ with slope $\\sqrt{3}$ passing through $F$ intersects the parabola $C$ at points $A$ and $B$. Then the chord length $|AB|=$?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;F: Point;Expression(C) = (y^2 = 6*x);Focus(C) = F;PointOnCurve(F, l);Slope(l) = sqrt(3);Intersection(l, C) = {A, B};IsChordOf(LineSegmentOf(A,B),C)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[47, 52]], [[2, 21], [53, 59]], [[60, 63]], [[64, 67]], [[24, 27], [29, 32]], [[2, 21]], [[2, 27]], [[28, 52]], [[33, 52]], [[47, 69]], [[53, 79]]]",
    "query_spans": "[[[72, 81]]]",
    "process": "Given that $ F\\left(\\frac{3}{2},0\\right) $, the line $ l: y = \\sqrt{3}\\left(x - \\frac{3}{2}\\right) $. Substituting into the parabola equation and simplifying yields: $ x^2 - 5x + \\frac{9}{4} = 0 $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, so $ x_{1} + x_{2} = 5 $. The directrix of the parabola is: $ x = -\\frac{3}{2} $. By the definition of a parabola: $ |AB| = |AF| + |BF| = x_{1} + \\frac{3}{2} + x_{2} + \\frac{3}{2} = x_{1} + x_{2} + 3 = 5 + 3 = 8 $."
  },
  {
    "text": "Let the left focus of $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ be $F_{1}$, and let $P(x_{0}, y_{0})$ be a point on the ellipse. Then the maximum value of $|P F_{1}|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;x0: Number;y0: Number;Coordinate(P) = (x0, y0);Expression(G) = (x^2/25 + y^2/16 = 1);LeftFocus(G) = F1;PointOnCurve(P, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, F1)))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 38], [71, 73]], [[52, 70]], [[43, 50]], [[52, 70]], [[52, 70]], [[52, 70]], [[1, 38]], [[1, 50]], [[53, 76]]]",
    "query_spans": "[[[78, 95]]]",
    "process": "Given the ellipse equation $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1^{1}$, then $a=5$, $b=4$, $c=3$, the left focus is $F_{1}$ with coordinates $(-3,0)$, then $\\lambda\\because-5\\leqslant x_{0}\\leqslant5$, hence when $x_{0}=5$, that is, when $P(x_{0},y_{0})$ is the right vertex of the ellipse $(5,0)$, the maximum value of $|PF_{1}|$ is $8$, hence the maximum value of $|PF_{1}|$ is $8$."
  },
  {
    "text": "Given the parabola $ C $: $ y = \\frac{1}{8} x^{2} $, the distance from a point $ M $ on the parabola to the focus is $ 5 $. Then, the distance from point $ M $ to the $ y $-axis is?",
    "fact_expressions": "C: Parabola;M: Point;Expression(C) = (y = x^2/8);PointOnCurve(M, C);Distance(M, Focus(C)) = 5",
    "query_expressions": "Distance(M, yAxis)",
    "answer_expressions": "2*sqrt(6)",
    "fact_spans": "[[[2, 31]], [[33, 37], [49, 53]], [[2, 31]], [[2, 37]], [[2, 47]]]",
    "query_spans": "[[[49, 63]]]",
    "process": "The equation of parabola C can be written as $x^{2}=8y$. Let $M(x_{0},y_{0})$. Since the distance from point M to the focus is 5, the distance from point M to the directrix $y=-2$ is 5, thus $y_{0}=3$. Substituting $y_{0}=3$ into $x^{2}=8y$ gives $|x_{0}|=2\\sqrt{6}$. Therefore, the distance from point M to the y-axis is $2\\sqrt{6}$."
  },
  {
    "text": "The foci of hyperbola $C$ are $F_{1}(-\\sqrt{5}, 0)$, $F_{2}(\\sqrt{5}, 0)$. A line passing through point $F_{2}$ intersects the right branch of the hyperbola at points $A$ and $B$. If $\\overrightarrow{A F_{2}}=3 \\overrightarrow{F_{2} B}$, $3|AB|=4|B F_{1}|$, then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;F1: Point;Coordinate(F1) = (-sqrt(5), 0);F2: Point;Coordinate(F2) = (sqrt(5), 0);Focus(C) = {F1, F2};PointOnCurve(F2, G);G: Line;Intersection(G, RightPart(C)) = {A, B};A: Point;B: Point;VectorOf(A, F2) = 3*VectorOf(F2, B);3*Abs(LineSegmentOf(A, B)) = 4*Abs(LineSegmentOf(B, F1))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/2 - y^2/3 = 1",
    "fact_spans": "[[[2, 8], [70, 73], [162, 165]], [[12, 33]], [[12, 33]], [[36, 56], [58, 66]], [[36, 56]], [[2, 56]], [[57, 69]], [[67, 69]], [[67, 87]], [[78, 81]], [[82, 85]], [[89, 142]], [[143, 161]]]",
    "query_spans": "[[[162, 170]]]",
    "process": "Let the equation of hyperbola $ C $ be: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$. Let $|F_{2}B|=x$. According to the given conditions and the definition of the hyperbola, we have $ a=x $, and $ BF_{1}\\bot BF_{2} $, $|BF_{1}|=3x$, $|F_{1}F_{2}|=2\\sqrt{5}$, then we can find $ a=\\sqrt{2} $. Using $ b^{2}=c^{2}-a^{2} $, we can find $ b^{2} $, and thus obtain the equation of hyperbola $ C $. Let the equation of hyperbola $ C $ be: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$. Let $|F_{2}B|=x$, then $|AF_{2}|=3x$, $|AB|=4x$, $|F_{1}F_{2}|=2\\sqrt{5}$. Since $ 3|AB|=4|BF_{1}| $, we have $|BF_{1}|=3x$. By the definition of the hyperbola: $|BF_{1}|-|BF_{2}|=2a=2x$, so $ a=x $. $|AF_{1}|=|AF_{2}|+2a=3x+2x=5x$. Since $|AF_{1}|^{2}=|AB|^{2}+|BF_{1}|^{2}$, we have $ BF_{1}\\bot BF_{2} $, so $|F_{1}F_{2}|^{2}=|BF_{2}|^{2}+|BF_{1}|^{2}$, that is, $ 20=x^{2}+9x^{2}=10x^{2} $, solving gives $ x=\\sqrt{2} $. Thus $ a=\\sqrt{2} $, so $ b^{2}=c^{2}-a^{2}=5-2=3 $. Therefore, the equation of hyperbola $ C $ is: $\\frac{x^{2}}{2}-\\frac{y^{2}}{3}=1$."
  },
  {
    "text": "Given the ellipse $ C $: $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ $ (a > b > 0) $, the coordinates of its right focus are $ (c, 0) $. The line $ y = \\frac{\\sqrt{6}}{3} b $ intersects the ellipse $ C $ at points $ A $ and $ B $. If $ OA \\perp OB $, then $ \\frac{b^{2}}{2 a c} $ equals?",
    "fact_expressions": "C: Ellipse;a: Number;b: Number;c:Number;G: Line;O: Origin;A: Point;B: Point;IsPerpendicular(LineSegmentOf(O, A), LineSegmentOf(O, B));a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y = b*(sqrt(6)/3));Coordinate(RightFocus(C)) = (c, 0);Intersection(G, C) = {A, B}",
    "query_expressions": "b^2/(2*a*c)",
    "answer_expressions": "sqrt(2)/4",
    "fact_spans": "[[[2, 59], [102, 107]], [[9, 59]], [[9, 59]], [[66, 74]], [[75, 101]], [[120, 135]], [[109, 112]], [[113, 116]], [[120, 135]], [[9, 59]], [[9, 59]], [[2, 59]], [[75, 101]], [[2, 74]], [[75, 118]]]",
    "query_spans": "[[[137, 161]]]",
    "process": "Since the line $ y = \\frac{\\sqrt{6}}{3}b $ intersects the ellipse $ C $ at points $ A $ and $ B $, we have $ A\\left(-\\frac{\\sqrt{3}}{3}a, \\frac{\\sqrt{6}}{3}b\\right) $, $ B\\left(\\frac{\\sqrt{3}}{3}a, \\frac{\\sqrt{6}}{3}b\\right) $. Since $ OA \\perp OB $, it follows that $ -\\frac{\\sqrt{3}}{3}a \\times \\frac{\\sqrt{3}}{3}a + \\frac{\\sqrt{6}}{3}b \\times \\frac{\\sqrt{6}}{3}b = 0 $, $ a^{2} = 2b^{2} $, $ \\frac{b^{2}}{2ac} = \\frac{1}{2}\\sqrt{(b^{2})^{2}} = \\frac{1}{2}\\sqrt{\\frac{b^{4}}{a^{2}(a^{2}-b^{2}})} = \\frac{\\sqrt{2}}{4} $."
  },
  {
    "text": "Given that the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has an eccentricity of $2$, and its foci coincide with the foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$. Then the equations of the asymptotes of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Ellipse;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2/16 + y^2/9 = 1);Eccentricity(G) = 2;Focus(G) = Focus(H)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*sqrt(3)*x",
    "fact_spans": "[[[2, 48], [107, 110]], [[5, 48]], [[5, 48]], [[60, 98]], [[2, 48]], [[60, 98]], [[2, 56]], [[2, 103]]]",
    "query_spans": "[[[107, 118]]]",
    "process": ""
  },
  {
    "text": "If the length of the minor axis of an ellipse is $6$, and the shortest distance from a focus to one endpoint of the major axis is $1$, then what is the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;Length(MinorAxis(G)) = 6;Min(Distance(Focus(G),OneOf(Endpoint(MajorAxis(G)))))=1",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "4/5",
    "fact_spans": "[[[1, 3], [33, 35]], [[1, 11]], [[1, 31]]]",
    "query_spans": "[[[33, 41]]]",
    "process": "According to the problem, we have b=3, a-c=1. Also, a^{2}=b^{2}+c^{2}, solving gives a=5, c=4, ∴ the eccentricity of the ellipse is e=\\frac{c}{a}=\\frac{4}{5}."
  },
  {
    "text": "It is known that the center of ellipse $G$ is at the origin of coordinates, the major axis lies on the $x$-axis, the eccentricity is $\\frac{\\sqrt{3}}{2}$, and the sum of the distances from a point on $G$ to the two foci of $G$ is $12$. Then the equation of ellipse $G$ is?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;OverlappingLine(MajorAxis(G),xAxis);Eccentricity(G) = sqrt(3)/2;D: Point;PointOnCurve(D, G);F1: Point;F2: Point;Focus(G) = {F1, F2};Distance(D, F1) + Distance(D, F2) = 12",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/36+y^2/9=1",
    "fact_spans": "[[[2, 7], [51, 54], [58, 61], [78, 83]], [[11, 15]], [[2, 15]], [[2, 24]], [[2, 49]], [], [[51, 57]], [], [], [[58, 66]], [[51, 76]]]",
    "query_spans": "[[[78, 88]]]",
    "process": ""
  },
  {
    "text": "Let the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ be $F_{1}$ and $F_{2}$, respectively, and let $P$ be a point on this hyperbola. If $P F_{1}$ is perpendicular to the $x$-axis and $\\cos \\angle P F_{2} F_{1}=\\frac{12}{13}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), xAxis);Cos(AngleOf(P, F2, F1)) = 12/13",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3/2",
    "fact_spans": "[[[1, 57], [87, 90], [157, 160]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[66, 73]], [[74, 81]], [[1, 81]], [[1, 81]], [[82, 85]], [[82, 93]], [[95, 111]], [[112, 153]]]",
    "query_spans": "[[[157, 166]]]",
    "process": "Since $ PF_{1} \\perp F_{1}F_{2} $, $ \\cos\\angle PF_{2}F_{1} = \\frac{12}{13} $, so $ |PF| = \\frac{b^{2}}{a} $, $ |PF_{2}| = \\frac{13b^{2}}{5a} $, $ |PF_{2}| - |PF_{1}| = 2a $. Therefore, $ \\frac{13b^{2}}{5a} - \\frac{b^{2}}{a} = 2a $, i.e., $ \\frac{8b^{2}}{5a} = 2a $, so $ 8b^{2} = 10a^{2} $, which gives $ 8(c^{2} - a^{2}) = 10a^{2} $, $ \\therefore e^{2} = \\frac{c^{2}}{a^{2}} = \\frac{9}{4} $, i.e., $ e = \\frac{3}{2} $, hence fill in $ \\frac{3}{2} $."
  },
  {
    "text": "A point $(5, m)$ on the parabola $y^{2}=2 p x(p>0)$ is at a distance of $6$ from the focus $F$. Let $P$ and $Q$ be moving points on the parabola and the circle $(x-3)^{2}+y^{2}=1$, respectively. Then the minimum value of $|P Q|+|P F|$ is?",
    "fact_expressions": "G: Parabola;p: Number;m:Number;H: Circle;I: Point;P: Point;Q: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (y^2 + (x - 3)^2 = 1);Coordinate(I) = (5, m);PointOnCurve(I,G);Focus(G)=F;Distance(I,F)=6;PointOnCurve(P,G);PointOnCurve(Q,H)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F)) + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 21], [57, 60]], [[3, 21]], [[24, 32]], [[61, 81]], [[24, 32]], [[47, 50]], [[51, 54]], [[35, 38]], [[3, 21]], [[0, 21]], [[61, 81]], [[24, 32]], [[0, 32]], [[0, 38]], [[24, 45]], [[47, 85]], [[47, 85]]]",
    "query_spans": "[[[87, 106]]]",
    "process": "Using the definition of a parabola, find the value of $ p $. Draw $ PN \\perp $ to the directrix of the parabola from point $ P $, intersecting the directrix at point $ N $. Let the directrix intersect the $ x $-axis at point $ M_{1} $. By the definition of the parabola, $ |PF| = |PN| $, so $ |PQ| + |PF| = |PQ| + |PN| \\geqslant |MM_{1}| - 1 $. The parabola $ C: y^{2} = 2px $ ($ p > 0 $) has its focus on the $ x $-axis and directrix $ x = -\\frac{p}{2} $. The distance from point $ (5, t) $ to the focus is $ d = 5 + \\frac{p}{2} = 6 $, so $ p = 2 $. Therefore, the equation of the parabola is $ y^{2} = 4x $. Draw $ PN \\perp $ to the directrix of the parabola from point $ P $, intersecting the directrix at point $ N $. The directrix intersects the $ x $-axis at point $ M_{1} $. As shown in the figure, the circle $ (x - 3)^{2} + y^{2} = 1 $ has center $ M(3, 0) $ and radius $ 1 $. By the definition of the parabola, $ |PF| = |PN| $, so $ |PQ| + |PF| = |PQ| + |PN| \\geqslant |MM_{1}| - 1 = 3 $. The answer is: $ 3 $"
  },
  {
    "text": "The left and right foci of the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ are $F_{1}$, $F_{2}$ respectively, and $P$ is any point on the ellipse $E$. The range of the maximum value of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ is $[2 c^{2}, 4 c^{2}]$, where $c=\\sqrt{a^{2}-b^{2}}$. Then, what is the range of the eccentricity $e$ of the ellipse?",
    "fact_expressions": "E: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;c:Number;e: Number;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(E) =F1;RightFocus(E)=F2;PointOnCurve(P,E);Range(Max(DotProduct(VectorOf(P,F1),VectorOf(P,F2))))=[2*c^2,4*c^2];c=sqrt(a^2-b^2);Eccentricity(E) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "[sqrt(5)/5, sqrt(3)/3]",
    "fact_spans": "[[[0, 57], [86, 91], [211, 213]], [[7, 57]], [[7, 57]], [[82, 85]], [[66, 73]], [[74, 81]], [[187, 209]], [[217, 220]], [[7, 57]], [[7, 57]], [[0, 57]], [[0, 81]], [[0, 81]], [[82, 95]], [[97, 184]], [[187, 209]], [[211, 220]]]",
    "query_spans": "[[[217, 227]]]",
    "process": "From the given conditions, we have $ F_{1}(-c,0) $, $ F_{2}(c,0) $. Let $ P(x,y) $ be any point on the ellipse $ E $. Since $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $, $ \\therefore x^{2} = \\frac{a^{2}(b^{2}-y^{2})}{b^{2}(b^{2}-y)^{2}} + y^{2} - c^{2} = a^{2} - c^{2} - \\frac{c^{2}y^{2}}{b^{2}} $, $ \\because -b \\leqslant y \\leqslant b $, $ \\therefore \\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}} = x^{2} + y^{2} - c^{2} = \\frac{a^{2}(b)}{b^{2}} $. Therefore, when $ y = 0 $, $ \\overrightarrow{PF}_{1} \\cdot \\overrightarrow{PF_{2}} $ reaches its maximum value $ a^{2} - c^{2} $, i.e., $ 2c^{2} \\leqslant a^{2} - c^{2} \\leqslant 4c^{2} $, $ \\therefore \\sqrt{3}c \\leqslant a \\leqslant \\sqrt{5}c $, $ \\therefore \\frac{\\sqrt{5}}{5} \\leqslant e \\leqslant \\frac{\\sqrt{3}}{3} $"
  },
  {
    "text": "If the curve represented by the equation $x^{2}-2 m y^{2}=4$ is an ellipse, then the range of real values for $m$ is?",
    "fact_expressions": "G: Ellipse;H: Curve;Expression(H) = (-2*m*y^2 + x^2 = 4);m: Real;H = G",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-oo, 0)&(Negation(m=-1/2))",
    "fact_spans": "[[[28, 30]], [[25, 27]], [[1, 27]], [[32, 37]], [[25, 30]]]",
    "query_spans": "[[[32, 44]]]",
    "process": "By the given condition, the equation $x^{2}-2my^{2}=4$ can be rewritten as $\\frac{x^{2}}{4}+\\frac{y^{2}}{-\\frac{2}{m}}=1$, so we have $-\\frac{2}{m}>0$, $-\\frac{2}{m}\\neq4$, solving yields $m<0$ and $m\\neq-\\frac{1}{2}$."
  },
  {
    "text": "Given the parabola $y=\\frac{1}{2} x^{2}$ has focus $F$ and directrix $l$, point $M$ lies on $l$, segment $MF$ intersects the parabola at point $N$. If $|MN| = \\sqrt{2}|NF|$, then $|MF| =$ ?",
    "fact_expressions": "M: Point;F: Point;G: Parabola;N: Point;l: Line;Expression(G) = (y = x^2/2);Focus(G) = F;Directrix(G) = l;PointOnCurve(M, l);Intersection(LineSegmentOf(M, F), G) = N;Abs(LineSegmentOf(M, N)) = sqrt(2)*Abs(LineSegmentOf(N, F))",
    "query_expressions": "Abs(LineSegmentOf(M, F))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[42, 46]], [[30, 33]], [[2, 26], [59, 62]], [[64, 68]], [[37, 41], [47, 50]], [[2, 26]], [[2, 33]], [[2, 40]], [[43, 51]], [[52, 68]], [[70, 92]]]",
    "query_spans": "[[[94, 102]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has eccentricity $e=\\frac{1}{2}$, $A$, $B$ are the left and right vertices of the ellipse, $P$ is a moving point on the ellipse distinct from $A$, $B$, and the angles of inclination of lines $PA$, $PB$ are $\\alpha$, $\\beta$ respectively, then $\\frac{\\cos (\\alpha+\\beta)}{\\cos (\\alpha-\\beta)}$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;e: Number;Eccentricity(G) = e;e = 1/2;A: Point;B: Point;LeftVertex(G) = A;RightVertex(G) = B;P: Point;PointOnCurve(P, G);Negation(P=A);Negation(P=B);alpha: Number;beta: Number;Inclination(LineOf(P, A)) = alpha;Inclination(LineOf(P, B)) = beta",
    "query_expressions": "Cos(alpha + beta)/Cos(alpha - beta)",
    "answer_expressions": "7",
    "fact_spans": "[[[2, 54], [84, 86], [96, 98]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[58, 73]], [[2, 73]], [[58, 73]], [[75, 78], [102, 105]], [[80, 83], [107, 110]], [[75, 91]], [[75, 91]], [[92, 95]], [[92, 113]], [[92, 113]], [[92, 113]], [[133, 141]], [[143, 150]], [[114, 150]], [[114, 150]]]",
    "query_spans": "[[[153, 204]]]",
    "process": "By the given conditions, A(-a,0), B(a,0), let P(x,y), then $\\tan\\alpha=\\frac{y}{x+a}$, $\\tan\\beta=\\frac{y}{x-a}$, $\\therefore\\tan\\alpha\\tan\\beta=\\frac{y}{x+a}\\cdot\\frac{y}{x-a}=\\frac{y^{2}}{x^{2}-a^{2}}$. Since the eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $e=\\frac{1}{2}$, $\\therefore\\frac{a^{2}-b^{2}}{a^{2}}=\\frac{1}{4}$, $\\therefore a^{2}=\\frac{4}{3}b^{2}$, $\\frac{x^{2}}{\\frac{4}{3}b^{2}}+\\frac{y^{2}}{b^{2}}=1$, $\\therefore\\frac{\\cos\\alpha\\cos\\beta-\\sin\\alpha\\sin\\beta}{\\cos\\alpha\\cos\\beta+\\sin\\alpha\\sin\\beta}=\\tan\\beta=\\frac{1+\\frac{3}{4}}{1-\\frac{3}{4}}=7$"
  },
  {
    "text": "Let the line $y = x - 1$ intersect the ellipse $\\frac{x^{2}}{2} + y^{2} = 1$ at points $A$ and $B$. What are the coordinates of the midpoint of segment $AB$?",
    "fact_expressions": "H: Line;Expression(H) = (y = x - 1);G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);A: Point;B: Point;Intersection(H, G) = {A, B}",
    "query_expressions": "Coordinate(MidPoint(LineSegmentOf(A, B)))",
    "answer_expressions": "(2/3, -1/3)",
    "fact_spans": "[[[1, 10]], [[1, 10]], [[11, 38]], [[11, 38]], [[41, 44]], [[45, 48]], [[1, 50]]]",
    "query_spans": "[[[52, 66]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Solving the system \n\\[\n\\begin{cases}\ny = x - 1 \\\\\n\\frac{x^{2}}{2} + y^{2} = 1\n\\end{cases}\n\\]\ngives $ 3x^{2} - 4x = 0 $, so $ x_{1} + x_{2} = \\frac{4}{3} $. Then the x-coordinate of the midpoint of segment $ AB $ is $ \\frac{x_{1} + x_{2}}{2} = \\frac{4}{6} = \\frac{2}{3} $. Substituting into $ y = x - 1 $, we get $ y = \\frac{2}{3} - 1 = -\\frac{1}{3} $. Therefore, the coordinates of the midpoint of segment $ AB $ are $ \\left( \\frac{2}{3}, -\\frac{1}{3} \\right) $."
  },
  {
    "text": "The hyperbola $x^{2}-y^{2}=1$ intersects the line $x+2 y+3=0$ at points $A$ and $B$, and the midpoint of segment $AB$ is $P$, with $O$ being the origin. Then the slope of line $OP$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;P: Point;O: Origin;A: Point;B: Point;Expression(G) = (x^2 - y^2 = 1);Expression(H) = (x + 2*y + 3 = 0);Intersection(G, H) = {A, B};MidPoint(LineSegmentOf(A,B))=P",
    "query_expressions": "Slope(LineOf(O,P))",
    "answer_expressions": "-2",
    "fact_spans": "[[[0, 18]], [[19, 32]], [[56, 59]], [[60, 63]], [[34, 37]], [[38, 41]], [[0, 18]], [[19, 32]], [[0, 43]], [[46, 59]]]",
    "query_spans": "[[[70, 82]]]",
    "process": "The hyperbola $ x^{2} - y^{2} = 1 $ and the line $ x + 2y + 3 = 0 $ are solved simultaneously, yielding \\begin{cases} \\\\ x - \\end{cases}, eliminating $ x $ gives $ 3y^{2} + 12y + 8 = 0 $, $+2v+3$. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ y_{1} + y_{2} = -4 $, $ x_{1} + x_{2} = -2(y_{1} + y_{2}) - 6 = 2 $, so the coordinates of $ P $ are $ (1, -2) $. Therefore, the slope of line $ OP $ is $ -2 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1$ with foci $F_{1}$ and $F_{2}$, point $M$ lies on the hyperbola such that $M F_{1} \\perp x$-axis. Find the distance from $F_{1}$ to the line $F_{2}M$.",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/6 - y^2/3 = 1);F1: Point;F2: Point;Focus(G) = {F1,F2};M: Point;PointOnCurve(M, G) = True;IsPerpendicular(LineSegmentOf(M, F1), xAxis) = True",
    "query_expressions": "Distance(F1, LineOf(F2, M))",
    "answer_expressions": "6/5",
    "fact_spans": "[[[2, 40], [65, 68]], [[2, 40]], [[44, 51], [90, 97]], [[52, 59]], [[2, 59]], [[60, 64]], [[60, 69]], [[70, 88]]]",
    "query_spans": "[[[90, 113]]]",
    "process": ""
  },
  {
    "text": "Let the circle centered at the vertex of the parabola $ C: y^{2} = 2px \\ (p > 0) $ intersect the parabola at points $ A $ and $ B $, and intersect the directrix of the parabola at points $ D $ and $ E $. Given that $ |AB| = 4\\sqrt{2} $, $ |DE| = 2\\sqrt{5} $, then the standard equation of the parabola is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;p>0;G: Circle;Vertex(C) = Center(G);Intersection(C, G) = {A, B};A: Point;B: Point;Intersection(G, Directrix(C)) = {D, E};D: Point;E: Point;Abs(LineSegmentOf(A, B)) = 4*sqrt(2);Abs(LineSegmentOf(D, E)) = 2*sqrt(5)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=8*x",
    "fact_spans": "[[[1, 27], [36, 39], [51, 54], [110, 113]], [[1, 27]], [[9, 27]], [[9, 27]], [[34, 35]], [[0, 35]], [[34, 49]], [[40, 43]], [[44, 47]], [[34, 67]], [[58, 61]], [[62, 65]], [[70, 88]], [[90, 108]]]",
    "query_spans": "[[[110, 120]]]",
    "process": "Let line AB intersect the x-axis at point M, and line DE intersect the x-axis at point N. Then M and N are the midpoints of segments AB and DE, respectively. Given |AB| = 4\\sqrt{2}, |DE| = 2\\sqrt{5}, we have |AM| = 2\\sqrt{2}, |DN| = \\sqrt{5}. Then |ON| = \\frac{p}{2}, x_{A} = \\frac{(2\\sqrt{2})^{2}}{2p} = \\frac{4}{p}. Since |OD| = |OA|, it follows that |ON|^{2} + |DN|^{2} = |OM|^{2} + |AM|^{2}, i.e., \\frac{p^{2}}{4} + 5 = \\frac{16}{p^{2}} + 8. Solving gives p = 4. Therefore, the standard equation of the parabola is y^{2} = 8x."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $A$ is a point on the hyperbola such that $A F_{2} \\perp x$-axis. If the inradius of $\\Delta A F_{1} F_{2}$ is $(\\sqrt{3}-1) a$, then its asymptotes' equations are?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;A: Point;F1:Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(A,G);IsPerpendicular(LineSegmentOf(A,F2),xAxis);Radius(InscribedCircle(TriangleOf(A,F1,F2)))=(sqrt(3)-1)*a",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(2)*x",
    "fact_spans": "[[[2, 58], [87, 90], [162, 163]], [[5, 58]], [[144, 160]], [[83, 86]], [[67, 74]], [[75, 82]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 82]], [[2, 82]], [[83, 93]], [[95, 113]], [[115, 160]]]",
    "query_spans": "[[[162, 170]]]",
    "process": "Since point A lies on the hyperbola and $ AF_{2} \\perp x $-axis, A is on the right branch of the hyperbola. By the definition of the hyperbola, we obtain $ |AF_{1}| \\cdot |AF_{2}| = 2a $. Let the inradius of right triangle $ \\triangle AF_{1}F_{2} $ be $ r $. Using equal areas, we have $ S_{\\triangle AF_{1}F_{2}} = \\frac{1}{2}|AF_{2}||F_{1}F_{2}| = \\frac{1}{2}r(|AF_{1}| + |AF_{2}| + |F_{1}F_{2}|) $. By the Pythagorean theorem, $ |AF_{2}|^{2} + |F_{1}F_{2}|^{2} = |AF_{1}|^{2} $, solving gives $ r = \\frac{|AF_{2}| + |F_{1}F_{2}| - |AF_{1}|}{2} = \\frac{2c - 2a}{2} = c - a = (\\sqrt{3} - 1)a \\Rightarrow c = \\sqrt{3}a $, thus $ b = \\sqrt{2}a $. Therefore, the asymptotes are $ y = \\pm \\sqrt{2}x $."
  },
  {
    "text": "The product of the distances from a point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ to its two foci is $m$. Then the maximum value of $m$ is?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/25 + y^2/9 = 1);PointOnCurve(P, G);F1:Point;F2:Point;Focus(G)={F1,F2};Distance(P, F1)*Distance(P,F2) = m;m:Number",
    "query_expressions": "Max(m)",
    "answer_expressions": "25",
    "fact_spans": "[[[0, 38]], [[41, 44]], [[0, 38]], [[0, 44]], [], [], [[0, 48]], [[0, 57]], [[54, 57], [59, 62]]]",
    "query_spans": "[[[59, 68]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, one of its asymptotes is $y=x$, and the distance from a focus to an asymptote is $\\sqrt{2}$. Then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Distance(OneOf(Focus(G)),OneOf(Asymptote(G)))=sqrt(2);Expression(OneOf(Asymptote(G))) = (y = x)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/2 - y^2/2 = 1",
    "fact_spans": "[[[2, 58], [101, 104]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 99]], [[2, 73]]]",
    "query_spans": "[[[101, 111]]]",
    "process": "Since the asymptote equation is y = x, we have a = b. The distance from one focus to an asymptote is \\sqrt{2}, so \\frac{|bc|}{\\sqrt{1+\\frac{b^{2}}{a^{2}}}}=b=\\sqrt{2}. Hence, the standard equation of the hyperbola is \\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1."
  },
  {
    "text": "If the asymptotes of a hyperbola are given by $y = \\pm 3x$, and the coordinates of one of its foci are $(\\sqrt{10}, 0)$, then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Coordinate(OneOf(Focus(G))) = (sqrt(10), 0);Expression(Asymptote(G)) = (y = pm*3*x)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/9 = 1",
    "fact_spans": "[[[1, 4], [23, 24], [52, 55]], [[23, 49]], [[1, 22]]]",
    "query_spans": "[[[52, 62]]]",
    "process": "From the asymptote equations of the hyperbola, we know that $ b/a = 3 $\\textcircled{1}. Since one of its foci is $ (\\sqrt{10}, 0) $, it follows that $ c = \\sqrt{10} $\\textcircled{2}. Also, $ c^2 = a^2 + b^2 $\\textcircled{3}. Solving \\textcircled{1}\\textcircled{2}\\textcircled{3} simultaneously, we obtain $ a^2 = 1 $, $ b^2 = 9 $. Therefore, the equation of the hyperbola is $ x^2 - y^2/9 = 1 $."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, respectively. If $P$ is a moving point on this ellipse, then the minimum value of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G)",
    "query_expressions": "Min(DotProduct(VectorOf(P, F1), VectorOf(P, F2)))",
    "answer_expressions": "-2",
    "fact_spans": "[[[19, 46], [59, 61]], [[54, 57]], [[1, 8]], [[9, 16]], [[19, 46]], [[1, 52]], [[1, 52]], [[54, 67]]]",
    "query_spans": "[[[69, 132]]]",
    "process": "P(x,y),F_{1}(-\\sqrt{3},0),F_{2}(\\sqrt{3},0),\\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=(-\\sqrt{3}-x,-y)(\\sqrt{3}-x,-y)=x^{2}-3+y^{2}=x^{2}-3+1-\\frac{x^{2}}{4}=\\frac{3}{4}x^{2}-2,x\\in[-2,2] so when x=0, the minimum value of \\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}} is -2, therefore fill in: -2"
  },
  {
    "text": "What is the distance from a point $P\\left(\\frac{1}{2}, 2\\right)$ on the parabola $y^{2}=8x$ to its focus?",
    "fact_expressions": "G: Parabola;P: Point;Expression(G) = (y^2 = 8*x);Coordinate(P) = (1/2, 2);PointOnCurve(P,G)",
    "query_expressions": "Distance(P, Focus(G))",
    "answer_expressions": "5/2",
    "fact_spans": "[[[0, 14]], [[17, 37]], [[0, 14]], [[17, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 45]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus $F(-c, 0)$, the right vertex $A(a, 0)$, the upper vertex $B(0, b)$, satisfying $\\overrightarrow{F B} \\cdot \\overrightarrow{A B}=0$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F: Point;A: Point;B: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (-c, 0);c:Number;Coordinate(A) = (a, 0);Coordinate(B) = (0, b);LeftFocus(G)=F;RightVertex(G)=A;UpperVertex(G)=B;DotProduct(VectorOf(F, B), VectorOf(A, B)) = 0",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(\\sqrt{5}-1)/2",
    "fact_spans": "[[[2, 54], [150, 152]], [[4, 54]], [[4, 54]], [[58, 68]], [[72, 81]], [[85, 94]], [[4, 54]], [[4, 54]], [[2, 54]], [[58, 68]], [[58, 68]], [[72, 81]], [[85, 94]], [[2, 68]], [[2, 81]], [[2, 94]], [[97, 148]]]",
    "query_spans": "[[[150, 158]]]",
    "process": "Using the coordinate formula of the dot product, we obtain the answer. From $\\overrightarrow{FB}\\cdot\\overrightarrow{AB}=0$, we get $(c,b)\\cdot(-a,b)=0$, that is, $ac=b^{2}=a^{2}-c^{2}$. Then $e^{2}+e-1=0$, solving yields $e=\\frac{\\sqrt{5}-1}{2}$ or $\\frac{\\sqrt{5}-1}{2}$ (discarded)."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the hyperbola $\\frac{x^{2}}{m^{2}}-\\frac{y^{2}}{n^{2}}=1$ $(m>0, n>0)$ have the same foci $F_{1}$, $F_{2}$, and $P$ is a common point of the ellipse and the hyperbola. The eccentricities of the ellipse and the hyperbola are $e_{1}$, $e_{2}$ respectively, $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$, and $e_{2} \\in[\\sqrt{2}, \\sqrt{3}]$, then the range of values for $e_{1}$ is?",
    "fact_expressions": "H: Ellipse;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;G: Hyperbola;Expression(G) = (-y^2/n^2 + x^2/m^2 = 1);m: Number;n: Number;m>0;n>0;F1: Point;F2: Point;Focus(H) = {F1, F2};Focus(G) = {F1, F2};P: Point;OneOf(Intersection(H, G)) = P;e1: Number;e2: Number;Eccentricity(H) = e1;Eccentricity(G) = e2;AngleOf(F1, P, F2) = pi/3;In(e2, [sqrt(2), sqrt(3)])",
    "query_expressions": "Range(e1)",
    "answer_expressions": "[sqrt(3)/3, sqrt(10)/5]",
    "fact_spans": "[[[2, 54], [137, 139], [150, 152]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[55, 111], [140, 143], [153, 156]], [[55, 111]], [[58, 111]], [[58, 111]], [[58, 111]], [[58, 111]], [[117, 124]], [[125, 132]], [[2, 132]], [[2, 132]], [[133, 136]], [[133, 149]], [[163, 170], [252, 259]], [[172, 179]], [[150, 179]], [[150, 179]], [[181, 217]], [[219, 250]]]",
    "query_spans": "[[[252, 266]]]",
    "process": "From the given conditions: |PF_{1}|-|PF_{2}|=2m, |PF_{1}|+|PF_{2}|=2a, F_{1}F_{2}=2c, a^{2}-b^{2}=m^{2}+n^{2}=c^{2}, solving yields |PF_{1}|=a+m, |PF_{2}|=a-m. By the law of cosines: \\cos\\angle F_{1}PF_{2}=\\frac{PF_{1}^{2}+PF_{2}^{2}-F_{1}F_{2}^{2}}{2\\cdot PF_{1}\\cdot PF_{2}}=\\frac{2a^{2}+2m^{2}-4c^{2}}{2a^{2}-2m^{2}}=\\frac{1}{2}, solving gives: a^{2}+3m^{2}=4c^{2}. Since \\frac{2}{2}12=c^{2}, solving yields: n^{2}=\\frac{1}{3}b^{2}, m^{2}=a^{2}-\\frac{4}{3}b^{2}. Because e_{2}=\\sqrt{1+\\frac{n^{2}}{m^{2}}}\\in[\\sqrt{2},\\sqrt{3}], that is"
  },
  {
    "text": "The vertex of the parabola $C$ is at the origin, and its focus lies on the $x$-axis. The line $y = x$ intersects the parabola $C$ at points $A$ and $B$. If $F(2,2)$ is the midpoint of $AB$, then the equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;O: Origin;Vertex(C) = O;PointOnCurve(Focus(C), xAxis);G: Line;Expression(G) = (y = x);A: Point;B: Point;Intersection(G, C) = {A, B};F: Point;Coordinate(F) = (2, 2);MidPoint(LineSegmentOf(A, B)) = F",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[2, 8], [34, 40], [72, 78]], [[14, 16]], [[2, 16]], [[2, 25]], [[26, 33]], [[26, 33]], [[42, 45]], [[46, 49]], [[26, 51]], [[53, 61]], [[53, 61]], [[53, 70]]]",
    "query_spans": "[[[72, 83]]]",
    "process": ""
  },
  {
    "text": "Given the line $l$: $3x - y - 1 = 0$ and the parabola $C$: $y^2 = 3x$ intersect at points $M$ and $N$, and $O$ is the origin, then the area of $\\triangle OMN$ is?",
    "fact_expressions": "C: Parabola;l: Line;O: Origin;M: Point;N: Point;Expression(C) = (y^2 = 3*x);Expression(l) = (3*x - y - 1 = 0);Intersection(l, C) = {M, N}",
    "query_expressions": "Area(TriangleOf(O, M, N))",
    "answer_expressions": "sqrt(5)/6",
    "fact_spans": "[[[21, 40]], [[2, 20]], [[52, 55]], [[42, 45]], [[46, 49]], [[21, 40]], [[2, 20]], [[2, 51]]]",
    "query_spans": "[[[62, 84]]]",
    "process": "Let line $ l: 3x - y - 1 = 0 $ intersect the x-axis at point $ A $, $ S = \\frac{1}{2} \\cdot |OA| \\cdot |y_{1} - y_{2}| $, solve by combining equations. Combining  \n\\[\n\\begin{cases}\ny^{2} = 3x, \\\\\n3x = y + 1,\n\\end{cases}\n\\]\nthen $ y^{2} - y - 1 = 0 $. Let $ M(x_{1}, y_{1}) $, $ N(x_{2}, y_{2}) $. Line $ l: 3x - y - 1 = 0 $ intersects the x-axis at point $ A $, then $ y_{1} + y_{2} = 1 $, $ y_{1} y_{2} = -1 $. Hence the area of $ \\triangle OMN $ is $ S = \\frac{1}{2} \\cdot |OA| \\cdot |y_{1} - y_{2}| = \\frac{1}{2} \\times \\frac{1}{3} \\times \\sqrt{1 + 4} = \\frac{\\sqrt{5}}{6} $."
  },
  {
    "text": "A line passing through the left focus $F$ of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ and not perpendicular to the $x$-axis intersects the ellipse at points $A$ and $B$. The perpendicular bisector of $AB$ intersects the $x$-axis at point $N$. Then $\\frac{|N F|}{|A B|}$=?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;N: Point;F: Point;Expression(G) = (x^2/9 + y^2/5 = 1);LeftFocus(G) = F;PointOnCurve(F, H);Negation(IsPerpendicular(xAxis, H));Intersection(H, G) = {A, B};Intersection(PerpendicularBisector(LineSegmentOf(A, B)), xAxis) = N",
    "query_expressions": "Abs(LineSegmentOf(N, F))/Abs(LineSegmentOf(A, B))",
    "answer_expressions": "1/3",
    "fact_spans": "[[[1, 38], [57, 59]], [[54, 56]], [[60, 63]], [[64, 67]], [[87, 91]], [[41, 44]], [[1, 38]], [[1, 44]], [[0, 56]], [[45, 56]], [[54, 69]], [[70, 91]]]",
    "query_spans": "[[[93, 116]]]",
    "process": "From the ellipse equation, the left focus is $ F(-2,0) $. Let the equation of line $ AB $ be $ y = k(x - 2) $. Substituting the equation of line $ AB $ into the ellipse equation, eliminating $ y $, and simplifying yields $ \\left( \\frac{1}{9} + \\frac{1}{5}k^{2} \\right)x^{2} - \\frac{4}{5}k^{2}x + \\frac{4}{5}k^{2} - 1 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, so $ x_{1} + x_{2} = \\frac{\\frac{4}{5}k^{2}}{\\frac{1}{9} + \\frac{1}{5}k^{2}} $, $ x_{1}x_{2} = \\frac{\\frac{4}{5}k^{2} - 1}{\\frac{1}{0} + \\frac{1}{5}k^{2}} $. By the chord length formula, $ |AB| = \\frac{2}{3}\\frac{1 + k^{2}}{9} + \\frac{1}{5}k^{2} $. Substituting $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $ into the ellipse equation and subtracting the two equations gives $ \\frac{1}{9}(x_{1} + x_{2})(x_{1} - x_{2}) + \\frac{1}{5}(y_{1} + y_{2})(y_{1} - y_{2}) = 0 $. Let the midpoint of $ AB $ be $ M(x_{0}, y_{0}) $, then $ y_{0}k = -\\frac{5}{9}x_{0} $. The perpendicular bisector of $ AB $ is $ y = -\\frac{1}{k}(x - |NF| = 2 - \\frac{4}{9}x_{0} = \\frac{2}{9}\\frac{1 + k^{2}}{1 + \\frac{1}{4}k^{2}} $, so $ \\frac{|NF|}{|AB|} = \\frac{1}{3} $. $ -\\frac{1}{k}(x - x_{0}) + y_{0} $, setting $ y = 0 $ gives $ x = ky_{0} + x_{0} = \\frac{4}{9}x_{0} $. So key points: 1. Chord problem of line intersecting ellipse; 2. Point difference method solving midpoint ☆ problem"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$, and $P$ is a point on the ellipse, if $S_{\\triangle F_{1} P F_{2}}=4$, then what is $\\angle F_{1} P F_{2}$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/8 + y^2/4 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);Area(TriangleOf(F1, P, F2)) = 4",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "ApplyUnit(90, degree)",
    "fact_spans": "[[[18, 55], [66, 68]], [[18, 55]], [[2, 9]], [[10, 17]], [[2, 61]], [[2, 61]], [[62, 65]], [[62, 71]], [[73, 104]]]",
    "query_spans": "[[[106, 131]]]",
    "process": "By the definition of the ellipse, $|PF_{1}|+|PF_{2}|=4\\sqrt{2}$, and $|F_{1}F_{2}|^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}||PF_{2}|\\cos\\angle F_{1}PF_{2}$. Also, $|F_{1}F_{2}|=4$, and using the triangle area formula gives the result. From the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$, we have $a=2\\sqrt{2}$, $b=2$, $c=2$. By the definition of the ellipse, $|PF_{1}|+|PF_{2}|=4\\sqrt{2}$, and $|F_{1}F_{2}|=4$. In $\\Delta F_{1}PF_{2}$, $|F_{1}F_{2}|^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}||PF_{2}|\\cos\\angle F_{1}PF_{2}$, so $16=(4\\sqrt{2})^{2}-2|PF_{1}||PF_{2}|-2|PF_{1}||PF_{2}|\\cos\\angle F_{1}PF_{2}$ ①. Also, $S_{\\Delta F_{1}PF_{2}}=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|\\sin\\angle F_{1}PF_{2}=4$ ②. Let $|PF_{1}|\\cdot|PF_{2}|=t$ $(t>0)$. Equation ① simplifies to: $(1+\\cos\\angle F_{1}PF_{2})t=8$ ③. Equation ② simplifies to: $t\\sin\\angle F_{1}PF_{2}=8$ ④. From ③ and ④: $\\sin\\angle F_{1}PF_{2}=\\cos\\angle F_{1}PF_{2}+1 \\Rightarrow \\sin\\angle F_{1}PF_{2}-\\cos\\angle F_{1}PF_{2}=1$, $\\sqrt{2}\\sin(\\angle F_{1}PF_{2}-45^{\\circ})=1 \\Rightarrow \\sin(\\angle F_{1}PF_{2}-45^{\\circ})=\\frac{\\sqrt{2}}{2}$. Since $0^{\\circ}<\\angle F_{1}PF_{2}<180^{\\circ}$, it follows that $\\angle F_{1}PF_{2}-45^{\\circ}=45^{\\circ}$, so $\\angle F_{1}PF_{2}=90^{\\circ}$."
  },
  {
    "text": "Given that the asymptotes of a hyperbola with foci on the $x$-axis are $y=\\pm \\frac{3}{4} x$, what is the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), xAxis);Expression(Asymptote(G)) = (y = pm*(3/4)*x)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/4",
    "fact_spans": "[[[11, 14], [45, 48]], [[2, 14]], [[11, 42]]]",
    "query_spans": "[[[45, 54]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{2}$ intersects the ellipse at points $A$ and $B$. If $F_{2}$ is the midpoint of segment $AB$, then the area of $\\triangle A F_{1} B$ is?",
    "fact_expressions": "G: Ellipse;H: Line;B: Point;A: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, H);Intersection(H, G) = {A, B};MidPoint(LineSegmentOf(A,B)) = F2",
    "query_expressions": "Area(TriangleOf(A, F1, B))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 29], [65, 67]], [[62, 64]], [[72, 75]], [[68, 71]], [[37, 44]], [[54, 61], [45, 52], [79, 86]], [[2, 29]], [[2, 52]], [[2, 52]], [[53, 64]], [[62, 77]], [[79, 97]]]",
    "query_spans": "[[[99, 125]]]",
    "process": "Since a line passing through $F_{2}$ intersects the ellipse at points $A$ and $B$, and $F_{2}$ is the midpoint of segment $AB$, $AB$ is perpendicular to the $x$-axis. According to the problem, $a=2$, $b=1$, $c=\\sqrt{a^{2}-b^{2}}=\\sqrt{3}$, so $F_{1}F_{2}=2c=2\\sqrt{3}$, $F_{1}(-\\sqrt{3},0)$, $F_{2}(\\sqrt{3},0)$. Let $A(\\sqrt{3},y_{1})$, $B(\\sqrt{3},y_{2})$, and $A$ is in the first quadrant, so $AB=|y_{1}-y_{2}|$. Substituting point $A(\\sqrt{3},y_{1})$ into the ellipse equation gives: $\\frac{3}{4}+y_{1}^{2}=1$, so $y_{1}=\\frac{1}{2}$. By symmetry, $y_{2}=-\\frac{1}{2}$, so $AB=|y_{1}-y_{2}|=1$. In $\\triangle AF_{1}B$, $F_{1}F_{2}\\bot AB$, so $S_{\\triangle AF_{1}B}=\\frac{1}{2}\\times F_{1}F_{2}\\times AB=\\sqrt{3}$."
  },
  {
    "text": "Given point $P$ is a moving point on the parabola $C$: $y^{2}=4x$, the focus of parabola $C$ is $F$, and point $A(3,1)$. Then the minimum value of $|PA|+|PF|$ is?",
    "fact_expressions": "C: Parabola;A: Point;P: Point;F: Point;Expression(C) = (y^2 = 4*x);Coordinate(A) = (3, 1);PointOnCurve(P, C);Focus(C) = F",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "4",
    "fact_spans": "[[[7, 26], [31, 37]], [[46, 55]], [[2, 6]], [[41, 44]], [[7, 26]], [[46, 55]], [[2, 30]], [[31, 44]]]",
    "query_spans": "[[[57, 76]]]",
    "process": "Let the projection of point P onto the directrix be D. Then, according to the definition of a parabola, |PF| = |PD|, and thus the problem is transformed into finding the minimum value of |PA| + |PD|. It can be inferred that |PA| + |PD| is minimized when points D, P, and A are collinear. The answer is obtained as follows: the directrix of the parabola C: y^{2} = 4x is x = -1. Let the projection of point P onto the directrix be D, as shown in the figure. Then, by the definition of the parabola, |PF| = |PD|. To minimize |PA| + |PF|, we need to minimize |PA| + |PD|. When D, P, and A are collinear, |PA| + |PD| is minimized, and its value is 3 - (-1) = 4."
  },
  {
    "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are given by $y=\\pm \\frac{2 \\sqrt{6}}{5} x$, and the hyperbola passes through the point $(5,3 \\sqrt{2})$. What is its focal distance?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Expression(Asymptote(G)) = (y = pm*x*2*sqrt(6)/5);H: Point;Coordinate(H) = (5, 3*sqrt(2));PointOnCurve(H,G) = True",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "7",
    "fact_spans": "[[[0, 56], [115, 116]], [[0, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[0, 93]], [[96, 113]], [[96, 113]], [[0, 113]]]",
    "query_spans": "[[[115, 120]]]",
    "process": "Given: The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ has asymptotes $y=\\pm\\frac{\\sqrt[2]{6}}{5}x$, and passes through the point $(5,3\\sqrt{2})$, that is, $\\left|\\frac{b}{a}=\\frac{2\\sqrt{6}}{\\frac{18}{a^{2}}-\\frac{18}{b^{2}}}\\right|=1^{\\frac{1}{1}}$. Solving gives: $\\begin{cases}a=\\frac{5}{2}\\\\b=\\sqrt{6}\\end{cases}$, so $c=\\sqrt{a^{2}+b^{2}}=\\frac{7}{2}$, therefore its focal distance is $7$."
  },
  {
    "text": "Given that the distance from point $P(2,0)$ to the asymptotes of the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is less than $\\sqrt{2}$, what is the range of values for the eccentricity of the hyperbola?",
    "fact_expressions": "E: Hyperbola;b: Number;a: Number;P: Point;a>0;b>0;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (2, 0);Distance(P, Asymptote(E))<sqrt(2)",
    "query_expressions": "Range(Eccentricity(E))",
    "answer_expressions": "(1, \\sqrt{2})",
    "fact_spans": "[[[12, 73], [94, 97]], [[19, 73]], [[19, 73]], [[2, 11]], [[19, 73]], [[19, 73]], [[12, 73]], [[2, 11]], [[2, 92]]]",
    "query_spans": "[[[94, 107]]]",
    "process": "From the condition we get: \\frac{2b}{\\sqrt{a2+b^{2}}}<\\sqrt{2}\\Rightarrow\\frac{2b}{c}<\\sqrt{2}, thus obtaining e^{2}<2, so the range of the hyperbola's eccentricity is (1,\\sqrt{2})"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ has a semi-focal length of $c$. If $b^{2}-4 a c<0$, then what is the range of its eccentricity?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;c: Number;HalfFocalLength(G) = c;-4*a*c + b^2 < 0",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1,2+sqrt(5))",
    "fact_spans": "[[[2, 59], [86, 87]], [[2, 59]], [[69, 84]], [[69, 84]], [[5, 59]], [[5, 59]], [[64, 67]], [[2, 67]], [[69, 84]]]",
    "query_spans": "[[[86, 98]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{y^{2}}{5}+\\frac{x^{2}}{4}=1$, $F_{1}$ and $F_{2}$ are the two foci of the ellipse, and $\\angle F_{1} P F_{2}=30^{\\circ}$, then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 + y^2/5 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2};AngleOf(F1, P, F2) = ApplyUnit(30, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "8 - 4*sqrt(3)",
    "fact_spans": "[[[6, 43], [64, 66]], [[48, 55]], [[2, 5]], [[56, 63]], [[6, 43]], [[2, 47]], [[48, 71]], [[73, 106]]]",
    "query_spans": "[[[108, 135]]]",
    "process": "From the ellipse $\\frac{y^{2}}{5}+\\frac{x^{2}}{4}=1$, we get $a=\\sqrt{5}$, $b=2$, then $2a=2\\sqrt{5}$, $c=\\sqrt{a^{2}-b^{2}}=1$, $\\therefore|PF_{1}|+|PF_{2}|=2a=2\\sqrt{5}$, $|F_{1}F_{2}|=2c=2$. By the cosine theorem, we have: $|F_{1}F_{2}^{2}|=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}||PF_{2}|\\cos30^{\\circ}$. $\\therefore4c^{2}=(|PF_{1}|+|PF_{2}|)^{2}-2|PF_{1}||PF_{2}|-\\sqrt{3}|PF_{1}||PF_{2}|$, that is $|PF|_{1}|_{1}|PF_{2}|=\\frac{16}{2+\\sqrt{3}}=16(2-\\sqrt{3})$. $\\therefore$ the area of $\\triangle F_{1}PF_{2}$ is $S=\\frac{1}{2}|PF_{1}||PF_{2}|\\sin30^{\\circ}=\\frac{1}{2}\\times16(2-\\sqrt{3})\\times\\frac{1}{2}=4(2-\\sqrt{3})=8-4\\sqrt{3}$."
  },
  {
    "text": "If a point $P$ on the ellipse $\\frac{x^{2}}{100}+\\frac{y^{2}}{36}=1$ has a distance of $6$ to the focus $F_{1}$, then what is the distance from point $P$ to the other focus $F_{2}$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2:Point;Expression(G) = (x^2/100 + y^2/36 = 1);PointOnCurve(P, G);OneOf(Focus(G)) =F1;OneOf(Focus(G)) = F2;Negation(F1= F2);Distance(P, F1) = 6",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 42]], [[45, 48], [69, 73]], [[51, 58]], [[79, 86]], [[2, 42]], [[2, 48]], [[2, 58]], [[2, 86]], [[2, 86]], [[45, 66]]]",
    "query_spans": "[[[69, 91]]]",
    "process": ""
  },
  {
    "text": "If a moving point $(x, y)$ varies on the curve $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, then what is the maximum value of $x^{2}+2 y$?",
    "fact_expressions": "G: Curve;b: Number;H: Point;b>0;Expression(G) = (x^2/4 + y^2/b^2 = 1);Coordinate(H) = (x1, y1);x1:Number;y1:Number;PointOnCurve(H,G)",
    "query_expressions": "Max(x1^2+2*y1)",
    "answer_expressions": "{(b^2/4+4)&(0,4],(2*b)&(b>4)}",
    "fact_spans": "[[[12, 58]], [[14, 58]], [[3, 11]], [[14, 58]], [[12, 58]], [[3, 11]], [[3, 11]], [[3, 11]], [[3, 59]]]",
    "query_spans": "[[[62, 79]]]",
    "process": "First, derive the functional relationship of $x^{2}+2y$ with respect to $y$ from the ellipse equation, paying special attention to $-b\\leqslant y\\leqslant b$, then use the method for finding maximum and minimum values of a quadratic function on a closed interval, and discuss by cases. [Detailed solution] Since $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, then $x^{2}+2y=-\\frac{4}{b^{2}}y^{2}+2y+4=-\\frac{4}{b^{2}}(y-\\frac{b^{2}}{4})^{2}+4+\\frac{b^{2}}{4}$ $(-b\\leqslant y\\leqslant b)$. Let $f(y)=-\\frac{4}{b^{2}}y^{2}+2y+4=-\\frac{4}{b^{2}}(y-\\frac{b^{2}}{4})^{2}+4+\\frac{b^{2}}{4}$, $(-b\\leqslant y\\leqslant b)$. When $\\frac{b^{2}}{4}\\leqslant b$, i.e., $0<b\\leqslant 4$, $f(y)_{\\max}=f(\\frac{b^{2}}{4})=4+\\frac{b^{2}}{4}$. When $\\frac{b^{2}}{4}>b$, i.e., $b>4$, the function $f(y)$ is increasing on $[-b,b]$, so $f(y)_{\\max}=f(b)=2b$. In summary, the maximum value of $x^{2}+2y$ is $\\begin{cases} \\frac{b^{2}}{4}+4, & 0<b\\leqslant 4 \\\\ 2b, & b>4 \\end{cases}$"
  },
  {
    "text": "Let $P$ be an arbitrary point on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and $F_{1}$ be one of the foci of the ellipse. Then, what is the range of values for $|PF_{1}|$?",
    "fact_expressions": "P: Point;G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;PointOnCurve(P, G);F1: Point;OneOf(Focus(G)) = F1",
    "query_expressions": "Range(Abs(LineSegmentOf(P, F1)))",
    "answer_expressions": "[a-sqrt(a^2-b^2),a+sqrt(a^2-b^2)]",
    "fact_spans": "[[[1, 4]], [[5, 57], [72, 74]], [[5, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[1, 63]], [[64, 71]], [[64, 79]]]",
    "query_spans": "[[[81, 98]]]",
    "process": ""
  },
  {
    "text": "The right focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $F_{1}$, and the right directrix is $l_{1}$. If the length of the chord passing through point $F_{1}$ and perpendicular to the $x$-axis is equal to the distance from point $F_{1}$ to $l_{1}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;l1: Line;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F1;RightDirectrix(G) = l1;f: LineSegment;PointOnCurve(F1, f);IsPerpendicular(f, xAxis);IsChordOf(f, G);Length(f) = Distance(F1, l1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[0, 52], [123, 125]], [[2, 52]], [[2, 52]], [[57, 64], [79, 87], [102, 110]], [[69, 76], [111, 118]], [[2, 52]], [[2, 52]], [[0, 52]], [[0, 64]], [[0, 76]], [], [[0, 97]], [[0, 97]], [[0, 97]], [[0, 121]]]",
    "query_spans": "[[[123, 131]]]",
    "process": "Let the chord AB passing through point $ F_{1} $ and perpendicular to the x-axis be given, and let $ A(x_{1},y_{1}) $, then $ x_{1}=c $. Since point A lies on the ellipse, $ \\frac{c^{2}}{a^{2}}+\\frac{y_{1}^{2}}{b^{2}}=1 $, so $ y_{1}^{2}=\\frac{b^{4}}{a^{2}} $, the chord length $ |AB|=\\frac{2b^{2}}{a} $. The distance from point $ F_{1} $ to $ l_{1} $ is $ \\frac{a^{2}}{c}-c=\\frac{b^{2}}{c} $. From the given condition, $ \\frac{2b^{2}}{a}=\\frac{b^{2}}{c} $, solving yields $ a=2c $, so the eccentricity of the ellipse $ e=\\frac{c}{a}=\\frac{1}{2} $."
  },
  {
    "text": "Given point $A(0,-2)$, line $l$: $y=2$, then the equation of the locus of the centers of the circles passing through point $A$ and tangent to line $l$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (0, -2);l: Line;Expression(l) = (y = 2);PointOnCurve(A, G) = True;IsTangent(l, G) = True;G: Circle",
    "query_expressions": "LocusEquation(Center(G))",
    "answer_expressions": "x^2 = -8y",
    "fact_spans": "[[[2, 12], [28, 32]], [[2, 12]], [[13, 25], [34, 39]], [[13, 25]], [[27, 43]], [[33, 43]], [[42, 43]]]",
    "query_spans": "[[[42, 53]]]",
    "process": "Let the center of the known circle be C(x, y). Since the circle C passing through point A is tangent to the line l: y = 2, we have |CA| = d, where d is the distance from point C to the line. Thus, the locus of C is a parabola with the origin as the vertex, A as the focus, and l as the directrix. Therefore, the equation of the locus of the center of the moving circle is x^{2} = -8y."
  },
  {
    "text": "Let the two foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ be $F_{1}$ and $F_{2}$, respectively. Point $P$ lies on the ellipse such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, and $\\tan \\angle P F_{1} F_{2}=\\frac{\\sqrt{3}}{3}$. Then the eccentricity of this ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Focus(G)={F1,F2};PointOnCurve(P, G);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Tan(AngleOf(P,F1,F2))=sqrt(3)/3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[1, 53], [82, 84], [198, 200]], [[3, 53]], [[3, 53]], [[77, 81]], [[61, 68]], [[69, 76]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 76]], [[77, 85]], [[87, 146]], [[148, 195]]]",
    "query_spans": "[[[198, 206]]]",
    "process": "In \\triangleF_{1}PF_{2}, \\because\\overrightarrow{PF}\\cdot\\overrightarrow{PF_{2}}=0, \\therefore\\angleF_{1}PF_{2}=90^{0} \\because\\tan\\anglePF_{1}F_{2}=\\frac{\\sqrt{3}}{3}, \\therefore\\anglePF_{1}F_{2}=30^{0}, let |F_{1}F_{2}|=2c, then |PF_{2}|=c, |PF_{1}|=\\sqrt{3}c, \\therefore2a=|PF_{2}|+|PF_{1}|=(1+\\sqrt{3})c, \\therefore e=\\frac{2c}{2a}=\\frac{2c}{(1+\\sqrt{3})c}=\\sqrt{3}-1"
  },
  {
    "text": "Given the parabola $y^{2}=4 x$, a line passing through the point $P(4 , 0)$ intersects the parabola at two points $A(x_{1} , y_{1})$ and $B(x_{2}, y_{2})$. Then the minimum value of $y_{1}^{2}+y_{2}^{2}$ is?",
    "fact_expressions": "C:Parabola;Expression(C)=(y^2=4*x);P:Point;Coordinate(P)=(4,0);L:Line;PointOnCurve(P,L);A:Point;B:Point;x1:Number;x2:Number;y1:Number;y2:Number;Coordinate(A)=(x1,y1);Coordinate(B)=(x2,y2);Intersection(L,C)={A,B}",
    "query_expressions": "Min(y1^2+y2^2)",
    "answer_expressions": "32",
    "fact_spans": "[[[2, 16], [33, 36]], [[2, 16]], [[18, 29]], [[19, 29]], [[30, 32]], [[17, 32]], [[39, 57]], [[60, 77]], [[39, 57]], [[60, 77]], [[81, 102]], [[81, 102]], [[39, 57]], [[60, 77]], [[30, 79]]]",
    "query_spans": "[[[81, 108]]]",
    "process": ""
  },
  {
    "text": "Let the directrix of the parabola $y^{2}=8x$ intersect the $x$-axis at point $Q$. If a line $l$ passing through point $Q$ has a common point with the parabola, then the range of the slope of line $l$ is?",
    "fact_expressions": "l: Line;G: Parabola;Q: Point;Expression(G) = (y^2 = 8*x);Intersection(Directrix(G), xAxis) = Q;PointOnCurve(Q, l);IsIntersect(l,G)",
    "query_expressions": "Range(Slope(l))",
    "answer_expressions": "[-1,1]",
    "fact_spans": "[[[37, 42], [52, 57]], [[1, 15], [43, 46]], [[25, 29], [32, 36]], [[1, 15]], [[1, 29]], [[31, 42]], [[37, 50]]]",
    "query_spans": "[[[52, 67]]]",
    "process": "The directrix of the parabola $ y^{2} = 8x $ is $ x = -2 $, so $ Q(-2,0) $. It is clear that the slope of line $ l $ must exist; thus, let the equation of line $ l $ be $ y = k(x + 2) $. Combining this with the parabola's equation and eliminating $ x $, we obtain $ ky^{2} - 8y + 16k = 0 $. When $ k = 0 $, line $ l $ intersects the parabola at one point; when $ k \\neq 0 $, from $ \\Delta = 64 - 64k^{2} \\geqslant 0 $, we solve to get $ -1 \\leqslant k \\leqslant 1 $, and $ k \\neq 0 $. In summary, $ -1 \\leqslant k \\leqslant 1 $."
  },
  {
    "text": "$P$ is a moving point on the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F$ is a focus of the ellipse $C$, the maximum value of $|P F|$ is $5$, and the minimum value is $1$. What is the length of the minor axis of the ellipse $C$?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;F: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, C);OneOf(Focus(C))=F;Max(Abs(LineSegmentOf(P, F))) = 5;Min(Abs(LineSegmentOf(P,F)))=1",
    "query_expressions": "Length(MinorAxis(C))",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[4, 61], [71, 76], [71, 76]], [[10, 61]], [[10, 61]], [[0, 3]], [[67, 70]], [[10, 61]], [[10, 61]], [[4, 61]], [[0, 66]], [[67, 81]], [[82, 97]], [[82, 105]]]",
    "query_spans": "[[[107, 118]]]",
    "process": "According to the positional relationship of the point on the ellipse, determine the positions where |PF| reaches its maximum and minimum values, then find a and c, and subsequently obtain the length of the minor axis. [Solution] F is a focus of ellipse C, so the maximum value of |PF| is a + c = 5, and the minimum value of |PF| is a - c = 1. Therefore, a = 3, c = 2. Thus, the length of the minor axis is 2b = 2\\sqrt{3^{2} - 2^{2}} = 2\\sqrt{5}."
  },
  {
    "text": "The distance from the point $(0,1)$ to the asymptote of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ is?",
    "fact_expressions": "G: Hyperbola;H: Point;Expression(G) = (x^2 - y^2/3 = 1);Coordinate(H) = (0, 1)",
    "query_expressions": "Distance(H, Asymptote(G))",
    "answer_expressions": "1/2",
    "fact_spans": "[[[9, 37]], [[0, 8]], [[9, 37]], [[0, 8]]]",
    "query_spans": "[[[0, 46]]]",
    "process": "\\because the asymptote equation is \\sqrt{3}x\\pmy=0, \\therefore the distance from the point (0,1) to the line is d=\\frac{|1|}{\\sqrt{3+1}}=\\frac{1}{2},"
  },
  {
    "text": "Given that one of the asymptotes of a hyperbola with foci on the $x$-axis has an inclination angle of $\\frac{\\pi}{6}$, and the distance from its focus to the asymptote is $2$, then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), xAxis);Inclination(OneOf(Asymptote(G))) = pi/6;Distance(Focus(G), Asymptote(G)) = 2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/12 - y^2/4 = 1",
    "fact_spans": "[[[11, 14], [43, 44], [60, 63]], [[2, 14]], [[11, 40]], [[43, 57]]]",
    "query_spans": "[[[60, 70]]]",
    "process": "Let the equation of a hyperbola with foci on the x-axis be: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$, and let the angle of inclination of one asymptote be $\\frac{\\pi}{6}$. Take the focus $F(c,0)$. Since the distance from the focus to the asymptote is $2$, we have:\n\\[\n\\begin{cases}\n\\frac{b}{a}=\\tan\\frac{\\pi}{6}=\\frac{\\sqrt{3}}{3}\\\\\nc^{2}=a^{2}+b^{2}\\\\\n\\frac{bc}{\\sqrt{a^{2}+b^{2}}}=2\n\\end{cases}\n\\]\nSolving gives $a=2\\sqrt{3}$, $b=2$. Therefore, the equation of the hyperbola is: $\\frac{x^{2}}{12}-\\frac{y^{2}}{4}=1$."
  },
  {
    "text": "Given $A(0 ,-4)$ , $B(3,2)$, the shortest distance from a point on the parabola $y^{2}=x$ to the line $AB$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (0, -4);B: Point;Coordinate(B) = (3, 2);G: Parabola;Expression(G) = (y^2 = x);D: Point;PointOnCurve(D, G) = True",
    "query_expressions": "Min(Distance(D, LineOf(A, B)))",
    "answer_expressions": "3*sqrt(5)/5",
    "fact_spans": "[[[2, 12]], [[2, 12]], [[14, 23]], [[14, 23]], [[24, 36]], [[24, 36]], [[38, 39]], [[24, 39]]]",
    "query_spans": "[[[38, 54]]]",
    "process": "The equation of line AB obtained from the two-point form is $ 2x - y - 4 = 0 $. Let an arbitrary point $ P $ on the parabola $ y^{2} = x $ have coordinates $ (t, t^{2}) $, then $ d = \\frac{|2t - t^{2} - 4|}{\\sqrt{5}} = \\frac{|t^{2} - 2t + 4|}{\\sqrt{5}} = \\frac{(t - 1)^{2} + 3}{\\sqrt{5}} \\geqslant \\frac{3}{\\sqrt{5}} = \\frac{3\\sqrt{5}}{5} $. The shortest distance from any point on the parabola $ y^{2} = x $ to the line AB is $ \\frac{\\sqrt[3]{5}}{5} $."
  },
  {
    "text": "Given a moving line $l$ passing through the origin intersects the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1$ at points $A$ and $B$, $D$ is the upper vertex of the ellipse $C$. If the slopes of lines $AD$ and $BD$ exist and are $k_{1}$, $k_{2}$ respectively, then $k_{1} k_{2}=$?",
    "fact_expressions": "l: Line;C: Ellipse;D: Point;A: Point;B: Point;k1: Number;k2: Number;Expression(C) = (x^2/3 + y^2/2 = 1);O: Origin;PointOnCurve(O, l);Intersection(l, C) = {A, B};UpperVertex(C) = D;Slope(LineOf(A, D)) = k1;Slope(LineOf(B, D)) = k2",
    "query_expressions": "k1*k2",
    "answer_expressions": "-2/3",
    "fact_spans": "[[[9, 12]], [[13, 50], [66, 71]], [[62, 65]], [[52, 55]], [[56, 59]], [[101, 108]], [[109, 116]], [[13, 50]], [[3, 5]], [[2, 12]], [[9, 61]], [[62, 75]], [[77, 116]], [[77, 116]]]",
    "query_spans": "[[[118, 133]]]",
    "process": "From the given information, $ D(0,\\sqrt{2}) $, let $ A(x_{1},y_{1}) $, $ B(-x_{1},-y_{1}) $, then $ k_{1}k_{2}=\\frac{y_{1}-\\sqrt{2}}{x_{1}} \\cdot \\frac{-y_{1}-\\sqrt{2}}{x_{1}} = \\frac{y_{1}^{2}-2}{x_{1}^{2}} $. Since $ A $ lies on the ellipse, $ \\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1 $, i.e., $ y_{1}^{2} = 2 - \\frac{2}{3}x_{1}^{2} $, so $ k_{1}k_{2} = -\\frac{2}{3} $."
  },
  {
    "text": "If there are two points $A$ and $B$ on the parabola $y^{2}=8x$, and the line $AB$ is perpendicular to the $x$-axis, $O$ is the coordinate origin, and the area of $\\triangle OAB$ is $8$, then the sum of the distances from points $A$ and $B$ to the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;O: Origin;Expression(G) = (y^2 = 8*x);PointOnCurve(A,G);PointOnCurve(B,G);IsPerpendicular(LineOf(A,B),xAxis);Area(TriangleOf(O, A, B)) = 8",
    "query_expressions": "Distance(A,Directrix(G))+Distance(B,Directrix(G))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 15], [88, 91]], [[19, 22], [79, 83]], [[23, 26], [84, 87]], [[43, 46]], [[1, 15]], [[1, 26]], [[1, 26]], [[28, 42]], [[53, 77]]]",
    "query_spans": "[[[79, 101]]]",
    "process": "Given that the directrix of the parabola is $ x = -2 $, and since $ AB $ is perpendicular to the $ x $-axis, assume without loss of generality that point $ A $ is above, let $ A\\left(\\frac{y_{0}^{2}}{8}, y_{0}\\right) $, $ B\\left(\\frac{y_{0}^{2}}{8}, -y_{0}\\right) $. The area of $ \\triangle OAB $ is $ \\frac{1}{2}|y_{0} - (-y_{0})| \\times \\frac{y_{0}^{2}}{8} = 8 $, so $ y_{0} = 4 $. Thus, $ A(2, 4) $, $ B(2, -4) $, then the sum of the distances from points $ A $ and $ B $ to the directrix of the parabola is $ 4 + 4 = 8 $."
  },
  {
    "text": "Given that moving circle $Q$ is externally tangent to circle $C_{1}$: $x^{2}+(y+4)^{2}=9$, and internally tangent to circle $C_{2}$: $x^{2}+(y-4)^{2}=9$, then the trajectory equation of the center of moving circle $Q$ is?",
    "fact_expressions": "Q: Circle;C1:Circle;C2:Circle;Expression(C1)=(x^2+(y+4)^2=9);Expression(C2)=(x^2+(y-4)^2=9);IsOutTangent(Q,C1);IsInTangent(Q,C2);Q1:Point;Center(Q)=Q1",
    "query_expressions": "LocusEquation(Q1)",
    "answer_expressions": "{(y^2/9-x^2/7=1)&(y>0)}",
    "fact_spans": "[[[2, 7], [74, 76]], [[8, 37]], [[41, 70]], [[8, 37]], [[41, 70]], [[2, 39]], [[2, 72]], [[78, 81]], [[74, 81]]]",
    "query_spans": "[[[78, 88]]]",
    "process": "According to the problem and the definition of a hyperbola, the trajectory of the center Q of the moving circle is the upper branch of a hyperbola with foci at $C_{1}$ and $C_{2}$. By finding the values of $a$, $b$, and $c$, the equation of the trajectory can be determined. Let the radius of the moving circle Q be $R$. Since the moving circle Q is externally tangent to the circle $C_{1}: x^{2}+(y+4)^{2}=9$ and internally tangent to the circle $C_{2}: x^{2}+(y-4)^{2}=9$, it follows that $|QC_{1}|=3+R$, $|QC_{2}|=R-3$. Therefore, $|QC_{1}|-|QC_{2}|=6<|C_{1}C_{2}|=8$. By the definition of a hyperbola, the trajectory of the center Q of the moving circle is the upper branch of a hyperbola with foci at $C_{1}$ and $C_{2}$, where $2a=6$, $2c=8$, solving gives $a=3$, $c=4$. Also, $b^{2}=c^{2}-a^{2}=16-9=7$. Thus, the equation of the trajectory of the center Q of the moving circle is $\\frac{y^{2}}{9}-\\frac{x^{2}}{7}=1$ $(y>0)$."
  },
  {
    "text": "An ellipse $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{3-m}=1$ has one focus at $(0,1)$, then $m$ equals?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (y^2/(3 - m) + x^2/m^2 = 1);Coordinate(OneOf(Focus(G))) = (0, 1)",
    "query_expressions": "m",
    "answer_expressions": "{-2, 1}",
    "fact_spans": "[[[0, 43]], [[58, 61]], [[0, 43]], [[0, 56]]]",
    "query_spans": "[[[58, 64]]]",
    "process": "Since the foci of the given ellipse lie on the y-axis, and $3 - m - m^{2} = 1$, that is, $m^{2} + m - 2 = 0$, solving gives $m = -2$ or $1$."
  },
  {
    "text": "Draw a line $l$ through the point $P(1,1)$ such that the midpoint of the chord intercepted by the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ is exactly $P$. Then the equation of the line $l$ is?",
    "fact_expressions": "l: Line;G: Ellipse;P: Point;Expression(G) = (x^2/9 + y^2/4 = 1);Coordinate(P) = (1, 1);PointOnCurve(P, l) ;MidPoint(InterceptChord(l, G)) = P",
    "query_expressions": "Expression(l)",
    "answer_expressions": "4*x + 9*y = 13",
    "fact_spans": "[[[11, 16], [19, 20], [73, 78]], [[21, 58]], [[1, 10], [68, 71]], [[21, 58]], [[1, 10]], [[0, 16]], [[19, 71]]]",
    "query_spans": "[[[73, 83]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$, $F_{2}$ respectively, and the eccentricity is $\\frac{1}{2}$. A line passing through $F_{2}$ intersects the ellipse $C$ at points $A$, $B$. If the perimeter of $\\Delta F_{1} A B$ is $8$, then the equation of the ellipse is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;Eccentricity(C) = 1/2;L: Line;PointOnCurve(F2,L) = True;Intersection(L,C) = {A,B};A: Point;B: Point;Perimeter(TriangleOf(F1,A,B)) = 8",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4+y^2/3=1",
    "fact_spans": "[[[2, 59], [116, 121], [162, 164]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[68, 75]], [[76, 83], [105, 112]], [[2, 83]], [[2, 83]], [[2, 102]], [[113, 115]], [[104, 115]], [[113, 132]], [[123, 126]], [[127, 130]], [[135, 160]]]",
    "query_spans": "[[[162, 168]]]",
    "process": "According to the definition of an ellipse, given that the perimeter of $\\triangle F_{1}AB$ is 8, we have $4a=8$, so $a=2$. Then, since the eccentricity is $\\frac{1}{2}$, i.e., $\\frac{c}{a}=\\frac{1}{2}$, we get $c=1$, and thus $b^{2}=a^{2}-c^{2}=3$. Therefore, the equation of the ellipse is $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$."
  },
  {
    "text": "Given that the line $x - m y - 2 = 0$ intersects the parabola $C$: $y^2 = \\frac{1}{2} x$ at points $A$ and $B$. $P$ is the midpoint of segment $AB$. A line through $P$ parallel to the $x$-axis intersects $C$ at point $Q$. If the circle with diameter $AB$ passes through $Q$, then $m = $?",
    "fact_expressions": "H: Line;Expression(H) = (-m*y + x - 2 = 0);m: Number;C: Parabola;Expression(C) = (y^2 = x/2);A: Point;B: Point;Intersection(H, C) = {A, B};P: Point;MidPoint(LineSegmentOf(A, B)) = P;L: Line;PointOnCurve(P, L);IsParallel(L, xAxis);Q: Point;Intersection(L, C) = Q;G: Circle;IsDiameter(LineSegmentOf(A, B), G);PointOnCurve(Q, G)",
    "query_expressions": "m",
    "answer_expressions": "pm*2",
    "fact_spans": "[[[2, 15]], [[2, 15]], [[114, 117]], [[16, 45], [86, 89]], [[16, 45]], [[47, 50]], [[51, 54]], [[2, 56]], [[57, 60], [73, 76]], [[57, 71]], [], [[72, 85]], [[72, 85]], [[90, 94], [109, 112]], [[72, 94]], [[106, 107]], [[96, 107]], [[106, 112]]]",
    "query_spans": "[[[114, 119]]]",
    "process": "Let the coordinates of AB, and solve the system of equations of the line and the parabola to find the sum of the two roots, thereby obtaining the coordinates of the midpoint P of AB. Find the coordinates of Q according to the given conditions, then compute the chord lengths |AB| and |PQ|, and finally determine the value of m. Let A(x_{1},y_{1}), B(x_{2},y_{2}), from \\begin{cases}x-my\\\\y2=\\frac{1}{2}x\\end{cases}-2=0, rearranging gives 2y^{2}-my-2=0, \\triangle=m^{2}+8>0, y_{1}+y_{2}=\\frac{m}{2}, y_{1}y_{2}=-1, so the midpoint P of AB is (\\frac{m^{2}}{4}+2,\\frac{m}{4}), then Q(\\frac{m^{2}}{8},\\frac{m}{4}), thus |PQ|=\\frac{m^{2}}{8}+2. Also |AB|=\\sqrt{1+m^{2}}|_{y_{1}}-y_{2}|=\\sqrt{1+m^{2}}\\sqrt{\\frac{m^{2}}{4}+4}, so \\sqrt{1+m^{2}}\\sqrt{\\frac{m^{2}}{4}+4}=2(\\frac{m^{2}}{8}+2), i.e., \\sqrt{1+m^{2}}=\\sqrt{\\frac{m^{2}}{4}+4}, solving yields m=\\pm2. This problem examines properties of the parabola and properties of a circle with a line segment as diameter, classified as a medium-difficulty problem."
  },
  {
    "text": "Given the parabola $C$: $x^{2}=2 y$ with two moving points $P$, $Q$ on it, and $|P Q|=5$, then the minimum distance from the midpoint of segment $P Q$ to the $x$-axis is?",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 2*y);P: Point;Q: Point;PointOnCurve(P, C);PointOnCurve(Q, C);Abs(LineSegmentOf(P, Q)) = 5",
    "query_expressions": "Min(Distance(MidPoint(LineSegmentOf(P, Q)), xAxis))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 21]], [[2, 21]], [[26, 29]], [[30, 33]], [[2, 33]], [[2, 33]], [[35, 44]]]",
    "query_spans": "[[[46, 69]]]",
    "process": "Let the focus of parabola $ C $ be $ F $, the projection of point $ P $ on the directrix $ y = -\\frac{1}{2} $ be $ P_{1} $, the projection of point $ Q $ on the line $ y = -\\frac{1}{2} $ be $ Q_{1} $, the midpoint of segment $ PQ $ be $ E $, and the distance from point $ E $ to the $ x $-axis be $ d $. Then $ |PP_{1}| + |QQ| = |PF| + |QF| \\geqslant |PQ| = 5d = \\frac{1}{2}(|PP_{1}| + |QQ_{1}|) - 0.5 $, $ \\therefore d \\geqslant 2 $, with equality if and only if $ |PF| + |QF| = |PQ| $, i.e., when points $ P $, $ F $, $ Q $ are collinear. Therefore, the minimum distance from the midpoint of segment $ PQ $ to the $ x $-axis is $ 2 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with left and right foci $F_{1}$ and $F_{2}$, a line passing through $F_{2}$ intersects the right branch of the hyperbola at points $A$ and $B$. If $\\overrightarrow{B F_{1}} \\cdot \\overrightarrow{B F_{2}}=0$ and $\\sin \\angle F_{1} A F_{2}=\\frac{3}{5}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;LeftFocus(G) = F1;RightFocus(G) = F2;F1: Point;F2: Point;H: Line;PointOnCurve(F2, H) = True;Intersection(H, RightPart(G)) = {A, B};B: Point;A: Point;DotProduct(VectorOf(B, F1), VectorOf(B, F2)) = 0;Sin(AngleOf(F1, A, F2)) = 3/5",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)/2",
    "fact_spans": "[[[2, 58], [93, 96], [213, 216]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 80]], [[2, 80]], [[65, 72]], [[73, 80], [82, 89]], [[90, 92]], [[81, 92]], [[90, 108]], [[103, 106]], [[99, 102]], [[110, 169]], [[171, 210]]]",
    "query_spans": "[[[213, 222]]]",
    "process": "Let $|AF_{1}|=m$. Since $\\overrightarrow{BF}_{1}\\cdot\\overrightarrow{BF}_{2}=0$ and $\\cos\\angle F_{1}AF_{2}=\\frac{4}{5}$, it follows that $|BF_{1}|=\\frac{3}{5}m$, $|AB|=\\frac{4}{5}m$. By the definition of the hyperbola, we have $|BF_{2}|=\\frac{3}{5}m-2a$, $|AF_{2}|=m-2a$. Since $|AF_{2}|+|BF_{2}|=|AB|$, then $\\frac{3}{5}m-2a+m-2a=\\frac{4}{5}m$. Solving gives $m=5a$. Therefore, in $\\triangle BF_{1}F_{2}$, $|BF_{1}|^{2}+|BF_{2}|^{2}=|F_{1}F_{2}|^{2}$, that is, $9a^{2}+a^{2}=4c^{2}$. Solving gives $e=\\frac{\\sqrt{10}}{2}$."
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{4}+y^{2}=1$ has left and right foci $F_{1}$ and $F_{2}$, respectively. Point $P$ lies on the ellipse such that $\\angle F_{1} P F_{2}=90^{\\circ}$. Then the area of $\\triangle F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 27], [56, 58]], [[35, 42]], [[51, 55]], [[43, 50]], [[0, 27]], [[0, 50]], [[0, 50]], [[51, 59]], [[61, 94]]]",
    "query_spans": "[[[96, 126]]]",
    "process": "From the definition of the ellipse and the Pythagorean theorem, we obtain |PF_{1}||PF_{2}|, and then use the triangle area formula to get the answer. By the definition of the ellipse, |PF_{1}| + |PF_{2}| = 2a = 4, so |PF_{1}|^{2} + |PF_{2}|^{2} + 2|PF_{1}||PF_{2}| = 4a^{2} = 16. By the Pythagorean theorem, |PF_{1}|^{2} + |PF_{2}|^{2} = 4c^{2} = 12. Therefore, |PF_{1}||PF_{2}| = 2(a^{2} - c^{2}) = 2b^{2} = 2, then the area S of \\triangle F_{1}PF_{2} is S = \\frac{1}{2}|PF_{1}||PF_{2}| = b^{2} ="
  },
  {
    "text": "There are two distinct points $P$ and $Q$ on the parabola $y^{2}=4x$ with focus $F$, satisfying $\\overrightarrow{PF} = \\lambda \\overrightarrow{FQ}$ ($\\lambda>1$). If the distance from the midpoint $M$ of segment $PQ$ to the directrix of the parabola is $\\frac{8}{3}$, then $\\lambda=$?",
    "fact_expressions": "G: Parabola;P: Point;Q: Point;F: Point;M: Point;Expression(G) = (y^2 = 4*x);Focus(G)=F;PointOnCurve(P,G);PointOnCurve(Q,G);Negation(P=Q);lambda:Number;lambda>1;VectorOf(P, F) = lambda*VectorOf(F, Q);MidPoint(LineSegmentOf(P,Q))=M;Distance(M, Directrix(G)) = 8/3",
    "query_expressions": "lambda",
    "answer_expressions": "3",
    "fact_spans": "[[[7, 21], [117, 120]], [[28, 31]], [[32, 35]], [[3, 6]], [[113, 116]], [[7, 21]], [[0, 21]], [[7, 35]], [[7, 35]], [[7, 35]], [[142, 151]], [[39, 101]], [[39, 101]], [[103, 116]], [[113, 140]]]",
    "query_spans": "[[[142, 153]]]",
    "process": "Method 1: The focus of the parabola $ y^{2}=4x $ is $ F(1,0) $, and the equation of the directrix is $ x=-1 $. Let $ P(x_{1},y_{1}) $, $ Q(x_{2},y_{2}) $. Given $ \\overrightarrow{PF}=\\lambda\\overrightarrow{FQ} $ ($ \\lambda>1 $), so points $ P $, $ Q $, $ F $ are collinear. Let the line equation be $ x=my+1 $, $ \\lambda=-\\frac{y_{1}}{y_{2}} $. Since the distance from the midpoint $ M $ of segment $ PQ $ to the directrix of the parabola is $ \\frac{8}{3} $, combining  \n$$\n\\begin{cases}\nx_{1}+x_{2}+1=\\frac{8}{3}, \\\\\nx_{1}+x_{2}=\\frac{10}{3}, \\therefore m(y_{1}+y_{2})=\\frac{4}{3} \\\\\nx=my+1 \\\\\ny^{2}=4x\n\\end{cases}\n$$\nEliminating $ x $ gives $ y^{2}-4my-4=0 $, $ \\textcircled{1} $, $ m $, so $ m(y_{1}+y_{2})=4m^{2}=\\frac{4}{3} $, $ m=\\pm\\frac{\\sqrt{3}}{3} $. Since $ y_{1}+y_{2}=4m $, $ \\therefore m(y_{1}+y_{2})=4m^{2}=\\frac{4}{3} $, $ m=\\pm\\frac{\\sqrt{3}}{3} $. When $ m=\\frac{\\sqrt{3}}{3} $, equation $ \\textcircled{1} $ becomes $ \\sqrt{3}y^{2}-4y-4\\sqrt{3}=0 $, solving gives $ y=-\\frac{2}{\\sqrt{3}} $ or $ y=2\\sqrt{3} $. Since $ \\lambda>1 $, $ \\therefore \\lambda=-\\frac{y_{1}}{y_{2}}=3 $. Similarly, when $ m=-\\frac{\\sqrt{3}}{3} $, $ \\lambda=3 $.  \n\nMethod 2: Without loss of generality, assume point $ P $ is in the first quadrant. Draw $ PA\\bot $ directrix at point $ A $, draw $ QB\\bot $ directrix at point $ B $, draw $ MC\\bot $ directrix at point $ C $. Since $ |MC|=\\frac{1}{2}(|PA|+|QB|)=\\frac{1}{2}(|PF|+|QF|)=\\frac{1}{2}|PQ| $, $ \\therefore |PQ|=\\frac{16}{3} $. Let the inclination angle of line $ PQ $ be $ \\theta $, $ |PQ|=\\frac{2p}{\\sin^{2}\\theta}=\\frac{4}{\\sin^{2}\\theta}=\\frac{16}{3} $, $ \\therefore \\theta=60^{\\circ} $, $ \\therefore \\frac{|PF|}{|QF|}=\\frac{p}{\\frac{p-\\cos\\theta}{1+\\cos\\theta}}=\\frac{1+\\cos\\theta}{1-\\cos\\theta}=3 $."
  },
  {
    "text": "What is the equation of the directrix of the parabola $y=-\\frac{1}{a} x^{2}$?",
    "fact_expressions": "G: Parabola;a: Number;Expression(G) = (y = -x^2/a)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=a/4",
    "fact_spans": "[[[0, 25]], [[3, 25]], [[0, 25]]]",
    "query_spans": "[[[0, 32]]]",
    "process": "From $ y = -\\frac{1}{a}x^2 $, we get $ x^{2} = -ay $, then the equation of the directrix is $ y = \\frac{a}{4} $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, its two asymptotes are the lines $l_{1}$ and $l_{2}$. A line $l$, passing through the right focus $F$ and perpendicular to $l_{1}$, intersects $l_{1}$ and $l_{2}$ at points $A$ and $B$ respectively, and $\\overrightarrow{F B}=2 \\overrightarrow{A F}$. Then the eccentricity of this hyperbola is?",
    "fact_expressions": "l1: Line;l2:Line;l:Line;G: Hyperbola;b: Number;a: Number;F: Point;B: Point;A: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Asymptote(G)={l1,l2};RightFocus(G)=F;PointOnCurve(F,l);IsPerpendicular(l,l1);Intersection(l1,l)=A;Intersection(l2,l)=B;VectorOf(F, B) = 2*VectorOf(A, F)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[67, 76], [100, 107], [116, 123]], [[78, 86], [125, 132]], [[108, 113]], [[2, 58], [192, 195]], [[5, 58]], [[5, 58]], [[93, 96]], [[137, 140]], [[133, 136]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 86]], [[2, 96]], [[87, 113]], [[97, 113]], [[108, 142]], [[108, 142]], [[144, 189]]]",
    "query_spans": "[[[192, 201]]]",
    "process": "The equations of the asymptotes of the hyperbola are $ y = \\pm\\frac{b}{a}x $. Without loss of generality, assume the equation of line $ l $ is $ y = -\\frac{a}{b}(x - c) $. From \n\\[\n\\begin{cases}\ny = -\\frac{b}{3}(x - c) \\\\\nb = -x(x - )\n\\end{cases}\n\\]\n, since $ \\frac{x - a}{a - b} = \\frac{x + b}{a} = \\frac{x}{a + b - a} = 2\\overline{2}(a - a)(a - b), -\\frac{a + b}{a - b} $, simplifying yields $ c^{2} = 2a^{2} - 2b^{2} $, that is, $ 3c^{2} = 4a^{2} $, hence $ e = \\frac{2\\sqrt{3}}{2} $,"
  },
  {
    "text": "Given that an ellipse is tangent to the $x$-axis and has foci at $F_{1}(1,1)$ and $F_{2}(5,2)$, what is the length of its major axis?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Coordinate(F1) = (1, 1);Coordinate(F2) = (5, 2);Focus(G)={F1,F2};IsTangent(xAxis,G)",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "5",
    "fact_spans": "[[[2, 4], [46, 47]], [[19, 31]], [[32, 44]], [[19, 31]], [[32, 44]], [[2, 44]], [[2, 11]]]",
    "query_spans": "[[[46, 52]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the focus of the parabola $y^{2}=8x$, point $A(2,2\\sqrt{3})$, and there exists a point $P$ on the parabola such that $|PA|+|PF|$ is minimized, then the coordinates of point $P$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;A: Point;Coordinate(A) = (2, 2*sqrt(3));P: Point;PointOnCurve(P, G);WhenMin(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(3/2, 2*sqrt(3))",
    "fact_spans": "[[[6, 20], [43, 46]], [[6, 20]], [[2, 5]], [[2, 23]], [[24, 42]], [[24, 42]], [[50, 53], [76, 80]], [[43, 53]], [[56, 74]]]",
    "query_spans": "[[[76, 85]]]",
    "process": "As shown in the figure, the parabola $ y^{2}=8x $ has focus $ F(2,0) $ and directrix line $ l: x=-2 $. Draw $ PB $ perpendicular from point $ P $ to the directrix of the parabola at point $ B $. By the definition of a parabola, $ |PF| = |PB| $, so $ |PA| + |PF| = |PA| + |PB| $. The sum $ |PA| + |PF| $ reaches its minimum value if and only if points $ A $, $ P $, and $ B $ are collinear. At this time, the equation of line $ PA $ is $ y=2\\sqrt{3} $. Solving the system of equations consisting of the line $ PA $ and the parabola:\n$$\n\\begin{cases}\ny=2\\sqrt{3} \\\\\ny^{2}=8x\n\\end{cases}\n$$\nyields\n$$\n\\begin{cases}\nx=\\frac{3}{2} \\\\\ny=2\\sqrt{3}\n\\end{cases}\n$$\nTherefore, the coordinates of point $ P $ are $ \\left(\\frac{3}{2}, 2\\sqrt{3}\\right) $."
  },
  {
    "text": "If $A$ is the vertex of the parabola $y=\\frac{1}{4} x^{2}$, and a line passing through the focus of the parabola intersects the parabola at points $B$ and $C$, then $\\overrightarrow{A B} \\cdot \\overrightarrow{A C}$ equals?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;C: Point;Expression(G) = (y = (1/4)*x^2);Vertex(G) = A;PointOnCurve(Focus(G), H);Intersection(H, G) = {B, C}",
    "query_expressions": "DotProduct(VectorOf(A, B), VectorOf(A, C))",
    "answer_expressions": "-3",
    "fact_spans": "[[[5, 29], [34, 37], [43, 46]], [[40, 42]], [[1, 4]], [[47, 50]], [[51, 54]], [[5, 29]], [[1, 32]], [[33, 42]], [[40, 56]]]",
    "query_spans": "[[[58, 110]]]",
    "process": "From the given conditions, we have A(0,0), and the focus of the parabola is (0,1), so the equation of line BC is y = kx + 1. Solving the system of equations  \n\\begin{cases}y=kx+1\\\\y=\\frac{1}{4}x^{2}\\end{cases}  \nwe obtain \\frac{1}{4}x^{2}-kx-1=0. Let A(x_{1},y_{1}), B(x_{1},y_{1}), then x_{1}+x_{2}=4k, x_{1}\\cdot x_{2}=-4, so y_{1}y_{2}=(kx_{1}+1)(kx_{2}+1)=k^{2}x_{1}x_{2}+k(x_{1}+x_{2})+1, thus \\overrightarrow{AB}\\cdot\\overrightarrow{AC}=x_{1}x_{2}+y_{1}y_{2}=4(1+k^{2})+4k^{2}+1=-3. Method: This problem uses the dot product of planar vectors to examine the intersection of a line and a parabola, and investigates the application of the equation method in studying the positional relationship between lines and conic sections, making it a medium-difficulty problem. The key to solving this problem lies in using the coordinate representation of the vector dot product to transform the problem into finding relationships among the coordinates of the intersection points of the line and the parabola, which is precisely the focus of analytic geometry. By organizing the system of equations and applying Vieta's formulas along with the line equation, the solution can be obtained."
  },
  {
    "text": "Given the equation of circle $C$ is $x^{2}+y^{2}-10x=0$, find the trajectory equation of the center $P$ of a moving circle that is tangent to the $y$-axis and externally tangent to circle $C$?",
    "fact_expressions": "C: Circle;Expression(C) = (-10*x + x^2 + y^2 = 0);H: Circle;P: Point;IsTangent(H, yAxis);IsOutTangent(C, H);Center(H) = P",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "{(y^2 = 20*x)&(x > 0), (y = 0)&(x < 0)}",
    "fact_spans": "[[[2, 6], [40, 44]], [[2, 29]], [[47, 49]], [[51, 54]], [[31, 49]], [[39, 49]], [[47, 54]]]",
    "query_spans": "[[[51, 60]]]",
    "process": "The equation $x^{2}+y^{2}-10x=0$ is rewritten as $(x-5)^{2}+y^{2}=25$. If the moving circle is on the right side of the $y$-axis, then the distance from the center of the moving circle to the fixed point $(5,0)$ is equal to the distance to the fixed line $x=-5$, and its trajectory is a parabola. The equation is $y^{2}=20x$ $(x>0)$. If the moving circle is on the left side of the $y$-axis, then the trajectory of the center of the moving circle is the negative $x$-axis, with equation $y=0$ $(x<0)$. In summary, the trajectory equation of the center $P$ of the moving circle is $y^{2}=20x$ $(x>0)$ or $y=0$ $(x<0)$."
  },
  {
    "text": "The equation of the directrix of the curve $y=-2 x^{2}$ is?",
    "fact_expressions": "G: Curve;Expression(G) = (y = -2*x^2)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y = 1/8",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": "From $ y = -2x^{2} $, we get $ x^{2} = -\\frac{y}{2} $. Hence, its directrix equation is: $ y = \\frac{1}{a} $,"
  },
  {
    "text": "The asymptotes of the hyperbola $E$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1(a>0, b>0)$ are tangent to the circle $C$: $(x-2)^{2}+y^{2}=1$, then the eccentricity of the hyperbola $E$ is?",
    "fact_expressions": "E: Hyperbola;Expression(E) = (-x^2/b^2 + y^2/a^2 = 1);a: Number;b: Number;a>0;b>0;C: Circle;Expression(C) = (y^2 + (x - 2)^2 = 1);IsTangent(Asymptote(E),C) = True",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 61], [94, 100]], [[0, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[66, 90]], [[66, 90]], [[0, 92]]]",
    "query_spans": "[[[94, 106]]]",
    "process": "Circle $ C: (x-2)^{2} + y^{2} = 1 $ has center $ (2,0) $ and radius $ 1 $. Hyperbola $ E: \\frac{y^{2}}{a^{2}} - \\frac{x^{2}}{b^{2}} = 1 $ $ (a>0, b>0) $ has asymptotes $ y = \\pm\\frac{a}{b}x $. Since the asymptotes of hyperbola $ E $ are tangent to circle $ C $, it follows that $ \\frac{2a}{\\sqrt{a^{2}+b^{2}}} = \\frac{2a}{c} = 1 $, then $ e = \\frac{c}{a} = 2 $."
  },
  {
    "text": "Let the parabola $C$: $x^{2}=4 y$ have focus $F$, and its directrix intersects the $y$-axis at point $M$. A line $l$ passing through point $F$ intersects the parabola $C$ at points $A$ and $B$. If $\\angle A M B=90^{\\circ}$, then $|A F|$=?",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 4*y);F: Point;Focus(C) = F;M: Point;Intersection(Directrix(C), yAxis) = M;A: Point;B: Point;l: Line;PointOnCurve(F, l);Intersection(l, C) = {A, B};AngleOf(A, M, B) = ApplyUnit(90, degree)",
    "query_expressions": "Abs(LineSegmentOf(A, F))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 20], [28, 29], [55, 61]], [[1, 20]], [[24, 27], [44, 48]], [[1, 27]], [[38, 42]], [[28, 42]], [[62, 65]], [[66, 69]], [[49, 54]], [[43, 54]], [[49, 71]], [[73, 98]]]",
    "query_spans": "[[[100, 109]]]",
    "process": "Analysis: First, assume the equation of line AB is $ y = kx + 1 $, then use $ \\overrightarrow{MA} \\cdot \\overrightarrow{MB} = 0 $ to find the value of $ k $, and finally find $ |AF| $. Assume the equation of line AB is $ y = kx + 1 $, combine with \n\\[\n\\begin{cases}\nx^{2} = 4y \\\\\ny = kx + 1\n\\end{cases}\n\\Rightarrow x^{2} - 4kx - 4 = 0.\n\\]\nLet $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, $ x_{1} > 0 $. Then $ x_{1} + x_{2} = 4k $, $ x_{1} \\cdot x_{2} = -4 $. From the problem, $ \\overrightarrow{MA} = (x_{1}, y_{1} + 1) $, $ \\overrightarrow{MB} = (x_{2}, y_{2} + 1) $. Since $ \\angle AMB = 90^{\\circ} $, we have $ \\overrightarrow{MA} \\cdot \\overrightarrow{MB} = x_{1}x_{2} + (y_{1} + 1)(y_{2} + 1) = x_{1}x_{2} + (kx_{1} + 2)(kx_{2} + 2) = (1 + k^{2})x_{1}x_{2} + 2k(x_{1} + x_{2}) + 4 = -4(1 + k^{2}) + 8k^{2} + 4 = 4k^{2} = 0 $, so $ k = 0 $. Thus $ x_{1} = 2 $, $ y_{1} = 1 $. Therefore $ A(2, 1) $, so $ |AF| = 2 $."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=8x$ be $F$, and the directrix be $l$. Then, the equation of the circle with center $F$ and tangent to $l$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;l: Line;Directrix(G) = l;H: Circle;Center(H) = F;IsTangent(l, H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x-2)^2+y^2=16",
    "fact_spans": "[[[1, 15]], [[1, 15]], [[19, 22], [32, 35]], [[1, 22]], [[26, 29], [41, 44]], [[1, 29]], [[47, 48]], [[31, 48]], [[40, 48]]]",
    "query_spans": "[[[47, 53]]]",
    "process": "Given that the focus is F(2,0) and the directrix l is x = -2, the radius of the circle is r = 4, so the equation of the circle is (x-2)^{2} + y^{2} = 16."
  },
  {
    "text": "The product of the distances from a point $P$ on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{25}=1$ to its two foci is $m$. Then the maximum value of $m$ is?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/9 + y^2/25 = 1);PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};m: Number;Distance(P, F1)*Distance(P, F2) = m",
    "query_expressions": "Max(m)",
    "answer_expressions": "25",
    "fact_spans": "[[[0, 38]], [[42, 45]], [[0, 38]], [[0, 45]], [], [], [[0, 49]], [[56, 59], [61, 64]], [[0, 59]]]",
    "query_spans": "[[[61, 70]]]",
    "process": "Let the two foci of the ellipse be $ F_{1} $ and $ F_{2} $, with $ |PF_{1}| + |PF_{2}| = 2a = 10 $. Then $ m = |PF_{1}||PF_{2}| \\leqslant \\left( \\frac{|PF_{1}| + |PF_{2}|}{2} \\right)^{2} = 25 $, with equality if and only if $ |PF_{1}| = |PF_{2}| = 5 $, that is, when point $ P $ is at the endpoint of the minor axis of the ellipse, $ m $ attains its maximum value of 25."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $x^{2}=8 y$ intersects it at points $A$ and $B$, and $O$ is the origin. If $|A F|=6$, then the area of $\\triangle O A B$ is?",
    "fact_expressions": "G: Parabola;H: Line;O: Origin;A: Point;B: Point;F: Point;Expression(G) = (x^2 = 8*y);PointOnCurve(F, H);Focus(G) = F;Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, F)) = 6",
    "query_expressions": "Area(TriangleOf(O, A, B))",
    "answer_expressions": "6*sqrt(2)",
    "fact_spans": "[[[1, 15], [25, 26]], [[22, 24]], [[39, 42]], [[29, 32]], [[33, 36]], [[18, 21]], [[1, 15]], [[0, 24]], [[1, 21]], [[22, 38]], [[50, 59]]]",
    "query_spans": "[[[62, 84]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), without loss of generality, assume A is in the first quadrant and B is in the second quadrant. For the parabola x^{2}=8y, p=4, F(0,2), |AF|=y_{1}+\\frac{p}{2}=y_{1}+2=6, y_{1}=4, x_{1}^{2}=8y_{1}=32, and since point A is in the first quadrant, \\therefore x_{1}=4\\sqrt{2}. \\because AB is a chord passing through the focus, \\therefore x_{1}x_{2}=-p^{2}, x_{2}=\\frac{-4^{2}}{4\\sqrt{2}}=-\\therefore S_{AOAB}=\\frac{1}{2}|OF||x_{1}-x_{2}|=\\frac{1}{2}\\times2\\times(4\\sqrt{2}+2\\sqrt{2})=6\\sqrt{2}"
  },
  {
    "text": "Given fixed point $F(1 , 0)$, moving point $P$ on the $y$-axis, point $M$ on the $x$-axis, and $\\overrightarrow{P M} \\cdot \\overrightarrow{P F}=0$. Extend $MP$ to point $N$ such that $|\\overrightarrow{P M}|=|\\overrightarrow{P N}|$. Then the trajectory equation of point $N$ is?",
    "fact_expressions": "M: Point;P: Point;F: Point;N: Point;Coordinate(F) = (1, 0);PointOnCurve(P, yAxis);PointOnCurve(M, xAxis);DotProduct(VectorOf(P, M), VectorOf(P, F)) = 0;PointOnCurve(N,OverlappingLine(LineSegmentOf(M,P)));Abs(VectorOf(P, M)) = Abs(VectorOf(P, N))",
    "query_expressions": "LocusEquation(N)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[29, 33]], [[17, 20]], [[4, 14]], [[101, 105], [157, 161]], [[4, 14]], [[15, 28]], [[29, 39]], [[41, 92]], [[93, 105]], [[108, 155]]]",
    "query_spans": "[[[157, 168]]]",
    "process": "Since $|\\overrightarrow{PM}|=|\\overrightarrow{PN}|$, $P$ is the midpoint of $MN$. Let $N(x,y)$, then $M(-x,0)$, $P(0,\\frac{y}{2})$. From $\\overrightarrow{PM}\\cdot\\overrightarrow{PF}=0$, we get $(-x,-\\frac{y}{2})\\cdot(1,-\\frac{y}{2})=0$, so $(-x)^{2}+(-\\frac{y}{2})\\cdot(-\\frac{y}{2})=0$, then $y^{2}=4x$, thus the trajectory equation of point $N$ is $y^{2}=4x$."
  },
  {
    "text": "The focal distance of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{m}=1$ is $10$, then the value of the real number $m$ is?",
    "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (x^2/9 - y^2/m = 1);FocalLength(G) = 10",
    "query_expressions": "m",
    "answer_expressions": "16",
    "fact_spans": "[[[0, 38]], [[48, 53]], [[0, 38]], [[0, 46]]]",
    "query_spans": "[[[48, 57]]]",
    "process": "The hyperbola \\frac{x^{2}}{9}-\\frac{y^{2}}{m}=1 has a focal distance of 10. Therefore, a=3, c=5, so m=25-9=16. Hence, the correct answer is 16."
  },
  {
    "text": "Given a hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{16}=1$ with a focus $F_{1}(5,0)$, and one of its asymptotes is $l$. Draw $F_{1} A \\perp l$ from $F_{1}$ intersecting $l$ at $A$. Then the distance from $A$ to the origin is?",
    "fact_expressions": "G: Hyperbola;m: Number;F1: Point;A: Point;l:Line;Expression(G) = (-y^2/16 + x^2/m = 1);Coordinate(F1) = (5, 0);OneOf(Focus(G)) = F1;OneOf(Asymptote(G))=l;PointOnCurve(F1, LineSegmentOf(F1, A));IsPerpendicular(LineSegmentOf(F1, A),l);Intersection(LineSegmentOf(F1, A), l) = A;O:Origin",
    "query_expressions": "Distance(A, O)",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 41]], [[5, 41]], [[46, 58], [72, 79]], [[102, 105], [107, 110]], [[67, 70], [98, 101]], [[2, 41]], [[46, 58]], [[2, 58]], [[2, 70]], [[71, 97]], [[80, 97]], [[80, 105]], [[111, 113]]]",
    "query_spans": "[[[107, 117]]]",
    "process": "Using the equation of the hyperbola and the coordinates of the foci, we obtain m=9; then the equation of the asymptote l can be written, and subsequently, the distance from point A to the origin is solved using geometric methods according to the given conditions. Let the asymptote l: y=\\frac{4}{3}x, then \\tan\\angleAOF_{1}=\\frac{4}{3}, so \\cos\\angleAOF_{1}=\\frac{3}{5}; since |OF_{1}|=5, it follows that |OA|=5\\cos\\angleAOF_{1}=3."
  },
  {
    "text": "If the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{a^{2}}=1$ and the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{2}=1$ have the same foci, then $a$=?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Ellipse;Expression(G) = (-y^2/2 + x^2/a^2 = 1);Expression(H) = (x^2/4 + y^2/a^2 = 1);Focus(G)=Focus(H)",
    "query_expressions": "a",
    "answer_expressions": "pm*1",
    "fact_spans": "[[[43, 85]], [[93, 96]], [[1, 42]], [[43, 85]], [[1, 42]], [[1, 91]]]",
    "query_spans": "[[[93, 98]]]",
    "process": ""
  },
  {
    "text": "Find a point $P$ on $y^{2}=-4x$ such that the sum of the distance from $P$ to the focus $F$ and the distance from $P$ to $A(-2,1)$ is minimized. Then what are the coordinates of this point?",
    "fact_expressions": "C: Curve;Expression(C) = (y^2 = -4*x);P: Point;PointOnCurve(P, C);F: Point;Focus(C) = F;A: Point;Coordinate(A) = (-2, 1);WhenMin(Distance(P, F) + Distance(P, A))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(-1/4,1)",
    "fact_spans": "[[[1, 13]], [[1, 13]], [[17, 20], [22, 23], [53, 54]], [[1, 20]], [[26, 29]], [[1, 29]], [[34, 43]], [[34, 43]], [[22, 50]]]",
    "query_spans": "[[[53, 59]]]",
    "process": "Given the parabola equation $ y^{2} = -4x $, we obtain $ 2p = 4 $, $ \\frac{p}{2} = 1 $, $ \\therefore $ the focus coordinates are $ F(-1,0) $, and the directrix equation is $ x = 1 $. Let point $ P $'s projection on the directrix be $ Q $, connect $ PQ $, then by the definition of a parabola, $ |PF| = |PQ| $. By plane geometry knowledge, when points $ A $, $ P $, $ Q $ are collinear, $ |PQ| + |PA| $ reaches its minimum value, thus $ |PF| + |PA| $ also reaches its minimum value. $ \\therefore $ when $ |PF| + |PA| $ takes the minimum value, the ordinate of point $ P $ is 1. Substituting $ P(x,1) $ into the parabola equation yields $ x = -\\frac{1}{4} $, $ \\therefore $ the coordinates of point $ P $ that minimize the sum of distances from $ P $ to $ A $ and $ F $ are $ (-\\frac{1}{4},1) $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and point $P$ is a point on the ellipse, if $|P F_{1}|=4$, then $S_{\\triangle P F_{1} F_{2}}$=?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/25 + y^2/9 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 4",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "3*sqrt(15)",
    "fact_spans": "[[[18, 61], [72, 74]], [[18, 61]], [[2, 9]], [[10, 17]], [[2, 66]], [[2, 66]], [[67, 71]], [[67, 78]], [[80, 93]]]",
    "query_spans": "[[[95, 126]]]",
    "process": "Given that $ a^{2} = 25 $, $ b^{2} = 9 $, so $ a = 5 $, $ b = 3 $, $ c = 4 $. Since point $ P $ is a point on the ellipse, if $ |PF_{1}| = 4 $, then $ |PF_{2}| = 2a - |PF_{1}| = 10 - 4 = 6 $. Because $ |F_{1}F_{2}| = 2c = 8 $, in $ \\triangle PF_{1}F_{2} $, $ \\cos\\angle F_{1}PF_{2} = \\frac{|PF_{2}|^{2} + |PF_{1}|^{2} - |F_{1}F_{2}|^{2}}{2|PF_{2}||PF_{1}|} $, so $ \\sin\\angle F_{1}PF_{2} = \\sqrt{1 - \\left(-\\frac{1}{4}\\right)^{2}} = \\frac{\\sqrt{15}}{4} $, $ |PF_{2}||PF_{1}| $, so $ S_{\\Delta PF_{1}F_{2}} = \\frac{1}{2}|PF_{2}||PF_{1}|\\sin\\angle F_{1}PF_{2} = \\frac{1}{2} \\times 6 \\times 4 \\times \\frac{\\sqrt{15}}{4} = 3\\sqrt{15} $."
  },
  {
    "text": "On the right branch of the hyperbola $x^{2}-\\frac{y^{2}}{15}=1$, a point $P$ draws tangents to circle $C_{1}$: $(x+4)^{2}+y^{2}=4$ and circle $C_{2}$: $(x-4)^{2}+y^{2}=1$, with tangent points $M$ and $N$ respectively. Then the minimum value of $|P M|^{2}-|P N|^{2}$ is?",
    "fact_expressions": "G: Hyperbola;C1: Circle;C2: Circle;P: Point;M: Point;N: Point;l1:Line;l2:Line;Expression(G) = (x^2 - y^2/15 = 1);Expression(C1) = ((x + 4)^2 + y^2 = 4);Expression(C2) = ((x - 4)^2 + y^2 = 1);PointOnCurve(P, RightPart(G));TangentOfPoint(P,C1)=l1;TangentOfPoint(P,C2)=l2;TangentPoint(l1,C1)=M;TangentPoint(l2,C2)=N;PointOnCurve(P,l1);PointOnCurve(P,l2)",
    "query_expressions": "Min((Abs(LineSegmentOf(P, M)))^2 - (Abs(LineSegmentOf(P, N)))^2)",
    "answer_expressions": "13",
    "fact_spans": "[[[1, 30]], [[43, 71]], [[72, 100]], [[36, 39]], [[109, 112]], [[113, 116]], [], [], [[1, 30]], [[43, 71]], [[72, 100]], [[0, 39]], [[0, 103]], [[0, 103]], [[0, 116]], [[0, 116]], [[0, 103]], [[0, 103]]]",
    "query_spans": "[[[118, 145]]]",
    "process": "Circle $ C_{1}: (x+4)^{2} + y^{2} = 4 $ has center $ (-4,0) $ and radius $ r_{1} = 2 $; circle $ C_{2}: (x-4)^{2} + y^{2} = 1 $ has center $ (4,0) $ and radius $ r_{2} = 1 $; let the left and right foci of the hyperbola $ x^{2} - \\frac{y^{2}}{15} = 1 $ be $ F_{1}(-4,0) $, $ F_{2}(4,0) $. Connecting $ PF_{1} $, $ PF_{2} $, $ F_{1}M $, $ F_{2}N $, we obtain $ |PM|^{2} - |PN|^{2} = (|PF_{1}|^{2} - r_{1}^{2}) - (|PF_{2}|^{2} - r_{2}^{2}) = |PF_{1}|^{2} - |PF_{2}|^{2} - 3 = (|PF_{1}| - |PF_{2}|)(|PF_{1}| + |PF_{2}|) - 3 = 2a(|PF_{1}| + |PF_{2}|) - 3 \\geqslant 2 \\cdot 2c - 3 = 2 \\cdot 8 - 3 = 13 $. The equality holds if and only if $ P $ is the right vertex, thus the minimum value is 13."
  },
  {
    "text": "Given $A(-1,0)$, $B(2,4)$, and the area of $\\triangle A B C$ is $10$, what is the trajectory equation of vertex $C$?",
    "fact_expressions": "A: Point;B: Point;C: Point;Coordinate(A) = (-1, 0);Coordinate(B) = (2, 4);Area(TriangleOf(A, B, C)) = 10",
    "query_expressions": "LocusEquation(C)",
    "answer_expressions": "{4*x-3*y-16=0, 4*x-3*y+24=0}",
    "fact_spans": "[[[2, 11]], [[13, 21]], [[52, 55]], [[2, 11]], [[13, 21]], [[23, 48]]]",
    "query_spans": "[[[52, 62]]]",
    "process": "The equation of line AB derived from the two-point form of a line is , that is, the length of segment AB is . Let the coordinates of point C be , then , that is,  or ."
  },
  {
    "text": "Given that point $A\\left(\\frac{\\sqrt{15}}{2}, \\frac{1}{2}\\right)$ lies on the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, and $F_{1}$, $F_{2}$ are the left and right foci of hyperbola $C$, if $\\overrightarrow{A F_{1}} \\cdot \\overrightarrow{A F_{2}}=0$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;A: Point;F1: Point;F2: Point;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (sqrt(15)/2, 1/2);PointOnCurve(A, C);LeftFocus(C) = F1;RightFocus(C) = F2;DotProduct(VectorOf(A, F1), VectorOf(A, F2)) = 0",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[42, 93], [113, 119], [188, 194]], [[50, 93]], [[50, 93]], [[2, 41]], [[97, 104]], [[105, 112]], [[42, 93]], [[2, 41]], [[2, 96]], [[97, 125]], [[97, 125]], [[127, 186]]]",
    "query_spans": "[[[188, 200]]]",
    "process": "From the given, $\\overrightarrow{AF}_{1}\\bot\\overrightarrow{AF_{2}}$, so $|F_{1}F_{2}|=2|AO|=4$, hence $c=2$. Also $\\frac{(\\sqrt{15}}{a})-\\frac{(\\frac{1}{2})^{2}}{b^{2}}=1'$, and $a^{2}=4-b^{2}$, so $b^{2}=1$, $a^{2}=3$. Therefore, the eccentricity of hyperbola $C$ is $e=\\frac{2\\sqrt{3}}{3}$."
  },
  {
    "text": "A moving point $M$ is such that its distance to the $x$-axis is 2 less than its distance to the point $F(0,2)$. Then the equation of the trajectory of the moving point $M$ is?",
    "fact_expressions": "M: Point;F: Point;Coordinate(F) = (0, 2);Distance(M, xAxis) = Distance(M, F) - 2",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[3, 6], [37, 40]], [[16, 25]], [[16, 25]], [[3, 32]]]",
    "query_spans": "[[[37, 47]]]",
    "process": "Let the coordinates of point M be M(x, y). According to the problem, we have |y| + 2 = \\sqrt{x^{2} + (y - 2)^{2}}. Simplifying gives x^{2} = 4y + 4|y|. When y < 0, x = 0. When y \\geqslant 0, x^{2} = 8y."
  },
  {
    "text": "Given the parabola equation $y^{2}=-4 \\sqrt{3} x$, and the line $l$ with equation $\\sqrt{3} x+y-5=0$, there is a moving point $A$ on the parabola. The distance from point $A$ to the $y$-axis is $m$, and the distance from point $A$ to line $l$ is $n$. Then the minimum value of $m+n$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -4*sqrt(3)*x);l: Line;Expression(l) = (sqrt(3)*x + y - 5 = 0);A: Point;PointOnCurve(A,G) = True;m: Number;n: Number;Distance(A, yAxis) = m;Distance(A,l) = n",
    "query_expressions": "Min(m + n)",
    "answer_expressions": "4-sqrt(3)",
    "fact_spans": "[[[2, 5], [59, 62]], [[2, 29]], [[30, 35], [89, 94]], [[30, 57]], [[67, 70], [71, 75]], [[58, 70]], [[84, 87]], [[98, 101]], [[71, 87]], [[71, 101]]]",
    "query_spans": "[[[103, 114]]]",
    "process": "According to the equation and definition of the parabola, the distance from moving point A to the y-axis is $ m $, and the distance to the directrix is $ m+\\sqrt{3} $. The distance from a point to the directrix equals the distance from the point to the focus. Let $ d $ be the distance from point A to the line $ \\sqrt{3}x+y-5=0 $. Then $ m+\\sqrt{3}+d $ is the sum of the distances from point A to the focus and to the line $ \\sqrt{3}x+y-5=0 $. From the given condition, the minimum distance occurs when the two distances are collinear, and the minimum value is the distance from point A to the line: $ 4-\\sqrt{3} $. Hence, the result is: $ 4-\\sqrt{3} $"
  },
  {
    "text": "Given an ellipse with foci $F_{1}(-2 , 0)$ and $F_{2}(2 , 0)$ that intersects the line $x+\\sqrt{3} y+4=0$ at exactly one point, what is the length of the major axis of the ellipse?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Coordinate(F1) = (-2, 0);Coordinate(F2) = (2, 0);Focus(G) = {F1, F2};H: Line;Expression(H) = (x + sqrt(3)*y + 4 = 0);NumIntersection(G, H) = 1",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "2*sqrt(7)",
    "fact_spans": "[[[39, 41], [72, 74]], [[3, 19]], [[21, 35]], [[3, 19]], [[21, 35]], [[2, 41]], [[42, 62]], [[42, 62]], [[39, 70]]]",
    "query_spans": "[[[72, 80]]]",
    "process": "According to the problem, let the equation of the ellipse be $\\frac{x^{2}}{b^{2}+4}+\\frac{y^{2}}{b^{2}}=1$ $(b>0)$. Substituting $x=-\\sqrt{3}y-4$ into the ellipse equation yields $4(b^{2}+1)y^{2}+8\\sqrt{3}b^{2}y-b^{4}+12b^{2}=0$. Since the ellipse and the line $x+\\sqrt{3}y+4=0$ have exactly one intersection point, the discriminant $\\Delta=(8\\sqrt{3}b^{2})^{2}-4\\times4(b^{2}+1)(-b^{4}+12b^{2})=0$, which simplifies to $(b^{2}+4)(b^{2}-3)=0$. Therefore, $b^{2}=3$, and the major axis length is $2\\sqrt{b^{2}+4}=2\\sqrt{7}$."
  },
  {
    "text": "Given the parabola $y=a x^{2}(a>0)$ with directrix $l$, where $l$ intersects the two asymptotes of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ at points $A$ and $B$ respectively, if $|A B|=4$, then $a$=?",
    "fact_expressions": "G: Hyperbola;H: Parabola;a: Number;A: Point;B: Point;l:Line;Expression(G) = (x^2/4 - y^2 = 1);a>0;Expression(H) = (y = a*x^2);Directrix(H)=l;L1:Line;L2:Line;Asymptote(G)={L1,L2};Intersection(l,L1)=A;Intersection(l,L2)=B;Abs(LineSegmentOf(A, B)) = 4",
    "query_expressions": "a",
    "answer_expressions": "1/4",
    "fact_spans": "[[[33, 61]], [[2, 21]], [[93, 96]], [[71, 74]], [[75, 78]], [[25, 28], [29, 32]], [[33, 61]], [[5, 21]], [[2, 21]], [[2, 28]], [], [], [[33, 67]], [[29, 80]], [[29, 80]], [[82, 91]]]",
    "query_spans": "[[[93, 98]]]",
    "process": "From the parabola $ y = ax^{2} $ ($ a > 0 $), the directrix of the parabola is $ y = -\\frac{1}{4a} $. From the hyperbola $ \\frac{x^{2}}{4} - y^{2} = 1 $, the asymptotes are $ y = \\pm\\frac{1}{2}x $. Since $ |AB| = 4 $, by the symmetry of the hyperbola, the intersection point of $ y = -\\frac{1}{4a} $ and $ y = -\\frac{1}{2}x $ is $ (2, -\\frac{1}{4a}) $. Substituting this point into $ y = -\\frac{1}{2}x $ gives $ -\\frac{1}{4a} = -\\frac{2}{2} $, so $ a = \\frac{1}{4} $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, if $b=\\sqrt{3} a$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b = sqrt(3)*a",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 53], [71, 77]], [[55, 69]], [[55, 69]], [[2, 53]], [[55, 69]]]",
    "query_spans": "[[[71, 83]]]",
    "process": "Since $ b = \\sqrt{3}a $, it follows that $ c^{2} = a^{2} + b^{2} = a^{2} + 3a^{2} = 4a^{2} $, hence $ c = 2a $. Therefore, the eccentricity is $ e = \\frac{c}{a} = 2 $. The answer is: ?"
  },
  {
    "text": "Given the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through the origin intersects $E$ at points $P$ and $Q$, such that $P F_{2} \\perp F_{2} Q$, $S_{\\Delta P F_{2} Q}=\\frac{1}{2} a^{2}$, and $|P F_{2}|+|F_{2} Q|=4$. Then the standard equation of $E$ is?",
    "fact_expressions": "E: Ellipse;b: Number;a: Number;G: Line;P: Point;F2: Point;Q: Point;F1: Point;O:Origin;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(E) = F1;RightFocus(E) = F2;PointOnCurve(O,G);Intersection(G, E) = {P, Q};IsPerpendicular(LineSegmentOf(P, F2), LineSegmentOf(F2, Q));Area(TriangleOf(P,F,Q2))=a^2/2;Abs(LineSegmentOf(F2, Q)) + Abs(LineSegmentOf(P, F2)) = 4",
    "query_expressions": "Expression(E)",
    "answer_expressions": "x^2/4 + y^2/2 = 1",
    "fact_spans": "[[[2, 59], [93, 96], [199, 202]], [[8, 59]], [[8, 59]], [[90, 92]], [[97, 100]], [[76, 83]], [[101, 104]], [[68, 75]], [[85, 89]], [[8, 59]], [[8, 59]], [[2, 59]], [[2, 83]], [[2, 83]], [[84, 92]], [[90, 106]], [[108, 131]], [[133, 173]], [[174, 197]]]",
    "query_spans": "[[[199, 209]]]",
    "process": "As shown in the figure: since $OP=OQ$, $OF_{1}=OF_{2}$, the quadrilateral $PF_{1}QF_{2}$ is a parallelogram, so $PF_{1}=QF_{2}$, $PF_{2}=QF_{1}$; also because $PF_{2}\\bot F_{2}Q$, the parallelogram $PF_{1}QF_{2}$ is a rectangle. Let $PF_{1}=m$, $PF_{2}=n$. From the given conditions we have\n\\[\n\\begin{cases}\nm+n=2a=4 \\\\\nm^{2}+n^{2}=4c^{2}\n\\end{cases}\n\\]\nSolving gives\n\\[\n\\begin{cases}\n\\frac{1}{2}m=\\frac{1}{2}a^{2} \\\\\nc=\\sqrt{2}\n\\end{cases}\n\\]\nThen $b^{2}=a^{2}-c^{2}=2$,"
  },
  {
    "text": "Given that point $F$ is the focus of the parabola $E$: $y^{2}=4 x$, and point $A(2, m)$ lies on parabola $E$, then $|A F|$=?",
    "fact_expressions": "E: Parabola;A: Point;F: Point;m: Number;Expression(E) = (y^2 = 4*x);Coordinate(A) = (2, m);Focus(E) = F;PointOnCurve(A, E)",
    "query_expressions": "Abs(LineSegmentOf(A, F))",
    "answer_expressions": "3",
    "fact_spans": "[[[7, 26], [41, 47]], [[30, 40]], [[2, 6]], [[31, 40]], [[7, 26]], [[30, 40]], [[2, 29]], [[30, 48]]]",
    "query_spans": "[[[50, 59]]]",
    "process": "Substituting A(2,m) into the parabola equation, we obtain m=\\pm2\\sqrt{2}, and the focus is (1,0), thus |AF|=\\sqrt{1+8}=3"
  },
  {
    "text": "It is known that line $l$ passes through the focus of parabola $C$ and is perpendicular to the axis of symmetry of $C$. Line $l$ intersects $C$ at points $A$ and $B$, with $|AB| = 12$. Point $P$ lies on the directrix of $C$. Then, the area of $\\triangle ABP$ is?",
    "fact_expressions": "l: Line;PointOnCurve(Focus(C),l);C: Parabola;IsPerpendicular(l,SymmetryAxis(C)) = True;Intersection(l, C) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(A, B)) = 12;P: Point;PointOnCurve(P, Directrix(C))",
    "query_expressions": "Area(TriangleOf(A, B, P))",
    "answer_expressions": "36",
    "fact_spans": "[[[2, 7], [30, 33]], [[2, 17]], [[8, 14], [20, 23], [34, 37], [66, 69]], [[2, 29]], [[30, 48]], [[39, 42]], [[43, 46]], [[49, 59]], [[62, 65]], [[62, 76]]]",
    "query_spans": "[[[78, 100]]]",
    "process": "Analysis: From |AB| = 12, we can obtain p, and thus derive the equation of the parabola and its directrix. Therefore, the height on side AB in triangle ABP is easily found. Without loss of generality, assume the parabola equation is y^{2} = 2px, |AB| = 2p = 12, p = 6, \\therefore the directrix equation is x = -3, the distance from P to line AB is 6, \\therefore S_{\\triangle ABP} = \\frac{1}{2} \\times 12 \\times 6 = 36."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $A$ the upper vertex of the ellipse, and $M$ the midpoint of $A F_{2}$. If $M F_{1} \\perp A F_{2}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;F2: Point;M: Point;F1: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;UpperVertex(G) = A;MidPoint(LineSegmentOf(A, F2)) = M;IsPerpendicular(LineSegmentOf(M, F1), LineSegmentOf(A, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[17, 69], [80, 82], [131, 133]], [[19, 69]], [[19, 69]], [[76, 79]], [[9, 16]], [[87, 90]], [[1, 8]], [[19, 69]], [[19, 69]], [[17, 69]], [[1, 75]], [[1, 75]], [[76, 86]], [[87, 103]], [[105, 128]]]",
    "query_spans": "[[[131, 139]]]",
    "process": "According to the problem, $\\triangle AF_{1}F_{2}$ is an isosceles triangle and $|AF_{1}|=|F_{1}F_{2}|$, that is, $a=2c$, then the answer is obtained. According to the problem, $\\triangle AF_{1}F_{2}$ is an isosceles triangle, and $|AF_{1}|=|F_{1}F_{2}|$, so $a=2c$, hence $e=\\frac{c}{a}=\\frac{1}{2}$."
  },
  {
    "text": "Given the parabola $E$: $y^{2}=4x$ with focus $F$ and directrix $l$, a line $m$ passing through $F$ intersects $E$ at points $A$ and $B$. The perpendicular bisector of $AF$ intersects $l$ and the $x$-axis at points $P$ and $Q$, respectively. If $\\angle AFP = \\angle AFQ$, then $|AB|=$?",
    "fact_expressions": "E: Parabola;Expression(E) = (y^2 = 4*x);F: Point;Focus(E) = F;l: Line;Directrix(E) = l;m: Line;PointOnCurve(F, m);A: Point;B: Point;Intersection(m, E) = {A, B};P: Point;Q: Point;Intersection(PerpendicularBisector(LineSegmentOf(A, F)), l) = P;Intersection(PerpendicularBisector(LineSegmentOf(A, F)), xAxis) = Q;AngleOf(A, F, P) = AngleOf(A, F, Q)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[2, 21], [47, 50]], [[2, 21]], [[25, 28], [37, 40]], [[2, 28]], [[32, 35], [76, 79]], [[2, 35]], [[41, 46]], [[36, 46]], [[52, 55]], [[56, 59]], [[41, 61]], [[85, 88]], [[89, 92]], [[62, 94]], [[62, 94]], [[96, 123]]]",
    "query_spans": "[[[126, 135]]]",
    "process": "\\because PQ \\text{ is the perpendicular bisector of } AF, \\angle AFP = \\angle AFQ, \\therefore PA = PF = FQ \\text{ in quadrilateral } PAQF, \\text{ diagonals } AF \\text{ and } PQ \\text{ are perpendicular}, \\therefore \\text{ quadrilateral } PAQF \\text{ is a rhombus} \\text{ by the definition of parabola}: AF = PA \\text{ hence } PA = AF = PF \\therefore \\triangle APF \\text{ is equilateral} \\text{ thus } \\angle AFP = 60^{\\circ} \\text{ hence } \\angle AFP = \\angle AFQ = 60^{\\circ} \\text{ therefore line } AB: y = \\sqrt{3}(x - 1) \\text{ substitute line } y = \\sqrt{3}(x - 1) \\text{ into the parabola} \\begin{cases} y = \\sqrt{3}(x - 1) \\\\ v^2 = 4x \\end{cases} \\text{ yields } 3x^{2} - 10x + 3 = 0, \\text{ let } A(x_{1}, y_{1}), B(x_{2}, y_{2}), \\text{ then } x_{1} + x_{2} = \\frac{10}{3}, |AB| = x_{1} + x_{2} + p = \\frac{16}{3}"
  },
  {
    "text": "Given real numbers $x$, $y$ satisfying $\\frac{x^{2}}{3}+y^{2}=1$, then the minimum value of $x+y$ is?",
    "fact_expressions": "x_: Real;y_: Real;x_^2/3 + y_^2 = 1",
    "query_expressions": "Min(x_ + y_)",
    "answer_expressions": "-2",
    "fact_spans": "[[[2, 7]], [[10, 13]], [[15, 40]]]",
    "query_spans": "[[[42, 53]]]",
    "process": "Since real numbers $x$, $y$ satisfy $\\frac{x^{2}}{3}+y^{2}=1$, let $x=\\sqrt{3}\\cos\\theta$, $y=\\sin\\theta$. Therefore, $x+y=\\sqrt{3}\\cos\\theta+\\sin\\theta=2\\sin(\\theta+\\frac{\\pi}{3})\\in[-2,2]$, so the minimum value of $x+y$ is $-2$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of an ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with focal distance $2$, and $P$ is a point on the ellipse $C$. A tangent line $l$ to the ellipse $C$ is drawn at point $P$. If the product of the distances from $F_{1}$ and $F_{2}$ to the tangent line $l$ is $4$, then what is the eccentricity of the ellipse $C$?",
    "fact_expressions": "F1: Point;F2: Point;Focus(C) = {F1, F2};FocalLength(C) = 2;C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;P: Point;PointOnCurve(P, C);l: Line;TangentOnPoint(P, C) = l;Distance(F1, l) * Distance(F2, l) = 4;PointOnCurve(P, l)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/5",
    "fact_spans": "[[[2, 9], [122, 129]], [[10, 17], [130, 137]], [[2, 87]], [[18, 82]], [[25, 82], [92, 97], [109, 114], [154, 159]], [[25, 82]], [[32, 82]], [[32, 82]], [[32, 82]], [[32, 82]], [[88, 91], [104, 108]], [[88, 102]], [[117, 120], [140, 143]], [[103, 120]], [[122, 152]], [[103, 120]]]",
    "query_spans": "[[[154, 165]]]",
    "process": ""
  },
  {
    "text": "Given that the foci of the ellipse lie on the $x$-axis, the length of the major axis $2a$ is $10$, and the eccentricity is $\\frac{3}{5}$, then the standard equation of the ellipse is?",
    "fact_expressions": "E: Ellipse;PointOnCurve(Focus(E), xAxis);a: Number;Length(MajorAxis(E)) = 2*a;2*a = 10;Eccentricity(E) = 3/5",
    "query_expressions": "Expression(E)",
    "answer_expressions": "x^2/16 + y^2/25 = 1",
    "fact_spans": "[[[2, 4], [47, 49]], [[2, 12]], [[16, 21]], [[2, 21]], [[16, 26]], [[2, 44]]]",
    "query_spans": "[[[47, 56]]]",
    "process": "The major axis of the ellipse is 10, and the eccentricity is \\frac{3}{5}, so a=5, c=3, then b=4. The standard equation of the ellipse with foci on the x-axis is \\frac{x2}{16}+\\frac{y^{2}}{25}=1"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2px$ ($p>0$) with directrix $l$, a line passing through $M(1, 0)$ with slope $\\sqrt{3}$ intersects $l$ at point $A$ and intersects $C$ at a point $B$. If $\\overrightarrow{AM}=\\overrightarrow{MB}$, then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;M: Point;A: Point;B: Point;l: Line;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(M) = (1, 0);Directrix(C) = l;PointOnCurve(M, G);Slope(G)=sqrt(3);Intersection(l, G) = A;OneOf(Intersection(G,C))=B;VectorOf(A, M) = VectorOf(M, B)",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 27], [77, 80]], [[137, 140]], [[62, 64]], [[37, 47]], [[71, 75]], [[86, 89]], [[31, 34], [65, 68]], [[9, 27]], [[2, 27]], [[37, 47]], [[2, 34]], [[36, 64]], [[48, 64]], [[62, 75]], [[62, 89]], [[92, 135]]]",
    "query_spans": "[[[137, 142]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ passes through the point $(\\sqrt{2}, \\sqrt{3})$, and has eccentricity $2$. Then the standard equation of this hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Point;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(G) = (sqrt(2), sqrt(3));PointOnCurve(G, C);Eccentricity(C) = 2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2/3 = 1",
    "fact_spans": "[[[0, 51], [87, 90]], [[8, 51]], [[8, 51]], [[52, 75]], [[0, 51]], [[52, 75]], [[0, 75]], [[0, 84]]]",
    "query_spans": "[[[87, 97]]]",
    "process": "Since the eccentricity of the hyperbola is 2, we have c=2a, so c^{2}=4a^{2}=a^{2}+b^{2}, which implies b^{2}=3a^{2}. Substituting the point (\\sqrt{2},\\sqrt{3}) into the hyperbola equation gives: \\frac{2}{a^{2}}-\\frac{3}{3a^{2}}=1, solving yields a^{2}=1, b^{2}=3. Therefore, the standard equation of the hyperbola is C: x^{2}-\\frac{y^{2}}{3}=1"
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=2 px(p>0)$, draw a line inclined at $45^{\\circ}$ to intersect the parabola at points $A$ and $B$. If the length of segment $AB$ is $8$, then $p=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);Inclination(H) = ApplyUnit(45, degree);A: Point;B: Point;Intersection(H, G) = {A, B};Length(LineSegmentOf(A, B)) = 8",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 21], [48, 51]], [[1, 21]], [[79, 82]], [[4, 21]], [[24, 27]], [[1, 27]], [[45, 47]], [[0, 47]], [[28, 47]], [[52, 55]], [[58, 61]], [[45, 63]], [[65, 77]]]",
    "query_spans": "[[[79, 84]]]",
    "process": ""
  },
  {
    "text": "If the curve represented by the equation $9 m x^{2}+y^{2}=9$ is an ellipse with foci on the $y$-axis, then the range of the constant $m$ is the interval?",
    "fact_expressions": "G: Ellipse;H: Curve;m: Number;Expression(H)=(9*m*x^2+y^2=9);H=G;PointOnCurve(Focus(G),yAxis)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(1/9,+oo)",
    "fact_spans": "[[[37, 39]], [[25, 27]], [[41, 46]], [[1, 27]], [[25, 39]], [[28, 39]]]",
    "query_spans": "[[[41, 55]]]",
    "process": "First simplify the equation to $\\frac{x^{2}}{m}+\\frac{y^{2}}{9}=1$, then according to the problem, obtain $\\begin{cases}\\frac{1}{m}>0\\\\\\frac{1}{m}<9\\end{cases}$, and solve the system of inequalities. $9mx^{2}+y^{2}=9\\Rightarrow\\frac{x^{2}}{1}+\\frac{y^{2}}{9}=1$. Since the equation $\\frac{x^{2}}{1}+\\frac{y^{2}}{9}=1$ represents an ellipse with foci on the $y$-axis, it follows that $\\begin{cases}\\frac{1}{m}>0\\\\\\frac{1}{m}<9\\end{cases}$, solving yields $m>\\frac{1}{9}$."
  },
  {
    "text": "Draw a tangent from the right focus $F$ of the hyperbola $M$: $\\frac{x^{2}}{3}-y^{2}=1$ to the circle $C$: $x^{2}+(y+1)^{2}=\\frac{1}{2}$. This tangent intersects the right branch of $M$ at points $A$ and $B$. Then $|A B|$=?",
    "fact_expressions": "M: Hyperbola;Expression(M) = (x^2/3 - y^2 = 1);F: Point;RightFocus(M) = F;C: Circle;Expression(C) = (x^2 + (y + 1)^2 = 1/2);A: Point;B: Point;L:Line;TangentOfPoint(F,C)=L;Intersection(L,RightPart(M))={A,B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[1, 34], [85, 88]], [[1, 34]], [[38, 41]], [[1, 41]], [[42, 77]], [[42, 77]], [[93, 96]], [[97, 100]], [76, 77], [0, 77], [79, 99]]",
    "query_spans": "[[[104, 113]]]",
    "process": "First, set the line equation. Using the condition that the line is tangent to the circle, find the line equation, then use the chord length formula to find the chord length |AB|. Since the line passes through the right focus of the hyperbola and is tangent to the circle, the slope of the line exists. Let the line equation be y−0=k(x−2). From the tangency condition between the line and the circle, we have \\frac{|2k-1|}{\\sqrt{1+k^{2}}}=\\frac{\\sqrt{2}}{2}, solving gives k=1 or k=\\frac{1}{7}. When k=\\frac{1}{7}, the slope of one asymptote of the hyperbola is \\frac{\\sqrt{3}}{3}, and since \\frac{1}{7}<\\frac{\\sqrt{3}}{3}, this line does not intersect the right branch of the hyperbola at two points, so it is discarded. Thus, the line equation is y=x−2. Solving simultaneously with the hyperbola equation, eliminating variables yields 2x^{2}−12x+15=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), then x_{1}+x_{2}=6, x_{1}x_{2}=\\frac{15}{2}, so |AB|=\\sqrt{1+k^{2}}|x_{1}-x_{2}|=\\sqrt{2}\\cdot\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\sqrt{2}\\times\\sqrt{6^{2}-4\\times\\frac{15}{2}}=2\\sqrt{3}"
  },
  {
    "text": "The product of the distances from a point $P$ on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{25}=1$ to its two foci is $m$. Then, when $m$ takes its maximum value, the coordinates of point $P$ are?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/9 + y^2/25 = 1);PointOnCurve(P, G);F1:Point;F2:Point;Focus(G)={F1,F2};m:Number;Distance(P, F1)*Distance(P,F2) = m;WhenMax(m)",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(3, 0), (-3, 0)",
    "fact_spans": "[[[0, 38]], [[42, 45], [71, 75]], [[0, 38]], [[0, 45]], [], [], [[0, 49]], [[56, 59], [62, 65]], [[0, 59]], [[61, 70]]]",
    "query_spans": "[[[71, 80]]]",
    "process": "Let P(x,y), then m = e|y - \\frac{25}{4}| \\cdot e|y + \\frac{25}{4} = ex\\left(\\frac{25}{4}\\right)^{2} - y^{2} \\leqslant e^{2}\\left(\\frac{25}{4}\\right)^{2}, with equality if and only if y = 0, at which point the coordinates of P are (3,0) or (-3,0)."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$, a line $l$ passes through the focus of the parabola $C$ and intersects the parabola at points $A$ and $B$. The circle with diameter $AB$ intersects the directrix of the parabola at the common point $M(-2,-2)$. Then the slope $k$ of line $l$ is?",
    "fact_expressions": "l: Line;C: Parabola;G: Circle;A: Point;B: Point;M: Point;k: Number;p:Number;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(M) = ((-1)*2, (-1)*2);Slope(l) = k;PointOnCurve(Focus(C),l);Intersection(l,C)={A,B};IsDiameter(LineSegmentOf(A,B),G);Intersection(Directrix(C),G)=M",
    "query_expressions": "k",
    "answer_expressions": "-2",
    "fact_spans": "[[[28, 33], [94, 99]], [[2, 28], [34, 40], [44, 47], [71, 74]], [[69, 70]], [[49, 52]], [[53, 56]], [[82, 92]], [[102, 105]], [[10, 28]], [[10, 28]], [[2, 28]], [[82, 92]], [[94, 105]], [[28, 43]], [[28, 58]], [[59, 70]], [[59, 92]]]",
    "query_spans": "[[[102, 107]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since -\\frac{p}{2} = -2 \\Rightarrow p = 4. As shown in the figure, draw perpendiculars AP and BQ from A and B to the directrix, with P and Q as the feet of the perpendiculars. Let G be the midpoint of AB, and draw GH perpendicular to the directrix from G, with H as the foot. Then AP // BQ // GH, so |AP| = |AF|, |BQ| = |BF|, and |AP| + |BQ| = |AF| + |BF| = |AB|. Also, |GH| = \\frac{1}{2}(|AP| + |BQ|) = \\frac{1}{2}|AB|. Therefore, the circle with AB as diameter is tangent to the directrix, and point M coincides with point H. The common point of the circle with AB as diameter and the directrix of the parabola is M(-2,-2), so \\frac{y_{1}+y_{2}}{2} = -2. Because \\begin{cases} \\\\ \\end{cases} 2px_{1} \\Rightarrow (y_{1}+y_{2})(y_{1}-y_{2}) = 2p(x_{1}-x_{2}). So \\frac{y_{1}}{x_{1}} - \\frac{y_{2}}{x_{2}} = \\frac{2p}{y_{1}+y_{2}} = ."
  },
  {
    "text": "What are the coordinates of the points on the parabola $y^{2}=12 x$ whose distance from the focus is equal to $9$?",
    "fact_expressions": "C:Parabola;P:Point;Expression(C) = (y^2 = 12*x);PointOnCurve(P,C);Distance(P,Focus(C))=9",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "{(6,6*sqrt(2)),(6,-6*sqrt(2))}",
    "fact_spans": "[[[0, 15]], [[28, 29]], [[0, 15]], [[0, 29]], [[0, 29]]]",
    "query_spans": "[[[28, 34]]]",
    "process": ""
  },
  {
    "text": "A hyperbola has an asymptote given by the line $x - 2y = 0$, and passes through the point $(\\sqrt{5}, -1)$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Line;I: Point;Expression(H) = (x - 2*y = 0);Coordinate(I) = (sqrt(5), -1);OneOf(Asymptote(G)) = H;PointOnCurve(I, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - 4*y^2 = 1",
    "fact_spans": "[[[0, 3], [42, 45]], [[10, 21]], [[24, 40]], [[10, 21]], [[24, 40]], [[0, 21]], [[0, 40]]]",
    "query_spans": "[[[42, 50]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ $(m>0)$ is $x+3 y=0$, then $m$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2 + x^2/m = 1);m: Number;m>0;Expression(OneOf(Asymptote(G))) = (x + 3*y = 0)",
    "query_expressions": "m",
    "answer_expressions": "9",
    "fact_spans": "[[[2, 35]], [[2, 35]], [[55, 58]], [[5, 35]], [[2, 53]]]",
    "query_spans": "[[[55, 60]]]",
    "process": "The asymptotes of \\frac{x^{2}}{m}-y^{2}=1(m>0) are given by: y=\\pm\\frac{1}{\\sqrt{n}}. Also, one asymptote of the hyperbola is x+3y=0, that is, y=-\\frac{1}{3}x. Therefore, \\sqrt{m}=3, m=9"
  },
  {
    "text": "Let the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ have left focus $F$. A line passing through point $F$ intersects the ellipse $C$ at points $A$ and $B$. The inclination angle of line $l$ is $60^{\\circ}$, and $AF=2 FB$. Then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/a^2+y^2/b^2 = 1);a: Number;b: Number;a>b;b>0;F: Point;LeftFocus(C) = F;H: Line;PointOnCurve(F, H);Intersection(H, C) = {A, B};l: Line;Inclination(l) = ApplyUnit(60, degree);LineSegmentOf(A, F) = 2*LineSegmentOf(F, B);A: Point;B: Point",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2/3",
    "fact_spans": "[[[1, 58], [76, 81], [133, 138]], [[1, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[63, 66], [68, 72]], [[1, 66]], [[73, 75]], [[67, 75]], [[73, 95]], [[96, 101]], [[96, 118]], [[121, 130]], [[84, 87]], [[90, 93]]]",
    "query_spans": "[[[133, 144]]]",
    "process": ""
  },
  {
    "text": "If the chord $AB$ of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ is bisected by the point $(1,1)$, then the equation of the line containing $AB$ is?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;C: Point;Expression(G) = (x^2/9 + y^2/4 = 1);Coordinate(C) = (1, 1);IsChordOf(LineSegmentOf(A,B), G) = True;MidPoint(LineSegmentOf(A, B)) = C",
    "query_expressions": "Expression(OverlappingLine(LineSegmentOf(A,B)))",
    "answer_expressions": "4*x+9*y-13=0",
    "fact_spans": "[[[1, 38]], [[40, 45]], [[40, 45]], [[46, 54]], [[1, 38]], [[46, 54]], [[1, 45]], [[40, 56]]]",
    "query_spans": "[[[58, 72]]]",
    "process": "Let the intersection points of the line and the ellipse be A(x_{1},y_{1}), B(x_{2},y_{2}). Since (1,1) is the midpoint of AB, we have x_{1}+x_{2}=2, y_{1}+y_{2}=2. Therefore, k_{AB}=\\frac{y}{x}\\frac{y_{1}-y_{2}}{x_{1}}=-\\frac{1}{9}. Hence, the equation of the required line is y-1=-\\frac{4}{9}(x-1), that is, 4x+9y-13=0."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$, point $M(x_{0}, 2 \\sqrt{2})$ with $(x_{0}>\\frac{p}{2})$ is a point on parabola $C$, a circle centered at point $M$ intersects the line $x=\\frac{p}{2}$ at points $E$ and $G$, if $\\sin \\angle M F G=\\frac{1}{3}$, then the equation of parabola $C$ is?",
    "fact_expressions": "C: Parabola;p: Number;Expression(C) = (y^2 = 2*p*x);p > 0;F: Point;Focus(C) = F;M: Point;x0: Number;x0 > p/2;Coordinate(M) = (x0, 2*sqrt(2));PointOnCurve(M, C);Z: Circle;Center(Z) = M;H: Line;Expression(H) = (x = p/2);E: Point;G: Point;Intersection(Z, H) = {E, G};Sin(AngleOf(M, F, G)) = 1/3",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2 = 4*x",
    "fact_spans": "[[[2, 28], [80, 86], [164, 170]], [[10, 28]], [[2, 28]], [[10, 28]], [[32, 35]], [[2, 35]], [[36, 79], [91, 95]], [[37, 79]], [[37, 79]], [[36, 79]], [[36, 89]], [[99, 100]], [[90, 100]], [[101, 118]], [[101, 118]], [[120, 123]], [[124, 127]], [[99, 129]], [[131, 162]]]",
    "query_spans": "[[[164, 175]]]",
    "process": "As shown in the figure, draw MN\\bot EG, with the foot of the perpendicular at N. From the given conditions, point M(x_{0},2\\sqrt{2}), (x_{0}>\\frac{p}{2}), lies on the parabola C, so 2px_{0}=8\\textcircled{1}. By the definition of a parabola, |MF|=x_{0}+\\frac{p}{2}. Since \\sin\\angle MFG=\\frac{1}{3}, it follows that |MN|=\\frac{1}{3}|MF|=\\frac{1}{3}(x_{0}+\\frac{p}{2}). Therefore, x_{0}-\\frac{p}{2}=\\frac{1}{3}(x_{0}+\\frac{p}{2}), solving gives x_{0}=p\\textcircled{2}. From \\textcircled{1}\\textcircled{2}, we solve to get x_{0}=p=-2 (discarded) or x_{0}=p=2. Hence, the equation of parabola C is y^{2}=4x."
  },
  {
    "text": "Given that the intersection point of an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ and the directrix of the parabola $y^{2}=2 p x$ $(p>0)$ has coordinates $(-\\frac{4}{3}, \\frac{8}{3})$, and that a common point $M$ of the hyperbola and the parabola has coordinates $(x_{0}, 4)$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;H: Parabola;Expression(H) = (y^2 = 2*(p*x));p: Number;p>0;Coordinate(Intersection(OneOf(Asymptote(G)), Directrix(H))) = (-4/3, 8/3);M: Point;x0: Number;Coordinate(M) = (x0, 4);OneOf(Intersection(G, H)) = M",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5 - y^2/20 = 1",
    "fact_spans": "[[[2, 58], [126, 129], [160, 163]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[65, 86], [130, 133]], [[65, 86]], [[68, 86]], [[68, 86]], [[2, 124]], [[139, 142]], [[146, 158]], [[139, 158]], [[126, 158]]]",
    "query_spans": "[[[160, 168]]]",
    "process": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ are given by $y=\\pm\\frac{b}{a}x$. The directrix of the parabola $y^{2}=2px$ is $x=-\\frac{p}{2}$. From the given conditions, we have $\\frac{p}{2}=\\frac{4}{3}$, so $p=\\frac{8}{3}$. Also, $\\frac{b}{a}=2$, so $b=2a$$\\textcircled{1}$. Given the coordinates of point $M$ as $(x_{0},4)$, we obtain $16=2px_{0}=\\frac{16}{3}x_{0}$, solving gives $x_{0}=3$. Substituting $M(3,4)$ into the hyperbola equation yields $\\frac{9}{a^{2}}-\\frac{16}{b^{2}}=1$$\\textcircled{2}$. Solving equations $\\textcircled{1}$ and $\\textcircled{2}$ gives $a=\\sqrt{5}$, $b=2\\sqrt{5}$, thus the equation of the hyperbola is $\\frac{x^{2}}{5}-\\frac{y^{2}}{20}=$"
  },
  {
    "text": "Let point $ P $ be on the right branch of the hyperbola $ x^{2} - \\frac{y^{2}}{48} = 1 $. From $ P $, draw tangents to circle $ C_{1} $: $ (x+7)^{2} + y^{2} = 4 $ and circle $ C_{2} $: $ (x-7)^{2} + y^{2} = 1 $, with tangent points $ M $ and $ N $ respectively. Then the minimum value of $ |PM|^{2} - |PN|^{2} $ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/48 = 1);P: Point;PointOnCurve(P, RightPart(G));C1:Circle;Expression(C1) = ((x + 7)^2 + y^2 = 4);C2:Circle;Expression(C2) = ((x - 7)^2 + y^2 = 1);M: Point;N: Point;l1:Line;l2:Line;TangentPoint(l1,C1)=M;TangentPoint(l2,C2)=N;TangentOfPoint(P,C1)=l1;TangentOfPoint(P,C2)=l2",
    "query_expressions": "Min(Abs(LineSegmentOf(P, M))^2 - Abs(LineSegmentOf(P, N))^2)",
    "answer_expressions": "25",
    "fact_spans": "[[[1, 30]], [[1, 30]], [[36, 39]], [[0, 39]], [[43, 71]], [[43, 71]], [[72, 100]], [[72, 100]], [[109, 112]], [[113, 116]], [], [], [[0, 116]], [[0, 116]], [[0, 103]], [[0, 103]]]",
    "query_spans": "[[[118, 145]]]",
    "process": "From the hyperbola equation, its foci are at $(\\pm7,0)$; the center of circle $C_{2}$ is $C_{2}(7,0)$, radius $r_{2}=1$; since $PM$, $PN$ are tangent lines to the two circles respectively, $\\therefore PM^{2}=PC_{1}^{2}-r_{1}^{2}=PC_{1}^{2}-4$, $PN^{2}=PC_{2}^{2}-r_{2}^{2}=PC_{2}^{2}-1$ $\\therefore PM^{2}-PN^{2}=PC_{1}^{2}-PC_{2}^{2}-3=(PC_{1}+PC_{2})(PC_{1}-PC_{2})-3$. Since $P$ is a point on the right branch of the hyperbola and the foci of the hyperbola are $C_{1}, C_{2}$, $\\therefore PC_{1}-PC_{2}=2$. Also, $PC_{1}+PC_{2}\\geqslant C_{1}C_{2}=14$ (equality holds when $P$ is the right vertex of the hyperbola) $\\therefore PM^{2}-PN^{2}=(PC_{1}+PC_{2})(PC_{1}-PC_{2})-3\\geqslant14\\times2-3=25$. Thus, the minimum value of $PM^{2}-PN^{2}$ is 25."
  },
  {
    "text": "Given the line $l$: $x - y + 1 = 0$ intersects the ellipse $C$: $\\frac{x^{2}}{2} + y^{2} = 1$ at points $A$ and $B$, then $|AB| =$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/2 + y^2 = 1);l: Line;Expression(l) = (x - y + 1 = 0);A: Point;B: Point;Intersection(l, C) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "4*sqrt(2)/3",
    "fact_spans": "[[[19, 51]], [[19, 51]], [[2, 18]], [[2, 18]], [[53, 56]], [[57, 60]], [[2, 62]]]",
    "query_spans": "[[[64, 73]]]",
    "process": "From \\begin{cases}x-y+1=0\\\\\\frac{x^{2}}{2}+y^{2}=1\\end{cases}, solving gives \\begin{cases}x=0\\\\y=1\\end{cases} or \\begin{cases}x=-\\frac{4}{3}\\\\y=-\\frac{1}{3}\\end{cases}. Without loss of generality, let A(0,1), B(-\\frac{4}{3},-\\frac{1}{3}). Therefore, |AB|=\\sqrt{(\\frac{4}{3})^{2}\\times2}=\\frac{4\\sqrt{2}}{3}"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0, b>0$), one of its asymptotes is tangent to the curve $y=1+\\ln x$. Then, the eccentricity of this hyperbola is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;H: Curve;Expression(H) = (y = ln(x) + 1);IsTangent(OneOf(Asymptote(C)),H) = True",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 63], [88, 91]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[70, 83]], [[70, 83]], [[2, 85]]]",
    "query_spans": "[[[88, 97]]]",
    "process": "Let the coordinates of the tangent point be $(t, \\ln t + 1)$. Differentiating the function $y = 1 + \\ln x$ gives $y = \\frac{1}{x}$. Therefore, the equation of the tangent line to the curve $y = 1 + \\ln x$ at $x = t$ is $y - (1 + \\ln t) = \\frac{1}{t}(x - t)$. Since this tangent line passes through the origin, we have $-1 - \\ln t = -1$, solving which yields $t = 1$. Thus, the slope of the tangent line is $\\frac{b}{a} = 1$, so the eccentricity of the hyperbola is $e = \\frac{c}{a} = \\sqrt{1 + \\left(\\frac{b}{a}\\right)^2} = \\sqrt{2}$."
  },
  {
    "text": "Given that the circle $x^{2}+y^{2}-6 x-7=0$ is tangent to the directrix of the parabola $y^{2}=2 p x$ $(p>0)$, then $p$=?",
    "fact_expressions": "G: Parabola;p: Number;H: Circle;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (-6*x + x^2 + y^2 - 7 = 0);IsTangent(H, Directrix(G))",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[25, 47]], [[54, 57]], [[2, 24]], [[28, 47]], [[25, 47]], [[2, 24]], [[2, 52]]]",
    "query_spans": "[[[54, 59]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a point on the left branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola, and $\\cos \\angle P F_{1} F_{2}=\\sin \\angle P F_{2} F_{1}=\\frac{\\sqrt{5}}{5}$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, LeftPart(G));LeftFocus(G) = F1;RightFocus(G) = F2;Cos(AngleOf(P,F1,F2))=Sin(AngleOf(P,F2,F1));Sin(AngleOf(P,F2,F1))=sqrt(5)/5",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[6, 62], [84, 87], [168, 171]], [[9, 62]], [[9, 62]], [[2, 5]], [[68, 75]], [[76, 83]], [[9, 62]], [[9, 62]], [[6, 62]], [[2, 67]], [[68, 92]], [[68, 92]], [[94, 166]], [[94, 166]]]",
    "query_spans": "[[[168, 176]]]",
    "process": "According to the problem, $\\sin\\angle PF_{1}F_{2} = \\cos\\angle PF_{2}F_{1} = \\frac{2\\sqrt{5}}{5}$, so $\\sin\\angle F_{1}PF_{2} = \\sin(\\angle PF_{1}F_{2} + \\angle PF_{2}F_{1}) = 1$. Based on the definition of hyperbola and the law of sines, $e = \\frac{2c}{2a} = \\frac{|F_{1}F_{2}|}{|PF_{2}| - |PF_{1}|} = \\frac{}{\\sin\\angle}$. The method for solving this is examined using the law of sines in solving triangles, testing the mathematical thinking method of reduction and transformation; it is a medium-difficulty problem."
  },
  {
    "text": "Ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ has left and right foci $F_{1}$, $F_{2}$ respectively. Point $P$ lies on the ellipse, $G$ lies on $OP$, $|OP|=4|OG|$. In $\\Delta P F_{1} F_{2}$, the incenter is $I$. If $\\overrightarrow{IG}=\\lambda \\overrightarrow{F_{1} F_{2}}$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;O: Origin;P: Point;F1: Point;F2: Point;G: Point;I: Point;a > b;b > 0;lambda:Number;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);PointOnCurve(G, LineSegmentOf(O, P));Abs(LineSegmentOf(O, P)) = 4*Abs(LineSegmentOf(O, G));Incenter(TriangleOf(P,F1,F2))=I;VectorOf(I, G) = lambda*VectorOf(F1, F2)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "1/3",
    "fact_spans": "[[[0, 57], [87, 89], [212, 217]], [[7, 57]], [[7, 57]], [[95, 100]], [[82, 86]], [[66, 73]], [[74, 81]], [[91, 94]], [[147, 150]], [[7, 57]], [[7, 57]], [[152, 210]], [[0, 57]], [[0, 81]], [[0, 81]], [[82, 90]], [[91, 101]], [[102, 116]], [[119, 150]], [[152, 210]]]",
    "query_spans": "[[[212, 223]]]",
    "process": "Let point $ P(x_{0},y_{0}) $. From $ |OP| = 4|OG| $, the coordinates of $ G $ can be expressed. From $ \\overrightarrow{IG} = \\lambda \\overrightarrow{F_{1}F_{2}} $, the vertical coordinate of point $ I $ can be obtained. Then, since $ I $ is the incenter of $ \\triangle PF_{1}F_{2} $, combining with the area of $ \\triangle PF_{1}F_{2} $, we obtain $ \\frac{1}{2}(|PF_{1}| + |PF_{2}| + |F_{1}F_{2}|) \\times \\frac{1}{4}|y_{0}| = \\frac{1}{2} \\times |F_{1}F_{2}| \\times |y_{0}| $, from which the eccentricity can be found. Solution: Let point $ P(x_{0},y_{0}) $, $ y_{0} \\neq 0 $. Since $ |OP| = 4|OG| $, $ G(\\frac{1}{4}x_{0},\\frac{1}{4}y_{0}) $, $ \\overrightarrow{1} $. Since the incenter of $ \\triangle PF_{1}F_{2} $ is $ I $, the inradius of $ \\triangle PF_{1}F_{2} $ is $ \\frac{1}{4}|y_{0}| $. Therefore, the area of $ \\triangle PF_{1}F_{2} $ is $ \\frac{1}{2}(|PF_{1}| + |PF_{2}| + |F_{1}F_{2}|) \\times \\frac{1}{4}|y_{0}| = \\frac{1}{2} \\times |F_{1}F_{2}| \\times |y_{0}| $, that is, $ \\frac{1}{2}(2a + 2c) \\times \\frac{1}{4}|y_{0}| = \\frac{1}{2} \\times 2c \\times |y_{0}| $. Simplifying yields $ a = 3c $, so the eccentricity $ e = \\frac{c}{a} = \\frac{1}{3} $."
  },
  {
    "text": "What is the focal distance of the ellipse $\\frac{x^{2}}{2}+\\frac{y^{2}}{4}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2/4 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 42]]]",
    "process": "According to the quantitative relationship among a, b, c in an ellipse, solve for it. The focal distance of the ellipse \\frac{x^{2}}{2}+\\frac{y^{2}}{4}=1 is 2c=2\\sqrt{a^{2}-b^{2}}=2\\sqrt{4-2}=2\\sqrt{2}"
  },
  {
    "text": "Given that the ellipse $C$ passes through the point $M(1,-\\frac{3}{2})$ and has foci at $F_{1}(-1,0)$ and $F_{2}(1,0)$, then the standard equation of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;M: Point;Coordinate(M) = (1, -3/2);PointOnCurve(M, C);F1: Point;F2: Point;Coordinate(F1) = (-1, 0);Coordinate(F2) = (1, 0);Focus(C) = {F1, F2}",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4+y^2/3=1",
    "fact_spans": "[[[24, 29], [65, 70]], [[3, 23]], [[3, 23]], [[2, 29]], [[35, 48]], [[51, 63]], [[35, 48]], [[51, 63]], [[24, 63]]]",
    "query_spans": "[[[65, 77]]]",
    "process": "First, obtain $a$ from the definition of the ellipse, then find $b$ to get the equation of the ellipse. According to the problem, $2a = \\sqrt{(-1-1)^{2}+(0+\\frac{3}{2})^{2}} + \\sqrt{(1-1)^{2}+(0+\\frac{3}{2})^{2}} = 4$, so $a = 2$, thus $b = \\sqrt{2^{2}-1^{2}} = \\sqrt{3}$, therefore the equation of the ellipse is $\\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1$."
  },
  {
    "text": "The length of the imaginary axis of the hyperbola $x^{2}-y^{2}=2$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 2)",
    "query_expressions": "Length(ImageinaryAxis(G))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[0, 18]], [[0, 18]]]",
    "query_spans": "[[[0, 24]]]",
    "process": ""
  },
  {
    "text": "The standard equation of an ellipse that has the same focal distance as the ellipse $\\frac{x^{2}}{63}+\\frac{y^{2}}{38}=1$ and passes through the center of the circle $x^{2}+y^{2}-6 x-8 y=0$ is?",
    "fact_expressions": "G: Ellipse;Expression(G)=(x^2/63+y^2/38=1);H: Circle;Expression(H)=(x^2+y^2-6*x-8*y=0);C:Ellipse;FocalLength(G) = FocalLength(C);PointOnCurve(Center(H),C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 40]], [[1, 40]], [[49, 73]], [[49, 73]], [[77, 79]], [[0, 79]], [[48, 79]]]",
    "query_spans": "[[[77, 86]]]",
    "process": ""
  },
  {
    "text": "The focal length of the hyperbola $3 x^{2}-y^{2}=3$ is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (3*x^2 - y^2 = 3)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 26]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ has its right focus at $F$, then the distance from $F$ to the asymptote of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;F: Point;Expression(C) = (x^2/9 - y^2/16 = 1);RightFocus(C) = F",
    "query_expressions": "Distance(F, Asymptote(C))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 46], [60, 66]], [[51, 54], [56, 59]], [[2, 46]], [[2, 54]]]",
    "query_spans": "[[[56, 75]]]",
    "process": ""
  },
  {
    "text": "Point $M$ is a point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, and $F_{1}$, $F_{2}$ are the two foci of the ellipse. Then the equation of the locus of the incenter of $\\Delta F_{1} M F_{2}$ is?",
    "fact_expressions": "E: Ellipse;Expression(E) = (x^2/9 + y^2/5 = 1);M: Point;PointOnCurve(M, E);F1: Point;F2: Point;Focus(E) = {F1, F2}",
    "query_expressions": "LocusEquation(Incenter(TriangleOf(F1, M, F2)))",
    "answer_expressions": "(x^2/4 + 5*y^2/4 = 1) & Negation(y=0)",
    "fact_spans": "[[[5, 43], [66, 68]], [[5, 43]], [[0, 4]], [[0, 47]], [[49, 56]], [[57, 64]], [[49, 73]]]",
    "query_spans": "[[[76, 108]]]",
    "process": ""
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{m-1}+\\frac{y^{2}}{m^{2}-4}=1$ represents a hyperbola with foci on the $x$-axis, what is the range of values for $m$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(m - 1) + y^2/(m^2 - 4) = 1);PointOnCurve(Focus(G), xAxis);m:Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(1,2)",
    "fact_spans": "[[[58, 61]], [[2, 61]], [[49, 61]], [[63, 66]]]",
    "query_spans": "[[[63, 73]]]",
    "process": "The focus is on the x-axis, then \\begin{cases}m-1>0\\\\m^2-4<0\\end{cases}>0, solving gives 1<m<2. Therefore, the range of m is (1,2)."
  },
  {
    "text": "Let a line $l$ pass through the focus $F$ of the parabola $y^{2}=4x$, intersecting the parabola at points $A$ and $B$. Denote by $P$ the intersection point of the tangents $l_{1}$, $l_{2}$ to the parabola at points $A$ and $B$, respectively. Then the minimum area of $\\Delta ABP$ is?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G)=F;PointOnCurve(F, l);Intersection(l, G) = {A, B};l1:Line;l2:Line;TangentOnPoint(A,G)=l1;TangentOnPoint(B,G)=l2;Intersection(l1,l2)=P",
    "query_expressions": "Min(Area(TriangleOf(A, B, P)))",
    "answer_expressions": "4",
    "fact_spans": "[[[22, 27]], [[1, 15], [28, 31], [44, 47]], [[33, 36], [48, 51]], [[37, 40], [52, 55]], [[81, 84]], [[18, 21]], [[1, 15]], [[1, 21]], [[0, 27]], [[22, 42]], [[61, 68]], [[70, 77]], [[44, 77]], [[44, 77]], [[61, 84]]]",
    "query_spans": "[[[86, 108]]]",
    "process": "From $ y^{2} = 4x $, we get: $ F(1,0) $. Let the equation of line $ AB $ be: $ x = my + 1 $, $ A\\left(\\frac{y_{1}^{2}}{4}, y_{1}\\right) $, $ B\\left(\\frac{y_{2}^{2}}{4}, y_{2}\\right) $, and $ y_{1} > 0 > y_{2} $. Then solving the system \n\\[\n\\begin{cases}\nx = my + 1 \\\\\ny^{2} = 4x\n\\end{cases}\n\\]\nwe obtain: $ (0, \\frac{1}{2}] $. Then: $ y_{1} + y_{2} = 4m $, $ y_{1}y_{2} = -4 $. From $ y^{2} = 4x $, we get: $ y = 2\\sqrt{x} $ ($ y > 0 $), $ y = -2\\sqrt{x} $ ($ y < 0 $). Then when $ y > 0 $, $ y' = \\frac{1}{\\sqrt{x}} $, so the slope of the tangent at point $ A $ is: $ \\frac{1}{\\sqrt{\\frac{y_{1}^{2}}{4}}} = \\frac{2}{y_{1}} $. Therefore, $ l_{1}: y - y_{1} = \\frac{2}{y_{1}}\\left(x - \\frac{y_{1}^{2}}{4}\\right) $. Similarly, we get: $ l_{2}: y - y_{2} = \\frac{2}{y_{2}}\\left(x - \\frac{y_{2}^{2}}{4}\\right) $. Then: $ P\\left(\\frac{y_{1}y_{2}}{4}, \\frac{y_{1} + y_{2}}{2}\\right) $, i.e., $ P(-1, 2m) $. $ k_{AB} = \\frac{1}{m} $, then $ k_{PF} \\cdot k_{AB} = -1 $. $ |AB| = \\sqrt{1 + m^{2}} \\cdot \\sqrt{(y_{1} + y_{2})^{2} - 4y_{1}y_{2}} = \\sqrt{1 + m^{2}} \\cdot \\sqrt{16m^{2} + 16} = 4(m^{2} + 1) $. Also, $ |AB| = \\sqrt{1 + m^{2}} \\cdot \\sqrt{(y_{1} + y_{2})^{2} - 4y_{1}y_{2}} $, $ |PF| = \\sqrt{(1 + 1)^{2} + (0 - 2m)^{2}} = 2\\sqrt{m^{2} + 1} $. Since $ m^{2} \\geqslant 0 $, $ S_{\\triangle ABP} \\geqslant 4 $. Therefore, the minimum area of $ \\triangle ABP $ is 4. Final answer to this problem: 4. This problem examines the comprehensive application of lines and parabolas. The key is to use derivatives to quickly find the slopes of tangents, thereby obtaining the coordinates of the intersection point of the two tangents, and then using chord length formula and distance formula to get the base and height of the triangle, thus constructing a functional relationship for the area. It belongs to a more difficult problem."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has a real axis length of $2$ and an eccentricity of $2$, then the coordinates of the left focus of hyperbola $C$ are?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Length(RealAxis(C)) = 2;Eccentricity(C) = 2",
    "query_expressions": "Coordinate(LeftFocus(C))",
    "answer_expressions": "(-2, 0)",
    "fact_spans": "[[[2, 64], [82, 88]], [[2, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[2, 72]], [[2, 80]]]",
    "query_spans": "[[[82, 96]]]",
    "process": ""
  },
  {
    "text": "Given $|AB|=4$, point $P$ moves in the plane containing $A$ and $B$ such that $|PA|+|PB|=6$. Then the maximum value of $|PA|$ is? The minimum value is?",
    "fact_expressions": "A: Point;B: Point;P: Point;Abs(LineSegmentOf(A, B)) = 4;Abs(LineSegmentOf(P,A))+Abs(LineSegmentOf(P,B))=6",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)));Max(Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "2;5",
    "fact_spans": "[[[15, 18]], [[21, 24]], [[10, 14]], [[1, 9]], [[33, 48]]]",
    "query_spans": "[[[50, 68]], [[50, 63]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the points on the parabola $y^{2}=x$ that are equidistant from its directrix and vertex are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = x);P:Point;Distance(P,Directrix(G))=Distance(P,Vertex(G));PointOnCurve(P,G)",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1/8, pm*(sqrt(2)/4))",
    "fact_spans": "[[[0, 12], [14, 15]], [[0, 12]], [[25, 26]], [[13, 26]], [[0, 26]]]",
    "query_spans": "[[[25, 31]]]",
    "process": "The directrix of the parabola $ y^{2} = x $ is $ x = -\\frac{1}{4} $; the vertex is $ (0,0) $. Let $ P(x_{0}, y_{0}) $ be the point on the parabola equidistant from the directrix and the vertex. Then we have  \n$$\n\\begin{cases}\ny_{0}^{2} = x_{0} \\\\\n\\sqrt{x_{0}^{2} + y_{0}^{2}} = x_{0} + \\frac{1}{4}\n\\end{cases}\n$$\nSolving this system yields  \n$$\n\\begin{cases}\nx_{0} = \\frac{1}{8} \\\\\ny_{0} = \\pm \\frac{\\sqrt{2}}{4}\n\\end{cases}\n$$\n$ \\therefore P\\left( \\frac{1}{8}, \\pm \\frac{\\sqrt{2}}{4} \\right) $."
  },
  {
    "text": "Draw a line $l$ through the focus of the parabola $y^{2}=4x$, intersecting the parabola at points $A$ and $B$. Let $M$ be the midpoint of $AB$. If $|AB|=8$, then what is the horizontal coordinate of point $M$?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;M: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G),l);Intersection(l,G) = {A,B};MidPoint(LineSegmentOf(A, B)) = M;Abs(LineSegmentOf(A, B)) = 8",
    "query_expressions": "XCoordinate(M)",
    "answer_expressions": "3",
    "fact_spans": "[[[19, 24]], [[1, 15], [26, 29]], [[30, 33]], [[34, 37]], [[49, 52], [65, 69]], [[1, 15]], [[0, 24]], [[19, 39]], [[40, 52]], [[54, 63]]]",
    "query_spans": "[[[65, 75]]]",
    "process": "From the parabola $ y^{2} = 4x $, the focus is $ F(1,0) $. If $ AB \\perp x $-axis, then $ |AB| = 2p = 4 $, which does not satisfy the condition, so it is discarded. Let the equation of line $ l $ be: $ my = x - 1 $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Solving the system\n$$\n\\begin{cases}\nmy = x - 1 \\\\\ny^{2} = 4x\n\\end{cases}\n$$\nyields $ y^{2} - 4my - 4 = 0 $. Therefore,\n$$\n|AB| = \\sqrt{(1 + m^{2})[(y_{1} + y_{2})^{2} - 4y_{1}y_{2}]} = \\sqrt{(1 + m^{2})(16m^{2} + 16)} = 8,\n$$\nwhich simplifies to $ m^{2} = 1 $, so $ m = \\pm 1 $. When $ m = 1 $, solving the system\n$$\n\\begin{cases}\ny = x - 1 \\\\\ny^{2} = 4x\n\\end{cases}\n$$\ngives $ x^{2} - 6x + 1 = 0 $, so $ x_{1} + x_{2} = 6 $, hence $ \\frac{x_{1} + x_{2}}{2} = 3 $. Similarly, when $ m = -1 $, $ \\frac{x_{1} + x_{2}}{2} = 3 $. Therefore, the horizontal coordinate of the midpoint of segment $ AB $ is 3. b female an for. 3"
  },
  {
    "text": "Given that point $M$ lies on the parabola $y^{2}=4x$, $F$ is the focus of the parabola, and $A$ lies on the circle $C$: $(x-4)^{2}+(y-1)^{2}=1$, then the minimum value of $|MA|+|MF|$ is?",
    "fact_expressions": "G: Parabola;C: Circle;M: Point;A: Point;F: Point;Expression(G) = (y^2 = 4*x);Expression(C) = ((x - 4)^2 + (y - 1)^2 = 1);PointOnCurve(M, G);Focus(G) = F;PointOnCurve(A, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F)))",
    "answer_expressions": "4",
    "fact_spans": "[[[7, 21], [30, 33]], [[41, 69]], [[2, 6]], [[37, 40]], [[26, 29]], [[7, 21]], [[41, 69]], [[2, 25]], [[26, 36]], [[37, 70]]]",
    "query_spans": "[[[72, 92]]]",
    "process": ""
  },
  {
    "text": "Point $P(2,0)$, point $Q$ lies on the curve $C$: $y^{2}=2x$, then the minimum value of $|PQ|$ is?",
    "fact_expressions": "C: Curve;P: Point;Q: Point;Expression(C) = (y^2 = 2*x);Coordinate(P) = (2, 0);PointOnCurve(Q, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[15, 33]], [[0, 9]], [[10, 14]], [[15, 33]], [[0, 9]], [[10, 34]]]",
    "query_spans": "[[[36, 49]]]",
    "process": "Let Q(x,y), using the distance formula between two points, we obtain |PQ| = \\sqrt{(x-1)^{2}+3}, x\\in[0,+\\infty); thus, when x=1, 1\\in[0,+\\infty), |PQ| = \\sqrt{3}."
  },
  {
    "text": "Find the length of the chord cut from the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$ by the line $y=x+1$?",
    "fact_expressions": "G: Hyperbola;H: Line;Expression(G) = (x^2 - y^2/4 = 1);Expression(H) = (y = x + 1)",
    "query_expressions": "Length(InterceptChord(H, G))",
    "answer_expressions": "8*sqrt(2)/3",
    "fact_spans": "[[[1, 29]], [[30, 39]], [[1, 29]], [[30, 39]]]",
    "query_spans": "[[[1, 45]]]",
    "process": "Solving the system of equations \\begin{cases}y=x+1\\\\x^{2}-\\frac{y^{2}}{4}=1\\end{cases}, we obtain $3x^{2}-2x-5=0$. Let the line $y=x+1$ intersect the hyperbola at points $A$ and $B$, where $A(x_{1},y_{1})$, $B(x_{2},y_{2})$. Then $x_{1}+x_{2}=\\frac{2}{3}$, $x_{1}x_{2}=-\\frac{5}{3}$. Using the chord length formula, we get $|AB|=\\sqrt{1+k^{2}}\\cdot|x_{2}-x_{1}|=\\sqrt{2}\\cdot\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\sqrt{2}\\cdot\\sqrt{\\frac{4}{9}+\\frac{20}{3}}=\\frac{8\\sqrt{2}}{3}$."
  },
  {
    "text": "An ellipse $\\frac{x^{2}}{a^{2}}+y^{2}=1$ $(a>1)$ has a focus at $F$, and a point $P$ lies on the ellipse such that $|OP|=|OF|$ ($O$ being the coordinate origin). Then the area $S$ of $\\triangle OPF$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2 + x^2/a^2 = 1);a: Number;a>1;F: Point;OneOf(Focus(G)) = F;P: Point;PointOnCurve(P, G);O: Origin;Abs(LineSegmentOf(O, P)) = Abs(LineSegmentOf(O, F));S: Number;Area((TriangleOf(O, P, F))) = S",
    "query_expressions": "S",
    "answer_expressions": "1/2*sqrt(a^2-1)",
    "fact_spans": "[[[0, 36], [51, 53]], [[0, 36]], [[2, 36]], [[2, 36]], [[42, 45]], [[0, 45]], [[46, 50]], [[46, 54]], [[70, 73]], [[56, 69]], [[99, 102]], [[81, 102]]]",
    "query_spans": "[[[99, 104]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola equation $y^{2}-4 x^{2}=1$, then the asymptote equations are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-4*x^2 + y^2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*2*x",
    "fact_spans": "[[[2, 5]], [[2, 25]]]",
    "query_spans": "[[[2, 34]]]",
    "process": "y^{2}-4x^{2}=1\\Rightarrow y^{2}-\\frac{x^{2}}{\\frac{1}{4}}=1\\Rightarrow a=1,b=\\frac{1}{2}, therefore the asymptotes of the hyperbola y^{2}-4x^{2}=1 are: y=\\pm\\frac{a}{b}x\\Rightarrow y=\\pm2x."
  },
  {
    "text": "The parabola $x^{2}=8 y$ has focus $F$ and directrix $l$. Let $P$ be a point on the parabola, and $P A \\perp l$, where $A$ is the foot of the perpendicular. If the angle of inclination of line $A F$ is $60^{\\circ}$, then what is $|P F|$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 8*y);F: Point;Focus(G) = F;l: Line;Directrix(G) = l;P: Point;PointOnCurve(P, G);A: Point;IsPerpendicular(LineSegmentOf(P, A), l);FootPoint(LineSegmentOf(P, A), l) = A;Inclination(LineOf(A, F)) = ApplyUnit(60, degree)",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "8/3",
    "fact_spans": "[[[0, 14], [32, 35]], [[0, 14]], [[17, 20]], [[0, 20]], [[24, 27]], [[0, 27]], [[28, 31]], [[28, 38]], [[53, 56]], [[39, 52]], [[39, 59]], [[62, 87]]]",
    "query_spans": "[[[90, 100]]]",
    "process": "In $\\triangle APF$, by the definition of the parabola, we have $|PA| = |PF|$. Since $|AF|\\sin60^{\\circ} = 4$, and $PA \\bot l$, then $PA \\parallel y$-axis. Also, $\\angle AFO = \\angle PAF = \\angle PFA = 30^{\\circ}$. Draw $PB \\bot AF$ from $P$ to $B$, then $|BF| = \\frac{1}{2}|AF| = \\frac{4}{\\sqrt{3}}$, so $|PF| = \\frac{|BF|}{\\cos30^{\\circ}} = \\frac{8}{3}$."
  },
  {
    "text": "The distance from point $M$ to point $F(4,0)$ is less than its distance to the line $l$: $x+6=0$ by $2$. Then, what equation does point $M$ satisfy?",
    "fact_expressions": "l: Line;F: Point;M:Point;Coordinate(F) = (4, 0);Expression(l)=(x+6=0);Distance(M,F)+2=Distance(M,l)",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "y^2=16*x",
    "fact_spans": "[[[20, 34]], [[5, 14]], [[0, 4], [43, 47], [18, 19]], [[5, 14]], [[20, 34]], [[0, 41]]]",
    "query_spans": "[[[43, 54]]]",
    "process": "Since the distance from point M to point F(4,0) is 2 less than its distance to the line l: x+6=0, the distance from point M to point F(4,0) is equal to its distance to the line m: x+4=0. By the definition of a parabola, the trajectory of point M is a parabola with focus at point F(4,0), where \\frac{p}{2}=4 \\therefore p=8. The equation satisfied by point M is y^{2}=16x."
  },
  {
    "text": "The directrix of the parabola $y^{2}=4 x$ intersects the $x$-axis at point $P$. A line with slope $k$ $(k>0)$ passing through point $P$ intersects the parabola at points $A$ and $B$. $F$ is the focus of the parabola. If $|F A|=2|F B|$, then the slope $k$ of line $A B$ is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;F: Point;P: Point;k: Number;k>0;Expression(G) = (y^2 = 4*x);Intersection(Directrix(G), xAxis) = P;Slope(LineOf(A,B)) = k;Intersection(H,G)={A,B};Focus(G) = F;Abs(LineSegmentOf(F, A)) = 2*Abs(LineSegmentOf(F, B));H:Line;PointOnCurve(P,H);Slope(H)=k",
    "query_expressions": "k",
    "answer_expressions": "2*sqrt(2)/3",
    "fact_spans": "[[[0, 14], [68, 71]], [[54, 57]], [[58, 61]], [[64, 67]], [[25, 29], [31, 35]], [[38, 46], [102, 105]], [[38, 46]], [[0, 14]], [[0, 29]], [[36, 49]], [[47, 63]], [[64, 74]], [[76, 90]], [46, 48], [30, 48], [36, 48]]",
    "query_spans": "[[[102, 107]]]",
    "process": "From the given, we know $ P(-1,0) $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. From the known condition $ |FA| = 2|FB| $, we get $ x_{1}+1 = 2(x_{2}+1) $, that is, $ x_{1} = 2x_{2} + 1 $ ①. Let the equation of line $ AB $ be $ y = kx + k $, and together with $ y^{2} = 4x $, we obtain: $ k^{2}x^{2} + (2k^{2}-4)x + k^{2} = 0 $. Then $ x_{1}x_{2} = 1 $ ②. From ① and ②, solving gives $ x_{2} = \\frac{1}{2} $, $ x_{1} = 2 $. Substituting $ x_{1} = 2 $ into $ y^{2} = 4x $, and knowing $ k > 0 $ implies $ y_{1} > 0 $, we solve to get $ A(2, 2\\sqrt{2}) $. Therefore, $ k_{AB} = k_{AP} = \\frac{2\\sqrt{2}-0}{2-(-1)} = \\frac{2\\sqrt{2}}{3} $."
  },
  {
    "text": "From a point $P$ on the parabola $y^{2}=4x$, draw a perpendicular to the directrix of the parabola, with foot of perpendicular at $M$, and $|PM|=5$. Let the focus of the parabola be $F$. Then $\\cos \\angle MPF = ?$",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G);L: Line;PointOnCurve(P, L);IsPerpendicular(L, Directrix(G));M: Point;FootPoint(L, Directrix(G)) = M;Abs(LineSegmentOf(P, M)) = 5;F: Point;Focus(G) = F",
    "query_expressions": "Cos(AngleOf(M, P, F))",
    "answer_expressions": "3/5",
    "fact_spans": "[[[1, 15], [22, 25], [50, 53]], [[1, 15]], [[18, 21]], [[1, 21]], [], [[0, 30]], [[0, 30]], [[34, 37]], [[0, 37]], [[39, 48]], [[57, 60]], [[50, 60]]]",
    "query_spans": "[[[62, 83]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, find the standard equation of a hyperbola that shares the same asymptotes as hyperbola $C$ and passes through the point $(3,3 \\sqrt{2})$.",
    "fact_expressions": "C: Hyperbola;H:Hyperbola;G: Point;Expression(C) = (x^2/4 - y^2/12 = 1);Coordinate(G) = (3, 3*sqrt(2));Asymptote(C)=Asymptote(H);PointOnCurve(G, H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2/3 - y^2/9 = 1",
    "fact_spans": "[[[2, 46], [47, 53]], [[80, 83]], [[62, 79]], [[2, 46]], [[62, 79]], [[46, 83]], [[61, 83]]]",
    "query_spans": "[[[80, 89]]]",
    "process": "Let the equation of the hyperbola sharing asymptotes with hyperbola $ C: \\frac{x^{2}}{4} - \\frac{y^{2}}{12} = 1 $ be: $ \\frac{x^{2}}{4} - \\frac{y^{2}}{12} = \\lambda $. Since the hyperbola passes through $ (3, 3\\sqrt{2}) $, then $ \\frac{3^{2}}{4} - \\frac{(3\\sqrt{2})^{2}}{12} = \\lambda $, which simplifies to $ \\frac{x^{2}}{3} - \\frac{y^{2}}{9} = 1 $."
  },
  {
    "text": "If one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1  (a>0)$ is given by $3 x-2 y=0$, then $a$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/9 + x^2/a^2 = 1);a: Number;a>0;Expression(OneOf(Asymptote(G))) = (3*x - 2*y = 0)",
    "query_expressions": "a",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 50]], [[1, 50]], [[72, 75]], [[4, 50]], [[1, 70]]]",
    "query_spans": "[[[72, 77]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, the inclination angle of one of its asymptotes is $60^{\\circ}$, and it has equal focal distance to the ellipse $\\frac{x^{2}}{5}+y^{2}=1$. Then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;Inclination(OneOf(Asymptote(C))) = ApplyUnit(60, degree);G: Ellipse;Expression(G) = (x^2/5 + y^2 = 1);FocalLength(C) = FocalLength(G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[2, 53], [113, 116]], [[2, 53]], [[10, 53]], [[10, 53]], [[2, 76]], [[79, 106]], [[79, 106]], [[2, 111]]]",
    "query_spans": "[[[113, 121]]]",
    "process": "From the inclination angle of the asymptotes, the slope is obtained as $\\frac{b}{a}=\\sqrt{3}$; combining with the focal distance, we can find the equation. From the ellipse's equation, the focal distance is 4. Then, from the hyperbola's asymptote equation, we get: $\\frac{b}{a}=\\tan60^{\\circ}=\\sqrt{3}$. According to the given condition, $a^{2}+b^{2}=4$, solving yields: $a^{2}=1$, $b^{2}=3$. Therefore, the equation of the hyperbola is: $x^{2}-\\frac{y^{2}}{3}=$"
  },
  {
    "text": "Given that points $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $x^{2} + 2 y^{2} = 2$, respectively, and point $P$ is a moving point on this ellipse, then the minimum value of $|\\overrightarrow{P F_{1}} + \\overrightarrow{P F_{2}}|$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 2*y^2 = 2);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(VectorOf(P, F1) + VectorOf(P, F2)))",
    "answer_expressions": "2",
    "fact_spans": "[[[23, 42], [56, 58]], [[23, 42]], [[2, 10]], [[13, 20]], [[2, 48]], [[2, 48]], [[50, 54]], [[50, 64]]]",
    "query_spans": "[[[66, 125]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$, a circle centered at the origin $O$ with radius equal to the focal distance of $C$ intersects the $x$-axis at points $A$ and $B$. $P$ is a common point of circle $O$ and $C$. If $|P A|=\\sqrt{3}|P B|$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;O: Origin;O1: Circle;P: Point;A: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Center(O1) = O;Radius(O1) = FocalLength(C);Intersection(O1, xAxis) = {A, B};OneOf(Intersection(O1, C)) = P;Abs(LineSegmentOf(P, A)) = sqrt(3)*Abs(LineSegmentOf(P, B))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(\\sqrt{3} + \\sqrt{7})/2",
    "fact_spans": "[[[2, 66], [77, 80], [113, 116], [147, 150]], [[10, 66]], [[10, 66]], [[68, 73]], [[87, 88], [108, 112]], [[104, 107]], [[94, 97]], [[98, 101]], [[10, 66]], [[10, 66]], [[2, 66]], [[67, 88]], [[77, 88]], [[87, 103]], [[104, 122]], [[124, 145]]]",
    "query_spans": "[[[147, 156]]]",
    "process": ""
  },
  {
    "text": "Given a hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with eccentricity $\\frac{\\sqrt{5}}{2}$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $M$ lies on an asymptote of hyperbola $C$, and $OM \\perp MF_{2}$, where $O$ is the origin. If $S \\triangle OMF_{2} = 16$, then the length of the real axis of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;O: Origin;M: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(C)=sqrt(5)/2;LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(M,OneOf(Asymptote(C)));IsPerpendicular(LineSegmentOf(O,M),LineSegmentOf(M,F2));Area(TriangleOf(O,M,F2))=16;F1:Point",
    "query_expressions": "Length(RealAxis(C))",
    "answer_expressions": "16",
    "fact_spans": "[[[27, 88], [117, 123], [194, 197]], [[34, 88]], [[34, 88]], [[155, 158]], [[113, 116]], [[105, 112]], [[34, 88]], [[34, 88]], [[27, 88]], [[2, 88]], [[27, 112]], [[27, 112]], [[113, 132]], [[134, 153]], [[165, 192]], [[97, 104]]]",
    "query_spans": "[[[194, 203]]]",
    "process": "Combining with the image, based on the relationship between the asymptotes and foci of the hyperbola, the solution can be obtained. As shown in the figure, from the given conditions we have F_{2}(c,0); without loss of generality, assume point M lies on the asymptote y=\\frac{b}{a}x. According to the conditions, |F_{2}M|=\\frac{bc}{\\sqrt{a^{2}+b^{2}}}=b, so |OM|=\\sqrt{c^{2}-b^{2}}=a. Given S_{\\triangle OMF2}=16, we obtain \\frac{1}{2}ab=16, that is, ab=32. Also, a^{2}+b^{2}=c^{2}, \\frac{c}{a}=\\frac{\\sqrt{5}}{2}. Therefore, a=8, b=4, c=4\\sqrt{5}, so the real axis length of hyperbola C is 16."
  },
  {
    "text": "If an asymptote of the hyperbola $C$: $m x^{2}-y^{2}=1$ is perpendicular to the line $l$: $y=-2 x-1$, then the focal length of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (m*x^2 - y^2 = 1);m: Number;l: Line;Expression(l) = (y = -2*x -1);IsPerpendicular(OneOf(Asymptote(C)), l)",
    "query_expressions": "FocalLength(C)",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[1, 26], [54, 60]], [[1, 26]], [[9, 26]], [[33, 50]], [[33, 50]], [[1, 52]]]",
    "query_spans": "[[[54, 65]]]",
    "process": "Since one asymptote of the hyperbola $ C: mx^{2} - y^{2} = 1 \\Leftrightarrow \\frac{x^{2}}{m} - y^{2} = 1 $ is perpendicular to the line $ l: y = -2x - 1 $, the slope of this asymptote $ \\frac{b}{a} = \\frac{1}{2} $, $ \\therefore \\frac{1}{4} = \\frac{b^{2}}{a^{2}} = \\frac{1}{m} = m $, $ \\therefore a = \\sqrt{\\frac{1}{m}} = 2 $, $ b = 1 $, $ \\therefore c = \\sqrt{a^{2} + b^{2}} = \\sqrt{5} $, $ 2c = 2\\sqrt{5} $."
  },
  {
    "text": "Given that $F$ is a focus of the ellipse $C$, $B$ is an endpoint of the minor axis, the extension of the line segment $BF$ intersects $C$ at point $D$, and $\\overrightarrow{BF} = 2 \\overrightarrow{FD}$,  \nthen the eccentricity of $C$ is?",
    "fact_expressions": "F: Point;OneOf(Focus(C)) = F;C: Ellipse;B: Point;OneOf(Endpoint(MinorAxis(C))) = B;Intersection(OverlappingLine(LineSegmentOf(B,F)), C) = D;D: Point;VectorOf(B, F) = 2*VectorOf(F, D)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[2, 5]], [[2, 16]], [[6, 11], [41, 44], [99, 102]], [[17, 20]], [[6, 28]], [[29, 49]], [[45, 49]], [[51, 96]]]",
    "query_spans": "[[[99, 108]]]",
    "process": ""
  },
  {
    "text": "Given an ellipse $\\frac{x^{2}}{m}+y^{2}=1$ ($m>0$) with foci on the $x$-axis and focal distance $\\sqrt{2}$, what is the value of $m$?",
    "fact_expressions": "G: Ellipse;m: Number;m>0;Expression(G) = (y^2 + x^2/m = 1);PointOnCurve(Focus(G), xAxis);FocalLength(G) = sqrt(2)",
    "query_expressions": "m",
    "answer_expressions": "3/2",
    "fact_spans": "[[[11, 43]], [[59, 62]], [[13, 43]], [[11, 43]], [[2, 43]], [[11, 57]]]",
    "query_spans": "[[[59, 66]]]",
    "process": "Since the ellipse with foci on the x-axis $\\frac{x^{2}}{m}+y^{2}=1$ $(m>0)$ has a focal distance of $\\sqrt{2}$, then $a^{2}=m$, $b^{2}=1$, $c=\\frac{\\sqrt{2}}{2}$, so $m=1+\\frac{1}{2}=\\frac{3}{2}$"
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-\\frac{y^{2}}{b^{2}}=1$ ($b>0$), and the circle $M$: $x^{2}+(y-3)^{2}=1$ tangent to one asymptote of $C$ at point $P$ ($P$ lies in the second quadrant). If the line $PM$ intersects the other asymptote of the hyperbola at point $S$, and intersects the $x$-axis at point $T$, then the length of $ST$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2/b^2 = 1);b: Number;b>0;M: Circle;Expression(M) = (x^2 + (y - 3)^2 = 1);J1: Line;J2: Line;OneOf(Asymptote(C)) = J1;OneOf(Asymptote(C)) = J2;Negation(J1 = J2);T: Point;TangentPoint(M, J1) = P;P: Point;Quadrant(P) = 2;Intersection(OverlappingLine(LineSegmentOf(P, M)), J2) = S;S: Point;Intersection(OverlappingLine(LineSegmentOf(P, M)), xAxis) = T",
    "query_expressions": "Length(LineSegmentOf(S, T))",
    "answer_expressions": "72/7",
    "fact_spans": "[[[2, 44], [71, 74], [111, 114]], [[2, 44]], [[10, 44]], [[10, 44]], [[45, 70]], [[45, 70]], [], [], [[71, 80]], [[111, 121]], [[71, 121]], [[135, 139]], [[45, 87]], [[83, 87], [88, 91]], [[88, 97]], [[101, 127]], [[123, 127]], [[101, 139]]]",
    "query_spans": "[[[141, 150]]]",
    "process": "The asymptotes of the hyperbola are given by: $ y = \\pm bx $, with the circle center at $ M(0,3) $. Since the asymptote is tangent to the circle, we have $ \\frac{|3|}{\\sqrt{1+b^{2}}} = 1 $, yielding $ b = 2\\sqrt{2} $. Because line $ PM $ is perpendicular to line $ OP $, the equation of line $ PM $ is $ y = \\frac{\\sqrt{2}}{4}x + 3 $. Also, the equation of line $ OS $ is $ y = 2\\sqrt{2}x $. Solving these equations simultaneously gives point $ T(-6\\sqrt{2}, 0) $, $ S\\left(\\frac{6}{7}\\sqrt{2}, \\frac{24}{7}\\right) $. Using the distance formula between two points, we obtain: $ |ST| = \\frac{72}{7} $."
  },
  {
    "text": "The line $l$ passes through the focus $F$ of the parabola $y^{2}=4x$ and intersects the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$, then $y_{1} y_{2}$=?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;F:Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (x1,y1);Coordinate(B) = (x2,y2);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};x1:Number;x2:Number;y1:Number;y2:Number",
    "query_expressions": "y1*y2",
    "answer_expressions": "-4",
    "fact_spans": "[[[0, 5]], [[6, 20], [27, 30]], [[31, 48]], [[49, 66]], [[23, 26]], [[6, 20]], [[31, 48]], [[49, 66]], [[6, 26]], [[0, 26]], [[0, 66]], [[31, 48]], [[49, 66]], [[31, 48]], [[49, 66]]]",
    "query_spans": "[[[68, 83]]]",
    "process": ""
  },
  {
    "text": "From point $P(-1,2)$, two tangents $PA$, $PB$ are drawn to the parabola $y^{2}=4x$. Let the points of tangency be $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$, with $y_{1}>y_{2}$. If line $AB$ intersects the $x$-axis at point $C$, then $\\frac{S_{\\triangle A P C}}{S_{\\triangle B P C}}$=? ($S_{\\triangle A P C}$, $S_{\\triangle B P C}$ are the areas of $\\triangle A P C$, $\\triangle B P C$, respectively)",
    "fact_expressions": "P: Point;Coordinate(P) = (-1, 2);G: Parabola;Expression(G) = (y^2 = 4*x);TangentPoint(LineSegmentOf(P, A), G) = A;TangentPoint(LineSegmentOf(P, B), G) = B;A: Point;B: Point;Coordinate(A) = (x1, y1);x1: Number;y1: Number;Coordinate(B) = (x2, y2);x2: Number;y2: Number;y1 >y2;Intersection(LineOf(A,B), xAxis) = C;C: Point;TangentOfPoint(P,G)={LineSegmentOf(P,A),LineSegmentOf(P,B)}",
    "query_expressions": "Area(TriangleOf(A,P,C))/Area(TriangleOf(B,P,C))",
    "answer_expressions": "3+2*sqrt(2)",
    "fact_spans": "[[[1, 11]], [[1, 11]], [[12, 26]], [[12, 26]], [[0, 83]], [[0, 83]], [[47, 64]], [[66, 83]], [[47, 64]], [[47, 64]], [[47, 64]], [[66, 83]], [[66, 83]], [[66, 83]], [[85, 98]], [[100, 118]], [[114, 118]], [0, 42]]",
    "query_spans": "[[[120, 260]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ has focus $F$ and directrix $l$, and the circle $\\odot C$: $(x-a)^{2}+(y-b)^{2}=16$ passes through point $F$ and is tangent to $l$. The chord length intercepted on the $x$-axis by $\\odot C$ is $4$. Then $a=?$",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;l: Line;Directrix(G) = l;C: Circle;a: Number;b: Number;Expression(C) = ((x-a)^2+(y-b)^2=16);PointOnCurve(F, C);IsTangent(l, C);Length(InterceptChord(xAxis, C)) = 4",
    "query_expressions": "a",
    "answer_expressions": "{1, 3}",
    "fact_spans": "[[[2, 23]], [[2, 23]], [[5, 23]], [[5, 23]], [[27, 30], [75, 78]], [[2, 30]], [[34, 37], [80, 83]], [[2, 37]], [[39, 73], [91, 100]], [[112, 115]], [[49, 73]], [[39, 73]], [[39, 78]], [[39, 85]], [[86, 110]]]",
    "query_spans": "[[[112, 117]]]",
    "process": "From the given conditions, the center of the circle $(a,b)$ lies on the parabola, so $b^{2}=2pa$; also, for a point on the parabola, the distance to the focus equals the distance to the directrix, thus $a=4-\\frac{p}{2}$. Therefore, $b^{2}=2p(4-\\frac{p}{2})=8p-p^{2}$. Since the chord length intercepted by the $\\odot C$ on the $x$-axis is 4, according to the property of a circle: the square of the distance from the center to the chord plus the square of half the chord length equals the square of the radius; hence $b^{2}+4=4^{2}$, so $b^{2}=12$. Thus, $8p-p^{2}=12$, that is, $p^{2}-8p+12=0$, so $p=2$ or $p=6$, therefore $a=3$ or $a=1$."
  },
  {
    "text": "What is the eccentricity of the ellipse $2 x^{2}+y^{2}=4$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (2*x^2 + y^2 = 4)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[0, 19]], [[0, 19]]]",
    "query_spans": "[[[0, 25]]]",
    "process": ""
  },
  {
    "text": "Given that the two foci of ellipse $E$ are $F_{1}$ and $F_{2}$ respectively, point $P$ lies on the ellipse, and $\\tan \\angle P F_{1} F_{2}=\\frac{1}{3}$, $\\tan \\angle P F_{2} F_{1}=-3$, then the eccentricity of ellipse $E$ is?",
    "fact_expressions": "E: Ellipse;F1: Point;F2: Point;Focus(E) = {F1, F2};P: Point;PointOnCurve(P, E);Tan(AngleOf(P, F1, F2)) = 1/3;Tan(AngleOf(P, F2, F1)) = -3",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(10)/5",
    "fact_spans": "[[[2, 7], [36, 38], [117, 122]], [[15, 22]], [[23, 30]], [[2, 30]], [[31, 35]], [[31, 41]], [[43, 83]], [[85, 115]]]",
    "query_spans": "[[[117, 128]]]",
    "process": "From the given conditions, since $\\tan\\angle PF_{1}F_{2}=\\frac{1}{3}$, it follows that $\\sin\\angle PF_{1}F_{2}=\\frac{\\sqrt{10}}{10}$; and since $\\tan\\angle PF_{2}F_{1}=-3$, it follows that $\\sin\\angle PF_{2}F_{1}=\\frac{3\\sqrt{10}}{10}$. Then, $\\tan\\angle F_{2}PF_{1}=\\tan(\\pi-\\angle PF_{1}F_{2}-\\angle PF_{2}F_{1})\\frac{PF_{F}_{2}+\\angle PF_{2}F_{1})}{\\angle PF_{1}F_{2}+\\tan\\angle PF_{2}F_{1}}=\\frac{4}{3}$. Solving gives $\\sin\\angle F_{2}PF_{1}=\\frac{4}{5}$. Then, in $\\triangle PF_{1}F_{2}$, by the law of sines, we obtain $\\frac{\\sin\\angle PF_{1}F_{2}+\\sin\\angle PF_{2}F}{F}$"
  },
  {
    "text": "Ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, a line $l$ with slope $k$ intersects the ellipse at points $M$, $N$, point $A$ is the midpoint of segment $MN$, the slope of line $OA$ ($O$ is the coordinate origin) is $k^{\\prime}$, then $k \\cdot k^{\\prime}$=?",
    "fact_expressions": "O: Origin;A: Point;M: Point;N: Point;l: Line;C: Ellipse;k: Number;k1:Number;Expression(C) = (x^2/9 + y^2/4 = 1);Slope(l)=k;Intersection(l, C) = {M, N};MidPoint(LineSegmentOf(M, N)) = A;Slope(LineOf(O,A))=k1",
    "query_expressions": "k*k1",
    "answer_expressions": "-4/9",
    "fact_spans": "[[[92, 95]], [[70, 74]], [[61, 65]], [[66, 69]], [[50, 55]], [[0, 42], [56, 58]], [[46, 49]], [[106, 118]], [[0, 42]], [[43, 55]], [[50, 69]], [[70, 84]], [[85, 118]]]",
    "query_spans": "[[[121, 143]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through the origin intersects $E$ at points $P$ and $Q$, such that $P F_{2} \\perp F_{2} Q$, $S_{\\Delta P F_{2} Q}=\\frac{1}{2} a^{2}$, and $|P F_{2}|+|F_{2} Q|=4$. Find the length of the minor axis of ellipse $E$.",
    "fact_expressions": "E: Ellipse;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(E) = F1;RightFocus(E) = F2;O: Origin;PointOnCurve(O, G);G: Line;Intersection(G, E) = {P, Q};P: Point;Q: Point;IsPerpendicular(LineSegmentOf(P, F2), LineSegmentOf(F2, Q));Area(TriangleOf(P, F2, Q)) = a^2/2;Abs(LineSegmentOf(F2, Q)) + Abs(LineSegmentOf(P, F2)) = 4",
    "query_expressions": "Length(MinorAxis(E))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 59], [93, 96], [199, 204]], [[2, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[68, 75]], [[76, 83]], [[2, 83]], [[2, 83]], [[85, 89]], [[84, 92]], [[90, 92]], [[90, 106]], [[97, 100]], [[101, 104]], [[108, 131]], [[133, 173]], [[174, 197]]]",
    "query_spans": "[[[199, 210]]]",
    "process": "Connect $PF_{1}$, $QF_{1}$, according to the symmetry of the ellipse, $F_{1}PF_{2}Q$ is a rectangle. From $|PF_{1}|+|PF_{2}|=2a=4$, we get $a=2$. From $|PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}=4c^{2}$, combined with $|PF_{1}|+|PF_{2}|=2a$, we obtain $2|PF_{1}||PF_{2}|=4(a^{2}-c^{2})=4b^{2}=4S_{\\triangle PF_{2}Q}=4\\times\\frac{1}{2}a^{2}=8$, $\\therefore b=\\sqrt{2}$, $\\therefore$ the minor axis length of the ellipse is $2b=2\\sqrt{2}$."
  },
  {
    "text": "Draw a perpendicular line from the lower focus $F_1$ of the hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0, b>0)$ to the $y$-axis, intersecting the hyperbola at points $A$ and $B$. If the circle with diameter $AB$ passes exactly through its upper focus $F_2$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;A: Point;B: Point;F1: Point;F2: Point;l:Line;a>0;b>0;Expression(G) = (-x^2/b^2 + y^2/a^2 = 1);LowerFocus(G) = F1;PointOnCurve(F1,l);IsPerpendicular(l,yAxis);Intersection(l,G)={A,B};IsDiameter(LineSegmentOf(A,B),H);PointOnCurve(F2,H);UpperFocus(G)=F2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1+sqrt(2)",
    "fact_spans": "[[[1, 57], [78, 81], [107, 108], [121, 124]], [[4, 57]], [[4, 57]], [[103, 104]], [[82, 85]], [[86, 89]], [[61, 68]], [[111, 118]], [], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 68]], [[0, 76]], [[0, 76]], [[0, 91]], [[93, 104]], [[103, 118]], [[107, 118]]]",
    "query_spans": "[[[121, 130]]]",
    "process": "Draw a perpendicular line from the lower focus $ F_{1} $ of the hyperbola $ \\frac{y^{2}}{a^{2}} - \\frac{x^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) to the $ y $-axis, intersecting the hyperbola at points $ A $ and $ B $. Then $ |AB| = \\frac{2b^{2}}{a} $. The circle with $ AB $ as diameter passes exactly through its upper focus $ F_{2} $, yielding: $ \\frac{b^{2}}{a} = 2c' \\cdot c^{2} - a^{2} - 2ac = 0 $, leading to $ e^{2} - 2e - 1 = 0 $. Solving gives $ e = 1 + \\sqrt{2} $, while $ e = 1 - \\sqrt{2} $ is discarded."
  },
  {
    "text": "Given the equation of the parabola is $y=2 a x^{2}$, and it passes through the point $(1,4)$, then the coordinates of the focus are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 2*(a*x^2));H: Point;Coordinate(H) = (1, 4);PointOnCurve(H, G);a: Number",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, 1/16)",
    "fact_spans": "[[[2, 5]], [[2, 22]], [[25, 33]], [[25, 33]], [[2, 33]], [[9, 22]]]",
    "query_spans": "[[[2, 41]]]",
    "process": "Substituting the point (1,4) into the parabola equation gives the value of a, allowing us to find the parabola equation and then the focus coordinates. The parabola y=2ax^{2} passes through the point (1,4), so we have 4=2a, solving gives a=2, then the parabola is y=4x^{2}, that is, x^{2}=\\frac{1}{4}y, and the focus coordinates are (0,\\frac{1}{16})"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, $A_{1}$, $A_{2}$ are the two endpoints of the major axis, and point $P$ is a point on the ellipse such that the slope of line $P A_{1}$ lies in the range $[1,2]$. Then, what is the range of the slope of line $P A_{2}$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);A1: Point;A2: Point;Endpoint(MajorAxis(C)) = {A1, A2};P: Point;PointOnCurve(P, C) = True;Range(Slope(LineOf(P, A1))) = [1, 2]",
    "query_expressions": "Range(Slope(LineOf(P, A2)))",
    "answer_expressions": "[-3/4,-3/8]",
    "fact_spans": "[[[2, 44], [77, 79]], [[2, 44]], [[47, 54]], [[56, 63]], [[2, 71]], [[72, 76]], [[72, 83]], [[87, 114]]]",
    "query_spans": "[[[116, 137]]]",
    "process": "First obtain $A_{1}(-2,0)$, $A_{2}(2,0)$, let $P(m,n)$, according to the problem calculate $k_{PA_{1}}\\cdot k_{PA_{2}}=-\\frac{3}{4}$, then based on $1\\leqslant k_{PA_{1}}\\leqslant 2$, the answer can be obtained. [Detailed explanation] From the problem: $a=2$, thus $A_{1}(-2,0)$, $A_{2}(2,0)$, let $P(m,n)$, since $k_{PA_{1}}=\\frac{n}{m+2}$, $k_{PA_{2}}=\\frac{n}{m-2}$, $k_{PA_{1}}\\cdot k_{PA_{2}}=\\frac{n}{m+2}_{\\frac{12-3m^{2}}{4}}=\\frac{n^{2}}{m^{2}-4}$, because $1\\leqslant k_{PA_{1}}\\leqslant 2$, so $-\\frac{3}{4}\\leqslant k_{PA_{2}}\\leqslant -\\frac{3}{8}$"
  },
  {
    "text": "Points $A$ and $B$ are two points on the parabola $C$: $y^{2}=2 p x(p>0)$, $F$ is the focus of parabola $C$. If $\\angle A F B=60^{\\circ}$, and the distance from the midpoint $D$ of $AB$ to the directrix of parabola $C$ is $d$, then the maximum value of $\\frac{d}{|A B|}$ is?",
    "fact_expressions": "C: Parabola;p: Number;A: Point;B: Point;F: Point;D: Point;d: Number;p>0;Expression(C) = (y^2 = 2*(p*x));PointOnCurve(A, C);PointOnCurve(B, C);Focus(C) = F;AngleOf(A, F, B) = ApplyUnit(60, degree);MidPoint(LineSegmentOf(A, B)) = D;Distance(D, Directrix(C)) = d",
    "query_expressions": "Max(d/Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "1",
    "fact_spans": "[[[9, 35], [43, 49], [93, 99]], [[17, 35]], [[0, 4]], [[5, 8]], [[39, 42]], [[89, 92]], [[106, 109]], [[17, 35]], [[9, 35]], [[0, 38]], [[0, 38]], [[39, 52]], [[54, 80]], [[82, 92]], [[89, 109]]]",
    "query_spans": "[[[111, 134]]]",
    "process": "Let AF = a, BF = b, then d = \\frac{a+b}{2}, AB^{2} = a^{2} + b^{2} - 2ab\\cos\\angle AFB = a^{2} + b^{2} - ab\\frac{1}{2}\\sqrt{1+\\frac{3ab}{a^{2}+b^{2}-ab}} \\leqslant \\frac{1}{2}\\sqrt{1+\\frac{3ab}{2ab-ab}} = 1, with equality if and only if a = b."
  },
  {
    "text": "On the parabola $y^{2}=2 p x$, the point with abscissa $4$ is at a distance of $5$ from the focus. Then the value of $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;Expression(G) = (y^2 = 2*(p*x));P: Point;XCoordinate(P) = 4;PointOnCurve(P, G);Distance(P, Focus(G)) = 5",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 17]], [[40, 43]], [[1, 17]], [[27, 28]], [[19, 28]], [[0, 28]], [[1, 38]]]",
    "query_spans": "[[[40, 47]]]",
    "process": ""
  },
  {
    "text": "$P$ is a moving point on the right branch of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, $F$ is the right focus of the hyperbola, and $A(3,1)$ is given. Then the minimum value of $|P A|+|P F|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1);P: Point;PointOnCurve(P, RightPart(G));F: Point;RightFocus(G) = F;A: Point;Coordinate(A) = (3, 1)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "sqrt(25)-2*sqrt(3)",
    "fact_spans": "[[[4, 32], [44, 47]], [[4, 32]], [[0, 3]], [[0, 39]], [[40, 43]], [[40, 51]], [[54, 62]], [[54, 62]]]",
    "query_spans": "[[[64, 83]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively, if there exists a point $P$ on the ellipse $C$ such that the perpendicular bisector of the segment $P F_{1}$ passes exactly through the focus $F_{2}$, then the range of the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C);PointOnCurve(F2, PerpendicularBisector(LineSegmentOf(P, F1))) = True",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "[1/3,1)",
    "fact_spans": "[[[20, 77], [85, 90], [130, 135]], [[20, 77]], [[27, 77]], [[27, 77]], [[27, 77]], [[27, 77]], [[2, 9]], [[10, 17], [121, 128]], [[2, 83]], [[2, 83]], [[93, 97]], [[85, 97]], [[100, 128]]]",
    "query_spans": "[[[130, 146]]]",
    "process": "Let $ P(x_{0},y_{0}) $, then $ \\frac{x_{0}^{2}}{a^{2}} + \\frac{y_{0}^{2}}{b^{2}} = 1 $. Then the midpoint $ M $ of segment $ PF_{1} $ is $ \\left( \\frac{x_{0}-c}{2}, \\frac{y_{0}}{2} \\right) $. Since the perpendicular bisector of segment $ PF_{1} $ passes exactly through focus $ F_{2} $, $ k_{PF_{1}}k_{F_{2}M} = \\frac{y_{0}}{x_{0}+c} \\cdot \\frac{\\frac{y_{0}}{2}-0}{\\frac{x_{0}-c}{2}-c} = -1 $, which simplifies to $ \\frac{y_{0}^{2}}{(x_{0}+c)(x_{0}-3c)} $. Thus, $ b^{2}\\left(1-\\frac{x_{0}^{2}}{a^{2}}\\right) + (x_{0}+c)(x_{0}-3c) = 0 $, which becomes $ c^{2}x_{0}^{2} - 2a^{2}cx_{0} + b^{2}a^{2} - 3a^{2}c^{2} = 0 $. Solving gives $ x_{0} = \\frac{a^{2}-2ac}{c} $. Since $ -a \\leqslant x_{0} \\leqslant a $, we have $ -a \\leqslant \\frac{a^{2}-2ac}{c} \\leqslant a $, and given $ 0 < e < 1 $, solving yields $ \\frac{1}{3} \\leqslant e \\leqslant 1 $. The range of eccentricity of ellipse $ C $ is: $ \\left[ \\frac{1}{2},"
  },
  {
    "text": "The distance from the right focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ to the line $\\sqrt{5} x-2 y=0$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;Expression(G) = (x^2/4 - y^2/5 = 1);Expression(H) = (sqrt(5)*x - 2*y = 0)",
    "query_expressions": "Distance(RightFocus(G),H)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[0, 38]], [[43, 63]], [[0, 38]], [[43, 63]]]",
    "query_spans": "[[[0, 68]]]",
    "process": "From $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, we obtain $a^{2}=4$, $b^{2}=5$, so $c^{2}=a^{2}+b^{2}=4+5=9$. Thus, the coordinates of the right focus are $(3,0)$. The distance from the point $(3,0)$ to the line $\\sqrt{5}x-2y=0$ is $d=\\frac{|\\sqrt{5}\\times3-2\\times0|}{\\sqrt{5+4}}=\\sqrt{5}$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$, $F_{2}$ respectively. $M$ is a point on the asymptote of $C$, and $|O F_{2}|=|O M|$, $|M F_{1}|-|M F_{2}|=2 \\sqrt{2} a$. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;M: Point;PointOnCurve(M,Asymptote(C)) ;O: Origin;Abs(LineSegmentOf(O, F2)) = Abs(LineSegmentOf(O, M));Abs(LineSegmentOf(M, F1)) - Abs(LineSegmentOf(M, F2)) = 2*sqrt(2)*a",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 63], [91, 94], [157, 160]], [[2, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[71, 78]], [[79, 86]], [[2, 86]], [[2, 86]], [[87, 90]], [[87, 100]], [[102, 120]], [[102, 120]], [[121, 155]]]",
    "query_spans": "[[[157, 166]]]",
    "process": "From $|OF_{2}|=|OM|=\\frac{1}{2}|F_{1}F_{2}|=c$, we obtain $F_{1}M\\bot F_{2}M$. Let $M(x_{0},\\frac{b}{a}x_{0})$, $\\overrightarrow{F_{1}M}\\cdot\\overrightarrow{F_{2}M}=(x_{0}+c,\\frac{b}{a}x_{0})\\cdot(x_{0}-c,\\frac{b}{a}x_{0})=0$, we get $x_{0}^{2}-c^{2}+\\frac{b^{2}}{a^{2}}x_{0}^{2}=0$, solving yields $x_{0}^{2}=a^{2}$, i.e., $M(a,b)$, $|MF_{1}|=\\sqrt{(a+c)^{2}+b^{2}}=\\sqrt{2c^{2}+2ac}$, $|MF_{2}|=\\sqrt{(a-c)^{2}+b^{2}}=\\sqrt{2c^{2}-2ac}$, so $|MF_{1}||MF_{2}|=\\sqrt{4c^{4}-4a^{2}c^{2}}$ $\\textcircled{1}$ while $||MF_{1}|-|MF_{2}||=2\\sqrt{2}a$, squaring both sides gives $|MF_{1}|^{2}+|MF_{2}|^{2}-2|MF_{1}||MF_{2}|=(2\\sqrt{2}a)^{2}$, and since $|MF_{1}|^{2}+|MF_{2}|^{2}=|F_{1}F_{2}|^{2}=(2c)^{2}$, thus $|MF_{1}||MF_{2}|=4a^{2}-2c^{2}$ $\\textcircled{2}$ from $\\textcircled{1}$ and $\\textcircled{2}$ we obtain $(2a^{2}-c^{2})^{2}=c^{4}-a^{2}c^{2}$, yielding $4a^{2}=3c^{2}$, so the eccentricity $e=\\frac{c}{a}=\\frac{2\\sqrt{3}}{3}$"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ and the parabola $y^{2}=8x$ share a common focus $F$, and $P$ is an intersection point of the two curves. If $|PF|=5$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;P: Point;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 8*x);OneOf(Focus(G))=F;Focus(H)=F;OneOf(Intersection(G,H))=P;Abs(LineSegmentOf(P, F)) = 5",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[2, 60], [115, 118]], [[5, 60]], [[5, 60]], [[63, 77]], [[99, 102]], [[85, 88]], [[5, 60]], [[5, 60]], [[2, 60]], [[63, 77]], [[2, 88]], [[2, 88]], [[90, 102]], [[104, 113]]]",
    "query_spans": "[[[115, 122]]]",
    "process": ""
  },
  {
    "text": "If the left and right foci of the hyperbola $E$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ are $F_{1}$ and $F_{2}$ respectively, and point $P$ lies on hyperbola $E$ with $|P F_{1}|=4$, then $|P F_{2}|$ equals?",
    "fact_expressions": "E: Hyperbola;P: Point;F1: Point;F2: Point;Expression(E) = (x^2/9 - y^2/16 = 1);LeftFocus(E) = F1;RightFocus(E) = F2;PointOnCurve(P, E);Abs(LineSegmentOf(P, F1)) = 4",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "10",
    "fact_spans": "[[[1, 45], [75, 81]], [[70, 74]], [[54, 61]], [[62, 69]], [[1, 45]], [[1, 69]], [[1, 69]], [[70, 82]], [[84, 97]]]",
    "query_spans": "[[[99, 113]]]",
    "process": "Obtain a=3 for the hyperbola, and by the definition of a hyperbola, ||PF_{1}|-|PF_{2}||=2a=6; substitute the given conditions and solve the equation to obtain the required value. For hyperbola E: \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1, a=3, and by the definition of a hyperbola, ||PF_{1}|-|PF_{2}||=2a=6. Given |PF_{1}|=4, we have |4-|PF_{2}||=6. Solving yields |PF_{2}|=10 (discarding -2)."
  },
  {
    "text": "The standard equation of an ellipse with a focus at $F_{1}(-2 \\sqrt{3}, 0)$ and the sum of the lengths of the major axis and minor axis equal to $12$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;Coordinate(F1) = (-2*sqrt(3), 0);OneOf(Focus(G))=F1;Length(MinorAxis(G))+Length(MajorAxis(G))=12",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16+y^2/4=1",
    "fact_spans": "[[[44, 46]], [[5, 28]], [[5, 28]], [[0, 46]], [[29, 46]]]",
    "query_spans": "[[[44, 53]]]",
    "process": "The solution process is omitted"
  },
  {
    "text": "If the equation $\\frac{y^{2}}{2-k}+\\frac{x^{2}}{|k|-3}=1$ represents a hyperbola with foci on the $y$-axis, and the semi-focal length of the hyperbola is $c$, then what is the range of values for $c$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(Abs(k) - 3) + y^2/(2 - k) = 1);PointOnCurve(Focus(G), yAxis);HalfFocalLength(G) = c;c: Number",
    "query_expressions": "Range(c)",
    "answer_expressions": "{sqrt(5), sqrt(5 - 2*k)}",
    "fact_spans": "[[[55, 58], [59, 62]], [[1, 58]], [[46, 58]], [[59, 70]], [[67, 70], [72, 75]]]",
    "query_spans": "[[[72, 82]]]",
    "process": ""
  },
  {
    "text": "The distance from the focus $F$ to the directrix of the parabola $C$: $x=2 p y^{2}(p>0)$ is $2$. A line passing through point $F$ intersects $C$ at points $A$ and $B$. The intersection point of the directrix of $C$ and the $x$-axis is $M$. If the area of $\\triangle M A B$ is $3 \\sqrt{2}$, then $\\frac{|A F|}{|B F|}=$?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;M: Point;A: Point;B: Point;F: Point;p>0;Expression(C) = (x = 2*(p*y^2));Focus(C) = F;Distance(F,Directrix(C)) = 2;PointOnCurve(F, G);Intersection(G, C) = {A, B};Intersection(Directrix(C), xAxis) = M;Area(TriangleOf(M, A, B)) = 3*sqrt(2)",
    "query_expressions": "Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(B, F))",
    "answer_expressions": "{2, 1/2}",
    "fact_spans": "[[[0, 26], [52, 55], [67, 70]], [[8, 26]], [[49, 51]], [[82, 85]], [[57, 60]], [[61, 64]], [[44, 48], [29, 32]], [[8, 26]], [[0, 26]], [[0, 32]], [[0, 42]], [[43, 51]], [[49, 66]], [[67, 85]], [[87, 120]]]",
    "query_spans": "[[[122, 145]]]",
    "process": "The parabola $ C: x = 2py^2 $ ($ p > 0 $) is rewritten in standard form as: $ y^{2} = \\frac{1}{2p}x $ ($ p > 0 $). Since the distance from the focus $ F $ of the parabola to the directrix is 2, $ \\frac{1}{4p} = 2 $, so $ p = \\frac{1}{8} $. Therefore, the equation of the parabola is $ y^2 = 4x $ ($ p > 0 $), and the focus is $ F(1, 0) $. Since a line passing through point $ F $ intersects $ C $ at points $ A $ and $ B $, let the equation of line $ AB $ be: $ x = my + 1 $. Combining this with the parabola's equation yields: $ y^{2} - 4my - 4 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, and assume without loss of generality that point $ A $ lies above the $ x $-axis and point $ B $ lies below the $ x $-axis. Then $ y_{1} + y_{2} = 4m $, $ y_{1}y_{2} = -4 $. Let $ d $ be the distance from point $ M $ to line $ AB $. Then $ d = \\frac{|1+1|}{\\sqrt{m^{2}+1}} = \\frac{2}{\\sqrt{m^{2}+1}} $. Therefore, $ S_{\\Delta MAB} = \\frac{1}{2} AB \\cdot d = \\frac{1}{2} \\times 4(1+m^{2}) \\times \\frac{2}{\\sqrt{m^{2}+1}} = 4\\sqrt{m^{2}+1} = 3\\sqrt{2} $. Solving gives: $ m^{2} = \\frac{1}{8} $, so $ m = \\pm \\frac{\\sqrt{2}}{4} $. When $ m = \\frac{\\sqrt{2}}{4} $, $ y_{1} + y_{2} = \\sqrt{2} $, $ y_{1}y_{2} = -4 $. Solving yields: $ |y_{1}| = 2\\sqrt{2} $, $ \\begin{cases} y_{2} = -\\sqrt{2} \\end{cases} $. At this time: $ \\begin{cases} x_{1} = 2 \\\\ x_{2} = \\frac{1}{2} \\end{cases} $. Therefore, $ |AF| = x_{1} + 1 = 3 $, $ |BF| = x_{2} + 1 = \\frac{3}{2} $, $ \\frac{|AF|}{|BF|} = 2 $. When $ m = -\\frac{\\sqrt{2}}{4} $, $ y_{1} + y_{2} = -\\sqrt{2} $, $ y_{1}y_{2} = -4 $. Solving yields: $ \\begin{cases} y_{1} = \\sqrt{2} \\\\ y_{2} = -2\\sqrt{2} \\end{cases} $. Therefore, $ |AF| = x_{1} + 1 = \\frac{3}{2} $, $ |BF| = x_{2} + 1 = 3 $, $ \\frac{|AF|}{|BF|} = \\frac{1}{2} $."
  },
  {
    "text": "The equation of the parabola is $y^{2}=8 x$, with focus $F$. A line $l$ passing through $F$ intersects the parabola at two points $A$ and $B$, whose coordinates are $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$. Then $x_{1} x_{2}$=? $y_{1} y_{2}$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);l: Line;A: Point;B: Point;F: Point;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};x1: Number;x2: Number;y1: Number;y2: Number",
    "query_expressions": "x1*x2;y1*y2",
    "answer_expressions": "4\n-16",
    "fact_spans": "[[[0, 3], [37, 40], [18, 19]], [[0, 17]], [[31, 36]], [[44, 47], [60, 77]], [[48, 51], [79, 96]], [[22, 25], [27, 30]], [[60, 77]], [[79, 96]], [[18, 25]], [[26, 36]], [[31, 51]], [[60, 77]], [[79, 96]], [[60, 77]], [[79, 96]]]",
    "query_spans": "[[[99, 114]], [[114, 129]]]",
    "process": ""
  },
  {
    "text": "Given the line $l$: $m x - y + 1 - m = 0$, $m \\in \\mathbb{R}$, if the line passes through the focus of the parabola $y^{2} = 8x$, then the length of the chord $AB$ intercepted by the circle $(x-1)^{2} + (y-1)^{2} = 6$ is $?$.",
    "fact_expressions": "l: Line;G: Parabola;H: Circle;m:Real;A: Point;B: Point;Expression(G) = (y^2 = 8*x);Expression(H) = ((x - 1)^2 + (y - 1)^2 = 6);Expression(l) = (-m + m*x - y + 1 = 0);PointOnCurve(Focus(G),l);Length(InterceptChord(l,H))=LineSegmentOf(A,B)",
    "query_expressions": "LineSegmentOf(A,B)",
    "answer_expressions": "2*sqrt(6)",
    "fact_spans": "[[[2, 23], [35, 37], [60, 62]], [[39, 53]], [[63, 87]], [[24, 33]], [[92, 99]], [[92, 99]], [[39, 53]], [[63, 87]], [[2, 23]], [[35, 56]], [[60, 99]]]",
    "query_spans": "[[[92, 101]]]",
    "process": "The focus of the parabola is (2,0), 2m-0+1-m=0, solving gives: m=-1; l: y=-x+2, the center (1,1) lies on the line y=-x+2, so AB=2R=2\\sqrt{6}"
  },
  {
    "text": "For any point $M$ on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, excluding the endpoints of the minor axis, the lines connecting $M$ to the two endpoints $B_{1}$, $B_{2}$ of the minor axis intersect the $x$-axis at points $N$ and $K$, respectively. Then what is the minimum value of $|O N|+|O K|$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>b;b>0;M: Point;PointOnCurve(M, G);Negation(M=Endpoint(MinorAxis(G)));B1: Point;B2: Point;Endpoint(MinorAxis(G)) = {B1, B2};N: Point;K: Point;Intersection(LineOf(M, B1), xAxis) = N;Intersection(LineOf(M, B2), xAxis) = K;O: Origin",
    "query_expressions": "Min(Abs(LineSegmentOf(O, K)) + Abs(LineSegmentOf(O, N)))",
    "answer_expressions": "2*a",
    "fact_spans": "[[[0, 53]], [[0, 53]], [[2, 53]], [[2, 54]], [[2, 54]], [[2, 54]], [[60, 63]], [[0, 63]], [[0, 72]], [[79, 86]], [[87, 94]], [[0, 94]], [[103, 107]], [[108, 111]], [[60, 111]], [[60, 111]], [[113, 126]]]",
    "query_spans": "[[[113, 132]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $x^{2}=4 y$ with focus $F$, and the intersection point of the directrix and the $y$-axis being $M$, let $N$ be a point on the parabola such that $|N F|=\\frac{\\sqrt{3}}{2}|M N|$. Then $\\angle N M F$?",
    "fact_expressions": "G: Parabola;N: Point;F: Point;M: Point;Expression(G) = (x^2 = 4*y);Focus(G) = F;Intersection(Directrix(G), yAxis) = M;PointOnCurve(N, G);Abs(LineSegmentOf(N, F)) = (sqrt(3)/2)*Abs(LineSegmentOf(M, N))",
    "query_expressions": "AngleOf(N, M, F)",
    "answer_expressions": "ApplyUnit(30,degree)",
    "fact_spans": "[[[2, 16], [45, 48]], [[41, 44]], [[20, 23]], [[35, 39]], [[2, 16]], [[2, 23]], [[2, 39]], [[41, 52]], [[54, 85]]]",
    "query_spans": "[[[88, 103]]]",
    "process": "Let the distance from N to the directrix be d. By the definition of a parabola, we have d = |NF|. According to the given condition, \\cos\\angleNMF=\\frac{d}{|MN|}=\\frac{\\sqrt{3}}{2}, so \\angleNMF=30^{\\circ}"
  },
  {
    "text": "Given that the center of ellipse $C$ is at the origin, and one focus has coordinates $(0,3)$. If the point $(4,0)$ lies on ellipse $C$, then the eccentricity of ellipse $C$ equals?",
    "fact_expressions": "C: Ellipse;G: Point;O: Origin;Center(C) = O;Coordinate(OneOf(Focus(C))) = (0, 3);Coordinate(G) = (4, 0);PointOnCurve(G, C)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "3/5",
    "fact_spans": "[[[2, 7], [42, 47], [50, 55]], [[33, 41]], [[11, 15]], [[2, 15]], [[2, 31]], [[33, 41]], [[33, 48]]]",
    "query_spans": "[[[50, 62]]]",
    "process": "Analysis: According to the geometric conditions of the ellipse, we have b=4, c=3, solve for a and the eccentricity. Specifically, since b=4, c=3, it follows that a=5, c=\\frac{3}{5}."
  },
  {
    "text": "Point $P$ lies on the ellipse $C_{1}$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, the left focus of $C_{1}$ is $F$, and point $Q$ lies on the circle $C_{2}$: $(x+2)^{2}+(y-4)^{2}=1$. Then the minimum value of $|P Q|-|P F|$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P,C1) = True;C1: Ellipse;Expression(C1) = (x^2/4 + y^2/3 = 1);F: Point;LeftFocus(C1) = F;C2: Circle;Expression(C2) = ((x + 2)^2 + (y - 4)^2 = 1);Q: Point;PointOnCurve(Q,C2) = True",
    "query_expressions": "Min(-Abs(LineSegmentOf(P, F)) + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "0",
    "fact_spans": "[[[0, 4]], [[0, 52]], [[5, 51], [53, 60]], [[5, 51]], [[65, 68]], [[53, 68]], [[74, 106]], [[74, 106]], [[69, 73]], [[69, 107]]]",
    "query_spans": "[[[109, 128]]]",
    "process": "Point $P$ lies on the ellipse $C_{1}: \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1$, with the left focus of the ellipse $F(-1,0)$ and the right focus $E(1,0)$. As shown in the figure, from the circle $C_{2}: (x+2)^{2} + (y-4)^{2} = 1$, we obtain $C_{2}(-2,4)$ and radius $1$. By the definition of the ellipse, $|PF| + |PE| = 4$, so $|PF| = 4 - |PE|$. Then $|PQ| - |PF| = |PQ| + |PE| - 4$. When points $C_{2}$, $Q$, $P$, $E$ are collinear, $|PQ| - |PF|$ reaches its minimum value. Thus, $(|PQ| - |PF|)_{\\min} = C_{2}E - 1 - 4 = \\sqrt{9+16} - 5 = 0$."
  },
  {
    "text": "The asymptotes of a hyperbola centered at the origin with coordinate axes as symmetry axes are given by $y=\\pm \\frac{\\sqrt {2}}{2} x$, and the hyperbola passes through the point $P(2,1)$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;O: Origin;P: Point;SymmetryAxis(G)=axis;Expression(Asymptote(G)) = (y =pm*(sqrt(2)/2)*x);PointOnCurve(P, G);Coordinate(P)=(2,1);Center(G)=O",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/2-y^2=1",
    "fact_spans": "[[[14, 17], [55, 58], [70, 73]], [[3, 5]], [[59, 68]], [[6, 17]], [[14, 53]], [[55, 68]], [[59, 68]], [[0, 17]]]",
    "query_spans": "[[[70, 78]]]",
    "process": ""
  },
  {
    "text": "If the equation $\\frac{x^{2}}{|k|-2}+\\frac{y^{2}}{3-k}=1$ represents a hyperbola, then the range of real values for $k$ is?",
    "fact_expressions": "G: Hyperbola;k: Real;Expression(G)=(x^2/(Abs(k) - 2) + y^2/(3 - k) = 1)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-2,2)+(3,+oo)",
    "fact_spans": "[[[46, 49]], [[51, 56]], [[1, 49]]]",
    "query_spans": "[[[51, 63]]]",
    "process": "From the given condition, we have $(|k|-2)(3-k)<0$, solving yields $k\\in(-2,2)\\cup(3,+\\infty)$, so the range of real values for $k$ is $(-2,2)\\cup(3,+\\infty)$."
  },
  {
    "text": "Given that point $A(-2,3)$ lies on the directrix of the parabola $C$: $y^{2}=2 p x  (p>0)$, and denoting the focus of $C$ as $F$, what is the slope of line $A F$?",
    "fact_expressions": "C: Parabola;p: Number;A: Point;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(A) = (-2, 3);PointOnCurve(A, Directrix(C));Focus(C) = F",
    "query_expressions": "Slope(LineOf(A,F))",
    "answer_expressions": "-3/4",
    "fact_spans": "[[[13, 41], [47, 50]], [[21, 41]], [[2, 12]], [[54, 57]], [[21, 41]], [[13, 41]], [[2, 12]], [[2, 45]], [[47, 57]]]",
    "query_spans": "[[[59, 71]]]",
    "process": "From the properties of the parabola, obtain the value of p, then determine the coordinates of F, and finally use the slope formula to find the answer. Let the parabola C: y^{2}=2px (p>0) have the directrix equation x=-\\frac{p}{2}. According to the given condition, -\\frac{p}{2}=-2, so p=4, thus F(2,0). Therefore, k_{AF}=\\frac{3-0}{-2-2}=-\\frac{3}{4}"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$ and directrix $l$, let $M$ be a point on $l$, and $Q$ be an intersection point of line $MF$ and $C$. If $\\overrightarrow{FM}=3\\overrightarrow{FQ}$, then $|\\overrightarrow{QF}|$=?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 8*x);F: Point;Focus(C) = F;l: Line;Directrix(C) = l;M: Point;PointOnCurve(M, l);Q: Point;OneOf(Intersection(LineOf(M, F), C)) = Q;VectorOf(F, M) = 3*VectorOf(F, Q)",
    "query_expressions": "Abs(VectorOf(Q, F))",
    "answer_expressions": "8/3",
    "fact_spans": "[[[2, 21], [60, 63]], [[2, 21]], [[25, 28]], [[2, 28]], [[32, 35], [41, 44]], [[2, 35]], [[37, 40]], [[37, 47]], [[48, 51]], [[48, 68]], [[69, 114]]]",
    "query_spans": "[[[116, 142]]]",
    "process": "The focus of parabola C is F(2,0), and the directrix is l: x = -2. Let point Q(x,y) and M(-2,t), then \\overrightarrow{FM} = (-4,t), \\overrightarrow{FQ} = (x-2,y). From \\overrightarrow{FM} = 3\\overrightarrow{FQ}, we get 3(x-2) = -4, so x = \\frac{2}{3}. Therefore, |\\overrightarrow{QF}| = \\frac{2}{3} + 2 = \\frac{8}{3}."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, draw a line $l$ through point $P(-1,0)$ with slope $k$ $(k>0)$ intersecting the parabola $C$ at points $A$ and $B$. The line $AF$ intersects the parabola $C$ at another point $M$, and the line $BF$ intersects the parabola $C$ at another point $N$. If $\\frac{|AF|}{|FM|}+\\frac{|BF|}{|FN|}=18$, then $k=$?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;F: Point;B: Point;P: Point;M: Point;N: Point;k:Number;k>0;Expression(C) = (y^2 = 4*x);Coordinate(P) = (-1, 0);Focus(C) = F;PointOnCurve(P, l);Slope(l) = k;Intersection(l, C) = {A, B};Intersection(LineOf(A,F), C) = M;Intersection(LineOf(B,F), C) = N;Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(F, M)) + Abs(LineSegmentOf(B, F))/Abs(LineSegmentOf(F, N)) = 18",
    "query_expressions": "k",
    "answer_expressions": "sqrt(5)/5",
    "fact_spans": "[[[53, 58]], [[2, 21], [59, 65], [85, 91], [107, 113]], [[68, 72]], [[25, 28]], [[73, 76]], [[30, 40]], [[95, 98]], [[117, 120]], [[44, 52], [168, 171]], [[44, 52]], [[2, 21]], [[30, 40]], [[2, 28]], [[29, 58]], [[41, 58]], [[53, 76]], [[77, 98]], [[99, 120]], [[122, 166]]]",
    "query_spans": "[[[168, 173]]]",
    "process": "According to the problem, the graph is symmetric about the x-axis, and points A, B, P are collinear, thus $\\frac{y_{1}}{x_{1}+1}=\\frac{y_{2}}{x_{2}+1}$. From the focal radius formula, $|AF|=x_{1}+1=|NF|$, $|BF|=x_{2}+1=|MF|$, $\\frac{|AF|}{|FM|}+\\frac{|BF|}{|FN|}=\\frac{y_{1}}{y_{2}}+\\frac{y_{2}}{y_{1}}=18$, $(y_{1}+y_{2})^{2}=20y_{1}y_{2}$. Then using Vieta's formulas, the conclusion can be drawn. [Detailed solution] According to the problem, the graph is symmetric about the x-axis, and points A, B, P are collinear, thus $\\frac{y_{1}}{x_{1}+1}=\\frac{y_{2}}{x_{2}+1}$. From the focal radius formula, $|AF|=x_{1}+1=|NF|$, $|BF|=x_{2}+1=|MF|$. $\\frac{|AF|}{|FM|}+\\frac{|BF|}{|FN|}=\\frac{y_{1}}{y_{2}}+\\frac{y_{2}}{y_{1}}=18$, $\\therefore(y_{1}+y_{2})^{2}=20y_{1}y_{2}$, $+1$' yields $ky^{2}\\cdot4y+4k=0$, $\\therefore k>0$, $\\therefore k=\\frac{\\sqrt{5}}{5}$"
  },
  {
    "text": "If the eccentricity of the hyperbola $x^{2}+k y^{2}=1$ is $2$, then the value of the real number $k$ is?",
    "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (k*y^2 + x^2 = 1);Eccentricity(G) = 2",
    "query_expressions": "k",
    "answer_expressions": "-1/3",
    "fact_spans": "[[[1, 21]], [[31, 36]], [[1, 21]], [[1, 29]]]",
    "query_spans": "[[[31, 40]]]",
    "process": "First, from the hyperbola equation, we can determine $ a $ and $ b $, then find the expression for $ c $. Using the eccentricity of 2, we solve for $ k $. According to the problem, since the eccentricity of the hyperbola $ x^{2} + ky^{2} = 1 $ is 2, we have $ a = 1 $, $ b = \\sqrt{\\frac{1}{k}} $, $ \\therefore c = \\sqrt{1 - \\frac{1}{k}} $, $ \\therefore \\frac{c}{a} = \\sqrt{1 - \\frac{1}{k}} = 2 $, $ k = -\\frac{1}{3} $,"
  },
  {
    "text": "Given that the center of the hyperbola is at the origin, one focus is $F_{1}(-\\sqrt{5}, 0)$, point $P$ lies on the hyperbola, and the midpoint of segment $P F_{1}$ has coordinates $(0 , 2)$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;F1: Point;O:Origin;P: Point;Center(G)=O;Coordinate(F1) = (-sqrt(5), 0);PointOnCurve(P, G);OneOf(Focus(G))=F1;Coordinate(MidPoint(LineSegmentOf(P,F1)))=(0,2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 5], [43, 46], [78, 81]], [[16, 37]], [[8, 10]], [[38, 42]], [[2, 10]], [[16, 37]], [[38, 47]], [[2, 37]], [[49, 75]]]",
    "query_spans": "[[[78, 87]]]",
    "process": ""
  },
  {
    "text": "A point $M$ on the parabola $y^{2}=\\frac{1}{4} x$ is at a distance of $1$ from the focus. Then the horizontal coordinate of point $M$ is?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = x/4);PointOnCurve(M, G);Distance(M, Focus(G)) = 1",
    "query_expressions": "XCoordinate(M)",
    "answer_expressions": "15/16",
    "fact_spans": "[[[0, 24]], [[27, 30], [42, 46]], [[0, 24]], [[0, 30]], [[0, 40]]]",
    "query_spans": "[[[42, 52]]]",
    "process": "From the given, we have $\\frac{p}{2}=\\frac{1}{16}\\Rightarrow x_{0}+\\frac{1}{16}=1\\Rightarrow x_{0}=\\frac{15}{16}$."
  },
  {
    "text": "If the endpoints of a line segment $AB$ of fixed length $5$ move along the parabola $y^{2}=4x$, then the shortest distance from the midpoint of segment $AB$ to the $y$-axis is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Length(LineSegmentOf(A,B)) = 5;PointOnCurve(Endpoint(LineSegmentOf(A, B)), G)",
    "query_expressions": "Min(Distance(MidPoint(LineSegmentOf(A, B)), yAxis))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[20, 34]], [[10, 15]], [[10, 15]], [[20, 34]], [[1, 15]], [[8, 35]]]",
    "query_spans": "[[[39, 61]]]",
    "process": "As shown in the figure, let $ F $ be the focus of the parabola $ y^{2} = 4x $, and $ M $ be the midpoint of $ AB $. Draw perpendiculars from $ A $, $ B $, and $ M $ to the directrix of the parabola, with feet at $ P $, $ Q $, and $ N $, respectively. In the right trapezoid $ APQB $, since $ |AM| = |BM| $, it follows that $ |MN| = \\frac{1}{2}(|PA| + |QB|) $. Also, $ |PA| = |AF| $, $ |QB| = |BF| $, so $ |MN| = \\frac{1}{2}(|AF| + |BF|) $. By the property of plane geometry, $ |AF| + |BF| \\geqslant |AB| = 5 $, with equality if and only if $ AB $ passes through the focus $ F $. When $ AB $ is a focal chord, $ |MN| $ attains its minimum value $ \\frac{5}{2} $, and the distance from point $ M $ to the $ y $-axis is shortest, with the shortest distance being $ \\frac{5}{2} - 1 = \\frac{3}{2} $."
  },
  {
    "text": "The ellipse $x^{2}+2 y^{2}=2$ intersects the line $y=x+m$ at two points $A$ and $B$, and $|A B|=\\frac{4 \\sqrt{2}}{3}$. Find the value of the real number $m$.",
    "fact_expressions": "G: Ellipse;H: Line;m: Real;A: Point;B: Point;Expression(G) = (x^2 + 2*y^2 = 2);Expression(H) = (y = m + x);Intersection(G, H) = {A, B};Abs(LineSegmentOf(A, B)) = (4*sqrt(2))/3",
    "query_expressions": "m",
    "answer_expressions": "pm*1",
    "fact_spans": "[[[0, 19]], [[20, 29]], [[74, 79]], [[31, 34]], [[35, 38]], [[0, 19]], [[20, 29]], [[0, 40]], [[43, 71]]]",
    "query_spans": "[[[74, 83]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). From \\begin{cases}x^{2}+2y^{2}=2\\\\y=x+m\\end{cases}, we have: 3x^{2}+4mx+2m^{2}-2=0; then x_{1}+x_{2}=-\\frac{4m}{3}, x_{1}x_{2}=\\frac{2m^{2}-2}{3}, |AB|=\\sqrt{2}|x_{1}-x_{2}|=\\sqrt{2}\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\sqrt{2}\\sqrt{(-\\frac{4m}{3})^{2}-\\frac{8m^{2}-8}{3}}=\\frac{4\\sqrt{2}}{3}, which gives: m^{2}=1, i.e., m=\\pm1;"
  },
  {
    "text": "Let $P$ and $Q$ be points on the circle $x^{2}+(y-6)^{2}=2$ and the ellipse $\\frac{x^{2}}{10}+y^{2}=1$, respectively. Then the maximum distance between points $P$ and $Q$ is?",
    "fact_expressions": "H: Circle;Expression(H) = (x^2 + (y - 6)^2 = 2);G: Ellipse;Expression(G) = (x^2/10 + y^2 = 1);P: Point;Q: Point;PointOnCurve(P, H);PointOnCurve(Q, G)",
    "query_expressions": "Max(Distance(P, Q))",
    "answer_expressions": "6*sqrt(2)",
    "fact_spans": "[[[11, 31]], [[11, 31]], [[32, 60]], [[32, 60]], [[1, 4], [65, 68]], [[5, 8], [69, 72]], [[1, 63]], [[1, 63]]]",
    "query_spans": "[[[65, 82]]]",
    "process": "According to the problem, the center of the circle is $ C(0,6) $ and the radius is $ r = \\sqrt{2} $. Let any point on the ellipse be $ Q(\\sqrt{10}\\cos\\theta, \\sin\\theta) $. The maximum distance from point $ Q $ to a point on the circle is $ |QC| + r $, which is $ \\sqrt{(\\sqrt{10}\\cos\\theta)^2 + (\\sin\\theta - 6)^2} + \\sqrt{2} $, simplifying to $ \\overline{16} + \\sqrt{2} $. When the axis of symmetry $ \\sin\\theta = \\frac{-12}{2 \\times (-9)} = -\\frac{2}{3} $, the maximum value is $ (-\\frac{2}{3}) + 46 + \\sqrt{2} = 5\\sqrt{2} + \\sqrt{2} = 6\\sqrt{2} $. This question mainly examines the positional relationship between a circle and an ellipse, the representation of the maximum distance from a point outside a circle to a point on the circle, and the mathematical thinking methods of trigonometric substitution, as well as reduction and transformation. Given the standard equation of a circle, its center and radius are determined. The maximum and minimum distances from a point outside the circle to points on the circle can be transformed into the distance from the external point to the center plus or minus the radius."
  },
  {
    "text": "If the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has a focus $F(c, 0)$, and the distance from the right vertex $A$ to one of the asymptotes is $\\frac{1}{2} c$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;c: Number;Coordinate(F) = (c, 0);Focus(G) = F;A: Point;RightVertex(G) = A;Distance(A,OneOf(Asymptote(G))) = c/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 57], [105, 108]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[61, 70]], [[87, 102]], [[61, 70]], [[1, 70]], [[74, 77]], [[1, 77]], [[74, 102]]]",
    "query_spans": "[[[105, 114]]]",
    "process": "Right vertex $ A(a,0) $, the equation of an asymptote $ y=\\frac{b}{a}x $, i.e., $ bx-ay=0 $. The distance from the right vertex $ A $ to the asymptote is $ h=\\frac{|ab|}{\\sqrt{a^{2}+b^{2}}}=\\frac{1}{2}c $. Squaring both sides gives $ \\frac{a^{2}b^{2}}{a^{2}+b^{2}}=\\frac{1}{4}c^{2} $. $ \\therefore a^{2}(c^{2}-a^{2})=\\frac{1}{4}c^{4} $, $ \\therefore c^{4}-4a^{2}c^{2}+4a^{4}=0 $, $ \\therefore (c^{2}-2a^{2})^{2}=0 $, $ \\therefore \\frac{c^{2}}{a^{2}}=2 $, $ \\therefore e=\\frac{c}{a}=\\sqrt{2} $"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$, with focus $F$ and directrix $l$. Let $P$ be a point on the parabola $C$ in the first quadrant. Draw a perpendicular from point $P$ to $l$, with foot of perpendicular at $Q$. When the perimeter of $\\triangle PFQ$ is $12$, what is the area of $\\triangle PFQ$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l: Line;Directrix(C) = l;P: Point;PointOnCurve(P, C);Quadrant(P) = 1;Z: Line;PointOnCurve(P, Z);IsPerpendicular(Z, l);Q: Point;FootPoint(Z, l) = Q;Perimeter(TriangleOf(P, F, Q)) = 12",
    "query_expressions": "Area(TriangleOf(P, F, Q))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[2, 21], [42, 48], [22, 23]], [[2, 21]], [[26, 29]], [[22, 29]], [[33, 36], [63, 66]], [[22, 36]], [[38, 41], [58, 62]], [[38, 56]], [[38, 56]], [], [[57, 69]], [[57, 69]], [[73, 76]], [[57, 76]], [[79, 104]]]",
    "query_spans": "[[[106, 128]]]",
    "process": "From $ y^{2} = 4x $, we obtain the focus $ F(1,0) $ and the directrix $ l: x = -1 $. As shown in the figure, let $ |PQ| = |PF| = a $. By the property of the parabola, $ |PF| \\geqslant |OF| = 1 $, so $ a > 1 $. Thus, $ P(a-1, 2\\sqrt{a-1}) $, $ Q(-1, 2\\sqrt{a-1}) $. $ \\therefore QF = \\sqrt{4 + (2\\sqrt{a-1})^{2}} $. Since the perimeter of $ \\triangle PFQ $ is 12, $ \\therefore 2a + 2\\sqrt{a} = 12 $. Solving gives $ a = 4 $. $ \\therefore |QF| = 4 $, $ \\therefore \\triangle PFQ $ is an equilateral triangle with side length 4. $ \\therefore $ The area of $ \\triangle PFQ $ is $ \\frac{\\sqrt{3}}{4} \\times 4^{2} = 4\\sqrt{3} $."
  },
  {
    "text": "If $F_{1}$ and $F_{2}$ are the two foci of an ellipse $C$, $P$ is a point on $C$, and $\\angle F_{1} P F_{2}=60^{\\circ}$, $|P F_{1}|=3|P F_{2}|$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;F1: Point;P: Point;F2: Point;Focus(C) = {F1, F2};PointOnCurve(P, C);AngleOf(F1, P, F2) = ApplyUnit(60, degree);Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(7)/4",
    "fact_spans": "[[[17, 22], [32, 35], [99, 102]], [[1, 8]], [[28, 31]], [[9, 16]], [[1, 27]], [[28, 38]], [[40, 74]], [[75, 97]]]",
    "query_spans": "[[[99, 108]]]",
    "process": "From the definition of the ellipse, |PF₁| + |PF₂| = 2a, and given |PF₁| = 3|PF₂|, solving yields |PF₁| = \\frac{3}{2}a, |PF₂| = \\frac{1}{2}a. Also, |F₁F₂| = 2c. In triangle PF₁F₂, by the cosine law, 4c^{2} = \\frac{9}{4}a^{2} + \\frac{1}{4}a^{2} - 2 \\times \\frac{3}{2}a \\times \\frac{1}{2}a \\times \\cos60^{\\circ}, solving gives \\frac{c}{a} = \\frac{\\sqrt{7}}{4}."
  },
  {
    "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) have one asymptote intersecting the parabola $y^{2}=x$ at a point with horizontal coordinate $x_{0}$. If $x_{0}>1$, then the range of the eccentricity $e$ of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Parabola;e: Number;x0:Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y^2 = x);XCoordinate(OneOf(Intersection(OneOf(Asymptote(C)),G)))=x0;Eccentricity(C) = e;x0>1",
    "query_expressions": "Range(e)",
    "answer_expressions": "(1,\\sqrt{2})",
    "fact_spans": "[[[1, 63], [112, 118]], [[9, 63]], [[9, 63]], [[70, 82]], [[122, 125]], [[92, 99]], [[9, 63]], [[9, 63]], [[1, 63]], [[70, 82]], [[1, 99]], [[112, 125]], [[101, 110]]]",
    "query_spans": "[[[122, 132]]]",
    "process": "The hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ has an asymptote $ y = \\frac{b}{a}x $. Solving the system $ \\begin{cases} y^{2} = x \\\\ y = \\frac{b}{a}x \\end{cases} $ by eliminating $ y $, we obtain $ \\frac{b^{2}}{a^{2}}x^{2} = x $. Since $ x_{0} > 1 $, it follows that $ \\frac{b^{2}}{a^{2}} < 1 $, so $ b^{2} < a^{2} $. Therefore, $ e^{2} = \\frac{c^{2}}{a^{2}} = \\frac{a^{2} + b^{2}}{a^{2}} < 2 $, hence $ 1 < e < \\sqrt{2} $."
  },
  {
    "text": "Given that point $P(m, n)$ is a moving point on the parabola $x^{2} = -8y$, then the minimum value of $\\sqrt{m^{2}+n^{2}+4n+4} + \\sqrt{m^{2}+n^{2}-4m+2n+5}$ is?",
    "fact_expressions": "P: Point;m: Number;n: Number;Coordinate(P) = (m, n);PointOnCurve(P, G);G: Parabola;Expression(G) = (x^2 = -8*y)",
    "query_expressions": "Min(sqrt(2*n - 4*m + m^2 + n^2 + 5) + sqrt(4*n + m^2 + n^2 + 4))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 12]], [[3, 12]], [[3, 12]], [[2, 12]], [[2, 32]], [[13, 28]], [[13, 28]]]",
    "query_spans": "[[[34, 95]]]",
    "process": "From the given conditions, $\\sqrt{m^{2}+n^{2}+4n+4}+\\sqrt{m^{2}+n^{2}-4m+2n+5}$ represents the sum of distances from point $P(m,n)$ to $F(0,-2)$ and point $A(2,-1)$. According to the definition of a parabola, the result can be obtained. From the parabola $x^{2}=-8y$, it follows that the focus is $F(0,-2)$ and the directrix is $l:y=2$. $\\sqrt{m^{2}+n^{2}+4n+4}+\\sqrt{m^{2}+n^{2}-4m+2n+5}=\\sqrt{m^{2}+(n+2)^{2}}+\\sqrt{(m-2)^{2}+(n+1)^{2}}$ has the geometric meaning of the sum of distances from point $P(m,n)$ to $F(0,-2)$ and point $A(2,-1)$. According to the definition of the parabola, the distance from point $P(m,n)$ to $2-(-1)=3$."
  },
  {
    "text": "The length of the real axis of the hyperbola $x^{2}-\\frac{y^{2}}{m}=1$ is twice the length of the imaginary axis. What is the value of $m$?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2 - y^2/m = 1);2*Length(ImageinaryAxis(G)) = Length(RealAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "1/4",
    "fact_spans": "[[[0, 28]], [[43, 46]], [[0, 28]], [[0, 41]]]",
    "query_spans": "[[[43, 50]]]",
    "process": "By the given condition, $ m > 0 $, the real semi-axis length is $ a = 1 $, the imaginary semi-axis length is $ b = \\sqrt{m} $. Also, $ 2a = 2 \\times 2b $, so $ 1 = 2\\sqrt{m} $, $ m = \\frac{1}{4} $."
  },
  {
    "text": "A line $ l $ passing through the focus $ F $ of the parabola $ E: y^{2} = 4x $ intersects $ E $ at two distinct points $ A(x_{1}, y_{1}) $ and $ B(x_{2}, y_{2}) $. Then the minimum value of $ \\frac{1}{x_{1}} + \\frac{4}{x_{2}} $ is?",
    "fact_expressions": "E: Parabola;Expression(E) = (y^2 = 4*x);F: Point;Focus(E) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Intersection(l, E) = {A, B}",
    "query_expressions": "Min(4/x2 + 1/x1)",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 19], [31, 34]], [[1, 19]], [[21, 24]], [[1, 24]], [[25, 30]], [[0, 30]], [[40, 58]], [[60, 78]], [[40, 58]], [[60, 78]], [[40, 58]], [[60, 78]], [[40, 58]], [[60, 78]], [[25, 78]]]",
    "query_spans": "[[[80, 121]]]",
    "process": ""
  },
  {
    "text": "The line $ l $: $ y = kx + 1 $ intersects the right branch of the hyperbola $ C $: $ 2x^2 - y^2 = 1 $ at two distinct points $ A $, $ B $. Then the range of real values for $ k $ is?",
    "fact_expressions": "l: Line;Expression(l) = (y = k*x + 1);k: Real;C: Hyperbola;Expression(C) = (2*x^2 - y^2 = 1);A: Point;B: Point;Negation(A=B);Intersection(l, RightPart(C)) = {A, B}",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-2, -\\sqrt{2})",
    "fact_spans": "[[[0, 16]], [[0, 16]], [[62, 67]], [[17, 42]], [[17, 42]], [[52, 55]], [[56, 59]], [[47, 59]], [[0, 59]]]",
    "query_spans": "[[[62, 74]]]",
    "process": "Substitute the equation of line $ l $, $ y = kx + 1 $, into the equation of hyperbola $ C $, $ 2x^2 - y^2 = 1 $. After simplifying, we obtain $ (k^2 - 2)x^2 + 2kx + 2 = 0 $. According to the problem, the line $ l $ and hyperbola $ C $ satisfy  \n\\[\n\\begin{matrix}\nk^2 - 2 \\neq 0, & A = 4k^2 - 8(k^2 - 2) > 0 \\\\\n-\\frac{-2^k}{k^2 - 2} > 0, & \\frac{k^2}{2} > 0,\n\\end{matrix}\n\\]  \nsolving which yields the range of $ k $ as $ -2 < k < -\\sqrt{2} $."
  },
  {
    "text": "If the hyperbola is $\\frac{x^{2}}{2}+\\frac{y^{2}}{\\lambda}=1$, then the focal distance of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Z) = (x^2/2 + y^2/lambda = 1);Z: Curve;Z = G;lambda: Number",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(2-lambda)",
    "fact_spans": "[[[1, 4], [49, 52]], [[5, 46]], [[5, 46]], [[1, 46]], [[5, 46]]]",
    "query_spans": "[[[49, 57]]]",
    "process": "From $\\frac{x^{2}}{2}+\\frac{y^{2}}{\\lambda}=1$, we get $\\frac{x^{2}}{2}-\\frac{y^{2}}{-\\lambda}=1$, therefore the focal distance of this hyperbola is $2\\sqrt{2-\\lambda}$."
  },
  {
    "text": "Given the ellipse $ C $: $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $), with left and right foci $ F_{1} $, $ F_{2} $ respectively. Points $ A $, $ B $ lie on the ellipse and satisfy $ \\overrightarrow{A F_{1}} = 2 \\overrightarrow{F_{1} B} $, $ \\overrightarrow{A F_{2}} \\cdot \\overrightarrow{A F_{1}} = 0 $. Then the eccentricity of ellipse $ C $ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;A: Point;B: Point;PointOnCurve(A, C);PointOnCurve(B, C);VectorOf(A, F1) = 2*VectorOf(F1, B);DotProduct(VectorOf(A, F2), VectorOf(A,F1)) = 0",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[2, 59], [93, 95], [217, 222]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[68, 75]], [[76, 83]], [[2, 83]], [[2, 83]], [[84, 88]], [[89, 92]], [[84, 96]], [[89, 96]], [[100, 154]], [[156, 215]]]",
    "query_spans": "[[[217, 228]]]",
    "process": "Let $|AF_{1}|=2m$ $(m>0)$. Since $\\overrightarrow{AF}_{1}=2\\overrightarrow{F_{1}B}$, it follows that $|BF_{1}|=m$. Because $\\overrightarrow{AF}_{2}\\cdot\\overrightarrow{AF_{1}}=0$ and $|F_{1}F_{2}|=2c$, we have $|AF_{2}|=\\sqrt{|F_{1}F_{2}|^{2}-|AF_{1}|^{2}}=2\\sqrt{c^{2}-m^{2}}$. Also, since $|BF_{2}|=\\sqrt{|AB|^{2}+|AF_{2}|^{2}}=\\sqrt{4c^{2}+5m^{2}}$, and $|AF_{1}|+|AF_{2}|=|BF_{1}|+|BF_{2}|=2a$, it follows that $2m+2\\sqrt{c^{2}-m^{2}}=m+\\sqrt{4c^{2}+5m^{2}}$. Hence, $m+2\\sqrt{c^{2}-m^{2}}=\\sqrt{4c^{2}+5m^{2}}$. Then, $m^{2}+4c^{2}-4m^{2}+4m\\sqrt{c^{2}-m^{2}}=4c^{2}+5m^{2}$, so $c^{2}=5m^{2}$, thus $c=\\sqrt{5}m$. Since $2a=2m+2\\sqrt{c^{2}-m^{2}}=6m$, we get $a=3m$, so $e=\\frac{c}{a}=\\frac{\\sqrt{5}m}{3m}=\\frac{\\sqrt{5}}{3}$."
  },
  {
    "text": "Given that both asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ are tangent to the circle $C$: $x^{2}+y^{2}-6 x+5=0$, and the right focus of the hyperbola coincides with the center of the circle $C$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;C: Circle;Expression(C) = (-6*x + x^2 + y^2 + 5 = 0);IsTangent(Asymptote(G),C) = True;Center(C) = RightFocus(G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5 - y^2/4 = 1",
    "fact_spans": "[[[2, 61], [99, 102], [117, 120]], [[2, 61]], [[5, 61]], [[5, 61]], [[5, 61]], [[5, 61]], [[69, 95], [107, 111]], [[69, 95]], [[2, 97]], [[99, 114]]]",
    "query_spans": "[[[117, 125]]]",
    "process": ""
  },
  {
    "text": "The left focus of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ is $F_{1}$, $M$ is a point on the ellipse, $N$ is the midpoint of $M F_{1}$, $O$ is the origin. If $|O N|=3$, then $|M F_{1}|=$?",
    "fact_expressions": "G: Ellipse;M: Point;F1: Point;O: Origin;N: Point;Expression(G) = (x^2/25 + y^2/9 = 1);LeftFocus(G)=F1;PointOnCurve(M,G);MidPoint(LineSegmentOf(M, F1)) = N;Abs(LineSegmentOf(O, N)) = 3",
    "query_expressions": "Abs(LineSegmentOf(M, F1))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 38], [55, 57]], [[51, 54]], [[43, 50]], [[79, 82]], [[62, 65]], [[0, 38]], [[0, 50]], [[51, 61]], [[62, 78]], [[87, 96]]]",
    "query_spans": "[[[98, 111]]]",
    "process": "The left focus of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ is $F_{1}$, as shown in the figure. Let the right focus be $F_{2}$, then $a=5$. Since $N$ is the midpoint of $MF_{1}$ and $O$ is the midpoint of $F_{1}F_{2}$, $|ON|=3$, hence $|MF_{2}|=2|ON|=6$. Also, $|MF_{1}|+|MF_{2}|=2a=10$, so $|MF_{1}|=4$."
  },
  {
    "text": "$P$ is a point on the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$, $F_{1}$ and $F_{2}$ are its left and right foci, respectively, and $|P F_{1}|=5$, then $|P F_{2}|$=?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 - y^2/9 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(P, F1)) = 5",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "9",
    "fact_spans": "[[[4, 42], [63, 64]], [[0, 3]], [[47, 54]], [[55, 62]], [[4, 42]], [[0, 46]], [[47, 70]], [[47, 70]], [[72, 85]]]",
    "query_spans": "[[[87, 100]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and $P$ is a point on the ellipse such that $\\angle F_{1}PF_{2}=90^{\\circ}$, find the minimum value of the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;P: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Min(Eccentricity(G))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[20, 72], [82, 84], [122, 124]], [[22, 72]], [[22, 72]], [[2, 9]], [[78, 81]], [[12, 19]], [[22, 72]], [[22, 72]], [[20, 72]], [[2, 77]], [[2, 77]], [[78, 87]], [[88, 120]]]",
    "query_spans": "[[[122, 133]]]",
    "process": "Since $\\angle F_{1}PF_{2}=90^{\\circ}$, it follows that $|PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}$. Since $|PF_{1}|+|PF_{2}|=2a$ and $|F_{1}F_{2}|=2c$, we can solve to get $|PF_{1}||PF_{2}|=2a^{2}-2c^{2}$. Because $2a=|PF_{1}|+|PF_{2}|\\geqslant2\\sqrt{|PF_{1}||PF_{2}|}=2\\sqrt{2a^{2}-2c^{2}}$, simplifying yields $2c^{2}\\geqslanta^{2}$, that is, $e^{2}=\\frac{c^{2}}{a^{2}}\\geqslant\\frac{1}{2}$, so $e\\geqslant\\frac{\\sqrt{2}}{2}$."
  },
  {
    "text": "Given the circle $C$: $(x-1)^{2}+(y-1)^{2}=2$ passes through the right focus $F$ and the upper vertex $B$ of the ellipse $\\Gamma$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$, then the eccentricity of the ellipse $\\Gamma$ is?",
    "fact_expressions": "C: Circle;Gamma:Ellipse;Expression(Gamma) = (x^2/a^2+y^2/b^2=1);Expression(C) = ((x - 1)^2 + (y - 1)^2 = 2);a:Number;b:Number;a>b;b>0;F:Point;B:Point;RightFocus(Gamma)=F;UpperVertex(Gamma)=B;PointOnCurve(F,C);PointOnCurve(B,C)",
    "query_expressions": "Eccentricity(Gamma)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 30]], [[32, 96], [112, 122]], [[32, 96]], [[2, 30]], [[44, 96]], [[44, 96]], [[44, 96]], [[44, 96]], [[100, 103]], [[107, 110]], [[32, 103]], [[32, 110]], [[2, 103]], [[2, 110]]]",
    "query_spans": "[[[112, 128]]]",
    "process": "In the equation $(x-1)^{2}+(y-1)^{2}=2$, let $y=0$ to get $x=1, 0$. Let $x=0$, to get $y=1, 0$. According to the problem, $c=1$, $b=1$, so $a=\\sqrt{2}$, $e=\\frac{c}{a}=\\frac{\\sqrt{2}}{2}$"
  },
  {
    "text": "The standard equation of the hyperbola with asymptotes $2 x-\\sqrt{3} y=0$ and $2 x+\\sqrt{3} y=0$, passing through the point $(6,6)$ is?",
    "fact_expressions": "Asymptote(G) = {J1, J2};J1: Line;J2:Line;Expression(J1) = (2*x - sqrt(3)*y = 0);Expression(J2) = (2*x + sqrt(3)*y = 0);G: Hyperbola;H: Point;Coordinate(H) = (6, 6);PointOnCurve(H, G) = True",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/12 = 1",
    "fact_spans": "[[[0, 56]], [[4, 22]], [[23, 41]], [[4, 22]], [[23, 41]], [[53, 56]], [[44, 52]], [[44, 52]], [[43, 56]]]",
    "query_spans": "[[[53, 63]]]",
    "process": "\\because the asymptotes are 2x\\pm\\sqrt{3}y=0, let the hyperbola equation be (2x+\\sqrt{3}y)(2x-\\sqrt{3}y)=\\lambda (\\lambda\\neq0), that is, 4x^{2}-3y^{2}=\\lambda. Substituting (6,6) gives 4\\times36-3\\times36=\\lambda, \\therefore\\lambda=36, \\therefore the standard equation of the hyperbola is \\frac{x^{2}}{9}-\\frac{y^{2}}{12}="
  },
  {
    "text": "The standard equation of a parabola whose axis of symmetry is the $y$-axis and whose focus lies on the line $3x - 4y - 8 = 0$ is?",
    "fact_expressions": "G: Parabola;SymmetryAxis(G) = yAxis;H: Line;Expression(H) = (3*x - 4*y - 8 = 0);PointOnCurve(Focus(G), H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 = -8y",
    "fact_spans": "[[[29, 32]], [[0, 32]], [[12, 27]], [[12, 27]], [[9, 32]]]",
    "query_spans": "[[[29, 39]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a point on the right branch of the hyperbola $x^{2}-\\frac{y^{2}}{15}=1$, and $M$, $N$ are points on the circles $(x+4)^{2}+y^{2}=9$ and $(x-4)^{2}+y^{2}=1$ respectively, then the maximum value of $|P M|-|P N|$ is?",
    "fact_expressions": "G: Hyperbola;H: Circle;C:Circle;P: Point;M: Point;N: Point;Expression(G) = (x^2 - y^2/15 = 1);Expression(H) = (y^2 + (x + 4)^2 = 9);Expression(C)=(y^2 + (x - 4)^2 = 1);PointOnCurve(P, RightPart(G));PointOnCurve(M, H);PointOnCurve(N, C)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "6",
    "fact_spans": "[[[6, 35]], [[52, 72]], [[73, 92]], [[2, 5]], [[42, 45]], [[46, 49]], [[6, 35]], [[52, 72]], [[73, 92]], [[2, 41]], [[42, 95]], [[42, 95]]]",
    "query_spans": "[[[97, 116]]]",
    "process": "From the hyperbola equation, it is known that the two foci of the hyperbola are exactly the centers of the two circles. If |PM| - |PN| attains its maximum value, then it suffices to compute |PM|_{\\max} - |PN|_{\\min}. This can be solved using the definition of a hyperbola and the positional relationship between a point and a circle. The hyperbola is given by x^{2} - \\frac{y^{2}}{15} = 1, and P is a point on the right branch of the hyperbola, so |PF_{1}| - |PF_{2}| = 2a = 2. The two foci of the hyperbola are (-4,0) and (4,0), which are exactly the centers of the two circles (x+4)^{2} + y^{2} = 9 and (x-4)^{2} + y^{2} = 1, respectively. Thus, the radii of the two circles are r_{1} = 3 and r_{2} = 1. Therefore, from geometric properties, we have |PM|_{\\max} = |PF_{1}| + r_{1} = |PF_{1}| + 3. Similarly, |PN|_{\\min} = |PF_{2}| - r_{2} = |PF_{2}| - 1. Hence, the maximum value of |PM| - |PN| is |PM|_{\\max} - |PN|_{\\min} = (|PF_{1}| + 3) - (|PF_{2}| - 1) = |PF_{1}| - |PF_{2}| + 4 = 2 + 4 = 6. Therefore, the maximum value of |PM| - |PN| is 6."
  },
  {
    "text": "If a point $P$ on the parabola $x^{2}=4 y$ is at a distance of $2$ from the focus, then what are the coordinates of point $P$?",
    "fact_expressions": "G: Parabola;P: Point;Expression(G) = (x^2 = 4*y);PointOnCurve(P, G);Distance(P, Focus(G)) = 2",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(pm*2,1)",
    "fact_spans": "[[[1, 15]], [[33, 37], [18, 21]], [[1, 15]], [[1, 21]], [[1, 31]]]",
    "query_spans": "[[[33, 42]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with right focus $F(1,0)$, point $B$ is the upper vertex of ellipse $C$, and the slope of line $FB$ is $-\\sqrt{3}$. Then the equation of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;RightFocus(C)=F;Coordinate(F)=(1,0);B: Point;UpperVertex(C)=B;Slope(LineOf(F,B)) = -sqrt(3)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4+y^2/3=1",
    "fact_spans": "[[[2, 59], [77, 82], [112, 117]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[63, 71]], [[2, 71]], [[63, 71]], [[72, 76]], [[72, 86]], [[88, 110]]]",
    "query_spans": "[[[112, 122]]]",
    "process": "According to the ellipse equation, B(0,b) is obtained; then from the slope of line FB, b is found. Given focus F(1,0), c=1, and thus a can be determined, leading to the ellipse equation. Since ellipse C: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0) has right focus F(1,0), and point B is the upper vertex of ellipse C, so B(0,b), c=1. Also, the slope of line FB is -\\sqrt{3}, so \\frac{b-0}{0-1}=-\\sqrt{3}, solving gives b=\\sqrt{3}, therefore a=\\sqrt{b^{2}+c^{2}}=2. Hence, the equation of ellipse C is \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, a line passing through point $F_{1}$ intersects the ellipse at points $A$ and $B$, the perimeter of $\\triangle A B F_{2}$ is $12$, and the eccentricity of ellipse $C$ is $\\frac{\\sqrt{2}}{3}$, then the equation of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;Focus(C) = {F1, F2};H: Line;PointOnCurve(F1, H);A: Point;B: Point;Intersection(H, C) = {A, B};Perimeter(TriangleOf(A, B, F2)) = 12;Eccentricity(C) = sqrt(2)/3",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/9+y^2/7=1",
    "fact_spans": "[[[18, 75], [94, 96], [138, 143], [170, 175]], [[18, 75]], [[25, 75]], [[25, 75]], [[25, 75]], [[25, 75]], [[2, 9], [82, 90]], [[10, 17]], [[2, 80]], [[91, 93]], [[81, 93]], [[98, 101]], [[102, 105]], [[91, 107]], [[108, 137]], [[138, 168]]]",
    "query_spans": "[[[170, 180]]]",
    "process": "From the perimeter of \\triangle ABF_{2} being 12, we know a=3; then according to the eccentricity, c=\\sqrt{2}, and thus the answer can be obtained. [Detailed solution] Since the perimeter of \\triangle ABF_{2} is 4a, it follows that a=3; from \\frac{c}{a}=\\frac{\\sqrt{2}}{3} we get c=\\sqrt{2}, hence b^{2}=a^{2}-c^{2}=7, so the equation of ellipse C is \\frac{x^{2}}{9}+\\frac{y^{2}}{7}="
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $y=3 x$, then its eccentricity is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;Expression(OneOf(Asymptote(G))) = (y = 3*x)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 48], [66, 67]], [[2, 48]], [[5, 48]], [[5, 48]], [[2, 64]]]",
    "query_spans": "[[[66, 72]]]",
    "process": ""
  },
  {
    "text": "If a parabola passes through the points $(-1, \\frac{1}{2})$, $(2,2)$, then what is the standard equation of this parabola?",
    "fact_expressions": "G: Parabola;H: Point;I: Point;Coordinate(H) = (-1, 1/2);Coordinate(I) = (2, 2);PointOnCurve(H, G);PointOnCurve(I,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2=2*y",
    "fact_spans": "[[[1, 4], [38, 41]], [[6, 26]], [[28, 35]], [[6, 26]], [[28, 35]], [[1, 27]], [[1, 35]]]",
    "query_spans": "[[[38, 48]]]",
    "process": "Given the coordinates of two points, the equation of the parabola can be set up and the unknown coefficient can be determined. Since the parabola passes through the points $(-1,\\frac{1}{2})$ and $(2,2)$, and thus lies in the first and second quadrants, assume the parabola equation is $x^{2}=2py$, where $p>0$. Substituting the point $(2,2)$ yields $4p=4$, so $p=1$. Therefore, the equation of the parabola is: $x^{2}=2y$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$, its directrix intersects the $x$-axis at point $M$. A line passing through its focus $F$ intersects the parabola at points $A$ and $B$. Let the slopes of lines $MA$ and $MB$ be $k_{1}$ and $k_{2}$, respectively. Then the minimum value of $\\frac{1}{k_{1}^{2}}+\\frac{1}{k_{2}^{2}}$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);Intersection(Directrix(C),xAxis)=M;A: Point;M: Point;B: Point;F:Point;Focus(C)=F;G:Line;Intersection(G, C) = {A, B};PointOnCurve(F, G);Slope(LineOf(M,A))=k1;Slope(LineOf(M,B))=k2;k1:Number;k2:Number",
    "query_expressions": "Min(1/(k2^2) + 1/(k1^2))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 21], [22, 23], [38, 39], [48, 51]], [[2, 21]], [[22, 36]], [[54, 57]], [[32, 36]], [[58, 61]], [[41, 44]], [[38, 44]], [[45, 47]], [[45, 63]], [[37, 47]], [[65, 103]], [[65, 103]], [[86, 93]], [[96, 103]]]",
    "query_spans": "[[[105, 152]]]",
    "process": ""
  },
  {
    "text": "What are the coordinates of the focus of the parabola $y=a x^{2}  (a \\neq 0)$?",
    "fact_expressions": "G: Parabola;a: Number;Negation(a=0);Expression(G) = (y = a*x^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,1/(4*a))",
    "fact_spans": "[[[0, 26]], [[3, 26]], [[3, 26]], [[0, 26]]]",
    "query_spans": "[[[0, 33]]]",
    "process": ""
  },
  {
    "text": "The right directrix of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ intersects the two asymptotes at points $A$ and $B$, the right focus is $F$, and $F A \\perp F B$. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F: Point;A: Point;B: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;IsPerpendicular(LineSegmentOf(F, A), LineSegmentOf(F, B));L1:Line;L2:Line;Asymptote(G)={L1,L2};Intersection(RightDirectrix(G),L1)=A;Intersection(RightDirectrix(G),L2)=B",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 46], [94, 97]], [[3, 46]], [[3, 46]], [[72, 75]], [[58, 61]], [[62, 65]], [[0, 46]], [[0, 75]], [[77, 92]], [-1, -1], [-1, -1], [0, 53], [[0, 67]], [[0, 67]]]",
    "query_spans": "[[[94, 103]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $x^{2}=2 p y(p>0)$ with focus $F$, directrix $l$, point $P(4, y_{0})$ lies on the parabola, $K$ is the intersection point of $l$ and the $y$-axis, and $|P K|=\\sqrt{2}|P F|$, then $y_{0}$=?",
    "fact_expressions": "G: Parabola;p: Number;P: Point;K: Point;F: Point;l: Line;p>0;y0:Number;Expression(G) = (x^2 = 2*p*y);Coordinate(P) = (4,y0);Focus(G) = F;Directrix(G) = l;PointOnCurve(P, G);Intersection(l,yAxis) = K;Abs(LineSegmentOf(P, K)) = sqrt(2)*Abs(LineSegmentOf(P, F))",
    "query_expressions": "y0",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 23], [52, 55]], [[5, 23]], [[37, 51]], [[57, 60]], [[26, 29]], [[33, 36], [61, 64]], [[5, 23]], [[97, 104]], [[2, 23]], [[37, 51]], [[2, 29]], [[2, 36]], [[37, 56]], [[57, 72]], [[74, 95]]]",
    "query_spans": "[[[97, 106]]]",
    "process": "Draw a perpendicular from point P to the directrix l, with foot of perpendicular at M, then |PM| = |PF|. In right triangle PKM, since |PK| = \\sqrt{2}|PF| = \\sqrt{2}|PM|, \\therefore PM = KM = 4, \\therefore y_{0} = 4 - \\frac{p}{x}. Substitute P(4, 44 - \\frac{p}{2}) into the parabola equation x^{2} = 2py, solving gives p = 4. \\therefore y_{0} = 4 - 2 = 2. The correct answer is 2. This problem mainly examines the definition of parabola and equation thinking, as well as computational and transformation abilities, classified as a medium-level question."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively. A line passing through $F_{1}$ intersects the left branch of hyperbola $C$ at points $A$ and $B$. Connect $A F_{2}$ and $B F_{2}$. In $\\triangle A B F_{2}$, $\\sin \\frac{\\angle A B F_{2}}{2}=\\frac{1}{4}$, $|A B|=|B F_{2}|$. Then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;A: Point;F2: Point;B: Point;F1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, G);Intersection(G,LeftPart(C))={A,B};Sin(AngleOf(A, B, F2)/2) = 1/4;Abs(LineSegmentOf(A, B)) = Abs(LineSegmentOf(B, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[20, 81], [99, 105], [233, 239]], [[27, 81]], [[27, 81]], [[96, 98]], [[110, 113]], [[10, 17]], [[114, 117]], [[2, 9], [88, 95]], [[27, 81]], [[27, 81]], [[20, 81]], [[2, 86]], [[2, 86]], [[87, 98]], [[96, 119]], [[168, 213]], [[214, 231]]]",
    "query_spans": "[[[233, 245]]]",
    "process": "Let $|BF_{1}|=m$, then by the definition of hyperbola, we have $|BF_{2}|=2a+m$, $|AF_{1}|=2a$, $|AF_{2}|=4a$. From $\\sin\\frac{\\angle ABF_{2}}{2}=\\frac{1}{4}$, we obtain $m=6a$. Then in $\\triangle BF_{1}F_{2}$, using the law of cosines, we can set up an equation to solve for the eccentricity. Let $|BF_{1}|=m$, then by the definition of hyperbola, we get $|BF_{2}|=2a+m$, $|AF_{1}|=|AB|-|BF_{1}|=|BF_{2}|-m=2a$, then $|AF_{2}|=4a$, so $\\sin\\frac{\\angle ABF_{2}}{2}=\\frac{2a}{2a+m}=\\frac{1}{4}$, solving gives $m=6a$, thus $|BF_{2}|=8a$. In $\\Delta BF_{1}F_{2}$, $|F_{1}F_{2}|^{2}=|BF_{1}|^{2}+|BF_{2}|^{2}-2|BF_{1}|\\cdot|BF_{2}|\\cos\\angle F_{1}BF_{2}$, i.e., $4c^{2}=36a^{2}+64a^{2}-2\\cdot6a\\cdot8a\\cdot(1-2\\sin^{2}\\frac{\\angle ABF_{2}}{2})$, solving yields $e=\\frac{c}{a}=2$."
  },
  {
    "text": "One focus of the hyperbola is $F_{2}(2 , 0)$, and the eccentricity is $e=2$. Then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;F2: Point;Coordinate(F2) = (2, 0);OneOf(Focus(G) )= F2;Eccentricity(G)=e;e:Number;e=2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[0, 3], [34, 37]], [[9, 23]], [[9, 23]], [[0, 23]], [[0, 32]], [[27, 32]], [[27, 32]]]",
    "query_spans": "[[[34, 44]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, a line passing through $F_{1}$ intersects the left branch of hyperbola $C$ at points $P$ and $Q$, satisfying $|P F_{2}|=2 c$ $(c=\\sqrt{a^{2}+b^{2}})$, and $\\frac{S_{\\Delta P F_{1}} F_{2}}{S_{\\triangle Q F_{1}} F_{2}}=\\frac{2}{3}$. Then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;P: Point;F2: Point;Q: Point;F1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, G);Intersection(G, LeftPart(C)) = {P, Q};Abs(LineSegmentOf(P, F2)) = 2*c;c:Number;c=sqrt(a^2+b^2);Area(TriangleOf(P,F1,F2))/Area(TriangleOf(Q,F1,F2)) = 2/3",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "7/5",
    "fact_spans": "[[[20, 82], [101, 107], [239, 245]], [[28, 82]], [[28, 82]], [[98, 100]], [[112, 115]], [[10, 17]], [[116, 119]], [[2, 9], [90, 97]], [[28, 82]], [[28, 82]], [[20, 82]], [[2, 88]], [[2, 88]], [[89, 100]], [[98, 121]], [[125, 162]], [[125, 162]], [[125, 162]], [[163, 237]]]",
    "query_spans": "[[[239, 251]]]",
    "process": "By the definition of a hyperbola, we have |PF_{1}| = |PF_{2}| - 2a = 2c - 2a. Since \\frac{S_{\\triangle PF_{1}F_{2}}}{S_{\\triangle OF_{1}F_{2}}} = \\frac{|PF_{1}|}{|QF_{1}|} = \\frac{2}{3}, it follows that |QF_{1}| = 3(c - a), hence |QF_{2}| = 2a + |QF_{1}| = 3c - a. By the law of cosines, \\cos\\angle PF_{1}F_{2} = \\frac{|PF_{1}|^{2} + |F_{1}F_{2}|^{2} - |PF_{2}|^{2}}{2|PF_{1}||F_{1}F_{2}|}, \\cos\\angle QF_{1}F_{2} = \\frac{|QF_{1}|^{2} + |F_{1}F_{2}|^{2} - |QF_{2}|^{2}}{2|QF_{1}||F_{1}F_{2}|}. Also, \\cos\\angle PF_{1}F_{2} = -\\cos\\angle QF_{1}F_{2}, so 5c^{2} - 12ac + 7a^{2} = 0, i.e., 5e^{2} - 12e + 7 = 0. Solving gives e = \\frac{7}{5} or e = 1 (discarded)."
  },
  {
    "text": "The minimum value of $\\sqrt{(\\frac{t^{2}}{6}-\\frac{3}{2})^{2}+t^{2}}+\\sqrt{(\\frac{t^{2}}{6}-3)^{2}+(t-2)^{2}}$ is?",
    "fact_expressions": "t:Number",
    "query_expressions": "Min(sqrt(t^2 + (t^2/6 - 3/2)^2) + sqrt((t - 2)^2 + (t^2/6 - 3)^2))",
    "answer_expressions": "9/2",
    "fact_spans": "[[[0, 90]]]",
    "query_spans": "[[[0, 96]]]",
    "process": "Let $ t = y $, then $ \\sqrt{\\left( \\frac{y^{2}}{6} - \\frac{3}{2} \\right)^{2} + y^{2}} + \\sqrt{\\left( \\frac{y^{2} - 3}{6} \\right)^{2} + (y - 2)^{2}} $ represents the sum of the distances from point $ P\\left( \\frac{y^{2}}{6}, y \\right) $ to point $ F\\left( \\frac{3}{2}, 0 \\right) $ and point $ A(3, 2) $. Since point $ P\\left( \\frac{y^{2}}{6}, y \\right) $ lies on the parabola $ C: y^{2} = 6x $, and $ F $ is the focus of $ C $, as shown in the figure: the distance to $ x = -\\frac{3}{2} $ is $ d_{1} $. Therefore, $ \\sqrt{\\left( \\frac{t^{2}}{6} - \\frac{3}{2} \\right)^{2} + t^{2}} + \\sqrt{\\left( \\frac{t^{2} - 3}{6} \\right)^{2} + (t - 2)^{2}} = |PF| + |PA| = d + |PA| \\geqslant d_{1} = 3 + \\frac{3}{2} = \\frac{9}{2} $."
  },
  {
    "text": "Given a point $P$ on the hyperbola $x^{2}-\\frac{y^{2}}{15}=1$ such that the distance from $P$ to one of its foci is $4$, then the distance from point $P$ to the other focus is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/15 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;OneOf(Focus(G)) = F1;OneOf(Focus(G)) = F2;Negation(F1=F2);Distance(P, F1) = 4",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 31], [38, 39]], [[2, 31]], [[34, 37], [54, 58]], [[2, 37]], [], [], [[38, 44]], [[38, 64]], [[38, 64]], [[34, 51]]]",
    "query_spans": "[[[38, 69]]]",
    "process": "\\because the hyperbola x^{2}-\\frac{y^{2}}{15}=1 has a=1, c=4, then the shortest distance from a point on the hyperbola to a focus is 3. If a point P on the hyperbola x^{2}-\\frac{y^{2}}{15}=1 has a distance of 4 to one of its foci, then the distance from point P to the other focus is equal to 4+2=6, or 4-2=2 (discarded),"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{2}=1$ and the parabola $y^{2}=2 p x  (p>0)$ intersect at points $A$ and $B$, and $|A B|=2$, then $p=$?",
    "fact_expressions": "H: Ellipse;Expression(H) = (x^2/8 + y^2/2 = 1);G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;A: Point;B: Point;Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 2",
    "query_expressions": "p",
    "answer_expressions": "1/4",
    "fact_spans": "[[[2, 39]], [[2, 39]], [[40, 63]], [[40, 63]], [[86, 89]], [[43, 63]], [[65, 68]], [[69, 72]], [[2, 74]], [[75, 84]]]",
    "query_spans": "[[[86, 91]]]",
    "process": "Let A(x,y) (x>0, y>0) according to the problem. Since |AB|=2, we have y=1. Substituting into the ellipse equation gives x=2. Substituting A(2,1) into the parabola equation yields p=\\frac{1}{4}. Therefore, the answer is \\frac{1}{4}."
  },
  {
    "text": "Given that the line $x = m y + 1$ always has common points with the ellipse $\\frac{x^{2}}{a} + \\frac{y^{2}}{2} = 1$, the range of values for $a$ is?",
    "fact_expressions": "G: Ellipse;a: Number;H: Line;m: Number;Expression(G) = (y^2/2 + x^2/a = 1);Expression(H) = (x = m*y + 1);IsIntersect(H, G)",
    "query_expressions": "Range(a)",
    "answer_expressions": "[1,+oo)&\\Negation(a=2)",
    "fact_spans": "[[[14, 51]], [[58, 61]], [[2, 13]], [[4, 13]], [[14, 51]], [[2, 13]], [[2, 56]]]",
    "query_spans": "[[[58, 68]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, respectively, and point $P$ lies on the ellipse such that $|\\overrightarrow{P F_{1}}+\\overrightarrow{P F_{2}}|=2 \\sqrt{3}$, then $\\angle F_{1} P F_{2}$=?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P,G);Abs(VectorOf(P, F1) + VectorOf(P, F2)) = 2*sqrt(3)",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "pi/2",
    "fact_spans": "[[[20, 47], [57, 59]], [[52, 56]], [[2, 9]], [[10, 17]], [[20, 47]], [[2, 51]], [[2, 51]], [[52, 60]], [[61, 125]]]",
    "query_spans": "[[[128, 152]]]",
    "process": "The ellipse $\\frac{x^{2}}{4}+y^{2}=1$ has its foci on the $x$-axis, $|PF_{1}|+|PF_{2}|=2a=4$, $|F_{1}F_{2}|^{2}=2\\sqrt{3}$. By the law of cosines: $|F_{1}F_{2}|^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}|\\cdot|PF_{2}|\\cos\\angle F_{1}PF_{2}=12$. Then $|\\overrightarrow{PF_{1}}+\\overrightarrow{PF_{2}}|^{2}=|\\overrightarrow{PF_{1}}|^{2}+2\\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}+|\\overrightarrow{PF_{2}}|^{2}=12$, $\\therefore 2|PF_{1}||PF_{2}|\\cos\\angle F_{1}PF_{2}=0$, then $\\cos\\angle F_{1}PF_{2}=0$, $\\therefore \\angle F_{1}PF_{2}=\\frac{\\pi}{2}$"
  },
  {
    "text": "Let $F$ be a focus of the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, point $P$ lies on $C$, $O$ is the origin, and $|O P|=|O F|$. Then the area of $\\triangle O P F$ is?",
    "fact_expressions": "C: Hyperbola;O: Origin;P: Point;F: Point;Expression(C) = (x^2/4 - y^2/5 = 1);OneOf(Focus(C)) = F;PointOnCurve(P, C);Abs(LineSegmentOf(O, P)) = Abs(LineSegmentOf(O, F))",
    "query_expressions": "Area(TriangleOf(O, P, F))",
    "answer_expressions": "5/2",
    "fact_spans": "[[[5, 48], [59, 62]], [[64, 67]], [[54, 58]], [[1, 4]], [[5, 48]], [[1, 53]], [[54, 63]], [[73, 86]]]",
    "query_spans": "[[[88, 110]]]",
    "process": "As shown in the figure, without loss of generality, let $ F $ be the right focus of the hyperbola $ C: \\frac{x^2}{4} - \\frac{y^{2}}{5} = 1 $, and $ P $ be a point in the first quadrant. From the hyperbola equation, we have $ a^{2} = 4 $, $ b^{2} = 5 $, then $ c = 3 $. Thus, the equation of the circle centered at $ O $ with radius 3 is $ x^{2} + y^{2} = 9 $. Solving the system  \n\\[\n\\begin{cases}\nx^{2} + y^{2} = 9 \\\\\n\\frac{x^{2}}{4} - \\frac{y^{2}}{5} = 1\n\\end{cases}\n\\]\nyields $ P\\left( \\frac{2\\sqrt{14}}{3}, \\frac{5}{3} \\right) $. Therefore,  \n$ S_{\\triangle OPF} = \\frac{1}{2} \\times 3 \\times \\frac{5}{3} = \\frac{5}{2} $.  \nAnswer is"
  },
  {
    "text": "Given that $F$ is a focus of the hyperbola $C$: $\\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1$, point $P$ lies on an asymptote of the hyperbola $C$, and $O$ is the origin. If $|O P|=|O F|$, then the area of $\\triangle O P F$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/5 - y^2/4 = 1);F: Point;OneOf(Focus(C)) = F;P: Point;PointOnCurve(P, Asymptote(C));O: Origin;Abs(LineSegmentOf(O, P)) = Abs(LineSegmentOf(O, F))",
    "query_expressions": "Area(TriangleOf(O, P, F))",
    "answer_expressions": "3",
    "fact_spans": "[[[6, 49], [60, 66]], [[6, 49]], [[2, 5]], [[2, 54]], [[55, 59]], [[55, 71]], [[72, 75]], [[82, 95]]]",
    "query_spans": "[[[97, 119]]]",
    "process": "a^{2}=5, b^{2}=4, c^{2}=a^{2}+b^{2}=9, c=3, without loss of generality, let F be the right focus, let P(x_{0},y_{0}) be in the first quadrant and on the line y=\\frac{2\\sqrt{5}}{5}x, so y_{0}=\\frac{2\\sqrt{5}}{5}x_{0}\\textcircled{1} solving \\textcircled{1}\\textcircled{2} gives y_{0}=2, thus the area of \\Delta OPF is \\frac{1}{2}\\times|OF|\\times|y_{0}|=\\frac{1}{2}\\times3\\times2=3"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the common foci of an ellipse and a hyperbola, $P$ is one of their common points, and $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$, then the minimum value of the sum of the squares of the eccentricities of the ellipse and the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;F1: Point;P: Point;F2: Point;Focus(G)={F1,F2};Focus(H)={F1,F2};OneOf(Intersection(G,H))=P;AngleOf(F1, P, F2) = pi/3",
    "query_expressions": "Min(Eccentricity(G)^2+Eccentricity(H)^2)",
    "answer_expressions": "1+(sqrt(3)/2)",
    "fact_spans": "[[[21, 24], [85, 88]], [[18, 20], [82, 84]], [[2, 9]], [[30, 33]], [[10, 17]], [[2, 29]], [[2, 29]], [[30, 42]], [[44, 80]]]",
    "query_spans": "[[[82, 102]]]",
    "process": "Let the semi-major axis of the ellipse and the semi-transverse axis of the hyperbola be $ a_{1} $, $ a_{2} $, respectively, and the semi-focal length be $ c $. Assume $ PF_{1} > PF_{2} $, then we have $ PF_{1} + PF_{2} = 2a_{1} $, $ PF_{1} - PF_{2} = 2a_{2} $. Solving gives $ PF_{1} = a_{1} + a_{2} $, $ PF_{2} = a_{1} - a_{2} $. By the law of cosines, $ F_{1}F_{2}^{2} = PF_{1}^{2} + PF_{2}^{2} - 2PF_{1}PF_{2}\\cos\\frac{\\pi}{3} $, simplifying yields $ 4c^{2} = a_{1}^{2} + 3a_{2}^{2} $, $ e_{1}^{2} + p_{2}^{2} = \\frac{c^{2}}{a_{1}^{2}} + \\frac{c^{2}}{a_{2}^{2}} = \\frac{a_{1}^{2} + 3a_{2}^{2}}{4a_{1}^{2}} + \\frac{a_{1}^{2} + 3a_{2}^{2}}{4a_{2}^{2}} = 1 + \\frac{3a_{2}^{2}}{4a_{1}^{2}} + \\frac{a_{1}^{2}}{4a_{2}^{2}} \\geqslant 1 + \\frac{\\sqrt{3}}{2} $. The minimum value of the sum of the squares of the eccentricities of the ellipse and hyperbola is $ 1 + \\frac{\\sqrt{3}}{2} $."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $y^{2}=2 x$, $A$ is a moving point on the parabola, and point $B(-1,0)$, then when $\\frac{2|A B|}{2|A F|+1}$ takes its maximum value, the value of $|A B|$ is?",
    "fact_expressions": "G: Parabola;B: Point;A: Point;F: Point;Expression(G) = (y^2 = 2*x);Coordinate(B) = (-1, 0);Focus(G) = F;PointOnCurve(A, G);WhenMax(2*Abs(LineSegmentOf(A, B))/(2*Abs(LineSegmentOf(A, F)) + 1))",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "sqrt(6)",
    "fact_spans": "[[[6, 20], [28, 31]], [[36, 46]], [[24, 27]], [[2, 5]], [[6, 20]], [[36, 46]], [[2, 23]], [[24, 35]], [[48, 79]]]",
    "query_spans": "[[[80, 91]]]",
    "process": "Let A(x, y), then |AB| = \\sqrt{(x+1)^{2} + y^{2}} = \\sqrt{x^{2} + 4x + 1}, and |AF| = x + \\frac{1}{2}. Therefore, (\\frac{2|AB|}{2|AF|+1})^{2} = \\frac{x^{2}+4x+1}{x^{2}+2x+1} = 1 + \\frac{2x}{x^{2}+2x+1} = 1 + \\frac{1}{2} + \\frac{1}{2x} + 1^{\\leqslant} 1 + \\frac{1}{2\\sqrt{\\frac{x}{2} \\cdot \\frac{1}{2}} + 1} = \\frac{3}{2}, with equality if and only if x = 1. Thus, \\frac{2|AB|}{2|AF|+1} attains its maximum when x = 1, at which point |AB| = \\sqrt{1 + 4 + 1} = \\sqrt{6}."
  },
  {
    "text": "Given that the asymptotes of the hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ are $y=\\pm \\frac{1}{2} x$, and it passes through the point $(4, \\sqrt{3})$, find the equation of the hyperbola $C_{1}$?",
    "fact_expressions": "C1: Hyperbola;H: Point;Expression(C1)=(-y^2/b^2 + x^2/a^2 = 1);Coordinate(H) = (4, sqrt(3));Expression(Asymptote(C1)) = (y = pm*(x/2));PointOnCurve(H,C1);a: Number;b: Number;a>0;b>0",
    "query_expressions": "Expression(C1)",
    "answer_expressions": "x^2/4-y^2=1",
    "fact_spans": "[[[2, 67], [116, 126]], [[98, 114]], [[2, 67]], [[98, 114]], [[2, 95]], [[2, 114]], [[2, 67]], [[2, 67]], [[2, 67]], [[2, 67]]]",
    "query_spans": "[[[116, 130]]]",
    "process": "The asymptotes of the hyperbola $ C_{1}: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) are given by $ y = \\pm \\frac{1}{2}x $, so $ \\frac{b}{a} = \\frac{1}{2} $. From the point $ (4, \\sqrt{3}) $, we have: $ \\frac{16}{a^{2}} - \\frac{3}{b^{2}} = 1 $. Solving the system  \n\\[\n\\begin{cases}\n\\frac{b}{a} = \\frac{1}{2} \\\\\n\\frac{16}{a^{2}} - \\frac{3}{b^{2}} = 1\n\\end{cases}\n\\]\nyields $ a = 2 $, $ b = 1 $. The equation of the hyperbola $ C_{1} $ is $ \\frac{x^{2}}{4} - y^{2} = 1 $."
  },
  {
    "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ have its right focus at $F$, and left and right vertices at $A_{1}$, $A_{2}$ respectively. A line $l$ passing through $F$ and parallel to one asymptote of the hyperbola $C$ intersects the other asymptote at point $P$. If $P$ lies exactly on the circle with diameter $A_{1}A_{2}$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(C) = F;A1: Point;A2: Point;LeftVertex(C) = A1;RightVertex(C) = A2;PointOnCurve(F, l) = True;L1: Line;L2: Line;OneOf(Asymptote(C)) = L1;OneOf(Asymptote(C)) = L2;Negation(L1 = L2);IsParallel(l, L1) = True;l: Line;Intersection(l, L2) = P;P: Point;G: Circle;PointOnCurve(P, G) = True;IsDiameter(LineSegmentOf(A1, A2), G) = True",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 63], [105, 111], [167, 173]], [[1, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[68, 71], [100, 103]], [[1, 71]], [[80, 87]], [[90, 97]], [[1, 97]], [[1, 97]], [[99, 125]], [], [], [[105, 117]], [[105, 132]], [[105, 132]], [[104, 125]], [[120, 125]], [[105, 138]], [[135, 138], [140, 143]], [[163, 164]], [[140, 165]], [[146, 164]]]",
    "query_spans": "[[[167, 179]]]",
    "process": "According to the problem, let point $ F(c,0) $. The line $ l $ passing through $ F $ and parallel to one asymptote of hyperbola $ C $ is: $ y = \\frac{b}{a}(x - c) $. Therefore,\n$$\n\\begin{cases}\ny = \\frac{b}{a}(x - c) \\\\\ny = -\\frac{b}{a}\n\\end{cases}\n\\Rightarrow\n\\begin{cases}\nx = \\frac{c}{2} \\\\\ny = \\frac{bc}{2a}\n\\end{cases}\n$$\nSince the coordinates of $ A_1 $, $ A_2 $ are $ (a,0) $, $ (-a,0) $ respectively, then $ P $ lies exactly on the circle with $ A_1A_2 $ as diameter, so $ |OP| = a $, i.e., $ 4a^4 = a^2c^2 + b^2c^2 = a^2c^2 + (c^2 - a^2)c^2 \\Rightarrow e = \\sqrt{2} $. Hence, fill in $ \\sqrt{2} $."
  },
  {
    "text": "From the left focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw two tangents to the circle $O$: $x^{2}+y^{2}=a^{2}$, with points of tangency $A$ and $B$. The left vertex of the hyperbola is $C$. If $\\angle A C B=120^{\\circ}$, then the equations of the asymptotes of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;O: Circle;A: Point;C: Point;B: Point;F: Point;l1: Line;l2: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(O) = (x^2 + y^2 = a^2);LeftFocus(G) = F;TangentOfPoint(F, O) = {l1, l2};TangentPoint(l1, O) = A;TangentPoint(l2, O) = B;LeftVertex(G) = C;AngleOf(A, C, B) = ApplyUnit(120, degree)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[1, 57], [107, 110], [147, 150]], [[4, 57]], [[4, 57]], [[65, 90]], [[99, 102]], [[114, 117]], [[103, 106]], [[61, 64]], [], [], [[4, 57]], [[4, 57]], [[1, 57]], [[65, 90]], [[1, 64]], [[0, 95]], [[0, 106]], [[0, 106]], [[107, 117]], [[119, 145]]]",
    "query_spans": "[[[147, 158]]]",
    "process": "According to the asymptotic equations of the hyperbola: $ y = \\pm\\frac{b}{a}x' $, combining with the given conditions we obtain $ \\angle AFO = 30^{\\circ} $, thus obtaining the relationship between $ a $ and $ c $, and further deriving the relationship between $ a $ and $ b $, thereby obtaining the answer. From the problem, the equation of the hyperbola is: $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $). Therefore, the asymptotes of the hyperbola are: $ y = \\pm\\frac{b}{a}x $. Since $ \\angle ACB = 120^{\\circ} $, and an inscribed angle is half the central angle, we get $ \\angle AFB = 60^{\\circ} $. By symmetry, $ \\angle AFO = 30^{\\circ} $. Hence, $ c = 2a $. Also, since $ b^{2} = c^{2} - a^{2} $, we have $ \\frac{b}{a} = \\sqrt{3} $. Therefore, the asymptotes of the hyperbola are: $ y = \\pm\\sqrt{3}x $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and $P$ is a point on the ellipse such that $\\angle F_{1} PF_{2}=90^{\\circ}$, then what is the area of $\\triangle P F_{1} F_{2}$?",
    "fact_expressions": "F1: Point;F2: Point;C: Ellipse;Expression(C) = (x^2/25 + y^2/9 = 1);Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C);AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Area(AngleOf(P, F1, F2))",
    "answer_expressions": "9",
    "fact_spans": "[[[2, 9]], [[10, 17]], [[18, 61], [71, 73]], [[18, 61]], [[2, 66]], [[67, 70]], [[67, 76]], [[78, 110]]]",
    "query_spans": "[[[112, 138]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ and the hyperbola $\\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1$, their intersection point in the first quadrant is $P$. What is the distance from point $P$ to the left focus of the ellipse?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;P: Point;Expression(G) = (x^2/6 - y^2/3 = 1);Expression(H) = (x^2/25 + y^2/16 = 1);Quadrant(P) = 1;Intersection(G, H) = P",
    "query_expressions": "Distance(P, LeftFocus(H))",
    "answer_expressions": "5 + sqrt(6)",
    "fact_spans": "[[[42, 80]], [[2, 41], [99, 101]], [[94, 98], [89, 92]], [[42, 80]], [[2, 41]], [[81, 92]], [[2, 92]]]",
    "query_spans": "[[[94, 109]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{4}=1$ is $e=\\frac{\\sqrt{2}}{2}$, then $m=$?",
    "fact_expressions": "G: Ellipse;m: Number;e: Number;Expression(G) = (y^2/4 + x^2/m = 1);Eccentricity(G) = e ;e = sqrt(2)/2",
    "query_expressions": "m",
    "answer_expressions": "{2, 8}",
    "fact_spans": "[[[0, 37]], [[65, 68]], [[41, 63]], [[0, 37]], [[0, 63]], [[41, 63]]]",
    "query_spans": "[[[65, 70]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the directrix of the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/8 + y^2/4 = 1)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x = pm*4",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 43]]]",
    "process": ""
  },
  {
    "text": "If the line $y=k x$ is an asymptote of the hyperbola $x^{2}-4 y^{2}=1$, then $k=$?",
    "fact_expressions": "E: Hyperbola;F: Line;k: Number;Expression(E) = (x^2 - 4*y^2 = 1);Expression(F) = (y = k*x);F = OneOf(Asymptote(E))",
    "query_expressions": "k",
    "answer_expressions": "pm*1/2",
    "fact_spans": "[[[11, 31]], [[1, 10]], [[39, 42]], [[11, 31]], [[1, 10]], [[1, 37]]]",
    "query_spans": "[[[39, 44]]]",
    "process": "Let $x^{2}-4y^{2}=0$, solving gives $y=\\pm\\frac{1}{2}x$, hence $k=\\pm\\frac{1}{2}$."
  },
  {
    "text": "The length of the imaginary axis of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{m}=1$ is $2$ times the length of the real axis, then the value of $m$ =?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/4 - y^2/m = 1);Length(ImageinaryAxis(G)) = 2*Length(RealAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "16",
    "fact_spans": "[[[0, 38]], [[53, 56]], [[0, 38]], [[0, 51]]]",
    "query_spans": "[[[53, 60]]]",
    "process": "From the hyperbola equation, we know $a^{2}=4$, $b^{2}=m$, $\\therefore a=2$, $b=\\sqrt{m}$. Since the length of the imaginary axis is twice that of the real axis, we have $\\sqrt{m}=4$, $\\therefore m=16$."
  },
  {
    "text": "Given an ellipse $\\frac{x^{2}}{a^{2}}+y^{2}=1$ $(a>1)$ with left and right foci $F_{1}$, $F_{2}$, the left vertex is $A$, the upper vertex is $B$, and points $M$, $N$ are arbitrary points on the ellipse. If the maximum area of $\\Delta M A B$ is $2(\\sqrt{2}+1)$, then the maximum value of $\\frac{3|N F_{1}| \\cdot|N F_{2}|}{|N F_{1}|+4|N F_{2}|}$ is?",
    "fact_expressions": "G: Ellipse;a: Number;M: Point;A: Point;B: Point;N: Point;F1: Point;F2: Point;a>1;Expression(G) = (y^2 + x^2/a^2 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;LeftVertex(G)=A;UpperVertex(G)=B;PointOnCurve(M, G);PointOnCurve(N, G);Max(Area(TriangleOf(M,A,B)))=2*(sqrt(2)+1)",
    "query_expressions": "Max((3*Abs(LineSegmentOf(N, F1))*Abs(LineSegmentOf(N, F2)))/(Abs(LineSegmentOf(N, F1)) + 4*Abs(LineSegmentOf(N, F2))))",
    "answer_expressions": "8/3",
    "fact_spans": "[[[24, 62], [88, 90]], [[26, 62]], [[79, 83]], [[67, 70]], [[75, 78]], [[84, 87]], [[3, 10]], [[11, 18]], [[26, 62]], [[24, 62]], [[2, 62]], [[2, 62]], [[24, 70]], [[24, 78]], [[79, 95]], [[79, 95]], [[97, 133]]]",
    "query_spans": "[[[135, 197]]]",
    "process": "By the given conditions, A(-a,0), B(0,1), then |AB| = \\sqrt{1+a^{2}}, k_{AB} = \\frac{1}{a}. Let line l be a tangent to the ellipse parallel to AB, with equation: y = \\frac{1}{a}x + m. From \\begin{cases} y = \\frac{1}{a}x + m \\\\ x^{2} + a^{2}y^{2} = a^{2} \\end{cases}, we obtain: 2x^{2} + 2mx + a^{2}m^{2} - a^{2} = 0; \\therefore \\triangle = (2am)^{2} - 4 \\times 2 \\times (a^{2}m^{2} - a^{2}) = 0, yielding m = \\pm\\sqrt{2}; according to the conditions take m = -\\sqrt{2}. The distance from tangent line l to line AB is: d = \\frac{\\sqrt{2}+1}{\\sqrt{1+\\frac{1}{a^{2}}}}; \\therefore the maximum area is (S_{ABM})_{\\max} = \\frac{1}{2}\\sqrt{a^{2+1}}{\\sqrt{2}+1}{\\sqrt{\\frac{1}{a^{2}}+1}} = 2(\\sqrt{2}+1), giving a = 4. Let |NF_{1}| = m, |NF_{2}| = n, then m+n = 8; then \\frac{3|NF_{1}|\\cdot|NF_{2}|}{|NF_{1}|+4|NF_{2}|} = \\frac{3mn}{m+4n} = \\frac{3}{\\frac{1}{n}+\\frac{4}{m}} = \\frac{3\\times8}{(m+n)(\\frac{1}{n}+\\frac{c}{7}}- = \\frac{3\\times8}{1+4+\\frac{m}{n}+\\frac{4n}{m}} \\leqslant \\frac{24}{9} = \\frac{8}{3}, with equality if and only if m = \\frac{8}{3};"
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $x^{2}-\\frac{y^{2}}{24}=1$, and let $P$ be a common point of the hyperbola and the ellipse $\\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1$. Then the area of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;G: Hyperbola;Expression(G) = (x^2 - y^2/24 = 1);P: Point;H: Ellipse;Expression(H) = (x^2/49 + y^2/24 = 1);OneOf(Intersection(G, H)) = P",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "24",
    "fact_spans": "[[[1, 8]], [[9, 16]], [[1, 52]], [[1, 52]], [[17, 46], [57, 60]], [[17, 46]], [[53, 56]], [[61, 100]], [[61, 100]], [[53, 106]]]",
    "query_spans": "[[[108, 135]]]",
    "process": "In the hyperbola $ x^{2} - \\frac{y^{2}}{24} = 1 $, $ |F_{1}F_{2}| = 10 $, and $ P $ is a common point of the hyperbola and the ellipse $ \\frac{x^{2}}{49} + \\frac{y^{2}}{24} = 1 $. Therefore, $ |PF_{1}| - |PF_{2}| = 2 $, $ |PF_{1}| + |PF_{2}| = 14 $, so $ |PF_{1}| = 8 $, $ |PF_{2}| = 6 $. Hence, $ \\triangle PF_{1}F_{2} $ is a right triangle with area 24."
  },
  {
    "text": "The coordinates of the focus of the parabola $x^{2}=-2 y$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = -2*y)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, -1/2)",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 22]]]",
    "process": "The focus coordinates of the parabola $x^{2}=-2py$ ($p>0$) are $(0,-\\frac{p}{2})$. Since in the parabola $x^{2}=-2y$, $2p=2$, solving gives $p=1$. Therefore, the focus coordinates of the parabola $x^{2}=-2y$ are $(0,-\\frac{1}{2})$."
  },
  {
    "text": "Given a point $P$ on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ such that the distances from $P$ to the two foci $F_{1}$ and $F_{2}$ are $6$ and $2$, respectively, and the point $M (\\frac{3}{2} , 0)$ is equidistant from the lines $PF_{1}$ and $PF_{2}$, then the equation of this hyperbola is?",
    "fact_expressions": "P: Point;F1: Point;G: Hyperbola;b: Number;a: Number;M: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(M) = (3/2, 0);PointOnCurve(P,G);Focus(G)={F1,F2};Distance(P,F1)=6;Distance(P,F2)=2;Distance(M,LineOf(P,F1))=Distance(M,LineOf(P,F2))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/5 = 1",
    "fact_spans": "[[[62, 65]], [[69, 76]], [[2, 59], [149, 152]], [[5, 59]], [[5, 59]], [[98, 120]], [[77, 84]], [[5, 59]], [[5, 59]], [[2, 59]], [[98, 120]], [[2, 65]], [[2, 84]], [[62, 97]], [[62, 97]], [[98, 146]]]",
    "query_spans": "[[[149, 157]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $y^{2}=m x$ coincides with the right focus of the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$, then the value of the real number $m$ is?",
    "fact_expressions": "G: Parabola;m: Real;H: Ellipse;Expression(G) = (y^2 = m*x);Expression(H) = (x^2/6 + y^2/2 = 1);Focus(G) = RightFocus(H)",
    "query_expressions": "m",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 15]], [[64, 69]], [[19, 56]], [[1, 15]], [[19, 56]], [[1, 62]]]",
    "query_spans": "[[[64, 73]]]",
    "process": "From the ellipse equation $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$, we know $a^{2}=6$, $b^{2}=2$, then $c^{2}=a^{2}-b^{2}=4$, so the coordinates of the right focus of the ellipse are $(2,0)$. The focus coordinates of the parabola $y^{2}=mx$ are $(\\frac{m}{4},0)$. Since the focus of the parabola coincides with the right focus of the ellipse, $\\frac{m}{4}=2$, thus $m=8$."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, respectively. If there exists a point $P$ on the ellipse such that $(\\overrightarrow{O P}+\\overrightarrow{O F_{2}}) \\cdot \\overrightarrow{P F_{2}}=0$ ($O$ is the origin), then the area of $\\triangle F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;O: Origin;Expression(G) = (x^2/4 + y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);DotProduct((VectorOf(O,P)+VectorOf(O,F2)),VectorOf(P,F2))=0",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[19, 46], [54, 56]], [[1, 8]], [[61, 64]], [[9, 16]], [[149, 152]], [[19, 46]], [[1, 52]], [[1, 52]], [[54, 64]], [[66, 148]]]",
    "query_spans": "[[[160, 190]]]",
    "process": "Let $ M $ be the midpoint of $ PF_{2} $. According to the dot product of vectors being 0, we obtain the positional relationship between $ OM $ and $ PF_{2} $. Then, combining the midline theorem of triangles and the Pythagorean theorem in right triangles, we solve for the value of $ |PF_{1}|\\cdot|PF_{2}| $, thus the area of $ \\triangle F_{1}PF_{2} $ can be determined. As shown in the figure: Let $ M $ be the midpoint of $ PF_{2} $. Since $ (\\overrightarrow{OP} + \\overrightarrow{OF}_{2}) \\cdot \\overrightarrow{PF_{2}} = 0 $, it follows that $ 2\\overrightarrow{OM} \\cdot \\overrightarrow{PF_{2}} = 0 $, so $ OM \\perp PF_{2} $. Since $ O $ and $ M $ are the midpoints of $ F_{1}F_{2} $ and $ PF_{2} $ respectively, we have $ OM \\parallel PF_{1} $, hence $ PF_{1} \\perp PF_{2} $. Therefore,\n\n$$\n\\begin{cases}\n|PF_{1}|^{2} + |PF_{2}|^{2} = |F_{1}F_{2}|^{2} = 12 \\\\\n|PF_{1}| + |PF_{2}| = 2a = 4\n\\end{cases}\n$$\n\nThus,\n$$\n|PF_{1}|\\cdot|PF_{2}| = \\frac{(|PF_{1}| + |PF_{2}|)^{2} - (|PF_{1}|^{2} + |PF_{2}|^{2})}{2} = 2\n$$\n\nHence,\n$$\nS_{\\Delta F_{1}PF_{2}} = \\frac{|PF_{1}||PF_{2}|}{2} = 1\n$$"
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively. Let point $P$ lie on the left branch of the hyperbola, $M$ be the midpoint of $P F_{1}$, and $O M \\perp P F_{1}$, with $2|P F_{1}|=|P F_{2}|$. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, LeftPart(G));M: Point;MidPoint(LineSegmentOf(P, F1)) = M;O: Origin;IsPerpendicular(LineSegmentOf(O, M), LineSegmentOf(P, F1)) = True;2*Abs(LineSegmentOf(P, F1)) = Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[19, 75], [87, 90], [158, 161]], [[19, 75]], [[22, 75]], [[22, 75]], [[22, 75]], [[22, 75]], [[1, 8]], [[9, 16]], [[1, 81]], [[1, 81]], [[82, 86]], [[82, 95]], [[96, 99]], [[96, 112]], [[114, 133]], [[114, 133]], [[134, 156]]]",
    "query_spans": "[[[158, 167]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ intersect the circle $(x-2)^{2}+y^{2}=1$, then the range of the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 + (x - 2)^2 = 1);IsIntersect(Asymptote(G),H)",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1,2*sqrt(3)/3)",
    "fact_spans": "[[[2, 58], [87, 90]], [[5, 58]], [[5, 58]], [[63, 83]], [[5, 58]], [[5, 58]], [[2, 58]], [[63, 83]], [[2, 85]]]",
    "query_spans": "[[[87, 101]]]",
    "process": "\\because the asymptotes of the hyperbola are bx\\pm ay=0, and they intersect the circle x^{2}+(y-2)^{2}=1, \\therefore the distance from the center of the circle to the asymptote is less than the radius, i.e., \\frac{2b}{\\sqrt{a^{2}+b^{2}}}, \\therefore 3b^{2}<a^{2}, \\therefore c^{2}=a^{2}+b^{2}<\\frac{4}{3}a^{2}, \\therefore e=\\frac{c}{a}<\\frac{2\\sqrt{3}}{3}<e<\\frac{2\\sqrt{3}}{3}"
  },
  {
    "text": "Given that $P$ is a point on the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, $F_{1}$ and $F_{2}$ are the two foci of the hyperbola, and $\\angle F_{1} P F_{2}=60^{\\circ}$, then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[6, 34], [55, 58]], [[6, 34]], [[2, 5]], [[2, 38]], [[39, 46]], [[47, 54]], [[39, 63]], [[65, 98]]]",
    "query_spans": "[[[100, 127]]]",
    "process": "Let point $ P $ be a point on the right branch of the hyperbola $ \\frac{x^{2}}{4} - y^{2} = 1 $, and $ |F_{1}F_{2}| = 2\\sqrt{5} $. By the definition of the hyperbola, we have $ |PF_{1}| - |PF_{2}| = 2a = 4 $. Then in $ \\triangle PF_{1}F_{2} $, by the cosine law:  \n$ |F_{1}F_{2}|^{2} = |PF_{1}|^{2} + |PF_{2}|^{2} - 2|PF_{1}||PF_{2}|\\cos\\angle F_{1}PF_{2} $,  \nthat is,  \n$ |F_{1}F_{2}|^{2} = (|PF_{1}| - |PF_{2}|)^{2} + 2|PF_{1}||PF_{2}| - 2|PF_{1}||PF_{2}|\\cos\\angle F_{1}PF_{2} $,  \nso  \n$ (2\\sqrt{5})^{2} = 4^{2} + 2|PF_{1}||PF_{2}| - 2|PF_{1}||PF_{2}|\\cos 60^{\\circ} $,  \nsolving gives $ |PF_{1}||PF_{2}| = 4 $. Then  \n$ S_{\\Delta F_{1}PF_{2}} = \\frac{1}{2}|PF_{1}||PF_{2}|\\cdot\\sin\\angle F_{1}PF_{2} = \\frac{1}{2} \\times 4 \\times \\frac{\\sqrt{3}}{2} = \\sqrt{3} $."
  },
  {
    "text": "The coordinates of the focus of the parabola $y^{2}=8 x$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(2,0)",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": "First determine the position of the focus, that is, the parabola $ y^{2} = 8x $ is a standard equation with the focus on the positive x-axis, then find $ p = 2 $, yielding the focus coordinates $ (2, 0 )$"
  },
  {
    "text": "The equation of a tangent line to the parabola $y=x^{2}$ is $6 x-y-9=0$, then the coordinates of the point of tangency are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2);Expression(OneOf(TangentOnPoint(P, G))) = (6*x - y - 9 = 0);P: Point",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(3, 9)",
    "fact_spans": "[[[0, 12]], [[0, 12]], [[0, 31]], []]",
    "query_spans": "[[[0, 39]]]",
    "process": "Find the derivative of y from the equation of the curve. Since one tangent line to the curve is given by 6x - y - 9 = 0, set the derivative equal to 6 and solve for x, which gives the x-coordinate of the point of tangency. Substitute the obtained x-value into the equation of the curve to find the corresponding y-coordinate, then write down the coordinates of the point of tangency. [Solution] From y = x^{2}, we get y' = 2x. Since the tangent line equation is 6x - y - 9 = 0, the slope of the tangent line is 6. Thus, y' = 2x = 6, solving gives x = 3. Substituting x = 3 into y = x^{2}, we get y = 9. Therefore, the coordinates of the point of tangency are (3, 9)."
  },
  {
    "text": "The left focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ is $F$, point $P$ is any point on the ellipse, $A(1,2)$, then the minimum value of $|P A|-|P F|$ is?",
    "fact_expressions": "G: Ellipse;A: Point;P: Point;F: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Coordinate(A) = (1, 2);LeftFocus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) - Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "-2",
    "fact_spans": "[[[0, 37], [51, 53]], [[59, 67]], [[46, 50]], [[42, 45]], [[0, 37]], [[59, 67]], [[0, 45]], [[46, 58]]]",
    "query_spans": "[[[69, 88]]]",
    "process": "Using the ellipse equation, find the left focus $ F(-1,0) $ and the right focus $ F_{1}(1,0) $. By the definition of the ellipse equation, obtain $ |PF| = 4 - |PF_{1}| $, thereby getting $ |PA| - |PF| = |PA| + |PF_{1}| - 4 $. To minimize $ |PA| - |PF| $, it suffices to minimize $ |PA| + |PF_{1}| $, and use geometric reasoning to find the minimum value. [Detailed solution] Since $ a^{2} = 4 $, $ b^{2} = 3 $, we have $ c^{2} = a^{2} - b^{2} = 1 $. Thus, $ F(-1,0) $, and the right focus $ F_{1}(1,0) $. As shown in the figure, let point $ P $ be any point on the ellipse, so $ |PF| + |PF_{1}| = 2a = 4 $, hence $ |PF| = 4 - |PF_{1}| $. Then $ |PA| - |PF| = |PA| - (4 - |PF_{1}|) = |PA| + |PF_{1}| - 4 $. To minimize $ |PA| - |PF| $, we only need to minimize $ |PA| + |PF_{1}| $. Clearly, when points $ A $, $ P $, and $ F_{1} $ are collinear, $ |PA| + |PF_{1}| $ is minimized. At this time, $ |PA| + |PF_{1}|_{\\min} = 2 $, thus $ |PA| - |PF|_{\\min} = |PA| + |PF_{1}| - 4 = -2 $."
  },
  {
    "text": "Let one asymptote of the hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a, b>0)$ be $y=-2x$, and one focus coincide with the focus of the parabola $x^{2}=4 \\sqrt{10} y$. Then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;a>0;b>0;Expression(G) = (-x^2/b^2 + y^2/a^2 = 1);Expression(H) = (x^2 = 4*(sqrt(10)*y));Expression(OneOf(Asymptote(G))) = (y = -2*x);OneOf(Focus(G)) = Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/8 - x^2/2 = 1",
    "fact_spans": "[[[1, 55], [109, 112]], [[4, 55]], [[4, 55]], [[77, 101]], [[4, 55]], [[4, 55]], [[1, 55]], [[77, 101]], [[1, 70]], [[1, 106]]]",
    "query_spans": "[[[109, 119]]]",
    "process": "The asymptotes of the hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a,b>0)$ are given by $y=\\pm\\frac{a}{b}x$. From the condition, we have $\\frac{a}{b}=2$. The focus of the parabola $x^{2}=4\\sqrt{10}y$ is $(0,\\sqrt{10})$, so we obtain $a^{2}+b^{2}=c^{2}=10$. Solving gives $a=2\\sqrt{2}$, $b=\\sqrt{2}$. Thus, the equation of the hyperbola is $\\frac{y^{2}}{8}-\\frac{x^{2}}{2}=1$."
  },
  {
    "text": "What is the eccentricity of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/5 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3/2",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 44]]]",
    "process": ""
  },
  {
    "text": "A moving circle is externally tangent to the fixed circle $F$: $(x+2)^{2}+y^{2}=1$ and tangent to the fixed line $l$: $x=1$. What is the equation of the trajectory of the center $M$ of this moving circle?",
    "fact_expressions": "l: Line;F: Circle;E: Circle;Expression(F) = (y^2 + (x + 2)^2 = 1);IsOutTangent(E,F);IsTangent(E, l);M: Point;Center(E) = M;Expression(l)=(x=1)",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "y^2 = -8*x",
    "fact_spans": "[[[36, 49]], [[5, 30]], [[2, 4], [54, 56]], [[5, 30]], [[2, 33]], [[2, 51]], [[59, 62]], [[54, 69]], [36, 48]]",
    "query_spans": "[[[59, 69]]]",
    "process": "First, use the positional relationships between circles and between lines and circles to find the geometric conditions of the moving point M, then determine the trajectory of the moving point M according to the definition of a parabola, and finally write the trajectory equation using the standard equation of a parabola. [Detailed Explanation] From the given conditions, the distance from point M to the fixed point F(-2,0) minus 1 is equal to the distance from M to the fixed line l: x=1; that is, the distance from point M to the fixed point F(-2,0) is equal to the distance from M to the fixed line l: x=2. According to the definition of a parabola, the trajectory of point M is a parabola with focus at F(-2,0) and directrix t: x=2, so the trajectory equation is y^{2}=-8x."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$, a line $l$ intersects the parabola at points $A$ and $B$. If the coordinates of the midpoint of segment $AB$ are $\\left(\\frac{1}{2},-1\\right)$, then what is the equation of line $l$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);l: Line;A: Point;B: Point;Intersection(l, C) = {A, B};Coordinate(MidPoint(LineSegmentOf(A, B))) = (1/2, -1)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "y=-2*x",
    "fact_spans": "[[[2, 21], [29, 32]], [[2, 21]], [[22, 28], [79, 84]], [[33, 36]], [[37, 40]], [[23, 42]], [[45, 76]]]",
    "query_spans": "[[[79, 89]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Since $ A $, $ B $ lie on the parabola, $ y_{1}^{2}=4x_{1} $, $ y_{2}^{2}=4x_{2} $. Subtracting these two equations gives: $ y_{1}^{2}-y_{2}^{2}=4(x_{1}-x_{2}) $, that is, $ 4(x_{1}-x_{2})=(y_{1}-y_{2})(y_{1}+y_{2}) $, so the slope of $ AB $ is $ k=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{4}{y_{1}+y_{2}} $. Since the midpoint of segment $ AB $ has coordinates $ \\left(\\frac{1}{2}, \\frac{y_{1}+y_{2}}{2}\\right) = -1 $, then $ y_{1}+y_{2}=-2 $. Therefore, $ k=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{4}{y_{1}+y_{2}}=-2 $. Thus, the slope of line $ l $ is $ -2 $, and the corresponding equation is $ y+1=-2\\left(x-\\frac{1}{2}\\right) $, that is, $ y=-2x $."
  },
  {
    "text": "Let point $P(x_{1}, y_{1})$ lie on the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, and point $Q(x_{2}, y_{2})$ lie on the line $x+4 y-4=0$. Then the minimum value of $|x_{1}-x_{2}|+|y_{1}-2 y_{2}|$ is?",
    "fact_expressions": "G: Ellipse;H: Line;P: Point;Q: Point;x1:Number;x2:Number;y1:Number;y2:Number;Expression(G) = (x^2/2 + y^2 = 1);Expression(H) = (x + 4*y - 4 = 0);Coordinate(P) = (x1, y1);Coordinate(Q) = (x2, y2);PointOnCurve(P, G);PointOnCurve(Q, H)",
    "query_expressions": "Min(Abs(x1 - x2) + Abs(y1 - 2*y2))",
    "answer_expressions": "2-sqrt(6)/2",
    "fact_spans": "[[[20, 47]], [[68, 81]], [[1, 19]], [[49, 67]], [[2, 19]], [[50, 67]], [[2, 19]], [[50, 67]], [[20, 47]], [[68, 81]], [[1, 19]], [[49, 67]], [[1, 48]], [[49, 82]]]",
    "query_spans": "[[[84, 121]]]",
    "process": "Approach 1: Let $ y = 2y_{2} $, then $ t: x + 2y - 4 = 0 $. From the line $ l: y = -\\frac{1}{2}x + 2 $, it is known that the vertical distance is minimized. Set the equation of the tangent line parallel to line $ l $ as $ y = -\\frac{1}{2}x + m $. Find the value of $ m $ when tangency occurs, and the answer can be obtained.  \nApproach 2: Similar to Approach 1, knowing the vertical distance is minimized, set the line parallel to line $ l $ and tangent to the ellipse at point $ P(x_{0}, y_{0}) $. The tangent line equation is easily known as $ \\frac{x_{0}x}{2} + y_{0}y = 1 $, with slope $ k = -\\frac{x_{0}}{2y_{0}} = -\\frac{1}{2} $, thus $ x_{0} = y_{0} $. Substitute into the ellipse equation to find the coordinates of point $ P $, and thereby obtain the answer.  \n\nSolution:  \nSolution 1: Let $ y = 2y_{2} $, then $ t: x + 2y - 4 = 0 $. From the line $ l: y = -\\frac{1}{2}x + 2 $, it is known that the vertical distance is minimized. Set the equation of the tangent line parallel to line $ l $ as $ y = -\\frac{1}{2}x + m $. Combine with the ellipse $ x^{2} + 2y^{2} - 2 = 0 $, yielding $ 3x^{2} - 4mx + 4m^{2} - 4 = 0 $. From $ \\Delta = 0 $, we get $ m = \\frac{\\sqrt{6}}{2} $. Hence, $ (|x_{1} - x_{2}| + |y_{1} - 2y_{2}|)_{\\min} = 2 - \\frac{\\sqrt{6}}{2} $.  \n\nSolution 2: Similar to Solution 1, knowing the vertical distance is minimized, set the line parallel to line $ l $ and tangent to the ellipse at point $ P(x_{0}, y_{0}) $. The tangent line equation is $ \\frac{x_{0}x}{2} + y_{0}y = 1 $, with slope $ k = -\\frac{x_{0}}{2y_{0}} = -\\frac{1}{2} $, therefore $ x_{0} = y_{0} $. Substituting into the ellipse equation gives $ x_{0} = y_{0} = \\frac{\\sqrt{6}}{3} $, so $ P(\\frac{\\sqrt{6}}{3}, \\frac{\\sqrt{6}}{3}) $, and point $ Q(\\frac{\\sqrt{6}}{3}, 2 - \\frac{\\sqrt{6}}{6}) $. Thus, $ (|x_{1} - x_{2}| + |y_{1} - 2y_{2}|)_{\\min} = 2 - \\frac{\\sqrt{6}}{2} $."
  },
  {
    "text": "Given that the center of the hyperbola is at the origin $O$, the foci lie on the $x$-axis, its conjugate axis has length $2$, and the focal distance is twice the distance between the two directrices,  \nthen the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;O: Origin;Center(G) = O;PointOnCurve(Focus(G), xAxis);Length(ImageinaryAxis(G)) = 2;l1: Line;l2: Line;Directrix(G) = {l1, l2};FocalLength(G) = 2*Distance(l1, l2)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 = 1",
    "fact_spans": "[[[2, 5], [24, 25], [53, 56]], [[9, 14]], [[2, 14]], [[2, 23]], [[24, 33]], [], [], [], [[24, 49]]]",
    "query_spans": "[[[53, 61]]]",
    "process": ""
  },
  {
    "text": "The distance from point $P(2,0)$ to the asymptotes of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Coordinate(P) = (2, 0)",
    "query_expressions": "Distance(P,Asymptote(G))",
    "answer_expressions": "8/5",
    "fact_spans": "[[[10, 49]], [[0, 9]], [[10, 49]], [[0, 9]]]",
    "query_spans": "[[[0, 58]]]",
    "process": "The asymptotes of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ are $y=\\pm\\frac{4}{3}x$, or $4x\\pm3y=0$. Then the distance from the point $(2,0)$ to $4x-3y=0$ is $d=\\frac{8}{\\sqrt{4^{2}+(-3)^{2}}}=\\frac{8}{5}$"
  },
  {
    "text": "If the line $y=m(x-1)$ and the curve $x^{2}-y^{2}=4$ have only one common point, then the real value of $m$ is?",
    "fact_expressions": "G: Line;Expression(G) = (y = m*(x - 1));H: Curve;Expression(H) = (x^2 - y^2 = 4);NumIntersection(G, H) = 1;m: Real",
    "query_expressions": "m",
    "answer_expressions": "{pm*1, pm*2*sqrt(3)/3}",
    "fact_spans": "[[[1, 13]], [[1, 13]], [[14, 31]], [[14, 31]], [[1, 38]], [[40, 45]]]",
    "query_spans": "[[[40, 49]]]",
    "process": "From the hyperbola $x^{2}-y^{2}=4$, we obtain the equations of its asymptotes as $y=\\pm x$.  \n① When the line $y=m(x-1)$ is parallel to the asymptotes $y=\\pm x$ of the hyperbola, that is, when $m=\\pm1$, the line and the hyperbola have exactly one common point, satisfying the condition.  \n② When the line $y=m(x-1)$ is tangent to the right branch, the line and the hyperbola have exactly one intersection point.  \nSolving the system of equations  \n\\[\n\\begin{cases}\ny=m(x-1)\\\\\nx^{2}-y^{2}=4\n\\end{cases}\n\\]\nwe obtain $(m^{2}-1)x^{2}-2m^{2}x+m^{2}+4=0$.  \nLet $\\Delta=4m^{4}-4(m^{2}-1)(m^{2}+4)=0$, then $3m^{2}=4$, solving gives $m=\\pm\\frac{2\\sqrt{3}}{3}$.  \nIn conclusion, the set of real values of $m$ is $\\left\\{\\pm1,\\pm\\frac{2\\sqrt{3}}{3}\\right\\}$."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. Then $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);A: Point;B: Point;Intersection(H, G) = {A, B};O: Origin",
    "query_expressions": "DotProduct(VectorOf(O, A), VectorOf(O, B))",
    "answer_expressions": "-3",
    "fact_spans": "[[[1, 15], [25, 28]], [[1, 15]], [[18, 21]], [[1, 21]], [[22, 24]], [[0, 24]], [[30, 33]], [[35, 38]], [[22, 40]], [[43, 92]]]",
    "query_spans": "[[[43, 94]]]",
    "process": ""
  },
  {
    "text": "From a point $P$ on the directrix $l$ of the parabola $C$: $x^{2}=4 y$, draw tangents $PA$ and $PB$ to $C$, with points of tangency $A$ and $B$ respectively. Let $Q$ be the midpoint of chord $AB$. Then the minimum value of $|P Q|$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 4*y);l: Line;Directrix(C) = l;P: Point;PointOnCurve(P, l);A: Point;B: Point;TangentOfPoint(P,C)={LineOf(P,A),LineOf(P,B)};TangentPoint(LineOf(P,A),C)=A;TangentPoint(LineOf(P,B),C)=B;Q: Point;IsChordOf(LineSegmentOf(A, B), C);MidPoint(LineSegmentOf(A, B)) = Q",
    "query_expressions": "Min(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 20], [33, 36]], [[1, 20]], [[23, 26]], [[1, 26]], [[29, 32]], [[23, 32]], [[58, 61]], [[62, 65]], [[0, 52]], [[0, 65]], [[0, 65]], [[77, 80]], [[33, 73]], [[68, 80]]]",
    "query_spans": "[[[82, 95]]]",
    "process": "x^{2}=4y\\Rightarrow y=\\frac{1}{4}x^{2}\\Rightarrow y'=\\frac{1}{2}x. Let A(x_{1},y_{1}), B(x_{2},y_{2}), P(m,-1), then x_{1}^{2}=4y_{1}, x_{2}^{2}=4y_{2}. Then the tangent line PA: y-y_{1}=\\frac{1}{2}x_{1}(x-x_{1})\\Rightarrow 2y-2y_{1}=x_{1}x-x_{1}^{2}\\Rightarrow 2y-2y_{1}=x_{1}x-4y_{1}\\Rightarrow 2y+2y_{1}=x_{1}x. Since tangent PA passes through P, \\therefore mx_{1}=2y_{1}-2. Similarly, mx_{2}=2y_{2}-2, \\therefore the equation of line AB is: mx=2y-2. From \\begin{cases} mx=2y-2 \\\\ x^{2}=4y \\end{cases}, we obtain y^{2}-(2+m^{2})y+1=0. Then y_{1}+y_{2}=2+m^{2}, x_{1}+x_{2}=\\frac{2y_{1}-2}{m}+\\frac{2y_{2}-2}{m}=\\frac{2(2+m^{2})-4}{m}=\\frac{2m^{2}}{m}=2m. Then Q(m,1+\\frac{m^{2}}{2}), then |PQ|=\\sqrt{(0)^{2}+(\\frac{m^{2}}{2}+2)^{2}}=\\frac{m^{2}}{2}+2\\geqslant 2, so the minimum value of |PQ| is 2. Hence the answer is: 2."
  },
  {
    "text": "The number of intersection points between the line $x-y+2=0$ and the curve $(x-1)(x-2)+(y-3)(y-4)=0$ is?",
    "fact_expressions": "G: Line;H: Curve;Expression(G) = (x - y + 2 = 0);Expression(H) = ((x - 2)*(x - 1) + (y - 4)*(y - 3) = 0)",
    "query_expressions": "NumIntersection(G, H)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 11]], [[12, 39]], [[0, 11]], [[12, 39]]]",
    "query_spans": "[[[0, 46]]]",
    "process": ""
  },
  {
    "text": "From a point $M$ on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, draw two tangents to the circle $x^{2}+y^{2}=2$, with points of tangency $A$ and $B$. The line passing through $A$ and $B$ intersects the $x$-axis and $y$-axis at points $P$ and $Q$, respectively. Then, the minimum area of triangle $POQ$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1);M: Point;PointOnCurve(M, G);H: Circle;Expression(H) = (x^2 + y^2 = 2);L1: Line;L2: Line;TangentOfPoint(M, H) = {L1, L2};A: Point;B: Point;TangentPoint(L1, H) = A;TangentPoint(L2, H) = B;L: Line;PointOnCurve(A, L);PointOnCurve(B, L);P: Point;Q: Point;Intersection(L, xAxis) = P;Intersection(L, yAxis) = Q;O: Origin",
    "query_expressions": "Min(Area(TriangleOf(P, O, Q)))",
    "answer_expressions": "2/3",
    "fact_spans": "[[[1, 38]], [[1, 38]], [[41, 44]], [[1, 44]], [[45, 61]], [[45, 61]], [], [], [[0, 66]], [[70, 73], [79, 82]], [[74, 77], [83, 86]], [[0, 77]], [[0, 77]], [[87, 89]], [[78, 89]], [[78, 89]], [[103, 106]], [[107, 110]], [[87, 110]], [[87, 110]], [[112, 129]]]",
    "query_spans": "[[[112, 137]]]",
    "process": "Let $ M(x_{0},y_{0}) $, the coordinates of point $ A $ be $ (x_{1},y_{1}) $, and the coordinates of point $ B $ be $ (x_{2},y_{2}) $. Since $ |MA|^{2}+|OA|^{2}=|OM|^{2} $, $ |MB|^{2}+|OB|^{2}=|OM|^{2} $, simplifying yields \n\\[\n\\begin{cases}\nx_{1}x_{0}+y_{1}y_{0}=2 \\\\\nx_{2}x_{0}+y_{2}y_{0}=2\n\\end{cases}\n\\]\nso $ x_{1}, x_{2} $ are two solutions of the equation $ x_{0}x+y_{0}y=2 $, hence the equation of line $ AB $ is $ x_{0}x+y_{0}y=2 $. Thus, $ P(-\\frac{2}{x_{0}},0) $, $ Q(0,-\\frac{2}{y_{0}}) $, and $ \\frac{x_{0}^{2}}{9}+\\frac{y_{0}^{2}}{4}=1 $. The area of $ \\triangle POQ $ is $ S=\\frac{|-\\frac{2}{x_{0}}|\\cdot|-\\frac{2}{y_{0}}|}{2}=\\frac{2}{|x_{0}y_{0}|} $, and $ 1=\\frac{x_{0}^{2}}{9}+\\frac{y_{0}^{2}}{4} \\geqslant 2\\sqrt{\\frac{x_{0}^{2}}{9}\\cdot\\frac{y_{0}^{2}}{4}}=\\frac{|x_{0}y_{0}|}{3} $, so $ |x_{0}y_{0}| \\leqslant 3 $, thus $ \\frac{2}{|x_{0}y_{0}|} \\geqslant \\frac{2}{3} $. Equality holds when $ |\\frac{x_{0}}{3}|=|\\frac{y_{0}}{2}| $, i.e., \n\\[\n\\begin{cases}\nx_{0}=\\pm\\frac{3\\sqrt{2}}{2} \\\\\ny_{0}=\\pm\\sqrt{2}\n\\end{cases}\n\\quad \\text{or} \\quad\n\\begin{cases}\nx_{0}=\\pm\\frac{3\\sqrt{2}}{2} \\\\\ny_{0}=\\pm\\sqrt{2}\n\\end{cases}\n\\]\nIn summary, the minimum area of $ \\triangle POQ $ is $ \\frac{2}{3} $."
  },
  {
    "text": "Given that a line passing through the focus $F$ of the parabola $C$: $y^2 = 4x$ intersects the parabola $C$ at points $P$ and $Q$, and intersects the circle $x^2 + y^2 - 2x = 0$ at points $M$ and $N$, where $P$ and $M$ lie in the first quadrant, then the minimum value of $\\frac{1}{|PM|} + \\frac{1}{|QN|}$ is?",
    "fact_expressions": "C: Parabola;G:Circle;H: Line;P: Point;M: Point;Q: Point;N: Point;F: Point;Expression(C) = (y^2 = 4*x);Expression(G) = (-2*x + x^2 + y^2 = 0);Focus(C) = F;PointOnCurve(F, H);Intersection(H, C) = {P, Q};Quadrant(P)=1;Quadrant(M)=1;Intersection(H,G)={M,N}",
    "query_expressions": "Min(1/Abs(LineSegmentOf(Q, N)) + 1/Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "2",
    "fact_spans": "[[[3, 22], [31, 37]], [[49, 69]], [[28, 30]], [[38, 41], [82, 85]], [[86, 89], [86, 89]], [[42, 45]], [[74, 77]], [[24, 27]], [[3, 22]], [[49, 69]], [[3, 27]], [[2, 30]], [[28, 47]], [[82, 95]], [[86, 95]], [[28, 79]]]",
    "query_spans": "[[[97, 136]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and $M$ is a point on the ellipse such that $M F_{1} \\perp x$-axis and $\\angle F_{1} M F_{2}=45^{\\circ}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;M: Point;F1: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Focus(G) = {F1, F2};PointOnCurve(M,G);IsPerpendicular(LineSegmentOf(M,F1),xAxis);AngleOf(F1,M,F2)=ApplyUnit(45,degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)-1",
    "fact_spans": "[[[18, 72], [80, 82], [80, 82]], [[20, 72]], [[20, 72]], [[76, 79]], [[2, 9]], [[10, 17]], [[20, 72]], [[20, 72]], [[18, 72]], [[2, 75]], [[76, 85]], [[85, 103]], [[104, 137]]]",
    "query_spans": "[[[139, 147]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2x$ with focus $F$, the line $l$ passing through $F$ and perpendicular to the $x$-axis intersects $C$ at points $A$ and $B$. What is the length of the chord intercepted on the $y$-axis by the circle with diameter $AB$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*x);F: Point;Focus(C) = F;l: Line;PointOnCurve(F,l);IsPerpendicular(l,xAxis);B: Point;A: Point;Intersection(l, C) = {A, B};G: Circle;IsDiameter(LineSegmentOf(A,B),G)",
    "query_expressions": "Length(InterceptChord(yAxis,G))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 21], [48, 51]], [[2, 21]], [[25, 28], [30, 33]], [[2, 28]], [[42, 47]], [[29, 47]], [[34, 47]], [[57, 60]], [[53, 56]], [[42, 62]], [[76, 77]], [[64, 77]]]",
    "query_spans": "[[[76, 90]]]",
    "process": "For the parabola $ C: y^{2} = 2x $, its focus is $ F\\left(\\frac{1}{2}, 0\\right) $. Letting $ x = \\frac{1}{2} $, we obtain $ y = \\pm 1 $, hence $ |AB| = 2 $, so the required circle's radius $ r = 1 $. The distance from the center $ F $ to the $ y $-axis is $ \\frac{1}{2} $, thus the chord length intercepted by the $ y $-axis on the circle with diameter $ AB $ is $ 2\\sqrt{r^{2} - d^{2}} = 2\\sqrt{1 - \\frac{1}{4}} = \\sqrt{3} $."
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/9 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(3/2)*x",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 47]]]",
    "process": ""
  },
  {
    "text": "Given that the equation $x^{2}+k y^{2}=2$ represents an ellipse with foci on the $y$-axis, what is the range of real values for $k$?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (k*y^2 + x^2 = 2);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(0,1)",
    "fact_spans": "[[[31, 33]], [[35, 40]], [[1, 33]], [[22, 33]]]",
    "query_spans": "[[[35, 47]]]",
    "process": "The solution process is omitted"
  },
  {
    "text": "The equation of the parabola with the right vertex of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ as its vertex and the left focus as its focus is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (x^2/16 - y^2/9 = 1);Vertex(H) = RightVertex(G);Focus(H) = LeftFocus(G)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2 = -36*(x - 4)",
    "fact_spans": "[[[1, 40]], [[54, 57]], [[1, 40]], [[0, 57]], [[0, 57]]]",
    "query_spans": "[[[54, 62]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[0, 27]], [[0, 27]]]",
    "query_spans": "[[[0, 33]]]",
    "process": ""
  },
  {
    "text": "Given that points $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $|F_{1} F_{2}|=4$, point $Q(2, \\sqrt{2})$ lies on ellipse $C$, and $P$ is a moving point on ellipse $C$, then the maximum value of $\\overrightarrow{P Q} \\cdot \\overrightarrow{P F_{1}}$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;Q: Point;F1: Point;F2: Point;P: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(Q) = (2, sqrt(2));LeftFocus(C) = F1;RightFocus(C) = F2;Abs(LineSegmentOf(F1, F2)) = 4;PointOnCurve(Q, C);PointOnCurve(P, C)",
    "query_expressions": "Max(DotProduct(VectorOf(P, Q), VectorOf(P, F1)))",
    "answer_expressions": "9/2",
    "fact_spans": "[[[19, 76], [119, 124], [130, 135]], [[26, 76]], [[26, 76]], [[101, 118]], [[2, 10]], [[11, 18]], [[126, 129]], [[26, 76]], [[26, 76]], [[19, 76]], [[101, 118]], [[2, 82]], [[2, 82]], [[83, 100]], [[101, 125]], [[126, 139]]]",
    "query_spans": "[[[141, 200]]]",
    "process": "From the focal length and point Q on the ellipse, the equation of the ellipse can be determined, yielding the coordinates of F_{1}. Let P(x, y) and combine with the coordinate representation of the dot product of vectors. According to the given conditions: c=2, \\begin{cases}2\\\\\\frac{\\sqrt{2}}{2}+\\frac{2}{b^{2}}=1\\\\a=b^{2}+c\\end{cases}, solving gives a^{2}=8, b^{2}=4. Find the maximum value of \\overrightarrow{PQ}\\cdot\\overrightarrow{PF}_{1}. Therefore, the equation of the ellipse is \\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1, and thus F_{1}(-2,0). Let P(x,y), then we have: x^{2}=8-2y^{2}, so \\overrightarrow{PQ}\\cdot\\overrightarrow{PF}_{1}=(2-x,\\sqrt{2}-y)(-2-x,-y)=x^{2}-4+y^{2}-\\sqrt{2}y=-(y+\\frac{\\sqrt{2}}{2})^{2}+\\frac{9}{2}. The maximum value of \\overrightarrow{PQ}\\cdot\\overrightarrow{PF}_{1} is \\frac{9}{2}, achieved if and only if y=-\\frac{\\sqrt{2}}{2}\\in[-2,2]."
  },
  {
    "text": "Let the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ have left and right foci $F_{1}$, $F_{2}$ respectively, and let $P$ be a point on $C$ such that $P F_{2} \\perp F_{1} F_{2}$, $\\angle P F_{1} F_{2}=30^{\\circ}$. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P, F2), LineSegmentOf(F1, F2));AngleOf(P, F1, F2) = ApplyUnit(30, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[1, 57], [86, 89], [158, 161]], [[1, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[66, 73]], [[74, 81]], [[1, 81]], [[1, 81]], [[82, 85]], [[82, 92]], [[93, 120]], [[123, 156]]]",
    "query_spans": "[[[158, 167]]]",
    "process": "Let $ PF_{2} = m $. Based on the property that in a right triangle the side opposite the $ 30^{\\circ} $ angle is half the hypotenuse, and using the Pythagorean theorem, derive $ PF_{1} $, $ PF_{2} $, and $ F_{1}F_{2} $. Then solve using the definition of an ellipse and the eccentricity formula. In right triangle $ \\triangle PF_{2}F_{1} $, let $ PF_{2} = m $. Since $ \\angle PF_{1}F_{2} = 30^{\\circ} $, it follows that $ PF_{1} = 2m $, $ F_{1}F_{2} = \\sqrt{(2m)^{2} - m^{2}} = \\sqrt{3}m $. Hence, $ e = \\frac{2c}{2a} = \\frac{F_{1}F_{2}}{PF_{1} + PF_{2}} = \\frac{\\sqrt{3}}{3} $."
  },
  {
    "text": "Ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$ has left and right foci $F_{1}$, $F_{2}$. A line perpendicular to the $x$-axis passing through $F_{2}$ intersects $C$ at points $A$, $B$. Line $F_{1} A$ intersects the $y$-axis at point $D$. If $B D \\perp F_{1} A$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;L: Line;PointOnCurve(F2, L) = True;IsPerpendicular(L, xAxis) = True;Intersection(L, C) = {A, B};A: Point;B: Point;Intersection(LineSegmentOf(F1, A), yAxis) = D;D: Point;IsPerpendicular(LineSegmentOf(B, D), LineSegmentOf(F1, A)) = True",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[0, 59], [98, 101], [158, 163]], [[0, 59]], [[7, 59]], [[7, 59]], [[7, 59]], [[7, 59]], [[65, 72]], [[73, 80], [82, 89]], [[0, 80]], [[0, 80]], [], [[81, 97]], [[81, 97]], [[81, 112]], [[103, 106]], [[107, 110]], [[113, 134]], [[130, 134]], [[136, 156]]]",
    "query_spans": "[[[158, 169]]]",
    "process": ""
  },
  {
    "text": "The equation of the hyperbola that shares the same asymptotes as $x^{2}-\\frac{y^{2}}{4}=1$ and passes through the point $M(2,2)$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/4 = 1);C: Hyperbola;Asymptote(C) = Asymptote(G);M: Point;Coordinate(M) = (2, 2);PointOnCurve(M, C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/3 - y^2/12 = 1",
    "fact_spans": "[[[1, 29]], [[1, 29]], [[48, 51]], [[0, 51]], [[38, 47]], [[38, 47]], [[37, 51]]]",
    "query_spans": "[[[48, 55]]]",
    "process": "Let the hyperbola equation be $x^{2}-\\frac{y^{2}}{4}=k$, $k\\neq0$. Since the hyperbola passes through the point $(2,2)$, we have $2^{2}-\\frac{2^{2}}{4}=k$, so $k=3$. Therefore, the required hyperbola equation is $x^{2}-\\frac{y^{2}}{4}=3$, or $\\frac{x^{2}}{3}-\\frac{y^{2}}{12}=1$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, $(a>0, b>0)$, with left and right foci $F_{1}$, $F_{2}$ respectively. If $C$ intersects the line $y=x$, and there exists a point $P$ on the hyperbola, not a vertex, such that $\\angle P F_{2} F_{1}=3 \\angle P F_{1} F_{2}$, then the range of eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;G: Line;Expression(G) = (y = x);IsIntersect(C, G);P: Point;PointOnCurve(P, C);Negation(P=Vertex(C));AngleOf(P, F2, F1) = 3*AngleOf(P, F1, F2)",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(\\sqrt{2}, 2)",
    "fact_spans": "[[[2, 64], [89, 92], [105, 108], [170, 173]], [[2, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[72, 79]], [[80, 87]], [[2, 87]], [[2, 87]], [[93, 100]], [[93, 100]], [[89, 103]], [[116, 119]], [[105, 119]], [[105, 119]], [[123, 168]]]",
    "query_spans": "[[[170, 184]]]",
    "process": ""
  },
  {
    "text": "Given that the line $y = kx - 4$ and the parabola $y^2 = 8x$ have exactly one common point, what is the set of real values of $k$ satisfying this condition?",
    "fact_expressions": "G: Parabola;H: Line;k: Real;Expression(G) = (y^2 = 8*x);Expression(H) = (y = k*x - 4);NumIntersection(H, G) = 1",
    "query_expressions": "Range(k)",
    "answer_expressions": "{0, -1/2}",
    "fact_spans": "[[[14, 28]], [[2, 13]], [[44, 49]], [[14, 28]], [[2, 13]], [[2, 37]]]",
    "query_spans": "[[[44, 56]]]",
    "process": "Solve the system \\begin{cases}y=kx-4\\\\y^{2}=8x\\end{cases}, eliminate $x$ to obtain $ky^{2}-8y-32=0$. When $k=0$, $-8y-32=0$, solving gives $y=-4$, in which case the line $y=kx-4$ and the parabola $y^{2}=8x$ have exactly one common point, satisfying the condition. When $k\\neq0$, then $\\triangle=64+128k=0$, solving gives $k=-\\frac{1}{2}$. In summary, $k=0$ or $-\\frac{1}{2}$, so the set of real values of $k$ satisfying the condition is $\\left\\{0,-\\frac{1}{2}\\right\\}$. Hence, the answer: $0,-\\frac{1}{2}$"
  },
  {
    "text": "The distance from the vertex of the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{4}=1$ to its asymptote is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/2 - y^2/4 = 1)",
    "query_expressions": "Distance(Vertex(G), Asymptote(G))",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[0, 38], [42, 43]], [[0, 38]]]",
    "query_spans": "[[[0, 52]]]",
    "process": "According to the problem, the vertex coordinates are $(\\pm\\sqrt{2},0)$, and the asymptotes are given by $y=\\pm\\sqrt{2}x$; then the distance between the vertex coordinates $(\\pm\\sqrt{2},0)$ is $d=\\frac{2\\sqrt{3}}{3}$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the line $l$ is a tangent to the circle $O$: $x^{2}+y^{2}=b^{2}$. If the inclination angle of line $l$ is $\\frac{\\pi}{3}$ and it passes exactly through the right vertex of the ellipse, then the eccentricity of the ellipse is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;l: Line;O: Circle;Expression(O) = (x^2 + y^2 = b^2);IsTangent(l, O);Inclination(l) = pi/3;PointOnCurve(RightVertex(C), l)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 59], [129, 131], [137, 139]], [[2, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[60, 65], [98, 103]], [[66, 91]], [[66, 91]], [[60, 96]], [[98, 123]], [[98, 135]]]",
    "query_spans": "[[[137, 144]]]",
    "process": "The right vertex of the ellipse is (a,0), so the equation of the line is y-0=\\sqrt{3}(x-a) \\therefore \\sqrt{3}x-y-\\sqrt{3}a=0. Since the line is tangent to the circle, we have \\frac{|-\\sqrt{3}a|}{2}=b \\therefore 3a^{2}=4b^{2}=4a^{2}-4c^{2} \\therefore e=\\frac{c}{a}=\\frac{1}{2}"
  },
  {
    "text": "Point $P$ is a point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, and $F_{1}$, $F_{2}$ are the left and right foci of the ellipse, respectively. If $\\angle F_{1} P F_{2}=60^{\\circ}$, then $|P F_{1}| \\cdot |P F_{2}|$=?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/9 + y^2/4 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[5, 42], [65, 67]], [[47, 54]], [[0, 4]], [[55, 62]], [[5, 42]], [[0, 46]], [[47, 72]], [[47, 72]], [[74, 107]]]",
    "query_spans": "[[[109, 138]]]",
    "process": ""
  },
  {
    "text": "The parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$ and directrix $l$. From a point $A$ on the parabola $C$, draw a perpendicular to $l$, with foot $B$. Let point $P(\\frac{9 p}{2}, 0)$. The lines $AF$ and $BP$ intersect at point $E$. If $|P F|=2|A F|$ and the area of $\\Delta P A E$ is $6 \\sqrt{3}$, then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;B: Point;P: Point;A: Point;E: Point;F: Point;l: Line;l1:Line;p>0;Expression(C) = (y^2 = 2*p*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(A,C);PointOnCurve(A,l1);IsPerpendicular(l,l1);FootPoint(l,l1)=B;Coordinate(P)=(9*p/2,0);Intersection(LineSegmentOf(A,F),LineSegmentOf(B,P))=E;Abs(LineSegmentOf(P, F)) = 2*Abs(LineSegmentOf(A, F));Area(TriangleOf(P, A, E)) = 6*sqrt(3)",
    "query_expressions": "p",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 25], [41, 47]], [[160, 163]], [[64, 67]], [[69, 91]], [[50, 53]], [[106, 110]], [[29, 32]], [[36, 39], [54, 57]], [], [[7, 25]], [[0, 25]], [[0, 32]], [[0, 39]], [[41, 53]], [[40, 60]], [[40, 60]], [[40, 67]], [[69, 91]], [[92, 110]], [[112, 126]], [[128, 158]]]",
    "query_spans": "[[[160, 165]]]",
    "process": "From the given conditions, as shown in the figure, according to the definition of a parabola, we have |AF| = |AB|. Since |PF| = 2|AF|, then |PF| = 2|AF| = 2|AB| = 4p, solving gives |AB| = 2p. Let A(x_{0}, y_{0}), then |AB| = 2p, x_{0} + \\frac{p}{2} = 2p, solving gives x_{0} = \\frac{3p}{2}. Since A(x_{0}, y_{0}) lies on the parabola y^{2} = 2px, we have y_{0}^{2} = 2px_{0}, solving gives y_{0} = \\sqrt{3}p. Because AB // x-axis and |PF| = 2|AB|, then \\frac{|PF|}{|AB|} = \\frac{|EF|}{|AE|} = 2, so |EF| = 2|AE|. Since the area of \\triangle PAE is 6\\sqrt{3}, we have \\frac{1}{3}S_{\\triangle APF} = S_{\\triangle APE}, that is, \\frac{1}{3} \\times \\frac{1}{2} \\times 4p \\times \\sqrt{3}p = 6\\sqrt{3}, solving gives p = 3. Therefore, the value of p is 3."
  },
  {
    "text": "Given that $M$ is a point on the parabola $x^{2}=4 y$, $F$ is its focus, and point $A$ lies on the circle $C$: $(x+1)^{2}+(y-5)^{2}=1$, then the minimum value of $|M A|+|M F|$ is?",
    "fact_expressions": "G: Parabola;C: Circle;M: Point;A: Point;F: Point;Expression(G) = (x^2 = 4*y);Expression(C) = ((x + 1)^2 + (y - 5)^2 = 1);PointOnCurve(M, G);PointOnCurve(A, C);Focus(G)=F",
    "query_expressions": "Min(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F)))",
    "answer_expressions": "5",
    "fact_spans": "[[[6, 20], [28, 29]], [[37, 65]], [[2, 5]], [[32, 36]], [[24, 27]], [[6, 20]], [[37, 65]], [[2, 23]], [[32, 66]], [[24, 31]]]",
    "query_spans": "[[[69, 87]]]",
    "process": "The directrix of the parabola is $ l: y = -1 $, $ |MA| + |MF| = |MA| + d_{A-l} \\geqslant d_{C-l} - r = 5 + 1 - 1 = 5 $"
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=8x$ is $F$, the intersection point of the directrix and the $x$-axis is $M$, and $N$ is a point on the parabola satisfying $2|NF|=|MN|$, then the distance from point $F$ to the line $MN$ is?",
    "fact_expressions": "G: Parabola;M: Point;N: Point;F: Point;Expression(G) = (y^2 = 8*x);Focus(G) = F;Intersection(Directrix(G),xAxis)=M;PointOnCurve(N,G);2*Abs(LineSegmentOf(N, F)) = Abs(LineSegmentOf(M, N))",
    "query_expressions": "Distance(F, LineOf(M,N))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[2, 16], [43, 46]], [[35, 38]], [[39, 42]], [[20, 23], [70, 74]], [[2, 16]], [[2, 23]], [[2, 38]], [[39, 50]], [[54, 68]]]",
    "query_spans": "[[[70, 87]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and $F_{1}$, $F_{2}$ are the two foci of this ellipse. If the inradius of $\\Delta F_{1} P F_{2}$ is $\\frac{1}{2}$, then the value of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 + y^2/3 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2};Radius(InscribedCircle(TriangleOf(F1, P, F2))) = 1/2",
    "query_expressions": "DotProduct(VectorOf(P, F1), VectorOf(P, F2))",
    "answer_expressions": "9/4",
    "fact_spans": "[[[6, 43], [65, 67]], [[48, 55]], [[2, 5]], [[56, 63]], [[6, 43]], [[2, 47]], [[48, 72]], [[74, 116]]]",
    "query_spans": "[[[118, 179]]]",
    "process": "Let $ P $ be a point in the first quadrant on the ellipse $ \\frac{x^2}{4} + \\frac{y^{2}}{3} = 1 $, $ S_{\\Delta PF_{1}F_{2}} = \\frac{1}{2}(|PF_{1}| + |PF_{2}| + |F_{1}F_{2}|) \\cdot \\frac{1}{2} = \\frac{2}{3} = \\frac{1}{2}y_{P} = y_{P} $. So $ y_{P} = \\frac{3}{2} $. $ \\overrightarrow{PF}_{1} \\cdot \\overrightarrow{PF_{2}} = (-1 - x_{p}, -y_{p}) \\cdot (1 - x_{P}, -y_{P}) = x_{p}^2 - 1 + y_{p}^2 $, then substitute using the ellipse equation, and finally obtain the result according to the boundedness of the ellipse. [Solution] For the ellipse $ \\frac{x^2}{4} + \\frac{y^{2}}{3} = 1 $, $ a = 2 $, $ b = \\sqrt{3} $, $ c = 1 $. According to the definition of ellipse, $ |PF_{1}| + |PF_{2}| = 4 $, $ |F_{1}F_{2}| = 2 $. Assume $ P $ is a point in the first quadrant on the ellipse $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $, $ S_{\\Delta PF_{1}F_{2}} = \\frac{1}{2}(|PF_{1}| + |PF_{2}| + |F_{1}F_{2}|) \\cdot \\frac{1}{2} = \\frac{2}{3} = \\frac{1}{2}y_{P} = y_{P} $. So $ y_{P} = \\frac{3}{2} $, then $ \\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}} = (-1 - x_p, -y_{p}) \\cdot (1 - x_{P}, -y_{P}) = x_{p}^2 - 1 + y_{p}^{2} = 4(1 - \\frac{y_{0}^{2}}{3}) - 1 + y_{p}^{2} = 3 - \\frac{y_{P}^{2}}{2} = \\frac{9}{4} $."
  },
  {
    "text": "The line $l$ with slope $\\frac{\\sqrt{3}}{3}$ passes through the focus of the parabola $y^{2}=2 px(p>0)$. If line $l$ is tangent to the circle $(x-2)^{2}+y^{2}=4$, then $p$=?",
    "fact_expressions": "l: Line;G: Parabola;p: Number;H: Circle;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (y^2 + (x - 2)^2 = 4);Slope(l) = sqrt(3)/3;PointOnCurve(Focus(G), l);IsTangent(l, H)",
    "query_expressions": "p",
    "answer_expressions": "12",
    "fact_spans": "[[[24, 29], [55, 60]], [[30, 50]], [[85, 88]], [[61, 81]], [[33, 50]], [[30, 50]], [[61, 81]], [[0, 29]], [[24, 53]], [[55, 83]]]",
    "query_spans": "[[[85, 90]]]",
    "process": "The line $ l $ with slope $ \\frac{\\sqrt{3}}{3} $ passes through the focus $ F\\left(\\frac{p}{2}, 0\\right) $ of the parabola $ C: y^{2} = 2px $ ($ p > 0 $). The equation of line $ l $ is $ y = \\frac{\\sqrt{3}}{3}\\left(x - \\frac{p}{2}\\right) $, or $ x - \\sqrt{3}y - \\frac{p}{2} = 0 $. Since line $ l $ is tangent to the circle $ M: (x - 2)^{2} + y^{2} = 4 $, whose center is $ (2, 0) $ and radius is $ 2 $, we have $ \\frac{\\left|2 - \\frac{p}{2}\\right|}{\\sqrt{3 + 1}} = 2 $. Solving this gives $ p = 12 $ or $ p = -4 $ (discarded)."
  },
  {
    "text": "A point $Q(1, m)$ on the parabola $y^{2}=2 p x$ is at a distance of $5$ from the focus of the parabola. Then the real number $m$=?",
    "fact_expressions": "G: Parabola;p: Number;Q: Point;m: Real;Expression(G) = (y^2 = 2*p*x);Coordinate(Q) = (1, m);PointOnCurve(Q, G);Distance(Q,Focus(G))=5",
    "query_expressions": "m",
    "answer_expressions": "pm*4",
    "fact_spans": "[[[0, 16], [29, 32]], [[3, 16]], [[19, 28]], [[43, 48]], [[0, 16]], [[19, 28]], [[0, 28]], [[19, 41]]]",
    "query_spans": "[[[43, 50]]]",
    "process": "From the given conditions, it is known that the focus of the parabola lies on the x-axis and p>0. Since the point Q(1,m) on the parabola y^{2}=2px is at a distance of 5 from the focus of the parabola, according to the focal radius formula, we obtain 1+\\frac{p}{2}=5, so p=8, i.e., y^{2}=16x. Since the point Q(1,m) lies on the parabola, m^{2}=16, hence m=\\pm4."
  },
  {
    "text": "Given that the line $l$ passes through the focus of the parabola $C$: $y^{2}=4x$, and intersects $C$ at points $A$ and $B$. The tangents to $C$ at points $A$ and $B$ are drawn respectively, intersecting at point $P$. Then, the trajectory equation of point $P$ is?",
    "fact_expressions": "l: Line;C: Parabola;Expression(C) = (y^2 = 4*x);PointOnCurve(Focus(C), l);A: Point;B: Point;Intersection(l, C) = {A, B};Z1: Line;Z2: Line;TangentOfPoint(A, C) = Z1;TangentOfPoint(B, C) = Z2;P: Point;Intersection(Z1, Z2) = P",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x=-1",
    "fact_spans": "[[[2, 7], [31, 34]], [[8, 27], [35, 38], [64, 67]], [[8, 27]], [[2, 30]], [[40, 43], [53, 57]], [[46, 49], [58, 61]], [[31, 51]], [], [], [[52, 70]], [[52, 70]], [[74, 78], [80, 84]], [[52, 78]]]",
    "query_spans": "[[[80, 91]]]",
    "process": "Without loss of generality, flip the parabola to $x^{2}=4y$, and let the equation of the flipped line $l$ be $y=kx+1$. Let the coordinates of the flipped points $A$ and $B$ be $(x_{1},y_{1})$, $(x_{2},y_{2})$, respectively. Then, solving the system\n\\[\n\\begin{cases}\nx^{2}=4y, \\\\\ny=kx+1\n\\end{cases}\n\\]\nyields $x^{2}-4kx-4=0$ $\\textcircled{1}$. It is easy to obtain the tangent to the parabola $x^{2}=4y$ at point $A$. Solving the system\n\\[\n\\begin{cases}\ny-\\frac{1}{4}x^{2}=\\frac{1}{2}x_{1}(x-x_{1}), \\\\\ny-\\frac{1}{2}x(x-x),\n\\end{cases}\n\\]\nsimilarly, the tangent to the parabola $x^{2}=4y$ at point $B$ is given by $y-\\frac{1}{4}x_{2}^{2}=\\frac{1}{2}x_{1}(x-x_{1}x_{1}x_{2})$. Furthermore, from $\\textcircled{1}$ we get $x_{1}x_{2}=-4$, so $y=-1$. Hence, the trajectory equation of the corresponding point $P$ on the original parabola $C$ is $x=$"
  },
  {
    "text": "The equation of the directrix of a parabola is $y=1$, then its standard equation is?",
    "fact_expressions": "G: Parabola;Expression(Directrix(G)) = (y = 1)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 = -4*y",
    "fact_spans": "[[[0, 3], [16, 17]], [[0, 14]]]",
    "query_spans": "[[[16, 23]]]",
    "process": "From the equation of the directrix, the direction of the parabola's opening and the value of p can be determined, then the standard equation of the parabola can be established. Since the directrix of the parabola is y=1, the parabola opens downward and the focus is at coordinates (0,-1); thus, the standard equation of the parabola is x^{2}=-4y."
  },
  {
    "text": "Given the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has its left focus at point $F$. A perpendicular is drawn from point $F$ to one of the asymptotes of $E$, with the foot of the perpendicular at point $H$. If $|F H|=2$, and the area of $\\triangle F O H$ is $3$ (where $O$ is the coordinate origin), then the standard equation of $E$ is?",
    "fact_expressions": "E: Hyperbola;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;LeftFocus(E) = F;L: Line;H: Point;PointOnCurve(F, L);IsPerpendicular(OneOf(Asymptote(E)), L);FootPoint(OneOf(Asymptote(E)), L) = H;O: Origin;Abs(LineSegmentOf(F, H)) = 2;Area(TriangleOf(F, O, H)) = 3",
    "query_expressions": "Expression(E)",
    "answer_expressions": "x^2/9 - y^2/4 = 1",
    "fact_spans": "[[[2, 63], [78, 81], [149, 152]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 71], [73, 77]], [[2, 71]], [], [[94, 98]], [[72, 90]], [[72, 90]], [[72, 98]], [[138, 141]], [[100, 109]], [[111, 135]]]",
    "query_spans": "[[[149, 159]]]",
    "process": "From the hyperbola equation, we obtain F(-c,0), and the asymptotes are given by y=\\pm\\frac{b}{a}x. Using the point-to-line distance formula, we get |FH|=b=2. Then, using the area of \\triangle FOH being 3, we can find |OH|=3. Next, by the Pythagorean theorem, we can determine the value of c. Then, using a^{2}=c^{2}-b^{2}, we can find the value of a, thus solving the problem. From the hyperbola equation, we have F(-c,0), and the asymptotes are y=\\pm\\frac{b}{a}x, i.e., bx\\pm ay=0. Therefore, the distance from point F(-c,0) to bx\\pm ay=0 is |FH|=\\frac{|-bc|}{\\sqrt{a^{2}+b^{2}}}=b=2. S_{\\Delta FOH}=\\frac{1}{2}\\times|FH|\\times|OH|=\\frac{1}{2}\\times2\\times|OH|=3, so |OH|=3, hence |OM|=3. Thus, a^{2}=c^{2}-b^{2}=13-4=9, so the equation of hyperbola E is: \\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1"
  },
  {
    "text": "If a moving circle $M$ is externally tangent to circle $C_{1}$: $(x+4)^{2}+y^{2}=2$, and internally tangent to circle $C_{2}$: $(x-4)^{2}+y^{2}=2$, then what is the trajectory equation of the center $M1$ of the moving circle?",
    "fact_expressions": "M: Circle;C1: Circle;C2: Circle;M1: Point;Expression(C1) = ((x + 4)^2 + y^2 = 2);Expression(C2) = ((x - 4)^2 + y^2 = 2);IsOutTangent(M, C1);IsInTangent(M, C2);Center(M) = M1",
    "query_expressions": "LocusEquation(M1)",
    "answer_expressions": "And((x^2/2 - y^2/14 = 1), (x>=sqrt(2)))",
    "fact_spans": "[[[1, 6], [72, 74]], [[7, 35]], [[40, 68]], [[76, 80]], [[7, 35]], [[40, 68]], [[1, 37]], [[1, 70]], [[72, 80]]]",
    "query_spans": "[[[76, 86]]]",
    "process": ""
  },
  {
    "text": "The parabola $x^{2}=ay$ passes through the point $A(1, \\frac{1}{4})$, then the distance from point $A$ to the focus of this parabola is?",
    "fact_expressions": "G: Parabola;a: Number;A: Point;Expression(G) = (x^2 = a*y);Coordinate(A) = (1, 1/4);PointOnCurve(A, G)",
    "query_expressions": "Distance(A, Focus(G))",
    "answer_expressions": "5/4",
    "fact_spans": "[[[0, 13], [42, 45]], [[3, 13]], [[14, 34], [36, 40]], [[0, 13]], [[14, 34]], [[0, 34]]]",
    "query_spans": "[[[36, 53]]]",
    "process": ""
  },
  {
    "text": "Given that a focus of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{3 m}=1$ is $(0 , 2)$, and the focal length of the ellipse $\\frac{y^{2}}{n}-\\frac{x^{2}}{m}=1$ is equal to $4$, then $n$=?",
    "fact_expressions": "G: Hyperbola;m: Number;H: Ellipse;n: Number;I: Point;Expression(G) = (-y^2/(3*m)+ x^2/m = 1);Expression(H) = (y^2/n - x^2/m = 1);Coordinate(I) = (0, 2);OneOf(Focus(G))= I;FocalLength(H) = 4",
    "query_expressions": "n",
    "answer_expressions": "5",
    "fact_spans": "[[[2, 42]], [[60, 95]], [[58, 95]], [[105, 108]], [[48, 57]], [[2, 42]], [[58, 95]], [[48, 57]], [[2, 57]], [[58, 103]]]",
    "query_spans": "[[[105, 110]]]",
    "process": "Given that one focus of the hyperbola is (0,2), we can find m = -1. Based on the ellipse equation with focal length 4, we can compute to obtain the result. Since the focus of the hyperbola is (0,2), the focus lies on the y-axis. Thus, the equation of the hyperbola is \\frac{y^{2}}{-3m}-\\frac{x^{2}}{m}=1, so a^{2}=-3m, b^{2}=-m, and therefore c^{2}=-3m-m=-4m=4. Solving gives m = -1. The equation of the ellipse is \\frac{y^{2}}{n}+x^{2}=1, with n>0. The focal length of the ellipse is 4, so c^{2}=n-1=4 or 1-n=4, solving gives n=5 or -3 (discarded). The answer is: 5"
  },
  {
    "text": "Given that point $P$ is a moving point on the ellipse $C$: $\\frac{x^{2}}{3}+y^{2}=1$, then the maximum distance from point $P$ to the line $x+y-3=0$ is?",
    "fact_expressions": "C: Ellipse;G: Line;P: Point;Expression(C) = (x^2/3 + y^2 = 1);Expression(G) = (x + y - 3 = 0);PointOnCurve(P, C)",
    "query_expressions": "Max(Distance(P, G))",
    "answer_expressions": "5*sqrt(2)/2",
    "fact_spans": "[[[7, 39]], [[49, 60]], [[2, 6], [44, 48]], [[7, 39]], [[49, 60]], [[2, 42]]]",
    "query_spans": "[[[44, 68]]]",
    "process": "Let the line $ l: y = -x + m $ parallel to $ x + y - 3 = 0 $ be tangent to $ C: \\frac{x^{2}}{3} + y^{2} = 1 $. Solve for the equation of $ l $ at this time, then the maximum distance from point $ P $ to the line $ x + y - 3 = 0 $ can be found using the distance formula between parallel lines. Let the line $ l: y = -x + m $ ($ m \\neq 3 $) parallel to $ x + y - 3 = 0 $. When $ t $ is tangent to the ellipse $ C $, we have:\n\\[\n\\begin{cases}\ny = -x + m \\\\\nx^{2} + 3y^{2} = 3\n\\end{cases}\n\\]\nThus, $ 4x^{2} - 6mx + 3m^{2} - 3 = 0 $, so $ \\Delta = 36m^{2} - 16(3m^{2} - 3) = 0 $, hence $ m = \\pm 2 $. Therefore, $ l: x + y - 2 = 0 $ or $ l: x + y + 2 = 0 $. Take $ l: x + y + 2 = 0 $. At this time, the distance between $ l: x + y + 2 = 0 $ and $ x + y - 3 = 0 $ is $ d = \\frac{|2 - (-3)|}{\\sqrt{1+1}} = \\frac{5\\sqrt{2}}{2} $. Thus, the maximum distance from point $ P $ to the line $ x + y - 3 = 0 $ is $ \\underline{5\\sqrt{2}} $."
  },
  {
    "text": "Given that point $P(2,2 \\sqrt{2})$ lies on the parabola $y^{2}=2 p x$, then the distance from point $P$ to the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;P: Point;Coordinate(P) = (2, 2*sqrt(2));PointOnCurve(P, G)",
    "query_expressions": "Distance(P, Directrix(G))",
    "answer_expressions": "3",
    "fact_spans": "[[[21, 37], [48, 51]], [[21, 37]], [[24, 37]], [[2, 20], [43, 47]], [[2, 20]], [[2, 40]]]",
    "query_spans": "[[[43, 58]]]",
    "process": "\\because point P(2,2\\sqrt{2}) lies on the parabola y^{2}=2px, \\therefore (2\\sqrt{2})^{2}=2p\\times2, solving gives p=2, \\therefore the focus of the parabola is at (1,0), and the directrix equation is x=-1. \\therefore the distance from point P to the directrix of the parabola is 2-(-1)=3"
  },
  {
    "text": "It is known that the parabola and the ellipse both pass through the point $M(1,2)$, they share a common focus on the $x$-axis, the axis of symmetry of the ellipse is the coordinate axis, and the vertex of the parabola is at the origin. Then the length of the major axis of the ellipse is?",
    "fact_expressions": "H:Ellipse;G:Parabola;M:Point;Coordinate(M)=(1,2);PointOnCurve(M,H);PointOnCurve(M,G);Focus(G)=Focus(H);PointOnCurve(Focus(G),xAxis);PointOnCurve(Focus(H),xAxis);SymmetryAxis(H)=axis;O:Origin;Vertex(G)=O",
    "query_expressions": "Length(MajorAxis(H))",
    "answer_expressions": "2+2*sqrt(2)",
    "fact_spans": "[[[6, 8], [37, 39], [63, 65]], [[2, 5], [49, 52]], [[11, 20]], [[11, 20]], [[2, 20]], [[2, 20]], [[22, 35]], [[22, 35]], [[22, 35]], [[37, 47]], [[56, 60]], [[48, 60]]]",
    "query_spans": "[[[63, 71]]]",
    "process": ""
  },
  {
    "text": "Given that $O$ is the coordinate origin, $\\odot O_{1}$: $x^{2}+y^{2}=4$, $\\odot O_{2}$: $x^{2}+y^{2}=1$, $A$ is a moving point on $\\odot O_{1}$, connect $O A$, the line segment $O A$ intersects $\\odot O_{2}$ at point $B$, draw a perpendicular from $A$ to the $x$-axis intersecting the $x$-axis at point $C$, draw a perpendicular from $B$ to $AC$ intersecting $AC$ at point $D$, then the trajectory equation of point $D$ is?",
    "fact_expressions": "O: Origin;O1: Circle;Expression(O1) = (x^2 + y^2 = 4);O2: Circle;Expression(O2) = (x^2 + y^2 = 1);A: Point;PointOnCurve(A, O1);B: Point;Intersection(LineSegmentOf(O, A), O2) = B;L: Line;IsPerpendicular(L, xAxis);PointOnCurve(A, L);C: Point;Intersection(L, xAxis) = C;D: Point;L1: Line;IsPerpendicular(L1, LineSegmentOf(A, C));PointOnCurve(B, L1);Intersection(L1, LineSegmentOf(A, C)) = D",
    "query_expressions": "LocusEquation(D)",
    "answer_expressions": "x^2/4+y^2=1",
    "fact_spans": "[[[2, 5]], [[11, 41], [80, 93]], [[11, 41]], [[43, 73], [114, 127]], [[43, 73]], [[75, 79], [134, 137]], [[76, 97]], [[128, 132], [157, 161]], [[106, 132]], [], [[133, 145]], [[133, 145]], [[151, 155]], [[133, 155]], [[175, 179], [181, 185]], [], [[156, 169]], [[156, 169]], [[156, 179]]]",
    "query_spans": "[[[181, 192]]]",
    "process": ""
  },
  {
    "text": "If the equation $\\frac{x^{2}}{|k|-3}+\\frac{y^{2}}{2-k}=-1$ represents a hyperbola with foci on the $y$-axis, then the range of its semi-focal length $c$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(Abs(k) - 3) + y^2/(2 - k) = -1);PointOnCurve(Focus(G), yAxis);c: Number;HalfFocalLength(G) = c",
    "query_expressions": "Range(c)",
    "answer_expressions": "(1, +\\infty)",
    "fact_spans": "[[[56, 59], [61, 62]], [[1, 59]], [[47, 59]], [[66, 69]], [[61, 69]]]",
    "query_spans": "[[[66, 76]]]",
    "process": "Since dividing both sides of the equation $\\frac{x^2}{|k|-3} + \\frac{y^{2}}{2-k} = -1$ by $-1$ simplifies it to $\\frac{y^{2}}{k-2} - \\frac{x^{2}}{k-3} = 1$, and because the equation represents a hyperbola with foci on the $y$-axis, we have $k-2 > 0$, $|k|-3 > 0$, i.e., $k > 3$. Thus, the equation can be written as: $\\frac{y^{2}}{k-2} - \\frac{x^{2}}{k-3} = 1$. Since $c^{2} = a^{2} + b^{2} = 2k - 5 > 1$, the range of the semi-focal length $c$ is $c > 1$, i.e., $c \\in (1, +\\infty)$."
  },
  {
    "text": "The equation of the ellipse with the foci of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ as vertices and the vertices as foci is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(G) = (x^2/4 - y^2/5 = 1);Focus(G)=Vertex(H);Vertex(G)=Focus(H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2/9+y^2/5=1",
    "fact_spans": "[[[1, 39]], [[52, 54]], [[1, 39]], [[0, 54]], [[0, 54]]]",
    "query_spans": "[[[52, 58]]]",
    "process": "From the properties of hyperbolas, we know that the hyperbola $ C: \\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1 $ has foci at $ (\\pm3,0) $ and vertices at $ (\\pm2,0) $. Therefore, the ellipse has vertices at $ (\\pm3,0) $ and foci at $ (\\pm2,0) $. Since $ b^{2}=a^{2}-c^{2}=5 $, the equation of the ellipse is $ \\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1 $."
  },
  {
    "text": "Given the parabola $ C $: $ y^{2} = 2 p x $ ($ p > 0 $), the distance from the focus $ F $ to the directrix $ l $ is $ 2 $. Let $ A(1,2) $, $ D(x_{0}, 0) $, where $ x_{0} > 0 $, and point $ B $ lies on the parabola $ C $. If $ \\angle A B D = \\angle A D B = 45^{\\circ} $, then $ |B F| $ = ?",
    "fact_expressions": "C: Parabola;p: Number;A: Point;D: Point;B: Point;F: Point;l: Line;x0:Number;p>0;x0>0;Expression(C) = (y^2 = 2*p*x);Coordinate(A) = (1, 2);Coordinate(D) = (x0, 0);Focus(C)=F;Directrix(C)=l;Distance(F, l) = 2;PointOnCurve(B, C);AngleOf(A,B,D)=AngleOf(A,D,B);AngleOf(A,D,B)=ApplyUnit(45,degree)",
    "query_expressions": "Abs(LineSegmentOf(B, F))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 28], [92, 98]], [[10, 28]], [[50, 59]], [[60, 74]], [[87, 91]], [[31, 34]], [[37, 40]], [[77, 86]], [[10, 28]], [[77, 86]], [[2, 28]], [[50, 59]], [[60, 74]], [[2, 34]], [[2, 40]], [[31, 47]], [[87, 99]], [[101, 139]], [[101, 139]]]",
    "query_spans": "[[[141, 150]]]",
    "process": "According to the problem, p=2, then the parabola C: y^{2}=4x. It is clear that AF\\bot x-axis. Draw BM\\bot AF from point B, with M as the foot of the perpendicular, then \\triangle ABM\\cong\\triangle DAF, so |BM|=|AF|=2, therefore the x-coordinate of point B is 1+2=3. By the definition of the parabola, |BF|=3+1=4."
  },
  {
    "text": "Let the lower and upper vertices of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$ be $B_{1}$ and $B_{2}$, respectively. If point $P$ is a point on the ellipse such that the slopes of lines $PB_{1}$ and $PB_{2}$ are $\\frac{1}{4}$ and $-1$, respectively, then the eccentricity of the ellipse is?",
    "fact_expressions": "P: Point;B1: Point;G: Ellipse;b: Number;a: Number;B2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LowerVertex(G)=B1;UpperVertex(G)=B2;PointOnCurve(P, G);Slope(LineOf(P,B1))=1/4;Slope(LineOf(P,B2))=-1",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[83, 87]], [[64, 71]], [[1, 55], [88, 90], [142, 144]], [[3, 55]], [[3, 55]], [[74, 81]], [[3, 55]], [[3, 55]], [[1, 55]], [[1, 81]], [[1, 81]], [[83, 94]], [[96, 140]], [[96, 140]]]",
    "query_spans": "[[[142, 150]]]",
    "process": ""
  },
  {
    "text": "What is the eccentricity of the hyperbola $x^{2}-y^{2}=2$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 18]], [[0, 18]]]",
    "query_spans": "[[[0, 24]]]",
    "process": "The standard equation of the hyperbola $x^{2}-y^{2}=2$ is $\\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1$, then $a^{2}=2$, $b^{2}=2$, so $c^{2}=2+2=4$, that is, $a=\\sqrt{2}$, $c=2$, then the eccentricity $e=\\frac{c}{a}=\\frac{2}{\\sqrt{2}}=\\sqrt{2}$,"
  },
  {
    "text": "As shown in the figure, for the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $B$ is the upper vertex of the ellipse $C$. If the radius of the circumcircle of $\\Delta B F_{1} F_{2}$ is $\\frac{2 b}{3}$, then the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;a: Number;b: Number;B: Point;F1: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C)=F1;RightFocus(C)=F2;UpperVertex(C)=B;Radius(CircumCircle(TriangleOf(B,F1,F2)))=(2*b)/3",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[3, 60], [89, 94], [147, 152]], [[9, 60]], [[9, 60]], [[85, 88]], [[69, 76]], [[77, 84]], [[9, 60]], [[9, 60]], [[3, 60]], [[3, 85]], [[3, 85]], [[85, 98]], [[100, 145]]]",
    "query_spans": "[[[147, 158]]]",
    "process": "From the given conditions, the circumcenter of $\\triangle BF_{1}F_{2}$ lies on the segment $OB$, $|OF_{1}|=c$, $|MF_{1}|=|BM|=\\frac{2b}{3}$, we obtain $|OM|=\\frac{1}{3}b$. In $\\triangle OMF_{1}$, by the Pythagorean theorem: $|MF_{1}|^{2}=|OM|^{2}+|OF_{1}|^{2}$, that is $(\\frac{2b}{3})^{2}=(\\frac{b}{3})^{2}+c^{2}$. Combining with $b^{2}=a^{2}-c^{2}$, we can solve. From the given conditions: the circumcenter of $\\triangle BF_{1}F_{2}$ lies on the segment $OB$, $|OF_{1}|=c$. Let the circumcenter be $M$, then $|OM|=|OB|-|BM|=b-\\frac{2}{3}b=\\frac{1}{3}b$. In $\\triangle OMF_{1}$, by the Pythagorean theorem: $|MF_{1}|^{2}=|OM|^{2}+|OF_{1}|^{2}$, that is $(\\frac{2b}{3})^{2}=(\\frac{b}{3})^{2}+c^{2}$. Therefore, $b^{2}=3c^{2}$, i.e., $a^{2}-c^{2}=3c^{2}$, so $a=2c$, hence $e=\\frac{c}{a}=\\frac{1}{2}$."
  },
  {
    "text": "Given that $P$ is a point on the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, distinct from points $A(-a, 0)$ and $B(a, 0)$, and the eccentricity of $E$ is $\\frac{\\sqrt{3}}{2}$, then the product of the slopes of lines $AP$ and $BP$ is?",
    "fact_expressions": "E: Ellipse;b: Number;a: Number;A: Point;P: Point;B: Point;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (-a, 0);Coordinate(B) = (a, 0);PointOnCurve(P, E);Negation(P=A);Negation(P=B);Eccentricity(E) = sqrt(3)/2",
    "query_expressions": "Slope(LineOf(A,P))*Slope(LineOf(B,P))",
    "answer_expressions": "-1/4",
    "fact_spans": "[[[6, 63], [93, 96]], [[13, 63]], [[13, 63]], [[66, 77]], [[2, 5]], [[80, 89]], [[13, 63]], [[13, 63]], [[6, 63]], [[66, 77]], [[80, 89]], [[2, 92]], [[2, 92]], [[2, 92]], [[93, 121]]]",
    "query_spans": "[[[123, 143]]]",
    "process": "Let $ P(x_{0},y_{0}) $, we have $ \\frac{x_{0}^{2}}{a^{2}} + \\frac{y_{0}^{2}}{b^{2}} = 1 $, and $ \\frac{c}{a} = \\frac{\\sqrt{3}}{2} $, obtain $ \\frac{b}{a} = k_{AP}k_{BP} = \\frac{y_{0}}{x_{0}+a} \\cdot \\frac{y_{0}}{x_{0}-a} = \\frac{y_{0}^{2}}{x_{0}^{2}-a^{2}} $, obtain $ \\frac{b}{a} = \\frac{1}{2} $."
  },
  {
    "text": "Given that $P$ is the upper vertex of the ellipse $C$: $\\frac{x^{2}}{4}+y^{2}=1$, and a line $l$ passing through the origin intersects $C$ at points $A$ and $B$. If the area of $\\triangle P A B$ is $\\sqrt{2}$, then what is the slope of $l$?",
    "fact_expressions": "P: Point;C: Ellipse;Expression(C) = (x^2/4 + y^2 = 1);UpperVertex(C) = P;O: Origin;PointOnCurve(O, l) = True;l: Line;Intersection(l, C) = {A, B};A: Point;B: Point;Area(TriangleOf(P, A, B)) = sqrt(2)",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*1/2",
    "fact_spans": "[[[2, 5]], [[6, 38], [53, 56]], [[6, 38]], [[2, 42]], [[44, 46]], [[43, 52]], [[47, 52], [101, 104]], [[47, 66]], [[57, 60]], [[61, 64]], [[68, 99]]]",
    "query_spans": "[[[101, 109]]]",
    "process": "Let the equation of line AB be: y = kx, A(x_{1}, y_{1}), B(-x_{1}, -y_{1}). From \\begin{cases}\\frac{x^{2}}{4}+y^{2}=1\\\\y=kx\\end{cases}, we obtain (1+4k^{2})x^{2}-4=0. Thus, x_{1}^{2}=\\frac{4}{1+4k^{2}}. Since P(0,1), it follows that S_{\\triangle PAB} = \\frac{1}{2}|OP|\\times2|x| = \\frac{2}{\\sqrt{1+4k^{2}}} = \\sqrt{2}. Solving gives k = \\pm\\frac{1}{2}."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the right branch of the hyperbola, and $|P F_{1}|=4|P F_{2}|$. Then the maximum value of the eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;e: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F1)) = 4*Abs(LineSegmentOf(P, F2));Eccentricity(G) = e",
    "query_expressions": "Max(e)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[2, 59], [89, 92], [122, 125]], [[5, 59]], [[5, 59]], [[84, 88]], [[68, 75]], [[76, 83]], [[128, 131]], [[5, 59]], [[5, 59]], [[2, 59]], [[2, 83]], [[2, 83]], [[84, 96]], [[98, 120]], [[122, 131]]]",
    "query_spans": "[[[128, 137]]]",
    "process": ""
  },
  {
    "text": "Given the line $l$: $2x - y - 1 = 0$ and the parabola $x^{2} = -4y$ intersect at points $A$ and $B$, then $|AB| = $?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;Expression(G) = (x^2 = -4*y);Expression(l) = (2*x - y - 1 = 0);Intersection(l, G) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "20",
    "fact_spans": "[[[2, 20]], [[21, 36]], [[38, 41]], [[42, 45]], [[21, 36]], [[2, 20]], [[2, 47]]]",
    "query_spans": "[[[49, 59]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $\\frac{y^{2}}{9}-x^{2}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2 + y^2/9 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*3*x",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 36]]]",
    "process": "Directly transform the equation $\\frac{y^{2}}{9}-x^{2}=0$ to obtain the answer; $\\because\\frac{y^{2}}{9}-x^{2}=0\\Rightarrow y=\\pm3x$,"
  },
  {
    "text": "If $0<a<\\sqrt{2}$, then the range of the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3-a^{2}}=1$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/(3 - a^2) + x^2/a^2 = 1);a: Number;0<a;a<sqrt(2)",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(\\sqrt{6}/2, +\\infty)",
    "fact_spans": "[[[17, 65]], [[17, 65]], [[1, 15]], [[1, 15]], [[1, 15]]]",
    "query_spans": "[[[17, 74]]]",
    "process": ""
  },
  {
    "text": "Given $F_{1}(-3,0)$, $F_{2}(3,0)$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and there exists a point $P$ on an asymptote of hyperbola $C$ such that $|P F_{1}|=2|P F_{2}|$, then the maximum value of $b$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;Coordinate(F1) = (-3, 0);Coordinate(F2) = (3, 0);LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, Asymptote(C));Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Max(b)",
    "answer_expressions": "12/5",
    "fact_spans": "[[[30, 91], [98, 104]], [[30, 91]], [[141, 144]], [[38, 91]], [[38, 91]], [[38, 91]], [[2, 15]], [[17, 29]], [[2, 15]], [[17, 29]], [[2, 97]], [[2, 97]], [[111, 115]], [[98, 115]], [[117, 139]]]",
    "query_spans": "[[[141, 150]]]",
    "process": ""
  },
  {
    "text": "If a point $M$ on the parabola $y^{2}=2x$ is at a distance of $\\frac{3}{2}$ from its focus $F$, and $O$ is the coordinate origin, then the area of $\\Delta MFO$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);M: Point;PointOnCurve(M, G) = True;F: Point;Focus(G) = F;Distance(M, F) = 3/2;O: Origin",
    "query_expressions": "Area(TriangleOf(M, F, O))",
    "answer_expressions": "sqrt(2)/4",
    "fact_spans": "[[[1, 15], [22, 23]], [[1, 15]], [[18, 21]], [[1, 21]], [[26, 29]], [[22, 29]], [[18, 46]], [[49, 52]]]",
    "query_spans": "[[[59, 78]]]",
    "process": "The focus of the parabola is $F\\left(\\frac{1}{2},0\\right)$. Since the distance from $M$ to the focus of the parabola is $\\frac{3}{2}$, the x-coordinate of point $M$ is $\\frac{3}{2}-\\frac{1}{2}=1$. Substituting into the parabolic equation gives $y=\\pm\\sqrt{2}$. Without loss of generality, let $M(1,\\sqrt{2})$. Then the area of triangle $MFO$ is $\\frac{1}{2}\\times\\frac{1}{2}\\times\\sqrt{2}=\\frac{\\sqrt{2}}{4}$."
  },
  {
    "text": "The focus of the curve $\\sqrt{(x-1)^{2}+y^{2}}=\\frac{\\sqrt{2}}{2}(2-x)$ is the focus of hyperbola $C$, and the point $(3, \\frac{2 \\sqrt{39}}{3})$ lies on $C$. Then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Curve;H: Point;Expression(G) = (sqrt(y^2 + (x - 1)^2) = (sqrt(2)/2)*(2 - x));Coordinate(H) = (3, (2*sqrt(39))/3);Focus(G) = Focus(C);PointOnCurve(H, C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "3*x^2-(3/2)*y^2=1",
    "fact_spans": "[[[54, 60], [100, 103], [94, 97]], [[0, 50]], [[64, 93]], [[0, 50]], [[64, 93]], [[0, 63]], [[64, 98]]]",
    "query_spans": "[[[100, 108]]]",
    "process": ": Simplifying \\sqrt{(x-1)^{2}+y^{2}}=\\frac{\\sqrt{2}}{2}(2-x) yields: \\frac{x^{2}}{2}+y^{2}=1. This equation represents an ellipse with foci at (-1,0) and (1,0). According to the problem, let the equation of hyperbola C be: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0), and c=1. Since the point (3,\\frac{2\\sqrt{39}}{3}) lies on C, substituting it into the equation gives: \\frac{9}{a^{2}}-\\frac{52}{b^{2}}=1. Also, a^{2}+b^{2}=c^{2}=1, solving yields: a^{2}=\\frac{1}{3}, b^{2}=\\frac{2}{3}. Therefore, the equation of hyperbola C is: 3x^{2}-\\frac{3}{7}y^{2}=1."
  },
  {
    "text": "The point $(2,3)$ lies on the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and the focal length of $C$ is $4$. What is its eccentricity?",
    "fact_expressions": "P: Point;Coordinate(P) = (2, 3);PointOnCurve(P, C);C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;FocalLength(C) = 4",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 10]], [[2, 10]], [[2, 73]], [[11, 72], [75, 78], [87, 88]], [[11, 72]], [[19, 72]], [[19, 72]], [[19, 72]], [[19, 72]], [[75, 85]]]",
    "query_spans": "[[[87, 94]]]",
    "process": "Since the point (2,3) lies on the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $, we have $ \\frac{4}{a^{2}} - \\frac{9}{b^{2}} = 1 $. Also, given that the focal distance of the hyperbola $ C $ is 4, we obtain $ c = 2 $, then $ c^{2} = a^{2} + b^{2} = 4 $. Solving the system of equations, we get $ a = 1 $, so the eccentricity of the hyperbola is $ e = \\frac{c}{a} = 2 $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the endpoints of the minor axis are $P(0, b)$, $Q(0,-b)$, and one endpoint of the major axis is $M$. Let $AB$ be a chord passing through the center of the ellipse and not lying on the coordinate axes. If the product of the slopes of $PA$ and $PB$ equals $-\\frac{1}{4}$, then what is the distance from $P$ to the line $QM$?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;Q: Point;M: Point;P: Point;A: Point;B: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (0,b);Coordinate(Q) = (0,-b);Endpoint(MinorAxis(G))={P,Q};OneOf(Endpoint(MajorAxis(G)))=M;IsChordOf(LineSegmentOf(A,B),G);PointOnCurve(Center(G), LineSegmentOf(A,B));Negation(OverlappingLine(LineSegmentOf(A,B),axis));Slope(LineSegmentOf(P,A))*Slope(LineSegmentOf(P,B))=-1/4",
    "query_expressions": "Distance(P,LineOf(Q,M))",
    "answer_expressions": "{(4*sqrt(5)/5)*b,(2*sqrt(5)/5)*a}",
    "fact_spans": "[[[2, 54], [101, 103]], [[4, 54]], [[4, 54]], [[69, 79]], [[88, 91]], [[59, 68], [153, 156]], [[93, 98]], [[93, 98]], [[4, 54]], [[4, 54]], [[2, 54]], [[59, 68]], [[70, 79]], [[2, 79]], [[2, 91]], [[93, 116]], [[93, 116]], [[93, 116]], [[118, 151]]]",
    "query_spans": "[[[153, 169]]]",
    "process": ""
  },
  {
    "text": "It is known that the center of a hyperbola is at the origin, one of its foci coincides with the focus of the parabola $y=\\frac{1}{8} x^{2}$, and the eccentricity is $2$. Then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;O: Origin;Expression(H) = (y = x^2/8);Center(G) = O;OneOf(Focus(G)) = Focus(H);Eccentricity(G)=2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 - x^2/3 = 1",
    "fact_spans": "[[[2, 5], [59, 62]], [[19, 43]], [[9, 11]], [[19, 43]], [[2, 11]], [[2, 48]], [[2, 56]]]",
    "query_spans": "[[[59, 69]]]",
    "process": "The standard equation of the parabola is $x^{2}=8y$, and its focus is at $(0,2)$. This point is also a focus of the hyperbola, so $c=2$. Given $e=\\frac{c}{a}=2$, then $a=1$, $b^{2}=c^{2}-a^{2}=3$. The equation of the hyperbola is $y^{2}-\\frac{x^{2}}{3}=1$."
  },
  {
    "text": "Given that line $l$ passes through the focus of the ellipse $\\frac{y^{2}}{2}+x^{2}=1$ and intersects the ellipse at points $P$ and $Q$, and the perpendicular bisector of segment $PQ$ intersects the $x$-axis at point $M$, then the maximum area of $\\Delta M P Q$ is?",
    "fact_expressions": "l: Line;G: Ellipse;P: Point;Q: Point;M: Point;Expression(G) = (x^2 + y^2/2 = 1);PointOnCurve(Focus(G), l);Intersection(l, G) = {P, Q};Intersection(PerpendicularBisector(LineSegmentOf(P,Q)),xAxis) = M",
    "query_expressions": "Max(Area(TriangleOf(M, P, Q)))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 7]], [[9, 36], [42, 44]], [[47, 50]], [[51, 54]], [[77, 81]], [[9, 36]], [[2, 39]], [[2, 56]], [[57, 81]]]",
    "query_spans": "[[[83, 105]]]",
    "process": ""
  },
  {
    "text": "Given that $A$ and $B$ are the intersection points of a line passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) with the parabola, $O$ is the origin, and satisfying $\\overrightarrow{A F}=2 \\overrightarrow{F B}$, $S \\triangle_{O A B}=\\frac{\\sqrt{2}}{3}|A B|$, then the standard equation of the parabola is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;F: Point;B: Point;O: Origin;Z: Line;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(F, Z);Intersection(Z, G) = {A, B};VectorOf(A, F) = 2*VectorOf(F, B);Area(TriangleOf(O, A, B)) = Abs(LineSegmentOf(A, B))*sqrt(2)/3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = 4*x",
    "fact_spans": "[[[11, 33], [42, 45], [155, 158]], [[14, 33]], [[2, 5]], [[35, 38]], [[6, 9]], [[49, 52]], [[39, 41]], [[14, 33]], [[11, 33]], [[11, 38]], [[10, 41]], [[2, 48]], [[61, 106]], [[108, 153]]]",
    "query_spans": "[[[155, 165]]]",
    "process": "First, let A(x_{1},y_{1}), B(x_{2},y_{2}), according to \\overrightarrow{AF}=2\\overrightarrow{FB}, we obtain y_{1}=-2y_{2}; from the property of the focal chord of a parabola, we get y_{1}y_{2}=-p^{2}, solving gives |y_{2}|=\\frac{\\sqrt{2}}{2}p, |y_{1}|=\\sqrt{2}p; then solve to get |BF|=\\frac{3}{4}p, |AF|=\\frac{3}{2}p, |AB|=\\frac{9}{4}p; combining with the given conditions in the problem, using the triangle area formula, we can find p=2, thereby obtaining the equation of the parabola. Solution: Let A(x_{1},y_{1}), B(x_{2},y_{2}), \\overrightarrow{AF}=2\\overrightarrow{FB}, then y_{1}=-2y_{2}; also from the property of the focal chord of a parabola, y_{1}y_{2}=-p^{2}, so -2y_{2}^{2} = -p^{2}, thus |y_{2}| = \\frac{\\sqrt{2}}{2}p, |y_{1}| = \\sqrt{2}p; obtain |BF|=\\frac{3}{4}p, |AF|=\\frac{3}{2}p, |AB|=\\frac{9}{4}p; S_{\\triangle OAB} = \\frac{1}{2} \\cdot \\frac{p}{2} \\cdot (|y_{1}| + |y_{2}|) = \\frac{3\\sqrt{2}}{8}p^{2} = \\frac{\\sqrt{2}}{3} \\cdot \\frac{9}{4}p, solving gives p=2, the standard equation of the parabola is y^{2}=4x. Analysis: This problem examines conic sections, involving knowledge points such as solving for the equation of a parabola, properties of the focal chord of a parabola, and the area of a triangle in a parabola, belonging to a simple problem."
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-4 y^{2}=1$, the line $l$ passing through the point $P(2,0)$ has exactly one common point with $C$. Then the equation of the line $l$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;P: Point;Expression(C) = (x^2 - 4*y^2 = 1);Coordinate(P) = (2, 0);PointOnCurve(P,l);NumIntersection(l,C)=1",
    "query_expressions": "Expression(l)",
    "answer_expressions": "{y=(1/2)*x - 1,y=-(1/2)*x+1}",
    "fact_spans": "[[[40, 45], [57, 62]], [[2, 28], [46, 49]], [[30, 39]], [[2, 28]], [[30, 39]], [[29, 45]], [[40, 55]]]",
    "query_spans": "[[[57, 67]]]",
    "process": ""
  },
  {
    "text": "If the hyperbola $\\frac{x^{2}}{36}-\\frac{y^{2}}{m}=1$ has eccentricity $e=\\frac{5}{3}$, then $m=$?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/36 - y^2/m = 1);e:Number;Eccentricity(G)=e;e=5/3",
    "query_expressions": "m",
    "answer_expressions": "64",
    "fact_spans": "[[[1, 40]], [[61, 64]], [[1, 40]], [[44, 59]], [[1, 59]], [[44, 59]]]",
    "query_spans": "[[[61, 66]]]",
    "process": ""
  },
  {
    "text": "If the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has eccentricity $3$, then the distance from the focus of the parabola $y=\\frac{1}{4} x^{2}$ to the asymptotes of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(C) = 3;G: Parabola;Expression(G) = (y = x^2/4)",
    "query_expressions": "Distance(Focus(G),Asymptote(C))",
    "answer_expressions": "1/3",
    "fact_spans": "[[[1, 62], [100, 106]], [[1, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[1, 70]], [[72, 96]], [[72, 96]]]",
    "query_spans": "[[[72, 114]]]",
    "process": "y=\\frac{1}{4}x^2 has focus (0,1). The eccentricity of the hyperbola equals 3, i.e., \\frac{c}{a}=3, c^{2}=a^{2}+b^{2}, \\frac{c^{2}}{a^{2}}=1+(\\frac{b}{a})^{2}, \\frac{b}{a}=2\\sqrt{2}. Then the asymptotes of the hyperbola are y=\\pm2\\sqrt{2}x. The distance from the focus (0,1) to the asymptote of the hyperbola is d=\\frac{1}{\\sqrt{1+8}}=\\frac{1}{3}."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=x$, a line passing through the focus of $C$ intersects $C$ at points $A$ and $B$, and the chord $AB$ has length $2$. What is the x-coordinate of the point where the perpendicular bisector of segment $AB$ intersects the $x$-axis?",
    "fact_expressions": "C: Parabola;G: Line;A: Point;B: Point;Expression(C) = (y^2 = x);PointOnCurve(Focus(C), G);Intersection(G, C) = {A, B};IsChordOf(LineSegmentOf(A, B), C);Length(LineSegmentOf(A, B)) = 2",
    "query_expressions": "Coordinate(Intersection(PerpendicularBisector(LineSegmentOf(A, B)), xAxis))",
    "answer_expressions": "5/4",
    "fact_spans": "[[[2, 19], [21, 24], [31, 34]], [[28, 30]], [[36, 39]], [[40, 43]], [[2, 19]], [[20, 30]], [[28, 45]], [[31, 53]], [[48, 58]]]",
    "query_spans": "[[[60, 84]]]",
    "process": "The focus of the parabola is $(\\frac{1}{4},0)$, so the line $AB$ can be set as: $x=ky+\\frac{1}{4}$ $(k\\neq0)$. Combining with $y^{2}=x$, eliminating $x$ gives $y^{2}-ky-\\frac{1}{4}=0$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $y_{1}+y_{2}=k$, $|AB|=x_{1}+x_{2}+\\frac{1}{2}=(ky_{1}+\\frac{1}{4})+(ky_{2}+\\frac{1}{4})=k^{2}+1=2$, so $k=\\pm1$. When $k=1$, $\\frac{y_{1}+y_{2}}{2}=\\frac{1}{2}$, so the midpoint coordinates of $AB$ are $(\\frac{3}{4},\\frac{1}{2})$, then the perpendicular bisector equation of $AB$ is $y-\\frac{1}{2}=-(x-\\frac{3}{4})$, and the x-coordinate of its intersection with the $x$-axis is $\\frac{5}{4}$. Similarly, when $k=-1$, the x-coordinate of the intersection point between the perpendicular bisector of segment $AB$ and the $x$-axis is $\\frac{5}{4}$."
  },
  {
    "text": "The line $l$ passes through the focus $F$ of the parabola $y^{2}=2 p x(p>0)$, and intersects the parabola at points $P$ and $Q$. From $P$ and $Q$, perpendiculars $PR$ and $QS$ are drawn to the directrix, with feet $R$ and $S$, respectively. If $|PF| = a$, $|QF| = b$, and $M$ is the midpoint of $RS$, then $|MF| =$?",
    "fact_expressions": "l: Line;G: Parabola;p: Number;P: Point;R: Point;Q: Point;S: Point;F: Point;M: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(F, l);Intersection(l,G)={P,Q};L1:Line;L2:Line;PointOnCurve(P,LineSegmentOf(P,R));PointOnCurve(Q,LineSegmentOf(Q,S));Directrix(G)={L1,L2};IsPerpendicular(L1,LineSegmentOf(P,R));IsPerpendicular(L2,LineSegmentOf(Q,S));FootPoint(L1,LineSegmentOf(P,R))=R;FootPoint(L2,LineSegmentOf(Q,S))=S;Abs(LineSegmentOf(P,F))=a;Abs(LineSegmentOf(Q, F)) = b;MidPoint(LineSegmentOf(R, S)) = M;a:Number;b:Number",
    "query_expressions": "Abs(LineSegmentOf(M, F))",
    "answer_expressions": "sqrt(a*b)",
    "fact_spans": "[[[0, 5]], [[6, 27], [36, 39]], [[9, 27]], [[40, 43], [51, 54]], [[87, 90]], [[44, 47], [55, 58]], [[91, 94]], [[30, 33]], [[124, 127]], [[9, 27]], [[6, 27]], [[6, 33]], [[0, 33]], [[0, 49]], [], [], [[36, 81]], [[36, 81]], [[36, 63]], [[36, 80]], [[36, 80]], [[36, 94]], [[36, 94]], [[97, 108]], [[110, 121]], [[124, 136]], [[97, 108]], [[110, 121]]]",
    "query_spans": "[[[138, 147]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, a line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. If $|F_{2} A|+|F_{2} B|=12$, then $|A B|$=?",
    "fact_expressions": "G: Ellipse;H: Line;F2: Point;A: Point;B: Point;F1: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(G) = {F1, F2};PointOnCurve(F1, H);Intersection(H, G) = {A,B};Abs(LineSegmentOf(F2, A)) + Abs(LineSegmentOf(F2, B)) = 12",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[18, 56], [74, 76]], [[71, 73]], [[10, 17]], [[77, 80]], [[81, 84]], [[2, 9], [2, 9]], [[18, 56]], [[2, 61]], [[62, 73]], [[71, 86]], [[88, 112]]]",
    "query_spans": "[[[114, 123]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x$ ($p>0$) passes through the point $P(1,4)$, the lines $PA$ and $PB$ intersect the parabola $C$ at points $A$ and $B$, respectively. If the sum of the slopes of lines $PA$ and $PB$ is zero, then what is the slope of line $AB$?",
    "fact_expressions": "C: Parabola;p: Number;P: Point;A: Point;B: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(P) = (1, 4);PointOnCurve(P, C);Intersection(LineOf(P, A), C) = A;Intersection(LineOf(P, B), C) = B;Slope(LineOf(P, A)) + Slope(LineOf(P, B)) = 0",
    "query_expressions": "Slope(LineOf(A, B))",
    "answer_expressions": "-2",
    "fact_spans": "[[[2, 28], [57, 63]], [[10, 28]], [[30, 39]], [[65, 69]], [[70, 73]], [[10, 28]], [[2, 28]], [[30, 39]], [[2, 39]], [[40, 73]], [[40, 73]], [[75, 96]]]",
    "query_spans": "[[[98, 110]]]",
    "process": ""
  },
  {
    "text": "Given: $A(3 , 0)$, $B(9 , 5)$, and $P$ is an arbitrary point on the right branch of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$. Then the minimum value of $|PA|+| PB |$ is?",
    "fact_expressions": "G: Hyperbola;A: Point;B: Point;P: Point;Expression(G) = (x^2/4 - y^2/5 = 1);Coordinate(A) = (3, 0);Coordinate(B) = (9, 5);PointOnCurve(P, RightPart(G))",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B)))",
    "answer_expressions": "9",
    "fact_spans": "[[[30, 68]], [[3, 13]], [[15, 25]], [[26, 29]], [[30, 68]], [[3, 13]], [[15, 25]], [[26, 76]]]",
    "query_spans": "[[[78, 97]]]",
    "process": ""
  },
  {
    "text": "Given a point $P$ on the parabola $y^{2}=16x$ such that the distance from $P$ to the $x$-axis is $12$, then the distance from $P$ to the focus $F$ is equal to?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 16*x);P: Point;PointOnCurve(P, G);Distance(P, xAxis) = 12;F: Point;Focus(G) = F",
    "query_expressions": "Distance(P, F)",
    "answer_expressions": "13",
    "fact_spans": "[[[2, 17]], [[2, 17]], [[21, 24], [39, 42]], [[2, 24]], [[21, 37]], [[45, 48]], [[2, 48]]]",
    "query_spans": "[[[39, 54]]]",
    "process": ""
  },
  {
    "text": "Point $P$ moves on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and points $Q$, $R$ move on the circles $(x+1)^{2}+y^{2}=1$ and $(x-1)^{2}+y^{2}=1$ respectively. Then the minimum value of $|P Q|+|P R|$ is?",
    "fact_expressions": "G: Ellipse;H: Circle;C:Circle;P: Point;Q: Point;R: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Expression(H) = (y^2 + (x + 1)^2 = 1);Expression(C) = (y^2 + (x - 1)^2 = 1);PointOnCurve(P, G);PointOnCurve(Q, H);PointOnCurve(R, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, Q)) + Abs(LineSegmentOf(P, R)))",
    "answer_expressions": "2",
    "fact_spans": "[[[5, 42]], [[58, 77]], [[78, 97]], [[0, 4]], [[46, 49]], [[50, 53]], [[5, 42]], [[58, 77]], [[78, 97]], [[0, 45]], [[46, 100]], [[46, 100]]]",
    "query_spans": "[[[102, 121]]]",
    "process": "Since the centers of the two circles $(x+1)^{2}+y^{2}=1$ and $(x-1)^{2}+y^{2}=1$ are $F_{1}(-1,0)$, $F_{2}(1,0)$, exactly the left and right foci of the ellipse, we have $PQ+PR\\geqslant PF_{1}-r_{1}+PF_{2}-r_{2}=2a-2=4-2=2$"
  },
  {
    "text": "Points $A$ and $B$ are two points on the right branch of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$. The distance from the midpoint of $AB$ to the $y$-axis is $4$. Then the maximum value of $AB$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/5 = 1);A: Point;B: Point;PointOnCurve(A, RightPart(G));PointOnCurve(B, RightPart(G));Distance(MidPoint(LineSegmentOf(A, B)), yAxis) = 4",
    "query_expressions": "Max(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[9, 47]], [[9, 47]], [[0, 4]], [[5, 8]], [[0, 53]], [[0, 53]], [[54, 73]]]",
    "query_spans": "[[[75, 86]]]",
    "process": ""
  },
  {
    "text": "The distance from the focus of the hyperbola $\\frac{x^{2}}{9}-y^{2}=1$ to its asymptote is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2 = 1)",
    "query_expressions": "Distance(Focus(G), Asymptote(G))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 40]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{6}-\\frac{y^{2}}{b^{2}}=1$ is $\\frac{\\sqrt{6}}{2}$, then $b^{2}=$?",
    "fact_expressions": "G: Hyperbola;b: Number;Expression(G) = (x^2/6 - y^2/b^2 = 1);Eccentricity(G) = sqrt(6)/2",
    "query_expressions": "b^2",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 44]], [[71, 78]], [[2, 44]], [[2, 69]]]",
    "query_spans": "[[[71, 80]]]",
    "process": "\\because the eccentricity of the hyperbola \\frac{x^{2}}{6}-\\frac{y^{2}}{b^{2}}=1 is \\frac{\\sqrt{6}}{2}, then \\sqrt{\\frac{b^{2}+6}{6}}=\\frac{\\sqrt{6}}{2}, solving gives b^{2}=3."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=9x$ is $F$, and the intersection point of its directrix with the $x$-axis is $C$. A line passing through $F$ intersects the parabola at points $A$ and $B$. If the distance from the midpoint of chord $AB$ to the directrix of the parabola is $18$, then the cosine of $\\angle ACF$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;C: Point;F: Point;Expression(G) = (y^2 = 9*x);Focus(G) = F;Intersection(Directrix(G), xAxis) = C;PointOnCurve(F, H);Intersection(H, G) = {A, B};IsChordOf(LineSegmentOf(A, B), G);Distance(MidPoint(LineSegmentOf(A, B)), Directrix(G)) = 18",
    "query_expressions": "Cos(AngleOf(A, C, F))",
    "answer_expressions": "2*sqrt(5)/5",
    "fact_spans": "[[[2, 16], [24, 25], [48, 51], [74, 77]], [[45, 47]], [[53, 56]], [[57, 60]], [[36, 39]], [[20, 23], [41, 44]], [[2, 16]], [[2, 23]], [[24, 39]], [[40, 47]], [[45, 62]], [[48, 70]], [[65, 87]]]",
    "query_spans": "[[[89, 109]]]",
    "process": "First, find the coordinates of the focus and the equation of the directrix. Let the equation of the line be $ y = k\\left(x - \\frac{9}{4}\\right) $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. By solving the system of equations of the line and the parabola, obtain $ x_{1} + x_{2} $, then use the property of the focal chord to find $ k^{2} $. Due to symmetry, without loss of generality, take $ k = \\frac{\\sqrt{3}}{3} $, thus find the coordinates of intersection points between the line and the parabola, thereby obtaining $ \\tan\\angle ACF $, and then compute using fundamental relationships of trigonometric functions of the same angle. Solution: The parabola $ y^{2} = 9x $ has focus $ \\left(\\frac{9}{4}, 0\\right) $ and directrix $ x = -\\frac{9}{4} $, so $ C\\left(-\\frac{9}{4}, 0\\right) $. According to the problem, the slope of the line passing through $ F $ exists. Let the equation of the line be $ y = k\\left(x - \\frac{9}{4}\\right) $. From \n\\[\n\\begin{cases}\ny^{2} = 9x \\\\\ny = k\\left(x - \\frac{9}{4}\\right)\n\\end{cases}\n\\]\neliminating $ y $, we get $ k^{2}x^{2} - \\left(9 + \\frac{9}{2}k^{2}\\right)x + \\frac{81}{16}k^{2} = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = \\frac{9}{2} + \\frac{9}{k^{2}} $. Since the distance from the midpoint of chord $ AB $ to the directrix of the parabola is 18, $ |AB| = 36 $. And $ |AB| = x_{1} + x_{2} + p $, so $ x_{1} + x_{2} = 36 - \\frac{9}{2} $. Thus $ \\frac{9}{2} + \\frac{9}{k^{2}} = 36 - \\frac{9}{2} $, solving gives $ k^{2} = \\frac{1}{3} $, so $ k = \\pm\\frac{\\sqrt{3}}{3} $. Due to symmetry, without loss of generality, take $ k = \\frac{\\sqrt{3}}{3} $. Then $ \\frac{1}{3}x^{2} - \\frac{21}{2}x + \\frac{27}{16} = 0 $, solving yields $ x_{1} = \\frac{63 + 36\\sqrt{3}}{4} $ or $ x_{2} = \\frac{63 - 36\\sqrt{3}}{4} $, so $ y_{1} = \\frac{18 + 9\\sqrt{3}}{2} $ or $ y_{2} = \\frac{9\\sqrt{3} - 18}{2} $. If $ A\\left(\\frac{63 + 36\\sqrt{3}}{4}, \\frac{18 + 9\\sqrt{3}}{2}\\right) $, then $ \\tan\\angle ACF = \\frac{\\frac{18 + 9\\sqrt{3}}{2}}{\\frac{63 + 36\\sqrt{3}}{4} + \\frac{9}{4}} = \\frac{1}{2} $, and \n\\[\n\\begin{cases}\n\\tan\\angle ACF = \\frac{\\sin\\angle ACF}{\\cos\\angle ACF} = \\frac{1}{2} \\\\\n\\sin^{2}\\angle ACF + \\cos^{2}\\angle ACF = 1\n\\end{cases}\n\\]\nsolving gives $ \\cos\\angle ACF = \\frac{2\\sqrt{5}}{5} $; if $ A\\left(\\frac{63 - 36\\sqrt{3}}{4}, \\frac{9\\sqrt{3} - 18}{2}\\right) $, then $ \\tan\\angle ACF = \\frac{\\frac{9\\sqrt{3} - 18}{2}}{\\frac{63 - 36\\sqrt{3}}{4} + \\frac{9}{4}} = \\frac{1}{2} $, similarly we get $ \\cos\\angle ACF = \\frac{2\\sqrt{5}}{5} $; therefore $ \\cos\\angle ACF = \\frac{2\\sqrt{5}}{5} $."
  },
  {
    "text": "If the ellipse $\\Gamma$: $\\frac{x^{2}}{16}+\\frac{y^{2}}{m}=1$ with foci on the $x$-axis has a focal distance of $2 \\sqrt{7}$, then what is the value of $m$?",
    "fact_expressions": "PointOnCurve(Focus(Gamma), xAxis);Gamma: Ellipse;Expression(Gamma) = (x^2/16 + y^2/m = 1);m: Number;FocalLength(Gamma) = 2*sqrt(7)",
    "query_expressions": "m",
    "answer_expressions": "9",
    "fact_spans": "[[[1, 58]], [[10, 58]], [[10, 58]], [[76, 79]], [[10, 74]]]",
    "query_spans": "[[[76, 83]]]",
    "process": "From the given, we have: $2c=2\\sqrt{7}$, so $c=\\sqrt{7}$. Also, $c^{2}=a^{2}-b^{2}=16-m=7$, therefore $m=9$."
  },
  {
    "text": "Given that point $M$ lies on an ellipse with foci $A$ and $B$, and point $C$ is a point in the plane of the ellipse satisfying the following two conditions: (1) $\\overrightarrow{M A}+\\overrightarrow{M B}=2 \\overrightarrow{M C}$; (2) $|\\overrightarrow{M A}|=2|\\overrightarrow{M B}|=2|\\overrightarrow{M C}|$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;M: Point;A: Point;B: Point;C: Point;PointOnCurve(M, G);Focus(G) = {A,B};VectorOf(M,A)+VectorOf(M,B)=2*VectorOf(M,C);Abs(VectorOf(M,A))=2*Abs(VectorOf(M,B));2*Abs(VectorOf(M,B))=2*Abs(VectorOf(M,C))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[19, 21], [29, 31], [202, 204]], [[2, 6]], [[8, 11]], [[12, 15]], [[23, 27]], [[2, 22]], [[7, 21]], [[55, 121]], [[127, 199]], [[127, 199]]]",
    "query_spans": "[[[202, 210]]]",
    "process": "Using the symmetry of the ellipse and $\\overrightarrow{MA}+\\overrightarrow{MB}=2\\overrightarrow{MC}$, we obtain: point $C$ coincides with the origin $O$. Let $MA=m$. Using the definition of the ellipse and $|\\overrightarrow{MA}|=2|\\overrightarrow{MB}|=2|\\overrightarrow{MC}|$, we get: $m=\\frac{4a}{3}$, $\\frac{m}{2}=\\frac{2a}{3}$. Then applying the law of cosines to angle $B$ in two triangles and setting up equations, after simplification we obtain: $9c^{2}=6a^{2}$, thus solving the problem. [Detailed explanation] According to the conditions, draw the figure as follows: Since $O$ is the midpoint of $AB$, we have $\\overrightarrow{MA}+\\overrightarrow{MB}=2\\overrightarrow{MO}$. Also $\\overrightarrow{MA}+\\overrightarrow{MB}=2\\overrightarrow{MC}$, so $C$ coincides with the origin $O$. Let $MA=m$, then $MB=\\frac{m}{2}$, $MO=\\frac{m}{2}$. From the definition of the ellipse, we get: $MA+MB=m+\\frac{m}{2}=2a$. Therefore, $m=\\frac{4a}{3}$, $\\frac{m}{2}=\\frac{2a}{3}$. In triangles $\\triangle OBM$ and $\\triangle ABM$, by the law of cosines: $\\cos B = \\frac{(\\frac{2a}{3})^{2}+(2c)^{2}-(\\frac{4a}{3})^{2}}{2\\times2c\\times\\frac{2a}{3}} = \\frac{(\\frac{2a}{3})^{2}+c^{2}-(\\frac{2a}{3})^{2}}{2\\times c\\times\\frac{2a}{3}}$. Simplifying yields: $9c^{2}=6a^{2}$. Hence, $e=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{\\frac{6}{9}}=\\frac{\\sqrt{6}}{3}$."
  },
  {
    "text": "Given that $P$ is a moving point on the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, point $M$ is a moving point on the circle $(x+5)^{2}+y^{2}=4$, and point $N$ is a moving point on the circle $(x-5)^{2}+y^{2}=1$, then the maximum value of $|P M|-|P N|$ is?",
    "fact_expressions": "G: Hyperbola;H: Circle;A: Circle;P: Point;M: Point;N: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Expression(H) = (y^2 + (x + 5)^2 = 4);Expression(A) = (y^2 + (x - 5)^2 = 1);PointOnCurve(P, G);PointOnCurve(M, H);PointOnCurve(N, A)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "9",
    "fact_spans": "[[[6, 45]], [[55, 75]], [[85, 105]], [[2, 5]], [[50, 54]], [[80, 84]], [[6, 45]], [[55, 75]], [[85, 105]], [[2, 49]], [[50, 79]], [[80, 109]]]",
    "query_spans": "[[[111, 130]]]",
    "process": "Let the centers of the two circles be $ C_{1}(-5,0) $, $ C_{2}(5,0) $, which are the left and right foci of a hyperbola. Therefore, $ |PM| - |PN| \\leqslant |PC_{1}| + 2 - (|PC_{2}| - 1) = 2a + 3 = 2 \\times 3 + 3 = 9 $."
  },
  {
    "text": "Point $P$ is a moving point on the parabola $y^{2}=4x$. Then the minimum value of the sum of the distance from point $P$ to the $y$-axis and the distance from point $P$ to point $A(2,3)$ is equal to?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (2, 3);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, yAxis) + Distance(P, A))",
    "answer_expressions": "sqrt(10) - 1",
    "fact_spans": "[[[5, 19]], [[42, 51]], [[0, 4], [25, 29], [37, 41]], [[5, 19]], [[42, 51]], [[0, 23]]]",
    "query_spans": "[[[25, 62]]]",
    "process": "As shown in the figure, F(1,0) and the line l: x = -1 are respectively the focus and directrix of the parabola y^{2} = 4x, and PD \\bot l. Let d be the distance from point P to the y-axis. By the property of the parabola: PF = PD = d + 1, therefore d = PF - 1. Thus, the sum of the distance from point P to the y-axis and the distance from point P to point A(2,3) is AP + d = AP + PF - 1. By the fact that the shortest path between two points is a straight line, AP + d reaches its minimum value when points A, P, F are collinear. At this time, AP + d = AF - 1 = \\sqrt{(2-1)^{2}+(3-0)^{2}} - 1 = \\sqrt{10} - 1. Hence, the distance from point P to the y-axis"
  },
  {
    "text": "Given that the line $x=t$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$. If there exists a point $C$ on the parabola such that $AC \\perp BC$, then the range of values for $t$ is?",
    "fact_expressions": "H: Line;Expression(H) = (x = t);t: Number;G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;B: Point;Intersection(H, G) = {A, B};C: Point;PointOnCurve(C, G) = True;IsPerpendicular(LineSegmentOf(A, C), LineSegmentOf(B, C)) = True",
    "query_expressions": "Range(t)",
    "answer_expressions": "[4,+oo)",
    "fact_spans": "[[[2, 9]], [[2, 9]], [[67, 70]], [[10, 24], [37, 40]], [[10, 24]], [[25, 28]], [[29, 32]], [[2, 34]], [[43, 47]], [[37, 47]], [[50, 65]]]",
    "query_spans": "[[[67, 77]]]",
    "process": "From the given conditions, A(t,2\\sqrt{t}), B(t,-2\\sqrt{t}); let C(m,2\\sqrt{m}) (m\\geqslant0). Since AC\\bot BC, we have \\overrightarrow{AC}\\cdot\\overrightarrow{BC}=0. Therefore, (m-t)^{2}+(2\\sqrt{m}-2\\sqrt{t})(2\\sqrt{m}+2\\sqrt{t})=m^{2}+(4-2t)m+t^{2}-4t=0. Solving this equation yields: m=t (discarded) or m=t-4. From m=t-4\\geqslant0, the range of t is [4,+\\infty)."
  },
  {
    "text": "Given that the focus of the parabola is $F(0,-\\frac{1}{2})$, and the point $P(1, t)$ lies on the parabola, then the distance between point $P$ and $F$ is?",
    "fact_expressions": "G: Parabola;F: Point;Focus(G) = F;Coordinate(F) = (0, -1/2);P: Point;Coordinate(P) = (1, t);t: Number;PointOnCurve(P, G)",
    "query_expressions": "Distance(P, F)",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 5], [40, 43]], [[9, 28], [51, 54]], [[2, 28]], [[9, 28]], [[29, 39], [46, 50]], [[29, 39]], [[30, 39]], [[29, 44]]]",
    "query_spans": "[[[46, 59]]]",
    "process": "The focus of the parabola is $ F(0, -\\frac{1}{2}) $, so the standard equation of the parabola is: $ x^{2} = -2y $. Since point $ P(1, t) $ lies on the parabola, $ 1 = -2t $, solving gives $ t = -\\frac{1}{2} $. Therefore, $ |PF| = |-\\frac{1}{2}| + \\frac{p}{2} = \\frac{1}{2} + \\frac{1}{2} = 1 $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $P$ is a point on the ellipse $C$, and $\\overrightarrow{P F_{1}} \\perp \\overrightarrow{P F_{2}}$. If the area of $\\Delta P F_{1} F_{2}$ is $9$, then $b=$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C);IsPerpendicular(VectorOf(P, F1), VectorOf(P, F2));Area(TriangleOf(P, F1, F2)) = 9",
    "query_expressions": "b",
    "answer_expressions": "3",
    "fact_spans": "[[[18, 75], [85, 90]], [[18, 75]], [[186, 189]], [[25, 75]], [[25, 75]], [[25, 75]], [[2, 9]], [[10, 17]], [[2, 80]], [[81, 84]], [[81, 93]], [[95, 152]], [[155, 184]]]",
    "query_spans": "[[[186, 191]]]",
    "process": "From the given conditions, |PF₁| + |PF₂| = 2a, and vector PF₁ is perpendicular to vector PF₂, so |PF₁|² + |PF₂|² = |F₁F₂|² = 4c². Therefore, (|PF₁| + |PF₂|)² - 2|PF₁||PF₂| = 4c². Thus, 2|PF₁||PF₂| = 4a² - 4c² = 4b², so |PF₁||PF₂| = 2b². Hence, SΔPF₁F₂ = ½|PF₁||PF₂| = ½ × 2b² = b² = 9, so b = 3."
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{a^{2}} +\\frac{y^{2}}{b^{2}}=1(a>b>0)$ satisfies $a \\leq \\sqrt{3} b$, with eccentricity $e$. Then the maximum value of $e^{2}$ is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;e: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a <= sqrt(3)*b;Eccentricity(G) = e",
    "query_expressions": "Max(e^2)",
    "answer_expressions": "2/3",
    "fact_spans": "[[[0, 53]], [[2, 53]], [[2, 53]], [[79, 82]], [[2, 53]], [[2, 53]], [[0, 53]], [[55, 74]], [[0, 82]]]",
    "query_spans": "[[[84, 97]]]",
    "process": ""
  },
  {
    "text": "The distance from any point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{t}=1$ to one of its foci is always greater than $1$. Then, the range of values for $t$ is?",
    "fact_expressions": "G: Ellipse;t: Number;P:Point;Expression(G)=(x^2/4 + y^2/t=1);PointOnCurve(P,G);Distance(P,OneOf(Focus(G)))>1",
    "query_expressions": "Range(t)",
    "answer_expressions": "(3,4)+(4,25/4)",
    "fact_spans": "[[[0, 37]], [[60, 63]], [], [[0, 37]], [[0, 42]], [[0, 58]]]",
    "query_spans": "[[[60, 70]]]",
    "process": "When $t>4$, the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{t}=1$ represents an ellipse with foci on the $y$-axis, then $a=\\sqrt{t}$, $b=2$, $c=\\sqrt{t-4}$. From the given condition we have $a-c=\\sqrt{t}-\\sqrt{t-4}>1$, solving gives $4<t<\\frac{25}{4}$; when $0<t<4$, the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{t}=1$ represents an ellipse with foci on the $x$-axis, then $a=2$, $b=\\sqrt{t}$, $c=\\sqrt{4-t}$. From the given condition we have $a-c=2-\\sqrt{4-t}>1$, solving gives $3<t<4$; combining the above, the range of real number $t$ is $(3,4)\\cup(4,\\frac{25}{4})$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has its right focus at $F(2,0)$, and the distance from point $F$ to its asymptote is $1$. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (2, 0);RightFocus(G) = F;Distance(F,Asymptote(G))=1",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 58], [77, 78], [90, 93]], [[5, 58]], [[5, 58]], [[63, 71], [72, 76]], [[5, 58]], [[5, 58]], [[2, 58]], [[63, 71]], [[2, 71]], [[72, 88]]]",
    "query_spans": "[[[90, 99]]]",
    "process": "By the given condition, c=2, the asymptotes are y=\\pm\\frac{b}{a}x'\\therefore\\begin{cases}a2+b^{2}=4\\\\\\frac{2b}{\\sqrt{a2+b^{2}}}=1'\\end{cases} solving gives b=1,a=\\sqrt{3},\\therefore_{e}=\\frac{c}{a}=\\frac{2\\sqrt{3}}{3}."
  },
  {
    "text": "Given that the coordinates of point $M$ are $(1,0)$, and points $A$ and $B$ lie on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$ such that $MA \\perp MB$, then the minimum value of $\\overrightarrow{MA} \\cdot \\overrightarrow{BA}$ is?",
    "fact_expressions": "G: Ellipse;M: Point;A: Point;B: Point;Expression(G) = (x^2/16 + y^2/4 = 1);Coordinate(M) = (1, 0);PointOnCurve(A, G);PointOnCurve(B, G);IsPerpendicular(LineSegmentOf(M, A), LineSegmentOf(M, B))",
    "query_expressions": "Min(DotProduct(VectorOf(M, A), VectorOf(B, A)))",
    "answer_expressions": "11/3",
    "fact_spans": "[[[30, 68]], [[2, 6]], [[20, 23]], [[25, 29]], [[30, 68]], [[2, 17]], [[20, 72]], [[20, 72]], [[76, 91]]]",
    "query_spans": "[[[93, 148]]]",
    "process": ""
  },
  {
    "text": "If the difference between the distance from point $M$ on the parabola $C$: $y^{2}=2 p x$ ($p>0$) to the focus $F$ and the distance from $M$ to the $y$-axis is $2$, then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;p>0;Expression(C) = (y^2 = 2*(p*x));PointOnCurve(M,C);Focus(C)=F;M:Point;F:Point;Distance(M,F)-Distance(M,yAxis)=2",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 27]], [[59, 62]], [[9, 27]], [[1, 27]], [[1, 33]], [[1, 39]], [[29, 33]], [[36, 39]], [[29, 57]]]",
    "query_spans": "[[[59, 64]]]",
    "process": "As shown in the figure, the parabola $ y^{2} = 2px $ ($ p > 0 $) has focus at $ F\\left(\\frac{p}{2}, 0\\right) $, and the directrix equation is $ l: x = -\\frac{p}{2} $. Draw $ MB \\perp l $, intersecting the $ y $-axis at point $ A $. Since for a point $ M $ on the parabola $ C $, the difference between the distance to the focus $ F $ and the distance to the $ y $-axis is 2, that is, $ |MF| - |MA| = 2 $. According to the definition of a parabola, we have $ |MF| = |MB| $, so $ |MF| - |MA| = |MB| - |MA| = |AB| = 2 $, thus $ \\frac{p}{2} = 2 $, solving gives $ p = 4 $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2x$, a line $l$ passing through the point $P(4, 1)$ intersects the parabola $C$ at points $A$ and $B$, with point $P$ being exactly the midpoint of $AB$. Let $F$ be the focus of the parabola. Then $|\\overrightarrow{AF}| + |\\overrightarrow{BF}| = $?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;P: Point;F: Point;Expression(C) = (y^2 = 2*x);Coordinate(P) = (4, 1);PointOnCurve(P, l);Intersection(l, C) = {A, B};MidPoint(LineSegmentOf(A, B)) = P;Focus(C) = F",
    "query_expressions": "Abs(VectorOf(A, F)) + Abs(VectorOf(B, F))",
    "answer_expressions": "9",
    "fact_spans": "[[[35, 40]], [[2, 20], [41, 47], [80, 83]], [[50, 53]], [[54, 57]], [[23, 34], [61, 65]], [[76, 79]], [[2, 20]], [[23, 34]], [[21, 40]], [[35, 59]], [[61, 75]], [[76, 86]]]",
    "query_spans": "[[[88, 137]]]",
    "process": "From the parabola $ C: y^{2} = 2x $, we know that $ 2p = 2 $, then $ \\frac{p}{2} = \\frac{1}{2} $. Therefore, the focus of the parabola $ C: y^{2} = 2x $ has coordinates $ F\\left(\\frac{1}{2}, 0\\right) $. As shown in the figure, draw $ AM $ perpendicular to the directrix intersecting it at $ M $, draw $ BN $ perpendicular to the directrix intersecting it at $ N $, and draw $ PK $ perpendicular to the directrix intersecting it at $ K $. By the definition of a parabola, we have $ |AM| + |BN| = |AF| + |BF| $. Since $ P(4,1) $ is the midpoint of segment $ AB $, and quadrilateral $ AMNB $ is a trapezoid, by the midline theorem of a trapezoid, we obtain $ \\frac{1}{2}(|AM| + |BN|) = |PK| = 4 + \\frac{1}{2} = \\frac{9}{2} $. Then $ |AM| + |BN| = 9 $, so $ |AF| + |BF| = 9 $."
  },
  {
    "text": "Given that $P$ is a point on the parabola $y^{2}=4x$, $A(2, 2)$ is a fixed point in the plane, and $F$ is the focus of the parabola, find the coordinates of point $P$ such that $|PA|+|PF|$ is minimized.",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G);A: Point;Coordinate(A) = (2, 2);F: Point;Focus(G) = F;WhenMin(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1, 2)",
    "fact_spans": "[[[6, 20], [48, 51]], [[6, 20]], [[2, 5], [56, 60]], [[2, 24]], [[25, 35]], [[25, 35]], [[44, 47]], [[44, 54]], [[66, 79]]]",
    "query_spans": "[[[56, 64]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$, the line $l$: $y=-\\sqrt{3} x+\\sqrt{3} a$ intersects the right branch of $C$ at points $A$ and $B$, and intersects the $y$-axis at point $M$. If $\\overrightarrow{A B}=-2 \\overrightarrow{A M}$, then the asymptotes of $C$ are?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;l: Line;Expression(l)=(y=-sqrt(3)*a+sqrt(3)*x);Intersection(l, RightPart(C)) = {A, B};A: Point;B: Point;Intersection(l, yAxis) = M;M: Point;VectorOf(A, B) = -2*VectorOf(A, M)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*sqrt(6)*x/2",
    "fact_spans": "[[[2, 66], [101, 104], [185, 188]], [[2, 66]], [[10, 66]], [[10, 66]], [[10, 66]], [[10, 66]], [[67, 100]], [[67, 100]], [[67, 121]], [[111, 115]], [[118, 121]], [[67, 134]], [[130, 134]], [[137, 183]]]",
    "query_spans": "[[[185, 196]]]",
    "process": "This problem examines the equation and geometric properties of a hyperbola. As shown in the figure, draw $BD\\bot x$-axis, with foot at $D$. The line $y=-\\sqrt{3}(x-a)$ passes through $A(a,0)$, i.e., the right vertex of $C$, and the inclination angle of line $l$ is $\\frac{2\\pi}{3}$, so $\\angle MAO = \\angle BAD = \\frac{\\pi}{3}$. In right triangle $\\triangle AOM$, $|OA|=a$, then $|AM|=2a$, $|OM|=\\sqrt{3}a$. Since $|AB|=2|AM|$ and $\\triangle MAO \\sim \\triangle BAD$, it follows that $|AD|=2a$, $|BD|=2\\sqrt{3}a$, thus $B(3a,-2\\sqrt{3}a)$. Therefore, $\\frac{9a^{2}}{a^{2}} - \\frac{12a^{2}}{b^{2}} = 1$, solving gives $\\left(\\frac{b}{a}\\right)^{2} = \\frac{3}{2}$, then the asymptotes of $C$ are given by"
  },
  {
    "text": "Given that $F$ is the left focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, $A(1,4)$, and $P$ is a moving point on the right branch of the hyperbola, then the minimum value of $|P F|+|P A|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/12 = 1);F: Point;LeftFocus(G) = F;A: Point;Coordinate(A) = (1, 4);P: Point;PointOnCurve(P, RightPart(G)) = True",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "9",
    "fact_spans": "[[[7, 46], [65, 68]], [[7, 46]], [[2, 5]], [[2, 50]], [[51, 59]], [[51, 59]], [[61, 64]], [[61, 74]]]",
    "query_spans": "[[[76, 95]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of the hyperbola are $3x \\pm 4y = 0$ and the coordinates of the foci are $(\\pm 5, 0)$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Coordinate(Focus(G)) = (pm*5, 0);Expression(Asymptote(G)) = ( 3*x+pm*4*y= 0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16 - y^2/9 = 1",
    "fact_spans": "[[[2, 5], [46, 49]], [[2, 44]], [[2, 27]]]",
    "query_spans": "[[[46, 54]]]",
    "process": "Analysis: First, use the asymptote equations of the hyperbola to set up the equation of the hyperbola. Then, use the coordinates of the foci to determine the relevant coefficients. Transform $3x\\pm4y=0$ into $\\frac{x}{4}\\pm\\frac{y}{3}=0$, and let the hyperbola with asymptotes $\\frac{x}{4}\\pm\\frac{y}{3}=0$ be given by $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=\\lambda$. Since the foci of this hyperbola are $(\\pm5,0)$, it follows that $16\\lambda+9\\lambda=25$, solving which gives $\\lambda=1$. Thus, the equation of the hyperbola is $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. Suppose a line $l$ passes through $F_{1}$, intersects the ellipse $C$ at points $A$ and $B$, and intersects the $y$-axis at point $P$. If $\\overrightarrow{F_{1} P}=\\frac{3}{2} \\overrightarrow{A F_{1}}$ and $\\angle P F_{1} F_{2}=30^{\\circ}$, then the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "F1: Point;F2: Point;C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;LeftFocus(C) = F1;RightFocus(C) = F2;l: Line;PointOnCurve(F1, l);Intersection(l, C) = {A, B};A: Point;B: Point;Intersection(l, yAxis) = P;P: Point;VectorOf(F1, P) = (3/2)*VectorOf(A, F1);AngleOf(P, F1, F2) = ApplyUnit(30, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[1, 8], [89, 96]], [[9, 16]], [[19, 76], [97, 102], [227, 232]], [[19, 76]], [[25, 76]], [[25, 76]], [[25, 76]], [[25, 76]], [[1, 82]], [[1, 82]], [[83, 88]], [[83, 96]], [[83, 112]], [[103, 106]], [[107, 110]], [[83, 123]], [[119, 123]], [[127, 190]], [[192, 225]]]",
    "query_spans": "[[[227, 238]]]",
    "process": "Let the coordinates of point A be $(x_{1},y_{1})$, $F_{1}(-c,0)$, $F_{2}(c,0)$, $\\overrightarrow{F_{1}P}=\\frac{3}{2}\\overrightarrow{AF_{1}}$, so we have $0-(-c)=\\frac{3}{2}(-c-x_{1})$, solving gives: $x_{1}=-\\frac{5}{3}c$. Since $\\angle PF_{1}F_{2}=30^{\\circ}$, the equation of the line is $y=\\frac{\\sqrt{3}}{3}(x+c)$, so point A is $(-\\frac{5}{3}c,-\\frac{2\\sqrt{3}}{9})$, thus $|AF_{1}|=\\sqrt{(-\\frac{5}{3}c+c)^{2}+\\frac{4}{27}c^{2}}=\\frac{4\\sqrt{3}}{9}c$, $|AF_{2}|=\\sqrt{(-\\frac{5}{3}c-c)^{2}+\\frac{4}{27}c^{2}}=\\frac{14\\sqrt{3}}{9}c$, so $2a=|PF_{1}|+|PF_{2}|=\\frac{18\\sqrt{3}}{9}c$, hence $e=\\frac{2c}{2a}=\\frac{1}{\\sqrt{3}}=\\frac{\\sqrt{3}}{3}$"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{7}=1$, with left and right foci $F_{1}$ and $F_{2}$ respectively. There is a point $P$ on hyperbola $C$. If $|P F_{1}|=5$, then $|P F_{2}|=$?",
    "fact_expressions": "C: Hyperbola;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/9 - y^2/7 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 5",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "11",
    "fact_spans": "[[[2, 45], [71, 77]], [[81, 84]], [[55, 62]], [[63, 70]], [[2, 45]], [[2, 70]], [[2, 70]], [[71, 84]], [[86, 99]]]",
    "query_spans": "[[[101, 114]]]",
    "process": "Since |PF_{1}| = 5 < a + c = 7, it follows that |PF_{2}| - |PF_{1}| = 2a = 6, hence |PF_{2}| = 11."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, one of its asymptotes has the equation $x-\\sqrt{2} y=0$. Let $P$ be a point on $C$ such that the minimum value of $|O P|$ equals $2$. Then the standard equation of this hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;O: Origin;P: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(C))) = (x - sqrt(2)*y = 0);PointOnCurve(P, C);Min(Abs(LineSegmentOf(O, P))) = 2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4 - y^2/2 = 1",
    "fact_spans": "[[[2, 63], [95, 98], [123, 126]], [[10, 63]], [[10, 63]], [[103, 110]], [[91, 94]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 89]], [[91, 101]], [[103, 119]]]",
    "query_spans": "[[[123, 133]]]",
    "process": "From the given conditions, we have \\frac{b}{a}=\\frac{1}{\\sqrt{2}}, a=2, hence a=2, b=\\sqrt{2}, so its standard equation is \\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1."
  },
  {
    "text": "A moving circle passes through the fixed point $(0,-2)$ and is externally tangent to the fixed circle $x^{2}+(y-2)^{2}=4$. What is the trajectory equation of the center of the moving circle?",
    "fact_expressions": "K: Circle;H: Point;Coordinate(H) = (0, -2);PointOnCurve(H, K);G: Circle;Expression(G) = (x^2 + (y - 2)^2 = 4);IsOutTangent(K, G)",
    "query_expressions": "LocusEquation(Center(K))",
    "answer_expressions": "(y^2-x^2/3=1)&(y<0)",
    "fact_spans": "[[[0, 2], [40, 42]], [[5, 13]], [[5, 13]], [[0, 13]], [[16, 35]], [[16, 35]], [[0, 38]]]",
    "query_spans": "[[[40, 51]]]",
    "process": "Let the radius of the moving circle M be r, and it is externally tangent to the fixed circle C at point T. Let B(0,-2). According to the problem, |MB| = |MT| = r, |CT| = 2. Since the moving circle M is externally tangent to the fixed circle C, |MC| - |MB| = |CT| = 2 (a constant). Also, 2 < |BC| = 4. By the definition of a hyperbola, the locus of the moving point M is one branch of a hyperbola with foci at points B and C, where 2a = 2. Moreover, 2c = |BC| = 4, so c = 2, a = 1. Therefore, b^{2} = 3. The trajectory equation of the center M of the moving circle is y_{2} - \\frac{x^{2}}{3} = 1 (y < 0)."
  },
  {
    "text": "Given that the distance from a moving point $M$ to the point $A(1,0)$ is equal to its distance to the line $x=1$, then the equation of the trajectory of point $M$ is?",
    "fact_expressions": "G: Line;A: Point;Expression(G) = (x = 1);Coordinate(A) = (1, 0);M: Point;Distance(M, A) = Distance(M, G)",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "y^2 = 4*x",
    "fact_spans": "[[[24, 31]], [[8, 17]], [[24, 31]], [[8, 17]], [[4, 7], [22, 23], [36, 40]], [[4, 34]]]",
    "query_spans": "[[[36, 47]]]",
    "process": ""
  },
  {
    "text": "Let $F$ be the focus of the parabola $y=-\\frac{1}{4} x^{2}$, and let line $l$, tangent to the parabola at point $P(-4,-4)$, intersect the $x$-axis at point $Q$. Then $\\angle P Q F$=?",
    "fact_expressions": "l: Line;G: Parabola;P: Point;Q: Point;F: Point;Expression(G) = (y = -x^2/4);Coordinate(P) = (-4, -4);Focus(G) = F;TangentPoint(l, G) = P;Intersection(l, xAxis) = Q",
    "query_expressions": "AngleOf(P, Q, F)",
    "answer_expressions": "pi/2",
    "fact_spans": "[[[53, 58]], [[5, 30], [35, 38]], [[41, 52]], [[65, 69]], [[1, 4]], [[5, 30]], [[41, 52]], [[1, 33]], [[34, 58]], [[53, 69]]]",
    "query_spans": "[[[71, 87]]]",
    "process": "The focus of the parabola $x^{2}=-4y$ is $F(0,-1)$. Let the equation of the line passing through the tangent point $P(-4,-4)$ be $y=k(x+4)-4$. Then, solving the system of equations\n\\[\n\\begin{cases}\ny=k(x+4)-4 \\\\\nx^{2}=-4y\n\\end{cases}\n\\]\nyields $x^{2}+4kx+16k-16=0$. Therefore, since the line is tangent to the parabola, we have $16k^{2}-4(16k-16)=0$, solving which gives $k=2$. Hence, the equation of the line passing through the tangent point $P(-4,-4)$ and tangent to the parabola is: $y=2x+4$. Thus, $Q(-2,0)$, and at this point $k_{PQ}=\\frac{-4}{-4-(-2)}=2$, $k_{QF}=\\frac{0-(-1)}{-2-0}=-\\frac{1}{2}$. Since $k_{PQ}\\cdot k_{QF}=-1$, it follows that $PQ\\perp QF$, so $\\angle PQF=\\frac{\\pi}{2}$."
  },
  {
    "text": "If the line $y=k x-1$ and the parabola $y^{2}=4 x$ have exactly one common point, then the real number $k$=?",
    "fact_expressions": "E: Parabola;F: Line;k: Real;Expression(E) = (y^2 = 4*x);Expression(F) = (y = k*x - 1);NumIntersection(F, E) = 1",
    "query_expressions": "k",
    "answer_expressions": "{-1, 0}",
    "fact_spans": "[[[13, 27]], [[1, 12]], [[38, 43]], [[13, 27]], [[1, 12]], [[1, 36]]]",
    "query_spans": "[[[38, 45]]]",
    "process": "When $ k=0 $, combining numerical and graphical analysis shows that the line and the parabola have one common point; when $ k\\neq0 $, solving the system of equations formed by the line equation and the parabola equation, using $ A=0 $, yields the solution and thus the answer. [Detailed Solution] When $ k=0 $, combining numerical and graphical analysis shows that the line and the parabola have one common point; when $ k\\neq0 $, combining the equations gives  \n$$\n\\begin{cases}\ny = kx - 1 \\\\\ny^2 = 4x\n\\end{cases}\n$$\nleading to $ y^{2} - \\frac{4}{k}y - \\frac{4}{k} = 0 $, hence $ A = \\frac{16}{l^{2}} + \\frac{16}{k} = 0 $, so $ k = -1 $. Thus, $ k = -1 $ or $ 0 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a, b>0)$ has eccentricity $\\frac{\\sqrt{6}}{2}$, then the minimum value of $\\frac{a^{2}+4}{b}$ is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = sqrt(6)/2",
    "query_expressions": "Min((a^2 + 4)/b)",
    "answer_expressions": "4*sqrt(2)",
    "fact_spans": "[[[2, 56]], [[84, 103]], [[84, 103]], [[5, 56]], [[5, 56]], [[2, 56]], [[2, 81]]]",
    "query_spans": "[[[84, 109]]]",
    "process": "From $\\frac{c}{a}=\\frac{\\sqrt{6}}{2}$, we obtain $a=\\sqrt{2}b$, so $\\frac{a2+4}{b}=\\frac{2b^{2}+4}{b}=2(b+\\frac{2}{b})\\geqslant2\\cdot2\\sqrt{2}$. Hence, the minimum value of $\\frac{a2+4}{b}$ is $4\\sqrt{2}$."
  },
  {
    "text": "Given the hyperbola $\\frac{y^{2}}{9}-\\frac{x^{2}}{a}=1$ has asymptotes with equations $3 x \\pm 2 y=0$, then the value of $a$ is?",
    "fact_expressions": "G: Hyperbola;a: Number;Expression(G) = (y^2/9 - x^2/a = 1);Expression(Asymptote(G)) = (3*x+pm*2*y=0)",
    "query_expressions": "a",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 39]], [[63, 66]], [[1, 39]], [[1, 61]]]",
    "query_spans": "[[[63, 70]]]",
    "process": "From the hyperbola $\\frac{y^{2}}{9}-\\frac{x^{2}}{a}=1$, its asymptotes are given by $3x\\pm\\sqrt{a}y=0$. Since the asymptotes of the hyperbola $\\frac{y^{2}}{9}-\\frac{x^{2}}{a}=1$ are $3x\\pm2y=0$, it follows that $\\sqrt{a}=2$, so $a=4$."
  },
  {
    "text": "Given the equation of the ellipse is $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{25}=1$ $(a>5)$, its two foci are $F_{1}$ and $F_{2}$, and $|F_{1} F_{2}|=8$. The chord $A B$ passes through $F_{1}$. Then, the perimeter of $\\triangle A B F_{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/25 + x^2/a^2 = 1);a: Number;a > 5;F1: Point;F2: Point;Focus(G) = {F1, F2};Abs(LineSegmentOf(F1, F2)) = 8;A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G) = True;PointOnCurve(F1, LineSegmentOf(A, B)) = True",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "4*sqrt(41)",
    "fact_spans": "[[[2, 4], [54, 55]], [[2, 53]], [[8, 53]], [[8, 53]], [[63, 70], [105, 112]], [[71, 78]], [[54, 78]], [[80, 97]], [[99, 104]], [[99, 104]], [[54, 104]], [[99, 112]]]",
    "query_spans": "[[[114, 140]]]",
    "process": "According to the problem, the equation of the ellipse is $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{25}=1$ $(a>5)$, and the focal distance is 8, which means $c=4$. It follows that $-25+a^{2}=16$, so $a^{2}=41$. Therefore, the perimeter of $4ABF_{2}$ is twice the sum of the distances from a point on the ellipse to the two foci, which is $4a=4\\sqrt{41}$."
  },
  {
    "text": "The line $l$: $x = m y + 2$ passes through the focus $F$ of the parabola $C$: $y^2 = 2 p x$ $(p > 0)$, and intersects the parabola at points $A$ and $B$. A line passing through the origin goes through the midpoint $D$ of chord $AB$, and intersects the parabola at point $E$ (distinct from the origin). Then the range of $\\frac{|O E|}{|O D|}$ is?",
    "fact_expressions": "l: Line;Expression(l) = (x = m*y + 2);m: Number;C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, C) = {A, B};Z: Line;O: Origin;PointOnCurve(O, Z);IsChordOf(LineSegmentOf(A, B), C);D: Point;MidPoint(LineSegmentOf(A, B)) = D;PointOnCurve(D, Z);E: Point;Negation(E=O);Intersection(Z, C) = E",
    "query_expressions": "Range(Abs(LineSegmentOf(O, E))/Abs(LineSegmentOf(O, D)))",
    "answer_expressions": "(2, +oo)",
    "fact_spans": "[[[0, 16]], [[0, 16]], [[7, 16]], [[18, 46], [54, 57], [94, 97]], [[18, 46]], [[26, 46]], [[26, 46]], [[49, 52]], [[18, 52]], [[0, 52]], [[60, 63]], [[64, 67]], [[0, 69]], [[74, 76]], [[71, 73], [106, 108]], [[70, 76]], [[54, 84]], [[87, 90]], [[79, 90]], [[74, 90]], [[99, 103]], [[99, 109]], [[74, 103]]]",
    "query_spans": "[[[111, 139]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{6}=1$, respectively, a line passing through $F_{2}$ intersects the right branch of the hyperbola at points $A$ and $B$ (where point $A$ lies in the first quadrant). A circle $C$ is inscribed in $\\Delta A F_{1} F_{2}$ with radius $r$. Then the range of values for $r$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/2 - y^2/6 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;L: Line;PointOnCurve(F2, L) = True;Intersection(L, RightPart(G)) = {A, B};A: Point;B: Point;Quadrant(A) = 1;C: Circle;IsInTangent(C, TriangleOf(A, F1, F2)) = True;Radius(C) = r;r: Number",
    "query_expressions": "Range(r)",
    "answer_expressions": "(\\sqrt{6}/3,\\sqrt{6})",
    "fact_spans": "[[[20, 58], [78, 81]], [[20, 58]], [[2, 9]], [[10, 17], [67, 74]], [[2, 65]], [[2, 65]], [[75, 77]], [[66, 77]], [[75, 95]], [[86, 89], [99, 103]], [[90, 93]], [[99, 109]], [[111, 115]], [[111, 140]], [[111, 147]], [[144, 147], [149, 152]]]",
    "query_spans": "[[[149, 159]]]",
    "process": "From the hyperbola equation: the real semi-axis length is $ a=\\sqrt{2} $, the imaginary semi-axis length is $ b=\\sqrt{6} $, $ F_{2}(c,0) $ and $ c=2\\sqrt{2} $. Let circle $ C $ be tangent to $ \\triangle AF_{1}F_{2} $ at points $ M $, $ N $, $ E $, respectively, as shown in the figure below. By the tangent properties of a circle: $ |AN|=|AM| $, $ |F_{1}N|=|F_{1}E| $, $ |F_{2}M|=|F_{2}E| $. From the definition of the hyperbola: $ |AF_{1}|-|AF_{2}|=2a=|F_{1}N|-|F_{2}M| $, that is, $ |F_{1}E|-|F_{2}E|=2a $. Let $ E(x_{0},0) $, then $ x_{0}+c-(c-x_{0})=2a $, solving gives: $ x_{0}=a $. From the tangent property, we know: both $ C $ and $ E $ have the same horizontal coordinate $ a $. From the property of the incircle of a triangle: $ CF_{2} $ is the angle bisector of $ \\angle AF_{2}F_{1} $. Let the inclination angle of line $ AB $ be $ \\theta $, then $ \\angle CF_{2}E=\\frac{\\pi-\\theta}{2} $. Since $ |EF_{2}|=c-a=\\sqrt{2} $, $ r=|CE|=|EF_{2}|\\cdot\\tan\\angle CF_{2}E=(c-a)\\tan(\\frac{\\pi-\\theta}{2})=\\sqrt{2}\\cdot\\frac{1}{\\tan\\frac{\\theta}{2}} $. Since the hyperbola $ \\frac{x^{2}}{2}-\\frac{y^{2}}{6}=1 $ has asymptotes: $ y=\\pm\\sqrt{3}x $, their inclination angles are $ \\frac{\\pi}{3} $ and $ \\frac{2\\pi}{3} $, respectively. Since line $ AB $ intersects the right branch of the hyperbola at two points $ A $, $ B $, the inclination angle $ \\theta $ of line $ AB $ lies in $ (\\frac{\\pi}{3},\\frac{2\\pi}{3}) $. Then $ \\frac{\\theta}{2}\\in(\\frac{\\pi}{6},\\frac{\\pi}{3}) $, so $ \\tan\\frac{\\theta}{2}\\in(\\frac{\\sqrt{3}}{3},\\sqrt{3}) $. Therefore, $ r=|CE|=\\sqrt{2}\\cdot\\frac{1}{\\tan\\frac{\\theta}{2}}\\in(\\frac{\\sqrt{6}}{3},\\sqrt{6}) $."
  },
  {
    "text": "A point $M$ on the parabola $y=4 x^{2}$ is at a distance of $1$ from the focus. What is the distance from point $M$ to the $x$-axis?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y = 4*x^2);PointOnCurve(M, G);Distance(M, Focus(G)) = 1",
    "query_expressions": "Distance(M, xAxis)",
    "answer_expressions": "15/16",
    "fact_spans": "[[[0, 14]], [[17, 20], [32, 36]], [[0, 14]], [[0, 20]], [[0, 30]]]",
    "query_spans": "[[[32, 46]]]",
    "process": "y=4x^{2} is transformed into x^{2}=\\frac{1}{4}y, so the focus of the parabola is F(0,\\frac{1}{16}). Let point M(x,y), then y+\\frac{1}{16}=1, thus y=\\frac{15}{16}, so the distance from point M to the x-axis is \\frac{15}{16}."
  },
  {
    "text": "The line $l$ passes through the focus $F$ of the parabola $y^{2}=4x$, and intersects the parabola at points $A$ and $B$. If $\\overrightarrow{A F}=3 \\overrightarrow{F B}$, then the slope of line $l$ is?",
    "fact_expressions": "l: Line;G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;PointOnCurve(F, l) = True;Intersection(l, G) = {A, B};A: Point;B: Point;VectorOf(A, F) = 3*VectorOf(F, B)",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*sqrt(3)",
    "fact_spans": "[[[0, 5], [91, 96]], [[6, 20], [28, 31]], [[6, 20]], [[23, 26]], [[6, 26]], [[0, 26]], [[0, 42]], [[33, 36]], [[37, 40]], [[44, 89]]]",
    "query_spans": "[[[91, 101]]]",
    "process": "As shown in the figure, when point A is in the first quadrant, let M and N be the projections of points A and B on the directrix of the parabola, respectively. Draw BK perpendicular to AM from point B. In triangle ABK, ∠BAK equals the inclination angle of line AB, and its tangent value is k. According to the definition of the parabola, let |BF| = n, then |AF| = |AM| = 3n, |BF| = |BN| = n, ∴ |AK| = 2n. In right triangle ABK, |BK| = \\sqrt{|AB|^{2}-|AK|^{2}} = 2\\sqrt{3}n, so \\tan\\angleBAK = \\frac{|BK|}{|AK|} = \\sqrt{3}, thus the slope k of line l is \\sqrt{3}. When A is in the fourth quadrant, similarly, the slope k of line l is -\\sqrt{3}. Therefore, the slope of the line is \\pm\\sqrt{3}."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $A$, $B$ are two points on the ellipse, the perpendicular bisector of segment $AB$ intersects the $x$-axis at point $P(x_{0}, 0)$, then the range of $x_{0}$ is (expressed in terms of $a$, $b$)?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>b;b>0;A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);x0: Number;P: Point;Coordinate(P) = (x0, 0);Intersection(PerpendicularBisector(LineSegmentOf(A, B)), xAxis) = P",
    "query_expressions": "Range(x0)",
    "answer_expressions": "(-(a^2-b^2)/a, (a^2-b^2)/a)",
    "fact_spans": "[[[2, 54], [65, 67]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[56, 59]], [[61, 64]], [[56, 71]], [[56, 71]], [[109, 116]], [[93, 107]], [[93, 107]], [[72, 107]]]",
    "query_spans": "[[[109, 137]]]",
    "process": "Let the coordinates of A and B be $(x_{1},y_{1})$ and $(x_{2},y_{2})$. Since the perpendicular bisector of segment AB intersects the x-axis, AB is not parallel to the y-axis, i.e., $x_{1} \\ne x_{2}$. Also, the intersection point is $P(x_{0},0)$, so $|PA|=|PB|$, that is, $(x_{1}-x_{0})^{2}+y_{1}=(x_{2}-x_{0})^{2}+y_{2} \\cdots\\cdots \\textcircled{1}$. Since A and B lie on the ellipse, $\\therefore y_{1}^{2}=b^{2}-\\frac{b^{2}}{a^{2}}x_{1}^{2}, y_{2}^{2}=b^{2}-\\frac{b^{2}}{a^{2}}x_{2}^{2}$. Substituting the above expressions into $\\textcircled{1}$, we get $2(x_{2}-x_{1})x_{0}=(x_{2}^{2}-x_{1}^{2})\\frac{a^{2}-b^{2}}{a^{2}} \\cdots\\cdots \\textcircled{2}$. Since $x_{1} \\ne x_{2}$, we obtain $x_{0}=\\frac{x_{1}+x_{2}}{2} \\cdot \\frac{a^{2}-b^{2}}{a^{2}} \\cdots\\cdots \\textcircled{3}$. Since $-a \\le x_{1} \\le a,\\ -a \\le x_{2} \\le a$, and $x_{1} \\ne x_{2}$, $\\therefore -2a < x_{1}+x_{2} < 2a$, $\\therefore -\\frac{a^{2}-b^{2}}{a} < x_{0} < \\frac{a^{2}-b^{2}}{a}$, i.e., the answer is $\\left(-\\frac{a^{2}-b^{2}}{a},\\frac{a^{2}-b^{2}}{a}\\right)$."
  },
  {
    "text": "Given the hyperbola $\\frac{y^{2}}{4}-\\frac{x^{2}}{5}=1$, what is the distance from the center $C$ of the circle $x^{2}+y^{2}-6 x+8=0$ to the asymptotes of the hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/5 + y^2/4 = 1);H: Circle;Expression(H) = (-6*x + x^2 + y^2 + 8 = 0);C: Point;Center(H) = C",
    "query_expressions": "Distance(C, Asymptote(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 40], [71, 74]], [[2, 40]], [[42, 64]], [[42, 64]], [[67, 70]], [[42, 70]]]",
    "query_spans": "[[[67, 82]]]",
    "process": "The center of the circle $x^{2}+y^{2}-6x+8=0$ is $(3,0)$. The asymptotes of the hyperbola $\\frac{y^{2}}{4}-\\frac{x^{2}}{5}=1$ are given by $y=\\frac{2}{\\sqrt{5}}x$, or $2x-\\sqrt{5}y=0$. Therefore, the distance from the center $C$ to the asymptote of the hyperbola is $\\frac{|6|}{\\sqrt{4+5}}=2$."
  },
  {
    "text": "The standard equation of an ellipse with a focal length of $8$, minor axis length of $6$, and foci on the $x$-axis is?",
    "fact_expressions": "G: Ellipse;FocalLength(G) = 8;Length(MinorAxis(G)) = 6;PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/25 + y^2/9 = 1",
    "fact_spans": "[[[25, 27]], [[0, 27]], [[7, 27]], [[16, 27]]]",
    "query_spans": "[[[25, 34]]]",
    "process": "According to the problem, the ellipse has a focal distance of 8 and a minor axis length of 6, so 2c=8, 2b=6. Solving gives c=4, b=3, then a=\\sqrt{9+16}=5. Since the foci of the ellipse lie on the x-axis, its standard equation is: \\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1"
  },
  {
    "text": "The parabola $y^{2}=4x$, $O$ is the origin, $A$ and $B$ are two moving points on the parabola such that $OA \\perp OB$. When the inclination angle of line $AB$ is $45^{\\circ}$, the area of $\\triangle AOB$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);O: Origin;A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);IsPerpendicular(LineSegmentOf(O, A), LineSegmentOf(O, B));Inclination(LineOf(A, B)) = ApplyUnit(45, degree)",
    "query_expressions": "Area(TriangleOf(A, O, B))",
    "answer_expressions": "8*sqrt(5)",
    "fact_spans": "[[[0, 14], [34, 37]], [[0, 14]], [[16, 20]], [[26, 29]], [[30, 33]], [[26, 42]], [[26, 42]], [[44, 59]], [[61, 85]]]",
    "query_spans": "[[[87, 107]]]",
    "process": ""
  },
  {
    "text": "Given that the minor axis of ellipse $E$ has length $6$, and the distance from focus $F$ to one endpoint of the major axis is $9$, then the eccentricity of ellipse $E$ equals?",
    "fact_expressions": "E: Ellipse;Length(MinorAxis(E))=6;F:Point;OneOf(Focus(E))=F;Distance(F,OneOf(Endpoint(MajorAxis(E))))=9",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "4/5",
    "fact_spans": "[[[2, 7], [39, 44]], [[2, 15]], [[18, 21]], [[2, 21]], [[2, 37]]]",
    "query_spans": "[[[39, 51]]]",
    "process": ""
  },
  {
    "text": "Given that vertices $B$ and $C$ of $\\triangle ABC$ lie on the ellipse $\\frac{x^{2}}{3}+y^{2}=1$, vertex $A$ is one of the foci of the ellipse, and the other focus of the ellipse lies on side $BC$, then what is the perimeter of $\\triangle ABC$?",
    "fact_expressions": "G: Ellipse;B: Point;C: Point;A: Point;Expression(G) = (x^2/3 + y^2 = 1);PointOnCurve(B, G);PointOnCurve(C, G);OneOf(Focus(G)) = A;F:Point;OneOf(Focus(G))=K;Negation(A = F);PointOnCurve(F, LineSegmentOf(B, C))",
    "query_expressions": "Perimeter(TriangleOf(A, B, C))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[28, 55], [63, 65], [72, 74]], [[20, 23]], [[24, 27]], [[59, 62]], [[28, 55]], [[20, 56]], [[20, 56]], [[59, 70]], [[72, 81]], [[59, 81]], [[59, 81]], [[72, 88]]]",
    "query_spans": "[[[90, 110]]]",
    "process": "By the definition of an ellipse, the sum of the distances from a point on the ellipse to the two foci equals the length of the major axis, 2a. Thus, the perimeter of $\\triangle ABC$ is $4a = 4\\sqrt{3}$. Therefore, the answer is $4\\sqrt{3}$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, its two asymptotes intersect the directrix of the parabola $y^{2}=2 p x(p>0)$ at points $A$ and $B$, respectively, and $O$ is the origin. If the eccentricity of the hyperbola is $2$, and the area of $\\triangle A O B$ is $\\sqrt{3}$, then $p=$?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;p: Number;A: Point;O: Origin;B: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);p>0;Expression(H) = (y^2 = 2*p*x);L1:Line;L2:Line;Asymptote(G)={L1,L2};Intersection(L1,Directrix(H))=A;Intersection(L2,Directrix(H))=B;Eccentricity(G)=2;Area(TriangleOf(A,O,B))=sqrt(3)",
    "query_expressions": "p",
    "answer_expressions": "3/2",
    "fact_spans": "[[[2, 58], [112, 115]], [[5, 58]], [[5, 58]], [[65, 86]], [[158, 161]], [[93, 96]], [[103, 106]], [[97, 100]], [[5, 58]], [[5, 58]], [[2, 58]], [[68, 86]], [[65, 86]], [], [], [[2, 64]], [[2, 102]], [[2, 102]], [[112, 123]], [[124, 156]]]",
    "query_spans": "[[[158, 163]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $3x^{2} - y^{2} = m$ with foci on the $x$-axis has a focal length of $4$. What is the value of the real number $m$?",
    "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (3*x^2 - y^2 = m);PointOnCurve(Focus(G), xAxis);FocalLength(G)=4",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[9, 29]], [[38, 43]], [[9, 29]], [[0, 29]], [[9, 36]]]",
    "query_spans": "[[[38, 47]]]",
    "process": "From the given conditions, we know: m > 0. Transforming the hyperbola equation into standard form and using the fact that the focal length is 4, we can find the value of the real number m. Given m > 0, the hyperbola equation can be rewritten as \\frac{x^{2}}{\\frac{m}{3}} - \\frac{y^{2}}{m} = 1. Also, the hyperbola 3x^{2} - y^{2} = m has a focal length of 4. Then \\frac{m}{3} + m = 2^{2}, solving gives m = 3."
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through $F_{1}$ intersects the two asymptotes at points $A$ and $B$. If $\\overrightarrow{F_{1}B} \\cdot \\overrightarrow {F_{2}B}=0$ and $\\overrightarrow{F_{1}A}=\\lambda \\overrightarrow{A B}$, then $\\lambda=?$",
    "fact_expressions": "l: Line;C: Hyperbola;F1: Point;B: Point;F2: Point;A: Point;L1:Line;L2:Line;Expression(C) = (x^2 - y^2/3 = 1);LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(F1,l);Asymptote(C)={L1,L2};Intersection(l,L1)=A;Intersection(l,L2)=B;DotProduct(VectorOf(F1,B),VectorOf(F2,B))=0;VectorOf(F1, A) = lambda*VectorOf(A, B);lambda:Number",
    "query_expressions": "lambda",
    "answer_expressions": "1",
    "fact_spans": "[[[69, 74]], [[2, 35]], [[44, 51], [61, 68]], [[88, 91]], [[52, 59]], [[84, 87]], [], [], [[2, 35]], [[2, 59]], [[2, 59]], [[60, 74]], [[2, 82]], [[2, 93]], [[2, 93]], [[95, 153]], [[156, 210]], [[212, 221]]]",
    "query_spans": "[[[212, 223]]]",
    "process": ""
  },
  {
    "text": "The line $l$ intersects the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ at points $A$ and $B$, and the midpoint of segment $AB$ lies on the line $x=-\\frac{1}{2}$. Then, the range of the intercept of line $l$ on the $y$-axis is?",
    "fact_expressions": "l: Line;G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);A: Point;B: Point;Intersection(l, G) = {A, B};H: Line;Expression(H) = (x = -1/2);PointOnCurve(MidPoint(LineSegmentOf(A, B)), H)",
    "query_expressions": "Range(Intercept(l, yAxis))",
    "answer_expressions": "(-oo, -sqrt(2)/2]+[sqrt(2)/2, +oo)",
    "fact_spans": "[[[0, 5], [78, 83]], [[6, 33]], [[6, 33]], [[36, 39]], [[40, 43]], [[0, 45]], [[57, 75]], [[57, 75]], [[46, 76]]]",
    "query_spans": "[[[78, 99]]]",
    "process": "Let the equation of line $ l $ be $ y = kx + m $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = -1 $. From  \n$$\n\\begin{cases}\ny = kx + m \\\\\nx^{2} + 2y^{2} = 2\n\\end{cases}\n$$\nwe obtain $ (2k^{2} + 1)x^{2} + 4kmx + 2m^{2} - 2 = 0 $. From $ \\Delta = 16k^{2}m^{2} - 4(2k^{2} + 1)(2m^{2} - 2) > 0 $, we get $ m^{2} < 2k^{2} + 1 $. Since $ x_{1} + x_{2} = \\frac{-4km}{1 + 2k^{2}} = -1 $, it follows that $ 4km = 2k^{2} + 1 $, so  \n$$\nm^{2} = \\frac{(1 + 2k^{2})^{2}}{16k^{2}} = \\frac{k^{2}}{4} + \\frac{1}{16k^{2}} + \\frac{1}{4} \\geqslant 2\\sqrt{\\frac{k^{2}}{4} \\cdot \\frac{1}{16k^{2}}} + \\frac{1}{4} = \\frac{1}{2},\n$$\nwith equality if and only if $ \\frac{k^{2}}{4} = \\frac{1}{16k^{2}} $, i.e., $ k = \\pm \\frac{\\sqrt{2}}{2} $, at which point $ \\Delta > 0 $ is satisfied. Therefore, $ m \\geqslant \\frac{\\sqrt{2}}{2} $ or $ m \\leqslant -\\frac{\\sqrt{2}}{2} $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the point $(4,2)$ lies on one of its asymptotes. Then its eccentricity equals?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;H: Point;Coordinate(H) = (4, 2);PointOnCurve(H, OneOf(Asymptote(G))) = True",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 58], [68, 69], [78, 79]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[59, 67]], [[59, 67]], [[59, 76]]]",
    "query_spans": "[[[78, 85]]]",
    "process": "The asymptote equation is $ y = \\frac{a}{b}x $. The point $ (4,2) $ satisfies the equation: $ 2 = \\frac{a}{b} \\times 4 $, $ \\frac{a}{b} = \\frac{1}{2} $. Also, $ e = \\frac{c}{a} = \\sqrt{\\frac{a^2 + b^{2}}{a^2}} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\frac{\\sqrt{5}}{2} $"
  },
  {
    "text": "The eccentricity of the hyperbola $x^{2}-2 y^{2}=1$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - 2*y^2 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[0, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 26]]]",
    "process": "First, rewrite the equation into standard form, find a and c, then use the eccentricity formula e=\\frac{c}{a} to find the eccentricity. Since x^{2}-2y^{2}=1, it follows that x^{2}-\\frac{y^{2}}{\\frac{1}{2}}=1. \\therefore a^{2}=1, \\frac{2}{b^{2}}=\\frac{1}{2}, so a=1. \\because c^{2}=a^{2}+b^{2}, \\therefore c^{2}=\\frac{3}{2}, so c=\\frac{\\sqrt{6}}{2}. \\therefore e=\\frac{c}{a}=\\frac{\\sqrt{6}}{2}"
  },
  {
    "text": "Given that $F_{1}$, $F_{2}$ are the common foci of the ellipse $M$: $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{2}=1$ and the hyperbola $N$: $\\frac{x^{2}}{n^{2}}-y^{2}=1$, and $P$ is one of their common points such that $P F_{1} \\perp F_{1} F_{2}$, then the product of the eccentricities of the ellipse $M$ and the hyperbola $N$ is?",
    "fact_expressions": "F1: Point;F2: Point;Focus(M) = {F1,F2};Focus(N) = {F1,F2};M: Ellipse;Expression(M) = (y^2/2 + x^2/m^2 = 1);m: Number;N: Hyperbola;Expression(N) = (-y^2 + x^2/n^2 = 1);n: Number;P: Point;OneOf(Intersection(M,N)) = P;IsPerpendicular(LineSegmentOf(P,F1), LineSegmentOf(F1,F2)) = True",
    "query_expressions": "Eccentricity(M)*Eccentricity(N)",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 9]], [[10, 17]], [[2, 107]], [[2, 107]], [[18, 64], [152, 157]], [[18, 64]], [[25, 64]], [[65, 102], [158, 164]], [[65, 102]], [[73, 102]], [[108, 111]], [[108, 120]], [[122, 149]]]",
    "query_spans": "[[[152, 172]]]",
    "process": "This problem can first use the fact that the ellipse and hyperbola share foci to obtain $ m^{2}-2=n^{2}+1=c^{2} $. Then set $ x_{p}=c $, substitute into the equations of the ellipse and hyperbola respectively, yielding $ y_{p}^{2}=\\frac{4}{m^{2}} $ and $ y_{p}^{2}=\\frac{1}{n^{2}} $, so $ \\frac{1}{n^{2}}=\\frac{4}{m^{2}} $. Finally, solve the system $ \\begin{cases}\\frac{1}{n^{2}}=\\frac{4}{m^{2}}\\\\m^{2}-2=n^{2}+1\\end{cases} $ to find $ m^{2} $, $ n^{2} $, and the eccentricities of the ellipse and hyperbola, thus obtaining the result. Since $ F_{1},F_{2} $ are common foci of the ellipse $ M:\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{2}=1 $ and the hyperbola $ N:\\frac{x^{2}}{n^{2}}-y^{2}=1 $, we have $ m^{2}-2=n^{2}+1=c^{2} $. Since $ P $ is a common point and $ PF_{1}\\bot F_{1}F_{2} $, set $ x_{p}=c $. Then $ \\frac{x_{p}^{2}}{m^{2}}+\\frac{y_{p}^{2}}{2}=1 $, $ \\frac{c^{2}}{m^{2}}+\\frac{y_{p}^{2}}{2}=1 $, $ y_{p}^{2}=2-\\frac{2c^{2}}{m^{2}}=2-\\frac{2(m^{2}-2)}{m^{2}} $, $ \\frac{x_{p}^{2}}{n^{2}}-y_{p}^{2}=1 $, $ \\frac{c^{2}}{n^{2}}-y_{p}^{2}=1 $, $ y_{p}^{2}=\\frac{c^{2}}{n^{2}}-1=\\frac{n^{2}+1}{n^{2}}-1=\\frac{1}{n^{2}} $, so $ \\frac{1}{n^{2}}=\\frac{4}{m^{2}} $. Solving the system $ \\begin{cases}\\frac{1}{n^{2}}=\\frac{4}{m^{2}}\\\\m^{2}-2=n^{2}+1\\end{cases} $ gives $ m^{2}=4 $, $ n^{2}=1 $. Then the ellipse $ M:\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1 $, the hyperbola $ N:x^{2}-y^{2}=1 $, $ c^{2}=2 $, hence $ e_{M}=\\frac{\\sqrt{2}}{2} $, $ e_{N}=\\frac{\\sqrt{2}}{1}=\\sqrt{2} $, $ e_{M}\\cdot e_{N}=\\frac{\\sqrt{2}}{2}\\times\\sqrt{2}=1 $."
  },
  {
    "text": "If the equation of the directrix of the parabola $x=a y^{2}$ is $x=1$, then the value of the real number $a$ is?",
    "fact_expressions": "G: Parabola;a: Real;Expression(G) = (x = a*y^2);Expression(Directrix(G)) = (x = 1)",
    "query_expressions": "a",
    "answer_expressions": "-1/4",
    "fact_spans": "[[[1, 15]], [[28, 33]], [[1, 15]], [[1, 26]]]",
    "query_spans": "[[[28, 37]]]",
    "process": "Since the parabola equation can be rewritten as $ y^{2} = \\frac{1}{a}x $, and its directrix is $ x = 1 $, we have $ -\\frac{1}{4a} = 1 $, solving which gives $ a = -\\frac{1}{4} $."
  },
  {
    "text": "It is known that the line $l$ passes through the focus $F$ of the parabola $C$: $y^{2}=4x$, and $l$ intersects $C$ at two points $A$ and $B$, where point $A$ lies in the fourth quadrant. If $\\overrightarrow{A F}=2 \\overrightarrow{F B}$, then the slope of line $l$ is?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;F: Point;B: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F,l);Intersection(l,C)={A,B};Quadrant(A)=4;VectorOf(A, F) = 2*VectorOf(F, B)",
    "query_expressions": "Slope(l)",
    "answer_expressions": "-2*sqrt(2)",
    "fact_spans": "[[[2, 7], [37, 40], [116, 121]], [[9, 28], [41, 44]], [[46, 49], [58, 62]], [[31, 34]], [[50, 53]], [[9, 28]], [[9, 35]], [[2, 34]], [[37, 55]], [[58, 67]], [[69, 114]]]",
    "query_spans": "[[[116, 126]]]",
    "process": "According to the given conditions, assume the equation of the line is y = k(x - 1). Combine the line equation with the parabola equation. Based on the conditions, derive the quantitative relationship between the x-coordinates of the intersection points, then use Vieta's formulas to find the x-coordinates of the intersection points, thus obtaining the result. According to the problem, the focus F(1, 0) of the parabola y^{2} = 4x. Let the equation of line l be y = k(x - 1). From \\begin{cases} y = k(x - 1) \\\\ y^{2} = 4x \\end{cases}, we get k^{2}x^{2} - 2(k^{2} + 2)x + k^{2} = 0. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}). \\therefore x_{1} + x_{2} = 2 + \\frac{4}{k^{2}}, x_{1} \\cdot x_{2} = 1. \\cdot \\overrightarrow{AF} = 2\\overrightarrow{FB}, and \\overrightarrow{AF} = (1 - x_{1}, -y_{1}), \\overrightarrow{FB} = (x_{2} - 1, y_{2}). Then 1 - x_{1} = 2(x_{2} - 1), i.e., 2x_{2} + x_{1} - 3 = 0. \\because x_{1} = \\frac{1}{x_{2}}, \\therefore 2x_{2} + \\frac{1}{x_{2}} - 3 = 0, solving gives x_{2} = 1 or x_{2} = \\frac{1}{2}. \\therefore x_{1} = 1 or x_{1} = 2. Also, x_{1} + x_{2} = 2 + \\frac{4}{k^{2}} > 2, so x_{1} = 2, x_{2} = \\frac{1}{2}, giving 2 + \\frac{4}{k^{2}} = 2 + \\frac{1}{2}, thus: k = \\pm 2\\sqrt{2}. Combining with the graph, we get k = -2\\sqrt{2}."
  },
  {
    "text": "Given points $A(4,0)$ and $B(2,2)$, and let $M$ be a moving point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, then the maximum value of $|M A| + |M B|$ is?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;M: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Coordinate(A) = (4, 0);Coordinate(B) = (2, 2);PointOnCurve(M, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, B)))",
    "answer_expressions": "10 + 2*sqrt(10)",
    "fact_spans": "[[[27, 65]], [[2, 11]], [[12, 20]], [[22, 26]], [[27, 65]], [[2, 11]], [[12, 20]], [[23, 69]]]",
    "query_spans": "[[[71, 92]]]",
    "process": ""
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{m-1}+\\frac{y^{2}}{2-m}=1$ represents an ellipse with foci on the $x$-axis, what is the range of values for $m$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(m - 1) + y^2/(2 - m) = 1);PointOnCurve(Focus(G), xAxis) = True;m: Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(3/2,2)",
    "fact_spans": "[[[54, 56]], [[2, 56]], [[45, 56]], [[58, 61]]]",
    "query_spans": "[[[58, 68]]]",
    "process": ""
  },
  {
    "text": "The ellipse $a x^{2}+b y^{2}=1(a>0, b>0)$ intersects the line $y=1-x$ at points $A$ and $B$. The slope of the line passing through the origin and the midpoint of segment $AB$ is $\\frac{\\sqrt{3}}{2}$. Then the value of $\\frac{b}{a}$ is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;B: Point;A: Point;a>0;b>0;Expression(G) = (a*x^2 + b*y^2 = 1);Expression(H) = (y = 1 - x);Intersection(G, H) = {A, B};L:Line;O:Origin;PointOnCurve(O,L);PointOnCurve(MidPoint(LineSegmentOf(A,B)),L);Slope(L)=sqrt(3)/2",
    "query_expressions": "b/a",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[0, 31]], [[2, 31]], [[2, 31]], [[32, 41]], [[47, 50]], [[43, 46]], [[2, 31]], [[2, 31]], [[0, 31]], [[32, 41]], [[0, 52]], [[67, 69]], [[54, 56]], [[53, 69]], [[53, 69]], [[67, 93]]]",
    "query_spans": "[[[95, 112]]]",
    "process": "Using the point difference method yields \\frac{b(y_{1}-y_{2})(y_{1}+y_{2})}{a(x_{1}-x_{2})(x_{1}+x_{2})}=-1, then combining with the values of k_{AB} and k_{OM}, simplification gives the value of \\frac{b}{a}. Let A(x_{1},y_{1}), B(x_{2},y_{2}), and let the midpoint of segment AB be M(\\frac{x_{1}+x_{2}}{2},\\frac{y_{1}+y_{2}}{2}); then ax_{1}^{2}+by_{1}^{2}=1, ax_{2}^{2}+by_{2}^{2}=1, so ax_{1}^{2}-ax_{2}^{2}=-(by_{1}^{2}-by_{2}^{2}), \\frac{by_{1}^{2}-by_{2}^{2}}{ax_{1}^{2}-ax_{2}^{2}}=-1 \\therefore \\frac{b(y_{1}-y_{2})(y_{1}+y_{2})}{a(x_{1}-x_{2})(x_{1}+x_{2})}=-1; k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-1, k_{OM}=\\frac{y_{1}+y_{2}}{x_{1}+x_{2}-0}=\\frac{y_{1}+y_{2}}{x_{1}+x_{2}}=\\frac{\\sqrt{3}}{2} \\therefore \\frac{b}{a}\\times(-1)\\times\\frac{\\sqrt{3}}{2}=-1 \\therefore \\frac{b}{a}=\\frac{2\\sqrt{3}}{3}"
  },
  {
    "text": "Find the equation of the locus of the center $P$ of a circle that is externally tangent to both circle $A$: $(x+5)^{2}+y^{2}=49$ and circle $B$: $(x-5)^{2}+y^{2}=1$.",
    "fact_expressions": "A: Circle;Expression(A) = (y^2 + (x + 5)^2 = 49);B:Circle;Expression(B)=((x-5)^2+y^2=49);H:Circle;IsOutTangent(H,A);IsOutTangent(H,B);P:Point;Center(H)=P",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "{(x^2/9-y^2/16=1)&(x>0)}",
    "fact_spans": "[[[2, 27]], [[2, 27]], [[28, 52]], [[28, 52]], [[56, 57]], [[1, 57]], [[1, 57]], [[60, 63]], [[56, 63]]]",
    "query_spans": "[[[60, 70]]]",
    "process": ""
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ are $F_{1}$ and $F_{2}$, respectively. A line passing through focus $F_{1}$ intersects the ellipse at points $A$ and $B$. If the area of the incircle of $\\Delta A B F_{2}$ is $\\pi$, and the coordinates of points $A$ and $B$ are $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$, respectively, then the value of $|y_{1}-y_{2}|$ is?",
    "fact_expressions": "G: Ellipse;L: Line;A: Point;B: Point;F2: Point;F1: Point;x1:Number;y1:Number;x2:Number;y2:Number;Expression(G) = (x^2/16 + y^2/9 = 1);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, L);Intersection(L, G) = {A, B};Area(InscribedCircle(TriangleOf(A,B,F2)))=pi",
    "query_expressions": "Abs(y1- y2)",
    "answer_expressions": "8*sqrt(7)/7",
    "fact_spans": "[[[0, 38], [77, 79]], [[73, 75]], [[80, 83], [123, 126]], [[84, 87], [127, 131]], [[54, 61]], [[46, 53], [65, 72]], [[139, 155]], [[139, 155]], [[156, 172]], [[156, 172]], [[0, 38]], [[123, 172]], [[123, 172]], [[0, 61]], [[0, 61]], [[62, 75]], [[73, 89]], [[91, 121]]]",
    "query_spans": "[[[174, 193]]]",
    "process": ""
  },
  {
    "text": "The line $y = kx + 2$ has exactly one common point with the hyperbola $x^2 - y^2 = 1$, then the value of $k$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;k: Number;Expression(G) = (x^2 - y^2 = 1);Expression(H) = (y = k*x + 2);NumIntersection(H, G) = 1",
    "query_expressions": "k",
    "answer_expressions": "{pm*1, pm*sqrt(5)}",
    "fact_spans": "[[[12, 30]], [[0, 11]], [[41, 44]], [[12, 30]], [[0, 11]], [[0, 39]]]",
    "query_spans": "[[[41, 48]]]",
    "process": "Consider the case when the line is tangent to the hyperbola and when the line is parallel to the asymptote. When the line is tangent to the hyperbola:  \n\\begin{cases}x^{2}-y^{2}=1\\\\y=kx+2\\end{cases},  \nthen $(1-k^{2})x^{2}-4kx-5=0$. Then $A=16k^{2}+20(1-k^{2})=20-4k^{2}=0$, solving gives $k=\\pm\\sqrt{5}$; when the line is parallel to the asymptote, the asymptote equations are $y=\\pm x$, hence $k=\\pm1$. In summary: $k=\\pm\\sqrt{5}$ or $k=\\pm1$."
  },
  {
    "text": "Given the parabola $E$: $y^{2}=2 p x$ ($p>0$) with focus $F$, the line $l$ passes through point $F$ and intersects the parabola at points $A$ and $B$, and intersects its directrix at point $C$ (with point $B$ between points $A$ and $C$). If $|B C|=3|B F|$ and $|A B|=9$, then $p=$?",
    "fact_expressions": "l: Line;E: Parabola;p: Number;B: Point;C: Point;F: Point;A: Point;p>0;Expression(E) = (y^2 = 2*(p*x));Focus(E) = F;PointOnCurve(F, l);Intersection(l, E) = {A, B};Intersection(l, Directrix(E)) = C;Between(B, A, C);Abs(LineSegmentOf(B, C)) = 3*Abs(LineSegmentOf(B, F));Abs(LineSegmentOf(A, B)) = 9",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[36, 41]], [[2, 28], [47, 50], [63, 64]], [[117, 120]], [[73, 77], [56, 59]], [[68, 72], [83, 86]], [[32, 35], [42, 46]], [[52, 55], [78, 82]], [[10, 28]], [[2, 28]], [[2, 35]], [[36, 46]], [[36, 61]], [[36, 72]], [[73, 87]], [[90, 104]], [[106, 115]]]",
    "query_spans": "[[[117, 122]]]",
    "process": "As shown in the figure: draw BD\\bot l from point B, with D as the foot of the perpendicular. Let the acute angle of inclination of line AB be \\alpha, then \\angle CBD = \\alpha. By the definition of the parabola, |BF| = |BD|. Therefore, \\cos\\alpha = \\frac{|BD|}{|BC|} = \\frac{|BF|}{|BC|} = \\frac{1}{3}, \\therefore \\sin\\alpha = \\sqrt{1 - \\cos^{2}\\alpha} = \\frac{2\\sqrt{2}}{3}, \\tan\\alpha = \\frac{\\sin\\alpha}{\\cos\\alpha} = 2\\sqrt{2}. It is also known that the focus of parabola E is F(\\frac{p}{2}, 0), so the equation of line AB is y = 2\\sqrt{2}(x - \\frac{p}{2}). Let points A(x_{1}, y_{1}) and B(x_{2}, y_{2}). Solving the system of equations of line AB and parabola E: \\begin{cases} y = 2\\sqrt{2}(x - \\frac{p}{2}) \\end{cases}, eliminating y and simplifying yields 4x^{2} - 5px + p^{2} = 0. By Vieta's formulas, x_{1} + x_{2} = \\frac{5p}{4}. From the definition of the parabola, |AB| = x_{1} + x_{2} + p = \\frac{9p}{4} = 9, solving gives p = 4."
  },
  {
    "text": "Given that $e_{1}$ and $e_{2}$ are the eccentricities of an ellipse and a hyperbola having common foci $F_{1}$ and $F_{2}$, point $P$ is a common point of the two curves, $O$ is the midpoint of $F_{1} F_{2}$, and $P O = F_{2} O$, then $\\frac{e_{1} e_{2}}{\\sqrt{e_{1}^{2}+e_{2}^{2}}}=$?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;F1: Point;F2: Point;e1:Number;e2:Number;P: Point;O: Origin;Eccentricity(G)=e2;Eccentricity(H)=e1;Focus(G)={F1,F2};Focus(H)={F1,F2};OneOf(Intersection(G,H))=P;MidPoint(LineSegmentOf(F1, F2)) = O;LineSegmentOf(P, O) = LineSegmentOf(F2, O)",
    "query_expressions": "(e1*e2)/sqrt(e1^2 + e2^2)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[47, 50]], [[44, 46]], [[28, 35]], [[36, 43]], [[2, 9]], [[11, 19]], [[55, 59]], [[70, 73]], [[2, 54]], [[2, 54]], [[22, 50]], [[22, 50]], [[55, 69]], [[70, 90]], [[92, 105]]]",
    "query_spans": "[[[107, 157]]]",
    "process": "As shown in the figure: since in $\\Delta F_{1}PF_{2}$, $PO = F_{2}O = F_{1}O$, $\\therefore F_{1}P \\bot F_{2}P$. Let the major axis length of the ellipse be $2a$, the real axis length of the hyperbola be $2a_{2}$, and the focal distance be $2c$. Then $x + y = 4c^{2}$"
  },
  {
    "text": "Given that line $m$ passes through the focus $F$ of the parabola $y^{2}=2 p x$ $(p>0)$, intersects the parabola at points $A$ and $B$, and intersects its directrix $l$ at point $C$. If $|A F|=6$ and $\\overrightarrow{C B}=2 \\overrightarrow{B F}$, then $p=$?",
    "fact_expressions": "m: Line;l:Line;G: Parabola;p: Number;A: Point;F: Point;C: Point;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(F, m);Intersection(m,G)={A,B};Intersection(m,l)=C;Abs(LineSegmentOf(A, F)) = 6;VectorOf(C, B) = 2*VectorOf(B, F);Directrix(G)=l",
    "query_expressions": "p",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 6]], [[55, 58]], [[7, 28], [37, 40], [52, 53]], [[125, 128]], [[41, 44]], [[31, 34]], [[59, 63]], [[45, 48]], [[10, 28]], [[7, 28]], [[7, 34]], [[1, 34]], [[1, 50]], [[1, 63]], [[66, 75]], [[77, 123]], [52, 57]]",
    "query_spans": "[[[125, 130]]]",
    "process": "Draw perpendiculars from A and B to the directrix l, with feet N and M respectively. Draw a perpendicular from F to AN, with foot D. According to \\overrightarrow{CB}=2\\overrightarrow{BF} and the definition of the parabola, we obtain \\angle DFA = \\angle MCB = 30^{\\circ}. Based on this, |AD| = 3. Then, using the definition of the parabola, p can be found. As shown in the figure: draw perpendiculars from A and B to the directrix l, with feet N and M respectively. Draw a perpendicular from F to AN, with foot D. Since \\overrightarrow{CB} = 2\\overrightarrow{BF}, it follows that |CB| = 2|BF|. Because |BF| = |BM|, we have |CB| = 2|BM|. Therefore, \\angle MCB = 30^{\\circ}, so \\angle DFA = 30^{\\circ}. In right triangle ADF, since |AF| = 6, we get |AD| = 3. Because |AN| = |AF| = 6 and |AN| = |AD| + p = 3 + p, it follows that 6 = 3 + p, so p = 3."
  },
  {
    "text": "The hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a, b>0)$ has left and right foci $F_{1}$ and $F_{2}$, respectively. The line $y=\\sqrt{3} b$ intersects the right branch of $C$ at point $P$. If $|P F_{1}|=2|P F_{2}|$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;LeftFocus(C) = F1;RightFocus(C) = F2;F1: Point;F2: Point;G: Line;Expression(G) = (y = sqrt(3)*b);Intersection(G, RightPart(C)) = P;P: Point;Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "3/2",
    "fact_spans": "[[[0, 58], [100, 103], [139, 145]], [[0, 58]], [[7, 58]], [[7, 58]], [[7, 58]], [[7, 58]], [[0, 82]], [[0, 82]], [[65, 72]], [[75, 82]], [[83, 99]], [[83, 99]], [[83, 113]], [[109, 113]], [[115, 137]]]",
    "query_spans": "[[[139, 151]]]",
    "process": "Substitute $ y = \\sqrt{3}b $ into the equation of $ C $ to obtain $ x = 2a $, so $ P(2a, \\sqrt{3}b) $, $ F_{1}(-c, 0) $, $ F_{2}(c, 0) $. From the definition of the hyperbola, we have: $ |PF_{1}| = 4a $, $ |PF_{2}| = 2a $. Therefore, $ \\sqrt{(2a + c)^{2} + 3b^{2}} = 4a $, $ \\sqrt{(2a - c)^{2} + 3b^{2}} = 2a $. Simplifying yields $ 8ac = 12a^{2} $, so $ 2c = 3a $. Thus, the eccentricity of the hyperbola is $ \\frac{3}{2} $."
  },
  {
    "text": "In $\\triangle P M F$, point $P$ is any point on the parabola $C$: $x^{2}=4 y$ other than the vertex, $F$ is the focus of the parabola $C$, $M(0,-1)$, and the real number $k$ satisfies $\\sin \\angle P F M=k \\sin \\angle P M F$. Then the maximum value of $k$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 4*y);P: Point;PointOnCurve(P, C);Negation(P=Vertex(C));M: Point;Coordinate(M) = (0, -1);F: Point;Focus(C) = F;k: Real;Sin(AngleOf(P, F, M)) = k*Sin(AngleOf(P, M, F))",
    "query_expressions": "Max(k)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[25, 44], [59, 65]], [[25, 44]], [[20, 24]], [[20, 54]], [[20, 54]], [[69, 78]], [[69, 78]], [[55, 58]], [[55, 68]], [[79, 84], [127, 130]], [[86, 125]]]",
    "query_spans": "[[[127, 136]]]",
    "process": "According to the definition of a parabola and the law of sines, we obtain \\frac{1}{k}=\\frac{|PB|}{|PM|}, which is then transformed into \\sin\\alpha=\\frac{1}{k}, where \\alpha is the inclination angle of the line PM; combining algebra and geometry completes the solution. Draw a perpendicular from point P to the directrix, with foot B. Then by the definition of a parabola, |PB|=|PF|. Also, since \\sin\\angle PFM=k\\sin\\angle PMF, in \\triangle PFM, by the law of sines we have |PM|=k|PF|, so |PM|=k|PB|, hence \\frac{1}{k}=\\frac{|PB|}{|PM|}. Let the inclination angle of line PM be \\alpha, then \\sin\\alpha=\\frac{1}{k}. When k reaches its maximum value, \\sin\\alpha is minimized, at which point line PM is tangent to the parabola. Let the equation of line PM be y=mx-1. Solving the system \\begin{cases}y=mx-1\\\\x^{2}=4y\\end{cases} yields x^{2}-4mx+4=0. From \\Delta=16m^{2}-16=0, we get m=\\pm1, so \\tan\\alpha=\\pm1, then \\sin\\alpha=\\frac{\\sqrt{2}}{2}, thus the maximum value of k is \\frac{1}{\\sin\\alpha}=\\sqrt{2}."
  },
  {
    "text": "It is known that a line with slope $1$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$. If $M$ is the circumcenter of $\\triangle OAB$ (where $O$ is the origin), then when $|AB|$ is maximized, $|AB|=$?",
    "fact_expressions": "G: Parabola;H: Line;O: Origin;A: Point;B: Point;M: Point;Expression(G) = (y^2 = 4*x);Intersection(H, G) = {A, B};Slope(H) = 1;Circumcenter(TriangleOf(O, A, B)) = M;WhenMax(Abs(LineSegmentOf(A, B)))",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "4*sqrt(10)",
    "fact_spans": "[[[12, 26]], [[9, 11]], [[64, 67]], [[28, 31]], [[32, 35]], [[60, 63]], [[12, 26]], [[9, 37]], [[2, 11]], [[39, 63]], [[75, 86]]]",
    "query_spans": "[[[87, 96]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has right focus $F$, point $A$ lies on ellipse $C$, the line $AF$ is tangent to the circle $C^{\\prime}$: $(x-\\frac{c}{2})^{2}+y^{2}=\\frac{b^{2}}{16}$ at point $B$, and $\\overrightarrow{A F}=4 \\overrightarrow{B F}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;C1:Circle;b: Number;a: Number;c: Number;F: Point;A: Point;B: Point;a>0;b>0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(C1) = (y^2 + (-c/2 + x)^2 = b^2/16);RightFocus(C) = F;PointOnCurve(A, C);TangentPoint(LineOf(A,F),C1)=B;VectorOf(A, F) = 4*VectorOf(B, F)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[2, 63], [77, 82], [206, 209]], [[92, 150]], [[9, 63]], [[9, 63]], [[106, 150]], [[68, 71]], [[72, 76]], [[153, 157]], [[9, 63]], [[9, 63]], [[2, 63]], [[92, 150]], [[2, 71]], [[72, 83]], [[84, 157]], [[159, 204]]]",
    "query_spans": "[[[206, 215]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the circle $O$: $x^{2}+y^{2}=b^{2}$, if there exists a point $P$ on $C$ such that two tangents drawn from $P$ to circle $O$ touch the circle at points $A$ and $B$ respectively, satisfying $\\angle A P B=60^{\\circ}$, then the range of eccentricity of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;O: Circle;a: Number;b: Number;Expression(C) = (x^2/a^2 + y^2/b^2 = 1);a>b;b>0;Expression(O) = (x^2 + y^2 = b^2);P: Point;PointOnCurve(P, C);L1: Line;L2: Line;TangentOfPoint(P, O) = {L1, L2};A: Point;B: Point;TangentPoint(L1, O) = A;TangentPoint(L2, O) = B;AngleOf(A, P, B) = ApplyUnit(60, degree)",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "[\\sqrt{3}/2, 1)",
    "fact_spans": "[[[2, 59], [88, 91], [160, 165]], [[60, 85], [108, 112]], [[9, 59]], [[9, 59]], [[2, 59]], [[9, 59]], [[9, 59]], [[60, 85]], [[94, 98], [103, 107]], [[88, 98]], [], [], [[102, 117]], [[123, 126]], [[127, 130]], [[102, 130]], [[102, 130]], [[133, 158]]]",
    "query_spans": "[[[160, 176]]]",
    "process": ""
  },
  {
    "text": "Draw a perpendicular line from a focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) to an asymptote. If the foot of the perpendicular lies exactly on the perpendicular bisector of the segment $OF$ ($O$ being the origin), then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;OneOf(Focus(G)) = F;Z: Line;PointOnCurve(F, Z);IsPerpendicular(Z, OneOf(Asymptote(G)));PointOnCurve(FootPoint(Z, OneOf(Asymptote(G))), PerpendicularBisector(LineSegmentOf(O, F)));O: Origin",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 58], [108, 111]], [[1, 58]], [[4, 58]], [[4, 58]], [[4, 58]], [[4, 58]], [[63, 66]], [[1, 66]], [], [[0, 75]], [[0, 75]], [[0, 106]], [[89, 92]]]",
    "query_spans": "[[[108, 117]]]",
    "process": "Without loss of generality, draw a perpendicular line to the asymptote $ y = \\frac{b}{a}x $, with foot at point $ A $. The focus is $ F(\\sqrt{a^{2}+b^{2}}, 0) $. Since the foot lies exactly on the perpendicular bisector of segment $ OF $ (where $ O $ is the origin), we have $ A\\left(\\frac{\\sqrt{a^{2}+b^{2}}}{2}, \\frac{b\\sqrt{a^{2}+b^{2}}}{2a}\\right) $. Then $ k_{AF} = \\frac{\\frac{b\\sqrt{a^{2}+b^{2}}}{2a} - 0}{\\frac{\\sqrt{a^{2}+b^{2}}}{2} - \\sqrt{a^{2}+b^{2}}} = -\\frac{b}{a} $. From $ OA \\perp FA $, we know $ k_{OA} \\cdot k_{FA} = -1 $, so $ \\frac{b}{a} \\cdot \\left(-\\frac{b}{a}\\right) = -1 $, hence $ a = b $. Therefore, the hyperbola is equilateral, and its eccentricity is $ \\sqrt{2} $."
  },
  {
    "text": "The standard equation of the hyperbola sharing foci with the ellipse $C$: $\\frac{y^{2}}{16}+\\frac{x^{2}}{12}=1$ and passing through the point $(1, \\sqrt{3})$ is?",
    "fact_expressions": "G: Hyperbola;C: Ellipse;H: Point;Expression(C) = (x^2/12 + y^2/16 = 1);Coordinate(H) = (1, sqrt(3));Focus(G)=Focus(C);PointOnCurve(H,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/2 - x^2/2 = 1",
    "fact_spans": "[[[67, 70]], [[1, 45]], [[50, 66]], [[1, 45]], [[50, 66]], [[0, 70]], [[49, 70]]]",
    "query_spans": "[[[67, 77]]]",
    "process": "From the right standard equation of the ellipse, it is known that 1.1 the foci are on the y-axis, and the foci can be found. Thus, the standard equation of the hyperbola can be set as detailed \\] \\because\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{4-a^{2}}=1(a>0), substitute the point (1,\\sqrt{3}) into it, the result can be obtained \\therefore the foci of the required hyperbola are F_{1}(0,-2), F_{2}(0,2), let the hyperbola equation be \\frac{y^{2}}{a2}-\\frac{x2}{4-a2}=1(a>0), substituting (1,\\sqrt{3}) gives \\frac{3}{a^{2}}-\\frac{1}{4-a^{2}}=1, solving yields a^{2}=2 or a^{2}=6 (discarded) \\therefore the standard equation of the hyperbola is \\frac{y^{2}}{2}-\\frac{x^{2}}{2}=1"
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, and $F_{1}$, $F_{2}$ are the left and right foci of the ellipse respectively, if $\\angle F_{1} P  F_{2}=60^{\\circ}$, then the area of $\\triangle F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/16 + y^2/9 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[6, 44], [66, 68]], [[48, 55]], [[2, 5]], [[56, 63]], [[6, 44]], [[2, 47]], [[48, 74]], [[48, 74]], [[76, 110]]]",
    "query_spans": "[[[112, 142]]]",
    "process": ""
  },
  {
    "text": "The equation of the hyperbola that has the same asymptotes as the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$ and passes through the point $(1,2 \\sqrt{3})$ is?",
    "fact_expressions": "G: Hyperbola;C:Hyperbola;H: Point;Expression(G) = (x^2 - y^2/4 = 1);Coordinate(H) = (1, 2*sqrt(3));PointOnCurve(H, C);Asymptote(C)=Asymptote(G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/8 - x^2/2 = 1",
    "fact_spans": "[[[1, 29]], [[56, 59]], [[38, 55]], [[1, 29]], [[38, 55]], [[37, 59]], [[0, 59]]]",
    "query_spans": "[[[56, 63]]]",
    "process": "Let the required hyperbola equation be $x^{2}-\\frac{y^{2}}{4}=\\lambda$. Since the hyperbola passes through the point $(1,2\\sqrt{3})$, we have $1-\\frac{12}{4}=\\lambda$, so $\\lambda=-2$. The hyperbola equation is $x^{2}-\\frac{y^{2}}{4}=-2$, which is equivalent to $\\frac{y^{2}}{8}-\\frac{x^{2}}{2}=1$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, the distance from a focus to an asymptote is $3$. Then, the length of the imaginary axis of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/4 - y^2/b^2 = 1);b: Number;b>0;Distance(Focus(C),Asymptote(C))=3",
    "query_expressions": "Length(ImageinaryAxis(C))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 54], [70, 73]], [[2, 54]], [[10, 54]], [[10, 54]], [[2, 68]]]",
    "query_spans": "[[[70, 79]]]",
    "process": "Since one focus of the hyperbola $ C:\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1 $ ($ b>0 $) is $ (\\sqrt{4+b^{2}},0) $, and the equation of one asymptote is $ bx+2y=0 $, the distance from the focus of the hyperbola $ C:\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1 $ ($ b>0 $) to the asymptote is 3. We obtain $ \\frac{b\\sqrt{4+b^{2}}}{\\sqrt{4+b^{2}}}=3 $, solving gives $ b=3 $, then the length of the imaginary axis of the hyperbola is $ 2b=6 $."
  },
  {
    "text": "If $F_{1}$ and $F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, and point $P(8, y_{0})$ lies on the hyperbola, then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 - y^2 = 1);Coordinate(P) = (8, y0);Focus(G) = {F1, F2};PointOnCurve(P, G);y0:Number",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "5*sqrt(3)",
    "fact_spans": "[[[17, 45], [66, 69]], [[51, 65]], [[1, 8]], [[9, 16]], [[17, 45]], [[51, 65]], [[1, 50]], [[51, 70]], [[52, 65]]]",
    "query_spans": "[[[72, 99]]]",
    "process": "According to the problem, the equation of the hyperbola is: $\\frac{x^{2}}{4}-y^{2}=1$, with foci on the $x$-axis and $c=\\sqrt{4+1}=\\sqrt{5}$, so the coordinates of the foci are $(\\pm\\sqrt{5},0)$, then $|F_{1}F_{2}|=2\\sqrt{5}$. Since point $P(8,y_{0})$ lies on the hyperbola, we have $\\frac{8^{2}}{4}-y_{0}^{2}=1$, solving gives $y_{0}=\\pm\\sqrt{15}$. Therefore, the area $S$ of quadrilateral $F_{1}PF_{2}$ is $S=\\frac{1}{2}\\times|y_{0}|\\times|F_{1}F_{2}|=5\\sqrt{3}$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Let $P$ be a point on the ellipse such that $\\angle F_{1}PF_{2}=60^{\\circ}$. Then the value of $P F_{1}^{2}+P F_{2}^{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 + y^2/3 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "LineSegmentOf(P, F1)^2 + LineSegmentOf(P, F2)^2",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 39], [71, 73]], [[48, 55]], [[67, 70]], [[58, 65]], [[2, 39]], [[2, 65]], [[2, 65]], [[67, 76]], [[79, 110]]]",
    "query_spans": "[[[112, 141]]]",
    "process": ""
  },
  {
    "text": "What is the length of the shortest chord passing through the focus $F$ of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/9 = 1);H:LineSegment;F: Point;IsChordOf(H,G);OneOf(Focus(G))=F;PointOnCurve(F,H);WhenMin(Length(H))",
    "query_expressions": "Length(H)",
    "answer_expressions": "9/2",
    "fact_spans": "[[[1, 39]], [[1, 39]], [], [[42, 45]], [[0, 47]], [[1, 45]], [[0, 47]], [[0, 51]]]",
    "query_spans": "[[[0, 54]]]",
    "process": "According to the simple properties of an ellipse and the equation of the ellipse, it can be directly concluded that, from the geometric properties of the ellipse, the shortest chord passing through the focus of the ellipse and perpendicular to the major axis has a length of $\\frac{2b^{2}}{a}=\\frac{18}{4}=\\frac{9}{2}$."
  },
  {
    "text": "What is the length of the real axis of the hyperbola $x^{2}-\\frac{y^{2}}{9}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/9 = 1)",
    "query_expressions": "Length(RealAxis(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 34]]]",
    "process": "Since $ a=1 $, the length of the transverse axis of the hyperbola is $ 2a=2 $; the answer to be filled in is $ 2 $."
  },
  {
    "text": "Given the hyperbola $E$: $ \\frac{x^{2}}{6}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ has asymptotes with equations $y=\\pm \\frac{\\sqrt{3}}{3} x$, then the focal length of $E$ is equal to?",
    "fact_expressions": "E: Hyperbola;b: Number;b>0;Expression(E) = (x^2/6 - y^2/b^2 = 1);Expression(Asymptote(E)) = (y = pm*(x*(sqrt(3)/3)))",
    "query_expressions": "FocalLength(E)",
    "answer_expressions": "4*sqrt(2)",
    "fact_spans": "[[[2, 54], [91, 94]], [[9, 54]], [[9, 54]], [[2, 54]], [[2, 89]]]",
    "query_spans": "[[[91, 100]]]",
    "process": "\\because the hyperbola E:\\frac{x^{2}}{6}-\\frac{y^{2}}{b^{2}}=1(b>0) has asymptotes given by y=\\pm\\frac{\\sqrt{3}}{3}x, \\therefore\\frac{b^{2}}{6}=(\\frac{\\sqrt{3}}{3})^{2}, i.e., b^{2}=2, \\therefore c^{2}=6+b^{2}=8, 2c=4\\sqrt{2}, \\therefore the focal length of E is 4\\sqrt{2}."
  },
  {
    "text": "Let the center of the ellipse be at the origin, with foci on the $x$-axis, and eccentricity $e=\\frac{\\sqrt{3}}{2}$. It is known that the maximum distance from the point $P(0, \\frac{3}{2})$ to a point on the ellipse is $\\sqrt{7}$. Then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;O:Origin;P: Point;e:Number;Center(G)=O;Coordinate(P) = (0, 3/2);PointOnCurve(Focus(G), xAxis);Eccentricity(G)=e;e=sqrt(3)/2;H:Point;PointOnCurve(H,G);Max(Distance(P,H))=sqrt(7)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4+y^2=1",
    "fact_spans": "[[[1, 3], [70, 72], [93, 95]], [[7, 11]], [[49, 69]], [[24, 46]], [[1, 11]], [[49, 69]], [[1, 20]], [[1, 46]], [[24, 46]], [], [[70, 75]], [[49, 91]]]",
    "query_spans": "[[[93, 102]]]",
    "process": "According to the problem, assume the standard equation of the ellipse is $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. Then $e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{a^{2}-b^{2}}{a^{2}}=1-\\frac{b^{2}}{a^{2}}=\\frac{3}{4}$, so $\\frac{b^{2}}{a^{2}}=\\frac{1}{4}$, i.e., $a=2b$. Let $(x,y)$ be a point on the ellipse and $d$ be the distance from this point to point $P$, then $d^{2}=x^{2}+(y-\\frac{3}{2})^{2} = \\frac{y^{2}}{b^{2}}+y^{2}-3y+\\frac{9}{4}=-3(y+\\frac{1}{2})^{2}+4b^{2}+3$ $(-b\\leqslant y \\leqslant b)$. If $-b>-\\frac{1}{2}$, i.e., $b<\\frac{1}{2}$, then when $y=-b$, $d^{2}$ attains its maximum value, thus $d$ attains its maximum value. Hence $(\\sqrt{7})^{2}=(b+\\frac{3}{2})^{2}$, solving gives $b=\\pm\\sqrt{7}-\\frac{3}{2}$, which contradicts $0<b<\\frac{1}{2}$. If $-b\\leqslant-\\frac{1}{2}$, i.e., $b\\geqslant\\frac{1}{2}$, then when $y=-\\frac{1}{2}$, $d^{2}$ attains its maximum value, thus $d$ attains its maximum value. So $4b^{2}+3=(\\sqrt{7})^{2}$, solving gives $b^{2}=1$, therefore $a^{2}=4$. Hence, the required standard equation of the ellipse is $\\frac{x^{2}}{4}+y^{2}=1$."
  },
  {
    "text": "Given point $A(5,0)$ and a moving point $P$ on the parabola $y^{2}=4x$, point $M$ lies on segment $PA$ such that $|PM|=3|MA|$. Then the equation of the trajectory of point $M$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (5, 0);G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G);M: Point;PointOnCurve(M, LineSegmentOf(P, A));Abs(LineSegmentOf(P, M)) = 3*Abs(LineSegmentOf(M, A))",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "y^2 = x - 15/4",
    "fact_spans": "[[[2, 11]], [[2, 11]], [[12, 26]], [[12, 26]], [[30, 34]], [[12, 34]], [[35, 39], [67, 71]], [[35, 48]], [[51, 65]]]",
    "query_spans": "[[[67, 78]]]",
    "process": "Let M(x,y), P(x_{0},y_{0}). Since |PM| = 3|MA| and point M lies on the line segment PA, we have \\overrightarrow{PM} = 3\\overrightarrow{MA}. Therefore, (x - x_{0}, y - y_{0}) = 3(5 - x, 0 - y), so \\begin{cases} x - x_{0} = 15 - \\\\ y - y_{0} = -3y \\end{cases} 15 - 3x \\quad p_{0}(x_{0} = 4x - 15, \\quad y_{0} = 4y). Since the moving point P lies on the parabola, we have y_{0}^{2} = 4x_{0}. Therefore, 16y^{2} = 4(4x - 15), which simplifies to y^{2} = x - \\frac{15}{4}."
  },
  {
    "text": "Given that the eccentricity $e$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $2$, then the minimum value of $\\frac{b^{2}+1}{3 a}$ at this time is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;e: Number;Eccentricity(G)=e;e = 2",
    "query_expressions": "Min((b^2 + 1)/(3*a))",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 58]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[62, 65]], [[2, 65]], [[62, 69]]]",
    "query_spans": "[[[73, 100]]]",
    "process": "From the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ with eccentricity $2$, we obtain $\\frac{c}{a}=2$, so $\\frac{a^{2}+b^{2}}{a^{2}}=1+\\frac{b^{2}}{a^{2}}=4$. Hence, $b^{2}=3a^{2}$, then $\\frac{3a^{2}+1}{3a}=a+\\frac{1}{3a}\\geqslant2\\sqrt{a\\cdot\\frac{1}{3a}}=\\frac{2\\sqrt{3}}{3}$, with equality holding if and only if $a=\\frac{1}{3a}$, that is, when $a=\\frac{\\sqrt{3}}{3}$."
  },
  {
    "text": "Given that the line $l_{1}$: $x+2 y+1=0$ is perpendicular to the line $l_{2}$: $4 x+a y-2=0$, what is the value of $a$?",
    "fact_expressions": "l1: Line;Expression(l1) = (x + 2*y + 1 = 0);l2: Line;Expression(l2) = (a*y + 4*x - 2 = 0);a: Number;IsPerpendicular(l1, l2)",
    "query_expressions": "a",
    "answer_expressions": "-2",
    "fact_spans": "[[[2, 24]], [[2, 24]], [[25, 49]], [[25, 49]], [[54, 57]], [[2, 51]]]",
    "query_spans": "[[[54, 61]]]",
    "process": "The lines $ l_{1}: x + 2y + 1 = 0 $ and $ l_{2}: 4x + ay - 2 = 0 $ are perpendicular, then: $ 1 \\times 4 + 2a = 0 $, solving gives: $ a = -2 $."
  },
  {
    "text": "What is the length of the major axis of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{25}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/25 = 1)",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "10",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 45]]]",
    "process": "The major axis of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{25}=1$ lies on the $y$-axis, $a=5$, so the length of the major axis is: $2a=2\\times5=10$"
  },
  {
    "text": "Given points $M(-3,0)$, $N(3,0)$, $B(1,0)$, circle $C$ is tangent to line $MN$ at point $B$. Two lines passing through $M$ and $N$ respectively and tangent to circle $C$ intersect at point $P$. Then the equation of the trajectory of point $P$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (-3, 0);N: Point;Coordinate(N) = (3, 0) ;B: Point;Coordinate(B) = (1, 0);C: Circle;TangentPoint(C, LineOf(M, N)) = B;l1: Line;l2: Line;TangentOfPoint(M,C)=l1;TangentOfPoint(N,C)=l2;Intersection(l1, l2) = P;P: Point",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(x^2-y^2/8=1)&(x>1)",
    "fact_spans": "[[[2, 12], [53, 56]], [[2, 12]], [[14, 22], [57, 60]], [[14, 22]], [[24, 32], [47, 51]], [[24, 32]], [[33, 37], [61, 65]], [[33, 51]], [], [], [[52, 71]], [[52, 71]], [[52, 78]], [[80, 84], [74, 78]]]",
    "query_spans": "[[[80, 91]]]",
    "process": "As shown in the figure, let lines MP and NP be tangent to circle C at points A and D, respectively. By the tangent segment theorem, |MA| = |MB| = 4, |ND| = |NB| = 2, |PA| = |PD|. Therefore, |PM| - |PN| = (|PA| + |MA|) - (|PD| + |ND|) = |MA| - |ND| = 4 - 2 = 2 < |MN|. Thus, the trajectory of point P is the right branch of a hyperbola with foci at M(-3,0) and N(3,0), and real axis length 2 (excluding the right vertex). Let the equation of the hyperbola be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1. Then 2a = 2, c = 3, so a = 1, b^{2} = 9 - 1 = 8. Hence, the trajectory equation of point P is x^{2}-\\frac{y^{2}}{8}=1 (x>1)."
  },
  {
    "text": "Let $A$ and $B$ be two distinct points on the parabola $x^{2}=4 y$. Then the minimum value of $|\\overrightarrow{O A}+\\overrightarrow{O B}|^{2}-|\\overrightarrow{A B}|^{2}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);Negation(A=B);O: Origin",
    "query_expressions": "Min(Abs(VectorOf(O, A) + VectorOf(O, B))^2 - Abs(VectorOf(A, B))^2)",
    "answer_expressions": "-16",
    "fact_spans": "[[[9, 23]], [[9, 23]], [[1, 4]], [[5, 8]], [[1, 29]], [[1, 29]], [[1, 29]], [[31, 107]]]",
    "query_spans": "[[[31, 113]]]",
    "process": "Since the slope of line AB exists, let AB: $ y = kx + t $. From  \n\\[\n\\begin{cases}\ny = kx + t \\\\\nx^{2} = 4y\n\\end{cases}\n\\]  \neliminating $ y $ gives $ x^{2} - 4kx - 4t = 0 $, and $ \\Delta = 16k^{2} + 16t > 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, and midpoint of AB be $ M(x_{0}, y_{0}) $. Then  \n\\[\n\\begin{cases}\nx_{1} + x_{2} = 4k \\\\\nx_{1}x_{2} = -4t\n\\end{cases}\n\\]  \n$ x_{0} = \\frac{x_{1} + x_{2}}{2} = 2k $, $ y_{0} = kx_{0} + t = 2k^{2} + t $. Therefore, $ M(2k, 2k^{2} + t) $, so  \n\\[\n|\\overrightarrow{OM}| = \\sqrt{4k^{4} + 4k^{2}t + 4k^{2} + t^{2}}\n\\]  \nAlso,  \n\\[\n|\\overrightarrow{AB}| = \\sqrt{1 + k^{2}} |x_{1} - x_{2}| = \\sqrt{(1 + k^{2})(16k^{2} + 16t)} = \\sqrt{16k^{4} + 16k^{2}t + 16k^{2} + 16t}\n\\]  \n\\[\n|\\overrightarrow{OA} + \\overrightarrow{OB}|^{2} - |\\overrightarrow{AB}|^{2} = 4|\\overrightarrow{OM}|^{2} - |\\overrightarrow{AB}|^{2} = 4t^{2} - 16t = 4(t - 2)^{2} - 16 \\geqslant -16\n\\]  \nequality holds when $ t = 2 $. Therefore, the minimum value of $ |\\overrightarrow{OA} + \\overrightarrow{OB}|^{2} - |\\overrightarrow{AB}|^{2} $ is $ -16 $."
  },
  {
    "text": "What are the two asymptotes of the hyperbola $3 x^{2}-y^{2}=8$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (3*x^2 - y^2 = 8)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[0, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 28]]]",
    "process": "Let $3x^{2}-y^{2}=0$, $y^{2}=3x^{2}$, $y=\\pm\\sqrt{3}x$, so the required asymptotes are $y=\\pm\\sqrt{3}x$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ shares the same foci as the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, and the asymptotes of hyperbola $C$ are given by $y=\\pm 2 x$, then the equation of hyperbola $C$ is ?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Ellipse;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2/9 + y^2/4 = 1);Focus(C) = Focus(G);Expression(Asymptote(C)) = (y = pm*(2*x))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2/4 = 1",
    "fact_spans": "[[[2, 53], [99, 105], [125, 131]], [[10, 53]], [[10, 53]], [[54, 91]], [[2, 53]], [[54, 91]], [[2, 97]], [[99, 123]]]",
    "query_spans": "[[[125, 137]]]",
    "process": ""
  },
  {
    "text": "Given that the line $y=k(x-2)$ $(k>0)$ intersects the parabola $y^{2}=8x$ at points $A$ and $B$, and $F$ is the focus of the parabola. If $|FA|=2|FB|$, then the value of $k$ is?",
    "fact_expressions": "G: Parabola;H: Line;F: Point;A: Point;B: Point;k:Number;Expression(G) = (y^2 = 8*x);k>0;Expression(H) = (y = k*(x - 2));Intersection(H, G) = {A, B};Focus(G) = F;Abs(LineSegmentOf(F, A)) = 2*Abs(LineSegmentOf(F, B))",
    "query_expressions": "k",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[20, 34], [51, 54]], [[2, 19]], [[47, 50]], [[37, 40]], [[41, 44]], [[75, 78]], [[20, 34]], [[4, 19]], [[2, 19]], [[2, 46]], [[47, 57]], [[59, 73]]]",
    "query_spans": "[[[75, 82]]]",
    "process": ""
  },
  {
    "text": "Given that $AB$ is a chord passing through the left focus $F_{1}$ of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and $|A F_{2}|+|B F_{2}|=12$, where $F_{2}$ is the right focus of the ellipse, then the length of chord $AB$ is?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2/25 + y^2/16 = 1);LeftFocus(G) = F1;PointOnCurve(F1, LineSegmentOf(A, B));IsChordOf(LineSegmentOf(A,B),G);Abs(LineSegmentOf(A, F2)) + Abs(LineSegmentOf(B, F2)) = 12;RightFocus(G) = F2",
    "query_expressions": "Length(LineSegmentOf(A,B))",
    "answer_expressions": "8",
    "fact_spans": "[[[9, 48], [97, 99]], [[2, 7]], [[2, 7]], [[89, 96]], [[51, 58]], [[9, 48]], [[9, 58]], [[2, 58]], [[2, 60]], [[62, 86]], [[89, 103]]]",
    "query_spans": "[[[106, 114]]]",
    "process": ""
  },
  {
    "text": "The standard equation of the hyperbola passing through the two points $(-3,6 \\sqrt{2})$, $(2,3 \\sqrt{3})$ is?",
    "fact_expressions": "G: Hyperbola;A: Point;B: Point;Coordinate(A) = (-3, 6*sqrt(2));Coordinate(B) = (2, 3*sqrt(3));PointOnCurve(A, G);PointOnCurve(B, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/9 = 1",
    "fact_spans": "[[[40, 43]], [[4, 21]], [[23, 39]], [[4, 21]], [[23, 39]], [[0, 43]], [[0, 43]]]",
    "query_spans": "[[[40, 50]]]",
    "process": "Let the hyperbola equation be $ mx^{2} - ny^{2} = 1 $, $ mn > 0 $. According to the conditions, we have\n\\[\n\\begin{cases}\n9m - 72n = 1 \\\\\n4m - 27n = 1\n\\end{cases}\n\\]\nSolving gives $ m = 1 $, $ n = \\frac{1}{9} $. Therefore, the standard equation of the required hyperbola is: $ x^{2} - \\frac{y^{2}}{9} = 1 $."
  },
  {
    "text": "The sum of the distances from two points $A$ and $B$ on the parabola $y^{2}=4x$ to the focus $F$ is $5$. Then, the horizontal coordinate of the midpoint of segment $AB$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);Distance(A, F) + Distance(B, F) = 5;F: Point;Focus(G) = F",
    "query_expressions": "XCoordinate(MidPoint(LineSegmentOf(A,B)))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[18, 21]], [[22, 25]], [[0, 25]], [[0, 25]], [[18, 40]], [[28, 31]], [[0, 31]]]",
    "query_spans": "[[[42, 58]]]",
    "process": "The parabola $ y^{2} = 4x $ has focus coordinates $ F(1,0) $, and the directrix equation is $ x = -1 $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. $ |AF| + |BF| = x_{1} + x_{2} + 2 = 5 $, so $ x_{1} + x_{2} = 3 $. The horizontal coordinate of the midpoint of segment $ AB $ is $ \\frac{x_{1} + x_{2}}{2} = \\frac{3}{2} $."
  },
  {
    "text": "The right focus of the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1$ $(a > b > 0)$ is $F$, and the intersection point of its right directrix with the $x$-axis is $A$. There exists a point $P$ on the ellipse such that the perpendicular bisector of segment $AP$ passes through point $F$. Then, the range of values for the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;F: Point;P: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;Intersection(RightDirectrix(G), xAxis) = A;PointOnCurve(P, G);PointOnCurve(F, PerpendicularBisector(LineSegmentOf(A, P)))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[1/2, 1)",
    "fact_spans": "[[[0, 53], [61, 62], [79, 81], [110, 112]], [[2, 53]], [[2, 53]], [[74, 77]], [[57, 60], [104, 108]], [[84, 88]], [[2, 53]], [[2, 53]], [[0, 53]], [[0, 60]], [[61, 77]], [[78, 88]], [[90, 108]]]",
    "query_spans": "[[[110, 122]]]",
    "process": ""
  },
  {
    "text": "If the foci of a hyperbola are $F_{1}(-4 , 0)$, $F_{2}(4 , 0)$, and the length of the real axis is equal to the length of the imaginary axis, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;Focus(G) = {F1, F2};Coordinate(F1) = (-4, 0);Coordinate(F2) = (4, 0);Length(RealAxis(G)) = Length(ImageinaryAxis(G))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 4], [52, 55]], [[8, 23]], [[26, 40]], [[1, 40]], [[8, 23]], [[26, 40]], [[1, 50]]]",
    "query_spans": "[[[52, 62]]]",
    "process": ""
  },
  {
    "text": "Given that the focus of the parabola $x^{2}=4 y$ is $F$, the intersection point of the directrix and the axis is $M$, and $N$ is an arbitrary point on the parabola satisfying $|N F|=\\lambda|M N|$, then the range of values for $\\lambda$ is?",
    "fact_expressions": "G: Parabola;N: Point;F: Point;M: Point;Expression(G) = (x^2 = 4*y);Focus(G) = F;Intersection(Directrix(G),SymmetryAxis(G))=M;PointOnCurve(N, G);lambda: Number;Abs(LineSegmentOf(N, F)) = lambda*Abs(LineSegmentOf(M, N))",
    "query_expressions": "Range(lambda)",
    "answer_expressions": "[sqrt(2)/2, 1]",
    "fact_spans": "[[[2, 16], [40, 43]], [[36, 39]], [[20, 23]], [[32, 35]], [[2, 16]], [[2, 23]], [[2, 35]], [[36, 49]], [[75, 84]], [[53, 73]]]",
    "query_spans": "[[[75, 91]]]",
    "process": "From the given conditions, F(0,1), M(0,-1). Let N(x_{0},y_{0}), then x_{0}^{2}=4y_{0} (y_{0}\\geqslant0). When |NF|=\\lambda|MN|, \\lambda=1; when y_{0}>0, \\lambda=1-\\frac{4}{y_{0}+\\frac{1}{y_{0}}+6}. Since y_{0}+\\frac{1}{y_{0}}\\geqslant2, it follows that \\sqrt{1-\\frac{4}{2+6}}\\leqslant\\lambda<1, i.e., \\frac{\\sqrt{2}}{2}\\leqslant\\lambda<1. In summary, \\lambda\\in\\left[\\frac{\\sqrt{2}}{2},1\\right)"
  },
  {
    "text": "If the asymptotes of a hyperbola are given by $y=\\pm 3 x$ and its focal distance is $2 \\sqrt{10}$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*3*x);FocalLength(G)=2*sqrt(10)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{(x^2-y^2/9=1),(y^2/9-x^2=1)}",
    "fact_spans": "[[[1, 4], [23, 24], [43, 46]], [[1, 22]], [[23, 41]]]",
    "query_spans": "[[[43, 51]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $ C $: $ y^{2} = 2 p x $ ($ p > 0 $) with directrix $ l_{1} $, which passes through a vertex of the hyperbola $ x^{2} - y^{2} = 1 $. Point $ P $ lies on the parabola $ C $, and the line $ l_{2} $: $ 3x - 4y + 27 = 0 $. Then, the minimum value of the sum of the distances from point $ P $ to the lines $ l_{1} $ and $ l_{2} $ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;l1: Line;Directrix(C) = l1;G: Hyperbola;Expression(G) = (x^2 - y^2 = 1);PointOnCurve(OneOf(Vertex(G)), l1);P: Point;PointOnCurve(P, C);l2: Line;Expression(l2) = (3*x - 4*y + 27 = 0)",
    "query_expressions": "Min(Distance(P, l1) + Distance(P, l2))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 28], [70, 76]], [[2, 28]], [[10, 28]], [[10, 28]], [[31, 38], [110, 119]], [[2, 38]], [[41, 59]], [[41, 59]], [[31, 64]], [[65, 69], [105, 109]], [[65, 77]], [[78, 103], [120, 127]], [[78, 103]]]",
    "query_spans": "[[[105, 138]]]",
    "process": "According to the problem, we calculate $ p = 2 $, obtain the parabolic equation and the focus coordinates, then transform the sum of distances from point $ P $ to lines $ l_{1} $ and $ l_{2} $ into the sum of the distance from point $ P $ to focus $ F $ and the distance to $ l_{2} $. Thus, the minimum value is the distance from point $ F $ to line $ l_{2} $. From the problem, the left vertex of the hyperbola $ x^{2} - y^{2} = 1 $ is $ (-1, 0) $, so $ -\\frac{p}{2} = -1 $, hence $ p = 2 $. Then the equation of the parabola is $ C: y^2 = 4x $, with focus $ F(1, 0) $. The sum of distances from point $ P $ to lines $ l_{1} $ and $ l_{2} $ can thus be transformed into the sum of the distance from point $ P $ to focus $ F $ and the distance to $ l_{2} $. Therefore, the minimum value of the sum of distances from point $ P $ to lines $ l_{1} $ and $ l_{2} $ is the distance from point $ F $ to line $ l_{2} $, $ d = \\frac{|3 + 27|}{\\sqrt{3^{2} + 4^{2}}} = 6 $."
  },
  {
    "text": "Given a moving point $P(x, y)$ on the ellipse $\\frac{x^{2}}{100}+\\frac{y^{2}}{64}=1$. If the coordinates of point $A$ are $(6,0)$, $|\\overrightarrow{A M}|=1$, and $\\overrightarrow{P M} \\cdot \\overrightarrow{A M}=0$, then the minimum value of $|\\overrightarrow{P M}|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;A: Point;M: Point;x1: Number;y1: Number;Expression(G) = (x^2/100 + y^2/64 = 1);Coordinate(A) = (6, 0);Coordinate(P) = (x1, y1);PointOnCurve(P, G);Abs(VectorOf(A, M)) = 1;DotProduct(VectorOf(P, M), VectorOf(A, M)) = 0",
    "query_expressions": "Min(Abs(VectorOf(P, M)))",
    "answer_expressions": "sqrt(15)",
    "fact_spans": "[[[14, 54]], [[4, 13]], [[57, 61]], [[155, 179]], [[4, 13]], [[4, 13]], [[14, 54]], [[57, 72]], [[4, 13]], [[4, 55]], [[74, 100]], [[102, 153]]]",
    "query_spans": "[[[155, 185]]]",
    "process": "\\because\\overrightarrow{PM}\\cdot\\overrightarrow{AM}=0,\\therefore PM\\bot AM,\\therefore|PM|=\\sqrt{|AP|^{2}-|AM|^{2}}, and \\because|\\overrightarrow{AM}|=1,\\therefore the smaller |AP| is, the smaller |PM| is. Let P(10\\cosx,8\\sinx), then |AP|^{2}=(10\\cosx-6)^{2}+(8\\sinx-0)^{2}=100\\cos^{2}x-120\\cosx+36+64\\sin^{2}x=36\\cos^{2}x-120\\cosx+100=(6\\cosx-10)^{2},\\therefore the minimum value of |AP| is \\sqrt{(6-10)^{2}}=4,\\therefore the minimum value of |PM| is: \\sqrt{4^{2}-1^{2}}=\\sqrt{15}"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through the right focus $F_{2}$ intersects the right branch of the hyperbola at points $A$ and $B$. If $|A B|=5$, then the perimeter of $\\triangle A B F_{1}$ is?",
    "fact_expressions": "l: Line;G: Hyperbola;A: Point;B: Point;F1: Point;F2: Point;Expression(G) = (x^2/16 - y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, l);Intersection(l, RightPart(G)) = {A, B};Abs(LineSegmentOf(A, B)) = 5",
    "query_expressions": "Perimeter(TriangleOf(A, B, F1))",
    "answer_expressions": "26",
    "fact_spans": "[[[79, 84]], [[2, 41], [85, 88]], [[92, 95]], [[96, 99]], [[50, 57]], [[58, 65], [71, 78]], [[2, 41]], [[2, 65]], [[2, 65]], [[66, 84]], [[79, 101]], [[103, 112]]]",
    "query_spans": "[[[115, 141]]]",
    "process": ""
  },
  {
    "text": "Given that the moving point $P$ travels along the curve $2 x^{2}-y=0$, then the equation of the trajectory of the midpoint of the line segment joining point $A(0,-1)$ and point $P$ is?",
    "fact_expressions": "G: Curve;Expression(G) = (2*x^2 - y = 0);P: Point;PointOnCurve(P, G);A: Point;Coordinate(A) = (0, -1)",
    "query_expressions": "LocusEquation(MidPoint(LineSegmentOf(A, P)))",
    "answer_expressions": "2*y=8*x^2-1",
    "fact_spans": "[[[8, 23]], [[8, 23]], [[4, 7], [39, 43]], [[4, 26]], [[28, 38]], [[28, 38]]]",
    "query_spans": "[[[28, 54]]]",
    "process": "Let the coordinates of the midpoint of AP be (x, y), then the coordinates of point P are (2x, 2y+1). Since point P moves on the curve 2x^{2}-y=0, we have 2(2x)^{2}-(2y+1)=0, which simplifies to 2y=8x^{2}-1."
  },
  {
    "text": "Given that $A$ and $B$ are any two points on the parabola $y^{2}=4x$ (the line $AB$ is not perpendicular to the $x$-axis), the perpendicular bisector of segment $AB$ intersects the $x$-axis at point $M(m, 0)$. Then the range of values for $m$ is?",
    "fact_expressions": "A: Point;B: Point;G: Parabola;M: Point;m:Number;Expression(G) = (y^2 = 4*x);Coordinate(M) = (m, 0);PointOnCurve(A,G);PointOnCurve(B,G);Negation(IsPerpendicular(LineOf(A,B),xAxis));Intersection(PerpendicularBisector(LineSegmentOf(A, B)), xAxis) = M",
    "query_expressions": "Range(m)",
    "answer_expressions": "(2, +oo)",
    "fact_spans": "[[[2, 5]], [[8, 11]], [[12, 26]], [[64, 74]], [[76, 79]], [[12, 26]], [[64, 74]], [[2, 31]], [[2, 31]], [[32, 46]], [[48, 74]]]",
    "query_spans": "[[[76, 86]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $x^{2}-y^{2}=1$, points $F_{1}$ and $F_{2}$ are its two foci, and point $P$ is a point on the hyperbola. If $P F_{1} \\perp P F_{2}$, then the value of $|P F_{1}|+|P F_{2}|$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2 = 1);Focus(G)={F1,F2};PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))",
    "query_expressions": "Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[2, 20], [38, 39], [49, 52]], [[44, 48]], [[21, 29]], [[30, 37]], [[2, 20]], [[21, 43]], [[44, 55]], [[57, 80]]]",
    "query_spans": "[[[82, 107]]]",
    "process": "According to the hyperbola equation $ x^{2}-y^{2}=1 $, we can obtain the focal distance $ F_{1}F_{2}=2\\sqrt{2} $. Since $ PF_{1}\\bot PF_{2} $, it follows that $ |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2} $. Combining with the definition of the hyperbola, we get $ |PF_{1}|\\cdot|PF_{2}|=\\pm2 $. Finally, by solving simultaneously and completing the square, we obtain $ (|PF_{1}|+|PF_{2}|)^{2}=12 $, thus the value of $ |PF_{1}|+|PF_{2}| $ is $ 2\\sqrt{3} $. $ \\because PF_{1}\\bot PF_{2} $, $ \\therefore |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2} $. $ \\because $ the hyperbola equation is $ x^{2}-y^{2}=1 $, $ \\therefore a^{2}=b^{2}=1 $, $ c^{2}=a^{2}+b^{2}=2 $, we obtain $ F_{1}F_{2}=2\\sqrt{2} $, $ \\therefore |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}=8 $. Also $ \\because P $ is a point on the hyperbola $ x^{2}-y^{2}=1 $, $ \\therefore |PF_{1}|\\cdot|PF_{2}|=\\pm2a=\\pm2 $, $ (|PF_{1}|\\cdot|PF_{2}|)^{2}=4 $. Therefore, $ (|PF_{1}|+|PF_{2}|)^{2}=2(|PF_{1}|^{2}+|PF_{2}|^{2})-(|PF_{1}|\\cdot|PF_{2}|)^{2}=12 $, $ \\therefore $ the value of $ |PF_{1}|+|PF_{2}| $ is $ 2\\sqrt{3} $."
  },
  {
    "text": "The inclination angle of the tangent line to the parabola $y=x^{2}$ at the point $M(\\frac{1}{2}, \\frac{1}{4})$ is equal to?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y = x^2);Coordinate(M) = (1/2, 1/4);PointOnCurve(M,G)",
    "query_expressions": "Inclination(TangentOfPoint(M, G))",
    "answer_expressions": "pi/4",
    "fact_spans": "[[[1, 13]], [[15, 45]], [[1, 13]], [[15, 45]], [[1, 45]]]",
    "query_spans": "[[[0, 55]]]",
    "process": "From the problem, first find the slope of the tangent line at point M, then write the equation of the line. From the problem, y' = 2x, so the slope of the line passing through point M is 1, thus the angle of inclination of the tangent line passing through point M is \\frac{\\pi}{4}."
  },
  {
    "text": "The parabola $y^{2}=2 x$ intersects a line passing through the focus at points $A$ and $B$, and $O$ is the origin. Then $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);H: Line;PointOnCurve(Focus(G), H)  ;Intersection(G, H) = {A, B};O: Origin;A: Point;B: Point",
    "query_expressions": "DotProduct(VectorOf(O, A), VectorOf(O, B))",
    "answer_expressions": "-3/4",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[19, 21]], [[0, 21]], [[0, 32]], [[33, 36]], [[23, 26]], [[27, 30]]]",
    "query_spans": "[[[41, 92]]]",
    "process": "(1) When line $ AB \\perp x $-axis, in $ y^{2} = 2x $, let $ x = \\frac{1}{2} $, then $ y = \\pm 1 $, so $ A\\left(\\frac{1}{2}, 1\\right) $, $ B\\left(\\frac{1}{2}, -1\\right) $, we get $ \\overline{OA} \\cdot \\overline{OB} = \\left(\\frac{1}{2}, 1\\right) \\cdot \\left(\\frac{1}{2}, -1\\right) = -\\frac{3}{4} $. (2) When line $ AB $ is not perpendicular to the $ x $-axis, let the equation of $ AB $ be: $ y = k\\left(x - \\frac{1}{2}\\right) $. From \n\\[\n\\begin{cases}\ny = k\\left(x - \\frac{1}{2}\\right), \\\\\ny^{2} = 2x\n\\end{cases}\n\\]\neliminating $ y $, we obtain $ k^{2}x^{2} - \\left(k^{2} + 2\\right)x + \\frac{1}{4}k^{2} = 0 $, clearly $ k \\neq 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = \\frac{k^{2} + 2}{k^{2}} $, $ x_{1} \\cdot x_{2} = \\frac{1}{4} $, we get \n\\[\n\\overrightarrow{OA} \\cdot \\overrightarrow{OB} = (x_{1}, y_{1}) \\cdot (x_{2}, y_{2}) = x_{1} \\cdot x_{2} + y_{1}y_{2} = x_{1} \\cdot x_{2} + k\\left(x_{1} - \\frac{1}{2}\\right) \\cdot k\\left(x_{2} - \\frac{1}{2}\\right) = (1 + k^{2})x_{1} \\cdot x_{2} - \\frac{k^{2}}{2}(x_{1} + x_{2}) + \\frac{1}{4}k^{2} = \\frac{1}{4}(1 + k^{2}) - \\frac{k^{2}}{2} \\cdot \\frac{k^{2} + 2}{k^{2}} + \\frac{1}{4}k^{2} = -\\frac{3}{4}.\n\\]\nIn conclusion (1) and (2), we have $ \\overline{OA} \\cdot \\overline{OB} = -\\frac{3}{4} $."
  },
  {
    "text": "If a hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ has an asymptote whose chord length intercepted by the circle $(x-2)^{2}+y^{2}=4$ is $2$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;G: Circle;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y^2 + (x - 2)^2 = 4);Length(InterceptChord(OneOf(Asymptote(C)),G))=2",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 65], [103, 109]], [[8, 65]], [[8, 65]], [[72, 92]], [[8, 65]], [[8, 65]], [[1, 65]], [[72, 92]], [[1, 102]]]",
    "query_spans": "[[[103, 115]]]",
    "process": ""
  },
  {
    "text": "The equation $\\frac{x^{2}}{k-3}+\\frac{y^{2}}{k+3}=1$ represents an ellipse; then the range of values for $k$ is?",
    "fact_expressions": "G: Ellipse;Expression(G)=(x^2/(k - 3) + y^2/(k + 3) = 1) ;k:Number",
    "query_expressions": "Range(k)",
    "answer_expressions": "(3,+oo)",
    "fact_spans": "[[[43, 45]], [[0, 45]], [[47, 50]]]",
    "query_spans": "[[[47, 57]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, its asymptotes are tangent to the parabola $y=x^{2}+1$. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;G: Parabola;Expression(G) = (y = x^2 + 1);IsTangent(Asymptote(C), G)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 63], [86, 89]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 82]], [[68, 82]], [[2, 84]]]",
    "query_spans": "[[[86, 95]]]",
    "process": ""
  },
  {
    "text": "If the chord of the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$ is bisected by the point $A(1 , 1)$, then what is the equation of the line on which this chord lies?",
    "fact_expressions": "G: Ellipse;H: LineSegment;A: Point;Expression(G) = (x^2/36 + y^2/9 = 1);Coordinate(A) = (1, 1);IsChordOf(H, G);MidPoint(H) = A",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "x + 4*y - 5 = 0",
    "fact_spans": "[[[2, 40]], [], [[42, 53]], [[2, 40]], [[42, 53]], [[2, 41]], [[2, 55]]]",
    "query_spans": "[[[2, 70]]]",
    "process": "Let the endpoints of the chord be $ M(x_{1},y_{1}) $, $ N(x_{2},y_{2}) $, then we have \n\\[\n\\begin{cases}\n\\frac{x_{1}^{2}}{36} + \\frac{y_{1}^{2}}{9} = 1 \\\\\n\\frac{x_{2}^{2}}{36} + \\frac{y_{2}^{2}}{9} = 1\n\\end{cases}\n\\]\nSubtracting the two equations gives \n\\[\n\\frac{x_{1}^{2} - x_{2}^{2}}{36} + \\frac{y_{1}^{2} - y_{2}^{2}}{9} = 0,\n\\]\nwhich simplifies to \n\\[\n\\frac{y_{1} - y_{2}}{x_{1} - x_{2}} \\cdot \\frac{y_{1} + y_{2}}{x_{1} + x_{2}} = -\\frac{1}{4},\n\\quad\n\\frac{y_{1} - y_{2}}{x_{1} - x_{2}} = -\\frac{1}{4}.\n\\]\nFrom the given conditions, $ x_{1} + x_{2} = 2 $, $ y_{1} + y_{2} = 2 $, so the slope of the line containing the midpoint chord is $ -\\frac{1}{4} $. Thus, the equation is $ y - 1 = -\\frac{1}{4}(x - 1) $, which simplifies to $ x + 4y - 5 = 0 $."
  },
  {
    "text": "Given that the eccentricity of a hyperbola is $\\sqrt{2}$, and it has the same foci as the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$, then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/8 + y^2/4 = 1);Eccentricity(G) = sqrt(2);Focus(G) = Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/2 - y^2/2 = 1",
    "fact_spans": "[[[2, 5], [69, 72]], [[23, 60]], [[23, 60]], [[2, 20]], [[2, 66]]]",
    "query_spans": "[[[69, 79]]]",
    "process": "First, find the coordinates of the foci of the ellipse, which gives the foci of the hyperbola as $(\\pm2,0)$. Let the equation of the hyperbola be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$. From $e=\\frac{2}{a}=\\sqrt{2}$, we get $a=\\sqrt{2}$. Then, using $b^{2}=c^{2}-a^{2}$, we can find the value of $b$, thus obtaining the standard equation of the hyperbola. In the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$, $c^{2}=8-4=4$, so $c=2$. Therefore, the foci are $(-2,0)$, $(2,0)$. Let the equation of the hyperbola be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{k^{2}}=1$. From the given condition, $e=\\frac{2}{a}=\\sqrt{2}$, so $a=\\sqrt{2}$. Thus, $b^{2}=c^{2}-a^{2}=4-2=2$. Hence, the equation of the hyperbola is $\\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $A$ lies on the hyperbola $C$, $A F_{2} \\perp F_{1} F_{2}$, the line $A F_{1}$ intersects the hyperbola $C$ at another point $B$, and $\\overrightarrow{F_{1} A}=4 \\overrightarrow{F_{1} B}$. Then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;A: Point;PointOnCurve(A, C);IsPerpendicular(LineSegmentOf(A, F2), LineSegmentOf(F1, F2));Intersection(LineOf(A, F1), C) = {A, B};B: Point;VectorOf(F1, A) = 4*VectorOf(F1, B)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 63], [93, 99], [141, 147], [211, 217]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[72, 79]], [[80, 87]], [[2, 87]], [[2, 87]], [[88, 92]], [[88, 100]], [[101, 128]], [[129, 155]], [[152, 155]], [[156, 209]]]",
    "query_spans": "[[[211, 223]]]",
    "process": "Since $AF_{2}\\bot F_{1}F_{2}$, assume the coordinates of point $A$ are $(c,m)$ $(m>0)$, and the coordinates of point $B$ are $(s,t)$, with $\\frac{c^{2}}{a^{2}}-\\frac{m^{2}}{b^{2}}=1$, solving gives $m=\\frac{b^{2}}{a}$. Also, from $\\overrightarrow{F_{1}A}=(2c,\\frac{b^{2}}{a})$, $\\overrightarrow{F_{1}B}=(s+c,y)$, we have $(2c,\\frac{b^{2}}{a})=4(s+c,y)$, solving gives $s=-\\frac{c}{2}$, $t=\\frac{b^{2}}{4a}$. Substituting the coordinates of point $B$ into the hyperbola equation, we get $\\frac{c^{2}}{4a^{2}}-\\frac{b^{2}}{16a^{2}}=1$, $\\frac{4c^{2}-b^{2}}{16a^{2}}=1$, $3c^{2}+a^{2}=16a^{2}$, solving gives $c=\\sqrt{5}a$. The eccentricity of hyperbola $C$ is $e=\\frac{c}{a}=\\sqrt{5}$."
  },
  {
    "text": "Given two fixed points $A(0,4)$, $B(0,1)$, a moving point $P$ satisfies $|P A|=2|P B|$, then the trajectory equation of point $P$ is?",
    "fact_expressions": "A: Point;B: Point;P: Point;Coordinate(A) = (0, 4);Coordinate(B) = (0, 1);Abs(LineSegmentOf(P, A)) = 2*Abs(LineSegmentOf(P, B))",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2+y^2=4",
    "fact_spans": "[[[6, 14]], [[17, 25]], [[28, 31], [51, 54]], [[6, 14]], [[17, 25]], [[33, 47]]]",
    "query_spans": "[[[51, 61]]]",
    "process": "Let the coordinates of point P be (x, y). Using the distance formula between two points, express the condition |PA| = 2|PB| in coordinate form, then simplify to obtain the solution. Let the coordinates of point P be (x, y). From |PA| = 2|PB|, we get  \n\\sqrt{x^{2}+(y-4)^{2}} = 2\\sqrt{x^{2}+(y-1)^{2}}.  \nSimplifying yields x^{2} + y^{2} = 4. Therefore, the trajectory equation of the moving point P is x^{2} + y^{2} = 4."
  },
  {
    "text": "It is known that point $P$ is a moving point on the parabola $y^{2}=4x$, the projection of point $P$ onto the $y$-axis is $M$, and the coordinates of point $A$ are $(4, a)$. Then, when $|a|>4$, the minimum value of $|PA|+|PM|$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G);M: Point;Projection(P, yAxis) = M;A: Point;a: Number;Coordinate(A) = (4, a);Abs(a)>4",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[7, 21]], [[7, 21]], [[2, 6], [26, 30]], [[2, 25]], [[40, 43]], [[26, 43]], [[44, 48]], [[52, 61]], [[44, 61]], [[64, 71]]]",
    "query_spans": "[[[73, 90]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $x^{2}=2 p y(p>0)$ with focus $F$, if a point $M(x_{0}, 3)$ on $C$ is at a distance of $6$ from the focus $F$, then the value of $x_{0}$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 2*p*y);p: Number;p>0;F: Point;Focus(C) = F;M: Point;Coordinate(M) = (x0, 3);x0: Number;PointOnCurve(M, C) = True;Distance(M, F) = 6",
    "query_expressions": "x0",
    "answer_expressions": "pm*6",
    "fact_spans": "[[[2, 28], [37, 40]], [[2, 28]], [[9, 28]], [[9, 28]], [[32, 35], [59, 62]], [[2, 35]], [[43, 56]], [[43, 56]], [[71, 78]], [[37, 56]], [[43, 69]]]",
    "query_spans": "[[[71, 82]]]",
    "process": "According to the problem, |MF| = 3 + \\frac{p}{2} = 6, solving gives p = 6, thus the parabola C: x^{2} = 12y; substituting M(x_{0}, 3) yields x_{0}^{2} = 36, then x_{0} = \\pm6."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{m}+\\frac{y^{2}}{m^{2}-2}=1$ represents an ellipse, then the range of real values for $m$ is?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G)=(y^2/(m^2 - 2) + x^2/m = 1 )",
    "query_expressions": "Range(m)",
    "answer_expressions": "(\\sqrt{2},2)+(2,+\\infty)",
    "fact_spans": "[[[46, 48]], [[50, 55]], [[1, 48]]]",
    "query_spans": "[[[50, 62]]]",
    "process": "The equation $\\frac{x^{2}}{m}+\\frac{y^{2}}{m^{2}-2}=1$ represents an ellipse, then $\\begin{cases}m>0\\\\m^{2}-2>0\\\\m\\neq m^{2}-2\\end{cases}$, solving gives $m>\\sqrt{2}$ and $m\\neq2$, that is, $m\\in(\\sqrt{2},2)\\cup(2,+\\infty)$"
  },
  {
    "text": "Let the focus of the parabola $x^{2}=4 y$ be $F$. A line $l$ passing through point $P(1 , 4)$ intersects the parabola at points $A$ and $B$, and point $P$ is exactly the midpoint of $AB$. Then $|\\overrightarrow{A F}|+|\\overrightarrow{B F}|$=?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;P: Point;F: Point;Expression(G) = (x^2 = 4*y);Coordinate(P) = (1, 4);Focus(G) = F;PointOnCurve(P, l);Intersection(l, G) = {A,B};MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Abs(VectorOf(A, F)) + Abs(VectorOf(B, F))",
    "answer_expressions": "10",
    "fact_spans": "[[[37, 42]], [[43, 46], [1, 15]], [[49, 52]], [[55, 58]], [[25, 36], [62, 66]], [[19, 22]], [[1, 15]], [[25, 36]], [[1, 22]], [[23, 42]], [[37, 60]], [[62, 75]]]",
    "query_spans": "[[[77, 127]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=4x$ with focus $F$ and directrix $l$, a line passing through point $F$ with slope $\\sqrt{3}$ intersects the parabola at point $M$ (where $M$ is in the first quadrant), $MN \\perp l$ with foot of perpendicular at $N$, and line $NF$ intersects the $y$-axis at point $D$. Then $|MD|$=?",
    "fact_expressions": "G: Parabola;H: Line;F: Point;N: Point;M: Point;D: Point;l: Line;Expression(G) = (y^2 = 4*x);Focus(G) = F;Directrix(G) = l;PointOnCurve(F, H);Slope(H) = sqrt(3);Intersection(H, G) = M;IsPerpendicular(LineSegmentOf(M, N),l);FootPoint(LineSegmentOf(M, N),l)=N;Intersection(LineOf(N,F), yAxis) = D;Quadrant(M)=1",
    "query_expressions": "Abs(LineSegmentOf(M, D))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[2, 16], [54, 57]], [[51, 53]], [[20, 23], [32, 36]], [[90, 93]], [[58, 62], [63, 66]], [[107, 111]], [[27, 30]], [[2, 16]], [[2, 23]], [[2, 30]], [[31, 53]], [[37, 53]], [[51, 62]], [[73, 86]], [[73, 93]], [[94, 111]], [[63, 71]]]",
    "query_spans": "[[[113, 122]]]",
    "process": "As shown in the figure, let the directrix intersect the x-axis at E. It is easy to know that F(1,0), EF=2. By the definition of the parabola, |MN|=|MF|. According to the problem, \\angle MFx=60^{\\circ}, \\therefore \\angle NMF=60^{\\circ}, \\therefore \\triangle NMF is an equilateral triangle, \\therefore \\angle NFE=60^{\\circ}, \\therefore |NM|=\\frac{|EF|}{\\cos60^{\\circ}}=2|FE|=4. Also, OD is the midline of \\triangle FEN, \\therefore MD is the height of the equilateral \\triangle NMF, \\therefore |MD|=2\\sqrt{3}"
  },
  {
    "text": "Given that an asymptote of the hyperbola $x^{2}-\\frac{y^{2}}{a}=1$ is perpendicular to the line $x-2 y+3=0$, then $a$=?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Line;Expression(G) = (x^2 - y^2/a = 1);Expression(H) = (x - 2*y + 3 = 0);IsPerpendicular(OneOf(Asymptote(G)), H)",
    "query_expressions": "a",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 30]], [[54, 57]], [[37, 50]], [[2, 30]], [[37, 50]], [[2, 52]]]",
    "query_spans": "[[[54, 59]]]",
    "process": ""
  },
  {
    "text": "A point $M$ on the parabola $y^{2}=2x$ has a distance of $1$ to the focus. Then the horizontal coordinate of point $M$ is?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = 2*x);PointOnCurve(M, G);Distance(M, Focus(G)) = 1",
    "query_expressions": "XCoordinate(M)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[0, 14]], [[17, 20], [32, 36]], [[0, 14]], [[0, 20]], [[0, 30]]]",
    "query_spans": "[[[32, 42]]]",
    "process": "\\because the parabola equation is y^{2}=2x, \\therefore the focus of the parabola is F(\\frac{1}{2},0), let point M(x_{0},y_{0}), we get =1, solving gives x_{0}=\\frac{1}{2} (discard negative),"
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$, respectively. The line $l$ with slope $1$ passing through $F_{1}$ intersects the right branch of $C$ at point $P$. If $\\angle{F }_{1} F_{2} P=90^{\\circ}$, then the eccentricity of the hyperbola is equal to?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;F2: Point;F1: Point;a>0;b>0;LeftFocus(C)= F1;RightFocus(C)= F2;l:Line;Slope(l)=1;PointOnCurve(F1,l);P:Point;Intersection(l,RightPart(C))=P;AngleOf(F1, F2, P)=ApplyUnit(90, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "1+sqrt(2)",
    "fact_spans": "[[[18, 81], [110, 113], [161, 164]], [[18, 81]], [[25, 81]], [[25, 81]], [[8, 15]], [[0, 7], [96, 103]], [[25, 81]], [[25, 81]], [[0, 87]], [[0, 87]], [[104, 109]], [[88, 109]], [[95, 109]], [[118, 122]], [[104, 122]], [[124, 159]]]",
    "query_spans": "[[[161, 171]]]",
    "process": ""
  },
  {
    "text": "The perimeter of the triangle formed by a point $M$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ and its two foci is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);M:Point;F1:Point;F2:Point;PointOnCurve(M,G);Focus(G)={F1,F2}",
    "query_expressions": "Perimeter(TriangleOf(M,F1,F2))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 38]], [[0, 38]], [[41, 44]], [], [], [[0, 44]], [[0, 49]]]",
    "query_spans": "[[[0, 59]]]",
    "process": "The solution process is omitted"
  },
  {
    "text": "The standard equation of a parabola with directrix $y=\\frac{2}{3}$ is?",
    "fact_expressions": "G: Parabola;Expression(Directrix(G))=(y=2/3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2=-(8/3)*y",
    "fact_spans": "[[[21, 24]], [[0, 24]]]",
    "query_spans": "[[[21, 31]]]",
    "process": ""
  },
  {
    "text": "If a focus of the hyperbola $x^{2}-\\frac{y^{2}}{m}=1$ is $(2,0)$, then $m=$?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2 - y^2/m = 1);Coordinate(OneOf(Focus(G))) = (2, 0)",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 29]], [[44, 47]], [[1, 29]], [[1, 42]]]",
    "query_spans": "[[[44, 49]]]",
    "process": "Since one focus of the hyperbola $x^{2}-\\frac{y^{2}}{m}=1$ is $(2,0)$, we have $1+m=2^{2}$, solving gives $m=3$. Therefore, $m=3$."
  },
  {
    "text": "Given that $M$ is a point on the parabola $C$: $y^{2}=2 p x(p>0)$, $F$ is the focus of $C$, and a perpendicular is drawn from $M$ to the directrix of $C$, with foot of perpendicular at $N$. If $\\angle M F O=120^{\\circ}$ ($O$ is the origin), and the perimeter of $\\triangle M N F$ is $12$, then $|N F|$=?",
    "fact_expressions": "C: Parabola;p: Number;M: Point;N: Point;F: Point;O: Origin;p>0;Expression(C) = (y^2 = 2*(p*x));PointOnCurve(M, C);Focus(C) = F;L:Line;PointOnCurve(M,L);IsPerpendicular(L,Directrix(C));FootPoint(L,Directrix(C))=N;AngleOf(M, F, O) = ApplyUnit(120, degree);Perimeter(TriangleOf(M, N, F)) = 12",
    "query_expressions": "Abs(LineSegmentOf(N, F))",
    "answer_expressions": "4",
    "fact_spans": "[[[6, 32], [40, 43], [52, 55]], [[14, 32]], [[2, 5], [48, 51]], [[65, 68]], [[36, 39]], [[98, 101]], [[14, 32]], [[6, 32]], [[2, 35]], [[36, 46]], [], [[47, 61]], [[47, 61]], [[47, 68]], [[70, 97]], [[108, 133]]]",
    "query_spans": "[[[135, 144]]]",
    "process": "From \\angle MFO=120^{\\circ}, we obtain \\angle FMN=60^{\\circ}, and further deduce that \\triangle FMN is an equilateral triangle, thus enabling the solution. As shown in the figure, since \\angle MFO=120^{\\circ}, it follows that \\angle FMN=60^{\\circ}. Also, since M is a point on the parabola C, we have |FM|=|MN|. Therefore, \\triangle FMN is an equilateral triangle. Given that the perimeter of \\triangle FMN is 12, we obtain |NF|=4."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ has focus $F$, directrix $l$, a line passing through point $F$ intersects the parabola at points $A$ and $B$ (with point $B$ in the first quadrant), and intersects the directrix $l$ at point $P$. If $\\overrightarrow{A F}=\\frac{1}{2} \\overrightarrow{F B}$, $\\overrightarrow{A P}=\\lambda \\overrightarrow{A F}$, then $\\lambda=$?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;A: Point;F: Point;B: Point;P: Point;l: Line;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;Directrix(G) = l;PointOnCurve(F,H);Intersection(H,G)={A,B};Quadrant(B)=1;Intersection(H, l) = P;VectorOf(A, F) = VectorOf(F, B)/2;VectorOf(A, P) = lambda*VectorOf(A, F);lambda:Number",
    "query_expressions": "lambda",
    "answer_expressions": "-3",
    "fact_spans": "[[[2, 23], [47, 50]], [[5, 23]], [[44, 46]], [[52, 55]], [[27, 30], [39, 43]], [[56, 59], [62, 66]], [[81, 85]], [[34, 37], [76, 79]], [[5, 23]], [[2, 23]], [[2, 30]], [[2, 37]], [[38, 46]], [[44, 61]], [[62, 71]], [[44, 85]], [[88, 143]], [[144, 195]], [[197, 206]]]",
    "query_spans": "[[[197, 208]]]",
    "process": "Draw AA ⊥ l from point A, with foot at A, and draw BB ⊥ l from point B, with foot at B. By the definition of a parabola, |AA| = |AF|, |BB| = |BF|. Assume |AF| = x. Since \\overrightarrow{AF} = \\frac{1}{2}\\overrightarrow{FB}, it follows that |FB| = 2x. Because \\triangle PAA \\sim \\triangle PBB, we have \\frac{|PA|}{|PB|} = \\frac{|AA|}{|BB|} = \\frac{|AF|}{|BF|} = \\frac{1}{2}. Thus, \\frac{|PA|}{|PA| + |AB|} = \\frac{|PA|}{|PA| + 3x} = \\frac{1}{2}, so |PA| = 3x. Therefore, \\frac{|PA|}{|AF|} = \\frac{3x}{x} = 3. Since \\overrightarrow{AP} and \\overrightarrow{AF} are in opposite directions, \\lambda = -3."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. The line passing through $F_{1}$ and perpendicular to the $x$-axis intersects the ellipse at points $A$ and $B$. The line $A F_{2}$ intersects the ellipse again at point $C$. If $S_{\\triangle A B C}=3 S_{\\triangle B C F_{2}}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;F2: Point;A: Point;B: Point;C: Point;F1: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(F1, H);IsPerpendicular(H,xAxis);Intersection(H, G) = {A, B};Intersection(LineOf(A,F2),G)={A,C};Area(TriangleOf(A, B, C)) = 3*Area(TriangleOf(B, C, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/5",
    "fact_spans": "[[[2, 54], [101, 103], [126, 128], [189, 191]], [[4, 54]], [[4, 54]], [[98, 100]], [[72, 79]], [[104, 107]], [[108, 111]], [[135, 138]], [[63, 70], [82, 89]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 79]], [[2, 79]], [[81, 100]], [[90, 100]], [[98, 113]], [[104, 138]], [[140, 187]]]",
    "query_spans": "[[[189, 197]]]",
    "process": "Let the left and right foci of the ellipse be $ F_{1}(-c,0) $, $ F_{2}(c,0) $. Substituting $ x = -c $ into the ellipse equation yields $ y = \\pm\\frac{b^{2}}{a} $, so we can set $ A(-c,\\frac{b^{2}}{a}) $, $ C(x,y) $. From $ S_{\\triangle ABC} = 3S_{\\triangle BCF_{2}} $, we get $ \\overrightarrow{AF_{2}} = 2\\overrightarrow{F_{2}C} $, that is, $ (2c,-\\frac{b^{2}}{a}) = 2(x-c,y) $, which gives $ x = 2c $, $ y = -\\frac{b^{2}}{2a} = 2y' $. Substituting into the ellipse equation yields $ \\frac{4c^{2}}{a^{2}} + \\frac{b^{2}}{4a^{2}} = 1 $. Using $ e = \\frac{c}{a} $, $ b^{2} = a^{2} - c^{2} $, we obtain $ 4e^{2} + \\frac{1}{4} - \\frac{1}{4}e^{2} = 1 $, solving gives $ e = \\frac{\\sqrt{5}}{5} $. Hence the answer is: $ \\frac{\\sqrt{5}}{6} $"
  },
  {
    "text": "Construct a circle centered at the right focus $F_{2}$ of an ellipse such that the circle passes through the center of the ellipse and intersects the ellipse at points $M$, $N$. If the line $MF_{1}$ passing through the left focus $F_{1}$ of the ellipse is tangent to the circle $F_{2}$, then what is the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;H: Circle;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;Center(H) = F2;PointOnCurve(Center(G), H);M: Point;N: Point;Intersection(H, G) = {M, N};PointOnCurve(F1, LineOf(M, F1));IsTangent(LineOf(M, F1), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[1, 3], [26, 28], [32, 34], [45, 47], [82, 84]], [[20, 21], [24, 25], [69, 77]], [[50, 57]], [[7, 14]], [[45, 57]], [[1, 14]], [[0, 21]], [[24, 30]], [[35, 39]], [[40, 43]], [[24, 43]], [[44, 68]], [[58, 80]]]",
    "query_spans": "[[[82, 90]]]",
    "process": ""
  },
  {
    "text": "The directrix $l$ of the parabola $x^{2}=2 p y(p>0)$ is intercepted by the circle $x^{2}+y^{2}-6 x-1=0$ to form a chord of length $4$. Then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;H: Circle;p>0;l:Line;Expression(G) = (x^2 = 2*p*y);Expression(H) = (-6*x + x^2 + y^2 - 1 = 0);Directrix(G)=l;Length(InterceptChord(l,H))=4",
    "query_expressions": "p",
    "answer_expressions": "2*sqrt(6)",
    "fact_spans": "[[[0, 21]], [[61, 64]], [[28, 50]], [[3, 21]], [[24, 27]], [[0, 21]], [[28, 50]], [[0, 27]], [[24, 59]]]",
    "query_spans": "[[[61, 66]]]",
    "process": "According to the problem, the center of the circle $(x-3)^{2}+y^{2}=10$ is at $(3,0)$, and the radius is $r=\\sqrt{10}$. The directrix of the parabola $x^{2}=2py$ $(p>0)$ is given by $l: y=-\\frac{p}{2}$. Since the chord length intercepted on the circle $x^{2}+y^{2}-6x-1=0$ by the directrix $l$ of the parabola $x^{2}=2py$ is 4, the distance from the center $(3,0)$ to the directrix $l$ is $\\frac{p}{2}=\\sqrt{10-4}=\\sqrt{6}$, solving gives $p=2\\sqrt{6}$."
  },
  {
    "text": "Given that point $P$ is a point on the ellipse $C$: $\\frac{x^{2}}{36}+\\frac{y^{2}}{20}=1$, and $AB$ is a diameter of the circle $M$: $x^{2}-8x+y^{2}=0$, then the range of values for $\\overrightarrow{PA} \\cdot \\overrightarrow{PB}$ is?",
    "fact_expressions": "C: Ellipse;M: Circle;A: Point;B: Point;P: Point;Expression(C) = (x^2/36 + y^2/20 = 1);Expression(M) = (y^2 + x^2 - 8*x = 0);PointOnCurve(P, C);IsDiameter(LineSegmentOf(A, B), M)",
    "query_expressions": "Range(DotProduct(VectorOf(P, A), VectorOf(P, B)))",
    "answer_expressions": "[-12,84]",
    "fact_spans": "[[[7, 51]], [[61, 86]], [[55, 60]], [[55, 60]], [[2, 6]], [[7, 51]], [[61, 86]], [[2, 54]], [[55, 89]]]",
    "query_spans": "[[[91, 147]]]",
    "process": "x^{2}-8x+y^{2}=0 can be rewritten as (x-4)^{2}+y^{2}=16. The center M(4,0) is the right focus of the ellipse C, so \\overrightarrow{PA}\\cdot\\overrightarrow{PB}=(\\overrightarrow{PM}+\\overrightarrow{MA})\\cdot(\\overrightarrow{PM}+\\overrightarrow{MB})=\\overrightarrow{PM}^{2}-\\overrightarrow{MA}^{2}=\\overrightarrow{PM}^{2}-16. Since P is a point on the ellipse C: \\frac{x^{2}}{36}+\\frac{y^{2}}{20}=1, |\\overrightarrow{PM}|\\in[2,10]. Therefore, the range of \\overrightarrow{PA}\\cdot\\overrightarrow{PB} is [-12,84]."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=64 x$, a line passing through $F$ with slope $1$ intersects $C$ at points $A$ and $B$, and $|F A| > |F B|$, then $\\frac{|F A|}{|F B|} = $?",
    "fact_expressions": "C: Parabola;G: Line;A: Point;B: Point;F: Point;Expression(C) = (y^2 = 64*x);Focus(C)=F;PointOnCurve(F,G);Intersection(G,C)={A,B};Abs(LineSegmentOf(F, A)) > Abs(LineSegmentOf(F, B));Slope(G)=1",
    "query_expressions": "Abs(LineSegmentOf(F, A))/Abs(LineSegmentOf(F, B))",
    "answer_expressions": "3+2*sqrt(2)",
    "fact_spans": "[[[6, 25], [44, 47]], [[41, 43]], [[48, 51]], [[53, 56]], [[2, 5], [30, 33]], [[6, 25]], [[2, 28]], [[29, 43]], [[41, 58]], [[60, 73]], [[34, 43]]]",
    "query_spans": "[[[75, 98]]]",
    "process": ""
  },
  {
    "text": "The point $P(8, 1)$ bisects a chord of the hyperbola $x^{2}-4 y^{2}=4$. Then the equation of the line on which this chord lies is?",
    "fact_expressions": "P: Point;Coordinate(P) = (8, 1);G: Hyperbola;Expression(G) = (x^2 - 4*y^2 = 4);H: LineSegment;IsChordOf(H, G);MidPoint(H) = P",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "2*x-y-15=0",
    "fact_spans": "[[[0, 11]], [[0, 11]], [[13, 33]], [[13, 33]], [], [[13, 37]], [[0, 37]]]",
    "query_spans": "[[[13, 51]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=16 x$, a line $l$ with inclination angle $\\frac{\\pi}{6}$ passes through the focus $F$ and intersects the parabola at points $A$ and $B$, $O$ is the origin. Then the area of $\\triangle A B O$ is?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;O: Origin;F:Point;Expression(C) = (y^2 = 16*x);Inclination(l)=pi/6;Focus(C)=F;PointOnCurve(F,l);Intersection(l,C) = {A, B}",
    "query_expressions": "Area(TriangleOf(A, B, O))",
    "answer_expressions": "64",
    "fact_spans": "[[[43, 48]], [[2, 22], [55, 58]], [[59, 62]], [[63, 66]], [[69, 72]], [[51, 54]], [[2, 22]], [[23, 48]], [[2, 54]], [[43, 54]], [[43, 68]]]",
    "query_spans": "[[[79, 101]]]",
    "process": "The focus of the parabola $ y^{2}=16x $ is $ F(4,0) $, the equation of line $ l $ is $ y=\\frac{\\sqrt{3}}{3}(x-4) $. The line $ AB $ is given by $ y=\\frac{\\sqrt{3}}{3}(x-4) $, or $ x=\\sqrt{3}y+4 $. Substituting into $ y^{2}=16x $ yields: $ y^{2}=16(\\sqrt{3}y+4) $, that is, $ y^{2}-16\\sqrt{3}y-64=0 $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, $ \\therefore y_{1}+y_{2}=16\\sqrt{3} $, $ y_{1}y_{2}=-64 $, $ \\therefore y_{1}-y_{2}=\\sqrt{(y_{1}+y_{2})^{2}-4y_{1}y_{2}}=\\sqrt{16^{2}\\times}=16\\sqrt{3} $, $ y_{1}y_{2}=-64 $,"
  },
  {
    "text": "Let the parabola $ C $: $ y^{2} = 2 p x $ ($ p > 0 $) have focus $ F $ and directrix $ l $. Let point $ A $ be a point on $ C $, and a circle centered at $ F $ with radius $ FA $ intersects $ l $ at points $ B $ and $ D $. If $ \\angle FBD = 30^{\\circ} $, and the area of $ \\triangle ABD $ is $ \\sqrt{3} $, then the equation of the parabola $ C $ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;l: Line;Directrix(C) = l;A: Point;PointOnCurve(A, C);G: Circle;Center(G) = F;Radius(G)=LineSegmentOf(F,A);B: Point;D: Point;Intersection(G, l) = {B, D};AngleOf(F, B, D) = ApplyUnit(30, degree);Area(TriangleOf(A, B, D)) = sqrt(3)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=sqrt(2)*x",
    "fact_spans": "[[[1, 27], [47, 50], [149, 155]], [[1, 27]], [[9, 27]], [[9, 27]], [[31, 34], [55, 58]], [[1, 34]], [[38, 41], [73, 76]], [[1, 41]], [[42, 46]], [[42, 53]], [[71, 72]], [[54, 72]], [[62, 72]], [[77, 80]], [[81, 84]], [[71, 86]], [[88, 113]], [[116, 147]]]",
    "query_spans": "[[[149, 160]]]",
    "process": "By the given condition, $ F\\left(\\frac{p}{2},0\\right) $, $ l $: and $ |BD| = 2r\\cos30^{\\circ} = 2\\sqrt{3}p $, by the definition of the parabola, the distance from $ A $ to $ l $ is $ d = |FA| = 2p $, $ \\therefore S_{\\triangle ABD} = \\frac{1}{2}d|BD| = \\sqrt{3} $, that is $ 2\\sqrt{3}p^{2} = \\sqrt{3} $, solving gives $ p = \\frac{\\sqrt{2}}{2} $, $ \\therefore $ the equation of the parabola is $ y^{2} = \\sqrt{2}x $."
  },
  {
    "text": "Given that the left focus of the ellipse is $F_{1}$ and the right focus is $F_{2}$. If there exists a point $P$ on the ellipse such that the line segment $P F_{2}$ is tangent to the circle with the minor axis of the ellipse as its diameter, and the point of tangency is the midpoint of the segment $P F_{2}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;H: Circle;F2: Point;P: Point;F1: Point;LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P, G);IsTangent(LineSegmentOf(P,F2),H);IsDiameter(MinorAxis(G),H);TangentPoint(LineSegmentOf(P,F2),H)=MidPoint(LineSegmentOf(P,F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[2, 4], [31, 33], [59, 61], [90, 92]], [[68, 69]], [[21, 28]], [[38, 41]], [[9, 16]], [[2, 16]], [[2, 28]], [[31, 41]], [[44, 69]], [[58, 69]], [[44, 87]]]",
    "query_spans": "[[[90, 98]]]",
    "process": "Since the line segment $ PF_{2} $ is tangent to the circle with the minor axis of the ellipse as its diameter, and the point of tangency is the midpoint $ M $ of the segment $ PF_{2} $, then $ OM \\parallel PF_{1} $, $ OM \\perp PF_{2} $, $ \\therefore PF_{1} \\perp PF_{2} $. Let $ |F_{1}F_{2}| = 2c $, $ |PF_{1}| = 2|OM| = 2b $. By the definition of the ellipse, $ |PF_{2}| = 2a - 2b $. By the Pythagorean theorem, $ 4b^{2} + (2a - 2b)^{2} = 4c^{2} $. Solving gives $ b^{2} = \\frac{4}{9}a^{2} $, $ c^{2} = \\frac{5}{9}a^{2} $. Therefore, the eccentricity of the ellipse is $ e = \\frac{c}{a} = \\frac{\\sqrt{5}}{3} $."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=4 x$, two mutually perpendicular lines $l_{1}$ and $l_{2}$ are drawn through $F$. The line $l_{1}$ intersects $C$ at points $A$ and $B$, and the line $l_{2}$ intersects $C$ at points $D$ and $E$. Then the minimum value of $|A B|+|D E|$ is?",
    "fact_expressions": "C: Parabola;l1:Line;l2:Line;A: Point;B: Point;D: Point;E: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F,l1);PointOnCurve(F,l2);IsPerpendicular(l1,l2);Intersection(l1, C) = {A, B};Intersection(l2, C) = {D, E}",
    "query_expressions": "Min(Abs(LineSegmentOf(A, B)) + Abs(LineSegmentOf(D, E)))",
    "answer_expressions": "4",
    "fact_spans": "[[[6, 25], [71, 74], [71, 74]], [[41, 50], [61, 70]], [[52, 60], [86, 95]], [[76, 79]], [[80, 83]], [[101, 104]], [[105, 108]], [[30, 33], [30, 33]], [[6, 25]], [[2, 28]], [[29, 60]], [[29, 60]], [[36, 60]], [[61, 85]], [[86, 110]]]",
    "query_spans": "[[[112, 131]]]",
    "process": "From the given conditions, the parabola $ C: y^2 = 4x $ has focus $ F: (1, 0) $ and directrix $ x = -1 $. Let the equation of line $ l_1 $ be $ y = k(x - 1) $. Since lines $ l_1 $ and $ l_2 $ are perpendicular, the slope of $ l_2 $ is $ -\\frac{1}{k} $. Solving the system of equations with the parabola:\n$$\n\\begin{cases}\ny = k(x - 1)\n\\end{cases}\n$$\nEliminating $ y $, we obtain $ k^2x^2 - (2k^2 + 4)x + k^2 = 0 $. Let points $ A(x_1, y_1) $, $ B(x_2, y_2) $, $ C(x_3, y_3) $, $ D(x_4, y_4) $. By the relationship between roots and coefficients, $ x_1 + x_2 = \\frac{2k^2 + 4}{k^2} $, similarly $ x_3 + x_4 = \\frac{2\\frac{1}{k^2} + 4}{\\frac{1}{k^2}} $. According to the property of a parabola, the distance from a point on the parabola to the focus equals the distance to the directrix. Therefore, $ |AB| = x_1 + 1 + x_2 + 1 $, similarly $ |DE| = x_3 + 1 + x_4 + 1 $. Hence,\n$$\n|AB| + |DE| = \\frac{2k^2 + 4}{k^2} + \\frac{2\\frac{1}{k^2} + 4}{\\frac{1}{k^2}} + 4 = 8 + \\frac{4}{k^2} + 4k^2 \\geqslant 8 + 2\\sqrt{4 \\times 4} = 16,\n$$\nwith equality if and only if $ k^2 = 1 $."
  },
  {
    "text": "Let an ellipse centered at the origin share common foci with the hyperbola $2 x^{2}-2 y^{2}=1$, and let their eccentricities be reciprocals of each other. Then the equation of the ellipse is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;O: Origin;Expression(G) = (2*x^2 - 2*y^2 = 1);Center(H) = O;Focus(H) = Focus(G);InterReciprocal(Eccentricity(G), Eccentricity(H))",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2/2 + y^2 = 1",
    "fact_spans": "[[[10, 32]], [[7, 9], [53, 56]], [[4, 6]], [[10, 32]], [[1, 9]], [[7, 38]], [[7, 50]]]",
    "query_spans": "[[[53, 61]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, one of its asymptotes passes through the center of the circle $C^{\\prime}$: $x^{2}+y^{2}-4x-6y=0$. Then, the equations of the asymptotes of the hyperbola $C$ are?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;C1: Circle;Expression(C1) = (x^2 + y^2 - 4*x - 6*y = 0);PointOnCurve(Center(C1), OneOf(Asymptote(C)))",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*(3/2)*x",
    "fact_spans": "[[[2, 63], [113, 119]], [[2, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[70, 108]], [[70, 108]], [[2, 111]]]",
    "query_spans": "[[[113, 127]]]",
    "process": "According to the general equation of a circle, solve for the center coordinates; solve according to the asymptote equation of the hyperbola. As per the problem, the circle $ C': (x-2)^{2}+(y-3)^{2}=13 $ has center $ (2,3) $. Thus, $ y=\\frac{b}{a}x $ passes through point $ (2,3) $, so $ \\frac{b}{a}=\\frac{3}{2} $. Hence, the asymptote equations of the hyperbola $ C $ are $ y=\\pm\\frac{3}{2}x $."
  },
  {
    "text": "Given that $O$ is the coordinate origin, the parabola $C$: $y^{2}=8x$ has focus $F$, $P$ is a point on $C$, $PF$ is perpendicular to the $x$-axis, and $Q$ is a point on the $x$-axis. If $P$ lies on the circle with diameter $OQ$, then the equation of this circle is?",
    "fact_expressions": "O: Origin;C: Parabola;Expression(C) = (y^2 = 8*x);F: Point;Focus(C) = F;P: Point;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P, F), xAxis);Q: Point;PointOnCurve(Q, xAxis);G: Circle;IsDiameter(LineSegmentOf(O, Q), G);PointOnCurve(P, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "(x-5)^2+y^2=25",
    "fact_spans": "[[[2, 5]], [[11, 30], [42, 45]], [[11, 30]], [[34, 37]], [[11, 37]], [[38, 41], [75, 78]], [[38, 48]], [[49, 61]], [[62, 65]], [[62, 73]], [[91, 92], [96, 97]], [[79, 92]], [[75, 93]]]",
    "query_spans": "[[[96, 102]]]",
    "process": "From the given conditions: \\because the focus of the parabola $ C: y^{2} = 8x $ is $ F(2,0) $, and $ PF $ is perpendicular to the x-axis, \\therefore the x-coordinate of point $ P $ is 2, \\therefore $ y^{2} = 16 $, that is, $ y = \\pm4 $. Hence, the coordinates of point $ P $ are $ (2,4) $ or $ (2,-4) $. Also, \\because $ Q $ is a point on the x-axis, and $ OQ $ is the diameter, with point $ P $ lying on the circle, let the center of the circle be $ (m,0) $, then we have $ (2-m)^{2} + 16 = m^{2} $, that is, $ 20 - 4m = 0 \\Rightarrow m = 5 $. Therefore, the equation of the circle is $ (x-5)^{2} + y^{2} = 25x^{2} $."
  },
  {
    "text": "If the foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ are $F_{1}$ and $F_{2}$, and a point $P$ on the ellipse satisfies $\\angle F_{1} P F_{2}=90^{\\circ}$, then the area of $\\triangle F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "9",
    "fact_spans": "[[[1, 39], [66, 68]], [[1, 39]], [[45, 52]], [[53, 60]], [[1, 60]], [[61, 65]], [[61, 71]], [[73, 106]]]",
    "query_spans": "[[[108, 138]]]",
    "process": "Using the definition of an ellipse to solve focal triangle problems is simple and fast [detailed solution] Let |PF_{1}| = r_{1}, |PF_{2}| = r_{2}, then from \\angle F_{1}PF_{2} = 90^{\\circ} and |F_{1}F_{2}| = 8, we get r_{1}^{2} + r_{2}^{2} = 64, and r_{1} + r_{2} = 10. It follows that r_{1}r_{2} = 18, so S_{\\triangle PF_{1}F_{2}} = \\frac{1}{2}r_{1}r_{2} = 9. Answer is 9"
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, and $M$, $N$ are two distinct points on the parabola, if $|MF| + |NF| = 5$, then the distance from the midpoint of segment $MN$ to the $y$-axis is?",
    "fact_expressions": "G: Parabola;M: Point;N: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G)=F;PointOnCurve(M,G);PointOnCurve(N,G);Negation(M=N);Abs(LineSegmentOf(M, F)) + Abs(LineSegmentOf(N, F)) = 5",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(M,N)), yAxis)",
    "answer_expressions": "3/2",
    "fact_spans": "[[[2, 16], [32, 35]], [[24, 27]], [[28, 31]], [[20, 23]], [[2, 16]], [[2, 23]], [[24, 42]], [[24, 42]], [[24, 42]], [[44, 59]]]",
    "query_spans": "[[[61, 81]]]",
    "process": "From the parabola equation $ y^{2} = 4x $, we obtain its directrix equation $ x = -1 $. Let $ M(x_{1}, y_{1}) $, $ N(x_{2}, y_{2}) $. By the definition of the parabola, $ |MF| + |NF| = x_{1} + 1 + x_{2} + 1 = 5 $, so $ x_{1} + x_{2} = 3 $. Therefore, the horizontal coordinate of the midpoint of segment $ MN $ is $ \\frac{x_{1} + x_{2}}{2} = \\frac{3}{2} $. Hence, the distance from the midpoint of segment $ MN $ to the $ y $-axis is $ \\frac{3}{2} $."
  },
  {
    "text": "If the directrix of the parabola $y^{2}=2 p x(p>0)$ passes through a focus of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, then the equation of the directrix of this parabola is?",
    "fact_expressions": "G: Parabola;p: Number;H: Ellipse;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (x^2/9 + y^2/5 = 1);PointOnCurve(OneOf(Focus(H)), Directrix(G))",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=-2",
    "fact_spans": "[[[1, 22], [72, 75]], [[4, 22]], [[27, 64]], [[4, 22]], [[1, 22]], [[27, 64]], [[1, 69]]]",
    "query_spans": "[[[72, 82]]]",
    "process": "The foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ are $F_{1}(-2,0)$, $F(2,0)$. Since the directrix of the parabola $y^{2}=2px$ ($p>0$) passes through one focus of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, the equation of the directrix of this parabola is $x=-2$."
  },
  {
    "text": "Given that the point $(1 , 1)$ is the midpoint of a chord of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$, then the equation of the line containing this chord is?",
    "fact_expressions": "G: Ellipse;H: LineSegment;Expression(G) = (x^2/4 + y^2/2 = 1);Coordinate(MidPoint(H)) = (1, 1);IsChordOf(H,G)",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "x+2*y-3=0",
    "fact_spans": "[[[13, 50]], [], [[13, 50]], [[2, 56]], [[13, 53]]]",
    "query_spans": "[[[13, 69]]]",
    "process": ""
  },
  {
    "text": "The equation of the trajectory of the midpoint of the line segment connecting any point on the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ to the right focus?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);P:Point;PointOnCurve(P,G)",
    "query_expressions": "LocusEquation(MidPoint(LineSegmentOf(P,RightFocus(G))))",
    "answer_expressions": "((2*x-1)^2)/2+y^2=1",
    "fact_spans": "[[[0, 27]], [[0, 27]], [], [[0, 32]]]",
    "query_spans": "[[[0, 47]]]",
    "process": ""
  },
  {
    "text": "The ellipse $C$: $\\frac{x^{2}}{a^{2}}+y^{2}=1$ $(a>0)$ has eccentricity $\\frac{\\sqrt{3}}{2}$, $F_{1}$, $F_{2}$ are the two foci of $C$, a line $l$ passing through $F_{1}$ intersects $C$ at points $A$, $B$, then the maximum value of $|A F_{2}|+|B F_{2}|$ is equal to?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2 + x^2/a^2 = 1);a: Number;a>0;Eccentricity(C) = sqrt(3)/2;F1: Point;F2: Point;Focus(C) = {F1, F2};l: Line;PointOnCurve(F1, l) = True;Intersection(l, C) = {A, B};A: Point;B: Point",
    "query_expressions": "Max(Abs(LineSegmentOf(A, F2)) + Abs(LineSegmentOf(B, F2)))",
    "answer_expressions": "7",
    "fact_spans": "[[[0, 41], [86, 89], [110, 113]], [[0, 41]], [[7, 41]], [[7, 41]], [[0, 66]], [[69, 76], [96, 103]], [[78, 85]], [[69, 94]], [[104, 109]], [[95, 109]], [[104, 124]], [[115, 118]], [[119, 122]]]",
    "query_spans": "[[[126, 154]]]",
    "process": "Since the eccentricity is $\\frac{\\sqrt{3}}{2}$, we have $\\frac{\\sqrt{a^{2}-1}}{a}=\\frac{\\sqrt{3}}{2}\\Rightarrow a=$ From the definition of the ellipse, $|AF_{2}|+|BF_{2}|+|AB|=4a=8$, so $|AF_{2}|+|BF_{2}|=8-|AB|$. By the focal chord property, when $AB\\bot x$-axis, $|AB|$ takes the minimum value $2\\times\\frac{b^{2}}{a}=1$. Therefore, the maximum value of $|AF_{2}|+|BF_{2}|$ equals $8-1=7$."
  },
  {
    "text": "Given that $F$ is the right focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $A$ is an endpoint of the minor axis of the ellipse, the line $AF$ intersects the ellipse at another point $B$, and $\\overrightarrow{AF}=2 \\overrightarrow{FB}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "F: Point;G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;RightFocus(G) = F;A: Point;OneOf(Endpoint(MinorAxis(G))) = A;Intersection(LineOf(A, F), G) = {A,B};B: Point;VectorOf(A, F) = 2*VectorOf(F, B)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[2, 5]], [[6, 58], [67, 69], [85, 87], [144, 146]], [[6, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[2, 62]], [[63, 66]], [[63, 76]], [[63, 95]], [[92, 95]], [[97, 142]]]",
    "query_spans": "[[[144, 152]]]",
    "process": "Let A(0,-b), F(c,0), draw BC\\bot y-axis, with foot at C, as shown in the figure: |x|=\\sqrt{b^{2}+c^{2}}=a. From \\overrightarrow{AF}=2\\overrightarrow{FB}, we get: \\frac{|\\overrightarrow{AF}|}{|\\overrightarrow{AB}|}=\\frac{c}{|\\overrightarrow{BC}|}=\\frac{2}{3}. \\therefore |\\overrightarrow{BC}|=\\frac{3}{2}c, that is: x_{B}=\\frac{3}{2}c. From the focal radius formula of the ellipse, we have: |\\overrightarrow{BF}|=a-ex_{B}. \\therefore \\frac{|\\overrightarrow{AF}|}{|\\overrightarrow{FB}|}=\\frac{a}{a-ex_{B}}=\\frac{a}{a-\\frac{c}{a}\\cdot\\frac{3c}{2}}=2. Rearranging yields: a^{2}=3c^{2}. \\because e^{2}=\\frac{1}{3}, i.e., e=\\frac{\\sqrt{3}}{3}. The correct answer to this problem: \\frac{\\sqrt{3}}{3}"
  },
  {
    "text": "It is known that the equation of the directrix of the parabola $y=ax^{2}$ is $y=-2$. What is the value of the real number $a$?",
    "fact_expressions": "G: Parabola;a: Real;Expression(G) = (y = a*x^2);Expression(Directrix(G)) = (y =-2)",
    "query_expressions": "a",
    "answer_expressions": "1/8",
    "fact_spans": "[[[2, 15]], [[29, 34]], [[2, 15]], [[2, 27]]]",
    "query_spans": "[[[29, 38]]]",
    "process": ""
  },
  {
    "text": "Given that the foci of the hyperbola lie on the $x$-axis, the eccentricity is $2$, $F_{1}$, $F_{2}$ are the left and right foci, $P$ is a point on the hyperbola such that $\\angle F_{1} P F_{2}=60^{\\circ}$, and $S_{\\triangle P F_{1} F_{2}}=12 \\sqrt{3}$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), xAxis) = True;Eccentricity(G) = 2;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G) = True;AngleOf(F1, P, F2) = ApplyUnit(60, degree);Area(TriangleOf(P, F1, F2)) = 12*sqrt(3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/12 = 1",
    "fact_spans": "[[[2, 5], [53, 56], [140, 143]], [[2, 14]], [[2, 22]], [[25, 32]], [[35, 42]], [[2, 48]], [[2, 48]], [[49, 52]], [[49, 59]], [[61, 95]], [[97, 138]]]",
    "query_spans": "[[[140, 150]]]",
    "process": "From the hyperbola focal triangle area formula $ S = b^{2} \\cot \\xrightarrow{\\angle F_{1}PF_{2}} $, we get $ b^{2} = 12 $. Therefore, $ c^{2} - a^{2} = 12 $. Also, $ e = \\frac{c}{a} = 2 $. Hence, $ c^{2} = 16 $, $ a^{2} = 4 $, $ b^{2} = 12 $, and the equation is $ \\frac{x^{2}}{4} - \\frac{y^{2}}{12} = 1 $."
  },
  {
    "text": "The coordinates of the foci of the hyperbola $\\frac{y^{2}}{3}-x^{2}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2 + y^2/3 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, pm*2)",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 35]]]",
    "process": ""
  },
  {
    "text": "Let the left and right foci of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ be $F_{1}$ and $F_{2}$, respectively, and let point $P$ be a point on the hyperbola located in the first quadrant such that the area of $\\triangle P F_{1} F_{2}$ is $6$. Then the coordinates of point $P$ are?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 - y^2/5 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P,G);Quadrant(P)=1;Area(TriangleOf(P, F1, F2)) = 6",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(6*sqrt(5)/5,2)",
    "fact_spans": "[[[1, 39], [69, 72]], [[64, 68], [120, 124]], [[48, 55]], [[56, 63]], [[1, 39]], [[1, 63]], [[1, 63]], [[64, 84]], [[64, 84]], [[86, 118]]]",
    "query_spans": "[[[120, 129]]]",
    "process": ""
  },
  {
    "text": "If the line $y=k(x-1)$ has no intersection points with the hyperbola $x^{2}-y^{2}=4$, then the range of values for $k$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;k:Number;Expression(G) = (x^2 - y^2 = 4);Expression(H) = (y=k*(x-1));NumIntersection(G,H)=0",
    "query_expressions": "Range(k)",
    "answer_expressions": "{(-oo,-2*sqrt(3)/3),(2*sqrt(3)/3,+oo)}",
    "fact_spans": "[[[15, 33]], [[2, 14]], [[39, 42]], [[15, 33]], [[2, 14]], [[2, 37]]]",
    "query_spans": "[[[39, 49]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with left and right foci $F_{1}$, $F_{2}$, eccentricity $\\frac{\\sqrt{3}}{3}$, a line $l$ passing through $F_{2}$ intersects $C$ at points $A$, $B$. If the perimeter of $\\triangle AF_{1} B$ is $4 \\sqrt{3}$, then the equation of $C$ is?",
    "fact_expressions": "l: Line;C: Ellipse;b: Number;a: Number;F1: Point;B: Point;F2: Point;A: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;Eccentricity(C)=sqrt(3)/3;PointOnCurve(F2, l);Intersection(l,C)={A,B};Perimeter(TriangleOf(A,F1,B)) = 4*sqrt(3)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/3+y^2/2=1",
    "fact_spans": "[[[115, 120]], [[2, 59], [121, 124], [174, 177]], [[9, 59]], [[9, 59]], [[65, 72]], [[129, 132]], [[73, 80], [107, 114]], [[125, 128]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 80]], [[2, 80]], [[2, 105]], [[106, 120]], [[115, 134]], [[136, 172]]]",
    "query_spans": "[[[174, 182]]]",
    "process": "From the given condition, |AF_{1}|+|AF_{2}|+|BF_{1}|+|BF_{2}|=4\\sqrt{3}, find the value of a, then use the eccentricity to find the values of c and b, thus obtaining the equation of the ellipse. By the definition of an ellipse, |AF_{1}|+|AF_{2}|=2a, |BF_{1}|+|BF_{2}|=2a. Since |AF_{1}|+|AF_{2}|+|BF_{1}|+|BF_{2}|=4\\sqrt{3}, it follows that 4a=4\\sqrt{3}, solving gives a=\\sqrt{3}. Also, since e=\\frac{c}{a}=\\frac{\\sqrt{3}}{3}, we have c=1, so b^{2}=a^{2}-c^{2}=2. Therefore, the equation of the ellipse is \\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1"
  },
  {
    "text": "What are the coordinates of the focus of the parabola $y=x^{2}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,1/4)",
    "fact_spans": "[[[0, 12]], [[0, 12]]]",
    "query_spans": "[[[0, 19]]]",
    "process": "From the given conditions, we have 2p=1, that is, p=\\frac{1}{2}, so the focus coordinates of the parabola y=x^{2} are (0,\\frac{1}{4})"
  },
  {
    "text": "Draw two tangents from a focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) to the circle $x^{2}+y^{2}=a^{2}$, with points of tangency $A$ and $B$. If $\\angle AOB=120^{\\circ}$ ($O$ is the origin), then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;G: Circle;Expression(G) = (x^2 + y^2 = a^2);L1: Line;L2: Line;A: Point;B: Point;TangentOfPoint(OneOf(Focus(C)), G) = {L1, L2};TangentPoint(L1, G) = A;TangentPoint(L2, G) = B;O: Origin;AngleOf(A, O, B) = ApplyUnit(120, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 62], [147, 153]], [[1, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[68, 88]], [[68, 88]], [], [], [[99, 102]], [[104, 107]], [[0, 93]], [[0, 107]], [[0, 107]], [[135, 138]], [[110, 134]]]",
    "query_spans": "[[[147, 159]]]",
    "process": ""
  },
  {
    "text": "The equation $\\frac{x^{2}}{k-4}+\\frac{y^{2}}{10-k}=1$ represents an ellipse with foci on the $x$-axis. Then, the range of real values for $k$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(k - 4) + y^2/(10 - k) = 1);k: Real;PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(7, 10)",
    "fact_spans": "[[[53, 55]], [[0, 55]], [[57, 62]], [[44, 55]]]",
    "query_spans": "[[[57, 69]]]",
    "process": "According to the type of the standard equation of an ellipse, we can set up the equation and obtain the result. Since the equation is $\\frac{x^{2}}{k-4}+\\frac{y^{2}}{10-k}$, solving gives $7<k<10$."
  },
  {
    "text": "The eccentricity of the hyperbola $C$: $\\frac{x^{2}}{4}-y^{2}=1$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/4 - y^2 = 1)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[0, 33]], [[0, 33]]]",
    "query_spans": "[[[0, 39]]]",
    "process": "From the standard equation of the hyperbola, we obtain a, b, c, and then find the eccentricity: $a^2=4$, $b^{2}=1$, $\\therefore c^{2}=a^{2}+b^{2}=5$, so the eccentricity is $e=\\frac{\\sqrt{5}}{2}$."
  },
  {
    "text": "Draw a perpendicular line from the right vertex of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ to the $x$-axis, intersecting an asymptote of $C$ at point $A$. If a circle centered at the right focus of $C$ with radius $2$ passes through points $A$ and $O$ ($O$ being the origin), then the standard equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;A: Point;O: Origin;Z: Line;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(RightVertex(C), Z);IsPerpendicular(Z, xAxis);Intersection(Z, OneOf(Asymptote(C))) = A;Center(G) = RightFocus(C);Radius(G) = 2;PointOnCurve(A, G);PointOnCurve(O, G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2/3 = 1",
    "fact_spans": "[[[1, 52], [65, 68], [84, 87], [127, 133]], [[10, 52]], [[10, 52]], [[102, 103]], [[77, 80], [105, 108]], [[109, 112], [116, 119]], [], [[1, 52]], [[0, 64]], [[0, 64]], [[0, 80]], [[83, 103]], [[95, 103]], [[102, 108]], [[102, 115]]]",
    "query_spans": "[[[127, 140]]]",
    "process": "According to the properties of the hyperbola, obtain the value of $ c $ and the coordinates of point $ A $. Using the distance formula between two points, derive $ |AO| $, then deduce that $ \\triangle AOF $ is an equilateral triangle with side length 2, $ c = 2a $, solve to get $ b = \\sqrt{3} $, thus obtain the equation of the hyperbola. According to the problem, $ |FO| = 2 \\Rightarrow c = 2 $, $ A(a, b) $, so $ |AO| = c $, and $ |FO| = |FA| = c = 2 $, therefore $ \\triangle AOF $ is an equilateral triangle with side length 2. Thus $ c = 2a \\Rightarrow a = 1 $, then obtain $ b = \\sqrt{3} $, so the equation of the hyperbola is: $ x^{2} - \\frac{y^{2}}{3} = 1 $."
  },
  {
    "text": "The line $l$ passes through the focus $F$ of the parabola $y^{2}=4x$, intersects the parabola at points $A$ and $B$, and intersects its directrix at point $C$. If $\\overrightarrow{A F}=\\frac{3}{2} \\overrightarrow{B C}$, then the slope of line $l$ is?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;F: Point;B: Point;C: Point;Expression(G) = (y^2 = 4*x);Focus(G)=F;PointOnCurve(F, l);Intersection(l,G)={A,B};Intersection(l,Directrix(G))=C;VectorOf(A, F) = (3/2)*VectorOf(B, C)",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*sqrt(3)",
    "fact_spans": "[[[0, 5], [112, 117]], [[6, 20], [28, 31], [44, 45]], [[33, 36]], [[23, 26]], [[37, 40]], [[49, 53]], [[6, 20]], [[6, 26]], [[0, 26]], [[0, 42]], [[0, 53]], [[55, 110]]]",
    "query_spans": "[[[112, 122]]]",
    "process": ""
  },
  {
    "text": "If the line $y=k x-2$ intersects the parabola $y^{2}=8 x$ at points $A$ and $B$, and the abscissa of the midpoint of segment $A B$ is $2$, then $|A B|$=?",
    "fact_expressions": "G: Parabola;H: Line;k: Number;A: Point;B: Point;Expression(G) = (y^2 = 8*x);Expression(H) = (y = k*x - 2);Intersection(H, G) = {A, B};XCoordinate(MidPoint(LineSegmentOf(A,B))) = 2",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "2*sqrt(15)",
    "fact_spans": "[[[13, 27]], [[1, 12]], [[3, 12]], [[29, 32]], [[33, 36]], [[13, 27]], [[1, 12]], [[1, 38]], [[40, 58]]]",
    "query_spans": "[[[60, 69]]]",
    "process": ""
  },
  {
    "text": "Draw a chord $AB$ of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ through the right focus $F$, such that $\\overrightarrow{F A}+\\overrightarrow{F B}=0$. If the circle with diameter $AB$ passes through the left vertex of the hyperbola $C$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(C) = F;A: Point;B: Point;PointOnCurve(F,LineSegmentOf(A,B)) = True;IsChordOf(LineSegmentOf(A,B),C) = True;VectorOf(F,A) + VectorOf(F,B) = 0;G: Circle;IsDiameter(LineSegmentOf(A,B),G) = True;PointOnCurve(LeftVertex(C),G) = True",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 62], [70, 76], [148, 154], [160, 166]], [[1, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[66, 69]], [[1, 69]], [[80, 85]], [[80, 85]], [[0, 85]], [[0, 85]], [[88, 133]], [[145, 146]], [[135, 146]], [[145, 158]]]",
    "query_spans": "[[[160, 172]]]",
    "process": "From the given conditions, F is the midpoint of segment AB. Since the hyperbola is symmetric about the x-axis, it follows that AB \\bot x-axis. Letting x = c, we find |AF|. Then, since the angle subtended by a diameter is a right angle, let M be the left vertex of hyperbola C; then |MF| = |AF|. Simplifying and using the eccentricity formula of the hyperbola, the required value is obtained. \\overrightarrow{FA} + \\overrightarrow{FB} = \\overrightarrow{0}, so F is the midpoint of segment AB. Due to symmetry of the hyperbola about the x-axis, AB \\bot x-axis. Let x = c, then y^{2} = b^{2}\\left(\\frac{c^{2}}{a^{2}} - 1\\right), so y = \\pm\\frac{b^{2}}{a}. Assume |AF| = \\frac{b^{2}}{a}. The circle with AB as diameter passes through the left vertex M(-a, 0) of hyperbola C, so triangle AMB is an isosceles right triangle, hence |MF| = |AF|, that is, a + c = \\frac{b^{2}}{a} = \\frac{c^{2} - a^{2}}{a}, which simplifies to c = 2a, then e = \\frac{c}{a} = 2"
  },
  {
    "text": "Given that the left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ lies on the directrix of the parabola $y^{2}=16x$, then $a=$?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;H:Parabola;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 16*x);PointOnCurve(LeftFocus(G),Directrix(H))",
    "query_expressions": "a",
    "answer_expressions": "sqrt(15)",
    "fact_spans": "[[[2, 39]], [[65, 68]], [[5, 39]], [[44, 59]], [[2, 39]], [[44, 59]], [[2, 63]]]",
    "query_spans": "[[[65, 70]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, the left focus is $F_{1}$, point $P$ is any point on the ellipse distinct from the vertices, $O$ is the origin, and point $M$ is the midpoint of segment $P F_{1}$. Then the perimeter of $\\Delta M O F_{1}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;M: Point;O: Origin;Expression(G) = (x^2/25 + y^2/16 = 1);LeftFocus(G) = F1;PointOnCurve(P, G);Negation(P=Vertex(G));MidPoint(LineSegmentOf(P, F1)) = M",
    "query_expressions": "Perimeter(TriangleOf(M, O, F1))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 41], [59, 61]], [[54, 58]], [[46, 53]], [[82, 86]], [[72, 75]], [[2, 41]], [[2, 53]], [[54, 71]], [[54, 71]], [[82, 101]]]",
    "query_spans": "[[[103, 126]]]",
    "process": ""
  },
  {
    "text": "Given point $P(0 , 1)$ and the parabola $y=x^{2}+2$, where $Q$ is a moving point on the parabola, then the minimum value of $|P Q|$ is?",
    "fact_expressions": "P: Point;Coordinate(P) = (0, 1);Q: Point;G: Parabola;Expression(G) = (y = x^2 + 2);PointOnCurve(Q, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 13]], [[2, 13]], [[31, 34]], [[14, 28], [35, 38]], [[14, 28]], [[31, 42]]]",
    "query_spans": "[[[44, 57]]]",
    "process": "Let the coordinates of point Q be (a, a^{2}+2), then |PQ|^{2}=a^{4}+3a^{2}+1. Clearly, when a=0, the minimum value of |PQ| is 1. Let the coordinates of point Q be (a, a^{2}+2), then |PQ|^{2}=a^{2}+(a^{2}+1)^{2}=a^{4}+3a^{2}+1. Hence, when a^{2}=0, i.e., a=0, |PQ|^{2} attains its minimum value of 1, so the minimum value of |PQ| is 1."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=8x$, $M$ is a point on $C$, and the extension of $FM$ intersects the $y$-axis at point $N$. If $M$ is the midpoint of $FN$, then $|FN|$=?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 8*x);F: Point;M: Point;N: Point;Focus(C) = F;PointOnCurve(M, C);Intersection(OverlappingLine(LineSegmentOf(F, M)), yAxis) = N;MidPoint(LineSegmentOf(F, N)) = M",
    "query_expressions": "Abs(LineSegmentOf(F, N))",
    "answer_expressions": "6",
    "fact_spans": "[[[6, 25], [33, 36]], [[6, 25]], [[2, 5]], [[29, 32], [62, 65]], [[55, 59]], [[2, 28]], [[29, 39]], [[40, 59]], [[62, 74]]]",
    "query_spans": "[[[76, 85]]]",
    "process": "As shown in the figure, without loss of generality, assume point $ M $ lies in the first quadrant. Let the directrix of the parabola intersect the $ x $-axis at point $ F' $. Draw $ MB \\bot l $ at point $ B $, and $ NA \\bot l $ at point $ A $. From the equation of the parabola, the directrix equation is $ x = -2 $, so $ AN = 2 $, $ FF' = 4 $. In the right trapezoid $ ANFF' $, the median line $ BM = \\frac{AN + FF'}{2} = 3 $. By the definition of the parabola, $ MF = MB = 3 $. According to the given conditions, $ MN = MF = 3 $. Hence, $ |FN| = |FM| + |NM| = 3 + 3 = 6 $."
  },
  {
    "text": "What is the length of the imaginary axis of the hyperbola $\\frac{x^{2}}{12}-\\frac{y^{2}}{4}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/12 - y^2/4 = 1)",
    "query_expressions": "Length(ImageinaryAxis(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 45]]]",
    "process": "According to the problem, the standard equation of the hyperbola is \\frac{x^{2}}{12}-\\frac{y^{2}}{4}=1, then b=2, so the length of the imaginary axis is 2b=4"
  },
  {
    "text": "The vertex of a parabola is at the origin, and its focus is one of the foci of the ellipse $x^{2}+2 y^{2}=8$. Then, the distance from the focus of this parabola to its directrix is equal to?",
    "fact_expressions": "G: Parabola;O: Origin;Vertex(G) = O;H: Ellipse;Expression(H) = (x^2 + 2*y^2 = 8);Focus(G) = OneOf(Focus(H))",
    "query_expressions": "Distance(Focus(G),Directrix(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 3], [42, 45], [49, 50]], [[7, 11]], [[0, 11]], [[15, 34]], [[15, 34]], [[0, 39]]]",
    "query_spans": "[[[42, 59]]]",
    "process": ""
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=4x$ with an inclination angle of $60^{\\circ}$ intersects the parabola at points $A$ and $B$. Circles with diameters $AF$ and $BF$ are tangent to the $y$-axis at points $M$ and $N$, respectively. Then $|MN|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;L: Line;PointOnCurve(F, L);Inclination(L) = ApplyUnit(60, degree);Intersection(L, G) = {A, B};A: Point;B: Point;C1: Circle;IsDiameter(LineSegmentOf(A, F), C1) = True;C2: Circle;IsDiameter(LineSegmentOf(B, F), C2) = True;TangentPoint(C1, yAxis) = M;TangentPoint(C2, yAxis) = N;M: Point;N: Point",
    "query_expressions": "Abs(LineSegmentOf(M, N))",
    "answer_expressions": "4*sqrt(3)/3",
    "fact_spans": "[[[1, 15], [42, 45]], [[1, 15]], [[18, 21]], [[1, 21]], [[39, 41]], [[0, 41]], [[22, 41]], [[39, 55]], [[46, 49]], [[50, 53]], [], [[56, 74]], [], [[56, 74]], [[56, 92]], [[56, 92]], [[85, 88]], [[89, 92]]]",
    "query_spans": "[[[94, 103]]]",
    "process": "According to the given conditions, AF and BF can be obtained using the focal radius formula, then |MN| can be found as the difference between the y-coordinates of the midpoints of AF and BF. By the focal radius formula, AF = \\frac{2}{1-\\cos60^{\\circ}} = 4, BF = \\frac{2}{1+\\cos60^{\\circ}} = \\frac{4}{3}. Also, |MN| equals the difference between the y-coordinates y_{1}, y_{2} of the midpoints of AF and BF. Moreover, y_{1} = \\frac{1}{2} \\times AF \\times \\sin60^{\\circ} = \\sqrt{3}, y_{2} = -\\frac{1}{2} \\times BF \\times \\sin60^{\\circ} = -\\frac{\\sqrt{3}}{3}. Hence, MN = |y_{1} - y_{2}| = \\sqrt{3} - (-\\frac{\\sqrt{3}}{3}) = \\frac{4\\sqrt{3}}{3}"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$, $F_{2}$ respectively, and the eccentricity is $e$. If there exists a point $P$ on the ellipse such that $\\frac{P F_{1}}{P F_{2}}=e$, then the range of values for the eccentricity $e$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;e: Number;Eccentricity(G) = e;P: Point;PointOnCurve(P, G);LineSegmentOf(P, F1)/LineSegmentOf(P, F2) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "[\\sqrt{2}-1, 1)",
    "fact_spans": "[[[2, 54], [87, 89]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[62, 69]], [[70, 77]], [[2, 77]], [[2, 77]], [[82, 85], [132, 135]], [[2, 85]], [[92, 96]], [[87, 96]], [[99, 126]]]",
    "query_spans": "[[[132, 142]]]",
    "process": ""
  },
  {
    "text": "Let line $l$ passing through the focus $F$ of the parabola $C$: $y^{2}=2px$ ($p>0$) intersect the parabola $C$ at points $A$ and $B$. If a point $M(-2,2)$ on the directrix of the parabola $C$ satisfies $\\overrightarrow{MA} \\cdot \\overrightarrow{MB}=0$, then the value of $|AB|$ is?",
    "fact_expressions": "l: Line;C: Parabola;p: Number;M: Point;A: Point;B: Point;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(M) = (-2, 2);Focus(C)=F;PointOnCurve(F, l);Intersection(l, C) = {A, B};PointOnCurve(M,Directrix(C));DotProduct(VectorOf(M,A),VectorOf(M,B))=0",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "10",
    "fact_spans": "[[[33, 38]], [[1, 27], [39, 45], [58, 64]], [[8, 27]], [[70, 79]], [[47, 50]], [[51, 54]], [[29, 32]], [[8, 27]], [[1, 27]], [[70, 79]], [[1, 32]], [[0, 38]], [[33, 56]], [[58, 79]], [[81, 132]]]",
    "query_spans": "[[[134, 145]]]",
    "process": "Given that the directrix of the known parabola is $ x = -2 $, so $ \\frac{p}{2} = 2 $, $ p = 4 $, the equation of the parabola is $ y^2 = 8x $, and the focus is $ F(2,0) $. Since line $ l $ passes through the focus, its equation can be written as $ l: x = ky + 2 $ ($ k \\neq 0 $). Because $ \\overrightarrow{MA} \\cdot \\overrightarrow{MB} = 0 $, point $ M $ lies on the circle with $ AB $ as diameter. Let points $ A(x_1, y_1) $, $ B(x_2, y_2) $. Then  \n$$\n\\begin{cases}\ny_1^2 = 8x_1 \\\\\ny_2^2 = 8x_2\n\\end{cases}\n\\quad\n\\frac{1}{2} = 8x_1\n$$\nSubtracting these two equations gives  \n$$\n\\frac{y_1 - y_2}{x_1 - x_2} = \\frac{8}{y_1 + y_2} = \\frac{1}{k}\n$$\nLet the midpoint of $ AB $ be $ Q(x_0, y_0) $, then  \n$$\ny_0 = \\frac{y_1 + y_2}{2} = 4k', \\quad x_0 = 4k^2 + 2\n$$\nSo $ Q(4k^2 + 2, 4k) $ is the center of the circle with $ AB $ as diameter. By the definition of a parabola, the radius of the circle is  \n$$\nr = \\frac{|AB|}{2} = \\frac{x_1 + x_2 + 4}{2} = \\frac{2x_0 + 4}{2} = 4k^2 + 4\n$$\nSince  \n$$\n|QM|^2 = (x_0 + 2)^2 + (y_0 - 2)^2 = (4k^2 + 4)^2 + (4k - 2)^2 = r^2\n$$\nThus  \n$$\n(4k^2 + 4)^2 + (4k - 2)^2 = (4k^2 + 4)^2\n$$\nSolving gives $ k = \\frac{1}{2} $. So  \n$$\n|AB| = 2r = 2\\left[4 \\times \\left(\\frac{1}{7}\\right)^2 + 4\\right]\n$$\nThe final answer is: 10"
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{20}=1$ lie on the $x$-axis, and the focal distance is $8$. What is the eccentricity of this ellipse?",
    "fact_expressions": "G: Ellipse;a: Number;Expression(G) = (y^2/20 + x^2/a^2 = 1);PointOnCurve(Focus(G), xAxis);FocalLength(G) = 8",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2/3",
    "fact_spans": "[[[0, 42], [61, 63]], [[2, 42]], [[0, 42]], [[0, 51]], [[0, 58]]]",
    "query_spans": "[[[61, 69]]]",
    "process": "Since the focal distance of the ellipse is 2c=8, c=4, and the foci of the ellipse are on the x-axis, it follows that a^{2}=20+4^{2}=36, a=6, so the eccentricity of the ellipse is \\frac{c}{a}=\\frac{4}{6}=\\frac{2}{3}."
  },
  {
    "text": "The equation of the line on which the chord of the parabola $y^{2}=8 x$ lies, with $(1,-1)$ as its midpoint, is?",
    "fact_expressions": "I: Point;Coordinate(I) = (1, -1);G: Parabola;Expression(G) = (y^2 = 8*x);L: LineSegment;MidPoint(L) = I;IsChordOf(L, G)",
    "query_expressions": "Expression(OverlappingLine(L))",
    "answer_expressions": "4*x+y-3=0",
    "fact_spans": "[[[1, 9]], [[1, 9]], [[13, 27]], [[13, 27]], [], [[0, 29]], [[13, 29]]]",
    "query_spans": "[[[13, 37]]]",
    "process": "This chord is not perpendicular to the X-axis. Let the endpoints of the chord of the parabola $ y^{2} = 8x $ with midpoint $ (1, -1) $ be $ A(x_{1}, y_{1}) $ and $ B(x_{2}, y_{2}) $. Then we have $ y_{1}^{2} = 8x_{1} $, $ y_{2}^{2} = 8x_{2} $. Subtracting these two equations gives $ (y_{1} + y_{2})(y_{1} - y_{2}) = 8(x_{1} - x_{2}) $. Therefore, $ k = \\frac{y_{1} - y_{2}}{x_{1} - x_{2}} = -4 $. Hence, the equation of the line is $ y + 1 = 4(x - 1) $, i.e., $ 4x + y - 3 = 0 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, the two foci are denoted as $F_{1}$, $F_{2}$, and $P$ is a point on the hyperbola such that $\\angle F_{1} P F_{2}=90^{\\circ}$. Then the value of $|\\overrightarrow{P F_{1}}| \\cdot|\\overrightarrow{P F_{2}}|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1);F1: Point;F2: Point;Focus(G) = {F1,F2};P: Point;PointOnCurve(P,G) = True;AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Abs(VectorOf(P, F1))*Abs(VectorOf(P, F2))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 30], [58, 61]], [[2, 30]], [[38, 45]], [[46, 53]], [[2, 53]], [[54, 57]], [[54, 64]], [[66, 99]]]",
    "query_spans": "[[[101, 161]]]",
    "process": "The hyperbola $\\frac{x^2}{4}-y^{2}=1$ has $a=2$, $b=1$, $c=\\sqrt{5}$. Without loss of generality, let $P$ be a point on the right branch of the hyperbola, $|PF_{1}|=m$, $|PF_{2}|=n$. Then $m-n=2a=4$,\\textcircled{1} by the law of cosines we obtain $4c^{2}=m^{2}+n^{2}-2mn\\cos90^{\\circ}=m^{2}+n^{2}=20$,\\textcircled{2} combining \\textcircled{1} and \\textcircled{2} yields $mn=2$."
  },
  {
    "text": "The standard equation of a hyperbola with an asymptote $y = x$ and passing through the point $(2, 4)$ is?",
    "fact_expressions": "G: Hyperbola;Expression(OneOf(Asymptote(G))) = (y = x);H: Point;Coordinate(H) = (2, 4);PointOnCurve(H, G) = True",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 - x^2 = 12",
    "fact_spans": "[[[27, 30]], [[0, 30]], [[16, 26]], [[16, 26]], [[15, 30]]]",
    "query_spans": "[[[27, 36]]]",
    "process": ""
  },
  {
    "text": "The ellipse $m x^{2}+y^{2}=m$ has its foci on the $x$-axis, with left and right foci denoted as $F_{1}$ and $F_{2}$, respectively. The line $A B$ passes through $F_{1}$ and intersects the ellipse at points $A$ and $B$. If $\\overrightarrow{A F_{1}}=2 \\overrightarrow{F_{1} B}$ and $A F_{2}$ is perpendicular to the $x$-axis, then $m=$?",
    "fact_expressions": "G: Ellipse;m: Number;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (m*x^2 + y^2 = m);PointOnCurve(Focus(G), xAxis);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, LineOf(A, B));Intersection(LineOf(A, B), G) = {A, B};VectorOf(A, F1) = 2*VectorOf(F1, B);IsPerpendicular(LineSegmentOf(A, F2), xAxis)",
    "query_expressions": "m",
    "answer_expressions": "4/5",
    "fact_spans": "[[[9, 28], [29, 30], [74, 76]], [[162, 165]], [[78, 81]], [[82, 85]], [[46, 53]], [[38, 45], [63, 70]], [[9, 28]], [[0, 28]], [[29, 53]], [[29, 53]], [[54, 70]], [[54, 87]], [[89, 142]], [[144, 160]]]",
    "query_spans": "[[[162, 167]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ has a focal distance of $4$, and the point $P(1, \\sqrt{3})$ lies on the asymptote of $C$, then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;Expression(C) = (-x^2/b^2 + y^2/a^2 = 1);Coordinate(P) = (1, sqrt(3));FocalLength(C) = 4;PointOnCurve(P, Asymptote(C))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/3 - x^2 = 1",
    "fact_spans": "[[[2, 53], [79, 82], [89, 92]], [[10, 53]], [[10, 53]], [[61, 78]], [[2, 53]], [[61, 78]], [[2, 60]], [[61, 87]]]",
    "query_spans": "[[[89, 97]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the focus of the parabola $y=\\frac{x^{2}}{4}$, the line $l$ passing through point $F$ with an inclination angle of $150^{\\circ}$ intersects the parabola at points $A$ and $B$, and $l_{1}$, $l_{2}$ are the tangent lines to the parabola at points $A$ and $B$ respectively, with $l_{1}$ and $l_{2}$ intersecting at point $C$, then $\\overrightarrow{C A} \\cdot \\overrightarrow{C B}$=?",
    "fact_expressions": "F: Point;Focus(G) = F;G: Parabola;Expression(G) = (y = x^2/4);l: Line;PointOnCurve(F, l);Inclination(l) = ApplyUnit(150, degree);A: Point;B: Point;Intersection(l, G) = {A, B};l1: Line;l2: Line;TangentOnPoint(A, G) = l1;TangentOnPoint(B, G) = l2;C: Point;Intersection(l1, l2) = C",
    "query_expressions": "DotProduct(VectorOf(C, A), VectorOf(C, B))",
    "answer_expressions": "0",
    "fact_spans": "[[[2, 5], [33, 37]], [[2, 31]], [[6, 28], [62, 65], [97, 100]], [[6, 28]], [[56, 61]], [[32, 61]], [[38, 61]], [[67, 70], [101, 104]], [[71, 74], [105, 108]], [[56, 76]], [[77, 84], [115, 122]], [[86, 93], [124, 131]], [[77, 114]], [[77, 114]], [[134, 138]], [[115, 138]]]",
    "query_spans": "[[[140, 191]]]",
    "process": "The focus of the parabola $ y = \\frac{x^2}{4} $ is $ F(0,1) $, then the line $ l: y = -\\frac{\\sqrt{3}}{3}x + 1 $.  \n$$\n\\begin{cases}\ny = \\frac{x^2}{4} \\\\\ny = -\\frac{\\sqrt{3}}{3}x + 1\n\\end{cases}\n\\Rightarrow 3x^2 + 4\\sqrt{3}x - 12 = 0.\n$$\nSolving gives  \n$$\n\\begin{cases}\nx = \\frac{2\\sqrt{3}}{3} \\\\\ny = \\frac{1}{3}\n\\end{cases}\n\\quad \\text{or} \\quad\n\\begin{cases}\nx = -2\\sqrt{3} \\\\\ny = 3\n\\end{cases}\n$$\nSince $ y = \\frac{1}{2}x $, the slope of $ l_1 $ is $ k_1 = \\frac{1}{2} \\times \\frac{2\\sqrt{3}}{3} = \\frac{\\sqrt{3}}{3} $, and the slope of $ l_2 $ is $ k_2 = \\frac{1}{2} \\times (-2\\sqrt{3}) = -\\sqrt{3} $.  \nSince $ k_1 \\cdot k_2 = -1 $, it follows that $ CA \\perp CB $, hence $ \\overrightarrow{CA} \\cdot \\overrightarrow{CB} = 0 $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, a line $l$ passing through $F_{1}$ intersects the left and right branches of the hyperbola at points $A$ and $B$ respectively. If $A F_{1}=2 a$, $\\angle F_{1} A F_{2}=\\frac{2 \\pi}{3}$, then $\\frac{S_{\\triangle A F_{1} F_{2}}}{S_{\\triangle A B F_{2}}}=$?",
    "fact_expressions": "F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;l: Line;PointOnCurve(F1,l) = True;Intersection(l,LeftPart(G)) = A;A: Point;B: Point;LineSegmentOf(A, F1) = 2*a;AngleOf(F1,A,F2) = 2*pi/3;Intersection(l,RightPart(G)) = B",
    "query_expressions": "Area(TriangleOf(A,F1,F2))/Area(TriangleOf(A,B,F2))",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 9], [80, 87]], [[10, 17]], [[2, 78]], [[2, 78]], [[18, 74], [94, 97]], [[18, 74]], [[21, 74]], [[21, 74]], [[21, 74]], [[21, 74]], [[88, 93]], [[79, 93]], [[88, 114]], [[105, 108]], [[109, 112]], [[116, 129]], [[132, 170]], [[88, 114]]]",
    "query_spans": "[[[172, 235]]]",
    "process": "Since |AF₂| - |AF₁| = 2a, |AF₁| = 2a, therefore |AF₂| = 4a. Since |BF₁| - |BF₂| = 2a, it follows that |BA| + |AF₁| - |BF₂| = 2a, so |BA| = |BF₂|. Since ∠F₂AF₁ = \\frac{2\\pi}{3}, therefore ∠F₂AB = \\frac{\\pi}{3}. ∵ F₂A = 4a ∴ AB = 4a. Thus, \\frac{S_{\\triangle AFF}}{S_{\\triangle ABF_{2}}} = \\frac{\\frac{1}{2}|AF_{1}||AF_{2}|\\sin\\frac{2\\pi}{3}}{\\frac{1}{2}|AB||AF_{2}|\\sin\\frac{\\pi}{3}} = \\frac{|AF_{1}|}{|AB|} = \\frac{2a}{4a} = \\frac{1}{2}."
  },
  {
    "text": "It is known that the center of the hyperbola is at the origin, the foci are on the $y$-axis, the focal distance is $16$, and the eccentricity is $\\frac{4}{3}$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;O: Origin;Center(G) = O;PointOnCurve(Focus(G), yAxis);FocalLength(G) = 16;Eccentricity(G)=4/3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/36 - x^2/28 = 1",
    "fact_spans": "[[[2, 5], [48, 51]], [[9, 11]], [[2, 11]], [[2, 20]], [[2, 28]], [[2, 46]]]",
    "query_spans": "[[[48, 56]]]",
    "process": ""
  },
  {
    "text": "Given point $M(3,2)$, $F$ is the focus of the parabola $y^{2}=2 x$, and point $P$ moves along this parabola. When $|P M|+|P F|$ takes the minimum value, what are the coordinates of point $P$?",
    "fact_expressions": "G: Parabola;M: Point;P: Point;F: Point;Expression(G) = (y^2 = 2*x);Coordinate(M) = (3, 2);Focus(G) = F;PointOnCurve(P, G);WhenMin(Abs(LineSegmentOf(P, F)) + Abs(LineSegmentOf(P, M)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2,2)",
    "fact_spans": "[[[16, 30], [40, 43]], [[2, 11]], [[34, 38], [67, 71]], [[12, 15]], [[16, 30]], [[2, 11]], [[12, 33]], [[34, 46]], [[47, 66]]]",
    "query_spans": "[[[67, 76]]]",
    "process": ""
  },
  {
    "text": "Given that the foci of hyperbolas $C_{1}$ and $C_{2}$ lie on the $x$-axis and $y$-axis respectively, both have asymptotes with equations $y = \\pm \\frac{1}{a} x$ ($a > 0$), and their eccentricities are $e_{1}$ and $e_{2}$ respectively, then the minimum value of $e_{1} e_{2}$ is?",
    "fact_expressions": "C1: Hyperbola;C2: Hyperbola;PointOnCurve(Focus(C1), xAxis);PointOnCurve(Focus(C2), yAxis);a: Number;a>0;Expression(Asymptote(C1)) = (y = pm*(x/a));Expression(Asymptote(C2)) = (y = pm*(x/a));e1: Number;e2: Number;Eccentricity(C1) = e1;Eccentricity(C2) = e2",
    "query_expressions": "Min(e1*e2)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 12]], [[13, 20]], [[2, 36]], [[2, 36]], [[44, 70]], [[44, 70]], [[2, 70]], [[2, 70]], [[77, 84]], [[86, 93]], [[2, 93]], [[2, 93]]]",
    "query_spans": "[[[95, 114]]]",
    "process": "Let the standard equation of hyperbola $ C_{1} $ be $ \\frac{x^{2}}{a_{1}^{2}} - \\frac{y^{2}}{b_{1}^{2}} = 1 $ $ (a_{1} > 0, b_{1} > 0) $, and the standard equation of hyperbola $ C_{2} $ be $ \\frac{y^{2}}{a_{2}^{2}} - \\frac{x^{2}}{b_{2}^{2}} = 1 $ $ (a_{2} > 0, b_{2} > 0) $. From the given conditions, we have $ \\frac{b_{1}}{a_{1}} = \\frac{a_{2}}{b_{2}} = \\frac{1}{a} $. Then,  \n$ e_{1} = \\frac{c_{1}}{a_{1}} = \\sqrt{\\frac{c_{1}^{2}}{a_{1}^{2}}} = \\sqrt{\\frac{a_{1}^{2} + b_{1}^{2}}{a_{1}^{2}}} = \\sqrt{1 + \\frac{b_{1}^{2}}{a_{1}^{2}}} = \\sqrt{1 + \\frac{1}{a^{2}}} $.  \nSimilarly, $ e_{2} = \\sqrt{1 + a^{2}} $. Therefore,  \n$ e_{1}e_{2} = \\sqrt{\\left(1 + \\frac{1}{a^{2}}\\right)(1 + a^{2})} = \\sqrt{a^{2} + \\frac{1}{a^{2}} + 2} \\geqslant \\sqrt{2\\sqrt{a^{2} \\cdot \\frac{1}{a^{2}}} + 2} = 2 $,  \nwith equality holding if and only if $ a = 1 $. Hence, the minimum value of $ e_{1}e_{2} $ is $ 2 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $P$ lies on the hyperbola such that $\\angle F_{1} P F_{2}=90^{\\circ}$. Then the distance from point $P$ to the $x$-axis is equal to?",
    "fact_expressions": "G: Hyperbola;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 - y^2/12 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P,G);AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Distance(P, xAxis)",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 41], [70, 73]], [[50, 57]], [[66, 69], [112, 116]], [[58, 65]], [[2, 41]], [[2, 65]], [[2, 65]], [[66, 74]], [[77, 110]]]",
    "query_spans": "[[[112, 127]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $C$ centered at the origin with coordinate axes as its axes of symmetry shares a common point $P(2,-1)$ with the circle $O$: $x^{2}+y^{2}=5$, and the tangent line to circle $O$ at point $P$ is parallel to one of the asymptotes of hyperbola $C$. Then, the length of the real axis of this hyperbola is?",
    "fact_expressions": "C: Hyperbola;O: Origin;Center(C) = O;SymmetryAxis(C) = axis;O1: Circle;Expression(O1) = (x^2 + y^2 = 5);P: Point;Intersection(C, O1) = P;Coordinate(P) = (2, -1);IsParallel(TangentOnPoint(P, O1), OneOf(Asymptote(C)))",
    "query_expressions": "Length(RealAxis(C))",
    "answer_expressions": "sqrt(15)",
    "fact_spans": "[[[14, 20], [71, 77], [88, 91]], [[3, 5]], [[0, 20]], [[6, 20]], [[21, 42], [57, 61]], [[21, 42]], [[62, 66], [46, 55]], [[14, 55]], [[46, 55]], [[57, 85]]]",
    "query_spans": "[[[88, 97]]]",
    "process": "First, find the slope of the tangent line to obtain the asymptote equation. From the asymptote, assume the hyperbola equation and substitute point P to determine the hyperbola equation, then find the length of the real axis. Given that the slope of OP is $k_{OP}=-\\frac{1}{2}$, the slope of the tangent line to circle O at point P is 2. Therefore, one asymptote of the hyperbola is $2x-y=0$. Thus, assume the hyperbola equation as $(2x-y)\\cdot(2x+y)=m$ ($m\\neq0$). Since point $P(2,-1)$ lies on the hyperbola, $m=[2\\times2-(-1)]\\cdot[2\\times2+(-1)]=15$. Hence, the hyperbola equation is $4x^{2}-y^{2}=15$, or $\\frac{4x^{2}}{15}-\\frac{y^{2}}{15}=1$, so $a^{2}=\\frac{15}{4}$, and the length of the real axis is $2a=\\sqrt{15}$."
  },
  {
    "text": "Given a point $Q$ on the parabola $y^{2}=5x$ such that the distance from $Q$ to the focus $F$ is $\\frac{25}{4}$, then the distance from the origin to the line $FQ$ is?",
    "fact_expressions": "G: Parabola;Q: Point;F: Point;O:Origin;Expression(G) = (y^2 = 5*x);PointOnCurve(Q, G);Focus(G) = F;Distance(Q, F) = 25/4",
    "query_expressions": "Distance(O, LineOf(F,Q))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 16]], [[19, 22]], [[25, 28]], [[48, 52]], [[2, 16]], [[2, 22]], [[2, 28]], [[19, 46]]]",
    "query_spans": "[[[48, 65]]]",
    "process": "From the given conditions, F(\\frac{5}{4},0), let point Q(x,y), and let d be the distance from the origin to the line FQ. Then x+\\frac{5}{4}=\\frac{25}{4}, solving gives x=5. Substituting x=5 into the parabola equation yields |y|=5. In triangle QOF, S_{\\triangle QOF}=\\frac{1}{2}OF\\times|y|=\\frac{1}{2}QF\\times d, that is, \\frac{5}{4}\\times5=\\frac{25}{4}\\times d, solving gives d=1."
  },
  {
    "text": "If point $P(2, -1)$ bisects a chord of the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{8}=1$, then the equation of the line containing the chord is (write the result in general form)?",
    "fact_expressions": "G: Ellipse;H: LineSegment;P: Point;Expression(G) = (x^2/12 + y^2/8 = 1);Coordinate(P) = (2, -1);IsChordOf(H, G);MidPoint(H) = P",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "4*x - 3*y - 11 = 0",
    "fact_spans": "[[[14, 52]], [], [[1, 12]], [[14, 52]], [[1, 12]], [[14, 56]], [[1, 56]]]",
    "query_spans": "[[[14, 78]]]",
    "process": ""
  },
  {
    "text": "Let the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ have eccentricity $\\sqrt{3}$, and suppose one of its directrices coincides with the directrix of the parabola $y^{2}=4 x$. Then the equation of the asymptotes of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 4*x);Eccentricity(G) = sqrt(3);OneOf(Directrix(G))=Directrix(H)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(2)*x",
    "fact_spans": "[[[1, 60], [77, 78], [106, 109]], [[4, 60]], [[4, 60]], [[84, 98]], [[4, 60]], [[4, 60]], [[1, 60]], [[84, 98]], [[1, 75]], [[77, 103]]]",
    "query_spans": "[[[106, 117]]]",
    "process": ""
  },
  {
    "text": "Given that the point $P(t, \\sqrt{3} t)$ lies on the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, if $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse, $O$ is the origin, and $|O P|=\\frac{\\sqrt{14}}{7}|F_{1} F_{2}|$, then the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;a: Number;b: Number;t:Number;P: Point;O: Origin;F1: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (t, sqrt(3)*t);PointOnCurve(P, C);LeftFocus(C) = F1;RightFocus(C) = F2;Abs(LineSegmentOf(O,P))=(sqrt(14)/7)*Abs(LineSegmentOf(F1,F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[22, 79], [100, 102], [160, 165]], [[28, 79]], [[28, 79]], [[3, 21]], [[2, 21]], [[108, 111]], [[82, 89]], [[90, 97]], [[28, 79]], [[28, 79]], [[22, 79]], [[2, 21]], [[2, 80]], [[82, 107]], [[82, 107]], [[117, 158]]]",
    "query_spans": "[[[160, 171]]]",
    "process": "Since $ P(t,\\sqrt{3}t) $ lies on the ellipse $ C $, we have $ \\frac{t^{2}}{a^{2}}+\\frac{3t^{2}}{b^{2}}=1 $, which implies $ t^{2}=\\frac{a^{2}b^{2}}{3a^{2}+b^{2}} $. From $ |OP|=\\frac{\\sqrt{14}}{7}|F_{1}F_{2}| $, we get $ t^{2}=\\frac{2}{7}c^{2}=\\frac{2a^{2}-2b^{2}}{7} $. Therefore, $ \\frac{a^{2}b^{2}}{3a^{2}+b^{2}}=\\frac{2a^{2}-2b^{2}}{7} $. Rearranging yields $ (6a^{2}+b^{2})(a^{2}-2b^{2})=0 $. Hence, $ a^{2}=2b^{2}=2a^{2}-2c^{2} $, i.e., $ a^{2}=2c^{2} $, $ e=\\frac{c}{a}=\\frac{\\sqrt{2}}{2} $."
  },
  {
    "text": "Given that $F_{1}$, $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, a line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. If $|F_{2} A|+|F_{2} B|=12$, then $|AB|$=?",
    "fact_expressions": "G: Ellipse;H: Line;F2: Point;A: Point;B: Point;F1: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(G) = {F1,F2};PointOnCurve(F1, H);Intersection(H, G) = {A,B};Abs(LineSegmentOf(F2, A)) + Abs(LineSegmentOf(F2, B)) = 12",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[19, 57], [75, 77]], [[72, 74]], [[11, 18]], [[78, 81]], [[84, 87]], [[2, 10], [64, 71]], [[19, 57]], [[2, 62]], [[63, 74]], [[72, 89]], [[91, 115]]]",
    "query_spans": "[[[117, 125]]]",
    "process": ""
  },
  {
    "text": "Given that one focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ coincides with the center of the circle $x^{2}+y^{2}-10 x=0$, and the eccentricity of the hyperbola is equal to $\\sqrt{5}$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (-10*x + x^2 + y^2 = 0);OneOf(Focus(G))=Center(H);Eccentricity(G)=sqrt(5)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5 - y^2/10 = 1",
    "fact_spans": "[[[2, 48], [82, 85], [105, 108]], [[5, 48]], [[5, 48]], [[54, 75]], [[2, 48]], [[54, 75]], [[2, 80]], [[82, 101]]]",
    "query_spans": "[[[105, 115]]]",
    "process": ""
  },
  {
    "text": "Let $P$ be an arbitrary point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{8}=1$, and let $AB$ be an arbitrary diameter of the circle $C$: $(x-1)^{2}+y^{2}=1$. Then the range of values of $\\overrightarrow{PA} \\cdot \\overrightarrow{PB}$ is?",
    "fact_expressions": "G: Ellipse;C: Circle;A: Point;B: Point;P: Point;Expression(G) = (x^2/9 + y^2/8 = 1);Expression(C) = (y^2 + (x - 1)^2 = 1);PointOnCurve(P, G);IsDiameter(LineSegmentOf(A,B),C)",
    "query_expressions": "Range(DotProduct(VectorOf(P, A), VectorOf(P, B)))",
    "answer_expressions": "[3,15]",
    "fact_spans": "[[[4, 41]], [[54, 78]], [[48, 53]], [[48, 53]], [[0, 3]], [[4, 41]], [[54, 78]], [[0, 47]], [[48, 85]]]",
    "query_spans": "[[[87, 143]]]",
    "process": "By the given condition, the center of the circle $ C(1,0) $ is the right focus of the ellipse, and the radius of the circle is 1. Since $ AB $ is any diameter of the circle $ C: (x-1)^{2}+y^{2}=1 $, we have $ \\overrightarrow{PA}\\cdot\\overrightarrow{PB}=(\\overrightarrow{PC}+\\overrightarrow{CA})\\cdot(\\overrightarrow{PC}+\\overrightarrow{CB})=(\\overrightarrow{PC}+\\overrightarrow{CA})\\cdot(\\overrightarrow{PC}-\\overrightarrow{CA})=\\overrightarrow{PC}^{2}-\\overrightarrow{CA}^{2}=|\\overrightarrow{PC}|^{2}-1 $. By the definition of the ellipse, $ |\\overrightarrow{PC}|\\in[a-c,a+c]=[2,4] $, so $ \\overrightarrow{PA}\\cdot\\overrightarrow{PB}=|\\overrightarrow{PC}|^{2}-1\\in[3,15] $."
  },
  {
    "text": "It is known that the hyperbola $C$ passes through the point $(2 \\sqrt{3},-1)$ and has the same asymptotes as the hyperbola $\\frac{x^{2}}{12}-\\frac{y^{2}}{6}=1$. Then, the standard equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Point;Coordinate(G) = (2*sqrt(3), -1);PointOnCurve(G, C);H:Hyperbola;Expression(H)=(x^2/12-y^2/6=1);Asymptote(C)=Asymptote(H)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/10 - y^2/5 = 1",
    "fact_spans": "[[[2, 8], [79, 85]], [[9, 27]], [[9, 27]], [[2, 27]], [[31, 70]], [[31, 70]], [[2, 77]]]",
    "query_spans": "[[[79, 92]]]",
    "process": "Let the required hyperbola equation be $\\frac{x^{2}}{12}-\\frac{y^{2}}{6}=k$, substitute the coordinates of the given point to solve. According to the problem, let the required hyperbola equation be $\\frac{x^{2}}{12}-\\frac{y^{2}}{6}=k$. Since the hyperbola passes through the point $(2\\sqrt{3}, \\sqrt{1})$, we have $\\frac{12}{12}-\\frac{1}{6}=k$, so $k=\\frac{5}{6}$. Thus, the hyperbola equation is $\\frac{x^{2}}{12}-\\frac{y^{2}}{6}=\\frac{5}{6}$, which simplifies to $\\frac{x^{2}}{10}-\\frac{y^{2}}{5}=1$."
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the hyperbola $x^{2}-\\frac{y^{2}}{24}=1$, and let $P$ be a point on the hyperbola such that $3|P F_{1}|=4|P F_{2}|$. Then what is the perimeter of $\\triangle P F_{1} F_{2}$?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2/24 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);3*Abs(LineSegmentOf(P, F1)) = 4*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Perimeter(TriangleOf(P, F1, F2))",
    "answer_expressions": "24",
    "fact_spans": "[[[17, 46], [56, 59]], [[52, 55]], [[1, 8]], [[9, 16]], [[17, 46]], [[1, 51]], [[52, 63]], [[65, 88]]]",
    "query_spans": "[[[90, 119]]]",
    "process": "The hyperbola $ x^{2}-\\frac{y^{2}}{24}=1 $ has $ a=1 $, $ c=\\sqrt{1+24}=5 $, with two foci $ F_{1}(-5,0) $, $ F_{2}(5,0) $, so $ |F_{1}F_{2}|=10 $. Given $ 3|PF_{1}|=4|PF_{2}| $, let $ |PF_{2}|=x $, then $ |PF_{1}|=\\frac{4}{3}x $. By the definition of a hyperbola, $ \\frac{4}{3}x - x = 2 $, solving gives $ x=6 $. Therefore, $ |PF_{1}|=8 $, $ |PF_{2}|=6 $, $ |F_{1}F_{2}|=10 $, and the perimeter of $ \\triangle PF_{1}F_{2} $ is $ |PF_{1}| + |PF_{2}| + |F_{1}F_{2}| = 8+6+10=24 $."
  },
  {
    "text": "Let the vertex of the parabola be $(2 , 0)$, and the directrix be $x = -1$. Then its focus coordinates are?",
    "fact_expressions": "G: Parabola;Coordinate(Vertex(G)) = (2, 0);Expression(Directrix(G)) = (x = -1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(5, 0)",
    "fact_spans": "[[[1, 4], [33, 34]], [[1, 19]], [[1, 31]]]",
    "query_spans": "[[[33, 41]]]",
    "process": ""
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ has left and right foci $F_{1}$, $F_{2}$ respectively, and $P$ is a point on the ellipse such that $\\angle P F_{1} F_{2}=\\frac{\\pi}{6}$, $\\angle P F_{2} F_{1}=\\frac{\\pi}{3}$. Then the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);AngleOf(P, F1, F2) = pi/6;AngleOf(P, F2, F1) = pi/3;Eccentricity(G) = e;e: Number",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[0, 45], [73, 75], [157, 159]], [[0, 45]], [[2, 45]], [[2, 45]], [[53, 60]], [[61, 68]], [[0, 68]], [[0, 68]], [[69, 72]], [[69, 78]], [[80, 116]], [[119, 155]], [[157, 166]], [[163, 166]]]",
    "query_spans": "[[[163, 168]]]",
    "process": "From $\\angle PF_{1}F_{2} = \\frac{\\pi}{6}$, $\\angle PF_{2}F_{1} = \\frac{\\pi}{3}$ we get: $PF_{1} \\perp PF_{2}$ and $|PF_{2}| = \\frac{1}{2}|F_{1}F_{2}| = c$. By the definition of an ellipse: $|PF_{1}| = 2a - |PF_{2}| = 2a - c$. Also, $|PF_{1}|^{2} + |PF_{2}|^{2} = |F_{1}F_{2}|^{2}$, that is: $(2a - c)^{2} + c^{2} = 4c^{2}$. Simplifying yields: $2a = (\\sqrt{3} + 1)c$, solving gives: $e = \\frac{c}{a} = \\sqrt{3} - 1$. The correct result for this problem: $\\sqrt{3} -$"
  },
  {
    "text": "It is known that one focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ coincides with the focus of the parabola $y^{2}=4 x$, and the eccentricity of the hyperbola is equal to $\\sqrt{5}$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 4*x);OneOf(Focus(G)) = Focus(H);Eccentricity(G) = sqrt(5)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "5*x^2-(5/4)*y^2=1",
    "fact_spans": "[[[2, 59], [86, 89], [108, 111]], [[5, 59]], [[5, 59]], [[65, 79]], [[5, 59]], [[5, 59]], [[2, 59]], [[65, 79]], [[2, 84]], [[86, 105]]]",
    "query_spans": "[[[108, 116]]]",
    "process": ""
  },
  {
    "text": "Let $P$ be a moving point on the curve $y^{2}=4x$. Then the minimum value of the sum of the distance from point $P$ to point $A(-1,2)$ and the distance from point $P$ to the line $x=-1$ is?",
    "fact_expressions": "E: Curve;A: Point;P: Point;Expression(E) = (y^2 = 4*x);Coordinate(A) = (-1, 2);PointOnCurve(P, E);l: Line;Expression(l) = (x = -1)",
    "query_expressions": "Min(Distance(P, A) + Distance(P, l))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[5, 18]], [[31, 41]], [[1, 4], [26, 30], [45, 49]], [[5, 18]], [[31, 41]], [[1, 24]], [[50, 56]], [[50, 56]]]",
    "query_spans": "[[[26, 67]]]",
    "process": "Since point P lies on the parabola, the distance d from point P to the line x = -1 is equal to PF, so d + PA = PF + PA, and the minimum value is clearly AF. [Detailed solution] As shown in the figure, point P is on the parabola y^{2} = 4x, so the distance d from point P to the line x = -1 is equal to PF; thus, d + PA = PF + PA, and the minimum occurs when points P, A, and F are collinear, which is 2\\sqrt{2}."
  },
  {
    "text": "The coordinates of the point on the parabola $y=2 x^{2}$ that is closest to the line $y=4 x-5$ are?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Expression(G) = (y = 2*x^2);Expression(H) = (y = 4*x - 5);PointOnCurve(P, G);WhenMin(Distance(P,H))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1,2)",
    "fact_spans": "[[[0, 14]], [[16, 27]], [[33, 34]], [[0, 14]], [[16, 27]], [[0, 34]], [[15, 34]]]",
    "query_spans": "[[[33, 39]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{y^{2}}{12}+\\frac{x^{2}}{3}=1$, what is the equation of the line passing through the point $P(-1,2)$ such that the chord of the ellipse is bisected by $P$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/3 + y^2/12 = 1);G: LineSegment;IsChordOf(G, C) = True;P: Point;Coordinate(P) = (-1, 2);PointOnCurve(P, G) = True;MidPoint(G) = P",
    "query_expressions": "Expression(OverlappingLine(G))",
    "answer_expressions": "2*x-y+4=0",
    "fact_spans": "[[[2, 45]], [[2, 45]], [], [[2, 68]], [[49, 59], [61, 64]], [[49, 59]], [[2, 68]], [[2, 68]]]",
    "query_spans": "[[[2, 77]]]",
    "process": "Let the line passing through point P(-1,2) intersect the ellipse at points A(x_{1},y_{1}) and B(x_{2},y_{2}). Substitute into the equation of the ellipse, subtract the two equations, and use the midpoint formula and slope formula to find the slope of line AB, then obtain the equation of the line and thus the answer. Let the line passing through point P(-1,2) intersect the ellipse at A(x_{1},y_{1}) and B(x_{2},y_{2}). Then from \\frac{y_{1}^{2}}{12}+\\frac{x_{1}^{2}}{3}=1, \\frac{y_{2}^{2}}{12}+\\frac{x_{2}^{2}}{3}=1, subtracting the two equations yields \\frac{(y_{1}-y_{2})(y_{1}+y_{2})}{12}+\\frac{(x_{1}-x_{2})(x_{1}+x_{2})}{3}=0. Since point P(-1,2) is the midpoint of A and B, we have x_{1}+x_{2}=-2, y_{1}+y_{2}=4. Therefore, k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{12(x_{1}+x_{2})}{3(y_{1}+y_{2})}=2. Hence, the equation of the line passing through point P(-1,2) and bisected by P is y-2=2(x+1), or 2x-y+4=0."
  },
  {
    "text": "Draw a line through the focus of the parabola $y^{2}=4x$ intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $x_{1}+x_{2}=12$, then $|AB|$ equals?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Line;PointOnCurve(Focus(G), H);A: Point;B: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Intersection(H, G) = {A, B};x1 + x2 = 12",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "14",
    "fact_spans": "[[[1, 15], [22, 25]], [[1, 15]], [[19, 21]], [[0, 21]], [[26, 43]], [[45, 62]], [[26, 43]], [[45, 62]], [[26, 43]], [[45, 62]], [[26, 43]], [[45, 62]], [[19, 64]], [[66, 82]]]",
    "query_spans": "[[[84, 94]]]",
    "process": ""
  },
  {
    "text": "The ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has left and right foci $F_{1}$, $F_{2}$, upper and lower vertices $B_{1}$, $B_{2}$, and right vertex $A$. The line $A B_{1}$ intersects $B_{2} F_{1}$ at point $D$. If $2|A B_{1}|=3|B_{1} D|$, then the eccentricity of $C$ is equal to?",
    "fact_expressions": "C: Ellipse;a: Number;b: Number;a>b;b>0;F1: Point;F2: Point;B1: Point;B2: Point;A: Point;D: Point;Expression(C) = (x^2/a^2+y^2/b^2=1);LeftFocus(C)=F1;RightFocus(C)=F2;UpperVertex(C)=B1;LowerVertex(C)=B2;RightVertex(C)=A;Intersection(LineOf(A,B1), LineOf(B2,F1)) = D;2*Abs(LineSegmentOf(A,B1)) = 3*Abs(LineSegmentOf(B1,D))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "1/4",
    "fact_spans": "[[[0, 56], [172, 175]], [[6, 56]], [[6, 56]], [[6, 56]], [[6, 56]], [[65, 72]], [[73, 80]], [[89, 96]], [[97, 104]], [[109, 112]], [[140, 144]], [[0, 56]], [[0, 80]], [[0, 80]], [[0, 104]], [[0, 104]], [[0, 112]], [[113, 144]], [[147, 170]]]",
    "query_spans": "[[[172, 182]]]",
    "process": "As shown in the figure: let D(x_{0},y_{0}), from 2|AB_{1}|=3|B_{1}D|, we get \\frac{AB_{1}}{AD}=\\frac{3}{5}. According to similar triangles: \\frac{a}{a-x_{0}}=\\frac{3}{5}=\\frac{b}{y_{0}}, solving gives x_{0}=\\frac{2}{3}a, y_{0}=\\frac{5}{3}b. The equation of line B_{2}F_{1} is: \\frac{x}{-c}+\\frac{y}{-b}=1. Substituting point D into it yields:"
  },
  {
    "text": "The distance from a focus of the hyperbola $C$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0, b>0)$ to one of its asymptotes is $\\sqrt{2} a$. Then, the equations of the asymptotes of hyperbola $C$ are?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;a>0;b>0;Expression(C) = (-x^2/b^2 + y^2/a^2 = 1);Distance(Focus(C), OneOf(Asymptote(C))) = sqrt(2)*a",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "x + pm*sqrt(2)*y = 0",
    "fact_spans": "[[[0, 61], [88, 94]], [[74, 86]], [[8, 61]], [[8, 61]], [[8, 61]], [[0, 61]], [[0, 86]]]",
    "query_spans": "[[[88, 102]]]",
    "process": "Let the semi-focal distance of the hyperbola be $ c $, then the coordinates of the foci are $ (0,\\pm c) $, and the asymptotes of the hyperbola are given by: $ ax\\pm by=0 $. Hence, the distance from the focus to the asymptote is $ \\frac{|bc|}{\\sqrt{a^{2}+b^{2}}}=b $, so $ b=\\sqrt{2}a $, thus the equations of the asymptotes are: $ x\\pm\\sqrt{2}y=0 $."
  },
  {
    "text": "Given the parabola $E$: $x^{2}=8 y$ with focus $F$, a line $l$ passing through $F$ intersects $E$ at points $A$ and $B$, and intersects the $x$-axis at point $C$. If $A$ is the midpoint of segment $C F$, then $|A B|$=?",
    "fact_expressions": "l: Line;E: Parabola;C: Point;F: Point;A: Point;B: Point;Expression(E) = (x^2 = 8*y);Focus(E) = F;PointOnCurve(F, l);Intersection(l, E) = {A, B};Intersection(l,xAxis)=C;MidPoint(LineSegmentOf(C,F)) = A",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "9",
    "fact_spans": "[[[34, 39]], [[2, 21], [40, 43]], [[62, 66]], [[25, 28], [30, 33]], [[45, 48], [69, 72]], [[49, 52]], [[2, 21]], [[2, 28]], [[29, 39]], [[34, 54]], [[34, 66]], [[69, 83]]]",
    "query_spans": "[[[85, 94]]]",
    "process": "According to the problem, the parabola $ E: x^{2} = 8y $ gives $ p = 4 $, with focus $ F(0, 2) $. Since $ A $ is the midpoint of segment $ CF $, we have $ (-2\\sqrt{2}, 1) $. Then $ k_{AF} = \\frac{2 - 1}{0 - (-2\\sqrt{2})} = \\frac{\\sqrt{2}}{4} $. Thus, the equation of line $ AF $ is $ y = \\frac{\\sqrt{2}}{4}x + 2 $. Solving the system of equations  \n$$\n\\begin{cases}\ny = \\frac{\\sqrt{2}}{4}x + 2 \\\\\nx^{2} = 8y\n\\end{cases}\n$$  \nwe obtain $ x^{2} - 2\\sqrt{2}x - 16 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = 2\\sqrt{2} $, and $ y_{1} + y_{2} = \\frac{\\sqrt{2}}{4}(x_{1} + x_{2}) + 4 = 5 $. Therefore, $ |AB| = y_{1} + y_{2} + p = 5 + 4 = 9 $."
  },
  {
    "text": "Given that the line $y = kx + 1$ intersects the curve $y^2 = 2x$ at only one point, find the value of the real number $k$.",
    "fact_expressions": "G: Line;k: Real;H: Curve;Expression(G) = (y = k*x + 1);Expression(H) = (y^2 = 2*x);NumIntersection(G, H) = 1",
    "query_expressions": "k",
    "answer_expressions": "{0, 1/2}",
    "fact_spans": "[[[2, 13]], [[35, 40]], [[14, 27]], [[2, 13]], [[14, 27]], [[2, 33]]]",
    "query_spans": "[[[35, 44]]]",
    "process": "Solving the system of the line equation and the parabola equation yields: $k^{2}x^{2}+(2k-2)x+1=0$.  \n① If $k=0$, then $y=1$, $x=\\frac{1}{2}$, which satisfies the condition;  \n② If $k\\neq0$, then $\\Delta=(2k-2)^{2}-4k^{2}=0$, solving gives $k=\\frac{1}{2}$.  \nIn conclusion, $k=0$ or $\\frac{1}{2}$."
  },
  {
    "text": "There are two points $A$ and $B$ on the parabola $y^{2}=4x$, and $|AB|=6$. What is the minimum distance from the midpoint $M$ of segment $AB$ to the $y$-axis?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);Abs(LineSegmentOf(A, B)) = 6;M: Point;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Min(Distance(M, yAxis))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[18, 21]], [[22, 25]], [[0, 25]], [[0, 25]], [[28, 37]], [[50, 53]], [[40, 53]]]",
    "query_spans": "[[[50, 65]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, when point $P$ moves on the ellipse $C$, what is the trajectory equation of the incenter $I$ of $\\Delta P F_{1} F_{2}$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C);I: Point;Incenter(TriangleOf(P, F1, F2)) = I",
    "query_expressions": "LocusEquation(I)",
    "answer_expressions": "(x^2+3*y^2=1)&Negation(y=0)",
    "fact_spans": "[[[18, 60], [72, 77]], [[18, 60]], [[2, 9]], [[10, 17]], [[2, 66]], [[2, 66]], [[67, 71]], [[67, 78]], [[107, 110]], [[82, 110]]]",
    "query_spans": "[[[107, 117]]]",
    "process": "Consider a more general problem: Let $ P $ be a moving point on the ellipse $ C: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $), $ F_{1}, F_{2} $ be the two foci of the ellipse, and $ I $ be the incenter of $ \\triangle PF_{1}F_{2} $. Find the equation of the trajectory of point $ I $.\n\nSolution 1: As shown in the figure, let the incircle $ l $ touch $ F_{1}F_{2} $ at point $ H $, with radius $ r $, and let $ F_{1}H = y $, $ F_{2}H = z $, $ PF_{1} = x+y $, $ PF_{2} = x+z $. The product of the slopes of lines $ IF_{1} $ and $ IF_{2} $: $ k_{IF_{1}} \\cdot k_{IF_{2}} = $. According to Heron's formula, the area of $ \\triangle PF_{1}F_{2} $ is $ (x+y+z) $. Thus we have $ k_{IF_{1}} \\cdot k_{IF_{2}} = -\\frac{x}{x+y+z} $. Then, according to the slope product definition of the ellipse, the trajectory of point $ I $ is an ellipse with $ F_{1}F_{2} $ as the major axis and eccentricity $ e $ satisfying $ e^{2}-1 = -\\frac{a-c}{a+c} $. Its standard equation is $ \\frac{x^{2}}{c^{2}} + \\frac{y^{2}}{\\frac{a-c}{a+c} \\cdot c^{2}} = 1 $ ($ y \\neq 0 $).\n\nSolution 2: Let $ P(a\\cos\\theta, b\\sin\\theta) $, then $ \\sin\\theta \\neq 0 $. The area of triangle $ PF_{1}F_{2} $: $ S = \\frac{1}{2} \\cdot 2c \\cdot |b\\sin\\theta| = \\frac{1}{2}(2c+2a) \\cdot r $, where $ r $ is the inradius. Solving gives $ r = \\frac{bc \\cdot |\\sin\\theta|}{a+c} = |y_{1}| $. On the other hand, by properties of the incircle and focal radius formulas: $ (c - x_{1}) - (x_{1} + c) = |PF_{1}| - |PF_{2}| = (a - c\\cos\\theta) - (a + c\\cos\\theta) $, yielding $ x_{1} = c\\cos\\theta $. Eliminating $ \\theta $, the trajectory equation of point $ I $ is: $ \\frac{x^{2}}{c^{2}} + \\frac{y^{2}}{\\frac{a-c}{a+c} \\cdot c^{2}} = 1 $ ($ y \\neq 0 $). In this problem, $ a = 2 $, $ c = 1 $, substituting into the above yields the trajectory equation: $ x^{2} + 3y^{2} = 1 $ ($ y \\neq 0 $)."
  },
  {
    "text": "Through one focus of the hyperbola $\\frac{(y-3)^{2}}{2}-\\frac{(x+2)^{2}}{6}=1$, a line perpendicular to the real axis is drawn, intersecting the two asymptotes of the hyperbola at points $A$ and $B$, respectively. Then the length of segment $AB$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;A: Point;B: Point;Expression(G) = (-(x + 2)^2/6 + (y - 3)^2/2 = 1);PointOnCurve(OneOf(Focus(G)),H);IsPerpendicular(H,RealAxis(G));L1:Line;L2:Line;Asymptote(G)={L1,L2};Intersection(L1,H)=A;Intersection(L2,H)=B",
    "query_expressions": "Length(LineSegmentOf(A,B))",
    "answer_expressions": "4*sqrt(6)",
    "fact_spans": "[[[1, 47], [63, 66]], [[59, 61]], [[76, 79]], [[80, 83]], [[1, 47]], [[0, 61]], [[0, 61]], [], [], [[63, 72]], [[59, 83]], [[59, 83]]]",
    "query_spans": "[[[86, 97]]]",
    "process": "Make the substitution $ y = y - 3 $, $ x = x + 2 $, the equation of the hyperbola becomes $ \\frac{y^{2}}{2} - \\frac{x^{2}}{6} = 1 $. Find the coordinates of the foci and the equations of the asymptotes of the hyperbola $ \\frac{y^{2}}{2} - \\frac{x^{2}}{6} = 1 $, then the length of segment $ AB $ can be determined. Making the substitution $ y = y - 3 $, $ x = x + 2 $, in the rectangular coordinate system $ xOy $, the equation of the hyperbola is $ \\frac{y^{2}}{2} - \\frac{x^{2}}{6} = 1 $, where $ a = \\sqrt{2} $, $ b = \\sqrt{6} $, $ c = 2\\sqrt{2} $. The asymptotes of the hyperbola are $ y = \\pm\\frac{a}{b}x = \\pm\\frac{\\sqrt{3}}{3}x $. The line passing through the focus $ (0, 2\\sqrt{2}) $ of the hyperbola $ \\frac{y^{2}}{2} - \\frac{x^{2}}{6} = 1 $ and perpendicular to the real axis has the equation $ y = 2\\sqrt{2} $. Substituting $ y = 2\\sqrt{2} $ into the asymptote equation $ y = \\frac{\\sqrt{3}}{3}x $ gives $ x = 2\\sqrt{6} $. Therefore, the length of segment $ AB $ is $ 2|x| = 4\\sqrt{6} $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{m+1}+\\frac{y^{2}}{m}=1$ has eccentricity $\\frac{1}{2}$, then the length of the minor axis of $C$ is?",
    "fact_expressions": "C: Ellipse;m: Number;Expression(C) = (x^2/(m + 1) + y^2/m = 1);Eccentricity(C) = 1/2",
    "query_expressions": "Length(MinorAxis(C))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[2, 46], [66, 69]], [[9, 46]], [[2, 46]], [[2, 64]]]",
    "query_spans": "[[[66, 75]]]",
    "process": "Using the eccentricity, the equation of the ellipse can be found, and then the length of the minor axis of the ellipse can be determined. [Detailed solution] From the given conditions, c^{2}=a^{2}-b^{2}=m+1-m=1, and \\because e=\\frac{c}{a}=\\frac{1}{\\sqrt{m+1}}=\\frac{1}{2}, solving gives m=3, \\therefore the equation of the ellipse is \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1, then the length of the minor axis of C is 2b=2\\sqrt{3}"
  },
  {
    "text": "Let the right vertex of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ be $A$, and the right focus be $F$. A line passing through point $F$ and parallel to one asymptote of the hyperbola intersects the other asymptote at point $B$. Then the area of $\\triangle AFB$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;A: Point;F: Point;B: Point;l1: Line;l2: Line;Expression(G) = (x^2/9 - y^2/16 = 1);RightVertex(G) = A;RightFocus(G) = F;PointOnCurve(F, H);IsParallel(H, l1);Intersection(H, l2) = B;OneOf(Asymptote(G))=L1;OneOf(Asymptote(G))=L2;Negation(L1=L2)",
    "query_expressions": "Area(TriangleOf(A, F, B))",
    "answer_expressions": "10/3",
    "fact_spans": "[[[1, 40], [64, 67]], [[76, 78]], [[45, 48]], [[53, 56], [58, 62]], [[87, 91]], [], [], [[1, 40]], [[1, 48]], [[1, 56]], [[57, 78]], [[63, 78]], [[63, 91]], [62, 70], [62, 82], [62, 82]]",
    "query_spans": "[[[93, 114]]]",
    "process": "Draw the graph as shown: based on the properties of the hyperbola, the equation of line $ l $ is $ y = -\\frac{4}{3}x $. Line $ BF $ is parallel to one asymptote, indicating $ k = \\frac{4}{3} $. Since it passes through $ F(5,0) $, the equation of line $ BF $ is $ y = \\frac{4}{3}(x - 5) $. Solving these two equations simultaneously gives the coordinates of $ B $ as $ \\left(\\frac{5}{2}, -\\frac{10}{3}\\right) $, while the coordinates of $ A $ are $ (3,0) $. Therefore, $ S_{AAFB} = \\frac{1}{2} \\cdot 2 \\cdot \\frac{10}{3} = \\frac{10}{3} $. [Note] This problem examines the properties of hyperbolas and the positional relationship between lines and hyperbolas, with medium difficulty."
  },
  {
    "text": "Given the hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. The point $M(x_{0}, y_{0})$ in the first quadrant lies on the asymptote of the hyperbola $C_{1}$, and $M F_{1} \\perp M F_{2}$. If the parabola $C_{2}$: $y^{2}=2 p x$ $(p>0)$ with focus at $F_{2}$ passes through point $M$, then the eccentricity of the hyperbola $C_{1}$ is?",
    "fact_expressions": "C1: Hyperbola;C2: Parabola;M: Point;F1: Point;F2: Point;x0:Number;y0:Number;p:Number;p>0;a:Number;b:Number;a>0;b>0;Expression(C1)=(-y^2/b^2 + x^2/a^2 = 1);Expression(C2)=(y^2=2*p*x);Coordinate(M) = (x0, y0);LeftFocus(C1) =F1;RightFocus(C1)=F2;Quadrant(M)=1;PointOnCurve(M,Asymptote(C1));IsPerpendicular(LineSegmentOf(M, F1), LineSegmentOf(M, F2));Focus(C2)=F2;PointOnCurve(M,C2)",
    "query_expressions": "Eccentricity(C1)",
    "answer_expressions": "2+sqrt(5)",
    "fact_spans": "[[[2, 66], [116, 126], [207, 217]], [[170, 199]], [[97, 115], [201, 205]], [[75, 82]], [[83, 90], [159, 166]], [[98, 115]], [[98, 115]], [[181, 199]], [[181, 199]], [[13, 66]], [[13, 66]], [[13, 66]], [[13, 66]], [[2, 66]], [[170, 199]], [[97, 115]], [[2, 90]], [[2, 90]], [[91, 115]], [[97, 131]], [[133, 156]], [[158, 199]], [[170, 205]]]",
    "query_spans": "[[[207, 223]]]",
    "process": ""
  },
  {
    "text": "Given that $M$ is a point on the parabola $y^{2}=6 x$, $F$ is the focus of the parabola, and $\\angle M F O=60^{\\circ}$, then $|F M|$=?",
    "fact_expressions": "G: Parabola;M: Point;F: Point;O: Origin;Expression(G) = (y^2 = 6*x);PointOnCurve(M, G);Focus(G) = F;AngleOf(M, F, O) = ApplyUnit(60, degree)",
    "query_expressions": "Abs(LineSegmentOf(F, M))",
    "answer_expressions": "2",
    "fact_spans": "[[[6, 20], [31, 34]], [[2, 5]], [[27, 30]], [[39, 64]], [[6, 20]], [[2, 26]], [[27, 37]], [[39, 64]]]",
    "query_spans": "[[[66, 75]]]",
    "process": "Let |FM| = m. From the equation of the parabola, we know: F(\\frac{3}{2}, 0). Since the parabola is symmetric about the horizontal axis, assume without loss of generality that M lies in the first quadrant. Draw MN \\bot OF with foot N. Since \\angle MFO = 60^{\\circ}, then NF = FM \\cdot \\cos 60^{\\circ} = \\frac{1}{2}m, MN = FM \\cdot \\sin 60^{\\circ} = \\frac{\\sqrt{3}}{2}m. Thus, the x-coordinate of M is: \\frac{3}{2} - \\frac{1}{2}m, and the y-coordinate is: \\frac{\\sqrt{3}}{2}m. Therefore, (\\frac{\\sqrt{3}}{2}m)^2 = 6(\\frac{3}{2} - \\frac{1}{2}m) \\Rightarrow m = 2 (discarding the negative value)."
  },
  {
    "text": "Draw a chord $AB$ of the parabola $y^2 = 8x$ passing through the point $Q(4, 1)$, such that $Q$ exactly bisects the chord $AB$. Then the equation of the line $AB$ is?",
    "fact_expressions": "Q: Point;Coordinate(Q) = (4, 1);G: Parabola;Expression(G) = (y^2 = 8*x);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);PointOnCurve(Q, LineSegmentOf(A, B)) ;MidPoint(LineSegmentOf(A,B)) = Q",
    "query_expressions": "Expression(LineOf(A, B))",
    "answer_expressions": "4*x - y - 15 = 0",
    "fact_spans": "[[[1, 12], [39, 42]], [[1, 12]], [[13, 27]], [[13, 27]], [[29, 34]], [[29, 34]], [[13, 34]], [[0, 34]], [[13, 44]]]",
    "query_spans": "[[[46, 58]]]",
    "process": "From the given conditions, when AB is perpendicular to the x-axis, it does not satisfy the requirements; therefore, the slope of line AB exists. Let A(x_{1},y_{1}), B(x_{2},y_{2}), then y_{1}^{2}=8x_{1}\\textcircled{1}, y_{2}^{2}=8x_{2}\\textcircled{2}. And x_{1}+x_{2}=8, y_{1}+y_{2}=2, \\textcircled{1}-\\textcircled{2} gives (y_{1}+y_{2})(y_{1}-y_{2})=8(x_{1}-x_{2}), that is, 2(y_{1}-y_{2})=8(x_{1}-x_{2}), namely \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=4, thus the slope k of line AB is 4. Hence, the equation of line AB is y=4(x-4)+1, i.e., 4x-y-15=0. [Key Point] This question mainly examines the positional relationship between a line and a parabola, the application of the point difference method, and aims to assess students' transformation ability and computational solving skills."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $P$ is a point on the hyperbola. If $|P F_{1}| +|P F_{2}| =10$, then $P F_{1} \\cdot P F_{2}$=?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 - y^2/16 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 10;PointOnCurve(P,G)",
    "query_expressions": "LineSegmentOf(P, F1)*LineSegmentOf(P, F2)",
    "answer_expressions": "0",
    "fact_spans": "[[[2, 41], [70, 73]], [[66, 69]], [[50, 57]], [[58, 65]], [[2, 41]], [[2, 65]], [[2, 65]], [[79, 105]], [[66, 77]]]",
    "query_spans": "[[[107, 132]]]",
    "process": ""
  },
  {
    "text": "If one of the asymptotes of the hyperbola $\\frac{x^{2}}{m^{2}}-y^{2}=1$ $(m>0)$ is given by $x+\\sqrt{3} y=0$, then $m=$?",
    "fact_expressions": "G: Hyperbola;m: Number;m>0;Expression(G) = (-y^2 + x^2/m^2 = 1);Expression(OneOf(Asymptote(G))) = (x + sqrt(3)*y = 0)",
    "query_expressions": "m",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[1, 38]], [[65, 68]], [[4, 38]], [[1, 38]], [[1, 63]]]",
    "query_spans": "[[[65, 70]]]",
    "process": "The asymptotes of the hyperbola $\\frac{x^{2}}{m^{2}}-y^{2}=1$ $(m>0)$ are given by $y=\\pm\\frac{1}{m}x$. Since one asymptote of the hyperbola $\\frac{x^{2}}{m^{2}}-y^{2}=1$ $(m>0)$ is $x+\\sqrt{3}y=0$, that is, $y=-\\frac{\\sqrt{3}}{3}x$, it follows that $\\frac{1}{m}=\\frac{\\sqrt{3}}{3}$, so $m=\\sqrt{3}$."
  },
  {
    "text": "If the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ are $F_{1}$ and $F_{2}$ respectively, the focus of the parabola $y^{2}=4 b x$ is $M$, and $|F_{1} M|=2|F_{2} M|$, then the eccentricity of this ellipse is?",
    "fact_expressions": "G: Parabola;b: Number;H: Ellipse;a: Number;F1: Point;M: Point;F2: Point;Expression(G) = (y^2 = 4*(b*x));a > b;b > 0;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(H) = F1;RightFocus(H) = F2;Focus(G) = M;Abs(LineSegmentOf(F1, M)) = 2*Abs(LineSegmentOf(F2, M))",
    "query_expressions": "Eccentricity(H)",
    "answer_expressions": "{sqrt(10)/10, 3*sqrt(10)/10}",
    "fact_spans": "[[[78, 94]], [[3, 53]], [[1, 53], [128, 130]], [[3, 53]], [[62, 69]], [[98, 101]], [[70, 77]], [[78, 94]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 77]], [[1, 77]], [[78, 101]], [[103, 125]]]",
    "query_spans": "[[[128, 136]]]",
    "process": ""
  },
  {
    "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{3}=1$ are given by? The eccentricity is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2/3 = 1)",
    "query_expressions": "Expression(Asymptote(G));Eccentricity(G)",
    "answer_expressions": "y=pm*x\nsqrt(2)",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 46]], [[0, 51]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y=\\frac{1}{4} x^{2}$ and the circle $C$: $(x-1)^{2}+(y-2)^{2}=r^{2}$ ($r>0$) have a common point $P$, and if the tangent to the parabola at point $P$ is also tangent to the circle $C$, then $r=$?",
    "fact_expressions": "G: Parabola;C: Circle;r: Number;P: Point;Expression(G) = (y = x^2/4);r>0;Expression(C) = ((x - 1)^2 + (y - 2)^2 = r^2);Intersection(C,G)=P;IsTangent(TangentOnPoint(P,G),C)",
    "query_expressions": "r",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 26], [73, 76]], [[27, 64], [86, 90]], [[95, 98]], [[68, 71], [77, 81]], [[2, 26]], [[32, 64]], [[27, 64]], [[2, 71]], [[73, 93]]]",
    "query_spans": "[[[95, 100]]]",
    "process": "Let point $ P(x_{0},\\frac{1}{4}x_0^{2}) $. Then from $ x^{2}=4y $, differentiating gives $ y'=\\frac{1}{2}x $, so the slope of the tangent to the parabola at point $ P $ is $ k=\\frac{1}{2}x_{0} $. Since the center of the circle $ (x-1)^{2}+(y-2)^{2}=r^{2} $ $ (r>0) $ has coordinates $ C(1,2) $, then $ k_{PC}=\\frac{\\frac{1}{4}x_{0}^{2}-2}{x_{0}-1} $. Therefore, $ k_{PC} \\cdot k = \\frac{\\frac{1}{4}x_{0}^{2}-2}{x_{0}-1} \\cdot \\frac{1}{2}x_{0} = -1 $. Solving gives $ x_{0}=2 $, so $ P(2,1) $. Hence, $ r = \\begin{matrix} \\\\ PC \\end{matrix} = \\sqrt{(1-2)^{2}+(2-1)^{2}} = \\sqrt{2} $."
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/4 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 44]]]",
    "process": "From the given, $a^{2}=16$, $b^{2}=4$, so $c^{2}=a^{2}-b^{2}=12^{\\circ}$, that is, $a=4$, $c=2\\sqrt{3}$, therefore, $e=\\frac{c}{a}=\\frac{\\sqrt{3}}{2}$"
  },
  {
    "text": "Draw tangents from the point $(1 , \\frac{1}{2})$ to the circle $x^{2}+y^{2}=1$, with points of tangency $A$ and $B$. If the line $AB$ passes exactly through the focus and the upper vertex of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}} = 1(a>b>0)$, then the equation of the ellipse is?",
    "fact_expressions": "P: Point;Coordinate(P) = (1, 1/2);H: Circle;Expression(H) = (x^2 + y^2 = 1);L1: Line;L2: Line;A: Point;B: Point;TangentOfPoint(P, H) = {L1, L2};TangentPoint(L1, H) = A;TangentPoint(L2, H) = B;G: Ellipse;Expression(G)=(y^2/b^2+x^2/a^2=1);b: Number;a: Number;a>b;b>0;PointOnCurve(Focus(G), LineOf(A, B));PointOnCurve(UpperVertex(G), LineOf(A, B))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5+y^2/4=1",
    "fact_spans": "[[[1, 21]], [[1, 21]], [[22, 38]], [[22, 38]], [], [], [[47, 50]], [[51, 54]], [[0, 41]], [[0, 54]], [[0, 54]], [[67, 121], [130, 132]], [[67, 121]], [[69, 121]], [[69, 121]], [[69, 121]], [[69, 121]], [[57, 124]], [[57, 128]]]",
    "query_spans": "[[[130, 136]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{12-m}=1$ $(0<m<12)$ has asymptotes with slope $k$, and $k^{2}>2$. Then the range of values for $m$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/(12 - m) + x^2/m = 1);m: Number;0 < m;m < 12;Slope(Asymptote(G)) = k;k: Number;k^2 > 2",
    "query_expressions": "Range(m)",
    "answer_expressions": "(0, 4)",
    "fact_spans": "[[[0, 49]], [[0, 49]], [[72, 75]], [[3, 49]], [[3, 49]], [[0, 59]], [[56, 59]], [[61, 70]]]",
    "query_spans": "[[[72, 82]]]",
    "process": ""
  },
  {
    "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ have an asymptote $y=\\sqrt{2} x$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(C))) = (y = sqrt(2)*x)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[1, 63], [86, 89]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[1, 63]], [[1, 84]]]",
    "query_spans": "[[[86, 95]]]",
    "process": "From the hyperbola equation $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, it follows that its foci lie on the $x$-axis. Since one of its asymptotes is $y=\\sqrt{2}x$, we have $\\frac{b}{a}=\\sqrt{2}$, and $e=\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{3}$."
  },
  {
    "text": "Given that $F$ is the right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $A$ is the right vertex of $C$, and $B$ is a point on $C$ such that $BF$ is perpendicular to the $x$-axis. If the slope of $AB$ is $3$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(C) = F;A: Point;RightVertex(C) = A;B: Point;PointOnCurve(B, C);IsPerpendicular(LineSegmentOf(B, F), xAxis);Slope(LineSegmentOf(A, B)) = 3",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[6, 67], [76, 79], [88, 91], [125, 128]], [[6, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[2, 5]], [[2, 71]], [[72, 75]], [[72, 83]], [[84, 87]], [[84, 94]], [[96, 108]], [[111, 123]]]",
    "query_spans": "[[[125, 134]]]",
    "process": "Solving the system  \n\\begin{cases}x=c\\\\a^{2}-\\frac{y^{2}}{b^{2}}=1,\\\\c^{2}=b^{2}+a2\\end{cases}  \nyields  \n\\begin{cases}x=c\\\\y=\\pm\\frac{b^{2}}{a}\\end{cases},  \nso |BF|=\\frac{b^{2}}{a}. According to the problem, \\frac{|BF|}{|AF|}=3, |AF|=c-a, thus \\frac{b^{2}}{c-a}=\\frac{c^{2}-a^{2}}{a(c-a)}=3. Rearranging gives c+a=3a, so c=2a. Therefore, the eccentricity of hyperbola C is 2."
  },
  {
    "text": "Given that the focal distance of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ is $2 \\sqrt{5}$, and the length of the real axis of the hyperbola is twice the length of the imaginary axis, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;FocalLength(G) = 2*sqrt(5);2*Length(ImageinaryAxis(G)) = Length(RealAxis(G))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4-y^2=1",
    "fact_spans": "[[[2, 59], [77, 80], [95, 98]], [[2, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[2, 75]], [[77, 93]]]",
    "query_spans": "[[[95, 103]]]",
    "process": "\\because the length of the real axis of the hyperbola is twice the length of the imaginary axis, \\therefore 2a = 2 \\times 2b, we obtain a = 2b, c = \\sqrt{a^{2} + b^{2}} = \\sqrt{5}b = \\sqrt{5}, \\therefore b = 1, a = 2, thus the equation of the hyperbola is \\frac{x^{2}}{4} - y^{2} = 1"
  },
  {
    "text": "The equation of the circle with its center at the focus of the parabola $y^{2}=8x$ and tangent to the line $3x - y + 4 = 0$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);Center(H) = Focus(G);H: Circle;L: Line;Expression(L) = (3*x - y + 4 = 0);IsTangent(L,H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x-2)^2+y^2=10",
    "fact_spans": "[[[1, 15]], [[1, 15]], [[0, 41]], [[40, 41]], [[24, 37]], [[24, 37]], [[23, 41]]]",
    "query_spans": "[[[40, 46]]]",
    "process": "The focus of $ y^{2}=8x $ is $ (2,0) $. The distance from the center $ (2,0) $ to the line is $ d=r=\\frac{|6+4|}{\\sqrt{3^{2}+1^{2}}}=\\sqrt{10} $, so the equation of the circle is: $ (x-2)^{2}+y^{2}=10 $."
  },
  {
    "text": "If the asymptotes of a hyperbola are given by $y=\\pm \\frac{3}{2} x$, then its eccentricity is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*(3/2)*x)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{sqrt(13)/2,sqrt(13)/3}",
    "fact_spans": "[[[1, 4], [34, 35]], [[1, 32]]]",
    "query_spans": "[[[34, 41]]]",
    "process": ""
  },
  {
    "text": "Given that $O$ is the coordinate origin, $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. A line passing through point $F_{1}$ with slope $\\frac{\\sqrt{3}}{3}$ intersects the ellipse $C$ at point $P$, and $|O P|=|O F_{2}|$. Then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "O: Origin;C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;G: Line;PointOnCurve(F1, G);Slope(G) = sqrt(3)/3;P: Point;Intersection(G, C) = P;Abs(LineSegmentOf(O, P)) = Abs(LineSegmentOf(O, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[2, 5]], [[29, 86], [130, 135], [163, 168]], [[29, 86]], [[36, 86]], [[36, 86]], [[36, 86]], [[36, 86]], [[11, 18], [94, 102]], [[19, 26]], [[11, 92]], [[11, 92]], [[127, 129]], [[93, 129]], [[103, 129]], [[137, 141]], [[127, 141]], [[143, 160]]]",
    "query_spans": "[[[163, 174]]]",
    "process": "Draw a sketch according to the given conditions. From knowledge of triangles, we obtain $ PF_{2}=c $, $ PF_{1}=\\sqrt{3}c $. Since $ PF_{2}+PF_{1}=2a $, the eccentricity of the ellipse can be found. As shown in the figure, by the given condition $ |OP|=|OF_{2}| $, so $ |OP|=|OF_{2}|=|OF_{1}| $. Since $ k_{PF_{1}}=\\frac{\\sqrt{3}}{3}=30^{\\circ} $, thus $ \\angle OPF_{1}=30^{\\circ} $, $ \\angle OPF_{2}=\\angle OF_{2}P=\\angle POF_{2}=60^{\\circ} $, therefore $ PF_{2}=c $, $ PF_{1}=\\sqrt{3}c $. Since $ PF_{2}+PF_{1}=2a $, so $ 2a=c+\\sqrt{3}c $, hence $ e=\\frac{c}{a}=\\frac{2}{1+\\sqrt{3}}=\\sqrt{3}-1 $."
  },
  {
    "text": "With the origin as the vertex and the left directrix of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ as the directrix, the parabola intersects the right directrix of the ellipse $C$ at points $A$, $B$. Then $|AB|=$?",
    "fact_expressions": "G: Parabola;C: Ellipse;A: Point;B: Point;O:Origin;Expression(C) = (x^2/4 + y^2/3 = 1);Vertex(G)=O;LeftDirectrix(C)=Directrix(G);Intersection(RightDirectrix(C),G)={A,B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16",
    "fact_spans": "[[[57, 60]], [[8, 49], [61, 66]], [[72, 75]], [[77, 80]], [[1, 3]], [[8, 49]], [[0, 60]], [[7, 60]], [[57, 82]]]",
    "query_spans": "[[[84, 92]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $x^{2}-m y^{2}=1$ has an asymptote $y=\\frac{2 x}{3}$. Then $m=$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-m*y^2 + x^2 = 1);m: Number;Expression(OneOf(Asymptote(G))) = (y=2*x/3)",
    "query_expressions": "m",
    "answer_expressions": "9/4",
    "fact_spans": "[[[0, 20], [21, 22]], [[0, 20]], [[48, 51]], [[21, 46]]]",
    "query_spans": "[[[48, 53]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$, point $P(1, a)(a>1)$ lies on the parabola. Two tangents are drawn from $P$ to the circle $(x-1)^{2}+y^{2}=1$, intersecting the parabola at points $A$ and $B$ respectively. If the slope of line $AB$ is $-1$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;p: Number;H: Circle;A: Point;B: Point;P: Point;a: Number;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (y^2 + (x - 1)^2 = 1);Coordinate(P) = (1, a);a > 1;PointOnCurve(P, G);l1: Line;l2: Line;TangentOfPoint(P, H) = {l1, l2};Intersection(l1, G) = A;Intersection(l2, G) = B;Slope(LineOf(A, B)) = -1",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = 4*x",
    "fact_spans": "[[[2, 23], [40, 43], [79, 82], [110, 113]], [[5, 23]], [[50, 70]], [[83, 87]], [[88, 91]], [[24, 39], [46, 49]], [[25, 39]], [[5, 23]], [[2, 23]], [[50, 70]], [[24, 39]], [[25, 39]], [[24, 44]], [], [], [[45, 75]], [[45, 91]], [[45, 91]], [[93, 108]]]",
    "query_spans": "[[[110, 118]]]",
    "process": "From the given conditions, the two tangents drawn from point P to the circle are symmetric with respect to the line x=1. Let A(x_{1},y_{1}), B(x_{2},y_{2}), P(x_{P},y_{P}), then k_{PA}=\\frac{y_{P}-y_{1}}{x_{P}-x_{1}}. Similarly, k_{PB}=\\frac{2p}{y_{P}}+y_{2}, k_{AB}=\\frac{2p}{y_{1}^{1}}+\\frac{2p}{x}. Since the two tangents are symmetric about the line x=1, we obtain y_{1}+y_{2}=-2y_{P}, so k_{AB}=\\frac{2p}{y_{1}+y_{2}}=\\frac{2p}{-2y_{P}}=-1. Hence y_{P}=p, P(1,p). Substituting into the parabola equation gives p^{2}=2p\\cdot1, so p=2. Therefore, the parabola equation is y^{2}=4x."
  },
  {
    "text": "The equation of the directrix of the parabola $y^{2}=2 x$ is? The focus of this parabola is $F$, and the point $M(x_{0} , y_{0})$ lies on this parabola, with $|MF|=\\frac{5}{2}$. Then $x_{0}$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);F: Point;Focus(G) = F;M: Point;Coordinate(M) = (x0, y0);x0: Number;y0: Number;PointOnCurve(M, G);Abs(LineSegmentOf(M, F)) = 5/2",
    "query_expressions": "Expression(Directrix(G));x0",
    "answer_expressions": "x = -1/2\n2",
    "fact_spans": "[[[0, 14], [22, 25], [54, 57]], [[0, 14]], [[29, 32]], [[22, 32]], [[33, 52]], [[33, 52]], [[80, 87]], [[34, 52]], [[33, 58]], [[60, 78]]]",
    "query_spans": "[[[0, 21]], [[80, 89]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is perpendicular to the line $x+2 y-1=0$, then the eccentricity of the curve is equal to?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x + 2*y - 1 = 0);IsPerpendicular(OneOf(Asymptote(G)), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 58], [82, 84]], [[5, 58]], [[5, 58]], [[65, 78]], [[5, 58]], [[5, 58]], [[2, 58]], [[65, 78]], [[2, 80]]]",
    "query_spans": "[[[82, 91]]]",
    "process": ""
  },
  {
    "text": "Given that $O$ is the coordinate origin, $B$ and $F$ are the upper vertex and the right focus of the ellipse $\\frac{x^{2}}{a}+\\frac{y^{2}}{b}=1(a>b>0)$, respectively. If $|O B|=|O F|$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;a: Number;b: Number;O: Origin;B: Point;F: Point;a > b;b > 0;Expression(G) = (y^2/b + x^2/a = 1);RightFocus(G) = F;UpperVertex(G) = B;Abs(LineSegmentOf(O, B)) = Abs(LineSegmentOf(O, F))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[21, 65], [91, 93]], [[23, 65]], [[23, 65]], [[2, 5]], [[11, 14]], [[15, 18]], [[23, 65]], [[23, 65]], [[21, 65]], [[11, 73]], [[11, 73]], [[75, 88]]]",
    "query_spans": "[[[91, 99]]]",
    "process": "Since O is the coordinate origin, B and F are respectively the upper vertex and the right focus of the ellipse $\\frac{x^{2}}{a}+\\frac{y^{2}}{b}=1$ $(a>b>0)$, and $|OB|=|OF|$, we obtain $b=c$. Also, $a=\\sqrt{b^{2}+c^{2}}=\\sqrt{2}c$, so the eccentricity of the ellipse is $e=\\frac{c}{a}=\\frac{\\sqrt{2}}{2}$."
  },
  {
    "text": "The distance from point $A$ to point $F(1,0)$ is equal to the distance from point $A$ to the line $x=-1$. If the trajectory of point $A$ and the line passing through point $P(-1,0)$ with slope $k$ have no intersection, then what is the range of values for $k$?",
    "fact_expressions": "G: Line;L:Line;F: Point;P: Point;A:Point;Expression(G) = (x = -1);Coordinate(F) = (1, 0);Coordinate(P) = (-1, 0);Distance(A,F)=Distance(A,G);PointOnCurve(P,L);Slope(L)=k;k:Number;NumIntersection(Locus(A),L)=0",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-oo,-1)+(1,+oo)",
    "fact_spans": "[[[21, 29]], [[62, 64]], [[7, 16]], [[44, 54]], [[2, 6], [35, 39]], [[21, 29]], [[7, 16]], [[44, 54]], [[2, 34]], [[43, 64]], [[55, 64]], [[58, 61], [70, 73]], [[35, 68]]]",
    "query_spans": "[[[70, 80]]]",
    "process": "Let point A(x,y). According to the given conditions, point A lies on a parabola with focus F(1,0) and directrix x=-1. Therefore, the trajectory of point A is y^{2}=4x. From the given conditions, the equation of the line passing through point P(-1,0) with slope k is y=k(x+1). By solving \\begin{cases}y^{2}=4x\\\\y=kx+k\\end{cases}, eliminating x gives ky^{2}-4y+4k=0. When k=0, it clearly does not satisfy the conditions. When k\\neq0, according to the conditions, for ky^{2}-4y+4k=0, the discriminant \\Delta=(-4)^{2}-4k\\cdot k<0, simplifying yields k^{2}-1>0, solving gives k<-1 or k>1. Therefore, the range of values for k is (-\\infty,-1)\\cup(1,+\\infty)."
  },
  {
    "text": "A moving point $P$ travels along the parabola $y=2 x^{2}$. Then the equation of the trajectory of the midpoint $M$ of the line segment joining $P$ and the fixed point $Q(3,0)$ is?",
    "fact_expressions": "G: Parabola;Q: Point;P: Point;M:Point;Expression(G) = (y = 2*x^2);Coordinate(Q) = (3, 0);PointOnCurve(P, G);MidPoint(LineSegmentOf(P,Q))=M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "y=4*x^2-12*x+9",
    "fact_spans": "[[[7, 21]], [[30, 38]], [[3, 6], [26, 27]], [[43, 46]], [[7, 21]], [[30, 38]], [[3, 24]], [[26, 46]]]",
    "query_spans": "[[[43, 53]]]",
    "process": "Let M(x,y), P(m,n). Using the midpoint coordinate formula, we obtain \\begin{cases}m=2x-3\\\\n=2y\\end{cases}. Substituting into n=2m^{2} and eliminating m,n gives the solution. [Detailed explanation] Let M(x,y), P(m,n). Since P moves on the parabola y=2x^{2}, we have n=2m^{2}, \\textcircled{1}. Also, since M is the midpoint of the line segment joining P and the fixed point Q(3,0), we have \\begin{cases}2x=3+m\\\\2y=0+n\\end{cases}, that is, \\begin{cases}m=2x-3\\\\n=2y\\end{cases} \\textcircled{2}. Substituting \\textcircled{2} into \\textcircled{1} and eliminating m,n yields: 2y=2(2x-3)^{2}. Simplifying gives y=4x^{2}-12x+9. Thus, the trajectory equation of M is y=4x^{2}-12x+9."
  },
  {
    "text": "There is a point $P$ on the parabola $y=2 x^{2}$ such that the sum of its distance to the point $A(1,3)$ and its distance to the focus is minimized. Then the coordinates of point $P$ are?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;Expression(G) = (y = 2*x^2);Coordinate(A) = (1, 3);PointOnCurve(P, G);WhenMin(Distance(P, A) + Distance(P, Focus(G)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1, 2)",
    "fact_spans": "[[[1, 15]], [[25, 34]], [[19, 22], [38, 39], [23, 24], [51, 55]], [[1, 15]], [[25, 34]], [[0, 22]], [[1, 49]]]",
    "query_spans": "[[[51, 60]]]",
    "process": "From $ y = 2x^2 $, we get $ x^2 = \\frac{1}{2}y $, so the directrix of the parabola is $ y = -\\frac{1}{8} $. Let the focus be $ F $. Substituting $ x = 1 $ into $ y = 2x^2 $, we obtain $ y = 2 < 3 $, therefore point $ A(1,3) $ lies inside the parabola $ y = 2x^2 $. By the definition of a parabola, the distance from any point $ P $ on the parabola to the focus $ F $ equals its distance to the directrix. Hence, the minimum value of $ |PA| + |PF| $ is the distance from point $ A(1,3) $ to the directrix. At this minimum, point $ P $ is the intersection of the parabola $ y = 2x^2 $ and the line perpendicular to the directrix passing through point $ A(1,3) $ (i.e., $ x = 1 $). Therefore, the $ x $-coordinate of point $ P $ is 1. Substituting into $ y = 2x^2 $, we get $ y = 2 $. Thus, the coordinates of point $ P $ are $ (1,2) $."
  },
  {
    "text": "Given that the line $2 x-\\sqrt{3} y=0$ is an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (2*x - sqrt(3)*y = 0);OneOf(Asymptote(G))=H",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(21)/3",
    "fact_spans": "[[[23, 80], [89, 92]], [[26, 80]], [[26, 80]], [[2, 22]], [[26, 80]], [[26, 80]], [[23, 80]], [[2, 22]], [[2, 86]]]",
    "query_spans": "[[[89, 100]]]",
    "process": "\\frac{b}{a}=\\frac{2}{\\sqrt{3}}\\Rightarrow\\frac{c}{a}=\\frac{\\sqrt{7}}{\\sqrt{3}}=\\frac{\\sqrt{21}}{3}"
  },
  {
    "text": "The coordinates of two fixed points are $A(-1,0)$, $B(2,0)$. A moving point $M$ satisfies the condition $\\angle M B A = 2 \\angle M A B$. What is the trajectory equation of the moving point $M$?",
    "fact_expressions": "A: Point;B: Point;M: Point;Coordinate(A) = (-1, 0);Coordinate(B) = (2, 0);AngleOf(M, B, A) = 2*AngleOf(M, A, B)",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "(3*x^2 - y^2 = 3)&(x > 1)",
    "fact_spans": "[[[9, 19]], [[21, 29]], [[30, 32], [68, 71]], [[9, 19]], [[21, 29]], [[36, 65]]]",
    "query_spans": "[[[68, 78]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, one asymptote is $l$, the line passing through point $F_{2}$ and parallel to $l$ intersects the hyperbola $C$ at point $M$. If $|M F_{1}|=2|M F_{2}|$, then the equation of asymptote $l$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;l: Line;G:Line;M: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;OneOf(Asymptote(C)) = l;PointOnCurve(F2, G);IsParallel(l, G);Intersection(G, C) = M;Abs(LineSegmentOf(M, F1)) = 2*Abs(LineSegmentOf(M, F2))",
    "query_expressions": "Expression(l)",
    "answer_expressions": "y = pm*2*x",
    "fact_spans": "[[[2, 63], [118, 124]], [[9, 63]], [[9, 63]], [[94, 97], [109, 112], [158, 161]], [[115, 117]], [[125, 129]], [[72, 79]], [[80, 87], [99, 107]], [[9, 63]], [[9, 63]], [[2, 63]], [[2, 87]], [[2, 87]], [[2, 97]], [[98, 117]], [[108, 117]], [[115, 129]], [[131, 153]]]",
    "query_spans": "[[[158, 166]]]",
    "process": "Let the semi-focal distance of the hyperbola be $ c $, then $ F_{2}(c,0) $. By the symmetry of the hyperbola, without loss of generality, let the equation of line $ l $ be $ y = \\frac{b}{a}x $. Then the equation of the line passing through point $ F_{2} $ and parallel to $ l $ is $ y = \\frac{b}{a}(x - c) $. From  \n\\[\n\\begin{cases}\ny = \\frac{b}{a}(x - c) \\\\\n\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1\n\\end{cases}\n\\]  \neliminating $ y $ and simplifying yields: $ 2cx - c^{2} = a^{2} $, solving gives the x-coordinate of point $ M $ as $ x = \\frac{a^{2} + c^{2}}{2c} $. Thus,  \n\\[\n|MF_{2}| = \\sqrt{1 + \\left(\\frac{b}{a}\\right)^{2}} \\cdot \\left|\\frac{a^{2} + c^{2}}{2c} - c\\right| = \\frac{c}{a} \\cdot \\frac{c^{2} - a^{2}}{2c} = \\frac{b^{2}}{2a}, \\quad |MF_{1}| = 2|MF_{2}| = \\frac{b^{2}}{a}.\n\\]  \nBy the definition of the hyperbola: $ |MF_{1}| - |MF_{2}| = 2a $, therefore we have $ \\frac{b^{2}}{2a} = 2a $, i.e., $ \\frac{b}{a} = 2 $, so the equations of the asymptotes are: $ y = \\pm 2x $."
  },
  {
    "text": "The coordinates of the focus of the parabola $y=\\frac{1}{8} x^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2/8)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,2)",
    "fact_spans": "[[[0, 24]], [[0, 24]]]",
    "query_spans": "[[[0, 31]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the hyperbola $C$: $\\frac{x^{2}}{8}-\\frac{y^{2}}{4}=1$? The equations of the asymptotes?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/8 - y^2/4 = 1)",
    "query_expressions": "Eccentricity(C);Expression(Asymptote(C))",
    "answer_expressions": "sqrt(6)/2; y=pm*(sqrt(2)/2)*x",
    "fact_spans": "[[[0, 43]], [[0, 43]]]",
    "query_spans": "[[[0, 48]], [[0, 56]]]",
    "process": "From the given conditions, $ a^{2}=8\\Rightarrow a=2\\sqrt{2} $, $ b^{2}=4\\Rightarrow b=2 $, $ \\therefore c=\\sqrt{a^{2}+b^{2}}=2\\sqrt{3} $, $ \\therefore $ eccentricity $ e=\\frac{c}{a}=\\frac{\\sqrt{6}}{2} $, asymptotes equations are $ y=\\pm\\frac{\\sqrt{2}}{2}x $"
  },
  {
    "text": "Given $N(2,0)$, and $M$ is a moving point on $y^{2}=8x$, then the minimum value of $|MN|$ is?",
    "fact_expressions": "M: Point;N: Point;Coordinate(N)=(2,0);G:Curve;Expression(G)=(y^2 = 8*x);PointOnCurve(M,G)",
    "query_expressions": "Min(Abs(LineSegmentOf(M, N)))",
    "answer_expressions": "2",
    "fact_spans": "[[[13, 16]], [[2, 10]], [[2, 10]], [[17, 28]], [[17, 28]], [[13, 32]]]",
    "query_spans": "[[[34, 46]]]",
    "process": "According to the problem, N(2,0) is the focus of the parabola $ y^{2} = 8x $. Let $ M(x_{1}, y_{1}) $ where $ x_{1} \\geqslant 0 $. Then $ |MN| = x + 2 \\geqslant 2 $, so the minimum value of $ |MN| $ is 2."
  },
  {
    "text": "Given that the line $y = kx - 1$ intersects the right branch of the hyperbola $x^2 - y^2 = 1$ at two distinct points, what is the range of values for $k$?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*x - 1);k: Number;G: Hyperbola;Expression(G) = (x^2 - y^2 = 1);P1: Point;P2: Point;Intersection(H,RightPart(G)) = {P1,P2};Negation(P1=P2)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(1,\\sqrt{2})",
    "fact_spans": "[[[2, 13]], [[2, 13]], [[43, 46]], [[14, 32]], [[14, 32]], [], [], [[2, 41]], [[2, 41]]]",
    "query_spans": "[[[43, 53]]]",
    "process": "From the system of equations formed by the line y = kx - 1 and the hyperbola x^{2} - y^{2} = 1, eliminating y gives (1 - k^{2})x^{2} + 2kx - 2 = 0, where 1 - k^{2} \\neq 0. Since this equation has two unequal roots greater than 1, it follows that 4 > 0, 1 < k < \\sqrt{2}, \\frac{-k}{1 - k^{2}} > (2 + 2k - 2)(1 - k^{2}) > 0."
  },
  {
    "text": "Given the parabola $C$: $x^{2}=4 y$ with focus $F$, a line $l$ passing through $F$ intersects $C$ at two distinct points $A$ and $B$. Then the minimum value of $(|A F|+1) \\cdot|B F|$ is?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;F: Point;B: Point;Expression(C) = (x^2 = 4*y);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Negation(A = B)",
    "query_expressions": "Min((Abs(LineSegmentOf(A, F)) + 1)*Abs(LineSegmentOf(B, F)))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[34, 39]], [[2, 20], [40, 43]], [[47, 50]], [[25, 28], [30, 33]], [[51, 54]], [[2, 20]], [[2, 28]], [[29, 39]], [[34, 56]], [[44, 56]]]",
    "query_spans": "[[[58, 86]]]",
    "process": "From the given conditions, F(0,1). Clearly, the slope of line l exists. Let the equation of line l be y = kx + 1, A(x_{1},y_{1}), B(x_{2},y_{2}). From \\begin{cases}x^{2}=4y,\\\\y=kx+1\\end{cases}, we obtain x^{2}-4kx-4=0, so x_{1}x_{2}=-4, thus y_{1}y_{2}=\\frac{x_{1}^{2}}{4}\\cdot\\frac{x_{2}^{2}}{4}=1. Therefore, (|AF|+1)\\cdot|BF|=(y_{1}+1+1)(y_{2}+1)=y_{1}y_{2}+y_{1}+2y_{2}+2\\geqslant2\\sqrt{y_{1}\\cdot2y_{2}}+3=2\\sqrt{2}+3, with equality holding if and only if y_{1}=\\sqrt{2}, y_{2}=\\frac{\\sqrt{2}}{2}."
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{5 a}+\\frac{y^{2}}{4 a^{2}+1}=1$ lie on the $x$-axis. Then the range of its eccentricity $e$ is?",
    "fact_expressions": "G: Ellipse;a: Number;e: Number;Expression(G) = (y^2/(4*a^2 + 1) + x^2/(5*a) = 1);PointOnCurve(Focus(G), xAxis);Eccentricity(G) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "(0, \\sqrt{5}/5]",
    "fact_spans": "[[[0, 47], [58, 59]], [[2, 47]], [[63, 66]], [[0, 47]], [[0, 56]], [[58, 66]]]",
    "query_spans": "[[[63, 73]]]",
    "process": "Since the foci of the ellipse lie on the x-axis, we have 5a > 4a^{2}+1. Solving gives a \\in (\\frac{1}{4},1). Also, e = \\sqrt{1-\\frac{4a^{2}+1}{5a}} = \\sqrt{1-\\frac{1}{5}(4a+\\frac{1}{a})}. The function y = 4a + \\frac{1}{a} is monotonically decreasing on the interval (\\frac{1}{4},\\frac{1}{2}) and monotonically increasing on the interval (\\frac{1}{2},1). When a = \\frac{1}{4}, y = 5; when a = \\frac{1}{2}, y = 4; when a = 1, y = 5. Hence, y = 4a + \\frac{1}{a} \\in [4,5), so 1 - \\frac{1}{5}(4a + \\frac{1}{a}) \\in (0,\\frac{1}{5}], and thus e \\in (0,\\frac{\\sqrt{5}}{5}]."
  },
  {
    "text": "If the focal distance of the hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ is $4$, then what is the equation of its asymptotes?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (-y^2 + x^2/m = 1);FocalLength(G) = 4",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(3)/3*x",
    "fact_spans": "[[[38, 39], [1, 29]], [[4, 29]], [[1, 29]], [[1, 36]]]",
    "query_spans": "[[[38, 46]]]",
    "process": "The hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ has a focal distance of 4, so we obtain $m+1=4$, hence $m=3$. According to the given condition, the foci of the hyperbola lie on the $x$-axis, thus the asymptotes are given by: $y=\\pm\\frac{b}{a}x$. Therefore, the equations of the asymptotes of the hyperbola are: $y=\\pm\\frac{\\sqrt{3}}{3}x$."
  },
  {
    "text": "If the distance from a point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ to the focus $F_{1}$ is $6$, then what is the distance from point $P$ to the other focus $F_{2}$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2:Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(P, G);OneOf(Focus(G)) = F1;OneOf(Focus(G))=F2;Distance(P, F1) = 6;Negation(F1=F2)",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 40]], [[43, 46], [65, 69]], [[49, 56]], [[75, 82]], [[1, 40]], [[1, 46]], [[1, 56]], [[1, 82]], [[43, 63]], [[1, 82]]]",
    "query_spans": "[[[65, 87]]]",
    "process": "By the definition of an ellipse, we can solve it: from the definition of an ellipse, $ PF_{1} + PF_{2} = 10 $, and $ PF_{1} = 6 $, $ \\therefore PF_{2} = 4 $."
  },
  {
    "text": "Draw a tangent from the left focus $F(-c, 0)$ $(c>0)$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(b>a>0)$ to the circle $x^{2}+y^{2}=a^{2}$, with point of tangency $E$. Extend $FE$ to intersect the parabola $y^{2}=4 c x$ at point $P$. Let $O$ be the origin. If $\\overrightarrow{O E}=\\frac{1}{2}(\\overrightarrow{O F}+\\overrightarrow{O P})$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;c: Number;I: Circle;F: Point;E: Point;O: Origin;P: Point;l1:Line;b > a;a > 0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 4*(c*x));Expression(I) = (x^2 + y^2 = a^2);Coordinate(F) = (-c, 0);LeftFocus(G) = F;TangentOfPoint(F, I) =l1;TangentPoint(l1,I)=E;Intersection(OverlappingLine(LineSegmentOf(F,E)), H) =P;VectorOf(O, E) = (VectorOf(O, F) + VectorOf(O, P))/2;c>0",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(1+sqrt(5))/2",
    "fact_spans": "[[[1, 54], [223, 226]], [[4, 54]], [[4, 54]], [[112, 128]], [[115, 128]], [[74, 94]], [[58, 73]], [[101, 104]], [[134, 137]], [[129, 133]], [], [[4, 54]], [[4, 54]], [[1, 54]], [[112, 128]], [[74, 94]], [[58, 73]], [[1, 73]], [[0, 97]], [[0, 104]], [[105, 133]], [[144, 221]], [[58, 73]]]",
    "query_spans": "[[[223, 232]]]",
    "process": "Since |OF| = c, |OE| = a, OE \\bot EF, it follows that |EF| = b. Because \\overrightarrow{OE} = \\frac{1}{2}(\\overrightarrow{OF} + \\overrightarrow{OP}), E is the midpoint of PF, so |PF| = 2b. Also, since O is the midpoint of FF, PF // EO, so |PF| = 2a. Since the equation of the parabola is y^{2} = 4cx, the focus of the parabola is at (c, 0), meaning the right focus of the parabola coincides with that of the hyperbola. Draw a vertical line l through F perpendicular to the x-axis, and draw PD \\bot l from point P. Then l is the directrix of the parabola, so PD = PF = 2a. Hence, the x-coordinate of point P is 2a - c. Let P(x, y). In right triangle PDF, PD^{2} + DF^{2} = PF^{2}, that is, 4a^{2} + y^{2} = 4b^{2}. Substituting y^2 = 4c(2a - c), we get 4a^{2} + 4c(2a - c) = 4(c^{2} - a^{2}). Solving gives e = \\frac{1 + \\sqrt{5}}{2}."
  },
  {
    "text": "If the foci of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$ lie on the $y$-axis and its eccentricity is $\\frac{2}{3}$, then $m=$?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/5 + y^2/m = 1);PointOnCurve(Focus(G), yAxis);Eccentricity(G)=2/3",
    "query_expressions": "m",
    "answer_expressions": "9",
    "fact_spans": "[[[1, 38]], [[67, 70]], [[1, 38]], [[1, 47]], [[1, 65]]]",
    "query_spans": "[[[67, 72]]]",
    "process": "Given that $ m > 5 $, so $ a^{2} = m $, $ b^{2} = 5 $, thus $ \\frac{c}{a} = \\sqrt{1 - \\left( \\frac{b}{a} \\right)^{2}} = \\sqrt{1 - \\frac{5}{m}} = \\frac{2}{3} $, solving gives $ m = 9 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, the left focus is $F$, and the eccentricity is $\\sqrt{2}$. If the line passing through points $F$ and $P(0 , 4)$ is parallel to one of the asymptotes of the hyperbola, then what is the equation of the hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;P: Point;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (0, 4);LeftFocus(G) = F;Eccentricity(G) = sqrt(2);PointOnCurve(F, H);PointOnCurve(P, H);IsParallel(H, OneOf(Asymptote(G)))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/8 - y^2/8 = 1",
    "fact_spans": "[[[2, 59], [109, 112], [120, 123]], [[5, 59]], [[5, 59]], [[104, 106]], [[91, 101]], [[64, 67], [87, 90]], [[5, 59]], [[5, 59]], [[2, 59]], [[91, 101]], [[2, 67]], [[2, 82]], [[85, 106]], [[85, 106]], [[104, 118]]]",
    "query_spans": "[[[120, 128]]]",
    "process": "Let point $ F(-c,0) $ ($ c>0 $). Since the eccentricity of this hyperbola is $ \\sqrt{2} $, we have $ \\frac{c}{a} = \\sqrt{2} $, $\\textcircled{1}$. The line passing through $ F $ and $ P(0,4) $ is parallel to one asymptote of the hyperbola, so $ k_{FP} = \\frac{4-0}{0-(-c)} = \\frac{b}{a} $, $\\textcircled{2}$. Solving $\\textcircled{1}$ and $\\textcircled{2}$ together, we obtain $ b = 2\\sqrt{2} $. Also, $ a^{2} + b^{2} = c^{2} $, that is, $ a^{2} + 8 = c^{2} $, $\\textcircled{3}$. Solving $\\textcircled{1}$ and $\\textcircled{3}$ together, we get $ a^{2} = 8 $, $ c^{2} = 16 $. Hence, the equation of the hyperbola is $ \\frac{x^{2}}{8} - \\frac{y^{2}}{8} = 1 $."
  },
  {
    "text": "Let $k$ be a real number. If the curve represented by the equation $\\frac{x^{2}}{k-4}-\\frac{y^{2}}{k+4}=1$ is a hyperbola, then what is the range of values for $k$?",
    "fact_expressions": "G: Hyperbola;H: Curve;k:Real;Expression(H) = (x^2/(k - 4) - y^2/(k + 4) = 1);H=G",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-\\infty,-4)+(4,+\\infty)",
    "fact_spans": "[[[56, 59]], [[53, 55]], [[1, 4], [61, 64]], [[9, 55]], [[53, 59]]]",
    "query_spans": "[[[61, 71]]]",
    "process": "From the given condition, we have $(k-4)(k+4)>0$, solving yields $k>4$ or $k<-4$."
  },
  {
    "text": "The parabola $C_{1}$: $y^{2}=2 p x(p>0)$ and the hyperbola $C_{2}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0, b>0)$ intersect at points $A$ and $B$. The parabola $C_{1}$ intersects the two asymptotes of $C_{2}$ at points $C$ and $D$, distinct from the origin. Moreover, $AB$ and $CD$ pass through the foci of $C_{2}$ and $C_{1}$, respectively. Then $\\frac{|A B|}{|C D|}$=?",
    "fact_expressions": "C2:Hyperbola;C1:Parabola;A: Point;B: Point;C: Point;D: Point;p:Number;p>0;a:Number;b:Number;a>0;b>0;L1:Line;L2:Line;O:Origin;Expression(C1)=(y^2=2*p*x);Expression(C2)=(x^2/a^2-y^2/b^2=1);Intersection(C1,C2)={A,B};Asymptote(C2)={L1,L2};Intersection(C1,L1)=C;Intersection(C1,L2)=D;Negation(O=C);Negation(O=D);PointOnCurve(Focus(C2),LineSegmentOf(A,B));PointOnCurve(Focus(C1),LineSegmentOf(C,D))",
    "query_expressions": "Abs(LineSegmentOf(A, B))/Abs(LineSegmentOf(C, D))",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[30, 97], [117, 124], [167, 174]], [[0, 29], [109, 116], [175, 182]], [[99, 102]], [[103, 106]], [[141, 144]], [[146, 150]], [[11, 29]], [[11, 29]], [[42, 97]], [[42, 97]], [[42, 97]], [[42, 97]], [], [], [[136, 138]], [[0, 29]], [[30, 97]], [[0, 108]], [[117, 130]], [[109, 150]], [[109, 150]], [[134, 150]], [[134, 150]], [[153, 185]], [[153, 185]]]",
    "query_spans": "[[[187, 210]]]",
    "process": "From the given, we have A(c,\\frac{b^{2}}{a}), B(c,-\\frac{b^{2}}{a}), C(\\frac{p}{2},p), D(\\frac{p}{2},-p), and point A lies on the parabola, point C lies on the asymptote, so (\\frac{b^{2}}{a})^{2}=2pc', p=\\frac{b}{a}\\times\\frac{p}{2}, thus b=2a, 8a=\\sqrt{5}p, so \\frac{|AB|}{|CD|}=\\frac{2\\frac{b^{2}}{2p}}{2p}=\\frac{\\sqrt{5}}{2}, hence the solution. This problem involves the positional relationship between a parabola and a hyperbola, deriving coordinate relations of intersection points, classified as a medium-level question."
  },
  {
    "text": "If the intersection point $A$, distinct from the origin $O$, of the parabola $C_{1}$: $y^{2}=4x$ and the parabola $C_{2}$: $x^{2}=2py$ $(p>0)$ is at a distance of $3$ from the focus of the parabola $C_{1}$, then what is the equation of the parabola $C_{2}$?",
    "fact_expressions": "C1: Parabola;Expression(C1) = (y^2 = 4*x);C2: Parabola;Expression(C2) = (x^2 = 2*p*y);p: Number;p>0;O: Origin;A: Point;Negation(A=O);Intersection(C1, C2) = A;Distance(A, Focus(C1)) = 3",
    "query_expressions": "Expression(C2)",
    "answer_expressions": "x^2=sqrt(2)*y",
    "fact_spans": "[[[1, 24], [71, 81]], [[1, 24]], [[25, 57], [93, 103]], [[25, 57]], [[37, 57]], [[37, 57]], [[59, 64]], [[67, 70]], [[57, 70]], [[1, 70]], [[67, 91]]]",
    "query_spans": "[[[93, 108]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of the hyperbola is $2$, and the foci are $(-4,0)$, $(4,0)$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;P:Point;D:Point;Focus(G)={P,D};Coordinate(P) = (-4, 0);Coordinate(D) = (4, 0);Eccentricity(G) = 2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/12 = 1",
    "fact_spans": "[[[2, 5], [36, 39]], [[17, 25]], [[27, 34]], [[2, 34]], [[17, 25]], [[27, 34]], [[2, 13]]]",
    "query_spans": "[[[36, 43]]]",
    "process": "From the given conditions, we have $ c=4, \\frac{c}{a}=2 \\therefore a=2 \\therefore b^{2}=12 $, the hyperbola equation is $ \\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1 $."
  },
  {
    "text": "Write the equation of a hyperbola with eccentricity $\\sqrt{5}$ and asymptotes given by $y=\\pm 2 x$.",
    "fact_expressions": "G: Hyperbola;Eccentricity(G) = sqrt(5);Expression(Asymptote(G)) = (y = pm*(2*x))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/4 = 1 (the answer is not unique)",
    "fact_spans": "[[[37, 40]], [[4, 40]], [[19, 40]]]",
    "query_spans": "[[[37, 44]]]",
    "process": "From the given conditions, we have $ e = \\frac{c}{a} = \\sqrt{\\frac{c^{2}}{a^{2}}} = \\sqrt{\\frac{a^{2}+b^{2}}{a^{2}}} = \\sqrt{1+\\left(\\frac{b}{a}\\right)^{2}} = \\sqrt{5} $, which gives $ \\frac{b}{a} = 2 $. The asymptotes of the hyperbola are $ y = \\pm 2x = \\pm \\frac{b}{a}x $, so the foci of the hyperbola lie on the x-axis. Therefore, the standard equation of the hyperbola satisfying the conditions is $ x^{2} - \\frac{y^{2}}{4} = 1 $ (the answer is not unique)."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, and point $P$ is a moving point on the line $x-\\sqrt{2} y+4=0$, then the maximum value of $\\sin \\angle F_{1} P F_{2}$ is?",
    "fact_expressions": "F1: Point;F2: Point;G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;H: Line;Expression(H) = (x - sqrt(2)*y + 4 = 0);PointOnCurve(P, H)",
    "query_expressions": "Max(Sin(AngleOf(F1, P, F2)))",
    "answer_expressions": "(2*sqrt(3)+sqrt(6))/6",
    "fact_spans": "[[[2, 9]], [[10, 17]], [[18, 46]], [[18, 46]], [[2, 52]], [[2, 52]], [[53, 57]], [[58, 78]], [[58, 78]], [[53, 82]]]",
    "query_spans": "[[[84, 117]]]",
    "process": "Let $ P(x,y) $, then using the inclination angle and the tangent of the difference of two angles, we obtain  \n$$\n\\frac{1-\\sin^{2}\\angle F_{1}PF_{2}}{\\sin^{2}\\angle F_{1}PF_{2}} = \\left(\\frac{3}{4}y + \\frac{3}{y} - 2\\sqrt{2}\\right)^{2} \\quad (y \\neq 0)\n$$  \nUsing the basic inequality, the maximum value of $ \\sin\\angle F_{2}PF_{1} $ can be found. Since $ F_{1}, F_{2} $ are the left and right foci of the hyperbola $ \\frac{x^{2}}{3} - y^{2} = 1 $, it follows that $ F_{1}(-2,0), F_{2}(2,0) $. Let $ P(x,y) $, then  \n$$\n|\\tan\\angle F_{2}PF_{1}| = |\\tan(\\angle PF_{2}x - \\angle PF_{1}x)| = \n$$  \nSimplifying, we get  \n$$\n|\\tan\\angle F_{2}PF_{1}| = \\left| \\frac{4y}{x^{2}+y^{2}-4} \\right| = \\left| \\frac{4y}{y^{2}+(\\sqrt{2}y-4)^{2}-4} \\right|\n$$  \nWhen $ y = 0 $, $ |\\tan\\angle F_{2}PF_{1}| = 0 $, hence $ \\sin\\angle F_{2}PF_{1} = 0 $; when $ y \\neq 0 $, at this time  \n$$\n\\frac{3}{4}y + \\frac{3}{y} - 2\\sqrt{2} \\geqslant 3 - 2\\sqrt{2} \\quad \\text{or} \\quad \\frac{3}{4}y + \\frac{3}{y} - 2\\sqrt{2}\n$$  \n$$\n\\frac{1-\\sin^{2}\\angle F_{1}PF_{2}}{\\sin^{2}\\angle F_{1}PF_{2}} = \\left[ \\frac{y^{2}+(\\sqrt{2}y-4)^{2}-4}{4y} \\right| = \\left( \\frac{3}{4}y + \\frac{3}{y} - 2\\sqrt{2} \\right)^{2}\n$$  \nHence  \n$$\n\\left( \\frac{3}{4}y + \\frac{3}{y} - 2\\sqrt{2} \\right)^{2} \\geqslant 17 - 12\\sqrt{2},\n$$  \nwith equality if and only if $ y = 2 $. Therefore  \n$$\n\\frac{1-\\sin^{2}\\angle F_{1}PF_{2}}{\\sin^{2}\\angle F_{1}PF_{2}} \\geqslant 17 - 12\\sqrt{2},\n$$  \nsolving yields  \n$$\n\\sin^{2}\\angle F_{1}PF_{2} \\leqslant \\frac{3+2\\sqrt{2}}{6}\n$$  \nSince $ \\sin\\angle F_{1}PF_{2} > 0 $, it follows that  \n$$\n\\sin\\angle F_{1}PF_{2} \\leqslant \\frac{2\\sqrt{3}+\\sqrt{6}}{6},\n$$  \nwith equality if and only if $ y = 2 $."
  },
  {
    "text": "It is known that the vertex of the parabola is at the origin, the focus is on the $y$-axis, and the distance from a point $M(m,-2)$ on the parabola to the focus is $4$. Then, what is the equation of the parabola?",
    "fact_expressions": "G: Parabola;M: Point;O: Origin;m: Number;Coordinate(M) = (m, -2);Vertex(G) = O;PointOnCurve(Focus(G), yAxis);PointOnCurve(M, G);Distance(M, Focus(G)) = 4",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 = -8*y",
    "fact_spans": "[[[2, 5], [22, 25], [49, 52]], [[28, 37]], [[9, 11]], [[28, 37]], [[28, 37]], [[2, 11]], [[2, 20]], [[22, 37]], [[22, 47]]]",
    "query_spans": "[[[49, 57]]]",
    "process": "According to the problem, the parabola equation can be set as $x^{2}=-2px$ $(p>0)$, so the distance from point $P$ on the parabola to $F$ is equal to $\\frac{p}{2}-(-2)=4$, $\\therefore p=4$, thus the parabola equation is $x^{2}=-8y$."
  },
  {
    "text": "The line passing through the focus $F$ of the known parabola $y^{2}=16x$ intersects the parabola at points $A$ and $B$. Then the minimum value of $|AF|+2|BF|$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 16*x);F: Point;Focus(G) = F;PointOnCurve(F, L) = True;Intersection(L, G) = {A, B};L: Line;A: Point;B: Point",
    "query_expressions": "Min(Abs(LineSegmentOf(A, F)) + 2*Abs(LineSegmentOf(B, F)))",
    "answer_expressions": "12+8*sqrt(2)",
    "fact_spans": "[[[3, 18], [28, 31]], [[3, 18]], [[21, 24]], [[3, 24]], [[0, 27]], [[25, 41]], [[25, 27]], [[32, 35]], [[36, 39]]]",
    "query_spans": "[[[43, 63]]]",
    "process": "The focus of the parabola $ y^{2} = 16x $ is $ F(4,0) $. Let the equation of the line be $ x = my + 4 $. Combining it with the parabola gives $ y^{2} - 16my - 64 = 0 $. By Vieta's formulas, we have: $ y_{1} + y_{2} = 16m $, $ y_{1} \\cdot y_{2} = -64 $, $ x_{1} + x_{2} = 16m^{2} + 8 $, $ x_{1} \\cdot x_{2} = 16 $. Since $ |AF| = |x_{1} + 4| $, $ |BF| = |x_{2} + 4| $, $ \\therefore \\frac{1}{|AF|} + \\frac{1}{|BF|} = \\frac{1}{4} $, $ \\therefore |AF|^{+}|BF| = 4' $, $ \\therefore |AF| + 2|BF| = 4(|AF| + 2|BF|)\\left(\\frac{1}{|AF|} + \\frac{1}{|BF|}\\right) = 4\\left(3 + \\frac{2|BF|}{|AF|} + \\frac{|AF|}{|BF|}\\right) \\geqslant 12 + 8\\sqrt{2} $. The equality holds if and only if $ |AF| = \\sqrt{2}|BF| $."
  },
  {
    "text": "Let the minimum distance from points on the parabola $y^{2}=2 p x(p>0)$ to the line $3 x+4 y+12=0$ be $1$. Then $p=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;A: Point;PointOnCurve(A, G);H: Line;Expression(H) = (3*x + 4*y + 12 = 0);Min(Distance(A, H)) = 1",
    "query_expressions": "p",
    "answer_expressions": "{21/8, 51/8}",
    "fact_spans": "[[[1, 22]], [[1, 22]], [[55, 58]], [[4, 22]], [], [[1, 25]], [[26, 42]], [[26, 42]], [[1, 53]]]",
    "query_spans": "[[[55, 60]]]",
    "process": ""
  },
  {
    "text": "The point $P(x, y)$ is a moving point on the ellipse $2 x^{2} + 3 y^{2} = 12$, then the maximum value of $x + 2 y$ is?",
    "fact_expressions": "G: Ellipse;P: Point;x1:Number;y1:Number;Expression(G) = (2*x^2 + 3*y^2 = 12);Coordinate(P) = (x1, y1);PointOnCurve(P, G)",
    "query_expressions": "Max(x1 + 2*y1)",
    "answer_expressions": "sqrt(22)",
    "fact_spans": "[[[11, 33]], [[0, 10]], [[1, 10]], [[1, 10]], [[11, 33]], [[0, 10]], [[0, 39]]]",
    "query_spans": "[[[41, 54]]]",
    "process": "\\frac{x^{2}}{6}+\\frac{y^{2}}{4}=1,\\begin{cases}x=\\sqrt{6}\\sin\\theta\\\\y=2\\cos\\theta\\end{cases}, so x+2y=\\sqrt{6}\\sin\\theta+4\\cos\\theta=\\sqrt{22}\\sin(\\theta+\\varphi), the maximum value of x+2y is \\sqrt{22}."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2 p x(p>0)$ be $F$. If the distance from $F$ to the line $y=\\sqrt{3} x$ is $\\sqrt{3}$, then $p=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(G) = F;H: Line;Expression(H) = (y = sqrt(3)*x);Distance(F, H) = sqrt(3)",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 22]], [[1, 22]], [[67, 70]], [[4, 22]], [[26, 29], [31, 34]], [[1, 29]], [[35, 51]], [[35, 51]], [[31, 65]]]",
    "query_spans": "[[[67, 72]]]",
    "process": "Find the coordinates of the focus of the parabola and solve using the distance formula. For the parabola \\( y^2 = 2px \\) (\\( p > 0 \\)), the focus is \\( F\\left(\\frac{p}{2}, 0\\right) \\). Since the distance from \\( F \\) to the line \\( y = \\sqrt{3}x \\) is \\( \\sqrt{3} \\), we have \\( \\frac{|\\sqrt{3}|}{\\sqrt{3}+1} = \\sqrt{3} \\), solving gives \\( p = 4 \\); the answer is: 4."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively, $P$ is a point on the ellipse, $\\angle F_{1} P F_{2}=60^{\\circ}$, and the circumradius and inradius of $\\Delta P F_{1} F_{2}$ are $R$ and $r$, respectively. If $R=3 r$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G)=(y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree);R:Number;r:Number;Radius(CircumCircle(TriangleOf(P,F1,F2)))=R;Radius(InscribedCircle(TriangleOf(P,F1,F2)))=r;R = 3*r;a: Number;b: Number;a>0;b>0",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3/5",
    "fact_spans": "[[[20, 70], [81, 83], [178, 180]], [[2, 9]], [[77, 80]], [[10, 17]], [[20, 70]], [[2, 76]], [[2, 76]], [[77, 86]], [[87, 120]], [[160, 163]], [[164, 167]], [[122, 167]], [[122, 167]], [[169, 176]], [[20, 70]], [[20, 70]], [[20, 70]], [[20, 70]]]",
    "query_spans": "[[[178, 186]]]",
    "process": "Let |PF_{1}|=m, |PF_{2}|=n, then m+n=2a. According to the problem, \\frac{1}{2}(2a+2c)r=S_{\\triangle PF_{1}F_{2}}=\\frac{\\sqrt{3}}{4}mn, so mn=\\frac{4(a+c)r}{\\sqrt{3}}. In \\triangle PF_{1}F_{2}, by the law of cosines, m^{2}+n^{2}-mn=4c^{2}, we obtain (m+n)^{2}-4c=3mm, get m=\\frac{4a-c^{2})}{x^{2}}, b+\\frac{\\sqrt{3}(a-c)}{3}x^{\\frac{3(a+c)}{3}=\\frac{1}{3}\\sin60^{\\circ}=\\frac{2a}{3\\sqrt{3}}}=\\frac{2c}{3}, get e=\\frac{c}{3}=\\frac{3}{3}. Therefore \\frac{\\sqrt{3}(a-c}{3}}"
  },
  {
    "text": "If one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1(a>0)$ passes through $(1, a)$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Point;a>0;Expression(G) = (-y^2/4 + x^2/a^2 = 1);Coordinate(H) = (1, a);PointOnCurve(H, OneOf(Asymptote(G)))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[1, 48], [68, 71]], [[57, 65]], [[57, 65]], [[4, 48]], [[1, 48]], [[57, 65]], [[1, 65]]]",
    "query_spans": "[[[68, 77]]]",
    "process": "According to the hyperbola's asymptote equation passing through point (1,a), substitute the point into the asymptote equation to find a, then the eccentricity can be determined. [Detailed solution] The asymptote equations of the hyperbola are y=\\pm\\frac{b}{a}x=\\pm\\frac{2}{a}x. Since the asymptote passes through point (1,a), i.e., the asymptote equation y=\\frac{2}{a}x passes through (1,a), substituting gives a=\\sqrt{2} or a=-\\sqrt{2} (discarded). Then c=\\sqrt{a^{2}+b^{2}}=\\sqrt{6}. Therefore, the eccentricity e=\\frac{c}{a}=\\frac{\\sqrt{6}}{\\sqrt{2}}=\\sqrt{3}"
  },
  {
    "text": "Given that $F(2,0)$ is the right focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. The line $l$: $y=-\\frac{1}{3} x+m$ intersects the ellipse $C$ at points $A$ and $B$, with midpoint $P$ of segment $AB$. The slope $k$ of line $OP$ is $k=1$. Then the equation of the ellipse $C$ is?",
    "fact_expressions": "F: Point;RightFocus(C) = F;Coordinate(F) = (2, 0);C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;l: Line;m: Number;Expression(l) = (y = m - x/3);A: Point;B: Point;Intersection(l, C) = {A, B};P: Point;MidPoint(LineSegmentOf(A, B)) = P;O: Origin;k: Number;Slope(LineOf(O, P)) = k;k = 1",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/6+y^2/2=1",
    "fact_spans": "[[[2, 10]], [[2, 72]], [[2, 10]], [[11, 68], [101, 106], [150, 155]], [[11, 68]], [[18, 68]], [[18, 68]], [[18, 68]], [[18, 68]], [[73, 100]], [[80, 100]], [[73, 100]], [[109, 112]], [[113, 116]], [[73, 118]], [[128, 131]], [[119, 131]], [[135, 140]], [[143, 148]], [[133, 148]], [[143, 148]]]",
    "query_spans": "[[[150, 160]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, and let the midpoint of $ A,B $ be $ P(x_{0},y_{0}) $. Then from the given conditions, $ k_{OP} = \\frac{y_{0}}{x_{0}} = 1 $, $ \\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = -\\frac{1}{3} $. Since $ A,B $ are two distinct points on the ellipse, we have\n$$\n\\begin{cases}\n\\frac{x_{1}^{2}}{a^{2}} + \\frac{y_{1}^{2}}{b^{2}} = 1 \\\\\n\\frac{x_{2}^{2}}{a^{2}} + \\frac{y_{2}^{2}}{b^{2}} = 1\n\\end{cases}\n$$\nSubtracting these two equations yields\n$$\n\\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{a^{2}} + \\frac{(y_{1}-y_{2})(y_{1}+y_{2})}{b^{2}} = 0,\n$$\nwhich simplifies to\n$$\n\\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = -\\frac{b^{2}}{a^{2}} \\cdot \\frac{x_{0}}{y_{0}},\n$$\ni.e., $ -\\frac{1}{3} = -\\frac{b^{2}}{a^{2}} \\cdot \\frac{x_{0}}{y_{0}} $. Given $ k_{OP} = \\frac{y_0}{x_0} = 1 $, it follows $ \\frac{x_0}{y_0} = 1 $, so $ -\\frac{1}{3} = -\\frac{b^{2}}{a^{2}} $, thus $ a^{2} = 3b^{2} $. Together with $ a^{2} = b^{2} + c^{2} $ and $ c = 2 $, solving gives: $ a^{2} = 6 $, $ b^{2} = 2 $. Therefore, the equation of ellipse $ C $ is $ \\frac{x^{2}}{6} + \\frac{y^{2}}{2} = 1 $."
  },
  {
    "text": "Given that $F$ is a focus of the ellipse $C$, $B$ is an endpoint of the minor axis, the extension of segment $BF$ intersects $C$ at point $D$, and $\\overrightarrow{B F}=2\\overrightarrow{F D}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;F: Point;OneOf(Focus(C)) = F;B: Point;OneOf(Endpoint(MinorAxis(C))) = B;D: Point;Intersection(OverlappingLine(LineSegmentOf(B, F)), C) = D;VectorOf(B, F) = 2*VectorOf(F, D)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[6, 11], [40, 43], [96, 99]], [[2, 5]], [[2, 16]], [[17, 20]], [[6, 28]], [[44, 48]], [[29, 48]], [[50, 94]]]",
    "query_spans": "[[[96, 105]]]",
    "process": ""
  },
  {
    "text": "It is known that the foci of hyperbola $C$ lie on the $y$-axis and its eccentricity is $2$. Write an equation for a curve $C$ satisfying these conditions.",
    "fact_expressions": "C: Hyperbola;PointOnCurve(Focus(C), yAxis);Eccentricity(C) = 2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2 - x^2/3 = 1",
    "fact_spans": "[[[2, 8], [35, 40]], [[2, 17]], [[2, 25]]]",
    "query_spans": "[[[35, 45]]]",
    "process": "Let the hyperbola equation be \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1. Since the eccentricity of the hyperbola is 2, it follows that b^{2}=3a^{2}. Thus, the hyperbola equation is any one of \\frac{y^{2}}{\\lambda}-\\frac{x^{2}}{3\\lambda}=1 (\\lambda>0), and we can take y^{2}-\\frac{x^{2}}{3}=1."
  },
  {
    "text": "If a focus of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{3}=1$ lies on the directrix of the parabola $y^{2}=8 x$, then $m$=?",
    "fact_expressions": "G: Parabola;H: Ellipse;m: Number;Expression(G) = (y^2 = 8*x);Expression(H) = (y^2/3 + x^2/m = 1);PointOnCurve(OneOf(Focus(H)), Directrix(G))",
    "query_expressions": "m",
    "answer_expressions": "7",
    "fact_spans": "[[[44, 58]], [[1, 38]], [[64, 67]], [[44, 58]], [[1, 38]], [[1, 62]]]",
    "query_spans": "[[[64, 69]]]",
    "process": "First, find the directrix of the parabola $ y^{2} = 8x $, thereby obtaining the value of $ c $, and then determine the value of $ m $. The directrix of the parabola $ y^{2} = 8x $ is the line $ x = -2 $. Since one focus of the ellipse $ \\frac{x^{2}}{m} + \\frac{y^{2}}{3} = 1 $ lies on the directrix of the parabola $ y^{2} = 8x $, it follows that $ c = 2 $. Therefore, $ m = a^{2} = b^{2} + c^{2} = 3 + 2^{2} = 7 $."
  },
  {
    "text": "Given that one asymptote of a hyperbola has the equation $x-\\sqrt{2} y=0$, and the hyperbola passes through the point $(1, \\frac{\\sqrt{6}}{2})$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(OneOf(Asymptote(G))) = (x-sqrt(2)*y=0);H: Point;Coordinate(H) = (1, sqrt(6)/2);PointOnCurve(H,G) = True",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 - x^2 / 2 = 1",
    "fact_spans": "[[[2, 5], [63, 66]], [[2, 31]], [[35, 61]], [[35, 61]], [[2, 61]]]",
    "query_spans": "[[[63, 72]]]",
    "process": "By the given condition, the equation of one asymptote of the hyperbola is $x - \\sqrt{2}y = 0$, so we can assume the equation of the hyperbola to be: $x^{2} - 2y^{2} = \\lambda$. Substituting the point $(1, \\frac{\\sqrt{6}}{2})$, we get: $\\lambda = 1 - 2 \\times \\frac{3}{2} = -2$. $\\therefore y^{2} - \\frac{x^{2}}{2} = 1$"
  },
  {
    "text": "If $AB$ is a moving chord on the parabola $y=2px$ $(p>0)$, and $|AB|=a$ $(a>2p)$, then the shortest distance from the midpoint $M$ of $AB$ to the $y$-axis is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;M: Point;a: Number;p>0;Expression(G) = (y = 2*(p*x));IsChordOf(LineSegmentOf(A, B), G);Abs(LineSegmentOf(A, B)) = a;a > 2*p;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Min(Distance(M, yAxis))",
    "answer_expressions": "(a - p)/2",
    "fact_spans": "[[[7, 24]], [[10, 24]], [[1, 6]], [[1, 6]], [[55, 58]], [[30, 46]], [[10, 24]], [[7, 24]], [[1, 28]], [[30, 46]], [[30, 46]], [[48, 58]]]",
    "query_spans": "[[[55, 70]]]",
    "process": ""
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ are $F_1$ and $F_2$, respectively. A line passing through $F_2$ with an inclination angle of $120^{\\circ}$ intersects the ellipse at a point $M$. If $MF_1$ is perpendicular to the $x$-axis, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;M: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);F1: Point;LeftFocus(G) = F1;F2: Point;RightFocus(G) = F2;PointOnCurve(F2, H);OneOf(Intersection(H,G)) = M;Inclination(H) = ApplyUnit(120, degree);IsPerpendicular(LineSegmentOf(M,F1),xAxis)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(-\\infty,-1)",
    "fact_spans": "[[[0, 52], [98, 100], [127, 129]], [[2, 52]], [[2, 52]], [[95, 97]], [[106, 109]], [[2, 52]], [[2, 52]], [[0, 52]], [[61, 65]], [[0, 70]], [[72, 76], [66, 70]], [[0, 70]], [[71, 97]], [[95, 109]], [[77, 95]], [[111, 125]]]",
    "query_spans": "[[[127, 135]]]",
    "process": ""
  },
  {
    "text": "The equation of the directrix of the parabola $y^{2}=16 x$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 16*x)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=-4",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 22]]]",
    "process": "Analysis: Using the directrix equation of the parabola $ y^{2} = 2px $ ($ p > 0 $), which is $ x = -\\frac{p}{2} $, we can obtain the directrix equation of the parabola $ y^{2} = 16x $. Since the directrix equation of the parabola $ y^{2} = 2px $ ($ p > 0 $) is $ x = -\\frac{p}{2} $, the directrix equation of the parabola $ y^{2} = 16x $ is $ x = -4 $."
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x$ coincides with the right focus of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, then the real number $p$=?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 2*(p*x));p: Real;G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1);Focus(H) = RightFocus(G)",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 17]], [[1, 17]], [[57, 62]], [[21, 49]], [[21, 49]], [[1, 55]]]",
    "query_spans": "[[[57, 64]]]",
    "process": ""
  },
  {
    "text": "Given the parabola equation $y^{2}=4x$, a line $l$ passes through the fixed point $M(-2,1)$ with slope $k$. When the line $l$ has only one common point with the parabola $y^{2}=4x$, the set of values for the slope $k$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);l: Line;M: Point;Coordinate(M) = (-2, 1);PointOnCurve(M, l);k: Number;Slope(l) = k;NumIntersection(l, G) = 1",
    "query_expressions": "Range(k)",
    "answer_expressions": "{0, -1, 1/2}",
    "fact_spans": "[[[2, 5], [52, 66]], [[52, 66]], [[20, 25], [46, 51]], [[28, 37]], [[28, 37]], [[20, 37]], [[41, 44], [77, 80]], [[20, 44]], [[46, 73]]]",
    "query_spans": "[[[77, 87]]]",
    "process": "According to the problem, let the equation of the line be: $ y = k(x + 2) + 1 $. Substituting into the parabola equation and simplifying yields $ k^{2}x^{2} + (4k^{2} + 2k - 4)x + 4k^{2} + 4k + 1 = 0 $ (*). The line and parabola have only one common point if and only if (*) has exactly one root.  \n① When $ k = 0 $, $ y = 1 $ satisfies the condition;  \n② When $ k \\neq 0 $, $ \\Delta = (4k^{2} + 2k \\cdot 4)^{2} \\cdot 4k^{2}(4k^{2} + 4k + 1) = $. Simplifying gives $ 2k^{2} + k \\cdot 1 = 0 $, solving yields $ k = \\frac{1}{4} $ or $ k = -1 $.  \nIn conclusion, $ k \\in \\left[-1, 0, \\frac{1}{2}\\right) $."
  },
  {
    "text": "The line $l$: $y=k(x+\\sqrt{2})$ intersects the curve $C$: $x^{2}-y^{2}=1$ $(x<0)$ at two points $P$ and $Q$. Then, what is the range of the inclination angle of the line $l$?",
    "fact_expressions": "l: Line;k: Number;Expression(l) = (y = k*(x + sqrt(2)));C: Curve;Expression(C) = ((x^2 - y^2 = 1)&(x<0));P: Point;Q: Point;Intersection(l, C) = {P, Q}",
    "query_expressions": "Range(Inclination(l))",
    "answer_expressions": "(pi/4, pi/2) + (pi/2, 3*pi/4)",
    "fact_spans": "[[[0, 24], [65, 70]], [[7, 24]], [[0, 24]], [[25, 52]], [[25, 52]], [[54, 57]], [[58, 61]], [[0, 63]]]",
    "query_spans": "[[[65, 81]]]",
    "process": "\\because the equation of curve C is x^{2}-y^{2}=1 (x<0), \\therefore curve C is the left branch of a hyperbola, and its asymptotes are given by y=\\pm x. \\because the line l: y=k(x+\\sqrt{2}) intersects curve C at two points P and Q, \\therefore the slope k of line l satisfies k<-1 or k>1. \\because the slope of line l exists, \\therefore the range of the inclination angle of line l is (\\frac{\\pi}{4},\\frac{\\pi}{2})\\cup(\\frac{\\pi}{2},\\frac{3\\pi}{4})."
  },
  {
    "text": "Let $F$ be the focus of the parabola $y^{2}=12 x$, and let $A$, $B$, $C$ be three points on this parabola such that $\\overrightarrow{F A}+\\overrightarrow{F B}+\\overrightarrow{F C}=\\overrightarrow{0}$. Then $|\\overrightarrow{F A}|+|\\overrightarrow{F B}|+|\\overrightarrow{F C}|$=?",
    "fact_expressions": "F: Point;G: Parabola;Expression(G) = (y^2 = 12*x);Focus(G) = F;A: Point;B: Point;C: Point;PointOnCurve(A, G) = True;PointOnCurve(B, G) = True;PointOnCurve(C, G) = True;VectorOf(F, A) + VectorOf(F, B) + VectorOf(F, C) = 0",
    "query_expressions": "Abs(VectorOf(F, A)) + Abs(VectorOf(F, B)) + Abs(VectorOf(F, C))",
    "answer_expressions": "18",
    "fact_spans": "[[[1, 4]], [[5, 20], [37, 40]], [[5, 20]], [[1, 23]], [[24, 27]], [[28, 31]], [[32, 35]], [[24, 44]], [[24, 44]], [[24, 44]], [[46, 129]]]",
    "query_spans": "[[[131, 203]]]",
    "process": "Parabola focus coordinates $F(3,0)$, directrix equation: $x=-3$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, $C(x_{3},y_{3})$. Since $\\overrightarrow{FA}+\\overrightarrow{FB}+\\overrightarrow{FC}=\\overrightarrow{0}$, point $F$ is the centroid of $\\triangle ABC$, so $x_{1}+x_{2}+x_{3}=9$. Then by the definition of the parabola, $|FA|=x_{1}+3$, $|FB|=x_{2}-(-3)=x_{2}+3$, $|FC|=x_{3}-(-3)=x_{3}+3$, therefore $|\\overrightarrow{FA}|+|\\overrightarrow{FB}|+|\\overrightarrow{FC}|=x_{1}+3+x_{2}+3+x_{3}+3=18$."
  },
  {
    "text": "The ellipse and hyperbola with common foci $F_{1}$, $F_{2}$ have eccentricities $e_{1}$, $e_{2}$ respectively. Point $A$ is a common point on both curves and satisfies $\\angle F_{1} A F_{2}=60^{\\circ}$. Then the value of $\\frac{1}{e_{1}^{2}}+\\frac{3}{e_{2}^{2}}$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;F1: Point;A: Point;F2: Point;e1:Number;e2:Number;Focus(G)={F1,F2};Focus(H)={F1,F2};Eccentricity(H)=e1;Eccentricity(G)=e2;OneOf(Intersection(H,G))=A;AngleOf(F1, A, F2) = ApplyUnit(60, degree)",
    "query_expressions": "3/e2^2 + 1/e1^2",
    "answer_expressions": "4",
    "fact_spans": "[[[24, 27]], [[21, 23]], [[5, 12]], [[52, 56]], [[13, 20]], [[34, 41]], [[44, 51]], [[0, 27]], [[0, 27]], [[21, 51]], [[21, 51]], [[52, 66]], [[70, 103]]]",
    "query_spans": "[[[105, 150]]]",
    "process": "Let A be a point in the first quadrant, |AF₁| = m, |AF₂| = n. By the definition of the ellipse, we have m + n = 2a; by the definition of the hyperbola, we have m - n = 2a, yielding m = a + a, n = a - a. Given ∠F₁AF₂ = 60°, we obtain m² + n² - 2mn cos60° = (2c)², that is, (a + a)² + (a - a)² - 2(a + a)(a - a) × 1/2 = 4c², simplifying to a² + 3a² = 4c², then 1/e² + 3/e² = 4."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, the line $l$ is tangent to the parabola $C$ at point $Q$, and $P$ is a point on $l$ (not coinciding with $Q$). If the circle with diameter $PQ$ passes through $F$, then the minimum value of $|PF|$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);G: Circle;P: Point;Q: Point;F: Point;l:Line;Focus(C) = F;Negation(P=Q);PointOnCurve(P,l);TangentPoint(l,C)=Q;IsDiameter(LineSegmentOf(P,Q),G);PointOnCurve(F,G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 21], [35, 41]], [[2, 21]], [[83, 84]], [[49, 52]], [[44, 48], [63, 66]], [[25, 28], [88, 91]], [[29, 34], [53, 56]], [[2, 28]], [[49, 69]], [[49, 60]], [[29, 48]], [[71, 84]], [[83, 91]]]",
    "query_spans": "[[[93, 106]]]",
    "process": "By the symmetry of the parabola, let $ Q(m, 2\\sqrt{m}) $, then $ k_{QF} = \\frac{2\\sqrt{m}}{m-1} $, so the equation of line $ PF $ is $ y = \\frac{1-m}{2\\sqrt{m}}(x-1) $. From $ y^2 = 4x $, take $ y = 2\\sqrt{x} $, $ y = \\frac{1}{\\sqrt{x}} $, so the equation of line $ l $ is $ y - 2\\sqrt{m} = \\frac{1}{\\sqrt{m}}(x - m) $. Solving the system \n\\[\n\\begin{cases}\ny = \\frac{1-m}{2\\sqrt{m}}(x-1) \\\\\ny - 2\\sqrt{m} = \\frac{1}{\\sqrt{m}}(x - m)\n\\end{cases}\n\\]\ngives the x-coordinate of point $ P $ as $ x = -1 $, so point $ P $ moves on the directrix of the parabola. When the coordinates of point $ P $ are $ (-1, 0) $, $ |PF| $ is minimized, and the minimum value is $ 2 $."
  },
  {
    "text": "It is known that the line $x - 2y + 2 = 0$ passes through a vertex and a focus of the ellipse $\\frac{y^{2}}{a^{2}} + \\frac{x^{2}}{b^{2}} = 1\\ (a > b > 0)$. Then, what is the eccentricity of this ellipse?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;a > b;b > 0;Expression(G) = (x^2/b^2 + y^2/a^2 = 1);Expression(H) = (x - 2*y + 2 = 0);PointOnCurve(OneOf(Focus(G)),H);PointOnCurve(OneOf(Vertex(G)),H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(5)/5",
    "fact_spans": "[[[17, 69], [84, 86]], [[19, 69]], [[19, 69]], [[2, 15]], [[19, 69]], [[19, 69]], [[17, 69]], [[2, 15]], [[2, 79]], [[2, 79]]]",
    "query_spans": "[[[84, 92]]]",
    "process": ""
  },
  {
    "text": "It is known that the center of the ellipse is at the origin of the coordinate system, the foci lie on the $x$-axis, and the sum of the distances from point $P(3 \\sqrt{2}, 4)$ on the ellipse to the two foci is $12$. Then, what is the standard equation of the ellipse?",
    "fact_expressions": "G: Ellipse;Center(G) = O;O: Origin;PointOnCurve(Focus(G), xAxis) = True;P: Point;Coordinate(P) = (3*sqrt(2), 4);PointOnCurve(P, G) = True;F1: Point;F2: Point;Focus(G) = {F1,F2};Distance(P,F1) + Distance(P,F2) = 12",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/36+y^2/32=1",
    "fact_spans": "[[[2, 4], [22, 24], [60, 62]], [[2, 12]], [[8, 12]], [[2, 21]], [[25, 44]], [[25, 44]], [[22, 44]], [], [], [[22, 48]], [[25, 58]]]",
    "query_spans": "[[[60, 69]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola $y^{2}=-12 x$ is $F$, and point $M$ lies on the parabola (point $M$ is in the second quadrant), with $|M F|=7$. Then the coordinates of point $M$ are?",
    "fact_expressions": "G: Parabola;M: Point;F: Point;Expression(G) = (y^2 = -12*x);Focus(G) = F;Abs(LineSegmentOf(M, F)) = 7;PointOnCurve(M,G);Quadrant(M)=2",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(-4,4*\\sqrt{3})",
    "fact_spans": "[[[0, 16], [29, 32]], [[24, 28], [34, 38], [57, 61]], [[20, 23]], [[0, 16]], [[0, 23]], [[46, 55]], [[24, 33]], [[34, 43]]]",
    "query_spans": "[[[57, 65]]]",
    "process": "Let M(x_{0},y_{0}), x_{0}<0, y_{0}>0. By the definition of the parabola, p=6MF=|x_{0}|+\\frac{p}{2}=7, solving gives x_{0}=-4, substituting yields y_{0}=4\\sqrt{3} as the answer. -4A,F_{3}"
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, a line $l$ passing through point $F$ intersects the parabola at points $A$ and $B$. Extending $FB$ meets the directrix at point $C$. If $|BC|=2|BF|$, then the value of $\\frac{|BF|}{|AF|}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, G) = {A, B};C: Point;Intersection(OverlappingLine(LineSegmentOf(F, B)), Directrix(G)) = C;Abs(LineSegmentOf(B, C)) = 2*Abs(LineSegmentOf(B, F))",
    "query_expressions": "Abs(LineSegmentOf(B, F))/Abs(LineSegmentOf(A, F))",
    "answer_expressions": "1/3",
    "fact_spans": "[[[2, 16], [36, 39]], [[2, 16]], [[20, 23], [25, 29]], [[2, 23]], [[30, 35]], [[24, 35]], [[40, 43]], [[44, 47]], [[30, 49]], [[61, 65]], [[36, 65]], [[67, 81]]]",
    "query_spans": "[[[83, 108]]]",
    "process": "As shown in the figure: From the given conditions, we have F(1,0), the directrix is the line x = -1. Draw AM and BN perpendicular to the directrix from A and B, meeting it at M and N respectively. Then |BF| = |BN|, |AF| = |AM|. Since |BC| = 2|BF|, it follows that |CF| = 4. Also, \\frac{2}{AM} = \\frac{|BN|}{|CA|} = \\frac{|BC|}{|CF|} = \\frac{2}{|CF|}|BN| = \\frac{4}{3}, so |BF| = \\frac{4}{3}, |BC| = \\frac{8}{4+|AF|}. Solving gives: |AF| = |AM| = 4, hence \\frac{|BF|}{|AF|} = \\frac{4}{4} = \\frac{1}{3}."
  },
  {
    "text": "Let the line $l$: $y = x - 1$ intersect the parabola $y^{2} = 2 p x$ ($p > 0$) at points $A$ and $B$. If the horizontal coordinate of the midpoint of chord $AB$ is $2$, then the value of $p$ is?",
    "fact_expressions": "l: Line;Expression(l) = (y = x - 1);G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;A: Point;B: Point;p>0;Intersection(l, G) = {A, B};XCoordinate(MidPoint(LineSegmentOf(A, B))) = 2;IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "p",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 15]], [[1, 15]], [[16, 37]], [[16, 37]], [[71, 74]], [[40, 43]], [[44, 47]], [[19, 37]], [[1, 49]], [[52, 68]], [[16, 57]]]",
    "query_spans": "[[[71, 78]]]",
    "process": ""
  },
  {
    "text": "Given two points $A$ and $B$ on the parabola $y^{2}=4x$ with focus $F$ such that $\\overrightarrow{A F}=3 \\overrightarrow{F B}$, then the distance from the midpoint of chord $AB$ to the directrix is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(A, G);PointOnCurve(B,G);VectorOf(A, F) = 3*VectorOf(F, B);IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A, B)), Directrix(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[10, 24]], [[28, 31]], [[33, 37]], [[3, 6]], [[10, 24]], [[3, 24]], [[10, 36]], [[10, 36]], [[39, 84]], [[10, 91]]]",
    "query_spans": "[[[87, 102]]]",
    "process": ""
  },
  {
    "text": "Given that one focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ has coordinates $(\\sqrt{3}, 0)$ and its eccentricity is $\\frac{\\sqrt{3}}{2}$, what is the equation of the ellipse $C$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(OneOf(Focus(C))) = (sqrt(3), 0);Eccentricity(C)=sqrt(3)/2;a: Number;b: Number",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4+y^2=1",
    "fact_spans": "[[[2, 47], [97, 102]], [[2, 47]], [[2, 70]], [[2, 95]], [[2, 47]], [[2, 47]]]",
    "query_spans": "[[[97, 107]]]",
    "process": "According to the problem: $ a^{2}-b^{2}=3 $, $ e=\\frac{c}{a}=\\frac{\\sqrt{3}}{a}=\\frac{\\sqrt{3}}{2} $, solving gives $ a=2 $, $ b=1 $, thus the ellipse equation is $ \\frac{x^{2}}{4}+y^{2}=1 $."
  },
  {
    "text": "Given that one of the directrices of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1  (a>0)$ is $x=\\frac{3}{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2 + x^2/a^2 = 1);a: Number;a>0;Expression(OneOf(Directrix(G))) = (x = 3/2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 41], [64, 67]], [[2, 41]], [[5, 41]], [[5, 41]], [[2, 62]]]",
    "query_spans": "[[[64, 73]]]",
    "process": ""
  },
  {
    "text": "Given that point $F$ is the focus of the parabola $y^{2}=2 p x$ $(p>0)$, and points $A(2, y_{1})$, $B(\\frac{1}{2}, y_{2})$ are on the parabola located in the first and fourth quadrants respectively, if $|A F|=20$, then $|y_{1}-y_{2}|=$?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B:Point;F: Point;p>0;y1:Number;y2:Number;Expression(G) = (y^2 = 2*(p*x));Coordinate(A) = (2, y1);Coordinate(B) = (1/2, y2);Focus(G) = F;PointOnCurve(A,G);PointOnCurve(B,G);Quadrant(A)=1;Quadrant(B)=4;Abs(LineSegmentOf(A, F)) = 20",
    "query_expressions": "Abs(y1 - y2)",
    "answer_expressions": "18",
    "fact_spans": "[[[7, 28], [74, 77]], [[10, 28]], [[32, 46]], [[48, 71]], [[2, 6]], [[10, 28]], [[102, 117]], [[102, 117]], [[7, 28]], [[32, 46]], [[48, 71]], [[2, 31]], [[32, 88]], [[32, 88]], [[32, 88]], [[32, 88]], [[90, 100]]]",
    "query_spans": "[[[102, 119]]]",
    "process": "According to the definition of a parabola, we find $ p $, obtain the equation of the parabola, and further determine the coordinates of points $ B $ and $ A $, thus obtaining the result. Since the distance from point $ A(2,y_{1}) $ to the focus is equal to the distance to the directrix $ x=-\\frac{p}{2} $, $ |AF|=2+\\frac{p}{2}=20 $, so $ p=36 $. Then the equation of the parabola is $ y^{2}=72x $. Substituting $ x=\\frac{1}{2} $ into the equation gives $ y=-6 $ or $ y=6 $ (discarded), so $ B(\\frac{1}{2},-6) $. Substituting $ x=2 $ into the equation gives $ y=12 $, $ y=-12 $ (discarded), so $ A(2,12) $. Then $ |y_{1}-y_{2}|=|12-(-6)|=18 $."
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the common foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the hyperbola $\\frac{x^{2}}{m^{2}}-\\frac{y^{2}}{n^{2}}=1$ $(m>0, n>0)$, and let $P$ be one of their common points such that $\\angle F_{1} P F_{2}=\\frac{2}{3} \\pi$. When the product of the eccentricities of these two curves is minimized, what is the equation of the asymptotes of the hyperbola?",
    "fact_expressions": "H: Ellipse;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;G: Hyperbola;Expression(G) = (-y^2/n^2 + x^2/m^2 = 1);m: Number;n: Number;m>0;n>0;F1: Point;F2: Point;Focus(H) = {F1, F2};Focus(G) = {F1, F2};P: Point;OneOf(Intersection(H, G)) = P;AngleOf(F1, P, F2) = (2/3)*pi;WhenMin(Eccentricity(H)*Eccentricity(G))",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(2)/2)*x",
    "fact_spans": "[[[17, 69]], [[17, 69]], [[19, 69]], [[19, 69]], [[19, 69]], [[19, 69]], [[202, 205], [70, 126]], [[70, 126]], [[73, 126]], [[73, 126]], [[73, 126]], [[73, 126]], [[1, 8]], [[9, 16]], [[0, 131]], [[0, 131]], [[132, 135]], [[132, 144]], [[146, 184]], [[186, 201]]]",
    "query_spans": "[[[202, 214]]]",
    "process": "According to the problem, let the focal length be $2c$, and $P$ be a point in the second quadrant. Based on the given conditions and combining the definitions of the ellipse and hyperbola, we derive $3a^{2}+m^{2}=4c^{2}$. Using the eccentricity formula and the basic inequality, we find the minimum value of the product of the eccentricities and the condition for achieving this extremum, yielding $\\frac{n^{2}}{m^{2}}=\\frac{1}{2}$, and then obtain the equations of the asymptotes. By the symmetry of the ellipse and hyperbola, we can assume $P$ is a point in the second quadrant, as shown in the figure. According to the problem, the major axis of the ellipse is $2a$, and the transverse axis of the hyperbola is $2m$. Let the focal length be $2c$. From the definition of the ellipse, $|PF_{1}|+|PF_{2}|=2a$; from the definition of the hyperbola, $|PF_{2}|-|PF_{1}|=2m$. Solving these simultaneously gives: $|PF_{1}|=a-m$, $|PF_{2}|=a+m$. Also, $\\angle F_{1}PF_{2}=\\frac{2}{3}\\pi$. By the law of cosines:  \n$$\n\\cos\\angle F_{1}PF_{2}=\\frac{|PF_{1}|^{2}+|PF_{2}|^{2}-|F_{1}F_{2}|^{2}}{2|PF_{1}|\\cdot|PF_{2}|}\n$$  \nThat is,  \n$$\n-\\frac{1}{2}=\\frac{(a-m)^{2}+(a+m)^{2}-4c^{2}}{2(a-m)(a+m)},\n$$  \nwhich simplifies to:  \n$$\n3a^{2}+m^{2}=4c^{2},\n$$  \nor  \n$$\n\\frac{3a^{2}+m^{2}}{4}=c^{2}.\n$$  \nThe eccentricity of the ellipse is $e_{1}=\\frac{c}{a}$, and the eccentricity of the hyperbola is $e_{2}=\\frac{c}{m}$. Then  \n$$\ne_{1}e_{2}=\\frac{c}{a}\\cdot\\frac{c}{m}=\\frac{c^{2}}{am}=\\frac{3a^{2}+m^{2}}{4am}=\\frac{3a}{4m}+\\frac{m}{4a}\\geqslant2\\sqrt{\\frac{3a}{4m}\\cdot\\frac{m}{4a}}=\\frac{\\sqrt{3}}{2}.\n$$  \nEquality holds if and only if $\\frac{3a}{4m}=\\frac{m}{4a}$, i.e., $m=\\sqrt{3}a$, meaning the product of the eccentricities of the two curves is minimized. From $3a^{2}+m^{2}=4c^{2}$, we get $m^{2}=2c^{2}$. Since $c^{2}=m^{2}+n^{2}$, it follows that $n^{2}=c^{2}$, so $\\frac{n^{2}}{m^{2}}=\\frac{1}{2}$. Therefore, the asymptotes of the hyperbola are: $y=\\pm\\frac{n}{m}x=\\pm\\frac{\\sqrt{2}}{2}x$."
  },
  {
    "text": "The length of the imaginary axis of the hyperbola $m x^{2}+y^{2}=1$ is twice the length of the real axis, then $m=$?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (m*x^2 + y^2 = 1);Length(ImageinaryAxis(G)) = 2*Length(RealAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "-1/4",
    "fact_spans": "[[[0, 20]], [[35, 38]], [[0, 20]], [[0, 33]]]",
    "query_spans": "[[[35, 40]]]",
    "process": "First, we should know that the condition for the equation $ ax^{2} + by^{2} = 1 $ to represent a hyperbola is $ ab < 0 $. Therefore, in this problem, we have $ m < 0 $, so in the hyperbola $ mx^{2} + y^{2} = 1 $, $ a = 1 $, $ b = \\sqrt{-\\frac{1}{m}} $. The condition that the length of the imaginary axis is twice the length of the real axis is $ b = 2a $, thus we obtain $ m = -\\frac{1}{4} $."
  },
  {
    "text": "Given that the vertex of a parabola is at the origin and the focus is at $(0,-2)$, then the standard equation of this parabola is?",
    "fact_expressions": "G: Parabola;O:Origin;Vertex(G)=O;Coordinate(Focus(G))=(0,-2)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 = -8y",
    "fact_spans": "[[[10, 13], [30, 33]], [[5, 9]], [[2, 13]], [[10, 27]]]",
    "query_spans": "[[[30, 40]]]",
    "process": "Analysis: Based on the focus coordinates of the parabola, determine the form of the parabola. Assume the parabola equation is $ x^{2} = -2py $ ($ p > 0 $), find the value of $ p $, and obtain the standard equation. Specifically, from the focus coordinates of the parabola, assume the parabola equation is $ x^{2} = -2py $ ($ p > 0 $). Since $ \\frac{p}{2} = 2 $, we have $ p = 4 $. Therefore, the parabola equation is $ x^{2} = -8y $."
  },
  {
    "text": "The hyperbola $C$ passes through the point $(2, 3)$, and one of its asymptotes is $y = -\\sqrt{3}x$. Then the standard equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Point;Coordinate(G) = (2, 3);PointOnCurve(G, C);Expression(OneOf(Asymptote(C)))=(y=-sqrt(3)*x)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[0, 6], [44, 50]], [[7, 17]], [[7, 17]], [[0, 17]], [[0, 42]]]",
    "query_spans": "[[[44, 57]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$ has its right focus at point $F$; a line passing through point $F$ and parallel to one asymptote of the hyperbola intersects the hyperbola at points $P$ and $M$ lies on the line $PF$ such that $\\overrightarrow{O M} \\cdot \\overrightarrow{P F}=0$, then $\\frac{|\\overrightarrow{P M}|}{|\\overrightarrow{P F}|}$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/4 = 1);F: Point;RightFocus(G) = F;H: Line;PointOnCurve(F, H);IsParallel(H, OneOf(Asymptote(G))) ;Intersection(H, G) = P;P: Point;M: Point;PointOnCurve(M, LineOf(P, F)) ;DotProduct(VectorOf(O, M), VectorOf(P, F)) = 0;O: Origin",
    "query_expressions": "Abs(VectorOf(P, M))/Abs(VectorOf(P, F))",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 30], [48, 51], [61, 64]], [[2, 30]], [[35, 38], [40, 44]], [[2, 38]], [[58, 60]], [[39, 60]], [[45, 60]], [[58, 70]], [[66, 70]], [[71, 74]], [[71, 83]], [[87, 138]], [[87, 138]]]",
    "query_spans": "[[[141, 198]]]",
    "process": "The hyperbola $ x^{2} - \\frac{y^{2}}{4} = 1 $, then $ a = 1 $, $ b = 2 $, $ c = \\sqrt{a^{2} + b^{2}} = \\sqrt{5} $, yielding $ F(\\sqrt{5}, 0) $, the asymptotes are $ y = \\pm 2x $. Let the line passing through point $ F $ and parallel to one asymptote of the hyperbola be $ y = 2(x - \\sqrt{5}) $. Substituting into the hyperbola equation gives $ P\\left( \\frac{3\\sqrt{5}}{5}, -\\frac{4\\sqrt{5}}{5} \\right) $. From the line $ OM: y = -\\frac{1}{2}x $ and the line $ y = 2(x - \\sqrt{5}) $, we obtain"
  },
  {
    "text": "Through the right focus $F_2$ of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0, b>0$), draw a perpendicular line to one asymptote of the hyperbola, with foot of perpendicular at $P$, intersecting the hyperbola at point $A$. If $\\overrightarrow{F_{2} P}=3 \\overrightarrow{F_{2} A}$, then the equation of the asymptotes of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;RightFocus(C) = F2;F2: Point;L: Line;PointOnCurve(F2, L) = True;IsPerpendicular(L, OneOf(Asymptote(C))) = True;FootPoint(L, OneOf(Asymptote(C))) = P;P: Point;Intersection(L, C) = A;A: Point;VectorOf(F2, P) = 3*VectorOf(F2, A)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*x/2",
    "fact_spans": "[[[1, 62], [73, 76], [93, 96], [159, 165]], [[1, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[1, 72]], [[65, 72]], [], [[0, 84]], [[0, 84]], [[0, 91]], [[88, 91]], [[0, 102]], [[98, 102]], [[104, 157]]]",
    "query_spans": "[[[159, 173]]]",
    "process": "Without loss of generality, assume that one asymptote of the hyperbola is given by $ y = \\frac{b}{a}x $. Find the coordinates of point $ P $. Using $ \\overrightarrow{F_{2}P} = 3\\overrightarrow{F_{2}A} $, find the coordinates of point $ A $. Substitute the coordinates of point $ A $ into the equation of hyperbola $ C $ to find the value of $ \\frac{c^{2}}{a^{2}} $, then solve for $ \\frac{b}{a} $, thus obtaining the asymptote equations of hyperbola $ C $. Solution: As shown in the figure, without loss of generality, assume one asymptote of the hyperbola is $ y = \\frac{b}{a}x $. Then the slope of the line $ F_{2}P $ is $ -\\frac{a}{b} $, and the equation of line $ F_{2}P $ is: $ y = -\\frac{a}{b}(x - c) $. Solve the system:\n$$\n\\begin{cases}\ny = \\frac{b}{a}x \\\\\ny = -\\frac{a}{b}(x - c)\n\\end{cases}\n$$\nSolving gives $ P\\left( \\frac{a^{2}}{c}, \\frac{ab}{c} \\right) $. Let $ A(x_{0}, y_{0}) $, from $ \\overrightarrow{F_{2}P} = 3\\overrightarrow{F_{2}A} $, we get $ \\left( \\frac{a^{2}}{c} - c, \\frac{ab}{c} \\right) = 3(x_{0} - c, y_{0}) $, so\n$$\n\\begin{cases}\n\\frac{a^{2}}{c} - c = 3(x_{0} - c) \\\\\n\\frac{ab}{c} = 3y_{0}\n\\end{cases}\n$$\nThus $ A\\left( \\frac{a^{2} + 2c^{2}}{3c}, \\frac{ab}{3c} \\right) $. Substituting into $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $, we obtain\n$$\n\\frac{(a^{2} + 2c^{2})^{2}}{9a^{2}c^{2}} - \\frac{a^{2}b^{2}}{9b^{2}c^{2}} = 1\n$$\nSimplifying yields $ \\frac{c^{2}}{a^{2}} = \\frac{5}{4} $, then $ \\frac{b^{2}}{a^{2}} = \\frac{c^{2}}{a^{2}} - 1 = \\frac{1}{4} $, $ \\therefore \\frac{b}{a} = \\frac{1}{2} $. Therefore, the asymptote equations of hyperbola $ C $ are $ y = \\pm \\frac{1}{2}x $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 px$ ($p>0$) with directrix $l$, a line passing through $M(1, 0)$ with slope $\\sqrt{3}$ intersects $l$ at point $A$ and intersects $C$ at a point $B$. If $\\overrightarrow{A M}=\\overrightarrow{M B}$, then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;M: Point;A: Point;B: Point;l: Line;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(M) = (1, 0);Directrix(C) = l;PointOnCurve(M, G);Slope(G) = sqrt(3);Intersection(G, l) = A;OneOf(Intersection(G, C)) = B;VectorOf(A, M) = VectorOf(M, B)",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 28], [77, 80]], [[136, 139]], [[62, 64]], [[37, 47]], [[71, 75]], [[86, 89]], [[65, 68], [32, 35]], [[10, 28]], [[2, 28]], [[37, 47]], [[2, 35]], [[36, 64]], [[48, 64]], [[62, 75]], [[62, 89]], [[91, 134]]]",
    "query_spans": "[[[136, 141]]]",
    "process": ""
  },
  {
    "text": "Given that $A$, $F$, and $P$ are the left vertex, right focus, and a moving point on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively, if $\\angle P F A=2 \\angle P A F$ always holds, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F: Point;A: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftVertex(G)=A;RightFocus(G)=F;PointOnCurve(P,RightPart(G));AngleOf(P,F,A)=2*AngleOf(P,A,F)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[16, 72], [124, 127]], [[19, 72]], [[19, 72]], [[10, 13]], [[6, 9]], [[2, 5]], [[19, 72]], [[19, 72]], [[16, 72]], [[2, 88]], [[2, 88]], [[2, 88]], [[90, 119]]]",
    "query_spans": "[[[124, 133]]]",
    "process": "A(-a,0), F(c,0), let P(x_{0},y_{0}) \\therefore k_{AP}=\\frac{y_{0}}{x_{0}+a}, k_{FP}=\\frac{y_{0}}{x_{0}-c}, \\because \\angle PFA = 2\\angle PAF, k_{AP} = \\tan \\angle PAF, k_{FP} = -\\tan \\angle PFA. \\therefore \\frac{y_{0}}{x_{0}-c} = \\frac{2 \\times \\frac{y_{0}}{x_{0}+a}}{\\left(\\frac{y_{0}}{x_{0}+a}\\right)^{2} - 1} = \\frac{2y_{0}(x_{0}+a)}{y_{0}^{2} - (x_{0}+a)^{2}} \\therefore y_{0}^{2} - x_{0}^{2} - 2a x_{0} - a^{2} = 2x_{0}^{2} + 2a x_{0} - 2c x_{0} - 2a c, i.e., y_{0}^{2} - 3x_{0}^{2} - (4a - 2c)x_{0} - a^{2} + 2a c = 0. Also, P(x_{0},y_{0}) lies on the hyperbola, \\therefore \\begin{cases} y = \\frac{b^{2}}{2}x - b^{2}, \\\\ \\frac{b^{2} - 3 - b^{2}}{b} = 2a - c = 0 \\text{ holds identically}. \\\\ b = -3 = 0 \\\\ 2a - 2c = 0 \\end{cases}. c = 2a, i.e., e = 2."
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ are $F_{1}(-c, 0)$ and $F_{2}(c, 0)$, respectively. If there exists a point $P$ on the ellipse such that $c \\cdot \\overrightarrow{P F_{2}}=a \\cdot \\overrightarrow{P F_{1}}$, then the range of values for the eccentricity of this ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;c: Number;F1: Point;F2: Point;P: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);c*VectorOf(P, F2) = a*VectorOf(P, F1)",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[\\sqrt{2} - 1, 1)",
    "fact_spans": "[[[0, 52], [93, 95], [177, 179]], [[2, 52]], [[2, 52]], [[61, 75]], [[61, 75]], [[78, 91]], [[98, 102]], [[2, 52]], [[2, 52]], [[0, 52]], [[61, 75]], [[78, 91]], [[0, 91]], [[0, 91]], [[93, 102]], [[105, 173]]]",
    "query_spans": "[[[177, 189]]]",
    "process": ""
  },
  {
    "text": "Given that one focus of the ellipse with equation $x^{2}+k y^{2}=5$ is $(0,2)$, then what is $k$=?",
    "fact_expressions": "G: Ellipse;H: Point;Coordinate(H) = (0, 2);Expression(G) = (k*y^2 + x^2 = 5);OneOf(Focus(G)) = H;k: Number",
    "query_expressions": "k",
    "answer_expressions": "5/9",
    "fact_spans": "[[[2, 4]], [[30, 37]], [[30, 37]], [[2, 24]], [[2, 37]], [[40, 43]]]",
    "query_spans": "[[[40, 45]]]",
    "process": "From the given conditions, we have $ a^{2} = \\frac{5}{k} $, $ b^{2} = 5 $, then $ a^{2} - b^{2} = \\frac{5}{k} - 5 = 4 $, thus obtaining the answer. The ellipse equation $ x^{2} + ky^{2} = 5 $ is converted into standard form: $ \\frac{x^{2}}{5} + \\frac{y^{2}}{k} = 1 $. One focus of the ellipse is $ (0, 2) $, so the focus lies on the $ y $-axis and $ c = 2 $. Therefore, $ a^{2} = \\frac{5}{1} $, $ b^{2} = 5 $, then $ a^{2} - b^{2} = \\frac{5}{k} - 5 = 4 $, solving gives $ k = \\frac{5}{9} $."
  },
  {
    "text": "What is the minimum distance from a point on the parabola $y=4 x^{2}$ to its directrix?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 4*x^2);P: Point;PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, Directrix(G)))",
    "answer_expressions": "1/16",
    "fact_spans": "[[[0, 14], [18, 19]], [[0, 14]], [[16, 17]], [[0, 17]]]",
    "query_spans": "[[[16, 30]]]",
    "process": "The standard equation of the parabola is $x^{2}=\\frac{1}{4}y$, and the equation of the directrix is $y=-\\frac{1}{16}$, yielding the minimum distance."
  },
  {
    "text": "The circle centered at the left focus $F(-c, 0)$ of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with radius $c$ intersects the left directrix of the ellipse at two distinct points. What is the range of values for the eccentricity of this ellipse?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);D:Circle;b: Number;a: Number;F: Point;a > b;b > 0;c:Number;Coordinate(F) = (-c, 0);LeftFocus(G) = F;Center(D)=F;Radius(D)=c;NumIntersection(H,LeftDirectrix(G))=2",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(\\sqrt{2}/2,1)",
    "fact_spans": "[[[1, 53], [80, 82], [96, 98]], [[1, 53]], [[78, 79]], [[3, 53]], [[3, 53]], [[57, 67]], [[3, 53]], [[3, 53]], [[71, 74]], [[57, 67]], [[1, 67]], [[0, 79]], [[71, 79]], [[78, 93]]]",
    "query_spans": "[[[96, 109]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ is on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola, respectively, and $I$ is the incenter of $\\Delta P F_{1} F_{2}$. If $S_{\\Delta I P F_{1}}=\\frac{\\sqrt{6}}{3} S_{\\Delta I F_{1} F_{2}}+S_{\\Delta I P F_{2}}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;P: Point;F1: Point;F2: Point;I: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, RightPart(G));LeftFocus(G) = F1;RightFocus(G) = F2;Incenter(TriangleOf(P, F1, F2)) = I;Area(TriangleOf(I, P, F1)) = (sqrt(6)/3)*Area(TriangleOf(I, F1, F2)) + Area(TriangleOf(I, P, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[7, 63], [87, 90], [216, 219]], [[10, 63]], [[10, 63]], [[2, 6]], [[69, 76]], [[77, 84]], [[96, 99]], [[10, 63]], [[10, 63]], [[7, 63]], [[2, 68]], [[69, 95]], [[69, 95]], [[96, 125]], [[128, 215]]]",
    "query_spans": "[[[216, 225]]]",
    "process": "Let the inradius of $\\triangle PF_{1}F_{2}$ be $r$, $r_{2}\\frac{\\sqrt[4m+n]{-y+yF_{2}-4P_{2}r}}{|.2-b}.|-|-x-\\frac{1}{x}$"
  },
  {
    "text": "The center of the ellipse is at the origin, the coordinates of the foci are $(0 , \\pm 4)$, and the length of the major axis is $10$. Then the equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;O:Origin;Center(G)=O;Coordinate(Focus(G))=(0,pm*4);Length(MajorAxis(G)) = 10",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9+y^2/25=1",
    "fact_spans": "[[[0, 2], [39, 41]], [[5, 7]], [[0, 7]], [[0, 26]], [[0, 36]]]",
    "query_spans": "[[[39, 46]]]",
    "process": "The solution process is omitted"
  },
  {
    "text": "Given the ellipse $E$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ with a vertex at $H(2,0)$, for points $P(t, 0)$ on the $x$-axis, there exists a point $M$ on the ellipse $E$ such that $M P \\perp M H$. Then the range of real values for $t$ is?",
    "fact_expressions": "E: Ellipse;Expression(E) = (x^2/4 + y^2/3 = 1);H: Point;Coordinate(H) = (2, 0);OneOf(Vertex(E)) = H;P: Point;PointOnCurve(P, xAxis) ;Coordinate(P) = (t, 0);t: Real;M: Point;PointOnCurve(M, E);IsPerpendicular(LineSegmentOf(M, P), LineSegmentOf(M, H))",
    "query_expressions": "Range(t)",
    "answer_expressions": "(-2, -1)",
    "fact_spans": "[[[2, 44], [78, 83]], [[2, 44]], [[50, 58]], [[50, 58]], [[2, 58]], [[67, 77]], [[61, 77]], [[67, 77]], [[110, 115]], [[86, 90]], [[78, 90]], [[93, 108]]]",
    "query_spans": "[[[110, 122]]]",
    "process": "Let M(x_{0},y_{0}) with -2<x_{0}<2, then \\frac{x_{0}^{2}}{4}+\\frac{y_{0}^{2}}{3}=1,\\textcircled{1}\\overrightarrow{MP}=(t-x_{0},-y_{0}),\\overrightarrow{MH}=(2-x_{0},-y_{0}). From MP\\bot MH we obtain \\overrightarrow{MP}\\cdot\\overrightarrow{MH}=0, i.e., (t-x_{0})(2-x_{0})+y_{0}^{2}=0,\\textcircled{2}. Eliminating y_{0} from \\textcircled{1} and \\textcircled{2}, we get t(2-x_{0})=-\\frac{1}{4}x_{0}^{2}+2x_{0}-3. Since x_{0}\\neq2, it follows that t=\\frac{1}{4}x_{0}-\\frac{3}{2}. Because -2<x_{0}<2, we have -2<t<-1. Therefore, the range of the real number is (-2,-1)."
  },
  {
    "text": "Given a point $P$ on the left branch of the hyperbola $\\frac{x^{2}}{25}-\\frac{y^{2}}{144}=1$ such that the distance from $P$ to the left focus is $16$, what is the distance from point $P$ to the right directrix?",
    "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2/25 - y^2/144 = 1);PointOnCurve(P,LeftPart(G));Distance(P, LeftFocus(G)) = 16",
    "query_expressions": "Distance(P, RightDirectrix(G))",
    "answer_expressions": "10",
    "fact_spans": "[[[2, 43]], [[48, 51], [65, 69]], [[2, 43]], [[2, 51]], [[2, 63]]]",
    "query_spans": "[[[2, 78]]]",
    "process": "Since the distance from a point P on the left branch of the hyperbola $\\frac{x^{2}}{25}-\\frac{y^{2}}{144}=1$ to the left focus is 16, the distance from this point to the right focus is $16+10=26$, and the eccentricity is $e=\\frac{c}{a}=\\frac{\\sqrt{25+144}}{5}=\\frac{13}{5}$. Let $d$ be the distance from point P to the right directrix. Then, by the second definition of the hyperbola, we have $\\frac{26}{d}=\\frac{13}{5}$, solving for $d$ gives $d=10$, that is, the distance from point P to the right directrix is 10."
  },
  {
    "text": "Given point $A(0,2)$, the focus of the parabola $y^{2}=2 p x(p>0)$ is $F$, the directrix is $l$, and the segment $F A$ intersects the parabola at point $B$. Draw a perpendicular from $B$ to $l$, with foot $M$. If $A M \\perp M F$, then the area $S$ of triangle $A F M$ is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;F: Point;M: Point;B: Point;l: Line;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(A) = (0, 2);Focus(G) = F;Intersection(LineSegmentOf(F, A), G) = B;Directrix(G) = l;l1: Line;PointOnCurve(B, l1);IsPerpendicular(l1, l);FootPoint(l1, l) = M;IsPerpendicular(LineSegmentOf(A, M), LineSegmentOf(M, F));S: Number;Area(TriangleOf(A, F, M)) = S",
    "query_expressions": "S",
    "answer_expressions": "3*sqrt(2)/4",
    "fact_spans": "[[[12, 33]], [[15, 33]], [[2, 11]], [[37, 40]], [[81, 84]], [[60, 64]], [[44, 47], [71, 74]], [[15, 33]], [[12, 33]], [[2, 11]], [[12, 40]], [[48, 64]], [[12, 47]], [[75, 77]], [[66, 77]], [[71, 77]], [[66, 84]], [[86, 101]], [[116, 119]], [[103, 119]]]",
    "query_spans": "[[[116, 121]]]",
    "process": "From the definition of a parabola, we know |BF| = |BM|, F(\\frac{p}{2},0). Then, from the properties of a right triangle, point B is the midpoint of AF. Using the midpoint coordinate formula, we find the coordinates of point B and substitute them into the parabola equation to solve for p. Based on S_{AAFM} = 2S_{ABMF}, the result can be calculated. [Detailed solution] As shown in the figure: \\because AM \\bot MF, \\therefore from the properties of a right triangle, point B is the midpoint of AF, \\therefore B(\\frac{p}{4},1). Substituting point B(\\frac{p}{4},1) into the parabola equation: y^{2} = 2px (p > 0), we get 1 = 2p \\times \\frac{p}{4}. Solving gives p = \\sqrt{2}, \\therefore B(\\frac{\\sqrt{2}}{4},1), \\therefore S_{AAFM} = 2S_{ABMF} = 2 \\times \\frac{1}{2} \\times 1 \\times (\\frac{\\sqrt{2}}{4} + \\frac{\\sqrt{2}}{2}) = \\frac{3\\sqrt{2}}{4}"
  },
  {
    "text": "The line $l$ passing through the focus $F$ of the parabola $y^{2}=4x$ intersects it at two points $A$ and $B$. What is the equation of the locus of the midpoint of chord $AB$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;PointOnCurve(F,l) = True;l: Line;Intersection(l,G) = {A,B};A: Point;B: Point;IsChordOf(LineSegmentOf(A,B),G) = True",
    "query_expressions": "LocusEquation(MidPoint(LineSegmentOf(A, B)))",
    "answer_expressions": "y^2=2*(x-1)",
    "fact_spans": "[[[1, 15], [27, 28]], [[1, 15]], [[17, 20]], [[1, 20]], [[0, 26]], [[21, 26]], [[21, 40]], [[31, 34]], [[35, 38]], [[21, 47]]]",
    "query_spans": "[[[43, 57]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\Gamma_{1}$: $x^{2}-\\frac{y^{2}}{b^{2}}=1$, its left and right foci are $F_{1}$ and $F_{2}$ respectively. Taking $O$ as the vertex and $F_{2}$ as the focus, a parabola $\\Gamma_{2}$ is constructed. If the hyperbola $\\Gamma_{1}$ and the parabola $\\Gamma_{2}$ intersect at point $P$, and $\\angle P F_{1} F_{2}=45^{\\circ}$, then what is the equation of the directrix of the parabola $\\Gamma_{2}$?",
    "fact_expressions": "Gamma1: Hyperbola;Gamma2: Parabola;P: Point;F1: Point;F2: Point;O: Origin;Expression(Gamma1)=(x^2-y^2/b^2=1);LeftFocus(Gamma1)=F1;RightFocus(Gamma1)=F2;Vertex(Gamma2)=O;Focus(Gamma2)=F2;Intersection(Gamma1,Gamma2)=P;AngleOf(P, F1, F2) = ApplyUnit(45, degree)",
    "query_expressions": "Expression(Directrix(Gamma2))",
    "answer_expressions": "x=-(\\sqrt{2}+1)",
    "fact_spans": "[[[2, 48], [109, 124]], [[91, 106], [125, 140], [183, 198]], [[142, 146]], [[57, 64]], [[65, 72], [80, 87]], [[74, 77]], [[2, 48]], [[2, 72]], [[2, 72]], [[73, 106]], [[80, 106]], [[109, 146]], [[148, 181]]]",
    "query_spans": "[[[183, 205]]]",
    "process": "Let $ F_{1}(-c,0), F_{2}(c,0) $, then the equation of the parabola is $ y^{2}=4cx $. The equation of line $ PF_{1} $ is $ y=x+c $, $ y^{2}=4cx $, $ y=x+c $, $ |PF_{2}|=|F_{1}F_{2}|=2c $, $ |PF_{1}|=2\\sqrt{2}c $. According to the definition of hyperbola, $ |PF_{1}|-|PF_{2}|=2a $, $ 2\\sqrt{2}c-2c=2 \\Rightarrow c=\\sqrt{2}+1 $. Therefore, the directrix of the parabola is $ x=-c=-(\\sqrt{2}+1) $."
  },
  {
    "text": "Given that $F$ is the right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the line $y=\\sqrt{3} x$ intersects the hyperbola $C$ at points $A$ and $B$. If the midpoints of $A F$ and $B F$ are $M$ and $N$ respectively, $O$ is the origin, and $\\overrightarrow{O M} \\cdot \\overrightarrow{O N}=0$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(C) = F;G: Line;Expression(G) = (y = sqrt(3)*x);A: Point;B: Point;Intersection(G, C) = {A, B};O: Origin;N: Point;M: Point;MidPoint(LineSegmentOf(A,F)) = M;MidPoint(LineSegmentOf(B,F)) = N;DotProduct(VectorOf(O,M),VectorOf(O,N)) = 0",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)+1",
    "fact_spans": "[[[6, 67], [89, 95], [197, 203]], [[6, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[2, 5]], [[2, 71]], [[72, 88]], [[72, 88]], [[96, 99]], [[100, 103]], [[72, 105]], [[134, 137]], [[130, 133]], [[126, 129]], [[107, 133]], [[107, 133]], [[144, 195]]]",
    "query_spans": "[[[197, 209]]]",
    "process": "By symmetry, the origin O is the midpoint of AB, and the midpoints of AF and BF are M and N respectively. ∴ OM ∥ BF, ON ∥ AF. ∴ quadrilateral ONFM is a parallelogram. ∵ $\\overrightarrow{OM} \\cdot \\overrightarrow{ON} = 0$, ∴ OM ⊥ ON. ∴ quadrilateral ONFM is a rectangle. ∴ FA ⊥ FB. ∴ OF = OA. Let A(x,y)  \n$$\n\\begin{cases}\n\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 \\\\\ny = \\sqrt{3}x\n\\end{cases}\n$$\nSolving gives  \n$$\n\\begin{cases}\nx^{2} = \\frac{a^{2}b^{2}}{b^{2} - 3a^{2}} \\\\\ny^{2} = \\frac{3a^{2}b^{2}}{b^{2} - 3a^{2}}\n\\end{cases}\n$$\n∴ $\\frac{a^{2}b^{2}}{4a^{2}(c^{2} - a^{2})} \\cdot \\frac{3a^{2}b^{2}}{b^{2} - 3a^{2}} = c^{2}$, i.e., $4a^{2}b^{2} = c^{2}(b^{2} - 3a^{2})$, $b^{2} = c^{2} - a^{2}$, $c^{4} - 8a^{2}c^{2} + 4a^{4} = 0$. Dividing both sides by $a^{4}$ yields $e^{4} - 8e^{2} + 4 = 0$, $e > 1$. ∴ $e^{2} = 4 + 2\\sqrt{3} = (\\sqrt{3} + 1)^{2}$. ∴ $e = \\sqrt{3} + 1$. Hence fill in: $\\sqrt{3} + 1$"
  },
  {
    "text": "If the equation $\\frac{x^{2}}{4}-\\frac{y^{2}}{k}=1$ represents an ellipse, then the range of real values for $k$ is?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G)=(x^2/4 - y^2/k = 1)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-\\infty,-4)+(-4,0)",
    "fact_spans": "[[[40, 42]], [[44, 49]], [[1, 42]]]",
    "query_spans": "[[[44, 56]]]",
    "process": "\\because the equation \\frac{x^2}{4}-\\frac{y^{2}}{k}=1 represents an ellipse \\therefore \\begin{cases}-k>0\\\\-k\\neq4\\end{cases}, solving gives k<0, and k\\neq-4, \\therefore the range of real number k is (-\\infty,-4)\\cup(-4,0)"
  },
  {
    "text": "Given that one asymptote of the hyperbola $x^{2}-\\frac{y^{2}}{\\lambda}=1$ (where $\\lambda>0$ is a constant) is $y=2x$, then $\\lambda=$?",
    "fact_expressions": "G: Hyperbola;lambda: Number;Expression(G) = (x^2 - y^2/lambda = 1);lambda>0;Expression(OneOf(Asymptote(G))) = (y = 2*x)",
    "query_expressions": "lambda",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 36]], [[67, 76]], [[2, 36]], [[39, 50]], [[2, 65]]]",
    "query_spans": "[[[67, 78]]]",
    "process": "From the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$ ($\\lambda>0$), the asymptotes are given by $y=\\pm\\sqrt{2}x$. Since one of the asymptotes is $y=2x$, it follows that $\\sqrt{\\lambda}=2$, solving gives $\\lambda=4$."
  },
  {
    "text": "What are the coordinates of the foci of the ellipse $\\frac{x^{2}}{m-2}+\\frac{y^{2}}{m+5}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(m - 2) + y^2/(m + 5) = 1);m: Number",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, pm*sqrt(7))",
    "fact_spans": "[[[0, 41]], [[0, 41]], [[2, 41]]]",
    "query_spans": "[[[0, 48]]]",
    "process": ""
  },
  {
    "text": "The curve represented by the equation $\\frac{x^{2}}{2-k}-\\frac{y^{2}}{1-k}=1$ is an ellipse. What is the range of real values for $k$?",
    "fact_expressions": "G: Ellipse;H: Curve;k: Real;Expression(H) = (x^2/(2 - k) - y^2/(1 - k) = 1);H=G",
    "query_expressions": "Range(k)",
    "answer_expressions": "{(1,2)&Negation(k=3/2)}",
    "fact_spans": "[[[47, 49]], [[44, 46]], [[51, 56]], [[0, 46]], [[44, 49]]]",
    "query_spans": "[[[51, 63]]]",
    "process": "From the given conditions, we have  \n\\begin{cases}2-k>0\\\\1-k<0\\\\2-k\\neqk-1\\end{cases},  \nsolving this yields $1<k<2$ and $k\\neq\\frac{3}{2}$."
  },
  {
    "text": "Given a hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a line through any point $P$ on the hyperbola parallel to the $x$-axis, intersecting the two asymptotes of the hyperbola at points $A$ and $B$. If $\\overrightarrow{P A} \\cdot \\overrightarrow{P B}=-\\frac{a^{2}}{4}$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;P: Point;A: Point;B: Point;a>0;b>0;Expression(G) = (-x^2/b^2 + y^2/a^2 = 1);PointOnCurve(P,G);PointOnCurve(P,H);IsParallel(xAxis,H);DotProduct(VectorOf(P, A), VectorOf(P, B)) = -a^2/4;L1:Line;L2:Line;Asymptote(G)={L1,L2};Intersection(H,L1)=A;Intersection(H,L2)=B",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[1, 57], [78, 81], [168, 171]], [[4, 57]], [[4, 57]], [[74, 76]], [[62, 65]], [[88, 91]], [[92, 95]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 65]], [[0, 76]], [[66, 76]], [[99, 165]], [-1, -1], [-1, -1], [76, 84], [[74, 97]], [[74, 97]]]",
    "query_spans": "[[[168, 177]]]",
    "process": ""
  },
  {
    "text": "If the distance from point $A(x_{0},-3)$ on the parabola $y^{2}=2 p x(p>0)$ to its focus is twice the distance from $A$ to the $y$-axis, then $p$ equals?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;A: Point;x0: Number;Coordinate(A) = (x0, -3);PointOnCurve(A, G);Distance(A, Focus(G)) = 2*Distance(A, yAxis)",
    "query_expressions": "p",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 22], [39, 40]], [[1, 22]], [[63, 66]], [[4, 22]], [[24, 38], [46, 49]], [[25, 38]], [[24, 38]], [[1, 38]], [[24, 61]]]",
    "query_spans": "[[[63, 69]]]",
    "process": "From the given condition, we have $2x_{0}=x_{0}+\\frac{p}{2}$, solving gives $x_{0}=\\frac{p}{2}$, so $A(\\frac{p}{2},-3)$. Substituting into $y^{2}=2px$ $(p>0)$, we get $(-3)^{2}=2p\\cdot\\frac{p}{2}$. Together with $p>0$, solving yields $p=3$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $A$ lies on the right branch of the hyperbola, and the segment $A F_{1}$ intersects the left branch at point $B$. If $A F_{2} \\perp B F_{2}$ and $|B F_{1}|=\\frac{1}{3}|A F_{2}|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F1: Point;A: Point;F2: Point;B: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(A, RightPart(G));Intersection(LineSegmentOf(A,F1),LeftPart(G))=B;IsPerpendicular(LineSegmentOf(A, F2), LineSegmentOf(B, F2));Abs(LineSegmentOf(B, F1)) = Abs(LineSegmentOf(A, F2))/3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(65)/5",
    "fact_spans": "[[[2, 58], [88, 91], [180, 183]], [[5, 58]], [[5, 58]], [[67, 74]], [[83, 87]], [[75, 82]], [[112, 116]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 82]], [[2, 82]], [[83, 96]], [[88, 116]], [[119, 142]], [[145, 177]]]",
    "query_spans": "[[[180, 189]]]",
    "process": ""
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$. A line perpendicular to the $x$-axis is drawn through $F_{1}$, intersecting the hyperbola at points $M$, $N$. Given $\\cos \\angle M F_{2} N = \\frac{7}{8}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;M: Point;F2: Point;N: Point;F1: Point;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;L:Line;PointOnCurve(F1, L);IsPerpendicular(L,xAxis);Intersection(L, C) = {M, N};Cos(AngleOf(M, F2, N)) = 7/8",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(15)/3",
    "fact_spans": "[[[16, 67], [92, 95], [144, 147]], [[24, 67]], [[24, 67]], [[97, 100]], [[8, 15]], [[101, 104]], [[0, 7], [76, 83]], [[16, 67]], [[0, 74]], [[0, 74]], [], [[75, 91]], [[75, 91]], [[75, 106]], [[107, 142]]]",
    "query_spans": "[[[144, 153]]]",
    "process": "From the given conditions: $ MF_{2} = 2a + \\frac{b^{2}}{a} $, $ \\cos\\alpha = \\frac{2c}{2a + \\frac{b^{2}}{a}} = \\frac{2e}{e^{2}+1} $, $ \\cos\\angle MF_{2}N = \\frac{7}{8} $, we obtain: $ 2\\left(\\frac{2}{e^{2}+1}\\right)^{2} - 1 = \\frac{7}{8} $, which leads to: $ \\sqrt{15}(e^{2}+1) = 8e $, solving gives $ e = \\frac{\\sqrt{15}}{3} $ or $ e = \\frac{\\sqrt{5}}{5} $ (discarded)."
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $\\frac{\\sqrt{3}}{2}$, and the length of the minor axis is $4$. Then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = sqrt(3)/2;Length(MinorAxis(G))=4",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16+y^2/4=1",
    "fact_spans": "[[[0, 52], [87, 89]], [[2, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[0, 52]], [[0, 77]], [[0, 85]]]",
    "query_spans": "[[[87, 94]]]",
    "process": "From the given conditions, we have $e=\\frac{c}{a}=\\frac{\\sqrt{3}}{2}, 2b=4 \\Rightarrow b=2$, $\\therefore \\begin{cases} c=\\frac{\\sqrt{3}}{2} \\\\ a^{2}=b^{2}+c^{2}=4+c^{2} \\end{cases} \\Rightarrow \\begin{cases} a=4 \\\\ c=2\\sqrt{3} \\end{cases}$, $\\therefore$ the standard equation of the ellipse is $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$."
  },
  {
    "text": "The length of the chord cut by the line $y=x$ from the curve $2 x^{2}+y^{2}=2$ is?",
    "fact_expressions": "G: Line;H: Curve;Expression(G) = (y = x);Expression(H) = (2*x^2 + y^2 = 2)",
    "query_expressions": "Length(InterceptChord(H,G))",
    "answer_expressions": "4*sqrt(3)/3",
    "fact_spans": "[[[0, 7]], [[8, 27]], [[0, 7]], [[8, 27]]]",
    "query_spans": "[[[0, 34]]]",
    "process": "Solving the system \\begin{cases}2x^2+y^{2}=2\\\\y=x\\end{cases}, we get $3x^{2}-2=0$, so $x_{1}+x_{2}=0$, $x_{1}x_{2}=-\\frac{2}{3}$, thus the chord length is $l=\\sqrt{1+k^{2}}\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\frac{4\\sqrt{3}}{3}$."
  },
  {
    "text": "If the point $(2,-1)$ lies on the parabola $x=2 p y^{2}$, then the standard equation of this parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x = 2*(p*y^2));p: Number;H: Point;Coordinate(H) = (2, -1);PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=(1/2)*x",
    "fact_spans": "[[[11, 27], [31, 34]], [[11, 27]], [[14, 27]], [[1, 10]], [[1, 10]], [[1, 28]]]",
    "query_spans": "[[[31, 41]]]",
    "process": "From the given condition, we have 2 = 2p \\times (-1)^{2}, solving gives p = 1, then y^{2} = \\frac{1}{2}x, thus the standard equation of the parabola is y^{2} = \\frac{1}{2}x"
  },
  {
    "text": "Given the circle $C$: $(x-1)^{2}+(y-1)^{2}=2$, the ellipse $\\Gamma$: $\\frac{x^{2}}{2}+y^{2}=1$, and a ray $l$ passing through the origin $O$ intersecting the circle $C$ and the ellipse $\\Gamma$ at points $M$ and $N$ respectively, where point $M$ is distinct from point $O$, then the maximum value of $|OM| \\cdot |ON|$ is?",
    "fact_expressions": "Gamma: Ellipse;C: Circle;O: Origin;M: Point;N: Point;l:Ray;Expression(Gamma)=(x^2/2 + y^2 = 1);Expression(C) = ((x - 1)^2 + (y - 1)^2 = 2);PointOnCurve(O, l);Intersection(l,C)=M;Intersection(l,Gamma)=N;Negation(M=O)",
    "query_expressions": "Max(Abs(LineSegmentOf(O, M))*Abs(LineSegmentOf(O, N)))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[31, 68], [90, 100]], [[2, 30], [85, 89]], [[71, 76], [119, 123]], [[102, 105], [112, 116]], [[106, 109]], [[77, 82]], [[31, 68]], [[2, 30]], [[70, 82]], [[77, 111]], [[77, 111]], [[112, 123]]]",
    "query_spans": "[[[126, 150]]]",
    "process": "Let the equation of the ray be $ y = kx $. By simultaneously solving the equations of the line with the ellipse and the circle, and combining with the chord length formula, we obtain the expression for $ |OM| \\cdot |ON| $ in terms of $ k $:  \n$ |OM| \\cdot |ON| = 2\\sqrt{\\frac{2(1+k)^{2}}{1+2k^{2}}} $.  \nThen, by substituting $ t = 1 + k $ and using the maximum value problem of a quadratic function, we can find the maximum value of $ |OM| \\cdot |ON| $.  \n\n**Detailed Solution:** Let the equation of the ray be $ y = kx $.  \nSolving simultaneously  \n$$\n\\begin{cases}\ny = kx \\\\\nx^{2} + 2y^{2} = 2\n\\end{cases}\n$$  \ngives $ x_{N} = \\pm\\sqrt{\\frac{2}{1+2k^{2}}} $.  \nSolving simultaneously  \n$$\n\\begin{cases}\ny = kx, \\\\\n(x-1)^{2} + (y-1)^{2} = 2\n\\end{cases}\n$$  \ngives $ x_{M} = \\frac{2+2k}{1+k^{2}} $.  \nTherefore,  \n$ |OM| \\cdot |ON| = \\sqrt{1+k^{2}}|x_{M}| \\times \\sqrt{1+k^{2}}|x_{N}| = 2\\sqrt{\\frac{2(1+k)^{2}}{1+2k^{2}}} $.  \nLet $ \\frac{2 - 4t + 3}{2} = \\frac{3}{2t^{2}} - \\frac{2}{t} + 1 = \\frac{3}{2}\\left(\\frac{1}{t} - \\frac{2}{3}\\right)\\frac{(1+k)^{2}}{1+2k^{2}} $, set $ t = 1 + k $, then  \n$ 2\\sqrt{\\frac{2(1+k)}{1+2k^{2}}} \\leqslant 2\\sqrt{3} $, equality holds when $ t = \\frac{3}{2} $, i.e., $ k = \\frac{1}{2} $. Hence, the maximum value of $ |OM| \\cdot |ON| $ is $ 2\\sqrt{3} $."
  },
  {
    "text": "The focus of the parabola $C$: $y^{2}=2 p x(p>0)$ is $A$, and the intersection point of its directrix with the $x$-axis is $B$. If there exists a point $M$ on the line $3 x+4 y+25=0$ such that $\\angle A M B=90^{\\circ}$, then the range of values for the real number $p$ is?",
    "fact_expressions": "C: Parabola;p: Real;G: Line;A: Point;M: Point;B: Point;p>0;Expression(C) = (y^2 = 2*p*x);Expression(G) = (3*x + 4*y + 25 = 0);Focus(C) = A;Intersection(Directrix(C), xAxis) = B;PointOnCurve(M, G);AngleOf(A, M, B) = ApplyUnit(90, degree)",
    "query_expressions": "Range(p)",
    "answer_expressions": "[10,+\\infty)",
    "fact_spans": "[[[0, 26], [34, 35]], [[106, 111]], [[53, 69]], [[30, 33]], [[72, 76]], [[46, 49]], [[8, 26]], [[0, 26]], [[53, 69]], [[0, 33]], [[34, 49]], [[53, 76]], [[79, 104]]]",
    "query_spans": "[[[106, 118]]]",
    "process": "From the given, $ A(\\frac{p}{2},0), B(-\\frac{p}{2},0) $. Since $ M $ lies on the line $ 3x + 4y + 25 = 0 $, let point $ M(x, \\frac{-3x - 25}{4}) $. Thus, $ \\overrightarrow{AM} = (x - \\frac{p}{2}, \\frac{-3x - 25}{4}) $, $ \\overrightarrow{BM} = (x + \\frac{p}{2}, \\frac{-3x - 25}{4}) $; also $ \\angle AMB = 90^{\\circ} $, so $ \\overrightarrow{AM} \\cdot \\overrightarrow{BM} = (x - \\frac{p}{2})(x + \\frac{p}{2}) + (\\frac{-3x - 25}{4})^2 = 0 $, which simplifies to $ 25x^{2} + 150x + 625 - 4p^{2} = 0 $; thus $ \\Delta \\geqslant 0 $, i.e., $ 150^{2} - 4 \\times 25 \\times (625 - 4p^{2}) \\geqslant 0 $, solving gives $ p \\geqslant 10 $ or $ p \\leqslant -10 $, and since $ p > 0 $, the range of $ p $ is $ [10, +\\infty) $."
  },
  {
    "text": "If the line $y=2x$ and the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ have no common points, then the range of the eccentricity of the hyperbola is?",
    "fact_expressions": "H: Line;Expression(H) = (y = 2*x);G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;NumIntersection(H, G) = 0",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1, \\sqrt{5}]",
    "fact_spans": "[[[1, 10]], [[1, 10]], [[11, 67], [75, 78]], [[11, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[1, 72]]]",
    "query_spans": "[[[75, 88]]]",
    "process": "Since the line y=2x and the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0) have no common points, analyze to obtain \\frac{b}{a}\\leqslant2, then find the range of e. [Detailed solution] \\because the asymptotes of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0) are y=\\pm\\frac{b}{a}x, and the line y=2x has no common points with the hyperbola, \\therefore \\frac{b}{a}\\leqslant2, i.e., e=\\sqrt{1+\\frac{b^{2}}{a^{2}}}\\leqslant\\sqrt{5}, and e>1,"
  },
  {
    "text": "A line passing through the focus of the parabola $x^{2}=2 y$ intersects the parabola at points $A$ and $B$. If the ordinate of the midpoint of segment $AB$ is $4$, then what is the length of segment $AB$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 2*y);H: Line;PointOnCurve(Focus(G), H) = True;A: Point;B: Point;Intersection(H, G) = {A, B};YCoordinate(MidPoint(LineSegmentOf(A,B))) = 4",
    "query_expressions": "Length(LineSegmentOf(A,B))",
    "answer_expressions": "9",
    "fact_spans": "[[[1, 15], [21, 24]], [[1, 15]], [[18, 20]], [[0, 20]], [[25, 28]], [[29, 32]], [[18, 34]], [[36, 53]]]",
    "query_spans": "[[[55, 67]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), then \\frac{y_{1}+y_{2}}{2}=4, that is, y_{1}+y_{2}=8, \\therefore|AB|=y_{1}+y_{2}+p=8+1=9."
  },
  {
    "text": "The left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are $F_{1}$ and $F_{2}$ respectively, and $P$ is a point on the right branch of the hyperbola. $PF_{2}$ is tangent to the circle $x^{2}+y^{2}=b^{2}$ at point $G$, and $G$ is the midpoint of $PF_{2}$. Then the eccentricity $e$ of this hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;H: Circle;P: Point;F2: Point;F1: Point;G: Point;e: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P,RightPart(C));Expression(H) = (x^2 + y^2 = b^2);TangentPoint(LineSegmentOf(P, F2), H) = G;MidPoint(LineSegmentOf(P, F2)) = G;Eccentricity(C) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[0, 56], [85, 88], [149, 152]], [[3, 56]], [[3, 56]], [[103, 123]], [[81, 84]], [[73, 80]], [[65, 72]], [[125, 129], [131, 134]], [[156, 159]], [[3, 56]], [[3, 56]], [[0, 56]], [[0, 80]], [[0, 80]], [[81, 93]], [[103, 123]], [[94, 129]], [[131, 146]], [[149, 159]]]",
    "query_spans": "[[[156, 161]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$ and directrix $l$, $P$ is a point on $l$, $Q$ is an intersection point of line $PF$ and $C$, and if $\\overrightarrow{FP}=4\\overrightarrow{FQ}$, then $|QO|=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 8*x);F: Point;Focus(C) = F;l: Line;Directrix(C) = l;P: Point;PointOnCurve(P, l);Q: Point;OneOf(Intersection(LineOf(P, F), C)) = Q;O: Origin;VectorOf(F, P) = 4*VectorOf(F, Q)",
    "query_expressions": "Abs(LineSegmentOf(Q, O))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 21], [60, 63]], [[2, 21]], [[25, 28]], [[2, 28]], [[32, 35], [41, 44]], [[2, 35]], [[37, 40]], [[37, 47]], [[48, 51]], [[48, 68]], [[117, 126]], [[70, 115]]]",
    "query_spans": "[[[117, 128]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $E$: $y^{2}=2 p x(p>0)$ has focus $F$. A line $l$ with slope $\\sqrt{3}$ passing through $F$ intersects the parabola $E$ at points $A$ and $B$. If $||A F|-| B F||=4$, then $p=$?",
    "fact_expressions": "l: Line;E: Parabola;p: Number;B: Point;F: Point;A: Point;p>0;Expression(E) = (y^2 = 2*p*x);Focus(E) = F;PointOnCurve(F, l);Slope(l) = sqrt(3);Intersection(l, E) = {A, B};Abs(Abs(LineSegmentOf(A,F))-Abs(LineSegmentOf(B,F)))=4",
    "query_expressions": "p",
    "answer_expressions": "3",
    "fact_spans": "[[[55, 60]], [[2, 28], [61, 67]], [[100, 103]], [[72, 75]], [[32, 35], [32, 35]], [[68, 71]], [[10, 28]], [[2, 28]], [[2, 35]], [[36, 60]], [[41, 60]], [[55, 77]], [[80, 98]]]",
    "query_spans": "[[[100, 105]]]",
    "process": "By the given condition, ||AF| - |BF|| = |x_{1} + \\frac{p}{2} - (x_{2} + \\frac{p}{2})| = |x_{1} - x_{2}| = 4. By solving the line and the parabola simultaneously, we obtain x_{1} = \\frac{p}{6}, x_{2} = \\frac{3p}{2}. Substituting into the equation gives the solution. According to the problem, the focus F(\\frac{p}{2}, 0), assume the line l: y = \\sqrt{3}(x - \\frac{p}{2}), A(x_{1}, y_{1}), B(x_{2}, y_{2}). Thus, ||AF| - |BF|| = |x_{1} + \\frac{p}{2} - (x_{2} + \\frac{p}{2})| = |x_{1} - x_{2}| = 4. Solving the system \\begin{cases} y^{2} = 2px \\\\ y = \\sqrt{3}(x - \\frac{p}{2}) \\end{cases}, we get 3x^{2} - 5px + \\frac{3p^{2}}{4} = (3x - \\frac{p}{2})(x - \\frac{3p}{2}) = 0. Solving yields: x_{1} = \\frac{p}{6}, x_{2} = \\frac{3p}{2}. Hence, |x_{1} - x_{2}| = \\frac{8p}{6} = 4, solving gives: p = 3"
  },
  {
    "text": "If $a>1$, then the range of the eccentricity $e$ of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2 + x^2/a^2 = 1);a: Number;e: Number;a > 1;Eccentricity(G) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "(1, \\sqrt{2})",
    "fact_spans": "[[[8, 40]], [[8, 40]], [[1, 6]], [[44, 47]], [[1, 6]], [[8, 47]]]",
    "query_spans": "[[[44, 54]]]",
    "process": "The eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ is $e=\\frac{c}{a}=\\frac{\\sqrt{a^{2}+1}}{a}=\\sqrt{1+\\frac{1}{a^{2}}}$. Given $a>1$, it follows that $1<e<\\sqrt{2}$."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. A line passing through point $F_{1}$ intersects the ellipse $E$ at points $A$ and $B$, with $|A F_{1}|=3|B F_{1}|$. If $\\cos \\angle A F_{2} B=\\frac{3}{5}$, then the eccentricity of ellipse $E$ is?",
    "fact_expressions": "E: Ellipse;b: Number;a: Number;G: Line;A: Point;F1: Point;B: Point;F2: Point;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(E) = F1;RightFocus(E) = F2;PointOnCurve(F1, G);Intersection(G, E) = {A, B};Abs(LineSegmentOf(A, F1)) = 3*Abs(LineSegmentOf(B, F1));Cos(AngleOf(A, F2, B)) = 3/5",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[19, 76], [96, 101], [173, 178]], [[26, 76]], [[26, 76]], [[93, 95]], [[102, 105]], [[1, 8], [84, 92]], [[106, 109]], [[9, 16]], [[26, 76]], [[26, 76]], [[19, 76]], [[1, 82]], [[1, 82]], [[83, 95]], [[93, 111]], [[112, 134]], [[136, 171]]]",
    "query_spans": "[[[173, 184]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{3}=1$ is $2$, then $m$=?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (-y^2/3 + x^2/m = 1);Eccentricity(G) = 2",
    "query_expressions": "m",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 38]], [[48, 51]], [[0, 38]], [[0, 46]]]",
    "query_spans": "[[[48, 53]]]",
    "process": ""
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=2 x$ is $F$, point $P$ is a moving point on the parabola, and there is a point $A(3 , 2)$, find the minimum value of $|P A|+|P F|$.",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 2*x);Coordinate(A) = (3, 2);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "7/2",
    "fact_spans": "[[[2, 16], [29, 32]], [[39, 50]], [[24, 28]], [[20, 23]], [[2, 16]], [[39, 50]], [[2, 23]], [[24, 36]]]",
    "query_spans": "[[[52, 70]]]",
    "process": "Substitute $x=3$ into the parabola equation $y^{2}=2x$, solving gives $y=\\pm\\sqrt{6}$. Since $\\sqrt{6}>2$, point $A(3,2)$ lies inside the opening of the parabola. The focus of the parabola is $F\\left(\\frac{1}{2},0\\right)$, and the directrix is $l: x=-\\frac{1}{2}$. Let $d$ be the distance from a point on the parabola to the directrix $l$, then $|PA|+|PF|=|PA|+d$. As shown in the figure, $|PA|+d$ is minimized when $PA\\perp l$, and the minimum value is $3-\\left(-\\frac{1}{2}\\right)=\\frac{7}{2}$. Therefore, the minimum value of $|PA|+|PF|$ is $\\frac{7}{2}$."
  },
  {
    "text": "The coordinates of the focus of the parabola $y=4 x^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 4*x^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, 1/16)",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": "From $ y = 4x^2 $, we get $ x^{2} = \\frac{1}{4}y $, so the focus of the parabola lies on the $ y $-axis, and $ 2p = \\frac{1}{4} $, $ \\frac{p}{2} = \\frac{1}{16} $. Therefore, the coordinates of the focus of the parabola are $ \\left(0, \\frac{1}{16}\\right) $."
  },
  {
    "text": "Given that the eccentricity of the ellipse $x^{2}+m y^{2}=1$ is $\\frac{\\sqrt{3}}{2}$, then $m$=?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (m*y^2 + x^2 = 1);Eccentricity(G) = sqrt(3)/2",
    "query_expressions": "m",
    "answer_expressions": "{4, 1/4}",
    "fact_spans": "[[[1, 20]], [[47, 50]], [[1, 20]], [[1, 45]]]",
    "query_spans": "[[[47, 52]]]",
    "process": "Analysis: Converting the ellipse into standard form gives $x^{2}+\\frac{y^{2}}{m}=1$, from which we obtain $a^{2}=1, b^{2}=\\frac{1}{m}$ or $a^{2}=\\frac{1}{m}, b^{2}=1$. Given that the eccentricity of the ellipse is $\\frac{\\sqrt{3}}{2}$, we have $e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{a^{2}-b^{2}}{a^{2}}=\\frac{3}{4}$, simplifying to $a^{2}=4b^{2}$. Thus, $\\frac{1}{m}=4$ or $\\frac{4}{m}=1$, so $m=4$ or $\\frac{1}{4}$. Converting the ellipse into standard form gives $x^{2}+\\frac{y^{2}}{m}=1$. Since the eccentricity of the ellipse is $\\frac{\\sqrt{3}}{2}$, $\\therefore e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{a^{2}-b^{2}}{a^{2}}=\\frac{3}{4}$, $a^{2}=4b^{2}$, $\\therefore \\frac{1}{m}=4$ or $\\frac{4}{m}=1$. Hence $m=4$ or $\\frac{1}{4}$."
  },
  {
    "text": "Given that the line $l$: $y = 4x + m$, passing through the focus of the parabola $C$: $y^2 = 2px$ $(p > 0)$, intersects the parabola $C$ at points $A$ and $B$, and $|AB| = \\frac{17}{4}$, then $p = $?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;l: Line;Expression(l) = (y=4*x+m);m: Number;Intersection(l, C) = {A, B};PointOnCurve(Focus(C), l) = True;A: Point;B: Point;Abs(LineSegmentOf(A, B)) = 17/4",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[3, 31]], [[3, 31]], [[94, 97]], [[11, 31]], [[35, 51]], [[35, 51]], [[42, 51]], [[35, 70]], [[2, 51]], [[61, 64]], [[65, 68]], [[72, 92]]]",
    "query_spans": "[[[94, 99]]]",
    "process": ""
  },
  {
    "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ are tangent to the circle $(x-2)^{2} + y^{2}=3$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;H: Circle;Expression(H) = (y^2 + (x - 2)^2 = 3);IsTangent(Asymptote(G), H) = True",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 47], [79, 82]], [[1, 47]], [[4, 47]], [[4, 47]], [[52, 74]], [[52, 74]], [[1, 76]]]",
    "query_spans": "[[[79, 88]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the foci of the conic section $x^{2}+\\frac{y^{2}}{10}=1$ are?",
    "fact_expressions": "H: ConicSection;Expression(H) = (x^2 + y^2/10 = 1)",
    "query_expressions": "Coordinate(Focus(H))",
    "answer_expressions": "(0,pm*3)",
    "fact_spans": "[[[0, 30]], [[0, 30]]]",
    "query_spans": "[[[0, 37]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{3}=1$ has left focus $F$, point $A$ is any point on the ellipse distinct from the vertices, $O$ is the origin, and point $B$ is the midpoint of segment $AF$. Then the perimeter of $\\triangle F O B$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/6 + y^2/3 = 1);F: Point;LeftFocus(G) = F;PointOnCurve(A, G) = True;Negation(A = Vertex(G));A: Point;O: Origin;MidPoint(LineSegmentOf(A, F)) = B;B: Point",
    "query_expressions": "Perimeter(TriangleOf(F, O, B))",
    "answer_expressions": "sqrt(3) + sqrt(6)",
    "fact_spans": "[[[2, 39], [53, 55]], [[2, 39]], [[44, 47]], [[2, 47]], [[48, 65]], [[48, 65]], [[48, 52]], [[66, 69]], [[77, 92]], [[77, 81]]]",
    "query_spans": "[[[94, 116]]]",
    "process": "From the ellipse $\\frac{x^2}{6} + \\frac{y^{2}}{3} = 1$, we obtain $a = \\sqrt{6}$, $b = \\sqrt{3}$, $c = \\sqrt{3}$. Let $E$ be the right focus of the ellipse, connect $AE$, then $|AE| + |AF| = 2a = 2\\sqrt{6}$. Since $O$ is the midpoint of $EF$, and point $B$ is the midpoint of segment $AF$, it follows that $|BO| = \\frac{1}{2}|AE|$. Therefore, $|BF| + |BO| = \\frac{1}{2}|AE| + \\frac{1}{2}|AF| = a$. Thus, the perimeter of $\\triangle FOB$ is: $|BF| + |BO| + |OF| = a + c = \\sqrt{3} + \\sqrt{6}$."
  },
  {
    "text": "The equation of the directrix of the parabola $y^{2}=8 x$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=-2",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": ""
  },
  {
    "text": "If the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the circle $x^{2}+y^{2}=\\left(\\frac{b}{2}+c\\right)^{2}$ ($c$ is the semi-focal distance of the ellipse) have four distinct intersection points, then what is the range of values for the eccentricity $e$ of the ellipse?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Circle;c: Number;e: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + y^2 = (b/2 + c)^2);Eccentricity(G) = e;HalfFocalLength(G)=c;NumIntersection(G,H)=4",
    "query_expressions": "Range(e)",
    "answer_expressions": "(\\sqrt{5}/5, 3/5)",
    "fact_spans": "[[[1, 53], [93, 95], [111, 113]], [[3, 53]], [[3, 53]], [[54, 88]], [[89, 92]], [[117, 120]], [[3, 53]], [[3, 53]], [[1, 53]], [[54, 88]], [[111, 120]], [[89, 99]], [[1, 109]]]",
    "query_spans": "[[[117, 127]]]",
    "process": "Since the ellipse and the circle have four foci, the diameter of the circle $2r = b + 2c$ lies between the lengths of the major and minor axes of the ellipse, that is, $2b < b + 2c < 2a$. From $2b < b + 2c$, we get $b < 2c$; squaring both sides and simplifying yields $\\frac{1}{5} < \\frac{c^{2}}{a^{2}} < 1$, i.e., $\\frac{\\sqrt{5}}{5} < e < 1^{\\textcircled{1}}$. From $b + 2c < 2a$, we get $b < 2a - 2c$; squaring both sides and simplifying gives $5e^{2} - 8e + 3 > 0$, solving which yields $0 < e < \\frac{3}{5}\\textcircled{2}$. Combining $\\textcircled{1}$ and $\\textcircled{2}$, we obtain $\\frac{\\sqrt{5}}{5} < e < \\frac{3}{5}$. Hence, fill in $\\frac{\\sqrt{5}}{5}, \\frac{3}{5}$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, a line $l$ passing through point $F_{1}$ intersects the left branch of hyperbola $E$ at points $P$ and $Q$. If $|P F_{1}|=3|F_{1} Q|$ and $F_{2} Q \\perp P Q$, then the eccentricity of $E$ is?",
    "fact_expressions": "l: Line;E: Hyperbola;b: Number;a: Number;F2: Point;Q: Point;P: Point;F1: Point;a>0;b>0;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(E) = F1;RightFocus(E) = F2;PointOnCurve(F1, l);Intersection(l, LeftPart(E)) = {P, Q};Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(F1, Q));IsPerpendicular(LineSegmentOf(F2, Q), LineSegmentOf(P, Q))",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(10)/2",
    "fact_spans": "[[[95, 100]], [[18, 79], [101, 107], [168, 171]], [[26, 79]], [[26, 79]], [[10, 17]], [[116, 119]], [[112, 115]], [[2, 9], [86, 94]], [[26, 79]], [[26, 79]], [[18, 79]], [[2, 84]], [[2, 84]], [[85, 100]], [[95, 121]], [[123, 145]], [[147, 166]]]",
    "query_spans": "[[[168, 177]]]",
    "process": "Let $|QF_{1}|=m$. According to the problem, $|PF_{1}|=3m$. By the definition of a hyperbola, we have: $|PF_{2}|=2a+3m$, $|QF_{2}|=2a+m$. Since $F_{2}Q\\perp PQ$, as shown in the figure, in right triangle $\\triangle PQF_{2}$, $PQ^{2}+QF_{2}^{2}=PF_{2}^{2}$, that is $(4m)^{2}+(2a+m)^{2}=(2a+3m)^{2}$. Simplifying and rearranging yields: $m=a$. Thus, $|QF_{1}|=a$, $|QF_{2}|=3a$. In right triangle $\\triangle QF_{1}F_{2}$, $|F_{1}F_{2}|=2c$, $QF_{1}^{2}+QF_{2}^{2}=F_{1}F_{2}^{2}$, then $a^{2}+(3a)^{2}=(2c)^{2}$, i.e., $e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{5}{2}$, solving gives $e=\\frac{\\sqrt{10}}{2}$. Therefore, the eccentricity of $E$ is $\\frac{\\sqrt{10}}{2}$."
  },
  {
    "text": "Given $P(2 , 1)$, $Q(3 ,-2)$, the standard equation of the hyperbola passing through points $P$ and $Q$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;Q: Point;Coordinate(P) = (2, 1);Coordinate(Q) = (3, -2);PointOnCurve(P, G);PointOnCurve(Q, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[37, 40]], [[2, 12], [27, 30]], [[14, 24], [31, 34]], [[2, 12]], [[14, 24]], [[25, 40]], [[25, 40]]]",
    "query_spans": "[[[37, 47]]]",
    "process": ""
  },
  {
    "text": "The asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is parallel to the line $x-y+3=0$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;H: Line;Expression(H) = (x - y + 3 = 0);IsParallel(OneOf(Asymptote(G)),H) = True",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 56], [79, 82]], [[0, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[63, 74]], [[63, 74]], [[0, 76]]]",
    "query_spans": "[[[79, 88]]]",
    "process": "From the given conditions, $\\frac{b}{a}=1\\Rightarrow b=a$, $c=\\sqrt{2}a$, $e=\\sqrt{2}$"
  },
  {
    "text": "The standard equation of an ellipse centered at the origin, with major axis length $8$ and directrix equations $x = \\pm 8$ is?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;Length(MajorAxis(G)) = 8;Expression(Directrix(G)) = (x = pm*8)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16+y^2/12=1",
    "fact_spans": "[[[29, 31]], [[3, 5]], [[0, 31]], [[6, 31]], [[14, 31]]]",
    "query_spans": "[[[29, 37]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, its two foci are $F_{1}$ and $F_{2}$ respectively, and $P$ is a point on the hyperbola such that $\\angle F_{1} P F_{2}=\\frac{\\pi}{2}$. Then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/16 - y^2/9 = 1);AngleOf(F1, P, F2) = pi/2;Focus(G) = {F1, F2};PointOnCurve(P, G)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "9",
    "fact_spans": "[[[2, 41], [69, 72]], [[49, 56]], [[65, 68]], [[57, 64]], [[2, 41]], [[77, 113]], [[2, 64]], [[65, 75]]]",
    "query_spans": "[[[115, 142]]]",
    "process": "According to the problem, the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ has foci $F_{1}(-5,0)$, $F_{2}(5,0)$, and $||PF_{1}|-|PF_{2}||=8$. Since $\\angle F_{1}PF_{2}=\\frac{\\pi}{2}$, it follows that $|F_{1}F_{2}|^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}=(|PF_{1}|-|PF_{2}|)^{2}+2|PF_{1}||PF_{2}|$, so $2|PF_{1}||PF_{2}|=10^{2}-8^{2}=36$. Solving gives $|PF_{1}||PF_{2}|=18$, therefore the area of $\\triangle F_{1}PF_{2}$ is $S=\\frac{1}{2}|PF_{1}||PF_{2}|=9$."
  },
  {
    "text": "Let the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ intersect the line $y=x$ at points $M$ and $N$. If there exists a point $P$ on the ellipse such that the product of the slopes of lines $MP$ and $NP$ is $-\\frac{4}{9}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;H: Line;Expression(H) = (y = x);M: Point;N: Point;Intersection(G, H) = {M, N};P: Point;PointOnCurve(P, G);Slope(LineOf(M, P))*Slope(LineOf(N, P)) = -4/9",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[1, 53], [76, 78], [123, 125]], [[1, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[54, 61]], [[54, 61]], [[64, 67]], [[68, 71]], [[1, 73]], [[81, 85]], [[76, 85]], [[88, 121]]]",
    "query_spans": "[[[123, 131]]]",
    "process": "The foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ lie on the $x$-axis. Let $P(x,y)$, $M(m,m)$, $N(-m,-m)$. Then the slopes of lines $MP$ and $NP$ are $\\frac{y-m}{x-m}$ and $\\frac{y+m}{x+m}$ respectively. Since the product of the slopes of lines $MP$ and $NP$ is $-\\frac{4}{9}$, we have $\\frac{y-m}{x-m} \\cdot \\frac{y+m}{x+m} = -\\frac{4}{9}$, that is, $\\frac{y^{2}-m^{2}}{x^{2}-m^{2}} = -\\frac{4}{9}$. Since $M$ and $P$ are points on the ellipse $C$, it follows that $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, $\\frac{m^{2}}{a^{2}}+\\frac{m^{2}}{b^{2}}=1$, so $\\frac{y^{2}-m^{2}}{x^{2}-m^{2}} = -\\frac{b^{2}}{a^{2}}$, hence $\\frac{b^{2}}{a^{2}} = \\frac{4}{9}$. Therefore, the eccentricity of the ellipse is $e = \\frac{c}{a} = \\sqrt{1-\\frac{b^{2}}{a^{2}}} = \\sqrt{1-\\frac{4}{9}} = \\frac{\\sqrt{5}}{3}$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, the focal distance is $2c$, and a line $y=-\\sqrt{3}(x-c)$ intersects the hyperbola at a point $P$ such that $\\angle P F_{2} F_{1}=2 \\angle P F_{1} F_{2}$. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;c: Number;FocalLength(C) = 2*c;G: Line;Expression(G) = (y = -sqrt(3)*(-c + x));P: Point;OneOf(Intersection(G, C)) = P;AngleOf(P, F2, F1) = 2*AngleOf(P, F1, F2)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)+1",
    "fact_spans": "[[[2, 63], [118, 121], [178, 181]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[72, 79]], [[80, 87]], [[2, 87]], [[2, 87]], [[91, 96]], [[2, 96]], [[97, 117]], [[97, 117]], [[126, 129]], [[97, 129]], [[131, 176]]]",
    "query_spans": "[[[178, 187]]]",
    "process": "As shown in the figure, the slope of line $ PF_{2} $ is $ k = -\\sqrt{3} $, so the corresponding inclination angle of the line is $ 120^{\\circ} $, that is, $ \\angle PF_{2}F_{1} = 60^{\\circ} $, then $ \\angle PF_{2}F_{1} = 2\\angle PF_{1}F_{2} = 60^{\\circ} $, $ \\angle PF_{1}F_{2} = 30^{\\circ} $, i.e., $ \\angle F_{1}PF_{2} = 90^{\\circ} $, $ \\therefore |PF_{2}| = c $, $ |PF_{1}| = \\sqrt{3}c $. From the definition of hyperbola, we have: $ ||PF_{1}| - |PF_{2}|| = 2a $, i.e., $ \\sqrt{3}c - c = 2a $, thus $ e = \\frac{c}{a} = \\frac{2(\\sqrt{3}+1)}{2} = \\sqrt{3}+1 $. Therefore, the eccentricity of the hyperbola is $ e = \\sqrt{3}+1 $."
  },
  {
    "text": "The vertex of the parabola is $O$, the focus is $F$, and $M$ is a moving point on the parabola. Then the maximum value of $\\frac{M O}{M F}$ is?",
    "fact_expressions": "G: Parabola;O: Origin;Vertex(G) = O;F: Point;Focus(G) = F;M: Point;PointOnCurve(M, G)",
    "query_expressions": "Max(LineSegmentOf(M, O)/LineSegmentOf(M, F))",
    "answer_expressions": "(2*sqrt(3))/3",
    "fact_spans": "[[[0, 3], [21, 24]], [[6, 9]], [[0, 9]], [[13, 16]], [[0, 16]], [[17, 20]], [[17, 28]]]",
    "query_spans": "[[[30, 53]]]",
    "process": ""
  },
  {
    "text": "If $A$ and $B$ are any two points on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, satisfying $\\overrightarrow{A O}=\\frac{1}{2} \\overrightarrow{A B}$, where $O$ is the origin, and point $M$ is any point on the hyperbola distinct from $A$ and $B$, such that the product of the slopes of lines $MA$ and $MB$ is $\\frac{1}{2}$, then the asymptotes of the hyperbola are?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;a > 0;b > 0;Expression(C) = (x^2/a^2 - y^2/b^2 = 1);A: Point;B: Point;PointOnCurve(A, C);PointOnCurve(B, C);VectorOf(A, O) = 1/2*VectorOf(A, B);O: Origin;M: Point;PointOnCurve(M, C);Negation(M = A);Negation(M = B);Slope(LineOf(M, A)) * Slope(LineOf(M, B)) = 1/2",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y = pm*sqrt(2)/2*x",
    "fact_spans": "[[[9, 67], [150, 153], [202, 205]], [[12, 67]], [[12, 67]], [[12, 67]], [[12, 67]], [[9, 67]], [[1, 4], [157, 160]], [[5, 8], [161, 164]], [[1, 73]], [[1, 73]], [[77, 132]], [[135, 138]], [[144, 148]], [[144, 169]], [[144, 169]], [[144, 169]], [[171, 200]]]",
    "query_spans": "[[[202, 213]]]",
    "process": ""
  },
  {
    "text": "Let $k \\in R$, the moving line $k x + y = 0$ passing through fixed point $A$ and the moving line $x - k y + 2 k = 0$ passing through fixed point $B$ intersect at point $M(x, y)$ $(x > 0)$. If $M B = 2 M A$, then the coordinates of point $M$ are?",
    "fact_expressions": "L1: Line;L2:Line;k: Real;M: Point;B: Point;A: Point;x1:Number;y1:Number;x1>0;Expression(L1) = (k*x + y = 0);Expression(L2) = (2*k - k*y + x = 0);Coordinate(M) = (x1, y1);PointOnCurve(A,L1);PointOnCurve(B,L2);LineSegmentOf(M, B) = 2*LineSegmentOf(M, A);Intersection(L1,L2)=M",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(4/5, 2/5)",
    "fact_spans": "[[[21, 30]], [[41, 54]], [[1, 10]], [[86, 90], [56, 71]], [[34, 37]], [[14, 17]], [[57, 71]], [[57, 71]], [[57, 71]], [[21, 30]], [[41, 54]], [[56, 71]], [[11, 30]], [[31, 54]], [[73, 84]], [[21, 71]]]",
    "query_spans": "[[[86, 95]]]",
    "process": "A(0,0), B(0,2), and two moving lines are perpendicular to each other, i.e., MA\\botMB, so \\begin{cases}(x,y)\\cdot(x,y-2)=0\\\\2\\sqrt{x^{2}+y^{2}}=\\sqrt{x^{2}+(y-2)^{2}}\\end{cases}\\Rightarrow\\begin{cases}x^2+y^2-2y=0\\\\3x^{2}+3y^{2}+4y-4=0\\end{cases}\\begin{cases}y=\\frac{2}{5}\\\\x=\\frac{4}{5}(\\because x>0)\\end{cases}, thus the coordinates of point M are (\\frac{4}{5},\\frac{2}{5})"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{7}=1$, a line $l$ passes through its left focus $F_{1}$, intersecting the left branch of the hyperbola at points $A$ and $B$, with $|A B|=4$. $F_{2}$ is the right focus of the hyperbola, and the perimeter of $\\triangle A B F_{2}$ is $20$. Find the value of $m$.",
    "fact_expressions": "l: Line;G: Hyperbola;Expression(G) = (-y^2/7 + x^2/m = 1);m: Number;A: Point;B: Point;F2: Point;F1: Point;LeftFocus(G)=F1;PointOnCurve(F1,l);Intersection(l,LeftPart(G))={A,B};Abs(LineSegmentOf(A, B)) = 4;Perimeter(TriangleOf(A, B, F2)) = 20;RightFocus(G)=F2",
    "query_expressions": "m",
    "answer_expressions": "9",
    "fact_spans": "[[[41, 46]], [[2, 40], [47, 48], [60, 63], [96, 99]], [[2, 40]], [[135, 138]], [[66, 69]], [[70, 73]], [[88, 95]], [[51, 58]], [[47, 58]], [[41, 58]], [[41, 75]], [[77, 86]], [[104, 133]], [[88, 103]]]",
    "query_spans": "[[[135, 142]]]",
    "process": "From the given condition, |AB| + |AF_{2}| + |BF_{2}| = 20. Since |AB| = 4, it follows that |AF_{2}| + |BF_{2}| = 16. According to the definition of a hyperbola, 2a = |AF_{2}| - |AF_{1}| = |BF_{2}| - |BF_{1}|, so 4a = |AF_{2}| + |BF_{2}| - (|AF_{1}| + |BF_{1}|) = 16 - 4 = 12, thus a = 3, and therefore m = a^{2} = 9"
  },
  {
    "text": "Given that $P$ is a point on the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, lines $l_{1}$ and $l_{2}$ are drawn through point $P$ parallel to the two asymptotes of the hyperbola, intersecting the $x$-axis at points $M$ and $N$, respectively. Then $|O M| \\cdot|O N|$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/5 = 1);P: Point;PointOnCurve(P, G);l1: Line;l2: Line;Z1: Line;Z2: Line;Asymptote(G) = {Z1, Z2};PointOnCurve(P, l1);PointOnCurve(P, l2);IsParallel(Z1, l1);IsParallel(Z2, l2);M: Point;N: Point;Intersection(l1, xAxis) = M;Intersection(l2, xAxis) = N;O: Origin",
    "query_expressions": "Abs(LineSegmentOf(O, M))*Abs(LineSegmentOf(O, N))",
    "answer_expressions": "4",
    "fact_spans": "[[[6, 44], [54, 57]], [[6, 44]], [[2, 5], [49, 53]], [[2, 47]], [[65, 72], [82, 91]], [[74, 81], [93, 100]], [], [], [[54, 61]], [[48, 81]], [[48, 81]], [[53, 81]], [[53, 81]], [[108, 111]], [[112, 115]], [[82, 117]], [[82, 117]], [[119, 137]]]",
    "query_spans": "[[[119, 139]]]",
    "process": "The slopes of the two asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ are \\_. $(x_{0},y_{0})$, then the equations of $l_{1}$ and $l_{2}$ are respectively $y-y_{0}=\\frac{\\sqrt{5}}{2}(x-x_{0})$, $y-y_{0}=-\\frac{\\sqrt{5}}{2}(x-x_{0})$. Therefore, the coordinates of $M$ and $N$ are $_{M(x_{0}-\\frac{2\\sqrt{5}}{5}},y_{0},0)$, $N(x_{0}+\\frac{2\\sqrt{}}{5}\\overrightarrow{AB}\\cdot\\overrightarrow{ON}=x_{0}-\\frac{2\\sqrt{5}}{5}B.xx_{0}+\\frac{2\\sqrt{5}}{5}y_{0}=$ Also, since point $P$ lies on the hyperbola, then $\\frac{x_{0}^{2}}{0}-\\frac{y_{0}^{2}}{6}=1$, so $|OM|\\cdot|ON|=4$."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $E$: $x^{2}+\\frac{y^{2}}{b^{2}}=1$ $(0<b<1)$, respectively. A line passing through point $F_{1}$ intersects the ellipse $E$ at points $A$ and $B$. If $|A F_{1}|=3|F_{1} B|$ and $A F_{2} \\perp x$-axis, then the equation of the ellipse $E$ is?",
    "fact_expressions": "E: Ellipse;G: Line;A: Point;F1: Point;B: Point;F2: Point;b:Number;0<b;b<1;Expression(E)=(x^2+y^2/b^2=1);LeftFocus(E)=F1;RightFocus(E)=F2;PointOnCurve(F1, G);Intersection(G, E) = {A, B};Abs(LineSegmentOf(A, F1)) = 3*Abs(LineSegmentOf(F1, B));IsPerpendicular(LineSegmentOf(A,F2),xAxis)",
    "query_expressions": "Expression(E)",
    "answer_expressions": "x^2+3*y^2/2=1",
    "fact_spans": "[[[19, 62], [82, 87], [142, 147]], [[79, 81]], [[88, 91]], [[1, 8], [70, 78]], [[92, 95]], [[9, 16]], [[26, 62]], [[26, 62]], [[26, 62]], [[19, 62]], [[1, 68]], [[1, 68]], [[69, 81]], [[79, 97]], [[99, 121]], [[122, 140]]]",
    "query_spans": "[[[142, 152]]]",
    "process": "Let $F_{1}(-c,0)$, $F_{2}(c,0)$. Since $AF_{2} \\perp x$-axis, we have $x_{A}=c$, substituting into the ellipse equation gives $A(c,b^{2})$. Let $B(x,y)$. Since $|AF_{1}|=3|F_{1}B|$, we get $\\overrightarrow{AF}_{1}=3\\overrightarrow{F_{1}B}$. Therefore $(-c-c,-b^{2})=3(x+c,y)$, solving yields $\\begin{cases}x=-\\frac{5}{3}c\\\\y=-\\frac{1}{3}b^{2}\\end{cases}$, so $B(-\\frac{5}{3}c,-\\frac{1}{3}b^{2})$. Since $B$ lies on the ellipse, substitute $B(-\\frac{5}{3}c,-\\frac{1}{3}b^{2})$ into the ellipse equation: $(-\\frac{5}{3}c)^{2}+\\frac{(-\\frac{1}{3}b^{2})^{2}}{b^{2}}=1$. Also $b^{2}+c^{2}=1$, solving gives $b^{2}=\\frac{2}{3}, c^{2}=\\frac{1}{3}$. Thus the ellipse equation is: $x^{2}+\\frac{3y^{2}}{2}=1$."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ with focus $F$, a line with slope $2 \\sqrt{2}$ passes through $F$ and intersects the parabola at points $A$ and $B$. $O$ is the origin. If $A$ lies in the first quadrant, then $\\frac{S_{\\triangle A F O}}{S_{\\triangle B F O}}$=?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;A: Point;F: Point;B: Point;O: Origin;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(F, H);Slope(H) = 2*sqrt(2);Intersection(H, G) = {A, B};Quadrant(A)=1",
    "query_expressions": "Area(TriangleOf(A,F,O))/Area(TriangleOf(B,F,O))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 23], [55, 58]], [[5, 23]], [[47, 49]], [[60, 63], [80, 83]], [[27, 30], [50, 53]], [[64, 67]], [[70, 73]], [[5, 23]], [[2, 23]], [[2, 30]], [[47, 53]], [[31, 49]], [[47, 69]], [[80, 88]]]",
    "query_spans": "[[[91, 142]]]",
    "process": ""
  },
  {
    "text": "The minor axis of the ellipse is $2$, and the major axis is twice the minor axis. What is the distance from the center of the ellipse to its directrix?",
    "fact_expressions": "G: Ellipse;Length(MinorAxis(G)) = 2;Length(MajorAxis(G)) = 2*Length(MinorAxis(G))",
    "query_expressions": "Distance(Center(G), Directrix(G))",
    "answer_expressions": "4/sqrt(3)",
    "fact_spans": "[[[0, 2], [23, 25], [29, 30]], [[0, 10]], [[0, 21]]]",
    "query_spans": "[[[23, 37]]]",
    "process": ""
  },
  {
    "text": "Given an ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{3}=1(a>0)$ with one focus at $F(-1,0)$, a line $l$ passing through point $F$ with slope $1$ intersects the ellipse at points $C$ and $D$. Then the length of segment $CD$ is?",
    "fact_expressions": "l: Line;G: Ellipse;a: Number;C: Point;D: Point;F: Point;a>0;Expression(G) = (y^2/3 + x^2/a^2 = 1);Coordinate(F) = (-1, 0);OneOf(Focus(G)) = F;PointOnCurve(F, l);Slope(l) = 1;Intersection(l, G) = {C, D}",
    "query_expressions": "Length(LineSegmentOf(C, D))",
    "answer_expressions": "24/7",
    "fact_spans": "[[[78, 83]], [[2, 48], [85, 87]], [[4, 48]], [[89, 92]], [[93, 96]], [[54, 63], [66, 70]], [[4, 48]], [[2, 48]], [[54, 63]], [[2, 63]], [[64, 83]], [[71, 83]], [[78, 98]]]",
    "query_spans": "[[[100, 111]]]",
    "process": "From the focus of the ellipse, we can obtain a, then solve simultaneously the line and ellipse equations, and use the chord length formula to get the result. Let C(x_{1},y_{1}), D(x_{2},y_{2}). Since one focus of the ellipse is F(-1,0), so a^{2}=b^{2}+c^{2}=3+1=4, then the equation of the ellipse is \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1. The equation of line l is: y=x+1.  \n\\begin{cases}\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1\\\\y=x+1\\end{cases} \\Rightarrow 7x^{2}+8x-8=0  \nx_{1}x_{2}=-\\frac{8}{7}, x_{1}+x_{2}=-\\frac{8}{7}  \nFrom |CD|=\\sqrt{1+k^{2}}\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}^{x}}"
  },
  {
    "text": "The equation of the directrix of the parabola $x=16 y^{2}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x = 16*y^2)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x = -1/64",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 22]]]",
    "process": "The parabola is given by $ y^{2} = \\frac{1}{16}x $, then $ 2p = \\frac{1}{16} $, solving for $ p $ gives: $ p = \\frac{1}{32} $, the equation of the directrix is: $ x = -\\frac{p}{2} = -\\frac{1}{64} $"
  },
  {
    "text": "Given that point $P$ is a point on the parabola $y^{2}=-8x$, let $d_{1}$ be the distance from $P$ to the directrix of this parabola, and $d_{2}$ be the distance from $P$ to the line $x+y-10=0$. Then the minimum value of $d_{1}+d_{2}$ is?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;d1:Number;d2:Number;Expression(G) = (y^2 = -8*x);Expression(H) = (x + y - 10 = 0);PointOnCurve(P, G);Distance(P,Directrix(G)) = d1;Distance(P,H)=d2",
    "query_expressions": "Min(d1+d2)",
    "answer_expressions": "6*sqrt(2)",
    "fact_spans": "[[[7, 22], [32, 35]], [[50, 62]], [[2, 6], [27, 30]], [[41, 48]], [[66, 73]], [[7, 22]], [[50, 62]], [[2, 25]], [[27, 48]], [[27, 73]]]",
    "query_spans": "[[[75, 94]]]",
    "process": ""
  },
  {
    "text": "The equation $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ represents an ellipse with foci on the $x$-axis. Then, the range of the real number $m$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/m = 1);PointOnCurve(Focus(G), xAxis);m: Real",
    "query_expressions": "Range(m)",
    "answer_expressions": "(0,4)",
    "fact_spans": "[[[48, 50]], [[0, 50]], [[39, 50]], [[52, 57]]]",
    "query_spans": "[[[52, 62]]]",
    "process": ""
  },
  {
    "text": "Given that point $F$ is the right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the line $l$ passing through the origin with an inclination angle of $\\frac{\\pi}{3}$ intersects the left and right branches of $C$ at points $A$ and $B$, respectively, and $\\overrightarrow{A F} \\cdot \\overrightarrow{B F}=0$. Then the eccentricity of $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;A: Point;F: Point;B: Point;O:Origin;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;PointOnCurve(O,l);Inclination(l)=pi/3;Intersection(l,LeftPart(C))=A;Intersection(l,RightPart(C))=B;DotProduct(VectorOf(A, F), VectorOf(B, F)) = 0",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)+1",
    "fact_spans": "[[[97, 102]], [[7, 68], [103, 106], [180, 183]], [[15, 68]], [[15, 68]], [[116, 119]], [[2, 6]], [[120, 123]], [[74, 76]], [[15, 68]], [[15, 68]], [[7, 68]], [[7, 72]], [[73, 102]], [[77, 102]], [[97, 125]], [[97, 125]], [[127, 178]]]",
    "query_spans": "[[[180, 189]]]",
    "process": "Let F be the left focus of the hyperbola, connect AF, BF. From $\\overrightarrow{AF}\\cdot\\overrightarrow{BF}=0$, we get $AF\\bot BF$, hence quadrilateral AFBF is a rectangle. By the definition of hyperbola: $BF-BF=2a$, i.e., $\\sqrt{3}c-c=2a$, $\\therefore e=\\sqrt{3}+1$"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, let the left focus be $F$, the right vertex be $A$, and the intersection point of the line $x=a$ and one asymptote of the hyperbola be $B$. If $\\angle B F A=30^{\\circ}$, then the equations of the asymptotes of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;LeftFocus(G) = F;A: Point;RightVertex(G) = A;H: Line;Expression(H) = (x = a);Intersection(H, OneOf(Asymptote(G))) = B;B: Point;AngleOf(B, F, A) = ApplyUnit(30, degree)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y+pm*sqrt(3)*x=0",
    "fact_spans": "[[[2, 58], [83, 86], [129, 132]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[63, 66]], [[2, 66]], [[71, 74]], [[2, 74]], [[75, 82]], [[75, 82]], [[75, 99]], [[96, 99]], [[102, 127]]]",
    "query_spans": "[[[129, 140]]]",
    "process": "According to the value of $\\tan\\angle BFA$, obtain an equation in terms of $a, b$, thereby find the slope of the asymptote, and thus get the answer. From the given conditions, we have $A(a,0)$, the asymptotes of the hyperbola are $ay\\pm bx=0$. Without loss of generality, let point $B$ be the intersection of the line $x=a$ and $y=\\frac{b}{a}x$, then the coordinates of point $B$ are $(a,b)$. Since $AB\\perp FA$, $\\angle BFA=30^{\\circ}$, so $\\tan\\angle BFA=\\frac{|AB|}{|FA|}=\\frac{b}{a+c}=\\frac{}{a+}$. Solving gives $\\frac{b}{a}=\\sqrt{3}$, so the equations of the asymptotes are $y\\pm\\sqrt{3}x=0$."
  },
  {
    "text": "The distance from the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ to its asymptote is half the distance from the right focus to the left vertex. Then the eccentricity $e$ of the hyperbola is $?$.",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;Distance(RightFocus(G), Asymptote(G)) = (1/2)*Distance(RightFocus(G), LeftVertex(G));e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "5/3",
    "fact_spans": "[[[0, 46], [70, 73]], [[0, 46]], [[3, 46]], [[3, 46]], [[0, 68]], [[77, 80]], [[70, 80]]]",
    "query_spans": "[[[77, 82]]]",
    "process": "Since the distance from the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ to its asymptote is $b$, it follows that $b=\\frac{c+a}{2}$, $\\sqrt{c^{2}-a^{2}}=\\frac{c+a}{2}$, $c-a=\\frac{c+a}{4}$, $3c=5a$, $e=\\frac{c}{a}=\\frac{5}{3}$."
  },
  {
    "text": "The distance from the focus of the parabola $y^{2}=4 x$ to the line $l$: $5 x+12 y+3=0$ is?",
    "fact_expressions": "l: Line;G: Parabola;Expression(G) = (y^2 = 4*x);Expression(l)=(5*x+12*y+3=0)",
    "query_expressions": "Distance(Focus(G),l)",
    "answer_expressions": "8/13",
    "fact_spans": "[[[18, 38]], [[0, 14]], [[0, 14]], [[18, 38]]]",
    "query_spans": "[[[0, 43]]]",
    "process": "From the parabola $ y^{2} = 4x $, the focus is $ F(1,0) $. Therefore, the distance from point $ F(1,0) $ to the line $ l: 5x + 12y + 3 = 0 $ is $ d = \\frac{|5 + 3|}{\\sqrt{5^{2} + 12^{2}}} = \\frac{8}{13} $."
  },
  {
    "text": "Given that point $P$ is a moving point on the parabola $y^{2}=2x$, then the minimum value of the sum of the distance from $P$ to the point $(0,2)$ and the distance from $P$ to the directrix of this parabola is?",
    "fact_expressions": "G: Parabola;H: Point;P: Point;Expression(G) = (y^2 = 2*x);Coordinate(H) = (0, 2);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, H) + Distance(P, Directrix(G)))",
    "answer_expressions": "sqrt(17)/2",
    "fact_spans": "[[[7, 21], [50, 53]], [[33, 41]], [[2, 6], [29, 32], [45, 48]], [[7, 21]], [[33, 41]], [[2, 27]]]",
    "query_spans": "[[[29, 66]]]",
    "process": "Analysis: First, find the coordinates of the focus of the parabola. Then, by the definition of the parabola, we have $ d = |PF| + |PA| \\geqslant |AF| $. Next, compute the value of $ |AF| $. According to the problem, let $ P' $ be the projection of point $ P $ on the directrix of the parabola, and let $ F $ be the focus of the parabola. Then $ F\\left(\\frac{1}{2}, 0\\right) $. By the definition of the parabola, the distance from $ P $ to the directrix is $ |PP'| = |PF| $. Then the sum of the distance from point $ P $ to point $ A(0, 2) $ and the distance from $ P $ to the directrix of the parabola is $ d = |PF| + |PA| \\geqslant |AF| = \\sqrt{\\left(\\frac{1}{2}\\right)^2 + 2^2} = \\frac{\\sqrt{17}}{2} $."
  },
  {
    "text": "An ellipse $5 x^{2}+k y^{2}=5$ has a focus at $(0,2)$. What is the value of the real number $k$?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (k*y^2 + 5*x^2 = 5);Coordinate(OneOf(Focus(G))) = (0, 2)",
    "query_expressions": "k",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 21]], [[37, 42]], [[0, 21]], [[0, 34]]]",
    "query_spans": "[[[37, 46]]]",
    "process": "From $x^{2}+\\frac{y^{2}}{\\frac{5}{k}}=1$ we get $\\frac{5}{k}=1+4$, so $k=1$."
  },
  {
    "text": "Let the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ have left and right foci $F_{1}$, $F_{2}$ respectively, and let $A$ be a point on the ellipse such that $A F_{2} \\perp F_{1} F_{2}$. If the distance from the origin $O$ to the line $A F_{1}$ is $\\frac{1}{3}|O F_{1}|$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;A: Point;F2: Point;O: Origin;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(A,G);IsPerpendicular(LineSegmentOf(A, F2), LineSegmentOf(F1, F2));Distance(O, LineOf(A,F1)) = Abs(LineSegmentOf(O, F1))/3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[1, 53], [82, 84], [163, 165]], [[3, 53]], [[3, 53]], [[62, 69]], [[78, 81]], [[70, 77]], [[117, 122]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 77]], [[1, 77]], [[78, 87]], [[88, 115]], [[117, 160]]]",
    "query_spans": "[[[163, 171]]]",
    "process": "Since $AF_{2}\\bot F_{1}F_{2}$, assume point $A(c,y_{A})$, where $y_{A}>0$. Substituting into the ellipse equation $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, we get $\\frac{c^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$. Solving gives $y_{A}^{2}=\\frac{b^{2}(a^{2}-c^{2})}{a^{2}}=\\frac{b^{4}}{a^{2}}$, so $y_{A}=\\frac{b^{2}}{a}$, i.e., $|AF_{2}|=\\frac{b^{2}}{a}$. Draw $OE\\bot AF_{1}$ from $O$. Since the distance from the origin $O$ to line $AF_{1}$ is $\\frac{1}{3}|OF_{1}|$, i.e., $|OE|=\\frac{1}{3}|OF_{1}|$, and by $\\triangle OEF_{1} \\sim \\triangle AF_{2}F_{1}$, we obtain $\\frac{|AF_{2}|}{|F_{1}F_{2}|}=\\frac{|OE|}{|EF_{1}|}=\\frac{1}{2\\sqrt{2}}$, that is, $\\frac{b^{2}}{2c}=\\frac{b^{2}}{2ac}=\\frac{1}{2\\sqrt{2}}$. Using $b^{2}=a^{2}-c^{2}$, simplifying yields $\\sqrt{2}c^{2}+ac-\\sqrt{2}a^{2}=0$, i.e., $\\sqrt{2}e^{2}+e-\\sqrt{2}=0$. Since $e>0$, solving gives $e=\\frac{\\sqrt{2}}{2}$, i.e., the eccentricity of the ellipse is $\\frac{\\sqrt{2}}{2}$."
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ satisfies $a \\leq \\sqrt{3} b$. If the eccentricity is $e$, then the minimum value of $e^{2}+\\frac{1}{e^{2}}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;a <= sqrt(3)*b;e: Number;Eccentricity(G) = e",
    "query_expressions": "Min(e^2 + 1/(e^2))",
    "answer_expressions": "7/6",
    "fact_spans": "[[[0, 52]], [[0, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[54, 73]], [[79, 82]], [[0, 82]]]",
    "query_spans": "[[[84, 113]]]",
    "process": "Since $a\\leqslant\\sqrt{3}b$ and $a^{2}=b^{2}+c^{2}$, it follows that $a^{2}\\leqslant3b^{2}=3a^{2}-3c^{2}$. Solving gives: $0<e^{2}\\leqslant\\frac{2}{3}$. Because the function $y=e^{2}+\\frac{1}{e^{2}}$ with respect to $e^{2}$ is monotonically decreasing on $(0,\\frac{2}{3}]$, when $e^{2}=\\frac{2}{3}$, $y=e^{2}+\\frac{1}{e^{2}}$ attains the minimum value $\\frac{2}{3}+\\frac{3}{2}=\\frac{7}{6}$. Therefore, the answer is: $\\frac{7}{6}$."
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{2 m-1}-\\frac{y^{2}}{m+2}=1$ represents a hyperbola, what is the range of values for $m$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(2*m - 1) - y^2/(m + 2) = 1);m: Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-oo, -2) + (1/2, +oo)",
    "fact_spans": "[[[47, 50]], [[2, 50]], [[52, 55]]]",
    "query_spans": "[[[52, 62]]]",
    "process": "From the given condition, (2m-1)(m+2)>0, so m>\\frac{1}{2} or m<-2. Therefore, the range of values for m is (-\\infty,-2)\\cup(\\frac{1}{2},+\\infty)"
  },
  {
    "text": "If the focal distance of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{m}=1$ is $10$, then $m=$?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/16 - y^2/m = 1);FocalLength(G) = 10",
    "query_expressions": "m",
    "answer_expressions": "9",
    "fact_spans": "[[[1, 40]], [[50, 53]], [[1, 40]], [[1, 48]]]",
    "query_spans": "[[[50, 55]]]",
    "process": "According to the problem, $ a^{2} = 16 $, $ b^{2} = m $, $ 2c = 10 $, so $ m = b^{2} = c^{2} - a^{2} = \\left( \\frac{10}{2} \\right)^{2} - 16 = 9 $."
  },
  {
    "text": "The distance from the focus $F$ of the parabola $x^{2}=\\frac{1}{2} y$ to its directrix $l$ is?",
    "fact_expressions": "G: Parabola;F: Point;l: Line;Expression(G) = (x^2 = y/2);Focus(G) = F;Directrix(G) = l",
    "query_expressions": "Distance(F, l)",
    "answer_expressions": "1/4",
    "fact_spans": "[[[0, 24], [31, 32]], [[27, 30]], [[34, 37]], [[0, 24]], [[0, 30]], [[31, 37]]]",
    "query_spans": "[[[27, 42]]]",
    "process": "Using the standard equation of a parabola, we can obtain $ p $, and thus determine the distance from the focus to the directrix. [Detailed solution] The parabola $ x^{2} = \\frac{1}{2}y $, hence $ p = \\frac{1}{4} $, that is, the distance from its focus to the directrix is $ \\frac{1}{4} $."
  },
  {
    "text": "Given that the distance from the focus of the parabola $y^{2}=8x$ to the asymptotes of the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is not greater than $\\sqrt{3}$, then the range of the eccentricity of hyperbola $E$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);E: Hyperbola;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Negation(Distance(Focus(G), Asymptote(E)) > sqrt(3))",
    "query_expressions": "Range(Eccentricity(E))",
    "answer_expressions": "(1,2]",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[20, 81], [103, 109]], [[20, 81]], [[28, 81]], [[28, 81]], [[28, 81]], [[28, 81]], [[2, 101]]]",
    "query_spans": "[[[103, 120]]]",
    "process": "From the parabola equation, the focus coordinates are obtained. From the hyperbola equation, the asymptotes are derived. Using the point-to-line distance formula and the given conditions, an inequality involving b and c is established. Combining this with the relationship among a, b, and c in a hyperbola, the range of eccentricity is found. For the parabola $ y^{2} = 8x $, the focus coordinates are $ (2, 0) $. The hyperbola $ E: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) has asymptotes $ bx \\pm ay = 0 $. According to the problem, the distance from point $ (2, 0) $ to the line $ bx \\pm ay = 0 $ is $ \\frac{2b}{\\sqrt{b^{2} + a^{2}}} = \\frac{2b}{c} \\leqslant \\sqrt{3} $. Thus, $ 4b^{2} \\leqslant 3c^{2} $, i.e., $ 4(c^{2} - a^{2}) \\leqslant 3c^{2} $, $ c^{2} \\leqslant 4a^{2} $, $ e = \\frac{c}{a} \\leqslant 2 $. Therefore, the range of eccentricity for hyperbola $ E $ is $ (1, 2] $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$ with right focus $F$, upper vertex $A$, and point $P$ a moving point on the ellipse. When the perimeter of $\\triangle P A F$ is maximized, what is the area of $\\triangle P A F$?",
    "fact_expressions": "C: Ellipse;P: Point;A: Point;F: Point;Expression(C) = (x^2/2 + y^2 = 1);RightFocus(C) = F;UpperVertex(C) = A;PointOnCurve(P, C);WhenMax(Perimeter(TriangleOf(P, A, F)))",
    "query_expressions": "Area(TriangleOf(P, A, F))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[2, 34], [57, 59]], [[51, 55]], [[47, 50]], [[39, 42]], [[2, 34]], [[2, 42]], [[2, 50]], [[51, 63]], [[64, 88]]]",
    "query_spans": "[[[89, 111]]]",
    "process": "PA+PF+AF=a+PA+PF=a+(2a-PF_{1})+PA (where F_{1} is the left focus)=3a+PA-PF_{1}\\leqslant3a+AF_{1}=4a=4\\sqrt{2}, with equality if and only if points A, F_{1}, P are collinear, at which point P(-\\frac{4}{3},-\\frac{1}{3}), so S_{\\DeltaAFP}=S_{\\DeltaAFF_{1}}+S_{\\DeltaPFF_{1}}=\\frac{1}{2}FF_{1}\\cdot|y_{A}-y_{P}|=\\frac{1}{2}\\times2\\times\\frac{4}{3}=\\frac{4}{3}"
  },
  {
    "text": "The center of the ellipse is at the origin, the focal distance is $4$, and one directrix is $x = -4$. Then the equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;FocalLength(G) = 4;Expression(OneOf(Directrix(G)))=(x = -4)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/8+y^2/4=1",
    "fact_spans": "[[[0, 2], [30, 32]], [[6, 8]], [[0, 8]], [[0, 15]], [[0, 27]]]",
    "query_spans": "[[[30, 37]]]",
    "process": "The focal distance of the ellipse is 4, so 2c=4, c=2. Since the directrix is x=-4, the foci of the ellipse lie on the x-axis, and -\\frac{a^{2}}{c}=-4, so a^{2}=4c=8, b^{2}=a^{2}-c^{2}=8-4=4. Therefore, the equation of the ellipse is \\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1."
  },
  {
    "text": "The equation of the hyperbola with asymptotes $3x-4y-2=0$ and $3x+4y-10=0$, and one vertex at $(2,4)$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;Z: Line;Expression(H) = (3*x - 4*y - 2 = 0);Expression(Z) = (3*x + 4*y - 10 = 0);Asymptote(G) = {H, Z};Coordinate(OneOf(Vertex(G))) = (2, 4)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "(y - 1)^2/9 - (x - 2)^2/16 = 1",
    "fact_spans": "[[[49, 52]], [[1, 16]], [[18, 31]], [[1, 16]], [[17, 31]], [[0, 52]], [[36, 52]]]",
    "query_spans": "[[[49, 56]]]",
    "process": "According to the property that the intersection point of the asymptotes of a hyperbola is the center of symmetry of the hyperbola, solve the system of equations to find the center of symmetry of the hyperbola, then use the position of the vertex and the properties of translation to solve. Since the intersection point of the asymptotes of the hyperbola is the center of symmetry, the coordinates of the center of symmetry of the hyperbola are the solution of the system of equations  \n\\begin{cases}3x-4y-2=0\\\\3x-4y-10=0\\end{cases},  \nsolving gives  \n\\begin{cases}x=2\\\\y=1\\end{cases}.  \nTherefore, the center of symmetry of the hyperbola is at (2,1). Translating this hyperbola 2 units to the left and 1 unit downward results in a hyperbola with its center of symmetry at the origin. Thus, the vertex (2,4) of this hyperbola, after translating 2 units to the left and 1 unit downward, becomes (0,3). Therefore, for the hyperbola with center of symmetry at the origin, the foci lie on the y-axis and $ a = 3 $. After such translation, the asymptotes of the hyperbola become $ 3x - 4y = 0 $ and $ 3x + 4y = 0 $, so we have $ \\frac{b}{a} = \\frac{4}{3} \\Rightarrow b = 4 $. Hence, the equation of the hyperbola with center of symmetry at the origin is $ \\frac{y^{2}}{9} - \\frac{x^{2}}{16} = 1 $. Therefore, the equation of the original hyperbola is:"
  },
  {
    "text": "Draw a perpendicular line from a focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) to one of its asymptotes. If the foot of the perpendicular lies exactly on the perpendicular bisector of the segment $OF$ ($O$ being the origin), then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;OneOf(Focus(G)) = F;L: Line;PointOnCurve(F, L);IsPerpendicular(L, OneOf(Asymptote(G)));O: Origin;PointOnCurve(FootPoint(L, OneOf(Asymptote(G))), PerpendicularBisector(LineSegmentOf(O, F)))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 58], [104, 107]], [[1, 58]], [[4, 58]], [[4, 58]], [[4, 58]], [[4, 58]], [[63, 66]], [[1, 66]], [], [[0, 75]], [[0, 75]], [[88, 91]], [[0, 102]]]",
    "query_spans": "[[[104, 113]]]",
    "process": ""
  },
  {
    "text": "Let $P$ be a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$. The distance from $P$ to the left focus $F_{1}$ is 4 times the distance from $P$ to the right focus $F_{2}$. Then $\\cos \\angle P F_{2} F_{1}$ = ?",
    "fact_expressions": "G: Ellipse;P: Point;F2: Point;F1: Point;Expression(G) = (x^2/25 + y^2/9 = 1);PointOnCurve(P, G);LeftFocus(G)=F1;RightFocus(G)=F2;Distance(P,F1)=4*Distance(P,F2)",
    "query_expressions": "Cos(AngleOf(P, F2, F1))",
    "answer_expressions": "1/8",
    "fact_spans": "[[[5, 43]], [[47, 51], [1, 4]], [[70, 77]], [[55, 62]], [[5, 43]], [[1, 46]], [[5, 62]], [[5, 77]], [[5, 84]]]",
    "query_spans": "[[[86, 115]]]",
    "process": "According to the definition of an ellipse, we have: PF_{1}+PF_{2}=2a. Combining with the given conditions, find PF_{1} and PF_{2}, then use the cosine law to obtain the result. [Detailed solution] From the definition of the ellipse, we get: a^{2}=25, b^{2}=9, so c^{2}=25-9=16, hence c=4, therefore PF_{1}+PF_{2}=10. F_{1}F_{2}=8. According to the problem, PF_{1}=4PF_{2}, solving gives: PF_{1}=8, PF_{2}=2. In \\triangle PF_{1}F_{2}, using the cosine law, \\cos\\angle PF_{2}F_{1}=\\frac{PF_{2}^{2}+F_{2}F_{1}^{2}-PF_{1}^{2}}{2PF_{2}\\cdot F_{2}F_{1}}=\\frac{4+64-64}{2\\times2\\times8}=\\frac{1}{8}"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the distance from its right vertex to one of its asymptotes is $\\frac{a}{2}$. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Distance(RightVertex(C), OneOf(Asymptote(C)))=a/2",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 63], [68, 69], [93, 96]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 91]]]",
    "query_spans": "[[[93, 102]]]",
    "process": "From the asymptote equations $ y = \\pm\\frac{b}{a}x $, the distance from the right vertex $ (a,0) $ to one of its asymptotes is $ a\\times\\frac{b}{c} $. Then $ a\\times\\frac{b}{c} = \\frac{a}{2} $, so $ c = 2b $. From $ a^{2} + b^{2} = c^{2} $, we get $ a = \\sqrt{3}b $, then the eccentricity $ e = \\frac{c}{a} = \\frac{2\\sqrt{3}}{3} $."
  },
  {
    "text": "The point $P(4,1)$ bisects a chord of the hyperbola $x^{2}-4 y^{2}=4$, then the equation of the line on which this chord lies is?",
    "fact_expressions": "G: Hyperbola;H: LineSegment;P: Point;Expression(G) = (x^2 - 4*y^2 = 4);Coordinate(P) = (4, 1);IsChordOf(H,G);MidPoint(H)=P",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "x-y-3=0",
    "fact_spans": "[[[11, 31]], [], [[0, 9]], [[11, 31]], [[0, 9]], [[11, 35]], [[0, 35]]]",
    "query_spans": "[[[11, 50]]]",
    "process": "Let the endpoints of the chord be A(x_{1},y_{1}), B(x_{2},y_{2}). Since the midpoint of AB is P(4,1), we have x_{1}+x_{2}=8, y_{1}+y_{2}=2. Substituting A(x_{1},y_{1}), B(x_{2},y_{2}) into the hyperbola x^{2}-4y^{2}=4, we obtain \\begin{cases}x_{1}^{2}-4y_{1}^{2}=4\\\\x_{2}^{2}-4y_{2}^{2}=4\\end{cases}, (x_{1}+x_{2})(x_{1}-x_{2})-4(y_{1}+y_{2})(y_{1}-y_{2})=0, (x_{1}-x_{2})-8(y_{1}-y_{2})=0, \\therefore k=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=1, \\therefore the equation of the line containing this chord is x-y-3=0"
  },
  {
    "text": "Given that one of the directrices of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ is $x=-\\frac{3}{2}$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(OneOf(Directrix(G)))=(x=-3/2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 39], [64, 67]], [[5, 39]], [[5, 39]], [[2, 39]], [[2, 61]]]",
    "query_spans": "[[[64, 73]]]",
    "process": "The hyperbola \\frac{x^{2}}{a^{2}}-y^{2}=1 (a>0) has a directrix given by x=-\\frac{a^{2}}{\\sqrt{1+a^{2}}}=-\\frac{3}{2}; solving yields a=\\sqrt{3}, c=2; the eccentricity of the hyperbola is \\frac{c}{a}=\\frac{2}{\\sqrt{3}}=\\frac{2\\sqrt{3}}{3}"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(b>a>0)$, the distance from a point $P$ on the hyperbola to the origin is $b$, and $\\sin \\angle P F_{2} F_{1} = \\frac{5}{3} \\sin \\angle P F_{1} F_{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F2: Point;F1: Point;O:Origin;b > a;a > 0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);Distance(P,O) = b;Sin(AngleOf(P, F2, F1)) = (5/3)*Sin(AngleOf(P, F1, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3",
    "fact_spans": "[[[20, 73], [79, 82], [168, 171]], [[95, 98]], [[23, 73]], [[84, 88]], [[10, 17]], [[2, 9]], [[89, 91]], [[23, 73]], [[23, 73]], [[20, 73]], [[2, 78]], [[79, 88]], [[84, 98]], [[100, 165]]]",
    "query_spans": "[[[168, 177]]]",
    "process": "Let $F_{1}$ be the left focus of the hyperbola, and $F_{2}$ be the right focus of the hyperbola. Since $\\sin\\angle PF_{2}F_{1} = \\frac{5}{3}\\sin\\angle PF_{1}F_{2}$, by the law of sines we have $|PF_{1}| = \\frac{5}{3}|PF_{2}|$. Because $|PF_{1}| - |PF_{2}| = 2a$, it follows that $|PF_{2}| = 3a$, $|PF_{1}| = 5a$. From the problem, it is clear that $|OP| = b$, $|OF_{1}| = |OF_{2}| = c$. Since $\\angle POF_{2} + \\angle POF_{1} = \\pi$, we have $\\cos\\angle POF_{2} + \\cos\\angle POF_{1} = 0$. Then $\\frac{b^{2}+c^{2}-(3a)^{2}}{2bc} + \\frac{b^{2}+c^{2}-(5a)^{2}}{2bc} = 0$. Simplifying and combining terms yields $b^{2}+c^{2}=17a^{2}$. Also, $b^{2}=c^{2}-a^{2}$, $\\therefore c^{2}=9a^{2}$, i.e., $\\frac{c}{a}=3$."
  },
  {
    "text": "Given the ellipse $\\frac{y^{2}}{4}+x^{2}=1$, let $P$ be an arbitrary point on the ellipse. Through $P$, draw lines parallel to $l_{1}$: $y=2x$ and $l_{2}$: $y=-2x$, respectively, intersecting lines $l_{2}$ and $l_{1}$ at points $M$ and $N$. Then the maximum value of $|MN|$ is?",
    "fact_expressions": "G: Ellipse;l1: Line;l2: Line;M: Point;N: Point;P: Point;Expression(G) = (x^2 + y^2/4 = 1);Expression(l1) = (y = 2*x);Expression(l2) = (y = -2*x);PointOnCurve(P, G);L1:Line;L2:Line;PointOnCurve(P, L1);PointOnCurve(P, L2);IsParallel(l1, L1);IsParallel(l2, L2);Intersection(L1, l2) = M;Intersection(L2, l1) = N",
    "query_expressions": "Max(Abs(LineSegmentOf(M, N)))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 29], [36, 38]], [[52, 68], [105, 112]], [[69, 86], [93, 102]], [[113, 116]], [[117, 120]], [[31, 35], [45, 48]], [[2, 29]], [[52, 68]], [[69, 86]], [[32, 43]], [], [], [[44, 91]], [[44, 91]], [[44, 91]], [[44, 91]], [[44, 120]], [[44, 120]]]",
    "query_spans": "[[[122, 134]]]",
    "process": "According to the problem, draw the schematic diagram as follows: Let M(x_{1},y_{1}), N(x_{2},y_{2}), P(x_{0},y_{0}), then y_{1}=-2x_{1}, y_{2}=2x_{2}. From the problem, quadrilateral PMON is a parallelogram, so  \n\\begin{cases}x_{1}+x_{2}=x_{0}+0=\\frac{1}{2}(y_{2}-y_{1})\\\\y_{1}+y_{2}=y_{0}+0=2(x_{2}-x_{1})\\end{cases},  \ni.e.,  \n\\begin{cases}x_{2}-x_{1}=\\frac{y_{0}}{2}\\\\y_{2}-y_{1}=2x_{0}\\end{cases}.  \nAlso, since P is an arbitrary point on the ellipse, \\frac{y_{0}^{2}}{4}+x_{0}^{2}=1 \\Rightarrow \\frac{y_{0}^{2}}{4}=1-x_{0}^{2}.  \n|MN|=\\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}=\\sqrt{\\frac{y_{0}^{2}}{4}+4x_{0}^{2}}=\\sqrt{4x_{0}^{2}+(1-x_{0}^{2})}=\\sqrt{3x_{0}^{2}+1}.  \nSince -1\\leqslant x_{0}\\leqslant 1, \\therefore 0\\leqslant x_{0}^{2}\\leqslant 1, thus by function properties: when x_{0}^{2}=1, |MN|_{\\max}=\\sqrt{3\\times1+1}=2."
  },
  {
    "text": "Let point $P$ be an arbitrary point on the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{a^{2}-1}=1(a>1)$, and let $AB$ be an arbitrary diameter of the circle $M$: $(x-1)^{2}+y^{2}=1$. If the maximum value of $\\overrightarrow{P A} \\cdot \\overrightarrow{P B}$ is $15$, then $a$=?",
    "fact_expressions": "C: Ellipse;a: Number;M: Circle;A: Point;B: Point;P: Point;a>1;Expression(C) = (y^2/(a^2 - 1) + x^2/a^2 = 1);Expression(M) = (y^2 + (x - 1)^2 = 1);PointOnCurve(P, C);IsDiameter(LineSegmentOf(A,B),M);Max(DotProduct(VectorOf(P, A), VectorOf(P, B))) = 15",
    "query_expressions": "a",
    "answer_expressions": "3",
    "fact_spans": "[[[5, 62]], [[168, 171]], [[75, 99]], [[69, 74]], [[69, 74]], [[0, 4]], [[12, 62]], [[5, 62]], [[75, 99]], [[0, 68]], [[69, 106]], [[108, 166]]]",
    "query_spans": "[[[168, 173]]]",
    "process": "By the properties of a circle combined with linear operations and dot product operations of planar vectors, we obtain $\\overrightarrow{PA}\\cdot\\overrightarrow{PB}=|\\overrightarrow{PM}|^{2}-1$. Then, by the properties of the ellipse, we get $|\\overrightarrow{PM}|^{2}-1\\leqslant(a+c)^{2}-1$, and thus the solution follows. The ellipse $C:\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{a^{2}-1}=1$ $(a>1)$ has foci at $(-1,0)$, $(1,0)$, and semi-focal length $c=1$. The circle $M:(x-1)^{2}+y^{2}=1$ has center $M(1,0)$ and radius $1$. Since $AB$ is the diameter of circle $M$, we have $\\overrightarrow{MA}=-\\overrightarrow{MB}$. Then, $\\overrightarrow{PA}\\cdot\\overrightarrow{PB}=(\\overrightarrow{PM}+\\overrightarrow{MA})\\cdot(\\overrightarrow{PM}+\\overrightarrow{MB})=(\\overrightarrow{PM}+\\overrightarrow{MA})\\cdot(\\overrightarrow{PM}-\\overrightarrow{MA})=\\overrightarrow{PM}^{2}-\\overrightarrow{MA}^{2}=|\\overrightarrow{PM}|^{2}-1$. Moreover, since $P$ is a point on the ellipse and $M$ is the right focus of the ellipse, we have $|\\overrightarrow{PM}|^{2}-1\\leqslant(a+c)^{2}-1$. The equality holds when $P$ is the left vertex of the ellipse $(-a,0)$. Thus, $(a+c)^{2}-1=15$, so $a+c=4$, that is, $a=3$."
  },
  {
    "text": "Given that $P$ is a point on the left branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $F_{1}$, $F_{2}$ are the left and right foci of the hyperbola, and $\\cos \\angle P F_{1} F_{2}=\\sin \\angle P F_{2} F_{1}=\\frac{\\sqrt{5}}{5}$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, LeftPart(G));LeftFocus(G) = F1;RightFocus(G) = F2;Cos(AngleOf(P, F1, F2)) = Sin(AngleOf(P, F2, F1));Sin(AngleOf(P, F2, F1)) = sqrt(5)/5",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[6, 62], [85, 88], [169, 172]], [[9, 62]], [[9, 62]], [[2, 5]], [[68, 75]], [[77, 84]], [[9, 62]], [[9, 62]], [[6, 62]], [[2, 67]], [[68, 93]], [[68, 93]], [[95, 167]], [[95, 167]]]",
    "query_spans": "[[[169, 177]]]",
    "process": ""
  },
  {
    "text": "Given an ellipse with foci $F_{1}(-2,0)$, $F_{2}(2,0)$ that has exactly one intersection point with the line $x + \\sqrt{3} y+ 4=0$, then the length of the major axis of the ellipse is?",
    "fact_expressions": "G: Ellipse;H: Line;F1: Point;F2: Point;Expression(H) = (x + sqrt(3)*y + 4 = 0);Coordinate(F1) = (-2, 0);Coordinate(F2) = (2, 0);Focus(G)={F1,F2};NumIntersection(H,G)=1",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "2*sqrt(7)",
    "fact_spans": "[[[33, 35], [69, 71]], [[36, 59]], [[2, 15]], [[17, 29]], [[36, 59]], [[2, 15]], [[17, 29]], [[2, 35]], [[33, 67]]]",
    "query_spans": "[[[69, 77]]]",
    "process": ""
  },
  {
    "text": "The equation of the ellipse with foci $F_{1}(-1 , 0)$, $F_{2}(1 , 0)$ and having a common point with the line $x-y+3=0$, for which the eccentricity is maximized, is?",
    "fact_expressions": "G: Ellipse;F1: Point;Coordinate(F1) = (-1, 0);F2: Point;Coordinate(F2) = (1, 0);Focus(G) = {F1, F2};H: Line;Expression(H) = (x - y + 3 = 0);IsIntersect(H, G);WhenMax(Eccentricity(G))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5+y^2/4=1",
    "fact_spans": "[[[53, 55], [63, 65]], [[1, 16]], [[1, 16]], [[18, 32]], [[18, 32]], [[0, 55]], [[37, 48]], [[37, 48]], [[36, 55]], [[57, 65]]]",
    "query_spans": "[[[63, 69]]]",
    "process": ""
  },
  {
    "text": "A line $l$ with slope $\\frac{1}{2}$ passes through the focus $F$ of the parabola $C$: $y^{2}=2 p x$ ($p>0$), intersecting the parabola at points $A$ and $B$. If the distance from the midpoint $D$ of chord $AB$ to the directrix of the parabola is $10$, then $p=$?",
    "fact_expressions": "l: Line;C: Parabola;p: Number;A: Point;B: Point;D: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C)=F;PointOnCurve(F,l);Slope(l)=1/2;Intersection(l,C)={A,B};IsChordOf(LineSegmentOf(A,B),C);MidPoint(LineSegmentOf(A,B))=D;F:Point;Distance(D, Directrix(C)) = 10",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[17, 22]], [[23, 49], [57, 60], [86, 89]], [[102, 105]], [[62, 65]], [[66, 69]], [[82, 85]], [[31, 49]], [[23, 49]], [[23, 55]], [[17, 55]], [[0, 22]], [[17, 71]], [[57, 79]], [[74, 85]], [[52, 55]], [[82, 100]]]",
    "query_spans": "[[[102, 107]]]",
    "process": ""
  },
  {
    "text": "Given that point $P(x_{0}, y_{0})$ lies on the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and $M$, $N$ are the left and right vertices of the hyperbola, respectively. If the product of the slopes of lines $PM$ and $PN$ is $\\frac{1}{3}$, then the equations of the asymptotes of this hyperbola are?",
    "fact_expressions": "P: Point;x0: Number;y0: Number;Coordinate(P) = (x0, y0);E: Hyperbola;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;PointOnCurve(P, E);M: Point;N: Point;LeftVertex(E) = M;RightVertex(E) = N;Slope(LineOf(P, M))*Slope(LineOf(P, N)) = 1/3",
    "query_expressions": "Expression(Asymptote(E))",
    "answer_expressions": "y=pm*(sqrt(3)*x/3)",
    "fact_spans": "[[[2, 20]], [[3, 20]], [[3, 20]], [[2, 20]], [[21, 81], [96, 99], [138, 141]], [[21, 81]], [[28, 81]], [[28, 81]], [[28, 81]], [[28, 81]], [[2, 85]], [[86, 89]], [[90, 93]], [[86, 104]], [[86, 104]], [[105, 135]]]",
    "query_spans": "[[[138, 149]]]",
    "process": ""
  },
  {
    "text": "The chord of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$ passing through $P(1 , 1)$ is exactly bisected by the point $P$. Then, what is the equation of the line containing this chord?",
    "fact_expressions": "G: Ellipse;H: LineSegment;P: Point;Expression(G) = (x^2/4 + y^2/2 = 1);Coordinate(P) = (1, 1);PointOnCurve(P,H);MidPoint(H)=P;IsChordOf(H,G)",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "x+2*y-3=0",
    "fact_spans": "[[[0, 37]], [], [[39, 49], [54, 58]], [[0, 37]], [[39, 49]], [[0, 51]], [[0, 60]], [[0, 51]]]",
    "query_spans": "[[[0, 73]]]",
    "process": ""
  },
  {
    "text": "Given a point $M$ on the ellipse $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{m^{2}-7}=1$ $(m>\\sqrt{7})$ whose distances to the two foci are $5$ and $3$, respectively, find the eccentricity of the ellipse.",
    "fact_expressions": "G: Ellipse;m: Number;M: Point;m>sqrt(7);Expression(G) = (y^2/(m^2 - 7) + x^2/m^2 = 1);PointOnCurve(M, G);F1:Point;F2:Point;Focus(G) ={F1,F2};Distance(M, F1) = 5;Distance(M,F2)=3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(7)/4",
    "fact_spans": "[[[2, 62], [89, 91]], [[4, 62]], [[65, 68]], [[4, 62]], [[2, 62]], [[2, 68]], [], [], [[2, 73]], [[2, 86]], [[2, 86]]]",
    "query_spans": "[[[89, 97]]]",
    "process": ""
  },
  {
    "text": "Given an ellipse $C$: $\\frac{x^{2}}{m}+\\frac{y^{2}}{3}=1$ $(m>0)$ with foci on the $x$-axis, its focal distance is $2 \\sqrt{2}$. Find the real number $m=?$",
    "fact_expressions": "C: Ellipse;m: Real;m>0;Expression(C) = (y^2/3 + x^2/m = 1);PointOnCurve(Focus(C) ,xAxis);FocalLength(C) = 2*sqrt(2)",
    "query_expressions": "m",
    "answer_expressions": "5",
    "fact_spans": "[[[11, 60], [61, 62]], [[79, 84]], [[17, 60]], [[11, 60]], [[2, 60]], [[61, 77]]]",
    "query_spans": "[[[79, 86]]]",
    "process": "Since the ellipse with foci on the x-axis, $\\frac{x^{2}}{m}+y^{2}=1$ $(m>0)$, has a focal distance of $2\\sqrt{2}$, then $a^{2}=m$, $b^{2}=3$, $c=\\sqrt{2}$, so $a^{2}=m=b^{2}+c^{2}=3+2=5$."
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=2 p x  (p>0)$, draw a line $l$ intersecting the parabola at points $M$, $N$ and intersecting the directrix of the parabola at point $P$. If $\\overrightarrow{P M}=2 \\overrightarrow{P F}$, then what is the inclination angle of line $l$?",
    "fact_expressions": "l: Line;G: Parabola;p: Number;P: Point;M: Point;F: Point;N: Point;p>0;Expression(G) = (y^2 = 2*p*x);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {M, N};Intersection(l,Directrix(G)) = P;VectorOf(P, M) = 2*VectorOf(P, F)",
    "query_expressions": "Inclination(l)",
    "answer_expressions": "{pi/3,2*pi/3}",
    "fact_spans": "[[[31, 36], [111, 116]], [[1, 24], [37, 40], [51, 54]], [[4, 24]], [[58, 62]], [[41, 45]], [[27, 30]], [[46, 49]], [[4, 24]], [[1, 24]], [[1, 30]], [[0, 36]], [[31, 49]], [[31, 61]], [[64, 109]]]",
    "query_spans": "[[[111, 122]]]",
    "process": "F(\\frac{p}{2},0), let A(-\\frac{p}{2},0), draw the directrix of the parabola through A(-\\frac{p}{2},0), then |AF|==. Draw MB perpendicular to the directrix at B, then MB//x axis. \\because\\overrightarrow{PM}=2\\overrightarrow{PF}, F is the midpoint of PM, so A(-\\frac{p}{2},0) is the midpoint of PB. AF is the midline of \\trianglePMB, |AF|=\\frac{1}{2}|MB| \\therefore|BM|=2p, i.e., |FM|=2p, \\therefore|PF|=2p=2|AF|, \\therefore\\angleAPF=\\frac{\\pi}{6}, \\angleAFP=\\frac{\\pi}{3}. The inclination angle of line l is \\frac{\\pi}{3} or \\frac{2\\pi}{3}"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$, the right vertex and right focus are denoted as $A$ and $F$, respectively. The intersection point of its left directrix with the $x$-axis is $B$. If $A$ is the midpoint of segment $B F$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;B: Point;F: Point;A: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightVertex(C) = A;RightFocus(C) = F;Intersection(LeftDirectrix(C), xAxis) = B ;MidPoint(LineSegmentOf(B, F)) = A",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2) + 1",
    "fact_spans": "[[[2, 64], [83, 84], [118, 124]], [[10, 64]], [[10, 64]], [[97, 100]], [[79, 82]], [[75, 78], [102, 105]], [[10, 64]], [[10, 64]], [[2, 64]], [[2, 82]], [[2, 82]], [[83, 100]], [[102, 116]]]",
    "query_spans": "[[[118, 130]]]",
    "process": "From the given conditions: B(-\\frac{a^{2}}{c},0), A(a,0), F(c,0), then 2a = c - \\frac{a}{c}, which implies e^{2} - 2e - 1 = 0. Solving gives e = \\sqrt{2} + 1."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ with right focus $F$, and let $A$ be a point on the ellipse in the first quadrant. Connect $AF$ and extend it to intersect the ellipse again at point $B$. Connect $AO$ (where $O$ is the origin) and extend it to intersect the ellipse again at point $C$. If $S_{\\triangle ABC}=3$, then the coordinates of point $A$ are?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);F: Point;RightFocus(G) = F;A: Point;PointOnCurve(A, G) = True;Quadrant(A) = 1;Intersection(OverlappingLine(LineSegmentOf(A, F)), G) = B;B: Point;Intersection(OverlappingLine(LineSegmentOf(A, O)), G) = C;C: Point;O: Origin;Area(TriangleOf(A, B, C)) = 3",
    "query_expressions": "Coordinate(A)",
    "answer_expressions": "(1,3/2)",
    "fact_spans": "[[[2, 39], [52, 54], [74, 76], [103, 105]], [[2, 39]], [[44, 47]], [[2, 47]], [[48, 51], [138, 142]], [[48, 62]], [[48, 62]], [[63, 81]], [[77, 81]], [[82, 111]], [[107, 111]], [[90, 93]], [[113, 136]]]",
    "query_spans": "[[[138, 147]]]",
    "process": "Find F(1,0), let the equation of AB be x = my + 1, combine with the ellipse equation, apply Vieta's formulas and the perfect square formula, and according to the problem we have S_{\\triangleABO} = S_{\\triangleAOF} + S_{\\triangleBOF} = \\frac{1}{2} \\cdot |OF| \\cdot |y_{1} - y_{2}| = \\frac{3}{2}, thus |y_{1} - y_{2}| = 3. Squaring both sides and using Vieta's formulas, solve the equation to get m = 0, then obtain the coordinates of A. According to the problem, F(1,0), let the equation of AB be x = my + 1, combining with the ellipse equation gives (4 + 3m^{2})y^{2} + 6my - 9 = 0, let A(x_{1},y_{1}), B(x_{2},y_{2}), \\frac{m}{3m^{2}}'^{'} y_{1}y_{2} = y_{2}^{2} - 4y_{1}y_{2} = \\frac{36m2}{(4+3m)^{2}}. Since O is the midpoint of AC and the area of \\triangleABC is 3, the area of \\triangleABO is \\frac{3}{2}. S_{\\triangleABO} = S_{\\triangleAOF} + S_{\\triangleBOF} = \\frac{1}{2} \\cdot |OF| \\cdot |y_{1} - y_{2}| = \\frac{3}{2}, so |y_{1} - y_{2}| = 3, leading to \\frac{36m^{2}}{(4+3m^{2})^{2}} + \\frac{36}{4+3m^{2}} = 9, simplifying to 9m^{4} + m^{2} = 0, hence m = 0, then AB \\bot x-axis, giving A(1,\\frac{3}{2})."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{4}=1$ $(a>2)$ has a focal distance of $2 \\sqrt{5}$, then the length of the major axis of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/4 + x^2/a^2 = 1);a: Number;a>2;FocalLength(C) = 2*sqrt(5)",
    "query_expressions": "Length(MajorAxis(C))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 53], [71, 76]], [[2, 53]], [[9, 53]], [[9, 53]], [[2, 69]]]",
    "query_spans": "[[[71, 82]]]",
    "process": "\\because a>2, so the foci of ellipse C are on the x-axis, and a=\\sqrt{b^{2}+c^{2}}=\\sqrt{4+5}=3. Therefore, the major axis length of ellipse C is 2a=6."
  },
  {
    "text": "The coordinates of the left focus of the ellipse $\\frac{x^{2}}{7}+\\frac{y^{2}}{5}=1$ are?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/7 + y^2/5 = 1)",
    "query_expressions": "Coordinate(LeftFocus(G))",
    "answer_expressions": "(-\\sqrt{2},0)",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 46]]]",
    "process": "From the ellipse equation, obtain $a^{2}=7$, $b^{2}=5$, then find the value of $c$. According to the problem, the ellipse $\\frac{x^{2}}{7}+\\frac{y^{2}}{5}=1$ gives $a^{2}=7$, $b^{2}=5$. Since $c^{2}=a^{2}-b^{2}=7-5=2$, it follows that $c=\\sqrt{2}$, so the coordinates of the left focus are $(-\\sqrt{2},0)$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with eccentricity $e=\\sqrt{3}$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through point $F_{1}$ intersects the left and right branches of hyperbola $C$ at points $A$ and $B$ respectively. If $|A B|=4$, $|A F_{2}|=|B F_{2}|$, then $|F_{1} F_{2}|$=?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;A: Point;B: Point;F2: Point;F1: Point;e:Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(C) = e;e=sqrt(3);LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(F1, G);Intersection(G,LeftPart(C))=A;Intersection(G,RightPart(C))=B;Abs(LineSegmentOf(A, B)) = 4;Abs(LineSegmentOf(A, F2)) = Abs(LineSegmentOf(B, F2))",
    "query_expressions": "Abs(LineSegmentOf(F1, F2))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[2, 63], [118, 124]], [[10, 63]], [[10, 63]], [[115, 117]], [[135, 138]], [[139, 142]], [[97, 104]], [[89, 96], [106, 114]], [[67, 79]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 79]], [[67, 79]], [[2, 104]], [[2, 104]], [[105, 117]], [[115, 144]], [[115, 144]], [[146, 155]], [[156, 177]]]",
    "query_spans": "[[[179, 196]]]",
    "process": "According to the problem, we have\n\\begin{cases}\n|BF_{1}|-|BF_{2}|=2a \\\\\n|AF_{2}|-|AF_{1}|=2a \\\\\n|BF|=|BF_{1}|=2a|=4\n\\end{cases}.\nSolve for $ a $ using the given information. Also, the eccentricity of hyperbola $ C $ is $ e=\\sqrt{3} $, so $ c=\\sqrt{3} $, that is, $ |F_{1}F_{2}|=2\\sqrt{3} $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{12}+\\frac{y^{2}}{4}=1$, let the line $l$: $y=x+m(m \\in R)$ intersect the ellipse $C$ at two distinct points $A$, $B$, and $|A B|=3 \\sqrt{2}$. If the point $P(x_{0}, 2)$ satisfies $|\\overrightarrow{P A}|=|\\overrightarrow{P B}|$, then $x_{0}$=?",
    "fact_expressions": "l: Line;C: Ellipse;P: Point;A: Point;B: Point;x0:Number;m:Real;Expression(C) = (x^2/12 + y^2/4 = 1);Coordinate(P) = (x0, 2);Expression(l) = (y = x + m);Intersection(l, C) = {A, B};Negation(A=B);Abs(LineSegmentOf(A, B)) = 3*sqrt(2);Abs(VectorOf(P, A)) = Abs(VectorOf(P, B))",
    "query_expressions": "x0",
    "answer_expressions": "{-1,-3}",
    "fact_spans": "[[[48, 71]], [[2, 45], [72, 77]], [[114, 128]], [[83, 86]], [[87, 90]], [[180, 187]], [[55, 71]], [[2, 45]], [[114, 128]], [[48, 71]], [[48, 90]], [[79, 90]], [[93, 111]], [[130, 177]]]",
    "query_spans": "[[[180, 189]]]",
    "process": "From \\begin{cases}y=x+m\\\\\\frac{x^{2}}{12}+\\frac{y^{2}}{4}=1\\end{cases}, we obtain $4x^{2}+6mx+3m^{2}-12=0$ \\textcircled{1}. Since line $l$ intersects ellipse $C$ at two distinct points $A$ and $B$, we have $\\Delta=36m^{2}-16(3m^{2}-12)>0$, which gives $m^{2}<16$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $x_{1}, x_{2}$ are the roots of equation \\textcircled{1}, so $x_{1}+x_{2}=-\\frac{3m}{2}$, $x_{1}x_{2}=\\frac{3m^{2}-12}{4}$. Thus, $|AB|=\\sqrt{1+k^{2}}|x_{1}-x_{2}|=\\sqrt{2}\\times\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\sqrt{2}\\times\\sqrt{-\\frac{3}{4}m^{2}+12}$. Given $|AB|=3\\sqrt{2}$, we get $-\\frac{3}{4}m^{2}+12=9$, solving yields $m=\\pm2$. According to the problem, point $P$ is the intersection point of the perpendicular bisector of segment $AB$ and the line $y=2$. Let the midpoint of $AB$ be $E(x_{0},y_{0})$, then $x_{0}=\\frac{x_{1}+x_{2}}{2}=-\\frac{3m}{4}$, $y_{0}=x_{0}+m=\\frac{m}{4}$. When $m=2$, $E(-\\frac{3}{2},\\frac{1}{2})$, setting $y=2$, we get $x_{0}=-3$. When $m=-2$, $E(\\frac{3}{2},-\\frac{1}{2})$. Then, the perpendicular bisector of segment $AB$ has equation $y+\\frac{1}{2}=-(x-\\frac{3}{2})$, i.e., $y=-x+1$. Setting $y=2$, we get $x_{0}=-1$. In conclusion, the value of $x_{0}$ is $-3$ or $-1$."
  },
  {
    "text": "If the line $y=k(x+1)$ $(k>0)$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$, and the projections of points $A$ and $B$ on the directrix of the parabola are $M$ and $N$ respectively, and if $|BN|=2|AM|$, then the value of $k$ is?",
    "fact_expressions": "G: Parabola;H: Line;B: Point;N: Point;A: Point;M: Point;k:Number;Expression(G) = (y^2 = 4*x);k>0;Expression(H) = (y = k*(x + 1));Intersection(H, G) = {A, B};Projection(A,Directrix(G))=M;Projection(B,Directrix(G))=N;Abs(LineSegmentOf(B, N)) = 2*Abs(LineSegmentOf(A, M))",
    "query_expressions": "k",
    "answer_expressions": "2*sqrt(2)/3",
    "fact_spans": "[[[19, 33], [57, 60]], [[1, 18]], [[40, 43], [51, 54]], [[74, 77]], [[36, 39], [47, 50]], [[70, 73]], [[95, 98]], [[19, 33]], [[3, 18]], [[1, 18]], [[1, 45]], [[47, 77]], [[47, 77]], [[79, 93]]]",
    "query_spans": "[[[95, 102]]]",
    "process": "Let D(-1,0), then the line y=k(x+1) must pass through D(-1,0). Let A(x_{A},y_{A}), B(x_{B},y_{B}), then from |BN|=2|AM| it follows that A is the midpoint of BD, so {x_{A}=\\frac{-1+x_{B}}{xx_{0}=1}, \\begin{cases}x_{A}=\\frac{1}{2},\\\\x_{n}=2\\end{cases}, then B(2,2\\sqrt{2}) substituted into the line y=k(x+1), we have 2\\sqrt{2}=k(2+1), x_{A}x_{B}=k=\\frac{2\\sqrt{2}}{2}"
  },
  {
    "text": "Given the parabola $x^{2}=8 y$ with focus $F$ and directrix $l$, let point $P$ be a point on the parabola, and draw $P Q \\perp l$ with foot $Q$. If $|P F|=4$, then $\\angle F Q P$=?",
    "fact_expressions": "G: Parabola;P: Point;Q: Point;F: Point;l: Line;Expression(G) = (x^2 = 8*y);Focus(G) = F;Directrix(G) = l;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, Q), l);FootPoint(LineSegmentOf(P, Q), l) = Q;Abs(LineSegmentOf(P, F)) = 4",
    "query_expressions": "AngleOf(F, Q, P)",
    "answer_expressions": "ApplyUnit(45, degree)",
    "fact_spans": "[[[2, 16], [32, 35]], [[38, 41]], [[59, 62]], [[20, 23]], [[27, 30]], [[2, 16]], [[2, 23]], [[2, 30]], [[32, 41]], [[42, 55]], [[42, 62]], [[64, 73]]]",
    "query_spans": "[[[75, 91]]]",
    "process": "Since the focus of the parabola $ x^{2}=8y $ is $ F(0,2) $, the directrix is $ l: y=-2 $. Draw $ PQ \\perp l $ from a point $ P $ on the parabola, with foot $ Q $. If $ |PF|=4 $, then point $ P(\\pm4,2) $, and triangle $ PQF $ is an isosceles triangle, so $ \\angle FQP=45^{\\circ} $."
  },
  {
    "text": "Given that an ellipse is tangent to the $x$-axis, with left and right foci at $F_{1}(1,1)$ and $F_{2}(5,2)$ respectively, what is the distance from the origin $O$ to its left directrix?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Coordinate(F1) = (1, 1);Coordinate(F2) = (5, 2);IsTangent(G, xAxis);LeftFocus(G) =F1;RightFocus(G)=F2;O:Origin",
    "query_expressions": "Distance(O,LeftDirectrix(G))",
    "answer_expressions": "sqrt(17)/17",
    "fact_spans": "[[[2, 4], [56, 57]], [[22, 34]], [[36, 48]], [[22, 34]], [[36, 48]], [[2, 11]], [[2, 48]], [[2, 48]], [[50, 55]]]",
    "query_spans": "[[[50, 65]]]",
    "process": "This problem exceeds the requirements of Jiangsu's college entrance examination mathematics. Students may treat it as casual entertainment. Since the ellipse in this problem is not in standard form, we can only solve it based on the definition of an ellipse. $k_{F_{1}F_{2}}=\\frac{2-1}{5-1}=\\frac{1}{4}$, so the equation of the line containing the minor axis of the ellipse is $y-\\frac{3}{2}=-4(x-3)$, or $4x+y-\\frac{27}{2}=0$. The distance from the origin $O$ to this line is $\\frac{27}{\\sqrt{4^{2}+1}}=\\frac{27}{34}\\sqrt{17}$. By the optical property of ellipses (in fact, all conic sections): a ray emitted from one focus reflects off the ellipse and passes through the other focus (or its extension). In this problem, the tangent line is the $x$-axis; let the point of tangency be $P(x,0)$. Then $k_{PF_{1}}=-k_{PF_{2}}$, thus $\\frac{-1}{x-1}=-\\frac{0-2}{x-5}$, solving gives $x=\\frac{7}{3}$. Therefore, $2a=|PF_{1}|+|PF_{2}|=5$, so $a=\\frac{5}{2}$. Also, $2c=|F_{1}F_{2}|=\\sqrt{17}$, hence $c=\\frac{\\sqrt{17}}{2}$. Thus, $\\frac{a^{2}}{c}=\\frac{25\\sqrt{17}}{34}$. Therefore, the distance from the origin to the left directrix should be $\\frac{27\\sqrt{17}}{34}-\\frac{25\\sqrt{17}}{34}$."
  },
  {
    "text": "If the two foci of an ellipse are $F_{1}(-4 , 0)$, $F_{2}(4 , 0)$, and point $P$ lies on the ellipse such that the maximum area of $\\Delta P F_{1} F_{2}$ is $12$, then the equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;P: Point;Coordinate(F1) = (-4, 0);Coordinate(F2) = (4, 0);Focus(G) = {F1,F2};PointOnCurve(P, G);Max(Area(TriangleOf(P, F1, F2))) = 12",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/25+y^2/9=1",
    "fact_spans": "[[[1, 3], [45, 47], [87, 89]], [[7, 22]], [[24, 39]], [[40, 44]], [[7, 22]], [[25, 39]], [[1, 39]], [[40, 48]], [[50, 84]]]",
    "query_spans": "[[[87, 94]]]",
    "process": ""
  },
  {
    "text": "Let $P$ be a moving point on the parabola $y^{2}=8x$, and let $M$ be the projection of $P$ onto the $y$-axis. Given point $A(4,6)$, then the minimum value of $|PA|+|PM|$ is?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;M: Point;Expression(G) = (y^2 = 8*x);Coordinate(A) = (4, 6);PointOnCurve(P, G);Projection(P, yAxis) = M",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "2*sqrt(10) - 2",
    "fact_spans": "[[[5, 19]], [[41, 50]], [[1, 4], [24, 27]], [[36, 40]], [[5, 19]], [[41, 50]], [[1, 23]], [[24, 40]]]",
    "query_spans": "[[[52, 71]]]",
    "process": "Given the focus $ F(2,0) $, the directrix $ x = -2 $, extend $ PM $ to intersect the directrix at point $ H $, then $ |PF| = |PH| $. Therefore, $ |PM| = |PH| - 2 = |PF| - 2 $, so $ |PM| + |PA| = |PF| + |PA| - 2 $. Thus, it suffices to find the minimum value of $ |PF| + |PA| $. Since point $ A $ lies outside the parabola, by the triangle inequality that the sum of any two sides is greater than the third side, we have $ |PF| + |PA| \\geqslant |FA| $. The minimum value $ |FA| $ is attained when point $ P $ is the intersection point of segment $ FA $ and the parabola. Using the distance formula between two points, $ |FA| = 2\\sqrt{10} $. Hence, the minimum value of $ |PA| + |PM| $ is $ 2\\sqrt{10} - 2 $."
  },
  {
    "text": "Given fixed points $A(-3,0)$, $B(3,0)$, lines $AM$ and $BM$ intersect at point $M$, and the product of their slopes is $-\\frac{1}{9}$, then the trajectory equation of the moving point $M$ is?",
    "fact_expressions": "M: Point;A: Point;B: Point;Coordinate(A) = (-3, 0);Coordinate(B) = (3, 0);Intersection(LineOf(A,M),LineOf(B,M))=M;Slope(LineOf(A,M))*Slope(LineOf(B,M))=-1/9",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "(x^2/9+y^2=1)&Negation(x=pm*3)",
    "fact_spans": "[[[41, 45], [73, 76]], [[4, 13]], [[14, 22]], [[4, 13]], [[14, 22]], [[24, 45]], [[47, 69]]]",
    "query_spans": "[[[73, 83]]]",
    "process": "Let point M(x, y), (x ≠ ±3). Since k_{MA} = \\frac{y}{x+3}, k_{MB} = \\frac{y}{x-3}, the trajectory equation of moving point M is \\frac{x^{2}}{9} + y^{2} = 1 (x ≠ ±3)."
  },
  {
    "text": "If the line $y=x+1$ intersects the parabola $x^{2}=2 y$ at points $A$ and $B$, then the coordinates of the midpoint of segment $AB$ are?",
    "fact_expressions": "A: Point;B: Point;G: Parabola;H: Line;Expression(G) = (x^2 = 2*y);Expression(H) = (y = x + 1);Intersection(H, G) = {A, B}",
    "query_expressions": "Coordinate(MidPoint(LineSegmentOf(A, B)))",
    "answer_expressions": "(1, 2)",
    "fact_spans": "[[[28, 31]], [[34, 37]], [[11, 25]], [[1, 10]], [[11, 25]], [[1, 10]], [[1, 39]]]",
    "query_spans": "[[[41, 54]]]",
    "process": "The solution process is omitted"
  },
  {
    "text": "The ellipse $m x^{2}+n y^{2}=1$ intersects the line $y=1-x$ at points $M$ and $N$. If the slope of the line connecting the origin $O$ and the midpoint $P$ of segment $M N$ is $\\frac{\\sqrt{2}}{2}$, then the value of $\\frac{m}{n}$ is?",
    "fact_expressions": "G: Ellipse;n: Number;m: Number;H: Line;M: Point;N: Point;Expression(G) = (m*x^2 + n*y^2 = 1);Expression(H) = (y = 1 - x);Intersection(G, H) = {M, N};O: Origin;P: Point;MidPoint(LineSegmentOf(M, N)) = P;Slope(LineSegmentOf(O, P)) = sqrt(2)/2",
    "query_expressions": "m/n",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[0, 21]], [[91, 104]], [[91, 104]], [[22, 31]], [[33, 36]], [[37, 40]], [[0, 21]], [[22, 31]], [[0, 42]], [[44, 49]], [[60, 63]], [[50, 63]], [[44, 89]]]",
    "query_spans": "[[[91, 108]]]",
    "process": "Let points $M(x_{1},y_{1})$, $N(x_{2},y_{2})$. Using the point difference method, we obtain $k_{MN}\\cdot k_{OP} = -\\frac{m}{n}$, and combining with the given conditions, we can find the value of $\\frac{m}{n}$. Let points $M(x_{1},y_{1})$, $N(x_{2},y_{2})$, then\n$$\n\\begin{cases}\nmx_{1}^{2} + ny_{1}^{2} = 1 \\\\\nmx_{2}^{2} + ny_{2}^{2} = 1\n\\end{cases}\n$$\nSubtracting the two equations gives\n$$\nm(x_{1}-x_{2})(x_{1}+x_{2}) + n(y_{1}-y_{2})(y_{1}+y_{2}) = 0,\n$$\nwhich simplifies to\n$$\n\\frac{y_{1}-y_{2}}{x_{1}-x_{2}} \\cdot \\frac{y_{1}+y_{2}}{x_{1}+x_{2}} = -\\frac{m}{n}.\n$$\nLet point $P(x_{0},y_{0})$, then $\\frac{y_{0}}{x_{0}} = \\frac{\\sqrt{2}}{2}$. Since\n$$\n\\begin{cases}\nx_{1}+x_{2} = 2x_{0} \\\\\ny_{1}+y_{2} = 2y_{0}\n\\end{cases},\n$$\nit follows that $k_{MN} \\cdot k_{OP} = -\\frac{m}{n}$, namely $-\\frac{m}{n} = (-1) \\times \\frac{\\sqrt{2}}{2}$. Therefore, $\\frac{m}{n} = \\frac{\\sqrt{2}}{2}$."
  },
  {
    "text": "Given that point $P(m, n)$ is a moving point on the parabola $y = -\\frac{1}{4} x^{2}$, find the minimum value of $\\sqrt{m^{2}+(n+1)^{2}}+\\sqrt{(m-4)^{2}+(n+5)^{2}}$.",
    "fact_expressions": "G: Parabola;P: Point;Expression(G) = (y = -x^2/4);Coordinate(P) = (m, n);PointOnCurve(P, G);m: Number;n: Number",
    "query_expressions": "Min(sqrt(m^2 + (n + 1)^2) + sqrt((m - 4)^2 + (n + 5)^2))",
    "answer_expressions": "6",
    "fact_spans": "[[[13, 38]], [[2, 12]], [[13, 38]], [[2, 12]], [[2, 42]], [[3, 12]], [[3, 12]]]",
    "query_spans": "[[[44, 101]]]",
    "process": "By the geometric meaning of $\\sqrt{m^{2}+(n+1)^{2}}+\\sqrt{(m-4)^{2}+(n+5)^{2}}$ and the definition of a parabola, this is transformed into the distance from the point $(4,-5)$ to the directrix $l$ of the parabola. From $y=-\\frac{1}{4}x^{2}$, we obtain $x^{2}=-4y$, so the focus of $y=-\\frac{1}{4}x^{2}$ is $F(0,-1)$, and the directrix is $l: y=1$. The geometric meaning of $\\sqrt{m^{2}+(n+1)^{2}}+\\sqrt{(m-4)^{2}+(n+5)^{2}}$ is the sum of the distances from point $P(m,n)$ to $F(0,-1)$ and point $A(4,-5)$. Therefore, the minimum value of $\\sqrt{m^{2}+(n+1)^{2}}+\\sqrt{(m-4)^{2}+(n+5)^{2}}$ is $1-(-5)=6$."
  },
  {
    "text": "If the center of the hyperbola is at the origin, the foci are on the coordinate axes, the eccentricity is $\\sqrt{2}$, and it passes through the point $(4, -\\sqrt{10})$, then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;O: Origin;Center(G) = O;PointOnCurve(Focus(G),axis) = True;Eccentricity(G) = sqrt(2);H: Point;Coordinate(H) = (4, -sqrt(10));PointOnCurve(H,G) = True",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/6 - y^2/6 = 1",
    "fact_spans": "[[[1, 4], [56, 59]], [[7, 9]], [[1, 9]], [[1, 17]], [[1, 32]], [[35, 53]], [[35, 53]], [[1, 53]]]",
    "query_spans": "[[[56, 66]]]",
    "process": "From $ e = \\sqrt{2} $, we get: $ a = b $. Let the equation be $ \\frac{x^{2}}{m} - \\frac{y^{2}}{m} = 1 $, substituting the point $ (4, -\\sqrt{10}) $, the solution can be obtained. According to the problem, $ e = \\sqrt{2} $, $ c^{2} = 2a^{2} $, we get: $ a = b $, let the equation be $ \\frac{x^{2}}{m} - \\frac{y^{2}}{m} = 1 $, substituting the point $ (4, -\\sqrt{10}) $, then $ \\frac{16}{m} - \\frac{10}{m} = 1 $, solving gives $ m = 6 $. $ \\frac{x^{2}}{6} - \\frac{y^{2}}{6} = $"
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $2x+y=0$, and one focus is $(\\sqrt{5}, 0)$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(H) = (sqrt(5), 0);Expression(OneOf(Asymptote(G))) = (2*x + y = 0);OneOf(Focus(G))=H",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/4 = 1",
    "fact_spans": "[[[2, 58], [97, 100]], [[5, 58]], [[5, 58]], [[80, 95]], [[5, 58]], [[5, 58]], [[2, 58]], [[80, 95]], [[2, 74]], [[2, 95]]]",
    "query_spans": "[[[97, 107]]]",
    "process": "Since the asymptote is $2x + y = 0$, we have  \n\\[\n\\begin{cases}\n\\frac{b}{a} = 2 \\\\\nc = \\sqrt{5} \\\\\nc^{2} = b^{2} + a^{2}\n\\end{cases}\n\\]  \nSolving gives $a = 1$, $b = 2$, so the standard equation of the hyperbola is $x^{2} - \\frac{y^{2}}{4} = 1$."
  },
  {
    "text": "The minimum distance from any point $P$ on the parabola $y^{2}=-4 x$ to the left vertex of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{15}=1$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -4*x);P: Point;PointOnCurve(P, G);H: Ellipse;Expression(H) = (x^2/16 + y^2/15 = 1)",
    "query_expressions": "Min(Distance(P, LeftVertex(H)))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 15]], [[0, 15]], [[19, 22]], [[0, 22]], [[23, 62]], [[23, 62]]]",
    "query_spans": "[[[19, 72]]]",
    "process": ""
  },
  {
    "text": "Let a line passing through the focus of the parabola $y^{2}=2 p x$ intersect the parabola at two points, and let the ordinates of the two points be $y_{1}$, $y_{2}$. Then $y_{1} y_{2}$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;H: Line;PointOnCurve(Focus(G), H);NumIntersection(H, G) = 2;A: Point;B: Point;Intersection(H, G) = {A, B};y1: Number;y2: Number;YCoordinate(A) = y1;YCoordinate(B) = y2",
    "query_expressions": "y1*y2",
    "answer_expressions": "-p^2",
    "fact_spans": "[[[2, 18], [27, 30]], [[2, 18]], [[5, 18]], [[24, 26]], [[1, 26]], [[24, 35]], [], [], [[24, 35]], [[46, 53]], [[55, 62]], [[24, 62]], [[24, 62]]]",
    "query_spans": "[[[64, 79]]]",
    "process": "Let the equation of the line be $x = my + \\frac{p}{2}$. Solving this simultaneously with the parabola $y^2 = 2px$ gives $y^2 - 2pmy - p^2 = 0$. Therefore, by Vieta's formulas, we obtain $y_1 y_2 = -p^2$."
  },
  {
    "text": "What is the equation of the right directrix of the hyperbola $x^{2}-y^{2}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 1)",
    "query_expressions": "Expression(RightDirectrix(G))",
    "answer_expressions": "x=3/2",
    "fact_spans": "[[[0, 18]], [[0, 18]]]",
    "query_spans": "[[[0, 26]]]",
    "process": "Since the right directrix equation of the hyperbola $\\frac{x}{3}-y^{2}=1$ is $x=\\frac{a^{2}}{c}$, and $a^{2}=3$, $b^{2}=1$, $c^{2}=3+1=4$, $c=2$, therefore $x=\\frac{3}{2}$."
  },
  {
    "text": "If point $P$ is a point on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ with foci $F_{1}$ and $F_{2}$, satisfying $P F_{1} \\perp P F_{2}$ and $|P F_{1}|=2|P F_{2}|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2};IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[26, 72], [127, 130]], [[29, 72]], [[29, 72]], [[1, 5]], [[7, 14]], [[15, 22]], [[26, 72]], [[1, 75]], [[6, 72]], [[78, 101]], [[103, 125]]]",
    "query_spans": "[[[127, 136]]]",
    "process": "From the definition of the hyperbola, we have |PF_{1}| - |PF_{2}| = 2a. Since |PF_{1}| = 2|PF_{2}|, it follows that |PF_{1}| = 4a, |PF_{2}| = 2a. Also, since PF_{1} \\perp PF_{2}, we have |PF_{1}|^{2} + |PF_{2}|^{2} = |F_{1}F_{2}|^{2}, that is, (4a)^{2} + (2a)^{2} = (2c)^{2}. Simplifying gives c^{2} = 5a^{2}, so e = \\frac{c}{a} = \\sqrt{5}."
  },
  {
    "text": "Given that the eccentricity of a parabola is $e$, and the coordinates of the focus are $(0, e)$, then the standard equation of the parabola is?",
    "fact_expressions": "G: Parabola;Coordinate(Focus(G)) = (0, e);Eccentricity(G) = e;e:Number",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2=4*y",
    "fact_spans": "[[[2, 5], [29, 32]], [[2, 27]], [[2, 13]], [[10, 13]]]",
    "query_spans": "[[[29, 39]]]",
    "process": "From e=1, the focus coordinates are (0,1), so the standard equation of the parabola is x^{2}=4y"
  },
  {
    "text": "If an asymptote of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1(b>0)$ passes through the point $P(1 , 2)$, then the value of $b$ is?",
    "fact_expressions": "G: Hyperbola;b: Number;P: Point;b>0;Expression(G) = (x^2/4 - y^2/b^2 = 1);Coordinate(P) = (1, 2);PointOnCurve(P,OneOf(Asymptote(G)))",
    "query_expressions": "b",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 48]], [[68, 71]], [[55, 66]], [[4, 48]], [[1, 48]], [[55, 66]], [[1, 66]]]",
    "query_spans": "[[[68, 75]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the right focus is $F$, the left vertex is $A$, and the endpoints of the imaginary axis are $B_{1}$ and $B_{2}$. If $F$, $A$, $B_{1}$, $B_{2}$ lie on the same circle, then the eccentricity of the hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(G) = F;A: Point;LeftVertex(G) = A;Endpoint(ImageinaryAxis(G)) = {B1, B2};B1: Point;B2: Point;PointOnCurve(F, H) ;PointOnCurve(A, H) ;PointOnCurve(B1, H) ;PointOnCurve(B2, H) ;H: Circle",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(1+sqrt(5))/2",
    "fact_spans": "[[[2, 59], [134, 137]], [[2, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[64, 67], [103, 106]], [[2, 67]], [[72, 75], [107, 110]], [[2, 75]], [[2, 101]], [[86, 93], [111, 118]], [[94, 101], [119, 126]], [[103, 132]], [[103, 132]], [[103, 132]], [[103, 132]], [[130, 131]]]",
    "query_spans": "[[[134, 144]]]",
    "process": "According to the problem, the center of the circle lies at the intersection of the perpendicular bisectors of AF and $ B_{1}B_{2} $. Since $ B_{1}, B_{2} $ are the endpoints of the imaginary axis, the perpendicular bisector is the x-axis. Therefore, the center lies at the midpoint $ C $ of AF, and AF is the diameter, so $ AB_{1} \\perp FB_{1} $. The coordinates of the points are: $ A(-a,0) $, $ B_{1}(0,b) $, $ F(c,0) $, thus $ \\overrightarrow{AB_{1}} = (a,b) $, $ \\overrightarrow{FB_{1}} = (-c,b) $. We obtain: $ -ac + b^{2} = 0 $, i.e., $ c^{2} - a^{2} - ac = 0 $. Dividing both sides by $ a^{2} $ gives: $ e^{2} - e - 1 = 0 $, with $ e > 1 $. Solving yields: $ e = \\frac{1 + \\sqrt{5}}{2} $"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the common foci of the ellipse $C_{1}$ and the hyperbola $C_{2}$, $P$ is a common point of $C_{1}$ and $C_{2}$, and $\\angle F_{1} P F_{2}=\\frac{1}{3} \\pi$, the eccentricities of the ellipse $C_{1}$ and the hyperbola $C_{2}$ are $e_{1}$ and $e_{2}$ respectively, then the maximum value of $\\frac{1}{e_{1}}+\\frac{1}{e_{2}}$ is?",
    "fact_expressions": "C1:Ellipse;C2:Hyperbola;F1: Point;P: Point;F2: Point;e1:Number;e2:Number;Focus(C1)={F1,F2};Focus(C2)={F1,F2};OneOf(Intersection(C1,C2))=P;AngleOf(F1, P, F2) = pi/3;Eccentricity(C1)=e1;Eccentricity(C2)=e2",
    "query_expressions": "Max(1/e2 + 1/e1)",
    "answer_expressions": "sqrt(3)*4/3",
    "fact_spans": "[[[18, 27], [48, 55], [110, 119]], [[28, 38], [56, 63], [120, 130]], [[2, 9]], [[44, 47]], [[10, 17]], [[137, 145]], [[147, 154]], [[2, 43]], [[2, 43]], [[44, 69]], [[71, 109]], [[110, 154]], [[110, 154]]]",
    "query_spans": "[[[156, 195]]]",
    "process": "Let the major axis of ellipse $ C_{1} $ be $ 2a_{1} $, the real axis of hyperbola $ C_{2} $ be $ 2a_{2} $, and the common focal distance be $ 2c $. Let $ |PF_{1}| = r_{1} $, $ |PF_{2}| = r_{2} $, without loss of generality assume $ r_{1} > r_{2} $, then we have $ r_{1} + r_{2} = 2a_{1} $, $ r_{1} - r_{2} = 2a_{2} $, $ r_{1} = a_{1} + a_{2} $, $ r_{2} = a_{1} - a_{2} $. Therefore, in $ \\triangle PF_{1}F_{2} $, combining the law of cosines we get substituting yields $ 4c^{2} = 2a_{1}^{2} + 2a_{2}^{2} - a_{1}^{2} + a_{2}^{2} = a_{1}^{2} + 3a_{2}^{2} $, so $ \\frac{1}{e_{1}^{2}} + \\frac{3}{e_{2}^{2}} = 4 $. Using the Cauchy-Schwarz inequality, the solution can be obtained. Let the major axis of ellipse $ C_{1} $ be $ 2a_{1} $, the real axis of hyperbola $ C_{2} $ be $ 2a_{2} $, the common focal distance be $ 2c $, let $ |PF_{1}| = r_{1} $, $ |PF_{2}| = r_{2} $, without loss of generality assume $ r_{1} > r_{2} $, then we have $ r_{1} + r_{2} = 2a_{1} $, $ r_{1} - r_{2} = 2a_{2} $, $ r_{1} = a_{1} + a_{2} $, $ r_{2} = a_{1} - a_{2} $. Given $ \\angle F_{1}PF_{2} = \\frac{1}{3}\\pi $, therefore in $ \\triangle PF_{1}F_{2} $, we have $ |F_{1}F_{2}|^{2} = r_{1}^{2} + r_{2}^{2} - 2r_{1}r_{2}\\cos\\angle F_{1}PF_{2} $. Substituting we get $ 4c^{2} = (a_{1} + a_{2})^{2} + (a_{1} - a_{2})^{2} - 2(a_{1} + a_{2})(a_{1} - a_{2}) \\times \\frac{1}{2} = 2a_{1}^{2} + 2a_{2}^{2} - a_{1}^{2} + a_{2}^{2} = a_{1}^{2} + 3a_{2}^{2} $, so $ \\frac{1}{e_{1}^{2}} + \\frac{3}{e_{2}^{2}} = 4 $. $ (\\frac{1}{e_{1}} + \\frac{1}{e_{2}})^{2} = (\\frac{1}{e_{1}} \\times 1 + \\frac{\\sqrt{3}}{e_{2}} \\times \\frac{1}{\\sqrt{3}})^{2} \\leqslant [(\\frac{1}{e_{1}})^{2} + (\\frac{\\sqrt{3}}{e_{2}})^{2}][1^{2} + (\\frac{1}{\\sqrt{3}})^{2}] = (\\frac{1}{e_{1}^{2}} + \\frac{3}{e_{2}^{2}}) \\times \\frac{4}{3} = \\frac{16}{3} $. Therefore $ \\frac{1}{e_{1}} + \\frac{1}{e_{2}} \\leqslant \\frac{4}{3}\\sqrt{3} $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the right branch of the hyperbola, and $|P F_{1}|=4|P F_{2}|$. Then the range of values for the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F1)) = 4*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1, 5/3]",
    "fact_spans": "[[[2, 58], [88, 91], [122, 125]], [[5, 58]], [[5, 58]], [[83, 87]], [[67, 74]], [[75, 82]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 82]], [[2, 82]], [[83, 95]], [[97, 119]]]",
    "query_spans": "[[[122, 136]]]",
    "process": ""
  },
  {
    "text": "The standard equation of a parabola with directrix $x+1=0$ is?",
    "fact_expressions": "G: Parabola;Expression(Directrix(G)) = (x + 1 = 0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = 4*x",
    "fact_spans": "[[[13, 16]], [[0, 16]]]",
    "query_spans": "[[[13, 22]]]",
    "process": "From the directrix equation of the parabola $x = -1$, it is known that the parabola opens to the right, where $\\frac{p}{2} = 1$, yielding $p = 2$, so the standard equation of the parabola is $y^{2} = 4x$."
  },
  {
    "text": "Given the hyperbola $W$: $ \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with left focus at $F(-c, 0)$, the line $l$: $ y=\\frac{\\sqrt{3}}{3}(x+c) $ intersects the left and right branches of $W$ at points $A$ and $B$, respectively, intersects the $y$-axis at point $C$, and $O$ is the origin. If $\\overrightarrow{O A} \\cdot \\overrightarrow{C F}=\\frac{c^{2}}{3}$, then the eccentricity of $W$ is?",
    "fact_expressions": "l: Line;W: Hyperbola;b: Number;a: Number;c:Number;F: Point;O: Origin;A: Point;C: Point;B: Point;a>0;b>0;Expression(W) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (-c, 0);LeftFocus(W) = F;Expression(l)=(y = (sqrt(3)/3)*(c + x));Intersection(l, LeftPart(W)) = A;Intersection(l, RightPart(W))=B;Intersection(l, yAxis)=C;DotProduct(VectorOf(O, A), VectorOf(C, F)) = c^2/3",
    "query_expressions": "Eccentricity(W)",
    "answer_expressions": "(\\sqrt{21}+\\sqrt{3})/3",
    "fact_spans": "[[[79, 113]], [[2, 63], [114, 117], [228, 231]], [[9, 63]], [[9, 63]], [[68, 78]], [[68, 78]], [[149, 153]], [[127, 130]], [[144, 148]], [[131, 134]], [[9, 63]], [[9, 63]], [[2, 63]], [[68, 78]], [[2, 78]], [[79, 113]], [[79, 136]], [[79, 136]], [[79, 148]], [[161, 226]]]",
    "query_spans": "[[[228, 237]]]",
    "process": "Use vector relationships to find the length of |\\overrightarrow{FA}|. Let the other focus of the hyperbola be F, then use the cosine law to find the value of |\\overrightarrow{AF}|, and use the definition of the hyperbola to find the eccentricity. [Detailed solution] From the given information, \\angle AFO = \\frac{\\pi}{6}, then |OC| = \\frac{\\sqrt{3}}{3}|FO| = \\frac{\\sqrt{3}c}{3}, |CF| = \\frac{2\\sqrt{3}}{3}c. \\overrightarrow{OA} \\cdot \\overrightarrow{CF} = (\\overrightarrow{OF} + \\overrightarrow{FA}) \\cdot \\overrightarrow{CF} = \\overrightarrow{OF} \\cdot \\overrightarrow{CF} + \\overrightarrow{FA} \\cdot \\overrightarrow{CF} = c \\cdot \\frac{2\\sqrt{3}c}{3} \\cdot \\frac{\\sqrt{3}}{2} - |\\overrightarrow{FA}| \\cdot \\frac{2\\sqrt{3}c}{3} = c^{2} - \\frac{2\\sqrt{3}}{3}|\\overrightarrow{FA}| = \\frac{c^{2}}{3}, solving gives |\\overrightarrow{FA}| = \\frac{\\sqrt{3}c}{3}, hence A is the midpoint of CF, then |AF| = \\frac{1}{2}|CF| = \\frac{\\sqrt{3}}{3}c. Let the right focus of W be F. In \\triangle AFF, by the cosine law, \\cos\\angle AFF = \\frac{2}{2|AF|} \\cdot |\\overrightarrow{FF}| = \\frac{\\sqrt{3}}{2}, solving gives |AF| = \\frac{\\sqrt{21}}{3}c. By the definition of the hyperbola, |-|AF| = \\frac{\\sqrt{21-\\sqrt{3}}}{3}c', then the eccentricity of W is e = \\frac{c}{a} = \\frac{2\\sqrt{3}}{\\sqrt{7}-1} = \\frac{\\sqrt{21}+\\sqrt{3}}{3}."
  },
  {
    "text": "The parabola $x^{2}=4(y-m)$ intersects the circle $x^{2}+y^{2}=1$ at point $P$ in the first quadrant, and the tangents to the two curves at $P$ are perpendicular to each other. Then $m=$?",
    "fact_expressions": "P: Point;m: Number;H: Circle;G: Parabola;Expression(G) = (x^2 = 4*(-m + y));Expression(H) = (x^2 + y^2 = 1);Quadrant(P) = 1;Intersection(H, G) = P;IsPerpendicular(TangentOnPoint(P, G), TangentOnPoint(P, H))",
    "query_expressions": "m",
    "answer_expressions": "(\\sqrt{2} - 1)/2",
    "fact_spans": "[[[42, 46], [49, 53]], [[66, 69]], [[18, 34]], [[0, 17]], [[0, 17]], [[18, 34]], [[37, 46]], [[0, 46]], [[48, 64]]]",
    "query_spans": "[[[66, 71]]]",
    "process": "Slight"
  },
  {
    "text": "Given that the distance from point $M$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ to the left focus $F_{1}$ is $3$, $N$ is the midpoint of $M F_{1}$, and $O$ is the coordinate origin, then $|O N|=$?",
    "fact_expressions": "G: Ellipse;M: Point;F1: Point;O: Origin;N: Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(M, G);LeftFocus(G)=F1;Distance(M, F1) = 3;MidPoint(LineSegmentOf(M, F1)) = N",
    "query_expressions": "Abs(LineSegmentOf(O, N))",
    "answer_expressions": "7/2",
    "fact_spans": "[[[2, 41]], [[43, 47]], [[51, 58]], [[85, 88]], [[68, 71]], [[2, 41]], [[2, 47]], [[2, 58]], [[43, 65]], [[68, 84]]]",
    "query_spans": "[[[95, 104]]]",
    "process": "Since the major axis of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ is 10, we have $a=5$, $2a=10$. By the definition of the ellipse, $|MF_{2}|=10-3=7$, and since $ON$ is the midline of triangle $AMF_{1}F_{2}$, it follows that $|ON|=\\frac{7}{2}$."
  },
  {
    "text": "If the curve $x^{2}-y^{2}=4$ and the line $y=k(x-2)+3$ have two distinct common points, then the range of real values for $k$ is?",
    "fact_expressions": "G: Line;H: Curve;k: Real;Expression(G) = (y = k*(x - 2) + 3);Expression(H) = (x^2 - y^2 = 4);NumIntersection(H, G) = 2",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-oo, 13/12)&(Negation(k = pm*1)",
    "fact_spans": "[[[19, 33]], [[1, 18]], [[44, 49]], [[19, 33]], [[1, 18]], [[1, 42]]]",
    "query_spans": "[[[44, 56]]]",
    "process": "Solve the system of equations by eliminating variables to obtain an equation in terms of $x$, paying attention to the discussion of literal coefficients. For the hyperbola and the line to have two distinct common points, the quadratic coefficient must be non-zero and the discriminant must be positive; solve the inequality and take the intersection. Solution:  \nSolve the system  \n\\[\n\\begin{cases}\nx^{2}-y^{2}=4 \\\\\ny=k(x-2)+3\n\\end{cases}\n\\]  \nEliminating $y$ gives  \n\\[\n(1-k^{2})x^{2}+2k(2k-3)x-(2k-3)^{2}-4=0\n\\]  \nWhen $1-k^{2}=0$, i.e., $k=\\pm1$, the condition is not satisfied.  \nWhen $1-k^{2}\\neq0$, i.e., $k\\neq\\pm1$,  \nsince the curve $x^{2}-y^{2}=4$ and the line $y=k(x-2)+3$ have two distinct common points,  \nwe have  \n\\[\n\\Delta=4k^{2}(2k-3)^{2}+4(1-k^{2})[(2k-3)^{2}+4]=-4(12k-13)>0\n\\]  \nSolving this inequality yields $k<\\frac{13}{12}$."
  },
  {
    "text": "Given points $A(0,2)$, $B(0,-2)$, $C(3,2)$, if a moving point $M(x, y)$ satisfies $|M A|+|A C|=|M B|+|B C|$, then the trajectory equation of point $M$ is?",
    "fact_expressions": "A: Point;B: Point;C: Point;M: Point;x: Number;y: Number;Coordinate(A) = (0, 2);Coordinate(B) = (0, -2);Coordinate(C) = (3, 2);Coordinate(M) = (x, y);Abs(LineSegmentOf(A, C)) + Abs(LineSegmentOf(M, A)) = Abs(LineSegmentOf(B, C)) + Abs(LineSegmentOf(M, B))",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "(y^2 - x^2/3 = 1)&(y <= -1)",
    "fact_spans": "[[[2, 11]], [[13, 22]], [[24, 32]], [[36, 45], [74, 78]], [[36, 45]], [[36, 45]], [[2, 11]], [[13, 22]], [[24, 32]], [[36, 45]], [[47, 72]]]",
    "query_spans": "[[[74, 85]]]",
    "process": "According to |MA| + |AC| = |MB| + |BC|, since |AC| and |BC| are constants, simplify first and then analyze the distance relationship satisfied by M. Let M(x, y). Because |MA| + |AC| = |MB| + |BC|, it follows that |MA| + 3 = |MB| + \\sqrt{3^{2} + [2 - (-2)]^{2}}, so |MA| - |MB| = 2. Thus, the locus of M(x, y) is the lower branch of a hyperbola with foci A(0, 2) and B(0, -2), where 2a = 2. Here, a = 1, c = 2. Hence, b^{2} = c^{2} - a^{2} = 3. Therefore, y^{2} - \\frac{x^{2}}{3} = 1 (y \\leqslant -1)."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has eccentricity $\\frac{\\sqrt{2}}{2}$, and passes through the point $P\\left(\\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{3}}{2}\\right)$. A moving line $l$: $y=k x+m$ intersects the ellipse $C$ at two distinct points $A$, $B$, and satisfies $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}=0$ ($O$ is the origin). Then $3 m^{2}-2 k^{2}$=?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Eccentricity(C) = sqrt(2)/2;P: Point;Coordinate(P) = (sqrt(2)/2, sqrt(3)/2);PointOnCurve(P,C) = True;l: Line;Expression(l) = (y = k*x + m);m: Number;k: Number;Intersection(l, C) = {A, B};A: Point;B: Point;Negation(A=B);DotProduct(VectorOf(O, A), VectorOf(O, B)) = 0;O: Origin",
    "query_expressions": "-2*k^2 + 3*m^2",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 59], [150, 155]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[2, 84]], [[87, 131]], [[87, 131]], [[2, 131]], [[135, 149]], [[135, 149]], [[234, 251]], [[234, 251]], [[135, 168]], [[161, 164]], [[165, 168]], [[156, 168]], [[170, 222]], [[223, 226]]]",
    "query_spans": "[[[234, 253]]]",
    "process": "\\because ellipse C: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 (a > b > 0) has eccentricity \\frac{\\sqrt{2}}{2}, \\therefore \\frac{c}{a} = \\frac{\\sqrt{2}}{2}, i.e., a^{2} = 2b^{2} \\textcircled{1}. Also, the ellipse passes through point P\\left(\\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{3}}{2}\\right), \\therefore \\frac{1}{2a^{2}} + \\frac{3}{4b^{2}} = 1 \\textcircled{2}. Solving \\textcircled{1} and \\textcircled{2} together gives a^{2} = 2, b^{2} = 1. \\therefore the equation of ellipse C is \\frac{x^{2}}{2} + y^{2} = 1. Substituting line l: y = kx + m into the ellipse equation and simplifying yields (2k^{2} + 1)x^{2} + 4kmx + 2m^{2} - 2 = 0. From the problem condition, \\Delta = 8(2k^{2} - m^{2} + 1) > 0. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}). Then \\begin{cases} x_{1} + x_{2} = \\frac{-4km}{2k^{2} + 1} \\\\ x_{1}x_{2} = \\frac{2m^{2} - 2}{2k^{2} + 1} \\end{cases}. \\because \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = 0, \\therefore x_{1}x_{2} + y_{1}y_{2} = x_{1}x_{2} + (kx_{1} + m)(kx_{2} + m) = (1 + k^{2})x_{1}x_{2} + km(x_{1} + x_{2}) + m^{2} = 0. Substituting the above expressions gives x_{1}x_{2} + y_{1}y_{2} = \\frac{(2m^{2} - 2)(1 + k^{2})}{2k^{2} + 1} + \\frac{-4k^{2}m^{2}}{2k^{2} + 1} + m^{2} = \\frac{3m^{2} - 2k^{2} - 2}{2k^{2} + 1} = 0. Then 3m^{2} - 2k^{2} = 2."
  },
  {
    "text": "Given that $F$ is the left focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, $A(1,4)$, and point $P$ is a moving point on the right branch of the hyperbola, then the minimum value of $|P F|+|P A|$ is?",
    "fact_expressions": "G: Hyperbola;A: Point;P: Point;F: Point;Expression(G) = (x^2/4 - y^2/12 = 1);Coordinate(A) = (1, 4);LeftFocus(G) = F;PointOnCurve(P, RightPart(G))",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "9",
    "fact_spans": "[[[6, 45], [65, 68]], [[50, 58]], [[59, 64]], [[2, 5]], [[6, 45]], [[50, 58]], [[2, 49]], [[60, 74]]]",
    "query_spans": "[[[76, 95]]]",
    "process": "Find the minimum value of |PF| + |PA|. Since F is the left focus and P lies on the right branch of the hyperbola, transform the problem: let E be the right focus. By the definition of a hyperbola, |PF| - |PE| = 2a, that is, |PF| = |PE| + 2a = |PE| + 4. Thus, |PF| + |PA| = |PA| + |PE| + 4. Clearly, |PA| + |PE| \\geqslant |AE| = 5, with equality when points A, P, and E are collinear. Therefore, the minimum value is 9."
  },
  {
    "text": "If one of the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is given by $y=-\\frac{x}{3}$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (y = -x/3)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)/3",
    "fact_spans": "[[[1, 57], [85, 88]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 82]]]",
    "query_spans": "[[[85, 94]]]",
    "process": ""
  },
  {
    "text": "Given that the focal distance of an ellipse is $8$ and the eccentricity is $0.8$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;FocalLength(G)=8;Eccentricity(G)=0.8",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/25+y^2/9=1,y^2/25+x^2/9=1}",
    "fact_spans": "[[[2, 4], [23, 25]], [[2, 11]], [[2, 21]]]",
    "query_spans": "[[[23, 32]]]",
    "process": "Let the equation of the ellipse be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ or $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$ $(a>b>0)$. From the given conditions, we have $\\begin{cases}2c=8\\\\e=\\frac{c}{a}=0.8\\end{cases}$, solving yields: $\\begin{cases}a=5\\\\b=3\\\\c=4\\end{cases}$. If the foci of the ellipse lie on the $x$-axis, then the standard equation of the ellipse is: $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$. If the foci of the ellipse lie on the $y$-axis, then the standard equation of the ellipse is: $\\frac{y^{2}}{25}+\\frac{x^{2}}{9}=1$."
  },
  {
    "text": "Given that $P$ is a point on the parabola $y^{2}=4x$, let $d_{1}$ be the distance from $P$ to the directrix, and $d_{2}$ be the distance from $P$ to the point $A(1, 4)$. Then the minimum value of $d_{1}+d_{2}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G);d1: Number;Distance(P, Directrix(G)) = d1;d2: Number;A: Point;Coordinate(A) = (1, 4);Distance(P, A) = d2",
    "query_expressions": "Min(d1 + d2)",
    "answer_expressions": "4",
    "fact_spans": "[[[6, 20]], [[6, 20]], [[2, 5], [25, 28], [45, 48]], [[2, 23]], [[35, 42]], [[6, 42]], [[64, 71]], [[49, 60]], [[49, 60]], [[45, 71]]]",
    "query_spans": "[[[73, 92]]]",
    "process": ""
  },
  {
    "text": "Given $A(1, \\sqrt{3})$, $F$ is the left focus of the ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, and point $P$ is a moving point on the ellipse $C$. Then the minimum value of $|P A|+|P F|$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (1, sqrt(3));C: Ellipse;Expression(C) = (x^2/9 + y^2/5 = 1);F: Point;LeftFocus(C) = F;P: Point;PointOnCurve(P, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 18]], [[2, 18]], [[25, 66], [76, 81]], [[25, 66]], [[20, 24]], [[21, 70]], [[71, 75]], [[71, 85]]]",
    "query_spans": "[[[87, 106]]]",
    "process": "Let the right focus of ellipse C be F. According to the problem, F(2,0). By the definition of an ellipse, |PF| + |PF| = 6 and ||PA| - |PF|| \\leqslant |AF| = \\sqrt{(2-1)^{2} + (\\sqrt{3})^{2}} = 2, that is, -2 \\leqslant |PA| - |PF| \\leqslant 2, so |PF| - 2 \\leqslant |PA| \\leqslant 2 + |PF|. Therefore, |PA| + |PF| \\geqslant |PF| + |PF| - 2 = 4, with equality if and only if point P is the intersection of the extension of segment FA and ellipse C. Hence, the minimum value of |PA| + |PF| is 4."
  },
  {
    "text": "The hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has an endpoint of its imaginary axis at point $B$, and $F_{1}$, $F_{2}$ are the left and right foci of the hyperbola, respectively. The line segment $B F_{2}$ intersects the hyperbola at point $A$, and $\\overrightarrow{B A}=2 \\overrightarrow{A F_{2}}$. Then, the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;F1: Point;F2: Point;B: Point;A: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);OneOf(Endpoint(ImageinaryAxis(C))) = B;LeftFocus(C) = F1;RightFocus(C) = F2;Intersection(LineSegmentOf(B, F2), C) = A;VectorOf(B, A) = 2*VectorOf(A, F2)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(10)/2",
    "fact_spans": "[[[0, 61], [88, 91], [110, 113], [172, 178]], [[8, 61]], [[8, 61]], [[72, 79]], [[80, 87]], [[68, 71]], [[115, 119]], [[8, 61]], [[8, 61]], [[0, 61]], [[0, 71]], [[72, 97]], [[72, 97]], [[98, 119]], [[121, 170]]]",
    "query_spans": "[[[172, 184]]]",
    "process": "Let A(x, y). Since the right focus is F(c, 0), point B(0, b), and segment BF intersects the right branch of hyperbola C at point A, $\\overrightarrow{BA} = 2\\overrightarrow{AF}_{2}$, therefore $x = \\frac{2c}{3}, y = \\frac{b}{3}$. Substituting into the hyperbola equation yields $\\frac{4}{9} \\times \\frac{c^{2}}{a^{2}} - \\frac{1}{9} = 1$, hence $e = \\frac{\\sqrt{10}}{2}$."
  },
  {
    "text": "Given that point $P$ lies on the parabola $y^{2}=4x$, then the minimum value of the sum of the distance from point $P$ to point $Q(2,-1)$ and the distance from point $P$ to the focus of the parabola is?",
    "fact_expressions": "G: Parabola;Q: Point;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(Q) = (2, -1);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P,Q)+Distance(P,Focus(G)))",
    "answer_expressions": "3",
    "fact_spans": "[[[7, 21], [49, 52]], [[30, 40]], [[2, 6], [25, 29], [44, 48]], [[7, 21]], [[30, 40]], [[2, 22]]]",
    "query_spans": "[[[25, 63]]]",
    "process": "[Divide Wood Thousand] Convert the distance from point P to the parabola's focus into the distance from point P to the directrix, then use the principle that the sum of two sides is greater than the third side to find the value. Given that the parabola's focus is (1,0) and the directrix is $x = -1$, let $d$ be the distance from point P on the parabola to the directrix. Then the sum of the distance from point P to point Q(2,-1) and the distance from point P to the parabola's focus is $d + PQ$. Thus, the minimum value of this sum occurs when the line segment $PQ$ is perpendicular to the directrix, and at this time the sum of distances is $|2 - (-1)| = 3$."
  },
  {
    "text": "Given that $F$ is the left focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and a line $l$ passing through the origin $O$ intersects the hyperbola $C$ at points $P$ and $Q$. If $|P F|=3|Q F|$ and $\\angle P F Q=60^{\\circ}$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;P: Point;F: Point;Q: Point;O:Origin;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;PointOnCurve(O, l);Intersection(l, C) = {P, Q};Abs(LineSegmentOf(P, F)) = 3*Abs(LineSegmentOf(Q, F));AngleOf(P, F, Q) = ApplyUnit(60, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(13)/2",
    "fact_spans": "[[[80, 85]], [[6, 67], [86, 92], [148, 154]], [[14, 67]], [[14, 67]], [[94, 97]], [[2, 5]], [[98, 101]], [[74, 79]], [[14, 67]], [[14, 67]], [[6, 67]], [[2, 71]], [[72, 85]], [[80, 103]], [[105, 119]], [[121, 146]]]",
    "query_spans": "[[[148, 160]]]",
    "process": "Let $ F_{2} $ be the right focus. By the symmetry of the hyperbola, $ PFQF_{2} $ is a parallelogram, so $ |PF_{2}| = |QF| $. Therefore, $ |PF| = 3|QF| = 3|PF_{2}| $. Also, $ |PF| - |PF_{2}| = 2a $, so $ 2|PF_{2}| = 2a $, $ |PF_{2}| = a $, then $ |PF| = 3a $. Since $ \\angle PFQ = 60^{\\circ} $, it follows that $ \\angle FPF_{2} = 120^{\\circ} $. By the law of cosines, $ |F_{1}F_{2}|^{2} = |PF|^{2} + |PF_{2}|^{2} - 2|PF||PF_{2}|\\cos120^{\\circ} $, so $ 4c^{2} = 9a^{2} + a^{2} - 2 \\times 3a \\times a \\times (-\\frac{1}{2}) $, $ 4c^{2} = 13a^{2} $, thus $ e = \\frac{c}{a} = \\frac{\\sqrt{13}}{2} $."
  },
  {
    "text": "Given that one focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ coincides with the center of the circle $x^{2}+y^{2}-10 x=0$, and the eccentricity of the hyperbola is equal to $\\sqrt{5}$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (-10*x + x^2 + y^2 = 0);OneOf(Focus(G)) = Center(H);Eccentricity(G) = sqrt(5)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5 - y^2/20 = 1",
    "fact_spans": "[[[2, 48], [82, 85], [105, 108]], [[5, 48]], [[5, 48]], [[54, 75]], [[2, 48]], [[54, 75]], [[2, 80]], [[82, 101]]]",
    "query_spans": "[[[105, 115]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a point on the parabola $y^{2}=4x$, and $F$ is the focus of this parabola, find the standard equation of the circle with $PF$ as diameter and passing through $(0,2)$.",
    "fact_expressions": "G: Parabola;H: Circle;I: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(I) = (0, 2);PointOnCurve(P, G);Focus(G) = F;IsDiameter(LineSegmentOf(P,F),H);PointOnCurve(I,H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "{(x-3/2)^2+(y+sqrt(2)^2=9/4),((x-3/2)^2+(y^2-sqrt(2))^2=9/4)}",
    "fact_spans": "[[[6, 20], [29, 32]], [[56, 57]], [[48, 55]], [[2, 5]], [[24, 27]], [[6, 20]], [[48, 55]], [[2, 23]], [[24, 35]], [[37, 57]], [[47, 57]]]",
    "query_spans": "[[[56, 64]]]",
    "process": "Let $ P\\left(\\frac{y_{0}^{2}}{4}, y_{0}\\right) $, from the problem we know $ F(1,0) $, and by the definition of a parabola, the diameter of the circle is $ |PF| = 1 + \\frac{y_{0}^{2}}{4} $, the center of the circle is $ \\left(\\frac{1}{2} + \\frac{y_{0}^{2}}{8}, \\frac{y_{0}}{2}\\right) $. From the problem we have $ \\sqrt{\\left(\\frac{1}{2} + \\frac{y_{0}^{2}}{8} - 2\\right)^{2} + \\left(\\frac{y_{0}}{2} - 0\\right)^{2}} = \\frac{1}{2}\\left(1 + \\frac{y_{0}^{2}}{4}\\right) $, solving gives $ y_{0} = \\pm 2\\sqrt{2} $, so the center of the circle is $ \\left(\\frac{3}{2}, \\pm \\sqrt{2}\\right) $, radius is $ \\frac{3}{2} $, thus the standard equation of the required circle is $ \\left(x - \\frac{3}{2}\\right)^{2} + \\left(y \\pm \\sqrt{2}\\right)^{2} = \\frac{9}{4} $."
  },
  {
    "text": "The two endpoints of a line segment $AB$ of length $2$ slide on the parabola $y^{2}=x$. What is the minimum distance from the midpoint $M$ of segment $AB$ to the $y$-axis?",
    "fact_expressions": "B: Point;A: Point;Length(LineSegmentOf(A, B)) = 2;Endpoint(LineSegmentOf(A,B))={A,B};G: Parabola;Expression(G) = (y^2 = x);PointOnCurve(A,G);PointOnCurve(B,G);M: Point;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Min(Distance(M, yAxis))",
    "answer_expressions": "3/4",
    "fact_spans": "[[[8, 13]], [[8, 13]], [[0, 13]], [[6, 18]], [[19, 31]], [[19, 31]], [[6, 32]], [[6, 32]], [[45, 48]], [[36, 48]]]",
    "query_spans": "[[[45, 61]]]",
    "process": "As shown in the figure, the focus of the parabola is $ F\\left(\\frac{1}{4},0\\right) $, and the directrix equation is $ x = -\\frac{1}{4} $. Let the projections of points $ A $, $ M $, $ B $ on the directrix be $ C $, $ P $, $ D $, respectively, and the projection of $ M $ on the $ y $-axis be $ N $. Given that $ AB = 2 $, the latus rectum of the parabola is $ 2p = 1 $. To minimize the distance from the midpoint $ M $ of $ AB $ to the $ y $-axis, $ |BD| + |AC| $ must be minimized, which means minimizing $ |AF| + |BF| $. In $ \\triangle ABF $, $ |AF| + |BF| \\geqslant |AB| $, with equality when $ A $, $ F $, $ B $ are collinear (since the shortest chord through the focus of a parabola is the latus rectum, and the range of lengths for focal chords is $ [1, +\\infty) $, and $ AB = 2 > 1 $, it is possible for $ AB $ to pass through the focus $ F $). Therefore, the distance from the midpoint $ M $ of $ AB $ to the $ y $-axis is minimized when segment $ AB $ passes through the focus $ F $, and the minimum value is $ \\frac{|AB|}{2} - \\frac{1}{4} = \\frac{3}{4} $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$ with right focus $F$, and the line $l$: $y=k(x-2)$ ($k \\neq 0$) intersecting the ellipse $C$ at points $A$ and $B$. Let $P$ be the midpoint of $AB$, $O$ the origin, and let the slopes of lines $OP$, $AF$, $BF$ be $k_{OP}$, $k_{AF}$, $k_{BF}$ respectively. If $k_{AF}=k_{BF} \\cdot k_{OP}$, then $k=?$",
    "fact_expressions": "C: Ellipse;F: Point;Expression(C) = (x^2/2 + y^2 = 1);RightFocus(C) = F;l: Line;Expression(l) = (y = k*(x - 2));k: Number;Negation(k=0);A: Point;B: Point;Intersection(l, C) = {A, B};P: Point;MidPoint(LineSegmentOf(A, B)) = P;O: Origin;k1: Number;k2: Number;k3: Number;Slope(LineSegmentOf(O, P)) = k1;Slope(LineSegmentOf(A, F)) = k2;Slope(LineSegmentOf(B, F)) = k3;k2 = k1*k3",
    "query_expressions": "k",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 34], [71, 76]], [[39, 42]], [[2, 34]], [[2, 42]], [[43, 70]], [[43, 70]], [[208, 211]], [[49, 70]], [[78, 81]], [[82, 85]], [[43, 87]], [[97, 100]], [[88, 100]], [[102, 105]], [[140, 150]], [[152, 161]], [[163, 173]], [[111, 173]], [[111, 173]], [[111, 173]], [[175, 206]]]",
    "query_spans": "[[[208, 213]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, $ P(x_{0},y_{0}) $, then $ k_{AF}=\\frac{y_{1}}{x_{1}-1} $, $ k_{BF}=\\frac{y_{2}}{x_{2}-1} $, $ k_{OP}=\\frac{y_{0}}{x_{0}} $. From \\begin{cases}\\frac{x^{2}}{2}+y_{1}^{2}=1\\\\\\frac{x^{2}}{2}+y_{2}^{2}=1\\end{cases}, we obtain $ \\frac{x_{1}^{2}}{2}-\\frac{x_{2}^{2}}{2}=-(y_{1}^{2}-y_{2}^{2}) $, i.e., $ \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{1}{2}\\cdot\\frac{x_{1}+x_{2}}{y_{1}+y_{2}} $, so $ k=-\\frac{1}{2}\\cdot\\frac{x_{0}}{y_{0}} $, hence $ k\\cdot k_{OP}=-\\frac{1}{2} $. Solving the system of equations \\begin{cases}y=k(x-2)\\\\\\frac{x^{2}}{2}+y^{2}=1\\end{cases} yields $ (1+2k^{2})x^{2}-8k^{2}x+8k^{2}-2=0 $, then $ x_{1}+x_{2}=\\frac{8k^{2}}{1+2k^{2}} $, $ x_{1}x_{2}=\\frac{8k^{2}-2}{1+2k^{2}} $. Since $ k_{AF}+k_{BF}=\\frac{y_{1}}{x_{1}-1}+\\frac{y_{2}}{x_{2}-1}=\\frac{y_{1}(x_{2}-1)+y_{2}(x_{1}-1)}{(x_{1}-1)(x_{2}-1)}=\\frac{2kx_{1}x_{2}-3k(x_{1}+x_{2})+4k}{x_{1}x_{2}-(x_{1}+x_{2})+1}=0 $, therefore $ k_{OP}=\\frac{k_{AF}}{k_{BF}}=-1 $, thus $ k=\\frac{1}{2} $."
  },
  {
    "text": "From a focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a perpendicular line $l$ to an asymptote, with foot of perpendicular at $M$. The line $l$ intersects the $y$-axis at point $E$. If $\\overrightarrow{F M}=3 \\overrightarrow{M E}$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;OneOf(Focus(G)) = F;l: Line;PointOnCurve(F, l) = True;IsPerpendicular(l, Asymptote(G)) = True;FootPoint(l, Asymptote(G)) = M;M: Point;Intersection(l, yAxis) = E;E: Point;VectorOf(F, M) = 3*VectorOf(M, E)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 57], [149, 152]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[62, 65]], [[1, 65]], [[72, 75], [85, 88]], [[0, 75]], [[0, 75]], [[0, 82]], [[79, 82]], [[85, 98]], [[94, 98]], [[101, 146]]]",
    "query_spans": "[[[149, 158]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 px$ coincides with the right focus of the hyperbola $\\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1$, then the value of $p$ is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;Expression(G) = (x^2/6 - y^2/3 = 1);Expression(H) = (y^2 = 2*(p*x));Focus(H) = RightFocus(G)",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[20, 58]], [[1, 16]], [[66, 69]], [[20, 58]], [[1, 16]], [[1, 64]]]",
    "query_spans": "[[[66, 73]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the foci of the ellipse $C$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and point $P$ lies on the ellipse $C$ with $\\angle F_{1} P F_{2}=60^{\\circ}$, find the area of $\\Delta F_{1} P F_{2}$.",
    "fact_expressions": "C: Ellipse;F1: Point;P: Point;F2: Point;Expression(C) = (x^2/25 + y^2/16 = 1);Focus(C) = {F1, F2};PointOnCurve(P, C);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "16*sqrt(3)/3",
    "fact_spans": "[[[18, 62], [71, 76]], [[2, 9]], [[66, 70]], [[10, 17]], [[18, 62]], [[2, 65]], [[66, 77]], [[78, 111]]]",
    "query_spans": "[[[113, 140]]]",
    "process": "From $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, we get $a^{2}=25$, $b^{2}=16$, then $a=5$, $b=4$, $c=\\sqrt{a^{2}-b^{2}}=3$. Let $|PF_{1}|=m$, $|PF_{2}|=n$, then $m+n=2a=10$. In $\\triangle F_{1}PF_{2}$, $\\angle F_{1}PF_{2}=60^{\\circ}$, by the cosine law we have $|F_{1}F_{2}|^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}||PF_{2}|\\cos\\angle F_{1}PF_{2}$, so $4c^{2}=m^{2}+n^{2}-2mn\\cos60^{\\circ}$, $36=(m+n)^{2}-3mn$, so $3mn=100-36=64$, thus $mn=\\frac{64}{3}$, so $S_{\\triangle F_{1}PF_{2}}=\\frac{1}{2}mn\\sin60^{\\circ}=\\frac{1}{2}\\times\\frac{64}{3}\\times\\frac{\\sqrt{3}}{2}=\\frac{16\\sqrt{3}}{3}$,"
  },
  {
    "text": "Given a parabola $C$ with vertex at the origin and focus on the $x$-axis, the line $x - y = 0$ intersects the parabola $C$ at points $A$ and $B$. If $P(2,2)$ is the midpoint of $AB$, then the equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;G: Line;A: Point;B: Point;P: Point;O:Origin;Expression(G) = (x - y = 0);Coordinate(P) = (2, 2);Vertex(C)=O;PointOnCurve(Focus(C), xAxis);Intersection(G, C) = {A, B};MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[9, 15], [35, 41], [73, 79]], [[25, 34]], [[43, 46]], [[47, 50]], [[54, 62]], [[3, 5]], [[25, 34]], [[54, 62]], [[2, 15]], [[9, 24]], [[25, 52]], [[54, 71]]]",
    "query_spans": "[[[73, 84]]]",
    "process": ""
  },
  {
    "text": "Given that the distance from a moving point $P$ to the fixed point $(2,0)$ is equal to its distance to the fixed line $l$: $x=-2$, what is the equation of the trajectory of point $P$?",
    "fact_expressions": "P: Point;K: Point;Coordinate(K) = (2, 0);l: Line;Expression(l) = (x = -2);Distance(P, K) = Distance(P, l)",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "y^2 = 8*x",
    "fact_spans": "[[[4, 7], [21, 22], [44, 48]], [[10, 17]], [[10, 17]], [[26, 37]], [[26, 37]], [[4, 42]]]",
    "query_spans": "[[[44, 54]]]",
    "process": "Let P(x,y). According to the definition of a parabola, the trajectory of point P is the parabola y^{2}=2px, and p=2. Its equation is y^{2}=8x."
  },
  {
    "text": "A hyperbola $x^{2}-m y^{2}=m$ $(m>0)$ has one asymptote perpendicular to $y=2x$, and its right focus is $F$. Then, the area of the circle centered at the origin with radius $|OF|$ is?",
    "fact_expressions": "G: Hyperbola;m: Number;H: Circle;O: Origin;F: Point;l: Line;m>0;Expression(G) = (-m*y^2 + x^2 = m);RightFocus(G)=F;IsPerpendicular(OneOf(Asymptote(G)),H);Center(H)=O;Radius(H)=Abs(LineSegmentOf(O,F));Expression(l)=(y=2*x)",
    "query_expressions": "Area(H)",
    "answer_expressions": "5*pi",
    "fact_spans": "[[[0, 25]], [[3, 25]], [[69, 70]], [[52, 54]], [[46, 49]], [[32, 39]], [[3, 25]], [[0, 25]], [[0, 49]], [[0, 41]], [[51, 70]], [[58, 70]], [[32, 39]]]",
    "query_spans": "[[[69, 75]]]",
    "process": "From $x^{2}-my^{2}=m$ $(m>0)$, we obtain: $\\frac{x^{2}}{m}-y^{2}=1$, so $a=\\sqrt{m}$, $b=1$, thus the asymptotes are $y=\\pm\\frac{b}{a}x=\\pm\\frac{1}{\\sqrt{m}}$. Since one asymptote of the hyperbola $x^{2}-my^{2}=m$ $(m>0)$ is perpendicular to $y=2x$, we have $-\\frac{1}{\\sqrt{m}}\\times2=-1$, which gives $m=4$. Therefore, $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{m+1}=\\sqrt{5}$, so the right focus is $F(\\sqrt{5},0)$, and $|OF|=\\sqrt{5}$. The area of the circle with radius $|OF|$ is $\\pi\\times(\\sqrt{5})^{2}=5\\pi$. The answer is $5\\pi$."
  },
  {
    "text": "The coordinates of the focus of the parabola $y=-8 x^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = -8*x^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,-1/32)",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 22]]]",
    "process": "From $ y = -8x^{2} $, we get $ x^{2} = -\\frac{1}{8}y $, $ p = \\frac{1}{16} $, it follows that the parabola opens downward, with focus at: $ (0, -\\frac{1}{32}) $"
  },
  {
    "text": "Given the parabola $y^{2}=4x$, two mutually perpendicular lines are drawn through the origin $O$, intersecting the parabola at points $A$ and $B$ respectively, both different from $O$. When the chord length obtained by the intersection of line $AB$ and the circle $C$: $(x-3)^{2}+y^{2}=4$ is shortest, what is the equation of line $AB$?",
    "fact_expressions": "G: Parabola;C: Circle;O: Origin;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Expression(C) = (y^2 + (x - 3)^2 = 4);l1: Line;l2: Line;PointOnCurve(O, l1);PointOnCurve(O, l2);IsPerpendicular(l1, l2);Negation(O = A);Negation(O = B);Intersection(l1, G) = A;Intersection(l2, G) = B;WhenMin(Length(InterceptChord(LineOf(A, B), C)))",
    "query_expressions": "Expression(LineOf(A, B))",
    "answer_expressions": "x = 4",
    "fact_spans": "[[[2, 16], [37, 40]], [[68, 92]], [[18, 23], [44, 48]], [[49, 53]], [[54, 57]], [[2, 16]], [[68, 92]], [], [], [[17, 33]], [[17, 33]], [[17, 33]], [[41, 53]], [[41, 57]], [[17, 57]], [[17, 57]], [[59, 102]]]",
    "query_spans": "[[[103, 115]]]",
    "process": "Let OA: $ y = kx $ ($ k \\neq 0 $), solving the system of equations, we get  \n\\[\n\\begin{cases}\ny = kx, \\\\\ny^2 = 4x,\n\\end{cases}\n\\]  \nobtaining $ A\\left(\\frac{4}{k^2}, \\frac{4}{k}\\right) $. Let OB: $ y = -\\frac{1}{k}x $, similarly we obtain $ B(4k^2, -4k) $. When $ k \\neq \\pm1 $,  \n$ k_{AB} = \\frac{\\frac{4}{k} + 4k}{\\frac{4}{12} - 4k^2} = \\frac{k}{1 - k^2} $,  \nso the equation of line AB is $ y + 4k = \\frac{k}{1 - k^2}(x - 4k^2) $, simplifying to $ y = \\frac{k}{1 - k^2}(x - 4) $, thus line AB always passes through the point $ (4, 0) $. It is easy to prove that when $ k = \\pm1 $, line AB also passes through the point $ (4, 0) $. Since $ (4 - 3)^2 + 0^2 = 1 < 4 $, the point $ (4, 0) $ lies inside circle C, and both $ (4, 0) $ and the center $ (3, 0) $ lie on the x-axis. Therefore, the chord length intercepted by line AB on circle C is shortest if and only if AB is perpendicular to the x-axis, at which time the equation of line AB is $ x = 4 $."
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$, what is the eccentricity of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2/3 = 1)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 35], [38, 44]], [[2, 35]]]",
    "query_spans": "[[[38, 50]]]",
    "process": "From the hyperbola $ C: x^{2} - \\frac{y^{2}}{3} = 1 $, we have $ a = 1 $, $ b = \\sqrt{3} $, then $ c = \\sqrt{a^{2} + b^{2}} = 2 $. Therefore, the eccentricity is $ e = \\frac{c}{a} = 2 $."
  },
  {
    "text": "It is known that a line $l$ with slope $\\sqrt{3}$ passes through the focus $F$ of the parabola $y^{2}=2 p x$ $(p>0)$, and intersects the parabola at points $A$ and $B$, with $|A B|=8$. Then the value of $p$ is?",
    "fact_expressions": "l: Line;G: Parabola;p: Number;A: Point;B: Point;F:Point;p>0;Expression(G) = (y^2 = 2*(p*x));Slope(l)=sqrt(3);PointOnCurve(F,l);Focus(G)=F;Intersection(l,G)={A,B};Abs(LineSegmentOf(A, B)) = 8",
    "query_expressions": "p",
    "answer_expressions": "3",
    "fact_spans": "[[[16, 21]], [[23, 45], [54, 57]], [[81, 84]], [[59, 62]], [[63, 66]], [[48, 51]], [[26, 45]], [[23, 45]], [[2, 21]], [[16, 51]], [[23, 51]], [[16, 68]], [[70, 79]]]",
    "query_spans": "[[[81, 88]]]",
    "process": "The focus of the parabola $ y^{2} = 2px $ is $ F\\left(\\frac{p}{2}, 0\\right) $. According to the problem, the equation of line $ l $ is $ y = \\sqrt{3}\\left(x - \\frac{p}{2}\\right) $. Combining this with the parabola equation $ y^{2} = 2px $, we obtain $ 3\\left(x - \\frac{p}{2}\\right)^{2} = 2px $, which simplifies to $ 3x^{2} - 5px + \\frac{3p^{2}}{4} = 0 $. Thus, $ x_{1} + x_{2} = \\frac{5p}{3} $, so $ |AB| = x_{1} + x_{2} + p = \\frac{5p}{3} + p = \\frac{8p}{3} = 8 $, hence $ p = 3 $."
  },
  {
    "text": "The asymptotes of the hyperbola $C$: $\\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1$ are given by? If the right focus of the hyperbola $C$ coincides with the focus of the parabola $y^{2}=2 px$, then the equation of the directrix of the parabola is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/2 - y^2/2 = 1);G: Parabola;Expression(G) = (y^2 = 2*p*x);p:Number;RightFocus(C) = Focus(G)",
    "query_expressions": "Expression(Directrix(C));Expression(Directrix(G))",
    "answer_expressions": "y=pm*x\nx=-2",
    "fact_spans": "[[[0, 43], [52, 58]], [[0, 43]], [[63, 78], [85, 88]], [[63, 78]], [[66, 78]], [[52, 83]]]",
    "query_spans": "[[[0, 51]], [[85, 95]]]",
    "process": ""
  },
  {
    "text": "Given that $A$ and $B$ are two points on the parabola $C$: $x^{2}=4 y$, and $M(1,-2)$, if $\\overrightarrow{A M}=\\overrightarrow{M B}$, then the equation of line $A B$ is?",
    "fact_expressions": "C: Parabola;B: Point;A: Point;M: Point;Expression(C) = (x^2 = 4*y);Coordinate(M) = (1, -2);PointOnCurve(A, C);PointOnCurve(B, C);VectorOf(A, M) = VectorOf(M, B)",
    "query_expressions": "Expression(LineOf(A, B))",
    "answer_expressions": "x-2*y-5=0",
    "fact_spans": "[[[10, 29]], [[6, 9]], [[2, 5]], [[34, 43]], [[10, 29]], [[34, 43]], [[2, 33]], [[2, 33]], [[45, 88]]]",
    "query_spans": "[[[90, 102]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}) and M(1,-2). Since \\overrightarrow{AM}=\\overrightarrow{MB}, we have \\begin{cases}x_{1}+x_{2}=\\\\y_{1}+y_{2}=\\end{cases}. Also, x_{1}^{2}=4y_{1}, x_{2}^{2}=4y_{2}, then x_{1}^{2}-x_{2}^{2}=4y_{1}-4y_{2}, yielding x_{1}+x_{2}==\\frac{4y_{1}-4y_{2}}{x_{1}-x_{2}}=2. The slope of line AB is k=\\frac{1}{2}, thus the equation of line AB is y+2=\\frac{1}{2}(x-1), simplified to x-2y-5=0"
  },
  {
    "text": "For any point $P$ on the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, the product of its distances to the two asymptotes is a constant $c$. What is $c$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1);PointOnCurve(P,G);P:Point;L1:Line;L2:Line;Asymptote(G)={L1,L2};Distance(P,L1)*Distance(P,L2)=c;c:Number",
    "query_expressions": "c",
    "answer_expressions": "4/5",
    "fact_spans": "[[[0, 28], [36, 37]], [[0, 28]], [[0, 35]], [[32, 35]], [], [], [[36, 43]], [[32, 54]], [[51, 54], [55, 58]]]",
    "query_spans": "[[[55, 60]]]",
    "process": "The asymptotes of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ are $y=\\frac{1}{2}x$ and $y=-\\frac{1}{2}x$. Take a point $P(2,0)$ on the hyperbola, then the distance from point $P$ to each asymptote is $d=\\frac{2}{\\sqrt{5}}$, so $c=\\frac{2}{\\sqrt{5}}\\cdot\\frac{2}{\\sqrt{5}}=\\frac{4}{5}$."
  },
  {
    "text": "Given that point $P$ lies on the parabola $y^{2}=4x$, then when the sum of the distance from point $P$ to point $Q(3, 4)$ and the distance from point $P$ to the directrix of the parabola is minimized, what are the coordinates of point $P$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G);Q: Point;Coordinate(Q) = (3, 4);WhenMin(Distance(P, Q)+Distance(P, Directrix(G)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "((3+sqrt(5))/2, sqrt(5)+1)",
    "fact_spans": "[[[7, 21], [51, 54]], [[7, 21]], [[2, 6], [26, 30], [46, 50], [68, 72]], [[2, 22]], [[31, 42]], [[31, 42]], [[25, 67]]]",
    "query_spans": "[[[68, 77]]]",
    "process": "From the equation of the parabola, it is known that its directrix is $ l: x = -1 $ and the focus is $ F(1, 0) $. Draw $ PM \\perp l $ from point $ P $, with foot $ M $, and connect $ FM $, then $ |PM| = |PF| $. Therefore, when points $ P $, $ Q $, and $ F $ are collinear, $ |PM| + |PQ| $ reaches the minimum value. At this time, the equation of line $ QF $ is $ y = 2x - 2 $. Solving the system of equations \n\\[\n\\begin{cases}\ny = 2x - 2 \\\\\ny^{2} = 4x\n\\end{cases}\n\\]\nyields $ y^{2} - 2y - 4 = 0 $. Solving gives $ y = \\sqrt{5} + 1 $. Substituting into the line equation, we get $ x = \\frac{3 + \\sqrt{5}}{2} $. Thus, the coordinates of point $ P $ are $ \\left( \\frac{3 + \\sqrt{5}}{2}, \\sqrt{5} + 1 \\right) $."
  },
  {
    "text": "It is known that an ellipse centered at the origin passes through the point $A(-3,0)$ and has eccentricity $e=\\frac{\\sqrt{5}}{3}$. Then, what is the standard equation of the ellipse?",
    "fact_expressions": "O: Origin;Center(G) = O;G: Ellipse;A: Point;Coordinate(A) = (-3, 0);PointOnCurve(A,G) = True;e:Number;Eccentricity(G) = e;e=sqrt(5)/3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{(x^2/9+y^2/4=1),(y^2/(81/4)+x^2/9=1)}",
    "fact_spans": "[[[5, 9]], [[2, 12]], [[10, 12], [52, 54]], [[13, 23]], [[13, 23]], [[10, 23]], [[28, 50]], [[10, 50]], [[28, 50]]]",
    "query_spans": "[[[52, 61]]]",
    "process": "Classify according to the location of the foci, then determine the values of a, b, c based on the point through which the ellipse passes and the value of eccentricity, thus the equation of the ellipse can be found. If the foci are on the x-axis, from the problem we know a=3; since the eccentricity of the ellipse is e=\\frac{\\sqrt{5}}{3}, then c=\\sqrt{5}, b=2, so the equation of the ellipse is \\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1; if the foci are on the y-axis, then b=3, a^{2}-c^{2}=9, and the eccentricity e=\\frac{c}{a}=\\frac{\\sqrt{5}}{3}, solving gives a^{2}=\\frac{81}{4}, so the equation of the ellipse is \\frac{y^{2}}{\\frac{81}{4}}+\\frac{x^{2}}{9}=1"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the two foci are $F_{1}(-1,0)$ and $F_{2}(1, 0)$, and the ellipse $C$ passes through the point $P\\left(\\frac{4}{3}, \\frac{1}{3}\\right)$. What is the equation of the ellipse $C$?",
    "fact_expressions": "C: Ellipse;a: Number;b: Number;F1: Point;F2: Point;P: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-1, 0);Coordinate(F2) = (1, 0);Coordinate(P) = (4/3, 1/3);Focus(C) = {F1, F2};PointOnCurve(P, C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/2 + y^2 = 1",
    "fact_spans": "[[[2, 59], [99, 104], [139, 144]], [[9, 59]], [[9, 59]], [[67, 80]], [[83, 97]], [[106, 137]], [[9, 59]], [[9, 59]], [[2, 59]], [[67, 80]], [[83, 97]], [[106, 137]], [[2, 97]], [[99, 137]]]",
    "query_spans": "[[[139, 149]]]",
    "process": "From the given conditions, we know that $ c = 1 $. According to the definition of an ellipse, $ 2a = |PF_{1}| + |PF_{2}| = 2\\sqrt{2} $, so $ a = \\sqrt{2} $, and thus $ b = 1 $. The equation of the ellipse is $ \\frac{x^{2}}{2} + y^{2} = 1 $."
  },
  {
    "text": "A line $ l $ with an inclination angle of $30^{\\circ}$ passes through the left focus $F_{1}$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, intersecting the hyperbola at points $A$ and $B$. The perpendicular bisector of segment $AB$ passes through the right focus $F_{2}$. Then the asymptotes of this hyperbola have the equation?",
    "fact_expressions": "l: Line;Inclination(l) = ApplyUnit(30, degree);G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, l);A: Point;B: Point;Intersection(l, G) = {A, B};PointOnCurve(F2, PerpendicularBisector(LineSegmentOf(A, B)))",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*x",
    "fact_spans": "[[[17, 22]], [[0, 22]], [[24, 80], [93, 96], [134, 137]], [[24, 80]], [[27, 80]], [[27, 80]], [[27, 80]], [[27, 80]], [[84, 91]], [[124, 131]], [[24, 91]], [[93, 131]], [[17, 91]], [[97, 100]], [[101, 104]], [[17, 106]], [[107, 131]]]",
    "query_spans": "[[[134, 145]]]",
    "process": "As shown in the figure, $MF_{2}$ is the perpendicular bisector of $\\triangle ABF_{2}$, so $AF_{2}=BF_{2}$, and $\\angle MF_{1}F_{2}=30^{\\circ}$, thus $MF_{2}=2c\\cdot\\sin30^{\\circ}=c$, $MF_{1}=2c\\cdot\\cos30^{\\circ}=\\sqrt{3}c$. By the definition of hyperbola, $BF_{1}-BF_{2}=2a$, $AF_{2}-AF_{1}=2a$, so $AB=BF_{1}-AF_{1}=BF_{2}+2a-(AF_{2}-2a)=4a$. Hence $MA=2a$, $AF_{2}=\\sqrt{MA^{2}+MF_{2}^{2}}=\\sqrt{4a^{2}+c^{2}}$, $AF_{1}=MF_{1}-MA=\\sqrt{3}c-2a$, from $AF_{2}-AF_{1}=2a$, we get $\\sqrt{4a^{2}+c^{2}}-(\\sqrt{3}c-2a)=2a$, thus $4a^{2}+c^{2}=3c^{2}$, i.e., $c=\\sqrt{2}a$, $b=\\sqrt{c^{2}-a^{2}}=a$, then the asymptotes are $y=\\pm x$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the right focus is $F_{2}$, $O$ is the coordinate origin, and $M$ is a point on the $y$-axis. Point $A$ is an intersection point of line $M F_{2}$ and ellipse $C$, and $|O A|=|O F_{2}|=2|O M|$. Then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;M: Point;F2: Point;O: Origin;A: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F2;PointOnCurve(M, yAxis);OneOf(Intersection(LineOf(M,F2),C))=A;Abs(LineSegmentOf(O,A))=Abs(LineSegmentOf(O,F2));Abs(LineSegmentOf(O,F2))=2*Abs(LineSegmentOf(O,M))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[2, 58], [109, 114], [147, 152]], [[8, 58]], [[8, 58]], [[80, 83]], [[63, 70]], [[71, 74]], [[92, 96]], [[8, 58]], [[8, 58]], [[2, 58]], [[2, 70]], [[80, 91]], [[92, 119]], [[121, 145]], [[121, 145]]]",
    "query_spans": "[[[147, 158]]]",
    "process": "Since $F_{2}$ is the right focus of the ellipse, $F_{2}(c,0)$, so $|OA|=|OF_{2}|=2|OM|=c$. Since $M$ is a point on the $y$-axis, let $M(0,\\frac{c}{2})$, $A(x_{0},y_{0})$ $(y_{0}\\neq0)$, clearly $\\frac{x_{0}^{2}}{a^{2}}+\\frac{y_{0}^{2}}{b^{2}}=1$. Since points $A$, $M$, $F_{2}$ are collinear, we have: $\\overrightarrow{AM}=\\lambda\\overrightarrow{MF_{2}} \\Rightarrow (-x_{0},\\frac{c}{2}-y_{0})=\\lambda(c,-\\frac{c}{2}) \\Rightarrow \\begin{cases}-x_{0}=\\lambda c\\\\\\frac{c}{2}-y_{0}=-\\frac{\\lambda c}{2}\\end{cases} \\Rightarrow x_{0}=c-2y_{0}$. Thus we obtain $x_{0}=c-2y_{0}=c-2\\cdot\\frac{4c}{5}=-\\frac{3c}{5}$. Substituting into $\\frac{x_{0}^{2}}{a^{2}}+\\frac{y_{0}^{2}}{b^{2}}=1$, we get $\\frac{\\frac{9c^{2}}{25}}{a^{2}}+\\frac{\\frac{16c^{2}}{25}}{b^{2}}=1 \\Rightarrow 9c^{2}b^{2}+16c^{2}a^{2}=25a^{2}b^{2} \\Rightarrow 9c^{2}(a^{2}-c^{2})+16c^{2}a^{2}=25a^{2}(a^{2}-c^{2})$. Simplifying yields: $9c^{4}-50a^{2}c^{2}+25a^{4}=0 \\Rightarrow 9e^{4}-50e^{2}+25=0 \\Rightarrow (9e^{2}-5)(e^{2}-5)=0$, solving gives $e^{2}=\\frac{5}{9}$, or $e^{2}=5$, but $0<e<1$, so $e^{2}=\\frac{5}{9}$, i.e., $e=\\frac{\\sqrt{5}}{3}$."
  },
  {
    "text": "Given the parabola $C$: $x^{2}=4 y$ with focus $F$, and let $M$ be a point on $C$. The extension of $F M$ intersects the $x$-axis at point $N$. If $M$ is the midpoint of $F N$, then $|F N|$=?",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 4*y);F: Point;Focus(C) = F;M: Point;PointOnCurve(M, C);N: Point;Intersection(OverlappingLine(LineSegmentOf(F, M)), xAxis) = N;MidPoint(LineSegmentOf(F, N)) = M",
    "query_expressions": "Abs(LineSegmentOf(F, N))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 21], [33, 36]], [[2, 21]], [[25, 28]], [[2, 28]], [[29, 32], [62, 65]], [[29, 39]], [[55, 59]], [[40, 59]], [[62, 74]]]",
    "query_spans": "[[[76, 85]]]",
    "process": "Let M(x_{0},y_{0}), N(x_{1},0). Since the focus of the parabola C: x^{2}=4y is F, then F(0,1), and M is the midpoint of FN, so x_{0}=\\frac{x_{1}}{2}, y_{0}=\\frac{1}{2}. Point M lies on the parabola, (\\frac{x_{1}}{2})^{2}=4\\times\\frac{1}{2}=2x^{2}=8, x_{1}=\\pm2\\sqrt{2}. |FM|=\\frac{1}{2}+1=\\frac{3}{2}, |FN|=2|FM|=3"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1$ $(0<b<2)$, with left and right foci $F_{1}$ and $F_{2}$ respectively, a line $l$ passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. If the maximum value of $|B F_{2}|+|A F_{2}|$ is $6$, then the value of $b$ is?",
    "fact_expressions": "l: Line;G: Ellipse;b: Number;B: Point;F2: Point;A: Point;F1: Point;0 < b;b < 2;Expression(G) = (x^2/4 + y^2/b^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, l);Intersection(l, G) = {A,B};Max(Abs(LineSegmentOf(A, F2)) + Abs(LineSegmentOf(B, F2))) = 6",
    "query_expressions": "b",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[86, 91]], [[92, 94], [2, 50]], [[139, 142]], [[100, 104]], [[67, 75]], [[95, 99]], [[58, 65], [78, 85]], [[4, 50]], [[4, 50]], [[2, 50]], [[2, 76]], [[2, 76]], [[77, 91]], [[86, 106]], [[108, 137]]]",
    "query_spans": "[[[139, 146]]]",
    "process": "From 0 < b < 2, it is known that the foci lie on the x-axis. Since line l passing through $ F_{1} $ intersects the ellipse at points A and B, $ |BF_{2}| + |AF_{2}| + |BF_{1}| + |AF_{1}| = 2a + 2a = 8 $. Hence, $ |BF_{2}| + |AF_{2}| = 8 - |AB| $. When AB is perpendicular to the x-axis, $ |AB| $ is minimized and $ |BF_{2}| + |AF_{2}| $ is maximized. At this time, $ |AB| = \\frac{2b^{2}}{a} $, so $ 6 = 8 - \\frac{2b^{2}}{a} $. Solving gives $ b = \\sqrt{2} $. Key point: properties of the ellipse."
  },
  {
    "text": "Given point $M(1,-1)$ and parabola $C$: $x^{2}=4 y$, a line $l$ with slope $k$ passing through the focus of $C$ intersects $C$ at points $A$ and $B$, $N$ is the midpoint of $A B$, and $2|M N|=|A B|$. Then the value of $k$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (1, -1);C: Parabola;Expression(C) = (x^2 = 4*y);k: Number;l: Line;Slope(l) = k;PointOnCurve(Focus(C), l);A: Point;B: Point;Intersection(l, C) = {A, B};N: Point;MidPoint(LineSegmentOf(A, B)) = N;2*Abs(LineSegmentOf(M, N)) = Abs(LineSegmentOf(A, B))",
    "query_expressions": "k",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 12]], [[2, 12]], [[13, 32], [34, 37], [54, 57]], [[13, 32]], [[99, 102], [44, 47]], [[48, 53]], [[41, 53]], [[33, 53]], [[59, 62]], [[63, 66]], [[48, 68]], [[69, 72]], [[69, 81]], [[83, 97]]]",
    "query_spans": "[[[99, 106]]]",
    "process": "Let points $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, the focus of parabola $C$ be $F(0,1)$, and the equation of line $l$ be $y=kx+1$. Solving the system \n\\[\n\\begin{cases}\ny=kx+1 \\\\\nx^{2}=4y\n\\end{cases}\n\\]\nyields $x^{2}-4kx-4=0$, $d=16k^{2}+16>0$. By Vieta's formulas, we have $x_{1}+x_{2}=4k$, $x_{1}x_{2}=-4$. Since $N$ is the midpoint of $AB$, and $2|MN|=|AB|$, then $|MN|=|AN|=|BN|$. Therefore, $\\angle AMN=\\angle NAM$, $\\angle BMN=\\angle NBM$, so $\\angle AMB=\\angle AMN+\\angle BMN=\\frac{1}{2}(\\angle AMN+\\angle BMN+\\angle NAM+\\angle NBM)=90^{\\circ}$. Hence, $MA \\perp MB$. While $\\overrightarrow{MA}=(x_{1}-1,y_{1}+1)=(x_{1}-1,kx_{1}+2)$, similarly $\\overrightarrow{MB}=(x_{2}-1,kx_{2}+2)$, thus \n\\[\n\\overrightarrow{MA} \\cdot \\overrightarrow{MB} = (x_{1}-1)(x_{2}-1)+(kx_{1}+2)(kx_{2}+2) = (k^{2}+1)x_{1}x_{2}+(2k-1)(x_{1}+x_{2})+5 = -4(k^{2}+1)+4k(2k-1)+5 = 4k^{2}-4k+1=(2k-1)^{2}=0,\n\\]\nsolving gives $k=\\frac{1}{2}$."
  },
  {
    "text": "The asymptotes of the hyperbola $y^{2}-\\frac{x^{2}}{4}=1$ are given by?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/4 + y^2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(1/2)*x",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 38]]]",
    "process": "According to the equation of the hyperbola, find the values of a and b, then find the equations of the asymptotes of the hyperbola. Given that the foci of the hyperbola $ y^{2}-\\frac{x^{2}}{4}=1 $ lie on the y-axis, and $ a=1 $, $ b=2 $, so the equations of the asymptotes of the hyperbola are $ y=\\pm\\frac{a}{b}x=\\pm\\frac{1}{2}x $."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=8x$, a line passing through $F$ with an inclination angle of $60^{\\circ}$ intersects the parabola at points $A$ and $B$. Let $AF=\\lambda FB$, and $|FA|>|FB|$, then $\\lambda=$?",
    "fact_expressions": "C: Parabola;G: Line;A: Point;F: Point;B: Point;lambda: Number;Expression(C) = (y^2 = 8*x);Focus(C) = F;PointOnCurve(F, G);Inclination(G) = ApplyUnit(60, degree);Intersection(G, C) = {A, B};LineSegmentOf(A, F) = LineSegmentOf(F, B)*lambda;Abs(LineSegmentOf(F, A)) > Abs(LineSegmentOf(F, B))",
    "query_expressions": "lambda",
    "answer_expressions": "3",
    "fact_spans": "[[[6, 25], [54, 57]], [[51, 53]], [[58, 62]], [[2, 5]], [[64, 67]], [[102, 111]], [[6, 25]], [[2, 28]], [[29, 53]], [[34, 53]], [[51, 69]], [[72, 87]], [[89, 100]]]",
    "query_spans": "[[[102, 113]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $A$ is the left vertex of $C$, and $B$ is a point on $C$ such that $BF$ is perpendicular to the $x$-axis. If the slope of line $AB$ is $1$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(C) = F;A: Point;LeftVertex(C) = A;B: Point;PointOnCurve(B, C);IsPerpendicular(LineSegmentOf(B, F), xAxis);Slope(LineOf(A, B)) = 1",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[6, 67], [76, 79], [88, 91], [127, 130]], [[6, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[2, 5]], [[2, 71]], [[72, 75]], [[72, 83]], [[84, 87]], [[84, 94]], [[96, 108]], [[111, 125]]]",
    "query_spans": "[[[127, 136]]]",
    "process": "From the given conditions, we have A(-a,0), F(c,0). Letting x=c, we obtain y=\\pmb\\sqrt{\\frac{c^{2}}{a^{2}}-1}=\\pm\\frac{b^{2}}{a}. Without loss of generality, assume B(c,\\frac{b^{2}}{a}). Since the slope of line AB is 1, we get: \\frac{b^{2}}{c+a}=1, thus b^{2}=a(c+a). Also, b^{2}=c^{2}-a^{2}=(c-a)(c+a), hence c-a=a, so c=2a, e=\\frac{c}{x}=2+------------------"
  },
  {
    "text": "Given point $A(3,0)$, the right focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{3}=1(a>0)$ is $F$. If the midpoint of segment $AF$ lies exactly on the ellipse $C$, then the length of the major axis of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;a: Number;A: Point;F: Point;a>0;Expression(C) = (y^2/3 + x^2/a^2 = 1);Coordinate(A) = (3, 0);RightFocus(C) = F;PointOnCurve(MidPoint(LineSegmentOf(A,F)), C)",
    "query_expressions": "Length(MajorAxis(C))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[12, 63], [86, 91], [94, 99]], [[19, 63]], [[2, 11]], [[68, 71]], [[18, 63]], [[12, 63]], [[2, 11]], [[12, 71]], [[73, 92]]]",
    "query_spans": "[[[94, 105]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola $y^{2}=2 p x(p>0)$ is $F$, and point $A(0 , 2)$. If the midpoint $B$ of segment $FA$ lies on the parabola, then the distance from $B$ to the focus of the parabola is?",
    "fact_expressions": "F: Point;A: Point;G: Parabola;p: Number;B: Point;p>0;Expression(G) = (y^2 = 2*p*x);Coordinate(A) = (0, 2);Focus(G) = F;MidPoint(LineSegmentOf(F, A))=B;PointOnCurve(B, G)",
    "query_expressions": "Distance(B, Focus(G))",
    "answer_expressions": "3*sqrt(2)/4",
    "fact_spans": "[[[25, 28]], [[29, 40]], [[0, 21], [66, 69], [66, 69]], [[3, 21]], [[51, 54], [61, 64]], [[3, 21]], [[0, 21]], [[29, 40]], [[0, 28]], [[42, 54]], [[51, 59]]]",
    "query_spans": "[[[61, 76]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the common foci of an ellipse and a hyperbola, $P$ is one of their common points, and $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$, with the eccentricities of the ellipse and hyperbola being $e_{1}$ and $e_{2}$ respectively, what is the minimum value of $e_{1}^{2}+2 e_{2}^{2}$?",
    "fact_expressions": "G: Hyperbola;C:Ellipse;F1: Point;P: Point;F2: Point;Focus(G) = {F1,F2};Focus(C) = {F1,F2};e1:Number;e2:Number;OneOf(Intersection(G,C)) = P;AngleOf(F1, P, F2) = pi/3;Eccentricity(C) = e1;Eccentricity(G) = e2",
    "query_expressions": "Min(e1^2 + 2*e2^2)",
    "answer_expressions": "7/4+sqrt(6)/2",
    "fact_spans": "[[[21, 24], [84, 87]], [[18, 20], [81, 83]], [[2, 9]], [[30, 33]], [[10, 17]], [[2, 29]], [[2, 29]], [[94, 101]], [[103, 110]], [[30, 42]], [[44, 80]], [[81, 110]], [[81, 110]]]",
    "query_spans": "[[[112, 141]]]",
    "process": "Let the ellipse equation be $\\frac{x^{2}}{a_{1}^{2}}+\\frac{y^{2}}{b_{1}^{2}}=1$, and the hyperbola equation be $\\frac{x^{2}}{a_{2}^{2}}-\\frac{y^{2}}{b_{2}^{2}}=1$. From the definitions, we have $|PF_{1}|+|PF_{2}|=2a_{1}$, $|PF_{1}|-|PF_{2}|=2a_{2}$, so $|PF_{1}|=a_{1}+a_{2}$, $|PF_{2}|=a_{1}-a_{2}$. In triangle $F_{1}PF_{2}$, by the law of cosines, we obtain $(2c)^{2}=(a_{1}+a_{2})^{2}+(a_{1}-a_{2})^{2}-2(a_{1}+a_{2})(a_{1}-a_{2})\\cos\\frac{\\pi}{3}$, which simplifies to $4c^{2}=a_{1}^{2}+3a_{2}^{2}$. Then, $\\frac{c^{2}}{a_{1}^{2}}+2\\frac{c^{2}}{a_{2}^{2}}=\\frac{1}{4}\\left(\\frac{a_{1}^{2}+3a_{2}^{2}}{a_{1}^{2}}+2\\frac{a_{1}^{2}+3a_{2}^{2}}{a_{2}^{2}}\\right)=\\frac{7}{4}+\\frac{1}{4}\\left(\\frac{3a_{2}^{2}}{a_{1}^{2}}+\\frac{2a_{1}^{2}}{a_{2}^{2}}\\right)\\geqslant\\frac{7}{4}+\\frac{1}{4}\\cdot2\\sqrt{6}=\\frac{7}{4}+\\frac{\\sqrt{6}}{2}$, with equality if and only if $\\frac{3a_{2}^{2}}{a_{1}^{2}}=\\frac{2a_{1}^{2}}{a_{2}^{2}}$."
  },
  {
    "text": "A point $P$ on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ has a distance of $3$ to one of its foci. What is the distance from point $P$ to the other focus?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/9 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;OneOf(Focus(G)) = F1;OneOf(Focus(G)) = F2;Negation(F1=F2);Distance(P, F1) = 3",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "5",
    "fact_spans": "[[[0, 38], [45, 46]], [[0, 38]], [[41, 44], [62, 66]], [[0, 44]], [], [], [[45, 51]], [[45, 72]], [[45, 72]], [[41, 59]]]",
    "query_spans": "[[[45, 78]]]",
    "process": "First, determine $ a = 4 $ from the given conditions; then, using the definition of an ellipse, obtain an equation involving the desired distance $ d $ to reach the conclusion. Let the desired distance be $ d $. From the problem, we have: $ a = 4 $. According to the definition of the ellipse: $ 2a = 3 + d \\Rightarrow d = 2a - 3 = 5 $."
  },
  {
    "text": "The equation of the directrix of the parabola $y^{2}=-x$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -x)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=1/4",
    "fact_spans": "[[[0, 13]], [[0, 13]]]",
    "query_spans": "[[[0, 20]]]",
    "process": ""
  },
  {
    "text": "For the equation $\\frac{x^{2}}{k-5}+\\frac{y^{2}}{10-k}=1$ to represent an ellipse with foci on the $y$-axis, what is the range of real values for $k$?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (x^2/(k - 5) + y^2/(10 - k) = 1);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(5, 7.5)",
    "fact_spans": "[[[52, 54]], [[56, 61]], [[0, 54]], [[44, 54]]]",
    "query_spans": "[[[56, 68]]]",
    "process": ""
  },
  {
    "text": "It is known that an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is perpendicular to the line $3 x+4 y-1=0$. Then, the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;H: Line;Expression(H) = (3*x + 4*y - 1 = 0);IsPerpendicular(OneOf(Asymptote(G)), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[2, 58], [85, 88]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[65, 80]], [[65, 80]], [[2, 82]]]",
    "query_spans": "[[[85, 94]]]",
    "process": "The asymptotes of the hyperbola are given by $ bx\\pm ay=0 $. Since one of them is perpendicular to the line $ 3x+4y-1=0 $, it follows that $ 3b-4a=0 $, i.e., $ 9(c^{2}-a^{2})=16a^{2} $, hence $ \\frac{c}{a}=\\frac{5}{2} $, so $ e=\\frac{5}{2} $. If] This problem examines the calculation of the eccentricity of a hyperbola. For such problems, it suffices to find a set of relations among $ a,b,c $. This problem is easy."
  },
  {
    "text": "Given two fixed points $B(-3,0)$, $C(3,0)$, and the perimeter of $\\triangle ABC$ equal to $16$, then the trajectory equation of vertex $A$ is?",
    "fact_expressions": "B: Point;C: Point;A: Point;Coordinate(B) = (-3, 0);Coordinate(C) = (3, 0);Perimeter(TriangleOf(A, B, C)) = 16",
    "query_expressions": "LocusEquation(A)",
    "answer_expressions": "(x^2/25+y^2/16=1)&Negation(y=0)",
    "fact_spans": "[[[5, 15]], [[17, 25]], [[58, 61]], [[5, 15]], [[17, 25]], [[28, 54]]]",
    "query_spans": "[[[58, 68]]]",
    "process": "Since points B(-3,0) and C(3,0), we have BC=6. Given that the perimeter of triangle ABC is 16, it follows that AB+AC+BC=16, so AB+AC=10>6. By the definition of an ellipse, the locus of vertex A is an ellipse with foci at points B(-3,0) and C(3,0) and major axis length 10. Thus, the standard equation of this ellipse is \\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1. However, note that vertex A cannot be collinear or coincide with the fixed points B(-3,0) and C(3,0), otherwise a triangle cannot be formed, so y\\neq0. Therefore, the answer should be \\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1(y\\neq0)."
  },
  {
    "text": "The eccentricity of a hyperbola with asymptotes $\\sqrt{2} x \\pm y=0$ and foci on the $x$-axis is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (sqrt(2)*x + pm*y = 0);PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[36, 39]], [[0, 39]], [[27, 39]]]",
    "query_spans": "[[[36, 45]]]",
    "process": "By the given condition $\\frac{b}{a}=\\sqrt{2}$, combined with $e=\\frac{c}{a}=\\frac{\\sqrt{a^{2}+b^{2}}}{a^{2}}=\\sqrt{1+(\\frac{b}{a})^{2}}$, the solution is obtained. Hence, the asymptotes are $y=\\pm\\frac{b}{a}x$, so $\\frac{b}{a}=\\sqrt{2}$, then $e=\\frac{c}{a}=\\frac{\\sqrt{a^{2+b^{2}}}{a^{2}}=\\sqrt{1+(\\frac{b}{a})^{2}}=\\sqrt{3}$"
  },
  {
    "text": "The vertices of $\\triangle ABC$ are $B(-4 , 0)$, $C(4 , 0)$, and the incenter of $\\triangle ABC$ lies on the line $x=1$. Then the trajectory equation of vertex $A$ is?",
    "fact_expressions": "H: Line;B: Point;C: Point;A: Point;Expression(H) = (x = 1);Coordinate(B) = (-4, 0);Coordinate(C) = (4, 0);PointOnCurve(Center(InscribedCircle(TriangleOf(A,B,C))),H)",
    "query_expressions": "LocusEquation(A)",
    "answer_expressions": "(x^2-y^2/15=1)&(x>1)",
    "fact_spans": "[[[67, 74]], [[18, 29]], [[32, 42]], [[79, 82]], [[67, 74]], [[18, 29]], [[32, 42]], [[45, 75]]]",
    "query_spans": "[[[79, 89]]]",
    "process": ""
  },
  {
    "text": "Given $F_{1}$, $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{3}-y^{2}=1$, point $P$ lies on $C$, $\\angle F_{1} P F_{2}=60^{\\circ}$, then the area of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "C: Hyperbola;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/3 - y^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[18, 51], [63, 66]], [[58, 62]], [[2, 9]], [[10, 17]], [[18, 51]], [[2, 57]], [[2, 57]], [[58, 67]], [[68, 101]]]",
    "query_spans": "[[[103, 133]]]",
    "process": "Analysis: From the hyperbola equation, obtain $ c $. Using the definition of the hyperbola and combining with the cosine law in triangle, complete the square to simplify and find $ |PF_{1}|\\cdot|PF_{2}| $, then use the triangle area formula to obtain the solution. Hyperbola $ C: \\frac{x^{2}}{3}-y^{2}=1 $, then $ a^{2}=3, b^{2}=1 $, so $ c^{2}=a^{2}+b^{2}=4 $. By the definition of hyperbola, $ ||PF_{1}|-|PF_{2}||=2a=2\\sqrt{3} $. Squaring both sides gives $ |PF_{1}|^{2}+|PF_{2}|^{2}=12+2|PF_{1}|\\cdot|PF_{2}| $, and $ |F_{1}F_{2}|=2c=4 $, $ \\angle F_{1}PF_{2}=60^{\\circ} $. By the cosine law, $ \\cos\\angle F_{1}PF_{2}=\\frac{|PF_{1}|^{2}+|PF_{2}|^{2}-|F_{1}F_{2}|^{2}}{2|PF_{1}|\\cdot|PF_{2}|}=\\frac{12+2|PF_{1}|\\cdot|PF_{2}|-16}{2|PF_{1}|\\cdot|PF_{2}|}=\\frac{1}{2} $. Solving gives: $ |PF_{1}|\\cdot|PF_{2}|=4 $, then $ S_{\\triangle PF_{1}F_{2}}=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|\\cdot\\sin\\angle 60^{\\circ}=\\frac{1}{2}\\times4\\times\\frac{\\sqrt{3}}{2}=\\sqrt{3} $."
  },
  {
    "text": "It is known that the directrix of the parabola $y^{2}=4x$ intersects the two asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ at points $A$ and $B$, respectively, and $|AB|=2\\sqrt{3}$. Find the eccentricity $e$ of the hyperbola?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Parabola;A: Point;B: Point;e: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 4*x);l1:Line;l2:Line;Intersection(Directrix(H),l1) = A;Intersection(Directrix(H),l2) = B;Asymptote(G)={l1,l2};Abs(LineSegmentOf(A, B)) = 2*sqrt(3);Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "2",
    "fact_spans": "[[[20, 66], [107, 110]], [[23, 66]], [[23, 66]], [[2, 16]], [[76, 79]], [[80, 83]], [[114, 117]], [[20, 66]], [[2, 16]], [], [], [[2, 85]], [[2, 85]], [[20, 72]], [[87, 105]], [[107, 117]]]",
    "query_spans": "[[[114, 119]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{m+1}-\\frac{y^{2}}{m}=1$ $(m>0)$ is $x-\\sqrt{3} y=0$, then $m$=?",
    "fact_expressions": "G: Hyperbola;m: Number;m>0;Expression(G) = (x^2/(m + 1) - y^2/m = 1);Expression(OneOf(Asymptote(G))) = (x - sqrt(3)*y = 0)",
    "query_expressions": "m",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 48]], [[75, 78]], [[5, 48]], [[2, 48]], [[2, 73]]]",
    "query_spans": "[[[75, 80]]]",
    "process": "The asymptotes of the hyperbola $\\frac{x^{2}}{m+1}-\\frac{y^{2}}{m}=1$ $(m>0)$ are given by $y=\\pm\\sqrt{\\frac{m}{m+1}}$. From the hyperbola $\\frac{x^{2}}{m^{2}}-y^{2}=1$ $(m>0)$, one asymptote is $x-\\sqrt{3}y=0$, that is, $y=\\frac{\\sqrt{3}}{3}x$. Therefore, $\\sqrt{\\frac{m}{m+1}}=\\frac{\\sqrt{3}}{3}$, so $m=\\frac{1}{2}$."
  },
  {
    "text": "Let point $P$ be a moving point on the ellipse $C$: $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$, $F$ be the right focus of $C$, and fixed point $A(2,1)$. Then the range of values of $|P A|+|P F|$ is?",
    "fact_expressions": "C: Ellipse;A: Point;P: Point;F: Point;Expression(C) = (x^2/8 + y^2/4 = 1);Coordinate(A) = (2, 1);PointOnCurve(P, C);RightFocus(C) = F",
    "query_expressions": "Range(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "[4*sqrt(2)-sqrt(17),4*sqrt(2)+sqrt(17)]",
    "fact_spans": "[[[6, 48], [57, 60]], [[67, 75]], [[1, 5]], [[53, 56]], [[6, 48]], [[67, 75]], [[1, 52]], [[53, 64]]]",
    "query_spans": "[[[77, 97]]]",
    "process": "\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1,F is the right focus of C, F(2,0), left focus F_{1}(-2,0), |PA|+|PF|=|PA|+2a-|PF_{1}|=4\\sqrt{2}+|PA|-|PF_{1}|, |PA|+|PF|=|PA|+2a-|PF_{1}|=4\\sqrt{2}+|PA|-|PF_{1}|, |PA|+|PF|\\in[4\\sqrt{2}-\\sqrt{17},4\\sqrt{2}+\\sqrt{17}]"
  },
  {
    "text": "Ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ has left and right foci $F_{1}$, $F_{2}$ respectively. Point $A$ lies on the ellipse, $\\overrightarrow{A F_{1}} \\cdot \\overrightarrow{A F_{2}}=0$, the line $A F_{2}$ intersects the ellipse at point $B$, $|\\overrightarrow{A B}|=|\\overrightarrow{A F}|$, then the eccentricity of the ellipse is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;A: Point;PointOnCurve(A, C) = True;DotProduct(VectorOf(A, F1), VectorOf(A, F2)) = 0;Intersection(LineOf(A, F2), C) = B;B: Point;Abs(VectorOf(A, B)) = Abs(VectorOf(A, F))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "{sqrt(6)-sqrt(3), sqrt(9-6*sqrt(2))}",
    "fact_spans": "[[[0, 57], [87, 89], [163, 165], [220, 222]], [[0, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[66, 73]], [[74, 81]], [[0, 81]], [[0, 81]], [[82, 86]], [[82, 90]], [[91, 150]], [[151, 170]], [[166, 170]], [[171, 218]]]",
    "query_spans": "[[[220, 228]]]",
    "process": "We can use the condition that triangle ABF_{1} is an isosceles right triangle, set the side lengths, find the quantitative relationship between the side lengths and a, b, then substitute this relationship into the equation expressed by the Pythagorean theorem to solve for the eccentricity. According to the problem, triangle ABF_{1} is an isosceles right triangle. Let AF_{1} = AB = x, then x + x + \\sqrt{2}x = 4a, solving gives x = (4 - 2\\sqrt{2})a, AF_{2} = (2\\sqrt{2} - 2)a. In triangle AF_{1}F_{2}, by the Pythagorean theorem, (AF_{1})^{2} + (AF_{2})^{2} = (2c)^{2}, so e^{2} = 9 - 6\\sqrt{2}, e = \\sqrt{6} - \\sqrt{3}"
  },
  {
    "text": "Given points $O(0,0)$, $M(1,0)$, and that there exists at least one point $P$ on the curve $C$: $(x-5)^{2}-(y-4)^{2}=t^{2}$ $(t>0)$ such that $|P M|=|P O|$, then the minimum value of $t$ is?",
    "fact_expressions": "O: Point;M: Point;P: Point;C:Curve;t:Number;t>0;Expression(C) = ((x-5)^2 - (y-4)^2 = t^2);Coordinate(O) = (0, 0);Coordinate(M) = (1, 0);PointOnCurve(P,C);Abs(LineSegmentOf(P, M)) = Abs(LineSegmentOf(P, O))",
    "query_expressions": "Min(t)",
    "answer_expressions": "9/2",
    "fact_spans": "[[[2, 11]], [[12, 20]], [[65, 68]], [[22, 58]], [[86, 89]], [[26, 58]], [[22, 58]], [[2, 11]], [[12, 20]], [[22, 68]], [[71, 84]]]",
    "query_spans": "[[[86, 95]]]",
    "process": ""
  },
  {
    "text": "Given that a focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{m}=1$ lies on the circle $x^{2}+y^{2}-4 x-5=0$, then the asymptotes of the hyperbola are given by?",
    "fact_expressions": "G: Hyperbola;m: Number;H: Circle;Expression(G) = (x^2/9 - y^2/m = 1);Expression(H) = (-4*x + x^2 + y^2 - 5 = 0);PointOnCurve(OneOf(Focus(G)),H)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "pm*3*y+4*x=0",
    "fact_spans": "[[[2, 40], [71, 74]], [[5, 40]], [[46, 68]], [[2, 40]], [[46, 68]], [[2, 69]]]",
    "query_spans": "[[[71, 82]]]",
    "process": "Convert the equation of the circle into standard form: (x-2)^{2}+y^{2}=9, the intersection points with the x-axis are (-1,0), (5,0). \\therefore 9+m=5^{2} \\Rightarrow m=16, thus the new asymptote equations are y=\\pm\\frac{4}{3}x, i.e., 4x\\pm3y=0, hence fill in: 4x\\pm3y=0."
  },
  {
    "text": "Given the line $l$: $y = k(x + \\frac{p}{2})$ $(k > 0)$ intersects the parabola $\\Gamma$: $y^2 = 2px$ at points $A$ and $B$, the intersection point of the directrix of the parabola $\\Gamma$ and the $x$-axis is $C$, and it satisfies $\\overrightarrow{AB} + \\overrightarrow{AC} = \\overrightarrow{0}$, then the value of $k$ is?",
    "fact_expressions": "l: Line;Gamma: Parabola;A: Point;B: Point;C: Point;k: Number;p: Number;k > 0;Expression(l) = (y = k*(p/2 + x));Expression(Gamma) = (y^2 = 2*p*x);Intersection(Gamma, l) = {A, B};Intersection(Directrix(Gamma), xAxis) = C;VectorOf(A, B) + VectorOf(A, C) = 0",
    "query_expressions": "k",
    "answer_expressions": "sqrt(2)*2/3",
    "fact_spans": "[[[2, 34]], [[35, 61], [74, 85]], [[64, 67]], [[68, 71]], [[97, 100]], [[168, 171]], [[48, 61]], [[9, 34]], [[2, 34]], [[35, 61]], [[2, 73]], [[74, 100]], [[104, 166]]]",
    "query_spans": "[[[168, 175]]]",
    "process": "Let points $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, let $ m = \\frac{1}{k} $, then the equation of line $ l $ is $ x = my - \\frac{p}{2} $. Let point $ C(-\\frac{p}{2}, 0) $. Solve the system of equations of line $ l $ and parabola $ T $:  \n$$\n\\begin{cases}\nx = my - \\frac{p}{2} \\\\\ny^2 = 2px\n\\end{cases}\n$$\nEliminating $ x $ gives $ y^2 - 2pmy + p^2 = 0 $.  \n$ \\Delta = 4m^2p^2 - 4p^2 > 0 $, since $ m > 0 $, solve to get $ m > 1 $.  \nBy Vieta's formulas: $ y_1 + y_2 = 2mp $, $ y_1 y_2 = p^2 $.  \nSince $ \\overrightarrow{AB} + \\overrightarrow{AC} = \\overrightarrow{0} $, then  \n$ (x_2 - x_1, y_2 - y_1) + (-\\frac{p}{2} - x_1, -y_1) = \\overrightarrow{0} $,  \nthus $ y_2 - 2y_1 = 0 $, i.e., $ y_2 = 2y_1 $.  \nTherefore, $ 2mp = y_1 + y_2 = 3y_1 $, so $ y_1 = \\frac{2mp}{3} $,  \n$ p^2 = y_1 y_2 = 2y_1^2 = \\frac{8m^2p^2}{9} $,  \nsolving gives $ m = \\frac{3\\sqrt{2}}{4} $, hence $ k = \\frac{1}{m} = \\frac{2\\sqrt{2}}{3} $."
  },
  {
    "text": "Given the parabola $C_{1}$: $x^{2}=2 p y(p>0)$ and the hyperbola $C_{2}$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1(a>0, b>0)$ have the same focus and intersect at point $A$ in the first quadrant, and $F$ is the lower focus of the hyperbola $C_{2}$. If the line $A F$ has exactly one common point with the parabola $C_{1}$, then the eccentricity of the hyperbola $C_{2}$ is?",
    "fact_expressions": "C2: Hyperbola;C1: Parabola;F: Point;A: Point;p: Number;p >0;Expression(C1) = (x^2 =2*p*y);Expression(C2) = (-x^2/b^2 + y^2/a^2 = 1);a: Number;b: Number;a > 0;b > 0;Focus(C1) = Focus(C2);Intersection(C1, C2) = A;Quadrant(A) = 1;LowerFocus(C2) = F;NumIntersection(LineOf(A, F), C1) = 1",
    "query_expressions": "Eccentricity(C2)",
    "answer_expressions": "sqrt(2) + 1",
    "fact_spans": "[[[33, 98], [121, 131], [166, 176]], [[2, 32], [145, 155]], [[117, 120]], [[112, 116]], [[14, 32]], [[14, 32]], [[2, 32]], [[33, 98]], [[45, 98]], [[45, 98]], [[45, 98]], [[45, 98]], [[2, 104]], [[2, 116]], [[106, 116]], [[117, 135]], [[137, 164]]]",
    "query_spans": "[[[166, 182]]]",
    "process": "According to the problem, we have $ p = 2c $. Let $ A(x_{0}, y_{0}) $, then $ y_{0} = \\frac{1}{4c} \\cdot x_{0}^{2} $. Find the slope and equation of line $ AF $, combine it with the parabola equation, and use $ \\Delta = 0 $ to find the coordinates of point $ A $. Then substitute into the hyperbola equation and combine with $ b^{2} = c^{2} - a^{2} $ to obtain a homogeneous equation in $ a $ and $ c $. Then from $ e = \\frac{c}{a} $, the eccentricity can be obtained.  \n\nParabola $ C_{1}: x^{2} = 2py $ ($ p > 0 $) has focus at $ (0, \\frac{p}{2}) $, hyperbola $ C_{2}: \\frac{y^{2}}{a^{2}} - \\frac{x^{2}}{b^{2}} = 1 $ has upper focus at $ (0, c) $, $ F(0, -c) $. Since the two curves share common foci, we get $ \\frac{p}{2} = c $, so $ p = 2c $. Thus, parabola $ C_{1}: x^{2} = 4cy $. Let $ A(x_{0}, y_{0}) $, then $ y_{0} = \\frac{1}{4c} \\cdot x_{0}^{2} $. From the condition, line $ AF $ is tangent to the parabola at point $ A $.  \n$ k_{AF} = \\frac{y_{0} + c}{x_{0}} = \\frac{\\frac{1}{4c} \\cdot x_{0}^{2} + c}{x_{0}} = \\frac{x_{0}}{4c} + \\frac{c}{x_{0}} $,  \nso the equation of line $ AF $ is: $ y - (-c) = \\left( \\frac{x_{0}}{4c} + \\frac{c}{x_{0}} \\right)(x - 0) $,  \nwhich simplifies to: $ y = \\left( \\frac{x_{0}}{4c} + \\frac{c}{x_{0}} \\right)x - c $. Substituting into $ x^{2} = 4cy $ gives  \n$ x^{2} = 4c \\left[ \\left( \\frac{x_{0}}{4c} + \\frac{c}{x_{0}} \\right)x - c \\right] $,  \ni.e., $ x^{2} - \\left( x_{0} + \\frac{4c^{2}}{x_{0}} \\right)x + 4c^{2} = 0 $.  \nFrom $ \\Delta = \\left( x_{0} + \\frac{4c^{2}}{x_{0}} \\right)^{2} - 4 \\times 4c^{2} = 0 $, we get  \n$ x_{0} + \\frac{4c^{2}}{x_{0}} = 4c $, i.e., $ x_{0}^{2} - 4cx_{0} + 4c^{2} = 0 $,  \nthus $ x_{0} = 2c $, so $ y_{0} = \\frac{1}{4c} \\cdot x_{0}^{2} = c $, hence $ A(2c, c) $.  \nSince point $ A(2c, c) $ lies on the hyperbola, we have  \n$ \\frac{c^{2}}{a^{2}} - \\frac{4c^{2}}{b^{2}} = 1 $, i.e., $ b^{2}c^{2} - 4a^{2}c^{2} = a^{2}b^{2} $.  \nSince $ b^{2} = c^{2} - a^{2} $, we get  \n$ (c^{2} - a^{2})c^{2} - 4a^{2}c^{2} = a^{2}(c^{2} - a^{2}) $,  \ni.e., $ c^{4} - 6a^{2}c^{2} + a^{4} = 0 $. Dividing both sides by $ a^{4} $ yields:  \n$ e^{4} - 6e^{2} + 1 = 0 $, solving gives $ e^{2} = 3 + 2\\sqrt{2} $ or $ e^{2} = 3 - 2\\sqrt{2} $ (discarded).  \nSince $ e > 1 $, we have $ e = \\sqrt{2} + 1 $."
  },
  {
    "text": "Given that an asymptote of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1$ ($b>0$) has the equation $y=2x$, then $b$=?",
    "fact_expressions": "G: Hyperbola;b: Number;b>0;Expression(G) = (x^2 - y^2/b^2 = 1);Expression(OneOf(Asymptote(G)))=(y=2*x)",
    "query_expressions": "b",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 39]], [[58, 61]], [[5, 39]], [[2, 39]], [[2, 56]]]",
    "query_spans": "[[[58, 63]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the left focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$, $AB$ is a chord of the ellipse $C$ passing through $F$, and the perpendicular bisector of $AB$ intersects the $x$-axis at point $P$. If $\\overrightarrow{A F}=2 \\overrightarrow{F B}$, and $P$ is the midpoint of $OF$, then the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;A: Point;B: Point;O: Origin;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;IsChordOf(LineSegmentOf(A,B),C);PointOnCurve(F,LineSegmentOf(A,B));Intersection(PerpendicularBisector(LineSegmentOf(A, B)), xAxis) = P;VectorOf(A, F) = 2*VectorOf(F, B);MidPoint(LineSegmentOf(O, F)) = P;F:Point;P:Point",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[6, 65], [76, 81], [173, 178]], [[13, 65]], [[13, 65]], [[70, 75]], [[70, 75]], [[163, 168]], [[13, 65]], [[13, 65]], [[6, 65]], [[2, 69]], [[70, 87]], [[70, 87]], [[88, 109]], [[112, 157]], [[159, 171]], [[2, 5], [82, 85]], [[105, 109], [159, 162]]]",
    "query_spans": "[[[173, 184]]]",
    "process": ""
  },
  {
    "text": "The asymptotes of the hyperbola $\\frac{y^{2}}{4}-x^{2}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2 + y^2/4 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*2*x",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 36]]]",
    "process": "From the asymptote formula of the hyperbola, we get y=\\pm\\frac{a}{b}x, y=\\pm2x"
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and let $O$ be the coordinate origin. Draw a perpendicular from $F_{2}$ to an asymptote of $C$, with foot of the perpendicular at $P$. If $|P F_{1}|=\\sqrt{6}|O P|$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F1: Point;O: Origin;F2: Point;l: Line;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F2,l);IsPerpendicular(l, OneOf(Asymptote(C)));FootPoint(l, OneOf(Asymptote(C)))=P;Abs(LineSegmentOf(P, F1)) = sqrt(6)*Abs(LineSegmentOf(O, P))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[17, 78], [103, 106], [152, 155]], [[24, 78]], [[24, 78]], [[119, 122]], [[1, 8]], [[85, 88]], [[95, 102], [9, 16]], [], [[24, 78]], [[24, 78]], [[17, 78]], [[1, 84]], [[1, 84]], [[94, 115]], [[94, 115]], [[94, 122]], [[125, 150]]]",
    "query_spans": "[[[152, 161]]]",
    "process": "As shown in the figure: so |PF_{2}| = b, then |OP| = a, so |PF_{1}| = \\sqrt{6}a. Therefore, \\frac{a^{2}+c^{2}-(\\sqrt{6}a)^{2}}{2ac} = -\\frac{a^{2}+c^{2}-b^{2}}{2ac}. Solving gives: c^{2} = 3a^{2} \\Rightarrow e = \\sqrt{3}"
  },
  {
    "text": "The line passing through point $P(8,1)$ intersects the hyperbola $x^{2}-4 y^{2}=4$ at points $A$ and $B$, and $P$ is the midpoint of segment $AB$. Then the equation of line $AB$ is?",
    "fact_expressions": "P: Point;Coordinate(P) = (8, 1);H: Line;PointOnCurve(P, H);G: Hyperbola;Expression(G) = (x^2 - 4*y^2 = 4);A: Point;B: Point;Intersection(H, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Expression(LineOf(A, B))",
    "answer_expressions": "2*x-y-15=0",
    "fact_spans": "[[[1, 10], [50, 53]], [[1, 10]], [[11, 13]], [[0, 13]], [[14, 34]], [[14, 34]], [[37, 40]], [[42, 45]], [[11, 47]], [[50, 63]]]",
    "query_spans": "[[[66, 77]]]",
    "process": ""
  },
  {
    "text": "It is known that a hyperbola passes through the point $(4 , \\frac{4 \\sqrt{7}}{3})$, and its asymptotes have equations $y=\\pm \\frac{4}{3} x$. A circle $C$ passes through a vertex and a focus of the hyperbola, and its center lies on the hyperbola. Then, what is the distance from the center of the circle to the center of the hyperbola?",
    "fact_expressions": "G: Hyperbola;C: Circle;H: Point;Coordinate(H) = (4, (4*sqrt(7))/3);PointOnCurve(H, G);Expression(Asymptote(G)) = (y = pm*(4/3)*x);PointOnCurve(OneOf(Vertex(G)),C);PointOnCurve(OneOf(Focus(G)),C);PointOnCurve(Center(C),G)",
    "query_expressions": "Distance(Center(C), Center(G))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[2, 5], [70, 73], [87, 90], [97, 100]], [[64, 68]], [[6, 35]], [[6, 35]], [[2, 35]], [[2, 63]], [[64, 78]], [[64, 83]], [[64, 91]]]",
    "query_spans": "[[[64, 108]]]",
    "process": ""
  },
  {
    "text": "Let $F$ be the focus of the parabola $y^{2}=x$. A line passing through $F$ intersects the parabola at points $A$ and $B$. Then the minimum value of $|A F|+3|B F|$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;Expression(G) = (y^2 = x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B}",
    "query_expressions": "Min(Abs(LineSegmentOf(A, F)) + 3*Abs(LineSegmentOf(B, F)))",
    "answer_expressions": "sqrt(3)/2+1",
    "fact_spans": "[[[5, 17], [29, 32]], [[26, 28]], [[34, 37]], [[1, 4], [22, 25]], [[38, 41]], [[5, 17]], [[1, 20]], [[21, 28]], [[26, 43]]]",
    "query_spans": "[[[45, 65]]]",
    "process": "lThe equation of line AB and the parabola equation are solved simultaneously, and the solution can be obtained using the relationship between the roots and coefficients of a quadratic equation and the basic inequality. From the parabola equation, the focus is F(\\frac{1}{4},0). According to the given conditions, obviously the line AB has a slope; let the equation of line AB be: y=k(x-\\frac{1}{4}). Let A(x_{1},y_{1}), B(x_{2},y_{2}). Solving the system \\begin{cases}y=k(x-\\frac{1}{4})\\\\y^{2}=x\\end{cases}, we obtain: k^{2}x^{2}-(\\frac{1}{2}k^{2}+1)x+\\frac{k^{2}}{16}=0, x_{1}x_{2}=\\frac{1}{16}. From the property of the parabola, |AF|=x_{1}+\\frac{1}{4}, |BF|=x_{2}+\\frac{1}{4}, so |AF|+3|BF|=x_{1}+3x_{2}+1\\geqslant2\\sqrt{3x_{1}x_{2}}+1=\\frac{\\sqrt{3}}{2}+1"
  },
  {
    "text": "Given that point $O$ is the coordinate origin, point $F$ is the focus of the parabola $C$: $y^{2}=2 p x$ ($p>0$), and the perpendicular bisector of segment $O F$ intersects the parabola $C$ at a point $A$, with $|O A|=3$, then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;O: Origin;F: Point;A: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;OneOf(Intersection(PerpendicularBisector(LineSegmentOf(O,F)), C)) = A;Abs(LineSegmentOf(O, A)) = 3",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[17, 43], [62, 68]], [[90, 93]], [[2, 6]], [[12, 16]], [[74, 77]], [[25, 43]], [[17, 43]], [[12, 46]], [[48, 77]], [[79, 88]]]",
    "query_spans": "[[[90, 95]]]",
    "process": "From the given condition, we have $ x_{A} = \\frac{p}{4} $. Substituting into $ y^{2} = 2px $, we get $ |y_{A}| = \\frac{\\sqrt{2}}{2}p $, so $ |OA| = \\sqrt{\\left( \\frac{p}{4} \\right)^{2} + \\left( \\frac{\\sqrt{2}}{2}p \\right)^{2}} = \\frac{3}{4}p = 3 $, hence $ p = 4 $."
  },
  {
    "text": "If the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{12}=1$ ($a>0$) has eccentricity $2$, then what is the equation of its asymptotes?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2/12 + x^2/a^2 = 1);Eccentricity(G) = 2",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[1, 49], [59, 60]], [[4, 49]], [[4, 49]], [[1, 49]], [[1, 57]]]",
    "query_spans": "[[[59, 67]]]",
    "process": "According to the problem, we have $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{1 + \\frac{12}{a^{2}}} = 2 $, solving gives $ a = 2 $, therefore the asymptotes of this hyperbola are $ y = \\pm\\frac{b}{a}x = \\pm\\sqrt{3}x $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, the left and right vertices are $A$ and $B$ respectively. Point $P$ lies on the hyperbola $C$, and the product of the slopes of lines $PA$ and $PB$ is $1$. Then, what is the focal length of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;b: Number;A: Point;P: Point;B: Point;b>0;Expression(C) = (x^2/4 - y^2/b^2 = 1);LeftVertex(C) = A;RightVertex(C) = B;PointOnCurve(P, C);Slope(LineOf(P, A)) * Slope(LineOf(P, B)) = 1",
    "query_expressions": "FocalLength(C)",
    "answer_expressions": "4*sqrt(2)",
    "fact_spans": "[[[2, 54], [76, 82], [111, 117]], [[10, 54]], [[63, 66]], [[71, 75]], [[67, 70]], [[10, 54]], [[2, 54]], [[2, 70]], [[2, 70]], [[71, 83]], [[85, 109]]]",
    "query_spans": "[[[111, 122]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, with left and right foci $F_{1}$, $F_{2}$ respectively, point $P$ lies on the ellipse, $O$ is the origin, $|F_{1} F_{2}|=\\sqrt{2}|P F_{1}|$, $\\angle F_{1} P F_{2}=120^{\\circ}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;F2: Point;P: Point;O: Origin;a>0;b>0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);Abs(LineSegmentOf(F1, F2)) = sqrt(2)*Abs(LineSegmentOf(P, F1));AngleOf(F1,P,F2)=ApplyUnit(120,degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(\\sqrt{10}-\\sqrt{2})/2",
    "fact_spans": "[[[2, 60], [90, 92], [175, 177]], [[4, 60]], [[4, 60]], [[69, 76]], [[77, 84]], [[85, 89]], [[94, 97]], [[4, 60]], [[4, 60]], [[2, 60]], [[2, 84]], [[2, 84]], [[85, 93]], [[103, 137]], [[139, 173]]]",
    "query_spans": "[[[175, 183]]]",
    "process": "Let the semi-focal distance of the ellipse be $ c $. Since $ |F_{1}F_{2}| = \\sqrt{2}|PF_{1}| $, then $ |PF_{1}| = \\sqrt{2}c $. By the definition of an ellipse, we have $ |PF_{2}| = 2a - \\sqrt{2}c $. In triangle $ \\triangle F_{1}PF_{2} $, by the law of cosines:  \n$ |F_{1}F_{2}|^{2} = |PF_{1}|^{2} + |PF_{2}|^{2} - 2|PF_{1}|\\cdot|PF_{2}|\\cos\\angle F_{1}PF_{2} $,  \nthat is,  \n$ 4c^{2} = (\\sqrt{2}c)^{2} + (2a - \\sqrt{2}c)^{2} - 2\\sqrt{2}c\\cdot(2a - \\sqrt{2}c)\\cdot\\left(-\\frac{1}{2}\\right) $,  \nsimplifying yields  \n$ c^{2} + \\sqrt{2}ac - 2a^{2} = 0 $,  \nthus we have  \n$ e^{2} + \\sqrt{2}e - 2 = 0 $, and since $ 0 < e < 1 $, solving gives  \n$ e = \\frac{\\sqrt{10} - \\sqrt{2}}{2} $,  \nso the eccentricity of the ellipse is $ \\underline{\\sqrt{10}-\\sqrt{2}} $."
  },
  {
    "text": "The length of the imaginary axis of the hyperbola $\\frac{x^{2}}{a}+y^{2}=1$ is $2$ times the length of the real axis, then $a$=?",
    "fact_expressions": "G: Hyperbola;a: Number;Expression(G) = (y^2 + x^2/a = 1);Length(ImageinaryAxis(G)) = 2*Length(RealAxis(G))",
    "query_expressions": "a",
    "answer_expressions": "-4",
    "fact_spans": "[[[0, 28]], [[43, 46]], [[0, 28]], [[0, 41]]]",
    "query_spans": "[[[43, 48]]]",
    "process": "Using the definitions of the lengths of the imaginary axis and the real axis, an equation can be established to find the standard form of the hyperbola \\frac{x^{2}}{a}+y^{2}=1 as y^{2}-\\frac{x^{2}}{a}=1. The length of the imaginary axis is 2\\sqrt{-a}<2. From the given condition, 2\\sqrt{-a}=4, \\therefore a=-4,"
  },
  {
    "text": "Given that the focus of the parabola $x^{2}=4 y$ is $F$, the directrix is $l$, $P$ is a point on $l$, and $Q$ is an intersection point of the line $P F$ and the parabola $C$. If $\\overrightarrow{P Q}=4 \\overrightarrow{F Q}$, then $|\\overrightarrow{F Q}|$=?",
    "fact_expressions": "C: Parabola;F: Point;P: Point;Q: Point;l: Line;Expression(C)=(x^2=4*y);Focus(C)=F;Directrix(C) = l;PointOnCurve(P, l);OneOf(Intersection(LineOf(P,F),C))=Q;VectorOf(P, Q) = 4*VectorOf(F, Q)",
    "query_expressions": "Abs(VectorOf(F, Q))",
    "answer_expressions": "8/3",
    "fact_spans": "[[[2, 16], [55, 61]], [[20, 23]], [[31, 35]], [[43, 46]], [[27, 30], [36, 39]], [[2, 16]], [[2, 23]], [[2, 30]], [[31, 42]], [[43, 66]], [[69, 114]]]",
    "query_spans": "[[[116, 142]]]",
    "process": "Let $ P(a,-1) $, $ Q(m,\\frac{m^{2}}{4}) $. According to $ \\overrightarrow{PQ} = 4\\overrightarrow{FQ} $, we can construct an equation to solve for $ m^{2} $. From the parabola equation, we know: $ F(0,1) $, directrix $ l: y = -1 $. Let $ P(a,-1) $, $ Q(m,\\frac{m^{2}}{4}) $, then $ \\overrightarrow{PQ} = (m-a, \\frac{m^{2}}{4}+1) $, $ \\overrightarrow{FQ} = (m, \\frac{m^{2}}{4}-1) $. Since $ \\overrightarrow{PQ} = 4\\overrightarrow{FQ} $, we have $ \\frac{m^{2}}{4} + 1 = m^{2} - 4 $, solving gives: $ m^{2} = \\frac{20}{3} $. By the definition of the parabola, we obtain: $ |QF| = \\frac{m^{2}}{4} + 1 = \\frac{5}{3} + 1 = \\frac{8}{3} $."
  },
  {
    "text": "Through the center of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, draw a line intersecting the ellipse at points $A$ and $B$, and let $F_{1}$ be a focus of the ellipse. Then the maximum area of triangle $F_{1} A B$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);H: Line;PointOnCurve(Center(G), H);A: Point;B: Point;Intersection(H, G) = {A, B};F1: Point;OneOf(Focus(G)) = F1",
    "query_expressions": "Max(Area(TriangleOf(F1, A, B)))",
    "answer_expressions": "12",
    "fact_spans": "[[[1, 40], [47, 49], [69, 71]], [[1, 40]], [[44, 46]], [[0, 46]], [[51, 54]], [[55, 58]], [[44, 60]], [[61, 68]], [[61, 74]]]",
    "query_spans": "[[[76, 98]]]",
    "process": "From $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ we get $a^{2}=25$, $b^{2}=16$, so $a=5$, $b=4$, $c=3$. Let $A(x_{1},y_{1})$, then $S_{\\Delta F_{1}AB}=S_{\\Delta AF_{1}O}+S_{\\Delta BF_{1}O}=2S_{\\Delta AF_{1}O}=2\\times\\frac{1}{2}|F_{1}O||y_{1}|=3|y_{1}|$. Since $A(x_{1},y_{1})$ lies on the ellipse, $|y_{1}|\\leqslant b=4$, so $S_{\\triangle AF_{1}B}\\leqslant 12$, that is, the maximum area of triangle $F_{1}AB$ is $12$."
  },
  {
    "text": "Given that the hyperbolas $C_{1}$ and $C_{2}$ have foci on the $x$-axis and $y$-axis respectively, eccentricities $e_{1}$ and $e_{2}$, and share the same asymptotes, then the minimum value of $e_{1} \\cdot e_{2}$ is?",
    "fact_expressions": "C1:Hyperbola;C2:Hyperbola;PointOnCurve(Focus(C1),xAxis);PointOnCurve(Focus(C2),yAxis);e1:Number;e2:Number;Eccentricity(C1)=e1;Eccentricity(C2)=e2;Asymptote(C1)=Asymptote(C2)",
    "query_expressions": "Min(e1*e2)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 12]], [[13, 20]], [[2, 36]], [[2, 36]], [[43, 50]], [[52, 59]], [[2, 59]], [[2, 59]], [[2, 66]]]",
    "query_spans": "[[[68, 93]]]",
    "process": ""
  },
  {
    "text": "The equation $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{(m-1)^{2}}=1$ represents an ellipse with foci on the $y$-axis. Then the range of real values for $m$ is?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (y^2/(m - 1)^2 + x^2/m^2 = 1);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-oo,0)+(0,1/2)",
    "fact_spans": "[[[60, 62]], [[64, 69]], [[0, 62]], [[51, 62]]]",
    "query_spans": "[[[64, 76]]]",
    "process": "According to the problem, we can obtain a system of inequalities about the real number $ m $, from which the range of values for the real number $ m $ can be solved. From the problem, we have\n\\[\n\\begin{cases}\nm^{2} > 0 \\\\\n(m-1)^{2} > 0 \\\\\n(m-1)^{2} > m^{2}\n\\end{cases}\n\\]\nSolving this gives $ m < \\frac{1}{2} $ and $ m \\neq 0 $. Therefore, the range of values for the real number $ m $ is $ (-\\infty, 0) \\cup (0, \\frac{1}{2}) $."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=4x$ be $F$. A line $l$ passing through point $F$ intersects the parabola at two points $A$ and $B$. If point $M(2, t)$ satisfies $\\overrightarrow{O M}=\\frac{1}{2}(\\overrightarrow{O A}+\\overrightarrow{O B})$, then $|A B|=$?",
    "fact_expressions": "l: Line;G: Parabola;M: Point;O: Origin;A: Point;B: Point;F: Point;t:Number;Expression(G) = (y^2 = 4*x);Coordinate(M) = (2, t);Focus(G)=F;PointOnCurve(F, l);Intersection(l, G) = {A, B};VectorOf(O, M)=(1/2)*(VectorOf(O, A) + VectorOf(O, B))",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "6",
    "fact_spans": "[[[29, 34]], [[1, 15], [35, 38]], [[53, 63]], [[65, 142]], [[44, 47]], [[48, 51]], [[19, 22], [24, 28]], [[54, 63]], [[1, 15]], [[53, 63]], [[1, 22]], [[23, 34]], [[29, 51]], [[65, 142]]]",
    "query_spans": "[[[144, 153]]]",
    "process": "The focus of the parabola $ y^{2} = 4x $ is $ F(1,0) $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Since the line $ AB $ passes through the focus $ F(1,0) $, $ |AB| = x_{1} + x_{2} + 2 $. Also, $ \\overrightarrow{OM} = \\frac{1}{2}(\\overrightarrow{OA} + \\overrightarrow{OB}) $, then $ M(2,t) $ is the midpoint of $ AB $, so $ |AB| = 4 + 2 = 6 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{b^{2}}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $P$ is an arbitrary point on the hyperbola. If the minimum value of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ is $-7$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 - y^2/b^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P,G);Min(DotProduct(VectorOf(P, F1), VectorOf(P, F2))) = (-1)*7",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "4/3",
    "fact_spans": "[[[2, 44], [73, 76], [73, 76]], [[5, 44]], [[69, 72]], [[53, 60]], [[61, 68]], [[2, 44]], [[2, 68]], [[2, 68]], [[69, 80]], [[82, 148]]]",
    "query_spans": "[[[151, 160]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $9 y^{2}-16 x^{2}=144$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-16*x^2 + 9*y^2 = 144)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*(4/3)*x",
    "fact_spans": "[[[0, 25]], [[0, 25]]]",
    "query_spans": "[[[0, 33]]]",
    "process": "Convert the hyperbola $9y^{2}-16x^{2}=144$ into standard form: $\\frac{y^{2}}{16}-\\frac{x^{2}}{9}=1$. Therefore, $a=4$ and $b=3$, so the asymptotes of the hyperbola are $y=\\pm\\frac{a}{b}x$, that is, $y=\\pm\\frac{4}{3}x$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of an ellipse, if there exists a point $P$ on the ellipse such that $\\angle F_{1} P F_{2}=2 \\theta$ ($0<\\theta<\\frac{\\pi}{2}$, $\\theta$ is a known constant), then the range of values for the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;theta:Number;Focus(G) = {F1, F2};PointOnCurve(P,G);AngleOf(F1,P,F2) = 2*theta;0 < theta;theta < pi/2",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[\\sin(\\theta),1)",
    "fact_spans": "[[[21, 23], [31, 33], [119, 121]], [[2, 9]], [[36, 40]], [[10, 17]], [[103, 111]], [[2, 28]], [[31, 40]], [[42, 74]], [[76, 100]], [[76, 100]]]",
    "query_spans": "[[[119, 131]]]",
    "process": "According to the geometric meaning of the ellipse, since $0<\\theta<\\frac{\\pi}{2}$, we have $0<2\\theta<\\pi$. The minimum eccentricity of the ellipse is $e=\\frac{c}{a}=\\sin\\theta$. According to the range of values for the ellipse's eccentricity, $e\\in[\\sin\\theta,1)$."
  },
  {
    "text": "Given that the line $x=-2$ intersects the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{21}=1$ at points $A$ and $B$, and the right focus of the ellipse is point $F$, then the perimeter of $\\triangle A B F$ is?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;F: Point;Expression(G) = (x^2/25 + y^2/21 = 1);Expression(H) = (x = -2);Intersection(H, G) = {A, B};RightFocus(G) = F",
    "query_expressions": "Perimeter(TriangleOf(A, B, F))",
    "answer_expressions": "20",
    "fact_spans": "[[[11, 50], [61, 63]], [[2, 10]], [[51, 54]], [[55, 58]], [[68, 72]], [[11, 50]], [[2, 10]], [[2, 60]], [[61, 72]]]",
    "query_spans": "[[[74, 96]]]",
    "process": "Ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{21}=1$, so $c^{2}=a^{2}-b^{2}=25-21=4$. The line $x=-2$ passes through the left focus $F_{1}$ of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{21}=1$, and the right focus of the ellipse is $F$. By the definition of the ellipse, the perimeter of $\\triangle ABF$ is $AF+BF+AB=AF+AF_{1}+BF+BF_{1}=4a=4\\times5=20$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with right focus $F$, and $O$ as the coordinate origin. A line passing through $F$ intersects the right branch of the hyperbola at points $A$ and $B$. Line $AO$ is extended to intersect the hyperbola $C$ again at point $P$. If $|A F|=2|B F|$ and $\\angle P F B=60^{\\circ}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(C) = F;O: Origin;G: Line;PointOnCurve(F, G);A: Point;B: Point;Intersection(G, RightPart(C)) = {A, B};P: Point;Intersection(OverlappingLine(LineSegmentOf(A, O)), C) = P;Abs(LineSegmentOf(A, F)) = 2*Abs(LineSegmentOf(B, F));AngleOf(P, F, B) = ApplyUnit(60, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(13)/3",
    "fact_spans": "[[[2, 63], [91, 94], [119, 125], [179, 182]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 71], [84, 87]], [[2, 71]], [[73, 76]], [[88, 90]], [[83, 90]], [[97, 100]], [[102, 105]], [[88, 107]], [[126, 130]], [[108, 130]], [[133, 147]], [[150, 175]]]",
    "query_spans": "[[[179, 188]]]",
    "process": "Let the left focus of hyperbola $ C $ be $ F $, connect $ AF $, $ BF $, let $ |BF| = t $, then $ |AF| = 2t $. So $ |AF| = 2a + 2t $, $ |BF| = 2a + t $. By symmetry, quadrilateral $ AFBP $ is a parallelogram, hence $ \\angle FAB = 60^{\\circ} $. In $ \\triangle FAB $, by the law of cosines, $ (2a + t)^{2} = (2a + 2t)^{2} + (3t)^{2} - 2 \\times (2a + 2t) \\times 3t \\times \\cos 60^{\\circ} $, solving yields $ t = \\frac{a}{3} $. Hence $ |AF| = \\frac{8a}{3} $, $ |BF| = \\frac{2a}{3} $. In $ \\triangle FAF $, by the law of cosines, $ 4c^{2} = \\frac{64a^{2}}{9} + \\frac{4a^{2}}{9} - 2 \\times \\frac{8a}{3} \\times \\frac{2a}{3} \\times \\cos 60^{\\circ} = \\frac{52a^{2}}{9} $, solving yields: $ \\sqrt{13} $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$, and $P$ a point on $C$. If $A(-2,0)$, what are the coordinates of point $P$ when $|PA|$ is maximized?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 8*x);F: Point;Focus(C) = F;P: Point;PointOnCurve(P, C);A: Point;Coordinate(A) = (-2, 0);WhenMax(Abs(LineSegmentOf(P, A)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2, 4)",
    "fact_spans": "[[[2, 21], [33, 36]], [[2, 21]], [[25, 28]], [[2, 28]], [[29, 32], [62, 66]], [[29, 39]], [[41, 50]], [[41, 50]], [[51, 62]]]",
    "query_spans": "[[[62, 71]]]",
    "process": "As shown in the figure, $\\frac{|PA|}{|PF|} = \\frac{|PA|}{|PN|} = \\frac{1}{\\cos\\angle APN} = \\frac{1}{\\cos\\angle PAF}$. When $\\frac{|PA|}{|PF|}$ reaches its maximum value, $\\cos\\angle APN$ must reach its minimum value, at which point $AP$ is tangent to the parabola. Let the equation of the tangent line be $y = k(x + 2)$. Combining  \n$$\n\\begin{cases}\ny = k(x + 2) \\\\\ny^2 = 8x\n\\end{cases}\n$$\nand eliminating $x$, we obtain $ky^2 - 8y + 16k = 0$. The discriminant is $\\Delta = 64 - 64k^2 = 0$, so $k^2 = 1$, solving gives $k = \\pm1$ (negative discarded). Thus, $y^2 - 8y + 16 = 0$, solving gives $y = 4$. Substituting $y = 4$ into the line $y = x + 2$, we get $x = 2$. Therefore, when $\\frac{|PA|}{|PF|}$ is maximized, the coordinates of point $P$ are $(2, 4)$."
  },
  {
    "text": "It is known that a line $AB$ perpendicular to the $x$-axis is drawn through the right focus $F_2$ of the ellipse $\\frac{x^{2}}{64}+\\frac{y^{2}}{16}=1$, intersecting the ellipse at points $A$ and $B$, and $F_1$ is the left focus of the ellipse. Then the perimeter of $\\Delta A F_{1} B$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/64 + y^2/16 = 1);F2: Point;RightFocus(G) =  F2;PointOnCurve(F2, LineOf(A, B)) = True;IsPerpendicular(LineOf(A, B), xAxis) = True;Intersection(LineOf(A, B), G) = {A, B};A: Point;B: Point;F1: Point;LeftFocus(G) = F1",
    "query_expressions": "Perimeter(TriangleOf(A, F1, B))",
    "answer_expressions": "32",
    "fact_spans": "[[[4, 43], [93, 95], [72, 74]], [[4, 43]], [[47, 54]], [[4, 54]], [[2, 70]], [[55, 70]], [[63, 84]], [[75, 78]], [[79, 82]], [[85, 92]], [[85, 99]]]",
    "query_spans": "[[[101, 124]]]",
    "process": "\\because F_{1},F_{2} are the two foci of the ellipse \\frac{x^{2}}{64}+\\frac{y^{2}}{16}=1, \\therefore by the definition of an ellipse, |AF_{1}|+|AF_{2}|=2a, |BF_{1}|+|BF_{2}|=2a. \\therefore the perimeter of quadrilateral AF_{1}BF_{2} is |AB|+|AF_{1}|+|BF_{1}|=|AF_{1}|+|AF_{2}|+|BF_{1}|+|BF_{2}|=4a=32."
  },
  {
    "text": "Let $P$ be an arbitrary point on the right branch of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ not lying on the $x$-axis, and let $F_{1}$, $F_{2}$ be the left and right foci respectively. The point of tangency of the incircle of $\\Delta P F_{1} F_{2}$ with the $x$-axis is $M(m, 0)$ where $\\frac{\\sqrt{3}}{3} b \\leq m \\leq 2 b$. Then the maximum value of the eccentricity of this hyperbola is?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;M: Point;P: Point;F1: Point;F2: Point;m:Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(M) = (m, 0);PointOnCurve(P, RightPart(C));Negation(PointOnCurve(P, xAxis));LeftFocus(C) = F1;RightFocus(C) = F2;TangentPoint(InscribedCircle(TriangleOf(P,F1,F2)),xAxis)=M;(sqrt(3)/3)*b<=m;m<=2*b",
    "query_expressions": "Max(Eccentricity(C))",
    "answer_expressions": "2",
    "fact_spans": "[[[5, 65], [190, 193]], [[12, 65]], [[12, 65]], [[140, 187]], [[1, 4]], [[81, 88]], [[89, 96]], [[140, 187]], [[12, 65]], [[12, 65]], [[5, 65]], [[140, 187]], [[1, 80]], [[1, 80]], [[5, 104]], [[5, 104]], [[105, 187]], [[140, 187]], [[140, 187]]]",
    "query_spans": "[[[190, 202]]]",
    "process": "By the definition of the hyperbola, |PF_{1}|-|PF_{2}|=2a=|MF_{1}|-|MF_{2}|. Since |MF_{1}|+|MF_{2}|=2c, therefore |MF_{1}|=c+a, so M is the right vertex of hyperbola C. Then _{m}=a\\in[\\frac{\\sqrt{3}}{3}b,2b=\\sqrt{1+(b)^{2}}\\in[\\frac{\\sqrt{5}}{3}"
  },
  {
    "text": "The equations of the two asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{4}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/4 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*x",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 49]]]",
    "process": "Given a=2, b=2, the equations of the two asymptotes of the hyperbola \\frac{x^{2}}{4}-\\frac{y^{2}}{4}=1 are y=\\pm\\frac{b}{a}x=\\pm x"
  },
  {
    "text": "If the line $m x+n y-3=0$ and the circle $x^{2}+y^{2}=3$ have no common points, then taking $(m, n)$ as the coordinates of point $P$, how many common points does a line passing through point $P$ have with the ellipse $\\frac{x^{2}}{7}+\\frac{y^{2}}{3}=1$?",
    "fact_expressions": "G: Ellipse;H: Circle;I: Line;n: Number;m: Number;P:Point;Expression(G) = (x^2/7 + y^2/3 = 1);Expression(H) = (x^2 + y^2 = 3);Expression(I) = (m*x + n*y - 3 = 0);Coordinate(P) = (m, n);NumIntersection(I,H)=0;L:Line;PointOnCurve(P,L)",
    "query_expressions": "NumIntersection(L,G)",
    "answer_expressions": "2",
    "fact_spans": "[[[69, 106]], [[17, 33]], [[1, 16]], [[41, 49]], [[41, 49]], [[59, 63], [50, 54]], [[69, 106]], [[17, 33]], [[1, 16]], [[40, 57]], [[1, 38]], [[66, 68]], [[58, 68]]]",
    "query_spans": "[[[66, 114]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. Let $P$ be a point on its right directrix with ordinate $\\sqrt{3} c$ ($c$ is the semi-focal length), and suppose $F_{1} F_{2}=F_{2} P$. Then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;F2: Point;P: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightDirectrix(G));c: Number;HalfFocalLength(G) = c;YCoordinate(P) = sqrt(3)*c;LineSegmentOf(F1, F2) = LineSegmentOf(F2, P)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[19, 73], [84, 85], [142, 144]], [[21, 73]], [[21, 73]], [[1, 8]], [[9, 16]], [[80, 83]], [[21, 73]], [[21, 73]], [[19, 73]], [[1, 79]], [[1, 79]], [[80, 117]], [[106, 109]], [[84, 115]], [[80, 117]], [[119, 140]]]",
    "query_spans": "[[[142, 150]]]",
    "process": ""
  },
  {
    "text": "Given that a circle passes through the right vertex and the right focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and the center of the circle lies on this ellipse, what is the distance from the center of the circle to the center of the ellipse?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);H: Circle;PointOnCurve(RightVertex(G), H) ;PointOnCurve(RightFocus(G), H);PointOnCurve(Center(H), G)",
    "query_expressions": "Distance(Center(H), Center(G))",
    "answer_expressions": "sqrt(57)/4",
    "fact_spans": "[[[4, 41], [54, 56], [63, 65]], [[4, 41]], [[2, 3]], [[2, 45]], [[2, 49]], [[2, 57]]]",
    "query_spans": "[[[2, 72]]]",
    "process": ""
  },
  {
    "text": "Let the focal length of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(b>a>0)$ be $2 c$, and let the line $l$ pass through points $A(a, 0)$ and $B(0, b)$. It is known that the distance from the origin to the line $l$ is $\\frac{\\sqrt{3}}{4} c$. Then, what is the eccentricity of the hyperbola?",
    "fact_expressions": "l: Line;G: Hyperbola;b: Number;a: Number;A: Point;B: Point;b > a;a > 0;c: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (a, 0);Coordinate(B) = (0, b);FocalLength(G) = 2*c;PointOnCurve(A, l);PointOnCurve(B, l);O: Origin;Distance(O, l) = c*(sqrt(3)/4)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[65, 70], [101, 106]], [[1, 54], [138, 143]], [[4, 54]], [[4, 54]], [[71, 81]], [[84, 93]], [[4, 54]], [[4, 54]], [[59, 64]], [[1, 54]], [[72, 81]], [[84, 93]], [[1, 64]], [[65, 81]], [[65, 95]], [[98, 100]], [[98, 133]]]",
    "query_spans": "[[[138, 154]]]",
    "process": "Find the equation of line AB as $ bx+ay-ab=0 $. Based on the distance from the origin to line $ l $, a homogeneous equation in terms of $ a $ and $ b $ can be obtained, allowing the value of $ \\frac{b}{a} $ to be determined. Thus, the eccentricity of the hyperbola can be found as $ e=\\sqrt{1+(\\frac{b}{a})^{2}} $, leading to the solution. [Detailed Solution] The equation of line AB is $ \\frac{x}{a}+\\frac{y}{b}=1 $, that is, $ bx+ay-ab=0 $. Then the distance from the origin to line $ l $ is $ d=\\frac{ab}{\\sqrt{a^{2}+b^{2}}}=\\frac{ab}{c}=\\frac{\\sqrt{3}}{4}c $. Therefore, $ 4ab=\\sqrt{3}c^{2}=\\sqrt{3}(a^{2}+b^{2}) $, so $ \\sqrt{3}\\cdot(\\frac{b}{a})^{2}-\\frac{4b}{a}+\\sqrt{3}=0 $. Solving yields $ \\frac{b}{a}=\\sqrt{3} $ or $ \\frac{b}{a}=\\frac{\\sqrt{3}}{3} $ (discarded), so $ e=\\frac{c}{a}=\\sqrt{1+(\\frac{b}{a})^{2}}=2 $. Hence, the answer is: $ 2 $."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. If $|AF|=3$, then $|BF|=$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, F)) = 3",
    "query_expressions": "Abs(LineSegmentOf(B, F))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[1, 15], [26, 29]], [[22, 24]], [[30, 33]], [[18, 21]], [[34, 37]], [[1, 15]], [[1, 21]], [[0, 24]], [[22, 39]], [[41, 50]]]",
    "query_spans": "[[[52, 61]]]",
    "process": ""
  },
  {
    "text": "If an ellipse and a hyperbola have the same foci $F_{1}$, $F_{2}$, with eccentricities $e_{1}$, $e_{2}$ respectively, and $P$ is a common point on both curves such that $\\angle F_{1} P F_{2}=90^{\\circ}$, then the value of $\\frac{1}{e_{1}^{2}}+\\frac{1}{e_{2}^{2}}$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;F1: Point;P: Point;F2: Point;Focus(H) = {F1, F2};Focus(G) = {F1, F2};Focus(G) = Focus(H);e2: Number;e1: Number;Eccentricity(H) = e1;Eccentricity(G) = e2;OneOf(Intersection(G, H)) = P;AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "1/(e2^2) + 1/(e1^2)",
    "answer_expressions": "2",
    "fact_spans": "[[[4, 7]], [[1, 3]], [[14, 21]], [[55, 58]], [[22, 29]], [[1, 29]], [[1, 29]], [[1, 14]], [[45, 52]], [[36, 43]], [[1, 53]], [[1, 53]], [[55, 68]], [[72, 105]]]",
    "query_spans": "[[[107, 152]]]",
    "process": "Let the ellipse be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, the hyperbola be $\\frac{x^{2}}{m^{2}}-\\frac{y^{2}}{n^{2}}=1$, and the focal distance be $2c$. Then $PF_{1}+PF_{2}=2a$, $|PF_{1}-PF_{2}|=2m$. Thus $(PF_{1}+PF_{2})^{2}+(PF_{1}-PF_{2})^{2}=2(PF_{1}^{2}+PF_{2}^{2})=4a^{2}+4m^{2}$. Also, $\\angle F_{1}PF_{2}=90^{\\circ}$, so by the Pythagorean theorem, $PF_{1}^{2}+PF_{2}^{2}=4c^{2}$. Then $4a^{2}+4m^{2}=8c^{2} \\Rightarrow a^{2}+m^{2}=2c^{2}$, hence $\\frac{1}{e_{1}^{2}}+\\frac{1}{e_{2}^{2}}=\\frac{a^{2}}{c^{2}}+\\frac{m^{2}}{c^{2}}=\\frac{a^{2}+m^{2}}{c^{2}}=2$."
  },
  {
    "text": "The standard equation of a parabola with vertex at the origin, symmetric about the $x$-axis, and passing through the point $(-4,4)$ is?",
    "fact_expressions": "G: Parabola;Vertex(G) = O;O: Origin;SymmetryAxis(G) = xAxis;H: Point;Coordinate(H) = (-4, 4);PointOnCurve(H, G) = True",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=-4*x",
    "fact_spans": "[[[26, 29]], [[0, 29]], [[3, 5]], [[6, 29]], [[16, 25]], [[16, 25]], [[15, 29]]]",
    "query_spans": "[[[26, 36]]]",
    "process": ""
  },
  {
    "text": "Given $|\\vec{a}|=1$, $|\\vec{b}+\\vec{a}|+|\\vec{b}-\\vec{a}|=4$, then the minimum value of $|\\vec{b}-\\frac{1}{4} \\vec{a}|$ is?",
    "fact_expressions": "a: Vector;b: Vector;Abs(a) = 1;Abs(b + a) + Abs(b - a) = 4",
    "query_expressions": "Min(Abs(b - a/4))",
    "answer_expressions": "3*sqrt(5)/4",
    "fact_spans": "[[[2, 15]], [[17, 56]], [[2, 15]], [[17, 56]]]",
    "query_spans": "[[[58, 95]]]",
    "process": "Let $ A(1,0) $, $ B(x,y) $, $ \\overrightarrow{OA} = \\overrightarrow{a} = (1,0) $, $ \\overrightarrow{OB} = \\overrightarrow{b} = (x,y) $. Then $ \\overrightarrow{b} + \\overrightarrow{a} = (x+1,y) $, $ \\overrightarrow{b} - \\overrightarrow{a} = (x-1,y) $. Because $ |\\overrightarrow{b} + \\overrightarrow{a}| + |\\overrightarrow{b} - \\overrightarrow{a}| = 4 $, so $ \\sqrt{(x+1)^{2} + y^{2}} + \\sqrt{(x-1)^{2} + y^{2}} = 4 > 2 $, that is: point $ B $ lies on an ellipse with foci at $ (-1,0) $, $ (1,0) $ and $ 2a = 4 $. Therefore, $ B(x,y) $ satisfies $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $. $ \\overrightarrow{b} - \\frac{1}{4}\\overrightarrow{a} = (x - \\frac{1}{4}, y) $, then $ (\\overrightarrow{b} - \\frac{1}{4}\\overrightarrow{a})^{2} = (x - \\frac{1}{4})^{2} + y^{2} = x^{2} - \\frac{1}{2}x + \\frac{1}{16} + 3 - \\frac{3}{4}x^{2} = \\frac{1}{4}(x-1)^{2} + \\frac{45}{16} $. Because $ -2 \\leqslant x \\leqslant 2 $, so when $ x = 1 $, $ |\\overrightarrow{b} - \\frac{1}{4}\\overrightarrow{a}| $ attains the minimum value of $ \\underline{3\\sqrt{5}} $."
  },
  {
    "text": "The line $5x + 4y - 1 = 0$ intersects the ellipse $C$: $\\frac{y^{2}}{a^{2}} + \\frac{x^{2}}{b^{2}} = 1$ $(a > b > 0)$ at points $M$ and $N$. Let $P$ be the midpoint of $MN$, and suppose the slope of the line $OP$ is $\\frac{5}{4}$, where $O$ is the origin. Find the eccentricity of the ellipse $C$?",
    "fact_expressions": "G: Line;Expression(G) = (5*x + 4*y - 1 = 0);C: Ellipse;Expression(C) = (x^2/b^2 + y^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Intersection(G, C) = {M, N};M: Point;N: Point;MidPoint(LineSegmentOf(M, N)) = P;P: Point;O: Origin;Slope(LineOf(O, P)) = 5/4",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "3/5",
    "fact_spans": "[[[0, 15]], [[0, 15]], [[16, 73], [135, 140]], [[16, 73]], [[23, 73]], [[23, 73]], [[23, 73]], [[23, 73]], [[0, 83]], [[74, 77]], [[78, 81]], [[85, 96]], [[93, 96]], [[125, 128]], [[97, 122]]]",
    "query_spans": "[[[135, 145]]]",
    "process": "Let $ M(x_{1},y_{1}) $, $ N(x_{1},y_{1}) $, and the midpoint of $ MN $ be $ P(x_{0},y_{0}) $. Then  \n$$\n\\begin{cases}\n\\frac{y_{2}^{2}}{a^{2}} + \\frac{x_{1}}{b^{2}} = 1 \\\\\n\\frac{y_{2}^{2}}{a^{2}} + \\frac{x_{2}}{b^{2}} = 1\n\\end{cases}\n$$\nAccording to the midpoint $ P $ of the intersecting chord, solve using the point difference method. Let $ M(x_{1},y_{1}) $, $ N(x_{1},y_{1}) $, and the midpoint of $ MN $ be $ P(x_{0},y_{0}) $. Then  \n$$\n\\begin{cases}\n\\frac{y_{2}}{a^{2}} + \\frac{x_{1}2}{b^{2}} = 1 \\\\\n\\frac{y_{2}^{2}}{a^{2}} + \\frac{x_{2}^{2}}{2} = 1\n\\end{cases}\n$$\nSubtracting the two equations gives:  \n$ b^{2}(y_{2}-y_{2}) + a^{2}(x_{1}2 - x_{2}2) = 0 $,  \ni.e.,  \n$ \\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = -\\frac{a^{2}}{b^{2}} \\left( \\frac{x_{1}+x_{2}}{y_{1}+y_{2}} \\right) $,  \ni.e.,  \n$ k_{MN} = -\\frac{a^{2}}{b^{2}} \\cdot \\frac{1}{k_{0}} $.  \nSince $ k_{MN} = -\\frac{5}{4} $, $ k_{OP} = \\frac{5}{4} $,  \nso $ \\frac{b^{2}}{a^{2}} = \\frac{16}{25} $, therefore  \n$ e = \\frac{c}{a} = \\sqrt{1 - \\frac{b^{2}}{a^{2}}} = \\frac{3}{5} $, b female peace for.3"
  },
  {
    "text": "Given that the length of the major axis of an ellipse is $3$ times the length of the minor axis, and both the major and minor axes lie on the coordinate axes, passing through the point $A(3 , 0)$, then the equation of the ellipse?",
    "fact_expressions": "G: Ellipse;A: Point;Coordinate(A) = (3, 0);Length(MajorAxis(G)) = 3 * Length(MinorAxis(G));OverlappingLine(MajorAxis(G), axis);OverlappingLine(MinorAxis(G), axis);PointOnCurve(A, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/9 + y^2 = 1, x^2/9 + y^2/81 = 1}",
    "fact_spans": "[[[2, 4], [43, 45]], [[30, 41]], [[30, 41]], [[2, 17]], [[2, 28]], [[2, 28]], [[2, 41]]]",
    "query_spans": "[[[43, 49]]]",
    "process": ""
  },
  {
    "text": "If one asymptote of a hyperbola centered at the origin and with foci on the coordinate axes is given by $x+3 y=0$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;O: Origin;Center(G) = O;PointOnCurve(Focus(G), axis);Expression(OneOf(Asymptote(G))) = (x + 3*y = 0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{sqrt(10)/3, sqrt(10)}",
    "fact_spans": "[[[15, 18], [39, 42]], [[4, 6]], [[1, 18]], [[7, 18]], [[15, 36]]]",
    "query_spans": "[[[39, 48]]]",
    "process": "According to the given condition: $a:b=1:3$ or $a:b=3:1$, so $a:c=1:\\sqrt{10}$ or $a:c=\\sqrt{10}:1$, therefore the eccentricity of the hyperbola is $\\frac{\\sqrt{10}}{2}$ or $\\sqrt{10}$."
  },
  {
    "text": "What is the length of the imaginary axis of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/9 = 1)",
    "query_expressions": "Length(ImageinaryAxis(G))",
    "answer_expressions": "6",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 44]]]",
    "process": "According to the geometric properties of hyperbolas, the length of the imaginary axis can be determined. For the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$), the length of the imaginary axis is $2b$. Therefore, the length of the imaginary axis of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$ is $6$."
  },
  {
    "text": "The standard equation of an ellipse with eccentricity $\\frac{\\sqrt{3}}{2}$ and passing through the point $(2 , 0)$ is?",
    "fact_expressions": "G: Ellipse;H: Point;Coordinate(H) = (2, 0);PointOnCurve(H, G);Eccentricity(G)=sqrt(3)/2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/4+y^2=1,y^2/16+x^2/4=1}",
    "fact_spans": "[[[38, 40]], [[27, 37]], [[27, 37]], [[26, 40]], [[0, 40]]]",
    "query_spans": "[[[38, 47]]]",
    "process": "According to the problem, the point (2,0) is a vertex of the ellipse. Let the semi-major axis be $a$, the semi-minor axis be $b$, and the semi-focal distance be $c$. Then $\\frac{c}{a}=\\frac{\\sqrt{3}}{2}$. When the foci of the ellipse lie on the x-axis, $a=2$, $c=\\sqrt{3}$, $b^{2}=a^{2}-c^{2}=1$, and the standard equation of the ellipse is $\\frac{x^{2}}{4}+y^{2}=1$. When the foci of the ellipse lie on the y-axis, $b=2$, and from $a^{2}-\\left(\\frac{\\sqrt{3}}{2}a\\right)^{2}=b^{2}=4$, we get $a^{2}=16$, so the standard equation of the ellipse is $\\frac{y^{2}}{16}+\\frac{x^{2}}{4}=1$. Therefore, the required standard equations of the ellipse are $\\frac{x^{2}}{4}+y^{2}=1$ or $\\frac{y^{2}}{16}+\\frac{x^{2}}{4}=1$."
  },
  {
    "text": "$P$ is a point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, and $F_{1}$, $F_{2}$ are the two foci. Then the difference between the maximum and minimum values of $|PF_{1}| \\cdot |PF_{2}|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 + y^2/4 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2}",
    "query_expressions": "Max(Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))) - Min(Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)))",
    "answer_expressions": "5",
    "fact_spans": "[[[4, 41]], [[0, 3]], [[45, 52]], [[53, 60]], [[4, 41]], [[0, 44]], [[4, 65]]]",
    "query_spans": "[[[67, 105]]]",
    "process": ""
  },
  {
    "text": "$P$ is a point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and $F_{1}$ and $F_{2}$ are the foci of the ellipse. Then the maximum value of $k=|P F_{1}| \\cdot|P F_{2}|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2/3 = 1);PointOnCurve(P, G);Focus(G)={F1,F2};Abs(LineSegmentOf(P,F1))*Abs(LineSegmentOf(P,F2))=k;k:Number",
    "query_expressions": "Max(k)",
    "answer_expressions": "4",
    "fact_spans": "[[[4, 41], [62, 64]], [[0, 3]], [[45, 52]], [[53, 60]], [[4, 41]], [[0, 44]], [[45, 67]], [[69, 97]], [[69, 97]]]",
    "query_spans": "[[[69, 103]]]",
    "process": ""
  },
  {
    "text": "If the circle $(x-2)^{2}+y^{2}=2$ is tangent to the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 + (x - 2)^2 = 2);IsTangent(H, Asymptote(G))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[22, 78], [86, 89]], [[25, 78]], [[25, 78]], [[1, 21]], [[25, 78]], [[25, 78]], [[22, 78]], [[1, 21]], [[1, 84]]]",
    "query_spans": "[[[86, 95]]]",
    "process": ""
  },
  {
    "text": "Given that $A B$ and $C D$ are two chords passing through the focus $F$ of the parabola $y^{2}=8 x$ and are perpendicular to each other, then the value of $\\frac{1}{|A F| \\cdot|B F|}+\\frac{1}{|C F| \\cdot|D F|}$ is?",
    "fact_expressions": "A: Point;B: Point;C: Point;D: Point;G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;PointOnCurve(F,LineSegmentOf(A,B));IsChordOf(LineSegmentOf(A,B), G) ;PointOnCurve(F, LineSegmentOf(C,D));IsChordOf(LineSegmentOf(C,D),G) ;IsPerpendicular(LineSegmentOf(A,B), LineSegmentOf(C,D))",
    "query_expressions": "1/(Abs(LineSegmentOf(C, F))*Abs(LineSegmentOf(D, F))) + 1/(Abs(LineSegmentOf(A, F))*Abs(LineSegmentOf(B, F)))",
    "answer_expressions": "1/16",
    "fact_spans": "[[[2, 7]], [[2, 7]], [[10, 15]], [[10, 15]], [[17, 31]], [[17, 31]], [[33, 36]], [[17, 36]], [[2, 36]], [[2, 44]], [[2, 36]], [[2, 44]], [[2, 41]]]",
    "query_spans": "[[[46, 105]]]",
    "process": "From the given conditions, the slopes of lines AB and CD must exist. Let AB be $ y = k(x - 2) $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Combining with the parabola equation yields $ k^{2}x^{2} - (4k^{2} + 8)x + 4k^{2} = 0 $. Since $ \\Delta = 64(1 + k^{2}) > 0 $, we have $ x_{1} + x_{2} = \\frac{4(k^{2} + 2)}{k^{2}} $, $ x_{1}x_{2} = 4 $. Moreover, $ |AF| = x_{1} + 2 $, $ |BF| = x_{2} + 2 $, so $ |AF| \\cdot |BF| = (x_{1} + 2)(x_{2} + 2) = x_{1}x_{2} + 2(x_{1} + x_{2}) + 4 = \\frac{16(k^{2} + 1)}{k^{2}} $. Since $ CD \\perp AB $, let $ CD $ be $ y = \\frac{2 - x}{k} $, $ C(x_{3}, y_{3}) $, $ D(x_{4}, y_{4}) $. Combining with the parabola gives $ x^{2} - (8k^{2} + 4)x + 4 = 0 $. Similarly, $ x_{3} + x_{4} = 8k^{2} + 4 $, $ x_{3}x_{4} = 4 $, thus $ |CF| \\cdot |DF| = 16(k^{2} + 1) $. In conclusion, $ \\frac{1}{1} $"
  },
  {
    "text": "Through any point $P$ on the parabola $y^{2}=4x$, draw a perpendicular to the $y$-axis, with foot $Q$. Point $M$ moves on the line $2x - y + 2 = 0$. Then the minimum value of $PQ + PM$ is?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Q: Point;M: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (2*x - y + 2 = 0);PointOnCurve(P,G);L:Line;PointOnCurve(P,L);IsPerpendicular(L,yAxis);FootPoint(L,yAxis)=Q;PointOnCurve(M, H)",
    "query_expressions": "Min(LineSegmentOf(P, M) + LineSegmentOf(P, Q))",
    "answer_expressions": "(4*sqrt(5)/5)-1",
    "fact_spans": "[[[1, 15]], [[45, 58]], [[20, 23]], [[35, 38]], [[41, 44]], [[1, 15]], [[45, 58]], [[1, 23]], [], [[0, 31]], [[0, 31]], [[0, 38]], [[41, 59]]]",
    "query_spans": "[[[61, 76]]]",
    "process": ""
  },
  {
    "text": "It is known that the center of the ellipse is at the origin, one focus coincides with the focus of the parabola $y^{2}=8x$, and the coordinates of one vertex are $(0, 2)$. Then the equation of this ellipse is?",
    "fact_expressions": "H: Ellipse;O: Origin;Center(H) = O;G: Parabola;Expression(G) = (y^2 = 8*x);OneOf(Focus(H)) = Focus(G);Coordinate(OneOf(Vertex(H))) = (0, 2)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 4], [55, 57]], [[8, 10]], [[2, 10]], [[16, 30]], [[16, 30]], [[2, 35]], [[2, 53]]]",
    "query_spans": "[[[55, 61]]]",
    "process": ""
  },
  {
    "text": "The vertex is at the origin, and a point $P(m,-2)$ on the parabola with focus on the $y$-axis is at a distance of $4$ from the focus $F$. Then $m = $?",
    "fact_expressions": "G: Parabola;P: Point;O: Origin;F:Point;Focus(G)=F;Coordinate(P) = (m,-2);Vertex(G) = O;PointOnCurve(Focus(G),yAxis);PointOnCurve(P,G);Distance(P,F)=4;m:Number",
    "query_expressions": "m",
    "answer_expressions": "pm*4",
    "fact_spans": "[[[15, 18]], [[21, 30]], [[3, 5]], [[33, 36]], [[15, 36]], [[21, 30]], [[0, 18]], [[6, 18]], [[15, 30]], [[21, 44]], [[46, 49]]]",
    "query_spans": "[[[46, 51]]]",
    "process": "Let the equation of the parabola be: $x^{2}=2py$. Since $P(m,-2)$ is a point on the parabola, $\\therefore p<0$; by the focal radius formula of the parabola, $|PF|=2-\\frac{p}{2}=4$, solving gives: $p=-4$, $\\therefore$ the equation of the parabola is: $x^{2}=-8y$. $\\therefore m^{2}=(-8)\\times(-2)=16$, solving gives: $m=\\pm4$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$, $F_{1}$, $F_{2}$ are its two foci, point $M$ lies on the hyperbola, and $\\angle F_{1} MF_{2}=120^{\\circ}$, then the area of $\\triangle F_{1} MF_{2}$ is?",
    "fact_expressions": "G: Hyperbola;F1: Point;M: Point;F2: Point;Expression(G) = (x^2/4 - y^2/9 = 1);Focus(G) = {F1, F2};PointOnCurve(M, G);AngleOf(F1, M, F2) = ApplyUnit(120, degree)",
    "query_expressions": "Area(TriangleOf(F1, M, F2))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 40], [57, 58], [68, 71]], [[41, 48]], [[63, 67]], [[49, 56]], [[2, 40]], [[41, 62]], [[63, 72]], [[74, 107]]]",
    "query_spans": "[[[109, 138]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. If there exists a point $P$ on the ellipse such that $|P F_{1}|=\\frac{3}{2} e|P F_{2}|$, then the range of the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);e: Number;Abs(LineSegmentOf(P,F1))=(3/2)*e*Abs(LineSegmentOf(P,F2));Eccentricity(G) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "[1/3,1)",
    "fact_spans": "[[[2, 54], [80, 82], [129, 131]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[63, 70]], [[71, 78]], [[2, 78]], [[2, 78]], [[87, 90]], [[80, 90]], [[135, 138]], [[92, 126]], [[129, 138]]]",
    "query_spans": "[[[135, 145]]]",
    "process": "According to the focal radius formula, simplifying $|PF_{1}|=\\frac{3}{2}e|PF_{2}|$ yields $e(x+\\frac{a^{2}}{c})=\\frac{3}{2}\\cdot e\\cdot(\\frac{a^{2}}{c}-x)$, solving for $x$ gives $x=\\frac{\\frac{3}{2}c-a}{e(\\frac{3}{2}e+1)}$. Based on the range of the ellipse's horizontal coordinate, we have $-a\\leqslant\\frac{\\frac{3}{2}c-a}{e(\\frac{3}{2}e+1)}\\leqslant a$. Dividing the inequality by $a$ results in $-1\\leqslant\\frac{\\frac{3}{2}e-1}{e(\\frac{3}{2}e+1)}\\leqslant1$, solving this yields $\\frac{1}{3}\\leqslante<1$. Thus, the range of eccentricity is $[\\frac{1}{3},1)$."
  },
  {
    "text": "Given point $P(0 , 2)$, the focus of the parabola $C$: $y^{2}=2 px(p>0)$ is $F$, the intersection point of segment $PF$ and parabola $C$ is $M$, and from $M$ a perpendicular is drawn to the directrix of the parabola, with foot $Q$. If $\\angle PQF=90^{\\circ}$, then $p=$?",
    "fact_expressions": "P: Point;F: Point;C: Parabola;p: Number;Q: Point;M: Point;l: Line;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(P) = (0, 2);Focus(C) = F;Intersection(LineSegmentOf(P, F), C) = M;PointOnCurve(M, l);IsPerpendicular(l, Directrix(C));FootPoint(l, Directrix(C)) = Q;AngleOf(P, Q, F) = ApplyUnit(90, degree)",
    "query_expressions": "p",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 13]], [[42, 45]], [[14, 38], [53, 59], [73, 76]], [[116, 119]], [[85, 88]], [[63, 66], [69, 72]], [], [[21, 38]], [[14, 38]], [[2, 13]], [[14, 45]], [[46, 66]], [[68, 81]], [[68, 81]], [[67, 88]], [[91, 114]]]",
    "query_spans": "[[[116, 121]]]",
    "process": ""
  },
  {
    "text": "Given the line $l$: $k x - y - 2k + 1 = 0$ intersects the ellipse $C_1$: $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1\\ (a > b > 0)$ at points $A$, $B$, and intersects the circle $C_2$: $(x - 2)^2 + (y - 1)^2 = 1$ at points $C$, $D$. If there exists $k \\in \\left[-\\frac{3}{2}, -1\\right]$ such that $\\overrightarrow{AC} = \\overrightarrow{DB}$, then the range of the eccentricity of the ellipse $C_1$ is?",
    "fact_expressions": "l: Line;Expression(l) = (-2*k + k*x - y + 1 = 0);C1: Ellipse;Expression(C1) = (y^2/b^2 + x^2/a^2 = 1);Intersection(l, C1) = {A, B};A: Point;B: Point;C2: Circle;Expression(C2) = ((x - 2)^2 + (y - 1)^2 = 1);Intersection(l, C2) = {C, D};C: Point;D: Point;k: Number;In(k, [-3/2, -1]);VectorOf(A, C) = VectorOf(D, B);b:Number;a>b;b>0",
    "query_expressions": "Range(Eccentricity(C1))",
    "answer_expressions": "[1/2, sqrt(2)/2]",
    "fact_spans": "[[[2, 24]], [[2, 24]], [[25, 88], [221, 230]], [[25, 88]], [[2, 99]], [[90, 93]], [[94, 97]], [[101, 133]], [[101, 133]], [[2, 144]], [[135, 138]], [[139, 142]], [[149, 173]], [[149, 173]], [[176, 219]], [[25, 88]], [[25, 88]], [[25, 88]]]",
    "query_spans": "[[[221, 241]]]",
    "process": "The line has been given, which is a parallelogram, showing that the line always passes through the fixed point (2,1). The circle has center (2,1) and radius 1, and points C, D are endpoints of the diameter. From the center, it follows that the midpoint of AB is (2,1). Let A, B, B, B. Then A, \\prime. Subtracting these two equations yields, from which we obtain, and thus we have the following; therefore, the eccentricity of the ellipse is"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A circle with diameter $F_{1} F_{2}$ intersects an asymptote of $C$ in the first quadrant at point $P$. If $|P F_{1}|=2 b$, then the equation of the asymptotes of $C$ is?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;G: Circle;F1: Point;F2: Point;P: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;IsDiameter(LineSegmentOf(F1,F2),G);Intersection(G,Asymptote(C))=P;Abs(LineSegmentOf(P, F1)) = 2*b;Quadrant(P)=1",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[2, 63], [108, 111], [146, 149]], [[10, 63]], [[10, 63]], [[106, 107]], [[72, 79]], [[80, 87]], [[123, 127]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 87]], [[2, 87]], [[88, 107]], [[106, 127]], [[129, 144]], [[106, 127]]]",
    "query_spans": "[[[146, 157]]]",
    "process": "Draw the graph. First, find the distance from the focus to the asymptote |NF₂| = b. Then draw PQ ⊥ F₁F₂. Using the equal area method, combining APF₁Q, it follows that ∠POQ = 60°, so the slope k_{OP} can be found, and then solved accordingly. As shown in the figure, draw PQ ⊥ F₁F₂. The focus of the hyperbola is F₂(c, 0). Let one asymptote of the hyperbola be y = \\frac{b}{a}x, then the distance from point F₂ to the asymptote is |NF₂| = \\frac{bc}{c} = b. AOPF₂ is an isosceles triangle, so the heights on the equal sides are also equal, hence |PQ| = b, then \\frac{|PQ|}{|PF_{1}|} = \\frac{1}{2} ⇒ ∠PF₁Q = 30°, and ∠POQ = 2∠PF₁Q = 60°, thus k_{OP} = \\sqrt{3}. Therefore, the asymptotes of the hyperbola"
  },
  {
    "text": "If a moving circle is externally tangent to both fixed circles $(x+5)^{2}+y^{2}=1$ and $(x-5)^{2}+y^{2}=49$, then the equation of the trajectory of the center of the moving circle is?",
    "fact_expressions": "G1: Circle;G2: Circle;Expression(G2) = (y^2 + (x + 5)^2 = 1);G3: Circle;Expression(G3) = (y^2 + (x - 5)^2 = 49);IsOutTangent(G1, G2) = True;IsOutTangent(G1, G3) = True",
    "query_expressions": "LocusEquation(Center(G1))",
    "answer_expressions": "(x^2/9-y^2/16=1)&(x<=-3)",
    "fact_spans": "[[[1, 3], [52, 54]], [[7, 26]], [[7, 26]], [[27, 47]], [[27, 47]], [[1, 50]], [[1, 50]]]",
    "query_spans": "[[[52, 63]]]",
    "process": "Let circle $ C_{1} $ be $ (x+5)^{2}+y^{2}=1 $, then the center is $ C_{1}(-5,0) $, radius $ r_{1}=1 $. Let circle $ C_{2} $ be $ (x-5)^{2}+y^{2}=49 $, then the center is $ C_{2}(5,0) $, radius $ r_{2}=7 $, and $ |C_{1}C_{2}|=10 $. Let the moving circle have center $ C $ and radius $ r $. Since the moving circle is externally tangent to both circle $ C_{1} $ and circle $ C_{2} $, it follows that $ |CC_{1}|=r+1 $, $ |CC_{2}|=7+r $. Thus, $ |CC_{2}|-|CC_{1}|=6<|C_{1}C_{2}|=10 $. Therefore, the trajectory of point $ C $ is the left branch of a hyperbola with foci at $ C_{1}(-5,0) $ and $ C_{2}(5,0) $. Hence, $ a=3 $, $ c=5 $, $ b=\\sqrt{c^{2}-a^{2}}=\\sqrt{16}=4 $. Thus, the equation of the trajectory of the center $ C $ of the moving circle is: $ \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1 $ $ (x\\leqslant-3) $."
  },
  {
    "text": "Given that the circle $x^{2}+y^{2}=4$ passes through the foci and the endpoints of the minor axis of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, then the standard equation of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Circle;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = 4);PointOnCurve(Focus(C), G);PointOnCurve(Endpoint(MinorAxis(C)), G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/8 + y^2/4 = 1",
    "fact_spans": "[[[19, 76], [86, 91]], [[26, 76]], [[26, 76]], [[2, 18]], [[26, 76]], [[26, 76]], [[19, 76]], [[2, 18]], [[2, 79]], [[2, 84]]]",
    "query_spans": "[[[86, 98]]]",
    "process": "By the given condition, the circle $x^{2}+y^{2}=4$ passes through the foci and the endpoints of the minor axis of the ellipse $C: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$, so we obtain $b=c=2$, hence $a^{2}=b^{2}+c^{2}=8$. Therefore, the standard equation of the ellipse $C$ is $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$."
  },
  {
    "text": "Given the parabola $ C $: $ y^{2} = 2 p x $ ($ p > 0 $) with focus $ F $, directrix $ l $: $ x = -1 $, point $ M $ lies on the parabola $ C $, and the projection of point $ M $ onto the line $ l $: $ x = -1 $ is $ A $. If the slope of line $ A F $ is $ -\\sqrt{3} $, then the area of $ \\Delta M A F $ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p>0;F: Point;Focus(C) = F;l: Line;Directrix(C) = l;Expression(l) = (x=-1);M: Point;PointOnCurve(M, C) = True;A: Point;Projection(M,l) = A;Slope(LineOf(A,F)) = -sqrt(3);p: Number",
    "query_expressions": "Area(TriangleOf(M, A, F))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[2, 28], [55, 61]], [[2, 28]], [[10, 28]], [[32, 35]], [[2, 35]], [[38, 49], [68, 81]], [[2, 49]], [[38, 49]], [[50, 54], [63, 67]], [[50, 62]], [[86, 89]], [[63, 89]], [[91, 113]], [[10, 28]]]",
    "query_spans": "[[[115, 134]]]",
    "process": "\\because the directrix of parabola C is x = -1, \\therefore the focus of parabola C is F(1,0), and parabola C: y^{2} = 4x. \\because point M lies on parabola C, point A lies on the directrix l, MA \\bot l, and the slope of line AF is k_{AF} = -\\sqrt{3}, so the inclination angle of line AF is \\frac{2\\pi}{3}. Let N be the intersection point of the directrix and the x-axis, then \\angle AFN = \\frac{\\pi}{3}, NF = 2. \\therefore \\angle MAF = \\angle AFN = \\frac{\\pi}{3}, |AF| = 4. Also |MF| = |AM|, \\therefore \\triangle MAF is an equilateral triangle, \\therefore S_{\\Delta AMF} = \\frac{1}{2} \\times 4 \\times 4 \\times \\frac{\\sqrt{3}}{2} = 4\\sqrt{3}"
  },
  {
    "text": "From the right focus $F$ of the hyperbola $\\frac{x^{2}}{5}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, draw a perpendicular to the asymptote, with foot of perpendicular at $P$. If the area of $\\triangle P O F$ is $\\sqrt{5}$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;P: Point;O: Origin;F: Point;H: Line;b>0;Expression(G) = (x^2/5 - y^2/b^2 = 1);RightFocus(G) = F;PointOnCurve(F, H);IsPerpendicular(Asymptote(G), H);FootPoint(Asymptote(G), H) = P;Area(TriangleOf(P, O, F)) = sqrt(5)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3*sqrt(5)/5",
    "fact_spans": "[[[1, 48], [106, 109]], [[4, 48]], [[66, 69]], [[72, 89]], [[52, 55]], [], [[4, 48]], [[1, 48]], [[1, 55]], [[0, 62]], [[0, 62]], [[0, 69]], [[72, 103]]]",
    "query_spans": "[[[106, 115]]]",
    "process": ""
  },
  {
    "text": "The left and right foci of the hyperbola $C$: $\\frac{x^{2}}{4}-y^{2}=1$ are $F_{1}$ and $F_{2}$, respectively. A line passing through $F_{1}$ intersects the left branch of the hyperbola at points $A$ and $B$. Then the minimum value of $|A F_{2}|+|B F_{2}|$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/4 - y^2 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;G: Line;PointOnCurve(F1, G) = True;Intersection(G, LeftPart(C)) = {A, B};A: Point;B: Point",
    "query_expressions": "Min(Abs(LineSegmentOf(A, F2)) + Abs(LineSegmentOf(B, F2)))",
    "answer_expressions": "9",
    "fact_spans": "[[[0, 33], [70, 73]], [[0, 33]], [[42, 49], [59, 66]], [[50, 57]], [[0, 57]], [[0, 57]], [[67, 69]], [[58, 69]], [[67, 85]], [[76, 79]], [[80, 83]]]",
    "query_spans": "[[[87, 114]]]",
    "process": "|AF_{2}|+|BF_{2}|=|AF_{1}|+2a+|BF_{1}|+2a=|AB|+4a\\geqslant2\\frac{b^{2}}{a}+4a=2\\times\\frac{1}{2}+8=9"
  },
  {
    "text": "Given the hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ and the hyperbola $C_{2}$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{16}=1$ have the same asymptotes, and the right focus of $C_{1}$ is $F(\\sqrt{5}, 0)$. Then $a=?$ $b=?$",
    "fact_expressions": "C1: Hyperbola;C2: Hyperbola;F: Point;a: Number;b: Number;a > 0;b > 0;Expression(C1) = (x^2/a^2 - y^2/b^2 = 1);Expression(C2) = (x^2/4 - y^2/16 = 1);Asymptote(C1) = Asymptote(C2);Coordinate(F) = (sqrt(5), 0);RightFocus(C1) = F",
    "query_expressions": "a;b",
    "answer_expressions": "1\n2",
    "fact_spans": "[[[2, 68], [126, 133]], [[69, 117]], [[138, 154]], [[157, 160]], [[162, 165]], [[14, 68]], [[14, 68]], [[2, 68]], [[69, 117]], [[2, 124]], [[138, 154]], [[126, 154]]]",
    "query_spans": "[[[157, 162]], [[162, 167]]]",
    "process": ""
  },
  {
    "text": "Given a point $M(4, y)$ on the parabola $y^{2}=2 p x$ ($p>0$), and its distance to the focus $F$ is $5$, then the area of $\\triangle O F M$ ($O$ is the origin) is?",
    "fact_expressions": "G: Parabola;p: Number;M: Point;O: Origin;F: Point;p>0;Expression(G) = (y^2 = 2*p*x);Coordinate(M) = (4, y1);Focus(G)=F;Distance(M,F)=5;y1:Number;PointOnCurve(M,G)",
    "query_expressions": "Area(TriangleOf(O,F,M))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 23]], [[5, 23]], [[27, 36], [37, 38]], [[75, 78]], [[41, 44]], [[5, 23]], [[2, 23]], [[27, 36]], [[2, 44]], [[37, 51]], [[27, 36]], [2, 36]]",
    "query_spans": "[[[53, 85]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ lies on the parabola $y^{2}=4x$, if the distance from point $P$ to the line $y=x$ is $4\\sqrt{2}$, then the coordinates of point $P$ are?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (y = x);PointOnCurve(P, G);Distance(P, H) = 4*sqrt(2)",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "{(4,-4),(16,8)}",
    "fact_spans": "[[[6, 20]], [[30, 37]], [[55, 59], [2, 5], [25, 29]], [[6, 20]], [[30, 37]], [[2, 23]], [[25, 53]]]",
    "query_spans": "[[[55, 63]]]",
    "process": "Let $ P(t^{2}, 2t) $, by the point-to-line distance formula we have $ \\frac{|t^{2}-2t|}{\\sqrt{2}} = 4\\sqrt{2} $. Solving gives: $ t^{2}-2t-8=0 $ or $ t^{2}-2t+8=0 $, so $ t=4 $ or $ t=-2 $. Hence $ P(16,8) $ or $ P(4,-4) $."
  },
  {
    "text": "Given an ellipse $C$ centered at the origin with foci on the $x$-axis, the maximum distance from a point on the ellipse to a focus is $3$, and the minimum distance is $1$. Then the standard equation of the ellipse is?",
    "fact_expressions": "C: Ellipse;O: Origin;Center(C) = O;PointOnCurve(Focus(C), xAxis);K: Point;PointOnCurve(K, C);Max(Distance(K, Focus(C))) = 3;Min(Distance(K, Focus(C))) = 1",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4 + y^2/3 = 1",
    "fact_spans": "[[[17, 22], [49, 51]], [[5, 7]], [[2, 22]], [[8, 22]], [[24, 25]], [[17, 25]], [[17, 39]], [[17, 47]]]",
    "query_spans": "[[[49, 58]]]",
    "process": "Let the standard equation of the ellipse be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. From the given conditions: $a+c=3$, $a-c=1$, $\\therefore a=2$, $c=1$, $\\therefore b^{2}=a^{2}-c^{2}=3$, $\\therefore$ the standard equation of the ellipse is $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$."
  },
  {
    "text": "Point $P$ is a moving point on the parabola $y^{2}=4x$. The minimum value of the sum of the distance from point $P$ to the point $(0,-1)$ and the distance from point $P$ to the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;H: Point;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(H) = (0, -1);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, H) + Distance(P, Directrix(G)))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[5, 19], [45, 48]], [[30, 40]], [[0, 4], [25, 29]], [[5, 19]], [[30, 40]], [[0, 23]]]",
    "query_spans": "[[[25, 61]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. A line passing through $F_{2}$ intersects the ellipse at points $A$ and $B$, and $\\overrightarrow{A F_{1}} \\cdot \\overrightarrow{A F_{2}}=0$, $\\overrightarrow{A F_{2}}=2 \\overrightarrow{F_{2} B}$. Then the eccentricity of ellipse $E$ is?",
    "fact_expressions": "E: Ellipse;b: Number;a: Number;G: Line;A: Point;F1: Point;F2: Point;B: Point;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(E) = F1;RightFocus(E) = F2;PointOnCurve(F2, G);Intersection(G, E) = {A, B};DotProduct(VectorOf(A, F1), VectorOf(A, F2)) = 0;VectorOf(A, F2) = 2*VectorOf(F2, B)",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[19, 76], [95, 97], [224, 229]], [[26, 76]], [[26, 76]], [[92, 94]], [[98, 101]], [[1, 8]], [[9, 16], [84, 91]], [[102, 105]], [[26, 76]], [[26, 76]], [[19, 76]], [[1, 82]], [[1, 82]], [[83, 94]], [[92, 107]], [[109, 168]], [[169, 222]]]",
    "query_spans": "[[[224, 235]]]",
    "process": "Let $|BF_{2}|=x$, then $AF_{2}=2x$. From the given condition, $(2a-2x)^{2}+(3x)^{2}=(2a-x)^{2}$, solving gives $x=\\frac{a}{3}$. Therefore, $|AF_{2}|=\\frac{2a}{3}$, $|AF_{1}|=\\frac{4a}{3}$, simplifying the equation $(\\frac{4a}{3})^{2}+(\\frac{2a}{3})^{2}=(2c)$ yields the solution. \\because\\overrightarrow{AF}_{2}=2\\overrightarrow{F_{2}B}, let $|BF_{2}|=x$, then $|AF_{2}|=2x$. By the definition of ellipse, we obtain $|AF_{1}|=2a-2x$, $|BF_{1}|=2a-x$. \\because\\overrightarrow{AF}\\cdot\\overrightarrow{AF_{2}}=0,\\thereforeAF_{1}\\botAF_{2}$. In right triangle $\\triangle AF_{1}B$, there is $(2a-2x)^{2}+(3x)^{2}=(2a-x)^{2}$. Solving gives $x=\\frac{a}{3}$. \\therefore|AF_{2}|=\\frac{2a}{3}, |AF|=\\frac{4a}{3}$. In right triangle $\\triangle AF_{1}F_{2}$, there is $(\\frac{4a}{3})^{2}+(\\frac{2a}{3})^{2}=(2c)^{2}$. Simplifying yields $\\frac{c^{2}}{a^{2}}=\\frac{5}{9}$, \\therefore e=\\frac{c}{a}=\\frac{\\sqrt{5}}{3}$."
  },
  {
    "text": "A hyperbola centered at the origin with foci on the $y$-axis has an asymptote passing through the point $(-3,6)$. What is its eccentricity?",
    "fact_expressions": "G: Hyperbola;H: Point;O: Origin;Coordinate(H) = (-3, 6);Center(G)=O;PointOnCurve(Focus(G), yAxis);PointOnCurve(H,OneOf(Asymptote(G)))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[15, 18], [37, 38]], [[26, 35]], [[3, 5]], [[26, 35]], [[0, 18]], [[6, 18]], [[15, 35]]]",
    "query_spans": "[[[37, 44]]]",
    "process": "By the given conditions, the standard equation of the hyperbola can be written as \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1, and its asymptotes are given by y=\\pm\\frac{a}{b}x. Then, -\\frac{a}{b}\\times(-3)=6, so a=2b. Therefore, e=\\frac{c}{a}=\\sqrt{\\frac{a^{2}+b^{2}}{a^{2}}}=\\frac{\\sqrt{5}}{2}"
  },
  {
    "text": "The ellipse centered at the origin with coordinate axes as its axes of symmetry passes through the focus of the parabola $x^{2}=8 y$ and the vertex of the hyperbola $\\frac{x^{2}}{16}-y^{2}=1$. Then the eccentricity of this ellipse is equal to?",
    "fact_expressions": "G: Hyperbola;H: Parabola;I: Ellipse;O: Origin;Expression(G) = (x^2/16 - y^2 = 1);Expression(H) = (x^2 = 8*y);Center(I) = O;SymmetryAxis(I) = axis;PointOnCurve(Focus(H), I);PointOnCurve(Vertex(G), I)",
    "query_expressions": "Eccentricity(I)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[38, 67]], [[20, 34]], [[16, 18], [73, 75]], [[3, 7]], [[38, 67]], [[20, 34]], [[0, 18]], [[8, 18]], [[16, 37]], [[16, 70]]]",
    "query_spans": "[[[73, 82]]]",
    "process": "The focus coordinates of the parabola $x^{2}=8y$ are $(0,2)$. The vertex coordinates of the hyperbola $\\frac{x^{2}}{16}-y^{2}=1$ are $(4,0)$, $(-4,0)$. According to the problem, the foci of the ellipse lie on the x-axis, given by $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. Then $a=4$, $b=2$, hence $c=\\sqrt{a^{2}-b^{2}}=2\\sqrt{3}$. Therefore, its eccentricity is $e=\\frac{c}{a}=\\frac{\\sqrt{3}}{2}$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$ and directrix $l$, $P$ is a point on $l$, $Q$ is an intersection point of line $PF$ and $C$, and if $\\overrightarrow{QP}=3\\overrightarrow{QF}$, then $|QF|=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l: Line;Directrix(C) = l;P: Point;PointOnCurve(P, l);Q: Point;OneOf(Intersection(LineOf(P, F), C)) = Q;VectorOf(Q, P) = 3*VectorOf(Q, F)",
    "query_expressions": "Abs(LineSegmentOf(Q, F))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 21], [60, 63]], [[2, 21]], [[25, 28]], [[2, 28]], [[32, 35], [41, 44]], [[2, 35]], [[37, 40]], [[37, 47]], [[48, 51]], [[48, 68]], [[70, 115]]]",
    "query_spans": "[[[117, 128]]]",
    "process": "Let the distance from Q to / be d. By the definition of the parabola, |QF| = d. Combining with \\overrightarrow{QP} = 3\\overrightarrow{QF}, we find the equation of line PF and solve it together with y^{2} = 4x, then solve using |QF| = x + \\frac{p}{2}. As shown in the figure: let the distance from Q to be d, then by the definition of the parabola we have |QF| = d, \\overrightarrow{QP} = 3\\overrightarrow{QF}, \\therefore |QP| = 3d. The slope of line PF is \\pm2\\sqrt{2}. \\because F(1,0), directrix l: x = -1, \\therefore the equation of line PF is y = \\pm2\\sqrt{2}(x - 1). Solving together with y^{2} = 4x gives x = \\frac{1}{2} (discarded) or x = 2. \\therefore |QF| = 2 + 1 = 3."
  },
  {
    "text": "The line $l$ intersects the parabola $y^{2}=8x$ at points $A$ and $B$, and $l$ passes through the focus $F$ of the parabola. Given $A(8,8)$, find the distance from the midpoint of segment $AB$ to the directrix.",
    "fact_expressions": "l: Line;G: Parabola;Expression(G) = (y^2 = 8*x);A: Point;B: Point;Intersection(l, G) = {A, B};F: Point;Focus(G) = F;PointOnCurve(F, l);Coordinate(A) = (8, 8)",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A, B)), Directrix(G))",
    "answer_expressions": "25/4",
    "fact_spans": "[[[0, 5], [33, 36]], [[6, 20], [38, 41]], [[6, 20]], [[22, 25], [50, 58]], [[26, 29]], [[0, 31]], [[44, 47]], [[38, 47]], [[33, 47]], [[50, 58]]]",
    "query_spans": "[[[38, 78]]]",
    "process": "Analyze: First, obtain the coordinates of the focus from the parabola equation. Let the coordinates of point B be (x_{B}, y_{B}), then derive the equation of line AB. Substituting point B into this equation yields the coordinates of point B. Then, according to the definition of the parabola, the answer can be obtained. From the given information, for the parabola y^{2}=8x, we have p=4. Let the coordinates of point B be (x_{B}, y_{B}). Since line AB passes through the focus F, the equation of line AB is y=\\frac{4}{3}(x-2). Substituting point (x_{B}, y_{B}) into this equation gives y_{B}=\\frac{4}{3}(x_{B}-2)=\\frac{4}{3}(\\frac{y^{2}}{8}-2). Solving yields y_{B}=-2, so x_{B}=\\frac{1}{2}. Therefore, the distance from the midpoint of segment AB to the directrix is \\frac{8+\\frac{1}{2}}{2}+2=\\frac{25}{4}."
  },
  {
    "text": "Given that a line $l$ passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$, and $|AF|=3$, then the distance from the origin $O$ to the line $l$ is equal to?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, F)) = 3;O:Origin",
    "query_expressions": "Distance(O, l)",
    "answer_expressions": "2*sqrt(2)/3",
    "fact_spans": "[[[24, 29], [64, 69]], [[3, 17], [31, 34]], [[35, 38]], [[20, 23]], [[39, 42]], [[3, 17]], [[3, 23]], [[2, 29]], [[24, 44]], [[45, 54]], [[56, 63]]]",
    "query_spans": "[[[56, 75]]]",
    "process": "First, from |AF| = x_{1} + \\frac{P}{2}, solve for x_{1}, then substitute into the parabola equation y^{2} = 4x to find y_{1}; finally, according to \\triangle AMF \\sim \\triangle OHF, the answer to this problem can be obtained. Let point A(x_{1}, y_{1}). Since the parabola equation is y^{2} = 4x, we have P = 2. Also, because |AF| = x_{1} + \\frac{P}{2} = 3, it follows that x_{1} = 2. Substituting into y^{2} = 4x gives y_{1} = \\pm 2\\sqrt{2}. Draw a perpendicular from point A to the x-axis, with foot M; draw a perpendicular from point O to line l, with foot H. It is easy to see that \\triangle AMF \\sim \\triangle OHF, so \\frac{OF}{AF} = \\frac{OH}{AM}, that is, \\frac{1}{3} = \\frac{OH}{2\\sqrt{2}}, yielding OH = \\frac{2\\sqrt{2}}{3}. Therefore, the distance from point O to the line equals: 7 This problem mainly examines comprehensive problems involving the intersection of a line and a parabola; using similar triangles to obtain proportional corresponding sides is the key to solving this problem."
  },
  {
    "text": "The coordinates of the focus of the parabola $y^{2}=12 \\sqrt{3} x$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 12*(sqrt(3)*x))",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(3*sqrt(3),0)",
    "fact_spans": "[[[0, 24]], [[0, 24]]]",
    "query_spans": "[[[0, 31]]]",
    "process": "Since 2p=12\\sqrt{3}, it follows that \\frac{p}{2}=3\\sqrt{3}; therefore, the focus of the parabola y^{2}=12\\sqrt{3}x is at (3\\sqrt{3},0)."
  },
  {
    "text": "Given that $P$ is a common point of the ellipse $\\frac{x^{2}}{a_{1}^{2}}+\\frac{y^{2}}{b_{1}^{2}}=1$ $(a_{1}>b_{1}>0)$ and the hyperbola $\\frac{x^{2}}{a_{2}^{2}}-\\frac{y^{2}}{b_{2}^{2}}=1$ $(a_{2}>0, b_{2}>0)$, $F_{1}$, $F_{2}$ are the common foci of the ellipse and the hyperbola, $e_{1}$, $e_{2}$ are the eccentricities of the ellipse and the hyperbola respectively, and if $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$, then the maximum value of $\\frac{1}{e_{1}}+\\frac{1}{e_{2}}$ is?",
    "fact_expressions": "G: Hyperbola;b2: Number;a2: Number;H: Ellipse;a1: Number;b1: Number;F1: Point;P: Point;F2: Point;a2>0;b2>0;Expression(G) = (-y^2/b2^2 + x^2/a2^2 = 1);a1 > b1;b1 > 0;Expression(H) = (y^2/b1^2 + x^2/a1^2 = 1);OneOf(Intersection(H, G)) = P;Focus(G) = {F1, F2};Focus(H) = {F1, F2};e1: Number;e2: Number;Eccentricity(H) = e1;Eccentricity(G) = e2;AngleOf(F1, P, F2) = pi/3",
    "query_expressions": "Max(1/e2 + 1/e1)",
    "answer_expressions": "4*sqrt(3)/3",
    "fact_spans": "[[[75, 147], [173, 176], [204, 207]], [[78, 147]], [[78, 147]], [[6, 74], [170, 172], [201, 203]], [[8, 74]], [[8, 74]], [[154, 161]], [[2, 5]], [[162, 169]], [[78, 147]], [[78, 147]], [[75, 147]], [[8, 74]], [[8, 74]], [[6, 74]], [[2, 153]], [[154, 181]], [[154, 181]], [[182, 189]], [[191, 198]], [[182, 211]], [[182, 211]], [[213, 249]]]",
    "query_spans": "[[[251, 290]]]",
    "process": "Let |PF_{1}| = m, |PF_{2}| = n. From the definition of the ellipse, we have: m + n = 2a_{1} (1). From the definition of the hyperbola, we have: |m - n| = 2a_{2} (2). Adding (1)^{2} and (2)^{2} gives: m^{2} + n^{2} = 2(a_{1}^{2} + a_{2}^{2}). Subtracting (2)^{2} from (1)^{2} gives: mn = a_{1}^{2} - a_{2}^{2}. By the law of cosines: (2c)^{2} = m^{2} + n^{2} - 2mn \\cdot \\cos\\angle F_{1}PF_{2}. Therefore, a_{1}^{2} + 3a_{2}^{2} = 4c^{2}. Let a_{1} = 2c \\cdot \\cos\\theta, a_{2} = \\frac{2\\sqrt{3}}{3}c \\cdot \\sin\\theta, so \\frac{1}{e_{1}} + \\frac{1}{e_{2}} = \\frac{a_{1}}{c} + \\frac{a_{2}}{c} = 2\\cos\\theta + \\frac{2\\sqrt{3}}{3}\\sin\\theta = \\frac{4\\sqrt{3}}{3}\\sin(\\theta + \\frac{\\pi}{3}). When \\theta + \\frac{\\pi}{3} = 2k\\pi + \\frac{\\pi}{2} (k \\in Z), the maximum value of \\frac{1}{e_{1}} + \\frac{1}{e_{2}} is \\frac{4\\sqrt{3}}{3}."
  },
  {
    "text": "The equation of the trajectory of a moving point whose product of distances to the two coordinate axes is $2$ is?",
    "fact_expressions": "P: Point;Distance(P, xAxis) * Distance(P, yAxis) = 2",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "{(x*y=2),(x*y=-2)}",
    "fact_spans": "[[[14, 16]], [[0, 16]]]",
    "query_spans": "[[[14, 22]]]",
    "process": "Let the moving point be M with coordinates (x, y), and the product of the distances from the moving point to the two coordinate axes equals 2, i.e., |x|×|y|=2. Removing the absolute values gives the trajectory equations of the moving point as: xy=2 or xy=-2."
  },
  {
    "text": "If the line $x+2 y-2=0$ intersects the ellipse $m x^{2}+n y^{2}=1$ at points $C$ and $D$, point $M$ is the midpoint of $CD$, the slope of line $OM$ ($O$ being the origin) is $\\frac{1}{2}$, and $O C \\perp O D$, then $m+n=$?",
    "fact_expressions": "Z: Line;Expression(Z) = (x + 2*y - 2 = 0);G: Ellipse;Expression(G) = (m*x^2 + n*y^2 = 1);m: Number;n: Number;C: Point;D: Point;Intersection(Z, G) = {C, D};M: Point;MidPoint(LineSegmentOf(C, D)) = M;O: Origin;Slope(LineOf(O, M)) = 1/2;IsPerpendicular(LineSegmentOf(O, C), LineSegmentOf(O, D))",
    "query_expressions": "m + n",
    "answer_expressions": "5/4",
    "fact_spans": "[[[1, 14]], [[1, 14]], [[15, 36]], [[15, 36]], [[17, 36]], [[17, 36]], [[38, 42]], [[43, 46]], [[1, 46]], [[48, 52]], [[48, 60]], [[68, 71]], [[61, 92]], [[94, 109]]]",
    "query_spans": "[[[111, 118]]]",
    "process": "Let $ C(x_{1},y_{1}) $, $ D(x_{2},y_{2}) $, $ M(x_{3},y_{3}) $, then $ mx_{1}^{2}+ny_{1}^{2}=1 $, $ mx_{2}^{2}+ny_{2}^{2}=1 $, subtracting the two equations gives: $ m(x_{1}^{2}-x_{2}^{2})+n(y_{1}^{2}-y_{2}^{2})=0 \\Rightarrow m(x_{1}+x_{2})+n(y_{1}+y_{2})k_{CD}=0 \\Rightarrow m(2x_{3})+n(2y_{3})(-\\frac{1}{2})=0 \\Rightarrow m+nk_{OM}(-\\frac{1}{2})=0 \\Rightarrow m+n\\times\\frac{1}{2}(-\\frac{1}{2})=0 \\Rightarrow n=4m $. From the line $ x+2y-2=0 $ and the ellipse $ mx^{2}+4my^{2}=1 $, eliminating $ x $ yields: $ 8y^{2}-8y+4-\\frac{1}{m}=0 $. Also $ OC\\bot OD \\Rightarrow x_{1}x_{2}+y_{1}y_{2}=0 \\Rightarrow 5y_{1}y_{2}-4(y_{1}+y_{2})+4=0 $, so $ 5\\times\\frac{4-\\frac{1}{m}}{8}-4\\times1+4=0 \\Rightarrow m=\\frac{1}{4} $. $ m+n=\\frac{5}{4} $."
  },
  {
    "text": "A line $m$ passing through the point $M(-2, 0)$ intersects the ellipse $\\frac{x^{2}}{2} + y^{2} = 1$ at points $P_{1}$ and $P_{2}$, with the midpoint of segment $P_{1}P_{2}$ being $P$. Let the slope of line $m$ be $k_{1}$ $(k_{1} \\neq 0)$, and the slope of line $OP$ be $k_{2}$. Then the value of $k_{1}k_{2}$ is?",
    "fact_expressions": "m: Line;G: Ellipse;O: Origin;P: Point;P1: Point;P2: Point;M: Point;k1: Number;k2: Number;Negation(k1 = 0);Expression(G) = (x^2/2 + y^2 = 1);Coordinate(M) = (-2, 0);PointOnCurve(M, m);Intersection(m, G) = {P1, P2};MidPoint(LineSegmentOf(P1, P2)) = P;Slope(m) = k1;Slope(LineOf(O, P)) = k2",
    "query_expressions": "k1*k2",
    "answer_expressions": "-1/2",
    "fact_spans": "[[[14, 19], [91, 96]], [[20, 47]], [[124, 129]], [[86, 89]], [[49, 56]], [[57, 64]], [[1, 13]], [[100, 121]], [[133, 140]], [[100, 121]], [[20, 47]], [[1, 13]], [[0, 19]], [[14, 66]], [[67, 89]], [[91, 121]], [[122, 140]]]",
    "query_spans": "[[[142, 159]]]",
    "process": "Let the equation of line $ l $ be: $ y = k_{1}(x + 2) $, $ P_{1}(x_{1}, y_{1}) $, $ P_{2}(x_{2}, y_{2}) $. From  \n\\[\n\\begin{cases}\ny = k_{1}(x + 2) \\\\\nx^{2} + 2y^{2} = 2\n\\end{cases}\n\\]\nwe obtain after simplification:  \n$ (1 + 2k_{1}^{2})x^{2} + 8k_{1}^{2}x + 8k_{1}^{2} - 1 = 0 $,  \nso $ x_{1} + x_{2} = \\frac{-8k_{1}^{2}}{1 + 2k_{1}^{2}} $, $ x_{1}x_{2} = \\frac{8k_{1}^{2} - 1}{1 + 2k_{1}^{2}} $.  \nThus,  \n$ y_{1} + y_{2} = k_{1}(x_{1} + 2) + k_{1}(x_{2} + 2) = k_{1}(x_{1} + x_{2} + 4) = \\frac{4k_{1}}{1 + 2k_{1}^{2}} $.  \nTherefore,  \n$ P\\left( \\frac{-4k_{1}^{2}}{1 + 2k_{1}^{2}}, \\frac{-2k_{1}}{1 + 2k_{1}^{2}} \\right) $,  \n$ k_{2} = \\frac{\\frac{-2k_{1}}{1 + 2k_{1}^{2}}}{\\frac{-4k_{1}^{2}}{1 + 2k_{1}^{2}}} $,  \nso $ k_{1}k_{2} = -\\frac{1}{2} $."
  },
  {
    "text": "The maximum distance from a point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ to the line $2 x-\\sqrt {3} y+3 \\sqrt {3}=0$ is?",
    "fact_expressions": "G: Ellipse;H: Line;P0: Point;Expression(G) = (x^2/9 + y^2/4 = 1);Expression(H) = (2*x - sqrt(3)*y + 3*sqrt(3) = 0);PointOnCurve(P0, G)",
    "query_expressions": "Max(Distance(P0, H))",
    "answer_expressions": "sqrt(21)",
    "fact_spans": "[[[0, 37]], [[41, 74]], [[39, 40]], [[0, 37]], [[41, 74]], [[0, 40]]]",
    "query_spans": "[[[39, 82]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ is $x+\\sqrt{3} y=0$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (x + sqrt(3)*y = 0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 39], [67, 70]], [[5, 39]], [[5, 39]], [[2, 39]], [[2, 64]]]",
    "query_spans": "[[[67, 76]]]",
    "process": "According to the standard equation of the hyperbola, write the equations of the asymptotes, compare with the given asymptote equations, find the value of $ a $, and finally determine the eccentricity of the hyperbola. The asymptote equations of $ \\frac{x^{2}}{a^{2}}-y^{2}=1 $ are $ y=\\pm\\frac{x}{a} \\Rightarrow x\\pm ay=0 $, so $ a=\\sqrt{3} $. Hence, the eccentricity is $ e=\\frac{c}{a}=\\sqrt{\\frac{a^{2}+1}{a^{2}}}=\\frac{2\\sqrt{3}}{3} $."
  },
  {
    "text": "Draw a line $l$ through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$), not perpendicular to the coordinate axes, intersecting the parabola at points $A$ and $B$. The perpendicular bisector of segment $AB$ intersects the $x$-axis at point $M$. Then $\\frac{|F M|}{|A B|}$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;l: Line;PointOnCurve(F, l);Negation(IsPerpendicular(l,axis));A: Point;B: Point;Intersection(l, G) = {A, B};M: Point;Intersection(PerpendicularBisector(LineSegmentOf(A, B)), xAxis) = M",
    "query_expressions": "Abs(LineSegmentOf(F, M))/Abs(LineSegmentOf(A, B))",
    "answer_expressions": "1/2",
    "fact_spans": "[[[1, 22], [44, 47]], [[1, 22]], [[4, 22]], [[4, 22]], [[25, 28]], [[1, 28]], [[37, 42]], [[0, 42]], [[29, 42]], [[48, 51]], [[52, 55]], [[37, 57]], [[75, 79]], [[58, 79]]]",
    "query_spans": "[[[81, 104]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. From the properties of the parabola, we know $ |AB| = x_{1} + x_{2} + p $. Since $ y_{1}^{2} = 2px_{1} $, $ y_{2}^{2} = 2px_{2} $, it follows that $ y_{1}^{2} - y_{2}^{2} = 2px_{1} - 2px_{2} $. From the problem, we know $ x_{1} \\ne x_{2} $. Therefore, $ \\underline{(y_{1}-y_{2})(y_{1}+y_{2})} = 2p \\frac{x_{1}-x_{2}}{y_{1}+y_{2}} $, i.e., $ k_{AB} = \\frac{2p}{y_{1}+y_{2}} $. Thus, the perpendicular bisector of segment $ AB $ has the equation $ y - \\frac{y_{1} + y_{2}}{2} = -\\frac{y_{1}+y_{2}}{2p} \\left( x - \\frac{x_{1}+x_{2}}{2} \\right) $. Let $ y = 0 $, then the horizontal coordinate of point $ M $ is $ x_{M} = p + \\frac{x_{1}+x_{2}}{2} $, then $ |FM| = \\left| x_{M} - \\frac{p}{2} \\right| = \\left| p + \\frac{x_{1}+x_{2}}{2} - \\frac{p}{2} \\right| = \\left| \\frac{x_{1}+x_{2}+p}{2} \\right| = \\frac{x_{1}+x_{2}+p}{2} $, so $ \\frac{|FM|}{|AB|} = \\frac{1}{2} $."
  },
  {
    "text": "If the line $y = -x + m$ and the curve $y = \\sqrt{5 - \\frac{1}{4} x^{2}}$ have only one common point, then the range of values for $m$ is?",
    "fact_expressions": "G: Line;m: Number;H: Curve;Expression(G) = (y = m - x);Expression(H) = (y = sqrt(5 - x^2/4));NumIntersection(G, H) = 1",
    "query_expressions": "Range(m)",
    "answer_expressions": "{[-2*sqrt(5),2*sqrt(5)),5}",
    "fact_spans": "[[[1, 11]], [[53, 56]], [[12, 44]], [[1, 11]], [[12, 44]], [[1, 51]]]",
    "query_spans": "[[[53, 63]]]",
    "process": ""
  },
  {
    "text": "The equation of the circle with its center at the focus of the parabola $y^{2}=2 x$ and tangent to the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;H: Circle;Expression(G) = (y^2 = 2*x);Focus(G)=Center(H);IsTangent(Directrix(G),H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x-1/2)^2+y^2=1",
    "fact_spans": "[[[1, 15], [24, 27]], [[33, 34]], [[1, 15]], [[0, 34]], [[23, 34]]]",
    "query_spans": "[[[33, 39]]]",
    "process": "The focus of the parabola $y^{2}=2x$ is $(\\frac{1}{2},0)$, the directrix is $x=-\\frac{1}{2}$, and the distance from the focus to the directrix is 1. Therefore, the center of the circle is $(\\frac{1}{2},0)$ and the radius is 1. Hence, the standard equation of the circle is $(x-\\frac{1}{2})^{2}+y^{2}=1$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{1}$ intersects the left branch of the hyperbola at points $A$ and $B$, with segment $AB$ of length $5$. If $a=4$, then what is the perimeter of $\\triangle A B F_{2}$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, H) = True;H: Line;Intersection(H, LeftPart(G)) = {A, B};A: Point;B: Point;Length(LineSegmentOf(A, B)) = 5;a = 4",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "26",
    "fact_spans": "[[[2, 58], [95, 98]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[67, 74], [84, 91]], [[75, 82]], [[2, 82]], [[2, 82]], [[83, 94]], [[92, 94]], [[92, 112]], [[103, 106]], [[107, 110]], [[113, 125]], [[128, 133]]]",
    "query_spans": "[[[136, 162]]]",
    "process": "As shown in the figure, by the definition of a hyperbola, we have |AF_{2}| - |AF_{1}| = 2a, |BF_{2}| - |BF_{1}| = 2a. Adding these two equations gives: |AF_{2}| + |BF_{2}| = 4a + |AF_{1}| + |BF_{1}| = 4a + AB. Therefore, the perimeter of \\triangle ABF_{2} is |AF_{2}| + |BF_{2}| + |AB| = 4a + 2AB = 4 \\times 4 + 2 \\times 5 = 26."
  },
  {
    "text": "What is the focal distance of the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{2}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2/2 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 43]]]",
    "process": "From the problem, we have $ c = \\sqrt{3+2} = \\sqrt{5} $, so the focal distance of the hyperbola is $ 2\\sqrt{5} $."
  },
  {
    "text": "The equation of a hyperbola centered at the origin, with one focus at $(-5,0)$ and one asymptote being the line $4x - 3y = 0$, is?",
    "fact_expressions": "O: Origin;Center(G) = O;H: Point;OneOf(Focus(G)) = H;Coordinate(H) = (-5, 0);G: Hyperbola;Z: Line;OneOf(Asymptote(G)) = Z;Expression(Z) = (4*x - 3*y = 0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/16 = 1",
    "fact_spans": "[[[3, 5]], [[0, 45]], [[11, 19]], [[6, 45]], [[11, 19]], [[42, 45]], [[28, 41]], [[22, 45]], [[28, 41]]]",
    "query_spans": "[[[42, 49]]]",
    "process": ""
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{2}=1$ are $F_{1}$ and $F_{2}$, and point $P$ lies on this ellipse. If $|P F_{1}|=4$, then what is the measure of $\\angle F_{1} P F_{2}$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 + y^2/2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 4",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "2*pi/3",
    "fact_spans": "[[[0, 37], [63, 65]], [[57, 61]], [[41, 48]], [[49, 56]], [[0, 37]], [[0, 56]], [[57, 66]], [[68, 81]]]",
    "query_spans": "[[[83, 110]]]",
    "process": "From the ellipse equation $\\frac{x^2}{9} + \\frac{y^{2}}{2} = 1$, we obtain $a=3$, $b=\\sqrt{2}$, $c=\\sqrt{7}$. According to the definition of an ellipse, $|PF_{1}| + |PF_{2}| = 2a = 6$, $|F_{1}F_{2}| = 2c = 2\\sqrt{7}$. We get $4 + |PF_{2}| = 2a = 6$, solving gives $|PF_{2}| = 2$. In triangle $F_{1}PF_{2}$, by the cosine law, $\\cos\\angle F_{1}PF_{2} = \\frac{|PF_{1}|^{2} + |PF_{2}|^{2} - |F_{1}F_{2}|^{2}}{2|PF_{1}|\\cdot|PF_{2}|} = \\frac{16+4-28}{2\\times4\\times2} = -\\frac{1}{2}$. Since $0 < \\angle F_{1}PF_{2} < \\pi$, it follows that $\\angle F_{1}PF_{2} = \\frac{2\\pi}{3}$."
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ has two foci $F_{1}$, $F_{2}$, and $P$ is a point on the hyperbola. When the area of $\\Delta F_{1} P F_{2}$ is $2$, what is the value of $|P F_{1}-P F_{2} |$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G) = True;Area(TriangleOf(F1, P, F2)) = 2",
    "query_expressions": "Abs(LineSegmentOf(P, F1) - LineSegmentOf(P, F2))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 28], [54, 57]], [[0, 28]], [[34, 41]], [[42, 49]], [[0, 49]], [[50, 53]], [[50, 60]], [[62, 92]]]",
    "query_spans": "[[[93, 117]]]",
    "process": ""
  },
  {
    "text": "Given the equation $(k-1) x^{2}+(9-k) y^{2}=1$, if this equation represents an ellipse, then the range of values for $k$ is?",
    "fact_expressions": "G: Ellipse;k:Number;Expression(G)=(x^2*(k - 1) + y^2*(9 - k) = 1)",
    "query_expressions": "Range(k)",
    "answer_expressions": "{(1, 5),(5, 9)}",
    "fact_spans": "[[[38, 40]], [[44, 47]], [[2, 42]]]",
    "query_spans": "[[[44, 53]]]",
    "process": "Since the equation is $(k-1)x^{2}+(9-k)y^{2}=1$, it follows that $\\frac{x^{2}}{k-1}+\\frac{y^{2}}{9-k}=1$. Therefore, we have\n\\begin{cases}\n\\frac{1}{k-1}>0 \\\\\n\\frac{1}{9-k}>0 \\\\\n\\frac{1}{k-1}\\neq\\frac{1}{9-k}\n\\end{cases}\nSolving gives $1<k<5$ or $5<k<9$."
  },
  {
    "text": "The line $m$ intersects the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ at points $P_{1}$ and $P_{2}$, with the midpoint of segment $P_{1} P_{2}$ being $P$. Let the slope of line $m$ be $k_{1}$ $(k_{1} \\neq 0)$, and the slope of line $O P$ be $k_{2}$. Then the value of $k_{1} \\cdot k_{2}$ is?",
    "fact_expressions": "m: Line;G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);P1: Point;P2: Point;Intersection(m, G) = {P1, P2};P: Point;MidPoint(LineSegmentOf(P1, P2)) = P;k1: Number;Negation(k1=0);Slope(m) = k1;O: Origin;k2: Number;Slope(LineOf(O, P)) = k2",
    "query_expressions": "k1*k2",
    "answer_expressions": "-1/2",
    "fact_spans": "[[[0, 5], [78, 83]], [[6, 33]], [[6, 33]], [[37, 45]], [[46, 53]], [[0, 53]], [[73, 76]], [[54, 76]], [[87, 108]], [[87, 108]], [[78, 108]], [[111, 116]], [[120, 127]], [[109, 127]]]",
    "query_spans": "[[[129, 152]]]",
    "process": "Analysis: Let the points be substituted into the equation of the ellipse, use the point difference method, and combine with the coordinates of the midpoint $P$ of segment $P_{1}P_{2}$ to obtain the answer. Specifically, let $P_{1}(x_{1},y_{1})$, $P_{2}(x_{2},y_{2})$, and the midpoint $P(x_{0},y_{0})$. Then $k_{1}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$, $k_{2}=\\frac{y_{0}}{x_{0}}=\\frac{y_{1}+y_{2}}{x_{1}+x_{2}}$. Substituting points $P_{1}(x_{1},y_{1})$ and $P_{2}(x_{2},y_{2})$ into the ellipse equation $\\frac{x^{2}}{2}+y^{2}=1$, we get $\\frac{x_{1}^{2}}{2}+y_{1}^{2}=1$, $\\frac{x_{2}^{2}}{2}+y_{2}^{2}=1$. Subtracting these two equations yields $\\frac{x_{1}^{2}-x_{2}^{2}}{2}+(y_{1}^{2}-y_{2}^{2})=0$, which simplifies to $\\frac{y_{1}^{2}-y_{2}^{2}}{x_{1}^{2}-x_{2}^{2}}=\\frac{(y_{1}-y_{2})(y_{1}+y_{2})}{(x_{1}-x_{2})(x_{1}+x_{2})}=-\\frac{1}{2}$, that is, $k_{1}k_{2}=-\\frac{1}{2}$."
  },
  {
    "text": "The equation of the circle centered at the right focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ and tangent to the asymptotes of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);H: Circle;Center(H) = RightFocus(G);IsTangent(Asymptote(G), H) = True",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x-5)^2+y^2=4^2",
    "fact_spans": "[[[1, 40], [50, 53]], [[1, 40]], [[60, 61]], [[0, 61]], [[49, 61]]]",
    "query_spans": "[[[60, 66]]]",
    "process": "According to the problem, a=3, b=4, c=5, so the asymptotes are y=\\pm\\frac{4}{3}x, the right focus is (5,0). The distance from the right focus to the asymptote y=\\frac{4}{3}x \\Rightarrow 4x-3y=0 is \\frac{4\\times5}{4^{2}+3^{2}}=4. The equation of the required circle is (x-5)^{2}+y^{2}=4^{2}."
  },
  {
    "text": "The asymptotes of the hyperbola are given by $y=\\pm \\frac{3}{2} x$, and the distance between the two vertices is $6$. Then its equation is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G))=(y=pm*((3/2)*x));F1:Point;F2:Point;Vertex(G)={F1, F2};Distance(F1,F2)=6",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/9 - 4*y^2/81 = 1, y^2/9 - x^2/4 = 1}",
    "fact_spans": "[[[0, 3], [45, 46]], [[0, 31]], [], [], [[0, 35]], [[0, 43]]]",
    "query_spans": "[[[45, 51]]]",
    "process": ""
  },
  {
    "text": "If $P$ is a moving point on the parabola $y^{2}=8x$, and point $Q$ moves on a circle centered at point $C(2,0)$ with radius equal to $1$, then the minimum value of $|PQ|+|PC|$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);P: Point;PointOnCurve(P, G);Q: Point;H: Circle;PointOnCurve(Q, H);C: Point;Coordinate(C) = (2, 0);Center(H) = C;Length(Radius(H)) = 1",
    "query_expressions": "Min(Abs(LineSegmentOf(P, C)) + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "3",
    "fact_spans": "[[[5, 19]], [[5, 19]], [[1, 4]], [[1, 23]], [[24, 28]], [[52, 53]], [[24, 54]], [[30, 39]], [[30, 39]], [[29, 53]], [[43, 53]]]",
    "query_spans": "[[[59, 78]]]",
    "process": "Since point C is the focus of the parabola, |PC| equals the distance d from point P to the directrix x = -2 of the parabola. Moreover, the distance from the center C to the directrix of the parabola is 4, then |PQ| + |PC| = |PQ| + d \\geqslant 3, with equality when point P is at the origin and Q is at (1,0). Hence, the minimum value of |PQ| + |PC| is 3."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. A circle with diameter $F_{1}F_{2}$ intersects an asymptote of the hyperbola $C$ at point $M$, where $|M F_{1}|>|M F_{2}|$, and the midpoint of segment $M F_{1}$ lies on the other asymptote. Then, the eccentricity of this hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;M: Point;l1: Line;l2: Line;F1: Point;F2: Point;G: Circle;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;IsDiameter(LineSegmentOf(F1, F2), G);OneOf(Asymptote(C)) = l1;OneOf(Asymptote(C)) = l2;Negation(l1 = l2);Intersection(G, l1) = M;Abs(LineSegmentOf(M, F1)) > Abs(LineSegmentOf(M, F2));PointOnCurve(MidPoint(LineSegmentOf(M, F1)), l2)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 63], [110, 116], [179, 182]], [[10, 63]], [[10, 63]], [[124, 128]], [], [], [[72, 79]], [[80, 87]], [[108, 109]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 87]], [[2, 87]], [[88, 109]], [[110, 122]], [[110, 175]], [[110, 175]], [[108, 128]], [[129, 150]], [[110, 176]]]",
    "query_spans": "[[[179, 188]]]",
    "process": "As shown in the figure: since $Q, O$ are the midpoints of $MF_{1}, F_{1}F_{2}$ respectively, $\\therefore OQ \\parallel F_{2}M$. Since $F_{1}F_{2}$ is the diameter of the circle, $\\therefore OQ \\perp MF_{1}$, so the equation of the line containing $MF_{1}$ is $y = \\frac{a}{b}(x + c)$. Solving the system  \n$$\n\\begin{cases}\ny = -\\frac{b}{a}x \\\\\ny = \\frac{a}{b}(x + c)\n\\end{cases}\n$$\ngives $Q\\left(-\\frac{a^{2}}{c}, \\frac{ab}{c}\\right)$. Using the midpoint formula, we find $M\\left(\\frac{c^{2} - 2a^{2}}{c}, \\frac{2ab}{c}\\right)$. Substituting into $y = \\frac{b}{a}x$ yields $c^{2} = 4a^{2}$, i.e., $e^{2} = 4$, so $e = 2$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through $F_{2}$ intersects the right branch of $C$ at points $A$ and $B$, such that $\\overrightarrow{A F_{2}}=2 \\overrightarrow{F_{2} B}$, and the perimeter of $\\triangle A B F_{1}$ is equal to 3 times the focal length. If $\\angle A F_{1} B>\\angle A B F_{1}$, then the range of the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F2, l) ;l: Line;Intersection(l, RightPart(C)) = {A, B};A: Point;B: Point;VectorOf(A, F2) = 2*VectorOf(F2, B);Perimeter(TriangleOf(A, B, F1)) = FocalLength(C) * 3;AngleOf(A, F1, B) > AngleOf(A, B, F1)",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(8/3,+oo)",
    "fact_spans": "[[[2, 64], [104, 107], [250, 253]], [[2, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[73, 80]], [[81, 88], [90, 97]], [[2, 88]], [[2, 88]], [[89, 103]], [[98, 103]], [[98, 120]], [[111, 114]], [[115, 118]], [[122, 175]], [[104, 211]], [[213, 248]]]",
    "query_spans": "[[[250, 264]]]",
    "process": "Let |BF₂| = m, then |AF₂| = 2m. Let |AF₁| = p, |BF₁| = q. By the definition of the hyperbola, we have  \n\\begin{cases}p - 2m = 2a,\\\\q - m = 2a,\\end{cases}  \nso p + q = 4a + 3m. Also, the perimeter of \\triangle ABF is p + q + 3m = 4a + 6m = 6c,  \nthus m = \\frac{3c - 2a}{3}, so |AB| = 3m = 3c - 2a, |AF₁| = p = 2a + 2m = \\frac{6c + 2a}{3}.  \nIf \\angle AF₁B > \\angle ABF₁, then |AB| > |AF₁|, that is, 3c - 2a > \\frac{6c + 2a}{2}, solving gives 3c > 8a, so e > \\frac{8}{3}."
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{|m|-1}+\\frac{y^{2}}{2-m}=1$ represents an ellipse with foci on the $y$-axis, what is the range of values for $m$?",
    "fact_expressions": "G: Ellipse;m:Number;Expression(G) = (x^2/(Abs(m) - 1) + y^2/(2 - m) = 1);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-\\infty,-1)+(1,3/2)",
    "fact_spans": "[[[56, 58]], [[60, 63]], [[2, 58]], [[47, 58]]]",
    "query_spans": "[[[60, 70]]]",
    "process": "Based on the characteristics of the equation of an ellipse with foci on the y-axis, list the inequality relations and solve them to obtain the result. From the given conditions, we have  \n\\begin{cases} |m|-1>0 \\\\ 2-m>|m|-1 \\end{cases}.  \nSolving $|m|-1>0$ gives $m>1$ or $m<-1$; solving $2-m>|m|-1$ gives $0<m<\\frac{3}{2}$ or $m<0$; combining these, the range of values for $m$ is $(-\\infty,-1$."
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{20}=1$, and point $P$ lies on the hyperbola. If the distance from point $P$ to focus $F_{1}$ is $9$, then the distance from point $P$ to focus $F_{2}$ equals?",
    "fact_expressions": "G: Hyperbola;F2: Point;P: Point;F1: Point;Expression(G) = (x^2/16 - y^2/20 = 1);Focus(G)={F1,F2};PointOnCurve(P,G);Distance(P,F1) = 9",
    "query_expressions": "Distance(P,F2)",
    "answer_expressions": "17",
    "fact_spans": "[[[19, 59], [69, 72]], [[10, 18], [107, 114]], [[76, 80], [76, 80], [76, 80]], [[1, 8], [83, 90]], [[19, 59]], [[1, 63]], [[64, 73]], [[76, 98]]]",
    "query_spans": "[[[100, 120]]]",
    "process": "Since $F_{1}F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{20}=1$, we have $||PF_{1}|-|PF_{2}||=2a=8$. Given that the distance from point $P$ to focus $F_{1}$ is 9, i.e., $PF_{1}=9$, solving yields $|PF_{2}|=1$ or $17$. Since the minimum value of the focal radius is $c-a=6-4=2$, it follows that $|PF|=17$."
  },
  {
    "text": "If the focus of the parabola $x^{2}=2 p y(p>0)$ lies on the line $2 x-y+3=0$, then $p$=?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;p>0;Expression(G) = (x^2=2*p*y);Expression(H) = (2*x-y+3=0);PointOnCurve(Focus(G), H)",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 22]], [[42, 45]], [[26, 39]], [[4, 22]], [[1, 22]], [[26, 39]], [[1, 40]]]",
    "query_spans": "[[[42, 47]]]",
    "process": "The focus of the parabola $x^{2}=2py$ ($p>0$) lies on the $y$-axis. Since the focus of the parabola $x^{2}=2py$ ($p>0$) lies on the line $2x-y+3=0$, the coordinates of the focus are $(0,3)$, so $\\frac{p}{2}=3$, solving gives $p=6$."
  },
  {
    "text": "Given $\\tan \\alpha = -2$, the focus of the parabola $y^2 = 2px$ ($p > 0$) is $F(-\\sin \\alpha \\cos \\alpha, 0)$, the line $l$ passes through point $F$ and intersects the parabola at points $A$ and $B$, and $|AB| = 4$. Then the distance from the midpoint of segment $AB$ to the line $x = -\\frac{1}{2}$ is?",
    "fact_expressions": "l: Line;G: Parabola;p: Number;H: Line;B: Point;A: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (x = -1/2);alpha:Number;Coordinate(F) = (-Sin(alpha)*Cos(alpha), 0);Tan(alpha) = -2;Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, B)) = 4",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A,B)), H)",
    "answer_expressions": "21/10",
    "fact_spans": "[[[77, 82]], [[19, 40], [90, 93]], [[22, 40]], [[127, 145]], [[99, 103]], [[95, 98]], [[84, 88], [44, 76]], [[22, 40]], [[19, 40]], [[127, 145]], [[2, 18]], [[44, 76]], [[2, 18]], [[19, 76]], [[77, 88]], [[77, 103]], [[105, 114]]]",
    "query_spans": "[[[116, 150]]]",
    "process": ""
  },
  {
    "text": "The distance between the two directrices of the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{4}=1$ is?",
    "fact_expressions": "G: Hyperbola;L1: Line;L2: Line;Expression(G) = (x^2/3 - y^2/4 = 1);Directrix(G) = {L1, L2}",
    "query_expressions": "Distance(L1, L2)",
    "answer_expressions": "6*sqrt(7)/7",
    "fact_spans": "[[[0, 38]], [], [], [[0, 38]], [[0, 43]]]",
    "query_spans": "[[[0, 49]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=-4 x$ with focus $F$ and point $A(3,-3)$, $P$ is a point on the parabola. Then the minimum value of $|P A|+|P F|$ is equal to?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -4*x);F: Point;Focus(G) = F;A: Point;Coordinate(A) = (3, -3);P: Point;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "5",
    "fact_spans": "[[[2, 17], [39, 42]], [[2, 17]], [[20, 23]], [[2, 23]], [[24, 34]], [[24, 34]], [[35, 38]], [[35, 45]]]",
    "query_spans": "[[[47, 67]]]",
    "process": "Since point A is outside the parabola and the focus of the parabola is F(-1,0), when points P, A, and F are collinear, |PA| + |PF| is minimized. At this time, |PA| + |PF| = |AF| = \\sqrt{4^{2}+3^{2}}=5."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the two foci are $F_{1}$ and $F_{2}$, the eccentricity is $e=\\frac{\\sqrt{2}}{2}$, and point $P$ is the upper vertex of the ellipse. If the area of $\\Delta P F_{1} F_{2}$ is $1$, then what are the coordinates of the right focus $F_{2}$?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;e:Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Focus(G)={F1,F2};Eccentricity(G)=e;e=sqrt(2)/2;UpperVertex(G)=P;Area(TriangleOf(P, F1, F2)) = 1;RightFocus(G)=F2",
    "query_expressions": "Coordinate(F2)",
    "answer_expressions": "(1,0)",
    "fact_spans": "[[[2, 54], [109, 111]], [[4, 54]], [[4, 54]], [[104, 108]], [[62, 69]], [[70, 77], [70, 77]], [[81, 103]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 77]], [[2, 103]], [[81, 103]], [[104, 115]], [[117, 146]], [[109, 158]]]",
    "query_spans": "[[[151, 163]]]",
    "process": "From the given \\begin{cases}c=\\frac{\\sqrt{2}}{2}\\\\\\frac{1}{2}b\\cdot2c=1\\\\a^{2}=b^{2}+c\\end{cases}, solving yields c=1, hence the coordinates of the right focus F_{2} are (1,0)."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{5}=1$ $(m>0)$ has an asymptote equation $\\sqrt{5} x+2 y=0$, the left focus is $F$, point $P$ moves on the right branch of the hyperbola, and point $Q$ moves on the circle $x^{2}+(y-4)^{2}=1$. Then the minimum value of $|P Q|+|P F|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/5 + x^2/m = 1);m: Number;m>0;Expression(OneOf(Asymptote(G))) = (sqrt(5)*x + 2*y = 0);F: Point;LeftFocus(G) = F;P: Point;PointOnCurve(P, RightPart(G));H: Circle;Expression(H) = (x^2 + (y - 4)^2 = 1);Q: Point;PointOnCurve(Q, H)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F)) + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 45], [86, 89]], [[2, 45]], [[5, 45]], [[5, 45]], [[2, 72]], [[77, 80]], [[2, 80]], [[81, 85]], [[81, 92]], [[100, 120]], [[100, 120]], [[95, 99]], [[95, 121]]]",
    "query_spans": "[[[125, 144]]]",
    "process": "From the hyperbola $\\frac{x^2}{m}-\\frac{y^{2}}{5}=1$ $(m>0)$, one asymptote equation is $\\sqrt{5}x+2y=0$, we obtain $\\frac{\\sqrt{5m}}{m}=\\frac{\\sqrt{5}}{2}$, solving gives $m=4$. Thus, $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, the coordinates of the left focus of the hyperbola are $F(-3,0)$, and the right focus is $F(3,0)$. By the definition of the hyperbola, $|PF|-|PF|=4$, i.e., $|PF|=4+|PF|$. From the circle $x^{2}+(y-4)^{2}=1$, we get the center $C(0,4)$, radius $r=1$, $|PQ|+|PF|=4+|PF|+|PQ|\\geqslant|QF|+4$, the problem reduces to finding the minimum value from point $F$ to the circle $x^{2}+(y-4)^{2}=1$, that is, $|QF|_{\\min}=|CF|-1=\\sqrt{(3-0)^{2}+(0-4)^{2}}-1=5-1=4$, so $(|PQ|+|PF|)_{\\min}=4+4=8$. Therefore, the minimum value of $|PO|+|PF|$ is $8$. The answer is: $8$."
  },
  {
    "text": "The equation $\\frac{x^{2}}{25-k}+\\frac{y^{2}}{16+k}=1$, then for $k \\in$? the equation represents a hyperbola.",
    "fact_expressions": "G: Hyperbola;Expression(G)=(x^2/(25-k)+y^2/(16+k)=1);k:Number",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-oo,-16) + (25,+oo)",
    "fact_spans": "[[[59, 62]], [[0, 62]], [[2, 43]]]",
    "query_spans": "[[[45, 53]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $P$ is a point on the ellipse such that $\\angle F_{1} P F_{2}=60^{\\circ}$. Then the area of $\\Delta P F_{2} F_{1}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F2: Point;F1: Point;Expression(G) = (x^2/9 + y^2/5 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(P, F2, F1))",
    "answer_expressions": "5*sqrt(3)/3",
    "fact_spans": "[[[2, 39], [67, 69]], [[63, 66]], [[55, 62]], [[47, 54]], [[2, 39]], [[2, 62]], [[2, 62]], [[63, 73]], [[75, 108]]]",
    "query_spans": "[[[110, 137]]]",
    "process": "According to the definition of an ellipse, we obtain the value of |PF_{1}| + |PF_{2}|. Then, given \\angle F_{1}PF_{2} = 60^{\\circ}, in \\triangle PF_{2}F_{1}, apply the cosine law to find |PF_{1}||PF_{2}|. Based on the triangle area formula, the result can be obtained. According to the definition of the ellipse, we have |PF_{1}| + |PF_{2}| = 2a = 6, and the focal distance of the ellipse is |F_{1}F_{2}| = 2\\sqrt{9 - 5} = 4. Given \\angle F_{1}PF_{2} = 60^{\\circ}, in \\triangle PF_{2}F_{1}, by the cosine law, we get \\cos\\angle F_{1}PF_{2} = \\frac{|PF_{1}|^{2} + |PF_{2}|^{2} - |F_{1}F_{2}|^{2}}{2|PF_{1}||PF_{2}|} = \\frac{1}{2}. Therefore, \\frac{(|PF_{1}| + |PF_{2}|)^{2} - 2|PF_{1}||PF_{2}| - |F_{1}F_{2}|^{2}}{2|PF_{1}||PF_{2}|} = \\frac{36 - |F_{1}F_{2}|^{2}}{2|PF_{1}||PF_{2}|} - \\frac{2|PF_{1}||PF_{2}|}{2|PF_{1}||PF_{2}|} = \\frac{1}{2}, so |PF_{1}||PF_{2}| = \\frac{20}{3}. Thus, the area of \\triangle PF_{2}F_{1} is S_{\\triangle PF_{1}F_{2}} = \\frac{1}{2}|PF_{1}||PF_{2}|\\sin\\angle F_{1}PF_{2} = \\frac{1}{2} \\times \\frac{20}{3} \\times \\frac{\\sqrt{3}}{2} = \\frac{5\\sqrt{3}}{3}."
  },
  {
    "text": "It is known that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $y=\\sqrt{3} x$, and one of its foci coincides with the focus of the parabola $y^{2}=16 x$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 16*x);Expression(OneOf(Asymptote(G)))=(y=sqrt(3)*x);OneOf(Focus(G))=Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/12 = 1",
    "fact_spans": "[[[2, 58], [83, 84], [113, 116]], [[5, 58]], [[5, 58]], [[90, 105]], [[5, 58]], [[5, 58]], [[2, 58]], [[90, 105]], [[2, 81]], [[83, 110]]]",
    "query_spans": "[[[113, 121]]]",
    "process": ""
  },
  {
    "text": "It is known that the focus of the parabola $y^{2}=2 p x$ coincides with the right focus of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$. Then the value of $p$ is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;Expression(G) = (x^2 - y^2/3 = 1);Expression(H) = (y^2 = 2*(p*x));Focus(H) = RightFocus(G)",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[22, 50]], [[2, 18]], [[58, 61]], [[22, 50]], [[2, 18]], [[2, 56]]]",
    "query_spans": "[[[58, 65]]]",
    "process": "Since the hyperbola is $x^{2}-\\frac{y^{2}}{3}=1$, the right focus of the hyperbola is $(2,0)$, so $\\frac{p}{2}=2$, $\\therefore p=4$"
  },
  {
    "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ have its right focus at $F$. Draw a perpendicular from $F$ to an asymptote of $C$, with foot of the perpendicular at $A$, and $|O A|=2|A F|$, where $O$ is the origin. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(C) = F;L: Line;PointOnCurve(F,L) = True;IsPerpendicular(L,OneOf(Asymptote(C))) = True;FootPoint(L,OneOf(Asymptote(C))) = A;O: Origin;Abs(LineSegmentOf(O, A)) = 2*Abs(LineSegmentOf(A, F));A: Point",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[1, 62], [76, 79], [123, 126]], [[1, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[67, 70], [72, 75]], [[1, 70]], [], [[71, 88]], [[71, 88]], [[71, 94]], [[113, 116]], [[96, 110]], [[91, 94]]]",
    "query_spans": "[[[123, 132]]]",
    "process": "From the given information, find the slope of the asymptote, obtaining \\frac{b}{a}; combining with c^{2}-a^{2}=b^{2}, after transformation the eccentricity can be found. According to the problem, \\tan\\angleAOF=\\frac{|AF|}{|OA|}=\\frac{|AF|}{2|AF|}=\\frac{1}{2}; the asymptote equation is y=\\frac{b}{a}x, \\therefore\\frac{b}{a}=\\frac{1}{2}, e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{a^{2}+b^{2}}{a^{2}}=\\frac{a^{2}+\\frac{a^{2}}{4}}{a^{2}}=\\frac{5}{4}, hence_{e}=\\frac{\\sqrt{5}}{2}"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, one of its asymptotes is perpendicular to the line $l$: $x+3 y+2022=0$. Then the eccentricity of $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(l) = (x + 3*y + 2022 = 0);IsPerpendicular(OneOf(Asymptote(C)), l)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[70, 91]], [[2, 63], [95, 98]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 63]], [[70, 91]], [[2, 93]]]",
    "query_spans": "[[[95, 104]]]",
    "process": "The slope of the line $ l: x + 3y + 2022 = 0 $ is $ -\\frac{1}{3} $, so the slope of the asymptote of the hyperbola perpendicular to the line $ l: x + 3y + 2022 = 0 $ is $ 3 $. Therefore, $ \\frac{b}{a} = 3 $, so $ e = \\frac{c}{a} = \\sqrt{\\frac{b^{2} + a^{2}}{a^{2}}} = \\sqrt{1 + (\\frac{b}{a})^{2}} = \\sqrt{10} $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, and line $l$ passing through $F_{1}$ intersects ellipse $C$ at points $A$ and $B$, then the perimeter of $\\triangle A B F_{2}$ is equal to?",
    "fact_expressions": "l: Line;C: Ellipse;A: Point;B: Point;F2: Point;F1: Point;Expression(C) = (x^2/9 + y^2/4 = 1);Focus(C) = {F1, F2};PointOnCurve(F1, l);Intersection(l, C) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "12",
    "fact_spans": "[[[75, 80]], [[18, 60], [81, 86]], [[88, 91]], [[92, 95]], [[10, 17]], [[2, 9], [67, 74]], [[18, 60]], [[2, 65]], [[66, 80]], [[75, 97]]]",
    "query_spans": "[[[99, 126]]]",
    "process": "According to the problem, the perimeter of $\\triangle ABF_{2}$ passing through focus $F_{1}$ is $4a$. $F_{1}, F_{2}$ are the foci of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{4}=1$. A line passing through focus $F_{1}$ intersects the ellipse at points $A, B$, then the perimeter of $\\triangle ABF_{2}$ is $AB + F_{2}A + F_{2}B = AF_{1} + F_{1}B + F_{2}A + F_{2}B = 4a = 4 \\times 3 = 12$. Hence, the answer is: $12$."
  },
  {
    "text": "A moving circle $P$ is internally tangent to two fixed circles $O_{1}$: $x^{2}+y^{2}=1$ and $O_{2}$: $(x-4)^{2}+y^{2}=9$. What is the equation of the trajectory of the center of the moving circle $P$?",
    "fact_expressions": "P: Circle;O1: Circle;Expression(O1) = (x^2 + y^2 = 1);O2: Circle;Expression(O2) = ((x - 4)^2 + y^2 = 9);IsInTangent(P, O1);IsInTangent(P, O2)",
    "query_expressions": "LocusEquation(Center(P))",
    "answer_expressions": "((x-2)^2-y^2/3=1)&(x>2)",
    "fact_spans": "[[[4, 7], [74, 77]], [[12, 36]], [[12, 36]], [[39, 66]], [[39, 66]], [[4, 69]], [[4, 69]]]",
    "query_spans": "[[[74, 87]]]",
    "process": "Let P(x,y), and let the radius of circle P be R. Since circle O₁: x² + y² = 1 has center O₁(0,0) and radius 1, and circle O₂: (x−4)² + y² = 9 has center O₂(4,0) and radius 3, and since circle P is internally tangent to both circles O₁ and O₂, it follows that |PO₁| = R−1, |PO₂| = R−3, so |PO₁| − |PO₂| = 2 < |O₁O₂| = 4. Therefore, the locus of P is the right branch of a hyperbola with center (2,0), where a=1, c=2, so b=√(4−1)=√3. Thus, the equation of the locus of P is: (x−2)² − y²⁄3 = 1 (x>2)."
  },
  {
    "text": "What is the distance from the focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1,(a>0, b>0)$ to its asymptote?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1)",
    "query_expressions": "Distance(Focus(G), Asymptote(G))",
    "answer_expressions": "b",
    "fact_spans": "[[[0, 57], [61, 62]], [[3, 57]], [[3, 57]], [[3, 57]], [[3, 57]], [[0, 57]]]",
    "query_spans": "[[[0, 70]]]",
    "process": "Since the hyperbola is $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, $(a>0,b>0)$, its foci are at $(\\pm c,0)$ and the equations of the asymptotes are $y=\\pm\\frac{b}{a}x$. Therefore, the distance from a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, $(a>0,b>0)$ to its asymptote is $\\frac{\\frac{b}{a}c}{\\sqrt{1+(\\frac{b}{a})}}=b$. Hence, the answer is $b$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$, $P$ is a point on $C$, $P Q \\perp x$-axis with foot $Q$, $F$ is the focus of $C$, and $O$ is the origin. If $\\angle P O Q=45^{\\circ}$, then $\\cos \\angle P F Q$=?",
    "fact_expressions": "C: Parabola;p: Number;P: Point;Q: Point;F: Point;O: Origin;p>0;Expression(C) = (y^2 = 2*(p*x));PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P, Q), xAxis);FootPoint(LineSegmentOf(P, Q), xAxis) = Q;Focus(C) = F ;AngleOf(P, O, Q) = ApplyUnit(45, degree)",
    "query_expressions": "Cos(AngleOf(P, F, Q))",
    "answer_expressions": "3/5",
    "fact_spans": "[[[2, 28], [35, 38], [68, 71]], [[10, 28]], [[30, 34]], [[60, 63]], [[64, 67]], [[75, 78]], [[10, 28]], [[2, 28]], [[31, 41]], [[42, 56]], [[42, 63]], [[64, 74]], [[83, 108]]]",
    "query_spans": "[[[110, 131]]]",
    "process": "Assume P is above the x-axis. Since ∠POQ = 45°, we can set line OP: y = x. From  \n\\begin{cases}y=x\\\\y^{2}=2px\\end{cases},  \nwe get x = y = 2p, ∴ P(2p, 2p), Q(2p, 0). Also F(\\frac{p}{2}, 0), ∴ \\cos\\angle PFQ = \\frac{|FQ|}{|PF|} = \\frac{|2p - \\frac{p}{2}|}{\\sqrt{(2p - \\frac{p}{2})^{2} + (2p)^{2}}} = \\frac{3}{5}."
  },
  {
    "text": "The locus of the centers of circles passing through the point $A(3,0)$ and tangent to the $y$-axis is?",
    "fact_expressions": "G: Circle;A: Point;Coordinate(A) = (3, 0);PointOnCurve(A,G);IsTangent(yAxis,G)",
    "query_expressions": "LocusEquation(Center(G))",
    "answer_expressions": "Parabola",
    "fact_spans": "[[[19, 20]], [[1, 10]], [[1, 10]], [[0, 20]], [[11, 20]]]",
    "query_spans": "[[[19, 28]]]",
    "process": "As shown in the figure, let point $ P $ be a point satisfying the given condition. It is not difficult to conclude that the distance from point $ P $ to point $ A $ is equal to the distance from point $ P $ to the $ y $-axis. Therefore, point $ P $ lies on a parabola with focus at point $ A $ and the $ y $-axis as the directrix. Hence, the locus of point $ P $ is a parabola."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{36}+\\frac{y^{2}}{27}=1$, point $P$ is a point on the ellipse $C$, $|P F_{1}|=8$. If $M$ is the midpoint of segment $P F_{1}$, then the length of segment $O M$ is?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;O: Origin;M: Point;F2: Point;Expression(C) = (x^2/36 + y^2/27 = 1);Focus(C) = {F1, F2};PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 8;MidPoint(LineSegmentOf(P,F1))=M",
    "query_expressions": "Length(LineSegmentOf(O,M))",
    "answer_expressions": "2",
    "fact_spans": "[[[18, 61], [72, 77]], [[67, 71]], [[2, 9]], [[118, 123]], [[96, 99]], [[10, 17]], [[18, 61]], [[2, 66]], [[67, 80]], [[81, 94]], [[96, 114]]]",
    "query_spans": "[[[116, 127]]]",
    "process": "F_{1},F_{2} are the two foci of the ellipse C: \\frac{x^{2}}{36}+\\frac{y^{2}}{27}=1, giving F_{1}(-3,0), F_{2}(3,0). a= point P is on the ellipse C, |PF_{1}|=8, then |PF_{2}|=4, M is the midpoint of segment PF_{1}, then the length of segment OM is: \\frac{1}{2}|PF_{2}|=2."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with right focus $F$, if the circle centered at $F$ given by $x^{2}+y^{2}-6x+5=0$ is tangent to the asymptotes of this hyperbola, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Circle;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (-6*x + x^2 + y^2 + 5 = 0);RightFocus(G) = F;Center(H)=F;IsTangent(Asymptote(G),H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3*sqrt(5)/5",
    "fact_spans": "[[[2, 58], [101, 104], [113, 116]], [[5, 58]], [[5, 58]], [[77, 99]], [[63, 66], [70, 73]], [[5, 58]], [[5, 58]], [[2, 58]], [[77, 99]], [[2, 66]], [[69, 99]], [[77, 110]]]",
    "query_spans": "[[[113, 122]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the right branch of the hyperbola such that $|P F_{1}|=4|P F_{2}|$. Then the range of the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F1)) = 4*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1, 5/3]",
    "fact_spans": "[[[2, 58], [87, 90], [119, 122]], [[5, 58]], [[5, 58]], [[83, 86]], [[67, 74]], [[75, 82]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 82]], [[2, 82]], [[83, 94]], [[95, 117]]]",
    "query_spans": "[[[119, 132]]]",
    "process": "Combining the given conditions with the definition of a hyperbola, we obtain \\begin{cases}|PF_{1}|=\\frac{8a}{3}\\\\|PF_{2}|=\\frac{2a}{3}\\end{cases}; then using the law of cosines, we get \\cos\\theta=\\frac{17a^{2}-9c^{2}}{8a^{2}}=\\frac{17}{8}-\\frac{9}{8}e^{2}. By finding the range of \\cos\\theta, the result can be determined. Let \\angle F_{1}PF_{2}=\\theta, from \\begin{matrix}|PF_{1}|=4|PF_{2}| & |PF_{1}|=\\frac{8a}{|PF_{1}|-|PF_{2}|=2a}\\\\|PF_{2}|=\\frac{2a}{3} & 2||PF_{2}|^{2}-|F_{1}F_{2}|^{2}=\\frac{17a^{2}-9c^{2}}{8a^{2}}=\\frac{17}{8}-\\frac{9}{8}e^{2}\\end{matrix}. Since \\theta\\in(0,\\pi], it follows that \\cos\\theta\\in[-1,1), i.e., -1\\leqslant\\frac{17}{8}-\\frac{9}{8}e^{2}<1. Also, since e>1, we have 1<e\\leqslant\\frac{5}{3}."
  },
  {
    "text": "If a point $P$ on the ellipse $\\frac{x^{2}}{144}+\\frac{y^{2}}{36}=1$ has a distance of $10$ to the focus $F_{1}$, then what is the distance from point $P$ to the other focus $F_{2}$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/144 + y^2/36 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Distance(P, F1) = 10;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2)",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "14",
    "fact_spans": "[[[2, 42]], [[2, 42]], [[45, 48], [70, 74]], [[2, 48]], [[51, 58]], [[80, 87]], [[2, 67]], [2, 55], [2, 84], [2, 84]]",
    "query_spans": "[[[70, 92]]]",
    "process": "From the ellipse $\\frac{x^{2}}{144}+\\frac{y^{2}}{36}=1$, we obtain $a=12$. By the definition of an ellipse, $|PF_{1}|+|PF_{2}|=2a=24$. Since the distance from a point $P$ on the ellipse $\\frac{x^{2}}{144}+\\frac{y^{2}}{36}=1$ to focus $F_{1}$ is equal to 10, the distance from point $P$ to the other focus $F_{2}$ is: $24-10=14$."
  },
  {
    "text": "Given that $F$ is the left focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $A$ is the right vertex, and $P$ is a point on the ellipse such that $PF \\perp x$-axis. If $|PF|=\\frac{3}{4}|AF|$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F: Point;A:Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F;RightVertex(G) = A;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P,F),xAxis);Abs(LineSegmentOf(P,F))=(3/4)*Abs(LineSegmentOf(A,F))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/4",
    "fact_spans": "[[[6, 58], [75, 77], [125, 127]], [[8, 58]], [[8, 58]], [[71, 74]], [[2, 5]], [[63, 66]], [[8, 58]], [[8, 58]], [[6, 58]], [[2, 62]], [[6, 70]], [[71, 81]], [[82, 96]], [[98, 122]]]",
    "query_spans": "[[[125, 133]]]",
    "process": "According to the geometric properties of the ellipse, |PF|=\\frac{b^{2}}{a}, |AF|=a-c, so \\frac{b^{2}}{a}=\\frac{3}{4}(a+c), that is, 4b^{2}=3a^{2}-3ac. Since b^{2}=a^{2}-c^{2}, we have 4(a^{2}-c^{2})=3a^{2}+3ac. Rearranging gives 4c^{2}+3ac-a^{2}=0. Dividing both sides by a^{2} yields: 4e^{2}+3e-1=0, so (4e-1)(e+1)=0. Since 0<e<1, we have e=\\frac{1}{4}."
  },
  {
    "text": "Given that points $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $x^{2}+2 y^{2}=2$, respectively, and point $P$ is a moving point on this ellipse, then the minimum value of $|\\overrightarrow{P F_{1}}+\\overrightarrow{P F}_{2}|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 + 2*y^2 = 2);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(P,G)",
    "query_expressions": "Min(Abs(VectorOf(P,F1)+VectorOf(P,F2)))",
    "answer_expressions": "2",
    "fact_spans": "[[[23, 42], [54, 56]], [[48, 52]], [[2, 11]], [[13, 20]], [[23, 42]], [[2, 48]], [[2, 48]], [[48, 62]]]",
    "query_spans": "[[[65, 124]]]",
    "process": "Let P(x,y), then x^{2}+2y^{2}=2. From the ellipse equation \\frac{x^{2}}{2}+y^{2}=1, we have a=\\sqrt{2}, b=1, c=1, \\therefore F_{1}(-1,0), F_{2}(1,0). \\therefore \\overrightarrow{PF_{1}}=(-1-x,-y), \\overrightarrow{PF_{2}}=(1-x,-y), \\therefore \\overrightarrow{PF_{1}}+\\overrightarrow{PF_{2}}=(-2x,-2y). \\because \\overrightarrow{PF}+\\overrightarrow{PF}, | the minimum value is 2."
  },
  {
    "text": "Let $F$ be the right focus of the hyperbola $\\Gamma$: $x^{2}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, $O$ the coordinate origin, and $P$, $Q$ the two intersection points of the circle with diameter $OF$ and the asymptotes of the hyperbola $\\Gamma$. If $|PQ|=|OF|$, then $b=$?",
    "fact_expressions": "F: Point;Gamma: Hyperbola;Expression(Gamma) = (x^2 - y^2/b^2 = 1);b: Number;b > 0;RightFocus(Gamma) = F;O: Origin;P: Point;Q: Point;Intersection(H, Asymptote(Gamma)) = {P, Q};H: Circle;IsDiameter(LineSegmentOf(O, F), H);Abs(LineSegmentOf(P, Q)) = Abs(LineSegmentOf(O, F))",
    "query_expressions": "b",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 4]], [[5, 52], [86, 97]], [[5, 52]], [[123, 126]], [[18, 52]], [[1, 56]], [[57, 60]], [[66, 69]], [[70, 73]], [[66, 105]], [[84, 85]], [[74, 85]], [[108, 121]]]",
    "query_spans": "[[[123, 128]]]",
    "process": "From the given, obtain the coordinates of point p and substitute into the asymptote equation. From |PQ| = |OF|, we get p(\\frac{c}{2},\\frac{c}{2}). Since point p lies on the asymptote y = \\frac{b}{a}x, \\therefore \\frac{c}{2} = \\frac{b}{a} \\cdot \\frac{c}{2} \\Rightarrow a = b. Given a = 1, \\therefore b = 1"
  },
  {
    "text": "If a chord of the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{3}=1$ is bisected by the point $(2,1)$, then what is the equation of the line on which this chord lies?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/12 + y^2/3 = 1);H: LineSegment;IsChordOf(H, G);I: Point;Coordinate(I) = (2, 1);MidPoint(H) = I",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "x+2*y-4=0",
    "fact_spans": "[[[1, 39]], [[1, 39]], [], [[1, 43]], [[44, 52]], [[44, 52]], [[1, 54]]]",
    "query_spans": "[[[1, 68]]]",
    "process": "Let the coordinates of the two endpoints of the chord be A(x_{1},y_{1}) and B(x_{2},y_{2}), then \\frac{x_{1}^{2}}{12}+\\frac{y_{1}^{2}}{3}=1, \\frac{x_{2}^{2}}{12}+\\frac{y_{2}^{2}}{3}=1. Subtracting these two equations yields \\underline{(x_{1}-x_{2})(x_{1}}-y_{2}). Thus, \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{x_{1}+x_{2}}{4(y_{1}+y_{2})}. The slope of the line containing chord AB is -\\frac{1}{2}, and the equation of the line is y-1=-\\frac{1}{2}(x-2). Simplifying gives x+2y-4=0, so the equation of the line containing the chord is x+2y-4=0."
  },
  {
    "text": "Find the standard equation of a parabola for which the distance from the focus to the directrix is $2$?",
    "fact_expressions": "G: Parabola;Distance(Focus(G), Directrix(G)) = 2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2 = pm*4*x, x^2 = pm*4*y}",
    "fact_spans": "[[[14, 17]], [[1, 17]]]",
    "query_spans": "[[[14, 23]]]",
    "process": "\\because the distance from the focus to the directrix is 2. By the definition of a parabola, we have p=2. Therefore, the standard equations of the required parabola are in the following four forms: y^{2}=4x, y^{2}=-4x, x^{2}=4y, x^{2}=-4y"
  },
  {
    "text": "Let $M$ be a point on the right branch of the hyperbola $\\frac{x^{2}}{25}-\\frac{y^{2}}{9}=1$, $F_{1}$ be the left focus, and $| M F_{1} |=18$. Let $N$ be the midpoint of segment $M F_{1}$, and $O$ be the coordinate origin. Then $| ON |$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/25 - y^2/9 = 1);M: Point;PointOnCurve(M, RightPart(G));F1: Point;LeftFocus(G) = F1;Abs(LineSegmentOf(M, F1)) = 18;N: Point;MidPoint(LineSegmentOf(M, F1)) = N;O: Origin",
    "query_expressions": "Abs(LineSegmentOf(O, N))",
    "answer_expressions": "4",
    "fact_spans": "[[[5, 44]], [[5, 44]], [[1, 4]], [[1, 51]], [[52, 59]], [[5, 63]], [[65, 81]], [[83, 86]], [[83, 101]], [[102, 105]]]",
    "query_spans": "[[[112, 122]]]",
    "process": ""
  },
  {
    "text": "Let the line $a x + b y + c = 0$ ($c \\neq 0$) intersect the parabola $y^{2} = 2x$ at points $P$ and $Q$, $F$ be the focus of the parabola, and lines $PF$, $QF$ intersect the parabola again at points $M$ and $N$, respectively. Then the equation of line $MN$ is?",
    "fact_expressions": "H: Line;G: Parabola;P: Point;Q: Point;F: Point;a: Number;b: Number;c: Number;Negation(c=0);M: Point;N: Point;Expression(H) = (a*x + b*y + c = 0);Expression(G) = (y^2 = 2*x);Intersection(H, G)={P,Q};Focus(G) = F;Intersection(LineOf(P, F), G) = M;Intersection(LineOf(Q, F), G) = N",
    "query_expressions": "Expression(LineOf(M , N))",
    "answer_expressions": "4*c*x-2*b*y+a=0",
    "fact_spans": "[[[1, 26]], [[27, 41], [57, 60], [78, 81]], [[43, 46]], [[47, 50]], [[53, 56]], [[3, 26]], [[3, 26]], [[3, 26]], [[3, 26]], [[82, 85]], [[86, 89]], [[1, 26]], [[27, 41]], [[1, 52]], [[53, 63]], [[64, 89]], [[64, 89]]]",
    "query_spans": "[[[91, 103]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. A line passing through $F_{2}$ intersects the ellipse at points $A$ and $B$, such that $A F_{1} \\perp A B$ and $\\overrightarrow{A B}=3 \\overrightarrow{F_{2} B}$. Then the eccentricity of ellipse $E$ is?",
    "fact_expressions": "E: Ellipse;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(E) = F1;RightFocus(E) = F2;G: Line;PointOnCurve(F2, G);A: Point;B: Point;Intersection(G, E) = {A, B};IsPerpendicular(LineSegmentOf(A, F1), LineSegmentOf(A, B));VectorOf(A, B) = 3*VectorOf(F2, B)",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[19, 76], [95, 97], [181, 186]], [[19, 76]], [[26, 76]], [[26, 76]], [[26, 76]], [[26, 76]], [[1, 8]], [[9, 16], [84, 91]], [[1, 82]], [[1, 82]], [[92, 94]], [[83, 94]], [[98, 101]], [[102, 105]], [[92, 107]], [[109, 128]], [[130, 179]]]",
    "query_spans": "[[[181, 192]]]",
    "process": "Let |AB| = 3m, |AF_{2}| = 2|F_{2}B|, so |AF_{2}| = 2m, |F_{2}B| = m, so |AF_{1}| = 2a - 2m, |BF_{1}| = 2a - m. \\because AF_{1} \\bot AB, then |BF_{1}|^{2} = |AF_{1}|^{2} + |AB|^{2}, i.e., (2a - m)^{2} = (3m)^{2} + (2a - 2m)^{2}, solving gives m = \\frac{1}{3}a. From |F_{1}F_{2}|^{2} = |AF_{1}|^{2} + |AF_{2}|^{2}, i.e., 4c^{2} = (2m)^{2} + (2a - 2m)^{2}, so 4c^{2} = \\frac{4}{9}a^{2} + \\frac{16}{9}a^{2}, then therefore, the eccentricity of this ellipse is \\frac{\\sqrt{5}}{2}."
  },
  {
    "text": "$P$ and $F$ are a point and the focus on the parabola $x^{2}=-4 y$, respectively. Given point $A(1,-2)$, find the coordinates of point $P$ such that $|P A|+|P F|$ is minimized.",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = -4*y);P: Point;PointOnCurve(P, G);F: Point;Focus(G) = F;A: Point;Coordinate(A) = (1, -2);WhenMin(Abs(LineSegmentOf(P, A))+Abs(LineSegmentOf(P, F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1, -1/4)",
    "fact_spans": "[[[10, 25]], [[10, 25]], [[0, 3], [66, 70]], [[0, 31]], [[4, 7]], [[0, 31]], [[34, 44]], [[34, 44]], [[47, 64]]]",
    "query_spans": "[[[66, 74]]]",
    "process": ""
  },
  {
    "text": "It is known that $F$ is the focus of the parabola $C$: $y^{2}=16x$, $M$ is a moving point on $C$, and point $A(4,1)$. Then the minimum value of $|MA|+|MF|$ is?",
    "fact_expressions": "C: Parabola;A: Point;M: Point;F: Point;Expression(C) = (y^2 = 16*x);Coordinate(A) = (4, 1);Focus(C) = F;PointOnCurve(M, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F)))",
    "answer_expressions": "8",
    "fact_spans": "[[[6, 26], [34, 37]], [[42, 51]], [[30, 33]], [[2, 5]], [[6, 26]], [[42, 51]], [[2, 29]], [[30, 41]]]",
    "query_spans": "[[[53, 72]]]",
    "process": "Substitute $ x = 4 $ into the parabola equation $ y^2 = 16x $, we get $ y = \\pm 8 $. Since $ 8 > 1 $, point $ A $ is inside the parabola $ C $. Draw a perpendicular line from point $ M $ to the directrix $ x = -4 $ of parabola $ C $, with foot point $ B $, so $ |MB| $. Therefore, $ |MA| + |MF| = |MA| + |MB| $. When $ AB $ is perpendicular to the line $ x = -4 $, and point $ M $ is the intersection point of segment $ AB $ and parabola $ C $, $ |MA| + |MF| $ attains its minimum value, which is $ 4 + 4 = 8 $."
  },
  {
    "text": "The line $l$ passes through the point $P(2 , 0)$ with slope $\\frac{4}{3}$, and intersects the parabola $y^{2}=2 x$ at two points $A$ and $B$. Let $M$ be the midpoint of segment $A B$. Then the coordinates of point $M$ are?",
    "fact_expressions": "l: Line;P: Point;Coordinate(P) = (2, 0);Slope(l) = 4/3;PointOnCurve(P, l) = True;G: Parabola;Expression(G) = (y^2 = 2*x);Intersection(l, G) = {A, B};B: Point;A: Point;MidPoint(LineSegmentOf(A, B)) = M;M: Point",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(41/16, 3/4)",
    "fact_spans": "[[[0, 5]], [[7, 18]], [[7, 18]], [[0, 35]], [[0, 18]], [[38, 52]], [[38, 52]], [[0, 66]], [[61, 64]], [[55, 58]], [[68, 82]], [[79, 82], [84, 88]]]",
    "query_spans": "[[[84, 93]]]",
    "process": ""
  },
  {
    "text": "If the asymptotes of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are tangent to the circle $(x-2)^{2}+y^{2}=1$, then the equations of the asymptotes of $C$ are?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y^2 + (x - 2)^2 = 1);IsTangent(G, Asymptote(C))",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y = pm*x*sqrt(3)/3",
    "fact_spans": "[[[1, 62], [91, 94]], [[9, 62]], [[9, 62]], [[67, 87]], [[9, 62]], [[9, 62]], [[1, 62]], [[67, 87]], [[1, 89]]]",
    "query_spans": "[[[91, 102]]]",
    "process": "The asymptote equation is: bx - ay = 0. Since the asymptote is tangent to the circle, 1 = \\frac{|2b - a \\times 0|}{\\sqrt{a^{2} + b^{2}}}, so \\frac{b}{a} = \\frac{\\sqrt{3}}{3}. Hence, the asymptote equation is y = \\pm\\frac{\\sqrt{3}}{2}x. Fill in y = \\pm\\frac{\\sqrt{3}}{2}x["
  },
  {
    "text": "Let $F$ be the focus of the parabola $C$: $y^{2}=16 x$. A line passing through $F$ with an inclination angle of $45^{\\circ}$ intersects $C$ at points $A$ and $B$. Then $|A B|$=?",
    "fact_expressions": "C: Parabola;G: Line;A: Point;B: Point;F: Point;Expression(C) = (y^2 = 16*x);Focus(C) = F;PointOnCurve(F, G);Inclination(G) = ApplyUnit(45, degree);Intersection(G, C) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "32",
    "fact_spans": "[[[5, 25], [54, 57]], [[51, 53]], [[58, 61]], [[62, 65]], [[1, 4], [30, 33]], [[5, 25]], [[1, 28]], [[29, 53]], [[34, 53]], [[51, 67]]]",
    "query_spans": "[[[69, 78]]]",
    "process": "\\because F is the focus of the parabola C: y^{2} = 16x, so F(4,0). Since a line passing through F with an inclination angle of 45^{\\circ} intersects C at points A and B, we can set the line l: y = x - 4. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), then:  \n\\begin{cases} y^{2} = 16x \\\\ y = x - 4 \\end{cases}  \nEliminating y gives: x^{2} - 24x + 16 = 0.  \n\\begin{cases} x_{1} + x_{2} = 24 \\end{cases}  \n\\therefore |_{AB|}^{x_{x_{2}} = 16} = \\sqrt{1 + k^{2}} \\sqrt{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}} = \\sqrt{2} \\sqrt{24^{2} - 4 \\times 16} = 32 (x_{1}x_{2} = 16), i.e., the chord length |AB| = 32"
  },
  {
    "text": "If the symmetry axes of an ellipse are the coordinate axes, the sum of the major axis length and minor axis length is $18$, and the focal distance is $6$, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;FocalLength(G) = 6;SymmetryAxis(G)=axis;Length(MajorAxis(G))+Length(MinorAxis(G))=18",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/15+y^2/16=1,x^2/16+y^2/25=1}",
    "fact_spans": "[[[1, 3], [35, 37]], [[1, 33]], [[1, 11]], [[1, 26]]]",
    "query_spans": "[[[35, 42]]]",
    "process": ""
  },
  {
    "text": "If the asymptotes of a hyperbola are given by $y = \\pm 3x$, and one focus is $(0, \\sqrt{10})$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (0, sqrt(10));Expression(Asymptote(G)) = (y = pm*(3*x));OneOf(Focus(G))=H",
    "query_expressions": "Expression(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 4], [46, 49]], [[28, 44]], [[28, 44]], [[1, 22]], [[1, 44]]]",
    "query_spans": "[[[46, 54]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ lies on the parabola $y^{2}=4x$, find the minimum value of the sum of the distance from point $P$ to line $L_{1}$: $4x-3y+6=0$ and the distance from point $P$ to line $L_{2}$: $x=-1$.",
    "fact_expressions": "G: Parabola;L1: Line;L2: Line;P: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(P, G);Expression(L1)=(4*x-3*y+6=0);Expression(L2)=(x=-1)",
    "query_expressions": "Min(Distance(P,L1)+Distance(P,L2))",
    "answer_expressions": "2",
    "fact_spans": "[[[7, 21]], [[29, 53]], [[58, 75]], [[2, 6], [2, 6]], [[7, 21]], [[2, 22]], [[29, 53]], [[58, 75]]]",
    "query_spans": "[[[24, 86]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the focus of the parabola $y=2 x^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 2*x^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,1/8)",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": "Since the parabola equation is $x^{2}=\\frac{1}{2}y$, the focus coordinates are $(0,\\frac{p}{2})$, and $p=\\frac{1}{4}$, so the focus coordinates are $(0,\\frac{1}{8})$."
  },
  {
    "text": "From point $M(2,-2p)$, draw two tangents to the parabola $x^{2}=2py$ $(p>0)$, with points of tangency $A$ and $B$. If the vertical coordinate of the midpoint of segment $AB$ is $6$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;B: Point;A: Point;M: Point;l1: Line;l2:Line;p>0;Expression(G) = (x^2=2*p*y);Coordinate(M) = (2,-2*p);TangentOfPoint(M,G)={l1,l2};TangentPoint(l1,G)=A;TangentPoint(l2,G)=B;YCoordinate(MidPoint(LineSegmentOf(A,B)))=6",
    "query_expressions": "p",
    "answer_expressions": "{1,2}",
    "fact_spans": "[[[14, 35]], [[75, 78]], [[50, 53]], [[46, 49]], [[1, 13]], [], [], [[17, 35]], [[14, 35]], [[1, 13]], [[0, 40]], [[0, 53]], [[0, 53]], [[55, 73]]]",
    "query_spans": "[[[75, 82]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Since $ y = \\frac{x^{2}}{2p} $, thus $ y' = \\frac{x}{p} $. Hence, $ MA: y = \\frac{x_{1}}{p}(x - x_{1}) + \\frac{x_{1}^{2}}{2p} = \\frac{x_{1}}{p}x - \\frac{x_{1}^{2}}{2p} $. Similarly, $ MB: y = \\frac{x_{2}}{p}x - \\frac{x_{2}^{2}}{2p} $. From \n$$\n\\begin{matrix}\ny = \\frac{x_{1}}{p}x - \\frac{x_{1}^{2}}{2p} & x_{1} + x_{2} \\\\\ny = \\frac{x_{2}}{p}x - \\frac{x_{2}^{2}}{2p} & \\text{and } y_{M}\n\\end{matrix}\n= \\frac{x_{M}}{2p}\n$$\nthus $ x_{1} + \\frac{x_{2}}{2} = 4\\frac{1}{2} $, and $ \\frac{x_{1}^{2}}{2p} + \\frac{x_{2}^{2}}{2p} = 12 $, hence $ x_{1}^{2} + x_{2}^{2} = 24p $, i.e., $ (x_{1} + x_{2})^{2} = 16 = 24p - 8p^{2} $, solving yields $ p = 1 $ or $ p = 2 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the right focus is $F$. If a line passing through point $F$ with an inclination angle of $60^{\\circ}$ intersects the right branch of the hyperbola at exactly one point, then what is the range of values for the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F:Point;RightFocus(G) = F;L: Line;PointOnCurve(F,L) = True;Inclination(L) = ApplyUnit(60,degree);NumIntersection(L,RightPart(G)) = 1",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[2,+oo)",
    "fact_spans": "[[[2, 58], [94, 97], [111, 114]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[63, 66], [69, 73]], [[2, 66]], [[91, 93]], [[68, 93]], [[74, 93]], [[91, 108]]]",
    "query_spans": "[[[111, 124]]]",
    "process": "According to the relationship between the line and the asymptote, we obtain $\\frac{b}{a}\\geqslant\\sqrt{3}$; then, by calculating the range of eccentricity, we get the answer. If a line $l$ passing through $F$ has exactly one intersection point with the right branch of the hyperbola, then the inclination angle of its asymptote with positive slope should be no less than the inclination angle of $l$. Given that the inclination angle of $l$ is $60^{\\circ}$, thus $\\frac{b}{a}>\\sqrt{3}$, hence $e=\\frac{c}{a}\\geqslant2$."
  },
  {
    "text": "Given that the minor axis of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has length $8$, the upper vertex is $A$, the left vertex is $B$, and $F_{1}$, $F_{2}$ are the left and right foci of the ellipse respectively, and the area of $\\Delta F_{1} A B$ is $4$. Let point $P$ be any point on the ellipse. Then the range of values of $\\frac{1}{|P F_{1}|}+\\frac{1}{|P F_{2}|}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Length(MinorAxis(G)) = 8;A: Point;UpperVertex(G) = A;B: Point;LeftVertex(G) = B;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;Area(TriangleOf(F1, A, B)) = 4;P: Point;PointOnCurve(P, G)",
    "query_expressions": "Range(1/Abs(LineSegmentOf(P, F2)) + 1/Abs(LineSegmentOf(P, F1)))",
    "answer_expressions": "[2/5, 5/8]",
    "fact_spans": "[[[2, 54], [97, 99], [138, 140]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[2, 62]], [[67, 70]], [[2, 70]], [[75, 78]], [[2, 78]], [[79, 86]], [[87, 94]], [[79, 105]], [[79, 105]], [[107, 132]], [[133, 137]], [[133, 146]]]",
    "query_spans": "[[[148, 196]]]",
    "process": "First, determine the values of $a$, $b$, and $c$ based on the area of $\\triangle F_{1}AB$ and the length of the minor axis, find the range of $|PF_{1}|$, then combine and simplify $\\frac{1}{|PF_{1}|} + \\frac{1}{|PF_{2}|}$ into a function in terms of $|PF_{1}|$, and use a quadratic function to find its maximum and minimum values, thus obtaining the range. Solution: From the given information, $2b = 8$, so $b = 4$. The area of $\\triangle F_{1}AB$ is $4$, $\\therefore \\frac{1}{2}(a - c)b = 4$, $\\therefore a - c = 2$. Also, $a^{2} - c^{2} = (a - c)(a + c) = b^{2} = 16$, hence $a + c = 8$, $\\therefore a = 5$, $c = 3$, $\\therefore \\frac{2a}{|PF_{1}|(2a - |PF_{1}|)} = \\frac{10}{-(|PF_{1}| - 5)^{2} + 25}$. Since $a - c \\leqslant |PF_{1}| \\leqslant a + c$, i.e., $2 \\leqslant |PF_{1}| \\leqslant 8$. Therefore, when $|PF_{1}| = 5$, $-(|PF_{1}| - 5)^{2} + 25$ reaches its maximum value of $25$; when $|PF_{1}| = 2$ or $8$, $-(|PF_{1}| - 5)^{2} + 25$ reaches its minimum value of $16$, i.e., $16 \\leqslant -(|PF_{1}| - 5)^{2} + 25 \\leqslant 25$. $\\therefore \\frac{10}{25} \\leqslant \\frac{1}{|PF_{1}|} + \\frac{1}{|PF_{2}|} \\leqslant \\frac{10}{16}$, i.e., $\\frac{2}{5} \\leqslant \\frac{1}{|PF_{1}|} + \\frac{1}{|PF_{2}|} \\leqslant \\frac{5}{8}$. Thus, the range of $\\frac{1}{|PF_{1}|} + \\frac{1}{|PF_{2}|}$ is $\\left[\\frac{2}{5}, \\frac{5}{8}\\right]$."
  },
  {
    "text": "Given that point $F_{1}$ is the left focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, a line $l$ passing through the origin intersects the ellipse at points $A$ and $B$, and $M$, $N$ are the midpoints of $A F_{1}$ and $B F_{1}$ respectively. If $O M \\perp O N$, then the range of the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;O: Origin;M: Point;N: Point;A: Point;F1: Point;B: Point;l:Line;LeftFocus(G) = F1;Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, F1)) = M;MidPoint(LineSegmentOf(B, F1)) = N;IsPerpendicular(LineSegmentOf(O, M), LineSegmentOf(O, N));PointOnCurve(O,l)",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[\\sqrt{2}/2,1)",
    "fact_spans": "[[[11, 63], [78, 80], [143, 145]], [[11, 63]], [[13, 63]], [[13, 63]], [[13, 63]], [[13, 63]], [[69, 71]], [[91, 94]], [[95, 98]], [[81, 84]], [[2, 10]], [[85, 88]], [[72, 77]], [[2, 67]], [[72, 90]], [[91, 124]], [[91, 124]], [[126, 141]], [[68, 77]]]",
    "query_spans": "[[[143, 154]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$, a line $l$ passes through the point $P(m, 0)$ $(m>0)$ and intersects $C$ at points $A$ and $B$. The line passing through point $A$ and the vertex $O$ of $C$ intersects the directrix of $C$ at point $D$. If $BD$ is parallel to the axis of symmetry of $C$, then $m=$?",
    "fact_expressions": "l: Line;C: Parabola;G: Line;B: Point;D: Point;P: Point;A: Point;O:Origin;Expression(C) = (y^2 = 8*x);Coordinate(P) = (m, 0);m:Number;m>0;PointOnCurve(P, l);Intersection(l, C) = {A, B};PointOnCurve(A,G);PointOnCurve(O,G);Vertex(C)=O;Intersection(G,Directrix(C))=D;IsParallel(LineSegmentOf(B,D),SymmetryAxis(C))",
    "query_expressions": "m",
    "answer_expressions": "2",
    "fact_spans": "[[[22, 27]], [[2, 21], [45, 48], [66, 69], [79, 82], [98, 101]], [[76, 78]], [[53, 56]], [[86, 90]], [[28, 43]], [[49, 52], [61, 65]], [[72, 75]], [[2, 21]], [[28, 43]], [[109, 112]], [[29, 43]], [[22, 43]], [[22, 58]], [[59, 78]], [[59, 78]], [[66, 75]], [[76, 90]], [[92, 107]]]",
    "query_spans": "[[[109, 114]]]",
    "process": "By the given condition, the directrix equation of parabola $ C $ is $ x = -2 $. Let $ A\\left(\\frac{y_0^2}{8}, y_0\\right) $ ($ y_0 \\neq 0 $), then the equation of line $ OA $ is $ y = \\frac{8}{y_0}x $. From \n$$\n\\begin{cases}\nx = -2, \\\\\ny = \\frac{8}{y_0}x,\n\\end{cases}\n$$\nwe obtain $ y_D = -\\frac{16}{y_0} $. The equation of line $ AB $ is $ y = \\frac{8y_0}{y_0^2 - 8m}(x - m) $. From \n$$\n\\begin{cases}\ny = \\frac{8y_0}{y_0^2 - 8m}(x - m), \\\\\ny^2 = 8x,\n\\end{cases}\n$$\nwe get $ y_0 y^2 - (y_0^2 - 8m)y - 8y_0 m = 0 $. Thus $ y_0 y_B = -8m $, yielding $ y_B = -\\frac{8m}{y_0} $. Since $ BD $ is parallel to the axis of symmetry of $ C $, we have $ y_B = y_D $, that is, $ -\\frac{16}{y_0} = -\\frac{8m}{y_0} $, hence $ m = 2 $."
  },
  {
    "text": "What is the focal distance of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{32}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/32 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "8",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 44]]]",
    "process": "From the ellipse equation $\\frac{x2}{16}+\\frac{y^{2}}{32}=1$, we get $a^{2}=32$, $b^{2}=16$, $\\therefore c^{2}=a^{2}-b^{2}=16$. $\\therefore c=4$, $2c=8$."
  },
  {
    "text": "If the asymptotes of the hyperbola $C$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0, b>0)$ are given by $y=\\pm \\frac{\\sqrt{5}}{2} x$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-x^2/b^2 + y^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Expression(Asymptote(C)) = (y = pm*(sqrt(5)*x)/2)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "3*sqrt(5)/5",
    "fact_spans": "[[[1, 62], [99, 102]], [[1, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[1, 97]]]",
    "query_spans": "[[[99, 108]]]",
    "process": "From the hyperbola equation $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$, it follows that its foci lie on the $y$-axis. Since one of its asymptotes is $y=\\frac{\\sqrt{5}}{2}x$, we have $\\frac{a}{b}=\\frac{\\sqrt{5}}{2}$, $\\frac{b}{a}=\\frac{\\sqrt[2]{5}}{5}$, and $e=\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\frac{3\\sqrt{5}}{5}$."
  },
  {
    "text": "Given a point $P(x, y)$ on the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{30}=1$ such that the distance from $P(x, y)$ to one of the foci of the hyperbola is $9$, what is the value of $x^{2}+y^{2}$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/30 = 1);P: Point;x1: Number;y1: Number;Coordinate(P) = (x1, y1);PointOnCurve(P, G);Distance(P, OneOf(Focus(G))) = 9",
    "query_expressions": "x1^2 + y1^2",
    "answer_expressions": "139",
    "fact_spans": "[[[2, 42], [55, 58]], [[2, 42]], [[45, 54]], [[45, 54]], [[45, 54]], [[45, 54]], [[2, 54]], [[45, 69]]]",
    "query_spans": "[[[71, 88]]]",
    "process": "Given $ e = \\frac{\\sqrt{46}}{4} $, then by the definition of the hyperbola we have $ ex - 4 = 9 $, that is, $ ex = 13 \\Rightarrow x = \\frac{13}{e} = \\frac{4 \\times 13}{\\sqrt{46}} $, $ x^{2} = \\frac{16 \\times 169}{46} = \\frac{8 \\times 169}{23} $. Substituting into the hyperbola equation yields $ 0(\\frac{1}{16} \\times \\frac{8 \\times 169}{23} - 1) = 30(\\frac{1}{2} \\times \\frac{169}{23} - 1) = \\frac{15 \\times 169}{23} - 30 $, hence $ 2 = \\frac{8 \\times 169}{23} + \\frac{15 \\times 169}{23} - 30 = 169 - 30 = 139 $, the answer to be filled is $ 139 $."
  },
  {
    "text": "Given that the asymptotes of a hyperbola are $y=\\pm \\frac{x}{2}$, and the foci lie on the coordinate axes with a focal distance of $10$, then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y=pm*x/2);PointOnCurve(Focus(G),axis) = True;FocalLength(G) = 10",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{(x^2/20-y^2/5=1),(y^2/5-x^2/20=1)}",
    "fact_spans": "[[[2, 5], [51, 54]], [[2, 31]], [[2, 40]], [[2, 48]]]",
    "query_spans": "[[[51, 59]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$, its two foci are $F_{1}$ and $F_{2}$. Point $P$ lies on this ellipse. If $|P F_{1}|-|P F_{2}|=2$, then the area of $\\Delta P_1 F_2 F_1$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2/2 = 1);Focus(G)={F1,F2};PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)) = 2",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 39], [68, 70]], [[62, 66]], [[45, 52]], [[53, 60]], [[2, 39]], [[2, 60]], [[62, 71]], [[72, 95]]]",
    "query_spans": "[[[98, 123]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{25}=1$ with left focus $F_{1}$, point $P$ lies on the right branch of the hyperbola such that $P F_{1}$ is tangent to the circle $x^{2}+y^{2}=16$ at point $N$, $M$ is the midpoint of segment $P F_{1}$, and $O$ is the origin. Then $|M N|-|M O|$=?",
    "fact_expressions": "G: Hyperbola;H: Circle;P: Point;F1: Point;M: Point;N: Point;O: Origin;Expression(G) = (x^2/16 - y^2/25 = 1);Expression(H) = (x^2 + y^2 = 16);LeftFocus(G) = F1;PointOnCurve(P, RightPart(G));TangentPoint(LineSegmentOf(P, F1), H) = N;MidPoint(LineSegmentOf(P, F1)) = M",
    "query_expressions": "Abs(LineSegmentOf(M, N)) - Abs(LineSegmentOf(M, O))",
    "answer_expressions": "-1",
    "fact_spans": "[[[2, 42], [60, 63]], [[80, 101]], [[56, 59]], [[47, 54]], [[109, 112]], [[104, 108]], [[128, 131]], [[2, 42]], [[80, 101]], [[2, 54]], [[56, 68]], [[70, 108]], [[109, 127]]]",
    "query_spans": "[[[138, 153]]]",
    "process": ""
  },
  {
    "text": "Given that the center of the ellipse is at the origin, one focus is $F(-2 \\sqrt{3}, 0)$, and the length of the major axis is twice the length of the minor axis, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;F: Point;O: Origin;Coordinate(F) = (-2*sqrt(3), 0);Center(G) = O;OneOf(Focus(G)) = F;Length(MajorAxis(G)) = 2*Length(MinorAxis(G))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16 + y^2/4 = 1",
    "fact_spans": "[[[2, 4], [52, 54]], [[16, 35]], [[8, 10]], [[16, 35]], [[2, 10]], [[2, 35]], [[2, 49]]]",
    "query_spans": "[[[52, 61]]]",
    "process": ""
  },
  {
    "text": "The right focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $F$, and the intersection point of its right directrix with the $x$-axis is $A$. There exists a point $P$ on the ellipse such that the perpendicular bisector of segment $AP$ passes through point $F$. Then the range of values for the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;a: Number;b: Number;A: Point;P: Point;F: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;Intersection(RightDirectrix(G), xAxis) = A;PointOnCurve(P,G);PointOnCurve(F,PerpendicularBisector(LineSegmentOf(A,P)))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[1/2,1)",
    "fact_spans": "[[[0, 52], [60, 61], [78, 80], [109, 111]], [[2, 52]], [[2, 52]], [[73, 76]], [[83, 87]], [[56, 59], [103, 107]], [[2, 52]], [[2, 52]], [[0, 52]], [[0, 59]], [[60, 76]], [[78, 87]], [[89, 107]]]",
    "query_spans": "[[[109, 121]]]",
    "process": "By the given condition, there exists a point P on the ellipse such that the perpendicular bisector of segment AP passes through point F. That is, the distance from F to P equals the distance from F to A. While |FA| = \\frac{a^{2}}{c} - c = \\frac{b^{2}}{c}, and |PF| \\in (a - c, a + c], thus \\frac{b^{2}}{c} \\in (a - c, a + c], i.e., ac - c^{2} < b^{2} \\leqslant ac + c^{2},  \n\\begin{cases} ac - c^{2} < a^{2} - c^{2} \\\\ a^{2} - c^{2} \\leqslant ac + c^{2} \\end{cases}  \n\\Rightarrow \\frac{c}{a} \\leqslant -1 \\text{ or } \\frac{c}{a} \\geqslant \\frac{1}{2}.  \nSince e \\in (0,1), it follows that e \\in [\\frac{1}{2}, 1)."
  },
  {
    "text": "Given that $P$ is a point on the parabola $C$: $x^{2}=m y$ ($m>0$), the distance from $P$ to the focus of $C$ is $1$, and the distance from $P$ to the $x$-axis is $\\frac{3}{4}$, then $m$=?",
    "fact_expressions": "C: Parabola;m: Number;P: Point;m>0;Expression(C) = (x^2 = m*y);PointOnCurve(P, C);Distance(P, Focus(C)) = 1;Distance(P, xAxis) = 3/4",
    "query_expressions": "m",
    "answer_expressions": "1",
    "fact_spans": "[[[6, 30], [39, 42]], [[76, 79]], [[2, 5], [34, 38]], [[13, 30]], [[6, 30]], [[2, 33]], [[34, 51]], [[34, 74]]]",
    "query_spans": "[[[76, 81]]]",
    "process": "The directrix of parabola C is given by $ y = -\\frac{m}{4} $. According to the definition of a parabola, the distance from point P to the directrix $ y = -\\frac{m}{4} $ equals 1, so $ \\left|0 - \\left(-\\frac{m}{4}\\right)\\right| = 1 - \\frac{3}{4} $, from which the value of $ m $ can be found. [Solution] From the parabola $ C: x^{2} = my $ ($ m > 0 $), we obtain the directrix of the parabola: $ y = -\\frac{m}{4} $. By the definition of a parabola, the distance from point P to the directrix $ y = -\\frac{m}{4} $ equals the distance to the focus. Thus, the distance from point P to the directrix $ y = -\\frac{m}{4} $ is 1. Since the distance from point P to the x-axis is $ \\frac{3}{4} $, i.e., the distance from point P to $ y = 0 $ is $ \\frac{3}{4} $, it follows that $ \\left|0 - \\left(-\\frac{m}{4}\\right)\\right| = 1 - \\frac{3}{4} $. Given $ m > 0 $, solving yields: $ m = 1 $."
  },
  {
    "text": "The line $y = kx - 2$ intersects the parabola $y^2 = 8x$ at points $A$ and $B$. If the horizontal coordinate of the midpoint of $AB$ is $2$, then the chord length $|AB|$ is?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*x - 2);k: Number;G: Parabola;Expression(G) = (y^2 = 8*x);Intersection(H, G) = {A, B};A: Point;B: Point;XCoordinate(MidPoint(LineSegmentOf(A, B))) = 2;IsChordOf(LineSegmentOf(A, B), G) = True",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "2*sqrt(15)",
    "fact_spans": "[[[0, 11]], [[0, 11]], [[2, 11]], [[12, 26]], [[12, 26]], [[0, 36]], [[27, 30]], [[31, 34]], [[39, 54]], [[12, 65]]]",
    "query_spans": "[[[58, 67]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. It is clear that $ k=0 $ does not satisfy the condition. Substituting the line $ y=kx-2 $ into the parabola equation yields $ k^{2}x^{2}-(4k+8)x+4=0 $, $ =(4k+8)^{2}-16k^{2}>0\\Rightarrow k>-1 $, $ x_{1}+x_{2}=\\frac{4k+8}{k^{2}} $, $ x_{1}x_{2}=\\frac{4}{k^{2}} $. Since the horizontal coordinate of the midpoint of $ AB $ is 2, we have $ \\begin{cases} x+x \\\\ \\frac{x+1}{2}=\\frac{2k+4}{k^{2}} \\end{cases}=2\\Rightarrow k=2 $, so $ \\begin{cases} x_{1}+x_{2}=4 \\\\ x_{1}x_{2}=1 \\end{cases} $, then $ |AB|=\\sqrt{1+4}|x_{1}-x_{2}|=\\sqrt{5}\\cdot\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=2\\sqrt{15} $"
  },
  {
    "text": "Given the line $l$: $y = x - 1$ intersects the parabola $C$: $y^{2} = 4x$ at two distinct points $A$, $B$, $M$ is the midpoint of $AB$, and the perpendicular bisector of segment $AB$ intersects the $x$-axis at point $N$, then the length of $MN$ is?",
    "fact_expressions": "l: Line;Expression(l) = (y = x - 1);C: Parabola;Expression(C) = (y^2 = 4*x);Intersection(l, C) = {A, B};A: Point;B: Point;M: Point;MidPoint(LineSegmentOf(A, B)) = M;Intersection(PerpendicularBisector(LineSegmentOf(A, B)), xAxis) = N;N: Point;Negation(A=B)",
    "query_expressions": "Length(LineSegmentOf(M, N))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 15]], [[2, 15]], [[16, 34]], [[16, 34]], [[2, 49]], [[42, 45]], [[46, 49]], [[50, 53]], [[50, 62]], [[63, 86]], [[82, 86]], [37, 48]]",
    "query_spans": "[[[88, 97]]]",
    "process": "From the equations of line $ l $ and the parabola, together with Vieta's formulas, we obtain $ x_{1}+x_{2}=6 $. Then, using the midpoint coordinate formula and the equation of line $ l $, we find $ M(3,2) $. Next, from the equation of the perpendicular bisector of segment $ AB $, we determine $ N(5,0) $. The distance between two points can then be calculated. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. From  \n$$\n\\begin{cases}\ny=x-1 \\\\\ny^{2}=4x\n\\end{cases}\n$$\nwe get $ x^{2}-6x+1=0 $, so $ x_{1}+x_{2}=6 $. Hence, $ M(3,2) $. The equation of the perpendicular bisector of segment $ AB $ is $ y-2=-(x-3) $, that is, $ y=-x+5 $. Thus, $ N(5,0) $.  \n$ MN=\\sqrt{(3-5)^{2}+(2-0)^{2}}=2\\sqrt{2} $"
  },
  {
    "text": "In the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, what is the slope of the line containing the chord for which point $M(-1,2)$ is the midpoint?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/9 = 1);H: LineSegment;IsChordOf(H, G);M: Point;Coordinate(M) = (-1, 2);MidPoint(H) = M",
    "query_expressions": "Slope(OverlappingLine(H))",
    "answer_expressions": "9/32",
    "fact_spans": "[[[0, 38]], [[0, 38]], [], [[0, 56]], [[41, 51]], [[41, 51]], [[0, 56]]]",
    "query_spans": "[[[0, 65]]]",
    "process": "Let the line intersect the ellipse at points A(x_{1},y_{1}), B(x_{2},y_{2}); substituting into the ellipse gives:  \n\\begin{cases}x_{1}^{2}+\\frac{y_{1}^{2}}{16}=1\\\\x_{2}^{2}+\\frac{y_{2}^{2}}{16}=1\\end{cases}  \nSubtracting the two equations and simplifying yields:  \n\\frac{x_{1}+x_{2}}{y_{1}+y_{2}}=-\\frac{16}{9}\\cdot\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}  \nSince M is the midpoint of AB, then \\frac{x_{1}+x_{2}}{2}=1, \\frac{y_{1}+y_{2}}{2}=1, hence \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{9}{32}, that is, the slope of the line is \\frac{9}{32}."
  },
  {
    "text": "Given that point $P(2,-3)$ lies on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and the distance between the two foci of the hyperbola is $4$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (2, -3);PointOnCurve(P, G);Focus(G)={F1, F2} ;Distance(F1, F2) = 4",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[13, 69], [73, 76], [92, 95]], [[16, 69]], [[16, 69]], [[2, 12]], [], [], [[16, 69]], [[16, 69]], [[13, 69]], [[2, 12]], [[2, 72]], [[73, 80]], [[73, 89]]]",
    "query_spans": "[[[92, 99]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left focus is $F$. The distance from point $F$ to an asymptote of the hyperbola $C$ is $\\frac{1}{2} a$. Then, what is the equation of the asymptotes of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;F: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;Distance(F, OneOf(Asymptote(C))) = a/2",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*(1/2)*x",
    "fact_spans": "[[[2, 63], [111, 117], [77, 83]], [[93, 108]], [[10, 63]], [[72, 76]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 71]], [[72, 108]]]",
    "query_spans": "[[[111, 125]]]",
    "process": "From the hyperbola equation, its asymptotes are given by: $ y = \\pm\\frac{b}{a}x $. Since $ F(-c,0) $, the distance from point $ F $ to one of the asymptotes of hyperbola $ C $ is $ d = \\frac{|bc|}{\\sqrt{a^{2}+b^{2}}} = b = \\frac{1}{2}a $, therefore $ \\frac{b}{a} = \\frac{1}{2} $. Thus, the asymptotes of hyperbola $ C $ are: $ y = \\pm\\frac{1}{2}x $."
  },
  {
    "text": "The distance from the focus of the parabola ${y}^{2}=8 x$ to the asymptotes of the hyperbola $\\frac{x^{2}}{8}-\\frac{y^{2}}{2}=1$ is?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 8*x);G: Hyperbola;Expression(G) = (x^2/8 - y^2/2 = 1)",
    "query_expressions": "Distance(Focus(H),Asymptote(G))",
    "answer_expressions": "(2*sqrt(5))/5",
    "fact_spans": "[[[0, 16]], [[0, 16]], [[20, 58]], [[20, 58]]]",
    "query_spans": "[[[0, 67]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, a line passing through point $F$ with slope $1$ intersects the parabola $C$ at points $A$ and $B$. On the circle $E$ with diameter $AB$, there exist points $P$ and $Q$ such that the circle with diameter $PQ$ passes through point $D(-2, t)$. Then the range of real values for $t$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;Z: Line;PointOnCurve(F, Z);Slope(Z) = 1;A: Point;B: Point;Intersection(Z, C) = {A, B};E: Circle;IsDiameter(LineSegmentOf(A, B), E);P: Point;Q: Point;PointOnCurve(P, E);PointOnCurve(Q, E);U: Circle;IsDiameter(LineSegmentOf(P, Q), U);D: Point;t: Real;Coordinate(D) = (-2, t);PointOnCurve(D, U)",
    "query_expressions": "Range(t)",
    "answer_expressions": "[-1, 3]",
    "fact_spans": "[[[2, 21], [45, 51]], [[2, 21]], [[25, 28], [30, 34]], [[2, 28]], [[42, 44]], [[29, 44]], [[35, 44]], [[53, 57]], [[58, 61]], [[42, 61]], [[74, 78]], [[62, 78]], [[81, 85]], [[86, 89]], [[74, 85]], [[74, 89]], [[102, 103]], [[92, 103]], [[104, 115]], [[117, 122]], [[104, 115]], [[102, 115]]]",
    "query_spans": "[[[117, 129]]]",
    "process": "From the given conditions, the equation of line AB is $ x = y + 1 $. Solving the system of equations  \n\\[\n\\begin{cases}\nx = y + 1 \\\\\ny^2 = 4x\n\\end{cases}\n\\]\nyields $ y^2 - 4y - 4 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ y_{1} + y_{2} = 4 $, $ y_{1}y_{2} = -4 $. Let $ E(x_{E}, y_{E}) $, then $ y_{E} = \\frac{y_{1} + y_{2}}{2} = 2 $, $ x_{E} = y_{E} + 1 = 3 $. Also, $ |AB| = x_{1} + x_{2} + 2 = y_{1} + 1 + y_{2} + 1 + 2 = 8 $. Therefore, circle $ E $ has center $ (3, 2) $ and radius 4. Thus, point $ D $ is always outside circle $ E $, and there exist points $ P, Q $ on circle $ E $ such that the circle with diameter $ PQ $ passes through point $ D(-2, t) $, i.e., there exist points $ P, Q $ on circle $ E $ such that $ DP \\perp DQ $. Let two lines through $ D $ be tangent to circle $ E $ at points $ P' $, $ Q' $. To satisfy the condition, we require $ \\angle P'DQ' \\geqslant \\frac{\\pi}{2} $, so  \n\\[\n\\frac{|EP'|}{|DE|} = \\frac{4}{\\sqrt{(3+2)^2 + (2-t)^2}} \\geqslant \\frac{\\sqrt{2}}{2}\n\\]  \nRearranging gives $ t^2 - 4t - 3 \\leqslant 0 $, solving yields $ 2 - \\sqrt{7} \\leqslant t \\leqslant 2 + \\sqrt{7} $. Hence, the range of real number $ t $ is $ [2 - \\sqrt{7}, 2 + \\sqrt{7}] $."
  },
  {
    "text": "The curve $3 x^{2}+k y^{2}=6$ represents an ellipse with foci on the $x$-axis. What is the range of real values for $k$?",
    "fact_expressions": "G: Ellipse;H: Curve;k: Real;Expression(H) = (k*y^2 + 3*x^2 = 6);H = G;PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(3, +oo)",
    "fact_spans": "[[[32, 34]], [[0, 21]], [[36, 41]], [[0, 21]], [[0, 34]], [[23, 34]]]",
    "query_spans": "[[[36, 48]]]",
    "process": "According to the problem, $3x^{2}+ky^{2}=6$ can be rewritten in standard form as $\\frac{x^{2}}{2}+\\frac{y^{2}}{\\frac{6}{k}}=1$. Since it represents an ellipse with foci on the $x$-axis, we have $2>\\frac{6}{k}>0$, solving this gives $k>3$; therefore, the range of real values for $k$ is $(3,+\\infty)$."
  },
  {
    "text": "Given that the line $l$ passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) intersects the parabola at points $A$ and $B$, and $|A F|=\\frac{3}{2}|B F|$, then the slope $k$ of the line $l$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;l: Line;PointOnCurve(F, l);Intersection(l, G) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(A, F)) = (3/2)*Abs(LineSegmentOf(B, F));k: Number;Slope(l) = k",
    "query_expressions": "k",
    "answer_expressions": "pm*2*sqrt(6)",
    "fact_spans": "[[[3, 24], [36, 39]], [[3, 24]], [[6, 24]], [[6, 24]], [[26, 29]], [[3, 29]], [[30, 35], [78, 83]], [[2, 35]], [[30, 50]], [[41, 44]], [[45, 48]], [[52, 76]], [[86, 89]], [[78, 89]]]",
    "query_spans": "[[[86, 91]]]",
    "process": "When $A(x_{0},y_{0})$ is in the first quadrant, the inclination angle of the line is $\\theta$, $\\theta\\in(0,\\frac{\\pi}{2})$, we obtain $y_{0}=|AF|\\sin\\theta$, $x_{0}=\\frac{p}{2}+|AF|\\cos\\theta$. Substituting into $y^{2}=2px$ $(p>0)$, we get $|AF|^{2}\\sin^{2}\\theta x^{2}-2p\\cos\\theta|AF|x-p^{2}=0$. Using the quadratic formula to solve for $|AF|$, $|BF|$, and then using $|AF|=\\frac{3}{2}|BF|$ to solve. Similarly, when point $A(x_{0},y_{0})$ is in the fourth quadrant, solve by symmetry of the parabola. [Solution] Let the inclination angle of the line be $\\theta$. When point $A(x_{0},y_{0})$ is in the first quadrant, from the given conditions: $\\theta\\in(0,\\frac{\\pi}{2})$, then $y_{0}=|AF|\\sin\\theta$, $x_{0}=\\frac{p}{2}+|AF|\\cos\\theta$. Substituting into $y^{2}=2px$ $(p>0)$, we obtain: $|AF|^{2}\\sin^{2}\\theta x^{2}-2p\\cos\\theta|AF|x-p^{2}=0$, so $|AF|=\\frac{p}{1-\\cos\\theta}$, similarly $|BF|=\\frac{p}{1+\\cos\\theta}$. Therefore, $\\frac{|AF|}{|BF|}=\\frac{1+\\cos\\theta}{1-\\cos\\theta}=\\frac{3}{2}$, solving gives $\\cos\\theta=\\frac{1}{5}$, then $\\sin\\theta=\\frac{2\\sqrt{6}}{5}$, so $k=\\tan\\theta=2\\sqrt{6}$. When point $A(x_{0},y_{0})$ is in the fourth quadrant, from the given conditions: $\\theta\\in(\\frac{\\pi}{2},\\pi)$, by symmetry of the parabola, $k=\\tan\\theta=-2\\sqrt{6}$."
  },
  {
    "text": "Draw a tangent line $FM$ (with $M$ as the point of tangency) from the right focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) to the circle $x^{2}+y^{2}=a^{2}$, intersecting the $y$-axis at point $P$. If $M$ is the midpoint of segment $FP$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "F: Point;P: Point;G: Hyperbola;a: Number;b: Number;H: Circle;M: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + y^2 = a^2);RightFocus(G) = F;TangentOfPoint(F, H) = LineOf(F, M);TangentPoint(LineOf(F,M),H)= M ;Intersection(LineOf(F, M), yAxis) = P;MidPoint(LineSegmentOf(F, P)) = M",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[61, 64]], [[108, 112]], [[1, 57], [129, 132]], [[4, 57]], [[4, 57]], [[65, 85]], [[96, 99], [114, 117]], [[4, 57]], [[4, 57]], [[1, 57]], [[65, 85]], [[1, 64]], [[0, 92]], [[65, 100]], [[88, 112]], [[114, 127]]]",
    "query_spans": "[[[129, 138]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $x^{2}-\\frac{y^{2}}{24}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the left branch of the hyperbola and satisfies: $|P F_{1}|=\\frac{3}{5}|F_{1} F_{2}|$. The area of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2/24 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, LeftPart(G));Abs(LineSegmentOf(P, F1)) = (3/5)*Abs(LineSegmentOf(F1, F2))",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "24",
    "fact_spans": "[[[2, 31], [60, 63]], [[55, 59]], [[39, 46]], [[47, 54]], [[2, 31]], [[2, 54]], [[2, 54]], [[55, 68]], [[73, 109]]]",
    "query_spans": "[[[112, 139]]]",
    "process": "According to the problem, $ a=1 $, $ c=5 $, so $ |PF_{1}| = \\frac{3}{5}|F_{1}F_{2}| = 6 $. By the definition of hyperbola, $ |PF_{2}| = 2a + |PF_{1}| = 8 $. By the law of cosines and since $ |F_{1}F_{2}| = 10 $, triangle $ PF_{1}F_{2} $ is a right triangle, so the area is $ \\frac{1}{2} \\times 6 \\times 8 = 24 $."
  },
  {
    "text": "The equation of the hyperbola that has the same foci as $x^{2}-\\frac{y^{2}}{4}=1$ and passes through the point $(2 , \\sqrt{3})$ is?",
    "fact_expressions": "G: Hyperbola;C:Curve;H: Point;Expression(C)=(x^2 - y^2/4 = 1);Coordinate(H) = (2, sqrt(3));Focus(G) = Focus(C);PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/2 - y^2/3 = 1",
    "fact_spans": "[[[53, 56]], [[1, 26]], [[35, 52]], [[1, 26]], [[35, 52]], [[0, 56]], [[34, 56]]]",
    "query_spans": "[[[53, 60]]]",
    "process": ""
  },
  {
    "text": "An ellipse $\\frac{x^{2}}{m^{2}} + \\frac{y^{2}}{3-m}=1$ has one focus at $(0,1)$, then what is the value of $m$?",
    "fact_expressions": "G: Ellipse;m: Number;H: Point;Expression(G) = (y^2/(3 - m) + x^2/m^2 = 1);Coordinate(H) = (0, 1);OneOf(Focus(G)) = H",
    "query_expressions": "m",
    "answer_expressions": "{-2,1}",
    "fact_spans": "[[[0, 45]], [[60, 63]], [[51, 58]], [[0, 45]], [[51, 58]], [[0, 58]]]",
    "query_spans": "[[[60, 66]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, a line $l$ passing through $F$ intersects $C$ at points $A$ and $B$. Perpendiculars are drawn from $A$ and $B$ to the directrix $Z$, with feet of perpendiculars $P$ and $Q$, respectively. If $|A F|=3|B F|$, then $|P Q|=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);A: Point;B: Point;F: Point;P: Point;Q: Point;Focus(C) = F;l: Line;PointOnCurve(F, l);Intersection(l, C) = {A, B};L1: Line;L2: Line;Z: Line;Directrix(C) = Z;PointOnCurve(A,L1);PointOnCurve(B,L2);IsPerpendicular(Z,L1);IsPerpendicular(Z,L2);FootPoint(Z,L1) = P;FootPoint(Z,L2) = Q;Abs(LineSegmentOf(A, F)) = 3*Abs(LineSegmentOf(B, F))",
    "query_expressions": "Abs(LineSegmentOf(P, Q))",
    "answer_expressions": "8*sqrt(3)/3",
    "fact_spans": "[[[1, 20], [39, 42]], [[1, 20]], [[43, 46], [56, 59]], [[47, 50], [60, 63]], [[24, 27], [29, 32]], [[77, 80]], [[81, 84]], [[1, 27]], [[33, 38]], [[28, 38]], [[33, 52]], [], [], [[66, 69]], [[39, 69]], [[53, 72]], [[53, 72]], [[53, 72]], [[53, 72]], [[53, 84]], [[53, 84]], [[87, 101]]]",
    "query_spans": "[[[103, 112]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the slope of one asymptote is $\\frac{3}{4}$, and the focal length is $10$. Then the equation of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Slope(OneOf(Asymptote(C))) = 3/4;FocalLength(C) = 10",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/16 - y^2/9 = 1",
    "fact_spans": "[[[2, 63], [96, 102]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 86]], [[2, 94]]]",
    "query_spans": "[[[96, 107]]]",
    "process": "From \\begin{cases}\\frac{b}{a}=\\frac{3}{4}\\\\2c=10\\\\c^{2}=a^{2}+b^{2}\\end{cases}, we obtain \\begin{cases}a=4\\\\b=3\\\\c=5\\end{cases}, hence the equation of the hyperbola is: \\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1"
  },
  {
    "text": "The hyperbola $C$ is centered at the origin, with foci on the $x$-axis, eccentricity $e=\\frac{\\sqrt{6}}{2}$, and the distance from its foci to the asymptotes is $1$. Then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;O: Origin;Center(C) = O;PointOnCurve(Focus(C), xAxis);e: Number;Eccentricity(C) = e;e = sqrt(6)/2;Distance(Focus(C), Asymptote(C)) = 1",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/2-y^2=1",
    "fact_spans": "[[[0, 6], [63, 66], [47, 48]], [[10, 12]], [[0, 12]], [[0, 21]], [[25, 47]], [[0, 47]], [[25, 47]], [[47, 61]]]",
    "query_spans": "[[[63, 71]]]",
    "process": ""
  },
  {
    "text": "Given that point $A$ is the intersection of the axis of symmetry and the directrix of the parabola $y=\\frac{1}{4} x^{2}$, point $F$ is the focus of this parabola, and point $P$ lies on the parabola such that $|P F|=m|P A|$. When $m$ takes its minimum value, point $P$ lies exactly on a hyperbola with foci at $A$ and $F$. Then, the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;P: Point;F: Point;A: Point;Expression(H) = (y = x^2/4);Intersection(SymmetryAxis(H),Directrix(H))=A;Focus(H) = F;PointOnCurve(P, H);Abs(LineSegmentOf(P, F)) = m*Abs(LineSegmentOf(P, A));WhenMin(m);m:Number;Focus(G)={A,F};PointOnCurve(P, G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)+1",
    "fact_spans": "[[[112, 115], [119, 122]], [[7, 31], [36, 37], [49, 52], [61, 64]], [[56, 60], [93, 97]], [[43, 47], [105, 108]], [[2, 6], [101, 104]], [[7, 31]], [[2, 42]], [[43, 55]], [[56, 65]], [[68, 82]], [[83, 92]], [[84, 87]], [[100, 115]], [[93, 116]]]",
    "query_spans": "[[[119, 128]]]",
    "process": "By combining algebra and geometry, when $ m $ takes its minimum value, the line $ PA $ is tangent to the parabola. The coordinates of point $ P $ can then be obtained. According to the definition of a hyperbola, the result follows. As shown in the figure: draw $ PB $ perpendicular to the directrix, intersecting at point $ B $, then $ |PF| = |PB| $. Therefore, $ m = \\frac{|PF|}{|PA|} = \\frac{|PB|}{|PA|} = \\sin\\angle PAB $. Hence, $ m $ is minimized when line $ PA $ is tangent to the parabola. Let the equation of line $ PA $ be: $ y = kx - 1 $. Then  \n$$\n\\begin{cases}\ny = kx - 1 \\\\\nx^{2} = 4y\n\\end{cases}\n\\Rightarrow x^{2} - 4kx + 4 = 0\n$$  \nSo $ \\Delta = 0 $, i.e., $ k = \\pm 1 $. Without loss of generality, let $ k = 1 $. Then we obtain $ P(2,1) $, so $ |PA| = 2\\sqrt{2} $, $ |PF| = 2 $. Then $ 2a = |PA| - |PF| \\Rightarrow a = \\sqrt{2} - 1 $. Thus, $ e = \\frac{c}{a} = \\sqrt{2} + 1 $."
  },
  {
    "text": "It is known that the directrix of the parabola $y^{2}=2 px(p>0)$ is tangent to the circle $(x-3)^{2}+y^{2}=16$. What is the value of $p$?",
    "fact_expressions": "G: Parabola;p: Number;H: Circle;p>0;Expression(G) = (y^2 = 2*p*x);Expression(H) = (y^2 + (x - 3)^2 = 16);IsTangent(Directrix(G),H)",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 22]], [[51, 54]], [[26, 47]], [[5, 22]], [[2, 22]], [[26, 47]], [[2, 49]]]",
    "query_spans": "[[[51, 58]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=2 p x$ with focus $F$, let $M$ be the intersection point of the directrix and the $x$-axis, and let $N$ be a point on the parabola satisfying $|M N|=2|N F|$. Then $\\angle N M F$=?",
    "fact_expressions": "G: Parabola;p: Number;M: Point;N: Point;F: Point;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;Intersection(Directrix(G), xAxis) = M;PointOnCurve(N, G);Abs(LineSegmentOf(M, N)) = 2*Abs(LineSegmentOf(N, F))",
    "query_expressions": "AngleOf(N, M, F)",
    "answer_expressions": "pi/3",
    "fact_spans": "[[[2, 18], [45, 48]], [[5, 18]], [[37, 40]], [[41, 44]], [[22, 25]], [[2, 18]], [[2, 25]], [[2, 40]], [[41, 52]], [[56, 70]]]",
    "query_spans": "[[[72, 88]]]",
    "process": "Draw MA perpendicular to the directrix from point M, meeting at point A. By the definition of a parabola, we have: |NF| = |NA|. Also, |MN| = 2|NF| = 2|NA|, so $\\sin\\angle AMN = \\frac{|AN|}{|MN|} = \\frac{1}{2}$. Therefore, $\\angle AMN = \\frac{\\pi}{6}$, so $\\angle NMF = \\frac{\\pi}{3}$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{25}=1$, the two foci are denoted as $F_{1}$ and $F_{2}$. A line passing through point $F_{1}$ intersects the ellipse at points $A$ and $B$. What is the perimeter of triangle $A B F_{2}$?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 + y^2/25 = 1);Focus(G) = {F1,F2};PointOnCurve(F1, H);Intersection(H, G) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "20",
    "fact_spans": "[[[2, 40], [77, 79]], [[74, 76]], [[80, 83]], [[84, 87]], [[48, 55], [65, 73]], [[56, 63]], [[2, 40]], [[2, 63]], [[64, 76]], [[74, 89]]]",
    "query_spans": "[[[91, 110]]]",
    "process": "According to the problem, the equation of the ellipse is $\\frac{x^{2}}{9}+\\frac{y^{2}}{25}=1$, so $a=5$, therefore the perimeter of triangle $ABF_{2}$ is $4a=20$."
  },
  {
    "text": "Let $F$ be the focus of the parabola $x^{2}=8 y$, and let points $A$, $B$, $C$ lie on this parabola. If $\\overrightarrow{F A}+\\overrightarrow{F B}+\\overrightarrow{F C}=\\overrightarrow{0}$, then $|\\overrightarrow{F A}|+|\\overrightarrow{F B}|+|\\overrightarrow{F C}|$=?",
    "fact_expressions": "G: Parabola;F: Point;A: Point;B: Point;C: Point;Expression(G) = (x^2 = 8*y);Focus(G) = F;PointOnCurve(A, G);PointOnCurve(B, G);PointOnCurve(C, G);VectorOf(F, A) + VectorOf(F, B) + VectorOf(F, C) = 0",
    "query_expressions": "Abs(VectorOf(F, A)) + Abs(VectorOf(F, B)) + Abs(VectorOf(F, C))",
    "answer_expressions": "12",
    "fact_spans": "[[[5, 19], [37, 40]], [[1, 4]], [[23, 27]], [[28, 31]], [[32, 35]], [[5, 19]], [[1, 22]], [[23, 41]], [[28, 41]], [[32, 41]], [[43, 126]]]",
    "query_spans": "[[[128, 200]]]",
    "process": "From the given conditions, the parabola's focus is $ F(0,2) $, and the directrix is $ y = -2 $. From the condition, $ F $ is the centroid of triangle $ ABC $, so $ 2 = \\frac{y_{1}+y_{2}+y_{3}}{3} $. From the definition of the parabola, we obtain the result. [Solution] From the given conditions, $ p = 4 $, the focus is $ F(0,2) $, and the directrix is $ y = -2 $. Since $ \\overrightarrow{FA} + \\overrightarrow{FB} + \\overrightarrow{FC} = \\overrightarrow{0} $, $ F $ is the centroid of triangle $ ABC $. Let the vertical coordinates of $ A $, $ B $, $ C $ be $ y_{1} $, $ y_{2} $, $ y_{3} $, respectively. $ \\therefore 2 = \\frac{y_{1}+y_{2}+y_{3}}{3} $, so $ y_{1}+y_{2}+y_{3} = 6 $. From the definition of the parabola, $ |\\overrightarrow{FA}| + |\\overrightarrow{FB}| + |\\overrightarrow{FC}| = (y_{1}+2) + (y_{2}+2) + (y_{3}+2) = 12 $. Hence, the answer is $ 12 $?"
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $x^{2}=8 y$, $P$ is a point on $C$, and $M(-4,3)$, then the minimum value of $|P F|+|P M|$ is?",
    "fact_expressions": "C: Parabola;M: Point;P: Point;F: Point;Expression(C) = (x^2 = 8*y);Coordinate(M) = (-4, 3);Focus(C) = F;PointOnCurve(P, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F)) + Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "5",
    "fact_spans": "[[[6, 25], [33, 36]], [[40, 49]], [[29, 32]], [[2, 5]], [[6, 25]], [[40, 49]], [[2, 28]], [[29, 39]]]",
    "query_spans": "[[[51, 70]]]",
    "process": "From the equation of the parabola, we obtain the focus of the parabola $ F(0,2) $. According to the given condition, $ M $ lies inside the parabola. Connect $ MF $. Draw a perpendicular from $ M $ to the directrix $ y = -2 $, intersecting the parabola at $ P $, with foot of perpendicular $ N $. By the definition of the parabola, $ |PF| = |PN| $. Therefore, $ |PF| + |PM| = |PM| + |PN| \\geqslant |MN| = 3 + 2 = 5 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, the left and right vertices are denoted by $A$ and $B$ respectively. Point $P$ lies on the hyperbola $C$. If $\\angle P B A = \\angle P A B + \\frac{\\pi}{2}$, then what is the focal distance of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/4 - y^2/b^2 = 1);b: Number;P: Point;B: Point;A: Point;b>0;LeftVertex(C) = A;RightVertex(C) = B;PointOnCurve(P, C);AngleOf(P, B, A) = pi/2 + AngleOf(P, A, B)",
    "query_expressions": "FocalLength(C)",
    "answer_expressions": "4*sqrt(2)",
    "fact_spans": "[[[2, 54], [76, 82], [128, 134]], [[2, 54]], [[10, 54]], [[71, 75]], [[67, 70]], [[63, 66]], [[10, 54]], [[2, 70]], [[2, 70]], [[71, 83]], [[85, 126]]]",
    "query_spans": "[[[128, 139]]]",
    "process": "From $\\angle PBA = \\angle PAB + \\frac{\\pi}{2}$, it follows that $k_{PA} \\cdot k_{PB} = 1$. Let $P(x_{0}, y_{0})$, then $\\frac{y_{0}}{x_{0}+2} \\cdot \\frac{y_{0}}{x_{0}-2} = \\frac{y_{0}^{2}}{x_{0}^{2}-4} = 1$. Since point $P$ lies on hyperbola $C$, $\\therefore \\frac{x_{0}^{2}}{4} - \\frac{y_{0}^{2}}{b^{2}} = 1$, $\\frac{y_{0}^{2}}{x_{0}^{2}-4} = \\frac{b^{2}}{4}$. $\\therefore \\frac{b^{2}}{4} = 1$, i.e., $b = 2$, then the focal distance is $2\\sqrt{4+4} = 4\\sqrt{2}$."
  },
  {
    "text": "$P$ is a moving point on the right branch of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and $M$, $N$ are moving points on the circles $(x+5)^{2}+y^{2}=4$ and $(x-5)^{2}+y^{2}=1$, respectively. Then the maximum value of $|P M|-|P N|$ is?",
    "fact_expressions": "G: Hyperbola;Z1: Circle;Z2: Circle;P: Point;M: Point;N: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Expression(Z1) = (y^2 + (x + 5)^2 = 4);Expression(Z2) = (y^2 + (x - 5)^2 = 1);PointOnCurve(P, RightPart(G));PointOnCurve(M, Z1);PointOnCurve(N, Z2)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "9",
    "fact_spans": "[[[4, 43]], [[61, 81]], [[82, 101]], [[0, 3]], [[51, 54]], [[55, 58]], [[4, 43]], [[61, 81]], [[82, 101]], [[0, 50]], [[51, 105]], [[51, 105]]]",
    "query_spans": "[[[107, 126]]]",
    "process": ""
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $C$: $y^{2}=2 p x$ ($p>0$) intersects the parabola at points $A$ and $B$. If $|A F|=4|B F|$ ($O$ is the origin), then $\\frac{|A F|}{|O F|}$=?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;A: Point;F: Point;B: Point;O: Origin;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(F, G);Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, F)) = 4*Abs(LineSegmentOf(B, F))",
    "query_expressions": "Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(O, F))",
    "answer_expressions": "5",
    "fact_spans": "[[[1, 27], [37, 40]], [[9, 27]], [[34, 36]], [[41, 44]], [[30, 33]], [[45, 48]], [[67, 70]], [[9, 27]], [[1, 27]], [[1, 33]], [[0, 36]], [[34, 50]], [[52, 66]]]",
    "query_spans": "[[[78, 101]]]",
    "process": ""
  },
  {
    "text": "Given that the major axis length of ellipse $M$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $4 \\sqrt{2}$, and it has the same eccentricity as the ellipse $\\frac{x^{2}}{2}+\\frac{y^{2}}{4}=1$. Find the equation of ellipse $M$?",
    "fact_expressions": "M: Ellipse;b: Number;a: Number;a > b;b > 0;Expression(M) = (y^2/b^2 + x^2/a^2 = 1);Length(MajorAxis(M))=4*sqrt(2);G:Ellipse;Expression(G)=(y^2/4 + x^2/2= 1);Eccentricity(M)=Eccentricity(G)",
    "query_expressions": "Expression(M)",
    "answer_expressions": "x^2/8+y^2/4=1",
    "fact_spans": "[[[2, 59], [125, 130]], [[8, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[2, 59]], [[2, 76]], [[79, 116]], [[79, 116]], [[2, 123]]]",
    "query_spans": "[[[125, 134]]]",
    "process": "Given that the eccentricity of the ellipse is \\frac{\\sqrt{2}}{2}, and for the unknown ellipse, 2a=4\\sqrt{2}, so a=2\\sqrt{2}, hence \\frac{c}{a}=\\frac{c}{2,b}=\\frac{\\sqrt{2}}{2}, c=2, b^{2}=a^{2}-c^{2}=4, therefore the required equation of the ellipse is \\frac{x^{2}}{0}+\\frac{y^{2}}{4}=1"
  },
  {
    "text": "The line $y = kx - 2$ intersects the parabola $y^2 = 8x$ at points $A$ and $B$, and the horizontal coordinate of the midpoint of $AB$ is $2$. Then the value of $k$ is?",
    "fact_expressions": "G: Parabola;H: Line;k: Number;A: Point;B: Point;Expression(G) = (y^2 = 8*x);Expression(H) = (y = k*x - 2);Intersection(H, G) = {A, B};XCoordinate(MidPoint(LineSegmentOf(A, B))) = 2",
    "query_expressions": "k",
    "answer_expressions": "2",
    "fact_spans": "[[[12, 26]], [[0, 11]], [[56, 59]], [[28, 31]], [[32, 35]], [[12, 26]], [[0, 11]], [[0, 37]], [[39, 54]]]",
    "query_spans": "[[[56, 63]]]",
    "process": ""
  },
  {
    "text": "If the line $y=kx+1$ intersects the curve $x=\\sqrt{y^{2}+1}$ at two distinct points, then what is the range of values for $k$?",
    "fact_expressions": "G: Line;k: Number;H: Curve;Expression(G) = (y = k*x + 1);Expression(H) = (x = sqrt(y^2 + 1));NumIntersection(G, H) = 2",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-sqrt(2), -1)",
    "fact_spans": "[[[1, 11]], [[42, 45]], [[12, 32]], [[1, 11]], [[12, 32]], [[1, 40]]]",
    "query_spans": "[[[42, 52]]]",
    "process": "Given that the curve $ x = \\sqrt{y^{2} + 1} $ is the right branch of a hyperbola, and the line $ y = kx + 1 $ always passes through the point $ (0,1) $. When the line intersects the curve at two distinct points, the system of equations has two unequal common solutions, both positive roots. By solving the system and simplifying, we obtain $ (1 - k^{2})x^{2} - 2kx - 2 = 0 $, then  \n$$\n\\begin{matrix}\n4k^{2} + 8(1 - k^{2}) > 0 & \\frac{2k}{1 - k^{2}} \\\\\n\\frac{x}{1 - k} > 0 &\n\\end{matrix}\n$$\nSolving gives $ -\\sqrt{2} < k < -1 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, one of its asymptotes intersects the circle $x^{2}+(y-2 \\sqrt{3})^{2}=4$ at points $A$ and $B$, and $|A B|=2$. Find the eccentricity of the hyperbola $C$.",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;A: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + (y - 2*sqrt(3))^2 = 4);Intersection(OneOf(Asymptote(C)),G) = {A, B};Abs(LineSegmentOf(A, B)) = 2",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 63], [124, 130]], [[10, 63]], [[10, 63]], [[70, 99]], [[102, 105]], [[106, 109]], [[10, 63]], [[10, 63]], [[2, 63]], [[70, 99]], [[2, 111]], [[113, 122]]]",
    "query_spans": "[[[124, 136]]]",
    "process": "The asymptotes are determined from the equations of the hyperbola and circle as $ y = \\pm\\frac{b}{a}x $, with the circle's center at $ (0, 2\\sqrt{3}) $ and radius $ r = 2 $. According to the geometric relationship among intersecting chords, radius, and distance from chord to center, we have $ r^2 - d^{2} = \\frac{|AB|^{2}}{4} $. Combining this with the parameter relations of the hyperbola, we can find its eccentricity. From the given conditions: the hyperbola's asymptotes are $ y = \\pm\\frac{b}{a}x $, the circle's center is $ (0, 2\\sqrt{3}) $, and radius $ r = 2 $. Therefore, the distance from the center to the asymptote is $ d = \\frac{|2\\sqrt{3}|}{\\sqrt{1+\\frac{b^{2}}{a^{2}}}} = \\frac{2\\sqrt{3}a}{\\sqrt{a^{2}+b^{2}}} $, and $ |AB| = 2 $, thus $ r^2 - d^{2} = 1 $, hence $ \\frac{12a^{2}}{a^{2}+b^{2}} = 3 $. Since $ a^{2} + b^{2} = c^{2} $, $ e = \\frac{c}{a} > 1 $, therefore $ e = 2 $."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{2-m}+\\frac{y^{2}}{1-m}=1(m \\in R)$ represents a hyperbola, then the focal length of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (x^2/(2 - m) + y^2/(1 - m) = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[53, 56], [59, 62]], [[3, 51]], [[1, 56]]]",
    "query_spans": "[[[59, 67]]]",
    "process": "Since the equation $\\frac{x^2}{2-m} + \\frac{y^2}{1-m} = 1$ ($m \\in \\mathbb{R}$) represents a hyperbola, $(2-m)(1-m) < 0$, solving gives $1 < m < 2$. Thus $a^{2} = 2 - m$, $b^{2} = m - 1$, and since $c^{2} = a^{2} + b^{2}$, we have $c^{2} = 2 - m + m - 1 = 1$, so the focal distance of the hyperbola is $2c = 2$;"
  },
  {
    "text": "What is the distance from the focus to the directrix of the parabola $y^{2}=2 x$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 25]]]",
    "process": "Using the standard equation of the parabola, we obtain $ p=1 $. Since the distance from the focus to the directrix is $ p $, the result follows. For the parabola $ y^2=2x $, the distance from the focus to the directrix is $ p $, and from the standard equation we get $ p=1 $."
  },
  {
    "text": "Through the right focus $F$ of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1$, draw a line perpendicular to the $x$-axis, intersecting the two asymptotes of hyperbola $C$ at points $A$ and $B$. Let $O$ be the origin. Then the minimum area of $\\Delta O A B$ is?",
    "fact_expressions": "C: Hyperbola;a: Number;O: Origin;A: Point;B: Point;F: Point;Expression(C) = (-y^2/4 + x^2/a^2 = 1);RightFocus(C) = F;L: Line;PointOnCurve(F, L);IsPerpendicular(L, xAxis);L1: Line;L2: Line;Asymptote(C) = {L1, L2};Intersection(L, L1) = A;Intersection(L, L2) = B",
    "query_expressions": "Min(Area(TriangleOf(O, A, B)))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 48], [69, 75]], [[9, 48]], [[92, 95]], [[82, 85]], [[86, 89]], [[52, 55]], [[1, 48]], [[1, 55]], [], [[0, 68]], [[0, 68]], [], [], [[69, 81]], [[0, 91]], [[0, 91]]]",
    "query_spans": "[[[102, 125]]]",
    "process": "Find the hyperbola's $ b, c $, find the asymptotes of the hyperbola, substitute $ x = c $ into the asymptote equations to obtain the coordinates of $ A, B $, find the area of $ \\triangle OAB $, and use the basic inequality to find the minimum value. [Detailed solution] For hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{4} = 1 $, we have $ b = 2 $, $ c^{2} = a^{2} + 4 $, ($ a > 0 $). Let $ F(c, 0) $. The asymptotes of the hyperbola are $ y = \\pm \\frac{2}{a}x $. Substituting $ x = c $ gives intersection points $ A(c, \\frac{2c}{a}) $, $ B(c, -\\frac{2c}{a}) $. Thus, the area of $ \\triangle OAB $ is $ S = \\frac{1}{2}c \\cdot \\frac{4c}{a} = 2 \\cdot \\frac{a^{2} + 4}{a} = 2(a + \\frac{4}{a}) \\geqslant 4\\sqrt{a \\cdot \\frac{4}{a}} = 8 $. The area of $ \\triangle OAB $ reaches the minimum value of 8 if and only if $ a = 2 $."
  },
  {
    "text": "If the eccentricity of the ellipse $x^{2}+\\frac{y^{2}}{m}=1$ with foci on the $x$-axis is $\\frac{\\sqrt{3}}{2}$, then what is the value of $m$?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2 + y^2/m = 1);PointOnCurve(Focus(G), xAxis);Eccentricity(G) = (sqrt(3)/2)",
    "query_expressions": "m",
    "answer_expressions": "1/4",
    "fact_spans": "[[[10, 37]], [[65, 68]], [[10, 37]], [[1, 37]], [[10, 63]]]",
    "query_spans": "[[[65, 72]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{16}-\\frac{y^{2}}{18}=1$, with left and right foci $F_{1}$ and $F_{2}$ respectively, $P$ a point on $C$, $\\overrightarrow{F_{1} Q}=\\overrightarrow{Q P}$, $O$ the origin. If $|P F_{1}|=10$, then $|O Q|$=?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/16 - y^2/18 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C);Q: Point;VectorOf(F1, Q) = VectorOf(Q, P);O: Origin;Abs(LineSegmentOf(P, F1)) = 10",
    "query_expressions": "Abs(LineSegmentOf(O, Q))",
    "answer_expressions": "9",
    "fact_spans": "[[[2, 47], [76, 79]], [[2, 47]], [[56, 63]], [[64, 71]], [[2, 71]], [[2, 71]], [[72, 75]], [[72, 82]], [[83, 130]], [[83, 130]], [[132, 135]], [[142, 156]]]",
    "query_spans": "[[[158, 167]]]",
    "process": "The left and right foci of the hyperbola $ C: \\frac{x^2}{16} - \\frac{y^{2}}{18} = 1 $ are $ F_{1}, F_{2} $, respectively. Therefore, $ ||PF_{1}| - |PF_{2}|| = 2a = 8 $. Since $ |PF_{1}| = 10 $, it follows that $ |PF_{2}| = 2 $ or $ |PF_{2}| = 18 $. But $ |PF_{2}| > c + a $, so $ |PF_{2}| = 2 $ is discarded. Hence, $ |PF_{2}| = 18 $. Since $ \\overrightarrow{F_{1}Q} = \\overrightarrow{QP} $, point $ Q $ is the midpoint of $ PF_{2} $. Thus, $ OQ = \\frac{1}{2}|PF_{2}| = 9 $."
  },
  {
    "text": "Given the equation of the ellipse is $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, then the length of the major axis of this ellipse is equal to?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1)",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "10",
    "fact_spans": "[[[48, 50], [2, 4]], [[2, 45]]]",
    "query_spans": "[[[48, 57]]]",
    "process": "The length of the major axis of the ellipse is equal to 2a=2\\times5=10"
  },
  {
    "text": "If a line $l$ passing through the focus of the parabola $y^{2}=4x$ is tangent to the circle $(x-4)^{2}+y^{2}=4$, then what is the slope of the line $l$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), l) = True;l: Line;H: Circle;Expression(H) = (y^2 + (x - 4)^2 = 4);IsTangent(l, H) = True",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*2*sqrt(5)/5",
    "fact_spans": "[[[3, 17]], [[3, 17]], [[1, 25]], [[20, 25], [50, 55]], [[26, 46]], [[26, 46]], [[20, 48]]]",
    "query_spans": "[[[50, 60]]]",
    "process": "The focus of the parabola is $ F(1,0) $. Let the equation of line $ l $ be $ y = k(x - 1) $, or $ kx - y - k = 0 $. Since line $ l $ is tangent to the circle $ (x - 4)^2 + y^2 = 4 $, we have $ \\frac{|4k - k|}{\\sqrt{k^2 + 1}} = 2 $. Solving gives $ k = \\pm \\frac{2\\sqrt{5}}{5} $."
  },
  {
    "text": "Point $P$ is on the circle $(x-2)^{2}+(y-5)^{2}=1$, and point $Q$ is on the parabola $y^{2}=4x$. What is the minimum value of the sum of the distance from point $Q$ to the line $x=-1$ and the distance from point $Q$ to point $P$?",
    "fact_expressions": "G: Parabola;H: Circle;I: Line;P: Point;Q: Point;Expression(G) = (y^2 = 4*x);Expression(H) = ((x - 2)^2 + (y - 5)^2 = 1);Expression(I) = (x = -1);PointOnCurve(P, H);PointOnCurve(Q, G)",
    "query_expressions": "Min(Distance(Q,I)+Distance(Q,P))",
    "answer_expressions": "sqrt(26)-1",
    "fact_spans": "[[[38, 52]], [[5, 29]], [[62, 70]], [[0, 4], [75, 79]], [[33, 37], [57, 61]], [[38, 52]], [[5, 29]], [[62, 70]], [[0, 32]], [[33, 55]]]",
    "query_spans": "[[[57, 90]]]",
    "process": "As shown in the figure, d = PQ + QM \\geqslant (AQ - 1) + QF \\geqslant AF = \\sqrt{26} - 1, so fill in \\sqrt{26} - 1"
  },
  {
    "text": "The distance from point $M$ to point $F(0, -2)$ is less than its distance to the line $l$: $y - 3 = 0$ by $1$. Then, what is the equation of the trajectory of point $M$?",
    "fact_expressions": "M: Point;l: Line;Expression(l) = (y - 3 = 0);Distance(M, F) = Distance(M, l) - 1;F: Point;Coordinate(F) = (0, -2)",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 4], [20, 21], [44, 48]], [[22, 35]], [[22, 35]], [[0, 42]], [[5, 16]], [[5, 16]]]",
    "query_spans": "[[[44, 55]]]",
    "process": ""
  },
  {
    "text": "Given that point $A(1,2)$ lies on the parabola $C$: $y^{2}=2 p x$ ($p>0$), a line passing through point $B(2,-2)$ intersects the parabola $C$ at points $P$ and $Q$. If the slopes of lines $AP$ and $AQ$ are $k_{1}$ and $k_{2}$ respectively, then $k_{1} \\cdot k_{2}$ equals?",
    "fact_expressions": "C: Parabola;p: Number;H: Line;A: Point;P: Point;B: Point;Q: Point;p>0;Expression(C) = (y^2 = 2*p*x);Coordinate(A) = (1, 2);Coordinate(B) = (2,-2);PointOnCurve(A, C);PointOnCurve(B, H);Intersection(H, C) = {P, Q};k1:Number;k2:Number;Slope(LineOf(A,P))=k1;Slope(LineOf(A,Q))=k2",
    "query_expressions": "k1*k2",
    "answer_expressions": "-4",
    "fact_spans": "[[[12, 37], [54, 60]], [[19, 37]], [[51, 53]], [[2, 11]], [[61, 64]], [[40, 50]], [[65, 68]], [[19, 37]], [[12, 37]], [[2, 11]], [[40, 50]], [[2, 38]], [[39, 53]], [[51, 70]], [[91, 98]], [[99, 106]], [[72, 106]], [[72, 106]]]",
    "query_spans": "[[[108, 131]]]",
    "process": "Substituting the coordinates of point A(1,2) into the equation of the parabola according to the given conditions yields the value of p, and then the equation of the parabola can be found. Let the equation of line PQ be set and combined with the parabola's equation to find the sum and product of the two roots. Then compute the product of the slopes of lines AP and AQ, which simplifies to the constant value -4. Substituting the coordinates of A(1,2) into the parabola's equation gives 4=2p, solving yields p=2. Therefore, the equation of the parabola is $ y^{2}=4x $. According to the conditions, the slope of line PQ is not zero. Thus, let the equation of line PQ be: $ x=m(y+2)+2 $, and let $ P(x_{1},y_{1}) $, $ Q(x_{2},y_{2}) $. Combining the line and parabola equations:  \n\\[\n\\begin{cases}\nx=m(y+2)+2 \\\\\ny^{2}=4x\n\\end{cases}\n\\]  \nRearranging yields: $ y^{2}-4my-8m-8=0 $, then $ y_{1}+y_{2}=4m $, $ y_{1}y_{2}=-8m-8 $. According to the conditions, $ k_{1} $, so $ k_{1}k_{2}=-4 $."
  },
  {
    "text": "Given the equation of ellipse $E$ is $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, $AB$ is a chord with an inclination angle of $135^{\\circ}$, and $M(2,1)$ is the midpoint of chord $AB$, then the eccentricity of ellipse $E$ is?",
    "fact_expressions": "E: Ellipse;A: Point;B: Point;M: Point;a:Number;b:Number;a>b;b>0;Coordinate(M) = (2, 1);Expression(E) = (y^2/b^2 + x^2/a^2 = 1);IsChordOf(LineSegmentOf(A,B),E);Inclination(LineSegmentOf(A,B))=ApplyUnit(135,degree);MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 7], [68, 69], [113, 118]], [[62, 67]], [[62, 67]], [[93, 101]], [[11, 61]], [[11, 61]], [[11, 61]], [[11, 61]], [[93, 101]], [[2, 61]], [[62, 91]], [[62, 91]], [[93, 111]]]",
    "query_spans": "[[[113, 124]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{25}=1$ has left focus $F_{1}$, point $P$ lies on the right branch of the hyperbola, and $P F_{1}$ is tangent to the circle $x^{2}+y^{2}=16$ at point $N$, $M$ is the midpoint of segment $P F_{1}$, $O$ is the origin, then $|M N|-|M O|$=?",
    "fact_expressions": "G: Hyperbola;H: Circle;P: Point;F1: Point;M: Point;N: Point;O: Origin;Expression(G) = (x^2/16 - y^2/25 = 1);Expression(H) = (x^2 + y^2 = 16);LeftFocus(G) = F1;PointOnCurve(P, RightPart(G));TangentPoint(LineSegmentOf(P,F1),H)=N;MidPoint(LineSegmentOf(P,F1))=M",
    "query_expressions": "Abs(LineSegmentOf(M, N)) - Abs(LineSegmentOf(M, O))",
    "answer_expressions": "-1",
    "fact_spans": "[[[2, 42], [60, 63]], [[80, 97]], [[55, 59]], [[47, 54]], [[105, 108]], [[100, 104]], [[124, 127]], [[2, 42]], [[80, 97]], [[2, 54]], [[55, 68]], [[70, 104]], [[105, 123]]]",
    "query_spans": "[[[134, 149]]]",
    "process": "Let F' be the right focus of the hyperbola, connect PF'. Since M and O are the midpoints of FP and FF' respectively, |MO| = \\frac{1}{2}|PF|, so |FN| = \\sqrt{|OF|^{2} - |ON|^{2}} = 5. By the definition of the hyperbola, |PF| - |PF'| = 8, hence |MN| - |MO| = -\\frac{1}{2}|PF'| + |MF| - |FT| = -1."
  },
  {
    "text": "Point $P$ is a moving point on the parabola $\\frac{1}{4} y^{2}=x$. Then the minimum value of the sum of the distance from point $P$ to point $A(0,-1)$ and the distance from point $P$ to the line $x=-1$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2/4 = x);P: Point;PointOnCurve(P, G) = True;A: Point;Coordinate(A) = (0, -1);H: Line;Expression(H) = (x = -1)",
    "query_expressions": "Min(Distance(P,A)+Distance(P,H))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[5, 29]], [[5, 29]], [[0, 4], [36, 40], [55, 59]], [[0, 34]], [[41, 51]], [[41, 51]], [[60, 68]], [[60, 68]]]",
    "query_spans": "[[[36, 78]]]",
    "process": ""
  },
  {
    "text": "What is the distance from the focus to the directrix of the parabola $y^{2}=-8 x$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -8*x)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 26]]]",
    "process": "Problem Analysis: According to the problem, we have p=2, so the distance from the focus to the directrix of the parabola y^{2}=-8x is 4"
  },
  {
    "text": "Given that the graph of the equation $x y=\\frac{1}{4}$ is a hyperbola, and the asymptotes of this hyperbola are the lines $x=0$, $y=0$, then the focal distance of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H1: Line;H2:Line;Expression(G)=(x*y = 1/4);Expression(H1) = (x = 0);Expression(H2) = (y = 0);Asymptote(G)={H1,H2}",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[25, 28], [31, 34], [57, 60]], [[41, 48]], [[50, 55]], [[2, 28]], [[41, 48]], [[50, 55]], [[31, 55]]]",
    "query_spans": "[[[57, 65]]]",
    "process": "The center of symmetry of $ xy=\\frac{1}{4} $ is $ (0,0) $, and the axes of symmetry are $ y=x $, $ y=-x $. Since the asymptotes of the hyperbola are the lines $ x=0 $, $ y=0 $, the hyperbola is equilateral, so $ a=b $. Solving the system $ \\begin{cases} y=x \\\\ xy=\\frac{1}{4} \\end{cases} $, we obtain $ (\\frac{1}{2},\\frac{1}{2}) $, $ (-\\frac{1}{2},-\\frac{1}{2}) $. $ \\therefore 2a=\\sqrt{1+1}=\\sqrt{2} $, solving gives $ a=\\frac{\\sqrt{2}}{2} $, hence $ b=\\frac{\\sqrt{2}}{2} $, $ c=1 $, so $ 2c=2 $, i.e., the focal distance is 2."
  },
  {
    "text": "The line $l$ passing through the focus $F$ of the parabola $C$: $y^{2}=4x$ intersects $C$ at points $A$ and $B$ (with $A$ and $B$ in the first and fourth quadrants, respectively). If $|AF|=3|FB|$, then the slope of $l$ is?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;F: Point;B: Point;Expression(C) = (y^2 = 4*x);Focus(C)=F;PointOnCurve(F,l);Intersection(l,C)={A,B};Quadrant(A)=1;Quadrant(B)=4;Abs(LineSegmentOf(A,F))=3*Abs(LineSegmentOf(F,B))",
    "query_expressions": "Slope(l)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[84, 87], [27, 32]], [[1, 20], [33, 36]], [[37, 40], [47, 51]], [[23, 26]], [[41, 44], [52, 55]], [[1, 20]], [[1, 26]], [[0, 32]], [[27, 46]], [[47, 64]], [[47, 64]], [[67, 81]]]",
    "query_spans": "[[[84, 92]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with right focus $F$, if a line passing through point $F$ with an inclination angle of $45^{\\circ}$ intersects the right branch of the hyperbola at exactly one point, then what is the range of values for the eccentricity of this hyperbola?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(C) = F;G: Line;PointOnCurve(F, G);Inclination(G) = ApplyUnit(45, degree);NumIntersection(G, RightPart(C)) = 1",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "[\\sqrt{2}, +\\infty)",
    "fact_spans": "[[[2, 63], [99, 102], [116, 119]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 71], [74, 78]], [[2, 71]], [[96, 98]], [[73, 98]], [[79, 98]], [[96, 113]]]",
    "query_spans": "[[[116, 130]]]",
    "process": "According to the problem, determine the relationship between the line and the asymptote, obtaining $\\frac{b}{a}\\geqslant1$, then compute the range of eccentricity to obtain the answer. Denote the line passing through point $F$ as $l$. Since the line passing through point $F$ intersects the right branch of the hyperbola at exactly one point, the inclination angle of the asymptote with positive slope should be no less than the inclination angle of $l$. Given that the inclination angle of $l$ is $45^{\\circ}$, thus $\\frac{b}{a}\\geqslant\\tan45^{\\circ}=1$. Hence, $e=\\frac{c}{a}=\\sqrt{\\frac{a^{2}+b^{2}}{a^{2}}}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}\\geqslant\\sqrt{2}$"
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $C$: $2 x^{2}-y^{2}=1$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (2*x^2 - y^2 = 1)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*sqrt(2)*x",
    "fact_spans": "[[[0, 25]], [[0, 25]]]",
    "query_spans": "[[[0, 33]]]",
    "process": "\\because the standard equation of the hyperbola 2x^{2}-y^{2}=1 is: \\frac{x^{2}}{2}-y^{2}=1, \\therefore a^{2}=\\frac{1}{2}, b^{2}=1, we obtain a=\\frac{\\sqrt{2}}{2}, b=1. Also \\because the asymptotes of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 are given by y=\\pm\\frac{b}{a}x, \\therefore the asymptotes of the hyperbola 2x^{2}-y^{2}=1 are y=\\pm\\sqrt{2}x"
  },
  {
    "text": "Given that $F$ is the focus of the parabola $y^{2}=2 p x(p>0)$, $A$ and $B$ are two points on the parabola, $A$ lies in the first quadrant, $B$ lies in the fourth quadrant, and satisfies $\\overrightarrow{A F}=4 \\overrightarrow{F B}$, then the slope of line $A B$ is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;F: Point;p>0;Expression(G) = (y^2 = 2*p*x);Focus(G) = F;PointOnCurve(A, G);PointOnCurve(B, G);Quadrant(A) = 1;Quadrant(B) = 4;VectorOf(A, F) = 4*VectorOf(F, B)",
    "query_expressions": "Slope(LineOf(A,B))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[6, 27], [39, 42]], [[9, 27]], [[31, 34], [46, 49]], [[35, 38], [55, 58]], [[2, 5]], [[9, 27]], [[6, 27]], [[2, 30]], [[31, 45]], [[31, 45]], [[46, 54]], [[55, 63]], [[66, 111]]]",
    "query_spans": "[[[113, 125]]]",
    "process": "Let the angle of inclination of line AB be $\\theta$. Draw perpendiculars from points A and B to the directrix, with feet of perpendiculars denoted as $A_{1}$ and $B_{1}$, respectively. Draw a perpendicular from F to $AA_{1}$, with foot of perpendicular denoted as D. Then, according to the property of parabola, we have: $|AF|=|AA_{1}|=|A_{1}D|+|AD|=p+|AF|\\cos\\theta$, so $|AF|=\\frac{p}{1-\\cos\\theta}$. Similarly, $|BF|=\\frac{p}{1+\\cos\\theta}$. Given $\\overrightarrow{AF}=4\\overrightarrow{FB}$, it follows that $\\frac{p}{1-\\cos\\theta}=\\frac{4p}{1+\\cos\\theta}\\Rightarrow\\cos\\theta=\\frac{3}{5}\\Rightarrow k=\\tan\\theta=\\frac{4}{3}$"
  },
  {
    "text": "It is known that one directrix of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ coincides with the directrix of the parabola $y^{2}=-6x$. Then, the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(H) = (y^2 = -6*x);OneOf(Directrix(G)) = Directrix(H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 40], [69, 72]], [[5, 40]], [[46, 61]], [[5, 40]], [[2, 40]], [[46, 61]], [[2, 66]]]",
    "query_spans": "[[[69, 78]]]",
    "process": ""
  },
  {
    "text": "The line passing through the right vertex of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with slope $2$ intersects the right branch of the hyperbola at two points. Find the range of the eccentricity of this hyperbola.",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;H: Line;PointOnCurve(RightVertex(G), H);Slope(H) = 2;NumIntersection(H, RightPart(G)) = 2",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1, \\sqrt{5})",
    "fact_spans": "[[[1, 57], [73, 76], [86, 89]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[68, 70]], [[0, 70]], [[61, 70]], [[68, 83]]]",
    "query_spans": "[[[86, 99]]]",
    "process": "The asymptotes of the hyperbola are given by $ y = \\pm\\frac{b}{a}x $. According to the problem, we have $ \\frac{b}{a} < 2 $, so the eccentricity $ e = \\frac{c}{a} = \\sqrt{\\frac{a^{2}+b^{2}}{a^{2}}} = \\sqrt{1+(\\frac{b}{a})^{2}} < \\sqrt{5} $. Therefore, the range of the eccentricity $ e $ is $ (1,\\sqrt{5}) $."
  },
  {
    "text": "The focus of the parabola $y^{2}=4 x$ is $F$. A line passing through $F$ intersects the parabola at points $A$ and $B$, satisfying $\\frac{A F}{B F}=4$. Point $O$ is the origin. Then the area of $\\Delta A O F$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);A: Point;B: Point;Intersection(H, G) = {A, B};LineSegmentOf(A, F)/LineSegmentOf(B, F) = 4;O: Origin",
    "query_expressions": "Area(TriangleOf(A, O, F))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 14], [30, 33]], [[0, 14]], [[18, 21], [23, 26]], [[0, 21]], [[27, 29]], [[22, 29]], [[35, 38]], [[39, 42]], [[27, 44]], [[48, 67]], [[68, 72]]]",
    "query_spans": "[[[77, 96]]]",
    "process": "From the given conditions, p=2, p=2, \\begin{cases}\\frac{AF}{BF}=4,\\\\\\frac{1}{AF}+\\frac{1}{BF}=\\frac{2}{2}=1,\\end{cases}. AF=5, A(4,4), S=\\frac{1}{2}\\cdot1\\cdot4=2, so the answer is 2."
  },
  {
    "text": "Let $A$ be a point on the parabola $C$: $y^{2}=2 p x(p>0)$, and let $F$ be the focus of $C$. Draw a perpendicular from point $A$ to the directrix of $C$, with foot $B$. If the angle of inclination of line $F B$ is $150^{\\circ}$, and the area of $\\triangle F A B$ is $\\frac{\\sqrt{3}}{18}$, then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;F: Point;B: Point;A: Point;p>0;Expression(C) = (y^2 = 2*p*x);PointOnCurve(A, Directrix(C));Focus(C) = F;PointOnCurve(A,C);IsPerpendicular(G,Directrix(C));FootPoint(G,Directrix(C))=B;Inclination(LineOf(F,B))=ApplyUnit(150, degree);Area(TriangleOf(F, A, B)) = sqrt(3)/18",
    "query_expressions": "p",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[5, 31], [35, 38], [35, 38]], [[141, 144]], [], [[42, 45]], [[65, 68]], [[1, 4], [47, 51]], [[13, 31]], [[5, 31]], [[1, 34]], [[35, 45]], [[1, 34]], [[46, 61]], [[46, 68]], [[70, 95]], [[97, 139]]]",
    "query_spans": "[[[141, 146]]]",
    "process": "Let the intersection point of the directrix and the x-axis be M. Since the inclination angle of line FB is 150^{\\circ}, it follows that \\angle MFB = 30^{\\circ}. In the right triangle FBAp, so_{BF} = \\frac{2\\sqrt{3}}{3}p. From AB // x-axis, we have \\angle ABF = \\angle MFB = 30^{\\circ}. By the definition of a parabola, AB = AF, so \\angle ABF = \\angle AFB = \\angle MFB = 30^{\\circ}. Therefore, \\angle BAF = 120^{\\circ}. Thus, in triangle ABF, by the cosine law, BF^{2} = AB^{2} + AF^{2} - 2AB \\cdot AF \\cos \\angle BAF = 3AB^{2}, so AB = AF = \\frac{2}{3}p. Hence, the area of \\triangle FAB is \\frac{1}{2} \\times \\frac{2}{3}p \\times \\frac{2}{3}p \\sin 120^{\\circ} = \\frac{\\sqrt{3}}{18}, so_{p} = \\frac{\\sqrt{2}}{2}."
  },
  {
    "text": "The asymptote of the hyperbola $tx^{2}-y^{2}-1=0$ is perpendicular to the line $2x+y+1=0$, then $t=?$",
    "fact_expressions": "G: Hyperbola;t: Number;H: Line;Expression(G) = (t*x^2 - y^2 - 1 = 0);Expression(H) = (2*x + y + 1 = 0);IsPerpendicular(OneOf(Asymptote(G)), H)",
    "query_expressions": "t",
    "answer_expressions": "1/4",
    "fact_spans": "[[[0, 21]], [[45, 48]], [[28, 41]], [[0, 21]], [[28, 41]], [[0, 43]]]",
    "query_spans": "[[[45, 50]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=2 p x$ with focus $F$, if points $A$ and $B$ lie on the parabola, $|\\overrightarrow{A B}|=6$, $\\overrightarrow{A F} \\cdot \\overrightarrow{B F}=0$, and the projection of the midpoint $M$ of segment $A B$ onto the directrix of the parabola is $N$, then the maximum value of $|\\overrightarrow{M N}|$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;F: Point;Focus(G) = F;A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);Abs(VectorOf(A, B)) = 6;DotProduct(VectorOf(A, F), VectorOf(B, F)) = 0;M: Point;MidPoint(LineSegmentOf(A, B)) = M;N: Point;Projection(M, Directrix(G)) = N",
    "query_expressions": "Max(Abs(VectorOf(M, N)))",
    "answer_expressions": "3*sqrt(2)",
    "fact_spans": "[[[2, 18], [37, 40], [138, 141]], [[2, 18]], [[5, 18]], [[22, 25]], [[2, 25]], [[27, 31]], [[32, 35]], [[27, 43]], [[32, 43]], [[44, 70]], [[72, 123]], [[134, 137]], [[124, 137]], [[149, 152]], [[134, 152]]]",
    "query_spans": "[[[154, 184]]]",
    "process": "As shown in the figure, let |AF| = a, |BF| = b, then |AB| = \\sqrt{a^{2}+b^{2}} = 6. Combining the relationship between the quadratic mean and the arithmetic mean, \\frac{a+b}{2} \\leqslant \\frac{\\sqrt{a^{2}+b^{2}}}{2}, with equality if and only if a = b. Therefore, \\frac{|MN|}{|AB|} = \\frac{\\frac{a+b}{2}}{\\sqrt{a^{2}+b^{2}}} \\leqslant \\frac{\\sqrt{2}}{2}, so |MN| \\leqslant \\frac{\\sqrt{2}}{2}|AB|, that is, the maximum value of |MN| is \\frac{\\sqrt{2}}{2}|AB| = 3\\sqrt{2}."
  },
  {
    "text": "If $F_{1}$ and $F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, and point $P$ lies on the hyperbola with its horizontal coordinate being $2 \\sqrt{2}$, then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 - y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);XCoordinate(P) = 2*sqrt(2)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[17, 45], [56, 59]], [[1, 8]], [[51, 55], [62, 66]], [[9, 16]], [[17, 45]], [[1, 50]], [[51, 60]], [[62, 83]]]",
    "query_spans": "[[[85, 112]]]",
    "process": "From $\\frac{x^{2}}{4}-y^{2}=1$, we have $a^{2}=4$, $b^{2}=1$, so $c^{2}=a^{2}+b^{2}=5$, $c=\\sqrt{5}$, thus $|F_{1}F_{2}|=2\\sqrt{5}$. Since the horizontal coordinate of point $P$ is $2\\sqrt{2}$, we get $\\frac{8}{4}-y^{2}=1$, so $|y|=1$. Therefore, the area of $\\triangle F_{1}PF_{2}$ is $\\frac{1}{2}\\times|F_{1}F_{2}|\\times|y|=\\frac{1}{2}\\times2\\sqrt{5}\\times1=\\sqrt{5}$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2x$, a line $l$ passing through the focus intersects $C$ at points $A$ and $B$. If the circle with diameter $AB$ is tangent to the directrix of $C$ at point $M(-\\frac{1}{2}, \\frac{1}{2})$, then the equation of line $l$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*x);l: Line;PointOnCurve(Focus(C), l);A: Point;B: Point;Intersection(l, C) = {A, B};G: Circle;IsDiameter(LineSegmentOf(A, B), G);M: Point;Coordinate(M) = (-1/2, 1/2);TangentPoint(G, Directrix(C)) = M",
    "query_expressions": "Expression(l)",
    "answer_expressions": "y=2*x-1",
    "fact_spans": "[[[2, 21], [32, 35], [60, 63]], [[2, 21]], [[26, 31], [101, 104]], [[2, 31]], [[37, 40]], [[41, 44]], [[26, 46]], [[58, 59]], [[48, 59]], [[68, 99]], [[68, 99]], [[58, 99]]]",
    "query_spans": "[[[101, 109]]]",
    "process": "When the line/slope does not exist, it is obviously not valid. When the slope of line $ l $ exists, let the equation of $ l $ be $ y = k(x - \\frac{1}{2}) $. Let $ A(x_{1}, y_{1}) $. Solve the system of equations:\n$$\n\\begin{cases}\ny = k(x - \\frac{1}{2}) \\\\\ny^2 = 2x\n\\end{cases}\n$$\nEliminating $ x $ and simplifying, we get $ ky^{2} - 2y - k = 0 $, $ \\Delta = 4 + 4k^{2} > 0 $. Thus, $ y_{1} + y_{2} = \\frac{2}{k} $, $ y_{1}y_{2} = -1 $. Substituting into $ C: y^{2} = 2x $, we obtain $ x_{1} + x_{2} = \\frac{2}{k^{2}} + 1 $, $ x_{1}x_{2} = \\frac{1}{4} $. Since the circle with diameter $ AB $ is tangent to the directrix of $ C $ at point $ M(-\\frac{1}{2}, \\frac{1}{2}) $, we have $ MA \\perp MB $, so $ \\overrightarrow{MA} \\cdot \\overrightarrow{MB} = 0 $, and $ \\overrightarrow{MA} = (x_{1} + \\frac{1}{2}, y_{1} - \\frac{1}{2}) $, $ \\overrightarrow{MB} = (x_{2} + \\frac{1}{2}, y_{2} - \\frac{1}{2}) $. Therefore,\n$$\n(x_{1} + \\frac{1}{2})(x_{2} + \\frac{1}{2}) + (y_{1} - \\frac{1}{2})(y_{2} - \\frac{1}{2}) = 0\n$$\nwhich simplifies to $ \\frac{1}{k^{2}} - \\frac{1}{k} + \\frac{1}{4} = 0 $. Solving gives $ k = 2 $. Hence, the equation of line $ l $ is $ y = 2x - 1 $."
  },
  {
    "text": "Given the ellipse $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$ $(a>b>0)$, the upper focus is $F$, the left and right vertices are $B_{1}$, $B_{2}$ respectively, the lower vertex is $A$, the lines $A B_{2}$ and $B_{1} F$ intersect at point $P$. If $\\overrightarrow{A P}=2 \\overrightarrow{A B_{2}}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/b^2 + y^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;F: Point;UpperFocus(G) = F;B1: Point;B2: Point;LeftVertex(G) = B1;RightVertex(G) = B2;A: Point;LowerVertex(G) = A;P: Point;Intersection(LineOf(A, B2), LineOf(B1, F)) = P;VectorOf(A, P) = 2*VectorOf(A, B2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/3",
    "fact_spans": "[[[2, 54], [177, 179]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[59, 62]], [[2, 62]], [[71, 78]], [[79, 86]], [[2, 86]], [[2, 86]], [[91, 94]], [[2, 94]], [[120, 124]], [[95, 124]], [[126, 175]]]",
    "query_spans": "[[[177, 185]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 px$ coincides with the right focus of the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;H: Ellipse;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (x^2/6 + y^2/2 = 1);Focus(G) = RightFocus(H)",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 16]], [[65, 68]], [[20, 57]], [[1, 16]], [[20, 57]], [[1, 63]]]",
    "query_spans": "[[[65, 72]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $x^{2}=4 y$ with focus $F$, let $P$ be a moving point on the parabola $C$, and point $Q(0,-1)$. Then the minimum value of $\\frac{|P F|}{|P Q|}$ is?",
    "fact_expressions": "C: Parabola;Q: Point;P: Point;F: Point;Expression(C) = (x^2 = 4*y);Coordinate(Q) = (0, -1);Focus(C)=F;PointOnCurve(P, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F))/Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 21], [32, 38]], [[43, 53]], [[28, 31]], [[24, 27]], [[2, 21]], [[43, 53]], [[2, 27]], [[28, 42]]]",
    "query_spans": "[[[55, 82]]]",
    "process": "From the given conditions, the focus is F(0,1) and the directrix has equation y = -1. Draw PM perpendicular to the directrix from point P, with M as the foot of the perpendicular. Then by the definition of a parabola, |PF| = |PM|, so \\frac{|PF|}{|PQ|} = \\frac{|PM|}{|PQ|} = \\sin\\angle PQM, where \\angle PQM is an acute angle. Therefore, when \\angle PQM is minimized, \\frac{|PF|}{|PQ|} is minimized. Hence, \\frac{|PF|}{|PQ|} is minimized when PQ is tangent to the parabola. Let the point of tangency be P(a,\\frac{a^{2}}{4}), then the slope of PQ is \\frac{\\frac{a^{2}}{4}+1}{a}, and the slope of the tangent line is \\frac{a}{2}. From \\frac{\\frac{a^{2}}{4}+1}{a} = \\frac{a}{2}, solving gives a = \\pm2, yielding P(\\pm2,1), |PM| = 2, |PQ| = 2\\sqrt{2}, thus \\sin\\angle PQM = \\frac{\\sqrt{2}}{2}."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has eccentricity $e=\\frac{\\sqrt{3}}{2}$, $A$, $B$ are the left and right vertices of the ellipse, and $P$ is a point on the ellipse distinct from $A$ and $B$. The inclinations of lines $PA$ and $PB$ are $\\alpha$, $\\beta$ respectively. Then the minimum value of $| \\tan \\alpha - \\tan \\beta |$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;e: Number;Eccentricity(G) = e;e = sqrt(3)/2;A: Point;B: Point;LeftVertex(G) = A;RightVertex(G) = B;P: Point;PointOnCurve(P, G);Negation(P=A);Negation(P=B);alpha: Number;beta: Number;Inclination(LineOf(P, A)) = alpha;Inclination(LineOf(P, B)) = beta",
    "query_expressions": "Min(Abs(Tan(alpha) - Tan(beta)))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 54], [90, 92], [103, 105]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[58, 80]], [[2, 80]], [[58, 80]], [[82, 85], [109, 112]], [[86, 89], [113, 116]], [[82, 98]], [[82, 98]], [[99, 102]], [[99, 119]], [[99, 119]], [[99, 119]], [[140, 148]], [[150, 157]], [[120, 157]], [[120, 157]]]",
    "query_spans": "[[[159, 193]]]",
    "process": "Let the coordinates of point $ P $ be given, find the expression for $ |\\tan\\alpha - \\tan\\beta| $, simplify it, and then find the minimum value of $ |\\tan\\alpha - \\tan\\beta| $. According to the problem, $ \\frac{c}{a} = \\frac{\\sqrt{3}}{2} = \\sqrt{1 - \\left(\\frac{b}{a}\\right)^2} $, so $ \\frac{b}{a} = \\frac{1}{2} $, $ a = 2b $. Let $ P(x_0, y_0) $ ($ x_0 \\neq 0 $), then $ \\frac{x_0^2}{a^2} + \\frac{y_0^2}{b^2} = 1 $, i.e., $ \\frac{x_0^2}{a^2} + \\frac{y_0^2}{\\frac{a^2}{4}} = 1 $, simplifying yields $ x_0^2 - 4y_0^2 = a^2 $ ①. Since $ A $, $ B $ are the left and right vertices of the ellipse, $ A(-a, 0) $, $ B(a, 0) $. Therefore,  \n$$\n|\\tan\\alpha - \\tan\\beta| = \\left| \\frac{y_0}{x_0 + a} - \\frac{y_0}{x_0 - a} \\right| = \\left| \\frac{-2ay_0}{x_0^2 - a^2} \\right| = \\left| \\frac{-2ay_0}{-4y_0^2} \\right| = \\left| \\frac{a}{2y_0} \\right|.\n$$\nSince $ |y_0| \\in (0, b] = \\left(0, \\frac{a}{2}\\right] $, when $ |y_0| = \\frac{a}{2} $, $ |\\tan\\alpha - \\tan\\beta| $ reaches its minimum value $ \\frac{a}{2 \\cdot \\frac{a}{2}} = 1 $."
  },
  {
    "text": "What is the equation of the tangent line to the parabola $y=x^{2}$ at the point $(-1,1)$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2);H: Point;Coordinate(H) = (-1, 1)",
    "query_expressions": "Expression(TangentOnPoint(H, G))",
    "answer_expressions": "2*x+y+1=0",
    "fact_spans": "[[[0, 12]], [[0, 12]], [[13, 22]], [[13, 22]]]",
    "query_spans": "[[[0, 30]]]",
    "process": "Directly find the derivative of the parabola at the point (-1,1), which is the slope of the tangent line. Use the point-slope form of the line equation to write the tangent equation, then convert it to the general form. From $ y = x^{2} $, we get: $ y' = 2x $, $ \\therefore y'|_{x=-1} = -2 $. Therefore, the tangent line equation to the parabola $ y = x^{2} $ at the point (-1,1) is $ y - 1 = -2(x + 1) $, that is, $ 2x + y + 1 = 0 $."
  },
  {
    "text": "A line passing through the center of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ intersects the ellipse at points $A$ and $B$, and $F_{1}$ is the right focus of the ellipse. Then the maximum area of $\\triangle A B F_{1}$ is?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;F1: Point;Expression(G) = (x^2/25 + y^2/9 = 1);PointOnCurve(Center(G), H);Intersection(H, G) = {A, B};RightFocus(G) = F1",
    "query_expressions": "Max(Area(TriangleOf(A, B, F1)))",
    "answer_expressions": "12",
    "fact_spans": "[[[1, 39], [46, 48], [68, 70]], [[43, 45]], [[50, 53]], [[54, 57]], [[60, 67]], [[1, 39]], [[0, 45]], [[43, 59]], [[60, 74]]]",
    "query_spans": "[[[76, 110]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of a hyperbola are $3x \\pm 4y = 0$, and its foci are the endpoints of the major axis of the ellipse $\\frac{x^2}{10} + \\frac{y^2}{5} = 1$, then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (3*x + pm*4*y = 0);H: Ellipse;Expression(H) = (x^2/10 + y^2/5 = 1);Focus(G) = Endpoint(MajorAxis(H))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "5*x^2/32-5*y^2/18=1",
    "fact_spans": "[[[2, 5], [28, 29], [79, 82]], [[2, 27]], [[33, 71]], [[33, 71]], [[28, 76]]]",
    "query_spans": "[[[79, 87]]]",
    "process": "\\because the asymptotes of the hyperbola are given by 3x\\pm4y=0, \\therefore assume the equation of the hyperbola is: \\frac{x^{2}}{16}-\\frac{y^{2}}{9}=\\lambda, (\\lambda\\neq0). The foci of the hyperbola are (\\pm\\sqrt{10},0), \\therefore \\lambda>0, and 16\\lambda+9\\lambda=10, so \\lambda=\\frac{2}{5}. \\therefore the equation of this hyperbola is \\frac{5x^{2}}{32}-\\frac{5y^{2}}{18}=1"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, and point $P(2 , 1)$ lies on the asymptote of $C$. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (2, 1);PointOnCurve(P, Asymptote(C))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 53], [67, 70], [77, 80]], [[10, 53]], [[10, 53]], [[55, 66]], [[2, 53]], [[55, 66]], [[55, 75]]]",
    "query_spans": "[[[77, 86]]]",
    "process": "From the equation of the hyperbola, it is known that the foci lie on the x-axis. Given that $ P(2,1) $ lies on the asymptote, we have $ \\frac{b}{a} = \\frac{1}{2} $, so $ a = 2b $, and thus $ c = \\sqrt{5}b $. Therefore, the eccentricity is $ e = \\frac{c}{a} = \\frac{1}{2}\\sqrt{5} $."
  },
  {
    "text": "It is known that the vertex of the parabola is at the origin, the focus is on the $y$-axis, and the distance from the point $P(m,-2)$ on the parabola to the focus is $4$. Then the value of the real number $m$ is?",
    "fact_expressions": "G: Parabola;Vertex(G) = O;O: Origin;PointOnCurve(Focus(G), yAxis);P: Point;Coordinate(P) = (m, -2);m: Real;PointOnCurve(P, G);Distance(P, Focus(G)) = 4",
    "query_expressions": "m",
    "answer_expressions": "pm*4",
    "fact_spans": "[[[2, 5], [21, 24]], [[2, 11]], [[9, 11]], [[2, 20]], [[26, 36]], [[26, 36]], [[48, 53]], [[21, 36]], [[21, 46]]]",
    "query_spans": "[[[48, 57]]]",
    "process": "Let the standard equation of the parabola be $ x^{2} = -2py $ ($ p > 0 $). Given that the distance from point $ P $ to the focus is 4, we have $ \\frac{p}{2} + 2 = 4 $, so $ p = 4 $, hence $ x^{2} = -8y $. Substituting point $ P(m, -2) $ into $ x^{2} = -8y $, we get $ m = \\pm 4 $."
  },
  {
    "text": "The focus of the parabola $y^{2}=2 px(p>0)$ is $F$, points $A$ and $B$ lie on the parabola, and $\\angle AFB=120^{\\circ}$. A perpendicular is drawn from the midpoint $M$ of chord $AB$ to the directrix $l$, with foot of perpendicular $M_{1}$. Then the maximum value of $\\frac{|M M_{1}|}{|AB|}$ is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;F: Point;M: Point;l:Line;M1:Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(A,G);PointOnCurve(B, G);AngleOf(A, F, B) = ApplyUnit(120, degree);IsChordOf(LineSegmentOf(A,B),G);MidPoint(LineSegmentOf(A,B))=M;Directrix(G)=l;L:Line;PointOnCurve(M,L);IsPerpendicular(l,L);FootPoint(l,L)=M1",
    "query_expressions": "Max(Abs(LineSegmentOf(M, M1))/Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 20], [38, 41]], [[3, 20]], [[28, 32]], [[33, 37]], [[24, 27]], [[77, 80]], [[84, 87]], [[94, 101]], [[3, 20]], [[0, 20]], [[0, 27]], [[28, 42]], [[34, 42]], [[44, 68]], [[38, 75]], [[71, 80]], [[38, 87]], [], [[69, 90]], [[69, 90]], [[69, 101]]]",
    "query_spans": "[[[103, 133]]]",
    "process": ""
  },
  {
    "text": "The minimum distance from a point $P$ on the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ to $M(3,0)$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1);P: Point;PointOnCurve(P, G);M: Point;Coordinate(M) = (3, 0)",
    "query_expressions": "Min(Distance(P, M))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 28]], [[0, 28]], [[31, 34]], [[0, 34]], [[35, 43]], [[35, 43]]]",
    "query_spans": "[[[31, 51]]]",
    "process": "Let $ P(x_{0},y_{0}) $, then $ x_{0}^{2}-\\frac{y_{0}^{2}}{3}=1^{2} $. Thus, $ |PM|=\\sqrt{(x_{0}-3)^{2}+y_{0}^{2}}=\\sqrt{4x_{0}^{2}-6x_{0}+6}=\\sqrt{4(x_{0}-\\frac{3}{4})^{2}+\\frac{15}{4}} $. Since $ |x_{0}|\\geqslant1 $, when $ x_{0}=1 $, $ |PM|_{\\min}=2 $. Therefore, the minimum distance from a point $ P $ on the hyperbola $ x^{2}-\\frac{y^{2}}{3}=1 $ to $ M(3,0) $ is 2."
  },
  {
    "text": "If the left and right foci of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ are $F_{1}$ and $F_{2}$ respectively, and point $P$ is a point on the ellipse such that $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$, then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "C: Ellipse;F1: Point;P: Point;F2: Point;Expression(C) = (x^2/4 + y^2/3 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);AngleOf(F1, P, F2) = pi/3",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[1, 43], [72, 74]], [[51, 58]], [[67, 71]], [[59, 66]], [[1, 43]], [[1, 66]], [[1, 66]], [[67, 78]], [[79, 115]]]",
    "query_spans": "[[[117, 144]]]",
    "process": "\\because ellipse C:\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1,\\therefore a=2,b=\\sqrt{3},c=1, and \\because P is a point on the ellipse, \\angle F_{1}PF_{2}=\\frac{\\pi}{3}, F_{1},F_{2} are the left and right foci, \\therefore |PF_{1}|+|PF_{2}|=2a=4, |F_{1}F_{2}|=2. \\therefore |F_{1}F_{2}|^{2}=(|=\\frac{2a}{1}|+|\\frac{\\pi}{2}|)^{2}|-\\angle,=4, \\therefore |PF_{1}|\\cdot|PF_{2}|=4, \\therefore S_{APF_{1}F_{2}}=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|\\sin\\frac{\\pi}{3}=\\frac{1}{2}\\times4\\times\\frac{\\sqrt{3}}{2}=\\sqrt{3}"
  },
  {
    "text": "Given that the circle $N$: $x^{2}+y^{2}=b^{2}$ passes exactly through the foci of the ellipse $M$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, then the eccentricity of the ellipse $M$ is?",
    "fact_expressions": "M: Ellipse;b: Number;a: Number;N: Circle;Expression(M) = (y^2/b^2 + x^2/a^2 = 1);Expression(N) = (x^2 + y^2 = b^2);PointOnCurve(Focus(M), N)",
    "query_expressions": "Eccentricity(M)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[31, 80], [85, 90]], [[37, 80]], [[37, 80]], [[2, 27]], [[31, 80]], [[2, 27]], [[2, 83]]]",
    "query_spans": "[[[85, 96]]]",
    "process": ""
  },
  {
    "text": "If a hyperbola shares common foci with the ellipse $4 x^{2}+y^{2}=64$ and their eccentricities are reciprocals of each other, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (4*x^2 + y^2 = 64);Focus(G) = Focus(H);InterReciprocal(Eccentricity(G), Eccentricity(H))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/36 - x^2/12 = 1",
    "fact_spans": "[[[2, 5], [47, 50]], [[6, 26]], [[6, 26]], [[2, 32]], [[34, 44]]]",
    "query_spans": "[[[47, 55]]]",
    "process": "The ellipse $4x^{2}+y^{2}=64$ can be rewritten as $\\frac{x^{2}}{16}+\\frac{y^{2}}{64}=1$, $\\therefore a^{2}=64$, $c^{2}=64-16=48$, $\\therefore$ the foci are $(0,\\pm4\\sqrt{3})$, eccentricity $e=\\frac{\\sqrt{3}}{2}$, then for the hyperbola the foci lie on the $y$-axis, $c=4\\sqrt{3}$, $e=\\frac{2}{\\sqrt{3}}$ $\\therefore a=6$, $b=2\\sqrt{3}$, hence the equation of the hyperbola is $\\frac{y^{2}}{36}-\\frac{x^{2}}{12}=1$."
  },
  {
    "text": "Let $a \\in R$. It is known that the directrix $l$ of the parabola $y^{2}=4 x$ is tangent to the circle $C$: $x^{2}+y^{2}+2 a x-2 \\sqrt{3} y=0$. Then $a=$?",
    "fact_expressions": "G: Parabola;C: Circle;a: Real;l: Line;Expression(G) = (y^2 = 4*x);Expression(C) = (-2*sqrt(3)*y + 2*a*x + x^2 + y^2 = 0);IsTangent(l,C);Directrix(G)=l",
    "query_expressions": "a",
    "answer_expressions": "-1",
    "fact_spans": "[[[13, 27]], [[34, 74]], [[78, 81], [1, 10]], [[30, 33]], [[13, 27]], [[34, 74]], [[30, 76]], [[13, 33]]]",
    "query_spans": "[[[78, 83]]]",
    "process": "The directrix $ l $ of the parabola $ y^{2}=4x $ has the equation $ x=-1 $. The standard equation of circle $ C $ is $ (x+a)^{2}+(y-\\sqrt{3})^{2}=a^{2}+3 $, with center $ C(-a,\\sqrt{3}) $ and radius $ \\sqrt{a^{2}+3} $. Since the line $ l $ is tangent to circle $ C $, then $ |-a+1|=\\sqrt{a^{2}+3} $, solving gives $ a=-1 $."
  },
  {
    "text": "The focus of the parabola $y^{2}=16 x$ coincides with one focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1$, then what is the length of the real axis of the hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/9 + x^2/a^2 = 1);a: Number;H: Parabola;Expression(H) = (y^2 = 16*x);Focus(H)=OneOf(Focus(G))",
    "query_expressions": "Length(RealAxis(G))",
    "answer_expressions": "2*sqrt(7)",
    "fact_spans": "[[[18, 60], [69, 72]], [[18, 60]], [[21, 60]], [[0, 15]], [[0, 15]], [[0, 67]]]",
    "query_spans": "[[[69, 77]]]",
    "process": "The focus of the parabola $ y^{2} = 16x $ is $ (4, 0) $, which coincides with one focus of the hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{9} = 1 $. We obtain $ a^{2} + 9 = 16 $, solving gives $ a = \\sqrt{7} $. Therefore, the real axis length of the hyperbola is: $ 2\\sqrt{7} $."
  },
  {
    "text": "The left and right foci of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the hyperbola such that $|P F_{1}| \\cdot |P F_{2}|=64$. Find the area $S$ of $\\Delta P F_{1} F_{2}$.",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1) ;LeftFocus(G) = F1;RightFocus(G) = F2;F1: Point;F2: Point;P: Point;PointOnCurve(P, G) = True;Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 64;Area(TriangleOf(P, F1, F2)) = S;S: Number",
    "query_expressions": "S",
    "answer_expressions": "16*sqrt(3)",
    "fact_spans": "[[[0, 39], [69, 72]], [[0, 39]], [[0, 63]], [[0, 63]], [[48, 55]], [[56, 63]], [[64, 68]], [[64, 73]], [[75, 104]], [[106, 134]], [[131, 134]]]",
    "query_spans": "[[[131, 136]]]",
    "process": "By the definition of a hyperbola, we have $||PF_{1}|-|PF_{2}||=2\\sqrt{9}=6$. By the law of cosines, $\\cos\\angle F_{1}PF_{2}=\\frac{|PF_{1}|^{2}+|PF_{2}|^{2}-|F_{1}F_{2}|^{2}}{2|PF_{1}|\\cdot|PF_{2}|}=\\frac{(||PF_{1}|-|PF_{2}||)^{2}+2|PF_{1}|\\cdot|PF_{2}|-|F_{1}F_{2}|^{2}}{2|PF_{1}|\\cdot|PF_{2}|}$. Given that $||PF_{1}|-|PF_{2}||=6$, $|PF_{1}|\\cdot|PF_{2}|=64$, and $|F_{1}F_{2}|=2\\sqrt{9+16}=10$, substituting yields $\\cos\\angle F_{1}PF_{2}=\\frac{36+128-100}{128}=\\frac{1}{2}$. Since $\\angle F_{1}PF_{2}\\in(0,\\pi)$, it follows that $\\angle F_{1}PF_{2}=\\frac{\\pi}{3}$. Hence, $S=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|\\sin\\angle F_{1}PF_{2}=\\frac{1}{2}\\times64\\times\\frac{\\sqrt{3}}{2}=16\\sqrt{3}$."
  },
  {
    "text": "It is known that the focus of the parabola $y^{2}=16 x$ coincides with one focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{12}=1(a>0)$. Then, what is the equation of the asymptotes of the hyperbola?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;a>0;Expression(G) = (-y^2/12 + x^2/a^2 = 1);Expression(H) = (y^2 = 16*x);Focus(H) = OneOf(Focus(G))",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*sqrt(3)*x",
    "fact_spans": "[[[21, 69], [78, 81]], [[24, 69]], [[2, 17]], [[24, 69]], [[21, 69]], [[2, 17]], [[2, 76]]]",
    "query_spans": "[[[78, 89]]]",
    "process": "\\because the focus of the parabola y^2 = 16x is (4,0) \\therefore a focus of the hyperbola is (4,0) \\because \\frac{x^{2}}{a^2} - \\frac{y^{2}}{12} = 1 (a > 0), according to the hyperbola c^{2} = a^{2} + b^{2}, i.e., a^{2} + 12 = 16, solving gives: a = 2. According to the asymptote equations of a hyperbola with foci on the x-axis: v = \\pm \\frac{b}{a}x \\therefore the asymptote equations of the hyperbola are: v = +\\sqrt{3}x"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$, if a chord $AB$ is bisected by $M(1,1)$, then what is the equation of the line containing $AB$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/4 = 1);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);M: Point;Coordinate(M) = (1, 1);MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Expression(OverlappingLine(LineSegmentOf(A, B)))",
    "answer_expressions": "x+4*y-5=0",
    "fact_spans": "[[[2, 40], [42, 43]], [[2, 40]], [[47, 51]], [[47, 51]], [[42, 51]], [[52, 60]], [[52, 60]], [[47, 62]]]",
    "query_spans": "[[[64, 77]]]",
    "process": ""
  },
  {
    "text": "Let the parabola $C$: $y^{2}=4x$ have focus $F$. A line $l$ with positive slope passes through the focus $F$, intersects the parabola $C$ at points $A$ and $B$, and intersects the directrix at point $Q$. If $\\overrightarrow{AB} = \\overrightarrow{BQ}$, then what is the slope of line $l$?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;Q: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;Slope(l)>0;PointOnCurve(F,l);Intersection(l,C)={A,B};Intersection(l,Directrix(C))=Q;VectorOf(A, B) = VectorOf(B, Q)",
    "query_expressions": "Slope(l)",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[118, 123], [33, 38]], [[1, 20], [46, 52]], [[53, 56]], [[57, 60]], [[67, 71]], [[41, 44], [23, 26]], [[1, 20]], [[1, 26]], [[27, 38]], [[33, 44]], [[33, 62]], [[33, 71]], [[73, 116]]]",
    "query_spans": "[[[118, 128]]]",
    "process": "Draw perpendiculars from A and B to the directrix l, with feet of perpendiculars A' and B' respectively. ∴ |AF| = |AA'|, |BF| = |BB'|. ∵ |AB| = |BQ|, then B is the midpoint of AQ. In △AQA, |BB'| = \\frac{1}{2}|AA'|, ∴ |BF| = \\frac{1}{2}|AF|. Let |BF| = a, then |BB'| = a, |AB| = |BQ| = 3a, ∴ |QB'| = 2\\sqrt{2}a, ∴ k = \\tan\\angle QBB' = \\frac{|QB'|}{|BB'|} = \\frac{2\\sqrt{2}a}{a} = 2\\sqrt{2}"
  },
  {
    "text": "The moving circle $M$ is externally tangent to the fixed circle $C$: $x^{2}+y^{2}+4 x=0$, and tangent to the line $l$: $x=2$. What equation does the center $(x, y)$ of the moving circle $M$ satisfy?",
    "fact_expressions": "M: Circle;C: Circle;Expression(C) = (x^2 + y^2 + 4*x = 0);IsOutTangent(M, C) = True;l: Line;Expression(l) = (x = 2);IsTangent(M, l) = True;Center(M) = M1;M1: Point;Coordinate(M1) = (x1, y1);x1: Number;y1: Number",
    "query_expressions": "LocusEquation(M1)",
    "answer_expressions": "y^2+12*x-12=0",
    "fact_spans": "[[[2, 5], [56, 59]], [[8, 32]], [[8, 32]], [[2, 35]], [[38, 50]], [[38, 50]], [[2, 52]], [[56, 70]], [[62, 70]], [[62, 70]], [[62, 70]], [[62, 70]]]",
    "query_spans": "[[[62, 77]]]",
    "process": "Let the coordinates of point M be (x, y), C(-2, 0), the radius of the moving circle be r, and the distance from the moving circle to the line be d. Then, according to the properties of two externally tangent circles and a line tangent to a circle, we have |MC| = 2 + r, d = r, |MC| - d = 2. That is, \\sqrt{(x+2)^{2}+y^{2}}-(2-x)=2, simplifying yields y^{2}+12x-12=0, \\therefore the trajectory equation of the center of the moving circle is y^{2}+12x-12=0."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $P$ is a point on the ellipse, $O$ is the origin, $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, and $|\\overrightarrow{P F_{1}}|=3|\\overrightarrow{P F_{2}}|$, find the eccentricity of the ellipse.",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;O: Origin;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Abs(VectorOf(P, F1)) = 3*Abs(VectorOf(P, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)/4",
    "fact_spans": "[[[20, 72], [82, 84], [219, 221]], [[22, 72]], [[22, 72]], [[78, 81]], [[2, 9]], [[10, 17]], [[89, 92]], [[22, 72]], [[22, 72]], [[20, 72]], [[2, 77]], [[2, 77]], [[78, 88]], [[99, 158]], [[160, 216]]]",
    "query_spans": "[[[219, 227]]]",
    "process": "\\because|\\overrightarrow{PF_{1}}|=3|\\overrightarrow{PF_{2}}|, and |\\overrightarrow{PF_{1}}|+|\\overrightarrow{PF_{2}}|=2a. \\therefore|PF_{2}|=\\frac{a}{2},|PF_{1}|=\\frac{3a}{2}. \\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=0, \\therefore\\overrightarrow{PF_{1}}\\bot\\overrightarrow{PF_{2}}, i.e., AF_{1}PF_{2} is a right triangle, \\therefore|PF_{2}|^{2}+|PF_{1}|^{2}=|F_{1}F_{2}|^{2}. \\therefore(\\frac{a}{2})^{2}+(\\frac{3a}{2})^{2}=(2c)^{2}. Solving gives: e=\\frac{c}{a}=\\frac{\\sqrt{10}}{4}"
  },
  {
    "text": "Given that one of the directrices of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{3}=1$ $(a>0)$ is the line $x=4$, what is the eccentricity of this ellipse?",
    "fact_expressions": "G: Ellipse;a: Number;a>0;Expression(G) = (y^2/3 + x^2/a^2 = 1);Expression(OneOf(Directrix(G))) = (x = 4)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 48], [65, 67]], [[4, 48]], [[4, 48]], [[2, 48]], [[2, 61]]]",
    "query_spans": "[[[65, 73]]]",
    "process": ""
  },
  {
    "text": "The line passing through the right focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$ and perpendicular to the $x$-axis intersects the ellipse at points $M$ and $N$. The circle with $MN$ as diameter passes exactly through the left focus. Then, the eccentricity of the ellipse is equal to?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Circle;L: Line;M: Point;N: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(RightFocus(G),L);IsPerpendicular(L,xAxis);Intersection(L,G) = {M, N};IsDiameter(LineSegmentOf(M,N),H);PointOnCurve(LeftFocus(G),H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 55], [71, 73], [103, 105]], [[3, 55]], [[3, 55]], [[94, 95]], [[68, 70]], [[75, 78]], [[79, 82]], [[3, 55]], [[3, 55]], [[1, 55]], [[0, 70]], [[60, 70]], [[68, 84]], [[85, 95]], [[71, 101]]]",
    "query_spans": "[[[103, 112]]]",
    "process": ""
  },
  {
    "text": "Let the ellipse be $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and point $P(1, \\frac{3}{2})$ lies on the ellipse. Find the equation of the tangent line to the ellipse at $P$?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Coordinate(P) = (1, 3/2);PointOnCurve(P, G)",
    "query_expressions": "Expression(TangentOnPoint(P, G))",
    "answer_expressions": "x + 2*y - 4 = 0",
    "fact_spans": "[[[1, 38], [60, 62], [66, 68]], [[39, 59], [69, 72]], [[1, 38]], [[39, 59]], [[39, 63]]]",
    "query_spans": "[[[66, 79]]]",
    "process": "Since the slope of the tangent line exists, let the equation of the tangent line be $ y - \\frac{3}{2} = k(x - 1) $. Substituting into the ellipse equation, simplifying, and setting the discriminant equal to zero, we can solve for $ k $, thus obtaining the tangent line equation.  \nSubstitute $ y - \\frac{3}{2} = k(x - 1) $ into $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $, we get  \n$ 3x^{2} + 4\\left[k(x - 1) + \\frac{3}{2}\\right]^{2} = 12 $.  \nSimplifying and rearranging gives  \n$ (3 + 4k^{2})x^{2} + (12k - 8k^{2})x + 4k^{2} - 12k - 3 = 0 $.  \nLet $ \\Delta = (12k - 8k^{2})^{2} - 4(3 + 4k^{2})(4k^{2} - 12k - 3) = 0 $.  \nSimplifying and rearranging yields  \n$ 36k^{2} + 36k + 9 = 0 $, i.e., $ 4k^{2} + 4k + 1 = 0 $,  \nsolving gives $ k = -\\frac{1}{2} $.  \nThus, the tangent line equation is $ y - \\frac{3}{2} = -\\frac{1}{2}(x - 1) $, i.e., $ x + 2y - 4 = 0 $."
  },
  {
    "text": "Given that an asymptote of a hyperbola has the equation $y=x$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(OneOf(Asymptote(G)))=(y=x)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 5], [2, 5]], [[2, 19]]]",
    "query_spans": "[[[22, 31]]]",
    "process": ""
  },
  {
    "text": "Let the line $y=a$ intersect the curves $y^{2}=x$ and $y=e^{x}$ at points $M$ and $N$, respectively. When the length of segment $MN$ is minimized, what is the value of $a$?",
    "fact_expressions": "M: Point;N: Point;G: Line;a: Number;H: Curve;C:Curve;Expression(G) = (y = a);Expression(H) = (y^2 = x);Expression(C) = (y = e^x);Intersection(G, H) = M;Intersection(G, C) = N;WhenMin(LineSegmentOf(M, N))",
    "query_expressions": "a",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[34, 38]], [[39, 42]], [[1, 8]], [[57, 60]], [[11, 22]], [[23, 32]], [[1, 8]], [[11, 22]], [[23, 32]], [[1, 42]], [[1, 42]], [[44, 57]]]",
    "query_spans": "[[[57, 64]]]",
    "process": ""
  },
  {
    "text": "If the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3^{2}}=1$ $(a>0)$ is $2$, then $a$=?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2/9 + x^2/a^2 = 1);Eccentricity(G) = 2",
    "query_expressions": "a",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[1, 52]], [[62, 65]], [[4, 52]], [[1, 52]], [[1, 60]]]",
    "query_spans": "[[[62, 67]]]",
    "process": "From the given condition, $_{e}=\\frac{\\sqrt{a^{2}+9}}{a}=2$, $(a>0)$, from which we can find the value of $a$ as $\\sqrt{3}$"
  },
  {
    "text": "The real axis length and the imaginary axis length are equal for a hyperbola centered at the origin with foci on the $x$-axis, and the distance from a focus to an asymptote is $\\sqrt{2}$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;O: Origin;Center(G) = O;PointOnCurve(Focus(G), xAxis);Length(RealAxis(G)) = Length(ImageinaryAxis(G));Distance(Focus(G), Asymptote(G)) = sqrt(2)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 = 1",
    "fact_spans": "[[[15, 18], [53, 56]], [[3, 5]], [[0, 18]], [[6, 18]], [[15, 28]], [[15, 51]]]",
    "query_spans": "[[[53, 60]]]",
    "process": "Since the lengths of the real axis and the imaginary axis of a hyperbola with foci on the x-axis are equal, the equation of the hyperbola can be set as x^{2}-y^{2}=a^{2}. Then the focus is C(\\sqrt{2}a,0), the asymptote is y=x, and d=\\frac{\\sqrt{2}a}{\\sqrt{2}}=\\sqrt{2}, yielding a=\\sqrt{2}. Therefore, the equation of the hyperbola is x^{2}-y^{2}=2. [Comment] This question mainly examines the equation and geometric properties of hyperbolas, aiming to test the mastery of basic knowledge, and belongs to a fundamental problem."
  },
  {
    "text": "What is the length of the major axis of the ellipse $x^{2}+2 y^{2}=4$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 2*y^2 = 4)",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 19]], [[0, 19]]]",
    "query_spans": "[[[0, 25]]]",
    "process": "The equation of the ellipse $x^{2}+2y^{2}=4$ is rewritten as: $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$. Let the semi-major axis length of the ellipse be $a$, then $a^{2}=4$, solving gives $a=2$. Therefore, the major axis length of the ellipse $x^{2}+2y^{2}=4$ is $4$."
  },
  {
    "text": "In the rectangular coordinate system $x O y$, given that the asymptotes of a hyperbola with foci on the $x$-axis are $x \\pm 2 y = 0$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), xAxis);Expression(Asymptote(G)) = (x+pm*2*y = 0);O:Origin",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[28, 31], [54, 57]], [[19, 31]], [[28, 51]], [0, 14]]",
    "query_spans": "[[[54, 63]]]",
    "process": ""
  },
  {
    "text": "A point $P$ on the coordinate plane is equidistant from the points $A\\left(\\frac{1}{2}, 0\\right)$, $B(a, 2)$, and the line $x = -\\frac{1}{2}$. If there is exactly one such point $P$, then the real number $a$ is?",
    "fact_expressions": "A: Point;B: Point;P:Point;a: Real;G:Line;Expression(G) = (x = -1/2);Coordinate(A) = (1/2, 0);Coordinate(B) = (a, 2);Distance(P,A)=Distance(P,B);Distance(P,A)=Distance(P,G)",
    "query_expressions": "a",
    "answer_expressions": "{1/2,-1/2}",
    "fact_spans": "[[[11, 32]], [[35, 44]], [[7, 10], [77, 81]], [[90, 95]], [[46, 64]], [[46, 64]], [[11, 32]], [[35, 44]], [[7, 70]], [[7, 70]]]",
    "query_spans": "[[[90, 99]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$, a line $l$ passing through point $F$ intersects $C$ at points $A$ and $B$, and intersects the directrix of $C$ at point $M$. If $F$ is the midpoint of $AM$, then $|AB|=$?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;M: Point;B: Point;F: Point;Expression(C) = (y^2 = 8*x);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Intersection(l, Directrix(C)) = M;MidPoint(LineSegmentOf(A, M)) = F",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "32/2",
    "fact_spans": "[[[34, 39]], [[2, 20], [40, 43], [55, 58]], [[44, 47]], [[62, 66]], [[48, 51]], [[24, 27], [29, 33], [68, 71]], [[2, 20]], [[2, 27]], [[28, 39]], [[34, 53]], [[34, 66]], [[68, 80]]]",
    "query_spans": "[[[82, 91]]]",
    "process": "As shown in the figure, from the parabola $ C: y^{2} = 8x $, we have $ p = 4 $. Since $ F $ is the midpoint of $ AM $, $ AE = 2FG = 2p = 8 $, then $ x_{A} = 8 - \\frac{p}{2} = 6 $. By the focal chord property, $ x_{A}x_{B} = \\frac{p^{2}}{4} = 4 $, so $ x_{B} = \\frac{4}{6} = \\frac{2}{3} $. Therefore, $ BF = x_{B} + \\frac{p}{2} = \\frac{2}{3} + 2 = \\frac{8}{3} $. Hence, $ |AB| = |AF| + |BF| = 8 + \\frac{8}{3} = \\frac{32}{3} $."
  },
  {
    "text": "Given that $F_{1}$ is the left focus of the ellipse $C$: $\\frac{x^{2}}{4}+y^{2}=1$, a line $l$ passing through point $F_{1}$ intersects the ellipse $C$ at points $A$ and $B$. If $\\overrightarrow{B F_{1}}=3 \\overrightarrow{F_{1} A}$, then the slope of line $l$ is?",
    "fact_expressions": "F1: Point;LeftFocus(C) = F1;C: Ellipse;Expression(C) = (x^2/4 + y^2 = 1);l: Line;PointOnCurve(F1, l);A: Point;B: Point;Intersection(l, C) = {A, B};VectorOf(B, F1) = 3*VectorOf(F1, A)",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*sqrt(2)",
    "fact_spans": "[[[2, 9], [48, 56]], [[2, 46]], [[10, 42], [63, 68]], [[10, 42]], [[57, 62], [135, 140]], [[47, 62]], [[69, 72]], [[73, 76]], [[57, 78]], [[80, 133]]]",
    "query_spans": "[[[135, 145]]]",
    "process": "According to the problem, we find $ F_{1}(-\\sqrt{3},0) $. Let the equation of line $ l $ be: $ y = k(x + \\sqrt{3}) $. By solving the system of equations of the line and the ellipse and eliminating $ y $, we obtain the horizontal coordinates of the intersection points. From $ \\overrightarrow{BF}_{1} = 3\\overrightarrow{F_{1}A} $, we get $ x_{A} + 3x_{B} + 4\\sqrt{3} = 0 $. Substituting the horizontal coordinates of the intersection points allows us to solve the problem. The ellipse $ C: \\frac{x^{2}}{4} + y^{2} = 1 $, then $ a^{2} = 4 $, $ b^{2} = 1 $, so $ c^{2} = a^{2} - b^{2} $, i.e., $ c = \\sqrt{3} $. Thus, $ F_{1}(-\\sqrt{3},0) $. According to the problem, the slope of the line exists. Let the slope of line $ l $ be $ k $, we get $ (1 + 4k^{2})x^{2} + 8\\sqrt{3}k^{2}x + 12k^{2} - 4 = 0 $. Let $ A(x_{A}, y_{A}) $, $ B(x_{B}, y_{B}) $. Since $ \\overrightarrow{BF}_{1} = 3\\overrightarrow{F_{1}A} $, we have $ -3x_{B} - 3\\sqrt{3} = x_{A} + \\sqrt{3} $, simplifying gives $ x_{A} + 3x_{B} + 4\\sqrt{3} = 0 $. From $ x_{A} = \\frac{-4\\sqrt{3}k^{2} + 2\\sqrt{k^{2}+}}{1+4k^{2}}\\sqrt{3}k^{2}-2\\sqrt{k^{2}+1} $, substituting into $ x_{A} + 3x_{B} + 4\\sqrt{3} = 0 $ yields $ \\frac{-16\\sqrt{3}k^{2}}{1+4}\\frac{1+4k^{2}}{4\\sqrt{k^{2}+1}} + 4\\sqrt{3} = 0 $, solving gives $ k = \\pm\\sqrt{2} $."
  },
  {
    "text": "The line passing through the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ and perpendicular to the $x$-axis intersects the hyperbola at points $A$ and $B$, and intersects the asymptotes of the hyperbola at points $C$ and $D$. If $|AB| \\geq \\frac{5}{13}|CD|$, then the range of the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;A: Point;B: Point;C: Point;D: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(RightFocus(G), H);IsPerpendicular(H,xAxis);Intersection(H, G) = {A, B};Intersection(H,Asymptote(G))={C,D};Abs(LineSegmentOf(A, B)) >= (5/13)*Abs(LineSegmentOf(C, D))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[13/12, +oo)",
    "fact_spans": "[[[1, 57], [73, 76], [89, 92], [141, 144]], [[4, 57]], [[4, 57]], [[70, 72]], [[78, 81]], [[82, 85]], [[98, 101]], [[102, 105]], [[4, 57]], [[4, 57]], [[1, 57]], [[0, 72]], [[62, 72]], [[70, 87]], [[70, 107]], [[108, 138]]]",
    "query_spans": "[[[141, 154]]]",
    "process": "From the given conditions: |AB| = \\frac{2b^{2}}{a}, and the asymptotes are y = \\pm\\frac{b}{a}x, so |CD| = \\frac{2bc}{a}. From \\frac{2b^{2}}{a} \\geqslant \\frac{5}{13} \\cdot \\frac{2bc}{a}, simplifying yields b \\geqslant \\frac{5}{13}c, that is, b^{2} \\geqslant \\frac{25}{169}c^{2}, therefore c^{2} - a^{2} \\geqslant \\frac{25}{169}c^{2}, thus (\\frac{c}{a})^{2} \\geqslant \\frac{169}{144}, then the hyperbola's eccentricity e = \\frac{c}{a} \\geqslant \\frac{13}{12}"
  },
  {
    "text": "Given that the intersection point $P$ of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ and an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ has an abscissa of $1$, then the eccentricity of the hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Ellipse;P: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2/4 + y^2/3 = 1);Intersection(H,OneOf(Asymptote(G)))=P;XCoordinate(P) = 1",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(13)/2",
    "fact_spans": "[[[40, 99], [128, 131]], [[43, 99]], [[43, 99]], [[2, 39]], [[109, 112], [114, 118]], [[43, 99]], [[43, 99]], [[40, 99]], [[2, 39]], [[2, 112]], [[114, 126]]]",
    "query_spans": "[[[128, 138]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $x^{2}+2 y^{2}=4$, what is the length of the chord that has $(1,1)$ as its midpoint?",
    "fact_expressions": "G: Ellipse;H: Point;Expression(G) = (x^2 + 2*y^2 = 4);Coordinate(H) = (1, 1);L: LineSegment;IsChordOf(L, G);MidPoint(L) = H",
    "query_expressions": "Length(L)",
    "answer_expressions": "sqrt(30)/3",
    "fact_spans": "[[[2, 21]], [[24, 31]], [[2, 21]], [[24, 31]], [], [[2, 36]], [[2, 36]]]",
    "query_spans": "[[[2, 41]]]",
    "process": "Let the coordinates of the endpoints of the chord be A(a,b) and B(2-a,2-b). Set up the system of equations \\begin{cases}a^{2}+2b^{2}=4\\\\(2-a)^{2}+2(2-b)^{2}=4\\end{cases}, solving yields \\frac{a=1+\\frac{\\sqrt{6}}{3}}{\\sqrt{[(1-\\frac{\\sqrt{6}}{3})-(1+\\frac{\\sqrt{6}}{3})]^{2}}+[(1+\\frac{\\sqrt{6}}{6})-(1-\\frac{\\sqrt{6}}{6})]=\\frac{\\sqrt{30}}{3}."
  },
  {
    "text": "$M$ is a point on the parabola $y^{2}=4x$, $F$ is the focus, and $MF=4$. A perpendicular is drawn from point $M$ to the directrix $l$, with foot of the perpendicular at $K$. Then the area of triangle $MFK$ is?",
    "fact_expressions": "G: Parabola;M: Point;F: Point;l: Line;K: Point;Z: Line;Expression(G) = (y^2 = 4*x);PointOnCurve(M, G);Focus(G) = F;LineSegmentOf(M, F) = 4;PointOnCurve(M, Z);Directrix(G) = l;IsPerpendicular(Z, l);FootPoint(Z, l) = K",
    "query_expressions": "Area(TriangleOf(M, F, K))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[4, 18]], [[0, 3], [40, 44]], [[22, 25]], [[47, 50]], [[57, 60]], [], [[4, 18]], [[0, 21]], [[4, 28]], [[30, 37]], [[39, 53]], [[4, 50]], [[39, 53]], [[39, 60]]]",
    "query_spans": "[[[62, 77]]]",
    "process": "From the given conditions, we have: MK = MF = 4, \\frac{p}{2} = 1, so point M(3, 2\\sqrt{3}), therefore s_{\\triangle MFC} = s_{\\triangle ABC} - s_{\\triangle HFx} = \\frac{1}{2} \\times (2 + 4) \\times 2\\sqrt{3} - \\frac{1}{2} \\times 2 \\times 2\\sqrt{3} = 4\\sqrt{3}"
  },
  {
    "text": "If the equation $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{a-6}=1$ represents a hyperbola, then the range of real values for $a$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/a^2 + y^2/(a-6) = 1);a: Real",
    "query_expressions": "Range(a)",
    "answer_expressions": "(-\\infty,0)+(0,6)",
    "fact_spans": "[[[47, 50]], [[2, 50]], [[52, 57]]]",
    "query_spans": "[[[52, 64]]]",
    "process": "According to the equation representing a hyperbola, we obtain $a^{2}(a-6)<0$, from which we solve for the range of values of $a$. Since the equation $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{a-6}=1$ represents a hyperbola, it follows that $a^{2}(a-6)<0$, so $\\begin{cases}a\\neq0\\\\a-6\\end{cases}=0$. Therefore, $a\\in(-\\infty,0)\\cup(0,6)$."
  },
  {
    "text": "The line $y=2b$ intersects the left and right branches of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ at points $B$ and $C$, respectively. $A$ is the right vertex, $O$ is the coordinate origin. If $\\angle AOC = \\angle BOC$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;A: Point;O: Origin;C: Point;B: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = 2*b);Intersection(H,LeftPart(G))=B;Intersection(H,RightPart(G))=C;RightVertex(G)=A;AngleOf(A, O, C) = AngleOf(B, O, C)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(19)/2",
    "fact_spans": "[[[10, 66], [134, 137]], [[13, 66]], [[13, 66]], [[0, 9]], [[86, 89]], [[94, 97]], [[80, 83]], [[76, 79]], [[13, 66]], [[13, 66]], [[10, 66]], [[0, 9]], [[0, 85]], [[0, 85]], [[10, 93]], [[104, 131]]]",
    "query_spans": "[[[134, 143]]]",
    "process": "\\because\\angleAOC=\\angleBOC,\\therefore\\angleAOC=60C.\\frac{2\\sqrt{3}}{3}b,2b. Substituting into the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1, we get \\frac{4}{a^{b}}-4=1'\\therefore b=\\frac{\\sqrt{15}}{2}a'\\therefore_{c}=\\sqrt{a^{2+b^{2}}}=\\frac{\\sqrt{19}}{2}a'c=\\vee\\therefore_{e}=\\frac{c}{a}=\\frac{\\sqrt{19}}{2},"
  },
  {
    "text": "The focal distance of the ellipse $\\frac{x^{2}}{m} + \\frac{y^{2}}{4}=1$ is $2$, then what is the value of $m$?",
    "fact_expressions": "E: Ellipse;m: Number;Expression(E) = (y^2/4 + x^2/m = 1);FocalLength(E) = 2",
    "query_expressions": "m",
    "answer_expressions": "{5, 3}",
    "fact_spans": "[[[0, 39]], [[49, 52]], [[0, 39]], [[0, 46]]]",
    "query_spans": "[[[49, 57]]]",
    "process": ""
  },
  {
    "text": "Draw a line $l$ through the focus of the parabola $y=4x^{2}$, intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $y_{1}+y_{2}=2$, then the length of segment $AB$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 4*x^2);l: Line;PointOnCurve(Focus(G), l);A: Point;B: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Intersection(l, G) = {A, B};y1 + y2 = 2",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "17/8",
    "fact_spans": "[[[2, 16], [27, 30]], [[2, 16]], [[20, 25]], [[0, 25]], [[31, 48]], [[50, 67]], [[31, 48]], [[50, 67]], [[31, 48]], [[50, 67]], [[31, 48]], [[50, 67]], [[20, 69]], [[71, 86]]]",
    "query_spans": "[[[88, 100]]]",
    "process": "Using the formula for a focal chord, the chord length can be directly obtained. [Detailed explanation] Since $ y = 4x^{2} $, i.e., $ x^{2} = \\frac{1}{4}y $, then $ p = \\frac{1}{8} $. According to the focal chord length formula: $ AB = y_{1} + y_{2} + p = 2 + \\frac{1}{8} = \\frac{17}{8} $,"
  },
  {
    "text": "The equation of the hyperbola that shares the same asymptotes as $\\frac{x^{2}}{2}-y^{2}=1$ and passes through the point $(4, \\sqrt{3})$ is?",
    "fact_expressions": "G: Hyperbola;C: Hyperbola;H: Point;Expression(G) = (x^2/2 - y^2 = 1);Coordinate(H) = (4, sqrt(3));PointOnCurve(H, C);Asymptote(C)=Asymptote(G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/10 - y^2/5 = 1",
    "fact_spans": "[[[1, 29]], [[56, 59]], [[38, 55]], [[1, 29]], [[38, 55]], [[37, 59]], [[0, 59]]]",
    "query_spans": "[[[56, 63]]]",
    "process": ""
  },
  {
    "text": "Given the fixed point $Q(1,0)$, and $P$ is a moving point on the parabola $C$: $y^{2}=8x$, then the minimum value of $|PQ|$ is?",
    "fact_expressions": "C: Parabola;Q: Point;P: Point;Expression(C) = (y^2 = 8*x);Coordinate(Q) = (1, 0);PointOnCurve(P, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "1",
    "fact_spans": "[[[19, 38]], [[4, 12]], [[15, 18]], [[19, 38]], [[4, 12]], [[14, 42]]]",
    "query_spans": "[[[44, 57]]]",
    "process": "\\because point P is a moving point on the parabola y^{2}=x, \\because let P(x,2\\sqrt{2x}), |PQ| takes the minimum value 1"
  },
  {
    "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{(a+3)^{2}}=1$ $(a>0)$ are $y=\\pm 2x$, then $a=$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/(a + 3)^2 + x^2/a^2 = 1);a: Number;a>0;Expression(Asymptote(G)) = (y = pm*2*x)",
    "query_expressions": "a",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 57]], [[2, 57]], [[77, 80]], [[5, 57]], [[2, 75]]]",
    "query_spans": "[[[77, 82]]]",
    "process": "From the asymptote equations of the hyperbola, we obtain $\\frac{a+3}{a}=2$ ($a>0$); solving this equation gives the result. [Detailed Solution] When $a>0$, the asymptote equations of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{(a+3)^{2}}=1$ are $y=\\pm\\frac{a+3}{a}x$. According to the problem, $\\frac{a+3}{a}=2$, solving yields $a=3$."
  },
  {
    "text": "If the left and right foci of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ are $F_{1}$ and $F_{2}$ respectively, point $M$ lies on the hyperbola, and the perimeter of $\\Delta M F_{1} F_{2}$ is $20$, then the area of $\\Delta M F_{1} F_{2}$ equals?",
    "fact_expressions": "G: Hyperbola;M: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 - y^2/5 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(M, G);Perimeter(TriangleOf(M, F1, F2)) = 20",
    "query_expressions": "Area(TriangleOf(M, F1, F2))",
    "answer_expressions": "10*sqrt(2)",
    "fact_spans": "[[[1, 39], [69, 72]], [[64, 68]], [[48, 55]], [[56, 63]], [[1, 39]], [[1, 63]], [[1, 63]], [[64, 73]], [[75, 105]]]",
    "query_spans": "[[[107, 135]]]",
    "process": "Assume point M lies on the right branch of the hyperbola. From the hyperbola equation, we have a^{2}=4, b^{2}=5, so c=\\sqrt{4+5}=3. Since |MF_{1}|+|MF_{2}|+2c=20, it follows that |MF_{1}|+|MF_{2}|=14. Also, because |MF_{1}|-|MF_{2}|=4, we get |MF_{1}|=9, |MF_{2}|=5. In \\triangle MF_{1}F_{2}, by the law of cosines, \\cos\\angle F_{1}MF_{2}=\\frac{9^{2}+5^{2}-6^{2}}{2\\times9\\times5}=\\frac{7}{9}, so \\sin\\angle F_{1}MF_{2}=\\frac{4\\sqrt{2}}{9}. Therefore, the area S of \\triangle MF_{1}F_{2} is S=\\frac{1}{2}\\times9\\times5\\times\\frac{4\\sqrt{2}}{9}=10\\sqrt{2}."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Let $M$ be a moving point on the ellipse distinct from the endpoints of the major axis, and let $I$ be the incenter of $\\Delta M F_{1} F_{2}$. Then $\\frac{\\overrightarrow{M I} \\cdot \\overrightarrow{M F_{2}}}{|\\overrightarrow{M F_{2}}|}$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;M: Point;Negation(M = Endpoint(MajorAxis(G)));PointOnCurve(M, G);I: Point;Incenter(TriangleOf(M, F1, F2)) = I",
    "query_expressions": "DotProduct(VectorOf(M, I), VectorOf(M, F2))/Abs(VectorOf(M, F2))",
    "answer_expressions": "sqrt(2)-1",
    "fact_spans": "[[[2, 29], [58, 60]], [[2, 29]], [[38, 45]], [[46, 53]], [[2, 53]], [[2, 53]], [[54, 57]], [[54, 70]], [[54, 70]], [[97, 100]], [[71, 100]]]",
    "query_spans": "[[[102, 192]]]",
    "process": "Using the definition of an ellipse, properties of tangents to a circle, and the definition of the incenter, combined with knowledge of solving right triangles, we can obtain: let the incircle of $\\triangle MF_{1}F_{2}$ be tangent to $\\triangle MF_{1}F_{2}$ at points $D$, $E$, $F$. Let $MD=u$, $DF_{1}=v$, $FF_{2}=t$, then $MD=MF=u$, $DF_{1}=EF_{1}=v$, $EF_{2}=FF_{2}=t$. By the definition of the ellipse, we have $MF_{1}+MF_{2}=2a=2\\sqrt{2}$, $F_{1}F_{2}=2c=2$, thus $2u+v+t=2\\sqrt{2}$, $v+t=2$. Furthermore, $\\frac{\\overrightarrow{MF}}{|\\overrightarrow{MF_{2}}|}=|M|\\cos\\angle IMF=|MF|=u=\\sqrt{2}-1$,"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has the directrix equation $x=-\\frac{1}{2}$. If there is a point $A$ on $C$ located in the first quadrant, and the distance from point $A$ to the focus of the parabola is $\\frac{5}{2}$, then the coordinates of point $A$ are?",
    "fact_expressions": "C: Parabola;p: Number;A: Point;p>0;Expression(C) = (y^2 = 2*p*x);Expression(Directrix(C)) = (x = -1/2);PointOnCurve(A,C);Quadrant(A)=1;Distance(A, Focus(C)) = 5/2",
    "query_expressions": "Coordinate(A)",
    "answer_expressions": "(2,2)",
    "fact_spans": "[[[2, 28], [52, 55], [75, 78]], [[10, 28]], [[59, 62], [70, 74], [99, 103]], [[10, 28]], [[2, 28]], [[2, 50]], [[52, 62]], [[59, 68]], [[70, 97]]]",
    "query_spans": "[[[99, 108]]]",
    "process": "Since the directrix equation of the parabola $ C: y^{2} = 2px $ ($ p > 0 $) is $ x = -\\frac{1}{2} $, $ -\\frac{p}{2} = -\\frac{1}{2} $, so $ p = 1 $, hence $ C: y^{2} = 2x $. Given that the distance from point A to the focus of the parabola is $ \\frac{5}{2} $, $ x_{A} + \\frac{1}{2} = \\frac{5}{2} $, thus $ x_{A} = 2 $. Since point A lies in the first quadrant, $ y_{A}^{2} = 2 \\times 2 $, $ y_{A} = 2 $, so $ A(2, 2) $."
  },
  {
    "text": "Let the origin of the coordinate system be $O$, and let the parabola $y^{2}=2 x$ intersect a line passing through the focus at points $A$ and $B$. Then $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}$=?",
    "fact_expressions": "G: Parabola;H: Line;O: Origin;A:Point;B: Point;Expression(G) = (y^2 = 2*x);Intersection(G, H) = {A,B};PointOnCurve(Focus(G), H)",
    "query_expressions": "DotProduct(VectorOf(O, A), VectorOf(O, B))",
    "answer_expressions": "-3/4",
    "fact_spans": "[[[10, 24]], [[29, 31]], [[6, 9]], [[33, 36]], [[37, 41]], [[10, 24]], [[10, 43]], [[10, 31]]]",
    "query_spans": "[[[45, 96]]]",
    "process": ""
  },
  {
    "text": "Given point $M(-1,1)$ and parabola $C$: $y^{2}=4x$, a line is drawn through the focus of $C$ intersecting $C$ at points $A$ and $B$. If $\\angle AMB=90^{\\circ}$, then the chord length $|AB|=$?",
    "fact_expressions": "C: Parabola;G: Line;M: Point;A: Point;B: Point;Expression(C) = (y^2 = 4*x);Coordinate(M) = (-1, 1);PointOnCurve(Focus(C), G);Intersection(G, C) = {A, B};AngleOf(A, M, B) = ApplyUnit(90, degree);IsChordOf(LineSegmentOf(A,B),C)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "5",
    "fact_spans": "[[[13, 32], [34, 37], [44, 47]], [[41, 43]], [[2, 12]], [[49, 52]], [[53, 56]], [[13, 32]], [[2, 12]], [[33, 43]], [[41, 58]], [[60, 85]], [[44, 96]]]",
    "query_spans": "[[[89, 98]]]",
    "process": "Let the slope of line AB be $k$, points $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then  \n$$\n\\begin{cases}\ny_{1}^{2}=4x_{1} \\\\\ny_{2}^{2}=4x_{2}\n\\end{cases}\n\\therefore y_{1}^{2}-y_{2}^{2}=4(x_{1}-x_{2}), \\therefore k=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{4}{y_{1}+y_{2}},\n$$\nlet the midpoint of AB be $M'(x_{0},y_{0})$, the focus of the parabola be $F$, draw perpendiculars from points $A$, $B$ to the directrix $x=-1$, with feet $A'$, then  \n$$\n|MM|=\\frac{1}{2}|AB|=\\frac{1}{2}(|AF|+|BF|)=\\frac{1}{2}(|AA|+|BB|).\n$$\n$\\because M'(x_{0},y_{0})$ is the midpoint of AB, $\\therefore M$ is the midpoint of $A'B'$, $\\therefore MM$ is parallel to the x-axis.  \n$\\therefore y_{1}+y_{2}=2$, $k=2$. Let the angle of inclination of line AB be $\\theta$, then:  \n$$\n\\tan\\theta=2, \\text{ hence } \\sin\\theta=\\frac{2}{\\sqrt{5}}, |AB|=\\frac{2p}{\\sin^{2}\\theta}=\\underline{4}=\\text{from the chord length formula we get:}=5\n$$"
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{2+\\lambda}-\\frac{y^{2}}{1+\\lambda}=1$ represents a hyperbola, what is the range of values for $\\lambda$?",
    "fact_expressions": "G: Hyperbola;lambda: Number;Expression(G) = (x^2/(lambda + 2) - y^2/(lambda + 1) = 1)",
    "query_expressions": "Range(lambda)",
    "answer_expressions": "(-oo,-2)+(-1,+oo)",
    "fact_spans": "[[[57, 60]], [[62, 71]], [[2, 60]]]",
    "query_spans": "[[[62, 78]]]",
    "process": ""
  },
  {
    "text": "If the circle $(x-2)^{2}+y^{2}=1$ is tangent to the asymptotes of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-y^{2}=1(a>0)$, what is the equation of the asymptotes of hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;a: Number;G: Circle;a>0;Expression(C) = (-y^2 + x^2/a^2 = 1);Expression(G) = (y^2 + (x - 2)^2 = 1);IsTangent(G, Asymptote(C))",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y = pm*(sqrt(3)/3)*x",
    "fact_spans": "[[[71, 77], [22, 63]], [[29, 63]], [[1, 21]], [[29, 63]], [[22, 63]], [[1, 21]], [[1, 69]]]",
    "query_spans": "[[[71, 85]]]",
    "process": "Hyperbola $ C:\\frac{x^{2}}{a^{2}}-y^{2}=1 $ ($ a>0 $) has asymptotes given by $ y=\\pm\\frac{1}{a}x $, or $ x\\pm ay=0 $. Since the circle $ (x-2)^{2}+y^{2}=1 $ is tangent to the line, we have $ \\frac{2}{\\sqrt{1+a^{2}}}=1 $, solving gives $ a^{2}=3 $, so the asymptote equations of the hyperbola are $ y= $"
  },
  {
    "text": "The distance from a moving point $P$ to the point $A(0,2)$ is 2 less than the distance from $P$ to the line $l$: $y=-4$. Then the equation of the trajectory of point $P$ is?",
    "fact_expressions": "l: Line;A: Point;P: Point;Coordinate(A) = (0, 2);Expression(l) = (y = -4);Distance(P, A) = Distance(P, l) - 2",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2 = 8*y",
    "fact_spans": "[[[20, 32]], [[6, 15]], [[2, 5], [41, 45]], [[6, 15]], [[20, 32]], [[2, 39]]]",
    "query_spans": "[[[41, 52]]]",
    "process": "Since the distance from the moving point P to the point A(0,2) is 2 less than its distance to the line l: y = -4, the distance from the moving point P to the point A(0,2) equals its distance to the line l: y = -2. By the definition of a parabola, the moving point P traces a parabola with focus A(0,2) and directrix y = -2, and the equation is: x^{2} = 8y"
  },
  {
    "text": "Let the origin of the coordinate system be $O$, and let the parabola $y^{2}=2 x$ intersect a line passing through the focus at points $A$ and $B$. Then $O A \\cdot O B$=?",
    "fact_expressions": "G: Parabola;H: Line;O: Origin;A: Point;B: Point;Expression(G) = (y^2 = 2*x);PointOnCurve(Focus(G),H);Intersection(G, H) = {A, B}",
    "query_expressions": "LineSegmentOf(O, A)*LineSegmentOf(O, B)",
    "answer_expressions": "-3/4",
    "fact_spans": "[[[10, 24]], [[29, 31]], [[6, 9]], [[33, 36]], [[37, 40]], [[10, 24]], [[10, 31]], [[10, 42]]]",
    "query_spans": "[[[44, 61]]]",
    "process": ""
  },
  {
    "text": "If the directrix of the parabola $y^{2}=2 p x(p>0)$ coincides with one of the directrices of the hyperbola $x^{2}-y^{2}=1$, then $p$=?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 2*(p*x));p: Number;p>0;G: Hyperbola;Expression(G) = (x^2 - y^2 = 1);Directrix(H) = OneOf(Directrix(G))",
    "query_expressions": "p",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 22]], [[1, 22]], [[53, 56]], [[4, 22]], [[26, 44]], [[26, 44]], [[1, 51]]]",
    "query_spans": "[[[53, 58]]]",
    "process": "The directrix of the parabola $ y^{2} = 2px $ ($ p > 0 $) is: $ x = -\\frac{p}{2} $, the left directrix of the hyperbola is: $ x = -\\frac{a^{2}}{c} = -\\frac{1}{\\sqrt{2}} $. According to the problem, $ -\\frac{p}{2} = -\\frac{1}{\\sqrt{2}} $, solving gives $ p = \\sqrt{2} $. Hence, the answer is $ \\sqrt{2} $."
  },
  {
    "text": "Given that $M(x_{0}, y_{0})$ is a moving point on the parabola $y=\\frac{1}{8} x^{2}$, and the coordinates of point $N$ are $(2 \\sqrt{3}, 0)$, then the minimum value of $y_{0}+|\\overrightarrow{M N}|$ is?",
    "fact_expressions": "G: Parabola;M: Point;N: Point;x0: Number;y0: Number;Coordinate(N) = (2*sqrt(3), 0);Coordinate(M) = (x0, y0);Expression(G) = (y = x^2/8);PointOnCurve(M, G)",
    "query_expressions": "Min(y0 + Abs(VectorOf(M, N)))",
    "answer_expressions": "2",
    "fact_spans": "[[[20, 44]], [[2, 19]], [[50, 54]], [[2, 19]], [[2, 19]], [[50, 75]], [[2, 19]], [[20, 44]], [[2, 48]]]",
    "query_spans": "[[[78, 114]]]",
    "process": "From the graph, $ y_{0}+|\\overrightarrow{MN}|=MN+MX $. Combining with the definition of a parabola, consider transforming it and comparing distances to two fixed points $ P $ and $ N $, to find the extremum. [Solution] The parabola $ y=\\frac{1}{8}x^{2} $ is rewritten in standard form as $ x^{2}=8y $, with focus $ P(0,2) $ and directrix $ y=-2 $. $ y_{0}+|\\overrightarrow{MN}|=MN+MX=MN+MQ-XQ=MN+PM-MQ=MN+PM-2 $. According to the principle that the shortest distance between two points is a straight line, we have $ MN+PM\\geqslant PN=4 $."
  },
  {
    "text": "If the focal distance of the ellipse $\\frac{x^{2}}{2 m}+\\frac{y^{2}}{5}=1$ is equal to $4$, then the real number $m$=?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (y^2/5 + x^2/(2*m) = 1);FocalLength(G) = 4",
    "query_expressions": "m",
    "answer_expressions": "{9/2,1/2}",
    "fact_spans": "[[[1, 40]], [[50, 55]], [[1, 40]], [[1, 48]]]",
    "query_spans": "[[[50, 57]]]",
    "process": "Since the focal distance of the ellipse $\\frac{x^{2}}{2m}+\\frac{y^{2}}{5}=1$ is equal to 4, we have $2\\sqrt{5-2m}=4$ or $2\\sqrt{2m-5}=4$, solving which gives $m=\\frac{9}{2}$ or $m=\\frac{1}{2}$."
  },
  {
    "text": "If the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{m}=1$ passes through the point $(\\sqrt{3}, \\sqrt{2})$, then the length of the minor axis of this ellipse is?",
    "fact_expressions": "G: Ellipse;m: Number;H: Point;Expression(G) = (x^2/9 + y^2/m = 1);Coordinate(H) = (sqrt(3), sqrt(2));PointOnCurve(H, G)",
    "query_expressions": "Length(MinorAxis(G))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[1, 38], [66, 68]], [[3, 38]], [[40, 63]], [[1, 38]], [[40, 63]], [[1, 63]]]",
    "query_spans": "[[[66, 74]]]",
    "process": "Since the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{m}=1$ passes through the point $(\\sqrt{3},\\sqrt{2})$, we have $\\frac{3}{9}+\\frac{2}{m}=1$. Solving this gives: $m=3$; therefore, the length of the minor axis of the ellipse is: $2\\sqrt{3}$"
  },
  {
    "text": "Given the ellipse $ C: \\frac{x^{2}}{4} + \\frac{y^{2}}{b^{2}} = 1 $ $ (0 < b < 2) $, the left focus of $ C $ is $ F $, $ M $ is a moving point on $ C $, and point $ N(0, \\sqrt{3}) $. If the maximum value of $ |MN| + |MF| $ is $ 6 $, then the eccentricity of $ C $ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/b^2 = 1);b: Number;0 < b;b < 2;F: Point;LeftFocus(C) = F;M: Point;PointOnCurve(M, C) = True;N: Point;Coordinate(N) = (0, sqrt(3));Max(Abs(LineSegmentOf(M, F)) + Abs(LineSegmentOf(M, N))) = 6",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 54], [67, 70], [117, 120]], [[2, 54]], [[8, 54]], [[8, 54]], [[8, 54]], [[59, 62]], [[2, 62]], [[63, 66]], [[63, 74]], [[75, 92]], [[75, 92]], [[94, 115]]]",
    "query_spans": "[[[117, 126]]]",
    "process": "Let the right focus be F. By the definition of the ellipse, |MF| = 4 - |MF|, |MN| + |MF| = |MN| - |MF| + 4 \\leqslant |NF| + 4, with equality if and only if points M, N, F are collinear. |NF| + 4 = 6. Given N(0,\\sqrt{3}), F(c,0), \\therefore c = 1, e = \\frac{c}{a} = \\frac{1}{2}"
  },
  {
    "text": "Draw a tangent from the left focus $F(-c, 0)$ $(c>0)$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(b>a>0)$ to the circle $x^{2}+y^{2}=a^{2}$, with point of tangency $E$. Extend $FE$ to intersect the parabola $y^{2}=4 c x$ at point $P$. Let $O$ be the origin. If $\\overrightarrow{O E}=\\frac{1}{2}(\\overrightarrow{O F}+\\overrightarrow{O P})$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;c: Number;D: Circle;F: Point;E: Point;O: Origin;P: Point;l1:Line;b > a;a > 0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 4*c*x);Expression(D) = (x^2 + y^2 = a^2);Coordinate(F) = (-c, 0);LeftFocus(G)=F;TangentOfPoint(F,D)=l1;TangentPoint(l1,D)=E;Intersection(OverlappingLine(LineSegmentOf(F,E)),H)=P;c>0;VectorOf(O,E)=(VectorOf(O,F)+VectorOf(O,P))/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(1+sqrt(5))/2",
    "fact_spans": "[[[1, 54], [224, 227]], [[4, 54]], [[4, 54]], [[113, 129]], [[58, 73]], [[74, 94]], [[58, 73]], [[101, 104]], [[135, 138]], [[130, 134]], [], [[4, 54]], [[4, 54]], [[1, 54]], [[113, 129]], [[74, 94]], [[58, 73]], [[1, 73]], [[0, 97]], [[0, 104]], [[105, 134]], [[58, 73]], [[145, 222]]]",
    "query_spans": "[[[224, 233]]]",
    "process": ""
  },
  {
    "text": "The asymptotes of the hyperbola are given by $y=\\pm 2 x$, and it has the same foci as the ellipse $\\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1$. What are the coordinates of the foci? What is the equation of the hyperbola?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/49 + y^2/24 = 1);Expression(Asymptote(G)) = (y = pm*2*x);Focus(G) = Focus(H)",
    "query_expressions": "Coordinate(Focus(G));Expression(G)",
    "answer_expressions": "(pm*5,0)\nx^2/5-y^2/20=1",
    "fact_spans": "[[[2, 5], [73, 74], [80, 83]], [[26, 65]], [[26, 65]], [[2, 23]], [[2, 71]]]",
    "query_spans": "[[[73, 80]], [[80, 88]]]",
    "process": ""
  },
  {
    "text": "If real numbers $x$, $y$ satisfy the condition $x^{2}-y^{2}=1$, then the range of $\\frac{1}{x^{2}}+\\frac{2 y}{x}$ is?",
    "fact_expressions": "x: Real;y: Real;x^2 - y^2 = 1",
    "query_expressions": "Range(1/x^2 + 2*y/x)",
    "answer_expressions": "(-2, 2)",
    "fact_spans": "[[[1, 7]], [[9, 12]], [[16, 31]]]",
    "query_spans": "[[[33, 71]]]",
    "process": ""
  },
  {
    "text": "The distance from the focus of the parabola $x^{2}=8 y$ to the line $x-\\sqrt{3} y=0$ is?",
    "fact_expressions": "G: Parabola;H: Line;Expression(G) = (x^2 = 8*y);Expression(H) = (x - sqrt(3)*y = 0)",
    "query_expressions": "Distance(Focus(G),H)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[0, 14]], [[18, 36]], [[0, 14]], [[18, 36]]]",
    "query_spans": "[[[0, 41]]]",
    "process": "The focus of the parabola \\( x^{2} = 8y \\) is \\( (0, 2) \\), and the distance to the line \\( x - \\sqrt{3}y = 0 \\) is \\( d = \\frac{|0 - 2\\sqrt{3}|}{2} = \\sqrt{3} \\)."
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{3}=1$ with foci on the $x$-axis is $\\frac{1}{2}$, then the value of the real number $m$ is?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (y^2/3 + x^2/m = 1);PointOnCurve(Focus(G), xAxis);Eccentricity(G) = (1/2)",
    "query_expressions": "m",
    "answer_expressions": "4",
    "fact_spans": "[[[9, 46]], [[66, 71]], [[9, 46]], [[0, 46]], [[9, 64]]]",
    "query_spans": "[[[66, 75]]]",
    "process": "From the given information, $ a^{2} = m $, $ b^{2} = 3 $. Since the ellipse's eccentricity is $ \\frac{1}{2} $, it follows that $ e = \\sqrt{1 - \\left( \\frac{b}{a} \\right)^{2}} = \\sqrt{1 - \\frac{3}{m}} = \\frac{1}{2} $. Solving gives $ m = 4 $."
  },
  {
    "text": "The maximum value of the difference between the distance from a point $P$ on the parabola $y^{2}=4x$ to the line $x=-1$ and the distance to the point $Q(2,2)$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G) = True;H: Line;Expression(H) = (x = -1);Q: Point;Coordinate(Q) = (2, 2)",
    "query_expressions": "Max(Distance(P, H) - Distance(P, Q))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[17, 20]], [[0, 20]], [[21, 29]], [[21, 29]], [[34, 43]], [[34, 43]]]",
    "query_spans": "[[[17, 54]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$, $P$ is a point on $C$, $|P F_{1}| = 3|P F_{2}|$, and $\\cos \\angle P F_{1} F_{2} = \\frac{2 \\sqrt{2}}{3}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;P: Point;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, F2));Cos(AngleOf(P, F1, F2)) = (2*sqrt(2))/3",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[18, 24], [34, 37], [115, 118]], [[30, 33]], [[2, 9]], [[10, 17]], [[2, 29]], [[2, 29]], [[30, 40]], [[41, 63]], [[65, 113]]]",
    "query_spans": "[[[115, 124]]]",
    "process": "By the given condition, |PF_{1}| - |PF_{2}| = 2a, and |PF_{1}| = 3|PF_{2}|, then |PF_{1}| = 3|PF_{2}| = 3a, while |F_{1}F_{2}| = 2c. Therefore, \\cos\\angle PF_{1}F_{2} = \\frac{|PF_{1}|^{2} + |F_{1}F_{2}|^{2} - |PF_{2}|^{2}}{2|PF_{1}||F_{1}F_{2}|} = \\frac{2a^{2} + c^{2}}{3ac} = \\frac{2\\sqrt{2}}{3}, then (\\sqrt{2}a - c)^{2} = 0, so \\sqrt{2}a = c, hence e = \\frac{c}{a} = \\sqrt{2}."
  },
  {
    "text": "The focus of the parabola $y=\\frac{1}{16} x^{2}$ coincides with the upper focus of the hyperbola $\\frac{y^{2}}{3}-\\frac{x^{2}}{m}=1$, then $m$=?",
    "fact_expressions": "G: Hyperbola;m: Number;H: Parabola;Expression(G) = (y^2/3 - x^2/m = 1);Expression(H) = (y = x^2/16);Focus(H) = UpperFocus(G)",
    "query_expressions": "m",
    "answer_expressions": "13",
    "fact_spans": "[[[29, 67]], [[75, 78]], [[0, 25]], [[29, 67]], [[0, 25]], [[0, 73]]]",
    "query_spans": "[[[75, 80]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola equation $x^{2}-\\frac{y^{2}}{3}=1$, then the equation of the line $l$ containing the chord with midpoint $A(2,1)$ is?",
    "fact_expressions": "l: Line ;H:LineSegment;G: Hyperbola;A: Point;Coordinate(A) = (2, 1);Expression(G) = (x^2 - y^2/3 = 1);IsChordOf(H, G);MidPoint(H) = A;OverlappingLine(H,l)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[50, 55]], [], [[2, 5]], [[35, 43]], [[35, 43]], [[2, 32]], [[2, 48]], [[2, 48]], [[2, 55]]]",
    "query_spans": "[[[50, 60]]]",
    "process": "Let the equation of line $ l $ be $ y - 1 = k(x - 2) $, that is, $ y = kx - 2k + 1 $. Let the two intersection points be $ (x_{1}, y_{1}) $ and $ (x_{2}, y_{2}) $, then $ x_{1} + x_{2} = 4 $. Consider the system \n\\[\n\\begin{cases}\nx^{2} - \\frac{y^{2}}{3} = 1 \\\\\ny = kx - 2k + 1\n\\end{cases}\n\\]\nEliminating $ y $ gives $ (3 - k^{2})x^{2} + 2k(2k - 1)x - (2k - 1)^{2} - 3 = 0 $. $ \\therefore $ The equation of line $ l $ is: $ y = 6x - 11 $, that is, $ 6x - y - 11 = 0 $."
  },
  {
    "text": "The standard equation of an ellipse that has the same eccentricity as the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ and passes through the point $(2,-\\sqrt{3})$ is?",
    "fact_expressions": "G: Ellipse;H: Point;C:Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);Coordinate(H) = (2, -sqrt(3));Eccentricity(G)=Eccentricity(C);PointOnCurve(H,C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "{x^2/8+y^2/6=1,3*y^2/25+4*y^2/25=1}",
    "fact_spans": "[[[1, 38]], [[48, 64]], [[65, 67]], [[1, 38]], [[48, 64]], [[0, 67]], [[47, 67]]]",
    "query_spans": "[[[65, 74]]]",
    "process": ""
  },
  {
    "text": "Let the left and right foci of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$ be $F_{1}$ and $F_{2}$, respectively. A line $l$ passing through $F_{1}$ intersects the left branch of the hyperbola at points $A$ and $B$. Then the minimum value of $|B F_{2}|+|A F_{2}|$ is?",
    "fact_expressions": "l: Line;G: Hyperbola;B: Point;F2: Point;A: Point;F1: Point;Expression(G) = (x^2/4 - y^2/2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, l);Intersection(l, LeftPart(G)) = {A, B}",
    "query_expressions": "Min(Abs(LineSegmentOf(A, F2)) + Abs(LineSegmentOf(B, F2)))",
    "answer_expressions": "10",
    "fact_spans": "[[[73, 78]], [[1, 39], [79, 82]], [[89, 92]], [[56, 63]], [[85, 88]], [[48, 55], [65, 72]], [[1, 39]], [[1, 63]], [[1, 63]], [[64, 78]], [[73, 94]]]",
    "query_spans": "[[[96, 123]]]",
    "process": "According to the definition of hyperbola, transform and solve accordingly. From the standard equation of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$, we obtain $a=2$. By the definition of hyperbola, $|AF_{2}|-|AF_{1}|=4$, $|BF_{2}|-|BF_{1}|=4$, so $|AF_{2}|-|AF_{1}|+|BF_{2}|-|BF_{1}|=8$. Since $|AF_{1}|+|BF_{1}|=|AB|$, when line $l$ passes through point $F_{1}$ and is perpendicular to the $x$-axis, $|AB|$ is minimized. Therefore, $(|AF_{2}|+|BF_{2}|)_{\\min}=|AB|_{\\min}+8=\\frac{2b^{2}}{a}+8=10$."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $y^{2}=x$, and $A$, $B$ are two points on the parabola such that $|A F|+|B F|=3$, then the distance from the midpoint of segment $A B$ to the $y$-axis is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;F: Point;Expression(G) = (y^2 = x);Focus(G) = F;PointOnCurve(A, G);PointOnCurve(B, G);Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F)) = 3",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A,B)), yAxis)",
    "answer_expressions": "5/4",
    "fact_spans": "[[[6, 18], [31, 34]], [[22, 25]], [[26, 29]], [[2, 5]], [[6, 18]], [[2, 21]], [[22, 38]], [[22, 38]], [[39, 54]]]",
    "query_spans": "[[[56, 76]]]",
    "process": "According to the equation of the parabola, find the equation of the directrix. Using the definition of a parabola, the distance from a point on the parabola to the focus equals the distance to the directrix. Set up an equation to find the x-coordinate of the midpoint of A and B. Find the distance from the midpoint of segment AB to the y-axis. From the given conditions, F(\\frac{1}{4},0), the equation of the directrix is: x=-\\frac{1}{4}. Let A(x_{1},y_{1}), B(x_{2},y_{2}), |AF|+|BF|=x_{1}+\\frac{1}{4}+x_{2}+\\frac{1}{4}=3, therefore x_{1}+x_{2}=3-\\frac{1}{2}=\\frac{5}{2}, the distance from the midpoint of segment AB to the y-axis is \\frac{x_{1}+x_{2}}{2}=\\frac{5}{4}"
  },
  {
    "text": "Given that a line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$, and $|A F|=4$, then $|B F|=$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, F)) = 4",
    "query_expressions": "Abs(LineSegmentOf(B, F))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[3, 17], [28, 31]], [[24, 26]], [[32, 35]], [[20, 23]], [[36, 39]], [[3, 17]], [[3, 23]], [[2, 26]], [[24, 41]], [[42, 51]]]",
    "query_spans": "[[[53, 62]]]",
    "process": ""
  },
  {
    "text": "The line $y=2x-1$ intersects the hyperbola $\\frac{x^{2}}{8}-\\frac{y^{2}}{4}=1$ at points $A$ and $B$. What are the coordinates of the midpoint of $AB$?",
    "fact_expressions": "G: Hyperbola;H: Line;A: Point;B: Point;Expression(G) = (x^2/8 - y^2/4 = 1);Expression(H) = (y = 2*x - 1);Intersection(H,G)={A,B}",
    "query_expressions": "Coordinate(MidPoint(LineSegmentOf(A, B)))",
    "answer_expressions": "(4/7,1/7)",
    "fact_spans": "[[[12, 50]], [[0, 11]], [[52, 55]], [[56, 59]], [[12, 50]], [[0, 11]], [[0, 61]]]",
    "query_spans": "[[[63, 75]]]",
    "process": ""
  },
  {
    "text": "The left focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $F$. A line $l$ with slope $1$ passing through $F$ intersects the ellipse at points $A$ and $B$, and $|AB|=8$. If the slope of the tangent to the ellipse at $A$ is $k_{1}$, the slope of line $OA$ is $k_{2}$ (where $O$ is the origin), and $k_{1} k_{2}=-\\frac{1}{4}$, then what is the focal distance of the ellipse?",
    "fact_expressions": "l: Line;G: Ellipse;b: Number;a: Number;L: Line;A: Point;O: Origin;B: Point;F: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F;PointOnCurve(F, l);Slope(l) = 1;Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, B)) = 8;IsTangent(L,G);PointOnCurve(A,L);Slope(L)=k1;k1:Number;Slope(LineOf(O,A))=k2;k2:Number;k1*k2=-1/4",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "10*sqrt(3)",
    "fact_spans": "[[[73, 78]], [[0, 52], [79, 81], [109, 111], [184, 186]], [[2, 52]], [[2, 52]], [], [[82, 85], [105, 108]], [[145, 148]], [[86, 89]], [[62, 65], [57, 60]], [[2, 52]], [[2, 52]], [[0, 52]], [[0, 60]], [[61, 78]], [[66, 78]], [[73, 91]], [[93, 102]], [[104, 114]], [[104, 114]], [[104, 124]], [[117, 124]], [[125, 142]], [[135, 142]], [[156, 182]]]",
    "query_spans": "[[[184, 191]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the right focus is $F_{2}$, points $M$ and $N$ lie on the same asymptote of the hyperbola, and $O$ is the origin. If the line $F_{2} M$ is parallel to the other asymptote of the hyperbola, $O F_{2} \\perp F_{2} N$, and $|F_{2} M|=\\frac{\\sqrt{3}}{2}|F_{2} N|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F2: Point;M: Point;O: Origin;N: Point;L1:Line;L2:Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F2;PointOnCurve(M,L1);PointOnCurve(N,L1);IsParallel(LineOf(F2,M),L2);IsPerpendicular(LineSegmentOf(O, F2), LineSegmentOf(F2, N));Abs(LineSegmentOf(F2, M)) = (sqrt(3)/2)*Abs(LineSegmentOf(F2, N));OneOf(Asymptote(G)) = L1;OneOf(Asymptote(G)) = L2;Negation(L1=L2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[2, 58], [80, 83], [117, 120], [195, 198]], [[5, 58]], [[5, 58]], [[63, 70]], [[71, 75]], [[92, 95]], [[76, 79]], [], [], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 70]], [[71, 91]], [[71, 91]], [[103, 127]], [[129, 152]], [[153, 192]], [78, 87], [115, 124], [115, 124]]",
    "query_spans": "[[[195, 204]]]",
    "process": "As shown in the figure, let the angle of inclination of the asymptote $ y = \\frac{b}{a}x $ be $ \\theta $. Then $ \\theta \\in (0, \\frac{\\pi}{2}) $, $ \\angle NMF_{2} = 2\\theta $, $ \\angle ONF_{2} = \\frac{\\pi}{2} - \\theta $. In $ \\triangle MNF_{2} $, by the sine law we have $ \\frac{NF_{2}}{MF_{2}} = \\frac{\\sin 2\\theta}{\\sin(\\frac{\\pi}{2} - \\theta)} $, solving yields $ \\sin\\theta = \\frac{\\sqrt{3}}{3} $, $ \\tan\\theta = \\frac{1}{\\sqrt{2}} $, i.e., $ \\frac{b}{a} = \\frac{1}{\\sqrt{2}} $. From $ c^{2} = a^{2} + b^{2} $ we obtain $ \\frac{c}{a} = \\frac{\\sqrt{6}}{2} $. $ \\therefore $ the eccentricity of this hyperbola is $ \\frac{\\sqrt{6}}{2} $."
  },
  {
    "text": "The focal distance of a hyperbola is equal to $2$ times the distance between the two directrices of the hyperbola. What is the eccentricity of the hyperbola?",
    "fact_expressions": "G: Hyperbola;l1: Line;l2: Line;Directrix(G) = {l1,l2};FocalLength(G) = Distance(l1,l2)*2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 3], [8, 11], [26, 29]], [], [], [[8, 16]], [[0, 24]]]",
    "query_spans": "[[[26, 35]]]",
    "process": ""
  },
  {
    "text": "The line passing through the left focus $F$ of the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$ intersects the ellipse $C$ at points $A$ and $B$. If the perpendicular bisector of segment $AB$ has exactly one intersection point $M$ with the $x$-axis and exactly one intersection point $N$ with the $y$-axis, then the range of $|M F|$ is?",
    "fact_expressions": "C: Ellipse;G: Line;B: Point;A: Point;M: Point;F: Point;N:Point;Expression(C) = (x^2/2 + y^2 = 1);LeftFocus(C)=F;PointOnCurve(F, G);Intersection(G, C) = {A, B};Intersection(PerpendicularBisector(LineSegmentOf(A,B)),xAxis)=M;Intersection(PerpendicularBisector(LineSegmentOf(A,B)),yAxis)=N",
    "query_expressions": "Range(Abs(LineSegmentOf(M, F)))",
    "answer_expressions": "(1/2,1)",
    "fact_spans": "[[[1, 33], [43, 48]], [[40, 42]], [[54, 57]], [[50, 53]], [[91, 94]], [[36, 39]], [[95, 98]], [[1, 33]], [[1, 39]], [[0, 42]], [[40, 59]], [[61, 98]], [[61, 98]]]",
    "query_spans": "[[[100, 114]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1  (b>0)$ has eccentricity $2$, then the distance from one of its foci to one of its asymptotes is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/b^2 = 1);b: Number;b>0;Eccentricity(G) = 2",
    "query_expressions": "Distance(OneOf(Focus(G)), OneOf(Asymptote(G)))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[2, 51], [61, 62]], [[2, 51]], [[5, 51]], [[5, 51]], [[2, 59]]]",
    "query_spans": "[[[61, 79]]]",
    "process": ""
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=2 px(p>0)$ is $F(2,0)$, draw a perpendicular line from point $A(3,2)$ to its directrix, intersecting the parabola at point $E$, then $|EF|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Coordinate(F) = (2, 0);Focus(G) = F;A: Point;Coordinate(A) = (3, 2);L: Line;PointOnCurve(A,L);IsPerpendicular(L, Directrix(G));E: Point;Intersection(L, G) = E",
    "query_expressions": "Abs(LineSegmentOf(E, F))",
    "answer_expressions": "5/2",
    "fact_spans": "[[[2, 22], [47, 48], [55, 58]], [[2, 22]], [[5, 22]], [[5, 22]], [[26, 34]], [[26, 34]], [[2, 34]], [[37, 46]], [[37, 46]], [], [[36, 53]], [[36, 53]], [[62, 65]], [[36, 65]]]",
    "query_spans": "[[[68, 76]]]",
    "process": "By the given conditions, the focus of the parabola is $ F(2,0) $, the equation of the directrix is $ x = -2 $, and $ p = 4 $, so the equation of the parabola is $ y^{2} = 8x $. Let the perpendicular from point $ A(3,2) $ to its directrix intersect the directrix at point $ G $, then $ G(-2,2) $. Let the coordinates of point $ E $ be $ E(x,2) $, then $ 4 = 8x $, solving gives $ x = \\frac{1}{2} $. By the definition of the parabola, $ |EF| = |EG| = |AG| - |AE| = 5 - \\frac{5}{2} = \\frac{5}{2} $."
  },
  {
    "text": "If $A$ and $B$ are the left and right foci of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, and $C$ is a point on this hyperbola such that $\\cos \\angle A C B = \\frac{3}{5}$, then what is the perimeter of $\\triangle A B C$?",
    "fact_expressions": "A: Point;B: Point;G: Hyperbola;Expression(G) = (x^2/4 - y^2/12 = 1);LeftFocus(G) = A;RightFocus(G) = B;C: Point;PointOnCurve(C, G) = True;Cos(AngleOf(A, C, B)) = 3/5",
    "query_expressions": "Perimeter(TriangleOf(A, B, C))",
    "answer_expressions": "24",
    "fact_spans": "[[[1, 4]], [[5, 8]], [[9, 48], [60, 63]], [[9, 48]], [[1, 54]], [[1, 54]], [[55, 58]], [[55, 66]], [[68, 99]]]",
    "query_spans": "[[[101, 123]]]",
    "process": ""
  },
  {
    "text": "Given that the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1$ $(a>0)$ is centered at the origin, and its right focus coincides with the focus of the parabola $y^{2}=16x$, then the eccentricity of this hyperbola equals?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;O: Origin;a>0;Expression(G) = (-y^2/9 + x^2/a^2 = 1);Expression(H) = (y^2 = 16*x);Center(G) = O;RightFocus(G) = Focus(H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "4*sqrt(7)/7",
    "fact_spans": "[[[2, 51], [85, 88]], [[5, 51]], [[62, 77]], [[55, 57]], [[5, 51]], [[2, 51]], [[62, 77]], [[2, 57]], [[2, 82]]]",
    "query_spans": "[[[85, 95]]]",
    "process": ""
  },
  {
    "text": "Let $P$ be an intersection point of ellipse $C_{1}$ and hyperbola $C_{2}$ sharing common foci $F_{1}$, $F_{2}$, with $P F_{1} \\perp P F_{2}$. The eccentricity of ellipse $C_{1}$ is $e_{1}$, the eccentricity of hyperbola $C_{2}$ is $e_{2}$. If $e_{2}=3 e_{1}$, then $e_{1}$=?",
    "fact_expressions": "C1: Ellipse;C2: Hyperbola;F1: Point;F2: Point;Focus(C1) = {F1, F2};Focus(C2) = {F1, F2};P: Point;OneOf(Intersection(C1, C2)) = P;IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));e1: Number;e2: Number;Eccentricity(C1) = e1;Eccentricity(C2) = e2;e2 = 3*e1",
    "query_expressions": "e1",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[26, 35], [77, 86]], [[36, 46], [99, 109]], [[10, 17]], [[18, 25]], [[5, 46]], [[5, 46]], [[1, 4]], [[1, 51]], [[53, 76]], [[91, 98], [140, 147]], [[114, 121]], [[77, 98]], [[99, 121]], [[123, 138]]]",
    "query_spans": "[[[140, 149]]]",
    "process": "Let $\\angle F_{1}AF_{2}=2\\theta$. According to the geometric properties of the ellipse, we have $S_{A}PF_{1}F_{2}=b_{1}^{2}\\tan\\theta=b_{1}^{2}$. Since $e_{1}=\\frac{c}{a_{1}}$, $\\therefore a_{1}=\\frac{c}{e_{1}}$, $\\therefore b_{1}^{2}=a_{1}^{2}-c^{2}=c^{2}\\left(\\frac{1}{e_{1}^{2}}-1\\right)$. According to the geometric properties of the hyperbola, we have $S_{A}PF_{1}F_{2}=\\frac{b_{2}^{2}}{\\tan\\theta}=b_{2}^{2}$. Since $e_{2}=\\frac{c}{a_{2}}$, $\\therefore a_{2}=\\frac{c}{e_{2}}$, $\\therefore b_{2}^{2}=c^{2}-a_{2}^{2}=c^{2}\\left(1-\\frac{1}{e_{2}^{2}}\\right)$. $\\therefore c^{2}\\left(\\frac{1}{e_{1}^{2}}-1\\right)=c^{2}\\left(1-\\frac{1}{e_{2}^{2}}\\right)$, i.e., $\\frac{1}{e_{1}^{2}}+\\frac{1}{e_{2}^{2}}=2$. Since $3e_{1}=e_{2}$, $\\therefore e_{1}=\\frac{\\sqrt{5}}{3}$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, the length of the real axis is $4$, and the eccentricity is $\\sqrt{3}$. Point $P$ is a point on the hyperbola such that $\\angle F_{1} P F_{2}=120^{\\circ}$. Then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;Length(RealAxis(C)) = 4;Eccentricity(C) = sqrt(3);P: Point;PointOnCurve(P, C);AngleOf(F1, P, F2) = ApplyUnit(120, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "(8/3)*sqrt(3)",
    "fact_spans": "[[[2, 63], [116, 119]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[72, 79]], [[80, 87]], [[2, 87]], [[2, 87]], [[2, 95]], [[2, 110]], [[111, 115]], [[111, 122]], [[123, 157]]]",
    "query_spans": "[[[159, 186]]]",
    "process": "Hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) has left and right foci $ F_{1}, F_{2} $, respectively, with eccentricity $ \\sqrt{3} $, so $ \\frac{c}{a} = \\sqrt{3} $, thus $ 1 + \\frac{b^{2}}{a^{2}} = 3 $, hence $ \\frac{b}{a} = \\sqrt{2} $. The real axis length of hyperbola $ C $ is 4, so $ a = 2 $, then $ b = 2\\sqrt{2} $, $ c = 2\\sqrt{3} $. Point $ P $ is a point on the hyperbola, $ \\angle F_{1}PF_{2} = 120^{\\circ} $. Let $ |PF_{1}| = m $, $ |PF_{2}| = n $. By the definition of the hyperbola, $ |m - n| = 2a = 4 $, so $ m^{2} + n^{2} - 2mn = 16 $, $ \\textcircled{1} $. Since $ \\angle F_{1}PF_{2} = 120^{\\circ} $, we have $ m^{2} + n^{2} - 2mn\\cos120^{\\circ} = 4c^{2} = 48 $, $ \\textcircled{2} $. Solving equations $ \\textcircled{1} $ and $ \\textcircled{2} $ together yields $ mn = \\frac{32}{3} $, then the area $ S $ of $ \\triangle PF_{1}F_{2} $ is $ \\frac{1}{2}mn\\sin120^{\\circ} = \\frac{8\\sqrt{3}}{3} $;"
  },
  {
    "text": "Let the ellipse $\\Gamma$: $\\frac{x^{2}}{a^{2}}+y^{2}=1$ $(a>1)$ have left vertex $A$. A line $l$ passing through point $A$ intersects $\\Gamma$ at another point $B$, and intersects the $y$-axis at point $C$. If $|O A|=|O C|$, $|A B|=|B C|$, then $a=$?",
    "fact_expressions": "Gamma: Ellipse;Expression(Gamma) = (y^2 + x^2/a^2 = 1);a: Number;a>1;LeftVertex(Gamma) = A;A: Point;PointOnCurve(A, l);l: Line;Intersection(l, Gamma) = {A,B};B: Point;Intersection(l, yAxis) = C;C: Point;Abs(LineSegmentOf(O, A)) = Abs(LineSegmentOf(O, C));Abs(LineSegmentOf(A, B)) = Abs(LineSegmentOf(B, C));O: Origin",
    "query_expressions": "a",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[1, 47], [67, 75]], [[1, 47]], [[129, 132]], [[13, 47]], [[1, 54]], [[51, 54], [56, 60]], [[55, 66]], [[61, 66]], [[56, 85]], [[82, 85]], [[61, 98]], [[94, 98]], [[100, 113]], [[114, 127]], [[100, 113]]]",
    "query_spans": "[[[129, 134]]]",
    "process": "[Points and weight] From the ellipse equation and given conditions, we know: A(-a,0), C(0,\\pm a); then find the coordinates of B, and since B lies on the ellipse, solve for parameter a. According to the problem, A(-a,0). If line l intersects the y-axis above the x-axis, from |OA| = |OC| we have: C(0,a). Since |AB| = |BC|, i.e., B is the midpoint of AC, therefore B(-\\frac{a}{2},\\frac{a}{2}). Also, since B lies on the ellipse F, \\frac{1}{4}+\\frac{a^{2}}{4}=1, solving gives a=\\sqrt{3}."
  },
  {
    "text": "It is known that the hyperbola $C$ has its center at the origin and foci on the $x$-axis, with asymptotes given by $2x \\pm 3y = 0$, and the focal distance is $2\\sqrt{13}$. Then the standard equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;O: Origin;Center(C) = O;PointOnCurve(Focus(C), xAxis);Expression(Asymptote(C)) = (2*x + pm*3*y = 0);FocalLength(C) = 2*sqrt(13)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/9 - y^2/4 = 1",
    "fact_spans": "[[[2, 8], [24, 25], [65, 71]], [[12, 14]], [[2, 14]], [[2, 23]], [[24, 46]], [[24, 63]]]",
    "query_spans": "[[[65, 78]]]",
    "process": "The hyperbola $ C $ is centered at the origin, with foci on the $ x $-axis, and its asymptotes are given by $ 2x \\pm 3y = 0 $. The focal distance is $ 2\\sqrt{13} $. Let the equation of the hyperbola be: $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ $ (a > 0, b > 0) $. It follows that $ \\frac{b}{a} = \\frac{2}{3} $, and $ c^{2} = 13 = a^{2} + b^{2} $. Solving gives $ a = 3 $, $ b = 2 $. The standard equation of the required hyperbola is: $ \\frac{x^{2}}{9} - \\frac{y^{2}}{4} = 1 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0 , b>0)$, one of its asymptotes is parallel to the line $l$: $3 x-y+10=0$. Then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;a: Number;b: Number;a>0;b>0;Expression(C) = (-x^2/b^2 + y^2/a^2 = 1);Expression(l)=(3*x-y+10=0);IsParallel(OneOf(Asymptote(C)),l)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(10)/3",
    "fact_spans": "[[[73, 92]], [[2, 64], [94, 100]], [[10, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[2, 64]], [[73, 92]], [[2, 92]]]",
    "query_spans": "[[[94, 106]]]",
    "process": "The hyperbola $ C: \\frac{y^{2}}{a^{2}} - \\frac{x^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) has asymptotes $ y = \\pm \\frac{a}{b}x $. Since one of them is parallel to the line $ l: 3x - y + 10 = 0 $, it follows that $ \\frac{a}{b} = 3 $. Therefore, the eccentricity of hyperbola $ C $ is $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{1 + \\frac{1}{9}} = \\frac{\\sqrt{10}}{3} $."
  },
  {
    "text": "Given a point on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ such that the product of its distances to the two asymptotes is $\\frac{3}{4}$, if the eccentricity of the hyperbola is $2$, then what is the length of the imaginary axis of the hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;H: Point;PointOnCurve(H, G);L1: Line;L2: Line;Asymptote(G) = {L1, L2};Distance(H, L1)*Distance(H, L2) = 3/4;Eccentricity(G) = 2",
    "query_expressions": "Length(ImageinaryAxis(G))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[2, 48], [78, 81], [91, 94]], [[2, 48]], [[5, 48]], [[5, 48]], [], [[2, 52]], [], [], [[2, 57]], [[2, 76]], [[78, 89]]]",
    "query_spans": "[[[91, 100]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0 , b>0)$, let $P$ be a moving point on the $x$-axis. The line $y=2x+m$ $(m \\neq 0)$ passing through $P$ intersects the hyperbola $C$ at exactly one point. Then, the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-x^2/b^2 + y^2/a^2 = 1);b: Number;a: Number;a>0;b>0;P: Point;PointOnCurve(P, xAxis);G: Line;Expression(G) = (y = m + 2*x);m: Number;Negation(m=0);PointOnCurve(P, G);NumIntersection(G, C) = 1",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 64], [124, 130], [108, 114]], [[2, 64]], [[9, 65]], [[9, 65]], [[9, 65]], [[9, 65]], [[66, 70], [82, 85]], [[66, 79]], [[86, 107]], [[86, 107]], [[88, 107]], [[88, 107]], [[80, 107]], [[86, 122]]]",
    "query_spans": "[[[124, 136]]]",
    "process": "That is, the asymptotes of the hyperbola are parallel to the line y=2x+m, so \\frac{a}{b}=2, and the required eccentricity is e=\\frac{c}{a}=\\sqrt{1+(\\frac{b}{a})^{2}}=\\frac{\\sqrt{5}}{2}"
  },
  {
    "text": "Given that a line passing through the focus $F$ of the parabola $y^{2}=-2x$ with slope $\\sqrt{3}$ intersects the parabola at points $A$ and $B$, then $\\frac{|A F| \\cdot|B F|}{|A B|}$=?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;Expression(G) = (y^2 = -2*x);Focus(G) = F;Slope(H) = sqrt(3);PointOnCurve(F, H);Intersection(H, G) = {A, B}",
    "query_expressions": "(Abs(LineSegmentOf(A, F))*Abs(LineSegmentOf(B, F)))/Abs(LineSegmentOf(A, B))",
    "answer_expressions": "1/2",
    "fact_spans": "[[[3, 18], [43, 46]], [[40, 42]], [[48, 51]], [[21, 24]], [[52, 55]], [[3, 18]], [[3, 24]], [[26, 42]], [[2, 24]], [[40, 57]]]",
    "query_spans": "[[[59, 93]]]",
    "process": "The coordinates of the focus $ F $ of the parabola $ y^{2} = -2x $ are $ \\left(-\\frac{1}{2}, 0\\right) $. The equation of the line with slope $ \\sqrt{3} $ passing through the focus is $ y = \\sqrt{3}\\left(x + \\frac{1}{2}\\right) $. Solving the system of equations together with the parabola gives  \n\\[\n\\begin{cases}\ny^{2} = -2x \\\\\ny = \\sqrt{3}\\left(x + \\frac{1}{2}\\right)\n\\end{cases}\n\\]\nSimplifying yields $ 12x^{2} + 20x + 3 = 0 $. Let the coordinates of the two intersection points be $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $."
  },
  {
    "text": "Given that $F$ is the lower focus of the ellipse $C$: $\\frac{y^{2}}{4}+\\frac{x^{2}}{3}=1$, point $P$ is an arbitrary point on the ellipse $C$, and the coordinates of point $Q$ are $(1,1)$, find the coordinates of point $P$ when the value of $|P Q|+|P F|$ is maximized.",
    "fact_expressions": "C: Ellipse;P: Point;Q: Point;F: Point;Expression(C) = (x^2/3 + y^2/4 = 1);Coordinate(Q) = (1, 1);LowerFocus(C)=F;PointOnCurve(P, C);WhenMax(Abs(LineSegmentOf(P,Q))+Abs(LineSegmentOf(P,F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(-3/2, 1)",
    "fact_spans": "[[[6, 48], [58, 63]], [[53, 57], [107, 111]], [[69, 73]], [[2, 5]], [[6, 48]], [[69, 84]], [[2, 52]], [[53, 68]], [[86, 105]]]",
    "query_spans": "[[[107, 116]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola $x^{2}=-10 y$ lies on the line $2 m x+m y+1=0$, then $m$=?",
    "fact_expressions": "G: Parabola;H: Line;m: Number;Expression(G) = (x^2 = -10*y);Expression(H) = (2*(m*x) + m*y + 1 = 0);PointOnCurve(Focus(G), H)",
    "query_expressions": "m",
    "answer_expressions": "2/5",
    "fact_spans": "[[[0, 16]], [[20, 37]], [[40, 43]], [[0, 16]], [[20, 37]], [[0, 38]]]",
    "query_spans": "[[[40, 45]]]",
    "process": "Analysis: Since the focus of the parabola is $ F(0, -\\frac{5}{2}) $, substituting into the line equation $ 2mx + my = -1 $ yields $ m = \\frac{2}{5} $. The answer to be filled in is $ \\frac{2}{5} $."
  },
  {
    "text": "Let the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ have left and right foci $F_{1}$, $F_{2}$. Draw a line perpendicular to the $x$-axis through $F_{2}$ intersecting $C$ at points $A$, $B$. Let line $F_{1}B$ intersect the $y$-axis at point $D$. If $AD \\perp F_{1}B$, then the eccentricity of ellipse $C$ is equal to?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;L: Line;PointOnCurve(F2, L) = True;IsPerpendicular(L, xAxis) = True;Intersection(L, C) = {A, B};A: Point;B: Point;Intersection(LineSegmentOf(F1, B), yAxis) = D;D: Point;IsPerpendicular(LineSegmentOf(A, D), LineSegmentOf(F1, B)) = True",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[1, 58], [97, 100], [155, 160]], [[1, 58]], [[7, 58]], [[7, 58]], [[7, 58]], [[7, 58]], [[64, 71]], [[72, 79], [81, 88]], [[1, 79]], [[1, 79]], [], [[80, 96]], [[80, 96]], [[80, 111]], [[102, 105]], [[106, 109]], [[112, 132]], [[128, 132]], [[134, 153]]]",
    "query_spans": "[[[155, 167]]]",
    "process": "Since OD is parallel to F_{2}B, D is the midpoint of F_{1}B, and since AD\\botF_{1}B, it follows that AF_{1}=AB=2AF_{2}. Let AF_{2}=m, then AF_{1}=2m, F_{1}F_{2}=\\sqrt{3}m, therefore e=\\frac{c}{a}=\\frac{2c}{2a}=\\frac{F_{1}F_{2}}{AF_{1}+AF_{2}}=\\frac{\\sqrt{3}m}{2m+m}=\\frac{\\sqrt{3}}{3}"
  },
  {
    "text": "The left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are $F_{1}$ and $F_{2}$ respectively. A circle with diameter $F_{1} F_{2}$ intersects the right branch of the hyperbola at a point $P$. If $\\angle P F_{2} F_{1}=2 \\angle P F_{1} F_{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Circle;F1: Point;F2: Point;P: Point;a>0;b>0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;IsDiameter(LineSegmentOf(F1, F2), H);OneOf(Intersection(H, RightPart(G))) = P;AngleOf(P, F2, F1) = 2*AngleOf(P, F1, F2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3) + 1",
    "fact_spans": "[[[0, 56], [101, 104], [165, 168]], [[3, 56]], [[3, 56]], [[99, 100]], [[65, 72]], [[73, 80]], [[112, 115]], [[3, 56]], [[3, 56]], [[0, 56]], [[0, 80]], [[0, 80]], [[81, 100]], [[99, 115]], [[117, 162]]]",
    "query_spans": "[[[165, 174]]]",
    "process": "By the given condition, $\\angle F_{1}PF_{2}=\\frac{\\pi}{2}$, $\\therefore |PF_{2}|=\\frac{1}{2}|F_{1}F_{2}|=c$, $|PF_{1}|=\\sqrt{3}c$ $\\therefore 2a=\\sqrt{3}c-c$ $\\therefore e=\\frac{2c}{\\sqrt{3}c-c}\\frac{\\sqrt{3}}{2}=\\sqrt{3}+1$"
  },
  {
    "text": "Given a point $A$ on the parabola $y^{2}=4 x$ such that the distance from $A$ to the focus $F$ is $5$, then the distance from $A$ to the origin $O$, $|O A|$, is $?$.",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;PointOnCurve(A, G);F: Point;Focus(G) = F;Distance(A, F) = 5;O: Origin;Distance(A, O) = Abs(LineSegmentOf(O, A))",
    "query_expressions": "Abs(LineSegmentOf(O, A))",
    "answer_expressions": "4*sqrt(2)",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[20, 23], [38, 41]], [[2, 23]], [[26, 29]], [[2, 29]], [[20, 36]], [[42, 47]], [[38, 57]]]",
    "query_spans": "[[[50, 59]]]",
    "process": "Since |AF| = 5, the distance from A to the parabola's directrix is also equal to 5. Because the focus has coordinates (1,0), the directrix equation is x = -1. The x-coordinate of point A is 5 - 1 = 4. Substituting into the parabola equation gives the y-coordinate as 4. Therefore, the distance from point A to the origin is \\sqrt{4^{2}+4^{2}}=4\\sqrt{2}."
  },
  {
    "text": "Given that $B$ is the intersection point of the left directrix of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ and the $x$-axis, and point $A(0 , b)$, if the point $P$ satisfying $\\overrightarrow{A P}=2 \\overrightarrow{A B}$ lies on the hyperbola, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;B: Point;Intersection(LeftDirectrix(G), xAxis) = B;A: Point;Coordinate(A) = (0, b);P: Point;VectorOf(A, P) = 2*VectorOf(A, B);PointOnCurve(P, G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[6, 62], [141, 144], [148, 151]], [[6, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[2, 5]], [[2, 74]], [[75, 86]], [[75, 86]], [[136, 140]], [[90, 135]], [[136, 145]]]",
    "query_spans": "[[[148, 157]]]",
    "process": ""
  },
  {
    "text": "Given that the line $y=k(x+3)$ $(k>0)$ intersects the parabola $C$: $y^{2}=12x$ at points $A$ and $B$, and $F$ is the focus of $C$. If $|FA|=3|FB|$, then the value of $k$ is?",
    "fact_expressions": "C: Parabola;G: Line;F: Point;A: Point;B: Point;k:Number;Expression(C) = (y^2 = 12*x);k>0;Expression(G) = (y = k*(x + 3));Intersection(G, C) = {A, B};Focus(C) = F;Abs(LineSegmentOf(F, A)) = 3*Abs(LineSegmentOf(F, B))",
    "query_expressions": "k",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[20, 40], [57, 60]], [[2, 19]], [[53, 56]], [[43, 46]], [[47, 50]], [[83, 86]], [[20, 40]], [[4, 19]], [[2, 19]], [[2, 52]], [[53, 63]], [[66, 80]]]",
    "query_spans": "[[[83, 91]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). It is clear that F(3,0). Combining the line y=k(x+3) (k>0) with the parabola C: y^{2}=12x yields k^{2}x^{2}+(6k^{2}-12)x+9k^{2}=0, (k>0). Therefore, x_{1}+x_{2}=\\frac{12}{k^{2}}-6\\textcircled{1}, x_{1}x_{2}=9\\textcircled{2}. Since |FA|=3|FB|, by the definition of the parabola, we have |FA|=x_{1}+3, |FB|=x_{2}+3. Thus, x_{1}+3=3(x_{2}+3)\\textcircled{3}, which simplifies to x_{1}=3x_{2}+6. Solving \\textcircled{1}\\textcircled{2}\\textcircled{3} together gives k=\\frac{\\sqrt{3}}{2}."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively, if there exists a point $P$ on the left branch of the hyperbola such that $\\frac{|P F_{2}|^{2}}{|P F_{1}|^{2}}=8 a$, then the range of the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F2: Point;F1: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, LeftPart(G));Abs(LineSegmentOf(P, F2))^2/Abs(LineSegmentOf(P, F1))^2 = 8*a",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1, 3]",
    "fact_spans": "[[[20, 76], [84, 87], [142, 145]], [[23, 76]], [[23, 76]], [[94, 97]], [[10, 17]], [[2, 9]], [[23, 76]], [[23, 76]], [[20, 76]], [[2, 82]], [[2, 82]], [[84, 97]], [[99, 140]]]",
    "query_spans": "[[[142, 156]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and a point $P$ on the ellipse satisfies $\\angle F_{1} P F_{2}=90^{\\circ}$, then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[18, 45], [53, 55]], [[2, 9]], [[56, 60]], [[10, 17]], [[18, 45]], [[2, 50]], [[53, 60]], [[62, 95]]]",
    "query_spans": "[[[97, 124]]]",
    "process": ""
  },
  {
    "text": "The right focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ is the focus of a parabola. Then, the standard equation of the parabola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);H: Parabola;RightFocus(G) = Focus(H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2=20*x",
    "fact_spans": "[[[0, 39]], [[0, 39]], [[44, 47], [52, 55]], [[0, 50]]]",
    "query_spans": "[[[52, 62]]]",
    "process": "From $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, we obtain: $a^{2}=9$, $b^{2}=16$, so $c^{2}=a^{2}+b^{2}=9+16=25$, hence $c=5$. Therefore, the right focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ is $(5,0)$. Let the standard equation of the parabola be $y^{2}=2px$ ($p>0$), with focus at $(\\frac{p}{2},0)$. Thus, $\\frac{p}{2}=5$, so $p=10$. Therefore, the required equation of the parabola is: $y^{2}=20x$."
  },
  {
    "text": "If the eccentricities of the hyperbolas $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ and $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=-1$ $(a>b>0)$ are $e_{1}$ and $e_{2}$ respectively, then as $a$ and $b$ vary, the minimum value of $e_{1}^{2}+e_{2}^{2}$ is?",
    "fact_expressions": "G: Hyperbola;C:Hyperbola;b: Number;a: Number;a>b;b>0;e1:Number;e2:Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(C)=(x^2/a^2-y^2/b^2=-1);Eccentricity(G)=e1;Eccentricity(C)=e2",
    "query_expressions": "Min((e1)^2+(e2)^2)",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 47]], [[48, 99]], [[128, 131]], [[124, 127]], [[48, 99]], [[48, 99]], [[106, 113]], [[114, 121]], [[1, 47]], [[48, 99]], [[1, 121]], [[1, 121]]]",
    "query_spans": "[[[135, 162]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $\\frac{2 \\sqrt{3}}{3}$, then the equations of the asymptotes of this hyperbola are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(G) = (2*sqrt(3))/3",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(3)/3)*x",
    "fact_spans": "[[[2, 58], [88, 91]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 85]]]",
    "query_spans": "[[[88, 99]]]",
    "process": "\\because the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0,b>0) has eccentricity \\frac{2\\sqrt{3}}{3}, that is, \\frac{c}{a}=\\frac{2\\sqrt{3}}{3}, so \\frac{c^{2}}{a^{2}}=\\frac{4}{3}, so \\frac{a^{2}+b^{2}}{a^{2}}=\\frac{4}{3}, hence \\frac{b^{2}}{a^{2}}=\\frac{1}{3}, therefore the equations of the asymptotes of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0,b>0) are y=\\pm\\frac{b}{a}x=\\pm\\frac{\\sqrt{3}}{3}x"
  },
  {
    "text": "Given that lines $AB$ and $AC$ are drawn from point $A(2,2)$ tangent to the circle $x^{2}+(y-2)^{2}=1$, and intersect the parabola $x^{2}=2y$ at points $B$ and $C$, respectively, then the equation of line $BC$ is?",
    "fact_expressions": "G: Parabola;H: Circle;A: Point;B: Point;C: Point;Expression(G) = (x^2 = 2*y);Expression(H) = (x^2 + (y - 2)^2 = 1);Coordinate(A) = (2, 2);PointOnCurve(A,LineOf(A,B));PointOnCurve(A,LineOf(A,C));IsTangent(LineOf(A,B),H);IsTangent(LineOf(A,C),H);Intersection(LineOf(A,C),G)={B,C}",
    "query_expressions": "Expression(LineOf(B,C))",
    "answer_expressions": "6*x+3*y+4=0",
    "fact_spans": "[[[54, 68]], [[29, 49]], [[3, 12]], [[69, 72]], [[73, 76]], [[54, 68]], [[29, 49]], [[3, 12]], [[2, 20]], [[2, 28]], [[13, 51]], [[23, 51]], [[13, 78]]]",
    "query_spans": "[[[80, 92]]]",
    "process": "Let $ B(x_{1},\\frac{x^{2}}{2}) $, $ C(x_{2},\\frac{x^{2}}{2}) $, use the point-slope form to find line $ AB $, then use the distance from the center of the circle to the line equal to the radius to obtain $ 6x_{1}+3y_{1}+4=0 $, similarly $ 6x_{2}+3y_{2}+4=0 $, i.e., solve. Solution: Let $ B(x_{1},\\frac{x_{2}^{2}}{2}) $, $ C(x_{2},\\frac{x_{2}^{2}}{2}) $, $ \\therefore l_{AB}:(x_{1}+2)x-2y-2x_{1}=0 $. Since the distance from the center to line $ AB $ is $ d=\\frac{|-4-2x|}{\\sqrt{(x_{1}+2)^{2}+4}}=1 $, $ \\therefore 3x_{1}^{2}+12x_{1}+8=0 \\Rightarrow 6x_{1}+3y_{1}+4=0 $, similarly $ 6x_{2}+3y_{2}+4=0 $, $ \\therefore l_{BC}:6x+3y+4=0 $."
  },
  {
    "text": "Given that the foci of the ellipse $x^{2} \\sin \\alpha - y^{2} \\cos \\alpha = 1$ $(0 \\leq \\alpha < 2\\pi)$ lie on the $y$-axis, what is the range of values for $\\alpha$?",
    "fact_expressions": "G: Ellipse;alpha: Number;alpha>=0;alpha<2*pi;Expression(G) = (x^2*Sin(alpha) - y^2*Cos(alpha) = 1);PointOnCurve(Focus(G),yAxis)",
    "query_expressions": "Range(alpha)",
    "answer_expressions": "(\\pi/2, 3\\pi/4)",
    "fact_spans": "[[[2, 64]], [[75, 83]], [[4, 64]], [[4, 64]], [[2, 64]], [[2, 73]]]",
    "query_spans": "[[[75, 90]]]",
    "process": "The ellipse $ x^2\\sin\\alpha - y^2\\cos\\alpha = 1 $ ($ 0 \\leqslant \\alpha < 2\\pi $) is converted into standard form, yielding $ \\frac{x^{2}}{\\sin\\alpha} + \\frac{y^{2}}{-\\cos\\alpha} = 1 $. Since its foci lie on the $ y $-axis, $ 0 < -\\cos\\alpha < \\sin\\alpha $, $ 0 \\leqslant \\alpha < 2\\pi $. Therefore, $ \\frac{\\pi}{2} < \\alpha < \\frac{3\\pi}{4} $."
  },
  {
    "text": "The length of the imaginary axis of the hyperbola $m x^{2}+y^{2}=1$ is $2$ times the length of the real axis, then $m$ equals?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (m*x^2 + y^2 = 1);m: Number;Length(ImageinaryAxis(G)) = 2*Length(RealAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "-1/4",
    "fact_spans": "[[[0, 20]], [[0, 20]], [[35, 38]], [[0, 33]]]",
    "query_spans": "[[[35, 41]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x$ coincides with the right focus of the hyperbola $\\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1$, then the value of $p$ is?",
    "fact_expressions": "G: Hyperbola;C: Parabola;p: Number;Expression(G) = (x^2/6 - y^2/3 = 1);Expression(C) = (y^2 = 2*p*x);Focus(C) = RightFocus(G)",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[21, 59]], [[1, 17]], [[67, 70]], [[21, 59]], [[1, 17]], [[1, 65]]]",
    "query_spans": "[[[67, 74]]]",
    "process": ""
  },
  {
    "text": "A point $P$ on the parabola $y^{2}=16 x$ is at a distance of $12$ from the $x$-axis. Then the distance $|P F|$ between $P$ and the focus $F$ is $?$.",
    "fact_expressions": "G: Parabola;P: Point;F: Point;Expression(G) = (y^2 = 16*x);PointOnCurve(P, G);Distance(P, xAxis) = 12;Focus(G)=F;Distance(P,F)=Abs(LineSegmentOf(P,F))",
    "query_expressions": "Abs(LineSegmentOf(P,F))",
    "answer_expressions": "13",
    "fact_spans": "[[[0, 15]], [[19, 22], [37, 40]], [[43, 46]], [[0, 15]], [[0, 22]], [[19, 35]], [[0, 46]], [[37, 57]]]",
    "query_spans": "[[[50, 59]]]",
    "process": "According to the problem, the vertical coordinate of point P satisfies $|y|=12$. Substituting into the parabolic equation yields $x=9$. The directrix of the parabola is $x=4$. By the definition of a parabola, the distance between point P and the focus F is $9+4=13$."
  },
  {
    "text": "It is known that a hyperbola, having the lines $y = \\pm 2x$ as asymptotes, passes through the intersection point of the lines $x + y - 3 = 0$ and $2x - y + 6 = 0$. Then the length of the real axis of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Line;Expression(H) = (y = pm*(2*x));Asymptote(G) = H;L1: Line;L2: Line;Expression(L1) = (x + y - 3 = 0);Expression(L2) = (2*x - y + 6 = 0);PointOnCurve(Intersection(L1, L2), G)",
    "query_expressions": "Length(RealAxis(G))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[21, 24], [55, 58]], [[3, 16]], [[3, 16]], [[2, 24]], [[27, 38]], [[39, 50]], [[27, 38]], [[39, 50]], [[21, 53]]]",
    "query_spans": "[[[55, 64]]]",
    "process": "The coordinates of the intersection point of the lines $x+y-3=0$ and $2x-y+6=0$ can be found by solving\n\\[\n\\begin{cases}\nx+y-3=0 \\\\\n2x-y+6=0\n\\end{cases}\n\\]\nyielding\n\\[\n\\begin{cases}\nx=-1 \\\\\ny=4\n\\end{cases}\n\\]\nso the intersection point is $(-1,4)$. The equation of a hyperbola with asymptotes $y=\\pm2x$ can be written as $4x^{2}-y^{2}=k$. Substituting the point $(-1,4)$ into this equation gives $k=-12$. Thus, the equation of the hyperbola is $y^{2}-4x^{2}=12$, or $\\frac{y^{2}}{12}-\\frac{x^{2}}{3}=1$, so $a=\\sqrt{12}=2\\sqrt{3}$, and the length of the real axis is $2a=4\\sqrt{3}$."
  },
  {
    "text": "Given that moving circle $M$ is externally tangent to circle $A$: $(x+4)^{2}+y^{2}=2$ and internally tangent to circle $B$: $(x-4)^{2}+y^{2}=2$, then the trajectory equation of the center of moving circle $M$ is?",
    "fact_expressions": "M: Circle;A:Circle;B:Circle;Expression(A)=((x+4)^2+y^2=2);Expression(B)=((x-4)^2+y^2=2);IsOutTangent(M,A);IsInTangent(M,B);M1:Point;Center(M)=M1",
    "query_expressions": "LocusEquation(M1)",
    "answer_expressions": "{(x^2/2 - y^2/14 = 1) & (x >= sqrt(2))}",
    "fact_spans": "[[[4, 7], [64, 66]], [[8, 32]], [[36, 60]], [[8, 32]], [[36, 60]], [[4, 34]], [[4, 62]], [[68, 71]], [[64, 71]]]",
    "query_spans": "[[[68, 78]]]",
    "process": "Given circle A: $(x+4)^{2}+y^{2}=2$, center $A(-4,0)$, radius $\\sqrt{2}$; circle B: $(x-4)^{2}+y^{2}=2$, center $B(4,0)$, radius $\\sqrt{2}$. Let the moving circle center $M$ have coordinates $(x,y)$ and radius $r$. Then $|MA|=r+\\sqrt{2}$, $|MB|=r-\\sqrt{2}$. Therefore, $|MA|-|MB|=r+\\sqrt{2}-r+\\sqrt{2}=2\\sqrt{2}<|AB|=8$. By the definition of a hyperbola, the locus of point $M$ is the right branch of a hyperbola with foci $A$ and $B$, where $2a=2\\sqrt{2}$, $a=\\sqrt{2}$, $c=4$, so $b^{2}=c^{2}-a^{2}=14$. The equation of the hyperbola is $\\frac{x^{2}}{2}-\\frac{y^{2}}{14}=1$ $(x\\geqslant\\sqrt{2})$."
  },
  {
    "text": "One asymptote of the hyperbola is $y=\\sqrt{3} x$, and the foci are $(-4,0)$ and $(4,0)$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Coordinate(Focus(G)) = (pm*4, 0);Expression(OneOf(Asymptote(G)))=(y=sqrt(3)*x)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/12 = 1",
    "fact_spans": "[[[0, 3], [46, 49]], [[0, 44]], [[0, 26]]]",
    "query_spans": "[[[46, 53]]]",
    "process": "The solution process is omitted"
  },
  {
    "text": "What is the maximum distance from a point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$ to the line $x+2 y-\\sqrt{2}=0$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/4 = 1);Z: Line;Expression(Z) = (x + 2*y - sqrt(2) = 0);K: Point;PointOnCurve(K, G)",
    "query_expressions": "Max(Distance(K, Z))",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[0, 38]], [[0, 38]], [[42, 62]], [[42, 62]], [[40, 41]], [[0, 41]]]",
    "query_spans": "[[[40, 69]]]",
    "process": "Let the line $x+2y+c=0$ be tangent to the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$. From $\\begin{cases}x+2y+c=0,\\\\\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1,\\end{cases}$ eliminating $x$ and simplifying yields $8y^{2}+4cy+c^{2}-16=0$. From $4=16(32-c^{2})=0$, we obtain $c=\\pm4\\sqrt{2}$. When $c=4\\sqrt{2}$, it satisfies the condition ($c=-4\\sqrt{2}$ is discarded). When $c=4\\sqrt{2}$, it satisfies the condition ($c=-4\\sqrt{2}$ is discarded). Thus, $x+2y+4\\sqrt{2}=0$ is tangent to the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$. The maximum distance from a point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$ to the line $x+2y-\\sqrt{2}=0$ is the distance between the two parallel lines $\\frac{|-\\sqrt{2}-4\\sqrt{2}|}{\\sqrt{12}+2^{2}}=\\sqrt{10}$."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=2 x$, $M$ is a point on $C$, and the extension of $F M$ intersects the $y$-axis at point $N$. If $M$ is the midpoint of $F N$, then $|F N|$=?",
    "fact_expressions": "C: Parabola;F: Point;M: Point;N: Point;Expression(C) = (y^2 = 2*x);Focus(C) = F;PointOnCurve(M, C);Intersection(OverlappingLine(LineSegmentOf(F, M)), yAxis) = N;MidPoint(LineSegmentOf(F, N)) = M",
    "query_expressions": "Abs(LineSegmentOf(F, N))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[6, 25], [33, 36]], [[2, 5]], [[29, 32], [62, 65]], [[55, 59]], [[6, 25]], [[2, 28]], [[29, 39]], [[40, 59]], [[62, 74]]]",
    "query_spans": "[[[76, 85]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 43]]]",
    "process": "In the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, $a=2$, $b=\\sqrt{3}$, $c=\\sqrt{a^{2}-b^{2}}=1$, therefore, the eccentricity of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ is $e=\\frac{c}{a}=\\frac{1}{2}$."
  },
  {
    "text": "The equations of the two asymptotes of a hyperbola are $4x \\pm 3y = 0$, and one directrix is $x = \\frac{9}{5}$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (4*x+pm*3*y = 0);Expression(OneOf(Directrix(G))) = (x = 9/5)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/16 = 1",
    "fact_spans": "[[[0, 3], [53, 56]], [[0, 26]], [[0, 50]]]",
    "query_spans": "[[[53, 60]]]",
    "process": ""
  },
  {
    "text": "An ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ has a focus $F_{1}$. If there exists a point $P$ on the ellipse such that the circle with the minor axis of the ellipse as its diameter is tangent to the segment $P F_{1}$ at the midpoint of this segment, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;OneOf(Focus(G)) = F1;P: Point;PointOnCurve(P, G) = True;H: Circle;IsDiameter(MinorAxis(G),H) ;TangentPoint(H,LineSegmentOf(P,F1)) = MidPoint(LineSegmentOf(P,F1))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[0, 52], [67, 69], [82, 84], [114, 116]], [[0, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[58, 65]], [[0, 65]], [[74, 78]], [[67, 78]], [[90, 91]], [[81, 91]], [[90, 112]]]",
    "query_spans": "[[[114, 122]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ lies on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, $F_{1}$ and $F_{2}$ are the foci, and $\\angle F_{1} P F_{2}=90^{\\circ}$, find the area of $\\Delta F_{1} P F_{2}$.",
    "fact_expressions": "P: Point;PointOnCurve(P, G);G: Ellipse;Expression(G) = (x^2/9 + y^2/5 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "5",
    "fact_spans": "[[[2, 6]], [[2, 48]], [[7, 44]], [[7, 44]], [[49, 56]], [[57, 64]], [[7, 67]], [[69, 102]]]",
    "query_spans": "[[[104, 131]]]",
    "process": "From the ellipse $\\frac{x^2}{9}+\\frac{y^{2}}{5}=1$, we know $|PF_{1}|+|PF_{2}|=2a=6$. Also, $\\angle F_{1}PF_{2}=90^{\\circ}$, so $|PF_{1}|^{2}+|PF_{2}|^{2}=(2c)^{2}=16$. Since $|PF_{1}|^{2}+|PF_{2}|^{2}=(|PF_{1}|+|PF_{2}|)^{2}-2|PF_{1}|\\cdot|PF_{2}|=16$, solving gives $|PF_{1}|\\cdot|PF_{2}|=10$. Therefore, the area of $\\triangle F_{1}PF_{2}$ is $S=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|=5$."
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, $P$ a moving point on the ellipse (not coinciding with the endpoints of the major axis), $I$ the incenter of $\\Delta P F_{1} F_{2}$, and line $P I$ intersecting the $x$-axis at point $D$. Then $\\frac{P I}{I D}$=?",
    "fact_expressions": "G: Ellipse;I: Point;P: Point;F1: Point;F2: Point;D: Point;Expression(G) = (x^2/25 + y^2/16 = 1);Focus(G) = {F1, F2};PointOnCurve(P,G);Negation(P=Endpoint(MajorAxis(G)));Incenter(TriangleOf(P, F1, F2)) = I;Intersection(LineOf(P,I), xAxis) = D",
    "query_expressions": "LineSegmentOf(P, I)/LineSegmentOf(I, D)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[17, 56], [66, 68]], [[86, 89]], [[62, 65]], [[1, 8]], [[9, 16]], [[129, 133]], [[17, 56]], [[1, 61]], [[62, 72]], [[62, 85]], [[86, 115]], [[116, 133]]]",
    "query_spans": "[[[135, 154]]]",
    "process": ""
  },
  {
    "text": "Given point $A(4,4)$, if the focus of the parabola $y^{2}=2 px$ coincides with the right focus of the ellipse $\\frac{x^{2}}{10}+\\frac{y^{2}}{6}=1$, and there is a point $M$ on this parabola whose projection on the $y$-axis is $N$, then the minimum value of $|MA|+|MN|$ is?",
    "fact_expressions": "G: Parabola;p: Number;H: Ellipse;A: Point;M: Point;N: Point;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (x^2/10 + y^2/6 = 1);Coordinate(A) = (4, 4);Focus(G) = RightFocus(H);PointOnCurve(M, G);Projection(M,yAxis)=N",
    "query_expressions": "Min(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, N)))",
    "answer_expressions": "2*sqrt(5)-2",
    "fact_spans": "[[[13, 28], [78, 81]], [[16, 28]], [[32, 70]], [[2, 11]], [[85, 88], [89, 90]], [[100, 103]], [[13, 28]], [[32, 70]], [[2, 11]], [[13, 76]], [[78, 88]], [[89, 103]]]",
    "query_spans": "[[[105, 122]]]",
    "process": "Since the coordinates of the right focus of the ellipse are (2,0), we have \\frac{p}{2}=2, \\therefore p=4, so the equation of the parabola is y^{2}=8x. Let d be the distance from M to the directrix of the parabola. Draw AN perpendicular to the directrix from point A, with N as the foot of the perpendicular. According to the definition of the parabola, |MA|+|MN|=|MA|+d-2\\geqslant|AN|-2=6-2=4"
  },
  {
    "text": "In $\\triangle A B C$, points $A$ and $B$ are the left and right foci of hyperbola $E$, respectively, and point $C$ lies on hyperbola $E$, satisfying $\\overrightarrow{A B} \\cdot \\overrightarrow{A C}=0$, $(\\overrightarrow{A B}+\\overrightarrow{A C}) \\cdot \\overrightarrow{B C}=0$. Then the eccentricity of hyperbola $E$ is?",
    "fact_expressions": "E: Hyperbola;A: Point;B: Point;C: Point;LeftFocus(E) = A;RightFocus(E) = B;PointOnCurve(C, E);DotProduct(VectorOf(A, B), VectorOf(A, C)) = 0;DotProduct((VectorOf(A, B) + VectorOf(A, C)),VectorOf(B, C)) = 0",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(2)+1",
    "fact_spans": "[[[31, 37], [49, 55], [188, 194]], [[20, 24]], [[25, 28]], [[44, 48]], [[20, 43]], [[20, 43]], [[44, 56]], [[59, 110]], [[112, 186]]]",
    "query_spans": "[[[188, 200]]]",
    "process": "Since $\\overrightarrow{AB}\\cdot\\overrightarrow{AC}=0$, it follows that $\\overrightarrow{AB}\\bot\\overrightarrow{AC}$, i.e., $AB\\bot AC$; thus $\\triangle ABC$ is a right triangle with $BC$ as the hypotenuse. From $(\\overrightarrow{AB}+\\overrightarrow{AC})\\cdot\\overrightarrow{BC}=0 \\Rightarrow (\\overrightarrow{AB}+\\overrightarrow{AC})\\cdot(\\overrightarrow{AC}-\\overrightarrow{AB})=0 \\Rightarrow \\overrightarrow{AC}^{2}=\\overrightarrow{AB}^{2} \\Rightarrow |\\overrightarrow{AC}|=|\\overrightarrow{AB}|$, clearly $\\triangle ABC$ is an isosceles right triangle with $BC$ as the hypotenuse. Since points $A$ and $B$ are the left and right foci of hyperbola $E$, let the equation of hyperbola $E$ be: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$). Therefore $AB=AC=2c$, so $BC=\\sqrt{AB^{2}+AC^{2}}=\\sqrt{(2c)^{2}+(2c)^{2}}=2\\sqrt{2}c$. By the definition of a hyperbola: $CB-CA=2a \\Rightarrow 2\\sqrt{2}c-2c=2a \\Rightarrow e=\\frac{c}{a}=\\frac{1}{\\sqrt{2}-1}=\\sqrt{2}+1$"
  },
  {
    "text": "If the equation $\\frac{x^{2}}{m+2}+\\frac{y^{2}}{2-m}=1$ represents an ellipse, then the range of real values for $m$ is?",
    "fact_expressions": "E:Ellipse;Expression(E)=(y^2/(2-m)+x^2/(m+2)=1);m:Real",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-2,0)+(0,2)",
    "fact_spans": "[[[44, 46]], [[1, 46]], [[48, 53]]]",
    "query_spans": "[[[48, 60]]]",
    "process": "From the equation $\\frac{x^2}{m+2} + \\frac{y^2}{2-m} = 1$ representing an ellipse, we obtain $\\begin{cases} 2 - m > 0 \\\\ m + 2 \\neq 2 - m \\end{cases}$. Solving gives $-2 < m < 2$ and $m \\neq 0$. Therefore, the range of real values for $m$ is: $(-2, 0) \\cup (0, 2)$."
  },
  {
    "text": "If $A$ and $B$ are the two vertices on the minor axis of the ellipse $E$: $\\frac{x^{2}}{m}+y^{2}=1$ ($m>1$), and point $P$ is any point on the ellipse distinct from $A$ and $B$, and if the product of the slopes of lines $AP$ and $BP$ is $-\\frac{m}{4}$, then $m=$?",
    "fact_expressions": "A: Point;B: Point;Vertex(MinorAxis(E)) = {A, B};E: Ellipse;Expression(E) = (y^2 + x^2/m = 1);m: Number;m>1;P: Point;PointOnCurve(P, E);Negation(P = A);Negation(P = B);Slope(LineOf(A, P))*Slope(LineOf(B, P)) = -m/4",
    "query_expressions": "m",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 4], [67, 70]], [[5, 8], [71, 74]], [[1, 56]], [[11, 48], [62, 64]], [[11, 48]], [[118, 121]], [[18, 48]], [[57, 61]], [[57, 79]], [[57, 79]], [[57, 79]], [[81, 116]]]",
    "query_spans": "[[[118, 123]]]",
    "process": "Let the coordinates of point P be (x, y), then $ y^{2} = 1 - \\frac{x^{2}}{m} $. According to the problem, $ k_{AP} \\cdot k_{BP} = \\frac{(y-1)(y+1)}{x^{2}} = \\frac{y^{2}-1}{x^{2}} = \\frac{(1 - \\frac{x^{2}}{m}) - 1}{x^{2}} = -\\frac{1}{m} = -\\frac{m}{4} $, solving gives $ m = 2 $. Answer: $ 2 $"
  },
  {
    "text": "The equation of the parabola with vertex at the origin, axis of symmetry along the coordinate axes, and passing through the point $(-8, -4)$ is?",
    "fact_expressions": "G: Parabola;H: Point;O: Origin;Coordinate(H) = (-8, -4);PointOnCurve(H, G);Vertex(G) = O;SymmetryAxis(G) = axis",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2 = -16*y, y^2 = -2*x}",
    "fact_spans": "[[[29, 32]], [[18, 28]], [[3, 5]], [[18, 28]], [[16, 32]], [[0, 32]], [[7, 32]]]",
    "query_spans": "[[[29, 37]]]",
    "process": ""
  },
  {
    "text": "The asymptotes of the hyperbola are $y=\\pm \\frac{3}{2} x$, then the eccentricity is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*(3/2)*x)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(13)/3",
    "fact_spans": "[[[0, 3]], [[0, 29]]]",
    "query_spans": "[[[0, 36]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $ C $: $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ $ (a > b > 0) $, the length of the major axis is $ \\frac{3 \\sqrt{5}}{5} $ times the length of the minor axis, and the point $ (-1, \\frac{2 \\sqrt{10}}{3}) $ lies on the ellipse $ C $. Then, the standard equation of the ellipse $ C $ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(G) = (-1, (2*sqrt(10))/3);Length(MajorAxis(C)) = (3*sqrt(5)/5)*Length(MinorAxis(C));PointOnCurve(G, C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/9 + y^2/5 = 1",
    "fact_spans": "[[[2, 59], [124, 129], [132, 137]], [[9, 59]], [[9, 59]], [[93, 123]], [[9, 59]], [[9, 59]], [[2, 59]], [[93, 123]], [[2, 91]], [[93, 130]]]",
    "query_spans": "[[[132, 144]]]",
    "process": "The length of the major axis is $\\frac{3\\sqrt{5}}{5}$ times the length of the minor axis, and the point $(-1,\\frac{2\\sqrt{10}}{3})$ lies on the ellipse $C$, leading to the system:  \n$$\n\\begin{cases}\n\\frac{a}{b}=\\frac{3\\sqrt{5}}{4} \\\\\n\\frac{1}{a^{2}}+\\frac{40}{b^{2}}=1\n\\end{cases}\n$$\nSolving gives  \n$$\n\\begin{cases}\na^{2}=9 \\\\\nb^{2}=5\n\\end{cases}\n$$\nThus, the standard equation of ellipse $C$ is $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$."
  },
  {
    "text": "The standard equation of the parabola passing through the point $P(-2, -4)$ is?",
    "fact_expressions": "G: Parabola;P: Point;Coordinate(P) = (-2, -4);PointOnCurve(P, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=-8*x, x^2=-y}",
    "fact_spans": "[[[14, 17]], [[1, 13]], [[1, 13]], [[0, 17]]]",
    "query_spans": "[[[14, 24]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has eccentricity $e=\\frac{5}{4}$, and its right focus is $F_{2}(5,0)$, then the equation of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;F2: Point;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F2) = (5, 0);e:Number;Eccentricity(C) = e;e = 5/4;RightFocus(C) = F2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/16 - y^2/9 = 1",
    "fact_spans": "[[[2, 53], [74, 75], [93, 99]], [[10, 53]], [[10, 53]], [[79, 91]], [[2, 53]], [[79, 91]], [[57, 72]], [[2, 72]], [[57, 72]], [[74, 91]]]",
    "query_spans": "[[[93, 104]]]",
    "process": "From the eccentricity, find a, then find b from b^{2}=c^{2}-a^{2}, obtaining the hyperbola equation. Since the right focus of the desired hyperbola is F_{2}(5,0) and the eccentricity is e=\\frac{c}{a}=\\frac{5}{4}, it follows that c=5, a=4, b^{2}=c^{2}-a^{2}=9. Therefore, the equation of the desired hyperbola is \\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{2}+\\frac{y^{2}}{4}=1$, what is the equation of the line containing the chord for which point $M(1,1)$ is the midpoint?",
    "fact_expressions": "G: Ellipse;H: LineSegment;M: Point;Expression(G) = (x^2/2 + y^2/4 = 1);Coordinate(M) = (1, 1);IsChordOf(H, G);MidPoint(H) = M",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "2*x+y-3=0",
    "fact_spans": "[[[2, 39]], [], [[42, 51]], [[2, 39]], [[42, 51]], [[2, 56]], [[2, 56]]]",
    "query_spans": "[[[2, 64]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, find $ x_{1}+x_{2} $, $ y_{1}+y_{2} $, then use the point difference method to find the slope of line $ AB $, and then find the equation of line $ AB $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ \\frac{x_{1}+x_{2}}{2}=1 $, $ \\frac{y_{1}+y_{2}}{2}= $. Since $ A $, $ B $ lie on the ellipse, we have $ \\frac{x^{2}}{2}+\\frac{y^{2}}{4}=1 $. Subtracting the two equations gives $ \\underline{(x_{1}-x_{2})(x_{1}+x_{2})}+ $. Therefore, $ K_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{4(x_{1}+x_{2})}{2(y_{1}+y_{2})}= =-2 $, thus the equation of line $ AB $ is $ y-1=-2(x-1) $, i.e., $ 2x+y-3=0 $."
  },
  {
    "text": "Given the hyperbola equation $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, points $F_{1}$ and $F_{2}$ are its left and right foci, respectively, point $A$ lies on the circle $x^{2}+(y-6)^{2}=4$, and point $M$ lies on the right branch of the hyperbola. Then the minimum value of $|M F_{1}|+|M A|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);F1: Point;F2:Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Circle;Expression(H) = (x^2 + (y - 6)^2 = 4);A: Point;PointOnCurve(A, H);M: Point;PointOnCurve(M, RightPart(G))",
    "query_expressions": "Min(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F1)))",
    "answer_expressions": "4+sqrt(61)",
    "fact_spans": "[[[2, 5], [63, 64], [103, 106]], [[2, 45]], [[46, 54]], [[55, 62]], [[46, 68]], [[46, 68]], [[73, 93]], [[73, 93]], [[69, 72]], [[69, 97]], [[98, 102]], [[98, 110]]]",
    "query_spans": "[[[112, 135]]]",
    "process": ""
  },
  {
    "text": "Given that point $F$ is the left focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, a line $l$ passing through the origin $O$ intersects the ellipse at points $P$ and $Q$, and point $M$ is a point on the ellipse $C$ distinct from $P$ and $Q$. The slopes of lines $MP$ and $MQ$ are $k_{1}$ and $k_{2}$ respectively, with $k_{1} \\cdot k_{2}=-\\frac{5}{9}$. If $|PF|=2|QF|$, then $\\cos \\angle PFQ = ?$",
    "fact_expressions": "l: Line;C: Ellipse;b: Number;a: Number;M: Point;P: Point;Q: Point;F: Point;O: Origin;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;PointOnCurve(O, l);Intersection(l, C) = {P, Q};PointOnCurve(M, C);Negation(M = P);Negation(M = Q);Slope(LineOf(M, P))=k1;Slope(LineOf(M, Q))=k2;k1*k2 = -5/9;Abs(LineSegmentOf(P, F)) = 2*Abs(LineSegmentOf(Q, F));k1:Number;k2:Number",
    "query_expressions": "Cos(AngleOf(P, F, Q))",
    "answer_expressions": "-1/4",
    "fact_spans": "[[[76, 81]], [[7, 64], [100, 105], [82, 84]], [[13, 64]], [[13, 64]], [[95, 99]], [[85, 88], [108, 111]], [[89, 92], [112, 115]], [[2, 6]], [[70, 75]], [[13, 64]], [[13, 64]], [[7, 64]], [[2, 68]], [[69, 81]], [[76, 94]], [[95, 118]], [[95, 118]], [[95, 118]], [[119, 155]], [[119, 155]], [[157, 189]], [[191, 205]], [[139, 146]], [[147, 155]]]",
    "query_spans": "[[[207, 228]]]",
    "process": "From the condition $k_{1}\\cdot k_{2}=-\\frac{5}{9}$, the relationship among $a$, $b$, and $c$ can be simplified. From $|PF|=2|QF|$ and combining with the definition of the ellipse, $|PF|$ and $|QF|$ can be found. Then $\\cos\\angle PFQ$ can be found using the cosine law. [Detailed solution] Let $M(x_{0},y_{0})$, $P(x_{1},y_{1})$, from the known conditions we have $Q(-x_{1},-y_{1})$, $\\frac{x_{1}^{2}}{a^{2}}+\\frac{y_{1}^{2}}{b^{2}}=1$, $\\frac{x_{0}^{2}}{a^{2}}+\\frac{y_{0}^{2}}{b^{2}}=1$, so $\\frac{x_{0}^{2}-x_{1}^{2}}{a^{2}}+\\frac{y_{0}^{2}-y_{1}^{2}}{b^{2}}=0$, so $\\frac{b^{2}}{a^{2}}+\\frac{y_{0}^{2}-y_{1}^{2}}{x_{0}^{2}-x_{1}^{2}}=0$. Since $k_{1}\\cdot k_{2}=-\\frac{5}{9}$, we have $\\frac{y_{0}-y_{1}}{x_{0}-x_{1}}\\cdot\\frac{y_{0}+y_{1}}{x_{0}+x_{1}}=-\\frac{5}{9}$, so $\\frac{y_{0}^{2}-y_{1}^{2}}{x_{0}^{2}-x_{1}^{2}}=-\\frac{5}{9}$, thus $\\frac{b^{2}}{a^{2}}=\\frac{5}{9}$, so $\\frac{a^{2}-c^{2}}{a^{2}}=\\frac{5}{9}$, hence $\\frac{c^{2}}{a^{2}}=\\frac{4}{9}$. Let the right focus of the ellipse be $F_{2}$. Since $PQ$ and $FF_{2}$ bisect each other, the quadrilateral $PFQF_{2}$ is a parallelogram. Therefore $|PF_{2}|=|QF|$. Also $|PF|=2|QF|$, $|PF|+|PF_{2}|=2a$, so $|PF|=\\frac{4}{3}a$, $|PF_{2}|=\\frac{2}{3}a$, $\\cos\\angle PFQ=-\\cos\\angle FPF_{2}=-\\frac{|PF|^{2}+|PF_{2}|^{2}-|FF_{2}|^{2}}{2|PF|\\cdot|PF_{2}|}$"
  },
  {
    "text": "Let the parabola $C$: $y^{2}=2 p x$ ($p>0$) have directrix $l$ and focus $F$. A circle centered at $(1, \\sqrt{2})$ passes through point $F$ and is tangent to $l$. Points $A$ and $B$ lie on $C$ and $l$ respectively, point $O$ is the origin, and $\\overrightarrow{A O}=3 \\overrightarrow{O B}$. Then $|A O|$=?",
    "fact_expressions": "C: Parabola;p: Number;G: Circle;H: Point;F: Point;A: Point;O: Origin;B: Point;l: Line;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(H) = (1, sqrt(2));Directrix(C) = l;Focus(C) = F;Center(G) = H;IsTangent(G, l);PointOnCurve(A, C);PointOnCurve(B, l);VectorOf(A, O) = 3*VectorOf(O, B);PointOnCurve(F, G)",
    "query_expressions": "Abs(LineSegmentOf(A, O))",
    "answer_expressions": "\\sqrt{21}/2",
    "fact_spans": "[[[1, 27], [88, 91]], [[9, 27]], [[63, 64]], [[43, 59]], [[65, 69], [38, 41]], [[77, 81]], [[98, 101]], [[82, 85]], [[31, 34], [71, 74], [92, 95]], [[9, 27]], [[1, 27]], [[43, 59]], [[1, 34]], [[1, 41]], [[42, 64]], [[63, 76]], [[77, 96]], [[77, 96]], [[105, 150]], [[63, 69]]]",
    "query_spans": "[[[152, 161]]]",
    "process": "From the definition of the parabola, we know that (1,\\sqrt{2}) lies on C, so (\\sqrt{2})^{2}=2p\\times1, solving gives p=1. Thus, y^{2}=2x. Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since \\overrightarrow{AO}=3\\overrightarrow{OB}, it follows that x_{A}=-3x_{B}=-3\\times(-\\frac{p}{2})=\\frac{3}{2}, then y_{A}^{2}=2\\times\\frac{3}{2}=3, |AO|=\\sqrt{x_{A}^{2}+y_{A}^{2}}=\\frac{\\sqrt{21}}{2}"
  },
  {
    "text": "There is a point $M$ on the parabola $y^{2}=2 p x(p>0)$, its horizontal coordinate is $3$, and its distance to the focus is $5$. Then the equation of this parabola is?",
    "fact_expressions": "G: Parabola;p: Number;M: Point;p>0;Expression(G) = (y^2 = 2*(p*x));PointOnCurve(M, G);XCoordinate(M)=3;Distance(M,Focus(G))=5",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=8*x",
    "fact_spans": "[[[0, 21], [53, 56]], [[3, 21]], [[25, 28], [29, 30], [39, 40]], [[3, 21]], [[0, 21]], [[0, 28]], [[29, 38]], [[0, 50]]]",
    "query_spans": "[[[53, 61]]]",
    "process": "It is given that the x-coordinate of point M is 3, and the distance from point M to the focus is 5, so the distance from point M to the directrix is also 5. Since the distance from point M to the directrix equals \\frac{1}{2}p+3=5\\Rightarrow p=4, the equation of the parabola is y^{2}=8x."
  },
  {
    "text": "The ellipse $C$ is centered at the origin, with foci on the $x$-axis. If the eccentricity of the ellipse $C$ is $\\frac{1}{2}$, and one of its vertices coincides exactly with the focus of the parabola $x^{2}=8 \\sqrt{3} y$, then what is the standard equation of the ellipse $C$?",
    "fact_expressions": "C: Ellipse;Center(C) = O;O: Origin;PointOnCurve(Focus(C), xAxis) = True;Eccentricity(C) = 1/2;G: Parabola;Expression(G) = (x^2 = 8*(sqrt(3)*y));OneOf(Vertex(C)) = Focus(G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/16+y^2/12=1",
    "fact_spans": "[[[0, 5], [22, 27], [48, 49], [85, 90]], [[0, 11]], [[9, 11]], [[0, 20]], [[22, 46]], [[57, 80]], [[57, 80]], [[48, 83]]]",
    "query_spans": "[[[85, 97]]]",
    "process": "The focus of the parabola $x^{2}=8\\sqrt{3}y$, which is $(0,2\\sqrt{3})$, is a vertex of the ellipse, so $b=2\\sqrt{3}$. Since $\\frac{c}{a}=\\frac{1}{2}$ and $a^{2}=12+c^{2}$, solving gives $a^{2}=16$, $c^{2}=4$. Hence, the standard equation of the ellipse is $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$."
  },
  {
    "text": "If one of the asymptotes of the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{9}=1$ has an inclination angle of $60^{0}$, then the eccentricity of the hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;a: Number;Expression(G)=(x^2/a-y^2/9=1);Inclination(OneOf(Asymptote(G)))=ApplyUnit(60,degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 39], [60, 63]], [[4, 39]], [[1, 39]], [[1, 58]]]",
    "query_spans": "[[[60, 70]]]",
    "process": "The slope of the asymptotes of the hyperbola is \\frac{3}{\\sqrt{a}}=\\sqrt{3}, then a=3, c^{2}=a+9=3+9=12, c=2\\sqrt{3}. The eccentricity of the hyperbola is e=\\frac{c}{a}=\\frac{2\\sqrt{3}}{\\sqrt{3}}=2."
  },
  {
    "text": "Given that the hyperbola $C$ passes through the origin $O$, and the coordinates of the two foci are $F_{1}(-1,0)$, $F_{2}(3,0)$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;O: Origin;PointOnCurve(O, C);F1: Point;F2: Point;Coordinate(F1) = (-1, 0);Coordinate(F2) = (3, 0);Focus(C) = {F1, F2}",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 8], [56, 59]], [[10, 17]], [[2, 17]], [[27, 40]], [[42, 54]], [[27, 40]], [[42, 54]], [[2, 54]]]",
    "query_spans": "[[[56, 65]]]",
    "process": "The hyperbola C passes through the origin O, with two foci at F_{1}(-1,0) and F_{2}(3,0). It follows that 2c=4, so c=2, and c-a=1, so a=1. Therefore, the eccentricity of the hyperbola is: e=\\frac{c}{a}=2."
  },
  {
    "text": "Given that point $F$ is the focus of the parabola $C$: $y^{2}=2px$ ($p>0$), fixed point $A(1,2)$ and moving point $P$ lie on the parabola $C$, point $B(2,0)$, then the maximum value of $\\frac{|PF|-1}{|PB|^{2}}$ is?",
    "fact_expressions": "C: Parabola;p: Number;A: Point;B: Point;P: Point;F: Point;p>0;Expression(C) = (y^2 = 2*p*x);Coordinate(A) = (1, 2);Coordinate(B) = (2, 0);Focus(C) = F;PointOnCurve(A,C);PointOnCurve(P,C)",
    "query_expressions": "Max((Abs(LineSegmentOf(P, F)) - 1)/Abs(LineSegmentOf(P, B))^2)",
    "answer_expressions": "1/4",
    "fact_spans": "[[[7, 33], [55, 61]], [[14, 33]], [[39, 47]], [[63, 72]], [[50, 53]], [[2, 6]], [[14, 33]], [[7, 33]], [[39, 47]], [[63, 72]], [[2, 36]], [[39, 62]], [[39, 62]]]",
    "query_spans": "[[[74, 107]]]",
    "process": "Since point A(1,2) lies on the parabola, we have 4 = 2p, so p = 2, and the equation of the parabola is y^{2} = 4x, with focus F(1,0). Let P(x,y) be a moving point, then y = \\pm2\\sqrt{x}; take y = 2\\sqrt{x}, so P(x,2\\sqrt{x}). According to the definition of the parabola, the length PF - 1 equals the x-coordinate of point P. Then \\frac{|PF|-1}{|PB|^{2}} = \\frac{1}{x+\\frac{4}{x}} \\leqslant \\frac{1}{2\\sqrt{x\\cdot\\frac{4}{3}}} = \\frac{1}{4}, with equality holding if and only if x = 2."
  },
  {
    "text": "It is known that the focus of the parabola $y^{2}=-8x$ coincides with the left focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-y^{2}=1$ ($a>0$). Find the eccentricity of the hyperbola.",
    "fact_expressions": "C: Hyperbola;a: Number;G: Parabola;a>0;Expression(C) = (-y^2 + x^2/a^2 = 1);Expression(G) = (y^2 = -8*x);Focus(G) = LeftFocus(C)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[21, 63], [71, 74]], [[28, 63]], [[2, 17]], [[28, 63]], [[21, 63]], [[2, 17]], [[2, 69]]]",
    "query_spans": "[[[71, 80]]]",
    "process": "In the parabola $ y^{2} = -8x $, $ 2p = 8 $, $ p = 4 $, the focus is $ (-2, 0) $. It is the left focus of the hyperbola, so for the hyperbola $ C: \\frac{x^{2}}{a^{2}} - y^{2} = 1 $ $ (a > 0) $, $ c = 2 $, $ a = \\sqrt{2 - 1^{2}} = \\sqrt{3} $, thus the eccentricity is $ e = \\frac{c}{a} = \\frac{2}{\\sqrt{3}} = \\frac{2\\sqrt{3}}{3} $."
  },
  {
    "text": "$F$ is the left focus of the ellipse, $P$ is a point on the ellipse, $P F \\perp x$-axis, $O P \\| A B$, ($O$ is the coordinate origin). Find the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;P: Point;F: Point;O:Origin;LeftFocus(G) = F;PointOnCurve(P, G);IsParallel(LineSegmentOf(O,P),LineSegmentOf(A,B));IsPerpendicular(LineSegmentOf(P,F),xAxis)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[4, 6], [15, 17], [62, 64]], [[11, 14]], [[0, 3]], [[51, 54]], [[0, 10]], [[11, 20]], [[36, 48]], [[21, 35]]]",
    "query_spans": "[[[62, 69]]]",
    "process": ""
  },
  {
    "text": "The line $l$: $x-y=0$ intersects the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ at points $A$ and $B$, and point $C$ is a moving point on the ellipse. Then the maximum area of $\\triangle A B C$ is?",
    "fact_expressions": "l: Line;G: Ellipse;A: Point;B: Point;C: Point;Expression(G) = (x^2/2 + y^2 = 1);Expression(l)=(x - y = 0);Intersection(l, G) = {A, B};PointOnCurve(C, G)",
    "query_expressions": "Max(Area(TriangleOf(A, B, C)))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 14]], [[15, 42], [59, 61]], [[44, 47]], [[48, 51]], [[54, 58]], [[15, 42]], [[0, 14]], [[0, 53]], [[54, 65]]]",
    "query_spans": "[[[67, 92]]]",
    "process": ""
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{100}+\\frac{y^{2}}{64}=1$ has foci $F_{1}$ and $F_{2}$. A point $P$ on the ellipse satisfies $\\angle F_{1} P F_{2}=60^{\\circ}$. Then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/100 + y^2/64 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 40], [60, 62]], [[0, 40]], [[44, 51]], [[52, 59]], [[0, 59]], [[64, 68]], [[60, 68]], [[70, 103]]]",
    "query_spans": "[[[105, 132]]]",
    "process": ""
  },
  {
    "text": "A line $l$ passing through the point $M(0,1)$ intersects the ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ at points $A$ and $B$, and $F_{1}$ is the left focus of the ellipse. When the perimeter of $\\triangle A B F_{1}$ is maximized, what is the equation of line $l$?",
    "fact_expressions": "l: Line;C: Ellipse;M: Point;A: Point;B: Point;F1: Point;Expression(C) = (x^2/9 + y^2/5 = 1);Coordinate(M) = (0, 1);PointOnCurve(M, l);LeftFocus(C)=F1;Intersection(l, C) = {A, B};WhenMax(Perimeter(TriangleOf(A,B,F1)))",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x+2*y-2=0",
    "fact_spans": "[[[11, 16], [113, 118]], [[17, 59], [78, 80]], [[1, 10]], [[60, 63]], [[64, 67]], [[70, 77]], [[17, 59]], [[1, 10]], [[0, 16]], [[70, 84]], [[11, 69]], [[85, 112]]]",
    "query_spans": "[[[113, 123]]]",
    "process": "Left focus $F_{1}(-2,0)$, right focus $F_{2}(2,0)$, then $|AF_{1}|=6-|AF_{2}|$, $|BF_{1}|=6-|BF_{2}|$, so $|AF_{1}|+|BF_{1}|+|AB|=12+|AB|-(|AF_{2}|+|BF_{2}|)$. Clearly, $|AF_{2}|+|BF_{2}|\\geqslant|AB|$, with equality if and only if points $A$, $B$, $F_{2}$ are collinear, that is, the perimeter of the triangle is minimized when line $l$ passes through $F_{2}$. At this time, the equation of the line is $x+2y-2=0$."
  },
  {
    "text": "Given that $A$ and $B$ are the left and right vertices of the hyperbola $C$: $\\frac{x^{2}}{m}-\\frac{y^{2}}{2}=1$, respectively, and $P(3,4)$ is a point on $C$, then the standard equation of the circumcircle of $\\triangle P A B$ is?",
    "fact_expressions": "C: Hyperbola;m: Number;P: Point;A: Point;B: Point;Expression(C) = (-y^2/2 + x^2/m = 1);Coordinate(P) = (3, 4);LeftVertex(C) = A;RightVertex(C) = B;PointOnCurve(P, C)",
    "query_expressions": "Expression(CircumCircle(TriangleOf(P,A,B)))",
    "answer_expressions": "x^2+(y-3)^2=10",
    "fact_spans": "[[[12, 55], [71, 74]], [[20, 55]], [[62, 70]], [[2, 5]], [[6, 9]], [[12, 55]], [[62, 70]], [[2, 61]], [[2, 61]], [[62, 77]]]",
    "query_spans": "[[[79, 107]]]",
    "process": "\\because P(3,4) is a point on C, \\frac{9}{m}-\\frac{16}{2}=1 gives m=1, then B(1,0), \\therefore k_{PB}=\\frac{4}{2}=2, the perpendicular bisector equation of PB is y=-\\frac{1}{2}(x-2)+2, let x=0, then y=3, let the circumcenter M be (0,t), then M(0,3), r=|MB|=\\sqrt{1+3^{2}}, \\therefore the standard equation of the circumcircle of \\triangle PAB is x^{2}+(y-3)^{2}=10"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{7}=1$, the line $L$ passes through its left focus $F_{1}$, intersecting the left branch of the hyperbola at points $A$ and $B$, with $|A B|=4$, and $F_{2}$ is the right focus. The perimeter of $\\triangle A B F_{2}$ is $20$. Then $m=?$",
    "fact_expressions": "G: Hyperbola;m: Number;L: Line;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (-y^2/7 + x^2/m = 1);LeftFocus(G) = F1;PointOnCurve(F1,L);Intersection(L,LeftPart(G))={A,B};Abs(LineSegmentOf(A, B)) = 4;RightFocus(G)=F2;Perimeter(TriangleOf(A, B, F2)) = 20",
    "query_expressions": "m",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 40], [47, 48], [60, 63]], [[130, 133]], [[41, 46]], [[66, 69]], [[70, 73]], [[87, 94]], [[51, 58]], [[2, 40]], [[47, 58]], [[41, 58]], [[41, 75]], [[77, 86]], [[60, 98]], [[99, 128]]]",
    "query_spans": "[[[130, 135]]]",
    "process": ""
  },
  {
    "text": "Let point $P$ be a point on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and let $F_{1}$, $F_{2}$ be the left and right foci of the ellipse, respectively. Let $I$ be the incenter of $\\triangle P F_{1} F_{2}$. If the area of $\\Delta P F_{1} F_{2}$ is 3 times the area of $\\Delta I F_{1} F_{2}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;I: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;Incenter(TriangleOf(P, F1, F2)) = I;Area(TriangleOf(P,F1,F2))=3*Area(TriangleOf(I,F1,F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[6, 58], [80, 82], [181, 183]], [[8, 58]], [[8, 58]], [[1, 5]], [[62, 69]], [[70, 77]], [[89, 92]], [[8, 58]], [[8, 58]], [[6, 58]], [[1, 61]], [[62, 88]], [[62, 88]], [[89, 121]], [[123, 178]]]",
    "query_spans": "[[[181, 189]]]",
    "process": "Using the inradius, $ S_{APF_{1}F_{2}} $ and $ S_{\\triangle IF_{1}F_{2}} $ can be expressed respectively. Using the ratio of the areas of the two triangles, we obtain $ a+c=3c $, thereby finding the eccentricity.  \n$ \\therefore S_{APF_{1}F_{2}} = S_{AIPF_{1}} + S_{\\Delta IPF_{2}} + S_{AIF_{1}F_{2}} = \\frac{1}{2}(|PF_{1}| + |PF_{2}| + |F_{1}F_{2}|) \\cdot r = (a+c) \\cdot r $,  \nand $ S_{\\Delta IF_{1}F_{2}} = \\frac{1}{2}|F_{1}F_{2}| \\cdot r = c \\cdot r $,  \n$ S_{APF_{1}F_{2}} = 3S_{\\Delta IF_{1}F_{2}} $  \n$ \\therefore a+c=3c $  \n$ \\therefore e = \\frac{c}{a} = \\frac{1}{2} $  \nThe correct answer: $ \\frac{1}{2} $"
  },
  {
    "text": "One focus of the ellipse $x^{2}+\\frac{y^{2}}{k}=1$ is $(0, \\sqrt{2})$, then $k=$?",
    "fact_expressions": "G: Ellipse;k: Number;Expression(G) = (x^2 + y^2/k = 1);Coordinate(F)=(0,sqrt(2));F:Point;OneOf(Focus(G))=F",
    "query_expressions": "k",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 27]], [[49, 52]], [[0, 27]], [[33, 47]], [[33, 47]], [[0, 47]]]",
    "query_spans": "[[[49, 54]]]",
    "process": "The ellipse $x^{2}+\\frac{y^{2}}{k}=1$ has a focus at $(0,\\sqrt{2})$, indicating that the focus lies on the $y$-axis, so $a^{2}=k$, $b^{2}=1$, $c^{2}=2$. From $a^{2}=b^{2}+c^{2}$, we obtain $a^{2}=3$, thus $k=3$. The answer is: 3."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1$ $(2>b>0)$, the line $x+y=1$ intersects the ellipse at points $P$ and $Q$. Let $M$ be the midpoint of segment $PQ$, $O$ be the coordinate origin, and $OP \\perp OQ$. Then the slope of line $OM$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/b^2 = 1);b: Number;2 > b;b > 0;H: Line;Expression(H) = (x + y = 1);Intersection(H, G) = {P, Q};P: Point;Q: Point;M: Point;MidPoint(LineSegmentOf(P, Q)) = M;O: Origin;IsPerpendicular(LineSegmentOf(O, P), LineSegmentOf(O, Q)) = True",
    "query_expressions": "Slope(LineOf(O, M))",
    "answer_expressions": "1/7",
    "fact_spans": "[[[2, 50], [61, 63]], [[2, 50]], [[4, 50]], [[4, 50]], [[4, 50]], [[51, 60]], [[51, 60]], [[52, 74]], [[65, 68]], [[69, 72]], [[87, 90]], [[76, 90]], [[91, 94]], [[101, 116]]]",
    "query_spans": "[[[118, 130]]]",
    "process": "Solve the system \\begin{cases}x+y-1=0\\\\\\frac{x^2}{4}+\\frac{y^{2}}{b^{2}}=1\\end{cases}, eliminate $ y $ and simplify to obtain $(4+b^{2})x^{2}-8x+4-4b^{2}=0$. $\\Delta=64-4(4+b^{2})(4-4b^{2})=16b^{4}+48b^{2}>0$ always holds. Let $P(x_{1},y_{1})$, $Q(x_{2},y_{2})$, then $M\\left(\\frac{x_{1}+x_{2}}{2},\\frac{y_{1}+y_{2}}{2}\\right)$. Thus, $x_{1}+x_{2}=\\frac{8}{4+b^{2}}$, $x_{1}x_{2}=\\frac{4-4b^{2}}{4+b^{2}}$. Since $OP\\perp OQ$, we have $\\overrightarrow{OP}\\cdot\\overrightarrow{OQ}=0$, i.e., $x_{1}x_{2}+y_{1}y_{2}=0$. Hence, $x_{1}x_{2}+(1-x_{1})(1-x_{2})=0$, so $2x_{1}x_{2}-(x_{1}+x_{2})+1=0$. Solving gives $b^{2}=\\frac{4}{7}$, $\\frac{y_{1}+y_{2}}{2}=\\frac{1-x_{1}+1-x_{2}}{2}=1-\\frac{x_{1}+x_{2}}{2}=1-\\frac{4}{4+b^{2}}=1-\\frac{4}{4+\\frac{4}{7}}=1-\\frac{4}{\\frac{32}{7}}=1-\\frac{7}{8}=\\frac{1}{8}<ERR>+$.1"
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ are $F_{1}$ and $F_{2}$, respectively. A line passing through focus $F_{1}$ intersects the ellipse at points $A$ and $B$. If the area of the incircle of $\\triangle A B F_{2}$ is $\\pi$, and the coordinates of points $A$ and $B$ are $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$, respectively, then the value of $|y_{2}-y_{1}|$ is?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;F2: Point;F1: Point;x1:Number;x2:Number;y1:Number;y2:Number;Expression(G) = (x^2/16 + y^2/9 = 1);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);LeftFocus(G) = F1;RightFocus(G) = F2;I:Line;PointOnCurve(F1, I);Intersection(I, G) = {A, B};Area(InscribedCircle(TriangleOf(A,B,F2)))=pi",
    "query_expressions": "Abs(y2-y1)",
    "answer_expressions": "8*sqrt(7)/7",
    "fact_spans": "[[[0, 38], [77, 79]], [[80, 83], [128, 132]], [[84, 87], [133, 137]], [[55, 62]], [[47, 54], [66, 73]], [[145, 161]], [[162, 178]], [[145, 161]], [[162, 178]], [[0, 38]], [[128, 178]], [[128, 178]], [[0, 62]], [[0, 62]], [[74, 76]], [[63, 76]], [[74, 89]], [[91, 126]]]",
    "query_spans": "[[[180, 199]]]",
    "process": ""
  },
  {
    "text": "A asymptote of the hyperbola $tx^{2}-y^{2}=1$ is perpendicular to the line $2 x+y+1=0$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;t: Number;l: Line;Expression(G) = (t*x^2 - y^2 = 1);Expression(l) = (2*x + y + 1 = 0);IsPerpendicular(OneOf(Asymptote(G)),l)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[0, 19], [44, 47]], [[3, 19]], [[26, 39]], [[0, 19]], [[26, 39]], [[0, 41]]]",
    "query_spans": "[[[44, 53]]]",
    "process": ""
  },
  {
    "text": "Given the equation of the ellipse is $\\frac{x^{2}}{16}+\\frac{y^{2}}{m^{2}}=1$, with foci on the $x$-axis, what is the range of values for $m$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/m^2 = 1);PointOnCurve(Focus(G), xAxis);m:Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-4,0)+(0,4)",
    "fact_spans": "[[[2, 4]], [[2, 48]], [[2, 57]], [[58, 61]]]",
    "query_spans": "[[[58, 68]]]",
    "process": "Since the foci of the ellipse lie on the x-axis, we have 0 < m^{2} < 16; solving this gives the result. Given the equation of the ellipse \\frac{x^{2}}{16} + \\frac{y^{2}}{m^{2}} = 1 and that the foci lie on the x-axis, we obtain 0 < m^{2} < 16, so -4 < m < 0 or 0 < m < 4."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$, and $P$, $Q$ are two points on the right branch such that the line $PQ$ passes through $F_{2}$, then the value of $|PF_{1}|+|QF_{1}|-|PQ|$ is?",
    "fact_expressions": "G: Hyperbola;Q: Point;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/2 - y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));PointOnCurve(Q, RightPart(G));PointOnCurve(F2, LineOf(P, Q))",
    "query_expressions": "Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(Q, F1)) - Abs(LineSegmentOf(P, Q))",
    "answer_expressions": "4*sqrt(2)",
    "fact_spans": "[[[18, 46]], [[57, 60]], [[53, 56]], [[2, 9]], [[10, 17], [76, 83]], [[18, 46]], [[2, 52]], [[2, 52]], [[18, 67]], [[18, 67]], [[68, 83]]]",
    "query_spans": "[[[85, 113]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has its right focus at $F(2,0)$. A line passing through point $F$ intersects the ellipse at points $A$ and $B$. If the midpoint of $AB$ has coordinates $(1, \\frac{1}{2})$, then the equation of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;RightFocus(C) = F;Coordinate(F) = (2, 0);G: Line;PointOnCurve(F, G);A: Point;B: Point;Intersection(G, C) = {A, B};Coordinate(MidPoint(LineSegmentOf(A, B))) = (1, 1/2)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "(3*x^2)/16+(3*y^2)/4=1",
    "fact_spans": "[[[2, 59], [82, 84], [126, 131]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[64, 72], [74, 78]], [[2, 72]], [[64, 72]], [[79, 81]], [[73, 81]], [[85, 88]], [[89, 92]], [[79, 94]], [[95, 124]]]",
    "query_spans": "[[[126, 136]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $m x^{2}+5 y^{2}=5 m$ $(m>0)$ has eccentricity $e=\\frac{\\sqrt{10}}{5}$, find $m=$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (m*x^2 + 5*y^2 = 5*m);m: Number;m>0;e: Number;Eccentricity(G) = e ;e= sqrt(10)/5",
    "query_expressions": "m",
    "answer_expressions": "{3, 25/3}",
    "fact_spans": "[[[2, 30]], [[2, 30]], [[60, 63]], [[4, 30]], [[35, 58]], [[2, 58]], [[35, 58]]]",
    "query_spans": "[[[60, 65]]]",
    "process": "From the ellipse $mx^{2}+5y^{2}=5m$, we rewrite it as: $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$. When $0<m<5$, $a=\\sqrt{5}$, $c=\\sqrt{5-m}$, $\\therefore e=\\frac{c}{a}=\\frac{\\sqrt{5-m}}{\\sqrt{5}}=\\frac{\\sqrt{10}}{5}$, solving gives $m=3$; when $m>5$, $a=\\sqrt{m}$, $c=\\sqrt{m-5}$, $\\cdots\\cdot s e=\\frac{c}{a}=\\frac{\\sqrt{m-5}}{\\sqrt{m}}=\\frac{\\sqrt{10}}{5}$, solving gives $m=\\frac{25}{3}$."
  },
  {
    "text": "Let a chord $AB$ of the parabola $y^{2}=4x$ have point $P(1,\\ 1)$ as its midpoint. Then, the slope of the line containing this chord is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G) = True;P: Point;Coordinate(P) = (1, 1);MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Slope(OverlappingLine(LineSegmentOf(A, B)))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 15]], [[1, 15]], [[19, 23]], [[19, 23]], [[1, 23]], [[24, 35]], [[24, 35]], [[19, 38]]]",
    "query_spans": "[[[1, 52]]]",
    "process": ""
  },
  {
    "text": "The standard equation of the parabola passing through the point $P(4, -2)$ is?",
    "fact_expressions": "G: Parabola;P: Point;Coordinate(P) = (4, -2);PointOnCurve(P, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{(y^2=x), (x^2=-8*y)}",
    "fact_spans": "[[[14, 17]], [[2, 13]], [[2, 13]], [[0, 17]]]",
    "query_spans": "[[[14, 24]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x$ has coordinates $(1,0)$, then $p=$? What is the equation of the directrix?",
    "fact_expressions": "G: Parabola;p: Number;Expression(G) = (y^2 = 2*(p*x));Coordinate(Focus(G)) = (1, 0)",
    "query_expressions": "p;Expression(Directrix(G))",
    "answer_expressions": "2\nx = -1",
    "fact_spans": "[[[1, 17]], [[32, 35]], [[1, 17]], [[1, 30]]]",
    "query_spans": "[[[32, 37]], [[1, 43]]]",
    "process": "Since the parabola equation is $y^{2}=2px$, the focus coordinates are $(\\frac{p}{2},0)$. Given that the focus coordinates are $(1,0)$, then $p=2$, and the directrix equation is $x=-1$."
  },
  {
    "text": "The coordinates of the foci of the ellipse $25 x^{2}+16 y^{2}=1$ are?",
    "fact_expressions": "G: Ellipse;Expression(G) = (25*x^2 + 16*y^2 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, pm*3/20)",
    "fact_spans": "[[[0, 23]], [[0, 23]]]",
    "query_spans": "[[[0, 30]]]",
    "process": "The standard equation of the ellipse is \\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1, so the foci lie on the y-axis, where a^{2}=\\frac{1}{16}, b^{2}=\\frac{1}{25}, thus c^{2}=a^{2}-b^{2}=\\frac{1}{16}-\\frac{1}{25}=\\frac{9}{400}, hence c=\\frac{3}{20}. Therefore, the coordinates of the foci of this ellipse are (0,\\pm\\frac{3}{20}). Answer: (0,\\pm\\frac{3}{20})."
  },
  {
    "text": "Given that points $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{2} + y^{2} = 1$, respectively. A line passing through $F_{2}$ with an inclination angle of $\\frac{\\pi}{4}$ intersects the ellipse at points $A$ and $B$. Then, the area of $\\Delta F_{1} A B$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Line;PointOnCurve(F2, H);Inclination(H) = pi/4;A: Point;B: Point;Intersection(H, G) = {A, B}",
    "query_expressions": "Area(TriangleOf(F1, A, B))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[21, 48], [87, 89]], [[21, 48]], [[2, 10]], [[11, 18], [56, 63]], [[2, 54]], [[2, 54]], [[84, 86]], [[55, 86]], [[64, 86]], [[90, 93]], [[94, 97]], [[84, 99]]]",
    "query_spans": "[[[101, 124]]]",
    "process": "From the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, we can obtain the left focus $F_{1}$ and the right focus $F_{2}$ of the ellipse. Drawing a line through $F_{2}$ with an inclination angle of $\\frac{\\pi}{4}$, we get the equation of line $AB$ as $y=x-1$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$. By solving the system of equations consisting of the line and the ellipse, we can transform it into a quadratic equation in $x$. Then, using the relationships between roots and coefficients, the chord length formula, the point-to-line distance formula, and the triangle area formula, we can solve the problem. [Solution] From the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, we obtain the left focus $F_{1}(-1,0)$ and the right focus $F_{2}(1,0)$. Therefore, the equation of line $AB$ is $y=x-1$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$. Solving the system\n$$\n\\begin{cases}\ny=x-1 \\\\\n\\frac{x^{2}}{2}+y^{2}=1\n\\end{cases}\n$$\nwe obtain $3x^{2}-4x=0$. Thus, $x_{1}+x_{2}=\\frac{4}{3}$, $x_{1}x_{2}=0$, \n$$\n|AB|=\\sqrt{(1+1^{2})[(x_{1}+x_{2})^{2}-4x_{1}x_{2}]} = \\sqrt{2\\left[\\left(\\frac{4}{3}\\right)^{2}-4\\times0\\right]} = \\frac{4\\sqrt{2}}{3}\n$$\nThe distance from point $F_{1}$ to line $AB$ is \n$$\nd = \\frac{|(-1)\\times1-0-1|}{\\sqrt{2}} = \\sqrt{2}\n$$\nTherefore,\n$$\nS_{\\Delta AF_{1}B} = \\frac{1}{2} \\cdot d \\cdot |AB| = \\frac{1}{2} \\times \\sqrt{2} \\times \\frac{4\\sqrt{2}}{3} = \\frac{4}{3}\n$$"
  },
  {
    "text": "Given that a point $A$ on the parabola $y^{2} = -4x$ is at a distance of $5$ from the focus, find the distance from $A$ to the origin.",
    "fact_expressions": "G: Parabola;A: Point;Expression(G) = (y^2 = -4*x);PointOnCurve(A, G);Distance(A, Focus(G)) = 5;O:Origin",
    "query_expressions": "Distance(A,O)",
    "answer_expressions": "4*sqrt(2)",
    "fact_spans": "[[[2, 17]], [[20, 23], [36, 39]], [[2, 17]], [[2, 23]], [[2, 34]], [[40, 44]]]",
    "query_spans": "[[[36, 49]]]",
    "process": "According to the definition of a parabola: the distance to the focus equals the distance to the directrix, the distance from point A to the directrix is 5, thus the x-coordinate of point A is -4. Therefore, the coordinates of point P are $(-4,\\pm4)$, and then use the distance formula."
  },
  {
    "text": "The minimum distance from points on the parabola $y = -x^2$ to the line $4x + 3y - 8 = 0$ is?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Expression(G) = (y = -x^2);Expression(H) = (4*x + 3*y - 8 = 0);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, H))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[0, 13]], [[17, 32]], [[15, 16]], [[0, 13]], [[17, 32]], [[0, 16]]]",
    "query_spans": "[[[15, 41]]]",
    "process": ""
  },
  {
    "text": "Given that the line $y = kx - 1$ and the hyperbola $x^2 - y^2 = 4$ have no common points, find the range of real values for $k$.",
    "fact_expressions": "H: Line;Expression(H) = (y = k*x - 1);k: Real;G: Hyperbola;Expression(G) = (x^2 - y^2 = 4);NumIntersection(H, G) = 0",
    "query_expressions": "Range(k)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 13]], [[2, 13]], [[39, 44]], [[14, 32]], [[14, 32]], [[2, 37]]]",
    "query_spans": "[[[39, 51]]]",
    "process": ""
  },
  {
    "text": "The minor axis has length $\\sqrt{5}$, and the eccentricity is $e = \\frac{2}{3}$. The foci of the ellipse are $F_{1}$ and $F_{2}$. A line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. Then, the perimeter of $\\triangle ABF_{2}$ is?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;F2: Point;F1: Point;e: Number;Focus(G) = {F1, F2};Length(MinorAxis(G)) = sqrt(5);Eccentricity(G) = e;e = 2/3;PointOnCurve(F1, H);Intersection(H, G) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "12",
    "fact_spans": "[[[34, 36], [68, 70]], [[65, 67]], [[71, 74]], [[75, 78]], [[48, 55]], [[40, 47], [57, 64]], [[18, 33]], [[34, 55]], [[0, 36]], [[15, 36]], [[18, 33]], [[56, 67]], [[65, 80]]]",
    "query_spans": "[[[82, 106]]]",
    "process": ""
  },
  {
    "text": "Given that the circle $x^{2}+y^{2}+4 x+3=0$ is tangent to the directrix of the parabola $y^{2}=2 p x$ $(p>0)$, then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;H: Circle;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (4*x + x^2 + y^2 + 3 = 0);IsTangent(H, Directrix(G))",
    "query_expressions": "p",
    "answer_expressions": "2,6",
    "fact_spans": "[[[25, 46]], [[53, 56]], [[2, 24]], [[28, 46]], [[25, 46]], [[2, 24]], [[2, 51]]]",
    "query_spans": "[[[53, 58]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of a hyperbola with foci on the $x$-axis are $3x \\pm 4y = 0$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), xAxis);Expression(Asymptote(G)) = (3*x + pm*4*y = 0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/4",
    "fact_spans": "[[[11, 14], [38, 41]], [[2, 14]], [[11, 36]]]",
    "query_spans": "[[[38, 47]]]",
    "process": "Let the equation of the hyperbola with foci on the x-axis be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a,b>0)$. Since the asymptotes of this hyperbola are $y=\\pm\\frac{3}{4}x$, it follows that $\\frac{b}{a}=\\frac{3}{4}$. For a hyperbola, $c^{2}=a^{2}+b^{2}$, so the eccentricity is $e=\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{1+(\\frac{3}{4})^{2}}=\\frac{5}{4}$. Therefore, the eccentricity of the hyperbola is $\\frac{5}{4}$."
  },
  {
    "text": "The center of a moving circle lies on the parabola $y^{2}=8 x$, and the moving circle is always tangent to the line $x+2=0$. Then the moving circle must pass through the point?",
    "fact_expressions": "C:Circle;G: Parabola;Expression(G) = (y^2 = 8*x);PointOnCurve(Center(C),G);L: Line;Expression(L) = (x + 2 = 0);IsTangent(C,L);P:Point;PointOnCurve(P,C)",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2,0)",
    "fact_spans": "[[[0, 2], [23, 25], [40, 42]], [[6, 20]], [[6, 20]], [[0, 21]], [[27, 36]], [[27, 36]], [[23, 38]], [[44, 45]], [[40, 45]]]",
    "query_spans": "[[[40, 46]]]",
    "process": "First, write the coordinates of the focus and the equation of the directrix from the standard equation of the parabola, then combine with the definition of the parabola to conclude that the focus must lie on the moving circle, thereby solving the problem. The focus of the parabola $ y^{2} = 8x $ is $ F(2,0) $, and the equation of the directrix is $ x + 2 = 0 $. Hence, the distance from the center of the circle to the line $ x + 2 = 0 $, which equals the radius, is equal to the distance from the center to the focus $ F $, so $ F $ lies on the circle."
  },
  {
    "text": "The hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has left and right foci $F_{1}(-c, 0)$, $F_{2}(c, 0)$ respectively. A line passing through $F_{1}$ with slope $\\sqrt{3}$ intersects the left and right branches of the hyperbola at points $P$ and $Q$ respectively. If $QP = QF_{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;c: Number;Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);LeftFocus(C) = F1;RightFocus(C) = F2;G: Line;PointOnCurve(F1, G);Slope(G) = sqrt(3);P: Point;Q: Point;Intersection(G, LeftPart(C)) = P;Intersection(G, RightPart(C)) = Q;LineSegmentOf(Q, P) = LineSegmentOf(Q, F2)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(1+sqrt(13))/2",
    "fact_spans": "[[[0, 61], [124, 127], [162, 165]], [[0, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[69, 83], [100, 107]], [[84, 98]], [[69, 83]], [[69, 83]], [[84, 98]], [[0, 98]], [[0, 98]], [[121, 123]], [[99, 123]], [[107, 123]], [[136, 140]], [[141, 144]], [[121, 144]], [[121, 144]], [[146, 159]]]",
    "query_spans": "[[[162, 171]]]",
    "process": "According to QP = QF₂, by the definition of hyperbola, PF₁ = 2a, PF₂ = 4a. Since the slope of the line is √3, it follows that ∠PF₁F₂ = π/3. In ΔPF₁F₂, applying the cosine law gives cos(π/3) = (4a² + 4c² - 16a²) / (2 × 2a × 2c) ⇒ e² - e - 3 = 0. ∴ e = (1 + √13)/2"
  },
  {
    "text": "Given the two foci of a hyperbola $F_{1}(-\\sqrt{10} , 0)$, $F_{2}(\\sqrt{10} , 0)$, and $M$ is a point on this hyperbola such that $|MF_{1}|-| MF_{2} |=6$, then the equation of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;F1: Point;F2: Point;Coordinate(F1) = (-sqrt(10), 0);Coordinate(F2) = (sqrt(10), 0);Focus(C) = {F1, F2};M: Point;PointOnCurve(M, C);Abs(LineSegmentOf(M, F1)) - Abs(LineSegmentOf(M, F2)) = 6",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/9 - y^2 = 1",
    "fact_spans": "[[[2, 5], [66, 69], [99, 102]], [[10, 33]], [[36, 58]], [[10, 33]], [[36, 58]], [[2, 58]], [[61, 64]], [[61, 73]], [[74, 97]]]",
    "query_spans": "[[[99, 107]]]",
    "process": ""
  },
  {
    "text": "If the coordinates of the focus of the parabola ${y}^{2}=ax$ are $(2 , 0)$, then the value of the real number $a$ is?",
    "fact_expressions": "G: Parabola;a: Real;Expression(G) = (y^2 = a*x);Coordinate(Focus(G)) = (2, 0)",
    "query_expressions": "a",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 16]], [[33, 38]], [[1, 16]], [[1, 31]]]",
    "query_spans": "[[[33, 42]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a moving point on the parabola $y^{2}=4x$, $F$ is the focus of the parabola, and a perpendicular is drawn from point $P$ to the line $x=-2$, with foot of the perpendicular at $M$. Point $Q$ is a moving point on the circle $C$: $(x+3)^{2}+(y-4)^{2}=1$. Then the minimum value of $|PM|+|PQ|$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P, G);G: Parabola;Expression(G) = (y^2 = 4*x);Focus(G) = F;F: Point;H: Line;Expression(H) = (x = -2);Z: Line;PointOnCurve(P, Z) ;IsPerpendicular(H, Z) ;FootPoint(H, Z) = M ;M: Point;Q: Point;PointOnCurve(Q, C) ;C: Circle;Expression(C) = ((x + 3)^2 + (y - 4)^2 = 1)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, M)) + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "4*sqrt(2)",
    "fact_spans": "[[[2, 5], [37, 41]], [[2, 24]], [[6, 20], [29, 32]], [[6, 20]], [[25, 35]], [[25, 28]], [[42, 48]], [[42, 48]], [], [[36, 51]], [[36, 51]], [[36, 58]], [[55, 58]], [[60, 64]], [[60, 97]], [[65, 93]], [[65, 93]]]",
    "query_spans": "[[[99, 118]]]",
    "process": "The focus of the parabola $ y^{2}=4x $ has coordinates $ (1,0) $, and the equation of the directrix is $ x=-1 $. By the definition of a parabola, $ |PM|=|PF|+1 $, hence $ |PM|+|PQ|=|PF|+1+|PQ|\\geqslant|QF|+1 $, where points $ Q $, $ P $, $ F $ are collinear. Point $ Q $ lies on a circle centered at $ (-3,4) $ with radius 1, thus $ |QF|\\geqslant|CF|-1=\\sqrt{(-3-1)^{2}+(4-0)^{2}}-1=4\\sqrt{2}-1 $, therefore $ |PM|+|PQ|\\geqslant4\\sqrt{2} $."
  },
  {
    "text": "1. Given the parabola $C: y^{2}=2 p x(p>0)$ has focus $F$, and a line $l$ passing through point $F$ with an inclination angle of $\\frac{\\pi}{6}$ intersects the parabola $C$ at point $A$ in the first quadrant and intersects the directrix of the parabola at point $B$. If $\\overrightarrow{A F}=\\lambda \\overrightarrow{A B}$, then $\\lambda=$?",
    "fact_expressions": "C: Parabola;p: Number;p > 0;Expression(C) = (y**2 = 2*p*x);F: Point;Focus(C) = F;l: Line;PointOnCurve(F, l);Inclination(l) = pi/6;A: Point;Quadrant(A) = 1;Intersection(l, C) = A;B: Point;Intersection(l, Directrix(C)) = B;lambda: Number;VectorOf(A, F) = lambda * VectorOf(A, B)",
    "query_expressions": "lambda",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[5, 30], [79, 87], [101, 104]], [[9, 30]], [[9, 30]], [[5, 30]], [[36, 39], [42, 47]], [[5, 39]], [[71, 77]], [[41, 77]], [[49, 77]], [[93, 98]], [[89, 98]], [[71, 98]], [[108, 113]], [[71, 113]], [[116, 169]], [[116, 169]]]",
    "query_spans": "[[[172, 183]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the focus of the parabola $y=-2 x^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = -2*x^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,-1/8)",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 22]]]",
    "process": "The equation $ y = -2x^{2} $, which is $ x^{2} = -\\frac{1}{2}y $, thus its focus coordinates are $ (0, -\\frac{1}{8}) $."
  },
  {
    "text": "The line $l$: $x = m y + 2$ passes through the focus $F$ of the parabola $C$: $y^2 = 2 p x$ ($p > 0$), and intersects the parabola at two points $A$ and $B$. A line passing through the origin goes through the midpoint $D$ of chord $AB$, and intersects the parabola at point $E$ (distinct from the origin). Then the range of $\\frac{|O D|}{|O E|}$ is?",
    "fact_expressions": "l: Line;Expression(l) = (x=m*y+2);m: Number;C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, C) = {A, B};G: Line;O: Origin;PointOnCurve(O, G);IsChordOf(LineSegmentOf(A, B), C);D: Point;MidPoint(LineSegmentOf(A, B)) = D;PointOnCurve(D, G);E: Point;Intersection(G, C) = E;Negation(E=O)",
    "query_expressions": "Range(Abs(LineSegmentOf(O, D))/Abs(LineSegmentOf(O, E)))",
    "answer_expressions": "(0, 1/2)",
    "fact_spans": "[[[0, 16]], [[0, 16]], [[7, 16]], [[18, 44], [52, 55], [92, 95]], [[18, 44]], [[26, 44]], [[26, 44]], [[47, 50]], [[18, 50]], [[0, 50]], [[58, 61]], [[62, 65]], [[0, 67]], [[72, 74]], [[69, 71], [104, 106]], [[68, 74]], [[52, 82]], [[85, 88]], [[77, 88]], [[72, 88]], [[97, 101]], [[72, 101]], [[97, 107]]]",
    "query_spans": "[[[109, 137]]]",
    "process": "The line $ l: x = my + 2 $ passes through $ (2,0) $, which is the focus $ F(2,0) $ of the parabola $ C: y^{2} = 2px $ ($ p > 0 $), so $ p = 4 $, and the equation of the parabola is $ y^{2} = 8x $. Solving the system \n\\[\n\\begin{cases}\ny^{2} = 8x \\\\\nx = my + 2\n\\end{cases}\n\\]\nyields $ y^{2} - 8my - 16 = 0 $. The discriminant $ \\triangle = 64m^{2} + 64 > 0 $ always holds. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, so $ y_{1} + y_{2} = 8m $, $ x_{1} + x_{2} = m(y_{1} + y_{2}) + 4 = 8m^{2} + 4 $. Thus, the midpoint $ D $ of chord $ AB $ is $ (4m^{2} + 2, 4m) $. Therefore, the equation of $ OD $ is $ y = \\frac{4m}{4m^{2} + 2}x $. From the problem, $ m \\neq 0 $. Solving the system with the parabola $ y^{2} = 8x $:\n\\[\n\\begin{cases}\ny^{2} = 8x \\\\\ny = \\frac{4m}{4m^{2} + 2}x\n\\end{cases}\n\\]\ngives $ y_{E} = \\frac{4(2m^{2} + 1)}{m} $. Then $ \\frac{|OD|}{|OE|} = \\frac{|y_{D}|}{|y_{E}|} = \\frac{m^{2}}{2m^{2} + 1} = \\frac{1}{2 + \\frac{1}{m^{2}}} $. Since $ m^{2} > 0 $, $ \\frac{1}{m^{2}} > 0 $, so $ 2 + \\frac{1}{m^{2}} > 2 $, hence $ 0 < \\frac{1}{2 + \\frac{1}{m^{2}}} < \\frac{1}{2} $, that is, $ 0 < \\frac{|OD|}{|OE|} < \\frac{1}{2} $."
  },
  {
    "text": "Given the circle $C$: $x^{2}+y^{2}-2 x-5 y+4=0$, using the intersection points of circle $C$ with the coordinate axes as a focus and a vertex of a hyperbola respectively, the standard equation of the hyperbola satisfying the above conditions is?",
    "fact_expressions": "G: Hyperbola;C: Circle;Expression(C) = (-5*y - 2*x + x^2 + y^2 + 4 = 0);P:Point;P1:Point;Intersection(axis,C)={P,P1};OneOf(Focus(G))=P;OneOf(Vertex(G))=P1",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 - x^2/15 = 1",
    "fact_spans": "[[[50, 53], [70, 73]], [[2, 33], [35, 39]], [[2, 33]], [], [], [[35, 46]], [[34, 61]], [[34, 61]]]",
    "query_spans": "[[[70, 80]]]",
    "process": "The problem requires finding the coordinates of the vertices and foci of a hyperbola to obtain the value of $ b $, then determining the standard equation of the hyperbola based on its characteristics. Given the circle $ C: x^{2}+y^{2}-2x-5y+4=0 $, its intersections with the coordinate axes are points $ A(0,1) $ and $ B(0,4) $. Therefore, the vertex of the required hyperbola is $ A(0,1) $, and the focus is $ B(0,4) $. Hence, $ a=1 $, $ c=4 $, and the focus lies on the $ y $-axis. $ \\therefore b=\\sqrt{c^{2}-a^{2}}=\\sqrt{15} $. Thus, the standard equation of the required hyperbola is $ y^{2}-\\frac{x^{2}}{15}=1 $."
  },
  {
    "text": "The standard equation of an ellipse passing through the point $A(-1,-2)$ and having the same foci as the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{9}=1$ is?",
    "fact_expressions": "G: Ellipse;A: Point;H:Ellipse;Expression(H) = (x^2/6 + y^2/9 = 1);Coordinate(A) = (-1, -2);PointOnCurve(A, G);Focus(G)=Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/3 + y^2/6 = 1",
    "fact_spans": "[[[57, 59]], [[1, 12]], [[14, 51]], [[14, 51]], [[1, 12]], [[0, 59]], [[13, 59]]]",
    "query_spans": "[[[57, 66]]]",
    "process": "Since in the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{9}=1$, $c^{2}=9-3=3$, assume the desired ellipse equation is $\\frac{y^{2}}{m}+\\frac{x^{2}}{m-3}=1$ $(m>3)$, substituting point $A(-1,-2)$ gives $\\frac{4}{m}+\\frac{1}{m-3}=1$, solving yields $m=6$ or $m=2$ (discarded), thus the desired ellipse equation is $\\frac{x^{2}}{3}+\\frac{y^{2}}{6}=1$."
  },
  {
    "text": "If one focus of the ellipse $a^{2} x^{2}-2 a y^{2}=2$ is $(-\\sqrt{3}, 0)$, then $a=$?",
    "fact_expressions": "G: Ellipse;a: Number;H: Point;Expression(G) = (a^2*x^2 - 2*a*y^2 = 2);Coordinate(H) = (-sqrt(3), 0);OneOf(Focus(G)) = H",
    "query_expressions": "a",
    "answer_expressions": "-2/3",
    "fact_spans": "[[[1, 28]], [[52, 55]], [[34, 50]], [[1, 28]], [[34, 50]], [[1, 50]]]",
    "query_spans": "[[[52, 57]]]",
    "process": "Transform the ellipse $ a^{2x}2 - 2ay2 = 2 $ into standard form: $ \\frac{x2}{\\frac{2}{a2}} + \\frac{y^{2}}{-\\frac{1}{a}} = 1 $. One focus of the ellipse is $ (-\\sqrt{3}, 0) $, so the ellipse has its foci on the x-axis.  \n$ \\therefore \\begin{cases} -\\frac{1}{a} > 0 \\\\ \\frac{2}{a^{2}} - (-\\frac{1}{a}) = \\sqrt{3} \\end{cases} $.  \n$ \\Rightarrow \\begin{cases} a < 0 \\\\ 3a^{2} - a - 2 = 0 \\end{cases} $, solving gives: $ a = -\\frac{2}{3} $"
  },
  {
    "text": "Given that $F$ is the right focus of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and $P$ is a point on $C$ such that $|P F|=\\frac{3}{2}$, then the coordinates of point $P$ are?",
    "fact_expressions": "C: Ellipse;P: Point;F: Point;Expression(C) = (x^2/4 + y^2/3 = 1);RightFocus(C) = F;PointOnCurve(P, C);Abs(LineSegmentOf(P, F)) = 3/2",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1, pm*3/2)",
    "fact_spans": "[[[6, 48], [57, 60]], [[87, 91], [53, 56]], [[2, 5]], [[6, 48]], [[2, 52]], [[53, 64]], [[66, 85]]]",
    "query_spans": "[[[87, 96]]]",
    "process": "By the given condition, $ F(1,0) $, let $ P(x,y) $, then"
  },
  {
    "text": "Given that the line $MN$ passes through the left focus $F$ of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, intersecting the ellipse at points $M$ and $N$, and the line $PQ$ passes through the origin $O$ parallel to $MN$, intersecting the ellipse at points $P$ and $Q$, then $\\frac{|P Q|^{2}}{|M N|}$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);F: Point;LeftFocus(G) = F;M: Point;N: Point;PointOnCurve(F, LineOf(M, N));Intersection(LineOf(M, N), G) = {M, N};P: Point;Q: Point;O: Origin;PointOnCurve(O, LineOf(P, Q));IsParallel(LineOf(P, Q), LineSegmentOf(M, N));Intersection(LineSegmentOf(P, Q), G) = {P, Q}",
    "query_expressions": "Abs(LineSegmentOf(P, Q))^2/Abs(LineSegmentOf(M, N))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[10, 37], [46, 48], [89, 91]], [[10, 37]], [[41, 44]], [[10, 44]], [[50, 53]], [[54, 57]], [[2, 44]], [[2, 59]], [[93, 96]], [[97, 100]], [[68, 73]], [[60, 73]], [[60, 81]], [[83, 102]]]",
    "query_spans": "[[[104, 131]]]",
    "process": "Specialize, let $ MN \\perp x $-axis, then $ |MN| = \\frac{2b^{2}}{a} = \\frac{2}{\\sqrt{2}} = \\sqrt{2} $, $ |PQ|^{2} = 4 $, $ \\frac{|PQ|^{2}}{|MN|} = \\frac{4}{\\sqrt{2}} = 2\\sqrt{2} $"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a perpendicular line from its right vertex $A$ to one asymptote, intersecting the other asymptote at point $B$. If $|O B|=\\sqrt{3}|O A|$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;O: Origin;B: Point;A: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightVertex(C) = A;L1: Line;L2: Line;L: Line;PointOnCurve(A, L);IsPerpendicular(L, L1);Intersection(L, L2) = B;Abs(LineSegmentOf(O, B)) = sqrt(3)*Abs(LineSegmentOf(O, A));OneOf(Asymptote(C))=L1;OneOf(Asymptote(C))=L2;Negation(L1=L2)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "{2*sqrt(3)/3, sqrt(3)}",
    "fact_spans": "[[[2, 63], [65, 66], [119, 122]], [[10, 63]], [[10, 63]], [[95, 116]], [[89, 93]], [[69, 72]], [[10, 63]], [[10, 63]], [[2, 63]], [[65, 72]], [], [], [], [[64, 81]], [[64, 81]], [[64, 93]], [[95, 116]], [2, 75], [2, 85], [2, 85]]",
    "query_spans": "[[[119, 128]]]",
    "process": "As shown in the figure, without loss of generality, assume point $ B $ lies on the line $ y = -\\frac{b}{a}x $. It is easy to obtain the equation of line $ AB $ as $ y = -\\frac{a}{b}(x - a) $. Solving the system of equations of lines $ OB $ and $ AB $:\n$$\n\\begin{cases}\ny = -\\frac{b}{a}x \\\\\ny = -\\frac{a}{b}(x - a)\n\\end{cases}\n$$\nyields\n$$\n\\begin{cases}\nx = \\frac{a^3}{a^2 - b^2} \\\\\ny = -\\frac{a^2b}{a^2 - b^2}\n\\end{cases}\n$$\nThus, the coordinates of $ B $ are $ \\left( \\frac{a^3}{a^2 - b^2}, -\\frac{a^2b}{a^2 - b^2} \\right) $. Since $ |OB| = \\sqrt{3}|OA| $, it follows that $ |OB|^2 = 3|OA|^2 $, i.e.,\n$$\n\\left( \\frac{a^3}{a^2 - b^2} \\right)^2 + \\left( -\\frac{a^2b}{a^2 - b^2} \\right)^2 = 3a^2\n$$\nSimplifying yields $ a^4 + a^2b^2 = 3(a^2 - b^2)^2 $, leading to $ a^2 = 3b^2 $ or $ 2a^2 = b^2 $. Therefore, $ e^2 = \\frac{4}{3} $ or $ e^2 = 3 $, hence $ e = \\frac{2\\sqrt{3}}{3} $ or $ e = \\sqrt{3} $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. The line $MN$ passes through $F_{2}$ and intersects the right branch of the hyperbola at points $M$ and $N$. If $\\cos \\angle F_{1} M N = \\cos \\angle F_{1} F_{2} M$ and $\\frac{|F_{1} M|}{|F_{1} N|} = \\frac{1}{2}$, then the eccentricity of the hyperbola is equal to?",
    "fact_expressions": "M: Point;N: Point;G: Hyperbola;b: Number;a: Number;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, LineOf(M, N));Intersection(LineOf(M,N),RightPart(G))={M,N};Cos(AngleOf(F1, M, N)) = Cos(AngleOf(F1, F2, M));Abs(LineSegmentOf(F1, M))/Abs(LineSegmentOf(F1, N)) = 1/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[108, 111]], [[112, 115]], [[2, 59], [101, 104], [214, 217]], [[5, 59]], [[5, 59]], [[68, 75]], [[76, 83], [91, 98]], [[5, 59]], [[5, 59]], [[2, 59]], [[2, 83]], [[2, 83]], [[84, 98]], [[84, 117]], [[119, 169]], [[170, 212]]]",
    "query_spans": "[[[214, 224]]]",
    "process": ""
  },
  {
    "text": "From the left focus $F$ of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, draw a tangent to the circle $x^{2}+y^{2}=9$, with the point of tangency $T$. Extend $FT$ to intersect the right branch of the hyperbola at point $P$. If $M$ is the midpoint of segment $FP$ and $O$ is the origin, then $|MO|-|MT|=$?",
    "fact_expressions": "G: Hyperbola;H: Circle;F: Point;P: Point;T: Point;M: Point;O: Origin;l: Line;Expression(G) = (x^2/9 - y^2/16 = 1);Expression(H) = (x^2 + y^2 = 9);LeftFocus(G) = F;TangentOfPoint(F, H) = l;TangentPoint(l, H) = T;Intersection(OverlappingLine(LineSegmentOf(F,T)),RightPart(G))=P;MidPoint(LineSegmentOf(F, P)) = M",
    "query_expressions": "Abs(LineSegmentOf(M, O)) - Abs(LineSegmentOf(M, T))",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 40], [83, 86]], [[48, 64]], [[44, 47]], [[89, 93]], [[71, 74]], [[95, 98]], [[110, 113]], [], [[1, 40]], [[48, 64]], [[1, 47]], [[0, 67]], [[0, 74]], [[75, 93]], [[95, 109]]]",
    "query_spans": "[[[120, 135]]]",
    "process": ""
  },
  {
    "text": "Given that the focus of the parabola $x^{2}=4 y$ is $F$, if a point $P$ on this parabola has a distance of $2$ from the $y$-axis, then $|P F|$=?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;Expression(G) = (x^2 = 4*y);Focus(G) = F;PointOnCurve(P, G);Distance(P, yAxis) = 2",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 16], [26, 29]], [[32, 35]], [[20, 23]], [[2, 16]], [[2, 23]], [[26, 35]], [[32, 47]]]",
    "query_spans": "[[[49, 58]]]",
    "process": "F(0,1), let P(x_{0},y_{0}), then |x_{0}|=2, y_{0}=\\frac{x_{0}^{2}}{4}=1 \\therefore |PF|=y_{0}+1=1+1=2."
  },
  {
    "text": "Given that the moving circle $C$ passes through the point $A(-2,0)$ and is internally tangent to the circle $M$: $(x-2)^{2}+y^{2}=64$, find the equation of the locus of the center of the moving circle $C$.",
    "fact_expressions": "C: Circle;M: Circle;A: Point;Coordinate(A) = (-2, 0);PointOnCurve(A, C);Expression(M) = ((x - 2)^2 + y^2 = 64);IsInTangent(C, M)",
    "query_expressions": "LocusEquation(Center(C))",
    "answer_expressions": "x^2/16 + y^2/12 = 1",
    "fact_spans": "[[[2, 7], [51, 56]], [[21, 46]], [[8, 18]], [[8, 18]], [[2, 18]], [[21, 46]], [[2, 49]]]",
    "query_spans": "[[[51, 65]]]",
    "process": ""
  },
  {
    "text": "The minimum distance from a point on the parabola $y^{2}=4 x$ to the line $y=x+3$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Line;Expression(H) = (y = x + 3);A: Point;PointOnCurve(A, G)",
    "query_expressions": "Min(Distance(A, H))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[18, 27]], [[18, 27]], [[16, 17]], [[0, 17]]]",
    "query_spans": "[[[0, 35]]]",
    "process": "Let the parabola be $ y^{2} = 4x $, and a point on it be $ \\left( \\frac{y_{0}^{2}}{4}, y_{0} \\right) $. The distance from this point to the line $ y = x + 3 $ is $ \\underline{|} \\frac{+3}{1} = \\frac{ \\left| \\frac{y_{0}}{2} - 1^{2} + 2 \\right| }{ \\sqrt{2} } $. When $ \\frac{y_{0}}{2} = 1 $, i.e., $ y_{0} = 2 $, the minimum value $ \\sqrt{2} $ is attained."
  },
  {
    "text": "Given the hyperbola $M$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the asymptotes of which are tangent to the circle $x^{2}+(y-2 b)^{2}=a^{2}$, find the eccentricity of the hyperbola $M$.",
    "fact_expressions": "M: Hyperbola;Expression(M) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;G: Circle;Expression(G) = (x^2 + (-2*b + y)^2 = a^2);IsTangent(Asymptote(M), G)",
    "query_expressions": "Eccentricity(M)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 63], [98, 104]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 94]], [[68, 94]], [[2, 96]]]",
    "query_spans": "[[[98, 110]]]",
    "process": "According to the problem, the hyperbola $ M: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) has its foci on the x-axis; thus, its asymptotes are given by $ y = \\pm\\frac{b}{a}x $, or equivalently $ bx \\pm ay = 0 $. The circle $ x^{2} + (y - 2b)^{2} = a^{2} $ has center $ (0, 2b) $ and radius $ r = a $. Since the asymptotes of the hyperbola $ M $ are tangent to the circle $ x^{2} + (y - 2b)^{2} = a^{2} $, it follows that $ d = \\frac{|2ab|}{\\sqrt{a^{2} + b^{2}}} = a $. Rearranging yields $ a^{2} = 3b^{2} $, so $ a = \\sqrt{3}b $. Then $ c = \\sqrt{a^{2} + b^{2}} = 2b $, and the eccentricity of the hyperbola is $ e = \\frac{c}{a} = \\frac{2b}{\\sqrt{3}b} = \\frac{2\\sqrt{3}}{3} $."
  },
  {
    "text": "Given the ellipse $ C $: $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ $ (a > b > 0) $, the right focus is $ F $. A line passing through the origin $ O $ intersects the ellipse at points $ A $ and $ B $. Let $ M $ and $ N $ be the midpoints of segments $ AF $ and $ BF $, respectively. If there exists a circle with diameter $ MN $ that passes exactly through the origin $ O $, then the range of values for the eccentricity of the ellipse is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Circle;H: Line;A: Point;F: Point;B: Point;M: Point;N: Point;O: Origin;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;PointOnCurve(O,H);Intersection(H,C)={A,B};MidPoint(LineSegmentOf(A,F))=M;MidPoint(LineSegmentOf(B,F))=N;IsDiameter(LineSegmentOf(M,N),G);PointOnCurve(O,G)",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(\\sqrt{2}/2,1)",
    "fact_spans": "[[[2, 59], [81, 83], [149, 151]], [[9, 59]], [[9, 59]], [[136, 137]], [[78, 80]], [[84, 87]], [[64, 67]], [[88, 91]], [[94, 97]], [[98, 101]], [[140, 147], [70, 77]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 67]], [[68, 80]], [[78, 93]], [[94, 122]], [[94, 122]], [[126, 137]], [[136, 147]]]",
    "query_spans": "[[[149, 162]]]",
    "process": "Let the equation of AB be $ y = kx $. Solve the system of equations to find the coordinates of A and B, then obtain the coordinates of M and N. From $ OM \\perp ON $, set up an equation to get an equation in terms of $ k $. Let this equation have solutions to derive the relationship among $ a, b, c $, thereby obtaining the range of eccentricity. Let the equation of line AB be $ y = kx $. Given $ C(c,0) $, and M, N are the midpoints of AF and BF respectively,  \n$ \\therefore M\\left(\\frac{ab}{2\\sqrt{a^{2}k^{2}+b^{2}}}+\\frac{c}{2},\\frac{abk}{2\\sqrt{a^{2}k^{2}+b^{2}}}\\right),\\ N\\left(\\frac{-ab}{2\\sqrt{a^{2}k^{2}+b^{2}}}+\\frac{c}{2},\\frac{-abk}{2\\sqrt{a^{2}k^{2}+b^{2}}}\\right) $.  \nSince the circle with diameter MN passes exactly through the origin O,  \n$ \\therefore OM \\perp ON $,  \n$ \\left(\\frac{ab}{2\\sqrt{a^{2}k^{2}+b^{2}}}+\\frac{c}{2}\\right)\\left(\\frac{-ab}{2\\sqrt{a^{2}k^{2}+b^{2}}}+\\frac{c}{2}\\right)+\\frac{abk}{2\\sqrt{a^{2}k^{2}+b^{2}}}\\cdot\\frac{-abk}{2\\sqrt{a^{2}k^{2}+b^{2}}}=0 $,  \ni.e.,  \n$ \\frac{c^{2}}{4}-\\frac{a^{2}b^{2}}{4(a^{2}k^{2}+b^{2})}-\\frac{a^{2}b^{2}k^{2}}{4(a^{2}k^{2}+b^{2})}=0 $,  \n$ \\therefore c^{2}(a^{2}k^{2}+b^{2})-a^{2}b^{2}-a^{2}b^{2}k^{2}=0 $,  \n$ \\therefore (a^{2}c^{2}-a^{2}b^{2})k^{2}=a^{2}b^{2}-b^{2}c^{2}=b^{4} $, i.e., $ a^{2}(c^{2}-b^{2})k^{2}=b^{4} $.  \nSince there exists a line AB satisfying the conditions such that $ OM \\perp ON $,  \n$ \\therefore $ the equation in $ k $, $ a^{2}(c^{2}-b^{2})k^{2}=b^{4} $, has solutions,  \n$ \\therefore c^{2}>b^{2} $, i.e., $ c^{2}>a^{2}-c^{2} $, $ \\therefore 2c^{2}>a^{2} $,  \n$ \\therefore \\frac{c^{2}}{a^{2}}>\\frac{1}{2} $, $ \\therefore e=\\frac{c}{a}>\\frac{\\sqrt{2}}{2} $. Also $ e<1 $,  \n$ \\therefore \\frac{\\sqrt{2}}{2}<e<1 $."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{|m|-1}-\\frac{y^{2}}{m-2}=1$ represents a hyperbola, then what is the range of real values for $m$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(Abs(m) - 1) - y^2/(m - 2) = 1);m: Real",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-1, 1)+(2, +oo)",
    "fact_spans": "[[[47, 50]], [[2, 50]], [[53, 58]]]",
    "query_spans": "[[[53, 65]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola $y^{2}=8 x$ is $F$. Let $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$ be two moving points on the parabola. If $x_{1}+x_{2}+4=\\frac{2 \\sqrt{3}}{3}|A B|$, then the maximum value of $\\angle A F B$ is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 8*x);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Focus(G) = F;x1:Number;y1:Number;x2:Number;y2:Number;PointOnCurve(A,G);PointOnCurve(B, G);x1 + x2 + 4 = (2*sqrt(3)/3)*Abs(LineSegmentOf(A, B))",
    "query_expressions": "Max(AngleOf(A, F, B))",
    "answer_expressions": "2*pi/3",
    "fact_spans": "[[[0, 14], [61, 64]], [[23, 40]], [[43, 60]], [[18, 21]], [[0, 14]], [[23, 40]], [[43, 60]], [[0, 21]], [[23, 40]], [[23, 41]], [[43, 60]], [[43, 60]], [[23, 70]], [[23, 70]], [[72, 114]]]",
    "query_spans": "[[[116, 136]]]",
    "process": "From the parabola focal radius formula, we have $ AF = x_{1} + 2 $, $ BF = x_{2} + 2 $, and obtain $ AF + BF = \\frac{2\\sqrt{3}}{3}AB $. Combining the law of cosines and the AM-GM inequality, the solution can be found. [Detailed Solution] From the parabola focal radius formula, we have $ AF = x_{1} + 2 $, $ BF = x_{2} + 2 $, so from $ x_{1} + x_{2} + 4 = \\frac{2\\sqrt{3}}{3}|AB| $, we get $ AF + BF = \\frac{2\\sqrt{3}}{3}AB $. Given $ \\frac{1}{2}x_{2} - |AF|\\cdot|BF| - \\frac{|FF^{2} + |BF|^{2}||AB|^{2}}{2|AF|} = \\frac{1}{4}\\frac{4}{|F|} = -\\frac{1}{2}|\\frac{2}{2|AF|} $, so the maximum value of $ \\angle AFB $ is $ \\frac{2}{3}\\pi $. Therefore, fill in $ \\frac{2}{3}\\pi $. [Note] This problem examines the parabola's focal radius formula combined with the law of cosines and the AM-GM inequality, aiming to test students' analytical reasoning ability, computational and simplification skills, and is a medium-difficulty problem."
  },
  {
    "text": "The focal length of the ellipse $C$: $2 x^{2}+y^{2}=2$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (2*x^2 + y^2 = 2)",
    "query_expressions": "FocalLength(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 24]], [[0, 24]]]",
    "query_spans": "[[[0, 29]]]",
    "process": "The standard equation of the ellipse is $x^{2}+\\frac{y^{2}}{2}=1$, then $a=\\sqrt{2}$, $b=1$, so $c=\\sqrt{2-1}=1$, $2c=2$, the focal distance is $2$."
  },
  {
    "text": "The eccentricity of the hyperbola $x^{2}-3 y^{2}=3$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - 3*y^2 = 3)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[0, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 26]]]",
    "process": "The hyperbola equation $x^{2}-3y^{2}=3$ is rewritten as $\\frac{x^{2}}{3}-y^{2}=1$, yielding $a=\\sqrt{3}$, $b=1 \\Rightarrow c=2$, so its eccentricity is $e=\\frac{c}{a}=\\frac{2\\sqrt{3}}{3}$."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. If there exists a point $P$ on the line $x=\\frac{a^{2}}{c}$ such that the perpendicular bisector of the segment $PF_{1}$ passes through the point $F_{2}$, then the range of the eccentricity of the ellipse is?",
    "fact_expressions": "F1: Point;F2: Point;G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;LeftFocus(G) = F1;RightFocus(G) = F2;H: Line;Expression(H) = (x = a^2/c);c: Number;P: Point;PointOnCurve(P, H);PointOnCurve(F2, PerpendicularBisector(LineSegmentOf(P, F1)))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 8]], [[9, 16], [125, 133]], [[19, 71], [135, 137]], [[19, 71]], [[21, 71]], [[21, 71]], [[21, 71]], [[21, 71]], [[1, 77]], [[1, 77]], [[80, 101]], [[80, 101]], [[82, 101]], [[104, 108]], [[79, 108]], [[110, 133]]]",
    "query_spans": "[[[135, 148]]]",
    "process": ""
  },
  {
    "text": "A point $P(2, m)$ on the parabola $y^{2}=2 p x(p>0)$ has a distance of $4$ to its focus $F$. Then $p=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;P: Point;PointOnCurve(P, G);Coordinate(P) = (2, m);m: Number;F: Point;Focus(G) = F;Distance(P, F) = 4",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 21], [34, 35]], [[0, 21]], [[49, 52]], [[3, 21]], [[24, 33]], [[0, 33]], [[24, 33]], [[24, 33]], [[37, 40]], [[34, 40]], [[24, 47]]]",
    "query_spans": "[[[49, 54]]]",
    "process": "Since the distance from a point P(2,m) on the parabola $ y^{2}=2px $ ($ p>0 $) to its focus F is 4, we have $ 2+\\frac{p}{2}=4 $. Solving this gives $ p=4 $."
  },
  {
    "text": "Draw a line $l$ through point $M(1,1)$ intersecting the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$ at points $A$ and $B$, such that $M$ is the midpoint of $AB$. Then the equation of line $l$ is?",
    "fact_expressions": "l: Line;G: Ellipse;A: Point;B: Point;M: Point;Expression(G) = (x^2/4 + y^2/9 = 1);Coordinate(M) = (1, 1);PointOnCurve(M, l);Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Expression(l)",
    "answer_expressions": "9*x + 4*y - 13 = 0",
    "fact_spans": "[[[12, 17], [80, 85]], [[18, 55]], [[56, 59]], [[60, 63]], [[2, 11], [67, 70]], [[18, 55]], [[2, 11]], [[0, 17]], [[12, 65]], [[67, 78]]]",
    "query_spans": "[[[80, 90]]]",
    "process": "By the point-difference method, $\\frac{x_{M}}{4}+\\frac{y_{M}}{9}\\cdot k_{AB}=0 \\therefore \\frac{1}{4}+\\frac{1}{9}\\cdot k_{AB}=0, k_{AB}=-\\frac{9}{4} \\therefore l: y-1=-\\frac{9}{4}(x-1), 9x+4y-13=0.$"
  },
  {
    "text": "Point $A$ is the intersection point (other than the origin) of the parabola $C_{1}$: $y^{2}=2 p x(p>0)$ and an asymptote of the hyperbola $C_{2}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$. If the distance from point $A$ to the directrix of the parabola $C_{1}$ is $\\frac{3}{2} p$, then the eccentricity of the hyperbola $C_{2}$ is?",
    "fact_expressions": "C1:Parabola;C2:Hyperbola;a:Number;b:Number;p:Number;a>0;b>0;p>0;O:Origin;A:Point;Expression(C1)=(y^2=2*p*x);Expression(C2)=(x^2/a^2-y^2/b^2=1);Intersection(C1,OneOf(Asymptote(C2)))=A;Negation(A=O);Distance(A,Directrix(C1))=(3/2)*p",
    "query_expressions": "Eccentricity(C2)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[5, 35], [127, 137]], [[36, 104], [161, 171]], [[48, 104]], [[48, 104]], [[17, 35]], [[48, 104]], [[48, 104]], [[17, 35]], [[117, 119]], [[0, 4], [122, 126]], [[5, 35]], [[36, 104]], [[0, 113]], [[0, 120]], [[122, 159]]]",
    "query_spans": "[[[161, 178]]]",
    "process": "Take one asymptote of the hyperbola: $ y = \\frac{b}{a}x $, solve the system $ \\begin{cases} y^{2} = 2px \\\\ y = \\frac{b}{a}x \\end{cases} $, we get $ \\begin{cases} x = \\frac{2pa}{b^{2}} \\\\ y = \\frac{2pa}{b} \\end{cases} $. The distance from point $ A $ to the directrix of the parabola is $ \\frac{3}{3}p y = \\frac{1}{2} \\therefore \\frac{p}{2} + \\frac{2pa^{2}}{b^{2}} = \\frac{3}{2}p $, simplifying to $ \\frac{a^{2}}{b^{2}} = \\frac{1}{2} \\therefore $ the eccentricity of hyperbola $ C_{2} $ is $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{3} $"
  },
  {
    "text": "If the hyperbola equation $x^{2}+m y^{2}=1$ has foci on the $x$-axis and focal distance $2 \\sqrt{5}$, then what is the value of $m$?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (m*y^2 + x^2 = 1);PointOnCurve(Focus(G), xAxis);FocalLength(G) = 2*sqrt(5)",
    "query_expressions": "m",
    "answer_expressions": "-1/4",
    "fact_spans": "[[[2, 5]], [[51, 54]], [[2, 24]], [[2, 32]], [[2, 49]]]",
    "query_spans": "[[[51, 58]]]",
    "process": "Knowing the foci lie on the x-axis implies m<0; converting to standard form gives x^{2}-\\frac{y^{2}}{-\\frac{1}{m}}=1. Given the focal distance is 2\\sqrt{5}, we obtain 1+(-\\frac{1}{m})=(\\frac{2\\sqrt{5}}{2})^{2}, solving yields m=-\\frac{1}{4}."
  },
  {
    "text": "If a point $P$ on the parabola $y^{2}=4 x$ is at a distance of $3$ from its directrix, then what is the distance from point $P$ to the $x$-axis?",
    "fact_expressions": "G: Parabola;P: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(P, G);Distance(P,Directrix(G)) = 3",
    "query_expressions": "Distance(P, xAxis)",
    "answer_expressions": "sqrt(2)*2",
    "fact_spans": "[[[1, 15], [22, 23]], [[18, 21], [34, 38]], [[1, 15]], [[1, 21]], [[18, 32]]]",
    "query_spans": "[[[34, 48]]]",
    "process": "According to the definition of a parabola, solve for the horizontal coordinate of point P and substitute into the parabolic equation. From the given conditions, the directrix of the parabola $ y^{2} = 4x $ is $ x = -1 $. Since the distance from P to its directrix is 3, we have $ x_{P} = 2 $. Substituting into the parabolic equation, we obtain $ y_{P} = \\pm 2\\sqrt{2} $. Therefore, the distance from point P to the x-axis is $ 2\\sqrt{2} $."
  },
  {
    "text": "If points $A(0,-4)$, $B(3, 2)$, then the shortest distance from a point on the parabola $x^{2}=y$ to the line $AB$ is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;P0: Point;Expression(G) = (x^2 = y);Coordinate(A) = (0, -4);Coordinate(B)=(3, 2);PointOnCurve(P0, G)",
    "query_expressions": "Min(Distance(P0, LineOf(A,B)))",
    "answer_expressions": "3*sqrt(5)/5",
    "fact_spans": "[[[24, 36]], [[1, 11]], [[12, 22]], [[38, 39]], [[24, 36]], [[1, 11]], [[12, 22]], [[24, 39]]]",
    "query_spans": "[[[38, 54]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a moving point on the parabola $y=\\frac{1}{12} x^{2}$, $M(0,3)$, $N(4,3)$, then the minimum value of $|P M|+|P N|$ is?",
    "fact_expressions": "G: Parabola;M: Point;N: Point;P: Point;Expression(G) = (y = x^2/12);Coordinate(M) = (0, 3);Coordinate(N) = (4, 3);PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, M)) + Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "6",
    "fact_spans": "[[[6, 31]], [[36, 44]], [[45, 53]], [[2, 5]], [[6, 31]], [[36, 44]], [[45, 53]], [[2, 35]]]",
    "query_spans": "[[[55, 74]]]",
    "process": "It is easy to see that M is the focus of the parabola $ y = \\frac{1}{12}x^2 $. Let the distance from P to the directrix $ l: y = -3 $ be $ d $, then $ |PM| + |PN| = d + |PN| $. The minimum value of $ d + |PN| $ is the distance from N to the directrix, so the minimum value of $ |PM| + |PN| $ is $ 3 + 3 = 6 $. Hence, the answer is: $ 6 $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively, a line passing through $F_{1}$ with slope $\\frac{1}{3}$ intersects the two asymptotes of the hyperbola at points $A$ and $B$, respectively. If $|F_{2} A|=|F_{2} B|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;F2: Point;A: Point;B: Point;F1: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, H);Slope(H) = 1/3;Intersection(H, l1)=A;Abs(LineSegmentOf(F2, A)) = Abs(LineSegmentOf(F2, B));l1:Line;l2:Line;Intersection(H, l2)=B;Asymptote(G)={l1,l2}",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[20, 77], [161, 164], [114, 117]], [[23, 77]], [[23, 77]], [[111, 113]], [[10, 17]], [[127, 130]], [[131, 134]], [[2, 9], [86, 93]], [[23, 77]], [[23, 77]], [[20, 77]], [[2, 84]], [[2, 84]], [[85, 113]], [[94, 113]], [[111, 136]], [[138, 159]], [], [], [[111, 136]], [[114, 123]]]",
    "query_spans": "[[[161, 170]]]",
    "process": "Let $ F_{1}(-c,0) $, find the equation of the line passing through $ F_{1} $ with slope $ \\frac{1}{3} $. Solve the system of this line and the asymptotes of the hyperbola to find the coordinates of points $ A $ and $ B $. Then, using $ |F_{2}A| = |F_{2}B| $, obtain the relationship between $ a $ and $ b $. Finally, use $ e = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} $ to find the eccentricity of the hyperbola.  \n**Solution**: Let $ F_{1}(-c,0) $, then the equation of the line passing through $ F_{1} $ with slope $ \\frac{1}{3} $ is $ y = \\frac{1}{3}(x + c) $. From  \n$$\n\\begin{cases}\ny = \\frac{1}{3}(x + c) \\\\\ny = \\frac{b}{a}x\n\\end{cases}\n$$\nwe get  \n$$\nx = \\frac{ac}{3b - a},\\quad y = \\frac{bc}{3b - a}.\n$$\nFrom  \n$$\n\\begin{cases}\ny = \\frac{1}{3}(x + c) \\\\\ny = -\\frac{b}{a}x\n\\end{cases}\n$$\nwe get  \n$$\nx = -\\frac{ac}{3b + a},\\quad y = \\frac{bc}{3b + a}.\n$$\nWithout loss of generality, assume $ A\\left( \\frac{ac}{3b - a}, \\frac{bc}{3b - a} \\right) $, $ B\\left( -\\frac{ac}{3b + a}, \\frac{bc}{3b + a} \\right) $. Since $ |F_{2}A| = |F_{2}B| $ and $ F_{2}(c, 0) $, we have  \n$$\n\\left( \\frac{ac}{3b - a} - c \\right)^2 + \\left( \\frac{bc}{3b - a} \\right)^2 = \\left( -\\frac{ac}{3b + a} - c \\right)^2 + \\left( \\frac{bc}{3b + a} \\right)^2.\n$$\nSimplifying yields $ a^{2} = 4b^{2} $, i.e., $ a = 2b $. Hence, the eccentricity of the hyperbola is  \n$$\ne = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{1 + \\frac{b^{2}}{4b^{2}}} = \\sqrt{\\frac{5}{4}} = \\frac{\\sqrt{5}}{2}.\n$$"
  },
  {
    "text": "The coordinates of the focus of the parabola $y^{2}=2 x$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(1/2,0)",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": "The focus lies on the positive x-axis, and p=1. Using the focus formula (\\frac{p}{2},0), write the coordinates of the focus. For the parabola y^{2}=2x, the focus lies on the positive x-axis, and p=1, \\therefore\\frac{p}{2}=\\frac{1}{2}, hence the focus coordinates are (\\frac{1}{2},0)"
  },
  {
    "text": "Let the right focus of the ellipse $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{n^{2}}=1  (m>0 , n>0)$ coincide with the focus of the parabola $y^{2}=8 x$, and the eccentricity is $\\frac{1}{2}$. Then the standard equation of this ellipse is?",
    "fact_expressions": "G: Parabola;H: Ellipse;m: Number;n: Number;Expression(G) = (y^2 = 8*x);m>0;n>0;Expression(H) = (y^2/n^2 + x^2/m^2 = 1);RightFocus(H) = Focus(G);Eccentricity(H)=(1/2)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2/16+y^2/12=1",
    "fact_spans": "[[[64, 78]], [[1, 59], [104, 106]], [[3, 59]], [[3, 59]], [[64, 78]], [[3, 59]], [[3, 59]], [[1, 59]], [[1, 83]], [[1, 101]]]",
    "query_spans": "[[[104, 113]]]",
    "process": ""
  },
  {
    "text": "The focal distance of the ellipse $\\frac{x^{2}}{a^{2}}+y^{2}=1$ is $2 \\sqrt{3}$, then the eccentricity of this ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2 + x^2/a^2 = 1);a: Number;FocalLength(G) = 2*sqrt(3)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[0, 31], [50, 52]], [[0, 31]], [[2, 31]], [[0, 47]]]",
    "query_spans": "[[[50, 58]]]",
    "process": "Since the ellipse's focal distance is $2c=2\\sqrt{3}$, $c=\\sqrt{3}$. The ellipse's foci lie on the x-axis, thus $a^{2}=1^{2}+\\sqrt{3}^{2}=4$, $a=2$, so the ellipse's eccentricity is $\\frac{c}{a}=\\frac{\\sqrt{3}}{2}$."
  },
  {
    "text": "Through a point $P$ on the circle $x^{2}+y^{2}=8$, draw a perpendicular to the $x$-axis, with foot $H$. Then the equation of the locus of the midpoint $M$ of the segment $PH$ is?",
    "fact_expressions": "G: Circle;P: Point;H: Point;L: Line;M: Point;Expression(G) = (x^2 + y^2 = 8);PointOnCurve(P, G);FootPoint(L, xAxis) = H;IsPerpendicular(L, xAxis);PointOnCurve(P, L);MidPoint(LineSegmentOf(P, H)) = M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2/8 + y^2/2 = 1",
    "fact_spans": "[[[1, 17]], [[20, 23]], [[35, 38]], [], [[50, 53]], [[1, 17]], [[1, 23]], [[0, 38]], [[0, 31]], [[0, 31]], [[40, 53]]]",
    "query_spans": "[[[50, 60]]]",
    "process": "Let M(x,y), H(x,0), then P(x,2y). Since P lies on the circle x^{2}+y^{2}=8, \\therefore x^{2+4y^{2}}=8, simplifying gives \\frac{x^{2}}{8}+\\frac{y^{2}}{2}=1 \\therefore"
  },
  {
    "text": "Given the hyperbola $\\Gamma$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the right vertex of $\\Gamma$ is $A$. A line parallel to the $x$-axis intersects $\\Gamma$ at points $B$ and $C$. Denote $\\overrightarrow{A B} \\cdot \\overrightarrow{A C}=m$. If the eccentricity of $\\Gamma$ is $\\sqrt{2}$, then the set of values of $m$ is?",
    "fact_expressions": "Gamma: Hyperbola;H: Line;A: Point;B: Point;C: Point;m: Number;a: Number;b: Number;a > 0;b > 0;Expression(Gamma) = (-y^2/b^2 + x^2/a^2 = 1);RightVertex(Gamma) = A;IsParallel(H, xAxis);Intersection(H, Gamma) = {B, C};DotProduct(VectorOf(A, B), VectorOf(A, C)) = m;Eccentricity(Gamma) = sqrt(2)",
    "query_expressions": "Range(m)",
    "answer_expressions": "{0}",
    "fact_spans": "[[[2, 68], [88, 96], [161, 169]], [[85, 87]], [[73, 76]], [[97, 100]], [[101, 104]], [[186, 189]], [[15, 68]], [[15, 68]], [[15, 68]], [[15, 68]], [[2, 68]], [[2, 76]], [[77, 87]], [[85, 106]], [[108, 159]], [[161, 184]]]",
    "query_spans": "[[[186, 197]]]",
    "process": "Since the eccentricity of r is \\sqrt{2}, we have a = b, so the hyperbola r: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 becomes x^{2} - y^{2} = a^{2}. Combining with the dot product formula of vectors, the conclusion can be obtained. Solution: \\because the eccentricity of r is \\sqrt{2}, \\therefore a = b, \\therefore the hyperbola r: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 becomes x^{2} - y^{2} = a^{2}. Let B(-x, y), C(x, y), A(a, 0). \\overrightarrow{AB} \\cdot \\overrightarrow{AC} = m = (-x - a, y) \\cdot (x - a, y) = a^{2} - x^{2} + y^{2} = 0 \\therefore m = 0"
  },
  {
    "text": "Let points $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, respectively. Let point $P$ be a point on the ellipse, and let point $M$ be the midpoint of $F_{1}P$. If $|OM|=3$, then what is the distance from point $P$ to the left focus of the ellipse?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;O: Origin;M: Point;F2: Point;Expression(G) = (x^2/25 + y^2/16 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);MidPoint(LineSegmentOf(F1, P)) = M;Abs(LineSegmentOf(O, M)) = 3",
    "query_expressions": "Distance(P, LeftFocus(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[20, 59], [71, 73], [111, 113]], [[1, 9]], [[66, 70], [106, 110]], [[95, 104]], [[77, 81]], [[10, 17]], [[20, 59]], [[1, 65]], [[1, 65]], [[66, 76]], [[77, 94]], [[95, 104]]]",
    "query_spans": "[[[106, 121]]]",
    "process": "From the given condition: OM is the midline of triangle $ PF_{1}F_{2} $, $ \\therefore |OM| = 3|PF_{2}| = 6 $. Also, $ \\because |PF_{1}| + |PF_{2}| = 2a = 10 $, $ \\therefore |PF|_{1} = 4 $."
  },
  {
    "text": "The line $y=k(x-1)$ intersects the parabola $y^{2}=4x$ at two points $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$. If $x_{1}+x_{2}=6$, then $|AB|$=?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*(x - 1));G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;B: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Intersection(H, G) = {A, B};x1 + x2 = 6;k: Number",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[0, 12]], [[0, 12]], [[13, 27]], [[13, 27]], [[29, 46]], [[48, 65]], [[29, 46]], [[48, 65]], [[29, 46]], [[48, 65]], [[29, 46]], [[48, 65]], [[0, 67]], [[69, 84]], [[0, 12]]]",
    "query_spans": "[[[86, 95]]]",
    "process": "Substituting $ y = k(x - 1) $ into $ y^2 = 4x $ yields $ k^2x^2 - (2k^2 + 4)x + k^2 = 0 $. According to the problem and by Vieta's formulas, we have: $ x_1 + x_2 = \\frac{2k^2 + 4}{k^2} = 6 $, $ x_1x_2 = 1 $. Solving gives $ k^2 = 1 $. Therefore, $ |AB| = \\sqrt{1 + k^2} \\cdot \\sqrt{(x_1 + x_2)^2 - 4x_1x_2} = \\sqrt{2} \\cdot \\sqrt{36 - 4} = \\sqrt{64} = 8 $."
  },
  {
    "text": "Let the moving circle $M$ be tangent to the $y$-axis and externally tangent to the circle $C$: $x^{2}+y^{2}-2 x=0$. Then, what is the trajectory equation of the center $M$ of the moving circle?",
    "fact_expressions": "M: Circle;IsTangent(M,yAxis);IsOutTangent(M,C);C:Circle;Expression(C)=(x^2+y^2-2*x=0);M1:Point;Center(M)=M1",
    "query_expressions": "LocusEquation(M1)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[3, 6], [46, 48]], [[3, 13]], [[3, 43]], [[15, 40]], [[15, 40]], [[50, 53]], [[46, 53]]]",
    "query_spans": "[[[50, 60]]]",
    "process": "Let M(x,y). \\sqrt{(x-1)^{2}+y^{2}}=1+|x| \\therefore x\\geqslant0, y^{2}=4x, x<0, y=0, i.e., the trajectory equation is y^{2}=4x or y=0 (x<0)"
  },
  {
    "text": "Given that hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ and hyperbola $C_{2}$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{16}=1$ have the same asymptotes, and the right focus of $C_{1}$ is $F(\\sqrt{5}, 0)$, then $a=?$ $b=?$",
    "fact_expressions": "C1: Hyperbola;Expression(C1) = (x^2/a^2 - y^2/b^2 = 1);b: Number;a: Number;C2: Hyperbola;Expression(C2) = (x^2/4 - y^2/16 = 1);Asymptote(C1) = Asymptote(C2);F: Point;Coordinate(F) = (sqrt(5), 0);RightFocus(C1) = F",
    "query_expressions": "a;b",
    "answer_expressions": "1\n2",
    "fact_spans": "[[[2, 57], [115, 122]], [[2, 57]], [[149, 152]], [[144, 147]], [[58, 106]], [[58, 106]], [[2, 113]], [[127, 143]], [[127, 143]], [[115, 143]]]",
    "query_spans": "[[[144, 149]], [[149, 154]]]",
    "process": "Hyperbola $ C_{1}: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ and $ C_{2}: \\frac{x^{2}}{4} - \\frac{y^{2}}{16} = 1 $ have the same asymptotes, then $ b = 2a $. The right focus of $ C_{1} $ is $ F(\\sqrt{5}, 0) $, so $ c = \\sqrt{5} $. For hyperbolas, $ 5 = a^{2} + b^{2} = 5a^{2} $, hence $ a = 1 $, $ b = 2 $. 【Exam Point Location】 This question tests the definition and simple geometric properties of hyperbolas, as well as students' mastery of basic textbook knowledge; it is an easy question."
  },
  {
    "text": "The chord $AB$ passing through the focus of the parabola $y^{2}=8x$ has a midpoint with horizontal coordinate $3$. Then $|AB|=$?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;Expression(G) = (y^2 = 8*x);IsChordOf(LineSegmentOf(A, B), G);PointOnCurve(Focus(G), LineSegmentOf(A, B));XCoordinate(MidPoint(LineSegmentOf(A, B))) = 3",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "10",
    "fact_spans": "[[[1, 15]], [[20, 24]], [[20, 24]], [[1, 15]], [[1, 24]], [[0, 24]], [[20, 34]]]",
    "query_spans": "[[[36, 45]]]",
    "process": "The solution process is omitted"
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, and $F_{1}$, $F_{2}$ are the two foci of the ellipse, with $\\angle F_{1} P F_{2}=60^{\\circ}$, then $|P F_{1}||P F_{2}|$=?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/9 + y^2/5 = 1);PointOnCurve(P, G);Focus(G)={F1,F2};AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "20/3",
    "fact_spans": "[[[6, 43], [64, 66]], [[47, 54]], [[2, 5]], [[55, 62]], [[6, 43]], [[2, 46]], [[47, 71]], [[73, 106]]]",
    "query_spans": "[[[108, 130]]]",
    "process": "From the given conditions, |PF_{1}| + |PF_{2}| = 2a = 6. By the law of cosines, |F_{1}F_{2}|^{2} = |PF_{1}|^{2} + |PF_{2}|^{2} - 2|PF_{1}||PF_{2}|\\cos60^{\\circ}. From the ellipse \\frac{x^{2}}{9} + \\frac{y^{2}}{5} = 1, we obtain F_{1}(-2,0) and F_{2}(2,0). From the given conditions, |PF_{1}| + |PF_{2}| = 2a = 6. By the law of cosines, |F_{1}F_{2}|^{2} = |PF_{1}|^{2} + |PF_{2}|^{2} - 2|PF_{1}||PF_{2}|\\cos60^{\\circ}. Thus, 16 = (|PF_{1}| + |PF_{2}|)^{2} - 3|PF_{1}||PF_{2}|, that is, 16 = 36 - 3|PF_{1}||PF_{2}|. Therefore, |PF_{1}||PF_{2}| = \\frac{20}{3}."
  },
  {
    "text": "A moving point $P$ has the property that its distance to the point $A(6,0)$ is twice its distance to the line $x=\\frac{3}{2}$. What is the trajectory equation of the moving point $P$?",
    "fact_expressions": "G: Line;A: Point;P: Point;Expression(G) = (x = 3/2);Coordinate(A) = (6, 0);Distance(P, A) = 2*Distance(P, G)",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2/9 - y^2/27 = 1",
    "fact_spans": "[[[22, 39]], [[7, 16]], [[3, 6], [50, 53], [20, 21]], [[22, 39]], [[7, 16]], [[3, 47]]]",
    "query_spans": "[[[50, 60]]]",
    "process": "Let P(x, y). According to the problem, \\sqrt{(x-6)^{2}+y^{2}}=2|x-\\frac{3}{2}|. Simplifying yields: \\frac{x^{2}}{9}-\\frac{y^{2}}{27}=1"
  },
  {
    "text": "Draw a line through the right focus $F$ of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ perpendicular to the $x$-axis, intersecting the ellipse at points $A$ and $B$. Then the area of the circle with diameter $AB$ is?",
    "fact_expressions": "G: Ellipse;H: Circle;L: Line;A: Point;B: Point;F: Point;Expression(G) = (x^2/16 + y^2/9 = 1);RightFocus(G) = F;PointOnCurve(F, L);IsPerpendicular(L,xAxis);Intersection(L, G) = {A, B};IsDiameter(LineSegmentOf(A,B),H)",
    "query_expressions": "Area(H)",
    "answer_expressions": "81*pi/16",
    "fact_spans": "[[[1, 39], [58, 60]], [[83, 84]], [[55, 57]], [[62, 65]], [[66, 69]], [[43, 46]], [[1, 39]], [[1, 46]], [[0, 57]], [[47, 57]], [[55, 71]], [[73, 84]]]",
    "query_spans": "[[[83, 89]]]",
    "process": "From the given condition, in $\\frac{x^2}{16} + \\frac{y^2}{9} = 1$, $c = \\sqrt{16 - 9} = \\sqrt{7}$, so $F(\\sqrt{7}, 0)$. When $x = \\sqrt{7}$, $y = \\pm 3\\sqrt{1 - \\frac{7}{16}} = \\pm \\frac{9}{4}$, so $|AB| = \\frac{9}{2}$. Therefore, the area of the circle with $AB$ as diameter is $\\pi \\times \\left(\\frac{9}{4}\\right)^2 = \\frac{81\\pi}{16}$."
  },
  {
    "text": "Through a focus $F$ of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw two tangents to the circle $x^{2}+y^{2}=a^{2}$, with points of tangency $A$ and $B$, respectively. If $\\angle AFB = 120^{\\circ}$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;A: Point;F: Point;B: Point;l1: Line;l2: Line;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = a^2);OneOf(Focus(C)) = F;TangentOfPoint(F, G) = {l1, l2};TangentPoint(l1,G)=A;TangentPoint(l2,G)=B;AngleOf(A, F, B) = ApplyUnit(120, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[1, 62], [141, 147]], [[9, 62]], [[9, 62]], [[71, 91]], [[103, 106]], [[67, 70]], [[107, 110]], [], [], [[9, 62]], [[9, 62]], [[1, 62]], [[71, 91]], [[1, 70]], [[0, 96]], [[0, 110]], [[0, 110]], [[112, 139]]]",
    "query_spans": "[[[141, 153]]]",
    "process": ""
  },
  {
    "text": "In the rectangular coordinate system $x O y$, the focus of the parabola $y^{2}=2 p x$ ($p>0$) is $F$. Let $M$ be a moving point on the parabola. Then the maximum value of $\\frac{|M O|}{|M F|}$ is?",
    "fact_expressions": "E: Parabola;p: Number;p>0;Expression(E) = (y^2 = 2*(p*x));F: Point;Focus(E) = F;M: Point;PointOnCurve(M, E);O: Origin",
    "query_expressions": "Max(Abs(LineSegmentOf(M, O))/Abs(LineSegmentOf(M, F)))",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[17, 38], [52, 55]], [[20, 38]], [[20, 38]], [[17, 38]], [[42, 45]], [[17, 45]], [[48, 51]], [[48, 59]], [[61, 82]]]",
    "query_spans": "[[[61, 88]]]",
    "process": ""
  },
  {
    "text": "The center of the circle $(x+1)^{2}+y^{2}=1$ is the focus of the parabola $y^{2}=p x$ ($p<0$), then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;H: Circle;p<0;Expression(G) = (y^2 = p*x);Expression(H) = (y^2 + (x + 1)^2 = 1);Center(H)=Focus(G)",
    "query_expressions": "p",
    "answer_expressions": "-4",
    "fact_spans": "[[[24, 43]], [[48, 51]], [[0, 20]], [[27, 43]], [[24, 43]], [[0, 20]], [[0, 46]]]",
    "query_spans": "[[[48, 53]]]",
    "process": "\\because the center of the circle is (-1,0), \\therefore \\frac{p}{4} = -1, \\therefore p = -4."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ with focus $F$, the intersection point of the directrix and the $x$-axis being $M$, and $N$ a point on the parabola such that $|N F|=\\frac{\\sqrt{3}}{2}|M N|$, then $\\angle N M F$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;M: Point;Intersection(Directrix(G), xAxis) = M;N: Point;PointOnCurve(N, G);Abs(LineSegmentOf(N, F)) = (sqrt(3)/2)*Abs(LineSegmentOf(M, N))",
    "query_expressions": "AngleOf(N, M, F)",
    "answer_expressions": "pi/6",
    "fact_spans": "[[[2, 23], [52, 55]], [[2, 23]], [[5, 23]], [[5, 23]], [[27, 30]], [[2, 30]], [[43, 46]], [[2, 46]], [[48, 51]], [[48, 59]], [[64, 95]]]",
    "query_spans": "[[[98, 114]]]",
    "process": "Analysis: Using the properties of a parabola, draw a perpendicular line from N to the directrix, intersecting the directrix at $N_{1}$, then $NN_{1}=NF$, so $\\cos\\angle NMF = \\cos\\angle N_{1}NM$. In the right triangle $\\triangle N_{1}NM$, $\\cos\\angle N_{1}NM$ can be expressed. Calculation yields the answer. Draw a perpendicular from N to the directrix, intersecting at $N_{1}$, then $\\cos\\angle NMF = \\cos\\angle N_{1}NM = \\frac{|NN_{1}|}{|MN|} = \\frac{|NF|}{|MN|} = \\frac{\\sqrt{3}}{2}$. Hence, $\\angle NMF = \\frac{\\pi}{6}$."
  },
  {
    "text": "Given that the hyperbola $C$ is centered at the origin, the point $F(\\sqrt{2} , 0)$ is a focus of the hyperbola $C$, and the line $l$ perpendicular to an asymptote is drawn through point $F$, with foot of perpendicular at $M$. The line $l$ intersects the $y$-axis at point $E$. If $FM=ME$, then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;O: Origin;Center(C) = O;F: Point;OneOf(Focus(C)) = F;Coordinate(F) = (sqrt(2), 0);l: Line;PointOnCurve(F,l);IsPerpendicular(Asymptote(C),l);M: Point;FootPoint(Asymptote(C),l) = M;E: Point;Intersection(l, yAxis) = E;LineSegmentOf(F, M) = LineSegmentOf(M, E)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2 = 1",
    "fact_spans": "[[[2, 8], [34, 40], [96, 99]], [[12, 14]], [[2, 14]], [[15, 33], [47, 51]], [[15, 45]], [[15, 33]], [[58, 61], [70, 75]], [[46, 61]], [[34, 61]], [[66, 69]], [[34, 69]], [[81, 85]], [[70, 85]], [[87, 94]]]",
    "query_spans": "[[[96, 104]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$ and the point $P(4,2)$, a line $l$ passes through point $P$ and intersects the ellipse at points $A$ and $B$. When point $P$ is exactly the midpoint of segment $AB$, what is the equation of line $l$?",
    "fact_expressions": "l: Line;G: Ellipse;A: Point;B: Point;P: Point;Expression(G) = (x^2/36 + y^2/9 = 1);Coordinate(P) = (4, 2);PointOnCurve(P, l);Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x + 2*y - 8 = 0",
    "fact_spans": "[[[51, 56], [99, 104]], [[2, 40], [64, 66]], [[68, 71]], [[72, 75]], [[41, 50], [58, 62], [80, 84]], [[2, 40]], [[41, 50]], [[51, 62]], [[51, 77]], [[80, 97]]]",
    "query_spans": "[[[99, 109]]]",
    "process": "From the given condition, $\\frac{16}{36}+\\frac{4}{9}<1$, we know that point $P(4,2)$ lies inside the ellipse. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $\\textcircled{1}\\textcircled{2}$. Since $P(4,2)$ is exactly the midpoint of segment $AB$, we have $x_{1}+x_{2}=8$, $y_{1}+y_{2}=4$. Subtracting $\\textcircled{1}$ and $\\textcircled{2}$ gives $\\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{36}+$, so $k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{1}{2}$. Therefore, the equation of the line is $y-2=-\\frac{1}{2}(x-4)$, that is, $x+2y-8=0$."
  },
  {
    "text": "Given the parabola equation $y^{2}=4x$, and the circle $C$: $(x-2)^{2}+y^{2}=1$. Points $A$ and $B$ lie on circle $C$, point $P$ lies on the parabola, and satisfy $|AB|=2$. Then the minimum value of $\\overrightarrow{PA} \\cdot \\overrightarrow{PB}$ is?",
    "fact_expressions": "G: Parabola;C: Circle;A: Point;B: Point;P: Point;Expression(C) = (y^2 + (x-2)^2 = 1);Expression(G) = (y^2 = 4*x);PointOnCurve(A, C);PointOnCurve(B, C);PointOnCurve(P, G);Abs(LineSegmentOf(A, B)) = 2",
    "query_expressions": "Min(DotProduct(VectorOf(P, A), VectorOf(P, B)))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 5], [66, 69]], [[21, 45], [55, 59]], [[46, 50]], [[51, 54]], [[61, 65]], [[21, 45]], [[2, 20]], [[46, 60]], [[51, 60]], [[61, 70]], [[74, 83]]]",
    "query_spans": "[[[85, 140]]]",
    "process": "\\because the circle's center is at C(2,0), radius r=1, |AB|=2, \\therefore AB is the diameter of the circle, C is the midpoint of AB. Connect PC, PA, PB. Let P(m^{2},2m). \\overrightarrow{B}=(\\overrightarrow{PA}\\cdot\\overrightarrow{PB}=(\\overrightarrow{CA}-\\overrightarrow{CP})\\cdot(\\overrightarrow{CB}-\\overrightarrow{CP})=(-\\overrightarrow{CB}-\\overrightarrow{CP})\\cdot(\\overrightarrow{CB}-\\overrightarrow{CP})=|\\overrightarrow{CP}|^{2}-|\\overrightarrow{CB}|^{2}=|\\overrightarrow{CP}|^{2}-1=(m^{2}-2)^{2}+4m^{2}-1=m^{4}+3\\geqslant3, with equality if and only if m=0."
  },
  {
    "text": "Given that line $l$ passes through the origin and does not coincide with any coordinate axis, and intersects the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ at points $P$ and $Q$. Point $B$ is an arbitrary point on the ellipse distinct from $P$ and $Q$. If the product of the slopes of lines $BP$ and $BQ$ is $-\\frac{1}{4}$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "l: Line;O: Origin;PointOnCurve(O, l);Negation(l = axis);C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;P: Point;Q: Point;Intersection(l, C) = {P, Q};B: Point;PointOnCurve(B, C);Negation(B = P);Negation(B = Q);Slope(LineOf(B, P)) * Slope(LineOf(B, Q)) = -1/4",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[2, 7], [27, 30]], [[10, 14]], [[2, 14]], [[2, 25]], [[31, 88], [106, 108], [160, 165]], [[31, 88]], [[38, 88]], [[38, 88]], [[38, 88]], [[38, 88]], [[91, 94], [111, 114]], [[95, 98], [115, 118]], [[27, 100]], [[101, 105]], [[101, 123]], [[101, 123]], [[101, 123]], [[125, 158]]]",
    "query_spans": "[[[160, 171]]]",
    "process": ""
  },
  {
    "text": "Given that the length of the imaginary axis of the hyperbola $x^{2}+m y^{2}=1$ is 2 times the length of the real axis, then the real number $m$=?",
    "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (m*y^2 + x^2 = 1);Length(ImageinaryAxis(G)) = 2*Length(RealAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "-1/4",
    "fact_spans": "[[[2, 22]], [[37, 42]], [[2, 22]], [[2, 35]]]",
    "query_spans": "[[[37, 44]]]",
    "process": "The hyperbola equation is transformed into standard form as $x^{2}-\\frac{y^{2}}{1}=1$, hence $a=1$, $b=\\sqrt{-\\frac{1}{m}}$. According to the problem, $b=2a$, that is, $\\sqrt{-\\frac{1}{m}}=2$, solving yields $m=-\\frac{1}{4}$."
  },
  {
    "text": "If the point $(-1,2)$ lies on the parabola $x=a y^{2}$, then the equation of the directrix of this parabola is?",
    "fact_expressions": "H: Point;Coordinate(H) = (-1, 2);PointOnCurve(H, G);G: Parabola;Expression(G) = (x = a*y^2);a: Number",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=1",
    "fact_spans": "[[[1, 10]], [[1, 10]], [[1, 26]], [[11, 25], [29, 32]], [[11, 25]], [[14, 25]]]",
    "query_spans": "[[[29, 39]]]",
    "process": "First, find the value of $ a $, then convert the given equation into the standard form of a parabola to obtain its directrix equation. [Detailed solution] From the given condition, $ -1 = a \\times 2^{2} $, so $ a = -\\frac{1}{4} $. Then, the equation of the parabola is $ x = -\\frac{1}{4}y^{2} $, which is equivalent to $ y^{2} = -4x $. Hence, its directrix equation is $ x = 1 $."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=8x$ is the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1$ $(a>0)$, find the equations of the asymptotes of the hyperbola.",
    "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;a>0;Expression(G) = (-y^2/3 + x^2/a^2 = 1);Expression(H) = (y^2 = 8*x);Focus(H) = RightFocus(G)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[20, 67], [73, 76]], [[23, 67]], [[2, 16]], [[23, 67]], [[20, 67]], [[2, 16]], [[2, 71]]]",
    "query_spans": "[[[73, 84]]]",
    "process": "The focus of the parabola $ y^{2} = 8x $ is $ (2, 0) $, so $ a^{2} + 3 = 2^{2} $, $ a = 1 $, therefore the asymptotes of the hyperbola are $ y = \\pm\\sqrt{3}x $."
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=4x$, draw a line intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $x_{1}+x_{2}=6$, then $|AB|=$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;x1:Number;y1:Number;x2:Number;y2:Number;Expression(G) = (y^2 = 4*x);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Focus(G)=F;PointOnCurve(F,H);Intersection(H,G) = {B, A};x1+x2=6;F:Point",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 15], [25, 28]], [[22, 24]], [[29, 46]], [[47, 64]], [[29, 46]], [[29, 46]], [[47, 64]], [[47, 64]], [[1, 15]], [[29, 46]], [[47, 64]], [[1, 21]], [[0, 24]], [[22, 66]], [[69, 84]], [[18, 21]]]",
    "query_spans": "[[[87, 96]]]",
    "process": "According to the problem, p=2, the equation of the directrix of the parabola is x=-1. \\because a line passing through the focus of the parabola y^{2}=4x intersects the parabola at two points A(x_{1},y_{1}) and B(x_{2},y_{2}), \\therefore |AB|=x_{1}+x_{2}+2. Since x_{1}+x_{2}=6, \\therefore |AB|=x_{1}+x_{2}+2=8"
  },
  {
    "text": "Let the line $x - 3y + m = 0$ ($m \\neq 0$) intersect the two asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ ($a > 0$, $b > 0$) at points $A$ and $B$, respectively. If point $P(m, 0)$ satisfies $|PA| = |PB|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;m: Number;P: Point;A: Point;B: Point;a>0;b>0;Negation(m=0);l1:Line;l2:Line;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (m + x - 3*y = 0);Coordinate(P) = (m, 0);Asymptote(G)={l1,l2};Intersection(H,l1)=A;Intersection(H,l2)=B;Abs(LineSegmentOf(P, A)) = Abs(LineSegmentOf(P, B))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[25, 81], [126, 129]], [[28, 81]], [[28, 81]], [[1, 24]], [[3, 24]], [[100, 109]], [[91, 94]], [[95, 98]], [[28, 81]], [[28, 81]], [[3, 24]], [], [], [[25, 81]], [[1, 24]], [[100, 109]], [[25, 87]], [[1, 98]], [[1, 98]], [[111, 124]]]",
    "query_spans": "[[[126, 135]]]",
    "process": "From the equation of the hyperbola, its asymptotes are given by $ y = \\frac{b}{a}x $ and $ y = -\\frac{b}{a}x $. Solving these respectively with the line $ x - 3y + m = 0 $, we obtain $ A\\left(\\frac{-am}{a-3b}, \\frac{-bm}{a-3b}\\right) $, $ B\\left(\\frac{-am}{a+3b}, \\frac{bm}{a+3b}\\right) $. Given $ |PA| = |PB| $, let the midpoint of $ AB $ be $ E $, then $ E\\left(\\frac{\\frac{-am}{a-3b} + \\frac{-am}{a+3b}}{2}, \\frac{-bm}{a+3b} + \\frac{bm}{2}\\right) $. Since $ PE $ is perpendicular to the line $ x - 3y + m = 0 $, it follows that $ 2a^{2} = 8b^{2} = 8(c^{2} - a^{2}) $, i.e., $ \\frac{c^{2}}{a^{2}} = \\frac{5}{4} $. Also, since $ e = \\frac{c}{a} > 1 $, we have $ e = \\frac{\\sqrt{5}}{2} $."
  },
  {
    "text": "The standard equation of a parabola with directrix $x=3$ is?",
    "fact_expressions": "G: Parabola;Expression(Directrix(G)) = (x = 3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = -12x",
    "fact_spans": "[[[11, 14]], [[0, 14]]]",
    "query_spans": "[[[11, 21]]]",
    "process": ""
  },
  {
    "text": "Given that the chord of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ is bisected by the point $(1,1)$, then the equation of the line on which this chord lies is?",
    "fact_expressions": "G: Ellipse;H: LineSegment;I: Point;Expression(G) = (x^2/9 + y^2/5 = 1);Coordinate(I) = (1, 1);IsChordOf(H,G);MidPoint(H)=I",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "5*x+9*y-14=0",
    "fact_spans": "[[[2, 39]], [], [[42, 50]], [[2, 39]], [[42, 50]], [[2, 41]], [[2, 52]]]",
    "query_spans": "[[[2, 66]]]",
    "process": "Given that the chord of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ is bisected by the point $(1,1)$, let the two endpoints of this chord be $A(x_{1},y_{1})$, $B(x_{2},y_{2})$. Then  \n$$\n\\begin{cases}\n\\frac{x_{1}+x_{2}}{2}=1 \\\\\n\\frac{y_{1}+y_{2}}{2}=1\n\\end{cases}\n\\quad \\text{yielding} \\quad\n\\begin{cases}\nx_{1}+x_{2}=2 \\\\\ny_{1}+y_{2}=2\n\\end{cases}\n$$\nSince points $A$ and $B$ lie on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, subtracting the two equations gives  \n$$\n\\frac{x_{1}^{2}-x_{2}^{2}}{9}+\\frac{y_{1}^{2}-y_{2}^{2}}{5}=0,\n\\quad \\text{that is,} \\quad\n\\frac{(y_{1}-y_{2})(y_{1}+y_{2})}{(x_{1}-x_{2})(x_{1}+x_{2})}=-\\frac{5}{9},\n$$\nthus the slope of line $AB$ is  \n$$\nk_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{5}{9}.\n$$\nTherefore, the equation of the line containing this chord is  \n$$\ny-1=-\\frac{5}{9}(x-1),\n\\quad \\text{or} \\quad\n5x+9y-14=0.\n$$"
  },
  {
    "text": "If two mutually perpendicular lines $l_1$, $l_2$ passing through point $P(1,1)$ intersect the $x$-axis and $y$-axis at points $A$ and $B$ respectively, then the equation of the locus of the midpoint $M$ of $AB$ is?",
    "fact_expressions": "l1:Line;l2:Line;P:Point;Coordinate(P)=(1,1);IsPerpendicular(l1,l2);PointOnCurve(P,l1);PointOnCurve(P,l2);Intersection(l2,yAxis)=B;Intersection(l1,xAxis)=A;MidPoint(LineSegmentOf(A,B))=M;A:Point;B:Point;M:Point",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x+y-1=0",
    "fact_spans": "[[[21, 28]], [[29, 36]], [[2, 11]], [[2, 11]], [[12, 36]], [[1, 36]], [[1, 36]], [[21, 59]], [[21, 59]], [[61, 70]], [[50, 53]], [[54, 57]], [[67, 70]]]",
    "query_spans": "[[[67, 77]]]",
    "process": "Let the equation of line $ l_{1} $ be $ y - 1 = k(x - 1) $, then the equation of line $ l_{2} $ is $ y - 1 = -\\frac{1}{k}(x - 1) $. Thus, the intersection point of line $ l_{1} $ with the x-axis is $ A(1 - \\frac{1}{k}, 0) $, and the intersection point of line $ l_{2} $ with the y-axis is $ B(0, 1 + \\frac{1}{k}) $. Let $ M(x, y) $ be the midpoint of $ AB $, then we have\n$$\n\\begin{cases}\nx = \\frac{1}{2}(1 - \\frac{1}{k}) \\\\\ny = \\frac{1}{2}(1 + \\frac{1}{k})\n\\end{cases}\n$$\nAdding the two equations to eliminate $ k $ gives $ x + y = 1 $, that is, $ x + y - 1 = 0 $. Therefore, the trajectory equation of the midpoint $ M $ of $ AB $ is $ x + y - 1 = 0 $."
  },
  {
    "text": "The standard equation of a circle centered at the focus of the parabola $y^{2}=-8 x$ and tangent to the directrix of this parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -8*x);H:Circle;Center(H)=Focus(G);IsTangent(Directrix(G),H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x+2)^2+y^2=16",
    "fact_spans": "[[[1, 16], [26, 29]], [[1, 16]], [[35, 36]], [[0, 36]], [[24, 36]]]",
    "query_spans": "[[[35, 43]]]",
    "process": "Since the focus of $ y^{2} = -8x $ is $ (-2, 0) $ and the directrix is $ x = 2 $, the center of the circle is $ (-2, 0) $ and the radius $ r = 4 $. Therefore, the standard equation of the circle is $ (x + 2)^{2} + y^{2} = 16 $."
  },
  {
    "text": "Given fixed points $A(-2,0)$, $B(2,0)$, and a moving point $P$ satisfying $|P A|+|P B|=6$, then the trajectory equation of point $P$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (-2, 0);B: Point;Coordinate(B) = (2, 0);P: Point;Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B)) = 6",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2/9+y^2/5=1",
    "fact_spans": "[[[4, 13]], [[4, 13]], [[15, 23]], [[15, 23]], [[26, 29], [48, 52]], [[31, 46]]]",
    "query_spans": "[[[48, 59]]]",
    "process": "\\because the moving point P satisfies |PA| + |PB| = 6 > |AB| = 4, \\therefore by the definition of an ellipse, the trajectory of the moving point P is an ellipse with foci at A(-2,0) and B(2,0). Then a = 3, c = 2, b = \\sqrt{5}, \\therefore the equation of the trajectory of the moving point P is \\frac{x^{2}}{9} + \\frac{y^{2}}{5} = 1"
  },
  {
    "text": "Given point $P(2,0)$, and a moving point $Q$ such that the circle with diameter $PQ$ is tangent to the $y$-axis. A perpendicular is drawn from point $P$ to the line $x+(m-1)y+2m-5=0$, with foot of perpendicular at $R$. Then the minimum value of $|QP|+|QR|$ is?",
    "fact_expressions": "P: Point;Coordinate(P) = (2, 0);Q: Point;G: Circle;IsDiameter(LineSegmentOf(P, Q), G);IsTangent(yAxis, G);H: Line;Expression(H) = (2*m + x + y*(m - 1) - 5 = 0);m: Number;Z: Line;PointOnCurve(P, Z);IsPerpendicular(Z, H);R: Point;FootPoint(Z, H) = R",
    "query_expressions": "Min(Abs(LineSegmentOf(Q, P)) + Abs(LineSegmentOf(Q, R)))",
    "answer_expressions": "(9-sqrt(5))/2",
    "fact_spans": "[[[2, 11], [39, 43]], [[2, 11]], [[14, 17]], [[29, 30]], [[19, 30]], [[29, 37]], [[44, 65]], [[44, 65]], [[46, 65]], [], [[38, 68]], [[38, 68]], [[72, 75]], [[38, 75]]]",
    "query_spans": "[[[77, 96]]]",
    "process": "From the condition that moving point Q satisfies the circle with QP as diameter being tangent to the y-axis, it follows that the distance from moving point Q to fixed point P equals the distance from moving point Q to the line x = -2. Hence, the trajectory of moving point Q is y^{2} = 8x.  \n\\begin{cases} From x + (m-1)y + 2m - 5 = 0 we get x - y - 5 + m(y + 2) = 0, \\\\ x - y - 5 = 0 \\\\ y = -2 \\end{cases}  \nSolving gives D(3, -2), so the line x + (m-1)y + 2m - 5 = 0 passes through fixed point D(3, -2). Drawing a perpendicular from P to the line x + (m-1)y + 2m - 5 = 0, with foot at R, then point R lies on the circle with PD as diameter. The diameter form equation is (x-2)(x-3) + y(y+2) = 0, which converts to standard form: (x-\\frac{5}{2})^{2} + (y+1)^{2} = \\frac{5}{4}, with center E(\\frac{5}{2}, -1) and radius r = \\frac{\\sqrt{5}}{2}. Draw QM perpendicular from Q to the directrix, with foot M, and draw EG perpendicular from E to the directrix, with foot G. Then |QP| + |QR| \\geqslant |QM| + |QE| - \\frac{\\sqrt{5}}{2} \\geqslant |EG| - \\frac{\\sqrt{5}}{2} = \\frac{9}{2} - \\frac{\\sqrt{5}}{2} = \\frac{9 - \\sqrt{5}}{2}"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ with foci $F_{1}$ and $F_{2}$, and a point $P(x_{0}, y_{0})$ satisfying $0<\\frac{x_{0}^{2}}{4}+\\frac{y_{0}^{2}}{3}<1$, what is the range of values for $|P F_{1}|+|P F_{2}|$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;Coordinate(P) = (x0, y0);x0: Number;y0: Number;0 < x0^2/4 + y0^2/3;x0^2/4 + y0^2/3 < 1",
    "query_expressions": "Range(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 39]], [[2, 39]], [[44, 51]], [[52, 59]], [[2, 59]], [[60, 78]], [[60, 78]], [[61, 78]], [[61, 78]], [[80, 125]], [[80, 125]]]",
    "query_spans": "[[[128, 156]]]",
    "process": ""
  },
  {
    "text": "Point $P$ moves on the hyperbola $x^{2}-y^{2}=1$, and $O$ is the origin. The equation of the trajectory of the midpoint $M$ of segment $PO$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P, G);G: Hyperbola;Expression(G) = (x^2 - y^2 = 1);O: Origin;M: Point;MidPoint(LineSegmentOf(P, O)) = M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "4*x^2-4*y^2=1",
    "fact_spans": "[[[0, 4]], [[0, 26]], [[5, 23]], [[5, 23]], [[27, 30]], [[45, 48]], [[36, 48]]]",
    "query_spans": "[[[45, 55]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, point $P(x_{0}, y_{0})$ is an arbitrary point on the line $b x-a y+2 a=0$. If the circle $(x-x_{0})^{2}+(y-y_{0})^{2}=2$ has no common points with the right branch of hyperbola $C$, then the range of the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;y0: Number;x0: Number;H: Line;P: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = ((x - x0)^2 + (y - y0)^2 = 2);Expression(H) = (2*a - a*y + b*x = 0);Coordinate(P) = (x0, y0);PointOnCurve(P, H);NumIntersection(G, RightPart(C))=0",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(1,\\sqrt{2}]",
    "fact_spans": "[[[2, 63], [140, 146], [156, 159]], [[10, 63]], [[10, 63]], [[107, 139]], [[65, 82]], [[65, 82]], [[83, 100]], [[64, 82]], [[10, 63]], [[10, 63]], [[2, 63]], [[107, 139]], [[83, 100]], [[64, 82]], [[64, 105]], [[107, 154]]]",
    "query_spans": "[[[156, 169]]]",
    "process": "From the given conditions, the distance between the lines $ bx - ay = 0 $ and $ bx - ay + 2a = 0 $ is greater than or equal to $ \\sqrt{2} $, leading to a homogeneous inequality in terms of $ a $ and $ c $, based on which the range of the eccentricity of the hyperbola can be determined as follows: The line $ bx - ay + 2a = 0 $ and the asymptote of the hyperbola (and point $ P(x_{0}, y_{0}) $ lies on the line $ bx - ay + 2a = 0 $; since the circle $ (x - x_{0})^{2} + (y - y_{0})^{2} = 2 $ has no common point with the right branch of hyperbola $ C $, the distance between the lines $ bx - ay + 2a = 0 $ and $ bx - ay = 0 $ is greater than or equal to $ \\sqrt{2} $, that is, $ \\frac{2a}{\\sqrt{b^{2} + a^{2}}} = \\frac{2a}{c} \\geqslant \\sqrt{2} $, $ \\therefore e = \\frac{c}{a} \\leqslant \\sqrt{2} $. Moreover, since $ e > 1 $, $ \\therefore 1 < e \\leqslant \\sqrt{2} $. Therefore, the range of the eccentricity of the hyperbola is $ (1, \\sqrt{2}] $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of ellipse $C$, with focal distance $4$. If $P$ is a point on ellipse $C$ and the perimeter of $\\Delta P F_{1} F_{2}$ is $14$, then the eccentricity $e$ of ellipse $C$ is?",
    "fact_expressions": "Focus(C) = {F1, F2};F1: Point;F2: Point;C: Ellipse;FocalLength(C) = 4;P: Point;PointOnCurve(P, C) = True;Perimeter(TriangleOf(P, F1, F2)) = 14;e: Number;Eccentricity(C) = e",
    "query_expressions": "e",
    "answer_expressions": "2/5",
    "fact_spans": "[[[2, 28]], [[2, 9]], [[10, 17]], [[18, 23], [42, 47], [84, 89]], [[18, 35]], [[38, 41]], [[38, 50]], [[52, 82]], [[93, 96]], [[84, 96]]]",
    "query_spans": "[[[93, 98]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, with left and right foci $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on ellipse $C$, and the segment $P F_{2}$ is tangent to the circle $x^{2}+y^{2}=b^{2}$ at point $Q$. If $Q$ is the midpoint of segment $P F_{2}$, and $e$ is the eccentricity of $C$, then the minimum value of $\\frac{a^{2}+e^{2}}{3 b}$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C) = True;G: Circle;Expression(G) = (x^2 + y^2 = b^2);TangentPoint(LineSegmentOf(P, F2), G) = Q;Q: Point;MidPoint(LineSegmentOf(P, F2)) = Q;e: Number;Eccentricity(C) = e",
    "query_expressions": "Min((a^2 + e^2)/(3*b))",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[2, 59], [88, 93], [159, 162]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[67, 74]], [[75, 82]], [[2, 82]], [[2, 82]], [[83, 87]], [[83, 94]], [[107, 127]], [[107, 127]], [[95, 134]], [[130, 134], [136, 139]], [[136, 154]], [[155, 158]], [[155, 166]]]",
    "query_spans": "[[[168, 199]]]",
    "process": "Connect $ PF_{1} $, $ OQ $. Since $ OQ $ is the midline, we have $ OQ \\parallel PF_{1} $, $ |OQ| = \\frac{1}{2}|PF_{1}| $. From the circle $ x^{2} + y^{2} = b^{2} $, we get $ |OQ| = b $ and $ |PF_{1}| = 2b $. By the definition of the ellipse, $ |PF_{1}| + |PF_{2}| = 2a $, so $ |PF_{2}| = 2a - 2b $. Since $ OQ \\perp PF_{2} $, we have $ PF_{1} \\perp PF_{2} $, thus $ (2b)^{2} + (2a - 2b)^{2} = (2c)^{2} $, which simplifies to $ b^{2} + a^{2} - 2ab + b^{2} = c^{2} = a^{2} - b^{2} $, leading to $ 2a = 3b $, so $ b = \\frac{2}{3}a $, $ c = \\sqrt{a^{2} - b^{2}} = \\frac{\\sqrt{5}}{3}a $. Hence, $ e = \\frac{c}{a} = \\frac{\\sqrt{5}}{3} $. Then $ \\frac{a^{2} + e^{2}}{3b} = \\frac{a^{2} + \\frac{5}{9}}{2a} = \\frac{1}{2}\\left(a + \\frac{5}{9a}\\right) \\geqslant \\frac{1}{2} \\cdot 2\\sqrt{a \\cdot \\frac{5}{9a}} = \\frac{\\sqrt{5}}{3} $, with equality when $ a = \\frac{\\sqrt{5}}{3} $. Therefore, the minimum value of $ \\frac{a^{2} + e^{2}}{3b} $ is $ \\frac{\\sqrt{5}}{3} $."
  },
  {
    "text": "The vertex is at the origin, passing through the center of circle $C$: $x^{2}+y^{2}-2 x+2 \\sqrt{2} y=0$, and the directrix is perpendicular to the $x$-axis. What is the equation of the parabola?",
    "fact_expressions": "G: Parabola;C: Circle;O: Origin;Expression(C) = (2*sqrt(2)*y - 2*x + x^2 + y^2 = 0);Vertex(G) = O;PointOnCurve(Center(C),G);IsPerpendicular(Directrix(G),xAxis)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=2*x",
    "fact_spans": "[[[60, 63]], [[8, 46]], [[3, 5]], [[8, 46]], [[0, 63]], [[6, 63]], [[50, 63]]]",
    "query_spans": "[[[60, 67]]]",
    "process": "The center of the circle is $(1, -\\sqrt{2})$. Therefore, the focus of the parabola lies on the positive $x$-axis. Let the equation be $y^2 = 2px$. Substituting the point $(1, -\\sqrt{2})$ gives $2 = 2p$, solving yields $p = 1$. Thus, the equation of the parabola is $y^2 = 2x$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, $A$ is the bottom vertex, connect $AF_{2}$ and extend it to intersect the ellipse at point $B$, then the length of $B F_{1}$ is?",
    "fact_expressions": "F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);A: Point;LowerVertex(G) = A;B: Point;Intersection(OverlappingLine(LineSegmentOf(A,F2)),G)=B",
    "query_expressions": "Length(LineSegmentOf(B, F1))",
    "answer_expressions": "5*sqrt(2)/3",
    "fact_spans": "[[[2, 9]], [[10, 17]], [[2, 51]], [[2, 51]], [[18, 45], [74, 76]], [[18, 45]], [[52, 55]], [[18, 59]], [[77, 81]], [[60, 81]]]",
    "query_spans": "[[[83, 95]]]",
    "process": "The ellipse $\\frac{x^{2}}{2}+y^{2}=1$ has $a=\\sqrt{2}, b=1, c=1$, so $F_{1}(-1,0), F_{2}(1,0), A(0,-1)$. The equation of line $AF_{2}$ is $y=x-1$. Substituting into the ellipse equation $\\frac{x^{2}}{2}+y^{2}=1$, we get $3x^{2}-4x=0$, solving gives $x=0$ or $\\frac{4}{3}$, thus $B(\\frac{4}{3},\\frac{1}{3})$. Then $|BF_{1}|=\\sqrt{(\\frac{4}{3}+1)^{2}+\\frac{1}{9}}=\\frac{5\\sqrt{2}}{3}$."
  },
  {
    "text": "One focus of the ellipse $x^{2}+k y^{2}=1$ is $(0,2)$, then the value of $k$ is?",
    "fact_expressions": "G: Ellipse;k: Number;H: Point;Expression(G)=(k*y^2+x^2=1);Coordinate(H) = (0, 2);OneOf(Focus(G))=H",
    "query_expressions": "k",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 19]], [[34, 37]], [[25, 32]], [[0, 19]], [[25, 32]], [[0, 32]]]",
    "query_spans": "[[[34, 41]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ lies on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and $F_{1}$, $F_{2}$ are the left and right foci of the ellipse respectively. It is known that $\\angle F_{1} P F_{2}=120^{\\circ}$ and $|P F_{1}|=2|P F_{2}|$. Find the eccentricity of the ellipse?",
    "fact_expressions": "P: Point;G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;PointOnCurve(P, G) = True;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;AngleOf(F1, P, F2) = ApplyUnit(120, degree);Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(7)/3",
    "fact_spans": "[[[2, 6]], [[7, 59], [82, 84], [153, 155]], [[7, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[2, 63]], [[64, 71]], [[72, 79]], [[64, 90]], [[64, 90]], [[93, 127]], [[129, 151]]]",
    "query_spans": "[[[153, 161]]]",
    "process": "\\because|PF_{1}|=2|PF_{2}|,|PF_{1}|+|PF_{2}|=2a\\therefore|PF_{2}|=\\frac{2a}{3},|PF_{1}|=\\frac{4a}{3}\\because\\angle F_{1}PF_{2}=120^{\\circ},\\cos\\angle F_{1}PF_{2}=\\frac{(\\frac{2a}{3})^{2}+(\\frac{4a}{3})^{2}-4c^{2}}{2\\cdot\\frac{2a}{3}\\cdot\\frac{4a}{3}} \\text{ solving gives } \\frac{c^{2}}{a^{2}}=\\frac{7}{9}\\therefore e=\\frac{c}{a}=\\frac{\\sqrt{7}}{3}"
  },
  {
    "text": "The focus of the parabola $y^{2}=4 x$ is $F$. A line passing through $F$ intersects the parabola at points $A$ and $B$, satisfying $\\frac{|A F|}{|B F|}=4$. Point $O$ is the origin. Then the area of $\\triangle A O F$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);A: Point;B: Point;Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(B, F)) = 4;O: Origin",
    "query_expressions": "Area(TriangleOf(A, O, F))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 14], [30, 33]], [[0, 14]], [[18, 21], [23, 26]], [[0, 21]], [[27, 29]], [[22, 29]], [[35, 38]], [[39, 42]], [[27, 44]], [[48, 71]], [[72, 76]]]",
    "query_spans": "[[[81, 103]]]",
    "process": "As shown in the figure, from $\\frac{|AF|}{|BF|}=4$ we get $x_{A}+1=4(x_{B}+1)$, and from $\\triangle ACF \\sim \\triangle BLDF$ we obtain $\\frac{|CF|}{|DF|}=\\frac{|AF|}{|BF|}$, that is, $\\frac{x_{A}-|OF|}{|OF|-x_{B}}=4$, which simplifies to $\\frac{x_{A}-1}{1-x_{B}}=4$. Solving gives $x_{A}=4$, $x_{B}=\\frac{1}{4}$. Therefore, the coordinates of point $A$ are $A(4,4)$ or $A(4,-4)$, and $_{\\Delta AOF}=\\frac{1}{2}\\times1\\times4=2$."
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and $Q$, $R$ are points on the circles $(x+4)^{2}+y^{2}=\\frac{1}{4}$ and $(x-4)^{2}+y^{2}=\\frac{1}{4}$ respectively, then the minimum value of $|PQ|+|PR|$ is?",
    "fact_expressions": "G: Ellipse;H: Circle;C:Circle;P: Point;Q: Point;R: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Expression(H) = (y^2 + (x + 4)^2 = 1/4);Expression(C) = (y^2 + (x - 4)^2 = 1/4);PointOnCurve(P, G);PointOnCurve(Q, H);PointOnCurve(R, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, Q)) + Abs(LineSegmentOf(P, R)))",
    "answer_expressions": "9",
    "fact_spans": "[[[6, 44]], [[58, 88]], [[89, 119]], [[2, 5]], [[48, 51]], [[52, 55]], [[6, 44]], [[58, 88]], [[89, 119]], [[2, 47]], [[48, 123]], [[48, 123]]]",
    "query_spans": "[[[125, 142]]]",
    "process": ""
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and point $P$ lies on the hyperbola such that $|PF_{1}| \\cdot |PF_{2}| = \\frac{64}{3}$. Then $\\angle F_{1} P F_{2}$ = ?",
    "fact_expressions": "F1: Point;F2: Point;G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 64/3",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "ApplyUnit(120,degree)",
    "fact_spans": "[[[0, 7]], [[8, 15]], [[16, 55], [65, 68]], [[16, 55]], [[0, 60]], [[61, 64]], [[61, 69]], [[72, 109]]]",
    "query_spans": "[[[111, 134]]]",
    "process": "\\cos\\angle F_{1}PF_{2}=\\frac{|PF_{1}|^{2}+|PF_{2}|^{2}-|F_{1}F_{2}|^{2}}{2|PF_{1}||PF_{2}|}=\\frac{(|PF_{1}|-|PF_{2}|)^{2}+2|PF_{1}||PF_{2}|-|F_{1}F_{2}|^{2}}{2|PF_{1}||PF_{2}|}=\\frac{6^{2}+2\\times\\frac{64}{3}-4\\times25}{2\\times\\frac{64}{3}}=-\\frac{1}{2}, \\text{ we get } \\angle F_{1}PF_{2}=120^{\\circ}"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. The point $M(1,2)$ lies on the right branch of the hyperbola $C$, and $F_{2}$ lies on the circle with diameter $M F_{1}$. Then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;M: Point;Coordinate(M) = (1, 2);PointOnCurve(M, RightPart(C)) = True;PointOnCurve(F2, G) = True;IsDiameter(LineSegmentOf(M, F1), G) = True;G: Circle",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)+1",
    "fact_spans": "[[[2, 63], [98, 104], [142, 148]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[72, 79]], [[80, 87], [111, 118]], [[2, 87]], [[2, 87]], [[88, 97]], [[88, 97]], [[88, 109]], [[111, 140]], [[119, 140]], [[135, 136]]]",
    "query_spans": "[[[142, 154]]]",
    "process": "Since $F_{2}$ lies on the circumference of the circle with segment $MF_{1}$ as diameter, $\\therefore MF_{2} \\bot F_{1}F_{2}$, hence $2 = \\frac{b^{2}}{a}$. Also, $\\because \\frac{1}{a^{2}} - \\frac{4}{b^{2}} = 1$, $\\therefore a = \\sqrt{2} - 1$, $\\therefore c = \\sqrt{a^{2} - b^{2}} = 1$, so $e = \\frac{c}{a} = \\frac{1}{\\sqrt{2} - 1} = \\sqrt{2} + 1$, therefore the answer is $\\sqrt{2} + 1$. Key point: Properties of hyperbola"
  },
  {
    "text": "Given the parabola equation $y=4 x^{2}$, what are the coordinates of the focus of the parabola?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 4*x^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, 1/16)",
    "fact_spans": "[[[2, 5], [21, 24]], [[2, 19]]]",
    "query_spans": "[[[21, 31]]]",
    "process": ""
  },
  {
    "text": "$F$ is the right focus of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and point $A(-1,1)$ is fixed. $M$ is a moving point on the ellipse. The minimum value of $\\frac{1}{2}|MA|+|MF|$ is?",
    "fact_expressions": "F: Point;RightFocus(C) = F;C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);A: Point;Coordinate(A) = (-1, 1);M: Point;PointOnCurve(M, C)",
    "query_expressions": "Min((1/2)*Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F)))",
    "answer_expressions": "5/2",
    "fact_spans": "[[[0, 3]], [[0, 50]], [[4, 46], [68, 70]], [[4, 46]], [[53, 62]], [[53, 62]], [[64, 67]], [[64, 74]]]",
    "query_spans": "[[[76, 106]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=x$ intersects with line $l$ at points $A$ and $B$, then the minimum value of $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}$ ($O$ is the origin) is?",
    "fact_expressions": "l: Line;C: Parabola;O: Origin;A: Point;B: Point;Expression(C) = (y^2 = x);Intersection(C, l) = {A, B}",
    "query_expressions": "Min(DotProduct(VectorOf(O,A),VectorOf(O,B)))",
    "answer_expressions": "-1/4",
    "fact_spans": "[[[20, 25]], [[2, 19]], [[89, 92]], [[28, 31]], [[32, 35]], [[2, 19]], [[2, 37]]]",
    "query_spans": "[[[39, 105]]]",
    "process": "By the given condition, let point $ A(m^{2},m) $, $ B(n^{2},n) $, then $ \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = m^{2}n^{2} + mn = (mn + \\frac{1}{2})^{2} - \\frac{1}{4} \\geqslant -\\frac{1}{4} $. Therefore, the minimum value of $ \\overrightarrow{OA} \\cdot \\overrightarrow{OB} $ is $ -\\frac{1}{4} $."
  },
  {
    "text": "Given that point $P(a, b)$ is a moving point on the curve $(x-2 y+2) \\cdot \\sqrt{12-3 x^{2}-4 y^{2}}=0$, then the range of values of $a^{2}+(b+\\frac{1}{4})^{2}$ is?",
    "fact_expressions": "G: Curve;P: Point;Expression(G) = (sqrt(-4*y^2 + 12 - 3*x^2)*(x - 2*y + 2) = 0);PointOnCurve(P,G);a: Number;b: Number;Coordinate(P) = (a, b)",
    "query_expressions": "Range(a^2 + (b + 1/4)^2)",
    "answer_expressions": "[5/4, 17/4]",
    "fact_spans": "[[[13, 60]], [[2, 12]], [[13, 60]], [[2, 64]], [[3, 12]], [[3, 12]], [[2, 12]]]",
    "query_spans": "[[[65, 99]]]",
    "process": "The equation $(x-2y+2)\\cdot\\sqrt{12-3x^{2}-4y^{2}}=0$ is equivalent to $\\begin{cases}x-2y+2=0\\\\12-3x^{2}-4y^{2}\\geqslant0\\end{cases}$ or $12-3x^{2}-4y^{2}=0$, i.e., $\\begin{cases}x-2y+2=0\\\\\\frac{x^{2}}{4}+\\frac{y^{2}}{3}\\leqslant1\\end{cases}$ or $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$. Therefore, the graph corresponding to $P$ is as shown (the line segment inside and on the ellipse). Let $Q(0,-\\frac{1}{4})$, then $|PQ|^{2}=a^{2}+(b+\\frac{1}{4})^{2}$. When $P$ lies on the ellipse, $|PQ|^{2}=4(1-\\frac{b^{2}}{3})+(b+\\frac{1}{4})^{2}=-\\frac{1}{3}(b-\\frac{3}{4})^{2}+\\frac{17}{4}$. Since $-\\sqrt{3}\\leqslant b \\leqslant \\sqrt{3}$, it follows that $(\\sqrt{3}-\\frac{1}{4})^{2}\\leqslant|PQ|^{2}\\leqslant\\frac{17}{4}$. When $P$ lies on the line segment inside the ellipse, $|PQ|^{2}\\geqslant\\left|\\frac{|0+\\frac{1}{2}+2|}{\\sqrt{1+4}}\\right|^{2}=\\frac{5}{4}$. Also, in this case $|PQ|^{2}<\\frac{17}{4}$, and since $\\frac{5}{4}=1.25<(\\sqrt{3}-\\frac{1}{4})^{2}$, we have $|PQ|^{2}\\in[\\frac{5}{4},\\frac{17}{4}]$."
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$, with left and right foci denoted as $F_{1}$ and $F_{2}$, and left and right vertices as $A$ and $B$, respectively. A line passing through point $F_{1}$ intersects the right branch of hyperbola $C$ at point $P$, such that $P F_{2} \\perp A F_{2}$. Then the area of the circumcircle of $\\triangle A B P$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2/3 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;A: Point;B: Point;LeftVertex(C) = A;RightVertex(C) = B;H: Line;PointOnCurve(F1, H);P: Point;Intersection(H, RightPart(C)) = P;IsPerpendicular(LineSegmentOf(P, F2), LineSegmentOf(A, F2))",
    "query_expressions": "Area(CircumCircle(TriangleOf(A, B, P)))",
    "answer_expressions": "5*pi",
    "fact_spans": "[[[2, 35], [89, 95]], [[2, 35]], [[44, 51], [77, 85]], [[52, 59]], [[2, 59]], [[2, 59]], [[67, 70]], [[71, 74]], [[2, 74]], [[2, 74]], [[86, 88]], [[76, 88]], [[100, 104]], [[86, 104]], [[106, 129]]]",
    "query_spans": "[[[131, 156]]]",
    "process": "From $ PF_{2} \\perp AF_{2} $, find the coordinates of point $ P $, yielding that $ \\triangle APF_{2} $ is an isosceles right triangle. Then, using the sine law, the circumradius of the triangle can be found, leading to the answer. Hyperbola $ C: x^{2} - \\frac{y^{2}}{3} = 1 $, so $ F_{1}(-2,0) $, $ F_{2}(2,0) $, $ A(-1,0) $, $ B(1,0) $. From $ PF_{2} \\perp AF_{2} $, we have $ x_{P} = 2 $. Since point $ P $ lies on the right branch of hyperbola $ C $, substitute into $ C: 2^{2} - \\frac{y_{P}^{2}}{3} = 1 $, obtaining $ y_{P}^{2} = 9 $, so $ P(2,\\pm3) $. Thus $ |PF_{2}| = 3 $, $ |AF_{2}| = a + c = 3 $, so $ \\triangle APF_{2} $ is an isosceles right triangle. Hence $ \\sin\\angle PAF_{2} = \\frac{\\sqrt{2}}{2} $, $ |PB| = \\sqrt{3^{2} + 1^{2}} = \\sqrt{10} $. By the sine law, the circumradius $ R $ of $ \\triangle ABP $ satisfies: $ 2R = \\frac{|PB|}{\\sin\\angle PAF_{2}} = \\frac{\\sqrt{10}}{\\frac{\\sqrt{2}}{2}} = 2\\sqrt{5} $, so $ R = \\sqrt{5} $, and thus $ S = \\pi R^{2} = 5\\pi $."
  },
  {
    "text": "In a plane, there is a line segment $AB$ of length $4$. A moving point $P$ satisfies $|PA| + |PB| = 6$. Then the range of values for $|PA|$ is?",
    "fact_expressions": "B: Point;A: Point;P: Point;Length(LineSegmentOf(A, B)) = 4;Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B)) = 6",
    "query_expressions": "Range(Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "[1,5]",
    "fact_spans": "[[[14, 19]], [[14, 19]], [[22, 25]], [[5, 19]], [[27, 42]]]",
    "query_spans": "[[[44, 58]]]",
    "process": "By the definition of an ellipse, the trajectory of point P is an ellipse with A and B as foci. Then, according to the geometric property of an ellipse, $ a - c \\leqslant |PA| \\leqslant a + c $, the result can be obtained. Since $ |PA| + |PB| > |AB| $, the trajectory of point P is an ellipse with A and B as foci, where $ a = 3 $, $ c = 2 $. By the geometric property of the ellipse, $ 3 - 2 \\leqslant |PA| \\leqslant 3 + 2 $, that is, $ 1 \\leqslant |PA| \\leqslant 5 $. Therefore, the range of $ |PA| $ is $ [1, 5] $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ with focus $F$, a line through $F$ intersects $C$ at points $A$ and $B$. From $A$ and $B$, perpendiculars are drawn to the directrix $l$ of $C$, with feet $A_{1}$ and $B_{1}$, respectively. It is known that the areas of $\\triangle A A_{1} F$ and $\\triangle B B_{1} F$ are $9$ and $1$, respectively. Then the area of $\\triangle A_{1} B_{1} F$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;G: Line;PointOnCurve(F, G);A: Point;B: Point;Intersection(G, C) = {A, B};l: Line;Directrix(C) = l;L1: Line;L2: Line;PointOnCurve(A, L1);PointOnCurve(B, L2);IsPerpendicular(L1, l);IsPerpendicular(L2, l);A1: Point;B1: Point;FootPoint(L1, l) = A1;FootPoint(L2, l) = B1;Area(TriangleOf(A, A1, F)) = 9;Area(TriangleOf(B, B1, F)) = 1",
    "query_expressions": "Area(TriangleOf(A1, B1, F))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 28], [44, 47], [69, 72]], [[2, 28]], [[10, 28]], [[10, 28]], [[32, 35], [37, 40]], [[2, 35]], [[41, 43]], [[36, 43]], [[48, 51], [59, 62]], [[52, 55], [63, 66]], [[41, 57]], [[75, 78]], [[69, 78]], [], [], [[58, 81]], [[58, 81]], [[58, 81]], [[58, 81]], [[85, 92]], [[93, 100]], [[58, 100]], [[58, 100]], [[103, 159]], [[103, 159]]]",
    "query_spans": "[[[161, 191]]]",
    "process": "Let the line $ AB: x = ty + \\frac{p}{2} $. From \n\\[\n\\begin{cases}\nx = ty + \\frac{p}{2} \\\\\ny^2 = 2px\n\\end{cases}\n\\]\nwe get $ y^2 = 2p(ty + \\frac{p}{2}) = 2pty + p^2 $. Rearranging gives: $ y^2 - 2pty - p^2 = 0 $. Let $ A(x_1, y_1) $, $ B(x_2, y_2) $, then $ y_1 y_2 = -p^2 $, hence $ x_{1}x_{2} = \\frac{(y_1 y_2)^2}{4p^2} = \\frac{p^2}{4} $, $ x_1 + x_2 = t \\times 2pt + p = p(2t^2 + 1) $. Also, $ S_{\\Delta AA_1 F} = \\frac{1}{2}(x_1 + \\frac{p}{2})y_1 = 9 $, $ S_{\\Delta BB_1 F} = \\frac{1}{2}(x_2 + \\frac{p}{2})(-y_2) = 1 $, so $ S_{\\triangle AA_1 F} \\cdot S_{\\triangle BB_1 F} = \\frac{1}{4}(x_1 + \\frac{p}{2})(x_2 + \\frac{p}{2})y_1(-y_2) = 9 $. Rearranging yields $ x_1 x_2 + \\frac{p}{2}(x_1 + x_2) + \\frac{p^2}{4} = \\frac{36}{p^2} $, i.e., $ \\frac{p^2}{2} + \\frac{p^2}{2} = \\frac{36}{b^2} $, hence $ p^4(t^2 + 1) = 36 $. While $ S_{\\Delta A_1 B_1 F} = \\frac{1}{2} \\times p \\times |y_1 - y_2| = \\frac{1}{2} \\times p \\times \\sqrt{4p^2 t^2 + 4p^2} = \\frac{1}{2} \\times p \\times |y_1 - y_2| = p\\sqrt{t^2 + 1} = 6 $."
  },
  {
    "text": "Given that the left and right foci of a hyperbola are $F_{1}(-4,0)$ and $F_{2}(4,0)$ respectively, and a point $P$ on the hyperbola satisfies $||P F_{1}|-| P F_{2}||=4$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;P: Point;Coordinate(F1) = (-4, 0);Coordinate(F2) = (4, 0);PointOnCurve(P,G);Abs(Abs(LineSegmentOf(P,F1))-Abs(LineSegmentOf(P,F2)))=4;LeftFocus(G) = F1;RightFocus(G) = F2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/12 = 1",
    "fact_spans": "[[[2, 5], [43, 46], [82, 85]], [[14, 27]], [[30, 42]], [[47, 51]], [[14, 27]], [[30, 42]], [[43, 51]], [[53, 80]], [[2, 42]], [[2, 42]]]",
    "query_spans": "[[[82, 92]]]",
    "process": "The left and right foci of the hyperbola are $F_{1}(-4,0)$ and $F_{2}(4,0)$ respectively. A point $P$ on the hyperbola satisfies $||PF_{1}|-|PF_{2}||=4$. According to the definition of a hyperbola, $||PF_{1}|-|PF_{2}||=4=2a$, solving gives $a=2$, $c=4$. Using $c^{2}=a^{2}+b^{2}$, we solve for $b$ to get $b=2\\sqrt{3}$, and the foci of the hyperbola lie on the $x$-axis. Thus, the equation is: $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$"
  },
  {
    "text": "If one of the directrices of the ellipse $C$: $\\frac{x^{2}}{m+1}+y^{2}=1$ is given by $x=-2$, then $m=$?  \nAt this time, the minimum distance between the fixed point $(\\frac{1}{2}, 0)$ and a moving point on the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;m: Number;P: Point;P0: Point;Expression(C) = (x^2/(m + 1) + y^2 = 1);Coordinate(P) = (1/2, 0);Expression(OneOf(Directrix(C))) = (x = -2);PointOnCurve(P0, C)",
    "query_expressions": "m;Min(Distance(P0, P))",
    "answer_expressions": "1\nsqrt(3)/2",
    "fact_spans": "[[[1, 35], [82, 87]], [[51, 54]], [[62, 81]], [[88, 90]], [[1, 35]], [[62, 81]], [[1, 49]], [[82, 90]]]",
    "query_spans": "[[[51, 57]], [[62, 98]]]",
    "process": ""
  },
  {
    "text": "What is the focal length of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/9 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 43]]]",
    "process": "The ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$ has $a=3$, $b=2$, so $c=\\sqrt{a^{2}-b^{2}}=\\sqrt{5}$, thus the focal distance of the ellipse is $2c=2\\sqrt{5}$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$ $(a>b>0)$, the two foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the ellipse $C$, the perimeter of $\\Delta P F_{1} F_{2}$ is $18$, and the minimum value of $|P F_{1}|$ is $1$. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/b^2 + y^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C);Perimeter(TriangleOf(P, F1, F2)) = 18;Min(Abs(LineSegmentOf(P, F1))) = 1",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "4/5",
    "fact_spans": "[[[2, 61], [90, 95], [150, 153]], [[2, 61]], [[9, 61]], [[9, 61]], [[9, 61]], [[9, 61]], [[69, 76]], [[77, 84]], [[2, 84]], [[85, 89]], [[85, 96]], [[97, 127]], [[129, 148]]]",
    "query_spans": "[[[150, 159]]]",
    "process": "According to the definition of the ellipse, 2a + 2c = 18 ⇒ a + c = 9. Since the minimum value of |PF_{1}| is 1, it follows that a - c = 1. Therefore, a = 5, c = 4."
  },
  {
    "text": "Given that the line $y = -x + 1$ intersects the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1$ $(a > b > 0)$ at points $A$ and $B$, and $OA \\perp OB$ ($O$ being the origin), if the eccentricity of the ellipse $e \\in \\left[\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right]$, then the maximum value of $a$ is?",
    "fact_expressions": "H: Line;Expression(H) = (y = 1 - x);G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;A: Point;B: Point;Intersection(H, G) = {A, B};O: Origin;IsPerpendicular(LineSegmentOf(O, A), LineSegmentOf(O, B));e: Number;Eccentricity(G) = e;In(e, [1/2, sqrt(3)/2])",
    "query_expressions": "Max(a)",
    "answer_expressions": "sqrt(10)/2",
    "fact_spans": "[[[2, 12]], [[2, 12]], [[13, 65], [106, 108]], [[13, 65]], [[15, 65]], [[154, 157]], [[15, 65]], [[15, 65]], [[68, 71]], [[72, 75]], [[2, 77]], [[95, 98]], [[79, 94]], [[112, 152]], [[106, 152]], [[112, 152]]]",
    "query_spans": "[[[154, 163]]]",
    "process": "Let $ A(x_{1},y_{1}), B(x_{2},y_{2}) $, from \n$$\n\\begin{cases}\ny = -x + 1 \\\\\n\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1\n\\end{cases}\n$$\nwe get \n$$\n(a^{2} + b^{2})x^{2} - 2a^{2}x + a^{2} - a^{2}b^{2} = 0\n$$\n$ \\Delta = 4a^{4} - 4(a^{2} + b^{2})(a^{2} - a^{2}b^{2}) > 0 $, $ a^{2} + b^{2} > 1 $,\n$$\n\\begin{cases}\nx_{1} + x_{2} = \\frac{2a^{2}}{a^{2} + b^{2}} \\\\\nx_{1}x_{2} = \\frac{a^{2} - a^{2}b^{2}}{a^{2} + b^{2}}\n\\end{cases}\n$$\nSince $ OA \\perp OB $, $ \\therefore \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = x_{1}x_{2} + y_{1}y_{2} = 0 $, i.e., \n$$\n2x_{1}x_{2} - (x_{1} + x_{2}) + 1 = 0\n$$\n$$\n\\frac{2(a^{2} - a^{2}b^{2})}{a^{2} + b^{2}} - \\frac{2a^{2}}{a^{2} + b^{2}} + 1 = 0\n$$\nSimplifying gives \n$$\na^{2} + b^{2} = 2a^{2}b^{2}\n$$\n$$\na^{2} + a^{2} - c^{2} = 2a^{2}(a^{2} - c^{2})\n$$\n$$\n2a^{2} - a^{2}e^{2} = 2a^{2}(a^{2} - a^{2}e^{2})\n$$\n$$\n2a^{2} = \\frac{2 - e^{2}}{1 - e^{2}} = 1 + \\frac{1}{1 - e^{2}}\n$$\nSince $ e \\in \\left[\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right] $, $ \\therefore 2a^{2} \\in \\left[\\frac{7}{3}, 5\\right] $, thus $ a_{\\text{max}} = \\sqrt{\\frac{5}{2}} = \\frac{\\sqrt{10}}{2} $\n\nKey point: Geometric properties of the ellipse"
  },
  {
    "text": "$P$ is a point on the parabola $y^{2}=4x$. If the distance from $P$ to the focus is $5$, then what are the coordinates of point $P$?",
    "fact_expressions": "G: Parabola;P: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(P, G);Distance(P, Focus(G))=5",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(4,\\pm 4)",
    "fact_spans": "[[[4, 18]], [[0, 3], [23, 26], [39, 43]], [[4, 18]], [[0, 21]], [[4, 36]]]",
    "query_spans": "[[[39, 48]]]",
    "process": ""
  },
  {
    "text": "The line passing through the left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ and perpendicular to the $x$-axis intersects the hyperbola at points $M$ and $N$. The circle with $MN$ as diameter passes exactly through the right vertex of the hyperbola. Then, the eccentricity of the hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Circle;M: Point;N: Point;Z: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(LeftFocus(G),Z) ;IsPerpendicular(Z, xAxis);Intersection(Z, G) = {M, N};IsDiameter(LineSegmentOf(M, N), H);PointOnCurve(RightVertex(G), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 57], [73, 76], [104, 107], [113, 116]], [[4, 57]], [[4, 57]], [[100, 101]], [[79, 82]], [[85, 88]], [[70, 72]], [[4, 57]], [[4, 57]], [[1, 57]], [[0, 72]], [[62, 72]], [[70, 90]], [[91, 101]], [[100, 111]]]",
    "query_spans": "[[[113, 123]]]",
    "process": ""
  },
  {
    "text": "Given that the center of the hyperbola is at the origin, one of its vertices is $A(\\sqrt{2}, 0)$, and the two asymptotes are tangent to the circle centered at $A$ with radius $1$, then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;O:Origin;Center(G)=O;H: Circle;A: Point;Coordinate(A) = (sqrt(2), 0);OneOf(Vertex(G))=A;l1:Line;l2:Line;Asymptote(G)={l1,l2};Center(H)=A;Radius(H)=1;IsTangent(l1,H);IsTangent(l2,H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/2 - y^2/2 = 1",
    "fact_spans": "[[[2, 5], [14, 15], [65, 68]], [[9, 13]], [[2, 13]], [[58, 59]], [[21, 37], [45, 48]], [[21, 37]], [[14, 37]], [], [], [[14, 43]], [[44, 59]], [[51, 59]], [[14, 62]], [[14, 62]]]",
    "query_spans": "[[[65, 75]]]",
    "process": "A hyperbola has a vertex at $ A(\\sqrt{2},0) $, so the foci lie on the x-axis, thus $ a = \\sqrt{2} $. The hyperbola can be written as $ \\frac{x^{2}}{2} - \\frac{y^{2}}{b^{2}} = 1 $. The asymptotes are $ bx \\pm \\sqrt{2}y = 0 $. Also, $ \\frac{|\\sqrt{2}b + 0|}{\\sqrt{2 + b^{2}}} = 1 $, solving gives $ b^{2} = 2 $. Therefore, the equation of the hyperbola is $ \\frac{x^{2}}{2} - \\frac{y^{2}}{2} = 1 $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{8}=1$, and point $P$ lies on the hyperbola such that $|P F_{1}|=2|P F_{2}|$, find the area of $\\Delta P F_{1} F_{2}$.",
    "fact_expressions": "F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;G: Hyperbola;Expression(G) = (x^2/4 - y^2/8 = 1);P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "8*sqrt(3)",
    "fact_spans": "[[[2, 9]], [[10, 17]], [[2, 61]], [[2, 61]], [[18, 56], [67, 70]], [[18, 56]], [[62, 66]], [[62, 71]], [[74, 96]]]",
    "query_spans": "[[[98, 125]]]",
    "process": "From the given condition, |PF_{1}| = 2|PF_{2}|. Also, |PF_{1}| - |PF_{2}| = 4, so |PF_{1}| = 8, |PF_{2}| = 4. Since |F_{1}F_{2}| = 4\\sqrt{3}, it follows that |PF_{1}|^{2} = |PF_{2}|^{2} + |F_{1}F_{2}|^{2}, so \\angle F_{1}F_{2}P = \\frac{\\pi}{2}. Therefore, S_{\\triangle PF_{1}F_{2}} = \\frac{1}{2} \\cdot |PF_{2}| \\cdot |F_{1}F_{2}| = \\frac{1}{2} \\times 4 \\times 4\\sqrt{3} = 8\\sqrt{3}."
  },
  {
    "text": "Given that the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1$ $(a>0)$ has eccentricity $2a$, then the asymptotes of this hyperbola are?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2/3 + x^2/a^2 = 1);Eccentricity(G) = 2*a",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[2, 49], [62, 65]], [[5, 49]], [[5, 49]], [[2, 49]], [[2, 59]]]",
    "query_spans": "[[[62, 73]]]",
    "process": "From the given condition, $ e = \\frac{\\sqrt{a^{2}+3}}{a} = 2a $ and $ a > 0 $, solving yields $ a = 1 $, then the asymptotic equations are $ y = \\pm\\sqrt{3}x $."
  },
  {
    "text": "A point $M$ on the parabola $x^{2}=\\frac{1}{2} y$ is at a distance of $2$ from the focus. What is the vertical coordinate of point $M$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = y/2);M: Point;PointOnCurve(M, G);Distance(M, Focus(G)) = 2",
    "query_expressions": "YCoordinate(M)",
    "answer_expressions": "15/8",
    "fact_spans": "[[[0, 24]], [[0, 24]], [[28, 31], [43, 47]], [[0, 31]], [[0, 41]]]",
    "query_spans": "[[[43, 53]]]",
    "process": "First, find the equation of the directrix of the parabola, then according to the definition of the parabola, convert the distance from point M to the focus being 2 into the distance from point M to the directrix being 2, thus the y-coordinate of point M can be found. The directrix of the parabola $ x^{2}=\\frac{1}{2}y $ is $ y=-\\frac{1}{8} $. Let the y-coordinate of point M be $ y $, then since a point M on the parabola $ x^{2}=\\frac{1}{4}y $ has a distance of 2 to the focus, according to the definition of the parabola, the distance from point M to the directrix is 2, therefore $ y+\\frac{1}{8}=2 $, so $ y=\\frac{15}{8} $, hence the y-coordinate of point M is $ \\frac{15}{8} $."
  },
  {
    "text": "Through the focus $F$ of the parabola $y = a^{2}$ $(a > 0)$, draw a straight line intersecting the parabola at points $P$ and $Q$. If the lengths of segments $PF$ and $FQ$ are $p$ and $q$ respectively, then $\\frac{1}{p} + \\frac{1}{q} = $?",
    "fact_expressions": "P: Point;F: Point;G: Parabola;a: Number;H: Line;Q: Point;a>0;Expression(G) = (y = a^2);Focus(G)=F;PointOnCurve(F, H);Intersection(H, G) = {P, Q};Length(LineSegmentOf(P, F)) = p;Length(LineSegmentOf(F, Q)) = q;p:Number;q:Number",
    "query_expressions": "1/q + 1/p",
    "answer_expressions": "4*a",
    "fact_spans": "[[[36, 39]], [[23, 26]], [[1, 20], [31, 34]], [[4, 20]], [[28, 30]], [[40, 43]], [[4, 20]], [[1, 20]], [[1, 26]], [[0, 30]], [[28, 45]], [[47, 70]], [[47, 70]], [[63, 66]], [[67, 70]]]",
    "query_spans": "[[[72, 99]]]",
    "process": ""
  },
  {
    "text": "Given that $A$ and $B$ are the two endpoints of the major axis of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{m^{2}}=1$ $(m>0)$, if there exists a point $M$ on $C$ such that $\\angle A M B=150^{\\circ}$, then the range of values for $m$ is?",
    "fact_expressions": "C: Ellipse;m: Number;A: Point;M: Point;B: Point;m>0;Expression(C) = (x^2/4 + y^2/m^2 = 1);Endpoint(MajorAxis(C)) = {A, B};PointOnCurve(M,C);AngleOf(A,M,B)=ApplyUnit(150,degree)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(0,4-2*sqrt(3)]+[4+2*sqrt(3),+oo)",
    "fact_spans": "[[[10, 61], [71, 74]], [[111, 114]], [[2, 5]], [[77, 81]], [[6, 9]], [[17, 61]], [[10, 61]], [[2, 69]], [[71, 81]], [[83, 109]]]",
    "query_spans": "[[[111, 121]]]",
    "process": "Divide into two cases: x-axis as the major axis and y-axis as the major axis, and find the range of $ m $ respectively. When point $ M $ is the endpoint of the minor axis of the ellipse, $ \\angle AMB $ is maximized. Use trigonometric functions to establish inequalities and obtain the required range. When the x-axis is the major axis, i.e., $ m < 2 $, $ A(-2,0) $, $ B(2,0) $. When point $ M $ is the intersection point of the y-axis and the ellipse, $ \\angle AMB $ is maximized. To ensure that there exists a point $ M $ on ellipse $ C $ satisfying $ \\angle AMB = 150^{\\circ} $, it must hold that $ \\angle AM'B \\geqslant 150^{\\circ} $, i.e., $ \\angle AM'O \\geqslant 75^{\\circ} $. Therefore, $ \\tan\\angle AMO = \\frac{2}{m} \\geqslant \\tan75^{\\circ} = \\frac{\\tan45^{\\circ}+\\tan30^{\\circ}}{1-\\tan45^{\\circ}\\tan30^{\\circ}} = 2+\\sqrt{3} $, hence $ 0 < m \\leqslant 4-2\\sqrt{3} $. When the y-axis is the major axis, i.e., $ m > 2 $, $ A(0,2) $, $ B(0,-2) $. When point $ M $ is the intersection point of the x-axis and the ellipse, $ \\angle AMB $ is maximized. To ensure that there exists a point $ M $ on ellipse $ C $ satisfying $ \\angle AMB = 150^{\\circ} $, it must hold that $ \\angle AM'B \\geqslant 150^{\\circ} $, i.e., $ \\angle AM'O \\geqslant 75^{\\circ} $. Therefore, $ \\tan\\angle AMO = \\frac{m}{2} \\geqslant \\tan75^{\\circ} = \\frac{\\tan45^{\\circ}+\\tan30^{\\circ}}{1-\\tan45^{\\circ}\\tan30^{\\circ}} = 2+\\sqrt{3} $, hence $ m \\geqslant 4+2\\sqrt{3} $. In conclusion, the range of $ m $ is $ (0,4-2\\sqrt{3}] \\cup [4+2\\sqrt{3},+\\infty) $."
  },
  {
    "text": "Given a parabola $C$ with vertex at the origin and focus on the $x$-axis, the line $x - y = 0$ intersects the parabola $C$ at points $A$ and $B$. If $P(2, 2)$ is the midpoint of $AB$, then the equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;G: Line;A: Point;B: Point;P: Point;O:Origin;Expression(G) = (x - y = 0);Vertex(C)=O;Coordinate(P) = (2, 2);PointOnCurve(Focus(C), xAxis);Intersection(G, C) = {A, B};MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[11, 17], [11, 17], [11, 17]], [[27, 36]], [[45, 48]], [[49, 52]], [[57, 67]], [[3, 7]], [[27, 36]], [[2, 17]], [[57, 67]], [[11, 26]], [[27, 54]], [[57, 76]]]",
    "query_spans": "[[[78, 89]]]",
    "process": ""
  },
  {
    "text": "The equation of the line containing the chord of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ that passes through the point $M(3,-1)$ and is bisected by the point $M$ is?",
    "fact_expressions": "G: Hyperbola;L: LineSegment;M: Point;Expression(G) = (x^2/4 - y^2 = 1);Coordinate(M) = (3, -1);IsChordOf(L,G);MidPoint(L)=M;PointOnCurve(M,L)",
    "query_expressions": "Expression(OverlappingLine(L))",
    "answer_expressions": "3*x+4*y-5=0",
    "fact_spans": "[[[20, 48]], [], [[1, 11], [13, 17]], [[20, 48]], [[1, 11]], [[20, 50]], [[12, 50]], [[0, 50]]]",
    "query_spans": "[[[20, 58]]]",
    "process": "Since the hyperbola is symmetric about the x-axis and point M does not lie on the x-axis, the required line is not parallel to the y-axis, i.e., its slope is a real number. Let the slope of the required line be $ a $, and the coordinates of the two intersection points with the hyperbola be $ (3+t, -1+at) $ and $ (3-t, -1-at) $. Substituting these coordinates into the hyperbola equation, we obtain:\n$$\n\\frac{(3+t)^{2}}{4} - (-1+at)^{2} = 1\n$$\n$$\n\\frac{(3-t)^{2}}{4} - (-1-at)^{2} = 1\n$$\nSubtracting these two equations gives:\n$$\n3t + 4at = 0\n$$\n$$\n\\therefore a = -\\frac{3}{4}\n$$\nTherefore, the required line equation is:\n$$\ny + 1 = -\\frac{3}{4}(x - 3)\n$$\nwhich simplifies to:\n$$\n3x + 4y - 5 = 0\n$$"
  },
  {
    "text": "Let point $Q$ be any point on the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$ other than the endpoints of the major axis, $F_{1}$ and $F_{2}$ be the two foci, and let moving point $P$ satisfy $\\overrightarrow{P F}_{1}+\\overrightarrow{P F_{2}}+\\overrightarrow{P Q}=\\overrightarrow{0}$. Then the trajectory equation of the moving point $P$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F2: Point;Q: Point;F1: Point;Expression(G) = (x^2/36 + y^2/9 = 1);PointOnCurve(Q, G);Negation(Q = Endpoint(MajorAxis(G)));Focus(G) = {F1, F2};VectorOf(P, F1) + VectorOf(P, F2) + VectorOf(P, Q) = 0",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(x^2/4 + y^2 = 1)&Negation(x = pm*2)",
    "fact_spans": "[[[6, 44]], [[79, 82], [179, 182]], [[65, 72]], [[1, 5]], [[57, 64]], [[6, 44]], [[1, 56]], [[1, 56]], [[6, 76]], [[84, 175]]]",
    "query_spans": "[[[179, 189]]]",
    "process": "Let $ P(x,y) $, $ Q(x_{0},y_{0}) $ ($ x_{0} \\neq \\pm6 $). From the problem, we know: $ F_{1}(-3\\sqrt{3},0) $, $ F_{2}(3\\sqrt{3},0) $. From $ \\overrightarrow{PF_{1}} + \\overrightarrow{PF_{2}} + \\overrightarrow{PQ} = \\overrightarrow{0} $, we have $ (-3\\sqrt{3}-x+3\\sqrt{3}-x+x_{0}-x, -y-y+y_{0}-y) = (0,0) $. Therefore, \n$$\n\\begin{cases}\nx_{0}=3x \\\\\ny_{0}=3y\n\\end{cases}\n\\quad (x \\neq \\pm2)\n$$\nAlso, $ \\frac{x_{0}^{2}}{36} + \\frac{y_{0}^{2}}{9} = 1 $, so $ \\frac{x^{2}}{4} + y^{2} = 1 $ ($ x \\neq \\pm2 $)."
  },
  {
    "text": "Given that $P$ is a moving point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$, two tangents are drawn from point $P$ to the circle $(x-1)^{2}+y^{2}=1$, with points of tangency $A$ and $B$. Then the minimum value of $|A B|$ is?",
    "fact_expressions": "P: Point;G: Ellipse;Expression(G) = (x^2/16 + y^2/4 = 1);PointOnCurve(P, G) ;H: Circle;Expression(H) = (y^2 + (x - 1)^2 = 1);l1: Line;l2: Line;TangentOfPoint(P, H) = {l1, l2};TangentPoint(l1, H) = A;TangentPoint(l2, H) = B;A: Point;B: Point",
    "query_expressions": "Min(Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "4*sqrt(22)/11",
    "fact_spans": "[[[2, 5], [52, 56]], [[6, 44]], [[6, 44]], [[2, 50]], [[57, 77]], [[57, 77]], [], [], [[51, 82]], [[51, 95]], [[51, 95]], [[88, 91]], [[92, 95]]]",
    "query_spans": "[[[97, 110]]]",
    "process": "Connect PC, intersecting AB at H, we obtain H as the midpoint of AB. Find the center and radius. Connect AC, BC, we obtain AC\\botPA, BC\\botPB. Using the Pythagorean theorem and the triangle area formula, we obtain |AB|. Let P(4\\cos\\theta,2\\sin\\theta), \\theta\\in[0,2\\pi]. Using the distance formula between two points and the square relation of the same angle, combined with completing the square and the method for finding the extremum of a quadratic function, the desired minimum value can be obtained. As shown in the figure, connect PC, intersecting AB at H, we obtain H as the midpoint. The circle (x-1)^{2}+y^{2}=1 has center C(1,0) and radius r=1. Connect AC, BC, we obtain AC\\botPA, BC\\botPB. Then |PA|=|PB|=\\sqrt{|PC|^{2}-1}, and |AB|=2|AH|=\\frac{2|PA|\\cdot|AC|}{|PC|}=\\frac{2\\sqrt{|PC|^{2}-1}}{|PC|}=2\\sqrt{1-\\frac{1}{|PC|}}. Let P(4\\cos\\theta,2\\sin\\theta), \\theta\\in[0,2\\pi], we obtain |PC|^{2}=(4\\cos\\theta-1)^{2}+(2\\sin\\theta)^{2}=12\\cos^{2}\\theta-8\\cos\\theta+5=12(\\cos\\theta-\\frac{1}{3})^{2}+\\frac{11}{3}. When \\cos\\theta=\\frac{1}{3}, |PC|^{2} attains the minimum value \\frac{11}{3}, at which time |AB| attains the minimum value 2\\sqrt{1-\\frac{3}{11}}=\\frac{4\\sqrt{22}}{11}."
  },
  {
    "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{k+8}+\\frac{y^{2}}{9}=1$ is $\\frac{1}{2}$, then what is the value of $k$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(k + 8) + y^2/9 = 1);k: Number;Eccentricity(G) = 1/2",
    "query_expressions": "k",
    "answer_expressions": "{4,-4/5}",
    "fact_spans": "[[[1, 40]], [[1, 40]], [[60, 63]], [[1, 58]]]",
    "query_spans": "[[[60, 67]]]",
    "process": ""
  },
  {
    "text": "The equations of the two asymptotes of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "x \\pm \\left(\\sqrt{3} \\cdot y\\right) = 0",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 39]]]",
    "process": "For the hyperbola \\frac{x^{2}}{3}-y^{2}=1, a=\\sqrt{3}, b=1. Therefore, the asymptotes of the hyperbola \\frac{x^{2}}{3}-y^{2}=1 are y=\\pm\\frac{b}{a}x=\\pm\\frac{\\sqrt{3}}{3}x, that is, x\\pm\\sqrt{3}y=0"
  },
  {
    "text": "The eccentricity $e$ of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ is?",
    "fact_expressions": "G: Ellipse;e: Number;Expression(G) = (x^2/25 + y^2/9 = 1);Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "4/5",
    "fact_spans": "[[[0, 38]], [[42, 45]], [[0, 38]], [[0, 45]]]",
    "query_spans": "[[[42, 47]]]",
    "process": "From $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, we get $a^{2}=25$, $b^{2}=9$, then $c^{2}=a^{2}-b^{2}=16$, $\\therefore a=5$, $c=4$, $\\therefore e=\\frac{c}{a}=\\frac{4}{5}$"
  },
  {
    "text": "A line $l$ passing through the point $M(4,1)$ intersects the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$ at points $A$ and $B$, and $M$ is the midpoint of $AB$. Then the equation of line $l$ is?",
    "fact_expressions": "l: Line;M: Point;Coordinate(M) = (4, 1);PointOnCurve(M, l);G: Hyperbola;Expression(G) = (x^2 - y^2/2 = 1);A: Point;B: Point;Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Expression(l)",
    "answer_expressions": "8*x-y-31=0",
    "fact_spans": "[[[12, 17], [72, 77]], [[2, 11], [58, 61]], [[2, 11]], [[1, 17]], [[18, 46]], [[18, 46]], [[47, 50]], [[51, 54]], [[12, 56]], [[58, 70]]]",
    "query_spans": "[[[72, 82]]]",
    "process": "Let $A(x_{1},y_{1}),B(x_{2},y_{2})\\Rightarrow\\begin{cases}x_{1}^2-\\frac{y_{1}^2}{2}=1\\\\x_{2}^2-\\frac{y_{2}^2}{2}=1\\end{cases}\\Rightarrow(x_{1}+x_{2})(x_{1}-x_{2})-\\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{2}=0\\Rightarrow(x_{1}+x_{2})-\\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{2(x_{1}-x_{2})}=0\\Rightarrow8-\\frac{2k}{2}=0\\Rightarrow k=8\\Rightarrow$ the equation of line $l$ is $8x-y-31=0$."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{m+2}-\\frac{y^{2}}{m +1}=1$ represents an ellipse, then the range of real number $m$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(m + 2) - y^2/(m + 1) = 1);m: Real",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-2,-3/2)+(-3/2,-1)",
    "fact_spans": "[[[45, 47]], [[1, 47]], [[49, 54]]]",
    "query_spans": "[[[49, 61]]]",
    "process": ""
  },
  {
    "text": "If the minimum distance $|MN|$ from a moving point $M(x_1, y_1)$ on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$ to a fixed point $N(m, 0)$ $(0<m<2)$ is $1$, then $m=$?",
    "fact_expressions": "G: Ellipse;M: Point;N: Point;m: Number;0 < m;m < 2;x1: Number;y1: Number;Expression(G) = (x^2/4 + y^2/2 = 1);Coordinate(M) = (x1, y1);Coordinate(N) = (m, 0);PointOnCurve(M, G);Distance(M, N) = Abs(LineSegmentOf(M, N));Min(Abs(LineSegmentOf(M, N))) = 1",
    "query_expressions": "m",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 38]], [[42, 53]], [[56, 74]], [[94, 97]], [[56, 74]], [[56, 74]], [[42, 53]], [[42, 53]], [[1, 38]], [[42, 53]], [[56, 74]], [[1, 53]], [[42, 84]], [[77, 92]]]",
    "query_spans": "[[[94, 99]]]",
    "process": "Find |MN|, express y² in terms of x using the ellipse equation, and use the quadratic function to find its minimum value equal to 1, then solve accordingly. M(x,y) lies on the ellipse \\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1, so y^{2}=2-\\frac{x^{2}}{2}, -2\\leqslant x\\leqslant 2, |MN|=\\sqrt{\\frac{1}{2}(x-2m)^{2}-m^{2}+2}, 0<m<2. When 0<m\\leqslant1, x=2m, |MN|_{\\min}=\\sqrt{-m^{2}+2}=1, m=1, discard the negative value; when 1<m<2, x=2, |MN|_{\\min}=2-m=1, m=1, discard"
  },
  {
    "text": "The minimum value of the sum of distances from a moving point $M$ on the parabola $y=-\\frac{1}{4} x^{2}$ to the two fixed points $(0,-1)$ and $(1,-3)$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = -x^2/4);M: Point;PointOnCurve(M,G) = True;P1: Point;Coordinate(P1) = (0, -1);P2: Point;Coordinate(P2) = (1, -3)",
    "query_expressions": "Min(Distance(M,P1)+Distance(M,P2))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 25]], [[0, 25]], [[29, 32]], [[0, 32]], [[36, 44]], [[36, 44]], [[46, 54]], [[46, 54]]]",
    "query_spans": "[[[29, 65]]]",
    "process": "From the given conditions, the intersection point is $ F(0,-1) $. Let $ A(1,-3) $. Draw $ AN $ perpendicular to the directrix with foot $ N $, and draw $ MH $ perpendicular to the directrix with foot $ H $. Then $ MA + MF = MA + MH \\geqslant AN = 3 + 1 = 4 $."
  },
  {
    "text": "Given that point $P$ lies on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1(a>0)$, one asymptote of the hyperbola is $3x-2y=0$, and points $F_{1}$, $F_{2}$ are the left and right foci of the hyperbola respectively. If $|P F_{1}|=5$, then $|P F_{2}|$=?",
    "fact_expressions": "G: Hyperbola;a: Number;P: Point;F1: Point;F2: Point;a>0;Expression(G) = (-y^2/9 + x^2/a^2 = 1);PointOnCurve(P, G);Expression(OneOf(Asymptote(G))) = (3*x - 2*y = 0);LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(P, F1)) = 5",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "9",
    "fact_spans": "[[[7, 54], [58, 61], [101, 104]], [[10, 54]], [[2, 6]], [[82, 90]], [[91, 98]], [[10, 54]], [[7, 54]], [[2, 57]], [[58, 81]], [[82, 110]], [[82, 110]], [[111, 124]]]",
    "query_spans": "[[[126, 139]]]",
    "process": "By the given condition, the hyperbola is $\\frac{x^{2}}{a^2}-\\frac{y^{2}}{9}=1$, so the equations of its asymptotes are $y=\\pm\\frac{3}{a}x$. Since one asymptote is given as $3x-2y=0$, or $y=\\frac{3}{2}x$, we have $\\frac{3}{a}=\\frac{3}{2}$, solving gives $a=2$. Thus, the equation of the hyperbola is $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$. We obtain $a=2$, $b=3$, $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{13}$. Given $|PF_{1}|=5$, we have $|PF_{1}|<a+c$, so point $P$ lies on the left branch. By the definition of the hyperbola, $|PF_{2}|-|PF_{1}|=2a$, thus $|PF_{2}|=|PF_{1}|+2a=5+4=9$."
  },
  {
    "text": "Given that the center of circle $C$ is the focus of the parabola $x^{2}=4 y$, and the line $4 x-3 y-2=0$ intersects circle $C$ at points $A$ and $B$ with $|A B|=6$, find the standard equation of circle $C$.",
    "fact_expressions": "G: Parabola;C:Circle;H: Line;A: Point;B: Point;Expression(G) = (x^2 = 4*y);Expression(H) = (4*x - 3*y - 2 = 0);Center(C)=Focus(G);Intersection(H, C) = {A, B};Abs(LineSegmentOf(A, B)) = 6",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2+(y-1)^2=10",
    "fact_spans": "[[[10, 24]], [[2, 6], [44, 48], [73, 77]], [[28, 43]], [[51, 54]], [[55, 58]], [[10, 24]], [[28, 43]], [[2, 27]], [[28, 60]], [[62, 71]]]",
    "query_spans": "[[[73, 84]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ has eccentricity $\\frac{2 \\sqrt{3}}{3}$, then its asymptotes are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(G) = 2*sqrt(3)/3",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(3)*x/3",
    "fact_spans": "[[[2, 59], [88, 89]], [[2, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[2, 86]]]",
    "query_spans": "[[[88, 96]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, one of its asymptotes is given by $y=\\frac{12}{5} x$. Then, the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(C))) = (y = (12/5)*x)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "13/5",
    "fact_spans": "[[[2, 63], [92, 95]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 90]]]",
    "query_spans": "[[[92, 101]]]",
    "process": "The asymptotes of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) are given by: $ y = \\pm \\frac{b}{a}x $. Therefore, $ \\frac{b}{a} = \\frac{12}{5} $, so the eccentricity of $ C $ is $ e = \\sqrt{1 + \\left( \\frac{b}{a} \\right)^{2}} = \\frac{13}{5} $."
  },
  {
    "text": "Given that the moving point $A$ travels on the circle $P$: $x^{2}+y^{2}=1$, and point $Q$ is the midpoint of the distance between the fixed point $B(-3,4)$ and point $A$, then the trajectory equation of point $Q$ is?",
    "fact_expressions": "P: Circle;B: Point;A: Point;Expression(P) = (x^2 + y^2 = 1);Coordinate(B) = (-3, 4);PointOnCurve(A, P);Q:Point;MidPoint(Distance(A,B))=Q",
    "query_expressions": "LocusEquation(Q)",
    "answer_expressions": "x^2+y^2+3*x-4*y+6=0",
    "fact_spans": "[[[8, 28]], [[39, 48]], [[4, 7], [49, 53]], [[8, 28]], [[39, 48]], [[2, 31]], [[32, 36], [60, 64]], [[32, 58]]]",
    "query_spans": "[[[60, 71]]]",
    "process": "Let Q(x, y), then A(2x+3, 2y-4). Substituting A into the equation of circle P gives: (2x+3)^{2}+(2y-4)^{2}=1, which simplifies to x^{2}+y^{2}+3x-4y+6=0."
  },
  {
    "text": "Given the circle $C$: $x^{2}+(y-3)^{2}=9$, chords $OP$ are drawn from the origin to the circle $C$. Find the equation of the locus of the midpoint $Q$ of chord $OP$.",
    "fact_expressions": "C: Circle;Expression(C) = (x^2 + (y - 3)^2 = 9);O: Origin;PointOnCurve(O, LineSegmentOf(O, P)) = True;P: Point;IsChordOf(LineSegmentOf(O, P), C) = True;Q: Point;MidPoint(LineSegmentOf(O, P)) = Q",
    "query_expressions": "LocusEquation(Q)",
    "answer_expressions": "(x^2+(y-3/2)^2=9/4)&Negation(y = 0)",
    "fact_spans": "[[[2, 27], [32, 36]], [[2, 27]], [[29, 31]], [[28, 43]], [[38, 43]], [[32, 43]], [[54, 57]], [[46, 57]]]",
    "query_spans": "[[[54, 64]]]",
    "process": "Let $ Q(x,y) $ ($ y \\neq 0 $), then $ P(2x,2y) $. Substituting into the equation of circle $ C $, we get $ (2x)^{2} + (2y - 3)^{2} = 9 $, so the trajectory equation of point $ Q $ is $ x^{2} + \\left(y - \\frac{3}{2}\\right)^{2} = \\frac{9}{4} $ ($ y \\neq 0 $)."
  },
  {
    "text": "The left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $F(-2,0)$, point $A(0, \\sqrt{5})$, and point $P$ is a moving point on the right branch of the hyperbola. The minimum perimeter of $\\triangle A P F$ is $8$. Then, the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F: Point;A: Point;P: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (-2, 0);Coordinate(A) = (0, sqrt(5));LeftFocus(G) = F;PointOnCurve(P, RightPart(G));Min(Perimeter(TriangleOf(A, P, F))) = 8",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 56], [94, 97], [134, 137]], [[3, 56]], [[3, 56]], [[61, 70]], [[71, 88]], [[89, 93]], [[3, 56]], [[3, 56]], [[0, 56]], [[61, 70]], [[71, 88]], [[0, 70]], [[89, 103]], [[105, 132]]]",
    "query_spans": "[[[134, 143]]]",
    "process": "Let the right focus of the hyperbola be $ F(2,0) $, and $ |AF| = 3 $, so the perimeter of $ \\triangle APF $ is $ l = |AF| + |PF| + |AP| = 3 + |PF| + |AP| $. By the definition of the hyperbola, $ |PF| - |PF| = 2a $, that is, $ |PF| = 2a + |PF| $, so $ l = 3 + 2a + |PF| + |AP| $. When points $ A $, $ P $, and $ F $ are collinear, the perimeter $ l $ reaches its minimum value. At this time, $ |PF| + |AP| = |AF| = 3 $, so $ 3 + 2a + 3 = 8 $, solving gives $ a = 1 $, thus $ e = \\frac{c}{a} = 2 $."
  },
  {
    "text": "Given the ellipse equation $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}}=1(a>b>0)$, when $a^{2}+\\frac{16}{b(a-b)}$ is minimized, the eccentricity of the ellipse $e$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>b;b>0;e: Number;Eccentricity(G) = e;WhenMin(a^2 + 16/(b*(a - b)))",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[2, 4], [91, 93]], [[2, 58]], [[6, 58]], [[6, 58]], [[6, 58]], [[6, 58]], [[97, 100]], [[91, 100]], [[59, 90]]]",
    "query_spans": "[[[97, 102]]]",
    "process": ""
  },
  {
    "text": "If the two foci of an ellipse are $F_{1}(-3,0)$, $F_{2}(3,0)$, and the chord $AB$ of the ellipse passes through point $F_{1}$, with the perimeter of $\\triangle A B F_{2}$ equal to $20$, then what is the standard equation of the ellipse?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;F1: Point;F2: Point;Coordinate(F1) = (-3, 0);Coordinate(F2) = (3, 0);Focus(G) = {F1, F2};IsChordOf(LineSegmentOf(A, B), G);PointOnCurve(F1, LineSegmentOf(A, B));Perimeter(TriangleOf(A, B, F2)) = 20",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/25 + y^2/16 = 1",
    "fact_spans": "[[[1, 3], [38, 40], [90, 92]], [[42, 47]], [[42, 47]], [[9, 22], [48, 56]], [[25, 37]], [[9, 22]], [[25, 37]], [[1, 37]], [[38, 47]], [[42, 56]], [[58, 88]]]",
    "query_spans": "[[[90, 99]]]",
    "process": ""
  },
  {
    "text": "The length of the major axis of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ is equal to?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1)",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 27]], [[0, 27]]]",
    "query_spans": "[[[0, 34]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, the intersection point of its left directrix with the $x$-axis is point $P$. Then, the distance from point $P$ to one of its asymptotes is?",
    "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2/4 - y^2/12 = 1);Intersection(LeftDirectrix(G), xAxis) = P",
    "query_expressions": "Distance(P, OneOf(Asymptote(G)))",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[2, 41]], [[54, 58], [60, 64]], [[2, 41]], [[2, 58]]]",
    "query_spans": "[[[2, 77]]]",
    "process": "a=2, c=4, the equation of the left directrix is x=-\\frac{a^{2}}{c}=-1, so P(-1,0). The asymptotes are given by \\sqrt{3}x\\pm y=0, so the distance from P to the asymptotes is d=\\frac{|\\sqrt{3}\\pm 0|}{\\sqrt{3+1}}=\\frac{\\sqrt{3}}{2}, fill in \\frac{\\sqrt{3}}{2}"
  },
  {
    "text": "It is known that the vertices of a hyperbola lie on the coordinate axes, the center is at the origin, and the asymptotes pass through the point $P(m, 2m)$ $(m \\neq 0)$. What is the eccentricity of the hyperbola?",
    "fact_expressions": "G: Hyperbola;P: Point;O:Origin;PointOnCurve(Vertex(G),axis);Center(G)=O;Coordinate(P) = (m, 2*m);Negation(m=0);PointOnCurve(P,Asymptote(G));m:Number",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{sqrt(5),sqrt(5)/2}",
    "fact_spans": "[[[2, 5], [48, 51]], [[24, 46]], [[16, 18]], [[2, 12]], [[2, 18]], [[24, 46]], [[25, 46]], [[2, 46]], [[25, 46]]]",
    "query_spans": "[[[48, 57]]]",
    "process": "Discuss in two cases according to whether the foci are on the x-axis or y-axis. Set up the hyperbola equation, find the asymptote equations, use the condition that the asymptotes pass through point $ P(m,2m) $ to determine the relationship between $ a $ and $ b $, then use $ c^{2}=a^{2}+b^{2} $ and $ e=\\frac{c}{a} $ to obtain the solution.\n\nWhen the foci are on the x-axis, let the hyperbola equation be $ \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 $ ($ a>0,b>0 $), the asymptote equations are $ y=\\pm\\frac{b}{a}x $. Since the asymptotes pass through point $ P(m,2m) $ ($ m\\neq0 $), we have $ 2m=\\frac{b}{a}m $, solving gives $ b=2a $. Thus, $ b^{2}=4a^{2} $, $ c^{2}=a^{2}+b^{2}=a^{2}+4a^{2}=5a^{2} $. The eccentricity of the hyperbola is $ e=\\frac{c}{a}=\\sqrt{5} $.\n\nWhen the foci are on the y-axis, let the hyperbola equation be $ \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1 $ ($ a>0,b>0 $), the asymptote equations are $ y=\\pm\\frac{a}{b}x $. Since the asymptotes pass through point $ P(m,2m) $ ($ m\\neq0 $), we have $ 2m=\\frac{a}{b}m $, solving gives $ b=\\frac{1}{2}a $. Thus, $ b^{2}=\\frac{1}{4}a^{2} $, $ c^{2}=a^{2}+b^{2}=a^{2}+\\frac{1}{4}a^{2}=\\frac{5}{4}a^{2} $, the eccentricity of the hyperbola is $ e=\\frac{c}{a}=\\frac{\\sqrt{5}}{2} $.\n\nIn conclusion, the eccentricity of the hyperbola is $ \\sqrt{5} $ or $ \\frac{\\sqrt{5}}{2} $."
  },
  {
    "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{2}+\\frac{y^{2}}{m}=1$ with focus on the $y$-axis is $e=\\frac{1}{2}$, then $m=$?",
    "fact_expressions": "PointOnCurve(Focus(G), yAxis);G: Ellipse;Expression(G) = (x^2/2 + y^2/m = 1);m: Number;e: Number;Eccentricity(G) = e;e = 1/2",
    "query_expressions": "m",
    "answer_expressions": "8/3",
    "fact_spans": "[[[1, 47]], [[10, 47]], [[10, 47]], [[68, 71]], [[51, 66]], [[10, 66]], [[51, 66]]]",
    "query_spans": "[[[68, 73]]]",
    "process": ""
  },
  {
    "text": "Let $P$ be a point on the parabola $y = x^{2}$. When the distance from point $P$ to the line $x - y + 2 = 0$ is minimized, what are the coordinates of point $P$?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Expression(G) = (y = x^2);Expression(H) = (x - y + 2 = 0);PointOnCurve(P, G);WhenMin(Distance(P,H))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[5, 17]], [[27, 38]], [[1, 4], [22, 26], [45, 49]], [[5, 17]], [[27, 38]], [[1, 20]], [[21, 44]]]",
    "query_spans": "[[[45, 54]]]",
    "process": ""
  },
  {
    "text": "A point $M$ on the parabola ${y }^{2}=4 x$ is at a distance $|M F|=4$ from the focus $F$ of the parabola. What is the horizontal coordinate $x$ of point $M$?",
    "fact_expressions": "G: Parabola;M: Point;F: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(M, G);Focus(G) = F;Distance(M, F) = Abs(LineSegmentOf(M, F));Abs(LineSegmentOf(M, F)) = 4;x_: Number;XCoordinate(M) = x_",
    "query_expressions": "x_",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 17], [25, 28]], [[19, 23], [48, 52]], [[31, 34]], [[0, 17]], [[0, 23]], [[25, 34]], [[20, 46]], [[37, 46]], [[56, 59]], [[48, 59]]]",
    "query_spans": "[[[56, 61]]]",
    "process": ""
  },
  {
    "text": "Draw a line with slope $2$ passing through the right focus of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$, intersecting the ellipse at points $A$ and $B$, and let $O$ be the coordinate origin. Find the length of chord $AB$?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;O: Origin;Expression(G) = (x^2/5 + y^2/4 = 1);PointOnCurve(RightFocus(G), H);Intersection(H,G) = {A,B};Slope(H)=2;IsChordOf(LineSegmentOf(A,B),G)",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "5*sqrt(5)/3",
    "fact_spans": "[[[1, 38]], [[52, 54]], [[59, 63]], [[65, 68]], [[71, 74]], [[1, 38]], [[0, 54]], [[52, 70]], [[45, 54]], [[55, 86]]]",
    "query_spans": "[[[82, 89]]]",
    "process": ""
  },
  {
    "text": "Find the standard equation of the circle passing through the left and right foci $F_{1}$, $F_{2}$ and the upper vertex $B_{2}$ of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);F1: Point;F2: Point;B2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;UpperVertex(G) = B2;H: Circle;PointOnCurve(F1,H) = True;PointOnCurve(F2,H) = True;PointOnCurve(B2,H) = True",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2+y^2=1",
    "fact_spans": "[[[3, 30]], [[3, 30]], [[35, 42]], [[43, 50]], [[54, 61]], [[3, 50]], [[3, 50]], [[3, 61]], [[62, 63]], [[1, 63]], [[1, 63]], [[1, 63]]]",
    "query_spans": "[[[62, 69]]]",
    "process": ""
  },
  {
    "text": "The product of the distances from a point $P$ on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{25}=1$ to its two foci is $m$. When $m$ takes its maximum value, what are the coordinates of point $P$?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/9 + y^2/25 = 1);PointOnCurve(P, G);F1:Point;F2:Point;Focus(G)={F1,F2};Distance(P,F1)*Distance(P,F2) = m;WhenMax(m);m:Number",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "{(3,0),(-3,0)}",
    "fact_spans": "[[[0, 38]], [[42, 45], [70, 74]], [[0, 38]], [[0, 45]], [], [], [[0, 49]], [[0, 59]], [[60, 69]], [[61, 64], [56, 59]]]",
    "query_spans": "[[[70, 79]]]",
    "process": "Using the definition of an ellipse and the basic inequality, the maximum value of $ m $ can be found. From the equality condition $^{\\frac{1}{2}}$, it follows that point $ P $ is at the endpoint of the minor axis of the ellipse, from which the coordinates of point $ P $ can be determined. Denote the two foci of the ellipse as $ F_{1} $ and $ F_{2} $. By the definition of the ellipse, we have $ |PF_{1}| + |PF_{2}| = 10 $. Applying the basic inequality, we get $ m = |PF_{1}| \\cdot |PF_{2}| \\leqslant \\left( \\frac{|PF_{1}| + |PF_{2}|}{2} \\right)^{2} = 25 $. The equality holds if and only if $ |PF_{1}| = |PF_{2}| = 5 $, in which case point $ P $ is at the endpoint of the minor axis of the ellipse. Therefore, the coordinates of point $ P $ are $ (-3, 0) $ or $ (3, 0) $."
  },
  {
    "text": "Given the parabola $x^{2}=4 y$ with focus $F$ and directrix $l$, let $P$ be a point on the parabola, and let $PA \\perp l$ intersecting $l$ at point $A$. When $\\angle AFO=30^{\\circ}$ ($O$ being the origin), $|PF|=$?",
    "fact_expressions": "G: Parabola;P: Point;A: Point;F: Point;O: Origin;l: Line;Expression(G) = (x^2 = 4*y);Focus(G) = F;Directrix(G) = l;PointOnCurve(P, G);PointOnCurve(P, LineSegmentOf(P,A));IsPerpendicular(LineSegmentOf(P, A),l);FootPoint(LineSegmentOf(P, A),l)=A;AngleOf(A, F, O) = ApplyUnit(30, degree)",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[2, 16], [36, 39]], [[32, 35], [44, 47]], [[62, 66]], [[20, 23]], [[94, 97]], [[27, 30]], [[2, 16]], [[2, 23]], [[2, 30]], [[32, 42]], [[43, 61]], [[48, 61]], [[48, 66]], [[68, 93]]]",
    "query_spans": "[[[106, 115]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{2}$ is the right focus of the ellipse $m x^{2} + y^{2} = 4 m$ $(0 < m < 1)$, point $A(0,2)$, and point $P$ is any point on the ellipse, and the minimum value of $|P A| - |P F_{2}|$ is $-\\frac{4}{3}$, then $m = $?",
    "fact_expressions": "G: Ellipse;m: Number;A: Point;P: Point;F2: Point;0 < m;m < 1;Expression(G) = (m*x^2 + y^2 = 4*m);Coordinate(A) = (0, 2);RightFocus(G) = F2;PointOnCurve(P, G);Min(Abs(LineSegmentOf(P, A)) - Abs(LineSegmentOf(P, F2))) = -4/3",
    "query_expressions": "m",
    "answer_expressions": "2/9",
    "fact_spans": "[[[10, 38], [58, 60]], [[107, 110]], [[43, 52]], [[53, 57]], [[2, 9]], [[12, 38]], [[12, 38]], [[10, 38]], [[43, 52]], [[2, 42]], [[53, 65]], [[68, 104]]]",
    "query_spans": "[[[107, 112]]]",
    "process": "From mx^{2}+y^{2}=4m, we obtain the standard equation of the ellipse \\frac{x^{2}}{4}+\\frac{y^{2}}{4m}=1. Since 0<m<1, the foci of the ellipse lie on the x-axis, a^{2}=4, b^{2}=4m, c^{2}=4-4m. Let the left focus of the ellipse be F_{1}, then |PF_{1}|+|PF_{2}|=4, F_{1}(-\\sqrt{4-4m},0). Then |PA|-|PF_{2}|=|PA|+|PF_{1}|-4\\geqslant|AF_{1}|-4=2\\sqrt{2-m}-4=-\\frac{4}{3}, solving gives m=\\frac{2}{9}."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2 p x(p>0)$ be $F$, and the directrix be $l$. A line passing through the focus intersects the parabola at points $A$ and $B$. Perpendiculars from $A$ and $B$ to $l$ have feet at $C$ and $D$, respectively. If $|A F|=3|B F|$ and the area of triangle $C D F$ is $\\sqrt{3}$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(G) = F;l: Line;Directrix(G) = l;L: Line;PointOnCurve(Focus(G), L);Intersection(L, G) = {A, B};A: Point;B: Point;L1: Line;L2: Line;PointOnCurve(A, L1);PointOnCurve(B, L2);IsPerpendicular(L1, l) ;IsPerpendicular(L2, l);FootPoint(L1, l) = C;FootPoint(L2, l) = D;C: Point;D: Point;Abs(LineSegmentOf(A, F)) = 3*Abs(LineSegmentOf(B, F));Area(TriangleOf(C, D, F)) = sqrt(3)",
    "query_expressions": "p",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 22], [45, 48]], [[1, 22]], [[132, 135]], [[4, 22]], [[26, 29]], [[1, 29]], [[33, 36], [70, 73]], [[1, 36]], [[40, 42]], [[1, 42]], [[40, 58]], [[49, 52], [62, 65]], [[53, 56], [66, 69]], [], [], [[59, 76]], [[59, 76]], [[59, 76]], [[59, 76]], [[59, 87]], [[59, 87]], [[80, 83]], [[84, 87]], [[90, 104]], [[106, 130]]]",
    "query_spans": "[[[132, 139]]]",
    "process": ""
  },
  {
    "text": "A line passing through the point $P(1,1)$ intersects the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1$ at points $A$ and $B$, and $\\overrightarrow{A P}=\\lambda \\overrightarrow{P B}$. The point $Q$ satisfies $\\overrightarrow{A Q}=-\\lambda \\overrightarrow{Q B}$. If $O$ is the origin, then the minimum length of the segment $O Q$ is?",
    "fact_expressions": "P: Point;Coordinate(P) = (1, 1);H: Line;PointOnCurve(P, H);G: Ellipse;Expression(G) = (x^2/3 + y^2/2 = 1);A: Point;B: Point;Intersection(H, G) = {A, B};lambda: Number;VectorOf(A, P) = lambda*VectorOf(P, B);Q: Point;VectorOf(A, Q) = -lambda*VectorOf(Q, B);O: Origin",
    "query_expressions": "Min(Length(LineSegmentOf(O, Q)))",
    "answer_expressions": "6*sqrt(13)/13",
    "fact_spans": "[[[1, 10]], [[1, 10]], [[11, 13]], [[0, 13]], [[14, 51]], [[14, 51]], [[53, 57]], [[58, 61]], [[11, 61]], [[63, 114]], [[63, 114]], [[115, 119]], [[121, 173]], [[175, 178]]]",
    "query_spans": "[[[185, 200]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "4/5",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 44]]]",
    "process": ""
  },
  {
    "text": "It is known that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $y=\\sqrt{3} x$, and one of its foci coincides with the focus of the parabola $y^{2}=16 x$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x);H: Parabola;Expression(H) = (y^2 = 16*x);OneOf(Focus(G)) = Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/12 = 1",
    "fact_spans": "[[[2, 58], [82, 83], [112, 115]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 81]], [[89, 104]], [[89, 104]], [[82, 109]]]",
    "query_spans": "[[[112, 120]]]",
    "process": "From the given, $\\sqrt{3}=\\frac{b}{a}$, $c=4$ $\\therefore e=\\frac{c}{a}=\\sqrt{1+(\\frac{b}{a})^{2}}=2$ $\\therefore a=2$ $\\therefore$ the equation of the hyperbola is $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>c>0, a^{2}=b^{2}+c^{2})$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. If a circle $F_{2}$ is drawn with center at $F_{2}$ and radius $b-c$, and a tangent is drawn from a point $P$ on the ellipse to this circle, touching it at point $T$, and the minimum value of $|P T|$ is $\\frac{\\sqrt{3}}{2}(a-c)$, then what is the range of values for the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;c: Number;H: Circle;F2: Point;P: Point;T: Point;F1: Point;l: Line;a > b;b > c;c > 0;a^2=b^2+c^2;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;Center(H) = F2;Radius(H) = b - c;PointOnCurve(P, G);TangentOfPoint(P, H) = l;TangentPoint(l, H) = T;Min(Abs(LineSegmentOf(P, T))) = (sqrt(3)/2)*(a - c)",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[3/5, \\sqrt{2}/2)",
    "fact_spans": "[[[2, 75], [131, 133], [193, 195]], [[112, 117]], [[166, 191]], [[112, 117]], [[121, 129], [141, 142]], [[91, 98], [101, 108]], [[136, 139]], [[149, 152]], [[83, 90]], [], [[4, 75]], [[4, 75]], [[4, 75]], [[4, 75]], [[2, 75]], [[2, 98]], [[2, 98]], [[100, 129]], [[112, 129]], [[131, 139]], [[130, 145]], [[130, 152]], [[154, 191]]]",
    "query_spans": "[[[193, 206]]]",
    "process": ""
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{10}+\\frac{y^{2}}{11}=1$ are? The length of the major axis is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/10 + y^2/11 = 1)",
    "query_expressions": "Coordinate(Focus(G));Length(MajorAxis(G))",
    "answer_expressions": "(pm*1,0)\n2*sqrt(11)",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 44]], [[0, 49]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. If a point $P$ on the ellipse satisfies $|P F_{1}|=2|P F_{2}|$, then $|P F_{1}|$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Abs(LineSegmentOf(P, F1))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 39], [65, 67]], [[2, 39]], [[48, 55]], [[56, 63]], [[2, 63]], [[2, 63]], [[69, 73]], [[65, 73]], [[75, 97]]]",
    "query_spans": "[[[99, 112]]]",
    "process": "According to the given condition, for the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, $|PF_{1}|+|PF_{2}|=2a=6$, and $|PF_{1}|=2|PF_{2}|$, so $\\frac{3}{2}|PF_{1}|=6$, therefore $|PF_{1}|=4$."
  },
  {
    "text": "Let a point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ have distance $4$ to the left focus $F$. If point $M$ satisfies $\\overrightarrow{O M}=\\frac{1}{2}(\\overrightarrow{O P}+\\overrightarrow{O F})$, then $|\\overrightarrow{O M}|$=?",
    "fact_expressions": "G: Ellipse;O: Origin;M: Point;P: Point;F: Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(P, G);LeftFocus(G) = F;Distance(P, F) = 4;VectorOf(O, M) = (VectorOf(O, F) + VectorOf(O, P))/2",
    "query_expressions": "Abs(VectorOf(O, M))",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 40]], [[147, 171]], [[62, 66]], [[43, 46]], [[50, 53]], [[1, 40]], [[1, 46]], [[1, 53]], [[43, 60]], [[68, 145]]]",
    "query_spans": "[[[147, 173]]]",
    "process": "From the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, we have $a=5$, $b=4$. Let the right focus of the ellipse be $F$. Then $|PF|+|PF|=10$. Also, the distance from point $P$ to the left focus $F$ is $4$. Therefore, $|PF|=6$. Since $\\overrightarrow{OM}=\\frac{1}{2}(\\overrightarrow{OP}+\\overrightarrow{OF})$, then $M$ is the midpoint of $PF$. Also, $O$ is the midpoint of $FF$. Hence, $|OM|=\\frac{1}{2}|PF|=3$, that is, $|\\overrightarrow{OM}|=3$."
  },
  {
    "text": "The vertex of the parabola is at the origin, and the focus lies on the $y$-axis. A point $P(m, 1)$ on the parabola is at a distance of $4$ from the focus. Find the standard equation of the parabola?",
    "fact_expressions": "G: Parabola;P: Point;O: Origin;Coordinate(P) = (m, 1);Vertex(G) = O;PointOnCurve(Focus(G), yAxis);PointOnCurve(P, G);Distance(P, Focus(G)) = 4;m:Number",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2=12*y",
    "fact_spans": "[[[0, 3], [19, 22], [46, 49]], [[25, 34]], [[7, 9]], [[25, 34]], [[0, 9]], [[0, 18]], [[19, 34]], [[19, 44]], [[25, 34]]]",
    "query_spans": "[[[46, 55]]]",
    "process": "According to the problem, the parabola equation can be set as $x^{2}=2px$ ($p>0$), so the distance from point $P(m,1)$ on the parabola to the focus is equal to $1-(-\\frac{p}{2})=4$, $\\therefore p=6$, thus the parabola equation is $x^{2}=12y$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ with directrix $l$, a line passing through $M(1,0)$ with slope $\\sqrt{3}$ intersects $l$ at point $A$ and intersects $C$ at a point $B$. If $\\overrightarrow{A M}=\\overrightarrow{M B}$, then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;M: Point;A: Point;B: Point;l: Line;p>0;Expression(C) = (y^2 = 2*p*x);Coordinate(M) = (1, 0);PointOnCurve(M,G);Slope(G)=sqrt(3);Intersection(G,l)=A;OneOf(Intersection(G,C))=B;VectorOf(A, M) = VectorOf(M, B);Directrix(C)=l",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 28], [75, 78]], [[136, 139]], [[60, 62]], [[37, 45]], [[69, 73]], [[84, 87]], [[32, 35], [32, 35]], [[10, 28]], [[2, 28]], [[37, 45]], [[36, 62]], [[46, 62]], [[60, 73]], [[60, 87]], [[91, 134]], [[2, 35]]]",
    "query_spans": "[[[136, 141]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a moving point on the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, $A$ is the left vertex of $C$, and $F$ is the right focus of $C$, then the minimum value of $\\overrightarrow{A P} \\cdot \\overrightarrow{F P}$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P, C);C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);A: Point;LeftVertex(C) = A;F: Point;RightFocus(C) = F",
    "query_expressions": "Min(DotProduct(VectorOf(A, P), VectorOf(F, P)))",
    "answer_expressions": "0",
    "fact_spans": "[[[2, 5]], [[2, 52]], [[6, 48], [57, 60], [69, 72]], [[6, 48]], [[53, 56]], [[53, 64]], [[65, 68]], [[65, 76]]]",
    "query_spans": "[[[78, 133]]]",
    "process": "Let $ P(x_{0},y_{0}) $ with $ -2 \\leqslant x_{0} \\leqslant 2 $, then we have $ \\overrightarrow{AP} = (x_{0}+2, y_{0}) $, $ \\overrightarrow{FP} = (x_{0}-1, y_{0}) $. Substituting into $ \\overrightarrow{AP} \\cdot \\overrightarrow{FP} $, we get $ \\overrightarrow{AP} \\cdot \\overrightarrow{FP} = x_{0}^{2} + x_{0} - 2 + y_{0}^{2} $. Since point $ P $ lies on the ellipse, we have $ y_{0}^{2} = -\\frac{3}{4}x_{0}^{2} + 3 $. Substituting back, $ \\overrightarrow{AP} \\cdot \\overrightarrow{FP} $ becomes a quadratic function in terms of $ x_{0} $, and its extremum can thus be found. From the problem, $ A(-2,0) $, $ F(1,0) $, let $ P(x_{0},y_{0}) $ with $ -2 \\leqslant x_{0} \\leqslant 2 $, then $ \\overrightarrow{AP} = (x_{0}+2, y_{0}) $, $ \\overrightarrow{FP} = (x_{0}-1, y_{0}) $, so $ \\overrightarrow{AP} \\cdot \\overrightarrow{FP} = (x_{0}+2)(x_{0}-1) + y_{0}^{2} = x_{0}^{2} + x_{0} - 2 + y_{0}^{2} $. Since point $ P $ lies on the ellipse, $ \\frac{x_{0}^{2}}{4} + \\frac{y_{0}^{2}}{3} = 1 $, which gives $ y_{0}^{2} = -\\frac{3}{4}x_{0}^{2} + 3 $. Then $ \\overrightarrow{AP} \\cdot \\overrightarrow{FP} = \\frac{1}{4}x_{0}^{2} + x_{0} + 1 $. When $ x_{0} = -2 $, the minimum value of $ \\overrightarrow{AP} \\cdot \\overrightarrow{FP} $ is $ 0 $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$, points $A$ and $B$ lie on the parabola $C$. If $|AB|=6$, then the minimum value of the horizontal coordinate of the midpoint $M$ of segment $AB$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);A: Point;B: Point;PointOnCurve(A, C);PointOnCurve(B, C);Abs(LineSegmentOf(A, B)) = 6;M: Point;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Min(XCoordinate(M))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 21], [31, 37]], [[2, 21]], [[22, 26]], [[27, 30]], [[22, 38]], [[27, 38]], [[40, 49]], [[60, 63]], [[51, 63]]]",
    "query_spans": "[[[60, 73]]]",
    "process": "When the slope of line AB does not exist, it is easy to find that the x-coordinate of M is \\frac{9}{4}; when the slope of line AB exists, assume the equation of the line is y=kx+m(k\\neq0), combine it with the parabola equation to obtain the form of Vieta's formulas, use the chord length formula to express |AB|, derive km=1-\\frac{9k^{4}}{4(1+k^{2})}, substitute into \\frac{x_{1}+x_{2}}{2}, and combine with the basic inequality to find its minimum value; combining both cases gives the final result. Solution: When the slope of line AB does not exist, let its equation be x=t(t>0), then |AB|=4\\sqrt{t}=6, solving yields: t=\\frac{9}{4}, therefore the x-coordinate of midpoint M of AB is \\frac{9}{4}; when the slope of line AB exists, let its equation be: y=kx+m(k\\neq0). Combining \\begin{cases}y=kx+m\\\\y^{2}=4x\\end{cases} gives: k^{2}x^{2}+(2km-4)x+m^{2}=0, \\therefore \\Delta=(2km-4)^{2}-4k^{2}m^{2}=16-16km>0, i.e., km<1, then x_{1}+x_{2}=\\frac{4-2km}{k^{2}}, x_{1}x_{2}=\\frac{m^{2}}{k^{2}}, \\therefore km=1-\\frac{9k^{4}}{4(1+k^{2})}, \\frac{x_{1}+x_{2}}{2}=\\frac{4-2km}{2k^{2}}=\\frac{4-2\\left(1-\\frac{9k^{4}}{4(1+k^{2})}\\right)}{2k^{2}}=\\frac{2+\\frac{9k^{4}}{2(1+k^{2})}}{2k^{2}}=\\frac{1}{k^{2}}+\\frac{9k^{2}}{4(1+k^{2})}, let t=\\frac{1}{k^{2}}>0, then \\frac{x_{1}+x_{2}}{2}=\\frac{4t^{2}+4t+9}{4t+4}=\\frac{4(t+1)^{2}-4(t+1)+9}{4(t+1)}=(t+1)+\\frac{9}{4(t+1)}-1, because t>0, \\therefore t+1>1, \\therefore (t+1)+\\frac{9}{4(t+1)}\\geqslant2\\sqrt{(t+1)\\cdot\\frac{9}{4(t+1)}}=3 (equality holds if and only if t+1=\\frac{9}{4(t+1)}, i.e., t=\\frac{1}{2}), \\therefore \\frac{x_{1}+x_{2}}{2}\\geqslant3-1=2, i.e., the minimum x-coordinate of midpoint M of AB is 2; in summary, the minimum x-coordinate of midpoint M of AB is 2."
  },
  {
    "text": "If the distance from a point $A(4, m)$ on the parabola $y^{2}=2 p x$ to the directrix is $6$, then $m=$?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;Expression(G) = (y^2 = 2*p*x);Coordinate(A) = (4, m);PointOnCurve(A, G);Distance(A,Directrix(G)) = 6;m:Number",
    "query_expressions": "m",
    "answer_expressions": "pm*4*sqrt(2)",
    "fact_spans": "[[[2, 18]], [[5, 18]], [[21, 30]], [[2, 18]], [[21, 30]], [[2, 30]], [[2, 40]], [[43, 46]]]",
    "query_spans": "[[[43, 48]]]",
    "process": ""
  },
  {
    "text": "Given the equation of the ellipse is $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{9}=1$, its two foci are $F_{1}$, $F_{2}$. If $|F_{1} F_{2}|=8$, and chord $AB$ passes through $F_{1}$, then the perimeter of $\\triangle ABF_{2}$ is?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;F2: Point;F1: Point;Expression(G)=(x^2/a^2+y^2/9=1);a:Number;Focus(G)={F1,F2};Abs(LineSegmentOf(F1, F2)) = 8;IsChordOf(LineSegmentOf(A,B),G);PointOnCurve(F1, LineSegmentOf(A, B))",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "20",
    "fact_spans": "[[[2, 4], [48, 49]], [[91, 95]], [[91, 95]], [[63, 70]], [[96, 103], [96, 103]], [[2, 47]], [[8, 47]], [[48, 70]], [[72, 89]], [[48, 95]], [[91, 103]]]",
    "query_spans": "[[[105, 129]]]",
    "process": "Analysis: From |F_{1}F_{2}|=8 \\Rightarrow c=4, and a^{2}=b^{2}+c^{2} \\Rightarrow a=5, it follows that AABF_{2}=4a=20. Specifically, from |F_{1}F_{2}|=8, we have 2c=8, so c=4. Also, a^{2}=b^{2}+c^{2}=25, thus a=5. Therefore, the perimeter of AABF_{2} is 4a=20."
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-y^{2}=-1$, a line $l$ passing through $P(2,1)$ intersects the hyperbola $C$ at points $A$ and $B$, and $P$ is the midpoint of chord $AB$. Then the equation of line $l$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2 = -1);P: Point;Coordinate(P) = (2, 1);l: Line;PointOnCurve(P, l);A: Point;B: Point;Intersection(l, C) = {A, B};IsChordOf(LineSegmentOf(A, B), C);MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Expression(l)",
    "answer_expressions": "2*x-y-3=0",
    "fact_spans": "[[[2, 26], [44, 50]], [[2, 26]], [[29, 37], [63, 66]], [[29, 37]], [[38, 43], [78, 83]], [[28, 43]], [[52, 55]], [[56, 59]], [[38, 61]], [[44, 73]], [[63, 76]]]",
    "query_spans": "[[[78, 88]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ x_{1}+x_{2}=4 $, $ y_{1}+y_{2}=2 $. Since $ A $, $ B $ lie on the hyperbola, $ \\therefore \\begin{cases} x_{1}^{2} \\\\ x_{2}^{2} \\end{cases} x^{2}-y^{2}=-1 $\\textcircled{1}$ 2-y_{2}^{2}=-1 $\\textcircled{2}$ \\textcircled{1}-\\textcircled{2} $ gives: $ (x_{1}+x_{2})(x_{1}-x_{2})=(y_{1}+y_{2})(y_{1}-y_{2}) $, i.e. $ 4(x_{1}-x_{2})=2(y_{1}-y_{2}) $, i.e. $ k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{4}{2}=2' \\therefore AB: y-1=2(x-2) $, namely $ 2x-y-3=0 $, from $ \\begin{cases} y=2x-3 \\\\ x^{2}-y2=-1 \\end{cases} \\Rightarrow 3x^{2}-12x+8=0 $, $ \\because \\triangle>0 $, hence $ 2x-y-3=0 $ intersects the hyperbola at two points satisfying the condition. Therefore, the equation of $ l $ is: $ 2x-y-3=0 $."
  },
  {
    "text": "Given the hyperbola $\\Gamma_{1}$: $x^{2}-\\frac{y^{2}}{b^{2}}=1$, its left and right foci are $F_{1}$ and $F_{2}$ respectively. Taking $O$ as the vertex and $F_{2}$ as the focus, a parabola $\\Gamma_{2}$ is constructed. If the hyperbola $\\Gamma_{1}$ and the parabola $\\Gamma_{2}$ intersect at point $P$, and $\\angle P F_{1} F_{2}=45^{\\circ}$, then the equation of the directrix of the parabola $\\Gamma_{2}$ is?",
    "fact_expressions": "Gamma1: Hyperbola;Expression(Gamma1) = (x^2 - y^2/b^2 = 1);F1: Point;F2: Point;LeftFocus(Gamma1) = F1;RightFocus(Gamma1) = F2;O: Origin;Vertex(Gamma2) = O;Focus(Gamma2) = F2;Gamma2: Parabola;Intersection(Gamma1, Gamma2) = P;P: Point;AngleOf(P, F1, F2) = ApplyUnit(45, degree)",
    "query_expressions": "Expression(Directrix(Gamma2))",
    "answer_expressions": "x=-\\sqrt{2}-1",
    "fact_spans": "[[[2, 48], [108, 123]], [[2, 48]], [[57, 64]], [[65, 72], [80, 87]], [[2, 72]], [[2, 72]], [[74, 77]], [[73, 106]], [[80, 106]], [[91, 106], [124, 139], [182, 197]], [[108, 145]], [[141, 145]], [[147, 180]]]",
    "query_spans": "[[[182, 204]]]",
    "process": "Let the intersection points of the hyperbola $ I_{1} $ have coordinates $ F_{1}(-c,0) $, $ F_{2}(c,0) $, then the equation of the parabola $ I_{2} $ is $ y^{2}=4cx $. Since $ \\angle PF_{1}F_{2}=45^{\\circ} $, the slope of line $ PF $ is 1, so the equation of line $ PF $ is $ y=x+c $. Solving the system  \n\\[\n\\begin{cases}\ny=x+c \\\\\ny^{2}=4cx\n\\end{cases}\n\\]\ngives $ x=c $, $ y=2c $, that is, $ P(c,2c) $. Therefore, $ PF_{2} \\perp F_{1}F_{2} $. From $ \\angle PF_{1}F_{2}=45^{\\circ} $, we obtain $ |PF_{2}| = |F_{1}F_{2}| = 2c $, $ |PF_{1}| = 2\\sqrt{2}c $. According to the definition of a hyperbola, $ |PF_{1}| - |PF_{2}| = 2a $, that is, $ 2\\sqrt{2}c - 2c = 2a $. Solving gives $ c = \\sqrt{2} + 1 $, so the directrix equation of the parabola is $ x = -\\sqrt{2} - 1 $."
  },
  {
    "text": "If the eccentricity of the curve represented by the equation $\\frac{x^{2}}{t-5}+\\frac{y^{2}}{t-1}=1$ is $\\sqrt {2}$, then $t=?$",
    "fact_expressions": "G: Curve;Expression(G)=(x^2/(t-5)+y^2/(t-1)=1);Eccentricity(G)=sqrt(2);t:Number",
    "query_expressions": "t",
    "answer_expressions": "3",
    "fact_spans": "[[[45, 47]], [[1, 47]], [[45, 63]], [[65, 68]]]",
    "query_spans": "[[[65, 70]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ with focus $F$, point $P$ is a point on the parabola $C$. A circle centered at $F$ with radius $p$ intersects $PF$ at point $Q$. A tangent line from point $P$ to the circle $F$ touches the circle at point $A$. If $|P A|=\\sqrt{3} p$ and the area of $\\triangle O P Q$ is $\\frac{\\sqrt{3}}{2}$, then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;F: Point;P: Point;O: Origin;Q: Point;A: Point;G:Circle;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(P, C);Center(G)=F;Radius(G)=p;Intersection(LineSegmentOf(P,F),G)=Q;TangentOfPoint(P,G)=L;L:Line;TangentPoint(L,G)=A;Abs(LineSegmentOf(P, A)) = sqrt(3)*p;Area(TriangleOf(O, P, Q)) = sqrt(3)/2",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 28], [41, 47]], [[62, 65], [165, 168]], [[52, 55], [87, 90]], [[81, 85], [36, 40]], [[122, 139]], [[75, 79]], [[97, 100]], [[66, 67], [86, 90]], [[10, 28]], [[2, 28]], [[2, 35]], [[36, 50]], [[51, 67]], [[59, 67]], [[66, 79]], [[80, 93]], [], [[80, 100]], [[102, 120]], [[122, 163]]]",
    "query_spans": "[[[165, 170]]]",
    "process": "From the given conditions, |PF| = 2p, then Q is the midpoint of segment PF, then the area of \\triangle OPF is \\sqrt{3}, then find the coordinates of point P to establish an equation for solving. Since |PA| = \\sqrt{3}p^{x}, |FA| = p, and PA \\bot FA, it follows that |PF| = 2p. Because |FQ| = p, Q is the midpoint of segment PF. Since the area of \\triangle OPQ is \\frac{\\sqrt{3}}{2}, the area of \\triangle OPF is \\sqrt{3}. From |PF| = 2p = x_{P} + \\frac{p}{2}, we obtain x_{P} = \\frac{3p}{2}, so y_{P} = \\pm\\sqrt{3}p. Therefore, S_{\\triangle OPF} = \\frac{1}{2} \\times \\frac{p}{2} \\times \\sqrt{3}p = \\sqrt{3}, solving gives p = 2."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, and the point where its directrix intersects the $x$-axis is $Q$. A line passing through point $F$ intersects the parabola at points $A$ and $B$. If the circle with diameter $QF$ passes through point $B$, then the value of $|AF|-|BF|$ is?",
    "fact_expressions": "G: Parabola;H: Circle;L: Line;Q: Point;F: Point;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;Intersection(Directrix(G), xAxis) = Q;PointOnCurve(F,L);Intersection(L, G) = {A, B};IsDiameter(LineSegmentOf(Q, F),H);PointOnCurve(B,H)",
    "query_expressions": "Abs(LineSegmentOf(A, F)) - Abs(LineSegmentOf(B, F))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 16], [49, 52], [24, 25]], [[76, 77]], [[46, 48]], [[36, 39]], [[20, 23], [41, 45]], [[54, 57]], [[58, 61], [78, 82]], [[2, 16]], [[2, 23]], [[24, 39]], [[40, 48]], [[46, 63]], [[66, 77]], [[76, 82]]]",
    "query_spans": "[[[84, 101]]]",
    "process": "Suppose $ k $ exists, let the equation of line $ AB $ be: $ y = k(x - 1) $, combining with the parabola $ y^2 = 4x $ gives $ k^2(x^2 - 2x + 1) = 4x $, i.e., $ k^2x^2 - (2k^2 + 4)x + k^2 = 0 $. Let the two intersection points be $ A(x_2, y_2) $, $ B(x_1, y_1) $. Since the circle with diameter $ QF $ passes through point $ B $, $ \\angle QBA = 90^\\circ $, so $ (x_1 - 2)(x_1 + 2) + y_1^2 = 0 $, thus $ x_1^2 + y_1^2 = 4 $, hence $ x_1^2 + 4x_1 - 1 = 0 $ ($ x_1 > 0 $), so $ x_1 = \\sqrt{5} - 2 $, $ \\cdot x_1x_2 = 1 $, therefore $ x_2 = \\sqrt{5} + 2 $, thus $ |AF| - |BF| = (x_2 + 1) - (x_1 + 1) = 4 $."
  },
  {
    "text": "Given that the hyperbola $\\Gamma$ passes through the point $P(2,2)$ and has the same asymptotes as the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$, then the standard equation of the hyperbola $\\Gamma$ is?",
    "fact_expressions": "Gamma: Hyperbola;P: Point;G:Hyperbola;Expression(G) = (x^2/2 - y^2 = 1);Coordinate(P) = (2, 2);PointOnCurve(P, Gamma);Asymptote(G) = Asymptote(Gamma)",
    "query_expressions": "Expression(Gamma)",
    "answer_expressions": "y^2/2 - x^2/4 = 1",
    "fact_spans": "[[[2, 13], [65, 76]], [[15, 24]], [[27, 55]], [[27, 55]], [[15, 24]], [[2, 24]], [[2, 63]]]",
    "query_spans": "[[[65, 83]]]",
    "process": "Let the equation of hyperbola $\\Gamma$ be $\\frac{x^{2}}{2}-y^{2}=\\lambda$. Substitute the coordinates of point $P$ into the equation of hyperbola $\\Gamma$ to find the value of $\\lambda$, thus obtaining the standard equation of hyperbola $T$. [Detailed solution] Since hyperbola $\\Gamma$ has the same asymptotes as the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$, let the equation of hyperbola $T$ be $\\frac{x^{2}}{2}-y^{2}=\\lambda$. Substituting the coordinates of point $P$ into the equation of hyperbola $T$ gives $\\lambda=\\frac{2^{2}}{2}-2^{2}=-2$. Therefore, the equation of hyperbola $T$ is $\\frac{x^{2}}{2}-y^{2}=-2$, which in standard form is $\\frac{y^{2}}{2}-\\frac{x^{2}}{4}=1$."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, the directrix is $l$, and $P$ is a point on the parabola. $PQ$ is perpendicular to the directrix $l$ and intersects $l$ at point $Q$. The chord length intercepted by $l$ on the circle with diameter $FQ$ is $2\\sqrt{3}$. Then the length of $PF$ is?",
    "fact_expressions": "G: Parabola;H: Circle;F: Point;Q: Point;P: Point;l: Line;Expression(G) = (y^2 = 4*x);Focus(G) = F;Directrix(G) = l;PointOnCurve(P,G);IsPerpendicular(LineSegmentOf(P,Q),l);Intersection(LineSegmentOf(P,Q),l)=Q;IsDiameter(LineSegmentOf(F,Q),H);Length(InterceptChord(l,H))=2*sqrt(3)",
    "query_expressions": "Length(LineSegmentOf(P, F))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 16], [31, 34]], [[74, 75]], [[20, 23]], [[59, 63]], [[37, 40]], [[27, 30], [51, 54], [76, 79]], [[2, 16]], [[2, 23]], [[2, 30]], [[31, 40]], [[43, 56]], [[42, 63]], [[65, 75]], [[74, 97]]]",
    "query_spans": "[[[99, 109]]]",
    "process": "The directrix of the parabola is given by $ x = -1 $, and the focus is $ F(1,0) $. Let $ P(x_{0},y_{0}) $, without loss of generality assume $ y_{0} > 0 $, then $ Q(-1,y_{0}) $, the midpoint of $ FQ $ has coordinates $ \\left(0,\\frac{y_{0}}{2}\\right) $. The chord length intercepted on line $ l $ by the circle with $ FQ $ as diameter is $ 2\\sqrt{3} $. Since $ l \\perp y $-axis, we have $ y_{0} = 2\\sqrt{3} $, so $ x_{0} = 3 $, hence $ |PF| = x_{0} + 1 = 3 + 1 = 4 $."
  },
  {
    "text": "Given the parabola $y^{2}=4x$, a line passing through the focus $F$ with slope $k$ intersects the parabola at points $A$ and $B$. Points $A^{\\prime}$ and $B^{\\prime}$ are the feet of the perpendiculars from $A$ and $B$ to the line $x=-1$, respectively. Then, the product of the slopes of lines $A^{\\prime}F$ and $B^{\\prime}F$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Line;F: Point;Focus(G) = F;PointOnCurve(F, H);k: Number;Slope(H) = k;A: Point;B: Point;Intersection(H, G) = {A, B};A1: Point;B1: Point;Z: Line;Expression(Z) = (x = -1);L1: Line;L2: Line;PointOnCurve(A, L1);PointOnCurve(B, L2);IsPerpendicular(L1, Z);IsPerpendicular(L2, Z);FootPoint(L1, Z) = A1;FootPoint(L2, Z) = B1",
    "query_expressions": "Slope(LineOf(A1, F))*Slope(LineOf(B1, F))",
    "answer_expressions": "-1",
    "fact_spans": "[[[2, 16], [35, 38]], [[2, 16]], [[31, 33]], [[20, 23]], [[2, 23]], [[17, 33]], [[27, 30]], [[24, 33]], [[39, 42], [82, 85]], [[43, 46], [86, 89]], [[31, 48]], [[49, 62]], [[64, 76]], [[90, 98]], [[90, 98]], [], [], [[76, 102]], [[76, 102]], [[76, 102]], [[76, 102]], [[49, 105]], [[49, 105]]]",
    "query_spans": "[[[107, 146]]]",
    "process": "Let the line $ l: x = my + 1 $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ A(-1, y_{1}) $, $ B'(-1, y_{2}) $, $ F(1, 0) $, so  \n$$\n\\begin{cases}\nx = my \\\\\ny^{2} = 4\n\\end{cases}\n= my +\n$$\nRearranging gives $ y^{2} - 4my - 4 = 0 $. Therefore, $ y_{1}y_{2} = -4 $. Also,  \n$$\nk_{AF} \\cdot k_{BF} = \\frac{y_{1}}{-2} \\times \\frac{y_{2}}{-2} = \\frac{y_{1}y_{2}}{4} = -1,\n$$\nso $ k_{AF} \\cdot k_{BF} = -1 $."
  },
  {
    "text": "Given two points $A(-2,0)$, $B(2,0)$, lines $AM$ and $BM$ intersect at point $M$, and the product of the slopes of these two lines is $-\\frac{3}{4}$. Then the trajectory equation of point $M$ is?",
    "fact_expressions": "M: Point;A: Point;B: Point;Coordinate(A) = (-2, 0);Coordinate(B) = (2, 0);Intersection(LineOf(A,M),LineOf(B,M))=M;Slope(LineOf(A,M))*Slope(LineOf(B,M))=-3/4",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "(x^2/4+y^2/3=1)&Negation(x=pm*2)",
    "fact_spans": "[[[42, 46], [75, 79]], [[4, 14]], [[16, 24]], [[4, 14]], [[16, 24]], [[25, 46]], [[25, 73]]]",
    "query_spans": "[[[75, 86]]]",
    "process": "Let point $ M(x, y) $. From the product of the slopes of lines $ AM $ and $ BM $ being $ \\frac{y}{x+2} \\cdot \\frac{y}{x-2} = -\\frac{3}{4} $ ($ x \\neq \\pm 2 $), simplifying yields $ 3x^{2} + 4y^{2} = 12 $, or $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $ ($ x \\neq \\pm 2 $). Therefore, the trajectory equation of point $ M $ is $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $ ($ x \\neq \\pm 2 $)."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=4 x$, $M$ is a point on $C$, and the extension of $F M$ intersects the $y$-axis at point $N$. If $F M=\\frac{1}{2} M N$, then $|F N|$=?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;M: Point;PointOnCurve(M, C);N: Point;Intersection(OverlappingLine(LineSegmentOf(F, M)), yAxis) = N;LineSegmentOf(F, M) = (1/2)*LineSegmentOf(M, N)",
    "query_expressions": "Abs(LineSegmentOf(F, N))",
    "answer_expressions": "5",
    "fact_spans": "[[[6, 25], [33, 36]], [[6, 25]], [[2, 5]], [[2, 28]], [[29, 32]], [[29, 39]], [[55, 59]], [[40, 59]], [[62, 83]]]",
    "query_spans": "[[[85, 94]]]",
    "process": "Given $ F(1,0) $, let $ M(x_{0},y_{0}) $, $ N(x,y) $. Then from $ \\overrightarrow{FM} = \\frac{1}{2}\\overrightarrow{MN} $, we have $ (x_{0}-1,y_{0}) = \\frac{1}{2}(x-x_{0},y-y_{0}) \\Rightarrow \\begin{cases} x = 3x_{0}-2 \\\\ y = 3y_{0} \\end{cases} $. According to the problem, $ x = 0 $, then $ x_{0} = \\frac{2}{3} $, $ y_{0} = \\pm\\sqrt{\\frac{8}{3}} = \\pm\\frac{\\sqrt{24}}{3} $, $ y = 3y_{0} = \\pm\\sqrt{24} $. Thus, $ |\\overrightarrow{FN}| = \\sqrt{(1-0)^{2} + (0 \\pm \\sqrt{24})^{2}} = 5 $."
  },
  {
    "text": "The distance from the right focus of the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$ to its asymptote is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/4 = 1)",
    "query_expressions": "Distance(RightFocus(G), Asymptote(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 41]]]",
    "process": "From the hyperbola equation, find the values of a, b, c, then obtain the coordinates of the right focus and the asymptote equations; next, use the point-to-line distance formula to solve. Solution: From $x^{2}-\\frac{y^{2}}{4}=1$, we get $a=1$, $b=2$, so $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{5}$. Thus, the coordinates of the right focus of the hyperbola are $(\\sqrt{5},0)$, and the asymptote equations are $y=\\pm\\frac{b}{a}x=\\pm2x$. Then, the distance from the right focus of the hyperbola to the asymptote is $d=\\frac{|2\\sqrt{5}|}{\\sqrt{1^{2}+2^{2}}}=2$."
  },
  {
    "text": "Given that one focus of the ellipse is $F_{1}(-3 , 0)$, the length of the major axis is $10$, and the center is at the origin, what is the eccentricity of this ellipse?",
    "fact_expressions": "G: Ellipse;F1: Point;Coordinate(F1) = (-3, 0);OneOf(Focus(G)) = F1;Length(MajorAxis(G)) = 10;O: Origin;Center(G) = O",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3/5",
    "fact_spans": "[[[2, 4], [45, 47]], [[10, 25]], [[10, 25]], [[2, 25]], [[2, 34]], [[38, 42]], [[2, 42]]]",
    "query_spans": "[[[45, 53]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, let the right focus be $F$. Draw a perpendicular from $F$ to an asymptote, with foot of perpendicular at $M$, where $M$ lies in the first quadrant. The line segment $MF$ intersects the hyperbola at point $N$. If $\\overrightarrow{MN}=\\frac{1}{2} \\overrightarrow{NF}$, then the eccentricity of the hyperbola equals?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;F: Point;M: Point;N: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;PointOnCurve(F,H);IsPerpendicular(H,OneOf(Asymptote(G)));FootPoint(H,OneOf(Asymptote(G)))=M;Quadrant(M)=1;Intersection(LineSegmentOf(M,F),G) = N;VectorOf(M,N)=VectorOf(N,F)/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 58], [105, 108], [105, 108]], [[5, 58]], [[5, 58]], [], [[63, 66], [68, 71]], [[84, 87], [84, 87]], [[109, 113]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 66]], [[67, 80]], [[2, 80]], [[2, 87]], [[88, 96]], [[97, 113]], [[116, 171]]]",
    "query_spans": "[[[173, 183]]]",
    "process": "Since MF is perpendicular to the asymptote $ y = \\frac{b}{a}x $, the slope of line MF is $ -\\frac{a}{b} $. Thus, the equation of line MF is $ y = -\\frac{a}{b}(x - c) $, and solving this with the asymptote $ y = \\frac{b}{a}x $ gives the intersection point $ M\\left( \\frac{a^{2}}{c}, \\frac{ab}{c} \\right) $. Given $ \\overrightarrow{MN} = \\frac{1}{2}\\overrightarrow{NF} $, we find $ N\\left( \\frac{c}{3} + \\frac{2a^{2}}{3c}, \\frac{2ab}{3c} \\right) $. Since $ N $ lies on the hyperbola, substituting its coordinates into the hyperbola equation yields an equation in $ a, b, c $, from which the eccentricity can be determined. From the given condition, MF is perpendicular to the asymptote $ y = \\frac{b}{a}x $, so the slope of MF is $ -\\frac{a}{b} $. Since $ F(c, 0) $, the equation of line MF is $ y = -\\frac{a}{b}(x - c) $. Solving simultaneously with $ y = \\frac{b}{a}x $:\n$$\n\\begin{cases}\ny = -\\frac{a}{b}(x - c) \\\\\ny = \\frac{b}{a}x\n\\end{cases}\n$$\ngives\n$$\n\\begin{cases}\nx = \\frac{a^{2}}{c} \\\\\ny = \\frac{ab}{c}\n\\end{cases}\n$$\nThus, $ M\\left( \\frac{a^{2}}{c}, \\frac{ab}{c} \\right) $. Using $ \\overrightarrow{MN} = \\frac{1}{2}\\overrightarrow{NF} $, we obtain $ N\\left( \\frac{c}{3} + \\frac{2a^{2}}{3c}, \\frac{2ab}{3c} \\right) $. Since $ N $ lies on the hyperbola,\n$$\n\\frac{\\left( \\frac{c}{3} + \\frac{2a^{2}}{3c} \\right)^{2}}{a^{2}} - \\frac{\\left( \\frac{2ab}{3c} \\right)^{2}}{b^{2}} = 1\n$$\nSimplifying yields $ c^{2} = 5a^{2} $, so $ e = \\frac{c}{a} = \\sqrt{5} $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>b>0)$ has focal length $2 c$, right vertex at point $A$, and the parabola $x^{2}=2 p y(p>0)$ has focus $F$. If the segment length intercepted by the hyperbola on the directrix of the parabola is $2 c$, and $|P A|=c$, then the asymptotes of the hyperbola have equations?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;FocalLength(G) = 2*c;c: Number;A: Point;RightVertex(G) = A;H: Parabola;Expression(H) = (x^2 = 2*p*y);p: Number;p>0;F: Point;Focus(H) = F;Length(InterceptChord(Directrix(H), G)) = 2*c;Abs(LineSegmentOf(P, A)) = c;P: Point",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*x",
    "fact_spans": "[[[2, 57], [105, 108], [139, 142], [139, 142]], [[2, 57]], [[5, 57]], [[5, 57]], [[5, 57]], [[5, 57]], [[2, 66]], [[61, 66]], [[71, 74]], [[2, 74]], [[75, 96], [109, 112]], [[75, 96]], [[78, 96]], [[78, 96]], [[100, 103]], [[75, 103]], [[105, 126]], [[128, 137]], [[128, 137]]]",
    "query_spans": "[[[139, 150]]]",
    "process": ""
  },
  {
    "text": "An equation of a circle whose center lies on the $x$-axis and is tangent to the asymptotes of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ can be?",
    "fact_expressions": "G: Hyperbola;H: Circle;Expression(G) = (x^2/3 - y^2 = 1);PointOnCurve(Center(H),xAxis);IsTangent(Asymptote(G),H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "((x-2*m)^2+y^2=m^2)&Negation(m=0)",
    "fact_spans": "[[[11, 39]], [[48, 49]], [[11, 39]], [[0, 49]], [[10, 49]]]",
    "query_spans": "[[[48, 56]]]",
    "process": "First, find the asymptotes of the hyperbola, then use symmetry to assume the center of the circle, and use the condition that the circle is tangent to the lines to obtain the radius, thereby obtaining the equation of the desired circle. The asymptotes of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ are given by: $y=\\pm\\frac{\\sqrt{3}}{3}x$. To make the circle tangent to both asymptotes, let the center of the circle be $(2m,0)$, $m\\neq0$, then the radius of the circle is: $\\frac{|\\frac{2\\sqrt{3}}{3}m|}{\\sqrt{1+\\frac{1}{3}}}=|m|$. Therefore, the equation of the desired circle is: $(x-2m)^{2}+y^{2}=m^{2}$, $m\\neq0$."
  },
  {
    "text": "Given that the foci of hyperbolas $C_{1}$ and $C_{2}$ lie on the $x$-axis and $y$-axis respectively, their asymptotes are given by $y = \\pm \\frac{1}{a} x$, and their eccentricities are $e_{1}$ and $e_{2}$. Then the minimum value of $e_{1} + e_{2}$ is?",
    "fact_expressions": "C1: Hyperbola;C2: Hyperbola;Expression(Asymptote(C1)) = (y = pm*(1/a)*x);Expression(Asymptote(C2)) = (y = pm*(1/a)*x);PointOnCurve(Focus(C1), xAxis);PointOnCurve(Focus(C2), yAxis);e1: Number;e2: Number;Eccentricity(C1) = e1;Eccentricity(C2) = e2;a:Number",
    "query_expressions": "Min(e1 + e2)",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 12]], [[13, 20]], [[2, 64]], [[2, 64]], [[2, 36]], [[2, 36]], [[71, 78]], [[80, 88]], [[2, 88]], [[2, 88]], [[2, 64]]]",
    "query_spans": "[[[91, 110]]]",
    "process": ""
  },
  {
    "text": "The line $y=k x-2 k$ intersects the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{4}=1$ at two distinct points; then the range of real values for $k$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;k: Real;Expression(G) = (x^2/3 - y^2/4 = 1);Expression(H) = (y = k*x - 2*k);NumIntersection(H, G) = 2",
    "query_expressions": "Range(k)",
    "answer_expressions": "Negation(k=pm*(2*sqrt(3)/3))",
    "fact_spans": "[[[14, 52]], [[0, 13]], [[62, 67]], [[14, 52]], [[0, 13]], [[0, 60]]]",
    "query_spans": "[[[62, 74]]]",
    "process": ""
  },
  {
    "text": "Ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ has its right vertex at $A$, and $P$ is a point on ellipse $C$, $O$ is the coordinate origin. Given that $\\angle P O A=60^{\\circ}$, and $O P \\perp A P$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;A: Point;RightVertex(C) = A;P: Point;PointOnCurve(P, C);O: Origin;AngleOf(P, O, A) = ApplyUnit(60, degree);IsPerpendicular(LineSegmentOf(O, P), LineSegmentOf(A, P))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(2*sqrt(5))/5",
    "fact_spans": "[[[0, 57], [70, 75], [135, 140]], [[0, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[62, 65]], [[0, 65]], [[66, 69]], [[66, 78]], [[79, 82]], [[91, 116]], [[118, 133]]]",
    "query_spans": "[[[135, 146]]]",
    "process": "From the given conditions, |OP| = |OA| \\cos 60^{\\circ} = \\frac{a}{2}, it is easy to obtain P\\left(\\frac{1}{4}a, \\frac{\\sqrt{3}}{4}a\\right); substituting into the ellipse equation yields: \\frac{1}{16} + \\frac{3a^{2}}{16b^{2}} = 1, hence a^{2} = 5b^{2} = 5(a^{2} - c^{2}), so the eccentricity e = \\frac{2\\sqrt{5}}{5}. Key point: geometric properties of ellipses and eccentricity."
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and a line $l$ passing through $P$ intersects the circle $x^{2}+y^{2}=4$ at points $A$ and $B$, then the maximum value of $|P A| \\cdot|P B|$ is?",
    "fact_expressions": "P: Point;G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);PointOnCurve(P, G);l: Line;PointOnCurve(P, l);H: Circle;Expression(H) = (x^2 + y^2 = 4);A: Point;B: Point;Intersection(l, H) = {A, B}",
    "query_expressions": "Max(Abs(LineSegmentOf(P, A))*Abs(LineSegmentOf(P, B)))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 5], [39, 42]], [[6, 33]], [[6, 33]], [[2, 37]], [[43, 48]], [[38, 48]], [[49, 65]], [[49, 65]], [[66, 69]], [[70, 73]], [[43, 75]]]",
    "query_spans": "[[[77, 101]]]",
    "process": "As shown in the figure, draw $ OC \\perp AB $ from $ O $, with the foot of the perpendicular at $ C $. It follows that $ C $ is the midpoint of $ AB $, so $ |PA| \\cdot |PB| = |AC|^{2} - |PC|^{2} $. Then, by the Pythagorean theorem, $ |AC|^{2} - |PC|^{2} = 4 - |OP|^{2} $. The maximum and minimum values can be found using the boundedness of the ellipse. As shown in the figure, draw $ OC \\perp AB $ from $ O $, with the foot of the perpendicular at $ C $. Since $ C $ is the midpoint of $ AB $, we have $ |PA| \\cdot |PB| = (|AC| - |PC|)(|BC| + |PC|) = (|AC| - |PC|)(|AC| + |PC|) = |AC|^{2} - |PC|^{2} $. In right triangle $ \\triangle PCO $, $ |OC|^{2} + |PC|^{2} = |OP|^{2} $. In right triangle $ \\triangle ACO $, $ |OC|^{2} + |AC|^{2} = |OA|^{2} = 4 $. Solving these equations together gives $ |AC|^{2} - |PC|^{2} = 4 - |OP|^{2} $. Let $ P(x, y) $, then $ \\frac{x^{2}}{4} + y^{2} = 1 $ ($ -2 \\leqslant x \\leqslant 2 $). Therefore, $ |OP|^{2} = x^{2} + y^{2} = x^{2} + 1 - \\frac{x^{2}}{4} = \\frac{3}{4}x^{2} + 1 $, so $ 1 \\leqslant |OP|^{2} \\leqslant 4 $, and thus $ 0 \\leqslant 4 - |OP|^{2} \\leqslant 3 $, which means $ 0 \\leqslant |PA| \\cdot |PB| \\leqslant 3 $. Hence, the maximum value of $ |PA| \\cdot |PB| $ is $ 3 $."
  },
  {
    "text": "Given that the length of the major axis of an ellipse is $3$ times the length of its minor axis and it passes through the point $P(3,0)$, then the standard equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;P: Point;Coordinate(P) = (3, 0);Length(MajorAxis(G))=3*Length(MinorAxis(G));PointOnCurve(P,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/9+y^2=1,y^2/81+x^2/9=1}",
    "fact_spans": "[[[2, 4], [31, 33]], [[19, 28]], [[19, 28]], [[2, 16]], [[2, 28]]]",
    "query_spans": "[[[31, 40]]]",
    "process": "If the foci of the ellipse are on the x-axis. \\because the ellipse passes through point P(3,0) \\therefore a=3, and the major axis length is 3 times the minor axis length, \\therefore b=1, \\therefore the equation of the ellipse at this time is: \\frac{x^{2}}{9}+y^{2}=1; if the foci of the ellipse are on the y-axis, then b=3, similarly we get a=9, \\therefore the equation of the ellipse is \\frac{y^{2}}{81}+\\frac{x^{2}}{9}=1."
  },
  {
    "text": "If the distance from the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ to one of its asymptotes is $\\frac{a}{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Distance(RightFocus(G),OneOf(Asymptote(G)))=a/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[1, 57], [88, 91]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 86]]]",
    "query_spans": "[[[88, 97]]]",
    "process": "From the given condition, the distance from the right focus $(c,0)$ to one of the asymptotes $y=\\frac{b}{a}x$ is $\\frac{|bc|}{\\sqrt{a^{2}+b^{2}}}=\\frac{a}{2}$. Simplifying yields $\\frac{b}{a}=\\frac{1}{2}$, then the eccentricity $e=\\sqrt{1+(\\frac{b}{a})^{2}}=\\frac{\\sqrt{5}}{2}$."
  },
  {
    "text": "If a point $P$ on the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{7}=1$ is at a distance of $1$ from the right focus, then what is the distance from point $P$ to the origin?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/7 = 1);P: Point;PointOnCurve(P, G);Distance(P, RightFocus(G)) = 1;O: Origin",
    "query_expressions": "Distance(P, O)",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 39]], [[1, 39]], [[42, 45], [58, 62]], [[1, 45]], [[1, 56]], [[63, 65]]]",
    "query_spans": "[[[58, 70]]]",
    "process": "According to the problem, $ a=3 $, $ c=4 $, $ c-a=1 $, hence $ P $ is the right vertex, and the distance to the origin is $ a=3 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A perpendicular is drawn from $F_{1}$ to one asymptote of the hyperbola, with foot of perpendicular at $P$. If $|P F_{2}|=\\frac{\\sqrt{39}}{3} a$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F2: Point;F1: Point;a>0;b>0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;L:Line;PointOnCurve(F1, L);IsPerpendicular(L, OneOf(Asymptote(G)));FootPoint(L, OneOf(Asymptote(G)))=P;Abs(LineSegmentOf(P, F2)) = a*(sqrt(39)/3)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 58], [92, 95], [148, 151]], [[5, 58]], [[5, 58]], [[107, 110]], [[75, 82]], [[67, 74], [84, 91]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 82]], [[2, 82]], [], [[83, 103]], [[83, 103]], [[83, 110]], [[112, 145]]]",
    "query_spans": "[[[148, 157]]]",
    "process": "From the given conditions, one asymptote of the curve is given by $ bx + ay = 0 $. Using the point-to-line distance formula, the value of $ |F_{1}P| $ can be found, and then the values of $ |OP| $ and the cosine of $ \\angle POF_{1} $ can be determined. Since $ \\angle POF_{1} = \\pi - \\angle POF_{2} $, applying the cosine theorem in $ \\triangle POF_{2} $ allows us to find the relationship between $ a $ and $ c $. [Detailed Solution] Analysis: From the given conditions, $ F_{1}(-c, 0) $, and one asymptote of the hyperbola is $ bx + ay = 0 $. Then $ |F_{1}P| = \\frac{|-bc|}{\\sqrt{a^{2}+b^{2}}} = b $. Let $ O $ be the coordinate origin, then $ |F_{1}O| = c $, so $ |OP| = a $, $ \\cos\\angle POF_{1} = \\frac{a}{c} $. In $ \\triangle PF_{2}O $, $ |F_{2}O| = c $. By the cosine theorem, $ |F_{2}P|^{2} = c^{2} + a^{2} - 2ac \\cdot \\cos\\angle POF_{2} = c^{2} + a^{2} - 2ac \\cdot \\cos(\\pi - \\angle POF_{1}) $. Therefore, $ \\frac{39}{9}a^{2} = c^{2} + a^{2} + 2ac \\cdot \\frac{a}{c} $, simplifying yields $ \\frac{4}{3}a^{2} = c^{2} $, so $ e = \\frac{c}{a} = \\frac{2\\sqrt{3}}{3} $."
  },
  {
    "text": "Given point $M(-\\frac{1}{2}, 1)$ and parabola $C$: $y^{2}=2 x$, a line passing through the focus of $C$ with slope $k$ intersects $C$ at points $A$ and $B$. If $\\angle A M B=90^{\\circ}$, then $k=?$",
    "fact_expressions": "C: Parabola;G: Line;M: Point;A: Point;B: Point;Expression(C) = (y^2 = 2*x);Coordinate(M) = (-1/2, 1);PointOnCurve(Focus(C),G);Slope(G)=k;k:Number;Intersection(G,C)={A,B};AngleOf(A,M,B)=ApplyUnit(90,degree)",
    "query_expressions": "k",
    "answer_expressions": "1",
    "fact_spans": "[[[24, 43], [63, 66], [46, 49]], [[60, 62]], [[2, 23]], [[68, 71]], [[72, 75]], [[24, 43]], [[2, 23]], [[45, 62]], [[53, 62]], [[56, 59], [105, 108]], [[60, 77]], [[78, 103]]]",
    "query_spans": "[[[105, 110]]]",
    "process": "Let the equation of line AB be $ y = k(x - \\frac{1}{2}) $, and combine it with the parabola equation  \n$$\n\\begin{cases}\ny = k(x - \\frac{1}{2}) \\\\\ny^2 = 2x\n\\end{cases}\n$$\nAccording to $ \\angle AMB = 90^{\\circ} $, solve by $ \\overrightarrow{MA} \\cdot \\overrightarrow{MB} = 0 $.  \nSince the focus of the parabola $ C: y^2 = 2x $ is $ F(\\frac{1}{2}, 0) $, let the equation of line AB be $ y = k(x - \\frac{1}{2}) $. Combining with the parabola equation  \n$$\n\\begin{cases}\ny = k(x - \\frac{1}{2}) \\\\\ny^2 = 2x\n\\end{cases}\n$$\nwe obtain $ k^2x^2 - (k^2 + 2)x + \\frac{1}{4}k^2 = 0 $. Let $ A(x_1, y_1) $, $ B(x_2, y_2) $, then  \n$ x_1 + x_2 = \\frac{k^2 + 2}{k^2} $, $ x_1 \\cdot x_2 = \\frac{1}{4} $.  \nThus,  \n$ y_1 + y_2 = k(x_1 + x_2 - 1) = \\frac{2}{k} $,  \n$ y_1 \\cdot y_2 = k^2(x_1 \\cdot x_2 - (x_1 + x_2) + \\frac{1}{4}) = -1 $,  \nso  \n$ \\overrightarrow{MA} = (x_1 + \\frac{1}{2}, y_1 - 1) $, $ \\overrightarrow{MB} = (x_2 + \\frac{1}{2}, y_2 - 1) $.  \nSince $ \\angle AMB = 90^{\\circ} $, we have  \n$ \\overrightarrow{MA} \\cdot \\overrightarrow{MB} = x_1 \\cdot x_2 + \\frac{1}{2}(x_1 + x_2) + \\frac{1}{4} + y_1 \\cdot y_2 - (y_1 + y_2) + 1 = \\frac{1}{4} + \\frac{k^2 + 2}{2k^2} + \\frac{1}{4} - 1 - \\frac{2}{k} + 1 = 0 $,  \nwhich simplifies to $ k^2 - 2k + 1 = 0 $, solving gives $ k = 1 $."
  },
  {
    "text": "Given that the ellipse $\\frac{x^{2}}{3 m^{2}}+\\frac{y^{2}}{5 n^{2}}=1$ and the hyperbola $\\frac{x^{2}}{2 m^{2}}-\\frac{y^{2}}{3 n^{2}}=1$ have common foci, what is the eccentricity of the hyperbola?",
    "fact_expressions": "G: Hyperbola;m: Number;n: Number;H: Ellipse;Expression(G) = (-y^2/(3*n^2) + x^2/(2*m^2) = 1);Expression(H) = (y^2/(5*n^2) + x^2/(3*m^2) = 1);Focus(H) = Focus(G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(19)/4",
    "fact_spans": "[[[52, 102], [111, 114]], [[4, 51]], [[4, 51]], [[2, 51]], [[52, 102]], [[2, 51]], [[2, 108]]]",
    "query_spans": "[[[111, 120]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ passes through the point $(1, \\sqrt{2})$, and the range of its minor axis length is $[3, \\sqrt{11}]$, then what is the range of the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;H: Point;Coordinate(H) = (1, sqrt(2));PointOnCurve(H, G) = True;Range(Length(MinorAxis(G))) = [3, sqrt(11)]",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[1/2, sqrt(3)/2]",
    "fact_spans": "[[[2, 54], [72, 73], [100, 102]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[55, 71]], [[55, 71]], [[2, 71]], [[72, 98]]]",
    "query_spans": "[[[100, 112]]]",
    "process": "According to the problem, the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ passes through the point $(1,\\sqrt{2})$, then $\\frac{1}{a^{2}}+\\frac{2}{b^{2}}=1$. The range of the minor axis length $2b$ is $[3,\\sqrt{11}]$, so $b^{2}\\in[\\frac{9}{4},\\frac{11}{4}]$, thus $e=\\frac{c}{a}=\\sqrt{\\frac{a^{2}-b^{2}}{a^{2}}}=\\sqrt{1-\\frac{b^{2}}{a^{2}}}=\\sqrt{1-(1-\\frac{2}{b^{2}})b^{2}}=\\sqrt{3-b^{2}}\\in[\\frac{1}{2},\\frac{\\sqrt{3}}{2}]$"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(b>a>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $P$ is an arbitrary point on the left branch of the hyperbola. When the maximum value of $\\frac{2|P F_{1}|}{|P F_{2}|^{2}}$ is $\\frac{1}{4 a}$, what is the range of values for the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;b > a;a > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, LeftPart(G));Max(2*Abs(LineSegmentOf(P, F1))/Abs(LineSegmentOf(P, F2))^2) = 1/(4*a)",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(\\sqrt{2}, 3]",
    "fact_spans": "[[[2, 55], [84, 87], [152, 155]], [[2, 55]], [[5, 55]], [[5, 55]], [[5, 55]], [[5, 55]], [[64, 71]], [[72, 79]], [[2, 79]], [[2, 79]], [[80, 83]], [[80, 94]], [[96, 149]]]",
    "query_spans": "[[[152, 166]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $O$ the origin, $P$ a point on $C$ such that $|O P|=|O F_{1}|$, and $P F_{2}$ parallel to one asymptote of $C$. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;P: Point;F2: Point;O: Origin;F1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Focus(C) = {F1, F2};PointOnCurve(P, C);Abs(LineSegmentOf(O, P)) = Abs(LineSegmentOf(O, F1));IsParallel(LineSegmentOf(P, F2), OneOf(Asymptote(C)))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[17, 78], [96, 99], [130, 133], [143, 146]], [[25, 78]], [[25, 78]], [[92, 95]], [[9, 16]], [[83, 86]], [[1, 8]], [[25, 78]], [[25, 78]], [[17, 78]], [[1, 82]], [[92, 100]], [[101, 118]], [[120, 141]]]",
    "query_spans": "[[[143, 152]]]",
    "process": "By the given condition, without loss of generality, assume that point $ P $ lies in the first quadrant. As shown in the figure, since $ |F_{1}F_{2}| = 2|OF_{1}| = 2|OP| $, the three points $ P, F_{1}, F_{2} $ lie on a circle centered at the origin $ O $ with radius $ |OP| $. Therefore, $ \\angle F_{1}PF_{2} = 90^{\\circ} $, so $ |PF_{1}|^{2} + |PF_{2}|^{2} = 4c^{2} $. Since $ PF_{2} $ is parallel to an asymptote of hyperbola $ C $, we have $ \\tan\\angle F_{1}F_{2}P = \\frac{b}{a} $. Given $ |F_{1}F_{2}| = 2c $, it follows that $ |PF_{1}| = 2b $, $ |PF_{2}| = 2a $. By the definition of a hyperbola, $ |PF_{1}| - |PF_{2}| = 2b - 2a = 2a $, thus $ |PF_{1}| = 4a $, $ |PF_{2}| = 2a $. Substituting into $ |PF_{1}|^{2} + |PF_{2}|^{2} = 4c^{2} $, we get $ 16a^{2} + 4a^{2} = 4c^{2} $, so $ e = \\sqrt{\\frac{c^{2}}{a^{2}}} = \\sqrt{5} $."
  },
  {
    "text": "The equation of hyperbola $C$ is $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, $l_{1}$, $l_{2}$ are its asymptotes, $F$ is the right focus. Draw a line $l\\|l_{2}$ through $F$, and let $l$ intersect hyperbola $C$ at $R$ and intersect $l_{1}$ at $M$. If $\\overrightarrow{F R}=\\lambda \\overrightarrow{F M}$, and $\\lambda \\in(\\frac{1}{2}, \\frac{3}{4})$, then the range of the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;F: Point;R: Point;M: Point;l1:Line;l2:Line;l:Line;lambda:Number;a:Number;b:Number;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a>0;b>0;Asymptote(C) = {l1,l2};RightFocus(C) = F;PointOnCurve(F, l);IsParallel(l,l2) = True;Intersection(l, C) = R;Intersection(l,l1) = M;VectorOf(F, R) = lambda*VectorOf(F, M);In(lambda,(1/2,3/4))",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(\\sqrt{2},2)",
    "fact_spans": "[[[0, 6], [116, 122], [83, 84], [236, 239]], [[88, 91], [97, 100]], [[123, 126]], [[136, 139]], [[65, 73], [128, 135]], [[74, 82]], [[112, 115]], [[196, 235]], [[10, 63]], [[10, 63]], [[0, 63]], [[10, 63]], [[10, 63]], [[66, 87]], [[83, 95]], [[96, 111]], [[101, 111]], [[112, 126]], [[112, 139]], [[142, 193]], [[196, 235]]]",
    "query_spans": "[[[236, 250]]]",
    "process": "According to the asymptotes, solve for $ M\\left(\\frac{c}{2}, -\\frac{bc}{2a}\\right) $. Let $ R(x, y) $, and given $ \\overrightarrow{FR} = \\lambda \\overrightarrow{FM} $, solve to get $ \\begin{cases} x = c - \\frac{c}{2} \\\\ y = -\\frac{bc}{2a} \\end{cases} $, substitute into the hyperbola equation and simplify to obtain $ e^{2} = \\frac{1}{1-2} \\in (2,4) $, yielding the answer. [Detailed Solution] The asymptotes of the hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ are: $ y = \\pm \\frac{b}{a}x $. Without loss of generality, assume $ l_{1}: y = -\\frac{b}{a}x $, $ l_{2}: y = \\frac{b}{a}x $, then $ l: y = \\frac{b}{a}(x - c) $. Solving $ l $ and $ l_{1} $ simultaneously gives $ M\\left(\\frac{c}{2}, -\\frac{bc}{2a}\\right) $. Let $ R(x, y) $, $ \\overrightarrow{FR} = \\lambda \\overrightarrow{FM} $, hence $ (x - c, y) = \\lambda\\left(-\\frac{c}{2}, -\\frac{bc}{2a}\\right) $, thus $ \\begin{cases} x = c - \\frac{c\\lambda}{2} \\\\ y = -\\frac{bc}{2a} \\end{cases} $. Substituting into the hyperbola equation yields: $ \\frac{\\left(c - \\frac{c\\lambda}{2}\\right)^{2}}{a^{2}} - \\frac{\\left(-\\frac{bc\\lambda}{2a}\\right)^{2}}{b^{2}} = 1 $, simplifying gives $ e^{2} = \\frac{1}{1 - \\frac{\\lambda^{2}}{2}} \\in (2,4) $, hence $ e \\in (\\sqrt{2}, 2) $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ with two foci $F_{1}$, $F_{2}$, and $P$ a moving point on the ellipse. If $AB$ is a diameter of the circle centered at point $P$ with radius $1$, then the range of values of $\\overrightarrow{F_{1} A} \\cdot \\overrightarrow{F_{1} B}+\\overrightarrow{F_{2} A} \\cdot \\overrightarrow{F_{2} B}$ is?",
    "fact_expressions": "G: Ellipse;H: Circle;A: Point;B: Point;F1: Point;F2: Point;P: Point;Expression(G) = (x^2/9 + y^2/4 = 1);Focus(G)={F1,F2};PointOnCurve(P,G);Center(H)=P;Radius(H)=1;IsDiameter(LineSegmentOf(A,B),H)",
    "query_expressions": "Range(DotProduct(VectorOf(F1, A), VectorOf(F1, B)) + DotProduct(VectorOf(F2, A), VectorOf(F2, B)))",
    "answer_expressions": "[16,26]",
    "fact_spans": "[[[2, 39], [65, 67]], [[95, 96]], [[73, 78]], [[73, 78]], [[45, 52]], [[53, 60]], [[61, 64], [80, 84]], [[2, 39]], [[2, 60]], [[61, 71]], [[79, 96]], [[88, 96]], [[73, 101]]]",
    "query_spans": "[[[103, 223]]]",
    "process": "From the linear operations of vectors, we obtain $\\overrightarrow{F_{1}A}\\cdot\\overrightarrow{F_{1}B}+\\overrightarrow{F_{2}A}\\cdot\\overrightarrow{F_{2}B}=PF_{1}^{2}+PF_{2}^{2}-2=|PF_{1}|^{2}+|PF_{2}|^{2}-2$. Combining the definition of the ellipse, we get $|PF_{1}|^{2}+|PF_{2}|^{2}-2=2(|PF_{1}|-3)^{2}+16$. Then, using the geometric properties of the ellipse, we obtain $|PF_{1}|\\in[3-\\sqrt{5},3+\\sqrt{5}]$. Combining this with the method for finding the range of a quadratic function yields the solution. [Detailed explanation] From the given conditions, we have $|\\overrightarrow{PA}|=|\\overrightarrow{PB}|=1$ and $\\overrightarrow{PA}=-\\overrightarrow{PB}$. Then,  \n$\\overrightarrow{F_{1}A}\\cdot\\overrightarrow{F_{1}B}=(\\overrightarrow{F_{1}P}+\\overrightarrow{PA})\\cdot(\\overrightarrow{F_{1}P}+\\overrightarrow{PB})=\\overrightarrow{F_{1}P}^{2}+\\overrightarrow{F_{1}P}\\cdot(\\overrightarrow{PA}+\\overrightarrow{PB})+\\overrightarrow{PA}\\cdot\\overrightarrow{PB}=\\overrightarrow{F_{1}P}^{2}-1$.  \nSimilarly, $\\overrightarrow{F_{2}A}\\cdot\\overrightarrow{F_{2}B}=\\overrightarrow{F_{2}P}^{2}-1$. Therefore,  \n$\\overrightarrow{F_{1}A}\\cdot\\overrightarrow{F_{1}B}+\\overrightarrow{F_{2}A}\\cdot\\overrightarrow{F_{2}B}=|\\overrightarrow{F_{1}P}|^{2}+|\\overrightarrow{F_{2}P}|^{2}-2=|PF_{1}|^{2}+|PF_{2}|^{2}-2$.  \nFrom the definition of the ellipse, we have $|PF_{1}|+|PF_{2}|=6$, so  \n$|PF_{1}|^{2}+|PF_{2}|^{2}-2=|PF_{1}|^{2}+(6-|PF_{1}|)^{2}-2=2(|PF_{1}|-3)^{2}+16$.  \nUsing the geometric properties of the ellipse, we obtain $|PF_{1}|\\in[3-\\sqrt{5},3+\\sqrt{5}]$, which implies $2(|PF_{1}|-3)^{2}+16\\in[16,26]$. Thus, the range of $\\overrightarrow{F_{1}A}\\cdot\\overrightarrow{F_{1}B}+\\overrightarrow{F_{2}A}\\cdot\\overrightarrow{F_{2}B}$ is $[16,26]$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, a line $l$ with slope $k$ passing through the point $P(2,0)$ intersects the ellipse at two distinct points $A$ and $B$, and $O$ is the origin. Then the maximum area $S$ of $\\triangle AOB$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);P: Point;Coordinate(P) = (2, 0);l: Line;PointOnCurve(P, l);k: Number;Slope(l) = k;A: Point;B: Point;Intersection(l, G) = {A, B};Negation(A=B);O: Origin;S: Number;Area(TriangleOf(A, O, B)) = S",
    "query_expressions": "Max(S)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 29], [54, 56]], [[2, 29]], [[31, 40]], [[31, 40]], [[48, 53]], [[30, 53]], [[44, 47]], [[41, 53]], [[63, 66]], [[67, 70]], [[48, 70]], [[58, 70]], [[71, 74]], [[100, 103]], [[81, 103]]]",
    "query_spans": "[[[100, 109]]]",
    "process": "Obviously, the line is not perpendicular to the y-axis; let its equation be $x = my + 2$. From  \n\\[\n\\begin{cases}\nx = my + 2 \\\\\nx^{2} + 2y^{2} = 2\n\\end{cases}\n\\]  \neliminating $x$ and simplifying yields:  \n\\[\n\\begin{cases}\n(m^{2}+2)y^{2} + 4my + 2 = 0, \\\\\na > 0 \\quad y = m - 2m \\\\\ny_{1}y_{2} = \\frac{-4m}{m^{2}+2}, \\\\\ny_{1}y_{2} = \\frac{2}{m^{2}+2}\n\\end{cases}\n\\]\nThen $|y_{1} - y_{2}| = \\sqrt{(y_{1} + y_{2})^{2} - 4y_{1}y_{2}} = \\frac{\\sqrt{8m^{2} - 16}}{m^{2} + 2}$. Thus,  \n\\[\nS = \\frac{1}{2}|OP||y_{1} - y_{2}| = |y_{1} - y_{2}| = \\frac{\\sqrt{8m^{2} - 16}}{m^{2} + 2} = \\frac{2\\sqrt{2}\\sqrt{m^{2} - 2}}{m^{2} + 2} = \\frac{\\sqrt{m^{2} - 2}}{\\sqrt{2}} \\cdot \\frac{2\\sqrt{2}}{m^{2} - 2} = \\frac{2}{\\sqrt{m^{2} - 2}},\n\\]  \nequality holds when $\\sqrt{m^{2} - 2} = \\frac{2}{\\sqrt{m^{2} - 2}}$, i.e., $m = \\pm\\sqrt{6}$, so the maximum area $S$ of $\\triangle AOB$ is $\\underline{\\sqrt{2}}$."
  },
  {
    "text": "Given that point $F$ is the focus of the parabola $x^{2}=4 y$, point $M(1,2)$, and point $P$ is an arbitrary point on the parabola, then the minimum value of $|P M|+|P F|$ is?",
    "fact_expressions": "G: Parabola;M: Point;P: Point;F: Point;Expression(G) = (x^2 = 4*y);Coordinate(M) = (1, 2);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F)) + Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "3",
    "fact_spans": "[[[7, 21], [40, 43]], [[25, 34]], [[35, 39]], [[2, 6]], [[7, 21]], [[25, 34]], [[2, 24]], [[35, 49]]]",
    "query_spans": "[[[51, 70]]]",
    "process": "As shown in the figure, draw a perpendicular from point P to the directrix of the parabola $ y = -1 $, with foot of perpendicular at Q. Connect MQ. Then $ |PM| + |PF| = |PM| + |PQ| \\geqslant |MQ| \\geqslant 2 + 1 = 3 $. The equality holds if and only if M, P, Q are collinear. Hence, the minimum value of $ |PM| + |PF| $ is 3."
  },
  {
    "text": "Given points $A$ and $B$ on the parabola $C$: $y^{2}=4x$, distinct from the origin $O$, and $OA \\perp OB$, then the minimum area of $\\triangle OAB$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);A: Point;B: Point;PointOnCurve(A, C);PointOnCurve(B, C);O: Origin;Negation(O=A);Negation(O=B);IsPerpendicular(LineSegmentOf(O, A), LineSegmentOf(O, B))",
    "query_expressions": "Min(Area(TriangleOf(O, A, B)))",
    "answer_expressions": "16",
    "fact_spans": "[[[11, 30]], [[11, 30]], [[2, 6]], [[7, 10]], [[2, 42]], [[2, 42]], [[34, 39]], [[2, 42]], [[2, 42]], [[44, 59]]]",
    "query_spans": "[[[61, 87]]]",
    "process": ""
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, the directrix is $l$, $P$ is a point on the parabola, $PA \\perp l$ at $A$, if the inclination angle of the line $AF$ is $120^{\\circ}$, then $|PA|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;l: Line;Directrix(G) = l;P: Point;PointOnCurve(P, G);A: Point;IsPerpendicular(LineSegmentOf(P, A), l);FootPoint(LineSegmentOf(P, A), l) = A;Inclination(LineOf(A, F)) = ApplyUnit(120, degree)",
    "query_expressions": "Abs(LineSegmentOf(P, A))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 16], [37, 40]], [[2, 16]], [[20, 23]], [[2, 23]], [[27, 30]], [[2, 30]], [[33, 36]], [[33, 43]], [[58, 61]], [[44, 57]], [[44, 61]], [[63, 88]]]",
    "query_spans": "[[[91, 100]]]",
    "process": "As shown in the figure, let the directrix of the parabola intersect the x-axis at point E, connect PF, with point F(1,0) and line l: x = -1. Since the inclination angle of line AF is 120^{\\circ}, it follows that \\angle AFE = 60^{\\circ}. Also, PA \\bot l at A, so PA // x-axis, yielding \\angle PAF = 60^{\\circ}. By the definition of the parabola, |PF| = |PA|, thus \\triangle PAF is an equilateral triangle, so |PA| = |AF| = 2|EF| = 4, therefore |PA| = 4."
  },
  {
    "text": "In $\\triangle A B C$, $A B=8$, $A C=6$, $\\angle B A C=90^{\\circ}$, with $B$ as one focus, an ellipse is constructed such that the other focus lies on side $A C$, and the ellipse passes through points $A$ and $C$. Then the eccentricity of this ellipse is?",
    "fact_expressions": "G: Ellipse;A: Point;C: Point;B: Point;LineSegmentOf(A, B) = 8;LineSegmentOf(A, C) = 6;AngleOf(B, A, C) = ApplyUnit(90, degree);OneOf(Focus(G)) = B;Z: Point;OneOf(Focus(G)) = Z;PointOnCurve(Z, LineSegmentOf(A, C));PointOnCurve(A, G);PointOnCurve(C, G);Negation(B=Z)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[76, 78], [82, 84], [100, 102], [115, 117]], [[103, 106]], [[107, 110]], [[67, 70]], [[20, 28]], [[30, 37]], [[40, 65]], [[66, 78]], [], [[82, 90]], [[66, 98]], [[100, 112]], [[100, 112]], [[66, 90]]]",
    "query_spans": "[[[115, 123]]]",
    "process": "As shown in the figure, denote the other focus as D, then $\\triangle ABD$ is also a right triangle. $\\because AB=8$, $AC=6$, $\\angle CAB=90^{\\circ}$, $\\therefore BC=10$. By the definition of ellipse: $AB+AD=CB+CD=\\frac{1}{2}(AB+BC+CA)=\\frac{1}{2}(8+6+10)=12$. $\\therefore$ the major axis length of the ellipse $2a=12$, $\\therefore a=6$. Let the focal distance of the ellipse be $2c$, i.e., $BD=2c$. By the definition of ellipse: $AD=2a-AB=12-8=4$. Also $\\because AD=\\sqrt{BD^{2}-AB^{2}}=\\sqrt{4c^{2}-8^{2}}$, $\\therefore 4=\\sqrt{4c^{2}-64}$, solving gives $c=2\\sqrt{5}$. $\\therefore$ eccentricity $e=\\frac{c}{a}=\\frac{\\sqrt{5}}{3}$. Hence fill: $\\frac{\\sqrt{5}}{3}$. [Note] This problem examines finding the eccentricity of an ellipse. When the condition that a point P lies on the ellipse appears in the problem, pay attention to using the definition $|PF_{1}|+|PF_{2}|=2a$; when finding the eccentricity of an ellipse, the key is to obtain the values of $a$, $c$ or establish a relationship between $a$ and $c$."
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the hyperbola $x^{2}-\\frac{y^{2}}{24}=1$, and let $P$ be a common point of this hyperbola and the ellipse $\\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1$. Then $\\angle F_{1} P F_{2}$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/24 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};H: Ellipse;Expression(H) = (x^2/49 + y^2/24 = 1);P: Point;OneOf(Intersection(G, H)) = P",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "pi/2",
    "fact_spans": "[[[17, 46], [57, 60]], [[17, 46]], [[1, 8]], [[9, 16]], [[1, 51]], [[61, 100]], [[61, 100]], [[52, 55]], [[52, 106]]]",
    "query_spans": "[[[108, 132]]]",
    "process": "Let P be the intersection point of the ellipse and hyperbola in the first quadrant, let PF_{1}=m, PF_{2}=n, then m+n=14, m-n=2, \\therefore m^{2}+n^{2}=100, mn=48, \\therefore \\cos\\angle F_{1}PF_{2}=\\frac{m^{2}+n^{2}-10^{2}}{2mn}=0 \\therefore \\angle F_{1}PF_{2}=\\frac{\\pi}{2}"
  },
  {
    "text": "Let a point $P$ on the parabola $y^{2}=8x$ be at a distance of $4$ from the $y$-axis. Then, the distance from point $P$ to the focus of the parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);P: Point;PointOnCurve(P, G) = True;Distance(P, yAxis) = 4",
    "query_expressions": "Distance(P, Focus(G))",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 15], [41, 44]], [[1, 15]], [[18, 21], [35, 39]], [[1, 21]], [[18, 33]]]",
    "query_spans": "[[[35, 51]]]",
    "process": ""
  },
  {
    "text": "The distance from a moving point $P$ to the line $x=8$ is twice the distance from $P$ to the point $A(2,0)$. Then the trajectory equation of the moving point $P$ is?",
    "fact_expressions": "G: Line;A: Point;P: Point;Expression(G) = (x = 8);Coordinate(A) = (2, 0);Distance(P, G) = Distance(P, A)*2",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "3*x^2+4*y^2=48",
    "fact_spans": "[[[8, 15]], [[21, 30]], [[4, 7], [19, 20], [42, 45]], [[8, 15]], [[21, 30]], [[4, 38]]]",
    "query_spans": "[[[42, 52]]]",
    "process": "Let the coordinates of the moving point P be (x, y). Then the distance from point P to the line x = 8 is |x - 8|, and the distance from point P to point A is \\sqrt{(x - 2)^{2} + y^{2}}. From the given condition, we have |x - 8| = 2\\sqrt{(x - 2)^{2} + y^{2}}. Simplifying yields 3x^{2} + 4y^{2} = 48. Therefore, the trajectory equation of the moving point P is 3x^{2} + 4y^{2} = 48."
  },
  {
    "text": "Given real numbers $x$, $y$ satisfying $\\sqrt{(x-2)^{2}+y^{2}}+\\sqrt{(x+2)^{2}+y^{2}}=6$, then the maximum value of $2 x+y$ equals?",
    "fact_expressions": "x: Real;y: Real;sqrt(y^2 + (x - 2)^2) + sqrt(y^2 + (x + 2)^2) = 6",
    "query_expressions": "Max(2*x + y)",
    "answer_expressions": "sqrt(41)",
    "fact_spans": "[[[2, 7]], [[10, 13]], [[15, 64]]]",
    "query_spans": "[[[66, 80]]]",
    "process": ""
  },
  {
    "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ have left and right foci $F_{1}$ and $F_{2}$, respectively. A circle centered at $F_{2}$ is tangent to both asymptotes of the hyperbola $C$, and this circle passes through the midpoint of the segment $O F_{2}$. Then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;H: Circle;F2: Point;O: Origin;F1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;Center(H)=F2;IsTangent(Asymptote(C),H);PointOnCurve(MidPoint(LineSegmentOf(O,F2)),H)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[1, 62], [103, 109], [137, 143]], [[8, 62]], [[8, 62]], [[99, 100], [119, 120]], [[79, 86], [88, 95]], [[123, 132]], [[71, 78]], [[8, 62]], [[8, 62]], [[1, 62]], [[1, 86]], [[1, 86]], [[87, 100]], [[99, 116]], [[119, 135]]]",
    "query_spans": "[[[137, 149]]]",
    "process": ""
  },
  {
    "text": "Given that the right focus of the hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ coincides exactly with the focus of the parabola $y^{2}=8x$, then $m$=?",
    "fact_expressions": "G: Hyperbola;m: Number;H: Parabola;Expression(G) = (-y^2 + x^2/m = 1);Expression(H) = (y^2 = 8*x);RightFocus(G) =Focus(H)",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 30]], [[56, 59]], [[37, 51]], [[2, 30]], [[37, 51]], [[2, 54]]]",
    "query_spans": "[[[56, 61]]]",
    "process": ""
  },
  {
    "text": "Let point $P$ be a moving point on the parabola $y^{2}=4x$, $F$ be the focus of the parabola, and point $B$ have coordinates $(4,2)$. Then the minimum value of $|PB|+|PF|$ is?",
    "fact_expressions": "E: Parabola;Expression(E) = (y^2 = 4*x);P: Point;PointOnCurve(P, E);F: Point;Focus(E) = F;B: Point;Coordinate(B) = (4, 2)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, B)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "5",
    "fact_spans": "[[[8, 11], [37, 40]], [[8, 23]], [[1, 6]], [[0, 30]], [[32, 35]], [[31, 43]], [[45, 50]], [[45, 63]]]",
    "query_spans": "[[[67, 87]]]",
    "process": ""
  },
  {
    "text": "The line $l$ passing through the focus $F$ of the parabola $C$: $y^{2}=2 p x$ ($p>0$) intersects $C$ at points $A$ and $B$. The projections of points $A$ and $B$ onto the directrix are $M$ and $N$, respectively. The area of $\\triangle A F M$ and the area of $\\Delta B F N$ are reciprocals of each other. Then, the area of $\\triangle M F N$ is?",
    "fact_expressions": "l: Line;C: Parabola;p: Number;A: Point;F: Point;M: Point;B: Point;N: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C)=F;PointOnCurve(F,l);Intersection(l,C) = {B, A};Projection(A,Directrix(C))=M;Projection(B,Directrix(C))=N;InterReciprocal(Area(TriangleOf(A,F,M)),Area(TriangleOf(B,F,N)))",
    "query_expressions": "Area(TriangleOf(M, F, N))",
    "answer_expressions": "2",
    "fact_spans": "[[[34, 39]], [[1, 27], [40, 43]], [[9, 27]], [[46, 49], [58, 61]], [[30, 33]], [[78, 81]], [[51, 54], [63, 66]], [[83, 86]], [[9, 27]], [[1, 27]], [[1, 33]], [[0, 39]], [[34, 56]], [[40, 86]], [[40, 86]], [[87, 129]]]",
    "query_spans": "[[[131, 153]]]",
    "process": "Let $\\angle MAF = \\theta$, $|AF| = a$, $|BF| = b$. By the definition of the parabola, we have $|AM| = a$, $|BN| = b$, and $180^{\\circ} - 2\\angle AFM + 180^{\\circ} - 2\\angle BFN = 180^{\\circ}$, hence $\\angle AFM + \\angle BFN = 90^{\\circ}$, so $\\angle MFO + \\angle NFO = 90^{\\circ}$, i.e., $MF \\perp NF$. Let $\\angle MAF = \\theta$, then by the law of cosines, $|MF|^{2} = 2a^{2}(1 - \\cos\\theta)$, $|NF|^{2} = 2b^{2}(1 + \\cos\\theta)$, $S_{\\Delta MAF} = \\frac{1}{2}a^{2}\\sin\\theta$, $S_{\\Delta NBF} = \\frac{1}{2}b^{2}\\sin\\theta$. Since the area of $\\triangle AFM$ and the area of $\\triangle BFN$ are reciprocals, we have $\\frac{1}{2}a^{2}\\sin\\theta \\cdot \\frac{1}{2}b^{2}\\sin\\theta = 1$, i.e., $a^{2}b^{2}\\sin^{2}\\theta = 4$, so $(S_{\\Delta MFN})^{2} = \\left(\\frac{1}{4}|MF|^{2}|NF|^{2}\\right) = a^{2}b^{2}\\sin^{2}\\theta = 4$. Therefore, the area of $AMFN$ is $2$."
  },
  {
    "text": "In the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, $A$ is a vertex on the major axis, $B$ is a vertex on the minor axis, $F_{1}$ and $F_{2}$ are the left and right foci respectively, and it satisfies $\\overrightarrow{A F_{1}} \\cdot \\overrightarrow{A F_{2}}+\\overrightarrow{B F_{1}} \\cdot \\overrightarrow{B F_{2}}=0$, then the eccentricity $e$=?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;F1: Point;F2: Point;B: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;OneOf(Vertex(MajorAxis(G)))=A;OneOf(Vertex(MinorAxis(G)))=B;Eccentricity(G)=e;e:Number;DotProduct(VectorOf(A, F1), VectorOf(A, F2)) + DotProduct(VectorOf(B, F1), VectorOf(B, F2)) = 0",
    "query_expressions": "e",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[1, 53]], [[3, 53]], [[3, 53]], [[55, 58]], [[79, 86]], [[87, 94]], [[67, 70]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 102]], [[1, 102]], [[1, 66]], [[1, 78]], [[1, 229]], [[226, 229]], [[106, 221]]]",
    "query_spans": "[[[226, 231]]]",
    "process": "Let A(a,0), B(0,b), F₁(-c,0), F₂(c,0).  \n\\overrightarrow{AF}_{1} \\cdot \\overrightarrow{AF_{2}} + \\overrightarrow{BF_{1}} \\cdot \\overrightarrow{BF_{2}} = 0,  \n(-c-a,0) \\cdot (c-a,0) + (-c,-b) \\cdot (c,-b) = 0,  \na^{2}-c^{2}+b^{2}-c^{2}=0,  \na^{2}-c^{2}+(a^{2}-c^{2})-c^{2}=0,  \n2a^{2}=3c^{2},  \n\\frac{c^{2}}{a^{2}}=\\frac{2}{3},  \n\\frac{c}{a}=\\frac{\\sqrt{6}}{3}"
  },
  {
    "text": "What is the eccentricity of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{16}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/16 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(7)/4",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 44]]]",
    "process": "According to the ellipse equation, combining the properties of the ellipse, find a, c, then the eccentricity of the ellipse can be found. Since the ellipse equation is $\\frac{x^{2}}{9}+\\frac{y^{2}}{16}=1$, so $\\begin{cases}a^{2}=16\\\\b^{2}=9\\end{cases}\\Rightarrow\\begin{cases}a=4\\\\b=3\\end{cases}\\Rightarrow c=\\sqrt{a^{2}-b^{2}}=\\sqrt{7}$, therefore the eccentricity $e=\\frac{c}{a}=\\frac{\\sqrt{7}}{4}$"
  },
  {
    "text": "The coordinates of the focus of the parabola $x^{2}=16 y$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 16*y)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, 4)",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 22]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $x^{2}-4 y^{2}=4$, respectively, and point $P$ lies on the right branch of this hyperbola such that $|P F_{1}|+|P F_{2}|=6$, then $\\cos \\angle F_{1} P F_{2}$=?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 - 4*y^2 = 4);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P,RightPart(G));Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 6",
    "query_expressions": "Cos(AngleOf(F1, P, F2))",
    "answer_expressions": "3/5",
    "fact_spans": "[[[20, 40], [53, 56]], [[47, 51]], [[2, 9]], [[10, 17]], [[20, 40]], [[2, 46]], [[2, 46]], [[47, 60]], [[62, 85]]]",
    "query_spans": "[[[87, 116]]]",
    "process": ""
  },
  {
    "text": "Given that the equation $(k^{2}-1) x^{2}+3 y^{2}=1$ represents an ellipse with foci on the $y$-axis, what is the range of values for $k$?",
    "fact_expressions": "G: Ellipse;k:Number;Expression(G) = (x^2*(k^2 - 1) + 3*y^2 = 1);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-\\infty,-2)+(2,+\\infty)",
    "fact_spans": "[[[41, 43]], [[45, 48]], [[2, 43]], [[32, 43]]]",
    "query_spans": "[[[45, 55]]]",
    "process": "The equation $(k^{2}-1)x^{2}+3y^{2}=1$ can be rewritten as $\\frac{x^{2}}{k^{2}-1}+\\frac{y^{2}}{3}=1$. Since the foci of the ellipse lie on the $y$-axis, we have\n\\[\n\\begin{cases}\nk^{2}-1>0 \\\\\n\\frac{1}{k^{2}-1}<\\frac{1}{3}\n\\end{cases}\n\\]\nSolving this gives $k>2$ or $k<-2$. Answer: $(-\\infty,-2)\\cup(2,+\\infty)$"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{a^{2}-4}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. If there exists a point $P$ on the ellipse such that $P F_{1} \\perp P F_{2}$, and the area of $\\triangle P F_{1} F_{2}$ is $2$, then the value of $a^{2}$ is?",
    "fact_expressions": "G: Ellipse;a: Number;P: Point;F1: Point;F2: Point;Expression(G) = (y^2/(a^2 - 4) + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P,G);IsPerpendicular(LineSegmentOf(P,F1),LineSegmentOf(P,F2));Area(TriangleOf(P, F1, F2)) = 2",
    "query_expressions": "a^2",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 49], [76, 78]], [[146, 153]], [[81, 85]], [[58, 65]], [[66, 73]], [[2, 49]], [[2, 73]], [[2, 73]], [[76, 85]], [[87, 110]], [[112, 144]]]",
    "query_spans": "[[[146, 157]]]",
    "process": "According to the definition of an ellipse, |PF_{1}| + |PF_{2}| = 2a. Since PF_{1} \\perp PF_{2}, triangle PF_{1}F_{2} is a right triangle. Therefore, (|PF_{1}|)^{2} + (|PF_{2}|)^{2} = (2c)^{2}. Also, since the area of triangle PF_{1}F_{2} is 2, we have \\frac{1}{2} \\cdot |PF_{1}| \\cdot |PF_{2}| = 2, so |PF_{1}| \\cdot |PF_{2}| = 4. Thus, (2a)^{2} = (|PF_{1}| + |PF_{2}|)^{2} = (|PF_{1}|)^{2} + (|PF_{2}|)^{2} + 2|PF_{1}| \\cdot |PF_{2}| = 4c^{2} + 8, which gives a^{2} - c^{2} = 2 = b^{2}. From \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{a^{2} - 4} = 1, we get b^{2} = a^{2} - 4, so a^{2} - 4 = 2, a^{2} = 6."
  },
  {
    "text": "The equation of a parabola with vertex at the origin, coordinate axes as axes of symmetry, and focus on $3 x-2 y-6=0$ is?",
    "fact_expressions": "G: Parabola;O: Origin;Vertex(G) = O;SymmetryAxis(G) = axis;C: Curve;Expression(C) = (3*x - 2*y - 6 = 0);PointOnCurve(Focus(G), C)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=8*x, x^2=-12*y}",
    "fact_spans": "[[[32, 35]], [[3, 5]], [[0, 35]], [[6, 35]], [[17, 30]], [[17, 30]], [[14, 35]]]",
    "query_spans": "[[[32, 39]]]",
    "process": "The vertex is at the origin, the coordinate axes are the axes of symmetry, and the focus lies on 3x-2y-6=0. Therefore, the coordinates of the focus of the parabola are (2,0) or (0,-3). Hence, the standard equation of the parabola is y^{2}=8x or x^{2}=-12y."
  },
  {
    "text": "It is known that the circle $(x+1)^{2}+(y-1)^{2}=5$ passes through the right focus $F$ and the upper vertex $B$ of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$. Then, the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Circle;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = ((x + 1)^2 + (y - 1)^2 = 5);F: Point;B: Point;RightFocus(C) = F;UpperVertex(C) = B;PointOnCurve(F, G);PointOnCurve(B, G)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(10)/10",
    "fact_spans": "[[[28, 87], [103, 108]], [[35, 87]], [[35, 87]], [[2, 26]], [[35, 87]], [[35, 87]], [[28, 87]], [[2, 26]], [[91, 94]], [[98, 101]], [[28, 94]], [[28, 101]], [[2, 94]], [[2, 101]]]",
    "query_spans": "[[[103, 114]]]",
    "process": "In the equation $(x+1)^{2}+(y-1)^{2}=5$, let $y=0$, we get $x=1,-3$. Let $x=0$, we get $y=-1,3$. According to the problem, $c=1$, $b=3$, so $a=\\sqrt{10}$, $e=\\frac{c}{a}=\\frac{\\sqrt{10}}{10}$"
  },
  {
    "text": "Given that $M$ is a moving point on the parabola $y^{2}=4x$, $F$ is the focus of this parabola, and there is a fixed point $A(3, 2)$, then the minimum value of $|MA|+|MF|$ is $?$.",
    "fact_expressions": "G: Parabola;A: Point;M: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (3, 2);PointOnCurve(M, G);Focus(G) = F",
    "query_expressions": "Min(Abs(LineSegmentOf(M,A))+Abs(LineSegmentOf(M,F)))",
    "answer_expressions": "4",
    "fact_spans": "[[[6, 20], [31, 34]], [[43, 53]], [[2, 5]], [[25, 28]], [[6, 20]], [[43, 53]], [[2, 24]], [[25, 37]]]",
    "query_spans": "[[[55, 73]]]",
    "process": "Let the projection of point M on the directrix be D. Then, according to the definition of the parabola, |MF| = |MD|, so minimizing |MA| + |MF| is equivalent to minimizing |MA| + |MD|. The minimum value of |MA| + |MD| occurs when points D, M, A are collinear, and this minimum value is 3 - (-1) = 4."
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=4x$, draw a line intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. Then $x_{1} x_{2}$=?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;x1:Number;y1:Number;x2:Number;y2:Number;Expression(G) = (y^2 = 4*x);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Focus(G)=F;PointOnCurve(F,H);Intersection(H,G) = {B, A};F:Point",
    "query_expressions": "x1*x2",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 15], [25, 28]], [[22, 24]], [[29, 47]], [[49, 66]], [[29, 46]], [[29, 46]], [[49, 66]], [[49, 66]], [[1, 15]], [[29, 47]], [[49, 66]], [[1, 21]], [[0, 24]], [[22, 66]], [[18, 21]]]",
    "query_spans": "[[[68, 83]]]",
    "process": ""
  },
  {
    "text": "Draw a chord $AB$ through the focus of the parabola $y^{2}=8x$. Points $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$, and $x_{1}+x_{2}=10$, then $|AB|=$?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;x1: Number;y1: Number;x2: Number;y2: Number;Expression(G) = (y^2 = 8*x);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);x1 + x2 = 10;PointOnCurve(Focus(G), LineSegmentOf(A, B));IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "14",
    "fact_spans": "[[[1, 15]], [[26, 44]], [[45, 62]], [[26, 44]], [[26, 44]], [[45, 62]], [[45, 62]], [[1, 15]], [[26, 44]], [[45, 62]], [[64, 80]], [[0, 25]], [[1, 25]]]",
    "query_spans": "[[[82, 91]]]",
    "process": ""
  },
  {
    "text": "Given that an asymptote of the hyperbola $x^{2}-y^{2}=1$ is tangent to the curve $y=\\frac{1}{3} x^{3}+a$, then the value of $a$ is?",
    "fact_expressions": "G: Hyperbola;H: Curve;a: Number;Expression(G) = (x^2 - y^2 = 1);Expression(H) = (y = a + x^3/3);IsTangent(OneOf(Asymptote(G)), H)",
    "query_expressions": "a",
    "answer_expressions": "{2/3,-2/3}",
    "fact_spans": "[[[2, 20]], [[27, 52]], [[56, 59]], [[2, 20]], [[27, 52]], [[2, 54]]]",
    "query_spans": "[[[56, 63]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{100}+\\frac{y^{2}}{64}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ is a point on the ellipse $C$, and $|P F_{1}|=8$, then $|P F_{2}|=$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/100 + y^2/64 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 8",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "12",
    "fact_spans": "[[[2, 47], [80, 85]], [[2, 47]], [[56, 63]], [[66, 73]], [[2, 73]], [[2, 73]], [[75, 79]], [[75, 89]], [[91, 104]]]",
    "query_spans": "[[[106, 119]]]",
    "process": "From the ellipse equation $ C: \\frac{x^{2}}{100} + \\frac{y^{2}}{64} = 1 $, $\\therefore a = 10$, $b = 8$, $\\because |PF_{1}| + |PF_{2}| = 2a = 20$, $|PF_{1}| = 8$ $\\therefore |PF_{2}| = 20 - 8 = 12$"
  },
  {
    "text": "Let $P$ be a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{5}=1$, and let $F_{1}$, $F_{2}$ be the two foci of the ellipse. If $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P, G);G: Ellipse;Expression(G) = (x^2/25 + y^2/5 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "5",
    "fact_spans": "[[[1, 4]], [[1, 46]], [[5, 43], [63, 65]], [[5, 43]], [[47, 54]], [[55, 62]], [[47, 70]], [[71, 130]]]",
    "query_spans": "[[[132, 159]]]",
    "process": "Given |PF_{1}||PF_{2}|=10, and since \\overrightarrow{PF_{1}}\\bot\\overrightarrow{PF_{2}}, the area of \\triangle F_{1}PF_{2} is obtained. From the ellipse equation, a=5, c=\\sqrt{25-5}=2\\sqrt{5}, so |PF_{1}|+|PF_{2}|=2a=10, |F_{1}F_{2}|=2c=4\\sqrt{5}. Since \\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=0, it follows that \\overrightarrow{PF_{1}}\\bot\\overrightarrow{PF_{2}}, hence |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}=80. Since (|PF_{1}|+|PF_{2}|)^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}+2|PF_{1}||PF_{2}|, solving gives |PF_{1}||PF_{2}|=10. Because \\overrightarrow{PF_{1}}\\bot\\overrightarrow{PF_{2}}, S_{\\triangle F_{1}PF_{2}}=\\frac{1}{2}|PF_{1}||PF_{2}|=5"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, the right focus is $F$, and $P$ is a point on the ellipse. Point $A(0,2 \\sqrt{3})$. When point $P$ moves on the ellipse, the maximum perimeter of $\\triangle A P F$ is?",
    "fact_expressions": "G: Ellipse;A: Point;P: Point;F: Point;Expression(G) = (x^2/9 + y^2/5 = 1);Coordinate(A) = (0, 2*sqrt(3));PointOnCurve(P, G);RightFocus(G)=F",
    "query_expressions": "Max(Perimeter(TriangleOf(A,P,F)))",
    "answer_expressions": "14",
    "fact_spans": "[[[2, 39], [52, 54], [83, 85]], [[58, 76]], [[48, 51], [78, 82]], [[44, 47]], [[2, 39]], [[58, 76]], [[48, 57]], [[2, 47]]]",
    "query_spans": "[[[90, 116]]]",
    "process": ""
  },
  {
    "text": "Point $P$ is a point on the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and $F_{1}$, $F_{2}$ are its left and right foci, respectively. If $|P F_{1}|=10$, then $|P F_{2}|=$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);P: Point;PointOnCurve(P, G) = True;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(P, F1)) = 10",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "{4,16}",
    "fact_spans": "[[[5, 44], [66, 67]], [[5, 44]], [[0, 4]], [[0, 47]], [[48, 55]], [[56, 63]], [[48, 72]], [[48, 72]], [[74, 88]]]",
    "query_spans": "[[[90, 103]]]",
    "process": "According to the definition of a hyperbola, we have ||PF_{1}|-|PF_{2}||=2a=6, that is, |10-|PF_{2}||=6, so 10-|PF_{2}|=\\pm6. Solving gives |PF_{2}|=4 or 16. [Note] This question mainly examines the definition of a hyperbola and the solution method for equations involving absolute values, and belongs to basic problems."
  },
  {
    "text": "Given that the point $(1,2)$ lies on the parabola $y^{2}=2 p x$, then the coordinates of the focus of this parabola are?",
    "fact_expressions": "H: Point;Coordinate(H) = (1, 2);G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;PointOnCurve(H, G)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(1,0)",
    "fact_spans": "[[[2, 10]], [[2, 10]], [[11, 27], [31, 34]], [[11, 27]], [[14, 27]], [[2, 28]]]",
    "query_spans": "[[[31, 41]]]",
    "process": "From the given condition, we have $2p=4$, solving gives $p=2$, thus the focus coordinates of the parabola are $(1,0)$."
  },
  {
    "text": "The foci of the ellipse are $F_{1}(-4,0)$, $F_{2}(4,0)$. A chord $AB$ passes through $F_{1}$, and the perimeter of $\\triangle A B F_{2}$ is $20$. Then the equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;Focus(G) = {F1, F2};F1: Point;F2: Point;Coordinate(F1) = (-4, 0);Coordinate(F2) = (4, 0);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G) = True;PointOnCurve(F1, LineSegmentOf(A, B)) = True;Perimeter(TriangleOf(A, B, F2)) = 20",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/25+y^2/9=1",
    "fact_spans": "[[[0, 2], [84, 86]], [[0, 34]], [[7, 20], [36, 43]], [[22, 34]], [[7, 20]], [[22, 34]], [[45, 50]], [[45, 50]], [[0, 50]], [[35, 50]], [[52, 81]]]",
    "query_spans": "[[[84, 91]]]",
    "process": "By the given condition, since the perimeter of $\\triangle ABF_{2}$ is $4a = 20$, we have $a = 5$. From the coordinates of the foci, we know $c = 4$, and it lies on the $x$-axis, so $b^{2} = a^{2} - c^{2} = 9$. Therefore, the standard equation of the ellipse is $\\frac{x^{2}}{2} + \\frac{y^{2}}{3} = 1$."
  },
  {
    "text": "Let $AB$ be the major axis of ellipse $\\Gamma$, point $C$ lies on $\\Gamma$, and $\\angle C B A=\\frac{\\pi}{4}$. If $AB=4$, $B C=\\sqrt{2}$, then the distance between the two foci of $\\Gamma$ is?",
    "fact_expressions": "Gamma: Ellipse;A: Point;B: Point;MajorAxis(Gamma) = LineSegmentOf(A, B);C: Point;PointOnCurve(C, Gamma);AngleOf(C, B, A) = pi/4;LineSegmentOf(A, B) = 4;LineSegmentOf(B, C) = sqrt(2);F1: Point;F2: Point;Focus(Gamma) = {F1, F2}",
    "query_expressions": "Distance(F1, F2)",
    "answer_expressions": "4*sqrt(6)/3",
    "fact_spans": "[[[6, 16], [25, 33], [90, 98]], [[1, 5]], [[1, 5]], [[1, 19]], [[20, 24]], [[20, 34]], [[36, 64]], [[66, 72]], [[74, 88]], [], [], [[90, 103]]]",
    "query_spans": "[[[90, 110]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{n}=1$ $(m>0, n>0)$ has an eccentricity of $2$, and one of its foci coincides with the focus of the parabola $y^{2}=16 x$, then $n=$?",
    "fact_expressions": "G: Hyperbola;m: Number;n: Number;m>0;n>0;H:Parabola;Expression(G) = (-y^2/n + x^2/m = 1);Eccentricity(G) = 2;Expression(H) =(y^2 = 16*x);OneOf(Focus(G))=Focus(H)",
    "query_expressions": "n",
    "answer_expressions": "12",
    "fact_spans": "[[[2, 50]], [[5, 50]], [[87, 90]], [[5, 50]], [[5, 50]], [[65, 80]], [[2, 50]], [[2, 58]], [[65, 80]], [[2, 85]]]",
    "query_spans": "[[[87, 92]]]",
    "process": ""
  },
  {
    "text": "The eccentricity $e$ of the hyperbola $9 x^{2}-16 y^{2}=144$ is?",
    "fact_expressions": "G: Hyperbola;e: Number;Expression(G) = (9*x^2 - 16*y^2 = 144);Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "5/4",
    "fact_spans": "[[[0, 25]], [[29, 32]], [[0, 25]], [[0, 32]]]",
    "query_spans": "[[[29, 34]]]",
    "process": "The hyperbola $9x^{2}-16y^{2}=144$ is equivalent to $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, where $a=4$, $b=3$, $c=\\sqrt{4^{2}+3^{2}}=5$, $e=\\frac{c}{a}=\\frac{5}{4}$"
  },
  {
    "text": "Given point $P$ is a moving point on the parabola $C$: $y=x^{2}$. A tangent line is drawn from point $P$ to the circle $M$: $x^{2}+(y-2)^{2}=1$, with the point of tangency being $A$. Then the minimum value of $\\overrightarrow{P A} \\cdot \\overrightarrow{P M}$ is?",
    "fact_expressions": "C: Parabola;M: Circle;P: Point;A: Point;L: Line;Expression(C) = (y = x^2);Expression(M) = (x^2 + (y-2)^2 = 1);PointOnCurve(P, C);TangentOfPoint(P, M) = L;TangentPoint(L,M) = A",
    "query_expressions": "Min(DotProduct(VectorOf(P, A), VectorOf(P, M)))",
    "answer_expressions": "3/4",
    "fact_spans": "[[[7, 24]], [[35, 60]], [[2, 6], [30, 34]], [[69, 72]], [], [[7, 24]], [[35, 60]], [[2, 28]], [[29, 65]], [[35, 72]]]",
    "query_spans": "[[[74, 129]]]",
    "process": "From the given, we have: $\\overrightarrow{PA}\\cdot\\overrightarrow{PM}=|\\overrightarrow{PA}|^{2}=|\\overrightarrow{PM}|^{2}-1$. Let point $P(x,x^{2})$, then $|\\overrightarrow{PM}|^{2}-1=x^{2}+(x^{2}-2)^{2}-1=x^{4}-3x^{2}+3=(x^{2}-\\frac{3}{2})^{2}+\\frac{3}{4}\\geqslant\\frac{3}{4}$. When $x^{2}=\\frac{3}{2}$, $\\overrightarrow{PA}\\cdot\\overrightarrow{PM}=|\\overrightarrow{PM}|^{2}-1$ reaches the minimum value $\\frac{3}{4}$. Hence, the answer is: 3"
  },
  {
    "text": "Given that the focal distance of an ellipse is $6$, and the sum of the distances from any point on the ellipse to the two foci is equal to $10$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;FocalLength(G) = 6;P: Point;PointOnCurve(P,G) = True;F1: Point;F2: Point;Focus(G) = {F1, F2};Distance(P,F1) + Distance(P,F2) = 10",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/25 + y^2/16 = 1, y^2/25 + x^2/16 = 1}",
    "fact_spans": "[[[2, 4], [13, 15], [36, 38]], [[2, 11]], [[17, 18]], [[13, 18]], [], [], [[13, 23]], [[13, 34]]]",
    "query_spans": "[[[36, 45]]]",
    "process": "According to the problem, the focal distance of the ellipse is 6, so we have 2c = 6, that is, c = 3. Also, the sum of the distances from any point on the ellipse to the two foci is equal to 10, so we get 2a = 10, that is, a = 5. Then b^{2} = a^{2} - c^{2} = 25 - 9 = 16. When the foci are on the x-axis, the equation of the ellipse is \\frac{x^{2}}{2}+\\frac{y^{2}}{6}=1. When the foci of the ellipse are on the y-axis, the equation of the ellipse is \\frac{y^{2}}{c}+\\frac{x^{2}}{}=1"
  },
  {
    "text": "Given the hyperbola $M$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, with left and right foci $F_{1}$, $F_{2}$ respectively, and $|F_{1} F_{2}|=2 c$. If there exists a point $P$ on the right branch of the hyperbola $M$ such that $\\frac{a}{\\sin \\angle P F_{1} F_{2}}=\\frac{3 c}{\\sin \\angle P F_{2} F_{1}}$, then the range of the eccentricity of hyperbola $M$ is?",
    "fact_expressions": "M: Hyperbola;b: Number;a: Number;F1: Point;F2: Point;P: Point;a>0;b>0;Expression(M) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(M)=F1;RightFocus(M)=F2;Abs(LineSegmentOf(F1, F2)) = 2*c;PointOnCurve(P,RightPart(M));a/Sin(AngleOf(P, F1, F2)) = (3*c)/Sin(AngleOf(P, F2, F1));c:Number",
    "query_expressions": "Range(Eccentricity(M))",
    "answer_expressions": "(1,(2+sqrt(7))/3)",
    "fact_spans": "[[[2, 63], [111, 117], [206, 212]], [[10, 63]], [[10, 63]], [[72, 79]], [[80, 88]], [[123, 127]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 88]], [[2, 88]], [[89, 108]], [[111, 127]], [[129, 204]], [[89, 108]]]",
    "query_spans": "[[[206, 223]]]",
    "process": ""
  },
  {
    "text": "Given that the hyperbola $D$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has asymptotes $y=\\pm \\sqrt{3} x$, and the left and right foci are $F_{1}$ and $F_{2}$ respectively. Let $P$ be an arbitrary point on the right branch of hyperbola $D$, then the range of $\\frac{|P F_{1}|-|P F_{2}|}{|P F_{1}|+|P F_{2}|}$ is?",
    "fact_expressions": "D: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(D) = (-y^2/b^2 + x^2/a^2 = 1);Expression(Asymptote(D)) = (y = pm*sqrt(3)*x);LeftFocus(D) = F1;RightFocus(D) = F2;PointOnCurve(P, RightPart(D))",
    "query_expressions": "Range((Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)))/(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2))))",
    "answer_expressions": "(0, 1/2]",
    "fact_spans": "[[[26, 87], [116, 122]], [[34, 87]], [[34, 87]], [[112, 115]], [[96, 103]], [[104, 111]], [[34, 87]], [[34, 87]], [[26, 87]], [[2, 87]], [[26, 111]], [[26, 111]], [[112, 129]]]",
    "query_spans": "[[[131, 189]]]",
    "process": ""
  },
  {
    "text": "Let the point on the parabola $y^{2}=4 p x(p>0)$ have an abscissa of $6$, and the distance from this point to the focus is $10$. Then $p=?$",
    "fact_expressions": "G: Parabola;p: Number;p>0;H:Point;Expression(G) = (y^2 = 4*(p*x));PointOnCurve(H, G);XCoordinate(H)=6;Distance(H, Focus(G)) = 10",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 22]], [[45, 48]], [[4, 22]], [[31, 32]], [[1, 22]], [[1, 32]], [[23, 32]], [[1, 43]]]",
    "query_spans": "[[[45, 50]]]",
    "process": "The distance from a point on the parabola to the focus is equal to the distance to the directrix. Hence, $10 = 6 + \\frac{p}{2} \\therefore p = 4$"
  },
  {
    "text": "Let $A B$ be the major axis of the ellipse, point $C$ lies on the ellipse, and $\\angle C B A=45^{\\circ}$. If $A B=4$, $B C=\\sqrt{2}$, then the focal distance of the ellipse is?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;C: Point;MajorAxis(G) = LineSegmentOf(A, B);PointOnCurve(C, G);AngleOf(C, B, A) = ApplyUnit(45, degree);LineSegmentOf(A, B) = 4;LineSegmentOf(B,C)=sqrt(2)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "4*sqrt(6)/3",
    "fact_spans": "[[[7, 9], [18, 20], [75, 77]], [[1, 6]], [[1, 6]], [[13, 17]], [[1, 12]], [[13, 21]], [[23, 48]], [[50, 57]], [[59, 73]]]",
    "query_spans": "[[[75, 82]]]",
    "process": "As shown in the figure, let the standard equation of the ellipse be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$. From the given conditions, $2a=4$, so $a=2$. $\\because\\angle CBA=\\frac{\\pi}{4}$, $BC=\\sqrt{2}$, we can set $C(y_{0}-2,y_{0})$, $\\because B(-2,0)$. $\\therefore\\overrightarrow{BC}=(y_{0},y_{0})$, $\\therefore|\\overrightarrow{BC}|=\\sqrt{2}y_{0}=\\sqrt{2}$, solving gives $y_{0}=1$, $\\therefore$ the coordinates of point $C$ are $C(-1,1)$, $\\because$ point $C$ lies on the ellipse, $\\therefore\\frac{(-1)^{2}}{4}+\\frac{1^{2}}{b^{2}}=1$, $\\therefore b^{2}=\\frac{4}{3}$, $\\therefore c^{2}=a^{2}-b^{2}=4-\\frac{4}{3}=\\frac{8}{3}$, $c=\\frac{2\\sqrt{6}}{3}$, $\\therefore$ the focal length of the ellipse is $\\frac{4\\sqrt{6}}{3}$."
  },
  {
    "text": "Let one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ intersect the parabola $y=x^{2}+1$ at exactly one point. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;H: Parabola;Expression(H) = (y = x^2 + 1);NumIntersection(OneOf(Asymptote(G)), H) = 1",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[1, 47], [77, 80]], [[1, 47]], [[4, 47]], [[4, 47]], [[54, 68]], [[54, 68]], [[1, 75]]]",
    "query_spans": "[[[77, 86]]]",
    "process": ""
  },
  {
    "text": "The line $l$ passes through the focus $F$ of a parabola and intersects the parabola at two points $A$ and $B$. If $\\overrightarrow{A F}=5 \\overrightarrow{F B}$, then the slope of line $l$ is?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;F: Point;B: Point;Focus(G) = F;PointOnCurve(F, l);Intersection(l,G)={A,B};VectorOf(A, F) = 5*VectorOf(F, B)",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*sqrt(5)/2",
    "fact_spans": "[[[0, 5], [82, 87]], [[7, 10], [19, 22]], [[24, 27]], [[13, 16]], [[28, 31]], [[7, 16]], [[0, 16]], [[0, 33]], [[35, 80]]]",
    "query_spans": "[[[82, 92]]]",
    "process": "According to the problem, the focus of the parabola $ y^{2} = 4x $ is $ F(1,0) $. Let the equation of line $ l $ be $ y = k(x - 1) $. From  \n$$\n\\begin{cases}\ny = k(x - 1) \\\\\ny^{2} = 4x\n\\end{cases}\n$$  \nwe obtain $ k^{2}x^{2} - 2(k^{2} + 2)x + k^{2} = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $.  \n$ \\therefore x_{1} + x_{2} = 2 + \\frac{4}{k^{2}} $, $ x_{1} \\cdot x_{2} = 1 $.  \n$ \\because \\overrightarrow{AF} = 5\\overrightarrow{FB} $, $ \\therefore x_{1} = 5x_{2} - 5 $, i.e., $ 5x_{2} + x_{1} - 6 = 0 $.  \n$ \\because x_{1} = \\frac{1}{x_{2}} $, $ \\therefore 5x_{2} + \\frac{1}{x_{2}} - 6 = 0 $. Solving gives $ x_{2} = 1 $ or $ x_{2} = \\frac{1}{5} $.  \n$ \\therefore x_{1} = 1 $ or $ x_{1} = 5 $. Also, $ x_{1} + x_{2} = 2 + \\frac{4}{k^{2}} > 2 $. Substituting $ x_{1} = 5 $ gives $ k = \\pm \\frac{\\sqrt{5}}{2} $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, let $F_{1}$ and $F_{2}$ be its left and right foci respectively, and $P$ be a point on the hyperbola such that $|P F_{1}|=7$. Then the value of $|P F_{2}|$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 - y^2/16 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(P, F1)) = 7",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "13",
    "fact_spans": "[[[2, 41], [75, 78], [63, 64]], [[71, 74]], [[44, 51]], [[53, 60]], [[2, 41]], [[71, 81]], [[44, 70]], [[44, 70]], [[83, 96]]]",
    "query_spans": "[[[98, 113]]]",
    "process": "From the equation, we have $a^{2}=9 \\therefore a=3$, $2a=6 \\therefore ||PF|-|PF_{2}||=6 \\therefore |PF_{2}|=13$"
  },
  {
    "text": "Given $A(-2 , 0)$, $B(2 , 0)$, and the perimeter of $\\triangle ABC$ equals $10$, then the trajectory equation of vertex $C$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (-2, 0);B: Point;Coordinate(B) = (2, 0);Perimeter(TriangleOf(A, B, C)) = 10;C: Point;Vertex(TriangleOf(A,B,C)) = C",
    "query_expressions": "LocusEquation(C)",
    "answer_expressions": "x^2/9+y^2/5=1&Negation(y=0)",
    "fact_spans": "[[[2, 13]], [[2, 13]], [[16, 26]], [[16, 26]], [[28, 52]], [[56, 59]], [[28, 59]]]",
    "query_spans": "[[[56, 66]]]",
    "process": ""
  },
  {
    "text": "The vertex of the parabola $y^{2}=4 x$ is $O$, point $A$ has coordinates $(2,0)$, and a line $l$ with inclination angle $\\frac{\\pi}{4}$ intersects the segment $O A$ (without passing through points $O$ and $A$) and intersects the parabola at points $P$ and $Q$. Then the range of $|P Q|$ is?",
    "fact_expressions": "l: Line;G: Parabola;O: Point;A: Point;P: Point;Q: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (2, 0);Vertex(G)=O;Inclination(l)=pi/4;IsIntersect(l,LineSegmentOf(O,A));Negation(PointOnCurve(O,l));Negation(PointOnCurve(A,l));Intersection(l,G)={P,Q}",
    "query_expressions": "Range(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "(4*sqrt(2),4*sqrt(6))",
    "fact_spans": "[[[58, 63]], [[0, 14], [91, 94]], [[18, 21], [78, 82]], [[22, 26], [83, 87]], [[95, 98]], [[99, 102]], [[0, 14]], [[22, 37]], [[0, 21]], [[38, 63]], [[58, 73]], [[58, 88]], [[58, 88]], [[58, 104]]]",
    "query_spans": "[[[106, 120]]]",
    "process": "According to the problem, the inclination angle of line $ l $ is $ \\frac{\\pi}{4} $, so we can assume the equation of line $ l $ is $ y = x - m $. Since line $ l $ intersects segment $ OA $, we have $ 0 < m < 2 $. Solving the system of equations  \n\\[\n\\begin{cases}\ny = x - m \\\\\ny^2 = 4x\n\\end{cases}\n\\]\nwe obtain $ x^2 - (2m + 4)x + m^2 = 0 $. Let $ P(x_1, y_1) $, $ Q(x_2, y_2) $, then $ x_1 + x_2 = 2m + 4 $, $ x_1 \\cdot x_2 = m^2 $. Therefore,  \n\\[\n|PQ| = \\sqrt{1 + 1^2} |x_1 - x_2| = \\sqrt{2} \\times \\sqrt{(x_1 + x_2)^2 - 4x_1 x_2} = 4\\sqrt{2} \\times \\sqrt{m + 1}\n\\]  \nThus, $ 4\\sqrt{2} < |PO| < 4\\sqrt{6} $."
  },
  {
    "text": "Let point $F$ be the right focus of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{8}=1$, $M$ moves on the ellipse, and point $P(1,-2)$. Then the maximum perimeter of $\\triangle M P F$ is?",
    "fact_expressions": "G: Ellipse;P: Point;M: Point;F: Point;Expression(G) = (x^2/9 + y^2/8 = 1);Coordinate(P) = (1, -2);RightFocus(G) = F;PointOnCurve(M, G)",
    "query_expressions": "Max(Perimeter(TriangleOf(M, P, F)))",
    "answer_expressions": "8 + 2*sqrt(2)",
    "fact_spans": "[[[5, 42], [51, 53]], [[57, 67]], [[47, 50]], [[0, 4]], [[5, 42]], [[57, 67]], [[0, 46]], [[47, 54]]]",
    "query_spans": "[[[69, 94]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of a hyperbola are $y=\\pm \\frac{\\sqrt{3}}{3} x$, then its eccentricity is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*x*(sqrt(3)/3))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{2*sqrt(3)/3, 2}",
    "fact_spans": "[[[2, 5], [42, 43]], [[2, 40]]]",
    "query_spans": "[[[42, 48]]]",
    "process": "If the foci of the hyperbola lie on the x-axis, then $\\frac{b}{a}=\\frac{\\sqrt{3}}{3}$, then $e=\\frac{c}{a}=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{\\frac{a^{2}+b^{2}}{a^{2}}}=\\sqrt{1+(\\frac{b}{a})^{2}}=\\frac{2\\sqrt{3}}{3}$;"
  },
  {
    "text": "For any point $M$ on the circle $x^{2}+y^{2}=4$, draw a perpendicular from $M$ to the $x$-axis, with foot at $N$. Then the equation of the locus of the midpoint of the segment $MN$ is?",
    "fact_expressions": "G: Circle;N: Point;M: Point;Expression(G) = (x^2 + y^2 = 4);PointOnCurve(M, G);L: Line;PointOnCurve(M, L);IsPerpendicular(L, xAxis);FootPoint(L, xAxis) = N",
    "query_expressions": "LocusEquation(MidPoint(LineSegmentOf(M, N)))",
    "answer_expressions": "x^2 + 4*y^2 = 4",
    "fact_spans": "[[[1, 17]], [[36, 39]], [[22, 25]], [[1, 17]], [[1, 25]], [], [[0, 32]], [[0, 32]], [[0, 39]]]",
    "query_spans": "[[[41, 58]]]",
    "process": ""
  },
  {
    "text": "The center is at the origin, one of the foci is $(-2,0)$, and it passes through the point $(2,3)$. Then the equation of the ellipse is?",
    "fact_expressions": "Center(G) = O;O: Origin;F: Point;Coordinate(F) = (-2, 0);OneOf(Focus(G)) = F;G: Ellipse;I: Point;Coordinate(I) = (2, 3);PointOnCurve(I, G) = True",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16+y^2/12=1",
    "fact_spans": "[[[0, 37]], [[3, 5]], [[13, 21]], [[13, 21]], [[6, 37]], [[35, 37]], [[24, 32]], [[24, 32]], [[23, 37]]]",
    "query_spans": "[[[35, 41]]]",
    "process": "According to the problem, the foci of the ellipse lie on the x-axis. Let the standard equation of the ellipse be \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1. Then 2a = |PF_{1}| + |PF_{2}| = \\sqrt{(2+2)^{2}+3^{2}} + \\sqrt{(2-2)^{2}+3^{2}} = 8. Solving gives a = 4. Since c = 2, it follows that b^{2} = a^{2} - c^{2} = 12. Therefore, the equation of the ellipse is \\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1"
  },
  {
    "text": "What is the length of the chord cut by the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ from a line passing through its right focus and perpendicular to the $x$-axis?",
    "fact_expressions": "G: Hyperbola;L:Line;Expression(G) = (x^2/4 - y^2 = 1);IsPerpendicular(L,xAxis);PointOnCurve(RightFocus(G),L)",
    "query_expressions": "Length(InterceptChord(L,G))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 30], [46, 49]], [[43, 45]], [[2, 30]], [[35, 45]], [[0, 45]]]",
    "query_spans": "[[[43, 56]]]",
    "process": "\\because the coordinates of the right focus of the hyperbola \\frac{x^{2}}{4}-y^{2}=1 are (\\sqrt{5},0), \\therefore the equation of the line passing through the right focus of the hyperbola \\frac{x^{2}}{4}-y^{2}=1 and perpendicular to the x-axis is x=\\sqrt{5}; substituting x=\\sqrt{5} into the hyperbola equation \\frac{x^{2}}{4}-y^{2}=1 yields y=\\pm\\frac{1}{2}, \\therefore the length of the chord intercepted by the hyperbola on the line is 1,"
  },
  {
    "text": "If point $P(4,1)$ lies on the parabola $C$: $x^{2}=2 p y$ $(p>0)$, and the focus of parabola $C$ is $F$, then $|P F|=$?",
    "fact_expressions": "C: Parabola;p: Number;P: Point;F: Point;p>0;Expression(C) = (x^2 = 2*(p*y));Coordinate(P) = (4, 1);PointOnCurve(P, C);Focus(C) = F",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "5",
    "fact_spans": "[[[10, 36], [40, 46]], [[17, 36]], [[1, 9]], [[50, 53]], [[17, 36]], [[10, 36]], [[1, 9]], [[1, 39]], [[40, 53]]]",
    "query_spans": "[[[55, 64]]]",
    "process": "Since point P(4,1) lies on the parabola C: x^{2}=2py (p>0), we have 4^{2}=2p\\times1, which gives p=8. Then |PF|=1+\\frac{8}{2}=5."
  },
  {
    "text": "The equation of the directrix of the parabola $x=8 y^{2}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x = 8*y^2)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=-1/32",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": "From the given conditions: the standard equation of the parabola is $ y^{2} = \\frac{x}{8} $, therefore the directrix equation is $ x = -\\frac{1}{32} $."
  },
  {
    "text": "The line $y=a(x+2)$ intersects the curve $x^{2}-y|y|=1$ at exactly $2$ points. What is the range of real values for $a$?",
    "fact_expressions": "G: Line;Expression(G) = (y = a*(x + 2));a: Real;H: Curve;Expression(H) = (x^2 - y*Abs(y) = 1);NumIntersection(G, H) = 2",
    "query_expressions": "Range(a)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 12]], [[0, 12]], [[40, 45]], [[13, 29]], [[13, 29]], [[0, 38]]]",
    "query_spans": "[[[40, 52]]]",
    "process": ""
  },
  {
    "text": "The equation of the ellipse passing through the point $(3, 2)$ and having the same foci as the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ is?",
    "fact_expressions": "G: Ellipse;H: Point;C:Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1);Coordinate(H) = (3, 2);PointOnCurve(H, C);Focus(G)=Focus(C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/15+y^2/10=1",
    "fact_spans": "[[[14, 51]], [[2, 12]], [[57, 59]], [[14, 51]], [[2, 12]], [[0, 59]], [[13, 59]]]",
    "query_spans": "[[[57, 64]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C_{1}$: $2 x^{2}-y^{2}=8$, the hyperbola $C_{2}$ satisfies: ($1$) $C_{1}$ and $C_{2}$ have the same asymptotes, ($2$) the focal distance of $C_{2}$ is twice the focal distance of $C_{1}$, ($3$) the foci of $C_{2}$ lie on the $y$-axis. Then the equation of $C_{2}$ is?",
    "fact_expressions": "C1: Hyperbola;Expression(C1) = (2*x^2 - y^2 = 8);C2: Hyperbola;Asymptote(C1) = Asymptote(C2);FocalLength(C2) = 2*FocalLength(C1);PointOnCurve(Focus(C2), yAxis) = True",
    "query_expressions": "Expression(C2)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 31], [50, 57], [90, 97]], [[2, 31]], [[32, 42], [58, 65], [78, 85], [110, 117], [128, 135]], [[50, 72]], [[78, 103]], [[110, 126]]]",
    "query_spans": "[[[128, 140]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x$ ($p>0$) with focus $F$, a line $l$ passing through the focus intersects the parabola $C$ at points $M$ and $N$. If $\\overrightarrow{M N}=4 \\overrightarrow{M F}$, then what is the slope of line $l$?",
    "fact_expressions": "l: Line;C: Parabola;p: Number;M: Point;N: Point;F: Point;p>0;Expression(C) = (y^2 = 2*p*x);Focus(C) = F;PointOnCurve(F,l);Intersection(l, C) = {M, N};VectorOf(M, N) = 4*VectorOf(M, F)",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*sqrt(3)",
    "fact_spans": "[[[39, 44], [110, 115]], [[2, 27], [45, 51]], [[9, 27]], [[52, 55]], [[56, 59]], [[31, 34]], [[9, 27]], [[2, 27]], [[2, 34]], [[35, 44]], [[39, 61]], [[63, 108]]]",
    "query_spans": "[[[110, 120]]]",
    "process": "Draw perpendiculars from M and N to the directrix, with feet Q and P respectively, as shown in the figure. Draw MH\\botNP at H. The quadrilateral PNMQ is a right trapezoid. Let |MF|=a, then from \\overrightarrow{MN}=4\\overrightarrow{MF} we obtain |NF|=3a. Also, |MQ|=|MF|=a, |NP|=|NF|=3a. Thus |NH|=3a-a=2a, and |MN|=a+3a=4a. Therefore, in the right triangle HNM, \\cos\\angleHNM=\\frac{2a}{4a}=\\frac{1}{2}, \\angleHNM=60^{\\circ}, that is, the inclination angle of line MN is 60^{\\circ}, and the slope is \\sqrt{3}. By symmetry, the slope could also be -\\sqrt{3}."
  },
  {
    "text": "It is known that the vertex of the parabola is at the origin, the focus lies on the $y$-axis, and a point $M(a,-4)$ ($a>0$) on the parabola is at a distance of $5$ from the focus $F$. Then, what is the standard equation of this parabola?",
    "fact_expressions": "G: Parabola;M: Point;O: Origin;F: Point;Coordinate(M) = (a,-4);Vertex(G) = O;PointOnCurve(Focus(G), yAxis);PointOnCurve(M, G);Focus(G) = F;Distance(M, F) = 5;a:Number;a>0",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 = -4*y",
    "fact_spans": "[[[2, 5], [23, 26], [60, 63]], [[29, 43]], [[9, 13]], [[46, 49]], [[29, 43]], [[2, 13]], [[2, 22]], [[23, 43]], [[23, 49]], [[23, 56]], [[29, 43]], [[29, 43]]]",
    "query_spans": "[[[60, 70]]]",
    "process": "According to the problem, the parabola opens downward, so its equation is set as $ x^{2} = -2py $. Since a point $ M(a, -4) $ (with $ a > 0 $) on the parabola has a distance of 5 to the focus $ F $, the distance from $ M $ to the directrix $ y = \\frac{p}{2} $ is also 5. Thus, $ \\frac{p}{2} + 4 = 5 \\Rightarrow p = 2 $. Therefore, the equation of the parabola is: $ x^{2} = -4y $."
  },
  {
    "text": "Given the parabola $y=2 ax^{2}  (a<0)$, what are the coordinates of its focus?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 2*a*x^2);a: Number;a<0",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,1/(8*a))",
    "fact_spans": "[[[2, 24], [25, 26]], [[2, 24]], [[5, 24]], [[5, 24]]]",
    "query_spans": "[[[25, 33]]]",
    "process": ""
  },
  {
    "text": "What is the length of the chord passing through the focus of the hyperbola $x^{2}-y^{2}=4$ and parallel to the imaginary axis?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 4);L:LineSegment;IsChordOf(L,G);PointOnCurve(Focus(G),L);IsParallel(L,ImageinaryAxis(G))",
    "query_expressions": "Length(L)",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 19]], [[1, 19]], [], [[1, 30]], [[0, 30]], [[0, 30]]]",
    "query_spans": "[[[29, 33]]]",
    "process": "Since $c^{2}=a^{2}+b^{2}=8$, the coordinates of the foci are $(\\pm2\\sqrt{2},0)$; taking $(2\\sqrt{2},0)$, the equation of the line parallel to the imaginary axis is $x=2\\sqrt{2}$, solving the system $\\begin{cases}x=2\\sqrt{2}\\\\x^{2}-y^{2}=4\\end{cases}$ yields $\\begin{cases}x=2\\sqrt{2}\\\\y=\\pm2\\end{cases}$, then the chord length is: $2-(-2)=4$."
  },
  {
    "text": "The distance from the focus to the directrix of the parabola $y^{2}=6 x$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 6*x)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 25]]]",
    "process": "The distance from the focus to the directrix of the parabola \\( y^{2} = 6x \\) is \\( p \\). From the standard equation, we obtain \\( p = 3 \\)."
  },
  {
    "text": "It is known that one focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ coincides with the focus of the parabola $y^{2}=4x$, and the eccentricity of the hyperbola is $\\sqrt{5}$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 4*x);OneOf(Focus(G)) = Focus(H);Eccentricity(G) = sqrt(5)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "5*x^2 - (5/4)*y^2 = 1",
    "fact_spans": "[[[2, 58], [85, 88], [107, 110]], [[5, 58]], [[5, 58]], [[64, 78]], [[5, 58]], [[5, 58]], [[2, 58]], [[64, 78]], [[2, 83]], [[85, 104]]]",
    "query_spans": "[[[107, 115]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\Gamma$: $\\frac{x^{2}}{3}-\\frac{y^{2}}{2}=1$, $F_{1}$, $F_{2}$ are the left and right foci of $\\Gamma$, point $P$ is a point on $\\Gamma$. If $|P F_{1}| \\cdot|P F_{2}|=4$, then the value of $\\overrightarrow{P F_{1}}\\cdot\\overrightarrow{P F_{2}}$ is?",
    "fact_expressions": "Gamma: Hyperbola;Expression(Gamma) = (x^2/3-y^2/2=1);F1: Point;F2: Point;LeftFocus(Gamma) = F1;RightFocus(Gamma) = F2;P: Point;PointOnCurve(P, Gamma);Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 4",
    "query_expressions": "DotProduct(VectorOf(P, F1), VectorOf(P, F2))",
    "answer_expressions": "0",
    "fact_spans": "[[[1, 49], [70, 78], [89, 97]], [[1, 49]], [[52, 59]], [[62, 69]], [[52, 83]], [[52, 83]], [[84, 88]], [[84, 100]], [[102, 130]]]",
    "query_spans": "[[[132, 191]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the focus of the parabola $y = x^{2}$, $M$ and $N$ are two points on this parabola, and $|M F| + |N F| = 3$, then the distance from the midpoint of segment $MN$ to the $x$-axis is?",
    "fact_expressions": "M: Point;N: Point;G: Parabola;F: Point;Expression(G) = (y = x^2);Focus(G) = F;PointOnCurve(M, G);PointOnCurve(N, G);Abs(LineSegmentOf(M, F)) + Abs(LineSegmentOf(N, F)) = 3",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(M, N)), xAxis)",
    "answer_expressions": "5/4",
    "fact_spans": "[[[22, 26]], [[27, 31]], [[6, 18], [33, 36]], [[2, 5]], [[6, 18]], [[2, 21]], [[22, 40]], [[22, 40]], [[41, 56]]]",
    "query_spans": "[[[58, 77]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, and $O$ being the origin. Point $A$ lies on the parabola $C$, and $|AO|=|AF|$. What is the length of segment $AF$?",
    "fact_expressions": "C: Parabola;A: Point;F: Point;O: Origin;Expression(C) = (y^2 = 4*x);Focus(C)=F;PointOnCurve(A, C);Abs(LineSegmentOf(A, O)) = Abs(LineSegmentOf(A, F))",
    "query_expressions": "Length(LineSegmentOf(A,F))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[2, 21], [43, 49]], [[38, 42]], [[25, 28]], [[29, 32]], [[2, 21]], [[2, 28]], [[38, 50]], [[52, 65]]]",
    "query_spans": "[[[67, 78]]]",
    "process": "Without loss of generality, assume point A is above the x-axis. Then from |OF| = 1, we have $ x_{A} = \\frac{1}{2} $. Therefore, $ y_{A} = \\sqrt{2} $, so $ A\\left(\\frac{1}{2}, \\sqrt{2}\\right) $. Thus, $ |AF| = |OA| = \\sqrt{\\frac{1}{4} + 2} = \\frac{3}{2} $."
  },
  {
    "text": "The point $(2 , 3)$ lies on the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, and the focal length of $C$ is $4$. Then its eccentricity is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (2, 3);PointOnCurve(P, C);FocalLength(C) = 4",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[13, 74], [76, 79], [88, 89]], [[20, 74]], [[20, 74]], [[2, 12]], [[20, 74]], [[20, 74]], [[13, 74]], [[2, 12]], [[2, 75]], [[76, 86]]]",
    "query_spans": "[[[88, 95]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $P$ is a point on the hyperbola $C$. If $|\\overrightarrow{P F_{1}}+\\overrightarrow{P F_{2}}|=\\sqrt{| \\overrightarrow{P F_{1}}|^{2}+|\\overrightarrow{P F_{2}}|^{2}}$, and $|\\overrightarrow{P F_{1}}|=2|\\overrightarrow{P F_{2}}|$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(P,C);Abs(VectorOf(P, F1) + VectorOf(P, F2)) = sqrt(Abs(VectorOf(P, F1))^2 + Abs(VectorOf(P, F2))^2);Abs(VectorOf(P, F1)) = 2*Abs(VectorOf(P, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 62], [91, 97], [285, 291]], [[9, 62]], [[9, 62]], [[87, 90]], [[71, 78]], [[79, 86]], [[9, 62]], [[9, 62]], [[2, 62]], [[2, 86]], [[2, 86]], [[87, 101]], [[103, 226]], [[227, 283]]]",
    "query_spans": "[[[285, 297]]]",
    "process": "Let the semi-focal distance of the hyperbola be $ c $. Since $ |\\overrightarrow{PF_{1}}+\\overrightarrow{PF_{2}}| = \\sqrt{|\\overrightarrow{PF}|^{2} + |\\overrightarrow{PF_{2}}|^{2}} $, $ |\\overrightarrow{PF}| = 2t $, $ |\\overrightarrow{PF_{2}}| = t $, then according to the definition of the hyperbola, we obtain $ |\\overrightarrow{PF_{1}}| - |\\overrightarrow{PF_{2}}| = t = 2a \\Rightarrow 2c = \\sqrt{t^{2} + (2t)^{2}} = \\sqrt{5}t \\Rightarrow e = \\sqrt{5} $"
  },
  {
    "text": "Let the ellipse $ C $: $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $) have left and right foci $ F_{1} $, $ F_{2} $. A line passing through point $ F_{1} $ intersects ellipse $ C $ at points $ A $, $ B $. If $ \\overrightarrow{A F_{1}} = \\frac{3}{2} \\overrightarrow{F_{1} B} $, $ \\angle A F_{2} B = 90^{\\circ} $, then the eccentricity of ellipse $ C $ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, G) = True;G: Line;Intersection(G, C) = {A, B};A: Point;B: Point;VectorOf(A, F1) = (3/2)*VectorOf(F1, B);AngleOf(A, F2, B) = ApplyUnit(90, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/5",
    "fact_spans": "[[[1, 58], [94, 99], [208, 213]], [[1, 58]], [[7, 58]], [[7, 58]], [[7, 58]], [[7, 58]], [[65, 72], [82, 90]], [[73, 80]], [[1, 80]], [[1, 80]], [[81, 93]], [[91, 93]], [[91, 111]], [[102, 105]], [[106, 109]], [[113, 176]], [[177, 206]]]",
    "query_spans": "[[[208, 219]]]",
    "process": "Let $AF_{1}=3t$, $F_{1}B=2t$, $F_{2}A=m$, $F_{2}B=n$, then $AB=5t$. By the definition of the ellipse, we have $m+3t=2a$, $n+2t=2a$, so $m=2a-3t$, $n=2a-2t$. Applying the Pythagorean theorem in $\\triangle ABF_{2}$ gives $(2a-3t)^{2}+(2a-2t)^{2}=25t^{2}$, solving yields $a=3t$, $2a=-t$ (discarded). Thus $m=3t$, $n=4t$. In $\\triangle AF_{1}F_{2}$, $\\cos\\angle F_{1}AF_{2}=\\frac{3}{5}$. Applying the law of cosines gives $2c=3t\\sqrt{1+1-2\\times1\\times1\\times\\frac{3}{5}}=\\frac{6t}{\\sqrt{5}}$, so $c=\\frac{3t}{\\sqrt{5}}$, that is, $\\frac{c}{a}=\\frac{1}{\\sqrt{5}}=\\frac{\\sqrt{5}}{5}$. Therefore, the answer is $\\frac{\\sqrt{5}}{5}$."
  },
  {
    "text": "Given that the standard equation of the hyperbola $C$ is $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and the distance from its focus $F(3,0)$ to an asymptote is equal to $\\sqrt{5}$, then the standard equation of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F: Point;Focus(C) = F;Coordinate(F) = (3, 0);Distance(F, Asymptote(C)) = sqrt(5)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4 - y^2/5 = 1",
    "fact_spans": "[[[2, 8], [69, 70], [101, 104]], [[2, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[72, 80]], [[69, 80]], [[72, 80]], [[69, 99]]]",
    "query_spans": "[[[101, 111]]]",
    "process": "Analysis: According to the standard equation of hyperbola $ C $, obtain the equations of the asymptotes of the hyperbola. Then, since the distance from focus $ F(3,0) $ to the asymptote equals $ \\sqrt{5} $, using the point-to-line distance formula, we get $ \\frac{3b}{c} = \\sqrt{5} $. Thus, $ b $ can be found. Then, using $ a^{2} = c^{2} - b^{2} $, the standard equation of the hyperbola can be determined. Detail: $ \\because $ the standard equation of hyperbola $ C $ is $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $), $ \\therefore $ the equations of the asymptotes of the hyperbola are $ y = \\pm\\frac{b}{a}x $, i.e., $ bx \\pm ay = 0 $. $ \\because $ the distance from focus $ F(3,0) $ to the asymptote equals $ \\sqrt{5} $, $ \\therefore d = \\frac{3b}{\\sqrt{b^{2} + a^{2}}} = \\sqrt{5} $, i.e., $ \\frac{3b}{c} = \\sqrt{5} $. $ \\because c = 3 $, $ \\therefore b = \\sqrt{5} $, $ \\therefore a^{2} = c^{2} - b^{2} = 4 $, $ \\therefore $ the standard equation of the hyperbola is $ \\frac{x^{2}}{4} - \\frac{y^{2}}{5} = 1 $."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2 p x(p>0)$ be $F$, and let its directrix intersect the $x$-axis at point $C$. Draw a chord $AB$ through point $F$. If $\\angle C B F=90^{\\circ}$, then $A F-B F$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(G) = F;Intersection(Directrix(G), xAxis) = C;C: Point;PointOnCurve(F, LineSegmentOf(A, B)) = True;IsChordOf(LineSegmentOf(A, B), G) = True;A: Point;B: Point;AngleOf(C, B, F) = ApplyUnit(90, degree)",
    "query_expressions": "LineSegmentOf(A, F) - LineSegmentOf(B, F)",
    "answer_expressions": "2*p",
    "fact_spans": "[[[1, 22], [30, 31], [51, 52]], [[1, 22]], [[4, 22]], [[4, 22]], [[26, 29], [46, 50]], [[1, 29]], [[30, 44]], [[40, 44]], [[45, 59]], [[51, 59]], [[54, 59]], [[54, 59]], [[61, 86]]]",
    "query_spans": "[[[88, 99]]]",
    "process": "As shown in the figure below, let $ BF = x $, draw a perpendicular from $ B $ to line $ l $, with foot $ H $. Then it is easy to obtain $ \\triangle ACFB \\sim \\triangle ABCH $. Thus, $ BC = |CF||BH| = px $. Also, since $ CF = |BC| + |BF| \\Rightarrow p^{2} = x^{2} + px \\Rightarrow x = \\frac{\\sqrt{5}-1}{2}p $. By the focal chord property of the parabola, $ \\frac{1}{|AF|} + \\frac{1}{|BP|} = \\frac{2}{P} $, so $ \\frac{1}{|AF|} = \\frac{3-\\sqrt{5}}{2p} \\Rightarrow |AF| = \\frac{3+\\sqrt{5}}{2}p $. Therefore, $ |AF| - |BF| = 2p $. Hence, fill in: $ 2p $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{m}=1$ $(m>0)$ with a focus at $(3,0)$, what is its asymptote equation?",
    "fact_expressions": "C: Hyperbola;m: Number;m>0;Expression(C) = (x^2/4 - y^2/m = 1);Coordinate(OneOf(Focus(C))) = (3, 0)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*(sqrt(5)/2)*x",
    "fact_spans": "[[[2, 50], [67, 68]], [[10, 50]], [[10, 50]], [[2, 50]], [[2, 65]]]",
    "query_spans": "[[[67, 75]]]",
    "process": "Since one focus of the hyperbola $ C:\\frac{x^{2}}{4}-\\frac{y^{2}}{m}=1 $ $(m>0)$ has coordinates $(3,0)$, we have $ a^{2}=4 $, $ c^{2}=9 $, $ b^{2}=m $, and since $ c^{2}=a^{2}+b^{2} $, it follows that $ m=5 $. Therefore, the equation of the hyperbola is $ \\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1 $, and thus the asymptotes of the hyperbola are $ y=\\pm\\frac{\\sqrt{5}}{2}x^{;} $"
  },
  {
    "text": "What is the equation of the directrix of the parabola $x^{2}=-y$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = -y)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=1/4",
    "fact_spans": "[[[0, 13]], [[0, 13]]]",
    "query_spans": "[[[0, 20]]]",
    "process": "From the parabola equation $x^{2} = -y$, it can be seen that the focus lies on the negative half of the $y$-axis. From the characteristics of the standard equation, we know $2p = 1$, $p = \\frac{1}{2}$, hence the directrix is $y = \\frac{1}{4}$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=6x$, a line $l$ passing through the focus $F$ of the parabola intersects the parabola at point $A$ and intersects the directrix of the parabola at point $B$. If $\\overrightarrow{FB}=3\\overrightarrow{FA}$, then the distance from point $A$ to the origin is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 6*x);F: Point;Focus(C) = F;l: Line;PointOnCurve(F,l) = True;Intersection(C,l) = A;Intersection(Directrix(C),l) = B;B: Point;A: Point;VectorOf(F, B) = 3*(VectorOf(F, A));O: Origin",
    "query_expressions": "Distance(A,O)",
    "answer_expressions": "sqrt(13)/2",
    "fact_spans": "[[[2, 21], [23, 26], [39, 42], [49, 52]], [[2, 21]], [[29, 32]], [[23, 32]], [[33, 38]], [[22, 38]], [[33, 47]], [[33, 60]], [[56, 60]], [[43, 47], [110, 114]], [[62, 108]], [[115, 117]]]",
    "query_spans": "[[[110, 122]]]",
    "process": "Parabola C, the directrix is perpendicular to the x-axis, with foot at D, |DF| = 3. Draw a perpendicular from A to the directrix, with foot at C. By the definition of a parabola: |AF| = |AC|. Therefore, 3|AF| = |FB|, so |AB| = 2|AC|. From similar triangles, |AC| = 2. Thus, the x-coordinate of point A is |AC| - \\frac{3}{2} = \\frac{1}{2}. Hence, the coordinates of point A are (\\frac{3}{2}, -\\sqrt{3}). The distance from point A to the origin is \\frac{\\sqrt{13}}{2}."
  },
  {
    "text": "Given the ellipse $\\Gamma$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ has its right focus at $F$, and a line passing through the origin $O$ intersects the ellipse $\\Gamma$ at points $A$ and $B$, then the range of $\\frac{1}{|A F|}+\\frac{1}{|B F|}$ is?",
    "fact_expressions": "Gamma: Ellipse;H: Line;A: Point;F: Point;B: Point;Expression(Gamma) = (x^2/4 + y^2/3 = 1);RightFocus(Gamma)=F;O: Origin;PointOnCurve(O, H);Intersection(H, Gamma) = {A, B}",
    "query_expressions": "Range(1/Abs(LineSegmentOf(B, F)) + 1/Abs(LineSegmentOf(A, F)))",
    "answer_expressions": "[1, 4/3]",
    "fact_spans": "[[[2, 49], [68, 78]], [[65, 67]], [[80, 83]], [[54, 57]], [[84, 87]], [[2, 49]], [[2, 57]], [[59, 64]], [[58, 67]], [[65, 89]]]",
    "query_spans": "[[[91, 131]]]",
    "process": "Using the definition of an ellipse, let |AF| = x \\in [1,3], then |BF| = 4 - x. Construct the function f(x) = \\frac{1}{x} + \\frac{1}{4 - x}, x \\in [1,3], and use derivatives to find its range. Take the left focus F of the ellipse, connect AF, BF, AF', BF'. It is easy to see that quadrilateral AFBF' is a parallelogram, so |AF| + |BF| = |AF| + |AF'| = 2a = 4. Let |AF| = x \\in [1,3], then |BF| = 4 - x, hence \\frac{1}{|AF|} + \\frac{1}{|BF|} = \\frac{1}{x} + \\frac{1}{4 - x}. Let f(x) = \\frac{1}{x} + \\frac{1}{4 - x}, x \\in [1,3], then f'(x) = -\\frac{1}{(x-4)^{2}} + \\frac{1}{x^{2}} = \\frac{x^{2} - (x-4)^{2}}{x^{2}(x-4)^{2}} = \\frac{8(x-2)}{x^{2}(x-4)^{2}}. It is clear that the function f(x) is monotonically decreasing on [1,2) and monotonically increasing on [2,3]. Therefore, f(x)_{\\max} = f(1) = f(3) = \\frac{4}{3}, f(x)_{\\min} = f(2) = 1. Thus, the range of \\frac{1}{|AF|} + \\frac{1}{|BF|} is [1, \\frac{4}{3}]."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{12}=1$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively, and $M$ is a point on the hyperbola. If $\\angle F_{1} M F_{2}=60^{\\circ}$, then the area of triangle $F_{1} M F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;F1: Point;M: Point;F2: Point;Expression(G) = (x^2/16 - y^2/12 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(M,G);AngleOf(F1, M, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(F1, M, F2))",
    "answer_expressions": "12*sqrt(3)",
    "fact_spans": "[[[2, 42], [71, 74]], [[51, 58]], [[67, 70]], [[59, 66]], [[2, 42]], [[2, 66]], [[2, 66]], [[67, 77]], [[79, 112]]]",
    "query_spans": "[[[114, 137]]]",
    "process": "According to the definition of hyperbola and the law of cosines, |MF_{1}||MF_{2}|=48, then calculate the area of \\triangle MF_{1}F_{2} using the triangle area formula. By the symmetry of the hyperbola, assume point M lies on the right branch of the hyperbola, so |MF_{1}|>|MF_{2}|, and |MF_{1}|-|MF_{2}|=2a=8, thus (|MF_{1}|-|MF_{2}|)^{2}=|MF_{1}|^{2}+|MF_{2}|^{2}-2|MF_{1}||MF_{2}|=64. Also, 4c^{2}=112=|MF_{1}|^{2}+|MF_{2}|^{2}-|MF_{1}||MF_{2}|. Subtracting these two equations yields |MF_{1}||MF_{2}|=48. Therefore, S_{\\Delta MF_{1}F_{2}}=\\frac{1}{2}|MF_{1}||MF_{2}|\\times\\sin60^{\\circ}=12\\sqrt{3}."
  },
  {
    "text": "Given that $P$ is a moving point on the parabola $y^{2}=4x$, $F$ is the focus of the parabola, and point $Q$ lies on the circle $C$: $(x-2)^{2}+(y-1)^{2}=1$, then the minimum value of $|PQ|+|PF|$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);C: Circle;P: Point;Q: Point;F: Point;Expression(C) = ((x - 2)^2 + (y - 1)^2 = 1);PointOnCurve(P, G);Focus(G) = F;PointOnCurve(Q, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F)) + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "2",
    "fact_spans": "[[[6, 20], [29, 32]], [[6, 20]], [[41, 69]], [[2, 5]], [[36, 40]], [[25, 28]], [[41, 69]], [[2, 24]], [[25, 35]], [[36, 70]]]",
    "query_spans": "[[[72, 91]]]",
    "process": "Draw a perpendicular line from point Q to the directrix $ x = -1 $ of the parabola, with foot of perpendicular at N, as shown in the figure. By the definition of the parabola, $ |PF| = |PN| $. When points P, Q, and N pass through the center of the circle, $ |PQ| + |PF| $ attains its minimum value. The center of the circle is $ (2, 1) $ and the radius is 1. Therefore, the minimum value of $ |PQ| + |PF| $ is: 2."
  },
  {
    "text": "$P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola, respectively, and $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$. The line $P F_{2}$ intersects the $y$-axis at point $A$. Then the inradius of $\\Delta A F_{1} P$ is?",
    "fact_expressions": "G: Hyperbola;F2: Point;P: Point;A: Point;F1: Point;Expression(G) = (x^2/4 - y^2/9 = 1);PointOnCurve(P, RightPart(G));LeftFocus(G) = F1;RightFocus(G) = F2;DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Intersection(LineOf(P,F2), yAxis) = A",
    "query_expressions": "Radius(InscribedCircle(TriangleOf(A,F1,P)))",
    "answer_expressions": "2",
    "fact_spans": "[[[4, 42], [66, 69]], [[56, 63]], [[0, 3]], [[154, 158]], [[48, 55]], [[4, 42]], [[0, 47]], [[48, 75]], [[48, 75]], [[77, 136]], [[137, 158]]]",
    "query_spans": "[[[160, 186]]]",
    "process": "Analysis: This problem first derives the relationship between side lengths based on the inradius of a right triangle, then combines the definition of a hyperbola and the symmetry of the graph to reach the conclusion. Details: \\because PF_{1}\\bot PF_{2}, the inradius of \\triangle APF_{1} is r \\therefore |PF_{1}|+|PA|\\cdot|AF_{1}|=2r, \\therefore |PF_{2}|+2a+|PA|\\cdot|AF_{1}|=2r \\therefore |AF_{2}|-|AF_{1}|=2x-4, \\because by symmetry of the graph: |AF_{2}|=|AF_{1}|, \\therefore r=2"
  },
  {
    "text": "It is known that the vertex of the parabola $C$ is at the origin, and the focus of the parabola coincides with the right focus of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$. Then the equation of the parabola $C$ is?",
    "fact_expressions": "O: Origin;Vertex(C) = O;C: Parabola;G: Ellipse;Expression(G) = (x^2/16 + y^2/7 = 1);Focus(C) = RightFocus(G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=12*x",
    "fact_spans": "[[[5, 7]], [[2, 14]], [[8, 14], [64, 70]], [[18, 56]], [[18, 56]], [[8, 62]]]",
    "query_spans": "[[[64, 75]]]",
    "process": "The ellipse has $a^{2}=16$, $b^{2}=7$, thus $c^{2}=a^{2}-b^{2}=9$, hence $c=3$, so the coordinates of the right focus of the ellipse are $(3,0)$, therefore $\\frac{p}{2}=3$, so $2p=12$, thus the equation of the parabola is $y^{2}=12x$."
  },
  {
    "text": "The coordinates of the two foci of an ellipse are $(-2,0)$ and $(2,0)$, and it passes through the point $(2 \\sqrt{3}, \\sqrt{3})$. What is the standard equation of this ellipse?",
    "fact_expressions": "G: Ellipse;Focus(G) = {H, I};H: Point;Coordinate(H) = (-2, 0);I: Point;Coordinate(I) = (2, 0);J: Point;Coordinate(J) = (2*sqrt(3), sqrt(3));PointOnCurve(J, G) = True",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16+y^2/12=1",
    "fact_spans": "[[[2, 4], [61, 63]], [[2, 29]], [[13, 21]], [[13, 21]], [[22, 29]], [[22, 29]], [[33, 58]], [[33, 58]], [[2, 58]]]",
    "query_spans": "[[[61, 70]]]",
    "process": "According to the problem, the coordinates of the two foci of the ellipse are (-2,0) and (2,0), so c=2. Let the equation of the ellipse be \\frac{x^{2}}{a2}+\\frac{y^{2}}{a^{2}-4}=1. Since the ellipse passes through the point (2\\sqrt{3},\\sqrt{3}), we have \\frac{12}{a^{2}}+\\frac{3}{a^{2}-4}=1. Solving this gives a=4. Therefore, the equation of the ellipse is \\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1"
  },
  {
    "text": "Given point $Q(2 \\sqrt{2}, 0)$ and a moving point $P(x_1, y_1)$ on the parabola $y = \\frac{x^{2}}{4}$, find the minimum value of $y + |P Q|$.",
    "fact_expressions": "G: Parabola;Q: Point;P: Point;x1: Number;y1: Number;Expression(G) = (y = x^2/4);Coordinate(Q) = (2*sqrt(2), 0);Coordinate(P) = (x1, y1);PointOnCurve(P, G)",
    "query_expressions": "Min(y1+Abs(LineSegmentOf(P,Q)))",
    "answer_expressions": "2",
    "fact_spans": "[[[22, 44]], [[2, 21]], [[48, 59]], [[48, 59]], [[48, 59]], [[22, 44]], [[2, 21]], [[48, 59]], [[22, 59]]]",
    "query_spans": "[[[61, 76]]]",
    "process": "Let the focus of the parabola be $ F(0,1) $. From the property of the parabola: $ y + |PQ| = |PF| - 1 + |PQ| $, so the minimum value of $ y + |PQ| $ is $ |FQ| - 1 = 2 $."
  },
  {
    "text": "Given the hyperbola $E$: $m x^{2}+n y^{2}=1$ $(n>0)$ has eccentricity $2$, then what is the equation of its asymptotes?",
    "fact_expressions": "E: Hyperbola;m: Number;n: Number;n>0;Expression(E) = (m*x^2 + n*y^2 = 1);Eccentricity(E) = 2",
    "query_expressions": "Expression(Asymptote(E))",
    "answer_expressions": "y=pm*(sqrt(3)/3)*x",
    "fact_spans": "[[[2, 34], [44, 45]], [[9, 34]], [[9, 34]], [[9, 34]], [[2, 34]], [[2, 42]]]",
    "query_spans": "[[[44, 52]]]",
    "process": "Transform the hyperbola into $\\frac{y^{2}}{n}-\\frac{y^{2}}{m}=1(m<0)$, obtain $a$, $b$, $c$, and combining with the eccentricity of 2, derive $3n=-m$, then find the asymptotes of the hyperbola. [Detailed solution] According to the problem, the hyperbola $E: mx^{2}+ny^{2}=1$ ($n>0$) can be rewritten as $\\frac{y^{2}}{n}-\\frac{y^{2}}{m}=1$ ($m<0$), from which we get $a=\\sqrt{n}$, $b=\\sqrt{-m}$, then $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{n-m}$. Since the eccentricity of hyperbola $E$ is 2, that is, $\\frac{c}{a}=\\frac{\\sqrt{n-m}}{\\sqrt{n}}=2$, solving gives $3n=-m$. Therefore, the equations of the asymptotes of the hyperbola are $y=\\pm\\frac{\\sqrt{n}}{\\sqrt{-m}}x=\\pm\\frac{\\sqrt{3}}{3}x$."
  },
  {
    "text": "Given that the line $y=2 x+b$ is tangent to the parabola $x^{2}=4 y$ at point $A$, and $F$ is the focus of the parabola, the line $A F$ intersects the parabola again at point $B$. Then $|B F|$=?",
    "fact_expressions": "G: Parabola;b: Number;A: Point;F: Point;B: Point;Expression(G) = (x^2 = 4*y);Expression(H) = (y = b + 2*x);TangentPoint(H, G) = A;Focus(G)=F;Intersection(LineOf(A,F),G)={A,B};H:Line",
    "query_expressions": "Abs(LineSegmentOf(B, F))",
    "answer_expressions": "5/4",
    "fact_spans": "[[[14, 28], [40, 43], [55, 58]], [[4, 13]], [[31, 35]], [[36, 39]], [[62, 65]], [[14, 28]], [[2, 13]], [[2, 35]], [[36, 46]], [[31, 65]], [[2, 13]]]",
    "query_spans": "[[[67, 76]]]",
    "process": "Let A(x_{0},\\frac{1}{4}x_{0}^{2}), x^{2}=4y \\Rightarrow y=\\frac{1}{4}x^{2} \\Rightarrow y'=\\frac{1}{2}x, the tangent line at point A is y=2x+b, so \\frac{1}{2}x_{0}=2 \\Rightarrow x_{0}=4 \\therefore A(4,4), since F is the focus of the parabola, F(0,1), thus the equation of line AF is \\frac{y-4}{4-1}=\\frac{x-4}{4-0} \\Rightarrow 3x-4y+4=0, then we have \\begin{cases}3x-4y+4=0\\\\x^{2}=4y\\end{cases} \\Rightarrow x^{2}-3x-4=0, solving gives x=4 or x=-1, therefore the coordinates of point B are: (-1,\\frac{1}{4}), the directrix of the parabola is y=-1, by the definition of parabola we get: |BF|=\\frac{1}{4}-(-1)=\\frac{5}{4}."
  },
  {
    "text": "The sum of the lengths of the real axis and the imaginary axis of a hyperbola equals $\\sqrt{2}$ times its focal distance, and the coordinates of one vertex are $(0,2)$. Then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Coordinate(OneOf(Vertex(G))) = (0, 2);Length(RealAxis(G)) + Length(ImageinaryAxis(G))=sqrt(2)*FocalLength(G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/4 - x^2/4 = 1",
    "fact_spans": "[[[0, 3], [15, 16], [49, 52]], [[0, 47]], [[0, 30]]]",
    "query_spans": "[[[49, 59]]]",
    "process": "One vertex of the hyperbola has coordinates (0,2), so the foci of the hyperbola lie on the y-axis, and a=2. The sum of the lengths of the real axis and the imaginary axis equals \\sqrt{2} times its focal distance, then 2a+2b=\\sqrt{2}\\cdot2c and c^{2}=a^{2}+b^{2}, that is, 2+b=\\sqrt{2}c, c^{2}=4+b^{2}. Solving yields b=^{n}2, c=2^{n}\\sqrt{2}. \\therefore The standard equation of the hyperbola is \\frac{y^{2}}{4}-\\frac{x^{2}}{4}="
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=x$ intersects the parabola at points $A$ and $B$, and the projections of $A$ and $B$ on the line $x=\\frac{1}{4}$ are $M$ and $N$, respectively. Then the measure of $\\angle M F N$ is?",
    "fact_expressions": "G: Parabola;H: Line;C:Line;M: Point;F: Point;N: Point;A: Point;B:Point;Expression(G) = (y^2 = x);Expression(C) = (x = 1/4);Focus(G)=F;PointOnCurve(F,H);Intersection(H,G) = {A,B};Projection(A,C)=M;Projection(B,C)=N",
    "query_expressions": "AngleOf(M, F, N)",
    "answer_expressions": "ApplyUnit(90,degree)",
    "fact_spans": "[[[1, 13], [23, 26]], [[20, 22]], [[50, 67]], [[74, 77]], [[16, 19]], [[78, 81]], [[27, 30], [27, 30]], [[33, 36], [33, 36]], [[1, 13]], [[50, 67]], [[1, 19]], [[0, 22]], [[20, 38]], [[40, 81]], [[40, 81]]]",
    "query_spans": "[[[83, 102]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{7}=1$, $A(3,0)$, $F(4,0)$, $O$ is the origin. A line $l$ passing through point $F$ intersects the hyperbola $C$ at points $M$ and $N$. If there exists a point $P$ on line $l$ such that $|\\overrightarrow{A P}+\\overrightarrow{O P}|=4$, then the minimum value of $|M N|$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;A: Point;P: Point;O: Origin;M: Point;N: Point;F: Point;Expression(C) = (x^2/9 - y^2/7 = 1);Coordinate(A)=(3, 0);Coordinate(F)=(4,0);PointOnCurve(F, l);Intersection(l, C) = {M, N};PointOnCurve(P,l);Abs(VectorOf(A,P)+VectorOf(O,P))=4",
    "query_expressions": "Min(Abs(LineSegmentOf(M, N)))",
    "answer_expressions": "6",
    "fact_spans": "[[[85, 90], [85, 90]], [[2, 45], [91, 97]], [[48, 56]], [[117, 121]], [[70, 73]], [[98, 101]], [[102, 105]], [[59, 68], [80, 84]], [[2, 45]], [[48, 56]], [[59, 68]], [[79, 90]], [[85, 107]], [[109, 121]], [[123, 170]]]",
    "query_spans": "[[[172, 185]]]",
    "process": "Let the midpoint of OA be N, then the coordinates of N are $(\\frac{3}{2},0)$. From the given condition, there exists a point P on line $l$ such that $|\\overrightarrow{AP}+\\overrightarrow{OP}|=4=2|\\overrightarrow{NP}|$, i.e., $|PN|=2$, meaning the distance from N to line $l$ is less than or equal to 2. When line $l$ intersects each of the left and right branches of the hyperbola at one point, by the geometric properties of the hyperbola, the minimum chord length $|MN|$ is $2a=6$, at which time line $l$ is the x-axis and the distance from N to $l$ is 0, satisfying the condition. When line $l$ intersects the right branch of the hyperbola at two points, the shorter the chord, the greater the absolute value of the slope of the line. When the slope does not exist, i.e., when MN is the latus rectum, $|MN|$ attains its minimum value $\\frac{2b^{2}}{a}=\\frac{14}{3}<6$, but at this time the distance from point M to line $l$ is $4-\\frac{3}{2}=\\frac{5}{2}>2$. When the slope of the line exists, the range of the slope satisfies $|k|>\\frac{b}{a}=\\frac{\\sqrt{7}}{3}$, and the equation of the line is $y=k(x-4)$, $k^{2}>\\frac{7}{9}$. Since the distance from N to line $l$ is less than or equal to 2, we have: $\\frac{|k(\\frac{3}{2}-4)|}{\\sqrt{1+k^{2}}}\\leqslant2$, solving gives $|k|\\leqslant\\frac{4}{3}$, $\\therefore k^{2}\\in(\\frac{7}{9},\\frac{16}{9}]$. Substituting the line equation $y=k(x-4)$ into the hyperbola equation and simplifying yields $(9k^{2}-7)x^{2}-72k^{2}x+9(16k^{2}+7)=0$, $\\Delta=(8\\times9k^{2})^{2}-4\\times9(9k^{2}-7)(16k^{2}+7)=4\\times9\\times49(k^{2}+1)$, it is easy to see that $9k^{2}-7>0$. Let the x-coordinates of M and N be $x_{1},x_{2}$ respectively, then $|x_{1}-x_{2}|=\\frac{42\\sqrt{k^{2}+1}}{9k^{2}-7}$, $|MN|=\\sqrt{1+k^{2}}|x_{1}-x_{2}|=\\frac{42(k^{2}+1)}{9k^{2}-7}=\\frac{42}{9}1+\\frac{16}{k^{2}-7}$, $k^{2}-\\frac{7}{9}\\in(0,1)$, $\\therefore |MN|\\geqslant\\frac{42}{9}(1+\\frac{16}{9})=\\frac{350}{27}>6$. In summary, the minimum value of $|MN|$ is 6."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$. A line $l$ with slope $\\sqrt{5}$ passing through $F$ intersects $C$ at points $M$ and $N$. If the ordinate of the midpoint of segment $MN$ is $\\sqrt{10}$, then the distance from $F$ to the directrix of $C$ is?",
    "fact_expressions": "l: Line;C: Parabola;p: Number;N: Point;M: Point;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(F, l);Slope(l) = sqrt(5);Intersection(l, C) = {M, N};YCoordinate(MidPoint(LineSegmentOf(M, N))) = sqrt(10)",
    "query_expressions": "Distance(F, Directrix(C))",
    "answer_expressions": "5*sqrt(2)",
    "fact_spans": "[[[55, 60]], [[2, 28], [61, 64], [108, 111]], [[10, 28]], [[70, 73]], [[66, 69]], [[32, 35], [37, 40], [104, 107]], [[10, 28]], [[2, 28]], [[2, 35]], [[36, 60]], [[41, 60]], [[55, 75]], [[77, 102]]]",
    "query_spans": "[[[104, 119]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Using the point difference method, we obtain \\sqrt{5}(y_{1}+y_{2})=2p. Finally, according to the fact that the y-coordinate of the midpoint of segment AB is \\sqrt{10}, the result can be found. [Detailed solution] Let A(x_{1},y_{1}), B(x_{2},y_{2}), then y_{1}=2px_{1}, y_{2}^{2}=2px_{2}. Subtracting the two equations gives y_{1}-y_{2}=2px_{1}-2px_{2}, i.e., (y_{1}-y_{2})(y_{1}+y_{2})=2p(x_{1}-x_{2}). Since points A and B lie on a line l with slope \\sqrt{5}, we have \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\sqrt{5}. Therefore, from (y_{1}-y_{2})(y_{1}+y_{2})=2p(x_{1}-x_{2}), we get \\sqrt{5}(y_{1}+y_{2})=2p. Since the y-coordinate of the midpoint of segment AB is \\sqrt{10}, we have y_{1}+y_{2}=2\\sqrt{10}. Then \\sqrt{5}\\times2\\sqrt{10}=2p, so p=5\\sqrt{2}. Therefore, the distance from F to the directrix of C is 5\\sqrt{2}."
  },
  {
    "text": "If the focus of the parabola $y^{2}=a x(a>0)$ coincides with one focus of the hyperbola $\\frac{x^{2}}{7}-\\frac{y^{2}}{2}=1$, then the equation of the parabola is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;a: Number;Expression(G) = (x^2/7 - y^2/2 = 1);a>0;Expression(H) = (y^2 = a*x);Focus(H) = OneOf(Focus(G))",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2 = 12*x",
    "fact_spans": "[[[24, 62]], [[1, 20], [72, 75]], [[4, 20]], [[24, 62]], [[4, 20]], [[1, 20]], [[1, 69]]]",
    "query_spans": "[[[72, 80]]]",
    "process": "First, find the coordinates of the foci of the hyperbola $\\frac{x^{2}}{7}-\\frac{y^{2}}{2}=1$ and the parabola $y^{2}=ax$ ($a>0$). Since one focus of the parabola coincides with a focus of the hyperbola, solve for $a$, and thus obtain the equation of the parabola. Solution: The foci of the hyperbola $\\frac{x^{2}}{7}-\\frac{y^{2}}{2}=1$ are $(-3,0)$ and $(3,0)$. The focus of the parabola $y^{2}=ax$ ($a>0$) is $(\\frac{a}{4},0)$. Since the focus of the parabola $y^{2}=ax$ ($a>0$) coincides with one focus of the hyperbola $\\frac{x^{2}}{7}-\\frac{y^{2}}{2}=1$, we have $\\frac{a}{4}=3$, so $a=12$. Therefore, the equation of the parabola is $y^{2}=12x$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a, b>0)$ and the parabola $y^{2}=8x$ share a common focus $F$, and $P$ is an intersection point of the two curves. If $|PF|=5$, then the asymptotes of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;P: Point;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 8*x);OneOf(Focus(G)) = F;Focus(H) = F;OneOf(Intersection(H,G))=P;Abs(LineSegmentOf(P, F)) = 5",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[2, 56], [109, 112]], [[5, 56]], [[5, 56]], [[57, 71]], [[93, 96]], [[78, 81]], [[5, 56]], [[5, 56]], [[2, 56]], [[57, 71]], [[2, 81]], [[2, 81]], [[83, 96]], [[98, 107]]]",
    "query_spans": "[[[109, 119]]]",
    "process": "The focus of the parabola $ y^{2}=8x $ is $ (2,0) $, and the directrix is $ x=-2 $, so $ c=2 $. Let $ P(x_{0},y_{0}) $. Since $ |PF|=5 $, we have $ x_{0}+2=5 $, hence $ x_{0}=3 $. Without loss of generality, assume $ P(3,2\\sqrt{6}) $. Therefore, $ \\frac{(2\\sqrt{6})^{2}}{b^{2}}=1 $, solving gives $ a=1 $, $ b=\\sqrt{3} $. Thus, the asymptotes of the hyperbola are $ v=+\\frac{b}{x}=+\\sqrt{3}x $."
  },
  {
    "text": "Given the hyperbola equation $x^{2}-4 y^{2}=4$, find the equation of the line $l$ containing the chord for which $M(4,1)$ is the midpoint.",
    "fact_expressions": "l: Line;G: Hyperbola;M: Point;Z: LineSegment;Coordinate(M) = (4, 1);Expression(G) = (x^2 - 4*y^2 = 4);IsChordOf(Z, G);MidPoint(Z) = M;OverlappingLine(Z, l)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x - y - 3 = 0",
    "fact_spans": "[[[44, 49]], [[2, 5]], [[29, 37]], [], [[29, 37]], [[2, 25]], [[2, 42]], [[2, 42]], [[2, 49]]]",
    "query_spans": "[[[44, 54]]]",
    "process": ""
  },
  {
    "text": "Given that the focus of the parabola $x^{2}=4 y$ is $F$, the intersection point of the directrix and the $y$-axis is $M$, $N$ is a point on the parabola, and $|N F|=\\frac{\\sqrt{3}}{2}|M N|$, then $\\angle {N M F}$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);F: Point;Focus(G) = F;Intersection(Directrix(G),yAxis) = M;M: Point;PointOnCurve(N,G) = True;N: Point;Abs(LineSegmentOf(N, F)) = (sqrt(3)/2)*Abs(LineSegmentOf(M, N))",
    "query_expressions": "AngleOf(N, M, F)",
    "answer_expressions": "ApplyUnit(30, degree)",
    "fact_spans": "[[[2, 16], [43, 46]], [[2, 16]], [[20, 23]], [[2, 23]], [[2, 38]], [[35, 38]], [[39, 50]], [[39, 42]], [[52, 83]]]",
    "query_spans": "[[[86, 104]]]",
    "process": ""
  },
  {
    "text": "Given that the sum of the distances from a point $P$ on an ellipse to the two foci $F_{1}$, $F_{2}$ is $20$, then the maximum value of $P F_{1} \\cdot P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;PointOnCurve(P,G);Distance(P,F1)+Distance(P,F2)=20;Focus(G)={F1,F2}",
    "query_expressions": "Max(LineSegmentOf(P, F1)*LineSegmentOf(P, F2))",
    "answer_expressions": "100",
    "fact_spans": "[[[2, 4]], [[7, 10]], [[14, 21]], [[22, 29]], [[2, 10]], [[2, 39]], [[2, 29]]]",
    "query_spans": "[[[41, 70]]]",
    "process": "According to the definition of an ellipse: PF_{1} + PF_{2} = 2a = 20. Combining with the basic inequality, we have: PF_{1} \\cdot PF_{2} \\leqslant \\left( \\frac{PF_{1} + PF_{2}}{2} \\right)^{2} = 10^{2} = 100. The equality holds if and only if: PF_{1} = PF_{2} = 10, at which point PF_{1} \\cdot PF_{2} reaches the maximum value of 100. Hence, the maximum value of PF_{1} \\cdot PF_{2} is 100."
  },
  {
    "text": "The line $y=x$ intersects the parabola $y^{2}=2 x$ at points $A$ and $B$. Then $|A B|=$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;Expression(G) = (y^2 = 2*x);Expression(H) = (y = x);Intersection(H, G) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[8, 22]], [[0, 7]], [[24, 27]], [[28, 31]], [[8, 22]], [[0, 7]], [[0, 33]]]",
    "query_spans": "[[[35, 44]]]",
    "process": "Solving the system \\begin{cases}y=x\\\\y^2=2x\\end{cases}, the two intersection points are A(0,0), B(2,2), hence |AB|=\\sqrt{(2-0)^{2}+(2-0)^{2}}=2\\sqrt{2}"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4-b^{2}}-\\frac{y^{2}}{b^{2}}=1(0<b<2)$ intersects the $x$-axis at points $A$ and $B$, and $C(0, b)$, then the maximum area of $\\triangle A B C$ is?",
    "fact_expressions": "G: Hyperbola;b: Number;C: Point;A: Point;B: Point;0 < b;b < 2;Expression(G) = (x^2/(4 - b^2) - y^2/b^2 = 1);Coordinate(C) = (0, b);Intersection(G, xAxis) = {A, B}",
    "query_expressions": "Max(Area(TriangleOf(A, B, C)))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 57]], [[5, 57]], [[74, 83]], [[64, 67]], [[68, 71]], [[5, 57]], [[5, 57]], [[2, 57]], [[74, 83]], [[2, 73]]]",
    "query_spans": "[[[85, 111]]]",
    "process": "\\because the hyperbola \\frac{x^2}{4-b^{2}}-\\frac{y^{2}}{b^{2}}=1 (0<b<2) intersects the x-axis at points A and B, \\therefore A(-\\sqrt{4-b^{2}},0), B(\\sqrt{4-b^{2}},0). \\because point C(0,b), \\therefore the area S of \\Delta ABC = \\frac{1}{2} \\times 2\\sqrt{4-b^{2}} \\times b = \\sqrt{4-b^{2}} \\times b = \\sqrt{(4-b^{2})b^{2}} \\leqslant \\frac{4-b^{2}+b^{2}}{2} = 2, with equality if and only if b=\\sqrt{2}, \\therefore the maximum area of \\Delta ABC is 2, hence the answer is ?"
  },
  {
    "text": "Given that the horizontal coordinate of the midpoint of chord $AB$ of the parabola $y^2 = 4x$ is $2$, then the maximum value of $|AB|$ is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;Expression(G) = (y^2 = 4*x);IsChordOf(LineSegmentOf(A,B),G);XCoordinate(MidPoint(LineSegmentOf(A,B)))=2",
    "query_expressions": "Max(Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 16]], [[18, 23]], [[18, 23]], [[2, 16]], [[2, 23]], [[18, 34]]]",
    "query_spans": "[[[36, 49]]]",
    "process": ""
  },
  {
    "text": "Suppose the distance between the directrix of a parabola $y^{2}=m x$ and the line $x=1$ is $3$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;m: Number;H: Line;Expression(G) = (y^2 = m*x);Expression(H) = (x = 1);Distance(Directrix(G),H) = 3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=8*x,y^2=-16*x}",
    "fact_spans": "[[[2, 16], [39, 42]], [[5, 16]], [[20, 27]], [[2, 16]], [[20, 27]], [[2, 36]]]",
    "query_spans": "[[[39, 47]]]",
    "process": "The lines at a distance of 3 from the line $x=1$ are $x=4$ and $x=-2$, and the directrix equation of the parabola $y^{2}=mx$ is $x=-\\frac{m}{4}$. Therefore, $-\\frac{m}{4}=-2$ or $-\\frac{m}{4}=4$, so $m=8$ or $m=-16$."
  },
  {
    "text": "Let the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ be $F_{1}$, $F_{2}$, and let $P$ be a point on this hyperbola such that $2|P F_{1}|=3|P F_{2}|$. If $\\angle F_{1} P F_{2}=60^{\\circ}$, then the asymptotes of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);2*Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, F2));AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*sqrt(6)*x",
    "fact_spans": "[[[1, 57], [85, 88], [154, 157]], [[4, 57]], [[4, 57]], [[80, 83]], [[64, 71]], [[72, 79]], [[4, 57]], [[4, 57]], [[80, 91]], [[1, 79]], [[1, 79]], [[1, 57]], [[93, 116]], [[118, 151]]]",
    "query_spans": "[[[154, 165]]]",
    "process": "2|PF_{1}|=3|PF_{2}|, |PF_{1}|-|PF_{2}|=2a, hence |PF_{1}|=6a, |PF_{2}|=4a. In \\triangle PF_{1}F_{2}, using the law of cosines: 4c^{2}=36a^{2}+16a^{2}-2\\cdot6a\\cdot4a\\cos60^{\\circ}, simplifying yields c=\\sqrt{7}a, and b=\\sqrt{c^{2}-a^{2}}=\\sqrt{(\\sqrt{7}a)^{2}-a^{2}}=\\sqrt{6}a, thus the asymptotes are: y=\\pm\\sqrt{6}x"
  },
  {
    "text": "Given the parabola $y^{2}=4 x$, $P$ is a point on the parabola, $F$ is the focus, and $|PF|=5$. Find the coordinates of point $P$.",
    "fact_expressions": "G: Parabola;P: Point;F: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(P, G);Abs(LineSegmentOf(P, F)) = 5;Focus(G) = F",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(4, 4), (4, -4)",
    "fact_spans": "[[[2, 16], [23, 26]], [[19, 22], [48, 52]], [[30, 33]], [[2, 16]], [[19, 29]], [[38, 46]], [[23, 36]]]",
    "query_spans": "[[[48, 57]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and point $P$ is a point on the right branch of the hyperbola such that $|P F_{2}|= | F_{1} F_{2} |$. Then the area of triangle $P F_{1} F_{2}$ equals?",
    "fact_expressions": "G: Hyperbola;P: Point;F2: Point;F1: Point;Expression(G) = (x^2/9 - y^2/16 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P,RightPart(G));Abs(LineSegmentOf(P,F2))=Abs(LineSegmentOf(F1,F2))",
    "query_expressions": "Area(TriangleOf(P,F1,F2))",
    "answer_expressions": "48",
    "fact_spans": "[[[2, 41], [70, 73]], [[65, 69]], [[57, 64]], [[49, 56]], [[2, 41]], [[2, 64]], [[2, 64]], [[65, 78]], [[80, 108]]]",
    "query_spans": "[[[110, 134]]]",
    "process": ""
  },
  {
    "text": "Let $F$ be the focus of the parabola $C$: $y^{2}=3 x$. A line passing through $F$ with an inclination angle of $30^{\\circ}$ intersects $C$ at points $A$ and $B$, and $O$ is the origin. Then the area of $\\triangle A B O$ is?",
    "fact_expressions": "C: Parabola;G: Line;A: Point;B: Point;O: Origin;F: Point;Expression(C) = (y^2 = 3*x);Focus(C) = F;PointOnCurve(F, G);Inclination(G) = ApplyUnit(30, degree);Intersection(G, C) = {A, B}",
    "query_expressions": "Area(TriangleOf(A, B, O))",
    "answer_expressions": "9/4",
    "fact_spans": "[[[5, 24], [53, 56]], [[50, 52]], [[57, 60]], [[61, 64]], [[67, 70]], [[1, 4], [29, 32]], [[5, 24]], [[1, 27]], [[28, 52]], [[33, 52]], [[50, 66]]]",
    "query_spans": "[[[77, 99]]]",
    "process": ""
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ has an eccentricity of $\\frac{\\sqrt{3}}{2}$, and intersects the line $x+2 y+8=0$ at two points $P$ and $Q$. If $|P Q|=\\sqrt{10}$, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;a: Number;b: Number;H: Line;P: Point;Q: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x + 2*y + 8 = 0);Eccentricity(G) = sqrt(3)/2;Intersection(G, H) = {P, Q};Abs(LineSegmentOf(P, Q)) = sqrt(10)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/36+y^2/9=1",
    "fact_spans": "[[[0, 52], [79, 81], [128, 130]], [[2, 52]], [[2, 52]], [[82, 95]], [[98, 101]], [[102, 105]], [[2, 52]], [[2, 52]], [[0, 52]], [[82, 95]], [[0, 77]], [[79, 107]], [[109, 126]]]",
    "query_spans": "[[[128, 134]]]",
    "process": "First, based on the eccentricity and $a^{2}=b^{2}+c^{2}$, obtain the multiple relationship between $a^{2}$ and $b^{2}$, then simplify the ellipse equation and combine it with the line equation. Use Vieta's formulas and the chord length formula to express $|PQ|$, thereby solving for the values of $a^{2}$ and $b^{2}$, so that the ellipse equation can be determined. Solution: Since $e=\\frac{\\sqrt{3}}{2}$, we have $\\frac{c^{2}}{a^{2}}=\\frac{a^{2}-b^{2}}{a^{2}}=\\frac{3}{4}$, thus $a^{2}=4b^{2}$. Therefore, the ellipse equation is $x^{2}+4y^{2}=a^{2}$. Let $P(x_{1},y_{1})$, $Q(x_{2},y_{2})$. Combining the ellipse equation with $x+2y+8=0$, we get\n\\[\n\\begin{cases}\nx+2y+8=0 \\\\\nx^{2}+4y^{2}=a^{2}\n\\end{cases}\n\\]\nEliminating $y$ gives $2x^{2}+16x+64-a^{2}=0$, so $x_{1}+x_{2}=-8$, $x_{1}x_{2}=32-\\frac{a^{2}}{2}$. From $\\Delta=256-8(64-a^{2})>0$, we get $a^{2}>32$. Using the chord length formula:\n\\[\n|PQ|=\\sqrt{1+k^{2}} \\cdot \\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}} = \\sqrt{1+\\frac{1}{4}} \\cdot \\sqrt{64-4\\left(32-\\frac{a^{2}}{2}\\right)} = \\sqrt{10},\n\\]\nSolving yields $a^{2}=36$, $b^{2}=9$, satisfying $a^{2}>32$. Hence, the ellipse equation is $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$."
  },
  {
    "text": "The distance from a point $P$ on the parabola $y^{2}=16x$ to the $x$-axis is $12$. Then the distance $|PF|$ between point $P$ and the focus $F$ is $?$.",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 16*x);P: Point;PointOnCurve(P, G);Distance(P, xAxis) = 12;F: Point;Focus(G) = F;Distance(P, F) = Abs(LineSegmentOf(P, F))",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "13",
    "fact_spans": "[[[0, 15]], [[0, 15]], [[18, 21], [36, 40]], [[0, 21]], [[18, 34]], [[43, 46]], [[0, 46]], [[36, 56]]]",
    "query_spans": "[[[50, 58]]]",
    "process": ""
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=2x$ is $F$, and a fixed point $A(4,2)$. If there exists a point $M$ on the parabola such that $|MA|+|MF|$ is minimized, then the minimum value is?",
    "fact_expressions": "G: Parabola;A: Point;M: Point;F: Point;Expression(G) = (y^2 = 2*x);Coordinate(A) = (4, 2);Focus(G) = F;PointOnCurve(M, G);WhenMin(Abs(LineSegmentOf(M,A))+Abs(LineSegmentOf(M,F)))",
    "query_expressions": "Min(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F)))",
    "answer_expressions": "9/2",
    "fact_spans": "[[[2, 16], [37, 40]], [[26, 34]], [[45, 48]], [[20, 23]], [[2, 16]], [[26, 34]], [[2, 23]], [[37, 48]], [[50, 65]]]",
    "query_spans": "[[[50, 72]]]",
    "process": "Using the definition of a parabola, the minimum value of |MA| + |MF| is the distance from point A to the directrix. As shown in the figure: let P be any point on the parabola, draw PP_{1} perpendicular to the directrix from P, and draw AA_{1} perpendicular to the directrix from A. By the definition of the parabola, we obtain: |PA| + |PF| = |PA| + |PP_{1}| \\geqslant |MA| + |MA_{1}| = |MA| + |MF| = \\frac{9}{2}. Therefore, there exists a point M on the parabola such that |MA| + |MF| is minimized, and the minimum value is \\frac{9}{2}."
  },
  {
    "text": "Draw a perpendicular line from the left focus $F_{1}$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ to the $x$-axis, intersecting the hyperbola at point $P$. Let $F_{2}$ be the right focus. If $\\angle F_{1} P F_{2}=45^{\\circ}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F1: Point;P: Point;F2: Point;H:Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(F1,H);IsPerpendicular(H,xAxis);Intersection(H,G)=P;AngleOf(F1, P, F2) = ApplyUnit(45, degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)+1",
    "fact_spans": "[[[1, 60], [80, 83], [138, 141]], [[4, 60]], [[4, 60]], [[64, 71]], [[84, 88]], [[90, 97]], [], [[4, 60]], [[4, 60]], [[1, 60]], [[1, 71]], [[80, 101]], [[0, 79]], [[0, 79]], [[0, 88]], [[103, 136]]]",
    "query_spans": "[[[138, 147]]]",
    "process": ""
  },
  {
    "text": "If the curve $\\frac{x^{2}}{k}+\\frac{y^{2}}{1+k}=1$ represents an ellipse, then the range of values for $k$ is?",
    "fact_expressions": "G: Ellipse;H: Curve;Expression(H) = (y^2/(k + 1) + x^2/k = 1);k: Number;H = G",
    "query_expressions": "Range(k)",
    "answer_expressions": "(0, +\\infty)",
    "fact_spans": "[[[42, 44]], [[1, 40]], [[1, 40]], [[46, 49]], [[1, 44]]]",
    "query_spans": "[[[46, 56]]]",
    "process": ""
  },
  {
    "text": "The parabola $C$: $y^{2}=4 x$ has focus $F$, point $P$ moves on parabola $C$, and point $A(-1,0)$. When $\\frac{|P F |}{|P A|}$ attains its minimum value, what is the equation of line $A P$?",
    "fact_expressions": "C: Parabola;A: Point;P: Point;F: Point;Expression(C) = (y^2 = 4*x);Coordinate(A) = (-1, 0);Focus(C) = F;PointOnCurve(P, C);WhenMin(Abs(LineSegmentOf(P,F))/Abs(LineSegmentOf(P,A)))",
    "query_expressions": "Expression(LineOf(A,P))",
    "answer_expressions": "{x+y+1=0,x-y+1=0}",
    "fact_spans": "[[[0, 19], [33, 39]], [[41, 51]], [[29, 32]], [[23, 26]], [[0, 19]], [[41, 51]], [[0, 26]], [[29, 40]], [[52, 82]]]",
    "query_spans": "[[[82, 94]]]",
    "process": "Let the coordinates of point P be (4t^{2}, 4t), F(1,0), A(-1,\\begin{matrix}0)&A(-1,0\\\\2=(4t^{2}-1)^{2}\\end{matrix}+16t^{2}=16t^{4}+8t^{2}+12=(4t^{2}+1)^{2}+16t^{2}=16t^{4}+24t^{2}+1)^{2}=\\frac{16t^{4}+8t^{2}+1}{16t^{4}+24t^{2}+1}=1-\\frac{16t^{2}}{16t^{4}+24t^{2}+1}=1-\\frac{16}{16t^{2}+\\frac{1}{2}+24}\\geqslant\\frac{5}{12}=1-\\frac{16}{40}=\\frac{3}{5}, with equality if and only if 16t^{2}=\\frac{1}{t^{2}}, i.e., t=\\pm\\frac{1}{2}, at which point the coordinates of P are (1,2) or (1,-2); the equation of line AP is then y=\\pm(x+1), i.e., x+y+1=0 or x-y+1=0"
  },
  {
    "text": "Given the line $l$: $y = kx + 1$ $(k \\in \\mathbb{R})$, if there always exists a point $M$ on line $l$ such that the product of the slopes of the lines connecting $M$ to points $A(-1,0)$ and $B(1,0)$ is $-3m$ $(m > 0)$, then the range of real values for $m$ is?",
    "fact_expressions": "l: Line;Expression(l) = (y = k*x + 1);k: Real;M: Point;PointOnCurve(M, l);A: Point;Coordinate(A) = (-1, 0);B: Point;Coordinate(B) = (1, 0);m: Real;m>0;Slope(LineSegmentOf(M, A))*Slope(LineSegmentOf(M, B)) = -3*m",
    "query_expressions": "Range(m)",
    "answer_expressions": "[1/3, +oo)",
    "fact_spans": "[[[2, 29], [31, 36]], [[2, 29]], [[9, 29]], [[40, 44]], [[31, 44]], [[47, 56]], [[47, 56]], [[58, 66]], [[58, 66]], [[87, 92]], [[74, 85]], [[40, 85]]]",
    "query_spans": "[[[87, 99]]]",
    "process": "First, find the trajectory equation of point M such that the product of the slopes of the lines connecting M to two points A(-1,0) and B(1,0) is -3m (m>0). Determine the curve from the equation, then use whether the fixed point through which line l passes lies inside or on the curve to determine the conclusion. Let M(x,y), then k_{PA}k_{PB}=\\frac{y}{x+1}\\times\\frac{y}{x-1}=\\frac{y^{2}}{x^{2}-1}=-3m. Rearranging gives x^{2}+\\frac{y^{2}}{3m}=1. According to the problem, the line l: y=kx+1 always has common points with the curve x^{2}+\\frac{y^{2}}{3m}=1. The line y=kx+l passes through the fixed point P(0,1). When 3m=1, the curve x^{2}+\\frac{y^{2}}{3m}=1 represents the circle x^{2}+y^{2}=1, which also passes through the fixed point P(0,1), satisfying the condition; when 3m>1, the curve x^{2}+\\frac{y^{2}}{3m}=1 represents an ellipse, and the fixed point P(0,1) lies inside the ellipse x^{2}+\\frac{y^{2}}{3m}=1, satisfying the condition, m>\\frac{1}{3}; when 0<3m<1, the curve x^{2}+\\frac{y^{2}}{3m}=1 represents an ellipse, and the fixed point P(0,1) lies outside the ellipse x^{2}+\\frac{y^{2}}{3m}=1; in this case, the line y=1 has no common points with the ellipse, which does not satisfy the condition. In summary, m\\geqslant\\frac{1}{3}."
  },
  {
    "text": "Given $A(-1,0)$, $B$ is a moving point on the circle $F$: $x^{2}-2 x+y^{2}-11=0$ (with $F_{1}$ as the center of the circle), the perpendicular bisector of segment $AB$ intersects $BF_{1}$ at point $P$. Then the trajectory equation of the moving point $P$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (-1, 0);F: Circle;Expression(F) = (y^2 + x^2 - 2*x - 11 = 0);F1: Point;Center(F) = F1;B: Point;PointOnCurve(B, F) = True;P: Point;Intersection(PerpendicularBisector(LineSegmentOf(A,B)),LineSegmentOf(B,F1)) = P",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2/3+y^2/2=1",
    "fact_spans": "[[[2, 11]], [[2, 11]], [[18, 46]], [[18, 46]], [[47, 54]], [[18, 58]], [[14, 17]], [[14, 62]], [[87, 91], [95, 98]], [[63, 91]]]",
    "query_spans": "[[[93, 105]]]",
    "process": "Connect PA, PB, PF, then |PB| = |PA|; rewrite x^{2}-2x+y^{2}-11=0 as (x-1)^{2}+y^{2}=12, so F(1,0), |BF|=2\\sqrt{3}. Thus |PA|+|PF|=|PB|+|PF|=|BF|=2\\sqrt{3}>|AF|=2. Hence the locus of P is an ellipse with foci A and F, where a=\\sqrt{3}, c=1. Therefore b^{2}=3-1=2, and the equation of the locus of P is \\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1."
  },
  {
    "text": "If the distance from a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ to the left directrix is $5$, then what is the distance from this point to the right focus?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);D: Point;PointOnCurve(D, G) = True;Distance(D, LeftDirectrix(G)) = 5",
    "query_expressions": "Distance(D, RightFocus(G))",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 39]], [[1, 39]], [[56, 57]], [[1, 42]], [[1, 53]]]",
    "query_spans": "[[[1, 66]]]",
    "process": "According to the second definition of an ellipse, the ratio of the distance from a point on the ellipse to the left focus and the distance to the left directrix is the eccentricity of the ellipse. Given $a=5$, $b=3$, we obtain $c=4$, so the eccentricity is $\\frac{c}{a}=\\frac{4}{5}$. Let $d_{1}$ be the distance from a point on the ellipse to the left focus, and $d_{2}$ the distance from this point to the right focus. Then $\\frac{d_{1}}{5}=\\frac{4}{5}$, so $d_{1}=4$. According to the first definition of an ellipse, $d_{1}+d_{2}=2a$, we get $d_{2}=6$."
  },
  {
    "text": "Given $A(-2,0)$, $B(1,0)$, $Q(6,0)$, if a moving point $P(x^{\\prime}, y^{\\prime})$ satisfies $|P A|=2|P B|$, and let the midpoint of segment $P Q$ be $M$, then the trajectory equation of point $M$ is?",
    "fact_expressions": "P: Point;Q: Point;A: Point;B: Point;Coordinate(A) = (-2, 0);Coordinate(B) = (1, 0);Coordinate(Q) = (6, 0);Coordinate(P) = (x1, y1);x1: Number;y1: Number;Abs(LineSegmentOf(P, A)) = 2*Abs(LineSegmentOf(P, B));MidPoint(LineSegmentOf(P, Q)) = M;M: Point",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "(x-4)^2 + y^2 = 1",
    "fact_spans": "[[[37, 64]], [[24, 33]], [[2, 11]], [[13, 22]], [[2, 11]], [[13, 22]], [[24, 33]], [[37, 64]], [[37, 64]], [[37, 64]], [[66, 80]], [[82, 96]], [[93, 96], [99, 103]]]",
    "query_spans": "[[[99, 110]]]",
    "process": "Since A(-2,0), B(1,0), Q(6,0), and a moving point P(x,y) satisfies |PA| = 2|PB|, we have \\sqrt{(x+2)^{2}+y^{2}} = 2\\sqrt{(x-1)^{2}+y^{2}}. Simplifying gives x^{2}+y^{2}-4x=0, which rearranges to (x-2)^{2}+y^{2}=4\\textcircled{1}. Let M(x,y); by the midpoint formula, we get \\begin{cases}x=\\frac{x+6}{2}\\\\y=\\frac{y}{2}\\end{cases}, i.e., \\begin{cases}x=2x-6\\\\y=2y\\end{cases}\\textcircled{2}. Substituting \\textcircled{2} into \\textcircled{1} yields (2x-8)^{2}+(2y)^{2}=4. Therefore, the trajectory equation of point M is (x-4)^{2}+y^{2}=1."
  },
  {
    "text": "The point $P(x, y)$ is a moving point on the ellipse $2x^{2}+3y^{2}=12$, then the maximum value of $x+2y$ is?",
    "fact_expressions": "P: Point;x0:Number;y0:Number;Coordinate(P) = (x0, y0);G: Ellipse;Expression(G) = (2*x^2 + 3*y^2 = 12);PointOnCurve(P, G)",
    "query_expressions": "Max(x0 + 2*y0)",
    "answer_expressions": "sqrt(22)",
    "fact_spans": "[[[0, 11]], [[1, 11]], [[1, 11]], [[0, 11]], [[12, 34]], [[12, 34]], [[0, 40]]]",
    "query_spans": "[[[42, 55]]]",
    "process": ""
  },
  {
    "text": "Through the left focus $F(-c, 0)$ $(c>0)$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a line $FE$ with inclination angle $\\frac{\\pi}{6}$ intersecting the right branch of the hyperbola at point $P$. If $\\overrightarrow{O E}=\\frac{1}{2}(\\overrightarrow{O F}+\\overrightarrow{O P})$ and $\\overrightarrow{O E} \\cdot \\overrightarrow{E F}=0$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;Coordinate(F) = (-c, 0);LeftFocus(G) = F;c: Number;c>0;PointOnCurve(F, LineOf(F, E));E: Point;Inclination(LineOf(F, E)) = pi/6;P: Point;Intersection(LineOf(F, E), RightPart(G)) = P;O: Origin;VectorOf(O, E) = (1/2)*(VectorOf(O, F) + VectorOf(O, P));DotProduct(VectorOf(O, E), VectorOf(E, F)) = 0",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)+1",
    "fact_spans": "[[[1, 57], [107, 110], [251, 254]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[61, 76]], [[61, 76]], [[1, 76]], [[61, 76]], [[61, 76]], [[0, 105]], [[198, 249]], [[78, 105]], [[113, 117]], [[98, 117]], [[119, 196]], [[119, 196]], [[198, 249]]]",
    "query_spans": "[[[251, 260]]]",
    "process": "Since $\\overrightarrow{OE}\\cdot\\overrightarrow{EF}=0$, we have $OE\\bot EF$. Given that $\\angle PFO = \\frac{\\pi}{6}$, it follows that $OE = \\frac{1}{2}OF = \\frac{1}{2}c$. $\\overrightarrow{OE} = \\frac{1}{2}(\\overrightarrow{OF} + \\overrightarrow{OP})$, hence $E$ is the midpoint of $PF$. Let the right focus be $F'$, then $O$ is the midpoint of $FF'$. Thus, $PF' = 2OE = c$. $\\overrightarrow{OE}\\cdot\\overrightarrow{EF}=0$, so $OE\\bot EF$, therefore $PF\\bot PF'$. Since $PF - PF' = 2a$, we have $PF = PF' + 2a = 2a + c$. In right triangle $\\triangle PFF'$, $PF^{2} + PF'^{2} = FF'^{2}$, that is $(2a + c)^{2} + c^{2} = 4c^{2}$. Therefore, the eccentricity $e = \\sqrt{3} + 1$."
  },
  {
    "text": "Given that $P$ is a moving point on the parabola $y^{2}=4x$, the projection of point $P$ onto the $y$-axis is $M$, and the coordinates of point $A$ are $(2, 3)$, then the minimum value of $|PA|+|PM|$ is?",
    "fact_expressions": "G: Parabola;P: Point;A: Point;M: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (2, 3);PointOnCurve(P, G);Projection(P, yAxis) = M",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "sqrt(10) - 1",
    "fact_spans": "[[[6, 20]], [[2, 5], [25, 29]], [[43, 47]], [[39, 42]], [[6, 20]], [[43, 60]], [[2, 24]], [[25, 42]]]",
    "query_spans": "[[[62, 81]]]",
    "process": "When $ x=2 $, $ y^{2}=4\\times2=8 $, so $ y=\\pm2\\sqrt{2} $, that is, $ |y|=2\\sqrt{2} $. Since $ 3>2\\sqrt{2} $, point A is outside the parabola. Extend PM to intersect the line $ x=-1 $ at point N. By the definition of the parabola, $ PN=|PM|+1=|PF| $. When points A, P, F are collinear, $ |PA|+|PF| $ is minimized, and at this time $ |PA|+|PF|=|AF| $. The focus has coordinates $ F(1,0) $, so $ |AF|=\\sqrt{(2-1)^{2}+3^{2}}=\\sqrt{10} $. Thus, the minimum value of $ |PM|+1+|PA| $ is $ \\sqrt{10} $, so the minimum value of $ |PM|+|PA| $ is $ \\sqrt{10}-1 $."
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{2}=1$ are $F_{1}$, $F_{2}$, point $P$ lies on the ellipse. If $|P F_{1}|=4$, then what is the measure of $\\angle F_{1} P F_{2}$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 + y^2/2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 4",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "ApplyUnit(120,degree)",
    "fact_spans": "[[[0, 37], [62, 64]], [[57, 61]], [[41, 48]], [[49, 56]], [[0, 37]], [[0, 56]], [[57, 65]], [[67, 80]]]",
    "query_spans": "[[[82, 109]]]",
    "process": "Since $a=3$, $b=\\sqrt{2}$, $\\because c=\\sqrt{7}$, $|PF_{1}|=4$. Also, since $|PF_{1}|+|PF_{2}|=6$, $\\therefore |PF_{2}|=2$. Therefore, in triangle $PF_{1}F_{2}$, $\\cos\\angle F_{1}PF_{2}=\\frac{4^{2}+2^{2}-(2\\sqrt{7})^{2}}{2\\times4\\times2}=-\\frac{1}{2}$, $\\therefore \\angle F_{1}PF_{2}=\\frac{2\\pi}{3}$. Hence, fill in $120^{0}$. This problem involves the application of the definition of an ellipse and knowledge of solving triangles."
  },
  {
    "text": "Given that an asymptote of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ passes through the point $(1,2)$, then what is $b$? What is its eccentricity?",
    "fact_expressions": "G: Hyperbola;b: Number;H: Point;b>0;Expression(G) = (x^2 - y^2/b^2 = 1);Coordinate(H) = (1, 2);PointOnCurve(H,OneOf(Asymptote(G)))",
    "query_expressions": "b;Eccentricity(G)",
    "answer_expressions": "2\nsqrt(5)",
    "fact_spans": "[[[2, 39], [61, 62]], [[56, 59]], [[46, 54]], [[5, 39]], [[2, 39]], [[46, 54]], [[2, 54]]]",
    "query_spans": "[[[56, 61]], [[61, 67]]]",
    "process": "From the given information: the asymptotes of the hyperbola are $ y = \\pm bx $. Since $ y = bx $ passes through the point $ (1, 2) $, we have $ b = 2 $. Therefore, $ a^{2} = 1 $, $ b^{2} = 4 $, $ c^{2} = 1 + 4 = 5 $, $ \\therefore a = 1 $, $ c = \\sqrt{5} $, $ \\therefore e = \\frac{c}{a} = \\sqrt{5} $."
  },
  {
    "text": "Given that the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{6}=1$ and the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{1}=1$ $(a>0, b>0)$ have the same foci, find the equations of the asymptotes of the hyperbola.",
    "fact_expressions": "G: Hyperbola;a: Number;b:Number;H: Ellipse;a>0;b>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(H) = (x^2/9 + y^2/6 = 1);Focus(G) = Focus(H)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(2)/2)*x",
    "fact_spans": "[[[40, 92], [99, 102]], [[43, 92]], [[43, 92]], [[2, 39]], [[43, 92]], [[43, 92]], [[40, 92]], [[2, 39]], [[2, 97]]]",
    "query_spans": "[[[99, 110]]]",
    "process": "\\because the ellipse and hyperbola share common foci, \\therefore 9-6=a^{2}+1, solving gives: a=\\sqrt{2}, \\therefore the asymptotes of the hyperbola are: y=\\pm\\frac{1}{5}x=\\pm\\frac{\\sqrt{2}}{2}x"
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{18}+\\frac{y^{2}}{9}=1$, and $F_{1}$, $F_{2}$ are the left and right foci of the ellipse, respectively. If the area of $\\Delta F_{1} PF_{2}$ is $3 \\sqrt{3}$, then the value of $|PF_{1}| \\cdot |PF_{2}|$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/18 + y^2/9 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;Area(TriangleOf(F1, P, F2)) = 3*sqrt(3)",
    "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "12",
    "fact_spans": "[[[6, 44], [66, 68]], [[48, 55]], [[2, 5]], [[56, 63]], [[6, 44]], [[2, 47]], [[48, 74]], [[48, 74]], [[76, 113]]]",
    "query_spans": "[[[115, 143]]]",
    "process": ""
  },
  {
    "text": "The equation of the hyperbola with foci at $(0, \\pm 3)$ and the same asymptotes as the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$ is?",
    "fact_expressions": "Coordinate(Focus(G1)) = (0, pm*3);G1: Hyperbola;G2: Hyperbola;Expression(G2) = (x^2/2 - y^2 = 1);Asymptote(G1) = Asymptote(G2)",
    "query_expressions": "Expression(G1)",
    "answer_expressions": "y^2/3 - x^2/6 = 1",
    "fact_spans": "[[[0, 57]], [[54, 57]], [[18, 46]], [[18, 46]], [[17, 57]]]",
    "query_spans": "[[[54, 61]]]",
    "process": "Hyperbola \\frac{x^{2}}{2}-x^{2}, where a^{2}=2,^{2}-'. The asymptote equations are y=+\\frac{1}{v-}v, so in the required hyperbola equation \\frac{a}{b}-\\frac{1}{v\\cdot}, and \\because c=3,a^{2},1,2-x^{2} \\therefore a^{2}=2,^{2}-, hyperbola equation is \\frac{y^{2}}{3}-\\frac{x^{2}}{6}"
  },
  {
    "text": "Given two fixed points $A(-1,0)$ and $B(1,0)$, a moving point $P(x, y)$ travels along the line $l$: $y=x+2$. The ellipse $C$ has foci at $A$ and $B$ and passes through point $P$. Then, the maximum eccentricity of ellipse $C$ is?",
    "fact_expressions": "l: Line;C: Ellipse;A: Point;B: Point;P: Point;Coordinate(A) = (-1, 0);Coordinate(B) = (1, 0);Coordinate(P) = (x1, y1);x1:Number;y1:Number;PointOnCurve(P,l);Focus(C)={A,B};PointOnCurve(P,C);Expression(l)=(y=x+2)",
    "query_expressions": "Max(Eccentricity(C))",
    "answer_expressions": "sqrt(10)/5",
    "fact_spans": "[[[36, 50]], [[54, 59], [79, 84]], [[5, 14], [60, 63]], [[15, 23], [64, 67]], [[26, 35], [73, 77]], [[5, 14]], [[15, 23]], [[26, 35]], [[26, 35]], [[26, 35]], [[26, 53]], [[54, 70]], [[54, 77]], [[36, 50]]]",
    "query_spans": "[[[79, 94]]]",
    "process": "Analysis: Draw the line $ y = x + 2 $. Let $ C $ be the symmetric point of $ A $ with respect to the line $ y = x + 2 $. Then $ 2a = |PA| + |PB| \\geqslant |CD| + |DB| = |BC| $, from which the maximum value of $ a $ can be obtained. Since $ c = 1 $, the eccentricity formula gives the following. From the problem, we know $ c = 1 $, eccentricity $ e = \\frac{c}{a} $. The ellipse $ C $ has foci at $ A $ and $ B $ and passes through point $ P $, so $ c = 1 $. Since $ P $ moves on the line $ y = x + 2 $, we have $ 2a = |PA| + |PB| $. Let $ C(m, n) $ be the symmetric point of $ A $ with respect to the line $ y = x + 2 $. Then solving the system \n\\[\n\\begin{cases}\n\\frac{n}{m+1} = -1 \\\\\n\\frac{1}{2}n = \\frac{1}{2}(m-1) + 2\n\\end{cases}\n\\]\nyields \n\\[\n\\begin{cases}\nm = -2 \\\\\nn = 1\n\\end{cases}\n\\],\nso $ C(-2, 1) $. Thus, $ 2a = |PA| + |PB| \\geqslant |CD| + |DB| = |BC| = \\sqrt{10} $, and at this time $ a $ has minimum value $ \\frac{\\sqrt{10}}{2} $, corresponding to the maximum value of eccentricity $ \\frac{1}{\\sqrt{10}} = \\frac{\\sqrt{10}}{5} $."
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x(p>0)$ is a focus of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, then $p$=?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;Expression(G) = (x^2 - y^2/3 = 1);p>0;Expression(H) = (y^2 = 2*(p*x));Focus(H) = OneOf(Focus(G))",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[26, 54]], [[1, 22]], [[61, 64]], [[26, 54]], [[4, 22]], [[1, 22]], [[1, 59]]]",
    "query_spans": "[[[61, 66]]]",
    "process": "From the given conditions, the focus of the parabola $(\\frac{p}{2},0)$ is the right focus of the hyperbola, and since the coordinates of the right focus of the hyperbola are $(2,0)$, the answer can be obtained. Because the focus of the parabola $y^{2}=2px$ $(p>0)$ is $(\\frac{p}{2},0)$, the right focus of the hyperbola is $(\\frac{p}{2},0)$. Also, since the coordinates of the right focus of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ are $(2,0)$, we have $\\frac{p}{2}=2$, solving gives $p=4$."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2 p x(p>0)$ be $F$, and the directrix be $l$. A line passing through the focus intersects the parabola at points $A$ and $B$. Perpendiculars from $A$ and $B$ to $l$ have feet $C$ and $D$, respectively. If $|A F|=2|B F|$ and the area of triangle $C D F$ is $\\sqrt{2}$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(G) = F;l: Line;Directrix(G) = l;H: Line;PointOnCurve(F,H);A: Point;B: Point;Intersection(H, G) = {A, B};L1: Line;L2: Line;PointOnCurve(A,L1);PointOnCurve(B,L2);IsPerpendicular(l,L1);IsPerpendicular(l,L2);C: Point;D: Point;FootPoint(l,L1) = C;FootPoint(l,L2) = D;Abs(LineSegmentOf(A, F)) = 2*Abs(LineSegmentOf(B, F));Area(TriangleOf(C, D, F)) = sqrt(2)",
    "query_expressions": "p",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[1, 22], [47, 50]], [[1, 22]], [[132, 135]], [[4, 22]], [[26, 29]], [[1, 29]], [[33, 36], [72, 75]], [[1, 36]], [[42, 44]], [[38, 44]], [[51, 54], [64, 67]], [[55, 58], [68, 71]], [[42, 60]], [], [], [[61, 78]], [[61, 78]], [[61, 78]], [[61, 78]], [[81, 84]], [[85, 88]], [[61, 88]], [[61, 88]], [[90, 104]], [[106, 130]]]",
    "query_spans": "[[[132, 139]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since line AB passes through focus F, we have y_{1}y_{2} = -p^{2} (without loss of generality, assume C is in the first quadrant). Also, from |AF| = 2|BF|, it follows that |y_{1}| = 2|y_{2}|, i.e., y_{1} = -2y_{2}. Therefore, -2y_{2}^{2} = -p^{2}, y_{1} = -2y_{2} = \\sqrt{2}p. Thus, S_{\\triangle CDF} = \\frac{1}{2}|y_{1} - y_{2}| \\times p = \\frac{1}{2} \\times \\frac{3\\sqrt{2}}{2}p^{2} = \\sqrt{2}. Solving gives p = \\frac{2\\sqrt{3}}{3}."
  },
  {
    "text": "Given that $F$ is the right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, a line $l$ passing through $F$ is perpendicular to one asymptote of $C$, with foot of perpendicular at $A$, and intersects the other asymptote of $C$ at point $B$, and $|A B|=\\sqrt{3} a$, then the eccentricity of $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;A: Point;B: Point;F: Point;a>0;b>0;L1:Line;L2:Line;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;PointOnCurve(F,l);IsPerpendicular(l,L1);FootPoint(l,L1)=A;Intersection(l,L2)=B;Abs(LineSegmentOf(A, B)) = sqrt(3)*a;OneOf(Asymptote(C)) = L1;OneOf(Asymptote(C)) = L2;Negation(L1=L2)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[77, 82], [104, 107]], [[6, 67], [83, 86], [108, 111], [146, 149]], [[14, 67]], [[14, 67]], [[98, 102]], [[120, 124]], [[2, 5], [73, 76]], [[14, 67]], [[14, 67]], [], [], [[6, 67]], [[2, 71]], [[72, 82]], [[77, 94]], [[77, 101]], [[104, 124]], [[126, 144]], [81, 90], [106, 115], [81, 115]]",
    "query_spans": "[[[146, 155]]]",
    "process": "According to the problem, let the asymptotes of the hyperbola be $ y = \\pm\\frac{b}{a}x $, and let $ F(c,0) $. Then the distance from point $ F(c,0) $ to the asymptote $ y = \\frac{b}{a}x $ is $ d = \\frac{|bc|}{\\sqrt{a^{2}+b^{2}}} = b $, that is, $ |AF| = b $, so $ |OA| = a $. As shown in the figure, since $ |AB| = \\sqrt{3}a $, we have $ \\tan\\angle AOB = \\frac{\\sqrt{3}a}{a} = \\sqrt{3} $. Let $ \\angle AOF = \\angle BOF = \\theta $, then $ \\tan 2\\theta = \\frac{2\\tan\\theta}{1-\\tan^{2}\\theta} = \\sqrt{3} $, solving gives $ \\tan\\theta = \\frac{\\sqrt{3}}{3} $, i.e., $ \\frac{b}{a} = \\frac{\\sqrt{3}}{3} $, so $ e = \\sqrt{1+\\left(\\frac{b}{a}\\right)^{2}} = \\frac{2\\sqrt{3}}{3} $. $ y: 2\\sqrt{3} $"
  },
  {
    "text": "The focus of the parabola $C$: $y^{2}=2 p x$ ($p>0$) is $F$, and the intersection point of its directrix with the $x$-axis is $A$. If there exists a point $M$ on the line $x+y+4=0$ such that $\\angle F M A=90^{\\circ}$, then the range of values for the real number $p$ is?",
    "fact_expressions": "C: Parabola;p: Real;G: Line;F: Point;M: Point;A: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Expression(G) = (x + y + 4 = 0);Focus(C) = F;Intersection(Directrix(C), xAxis) = A;PointOnCurve(M, G);AngleOf(F, M, A) = ApplyUnit(90, degree)",
    "query_expressions": "Range(p)",
    "answer_expressions": "[4*sqrt(2), +oo)",
    "fact_spans": "[[[0, 26], [34, 35]], [[101, 106]], [[53, 64]], [[30, 33]], [[67, 71]], [[46, 49]], [[8, 26]], [[0, 26]], [[53, 64]], [[0, 33]], [[34, 49]], [[53, 71]], [[74, 99]]]",
    "query_spans": "[[[101, 113]]]",
    "process": "From the given conditions, $F(\\frac{p}{2},0)$, $A(-\\frac{p}{2},0)$. Since $M$ lies on the line $x+y+4=0$, let point $M(x,-x-4)$. Therefore, $\\overrightarrow{AM}=(x+\\frac{p}{2},-x-4)$, $\\overrightarrow{FM}=(x-\\frac{p}{2},-x-4)$. Also, $\\angle FMA = 90^{\\circ}$, so $\\overrightarrow{AM} \\cdot \\overrightarrow{FM} = (x+\\frac{p}{2})(x-\\frac{p}{2}) + (-x-4)^{2} = 0$, which gives $2x^{2}+8x+16-\\frac{p^{2}}{4}=0$. Thus, $A = 8^{2}-4\\times2\\times(16-\\frac{p^{2}}{4}) = 2p^{2}-64 \\geqslant 0$. Solving yields $p \\leqslant -4\\sqrt{2}$ or $p \\geqslant 4\\sqrt{2}$. Since $p > 0$, the range of $p$ is $[4\\sqrt{2},+\\infty)$."
  },
  {
    "text": "If point $P$ lies on the curve $C_{1}$: $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, point $Q$ lies on the curve $C_{2}$: $(x-5)^{2}+y^{2}=1$, and point $R$ lies on the curve $C_{3}$: $(x+5)^{2}+y^{2}=1$, then the maximum value of $|PQ|-| PR |$ is?",
    "fact_expressions": "C1: Curve;C2: Curve;C3: Curve;P: Point;Q: Point;R: Point;Expression(C1) = (x^2/16 - y^2/9 = 1);Expression(C2) = (y^2 + (x - 5)^2 = 1);Expression(C3) = (y^2 + (x + 5)^2 = 1);PointOnCurve(P, C1);PointOnCurve(Q, C2);PointOnCurve(R, C3)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, Q)) - Abs(LineSegmentOf(P, R)))",
    "answer_expressions": "10",
    "fact_spans": "[[[6, 52]], [[59, 88]], [[95, 124]], [[1, 5]], [[54, 58]], [[90, 94]], [[6, 52]], [[59, 88]], [[95, 124]], [[1, 53]], [[54, 89]], [[90, 125]]]",
    "query_spans": "[[[127, 146]]]",
    "process": ""
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{k+1}+\\frac{y^{2}}{3-k}=1$ $(k \\in \\mathbb{R})$ represents an ellipse with foci on the $x$-axis, what is the range of values for $k$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(k + 1) + y^2/(3 - k) = 1);PointOnCurve(Focus(G), xAxis);k:Real",
    "query_expressions": "Range(k)",
    "answer_expressions": "(1,3)",
    "fact_spans": "[[[63, 65]], [[2, 65]], [[54, 65]], [[67, 70]]]",
    "query_spans": "[[[67, 77]]]",
    "process": "\\begin{cases}k+1>0\\\\3-k>0\\\\k+1>3-k\\end{cases}, solving yields 1<k<3, hence fill in: 1<k<3"
  },
  {
    "text": "A line $l$ passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) intersects the parabola at points $A$ and $B$, and intersects the directrix at point $C$. If $C B=2B F$, then the slope of line $A B$ is?",
    "fact_expressions": "G: Parabola;p: Number;I: Line;B: Point;F: Point;A: Point;C: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(F, I);Intersection(I, G) = {A, B};Intersection(I,Directrix(G)) = C;LineSegmentOf(C, B) = 2 * LineSegmentOf(B, F)",
    "query_expressions": "Slope(LineOf(A, B))",
    "answer_expressions": "pm*sqrt(3)",
    "fact_spans": "[[[1, 22], [35, 38]], [[4, 22]], [[29, 34]], [[43, 46]], [[25, 28]], [[39, 42]], [[53, 57]], [[4, 22]], [[1, 22]], [[1, 28]], [[0, 34]], [[29, 48]], [[29, 57]], [[60, 70]]]",
    "query_spans": "[[[72, 84]]]",
    "process": ""
  },
  {
    "text": "The left focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $F$. A line $l$ passing through point $F$ with slope $\\frac{\\sqrt{2}}{2}$ intersects the ellipse at points $A$ and $B$ (with point $B$ in the first quadrant), and intersects the $y$-axis at point $E$. If $\\overrightarrow{A F}=\\overrightarrow{E B}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "l: Line;G: Ellipse;b: Number;a: Number;A: Point;F: Point;E: Point;B: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F;PointOnCurve(F, l);Slope(l) = sqrt(2)/2;Intersection(l, G) = {A, B};Quadrant(B) = 1;Intersection(l, yAxis) = E;VectorOf(A, F) = VectorOf(E, B)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[91, 96]], [[0, 53], [97, 99], [180, 182]], [[2, 52]], [[2, 52]], [[101, 104]], [[57, 60], [62, 66]], [[129, 133]], [[105, 108], [111, 115]], [[2, 52]], [[2, 52]], [[0, 52]], [[0, 60]], [[61, 96]], [[67, 96]], [[91, 110]], [[111, 120]], [[91, 133]], [[135, 178]]]",
    "query_spans": "[[[180, 188]]]",
    "process": ""
  },
  {
    "text": "Given point $A(4,0)$, the parabola $C$: $y^{2}=2 p x$ $(0<p<4)$ has directrix $l$, point $P$ lies on $C$, draw $P H \\perp l$ at $H$, and $|P H|=|P A|$, $\\angle A P H=120^{\\circ}$, then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;A: Point;P: Point;H: Point;0 < p;p < 4;Expression(C) = (y^2 = 2*(p*x));Coordinate(A) = (4, 0);l: Line;Directrix(C) = (l);PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P, H), l);FootPoint(LineSegmentOf(P, H), l) = H;Abs(LineSegmentOf(P, H)) = Abs(LineSegmentOf(P, A));AngleOf(A, P, H) = ApplyUnit(120, degree)",
    "query_expressions": "p",
    "answer_expressions": "8/5",
    "fact_spans": "[[[12, 42], [55, 58]], [[124, 127]], [[2, 11]], [[50, 54]], [[75, 78]], [[20, 42]], [[20, 42]], [[12, 42]], [[2, 11]], [[46, 49]], [[12, 49]], [[50, 59]], [[61, 74]], [[61, 78]], [[80, 93]], [[96, 122]]]",
    "query_spans": "[[[124, 129]]]",
    "process": "Let the focus be F, then \\angle PAF = \\frac{\\pi}{3}, x_{P} = \\frac{x_{P} + \\frac{p}{2}}{2} + \\frac{p}{2} \\Rightarrow x_{P} = \\frac{3p}{2}, so"
  },
  {
    "text": "Given a line $l$ with non-zero slope passing through the left focus $F$ of the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$, intersecting the ellipse at points $A$ and $B$. A point $M$ on the $y$-axis satisfies $|M A|=|M B|$. Then the range of $\\frac{|F M|}{|A B|}$ is?",
    "fact_expressions": "l: Line;Negation(Slope(l) = 0);C: Ellipse;Expression(C) = (x^2/2 + y^2 = 1);F: Point;LeftFocus(C) = F;PointOnCurve(F, l) = True;Intersection(l, C) = {A, B};A: Point;B: Point;PointOnCurve(M, yAxis) = True;M: Point;Abs(LineSegmentOf(M, A)) = Abs(LineSegmentOf(M, B))",
    "query_expressions": "Range(Abs(LineSegmentOf(F, M))/Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "(\\sqrt{2}/4,\\sqrt{2}/2]",
    "fact_spans": "[[[10, 15]], [[2, 15]], [[16, 48], [57, 59]], [[16, 48]], [[52, 55]], [[16, 55]], [[10, 55]], [[10, 69]], [[60, 63]], [[64, 67]], [[70, 80]], [[76, 80]], [[82, 95]]]",
    "query_spans": "[[[97, 125]]]",
    "process": "From the given conditions, point M is the intersection point of the perpendicular bisector of segment AB and the y-axis. Since F(-1,0), we can assume the equation of line l is x = my - 1 (m ≠ 0). Let A(x₁, y₁), B(x₂, y₂). Solving the system of equations  \n\\begin{cases}x=my-1\\\\\\frac{x^{2}}{2}+y^{2}=1\\end{cases}  \nwe obtain (m² + 2)y² - 2my - 1 = 0, then y₁ + y₂ = \\frac{2m}{m^{2}+2}, y₁y₂ = \\frac{-1}{m^{2}+2}. Thus, the midpoint coordinates of segment AB are (\\frac{-2}{m^{2}+2}, \\frac{m}{m^{2}+2}). |AB| = \\frac{3}{1m+1} \\in (\\frac{\\sqrt{2}}{4}, \\frac{\\sqrt{2}}{2})"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with a focus at $F(2,0)$, and the distance from $F$ to an asymptote of the hyperbola $C$ is $1$, then the equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F: Point;Coordinate(F) = (2, 0);OneOf(Focus(C)) = F;Distance(F, Asymptote(C)) = 1",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/3-y^2=1",
    "fact_spans": "[[[2, 63], [82, 88], [101, 107]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 76], [78, 81]], [[68, 76]], [[2, 76]], [[78, 99]]]",
    "query_spans": "[[[101, 112]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$, respectively, and point $P$ is any point on the ellipse, then the range of $\\frac{||PF_{1}|-|PF_{2}||}{|PF_{1}|}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/8 + y^2/4 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G)",
    "query_expressions": "Range(Abs(Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)))/Abs(LineSegmentOf(P, F1)))",
    "answer_expressions": "[0, 2*sqrt(2)+2]",
    "fact_spans": "[[[20, 57], [69, 71]], [[20, 57]], [[2, 9]], [[10, 17]], [[2, 63]], [[2, 63]], [[64, 68]], [[64, 77]]]",
    "query_spans": "[[[79, 122]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y=\\frac{1}{2} x^2$ has focus $F$, and $A$, $B$ are two points on the parabola such that $|A F|+|B F|=n$. If the perpendicular bisector of segment $A B$ intersects the $y$-axis at point $C(0,4)$, then $n=$?",
    "fact_expressions": "G: Parabola;B: Point;A: Point;C: Point;F: Point;Expression(G) = (y = x^2/2);Coordinate(C) = (0, 4);Focus(G)=F;PointOnCurve(A,G);PointOnCurve(B,G);Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F)) = n;Intersection(PerpendicularBisector(LineSegmentOf(A,B)), yAxis) = C;n:Number",
    "query_expressions": "n",
    "answer_expressions": "7",
    "fact_spans": "[[[2, 24], [40, 43]], [[36, 39]], [[32, 35]], [[87, 95]], [[28, 31]], [[2, 24]], [[87, 95]], [[2, 31]], [[32, 46]], [[32, 46]], [[48, 63]], [[65, 95]], [[97, 100]]]",
    "query_spans": "[[[97, 102]]]",
    "process": "Let points $ A(2t_{1},2t_{1}^{2}) $, $ B(2t_{2},2t_{2}^{2}) $, find the equation of the perpendicular bisector of segment $ AB $. Since this line passes through point $ C $, we obtain $ t_{1}^{2}+t_{2}^{2}=3 $. Then, using the definition of a parabola, the value of $ n $ can be found. Let points $ A(2t_{1},2t_{1}^{2}) $, $ B(2t_{2},2t_{2}^{2}) $, then $ k_{AB}=t_{1}+t_{2} $, the midpoint $ M $ of segment $ AB $ is $ (t_{1}+t_{2},t_{1}^{2}+t_{2}^{2}) $. Thus, the equation of the perpendicular bisector of $ AB $ is $ y=-\\frac{1}{t_{1}+t_{2}}(x-t_{1}-t_{2})+t_{1}^{2}+t_{2}^{2} $. This line passes through point $ (0,4) $, thus $ t_{1}^{2}+t_{2}^{2}+1=4 $, hence $ t_{1}^{2}+t_{2}^{2}=3 $. Therefore, $ n=|AF|+|BF|=2t_{1}^{2}+\\frac{1}{2}+2t_{2}^{2}+\\frac{1}{2}=2t_{1}^{2}+2t_{2}^{2}+1=2\\times3+1=7 $."
  },
  {
    "text": "The maximum area of a triangle with vertices at a point on the ellipse and the two foci is $1$. Then, the minimum length of the major axis is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;PointOnCurve(P, G);Focus(G) = {F1, F2};Max(Area(TriangleOf(P, F1, F2))) = 1",
    "query_expressions": "Min(Length(MajorAxis(G)))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[1, 3]], [], [], [], [[1, 6]], [[1, 11]], [[0, 27]]]",
    "query_spans": "[[[1, 38]]]",
    "process": ""
  },
  {
    "text": "The equation of the hyperbola centered at the vertex of the parabola $y^{2}=8 x$, with the focus as the right focus, and asymptotes $y=\\pm \\sqrt{3} x$ is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(H) = (y^2 = 8*x);Expression(Asymptote(G)) = (y = pm*sqrt(3)*x);Center(G) = Vertex(H);Focus(H) = RightFocus(G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/3 = 1",
    "fact_spans": "[[[54, 57]], [[1, 15]], [[1, 15]], [[30, 57]], [[0, 57]], [[0, 57]]]",
    "query_spans": "[[[54, 61]]]",
    "process": "Let the equation of the hyperbola be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0, b>0). According to the given conditions, we obtain b=\\sqrt{3}a. Combining with c^{2}=a^{2}+b^{2}, solve for the values of a and b, then obtain the equation of the hyperbola. Solution: The vertex of the parabola y^{2}=8x is at the origin, and the focus is F(2,0). According to the problem, assume the hyperbola's equation is \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0, b>0). Since the asymptotes of the hyperbola are y=\\pm\\sqrt{3}x, that is, \\frac{b}{a}=\\sqrt{3}, we get b=\\sqrt{3}a. Also from c^{2}=a^{2}+b^{2}, we obtain 4a^{2}=2^{2}, solving gives a=1, thus b=\\sqrt{3}. Therefore, the equation of the hyperbola is x^{2}-\\frac{y^{2}}{3}=1"
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. Point $P$ lies on the ellipse such that $P F_{1} \\perp P F_{2}$ and $|P F_{1}| \\cdot |P F_{2}|=2$. If $a=2 b$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;LeftFocus(G) = F1;RightFocus(G) = F2;F1: Point;F2: Point;PointOnCurve(P, G) = True;P: Point;IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2)) = True;Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 2;a = 2*b",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4+y^2=1",
    "fact_spans": "[[[19, 71], [83, 85], [152, 154]], [[19, 71]], [[21, 71]], [[21, 71]], [[21, 71]], [[21, 71]], [[1, 77]], [[1, 77]], [[1, 8]], [[9, 16]], [[78, 86]], [[78, 82]], [[88, 111]], [[113, 141]], [[143, 150]]]",
    "query_spans": "[[[152, 161]]]",
    "process": "\\because a=2b, a^{2}=b^{2}+c^{2}, \\therefore c^{2}=3b^{2} \\text{ and } PF_{1}\\bot PF_{2}, \\therefore |PF_{1}|^{2}+|PF_{2}|^{2}=(2c)^{2}=12b^{2} \\text{ by the definition of ellipse we know } |PF_{1}|+|PF_{2}|=2a=4b, (|PF_{1}|+|PF_{2}|)^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}+2|PF_{1}|\\cdot|PF_{2}|=12b^{2}+4=16b^{2}, \\therefore b=1, a=2 \\text{ therefore, the required standard equation of the ellipse is } \\frac{x^{2}}{4}+y^{2}=1"
  },
  {
    "text": "It is known that the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1$ $(a>0)$ has its center at the origin, and its right focus coincides with the focus of the parabola $y^{2}=16x$. Then, the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;O:Origin;a>0;Expression(G) = (-y^2/9 + x^2/a^2 = 1);Expression(H) = (y^2 = 16*x);RightFocus(G) = Focus(H);Center(G)=O",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "4*sqrt(7)/7",
    "fact_spans": "[[[2, 49], [82, 85]], [[5, 49]], [[59, 74]], [[52, 54]], [[5, 49]], [[2, 49]], [[59, 74]], [[2, 79]], [[2, 54]]]",
    "query_spans": "[[[82, 91]]]",
    "process": ""
  },
  {
    "text": "If point $P$ lies on the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, $F_{1}$ and $F_{2}$ are the two foci of the ellipse, and $\\angle F_{1} P F_{2}=90^{\\circ}$, then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/2 + y^2 = 1);PointOnCurve(P, G);Focus(G)={F1,F2};AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[6, 33], [53, 55]], [[35, 42]], [[1, 5]], [[43, 50]], [[6, 33]], [[1, 34]], [[35, 59]], [[61, 94]]]",
    "query_spans": "[[[96, 123]]]",
    "process": "(Analysis) From the ellipse equation and definition, we can find |F_{1}F_{2}| and |PF_{1}|+|PF_{2}|. Using the Pythagorean theorem, set up an equation involving |PF_{1}|\\cdot|PF_{2}| to solve for |PF_{1}|\\cdot|PF_{2}|, then find the triangle area. From the ellipse equation: a=\\sqrt{2}, b=1 \\therefore c=\\sqrt{a^{2}-b^{2}}=1 \\therefore |F_{1}F_{2}|=2c=2. From the ellipse definition: |PF_{1}|+|PF_{2}|=2a=2\\sqrt{2}. \\because \\angle F_{1}PF_{2}=90^{\\circ} \\therefore |PF_{1}|^{2}+|PF_{2}|^{2}=(|PF_{1}|+|PF_{2}|)^{2}-2|PF_{1}|\\cdot|PF_{2}|=|F_{1}F_{2}|^{2}. That is, 8-2|PF_{1}|\\cdot|PF_{2}|=4, solving gives: |PF_{1}|\\cdot|PF_{2}|=2 \\therefore S_{AF_{1}PF_{2}}=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|=1"
  },
  {
    "text": "It is known that the vertex of the parabola $C$ is at the origin of the coordinate system, and its focus coincides with the right focus of the hyperbola $\\frac{x^{2}}{7}-\\frac{y^{2}}{2}=1$. Then the equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;Vertex(C) = O;O: Origin;G: Hyperbola;Expression(G) = (x^2/7 - y^2/2 = 1);Focus(C) = RightFocus(G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=12*x",
    "fact_spans": "[[[2, 8], [66, 72]], [[2, 16]], [[12, 16]], [[20, 58]], [[20, 58]], [[2, 64]]]",
    "query_spans": "[[[66, 77]]]",
    "process": "Since for the hyperbola $\\frac{x^{2}}{7}-\\frac{y^{2}}{2}=1$, $a^{2}=7$, $b^{2}=2$, so $c=3$, thus its right focus is $(3,0)$, therefore the equation of the parabola $C$ is $y^{2}=12x$."
  },
  {
    "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{4}=1$ is $\\frac{1}{2}$, then what is $m$?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (y^2/4 + x^2/m = 1);Eccentricity(G) = 1/2",
    "query_expressions": "m",
    "answer_expressions": "{3,16/3}",
    "fact_spans": "[[[1, 38]], [[58, 61]], [[1, 38]], [[1, 56]]]",
    "query_spans": "[[[58, 63]]]",
    "process": ""
  },
  {
    "text": "Let the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{3}=1$ have two foci $F_{1}$, $F_{2}$ both on the $x$-axis, and let $P$ be a point on this ellipse in the first quadrant such that $\\frac{\\sin \\angle P F_{1} F_{2}+\\sin \\angle P F_{2} F_{1}}{\\sin \\angle F_{1} P F_{2}}=2$. Then the value of the positive number $m$ is?",
    "fact_expressions": "G: Ellipse;m: Number;m>0;P: Point;F1: Point;F2: Point;Expression(G) = (y^2/3 + x^2/m = 1);Focus(G)={F1,F2};PointOnCurve(F1,xAxis);PointOnCurve(F2,xAxis);Quadrant(P)=1;PointOnCurve(P, G);(Sin(AngleOf(P, F1, F2)) + Sin(AngleOf(P, F2, F1)))/(Sin(AngleOf(F1, P, F2)))= 2",
    "query_expressions": "m",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 38], [76, 78]], [[175, 180]], [[175, 180]], [[66, 69]], [[43, 50]], [[51, 58]], [[1, 38]], [[1, 58]], [[43, 65]], [[43, 65]], [[66, 75]], [[66, 82]], [[84, 173]]]",
    "query_spans": "[[[175, 184]]]",
    "process": "By the Law of Sines, we have: \\frac{\\sin\\angle PF_{1}F_{2} + \\sin\\angle PF_{2}F_{1}}{\\sin\\angle F_{1}PF_{2}} = \\frac{PF_{1} + PF_{2}}{F_{1}F_{2}} = 2, that is: \\frac{2a}{2c} = \\frac{a}{c} = 2. Combining with the definition of the ellipse, we have: \\frac{m}{m-3} = 4, solving gives: m = 4"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, a line $l$ with slope $2$ intersects $C$ at points $A$ and $B$. If $|AF|+|BF|=5$, then the equation of line $l$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l: Line;Slope(l) = 2;Intersection(l, C) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F)) = 5",
    "query_expressions": "Expression(l)",
    "answer_expressions": "2*x-y-2=0",
    "fact_spans": "[[[2, 21], [42, 45]], [[2, 21]], [[25, 28]], [[2, 28]], [[36, 41], [75, 80]], [[29, 41]], [[36, 56]], [[49, 52]], [[53, 56]], [[58, 73]]]",
    "query_spans": "[[[75, 85]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$), let the right focus be $F$. A circle with diameter $OF$ (where $O$ is the origin) intersects the two asymptotes of the hyperbola at points $A$ and $B$ respectively (distinct from the origin). If the area of $\\triangle OAB$ equals $\\frac{1}{8}ab$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;O: Origin;F: Point;A: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;IsDiameter(LineSegmentOf(O, F), G);l1: Line;l2: Line;Asymptote(C) = {l1, l2};Intersection(G, l1) = A;Intersection(G, l2) = B;Negation(A = O);Negation(B = O);Area(TriangleOf(O, A, B)) = (a*b)/8",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 64], [95, 98], [167, 173]], [[10, 64]], [[10, 64]], [[93, 94]], [[80, 83], [120, 122]], [[69, 72]], [[107, 110]], [[111, 114]], [[10, 64]], [[10, 64]], [[2, 64]], [[2, 72]], [[73, 94]], [], [], [[95, 103]], [[93, 116]], [[93, 116]], [[107, 123]], [[107, 123]], [[126, 165]]]",
    "query_spans": "[[[167, 179]]]",
    "process": "Connect AB intersecting the x-axis at D, we obtain |OA| = a, |AF| = b. From Rt\\triangle AOD \\sim Rt\\triangle FOA, we get |OD| = \\frac{a^{2}}{c}, |AD| = \\frac{ab}{c}. Then, based on the area, an equation in terms of a and c can be established to solve for the eccentricity. Connect AB intersecting the x-axis at D, connect AF, then OA \\bot AF, |OA| = a, |AF| = b. Since Rt\\triangle AOD \\sim Rt\\triangle FOA, it follows that \\frac{|OD|}{|OA|} = \\frac{|AD|}{|AF|} = \\frac{|OA|}{|OF|}, i.e., \\frac{|OD|}{a} = \\frac{|AD|}{b} = \\frac{a}{c}, so |OD| = \\frac{a^{2}}{c}, |AD| = \\frac{ab}{c}. Because S_{\\triangle OAD} = \\frac{1}{2} \\times |OD| \\times 2|AD| = \\frac{a^{3}b}{c^{2}} = \\frac{1}{8}ab, thus e^{2} = \\frac{c^{2}}{a^{2}} = 8, i.e., e = 2\\sqrt{2}."
  },
  {
    "text": "The focal length of the hyperbola $2 x^{2}-3 y^{2}=6$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (2*x^2 - 3*y^2 = 6)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[0, 22]], [[0, 22]]]",
    "query_spans": "[[[0, 27]]]",
    "process": "The standard equation of the hyperbola $2x^{2}-3y^{2}=6$ is $\\frac{x^{2}}{3}-\\frac{y^{2}}{2}=1$, therefore, the focal distance of this hyperbola is $2\\sqrt{2+3}$."
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$ with right focus $F$, a line $l$ passing through point $F$ intersects the hyperbola at points $P$ and $Q$. If the circle with diameter $PQ$ passes through a fixed point $M$, then $|M F|$=?",
    "fact_expressions": "l: Line;C: Hyperbola;G: Circle;Q: Point;P: Point;M: Point;F: Point;Expression(C) = (x^2 - y^2/3 = 1);RightFocus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {P, Q};IsDiameter(LineSegmentOf(P,Q),G);PointOnCurve(M, G)",
    "query_expressions": "Abs(LineSegmentOf(M, F))",
    "answer_expressions": "3",
    "fact_spans": "[[[50, 55]], [[2, 35], [56, 59]], [[85, 86]], [[66, 69]], [[62, 65]], [[89, 92]], [[40, 43], [45, 49]], [[2, 35]], [[2, 43]], [[44, 55]], [[50, 71]], [[73, 86]], [[85, 92]]]",
    "query_spans": "[[[94, 103]]]",
    "process": "The coordinates of point F are (2,0). The equation of the hyperbola can be rewritten as $3x^{2}-y^{2}=3$. $\\textcircled{1}$ When the slope of line $l$ does not exist, the coordinates of points P and Q are $(2,3)$ and $(2,-3)$ respectively. At this time, the equation of the circle with segment PQ as diameter is $(x-2)^{2}+y^{2}=9$. $\\textcircled{2}$ When the slope of line $l$ exists, let the coordinates of points P and Q be $(x_{1},y_{1})$, $(x_{2},y_{2})$, denote the left vertex of hyperbola C as $A(-1,0)$, and the equation of line $l$ as $y=k(x-2)$. Solving the system of equations\n$$\n\\begin{cases}\n3x^{2}-y^{2}=3 \\\\\ny=k(x-2)\n\\end{cases}\n$$\neliminating $y$ and simplifying gives $(3-k^{2})x^{2}+4k^{2}x-(3+4k^{2})=0$. \n$$\n\\begin{cases}\n3-k^{2}\\neq0 \\\\\n4=16k^{4}+4(3-k^{2})(3+4k^{2})=36(1+k^{2})>0\n\\end{cases}\n$$\nthat is, when $k\\neq\\pm\\sqrt{3}$,\n$$\n\\begin{cases}\nx_{1}+1 \\\\\nx_{1}+x_{2}\n\\end{cases}\n(x_{1}-2)(x_{2}-2)=k^{2}[x_{1}x_{2}-2(x_{1}+x_{2})+4],\\frac{+4k^{2}}{2-3}-\\frac{8k}{k^{2}}\n\\overrightarrow{AO}=(x_{2}+1,y_{1}\\overrightarrow{|P}\\cdot\\overrightarrow{AO}=(x_{1}+1)(x_{0}+1)+y_{1}=x_{1}x_{2}[(x_{1}+x_{2})+y_{1}y_{2}+1+1:=03+\\frac{x}{k^{2}-3}\n$$\nThus, the circle with segment PO as diameter passes through the fixed point $M(-1,0)$, $|MF|=3$."
  },
  {
    "text": "Given that the two endpoints $A$ and $B$ of a line segment $AB$ of length $4$ slide along the parabola $y = 2x^2$, if $M$ is the midpoint of the segment $AB$, then the shortest distance from point $M$ to the $x$-axis is?",
    "fact_expressions": "Length(LineSegmentOf(A, B)) = 4;A: Point;B: Point;Endpoint(LineSegmentOf(A, B)) = {A, B};G: Parabola;Expression(G) = (y = 2*x^2);PointOnCurve(A, G);PointOnCurve(B, G);M: Point;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Min(Distance(M, xAxis))",
    "answer_expressions": "15/8",
    "fact_spans": "[[[2, 15]], [[20, 23]], [[24, 27]], [[8, 27]], [[29, 43]], [[29, 43]], [[20, 44]], [[20, 44]], [[48, 51], [64, 68]], [[48, 62]]]",
    "query_spans": "[[[64, 80]]]",
    "process": "Let the focus of the parabola $ y = 2x^2 $ be $ F $. Draw perpendiculars from points $ A $, $ B $, $ M $ to the directrix $ y = -\\frac{1}{8} $ of the parabola $ y = 2x^2 $, with feet of perpendiculars being $ A_1 $, $ B_1 $, $ M_1 $, respectively. Then $ |MM_1| = \\frac{|AA_1| + |BB_1|}{2} = \\frac{|AF| + |BF|}{2} \\geqslant \\frac{1}{2}|AB| = 2 $, with equality holding if and only if points $ A $, $ B $, $ F $ are collinear. Therefore, when chord $ AB $ passes through the focus $ F $ of the parabola, $ |MM_1| $ attains the minimum value $ 2 $. At this time, the minimum distance from point $ M $ to the $ x $-axis is $ 2 - \\frac{1}{8} = \\frac{15}{8} $."
  },
  {
    "text": "Given that $P$ is a point on the ellipse $C$: $\\frac{x^{2}}{4}+y^{2}=1$, and points $F_{1}$, $F_{2}$ are its left and right foci respectively, with $|P F_{1}|=3|P F_{2}|$, then $|P F_{1}|$=?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/4 + y^2 = 1);PointOnCurve(P, C);LeftFocus(C) = F1;RightFocus(C) = F2;Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Abs(LineSegmentOf(P, F1))",
    "answer_expressions": "3",
    "fact_spans": "[[[6, 37], [58, 59]], [[2, 5]], [[41, 49]], [[50, 57]], [[6, 37]], [[2, 40]], [[41, 63]], [[41, 63]], [[64, 86]]]",
    "query_spans": "[[[88, 101]]]",
    "process": "From the given condition, we have |PF_{1}| + |PF_{2}| = 4. Since |PF_{1}| = 3|PF_{2}|, it follows that 4|PF_{2}| = 4, so |PF_{2}| = 1, and therefore |PF_{1}| = 3."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=x$ with a point $M(1,-1)$, points $A$ and $B$ are two moving points on the parabola $C$ such that $\\overrightarrow{M A} \\cdot \\overrightarrow{M B}=0$. Then, the maximum value of the distance from point $M$ to the line $AB$ is?",
    "fact_expressions": "C: Parabola;B: Point;A: Point;M: Point;Expression(C) = (y^2 = x);Coordinate(M) = (1, -1);PointOnCurve(M,C);PointOnCurve(A, C);PointOnCurve(B, C);DotProduct(VectorOf(M, A), VectorOf(M, B)) = 0",
    "query_expressions": "Max(Distance(M,LineOf(A,B)))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 19], [41, 47]], [[37, 40]], [[32, 36]], [[22, 31], [107, 111]], [[2, 19]], [[22, 31]], [[2, 31]], [[32, 52]], [[32, 52]], [[54, 105]]]",
    "query_spans": "[[[107, 128]]]",
    "process": "Let the equation of line AB be $x = my + n$, with $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$. Combining the equation of line AB and the parabola equation, we have\n\\[\n\\begin{cases}\ny^{2} = my + n \\\\\nx = my + n\n\\end{cases}\n\\]\ni.e., $y^{2} = my + n$, or $y^{2} - my - n = 0$. Since line AB intersects the parabola at two points, we have $y_{1}y_{2} = -n$, $y_{1} + y_{2} = m$, and $\\Delta = m^{2} + 4n > 0$. Given that $\\overrightarrow{MA} \\cdot \\overrightarrow{MB} = 0$, it follows that $(x_{1} - 1)(x_{2} - 1) + (y_{1} + 1)(y_{2} + 1) = 0$, i.e., $(y_{1}^{2} - 1)(y_{2} - 1) + (y_{2} + 1) = 0$, $(y_{1} + 1)(y_{2} + 1)(y_{1} - 1)(y_{2} - 1) + 1 = 0$. Solving gives $(y_{1} + 1)(y_{2} + 1) = 0$ or $(y_{1} - 1)(y_{2} - 1) + 1 = 0$. Simplifying yields $-n + m + 1 = 0$ or $-n - m + 2 = 0$. Since $\\Delta = m^{2} + 4n > 0$, we have $-n - m + 2 = 0$, so $n = 2 - m$. Thus, the equation of line AB is $x = my + 2 - m$, or $x - 2 = m(y - 1)$, hence line AB passes through the fixed point $C(2, 1)$. When MC is perpendicular to line AB, the distance from point M to line AB reaches its maximum value, which is $\\sqrt{(2 - 1)^{2} + (1 + 1)^{2}} = \\sqrt{5}$."
  },
  {
    "text": "What is the standard equation of the ellipse passing through two points $P_{1}(\\frac{1}{3}, \\frac{1}{3})$, $P_{2}(0,-\\frac{1}{2})$?",
    "fact_expressions": "G: Ellipse;P1: Point;P2: Point;Coordinate(P1) = (1/3, 1/3);Coordinate(P2) = (0, -1/2);PointOnCurve(P1,G);PointOnCurve(P2,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "5*x^2+4*y^2=1",
    "fact_spans": "[[[63, 65]], [[4, 37]], [[39, 62]], [[4, 37]], [[39, 62]], [[0, 65]], [[0, 65]]]",
    "query_spans": "[[[63, 72]]]",
    "process": "Let the equation be $mx^{2}+ny^{2}=1$. Substituting $(\\frac{1}{3},\\frac{1}{3})$ and $(0,\\frac{1}{2})$ gives $\\frac{1}{9}m+\\frac{1}{9}n=1$, $\\frac{1}{4}n=1$. Solving yields $m=5$, $n=4$, so the equation is $5x^{2}+4y^{2}=1$."
  },
  {
    "text": "If a point on the hyperbola $\\frac{x^{2}}{25}-\\frac{y^{2}}{9}=1$ has a distance of $12$ to one focus, then what is its distance to the other focus?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/25 - y^2/9 = 1);P:Point;F1:Point;F2:Point;PointOnCurve(P,G);OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2);Distance(P,F1)=12",
    "query_expressions": "Distance(P,F2)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 40]], [[1, 40]], [[58, 59]], [], [], [[1, 43]], [[1, 48]], [[1, 65]], [[1, 65]], [[1, 56]]]",
    "query_spans": "[[[1, 70]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola $x^{2}=2 m y(m>0)$ is $F$, and its directrix intersects the hyperbola $\\frac{x^{2}}{m^{2}}-\\frac{y^{2}}{n^{2}}=1(n>0)$ at two points $A$, $B$. If $\\angle A F B=120^{\\circ}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;m: Number;n: Number;H: Parabola;A: Point;F: Point;B: Point;n>0;Expression(G) = (-y^2/n^2 + x^2/m^2 = 1);m>0;Expression(H) = (x^2 = 2*(m*y));Focus(H) = F;Intersection(Directrix(H),G)={A,B};AngleOf(A, F, B) = ApplyUnit(120, degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3*sqrt(2)/4",
    "fact_spans": "[[[33, 84], [126, 129]], [[36, 84]], [[36, 84]], [[0, 21], [29, 30]], [[89, 92]], [[25, 28]], [[93, 96]], [[36, 84]], [[33, 84]], [[3, 21]], [[0, 21]], [[0, 28]], [[29, 96]], [[98, 124]]]",
    "query_spans": "[[[126, 135]]]",
    "process": "The directrix of the given parabola is $ y = -\\frac{m}{2} $, and the focus is $ F(0, \\frac{m}{2}) $. Substituting $ y = -\\frac{m}{2} $ into the hyperbola equation yields $ x = \\pm \\frac{m}{2n} \\sqrt{m^{2} + 4n^{2}} $. Since $ \\angle AFB = 120^{\\circ} $, it follows that $ \\underline{\\frac{m}{2n} \\sqrt{m^{2} + 4n^{2}}} = \\tan 60^{\\circ} = \\sqrt{3} $, $ m^{2} = 8n^{2} = 8(c^{2} - m^{2}) $, so $ _{e} = \\frac{c}{m} = \\frac{3\\sqrt{2}}{4} $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ with focus $F$ and directrix $l$, points $A$ and $B$ are two moving points on the parabola $C$ such that $A F \\perp A B$, $\\angle A B F=30^{\\circ}$. Let point $M$ be the midpoint of segment $A B$, and let point $N$ be the projection of $M$ onto the directrix $l$. Then the value of $\\frac{|M N|}{|A B|}$ is?",
    "fact_expressions": "C: Parabola;p: Number;B: Point;A: Point;F: Point;M: Point;N: Point;l: Line;p>0;Expression(C) = (y^2 = 2*p*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(A,C);PointOnCurve(B, C);IsPerpendicular(LineSegmentOf(A,F),LineSegmentOf(A,B));AngleOf(A,B,F)=ApplyUnit(30,degree);MidPoint(LineSegmentOf(A,B))=M;Projection(M,l)=N",
    "query_expressions": "Abs(LineSegmentOf(M, N))/Abs(LineSegmentOf(A, B))",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[2, 28], [54, 60]], [[10, 28]], [[50, 53]], [[45, 48]], [[32, 35]], [[122, 125]], [[136, 140]], [[39, 42], [128, 131]], [[10, 28]], [[2, 28]], [[2, 35]], [[2, 43]], [[44, 66]], [[44, 66]], [[68, 83]], [[85, 110]], [[112, 125]], [[122, 140]]]",
    "query_spans": "[[[142, 167]]]",
    "process": "As shown in the figure, draw $BE\\bot l$, $AD\\bot l$. By the definition of the parabola, we have $|AF|=|AD|$, $|BF|=|BE|$. In trapezoid $ABED$, $2|MN|=|AD|+|BE|=a+b$. Since $AF\\bot AB$, $\\angle ABF=30^{\\circ}$, it follows that $b=2a$, then $|MN|=\\frac{3a}{2}$. Also, $|AB|=\\sqrt{b^{2}-a^{2}}=\\sqrt{3}a$. Hence, $\\frac{|MN|}{|AB|}=\\frac{\\frac{3a}{2}}{\\sqrt{3}a}=\\frac{\\sqrt{3}}{2}$."
  },
  {
    "text": "The point $A$ on the parabola $y^{2}=2x$ is at a distance of $1$ from its focus. Then, the distance from point $A$ to the $y$-axis is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);A: Point;PointOnCurve(A, G);Distance(A, Focus(G)) = 1",
    "query_expressions": "Distance(A, yAxis)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[0, 14], [21, 22]], [[0, 14]], [[16, 20], [33, 37]], [[0, 20]], [[16, 31]]]",
    "query_spans": "[[[33, 47]]]",
    "process": "The equation of the directrix of the parabola $ y^{2} = 2x $ is $ x = -\\frac{1}{2} $. The distance from point A to its focus is 1, which means the distance to $ x = -\\frac{1}{2} $ is 1, so the distance to the y-axis is $ \\frac{1}{2} $."
  },
  {
    "text": "Given point $P$ is any point on the parabola $C$: $y^{2}=4x$, and point $A(3,0)$, then the minimum value of $|PA|$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);P: Point;A: Point;Coordinate(A) = (3, 0);PointOnCurve(P, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[7, 26]], [[7, 26]], [[2, 6]], [[32, 41]], [[32, 41]], [[2, 31]]]",
    "query_spans": "[[[44, 57]]]",
    "process": "Let the coordinates of an arbitrary point on the parabola be $(\\frac{y^{2}}{4}, y)$. By the distance formula between two points, we have $|PA| = \\sqrt{(\\frac{y^{2}}{4}-3)^{2}+y^{2}} = \\sqrt{\\frac{1}{16}y^4 - \\frac{1}{2}y^2 + 9}$. When $y^{2} = -\\frac{\\frac{1}{2}}{-2\\times\\frac{1}{16}} = 4$, that is, when $y = \\pm2$, $|PA|$ attains its minimum value $\\sqrt{8} = 2\\sqrt{2}$. This question primarily examines the method for finding the minimum distance from a point on a parabola to a fixed point, the distance formula between two points, and the method for finding the minimum value of a quadratic function, and belongs to a medium-difficulty problem. Since the point on the parabola is a moving point, we must set the coordinates of an arbitrary point on the parabola. When setting the coordinates, by using the equation of the parabola, we can set only one coordinate and express the other coordinate in terms of it, thereby reducing the number of unknowns."
  },
  {
    "text": "Given that $P$ is any point on the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{4}=1$ different from the left vertex $A$ and the right vertex $B$, denote the slopes of lines $PA$ and $PB$ as $k_{1}$, $k_{2}$, respectively. Then the value of $k_{1} \\cdot k_{2}$ is?",
    "fact_expressions": "P: Point;A: Point;G: Ellipse;B: Point;Expression(G) = (x^2/12 + y^2/4 = 1);PointOnCurve(P,G);LeftVertex(G)=A;RightVertex(G)=B;Negation(P=A);Negation(P=B);k1:Number;k2:Number;Slope(LineOf(P, A)) = k1;Slope(LineOf(P, B)) = k2",
    "query_expressions": "k1*k2",
    "answer_expressions": "-1/3",
    "fact_spans": "[[[2, 5]], [[51, 54]], [[6, 44]], [[58, 61]], [[6, 44]], [[2, 66]], [[6, 54]], [[6, 61]], [[2, 54]], [[2, 61]], [[85, 92]], [[94, 101]], [[68, 101]], [[68, 101]]]",
    "query_spans": "[[[104, 127]]]",
    "process": "Let P(x,y), A(-2\\sqrt{3},0), B(-2\\sqrt{3},0), then k_{1}=\\frac{y}{x+2\\sqrt{3}}, k_{2}=k_{1}k_{2}=\\frac{y}{x+2\\sqrt{3}}\\cdot\\frac{y}{x-2\\sqrt{3}}=\\frac{y^{2}}{x^{2}-12}, \\cdots\\cdots\\textcircled{1} Since P lies on the ellipse, \\frac{x^{2}}{12}+\\frac{y^{2}}{4}=1, that is, y^{2}=\\frac{12-x^{2}}{3} \\cdots\\cdots\\textcircled{2} Substituting \\textcircled{2} into \\textcircled{1}, we get k_{1}k_{2}=\\frac{y^{2}}{x^{2}-12}=-\\frac{1}{3}"
  },
  {
    "text": "$P$ is a point on the left branch of the hyperbola $x^{2}-y^{2}=16$, and $F_{1}$, $F_{2}$ are the left and right foci, respectively. Then $|P F_{1}|-|P F_{2}|$=?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2 = 16);PointOnCurve(P, LeftPart(G));LeftFocus(G) = F1;RightFocus(G) = F2",
    "query_expressions": "Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "-8",
    "fact_spans": "[[[4, 23]], [[0, 3]], [[30, 37]], [[38, 45]], [[4, 23]], [[0, 29]], [[4, 53]], [[4, 53]]]",
    "query_spans": "[[[55, 78]]]",
    "process": "The equation of the hyperbola is \\frac{x^{2}}{16}-\\frac{y^{2}}{16}=1, \\therefore a=4, \\therefore ||PF_{1}|-|PF_{2}||=2a=8, \\because P lies on the left branch of the hyperbola and F is the left focus, hence |PF|-|PF_{2}|=-8."
  },
  {
    "text": "Through the right focus $F$ of the hyperbola $x^{2}-y^{2}=4$, draw a line with an inclination angle of $105^{\\circ}$, intersecting the hyperbola at points $P$ and $Q$. Then the value of $|F P| \\cdot|F Q|$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;F: Point;P: Point;Q: Point;Expression(G) = (x^2 - y^2 = 4);RightFocus(G) = F;PointOnCurve(F, H);Inclination(H) = ApplyUnit(105, degree);Intersection(H,G) = {P,Q}",
    "query_expressions": "Abs(LineSegmentOf(F, P))*Abs(LineSegmentOf(F, Q))",
    "answer_expressions": "8*sqrt(3)/3",
    "fact_spans": "[[[1, 19], [49, 52]], [[45, 47]], [[23, 26]], [[53, 56]], [[57, 60]], [[1, 19]], [[1, 26]], [[0, 47]], [[27, 47]], [[45, 62]]]",
    "query_spans": "[[[64, 86]]]",
    "process": "By the given condition, $ F(2\\sqrt{2},0) $, the slope of the line $ k = \\tan105^{\\circ} = -(2+\\sqrt{3}) $, so the equation of the line is $ y = -(2+\\sqrt{3})(x-2\\sqrt{2}) $. Substituting into $ x^{2}-y^{2}=4 $, we obtain $ (6+4\\sqrt{3})x^{2}-4\\sqrt{2}(7+4\\sqrt{3})x+60+32\\sqrt{3}=0 $. Let $ P(x_{1},y_{1}) $, $ Q(x_{2},y_{2}) $, then $ x_{1}+x_{2} = \\frac{4\\sqrt{2}(7+4\\sqrt{3})}{6+4\\sqrt{3}} $, $ x_{1}x_{2} = \\frac{60+32\\sqrt{3}}{6+4\\sqrt{3}} $. Also, $ |FP| = \\sqrt{1+k^{2}}|x_{1}-2\\sqrt{2}| $, $ |FQ| = \\sqrt{1+k^{2}}|x_{2}-2\\sqrt{2}| $. Therefore, $ |FP|\\cdot|FQ| = (1+k^{2})|x_{1}x_{2}-2\\sqrt{2}(x_{1}+x_{2})+8| = (8+4\\sqrt{3})\\cdot\\left|\\frac{60+32\\sqrt{3}}{6+4\\sqrt{3}}-\\frac{16(7+4\\sqrt{3})}{6+4\\sqrt{3}}+8\\right| = \\frac{(8+4\\sqrt{3})(+4)}{6+4\\sqrt{3}} = \\frac{8\\sqrt{3}}{3} $."
  },
  {
    "text": "It is known that the center of the hyperbola is at the origin, and its eccentricity is $\\sqrt{3}$. If one of its directrices coincides with the directrix of the parabola $y^{2}=4x$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Center(G) = O;O: Origin;Eccentricity(G) = sqrt(3);H: Parabola;Expression(H) = (y^2 = 4*x);OneOf(Directrix(G)) = Directrix(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/3 - y^2/6 = 1",
    "fact_spans": "[[[2, 5], [29, 30], [58, 61]], [[2, 11]], [[9, 11]], [[2, 27]], [[36, 50]], [[36, 50]], [[29, 55]]]",
    "query_spans": "[[[58, 66]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, point $P$ is a moving point on the hyperbola $C$. If $P F_{1}=2 PF_{2}$, $\\angle F_{1} PF_{2}=60^{\\circ}$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Focus(C) = {F1, F2};PointOnCurve(P, C);LineSegmentOf(P, F1) = 2*LineSegmentOf(P, F2);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[18, 80], [89, 95], [156, 162]], [[26, 80]], [[26, 80]], [[84, 88]], [[2, 9]], [[10, 17]], [[26, 80]], [[26, 80]], [[18, 80]], [[2, 83]], [[84, 99]], [[101, 119]], [[121, 154]]]",
    "query_spans": "[[[156, 168]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ has a focus at $F(3 , 0)$, then $m$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2 + x^2/m = 1);m: Number;F: Point;Coordinate(F) = (3, 0);OneOf(Focus(G)) = F",
    "query_expressions": "m",
    "answer_expressions": "8",
    "fact_spans": "[[[0, 28]], [[0, 28]], [[45, 48]], [[33, 43]], [[33, 43]], [[0, 43]]]",
    "query_spans": "[[[45, 50]]]",
    "process": ""
  },
  {
    "text": "Given point $A(0,2)$, the focus of the parabola $y^{2}=2 p x(p>0)$ is $F$, the directrix is $l$, the line segment $FA$ intersects the parabola at point $B$, and a perpendicular is drawn from point $B$ to the directrix $l$, with foot of the perpendicular at $M$. If $AM \\perp MF$, then the standard equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;A: Point;F: Point;M: Point;l: Line;B: Point;H: Line;p: Number;Coordinate(A) = (0, 2);Expression(C) = (y^2 = 2*p*x);p > 0;Focus(C) = F;Directrix(C) = l;Intersection(LineSegmentOf(F,A), C) = B;PointOnCurve(B, H);IsPerpendicular(H,l);FootPoint(H,l) = M;IsPerpendicular(LineSegmentOf(A, M), LineSegmentOf(M, F))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2 = 2*sqrt(2)*x",
    "fact_spans": "[[[12, 33], [56, 59], [106, 112]], [[2, 11]], [[37, 40]], [[83, 86]], [[44, 47], [73, 76]], [[60, 64], [66, 70]], [], [[15, 33]], [[2, 11]], [[12, 33]], [[15, 33]], [[12, 40]], [[12, 47]], [[48, 64]], [[65, 79]], [[65, 79]], [[65, 86]], [[89, 104]]]",
    "query_spans": "[[[106, 119]]]",
    "process": "From the definition of the parabola, we have |BM| = |BF|, F(\\frac{p}{2}, 0). Since AM \\perp MF, point B is the midpoint of segment FA. Given point A(0, 2), it follows that B(\\frac{p}{4}, 1). Since point B lies on the parabola, we have 1 = 2p \\times \\frac{p}{4}, solving gives p = \\sqrt{2}. Therefore, the standard equation of parabola C is y^{2} = 2\\sqrt{2}x."
  },
  {
    "text": "Given the ellipse $C_{1}$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ and the circle $C_{2}$: $x^{2}+y^{2}=\\frac{3 b^{2}}{4}$, if there does not exist a point $P$ on the ellipse $C_{1}$ such that the two tangents drawn from point $P$ to the circle $C_{2}$ are perpendicular to each other, then the range of eccentricity of the ellipse $C_{1}$ is?",
    "fact_expressions": "C1:Ellipse;C2:Circle;P: Point;L1: Line;L2: Line;Expression(C1)=(y^2/b^2 + x^2/a^2 = 1);Expression(C2)=(x^2 + y^2 = (3*b^2)/4);Negation(PointOnCurve(P,C1));TangentOfPoint(P,C2)={L1,L2};IsPerpendicular(L1,L2);a:Number;b:Number;a>b;b>0",
    "query_expressions": "Range(Eccentricity(C1))",
    "answer_expressions": "(0, \\sqrt{3}/3)",
    "fact_spans": "[[[2, 63], [108, 117], [155, 164]], [[64, 105], [136, 144]], [[121, 125], [129, 133]], [], [], [[2, 63]], [[64, 105]], [[108, 125]], [[128, 149]], [[128, 153]], [[2, 63]], [[2, 63]], [[2, 63]], [[2, 63]]]",
    "query_spans": "[[[155, 175]]]",
    "process": "As shown in the figure, let two lines passing through point P be tangent to circle $ C_{2} $ at points M and N respectively. Since the two tangents are perpendicular to each other, we have $ |OP| = \\frac{\\sqrt{6}}{2}b $. From the given condition $ |OP| > a $, it follows that $ \\frac{b}{a} > \\frac{\\sqrt{6}}{3} $. Also, since $ e = \\sqrt{1 - \\left( \\frac{b}{a} \\right)^{2}} $, the result can be obtained. As shown in the figure, let two lines passing through point P be tangent to circle $ C_{2} $ at points M and N respectively. From the fact that the two tangents are perpendicular, we have $ |OP| = \\sqrt{2} \\times \\frac{\\sqrt{3}}{2}b = \\frac{\\sqrt{6}}{2}b $. Moreover, since there is no point P on ellipse $ C_{1} $ such that the two tangents from P to circle $ C_{2} $ are perpendicular, we have $ |OP| > a $, which implies $ \\frac{\\sqrt{6}}{2}b > a $. Hence, $ \\frac{b}{a} > \\frac{\\sqrt{6}}{3} $. Therefore, the eccentricity of ellipse $ C_{1} $ is $ e = \\frac{c}{a} = \\frac{\\sqrt{a^{2} - b^{2}}}{a} = \\sqrt{1 - \\left( \\frac{b}{a} \\right)^{2}} < \\sqrt{1 - \\left( \\frac{\\sqrt{6}}{3} \\right)^{2}} = \\frac{\\sqrt{3}}{3} $. Since $ e > 0 $, we have $ 0 < e < \\frac{\\sqrt{3}}{3} $."
  },
  {
    "text": "Given fixed point $A(4,0)$ and a moving point $B$ on the curve $x^{2}+y^{2}=8$, what is the trajectory equation of the midpoint $P$ of segment $A B$?",
    "fact_expressions": "G: Curve;A: Point;B: Point;P:Point;Expression(G) = (x^2 + y^2 = 8);Coordinate(A) = (4, 0);PointOnCurve(B, G);MidPoint(LineSegmentOf(A,B))=P",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(x-2)^2+y^2=2",
    "fact_spans": "[[[13, 30]], [[4, 12]], [[34, 37]], [[49, 52]], [[13, 30]], [[4, 12]], [[13, 37]], [[39, 52]]]",
    "query_spans": "[[[49, 59]]]",
    "process": "Let the coordinates of point P be (x, y) and point B be (a, b). Since point P is the midpoint of segment AB, we have  \n\\begin{cases}x=\\frac{4+a}{2}\\\\y=\\frac{0+b}{2}\\end{cases}  \nSolving gives  \n\\begin{cases}a=2x-4\\\\b=2y\\end{cases}.  \nSubstituting the coordinates of point B into the curve equation yields (2x-4)^{2}+(2y)^{2}=8. Simplifying gives (x-2)^{2}+y^{2}=2. Therefore, the trajectory equation of point P is (x-2)^{2}+y^{2}=2."
  },
  {
    "text": "It is known that the lower focus of the hyperbola $\\frac{y^{2}}{3}-\\frac{x^{2}}{6}=1$ coincides with the focus of the parabola $y=a x^{2}$, then $a$=?",
    "fact_expressions": "G: Hyperbola;H: Parabola;a: Number;Expression(G) = (-x^2/6 + y^2/3 = 1);Expression(H) = (y = a*x^2);LowerFocus(G) = Focus(H)",
    "query_expressions": "a",
    "answer_expressions": "-1/12",
    "fact_spans": "[[[2, 40]], [[45, 59]], [[66, 69]], [[2, 40]], [[45, 59]], [[2, 64]]]",
    "query_spans": "[[[66, 71]]]",
    "process": "From the hyperbola $\\frac{y^2}{3}-\\frac{x^{2}}{6}=1$, we obtain the lower focus $F(0,-3)$. From the parabola $y=ax^{2}$, we can rewrite it as $x^{2}=\\frac{1}{a}y$, whose focus is $F_{1}(0,\\frac{1}{4a})$. Since $F$ and $F_{1}$ coincide, we have $\\frac{1}{4a}=-3$, solving which gives $a=-\\frac{1}{12}$."
  },
  {
    "text": "If the hyperbola $\\frac{y^{2}}{16}-\\frac{x^{2}}{m}=1$ has eccentricity $e=2$, then $m=$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (y^2/16 - x^2/m = 1);m: Number;e: Number;Eccentricity(G) = e;e = 2",
    "query_expressions": "m",
    "answer_expressions": "48",
    "fact_spans": "[[[1, 40]], [[1, 40]], [[51, 54]], [[44, 49]], [[1, 49]], [[44, 49]]]",
    "query_spans": "[[[51, 56]]]",
    "process": "From the equation, we have $a^{2}=16$, $b^{2}=m$, $\\therefore c^{2}=16+m$, $\\therefore \\frac{16+m}{16}=4$, $\\therefore m=48$."
  },
  {
    "text": "Given that the focal distance of an ellipse is $8$, and the sum of the distances from a point on the ellipse to the two foci is equal to $16$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;FocalLength(G) = 8;PointOnCurve(P,G);Focus(G)={F1,F2};Distance(P,F1)+Distance(P,F2) = 16",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/64 + y^2/48 = 1, x^2/48 + y^2/64 = 1}",
    "fact_spans": "[[[2, 4], [12, 14], [36, 38]], [[17, 18]], [], [], [[2, 11]], [[12, 18]], [[12, 23]], [[12, 34]]]",
    "query_spans": "[[[36, 45]]]",
    "process": "From the given conditions, 2c = 8, 2a = 16, then a = 8, c = 4, and b^{2} = a^{2} - c^{2} = 48, so the standard equation of the ellipse is \\frac{x^{2}}{64} + \\frac{y^{2}}{48} = 1 or \\frac{y^{2}}{64} + \\frac{x^{2}}{48} = 1"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, let the right focus be $F$. From point $F$, a perpendicular is drawn to one asymptote of the hyperbola, with foot of the perpendicular at $M$, intersecting the other asymptote at $N$. If $7 \\overrightarrow {F M}=3 \\overrightarrow{F N}$, then the equations of the asymptotes of the hyperbola are?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;F: Point;RightFocus(C) = F;L1: Line;L2: Line;OneOf(Asymptote(C))=L1;OneOf(Asymptote(C))=L2;Negation(L1=L2);L: Line;M: Point;PointOnCurve(F, L);IsPerpendicular(L, L1);FootPoint(L, L1) = M;N: Point;Intersection(L, L2) = N;7*VectorOf(F, M) = 3*VectorOf(F, N)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*(sqrt(10)/2)*x",
    "fact_spans": "[[[2, 53], [68, 71], [152, 155]], [[2, 53]], [[10, 53]], [[10, 53]], [[58, 61], [63, 67]], [[2, 61]], [], [], [[68, 77]], [[68, 95]], [[68, 95]], [], [[84, 87]], [[62, 80]], [[62, 80]], [[62, 87]], [[96, 99]], [[62, 99]], [[101, 149]]]",
    "query_spans": "[[[152, 163]]]",
    "process": "By the given condition, the right focus of the hyperbola is $ F(c,0) $. Let the equation of one asymptote $ OM $ be $ y = \\frac{b}{a}x $, then the equation of the other asymptote $ ON $ is $ y = -\\frac{b}{a}x $. Let $ M(m, \\frac{bm}{a}) $, $ N(n, -\\frac{bn}{a}) $. Since $ 7\\overrightarrow{FM} = 3\\overrightarrow{FN} $, we have $ 7(m - c, \\frac{bm}{a}) = 3(n - c, -\\frac{bn}{a}) $, thus  \n$$\n\\begin{cases}\n7(m - c) = 3(n - c) \\\\\n\\frac{7bm}{a} = -\\frac{3bn}{a}\n\\end{cases}\n$$\nSolving gives  \n$$\n\\begin{cases}\nm = \\frac{2c}{7} \\\\\nn = -\\frac{2c}{3}\n\\end{cases}\n$$\nTherefore, the coordinates of point $ M $ are $ (\\frac{2c}{7}, \\frac{2bc}{7a}) $. Since $ OM \\perp FM $, we have  \n$$\nk_{OM} \\cdot k_{FM} = \\frac{b}{a} \\times \\frac{\\frac{2bc}{7a}}{\\frac{2c}{7} - c} = -1\n$$\nSimplifying yields $ \\frac{b^{2}}{a^{2}} = \\frac{5}{2} $. Hence, the equations of the asymptotes of the hyperbola are $ y = \\pm \\frac{b}{a}x = \\pm \\frac{\\sqrt{10}}{2}x $. Answer: $ y = \\pm \\frac{\\sqrt{10}}{2}x $"
  },
  {
    "text": "If $P$ is a point on the parabola $C$: $y^{2}=2 x$, $F$ is the focus of the parabola $C$, and point $A\\left(\\frac{7}{2}, 2\\right)$, then when $P A+P F$ takes the minimum value, the coordinates of point $P$ are?",
    "fact_expressions": "P: Point;C: Parabola;Expression(C) = (y^2 = 2*x);PointOnCurve(P, C) = True;F: Point;Focus(C) = F;A: Point;Coordinate(A) = (7/2, 2);WhenMin(LineSegmentOf(P, A) + LineSegmentOf(P, F))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2, 2)",
    "fact_spans": "[[[1, 4], [80, 84]], [[5, 24], [32, 38]], [[5, 24]], [[1, 27]], [[28, 31]], [[28, 41]], [[42, 62]], [[42, 62]], [[64, 78]]]",
    "query_spans": "[[[80, 89]]]",
    "process": "Draw a schematic diagram. From point P, draw a perpendicular line PB to the directrix of the parabola, with foot at B. From the figure, it can be seen that PA + PF reaches its minimum value when PA is perpendicular to the directrix of the parabola. Substitute $ y = 2 $ into $ y^2 = 2x $, yielding $ x = 2 $. At this time, the coordinates of point P are $ (2, 2) $."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. A line passing through $F_{2}$ intersects the ellipse at points $A$ and $B$. If $\\overrightarrow{A F_{1}} \\cdot \\overrightarrow{A F_{2}}=0$ and $\\overrightarrow{A F_{2}}=3 \\overrightarrow{F_{2} B}$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;A: Point;F1: Point;F2: Point;B: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F2, G);Intersection(G, C) = {A, B};DotProduct(VectorOf(A, F1), VectorOf(A, F2)) = 0;VectorOf(A, F2) = 3*VectorOf(F2, B)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[17, 74], [93, 95], [225, 230]], [[24, 74]], [[24, 74]], [[90, 92]], [[97, 100]], [[1, 8]], [[9, 16], [82, 89]], [[101, 104]], [[24, 74]], [[24, 74]], [[17, 74]], [[1, 80]], [[1, 80]], [[81, 92]], [[90, 106]], [[108, 167]], [[170, 223]]]",
    "query_spans": "[[[225, 236]]]",
    "process": "According to the problem and the definition of an ellipse, let $ F_{2}B = x $, then the lengths of the other sides can be expressed. Using the Pythagorean theorem, $ x $ can be found. In right triangle $ \\triangle AF_{1}F_{2} $, using the Pythagorean theorem, the relationship between $ a $ and $ c $ can be obtained, and substituting into the formula yields the answer. From the problem, $ AF_{1} \\perp AF_{2} $, and $ |\\overrightarrow{AF_{2}}| = 3|\\overrightarrow{F_{2}B}| $, as shown in the figure: let $ F_{2}B = x $, then $ AF_{2} = 3x $. Therefore, according to the definition of the ellipse, $ AF_{1} = 2a - 3x $, $ BF_{1} = 2a - x $. Since $ AF_{1} \\perp AF_{2} $, it follows that $ (2a - x)^{2} = (2a - 3x)^{2} + (3x + x)^{2} $. Solving gives $ x = \\frac{a}{3} $. Thus, $ AF_{1} = 2a - 3x = a $, $ AF_{2} = 3x = a $. In right triangle $ \\triangle AF_{1}F_{2} $, $ AF_{1}^{2} + AF_{2}^{2} = F_{1}F_{2}^{2} $, i.e., $ a^{2} + a^{2} = (2c)^{2} $. Therefore, $ e = \\frac{c}{a} = \\sqrt{\\frac{c^{2}}{a^{2}}} = \\sqrt{\\frac{1}{2}} = \\frac{\\sqrt{2}}{2} $."
  },
  {
    "text": "Given the line $y=\\frac{1}{2} x$ intersects the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1$ at points $A$ and $B$, and $P$ is a point on the hyperbola distinct from $A$ and $B$. When the slopes $k_{P A}$, $k_{P B}$ of lines $P A$ and $P B$ exist, $k_{P A} \\cdot k_{P B}$=?",
    "fact_expressions": "G: Hyperbola;H: Line;A: Point;P: Point;B: Point;k1: Number;k2: Number;Expression(G) = (x^2/9 - y^2/4 = 1);Expression(H) = (y = x/2);Intersection(H, G) = {A, B};PointOnCurve(P, G);Negation(A = P);Negation(B = P);Slope(LineOf(P,A))=k1;Slope(LineOf(P,B))=k2",
    "query_expressions": "k1*k2",
    "answer_expressions": "4/9",
    "fact_spans": "[[[22, 60], [76, 79]], [[2, 21]], [[62, 65], [83, 86]], [[72, 75]], [[66, 69], [87, 90]], [[112, 121]], [[124, 133]], [[22, 60]], [[2, 21]], [[2, 71]], [[72, 92]], [[72, 92]], [[72, 92]], [[94, 133]], [[94, 133]]]",
    "query_spans": "[[[137, 162]]]",
    "process": "Solving the system \\begin{cases}y=\\frac{1}{2}x\\\\\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1\\end{cases} yields \\frac{7}{144}x^{2}=1\\Rightarrow x=\\pm\\frac{12\\sqrt{7}}{7}, let A(\\frac{12\\sqrt{7}}{7},\\frac{6\\sqrt{7}}{7}), B(-\\frac{12\\sqrt{7}}{7},-\\frac{6\\sqrt{7}}{7}), since P is a point on the hyperbola different from A, B, let P(x,y) satisfy \\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1, k_{PA}=\\frac{y-\\frac{6\\sqrt{7}}{7}}{x-\\frac{12\\sqrt{7}}{7}}, k_{PB}=\\frac{y+\\frac{6\\sqrt{7}}{7}}{x+\\frac{12\\sqrt{7}}{7}} \\therefore k_{PA}k_{PB}=\\frac{y-\\frac{6\\sqrt{7}}{7}}{x-\\frac{12\\sqrt{7}}{7}}\\frac{y+\\frac{6\\sqrt{7}}{7}}{x+\\frac{12\\sqrt{7}}{7}}\\frac{y^{2}-\\frac{36}{7}}{x^{2}-\\frac{144}{7}}=\\frac{x^{2}\\frac{4}{9}-4-\\frac{36}{x^{2}-\\frac{144}{7}}=\\frac{4}{9}, hence fill in \\frac{4}{9}"
  },
  {
    "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $\\frac{\\sqrt{3}}{2}$, then the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ are?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Ellipse;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a > b;b > 0;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);Eccentricity(H) = sqrt(3)/2",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*x/2",
    "fact_spans": "[[[80, 126]], [[3, 53]], [[3, 53]], [[1, 53]], [[80, 126]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 78]]]",
    "query_spans": "[[[80, 134]]]",
    "process": ""
  },
  {
    "text": "If point $P$ is any point on the ellipse $C$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and points $A$, $B$ are the upper and lower vertices of ellipse $C$ respectively, and if the angles of inclination of lines $P A$, $P B$ are $\\alpha$, $\\beta$ respectively, then $\\frac{\\cos (\\alpha-\\beta)}{\\cos (\\alpha+\\beta)}=$?",
    "fact_expressions": "C: Ellipse;P: Point;A: Point;B: Point;alpha: Number;beta: Number;Expression(C) = (x^2/25 + y^2/16 = 1);PointOnCurve(P, C);UpperVertex(C) = A;LowerVertex(C) = B;Inclination(LineOf(P, A)) = alpha;Inclination(LineOf(P, B)) = beta",
    "query_expressions": "Cos(alpha - beta)/Cos(alpha + beta)",
    "answer_expressions": "9/41",
    "fact_spans": "[[[6, 50], [67, 72]], [[1, 5]], [[56, 60]], [[61, 64]], [[101, 109]], [[111, 120]], [[6, 50]], [[1, 55]], [[56, 77]], [[56, 77]], [[79, 119]], [[79, 119]]]",
    "query_spans": "[[[122, 173]]]",
    "process": "First, use the angle sum and difference formulas along with the relationships of trigonometric functions of the same angle to simplify $\\frac{\\cos(\\alpha-\\beta)}{\\cos(\\alpha+\\beta)}$ into $\\frac{1+\\tan\\alpha\\tan\\beta}{1-\\tan\\alpha\\tan\\beta}$. Then set point $P$, find the slopes of lines $PA$ and $PB$, and substitute the ellipse equation to simplify and solve. Solution: From $\\frac{\\cos(\\alpha-\\beta)}{\\cos(\\alpha+\\beta)} = \\frac{\\cos\\alpha\\cos\\beta + \\sin\\alpha\\sin\\beta}{\\cos\\alpha\\cos\\beta - \\sin\\alpha\\sin\\beta} = \\frac{1+\\tan\\alpha\\tan\\beta}{1-\\tan\\alpha\\tan\\beta}$, let point $P(x,y)$, given $A(0,4)$, $B(0,-4)$, the inclination angles of lines $PA$ and $PB$ are $\\alpha$, $\\beta$ respectively, so $\\tan\\alpha = \\frac{y-4}{x}$, $y+4$, so $\\frac{c}{c}$, and since $\\frac{x^{2}}{25} + \\frac{y^{2}}{16} = 1$, we get $x^{2} = (1 - \\frac{\\cos(\\alpha-\\beta)}{\\cos(\\alpha+\\beta)} = \\frac{x^{2}+y^{2}-16}{x^{2}-y^{2}+16} = \\frac{9}{41}$."
  },
  {
    "text": "The length of the major axis of an ellipse is $2$ times the length of the minor axis, and it passes through the point $(2,0)$. Then the standard equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;H: Point;Coordinate(H) = (2, 0);Length(MajorAxis(G)) = 2*Length(MinorAxis(G));PointOnCurve(H,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/4+y^2=1,y^2/16+x^2/4=1}",
    "fact_spans": "[[[0, 2], [30, 32]], [[19, 27]], [[19, 27]], [[0, 15]], [[0, 27]]]",
    "query_spans": "[[[30, 38]]]",
    "process": "Discuss two cases: the focus on the x-axis and the focus on the y-axis. According to the problem, $ a=2b $. <sup>①</sup> Let the standard equation of the ellipse be $ \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 $, then $ \\begin{cases}a=2b\\\\a=2\\end{cases} $, yielding $ \\begin{cases}a^{2}=4\\\\12-1\\end{cases} $, thus the standard equation of the ellipse is $ \\frac{x^{2}}{4}+y^{2}=1 $; <sup>②</sup> Let the standard equation of the ellipse be $ \\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1 $, then $ \\begin{cases}a=2\\\\a=2b\\end{cases} $, yielding $ \\begin{cases}\\frac{2}{a}=1\\\\b=2\\end{cases} $, $ \\frac{1}{2}=4 $, thus the standard equation of the ellipse is $ \\frac{y^{2}}{16}+\\frac{x^{2}}{4}=1 $."
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, with $F_{1}$ and $F_{2}$ being its left and right foci respectively, the segment $F_{2} A$ is perpendicular to the line $y=\\frac{b}{a} x$, with foot of perpendicular at point $A$, intersecting the hyperbola at point $B$. If $\\overrightarrow{F_{2} B}=\\overrightarrow{B A}$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Line;Expression(H) = (y = x*(b/a));IsPerpendicular(LineSegmentOf(F2, A), H);A: Point;FootPoint(LineSegmentOf(F2, A), H) = A;B: Point;Intersection(LineSegmentOf(F2, A), G) = B;VectorOf(F2, B) = VectorOf(B, A)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 56], [124, 127], [185, 188], [75, 76]], [[0, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[58, 65]], [[67, 74]], [[58, 81]], [[58, 81]], [[95, 114]], [[95, 114]], [[82, 114]], [[118, 122]], [[82, 122]], [[129, 133]], [[82, 133]], [[135, 182]]]",
    "query_spans": "[[[185, 194]]]",
    "process": ""
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1$ are $F_{1}$ and $F_{2}$, and point $P$ lies on the ellipse. If $|P F_{1}|=6$, then the area of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/49 + y^2/24 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 6",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "24",
    "fact_spans": "[[[0, 39], [64, 66]], [[59, 63]], [[43, 50]], [[51, 58]], [[0, 39]], [[0, 58]], [[59, 67]], [[69, 82]]]",
    "query_spans": "[[[84, 111]]]",
    "process": "From the ellipse $\\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1$, we obtain $a=7$, $b=2\\sqrt{6}$, $c=\\sqrt{a^{2}-b^{2}}=5$, so $|F_{1}F_{2}|=2c=10$. Since $|PF_{1}|+|PF_{2}|=2a=14$ and $|PF_{1}|=6$, then $|PF_{2}|=8$. Thus, $|PF_{1}|^{2}+|PF_{2}|^{2}=100=|F_{1}F_{2}|^{2}$, so $\\angle F_{1}PF_{2}=90^{\\circ}$. The area of $\\Delta PF_{1}F_{2}$ is $S_{\\triangle PF_{1}F_{2}}=\\frac{1}{2}|PF_{1}||PF_{2}|=24$."
  },
  {
    "text": "Given the hyperbola $x^2 - y^{2} = 4$, the line $l$: $y = k(x - 1)$ intersects the hyperbola at exactly one point. Find all possible values of $k$.",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 4);l: Line;Expression(l) = (y = k*(x - 1));k: Number;NumIntersection(l, G) = 1",
    "query_expressions": "k",
    "answer_expressions": "{pm*1, pm*2*sqrt(3)/3}",
    "fact_spans": "[[[2, 18], [37, 40]], [[2, 18]], [[19, 35]], [[19, 35]], [[49, 52]], [[19, 47]]]",
    "query_spans": "[[[49, 54]]]",
    "process": ""
  },
  {
    "text": "What is the eccentricity of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 45]]]",
    "process": "\\because from the problem we know a=3, b=4 \\therefore c=\\sqrt{3^{2}+4^{4}}=5 \\therefore eccentricity e=\\frac{c}{a}=\\frac{5}{3}"
  },
  {
    "text": "Let the two foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ be $F_{1}$, $F_{2}$, and let $P$ be any point on the ellipse other than the endpoints of the major axis. In $\\triangle P F_{1} F_{2}$, denote $\\angle F_{1} P F_{2}=\\alpha$, $\\angle P F_{1} F_{2}=\\beta$, $\\angle F_{1} F_{2} P=\\gamma$. Then the eccentricity $e$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Focus(G) = {F1,F2};F1: Point;F2: Point;P: Point;PointOnCurve(P,G) = True;Negation(P = Endpoint(MajorAxis(G)));alpha: Number;AngleOf(F1,P,F2) = alpha;beta: Number;AngleOf(P,F1,F2) = beta;gamma: Number;AngleOf(F1,F2,P) = gamma;e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "Sin(alpha)/(Sin(gamma)+Sin(beta))",
    "fact_spans": "[[[1, 53], [79, 81]], [[1, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[1, 74]], [[59, 66]], [[67, 74]], [[75, 78]], [[75, 93]], [[75, 93]], [[123, 152]], [[123, 152]], [[153, 182]], [[153, 182]], [[184, 213]], [[184, 213]], [[218, 221]], [[79, 221]]]",
    "query_spans": "[[[218, 223]]]",
    "process": "Let |PF_{1}| = m, |PF_{2}| = n. By the law of sines: \\frac{m}{\\siny} = \\frac{n}{\\sin\\beta} = \\frac{2c}{\\sin\\alpha}, it follows that: \\frac{m+n}{\\siny+\\sin\\beta} = \\frac{2c}{\\sin\\alpha}. Since m+n=2a, \\therefore e = \\frac{c}{a} = \\frac{\\sin\\alpha}{\\siny+\\sin\\beta}"
  },
  {
    "text": "Draw a line through the focus $F$ of the parabola $x^{2}=4 y$ perpendicular to the $y$-axis, intersecting the parabola at point $P$. Then the equation of the tangent line $l$ to the parabola at point $P$ is?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;F: Point;Expression(G) = (x^2 = 4*y);Focus(G)=F;PointOnCurve(F,H);IsPerpendicular(H,yAxis);Intersection(H, G) = P;TangentOnPoint(P, G)=l;l:Line",
    "query_expressions": "Expression(l)",
    "answer_expressions": "{x-y-1=0, x+y+1=0}",
    "fact_spans": "[[[1, 15], [33, 36], [45, 48]], [[30, 32]], [[39, 43], [49, 53]], [[18, 21]], [[1, 15]], [[1, 21]], [[0, 32]], [[22, 32]], [[30, 43]], [[45, 60]], [[57, 60]]]",
    "query_spans": "[[[57, 65]]]",
    "process": ""
  },
  {
    "text": "It is known that the line $l$ passes through the focus $F$ of the parabola $y^{2}=2 p x(p>0)$, intersecting the parabola at points $A$ and $B$. If $|A B|=7$, and the distance from the midpoint of segment $A B$ to the line $x=-3$ is $5$, then what is the value of $p$?",
    "fact_expressions": "l: Line;G: Parabola;p: Number;H: Line;B: Point;A: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (x = -3);Focus(G) = F;PointOnCurve(F, l);Intersection(l,G)={A,B};Abs(LineSegmentOf(A, B)) = 7;Distance(MidPoint(LineSegmentOf(A,B)), H) = 5",
    "query_expressions": "p",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 7]], [[9, 30], [38, 41]], [[93, 96]], [[76, 84]], [[47, 50]], [[43, 46]], [[33, 36]], [[12, 30]], [[9, 30]], [[76, 84]], [[9, 36]], [[2, 36]], [[2, 52]], [[54, 63]], [[65, 91]]]",
    "query_spans": "[[[93, 99]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since the midpoint coordinates of AB are (\\frac{x_{1}+x_{2}}{2},\\frac{y_{1}+y_{2}}{2}), and the distance from the midpoint of segment AB to the line x=-3 is 5, then \\frac{x_{1}+x_{2}}{2}=-3+5=2, x_{1}+x_{2}=4. Also |AB|=x_{1}+\\frac{p}{2}+x_{2}+\\frac{p}{2}=x_{1}+x_{2}+p=4+p=7, so p=3."
  },
  {
    "text": "What is the focal distance of the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/4 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 33]]]",
    "process": "Given $a^{2}=1$, $b^{2}=4$, so $c^{2}=5$, therefore the focal distance is $2\\sqrt{5}$."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, the intersection point of the directrix and the $x$-axis is $M$, and $N$ is a point on the parabola, if $|NF|=\\frac{\\sqrt{3}}{2}|MN|$, then $\\angle NMF=$?",
    "fact_expressions": "G: Parabola;N: Point;F: Point;M: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;Intersection(Directrix(G), xAxis) = M;PointOnCurve(N, G);Abs(LineSegmentOf(N, F)) = (sqrt(3)/2)*Abs(LineSegmentOf(M, N))",
    "query_expressions": "AngleOf(N, M, F)",
    "answer_expressions": "pi/6",
    "fact_spans": "[[[2, 16], [43, 46]], [[39, 42]], [[20, 23]], [[35, 38]], [[2, 16]], [[2, 23]], [[2, 38]], [[39, 50]], [[54, 85]]]",
    "query_spans": "[[[86, 102]]]",
    "process": ""
  },
  {
    "text": "Given that $A$ and $B$ are the common vertices of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$. $P$ is a moving point on the hyperbola, $M$ is a moving point on the ellipse ($P$, $M$ are distinct from $A$, $B$), and satisfy $\\overrightarrow{A P}+\\overrightarrow{B P}=\\lambda(\\overrightarrow{A M}+\\overrightarrow{B M})$, where $\\lambda \\in \\mathbb{R}$. Let the slopes of lines $A P$, $B P$, $A M$, $B M$ be denoted by $k_{1}$, $k_{2}$, $k_{3}$, $k_{4}$ respectively. If $k_{1}+k_{2}=5$, then $k_{3}+k_{4}$=?",
    "fact_expressions": "G: Hyperbola;b1: Number;a1: Number;H: Ellipse;A: Point;P: Point;B: Point;M: Point;k1:Number;k2:Number;k3:Number;k4:Number;a1>0;b1>0;Expression(G) = (-y^2/b1^2 + x^2/a1^2 = 1);a> b;b> 0;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);Vertex(G)={A,B};Vertex(H)={A,B};PointOnCurve(P,G);PointOnCurve(M,H);Negation(P=A);Negation(P=B);Negation(M=A);Negation(M=B);lambda:Real;VectorOf(A, P) + VectorOf(B, P) = lambda*(VectorOf(A, M) + VectorOf(B, M));Slope(LineOf(A,P))=k1;Slope(LineOf(B,P))=k2;Slope(LineOf(A,M))=k3;Slope(LineOf(B,M))=k4;k1+k2=5;a:Number;b:Number",
    "query_expressions": "k3 + k4",
    "answer_expressions": "-5",
    "fact_spans": "[[[63, 119], [129, 132]], [[66, 119]], [[66, 119]], [[10, 62], [141, 143]], [[2, 5], [159, 162]], [[125, 128], [148, 151]], [[6, 9], [163, 167]], [[137, 140], [153, 156]], [[324, 331]], [[333, 340]], [[341, 349]], [[351, 358]], [[66, 119]], [[66, 119]], [[63, 119]], [[12, 62]], [[12, 62]], [[10, 62]], [[2, 124]], [[2, 124]], [[125, 136]], [[137, 147]], [[148, 167]], [[148, 167]], [[148, 167]], [[148, 167]], [[269, 284]], [[172, 266]], [[286, 358]], [[286, 358]], [[286, 358]], [[286, 358]], [[360, 375]], [[12, 62]], [[12, 62]]]",
    "query_spans": "[[[377, 392]]]",
    "process": ""
  },
  {
    "text": "The slope of the line passing through the lower focus of the ellipse $\\frac{x^{2}}{2}+\\frac{y^{2}}{3}=1$ and tangent to the circle $x^{2}+y^{2}-3 x+y+\\frac{3}{2}=0$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2/3 = 1);L: Line;PointOnCurve(LowerFocus(G), L);H: Circle;Expression(H) = (y - 3*x + x^2 + y^2 + 3/2 = 0);IsTangent(L, H)",
    "query_expressions": "Slope(L)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 38]], [[1, 38]], [[82, 84]], [[0, 84]], [[45, 79]], [[45, 79]], [[44, 84]]]",
    "query_spans": "[[[82, 89]]]",
    "process": ""
  },
  {
    "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ are given by $y=\\pm \\frac{1}{2}x$, then what is the value of $b$?",
    "fact_expressions": "G: Hyperbola;b: Number;b>0;Expression(G) = (x^2/4 - y^2/b^2 = 1);Expression(Asymptote(G)) = (y = pm*(x/2))",
    "query_expressions": "b",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 49]], [[78, 81]], [[4, 49]], [[1, 49]], [[1, 76]]]",
    "query_spans": "[[[78, 84]]]",
    "process": ""
  },
  {
    "text": "If the curve $|y|=2^{x}+1$ and the line $y=b$ have no common points, then the range of values for $b$ is?",
    "fact_expressions": "G: Line;b: Number;H: Curve;Expression(G) = (y = b);Expression(H) = (Abs(y) = 2^x + 1);NumIntersection(H,G)=0",
    "query_expressions": "Range(b)",
    "answer_expressions": "[-1,1]",
    "fact_spans": "[[[17, 24]], [[31, 34]], [[1, 16]], [[17, 24]], [[1, 16]], [[1, 29]]]",
    "query_spans": "[[[31, 41]]]",
    "process": ""
  },
  {
    "text": "Given that point $M(-3 , 2)$ is a fixed point in the coordinate plane, if the focus of the parabola $y^{2}=2 x$ is $F$, and point $Q$ is a moving point on this parabola, then the minimum value of $|M Q|-|Q F|$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (-3, 2);G: Parabola;Expression(G) = (y^2 = 2*x);F: Point;Focus(G) = F;Q: Point;PointOnCurve(Q, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(M, Q)) - Abs(LineSegmentOf(Q, F)))",
    "answer_expressions": "5/2",
    "fact_spans": "[[[2, 14]], [[2, 14]], [[25, 39], [53, 56]], [[25, 39]], [[43, 46]], [[25, 46]], [[47, 51]], [[47, 61]]]",
    "query_spans": "[[[63, 82]]]",
    "process": "The directrix of the parabola is given by $x = -\\frac{1}{2}$. When $MQ \\parallel x$-axis, $|MQ| - |QF|$ attains its minimum value. At this time, substituting the $y$-coordinate of point $Q$, $y = 2$, into the parabola equation $y^2 = 2x$, we obtain the $x$-coordinate of $Q$ as $x = 2$. Then, $|QM| - |QF| = |2 + 3| - |2 + \\frac{1}{2}| = \\frac{5}{2}$."
  },
  {
    "text": "Given that point $A$ is the intersection of the axis of symmetry and the directrix of the parabola $x^{2}=4 y$, point $B$ is the focus of the parabola, and point $P$ lies on the parabola such that when $PA$ is tangent to the parabola, point $P$ lies exactly on the hyperbola with foci at $A$ and $B$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "H: Parabola;Expression(H) = (x^2 = 4*y);A: Point;Intersection(SymmetryAxis(H), Directrix(H)) = A;B: Point;Focus(H) = B;P: Point;PointOnCurve(P, H);IsTangent(LineSegmentOf(P, A), H);G: Hyperbola;PointOnCurve(P, G);Focus(G) = {A, B}",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)+1",
    "fact_spans": "[[[7, 21], [37, 40], [49, 52], [62, 65]], [[7, 21]], [[2, 6], [77, 80]], [[2, 31]], [[32, 36], [81, 84]], [[32, 43]], [[44, 48], [69, 73]], [[44, 53]], [[56, 67]], [[88, 91], [94, 97]], [[69, 92]], [[76, 91]]]",
    "query_spans": "[[[94, 103]]]",
    "process": "Without loss of generality, assume point P is in the first quadrant, as shown in the figure, A(0,-1), B(0,1). Draw a perpendicular from point P to the directrix, with foot N. Since line PA is tangent to the parabola, let the equation of line AP be y = kx - 1. Solving simultaneously:\n\\begin{cases}y=kx-1\\\\x^{2}=4y\\end{cases},\nwe obtain x^{2}-4kx+4=0. The discriminant Δ=16k^{2}-16=0, ∴k=1 (discarding k=-1). ∴P(2,1). ∴The length of the transverse axis of the hyperbola is |PA| - |PB| = 2(\\sqrt{2}-1), so a=\\sqrt{2}-1, c=1. ∴The eccentricity of the hyperbola is e=\\frac{c}{a}=\\frac{1}{\\sqrt{2}-1}=\\sqrt{2}+"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has an eccentricity of $2$. A line passing through the right focus of the hyperbola and perpendicular to the $x$-axis is intersected by the hyperbola forming a chord of length $m$. Then $\\frac{m}{b}$=?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = 2;PointOnCurve(RightFocus(G),H);IsPerpendicular(H,xAxis);Length(InterceptChord(H,G))=m;m:Number",
    "query_expressions": "m/b",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[2, 58], [68, 71], [68, 71]], [[5, 58]], [[5, 58]], [[84, 86]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 66]], [[67, 86]], [[76, 86]], [[84, 99]], [[96, 99]]]",
    "query_spans": "[[[101, 116]]]",
    "process": "Let the focal distance of the hyperbola be $2c$, then $\\frac{c}{a}=2$, i.e., $c=2a$, $\\therefore b=\\sqrt{3}a$. Substituting $x=c=2a$ into the hyperbola equation yields $y=\\pm3a$, $\\therefore m=6a$, $\\therefore \\frac{m}{b}=\\frac{6a}{\\sqrt{3}a}=2\\sqrt{3}$."
  },
  {
    "text": "It is known that the center of ellipse $G$ is at the origin of the coordinate system, its major axis lies on the $y$-axis, its eccentricity is $\\frac{\\sqrt{3}}{2}$, and the sum of the distances from a point on $G$ to the two foci of $G$ is $12$. Then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;OverlappingLine(MajorAxis(G), yAxis);Eccentricity(G) = sqrt(3)/2;H: Point;PointOnCurve(H, G);F1: Point;F2: Point;Focus(G) = {F1, F2};Distance(H, F1) + Distance(H, F2) = 12",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/36 + x^2/9 = 1",
    "fact_spans": "[[[2, 7], [51, 54], [58, 61], [78, 80]], [[11, 15]], [[2, 15]], [[2, 24]], [[2, 49]], [[56, 57]], [[51, 57]], [], [], [[58, 66]], [[51, 76]]]",
    "query_spans": "[[[78, 85]]]",
    "process": "The eccentricity is given as $\\frac{\\sqrt{3}}{2}$, so $\\frac{c}{a} = \\frac{\\sqrt{3}}{2}$. The sum of the distances from a point on $C$ to the two foci is 12, hence $2a = 12$, solving gives $a = 6$, $c = 3\\sqrt{3}$, then $b = 3$. Therefore, the standard equation of the ellipse $C$ is $\\frac{y^{2}}{36} + \\frac{x^{2}}{9} = 1$."
  },
  {
    "text": "Given the ellipse $\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the ellipse, $O$ is the coordinate origin, and $\\angle F_{1} P F_{2}=\\angle F_{1} O P=\\frac{2 \\pi}{3}$. Find the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;P: Point;F2: Point;O: Origin;a > b;b > 0;Expression(G) = (y/b^2 + x/a^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P,G);AngleOf(F1,P,F2)=AngleOf(F1,O,P);AngleOf(F1,O,P)=2*pi/3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(\\sqrt{10}-\\sqrt{2})/2",
    "fact_spans": "[[[2, 48], [78, 80], [148, 150]], [[4, 48]], [[4, 48]], [[57, 64]], [[73, 77]], [[65, 72]], [[82, 85]], [[4, 48]], [[4, 48]], [[2, 48]], [[2, 72]], [[2, 72]], [[73, 81]], [[91, 146]], [[91, 146]]]",
    "query_spans": "[[[148, 156]]]",
    "process": ""
  },
  {
    "text": "Given the equation of the ellipse is $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$. If $C$ is a point on the ellipse, $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse respectively, and $|C F_{1}|=2$, then $|C F_{2}|$=?",
    "fact_expressions": "G: Ellipse;C: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(C, G);LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(C, F1)) = 2",
    "query_expressions": "Abs(LineSegmentOf(C, F2))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 4], [51, 53], [75, 77]], [[47, 50]], [[57, 64]], [[65, 72]], [[2, 45]], [[47, 56]], [[57, 83]], [[57, 83]], [[86, 99]]]",
    "query_spans": "[[[101, 114]]]",
    "process": "The ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ has $a=5$. By the definition of an ellipse, $|CF_{1}|+|CF_{2}|=2a=10$. Given $|CF|=2$, we get $|CF_{2}|=10-2=8$. Hence the answer is: $8$."
  },
  {
    "text": "The ellipse passing through the point $(3, -2)$ and having the same foci as $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ is?",
    "fact_expressions": "G: Ellipse;H: Point;Coordinate(H) = (3,-2);C:Curve;Expression(C)=(x^2/9+y^2/4=1);Focus(C)=Focus(G);PointOnCurve(H,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/15 + y^2/10 = 1",
    "fact_spans": "[[[54, 56]], [[1, 11]], [[1, 11]], [[13, 48]], [[13, 48]], [[12, 56]], [[0, 56]]]",
    "query_spans": "[[[54, 58]]]",
    "process": ""
  },
  {
    "text": "There is a point $P$ on the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ such that the distance from $P$ to the left directrix is $\\frac{16}{5}$. Then, the distance from $P$ to the right focus is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);P: Point;PointOnCurve(P, G);Distance(P, LeftDirectrix(G)) = 16/5",
    "query_expressions": "Distance(P, RightFocus(G))",
    "answer_expressions": "34/3",
    "fact_spans": "[[[0, 39]], [[0, 39]], [[43, 46], [70, 73]], [[0, 46]], [[0, 68]]]",
    "query_spans": "[[[0, 82]]]",
    "process": ""
  },
  {
    "text": "The equation of the locus of the midpoint $M$ of the segment $PQ$, where $P$ is any point on the circle $x^{2}+y^{2}=8$ and $Q$ is the foot of the perpendicular from $P$ to the $x$-axis, is?",
    "fact_expressions": "G: Circle;Q: Point;P: Point;M:Point;Expression(G) = (x^2 + y^2 = 8);PointOnCurve(P,G);L:Line;PointOnCurve(P,L);IsPerpendicular(L,xAxis);FootPoint(L,xAxis)=Q;MidPoint(LineSegmentOf(P,Q))=M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2/8+y^2/2=1",
    "fact_spans": "[[[1, 17]], [[36, 39]], [[22, 25]], [[51, 54]], [[1, 17]], [[1, 25]], [], [[0, 32]], [[0, 32]], [[0, 39]], [[41, 54]]]",
    "query_spans": "[[[51, 61]]]",
    "process": "Using the midpoint coordinate formula, determine the relationship between the coordinates of P and M. Substitute the coordinates of P into the equation of the circle to obtain the trajectory equation of M. Let M(x, y), Q(x, 0), then P(x, 2y). Since P lies on the circle x^{2}+y^{2}=8, we have x^{2}+4y^{2}=8. Rearranging gives \\frac{x^{2}}{8}+\\frac{y^{2}}{2}=1."
  },
  {
    "text": "It is known that one focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is the center of the circle $x^{2}+y^{2}-6x+8=0$, and the length of the minor axis is $8$. Then, what is the left vertex of the ellipse?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Circle;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (-6*x + x^2 + y^2 + 8 = 0);Center(H) = OneOf(Focus(G));Length(MinorAxis(G)) = 8",
    "query_expressions": "Coordinate(LeftVertex(G))",
    "answer_expressions": "(-5, 0)",
    "fact_spans": "[[[2, 54], [96, 98]], [[4, 54]], [[4, 54]], [[60, 82]], [[4, 54]], [[4, 54]], [[2, 54]], [[60, 82]], [[2, 85]], [[2, 94]]]",
    "query_spans": "[[[96, 104]]]",
    "process": "\\because the standard equation of the circle is (x-3)^{2}+y^{2}=1, \\therefore the center coordinates are (3,0), \\therefore c=3. Also b=4, \\therefore a=\\sqrt{b^{2}+c^{2}}=5 \\because the foci of the ellipse lie on the c-axis. \\cdot the vertices of the ellipse are (-5,0)."
  },
  {
    "text": "Given that the directrix of the parabola $y^{2}=2 x$ is also a directrix of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{3}=1$, then the equations of the two asymptotes of this hyperbola are?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 2*x);G: Hyperbola;Expression(G) = (-y^2/3 + x^2/m = 1);m: Number;Directrix(H) = OneOf(Directrix(G))",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[21, 59], [67, 70]], [[21, 59]], [[24, 59]], [[2, 64]]]",
    "query_spans": "[[[67, 80]]]",
    "process": "Since the directrix of the parabola $ y^{2} = 2x $ is $ x = -\\frac{1}{2} $, and one left directrix of the hyperbola $ \\frac{x^{2}}{m} - \\frac{y^{2}}{3} = 1 $ is $ x = -\\frac{m}{\\sqrt{m+3}} $ ($ m > 0 $), we have $ \\frac{1}{2} = \\frac{m}{\\sqrt{m+3}} $. Solving this gives $ m = 1 $. Therefore, the equation of the hyperbola is $ x^{2} - \\frac{y^{2}}{3} = 1 $, and its asymptotes are $ y = \\pm\\sqrt{3}x $."
  },
  {
    "text": "If the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ passes through the focus of the parabola $y^{2}=8 x$, and has the same foci as the hyperbola $x^{2}-y^{2}=1$, then the equation of this ellipse is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;C:Ellipse;b: Number;a: Number;Expression(G) = (x^2 - y^2 = 1);Expression(H) = (y^2 = 8*x);Expression(C) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(Focus(H),C);Focus(G)=Focus(C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4 + y^2/2 = 1",
    "fact_spans": "[[[67, 85]], [[47, 61]], [[1, 46], [94, 96]], [[3, 46]], [[3, 46]], [[67, 85]], [[47, 61]], [[1, 46]], [[1, 64]], [[1, 91]]]",
    "query_spans": "[[[94, 101]]]",
    "process": "Since the ellipse passes through the focus of the parabola at (2,0), and has foci at F_{1}(-\\sqrt{2},0), F_{2}(\\sqrt{2},0), it follows that a=2, c=\\sqrt{2}, \\therefore b^{2}=a^{2}-c^{2}=2, \\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1."
  },
  {
    "text": "Given an ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{4}=1$ with one focus at $(2,0)$, what is the eccentricity of $C$?",
    "fact_expressions": "C: Ellipse;a: Number;G: Point;Expression(C) = (y^2/4 + x^2/a^2 = 1);Coordinate(G) = (2, 0);OneOf(Focus(C)) = G",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 48], [63, 66]], [[9, 48]], [[54, 61]], [[2, 48]], [[54, 61]], [[2, 61]]]",
    "query_spans": "[[[63, 72]]]",
    "process": "By the given condition, one focus of the ellipse $ C: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{4} = 1 $ is $ (2,0) $, so $ c = 2 $. From $ c^{2} = a^{2} - b^{2} $, we obtain $ a^{2} - 4 = 4 $, solving gives $ a = 2\\sqrt{2} $. Therefore, the eccentricity of ellipse $ C $ is $ e = \\frac{c}{a} = \\frac{2}{2\\sqrt{2}} = \\frac{\\sqrt{2}}{2} $."
  },
  {
    "text": "If the point $P(m, 2 \\sqrt{3})$ lies on the parabola $y^{2}=4 x$ with focus $F$, then $|P F|$=?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(P) = (m, 2*sqrt(3));PointOnCurve(P, G);Focus(G) = F;m:Number",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "4",
    "fact_spans": "[[[29, 43]], [[1, 20]], [[22, 25]], [[29, 43]], [[1, 20]], [[1, 44]], [[21, 43]], [[2, 20]]]",
    "query_spans": "[[[46, 55]]]",
    "process": "Analysis: According to the given conditions, first find the coordinates of point P, then use the definition of the parabola to solve for |PF|. Since point P(m, 2\\sqrt{3}) lies on the parabola y^{2}=4x, \\therefore (2\\sqrt{3})^{2}=4m, solving gives m=3, \\therefore the coordinates of point P are (3, 2\\sqrt{3}). The directrix equation of the parabola is x=-1, \\therefore |PF|=3+1=4"
  },
  {
    "text": "Given the ellipse $x^{2}+\\frac{y^{2}}{b^{2}}=1$ $(0<b<1)$ with left and right foci $F_{1}$, $F_{2}$, a parabola is drawn with vertex $O$ and focus $F_{2}$, intersecting the ellipse at point $P$, and $\\angle P F_{1} F_{2}=45^{\\circ}$. Then the equation of the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;H: Ellipse;b: Number;P: Point;F1: Point;F2: Point;O:Origin;0 < b;b < 1;Expression(H) = (x^2 + y^2/b^2 = 1);LeftFocus(H) = F1;RightFocus(H)=F2;Vertex(G) = O;Focus(G)=F2;Intersection(G, H) = P;AngleOf(P, F1, F2) = ApplyUnit(45, degree)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x = 1-\\sqrt{2}",
    "fact_spans": "[[[85, 88], [132, 135]], [[2, 40], [89, 91]], [[4, 40]], [[92, 95]], [[47, 54]], [[57, 64], [74, 81]], [[67, 70]], [[4, 40]], [[4, 40]], [[2, 40]], [[2, 64]], [[2, 64]], [[66, 88]], [[74, 88]], [[85, 95]], [[97, 130]]]",
    "query_spans": "[[[132, 142]]]",
    "process": "Let $ F_{1}(-c,0), F_{2}(c,0) $, then the parabola $ y^{2} = 4cx $, the line $ PF_{1}: y = x + c $. Solving the system of equations \n\\[\n\\begin{cases}\ny^2 = 4cx \\\\\ny = x + c\n\\end{cases}\n\\]\nyields $ x = c, y = 2c $. Therefore, the coordinates of point $ P $ are $ (c, 2c) $, so $ PF_{2} \\perp F_{1}F_{2} $. Also, $ PF_{2} = F_{2}F_{1} = 2c $, so $ PF_{1} = 2\\sqrt{2}c $. Hence, $ PF_{1} = 2\\sqrt{2}c $, and $ PF_{1} + PF_{2} = (2 + 2\\sqrt{2})c = 2a = 2 $, then $ c = \\sqrt{2} - 1 $. Therefore, the directrix equation of the parabola is: $ x = -c = 1 - \\sqrt{2} $."
  },
  {
    "text": "Given that $P$ is a point on the hyperbola $\\frac{x^{2}}{64}-\\frac{y^{2}}{36}=1$, and $F_{1}$, $F_{2}$ are the two foci of the hyperbola. If $|PF_{1}|=17$, then the value of $|PF_{2}|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/64 - y^2/36 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};Abs(LineSegmentOf(P, F1)) = 17",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "33",
    "fact_spans": "[[[6, 46], [67, 70]], [[6, 46]], [[2, 5]], [[2, 49]], [[50, 57]], [[59, 66]], [[50, 75]], [[77, 90]]]",
    "query_spans": "[[[92, 106]]]",
    "process": "By the definition of hyperbola, ||PF₁| - |PF₂|| = 2√64 = 16, so |PF₂| = 16 + |PF₁| = 16 + 17 = 33 or |PF₂| = 1 (discarded),"
  },
  {
    "text": "Given $0<\\alpha<\\frac{\\pi}{2}$, the equation $x^{2} \\sin \\alpha+y^{2} \\cos \\alpha=1$ represents an ellipse with foci on the $y$-axis. What is the range of $\\alpha$?",
    "fact_expressions": "G: Ellipse;alpha:Number;0<alpha;alpha<pi/2;Expression(G) = (x^2*Sin(alpha) + y^2*Cos(alpha) = 1);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(alpha)",
    "answer_expressions": "(\\pi/4,\\pi/2)",
    "fact_spans": "[[[79, 81]], [[83, 91]], [[2, 26]], [[2, 26]], [[27, 81]], [[70, 81]]]",
    "query_spans": "[[[83, 97]]]",
    "process": ""
  },
  {
    "text": "The distance from the focus of the parabola $y^{2}=8 x$ to the asymptotes of the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1$ is?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 8*x);G: Hyperbola;Expression(G) = (x^2/2 - y^2/2 = 1)",
    "query_expressions": "Distance(Focus(H), Asymptote(G))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[18, 56]], [[18, 56]]]",
    "query_spans": "[[[0, 65]]]",
    "process": "Analysis: From the given conditions, the focus of the parabola is found to be (2,0), and the asymptotes of the hyperbola are found to be x\\pmy=0. Using the point-to-line distance formula, the required distance can be obtained. Specifically, from the parabola y^{2}=8x, its focus is (2,0). Also, the asymptotes of the hyperbola \\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1 are given by x\\pmy=0. Therefore, the required distance is d=\\frac{2}{\\sqrt{2}}=\\sqrt{2}"
  },
  {
    "text": "Given that the line $l$ with slope $k$ $(k>0)$ passes through the focus $F$ of the parabola $y^{2}=2 p x$ $(p>0)$ and intersects the parabola at points $A$ and $B$, and $|A F|=\\frac{3}{2}|B F|$, then the slope of line $l$ is?",
    "fact_expressions": "l: Line;G: Parabola;p: Number;A: Point;F: Point;B: Point;p>0;k:Number;k>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;Slope(l)=k;PointOnCurve(F, l);Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, F)) = (3/2)*Abs(LineSegmentOf(B, F))",
    "query_expressions": "Slope(l)",
    "answer_expressions": "2*sqrt(6)",
    "fact_spans": "[[[42, 47], [90, 95]], [[3, 24], [48, 51]], [[6, 24]], [[53, 56]], [[26, 29]], [[57, 60]], [[6, 24]], [[33, 41]], [[33, 41]], [[3, 24]], [[3, 29]], [[30, 47]], [[2, 47]], [[42, 62]], [[64, 88]]]",
    "query_spans": "[[[90, 100]]]",
    "process": "Let the inclination angle of the line be $\\theta$. As shown in the figure, since $|AM|=|AF|$, $\\cos\\theta=\\frac{|FQ|}{|AF|}=\\frac{|AF|-p}{|AF|}$, $\\therefore|AF|=\\frac{p}{1-\\cos\\theta}$. Similarly, $|BF|=\\frac{1}{1}\\frac{p}{+\\cos\\theta}$. $\\therefore\\frac{|AF|}{|BF|}=\\frac{1+\\cos\\theta}{1-\\cos\\theta}=\\frac{3}{2}$. Solving gives $\\cos\\theta=\\frac{1}{5}$, so $k=\\tan\\theta=2\\sqrt{6}$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$ with foci $F_{1}$ and $F_{2}$, and a point $P(x_{0}, y_{0})$ satisfying $0<\\frac{x_{0}^{2}}{2}+y_{0}^{2}<1$, then the range of $|P F_{1}|+|P F_{2}|$ is? The number of common points between the line $\\frac{x_{0} x}{2}+y_{0} y=1$ and the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;H: Line;x0: Number;y0: Number;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/2 + y^2 = 1);Expression(H) = (y*y0 + (x*x0)/2 = 1);Coordinate(P) = (x0, y0);Focus(C) = {F1, F2};0 < x0^2/2 + y0^2",
    "query_expressions": "Range(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)));NumIntersection(H, C)",
    "answer_expressions": "[2, 2*sqrt(2))\n0",
    "fact_spans": "[[[2, 34], [174, 179]], [[142, 173]], [[58, 76]], [[58, 76]], [[57, 76]], [[39, 46]], [[48, 56]], [[2, 34]], [[142, 173]], [[57, 76]], [[2, 56]], [[78, 112]]]",
    "query_spans": "[[[114, 142]], [[142, 187]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ with right focus $F$, point $P$ lies on ellipse $C$, and point $A(5,8)$. Then the minimum value of $|P A|-|P F|$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);F: Point;RightFocus(C) = F;P: Point;PointOnCurve(P, C);A: Point;Coordinate(A) = (5, 8)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) - Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 44], [57, 62]], [[2, 44]], [[48, 51]], [[2, 51]], [[52, 56]], [[52, 63]], [[65, 74]], [[65, 74]]]",
    "query_spans": "[[[76, 95]]]",
    "process": "From the definition of the ellipse, |PF| = 4 - |PF₁|, thus transforming |PA| - |PF| into |PA| + |PF₁| - 4. It is analyzed that when points P, A, and F₁ are collinear, |PA| + |PF₁| reaches its minimum value, allowing the result to be obtained. From the definition of the ellipse: |PF₁| + |PF| = 4, so |PF| = 4 - |PF₁|. Therefore, |PA| - |PF| = |PA| - (4 - |PF₁|) = |PA| + |PF₁| - 4. The minimum value of |PA| + |PF₁| occurs when P, A, and F₁ are collinear; hence, |PA| - |PF| ≥ |AF₁| - 4. Given F₁(-1, 0), it follows that |AF₁| = √((5 + 1)² + 8²) = 10. Thus, |PA| - |PF| ≥ 6, and therefore the minimum value of |PA| - |PF| is 6."
  },
  {
    "text": "Given that $F_{1}(0, -2)$ and $F_{2}(0, 2)$ are the two foci of an ellipse, and a chord $AB$ of the ellipse passes through $F_{2}$. If the perimeter of $\\triangle A F_{1} B$ is $16$, then what is the standard equation of the ellipse?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;F1: Point;F2: Point;Coordinate(F1) = (0, -2);Coordinate(F2) = (0, 2);Focus(G) = {F1, F2};PointOnCurve(F2, LineSegmentOf(A, B));IsChordOf(LineSegmentOf(A, B), G);Perimeter(TriangleOf(A, F1, B)) = 16",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/16 + x^2/12 = 1",
    "fact_spans": "[[[32, 34], [92, 94], [49, 51]], [[53, 58]], [[53, 58]], [[2, 16]], [[17, 31], [41, 48]], [[2, 16]], [[17, 31]], [[2, 39]], [[40, 58]], [[49, 58]], [[60, 89]]]",
    "query_spans": "[[[92, 101]]]",
    "process": ""
  },
  {
    "text": "Given point $A(4,0)$ and the focus $F$ of the parabola $y^{2}=4x$, if a point $P$ on the parabola satisfies $|PA|=2|PF|$, then the horizontal coordinate of $P$ is?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (4, 0);Focus(G)=F;PointOnCurve(P,G);Abs(LineSegmentOf(P,A))=2*Abs(LineSegmentOf(P,F))",
    "query_expressions": "XCoordinate(P)",
    "answer_expressions": "2*sqrt(2)-2",
    "fact_spans": "[[[12, 26], [34, 37]], [[2, 11]], [[39, 43], [61, 64]], [[29, 32]], [[12, 26]], [[2, 11]], [[12, 32]], [[34, 43]], [[45, 59]]]",
    "query_spans": "[[[61, 70]]]",
    "process": "The focus of the parabola is F(1,0). Let P(\\frac{a^{2}}{4},a). Using the distance formula between two points, we get \\sqrt{(\\frac{a^{2}}{4}-4)^{2}+a^{2}}=2\\sqrt{(\\frac{a^{2}}{4}-1)^{2}}+, yielding \\frac{a^{2}}{4}=2\\sqrt{2}-2."
  },
  {
    "text": "The distance between two fixed points $A$ and $B$ on a plane is $4$. A moving point $P$ satisfies $PA - PB = 2$. What is the minimum value of the distance from point $P$ to the midpoint of $AB$?",
    "fact_expressions": "A: Point;B: Point;Distance(A, B) = 4;P: Point;LineSegmentOf(P, A) - LineSegmentOf(P, B) = 2",
    "query_expressions": "Min(Distance(P, MidPoint(LineSegmentOf(A, B))))",
    "answer_expressions": "1",
    "fact_spans": "[[[6, 9]], [[10, 13]], [[6, 21]], [[25, 28], [44, 48]], [[30, 41]]]",
    "query_spans": "[[[44, 65]]]",
    "process": "\\because the distance between two fixed points A and B on a plane is 4, and a moving point P satisfies PA - PB = 2 (2 < 4), \\therefore the trajectory of point P is the right branch of a hyperbola with foci at A and B, and 2a = 2, a = 1; hence, the minimum distance from point P to the midpoint of AB (i.e., the origin) is a."
  },
  {
    "text": "A point $M(x_{0}, y_{0})$ on the parabola $y^{2}=8x$ is at a distance of $6$ from its focus. Then, what is the distance from point $M$ to the origin $O$?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = 8*x);Coordinate(M) = (x0, y0);PointOnCurve(M, G);Distance(M, Focus(G)) = 6;O:Origin;x0:Number;y0:Number",
    "query_expressions": "Distance(M, O)",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[0, 14], [36, 37]], [[17, 35], [48, 52]], [[0, 14]], [[17, 35]], [[0, 35]], [[17, 46]], [[53, 60]], [[17, 35]], [[17, 35]]]",
    "query_spans": "[[[48, 65]]]",
    "process": "According to the problem, the directrix equation of the parabola $ y^{2} = 8x $ is $ x = -2 $. If a point $ M(x_{0}, y_{0}) $ on the parabola $ y^{2} = 8x $ has a distance of 6 to its focus, then its distance to the directrix is also 6. Thus, $ x_{0} - (-2) = 6 $, solving gives: $ x_{0} = 4 $. Since $ M $ lies on the parabola, $ y^{2} = 8x_{0} = 32 $. Then the distance from $ M $ to the origin $ O $ is $ d = \\sqrt{x_{0}^{2} + y_{0}^{2}} = \\sqrt{48} = 4\\sqrt{3} $."
  },
  {
    "text": "Draw a perpendicular line from a focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ to one of its asymptotes. If the foot of the perpendicular lies exactly on the perpendicular bisector of the segment $OF$ ($O$ being the origin), then the eccentricity of the hyperbola is?",
    "fact_expressions": "O: Origin;F: Point;G: Hyperbola;b: Number;a: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);OneOf(Focus(G))=F;L:Line;PointOnCurve(F,L);IsPerpendicular(L,OneOf(Asymptote(G)));PointOnCurve(FootPoint(L,OneOf(Asymptote(G))),PerpendicularBisector(LineSegmentOf(O,F)))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[77, 80]], [[52, 55]], [[1, 47], [93, 96]], [[4, 47]], [[4, 47]], [[1, 47]], [[1, 55]], [], [[0, 64]], [[0, 64]], [[0, 91]]]",
    "query_spans": "[[[93, 102]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is an arbitrary point on the parabola $y^{2}=4x$ distinct from the vertex, draw a perpendicular from point $P$ to the $y$-axis, with foot of perpendicular at point $Q$. Let point $M(7,6)$. When the sum of the lengths of segments $PM$ and $PQ$ is minimized, what are the coordinates of point $P$?",
    "fact_expressions": "G: Parabola;M: Point;P: Point;Q: Point;Expression(G) = (y^2 = 4*x);Coordinate(M) = (7, 6);PointOnCurve(P, G);Negation(P=Vertex(G));L:Line;PointOnCurve(P,L);IsPerpendicular(L,yAxis);FootPoint(L,yAxis)=Q;WhenMin(Length(LineSegmentOf(P,M))+Length(LineSegmentOf(P,Q)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(3+2*sqrt(2),2+2*sqrt(2))",
    "fact_spans": "[[[6, 20]], [[54, 63]], [[2, 5], [33, 37], [90, 94]], [[49, 53]], [[6, 20]], [[54, 63]], [[2, 31]], [[2, 31]], [], [[32, 45]], [[32, 45]], [[32, 53]], [[65, 89]]]",
    "query_spans": "[[[90, 99]]]",
    "process": "The focus $ F $ of the parabola $ y^{2} = 4x $ has coordinates $ (1,0) $. According to the given condition, $ |PQ| = |PF| - 1 $, so $ |PM| + |PQ| = |PM| + |PF| - 1 $, hence $ (|PM| + |PQ|)_{\\min} = (|PM| + |PF|)_{\\min} - 1 $. The equation of line $ MF $ is $ y = x - 1 $. Solving \n\\[\n\\begin{cases}\ny = x - 1, \\\\\ny = 4x,\n\\end{cases}\n\\]\nwe get \n\\[\n\\begin{cases}\nx = 3 - 2\\sqrt{2} \\\\\ny = 2 - 2\\sqrt{2}\n\\end{cases}\n\\quad \\text{or} \\quad\n\\begin{cases}\nx = 3 + 2\\sqrt{2} \\\\\ny = 2 + 2\\sqrt{2}\n\\end{cases}.\n\\]\nFrom graphical analysis, the $ x $-coordinate of point $ P $ is greater than $ 1 $, so the required coordinates of point $ P $ are $ (3 + 2\\sqrt{2}, 2 + 2\\sqrt{2}) $."
  },
  {
    "text": "The coordinates of the focus of the parabola $2 y^{2}+x=0$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (x + 2*y^2 = 0)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(-1/8,0)",
    "fact_spans": "[[[0, 16]], [[0, 16]]]",
    "query_spans": "[[[0, 23]]]",
    "process": "Since $ y^{2} = -\\frac{1}{2}x $, the coordinates of the focus are $ (-\\frac{1}{8}, 0) $."
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{m-1}+\\frac{y^{2}}{2-m}=1$ represents an ellipse with foci on the $y$-axis, what is the range of values for $m$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(m - 1) + y^2/(2 - m) = 1);PointOnCurve(Focus(G), yAxis);m: Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(1,3/2)",
    "fact_spans": "[[[54, 56]], [[2, 56]], [[45, 56]], [[58, 61]]]",
    "query_spans": "[[[58, 68]]]",
    "process": "From the given condition, we have $2 - m > m - 1 > 0$, therefore $1 < m < \\frac{3}{2}$. Thus, the range of values for $m$ is $(1, \\frac{3}{2})$."
  },
  {
    "text": "It is known that the foci of the ellipse lie on the $x$-axis, one vertex is at $A(0,-1)$, and the distance from its right focus to the line $x - y + 2\\sqrt{2} = 0$ is $3$. Then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;Expression(H) = (x - y + 2*sqrt(2) = 0);Coordinate(A) = (0, -1);OneOf(Vertex(G))=A;PointOnCurve(Focus(G),xAxis);Distance(RightFocus(G),H)=3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/3+y^2=1",
    "fact_spans": "[[[2, 4], [29, 30], [63, 65]], [[34, 54]], [[19, 28]], [[34, 54]], [[19, 28]], [[2, 28]], [[2, 13]], [[29, 61]]]",
    "query_spans": "[[[63, 70]]]",
    "process": "According to the problem, the ellipse equation is in standard form, $ b=1 $, the right focus is at $ (c,0) $, and its distance to the given line is $ \\frac{|c+2\\sqrt{2}|}{\\sqrt{2}}=3 $, so $ c=\\sqrt{2} $, thus $ a^{2}=b^{2}+c^{2}=3 $, and the equation of the ellipse is $ \\frac{x^{2}}{3}+y^{2}=1 $."
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$, the left focus is $F_{1}$, the vertex $Q(0,2 \\sqrt{3})$, and $P$ is a moving point on the right branch of the hyperbola $C$. Then the minimum value of $|P F_{1}|+|P Q|$ is equal to?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2/3 = 1);F1: Point;LeftFocus(C) = F1;Q: Point;Vertex(C) = Q;Coordinate(Q) = (0, 2*sqrt(3));P: Point;PointOnCurve(P, RightPart(C))",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 35], [74, 80]], [[2, 35]], [[40, 47]], [[2, 47]], [[50, 67]], [[2, 67]], [[50, 67]], [[70, 73]], [[70, 86]]]",
    "query_spans": "[[[88, 112]]]",
    "process": "Combining the given conditions, draw the graph: according to the properties of the hyperbola, we know |PF_{1}|-|PF_{2}|=2a=2, thus |PF_{1}|=|PF_{2}|+2, so |PF_{1}|+|PQ|=|PF_{2}|+|PQ|+2\\geqslant|QF_{2}|+2, and O(0,2\\sqrt{3}), F_{2}(2,0), therefore |OF_{2}|=\\sqrt{2^{2}+(2\\sqrt{3})^{2}}=4, hence the minimum value is 6."
  },
  {
    "text": "The distance from point $A$ on the parabola $y^{2}=2 p x(p>0)$ to the focus $F$ is $2$. A circle with diameter $A F$ intersects the $y$-axis at point $H(0,1)$. Then $p=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;A: Point;PointOnCurve(A, G) = True;F: Point;Focus(G) = F;Distance(A, F) = 2;IsDiameter(LineSegmentOf(A, F), V) = True;V: Circle;Intersection(V, yAxis) = H;H: Point;Coordinate(H) = (0, 1)",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 21]], [[0, 21]], [[68, 71]], [[3, 21]], [[22, 26]], [[0, 26]], [[29, 32]], [[0, 32]], [[22, 38]], [[39, 50]], [[49, 50]], [[49, 66]], [[57, 66]], [[57, 66]]]",
    "query_spans": "[[[68, 73]]]",
    "process": "Method 1: Given $ F\\left(\\frac{p}{2}, 0\\right) $, and the distance from point $ A $ to focus $ F $ is 2, so the x-coordinate of point $ A $ is $ 2 - \\frac{p}{2} $. Thus, the y-coordinate of point $ A $ satisfies $ y^2 = 2p\\left(2 - \\frac{p}{2}\\right) = 4p - p^{20} $; that is, $ \\overrightarrow{HF} = \\left(\\frac{p}{2}, -1\\right) $, $ \\overrightarrow{HA} = \\left(2 - \\frac{p}{2}, y - 1\\right) $. From $ \\overrightarrow{HF} \\cdot \\overrightarrow{HA} = 0 $, we obtain $ \\frac{4p - p^2}{4} - y + 1 = 0 $. Then, solving together with $ \\textcircled{1} $, we get $ y = 2 $, and substituting back into $ \\textcircled{1} $ gives $ p = 2 $. Method 2: Let $ A(x_0, y_0) $, $ F\\left(\\frac{p}{2}, 0\\right) $. By the focal radius formula, $ x_0 + \\frac{p}{2} = 2 $. Then, the distance from the midpoint of segment $ AF $ to the y-axis is $ d = \\frac{x_0 + \\frac{p}{2}}{2} = \\frac{2}{2} = 1 $. Therefore, the circle with diameter $ |AF| $ is tangent to the y-axis. According to the problem, the point of tangency is $ H(0, 1) $, so the y-coordinate of point $ A $ is 2, and the x-coordinate is $ 2 - \\frac{p}{2} $. Then, $ 2p\\left(2 - \\frac{p}{2}\\right) = 4 $, solving gives: $ p = 2 $."
  },
  {
    "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $\\frac{\\sqrt{3}}{2}$, then what is the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$?",
    "fact_expressions": "H: Ellipse;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Eccentricity(H) = sqrt(3)/2;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[1, 53]], [[1, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[1, 78]], [[80, 126]], [[80, 126]]]",
    "query_spans": "[[[80, 132]]]",
    "process": "From the given condition, we have $\\frac{\\sqrt{a^{2}-b^{2}}}{a}=\\frac{\\sqrt{3}}{2}$, then $a=2b$, and the eccentricity of the hyperbola is $\\frac{\\sqrt{a^{2}+b^{2}}}{a}=\\frac{\\sqrt{5}b}{2b}=\\frac{\\sqrt{5}}{2}$."
  },
  {
    "text": "The midpoint of the chord $AB$ of the ellipse $x^{2}+2 y^{2}=4$ is $(-1,-1)$. What is the length of the chord $AB$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 2*y^2 = 4);IsChordOf(LineSegmentOf(A, B), G);Coordinate(MidPoint(LineSegmentOf(A, B))) = (-1, -1);A:Point;B:Point",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "sqrt(30)/3",
    "fact_spans": "[[[0, 19]], [[0, 19]], [[0, 26]], [[21, 39]], [[0, 26]], [[0, 26]]]",
    "query_spans": "[[[42, 51]]]",
    "process": "Let the midpoint of A(x_{1},y_{1}), B(x_{2},y_{2}) be (-1,-1), so x_{1}+x_{2}=-2, y_{1}+y_{2}=-2, x_{1}^{2}+2y_{1}^{2}=4, x_{2}^{2}+2y_{2}^{2}=4. Subtracting the two equations: x_{1}^{2}-x_{2}^{2}+2y_{1}^{2}-2y_{2}^{2}=0, (x_{1}+x_{2})(x_{1}-x_{2})+2(y_{1}+y_{2})(y_{1}-y_{2})=0, (x_{1}-x_{2})+2(y_{1}-y_{2})=0, thus: k_{AB}=-\\frac{1}{2}. The equation of the line containing chord AB is y=-\\frac{1}{2}x-\\frac{3}{2}. Solving simultaneously with x^{2}+2y^{2}=4 and simplifying yields: 3x^{2}+6x+1=0, x_{1}+x_{2}=-2, x_{1}x_{2}=\\frac{1}{3}. Therefore |AB|=\\sqrt{1+\\frac{1}{4}}\\times\\sqrt{4-\\frac{4}{3}}=\\frac{\\sqrt{30}}{3}"
  },
  {
    "text": "If the foci of the ellipse $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$ lie on the $y$-axis, and tangents are drawn from the point $(1, \\frac{1}{2})$ to the circle $x^{2}+y^{2}=1$, with points of tangency $A$ and $B$, respectively, such that the line $AB$ passes exactly through the upper focus and the right vertex of the ellipse, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Circle;B: Point;A: Point;P:Point;L1:Line;L2:Line;Expression(G) = (x^2/b^2 + y^2/a^2 = 1);Expression(H) = (x^2 + y^2 = 1);Coordinate(P) = (1, 1/2);PointOnCurve(Focus(G), yAxis);TangentOfPoint(P,H)={L1,L2};TangentPoint(L1,H)=A;TangentPoint(L2,H)=B;PointOnCurve(UpperFocus(G),LineOf(A,B));PointOnCurve(RightVertex(G),LineOf(A,B))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 + y^2/5 = 1",
    "fact_spans": "[[[1, 46], [121, 123], [121, 123]], [[3, 46]], [[3, 46]], [[77, 93]], [[106, 109]], [[102, 105]], [[57, 76]], [], [], [[1, 46]], [[77, 93]], [[57, 76]], [[1, 55]], [[56, 96]], [[56, 109]], [[56, 109]], [[110, 127]], [[110, 131]]]",
    "query_spans": "[[[133, 140]]]",
    "process": "Let the tangent lines to the circle $x^{2}+y^{2}=1$ passing through the point $(1,\\frac{1}{2})$ be denoted by $l$. By case analysis, find the points of tangency $A$ and $B$ where line $l$ touches the circle, obtain the equation of line $AB$, thereby finding the intersection points of line $AB$ with the $x$-axis and $y$-axis, yielding the right focus and upper vertex of the ellipse, and then determine the equation of the ellipse. Detailed solution: Let the tangent lines to the circle $x^{2}+y^{2}=1$ passing through the point $(1,\\frac{1}{2})$ be $y-\\frac{1}{2}=k(x-1)$, i.e., $kx-y-k+\\frac{1}{2}=0$. When line $l$ is perpendicular to the $x$-axis, $k$ does not exist, and the equation of the line is $x=1$, which is tangent to the circle $x^{2}+y^{2}=1$ at point $A(1,0)$. When line $l$ is not perpendicular to the $x$-axis, the distance from the origin to line $l$ is $d=\\frac{|-k+\\frac{1}{2}|}{\\sqrt{1+k^{2}}}=1$, solving gives $k=-\\frac{3}{4}$. At this time, the equation of line $l$ is $y=-\\frac{3}{4}x+\\frac{5}{4}$, and this line is tangent to the circle $x^{2}+y^{2}=1$ at point $B(\\frac{3}{5},\\frac{4}{5})$. Therefore, the slope of line $AB$ is $k_{1}=\\frac{0-\\frac{4}{5}}{1-\\frac{3}{5}}=-2$, and the equation of line $AB$ is $y=-2(x-1)$. Thus, line $AB$ intersects the $x$-axis at point $A(1,0)$ and the $y$-axis at point $C(0,2)$. For the ellipse $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$, the right focus is $(1,0)$ and the upper vertex is $(0,2)$, so $c=1$, $b=2$, giving $a^{2}=b^{2}+c^{2}=5$. Hence, the standard equation of the ellipse is $\\frac{y^{2}}{5}+x^{2}=1$."
  },
  {
    "text": "What are the coordinates of the focus of the parabola $y=a x^{2}(a<0)$?",
    "fact_expressions": "G: Parabola;a: Number;a<0;Expression(G) = (y = a*x^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,1/(4*a))",
    "fact_spans": "[[[0, 19]], [[3, 19]], [[3, 19]], [[0, 19]]]",
    "query_spans": "[[[0, 26]]]",
    "process": "The standard equation of the parabola is $x^{2}=\\frac{1}{a}y$, therefore, the coordinates of the focus of this parabola are $(0,\\frac{1}{4a})$."
  },
  {
    "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ is $\\frac{\\sqrt{3}}{2}$, then what is the eccentricity of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$?",
    "fact_expressions": "H: Ellipse;Expression(H) = (x^2/4 + y^2/b^2 = 1);Eccentricity(H) = sqrt(3)/2;b: Number;b>0;G: Hyperbola;Expression(G) = (x^2/4 - y^2/b^2 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2, sqrt(5)",
    "fact_spans": "[[[1, 47]], [[1, 47]], [[1, 72]], [[3, 47]], [[3, 47]], [[74, 116]], [[74, 116]]]",
    "query_spans": "[[[74, 122]]]",
    "process": "From the given condition, \\frac{4-b^{2}}{4}=\\frac{3}{4} or \\frac{b^{2}-4}{b^{2}}=\\frac{3}{4}, we obtain b^{2}=1 or b^{2}=16. When b^{2}=1, the eccentricity of the hyperbola \\frac{x^{2}}{4}-y^{2}=1 is \\frac{\\sqrt{5}}{2}; when b^{2}=16, the eccentricity of the hyperbola \\frac{x^{2}}{4}-\\frac{y^{2}}{16}=1 is \\sqrt{5}."
  },
  {
    "text": "What is the focal distance of the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/2 - y^2 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 33]]]",
    "process": "The focal distance of the hyperbola \\frac{x^{2}}{2}-y^{2}=1 is 2c=2\\sqrt{a^{2+b^{2}}}=2\\sqrt{3}"
  },
  {
    "text": "Given the two foci of a hyperbola $F_1(-\\sqrt{10}, 0)$, $F_2(\\sqrt{10}, 0)$, and $P$ is a point on this hyperbola such that $\\overrightarrow{PF_1} \\cdot \\overrightarrow{PF_2}=0$, $|PF_1| \\cdot |PF_2|=2$, then the equation of this hyperbola is?",
    "fact_expressions": "E: Hyperbola;F1: Point;F2: Point;Coordinate(F1) = (-sqrt(10), 0);Coordinate(F2) = (sqrt(10), 0);P: Point;PointOnCurve(P, E);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Abs(LineSegmentOf(P, F1)) * Abs(LineSegmentOf(P, F2)) = 2;Focus(G)={F1,F2}",
    "query_expressions": "Expression(E)",
    "answer_expressions": "x**2/9-y**2=1",
    "fact_spans": "[[[2, 5], [69, 72], [167, 170]], [[11, 34]], [[37, 60]], [[11, 34]], [[37, 60]], [[63, 66]], [[63, 76]], [[79, 132]], [[134, 163]], [2, 59]]",
    "query_spans": "[[[165, 175]]]",
    "process": ""
  },
  {
    "text": "Let $F$ be the focus of the parabola $y=-\\frac{1}{4} x^{2}$, and let line $l$, tangent to the parabola at point $P(-4,-4)$, intersect the $x$-axis at point $Q$. Then the value of $\\angle PQF$ is?",
    "fact_expressions": "l: Line;G: Parabola;P: Point;Q: Point;F: Point;Expression(G) = (y = -x^2/4);Coordinate(P) = (-4, -4);Focus(G) = F;TangentPoint(l,G)=P;Intersection(l,xAxis)=Q",
    "query_expressions": "AngleOf(P, Q, F)",
    "answer_expressions": "pi/2",
    "fact_spans": "[[[54, 59]], [[5, 30], [35, 38]], [[41, 53]], [[68, 71]], [[1, 4]], [[5, 30]], [[41, 53]], [[1, 33]], [[34, 59]], [[54, 71]]]",
    "query_spans": "[[[73, 89]]]",
    "process": ""
  },
  {
    "text": "The equation of an ellipse that has the same foci as the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ and has a major axis length of $4 \\sqrt{5}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1);E: Ellipse;Focus(G) = Focus(E);Length(MajorAxis(E)) = 4*sqrt(5)",
    "query_expressions": "Expression(E)",
    "answer_expressions": "x^2/20 + y^2/15 = 1",
    "fact_spans": "[[[1, 38]], [[1, 38]], [[62, 64]], [[0, 64]], [[45, 64]]]",
    "query_spans": "[[[62, 68]]]",
    "process": "The semi-focal distance of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ is $\\sqrt{9-4}=\\sqrt{5}$, so the required ellipse has $c=\\sqrt{5}$, and the foci lie on the $x$-axis. Since the major axis length of the required ellipse is $2a=4\\sqrt{5}$, then $a=2\\sqrt{5}$, so $b^{2}=20-5=15$. Therefore, the equation of the required ellipse is $\\frac{x^{2}}{20}+\\frac{y^{2}}{15}=1$. Hence, fill in: $\\frac{x^{2}}{20}+\\frac{y^{2}}{15}=1$."
  },
  {
    "text": "Let $m$, $b$ be real numbers. Given that the ellipse $\\frac{x^{2}}{10}+\\frac{y^{2}}{m}=1$ passing through the point $P\\left(\\frac{\\sqrt{10}}{3}, \\frac{8}{3}\\right)$ and the hyperbola $x^{2}-\\frac{y^{2}}{b}=1$ have the same foci, then $b=$?",
    "fact_expressions": "m: Real;b: Real;P: Point;Coordinate(P) = (sqrt(10)/3, 8/3);H: Ellipse;Expression(H) = (x^2/10 + y^2/m = 1);PointOnCurve(P, H);G: Hyperbola;Expression(G) = (x^2 - y^2/b = 1);Focus(H) = Focus(G)",
    "query_expressions": "b",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 4]], [[7, 10], [132, 135]], [[18, 56]], [[18, 56]], [[57, 95]], [[57, 95]], [[16, 95]], [[96, 124]], [[96, 124]], [[57, 130]]]",
    "query_spans": "[[[132, 137]]]",
    "process": "Since point $ P\\left(\\frac{\\sqrt{10}}{3},\\frac{8}{3}\\right) $ lies on the ellipse $ \\frac{x^2}{10} + \\frac{y^2}{m} = 1 $, we have $ \\frac{(\\sqrt{10})^2}{10} + \\frac{(8)^2}{m} = 1 $, solving gives $ m = 8 $. Therefore, the equation of the ellipse is $ \\frac{x^2}{10} + \\frac{y^2}{8} = 1 $. Since the ellipse $ \\frac{x^2}{10} + \\frac{y^2}{8} = 1 $ and the hyperbola $ x^2 - \\frac{y^2}{b} = 1 $ have the same foci, it follows that $ 10 - 8 = 1 + b $, solving gives $ b = 1 $."
  },
  {
    "text": "The coordinates of the foci of the hyperbola $x^{2}-\\frac{y^{2}}{9}=1$ are? The coordinates of the vertices are? The length of the real axis is? The length of the imaginary axis is? The equations of the asymptotes are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/9 = 1)",
    "query_expressions": "Coordinate(Focus(G));Coordinate(Vertex(G));Length(RealAxis(G));Length(ImageinaryAxis(G));Expression(Asymptote(G))",
    "answer_expressions": "(pm*sqrt(10), 0)\n(pm*1, 0)\n2\n6\ny=pm*3*x",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 35]], [[0, 41]], [[0, 46]], [[0, 51]], [[0, 58]]]",
    "process": "It is known that the foci of the hyperbola lie on the x-axis, $a^{2}=1$, $b^{2}=9$, $c^{2}=10$, so the coordinates of the foci are $(\\pm\\sqrt{10},0)$, the vertices are $(\\pm1,0)$, the length of the transverse axis is $2a=2$, the length of the conjugate axis is $2b=6$, and the equations of the asymptotes are $y=\\pm\\frac{b}{a}x=\\pm3x$."
  },
  {
    "text": "If the directrix of the parabola $y^{2}=2 p x$ passes through the left focus of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, then the real number $p$=?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Real;Expression(G) = (x^2 - y^2/3 = 1);Expression(H) = (y^2 = 2*(p*x));PointOnCurve(LeftFocus(G), Directrix(H))",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[22, 50]], [[1, 17]], [[56, 61]], [[22, 50]], [[1, 17]], [[1, 54]]]",
    "query_spans": "[[[56, 63]]]",
    "process": "The right focus of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ is $(-2,0)$, so the directrix of the parabola $y^{2}=2px$ is $x=-2$, $\\therefore -\\frac{p}{2}=-2$, $\\therefore p=4$,"
  },
  {
    "text": "Given the parabola $C$: $x^{2}=2 p y(p>0)$, the focus of which is the center of the circle $x^{2}+(y-1)^{2}=2$. A line passing through the focus of the parabola $C$ and having an inclination angle of $60^{\\circ}$ intersects the parabola $C$ at points $A$ and $B$. Then $|A B|$=?",
    "fact_expressions": "C: Parabola;p: Number;G: Circle;H: Line;A: Point;B: Point;p>0;Expression(C) = (x^2 = 2*p*y);Expression(G) = (x^2 + (y - 1)^2 = 2);Center(G)=Focus(C);PointOnCurve(Focus(C),H);Inclination(H)=ApplyUnit(60,degree);Intersection(H,C)={A,B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16",
    "fact_spans": "[[[2, 28], [59, 65], [89, 95]], [[10, 28]], [[32, 52]], [[86, 88]], [[96, 99]], [[100, 103]], [[10, 28]], [[2, 28]], [[32, 52]], [[2, 55]], [[57, 88]], [[69, 88]], [[86, 105]]]",
    "query_spans": "[[[107, 116]]]",
    "process": "The center of the circle $x^{2}+(y-1)^{2}=2$ is $(0,1)$, so $\\frac{p}{2}=1$, that is, $p=2$, hence the parabola $C: x^{2}=4y$. Since the inclination angle of the line is $60^{\\circ}$, the slope $k=\\sqrt{3}$, thus the equation of line $AB$: $y=\\sqrt{3}x+1$. Solving the system of equations of the line and the parabola $\\begin{cases}x^{2}=4y\\\\y=\\sqrt{3}x+1\\end{cases}$, we obtain: $x^{2}-4\\sqrt{3}x-4=0$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $x_{1}+x_{2}=4\\sqrt{3}$, $x_{1}x_{2}=-4$, then $|AB|=\\sqrt{1+3}\\cdot\\sqrt{(4\\sqrt{3})^{2}-4\\times(-4)}=16$."
  },
  {
    "text": "Given that $AB$ is a chord of the parabola $x^{2}=y$, and if the equation of the perpendicular bisector of this chord is $y=-x+3$, then what is the equation of the line on which the chord $AB$ lies?",
    "fact_expressions": "IsChordOf(LineSegmentOf(A, B), G) = True;G: Parabola;Expression(G) = (x^2 = y);Expression(PerpendicularBisector(LineSegmentOf(A, B))) = (y = -x + 3);A: Point;B: Point",
    "query_expressions": "Expression(OverlappingLine(LineSegmentOf(A, B)))",
    "answer_expressions": "x-y+2=0",
    "fact_spans": "[[[2, 22]], [[8, 20]], [[8, 20]], [[2, 45]], [[2, 7]], [[2, 7]]]",
    "query_spans": "[[[48, 62]]]",
    "process": "Let points A(x_{1},y_{1}) and B(x_{2},y_{2}), and let the midpoint of segment AB be M(x_{0},y_{0}). From \\begin{cases}x_{1}^{2}=y_{1}\\\\x_{2}^{2}=y_{2}\\end{cases}, subtracting the two equations gives x_{1}^{2}-x_{2}^{2}=y_{1}-y_{2}, i.e., (x_{1}-x_{2})(x_{1}+x_{2})=y_{1}-y_{2}. \\therefore \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=x_{1}+x_{2}=2x_{0}'. Since the equation of the perpendicular bisector of segment AB is y=-x+3, the slope of line AB is \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=2x_{0}=1, \\therefore x_{0}=\\frac{1}{2}, then y_{0}=-x_{0}+3=\\frac{5}{2}, so the coordinates of point M are (\\frac{1}{2},\\frac{5}{2}). Also, since the slope of line AB is 1, the equation of the chord AB is y-\\frac{5}{2}=x-\\frac{1}{2}, i.e., x-y+2=0."
  },
  {
    "text": "Given that point $P$ is a moving point on the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1  (b>0)$, and the minimum value of $|OP|$ is $1$, where $O$ is the origin, then $b$=?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/b^2 = 1);b: Number;b>0;P: Point;PointOnCurve(P, C);O: Origin;Min(Abs(LineSegmentOf(O, P))) = 1",
    "query_expressions": "b",
    "answer_expressions": "1",
    "fact_spans": "[[[7, 60]], [[7, 60]], [[95, 98]], [[14, 60]], [[2, 6]], [[2, 64]], [[84, 87]], [[66, 81]]]",
    "query_spans": "[[[95, 100]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, the focal distance is $2c$, and the line $y=\\frac{\\sqrt{3}}{3}(x+c)$ intersects the hyperbola at a point $P$ such that $\\angle P F_{2} F_{1}=2 \\angle P F_{1} F_{2}$. Find the eccentricity $e$ of the hyperbola $C$.",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;c: Number;P: Point;F2: Point;F1: Point;e: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y = (sqrt(3)/3)*(c + x));LeftFocus(C) = F1;RightFocus(C) = F2;FocalLength(C) = 2*c;AngleOf(P, F2, F1) = 2*AngleOf(P, F1, F2);OneOf(Intersection(G, C)) = P;Eccentricity(C) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3) + 1",
    "fact_spans": "[[[2, 63], [127, 130], [187, 193]], [[10, 63]], [[10, 63]], [[97, 126]], [[91, 96]], [[135, 138]], [[80, 87]], [[72, 79]], [[197, 200]], [[10, 63]], [[10, 63]], [[2, 63]], [[97, 126]], [[2, 87]], [[2, 87]], [[2, 96]], [[140, 185]], [[97, 138]], [[187, 200]]]",
    "query_spans": "[[[197, 202]]]",
    "process": "It is easy to see that $\\angle PF_{1}F_{2}$ is the inclination angle of the line. Then, according to $\\angle PF_{2}F_{1} = 2\\angle PF_{1}F_{2}$, we obtain $PF_{2} = c$, $PF_{1} = \\sqrt{3}c$. Then use the definition of hyperbola to solve. Since the line $y = \\frac{\\sqrt{3}}{3}(x + c)$ intersects the hyperbola at a point $P$, so $\\angle PF_{1}F_{2} = 30^{\\circ}$. Because $\\angle PF_{2}F_{1} = 2\\angle PF_{1}F_{2}$, so $\\angle PF_{2}F_{1} = 60^{\\circ}$, $\\angle F_{1}PF_{2} = 90^{\\circ}$. Therefore, $PF_{2} = c$, $PF_{1} = \\sqrt{3}c$. By the definition of hyperbola, $PF_{1} - PF_{2} = 2a$, that is, $\\sqrt{3}c - c = 2a$, solving gives $e = \\sqrt{3} + 1$."
  },
  {
    "text": "The equation of the hyperbola with focus $(0,6)$ and having the same asymptotes as the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$ is?",
    "fact_expressions": "G: Hyperbola;H: Point;C:Hyperbola;Expression(G) = (x^2/2 - y^2 = 1);Coordinate(H) = (0, 6);Focus(C)=H;Asymptote(C)=Asymptote(G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/12 - x^2/24 = 1",
    "fact_spans": "[[[13, 41]], [[3, 10]], [[49, 52]], [[13, 41]], [[3, 10]], [[0, 52]], [[12, 52]]]",
    "query_spans": "[[[49, 56]]]",
    "process": "From \\frac{x^{2}}{2}-y^{2}=1, the asymptotes of the hyperbola are y=\\pm\\frac{\\sqrt{2}}{2}. Let the equation of the hyperbola be: \\frac{x^{2}}{2}-y^{2}=\\lambda (\\lambda<0). \\therefore \\frac{x^{2}}{22}-\\frac{y^{2}}{2}=1 \\therefore -\\lambda-2\\lambda=36, \\therefore \\lambda=-12. Thus, the equation of the hyperbola is \\frac{y^{2}}{12}-\\frac{x^{2}}{24}=1. Answer: \\frac{y^{2}}{12}-\\frac{x^{2}}{24}=1."
  },
  {
    "text": "It is known that a line $l$ parallel to the $x$-axis intersects the parabola $x^{2}=4y$ at points $A$ and $B$, and $|AB|=8$. Then the equation of $l$ is?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;Expression(G) = (x^2 = 4*y);IsParallel(xAxis, l);Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, B)) = 8",
    "query_expressions": "Expression(l)",
    "answer_expressions": "y=4",
    "fact_spans": "[[[10, 15], [53, 56]], [[16, 30]], [[31, 34]], [[35, 38]], [[16, 30]], [[2, 15]], [[10, 40]], [[42, 51]]]",
    "query_spans": "[[[53, 61]]]",
    "process": "First draw the graph. From |AB| = 8, the x-coordinate of point B can be found. Substituting into the parabola equation gives $ y_{B} $, thus allowing to solve for the equation of line $ l $. As shown in the figure, $ |AB| = 8 \\Rightarrow x_{B} = 4 $, substituting $ x_{B} = 4 $ into $ x^{2} = 4y $ yields $ y_{B} = 4 $, then the equation of line $ l $ is $ y = 4 $."
  },
  {
    "text": "Given the equation of the ellipse is $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{25}=1$ $(a>5)$, its two foci are $F_{1}$ and $F_{2}$, and $|F_{1} F_{2}|=8$. The chord $AB$ (a line segment between any two points on the ellipse) passes through point $F_{1}$. Then, what is the perimeter of $\\triangle ABF_{2}$?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;F1: Point;F2: Point;a:Number;a>5;Expression(G)=(x^2/a^2+y^2/25=1);Focus(G)={F1,F2};PointOnCurve(A,G);PointOnCurve(B,G);PointOnCurve(F1,LineSegmentOf(A,B));Abs(LineSegmentOf(F1, F2)) = 8;IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "Perimeter(TriangleOf(A,B,F2))",
    "answer_expressions": "4*sqrt(41)",
    "fact_spans": "[[[2, 4], [55, 56], [108, 110]], [[103, 107]], [[103, 107]], [[64, 71], [121, 129]], [[74, 81]], [[8, 55]], [[8, 55]], [[2, 55]], [[55, 81]], [[103, 119]], [[103, 119]], [[103, 129]], [[84, 101]], [[55, 107]]]",
    "query_spans": "[[[131, 155]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, respectively, and let $P$ be a point on the ellipse. Let point $A$ be the incenter of $\\Delta P F_{1} F_{2}$, and let the extension of segment $P A$ intersect segment $F_{1} F_{2}$ at point $B$. Then $\\frac{|B A|}{|A P|}$=?",
    "fact_expressions": "C: Ellipse;A: Point;P: Point;F1: Point;F2: Point;B: Point;Expression(C) = (x^2/9 + y^2/5 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);Incenter(TriangleOf(P, F1, F2)) = A;Intersection(OverlappingLine(LineSegmentOf(P,A)),LineSegmentOf(F1,F2)) = B",
    "query_expressions": "Abs(LineSegmentOf(B, A))/Abs(LineSegmentOf(A, P))",
    "answer_expressions": "2/3",
    "fact_spans": "[[[19, 61], [72, 74]], [[78, 82]], [[68, 71]], [[1, 8]], [[9, 16]], [[137, 141]], [[19, 61]], [[1, 67]], [[1, 67]], [[68, 77]], [[78, 108]], [[109, 141]]]",
    "query_spans": "[[[143, 166]]]",
    "process": "Connect $AF_{1}$, $B_{1}F$, according to the property of the angle bisector, we obtain $\\frac{BF_{1}}{PF_{1}}=\\frac{BA}{PA}=\\frac{BF_{2}}{PF_{2}}$; by the equal sum property, we get $\\frac{BA}{PA}=\\frac{BF_{1}+BF_{2}}{PF_{1}+PP_{2}}=\\frac{2c}{2a}=\\frac{c}{a}=\\frac{2}{3}$."
  },
  {
    "text": "Draw a line through the focus of the ellipse $x^{2}+2 y^{2}=2$ with an inclination angle of $45^{\\circ}$, intersecting the ellipse at points $A$ and $B$. The center of the ellipse is $O$. Then the area of $\\Delta A O B$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 2*y^2 = 2);H: Line;PointOnCurve(Focus(G), H);Inclination(H) = ApplyUnit(45, degree);A: Point;B: Point;Intersection(H, G) = {A, B};O: Origin;Center(G) = O",
    "query_expressions": "Area(TriangleOf(A, O, B))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 20], [46, 48], [60, 62]], [[1, 20]], [[43, 45]], [[0, 45]], [[26, 45]], [[50, 53]], [[54, 57]], [[43, 59]], [[66, 69]], [[60, 69]]]",
    "query_spans": "[[[71, 90]]]",
    "process": ""
  },
  {
    "text": "Given that line $l$ intersects the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$ at points $P$ and $Q$ (the slope of line $PQ$ is greater than $0$), and $OP \\perp OQ$, if the area of $\\triangle OPQ$ is $\\frac{2\\sqrt{3}}{5}$, then what is the equation of line $PQ$?",
    "fact_expressions": "l: Line;C: Ellipse;Expression(C) = (x^2/2 + y^2 = 1);P: Point;Q: Point;Intersection(l, C) = {P, Q};Slope(LineOf(P, Q))>0;O: Origin;IsPerpendicular(LineSegmentOf(O, P), LineSegmentOf(O, Q));Area(TriangleOf(O, P, Q)) = (2*sqrt(3))/5",
    "query_expressions": "Expression(LineOf(P, Q))",
    "answer_expressions": "{y=sqrt(2)*x + pm*sqrt(2), y=(sqrt(2)/4)*x+pm*(sqrt(3)/2)}",
    "fact_spans": "[[[2, 7]], [[8, 40]], [[8, 40]], [[42, 45]], [[46, 49]], [[2, 51]], [[53, 68]], [[71, 86]], [[71, 86]], [[88, 131]]]",
    "query_spans": "[[[133, 145]]]",
    "process": "Let the line be $ y = kx + m $ ($ k > 0 $), combine with the ellipse equation, using $ OP \\bot OQ $, we obtain $ 3m^{2} = 2(k^{2} + 1) $, calculate the distance from the origin to the line and solve to get $ m^{2} = k^{2} = 2 $, then write the equation of the line. Let the line be $ y = kx + m $ ($ k > 0 $), combine with the ellipse equation $ \\frac{x^{2}}{2} + y^{2} = 1 $, eliminate variables to get: $ (2k^{2} + 1)x^{2} + 4kmx + 2m^{2} - 2 = 0 $. When $ \\Delta > 0 $, $ x_{1} + x_{2} = -\\frac{4km}{2k^{2} + 1} $, $ x_{1} \\cdot x_{2} = \\frac{2m^{2} - 2}{2k^{2} + 1} $. Since $ OP \\bot OQ $, we have $ y_{1}y_{2} + x_{1}x_{2} = 0 $, simplifying yields $ 3m^{2} = 2(k^{2} + 1) $ $\\textcircled{1}$. Also, the distance from the origin to the line is $ d = \\frac{|m|}{\\sqrt{1 + k^{2}}} = \\frac{2}{\\sqrt{5}} $, $ |PQ| = \\sqrt{1 + k^{2}} \\cdot \\sqrt{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}} = \\frac{2\\sqrt{3}}{\\sqrt{5}} $, solving gives $ m^{2} = 2 $ or $ m^{2} = \\frac{3}{4} $. When $ m^{2} = 2 $, $ k^{2} = 2 $, since $ k > 0 $, we have $ k = \\sqrt{2} $, $ m = \\pm\\sqrt{2} $. When $ m^{2} = \\frac{3}{4} $, $ k^{2} = \\frac{1}{8} $, so $ k = \\frac{\\sqrt{2}}{4} $, $ m = \\pm\\frac{\\sqrt{3}}{2} $. After verification, both satisfy $ \\Delta > 0 $. Therefore, the required line equations are $ y = \\sqrt{2}x \\pm \\sqrt{2} $ or $ y = \\frac{\\sqrt{2}}{4}x \\pm \\frac{\\sqrt{3}}{2} $. [Note] This problem mainly examines the equation of an ellipse, the relationship between a line and an ellipse, chord length formula, and distance from a point to a line, classified as a difficult problem."
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1$ $(a>0)$ is $y=-2x$, then $a$=?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2/4 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (y = -2*x)",
    "query_expressions": "a",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 49]], [[66, 69]], [[5, 49]], [[2, 49]], [[2, 64]]]",
    "query_spans": "[[[66, 71]]]",
    "process": "The hyperbola $\\frac{x^{2}}{a2}-\\frac{y^{2}}{4}=1$ ($a>0$) has asymptotes $y=\\pm\\frac{2x}{a}$, so $\\frac{2}{a}=2$, solving gives $a=1$."
  },
  {
    "text": "Through the point $P\\left(\\frac{\\sqrt{10}}{2}, 0\\right)$, draw a line intersecting the curve $x^{2}+12 y^{2}=1$ at points $M$ and $N$. Then the minimum value of $|P M| \\cdot|P N|$ is equal to?",
    "fact_expressions": "G: Line;H: Curve;P: Point;M: Point;N: Point;Expression(H) = (x^2 + 12*y^2 = 1);Coordinate(P) = (sqrt(10)/2, 0);PointOnCurve(P, G);Intersection(G, H) = {M, N}",
    "query_expressions": "Min(Abs(LineSegmentOf(P, M))*Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "19/20",
    "fact_spans": "[[[30, 32]], [[33, 53]], [[1, 29]], [[55, 59]], [[60, 63]], [[33, 53]], [[1, 29]], [[0, 32]], [[30, 63]]]",
    "query_spans": "[[[65, 90]]]",
    "process": ""
  },
  {
    "text": "Given point $M(3,2)$, $F$ is the focus of the parabola $y^{2}=2x$, and point $P$ moves along this parabola. When the perimeter of $\\Delta PMF$ is minimized, what are the coordinates of point $P$?",
    "fact_expressions": "M: Point;Coordinate(M) = (3, 2);F: Point;Focus(G) = F;G: Parabola;Expression(G) = (y^2 = 2*x);P: Point;PointOnCurve(P, G);WhenMin(Perimeter(TriangleOf(P, M, F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2, 2)",
    "fact_spans": "[[[2, 11]], [[2, 11]], [[14, 17]], [[14, 35]], [[18, 32], [42, 45]], [[18, 32]], [[36, 40], [71, 75]], [[36, 48]], [[49, 70]]]",
    "query_spans": "[[[71, 80]]]",
    "process": ""
  },
  {
    "text": "It is known that the focus of the parabola $y^{2}=8x$ is exactly the right focus of the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{2}=1$ $(a>0)$. Then, the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;a>0;Expression(G) = (-y^2/2 + x^2/a = 1);Expression(H) = (y^2 = 8*x);Focus(H)=RightFocus(G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[22, 65], [72, 75]], [[25, 65]], [[2, 16]], [[25, 65]], [[22, 65]], [[2, 16]], [[2, 69]]]",
    "query_spans": "[[[72, 81]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, $A$ is a moving point on the right branch of $C$, the incenter of $\\Delta A F_{1} F_{2}$ is $D$, and the radius $r \\in(0,1]$. Then the range of $|F_{1} D|$ is?",
    "fact_expressions": "C: Hyperbola;A: Point;F1: Point;F2: Point;D: Point;Expression(C) = (x^2 - y^2/3 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(A, RightPart(C));Center(InscribedCircle(TriangleOf(A,F1,F2)))=D;r:Number;Radius(InscribedCircle(TriangleOf(A,F1,F2)))=r;In(r,(0,1])",
    "query_expressions": "Range(Abs(LineSegmentOf(F1, D)))",
    "answer_expressions": "(3, \\sqrt{10}]",
    "fact_spans": "[[[2, 34], [62, 65]], [[58, 61]], [[42, 49]], [[50, 57]], [[102, 105]], [[2, 34]], [[2, 57]], [[2, 57]], [[58, 71]], [[72, 105]], [[108, 120]], [[72, 120]], [[108, 120]]]",
    "query_spans": "[[[122, 140]]]",
    "process": "By combining numerical and geometric analysis, it can be seen that the point of tangency between the circle and $F_{1}F_{2}$ is the right vertex, so $|F_{1}D|^{2}=(a+c)^{2}+r^{2}=9+r^{2}$, thus obtaining the solution. According to the problem, $F_{1}(-2,0)$, $F_{2}(2,0)$, let the incircle of $\\triangle AF_{1}F_{2}$ be tangent to $AF_{1}$, $AF_{2}$ at points $A_{1}$, $B_{1}$ respectively, and tangent to $F_{1}F_{2}$ at point $P$, then $|AA_{1}|=|AB_{1}|$, $|F_{1}A_{1}|=|F_{1}P|$, $|F_{2}B_{1}|=|F_{2}P|$. Since point $A$ lies on the right branch of the hyperbola, $\\therefore |F_{1}A|-|F_{2}A|=2a=2$, $\\therefore |PF_{1}|-|PF_{2}|=2a=2$, while $|F_{1}P|+|F_{2}P|=2c=4$. Let the coordinates of point $P$ be $(x,0)$, then from $|F_{1}A|-|F_{2}A|=2a=2$, we get $(x+c)-(c-x)=2a$, solving gives $x=a=1$. The point of tangency between the circle and $F_{1}F_{2}$ is the right vertex, so $|F_{1}D|^{2}=(a+c)^{2}+r^{2}=9+r^{2}$, therefore $|F_{1}D|\\in(3,\\sqrt{10}]$"
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and $P$ is a point on the ellipse. If the area of $\\triangle P F_{1} F_{2}$ is $1$, $\\tan \\angle P F_{1} F_{2}=\\frac{1}{2}$, $\\tan \\angle P F_{2} F_{1}=-2$, then $a=?$",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);Area(TriangleOf(P, F1, F2)) = 1;Tan(AngleOf(P, F1, F2)) = 1/2;Tan(AngleOf(P, F2, F1)) = -2",
    "query_expressions": "a",
    "answer_expressions": "sqrt(15)/2",
    "fact_spans": "[[[16, 68], [78, 80]], [[16, 68]], [[18, 68]], [[193, 196]], [[18, 68]], [[18, 68]], [[0, 7]], [[8, 15]], [[0, 73]], [[74, 77]], [[74, 84]], [[87, 119]], [[121, 160]], [[161, 191]]]",
    "query_spans": "[[[193, 198]]]",
    "process": "Assume point P is below the x-axis. According to $\\tan\\angle PF_{1}F_{2}=\\frac{1}{2}$, $\\tan\\angle PF_{2}F_{1}=-2$, the slopes and equations of lines $PF_{1}$ and $PF_{2}$ can be obtained. Solving the system of equations gives the coordinates of P. Using the area, we can calculate $c=\\frac{\\sqrt{3}}{2}$. Substituting the coordinates of P into the ellipse equation, combined with $b^{2}=a^{2}-c^{2}$, solving the system yields $a^{2}$; discarding one value based on $a>c$ gives the answer. [Solution] Assume point P is below the x-axis, as shown in the figure: since $\\tan\\angle PF_{1}F_{2}=\\frac{1}{2}$, then $k_{PF_{1}}=-\\frac{1}{2}$, the equation of line $PF_{1}$ is: $y=-\\frac{1}{2}(x+c)$; since $\\tan\\angle PF_{2}F_{1}=-2$, then $k_{PF_{2}}=-2$, the equation of line $PF_{2}$ is: $y=-2(x-c)$. Solve the system\n$$\n\\begin{cases}\ny=-\\frac{1}{2}(x+c) \\\\\ny=-2(x-c)\n\\end{cases}\n$$\nto get\n$$\n\\begin{cases}\nx=\\frac{5}{3}c \\\\\ny=-\\frac{4}{3}c\n\\end{cases}\n$$\ni.e., $P(\\frac{5}{3}c,-\\frac{4}{3}c)$. Since the area of $\\triangle PF_{1}F_{2}$ is 1, we have $\\frac{1}{2}\\times2c\\times\\frac{4}{3}c=1$, so $c=\\frac{\\sqrt{3}}{2}$, thus $P(\\frac{5\\sqrt{3}}{6},-\\frac{2\\sqrt{3}}{3})$. Substitute $P(\\frac{5\\sqrt{3}}{6},-\\frac{2\\sqrt{3}}{3})$ into $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, obtain $\\frac{(\\frac{5\\sqrt{3}}{6})^{2}}{a^{2}}+\\frac{(-\\frac{2\\sqrt{3}}{3})^{2}}{b^{2}}=1$, so $\\frac{75}{36a^{2}}+\\frac{12}{9b^{2}}=1$, simplify to $\\frac{25}{12a^{2}}+\\frac{4}{3b^{2}}=1$. Using $b^{2}=a^{2}-c^{2}=a^{2}-\\frac{3}{4}$, substitute to get $\\frac{25}{12a^{2}}+\\frac{4}{3(a^{2}-\\frac{3}{4})}=1$, multiply both sides by $12a^{2}(a^{2}-\\frac{3}{4})$: $25(a^{2}-\\frac{3}{4})+16a^{2}=12a^{2}(a^{2}-\\frac{3}{4})$, expand: $25a^{2}-\\frac{75}{4}+16a^{2}=12a^{4}-9a^{2}$, rearrange: $12a^{4}-50a^{2}+\\frac{75}{4}=0$. So $(3a^{2}-\\frac{5}{4})(4a^{2}-15)=0$, solve to get $a^{2}=\\frac{5}{12}$ or $a^{2}=\\frac{15}{4}$, since $a^{2}>c^{2}=\\frac{3}{4}$, discard $a^{2}=\\frac{5}{12}$, so $a^{2}=\\frac{15}{4}$, hence $a=\\frac{\\sqrt{15}}{2}$."
  },
  {
    "text": "Given two fixed points $F_{1}(5,0)$, $F_{2}(-5,0)$, the absolute value of the difference in distances from any point $P$ on the curve to $F_{1}$ and $F_{2}$ is $6$. Then the equation of this curve is?",
    "fact_expressions": "G: Curve;F1: Point;F2: Point;P: Point;Coordinate(F1)=(5,0);Coordinate(F2)=(-5,0);PointOnCurve(P,G);Abs(Distance(P,F1)-Distance(P,F2))=6",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/16 = 1",
    "fact_spans": "[[[33, 35], [73, 75]], [[5, 17], [42, 49]], [[19, 32], [50, 57]], [[37, 41]], [[5, 17]], [[19, 32]], [[33, 41]], [[37, 70]]]",
    "query_spans": "[[[72, 80]]]",
    "process": "\\because|F_{1}F_{2}|=10,||PF_{1}|-|PF_{2}||=6,\\thereforeP point's trajectory is a hyperbola with foci F_{1},F_{2} and real axis length 6, so 2a=6, a=3; also c=5,\\thereforeb^{2}=c^{2}-a^{2}=16,\\therefore the curve equation is \\frac{x^{2}}{q}-\\frac{y^{2}}{16}=1"
  },
  {
    "text": "The center of the hyperbola is at the origin, the eccentricity is $2$, and the coordinates of one focus are $(0, 2)$. Then what is the equation of this hyperbola?",
    "fact_expressions": "G: Hyperbola;O:Origin;Center(G)=O;Eccentricity(G)=2;Coordinate(OneOf(Focus(G))) = (0, 2)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 - x^2/3 = 1",
    "fact_spans": "[[[0, 3], [43, 46]], [[7, 11]], [[0, 11]], [[0, 20]], [[0, 40]]]",
    "query_spans": "[[[43, 50]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $E$ is centered at the origin with an eccentricity of $2$. If one of its vertices coincides exactly with the focus of the parabola $y^{2}=8 x$, then the length of the imaginary axis of $E$ is equal to?",
    "fact_expressions": "E: Hyperbola;Center(E) = O;O: Origin;Eccentricity(E) = 2;OneOf(Vertex(E)) = Focus(G);G: Parabola;Expression(G) = (y^2 = 8*x)",
    "query_expressions": "Length(ImageinaryAxis(E))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[0, 6], [21, 22], [49, 52]], [[0, 11]], [[9, 11]], [[0, 19]], [[21, 47]], [[30, 44]], [[30, 44]]]",
    "query_spans": "[[[49, 59]]]",
    "process": "The focus of the parabola has coordinates (2,0), so in the hyperbola, a=2. Also, e=\\frac{c}{a}=2, thus c=4, b=\\sqrt{c^{2}-a^{2}}=2\\sqrt{3}, 2b=4\\sqrt{3}"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$, its directrix intersects the $x$-axis at point $M$. A line passing through $M$ intersects $C$ at points $A$ and $B$, such that $|MA|=|AB|$. If $F$ is the focus of $C$, then the area of $\\triangle ABF$ is?",
    "fact_expressions": "C: Parabola;G: Line;A: Point;B: Point;F: Point;M: Point;Expression(C) = (y^2 = 4*x);Intersection(Directrix(C), xAxis) = M;PointOnCurve(M, G);Intersection(G, C) = {A, B};Abs(LineSegmentOf(M, A)) = Abs(LineSegmentOf(A, B));Focus(C) = F",
    "query_expressions": "Area(TriangleOf(A, B, F))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 21], [45, 48], [79, 82]], [[42, 44]], [[49, 52]], [[53, 56]], [[75, 78]], [[31, 35], [37, 41]], [[2, 21]], [[2, 35]], [[36, 44]], [[42, 58]], [[60, 73]], [[75, 85]]]",
    "query_spans": "[[[87, 109]]]",
    "process": "From the problem, we know M(-1,0). The slope of line AB exists and is non-zero. According to the symmetry of the parabola, assume without loss of generality that the equation of line AB is x = my - 1 (m > 0). Substituting x = my - 1 into y^{2} = 4x gives y^{2} - 4my + 4 = 0. Let A(x_{1},y_{1}), B(x_{2},y_{2}). Then when \\triangle > 0, y_{1} + y_{2} = 4m, y_{1} \\cdot y_{2} = 4. Also |MA| = |AB|, \\therefore y_{2} = 2y_{1}, \\therefore y_{1} = \\sqrt{2}, y_{2} = 2\\sqrt{2}, \\therefore S_{\\triangle ABF} = S_{\\Delta BMF} - S_{\\Delta AMF} = \\frac{1}{2} \\cdot |AF| \\cdot (y_{2} - y_{1}) = \\frac{1}{2} \\times 2 \\times (2\\sqrt{2} - \\sqrt{2}) = \\sqrt{2}."
  },
  {
    "text": "If the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ are $F_{1}$ and $F_{2}$ respectively, and the focus of the parabola $y^{2}=2 b x$ is $F$, and if $\\overrightarrow{F_{1} F}=3 \\overrightarrow{F F_{2}}$, then the eccentricity of this ellipse is?",
    "fact_expressions": "G: Parabola;b: Number;H: Ellipse;a: Number;F1: Point;F: Point;F2: Point;Expression(G) = (y^2 = 2*(b*x));a > b;b > 0;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(H) = F1;RightFocus(H) = F2;Focus(G) = F;VectorOf(F1, F) = 3*VectorOf(F, F2)",
    "query_expressions": "Eccentricity(H)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[78, 94]], [[3, 53]], [[1, 53], [159, 161]], [[3, 53]], [[62, 69]], [[98, 101]], [[70, 77]], [[78, 94]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 77]], [[1, 77]], [[78, 101]], [[103, 156]]]",
    "query_spans": "[[[159, 167]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively. A perpendicular is drawn from $F_{1}$ to the asymptote $y=\\frac{b}{a}x$ of hyperbola $C$, with foot of perpendicular at $P$, intersecting the left branch of hyperbola $C$ at point $Q$. If $OQ \\parallel PF_{2}$ ($O$ being the origin), then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "F1: Point;F2: Point;C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, L) = True;J: Line;OneOf(Asymptote(C)) = J;Expression(J) = (y = b*x/a);IsPerpendicular(L, J) = True;L: Line;FootPoint(L, J) = P;P: Point;Intersection(L, LeftPart(C)) = Q;Q: Point;IsParallel(LineSegmentOf(O, Q), LineSegmentOf(P, F2)) = True;O: Origin",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 9], [90, 97]], [[10, 17]], [[20, 82], [98, 104], [138, 144], [183, 189]], [[20, 82]], [[28, 82]], [[28, 82]], [[28, 82]], [[28, 82]], [[2, 88]], [[2, 88]], [[89, 128]], [[108, 125]], [[98, 125]], [[98, 125]], [[89, 128]], [], [[89, 135]], [[132, 135]], [[89, 153]], [[149, 153]], [[155, 171]], [[172, 175]]]",
    "query_spans": "[[[183, 195]]]",
    "process": "Since $ OQ \\parallel PF_{2} $, and $ O $ is the midpoint of $ F_{1}F_{2} $, it follows that $ Q $ is the midpoint of $ F_{1}P $. Since $ PF_{1} \\perp PO $, the distance from point $ F_{1}(-c,0) $ to the asymptote $ y = \\frac{b}{a}x $ is $ |PF_{1}| = \\frac{|-bc|}{\\sqrt{b^{2}+a^{2}}} = b $. Also, $ |F_{1}O| = c $, so $ \\cos\\angle PF_{1}O = \\frac{b}{c} $. Connecting $ QF_{2} $, it is clear that $ |QF_{1}| = \\frac{1}{2}|PF_{1}| = \\frac{b}{2} $, then by the definition of the hyperbola, $ |QF_{2}| = |QF_{1}| + 2a = 2a + \\frac{b}{2} $. In $ \\triangle QF_{1}F_{2} $, by the law of cosines, $ \\cos\\angle QF_{1}F_{2} = \\frac{\\frac{b^{2}}{4} + 4c^{2} - (2a + \\frac{b}{2})^{2}}{2 \\times \\frac{b}{2} \\times 2c} = \\frac{b}{c} $. Simplifying yields $ b = a $, so the eccentricity of hyperbola $ C $ is $ e = \\frac{c}{a} = \\sqrt{\\frac{a^{2}+b^{2}}{a^{2}}} = \\sqrt{2} $."
  },
  {
    "text": "Given that the left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ coincides with the focus of the parabola $y^{2}=-12x$, and the asymptotes of the hyperbola are given by $y=\\pm \\sqrt{2}x$, find the real number $a=?$",
    "fact_expressions": "G: Hyperbola;b: Number;a: Real;H: Parabola;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = -12*x);LeftFocus(G) = Focus(H);Expression(Asymptote(G)) = (y = pm*(sqrt(2)*x))",
    "query_expressions": "a",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 58], [83, 86]], [[5, 58]], [[113, 118]], [[63, 79]], [[5, 58]], [[5, 58]], [[2, 58]], [[63, 79]], [[2, 82]], [[83, 111]]]",
    "query_spans": "[[[113, 120]]]",
    "process": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$, the left focus is the focus of the parabola $y^{2}=-12x$, the focus is: $(3,0)$. The asymptotes of the hyperbola are $y=\\pm\\sqrt{2}x$, $\\frac{b}{a}=\\sqrt{2}$, $c=3\\Rightarrow a^{2}=3\\Rightarrow a=\\sqrt{3}$."
  },
  {
    "text": "The ellipse, centered at the origin with foci $F_{1}$ and $F_{2}$ on the $x$-axis, passes through the point $(0 , \\sqrt{5})$ and has eccentricity $\\frac{2}{3}$. A line is drawn through $F_{1}$ intersecting the ellipse at points $A$ and $B$. Then, the perimeter of $\\triangle A B F_{2}$ is?",
    "fact_expressions": "G: Ellipse;H: Line;P: Point;A: Point;B: Point;F2: Point;F1: Point;O:Origin;Coordinate(P) = (0, sqrt(5));PointOnCurve(P,G);Eccentricity(G)=2/3;Center(G)=O;Focus(G)={F1,F2};PointOnCurve(F1,xAxis);PointOnCurve(F2,xAxis);PointOnCurve(F1,H);Intersection(H, G) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "12",
    "fact_spans": "[[[37, 39], [87, 89]], [[84, 86]], [[1, 18]], [[90, 93]], [[94, 97]], [[66, 73]], [[57, 65], [75, 83]], [[42, 44]], [[1, 18]], [[0, 39]], [[19, 39]], [[37, 44]], [[37, 73]], [[45, 73]], [[45, 73]], [[74, 86]], [[84, 99]]]",
    "query_spans": "[[[101, 127]]]",
    "process": ""
  },
  {
    "text": "An ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{3}=1$ has a focus $F_{1}$, and point $P$ lies on the ellipse. If the midpoint $M$ of segment $P F_{1}$ lies on the $y$-axis, then what is the $y$-coordinate of point $M$?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;M: Point;Expression(G) = (x^2/12 + y^2/3 = 1);OneOf(Focus(G)) = F1;PointOnCurve(P, G);MidPoint(LineSegmentOf(P,F1))=M;PointOnCurve(M, yAxis)",
    "query_expressions": "YCoordinate(M)",
    "answer_expressions": "pm*(sqrt(3)/4)",
    "fact_spans": "[[[0, 38], [57, 59]], [[44, 51]], [[52, 56]], [[87, 91], [76, 79]], [[0, 38]], [[0, 51]], [[52, 60]], [[62, 79]], [[76, 85]]]",
    "query_spans": "[[[87, 97]]]",
    "process": "Since the midpoint $M$ of segment $PF_{1}$ lies on the $y$-axis and $O$ is the midpoint of segment $F_{1}F_{2}$, $OM$ is the midline of $\\triangle PF_{1}F_{2}$. Find the coordinates of point $P$, then substitute to obtain the solution. $\\because$ the midpoint $M$ of segment $PF_{1}$ lies on the $y$-axis and $O$ is the midpoint of segment $F_{1}F_{2}$, $\\therefore$ $OM$ is the midline of $\\triangle PF_{1}F_{2}$, $\\therefore$ $PF_{2} \\perp x$-axis, $\\therefore$ the $x$-coordinate of point $P$ is $3$ or $-3$, $\\because$ point $P$ lies on the ellipse, $\\therefore$ $\\frac{9}{12} + \\frac{y^{2}}{3} = 1$, that is, $y^{2} = \\frac{3}{4}$, $\\therefore$ $y = \\pm \\frac{\\sqrt{3}}{2}$, $\\therefore$ the $y$-coordinate of point $M$ is $\\pm \\frac{\\sqrt{3}}{4}$."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ with focus $F$, a line passing through the focus $F$ with slope $\\frac{1}{3}$ intersects the parabola at points $A$ and $B$, and $O$ is the origin. Then $\\cos \\angle A O B$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(G) = F;H: Line;A: Point;B: Point;O: Origin;PointOnCurve(F, H);Slope(H) = 1/3;Intersection(H, G) = {A,B}",
    "query_expressions": "Cos(AngleOf(A, O, B))",
    "answer_expressions": "-3/13",
    "fact_spans": "[[[2, 23], [58, 61]], [[2, 23]], [[5, 23]], [[5, 23]], [[27, 30], [34, 37]], [[2, 30]], [[55, 57]], [[64, 67]], [[68, 71]], [[74, 77]], [[31, 57]], [[38, 57]], [[55, 73]]]",
    "query_spans": "[[[84, 105]]]",
    "process": "Find the focus of the parabola, set up the equation of line AB and the coordinates of points A and B, combine with the parabola equation, apply Vieta's formulas and the coordinate representation of the dot product of vectors along with the angle formula to compute the required value. The parabola $ y^{2} = 2px $ ($ p > 0 $) has focus $ F\\left(\\frac{p}{2}, 0\\right) $. Let $ AB: y = \\frac{1}{3}\\left(x - \\frac{p}{2}\\right) $, combining with $ y^{2} = 2px $, we obtain $ 4x^{2} - 76px + p^{2} = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = 19p $, $ x_{1}x_{2} = \\frac{p^{2}}{4} $. Then $ \\overrightarrow{OA} \\cdot \\overrightarrow{O} \\frac{(x_{1} + y_{1}^{2})(x_{2}^{2} + y_{2}^{2})}{p^{2+2p(x_{1}+x_{2})}} = \\sqrt{\\frac{p^{2} + 2px}{4}\\left(\\frac{p^{2}}{4} + \\right.} = \\sqrt{x_{1}x_{2}(x_{1}x_{2} + 4p^{2+2p(x_{1}+x_{2})}} = \\sqrt{\\frac{p^{2}}{4}\\left(\\frac{p}{4}\\right.} \\frac{\\sqrt{\\overrightarrow{OA} \\cdot \\overrightarrow{OB}} = \\frac{-\\frac{3}{4}p^{2}}{\\frac{3}{13}} $"
  },
  {
    "text": "Given that the slope of a line with inclination angle $\\alpha$ is equal to the eccentricity of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, then $\\sin (\\pi-\\alpha)$=?",
    "fact_expressions": "alpha: Number;Inclination(H) = alpha;H: Line;G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1);Slope(H) = Eccentricity(G)",
    "query_expressions": "Sin(-alpha + pi)",
    "answer_expressions": "2*sqrt(5)/5",
    "fact_spans": "[[[6, 14]], [[2, 17]], [[15, 17]], [[22, 50]], [[22, 50]], [[15, 54]]]",
    "query_spans": "[[[56, 77]]]",
    "process": "From $x^{2}-\\frac{y^{2}}{3}=1$, the eccentricity of the hyperbola is known to be $e=2$, that is, $\\tan\\alpha=e=2$, and the angle of inclination $\\alpha\\in[0,\\pi)$, so $\\sin\\alpha=\\frac{2\\sqrt{5}}{5}$, then $\\sin(\\pi-\\alpha)=\\sin\\alpha=\\frac{2\\sqrt{5}}{5}$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, and $O$ as the origin. Point $P$ lies on the parabola $C$, and $PF \\perp OF$. Then $|\\overrightarrow{OF} - \\overrightarrow{PF}| = $?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C)=F;O: Origin;P: Point;PointOnCurve(P,C);IsPerpendicular(LineSegmentOf(P,F),LineSegmentOf(O,F))",
    "query_expressions": "Abs(VectorOf(O,F)-VectorOf(P,F))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 21], [44, 50]], [[2, 21]], [[25, 28]], [[2, 28]], [[29, 32]], [[39, 43]], [[39, 51]], [[54, 69]]]",
    "query_spans": "[[[72, 119]]]",
    "process": ""
  },
  {
    "text": "What is the standard equation of a hyperbola that shares foci with the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ and has eccentricity $e=\\frac{4}{3}$?",
    "fact_expressions": "H: Ellipse;Expression(H) = (x^2/25 + y^2/9 = 1);Focus(H) = Focus(G);G: Hyperbola;e: Number;Eccentricity(G) = e;e = 4/3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/7 = 1",
    "fact_spans": "[[[1, 39]], [[1, 39]], [[0, 68]], [[65, 68]], [[49, 64]], [[46, 68]], [[49, 64]]]",
    "query_spans": "[[[65, 74]]]",
    "process": ""
  },
  {
    "text": "The focal distance of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ is equal to?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 43]]]",
    "process": ""
  },
  {
    "text": "Let a line passing through the left focus $F$ of a hyperbola and perpendicular to the $x$-axis intersect the hyperbola at points $A$ and $B$. From $A$ and $B$, perpendiculars are drawn to the same asymptote of the hyperbola, with feet at $P$ and $Q$, respectively. If $|AP| + |BQ| = 2a$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;F: Point;LeftFocus(G) = F;H: Line;PointOnCurve(F, H);IsPerpendicular(H, xAxis);A: Point;B: Point;Intersection(H, G) = {A, B};Z1: Line;Z2: Line;K: Line;OneOf(Asymptote(G)) = K;PointOnCurve(A, Z1);PointOnCurve(B, Z2);IsPerpendicular(Z1, K);IsPerpendicular(Z2, K);P: Point;Q: Point;FootPoint(Z1, K) = P;FootPoint(Z2, K) = Q;Abs(LineSegmentOf(A, P)) + Abs(LineSegmentOf(B, Q)) = 2*a",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 4], [22, 25], [48, 51], [97, 100]], [[7, 10]], [[1, 10]], [[19, 21]], [[0, 21]], [[11, 21]], [[27, 30], [38, 41]], [[31, 34], [42, 45]], [[19, 36]], [], [], [], [[48, 58]], [[37, 61]], [[37, 61]], [[37, 61]], [[37, 61]], [[67, 70]], [[71, 74]], [[37, 74]], [[37, 74]], [[77, 94]]]",
    "query_spans": "[[[97, 106]]]",
    "process": "Let the hyperbola equation be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1, the left focus (-c,0), one asymptote bx-ay=0, \\frac{(-c)^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1\\Rightarrow y=\\pm\\frac{b^{2}}{a}. Without loss of generality, let A(-c,\\frac{b^{2}}{a}), B(-c,-\\frac{b^{2}}{a}), then |AP|=\\frac{|-bc-b^{2}|}{c}=\\frac{b^{2}+bc}{c}, |BP|=\\frac{|-bc+b^{2}|}{c}=\\frac{bc-b^{2}}{c}. According to the condition, |AP|+|BQ|=2a, i.e., \\frac{b^{2}+bc}{c}+\\frac{bc-b^{2}}{c}=2b=2a, a=b. The hyperbola is an equilateral hyperbola, and the eccentricity is \\sqrt{2}."
  },
  {
    "text": "The directrix of the parabola $y^{2}=2 p x(p>0)$ is tangent to the circle $x^{2}+y^{2}+2 x=0$, then $p$=?",
    "fact_expressions": "G: Parabola;p: Number;H: Circle;p>0;Expression(G) = (y^2 = 2*p*x);Expression(H) = (2*x + x^2 + y^2 = 0);IsTangent(Directrix(G),H)",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 21]], [[49, 52]], [[25, 45]], [[0, 21]], [[25, 45]], [[0, 47]], [[0, 47]]]",
    "query_spans": "[[[49, 54]]]",
    "process": "From the given conditions, the equation of the directrix of the parabola \\( y^{2} = 2px \\) (\\( p > 0 \\)) is \\( x = -\\frac{p}{2} \\). The center of the circle \\( x^{2} + y^{2} + 2x = 0 \\) is at \\( (-1, 0) \\), and its radius is \\( r = 1 \\). Since the directrix \\( x = -\\frac{p}{2} \\) is tangent to the circle \\( x^{2} + y^{2} + 2x = 0 \\), it follows that \\( -\\frac{p}{2} - (-1) = 1 \\). Solving this gives \\( p = 4 \\)."
  },
  {
    "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $e_{1}$, and the eccentricity of the hyperbola $\\frac{x^{2}}{b^{2}}-\\frac{y^{2}}{a^{2}}=1$ is $e_{2}$. Then the minimum value of $e_{1}+e_{2}$ is?",
    "fact_expressions": "a: Number;b: Number;e1: Number;e2: Number;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = e1;Z: Hyperbola;Expression(Z) = (x^2/b^2 - y^2/a^2 = 1);Eccentricity(Z) = e2",
    "query_expressions": "Min(e1 + e2)",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[3, 46]], [[3, 46]], [[51, 58]], [[110, 117]], [[0, 46]], [[0, 46]], [[0, 58]], [[59, 105]], [[59, 105]], [[59, 117]]]",
    "query_spans": "[[[119, 138]]]",
    "process": "From the equation of the hyperbola, we know that $ e_{1} = \\frac{c}{a} $, $ e_{2} = \\frac{c}{b} $, so $ e_{1} + e_{2} = \\frac{c}{a} + \\frac{c}{b} = \\frac{c(a+b)}{ab} $. Since $ \\left( \\frac{4c}{a+b} \\right)^{2} = \\frac{16(a^{2}+b^{2})}{a^{2}+b^{2}+2ab} \\geqslant \\frac{16(a^{2}+b^{2})}{2(a^{2}+b^{2})} = 8 $, the minimum value of $ e_{1} + e_{2} $ is $ \\sqrt{8} = 2\\sqrt{2} $."
  },
  {
    "text": "The point $M$ on the parabola $x^{2}=2 y$ is at a distance $|M F|=\\frac{5}{2}$ from its focus $F$. What are the coordinates of point $M$?",
    "fact_expressions": "G: Parabola;M: Point;F: Point;Expression(G) = (x^2 = 2*y);PointOnCurve(M, G);Focus(G) = F;Distance(M, F) = Abs(LineSegmentOf(M, F)) ;Abs(LineSegmentOf(M, F)) = 5/2",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(pm*2,2)",
    "fact_spans": "[[[0, 14], [21, 22]], [[16, 20], [52, 56]], [[24, 27]], [[0, 14]], [[0, 20]], [[21, 27]], [[16, 49]], [[30, 49]]]",
    "query_spans": "[[[52, 61]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, with left and right foci $F_{1}$ and $F_{2}$ respectively, and the parabola $C_{2}$ whose vertex is at the origin and whose directrix coincides with the left directrix of the hyperbola $C_{1}$. If the intersection point $P$ of the hyperbola $C_{1}$ and the parabola $C_{2}$ satisfies $P F_{2} \\perp F_{1} F_{2}$, then the eccentricity of the hyperbola $C_{1}$ is?",
    "fact_expressions": "C1: Hyperbola;C2: Parabola;P: Point;F2: Point;F1: Point;O: Origin;Expression(C1)=(-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C1)=F1;RightFocus(C1)=F2;Vertex(C2) = O;Directrix(C2)=LeftDirectrix(C1);Intersection(C1,C2)=P;IsPerpendicular(LineSegmentOf(P, F2), LineSegmentOf(F1, F2));a: Number;b: Number;a > 0;b > 0",
    "query_expressions": "Eccentricity(C1)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 67], [114, 124], [132, 142], [190, 200]], [[92, 102], [109, 110], [143, 153]], [[156, 159]], [[84, 91]], [[76, 83]], [[106, 108]], [[2, 67]], [[2, 91]], [[2, 91]], [[92, 108]], [[109, 130]], [[132, 159]], [[161, 188]], [[14, 67]], [[14, 67]], [[14, 67]], [[14, 67]]]",
    "query_spans": "[[[190, 206]]]",
    "process": "First, set the parabola equation as $ y^{2} = 2px $ ($ p > 0 $), then according to the given conditions, obtain the relationship among $ p $, $ a $, and $ c $. Combine the parabola equation with the hyperbola equation to get a quadratic equation in terms of $ x $. According to the conditions, substitute $ x = c $ and $ p = \\frac{2a^{2}}{c} $ into the equation and simplify to find the eccentricity $ e $ of the hyperbola. [Detailed solution] Let the parabola equation be $ y^{2} = 2px $ ($ p > 0 $). From the given conditions, we have $ -\\frac{p}{2} = -\\frac{a^{2}}{c} $, therefore $ p = \\frac{2a^{2}}{c} $. Combining the parabola equation with the hyperbola equation yields $ \\frac{x^{2}}{a^{2}} - \\frac{2px}{b^{2}} = 1 $ (*). Since $ PF_{2} \\perp F_{1}F_{2} $, the x-coordinate of point $ P $ is $ x = c $, so $ x = c $ is a root of equation (*). Therefore, substituting $ x = c $ and $ p = \\frac{2a^{2}}{c} $ into (*) and simplifying gives $ e^{4} - 2e^{2} - 3 = 0 $, solving which yields $ e^{2} = 3 $ or $ -1 $ (discarded), thus $ e = \\sqrt{3} $."
  },
  {
    "text": "The equation of one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1$ $(a>0)$ is $y=2x$, then the value of $a$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/4 + x^2/a^2 = 1);a: Number;a>0;Expression(OneOf(Asymptote(G))) = (y = 2*x)",
    "query_expressions": "a",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 47]], [[0, 47]], [[65, 68]], [[3, 47]], [[0, 63]]]",
    "query_spans": "[[[65, 72]]]",
    "process": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1$ ($a>0$) has an asymptote with equation $y=2x$, from which we obtain: $\\frac{2}{a}=2$, solving gives $a=1$."
  },
  {
    "text": "Given $A_{1}(-3,0)$, $A_{2}(3,0)$ are the left and right vertices of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$. There exists a point $P$ on an asymptote of hyperbola $C$ such that $|P A_{1}|=2|P A_{2}|$. Then the maximum value of $b$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;A1: Point;A2: Point;P: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftVertex(C)=A1;RightVertex(C)=A2;Coordinate(A1) = (-3, 0);Coordinate(A2) = (3, 0);PointOnCurve(P,Asymptote(C));Abs(LineSegmentOf(P,A1))=2*Abs(LineSegmentOf(P,A2))",
    "query_expressions": "Max(b)",
    "answer_expressions": "4",
    "fact_spans": "[[[31, 91], [98, 104]], [[142, 145]], [[38, 91]], [[2, 15]], [[18, 30]], [[113, 116]], [[38, 91]], [[38, 91]], [[31, 91]], [[2, 97]], [[2, 97]], [[2, 15]], [[18, 30]], [[98, 116]], [[118, 140]]]",
    "query_spans": "[[[142, 151]]]",
    "process": "According to the problem, a=3. Without loss of generality, by symmetry, assume the asymptote is y=\\frac{b}{3}x, and let P(3m,bm) satisfy |PA_{1}|=2|PA_{2}|. Then (3m+3)^{2}+(bm)^{2}=4(3m-3)^{2}+4(bm)^{2}. Simplifying yields: (27+3b^{2})m^{2}-90m+27=0. The discriminant d=90^{2}-4\\times27\\times(27+3b^{2})\\geqslant0, solving gives b\\leqslant4."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=8x$ be $F$, and let $P$ lie on this parabola such that $|PF|=5$. Then the coordinates of point $P$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;P: Point;PointOnCurve(P, G) = True;Abs(LineSegmentOf(P, F)) = 5",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(3,2*sqrt(6)), (3,-2*sqrt(6))",
    "fact_spans": "[[[1, 15], [28, 31]], [[1, 15]], [[19, 22]], [[1, 22]], [[23, 26], [44, 48]], [[23, 32]], [[33, 42]]]",
    "query_spans": "[[[44, 53]]]",
    "process": "Let the horizontal coordinate of point P be x. The directrix equation of the parabola y^{2}=8x is x=-2. Since point P lies on the parabola and |PF|=5, ∴ x+2=5 ∴ x=3. Since point P lies on the parabola, ∴ y^{2}=24 ∴ y=\\pm2\\sqrt{6}. ∴ The coordinates of point P are (3,2\\sqrt{6}) or (3,-2\\sqrt{6})."
  },
  {
    "text": "Given that point $p$ is a moving point on the parabola $x = \\frac{1}{4} y^{2}$, what is the minimum value of the sum of the distance from point $p$ to point $A(0,-1)$ and the distance from point $p$ to the line $x = -1$?",
    "fact_expressions": "p: Point;G: Parabola;Expression(G) = (x = y^2/4);PointOnCurve(p, G);A: Point;Coordinate(A) = (0, -1);H: Line;Expression(H) = (x = -1)",
    "query_expressions": "Min(Distance(p,A)+Distance(p,H))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 6], [38, 42], [57, 61]], [[7, 31]], [[7, 31]], [[2, 36]], [[43, 53]], [[43, 53]], [[62, 70]], [[62, 70]]]",
    "query_spans": "[[[38, 81]]]",
    "process": ""
  },
  {
    "text": "Given that the point $(3, \\sqrt{15})$ lies on the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{12}=1$ $(a>0)$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;a: Number;G: Point;a>0;Expression(C) = (-y^2/12 + x^2/a^2 = 1);Coordinate(G) = (3, sqrt(15));PointOnCurve(G, C)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[20, 73], [76, 82]], [[28, 73]], [[2, 19]], [[28, 73]], [[20, 73]], [[2, 19]], [[2, 74]]]",
    "query_spans": "[[[76, 88]]]",
    "process": "From the given condition, we have \\frac{9}{a^{2}}-\\frac{15}{12}=1, solving gives a=2, so c=\\sqrt{4+12}=4, hence the eccentricity of hyperbola C is \\frac{4}{3}=2."
  },
  {
    "text": "Given that the length of the real axis of a hyperbola is $2$, one focus is $F(1,0)$, and it always passes through the origin, what is the equation of the trajectory of the center of this hyperbola?",
    "fact_expressions": "G: Hyperbola;F: Point;Coordinate(F) = (1, 0);OneOf(Focus(G)) = F;Length(RealAxis(G)) = 2;O: Origin;PointOnCurve(O, G)",
    "query_expressions": "LocusEquation(Center(G))",
    "answer_expressions": "x^2 + y^2 - x - 2 = 0",
    "fact_spans": "[[[2, 5], [34, 37]], [[18, 26]], [[18, 26]], [[2, 26]], [[2, 12]], [[29, 31]], [[2, 31]]]",
    "query_spans": "[[[34, 46]]]",
    "process": ""
  },
  {
    "text": "The standard equation of an ellipse with a major axis length of $4$, minor axis length of $2 \\sqrt{3}$, and foci on the $x$-axis is?",
    "fact_expressions": "G: Ellipse;Length(MajorAxis(G))=4;Length(MinorAxis(G))=2*sqrt(3);PointOnCurve(Focus(G),xAxis)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4+y^2/3=1",
    "fact_spans": "[[[0, 2]], [[0, 10]], [[0, 27]], [[0, 35]]]",
    "query_spans": "[[[0, 43]]]",
    "process": "According to the problem, assume the ellipse equation $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, so $\\begin{cases}2a=4\\\\2b=2\\sqrt{3}\\end{cases} \\Rightarrow \\begin{cases}b=\\sqrt{1}\\end{cases}$, therefore the equation of the ellipse is: $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$."
  },
  {
    "text": "The ellipse passes through the point $(0, \\sqrt{5})$ and has an eccentricity of $\\frac{2}{3}$. The center of the ellipse is at the origin, and the two foci on the $x$-axis are $F_{1}$ and $F_{2}$. A line is drawn through $F_{1}$ intersecting the ellipse at points $A$ and $B$. Then, what is the perimeter of $\\triangle A B F_{2}$?",
    "fact_expressions": "G: Ellipse;H: Line;O:Origin;A: Point;B: Point;F2: Point;F1: Point;P:Point;Coordinate(P) = (0, sqrt(5));PointOnCurve(P,G);Eccentricity(G)=2/3;Center(G)=O;PointOnCurve(F1,xAxis);PointOnCurve(F2,xAxis);Focus(G)={F1,F2};PointOnCurve(F1,H);Intersection(H, G) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "12",
    "fact_spans": "[[[37, 39], [86, 88]], [[83, 85]], [[42, 44]], [[89, 92]], [[93, 96]], [[65, 72]], [[57, 64], [75, 82]], [[1, 18]], [[1, 18]], [[0, 39]], [[19, 39]], [[37, 44]], [[45, 72]], [[45, 72]], [[37, 72]], [[74, 85]], [[83, 98]]]",
    "query_spans": "[[[100, 126]]]",
    "process": "Assume the standard equation of the ellipse is \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0). Since the minor axis length is 2\\sqrt{5} and the eccentricity e=\\frac{2}{3}, we have b=\\sqrt{5}, \\frac{c}{a}=\\frac{2}{3}, a^{2}=b^{2}+c^{2}. Solving gives a=3. Therefore, the perimeter of \\triangle ABF_{2} = |AF_{1}|+|AB|+|BF_{1}| = 4a = 12."
  },
  {
    "text": "The moving point $M$ is such that its distance to the point $(-1,0)$ equals its distance to the line $x=1$. Then, the equation of the trajectory of point $M$ is?",
    "fact_expressions": "M: Point;H: Point;Coordinate(H) = (-1, 0);L: Line;Expression(L) = (x = 1);Distance(M, H) = Distance(M, L)",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "y^2=-4*x",
    "fact_spans": "[[[2, 5], [17, 20], [33, 37]], [[6, 15]], [[6, 15]], [[21, 28]], [[21, 28]], [[2, 31]]]",
    "query_spans": "[[[33, 44]]]",
    "process": "In the Cartesian coordinate system xOy, the locus of a moving point M that is equidistant from the point (-1,0) and the line x=1 is a parabola with focus at point A(-1,0) and directrix the line x=1, p=-2, hence the equation of the parabola is y^{2}=-4x;"
  },
  {
    "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ are $y=\\pm 2 x$, what is the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(Asymptote(G)) = (y = pm*(2*x))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 48], [70, 73]], [[5, 48]], [[5, 48]], [[2, 48]], [[2, 66]]]",
    "query_spans": "[[[70, 79]]]",
    "process": ""
  },
  {
    "text": "If the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has eccentricity $\\sqrt{10}$, then the equations of the asymptotes of the hyperbola $C$ are?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;a > 0;b > 0;Expression(C) = (x^2/a^2 - y^2/b^2 = 1);Eccentricity(C) = sqrt(10)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y = pm*3*x",
    "fact_spans": "[[[1, 62], [80, 86]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[1, 62]], [[1, 78]]]",
    "query_spans": "[[[80, 94]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $3x+y=0$, what is the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (3*x + y = 0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[2, 58], [79, 82]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 76]]]",
    "query_spans": "[[[79, 88]]]",
    "process": "By the given condition, one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $3x+y=0$, so $b=3a$, thus $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{10}a$, therefore $e=\\frac{c}{a}=\\sqrt{10}$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has an eccentricity of $\\frac{\\sqrt{3}}{2}$. A line passing through the right focus $F$ with slope $k$ $(k>0)$ intersects $C$ at points $A$ and $B$. If $\\overrightarrow{AF}=3 \\overrightarrow{F B}$, then $k=$?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;A: Point;F: Point;B: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Eccentricity(C) = sqrt(3)/2;RightFocus(C)=F;PointOnCurve(F, G);Slope(G)=k;k:Number;k>0;Intersection(G, C) = {A,B};VectorOf(A, F) = 3*VectorOf(F, B)",
    "query_expressions": "k",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 59], [110, 113]], [[9, 59]], [[9, 59]], [[106, 109]], [[116, 119]], [[90, 93]], [[121, 124]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 85]], [[2, 93]], [[86, 108]], [[94, 109]], [[174, 177]], [[97, 105]], [[106, 126]], [[128, 172]]]",
    "query_spans": "[[[174, 179]]]",
    "process": ""
  },
  {
    "text": "$P$ is a point on the right branch of the hyperbola $x^{2}-\\frac{y^{2}}{15}=1$, and $M$, $N$ are points on the circles $(x+4)^{2}+y^{2}=4$ and $(x-4)^{2}+y^{2}=1$, respectively. Then the maximum value of $|PM|-|PN|$ is?",
    "fact_expressions": "G: Hyperbola;H: Circle;Z: Circle;P: Point;M: Point;N: Point;Expression(G) = (x^2 - y^2/15 = 1);Expression(H) = (y^2 + (x + 4)^2 = 4);Expression(Z) = (y^2 + (x - 4)^2 = 1);PointOnCurve(P, RightPart(G));PointOnCurve(M, H);PointOnCurve(N, Z)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "5",
    "fact_spans": "[[[4, 33]], [[49, 69]], [[70, 89]], [[0, 3]], [[39, 42]], [[43, 46]], [[4, 33]], [[49, 69]], [[70, 89]], [[0, 38]], [[39, 92]], [[39, 92]]]",
    "query_spans": "[[[94, 113]]]",
    "process": ""
  },
  {
    "text": "Point $M$ is the intersection of the axis of symmetry and the directrix of the parabola $C$: $x^{2}=2 p y(p>0)$, point $F$ is the focus of the parabola $C$, and point $P$ lies on the parabola $C$. In $\\triangle F P M$, $\\sin \\angle P F M=\\lambda \\sin \\angle P M F$. Then the maximum value of $\\lambda$ is?",
    "fact_expressions": "C: Parabola;p: Number;F: Point;P: Point;M: Point;p>0;Expression(C) = (x^2 = 2*(p*y));Intersection(SymmetryAxis(C),Directrix(C))=M;Focus(C) = F;PointOnCurve(P, C);Sin(AngleOf(P, F, M)) = lambda*Sin(AngleOf(P, M, F));lambda:Number",
    "query_expressions": "Max(lambda)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[5, 31], [47, 53], [62, 68]], [[13, 31]], [[42, 46]], [[57, 61]], [[0, 4]], [[13, 31]], [[5, 31]], [[0, 41]], [[42, 56]], [[57, 69]], [[90, 135]], [[137, 146]]]",
    "query_spans": "[[[137, 152]]]",
    "process": "By the law of sines, |PM| = \\lambda|PF|. According to the definition of the parabola, \\frac{1}{\\lambda} = \\frac{|PB|}{|PM|}, i.e., \\sin\\alpha = \\frac{1}{2}. When \\lambda reaches its maximum value, \\sin\\alpha is minimized, and at this moment the line PM is tangent to the parabola. Substituting the line equation into the parabola equation and setting A = 0, we solve for k and thus obtain the maximum value of \\lambda. As shown in the figure, draw a perpendicular from point P to the directrix, with foot B. Then by the definition of the parabola, |PF| = |PB|. Given \\sin\\angle PFM = \\lambda \\sin\\angle PMF, by the law of sines in \\triangle PFM: |PM| = \\lambda|PF|. Hence |PM| = \\lambda|PB|, so \\frac{1}{\\lambda} = \\frac{|PB|}{|PM|}. Let the angle of inclination of PM be \\alpha, then \\sin\\alpha = \\frac{1}{\\lambda}. When \\lambda reaches its maximum value, \\sin\\alpha is minimized, and at this moment the line PM is tangent to the parabola. Let the equation of line PM be y = kx - \\frac{p}{2}, then\n\\begin{cases}\nx^{2} = 2py \\\\\ny = kx - \\frac{p}{2}\n\\end{cases}\nwhich gives x^{2} - 2pkx + p^{2} = 0. Thus A = 4p^{2}k^{2} - 4p^{2} = 0, so k = \\pm1, i.e., \\tan\\alpha = \\pm1, then \\sin\\alpha = \\frac{\\sqrt{2}}{2}. Therefore, the maximum value of \\lambda is \\frac{1}{\\sin\\alpha} = \\sqrt{2}."
  },
  {
    "text": "The two asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ are perpendicular to each other, then its eccentricity is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;L1: Line;L2: Line;Asymptote(G) = {L1, L2};IsPerpendicular(L1, L2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 46], [59, 60]], [[0, 46]], [[3, 46]], [[3, 46]], [], [], [[0, 52]], [[0, 56]]]",
    "query_spans": "[[[59, 66]]]",
    "process": ""
  },
  {
    "text": "If the equation of ellipse $C$ is $\\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1$, then its eccentricity is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/3 + y^2/4 = 1)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[1, 6], [47, 48]], [[1, 45]]]",
    "query_spans": "[[[47, 53]]]",
    "process": "From the ellipse equation, we obtain: a=2, b=\\sqrt{3}, c=1, e=\\frac{c}{a}=\\frac{1}{2}"
  },
  {
    "text": "If the circle $(x-2)^{2}+y^{2}=1$ is tangent to the asymptotes of the hyperbola $C$: $\\frac{y^{2}}{m^{2}}-\\frac{x^{2}}{9}=1$ $(m>0)$, then the equations of the asymptotes of hyperbola $C$ are?",
    "fact_expressions": "C: Hyperbola;m: Number;G: Circle;m>0;Expression(C) = (-x^2/9 + y^2/m^2 = 1);Expression(G) = (y^2 + (x - 2)^2 = 1);IsTangent(G, Asymptote(C))",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*(sqrt(3)/3)*x",
    "fact_spans": "[[[22, 73], [81, 87]], [[29, 73]], [[1, 21]], [[29, 73]], [[22, 73]], [[1, 21]], [[1, 79]]]",
    "query_spans": "[[[81, 95]]]",
    "process": "【Topic】Equation of tangent lines to a circle, simple properties of hyperbolas. The asymptotes of the hyperbola are: $ y = \\pm\\frac{m}{3}x $. The circle $ (x-2^{2}) + y^{2} = 1 $ has center $ (2, 0) $ and radius $ 1 $. Since they are tangent, $ d = \\frac{|2m|}{\\sqrt{m^{2}+9}} = 1 \\Rightarrow m = \\sqrt{3} $ ($ m > 0 $). Therefore, the asymptotes of hyperbola $ C $ are: $ y = \\pm\\frac{\\sqrt{3}}{3}x $."
  },
  {
    "text": "The focal distance of the ellipse $\\frac{x^{2}}{16-m}+\\frac{y^{2}}{m-2}=1$ is $4$, then the value of $m$ is?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/(16 - m) + y^2/(m - 2) = 1);FocalLength(G) = 4",
    "query_expressions": "m",
    "answer_expressions": "{7,11}",
    "fact_spans": "[[[0, 42]], [[51, 54]], [[0, 42]], [[0, 49]]]",
    "query_spans": "[[[51, 58]]]",
    "process": "In the ellipse $\\frac{x^2}{16-m}+\\frac{y^{2}}{m-2}=1$, from the given condition we have $2c=4$, solving gives $c=2$. If the foci of the ellipse lie on the $x$-axis, we obtain\n\\[\n\\begin{cases}\n16-m>0 \\\\\nm-2>0 \\\\\n(16-m)-(m-2)=c^{2}=4\n\\end{cases}\n\\]\nSolving $16-m>0$ yields $m=7$; if the foci of the ellipse lie on the $y$-axis, $16-m>0$ gives\n\\[\n\\begin{cases}\n16-m>0 \\\\\nm-2>0 \\\\\n(m-2)-(16-m)=c^{2}=4\n\\end{cases}\n\\]\nSolving yields $m=11$. Therefore, $m=7$ or $11$."
  },
  {
    "text": "The distance from a point $M$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ to the left focus $F_{1}$ is $4$. What is the distance from $M$ to the right focus $F_{2}$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);M: Point;PointOnCurve(M, G);F1: Point;LeftFocus(G) = F1;Distance(M, F1) = 4;F2: Point;RightFocus(G) = F2",
    "query_expressions": "Distance(M, F2)",
    "answer_expressions": "6",
    "fact_spans": "[[[0, 38]], [[0, 38]], [[41, 44], [64, 67]], [[0, 44]], [[48, 55]], [[0, 55]], [[41, 62]], [[71, 78]], [[0, 78]]]",
    "query_spans": "[[[64, 83]]]",
    "process": ""
  },
  {
    "text": "Given that the line $y = kx + 1$ intersects the right branch of the hyperbola $3x^2 - y^2 = 3$ at two distinct points, what is the range of values for $k$?",
    "fact_expressions": "G: Hyperbola;H: Line;k: Number;Expression(G) = (3*x^2 - y^2 = 3);Expression(H) = (y = k*x + 1);NumIntersection(H, RightPart(G)) = 2",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-2, -\\sqrt{3})",
    "fact_spans": "[[[14, 34]], [[2, 13]], [[47, 50]], [[14, 34]], [[2, 13]], [[2, 45]]]",
    "query_spans": "[[[47, 57]]]",
    "process": "\\begin{cases}y=kx+1\\\\3x^{2}-y^{2}=3\\end{cases} eliminating y and simplifying gives (3-k^{2})x^{2} according to the problem we have x_{1}+x_{2}=\\frac{2}{2k}+16(3-k^{2})>0 2<k<-\\sqrt{\\begin{matrix}3-k&"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, the left focus is point $F_{1}$, the right focus is point $F_{2}$, and point $M(x, y)$ $(x \\neq \\pm 5)$ is a moving point on hyperbola $C$. Then the range of the product of the slopes of lines $M F_{1}$ and $M F_{2}$, $k_{M F_{1}} \\cdot k_{M F_{2}}$, is?",
    "fact_expressions": "C: Hyperbola;M: Point;F1: Point;F2: Point;x1:Number;y1:Number;Negation(x1=pm*5);k1:Number;k2:Number;Expression(C) = (x^2/9 - y^2/16 = 1);Coordinate(M) = (x1, y1);LeftFocus(C) =F1;RightFocus(G)=F2;PointOnCurve(M,C);Slope(LineOf(M,F1))=k1;Slope(LineOf(M,F2))=k2",
    "query_expressions": "k1*k2",
    "answer_expressions": "(-\\infty,0] \\cup (16/9,+\\infty)",
    "fact_spans": "[[[2, 46], [99, 105]], [[74, 98]], [[51, 59]], [[65, 73]], [[75, 98]], [[75, 98]], [[75, 98]], [[136, 167]], [[136, 167]], [[2, 46]], [[74, 98]], [[2, 59]], [[2, 73]], [[74, 109]], [[110, 167]], [[110, 167]]]",
    "query_spans": "[[[136, 174]]]",
    "process": "Because $k_{MF_{1}}\\cdot k_{MF_{6}} = \\frac{y}{x+5} \\cdot \\frac{y}{x-5} = \\frac{\\frac{16}{9}(x^{2}-9)}{x^{2}-25} = \\frac{16}{9}\\left(1 + \\frac{16}{x^{2}-25}\\right) (-\\infty, 0] \\cup \\left(\\frac{16}{0}, +\\infty\\right)$"
  },
  {
    "text": "Let $F_{1}$ be the left focus of the ellipse $C$: $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$, $M$ any point on the ellipse, and $P$ the midpoint of the segment $F_{1} M$. Then the equation of the trajectory of the moving point $P$ is?",
    "fact_expressions": "C: Ellipse;F1: Point;M: Point;P: Point;Expression(C) = (x^2/16 + y^2/12 = 1);LeftFocus(C) = F1;PointOnCurve(M, C);MidPoint(LineSegmentOf(F1,M)) = P",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(x+1)^2/4+y^2/3=1",
    "fact_spans": "[[[9, 53], [62, 64]], [[1, 8]], [[58, 61]], [[70, 73], [92, 95]], [[9, 53]], [[1, 57]], [[58, 69]], [[70, 88]]]",
    "query_spans": "[[[92, 103]]]",
    "process": "For the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$, its left focus $F_{1}$ has coordinates $(-2,0)$. Let the coordinates of points $M$ and $P$ be $(x_{0},y_{0})$ and $(x,y)$ respectively. Since point $P$ is the midpoint of segment $F_{1}M$, we have $x=\\frac{-2+x_{0}}{2}$, $y=\\frac{y_{0}}{2}$, that is, $x_{0}=2x+2$, $y_{0}=2y$. Since point $M$ lies on the ellipse, it follows that $\\frac{x_{0}^{2}}{16}+\\frac{y_{0}^{2}}{12}=1$, i.e., $\\frac{(2x+2)^{2}}{16}+\\frac{4y^{2}}{12}=1$. Simplifying yields: $\\frac{(x+1)^{2}}{4}+\\frac{y^{2}}{3}=1$."
  },
  {
    "text": "Let the left and right vertices of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ be $A$ and $B$, respectively. Point $P$ lies on the hyperbola and is distinct from $A$ and $B$. Let $O$ be the origin. If the product of the slopes of lines $PA$ and $PB$ is $\\frac{7}{9}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;A: Point;B: Point;LeftVertex(G) = A;RightVertex(G) = B;P: Point;PointOnCurve(P, G);Negation(P=A);Negation(P=B);O: Origin;Slope(LineOf(P, A))*Slope(LineOf(P, B)) = 7/9",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "4/3",
    "fact_spans": "[[[1, 47], [67, 70], [130, 133]], [[1, 47]], [[4, 47]], [[4, 47]], [[54, 57], [75, 78]], [[58, 61], [79, 82]], [[1, 61]], [[1, 61]], [[62, 66]], [[62, 71]], [[62, 84]], [[62, 84]], [[85, 88]], [[95, 128]]]",
    "query_spans": "[[[130, 139]]]",
    "process": "By the symmetry of the hyperbola, A and B are symmetric with respect to the origin. Let A(x_{1},y_{1}), B(-x_{1},-y_{1}), P(x,y), then \\frac{x_{1}^{2}}{a^{2}}-\\frac{y_{1}^{2}}{b^{2}}=1, and the hyperbola is \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1. Therefore, k_{PA}\\cdot k_{PB}=\\frac{y_{1}-y}{x_{1}-x}\\cdot\\frac{y_{1}+y}{x_{1}+x}=\\frac{b^{2}}{a^{2}}=\\frac{7}{9}. Hence, the eccentricity of the hyperbola is e=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\frac{4}{3}."
  },
  {
    "text": "Given that point $P$ lies on the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$, and $F_{1}$, $F_{2}$ are the foci of this hyperbola. If $\\angle F_{1} PF_{2}=60^{\\circ}$, then the area of $\\Delta F_{1} PF_{2}$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/3 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "3*sqrt(3)",
    "fact_spans": "[[[7, 45], [68, 71]], [[7, 45]], [[2, 6]], [[2, 48]], [[49, 56]], [[58, 66]], [[49, 74]], [[76, 108]]]",
    "query_spans": "[[[110, 136]]]",
    "process": ""
  },
  {
    "text": "The tangent line $M T$ to the circle $x^{2}+y^{2}=9$ passes through the left focus $F$ of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{12}=1$, where $T$ is the point of tangency, $M$ is the intersection point of the tangent line and the right branch of the hyperbola, $P$ is the midpoint of $M F$, and $O$ is the origin. Then $|P O|-|P T|$=?",
    "fact_expressions": "G: Hyperbola;H: Circle;M: Point;T: Point;F: Point;P: Point;O: Origin;Expression(G) = (x^2/9 - y^2/12 = 1);Expression(H) = (x^2 + y^2 = 9);IsTangent(LineOf(M,T),H);LeftFocus(G)=F;PointOnCurve(F,LineOf(M,T));TangentPoint(LineOf(M,T),H)=T;Intersection(LineOf(M,N),RightPart(G))=M;MidPoint(LineSegmentOf(M, F)) = P",
    "query_expressions": "Abs(LineSegmentOf(P, O)) - Abs(LineSegmentOf(P, T))",
    "answer_expressions": "2*sqrt(3)-3",
    "fact_spans": "[[[25, 64], [88, 91]], [[0, 16]], [[81, 84]], [[74, 77]], [[68, 71]], [[97, 100]], [[110, 113]], [[25, 64]], [[0, 16]], [[0, 24]], [[25, 71]], [[19, 71]], [[19, 80]], [[81, 96]], [[97, 109]]]",
    "query_spans": "[[[120, 135]]]",
    "process": "Let the right focus be F, |FF| = \\sqrt{OF^{2} - |OT|} = b = |P| = |PF| - |FF| = \\frac{1}{2}|MF| - b, |PO| = \\frac{1}{2}|PF| \\Rightarrow |PO| - |PF| = b - \\frac{1}{2}|MF| - |MF| = b - a = 2\\sqrt{3} - 3"
  },
  {
    "text": "The distance from the focus of a hyperbola to its asymptote equals the length of the real axis; then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Distance(Focus(G),Asymptote(G)) =  Length(RealAxis(G))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[0, 3], [20, 23]], [[0, 18]]]",
    "query_spans": "[[[20, 29]]]",
    "process": "\\because the distance from focus F(c,0) to the asymptote y=\\frac{b}{a}x equals the length of the real axis. \\frac{bc}{\\sqrt{a^{2}+b^{2}}}=2a, \\therefore b=2a, \\therefore e^{2}=\\frac{c^{2}}{a^{2}}=1+\\frac{b^{2}}{a^{2}}=5 \\therefore e=\\sqrt{5}"
  },
  {
    "text": "If the focus of the parabola $y^{2}=8 a x$ coincides with the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 8*(a*x));G: Hyperbola;Expression(G) = (-y^2 + x^2/a^2 = 1);a: Number;Focus(H) = RightFocus(G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 17]], [[1, 17]], [[21, 53], [61, 64]], [[21, 53]], [[4, 17]], [[1, 59]]]",
    "query_spans": "[[[61, 70]]]",
    "process": "The focus of the parabola $ y^{2} = 8ax $ is at $ (2a, 0) $, and the right focus of the hyperbola $ \\frac{x^{2}}{a^{2}} - y^{2} = 1 $ is at $ (\\sqrt{a^{2}+1}, 0) $. Then $ 2a = \\sqrt{a^{2}+1} $, yielding $ a^{2} = \\frac{1}{3} $, $ e = \\sqrt{1 + \\frac{1}{a^{2}}} = 2 $;"
  },
  {
    "text": "It is known that a hyperbola passes through the point $(2 \\sqrt{3}, 2)$ and has asymptotes with equations $y=\\pm \\frac{\\sqrt{2}}{2} x$. Then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (2*sqrt(3), 2);PointOnCurve(H, G);Expression(Asymptote(G)) = (y = pm*x*(sqrt(2)/2))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/2 = 1",
    "fact_spans": "[[[2, 5], [63, 66]], [[6, 24]], [[6, 24]], [[2, 24]], [[2, 60]]]",
    "query_spans": "[[[63, 73]]]",
    "process": "When the foci are on the x-axis, let the hyperbola equation be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1, \\therefore \\frac{12}{a^{2}}-\\frac{4}{b^{2}}=1, at this time the asymptote equations are y=\\pm\\frac{b}{a}x=\\pm\\frac{\\sqrt{2}}{2}x', \\therefore a^{2}=4, b^{2}=2, \\therefore the hyperbola equation is \\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1. When the foci are on the y-axis, let the hyperbola equation be \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1, \\therefore \\frac{4}{a^{2}}-\\frac{12}{b^{2}}=1, at this time the asymptote equations are y=\\pm\\frac{a}{b}x=\\pm\\frac{\\sqrt{2}}{2}x', \\therefore a2<0, discard."
  },
  {
    "text": "Given that the line $y = kx + 1$ intersects the ellipse $\\frac{x^{2}}{4} + y^{2} = 1$ at points $A$ and $B$, if the x-coordinate of the midpoint of $AB$ is $1$, then the value of $k$ is?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*x + 1);k: Number;G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);A: Point;B: Point;Intersection(H, G) = {A, B};XCoordinate(MidPoint(LineSegmentOf(A, B))) = 1",
    "query_expressions": "k",
    "answer_expressions": "-1/2",
    "fact_spans": "[[[2, 13]], [[2, 13]], [[72, 75]], [[14, 41]], [[14, 41]], [[44, 47]], [[48, 51]], [[2, 53]], [[55, 70]]]",
    "query_spans": "[[[72, 79]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Substituting $ y = kx + 1 $ into $ \\frac{x^{2}}{4} + y^{2} = 1 $ gives $ (1+4k^{2})x^{2} + 8kx = 0 $. $ x_{1} + x_{2} = -\\frac{8k}{1+4k^{2}} $. Since the horizontal coordinate of the midpoint of $ AB $ is 1, $ -\\frac{4k}{1+4k^{2}} = 1 $. Solving yields $ k = -\\frac{1}{2} $."
  },
  {
    "text": "The point $A$ on the parabola $x^{2}=2 p y(p>0)$ with ordinate $4$ has a distance of $5$ from its focus $F$. Then, what is the distance from point $A$ to the origin?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 2*(p*y));p: Number;p>0;PointOnCurve(A,G) = True;A: Point;YCoordinate(A) = 4;F: Point;Focus(G) = F;Distance(A,F) = 5;O: Origin",
    "query_expressions": "Distance(A,O)",
    "answer_expressions": "4*sqrt(2)",
    "fact_spans": "[[[0, 21], [35, 36]], [[0, 21]], [[3, 21]], [[3, 21]], [[0, 34]], [[30, 34], [51, 55]], [[22, 34]], [[38, 41]], [[35, 41]], [[30, 48]], [[56, 58]]]",
    "query_spans": "[[[51, 63]]]",
    "process": "According to the problem, the directrix of the parabola $ x^{2} = 2py $ is $ y = -\\frac{p}{2} $. If the distance from the point with ordinate 4 to the focus is 5, then the distance from this point to the directrix is also 5. Thus, $ 4 - \\left(-\\frac{p}{2}\\right) = 5 $, solving which gives $ p = 2 $. The standard equation of the parabola is $ x^{2} = 4y $, so point $ A(\\pm4, 4) $, and therefore $ OA = \\sqrt{(\\pm4)^{2} + 4^{2}} = 4\\sqrt{2} $."
  },
  {
    "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ are tangent to the circle $x^{2}+(y-4)^{2}=4$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "S: Hyperbola;Expression(S) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;C: Circle;Expression(C) = (x^2 + (y - 4)^2 = 4);IsTangent(Asymptote(S),C)",
    "query_expressions": "Eccentricity(S)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 58], [87, 90]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[63, 83]], [[63, 83]], [[2, 85]]]",
    "query_spans": "[[[87, 96]]]",
    "process": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0,b>0)$ are given by $y=\\pm\\frac{b}{a}x$, i.e., $bx\\pm ay=0$. The circle $x^{2}+(y-4)^{2}=4$ has center $(0,4)$ and radius $2$. Since the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0,b>0)$ are tangent to the circle $x^{2}+(y-4)^{2}=4$, it follows that $\\frac{|4a|}{\\sqrt{a^{2}+b^{2}}}=2$, i.e., $\\frac{4a}{c}=2$, yielding $c=2a$, so $e=\\frac{c}{a}=2$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, a line passing through the left focus $F_{1}$ intersects the ellipse at points $A$ and $B$, satisfying $|A B|=|B F_{2}|=4|B F_{1}|$. What is the eccentricity of this ellipse?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;A: Point;B: Point;F2: Point;F1: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A,B))=Abs(LineSegmentOf(B,F2));Abs(LineSegmentOf(B,F2))=4*Abs(LineSegmentOf(B,F1))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)/5",
    "fact_spans": "[[[18, 70], [92, 94], [92, 94]], [[20, 70]], [[20, 70]], [[89, 91]], [[95, 98]], [[99, 102]], [[10, 17]], [[2, 9], [81, 88]], [[20, 70]], [[20, 70]], [[18, 70]], [[2, 76]], [[2, 76]], [[77, 91]], [[89, 104]], [[108, 136]], [[108, 136]]]",
    "query_spans": "[[[138, 146]]]",
    "process": "First, obtain |AB|, |BF₂|, |AF₂| according to the definition of the ellipse, then use the cosine law to set up an equation and solve for the eccentricity. Since |BF₂| = 4|BF₁| and |BF₂| + |BF₁| = 2a, it follows that |BF₂| = 4|BF₁| = \\frac{8a}{5}, \\therefore |AB| = \\frac{8a}{5}, |AF₁| = \\frac{6}{x}\\frac{a}{5}, |AF₂| = 2a - \\frac{6a}{5} = \\frac{4a}{5}. Therefore, \\cos\\angle ABF_{2} = \\frac{(\\frac{8}{5}a)^{2} + (\\frac{2}{5}a)^{2}}{2 \\times \\frac{8}{5}a \\times \\frac{2}{5}a} = \\frac{(\\frac{8}{5}a)^{2} + (\\frac{8}{5}a)^{2} - (\\frac{4}{5}a)}{2 \\times \\frac{8}{5}a \\times \\frac{8}{5}a}, \\therefore 2a^{2} = 5c^{2}, \\therefore e = \\frac{\\sqrt{10}}{5}."
  },
  {
    "text": "What is the equation of the directrix of the parabola $y=2 x^{2}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 2*x^2)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y = -1/8",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": "First, convert the parabola equation into standard form and find the value of p. From y=2x^{2}, the parabola equation becomes x^{2}=\\frac{1}{2}y, so p=\\frac{1}{4}; thus, the directrix of the parabola y=2x^{2} is y=-\\frac{p}{2}=-\\frac{1}{8}."
  },
  {
    "text": "Given point $A(3,-1)$, $F$ is the focus of the parabola $y^{2}=4x$, and $M$ is any point on the parabola. Then the minimum value of $|MF|+|MA|$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (3, -1);F: Point;Focus(G) = F;G: Parabola;Expression(G) = (y^2 = 4*x);M: Point;PointOnCurve(M, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F)))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 12]], [[2, 12]], [[14, 17]], [[14, 35]], [[18, 32], [40, 43]], [[18, 32]], [[36, 39]], [[36, 48]]]",
    "query_spans": "[[[50, 69]]]",
    "process": "Analysis: Let $ M(x_{0},y_{0}) $. Draw a perpendicular line from point $ M $ to the directrix $ x = -1 $, with foot of perpendicular $ H $. By the definition of a parabola, $ |MF| = |MH| $. Thus, the problem is transformed into finding the minimum value of $ |MA| + |MH| $. From the figure, it can be seen that $ |MA| + |MH| $ reaches its minimum if and only if points $ M $, $ F $, and $ H $ are collinear. The minimum value is $ |AH| = 3 - (-1) = 4 $. The answer is $ 4 $."
  },
  {
    "text": "What are the coordinates of the focus of the parabola $y=a x^{2}$?",
    "fact_expressions": "G: Parabola;a: Number;Expression(G) = (y = a*x^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,1/(4*a))",
    "fact_spans": "[[[0, 14]], [[3, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": "First, rewrite the parabola equation into standard form, and then use the properties of the parabola to find the coordinates of the focus. When $ a > 0 $, rearranging the parabola equation gives $ x^{2} = \\frac{1}{a}y' $, so $ p = \\frac{1}{2a} $. For the parabola $ x^{2} = 2py $ ($ p > 0 $), the focus is at $ (0, \\frac{p}{2}) $, hence the desired focus coordinates are $ (0, \\frac{1}{4a}) $. The same result can be obtained when $ a < 0 $."
  },
  {
    "text": "Given that the line $x=t$ intersects the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ at points $P$ and $Q$, and point $F$ is the left focus of the ellipse, find the value of $t$ when $\\overrightarrow{F P} \\cdot \\overrightarrow{F Q}$ takes its minimum value.",
    "fact_expressions": "H: Line;Expression(H) = (x = t);t: Number;G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);P: Point;Q: Point;Intersection(H, G) = {P, Q};F: Point;LeftFocus(G) = F;WhenMin(DotProduct(VectorOf(F, P), VectorOf(F, Q)))",
    "query_expressions": "t",
    "answer_expressions": "-50/17",
    "fact_spans": "[[[2, 9]], [[2, 9]], [[130, 133]], [[10, 48], [67, 69]], [[10, 48]], [[50, 53]], [[54, 57]], [[2, 59]], [[61, 65]], [[61, 73]], [[76, 130]]]",
    "query_spans": "[[[130, 137]]]",
    "process": "It is easy to see that the left focus of the ellipse is $F(-4,0)$. By symmetry, let $P(t,y_{0})$, $Q(t,-y_{0})$, and $-5<t<5$, then $\\overrightarrow{FP}=(t+4,y_{0})$, $\\overrightarrow{FQ}=(t+4,-y_{0})$, so $\\overrightarrow{FP}\\cdot\\overrightarrow{FQ}=(t+4,y_{0})\\cdot(t+4,-y_{0})=(t+4)^{2}-y_{0}^{2}$. Since $y_{0}^{2}=9(1-\\frac{t^{2}}{25})=9-\\frac{9}{25}t^{2}$, we have $\\overrightarrow{FP}\\cdot\\overrightarrow{FQ}=(t+4)^{2}-y_{0}^{2}=\\frac{34}{25}t^{2}+8t+7$. Therefore, when $t=-\\frac{50}{17}$, $\\overrightarrow{FP}\\cdot\\overrightarrow{FQ}$ reaches its minimum value. Hence, fill in $-\\frac{50}{17}$. (This problem mainly examines the standard equation of an ellipse, the dot product of vectors, and the maximum/minimum value of a quadratic function, belonging to a medium-difficulty question.)"
  },
  {
    "text": "A line $l$ passing through the point $M(0,1)$ intersects the ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ at points $A$ and $B$, and $F_{1}$ is the left focus of the ellipse. When the perimeter of $\\triangle A B F_{1}$ is maximized, what is the slope of line $l$?",
    "fact_expressions": "M: Point;Coordinate(M) = (0, 1);l: Line;PointOnCurve(M, l);C: Ellipse;Expression(C) = (x^2/9 + y^2/5 = 1);A: Point;B: Point;Intersection(l, C) = {A, B};F1: Point;LeftFocus(C) = F1;WhenMax(Perimeter(TriangleOf(A, B, F1)))",
    "query_expressions": "Slope(l)",
    "answer_expressions": "-1/2",
    "fact_spans": "[[[1, 10]], [[1, 10]], [[11, 16], [113, 118]], [[0, 16]], [[17, 59], [78, 80]], [[17, 59]], [[60, 63]], [[64, 67]], [[11, 69]], [[70, 77]], [[70, 84]], [[85, 112]]]",
    "query_spans": "[[[113, 123]]]",
    "process": "From the given conditions, the left and right foci are respectively $F_{1}(-2,0)$, $F_{2}(2,0)$, then $AF_{1}=6-AF_{2}$, $BF_{1}=6-BF_{2}$. $\\therefore$ The perimeter $L$ of $\\Delta ABF_{1}$ is $L=AF_{1}+BF_{1}+AB=12+AB-(AF_{2}+BF_{2})$. Also, since $AF_{2}+BF_{2}\\geqslant AB$, with equality if and only if $A$, $B$, $F_{2}$ are collinear, $\\therefore L_{\\max}=12$, i.e., the line $l$ passes through $M(0,1)$, $F_{2}(2,0)$, so $k_{1}=-\\frac{1}{2}$."
  },
  {
    "text": "Given that a line $l$ with slope $k$ intersects the hyperbola $C$: $\\frac{x^{2}}{2}-\\frac{y^{2}}{3}=1$ at points $A$ and $B$, if the midpoint of segment $AB$ is $M(2,1)$, then the value of $k$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;k:Number;B: Point;A: Point;M: Point;Expression(C) = (x^2/2 - y^2/3 = 1);Coordinate(M) = (2, 1);Slope(l)=k;Intersection(l,C) = {A, B};MidPoint(LineSegmentOf(A,B))=M",
    "query_expressions": "k",
    "answer_expressions": "3",
    "fact_spans": "[[[9, 14]], [[15, 58]], [[5, 8], [92, 95]], [[65, 68]], [[61, 64]], [[82, 90]], [[15, 58]], [[82, 90]], [[2, 14]], [[9, 70]], [[71, 90]]]",
    "query_spans": "[[[92, 99]]]",
    "process": "Let the equation of the line passing through point M(2,1) be y = k(x - 2) + 1 or x = 2; when k exists, combining gives (3 - 2k^{2})x^{2} + (8k^{2} - 4k)x - 8k^{2} + 8k - 8 = 0. When the line intersects the hyperbola at two distinct points, it must have _{\\triangle} = (8k^{2} - 4k)^{2} - 4(3 - 2k^{2})(-8k^{2} + 8k - 8) > 0. Also, the two distinct roots of the equation are the x-coordinates of intersection points A and B, and P is the midpoint of segment AB, \\therefore x_{1} + x_{2} = 4, i.e.: \\frac{(8k^{2} - 4k)}{3 + 2k^{2}} = 4 \\Rightarrow k = 3. Upon verification, satisfies \\triangle > 0. When x = 2, the chord's midpoint lies on the x-axis, which does not satisfy the condition. \\therefore k = 3, the final answer is 3"
  },
  {
    "text": "In triangle $\\triangle A B C$, $A B=2$ and $A C=2 B C$, then the maximum area of triangle $A B C$ is?",
    "fact_expressions": "A: Point;B: Point;C: Point;LineSegmentOf(A,B)=2;LineSegmentOf(A,C)=2*LineSegmentOf(B,C)",
    "query_expressions": "Max(Area(TriangleOf(A,B,C)))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[30, 41]], [[30, 41]], [[30, 41]], [[22, 29]], [[30, 41]]]",
    "query_spans": "[[[43, 61]]]",
    "process": ""
  },
  {
    "text": "Draw a line through the right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0, b>0)$ that is parallel to one of its asymptotes, intersecting $C$ at point $P$. If the horizontal coordinate of point $P$ is $2 a$, then what is the eccentricity of $C$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;G: Line;PointOnCurve(RightFocus(C), G);IsParallel(G, Asymptote(C));P: Point;Intersection(G, C) = P;XCoordinate(P) = 2*a",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2+sqrt(3)",
    "fact_spans": "[[[1, 64], [83, 86], [111, 114], [72, 73]], [[1, 64]], [[9, 64]], [[9, 64]], [[9, 64]], [[9, 64]], [[79, 81]], [[0, 81]], [[71, 81]], [[87, 91], [94, 98]], [[79, 91]], [[94, 108]]]",
    "query_spans": "[[[111, 120]]]",
    "process": "The right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $(c,0)$. Assume the line drawn is parallel to the asymptote $y=\\frac{b}{a}x$ of the hyperbola, with equation $y=\\frac{b}{a}(x-c)$. Substituting into $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, the x-coordinate of point $P$ is found to be $x=\\frac{a2+c^{2}}{2c}$. From $\\frac{a2+c^{2}}{2c}=2a$, we obtain $(\\frac{c}{a})^{2}-4\\frac{c}{a}+1=0$. Solving gives $\\frac{c}{a}=2+\\sqrt{3}$, $\\frac{c}{a}=2-\\sqrt{3}$ (discarded, since the eccentricity $\\frac{c}{a}>1$). Thus, the eccentricity of the hyperbola is $2+\\sqrt{3}$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. The circle with diameter $F_{1}F_{2}$ intersects the ellipse at point $P\\left(\\frac{3 \\sqrt{5}}{5}, \\frac{4 \\sqrt{5}}{5}\\right)$. Then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Circle;IsDiameter(LineSegmentOf(F1, F2), H);P: Point;Coordinate(P) = ((3*sqrt(5))/5, (4*sqrt(5))/5);Intersection(H, G) = P",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9+y^2/4=1",
    "fact_spans": "[[[2, 54], [102, 104], [155, 157]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[63, 70]], [[71, 78]], [[2, 78]], [[2, 78]], [[100, 101]], [[80, 101]], [[106, 153]], [[106, 153]], [[100, 153]]]",
    "query_spans": "[[[155, 162]]]",
    "process": "According to the problem, |PO| = \\sqrt{\\frac{9}{5}+\\frac{16}{5}} = \\sqrt{5} = c, hence F_{1}(-\\sqrt{5},0), F_{2}(\\sqrt{5},0). \\therefore |PF_{1}| + |PF_{2}| = \\sqrt{(-\\sqrt{5}-\\frac{3\\sqrt{5}}{5})^{2}+(\\frac{4\\sqrt{5}}{5})^{2}} + \\sqrt{(\\sqrt{5}-\\frac{3\\sqrt{5}}{5})^{2}} + (\\frac{4\\sqrt{5}}{5})^{2} = 4 + 2 = 6 = 2a, a = 3, b = \\sqrt{a^{2}-c^{2}} = . Therefore, the equation of the ellipse is \\frac{x^{2}}{9} + \\frac{y^{2}}{4} = 1"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ with foci $F_{1}$ and $F_{2}$, and a point $P(x_{0}, y_{0})$ satisfying $0<\\frac{x_{0}^{2}}{4}+\\frac{y_{0}^{2}}{3}<1$, then the range of $|P F_{1}|+|P F_{2}|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Coordinate(P) = (x0, y0);Focus(G)={F1,F2};x0:Number;y0:Number;0<x0^2/4+y0^2/3;x0^2/4+y0^2/3<1",
    "query_expressions": "Range(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)))",
    "answer_expressions": "[2,4)",
    "fact_spans": "[[[2, 39]], [[60, 78]], [[44, 51]], [[52, 59]], [[2, 39]], [[60, 78]], [[2, 59]], [[61, 78]], [[61, 78]], [[80, 125]], [[80, 125]]]",
    "query_spans": "[[[127, 155]]]",
    "process": "From the given conditions, a=2, |F_{1}F_{2}|=2c=2. Since 0<\\frac{x^{2}}{4}+\\frac{y^{2}}{3}<1, the point P(x_{0},y_{0}) lies inside the ellipse and does not coincide with the origin O. Therefore, |F_{1}F_{2}|\\leqslant|PF_{1}|+|PF_{2}|<2a, so 2\\leqslant|PF_{1}|+|PF_{2}|<4, hence |PF_{1}|+|PF_{2}|\\in[2,4)."
  },
  {
    "text": "An ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{4}=1$ has a focus at $(0,1)$, then $m$=?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (y^2/4 + x^2/m = 1);Coordinate(OneOf(Focus(G))) = (0, 1)",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 37]], [[52, 55]], [[0, 37]], [[0, 50]]]",
    "query_spans": "[[[52, 57]]]",
    "process": "From the problem, we have 4 - m = 1, solve for m. \\because one focus is (0,1), \\therefore the focus lies on the y-axis \\therefore 4 - m = 1, solving gives m = 3."
  },
  {
    "text": "What is the length of the major axis of the ellipse $x^{2}+4 y^{2}=16$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 4*y^2 = 16)",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "8",
    "fact_spans": "[[[0, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 26]]]",
    "process": "From the given problem, the standard equation of the ellipse is \\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1, then a^{2}=16, so a=4, thus the length of the major axis is 8."
  },
  {
    "text": "From a moving point $M$ on the ellipse $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1\\ (a>b>0)$, draw two tangent lines to the circle $x^{2}+y^{2}=\\frac{b^{2}}{2}$, with points of tangency $P$ and $Q$. Let the line $PQ$ intersect the $x$-axis and $y$-axis at points $E$ and $F$, respectively. Then the minimum area of $\\Delta EOF$ is?",
    "fact_expressions": "D: Ellipse;a: Number;b: Number;a > b;b > 0;Expression(D) = (x^2/b^2 + y^2/a^2 = 1);M: Point;PointOnCurve(M, D);C: Circle;Expression(C) = (x^2 + y^2 = b^2/2);l1, l2: Line;TangentOfPoint(M, C) = {l1, l2};P: Point;Q: Point;TangentPoint(l1, C) = P;TangentPoint(l2, C) = Q;E: Point;F: Point;Intersection(LineOf(P, Q), xAxis) = E;Intersection(LineOf(P, Q), yAxis) = F;O: Origin",
    "query_expressions": "Min(Area(TriangleOf(E, O, F)))",
    "answer_expressions": "b^3/(4*a)",
    "fact_spans": "[[[1, 55]], [[3, 55]], [[3, 55]], [[3, 55]], [[3, 55]], [[1, 55]], [[59, 62]], [[1, 62]], [[63, 93]], [[63, 93]], [[96, 98]], [[0, 98]], [[102, 105]], [[106, 109]], [[0, 109]], [[0, 109]], [[133, 136]], [[137, 140]], [[110, 140]], [[110, 140]], [[142, 156]]]",
    "query_spans": "[[[142, 164]]]",
    "process": "Let M(x_{0},y_{0}), P(x_{1},y_{1}), Q(x_{2},y_{2}), and let the equations of lines MP and MQ be x_{1}x + y_{1}y = \\frac{b^{2}}{2}, x_{2}x + y_{2}y = \\frac{b^{2}}{2}. Since point M lies on MP and MQ, we have x_{1}x_{0} + y_{1}y_{0} = \\frac{b^{2}}{2}, x_{2}x_{0} + y_{2}y_{0} = \\frac{b^{2}}{2}. It follows that the coordinates of points P and Q satisfy the equation x_{0}x + y_{0}y = \\frac{b^{2}}{2}, so the equation of line PQ is x_{0}x + y_{0}y = \\frac{b^{2}}{2}. Thus, the intersections of line PQ with the x-axis and y-axis are E(\\frac{b^{2}}{2x_{0}}, 0) and F(0, \\frac{b^{2}}{2y_{0}}), respectively. Therefore, the area of AEOF is S_{AEOF} = \\frac{1}{2}|OE||OF| = \\frac{b^{4}}{8|x_{0}y_{0}|}. Since b^{2}y_{0}^{2} + a^{2}x_{0}^{2} = a^{2}b^{2}, and b^{2}x_{0}^{2} + a^{2}y_{0}^{2} \\geqslant 2ab|x_{0}y_{0}|, it follows that |x_{0}y_{0}| \\leqslant \\frac{ab}{2}. Hence, S_{AEOF} = \\frac{b^{4}}{8|x_{0}y_{0}|} \\geqslant \\frac{b^{3}}{4a}, with equality if and only if b^{2}y_{0}^{2} = a^{2}x_{0}^{2} = \\frac{a^{2}b^{2}}{2}, when the area of AEOF attains the minimum value \\frac{b^{3}}{4a}."
  },
  {
    "text": "Given the parabola $y^{2}=8 x$, $F$ is its focus, and $P$ is any point on the parabola. Then the equation of the locus of the midpoint of the segment $P F$ is?",
    "fact_expressions": "G: Parabola;F: Point;P: Point;Expression(G) = (y^2 = 8*x);Focus(G)=F;PointOnCurve(P, G)",
    "query_expressions": "LocusEquation(MidPoint(LineSegmentOf(P,F)))",
    "answer_expressions": "y^2=4*x-4",
    "fact_spans": "[[[2, 16], [23, 24], [31, 34]], [[19, 22]], [[27, 30]], [[2, 16]], [[19, 26]], [[27, 39]]]",
    "query_spans": "[[[41, 57]]]",
    "process": ""
  },
  {
    "text": "The ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$ has left and right foci $F_{1}$ and $F_{2}$, respectively. Point $P$ lies on ellipse $C$, and it is given that $|P F_{1}|=3$. Then $|P F_{2}|=$?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/4 + y^2/2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 3",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 42], [72, 77]], [[67, 71]], [[51, 58]], [[59, 66]], [[0, 42]], [[0, 66]], [[0, 66]], [[67, 78]], [[81, 94]]]",
    "query_spans": "[[[96, 109]]]",
    "process": "According to the definition of an ellipse, given the ellipse $ C: \\frac{x^{2}}{4} + \\frac{y^{2}}{2} = 1 $, then $ a^{2} = 4 $, so $ a = 2 $. By the definition of an ellipse, we have $ |PF_{1}| + |PF_{2}| = 2a = 4 $. Since $ |PF_{1}| = 3 $, therefore $ |PF_{2}| = 1 $."
  },
  {
    "text": "If the line passing through the two points $A(a, 0)$ and $B(0, a)$ does not intersect the parabola $y=x^{2}-2 x-3$, then what is the range of real values for $a$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;a: Real;Expression(G) = (y = x^2 - 2*x - 3);Coordinate(A) = (a, 0);Coordinate(B) = (0, a);PointOnCurve(A,H);PointOnCurve(B,H);NumIntersection(H,G)=0",
    "query_expressions": "Range(a)",
    "answer_expressions": "(-\\infty,-13/4)",
    "fact_spans": "[[[28, 46]], [[25, 27]], [[5, 14]], [[15, 24]], [[53, 58]], [[28, 46]], [[5, 14]], [[15, 24]], [[2, 27]], [[2, 27]], [[25, 50]]]",
    "query_spans": "[[[53, 65]]]",
    "process": "Mainly examines the positional relationship between a line and a parabola, and the application of transformation and reduction thinking. The equation of the line passing through AB is $ y = -x + a $. Substituting into $ y = x^{2} - 2x - 3 $ gives: $ -x + a = x^{2} - 2x - 3 $, so $ x^{2} - x - 3 - a = 0 $. Since the line and the parabola have no intersection points, the equation has no solution, meaning the discriminant $ = 1 - 4(-3 - a) < 0 $, thus $ a < -\\frac{13}{4} $. Extension of thought: Transforming the study of the positional relationship between a line and a parabola into determining the number of solutions of a quadratic equation in one variable is a typical example of \"algebraization\" of geometric problems."
  },
  {
    "text": "What is the minimum distance from a point $P(x, y)$ on the parabola $y^{2}=x$ to the point $\\left(\\frac{9}{4}, 0\\right)$?",
    "fact_expressions": "G: Parabola;H: Point;P: Point;Expression(G) = (y^2 = x);Coordinate(H) = (9/4, 0);Coordinate(P) = (x1, y1);x1:Number;y1:Number;PointOnCurve(P,G)",
    "query_expressions": "Min(Distance(P, H))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 12]], [[24, 43]], [[13, 23]], [[0, 12]], [[24, 43]], [[13, 23]], [[14, 23]], [[14, 23]], [[0, 23]]]",
    "query_spans": "[[[13, 51]]]",
    "process": ""
  },
  {
    "text": "If point $M(-2,8)$ lies on the directrix of the parabola $y^{2}=2 p x$, then the value of the real number $p$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (-2, 8);G: Parabola;Expression(G) = (y^2 = 2*p*x);PointOnCurve(M, Directrix(G));p: Real",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 11]], [[1, 11]], [[12, 28]], [[12, 28]], [[1, 32]], [[34, 39]]]",
    "query_spans": "[[[34, 43]]]",
    "process": "The equation of the directrix of the parabola is $x = -\\frac{p}{2} = -2$, so $p = 4$."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, and a line passing through the point $(2,0)$ intersects the parabola at points $A$ and $B$, then the minimum value of $\\frac{1}{|AF|}+\\frac{1}{|BF|}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;I: Point;Coordinate(I) = (2, 0);H: Line;PointOnCurve(I, H) = True;Intersection(H, G) = {A, B};A: Point;B: Point",
    "query_expressions": "Min(1/Abs(LineSegmentOf(B, F)) + 1/Abs(LineSegmentOf(A, F)))",
    "answer_expressions": "2/3",
    "fact_spans": "[[[2, 16], [37, 40]], [[2, 16]], [[20, 23]], [[2, 23]], [[25, 33]], [[25, 33]], [[34, 36]], [[24, 36]], [[34, 50]], [[41, 44]], [[45, 48]]]",
    "query_spans": "[[[52, 91]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), then \\frac{1}{|AF|}+\\frac{1}{|BF|}=\\frac{1}{x_{1}+} Let the line AB: x=my+2 (m\\neq0). Solving the system \\begin{cases}x=my+2\\\\y^{2}=4x\\end{cases}, eliminating x and simplifying yields y^{2}-4my-8=0, \\triangle=16m^{2}+32>0, y_{1}+y_{2}=4m, y_{1}y_{2}=-8, so x_{1}+x_{2}=my_{1}+2+my_{2}+2=m(y_{1}+y_{2})+4=4m^{2}+4, \\frac{(y_{1}+y_{2})^{2}-2y_{1}y_{2}}{4}= \\frac{(4m)^{2}-2(-8)}{4}= \\frac{16m^{2}+16}{4}=4m^{2}+4, x_{1}x_{2}=(my_{1}+2)(my_{2}+2)=m^{2}y_{1}y_{2}+2m(y_{1}+y_{2})+4=m^{2}(-8)+2m(4m)+4=-8m^{2}+8m^{2}+4=4, thus \\frac{1}{|AF|}+\\frac{1}{|BF|}=\\frac{x_{1}+x_{2}+2}{x_{1}x_{2}+x_{1}+x_{2}+1}=\\frac{4m^{2}+4+2}{4+4m^{2}+4}=\\frac{4m^{2}+6}{4m^{2}+8}=\\frac{2(2m^{2}+3)}{4(m^{2}+2)}=\\frac{2m^{2}+3}{2(m^{2}+2)}, but correction from original: \\frac{4m^{2}+6}{4m^{2}+9}, then \\frac{6}{5+3}=\\frac{1}{1+\\frac{3}{4m^{2}+}} \\frac{1}{1+\\frac{3}{6}}=\\frac{2}{3}, equality holds if and only if m=0, therefore the minimum value of \\frac{1}{|AF|}+\\frac{1}{|BF|} is \\frac{2}{3}."
  },
  {
    "text": "Given the equation of the ellipse is $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{25}=1$ $(a>5)$, its two foci are $F_{1}$ and $F_{2}$, and $|F_{1} F_{2}|=8$. A chord $A B$ (a line segment between any two points on the ellipse) passes through point $F_{1}$. Then, what is the perimeter of $\\triangle A B F_{2}$?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;F2: Point;F1: Point;a: Number;a > 5;Expression(G) = (y^2/25 + x^2/a^2 = 1);Abs(LineSegmentOf(F1, F2)) = 8;Focus(G) = {F1, F2};IsChordOf(LineSegmentOf(A, B), G);PointOnCurve(F1, LineSegmentOf(A, B))",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "4*sqrt(41)",
    "fact_spans": "[[[2, 4], [105, 108], [54, 55]], [[99, 104]], [[99, 104]], [[71, 78]], [[63, 70]], [[8, 53]], [[8, 53]], [[2, 53]], [[80, 97]], [[54, 78]], [[98, 117]], [[98, 126]]]",
    "query_spans": "[[[129, 155]]]",
    "process": ""
  },
  {
    "text": "If the distance from one focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ to one asymptote is equal to $\\frac{1}{4}$ of the focal distance, then what is the equation of the asymptotes of this hyperbola?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Distance(OneOf(Focus(G)),OneOf(Asymptote(G)))=(1/4)*FocalLength(G)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(3)/3)*x",
    "fact_spans": "[[[1, 57], [92, 95]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 89]]]",
    "query_spans": "[[[91, 103]]]",
    "process": "Using the point-to-line distance formula, calculate the distance from the focus to the asymptote, then solve for the value of $\\frac{b}{a}$ based on the condition that this distance equals $\\frac{1}{4}$ of the focal length, thus obtaining the asymptotic equations of the hyperbola. [Solution] Since the distance $d$ from the focus to the asymptote is $d = \\frac{|bc|}{\\sqrt{a^{2}+b^{2}}} = b$, it follows that $b = \\frac{1}{4} \\cdot 2c = \\frac{c}{2}$, so $c^{2} = 4b^{2} = a^{2} + b^{2}$, hence $\\frac{b}{a} = \\frac{\\sqrt{3}}{3}$, therefore the asymptotic equations are: $y = \\pm \\frac{\\sqrt{3}}{3}x$."
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $9 x^{2}-y^{2}=-1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (9*x^2 - y^2 = -1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "3*x+pm*y=0",
    "fact_spans": "[[[0, 21]], [[0, 21]]]",
    "query_spans": "[[[0, 29]]]",
    "process": ""
  },
  {
    "text": "The asymptotes of the hyperbola $x^{2}-y^{2}=m(m \\neq 0)$ are?",
    "fact_expressions": "G: Hyperbola;m: Number;Negation(m=0);Expression(G) = (x^2 - y^2 = m)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*x",
    "fact_spans": "[[[0, 28]], [[3, 28]], [[3, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 36]]]",
    "process": "Since the hyperbola equation is $x^{2}-y^{2}=m$ $(m\\neq0)$, let $m=0$, we get $x^{2}-y^{2}=0$, i.e., $y=\\pm x$."
  },
  {
    "text": "Given the ellipse $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$ $(a > b > 0)$ has eccentricity $\\frac{\\sqrt{2}}{2}$, and the three vertices of triangle $ABC$ lie on the ellipse. Let the midpoints of its three sides $AB$, $BC$, $AC$ be $D$, $E$, $F$, respectively, and the slopes of the lines containing the three sides be $k_1$, $k_2$, $k_3$ ($k_1 k_2 k_3 \\neq 0$). If the sum of the slopes of the lines $OD$, $OE$, $OF$ is $-1$ ($O$ being the origin), then $\\frac{1}{k_1} + \\frac{1}{k_2} + \\frac{1}{k_3} = $?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;Eccentricity(G) = sqrt(2)/2;A: Point;B: Point;C: Point;PointOnCurve(Vertex(TriangleOf(A, B, C)), G);D: Point;E: Point;F: Point;MidPoint(LineSegmentOf(A, B)) = D;MidPoint(LineSegmentOf(B, C)) = E;MidPoint(LineSegmentOf(A, C)) = F;k1: Number;k2: Number;k3: Number;Slope(OverlappingLine(LineSegmentOf(A, B))) = k1;Slope(OverlappingLine(LineSegmentOf(B, C))) = k2;Slope(OverlappingLine(LineSegmentOf(A, C))) = k3;Negation(k1*k2*k3=0);O: Origin;Slope(LineOf(O, D)) + Slope(LineOf(O, E)) + Slope(LineOf(O, F)) = -1",
    "query_expressions": "1/k1 + 1/k2 + 1/k3",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 50], [91, 93]], [[2, 50]], [[4, 50]], [[4, 50]], [[4, 50]], [[4, 50]], [[2, 75]], [[79, 84]], [[79, 84]], [[79, 84]], [[76, 94]], [[121, 124]], [[125, 128]], [[129, 132]], [[96, 132]], [[96, 132]], [[96, 132]], [[147, 154]], [[155, 162]], [[163, 170]], [[101, 170]], [[101, 170]], [[101, 170]], [[171, 197]], [[230, 233]], [[200, 229]]]",
    "query_spans": "[[[241, 292]]]",
    "process": ""
  },
  {
    "text": "When the line $k x - y + 3 = 0$ is tangent to the ellipse $\\frac{x^{2}}{16} + \\frac{y^{2}}{4} = 1$, $k = $?",
    "fact_expressions": "G: Ellipse;H: Line;k: Number;Expression(G) = (x^2/16 + y^2/4 = 1);Expression(H) = (k*x - y + 3 = 0);IsTangent(H,G)",
    "query_expressions": "k",
    "answer_expressions": "pm*sqrt(5)/4",
    "fact_spans": "[[[15, 53]], [[1, 14]], [[57, 60]], [[15, 53]], [[1, 14]], [[1, 55]]]",
    "query_spans": "[[[57, 62]]]",
    "process": "Solving the line equation together with the ellipse equation yields (1+4k^{2})x^{2}+24kx+20=0. \\therefore A=0. \\therefore (24k)^{2}-4(1+4k^{2})\\times20=0. \\therefore k=\\pm\\frac{\\sqrt{5}}{4}"
  },
  {
    "text": "$F$ is the focus of the parabola $C$: $y^{2}=4x$, $P$ is a point on $C$ located in the first quadrant, the projection of point $P$ onto the directrix of $C$ is $Q$, and $|PQ|=2$. Then the equation of the circumcircle of $\\triangle PQF$ is?",
    "fact_expressions": "C: Parabola;P: Point;Q: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;Quadrant(P) = 1;PointOnCurve(P, C);Projection(P, Directrix(C)) = Q;Abs(LineSegmentOf(P, Q)) = 2",
    "query_expressions": "Expression(CircumCircle(TriangleOf(P,Q,F)))",
    "answer_expressions": "x^2+(y-1)^2=2",
    "fact_spans": "[[[4, 23], [31, 34], [51, 54]], [[27, 30], [46, 50]], [[62, 65]], [[0, 3]], [[4, 23]], [[0, 26]], [[27, 45]], [[27, 45]], [[46, 65]], [[67, 76]]]",
    "query_spans": "[[[78, 103]]]",
    "process": "From the given information, it can be determined that $\\triangle FPQ$ is a right triangle, so the circumcenter of $\\triangle PQF$ is the midpoint of $FQ$. The equation of the circle can be written once the center and radius are found. From the parabola equation, the focus is $F(1,0)$ and the directrix is $x = -1$. Since $|PQ| = 2$, we have $x_{P} + 1 = 2$, so $x_{P} = 1$, then $y_{P} = 2$. Therefore, $P(1,2)$, $Q(-1,2)$. Hence, $FP \\perp PQ$, so $\\triangle FPQ$ is a right triangle. Thus, the circumcenter of $\\triangle PQF$ is the midpoint of $FQ$, i.e., the center is $(0,1)$, and the radius is $\\frac{1}{2}|FQ| = \\sqrt{2}$. Therefore, the equation of the circumcircle of $\\triangle PQF$ is $x^{2} + (y-1)^{2} = 2$."
  },
  {
    "text": "Given the parabola $E$: $y^{2}=4x$ with focus $F$ and directrix $l$, where $l$ intersects the $x$-axis at point $T$. Let $A$ be a point on $E$, $AA_{1}$ perpendicular to $l$ with foot of perpendicular $A_{1}$, and $A_{1}F$ intersecting the $y$-axis at point $S$. If $ST \\parallel AF$, then $|AF|=$?",
    "fact_expressions": "E: Parabola;A: Point;A1: Point;F: Point;S: Point;T: Point;l: Line;Expression(E) = (y^2 = 4*x);Focus(E) = F;Directrix(E) = l;Intersection(l, xAxis) = T;PointOnCurve(A, E);IsPerpendicular(LineSegmentOf(A, A1), l);FootPoint(LineSegmentOf(A, A1), l) = A1;Intersection(LineSegmentOf(A1, F), yAxis) = S;IsParallel(LineSegmentOf(S, T), LineSegmentOf(A, F))",
    "query_expressions": "Abs(LineSegmentOf(A, F))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 21], [56, 59]], [[52, 55]], [[82, 89]], [[25, 28]], [[107, 111]], [[47, 51]], [[32, 35], [38, 41], [75, 78]], [[2, 21]], [[2, 28]], [[2, 36]], [[38, 51]], [[52, 62]], [[63, 78]], [[63, 89]], [[92, 111]], [[113, 125]]]",
    "query_spans": "[[[127, 136]]]",
    "process": "Let A(x_{1},y_{1}), A_{1}(-1,y_{1}), F(1,0) K_{A_{1}F}=-\\frac{y_{1}}{2}_{y_{1}} l_{A_{1}F}: y hence s(0,\\frac{y_{1}}{2}) K_{ST}=\\frac{y_{1}^{2}}{2}=K_{AF}=\\frac{y_{1}}{x_{1}-1} \\therefore x_{1}=3, then |AF|=x_{1}+1=4"
  },
  {
    "text": "Given that the fixed point $F$ through which the line $(1+4k)x - (2-3k)y - (3+12k) = 0$ $(k \\in \\mathbb{R})$ passes is exactly one focus of the ellipse $C$, and the maximum distance from any point on the ellipse $C$ to the point $F$ is $8$. Then the standard equation of the ellipse $C$ is?",
    "fact_expressions": "G: Line;Expression(G)=((1+4*k)*x - (2-3*k)*y - (3+12*k) = 0);k: Real;F: Point;OneOf(Focus(C))=F;PointOnCurve(F,G);C: Ellipse;P: Point;PointOnCurve(P,C);Max(Distance(P,F)) = 8",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/25+y^2/16=1",
    "fact_spans": "[[[2, 45]], [[2, 45]], [[4, 45]], [[51, 54], [79, 83]], [[51, 67]], [[2, 54]], [[57, 62], [70, 75], [95, 100]], [[77, 78]], [[70, 78]], [[77, 92]]]",
    "query_spans": "[[[95, 107]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has a major axis length of $4$ and an eccentricity of $\\frac{\\sqrt{3}}{2}$, then the equation of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Length(MajorAxis(C)) = 4;Eccentricity(C) = sqrt(3)/2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "{x^2/4+y^2=1, y^2/4+x^2=1}",
    "fact_spans": "[[[2, 62], [97, 102]], [[2, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[2, 70]], [[2, 95]]]",
    "query_spans": "[[[97, 107]]]",
    "process": "If the foci of ellipse C lie on the x-axis, then $a=2$, $\\frac{c}{a}=\\frac{\\sqrt{3}}{2}$, so $c=\\sqrt{3}$, $b=\\sqrt{a^{2}-c^{2}}=1$; in this case, the equation of ellipse C is $\\frac{x^{2}}{4}+y^{2}=1$. If the foci of ellipse C lie on the y-axis, then $b=2$, $\\frac{c}{b}=\\frac{\\sqrt{3}}{2}$, so $c=\\sqrt{3}$, $a=\\sqrt{b^{2}-c^{2}}=1$; in this case, the equation of ellipse C is $\\frac{y^{2}}{4}+x^{2}=1$."
  },
  {
    "text": "It is known that the foci of ellipse $C$ lie on the coordinate axes, and the ellipse passes through points $A(-\\sqrt{3},-2)$ and $B(-2 \\sqrt{3}, 1)$. What is the standard equation of ellipse $C$?",
    "fact_expressions": "C: Ellipse;PointOnCurve(Focus(C), axis) = True;A: Point;Coordinate(A) = (-sqrt(3), -2);B: Point;Coordinate(B) = (-2*sqrt(3), 1);PointOnCurve(A, C) = True;PointOnCurve(B, C) = True",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/15 + y^2/5 = 1",
    "fact_spans": "[[[2, 7], [60, 65]], [[2, 15]], [[19, 36]], [[19, 36]], [[37, 56]], [[37, 56]], [[2, 58]], [[2, 58]]]",
    "query_spans": "[[[60, 72]]]",
    "process": "Let the desired ellipse equation be: $mx^{2}+ny^{2}=1$ $(m>0,n>0,m\\neq n)$. Substituting the coordinates of points A and B into the equation gives:  \n$$\n\\begin{cases}\n3m+4n=1 \\\\\n12m+n=1\n\\end{cases}\n$$\nSolving yields  \n$$\n\\begin{cases}\nm=\\frac{1}{15} \\\\\nn=\\frac{1}{6}\n\\end{cases}\n$$\nThe standard equation of the desired ellipse is: $\\frac{x^{2}}{4}+\\frac{y^{2}}{4}=1$"
  },
  {
    "text": "Given the parabola $T$: $y^{2}=2 p x$ ($p>0$), the chord length intercepted on the directrix of this parabola by the circle $C$: $x^{2}+y^{2}-4 y-4=0$ is $4$. Then, the equation of the parabola $T$ is?",
    "fact_expressions": "T: Parabola;p: Number;C: Circle;p>0;Expression(T) = (y^2 = 2*(p*x));Expression(C) = (-4*y + x^2 + y^2 - 4 = 0);Length(InterceptChord(Directrix(T), C)) = 4",
    "query_expressions": "Expression(T)",
    "answer_expressions": "y^2 = 8*x",
    "fact_spans": "[[[2, 28], [70, 76]], [[10, 28]], [[32, 59]], [[10, 28]], [[2, 28]], [[32, 59]], [[2, 68]]]",
    "query_spans": "[[[70, 81]]]",
    "process": "Convert the equation of circle C into standard form: $x^{2}+(y-2)^{2}=8$, then the center of the circle is $C(0,2)$ and the radius is $r=2\\sqrt{2}$. Since the directrix equation of parabola T is $x=-\\frac{p}{2}$, the distance from the center to the directrix is $d=\\frac{p}{2}$. Given the radius of circle C is $r=2\\sqrt{2}$ and the chord length is 4, we have $\\left(\\frac{p}{2}\\right)^{2}+2^{2}=(2\\sqrt{2})^{2}$, solving gives $p=4$. Thus, the equation of parabola T is $y^{2}=8x$."
  },
  {
    "text": "The focus of the parabola $x^{2}=4 y$ is $F$, $P$ is a point on the parabola, and $O$ is the origin. The circumcircle of $\\triangle O P F$ is tangent to the directrix of the parabola. Then the radius of this circumcircle is?",
    "fact_expressions": "G: Parabola;O: Origin;P: Point;F: Point;Expression(G) = (x^2 = 4*y);Focus(G)=F;PointOnCurve(P,G);IsTangent(CircumCircle(TriangleOf(O,P,F)),Directrix(G))",
    "query_expressions": "Radius(CircumCircle(TriangleOf(O,P,F)))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[0, 14], [26, 29], [64, 67]], [[33, 36]], [[22, 25]], [[18, 21]], [[0, 14]], [[0, 21]], [[22, 32]], [[42, 72]]]",
    "query_spans": "[[[42, 83]]]",
    "process": "Find the focus and directrix equation of the parabola. Since the circumcenter lies on the perpendicular bisector of segment OF, the y-coordinate of the center is \\frac{1}{2}. Then, using the condition for tangency between a line and a circle: d=r, the required radius can be calculated. Detailed solution: The focus of the parabola x^{2}=4y is F(0,1), and the directrix equation of the parabola is y=-1. Let the circumcenter C of \\triangle OPF be (m,n) with radius r. Since C lies on the perpendicular bisector of segment OF, we have n=\\frac{1}{2}. Since the circumcircle is tangent to the directrix, we obtain n+1=r, thus r=\\frac{3}{2}."
  },
  {
    "text": "The standard equation of an ellipse with eccentricity $e=\\frac{1}{2}$ and a focus at $F(0,-3)$ is?",
    "fact_expressions": "G: Ellipse;e: Number;Eccentricity(G) = e;e = 1/2;F: Point;OneOf(Focus(G)) = F;Coordinate(F) = (0, -3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/27+y^2/36=1",
    "fact_spans": "[[[36, 38]], [[3, 18]], [[0, 38]], [[3, 18]], [[26, 35]], [[21, 38]], [[26, 35]]]",
    "query_spans": "[[[36, 44]]]",
    "process": ""
  },
  {
    "text": "Given a line with slope $-\\frac{1}{3}$ intersecting the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{7}=1$ at two distinct points $A$ and $B$, and a point $M$ on the $y$-axis satisfying $|M A|=|M B|$, what is the range of possible values for the $y$-coordinate of point $M$?",
    "fact_expressions": "G: Ellipse;H: Line;M: Point;A: Point;B: Point;Expression(G) = (x^2/9 + y^2/7 = 1);Slope(H)= -1/3;Intersection(H,G)={A,B};Negation(A=B);PointOnCurve(M,yAxis);Abs(LineSegmentOf(M,A))=Abs(LineSegmentOf(M,B))",
    "query_expressions": "Range(YCoordinate(M))",
    "answer_expressions": "(-sqrt(2)/2, sqrt(2)/2)",
    "fact_spans": "[[[23, 60]], [[20, 22]], [[76, 79], [105, 109]], [[68, 71]], [[72, 75]], [[23, 60]], [[2, 22]], [[20, 75]], [[63, 75]], [[76, 87]], [[90, 103]]]",
    "query_spans": "[[[105, 120]]]",
    "process": "Let the equation of line AB be $ y = -\\frac{1}{3}x $. From  \n$$\n\\begin{cases}\ny = -\\frac{1}{3}x + t \\\\\n\\frac{x^{2}}{9} + \\frac{y^{2}}{7} = 1\n\\end{cases}\n$$  \neliminate $ y $ and simplify to obtain $ 8x^{2} - 6tx + 9t^{2} - 63 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then  \n$ x_{1} + x_{2} = \\frac{3}{4}t $, $ x_{1} \\cdot x_{2} = \\frac{9t^{2} - 63}{8} $.  \n$ \\Delta = 36t^{2} - 32(9t^{2} - 63) > 0 $, solving gives $ -2\\sqrt{2} < t < 2\\sqrt{2} $.  \n$ \\frac{x_{1} + x_{2}}{2} = \\frac{3}{8}t $,  \n$ \\frac{y_{1} + y_{2}}{2} = \\frac{-\\frac{1}{3}(x_{1} + x_{2}) + 2t}{2} = -\\frac{1}{6}(x_{1} + x_{2}) + t = -\\frac{1}{6} \\times \\frac{3}{4}t + t = \\frac{7}{8}t $.  \nThe equation of the perpendicular bisector of AB is $ y - \\frac{7}{8}t = 3(x - \\frac{3}{8}t) $.  \nLet $ x = 0 $, we get $ y = -\\frac{1}{4}t $, that is, the range of the ordinate of point M is $ \\left( -\\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{2}}{2} \\right) $."
  },
  {
    "text": "Given the parabola $C$ with equation $y^{2}=2 p x  (p>0)$, a line segment $AB$ of length $4p$ has its endpoints $A$ and $B$ moving on the parabola $C$. Then, the minimum distance from the midpoint $M$ of segment $AB$ to the $y$-axis is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;A: Point;B: Point;Length(LineSegmentOf(A, B)) = 4*p;Endpoint(LineSegmentOf(A, B)) = {A, B};PointOnCurve(A, C);PointOnCurve(B, C);M: Point;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Min(Distance(M, yAxis))",
    "answer_expressions": "(3/2)*p",
    "fact_spans": "[[[2, 8], [64, 70]], [[2, 32]], [[12, 32]], [[12, 32]], [[56, 59]], [[60, 63]], [[35, 51]], [[44, 63]], [[56, 71]], [[60, 71]], [[85, 88]], [[75, 88]]]",
    "query_spans": "[[[85, 101]]]",
    "process": "l: x = -\\frac{p}{2}, respectively draw AC \\bot l, BD \\bot l, MH \\bot l from A, B, M, with feet of perpendiculars C, D, H. To find the minimum distance from M to the y-axis, first find the minimum distance d from M to the directrix of the parabola using the definition of parabola, then the desired distance is d - \\frac{p}{2}. [Solution] From the given conditions, the directrix of the parabola is l: x = -\\frac{p}{2}. Draw AC \\bot l, BD \\bot l, MH \\bot l from A, B, M, with feet of perpendiculars C, D, H. In the right trapezoid ABDC, MH = \\underline{AC + BD}. By the definition of parabola, AC = AF, BD = BF (where F is the focus of the parabola). Thus, MH = \\frac{AF + BF}{2} \\geqslant \\frac{AB}{2} = 2p. Therefore, the minimum distance from the midpoint M of AB to the directrix of the parabola is 2p. Hence, the shortest distance from the midpoint M of segment AB to the y-axis is 2p - \\frac{p}{2} = \\frac{3p}{2}."
  },
  {
    "text": "If the eccentricity of the ellipse $x^{2}+\\frac{y^{2}}{m}=1$ is $\\frac{\\sqrt{3}}{2}$, then what is the value of $m$?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2 + y^2/m = 1);Eccentricity(G) = sqrt(3)/2",
    "query_expressions": "m",
    "answer_expressions": "{4,1/4}",
    "fact_spans": "[[[1, 28]], [[55, 58]], [[1, 28]], [[1, 53]]]",
    "query_spans": "[[[55, 62]]]",
    "process": ""
  },
  {
    "text": "Let the ellipse be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{3}=1$ $(a>\\sqrt{3})$, with right focus $F$ and right vertex $A$. Given that $\\frac{1}{|O F|}+\\frac{1}{|O A|}=\\frac{3 e}{|F A|}$, where $O$ is the coordinate origin and $e$ is the eccentricity of the ellipse, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/3 + x^2/a^2 = 1);a: Number;a>sqrt(3);F: Point;RightFocus(G) = F;A: Point;RightVertex(G) = A;e: Number;1/Abs(LineSegmentOf(O, F)) + 1/Abs(LineSegmentOf(O, A)) = (3*e)/Abs(LineSegmentOf(F, A));O: Origin;Eccentricity(G) = e",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4+y^2/3=1",
    "fact_spans": "[[[1, 54], [140, 142], [148, 150]], [[1, 54]], [[3, 54]], [[3, 54]], [[59, 62]], [[1, 62]], [[67, 70]], [[1, 70]], [[136, 139]], [[73, 124]], [[127, 130]], [[136, 146]]]",
    "query_spans": "[[[148, 155]]]",
    "process": "From $\\frac{1}{|OF|}+\\frac{1}{|OA|}=\\frac{3e}{|FA|}$, we get $\\frac{1}{c}+\\frac{1}{a}=\\frac{3c}{a(a-c)}$, simplifying yields $a^{2}=4c^{2}$. Also $a^{2}-c^{2}=b^{2}=3$, so $c^{2}=1$, hence $a^{2}=4$, therefore the equation of the ellipse is $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$."
  },
  {
    "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ have left and right foci $F_{1}$ and $F_{2}$, respectively. If there exists a point $P$ on the right branch of the hyperbola such that $|P F_{1}|=2|P F_{2}|$, then the range of the eccentricity $e$ of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;e: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, RightPart(C));Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2));Eccentricity(C) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "(1, 3]",
    "fact_spans": "[[[1, 62], [89, 92], [130, 136]], [[8, 62]], [[8, 62]], [[100, 103]], [[71, 78]], [[79, 86]], [[140, 143]], [[8, 62]], [[8, 62]], [[1, 62]], [[1, 86]], [[1, 86]], [[89, 103]], [[106, 128]], [[130, 143]]]",
    "query_spans": "[[[140, 150]]]",
    "process": "Using the definition of the hyperbola, we obtain |PF_{2}| = 2a. Then, from |PF_{2}| \\geqslant c - a and e > 1, we can find the range of the eccentricity e of hyperbola C. [Solution] By the definition of the hyperbola, |PF_{1}| - |PF_{2}| = 2|PF_{2}| - |PF_{2}| = |PF_{2}| = 2a. Since |PF_{2}| = 2a \\geqslant c - a, it follows that c \\leqslant 3a. Because e > 1, we have 1 < e \\leqslant 3. Therefore, the range of the eccentricity e of hyperbola C is (1, 3]."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, from the right focus $F$ of hyperbola $C$, draw a perpendicular to an asymptote of $C$, with foot of perpendicular at $M$. Extend $F M$ to intersect the $y$-axis at point $P$, and $|F M|=4|P M|$. Then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;F: Point;M: Point;P: Point;L:Line;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C)=F;PointOnCurve(F,L);IsPerpendicular(Asymptote(C),L);FootPoint(Asymptote(C),L)=M;Intersection(OverlappingLine(LineSegmentOf(F,M)),yAxis)=P;Abs(LineSegmentOf(F, M)) = 4*Abs(LineSegmentOf(P, M))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 63], [65, 71], [79, 82], [133, 139]], [[10, 63]], [[10, 63]], [[75, 78]], [[93, 96]], [[111, 115]], [], [[10, 63]], [[10, 63]], [[2, 63]], [[65, 78]], [[64, 89]], [[64, 89]], [[64, 96]], [[97, 115]], [[117, 131]]]",
    "query_spans": "[[[133, 145]]]",
    "process": "The hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) has asymptotes given by $ y = \\frac{b}{a}x $. The right focus is $ F(c, 0) $. The line passing through $ F $ and perpendicular to the asymptote is $ y = -\\frac{a}{b}(x - c) $. Solving the system  \n$$\n\\begin{cases}\ny = \\frac{b}{a}x \\\\\ny = -\\frac{a}{b}(x - c)\n\\end{cases}\n$$\ngives: $ x_{M} = \\frac{a^{2}}{c}, y_{M} = \\frac{ab}{c} $. In $ y = -\\frac{a}{b}(x - c) $, setting $ x = 0 $ gives: $ y_{p} = \\frac{ac}{b} $. Since $ |FM| = 4|PM| $, it follows that $ \\overrightarrow{FM} = 4\\overrightarrow{MP} $. Therefore,  \n$ \\frac{a^{2}}{c} - c = 4\\left(0 - \\frac{a^{2}}{c}\\right) $.  \nRearranging yields: $ 5a^{2} = c^{2} $, so $ e^{2} = 5 $, hence $ e = \\sqrt{5} $. Thus, the eccentricity of hyperbola $ C $ is $ \\sqrt{5} $."
  },
  {
    "text": "Given a point $M$ on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$ such that the distance from $M$ to the left focus $F_{1}$ is $6$, and $N$ is the midpoint of $M F_{1}$, then $|O N|$=?",
    "fact_expressions": "G: Ellipse;M: Point;F1: Point;O: Origin;N: Point;Expression(G) = (x^2/16 + y^2/12 = 1);PointOnCurve(M,G);LeftFocus(G)=F1;Distance(M,F1)=6;MidPoint(LineSegmentOf(M,F1))=N",
    "query_expressions": "Abs(LineSegmentOf(O, N))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 41]], [[44, 47]], [[51, 58]], [[85, 92]], [[67, 70]], [[2, 41]], [[2, 47]], [[2, 58]], [[44, 66]], [[67, 83]]]",
    "query_spans": "[[[85, 94]]]",
    "process": "According to the problem, $ a=4 $, $ b=2\\sqrt{3} $, $ c=2 $. Let $ F_{2} $ be the right focus of the ellipse, then $ |MF_{2}| = 2a - |MF_{1}| = 2 $. Since $ N $ is the midpoint of $ MF_{1} $ and $ O $ is the midpoint of $ F_{1}F_{2} $, it follows that $ ON \\parallel MF_{2} $ and $ |ON| = \\frac{1}{2}|MF_{2}| = 1 $."
  },
  {
    "text": "If two vertices of $\\triangle A B C$ are $B(0,-3)$, $C(0,3)$, and the perimeter is $16$, then the trajectory equation of the third vertex $A$ is?",
    "fact_expressions": "Vertex(TriangleOf(A, B, C)) = {A, B, C};B: Point;Coordinate(B) = (0, -3);C: Point;Coordinate(C) = (0, 3);Perimeter(TriangleOf(A, B, C)) = 16;A: Point",
    "query_expressions": "LocusEquation(A)",
    "answer_expressions": "(y^2/25+x^2/16=1)&Negation(x = 0)",
    "fact_spans": "[[[1, 61]], [[23, 32]], [[23, 32]], [[35, 43]], [[35, 43]], [[1, 51]], [[58, 61]]]",
    "query_spans": "[[[58, 68]]]",
    "process": "Since two vertices of $\\triangle ABC$ are $B(0,-3)$ and $C(0,3)$, we have $|BC|=6$. Since the perimeter of the triangle is $16$, that is, $|AB|+|AC|+|BC|=16$, it follows that $|AB|+|AC|=10 > |BC|=6$. By the definition of an ellipse: a moving point $A$ has a constant sum of distances to two fixed points $B(0,-3)$ and $C(0,3)$, and this sum is greater than the distance between the two fixed points. Therefore, the locus of point $A$ is an ellipse with foci at $B(0,-3)$ and $C(0,3)$, and $2a=10$. Thus, $c=3$, $a=5$, $b=\\sqrt{a^{2}-c^{2}}=\\sqrt{5^{2}-3^{2}}=4$. The equation of the ellipse is: $\\frac{y^{2}}{25}+\\frac{x^{2}}{16}=1$. Since points $A$, $B$, and $C$ are not collinear, point $A$ cannot lie on the $y$-axis. Therefore, the trajectory equation of vertex $A$ is: $\\frac{y^{2}}{25}+\\frac{x^{2}}{16}=1$ $(x\\neq0)$."
  },
  {
    "text": "Let the left and right vertices of the hyperbola $x^{2}-y^{2}=6$ be $A_{1}$ and $A_{2}$, respectively, and let $P$ be a point on the right branch of the hyperbola located in the first quadrant. The slopes of the lines $P A_{1}$ and $P A_{2}$ are $k_{1}$ and $k_{2}$, respectively. Then the value of $k_{1} k_{2}$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;A1: Point;A2:Point;Expression(G) = (x^2 - y^2 = 6);LeftVertex(G)=A1;RightVertex(G)=A2;PointOnCurve(P,RightPart(G));Quadrant(P)=1;k1:Number;k2:Number;Slope(LineOf(P,A1))=k1;Slope(LineOf(P,A2))=k2",
    "query_expressions": "k1*k2",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 19], [47, 50]], [[43, 46]], [[27, 34]], [[35, 42]], [[1, 19]], [[1, 42]], [[1, 42]], [[43, 55]], [[43, 63]], [[92, 99]], [[101, 108]], [[64, 108]], [[64, 108]]]",
    "query_spans": "[[[110, 127]]]",
    "process": "The left and right vertices of the hyperbola $x^{2}-y^{2}=6$ are $A_{1}(\\sqrt{6},0)$ and $A_{2}(-\\sqrt{6},0)$, respectively. Let $p(x,y)$, $x>0$, $y>0$, then $k_{1}k_{2}=\\frac{y}{x+\\sqrt{6}}\\cdot\\frac{y}{x-\\sqrt{6}}=\\frac{y^{2}}{x^{2}-6}=1$"
  },
  {
    "text": "The equation $\\frac{x^{2}}{m^{2}-2 m}+\\frac{y^{2}}{3}=1$ represents an ellipse with foci on the $y$-axis. Then, the range of values for $m$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(m^2 - 2*m) + y^2/3 = 1);m: Number;PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-1, 0) + (2, 3)",
    "fact_spans": "[[[56, 58]], [[0, 58]], [[60, 63]], [[47, 58]]]",
    "query_spans": "[[[60, 70]]]",
    "process": ""
  },
  {
    "text": "Given point $A(-3,-\\sqrt{5})$, $P$ is a moving point on the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and $B$ is a moving point on the circle $C_{1}$: $(x-1)^{2}+y^{2}=1$. Then the maximum value of $|P B|-|P A|$ is?",
    "fact_expressions": "C: Ellipse;C1: Circle;A: Point;P: Point;B: Point;Expression(C) = (x^2/4 + y^2/3 = 1);Expression(C1)=((x-1)^2+y^2=1);Coordinate(A) = (-3, -sqrt(5));PointOnCurve(P, C);PointOnCurve(B,C1)",
    "query_expressions": "Max(-Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B)))",
    "answer_expressions": "2",
    "fact_spans": "[[[27, 69]], [[78, 106]], [[2, 20]], [[23, 26]], [[74, 77]], [[27, 69]], [[78, 106]], [[2, 20]], [[22, 73]], [[74, 110]]]",
    "query_spans": "[[[112, 131]]]",
    "process": "From the ellipse $ C: \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $, we obtain $ a = 2 $, $ b = \\sqrt{3} $, $ c = 1 $. Let the right focus be $ F(-1, 0) $. Since $ P $ is a moving point on the ellipse $ C: \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $, and $ B $ is a moving point on the circle $ C_{1}: (x - 1)^{2} + y^{2} = 1 $, then $ |PB| - |PA| \\leqslant 1 + |PF| - |PA| = 1 + 2a - |PF| - |PA| = 5 - (|PF| + |PA|) $. Since $ |PF| + |PA| \\geqslant |AF| = \\sqrt{(-1 + 3)^{2} + (0 + \\sqrt{5})^{2}} = 3 $, with equality if and only if $ A $, $ P $, $ F $ are collinear, $ |PB| - |PA| < 5 - (|PF| + |PA|) < 2b $, but this is incorrect."
  },
  {
    "text": "Given the parabola $C$: $y=\\frac{1}{8} x^{2}$ with focus $F$, point $P$ lies on $C$, and $|P F|=10$. Then the area of $\\triangle P O F$ (where $O$ is the origin) is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y = x^2/8);F: Point;Focus(C) = F;P: Point;PointOnCurve(P, C);Abs(LineSegmentOf(P, F)) = 10;O: Origin",
    "query_expressions": "Area(TriangleOf(P, O, F))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 31], [44, 47]], [[2, 31]], [[35, 38]], [[2, 38]], [[39, 43]], [[39, 48]], [[50, 60]], [[82, 85]]]",
    "query_spans": "[[[62, 96]]]",
    "process": ""
  },
  {
    "text": "It is known that the line $x - 2y + 2 = 0$ passes through a vertex and a focus of the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1$ $(a > b > 0)$. Then the equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x - 2*y + 2 = 0);PointOnCurve(OneOf(Vertex(G)),H);PointOnCurve(OneOf(Focus(G)),H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5+y^2=1",
    "fact_spans": "[[[17, 69], [84, 86]], [[19, 69]], [[19, 69]], [[2, 15]], [[19, 69]], [[19, 69]], [[17, 69]], [[2, 15]], [[2, 74]], [[2, 79]]]",
    "query_spans": "[[[84, 91]]]",
    "process": "Find the vertices and foci of the ellipse, then obtain a and b, so that the equation of the ellipse can be determined. For the line x-2y+2=0, when x=0, y=1; when y=0, x=-2. Then in the ellipse, b=1, c=2, so a^{2}=b^{2}+c^{2}=5. Therefore, the equation of the ellipse is \\frac{x^{2}}{5}+y^{2}=1."
  },
  {
    "text": "Given that on the parabola $y^{2}=2 px(p>0)$, the point with abscissa $1$ has equal distance to the vertex and to the directrix, then the equation of this parabola is?",
    "fact_expressions": "G: Parabola;p: Number;p>0;Expression(G) = (y^2 = 2*(p*x));P: Point;PointOnCurve(P, G);XCoordinate(P) = 1;Distance(P, Vertex(G)) = Distance(P, Directrix(G))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = 8*x",
    "fact_spans": "[[[2, 22], [50, 53]], [[5, 22]], [[5, 22]], [[2, 22]], [[31, 32]], [[2, 32]], [[23, 32]], [[2, 47]]]",
    "query_spans": "[[[50, 58]]]",
    "process": ""
  },
  {
    "text": "The equation of the directrix of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x = pm*1/2",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 35]]]",
    "process": "From the hyperbola equation, we know: a=1, c=\\sqrt{1+3}=2, the foci lie on the x-axis, \\therefore the equations of the directrices are x=\\pm\\frac{a^2}{c}="
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(b>a>0)$, the right focus is $F$, and $O$ is the origin. If there exists a line $l$ passing through point $F$ intersecting the right branch of hyperbola $C$ at points $A$ and $B$ such that $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}=0$, then the range of eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;O: Origin;B: Point;A: Point;F: Point;e: Number;b > a;a > 0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;PointOnCurve(F, l);Intersection(l, RightPart(C)) = {A, B};DotProduct(VectorOf(O, A), VectorOf(O, B)) = 0;Eccentricity(C) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "[(sqrt(5)+1)/2,sqrt(3))",
    "fact_spans": "[[[81, 86]], [[2, 60], [92, 98], [167, 170]], [[9, 60]], [[9, 60]], [[69, 72]], [[106, 109]], [[102, 105]], [[65, 68], [87, 91]], [[174, 177]], [[9, 60]], [[9, 60]], [[2, 60]], [[2, 68]], [[81, 91]], [[81, 111]], [[114, 165]], [[167, 177]]]",
    "query_spans": "[[[174, 184]]]",
    "process": "Let $ F(c,0) $, $ A(x_{1},y_{2}) $, $ B(x_{2},y_{2}) $, the equation of line $ l $ is $ x=ty+c $. From  \n$$\n\\begin{cases}\nx=ty+c \\\\\n\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 \\\\\nc=a^{2}+b^{2}\n\\end{cases}\n$$\nwe obtain by simplification $ (b^{2}t^{2}-a^{2})y^{2}+2tb^{2}cy+b^{4}=0 $. Since the line intersects the right branch of hyperbola $ C $ at two points $ A $ and $ B $, we have  \n$ \\Delta=(2tb^{2}c)^{2}-4(b^{2}t^{2}-a^{2})b^{4}>0 $, and $ y_{1}y_{2}=\\frac{b^{4}}{b^{2}t^{2}-a^{2}}<0 $. Solving these two inequalities yields $ 0\\leqslant t^{2}<\\frac{a^{2}}{b^{2}} $.  \nBecause $ \\overrightarrow{OA}\\cdot\\overrightarrow{OB}=x_{1}x_{2}+y_{1}y_{2}=(ty_{1}+c)(ty_{2}+c)+y_{1}y_{2}=(t^{2}+1)y_{1}y_{2}+ct\\cdot(y_{1}+y_{2})+c^{2} $, simplifying gives $ \\overline{C} $. Since $ \\overrightarrow{OA}\\cdot\\overrightarrow{OB}=0 $, we have $ b^{2}t^{2}-a^{2} $, i.e.,  \n$ (t^{2}+1)b^{4}-2t^{2}b^{2}c^{2}+c^{2}(b^{2}t^{2}-a^{2})=0 $, which simplifies to  \n$ t^{2}\\frac{2c^{2}-b^{4}}{(b^{2}-c^{2})}=\\frac{b^{4}-a^{2}c^{2}}{b^{2}a^{2}} $. From $ 0\\leqslant t^{2}<\\frac{a^{2}}{b^{2}} $, we get  \n$$\n\\begin{cases}\nb^{4}-a^{2}c^{2}\\geqslant0 \\\\\n\\frac{b^{4}-a^{2}c^{2}}{b^{2}a^{2}}<\\frac{a^{2}}{b^{2}}\n\\end{cases}\n$$\nSolving yields $ \\frac{\\sqrt{5}+1}{2}\\leqslant e<\\sqrt{3} $. Therefore, the range of eccentricity of the hyperbola is $ \\left[\\frac{\\sqrt{5}+1}{2},\\sqrt{3}\\right) $. This problem examines the relationship between a line and a hyperbola, and finding the eccentricity using properties of hyperbolas; it is relatively difficult."
  },
  {
    "text": "Given that the two foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ are $F_{1}$ and $F_{2}$, and point $M$ is a point on the ellipse such that $|M F_{1}|-|M F_{2}|=2$, then the area of $\\Delta F_{1} F_{2} M$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;M: Point;Expression(G) = (x^2/9 + y^2/4 = 1);Focus(G) ={F1,F2};PointOnCurve(M, G);Abs(LineSegmentOf(M, F1)) - Abs(LineSegmentOf(M, F2)) = 2",
    "query_expressions": "Area(TriangleOf(F1, F2, M))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 39], [66, 68]], [[45, 52]], [[53, 60]], [[61, 65]], [[2, 39]], [[2, 60]], [[61, 71]], [[73, 96]]]",
    "query_spans": "[[[98, 125]]]",
    "process": "From the definition of an ellipse, |MF_{1}| + |MF_{2}| = 6, and |MF_{1}| - |MF_{2}| = 2. Solving the system of equations  \n\\begin{cases}|MF_{1}|+|MF_{2}|=6\\\\|MF_{1}|-|MF_{2}|=2\\end{cases},  \nwe obtain \\begin{cases}|MF_{1}|=4\\\\|MF_{2}|=2\\end{cases}.  \nAlso, |F_{1}F_{2}| = 2\\sqrt{5}, so |MF_{1}|^{2} + |MF_{2}|^{2} = |F_{1}F_{2}|^{2}.  \nTherefore, \\triangle F_{1}F_{2}M is a right triangle with |MF_{1}| and |MF_{2}| as the legs.  \nThus, the area of \\triangle F_{1}F_{2}M is \\frac{1}{2} \\cdot |MF_{1}| \\cdot |MF_{2}| = \\frac{1}{2} \\times 4 \\times 2 = 4.  \nHence, the answer is: 4.  \n(This question mainly examines the definition and simple properties of an ellipse, and belongs to basic problems.)"
  },
  {
    "text": "Let $M$ and $N$ be the two endpoints of the transverse axis of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and let $Q$ be a point on the hyperbola (distinct from $M$ and $N$), such that $\\tan \\angle Q M N \\cdot \\tan \\angle Q N M = -2$. Then the equations of the asymptotes of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;Q: Point;M: Point;N: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Endpoint(RealAxis(G)) = {M, N};PointOnCurve(Q, G);Negation(Q = M);Negation(Q = N);Tan(AngleOf(Q, M, N))*Tan(AngleOf(Q, N, M)) = -2",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*sqrt(2)*x",
    "fact_spans": "[[[9, 65], [77, 80], [147, 150]], [[12, 65]], [[12, 65]], [[73, 76]], [[1, 4], [87, 90]], [[5, 8], [91, 94]], [[12, 65]], [[12, 65]], [[9, 65]], [[1, 72]], [[73, 84]], [[73, 97]], [[73, 97]], [[99, 145]]]",
    "query_spans": "[[[147, 158]]]",
    "process": "Let Q(m,n) (m≠±a), then tan∠QMN = \\frac{n}{m+a}, tan∠QNM = \\frac{n}{a−m}. Therefore, tan∠QMN⋅tan∠QNM = \\frac{n^{2}}{a^{2}−m^{2}} = −2. Since Q lies on the hyperbola, we have \\frac{m^{2}}{a^{2}}−\\frac{n^{2}}{b^{2}} = 1, so m^{2} = a^{2}(1+\\frac{n^{2}}{b^{2}}). It follows that tan∠QMN⋅tan∠QNM = −\\frac{b^{2}}{a^{2}} = −2, thus \\frac{b^{2}}{a^{2}} = 2. Hence, the hyperbola NH^{^{2}} has asymptotes given by y = ±\\sqrt{2}x."
  },
  {
    "text": "The equation of the hyperbola that has the same foci as the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$ and passes through the point $P(2,1)$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;C:Hyperbola;Expression(G) = (x^2/4 - y^2/2 = 1);Coordinate(P) = (2, 1);Focus(G)=Focus(C);PointOnCurve(P,C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/3 - y^2/3 = 1",
    "fact_spans": "[[[1, 39]], [[46, 55]], [[56, 59]], [[1, 39]], [[46, 55]], [[0, 59]], [[45, 59]]]",
    "query_spans": "[[[56, 64]]]",
    "process": "According to the problem, let the required hyperbola be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$. Since the two curves have the same foci, $a^{2}+b^{2}=c^{2}=4+2=6$ $\\textcircled{1}$. Also, point $P(2,1)$ lies on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, so $\\frac{4}{a^{2}}-\\frac{1}{b^{2}}=1$ $\\textcircled{2}$. Solving $\\textcircled{1}$ and $\\textcircled{2}$ simultaneously gives $a^{2}=b^{2}=3$. Therefore, the equation of the required hyperbola is $\\frac{x^{2}}{3}-\\frac{y^{2}}{3}=1$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively, and $P$ is a point on the asymptote of hyperbola $C$ in the first quadrant. Let $|P F_{1}|=\\lambda|P F_{2}|$. If the maximum value of $\\lambda$ is $\\sqrt{2}$, then what is the equation of the asymptotes of hyperbola $C$?",
    "fact_expressions": "LeftFocus(C) = F1;RightFocus(C) = F2;F1: Point;F2: Point;C: Hyperbola;Expression(C) = (x^2/a^2 - y^2/b^2 = 1);a: Number;b: Number;a>0;b>0;P: Point;Quadrant(P) = 1;PointOnCurve(P, Asymptote(C)) = True;Abs(LineSegmentOf(P, F1)) = lambda*Abs(LineSegmentOf(P, F2));lambda: Number;Min(lambda) = sqrt(2)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*2*sqrt(2)*x",
    "fact_spans": "[[[2, 88]], [[2, 88]], [[2, 9]], [[10, 17]], [[20, 81], [169, 175], [98, 104]], [[20, 81]], [[28, 81]], [[28, 81]], [[28, 81]], [[28, 81]], [[89, 92]], [[89, 98]], [[89, 111]], [[113, 141]], [[143, 152]], [[143, 167]]]",
    "query_spans": "[[[169, 183]]]",
    "process": "According to the problem, the asymptote of the hyperbola passing through the first quadrant has the equation $ y = \\frac{b}{a}x $; thus, let $ P\\left(x, \\frac{bx}{a}\\right) $, $ x > 0 $,  \n$ \\lambda^{2} = \\frac{|PF_{1}|^{2}}{|PF_{2}|^{2}} = \\frac{(x+c)^{2} + \\frac{b^{2}x^{2}}{a^{2}}}{(x-c)^{2} + \\frac{b^{2}x^{2}}{a^{2}}} = 1 + \\frac{4}{a^{2}x + \\frac{c}{x}} \\geq 1 + \\frac{4}{c} - 2 $, with equality if and only if “$ x = a $”, at which point the eccentricity $ e = 3 $, $ 1 + \\left(\\frac{b}{a}\\right)^{2} = 9 $, $ \\frac{b}{a} = 2\\sqrt{2} $. The asymptotes of curve $ C $ are $ y = \\pm 2\\sqrt{2}x $."
  },
  {
    "text": "If the focal distance of the hyperbola $\\frac{y^{2}}{5}-x^{2}=m(m>0)$ equals its eccentricity, then $m$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2 + y^2/5 = m);m: Number;m>0;FocalLength(G) = Eccentricity(G)",
    "query_expressions": "m",
    "answer_expressions": "1/20",
    "fact_spans": "[[[1, 34]], [[1, 34]], [[44, 47]], [[4, 34]], [[1, 42]]]",
    "query_spans": "[[[44, 49]]]",
    "process": "Analysis: First, convert the hyperbola equation into the standard form $\\frac{y^{2}}{5m}-\\frac{x^{2}}{m}=1$. Since $2c=\\frac{c}{a}$, it follows that $a=\\frac{1}{2}$. According to the standard form of the hyperbola where $a^{2}=5m$, the value of $m$ can be determined. Detailed solution: Because $m>0$, the equation of the hyperbola can be rewritten in standard form as $\\frac{y^{2}}{5m}-\\frac{x^{2}}{m}=1$. Given that the focal length equals the eccentricity, $2c=\\frac{c}{a}$, so $a=\\frac{1}{2}$. From the standard equation, $a^{2}=5m$, thus $m=\\frac{a^{2}}{5}=\\frac{1}{20}$. Therefore, the value of $m$ is $\\frac{1}{20}$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x$, with focus $F(1,0)$. A line passing through $F$ intersects $C$ at points $A$ and $B$. The tangent to $C$ at $A$ intersects the directrix of $C$ at point $P$. If $|P F|=\\sqrt{5}$, then $|A B|=$?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;F: Point;P: Point;A: Point;B: Point;Expression(C) = (y^2 = 2*p*x);Coordinate(F) = (1, 0);Focus(C) = F;PointOnCurve(F,G);Intersection(G, C) = {A, B};Intersection(TangentOnPoint(A,C),Directrix(C))=P;Abs(LineSegmentOf(P, F)) = sqrt(5)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "5",
    "fact_spans": "[[[2, 23], [44, 47], [59, 62], [71, 74]], [[10, 23]], [[41, 43]], [[37, 40], [27, 35]], [[79, 83]], [[63, 66], [49, 52]], [[53, 56]], [[2, 23]], [[27, 35]], [[2, 35]], [[36, 43]], [[41, 58]], [[59, 83]], [[85, 101]]]",
    "query_spans": "[[[103, 112]]]",
    "process": "Since the focus is $ F(1,0) $, the parabola is $ C: y^{2} = 4x $. Without loss of generality, assume $ P $ lies in the second quadrant. Since $ |PF| = \\sqrt{5} $, we have $ P(-1,1) $. Let the equation of line $ AB $ be $ x = my + 1 $, and by substituting into $ y^{2} = 4x $ and eliminating $ x $, we obtain $ y^{2} - 4my - 4 = 0 $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ y_{1} + y_{2} = 4m $, $ y_{1}y_{2} = -4 $, so $ |AB| = \\sqrt{(1+m^{2})(16m^{2}+16)} = 4(m^{2}+1) $. Let the tangent line $ PA $ have equation $ x = ny + t $, and by substituting into $ y^{2} = 4x $ and eliminating $ x $, we get $ y^{2} - 4ny - 4t = 0 $. Since $ \\triangle = 16n^{2} + 16t = 0 $, it follows that $ n^{2} + t = 0 $, so $ y^{2} - 4ny + 4n^{2} = 0 $, $ y = 2n $, i.e., $ y_{1} = 2n $, $ y_{2} = -\\frac{2}{n} $, $ m = \\frac{n}{2} - \\frac{1}{2n} $, thus $ P(-1, n - \\frac{1}{n}) $. Therefore, $ n - \\frac{1}{n} = 1 $, $ m = \\frac{1}{2} $, $ |AB| = 4(m^{2}+1) = 5 $."
  },
  {
    "text": "A line passing through the left focus $F$ of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$ intersects the left branch of the hyperbola at points $M$ and $N$, and $F_{2}$ is its right focus. Then the value of $| M F_{2}|+|NF_{2}|-|M N|$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;M: Point;F2: Point;N: Point;F: Point;Expression(G) = (x^2/4 - y^2/3 = 1);LeftFocus(G)=F;PointOnCurve(F, H);Intersection(H,LeftPart(G)) = {M, N};RightFocus(G) = F2",
    "query_expressions": "Abs(LineSegmentOf(M, F2)) + Abs(LineSegmentOf(N, F2)) - Abs(LineSegmentOf(M, N))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 39], [49, 52], [74, 75]], [[46, 48]], [[56, 59]], [[66, 73]], [[60, 63]], [[42, 45]], [[1, 39]], [[1, 45]], [[0, 48]], [[46, 65]], [[66, 78]]]",
    "query_spans": "[[[80, 111]]]",
    "process": ""
  },
  {
    "text": "Given that $F(2,0)$ is the right focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the length of the chord passing through $F$ and perpendicular to the $x$-axis is $6$. If $A(-2, \\sqrt{2})$, and point $M$ is any point on the ellipse, then the maximum value of $|M F|+|M A|$ is?",
    "fact_expressions": "F: Point;Coordinate(F) = (2, 0);RightFocus(G) = F;G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;L: LineSegment;PointOnCurve(F,L) = True;IsPerpendicular(L,xAxis) = True;IsChordOf(L,G) = True;Length(L) = 6;A: Point;Coordinate(A) = (-2, sqrt(2));M: Point;PointOnCurve(M, G) = True",
    "query_expressions": "Max(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F)))",
    "answer_expressions": "8+sqrt(2)",
    "fact_spans": "[[[2, 10], [69, 72]], [[2, 10]], [[2, 67]], [[11, 63], [116, 118]], [[11, 63]], [[13, 63]], [[13, 63]], [[13, 63]], [[13, 63]], [], [[11, 82]], [[11, 82]], [[11, 82]], [[11, 89]], [[92, 109]], [[92, 109]], [[111, 115]], [[111, 122]]]",
    "query_spans": "[[[124, 143]]]",
    "process": "Let the left focus of the ellipse be F. Given that the focus of the ellipse is F(2,0), then c=2. Also, the length of the chord passing through F and perpendicular to the x-axis is 6, that is, \\frac{b^{2}}{a}=6, then \\frac{a^{2}-c^{2}}{a}=\\frac{a^{2}-4}{a}=6. Solving gives a=4, so b^{2}=a^{2}-c^{2}=12. Also, |MA|+|MF|=|MA|+8-|MF|=8+|MA|-|MF|. When points M, A, and F are collinear, the maximum value is attained, at which point |MA|-|MF|=\\sqrt{2}. Therefore, the maximum value of |MA|+|MF| is 8+\\sqrt{2}."
  },
  {
    "text": "The length of the imaginary axis of the hyperbola $m x^{2}-y^{2}=1$ is $2$ times the length of the real axis, then $m$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (m*x^2 - y^2 = 1);m: Number;Length(ImageinaryAxis(G)) = 2*Length(RealAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 20]], [[0, 20]], [[35, 38]], [[0, 33]]]",
    "query_spans": "[[[35, 40]]]",
    "process": "From the given conditions, \\begin{cases}m>0\\\\m=\\frac{1}{4}\\end{cases}, we obtain $ m=4 $."
  },
  {
    "text": "The distance from a moving point $P$ in the plane to the point $(1,0)$ is less than its distance to the line $x=-3$ by $2$. Then the equation of the trajectory of point $P$ is?",
    "fact_expressions": "G: Line;H: Point;P: Point;Expression(G) = (x =-3);Coordinate(H) = (1, 0);Distance(P, H) = Distance(P, G) - 2",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[22, 30]], [[9, 17]], [[5, 8], [39, 43]], [[22, 30]], [[9, 17]], [[5, 37]]]",
    "query_spans": "[[[39, 50]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/3 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(7)/2",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 44]]]",
    "process": "\\because the equation of the hyperbola is \\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1 \\therefore a=2, b=\\sqrt{3} \\therefore c=\\sqrt{a^{2}+b^{2}}=\\sqrt{7} \\therefore_{e}=\\frac{c}{a}=\\frac{\\sqrt{7}}{2}"
  },
  {
    "text": "Given that a line passes through the focus $F$ of the parabola $y^2 = 4x$, and intersects the parabola at points $A$ and $B$. A point $C$ on the directrix of the parabola satisfies $\\overrightarrow{CB} = 2 \\overrightarrow{BF}$. Then $|AF| =$?",
    "fact_expressions": "G: Parabola;H: Line;C: Point;B: Point;F: Point;A: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H,G)={A,B};PointOnCurve(C,Directrix(G));VectorOf(C,B)=2*VectorOf(B,F)",
    "query_expressions": "Abs(LineSegmentOf(A, F))",
    "answer_expressions": "4",
    "fact_spans": "[[[6, 20], [30, 33], [45, 48]], [[2, 4]], [[55, 58]], [[38, 41]], [[23, 26]], [[34, 37]], [[6, 20]], [[6, 26]], [[2, 26]], [[2, 43]], [[45, 58]], [[60, 105]]]",
    "query_spans": "[[[108, 117]]]",
    "process": "\\because\\overrightarrow{CB}=2\\overrightarrow{BF},\\therefore C is the intersection point of line AB and the directrix. Draw perpendiculars AN and BM from A and B to the directrix, with N and M as the feet of the perpendiculars. Let K be the intersection point of the directrix and the x-axis, as shown in the figure. \\because |BM|=|BF|,\\therefore |CB|=2|BM|,\\therefore \\angle MNB=\\frac{\\pi}{6}. The equation of the parabola is y^{2}=4x, then p=2, so |KF|=2, therefore |CF|=2|KF|=4. Also |CA|=2|AN|, and |AN|=|AF|, so |AF|=|CF|=4."
  },
  {
    "text": "The standard equation of a parabola with the directrix line $x+3=0$ is?",
    "fact_expressions": "H: Line;Expression(H) = (x + 3 = 0);G: Parabola;Directrix(G) = H",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=12*x",
    "fact_spans": "[[[1, 10]], [[1, 10]], [[14, 17]], [[0, 17]]]",
    "query_spans": "[[[14, 24]]]",
    "process": ""
  },
  {
    "text": "$P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and $M$, $N$ are points on the circles $(x+5)^{2}+y^{2}=4$ and $(x-5)^{2}+y^{2}=1$, respectively. Then the maximum value of $|P M|-|P N|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);P: Point;PointOnCurve(P, RightPart(G));H: Circle;Expression(H) = (y^2 + (x + 5)^2 = 4);K:Circle;Expression(K) = (y^2 + (x - 5)^2 = 1);M: Point;N: Point;PointOnCurve(M, H);PointOnCurve(N, K)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "9",
    "fact_spans": "[[[4, 43]], [[4, 43]], [[0, 3]], [[0, 49]], [[60, 80]], [[60, 80]], [[81, 100]], [[81, 100]], [[50, 53]], [[54, 57]], [[50, 103]], [[50, 103]]]",
    "query_spans": "[[[105, 124]]]",
    "process": ""
  },
  {
    "text": "The line $x - y - 2 = 0$ intersects the parabola $y^{2} = 2 p x$ ($p > 0$) at points $A$ and $B$. If the segment $AB$ is bisected by the point $M(4, 2)$, then the equation of the directrix of the parabola is?",
    "fact_expressions": "H: Line;Expression(H) = (x - y - 2 = 0);G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;A: Point;B: Point;Intersection(H, G) = {A, B};M: Point;Coordinate(M) = (4, 2);MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=-1",
    "fact_spans": "[[[0, 11]], [[0, 11]], [[12, 33], [67, 70]], [[12, 33]], [[15, 33]], [[15, 33]], [[35, 38]], [[39, 42]], [[0, 44]], [[54, 63]], [[54, 63]], [[46, 65]]]",
    "query_spans": "[[[67, 77]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since the segment AB is bisected by point M(4,2), we have y_{1}+y_{2}=4. Also, y_{1}^{2}=2px_{1}, y_{2}^{2}=2px_{2}, so (y_{1}+y_{2})(y_{1}-y_{2})=2p(x_{1}-x_{2}). According to the problem, the slope of line l exists and equals 1, so x_{1}\\neq x_{2}, thus 4\\times\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=2p, i.e., 4\\times1=2p, so p=2. Hence, the directrix equation of the parabola is x=-1."
  },
  {
    "text": "The distance from a point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ to its left focus $F_{1}$ is $6$, then what is the distance from point $P$ to the right directrix of the ellipse?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;Distance(P, F1) = 6",
    "query_expressions": "Distance(P, RightDirectrix(G))",
    "answer_expressions": "20/3",
    "fact_spans": "[[[0, 39], [46, 47], [72, 74]], [[42, 45], [67, 71]], [[51, 58]], [[0, 39]], [[0, 45]], [[46, 58]], [[42, 65]]]",
    "query_spans": "[[[67, 82]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ passes through the point $P(2, y_{0})$, $F$ is the focus of the parabola, and $|P F|=4$, then the value of $y_{0}$ is?",
    "fact_expressions": "C: Parabola;p: Number;P: Point;F: Point;p>0;y0: Number;Expression(C) = (y^2 = 2*(p*x));Coordinate(P) = (2, y0);PointOnCurve(P, C);Focus(C) = F;Abs(LineSegmentOf(P, F)) = 4",
    "query_expressions": "y0",
    "answer_expressions": "pm*4",
    "fact_spans": "[[[2, 28], [51, 54]], [[10, 28]], [[30, 45]], [[47, 50]], [[10, 28]], [[70, 77]], [[2, 28]], [[30, 45]], [[2, 44]], [[47, 57]], [[59, 68]]]",
    "query_spans": "[[[70, 81]]]",
    "process": "\\because the parabola C: y^{2}=2px (p>0) passes through the point P(2,y_{0}), F is the focus of the parabola, and |PF|=4. \\therefore by the definition of the parabola, we obtain \\frac{p}{2}+2=4, solving gives p=4, \\therefore y^{2}=8x, \\because the x-coordinate of P is 2, \\therefore y_{0}^{2}=2\\times8=16, solving gives y_{0}=\\pm4"
  },
  {
    "text": "If the foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ lie on the $x$-axis, and tangents are drawn from the point $(2 , 1)$ to the circle $x^{2}+y^{2}=4$, with points of tangency $A$ and $B$, respectively, such that the line $AB$ passes exactly through the right focus and the upper vertex of the ellipse, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Circle;P:Point;B: Point;A: Point;L1:Line;L2:Line;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + y^2 = 4);PointOnCurve(Focus(G), xAxis);Coordinate(P)=(2,1);TangentOfPoint(P,H)={L1,L2};TangentPoint(L1,H)=A;TangentPoint(L2,H)=B;PointOnCurve(RightFocus(G),LineOf(A,B));PointOnCurve(UpperVertex(G),LineOf(A,B))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/20+y^2/16=1",
    "fact_spans": "[[[1, 46], [112, 114], [112, 114]], [[3, 46]], [[3, 46]], [[68, 84]], [[57, 67]], [[97, 100]], [[93, 96]], [], [], [[1, 46]], [[68, 84]], [[1, 55]], [[57, 67]], [[56, 87]], [[56, 100]], [[56, 100]], [[101, 118]], [[101, 122]]]",
    "query_spans": "[[[124, 130]]]",
    "process": ""
  },
  {
    "text": "Given that the lines connecting point $P$ on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{8}=1$ to the two foci are perpendicular to each other, what are the coordinates of point $P$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/8 = 1);P: Point;PointOnCurve(P, G) = True;F1: Point;F2: Point;Focus(G) = {F1, F2};IsPerpendicular(LineSegmentOf(P,F1), LineSegmentOf(P, F2)) = True",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(0,pm*2*sqrt(2))",
    "fact_spans": "[[[2, 40]], [[2, 40]], [[42, 46], [59, 63]], [[2, 46]], [], [], [[2, 50]], [[2, 57]]]",
    "query_spans": "[[[59, 68]]]",
    "process": "Let $ P(x,y) $, $ c^{2} = a^{2} - b^{2} = 16 - 8 = 8 $, $ F_{1}(-2\\sqrt{2},0) $, $ F_{2}(2\\sqrt{2},0) $, $ \\overrightarrow{PF_{1}} = (-2\\sqrt{2}-x,-y) $, $ \\overrightarrow{PF_{2}} = (2\\sqrt{2}-x,-y) $. Since $ \\overrightarrow{PF_{1}} \\bot \\overrightarrow{PF_{2}} $, $ \\therefore \\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}} = 0 \\Rightarrow (-2\\sqrt{2}-x)(2\\sqrt{2}-x) + y^{2} = 0 $, i.e., $ x^{2} + y^{2} - 8 = 0 $, $\\textcircled{1}$ $ \\because \\frac{x^{2}}{16} + \\frac{y^{2}}{8} = 1 $ $\\textcircled{2} $. From $\\textcircled{1}$ and $\\textcircled{2}$ we get $ x^{2} = 0 $, $ y^{2} = 8 $, i.e., $ x = 0 $, $ y = \\pm 2\\sqrt{2} $. Therefore, the coordinates of point $ P $ are $ (0, \\pm 2\\sqrt{2}) $."
  },
  {
    "text": "The eccentricity of the ellipse $x^{2}+4 y^{2}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 4*y^2 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[0, 19]], [[0, 19]]]",
    "query_spans": "[[[0, 25]]]",
    "process": ""
  },
  {
    "text": "The line $l$ passing through point $P(1,1)$ intersects the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ at points $A$ and $B$, and $\\overrightarrow{A P}=\\lambda \\overrightarrow{P B}$. Point $Q$ satisfies $\\overrightarrow{A Q}=-\\lambda \\overrightarrow{Q B}$. If $O$ is the origin, then the minimum value of $|O Q|$ is?",
    "fact_expressions": "l: Line;G: Ellipse;P: Point;A: Point;B: Point;Q: Point;O: Origin;Expression(G) = (x^2/4 + y^2/3 = 1);Coordinate(P) = (1, 1);PointOnCurve(P, l);Intersection(l, G) = {A, B};VectorOf(A,P)=lambda*VectorOf(P,B);lambda:Number;VectorOf(A,Q)=-lambda*VectorOf(Q,B)",
    "query_expressions": "Min(Abs(LineSegmentOf(O, Q)))",
    "answer_expressions": "12/5",
    "fact_spans": "[[[11, 16]], [[17, 54]], [[1, 10]], [[56, 60]], [[61, 64]], [[117, 121]], [[177, 180]], [[17, 54]], [[1, 10]], [[0, 16]], [[11, 64]], [[66, 117]], [[123, 175]], [[123, 175]]]",
    "query_spans": "[[[187, 200]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, $ Q(m,n) $, then \n\\[\n\\begin{cases}\nx_{1}+\\lambda x_{2}=1+\\lambda \\\\\nx_{1}-\\lambda x_{2}=m(1-\\lambda)\n\\end{cases}\n\\]\nThus $ x_{1}^{2}-(\\lambda x_{2})^{2}=m(1-\\lambda^{2}) $, similarly $ y_{1}^{2}-(\\lambda y_{2})^{2}=n(1-\\lambda^{2}) $, thus we can obtain \n\\[\n\\left(\\frac{x^{2}}{4}+\\frac{y^{2}}{3}\\right)-2^{2}\\left(\\frac{x^{2}}{4}+\\frac{y^{2}}{3}\\right)=(1-\\lambda^{2})\\left(\\frac{m}{4}+\\frac{n}{3}\\right).\n\\]\nThat is $ \\frac{m}{4}+\\frac{n}{3}=1 $, so the trajectory of point $ Q $ is a straight line, $ |OQ|_{\\min} $ is the distance from the origin to the line. Therefore, [Note] Common types and solving strategies for coordinate representation problems of collinear planar vectors: (1) Using the condition of two vectors being collinear to find vector coordinates. Generally, when finding a vector collinear with a known vector $ \\mathbf{a} $, one can set the desired vector as $ \\lambda \\mathbf{a} $ ($ \\lambda \\in \\mathbb{R} $), then combine other conditions to form an equation in $ \\lambda $, solve for $ \\lambda $, and substitute back into $ \\lambda \\mathbf{a} $ to obtain the required vector. (2) Using two collinear vectors to find parameters. If two vectors are known to be collinear and certain parameters need to be found, it is convenient to use the necessary and sufficient condition for $ \\mathbf{a}=(x_{1},y_{1}) $, $ \\mathbf{b}=(x_{2},y_{2}) $ to be parallel: $ x_{1}y_{2}=x_{2}y_{1} $."
  },
  {
    "text": "Given that the moving line $l$ intersects the ellipse $C$: $\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1$ at two distinct points $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$, and the area of $\\triangle O P Q$ is $S_{\\triangle O P Q}=\\frac{\\sqrt{6}}{2}$, where $O$ is the origin, then $x_{1}^{2}+x_{2}^{2}$=?",
    "fact_expressions": "l: Line;C: Ellipse;P: Point;Q: Point;O: Origin;Expression(C) = (x^2/3 + y^2/2 = 1);Coordinate(P) = (x1,y1);Coordinate(Q) = (x2,y2);Intersection(l, C) = {P,Q};Negation(P=Q);Area(TriangleOf(O, P, Q)) = sqrt(6)/2;x1:Number;y1:Number;x2:Number;y2:Number",
    "query_expressions": "(x1)^2+(x2)^2",
    "answer_expressions": "3",
    "fact_spans": "[[[5, 8]], [[9, 51]], [[53, 70]], [[73, 90]], [[159, 162]], [[9, 51]], [[53, 70]], [[73, 90]], [[5, 94]], [[53, 94]], [[96, 156]], [[53, 70]], [[53, 70]], [[73, 90]], [[73, 90]]]",
    "query_spans": "[[[169, 192]]]",
    "process": "When the slope of line $ l $ does not exist, points $ P $ and $ Q $ are symmetric with respect to the $ x $-axis, so $ x_{2}=x_{1} $, $ y_{2}=-y_{1} $. Since $ P(x_{1},y_{1}) $ lies on the ellipse, $ \\frac{x_{1}^{2}}{3}+\\frac{y_{1}^{2}}{2}=1 $. Also, $ S_{\\triangle OPQ}=\\frac{\\sqrt{6}}{2} $, so $ |x_{1}||y_{1}|=\\frac{\\sqrt{6}}{2} $. Hence, $ |x_{1}|=\\frac{\\sqrt{6}}{2} $, $ |y_{1}|=1 $, so $ x_{1}^{2}+x_{2}^{2}=3 $. When the slope of line $ l $ exists, let the equation of line $ l $ be $ y=kx+m $, where $ m\\neq0 $. Substituting into the ellipse $ C: \\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1 $, we obtain $ (2+3k^{2})x^{2}+6kmx+3(m^{2}-2)=0 $. From $ \\Delta=36k^{2}m^{2}-12(2+3k^{2})(m^{2}-2)>0 $, we solve to get $ 3k^{2}+2>m^{2} $. Thus, $ 3k^{2}+2=2m^{2} $ and satisfies $ 3k^{2}+2>m^{2} $. Therefore, $ x_{1}^{2}+x_{2}^{2}=(x_{1}+x_{2})^{2}-2x_{1}x_{2}=\\left(\\frac{-6km}{2+3k^{2}}\\right)^{2}-2\\times\\frac{3m^{2}-6}{2+3k^{2}}=3 $. In conclusion, $ x_{1}^{2}+x_{2}^{2}=3 $."
  },
  {
    "text": "Parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$, $M$ is a point on parabola $C$, $O$ is the coordinate origin. If the circumcircle of $\\Delta O F M$ is tangent to the directrix of parabola $C$, and the area of this circle is $9 \\pi$, then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;O: Origin;F: Point;M: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C)=F;PointOnCurve(M, C);IsTangent(CircumCircle(TriangleOf(O,F,M)),Directrix(C));Area(CircumCircle(TriangleOf(O,F,M))) = 9*pi",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 26], [38, 44], [77, 83]], [[105, 108]], [[48, 51]], [[30, 33]], [[34, 37]], [[8, 26]], [[0, 26]], [[0, 33]], [[34, 47]], [[58, 88]], [[58, 103]]]",
    "query_spans": "[[[105, 110]]]",
    "process": "According to the properties of a circle and the definition of a parabola, set up equations to solve. \\because the circumcircle of $4OFM$ is tangent to the directrix of parabola $C$, \\therefore the distance from the center of the circumcircle of $AOFM$ to the directrix equals the radius of the circle. \\because the area of the circle is $9\\pi$, \\therefore the radius of the circle is $3$. Also, \\because the center lies on the perpendicular bisector of $OF$, $OF = \\frac{p}{2}$, \\therefore $\\frac{p}{2} + \\frac{p}{4} = 3$, $p = 4$."
  },
  {
    "text": "If the line $l$: $y=\\sqrt{3}(x-1)$ intersects the right branch of the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ at two distinct points, then the range of the slope $k$ of the asymptotes of hyperbola $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;b>0;Expression(C) = (x^2/4 - y^2/b^2 = 1);Expression(l) = (y = sqrt(3)*(x -1 ));NumIntersection(l, RightPart(C)) = 2;k: Number;Slope(Asymptote(C)) = k",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-sqrt(3), -3/2) + (3/2, sqrt(3))",
    "fact_spans": "[[[1, 25]], [[26, 80], [92, 98]], [[34, 80]], [[34, 80]], [[26, 80]], [[1, 25]], [[1, 91]], [[104, 107]], [[92, 107]]]",
    "query_spans": "[[[104, 114]]]",
    "process": "According to the problem, solve the system of equations  \n\\begin{cases}y=\\sqrt{3}(x-1)\\\\\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1\\end{cases},  \nsimplifying yields (b^{2}-12)x^{2}+24x-12-4b^{2}=0. To ensure that the line l: y=\\sqrt{3}(x-1) intersects the right branch of the hyperbola C: \\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1 (b>0) at two distinct points, the following must hold:  \n\\begin{cases}a=24^{2}+4\\times(b^{2}-12)\\\\x_{1}+x_{2}=\\frac{-24}{b^{2}-12}>0\\\\x_{1}x_{2}=\\frac{-12-4b^{2}}{2}>0\\end{cases}\\times(b^{2}-12)(12+4b^{2})>0.  \nSince b>0, solving gives 3<b<F.2. Also, the equations of the asymptotes of the hyperbola are y=\\pm\\frac{b}{2}x. Therefore, the range of the slope k of the asymptotes of hyperbola C is (-\\sqrt{3},-\\frac{3}{2})\\cup(\\frac{3}{2},\\sqrt{3})."
  },
  {
    "text": "The eccentricity of the hyperbola $C$: $\\frac{x^{2}}{4}-y^{2}=1$ is? The equations of the asymptotes are?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/4 - y^2 = 1)",
    "query_expressions": "Eccentricity(C);Expression(Asymptote(C))",
    "answer_expressions": "sqrt(5)/2\ny=pm*(1/2)*x",
    "fact_spans": "[[[0, 33]], [[0, 33]]]",
    "query_spans": "[[[0, 39]], [[0, 46]]]",
    "process": "a^{2}=4,b^{2}=1\\thereforec^{2}=a^{2}+b^{2}=5, so the eccentricity e=\\frac{c}{a}=\\frac{\\sqrt{5}}{2}, the asymptotes are y=\\pm\\frac{b}{a}x=\\pm\\frac{1}{2}x,"
  },
  {
    "text": "Given that $P(x, y)$ is a moving point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, then the maximum value of $\\frac{4 x}{5}+\\frac{3 y}{4}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/25 + y^2/16 = 1);Coordinate(P) = (x1, y1);PointOnCurve(P, G);x1:Number;y1:Number",
    "query_expressions": "Max((4*x1)/5 + (3*y1)/4)",
    "answer_expressions": "5",
    "fact_spans": "[[[12, 51]], [[2, 11]], [[12, 51]], [[2, 11]], [[2, 55]], [[2, 11]], [[2, 11]]]",
    "query_spans": "[[[57, 92]]]",
    "process": "Since point P(x, y) is a moving point on the ellipse \\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1, we can set P(5\\cos\\theta,4\\sin\\theta). Hence, \\frac{4x}{5}+\\frac{3y}{4}=4\\cos\\theta+3\\sin\\theta=5\\sin(\\theta+\\varphi), where \\sin\\varphi=\\frac{4}{5}, \\cos\\varphi=\\frac{3}{5}. Therefore, the maximum value of \\frac{4x}{6}+\\frac{3y}{4} is 5."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, the line $l$ passing through point $F$ with slope $\\sqrt{2}$ intersects the parabola at points $A$ and $B$. The circle with diameter $AB$ intersects the $y$-axis at points $M$ and $N$. Let $H$ be the midpoint of segment $AB$. Then the value of $\\tan \\angle HMN$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;l: Line;PointOnCurve(F, l);Slope(l) = sqrt(2);A: Point;B: Point;Intersection(l, G) = {A, B};S: Circle;IsDiameter(LineSegmentOf(A, B), S) = True;M: Point;N: Point;Intersection(S, yAxis) = {M, N};H: Point;MidPoint(LineSegmentOf(A,B)) = H",
    "query_expressions": "Tan(AngleOf(H, M, N))",
    "answer_expressions": "2*sqrt(5)/5",
    "fact_spans": "[[[2, 16], [50, 53]], [[2, 16]], [[20, 23], [25, 29]], [[2, 23]], [[44, 49]], [[24, 49]], [[30, 49]], [[54, 57]], [[58, 61]], [[44, 63]], [[76, 77]], [[64, 77]], [[83, 86]], [[87, 90]], [[76, 92]], [[105, 108]], [[94, 108]]]",
    "query_spans": "[[[110, 133]]]",
    "process": "As shown in the figure, the focus of the parabola is $ F(1,0) $, and the line $ l: y = \\sqrt{2}(x - 1) $ gives: $ x^2 - 4x + 1 = 0 $. Let the horizontal coordinates of $ A $ and $ B $ be $ x_{1} $, $ x_{2} $, respectively, then $ x_{1} + x_{2} = 4 $, $ \\therefore |AB| = x_{1} + 1 + x_{2} + 1 = 6 $, $ \\therefore |HA| = |HM| = \\frac{1}{2}|AB| = 3 $. Since $ x_{H} = \\frac{x_{1} + x_{2}}{2} = 2 $, let $ C $ be the midpoint of $ MN $, then $ CH \\perp MN $, $ |MC| = \\sqrt{|MH|^{2} - |GC|^{2}} = \\sqrt{3^{2} - 2^{2}} = \\sqrt{5} $, $ \\therefore \\tan\\angle HMN = \\tan\\angle HMC = \\frac{|CH|}{|MC|} = \\frac{2}{\\sqrt{5}} = \\frac{2\\sqrt{5}}{5} $,"
  },
  {
    "text": "The coordinates of the focus of the parabola $x^{2}=y$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = y)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,1/4)",
    "fact_spans": "[[[0, 12]], [[0, 12]]]",
    "query_spans": "[[[0, 19]]]",
    "process": "Since the focus of $x^{2}=2py$ ($p>0$) is $(0,\\frac{p}{2})$, the focus of the parabola $x^{2}=y$ is $(0,\\frac{1}{4})$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1$ $(0<b<2)$, with left and right foci $F_{1}$ and $F_{2}$ respectively, a line $l$ passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. If the maximum value of $|B F_{2}|+| A F_{2} |$ is $5$, then the value of $b$ is?",
    "fact_expressions": "l: Line;G: Ellipse;b: Number;B: Point;F2: Point;A: Point;F1: Point;0 < b;b < 2;Expression(G) = (x^2/4 + y^2/b^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(F1, l);Intersection(l, G) = {A, B};Max(Abs(LineSegmentOf(A, F2)) + Abs(LineSegmentOf(B, F2))) = 5",
    "query_expressions": "b",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[83, 88]], [[2, 50], [89, 91]], [[136, 139]], [[96, 99]], [[66, 73]], [[92, 95]], [[58, 65], [75, 82]], [[4, 50]], [[4, 50]], [[2, 50]], [[2, 73]], [[2, 73]], [[74, 88]], [[83, 101]], [[103, 134]]]",
    "query_spans": "[[[136, 143]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x(p>0)$ is exactly the right focus of the hyperbola $\\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1$, then $p$=?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;Expression(G) = (x^2/5 - y^2/4 = 1);p>0;Expression(H) = (y^2 = 2*(p*x));Focus(H) = RightFocus(G)",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[28, 66]], [[1, 22]], [[72, 75]], [[28, 66]], [[4, 22]], [[1, 22]], [[1, 70]]]",
    "query_spans": "[[[72, 77]]]",
    "process": "The focus of the parabola $ y^{2}=2px $ ($ p>0 $) has coordinates $ \\left(\\frac{p}{2},0\\right) $. For the hyperbola $ \\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1 $, $ a^{2}=5 $, $ b^{2}=4 $, $ \\therefore c=\\sqrt{a^{2}+b^{2}}=3 $, $ \\therefore $ the right focus of the hyperbola $ \\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1 $ is $ (3,0) $, then $ \\frac{p}{2}=3 $, yielding $ p=6 $."
  },
  {
    "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{16}+\\frac{y^{2}}{m}=1$ is $\\frac{5}{4}$, then $m=$?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/16 + y^2/m = 1);Eccentricity(G) = 5/4",
    "query_expressions": "m",
    "answer_expressions": "-9",
    "fact_spans": "[[[0, 39]], [[59, 62]], [[0, 39]], [[0, 57]]]",
    "query_spans": "[[[59, 64]]]",
    "process": "The hyperbola $\\frac{x^{2}}{16}+\\frac{y^{2}}{m}=1$, then $a^{2}=16$, $b^{2}=-m$, $\\therefore c^{2}=16-m$ $\\because e=\\frac{5}{4}$ $\\therefore e^{2}=\\frac{16-m}{16}=\\frac{25}{16}$ $\\therefore m=-9$"
  },
  {
    "text": "Given that the standard equation of ellipse $C$ is $\\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1$, what is the focal length of ellipse $C$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/49 + y^2/24 = 1)",
    "query_expressions": "FocalLength(C)",
    "answer_expressions": "10",
    "fact_spans": "[[[2, 7], [52, 57]], [[2, 50]]]",
    "query_spans": "[[[52, 62]]]",
    "process": "Given the standard equation of ellipse C is \\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1, a^{2}=49, b^{2}=24, so c^{2}=a^{2}-b^{2}=49-24=25, so c=5, therefore the focal distance of ellipse C is 2c=10"
  },
  {
    "text": "If $\\frac{x^{2}}{1+m}+\\frac{y^{2}}{1-m}=1$ represents a hyperbola, then the range of values for $m$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(m + 1) + y^2/(1 - m) = 1);m: Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-oo, -1) + (1, +oo)",
    "fact_spans": "[[[42, 45]], [[1, 45]], [[47, 50]]]",
    "query_spans": "[[[47, 57]]]",
    "process": "Analysis: According to the concept of a hyperbola, solve the inequality $(1+m)(1-m)<0$. Detailed solution:  \n$\\because \\frac{x^{2}}{1+m} + \\frac{y^{2}}{1-m} = 1$ represents a hyperbola,  \n$\\therefore (1+m)(1-m) < 0$, solving yields $m < -1$ or $m > 1$.  \n$\\therefore$ The range of $m$ is $m < -1$ or $m > 1$."
  },
  {
    "text": "If the curve $C$: $x y+3 x+k y+2=0$, when $k$=? does the curve pass through the point $(2, -1)$",
    "fact_expressions": "C: Curve;k: Number;G: Point;Expression(C) = (k*y + x*y + 3*x + 2 = 0);Coordinate(G) = (2, -1);PointOnCurve(G, C)",
    "query_expressions": "k",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 25], [34, 36]], [[27, 30]], [[38, 48]], [[1, 25]], [[38, 48]], [[34, 48]]]",
    "query_spans": "[[[27, 32]]]",
    "process": "Since the curve passes through the point (2, -1), substituting x=2, y=-1 into the equation gives 2\\times(-1)+3\\times2-k+2=0, solving for k yields k=6. 【Analysis】This problem examines the relationship between a curve's equation and the equation of a curve, belonging to basic problems. For such problems, given a point on the curve, to determine the parameter in the equation, simply substitute the coordinates of the point and calculate accurately."
  },
  {
    "text": "If the line $l$ passes through a focus of the hyperbola $\\frac{y^{2}}{16}-\\frac{x^{2}}{8}=1$ and is perpendicular to the real axis, intersecting the hyperbola at points $A$ and $B$, then the length of segment $AB$ is?",
    "fact_expressions": "l: Line;G: Hyperbola;B: Point;A: Point;Expression(G) = (-x^2/8 + y^2/16 = 1);PointOnCurve(OneOf(Focus(G)),l);IsPerpendicular(l,RealAxis(G));Intersection(l, G) = {A, B}",
    "query_expressions": "Length(LineSegmentOf(A,B))",
    "answer_expressions": "4",
    "fact_spans": "[[[55, 60]], [[3, 42], [61, 64]], [[70, 73]], [[66, 69]], [[3, 42]], [[1, 60]], [[3, 60]], [[55, 75]]]",
    "query_spans": "[[[77, 88]]]",
    "process": "The hyperbola $\\frac{y^{2}}{16}-\\frac{x^{2}}{8}=1$ has $a=4$, $b=2\\sqrt{2}$, $c=\\sqrt{a^{2}+b^{2}}=2\\sqrt{6}$, yielding a focus at $(0,2\\sqrt{6})$, and the line $l: y=2\\sqrt{6}$. Substituting into the hyperbola equation gives $\\frac{24}{16}-\\frac{x^{2}}{8}=1$, solving yields $x=\\pm2$, then $|AB|=4$,"
  },
  {
    "text": "Given that the distance from a moving point $M$ to the right focus of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ is equal to the distance from $M$ to the line $x=-4$, then the equation of the trajectory of point $M$ is?",
    "fact_expressions": "G: Ellipse;H: Line;M: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Expression(H) = (x = -4);Distance(M, RightFocus(G)) = Distance(M, H)",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "y^2 = 16*x",
    "fact_spans": "[[[8, 46]], [[55, 63]], [[4, 7], [72, 75]], [[8, 46]], [[55, 63]], [[4, 68]]]",
    "query_spans": "[[[72, 82]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ with left and right foci $F_{1}$, $F_{2}$, points $A$, $B$ lie on the ellipse ($A$, $B$ may coincide) and satisfy $\\overrightarrow{F_{1} A}=\\lambda \\overrightarrow{F_{2} B}(\\lambda \\in[\\frac{1}{3}, 3])$. Then the range of $|F_{1} A|+|F_{2} B|$ is?",
    "fact_expressions": "C: Ellipse;F1: Point;A: Point;F2: Point;B: Point;lambda: Number;Expression(C) = (x^2/4 + y^2/3 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(A, C);PointOnCurve(B, C);VectorOf(F1, A) = lambda*VectorOf(F2, B);In(lambda, [1/3, 3])",
    "query_expressions": "Range(Abs(LineSegmentOf(F1, A)) + Abs(LineSegmentOf(F2, B)))",
    "answer_expressions": "[3, 4]",
    "fact_spans": "[[[2, 44], [76, 78]], [[50, 57]], [[66, 71]], [[58, 65]], [[72, 75]], [[95, 183]], [[2, 44]], [[2, 65]], [[2, 65]], [[67, 79]], [[67, 79]], [[95, 183]], [[95, 183]]]",
    "query_spans": "[[[185, 213]]]",
    "process": "Draw the figure, and use the symmetry of the ellipse to transform the problem into finding the range of chord lengths passing through the foci; combine this with a, b, c and geometric properties to solve. As shown in the figure, from $\\overrightarrow{F_{1}A}=\\lambda\\overrightarrow{F_{2}B}$ ($\\lambda\\in[\\frac{1}{3},3]$), it follows that the slopes of lines $AB'$ and $BF_{2}$ are equal. By the symmetry of the ellipse, $|BF_{2}|=|B'F_{1}|$. When points $A$ and $B$ are both at the left vertex, $|F_{1}A|=a-c=1$, $|F_{2}B|=a+c=3$, so $\\lambda=\\frac{1}{3}$. Similarly, when $A$ and $B$ are both at the right vertex, $\\lambda=3$. Thus, the problem is transformed into finding the range of chord lengths passing through the foci of the ellipse. From geometric properties, the shortest chord length is the latus rectum $\\frac{2b^{2}}{a}=3^{n}$, and the longest chord length is the major axis $2a=4$. Therefore, the range of $|F_{1}A|+|F_{2}B|$ is $[3,4]$."
  },
  {
    "text": "Given that the distance from a vertex of the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{m}=1$ $(m>0)$ to its asymptote is $1$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;m: Number;m>0;Expression(G) = (x^2/2 - y^2/m = 1);Distance(OneOf(Vertex(G)), Asymptote(G)) = 1",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 45], [63, 66]], [[5, 45]], [[5, 45]], [[2, 45]], [[2, 61]]]",
    "query_spans": "[[[63, 72]]]",
    "process": "By the given condition, one asymptote of the hyperbola is $\\sqrt{m}x-\\sqrt{2}y=0$, and one vertex is $(\\sqrt{2},0)$. Then, using the distance from a point to a line, we obtain $\\frac{|\\sqrt{m}\\cdot\\sqrt{2}|}{\\sqrt{m+2}}=1$, solving gives $m=2$. Therefore, the eccentricity $e=\\sqrt{1+\\frac{m}{2}}=\\sqrt{2}$"
  },
  {
    "text": "The line $y = x + m$ intersects the curve $y = \\sqrt{3 - \\frac{3x^2}{4}}$ at two points; then the range of values for $m$ is?",
    "fact_expressions": "G: Line;m: Number;H: Curve;Expression(G) = (y = m + x);Expression(H) = (y = sqrt(3 - 3*x^2/4));NumIntersection(G, H) = 2",
    "query_expressions": "Range(m)",
    "answer_expressions": "[2, sqrt(7))",
    "fact_spans": "[[[0, 9]], [[50, 53]], [[10, 42]], [[0, 9]], [[10, 42]], [[0, 48]]]",
    "query_spans": "[[[50, 60]]]",
    "process": "According to the problem, the curve $ y = \\sqrt{3 - \\frac{3x^{2}}{4}} $ can be rewritten as $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $, ($ y \\geqslant 0 $), which represents the upper half of the ellipse $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $, as shown in the figure. If  \n$$\n\\begin{cases}\ny = x + m \\\\\n\\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1\n\\end{cases}\n$$\nthen it follows: $ 7x^{2} + 8mx + 4m^{2} - 12 = 0 $,  \n$ \\triangle = 64m^{2} - 4 \\times 7 \\times (4m^{2} - 12) = 0 $,  \nsolving gives $ m = \\pm \\sqrt{7} $.  \nFrom the graph, when $ m = \\sqrt{7} $, the line $ y = x + m $ is tangent to the upper half of the ellipse $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $. Meanwhile, when $ m = 2 $, the line $ y = x + 2 $ intersects the upper half of the ellipse $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $ at two points. In conclusion: when the line $ y = x + m $ and the curve $ y = \\sqrt{3 - \\frac{3x^{2}}{4}} $ have two common points, it must hold that $ 2 \\leqslant m < \\sqrt{7} $, i.e., the range of $ m $ is $ [2, \\sqrt{7}) $."
  },
  {
    "text": "The parabola $y^{2}=2 p x(p>0)$ passes through the center of the circle $x^{2}+y^{2}-4 x+8 y+19=0$, and point $A(3, m)$ lies on the parabola. Then the distance from $A$ to the focus $F$ of the parabola is?",
    "fact_expressions": "G: Parabola;p: Number;H: Circle;m: Number;A: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (8*y - 4*x + x^2 + y^2 + 19 = 0);Coordinate(A) = (3, m);Focus(G) = F;PointOnCurve(A, G);PointOnCurve(Center(H), G)",
    "query_expressions": "Distance(A, F)",
    "answer_expressions": "5",
    "fact_spans": "[[[0, 21], [63, 66], [75, 78]], [[3, 21]], [[22, 49]], [[53, 62]], [[53, 62], [71, 74]], [[80, 83]], [[3, 21]], [[0, 21]], [[22, 49]], [[53, 62]], [[75, 83]], [[53, 69]], [[0, 52]]]",
    "query_spans": "[[[71, 88]]]",
    "process": "Find the coordinates of the center of the circle, then obtain the equation of the parabola, followed by the equation of the directrix of the parabola. Combining with the definition of a parabola, find the distance from point A to the focus F of the parabola. The center of the circle $x^{2}+y^{2}-4x+8y+19=0$ is $(-\\frac{-4}{2},-\\frac{8}{2})$, that is, $(2,-4)$. Substituting into the parabola equation gives $(-4)^{2}=2p\\times2\\Rightarrow p=4$, so the equation of the parabola is $y^{2}=8x$, and its directrix equation is $x=-2$. For point $A(3,m)$, the distance from A to the focus F of the parabola equals the distance from A to the directrix of the parabola, which is $3+2=5$."
  },
  {
    "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{m-1}-\\frac{y^{2}}{m+1}=1$ is $\\frac{3}{2}$, then the value of the real number $m$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(m - 1) - y^2/(m + 1) = 1);m: Real;Eccentricity(G) = 3/2",
    "query_expressions": "m",
    "answer_expressions": "pm*9",
    "fact_spans": "[[[0, 42]], [[0, 42]], [[62, 67]], [[0, 60]]]",
    "query_spans": "[[[62, 71]]]",
    "process": "According to the problem, when the foci are on the x-axis, $ e^{2} = 1 + \\frac{b^{2}}{a^{2}} = 1 + \\frac{m+1}{m-1} = \\frac{9}{4} $, so $ m = 9 $; when the foci are on the y-axis, $ e^{2} = 1 + \\frac{b^{2}}{a^{2}} = 1 + \\frac{-(m-1)}{-(m+1)} = \\frac{9}{4} $, so $ m = -9 $. In conclusion, the values of $ m $ are $ 9, -9 $. This question mainly examines the eccentricity of a hyperbola, the coordinate axis on which the foci of the hyperbola lie, and the mathematical thinking method of case analysis. It is a basic problem."
  },
  {
    "text": "Point $P$ lies on the curve $\\frac{x^{2}}{2}-y^{2}=1$, point $Q$ lies on the curve $x^{2}+(y-3)^{2}=4$, the midpoint of segment $PQ$ is $M$, and $O$ is the origin. Then the minimum length of segment $OM$ is?",
    "fact_expressions": "P: Point;G1: Curve;Expression(G1) = (x^2/2 - y^2 = 1);PointOnCurve(P, G1);Q: Point;G2: Curve;Expression(G2) = (x^2 + (y - 3)^2 = 4);PointOnCurve(Q, G2);M: Point;MidPoint(LineSegmentOf(P, Q)) = M;O: Origin",
    "query_expressions": "Min(Length(LineSegmentOf(O, M)))",
    "answer_expressions": "sqrt(2)-1",
    "fact_spans": "[[[0, 4]], [[5, 32]], [[5, 32]], [[0, 33]], [[34, 38]], [[39, 60]], [[39, 60]], [[34, 61]], [[73, 76]], [[62, 76]], [[77, 80]]]",
    "query_spans": "[[[87, 101]]]",
    "process": "Let M(x,y), P(x_{1},y_{1}), then Q(2x-x_{1},2y-y_{1}), then (2x-x_{1})^{2}+(2y-y_{1}-3)^{2}=4, which can be rewritten as (x-\\frac{x_{1}}{2})^{2}+(y-\\frac{y_{1}}{2}-\\frac{3}{2})^{2}=1. Let C(\\frac{x_{1}}{2},\\frac{y_{1}}{2}+\\frac{3}{2}), then |OM|\\geqslant|OC|-1, |OC|^{2}=\\frac{x_{1}^{2}}{4}+\\frac{(y_{1}+3)^{2}}{4}=\\frac{3y_{1}^{2}+6y_{1}+11}{4}=\\frac{3(y_{1}+1)^{2}+8}{4}\\geqslant2, |OC|\\geqslant\\sqrt{2}, |OM|\\geqslant\\sqrt{2}-1, that is, the minimum length of segment OM is \\sqrt{2}-1, \\frac{4}{"
  },
  {
    "text": "The center of the ellipse is at the origin, the foci are on the coordinate axes, and the two vertices are $(3 , 0)$, $(0 , \\sqrt{2})$. Then the equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;Center(G) = O;O: Origin;PointOnCurve(Focus(G), axis);D1: Point;D2: Point;Coordinate(D1) = (3, 0);Coordinate(D2) = (0, sqrt(2));Vertex(G)={D1,D2}",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9+y^2/2=1",
    "fact_spans": "[[[0, 2], [55, 57]], [[0, 10]], [[6, 10]], [[0, 18]], [[25, 34]], [[36, 52]], [[25, 34]], [[36, 52]], [[0, 52]]]",
    "query_spans": "[[[55, 62]]]",
    "process": ""
  },
  {
    "text": "The eccentricities of confocal ellipse and hyperbola are $e_{1}$, $e_{2}$ respectively. If the minor axis length of the ellipse is $2$ times the imaginary axis length of the hyperbola, then the maximum value of $\\frac{1}{e_{1}}+\\frac{2}{e_{2}}$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;e1:Number;e2:Number;Focus(G)=Focus(H);Eccentricity(H)=e1;Eccentricity(G)=e2;Length(MinorAxis(H))=2*Length(ImageinaryAxis(G))",
    "query_expressions": "Max(2/e2 + 1/e1)",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[7, 10], [43, 46]], [[4, 6], [36, 38]], [[17, 24]], [[27, 34]], [[0, 10]], [[4, 34]], [[4, 34]], [[36, 55]]]",
    "query_spans": "[[[57, 96]]]",
    "process": "Let the semi-focal length of the ellipse and hyperbola be $ c $, and the length of the imaginary axis of the hyperbola be $ 2b_{2} $, then the minor axis length of the ellipse is $ 2b_{1} = 4b_{2} $, that is, $ b_{1} = 2b_{2} $, so $ b_{1}^{2} = 4b_{2}^{2} $, i.e., $ a_{1}^{2} - c^{2} = 4(c^{2} - a_{2}^{2}) $, i.e., $ a_{1}^{2} + 4a_{2}^{2} = 5c^{2} $, i.e., $ \\left(\\frac{a_{1}}{c}\\right)^{2} + 4\\left(\\frac{a_{2}}{c}\\right)^{2} = 5 $, i.e., $ \\frac{1}{e_{1}^{2}} + \\frac{4}{e_{2}^{2}} = 5 $, then $ \\left(\\frac{1}{e_{1}} + \\frac{2}{e_{2}}\\right)^{2} = \\frac{1}{e_{1}^{2}} + \\frac{4}{e_{2}^{2}} + 2 \\cdot \\frac{1}{e_{1}} \\cdot \\frac{2}{e_{2}} \\leqslant 2\\left(\\frac{1}{e_{1}^{2}} + \\frac{4}{e_{2}^{2}}\\right) = 10 $ (equality holds if and only if $ \\frac{1}{e_{1}^{2}} = \\frac{4}{e_{2}^{2}} $, i.e., $ e_{2} = 2e_{1} $), i.e., $ \\frac{1}{e_{1}} + \\frac{2}{e_{2}} \\leqslant \\sqrt{10} $, i.e., the maximum value of $ \\frac{1}{e_{1}} + \\frac{2}{e_{2}} $ is $ \\sqrt{10} $."
  },
  {
    "text": "The line $y = x + 1$ intersects the ellipse $m x^{2} + n y^{2} = 1$ $(m > n > 0)$ at points $A$ and $B$. If the x-coordinate of the midpoint of chord $AB$ is $-\\frac{1}{3}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "H: Line;Expression(H) = (y = x + 1);G: Ellipse;Expression(G) = (m*x^2 + n*y^2 = 1);m: Number;n: Number;m > n;n > 0;Intersection(H, G) = {A, B};A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G) = True;XCoordinate(MidPoint(LineSegmentOf(A, B))) = -1/3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[0, 9]], [[0, 9]], [[10, 37], [83, 85]], [[10, 37]], [[12, 37]], [[12, 37]], [[12, 37]], [[12, 37]], [[0, 51]], [[40, 43]], [[46, 49]], [[10, 58]], [[54, 81]]]",
    "query_spans": "[[[83, 92]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a moving point on the parabola $y^{2}=4x$, and $Q(0,3)$ is a fixed point, what is the minimum value of the sum of the distance from point $P$ to point $Q$ and the distance from point $P$ to the directrix of the parabola?",
    "fact_expressions": "G: Parabola;Q: Point;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(Q) = (0, 3);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P,Q)+Distance(P,Directrix(G)))",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[6, 20], [57, 60]], [[28, 36], [44, 48]], [[2, 5], [39, 43], [52, 56]], [[6, 20]], [[28, 36]], [[2, 25]]]",
    "query_spans": "[[[39, 74]]]",
    "process": "The focus of the parabola $ y^{2} = 4x $ is $ F(1,0) $. According to the definition of a parabola, the distance from point $ P $ to the directrix is equal to the distance from point $ P $ to the focus. Let the distance from point $ P $ to the directrix of the parabola be $ d $, so $ |PQ| + d = |PQ| + |PF| $. It follows that when points $ P $, $ Q $, and $ F $ are collinear, the sum of the distance from point $ P $ to point $ Q $ and the distance from point $ P $ to the directrix is minimized. Therefore, the minimum value is $ |PF| = \\sqrt{1^{2} + 3^{2}} = \\sqrt{10} $."
  },
  {
    "text": "The hyperbola with foci on the $x$-axis passes through the point $P(4 \\sqrt{2},-3)$, and the lines connecting $Q(0,5)$ to the two foci are perpendicular to each other. Then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;P: Point;Q: Point;F1: Point;F2: Point;Coordinate(P) = (4*sqrt(2), -3);Coordinate(Q) = (0, 5);PointOnCurve(Focus(G), xAxis);PointOnCurve(P, G);Focus(G) = {F1, F2};IsPerpendicular(LineSegmentOf(Q, F1), LineSegmentOf(Q, F2))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16 - y^2/9 = 1",
    "fact_spans": "[[[9, 12], [57, 60]], [[14, 33]], [[35, 43]], [], [], [[14, 33]], [[35, 43]], [[0, 12]], [[9, 33]], [[9, 47]], [[9, 54]]]",
    "query_spans": "[[[57, 67]]]",
    "process": "Let the foci be $F_{1}(-c,0)$, $F_{2}(c,0)$ $(c>0)$, then from $QF_{1}\\bot QF_{2}$, we get $k_{QF_{1}}\\cdot k_{QF_{2}}=-1$, $\\therefore \\frac{5}{c}\\cdot\\frac{5}{2}\\frac{x}{x}=-1$, $\\therefore c=5$. $\\because$ the hyperbola passes through $(4\\sqrt{2},-3)$, $\\therefore \\frac{32}{a^{2}}-\\frac{9}{b^{2}}=1$, and $\\because c^{2}=a^{2}+b^{2}=25$, $\\therefore a^{2}=16$, $b^{2}=9$, $\\therefore$ the standard equation of the hyperbola is $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$."
  },
  {
    "text": "Given the parabola $E$: $y^{2}=2 p x(p>0)$ with focus $F$, a line $L$ passing through point $F$ intersects the parabola $E$ at points $A$ and $B$, and intersects the circle $(x-\\frac{p}{2})^{2}+y^{2}=p^{2}$ at points $C$ and $D$. If $|A B|=3|C D|$, then what is the slope of line $L$?",
    "fact_expressions": "E: Parabola;p: Number;p > 0;Expression(E) = (y^2 = 2*p*x);F: Point;Focus(E) = F;L: Line;PointOnCurve(F, L);A: Point;B: Point;Intersection(L, E) = {A, B};H: Circle;Expression(H) = ((x-p/2)^2 + y^2 = p^2);C: Point;D: Point;Abs(LineSegmentOf(A, B)) = 3*Abs(LineSegmentOf(C, D))",
    "query_expressions": "Slope(L)",
    "answer_expressions": "pm*sqrt(2)/2",
    "fact_spans": "[[[2, 28], [48, 54]], [[10, 28]], [[10, 28]], [[2, 28]], [[32, 35], [37, 41]], [[2, 35]], [[42, 47], [69, 74], [140, 145]], [[36, 47]], [[56, 59]], [[62, 65]], [[36, 67]], [[75, 109]], [[75, 109]], [[111, 114]], [[116, 120]], [[124, 138]]]",
    "query_spans": "[[[140, 150]]]",
    "process": ""
  },
  {
    "text": "The distance from the right focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ to the line $y = \\sqrt{3} x$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);H: Line;Expression(H) = (y = sqrt(3)*x)",
    "query_expressions": "Distance(RightFocus(G), H)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[0, 37]], [[0, 37]], [[42, 60]], [[42, 60]]]",
    "query_spans": "[[[0, 65]]]",
    "process": "Since the equation of the ellipse is $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, so $c^{2}=a^{2}-b^{2}=1$, thus the coordinates of the right focus are $(1,0)$. The line equation in general form is $\\sqrt{3}x-y=0$. By the point-to-line distance formula, we get $d=\\frac{|\\sqrt{3}\\times1|}{\\sqrt{(\\sqrt{3})^{2}+(-1)^{2}}}=\\frac{\\sqrt{3}}{2}$"
  },
  {
    "text": "Let the lower and upper vertices of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ be $B_1$ and $B_2$, respectively. If point $P$ is a point on the ellipse such that the slopes of lines $PB_1$ and $PB_2$ are $\\frac{1}{4}$ and $-1$, respectively, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;B1: Point;B2:Point;P: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LowerVertex(G)=B1;UpperVertex(G)=B2;PointOnCurve(P, G);Slope(LineOf(P,B1))=1/4;Slope(LineOf(P,B2))=-1",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[1, 53], [82, 84], [133, 135]], [[3, 53]], [[3, 53]], [[62, 68]], [[70, 75]], [[77, 81]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 75]], [[1, 75]], [[77, 88]], [[90, 131]], [[90, 131]]]",
    "query_spans": "[[[133, 141]]]",
    "process": ""
  },
  {
    "text": "Given that the left vertex of the ellipse $3 x^{2}+4 y^{2}=12$ is $A$, and the upper vertex is $B$, then $|A B|=$?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;Expression(G) = (3*x^2 + 4*y^2 = 12);LeftVertex(G) = A;UpperVertex(G) = B",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "sqrt(7)",
    "fact_spans": "[[[2, 24]], [[29, 32]], [[37, 40]], [[2, 24]], [[2, 32]], [[2, 40]]]",
    "query_spans": "[[[42, 51]]]",
    "process": "The ellipse $3x^{2}+4y^{2}=12$ is converted to the standard equation $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$. Since the left vertex is $A(-2,0)$ and the upper vertex is $B(0,\\sqrt{3})$, $\\left|AB\\right|=\\sqrt{4+3}=\\sqrt{7}$."
  },
  {
    "text": "If the left focus of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{p^{2}}=1$ lies on the directrix of the parabola $y^{2}=2 p x$, then the value of $p$ is?",
    "fact_expressions": "H: Ellipse;Expression(H) = (x^2/5 + y^2/p^2 = 1);G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;PointOnCurve(LeftFocus(H), Directrix(G)) = True",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 42]], [[1, 42]], [[47, 63]], [[47, 63]], [[69, 72]], [[1, 67]]]",
    "query_spans": "[[[69, 76]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{100}+\\frac{y^{2}}{36}=1$, if the distance from $P$ to the right directrix of the ellipse is $\\frac{17}{2}$, then what is the distance from point $P$ to the right focus?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/100 + y^2/36 = 1);P: Point;PointOnCurve(P, G);Distance(P, RightDirectrix(G)) = 17/2",
    "query_expressions": "Distance(P, RightFocus(G))",
    "answer_expressions": "34/5",
    "fact_spans": "[[[6, 46], [56, 58]], [[6, 46]], [[2, 5], [52, 55], [81, 85]], [[2, 50]], [[52, 79]]]",
    "query_spans": "[[[56, 93]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the points on the parabola $y^{2}=12 x$ whose distance from the focus is equal to $9$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 12*x);P:Point;PointOnCurve(P,G);Distance(P,Focus(G))=9",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "{(6,6*sqrt(2)),(6,-6*sqrt(2))}",
    "fact_spans": "[[[0, 15], [17, 18]], [[0, 15]], [[29, 30]], [[0, 30]], [[16, 30]]]",
    "query_spans": "[[[29, 35]]]",
    "process": "According to the property that the distance from a point on a parabola to its focus equals the distance from the point to the directrix, the x-coordinate of the desired point can be obtained, thus leading to the conclusion. The directrix of the parabola $ y^{2} = 12x $ is $ x = -3 $. Since the distance from a point on the parabola $ y^{2} = 12x $ to the focus is 9, based on the property that the distance from a point on the parabola to the focus equals the distance from the point to the directrix, the x-coordinate of the desired point is 6. Substituting into the parabolic equation yields $ y^{2} = 72 $, so $ y = \\pm 6\\sqrt{2} $. Therefore, the coordinates of the desired point are $ (6, \\pm 6\\sqrt{2}) $."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2 px(p>0)$ be $F$, and point $A(0 , 2)$. If the midpoint $B$ of segment $F A$ lies on the parabola, then the distance from $B$ to the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;p: Number;F: Point;A: Point;B: Point;p>0;Expression(G) = (y^2 = 2*p*x);Coordinate(A) = (0, 2);Focus(G) = F;MidPoint(LineSegmentOf(F,A))=B;PointOnCurve(B, G)",
    "query_expressions": "Distance(B, Directrix(G))",
    "answer_expressions": "3*sqrt(2)/4",
    "fact_spans": "[[[1, 21], [56, 59], [67, 70]], [[4, 21]], [[25, 28]], [[29, 40]], [[52, 55], [62, 65]], [[4, 21]], [[1, 21]], [[29, 40]], [[1, 28]], [[42, 55]], [[52, 60]]]",
    "query_spans": "[[[62, 77]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ pass through the point $P(1 , \\frac{4}{3})$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (1, 4/3);PointOnCurve(P, Asymptote(G))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[2, 61], [90, 93]], [[5, 61]], [[5, 61]], [[66, 87]], [[5, 61]], [[5, 61]], [[2, 61]], [[66, 87]], [[2, 87]]]",
    "query_spans": "[[[90, 99]]]",
    "process": ""
  },
  {
    "text": "If the curve $y=\\sqrt{|x^{2}-9|}$ and the line $x+y-m=0$ have one intersection point, then the range of real values for $m$ is?",
    "fact_expressions": "G: Line;m: Real;H: Curve;Expression(G) = (-m + x + y = 0);Expression(H) = (y = sqrt(Abs(x^2 - 9)));NumIntersection(H, G)=1",
    "query_expressions": "Range(m)",
    "answer_expressions": "{-3}+[0，3)+(3*sqrt(2),+oo)",
    "fact_spans": "[[[24, 35]], [[42, 47]], [[1, 23]], [[24, 35]], [[1, 23]], [[1, 40]]]",
    "query_spans": "[[[42, 54]]]",
    "process": ""
  },
  {
    "text": "It is known that the angle of inclination of an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $45^{\\circ}$, and the hyperbola passes through the point $(3,1)$. Then the focal distance of the hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Inclination(OneOf(Asymptote(G))) = ApplyUnit(45, degree);H: Point;Coordinate(H) = (3, 1);PointOnCurve(H, G)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 58], [94, 97]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 81]], [[84, 92]], [[84, 92]], [[2, 92]]]",
    "query_spans": "[[[94, 103]]]",
    "process": "The asymptotes of the hyperbola are given by $ y = \\pm\\frac{b}{a}x $. From the problem, we have $ \\frac{b}{a} = 1 $, so $ b = a $. Therefore, the standard equation of the hyperbola is $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{a^{2}} = 1 $. Substituting the point $ (3,1) $ into this equation gives $ \\frac{9}{a^{2}} - \\frac{1}{a^{2}} = 1 $, yielding $ a = b = 2\\sqrt{2} $. Thus, the focal distance of the hyperbola is $ 2\\sqrt{a^{2}+b^{2}} = 2\\times4 = 8 $."
  },
  {
    "text": "Given the parabola $y^{2}=4 x$, line $l$ passes through the fixed point $(-1,0)$. When line $l$ has only one common point with the parabola, what is the slope of line $l$?",
    "fact_expressions": "l: Line;G: Parabola;H: Point;Expression(G) = (y^2 = 4*x);Coordinate(H) = (-1, 0);PointOnCurve(H,l);NumIntersection(l, G) = 1",
    "query_expressions": "Slope(l)",
    "answer_expressions": "{0,pm*1}",
    "fact_spans": "[[[17, 25], [55, 60], [37, 42]], [[2, 16], [43, 46]], [[28, 36]], [[2, 16]], [[28, 36]], [[18, 36]], [[37, 53]]]",
    "query_spans": "[[[55, 65]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the foci of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(pm*5,0)",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 46]]]",
    "process": "From the hyperbola equation $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, we have $a=3$, $b=4$, $c=\\sqrt{a^{2}+b^{2}}=5$, so the coordinates of the hyperbola's foci are $(\\pm5,0)$,"
  },
  {
    "text": "If the curve $\\frac{x^{2}}{4+k}+\\frac{y^{2}}{1-k}=1$ represents a hyperbola, then the range of values for $k$ is?",
    "fact_expressions": "G: Hyperbola;H: Curve;k: Number;Expression(H) = (x^2/(k + 4) + y^2/(1 - k) = 1);H = G",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-oo, -4)+(1, +oo)",
    "fact_spans": "[[[44, 47]], [[1, 42]], [[49, 52]], [[1, 42]], [[1, 47]]]",
    "query_spans": "[[[49, 59]]]",
    "process": "\\because the curve \\frac{x^{2}}{4+k}+\\frac{y^{2}}{1-k}=1 represents a hyperbola, \\therefore (4+k)(1-k)<0, solving gives k<-4 or k>1"
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{m^{2}+n}-\\frac{y^{2}}{3 m^{2}-n}=1$ represents a hyperbola, and the distance between the two foci of this hyperbola is $4$, then the range of values for $n$ is?",
    "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;m: Number;n: Number;Expression(G) = (x^2/(m^2 + n) - y^2/(3*m^2 - n) = 1);Focus(G) = {F1, F2};Distance(F1, F2) = 4",
    "query_expressions": "Range(n)",
    "answer_expressions": "(-1, 3)",
    "fact_spans": "[[[55, 58], [61, 64]], [], [], [[4, 53]], [[77, 80]], [[2, 58]], [[61, 67]], [[61, 75]]]",
    "query_spans": "[[[77, 87]]]",
    "process": ""
  },
  {
    "text": "If the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{m}=1$ has the same foci as the hyperbola $x^{2}-8 y^{2}=8$, then what is the value of $m$?",
    "fact_expressions": "H: Ellipse;Expression(H) = (x^2/12 + y^2/m = 1);m: Number;G: Hyperbola;Expression(G) = (x^2 - 8*y^2 = 8);Focus(H) =Focus(G)",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 39]], [[1, 39]], [[67, 70]], [[40, 60]], [[40, 60]], [[1, 65]]]",
    "query_spans": "[[[67, 74]]]",
    "process": "Transforming the hyperbola equation into standard form gives: \\frac{x^{2}}{8}-y^{2}=1, so the coordinates of the foci of the hyperbola are (\\pm3,0). Since the ellipse and the hyperbola have the same foci, from the equation of the ellipse we obtain: m=12-9=3"
  },
  {
    "text": "A line $ l $ with slope $ 2 $ intersects the parabola $ y^{2} = 2 p x $ ($ p > 0 $) at points $ A $ and $ B $. If the midpoint of $ A $ and $ B $ is $ M(2,1) $, then the value of $ p $ is?",
    "fact_expressions": "l: Line;G: Parabola;p: Number;M: Point;A: Point;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(M) = (2, 1);Slope(l)= 2;Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A,B))=M",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[7, 12]], [[13, 34]], [[71, 74]], [[61, 69]], [[37, 40], [48, 51]], [[41, 44], [52, 55]], [[16, 34]], [[13, 34]], [[61, 69]], [[0, 12]], [[7, 46]], [[48, 69]]]",
    "query_spans": "[[[71, 78]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}) be such that \\begin{cases}y_{1}=2px_{1}\\\\y_{2}=2px_{2}\\end{cases}, then y_{1}-y_{2}^{2}=2p(x_{1}-x_{2}), i.e., (y_{1}-y_{2})(y_{1}+y_{2})=2p(x_{1}-x_{2}). Since the slope of line l is 2, and the midpoint of points A and B is M(2,1), it follows that \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{2p}{v_{+}v}, i.e., \\frac{2p}{2}=2, solving gives D=2, the answer is."
  },
  {
    "text": "Given the hyperbola equation $\\frac{x^{2}}{16}-\\frac{y^{2}}{8}=1$, point $P$ lies on the hyperbola and its distance to one focus $F_{1}$ is $10$, with the other focus being $F_{2}$. Point $N$ is the midpoint of $P F_{1}$. Then the magnitude of $O N$ ($O$ is the coordinate origin) is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/8 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;OneOf(Focus(G)) = F1;OneOf(Focus(G)) = F2;Negation(F1=F2);Distance(P, F1) = 10;N: Point;MidPoint(LineSegmentOf(P, F1)) = N;O: Origin",
    "query_expressions": "LineSegmentOf(O, N)",
    "answer_expressions": "1, 9",
    "fact_spans": "[[[2, 5], [51, 54]], [[2, 45]], [[46, 50]], [[46, 55]], [[64, 71]], [[86, 93]], [[51, 71]], [[51, 93]], [[51, 93]], [[46, 79]], [[94, 98]], [[94, 111]], [[123, 126]]]",
    "query_spans": "[[[113, 135]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=2 m x(m>0)$ with focus $F$, a line passing through the focus $F$ intersects the parabola at points $A$ and $B$. The equation of the circle with diameter $A B$ is $x^{2}+y^{2}-2 x-2 t y+t^{2}-15=0$. Then $m$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(m*x));m: Number;m>0;F: Point;Focus(G) = F;L: Line;PointOnCurve(F, L) = True;Intersection(L, G) = {A, B};A: Point;B: Point;H: Circle;IsDiameter(LineSegmentOf(A,B),H) = True;Expression(H) = (x^2+y^2-2*x-2*t*y+t^2-15=0);t: Number",
    "query_expressions": "m",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 23], [41, 44]], [[2, 23]], [[106, 109]], [[5, 23]], [[27, 30], [34, 37]], [[2, 30]], [[38, 40]], [[31, 40]], [[38, 54]], [[45, 48]], [[49, 52]], [[65, 66]], [[55, 66]], [[65, 104]], [[70, 104]]]",
    "query_spans": "[[[106, 111]]]",
    "process": "By the given condition, the equation of the circle with AB as diameter is $(x-1)^{2}+(y-t)^{2}=16$, so the center coordinates are $(1,t)$ and the radius is 4. Therefore, the x-coordinate of the midpoint of chord AB is 1, and $|AB|=8$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $x_{1}+x_{2}=2$. From the definition of the parabola, we obtain $|AB|=x_{1}+x_{2}+m=8$, so $m=6$."
  },
  {
    "text": "Let point $P$ be a moving point on the parabola $C$: $y^{2}=4x$, $F$ be the focus of the parabola, and $O$ be the origin. Then the maximum value of $\\frac{|OP|}{|PF|}$ is?",
    "fact_expressions": "C: Parabola;O: Origin;P: Point;F: Point;Expression(C) = (y^2 = 4*x);PointOnCurve(P, C);Focus(C) = F",
    "query_expressions": "Max(Abs(LineSegmentOf(O, P))/Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[6, 25], [34, 37]], [[41, 44]], [[1, 5]], [[30, 33]], [[6, 25]], [[1, 29]], [[30, 40]]]",
    "query_spans": "[[[51, 78]]]",
    "process": "Let point P(x,y), then y^{2}=4x, so x\\geqslant0, we can obtain \\frac{|OP|}{|PF|}=\\sqrt{1+\\frac{2}{x+1}-\\frac{3}{(x+1)}}, use the basic properties of quadratic functions to find the maximum value of the quadratic function y=-3t^{2}+2t+1, thus obtaining the maximum of \\frac{|C}{|F|}\\frac{7P|}{F|}. Let point P(x,y), then y^{2}=4x, so x\\geqslant0. The directrix equation of parabola C is x=-1. By the definition of a parabola, we have |PF|. Therefore, \\frac{|OP|}{|PF|}=\\frac{\\sqrt{x^{2}+y^{2}}}{x+1}=\\sqrt{\\frac{x^{2}+4x}{x^{2}+2x+}}. Since x\\geqslant0, let t=\\frac{1}{x+1}\\in(0,1], y=-3t^{2}+2t+1=--. When and only when t=\\frac{1}{3}, the function y=-3t^{2}+2t+1 reaches the maximum value \\frac{4}{3}. Therefore, \\frac{1}{1}\\frac{2}{10P|}=\\frac{4}{3}."
  },
  {
    "text": "Let point $P$ be a point on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$, and let $F_{1}$, $F_{2}$ be the left and right foci of the hyperbola, respectively. Let $I$ be the incenter of $\\triangle P F_{1} F_{2}$. If $2(S_{\\triangle P F_{1} I}-S_{\\triangle P F_{2} I})=S_{\\triangle F_{1} F_{2} I}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;I: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;Incenter(TriangleOf(P, F1, F2)) = I;2*(Area(TriangleOf(P, F1, I)) - Area(TriangleOf(P, F2, I))) = Area(TriangleOf(F1, F2, I))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[6, 63], [85, 88], [212, 215]], [[9, 63]], [[9, 63]], [[1, 5]], [[67, 74]], [[75, 82]], [[95, 98]], [[9, 63]], [[9, 63]], [[6, 63]], [[1, 66]], [[67, 94]], [[67, 94]], [[95, 127]], [[129, 209]]]",
    "query_spans": "[[[212, 221]]]",
    "process": "As shown in the figure, let the incircle $ I $ of $ \\triangle PF_{1}F_{2} $ be tangent to the three sides $ F_{1}F_{2} $, $ PF_{1} $, $ PF_{2} $ at points $ E $, $ F $, $ G $ respectively. Connect $ IE $, $ IF $, $ IG $, then $ IE \\perp F_{1}F_{2} $, $ IF \\perp PF_{1} $, $ IG \\perp PF_{2} $, which are the altitudes of $ \\triangle IF_{1}F_{2} $, $ \\triangle IPF_{1} $, $ \\triangle IPF_{2} $, respectively. $ \\therefore S_{\\triangle IPF_{1}} = \\frac{1}{2} \\times |PF_{1}| \\times |IF| = \\frac{r}{2}|PF_{1}| $, $ S_{\\triangle IPF_{2}} = \\frac{1}{2} \\times |PF_{2}| \\times |IG| = \\frac{r}{2}|PF_{2}| $, $ S_{\\triangle IF_{1}F_{2}} = \\frac{1}{2} \\times |F_{1}F_{2}| \\times |IE| = \\frac{r}{2}|F_{1}F_{2}| $, where $ r $ is the inradius of $ \\triangle PF_{1}F_{2} $. Since $ S_{\\triangle IPF_{1}} = S_{\\triangle IPF_{2}} + \\frac{1}{2}S_{\\triangle IF_{1}F_{2}} $, $ \\therefore \\frac{r}{2}|PF_{1}| = \\frac{r}{2}|PF_{2}| + \\frac{r}{4}|F_{1}F_{2}| $, dividing both sides by $ \\frac{r}{2} $ yields: $ |PF_{1}| = |PF_{2}| + \\frac{1}{3}|F_{1}F_{2}| $. According to the definition of a hyperbola, $ |PF_{1} - PF_{2}| = 2a $, $ \\frac{1}{3}|F_{1}F_{2}| = c $. $ \\therefore 2a = c $, so the eccentricity is $ e = \\frac{c}{a} = 2 $."
  },
  {
    "text": "Given that the line $l$ passing through point $P(1, 0)$ with an inclination angle of $60^{\\circ}$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$, then the chord length $|AB|$=?",
    "fact_expressions": "l: Line;G: Parabola;P: Point;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Coordinate(P) = (1, 0);Inclination(l) = ApplyUnit(60, degree);Intersection(l, G) = {A, B};IsChordOf(LineSegmentOf(A, B), G);PointOnCurve(P, l)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[32, 37]], [[38, 54]], [[3, 14]], [[56, 59]], [[60, 63]], [[38, 54]], [[3, 14]], [[15, 37]], [[32, 65]], [[32, 76]], [[2, 37]]]",
    "query_spans": "[[[69, 78]]]",
    "process": ""
  },
  {
    "text": "If the chord length intercepted by the circle with center $(-2,1)$ on the directrix of the parabola $y^{2}=6 x$ is equal to $\\sqrt{3}$, then what is the radius of the circle?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 6*x);H: Circle;P: Point;Coordinate(P) = (-2, 1);Center(H) = P;Length(InterceptChord(Directrix(G), H)) = sqrt(3)",
    "query_expressions": "Radius(H)",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 15]], [[1, 15]], [[31, 32], [53, 54]], [[22, 30]], [[22, 30]], [[19, 32]], [[1, 49]]]",
    "query_spans": "[[[53, 59]]]",
    "process": "The directrix of the parabola $ y^{2} = 6x $ is $ x = -\\frac{3}{2} $. The distance from the center $ (-2, 1) $ to it equals $ -\\frac{3}{2} - (-2) = \\frac{1}{2} $. Given the chord length is $ \\sqrt{3} $, the radius of the circle is $ \\sqrt{(\\frac{1}{2})^{2}} $."
  },
  {
    "text": "The point $M(4, y)$ on the parabola $y^{2}=2 p x(p>0)$ is at a distance of $5$ from the focus $F$, and $O$ is the origin. Find $OM=?$",
    "fact_expressions": "G: Parabola;p: Number;O: Origin;M: Point;F: Point;y1:Number;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(M) = (4, y1);PointOnCurve(M, G);Focus(G) = F;Distance(M, F) = 5",
    "query_expressions": "LineSegmentOf(O, M)",
    "answer_expressions": "4*sqrt(2)",
    "fact_spans": "[[[0, 21]], [[3, 21]], [[48, 52]], [[23, 33]], [[36, 39]], [[24, 33]], [[3, 21]], [[0, 21]], [[23, 33]], [[0, 33]], [[0, 39]], [[23, 46]]]",
    "query_spans": "[[[59, 66]]]",
    "process": "4+\\frac{p}{2}=5, p=2, the equation of the parabola is y^{2}=4x, from y^{2}=4\\times4=16 we get y=\\pm4,"
  },
  {
    "text": "The parabola with vertex at the origin and coordinate axes as axes of symmetry passes through the point $(-1 , 2)$, then its equation is?",
    "fact_expressions": "G: Parabola;H: Point;Coordinate(H) = (-1, 2);O: Origin;Vertex(G) = O;SymmetryAxis(G) = axis;PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2 = -4*x, x^2 = y/2}",
    "fact_spans": "[[[14, 17], [32, 33]], [[19, 30]], [[19, 30]], [[3, 5]], [[0, 17]], [[6, 17]], [[14, 30]]]",
    "query_spans": "[[[32, 38]]]",
    "process": "When the focus of the parabola lies on the negative x-axis, let its equation be $ y^{2} = -2px $ ($ p > 0 $). Substituting $ (-1, 2) $, we get $ p = 2 $, so the equation is $ y^{2} = -4x $. When the focus of the parabola lies on the positive y-axis, let its equation be $ x^{2} = 2py $ ($ p > 0 $). Substituting $ (-1, 2) $, we get $ p = \\frac{1}{4} $, so the equation is $ x^{2} = \\frac{1}{2}y $."
  },
  {
    "text": "The focus of the parabola $y^{2}=2 p x(p>0)$ is $F$, the directrix is $l$, and $A$, $B$ are two moving points on the parabola such that $\\angle A F B=\\frac{\\pi}{2}$. Let $M$ be the midpoint of segment $A B$, and let $N$ be the projection of $M$ onto $l$. Then the maximum value of $\\frac{|M N|}{|A B|}$ is?",
    "fact_expressions": "G: Parabola;p: Number;B: Point;A: Point;F: Point;M: Point;N: Point;l: Line;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;Directrix(G) = l;PointOnCurve(A, G);PointOnCurve(B, G);AngleOf(A, F, B) = pi/2;MidPoint(LineSegmentOf(A, B)) = M;Projection(M, l) = N",
    "query_expressions": "Max(Abs(LineSegmentOf(M, N))/Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[0, 21], [47, 50]], [[3, 21]], [[43, 46]], [[38, 41]], [[25, 28]], [[101, 104]], [[113, 116]], [[32, 35], [105, 108]], [[3, 21]], [[0, 21]], [[0, 28]], [[0, 35]], [[37, 56]], [[37, 56]], [[60, 88]], [[91, 104]], [[101, 116]]]",
    "query_spans": "[[[118, 147]]]",
    "process": "By the definition of a parabola, we obtain $|MN|=\\frac{1}{2}(|AF|+|BF|)$. Combining with the Pythagorean theorem, we get $\\frac{|MN|^{2}}{|AB|^{2}}=\\frac{\\frac{1}{4}(|AF|+|BF|)^{2}}{|AF|^{2}+|BF|^{2}}$. After appropriate manipulation and applying the basic inequality, the desired maximum value can be found. [Solution] Draw $AG\\bot l$ from point $A$, and draw $BE\\bot l$ from point $B$. By the property of the parabola, we know $|AG|=|AF|$, $|BE|=|BF|$. Since $M$ is the midpoint of $AB$, $MN$ is the median line of trapezoid $AGEB$, so $|MN|=\\frac{1}{2}(|AG|+|BE|)=\\frac{1}{2}(|AF|+|BF|)$. In $\\triangle ABF$, $|AB|=\\sqrt{|AF|^{2}+|BF|^{2}}=\\sqrt{|AF|^{2}+|BF|^{2}}$, then $\\frac{|MN|^{2}}{|AB|^{2}}=\\frac{\\frac{1}{4}(|AF|+|BF|)^{2}}{|AF|^{2}+|BF|^{2}}=\\frac{1}{4}\\left(1+\\frac{2|AF||BF|}{|AF|^{2}+|BF|^{2}}\\right)\\leqslant\\frac{1}{4}\\left(1+\\frac{2|AF||BF|}{2|AF||BF|}\\right)=\\frac{1}{2}$, with equality if and only if $|AF|=|BF|$. Therefore, the maximum value of $\\frac{|MN|}{|AB|}$ is $\\frac{\\sqrt{2}}{2}$."
  },
  {
    "text": "Given the parabola $C$: $x^{2}=2py$ ($p>0$) with focus $F$, a line passing through point $F$ intersects the parabola $C$ at points $M$ and $N$, and $|MN|=8$. Then, the distance from the midpoint of segment $MN$ to the directrix of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;Expression(C)=(x^2 = 2*(p*y));F:Point;Focus(C)=F;M: Point;N: Point;L:Line;PointOnCurve(F,L);Intersection(L, C) = {M, N};Abs(LineSegmentOf(M, N)) = 8;p:Number;p>0",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(M,N)),Directrix(C))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 28], [45, 51], [87, 93]], [[2, 28]], [[37, 41], [32, 35]], [[2, 35]], [[53, 56]], [[57, 60]], [[42, 44]], [[36, 44]], [[42, 62]], [[64, 73]], [[9, 28]], [[9, 28]]]",
    "query_spans": "[[[76, 101]]]",
    "process": "Draw perpendiculars from points M and N to the directrix of the parabola C, with feet of the perpendiculars P and Q, respectively. By the definition of a parabola, |MP| = |MF|, |NQ| = |NF|, so |MP| + |NQ| = |MN| = 8. The distance from the midpoint of segment MN to the directrix of parabola C is the length of the median line of trapezoid MNQP, which is \\frac{1}{2}\\times(|MP|+|NQ|)=4"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ has equal lengths of the real axis and the imaginary axis, then the asymptote equations of $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ are?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Length(RealAxis(C)) = Length(ImageinaryAxis(C))",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y = pm*x",
    "fact_spans": "[[[2, 64], [76, 135]], [[10, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[2, 64]], [[2, 74]]]",
    "query_spans": "[[[76, 143]]]",
    "process": "According to the problem, 2a = 2b, that is, a = b, so the asymptotes of the hyperbola C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 (a > 0, b > 0) are given by y = \\pm\\frac{b}{a}x = \\pm x."
  },
  {
    "text": "If the asymptotes of a hyperbola are given by $y=\\pm \\frac{1}{3} x$, and it passes through the point $(6, \\sqrt{3})$, then what is the equation of the hyperbola?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (6, sqrt(3));Expression(Asymptote(G)) = (y = pm*(x/3));PointOnCurve(H,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9-y^2=1",
    "fact_spans": "[[[1, 4], [54, 57]], [[36, 52]], [[36, 52]], [[1, 32]], [[1, 52]]]",
    "query_spans": "[[[54, 62]]]",
    "process": "According to the problem, the asymptotes of the hyperbola are $ y = \\pm\\frac{1}{3}x $, so the standard equation of the hyperbola can be written as $ \\frac{x^{2}}{9} - y^{2} = \\lambda $ ($ \\lambda \\neq 0 $). Since the hyperbola passes through the point $ (6, \\sqrt{3}) $, substituting the point $ (6, \\sqrt{3}) $ into the equation yields $ \\lambda = $. Therefore, the equation of this hyperbola is $ \\frac{x^{2}}{9} - y^{2} = 1 $."
  },
  {
    "text": "Given that $F$ is the right focus of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{8}=1$. If $P$ is a point on the left branch of $C$ and $A(0,6 \\sqrt{6})$ is a point on the $y$-axis, then the minimum area of $\\Delta A P F$ is?",
    "fact_expressions": "C: Hyperbola;A: Point;P: Point;F: Point;Expression(C) = (x^2 - y^2/8 = 1);Coordinate(A) = (0, 6*sqrt(6));RightFocus(C) = F;PointOnCurve(P, LeftPart(C));PointOnCurve(A, yAxis)",
    "query_expressions": "Min(Area(TriangleOf(A, P, F)))",
    "answer_expressions": "9*sqrt(6)+6",
    "fact_spans": "[[[6, 39], [49, 52]], [[59, 76]], [[45, 48]], [[2, 5]], [[6, 39]], [[59, 76]], [[2, 43]], [[45, 58]], [[59, 84]]]",
    "query_spans": "[[[86, 108]]]",
    "process": "First, find the foci of the hyperbola, the equation of line AF, and the length of AF; let the line $ y = -2\\sqrt{6}x + t $ be tangent to the hyperbola, with the point of tangency on the left branch. Combine the hyperbola equation, eliminate $ y $, and set the discriminant to 0 to solve for $ t $. Then, using the distance formula between parallel lines, obtain the minimum area of the triangle. The right focus of hyperbola $ C: x^2 - \\frac{y^2}{8} = 1 $ is $ (3, 0) $. Given $ A(0, 6\\sqrt{6}) $, the equation of line AF is $ y = -2\\sqrt{6}x + 6\\sqrt{6} $, and $ |AF| = \\sqrt{9 + (6\\sqrt{6})^2} = 15 $. Let the line $ y = -2\\sqrt{6}x + t $ be tangent to the hyperbola, with the point of tangency on the left branch. Combining  \n$$\n\\begin{cases}\ny = t - 2\\sqrt{6}x \\\\\n8x^2 - y^2 = 8\n\\end{cases}\n$$  \ngives $ 16x^2 - 4\\sqrt{6}tx + t^2 + 8 = 0 $. From $ \\Delta = 96t^2 - 4 \\times 16(t^2 + 8) = 0 $, solving yields $ t = -4 $ (discard $ t = 4 $). Then, the distance from point $ P $ to line AF is $ d = \\frac{|6\\sqrt{6} + 4|}{\\sqrt{1 + 24}} = \\frac{4 + 6\\sqrt{6}}{5} $. Thus, the minimum area of $ \\triangle APF $ is $ \\frac{1}{2}d \\cdot |AF| = \\frac{1}{2} \\times \\frac{4 + 6\\sqrt{6}}{5} \\times 15 = 6 + 9\\sqrt{6} $."
  },
  {
    "text": "If the asymptotes of a hyperbola are given by $y = \\pm 2x$, and one of its foci is $(0, \\sqrt{5})$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*2*x);Coordinate(OneOf(Focus(G))) = (0, sqrt(5))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/4-x^2=1",
    "fact_spans": "[[[1, 4], [23, 24], [47, 50]], [[1, 22]], [[23, 45]]]",
    "query_spans": "[[[47, 55]]]",
    "process": "It is known from the problem that the hyperbola has its foci on the y-axis, and \\begin{cases}\\frac{a}{b}=2\\\\a^{2}+b^{2}=5\\end{cases}, solving gives \\begin{cases}a=2\\\\b=1\\end{cases}, so the equation of the hyperbola is \\frac{y^{2}}{4}-x^{2}=1"
  },
  {
    "text": "The moving circle $M$ is externally tangent to the circle $C_{1}$: $(x+1)^{2}+y^{2}=1$, and internally tangent to the circle $C_{2}$: $(x-1)^{2}+y^{2}=25$. Then, the trajectory equation of the center $M$ of the moving circle is?",
    "fact_expressions": "M: Circle;M1:Point;C1:Circle;C2:Circle;Expression(C1)=((x+1)^2+y^2=1);Expression(C2)=((x-1)^2+y^2=25);IsOutTangent(M,C1);IsInTangent(M,C2);Center(M)=M1",
    "query_expressions": "LocusEquation(M1)",
    "answer_expressions": "x^2/9+y^2/8=1",
    "fact_spans": "[[[2, 5], [71, 73]], [[75, 78]], [[6, 34]], [[38, 67]], [[6, 34]], [[38, 67]], [[2, 36]], [[2, 69]], [[71, 78]]]",
    "query_spans": "[[[75, 85]]]",
    "process": ""
  },
  {
    "text": "Point $M$ is a point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and $F_{1}$, $F_{2}$ are the left and right foci of the ellipse, respectively. Find the coordinates of point $M$ satisfying $|MF_{1}|=3|MF_{2}|$.",
    "fact_expressions": "G: Ellipse;M: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2/3 = 1);PointOnCurve(M, G);LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(M, F1)) = 3*Abs(LineSegmentOf(M, F2))",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(pm*2, 0)",
    "fact_spans": "[[[5, 42], [65, 67]], [[0, 4], [97, 101]], [[47, 54]], [[55, 62]], [[5, 42]], [[0, 46]], [[47, 71]], [[47, 71]], [[75, 95]]]",
    "query_spans": "[[[97, 105]]]",
    "process": ""
  },
  {
    "text": "Given that the ellipse is centered at the origin, with coordinate axes as its axes of symmetry, and passes through the two points $(-\\frac{3}{2}, \\frac{5}{2})$, $(\\sqrt{3}, \\sqrt{5})$, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;H: Point;I: Point;O: Origin;Coordinate(H) = (-3/2, 5/2);Coordinate(I) = (sqrt(3), sqrt(5));Center(G) = O;SymmetryAxis(G)=axis;PointOnCurve(H,G);PointOnCurve(I,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/10 + x^2/6 = 1",
    "fact_spans": "[[[2, 4], [79, 81]], [[25, 54]], [[55, 77]], [[8, 10]], [[25, 54]], [[55, 77]], [[2, 10]], [[2, 19]], [[2, 77]], [[2, 77]]]",
    "query_spans": "[[[79, 85]]]",
    "process": "Let the equation of the ellipse be $mx^{2}+ny^{2}=1$ ($m>0$, $n>0$, and $m\\neq n$). The ellipse passes through the points $(-\\frac{3}{2},\\frac{5}{2})$, $(\\sqrt{3},\\sqrt{5})$, then  \n$$\n\\begin{cases}\n\\frac{9}{4}m+\\frac{25}{4}n=1 \\\\\n3m+5n=1\n\\end{cases}\n$$\nSolving gives  \n$$\n\\begin{cases}\nm=\\frac{1}{6} \\\\\nn=\\frac{1}{10}\n\\end{cases},\n$$\nso the required ellipse equation is $\\frac{y^{2}}{10}+\\frac{x^{2}}{6}=1$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, $A$ is the top vertex of the ellipse. A line is drawn from point $F_{1}$ such that $F_{1} D \\perp A F_{2}$, with $D$ being the foot of the perpendicular. If $|F_{2} D|=\\frac{1}{2}|F_{1} F_{2}|$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;F1: Point;D: Point;A: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) =F1;RightFocus(C)=F2;UpperVertex(C)=A;PointOnCurve(F1,LineSegmentOf(F1,D));IsPerpendicular(LineSegmentOf(F1,D),LineSegmentOf(A,F2));FootPoint(LineSegmentOf(F1,D),LineSegmentOf(A,F2))=D;Abs(LineSegmentOf(F2, D)) = Abs(LineSegmentOf(F1, F2))/2",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 59], [88, 90], [174, 179]], [[9, 59]], [[9, 59]], [[68, 75], [95, 103]], [[131, 134]], [[84, 87]], [[76, 83]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 83]], [[2, 83]], [[84, 93]], [[94, 127]], [[104, 127]], [[104, 134]], [[136, 172]]]",
    "query_spans": "[[[174, 185]]]",
    "process": "In Rt△F₁DF₂, if |F₂D| = \\frac{1}{2}|F₁F₂|, then ∠DF₁F₂ = 30^{\\circ}, hence ∠F₁F₂D = 90^{\\circ} - ∠DF₁F₂ = 90^{\\circ} - 30^{\\circ} = 60^{\\circ}. Also obviously |AF₁| = |AF₂|, so △AF₁F₂ is an equilateral triangle, hence |AF₁| = |F₁F₂|, i.e., a = 2c, thus \\frac{c}{a} = \\frac{1}{2}, i.e., the eccentricity of ellipse C is \\frac{1}{2}"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{b^{2}}=1$ $(3>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, $O$ is the coordinate origin, $P$ is a point on the ellipse, the extension of $P F_{2}$ intersects the ellipse again at point $A$, if $|O F_{1}|=|O A|$, and the area of triangle $\\triangle O F_{1} A$ is $2$, then $|P F_{2}|$=?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/9 + y^2/b^2 = 1);b: Number;3 > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;O: Origin;PointOnCurve(P, C) = True;P: Point;Intersection(OverlappingLine(LineSegmentOf(P,F2)),C) = A;A: Point;Abs(LineSegmentOf(O, F1)) = Abs(LineSegmentOf(O, A));Area(TriangleOf(O,F1,A)) = 2",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "{1,4/5}",
    "fact_spans": "[[[2, 55], [93, 95], [111, 113]], [[2, 55]], [[9, 55]], [[9, 55]], [[9, 55]], [[64, 71]], [[72, 79]], [[2, 79]], [[2, 79]], [[80, 83]], [[89, 98]], [[89, 92]], [[99, 119]], [[115, 119]], [[121, 138]], [[140, 168]]]",
    "query_spans": "[[[170, 183]]]",
    "process": "Since |OF₁| = |OA|, it follows that ∠F₂AF₁ = 90°, so the area S of △OF₁A is S = ½ × ½ |AF₁|·|AF₂| = 2, hence |AF₁|·|AF₂| = 8. By the definition of an ellipse, |AF₁| + |AF₂| = 6, so |AF₁| = 2 and |AF₂| = 4 or |AF₁| = 4 and |AF₂| = 2. Let |PF₂| = n, then |PF₁| = 6 − n. When  \n\\begin{cases} |AF₁| = 2 \\\\ |AF₂| = 4 \\end{cases},  \nby the Pythagorean theorem, |AF₁|² + |AP|² = |PF₁|², i.e., 2² + (4 + n)² = (6 − n)², solving gives n = ⁴⁄₅; when |AF₁| = 4 and |AF₂| = 2, by the Pythagorean theorem, 4² + (2 + n)² = (6 − n)², solving gives n = 1. In summary, |PF₂| = 1 or ⁴⁄₅."
  },
  {
    "text": "The equations of the asymptotes of the hyperbola are $3 x \\pm 4 y=0$, then the eccentricity of the hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (pm*4*y + 3*x = 0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 3], [27, 30]], [[0, 25]]]",
    "query_spans": "[[[27, 37]]]",
    "process": ""
  },
  {
    "text": "If points $O$ and $F(-2,0)$ are respectively the center and left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1(a>0)$, and point $P$ is any point on the right branch of the hyperbola, then the range of values of $\\overrightarrow{O P} \\cdot \\overrightarrow{F P}$ is?",
    "fact_expressions": "G: Hyperbola;a: Number;F: Point;O: Origin;P: Point;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Coordinate(F) = (-2, 0);PointOnCurve(P, RightPart(G));Center(G)=O;LeftFocus(G)=F",
    "query_expressions": "Range(DotProduct(VectorOf(O, P), VectorOf(F, P)))",
    "answer_expressions": "[3+2*\\sqrt{3},+\\infty)",
    "fact_spans": "[[[19, 56], [69, 72]], [[22, 56]], [[6, 16]], [[1, 5]], [[64, 68]], [[22, 56]], [[19, 56]], [[6, 16]], [[64, 80]], [[1, 63]], [[1, 63]]]",
    "query_spans": "[[[82, 138]]]",
    "process": "From F(-2,0), we get a=\\sqrt{3}; let P(x,y), x\\geqslant\\sqrt{3}. Also, \\overrightarrow{OP}\\cdot\\overrightarrow{FP}=x(x+2)+y^{2}. Given \\frac{x^{2}}{3}-y^{2}=1, then y^{2}=\\frac{x^{2}}{3}-1. Thus, \\overrightarrow{OP}\\cdot\\overrightarrow{FP}=x(x+2)+y^{2}=\\frac{4}{3}x^{2}+2x-1. When x\\geqslant\\sqrt{3}, the function is monotonically increasing; therefore, the range of \\overrightarrow{OP}\\cdot\\overrightarrow{FP} is [3+2\\sqrt{3},+\\infty)."
  },
  {
    "text": "Draw a line $l$ through the right focus $F$ of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$, intersecting the right branch of the hyperbola at point $A$. If $l$ intersects the left branch of the hyperbola at another point, then the range of the length of segment $A F$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;A: Point;F: Point;Expression(C) = (x^2 - y^2/3 = 1);RightFocus(C) = F;PointOnCurve(F, l);Intersection(l, RightPart(C)) = A;NumIntersection(l, LeftPart(C)) = 1",
    "query_expressions": "Range(Length(LineSegmentOf(A, F)))",
    "answer_expressions": "[1, 3/2)",
    "fact_spans": "[[[42, 47], [63, 66]], [[1, 34], [49, 52], [67, 70]], [[57, 61]], [[38, 41]], [[1, 34]], [[1, 41]], [[0, 47]], [[42, 61]], [[63, 80]]]",
    "query_spans": "[[[82, 98]]]",
    "process": "From the hyperbola equation, we obtain the asymptotes of the hyperbola: $ y = \\pm\\sqrt{3}x $. Combining the properties of the hyperbola, it is known that when the slope $ k $ of line $ l $ lies in $ [-\\sqrt{3}, \\sqrt{3}] $, the condition is satisfied. Consider the critical cases: when the slope of the line is 0, $ AF = 1 $; when the slope of the line is $ \\sqrt{3} $, the line equation is: $ y - 0 = \\sqrt{3}(x - 2) $. Solving simultaneously with the tangent line equation gives: $ 3x^{2} - 3(x - 2)^{2} = 3 $, then: $ x = \\frac{5}{4} $, $ y = -\\frac{3}{4}\\sqrt{3} $. At this point, the distance between points $ A(\\frac{5}{4}, -\\frac{3}{4}\\sqrt{3}) $ and $ F(2, 0) $ is: $ (\\frac{5}{4} - 2)^{2} + (-\\frac{3}{4}\\sqrt{3} - 0)^{2} = \\frac{3}{2} $. In summary, the range of length of segment $ AF $ is $ [1, \\frac{3}{2}) $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through $F_{1}$ is tangent to the circle $x^{2}+y^{2}=a^{2}$ at point $T$, and intersects the right branch of the hyperbola $C$ at point $P$. If $\\overrightarrow {F_{1} P}=3 \\overrightarrow{F_{1} T}$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;G: Circle;F1: Point;T: Point;P: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = a^2);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, l);TangentPoint(l, G) = T;Intersection(l, RightPart(C)) = P;VectorOf(F1, P) = 3*VectorOf(F1, T)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(13)/2",
    "fact_spans": "[[[98, 103], [133, 138]], [[2, 64], [139, 145], [213, 219]], [[10, 64]], [[10, 64]], [[104, 124]], [[73, 80], [90, 97]], [[127, 131]], [[150, 154]], [[81, 88]], [[10, 64]], [[10, 64]], [[2, 64]], [[104, 124]], [[2, 88]], [[2, 88]], [[89, 103]], [[98, 131]], [[133, 154]], [[156, 210]]]",
    "query_spans": "[[[213, 225]]]",
    "process": "As shown in the figure, from the given conditions, |OF_{1}| = |OF_{2}| = c, |OT| = a, then |F_{1}T| = b. Since \\overrightarrow{F_{1}P} = 3\\overrightarrow{F_{1}T}, therefore |TP| = 2b, |F_{1}P| = 3b. Also, since |PF_{1}| - |PF_{2}| = 2a, then |PF_{2}| = 3b - 2a. Draw F_{2}M perpendicular to OT, obtaining F_{2}M = 2a, |TM| = b, then |PM| = b. In the right triangle \\triangle MPF_{2}, |PM|^{2} + |MF_{2}|^{2} = |PF_{2}|^{2}, that is, b^{2} + (2a)^{2} = (3b - 2a)^{2}, yielding 2b = 3a. Also, since c^{2} = a^{2} + b^{2}, then c^{2} = a^{2} + \\frac{9}{4}a^{2}, i.e., 4c^{2} = 13a^{2}. Therefore, e = \\frac{\\sqrt{13}}{2}."
  },
  {
    "text": "Given that point $P$ lies on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$, and its left and right foci are $F_{1}$ and $F_{2}$ respectively, if the circumradius of $\\Delta F_{1} P F_{2}$ is $4$, then what is the area of $\\Delta F_{1} P F_{2}$?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/16 + y^2/4 = 1);PointOnCurve(P, G);LeftFocus(G)=F1;RightFocus(G)=F2;Radius(CircumCircle(TriangleOf(F1,P,F2)))=4",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "{4*sqrt(3)/3,4*sqrt(3)}",
    "fact_spans": "[[[7, 45], [49, 50]], [[58, 65]], [[2, 6]], [[66, 73]], [[7, 45]], [[2, 48]], [[49, 73]], [[49, 73]], [[75, 107]]]",
    "query_spans": "[[[109, 136]]]",
    "process": "From the given, $ F_{1}F_{2}=2c=2\\sqrt{a^{2}-b^{2}}=4\\sqrt{3} $. By the sine law, $ \\frac{F_{1}F_{2}}{\\sin\\angle F_{1}PF_{2}} = 2R $. Since $ F_{1}F_{2}=4\\sqrt{3} $ and $ R=4 $, substituting gives $ \\sin\\angle F_{1}PF_{2}=\\frac{\\sqrt{3}}{2} $, hence $ \\angle F_{1}PF_{2}=\\frac{\\pi}{3} $ or $ \\frac{2\\pi}{3} $. When point $ P $ is at the upper vertex, $ \\angle F_{1}PF_{2}=\\frac{2\\pi}{3} $, so both $ \\angle F_{1}PF_{2}=\\frac{\\pi}{3} $ and $ \\frac{2\\pi}{3} $ are possible. Thus, when $ \\angle F_{1}PF_{2}=\\frac{\\pi}{3} $, $ S_{AF_{1}PF_{2}}=b^{2}\\tan\\frac{\\angle F_{1}PF_{2}}{2}=4\\cdot\\frac{\\sqrt{3}}{3}=\\frac{4\\sqrt{3}}{3} $. When $ \\angle F_{1}PF_{2}=\\frac{2\\pi}{3} $, $ S_{\\Delta F_{1}PF_{2}}=b^{2}\\tan\\frac{\\angle F_{1}PF_{2}}{2}=4\\sqrt{3} $. Therefore, the answer is $ \\frac{4\\sqrt{3}}{3} $ or $ 4\\sqrt{3} $. [Note: Relating the circumradius suggests using the sine law; to find the area of the focal triangle, use $ S_{AF_{1}PF_{2}}=b^{2}\\tan^{\\frac{\\angle}{-}} $]"
  },
  {
    "text": "What is the eccentricity of the hyperbola $4 x^{2}-5 y^{2}=-20$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (4*x^2 - 5*y^2 = -20)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3/2",
    "fact_spans": "[[[0, 24]], [[0, 24]]]",
    "query_spans": "[[[0, 30]]]",
    "process": "The solution process is omitted"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{4}+y^{2}=1$, and point $P$ lies on the ellipse $C$ such that $\\angle F_{1} P F_{2}=60^{\\circ}$, then $|P F_{1}| \\cdot|P F_{2}|$=?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[18, 50], [62, 67]], [[18, 50]], [[2, 9]], [[10, 17]], [[2, 56]], [[2, 56]], [[57, 61]], [[57, 68]], [[69, 102]]]",
    "query_spans": "[[[104, 132]]]",
    "process": "From the definition of the ellipse, we have |PF_{1}| + |PF_{2}| = 4. Using the law of cosines, we obtain |PF_{1}|^{2} + |PF_{2}|^{2} - 2|PF_{1}||PF_{2}|\\cos60^{\\circ} = |F_{1}F_{2}|^{2}. Thus, (|PF_{1}| + |PF_{2}|)^{2} - 3|PF_{1}|\\cdot|PF_{2}| = |F_{1}F_{2}|^{2} = 12. Solving gives 3|PF_{1}||PF_{2}| = 4, that is, |PF_{1}|\\cdot|PF_{2}| = \\frac{4}{3}."
  },
  {
    "text": "If the foci of the ellipse $\\frac{y^{2}}{m}+x^{2}=1$ lie on the $y$-axis, and the major axis length is twice the minor axis length, then $m=?$",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + y^2/m = 1);m: Number;PointOnCurve(Focus(G), yAxis);Length(MajorAxis(G)) = 2*Length(MinorAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 28]], [[1, 28]], [[53, 56]], [[1, 37]], [[1, 51]]]",
    "query_spans": "[[[53, 58]]]",
    "process": "Since the foci of the ellipse $\\frac{y^{2}}{m}+x^{2}=1$ lie on the $y$-axis, we have $m>1$. Because the major axis length is twice the minor axis length, we have $2\\sqrt{m}=2\\times2\\times1\\Rightarrow m=4$,"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$, a chord with midpoint $(1,1)$ intersects the parabola $C$ at points $M$ and $N$. If $|M N|=\\sqrt{15}$, then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;M: Point;N: Point;H:LineSegment;p>0;Expression(C) = (y^2 = 2*p*x);Coordinate(P) = (1, 1);IsChordOf(H,C);Intersection(H, C) = {M, N};Abs(LineSegmentOf(M, N)) = sqrt(15);MidPoint(H)=P;P:Point",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 28], [44, 50]], [[82, 85]], [[52, 55]], [[56, 59]], [], [[10, 28]], [[2, 28]], [[30, 38]], [[2, 43]], [[29, 61]], [[63, 80]], [[2, 43]], [[30, 38]]]",
    "query_spans": "[[[82, 87]]]",
    "process": "Let the line pass through points M(x_{1},y_{1}), N(x_{2},y_{2}), then ^{k}MN^{=} so the equation of line MN is y-1=p(x-1), that is, l_{MN}: y=px+(1-p). From \\begin{cases}y=px+(1-p)\\\\y2=2px\\end{cases}, eliminating y and simplifying yields p^{2x2}-2p^{2}x+(p^{2}-2p+1)=0. \\triangle>0 \\Rightarrow p>=\\sqrt{1+p^{2}\\times2\\times\\frac{\\sqrt{2p-1}}{p}}=\\sqrt{15}. Squaring both sides and simplifying gives: 8p^{3}-19p^{2}+8p-4=0. (19p^{2}-8p-60)(p^{2}+2p+4)-(1+\\frac{1}{2}. Solving yields p=2."
  },
  {
    "text": "Given circle $C_{1}$: $x^{2}+(y-2)^{2}=4$, parabola $C_{2}$: $y^{2}=2 p x(p>0)$, $C_{1}$ and $C_{2}$ intersect at points $A$ and $B$, $|A B|=\\frac{8 \\sqrt{5}}{5}$, then the equation of parabola $C_{2}$ is?",
    "fact_expressions": "C2: Parabola;p: Number;C1: Circle;A: Point;B: Point;p > 0;Expression(C2) = (y^2 = 2*(p*x));Expression(C1) = (x^2 + (y - 2)^2 = 4);Intersection(C1, C2) = {A, B};Abs(LineSegmentOf(A, B)) = 8*sqrt(5)/5",
    "query_expressions": "Expression(C2)",
    "answer_expressions": "y^2 = 32*x/5",
    "fact_spans": "[[[32, 62], [72, 79], [122, 132]], [[44, 62]], [[2, 31], [63, 71]], [[82, 85]], [[86, 89]], [[44, 62]], [[32, 62]], [[2, 31]], [[64, 91]], [[92, 120]]]",
    "query_spans": "[[[122, 137]]]",
    "process": "According to the chord length formula for the intersection of a line and a circle, we have $2\\sqrt{R^{2}-d^{2}}=2\\sqrt{4-d^{2}}=\\frac{8\\sqrt{5}}{5}$, solving gives $d=\\frac{2}{5}\\sqrt{5}$. Let the equation of line $AB$ be $y=kx$, the distance from the center $(0,2)$ to the line is $d=\\frac{2}{\\sqrt{1+k^{2}}}=\\frac{2\\sqrt{5}}{5}$, solving gives $k=-2$ (discarded) or $k=2$. $\\begin{cases}y=2x\\\\x^{2}+(y-2)^{2}=4\\end{cases}$, solving gives $\\begin{cases}x=0\\\\y=0\\end{cases}$ or $\\begin{cases}x=\\frac{8}{5}\\\\y=\\frac{16}{4}\\end{cases}$, substituting into the parabola equation $(\\frac{16}{5})^{2}=2p\\times\\frac{8}{5}$, solving gives $2p=\\frac{32}{5}$, so the parabola equation is $y^{2}=\\frac{32}{5}x$."
  },
  {
    "text": "The distance from a point $P$ on the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ to the point $(5,0)$ is $9$. Then, what is the distance from $P$ to $(-5,0)$?",
    "fact_expressions": "G: Hyperbola;H: Point;I: Point;P: Point;Expression(G) = (x^2/16 - y^2/9 = 1);Coordinate(H) = (5, 0);Coordinate(I) = (-5, 0);PointOnCurve(P, G);Distance(P, H) = 9",
    "query_expressions": "Distance(P, I)",
    "answer_expressions": "{17,1}",
    "fact_spans": "[[[0, 39]], [[46, 54]], [[67, 75]], [[41, 45], [63, 66]], [[0, 39]], [[46, 54]], [[67, 75]], [[0, 45]], [[41, 61]]]",
    "query_spans": "[[[63, 79]]]",
    "process": ""
  },
  {
    "text": "Given that the circle $x^{2}+y^{2}-9=0$ is tangent to the directrix of the parabola $y^{2}=2 px  (p>0)$, then $p$=?",
    "fact_expressions": "H: Circle;Expression(H) = (x^2 + y^2 - 9 = 0);G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;IsTangent(H, Directrix(G))",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 20]], [[2, 20]], [[21, 43]], [[21, 43]], [[50, 53]], [[24, 43]], [[2, 48]]]",
    "query_spans": "[[[50, 55]]]",
    "process": ""
  },
  {
    "text": "The two foci of a hyperbola are given as $F_{1}(-\\sqrt{5}, 0)$, $F_{2}(\\sqrt{5}, 0)$, $P$ is a point on the hyperbola such that $P F_{1} \\perp P F_{2}$, and $|P F_{1} |\\cdot| P F_{2} |=2$. Then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;F1: Point;Coordinate(F1) = (-sqrt(5), 0);F2: Point;Coordinate(F2) = (sqrt(5), 0);Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4-y^2=1",
    "fact_spans": "[[[2, 5], [65, 68], [131, 134]], [[13, 34]], [[13, 34]], [[37, 58]], [[37, 58]], [[2, 58]], [[61, 64]], [[61, 72]], [[74, 97]], [[99, 129]]]",
    "query_spans": "[[[131, 141]]]",
    "process": "Since the two foci of the hyperbola are $ F_{1}(-\\sqrt{5},0) $ and $ F_{2}(\\sqrt{5},0) $, the foci lie on the x-axis and $ c = \\sqrt{5} $. Since triangle $ PF_{1}F_{2} $ is a right triangle, $ |PF_{1}|^{2} + |PF_{2}|^{2} = 4c^{2} = 20 $. Thus, $ (|PF_{1}| - |PF_{2}|)^{2} + 2|PF_{1}|\\cdot|PF_{2}| = 20 $. By the definition of the hyperbola, $ (2a)^{2} + 4 = 20 $, so $ a^{2} = 4 $, hence $ b^{2} = 5 - 4 = 1 $. Therefore, the equation of the hyperbola is $ \\frac{x^{2}}{4} - y^{2} = 1 $."
  },
  {
    "text": "Let line $l$ pass through a focus of hyperbola $C$ and be perpendicular to one of the symmetry axes of $C$. The line $l$ intersects $C$ at points $A$ and $B$, and $|AB|$ is twice the length of the real axis of $C$. Then the eccentricity of $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;A: Point;B: Point;PointOnCurve(OneOf(Focus(C)),l);IsPerpendicular(l,OneOf(SymmetryAxis(C)));Intersection(l, C) = {A,B};Abs(LineSegmentOf(A,B))=2*Length(RealAxis(C))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[1, 6], [33, 36]], [[7, 13], [21, 24], [21, 24], [21, 24], [60, 63]], [[42, 45]], [[46, 49]], [[1, 18]], [[1, 32]], [[33, 51]], [[52, 72]]]",
    "query_spans": "[[[74, 83]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of an ellipse with foci on the $x$-axis is $\\frac{1}{2}$, and its major axis length equals the radius of the circle $x^{2}+y^{2}-2 x-15=0$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;PointOnCurve(Focus(G), xAxis);Eccentricity(G) = 1/2;Length(MajorAxis(G)) = Radius(H);H: Circle;Expression(H) = (-2*x + x^2 + y^2 - 15 = 0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4+y^2/3=1",
    "fact_spans": "[[[11, 13], [32, 33], [67, 69]], [[2, 13]], [[11, 31]], [[32, 65]], [[39, 62]], [[39, 62]]]",
    "query_spans": "[[[67, 76]]]",
    "process": "\\because the standard equation of the circle x^{2}+y^{2}-2x-15=0 is (x-1)^{2}+y^{2}=16, \\therefore the radius of the circle x^{2}+y^{2}-2x-15=0 is 4, \\therefore 2a=4, a=2 \\because e=\\frac{c}{a}=\\frac{1}{2}, \\therefore c=1, b^{2}=a^{2}-c^{2}=3 \\therefore the standard equation of the ellipse is \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1"
  },
  {
    "text": "The distance from the focus of the parabola $x^{2}=-2 p y(p>0)$ to the line $y=2$ is $5$, then $p=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = -2*p*y);p: Number;p>0;H: Line;Expression(H) = (y = 2);Distance(Focus(G), H) = 5",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[0, 22]], [[0, 22]], [[42, 45]], [[3, 22]], [[26, 33]], [[26, 33]], [[0, 40]]]",
    "query_spans": "[[[42, 47]]]",
    "process": "From the given condition, we have $2 + \\frac{p}{2} = 5$, $\\therefore p = 6$. Fill in $6$."
  },
  {
    "text": "The maximum area of a triangle with vertices at a point on the ellipse and the two foci is $2$. Then, the minimum length of the major axis of the ellipse is?",
    "fact_expressions": "G: Ellipse;P:Point;F1:Point;F2:Point;PointOnCurve(P,G);Focus(G)={F1,F2};Max(Area(TriangleOf(P,F1,F2)))=2",
    "query_expressions": "Min(Length(MajorAxis(G)))",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 3], [31, 33]], [], [], [], [[1, 6]], [[1, 11]], [[0, 29]]]",
    "query_spans": "[[[31, 42]]]",
    "process": "From the given condition, when the point on the ellipse is at the endpoint of the minor axis, the area of the triangle reaches its maximum value, i.e., $ bc = 2 $. $\\therefore a^{2} = b^{2} + c^{2} \\geqslant 2bc = 4$, $\\therefore a \\geqslant 2$, with equality if and only if $ b = c = \\sqrt{2} $. $\\therefore 2a \\geqslant 4$, that is, the minimum length of the major axis of the ellipse is 4."
  },
  {
    "text": "The left focus of the hyperbola $\\frac{x^{2}}{3}-\\frac{16 y^{2}}{p^{2}}=1(p>0)$ lies on the directrix of the parabola $y^{2}=2 p x$. Then, the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - 16*y^2/p^2 = 1);p: Number;p>0;H: Parabola;Expression(H) = (y^2 = 2*p*x);PointOnCurve(LeftFocus(G), Directrix(H))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[0, 50], [78, 81]], [[0, 50]], [[3, 50]], [[3, 50]], [[55, 71]], [[55, 71]], [[0, 75]]]",
    "query_spans": "[[[78, 87]]]",
    "process": "Let the positive number $ a $ denote the real semi-axis of the hyperbola equation, and the positive number $ b $ denote the imaginary semi-axis. From the hyperbola equation, we know $ a^{2}=3 $, $ b^{2}=\\frac{p^{2}}{16} $. Thus, the focal semi-distance of the hyperbola is $ c=\\sqrt{a^{2}+b^{2}}=\\sqrt{3} $. From the directrix equation of the parabola $ x=-\\frac{p}{2} $, we obtain $ -c=-\\frac{16}{2} $, $ \\frac{p}{2}=\\sqrt{a^{2}+b^{2}}=\\sqrt{3+\\frac{p^{2}}{16}} $, solving gives $ p=4 $. Hence, the eccentricity of the hyperbola is $ e=\\frac{c}{a}=\\frac{\\sqrt{3+\\frac{4^{2}}{16}}}{\\sqrt{3}}=\\frac{2\\sqrt{3}}{3} $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the right focus is $F$. A circle with diameter $OF$ ($O$ being the origin) intersects the two asymptotes of the hyperbola at points $A$ and $B$ respectively (distinct from the origin). If the area of $\\triangle OAB$ equals $\\frac{1}{4}ab$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(C) = F;IsDiameter(LineSegmentOf(O, F), G) = True;G: Circle;O: Origin;L1: Line;L2: Line;Asymptote(C) = {L1, L2};Intersection(G, L1) = A;Intersection(G, L2) = B;Negation(A = O);Negation(B = O);A: Point;B: Point;Area(TriangleOf(O, A, B)) = a*b/4",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 63], [95, 98], [167, 173]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 71]], [[2, 71]], [[72, 94]], [[93, 94]], [[79, 82], [121, 123]], [], [], [[95, 103]], [[93, 116]], [[93, 116]], [[107, 124]], [[107, 124]], [[107, 110]], [[111, 114]], [[126, 165]]]",
    "query_spans": "[[[167, 179]]]",
    "process": "Let the circle with diameter OF (O being the coordinate origin) intersect the line $ y = \\frac{b}{a}x $ at point A. It is clear that point F has coordinates $ (c, 0) $. The equation of the circle with diameter OF is $ \\left(x - \\frac{c}{2}\\right)^2 + y^2 = \\frac{c^2}{4} $, or equivalently $ x^2 + y^2 - cx = 0 $. Solving the system \n\\[\n\\begin{cases}\ny = \\frac{b}{a}x \\\\\nx^2 + y^2 - cx = 0 \\\\\nx > 0\n\\end{cases}\n\\]\nyields \n\\[\n\\begin{cases}\nx = \\frac{a^2}{c} \\\\\ny = \\frac{ab}{c}\n\\end{cases}\n\\],\nso point A is $ \\left(\\frac{a^2}{c}, \\frac{ab}{c}\\right) $. Similarly, point B is $ \\left(\\frac{a^2}{c}, -\\frac{ab}{c}\\right) $. Therefore,\n$ S_{\\triangle OAB} = \\frac{1}{2} \\times \\frac{a^2}{c} \\times \\frac{2ab}{c} = \\frac{a^3b}{c^2} = \\frac{1}{4}ab $. Rearranging gives $ \\frac{c^2}{a^2} = 4 $. Hence, the eccentricity of hyperbola C is $ e = \\frac{c}{a} = 2 $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1$, with left and right foci $F_{1}$, $F_{2}$ respectively, and $P$ a point on the ellipse $C$ in the first quadrant satisfying $P F_{1} \\perp P F_{2}$. Extend $P F_{2}$ to intersect the ellipse $C$ at point $Q$. Then the inradius of $\\triangle F_{1} P Q$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/49 + y^2/24 = 1);LeftFocus(C) = F1;F1: Point;RightFocus(C) = F2;F2: Point;P: Point;PointOnCurve(P,C) = True;Quadrant(P) = 1;IsPerpendicular(LineSegmentOf(P,F1),LineSegmentOf(P,F2)) = True;Intersection(OverlappingLine(LineSegmentOf(P,F2)),C) = Q;Q: Point",
    "query_expressions": "Radius(InscribedCircle(TriangleOf(F1,P,Q)))",
    "answer_expressions": "12/5",
    "fact_spans": "[[[2, 46], [75, 80], [129, 134]], [[2, 46]], [[2, 70]], [[55, 62]], [[2, 70]], [[63, 70]], [[71, 74]], [[71, 90]], [[71, 90]], [[93, 116]], [[117, 139]], [[135, 139]]]",
    "query_spans": "[[[141, 170]]]",
    "process": "By the properties of the ellipse, the inradius of right triangle $\\triangle F_{1}PQ$ is $r = \\frac{1}{2}(F_{1}P + PQ - F_{1}Q) = \\frac{1}{2}(PF_{1} + PF_{2} + QF_{2} - QF_{1}) = QF_{2}$. In $\\triangle F_{1}PF_{2}$, $F_{1}F_{2} = 2 \\times \\sqrt{49 - 24} = 10$, $PF_{1} + PF_{2} = 14$. Since $\\angle F_{1}PF_{2} = \\frac{\\pi}{2}$, it follows that $PF_{1}^{2} + PF_{2}^{2} = F_{1}F_{2}^{2}$. Thus, $PF_{1} = 8$, $PF_{2} = 6$, and we obtain $\\cos \\angle PF_{2}F_{1} = \\frac{3}{5}$. Therefore, in $\\triangle F_{1}QF_{2}$, $QF_{1}^{2} = F_{1}F_{2}^{2} + QF_{2}^{2} - 2QF_{2} \\cdot F_{1}F_{2} \\cdot \\cos \\angle QF_{2}F_{1}$. We get $(14 - QF_{2})^{2} = 100 + QF_{2}^{2} - 2QF_{2} \\cdot 10 \\cdot (-\\frac{3}{5})$, solving yields: $QF_{2} = \\frac{12}{5}$. Then the inradius of $\\triangle F_{1}PQ$ is $\\frac{12}{5}$."
  },
  {
    "text": "Given two lines passing through points $A(-1,0)$ and $B(1,0)$ respectively intersect at point $P$, if the product of the slopes of lines $PA$ and $PB$ is $-1$, then what is the trajectory equation of the moving point $P$?",
    "fact_expressions": "A: Point;Coordinate(A) = (-1, 0);B: Point;Coordinate(B) = (1, 0);L1: Line;L2: Line;PointOnCurve(A, L1) = True;PointOnCurve(B, L2) = True;Intersection(L1, L2) = P;P: Point;Slope(LineOf(P, A))*Slope(LineOf(P, B)) = -1",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(x^2+y^2=1)&Negation(y=0)",
    "fact_spans": "[[[5, 15]], [[5, 15]], [[16, 25]], [[16, 25]], [], [], [[2, 30]], [[2, 30]], [[2, 37]], [[33, 37], [66, 69]], [[39, 62]]]",
    "query_spans": "[[[66, 76]]]",
    "process": "Let point P(x, y). Since the product of the slopes of lines PA and PB is -1, we have k_{PA} \\cdot k_{PB} = -1, that is, \\frac{y}{x-1} \\cdot \\frac{y}{x+1} = -1. Rearranging gives: x^{2} + y^{2} = 1 (y \\neq 0). Therefore, the trajectory equation of the moving point P is x^{2} + y^{2} = 1 (y \\neq 0)."
  },
  {
    "text": "The line $ l $ with slope $ 2 $ passes through the right focus of the hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0 $, $ b > 0 $) and intersects both the left and right branches of the hyperbola. What is the range of the eccentricity $ e $ of the hyperbola?",
    "fact_expressions": "l: Line;G: Hyperbola;b: Number;a: Number;e: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Slope(l)=2;PointOnCurve(RightFocus(G),l);IsIntersect(l,LeftPart(G));IsIntersect(l,RightPart(G));Eccentricity(G) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "(\\sqrt{5}, +\\infty)",
    "fact_spans": "[[[7, 12]], [[13, 69], [75, 78], [89, 92]], [[16, 69]], [[16, 69]], [[96, 99]], [[16, 69]], [[16, 69]], [[13, 69]], [[0, 12]], [[7, 73]], [[7, 87]], [[7, 87]], [[89, 99]]]",
    "query_spans": "[[[96, 105]]]",
    "process": "According to the problem and the given figure, the slope $\\frac{b}{a}$ of one asymptote of the hyperbola is greater than $2$, that is, $\\frac{b}{a}>2$. Therefore, the eccentricity of the hyperbola $e=\\frac{c}{a}=\\sqrt{1+(\\frac{b}{a})^{2}}>\\sqrt{5}$."
  },
  {
    "text": "Given that the coordinates of the focus of a parabola are $(0,3)$, then the standard equation of the parabola is?",
    "fact_expressions": "G: Parabola;Coordinate(Focus(G)) = (0, 3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 = -12 \\cdot y",
    "fact_spans": "[[[2, 5], [20, 23]], [[2, 18]]]",
    "query_spans": "[[[20, 30]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the right focus of the hyperbola $x^{2}-2 y^{2}=6$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - 2*y^2 = 6)",
    "query_expressions": "Coordinate(RightFocus(G))",
    "answer_expressions": "(3,0)",
    "fact_spans": "[[[0, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 28]]]",
    "process": "Convert the equation of the hyperbola into standard form and then determine the coordinates of the foci. Converting the equation of the hyperbola into standard form gives: \\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1, then a^{2}=6, b^{2}=3, c^{2}=a^{2}+b^{2}=9, so c=3, therefore the coordinates of the right focus are (3,0)."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through point $F_{1}$ intersects the ellipse, and the shortest chord intercepted by the ellipse satisfies $|MN|=\\frac{32}{5}$, and the perimeter of $\\Delta MF_{2}N$ is $20$. Then the eccentricity $e$ of the ellipse equals?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Line;PointOnCurve(F1, H);IsIntersect(H, G);M: Point;N: Point;Min(Length(InterceptChord(H, G))) = Abs(LineSegmentOf(M, N));Abs(LineSegmentOf(M, N)) = 32/5;Perimeter(TriangleOf(M, F2, N)) = 20;e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "3/5",
    "fact_spans": "[[[2, 47], [85, 87], [91, 93], [148, 150]], [[2, 47]], [[4, 47]], [[4, 47]], [[56, 63], [73, 81]], [[64, 71]], [[2, 71]], [[2, 71]], [[82, 84]], [[72, 84]], [[82, 89]], [[101, 120]], [[101, 120]], [[82, 120]], [[101, 120]], [[122, 146]], [[154, 157]], [[148, 157]]]",
    "query_spans": "[[[154, 160]]]",
    "process": ""
  },
  {
    "text": "It is known that the ellipse $E$ has its center at the coordinate origin and an eccentricity of $\\frac{1}{2}$. The right focus of $E$ coincides with the focus of the parabola $C$: $y^{2}=8x$. Points $A$ and $B$ are the two intersection points of the directrix of $C$ and the ellipse $E$. Then $|AB|=$?",
    "fact_expressions": "C: Parabola;E: Ellipse;A: Point;B: Point;O:Origin;Expression(C) = (y^2 = 8*x);Center(E)=O;Eccentricity(E)=1/2;RightFocus(E) = Focus(C);Intersection(Directrix(C),E)={A,B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "6",
    "fact_spans": "[[[42, 61], [75, 78]], [[2, 7], [34, 37], [82, 87]], [[67, 70]], [[71, 74]], [[11, 15]], [[42, 61]], [[2, 15]], [[2, 33]], [[34, 66]], [[67, 92]]]",
    "query_spans": "[[[94, 103]]]",
    "process": "The focus of the parabola is (2,0), and the equation of the directrix is x = -2. Thus, for the ellipse, c = 2. Since \\frac{c}{a} = \\frac{1}{2}, it follows that a = 4, b^{2} = 12. The equation of the ellipse is \\frac{x^{2}}{16} + \\frac{y^{2}}{12} = 1. Substituting x = -2 into the ellipse equation gives y = \\pm3, hence |AB| = 6."
  },
  {
    "text": "$\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G)=(x^2/16 - y^2/9 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/4",
    "fact_spans": "[[[39, 42]], [[0, 42]]]",
    "query_spans": "[[[39, 48]]]",
    "process": "From the hyperbola equation $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, we have $a=4$, $b=3$, so $c=\\sqrt{a^{2}+b^{2}}=5$. Therefore, the eccentricity of the hyperbola is $e=\\frac{c}{a}=\\frac{5}{4}$."
  },
  {
    "text": "Given a parabola $C$: $x^{2}=8y$, draw tangents $PA$ and $PB$ from an arbitrary point $P$ on the directrix of the parabola, with points of tangency $A$ and $B$, respectively. Then the minimum value of the sum of the distances from point $A$ to the directrix and from point $B$ to the directrix is?",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 8*y);P: Point;PointOnCurve(P, Directrix(C));A: Point;B: Point;TangentOfPoint(P, C) = {LineOf(P, A), LineOf(P, B)};TangentPoint(LineOf(P, A), C) = A;TangentPoint(LineOf(P, B), C) = B",
    "query_expressions": "Min(Distance(A, Directrix(C))+ Distance(B, Directrix(C)))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 20], [32, 35]], [[1, 20]], [[28, 31]], [[1, 31]], [[56, 59], [65, 69]], [[60, 63], [76, 80]], [[0, 50]], [[0, 63]], [[0, 63]]]",
    "query_spans": "[[[32, 94]]]",
    "process": "Let $ A(x_{1},\\frac{x_{1}^2}{8}), B(x_{2},\\frac{x_{2}^2}{8}) $, from $ x^{2}=8y $ we get $ y=\\frac{x^{2}}{8} $, according to the geometric meaning of derivative, obtain the equations of the two tangent lines, solve the system to find the coordinates of point $ P $, then convert the distance to the directrix into the distance to the focus, the minimum distance occurs when the three points are collinear, and then find the minimum value. Solution: Let $ A(x_{1},\\frac{x_{1}^2}{8}), B(x_{2},\\frac{x_{2}^2}{8}) $, from $ x^{2}=8y $ we get $ y=\\frac{x^{2}}{8} $, so $ y'=\\frac{x}{4} $, thus the equations of lines $ PA $ and $ PB $ are respectively:  \n$ y-\\frac{x_{1}^2}{8}=\\frac{x_{1}}{4}(x-x_{1}) $,  \n$ y-\\frac{x_{2}^2}{8}=\\frac{x_{2}}{4}(x-x_{2}) $.  \nSolving the system  \n$$\n\\begin{cases}\ny-\\frac{x_{1}^2}{8}=\\frac{x_{1}}{4}(x-x_{1}) \\\\\ny-\\frac{x_{2}^2}{8}=\\frac{x_{2}}{4}(x-x_{2})\n\\end{cases}\n$$\nyields  \n$$\n\\begin{cases}\nx=\\frac{x_{1}+x_{2}}{2} \\\\\ny=\\frac{x_{1}x_{2}}{8}\n\\end{cases}\n$$\ni.e., $ P\\left(\\frac{x_{1}+x_{2}}{2},\\frac{x_{1}x_{2}}{8}\\right) $. Since $ P $ lies on the directrix, $ \\frac{x_{1}x_{2}}{8}=-2 $, so $ x_{1}x_{2}=-16 $. Let the equation of line $ AB $ be $ y=kx+m $, substituting into the parabola's equation gives: $ x^{2}-8kx-8m=0 $, yielding $ x_{1}x_{2}=-8m $, so $ m=2 $, i.e., the line always passes through the point $ (0,2) $, which is the focus $ (0,2) $, thus the equation of line $ AB $ is $ y=kx+2 $. Substituting into the parabola's equation: $ x^{2}-8kx-16=0 $, $ x_{1}+x_{2}=8k $, so $ y_{1}+y_{2}=k(x_{1}+x_{2})+4=8k^{2}+4 $. The sum of the distances from point $ A $ to the directrix and from point $ B $ to the directrix equals $ AF+BF=y_{1}+y_{2}+4=8k^{2}+8\\geqslant8 $. Therefore, when $ k=0 $, the sum of distances is minimized and equals 8, at this time line $ AB $ is parallel to the $ x $-axis."
  },
  {
    "text": "The real axis endpoints of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$ are $M$ and $N$. A point $P$, distinct from $M$ and $N$, lies on this hyperbola. Then the product of the slopes of $PM$ and $PN$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;M: Point;N: Point;Expression(G) = (x^2/4 - y^2/3 = 1);Negation(P=M);Negation(P=N);Endpoint(RealAxis(G)) = {M, N};PointOnCurve(P, G)",
    "query_expressions": "Slope(LineSegmentOf(P, M))*Slope(LineSegmentOf(P, N))",
    "answer_expressions": "3/4",
    "fact_spans": "[[[0, 38], [69, 72]], [[63, 67]], [[44, 47], [55, 58]], [[48, 51], [59, 62]], [[0, 38]], [[52, 67]], [[52, 67]], [[0, 51]], [[63, 73]]]",
    "query_spans": "[[[76, 94]]]",
    "process": "The real axis endpoints of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$ are $M$ and $N$, so we have $M(-2,0)$, $N(2,0)$. Let the coordinates of point $P$ be $(x,y)$, then $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$, $y=\\frac{3}{4}(x^{2}-4)$, we get $k_{PM}\\cdot k_{PN}=\\frac{y}{x+2}\\cdot\\frac{y}{x-2}=\\frac{y^{2}}{x^{2}-4}$. Substituting $y=\\frac{3}{4}(x^{2}-4)$ yields $k_{PM}\\cdot k_{PN}=\\frac{3}{4}$. Hence, the answer is $\\frac{3}{4}$.  \n【Analysis】This problem mainly examines the properties of hyperbolas and the slopes of lines. Mastering the properties of hyperbolas and applying them flexibly is the key to solving the problem."
  },
  {
    "text": "The line $\\sqrt{3} x - y = 0$ is an asymptote of the hyperbola $x^{2} - \\frac{y^{2}}{b^{2}} = 1$ $(b > 0)$, then the value of $b$ is?",
    "fact_expressions": "G: Hyperbola;b: Number;H: Line;b>0;Expression(G) = (x^2 - y^2/b^2 = 1);Expression(H) = (sqrt(3)*x - y = 0);OneOf(Asymptote(G))=H",
    "query_expressions": "b",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[19, 56]], [[64, 67]], [[0, 18]], [[22, 56]], [[19, 56]], [[0, 18]], [[0, 62]]]",
    "query_spans": "[[[64, 71]]]",
    "process": "From the hyperbola equation, it follows that the asymptotes of the hyperbola satisfy: $x^{2}-\\frac{y^{2}}{b^{2}}=0$, which simplifies to: $y=\\pm bx$, i.e., $bx\\pm y=0$. Thus, one asymptote of the hyperbola is: $bx-y=0$. According to the given conditions, we obtain: $b=\\sqrt{3}$."
  },
  {
    "text": "The equation $\\frac{x^{2}}{2-a}+\\frac{y^{2}}{a+1}=1$ represents an ellipse, then $a \\in$?",
    "fact_expressions": "G: Ellipse;Expression(G)=(x^2/(2-a)+y^2/(a+1)=1);a:Number",
    "query_expressions": "Range(a)",
    "answer_expressions": "{(-1,1/2)+(1/2,2)}",
    "fact_spans": "[[[43, 45]], [[0, 45]], [[2, 41]]]",
    "query_spans": "[[[47, 55]]]",
    "process": "Since the equation $\\frac{x^2}{2-a}+\\frac{y^{2}}{a+1}=1$ represents an ellipse, we have $\\begin{cases}2-a>0\\\\a+1>0\\\\2-a\\neqa+1\\end{cases}\\Rightarrowa\\in(-1,\\frac{1}{2})\\cup(\\frac{1}{2},2)$"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, and directrix $l$. If point $A(0,-3)$, and point $P$ is a point on parabola $C$, with $PM \\perp l$ at $M$, then the minimum value of $|PM|+|PA|$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l: Line;Directrix(C) = l;A: Point;Coordinate(A) = (0, -3);P: Point;PointOnCurve(P, C);M: Point;IsPerpendicular(LineSegmentOf(P, M), l);FootPoint(LineSegmentOf(P, M), l) = M",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[9, 28], [56, 62]], [[9, 28]], [[5, 8]], [[2, 28]], [[32, 37]], [[9, 37]], [[40, 50]], [[40, 50]], [[51, 55]], [[51, 65]], [[81, 84]], [[67, 80]], [[67, 84]]]",
    "query_spans": "[[[87, 106]]]",
    "process": ""
  },
  {
    "text": "Given that $P(m, n)$ is a moving point on the ellipse $x^{2}+\\frac{y^{2}}{2}=1$, then the range of values of $m^{2}+n^{2}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2 + y^2/2 = 1);Coordinate(P) = (m, n);PointOnCurve(P, G);m:Number;n:Number",
    "query_expressions": "Range(m^2 + n^2)",
    "answer_expressions": "[1,2]",
    "fact_spans": "[[[12, 39]], [[2, 11]], [[12, 39]], [[2, 11]], [[2, 45]], [[2, 11]], [[2, 11]]]",
    "query_spans": "[[[47, 67]]]",
    "process": "Since P(m,n) is a moving point on the ellipse \\( x^{2} + \\frac{y^{2}}{2} = 1 \\), we have \\( m^{2} + \\frac{n^{2}}{2} = 1 \\), that is, \\( n^{2} = 2 - 2m^{2} \\), so \\( m^{2} + n^{2} = 2 - m^{2} \\). Also, \\( -1 < m < 1 \\), \\( 0 < m^{2} < 1 \\), so \\( 1 < 2 - m^{2} < 2 \\), that is, \\( 1 < m^{2} + n^{2} < 2 \\). Final answer: one 1.21"
  },
  {
    "text": "The standard equation of an ellipse centered at the origin, with foci on the $x$-axis, major axis length $18$, and focal distance $6$ is?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;PointOnCurve(Focus(G),xAxis);Length(MajorAxis(G))=18;FocalLength(G) = 6",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/81+y^2/72=1",
    "fact_spans": "[[[32, 34]], [[3, 7]], [[0, 34]], [[8, 34]], [[16, 34]], [[25, 34]]]",
    "query_spans": "[[[32, 41]]]",
    "process": "According to the problem, the foci of the ellipse lie on the x-axis. Given that the major axis length is 18 and the focal distance is 6, we have a=9, c=3, then b^{2}=a^{2}-c^{2}=81-9=72. Hence, the standard equation of the ellipse is: \\frac{x^{2}}{81}+\\frac{y^{2}}{72}=1."
  },
  {
    "text": "Given that point $F(-1,0)$ is a focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+y^{2}=1$ $(a>0)$, point $M$ is an arbitrary point on the ellipse $C$, and point $N(3,2)$. When $|MN|+|MF|$ attains its maximum value, what is the slope of line $MN$?",
    "fact_expressions": "C: Ellipse;a: Number;N: Point;M: Point;F: Point;a>0;Expression(C) = (y^2 + x^2/a^2 = 1);Coordinate(F) = (-1, 0);Coordinate(N) = (3, 2);OneOf(Focus(C))=F;PointOnCurve(M,C);WhenMax(Abs(LineSegmentOf(M, F)) + Abs(LineSegmentOf(M, N)))",
    "query_expressions": "Slope(LineOf(M,N))",
    "answer_expressions": "1",
    "fact_spans": "[[[13, 54], [65, 70]], [[20, 54]], [[76, 85]], [[60, 64]], [[2, 12]], [[20, 54]], [[13, 54]], [[2, 12]], [[76, 85]], [[2, 59]], [[60, 75]], [[87, 105]]]",
    "query_spans": "[[[106, 118]]]",
    "process": ""
  },
  {
    "text": "When the eccentricity of the hyperbola $\\frac{x^{2}}{t}-\\frac{y^{2}}{t^{2}-t+4}=1$ $(t>0)$ is minimized, what are the equations of the two asymptotes of the hyperbola?",
    "fact_expressions": "G: Hyperbola;t: Number;t>0;Expression(G) = (-y^2/(t^2 - t + 4) + x^2/t = 1);WhenMin(Eccentricity(G))",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*sqrt(3)*x",
    "fact_spans": "[[[1, 52], [61, 64]], [[4, 52]], [[4, 52]], [[1, 52]], [[1, 59]]]",
    "query_spans": "[[[61, 74]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through $F_{2}$ and perpendicular to the $x$-axis intersects the two asymptotes of the hyperbola at points $A$ and $B$ respectively, with $|AB|=3\\sqrt{5}$. Given $M(4,1)$, and a moving point $P(x, y)$ on the hyperbola, find the minimum value of $|PM|+|PF_{2}|$.",
    "fact_expressions": "l: Line;G: Hyperbola;b: Number;M: Point;P: Point;A: Point;B: Point;F1:Point;F2: Point;l1:Line;l2:Line;x1:Number;y1:Number;Expression(G) = (x^2/4 - y^2/b^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(F2,l);IsPerpendicular(l,xAxis);Asymptote(G)={l1,l2};Intersection(l,l1)=A;Intersection(l,l2)=B;Coordinate(M) = (4, 1);Coordinate(P) = (x1, y1);Abs(LineSegmentOf(A, B)) = 3*sqrt(5);PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F2)) + Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[87, 92]], [[2, 44], [93, 96], [158, 161]], [[5, 44]], [[137, 145]], [[148, 157]], [[106, 109]], [[110, 113]], [[53, 60]], [[63, 70], [63, 70]], [], [], [[148, 157]], [[148, 157]], [[2, 44]], [[2, 70]], [[2, 70]], [[70, 92]], [[79, 92]], [[93, 102]], [[87, 115]], [[87, 115]], [[137, 145]], [[148, 157]], [[116, 134]], [[148, 162]]]",
    "query_spans": "[[[164, 187]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ is on the left branch of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola, respectively, and $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$. If the midpoint $N$ of $P F_{2}$ lies in the first quadrant and $N$ lies on one of the asymptotes of the hyperbola, then the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F2: Point;F1: Point;N: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, LeftPart(C));LeftFocus(C) = F1;RightFocus(C) = F2;DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;MidPoint(LineSegmentOf(P, F2)) = N;Quadrant(N) = 1;PointOnCurve(N, OneOf(Asymptote(C)))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[7, 68], [90, 93], [188, 191], [200, 203]], [[15, 68]], [[15, 68]], [[2, 6]], [[82, 89]], [[74, 81]], [[174, 177], [184, 187]], [[15, 68]], [[15, 68]], [[7, 68]], [[2, 73]], [[74, 99]], [[74, 99]], [[101, 160]], [[162, 177]], [[174, 182]], [[184, 198]]]",
    "query_spans": "[[[200, 209]]]",
    "process": "By the given condition, let |PF₁| = m, |PF₂| = n. According to the definition of hyperbola, we have n - m = 2a ①; let F₁(-c,0), F₂(c,0). Since \\overrightarrow{PF₁} \\cdot \\overrightarrow{PF₂} = 0, triangle F₁PF₂ is a right triangle with right angle at P, so m² + n² = 4c² ②; the equation of line ON is y = (b/a)x. From the condition, in right triangle ONF₂, |ON| = (1/2)m, |NF₂| = (1/2)n, thus (b/a) = (n/m) ③; from ① and ③, we obtain m = (2a²)/(b - a), n = (2ab)/(b - a). Substituting into ② gives (4a⁴)/((b - a)²) + (4a²b²)/((b - a)²) = 4c². Using c² = a² + b², this simplifies to a² = (b - a)², yielding b = 2a, c = √(a² + b²) = √5 a, then e = c/a = √5"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{7}=1$ has left focus $F_{1}$, and point $P$ lies on the right branch of the hyperbola. If the segment $P F_{1}$ intersects the circle $x^{2}+y^{2}=16$ at point $M$, and $\\overrightarrow{F_{1} M}=\\overrightarrow{M P}$, then the slope of line $P F_{1}$ is?",
    "fact_expressions": "G: Hyperbola;H: Circle;F1: Point;P: Point;M: Point;Expression(G) = (x^2/9 - y^2/7 = 1);Expression(H) = (x^2 + y^2 = 16);LeftFocus(G) = F1;PointOnCurve(P,RightPart(G));Intersection(LineSegmentOf(P,F1),H) = M;VectorOf(F1, M) = VectorOf(M, P)",
    "query_expressions": "Slope(LineOf(P,F1))",
    "answer_expressions": "pm*sqrt(15)/7",
    "fact_spans": "[[[2, 40], [58, 61]], [[79, 96]], [[45, 52]], [[53, 57]], [[99, 103]], [[2, 40]], [[79, 96]], [[2, 52]], [[53, 65]], [[67, 103]], [[105, 152]]]",
    "query_spans": "[[[154, 170]]]",
    "process": "Let the right focus of the hyperbola be $ F_{2} $, connect $ PF_{2} $, $ MO $. From $ \\overrightarrow{F_{1}M} = \\overrightarrow{MP} $, we obtain that $ M $ is the midpoint of segment $ PF_{1} $. Then we get $ PF_{2} = 8 $. By the definition of the hyperbola, we obtain $ PF_{1} = 14 $. Then in $ \\triangle OMF_{1} $, use the cosine law to compute $ \\cos\\angle OF_{1}M $, and then compute $ \\tan\\angle OF_{1}M $. Let the right focus of the hyperbola be $ F_{2} $, connect $ PF_{2} $, $ MO $. From $ \\overrightarrow{F_{1}M} = \\overrightarrow{MP} $, we obtain that $ M $ is the midpoint of segment $ PF $, so $ PF_{2} = 2MO = 8 $. By the definition of the hyperbola, $ PF_{1} - PF_{2} = 2a = 6 $, so $ PF_{1} = 14 $, thus $ MF_{1} = 7 $. In $ \\triangle OMF_{1} $, by the cosine law, $ \\cos\\angle OF_{1}M = \\frac{49 + 16 - 16}{2 \\times 7 \\times 4} = \\frac{7}{8} $. Therefore, $ \\tan\\angle OF_{1}M = \\frac{\\sqrt{15}}{7} $. Hence, the slope of line $ PF_{1} $ is $ \\pm \\frac{\\sqrt{15}}{7} $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with right focus $F(c, 0)$, point $P$ lies on the left branch of hyperbola $C$. If line $FP$ is tangent to circle $E$: $(x-\\frac{c}{3})^{2}+y^{2}=\\frac{b^{2}}{9}$ at point $M$ and $\\overrightarrow{P M}=2 \\overrightarrow{M F}$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;E: Circle;c: Number;P: Point;F: Point;M: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(E) = (y^2 + (-c/3 + x)^2 = b^2/9);Coordinate(F) = (c, 0);RightFocus(C) = F;PointOnCurve(P, LeftPart(C));TangentPoint(LineOf(F, P), E) = M;VectorOf(P, M) = 2*VectorOf(M, F)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 63], [83, 89], [206, 212]], [[10, 63]], [[10, 63]], [[103, 151]], [[68, 77]], [[78, 82]], [[68, 77]], [[154, 158]], [[10, 63]], [[10, 63]], [[2, 63]], [[103, 151]], [[68, 77]], [[2, 77]], [[78, 93]], [[95, 158]], [[159, 204]]]",
    "query_spans": "[[[206, 219]]]",
    "process": "Let the left focus of hyperbola $ C $ be $ F_{1} $. Since the center of the circle is $ E\\left(\\frac{c}{3}, 0\\right) $, we have $ |F_{1}E| = 2|EF| $. Also, since $ \\overrightarrow{PM} = 2\\overrightarrow{MF} $, it follows that $ EM \\parallel PF_{1} $ and $ |PF_{1}| = 3|EM| = b $. By the definition of the hyperbola, $ |PF| = 2a + b $, and $ PF_{1} \\perp PF $. In right triangle $ \\triangle F_{1}PF $, we have $ |F_{1}F|^{2} = |F_{1}P|^{2} + |FP|^{2} \\Rightarrow (2c)^{2} = b^{2} + (2a + b)^{2} \\Rightarrow b = 2a \\Rightarrow e = \\frac{c}{a} = \\sqrt{5} $."
  },
  {
    "text": "A line passing through the focus of the parabola $y^{2}=8x$ intersects the parabola at two points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $x_{1}+x_{2}=6$, then $|AB|$ equals?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);H: Line;PointOnCurve(Focus(G), H);A: Point;B: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Intersection(H, G) = {A,B};x1 + x2 =6",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "10",
    "fact_spans": "[[[1, 15], [22, 25]], [[1, 15]], [[19, 21]], [[0, 21]], [[26, 44]], [[46, 65]], [[26, 44]], [[46, 65]], [[26, 44]], [[46, 65]], [[26, 44]], [[47, 65]], [[19, 67]], [[70, 85]]]",
    "query_spans": "[[[88, 98]]]",
    "process": "In the parabola \\( y^{2} = 8x \\), \\( p = 4 \\), the focus is \\( F(2,0) \\), and since the line \\( AB \\) passes through the focus \\( F(2,0) \\), according to the definition of the parabola, we have \\( |AB| = |AF| + |FB| = \\left(x_{1} + \\frac{p}{2}\\right) + \\left(x_{2} + \\frac{p}{2}\\right) = x_{1} + x_{2} + p = 6 + 4 = 10 \\)."
  },
  {
    "text": "The standard equation of an ellipse passing through the point $(\\sqrt{5}, \\sqrt{3})$ and having the same foci as the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ is?",
    "fact_expressions": "G: Ellipse;H: Point;C:Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);Coordinate(H) = (sqrt(5), sqrt(3));Focus(G)=Focus(C);PointOnCurve(H,C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/20+y^2/4=1",
    "fact_spans": "[[[27, 65]], [[1, 24]], [[71, 73]], [[27, 65]], [[1, 24]], [[26, 73]], [[0, 73]]]",
    "query_spans": "[[[71, 80]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$, a line passing through the focus $F$ intersects the parabola at points $M$ and $N$ (point $M$ is in the first quadrant). If the slope of the line $MN$ is $\\sqrt{3}$, and the horizontal coordinate of point $M$ is $6$, then $p=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(G) = F;H: Line;PointOnCurve(F,H) = True;Intersection(H,G) = {M,N};M: Point;N: Point;Quadrant(M) = 1;Slope(LineOf(M,N)) = sqrt(3);XCoordinate(M) = 6",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 23], [34, 37]], [[2, 23]], [[97, 100]], [[5, 23]], [[27, 30]], [[2, 30]], [[31, 33]], [[24, 33]], [[31, 48]], [[39, 42], [49, 53], [83, 87]], [[43, 46]], [[49, 58]], [[61, 82]], [[83, 95]]]",
    "query_spans": "[[[97, 102]]]",
    "process": "From the given conditions, the focus of the parabola $ y^{2} = 2px $ ($ p > 0 $) lies on the $ x $-axis, and the equation of the directrix is $ x = -\\frac{p}{2} $. Let $ M(x_{0}, y_{0}) $, then $ |MF| = x_{0} + \\frac{p}{2} = 6 + \\frac{p}{2} $. Let the angle of inclination of line $ MN $ be $ \\alpha $, then $ \\tan\\alpha = \\sqrt{3} $. Since $ \\alpha \\in [0, \\pi) $, it follows that $ \\cos\\alpha = \\frac{1}{2} $. Therefore, $ |MF| = \\frac{x_{0} - \\frac{p}{2}}{\\cos\\alpha} = 2(6 - \\frac{p}{2}) = 6 + \\frac{p}{2} $. Solving gives $ p = 4 $."
  },
  {
    "text": "Given that $F_{1}$, $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and $P$ is a point on the ellipse $C$ such that $\\overrightarrow {P F_{1}} \\perp \\overrightarrow {P F_{2}}$. If the area of $\\Delta P F_{1} F_{2}$ is $9$, then $b$=?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Focus(C)={F1,F2};PointOnCurve(P, C);IsPerpendicular(VectorOf(P, F2),VectorOf(P, F1));Area(TriangleOf(P, F1, F2)) = 9",
    "query_expressions": "b",
    "answer_expressions": "3",
    "fact_spans": "[[[19, 75], [85, 90]], [[189, 192]], [[25, 75]], [[81, 84]], [[2, 9]], [[11, 18]], [[25, 75]], [[25, 75]], [[19, 75]], [[2, 80]], [[81, 94]], [[96, 155]], [[158, 187]]]",
    "query_spans": "[[[189, 194]]]",
    "process": ""
  },
  {
    "text": "Through the focus of the parabola $y^{2}=a x$, a line is drawn intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $x_{1}+x_{2}=8$ and $|A B|=10$, then $a=$?",
    "fact_expressions": "G: Parabola;a: Number;H: Line;A: Point;B: Point;x1: Number;x2: Number;y1: Number;y2: Number;Expression(G) = (y^2 = a*x);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);PointOnCurve(Focus(G), H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 10;x1 + x2 = 8",
    "query_expressions": "a",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 15], [22, 25]], [[96, 99]], [[19, 21]], [[26, 43]], [[46, 63]], [[26, 43]], [[46, 63]], [[26, 43]], [[46, 63]], [[1, 15]], [[26, 43]], [[46, 63]], [[0, 21]], [[19, 65]], [[84, 94]], [[68, 83]]]",
    "query_spans": "[[[96, 101]]]",
    "process": "Test question analysis: \\because|AB|=x_{1}+x_{2}+p\\therefore solve for P=2, i.e. a=4"
  },
  {
    "text": "It is known that the center of the hyperbola lies at the origin of the coordinate system, the foci are on the $y$-axis, and the eccentricity is $2$. Then, what are the asymptote equations of hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;O:Origin;Center(C)=O;PointOnCurve(Focus(C), yAxis);Eccentricity(C)=2",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "sqrt(3)*x+pm*3*y=0",
    "fact_spans": "[[[2, 5], [32, 38]], [[9, 13]], [[2, 13]], [[2, 22]], [[2, 30]]]",
    "query_spans": "[[[32, 46]]]",
    "process": "From $ e = \\frac{c}{a} = 2 $, we get $ \\frac{a^2 + b^{2}}{a^{2}} = 4 $, solving yields $ \\frac{b}{a} = \\sqrt{3} $. Since the foci of the hyperbola lie on the y-axis, the asymptotes of the hyperbola are $ y = \\pm\\frac{a}{b}x $, that is, $ y = \\pm\\frac{\\sqrt{3}}{3}x $."
  },
  {
    "text": "If the line $ax - y + 1 = 0$ passes through the focus of the parabola $y^{2} = 4x$, then the real number $a$ = ?",
    "fact_expressions": "G: Parabola;H: Line;a: Real;Expression(G) = (y^2 = 4*x);Expression(H) = (a*x - y + 1 = 0);PointOnCurve(Focus(G), H)",
    "query_expressions": "a",
    "answer_expressions": "-1",
    "fact_spans": "[[[15, 29]], [[1, 13]], [[34, 39]], [[15, 29]], [[1, 13]], [[1, 32]]]",
    "query_spans": "[[[34, 41]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ is a point on the ellipse such that $\\angle P F_{1} F_{2}=30^{\\circ}$, $\\angle PF_{2} F_{1}=60^{\\circ}$. Then the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G) = True;P: Point;AngleOf(P, F1, F2) = ApplyUnit(30, degree);AngleOf(P, F2, F1) = ApplyUnit(60, degree);e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[2, 47], [76, 78], [153, 155]], [[2, 47]], [[4, 47]], [[4, 47]], [[55, 62]], [[63, 70]], [[2, 70]], [[2, 70]], [[71, 81]], [[71, 75]], [[83, 117]], [[119, 151]], [[159, 162]], [[153, 162]]]",
    "query_spans": "[[[159, 164]]]",
    "process": ""
  },
  {
    "text": "The distance between the two foci of the ellipse $\\frac{x^{2}}{3}+y^{2}=1$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Expression(G) = (x^2/3 + y^2 = 1);Focus(G) = {F1, F2}",
    "query_expressions": "Distance(F1, F2)",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[0, 27]], [], [], [[0, 27]], [[0, 30]]]",
    "query_spans": "[[[0, 37]]]",
    "process": "Find the focal distance of the ellipse. From the given condition, we have $c^{2}=3-1=2$, $\\therefore c=\\sqrt{2}$, $\\therefore 2c=2\\sqrt{2}$."
  },
  {
    "text": "Given that the distance from a moving point $M(x, y)$ to the fixed point $(2,0)$ is less by $1$ than its distance to the line $x=-3$, then the equation of the trajectory of point $M$ is?",
    "fact_expressions": "G: Line;H: Point;M: Point;x1: Number;y1: Number;Expression(G) = (x = -3);Coordinate(H) = (2, 0);Coordinate(M) = (x1, y1);Distance(M, H) = Distance(M, G) - 1",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "y^2 = 8*x",
    "fact_spans": "[[[28, 36]], [[16, 23]], [[4, 13], [47, 50]], [[4, 13]], [[4, 13]], [[28, 36]], [[16, 23]], [[4, 13]], [[4, 43]]]",
    "query_spans": "[[[47, 57]]]",
    "process": "\\because the distance from the moving point M(x,y) to the fixed point (2,0) is 1 less than the distance to the line x=-3, \\therefore the distance from the moving point M(x,y) to the fixed point (2,0) is equal to the distance to the line x=-2. According to the definition of a parabola, the locus of the moving point M is a parabola, with equation y^{2}=8x."
  },
  {
    "text": "Let the right focus of the ellipse $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{n^{2}}=1  (m>0 , n>0)$ be $(2,0)$, and the eccentricity be $\\frac{\\sqrt{2}}{2}$. Then the equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/n^2 + x^2/m^2 = 1);m: Number;n: Number;m>0;n>0;RightFocus(G) = Z;Eccentricity(G) = sqrt(2)/2;Z: Point;Coordinate(Z) = (2,0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/8+y^2/4=1",
    "fact_spans": "[[[1, 59], [99, 101]], [[1, 59]], [[3, 59]], [[3, 59]], [[3, 59]], [[3, 59]], [[1, 71]], [[1, 96]], [[64, 71]], [[64, 71]]]",
    "query_spans": "[[[99, 106]]]",
    "process": "Given $ c=2 $, $ e=\\frac{c}{a}=\\frac{\\sqrt{2}}{2} $, together with $ a^{2}=b^{2}+c^{2} $, we can solve for $ m $ and $ n $ in this problem. The right focus of the ellipse is $ (2,0) $, so $ m^{2}-n^{2}=4 $, $ e=\\frac{\\sqrt{2}}{2}=\\frac{2}{m} $, hence $ m=2\\sqrt{2} $. Substituting into $ m^{2}-n^{2}=4 $, we get $ n^{2}=4 $. Therefore, the equation of the ellipse is $ \\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1 $."
  },
  {
    "text": "The distance from point $M(m, \\frac{1}{2})$ on the parabola $y = a x^{2}$ $(a > 0)$ to its directrix $l$ is $1$. Then the value of $a$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = a*x^2);a: Number;a>0;M: Point;Coordinate(M) = (m, 1/2);m: Number;PointOnCurve(M, G);l: Line;Directrix(G) = l;Distance(M, l) = 1",
    "query_expressions": "a",
    "answer_expressions": "1/2",
    "fact_spans": "[[[0, 21], [43, 44]], [[0, 21]], [[58, 61]], [[3, 21]], [[22, 42]], [[22, 42]], [[23, 42]], [[0, 42]], [[46, 49]], [[43, 49]], [[22, 56]]]",
    "query_spans": "[[[58, 65]]]",
    "process": "The parabola $ y = ax^2 $ ($ a > 0 $), or equivalently $ x^2 = \\frac{1}{a}y $ ($ a > 0 $), has the directrix equation $ y = -\\frac{1}{4a} $. For a point $ M(m, \\frac{1}{2}) $ on the parabola $ y = ax^2 $ ($ a > 0 $), the distance from $ M $ to its directrix $ l $ is 1, so we have $ \\frac{1}{2} + \\frac{1}{4a} = 1 $, which gives $ a = \\frac{1}{2} $."
  },
  {
    "text": "Given the focus $F$ of the parabola $y^{2}=4x$, and a point $A$ on the parabola whose distance to the $y$-axis is $3$, then $|AF|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;A: Point;PointOnCurve(A, G);Distance(A, yAxis) = 3",
    "query_expressions": "Abs(LineSegmentOf(A, F))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 16], [24, 27]], [[2, 16]], [[19, 22]], [[2, 22]], [[30, 33]], [[24, 33]], [[30, 44]]]",
    "query_spans": "[[[46, 55]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$, $A$, $B$ are its left and right vertices, respectively. A moving point $M$ satisfies $MB \\perp AB$. Connect $AM$ intersecting the ellipse at point $P$. On the $x$-axis, there exists a fixed point $Q$ distinct from points $A$ and $B$, such that the circle with diameter $MP$ passes through the intersection point of lines $BP$ and $MQ$. Then the coordinates of point $Q$ are?",
    "fact_expressions": "G: Ellipse;H: Circle;B: Point;P: Point;A: Point;M: Point;Q: Point;Expression(G) = (x^2/4 + y^2/2 = 1);LeftVertex(G)=A;RightVertex(G)=B;IsPerpendicular(LineSegmentOf(M, B),LineSegmentOf(A, B));Intersection(LineSegmentOf(A,M),G)=P;PointOnCurve(Q,xAxis);Negation(Q = B);Negation(Q=A);IsDiameter(LineSegmentOf(M,P),H);PointOnCurve(Intersection(LineSegmentOf(B,P),LineSegmentOf(M,Q)),H)",
    "query_expressions": "Coordinate(Q)",
    "answer_expressions": "(0, 0)",
    "fact_spans": "[[[2, 39], [52, 53], [90, 92]], [[131, 132]], [[48, 51], [112, 115]], [[93, 97]], [[42, 46], [107, 111]], [[61, 64]], [[118, 121], [152, 156]], [[2, 39]], [[42, 58]], [[42, 58]], [[66, 81]], [[82, 97]], [[98, 121]], [[105, 121]], [[105, 121]], [[122, 132]], [[131, 150]]]",
    "query_spans": "[[[152, 161]]]",
    "process": "Let M(2,t), then AM: y = \\frac{t}{4}(x+2). Combining with the ellipse equation and eliminating y yields (t^{2}+8)x^{2}+4t^{2}x+4t^{2}-32=0, so x_{P} = \\frac{16-2t^{2}}{t^{2}+8}, y_{P} = \\frac{8t}{t^{2}+8}. Therefore, k_{BP} = \\frac{\\frac{8t}{t^{2}+8}}{\\frac{16-2t^{2}}{t^{2}+8}-2} = -\\frac{2}{t}, i.e., k_{BP}k_{OM} = -1. The coordinates of point Q are O(0,0)."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. A circle with diameter $F_{1}F_{2}$ intersects an asymptote of $C$ at point $P$ (where $P$ lies in the first quadrant). If the midpoint $Q$ of segment $PF_{1}$ lies on the other asymptote of $C$, then the eccentricity $e$ of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;F1: Point;F2: Point;P: Point;e: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;Eccentricity(C) = e;IsDiameter(LineSegmentOf(F1, F2), G);L1: Line;L2: Line;Intersection(G, L1) = P;Quadrant(P) = 1;Q: Point;MidPoint(LineSegmentOf(P, F1)) = Q;PointOnCurve(Q, L2);OneOf(Asymptote(G)) = L1;OneOf(Asymptote(G)) = L2;Negation(L1=L2)",
    "query_expressions": "e",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 63], [110, 113], [156, 159], [169, 172]], [[10, 63]], [[10, 63]], [[108, 109]], [[72, 79]], [[80, 87]], [[120, 125], [126, 129]], [[176, 179]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 87]], [[2, 87]], [[169, 179]], [[88, 109]], [], [], [[108, 125]], [[126, 135]], [[152, 155]], [[138, 155]], [[152, 167]], [108, 116], [154, 163], [108, 163]]",
    "query_spans": "[[[176, 181]]]",
    "process": "From the figure, it can be seen that OQ is the perpendicular bisector of segment F_{1}P, and OP is the median to the hypotenuse of right triangle F_{1}PF_{2}, \\therefore |OP| = c and \\angle F_{1}OQ = \\angle POQ = \\angle POF_{2} = 60^{\\circ}, \\therefore \\frac{b}{a} = \\tan 60^{\\circ} = \\sqrt{3}, so e = 2"
  },
  {
    "text": "The chord $AB$ of the parabola $y^{2}=8x$ is perpendicular to the $x$-axis, and $|AB|=4\\sqrt{6}$. Then, what is the distance from $AB$ to the focus?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;Expression(G) = (y^2 = 8*x);IsChordOf(LineSegmentOf(A,B),G);IsPerpendicular(LineSegmentOf(A,B),xAxis);Abs(LineSegmentOf(A,B))=4*sqrt(6)",
    "query_expressions": "Distance(LineSegmentOf(A,B),Focus(G))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 14]], [[51, 56]], [[51, 56]], [[0, 14]], [[0, 30]], [[16, 30]], [[32, 49]]]",
    "query_spans": "[[[0, 64]]]",
    "process": ""
  },
  {
    "text": "In $\\triangle ABC$, $|\\overrightarrow{B C}|=4$, the incircle of $\\triangle ABC$ touches $BC$ at point $D$, and $|\\overrightarrow{B D}|-|\\overrightarrow{C D}|=2 \\sqrt{2}$, then the equation of the locus of vertex $A$ is?",
    "fact_expressions": "B: Point;C: Point;A: Point;D: Point;Abs(VectorOf(B, C)) = 4;TangentPoint(InscribedCircle(TriangleOf(A,B,C)),LineSegmentOf(B,C))=D;Abs(VectorOf(B, D)) - Abs(VectorOf(C, D)) = 2*sqrt(2)",
    "query_expressions": "LocusEquation(A)",
    "answer_expressions": "{(x^2/2-y^2/2=1)&(x>sqrt(2))}",
    "fact_spans": "[[[18, 44]], [[18, 44]], [[138, 141]], [[70, 74]], [[18, 44]], [[45, 74]], [[76, 134]]]",
    "query_spans": "[[[138, 148]]]",
    "process": ""
  },
  {
    "text": "The equation of the locus of points equidistant from the $x$-axis and the point $F(0,4)$ is?",
    "fact_expressions": "F: Point;G: Point;Coordinate(F) = (0, 4);Distance(G,xAxis) = Distance(F,G)",
    "query_expressions": "LocusEquation(G)",
    "answer_expressions": "y=(1/8)*x^2+2",
    "fact_spans": "[[[10, 19]], [[25, 26]], [[10, 19]], [[0, 26]]]",
    "query_spans": "[[[25, 33]]]",
    "process": ""
  },
  {
    "text": "The point $P$ lies on the curve represented by the equation $\\sqrt{(x-5)^{2}+y^{2}}-\\sqrt{(x+5)^{2}+y^{2}}=8$, and the ordinate of point $P$ is $3$. Then its abscissa is?",
    "fact_expressions": "G: Curve;P: Point;Expression(G) = (sqrt((x - 5)^2 + y^2) - sqrt((x + 5)^2 + y^2) = 8);PointOnCurve(P, G);YCoordinate(P) = 3",
    "query_expressions": "XCoordinate(P)",
    "answer_expressions": "-4*sqrt(2)",
    "fact_spans": "[[[60, 62]], [[0, 4], [66, 70], [80, 81]], [[5, 62]], [[0, 65]], [[66, 78]]]",
    "query_spans": "[[[80, 86]]]",
    "process": "Take points $F_{1}(-5,0)$, $F_{2}(5,0)$, then $|F_{1}F_{2}|=10$. From the given condition, $|PF_{2}|-|PF_{1}|=8<10$, so the locus of point $P$ is the left branch of a hyperbola with foci at $F_{1}(-5,0)$, $F_{2}(5,0)$ and $a=4$. Therefore, the equation of the locus of point $P$ is $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ $(x<0)$. Let point $P(x_{0},3)$ $(x_{0}<0)$, then $\\frac{x_{0}^{2}}{16}-\\frac{3^{2}}{9}=1$, solving gives $x_{0}=-4\\sqrt{2}$."
  },
  {
    "text": "The standard equation of an ellipse sharing foci with the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ and passing through the point $(4,0)$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1);H: Point;Coordinate(H) = (4, 0);Z: Ellipse;Focus(G) = Focus(Z);PointOnCurve(H, Z)",
    "query_expressions": "Expression(Z)",
    "answer_expressions": "x^2/16 + y^2/11 = 1",
    "fact_spans": "[[[1, 38]], [[1, 38]], [[44, 52]], [[44, 52]], [[53, 55]], [[0, 55]], [[43, 55]]]",
    "query_spans": "[[[53, 62]]]",
    "process": "Let the equation of the ellipse be \\frac{x^{2}}{9+k}+\\frac{y^{2}}{4+k}=1. Substitute the point (4,0) to solve, thus the standard equation of the ellipse is \\frac{x^{2}}{16}+\\frac{y^{2}}{11}=1."
  },
  {
    "text": "Given that $AB$ is a chord of the parabola $y^2 = 4x$ passing through its focus $F$, and $O$ is the origin, then $\\overrightarrow{OA} \\cdot \\overrightarrow{OB}$ = ?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;O: Origin;Expression(G) = (y^2 = 4*x);IsChordOf(LineSegmentOf(A,B),G);PointOnCurve(F,LineSegmentOf(A,B));Focus(G)=F;F: Point",
    "query_expressions": "DotProduct(VectorOf(O, A), VectorOf(O, B))",
    "answer_expressions": "-3",
    "fact_spans": "[[[9, 23]], [[2, 7]], [[2, 7]], [[31, 34]], [[9, 23]], [[2, 30]], [[2, 30]], [[9, 28]], [25, 27]]",
    "query_spans": "[[[39, 90]]]",
    "process": ""
  },
  {
    "text": "Through a point $M$ (not the vertex) on the parabola $C$: $y = x^{2}$, draw a tangent line to $C$ intersecting the $x$-axis and $y$-axis at points $A$ and $B$, respectively. Then $\\frac{|M A|}{|M B|} = $?",
    "fact_expressions": "C: Parabola;Expression(C) = (y = x^2);M: Point;PointOnCurve(M,C) = True;Intersection(L1 , xAxis) = A;Intersection(L1 , yAxis) = B;A: Point;B: Point;L1: Line;TangentOnPoint(M,C) = L1;Negation(M = Vertex(C))",
    "query_expressions": "Abs(LineSegmentOf(M, A)) / Abs(LineSegmentOf(M, B))",
    "answer_expressions": "1/2",
    "fact_spans": "[[[1, 18], [31, 34]], [[1, 18]], [[22, 25]], [[1, 30]], [[0, 60]], [[0, 60]], [[51, 54]], [[55, 58]], [], [[0, 37]], [[0, 30]]]",
    "query_spans": "[[[62, 85]]]",
    "process": "Using derivatives to find the equation of the tangent line, obtain the coordinates of two points, and the result follows. From $ y = x^{2} $, we have $ y' = 2x $. Let point $ M(x_{0}, x_{0}^{2}) $ ($ x_{0} \\neq 0 $), then the slope of the tangent line to curve $ C $ at $ M $ is $ k = 2x_{0} $. Thus, the equation of the tangent line to curve $ C $ at $ M $ is: $ y - x_{0}^{2} = 2x_{0}(x - x_{0}) $, that is, $ y = 2x_{0}x - x_{0}^{2} $. Therefore, $ A\\left(\\frac{x_{0}}{2}, 0\\right) $, $ B(0, -x_{0}^{2}) $. From the coordinates of points $ M $, $ A $, $ B $, it follows that point $ A $ is the midpoint of $ BM $, so $ \\frac{|MA|}{|MB|} = \\frac{1}{2} $."
  },
  {
    "text": "Through any point $M$ on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, draw a line parallel to the real axis intersecting the asymptotes at points $P$ and $Q$. Then $M P \\cdot M Q = $?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;M: Point;PointOnCurve(M, G);H: Line;PointOnCurve(M, H);IsParallel(H, RealAxis(G));P: Point;Q: Point;Intersection(H, Asymptote(G)) = {P, Q}",
    "query_expressions": "LineSegmentOf(M, P)*LineSegmentOf(M, Q)",
    "answer_expressions": "a^2",
    "fact_spans": "[[[2, 59]], [[2, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[63, 66]], [[2, 66]], [[74, 76]], [[0, 76]], [[2, 76]], [[83, 86]], [[87, 90]], [[2, 92]]]",
    "query_spans": "[[[94, 111]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with left and right foci $F_{1}(-c, 0)$, $F_{2}(c, 0)$ respectively, if there exists a point $P$ on the ellipse such that $\\frac{a}{\\sin \\angle P F_{1} F_{2}}=\\frac{c}{\\sin \\angle P F_{2} F_{1}}$, then the range of the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;c: Number;Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);a/Sin(AngleOf(P, F1, F2)) = c/Sin(AngleOf(P, F2, F1))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(\\sqrt{2}-1,1)",
    "fact_spans": "[[[2, 54], [94, 96], [181, 183]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[63, 77]], [[79, 92]], [[63, 77]], [[63, 77]], [[79, 92]], [[2, 92]], [[2, 92]], [[101, 104]], [[94, 104]], [[105, 178]]]",
    "query_spans": "[[[181, 194]]]",
    "process": "In $\\triangle PF_{1}F_{2}$, by the law of sines: $\\frac{|PF_{2}|}{\\sin\\angle PF_{1}F_{2}} = \\frac{|PF_{1}|}{\\sin\\angle PF_{2}F_{1}}$, then from the given condition: $\\frac{|PF_{2}|}{a} = \\frac{|PF_{1}|}{c}$, i.e., $a|PF_{1}| = c|PF_{2}|$. Let the point be $(x_{0}, y_{0})$. By the focal radius formulas, we have: $|PF_{1}| = a + ex_{0}$, $|PF_{2}| = a - ex_{0}$. Then $a(a + ex_{0}) = c(a - ex_{0})$. Solving gives: $x_{0} = \\frac{a(a - c)}{e(-a + c)} = \\frac{a(e - 1)}{e(e + 1)}$. From the geometric property of the ellipse, we know: $x_{0} > -a$, so $\\frac{a(e - 1)}{e(e + 1)} > -a$. Rearranging yields $e^{2} + 2e - 1 > 0$, solving gives: $e < \\sqrt{2} - 1$ or $e > \\sqrt{2} - 1$. Since $e \\in (0, 1)$, the eccentricity of the ellipse is: $e \\in (\\sqrt{2} - 1, 1)$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Points $M$ and $N$ are moving points on the asymptote and the left branch of the hyperbola, respectively. Then the minimum value of $|M N|+|N F_{2}|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;M: Point;N: Point;PointOnCurve(M, Asymptote(G));PointOnCurve(N, LeftPart(G))",
    "query_expressions": "Min(Abs(LineSegmentOf(M, N)) + Abs(LineSegmentOf(N, F2)))",
    "answer_expressions": "1+2*sqrt(3)",
    "fact_spans": "[[[2, 30], [71, 74]], [[2, 30]], [[40, 47]], [[48, 55]], [[2, 55]], [[2, 55]], [[56, 60]], [[61, 64]], [[56, 80]], [[56, 80]]]",
    "query_spans": "[[[82, 106]]]",
    "process": "According to the problem, $ F_{1}(-2,0) $, $ F_{2}(2,0) $, without loss of generality, take one asymptote as $ y = \\frac{\\sqrt{3}}{3}x $. By the definition of a hyperbola, $ |NF_{2}| - |NF_{1}| = 2a = 2\\sqrt{3} $. Therefore, $ |NF_{2}| = 2\\sqrt{3} + |NF_{1}| $, then $ |MN| + |NF_{2}| = |MN| + |NF_{1}| + 2\\sqrt{3} $. Thus, when points $ M $, $ N $, $ F_{1} $ are collinear and $ F_{1}M $ is perpendicular to the asymptote $ y = \\frac{\\sqrt{3}}{3}x $, $ |MN| + |NF_{2}| $ reaches its minimum value. At this time, the distance from $ F_{1} $ to the asymptote is $ d $, and the minimum value is: $ d + 2\\sqrt{3} = 1 + 2\\sqrt{3} $."
  },
  {
    "text": "Let the parabola be $y^{2}=4 x$, and let a line $l$ passing through the point $(1,0)$ intersect the parabola at points $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$. Then $x_{1} x_{2}+y_{1} y_{2}$=?",
    "fact_expressions": "l: Line;G: Parabola;H: Point;P: Point;Q: Point;x1: Number;x2: Number;y1: Number;y2: Number;Expression(G) = (y^2 = 4*x);Coordinate(H) = (1, 0);Coordinate(P) = (x1, y1);Coordinate(Q) = (x2, y2);PointOnCurve(H, l);Intersection(l, G) = {P, Q}",
    "query_expressions": "x1*x2 + y1*y2",
    "answer_expressions": "-3",
    "fact_spans": "[[[27, 32]], [[1, 4], [33, 36]], [[18, 26]], [[38, 55]], [[57, 75]], [[38, 55]], [[58, 75]], [[38, 55]], [[58, 75]], [[1, 16]], [[18, 26]], [[38, 55]], [[58, 75]], [[17, 32]], [[27, 77]]]",
    "query_spans": "[[[79, 106]]]",
    "process": "Let the equation of line $ l $ be $ x = ty + 1 $, then  \n$$\n\\begin{cases}\nx = ty + 1 \\\\\ny = 4x\n\\end{cases}\n$$\nwe obtain $ y^{2} - 4ty - 4 = 0 \\cdot y_{1}y_{2} = -4 $. Also, since $ y_{1}^{2} = 4x_{1} $, $ y_{2}^{2} = 4x_{2} $,  \n$ \\therefore x_{1}x_{2} = \\frac{y_{1}^{2}y_{2}^{2}}{16} = 1 $, so $ x_{1}x_{2} + y_{1}y_{2} = 1 - 4 = -3 $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus is $F$, the right vertex is $A$, and point $B$ lies on the ellipse such that $B F \\perp x$-axis. The line $A B$ intersects the $y$-axis at point $P$. If $\\overrightarrow{A P}=2 \\overrightarrow{P B}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;B: Point;F: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F;RightVertex(G) = A;PointOnCurve(B, G);IsPerpendicular(LineSegmentOf(B,F),xAxis);Intersection(LineOf(A,B),yAxis)=P;VectorOf(A,P)=2*VectorOf(P,B);P:Point",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 54], [76, 78], [163, 165]], [[4, 54]], [[4, 54]], [[67, 70]], [[71, 75]], [[59, 62]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 62]], [[2, 70]], [[71, 79]], [[81, 95]], [[96, 113]], [[116, 161]], [[109, 113]]]",
    "query_spans": "[[[163, 171]]]",
    "process": "As shown in the figure, since $ BF \\perp x $-axis, we have $ x_{B} = -c $, $ y_{B} = \\frac{b^{2}}{a} $; let point $ P(0, t) $, because $ \\overrightarrow{AP} = 2\\overrightarrow{PB} $, so $ (-a, t) = 2(-c, \\frac{b^{2}}{a} - t) $, we obtain $ a = 2c $; therefore $ e = \\frac{c}{a} = \\frac{1}{2} $."
  },
  {
    "text": "Given that the line $l$: $y = kx + t$ is tangent to the circle $x^{2} + (y+1)^{2} = 1$ and intersects the parabola $C$: $x^{2} = 4y$ at two distinct points $M$, $N$, what is the range of real values for $t$?",
    "fact_expressions": "l: Line;C: Parabola;G: Circle;M: Point;N: Point;t: Real;k: Number;Expression(C) = (x^2 = 4*y);Expression(G) = (x^2 + (y + 1)^2 = 1);Expression(l) = (y = k*x + t);Intersection(l, C) = {M, N};IsTangent(l, G);Negation(M = N)",
    "query_expressions": "Range(t)",
    "answer_expressions": "{(-oo, -3), (0, +oo)}",
    "fact_spans": "[[[2, 18]], [[43, 62]], [[19, 39]], [[69, 72]], [[73, 76]], [[79, 84]], [[8, 18]], [[43, 62]], [[19, 39]], [[2, 18]], [[2, 76]], [[2, 41]], [[64, 76]]]",
    "query_spans": "[[[79, 91]]]",
    "process": "Since the line is tangent to the circle, $\\frac{|t+1|}{\\sqrt{1+k^{2}}}=1\\Rightarrow k^{2}=t^{2}+2t$. Substituting the line equation into the parabola equation and simplifying yields $x^{2}-4kx-4t=0$. Then, from $\\Delta=16k^{2}+16t=16(t^{2}+2t)+16t>0$, we obtain $t>0$ or $t<-3$."
  },
  {
    "text": "Let the foci of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ be $F_{1}$, $F_{2}$, and let $P$ be a point on this hyperbola. If $|P F_{1}|=7$, then $|P F_{2}|$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);F1: Point;F2: Point;Focus(G) = {F1,F2};P: Point;PointOnCurve(P,G) = True;Abs(LineSegmentOf(P, F1)) = 7",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "13",
    "fact_spans": "[[[1, 40], [65, 68]], [[1, 40]], [[44, 51]], [[52, 59]], [[1, 59]], [[60, 63]], [[60, 72]], [[74, 87]]]",
    "query_spans": "[[[89, 102]]]",
    "process": "According to the hyperbola definition ||PF_{1}|-|PF_{2}||=2a, solve as follows. From the definition of the hyperbola, ||PF_{1}|-|PF_{2}||=2a=6, and since |PF_{1}|=7, then |PF_{2}|=1 or |PF_{2}|=13. Upon verification, |PF_{2}|=1<c-a is discarded, so |PF_{2}|=13."
  },
  {
    "text": "Let a line with slope $1$ pass through the focus $F$ of the parabola ${y}^{2}=ax$ ($a>0$), and intersect the $y$-axis at point $A$. If the area of $\\triangle OAF$ ($O$ being the origin) is $8$, then the value of $a$ is?",
    "fact_expressions": "L: Line;Slope(L) = 1;C: Parabola;a:Number;a>0;Expression(C) = (y**2=a*x);PointOnCurve(F, L);A: Point;A = Intersection(L, yAxis);F: Point;F = Focus(C);O: Origin;Area(TriangleOf(O, A, F)) = 8",
    "query_expressions": "a",
    "answer_expressions": "16",
    "fact_spans": "[[[8, 10]], [[1, 10]], [[11, 33]], [[88, 91]], [[14, 33]], [[11, 33]], [[8, 39]], [[48, 52]], [[8, 52]], [[36, 39]], [[11, 39]], [[70, 73]], [[54, 86]]]",
    "query_spans": "[[[88, 95]]]",
    "process": "According to the problem, $ F\\left(\\frac{a}{4},0\\right) $, the line $ l $ is $ y = x - \\frac{a}{4} $, so $ A\\left(0,-\\frac{a}{4}\\right) $, the area of $ \\triangle OAF $ is $ \\frac{1}{2} \\times \\frac{a}{4} \\times \\frac{a}{4} = 8 $. Solving gives $ a = \\pm 16 $, and according to the problem, only $ a = 16 $ is valid."
  },
  {
    "text": "Given that point $F$ is the focus of the parabola $y^{2}=8x$, a line passing through $F$ intersects the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. Then $x_{1} x_{2}-y_{1} y_{2}$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);A: Point;B: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Intersection(H, G) = {A, B}",
    "query_expressions": "x1*x2 - y1*y2",
    "answer_expressions": "20",
    "fact_spans": "[[[7, 21], [33, 36]], [[7, 21]], [[2, 6], [26, 29]], [[2, 24]], [[30, 32]], [[25, 32]], [[37, 54]], [[57, 74]], [[37, 54]], [[57, 74]], [[37, 54]], [[57, 74]], [[37, 54]], [[57, 74]], [[30, 74]]]",
    "query_spans": "[[[76, 103]]]",
    "process": "The focus of the parabola has coordinates $ F(2,0) $. When the slope $ k $ of the line passing through $ F $ does not exist, $ x_{1} = x_{2} = 2 $, then $ y^{2} = 8 \\times 2 = 16 $, so $ y = \\pm 4 $, thus $ y_{1}y_{2} = -4 \\times 4 = -16 $, and $ x_{1}x_{2} - y_{1}y_{2} = 2 \\times 2 - (-16) = 4 + 16 = 20 $. When the slope $ k $ of the line passing through $ F $ exists, the equation of the line is $ y = k(x - 2) $. Combining with the equation $ y^{2} = 8x $ gives $ ky^{2} - 8y - 16k = 0 $, so $ y_{1}y_{2} = \\frac{-16k}{k} = -16 $, and $ x_{1}x_{2} = \\frac{y_{1}^{2}}{8} \\times \\frac{y_{2}^{2}}{8} = \\frac{16 \\times 16}{8 \\times 8} = 4 $, then $ x_{1}x_{2} - y_{1}y_{2} = 4 + 16 = 20 $."
  },
  {
    "text": "Given that the line $y = x - 4$ intersects the parabola $y^2 = 2px$ ($p > 0$) at points $A$ and $B$, and $O$ is the origin such that $OA \\perp OB$, find $p = ?$",
    "fact_expressions": "G: Parabola;p: Number;H: Line;O: Origin;A: Point;B: Point;p>0;Expression(G) = (y^2 = 2*p*x);Expression(H) = (y = x - 4);Intersection(H, G) = {A, B};IsPerpendicular(LineSegmentOf(O,A),LineSegmentOf(O,B))",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[12, 33]], [[71, 74]], [[2, 11]], [[45, 48]], [[35, 38]], [[39, 42]], [[15, 33]], [[12, 33]], [[2, 11]], [[2, 44]], [[54, 69]]]",
    "query_spans": "[[[71, 76]]]",
    "process": "Solve the system of equations formed by the line and the parabola, use the perpendicular relationship between the lines, and directly apply the relationship between roots and coefficients. Let A(x_{1},y_{1}), B(x_{2},y_{2}). Solve the system:  \n\\begin{cases}y=x-4\\\\y^{2}=2px\\end{cases} \\Rightarrow y^{2}=2p(y+4),  \nwe obtain y^{2}-2py-8p=0, then  \n\\begin{cases}y_{1}y_{2}=-8p\\\\y_{1}+y_{2}=2p\\end{cases}.  \n\\because OA\\bot OB, \\therefore x_{1}x_{2}+y_{1}y_{2}=0.  \nThus, x_{1}x_{2}=\\frac{y_{1}^{2}}{2p}\\cdot\\frac{y_{2}^{2}}{2p}=16,\\ 16+(-8p)=0 \\Rightarrow p=2"
  },
  {
    "text": "A line passing through the right focus $F_{2}$ of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ intersects the ellipse at points $M$ and $N$. Let $|F_{2} M|=m$, $|F_{2} N|=n$, then $\\frac{m n}{m+n}$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);F2: Point;RightFocus(G) = F2;H: Line;PointOnCurve(F2, H);M: Point;N: Point;Intersection(H, G) = {M, N};m: Number;n: Number;Abs(LineSegmentOf(F2, M)) = m;Abs(LineSegmentOf(F2, N)) = n",
    "query_expressions": "(m*n)/(m + n)",
    "answer_expressions": "3/4",
    "fact_spans": "[[[1, 38], [53, 55]], [[1, 38]], [[42, 49]], [[1, 49]], [[50, 52]], [[0, 52]], [[56, 59]], [[60, 63]], [[50, 65]], [[67, 80]], [[81, 94]], [[67, 80]], [[81, 94]]]",
    "query_spans": "[[[96, 115]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a moving point on the parabola $x^{2}=4 y$, the projection of point $P$ onto the $x$-axis is $M$, and the coordinates of point $A$ are $(2,0)$, then the minimum value of $|PA|+| PM|$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);P: Point;PointOnCurve(P, G);M: Point;Projection(P, xAxis) = M;A: Point;Coordinate(A) = (2, 0)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "sqrt(5)-1",
    "fact_spans": "[[[6, 20]], [[6, 20]], [[2, 5], [25, 29]], [[2, 24]], [[39, 42]], [[25, 42]], [[43, 47]], [[43, 58]]]",
    "query_spans": "[[[60, 78]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ has two foci at $F_{1}(-\\frac{{\\sqrt{3}}}{2} , 0)$ and $F_{2}(\\frac{\\sqrt {3}}{2} , 0)$, point $P$ lies on the hyperbola in the first quadrant, and $\\tan \\angle PF_{1} F_{2}=\\frac{1}{2}$, $\\tan \\angle PF_{2} F_{1}=-2$. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F1: Point;F2: Point;P: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-sqrt(3)/2, 0);Coordinate(F2) = (sqrt(3)/2, 0);Focus(G) = {F1,F2};Quadrant(P) = 1;PointOnCurve(P, G);Tan(AngleOf(P,F1,F2))=1/2;Tan(AngleOf(P, F2, F1)) = -2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3*sqrt(5)/5",
    "fact_spans": "[[[2, 61], [147, 150], [227, 230]], [[5, 61]], [[5, 61]], [[67, 101]], [[104, 136]], [[137, 141]], [[5, 61]], [[5, 61]], [[2, 61]], [[67, 101]], [[104, 136]], [[2, 136]], [[137, 153]], [[137, 153]], [[155, 193]], [[196, 225]]]",
    "query_spans": "[[[227, 236]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A circle with diameter $F_{1}F_{2}$ intersects an asymptote of the hyperbola at point $(3,4)$. Then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Circle;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;IsDiameter(LineSegmentOf(F1, F2), H);Coordinate(OneOf(Intersection(H, Asymptote(G)))) = (3, 4)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/16 = 1",
    "fact_spans": "[[[2, 59], [104, 107], [126, 129]], [[5, 59]], [[5, 59]], [[102, 103]], [[68, 75]], [[76, 83]], [[5, 59]], [[5, 59]], [[2, 59]], [[2, 83]], [[2, 83]], [[84, 103]], [[102, 123]]]",
    "query_spans": "[[[126, 134]]]",
    "process": "The radius of the circle is $ c $, and since the point $ (3,4) $ lies on the asymptote, we have $ \\frac{3b}{a} = 4 $. Thus, $ a $ and $ b $ can be determined, giving the hyperbola equation. According to the problem, the radius of the circle is 5, and the point $ (3,4) $ lies on the asymptote $ y = \\frac{b}{a}x $ passing through the first and third quadrants. Therefore, we have\n$$\n\\begin{cases}\na^{2} + b^{2} = 25 \\\\\n4 = 3 \\times \\frac{b}{a}\n\\end{cases}\n$$\nSolving this system yields\n$$\n\\begin{cases}\na = 3 \\\\\nb = 4\n\\end{cases}\n$$\nThus, the equation of the hyperbola is $ \\frac{x^{2}}{9} - \\frac{y^{2}}{16} = 1 $."
  },
  {
    "text": "Point $P$ lies on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, with left and right foci $F_{1}$ and $F_{2}$ respectively. The line $P F_{1}$ is tangent to the circle centered at the origin $O$ with radius $a$ at point $A$. The perpendicular bisector of segment $P F_{1}$ passes exactly through point $F_{2}$. Then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;A:Point;F1: Point;P: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, RightPart(G));LeftFocus(G) = F1;RightFocus(G) = F2;Center(H)=O;O:Origin;Radius(H)=a;TangentPoint(LineOf(P,F1),H)=A;PointOnCurve(F2,PerpendicularBisector(LineSegmentOf(P,F1)))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[5, 62], [67, 68], [163, 166]], [[8, 62]], [[116, 119]], [[123, 124]], [[127, 131]], [[76, 83]], [[0, 4]], [[84, 91], [152, 160]], [[8, 62]], [[8, 62]], [[5, 62]], [[0, 66]], [[67, 91]], [[67, 91]], [[104, 124]], [[105, 112]], [[116, 124]], [[92, 131]], [[132, 160]]]",
    "query_spans": "[[[163, 172]]]",
    "process": ""
  },
  {
    "text": "$AB$ is a line segment of length $4$ in the plane, $P$ is a moving point in the plane such that $|PA| + |PB| = 6$, and $M$ is the midpoint of $AB$. Then the range of values of $|PM|$ is?",
    "fact_expressions": "A: Point;B: Point;Length(LineSegmentOf(A, B)) = 4;P: Point;Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B)) = 6;M: Point;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Range(Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "[\\sqrt{5}, 3]",
    "fact_spans": "[[[0, 5]], [[0, 5]], [[0, 20]], [[21, 24]], [[34, 49]], [[51, 54]], [[51, 63]]]",
    "query_spans": "[[[65, 79]]]",
    "process": "By the given condition, |PA| + |PB| > |AB|, so the trajectory of P is an ellipse with foci at A and B and major axis length 6. If M(0,0), A(-2,0), B(2,0), then the equation of the trajectory of P is \\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1. Therefore, the range of |PM| is [b,a], that is, \\sqrt{5},3"
  },
  {
    "text": "If an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1(a>0)$ is perpendicular to the line $y=\\frac{3}{2} x$, then the length of the real axis of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Line;a>0;Expression(G) = (-y^2/4 + x^2/a^2 = 1);Expression(H) = (y = (3/2)*x);IsPerpendicular(OneOf(Asymptote(G)), H)",
    "query_expressions": "Length(RealAxis(G))",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 48], [79, 82]], [[4, 48]], [[55, 74]], [[4, 48]], [[1, 48]], [[55, 74]], [[1, 76]]]",
    "query_spans": "[[[79, 88]]]",
    "process": "\\because the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1 (a>0) has one asymptote perpendicular to the line y=\\frac{3}{2}x, \\therefore the asymptotes of the hyperbola are given by ay=\\pm2x. \\therefore -\\frac{2}{a}\\cdot\\frac{3}{2}=-1, yielding a=3, \\therefore 2a=6."
  },
  {
    "text": "If the two foci of an ellipse are $F_{1}(-4,0)$ and $F_{2}(4,0)$, and point $P$ lies on the ellipse such that the maximum area of triangle $P F_{1} F_{2}$ is $12$, then what is the equation of this ellipse?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Coordinate(F1) = (-4, 0);Coordinate(F2) = (4, 0);Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);Max(Area(TriangleOf(P, F1, F2))) = 12",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/25+y^2/9=1",
    "fact_spans": "[[[1, 3], [44, 46], [82, 84]], [[10, 23]], [[25, 38]], [[10, 23]], [[25, 38]], [[1, 38]], [[39, 43]], [[39, 47]], [[49, 79]]]",
    "query_spans": "[[[82, 88]]]",
    "process": "According to the problem, $ c = 4 $, $ 2c = 8 $, the foci of the ellipse lie on the x-axis, and the maximum area of triangle $ PF_{1}F_{2} $ is $ \\frac{1}{2} \\times 8 \\times b = 12 \\Rightarrow b = 3 $, so $ a = \\sqrt{b^{2} + c^{2}} = \\sqrt{9 + 16} = 5 $. Therefore, the equation of the ellipse is $ \\frac{x^{2}}{25} + \\frac{y^{2}}{9} = 1 $."
  },
  {
    "text": "Find the equation of the hyperbola with foci at the foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{8}=1$ and passing through the point $(2, \\frac{3 \\sqrt{5}}{2})$.",
    "fact_expressions": "G: Hyperbola;H: Ellipse;P: Point;Expression(H) = (x^2/9 + y^2/8 = 1);Coordinate(P) = (2, 3*sqrt(5)/2);PointOnCurve(P, G);Focus(G)=Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "4*x^2-4*y^2/3=1",
    "fact_spans": "[[[77, 80]], [[2, 39]], [[48, 76]], [[2, 39]], [[48, 76]], [[47, 80]], [[1, 80]]]",
    "query_spans": "[[[77, 84]]]",
    "process": "From the standard equation of the ellipse, it is known that the foci of the ellipse lie on the x-axis. Let the standard equation of the hyperbola be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$. According to the given conditions, we have  \n$\\begin{cases}a^{2}+b^{2}=1,\\\\\\frac{4}{a^{2}}-\\frac{45}{4b^{2}}=1,\\end{cases}$  \n$\\begin{cases}a^{2}=\\frac{1}{4},\\\\b^{2}=\\frac{3}{4}\\end{cases}$ or $\\begin{cases}a^{2}=16,\\\\b^{2}=-15\\end{cases}$ (discarded as it does not satisfy the conditions),  \n$\\therefore$ the equation of the hyperbola is $4x^{2}-\\frac{4y^{2}}{3}=1$."
  },
  {
    "text": "The right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $F$, and $B$ is a point on its left branch. The segment $BF$ intersects an asymptote of the hyperbola at $A$, and $(\\overrightarrow{O F}-\\overrightarrow{O B}) \\cdot \\overrightarrow{O A}=0$, $2 \\overrightarrow{O A}=\\overrightarrow{O B}+\\overrightarrow{O F}$, where $O$ is the coordinate origin. Then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;B: Point;F: Point;O: Origin;A: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;PointOnCurve(B,LeftPart(G));Intersection(LineSegmentOf(B,F),OneOf(Asymptote(G)))=A;DotProduct((-VectorOf(O, B) + VectorOf(O, F)),VectorOf(O, A)) = 0;2*VectorOf(O, A) = VectorOf(O, B) + VectorOf(O, F)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[0, 56], [69, 70], [84, 87], [258, 261]], [[3, 56]], [[3, 56]], [[65, 68]], [[61, 64]], [[247, 250]], [[96, 99]], [[3, 56]], [[3, 56]], [[0, 56]], [[0, 64]], [[65, 75]], [[76, 99]], [[102, 176]], [[177, 243]]]",
    "query_spans": "[[[258, 267]]]",
    "process": "Without loss of generality, assume point B lies in the second quadrant. From the given conditions, OA perpendicularly bisects segment BF. Let F(c,0), B(m,n), then \\frac{n-0}{m-c}=\\frac{a}{b}, and \\frac{n}{2}=\\frac{b}{a}\\frac{m+c}{2}, yielding m=\\frac{a^{2}-b^{2}}{c}, n=\\frac{2ab}{c}. Substituting into the hyperbola equation gives \\frac{(a^{2}-b^{2})^{2}}{c^{2}a^{2}}-\\frac{4a^{2}b^{2}}{c^{2}b^{2}}=1. Since b^{2}=c^{2}-a^{2}, simplifying and rearranging yields c^{2}=5a^{2}. Therefore, the eccentricity of the hyperbola is e=\\frac{c}{a}=\\sqrt{5}."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, with left and right foci $F_{1}$, $F_{2}$ respectively. The line passing through $F_{2}(1,0)$ with slope $1$ intersects the ellipse at points $A$, $B$. If the area of triangle $F_{1} A B$ equals $\\sqrt{2} b^{2}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;Coordinate(F2) = (1, 0);PointOnCurve(F2, G) = True;Slope(G) = 1;G: Line;Intersection(G, C) = {A, B};A: Point;B: Point;Area(TriangleOf(F1, A, B)) = sqrt(2)*b^2",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[2, 59], [157, 159], [107, 109]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[67, 74]], [[75, 82], [84, 96]], [[2, 82]], [[2, 82]], [[84, 96]], [[83, 106]], [[97, 106]], [[104, 106]], [[104, 117]], [[110, 113]], [[114, 117]], [[119, 154]]]",
    "query_spans": "[[[157, 165]]]",
    "process": "From the given condition, the equation of line AB is $x = y + 1$. Substituting into the ellipse equation yields: $(a^{2} + b^{2})y^{2} + 2b^{2}y + b^{2} - a^{2}b^{2} = 0$. Let points $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$, then $y_{1} + y_{2} = \\frac{-2b^{2}}{a^{2} + b^{2}}$, $y_{1}y_{2} = \\frac{b^{2} - a^{2}b^{2}}{a^{2} + b^{2}}$. Given $S_{AF_{1}AB} = \\frac{1}{2} \\times |F_{1}F_{2}| \\times |y_{1} - y_{2}| = \\sqrt{2}b^{2}$, and $a^{2} - b^{2} = 1$, solve for $a$, then find the eccentricity. Solution: From the problem, the equation of line AB is $x = y + 1$, substitute into $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1$ to eliminate $x$, obtaining: $(a^{2} + b^{2})y^{2} + 2b^{2}y + b^{2} - a^{2}b^{2} = 0$. Let points $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$, then $y_{1}$, $\\therefore |y_{1} - y_{2}| = \\sqrt{(y_{1} + y_{2})^{2} - 4y_{1}y_{2}} = \\sqrt{\\left(\\frac{-2b^{2}}{a^{2} + b^{2}}\\right)^{2} - 4 \\cdot \\frac{b^{2} - a^{2}b^{2}}{a^{2} + b^{2}}} = \\frac{2ab\\sqrt{a^{2} + b^{2} - 1}}{a^{2} + b^{2}}$. While $S_{AF_{1}AB} = \\frac{1}{2} \\times |F_{1}F_{2}| \\times |y_{1} - y_{2}| = \\frac{1}{2} \\cdot 2\\sqrt{a^{2} - b^{2}} \\cdot \\frac{2ab\\sqrt{a^{2} + b^{2} - 1}}{a^{2} + b^{2}} = \\frac{2ab\\sqrt{a^{2} + b^{2} - 1}\\sqrt{a^{2} - b^{2}}}{a^{2} + b^{2}} = \\sqrt{2}b^{2}$, and $a^{2} - b^{2} = 1$, solving gives: $a = \\frac{\\sqrt{3} + 1}{\\sqrt{2}}$, so the eccentricity $e = \\sqrt{1 - \\frac{b^{2}}{a^{2}}} = \\sqrt{3} - 1$."
  },
  {
    "text": "Given that the center of the ellipse is at the origin, one focus is $F(-2 \\sqrt{3} , 0)$, and the length of the major axis is twice the length of the minor axis, then the standard equation of the ellipse is ?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;F: Point;OneOf(Focus(G)) = F;Coordinate(F) = (-2*sqrt(3), 0);Length(MajorAxis(G)) = 2*Length(MinorAxis(G))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16+y^2/9=1",
    "fact_spans": "[[[2, 4], [52, 54]], [[7, 9]], [[2, 9]], [[15, 35]], [[2, 35]], [[15, 35]], [[2, 49]]]",
    "query_spans": "[[[52, 62]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse equation $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, points $F_{1}(2,0)$, $A(1,1)$, and $P$ is any point on the ellipse. Then the range of values for $|P A|+|P F_{1}|$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;A: Point;P: Point;Coordinate(F1) = (2, 0);Coordinate(A) = (1, 1);Expression(G) = (x^2/9 + y^2/5 = 1);PointOnCurve(P, G)",
    "query_expressions": "Range(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F1)))",
    "answer_expressions": "[6-sqrt(10),6+sqrt(10)]",
    "fact_spans": "[[[2, 4], [73, 75]], [[42, 55]], [[57, 66]], [[69, 72]], [[42, 55]], [[57, 66]], [[2, 41]], [[69, 80]]]",
    "query_spans": "[[[82, 106]]]",
    "process": ""
  },
  {
    "text": "If the distance from a point $(x_{0}, y_{0})$ on the parabola $x^{2}=24 y$ to the focus is 4 times the distance from that point to the $x$-axis, then $y_{0}$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 24*y);x0: Number;y0: Number;H: Point;Coordinate(H) = (x0, y0);PointOnCurve(H, G);Distance(H, Focus(G)) = 4*Distance(H, xAxis)",
    "query_expressions": "y0",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 16]], [[1, 16]], [[19, 35]], [[58, 65]], [[19, 35], [43, 44]], [[19, 35]], [[1, 35]], [[1, 56]]]",
    "query_spans": "[[[58, 67]]]",
    "process": "The equation of the directrix of the parabola is $ y = -6 $. The distance from a point to the focus is equal to the distance from the point to the directrix, so $ y_{0} + 6 = 4y_{0} $. Solving gives $ y_{0} = 2 $. Hence, fill in $ 2 $."
  },
  {
    "text": "The focus of the parabola $y^{2}=4 x$ is $F$. A line passing through point $F$ intersects the parabola at points $A$ and $B$, and $A F=2 B F$. What are the coordinates of point $A$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};LineSegmentOf(A, F) = 2*LineSegmentOf(B, F)",
    "query_expressions": "Coordinate(A)",
    "answer_expressions": "(5,2*sqrt(2)), (5,-2*sqrt(2))",
    "fact_spans": "[[[0, 14], [31, 34]], [[28, 30]], [[35, 38], [59, 63]], [[18, 21], [23, 27]], [[39, 42]], [[0, 14]], [[0, 21]], [[22, 30]], [[28, 44]], [[46, 57]]]",
    "query_spans": "[[[59, 68]]]",
    "process": ""
  },
  {
    "text": "The asymptotes of the hyperbola $C$: $x^{2}-y^{2}=1$ are given by? If the right vertex of hyperbola $C$ is $A$, and a line $l$ passing through $A$ intersects the two asymptotes of hyperbola $C$ at points $P$ and $Q$, such that $\\overrightarrow{P A}=2 \\overrightarrow{A Q}$, then the slope of line $l$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;P: Point;A: Point;Q: Point;Expression(C) = (x^2 - y^2 = 1);RightVertex(C)=A;PointOnCurve(A, l);Intersection(l,J2)=Q;VectorOf(P, A) = 2*VectorOf(A, Q);J1:Line;J2:Line;Asymptote(C)={J1,J2}",
    "query_expressions": "Expression(Asymptote(C));Slope(l)",
    "answer_expressions": "x+pm*y=0;pm*3",
    "fact_spans": "[[[130, 135], [52, 57]], [[0, 23], [32, 38], [58, 64]], [[72, 75]], [[43, 46], [48, 51]], [[76, 79]], [[0, 23]], [[32, 46]], [[47, 57]], [[52, 81]], [[83, 128]], [-1, -1], [-1, -1], [58, 69]]",
    "query_spans": "[[[0, 31]], [[130, 140]]]",
    "process": "This problem mainly examines the properties of hyperbolas and their applications. [Analysis] Write the equations of line $ l $ and the asymptotes of the hyperbola, then find the coordinates of points $ P $ and $ Q $, and use $ \\overrightarrow{PA} = 2\\overrightarrow{AQ} $ to find the corresponding value of $ k $. Replacing $ 1 $ in the equation $ x^2 - y^2 = 1 $ with $ 0 $ gives $ x^2 - y^2 = 0 $, which simplifies to the asymptote equations $ x \\pm y = 0 $. From $ x^2 - y^2 = 1 $, we get point $ A(1, 0) $; let the equation of line $ l $ be $ y = k(x - 1) $.  \n① When $ k > 0 $, solving  \n$$\n\\begin{cases}\ny = k(x - 1) \\\\\ny = x\n\\end{cases}\n$$\ngives $ x = \\frac{k}{k - 1} $, $ y = \\frac{k}{k - 1} $, so $ P\\left( \\frac{k}{k - 1}, \\frac{k}{k - 1} \\right) $, then $ \\overrightarrow{PA} = \\left( 1 - \\frac{k}{k - 1}, -\\frac{k}{k - 1} \\right) $. Similarly, we obtain $ Q\\left( \\frac{k}{k + 1}, -\\frac{k}{k + 1} \\right) $, then $ \\overrightarrow{AQ} = \\left( \\frac{k}{k + 1} - 1, -\\frac{k}{k + 1} \\right) $. Solving $ 1 - \\frac{k}{k - 1} = 2\\left( \\frac{k}{k + 1} - 1 \\right) $ yields $ k = 3 $.  \n② When $ k < 0 $, solving  \n$$\n\\begin{cases}\ny = k(x - 1) \\\\\ny = -x\n\\end{cases}\n$$\ngives $ x = \\frac{k}{k - 1} $, $ y = \\frac{k}{k - 1} $, so $ Q\\left( \\frac{k}{k - 1}, \\frac{k}{k - 1} \\right) $, then $ \\overrightarrow{AQ} = \\left( \\frac{k}{k - 1} - 1, \\frac{k}{k - 1} \\right) $. Similarly, we obtain $ P\\left( \\frac{k}{k + 1}, -\\frac{k}{k + 1} \\right) $, then $ \\overrightarrow{PA} = \\left( 1 - \\frac{k}{k + 1}, \\frac{k}{k + 1} \\right) $. From $ \\overrightarrow{PA} = 2\\overrightarrow{AQ} $, we have $ 1 - \\frac{k}{k + 1} = 2\\left( \\frac{k}{k - 1} - 1 \\right) $, solving gives $ k = -3 $.  \nFrom ① and ②, the slope of the line is $ k = \\pm 3 $.  \nComment: Considering the symmetry of the hyperbola, it is necessary to discuss the parameter $ k $ in two cases: $ k > 0 $ and $ k < 0 $. Transforming the vector relation $ \\overrightarrow{PA} = 2\\overrightarrow{AQ} $ into coordinate relations is also one of the keys to solving this problem."
  },
  {
    "text": "Given a point $M$ on the parabola $y^{2}=2 p x(p>0)$ such that the distance from $M$ to the focus $F$ is equal to $2 p$, what is the slope of the line $M F$?",
    "fact_expressions": "G: Parabola;p: Number;M: Point;F: Point;p>0;Expression(G) = (y^2 = 2*p*x);PointOnCurve(M, G);Focus(G) = F;Distance(M, F) = 2*p",
    "query_expressions": "Slope(LineOf(M,F))",
    "answer_expressions": "{sqrt(3),-sqrt(3)}",
    "fact_spans": "[[[2, 23]], [[40, 45]], [[26, 29]], [[32, 35]], [[5, 23]], [[2, 23]], [[2, 29]], [[2, 35]], [[26, 45]]]",
    "query_spans": "[[[48, 60]]]",
    "process": "Since the distance |MF| = 2p from a point M on the parabola y^{2} = 2px (p > 0) to the focus F is given, we have x_{M} + \\frac{p}{2} = 2p, so x_{M} = \\frac{3}{2}p, and thus y_{M} = \\pm\\sqrt{3}p. Therefore, the coordinates of point M are: (\\frac{3}{2}p, \\pm\\sqrt{3}p). When the coordinates of point M are (\\frac{3}{2}p, \\sqrt{3}p), the slope of line MF is \\frac{\\sqrt{3}p - 0}{\\frac{3}{2}p - \\frac{p}{2}} = \\sqrt{3}. When the coordinates of point M are (\\frac{3}{2}p, -\\sqrt{3}p), the slope of line MF is \\frac{-\\sqrt{3}p - 0}{\\frac{3}{2}p - \\frac{p}{2}} = -\\sqrt{3}. In conclusion, the slope of line MF is \\sqrt{3} or -\\sqrt{3}."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, with two foci $F_{1}$ and $F_{2}$, and a point $P$ on the hyperbola $C$. If $|P F_{1}|=10$, then $|P F_{2}|$=?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/16 - y^2/9 = 1);F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 10",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "{18, 2}",
    "fact_spans": "[[[2, 46], [73, 79]], [[2, 46]], [[53, 60]], [[61, 68]], [[2, 68]], [[69, 72]], [[69, 82]], [[84, 98]]]",
    "query_spans": "[[[100, 113]]]",
    "process": "From $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, we get $a^{2}=16$, $b^{2}=9$, then $a=4$, $b=3$, $c=\\sqrt{a^{2}+b^{2}}=5$. Since the two foci of hyperbola $C: \\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ are $F_{1}$ and $F_{2}$, and $P$ is a point on hyperbola $C$ with $|PF_{1}|=10$, so $||PF_{2}|-|PF_{1}||=2a=8$, that is, $||PF_{2}|-10|=8$, thus $|PF_{2}|=18$ or $|PF_{2}|=2$. Since $|PF_{1}|=10>a+c=9$, both $|PF_{2}|=18$ and $|PF_{2}|=2$ satisfy the conditions."
  },
  {
    "text": "The coordinates of the foci of the ellipse $m x^{2}+n y^{2}+m n=0(m<n<0)$ are?",
    "fact_expressions": "G: Ellipse;m: Number;n: Number;m < n;n < 0;Expression(G) = (m*n + m*x^2 + n*y^2 = 0)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,pm*sqrt(n-m))",
    "fact_spans": "[[[0, 32]], [[2, 32]], [[2, 32]], [[2, 32]], [[2, 32]], [[0, 32]]]",
    "query_spans": "[[[0, 39]]]",
    "process": "The equation of the ellipse $ mx^{2} + ny^{2} + mn = 0 $ ($ m < n < 0 $) is converted into standard form as: $ \\frac{x^{2}}{n} + \\frac{y^{2}}{m} = 1 $. Since $ m < n < 0 $, we have $ -m > -n > 0 $, therefore the foci of the ellipse lie on the $ y $-axis. Thus, $ c^{2} = -m + n $. Since the foci of the ellipse are on the $ y $-axis, the coordinates of the foci are: $ (0, -\\sqrt{n - m}) $, $ (0, \\sqrt{n - m}) $."
  },
  {
    "text": "Given that point $P$ is any point on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ except the vertices, $F_{1}$ and $F_{2}$ are the left and right foci respectively, $c$ is the semi-focal length, and the incircle of $\\Delta P F_{1} F_{2}$ touches $F_{1} F_{2}$ at point $M$, then $|F_{1} M| \\cdot|F_{2} M|$=?",
    "fact_expressions": "P: Point;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;PointOnCurve(P, G);Negation(Vertex(G) = P);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;c: Number;HalfFocalLength(G) = c;TangentPoint(InscribedCircle(TriangleOf(P, F1, F2)), LineSegmentOf(F1, F2)) = M;M: Point",
    "query_expressions": "Abs(LineSegmentOf(F1, M))*Abs(LineSegmentOf(F2, M))",
    "answer_expressions": "2/3",
    "fact_spans": "[[[2, 6]], [[7, 53]], [[7, 53]], [[10, 53]], [[10, 53]], [[2, 63]], [[2, 63]], [[64, 71]], [[72, 79]], [[7, 87]], [[7, 87]], [[88, 91]], [[7, 95]], [[96, 142]], [[138, 142]]]",
    "query_spans": "[[[144, 172]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, the line $MN$ passes through the focus $F$ and intersects the parabola $C$ at points $M$ and $N$. Point $P$ lies on the directrix $l$ of the parabola $C$ such that $PF \\perp MN$. Connect $PM$ intersecting the $y$-axis at point $Q$. Draw $QD \\perp MF$ meeting $MF$ at point $D$. If $|MD|=2|FN|$, then $|MF|=$?",
    "fact_expressions": "C: Parabola;M: Point;N: Point;P: Point;F: Point;Q: Point;D: Point;l:Line;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F, LineOf(M,N));Intersection(LineOf(M,N),C) = {M,N};PointOnCurve(P,l);Directrix(C)=l;IsPerpendicular(LineSegmentOf(P,F),LineSegmentOf(M,N));Intersection(LineSegmentOf(P,M),yAxis)=Q;PointOnCurve(Q,LineSegmentOf(Q,D));IsPerpendicular(LineSegmentOf(Q,D),LineSegmentOf(M,F));Abs(LineSegmentOf(M, D)) = 2*Abs(LineSegmentOf(F, N));FootPoint(LineSegmentOf(Q,D),LineSegmentOf(M,F))=D",
    "query_expressions": "Abs(LineSegmentOf(M, F))",
    "answer_expressions": "sqrt(3)+2",
    "fact_spans": "[[[2, 21], [43, 49], [65, 71]], [[51, 54]], [[55, 58]], [[61, 64]], [[24, 27], [38, 41]], [[109, 113], [115, 118]], [[135, 139]], [[73, 76]], [[2, 21]], [[2, 27]], [[28, 41]], [[28, 60]], [[61, 79]], [[65, 76]], [[80, 95]], [[98, 113]], [[114, 134]], [[119, 134]], [[141, 155]], [[119, 139]]]",
    "query_spans": "[[[157, 166]]]",
    "process": "Let $ M(x_{1},y_{1}) $, $ N(x_{2},y_{2}) $, and the equation of line $ MN $ be $ y = k(x - 1) $. Combining  \n$$\n\\begin{cases}\ny = k(x - 1) \\\\\ny^2 = 4x\n\\end{cases}\n$$  \ngives $ k^{2}x^{2} - (2k^{2} + 4)x + k^{2} = 0 $, so $ x_{1} + x_{2} = 2 + \\frac{4}{k^{2}} $\\textcircled{1}, $ x_{1}x_{2} = 1 $\\textcircled{2}. Since $ |MD| = 2|FN| $, we have $ |MD| = 2(x_{2} + 1) $. From the problem, $ QD \\parallel PF $, so $ \\frac{|MD|}{|MF|} = \\frac{|MQ|}{|MP|} = \\frac{|MB|}{|MA|} $, then $ \\frac{2(x_{2} + 1)}{x_{1} + 1} = \\frac{x_{1}}{x_{1} + 1} $. Simplifying gives $ x_{2} = \\frac{1}{2}x_{1} - 1 $\\textcircled{3}. Combining \\textcircled{1}\\textcircled{2}\\textcircled{3}, solving yields $ k^{2} = \\frac{4\\sqrt{3} + 4}{3} $, $ x_{1} = \\sqrt{3} + 1 $, so $ |MF| = x_{1} + 1 = \\sqrt{3} + 2 $."
  },
  {
    "text": "Given that the upper focus of the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1$ is $F$, and the lines $x+y+1=0$ and $x+y-1=0$ intersect the ellipse at points $A$, $B$, $C$, $D$, then $A F+B F+C F+D F$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/3 + y^2/4 = 1);F: Point;UpperFocus(G) = F;Z1: Line;Z2: Line;Expression(Z1) = (x + y + 1 = 0);Expression(Z2) = (x + y - 1 = 0);A: Point;B: Point;C: Point;D: Point;Intersection(Z1, G) = {A, B};Intersection(Z2, G) = {C, D}",
    "query_expressions": "LineSegmentOf(A, F) + LineSegmentOf(B, F) + LineSegmentOf(C, F) + LineSegmentOf(D, F)",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 39], [70, 72]], [[2, 39]], [[44, 47]], [[2, 47]], [[48, 59]], [[60, 69]], [[48, 59]], [[60, 69]], [[75, 79]], [[81, 84]], [[86, 89]], [[91, 94]], [[48, 94]], [[48, 94]]]",
    "query_spans": "[[[97, 116]]]",
    "process": "Substitute the line $x + y + 1 = 0$ into the ellipse $\\frac{x^{2}}{3} + \\frac{y^{2}}{4} = 1$, and simplify to obtain $7x^{2} + 6x - 9 = 0$. Let $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$, then $x_{1} + x_{2} = -\\frac{6}{7}$, $x_{1}x_{2} = -\\frac{9}{7}$, $\\therefore AB = \\sqrt{(1+1)(-\\frac{6}{7})^{2} - 4 \\times (-\\frac{99}{7})} = \\frac{24}{7}$. Similarly, we get $CD = CF + DF = \\frac{24}{7}$. $\\because AF + BF + AB = 4a = 8$, $\\therefore AF + BF = 8 - AB = 8 - \\frac{24}{7}$, $\\therefore AF + BF + CF + DF = (8 - \\frac{24}{4}) + \\frac{24}{7} = 8$."
  },
  {
    "text": "If the focus of the parabola $x^{2}=2 p y(p>0)$ lies on the circle $x^{2}+y^{2}+2 x-1=0$, then the equation of the directrix of this parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 2*p*y);p: Number;p > 0;H: Circle;Expression(H) = (2*x + x^2 + y^2 - 1 = 0);PointOnCurve(Focus(G), H) = True",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=-1",
    "fact_spans": "[[[1, 22], [53, 56]], [[1, 22]], [[4, 22]], [[4, 22]], [[26, 48]], [[26, 48]], [[1, 49]]]",
    "query_spans": "[[[53, 63]]]",
    "process": "Find the coordinates of the intersection point of the circle $x^{2}+y^{2}+2x-1=0$ with the positive $y$-axis, which gives the focus coordinates of the parabola, then the answer can be obtained. From $x^{2}+y^{2}+2x\\cdot1=0$, set $x=0$, we get $y^{2}=1$, so $y=\\pm1$. Since the focus of the parabola $x^{2}=2py$ ($p>0$) lies on the circle $x^{2}+y^{2}+2x\\cdot1=0$, the focus coordinates of the parabola $x^{2}=2py$ ($p>0$) are $(0,1)$, thus $\\frac{p}{2}=1$. Therefore, the directrix equation of the parabola $x^{2}=2py$ ($p>0$) is $y=-\\frac{p}{2}=-1$."
  },
  {
    "text": "The minimum value of the sum of distances from a moving point $M$ on the parabola $y=-\\frac{1}{4} x^{2}$ to two fixed points $F(0, -1)$ and $E(1, -3)$ is?",
    "fact_expressions": "G: Parabola;F: Point;E: Point;M: Point;PointOnCurve(M, G);Expression(G) = (y = -x^2/4);Coordinate(F) = (0, -1);Coordinate(E) = (1, -3)",
    "query_expressions": "Min(Distance(M, F)+Distance(M, E))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 25]], [[36, 46]], [[47, 57]], [[29, 32]], [[0, 32]], [[0, 25]], [[36, 46]], [[47, 57]]]",
    "query_spans": "[[[29, 68]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with left and right foci $F_{1}$, $F_{2}$, and a point $P$ on an asymptote of the hyperbola $C$ such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$. If the line $P F_{1}$ is tangent to the circle $x^{2}+y^{2}=a^{2}$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;F1: Point;P: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = a^2);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, Asymptote(C));DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;IsTangent(LineOf(P,F1),G)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 63], [90, 96], [201, 207]], [[10, 63]], [[10, 63]], [[177, 197]], [[69, 76]], [[85, 89]], [[77, 84]], [[10, 63]], [[10, 63]], [[2, 63]], [[177, 197]], [[2, 84]], [[2, 84]], [[85, 103]], [[104, 163]], [[165, 199]]]",
    "query_spans": "[[[201, 213]]]",
    "process": "Draw the graph, let $ PF_{1} $ be tangent to the circle $ x^{2}+y^{2}=a^{2} $ at point $ E $, analyze that $ \\angle POF_{2}=\\frac{\\pi}{3} $, then the value of $ \\frac{b}{a} $ can be found, and further the eccentricity of hyperbola $ C $ is $ e=\\sqrt{1+(\\frac{b}{a})^{2}} $, thus the solution is obtained. As shown in the figure below, let $ PF_{1} $ be tangent to the circle $ x^{2}+y^{2}=a^{2} $ at point $ E $, then $ |OE|=a $, $ OE\\bot PF_{1} $, so $ OE/\\!/PF_{2} $. Since $ O $ is the midpoint of $ F_{1}F_{2} $, then $ E $ is the midpoint of $ PF_{1} $, $ \\therefore |PF_{2}|=2|OE|=2a $. From the property of right triangle, we get $ |OF_{1}|=|OP| $. Since $ E $ is the midpoint of $ PF_{1} $, then $ \\angle EOF_{1}=\\angle POE $. Because the two asymptotes of the hyperbola are symmetric about the $ y $-axis, we have $ \\angle POF_{2}=\\angle EOF_{1} $. Therefore, $ \\angle EOF_{1}=\\angle POE=\\angle POF_{2} $, then $ \\angle EOF_{1}+\\angle POE+\\angle POF_{2}=3\\angle POF_{2}=\\pi $. So, $ \\angle POF_{2}=\\frac{\\pi}{3} $, then $ \\frac{b}{a}=\\tan\\frac{\\pi}{3}=\\sqrt{3} $. Therefore, the eccentricity of hyperbola $ C $ is $ e=\\frac{c}{a}=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{\\frac{a^{2}+b^{2}}{a^{2}}}=\\sqrt{1+(\\frac{b}{a})^{2}}=2 $."
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x$ coincides with the left focus of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;H: Ellipse;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (x^2/9 + y^2/5 = 1);Focus(G) = LeftFocus(H)",
    "query_expressions": "p",
    "answer_expressions": "-4",
    "fact_spans": "[[[1, 17]], [[66, 69]], [[21, 58]], [[1, 17]], [[21, 58]], [[1, 64]]]",
    "query_spans": "[[[66, 73]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2x$ with focus $F$ and directrix $l$, let point $A \\in l$, and segment $AF$ intersects the parabola $C$ at point $B$. If $\\overrightarrow{FA}=3\\overrightarrow{FB}$, then $|\\overrightarrow{AF}|$=?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*x);F: Point;Focus(C) = F;l: Line;Directrix(C) = l;A: Point;In(A, l);Intersection(LineSegmentOf(A,F), C) = B;B: Point;VectorOf(F, A) = 3*VectorOf(F, B)",
    "query_expressions": "Abs(VectorOf(A, F))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 21], [55, 61]], [[2, 21]], [[25, 28]], [[2, 28]], [[32, 35]], [[2, 35]], [[36, 46]], [[36, 46]], [[47, 66]], [[62, 66]], [[68, 113]]]",
    "query_spans": "[[[115, 141]]]",
    "process": "From the given condition, $x_{B}=\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}$, $\\therefore y_{B}^{2}=2\\times\\frac{1}{6}=\\frac{1}{3}$"
  },
  {
    "text": "The equation of the directrix of the hyperbola $\\frac{y^{2}}{16}-\\frac{x^{2}}{9}=1$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/9 + y^2/16 = 1)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=pm*16/5",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 46]]]",
    "process": ""
  },
  {
    "text": "It is known that the focus of the parabola $y = m x^{2}$ ($m > 0$) coincides with one focus of the ellipse $\\frac{4 y^{2}}{9} + \\frac{x^{2}}{2} = 1$, then $m$ = ?",
    "fact_expressions": "G: Parabola;m: Number;H: Ellipse;m>0;Expression(G) = (y = m*x^2);Expression(H) = (x^2/2 + (4*y^2)/9 = 1);Focus(G) = OneOf(Focus(H))",
    "query_expressions": "m",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 21]], [[73, 76]], [[25, 64]], [[5, 21]], [[2, 21]], [[25, 64]], [[2, 71]]]",
    "query_spans": "[[[73, 78]]]",
    "process": "The equation of the parabola $ y = mx^{2} $ ($ m > 0 $) can be rewritten in standard form as $ x^{2} = \\frac{1}{m}y $, so its focus is $ \\left(0, \\frac{1}{4m}\\right) $. Since the focus of the parabola $ y = mx^{2} $ ($ m > 0 $) coincides with one focus of the ellipse $ \\frac{4y^{2}}{9} + \\frac{x^{2}}{2} = 1 $, it follows that $ \\frac{9}{4} - 2 = \\left(\\frac{1}{4m}\\right)^{2} $. Solving gives $ m = \\frac{1}{2} $. The answer to be filled in is $ \\frac{1}{2} $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus is $F$, and the lower vertex is $A$. If a line parallel to $AF$ with a $y$-axis intercept of $3-\\sqrt{2}$ is tangent to the circle $x^{2}+(y-3)^{2}=1$, then the eccentricity of this ellipse is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Circle;A: Point;F: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + (y - 3)^2 = 1);LeftFocus(C) = F;LowerVertex(C) = A;L:Line;IsParallel(LineSegmentOf(A,F),L);Intercept(L,yAxis)=(3-sqrt(2));IsTangent(L,G)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 59], [137, 139]], [[9, 59]], [[9, 59]], [[112, 132]], [[72, 75]], [[64, 67]], [[9, 59]], [[9, 59]], [[2, 59]], [[112, 132]], [[2, 67]], [[2, 75]], [[109, 111]], [[78, 111]], [[87, 111]], [[109, 134]]]",
    "query_spans": "[[[137, 145]]]",
    "process": "Let $ l: y = kx + 3 - \\sqrt{2} $ $ \\left( k = \\frac{-b}{c} < 0 \\right) $, $ \\therefore \\frac{\\sqrt{2}}{\\sqrt{1 + k^{2}}} = 1 $, $ \\therefore k = -1 $, $ \\frac{b}{c} = 1 $, $ e = \\frac{\\sqrt{2}}{2} $"
  },
  {
    "text": "Given the equation of the parabola $x=\\frac{1}{36} y^{2}$, then the equation of the directrix of this parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x = y^2/36)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=-9",
    "fact_spans": "[[[2, 5], [33, 36]], [[2, 30]]]",
    "query_spans": "[[[33, 43]]]",
    "process": "x=\\frac{1}{36}y^{2}, the focus is on the x-axis, and \\frac{p}{2}=9, \\therefore the directrix equation of the parabola is x=-9"
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $4 x^{2}-9 y^{2}=-36$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (4*x^2 - 9*y^2 = -36)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(2/3)*x",
    "fact_spans": "[[[0, 24]], [[0, 24]]]",
    "query_spans": "[[[0, 32]]]",
    "process": "Convert $4x^{2}-9y^{2}=-36$ into the standard form $\\frac{y^{2}}{4}-\\frac{x^{2}}{9}=1$, we get $a^{2}=4$, $b^{2}=9$, so the asymptotes are: $y=\\pm\\frac{a}{b}x=\\pm\\frac{2}{3}x$"
  },
  {
    "text": "The hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has a semi-focal distance of $c$. If the hyperbola $E$ and the circle $(x-c)^{2}+y^{2}=9 a^{2}$ have exactly three common points, then the eccentricity of $E$ is?",
    "fact_expressions": "E: Hyperbola;b: Number;a: Number;G: Circle;c: Number;a>0;b>0;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y^2 + (-c + x)^2 = 9*a^2);HalfFocalLength(E) = c;NumIntersection(E, G)=3",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 61], [71, 77], [113, 116]], [[8, 61]], [[8, 61]], [[78, 104]], [[66, 69]], [[8, 61]], [[8, 61]], [[0, 61]], [[78, 104]], [[0, 69]], [[71, 111]]]",
    "query_spans": "[[[113, 122]]]",
    "process": "\\because the hyperbola E and the circle: (x-c)^{2}+y^{2}=9a^{2} have exactly three common points, \\therefore the circle: (x-c)^{2}+y^{2}=9a^{2} passes exactly through the left vertex of the hyperbola, \\therefore 3a=a+c, \\therefore e=\\frac{c}{a}=2,"
  },
  {
    "text": "Given that $P$ is a moving point on the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, $F_{1}$ and $F_{2}$ are the left and right foci of ellipse $C$, $O$ is the coordinate origin, and $d$ is the distance from $O$ to the tangent line of ellipse $C$ at point $P$. If $|P F_{1}| \\cdot|P F_{2}|=\\frac{24}{7}$, then $d=$?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;O: Origin;d:Number;Expression(C)=(x^2/4+y^2/3=1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P,C);Distance(O,TangentOnPoint(P,C))=d;Abs(LineSegmentOf(P,F1))*Abs(LineSegmentOf(P,F2))=24/7",
    "query_expressions": "d",
    "answer_expressions": "sqrt(14)/2",
    "fact_spans": "[[[6, 48], [70, 75], [95, 100]], [[2, 5], [101, 105]], [[54, 61]], [[62, 69]], [[82, 85], [91, 94]], [[158, 161], [112, 115]], [[6, 48]], [[54, 81]], [[54, 81]], [[2, 53]], [[91, 115]], [[117, 156]]]",
    "query_spans": "[[[158, 163]]]",
    "process": "Let $|PF_{1}|=m$, $|PF_{2}|=n$, then $\\begin{cases}m+n=4\\\\mn=\\frac{24}{7}\\end{cases}$. Without loss of generality, assume $P$ is in the first quadrant, then $|PF_{1}|=2+\\frac{2}{\\sqrt{7}}$, $|PF_{2}|=2-\\frac{2}{\\sqrt{7}}$. Hence, the circle centered at $F_{1}$ with radius $PF_{1}$ is: $(x+1)^{2}+y^{2}=(2+\\frac{2}{\\sqrt{7}})^{2}$, $\\textcircled{1}$; the circle centered at $F_{2}$ with radius $PF_{2}$ is: $(x-1)^{2}+y^{2}=(2-\\frac{2}{\\sqrt{7}})^{2}$, $\\textcircled{2}$. Subtracting $\\textcircled{1}-\\textcircled{2}$ gives: $x=\\frac{4}{\\sqrt{7}}$, substituting into the ellipse equation yields: $y=\\frac{3}{\\sqrt{7}}$, hence $P(\\frac{4}{\\sqrt{7}},\\frac{3}{\\sqrt{7}})$. When $y>0$, from $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ we get $y=(3-\\frac{3}{4}x^{2})^{\\frac{1}{2}}$, thus $y'=\\frac{1}{2}(3-\\frac{3}{4}x^{2})^{-\\frac{1}{2}}\\cdot(-\\frac{3}{2}x)$. Therefore, the slope of the tangent line to the ellipse at $P$ is $k=\\frac{1}{2}(3-\\frac{3}{4}\\cdot\\frac{16}{7})^{\\frac{1}{2}}\\cdot(-\\frac{3}{2}\\cdot\\frac{4}{\\sqrt{7}})=-1$. Thus, the tangent line equation is: $y-\\frac{3}{\\sqrt{7}}=-(x-\\frac{4}{\\sqrt{7}})$, i.e., $x+y-\\sqrt{7}=0$. Therefore, the distance from the origin $O$ to the tangent line is $d=\\frac{\\sqrt{7}}{\\sqrt{2}}=\\frac{\\sqrt{14}}{2}$."
  },
  {
    "text": "What is the length of the major axis of the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/3 + y^2/4 = 1)",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 43]]]",
    "process": "Since the foci of the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1$ lie on the vertical axis, $a^{2}=4\\Rightarrow a=2$, therefore the major axis length of the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1$ is $2a=4$."
  },
  {
    "text": "The distance from the vertex of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ to its asymptote is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1)",
    "query_expressions": "Distance(Vertex(G), Asymptote(G))",
    "answer_expressions": "2*sqrt(5)/5",
    "fact_spans": "[[[0, 28], [32, 33]], [[0, 28]]]",
    "query_spans": "[[[0, 42]]]",
    "process": "Without loss of generality, let the vertex be (2,0), and one asymptote be y=\\frac{1}{2}x, that is, x-2y=0. The distance from the point to the line is d=\\frac{2}{\\sqrt{5}}=\\frac{2\\sqrt{5}}{5}."
  },
  {
    "text": "The point on the parabola $x^{2}=\\frac{1}{4} y$ that is closest to the line $y=4 x-5$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = y/4);H: Line;Expression(H) = (y = 4*x - 5);P: Point;PointOnCurve(P, G);WhenMin(Distance(P, H))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1/2, 1)",
    "fact_spans": "[[[0, 24]], [[0, 24]], [[26, 37]], [[26, 37]], [[43, 44]], [[0, 44]], [[25, 44]]]",
    "query_spans": "[[[43, 46]]]",
    "process": ""
  },
  {
    "text": "Given the circle $(x+1)^{2}+y^{2}=4$ and the parabola $y^{2}=m x(m \\neq 0)$, their directrix intersects at points $A$ and $B$, and $|A B|=2 \\sqrt{3}$. Then the value of $m$ is?",
    "fact_expressions": "H: Circle;Expression(H) = (y^2 + (x + 1)^2 = 4);G: Parabola;Expression(G) = (y^2 = m*x);m: Number;Negation(m = 0);Intersection(H, Directrix(G)) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(A, B)) = 2*sqrt(3)",
    "query_expressions": "m",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 22]], [[2, 22]], [[23, 47]], [[23, 47]], [[83, 86]], [[26, 47]], [[2, 61]], [[52, 55]], [[56, 59]], [[63, 81]]]",
    "query_spans": "[[[83, 90]]]",
    "process": "Since the directrix of the parabola $ y^{2} = mx $ ($ m \\neq 0 $) is $ x = -\\frac{m}{4} $, the distance from the center $ (-1, 0) $ to the line $ x = -\\frac{m}{4} $ is $ |-1 + \\frac{m}{4}| $. Also, $ |AB| = 2\\sqrt{3} $, so $ 4 = \\left(|-1 + \\frac{m}{4}|\\right)^{2} + \\left(\\frac{|AB|}{2}\\right)^{2} $. Since $ m \\neq 0 $, it follows that $ m = 8 $."
  },
  {
    "text": "Point $A(x_{0} , y_{0})$ lies on the right branch of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{32}=1$. If the distance from point $A$ to the left focus is equal to $4 x_{0}$, then $x_{0}=$?",
    "fact_expressions": "A: Point;x0: Number;y0: Number;Coordinate(A) = (x0, y0);PointOnCurve(A, RightPart(G));G: Hyperbola;Expression(G) = (x^2/4 - y^2/32 = 1);Distance(A, LeftFocus(G)) = 4*x0",
    "query_expressions": "x0",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 19], [65, 69]], [[89, 96]], [[1, 19]], [[0, 19]], [[0, 63]], [[20, 59]], [[20, 59]], [[20, 87]]]",
    "query_spans": "[[[89, 98]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{6}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through $F_{2}$ intersects the right branch of the hyperbola at points $A$ and $B$. Then, the minimum value of $|A F_{1}|+|B F_{1}|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/6 = 1);A: Point;F1: Point;B: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;l:Line;PointOnCurve(F2, l);Intersection(l, RightPart(G)) = {A, B}",
    "query_expressions": "Min(Abs(LineSegmentOf(A, F1)) + Abs(LineSegmentOf(B, F1)))",
    "answer_expressions": "16",
    "fact_spans": "[[[2, 40], [80, 83]], [[2, 40]], [[87, 90]], [[48, 55]], [[91, 94]], [[56, 63], [66, 73]], [[2, 63]], [[2, 63]], [[74, 79]], [[65, 79]], [[74, 96]]]",
    "query_spans": "[[[98, 125]]]",
    "process": "From the standard equation of the hyperbola, we obtain: $ a=3 $. Then, by the definition of the hyperbola, we have: $ |AF_{1}|-|AF_{2}|=2a=6 $, $ |BF_{1}|-|BF_{2}|=2a=6 $. Thus, $ |AF_{1}|+|BF_{1}|-(|AF_{2}|+|BF_{2}|)=12 $. Based on the positional characteristics of points $ A $ and $ B $, combined with the minimum latus rectum, we obtain the answer. From the hyperbola $ \\frac{x^{2}}{9}-\\frac{y^{2}}{6}=1 $, we get: $ a=3 $, $ b=\\sqrt{6} $. By the definition of the hyperbola, we have: $ |AF_{1}|-|AF_{2}|=2a=6 $ $\\cdots\\textcircled{1}$, $ |BF_{1}|-|BF_{2}|=2a=6 $ $\\cdots\\textcircled{2} $. Therefore, adding $\\textcircled{1}$ and $\\textcircled{2}$ gives: $ |AF_{1}|+|BF_{1}|-(|AF_{2}|+|BF_{2}|)=12 $. Since the line $ l $ passing through $ F_{2} $ intersects the right branch of the hyperbola at points $ A $ and $ B $, we have $ |AF_{2}|+|BF_{2}|=|AB| $. When $ |AB| $ is the latus rectum of the hyperbola, $ |AB| $ is minimized. Hence, $ [|AF_{1}|+|BF_{1}|-(|AF_{2}|+|BF_{2}|)]_{\\min}=|AF_{1}|+|BF_{1}|-|AB|=12 $. So $ |AF_{1}|+|BF_{1}|=|AB|+12 \\geqslant \\frac{2b^{2}}{a}+12 = \\frac{2\\times6}{3}+12=16 $."
  },
  {
    "text": "What is the length of the shortest chord passing through the focus $F$ of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/12 = 1);H:LineSegment;PointOnCurve(F,H);Focus(G)=F;IsChordOf(H,G);WhenMin(Length(H));F:Point",
    "query_expressions": "Length(H)",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 40]], [[1, 40]], [], [[0, 48]], [[1, 46]], [[0, 48]], [[0, 53]], [[43, 46]]]",
    "query_spans": "[[[0, 56]]]",
    "process": "Let $ F(2,0) $. Among all chords passing through the focus $ F $, the chord perpendicular to the $ x $-axis is the shortest. When $ x = 2 $, $ y = \\pm 3 $, so the shortest chord length is $ 6 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$, the right focus is $F$, points $A$ and $B$ lie on the two asymptotes of $C$ respectively, $A F \\perp x$-axis, $A B \\perp O B$, $B F / / O A$ ($O$ is the origin), if the focal length of $C$ is $4$, then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;A: Point;F: Point;B:Point;O:Origin;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;L1:Line;L2:Line;Asymptote(C)={L1,L2};PointOnCurve(A,L1);PointOnCurve(B,L2);IsPerpendicular(LineSegmentOf(A,F),xAxis);IsPerpendicular(LineSegmentOf(A,B),LineSegmentOf(O,B));IsParallel(LineSegmentOf(B,F),LineSegmentOf(O,A));FocalLength(C)=4",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/3-y^2=1",
    "fact_spans": "[[[2, 64], [84, 87], [153, 156], [165, 168]], [[10, 64]], [[10, 64]], [[73, 77]], [[69, 72]], [[78, 81]], [[142, 145]], [[10, 64]], [[10, 64]], [[2, 64]], [[2, 72]], [], [], [[84, 93]], [[73, 94]], [[73, 94]], [[95, 109]], [[110, 125]], [[128, 141]], [[153, 163]]]",
    "query_spans": "[[[165, 173]]]",
    "process": "From the given condition: $ c=2 $. Since $ AF \\perp x $-axis, let $ A(c,\\frac{bc}{a}) $. Also, $ BF \\parallel OA $, so the equation of line $ BF $ is $ y=\\frac{b}{a}(x-c) $.  \n$$\n\\begin{cases}\ny=\\frac{b}{a}(x-c) \\\\\ny=-\\frac{b}{a}x\n\\end{cases}\n\\Rightarrow\n\\begin{cases}\nx=\\frac{c}{2} \\\\\ny=-\\frac{bc}{2a}\n\\end{cases}\n$$\nThen $ B(\\frac{c}{2},-\\frac{bc}{2a}) $. So $ \\overrightarrow{AB}=(-\\frac{c}{2},-\\frac{3bc}{2a}) $, $ \\overrightarrow{OB}=(\\frac{c}{2},-\\frac{bc}{2a}) $. Since $ AB \\perp OB $, we have $ \\overrightarrow{AB} \\cdot \\overrightarrow{OB}=0 $, then  \n$$\n-\\frac{c^{2}}{4}+\\frac{3b^{2}c^{2}}{4a^{2}}=0\n$$\nSo $ 3b^{2}=a^{2}=c^{2}-b^{2} \\Rightarrow b^{2}=1 $. Hence $ a^{2}=3 $, then the equation of $ C $ is $ \\frac{x^{2}}{3}-y^{2}=1 $."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, the directrix is $l$, point $A$ is a point on the parabola, and point $B$ is the projection of point $A$ onto the directrix $l$. If the distance from point $B$ to the line $AF$ is $2\\sqrt{3}$, then what is the length of $AF$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;l: Line;Directrix(G) = l;A: Point;PointOnCurve(A, G);B: Point;Projection(A, l) = B;Distance(B, LineOf(A, F)) = 2*sqrt(3)",
    "query_expressions": "Length(LineSegmentOf(A, F))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 16], [36, 39]], [[2, 16]], [[20, 23]], [[2, 23]], [[27, 30], [56, 59]], [[2, 30]], [[31, 35], [49, 53]], [[31, 43]], [[44, 48], [65, 69]], [[44, 63]], [[65, 92]]]",
    "query_spans": "[[[94, 102]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus is $F$, the right vertex is $A$, and the upper vertex is $B$. If the distance from point $F$ to the line $AB$ is $\\frac{5 \\sqrt{14}}{14} b$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;B: Point;F: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F;RightVertex(G)=A;UpperVertex(G)=B;Distance(F,LineOf(A,B))=(5*sqrt(14)/14)*b",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2/3",
    "fact_spans": "[[[2, 54], [126, 128]], [[4, 54]], [[4, 54]], [[67, 70]], [[75, 78]], [[59, 62], [81, 85]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 62]], [[2, 70]], [[2, 78]], [[81, 123]]]",
    "query_spans": "[[[126, 134]]]",
    "process": "The equation of line AB is bx + ay - ab = 0. The distance from point F(-c, 0) to line AB is \\frac{|-bc - ab|}{\\sqrt{a^{2} + b^{2}}} = \\frac{5\\sqrt{14}}{14}b \\frac{(a + c)^{2}}{a^{2} + b^{2}} = \\frac{25}{14} \\therefore a^{2} + c^{2} + 2ac = \\frac{25}{14}(2a^{2} - c^{2}) \\therefore (2a - 3c)(18a + 13c) = 0 \\therefore e = \\frac{c}{a} = \\frac{2}{3}"
  },
  {
    "text": "The right focus of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ is $F$. What is the standard equation of the parabola with focus at point $F$?",
    "fact_expressions": "G: Parabola;H: Ellipse;F: Point;Expression(H) = (x^2/4 + y^2 = 1);RightFocus(H) = F;Focus(G) = F",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = 4*sqrt(3)*x",
    "fact_spans": "[[[45, 48]], [[0, 27]], [[32, 35], [37, 41]], [[0, 27]], [[0, 35]], [[36, 48]]]",
    "query_spans": "[[[45, 55]]]",
    "process": ""
  },
  {
    "text": "The line $y = kx - 2$ and the hyperbola $x^2 - y^2 = 1$ have exactly one common point; then $k = $?",
    "fact_expressions": "G: Hyperbola;H: Line;k: Number;Expression(G) = (x^2 - y^2 = 1);Expression(H) = (y = k*x - 2);NumIntersection(H, G) = 1",
    "query_expressions": "k",
    "answer_expressions": "pm*1, pm*sqrt(5)",
    "fact_spans": "[[[12, 30]], [[0, 11]], [[41, 44]], [[12, 30]], [[0, 11]], [[0, 39]]]",
    "query_spans": "[[[41, 46]]]",
    "process": "Solving the system \\begin{cases}y=kx-2\\\\x^{2}-y^{2}=1\\end{cases}, we obtain (1-k^{2})x^{2}+4kx-5=0\\textcircled{1} When 1-k^{2}=0, we get k=\\pm1; at this time, the line is parallel to the asymptotes of the hyperbola, and the line intersects the hyperbola at exactly one point, satisfying the condition; \\textcircled{2} When 1-k^{2}\\neq0, since the line and the hyperbola have exactly one common point, we have \\triangle=16k^{2}+20(1-k^{2})=0, solving gives k=\\pm\\sqrt{5}, which satisfies the condition. In conclusion, k=\\pm1,\\pm\\sqrt{5}."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, the left focus is $F_{1}$, and $M$ is any point on the ellipse $C$. Then the minimum value of $|M F_{1}|$ is?",
    "fact_expressions": "C: Ellipse;M: Point;F1: Point;Expression(C) = (x^2/4 + y^2/3 = 1);LeftFocus(C)=F1;PointOnCurve(M,C)",
    "query_expressions": "Min(Abs(LineSegmentOf(M, F1)))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 44], [61, 66]], [[57, 60]], [[49, 56]], [[2, 44]], [[2, 56]], [[57, 71]]]",
    "query_spans": "[[[73, 90]]]",
    "process": ""
  },
  {
    "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ $(m<4)$ is $\\frac{1}{2}$, then $m=$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/m = 1);m: Number;m<4;Eccentricity(G) = 1/2",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 43]], [[1, 43]], [[63, 66]], [[3, 43]], [[1, 61]]]",
    "query_spans": "[[[63, 68]]]",
    "process": "From the given, we have $ a^2 = 4 $, $ b^{2} = m $, find $ c $, and $ m $ can be found from the eccentricity. According to the problem, $ 0 < m < 4 $, $ e = \\underline{\\sqrt{4 - m}} = \\frac{1}{2} $, $ m = 3 $."
  },
  {
    "text": "It is known that $F$ is the focus of the parabola $C$: $y^{2}=4x$. A line passing through $F$ with slope $\\sqrt{3}$ intersects $C$ at points $A$ and $B$. Given $|FA| > |FB|$, then the value of $\\frac{|FA|}{|FB|}$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;G: Line;PointOnCurve(F, G);Slope(G) = sqrt(3);A: Point;B: Point;Intersection(G, C) = {A, B};Abs(LineSegmentOf(F, A)) > Abs(LineSegmentOf(F, B))",
    "query_expressions": "Abs(LineSegmentOf(F, A))/Abs(LineSegmentOf(F, B))",
    "answer_expressions": "3",
    "fact_spans": "[[[6, 25], [51, 54]], [[6, 25]], [[2, 5], [30, 33]], [[2, 28]], [[48, 50]], [[29, 50]], [[34, 50]], [[55, 58]], [[59, 62]], [[48, 64]], [[67, 80]]]",
    "query_spans": "[[[82, 106]]]",
    "process": ""
  },
  {
    "text": "The standard equation of a hyperbola with an asymptote $3x + 4y = 0$ and a focus at $(4, 0)$ is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (4, 0);OneOf(Focus(G)) = H;Expression(OneOf(Asymptote(G)))=(3*x + 4*y=0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/(256/25) - y^2/(144/25) = 1",
    "fact_spans": "[[[34, 37]], [[26, 33]], [[26, 33]], [[22, 37]], [[0, 37]]]",
    "query_spans": "[[[34, 43]]]",
    "process": ""
  },
  {
    "text": "Given two points $M(-2,0)$, $N(2,0)$, a point $P$ satisfies $\\overrightarrow{P M} \\cdot \\overrightarrow{P N}=12$, then the equation of the trajectory of point $P$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (-2, 0);N: Point;Coordinate(N) = (2, 0);P: Point;DotProduct(VectorOf(P, M), VectorOf(P, N)) = 12",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2+y^2=16",
    "fact_spans": "[[[4, 13]], [[4, 13]], [[15, 23]], [[15, 23]], [[25, 29], [86, 90]], [[31, 83]]]",
    "query_spans": "[[[86, 97]]]",
    "process": "Let P(x,y), then \\overrightarrow{PM}=(-2-x,-y), \\overrightarrow{PN}=(2-x,-y). \\therefore \\overrightarrow{PM}\\cdot\\overrightarrow{PN}=(-2-x)(2-x)+(-y)^{2}=x^{2}-4+y^{2}=12, i.e. x^{2}+y^{2}=16. \\therefore The trajectory equation of point P is x^{2}+y^{2}=16."
  },
  {
    "text": "Construct an ellipse with foci $F_{1}$ and $F_{2}$. The sum of the distances from a point $P_{1}$ on the ellipse to $F_{1}$ and $F_{2}$ is $10$. Another point $P_{2}$ on the ellipse satisfies $P_{2}F_{1}=P_{2}F_{2}$. Then $P_{2}F_{1}$=?",
    "fact_expressions": "G: Ellipse;P2: Point;P1:Point;F1: Point;F2: Point;Focus(G)={F1,F2};PointOnCurve(P1,G);Distance(P1,F1)+Distance(P1,F2)=10;PointOnCurve(P2,G);LineSegmentOf(P2,F1)=LineSegmentOf(P2,F2)",
    "query_expressions": "LineSegmentOf(P2, F1)",
    "answer_expressions": "5",
    "fact_spans": "[[[20, 22], [23, 25], [62, 64]], [[68, 75]], [[28, 35]], [[1, 8], [36, 43]], [[9, 16], [9, 16]], [[1, 22]], [[23, 35]], [[28, 61]], [[62, 75]], [[77, 101]]]",
    "query_spans": "[[[103, 118]]]",
    "process": "According to the definition of an ellipse, the length relationship between line segments can be derived, thus obtaining the answer. Since point P lies on the ellipse, $ P_{2}F_{1} + P_{2}F_{2} = 10 $, and $ P_{2}F_{1} = P_{2}F_{2} $, so $ P_{2}F_{1} = 5 $."
  },
  {
    "text": "If the line segment $x+y=1(-1 \\leqslant x \\leqslant 1)$ and the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=k(k>0)$ have no intersection points, then the range of values for the real number $k$ is?",
    "fact_expressions": "G: Ellipse;k: Real;H: LineSegment;k>0;Expression(G) = (x^2/3 + y^2/2 = k);Expression(H) = And((x + y = 1), (-1 <= x), (x<= 1));NumIntersection(H,G)=0",
    "query_expressions": "Range(k)",
    "answer_expressions": "{(0, 1/5), (7/3, +oo)}",
    "fact_spans": "[[[39, 81]], [[87, 92]], [[1, 38]], [[41, 81]], [[39, 81]], [[1, 38]], [[1, 85]]]",
    "query_spans": "[[[87, 99]]]",
    "process": "\\because the line segment \\( x + y = 1 \\) \\((-1 \\leqslant x \\leqslant 1)\\) and the ellipse \\( \\frac{x^{2}}{3} + \\frac{y^{2}}{2} = k \\) \\((k > 0)\\) have no intersection points,  \n\\therefore the line segment \\( x + y = 1 \\) \\((-1 \\leqslant x \\leqslant 1)\\) lies either inside or outside the ellipse.  \nWhen the line segment \\( x + y = 1 \\) \\((-1 \\leqslant x \\leqslant 1)\\) lies inside the ellipse,  \n\\[\n\\begin{cases}\n\\frac{1}{3} < k \\\\\n\\frac{1}{3} + \\frac{4}{2} < k\n\\end{cases}\n\\]\n\\therefore \\( k > \\frac{7}{3} \\).  \nWhen the line segment \\( x + y = 1 \\) \\((-1 \\leqslant x \\leqslant 1)\\) lies outside the ellipse, the segment contains the portion of the line in the first quadrant, and since the center of the ellipse is the origin, the line containing the segment has no common point with the ellipse.  \nSubstituting \\( y = 1 - x \\) into \\( \\frac{x^{2}}{3} + \\frac{y^{2}}{2} = k \\), we obtain \\( 5x^{2} - 6x - 6k + 3 = 0 \\),  \n\\therefore \\( \\Delta = 36 - 20(-6k + 3) < 0 \\), and since \\( k > 0 \\), \\therefore \\( 0 < k < \\frac{1}{5} \\).  \nIn summary, \\( 0 < k < \\frac{1}{5} \\) or \\( k > \\frac{7}{3} \\)."
  },
  {
    "text": "The foci of the hyperbola lie on the $x$-axis, the length of the real axis is $4$, and the eccentricity is $3$. Then, the standard equation of this hyperbola is? The asymptote equations are?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), xAxis);Length(RealAxis(G)) = 4;Eccentricity(G) = 3",
    "query_expressions": "Expression(G);Expression(Asymptote(G))",
    "answer_expressions": "x^2/4 - y^2/32 = 1  \ny = pm*2*sqrt(2)*x",
    "fact_spans": "[[[0, 3], [31, 34]], [[0, 12]], [[0, 20]], [[0, 28]]]",
    "query_spans": "[[[31, 41]], [[31, 48]]]",
    "process": "Let the standard equation of the hyperbola be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$. Then $2a=4$, so $a=2$. $e=\\frac{c}{a}=3$, thus $c=6$, $b=4\\sqrt{2}$. Therefore, the standard equation of the hyperbola is $\\frac{x^{2}}{4}-\\frac{y^{2}}{32}=1$, and the asymptotes are $y=\\pm\\frac{b}{a}x=\\pm2\\sqrt{2}x$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and point $P$ is a point on the ellipse in the first quadrant such that $P F_{1} \\perp P F_{2}$, then $|P F_{2}|=$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;Quadrant(P) = 1;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "2-sqrt(2)",
    "fact_spans": "[[[18, 45], [57, 59]], [[52, 56]], [[2, 9]], [[10, 17]], [[18, 45]], [[2, 51]], [[2, 51]], [[52, 67]], [[52, 67]], [[69, 92]]]",
    "query_spans": "[[[94, 107]]]",
    "process": ""
  },
  {
    "text": "If the equation $\\frac{x^{2}}{m}+\\frac{y^{2}}{1-m}=1$ represents an ellipse with foci on the $y$-axis, then the range of real values for $m$ is?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (y^2/(1 - m) + x^2/m = 1);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(0,1/2)",
    "fact_spans": "[[[51, 53]], [[55, 60]], [[1, 53]], [[42, 53]]]",
    "query_spans": "[[[55, 67]]]",
    "process": "From the given condition, the equation $\\frac{x^{2}}{m} + \\frac{y^{2}}{1-m} = 1$ represents an ellipse with foci on the $y$-axis, so we have $1 - m > m > 0$, solving this yields: $0 < m < \\frac{1}{2}$. Therefore, the range of real number $m$ is: $(0, \\frac{1}{2})$."
  },
  {
    "text": "Given that $F$ is the left focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, point $A(1 , \\sqrt{5})$, and $P$ is a moving point on the right branch of the hyperbola, then the minimum value of $|P F|+| P A |$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/5 = 1);F: Point;LeftFocus(G) = F;A: Point;Coordinate(A) = (1, sqrt(5));P: Point;PointOnCurve(P, RightPart(G))",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "7",
    "fact_spans": "[[[6, 44], [74, 77]], [[6, 44]], [[2, 5]], [[2, 48]], [[49, 68]], [[49, 68]], [[70, 73]], [[70, 83]]]",
    "query_spans": "[[[85, 106]]]",
    "process": "Let the right focus of the hyperbola $\\frac{x^2}{4}-\\frac{y^{2}}{5}=1$ be $F_{1}$. Since $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{4+5}=3$, the coordinates of $F_{1}$ are $(3,0)$. Because $P$ is a moving point on the right branch of the hyperbola, $|PF|-|PF_{1}|=2a=4$, therefore $|PF|+|PA|=|PF_{1}|+|PA|+4$. Clearly, when $F_{1}$, $P$, and $A$ are collinear, $|PF|+|PA|$ reaches its minimum value, which is $|F_{A}|+4=\\sqrt{(3-1)^{2}+(0-\\sqrt{5})^{2}}+4=3+4=7$."
  },
  {
    "text": "Given that the distance from the point $P(1, y_{0})$ $(y_{0}>0)$ on the parabola $C$: $y^{2}=2 p x$ $(p>0)$ to the focus is $2$, then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;P: Point;p>0;y0:Number;Expression(C) = (y^2 = 2*(p*x));Coordinate(P) = (1, y0);PointOnCurve(P, C);Distance(P, Focus(C)) = 2;y0>0",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 30]], [[69, 72]], [[32, 57]], [[10, 30]], [[33, 57]], [[2, 30]], [[32, 57]], [[2, 57]], [[2, 67]], [[33, 57]]]",
    "query_spans": "[[[69, 74]]]",
    "process": "\\because the distance from the point P(1,y_{0}) (y_{0}>0) on the parabola C: y^{2}=2px (p>0) to the focus is 2, \\therefore by the definition of a parabola, 1+\\frac{p}{2}=2, solving gives p=2."
  },
  {
    "text": "If the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1$ $(a>0)$ is $2$, then what is the value of $a$?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2/9 + x^2/a^2 = 1);Eccentricity(G) = 2",
    "query_expressions": "a",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[1, 48]], [[58, 61]], [[4, 48]], [[1, 48]], [[1, 56]]]",
    "query_spans": "[[[58, 64]]]",
    "process": ""
  },
  {
    "text": "Let $F$ be the focus of the parabola $C$: $y^{2}=4 x$, and let the curve $y=\\frac{k}{x}(k>0)$ intersect $C$ at point $P$. If $P F \\perp x$-axis, then $k=$?",
    "fact_expressions": "C: Parabola;G: Curve;k: Number;P: Point;F: Point;Expression(C) = (y^2 = 4*x);k>0;Expression(G) = (y = k/x);Focus(C) = F;Intersection(G, C) = P;IsPerpendicular(LineSegmentOf(P,F),xAxis)",
    "query_expressions": "k",
    "answer_expressions": "2",
    "fact_spans": "[[[5, 24], [51, 54]], [[28, 50]], [[79, 82]], [[56, 60]], [[1, 4]], [[5, 24]], [[30, 50]], [[28, 50]], [[1, 27]], [[28, 61]], [[63, 77]]]",
    "query_spans": "[[[79, 84]]]",
    "process": ""
  },
  {
    "text": "It is known that one focus of a hyperbola coincides with the focus of the parabola $x=-\\frac{1}{8} y^{2}$, and the eccentricity of the hyperbola is $2$. What is the equation of the asymptotes of the hyperbola?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(H) = (x = (-1/8)*y^2);OneOf(Focus(G)) = Focus(H);Eccentricity(G) = 2",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[2, 5], [43, 46], [57, 60]], [[11, 36]], [[11, 36]], [[2, 41]], [[43, 54]]]",
    "query_spans": "[[[57, 68]]]",
    "process": ""
  },
  {
    "text": "What is the length of the imaginary axis of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1)",
    "query_expressions": "Length(ImageinaryAxis(G))",
    "answer_expressions": "6",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 45]]]",
    "process": ""
  },
  {
    "text": "Let $P$ be a point on the parabola $y^{2}=2 x$, and $A(a, 0)$ where $0<a<1$. Then the minimum value of $|P A|$ is?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;a: Number;0 < a;a < 1;Expression(G) = (y^2 = 2*x);Coordinate(A) = (a, 0);PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "a",
    "fact_spans": "[[[5, 19]], [[24, 42]], [[1, 4]], [[24, 42]], [[24, 42]], [[24, 42]], [[5, 19]], [[24, 42]], [[1, 23]]]",
    "query_spans": "[[[44, 57]]]",
    "process": "By the given condition, the distance from point P(x, y) to point A(a, 0) (a ∈ ℝ) is d = √((x - a)² + y²). Since y² = 2x, ∴ d = √((x - a)² + y²) = √((x - a)² + 2x) = √([x - (a - 1)]² + 2a - 1). Since 0 < a < 1, ∴ when x = 0, the minimum value is a."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, the focal distance is $4$, point $P$ is on the right branch of the hyperbola such that $P F_{1} \\perp P F_{2}$, and $\\overrightarrow{F_{1} P} \\cdot \\overrightarrow{F_{1} O}=6$. Then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;O: Origin;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;FocalLength(G) = 4;PointOnCurve(P, RightPart(G));IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));DotProduct(VectorOf(F1, P), VectorOf(F1, O)) = 6",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)+1",
    "fact_spans": "[[[2, 59], [94, 97], [192, 195]], [[5, 59]], [[5, 59]], [[89, 93]], [[66, 73]], [[74, 81]], [[130, 189]], [[5, 59]], [[5, 59]], [[2, 59]], [[2, 81]], [[2, 81]], [[2, 88]], [[89, 102]], [[104, 128]], [[130, 189]]]",
    "query_spans": "[[[192, 201]]]",
    "process": "Let $ P((x,y), F_{1}(-2,0) $, then $ \\overrightarrow{F_{1}P} \\cdot \\overrightarrow{F_{1}O} = (x+2,y) \\cdot (2,0) = 2(x+2) = 6 $, so $ x = 1 $. Without loss of generality, let $ P(1, \\frac{b\\sqrt{1-a^{2}}}{a}) $, then from $ PF_{1} \\perp PF_{2} $, we get $ \\overrightarrow{F_{1}P} \\cdot \\overrightarrow{F_{2}P} = -3 + \\frac{b^{2}(1-a^{2})}{a^{2}} = 0 $, that is $ (4-a^{2})(1-a^{2}) = 3a^{2} $, solving gives $ a = \\sqrt{3}-1 $, then the eccentricity of this hyperbola is $ e = \\frac{c}{a} = \\frac{2}{\\sqrt{3}-1} = \\sqrt{3}+1 $."
  },
  {
    "text": "The two foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$ are $F_{1}$ and $F_{2}$, respectively. Let $P$ be a point on the ellipse such that $|P F_{2}|=\\frac{\\sqrt{3}}{2}|P F_{1}|$. Then the maximum value of $\\angle P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F2: Point;F1: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Focus(G)={F1,F2};PointOnCurve(P,G);Abs(LineSegmentOf(P, F2)) = (sqrt(3)/2)*Abs(LineSegmentOf(P, F1))",
    "query_expressions": "Max(AngleOf(P, F1, F2))",
    "answer_expressions": "pi/3",
    "fact_spans": "[[[0, 54], [82, 84]], [[2, 54]], [[2, 54]], [[78, 81]], [[70, 77]], [[62, 69]], [[2, 54]], [[2, 54]], [[0, 54]], [[0, 77]], [[78, 87]], [[89, 129]]]",
    "query_spans": "[[[134, 162]]]",
    "process": "From the definition of the ellipse, we have |PF_{1}| + |PF_{2}| = 2a. Given |PF_{2}| = \\frac{\\sqrt{3}}{2}|PF_{1}|, we obtain |PF_{1}| = 4(2 - \\sqrt{3})a, |PF_{2}| = 2\\sqrt{3}(2 - \\sqrt{3})a. In \\triangle PF_{1}F_{2}, |F_{1}F_{2}| = 2c. 2 = \\frac{|PF_{1}|^{2} + |F_{1}F_{2}|^{2} - |PF_{2}|^{2}}{2|PF_{1}||F_{1}F_{2}|} = \\frac{4c^{2} + [4(2 - \\sqrt{3})a]}{4c \\cdot 4(2 - \\sqrt{3})a}(2 - \\sqrt{3})a]\\frac{\\sqrt{3})a}{c} \\geqslant 2 \\times \\frac{1}{4} = \\frac{1}{2}. The equality holds if and only if c = (2 - \\sqrt{3})a, so the maximum value of \\angle PF_{1}F_{2} is \\frac{\\pi}{3}."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, its two foci are $F_{1}(-c, 0)$, $F_{2}(c, 0)(c>0)$. Draw a perpendicular line from point $M(3 c, 0)$ to an asymptote of this hyperbola, with foot of perpendicular at $P$. If $|O P|=5 b$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;c:Number;c>0;Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);Focus(C)={F1,F2};M: Point;Coordinate(M) = (3*c, 0);l1: Line;PointOnCurve(M, l1);IsPerpendicular(l1,Asymptote(C));P: Point;FootPoint(l1,Asymptote(C))=P;O: Origin;Abs(LineSegmentOf(O, P)) = 5*b",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(34)/5",
    "fact_spans": "[[[2, 63], [121, 124], [152, 158]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[71, 85]], [[87, 105]], [[87, 105]], [[87, 105]], [[71, 85]], [[87, 105]], [[2, 105]], [[107, 119]], [[107, 119]], [], [[106, 130]], [[106, 130]], [[134, 137]], [[106, 137]], [[139, 150]], [[139, 150]]]",
    "query_spans": "[[[152, 164]]]",
    "process": "The distance from $F_{2}(c,0)$ to an asymptote line $ay-bx=0$ of the hyperbola is: $d=\\frac{|0\\times a-b\\times c|}{\\sqrt{a^{2}+b^{2}}}=b$. As shown in the figure, draw $F_{2}N\\bot OP$ intersecting $OP$ at point $N$, then $NF_{2}//MP$, we obtain $\\frac{|OF_{2}|}{|OM|}=\\frac{|NF_{2}|}{|PM|}=\\frac{c}{3c}=\\frac{1}{3}$, so $|PM|=3b$, and $|OP|=5b$. In the right triangle $OPM$, $\\tan\\angle POM=\\frac{b}{a}=\\frac{|PM|}{|PO|}=\\frac{3b}{5b}=\\frac{3}{5}$, thus $\\frac{b}{a}=\\frac{3}{5}$, so $e=\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{1+\\frac{9}{25}}=\\frac{\\sqrt{34}}{5}$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$, let the right focus be $F$, the right vertex be $A$, and one endpoint of the imaginary axis be $B$. If the distance from point $F$ to the line $AB$ is $\\frac{b}{3}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(G) = F;A: Point;RightVertex(G) = A;B: Point;OneOf(Endpoint(ImageinaryAxis(G))) = B;Distance(F, LineOf(A, B)) = b/3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3/2",
    "fact_spans": "[[[2, 61], [122, 125]], [[2, 61]], [[5, 61]], [[5, 61]], [[5, 61]], [[5, 61]], [[66, 69], [91, 95]], [[2, 69]], [[74, 77]], [[2, 77]], [[86, 89]], [[2, 89]], [[91, 120]]]",
    "query_spans": "[[[122, 131]]]",
    "process": "From the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$, we obtain: the right focus $F(c,0)$, the right vertex $A(a,0)$, the upper endpoint of the imaginary axis $B(0,b)$, and $a^{2}=b^{2}+c^{2}$. Then the line $AB$: $\\frac{x}{a}+\\frac{y}{b}=1$. Given that the distance from point $F$ to line $AB$ is $\\frac{b}{3}$, we have: $\\frac{|\\frac{c}{a}-1|}{\\sqrt{(\\frac{1}{a})^{2}+(\\frac{1}{b})^{2}}}=\\frac{b}{3}$. Simplifying yields: $3a=2c$. Therefore, the eccentricity $e=\\frac{c}{a}=\\frac{3}{2}$."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $x^{2}=8 y$ intersects the parabola at points $A$ and $B$, and $O$ is the origin. If $\\frac{|A F|}{|B F|}=\\frac{1}{2}$, then the area of $\\triangle A O B$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 8*y);F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);A: Point;B: Point;Intersection(H, G) = {A, B};O: Origin;Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(B, F)) = 1/2",
    "query_expressions": "Area(TriangleOf(A, O, B))",
    "answer_expressions": "6*sqrt(2)",
    "fact_spans": "[[[1, 15], [26, 29]], [[1, 15]], [[18, 21]], [[1, 21]], [[22, 24]], [[0, 24]], [[30, 33]], [[34, 37]], [[22, 39]], [[40, 43]], [[51, 84]]]",
    "query_spans": "[[[86, 108]]]",
    "process": "It is easy to see that the slope of line AB exists, denoted as y = kx + 2. From \\begin{cases}x^{2}=8y,\\\\y=kx+2\\end{cases}, we obtain x^{2}-8kx-16=0. \\therefore x_{A}x_{B}=-16. Also, \\because \\frac{|AF|}{|BF|}=\\frac{1}{2}, \\therefore x_{B}=-2x_{A}, \\therefore \\begin{cases}x_{A}=-2\\sqrt{2},\\\\x_{B}=4\\sqrt{2}\\end{cases}. Or x_{B}=-4\\sqrt{2}, then S_{AOAB}=\\frac{1}{2}|OF||x_{B}-x_{A}|=6\\sqrt{2}."
  },
  {
    "text": "It is known that the line $y = kx - 2$ intersects the parabola $y^2 = 8x$ at two distinct points $A$, $B$, and the x-coordinate of the midpoint of $AB$ is $2$. Then the value of $k$ is?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*x - 2);k: Number;G: Parabola;Expression(G) = (y^2 = 8*x);A: Point;B: Point;Negation(A=B);Intersection(H, G) = {A, B};XCoordinate(MidPoint(LineSegmentOf(A, B))) = 2",
    "query_expressions": "k",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 12]], [[2, 12]], [[62, 65]], [[13, 27]], [[13, 27]], [[34, 38]], [[41, 44]], [[29, 44]], [[2, 44]], [[46, 60]]]",
    "query_spans": "[[[62, 69]]]",
    "process": "\\because the line y=kx-2 intersects the parabola y^{2}=8x at two distinct points A and B, \\therefore k\\neq0. From \\begin{cases}y=kx-2\\\\y^{2}=8x\\end{cases}, we obtain k^{2}x^{2}-4kx-8x+4=0, \\therefore x_{1}+x_{2}=\\frac{4k+8}{k^{2}}. Since the horizontal coordinate of the midpoint of AB is 2, \\therefore x_{1}+x_{2}=\\frac{4k+8}{k^{2}}=4. Solving gives: k=-1 or k=2. Checking, when k=-1, the equation k^{2}x^{2}-4kx-8x+4=0 has only one solution, meaning points A and B coincide, \\therefore k\\neq-1, \\therefore k=2. Hence the answer: 2"
  },
  {
    "text": "Given that $F$ is the left focus of the hyperbola $C$: $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, $P$ is a point on the right branch of the hyperbola, and point $A$ has coordinates $(2,4)$, find the distance from point $F$ to line $AP$ when the perimeter of $\\triangle APF$ is minimized.",
    "fact_expressions": "C: Hyperbola;A: Point;P: Point;F: Point;Expression(C) = (x^2/16 - y^2/9 = 1);Coordinate(A) = (2, 4);LeftFocus(C) = F;PointOnCurve(P, RightPart(C));WhenMin(Perimeter(TriangleOf(A, P, F)))",
    "query_expressions": "Distance(F, LineOf(A,P))",
    "answer_expressions": "8",
    "fact_spans": "[[[6, 50], [59, 62]], [[68, 72]], [[55, 58]], [[2, 5], [109, 113]], [[6, 50]], [[68, 83]], [[2, 54]], [[55, 67]], [[85, 108]]]",
    "query_spans": "[[[109, 126]]]",
    "process": "The left focus of the hyperbola is $ F(-5,0) $, and let the right focus be $ F'(5,0) $. According to the given conditions, the perimeter of $ \\triangle APF $ is $ |AF| + |PF| + |AP| = |AF| + 8 + |PF'| + |AP| \\geqslant |AF| + 8 + |PF'| $, where equality holds when points $ A $, $ P $, and $ F' $ are collinear, i.e., the perimeter of $ \\triangle APF $ is minimized. At this time, the equation of line $ AP $ is $ 4x + 3y - 20 = 0 $. Therefore, the distance from point $ F $ to line $ AP $ is $ \\underline{|4\\times(-5)+0-20|} = 8 $."
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-\\frac{y^{2}}{4}=1$, what is the equation of the asymptotes? What is the eccentricity $e$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2/4 = 1);Eccentricity(C) = e ;e: Number",
    "query_expressions": "Expression(Asymptote(C));e",
    "answer_expressions": "pm*y+2*x=0;sqrt(5)",
    "fact_spans": "[[[2, 34]], [[2, 34]], [[2, 49]], [[46, 49]]]",
    "query_spans": "[[[2, 43]], [[46, 51]]]",
    "process": "From the given information, the foci of the hyperbola lie on the x-axis, so its asymptotes are given by $ y = \\pm\\frac{b}{a}x $, and the eccentricity is $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} $. It is known that the foci of the hyperbola lie on the x-axis, with $ a = 1 $, $ b = 2 $, $ c = \\sqrt{5} $. Thus, its asymptotes are $ y = \\pm\\frac{b}{a}x = \\pm2x $, or $ 2x \\pm y = 0 $, and the eccentricity is $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{5} $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $A$ is a point on the left branch of the hyperbola. If line $A F_{1}$ is parallel to the line $y=\\frac{b}{a} x$ and the perimeter of $\\Delta A F_{1} F_{2}$ is $9 a$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;F1: Point;A: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = x*(b/a));LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(A, LeftPart(G));IsParallel(LineOf(A,F1),H);Perimeter(TriangleOf(A,F1,F2))=9*a",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 58], [88, 91], [166, 169]], [[5, 58]], [[5, 58]], [[111, 130]], [[67, 74]], [[83, 87]], [[75, 82]], [[5, 58]], [[5, 58]], [[2, 58]], [[111, 130]], [[2, 82]], [[2, 82]], [[83, 97]], [[99, 132]], [[133, 164]]]",
    "query_spans": "[[[166, 175]]]",
    "process": "By the definition of a hyperbola, |AF₂| - |AF| = 2a, and |AF₂| + |AF₁| = 9a - 2c. Solving gives |AF₂| = (11a - 2c)/2, |AF₁| = (7a - 2c)/2. Since line AF₁ is parallel to the line y = (b/a)x, tan∠AF₁F₂ = b/a, hence cos∠AF₁F₂ = a/c. By the law of cosines: cos∠AF₁F₂ = a/c = (|AF₁|² + 4c² - |AF₂|²)/(2|AF₁|·2c), that is, 1/e = (-18 + 4e + 4e²)/(14e - 4e²). Simplifying yields e² + 2e - 8 = 0, solving gives e = 2 or e = -4 (discarded). This problem mainly examines the definition of a hyperbola, the law of cosines, and the eccentricity of a hyperbola, and is considered difficult."
  },
  {
    "text": "Given an ellipse centered at the origin with foci at $(0, \\pm 5 \\sqrt{2})$, the x-coordinate of the midpoint of the chord intercepted by the line $3x - y - 2 = 0$ is $\\frac{1}{2}$. Then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;Center(G) = O;O: Origin;Coordinate(Focus(G)) = (0, pm*5*sqrt(2));H: Line;Expression(H) = (3*x - y - 2 = 0);XCoordinate(MidPoint(InterceptChord(H, G))) = 1/2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/75 + x^2/25 = 1",
    "fact_spans": "[[[35, 37], [79, 81]], [[2, 37]], [[5, 7]], [[8, 37]], [[38, 51]], [[38, 51]], [[35, 76]]]",
    "query_spans": "[[[79, 86]]]",
    "process": "Let the equation of the ellipse be $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$ $(a>b>0)$, then $a^{2}=b^{2}+c^{2}=b^{2}+50$ $\\textcircled{1}$. Let the endpoints of the chord formed by the intersection of the line $3x-y-2=0$ and the ellipse be $A(x_{1},y_{1})$, $B(x_{2},y_{2})$. Then  \n$$\n\\begin{cases}\nb^{2}y_{1}^{2}+a^{2}x_{1}^{2}=a^{2}b^{2} \\\\\nb^{2}y_{2}^{2}+a^{2}x_{2}^{2}=a^{2}b^{2}\n\\end{cases}\n$$\n$$\n\\therefore b^{2}(y_{1}-y_{2})(y_{1}+y_{2})+a^{2}(x_{1}-x_{2})(x_{1}+x_{2})=0\n$$\nThe abscissa of the midpoint of the chord is $\\frac{1}{2}$, then the ordinate is $-\\frac{1}{2}$, that is, $x_{1}+x_{2}=2\\times\\frac{1}{2}=1$, $y_{1}+y_{2}=2\\times(-\\frac{1}{2})=-1$, $\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=3$.  \n$$\n\\therefore b^{2}\\times3\\times(-1)+a^{2}\\times1=0, \\text{ i.e. } a^{2}=3b^{2} \\textcircled{2}\n$$\nSolving $\\textcircled{1}$ and $\\textcircled{2}$ together gives: $a^{2}=75$, $b^{2}=25$.  \nHence, the equation of the ellipse is $\\frac{y^{2}}{75}+\\frac{x^{2}}{25}=1$."
  },
  {
    "text": "If two mutually perpendicular lines $l_{1}$, $l_{2}$ passing through the left and right foci $F_{1}$, $F_{2}$ of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ respectively intersect at a point on the ellipse, then the range of the eccentricity of this ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, l1) = True;PointOnCurve(F2, l2) = True;IsPerpendicular(l1, l2) = True;l1: Line;l2: Line;PointOnCurve(Intersection(l1, l2), G) = True",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[\\sqrt{2}/2,1)",
    "fact_spans": "[[[4, 56], [111, 113], [117, 119]], [[4, 56]], [[6, 56]], [[6, 56]], [[6, 56]], [[6, 56]], [[62, 69]], [[70, 77]], [[4, 77]], [[4, 77]], [[1, 107]], [[1, 107]], [[81, 107]], [[88, 97]], [[100, 107]], [[88, 114]]]",
    "query_spans": "[[[117, 130]]]",
    "process": "Let the intersection point of the two lines be $ M $, $ F_{1}(-c,0) $, $ F_{2}(c,0) $, and the coordinate origin be $ O $. By the definition of the ellipse, we have $ |MF_{1}| + |MF_{2}| = 2a $. Since $ MF_{1} \\perp MF_{2} $, it follows that $ |MF_{1}|^{2} + |MF_{2}|^{2} = |F_{1}F_{2}|^{2} = 4c^{2} $. By the AM-QM inequality, we obtain $ \\frac{|MF_{1}| + |MF_{2}|}{2} \\leqslant \\sqrt{\\frac{|MF_{1}|^{2} + |MF_{2}|^{2}}{2}} $, i.e., $ a \\leqslant \\sqrt{2}c $, with equality holding if and only if $ |MF_{1}| = |MF_{2}| = a $. Thus, $ e = \\frac{c}{a} \\geqslant \\frac{\\sqrt{2}}{2} $, and since $ e < 1 $, we have $ \\frac{\\sqrt{2}}{2} \\leqslant e < 1 $."
  },
  {
    "text": "Given that both asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) are tangent to the circle $C$: $(x-3)^{2}+y^{2}=4$, then the eccentricity of this hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;C: Circle;Expression(C) = (y^2 + (x - 3)^2 = 4);l1:Line;l2:Line;Asymptote(G)={l1,l2};IsTangent(l1,C);IsTangent(l2,C)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3*sqrt(5)/5",
    "fact_spans": "[[[2, 58], [95, 98]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[66, 90]], [[66, 90]], [-1, -1], [-1, -1], [2, 61], [2, 89], [2, 89]]",
    "query_spans": "[[[95, 105]]]",
    "process": ""
  },
  {
    "text": "Given that a focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{5}=1$ ($a>0$) coincides with the focus of the parabola $y^{2}=12x$, then $a=$? If a point $P$ on the hyperbola is at a distance of $2$ from $F$, then what is the distance from point $P$ to the other focus of the hyperbola?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;F: Point;P: Point;F2: Point;a>0;Expression(G) = (-y^2/5 + x^2/a^2 = 1);Expression(H) = (y^2 = 12*x);OneOf(Focus(G)) = F;OneOf(Focus(G)) = F2;Negation(F = F2);Focus(H) = F;PointOnCurve(P,G);Distance(P,F) = 2",
    "query_expressions": "Distance(P, F2);a",
    "answer_expressions": "6;2",
    "fact_spans": "[[[2, 49], [85, 88], [113, 116]], [[80, 83]], [[58, 73]], [[54, 57], [95, 98]], [[108, 112], [91, 94]], [], [[5, 49]], [[2, 49]], [[58, 73]], [[2, 57]], [[113, 122]], [[54, 122]], [[54, 78]], [[85, 94]], [[91, 105]]]",
    "query_spans": "[[[108, 127]], [[80, 85]]]",
    "process": ""
  },
  {
    "text": "What is the length of the major axis of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1)",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[0, 27]], [[0, 27]]]",
    "query_spans": "[[[0, 33]]]",
    "process": "According to the problem, in the ellipse given by the equation $\\frac{x^{2}}{2}+y^{2}=1$, where $a=\\sqrt{2}$, the length of its major axis is $2a=2\\sqrt{2}$."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$, and intersects the line $x=-1$ at point $P$. If $\\overrightarrow{P A}=\\lambda \\overrightarrow{A F}$, $\\overrightarrow{P B}=\\mu \\overrightarrow{B F}$ ($\\lambda, \\mu \\in \\mathbb{R}$), then $\\lambda+\\mu=$?",
    "fact_expressions": "G: Parabola;H: Line;C:Line;P: Point;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 4*x);Expression(C) = (x = -1);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Intersection(H,C)=P;VectorOf(P, A) = lambda*VectorOf(A, F);VectorOf(P, B) = mu*VectorOf(B, F);lambda:Real;mu:Real",
    "query_expressions": "lambda + mu",
    "answer_expressions": "0",
    "fact_spans": "[[[1, 15], [27, 30]], [[22, 24]], [[42, 50]], [[51, 55]], [[31, 34]], [[18, 21]], [[35, 38]], [[1, 15]], [[42, 50]], [[1, 21]], [[0, 24]], [[22, 40]], [[22, 55]], [[57, 108]], [[110, 177]], [[110, 177]], [[110, 177]]]",
    "query_spans": "[[[179, 194]]]",
    "process": ""
  },
  {
    "text": "The line $l$ passes through the intersection point of the parabola $y=x^{2}-3 x+1$ and the $y$-axis, and is parallel to the line $x+2 y=0$. Then the equation of line $l$ is?",
    "fact_expressions": "l: Line;G: Parabola;H: Line;Expression(G) = (y = x^2 - 3*x + 1);Expression(H) = (x + 2*y = 0);PointOnCurve(Intersection(G, yAxis), l);IsParallel(l, H)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x + 2*y - 2 = 0",
    "fact_spans": "[[[0, 5], [51, 56]], [[7, 25]], [[36, 47]], [[7, 25]], [[36, 47]], [[0, 33]], [[0, 49]]]",
    "query_spans": "[[[51, 61]]]",
    "process": "The intersection with the y-axis is (0,1). Let the line be $x + 2y + c = 0$. Substituting (0,1) gives $c = -2$, so the equation of the line is $x + 2y - 2 = 0$."
  },
  {
    "text": "If the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{n}=1(m>n>0)$ and the curve $x^{2}+y^{2}=m-n$ have no intersection points, then the range of the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "G: Ellipse;m: Number;n: Number;H: Curve;e: Number;m > n;n > 0;Expression(G) = (y^2/n + x^2/m = 1);Expression(H) = (x^2 + y^2 = m - n);NumIntersection(G, H)=0;Eccentricity(G) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "(0, \\sqrt{2}/2)",
    "fact_spans": "[[[1, 45], [70, 72]], [[3, 45]], [[3, 45]], [[46, 65]], [[76, 79]], [[3, 45]], [[3, 45]], [[1, 45]], [[46, 65]], [[1, 68]], [[70, 79]]]",
    "query_spans": "[[[76, 86]]]",
    "process": "By the given condition, $\\sqrt{m-n}<\\sqrt{n}$, that is, $m<2n$, so $n<m<2n$, $\\frac{1}{2}<\\frac{n}{m}<1$, therefore $e=\\frac{c}{a}=\\frac{\\sqrt{m-n}}{\\sqrt{m}}=\\sqrt{\\frac{m-n}{m}}=\\sqrt{1-\\frac{n}{m}}\\in(0,\\frac{\\sqrt{2}}{2})$"
  },
  {
    "text": "The standard equation of the ellipse with focus $F(0,5 \\sqrt{2})$ and for which the x-coordinate of the midpoint of the chord intercepted by the line $y=2 x-1$ is $\\frac{2}{7}$ is?",
    "fact_expressions": "C: Ellipse;H: Line;Expression(H) = (y = 2*x - 1);Coordinate(F) = (0, 5*sqrt(2)) ;XCoordinate(MidPoint(InterceptChord(H,C)))=2/7;Focus(C) =F;F:Point",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/25 + y^2/75 = 1",
    "fact_spans": "[[[59, 61]], [[23, 34]], [[23, 34]], [[3, 20]], [[22, 61]], [[0, 61]], [[3, 20]]]",
    "query_spans": "[[[59, 68]]]",
    "process": "Let the endpoints of the chord cut by the ellipse on the line be $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Using the \"point difference method\", establish the relationship among the coordinates of the midpoint of the chord, the slope of the line containing the chord, and $ a, b $ in the ellipse equation. Combining $ c=5\\sqrt{2} $, the conclusion can be obtained. Let the required ellipse equation be $ \\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1 $ $ (a>b>0) $, and the endpoints of the chord cut by the line on the ellipse be $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Substituting the coordinates of points $ A $ and $ B $ into the ellipse equation gives\n$$\n\\begin{cases}\nx+x_{2},y_{1}+y_{2} \\\\\n\\frac{y}{a}+\\frac{x^{2}}{b^{2}}=1 \\\\\n\\frac{y^{2}}{2}+\\frac{x^{2}}{2}=1\n\\end{cases}\n$$\nSubtracting the two equations and simplifying yields\n$$\n\\frac{a^{2}}{b^{2}}=-\\frac{y_{1}}{x_{1}-x_{2}}\\frac{y_{1}+y_{2}}{1+x_{2}}=-2\\frac{-\\frac{6}{7}}{4}=3,\n$$\nso $ a^{2}=3b^{2} $. Also, $ c^{2}=a^{2}-b^{2}=50 $, so $ a^{2}=75 $, $ b^{2}=25 $. Therefore, the standard equation of the required ellipse is $ \\frac{y^{2}}{75}+\\frac{x^{2}}{25}=1 $."
  },
  {
    "text": "It is known that $F$ is the right focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $A$ is the left vertex of the ellipse $C$, and $P$ is a point on the ellipse $C$ such that $PF$ is perpendicular to the $x$-axis. If the slope of the line $AP$ is $\\frac{\\sqrt{3}}{3}$, then the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;A: Point;P: Point;F: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;LeftVertex(C) = A;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P,F),xAxis);Slope(LineOf(A,P)) = sqrt(3)/3",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(3-sqrt(3))/3",
    "fact_spans": "[[[6, 63], [72, 77], [86, 91], [143, 148]], [[12, 63]], [[12, 63]], [[68, 71]], [[82, 85]], [[2, 5]], [[12, 63]], [[12, 63]], [[6, 63]], [[2, 67]], [[68, 81]], [[82, 94]], [[96, 108]], [[110, 141]]]",
    "query_spans": "[[[143, 154]]]",
    "process": "Let the angle of inclination of line AP be $\\theta$. In right triangle $PAF$, from the given conditions we have $\\tan\\theta = \\frac{b^{2}}{a+c} = \\frac{\\sqrt{3}}{3}$. Rearranging gives $3b^{2} = \\sqrt{3}(a^{2}+ac)$, that is, $3(a^{2}-c^{2}) = \\sqrt{3}(a^{2}+ac)$. This leads to $3e^{2} + \\sqrt{3}e - 3 + \\sqrt{3} = 0$. Solving yields $e = -1$ (discarded), $e = \\frac{3-\\sqrt{3}}{3}$."
  },
  {
    "text": "Let $F$ be the focus of the parabola $C$: $y^{2}=4 x$. A line $l$ passing through $F$ intersects the parabola $C$ at points $A$ and $B$. When $|A B|=6$, what is the length of the chord obtained by the intersection of the circle with diameter $A B$ and the $y$-axis?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, B)) = 6;G: Circle;IsDiameter(LineSegmentOf(A, B), G)",
    "query_expressions": "Length(InterceptChord(yAxis, G))",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[5, 24], [39, 45]], [[5, 24]], [[1, 4], [29, 32]], [[1, 27]], [[33, 38]], [[28, 38]], [[46, 49]], [[50, 53]], [[33, 55]], [[57, 66]], [[78, 79]], [[68, 79]]]",
    "query_spans": "[[[78, 92]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), \\therefore |AB| = x_{1} + 1 + x_{2} + 1 = 6 \\Rightarrow x_{1} + x_{2} = 4, \\therefore the distance from the center of the circle with AB as diameter to the chord formed by intersection with the y-axis is 2, \\therefore the required chord length is 2\\sqrt{3^{2} - 2^{2}} = 2\\sqrt{5}"
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ with focus $F$, a line passing through point $F$ intersects the parabola at points $A$ and $B$, and $|F A||F B|=6$, $|A B|=6$. Then $p=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);Intersection(H, G) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(F, A))*Abs(LineSegmentOf(F, B)) = 6;Abs(LineSegmentOf(A, B)) = 6",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 23], [40, 43]], [[2, 23]], [[82, 85]], [[5, 23]], [[27, 30], [32, 36]], [[2, 30]], [[37, 39]], [[31, 39]], [[37, 53]], [[44, 47]], [[48, 51]], [[55, 69]], [[71, 80]]]",
    "query_spans": "[[[82, 87]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, the equation of the line is $ y = k\\left(x - \\frac{p}{2}\\right) $. From the given conditions, $ k $ exists. Combining the system:  \n$$\n\\begin{cases}\ny = k\\left(x - \\frac{p}{2}\\right) \\\\\n2 - 2ny\n\\end{cases}\n$$\nSimplifying yields:  \n$ k^{2}x^{2} - (pk^{2} + 2p)x + \\frac{1}{4}k^{2}p^{2} = 0 $,  \nthen $ x_{1} + x_{2} = \\frac{pk^{2} + 2p}{k^{2}} $, $ x_{1}x_{2} = \\frac{1}{4}p^{2} $. By the definition of the parabola, $ AF = x_{1} + \\frac{p}{2} $, $ BF = x_{2} + \\frac{p}{2} $. Substituting into Vieta's formulas gives:  \n$ \\frac{n^{2}}{4} + \\frac{p}{2}(6 - p) + \\frac{p^{2}}{4} = 6 $, solving yields $ p = 2 $."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x  (p>0)$, the focus is $F$, $O$ is the origin, and $M$ is a point on the parabola such that $|M F|=4|O F|$, and the area of $\\Delta M F O$ is $4 \\sqrt{3}$. Then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;O: Origin;M: Point;PointOnCurve(M, G);Abs(LineSegmentOf(M, F)) = 4*Abs(LineSegmentOf(O, F));Area(TriangleOf(M, F, O)) = 4*sqrt(3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=8*x",
    "fact_spans": "[[[2, 25], [46, 49], [103, 106]], [[2, 25]], [[5, 25]], [[5, 25]], [[29, 32]], [[2, 32]], [[33, 36]], [[42, 45]], [[42, 52]], [[54, 68]], [[70, 100]]]",
    "query_spans": "[[[103, 111]]]",
    "process": ""
  },
  {
    "text": "Let the line $y=k(x+3)$ intersect the parabola $y=ax^{2}$ at points $A(x_{1} , y_{1})$ and $B(x_{2} , y_{2})$. Then the value of $\\frac{1}{x_{1}}+\\frac{1}{x_{2}}$ is?",
    "fact_expressions": "G: Parabola;H: Line;a: Number;A: Point;B: Point;k: Number;x1: Number;y1: Number;x2: Number;y2: Number;Expression(G) = (y = a*x^2);Expression(H) = (y = k*(x + 3));Coordinate(B) = (x2, y2);Coordinate(A) = (x1, y1);Intersection(H, G) = {A, B}",
    "query_expressions": "1/x2 + 1/x1",
    "answer_expressions": "-1/3",
    "fact_spans": "[[[14, 27]], [[1, 13]], [[17, 27]], [[29, 47]], [[48, 66]], [[3, 13]], [[29, 47]], [[29, 47]], [[48, 66]], [[48, 66]], [[14, 27]], [[1, 13]], [[48, 66]], [[29, 47]], [[1, 68]]]",
    "query_spans": "[[[70, 107]]]",
    "process": ""
  },
  {
    "text": "A line $ l $ with slope $ -\\frac{1}{3} $ intersects the ellipse $ C $: $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $) at points $ A $ and $ B $. The midpoint of segment $ AB $ has coordinates $ (1,1) $. What is the eccentricity of ellipse $ C $?",
    "fact_expressions": "l: Line;C: Ellipse;b: Number;a: Number;A: Point;B: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Slope(l) = -1/3;Intersection(l, C) = {A, B};Coordinate(MidPoint(LineSegmentOf(A,B))) = (1, 1)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[18, 23]], [[24, 83], [118, 123]], [[31, 83]], [[31, 83]], [[86, 89]], [[90, 93]], [[31, 83]], [[31, 83]], [[24, 83]], [[0, 23]], [[18, 95]], [[96, 116]]]",
    "query_spans": "[[[118, 130]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ \\frac{x_{1}^{2}}{a^{2}}+\\frac{y_{1}^{2}}{b^{2}}=1 $\\textcircled{1}, $ \\frac{x_{2}^{2}}{a^{2}}+\\frac{y_{2}^{2}}{b^{2}}=1 $\\textcircled{2}. Since $ (1,1) $ is the midpoint of segment $ AB $, $ \\frac{1}{2}(x_{1}+x_{2})=1 $, $ \\frac{1}{2}(y_{1}+y_{2})=1 $. Since the equation of line $ AB $ is $ y=-\\frac{1}{3}(x-1)+1 $, $ y_{1}-y_{2}=-\\frac{1}{3}(x_{1}-x_{2}) $. Subtracting equations \\textcircled{1} and \\textcircled{2} gives: $ \\frac{1}{a^{2}}(x_{1}^{2}-x_{2}^{2})+\\frac{1}{b^{2}}(y_{1}^{2}-y_{2}^{2})=0 $, $ \\therefore \\frac{1}{a^{2}}(x_{1}-x_{2})(x_{1}+x_{2})+\\frac{1}{b^{2}}(y_{1}-y_{2})(y_{1}+y_{2})=0 $, $ \\therefore 2\\times\\frac{1}{a^{2}}(x_{1}-x_{2})+2\\times\\frac{1}{b^{2}}(y_{1}-y_{2})=0 $, $ \\therefore \\frac{b^{2}}{a^{2}}=\\frac{1}{3} $, $ \\therefore e^{2}=1-\\frac{b^{2}}{a^{2}}=\\frac{2}{3} $, $ e=\\frac{\\sqrt{6}}{3} $"
  },
  {
    "text": "The number of common points between the line $y=x+3$ and the curve $\\frac{y^{2}}{9}-\\frac{x|x|}{4}=1$ is?",
    "fact_expressions": "G: Line;H: Curve;Expression(G) = (y = x + 3);Expression(H) = (-x*Abs(x)/4 + y^2/9 = 1)",
    "query_expressions": "NumIntersection(G, H)",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 9]], [[10, 46]], [[0, 9]], [[10, 46]]]",
    "query_spans": "[[[0, 55]]]",
    "process": "When $ x \\geqslant 0 $, the equation of the curve $ \\frac{y^{2}}{9} - \\frac{x|x|}{4} = 1 $ is $ \\frac{y^{2}}{9} - \\frac{x^{2}}{4} = 1 $. When $ x < 0 $, the equation of the curve $ \\frac{y^{2}}{9} - \\frac{x|x|}{4} = 1 $ is $ \\frac{y^{2}}{9} + \\frac{x^{2}}{4} = 1 $. The graph of $ |x| = 1 $ is shown in the figure on the right. Drawing the graph of the line $ v = x + 3 $ in the coordinate system, the number of intersection points between the line and the curve is 3."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2x$ with focus $F$, point $P$ lies on parabola $C$ such that $|PF|=a$ $(a>\\frac{1}{2})$. Let the directrix of parabola $C$ be $l$. Draw $PM \\perp l$ from point $P$, with foot of perpendicular at $M$. If the area of the incircle of $\\Delta MPF$ is $\\frac{(3-\\sqrt{5})a\\pi}{4}$, then $a=?$",
    "fact_expressions": "C: Parabola;F: Point;Expression(C) = (y^2 = 2*x);Focus(C) = F;P: Point;PointOnCurve(P, C);a: Number;Abs(LineSegmentOf(P, F)) = a;a > 1/2;l: Line;Directrix(C) = l;M: Point;PointOnCurve(P,LineSegmentOf(P,M));IsPerpendicular(LineSegmentOf(P,M),l);FootPoint(LineSegmentOf(P,M),l) = M;Area(InscribedCircle(TriangleOf(M,P,F))) = (3-sqrt(5))*a*pi/4",
    "query_expressions": "a",
    "answer_expressions": "5/2",
    "fact_spans": "[[[2, 21], [34, 40], [69, 75]], [[25, 28]], [[2, 21]], [[2, 28]], [[29, 33], [84, 88]], [[29, 41]], [[165, 168]], [[43, 67]], [[43, 67]], [[79, 82]], [[69, 82]], [[106, 109]], [[83, 102]], [[89, 102]], [[89, 109]], [[111, 163]]]",
    "query_spans": "[[[165, 170]]]",
    "process": "From the given conditions, we have $ F\\left(\\frac{1}{2},0 \\right) $, line $ l: x = -\\frac{1}{2} $. Without loss of generality, assume point $ P(x_0, y_0) $ lies in the first quadrant, so $ x_0 > 0 $, $ y_0 > 0 $. Since $ |PF| = a $ ($ a > \\frac{1}{2} $), it follows that $ |PM| = |PF| = x_0 + \\frac{1}{2} = a $, so $ y_0 = \\sqrt{2x_0} = \\sqrt{2a - 1} $. Then $ \\frac{1}{2}(|PM| + |PF| + |MF|)r = \\frac{1}{2}y_0 \\cdot |PM| $, i.e., $ (2a + \\sqrt{2a})r = a\\sqrt{2a - 1} $, so $ r = \\frac{a\\sqrt{2a - 1}}{2a + \\sqrt{2a}} = \\frac{\\sqrt{a} \\cdot \\sqrt{2a - 1}}{\\sqrt{2}(\\sqrt{2a} + 1)} $. Thus, the area of the incircle of $ \\triangle MPF $ is $ \\pi r^2 = \\frac{\\pi a(2a - 1)}{2(\\sqrt{2a} + 1)^2} = \\frac{\\pi a(\\sqrt{2a} - 1)}{2(\\sqrt{2a} + 1)} $. Also, the area of the incircle of $ \\triangle MPF $ is $ \\frac{(3 - \\sqrt{5})a\\pi}{4} $, so $ \\frac{\\pi a(\\sqrt{2a} - 1)}{2(\\sqrt{2a} + 1)} = \\frac{(3 - \\sqrt{5})a\\pi}{4} $, i.e., $ \\frac{\\sqrt{2a} - 1}{\\sqrt{2a} + 1} = \\frac{3 - \\sqrt{5}}{2} $. Solving gives $ \\sqrt{2a} = \\sqrt{5} $, so $ a = \\frac{5}{2} $."
  },
  {
    "text": "The line $x+2 y+\\sqrt{5}=0$ intersects the circle $x^{2}+y^{2}=2$ at points $A$ and $B$, and $O$ is the origin. Then $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}$=?",
    "fact_expressions": "G: Circle;H: Line;O: Origin;A: Point;B: Point;Expression(G) = (x^2 + y^2 = 2);Expression(H) = (x + 2*y + sqrt(5) = 0);Intersection(H, G) = {A, B}",
    "query_expressions": "DotProduct(VectorOf(O, A), VectorOf(O, B))",
    "answer_expressions": "0",
    "fact_spans": "[[[21, 37]], [[0, 20]], [[50, 53]], [[40, 43]], [[44, 47]], [[21, 37]], [[0, 20]], [[0, 49]]]",
    "query_spans": "[[[58, 109]]]",
    "process": ""
  },
  {
    "text": "Let the semi-focal length of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ be $c$, and the distance from the origin to the line $b x+a y=a b$ equals $\\frac{1}{4} c+1$. Then the minimum value of $c$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;HalfFocalLength(G) = c;c: Number;H: Line;Expression(H) = (a*y + b*x = a*b);O: Origin;Distance(O, H) = c/4 + 1",
    "query_expressions": "Min(c)",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 57]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 64]], [[61, 64], [109, 112]], [[70, 85]], [[70, 85]], [[65, 69]], [[65, 107]]]",
    "query_spans": "[[[109, 118]]]",
    "process": "According to the problem, $\\frac{ab}{\\sqrt{a^{2}+b^{2}}}=\\frac{ab}{c}=\\frac{c}{4}+1$, $\\therefore ab=\\frac{1}{4}c^{2}+c$, $\\because ab\\leqslant\\frac{a^{2}+b^{2}}{2}=\\frac{c^{2}}{2}$, $\\therefore \\frac{1}{4}c^{2}+c\\leqslant\\frac{c^{2}}{2}$, solving gives $c\\geqslant4$ or $c\\leqslant0$ (discarded)"
  },
  {
    "text": "If the chord of the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$ is bisected by the point $(4,2)$, then what is the slope of the line containing this chord?",
    "fact_expressions": "G: Ellipse;H: LineSegment;I: Point;Expression(G) = (x^2/36 + y^2/9 = 1);Coordinate(I) = (4, 2);IsChordOf(H, G);MidPoint(H) = I",
    "query_expressions": "Slope(OverlappingLine(H))",
    "answer_expressions": "-1/2",
    "fact_spans": "[[[1, 39]], [], [[42, 50]], [[1, 39]], [[42, 50]], [[1, 41]], [[1, 52]]]",
    "query_spans": "[[[1, 65]]]",
    "process": ""
  },
  {
    "text": "Point $A(3 , 2)$ is a fixed point, point $F$ is the focus of the parabola $y^{2}=4 x$, and point $P$ moves along the parabola $y^{2}=4 x$. If $|PA|+|PF|$ attains its minimum value, then the coordinates of point $P$ are?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (3, 2);Focus(G) = F;PointOnCurve(P, G);WhenMin(Abs(LineSegmentOf(P, A))+Abs(LineSegmentOf(P, F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1, 2)",
    "fact_spans": "[[[20, 34], [43, 57]], [[0, 11]], [[38, 42], [80, 84]], [[15, 19]], [[20, 34]], [[0, 11]], [[15, 37]], [[38, 58]], [[62, 78]]]",
    "query_spans": "[[[80, 89]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, a perpendicular line $F H$ is drawn from point $F$ to an asymptote (point $H$ being the foot of the perpendicular), intersecting the right branch of the hyperbola at point $A$. If $A$ is the midpoint of segment $F H$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Point;F: Point;A: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;PointOnCurve(F,LineSegmentOf(F,H));IsPerpendicular(LineSegmentOf(F,H),Asymptote(G));FootPoint(LineSegmentOf(F,H),Asymptote(G))=H;Intersection(LineSegmentOf(F,H),RightPart(C))=A;MidPoint(LineSegmentOf(F,H))=A",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[6, 62], [96, 99], [125, 128]], [[9, 62]], [[9, 62]], [[85, 89]], [[2, 5], [68, 72]], [[103, 107], [109, 112]], [[9, 62]], [[9, 62]], [[6, 62]], [[2, 66]], [[67, 84]], [[6, 84]], [[6, 93]], [[79, 107]], [[109, 123]]]",
    "query_spans": "[[[125, 134]]]",
    "process": "By the given condition, assume that from the right focus $ F $, a perpendicular line $ FH $ is drawn to the asymptote $ y = \\frac{b}{a}x $. The equation of $ FH $ is $ y = -\\frac{a}{b}(x - c) $. Solving the system of equations  \n\\[\n\\begin{cases}\ny = -\\frac{a}{b}(x - c) \\\\\ny = \\frac{b}{a}x\n\\end{cases}\n\\]\nyields $ x = \\frac{a^{2}}{c} $, $ y = \\frac{ab}{c} $, so the coordinates of point $ H $ are $ \\left( \\frac{a^{2}}{c}, \\frac{ab}{c} \\right) $. Using the midpoint formula, point $ A $ is $ \\left( \\frac{a^{2} + c^{2}}{2c}, \\frac{ab}{2c} \\right) $. Substituting into the hyperbola equation gives  \n\\[\n\\frac{(a^{2} + c^{2})^{2}}{4a^{2}c^{2}} - \\frac{a^{2}}{4c^{2}} = 1\n\\]  \nSimplifying yields $ \\frac{c^{2}}{a^{2}} = 2 $, i.e., $ e^{2} = 2 $, so $ e = \\sqrt{2} $."
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $4 x^{2}-y^{2}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (4*x^2 - y^2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*2*x",
    "fact_spans": "[[[0, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 28]]]",
    "process": "Given the hyperbola $4x^{2} - y^{2} = 1$, let $4x^{2} - y^{2} = 0$; the asymptotes are then given by: $y = \\pm 2x$"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(b>a>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through the right focus and parallel to one asymptote intersects the hyperbola at point $A$. If the inradius of $\\Delta A F_{1} F_{2}$ is $\\frac{b}{4}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;I: Line;A: Point;F1: Point;F2: Point;b > a;a > 0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(RightFocus(G), I);IsParallel(I, OneOf(Asymptote(G)));Intersection(I, G) = A;Radius(InscribedCircle(TriangleOf(A, F1, F2))) = b/4",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[2, 55], [98, 101], [152, 155]], [[5, 55]], [[5, 55]], [[95, 97]], [[102, 106]], [[65, 72]], [[73, 80]], [[5, 55]], [[5, 55]], [[2, 55]], [[2, 80]], [[2, 80]], [[2, 97]], [[2, 97]], [[95, 106]], [[108, 150]]]",
    "query_spans": "[[[152, 161]]]",
    "process": "The left and right foci of the hyperbola are $F_{1}(-c,0)$, $F_{2}(c,0)$, respectively. Let an asymptote of the hyperbola be $y=\\frac{b}{a}x$. Then the equation of line $AF_{2}$ is $y=\\frac{b}{a}(x-c)$. Solving this together with the hyperbola equation $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(b>a>0)$, we obtain the coordinates of point $A$: $A\\left(\\frac{c^{2}+a^{2}}{2c},\\frac{b(a^{2}-c^{2})}{2ac}\\right)$. Let $|AF_{1}|=m$, $|AF_{2}|=n$. Using the equal area method for triangles, we have $\\frac{1}{2}\\cdot\\frac{b}{4}(m+n+2c)=\\frac{1}{2}\\cdot2c\\cdot\\frac{b(c^{2}-a^{2})}{2ac}$, which simplifies to $m+n=\\frac{4c^{2}}{a}-4a-2c$ $\\textcircled{1}$. By the definition of a hyperbola, $m-n=2a$ $\\textcircled{2}$. In triangle $AF_{1}F_{2}$, $n\\sin\\theta=\\frac{b(c^{2}-a^{2})}{2ac}$, where $\\theta$ is the inclination angle of line $AF_{2}$. Since $\\tan\\theta=\\frac{b}{a}$ and $\\sin^{2}\\theta+\\cos^{2}\\theta=1$, we get $\\sin\\theta=\\frac{b}{\\sqrt{a^{2}+b^{2}}}=\\frac{b}{c}$. Thus, $n=\\frac{c^{2}-a^{2}}{2a}$ $\\textcircled{3}$. From equations $\\textcircled{1}$, $\\textcircled{2}$, and $\\textcircled{3}$, simplifying yields $3c^{2}-2ac-5a^{2}=0$, or $(3c-5a)(c+a)=0$. Therefore, $3c=5a$, and hence $e=\\frac{c}{a}=\\frac{5}{3}$."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=9x$ is $F$, its directrix intersects the $x$-axis at point $M$, and $N$ is a point on the parabola satisfying $\\sqrt{6}|NF|=2|MN|$, then the distance from point $F$ to the line $MN$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 9*x);F: Point;Focus(G) = F;Intersection(Directrix(G), xAxis) = M;M: Point;PointOnCurve(N, G) = True;N: Point;sqrt(6)*Abs(LineSegmentOf(N, F)) = 2*Abs(LineSegmentOf(M, N))",
    "query_expressions": "Distance(F, LineOf(M, N))",
    "answer_expressions": "3*sqrt(3)/2",
    "fact_spans": "[[[2, 16], [24, 25], [44, 47]], [[2, 16]], [[20, 23], [79, 83]], [[2, 23]], [[24, 39]], [[35, 39]], [[40, 51]], [[40, 43]], [[55, 77]]]",
    "query_spans": "[[[79, 96]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{2}=1$ has foci $F_{1}$ and $F_{2}$, respectively. A point $P$ lies on the ellipse such that $|P F_{1}|=2|P F_{2}|$. Then, the area of triangle $F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 + y^2/2 = 1);Focus(G)={F1,F2};PointOnCurve(P,G);Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Area(TriangleOf(F1,P,F2))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 39], [68, 70]], [[63, 67]], [[45, 52]], [[55, 62]], [[2, 39]], [[2, 61]], [[63, 71]], [[73, 95]]]",
    "query_spans": "[[[97, 120]]]",
    "process": ""
  },
  {
    "text": "Point $A(x, y)$ lies on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$. What is the maximum value of $x-2y$?",
    "fact_expressions": "G: Ellipse;A: Point;x1:Number;y1:Number;Expression(G) = (x^2/9 + y^2/4 = 1);Coordinate(A) = (x1, y1);PointOnCurve(A, G)",
    "query_expressions": "Max(x1 - 2*y1)",
    "answer_expressions": "5",
    "fact_spans": "[[[11, 48]], [[0, 10]], [[1, 10]], [[1, 10]], [[11, 48]], [[0, 10]], [[0, 51]]]",
    "query_spans": "[[[53, 65]]]",
    "process": ""
  },
  {
    "text": "Through the focus of the parabola $y^{2}=4x$, draw a straight line with inclination angle $\\frac{\\pi}{4}$ intersecting the parabola at points $P$ and $Q$, and let $O$ be the origin. Then the area of $\\triangle POQ$ is?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;O: Origin;Q: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), H);Inclination(H)=pi/4;Intersection(H, G) = {P, Q}",
    "query_expressions": "Area(TriangleOf(P, O, Q))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[1, 15], [43, 46]], [[40, 42]], [[47, 50]], [[57, 60]], [[51, 54]], [[1, 15]], [[0, 42]], [[20, 42]], [[40, 56]]]",
    "query_spans": "[[[67, 89]]]",
    "process": "Let P(x_{1},y_{1}), Q(x_{2},y_{2}), then S=\\frac{1}{2}|OF||y_{1}-y_{2}|. The line passing through the focus (1,0) of the parabola y^{2}=4x with inclination angle \\frac{\\pi}{4} is x-y-1=0, i.e., x=1+y. Substituting x=1+y into y^{2}=4x yields: y^{2}=4(1+y), i.e., y^{2}-4y-4=0, so y_{1}+y_{2}=4, y_{1}y_{2}=-4, \\therefore |y_{1}-y_{2}|=4\\sqrt{2}. Thus, S=\\frac{1}{2}|OF||y_{1}-y_{2}|=\\frac{1}{2}\\times1\\times4\\sqrt{2}=2\\sqrt{2}"
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{y^{2}}{9}+\\frac{x^{2}}{4}=1$, the two foci of the ellipse are $F_{1}$ and $F_{2}$, and $\\cos \\angle F_{1} P F_{2}=\\frac{1}{3}$, then the distance from point $P$ to the $y$-axis is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 + y^2/9 = 1);PointOnCurve(P, G);Focus(G) = {F1,F2};Cos(AngleOf(F1, P, F2)) = 1/3",
    "query_expressions": "Distance(P, yAxis)",
    "answer_expressions": "2*sqrt(10)/5",
    "fact_spans": "[[[6, 43], [47, 49]], [[57, 64]], [[115, 119], [2, 5]], [[65, 72]], [[6, 43]], [[2, 46]], [[47, 72]], [[74, 113]]]",
    "query_spans": "[[[115, 129]]]",
    "process": "From the ellipse $\\frac{y^{2}}{9}+\\frac{x^{2}}{4}=1$, we obtain $|PF_{1}|+|PF_{2}|=6$, $|F_{1}F_{2}|=2\\sqrt{5}$, $e=\\frac{c}{a}=\\frac{\\sqrt{5}}{3}$, $\\therefore\\cos\\angle F_{1}PF_{2}=\\frac{|PF_{1}|^{2}+|PF_{2}|^{2}-|F_{1}F_{2}|^{2}}{2|PF_{1}||PF_{2}|}=\\frac{16}{2|PF_{1}||PF_{2}|}-1=\\frac{1}{3}$, we get $|PF_{1}|\\cdot|PF_{2}|=6$. Without loss of generality, assume $|PF_{1}|>|PF_{2}|$, the abscissa of $P$ is $x$, $x>0$, $\\therefore|PF_{1}|=3+\\sqrt{3}$, $|PF_{2}|=3-\\sqrt{3}$, $3+\\sqrt{3}=3+\\frac{\\sqrt{5}}{3}y$, solving gives $y=\\frac{3\\sqrt{15}}{5}$, at this time $x=\\pm\\frac{2\\sqrt{10}}{5}$"
  },
  {
    "text": "A point $P$ on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ is at a distance of $\\frac{3}{2}$ from the right focus $F$. What is the distance from $P$ to the left directrix?",
    "fact_expressions": "G: Ellipse;P: Point;F: Point;Expression(G) = (x^2/4 + y^2 = 1);PointOnCurve(P, G);RightFocus(G)=F;Distance(P, F) = 3/2",
    "query_expressions": "Distance(P,LeftDirectrix(G))",
    "answer_expressions": "5*sqrt(3)/3",
    "fact_spans": "[[[0, 27]], [[30, 33], [60, 63]], [[37, 40]], [[0, 27]], [[0, 33]], [[0, 40]], [[30, 57]]]",
    "query_spans": "[[[0, 72]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. If there exists a point $P$ on $C$ such that $PF_{1} \\perp PF_{2}$ and $\\angle P F_{1} F_{2}=30^{\\circ}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));AngleOf(P, F1, F2) = ApplyUnit(30, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[18, 75], [83, 86], [154, 157]], [[18, 75]], [[25, 75]], [[25, 75]], [[25, 75]], [[25, 75]], [[1, 8]], [[10, 17]], [[1, 80]], [[91, 94]], [[82, 94]], [[96, 117]], [[119, 152]]]",
    "query_spans": "[[[154, 163]]]",
    "process": ""
  },
  {
    "text": "If the line $m x - n y = 4$ has no intersection points with the circle $x^{2} + y^{2} = 4$, then the number of intersection points between the line passing through point $P(m, n)$ and the ellipse $\\frac{x^{2}}{9} + \\frac{y^{2}}{4} = 1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G)=(x^2/9+y^2/4=1);H:Line;Expression(H)=(m*x-n*y=4);m:Number;n:Number;C:Circle;Expression(C)=(x^2+y^2=4);NumIntersection(H,C)=0;P:Point;Coordinate(P)=(m,n);L:Line;PointOnCurve(P,L)",
    "query_expressions": "NumIntersection(L,G)",
    "answer_expressions": "2",
    "fact_spans": "[[[52, 89]], [[52, 89]], [[1, 14]], [[1, 14]], [[3, 14]], [[3, 14]], [[15, 31]], [[15, 31]], [[1, 35]], [[38, 48]], [[38, 48]], [[49, 51]], [[37, 51]]]",
    "query_spans": "[[[49, 96]]]",
    "process": "First, from the given conditions we obtain m^{2}+n^{2}<4, thereby concluding that P(m,n) lies inside the ellipse. The number of intersection points between the line and the ellipse is determined as follows: Given that the distance d from the center (0,0) to the line mx-ny-4=0 is d=\\frac{4}{\\sqrt{m^{2}+n^{2}}}>2, simplifying yields: m^{2}+n^{2}<4. Thus, P(m,n) lies within a circle centered at (0,0) with radius 2. Moreover, since the ellipse \\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1 has a=3, b=2, P(m,n) lies inside the ellipse. Therefore, the line passing through point P(m,n) intersects the ellipse \\frac{x^{2}}{}+\\frac{y^{2}}{}=1 at 2 points. Hence, the answer is: ? Qing. This question primarily examines the relationship between lines and ellipses, points and ellipses, as well as the relationship between lines and circles, and belongs to a simple problem."
  },
  {
    "text": "The hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1(a>0)$ has an asymptote whose angle of inclination is $60^{\\circ}$, and $F_{1}$, $F_{2}$ are the left and right foci. If the line $x=2$ intersects the hyperbola $C$ at point $P$, then the perimeter of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "C: Hyperbola;a: Number;G: Line;P: Point;F1: Point;F2: Point;a>0;Expression(C) = (-y^2/3 + x^2/a^2 = 1);Expression(G) = (x = 2);Inclination(OneOf(Asymptote(C))) = ApplyUnit(60, degree);LeftFocus(C) =F1;RightFocus(C)=F2;Intersection(G, C) = P",
    "query_expressions": "Perimeter(TriangleOf(P, F1, F2))",
    "answer_expressions": "12",
    "fact_spans": "[[[0, 52], [111, 117]], [[8, 52]], [[103, 110]], [[119, 123]], [[78, 86]], [[88, 95]], [[8, 52]], [[0, 52]], [[103, 110]], [[0, 75]], [[0, 101]], [[0, 101]], [[103, 123]]]",
    "query_spans": "[[[125, 152]]]",
    "process": "Since one asymptote of the hyperbola $ C:\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1 $ ($ a>0 $) has an inclination angle of $ 60^{\\circ} $, then $ \\frac{\\sqrt{3}}{a}=\\tan60^{\\circ}=\\sqrt{3} $, yielding $ a=1 $. Therefore, the focal distance of the hyperbola $ C $ is $ |F_{1}F_{2}|=2\\sqrt{a^{2}+3}=4 $. Let point $ P $ be a point in the first quadrant. Solving the system $ \\begin{cases}x=2\\\\x^{2}-\\frac{y^{2}}{3}=1\\end{cases} $, since $ y>0 $, we obtain $ \\begin{cases}x=2\\\\y=3\\end{cases} $. It is clear that $ F_{1}(-2,0) $, $ F_{2}(2,0) $, thus $ |PF_{1}|=\\sqrt{(2+2)^{2}+3^{2}}=5 $, $ |PF_{2}|=\\sqrt{(2-2)^{2}+3^{2}}=3 $. Hence, the perimeter of $ \\triangle PF_{1}F_{2} $ is $ |PF_{1}|+|PF_{2}|+|F_{1}F_{2}|=5+3+4=12 $."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=2 p x(p>0)$, a line with slope $1$ passing through $F$ intersects the parabola $C$ at points $A$ and $B$, and $|F A|>|F B|$, then $\\frac{|F A|}{|F B|}$=?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;F: Point;A: Point;B: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(F, G);Slope(G) = 1;Intersection(G, C) = {A, B};Abs(LineSegmentOf(F, A)) > Abs(LineSegmentOf(F, B))",
    "query_expressions": "Abs(LineSegmentOf(F, A))/Abs(LineSegmentOf(F, B))",
    "answer_expressions": "3+2*sqrt(2)",
    "fact_spans": "[[[6, 32], [51, 57]], [[14, 32]], [[48, 50]], [[2, 5], [37, 40]], [[58, 61]], [[62, 65]], [[14, 32]], [[6, 32]], [[2, 35]], [[36, 50]], [[41, 50]], [[48, 67]], [[69, 82]]]",
    "query_spans": "[[[84, 107]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}) be given by \\begin{cases}y=x-\\frac{p}{2}\\\\y^{2}=2px\\end{cases}, then we obtain x^{2}-3px+\\frac{p^{2}}{4}=0, (x_{1}>x_{2}), x_{1}=\\frac{3+2\\sqrt{2}}{2}p, x_{2}=\\frac{3-2\\sqrt{2}}{2}p. Therefore, by the definition of the parabola, \\frac{|FA|}{|FB|}=\\frac{x_{1}+\\frac{p}{2}}{x_{2}+\\frac{p}{2}}=3+2\\sqrt{2}"
  },
  {
    "text": "Given that point $M$ is a point on the parabola $C$: $y^{2}=8x$, and $F$ is the focus of parabola $C$, what is the length of the chord cut from the circle centered at $M$ with radius $|MF|=4$ by the line $x=-1$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 8*x);G: Circle;F:Point;Focus(C) = F;H: Line;M: Point;Center(G)=M;PointOnCurve(M, C);Expression(H) = (x = -1);Radius(G)=Abs(LineSegmentOf(M,F));Abs(LineSegmentOf(M,F))=4",
    "query_expressions": "Length(InterceptChord(H,G))",
    "answer_expressions": "2*sqrt(7)",
    "fact_spans": "[[[7, 26], [34, 40]], [[7, 26]], [[66, 67]], [[30, 33]], [[30, 43]], [[68, 76]], [[2, 6], [46, 49]], [[45, 67]], [[2, 29]], [[68, 76]], [[53, 67]], [[53, 62]]]",
    "query_spans": "[[[66, 83]]]",
    "process": "As shown in the figure, according to the properties of the parabola, we have MF = MQ = 4 = 2 + x_{M}, yielding M(2,4). Therefore, the distance from M to x = 1 is 3. According to the chord length formula for a line and a circle, the chord length is: 2.\\sqrt{16 - 9} = 2\\sqrt{7}."
  },
  {
    "text": "Given that the equation of the ellipse is $x^{2}+2 y^{2}-4=0$, then the equation of the line containing the chord with midpoint $M(1,1)$ is?",
    "fact_expressions": "G: Ellipse;H: LineSegment;M: Point;Coordinate(M) = (1, 1);Expression(G) = (x^2 + 2*y^2 - 4 = 0);IsChordOf(H, G);MidPoint(H) = M",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "x+2*y-3=0",
    "fact_spans": "[[[2, 4]], [], [[30, 38]], [[30, 38]], [[2, 27]], [[2, 43]], [[2, 43]]]",
    "query_spans": "[[[2, 51]]]",
    "process": "Let the endpoints of the chord be A(x_{1},y_{1}) and B(x_{2},y_{2}). Since point M is the midpoint of chord AB, we have \\begin{cases}x_{1}+x_{2}=2\\\\y_{1}+y_{2}=2\\end{cases}. Substituting the coordinates of points A and B into the ellipse equation gives \\begin{cases}x_{1}^{2}+2y_{1}^{2}=4\\\\x_{2}^{2}+2y_{2}^{2}=4\\end{cases}. Using the point difference method, the slope of the line containing chord AB can be found. Finally, using the point-slope form, the equation of the line containing chord AB can be obtained. Let the endpoints of the chord be A(x_{1},y_{1}) and B(x_{2},y_{2}). Since point M is the midpoint of chord AB, then \\begin{cases}\\frac{x_{1}+x_{2}}{2}=1\\\\\\frac{y_{1}+y_{2}}{2}=1\\end{cases}, which yields \\begin{cases}x_{1}+x_{2}=2\\\\y_{1}+y_{2}=2\\end{cases}. Substituting the coordinates of points A and B into the ellipse equation gives \\begin{cases}x_{1}^{2}+2y_{1}^{2}=4\\\\x_{2}^{2}+2y_{2}^{2}=4\\end{cases}. Subtracting these two equations gives (x_{1}^{2}-x_{2}^{2})+2(y_{1}^{2}-y_{2}^{2})=0, that is, (x_{1}+x_{2})(x_{1}-x_{2})+2(y_{1}+y_{2})(y_{1}-y_{2})=0, which leads to 2(x_{1}-x_{2})+4(y_{1}-y_{2})=0. Therefore, the slope of line AB is \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{1}{2}. Hence, the equation of the line containing chord AB is y-1=-\\frac{1}{2}(x-1), that is, x+2y-3=0."
  },
  {
    "text": "Given the parabola $C$: $x^{2}=2 p y(p>0)$, its directrix intersects the coordinate axis at point $A$. If a line passing through point $A$ is tangent to the parabola $C$ at point $B$, and $|A B|=2$, then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;A: Point;B: Point;p>0;Expression(C) = (x^2 = 2*p*y);Intersection(Directrix(C),axis)=A;PointOnCurve(A,G);TangentPoint(G,C)=B;Abs(LineSegmentOf(A, B))=2",
    "query_expressions": "p",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 28], [52, 58]], [[78, 81]], [[49, 51]], [[37, 41], [44, 48]], [[61, 65]], [[10, 28]], [[2, 28]], [[2, 41]], [[43, 51]], [[49, 65]], [[67, 76]]]",
    "query_spans": "[[[78, 83]]]",
    "process": "From the given conditions, the directrix of parabola $ C $ is $ y = -\\frac{p}{2} $, point $ A(0, -\\frac{p}{2}) $, and the slope $ k $ of the tangent line must exist. Let the equation of the tangent line be $ y = kx - \\frac{p}{2} $, and the point of tangency be $ B(x_0, y_0) $. Solving simultaneously the equations of parabola $ C $ and the tangent line:\n$$\n\\begin{cases}\nx^2 = 2py, \\\\\ny = kx - \\frac{p}{2},\n\\end{cases}\n$$\neliminating $ y $ gives $ x^2 - 2pkx + p^2 = 0 $. From $ \\Delta = 4p^2k^2 - 4p^2 = 0 $, we solve to get $ k = \\pm1 $. When $ k = 1 $, then $ x^2 - 2px + p^2 = 0 $, yielding $ x_0 = p $, so $ y_0 = \\frac{p}{2} $. Since $ |AB| = 2 $, we have $ x_0^2 + (y_0 + \\frac{p}{2})^2 = 4 $, solving gives $ p = \\sqrt{2} $. When $ k = -1 $, similarly we obtain $ p = \\sqrt{2} $."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=8x$ be $F$, the directrix be $l$, and let $P$ be a point on the parabola with $PA \\perp l$, where $A$ is the foot of the perpendicular. If the angle of inclination of the line $AF$ is $\\frac{2\\pi}{3}$, then $|PF|=$?",
    "fact_expressions": "G: Parabola;A: Point;F: Point;P: Point;l: Line;Expression(G) = (y^2 = 8*x);Focus(G) = F;Directrix(G)=l;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P,A),l);FootPoint(LineSegmentOf(P,A),l)=A;Inclination(LineOf(A,F)) = (2*pi)/3",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 15], [35, 38]], [[56, 60]], [[19, 22]], [[30, 34]], [[26, 29]], [[1, 15]], [[1, 22]], [[1, 29]], [[31, 41]], [[42, 55]], [[42, 63]], [[67, 96]]]",
    "query_spans": "[[[100, 109]]]",
    "process": "According to the problem, by setting P(m,n) and using the fact that △APF is an isosceles triangle and a right triangle, find m, then solve using the definition of a parabola. From the problem: the focus of the parabola y^{2}=8x is F(2,0); the directrix of the parabola y^{2}=8x is x=-2. Without loss of generality, let P(m,n), then 8m=n^{2}. ∴|PA|=2+m, |FA|=\\sqrt{4^{2}+n^{2}}. From the definition of the parabola: |PF|=|PA|=2+m, ∴△APF is an isosceles triangle. Also, ∠AFx=\\frac{2\\pi}{3}, ∴|FA|\\cos60^{\\circ}=4, ∴|FA|=8. That is, \\sqrt{4^{2}+n^{2}}=8, n^{2}=48. Hence: 8m=48, solving gives: m=6, |PF|=2+6=8"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, the left focus is $F$, and $P$ is a point on the right branch of the hyperbola. If the midpoint of $FP$ lies on the circle with radius $|OF|$, then the horizontal coordinate of $P$ is?",
    "fact_expressions": "G: Hyperbola;F: Point;P: Point;O: Origin;H:Circle;Expression(G) = (x^2/4 - y^2/5 = 1);LeftFocus(G)=F;PointOnCurve(P,RightPart(G));Radius(H)=Abs(LineSegmentOf(O,F));PointOnCurve(MidPoint(LineSegmentOf(F,P)),H)",
    "query_expressions": "XCoordinate(P)",
    "answer_expressions": "16/3",
    "fact_spans": "[[[2, 40], [52, 55]], [[44, 47]], [[48, 51], [87, 90]], [[72, 79]], [[83, 84]], [[2, 40]], [[2, 47]], [[48, 60]], [[71, 84]], [[62, 85]]]",
    "query_spans": "[[[87, 96]]]",
    "process": "Let M be the midpoint of PF, and F be the right focus of the hyperbola. It is easy to know that $ a=2 $, $ b=\\sqrt{5} $, $ c=3 $. Since M is the midpoint of PF, $ |PF| = 2|OM| = 2|OF| = 6 $. By the definition of the hyperbola, $ |PF| = |PF| + 4 = 10 $. Connect MF', then $ \\angle FMF' = \\frac{\\pi}{2} $, $ |MF| = 5 $, so $ \\cos\\angle MFF = \\frac{5}{6} $. Therefore, $ x_{P} + 3 = 10\\cos\\angle MFF = \\frac{25}{3} $, that is, $ x_{P} = \\frac{25}{3} - 3 = \\frac{16}{3} $."
  },
  {
    "text": "The standard equation of a hyperbola with eccentricity $\\sqrt{3}$ and passing through $(-\\sqrt{3}, 2)$ is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (-sqrt(3), 2);PointOnCurve(H, G);Eccentricity(G)=sqrt(3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2-y^2/2=1,y^2/(5/2)-x^2/5=1}",
    "fact_spans": "[[[35, 38]], [[18, 34]], [[18, 34]], [[16, 38]], [[0, 38]]]",
    "query_spans": "[[[35, 45]]]",
    "process": "When the foci of the hyperbola are on the x-axis, let the equation be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1, then \\frac{c}{a}=\\sqrt{3}, then b=\\sqrt{2}a^{n}, \\frac{3}{a^{2}}-\\frac{4}{2a^{2}}=1, solving gives a^{2}=1, b^{2}=2. \\therefore the standard equation of the required hyperbola is x^{2}-\\frac{y}{-=1. When the foci of the hyperbola are on the y-axis, let the equation be \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1, then \\frac{c}{a}=\\sqrt{3}, then b=\\sqrt{2}a, \\frac{4}{a^{2}}-\\frac{3}{2a^{2}}=1, solving gives a^{2}=\\frac{5}{2}, b^{2}=5. \\therefore the standard equation of the required hyperbola is \\frac{y^{2}}{5}-\\frac{x^{2}}{5}=1"
  },
  {
    "text": "The eccentricity $e \\in (1,2)$ of the hyperbola $\\frac{x^{2}}{4}+\\frac{y^{2}}{k}=1$, then the range of real number $k$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 + y^2/k = 1);k: Real;e:Number;Eccentricity(G)=e;In(e, (1, 2))",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-12,0)",
    "fact_spans": "[[[0, 38]], [[0, 38]], [[56, 61]], [[42, 54]], [[0, 54]], [[42, 54]]]",
    "query_spans": "[[[56, 68]]]",
    "process": "The hyperbola equation can be rewritten as \\frac{x^{2}}{4}-\\frac{y^{2}}{-k}=1, then a^{2}=4, b^{2}=-k, c^{2}=4-k, e=\\frac{c}{a}=\\frac{\\sqrt{4-k}}{2}. Since e\\in(1,2), that is, 1<\\frac{\\sqrt{4-k}}{2}<2, solving gives -12<k<0."
  },
  {
    "text": "If $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{4}-y^{2}=1$ $(a>0, b>0)$, and point $P$ lies on the hyperbola $C$, if $\\angle F_{1} P F_{2}=120^{\\circ}$, then what is the distance from $P$ to the $x$-axis?",
    "fact_expressions": "F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;C: Hyperbola;Expression(C) = (x^2/4 - y^2 = 1);a: Number;b: Number;a>0;b>0;P: Point;PointOnCurve(P, C) = True;AngleOf(F1, P, F2) = ApplyUnit(120, degree)",
    "query_expressions": "Distance(P, xAxis)",
    "answer_expressions": "sqrt(15)/15",
    "fact_spans": "[[[1, 8]], [[9, 16]], [[1, 66]], [[1, 66]], [[17, 60], [72, 78]], [[17, 60]], [[25, 60]], [[25, 60]], [[25, 60]], [[25, 60]], [[67, 71], [117, 120]], [[67, 79]], [[81, 115]]]",
    "query_spans": "[[[117, 130]]]",
    "process": "According to the cosine law, 20=16+3PF_{1}\\cdot PF, calculate PF_{1}\\cdot PF=\\frac{4}{3}, then use the area formula to get \\frac{1}{2}\\times2ch=\\frac{1}{2}PF_{1}\\cdot PF_{2}\\sin120^{\\circ}, compute to obtain the answer. Solution: According to the cosine law, F_{1}F_{2}^{2}=PF_{1}^{2}+PF_{2}^{2}-2PF_{1}PF_{2}\\cos120^{\\circ} \\therefore 20=(PF_{1}-PF_{2})^{2}+3PF_{1}\\cdot PF_{2}=16+3PF_{1}\\cdot PF Hence PF_{1}\\cdot PF=\\frac{4}{3} S_{\\triangle PF_{1}F_{2}}=\\frac{1}{2}\\times2ch=\\frac{1}{2}PF_{1}\\cdot PF_{2}\\sin120^{\\circ}=\\frac{\\sqrt{3}}{3} \\therefore h=\\frac{\\sqrt{15}}{15}"
  },
  {
    "text": "Given the two foci of a hyperbola $F_{1}(-\\sqrt{10} , 0)$, $F_{2}(\\sqrt{10} , 0)$, and $M$ is a point on this hyperbola satisfying $\\overrightarrow{M F_{1}} \\cdot \\overrightarrow{M F_{2}}=0$, $|\\overrightarrow{M F_{1}}| \\cdot |\\overrightarrow{M F_{2}}|=2$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;M: Point;Coordinate(F1) = (-sqrt(10), 0);Coordinate(F2) = (sqrt(10), 0);Focus(G) = {F1,F2};PointOnCurve(M, G);DotProduct(VectorOf(M, F1), VectorOf(M, F2)) = 0;Abs(VectorOf(M, F1))*Abs(VectorOf(M, F2)) = 2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9-y^2=1",
    "fact_spans": "[[[2, 5], [66, 69], [204, 207]], [[11, 34]], [[37, 59]], [[61, 64]], [[11, 34]], [[37, 59]], [[2, 59]], [[61, 73]], [[78, 137]], [[139, 201]]]",
    "query_spans": "[[[204, 212]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through point $F_{1}$ intersects the ellipse at points $M$ and $N$. Then, the perimeter of $\\Delta F_{2} M N$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/5 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;l: Line;PointOnCurve(F1, l);M: Point;N: Point;Intersection(l, G) = {M, N}",
    "query_expressions": "Perimeter(TriangleOf(F2, M, N))",
    "answer_expressions": "12",
    "fact_spans": "[[[2, 39], [79, 81]], [[2, 39]], [[47, 54], [64, 72]], [[55, 62]], [[2, 62]], [[2, 62]], [[73, 78]], [[63, 78]], [[82, 85]], [[86, 89]], [[73, 91]]]",
    "query_spans": "[[[93, 116]]]",
    "process": "Since the line $l$ passing through point $F_{1}$ intersects the ellipse at points $M$ and $N$, by the definition of the ellipse, we have: $|MF_{1}|+|MF_{2}|=2a=6$, $|NF_{1}|+|NF_{2}|=2a=6$. Therefore, the perimeter of $\\triangle F_{2}MN$ is $(|MF_{1}|+|MF_{2}|)+(|NF_{1}|+|NF_{2}|)=12$."
  },
  {
    "text": "The equation $\\frac{x^{2}}{9-k}+\\frac{y^{2}}{k-3}=1$ represents a hyperbola, then $k$ satisfies?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(9 - k) + y^2/(k - 3) = 1);k: Number",
    "query_expressions": "k",
    "answer_expressions": "(-oo,3)+(9,+oo)",
    "fact_spans": "[[[43, 46]], [[0, 46]], [[48, 51]]]",
    "query_spans": "[[[48, 54]]]",
    "process": "The solution process is omitted"
  },
  {
    "text": "Given the parabola equation $y^{2}=4 \\sqrt{2} x$, then the coordinates of the focus of the parabola are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*(sqrt(2)*x))",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(\\sqrt{2},0)",
    "fact_spans": "[[[2, 5], [29, 32]], [[2, 27]]]",
    "query_spans": "[[[29, 39]]]",
    "process": "The coordinates of the focus of the parabola are $(\\frac{4\\sqrt{2}}{4},0)=(\\sqrt{2},0)$"
  },
  {
    "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{2}+\\frac{y^{2}}{m}=1$ with foci on the $x$-axis is $\\frac{1}{2}$, then what is the value of $m$?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/2 + y^2/m = 1);PointOnCurve(Focus(G), xAxis);Eccentricity(G) = 1/2",
    "query_expressions": "m",
    "answer_expressions": "3/2",
    "fact_spans": "[[[10, 47]], [[67, 70]], [[10, 47]], [[1, 47]], [[10, 65]]]",
    "query_spans": "[[[67, 74]]]",
    "process": ""
  },
  {
    "text": "A moving point on the parabola $y^{2}=8 x$ is $P$, with focus $F$. Let circle $M$ be the circle with diameter $P F$. When the area of circle $M$ is minimized, what is the equation of circle $M$?",
    "fact_expressions": "G: Parabola;M: Circle;P: Point;F: Point;Expression(G) = (y^2 = 8*x);PointOnCurve(P,G);Focus(G) = F;IsDiameter(LineSegmentOf(P,F),M);WhenMin(Area(M))",
    "query_expressions": "Expression(M)",
    "answer_expressions": "x^2+y^2-2*x=0",
    "fact_spans": "[[[0, 14]], [[42, 46], [39, 40], [48, 52], [59, 63]], [[19, 22]], [[25, 28]], [[0, 14]], [[0, 22]], [[0, 28]], [[29, 40]], [[47, 58]]]",
    "query_spans": "[[[59, 68]]]",
    "process": "By the given condition, the focus of the parabola $ y^{2} = 8x $ is $ F(2,0) $, and let point $ P(x,y) $. By the definition of a parabola, $ |PF| = x + 2 $, so the area of circle $ M $ is $ s = \\pi \\left( \\frac{|PF|}{2} \\right)^{2} = \\pi \\left( \\frac{x+2}{2} \\right)^{2} $. When $ x = 0 $, the area of circle $ M $ is minimized; at this time, point $ P(0,0) $, $ |PF| = 2 $, so the center of the circle $ M(1,0) $, radius $ r = \\frac{|PF|}{2} = 1 $. Therefore, circle $ M: (x-1)^{2} + y^{2} = 1 $, or $ x^{2} + y^{2} - 2x = 0 $."
  },
  {
    "text": "The equation $\\frac{x^{2}}{a-5}+\\frac{y^{2}}{7-a}=1$ represents an ellipse; then the range of real values for $a$ is?",
    "fact_expressions": "E: Ellipse;a: Real;Expression(E)  = (x**2/(a - 5) + y**2/(7 - a) = 1)",
    "query_expressions": "Range(a)",
    "answer_expressions": "(5,6)+(6,7)",
    "fact_spans": "[[[43, 45]], [[47, 52]], [[0, 45]]]",
    "query_spans": "[[[47, 59]]]",
    "process": ""
  },
  {
    "text": "The length of the chord cut by the line $y=2x+1$ from the parabola $y^{2}=12x$ is equal to?",
    "fact_expressions": "G: Parabola;H: Line;Expression(G) = (y^2 = 12*x);Expression(H) = (y = 2*x + 1)",
    "query_expressions": "Length(InterceptChord(H,G))",
    "answer_expressions": "sqrt(15)",
    "fact_spans": "[[[0, 15]], [[16, 27]], [[0, 15]], [[16, 27]]]",
    "query_spans": "[[[0, 34]]]",
    "process": "Let the line intersect the parabola at points A and B, where A(x_{1},y_{1}), B(x_{2},y_{2}). Solving the system \\begin{cases}y^{2}=12x\\\\y=2x+1\\end{cases} \\Rightarrow 4x^{2}-8x+1=0, c_{2}=2, x_{1}x_{2}=\\frac{1}{4}. This question mainly examines chord length problems in the context of the positional relationship between a line and a parabola, while also testing students' computational ability, and is considered an easy problem."
  },
  {
    "text": "If on the parabola $y^{2}=2 p x$, the distance from any point to the point $(1,0)$ is equal to the distance to the line $x=-1$, then $p$=?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;I: Point;P:Point;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (x = -1);Coordinate(I) = (1, 0);PointOnCurve(P,G);Distance(P,I)=Distance(P,H)",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 17]], [[51, 54]], [[36, 44]], [[23, 31]], [], [[1, 17]], [[36, 44]], [[23, 31]], [[1, 22]], [[1, 49]]]",
    "query_spans": "[[[51, 56]]]",
    "process": "From the definition of the parabola, we obtain $\\frac{p}{2}=1$, solving gives $p=2$."
  },
  {
    "text": "Given that a focus of the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{2}=1$ has coordinates $(-\\sqrt{3}, 0)$, what is its asymptote equation?",
    "fact_expressions": "G: Hyperbola;a: Number;Expression(G) = (-y^2/2 + x^2/a = 1);Coordinate(OneOf(Focus(G))) = (-sqrt(3), 0)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(2)*x",
    "fact_spans": "[[[2, 40], [66, 67]], [[5, 40]], [[2, 40]], [[2, 64]]]",
    "query_spans": "[[[66, 74]]]",
    "process": ""
  },
  {
    "text": "The equation of the hyperbola that has the same foci as the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ and passes through the point $P(2 , 1)$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;P: Point;Expression(H) = (x^2/4 + y^2 = 1);Coordinate(P) = (2, 1);Focus(H) = Focus(G);PointOnCurve(P, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/2 - y^2 = 1",
    "fact_spans": "[[[48, 51]], [[1, 28]], [[36, 47]], [[1, 28]], [[36, 47]], [[0, 51]], [[35, 51]]]",
    "query_spans": "[[[48, 55]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$ has its right focus at $F$, point $P$ lies on one of the asymptotes of hyperbola $C$, and $O$ is the origin. If $|O P|=2|O F|$, then the area of $\\triangle P F O$ is?",
    "fact_expressions": "C: Hyperbola;P: Point;F: Point;O: Origin;Expression(C) = (x^2/4 - y^2/2 = 1);RightFocus(C) = F;PointOnCurve(P, OneOf(Asymptote(C)));Abs(LineSegmentOf(O, P)) = 2*Abs(LineSegmentOf(O, F))",
    "query_expressions": "Area(TriangleOf(P, F, O))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[0, 43], [57, 63]], [[52, 56]], [[48, 51]], [[71, 74]], [[0, 43]], [[0, 51]], [[52, 70]], [[81, 95]]]",
    "query_spans": "[[[97, 119]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$. If there exists a point $P$ on $C$ such that $PF_{1} \\perp P F_{2}$ and $\\angle P F_{1} F_{2}=30^{\\circ}$, then the eccentricity of $C$ is?",
    "fact_expressions": "F1: Point;F2: Point;C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C) = True;IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2)) = True;AngleOf(P, F1, F2) = ApplyUnit(30, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 8]], [[9, 16]], [[17, 81], [90, 93], [163, 166]], [[17, 81]], [[25, 81]], [[25, 81]], [[25, 81]], [[25, 81]], [[1, 86]], [[98, 101]], [[89, 101]], [[104, 126]], [[128, 161]]]",
    "query_spans": "[[[163, 172]]]",
    "process": ""
  },
  {
    "text": "If the directrix of the parabola $y^{2}=2 p x(p>0)$ passes through an intersection point of the line $y=x+1$ and the coordinate axes, then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (y = x + 1);PointOnCurve(OneOf(Intersection(H,axis)),Directrix(G))",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 22]], [[47, 50]], [[27, 36]], [[4, 22]], [[1, 22]], [[27, 36]], [[1, 45]]]",
    "query_spans": "[[[47, 52]]]",
    "process": "The directrix of the parabola $ y^{2} = 2px $ ($ p > 0 $) is $ x = -\\frac{p}{2} $. Therefore, it passes through the intersection point of the line $ y = x + 1 $ and the coordinate axis, which is $ (-1, 0) $. Hence, $ -\\frac{p}{2} = -1 $, so $ p = 2 $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{20}+\\frac{y^{2}}{16}=1$, respectively, and point $P$ is any point on the ellipse $C$. Let $m= |\\overrightarrow{P F_{1}}| \\cdot |\\overrightarrow {P F_{2}}|$, then the maximum value of $m$ is?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/20 + y^2/16 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);m:Number;Abs(VectorOf(P,F1))*Abs(VectorOf(P,F2))=m",
    "query_expressions": "Max(m)",
    "answer_expressions": "20",
    "fact_spans": "[[[20, 64], [75, 80]], [[70, 74]], [[2, 9]], [[10, 17]], [[20, 64]], [[2, 69]], [[2, 69]], [[70, 85]], [[154, 157]], [[87, 152]]]",
    "query_spans": "[[[154, 163]]]",
    "process": "From the definition of an ellipse, we know that |PF_{1}| + |PF_{2}| = 2a = 4\\sqrt{5}, so m = |PF_{1}| \\cdot |PF_{2}| \\leqslant \\left( \\frac{|PF_{1}| + |PF_{2}|}{2} \\right)^{2} = 20, with equality holding if and only if |PF_{1}| = |PF_{2}| = 2\\sqrt{5}. Therefore, the maximum value of m is 20."
  },
  {
    "text": "Draw a line through the focus of the parabola $y^{2}=4x$ intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $x_{1}+x_{2}=6$, then $|AB|$=?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);PointOnCurve(Focus(G), H);Intersection(H, G) = {A,B};x1+x2=6;x1:Number;x2:Number;y1:Number;y2:Number",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 15], [22, 25]], [[19, 21]], [[26, 43]], [[46, 63]], [[1, 15]], [[26, 43]], [[46, 63]], [[0, 21]], [[19, 65]], [[68, 83]], [[26, 43]], [[46, 63]], [[26, 43]], [[46, 63]]]",
    "query_spans": "[[[86, 95]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $x^{2}=8 y$, $P$ is a point on $C$, and $M(-4 , 3)$, then the minimum perimeter of $\\triangle P M F$ is?",
    "fact_expressions": "C: Parabola;M: Point;P: Point;F: Point;Expression(C) = (x^2 = 8*y);Coordinate(M) = (-4, 3);Focus(C) = F;PointOnCurve(P, C)",
    "query_expressions": "Min(Perimeter(TriangleOf(P, M, F)))",
    "answer_expressions": "5+sqrt(17)",
    "fact_spans": "[[[6, 25], [33, 36]], [[40, 51]], [[29, 32]], [[2, 5]], [[6, 25]], [[40, 51]], [[2, 28]], [[29, 39]]]",
    "query_spans": "[[[53, 78]]]",
    "process": "As shown in the figure, let F be the focus of the parabola C: x^{2}=8y, P a point on C, and M(-4,3). The focus of the parabola C: x^{2}=8y is F(0,2), and the equation of the directrix is y=-2. Draw a perpendicular from P to the directrix, with foot Q. Then |PF|=|PQ|. |PM|+|PF|=|PM|+|PQ|\\geqslant|MQ|=5, with equality if and only if points M, P, Q are collinear. Therefore, the minimum perimeter of \\triangle PMF is 5+\\sqrt{(-4)^{2}+(3-2)^{2}}=5+\\sqrt{17}."
  },
  {
    "text": "The line $l$ passing through the focus of the parabola $E$: $y^{2}=4x$ intersects $E$ at points $A$ and $B$, and intersects the directrix of $E$ at point $C$. If point $A$ lies in the first quadrant and $B$ is the midpoint of $AC$, then the slope of line $l$ is equal to?",
    "fact_expressions": "l: Line;E: Parabola;A: Point;C: Point;B: Point;Expression(E) = (y^2 = 4*x);PointOnCurve(Focus(E), l);Intersection(l, E) = {A, B};Intersection(l,Directrix(E))=C;Quadrant(A)=1;MidPoint(LineSegmentOf(A, C)) = B",
    "query_expressions": "Slope(l)",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[25, 30], [89, 94]], [[2, 21], [48, 51], [31, 34]], [[37, 40], [63, 67]], [[56, 60]], [[41, 44], [75, 78]], [[2, 21]], [[0, 30]], [[25, 46]], [[25, 60]], [[63, 73]], [[75, 87]]]",
    "query_spans": "[[[89, 100]]]",
    "process": "Let the slope of line $ l $ be $ k $, so the equation of the line is $ y = k(x - 1) $. By solving the system of equations and using the relationship between roots and coefficients, we obtain $ x_{1}x_{2} = 1 $. Then, since $ B $ is the midpoint of $ A $ and $ C $, we get $ x_{1} = 2x_{2} + 1 $. Solving the system of equations, we find $ B\\left(\\frac{1}{2}, -\\sqrt{2}\\right) $. Then, using the slope formula, the solution can be obtained. [Detailed explanation] As shown in the figure, for the parabola $ E: y^{2} = 4x $, the focus coordinates are $ F(1, 0) $, and the directrix equation is $ x = -1 $. Let the slope of line $ l $ be $ k $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, $ C(-1, y_{3}) $. Thus, the equation of the line is $ y = k(x - 1) $, $ \\cdots\\cdots\\cdots\\cdots\\textcircled{1} $. Solve the system of equations:\n$$\n\\begin{cases}\ny = k(x - 1) \\\\\ny^{2} = 4x\n\\end{cases}\n$$\nRearranging gives $ k^{2}x - (2k^{2} + 4)x + k^{2} = 0 $, so $ x_{1}x_{2} = 1 $. Since $ B $ is the midpoint of $ A $ and $ C $, we have $ x_{2} = \\frac{x_{1} + (-1)}{2} $, i.e., $ x_{1} = 2x_{2} + 1 $, $ \\cdots\\cdots\\cdots\\cdots\\textcircled{2} $. Solving $ \\textcircled{1} $ and $ \\textcircled{2} $ together, we obtain $ x_{2} = \\frac{1}{2} $. Substituting into the parabola's equation, we find $ y_{2} = -\\sqrt{2} $, so $ B\\left(\\frac{1}{2}, -\\sqrt{2}\\right) $. Therefore, the slope of line $ BF $ is $ k = \\frac{0 - (-\\sqrt{2})}{1 - \\frac{1}{2}} = 2\\sqrt{2} $, i.e., the slope of line $ l $ is $ 2\\sqrt{2} $."
  },
  {
    "text": "The distance from a point $P$ on the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ to the point $(5,0)$ is $8.5$. Then, what is the distance from point $P$ to the point $(-5,0)$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1);P: Point;PointOnCurve(P, G);A: Point;Coordinate(A) = (5, 0);Distance(P, A) = 8.5;B: Point;Coordinate(B) = (-5, 0)",
    "query_expressions": "Distance(P, B)",
    "answer_expressions": "16.5",
    "fact_spans": "[[[0, 39]], [[0, 39]], [[41, 45], [65, 69]], [[0, 45]], [[46, 54]], [[46, 54]], [[41, 63]], [[70, 79]], [[70, 79]]]",
    "query_spans": "[[[65, 84]]]",
    "process": "Let the two foci of the hyperbola be $ F_{1}(-5,0) $ and $ F_{2}(5,0) $. By the definition of a hyperbola, $ ||PF_{1}|-|PF_{2}|| = 8 $, so $ |PF_{1}| = 16.5 $ or $ |PF_{1}| = 0.5 $. The shortest distance from a point on the left branch of the hyperbola to the left focus is 1, thus $ |PF_{1}| = 0.5 $ does not satisfy the conditions. In fact, when solving such problems, one should flexibly apply the definition of the hyperbola to analyze the existence of point $ P $, and then proceed to solve. In this problem, since the distance from the left vertex to the right focus is $ 9 > 8.5 $, point $ P $ must lie on the right branch, hence the required $ |PF_{1}| = 16.5 $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$, respectively, and $P$ is a point on the ellipse $C$ such that $P F_{1} \\perp P F_{2}$, then the area of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/16 + y^2/4 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "4",
    "fact_spans": "[[[20, 63], [74, 79]], [[70, 73]], [[2, 9]], [[10, 17]], [[20, 63]], [[2, 69]], [[2, 69]], [[70, 83]], [[85, 108]]]",
    "query_spans": "[[[110, 137]]]",
    "process": "From the ellipse equation, we know $ a=4 $, $ b=2 $, $ c^{2}=12 $. Since $ PF_{1} \\perp PF_{2} $, it follows that  \n$$\n\\begin{cases}\n|PF_{1}|^{2}+|PF_{2}|^{2}=4c^{2}=48 \\\\\n|PF_{1}|+|PF_{2}|=2a=8\n\\end{cases}\n$$\nThus, $ |PF_{1}||PF_{2}|=8 $, so $ S_{\\triangle PF_{1}F_{2}} = \\frac{1}{2}|PF_{1}||PF_{2}|=4 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0, b>0)$, one of its asymptotes has the equation $3x - 2y = 0$. What is the eccentricity of this hyperbola?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-x^2/b^2 + y^2/a^2 = 1);Expression(OneOf(Asymptote(C))) = (3*x - 2*y = 0)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(13)/3",
    "fact_spans": "[[[2, 64], [88, 91]], [[9, 64]], [[9, 64]], [[9, 64]], [[9, 64]], [[2, 64]], [[2, 85]]]",
    "query_spans": "[[[88, 97]]]",
    "process": "From the standard equation of the hyperbola, the asymptotes are given by $ y = \\pm\\frac{a}{b}x $. Combining this with the equation of one asymptote of the hyperbola $ 3x - 2y = 0 $, we obtain $ \\frac{b}{a} = \\frac{2}{3} $. Finally, using the eccentricity formula of the hyperbola $ e = \\sqrt{1 + \\left(\\frac{b}{a}\\right)^2} $, the result can be found. [Detailed solution] From the given, the hyperbola $ C: \\frac{y^2}{a^2} - \\frac{x^2}{b^2} = 1 $ ($ a > 0, b > 0 $) has asymptotes $ y = \\pm\\frac{a}{b}x $. Since one asymptote is $ 3x - 2y = 0 $, i.e., $ y = \\frac{3}{2}x $, we get $ \\frac{a}{b} = \\frac{3}{2} $, then $ \\frac{b}{a} = \\frac{2}{3} $, thus $ e = \\sqrt{1 + \\left(\\frac{b}{a}\\right)^2} = \\sqrt{1 + \\frac{4}{9}} = \\frac{\\sqrt{13}}{3} $. Therefore, the eccentricity of the hyperbola is $ \\frac{\\sqrt{13}}{3} $."
  },
  {
    "text": "Let the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ be $F_{1}$ and $F_{2}$, respectively. A line passing through $F_{1}$ intersects the left and right branches of the hyperbola at points $M$ and $N$. If the circle with diameter $MN$ passes through point $F_{2}$ and $|M F_{2}|=|N F_{2}|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, L) = True;Intersection(L, LeftPart(G)) = M;Intersection(L, RightPart(G)) = N;L: Line;M: Point;N: Point;IsDiameter(LineSegmentOf(M, N), H) = True;H: Circle;PointOnCurve(F2, H) = True;Abs(LineSegmentOf(M, F2)) = Abs(LineSegmentOf(N, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[1, 57], [99, 102], [163, 166]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[65, 72], [85, 93]], [[74, 82], [131, 139]], [[1, 81]], [[1, 81]], [[84, 96]], [[94, 115]], [[94, 115]], [[94, 96]], [[107, 111]], [[112, 115]], [[118, 129]], [[128, 129]], [[128, 139]], [[140, 161]]]",
    "query_spans": "[[[163, 172]]]",
    "process": "As shown in the figure, let H be the midpoint of segment MN, and triangle $AHF_{1}F_{2}$ be a right triangle. Let $|MF_{2}| = |NF_{2}| = m$, then $|MN| = \\sqrt{2}m$. By the definition of hyperbola, we have $|MF_{2}| - |MF_{1}| = 2a$, $|NF_{1}| - |NF_{2}| = 2a$. Also, $|NF_{1}| = |MN| + |MF_{1}|$, so $|MN| = 4a$, thus $\\sqrt{2}m = 4a$, then $m = 2\\sqrt{2}a$, so $|MF_{1}| = |MF_{2}| - 2a = 2\\sqrt{2}a - 2a$. Therefore, $|HF_{1}| = |MF_{1}| + |MH| = 2\\sqrt{2}a$, $|HF_{2}| = \\frac{1}{2}|MN| = 2a$. In right triangle $AHF_{1}F_{2}$, by Pythagorean theorem we get $(2\\sqrt{2}a)^{2} + 4a^{2} = 4c^{2}$, that is $3a^{2} = c^{2}$, so $c = \\sqrt{3}a$. Thus, eccentricity $e = \\frac{c}{a} = \\sqrt{3}$."
  },
  {
    "text": "If the focus of the parabola $C$: $y^{2}=2 p x(p>0)$ lies on the line $l$: $x+2 y-3=0$, then $p$=?",
    "fact_expressions": "l: Line;C: Parabola;p: Number;p>0;Expression(C) = (y^2 = 2*(p*x));Expression(l)=(x + 2*y - 3 = 0);PointOnCurve(Focus(C),l)",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[31, 49]], [[1, 27]], [[52, 55]], [[9, 27]], [[1, 27]], [[31, 49]], [[1, 50]]]",
    "query_spans": "[[[52, 57]]]",
    "process": "From the given conditions, the coordinates of the focus $F$ of the parabola $C$ are $(\\frac{p}{2},0)$, then $\\frac{p}{2}-3=0$, solving gives $p=6$."
  },
  {
    "text": "Let $m$ be a positive real number. If the focal distance of the ellipse $\\frac{x^{2}}{m^{2}+16}+\\frac{y^{2}}{9}=1$ is $8$, then $m=$?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (x^2/(m^2 + 16) + y^2/9 = 1);FocalLength(G) = 8;m>0",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[11, 55]], [[64, 67], [1, 4]], [[11, 55]], [[11, 62]], [[1, 8]]]",
    "query_spans": "[[[64, 69]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the focus of the parabola $y=-\\frac{1}{8} x^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = -x^2/8)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,-2)",
    "fact_spans": "[[[0, 25]], [[0, 25]]]",
    "query_spans": "[[[0, 32]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively, and $P$ is a point on $E$ such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$ and the area of $\\Delta P F_{1} F_{2}$ is $\\frac{1}{2} a b$, then the eccentricity of the ellipse $E$ is?",
    "fact_expressions": "E: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(E) = F1;RightFocus(E) = F2;PointOnCurve(P, E);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Area(TriangleOf(P, F1, F2)) = (a*b)/2",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[20, 77], [88, 91], [202, 207]], [[183, 200]], [[183, 200]], [[84, 87]], [[2, 9]], [[10, 17]], [[27, 77]], [[27, 77]], [[20, 77]], [[2, 83]], [[2, 83]], [[84, 94]], [[96, 155]], [[157, 200]]]",
    "query_spans": "[[[202, 213]]]",
    "process": "Because $\\overrightarrow{PF}_{1}\\cdot\\overrightarrow{PF_{2}}=0$, so $PF_{1}\\bot PF_{2}$, so $|PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}=4c^{2}$. Because $|PF_{1}|+|PF_{2}|=2a$, so $(|PF_{1}|+|PF_{2}|)^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}+2|PF_{1}|\\cdot|PF_{2}|=4a^{2}$. So $2|PF_{1}|\\cdot|PF_{2}|=4a^{2}-4c^{2}=4b^{2}$, so $S_{\\triangle F_{1}PF_{2}}=\\frac{|PF_{1}|\\cdot|PF_{2}|}{2}=b^{2}$. So $b^{2}=\\frac{1}{2}ab$, so $a=2b=2\\sqrt{a^{2}-c^{2}}$. So $e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{3}{4}$ and $e\\in(0,1)$, so $e=\\frac{\\sqrt{3}}{2}$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1(a>0)$ has a focus at $F(2,0)$, then what is the equation of the asymptotes of hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;a: Number;F: Point;a>0;Expression(C) = (-y^2/3 + x^2/a^2 = 1);Coordinate(F) = (2, 0);OneOf(Focus(C)) = F",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[2, 54], [71, 77]], [[10, 54]], [[60, 68]], [[10, 54]], [[2, 54]], [[60, 68]], [[2, 68]]]",
    "query_spans": "[[[71, 85]]]",
    "process": "According to the definition of hyperbola, obtain the value of c, then find the value of a, and directly write out the asymptote equations to get the answer. [Detailed solution] According to the given conditions, c = 2. Using the property of hyperbola $a^2 + b^{2} = c^{2}$, it is clear that $a^{2} + 3 = 4$, so $a = 1$. The asymptote equations of the hyperbola are $y = \\pm\\frac{b}{a}x$, $\\therefore y = \\pm\\sqrt{3}x$"
  },
  {
    "text": "The equation of a parabola with vertex at the origin and focus $F(0 , 5)$ is?",
    "fact_expressions": "G: Parabola;O: Origin;Vertex(G) = O;F: Point;Coordinate(F) = (0, 5);Focus(G) = F",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2=20*y",
    "fact_spans": "[[[20, 23]], [[3, 5]], [[0, 23]], [[9, 19]], [[9, 19]], [[6, 23]]]",
    "query_spans": "[[[20, 27]]]",
    "process": ""
  },
  {
    "text": "The line $l$ passes through the right focus of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{4}=1$ and intersects the right branch of the hyperbola at points $A$ and $B$, with $|A B|=4$. Then, the perimeter of the triangle formed by $A$, $B$, and the left focus of the hyperbola is?",
    "fact_expressions": "l: Line;PointOnCurve(RightFocus(G), l);G: Hyperbola;Expression(G) = (x^2/16 - y^2/4 = 1);A: Point;B: Point;Intersection(l, RightPart(G)) = {A, B};Abs(LineSegmentOf(A, B)) = 4;F: Point;LeftFocus(G) = F",
    "query_expressions": "Perimeter(TriangleOf(A, B, F))",
    "answer_expressions": "24",
    "fact_spans": "[[[0, 5]], [[0, 49]], [[6, 45], [51, 54], [88, 91]], [[6, 45]], [[59, 62], [80, 83]], [[63, 66], [84, 87]], [[0, 68]], [[69, 78]], [], [[88, 95]]]",
    "query_spans": "[[[80, 105]]]",
    "process": "From the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{4}=1$, we get $a=4$. Let the left focus of the hyperbola be $F_{1}$ and the right focus be $F_{2}$. By the definition of the hyperbola, we have $|AF_{1}|-|AF_{2}|=2a=8$ and $|BF_{1}|-|BF_{2}|=2a=8$. Since $|AB|=4$, it follows that $|AF_{2}|+|BF_{2}|=4$. Therefore, $|AF_{1}|+|BF_{1}|=20$, and thus the perimeter of the triangle is $24$."
  },
  {
    "text": "Given that the equation $(m-2) x^{2}+m y^{2}=1$ represents a hyperbola, what is the range of values for $m$?",
    "fact_expressions": "G: Hyperbola;Expression(G)=((m-2)*x^2+m*y^2=1);m:Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(0,2)",
    "fact_spans": "[[[29, 32]], [[2, 32]], [[34, 37]]]",
    "query_spans": "[[[34, 44]]]",
    "process": "Since the equation $(m-2)x^{2}+my^{2}=1$ represents a hyperbola, we have $m(m-2)<0$, that is, $0<m<2$. Therefore, the range of values for $m$ is $(0,2)$."
  },
  {
    "text": "The left and right vertices of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ are $A$ and $B$, respectively, and $P$ is an arbitrary point on the ellipse. Then the product of the slopes of line $PA$ and line $PB$ is equal to?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/9 = 1);A: Point;B: Point;LeftVertex(G) = A;RightVertex(G) = B;P: Point;PointOnCurve(P, G)",
    "query_expressions": "Slope(LineOf(P, A))*Slope(LineOf(P, B))",
    "answer_expressions": "-9/16",
    "fact_spans": "[[[0, 38], [59, 61]], [[0, 38]], [[47, 50]], [[51, 54]], [[0, 54]], [[0, 54]], [[55, 58]], [[55, 66]]]",
    "query_spans": "[[[68, 91]]]",
    "process": "According to the properties of an ellipse, the product of the slopes is constant; substitute the values to calculate. [Detailed solution] According to the properties of an ellipse, $k_{PA} \\cdot k_{PB} = -\\frac{b^{2}}{a^{2}} = -\\frac{9}{16}$"
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=8 x$, $P$ is a point on $C$, $O$ is the origin, and $|P F|=6$, then $|O P|=$?",
    "fact_expressions": "C: Parabola;P: Point;F: Point;O: Origin;Expression(C) = (y^2 = 8*x);Focus(C) = F;PointOnCurve(P, C);Abs(LineSegmentOf(P, F)) = 6",
    "query_expressions": "Abs(LineSegmentOf(O, P))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[6, 25], [33, 36]], [[29, 32]], [[2, 5]], [[40, 43]], [[6, 25]], [[2, 28]], [[29, 39]], [[50, 59]]]",
    "query_spans": "[[[61, 70]]]",
    "process": "Let $ P(x_{P}, y_{P}) $. By the property of the parabola, $ |PF| = x_{P} + 2 = 6 $, so $ x_{P} = 4 $. Also, $ y_{P}^{2} = 8x_{P} = 32 $, so $ |OP| = \\sqrt{x_{P}^{2} + y_{P}^{2}} = \\sqrt{48} = 4\\sqrt{3} $."
  },
  {
    "text": "Let the left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ be $F$, and let the left and right vertices be $A$ and $B$, respectively. A line passing through $F$ and perpendicular to the $x$-axis intersects the hyperbola at points $P$ and $Q$. If $A P \\perp B Q$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;A: Point;P: Point;B: Point;Q: Point;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F;LeftVertex(G)=A;RightVertex(G)=B;PointOnCurve(F,H);IsPerpendicular(H,xAxis);Intersection(H,G)={P,Q};IsPerpendicular(LineSegmentOf(A, P), LineSegmentOf(B, Q))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 57], [100, 103], [133, 136]], [[4, 57]], [[4, 57]], [[97, 99]], [[74, 77]], [[105, 108]], [[79, 82]], [[109, 112]], [[62, 65]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 65]], [[1, 82]], [[1, 82]], [[84, 99]], [[89, 99]], [[97, 114]], [[116, 131]]]",
    "query_spans": "[[[133, 142]]]",
    "process": "Find the coordinates of P and Q. Since AP ⊥ BQ, we have k_{AP} \\cdot k_{BQ} = -1, leading to c^{4} - 3a^{2}c^{2} + 2a^{4} = 0, i.e., e^{4} - 3e^{2} + 2 = 0, which can then be solved. [Detailed solution] ∵ PQ ⊥ x-axis, substitute x = -c into the hyperbola to get y = ±\\frac{b^{2}}{a}; without loss of generality, let P(-c, \\frac{b^{2}}{a}), Q(-c, -\\frac{b^{2}}{a}); ∵ A(-a, 0), B(a, 0); ∵ AP ⊥ BQ, ∴ k_{AP} \\cdot k_{BQ} = -1, i.e., \\frac{\\frac{b^{2}}{a}}{-c+a} \\cdot \\frac{-\\frac{b^{2}}{a}}{c-a} = -1, i.e., b^{4} = a^{2}c^{2} - a^{4}, thus c^{4} - 3a^{2}c^{2} + 2a^{4} = 0, ∴ e^{4} - 3e^{2} + 2 = 0; solving gives e^{2} = 1 (discarded) or e^{2} = 2, ∴ e = \\sqrt{2}."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{3}+y^{2}=1$ with two foci $F_{1}$, $F_{2}$, and a point $P(x_{0},y_{0})$ satisfying $0<\\frac{x_{0}^{2}}{3} + y_{0}^{2}<1$, what is the range of values for $|P F_{1}|+|P F_{2}|$? How many common points are there between the line $\\frac{x_{0} x}{3} + y_{0} y=1$ and the ellipse $C$?",
    "fact_expressions": "C: Ellipse;G: Line;y0: Number;x0: Number;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/3 + y^2 = 1);Expression(G) = (y*y0 + (x*x0)/3 = 1);Coordinate(P) = (x0, y0);Focus(C) = {F1,F2};0<x0^2/3+y0^2;x0^2/3+y0^2<1",
    "query_expressions": "Range(Abs(LineSegmentOf(P,F1))+Abs(LineSegmentOf(P,F2)));NumIntersection(G,C)",
    "answer_expressions": "[2*sqrt(2),2*sqrt(3))\n0",
    "fact_spans": "[[[2, 34], [180, 185]], [[146, 179]], [[60, 76]], [[60, 76]], [[59, 76]], [[40, 47]], [[49, 57]], [[2, 34]], [[146, 179]], [[59, 76]], [[2, 57]], [[78, 115]], [[78, 115]]]",
    "query_spans": "[[[118, 146]], [[146, 194]]]",
    "process": ""
  },
  {
    "text": "For a moving point $P$ on the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{27}=1$, tangents are drawn to circle $C_{1}$: $(x+3)^{2}+y^{2}=4$ and circle $C_{2}$: $(x-3)^{2}+y^{2}=1$, with points of tangency $M$ and $N$ respectively. Then the minimum value of $|P M|^{2}+2|P N|^{2}$ is?",
    "fact_expressions": "G: Ellipse;C1:Circle;C2:Circle;P: Point;M: Point;N: Point;L1:Line;L2:Line;Expression(G) = (x^2/36 + y^2/27 = 1);Expression(C1)=((x+3)^2+y^2=4);Expression(C2)=((x-3)^2+y^2=1);PointOnCurve(P,G);TangentOfPoint(P,C1)=L1;TangentOfPoint(P,C2)=L2;TangentPoint(L1,C1)=M;TangentPoint(L2,C2)=N",
    "query_expressions": "Min(Abs(LineSegmentOf(P, M))^2 + 2*Abs(LineSegmentOf(P, N))^2)",
    "answer_expressions": "90",
    "fact_spans": "[[[1, 40]], [[50, 78]], [[79, 107]], [[44, 47]], [[116, 119]], [[120, 123]], [], [], [[1, 40]], [[50, 78]], [[79, 107]], [[1, 47]], [[0, 110]], [[0, 110]], [[0, 123]], [[0, 123]]]",
    "query_spans": "[[[125, 153]]]",
    "process": "\\because a=6, b=3\\sqrt{3}, c=\\sqrt{a^{2}-b^{2}}=3, it is clear that C_{1}(-3,0) and C_{2}(3,0) are the two foci of the ellipse. 2|PN|^{2}=|PC_{1}|^{2}-4+2(|PC_{2}|^{2}-1)=|PC_{1}|^{2}+2|PC_{2}|^{2}-6. According to the definition of the ellipse, |PC_{1}|+|PC_{2}|=2a=12. Let |PC_{2}|=t, then a-c\\leqslant t\\leqslant a+c, i.e., 3\\leqslant t\\leqslant 9. Then |PM|^{2}+2|PN|^{2}=(12-t)^{2}+2t^{2}-6=3t^{2}-24t+138=3(t^{2}-8t+46). When t=4, |PM|^{2}+2|PN|^{2} attains the minimum value 90."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, the intersection point of the directrix and the $x$-axis is $M$, and $N$ is a point on the parabola satisfying $|NF|=\\frac{\\sqrt{3}}{2}|MN|$, then $\\angle NMF=$?",
    "fact_expressions": "G: Parabola;N: Point;F: Point;M: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;Intersection(Directrix(G), xAxis) = M;PointOnCurve(N, G);Abs(LineSegmentOf(N, F)) = (sqrt(3)/2)*Abs(LineSegmentOf(M, N))",
    "query_expressions": "AngleOf(N, M, F)",
    "answer_expressions": "pi/6",
    "fact_spans": "[[[2, 16], [43, 46]], [[39, 42]], [[20, 23]], [[35, 38]], [[2, 16]], [[2, 23]], [[2, 38]], [[39, 50]], [[54, 84]]]",
    "query_spans": "[[[86, 100]]]",
    "process": ""
  },
  {
    "text": "The focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is the center of the circle $M$: $(x-3)^{2}+y^{2}=1$, and the major axis length of $C$ is $10$. Then the eccentricity of this ellipse is equal to?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;M: Circle;Expression(M) = (y^2 + (x - 3)^2 = 1);OneOf(Focus(C)) = Center(M);Length(MajorAxis(C)) = 10",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "3/5",
    "fact_spans": "[[[0, 57], [92, 95], [107, 109]], [[0, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[63, 87]], [[63, 87]], [[0, 90]], [[92, 104]]]",
    "query_spans": "[[[107, 116]]]",
    "process": "From the equation of circle M, we obtain the center M(3,0), so according to the problem, c=3. Given 2a=10, we have a=5, so the eccentricity of the ellipse is e=\\frac{c}{a}=\\frac{3}{8}. The answer is: \\frac{3}{3}"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{m^{2}-1}+\\frac{y^{2}}{m^{2}}=1$ $(m>0)$, its two foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the ellipse, and the maximum area of $\\Delta P F_{1} F_{2}$ is $\\sqrt{3}$. Then, the length of the minor axis of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/(m^2 - 1) + y^2/m^2 = 1);m: Number;m>0;F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C);Max(Area(TriangleOf(P, F1, F2))) = sqrt(3)",
    "query_expressions": "Length(MinorAxis(C))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[2, 59], [88, 90], [136, 141]], [[2, 59]], [[9, 59]], [[9, 59]], [[67, 74]], [[75, 82]], [[2, 82]], [[83, 87]], [[83, 93]], [[95, 134]]]",
    "query_spans": "[[[136, 147]]]",
    "process": "From the equation of the ellipse, it is known that the foci $ F_{1}, F_{2} $ of the ellipse lie on the y-axis, and $ |F_{1}F_{2}| = 2\\sqrt{m^{2} - (m^{2} - 1)} $. According to the problem, when point $ P $ is the left or right vertex of the ellipse $ C $, the area of $ \\triangle PF_{1}F_{2} $ is maximized, and $ \\frac{1}{2}|F_{1}F_{2}|\\sqrt{m^{2} - 1} = \\sqrt{3} $. Solving gives $ m = 2 $, so the length of the minor axis of ellipse $ C $ is $ 2\\sqrt{m^{2} - 1} = 2\\sqrt{3} $."
  },
  {
    "text": "The line $l$ passing through the origin intersects the left and right branches of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ at points $A$ and $B$, respectively. $F(-\\sqrt{3}, 0)$ is the left focus of the hyperbola. If $|FA|+|FB|=4$, $\\overrightarrow{FA} \\cdot \\overrightarrow{FB}=0$, then the equation of the hyperbola is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;F: Point;A: Point;B: Point;O: Origin;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (-sqrt(3), 0);LeftFocus(C) = F;Intersection(l,LeftPart(C)) = A;Intersection(l,RightPart(C)) = B;Abs(LineSegmentOf(F, A)) + Abs(LineSegmentOf(F, B)) = 4;DotProduct(VectorOf(F, A), VectorOf(F,B)) = 0;PointOnCurve(O,l)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/2-y^2=1",
    "fact_spans": "[[[4, 9]], [[10, 71], [109, 112], [189, 192]], [[17, 71]], [[17, 71]], [[91, 108]], [[81, 84]], [[85, 88]], [[1, 3]], [[17, 71]], [[17, 71]], [[10, 71]], [[91, 108]], [[91, 116]], [[4, 90]], [[4, 90]], [[118, 133]], [[135, 186]], [[0, 9]]]",
    "query_spans": "[[[189, 197]]]",
    "process": "Let |FB| = x, then |FA| = 4 - x. Since the line l passing through the origin intersects the left and right branches of the hyperbola C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 (a > 0, b > 0) at points A and B respectively, and F(-\\sqrt{3}, 0) is the left focus of the hyperbola, let F_{1}(\\sqrt{3}, 0) be the right focus of the hyperbola. Given \\overrightarrow{FA} \\cdot \\overrightarrow{FB} = 0, by the symmetry of the hyperbola, the quadrilateral FAF_{1}B is a rectangle. Therefore, x^{2} - 4x + 2 = 0, so x = 2 \\pm \\sqrt{2}. |FB| = 2 + \\sqrt{2}, |FA| = 2 - \\sqrt{2}. Thus, 2a = |FB| - |FA| = 2\\sqrt{2}, so a = \\sqrt{2}, therefore b = 1. Hence, the equation of the hyperbola is \\frac{x^{2}}{2} - y^{2} = 1."
  },
  {
    "text": "Given that the center of the circle $(x-a)^{2}+(y-b)^{2}=r^{2}$ is the focus of the parabola $y^{2}=4x$, and that the circle is tangent to the line $3x+4y+2=0$, then the equation of the circle is?",
    "fact_expressions": "H: Circle;Expression(H) = ((-a + x)^2 + (-b + y)^2 = r^2);b: Number;a: Number;r: Number;G: Parabola;Expression(G) = (y^2 = 4*x);Center(H) = Focus(G);L: Line;Expression(L) = (3*x + 4*y + 2 = 0);IsTangent(H, L)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x-1)^2+y^2=1",
    "fact_spans": "[[[2, 30], [74, 75]], [[2, 30]], [[3, 30]], [[3, 30]], [[3, 30]], [[34, 48]], [[34, 48]], [[2, 51]], [[54, 69]], [[54, 69]], [[2, 71]]]",
    "query_spans": "[[[74, 80]]]",
    "process": "The focus of the parabola \\( y^{2} = 4x \\) is \\( (1,0) \\). The distance from this point to the line \\( 3x + 4y + 2 = 0 \\) is \\( r = \\frac{|3 + 0 + 2|}{\\sqrt{3^{2} + 4^{2}}} = 1 \\), so the equation of the circle is \\( (x - 1)^{2} + y^{2} = 1 \\)."
  },
  {
    "text": "Let $P$ be an arbitrary point on the parabola $y^{2}=4x$, and let $Q$ be the projection of $P$ onto the $y$-axis. Given point $M(4, 5)$, find the minimum value of the sum of the lengths $PQ$ and $PM$.",
    "fact_expressions": "G: Parabola;P: Point;Q: Point;M: Point;Expression(G) = (y^2 = 4*x);Coordinate(M) = (4, 5);PointOnCurve(P, G);Projection(P, yAxis) = Q",
    "query_expressions": "Min(Length(LineSegmentOf(P, Q)) + Length(LineSegmentOf(P, M)))",
    "answer_expressions": "sqrt(34) - 1",
    "fact_spans": "[[[4, 18]], [[0, 3], [24, 27]], [[37, 40]], [[41, 52]], [[4, 18]], [[41, 52]], [[0, 23]], [[24, 40]]]",
    "query_spans": "[[[54, 73]]]",
    "process": ""
  },
  {
    "text": "If the distance from point $P$ to point $F(4,0)$ is less than its distance to the line $x+5=0$ by $1$, then the equation of the locus of moving point $P$ is?",
    "fact_expressions": "G: Line;F: Point;P: Point;Expression(G) = (x + 5 = 0);Coordinate(F) = (4, 0);Distance(P, F) = Distance(P, G) - 1",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "y^2=16*x",
    "fact_spans": "[[[21, 30]], [[6, 15]], [[1, 5], [19, 20], [41, 44]], [[21, 30]], [[6, 15]], [[1, 37]]]",
    "query_spans": "[[[41, 51]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-\\frac{y^{2}}{m}=1$ $(m>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{2}$ intersects the right branch of the hyperbola $C$ at points $P$ and $Q$. If $P Q \\perp P F_{1}$ and $|P F_{1}|=|P Q|$, then $m=$?",
    "fact_expressions": "C: Hyperbola;m: Number;G: Line;P: Point;Q: Point;F1: Point;F2:Point;m>0;Expression(C) = (x^2 - y^2/m = 1);LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(F2, G);Intersection(G,RightPart(C)) = {P, Q};IsPerpendicular(LineSegmentOf(P, Q), LineSegmentOf(P, F1));Abs(LineSegmentOf(P, F1)) = Abs(LineSegmentOf(P, Q))",
    "query_expressions": "m",
    "answer_expressions": "4 - 2*sqrt(2)",
    "fact_spans": "[[[2, 40], [81, 87]], [[144, 147]], [[78, 80]], [[93, 96]], [[97, 100]], [[49, 56]], [[59, 66], [70, 77]], [[9, 40]], [[2, 40]], [[2, 67]], [[2, 67]], [[69, 80]], [[78, 102]], [[104, 123]], [[125, 142]]]",
    "query_spans": "[[[144, 149]]]",
    "process": ""
  },
  {
    "text": "The equation of a parabola with the vertex at the origin and one directrix of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ as its directrix is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (x^2/3 - y^2 = 1);O: Origin;Directrix(H) = OneOf(Directrix(G));Vertex(H) = O",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2 = pm*6*x",
    "fact_spans": "[[[1, 29]], [[44, 47]], [[1, 29]], [[41, 43]], [[0, 47]], [[38, 47]]]",
    "query_spans": "[[[44, 51]]]",
    "process": ""
  },
  {
    "text": "If a point $P$ on the ellipse $\\frac{x^{2}}{100}+\\frac{y^{2}}{16}=1$ has a distance of $6$ to the left focus, then what is the distance from point $P$ to the right focus?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/100 + y^2/16 = 1);PointOnCurve(P, G);Distance(P, LeftFocus(G)) = 6",
    "query_expressions": "Distance(P, RightFocus(G))",
    "answer_expressions": "14",
    "fact_spans": "[[[2, 42]], [[45, 48], [62, 66]], [[2, 42]], [[2, 48]], [[2, 59]]]",
    "query_spans": "[[[2, 75]]]",
    "process": "Let the left and right foci of the ellipse be \\( F_{1} \\) and \\( F_{2} \\). From the problem, we have \\( a = 10 \\), \\( |PF_{1}| = 6 \\). By the definition of the ellipse, \\( |PF_{1}| + |PF_{2}| = 2a \\), so \\( |PF_{2}| = 2a - |PF_{1}| = 20 - 6 = 14 \\)."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{64}-\\frac{y^{2}}{36}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. The line $l$ passes through point $F_{1}$ and intersects the left branch of the hyperbola at points $A$ and $B$, with $|A B|=9$. Then, the perimeter of $\\triangle A B F_{2}$ is?",
    "fact_expressions": "l: Line;G: Hyperbola;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2/64 - y^2/36 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, l);Intersection(l,LeftPart(G))={A, B};Abs(LineSegmentOf(A, B)) = 9",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "50",
    "fact_spans": "[[[67, 72]], [[2, 42], [82, 85]], [[89, 92]], [[93, 96]], [[59, 66]], [[51, 58], [73, 81]], [[2, 42]], [[2, 66]], [[2, 66]], [[67, 81]], [[67, 98]], [[100, 109]]]",
    "query_spans": "[[[111, 137]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{6}=1$ has an asymptote with equation $y=3x$. What is the value of the real number $m$?",
    "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (-y^2/6 + x^2/m = 1);Expression(OneOf(Asymptote(G))) = (y = 3*x)",
    "query_expressions": "m",
    "answer_expressions": "2/3",
    "fact_spans": "[[[0, 38]], [[56, 61]], [[0, 38]], [[0, 54]]]",
    "query_spans": "[[[56, 65]]]",
    "process": "Since the equation represents a hyperbola, we have $ m > 0 $. The asymptotes of the hyperbola are given by $ y = \\pm\\sqrt{\\frac{6}{m}}x $, so $ \\sqrt{\\frac{6}{m}} = 3 $. Solving gives $ m = \\frac{2}{3} $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ with focus $F$, if the parabola $C$ intersects the circle $O$: $x^{2}+y^{2}=12$ at points $P$ and $Q$, and $|P Q|=4 \\sqrt{2}$, then the area of $\\Delta P F O$ is ($O$ being the origin)?",
    "fact_expressions": "C: Parabola;p: Number;O: Origin;H:Circle;P: Point;F: Point;Q: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Expression(H) = (x^2 + y^2 = 12);Focus(C) = F;Intersection(C, H) = {P, Q};Abs(LineSegmentOf(P, Q)) = 4*sqrt(2)",
    "query_expressions": "Area(TriangleOf(P,F,O))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 28], [37, 43]], [[10, 28]], [[118, 121]], [[44, 66]], [[68, 71]], [[32, 35]], [[72, 75]], [[10, 28]], [[2, 28]], [[44, 66]], [[2, 35]], [[37, 77]], [[79, 97]]]",
    "query_spans": "[[[99, 128]]]",
    "process": "Without loss of generality, assume point P is in the first quadrant. Since the parabola C intersects the circle O: $x^{2}+y^{2}=12$ at points P and Q, and $|PQ|=4\\sqrt{2}$, then $y_{P}=2\\sqrt{2}$. Substituting into $x^{2}+y^{2}=12$, we solve to get $x_{P}=2$, hence $P(2,2\\sqrt{2})$. Substituting into the parabola C: $y^{2}=2px$ ($p>0$), we solve to get $p=2$, thus $S_{\\triangle PFO}=\\frac{1}{2}\\times2\\sqrt{2}\\times1=\\sqrt{2}$."
  },
  {
    "text": "Let $O$ be the origin, $P$ be an arbitrary point on the parabola $y^{2}=2 p x(p>0)$ with focus $F$, and $M$ be a point on the segment $P F$ such that $|P M|=3|M F|$. Then the maximum value of the slope of the line $O M$ is?",
    "fact_expressions": "G: Parabola;p: Number;O: Origin;M: Point;P: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(P, G);PointOnCurve(M, LineSegmentOf(P, F));Abs(LineSegmentOf(P, M)) = 3*Abs(LineSegmentOf(M, F))",
    "query_expressions": "Max(Slope(LineOf(O, M)))",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[22, 43]], [[25, 43]], [[1, 4]], [[49, 52]], [[10, 13]], [[15, 18]], [[25, 43]], [[22, 43]], [[14, 43]], [[10, 48]], [[49, 63]], [[65, 79]]]",
    "query_spans": "[[[81, 97]]]",
    "process": "From the given conditions, point $F(\\frac{p}{2},0)$, $p>0$, let $P(\\frac{y_{0}^{2}}{2p},y_{0})$ $(y_{0}>0)$, from $|PM|=3|MF|$ we obtain $\\overrightarrow{PF}=4\\overrightarrow{MF}$, then $\\overrightarrow{MF}=(\\frac{p}{8}-\\frac{y_{0}^{2}}{8p},-\\frac{y_{0}}{4})$, $\\therefore$ point $M(\\frac{3p}{8}+\\frac{y_{0}^{2}}{8p},\\frac{y_{0}}{4})$, $\\therefore k_{OM}=\\frac{\\frac{y_{0}}{4}}{\\frac{3p}{8}+\\frac{y_{0}^{2}}{8p}}=\\frac{1}{2y_{0}+\\frac{y_{0}}{2p}}\\leqslant\\frac{1}{2\\sqrt{\\frac{3p}{2y_{0}}\\cdot\\frac{y_{0}}{2p}}}=\\frac{\\sqrt{3}}{3}$, equality holds if and only if $\\frac{3p}{2y_{0}}=\\frac{y_{0}}{2p}$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$, $F_{2}$ respectively, and the eccentricity is $e$. If there exists a point $P$ on the ellipse such that $\\frac{PF_{1}}{PF_{2}}=e$, then the range of values for the eccentricity $e$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;e: Number;Eccentricity(G) = e;P: Point;PointOnCurve(P, G);LineSegmentOf(P, F1)/LineSegmentOf(P, F2) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "[\\sqrt{2}-1, 1)",
    "fact_spans": "[[[2, 54], [87, 89]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[62, 69]], [[70, 77]], [[2, 77]], [[2, 77]], [[82, 85], [130, 133]], [[2, 85]], [[92, 96]], [[87, 96]], [[99, 124]]]",
    "query_spans": "[[[130, 140]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{n}+\\frac{y^{2}}{12-n}=-1$ is $\\sqrt{3}$, then $n$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (y^2/(12 - n) + x^2/n = -1);n: Number;Eccentricity(G) = sqrt(3)",
    "query_expressions": "n",
    "answer_expressions": "{-12,24}",
    "fact_spans": "[[[2, 44]], [[2, 44]], [[61, 64]], [[2, 59]]]",
    "query_spans": "[[[61, 66]]]",
    "process": "First, standardize the hyperbola, then discuss the corresponding $ a, b $ when the foci are on the $ x $-axis and $ y $-axis respectively. The eccentricity can be found using $ e = \\frac{c}{a} $. [Detailed solution] From the given, $ \\frac{x^{2}}{n} + \\frac{y^{2}}{12 - n} = -1 \\Rightarrow \\frac{y^{2}}{n - 12} - \\frac{x^{2}}{n} = 1 $. Since $ \\frac{y^{2}}{n - 12} - \\frac{x^{2}}{n} = 1 $ represents a hyperbola, $ (n - 12) \\cdot n > 0 \\Rightarrow n < 0 $ or $ n > 12 $. When the focus is on the $ x $-axis,  \n$$\n\\begin{cases}\na^{2} = -n \\\\\nb^{2} = 12 - n \\\\\nc^{2} = a^{2} + b^{2}\n\\end{cases}\n\\Rightarrow c^{2} = 12 - 2n,\n$$\nsince $ e = \\sqrt{3} \\Rightarrow e^{2} = 3 $, so when the focus is on the $ y $-axis,  \n$$\n\\begin{cases}\nc = a + b\n\\end{cases}.\n\\begin{cases}\na = n - 12 \\\\\nb^{2} = n \\\\\nc = a + b^{2}\n\\end{cases}\n3 = \\frac{2n - 12}{n - 12} \\Rightarrow n = 24.\n$$\nIn conclusion, the values of $ n $ are $ -12 $ or $ 24 $."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2 p x(p>0)$ be $F$, and let a point $M(3, y_{0})$ on the parabola be at a distance of $6$ from $F$. Then $y_{0}=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;M: Point;y0: Number;Coordinate(M) = (3, y0);PointOnCurve(M, G);Distance(M, F) = 6",
    "query_expressions": "y0",
    "answer_expressions": "pm*6",
    "fact_spans": "[[[1, 22], [30, 33]], [[1, 22]], [[4, 22]], [[4, 22]], [[26, 29], [50, 53]], [[1, 29]], [[36, 49]], [[62, 69]], [[36, 49]], [[30, 49]], [[36, 60]]]",
    "query_spans": "[[[62, 71]]]",
    "process": "Since the focus of the parabola $ y^{2}=2px $ ($ p>0 $) is $ F $, and the distance from a point $ M(3,y_{0}) $ on the parabola to $ F $ is 6, we have $ 3+\\frac{p}{2}=6 $. Solving gives $ p=6 $. Thus, the equation of the parabola is $ y^{2}=12x $. Therefore, $ y_{0}^{2}=12\\times3 $, yielding $ y_{0}=\\pm6 $."
  },
  {
    "text": "Given that the eccentricity of the ellipse $m x^{2}+4 y^{2}=1$ is $\\frac{\\sqrt{2}}{2}$, then the real number $m$ equals?",
    "fact_expressions": "G: Ellipse;Expression(G) = (m*x^2 + 4*y^2 = 1);m: Real;Eccentricity(G) = sqrt(2)/2",
    "query_expressions": "m",
    "answer_expressions": "{2,8}",
    "fact_spans": "[[[2, 23]], [[2, 23]], [[50, 55]], [[2, 48]]]",
    "query_spans": "[[[50, 58]]]",
    "process": "From the ellipse equation $ mx^{2}+4y^{2}=1 $, we obtain $ \\frac{x^{2}}{m}+\\frac{y^{2}}{4}=1 $ ($ m\\neq0 $) ① If the foci are on the x-axis, then $ m<4 $, so $ a^{2}=\\frac{1}{m} $, $ b^{2}=\\frac{1}{4} $, thus $ c^{2}=a^{2}-b^{2}=\\frac{1}{m}-\\frac{1}{4} $, therefore $ \\frac{c^{2}}{a^{2}}=\\frac{\\frac{1}{m}-\\frac{1}{4}}{\\frac{1}{m}}=\\frac{1}{2} $, which gives $ m=2 $. ② If the foci are on the y-axis, then $ m>4 $, so $ a^{2}=\\frac{1}{4} $, $ b^{2}=\\frac{1}{m} $, thus $ c^{2}=a^{2}-b^{2}=\\frac{1}{4}-\\frac{1}{m} $, therefore $ \\frac{c^{2}}{a^{2}}=\\frac{\\frac{1}{4}-\\frac{1}{m}}{\\frac{1}{4}}=\\frac{1}{2} $, which gives $ m=8 $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, a line $l$ passing through point $F$ with slope $1$ intersects the parabola $C$ at points $M$ and $N$, and $b=\\frac{|MF|+|NF|}{2}$. If the horizontal coordinate of the intersection point between the perpendicular bisector of segment $MN$ and the $x$-axis is $a$, then the value of $a-b$ is?",
    "fact_expressions": "l: Line;C: Parabola;M: Point;N: Point;F: Point;a:Number;b:Number;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F, l);Slope(l) = 1;Intersection(l, C) = {M, N};b = (Abs(LineSegmentOf(M, F)) + Abs(LineSegmentOf(N, F)))/2;XCoordinate(Intersection(PerpendicularBisector(LineSegmentOf(M,N))), xAxis) = a",
    "query_expressions": "a - b",
    "answer_expressions": "1",
    "fact_spans": "[[[42, 47]], [[2, 21], [48, 54]], [[55, 58]], [[59, 62]], [[25, 28], [30, 34]], [[117, 120]], [[65, 90]], [[2, 21]], [[2, 28]], [[29, 47]], [[35, 47]], [[42, 64]], [[65, 90]], [[92, 120]]]",
    "query_spans": "[[[122, 131]]]",
    "process": ""
  },
  {
    "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{12}-\\frac{y^{2}}{9}=-1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/12 - y^2/9 = -1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(3)/2)*x",
    "fact_spans": "[[[0, 40]], [[0, 40]]]",
    "query_spans": "[[[0, 48]]]",
    "process": ""
  },
  {
    "text": "The equation of the circle centered at the upper focus of the hyperbola $y^{2}-\\frac{x^{2}}{3}=1$ and tangent to the asymptotes of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/3 + y^2 = 1);H: Circle;Center(H) = UpperFocus(G);IsTangent(Asymptote(G), H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2+(y-2)^2=3",
    "fact_spans": "[[[1, 29], [39, 42]], [[1, 29]], [[49, 50]], [[0, 50]], [[37, 50]]]",
    "query_spans": "[[[49, 55]]]",
    "process": "From the given conditions, a=1, b=\\sqrt{3}, then c=2, the upper focus F(0,2) is the center of the circle, and the distance from F to the asymptote is r=b=\\sqrt{3}; thus, the circle is x^{2}+(y-2)^{2}=3."
  },
  {
    "text": "Point $A(x_{0}, y_{0})$ lies on the right branch of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{32}=1$. If the distance from point $A$ to the right focus is equal to $2 x_{0}$, then $x_{0}=?$",
    "fact_expressions": "G: Hyperbola;A: Point;x0:Number;y0:Number;Expression(G) = (x^2/4 - y^2/32 = 1);Coordinate(A) = (x0, y0);PointOnCurve(A, RightPart(G));Distance(A,RightFocus(G)) = 2*x0",
    "query_expressions": "x0",
    "answer_expressions": "2",
    "fact_spans": "[[[19, 58]], [[0, 18]], [[90, 97]], [[1, 18]], [[19, 58]], [[0, 18]], [[0, 62]], [[19, 87]]]",
    "query_spans": "[[[90, 99]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$, a line $l$ passing through the focus $F$ intersects the parabola $C$ at points $A$ and $B$. If the projections on the $y$-axis of the midpoints of segments $AF$ and $BF$ are $P$ and $Q$ respectively, and $|PQ|=4$, then the equation of line $l$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l: Line;PointOnCurve(F, l) = True;Intersection(l, C) = {A, B};A: Point;B: Point;Projection(MidPoint(LineSegmentOf(A, F)), yAxis) = P;P: Point;Projection(MidPoint(LineSegmentOf(B, F)), yAxis) = Q;Q: Point;Abs(LineSegmentOf(P, Q)) = 4",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x \\pm \\sqrt{3} y - 1 = 0",
    "fact_spans": "[[[2, 21], [35, 41]], [[2, 21]], [[25, 28]], [[2, 28]], [[29, 34], [103, 108]], [[22, 34]], [[29, 52]], [[43, 46]], [[47, 50]], [[54, 90]], [[83, 86]], [[54, 90]], [[87, 90]], [[92, 101]]]",
    "query_spans": "[[[103, 113]]]",
    "process": ""
  },
  {
    "text": "Point $A(x_{0} , y_{0})$ lies on the right branch of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{32}=1$. If the distance from point $A$ to the right focus is equal to $2 x_{0}$, then $x_{0}=$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/32 = 1);A: Point;x0: Number;y0: Number;Coordinate(A) = (x0, y0);PointOnCurve(A, RightPart(G));Distance(A, RightFocus(G)) = 2*x0",
    "query_expressions": "x0",
    "answer_expressions": "2",
    "fact_spans": "[[[20, 59]], [[20, 59]], [[0, 19], [65, 69]], [[89, 96]], [[1, 19]], [[0, 19]], [[0, 63]], [[20, 87]]]",
    "query_spans": "[[[89, 98]]]",
    "process": ""
  },
  {
    "text": "The line $m$ passing through the point $M(-2,0)$ intersects the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ at points $P_{1}$ and $P_{2}$, with the midpoint of segment $P_{1} P_{2}$ being $P$. Let the slope of line $m$ be $k_{1}$ and the slope of line $O P$ be $k_{2}$. What is the value of $k_{1} k_{2}$?",
    "fact_expressions": "M: Point;Coordinate(M) = (-2, 0);m: Line;PointOnCurve(M, m) = True;G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);Intersection(m, G) = {P1, P2};P1: Point;P2: Point;MidPoint(LineSegmentOf(P1,P2)) = P;P: Point;Slope(m) = k1;k1: Number;O: Origin;Slope(LineOf(O,P)) = k2;k2: Number",
    "query_expressions": "k1*k2",
    "answer_expressions": "-1/2",
    "fact_spans": "[[[1, 11]], [[1, 11]], [[12, 17], [87, 92]], [[0, 17]], [[18, 45]], [[18, 45]], [[12, 62]], [[47, 54]], [[55, 62]], [[63, 85]], [[82, 85]], [[87, 103]], [[96, 103]], [[106, 111]], [[104, 122]], [[115, 122]]]",
    "query_spans": "[[[124, 141]]]",
    "process": ""
  },
  {
    "text": "The line $l$ passes through the focus $F(1,0)$ of the parabola $C$: $y^{2}=2 p x(p>0)$ and intersects $C$ at points $A$ and $B$. Then $\\frac{1}{|A F|}+\\frac{1}{|B F|}$=?",
    "fact_expressions": "l: Line;C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Coordinate(F) = (1, 0);Focus(C) = F;PointOnCurve(F, l) = True;Intersection(l, C) = {A, B};A: Point;B: Point",
    "query_expressions": "1/Abs(LineSegmentOf(B, F)) + 1/Abs(LineSegmentOf(A, F))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 5]], [[6, 32], [45, 48]], [[6, 32]], [[14, 32]], [[14, 32]], [[35, 43]], [[35, 43]], [[6, 43]], [[0, 43]], [[0, 59]], [[50, 53]], [[54, 57]]]",
    "query_spans": "[[[61, 96]]]",
    "process": "From the given, the focus of the parabola $ C: y^{2} = 2px $ ($ p > 0 $) is $ F(1, 0) $, so $ \\frac{p}{2} = 1 $, hence $ p = 2 $. Therefore, the equation of parabola $ C $ is: $ y^{2} = 4x $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. By the definition of the parabola: $ |AF| = x_{1} + 1 $, $ |BF| = x_{2} + 1 $. When the slope does not exist, $ x_{1} = x_{2} = 1 $, so: $ \\frac{1}{|AF|} + \\frac{1}{|BF|} = \\frac{1}{x_{1}+1} + \\frac{1}{x_{2}+1} = \\frac{1}{2} + \\frac{1}{2} = 1 $. When the slope exists, let the slope of line $ l $ be $ k $ ($ k \\neq 0 $), then the equation of the line is: $ y = k(x - 1) $. Solving simultaneously \n\\[\n\\begin{cases}\ny = k(x - 1) \\\\\ny^{2} = 4x\n\\end{cases}\n\\]\nyields: $ k^{2}x^{2} - 2(k^{2} + 2)x + k^{2} = 0 $, $ \\frac{1}{x_{1}+1} + \\frac{1}{x_{2}+1} = \\frac{x_{1} + x_{2} + 2}{x_{1}x_{2} + x_{1} + x_{2} + 1} = \\frac{x_{1} + x_{2} + 2}{x_{1} + x_{2} + 2} = 1 $."
  },
  {
    "text": "If the left and right foci of the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ are $F_{1}$, $F_{2}$ respectively, $P$ is a point on the right branch of $E$, $|P F_{1}|=|F_{1} F_{2}|$, $\\angle P F_{1} F_{2}=30^\\circ$, and the area of $\\triangle P F_{1} F_{2}$ is $2$, then $a=?$",
    "fact_expressions": "E: Hyperbola;a: Number;b: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(E) = F1;RightFocus(E) = F2;PointOnCurve(P, RightPart(E));Abs(LineSegmentOf(P, F1)) = Abs(LineSegmentOf(F1, F2));AngleOf(P, F1, F2) = ApplyUnit(30, degree);Area(TriangleOf(P, F1, F2)) = 2",
    "query_expressions": "a",
    "answer_expressions": "sqrt(2) + 1 - sqrt(3)",
    "fact_spans": "[[[1, 61], [90, 93]], [[188, 191]], [[8, 61]], [[86, 89]], [[70, 77]], [[78, 85]], [[8, 61]], [[8, 61]], [[1, 61]], [[1, 85]], [[1, 85]], [[86, 98]], [[99, 124]], [[127, 153]], [[154, 186]]]",
    "query_spans": "[[[188, 193]]]",
    "process": "|PF_{1}|=|F_{1}F_{2}|=2c, \\angle PF_{1}F_{2}=30^{\\circ}, the area of \\triangle PF_{1}F_{2} is 2, so we have \\frac{1}{2}\\times4c^{2}\\times\\sin30^{\\circ}=2, solving gives c=\\sqrt{2}, P(\\sqrt{6}-2,\\sqrt{2}), substituting into the hyperbola equation yields \\frac{(\\sqrt{6}-2)^{2}}{a^{2}}-\\frac{2}{2-a_{2}}=1, solving gives a=\\sqrt{2}+1-\\sqrt{3}. This problem examines the focal triangle of a hyperbola; the key to solving it is obtaining |PF_{1}|=2c, then combining with the area of the triangle to solve"
  },
  {
    "text": "The distance from the center $C$ of the circle $x^{2}-2 x+y^{2}=0$ to the directrix $l$ of the parabola $y^{2}=4 x$ is?",
    "fact_expressions": "H: Circle;Expression(H) = (y^2 + x^2 - 2*x = 0);C: Point;Center(H) = C;G: Parabola;Expression(G) = (y^2 = 4*x);l: Line;Directrix(G) = l",
    "query_expressions": "Distance(C, l)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 20]], [[0, 20]], [[23, 26]], [[0, 26]], [[27, 41]], [[27, 41]], [[44, 47]], [[27, 47]]]",
    "query_spans": "[[[23, 52]]]",
    "process": "The center of the circle $x^{2}-2x+y^{2}=0$ is $C(1,0)$, the directrix of the parabola $y^{2}=4x$ is $l:x=-1$, $d=|1-(-1)|=2$."
  },
  {
    "text": "Given that the two foci of a hyperbola are $F_{1}(-3 , 0)$ and $F_{2}(3 , 0)$, and one asymptote has the equation $y=\\sqrt{2} x$, then its standard equation is?",
    "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;Coordinate(F1) = (-3, 0);Coordinate(F2) = (3, 0);Focus(G) = {F1, F2};Expression(OneOf(Asymptote(G))) = (y = sqrt(2)*x)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/3 - y^2/6 = 1",
    "fact_spans": "[[[2, 5], [70, 71]], [[13, 28]], [[31, 45]], [[13, 28]], [[31, 45]], [[2, 45]], [[2, 68]]]",
    "query_spans": "[[[70, 78]]]",
    "process": "Let the equation of the hyperbola be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$. According to the given conditions, we have\n\\[\n\\begin{cases}\na^{2}+b^{2}=9 \\\\\n\\frac{b}{a}=\\sqrt{2}\n\\end{cases}\n\\]\nSolving this system yields\n\\[\n\\begin{cases}\na^{2}=3 \\\\\nb^{2}=6\n\\end{cases}\n\\]\nThus, the equation of the hyperbola is $\\frac{x^{2}}{3}-\\frac{y^{2}}{6}=1$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are denoted by $F_{1}$ and $F_{2}$, respectively. If there exists a point $P$ on the right branch of the hyperbola such that $|P F_{1}|=3|P F_{2}|$, then the range of the eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, F2));e: Number;Eccentricity(G) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "(1, 2]",
    "fact_spans": "[[[2, 61], [88, 91], [129, 132]], [[2, 61]], [[5, 61]], [[5, 61]], [[5, 61]], [[5, 61]], [[70, 77]], [[78, 85]], [[2, 85]], [[2, 85]], [[99, 102]], [[87, 102]], [[105, 127]], [[136, 139]], [[129, 139]]]",
    "query_spans": "[[[136, 146]]]",
    "process": "Let the x-coordinate of point P be x. Since |PF₁| = 3|PF₂|, and P lies on the right branch of the hyperbola (x ≥ a), therefore according to the second definition of hyperbola, we have e(x + \\frac{a^2}{c}) = 3e(x - \\frac{a^{2}}{c}). Thus ex = 2a. Since x ≥ a, it follows that ex ≥ ea, i.e., 2a ≥ ea. Therefore e ≤ 2. Also, since e > 1, we have 1 < e ≤ 2."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has asymptotes with equations $y=\\pm \\frac{1}{2} x$, and the hyperbola passes through the point $(\\frac{10}{3}, \\frac{4}{3})$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;Expression(Asymptote(G)) = (y = pm*(x/2));H: Point;Coordinate(H) = (10/3, 4/3);PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4-y^2=1",
    "fact_spans": "[[[2, 58], [88, 91], [124, 127]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 86]], [[92, 122]], [[92, 122]], [[88, 122]]]",
    "query_spans": "[[[124, 134]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a moving point on the right branch of the hyperbola $x^{2}-y^{2}=1$, if the distance from point $P$ to the line $y=x+2$ is always greater than $m$, then what is the range of real values for $m$?",
    "fact_expressions": "G: Hyperbola;H: Line;P: Point;m: Real;Expression(G) = (x^2 - y^2 = 1);Expression(H) = (y = x + 2);PointOnCurve(P, RightPart(G));Distance(P, H) > m",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-oo, sqrt(2)]",
    "fact_spans": "[[[6, 24]], [[39, 48]], [[2, 5], [34, 38]], [[53, 56], [61, 66]], [[6, 24]], [[39, 48]], [[2, 32]], [[34, 56]]]",
    "query_spans": "[[[61, 73]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{y^{2}}{4}-\\frac{x^{2}}{m}=1$ is $y=2x$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (y^2/4 - x^2/m = 1);Expression(OneOf(Asymptote(G))) = (y = 2*x)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 40], [58, 61]], [[5, 40]], [[2, 40]], [[2, 56]]]",
    "query_spans": "[[[58, 67]]]",
    "process": "From the given conditions, we have $ m > 0 $, and the hyperbola's foci lie on the y-axis with $ a = 2 $, $ b = \\sqrt{m} $. The asymptotes of the hyperbola are $ y = \\pm\\frac{a}{b}x = \\pm\\frac{2}{\\sqrt{m}}x $, so $ \\frac{2}{\\sqrt{m}} = 2 $, $ m = 1 $, $ b = 1 $, $ c = \\sqrt{a^{2} + b^{2}} = \\sqrt{5} $. Therefore, the eccentricity is $ e = \\frac{c}{a} = \\frac{\\sqrt{5}}{2} $."
  },
  {
    "text": "Given the circle $C$: $(x-a)^{2}+(y-b)^{2}=r^{2}$ $(b>0)$, with its center on the parabola $y^{2}=4x$, passing through the point $A(3,0)$, and tangent to the directrix of the parabola, then the equation of circle $C$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);C: Circle;Expression(C) = ((-a + x)^2 + (-b + y)^2 = r^2);r: Number;b: Number;a: Number;A: Point;b>0;PointOnCurve(A,C);Coordinate(A) = (3, 0);PointOnCurve(Center(C),G);IsTangent(C,Directrix(G))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "(x-2)^2+(y-2*sqrt(2))^2=9",
    "fact_spans": "[[[43, 57], [73, 76]], [[43, 57]], [[2, 39], [83, 87]], [[2, 39]], [[7, 39]], [[7, 39]], [[7, 39]], [[61, 70]], [[7, 39]], [[2, 70]], [[61, 70]], [[2, 58]], [[2, 81]]]",
    "query_spans": "[[[83, 92]]]",
    "process": "The directrix of the parabola $ y^{2} = 4x $ is $ x = -1 $, so $ r = |a - (-1)| = |a + 1| $. Since the circle passes through the point $ A(3,0) $, we have $ (3 - a)^{2} + (0 - b)^{2} = (a + 1)^{2} $. The center lies on the parabola $ y^{2} = 4x $, so $ b^{2} = 4a $. Solving the system of equations simultaneously gives $ a = 2 $, $ b = 2\\sqrt{2} $. Therefore, the equation of the required circle is $ (x - 2)^{2} + (y - 2\\sqrt{2})^{2} = 9 $."
  },
  {
    "text": "Given that the line $l$ intersects the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ at points $A$ and $B$, if point $P(6,3)$ is the midpoint of segment $AB$, then the equation of line $l$ is?",
    "fact_expressions": "l: Line;G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1);A: Point;B: Point;Intersection(l, G) = {A, B};P: Point;Coordinate(P) = (6, 3);MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Expression(l)",
    "answer_expressions": "A straight line does not exist",
    "fact_spans": "[[[2, 7], [72, 77]], [[8, 36]], [[8, 36]], [[39, 42]], [[43, 46]], [[2, 48]], [[50, 59]], [[50, 59]], [[50, 70]]]",
    "query_spans": "[[[72, 82]]]",
    "process": "Suppose the coordinates of A and B, using the point difference method, then according to the midpoint coordinate formula, the slope of line $ l $ can be obtained, and then a simple judgment leads to the result. According to the problem, let the slope of line $ l $ be $ k $, $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Then  \n$$\n\\begin{cases}\n\\frac{x_{1}^{2}}{4}-y_{1}^{2}=1 \\\\\n\\frac{x_{2}^{2}}{4}-y_{2}^{2}=1\n\\end{cases}\n\\Rightarrow \\frac{(x_{1}-x_{2})(x_{1}+x_{2})}{4}=(y_{1}-y_{2})(y_{1}+y_{2})\n$$\nSo  \n$$\n\\frac{y_{1}-y_{2}}{x_{1}-x_{2}} \\cdot \\frac{y_{1}+y_{2}}{x_{1}+x_{2}} = \\frac{1}{4},\n$$\nand  \n$$\n\\begin{cases}\nx_{1}+x_{2}=12 \\\\\ny_{1}+y_{2}=6\n\\end{cases}\n$$\nSo  \n$$\nk \\cdot \\frac{6}{12} = \\frac{1}{4} \\Rightarrow k = \\frac{1}{2}\n$$\nThus, the equation of line $ l $ is $ y-3 = \\frac{1}{2}(x-6) $, that is, $ y = \\frac{1}{2}x $. But $ y = \\frac{1}{2}x $ is an asymptote of the hyperbola, hence such line does not exist."
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1$ $(a > b > 0)$ with foci $F_{1}$ and $F_{2}$, if $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}} = 0$, $\\tan \\angle P F_{1} F_{2} = \\frac{1}{2}$, then the eccentricity of this ellipse is?",
    "fact_expressions": "P: Point;PointOnCurve(P,G) = True;G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Focus(G) = {F1,F2};F1: Point;F2: Point;DotProduct(VectorOf(P,F1),VectorOf(P,F2))=0;Tan(AngleOf(P,F1,F2)) = 1/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[2, 5]], [[2, 84]], [[26, 80], [190, 192]], [[26, 80]], [[28, 80]], [[28, 80]], [[28, 80]], [[28, 80]], [[6, 80]], [[7, 14]], [[15, 22]], [[86, 145]], [[147, 186]]]",
    "query_spans": "[[[190, 198]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and point $P$ lies on the ellipse such that $\\overrightarrow{F_{1} P} \\cdot \\overrightarrow{P F_{2}}=0$. Then the area of $\\triangle F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);DotProduct(VectorOf(F1, P), VectorOf(P, F2)) = 0",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[17, 44], [55, 57]], [[1, 8]], [[50, 54]], [[9, 16]], [[17, 44]], [[1, 49]], [[50, 58]], [[60, 119]]]",
    "query_spans": "[[[121, 151]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the right focus is $F$, $O$ is the coordinate origin. A circle centered at $F$ with radius $2 \\sqrt{3} a$ intersects one asymptote of the hyperbola $C$ at points $P$ and $Q$, and $\\overrightarrow{F P} \\cdot \\overrightarrow{F Q}=-6 a^{2}$. If $\\overrightarrow{O P}=\\lambda \\overrightarrow{O Q}$, then $\\lambda=?$",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(C) = F;O: Origin;Center(G) = F;Radius(G) = 2*sqrt(3)*a;G: Circle;Intersection(G, OneOf(Asymptote(C))) = {P, Q};P: Point;Q: Point;DotProduct(VectorOf(F, P), VectorOf(F, Q)) = -6*a^2;VectorOf(O, P) = lambda*VectorOf(O, Q);lambda: Number",
    "query_expressions": "lambda",
    "answer_expressions": "-2, -1/2",
    "fact_spans": "[[[2, 63], [109, 115]], [[2, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[68, 71], [82, 85]], [[2, 71]], [[72, 75]], [[81, 108]], [[89, 108]], [[107, 108]], [[107, 132]], [[123, 126]], [[127, 130]], [[134, 192]], [[195, 246]], [[248, 257]]]",
    "query_spans": "[[[248, 259]]]",
    "process": "As shown in the figure: \\because|\\overrightarrow{FP}|=|\\overrightarrow{FQ}|=r=2\\sqrt{3}a,\\therefore2\\sqrt{3}a\\times2\\sqrt{3}a\\times\\cos\\angle QFP=-6a^{2},\\angle QFP=120^{\\circ}. Draw FM\\bot PQ from point F, then: |FM|=\\sqrt{3}a, |PM|=3a. The asymptote equations are: bx-ay=0. The focus coordinates are F(c,0), then: |FM|=\\frac{|bc|}{\\sqrt{a^{2}+b^{2}}}=b. Rearranging gives: b=\\sqrt{3}a. Thus: 3a^{2}=b^{2}=c^{2}-a^{2}, \\therefore c=2a. OM=\\sqrt{(2a)^{2}-(\\sqrt{3}a)^{2}}=a. Hence: |OP|=4a, |OQ|=2a, \\therefore\\lambda=\\frac{|\\overrightarrow{OP}|}{|\\overrightarrow{OQ}|}=-2 or -\\frac{1}{2}"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$, $F_{2}$ respectively. Let $P$ be a point on $C$ distinct from the left and right vertices, $M$ be the incenter of $\\Delta P F_{1} F_{2}$, and suppose $5 \\overrightarrow{M F_{1}}+3 \\overrightarrow{M F_{2}}+3 \\overrightarrow{M P}=\\overrightarrow{0}$. Then the eccentricity of the ellipse is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C) = True;Negation(P = LeftVertex(C));Negation(P = RightVertex(C));M: Point;Incenter(TriangleOf(P, F1, F2)) = M;5*VectorOf(M, F1) + 3*VectorOf(M, F2) + 3*VectorOf(M, P) = 0",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "3/8",
    "fact_spans": "[[[2, 59], [87, 90], [231, 233]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[67, 74]], [[75, 82]], [[2, 82]], [[2, 82]], [[83, 86]], [[83, 100]], [[83, 100]], [[83, 100]], [[101, 104]], [[101, 129]], [[131, 228]]]",
    "query_spans": "[[[231, 239]]]",
    "process": "Let $ P(x_{0},y_{0}) $, $ M(x,y) $, we have $ F_{1}(-c,0) $, $ F_{2}(c,0) $. Then $ \\overrightarrow{MF_{1}}=(-c-x,-y) $, $ \\overrightarrow{MF_{2}}=(c-x,-y) $, $ \\overrightarrow{MP}=(x_{0}-x,y_{0}-y) $. Since $ 5\\overrightarrow{MF_{1}}+3\\overrightarrow{MF_{2}}+3\\overrightarrow{MP}=\\overrightarrow{0} $, it follows that $ 5(-c-x,-y)+3(c-x,-y)+3(x_{0}-x,y_{0}-y)=(0,0) $. Then we obtain $ x=\\frac{3x_{0}-2c}{2} $, $ y=\\frac{3y_{0}}{11} $, and the inradius of $ \\triangle PF_{1}F_{2} $ is $ \\frac{3|y_{0}|}{11} $. By the definition of an ellipse, $ |PF_{1}|+|PF_{2}|=2a $, and $ |F_{1}F_{2}|=2c $. Thus, $ S_{\\Delta PF_{1}F_{2}} = \\frac{1}{2}(|PF_{1}|+|PF_{2}|+|F_{1}F_{2}|)\\cdot\\frac{3|y_{0}|}{11} = \\frac{1}{2}|F_{1}F_{2}|\\cdot|y_{0}| $, i.e., $ \\frac{1}{2}(2a+2c)\\cdot\\frac{3|y_{0}|}{11} = \\frac{1}{2}2c\\cdot|y_{0}| $. Then we get $ 3a=8c $, so the eccentricity is $ \\frac{c}{a} = \\frac{3}{8} $."
  },
  {
    "text": "If a focus of the hyperbola $\\frac{x^{2}}{3} + \\frac{y^{2}}{p}=1$ lies on the directrix of the parabola $y^{2}=2 p x$, then the value of $p$ is?",
    "fact_expressions": "G: Hyperbola;p: Number;H: Parabola;Expression(G) = (x^2/3 + y^2/p = 1);Expression(H) = (y^2 = 2*(p*x));PointOnCurve(OneOf(Focus(G)), Directrix(H))",
    "query_expressions": "p",
    "answer_expressions": "-6",
    "fact_spans": "[[[1, 41]], [[69, 72]], [[47, 63]], [[1, 41]], [[47, 63]], [[1, 67]]]",
    "query_spans": "[[[69, 76]]]",
    "process": ""
  },
  {
    "text": "The sum of the distances from points on curve $C$ to $F_{1}(0,-1)$ and $F_{2}(0,1)$ is $4$. Then the equation of curve $C$ is?",
    "fact_expressions": "C: Curve;F1: Point;F2: Point;P: Point;PointOnCurve(P, C);Coordinate(F1) = (0, -1);Coordinate(F2) = (0, 1);Distance(P, F1) + Distance(P, F2) = 4",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/4 + x^2/3 = 1",
    "fact_spans": "[[[0, 5], [47, 52]], [[9, 22]], [[24, 36]], [[7, 8]], [[0, 8]], [[9, 22]], [[24, 36]], [[7, 45]]]",
    "query_spans": "[[[47, 57]]]",
    "process": "From the given conditions, the sum of the distances from any point on curve C to $ F_{1}(0,-1) $ and $ F_{2}(0,1) $ is 4. Therefore, by the definition of an ellipse, this curve is an ellipse. Since the foci are $ F_{1}(0,-1) $ and $ F_{2}(0,1) $, the foci of the ellipse lie on the y-axis. Moreover, $ a=2 $, $ c=1 $, so $ b=\\sqrt{3} $. Thus, the equation of the ellipse is: $ \\frac{y^{2}}{4} + \\frac{x^{2}}{3} = 1 $."
  },
  {
    "text": "Given real numbers $x$, $y$ satisfying $x|x| = y|y| + 1$, then the range of $x^{2}+y^{2}-2xy$ is?",
    "fact_expressions": "x_:Real;y_:Real;x_*Abs(x_)=y_*Abs(y_)+1",
    "query_expressions": "Range(-2*x_^2+y_^2+x_^2+y_^2)",
    "answer_expressions": "(0, 2]",
    "fact_spans": "[[[1, 6]], [[8, 11]], [[13, 26]]]",
    "query_spans": "[[[28, 54]]]",
    "process": "From $ x|x| = y|y| + 1 $, when $ x \\geqslant 0, y \\geqslant 0 $, we get $ x^{2} - y^{2} = 1 $, representing the part of a hyperbola with asymptotes $ y = x $ and foci on the $ x $-axis located in the first quadrant (including the coordinate axes). When $ x > 0, y < 0 $, we get $ x^{2} + y^{2} = 1 $, representing the portion in the fourth quadrant of a circle centered at the origin with radius 1. When $ x \\leqslant 0, y > 0 $, we get $ x^{2} + y^{2} = -1 $, which does not represent any graph. When $ x \\leqslant 0, y < 0 $, we get $ y^{2} - x^{2} = 1 $, representing the part of a hyperbola with asymptotes $ y = x $ and foci on the $ y $-axis located in the third quadrant (including the coordinate axes). The graph is shown as follows. $ x^{2} + y^{2} - 2xy = (x - y)^{2} $. Let $ z = x - y $, then $ y = x - z $. From the graph, when the line $ y = x - z $ is tangent to $ \\begin{cases} x^{2} + y^{2} = 1 \\\\ x > 0, y < 0 \\end{cases} $, $ z^{2} $ is maximized. At this time, $ \\frac{|-z|}{\\sqrt{1+1}} = 1 $, so $ z^{2} = 2 $, that is, the maximum value of $ x^{2} + y^{2} - 2xy $ is 2. When the line $ y = x - z $ approaches infinitely close to the asymptote $ y = x $ of the hyperbola, $ z $ tends to 0. Therefore, the range of $ x^{2} + y^{2} - 2xy $ is $ (0, 2] $."
  },
  {
    "text": "The equation of the hyperbola passing through the points $(-\\sqrt{2}, \\sqrt{3})$, $(\\frac{\\sqrt{5}}{3}, \\sqrt{2})$ is?",
    "fact_expressions": "G: Hyperbola;H: Point;I: Point;Coordinate(H) = (-sqrt(2), sqrt(3));Coordinate(I) = (sqrt(5)/3, sqrt(2));PointOnCurve(H,G);PointOnCurve(I,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[61, 64]], [[2, 26]], [[28, 60]], [[2, 26]], [[28, 60]], [[0, 64]], [[0, 64]]]",
    "query_spans": "[[[61, 68]]]",
    "process": ""
  },
  {
    "text": "Given point $M(0,2)$, a line $AB$ passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. If $\\angle AMF = \\frac{\\pi}{2}$, then the coordinates of point $B$ are?",
    "fact_expressions": "M: Point;Coordinate(M) = (0, 2);G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;A: Point;B: Point;PointOnCurve(F, LineOf(A, B));Intersection(LineOf(A, B), G) = {A, B};AngleOf(A, M, F) = pi/2",
    "query_expressions": "Coordinate(B)",
    "answer_expressions": "(1/4, -1)",
    "fact_spans": "[[[2, 11]], [[2, 11]], [[13, 27], [42, 45]], [[13, 27]], [[30, 33]], [[13, 33]], [[46, 49]], [[50, 53], [87, 91]], [[12, 41]], [[34, 55]], [[57, 85]]]",
    "query_spans": "[[[87, 95]]]",
    "process": "From the parabola equation we get: F(1,0). Let the equation of line AB be: x = my + 1. Let A(x_{1},y_{1}), B(x_{2},y_{2}).  \nSolving the system  \n\\begin{cases}x=my+1\\\\y^{2}=4x\\end{cases}  \nwe obtain: y^{2}-4my-4=0.  \n\\therefore y_{1}y_{2}=-4.  \n\\because \\angle AMF = \\frac{\\pi}{2},  \n\\therefore \\overrightarrow{AM} \\cdot \\overrightarrow{MF} = 0.  \nAlso, \\overrightarrow{AM} = (-x_{1}, 2-y_{1}), \\overrightarrow{MF} = (1, -2),  \n4 + 2v = 0.  \nAnd y_{1}^{2} = 4x_{1}.  \n\\because y_{1}y_{2} -----  \nAlso y_{2}^{2} = 4x_{2},  \n\\therefore B(\\frac{1}{4}, -1).  \nThe correct answer to this problem: (\\frac{1}{4}, -1)"
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2 x$ be $F$, and let a line passing through point $F$ intersect the parabola at points $A$ and $B$. Then the minimum value of $A F+4 B F$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 2*x);Focus(G) = F;PointOnCurve(F,H);Intersection(H,G)={A,B}",
    "query_expressions": "Min(LineSegmentOf(A, F) + 4*LineSegmentOf(B, F))",
    "answer_expressions": "4.5",
    "fact_spans": "[[[1, 15], [33, 36]], [[29, 31]], [[37, 40]], [[19, 22], [24, 28]], [[41, 44]], [[1, 15]], [[1, 22]], [[23, 31]], [[29, 46]]]",
    "query_spans": "[[[48, 65]]]",
    "process": "According to the problem, the focus of the parabola has coordinates: F(1,0). A line passing through the focus intersects the parabola y^{2}=2x at two points, and the slope of the line must exist. Let the equation of the line passing through the focus F(1,0) and intersecting the parabola at points A(x_{1},y_{1}) and B(x_{2},y_{2}) be: y=k(x-1). Substituting into y^{2}=2x, it simplifies to: k^{2}x^{2}-(k^{2}+2)x+\\frac{1}{4}k^{2}=0. By Vieta's formulas, we obtain: x_{1}x_{2}=\\frac{1}{4}. According to the definition of the parabola: AF+4BF=x_{1}+\\frac{1}{2}+4(x_{2}+\\frac{1}{2})=x_{1}+4x_{2}+\\frac{5}{2}\\geqslant2\\sqrt{4x_{1}x_{2}}+\\frac{5}{2}=2\\sqrt{4\\times\\frac{1}{4}}+\\frac{5}{2}=\\frac{9}{2} (equality holds if and only if \"x_{1}=4x_{2}=1\"). Therefore, the minimum value of AF+4BF is 4.5."
  },
  {
    "text": "The focus of the parabola $y^{2}=a x$ is $(-1,0)$, then $a$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = a*x);a: Number;Coordinate(Focus(G)) = (-1, 0)",
    "query_expressions": "a",
    "answer_expressions": "-4",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[28, 31]], [[0, 26]]]",
    "query_spans": "[[[28, 33]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 35]]]",
    "process": "e=\\frac{c}{a}=\\frac{\\sqrt{4+1}}{2}=\\frac{\\sqrt{5}}{2}.【Topic Location】Hyperbola and its eccentricity"
  },
  {
    "text": "If a hyperbola shares the same asymptotes as $\\frac{x^{2}}{8}-y^{2}=1$ and has the same foci as the ellipse $\\frac{x^{2}}{20}+\\frac{y^{2}}{2}=1$, then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;C:Curve;Expression(H) = (x^2/20 + y^2/2 = 1);Expression(C)= (x^2/8 - y^2 = 1);Asymptote(G)=Asymptote(C);Focus(G) = Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16 - y^2/2 = 1",
    "fact_spans": "[[[3, 6], [88, 91]], [[41, 79]], [[7, 32]], [[41, 79]], [[7, 32]], [[3, 38]], [[3, 85]]]",
    "query_spans": "[[[88, 96]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{4 m^{2}}=1(m>0)$ are $y=\\pm 2 x$, then $m$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/(4*m^2) + x^2/m = 1);m: Number;m>0;Expression(Asymptote(G)) = (y = pm*(2*x))",
    "query_expressions": "m",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 51]], [[2, 51]], [[71, 74]], [[5, 51]], [[2, 69]]]",
    "query_spans": "[[[71, 76]]]",
    "process": "Since $a^{2}=m$, $b^{2}=4m^{2}$, so $a=\\sqrt{m}$, $b=2m$, the asymptote equations are $v=\\pm2\\sqrt{m}x$. Then $\\frac{b}{a}=2\\sqrt{m}=2$, solving gives $m=1$. The answer is 1."
  },
  {
    "text": "Given that the distance from a point on the parabola $y^{2}=4x$ to the focus is $5$, what are the coordinates of this point?",
    "fact_expressions": "G: Parabola;P:Point;Expression(G) = (y^2 = 4*x);PointOnCurve(P,G);Distance(P,Focus(G))=5",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(4,\\pm 4)",
    "fact_spans": "[[[2, 16]], [[31, 32]], [[2, 16]], [[2, 19]], [[2, 29]]]",
    "query_spans": "[[[31, 37]]]",
    "process": ""
  },
  {
    "text": "Through the right focus $F_{2}$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a, b>0$), a line perpendicular to the $x$-axis intersects an asymptote at a point $P$ in the first quadrant. $F_{1}$ is the left focus, and the angle of inclination of the line $F_{1}P$ is $\\frac{\\pi}{4}$. Then the eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;F2: Point;a>0;b>0;RightFocus(G) = F2;H: Line;PointOnCurve(F2,H) = True;IsPerpendicular(H,xAxis) = True;P: Point;Intersection(H,Asymptote(G)) = P;Quadrant(P) = 1;F1: Point;LeftFocus(G) = F1;Inclination(LineOf(F1,P)) = pi/4;e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[1, 55], [139, 142]], [[1, 55]], [[4, 55]], [[4, 55]], [[59, 66]], [[4, 55]], [[4, 55]], [[1, 66]], [[75, 77]], [[0, 77]], [[67, 77]], [[90, 93]], [[0, 93]], [[83, 93]], [[94, 101]], [[0, 105]], [[106, 137]], [[146, 149]], [[139, 149]]]",
    "query_spans": "[[[146, 151]]]",
    "process": "First, find the coordinates of point P, then obtain the relationship between a and c from the slope formula, and finally get the answer using the eccentricity formula. From the given conditions, we can find: $ P(c,\\frac{bc}{a}) $, then $ k_{F_{1}P} = \\tan\\frac{\\pi}{4} = \\frac{bc}{2c} = \\frac{b}{2a} \\Rightarrow b = 2a \\Rightarrow c^{2} - a^{2} = 4a^{2} $. Therefore, $ c^{2} = 5a^{2} $, yielding $ e^{2} = 5 \\Rightarrow e = \\sqrt{5} $."
  },
  {
    "text": "Draw a chord $AB$ of the parabola $y^{2}=8x$ passing through $M(1, 0)$. If $|AB|=\\frac{8 \\sqrt{10}}{3}$, then the inclination angle of the line $AB$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);M: Point;Coordinate(M) = (1, 0);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);PointOnCurve(M, LineSegmentOf(A, B));Abs(LineSegmentOf(A, B)) = (8*sqrt(10))/3",
    "query_expressions": "Inclination(LineOf(A, B))",
    "answer_expressions": "{ApplyUnit(60, degree), ApplyUnit(120, degree)}",
    "fact_spans": "[[[12, 26]], [[12, 26]], [[1, 11]], [[1, 11]], [[28, 33]], [[28, 33]], [[12, 33]], [[0, 33]], [[35, 64]]]",
    "query_spans": "[[[66, 79]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is an arbitrary point on the ellipse $\\frac{x^{2}}{4}+y=1$, and $F_{1}$, $F_{2}$ are the two foci of the ellipse. Then the minimum value of $|P F_{1}|^{2}+|P F_{2}|^{2}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y = 1);PointOnCurve(P, G);Focus(G) = {F1, F2}",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F1))^2 + Abs(LineSegmentOf(P, F2))^2)",
    "answer_expressions": "8",
    "fact_spans": "[[[6, 29], [51, 53]], [[2, 5]], [[35, 42]], [[43, 50]], [[6, 29]], [[2, 34]], [[35, 58]]]",
    "query_spans": "[[[60, 95]]]",
    "process": "Using the important inequality and combining with the definition of an ellipse, we can directly solve it. From $\\frac{|PF_{1}|+|PF_{2}|}{2}\\leqslant\\sqrt{\\frac{|PF_{1}|^{2}+|PF_{2}|^{2}}{2}}\\Rightarrow\\frac{|PF_{1}|^{2}+|PF_{2}|^{2}}{2}\\geqslanta^{2}\\Rightarrow|PF_{1}|^{2}+|PF_{2}|^{2}\\geqslant2a^{2}=8$ (equality holds if and only if $|PF_{1}|=|PF_{2}|$)"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{4}+y^{2}=1$, a line passing through $F_{1}$ intersects the ellipse $C$ at points $A$ and $B$. Then, what is the perimeter of $\\triangle A B F_{2}$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;G: Line;PointOnCurve(F1, G);A: Point;B: Point;Intersection(G, C) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "8",
    "fact_spans": "[[[18, 50], [68, 73]], [[18, 50]], [[2, 9], [57, 64]], [[10, 17]], [[2, 55]], [[2, 55]], [[65, 67]], [[56, 67]], [[74, 77]], [[78, 81]], [[65, 83]]]",
    "query_spans": "[[[85, 111]]]",
    "process": "Since points A and B lie on the ellipse, |AF₁| + |AF₂| = 2a = 4, |BF₁| + |BF₂| = 2a = 4, so the perimeter of △ABF₂ is |AB| + |AF₂| + |BF₂| = |AF₁| + |AF₂| + |BF| + |BF₂| = 8."
  },
  {
    "text": "Given circle $C_{1}$: $(x+3)^{2}+y^{2}=1$ and circle $C_{2}$: $(x-3)^{2}+y^{2}=9$, a moving circle $M$ is externally tangent to both circle $C_{1}$ and circle $C_{2}$. Then the trajectory equation of the center $M$ of the moving circle is?",
    "fact_expressions": "M: Circle;C1: Circle;C2: Circle;Expression(C1) = ((x + 3)^2 + y^2 = 1);Expression(C2) = ((x - 3)^2 + y^2 = 9);IsOutTangent(M, C1);IsOutTangent(M, C2);M1: Point;Center(M) = M1",
    "query_expressions": "LocusEquation(M1)",
    "answer_expressions": "(x^2 - y^2/8 = 1)&(x <= -1)",
    "fact_spans": "[[[62, 65]], [[2, 30], [68, 76]], [[31, 59], [77, 85]], [[2, 30]], [[31, 59]], [[62, 87]], [[62, 87]], [[94, 97]], [[89, 97]]]",
    "query_spans": "[[[94, 104]]]",
    "process": "As shown in the figure, let the moving circle M be externally tangent to circle $ C_{1} $ and circle $ C_{2} $ at points A and B, respectively. According to the condition for two circles being externally tangent, we have $ |MC_{1}| - |AC_{1}| = |MA| $, $ |MC_{2}| - |BC_{2}| = |MB| $. Since $ |MA| = |MB| $, it follows that $ |MC_{1}| - |AC_{1}| = |MC_{2}| - |BC_{2}| $, i.e., $ |MC_{2}| - |MC_{1}| = |BC_{2}| - |AC_{1}| = 2 $. Thus, the difference of distances from point M to the two fixed points $ C_{2} $, $ C_{1} $ is a constant and less than $ |C_{1}C_{2}| = 6 $. By the definition of a hyperbola, the trajectory of the moving point M is the left branch of a hyperbola, where $ a = 1 $, $ c = 3 $, then $ b^{2} = 8 $. Hence, the equation of the trajectory of point M is $ x^{2} - \\frac{y^{2}}{8} = 1 $ $ (x \\leqslant -1) $."
  },
  {
    "text": "Given that $F$ is the left focus of the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$, $P$ is an arbitrary point on the ellipse $C$, and the coordinates of point $Q$ are $(4,3)$, then the maximum value of $|P Q|+|P F|$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/2 + y^2 = 1);F: Point;LeftFocus(C) = F;P: Point;PointOnCurve(P, C);Q: Point;Coordinate(Q) = (4, 3)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, F)) + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "5*sqrt(2)",
    "fact_spans": "[[[6, 38], [47, 52]], [[6, 38]], [[2, 5]], [[2, 42]], [[43, 46]], [[43, 57]], [[58, 62]], [[58, 73]]]",
    "query_spans": "[[[75, 94]]]",
    "process": "\\because point F is the left focus of the ellipse \\frac{x^{2}}{2}+y^{2}=1, \\therefore F(-1,0), let the right focus of the ellipse be F(1,0). \\because point P is any point on the ellipse C, and the coordinates of point Q are (4,3), |PQ|+|PF|=|PQ|+2\\sqrt{2}-|PF|=2\\sqrt{2}+|PQ|-|PF|. Also, \\because |PQ|-|PF|\\leqslant|QF|=3\\sqrt{2}, \\therefore |PQ|+|PF|\\leqslant5\\sqrt{2}, that is, the maximum value of |PQ|+|PF| is 5\\sqrt{2}, at which point Q, F, P are collinear."
  },
  {
    "text": "The parabola $C$: $y^{2}=4 x$ has focus $F$ and directrix $l$. Let $M$ be a point on $C$ in the first quadrant, and point $N$ lies on $l$. Given that $M F \\perp N F$, $|M F|=5$, find the coordinates of point $P$, the intersection of line $M N$ with the $y$-axis.",
    "fact_expressions": "C: Parabola;M: Point;N: Point;F: Point;l: Line;Expression(C) = (y^2 = 4*x);Focus(C) = F;Directrix(C) = l;Quadrant(M) = 1;PointOnCurve(M, C);PointOnCurve(N, l);IsPerpendicular(LineSegmentOf(M, F), LineSegmentOf(N, F));Abs(LineSegmentOf(M, F)) = 5;Intersection(LineOf(M,N),yAxis)=P;P:Point",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(0, 2)",
    "fact_spans": "[[[0, 19], [40, 43]], [[36, 39]], [[54, 58]], [[23, 26]], [[30, 34], [59, 62]], [[0, 19]], [[0, 26]], [[0, 34]], [[35, 53]], [[35, 53]], [[54, 63]], [[66, 81]], [[83, 92]], [[94, 111]], [[108, 111]]]",
    "query_spans": "[[[108, 116]]]",
    "process": "Let $ M(x_{0},y_{0}) $, then $ x_{0}+1=5 $, $ x_{0}=4 $, from $ x_{0}=\\frac{y_{0}^{2}}{4} $ we get $ y_{0}=4 $. Let $ M_{1} $ be the projection of $ M $ on $ l $, by the definition of parabola, $ |MF|=|MM_{1}| $. Since $ MF\\bot NF $, we have $ \\triangle MFN \\cong \\triangle MM_{1}N $, hence $ MN $ perpendicularly bisects $ M_{1}F $, and the line $ MN $ passes through the midpoint of segment $ M_{1}F $. Because $ l \\parallel y $-axis, the midpoint of $ M_{1}F $ lies on the $ y $-axis. Given $ M_{1}(-1,4) $, $ F(1,0) $, the coordinates of point $ P $ are $ (0,2) $."
  },
  {
    "text": "If point $P$ is a point on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ with foci $F_{1}$ and $F_{2}$, satisfying $P F_{1} \\perp P F_{2}$ and $P F_{1}=2 P F_{2}$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));LineSegmentOf(P, F1) = 2*LineSegmentOf(P, F2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[26, 72], [125, 128]], [[26, 72]], [[29, 72]], [[29, 72]], [[1, 5]], [[1, 75]], [[7, 14]], [[15, 22]], [[6, 72]], [[78, 101]], [[103, 122]]]",
    "query_spans": "[[[125, 134]]]",
    "process": "Let $|PF_{2}| = x$, then $PF_{1} = 2x$. Since point $P$ is a point on the hyperbola $\\frac{x^{2}}{a^{2}} - \\frac{x^{2}}{b^{2}} = 1$ with foci $F_{1}, F_{2}$ satisfying $PF_{1} \\perp PF_{2}$, we have $x^{2} + 4x^{2} = 4c^{2}$, solving gives $x = \\frac{2\\sqrt{5}}{5}c$, so $|PF_{2}| = \\frac{2\\sqrt{5}}{5}c$, $|PF_{1}| = \\frac{4\\sqrt{5}}{5}c$. By the definition of hyperbola, $a = \\frac{1}{2}(|PF_{1}| - |PF_{2}|) = \\frac{\\sqrt{5}}{5}c$, therefore the eccentricity of the hyperbola $e = \\frac{c}{a} = \\frac{c}{\\frac{\\sqrt{5}}{5}c} = \\sqrt{5}$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has eccentricity $\\sqrt{3}$, $A$, $B$ are the left and right vertices, point $P$ is an arbitrary point on hyperbola $C$ in the first quadrant, point $O$ is the origin, and the slopes of lines $PA$, $PB$, $PO$ are $k_{1}$, $k_{2}$, $k_{3}$ respectively. Let $m=k_{1} k_{2} k_{3}$, then the range of $m$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;A: Point;P: Point;B: Point;O: Origin;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(C) = sqrt(3);LeftVertex(C)=A;RightVertex(C)=B;PointOnCurve(P, C);Quadrant(P)=1;Slope(LineOf(P,A))=k1;Slope(LineOf(P,B))=k2;Slope(LineOf(P,O))=k3;k1:Number;k2:Number;k3:Number;m:Number;m=k1*k2*k3",
    "query_expressions": "Range(m)",
    "answer_expressions": "(0,2*sqrt(2))",
    "fact_spans": "[[[2, 63], [102, 108]], [[10, 63]], [[10, 63]], [[81, 84]], [[97, 101]], [[87, 90]], [[119, 123]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 78]], [[2, 96]], [[2, 96]], [[97, 118]], [[97, 118]], [[130, 184]], [[130, 184]], [[130, 184]], [[157, 164]], [[166, 174]], [[177, 184]], [[209, 212]], [[186, 207]]]",
    "query_spans": "[[[209, 219]]]",
    "process": ""
  },
  {
    "text": "The line $l$ passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. If $|AF|=4$, then the area of $\\triangle OAB$ ($O$ being the origin) is?",
    "fact_expressions": "l: Line;G: Parabola;O: Origin;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G)=F;PointOnCurve(F, l);Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, F)) = 4",
    "query_expressions": "Area(TriangleOf(O,A,B))",
    "answer_expressions": "4*sqrt(3)/3",
    "fact_spans": "[[[22, 27]], [[1, 15], [28, 31]], [[73, 76]], [[33, 36]], [[37, 40]], [[18, 21]], [[1, 15]], [[1, 21]], [[0, 27]], [[22, 42]], [[44, 53]]]",
    "query_spans": "[[[55, 87]]]",
    "process": "The coordinates of point A can be found using the focal radius formula. After finding the equation of line AB, solving it together with the parabola equation gives the coordinates of point B; then the area of triangle can be obtained by splitting it into two triangles. From the given condition, F(1,0). Assume A(x_{1},y_{1}) lies in the first quadrant, |AF|=x_{1}+1=4, x_{1}=3, y_{1}=2\\sqrt{3}. Let B(x_{2},y_{2}), k_{AB}=\\frac{2\\sqrt{3}}{3-1}=\\sqrt{3}, \\therefore AB: y=\\sqrt{3}(x-1). Solving the system of equations \\begin{cases}y=\\sqrt{3}(x-1)\\end{cases}, simplifying yields 3x^{2}-10x+3=0, solving gives x_{2}=\\frac{1}{3}, y_{2}=-\\frac{2\\sqrt{3}}{3}. S_{\\DeltaOAB}=\\frac{1}{2}|OF|\\cdot|y_{1}|+\\frac{1}{2}|OF|\\cdot|y_{2}|=\\frac{4\\sqrt{3}}{3}."
  },
  {
    "text": "Given a point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ such that the distance from $P$ to the right directrix is $10$, then the distance from point $P$ to its left focus is?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(P, G);Distance(P, RightDirectrix(G)) = 10",
    "query_expressions": "Distance(P, LeftFocus(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 41], [66, 67]], [[44, 47], [61, 65]], [[2, 41]], [[2, 47]], [[2, 59]]]",
    "query_spans": "[[[61, 76]]]",
    "process": ""
  },
  {
    "text": "Let $A(x_{1} , y_{1})$ , $B(x_{2} , y_{2})$ be two points on the parabola $y=2 x^{2}$, and line $l$ be the perpendicular bisector of $AB$. The line $l$ passes through the focus $F$ of the parabola if and only if $x_{1}+x_{2}$ takes what value?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;F: Point;x1: Number;x2: Number;y1: Number;y2: Number;Expression(G) = (y = 2*x^2);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);PointOnCurve(A, G);PointOnCurve(B, G);PerpendicularBisector(LineSegmentOf(A, B)) = l;Focus(G) = F;PointOnCurve(F, l)",
    "query_expressions": "x1 + x2",
    "answer_expressions": "{(5/4, +oo), 0}",
    "fact_spans": "[[[60, 65], [101, 106]], [[41, 55], [107, 110]], [[1, 20]], [[22, 40]], [[113, 116]], [[1, 20]], [[22, 40]], [[1, 20]], [[22, 40]], [[41, 55]], [[1, 20]], [[22, 40]], [[1, 59]], [[1, 59]], [[60, 77]], [[107, 116]], [[101, 116]]]",
    "query_spans": "[[[83, 99]]]",
    "process": ""
  },
  {
    "text": "Given that the conic section $C$ passes through the point $(2,1)$ and has an eccentricity of $\\sqrt{2}$, what is the standard equation of the curve $C$?",
    "fact_expressions": "C:ConicSection;H: Point;Coordinate(H) = (2, 1);PointOnCurve(H, C);Eccentricity(C) = sqrt(2)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/3 - y^2/3 = 1",
    "fact_spans": "[[[2, 9], [35, 40]], [[10, 18]], [[10, 18]], [[2, 18]], [[2, 33]]]",
    "query_spans": "[[[35, 47]]]",
    "process": "Since the eccentricity of the conic section C is \\sqrt{2}, the curve C is a hyperbola. Let its real semi-axis length be a, imaginary semi-axis length be b, and semi-focal distance be c. Then from e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{a^{2}+b^{2}}{a^{2}}=1+\\frac{b^{2}}{a^{2}}=2 we obtain: a=b, meaning the curve C is an equilateral hyperbola. Assume its equation is x^{2}-y^{2}=\\lambda (\\lambda\\neq0). Since the point (2,1) lies on curve C, then \\lambda=3. Thus, the equation of curve C is: x^{2}-y^{2}=3. Therefore, the standard equation of curve C is \\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1."
  },
  {
    "text": "If the line $l$ passing through the point $P(1,1)$ with slope $k$ has only one common point with the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$, then $k=$?",
    "fact_expressions": "l: Line;G: Hyperbola;P: Point;Expression(G) = (x^2 - y^2/4 = 1);Coordinate(P) = (1, 1);PointOnCurve(P, l);Slope(l) = k;NumIntersection(l, G) = 1;k: Number",
    "query_expressions": "k",
    "answer_expressions": "{5/2, pm*2}",
    "fact_spans": "[[[19, 24]], [[25, 53]], [[2, 11]], [[25, 53]], [[2, 11]], [[1, 24]], [[12, 24]], [[19, 60]], [[62, 65]]]",
    "query_spans": "[[[62, 67]]]",
    "process": "From the given conditions, we have $ l: y = k(x - 1) + 1 $. Substituting into the hyperbola equation yields $ (4 - k^{2})x^{2} - 2(k - k^{2})x - k^{2} + 2k - 5 = 0 $. When $ 4 - k^{2} = 0 $, i.e., $ k = \\pm 2 $, the line $ l $ is parallel to the asymptotes of the hyperbola, and thus the line intersects the hyperbola at only one point. When $ 4 - k^{2} \\neq 0 $, $ A = 4(k - k^{2})^{2} - 4(4 - k^{2})(-k^{2} + 2k - 5) = 0 $, solving gives $ k = \\frac{5}{2} $. In conclusion, when $ k = \\frac{5}{2} $ or $ k = \\pm 2 $, the line intersects the hyperbola at only one point."
  },
  {
    "text": "Given the hyperbola $C$: $4 y^{2}-9 x^{2}=-36$, then  \n($1$) The length of the real axis of hyperbola $C$ is ?, the length of the imaginary axis is ?  \n($2$) The coordinates of the foci of hyperbola $C$ are ?, the eccentricity is ?  \n($3$) The equations of the asymptotes of hyperbola $C$ are?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-9*x^2 + 4*y^2 = (-1)*36)",
    "query_expressions": "Length(RealAxis(C));Length(ImageinaryAxis(C));Coordinate(Focus(C));Eccentricity(C);Expression(Asymptote(C))",
    "answer_expressions": "4\n6\n{(sqrt(13),0),(-sqrt(13),0)}\nsqrt(13)/2\ny=pm*3*x/2",
    "fact_spans": "[[[2, 31], [40, 46], [66, 72], [93, 99]], [[2, 31]]]",
    "query_spans": "[[[40, 52]], [[40, 59]], [[66, 79]], [[66, 86]], [[93, 107]]]",
    "process": "Transform the hyperbola equation into the standard form $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$, we can see that the semi-transverse axis length is $a=\\sqrt{4}=2$, and the semi-conjugate axis length is $b=\\sqrt{9}=3$. (1) Since $a=2$, $b=3$, the transverse axis length of hyperbola $C$ is $2a=4$, and the conjugate axis length is $2b=6$. (2) Since $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{4+9}=\\sqrt{13}$, the foci coordinates of hyperbola $C$ are $(\\sqrt{13},0)$, $(-\\sqrt{13},0)$. Because $a=2$, $c=\\sqrt{13}$, the eccentricity of hyperbola $C$ is $e=\\frac{c}{a}=\\frac{\\sqrt{13}}{2}$. (3) Let $4y^{2}-9x^{2}=0$, simplifying gives $y=\\pm\\frac{3}{2}x$, thus the asymptotes equations of hyperbola $C$ are $y=\\pm\\frac{3}{2}x$. [Comment] This problem examines the hyperbola equation and tests simple geometric properties of hyperbolas, belonging to basic problems."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with left focus $F_{1}(-c , 0)$, right focus $F_{2}(c, 0)$, if there exists a point $P$ on the ellipse such that $|P F_{1}|=2 c$, $\\angle F_{1} P F_{2}=30^{\\circ}$, then the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;c: Number;F1: Point;F2: Point;P: Point;e: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 2*c;AngleOf(F1, P, F2) = ApplyUnit(30, degree);Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "(\\sqrt{3} - 1)/2",
    "fact_spans": "[[[2, 54], [92, 94], [156, 158]], [[4, 54]], [[4, 54]], [[58, 73]], [[58, 73]], [[77, 90]], [[99, 102]], [[162, 165]], [[4, 54]], [[4, 54]], [[2, 54]], [[58, 73]], [[77, 90]], [[2, 73]], [[2, 90]], [[92, 102]], [[104, 119]], [[120, 153]], [[156, 165]]]",
    "query_spans": "[[[162, 167]]]",
    "process": "From the definition of the ellipse, we have 2a = |PF₁| + |PF₂|. Given |PF₁| = 2c, it follows that |PF₂| = 2a - 2c. In triangle F₁PF₂, by the law of cosines, we obtain cos∠F₁PF₂ = cos30° = (|PF₁|² + |PF₂|² - |F₁F₂|²) / (2|PF₁|·|PF₂|) = (4c² + (2a - 2c)² - 4c²) / (2·2c·(2a - 2c)). Simplifying yields c = √3(a - c)."
  },
  {
    "text": "Given a line $L$ with slope $-\\frac{1}{2}$ intersecting the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ at points $A$ and $B$, if point $P(2, 1)$ is the midpoint of $AB$, then the eccentricity of $C$ equals?",
    "fact_expressions": "L: Line;Slope(L) = -1/2;C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;A: Point;B: Point;Intersection(L, C) = {A, B};P: Point;Coordinate(P) = (2, 1);MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[20, 25]], [[2, 25]], [[26, 83], [117, 120]], [[26, 83]], [[33, 83]], [[33, 83]], [[33, 83]], [[33, 83]], [[84, 87]], [[89, 92]], [[20, 94]], [[96, 107]], [[96, 107]], [[96, 115]]]",
    "query_spans": "[[[117, 127]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ \\frac{x_{1}^{2}}{a^{2}} + \\frac{y_{1}^{2}}{b^{2}} = 1 $, $ \\frac{x_{2}^{2}}{a^{2}} + \\frac{y_{2}^{2}}{b^{2}} = 1 $. Since a line with slope $ -\\frac{1}{2} $ intersects the ellipse $ C: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $) at points $ A $, $ B $, and point $ P(2,1) $ is the midpoint of $ AB $, we have $ x_{1} + x_{2} = 4 $, $ y_{1} + y_{2} = 2 $, $ \\frac{y_{1} - y_{2}}{x_{1} - x_{2}} = -\\frac{1}{2} $. Then it follows that $ \\frac{4}{a^{2}} + (-\\frac{1}{2}) \\times \\frac{2}{b^{2}} = 0 $, $ \\therefore a = 2b $, $ \\therefore c = \\sqrt{3}b $, $ \\therefore e = \\frac{c}{a} = \\frac{\\sqrt{3}}{2} $."
  },
  {
    "text": "If an ellipse and a hyperbola have the same foci $F_{1}$, $F_{2}$, point $P$ is a common point of the two curves, and $\\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}}=0$, where $e_{1}$, $e_{2}$ are their eccentricities respectively, then the value of $\\frac{1}{e_{1} ^2}+\\frac{1}{e_{2}^2}$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;P: Point;F1: Point;F2: Point;e1:Number;e2:Number;Focus(G)={F1,F2};Focus(H)={F1,F2};OneOf(Intersection(G,H))=P;DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Eccentricity(H)=e1;Eccentricity(G)=e2",
    "query_expressions": "1/(e1^2) + 1/(e2^2)",
    "answer_expressions": "2",
    "fact_spans": "[[[4, 7]], [[1, 3]], [[28, 32]], [[12, 19]], [[20, 27]], [[104, 111]], [[112, 119]], [[1, 27]], [[1, 27]], [[28, 43]], [[46, 103]], [[104, 128]], [[104, 128]]]",
    "query_spans": "[[[130, 172]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the directrix of the parabola $y^{2}=-12 x$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -12*x)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=3",
    "fact_spans": "[[[0, 16]], [[0, 16]]]",
    "query_spans": "[[[0, 22]]]",
    "process": "2p=12, p=6, the focus is (-3,0), therefore the directrix is x=3."
  },
  {
    "text": "The focal length of the hyperbola $C$: $\\frac{x^{2}}{3}-y^{2}=1$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/3 - y^2 = 1)",
    "query_expressions": "FocalLength(C)",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 33]], [[0, 33]]]",
    "query_spans": "[[[0, 38]]]",
    "process": "Directly use the focal length formula to obtain the answer. [Detailed solution] Hyperbola C: \\frac{x^{2}}{3}-y^{2}=1, c^{2}=a^{2}+b^{2}=4, c=2, the focal length is 2c=4"
  },
  {
    "text": "It is known that the line $x+3y-7=0$ intersects the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{b^{2}}=1$ $(0<b<3)$ at two points $A$ and $B$. The two foci of the ellipse are $F_{1}$ and $F_{2}$, and the midpoint of segment $AB$ is $C(1,2)$. Then the area of $\\Delta C F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;b: Number;H: Line;B: Point;A: Point;C: Point;F1: Point;F2: Point;0 < b;b < 3;Expression(G) = (x^2/9 + y^2/b^2 = 1);Expression(H) = (x + 3*y - 7 = 0);Coordinate(C) = (1, 2);Intersection(H, G) = {A, B};Focus(G)={F1,F2};MidPoint(LineSegmentOf(A,B)) = C",
    "query_expressions": "Area(TriangleOf(C, F1, F2))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[16, 64], [77, 79]], [[18, 64]], [[2, 15]], [[71, 74]], [[67, 70]], [[114, 122]], [[87, 94]], [[95, 102]], [[18, 64]], [[18, 64]], [[16, 64]], [[2, 15]], [[114, 122]], [[2, 76]], [[77, 102]], [[103, 122]]]",
    "query_spans": "[[[124, 151]]]",
    "process": "Using the point difference method, combined with the given conditions, we can find $ b^{2} $. Based on the relationship among $ a $, $ b $, and $ c $, we can determine the value of $ c $, and substitute into the formula to obtain the required result. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Since the slope of line $ AB $ is $ -\\frac{1}{3} $, then  \n$$\n\\begin{cases}\n\\frac{x^{2}}{9}+\\frac{y^{2}}{b^{2}}=1 \\\\\n\\frac{x^{2}}{9}+\\frac{y^{2}}{b^{2}}=1\n\\end{cases}\n$$\nSubtracting the two equations gives  \n$$\n\\frac{y_{1}-y_{2}}{x_{2}} = -\\frac{b^{2}(x_{1}+x_{2})}{9(y_{1})}\n$$\nThe midpoint of segment $ AB $ is $ C(1,2) $, so  \n$$\n-\\frac{2b^{2}}{9\\times4} = -\\frac{1}{3}\n$$\nSolving yields $ b^{2}=6 $. Since $ a=3 $, we have $ c=\\sqrt{3} $. Hence, the area of $ \\triangle CF_{1}F_{2} $ is  \n$$\n\\frac{1}{2}\\times2\\sqrt{3}\\times2=2\\sqrt{3}\n$$"
  },
  {
    "text": "Given that the circle $x^{2}+y^{2}-2 x-3=0$ is tangent to the directrix of the parabola $y=2 p x^{2}  (p>0)$, then $p$=?",
    "fact_expressions": "G: Parabola;p: Number;H: Circle;p>0;Expression(G) = (y = 2*(p*x^2));Expression(H) = (-2*x + x^2 + y^2 - 3 = 0);IsTangent(H, Directrix(G))",
    "query_expressions": "p",
    "answer_expressions": "1/16",
    "fact_spans": "[[[25, 48]], [[55, 58]], [[2, 24]], [[28, 48]], [[25, 48]], [[2, 24]], [[2, 53]]]",
    "query_spans": "[[[55, 60]]]",
    "process": "Circle $x^{2}+y^{2}-2x-3=0\\Rightarrow(x-1)^{2}+y^{2}=4$, center $(1,0)$, radius $r=2$. Parabola $y=2px^{2}\\Rightarrow x^{2}=\\frac{1}{2p}y$, directrix equation $y=-\\frac{1}{8p}$. Since the circle $(x-1)^{2}+y^{2}=4$ is tangent to $y=-\\frac{1}{8p}$, it follows that $\\frac{1}{8p}=2$, solving gives $p=\\frac{1}{16}$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F$ is the right focus of the ellipse, and $AB$ is a chord passing through the center $O$ of the ellipse. Then the maximum area of $\\triangle ABF$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;RightFocus(G) = F;A: Point;B: Point;O: Origin;Center(G) = O;PointOnCurve(O, LineSegmentOf(A, B));IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "Max(Area(TriangleOf(A, B, F)))",
    "answer_expressions": "b*sqrt(a^2-b^2)",
    "fact_spans": "[[[2, 56], [61, 63], [75, 77]], [[2, 56]], [[4, 56]], [[4, 56]], [[4, 56]], [[4, 56]], [[57, 60]], [[57, 67]], [[68, 73]], [[68, 73]], [[79, 82]], [[75, 82]], [[68, 84]], [[68, 84]]]",
    "query_spans": "[[[86, 111]]]",
    "process": "The area of 4ABF is equal to the sum of the areas of 4AOF and BFOF. Let the distance from point A to the x-axis be h. Since AB is a chord passing through the center of the ellipse, the distance from point B to the x-axis is also h. Therefore, the areas of AAOF and BBOF are equal, so S_{AABF} = \\frac{1}{2}c \\times 2h = ch. Since the maximum value of h is b, the maximum area of 4ABF is bc = b\\sqrt{a^{2}-b^{2}}."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $x^{2}=4 y$, $P$ is a moving point on the parabola, and the coordinates of point $A$ are $(0,-1)$, then the maximum value of $\\frac{\\sqrt{2}|P A|+|P F|}{|P F|}$ is?",
    "fact_expressions": "G: Parabola;P: Point;A: Point;F: Point;Expression(G) = (x^2 = 4*y);Coordinate(A) = (0, -1);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Max((sqrt(2)*Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))/Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "3",
    "fact_spans": "[[[6, 20], [28, 31]], [[24, 27]], [[37, 41]], [[2, 5]], [[6, 20]], [[37, 53]], [[2, 23]], [[24, 35]]]",
    "query_spans": "[[[55, 96]]]",
    "process": "\\frac{\\sqrt{2}|PA|+|PF|}=\\sqrt{2}\\frac{|PA|}{|PF|}+1' From the definition of the parabola, |PF| equals the distance from P to the directrix y=-1. Let \\theta be the angle between line PA and the directrix. Then \\sqrt{2}\\frac{|PA|}{|PF|}+1=\\frac{\\sqrt{2}}{\\sin\\theta}+1'\\textcircled{1} If the slope of PA does not exist, then the original expression = \\sqrt{2}+1\\textcircled{2} If the slope of PA exists, when PA is tangent to the parabola, \\sin\\theta is minimized. Let the equation of line PA be y=kx-1. Solving simultaneously \\begin{cases}y=kx-1\\\\x^{2}=4y\\end{cases}, we obtain x^{2}-4kx+4=0. From \\triangle=0, we get |k|=1, so |\\tan\\theta|=1, hence \\sin\\theta=\\frac{\\sqrt{2}}{2}. At this point, \\sqrt{2}\\frac{|PA|}{|PF|}+1=\\frac{\\sqrt{2}}{\\frac{\\sqrt{2}}{2}}+1=3"
  },
  {
    "text": "If point $P(2,-1)$ is the midpoint of chord $A B$ of the circle $(x-1)^{2}+y^{2}=25$, then the equation of line $A B$ is?",
    "fact_expressions": "G: Circle;Expression(G) = (y^2 + (x - 1)^2 = 25);P: Point;Coordinate(P) = (2, -1);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Expression(LineOf(A, B))",
    "answer_expressions": "x-y-3=0",
    "fact_spans": "[[[12, 33]], [[12, 33]], [[1, 11]], [[1, 11]], [[35, 40]], [[35, 40]], [[12, 40]], [[1, 43]]]",
    "query_spans": "[[[45, 57]]]",
    "process": ""
  },
  {
    "text": "Given that the focus of the parabola is $F$, the intersection point of the directrix and the $x$-axis is $M$, $N$ is a point on the parabola, and $|N F|=\\frac{2}{3}|M N|$, then $\\tan \\angle N M F$=?",
    "fact_expressions": "G: Parabola;Focus(G) = F;F: Point;Intersection(Directrix(G),xAxis) = M;M: Point;PointOnCurve(N,G) = True;N: Point;Abs(LineSegmentOf(N, F)) = (2/3)*Abs(LineSegmentOf(M, N))",
    "query_expressions": "Tan(AngleOf(N, M, F))",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 5], [32, 35]], [[2, 12]], [[9, 12]], [[2, 27]], [[24, 27]], [[28, 39]], [[28, 31]], [[43, 67]]]",
    "query_spans": "[[[69, 90]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of a hyperbola has the equation $y=2x$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(OneOf(Asymptote(G))) = (y = 2*x)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{sqrt(5),sqrt(5)/2}",
    "fact_spans": "[[[2, 5], [23, 26]], [[2, 21]]]",
    "query_spans": "[[[23, 32]]]",
    "process": "If the hyperbola equation is \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1, since one of its asymptotes is y=2x, then \\frac{b}{a}=2. Therefore, e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{a^{2}+b^{2}}{a^{2}}=1+\\frac{b^{2}}{a^{2}}=5, e=\\sqrt{5}. If the hyperbola equation is \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{\\frac{x}{12}}=1, since one of its asymptotes is y=2x, then \\frac{a}{b}=2. Therefore, e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{a^{2}+b^{2}}{a^{2}}=1+\\frac{b^{2}}{a^{2}}=\\frac{5}{4}, e=\\frac{\\sqrt{5}}{2}."
  },
  {
    "text": "If the sum of the distances from a point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ to its two foci is $m-3$, then what is the eccentricity of this ellipse?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/m = 1);m: Number;K: Point;PointOnCurve(K, G);F1: Point;F2: Point;Focus(G) = {F1, F2};Distance(K, F1)+Distance(K, F2) = m-3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[1, 38], [60, 62]], [[1, 38]], [[52, 57]], [], [[1, 41]], [], [], [[1, 46]], [[1, 57]]]",
    "query_spans": "[[[60, 68]]]",
    "process": "When m<4, by the definition of ellipse we have m-3=4, solving gives m=7, which does not satisfy the condition. When m>4, by the definition of ellipse we have m-3=2\\sqrt{m}, solving gives m=9, so e=\\frac{c}{a}=\\frac{\\sqrt{9-4}}{3}=\\frac{\\sqrt{5}}{3}, hence fill in \\frac{\\sqrt{5}}{3}."
  },
  {
    "text": "It is known that the focus of the parabola $y^{2}=8x$ coincides with the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$. Then, the asymptotes of the hyperbola are given by?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 8*x);G: Hyperbola;Expression(G) = (-y^2 + x^2/a^2 = 1);a: Number;Focus(H) = RightFocus(G)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "x \\pm \\sqrt{3} y = 0",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[20, 52], [60, 63]], [[20, 52]], [[23, 52]], [[2, 58]]]",
    "query_spans": "[[[60, 71]]]",
    "process": "Since the focus of the parabola $ y^{2} = 8x $ is $ (2,0) $, we have $ a^{2} + 1^{2} = 2^{2} $, then $ a^{2} = 3 $, so the asymptotes are: $ y = \\pm\\frac{b}{a}x = \\pm\\frac{\\sqrt{3}}{3}x $, that is, $ x \\pm \\sqrt{3}y = 0 $."
  },
  {
    "text": "Draw a line $l$ with slope $k$ through the focus $F$ of the parabola $C$: $y^{2}=2 p x$ ($p>0$). The line $l$ intersects $C$ at points $A$ and $B$. If $|A B|=\\frac{5}{2} p$, then $k=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;l: Line;PointOnCurve(F,l) = True;Slope(l) = k;k: Number;Intersection(l, C) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(A, B)) = (5/2)*p",
    "query_expressions": "k",
    "answer_expressions": "pm*2",
    "fact_spans": "[[[1, 27], [53, 56]], [[1, 27]], [[9, 27]], [[9, 27]], [[30, 33]], [[1, 33]], [[41, 46], [49, 52]], [[0, 46]], [[34, 46]], [[37, 40], [92, 95]], [[49, 67]], [[58, 61]], [[62, 65]], [[69, 90]]]",
    "query_spans": "[[[92, 97]]]",
    "process": "It is clear from the problem that $k \\neq 0$. Let $l: y = k\\left(x - \\frac{p}{2}\\right)$. Solving the system \n\\[\n\\begin{cases}\ny = k\\left(x - \\frac{p}{2}\\right) \\\\\ny^2 = 2px,\n\\end{cases}\n\\]\neliminating variables gives $k^2x^2 - (k^2 + 2}px + \\frac{k^2p^2}{4} = 0$. Let $A(x_1, y_1)$, $B(x_2, y_2)$, then $x_1 + x_2 = \\frac{k^2 + 2}{k^2}p$, so $|AB| = x_1 + x_2 + p = \\frac{2k^2 + 2}{k^2}p$. Therefore, $\\frac{2k^2 + 2}{k^2} = \\frac{5}{2}$, solving gives $k = \\pm 2$."
  },
  {
    "text": "Given that point $P$ is a moving point on the ellipse $C$: $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$, $F_{1}$ and $F_{2}$ are the left and right foci respectively, and $O$ is the origin, then the range of $\\frac{|{|PF_{1}|-|PF_{2}|}|}{|OP|}$ is?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;O: Origin;Expression(C) = (x^2/8 + y^2/4 = 1);PointOnCurve(P, C);LeftFocus(C) = F1;RightFocus(C) = F2",
    "query_expressions": "Range(Abs((Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2))))/Abs(LineSegmentOf(O, P)))",
    "answer_expressions": "[0, \\sqrt{2}]",
    "fact_spans": "[[[7, 49]], [[2, 6]], [[54, 61]], [[64, 71]], [[80, 83]], [[7, 49]], [[2, 53]], [[7, 79]], [[7, 79]]]",
    "query_spans": "[[[90, 134]]]",
    "process": ""
  },
  {
    "text": "The line passing through the left focus $F$ of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ also passes through the upper vertex $B$ of $C$, and intersects the ellipse at another point $A$. If $|B F|=3|A F|$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;LeftFocus(C) = F;G: Line;PointOnCurve(F, G);B: Point;UpperVertex(C) = B;PointOnCurve(B, G);A: Point;Intersection(G, C) = {A,B};Abs(LineSegmentOf(B, F)) = 3*Abs(LineSegmentOf(A, F))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[1, 58], [69, 72], [82, 84], [112, 115]], [[1, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[62, 65]], [[1, 65]], [[66, 68]], [[0, 68]], [[76, 79]], [[69, 79]], [[66, 79]], [[91, 94]], [[66, 94]], [[96, 110]]]",
    "query_spans": "[[[112, 121]]]",
    "process": "From the given conditions, we have B(0,b), F(-c,0). From |BF| = 3|AF|, we obtain point A(-\\frac{4}{3}c, -\\frac{b}{3}). Since point A lies on the ellipse, it satisfies: \\frac{(-\\frac{4}{3}c)^{2}}{a^{2}} + \\frac{(-\\frac{b}{3})^{2}}{b^{2}} = 1. Simplifying yields: \\frac{16}{9} \\cdot \\frac{c^{2}}{a^{2}} = \\frac{8}{9}, \\therefore e^{2} = \\frac{c^{2}}{a^{2}} = \\frac{1}{2}, e = \\frac{\\sqrt{2}}{2}"
  },
  {
    "text": "Let point $P$ be an intersection point of ellipse $C_{1}$ and hyperbola $C_{2}$ that share common foci $F_{1}$, $F_{2}$, and $P F_{1} \\perp P F_{2}$. The eccentricity of ellipse $C_{1}$ is $e_{1}$, the eccentricity of hyperbola $C_{2}$ is $e_{2}$. If $e_{2}=3 e_{1}$, then $e_{1}$=?",
    "fact_expressions": "P: Point;F1: Point;F2: Point;C1: Ellipse;C2: Hyperbola;Focus(C1) = {F1, F2};Focus(C2) = {F1, F2};OneOf(Intersection(C1, C2)) = P;IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2)) = True;e1: Number;e2: Number;Eccentricity(C1) = e1;Eccentricity(C2) = e2;e2 = 3*e1",
    "query_expressions": "e1",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[1, 5]], [[11, 18]], [[19, 26]], [[27, 36], [78, 87]], [[37, 47], [100, 110]], [[6, 47]], [[6, 47]], [[1, 52]], [[54, 77]], [[92, 99], [141, 148]], [[115, 122]], [[78, 99]], [[100, 122]], [[124, 139]]]",
    "query_spans": "[[[141, 150]]]",
    "process": "Without loss of generality, let $F_{1}, F_{2}$ be the left and right foci, respectively, the semi-major axis of the ellipse be $a_{1}$, the semi-transverse axis of the hyperbola be $a_{2}$, and $P$ be the intersection point of the ellipse and the hyperbola in the first quadrant. According to the definitions of the ellipse and hyperbola, we set up the equations to solve for $|PF_{1}|$ and $|PF_{2}|$. Using the Pythagorean theorem and combining with the eccentricity formulas of the ellipse and hyperbola, we obtain $\\frac{1}{e_{1}^{2}} + \\frac{1}{e_{2}^{2}} = 2$. Then, given $e_{2} = 3e_{1}$, we can solve for the result.  \n\n**Detailed solution:** Without loss of generality, let $F_{1}, F_{2}$ be the left and right foci, respectively, the semi-major axis of the ellipse be $a_{1}$, the semi-transverse axis of the hyperbola be $a_{2}$, and $P$ be the intersection point of the ellipse and the hyperbola in the first quadrant. Then, according to the definitions of the ellipse and hyperbola, we have  \n$$\n\\begin{cases}\n|PF_{1}| + |PF_{2}| = 2a_{1} \\\\\n|PF_{1}| - |PF_{2}| = 2a_{2}\n\\end{cases}\n$$\nSolving gives $|PF_{1}| = a_{1} + a_{2}$, $|PF_{2}| = a_{1} - a_{2}$. Since $PF_{1} \\perp PF_{2}$, it follows that $|PF_{1}|^{2} + |PF_{2}|^{2} = |F_{1}F_{2}|^{2}$, i.e.,  \n$$\n(a_{1} + a_{2})^{2} + (a_{1} - a_{2})^{2} = (2c)^{2}\n$$\nSimplifying yields $a_{1}^{2} + a_{2}^{2} = 2c^{2}$, so  \n$$\n\\left(\\frac{a_{1}}{c}\\right)^{2} + \\left(\\frac{a_{2}}{c}\\right)^{2} = 2\n$$\nthat is,  \n$$\n\\frac{1}{e_{1}^{2}} + \\frac{1}{e_{2}^{2}} = 2\n$$\nMoreover, since $e_{2} = 3e_{1}$, we have $e_{1}^{2} = \\frac{5}{9}$, hence $e_{1} = \\frac{\\sqrt{5}}{3}$."
  },
  {
    "text": "The distance from the focus $F$ of the parabola $y=\\frac{1}{4} x^{2}$ to the asymptote of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;F: Point;Expression(G) = (x^2/4 - y^2 = 1);Expression(H) = (y = x^2/4);Focus(H) = F",
    "query_expressions": "Distance(F, Asymptote(G))",
    "answer_expressions": "(2/5)*sqrt(5)",
    "fact_spans": "[[[31, 59]], [[0, 24]], [[27, 30]], [[31, 59]], [[0, 24]], [[0, 30]]]",
    "query_spans": "[[[0, 67]]]",
    "process": "First determine the position of the focus of the parabola, thereby obtaining the coordinates of the focus of the parabola. Then, using the given conditions, find the asymptote equations of the hyperbola, and substitute into the point-to-line distance formula to obtain the result. The focus of the parabola $ y = \\frac{1}{4}x^{2} $ lies on the y-axis, and $ p = 2 $, $ \\therefore $ the focus coordinates of the parabola $ y = \\frac{1}{4}x^{2} $ are $ (0,1) $. From the problem: the asymptote equations of the hyperbola $ \\frac{x^{2}}{4} - y^{2} = 1 $ are $ x \\pm 2y = 0 $. $ \\therefore $ the distance $ d $ from $ F $ to its asymptote is $ d = \\frac{2}{\\sqrt{1+4}} \\frac{2}{5}\\sqrt{5} $."
  },
  {
    "text": "Given that the equation of the parabola $\\Gamma$ is $y^{2}=6 x$, and the line $l$ passes through the focus of the parabola and has an inclination angle of $120^{\\circ}$, then the length of the chord intercepted on the line $l$ by the parabola $\\Gamma$ is?",
    "fact_expressions": "Gamma: Parabola;Expression(Gamma) = (y^2 = 6*x);l: Line;PointOnCurve(Focus(Gamma), l);Inclination(l) = ApplyUnit(120, degree)",
    "query_expressions": "Length(InterceptChord(l, Gamma))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 13], [36, 39], [62, 73]], [[2, 28]], [[29, 34], [74, 79]], [[29, 42]], [[29, 60]]]",
    "query_spans": "[[[62, 86]]]",
    "process": "From $ y^{2} = 6x $, it is known that the directrix of the parabola is $ x = -1.5 $, the focus is $ F(1.5, 0) $, and $ p = 3 $. Since the inclination angle of line $ l $ is $ 120^{\\circ} $, and line $ l $ passes through point $ F(1.5, 0) $, the equation of line $ l $ is $ y - 0 = -\\tan 60^{\\circ}(x - 1.5) $, that is, $ y = -\\sqrt{3}x + \\frac{3\\sqrt{3}}{2} $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Solving simultaneously the equations of the line and the parabola:\n$$\n\\begin{cases}\ny^{2} = 6x \\\\\ny = -\\sqrt{3}x + \\frac{3\\sqrt{3}}{2}\n\\end{cases}\n$$\nSimplifying and rearranging yields $ 4x^{2} - 20x + 9 = 0 $, so $ x_{1} + x_{2} = 5 $. By the definition of the parabola, $ |AB| = p + x_{1} + x_{2} = 8 $."
  },
  {
    "text": "A point $A$ on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ has a distance of $\\frac{5}{2}$ to the left focus. Then, what is the distance from point $A$ to the right directrix?",
    "fact_expressions": "G: Ellipse;A: Point;Expression(G) = (x^2/4 + y^2/3 = 1);PointOnCurve(A, G);Distance(A, LeftFocus(G)) = 5/2",
    "query_expressions": "Distance(A,RightDirectrix(G))",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 37]], [[40, 43], [66, 70]], [[0, 37]], [[0, 43]], [[0, 64]]]",
    "query_spans": "[[[0, 79]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. $P$ is a point on the right branch of the hyperbola such that $|P F_{1}| \\cdot|P F_{2}|=3 a^{2}$. Then the range of the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, RightPart(C));Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 3*a^2",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(1, 2]",
    "fact_spans": "[[[2, 63], [92, 95], [139, 145]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[72, 79]], [[80, 87]], [[2, 87]], [[2, 87]], [[88, 91]], [[88, 101]], [[103, 137]]]",
    "query_spans": "[[[139, 156]]]",
    "process": "Let the focal distance of hyperbola C be $2c$. From the given conditions, we have  \n$$\n\\begin{cases}\n|PF_{1}|-|PF_{2}|=2a \\\\\n|PF_{1}|\\cdot|PF_{2}|=3a^{2}\n\\end{cases}\n$$\nSolving gives $|PF_{1}|=3a$, $|PF_{2}|=a$. From $|PF_{2}|\\geqslant c-a$ we obtain $a\\geqslant c-a$, then $\\frac{c}{a}\\leqslant 2$, that is, the range of the eccentricity of hyperbola C is $(1,2)$."
  },
  {
    "text": "Given that one asymptote of the hyperbola $C$ has the equation $x+2y=0$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(OneOf(Asymptote(C))) = (x + 2*y = 0)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "{sqrt(5)/2, sqrt(5)}",
    "fact_spans": "[[[2, 8], [28, 34]], [[2, 26]]]",
    "query_spans": "[[[28, 40]]]",
    "process": "When the foci of hyperbola C are on the x-axis, then $\\frac{b}{a}=\\frac{1}{2}$, and since $c^{2}=a^{2}+b^{2}$, it follows that $\\frac{c^{2}}{a^{2}}=\\frac{5}{4}$, $e=\\frac{\\sqrt{5}}{2}$. When the foci of hyperbola C are on the y-axis, then $\\frac{a}{b}=\\frac{1}{2}$, and since $c^{2}=a^{2}+b^{2}$, it follows that $\\frac{c^{2}}{a^{2}}=5$, $e=\\sqrt{5}$."
  },
  {
    "text": "Given the ellipse $\\frac{y^{2}}{9}+\\frac{x^{2}}{5}=1$, the upper focus is $F$, $M$ is a point on the ellipse, and point $A(2 \\sqrt{3}, 0)$. When point $M$ moves on the ellipse, the maximum value of $|M A|+|M F|$ is?",
    "fact_expressions": "G: Ellipse;A: Point;M: Point;F: Point;Expression(G) = (x^2/5 + y^2/9 = 1);Coordinate(A) = (2*sqrt(3), 0);UpperFocus(G) = F;PointOnCurve(M, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F)))",
    "answer_expressions": "10",
    "fact_spans": "[[[2, 39], [52, 54]], [[58, 77]], [[48, 51], [79, 83]], [[44, 47]], [[2, 39]], [[58, 77]], [[2, 47]], [[48, 57]]]",
    "query_spans": "[[[91, 110]]]",
    "process": "Let the lower focus of the ellipse be $ F $. By the definition of the ellipse: $ |MF| = 6 - |MF| $, and using $ |MA| - |MF'| \\leqslant |AF'| $, the maximum value of $ |MA| + |MF| $ can be obtained. As shown in the figure, let the lower focus of the ellipse be $ F' $.  \n$ \\frac{y^{2}}{9} + \\frac{x^{2}}{5} = 1 \\therefore a = 3, 2a = 6 $, again $ \\because |MF| + |MF| = 2a = 6 $, i.e., $ |MF| = 6 - |MF| $,  \n$ \\therefore |MA| + |MF| = |MA| - |MF| + 6 $,  \nagain $ \\because |MA| - |MF| \\leqslant |AF'| $. Equality holds if and only if $ A, F, M $ are collinear and $ F $ lies on segment $ AM $,  \n$ \\because c = \\sqrt{a^{2} - b^{2}} = \\sqrt{9 - 5} = 2, \\cdot F'(0, -2) $  \n$ \\therefore |AF'| = \\sqrt{(2\\sqrt{3})^{2} + 2^{2}} = 4 $  \n$ \\therefore |MA| - |MF| \\leqslant 4 $,  \n$ \\therefore $ the maximum value of $ |MA| + |MF| $ is $ 4 + 6 = 10 $"
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, $F_{1}$ and $F_{2}$ are the two foci of the ellipse, and when $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$, then the area of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2};AngleOf(F1,P,F2)=pi/3",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[6, 33], [54, 56]], [[2, 5]], [[38, 45]], [[46, 53]], [[6, 33]], [[2, 37]], [[38, 61]], [[63, 99]]]",
    "query_spans": "[[[102, 129]]]",
    "process": "As shown in the figure, $2a=4$, $c=\\sqrt{a^{2}-b^{2}}=\\sqrt{3}$, $\\therefore|PF_{1}|+|PF_{2}|=2a=4$. By the cosine law: $|F_{1}F_{2}|^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}||PF_{2}|\\cos60^{\\circ}$. $\\therefore4c^{2}=(|PF_{1}|+|PF_{2}|)^{2}-3|PF_{1}||PF_{2}|$, that is, $|PF_{1}||PF_{2}|=\\frac{4}{3}$. $\\therefore$ The area $S$ of $\\Delta F_{1}PF_{2}$ is $S=\\frac{1}{2}|PF_{1}||PF_{2}|\\sin60^{\\circ}=\\frac{1}{2}\\times\\frac{4}{3}\\times\\frac{\\sqrt{3}}{2}=\\frac{\\sqrt{3}}{3}$."
  },
  {
    "text": "Given that $(4 , 2)$ is the midpoint of the segment of line $l$ intercepted by the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$, then the equation of $l$ is?",
    "fact_expressions": "l: Line;G: Ellipse;Expression(G) = (x^2/36 + y^2/9 = 1);Coordinate(MidPoint(InterceptChord(l,G))) = (4, 2)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x+2*y-8=0",
    "fact_spans": "[[[12, 17], [67, 70]], [[18, 56]], [[18, 56]], [[2, 65]]]",
    "query_spans": "[[[67, 75]]]",
    "process": ""
  },
  {
    "text": "What is the eccentricity of the hyperbola $x^{2}-\\frac{y^{2}}{35}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/35 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "6",
    "fact_spans": "[[[0, 29]], [[0, 29]]]",
    "query_spans": "[[[0, 35]]]",
    "process": "From the hyperbola $x^{2}-\\frac{y^{2}}{35}=1$, we have $a=1$, $b=\\sqrt{35}$, so $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{36}=6$, therefore the eccentricity $e=\\frac{c}{a}=6$."
  },
  {
    "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{9}-y^{2}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "pm*x-3*y=0",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 36]]]",
    "process": "From the hyperbola equation, it is known that the foci of the hyperbola lie on the horizontal axis, $ a=3 $, $ b=1 $, so the asymptotes of the hyperbola $ \\frac{x^{2}}{9}-y^{2}=1 $ are: $ y=\\pm\\frac{b}{a}x \\Rightarrow y=\\pm\\frac{1}{3}x \\Rightarrow \\pm x-3y=0 $"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $AB$ is a chord passing through $F_{2}$ on the right branch, $\\overrightarrow{A F_{2}}=\\frac{3}{4} \\overrightarrow{A B}$ and $|A F_{1}|=|A F_{2}|+\\frac{1}{2}|A B|$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;A: Point;B: Point;F2: Point;F1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;IsChordOf(LineSegmentOf(A,B),RightPart(C));PointOnCurve(F2,LineSegmentOf(A,B));VectorOf(A,F2)=(3/4)*VectorOf(A,B);Abs(LineSegmentOf(A,F1))=Abs(LineSegmentOf(A,F2))+(1/2)*Abs(LineSegmentOf(A,B))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(10)/2",
    "fact_spans": "[[[22, 83], [212, 215]], [[30, 83]], [[30, 83]], [[90, 95]], [[90, 95]], [[12, 19], [100, 107]], [[2, 9]], [[30, 83]], [[30, 83]], [[22, 83]], [[2, 89]], [[2, 89]], [[22, 111]], [[90, 111]], [[112, 171]], [[172, 210]]]",
    "query_spans": "[[[212, 221]]]",
    "process": "As shown in the figure, in the hyperbola, |AF₁| - |AF₂| = 2a, |AF₁| = |AF₂| + (1/2)|AB|, ∴ |AB| = 4a, |BF₂| = a. Also, by the definition of hyperbola, |AF| = 5a, |BF₁| = 3a. Hence in △AF₁B, |AF| = 5a, |BF₁| = 3a, |AB| = 4a, ∴ ∠ABF₂ = 90°. In △BF₁F₂, |BF₁| = 3a, |BF₂| = a, |F₁F₂| = 2c, hence (3a)² + a² = (2c)² ∴ e² = c²/a² = 5/2, e = √10 / (√10 / {})"
  },
  {
    "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-y^{2}=1  (a>0)$ intersect the line $l$: $y + x = 1$ at two distinct points $A$, $B$. Let point $P$ be the intersection of line $l$ and the $y$-axis, and $PA = \\frac{5}{12} PB$. Then $a = $?",
    "fact_expressions": "l: Line;C: Hyperbola;a: Number;P: Point;A: Point;B: Point;a>0;Expression(C) = (-y^2 + x^2/a^2 = 1);Negation(A=B);Intersection(C, l) = {A, B};Expression(l)=(y+x=1);Intersection(l,yAxis)=P;LineSegmentOf(P, A) = (5/12)*LineSegmentOf(P, B)",
    "query_expressions": "a",
    "answer_expressions": "17/13",
    "fact_spans": "[[[45, 59], [75, 80]], [[1, 44]], [[115, 118]], [[88, 91]], [[66, 69]], [[70, 73]], [[8, 44]], [[1, 44]], [[62, 73]], [[1, 73]], [[45, 59]], [[75, 91]], [[93, 113]]]",
    "query_spans": "[[[115, 120]]]",
    "process": ""
  },
  {
    "text": "Let the semi-focal length of the hyperbola be $c$, the distance between the two directrices be $d$, and $c = d$. Then the eccentricity $e$ of the hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;c: Number;HalfFocalLength(G) = c;l1: Line;l2: Line;Directrix(G) = {l1,l2};Distance(l1,l2) = d;d: Number;c = d;e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 4], [34, 37]], [[9, 12]], [[1, 12]], [], [], [[1, 17]], [[1, 25]], [[22, 25]], [[27, 32]], [[41, 44]], [[34, 44]]]",
    "query_spans": "[[[41, 47]]]",
    "process": ""
  },
  {
    "text": "Given point $Q(2 \\sqrt{2}, 0)$, and point $P(x_{0}, y_{0})$ is a moving point on the parabola $y=\\frac{1}{4} x^{2}$, then the minimum value of $y_{0}+|P Q|$ is?",
    "fact_expressions": "Q: Point;Coordinate(Q) = (2*sqrt(2), 0);P: Point;x0: Number;y0: Number;Coordinate(P) = (x0, y0);G: Parabola;Expression(G) = (y = x^2/4);PointOnCurve(P, G)",
    "query_expressions": "Min(y0 + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 21]], [[2, 21]], [[22, 40]], [[23, 40]], [[23, 40]], [[22, 40]], [[41, 65]], [[41, 65]], [[22, 69]]]",
    "query_spans": "[[[71, 90]]]",
    "process": "By the given condition, the focus of the parabola is $ F(0,1) $, so $ y_{0} = |PF| - 1 $, hence $ y_{0} + |PQ| = |PF| + |PQ| - 1 $. Also, $ |PF| + |PQ| - 1 \\geqslant |FQ| - 1 = \\sqrt{(2\\sqrt{2})^{2} + 1} - 1 = 2 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Let $P$ be a point on the right branch of the hyperbola such that $(\\overrightarrow{O P}+\\overrightarrow{O F_{2}}) \\cdot \\overrightarrow{P F_{2}}=0$ ($O$ is the origin), and $2|\\overrightarrow{P F_{1}}|=3|\\overrightarrow{P F_{2}}|$. Find the eccentricity of the hyperbola $C$.",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;O: Origin;P: Point;F2: Point;F1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) =F1;RightFocus(C)=F2;PointOnCurve(P,RightPart(C));DotProduct((VectorOf(O, F2) + VectorOf(O, P)),VectorOf(P, F2) )= 0;2*Abs(VectorOf(P, F1)) = 3*Abs(VectorOf(P, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(13)",
    "fact_spans": "[[[2, 63], [93, 96], [258, 264]], [[10, 63]], [[10, 63]], [[187, 190]], [[88, 91]], [[80, 87]], [[72, 79]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 87]], [[2, 87]], [[88, 101]], [[103, 186]], [[199, 256]]]",
    "query_spans": "[[[258, 270]]]",
    "process": "\\because(\\overrightarrow{OP}+\\overrightarrow{OF})\\cdot\\overrightarrow{PF_{2}}=0,\\therefore\\triangle OPF_{2} is an isosceles triangle and |OP|=|OF_{2}|, also |OF_{1}|=|OF_{2}| \\therefore|OP|=|OF_{1}|=|OF_{2}|,\\therefore PF_{1}\\bot PF_{2}. Also |PF_{1}|-|PF_{2}|=2a, 2|PF_{1}|=3|PF_{2}| \\therefore|PF_{1}|=6a,|PF_{2}|=4a, then (6a)^{2}+(4a)^{2}=(2c)^{2}, we obtain \\frac{c^{2}}{a^{2}}=13.\\therefore the eccentricity of hyperbola C is \\sqrt{13}"
  },
  {
    "text": "Given the ellipse $E$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, with left and right foci $F_{1}$ and $F_{2}$ respectively, a fixed point $A(2, \\sqrt{5})$, and a moving point $P$ on the ellipse $E$, then the maximum value of $|P A|+|P F_{1}|$ is?",
    "fact_expressions": "E: Ellipse;A: Point;P: Point;F1: Point;F2: Point;Expression(E) = (x^2/25 + y^2/9 = 1);Coordinate(A) = (2, sqrt(5));LeftFocus(E) = F1;RightFocus(E) = F2;PointOnCurve(P, E)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F1)))",
    "answer_expressions": "13",
    "fact_spans": "[[[2, 45], [94, 99]], [[72, 88]], [[89, 93]], [[54, 61]], [[62, 69]], [[2, 45]], [[72, 88]], [[2, 69]], [[2, 69]], [[89, 103]]]",
    "query_spans": "[[[105, 128]]]",
    "process": "From the given conditions, we have $ F_{2}(4,0) $, then $ |AF_{2}| = \\sqrt{2^{2} + (\\sqrt{5})^{2}} = 3 $. Since $ |PA| - |PF_{2}| \\leqslant |AF_{2}| = 3 $. Because point $ P $ lies on the ellipse $ E $, we have $ |PF_{1}| + |PF_{2}| = 2a = 10 $, so $ |PF_{1}| = 10 - |PF_{2}| $. Hence, $ |PA| + |PF_{1}| = 10 + |PA| - |PF_{2}| \\leqslant 13 $. Therefore, the maximum value of $ |PA| + |PF_{1}| $ is 13."
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ satisfies $a \\leq  \\sqrt{3} b$, with eccentricity $e$. Then the maximum value of $e^{2}$ is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Eccentricity(G)=e;a <=sqrt(3)*b;e:Number",
    "query_expressions": "Max(e^2)",
    "answer_expressions": "2/3",
    "fact_spans": "[[[0, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[0, 52]], [[0, 82]], [[54, 74]], [[79, 82]]]",
    "query_spans": "[[[84, 97]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$ and directrix $l$, $P$ is a point on the directrix $l$, and $Q$ is an intersection point of line $PF$ and parabola $C$. If $\\overrightarrow{PF}=4\\overrightarrow{QF}$, then $|QF|$=?",
    "fact_expressions": "C: Parabola;P: Point;F: Point;Q: Point;l:Line;Expression(C) = (y^2 = 8*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(P, l);OneOf(Intersection(LineOf(P,F),C))=Q;VectorOf(P, F) = 4*VectorOf(Q, F)",
    "query_expressions": "Abs(LineSegmentOf(Q, F))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 21], [63, 69]], [[38, 41]], [[25, 28]], [[51, 54]], [[32, 35], [44, 47]], [[2, 21]], [[2, 28]], [[2, 35]], [[38, 50]], [[51, 74]], [[76, 121]]]",
    "query_spans": "[[[123, 132]]]",
    "process": "Let P(-2, t), Q(x_{0}, y_{0}), since \\overrightarrow{PF} = 4\\overrightarrow{OF}, F(2, 0), then (4, -t) = 4(2 - x_{0}, -y_{0}), so 4 = 8 - 4x_{0}, solving gives x_{0} = 1, so |OF| = x_{0} + 2 = 3. The answer is: 3"
  },
  {
    "text": "The standard equation of an ellipse with a major axis length of $6$, a focal distance of $2 \\sqrt{3}$, and foci on the $x$-axis is?",
    "fact_expressions": "G: Ellipse;Length(MajorAxis(G)) = 6;FocalLength(G) = 2*sqrt(3);PointOnCurve(Focus(G),xAxis)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9+y^2/6=1",
    "fact_spans": "[[[33, 35]], [[0, 35]], [[8, 35]], [[24, 35]]]",
    "query_spans": "[[[33, 42]]]",
    "process": "From the given conditions, we know that 2a=6, 2c=2\\sqrt{3}, and the result can be calculated accordingly. The length of the major axis is 6, the focal distance is 2\\sqrt{3}, and the foci lie on the x-axis, so 2a=6, 2c=2\\sqrt{3}. Solving gives: a=3, c=\\sqrt{3}. From a^{2}=b^{2}+c^{2}, it follows that b^{2}=6. Therefore, the standard equation of the ellipse is \\frac{x^{2}}{9}+\\frac{y^{2}}{6}=1"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{3}=1$, and point $P$ lies on the ellipse. If the midpoint of segment $P F_{1}$ lies on the $y$-axis, then the value of $|P F_{1}|$ is?",
    "fact_expressions": "F1: Point;F2: Point;Focus(G) = {F1, F2};G: Ellipse;Expression(G) = (x^2/12 + y^2/3 = 1);P: Point;PointOnCurve(P, G);PointOnCurve(MidPoint(LineSegmentOf(P, F1)), yAxis)",
    "query_expressions": "Abs(LineSegmentOf(P, F1))",
    "answer_expressions": "7*sqrt(3)/2",
    "fact_spans": "[[[2, 9]], [[10, 17]], [[2, 61]], [[18, 56], [67, 69]], [[18, 56]], [[62, 66]], [[62, 70]], [[73, 93]]]",
    "query_spans": "[[[95, 110]]]",
    "process": "From the given conditions, we obtain that $ PF_{2} $ is parallel to the $ y $-axis, and then combining the ellipse equation and the definition of the ellipse, we can organize and calculate to obtain the final result. Since the origin $ O $ is the midpoint of $ F_{1}F_{2} $, $ \\therefore PF_{2} $ is parallel to the $ y $-axis, i.e., $ PF_{2} $ is perpendicular to the $ x $-axis. $ \\because c=3 $, $ \\therefore |F_{1}F_{2}|=6 $. Let $ |PF_{1}|=x $, then according to the definition of the ellipse, $ |PF_{2}|=4\\sqrt{3}-x $. $ \\therefore (4\\sqrt{3}-x)^{2}+36=x^{2} $, solving gives $ x=\\frac{7\\sqrt{3}}{2} $"
  },
  {
    "text": "Given the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $A$ lies on the left branch of the hyperbola $E$, and $\\angle F_{1} A F_{2}=120^{\\circ}$, $|A F_{2}|=3|A F_{1}|$. Then the eccentricity of the hyperbola $E$ is?",
    "fact_expressions": "E: Hyperbola;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(E) = F1;RightFocus(E) = F2;A: Point;PointOnCurve(A, LeftPart(E));AngleOf(F1, A, F2) = ApplyUnit(120, degree);Abs(LineSegmentOf(A, F2)) = 3*Abs(LineSegmentOf(A, F1))",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(13)/2",
    "fact_spans": "[[[2, 63], [93, 99], [164, 170]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[72, 79]], [[80, 87]], [[2, 87]], [[2, 87]], [[88, 92]], [[88, 103]], [[105, 139]], [[140, 162]]]",
    "query_spans": "[[[164, 176]]]",
    "process": "Since point A lies on the left branch of hyperbola E, |AF₂| - |AF₁| = 2a. Given ∠F₁AF₂ = 120° and |AF₂| = 3|AF₁|, it follows that |AF₂| = 3|AF₁| = 3a. By the law of cosines, 4c² = 9a² + a² - 2 × a × 3a cos120°, so 4c² = 13a². Solving gives e = \\frac{\\sqrt{13}}{2}."
  },
  {
    "text": "If the center of a circle is the focus of the parabola $x^{2}=4 y$, and the circle is tangent to the line $y=x+3$, then the standard equation of the circle is?",
    "fact_expressions": "G: Parabola;H: Circle;L: Line;Center(H)=Focus(G);Expression(G) = (x^2 = 4*y);Expression(L) = (y = x + 3);IsTangent(L,H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2+(y-1)^2=2",
    "fact_spans": "[[[8, 22]], [[3, 4], [28, 29], [44, 45]], [[30, 39]], [[3, 25]], [[8, 22]], [[30, 39]], [[28, 41]]]",
    "query_spans": "[[[44, 52]]]",
    "process": "The focus of the parabola is (0,1), so the center of the circle is (0,1). The radius of the circle is $ R = \\frac{|0 - 1 + 3|}{\\sqrt{2}} = \\sqrt{2} $, hence the equation of the circle is: $ x^{2} + (y - 1)^{2} = 2 $."
  },
  {
    "text": "The line passing through the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ with an inclination angle of $60^{\\circ}$ intersects the hyperbola at exactly one point. Then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(RightFocus(G), H);Inclination(H)=ApplyUnit(60,degree);NumIntersection(H,G)=1",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 59], [84, 87], [98, 101]], [[5, 59]], [[5, 59]], [[81, 83]], [[5, 59]], [[5, 59]], [[2, 59]], [[0, 83]], [[64, 83]], [[81, 95]]]",
    "query_spans": "[[[98, 107]]]",
    "process": "This indicates that the straight line with an inclination angle of $60^{\\circ}$ is parallel to one asymptote, thus allowing us to derive a relationship to find the eccentricity. The line passing through the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ with an inclination angle of $60^{\\circ}$ intersects the hyperbola at exactly one point. Therefore, according to the geometric properties of the hyperbola, the given line must be parallel to one asymptote of the hyperbola $y=\\frac{b}{a}x$, so $\\frac{b}{a}=\\tan60^{\\circ}=\\sqrt{3}$, that is, $b=\\sqrt{3}a$, hence $c=\\sqrt{a^{2}+b^{2}}=2a$, and therefore $e=\\frac{c}{a}=2$."
  },
  {
    "text": "Given a hyperbola centered at the origin with foci on the $x$-axis, the eccentricity is $e = \\frac{\\sqrt{6}}{2}$, and the distance from a focus to an asymptote is $\\sqrt{2}$. Then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;O: Origin;e:Number;Center(G)=O;PointOnCurve(Focus(G), xAxis);Eccentricity(G) = e;e = sqrt(6)/2;Distance(Focus(G), Asymptote(G))=sqrt(2)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/2 = 1",
    "fact_spans": "[[[17, 20], [71, 74], [47, 48]], [[5, 7]], [[24, 46]], [[2, 20]], [[8, 20]], [[17, 46]], [[24, 46]], [[47, 68]]]",
    "query_spans": "[[[71, 79]]]",
    "process": "First, from the eccentricity $ e = \\frac{\\sqrt{6}}{2} $, we obtain the asymptotes' equations as $ y = \\pm \\frac{\\sqrt{2}}{2}x $. From the distance from a focus to an asymptote being $ \\sqrt{2} $, we get $ c = \\sqrt{6} $, thus $ a = 2 $, $ b = \\sqrt{2} $, and then obtain the standard equation of the hyperbola. Since the eccentricity of the hyperbola is $ e = \\frac{\\sqrt{6}}{2} $, it follows that $ \\frac{c^{2}}{a^{2}} = \\frac{a^{2}+b^{2}}{a^{2}} = 1 + \\frac{b^{2}}{a^{2}} = \\frac{3}{2} $. Therefore, $ \\frac{b}{a} = \\frac{\\sqrt{2}}{2} $, so the asymptotes' equations of the hyperbola are $ y = \\pm \\frac{\\sqrt{2}}{2}x $. The distance $ d $ from the point $ (c,0) $ to one asymptote line $ y = \\frac{\\sqrt{2}}{2}x $ is $ d = \\frac{\\sqrt{2}c}{\\sqrt{2+4}} = \\sqrt{2} $, solving gives $ c = \\sqrt{6} $. Hence, $ a = 2 $, $ b = \\sqrt{2} $, and the equation of the hyperbola is $ \\frac{x^{2}}{4} - \\frac{y^{2}}{2} = 1 $."
  },
  {
    "text": "If the equation $x^{2}-k y^{2}=2$ represents an ellipse with foci on the $y$-axis, then the range of real values for $k$ is?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (-k*y^2 + x^2 = 2);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-1, 0)",
    "fact_spans": "[[[32, 34]], [[37, 42]], [[2, 34]], [[23, 34]]]",
    "query_spans": "[[[37, 49]]]",
    "process": "According to the problem, the equation $ x^{2} - ky^{2} = 2 $ can be rewritten as $ \\frac{x^{2}}{2} + \\frac{y^{2}}{-\\frac{2}{k}} = 1 $. Since the equation $ x^{2} - ky^{2} = 2 $ represents an ellipse with foci on the $ y $-axis, it follows that $ -\\frac{2}{k} > 2 $, solving which yields $ -1 < k < 0 $. Therefore, the range of real values for $ k $ is $ (-1, 0) $."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, respectively, $P$ be any point on the ellipse, and the coordinates of point $M$ be $(6,4)$. Then the maximum value of $|P M|+|P F_{1}|$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);M: Point;Coordinate(M) = (6, 4)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "15",
    "fact_spans": "[[[19, 58], [69, 71]], [[19, 58]], [[1, 8]], [[9, 16]], [[1, 64]], [[1, 64]], [[65, 68]], [[65, 75]], [[76, 80]], [[76, 91]]]",
    "query_spans": "[[[93, 116]]]",
    "process": "From the ellipse equation, we have $a^2=25$, $b^{2}=16$, $\\therefore c^{2}=9$, $\\therefore a=5$, $c=3$, the coordinates of the two foci are $(\\pm3,0)$. By the definition of an ellipse, $|PM|+|PF|=|PM|+2a-|PF_{2}|=|PM|-|PF_{2}|+10$. Combining the triangle inequality, we know $|PM|-|PF_{2}|\\leqslant|MF_{2}|=5$, so $|PM|-|PF_{2}|+10\\leqslant15$, the maximum value is 15. Key point: Application of the ellipse equation and its definition"
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{2-k}+\\frac{y^{2}}{3+k}=1$ represents an ellipse, what is the range of real values for $k$?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G)=(x^2/(2 - k) + y^2/(k + 3) = 1)",
    "query_expressions": "Range(k)",
    "answer_expressions": "{(-3,2)&(Negation(k=-1/2))}",
    "fact_spans": "[[[45, 47]], [[49, 54]], [[2, 47]]]",
    "query_spans": "[[[49, 61]]]",
    "process": "Since $\\frac{x^2}{2-k}+\\frac{y^{2}}{3+k}=1$ represents an ellipse, we have $\\begin{cases}2-k>0\\\\3+k>0\\\\2-k\\neq3+k\\end{cases}\\Rightarrow-3<k<2$ and $k\\neq-\\frac{1}{2}$"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of a hyperbola, a line passing through $F_{2}$ and perpendicular to the real axis intersects the hyperbola at points $P$ and $Q$, and $\\angle P F_{1} Q = 60^{\\circ}$, then the eccentricity $e =$?",
    "fact_expressions": "G: Hyperbola;H: Line;P: Point;F1: Point;Q: Point;F2: Point;e:Number;Focus(G) = {F1, F2};PointOnCurve(F2, H);IsPerpendicular(H,RealAxis(G));Intersection(H, G) = {P, Q};AngleOf(P, F1, Q) = ApplyUnit(60, degree);Eccentricity(G)=e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[18, 21], [44, 47]], [[41, 43]], [[48, 51]], [[2, 9]], [[52, 55]], [[10, 17], [27, 34]], [[92, 95]], [[2, 25]], [[26, 43]], [[18, 43]], [[41, 57]], [[58, 87]], [[44, 95]]]",
    "query_spans": "[[[92, 97]]]",
    "process": "Let the hyperbola equation be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$. When $x=c$, $\\frac{y^{2}}{b^{2}}=\\frac{c^{2}}{a^{2}}-1$, $y=\\pm\\frac{b^{2}}{a}$, so $|PQ|=\\frac{2b^{2}}{a}$. Also $\\angle PF_{1}Q=60^{\\circ}$, so $2c=\\sqrt{3}\\cdot\\frac{b^{2}}{a}$, i.e., $\\sqrt{3}e^{2}-2e-\\sqrt{3}=0$, solving gives $e=\\sqrt{3}$."
  },
  {
    "text": "If a point $P$ on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ has a distance of $2$ to the left focus $F$, then what is the distance from $P$ to the right directrix?",
    "fact_expressions": "G: Ellipse;P: Point;F: Point;Expression(G) = (x^2/9 + y^2/5 = 1);PointOnCurve(P, G);LeftFocus(G) = F;Distance(P, F) = 2",
    "query_expressions": "Distance(P, RightDirectrix(G))",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 38]], [[41, 44], [60, 63]], [[48, 51]], [[1, 38]], [[1, 44]], [[1, 51]], [[41, 58]]]",
    "query_spans": "[[[1, 72]]]",
    "process": "According to the problem, a=3, c=2. Let the right focus of the ellipse be F, so |PF|+|PF|=2a=6, hence |PF|=4. Let d be the distance from P to the right directrix, so \\frac{|PF|}{d}=e=\\frac{2}{3}, solving gives d=6"
  },
  {
    "text": "Given that $P$ moves on the circle $(x-1)^{2}+y^{2}=1$, and $Q$ moves on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, then the minimum value of $|P Q|$ is?",
    "fact_expressions": "G: Ellipse;H: Circle;P: Point;Q: Point;Expression(G) = (x^2/9 + y^2/4 = 1);Expression(H) = (y^2 + (x - 1)^2 = 1);PointOnCurve(P, H);PointOnCurve(Q, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "4*sqrt(5)/5-1",
    "fact_spans": "[[[34, 71]], [[6, 26]], [[2, 5]], [[30, 33]], [[34, 71]], [[6, 26]], [[2, 27]], [[30, 72]]]",
    "query_spans": "[[[76, 88]]]",
    "process": ""
  },
  {
    "text": "Given that point $A$ lies on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, point $O$ is the origin, and the slope of line $OA$ is $\\frac{\\sqrt{3}}{3}$. If the perpendicular bisector of segment $OA$ passes through a vertex of the hyperbola, then the equations of the asymptotes of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;A: Point;PointOnCurve(A, G);O: Origin;Slope(LineOf(O, A)) = sqrt(3)/3;PointOnCurve(Vertex(G), PerpendicularBisector(LineSegmentOf(O, A)))",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*((sqrt(15)/5)*x)",
    "fact_spans": "[[[7, 63], [123, 126], [131, 134]], [[7, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 6]], [[2, 64]], [[65, 69]], [[75, 106]], [[108, 129]]]",
    "query_spans": "[[[131, 142]]]",
    "process": "Without loss of generality, assume point A is in the first quadrant. At this time, the perpendicular bisector of segment OA passes through the right vertex B(a,0) of the hyperbola as shown in the figure. Connect AB, then |AB| = |OB| = a. Given that the slope of line OA is \\frac{\\sqrt{3}}{3}, the inclination angle of line OA is 30^{\\circ}. Therefore, the inclination angle of line AB is 60^{\\circ}, so point A is \\left(\\frac{3}{2}a,\\frac{\\sqrt{3}}{2}a\\right). Since point A lies on the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1, we obtain \\frac{9}{4}-\\frac{3a^{2}}{4b^{2}}=1, which gives 3a^{2}=5b^{2}, i.e., \\frac{b}{a}=\\frac{\\sqrt{15}}{5}. Hence, the asymptotes of the hyperbola are y=\\pm\\frac{\\sqrt{15}}{5}x."
  },
  {
    "text": "Given that point $F$ is the right focus of the hyperbola $C$: $\\frac{x^{2}}{5}-\\frac{y^{2}}{3}=1$, then the distance from point $F$ to an asymptote of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;F: Point;Expression(C) = (x^2/5 - y^2/3 = 1);RightFocus(C) = F",
    "query_expressions": "Distance(F,OneOf(Asymptote(C)))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[7, 50], [61, 67]], [[2, 6], [2, 6]], [[7, 50]], [[2, 54]]]",
    "query_spans": "[[[56, 78]]]",
    "process": "The hyperbola $ C: \\frac{x^{2}}{5}-\\frac{y^{2}}{3}=1 $ has $ a^{2}=5 $, $ b^{2}=3 $, so $ F(2\\sqrt{2},0) $. Let one asymptote of the hyperbola be $ y=\\frac{\\sqrt{3}}{\\sqrt{5}}x $, i.e., $ \\sqrt{3}x-\\sqrt{5}y=0 $. Then the distance from $ F $ to the asymptote is $ d=\\frac{\\sqrt{3}\\times2\\sqrt{2}}{2\\sqrt{2}}=\\sqrt{3} $."
  },
  {
    "text": "Given the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, what is the distance from its focus to the asymptote?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1)",
    "query_expressions": "Distance(Focus(G),Asymptote(G))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 30], [33, 34]], [[2, 30]]]",
    "query_spans": "[[[33, 46]]]",
    "process": "By the given condition, for the hyperbola $ x^{2} - \\frac{y^{2}}{3} = 1 $, we have $ a = 1 $, $ b = \\sqrt{3} $, then $ c = \\sqrt{a^{2} + b^{2}} = 2 $. Thus, the asymptotes of the hyperbola are $ y = \\pm\\sqrt{3}x $, or $ \\pm\\sqrt{3}x - y = 0 $, and the coordinates of the foci are $ F(\\pm2, 0) $. Therefore, the distance from a focus to an asymptote is $ d = \\frac{2\\sqrt{3}}{\\sqrt{(\\sqrt{3})^{2} + (-1)^{2}}} = \\sqrt{3} $."
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{m+2}=1$ represents an ellipse with foci on the $x$-axis, what is the range of values for $m$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/(m + 2) + x^2/m^2 = 1);PointOnCurve(Focus(G), xAxis);m: Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-2, -1) + (2, +oo)",
    "fact_spans": "[[[56, 58]], [[4, 58]], [[47, 58]], [[60, 63]]]",
    "query_spans": "[[[60, 70]]]",
    "process": "From the given condition, we have $ m^{2} > m + 2 > 0 $, solving this yields the result. According to the problem, the equation $ \\frac{x^2}{m^{2}} + \\frac{y^{2}}{m+2} = 1 $ represents an ellipse with foci on the x-axis, so $ m^{2} > m + 2 > 0 $. Solving gives $ m > 2 $ or $ -2 < m < -1 $. Therefore, the range of $ m $ is $ (-2, -1) \\cup (2, +\\infty) $."
  },
  {
    "text": "If a line $l$ passing through the origin $O$ intersects the circle $x^{2}+y^{2}-4 y+3=0$ at two distinct points $A$ and $B$, then the equation of the trajectory of the midpoint $M$ of chord $AB$ is?",
    "fact_expressions": "l: Line;G: Circle;A: Point;B: Point;O: Origin;M: Point;Expression(G) = (-4*y + x^2 + y^2 + 3 = 0);PointOnCurve(O, l);Intersection(l, G) = {A, B};IsChordOf(LineSegmentOf(A, B), G);MidPoint(LineSegmentOf(A,B)) = M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "(x^2 + (y-1)^2 = 1)&(3/2 < y <= 2)",
    "fact_spans": "[[[11, 16]], [[17, 39]], [[47, 50]], [[51, 54]], [[3, 10]], [[65, 68]], [[17, 39]], [[1, 16]], [[11, 54]], [[17, 62]], [[57, 68]]]",
    "query_spans": "[[[65, 75]]]",
    "process": "Let the equation of line l be y = kx. Points A(x₁, y₁) and B(x₂, y₂) together with the circle form a system of equations. Eliminating y yields: (1 + k²)x² - 4kx + 3 = 0. From Δ = 16k² - 4(1 + k²) × 3 > 0, we obtain k² > 3. By Vieta's formulas, we have x₁ + x₂ = \\frac{4k}{1 + k²}. Therefore, the parametric equations for the trajectory C of the midpoint M of segment AB are:\n\\begin{cases} x = \\frac{2k}{1 + k²} \\\\ y = \\frac{2k²}{1 + k²} \\end{cases},\nwhere k² > 3. Hence, the equation of the trajectory C of the midpoint M of segment AB is: x² + (y - 1)² = 1, where \\frac{3}{2} < y \\leqslant 2."
  },
  {
    "text": "If the focal distance of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is equal to $\\sqrt{3}$ times the length of the real axis, then the asymptotes of $C$ have equations?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;FocalLength(C) = sqrt(3)*Length(RealAxis(C))",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*sqrt(2)*x",
    "fact_spans": "[[[1, 62], [84, 87]], [[1, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[1, 82]]]",
    "query_spans": "[[[84, 95]]]",
    "process": ""
  },
  {
    "text": "If the focal distance of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{m}=1$ is $6$, then the value of $m$ is?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/4 - y^2/m = 1);FocalLength(G) = 6",
    "query_expressions": "m",
    "answer_expressions": "5",
    "fact_spans": "[[[1, 39]], [[48, 51]], [[1, 39]], [[1, 46]]]",
    "query_spans": "[[[48, 55]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $M$: $x^{2}=2 p y(p>0)$ with focus $F$, a line $l$ passing through point $F$ with slope $\\frac{5}{12}$ intersects the parabola $M$ at points $A$ and $B$ (where point $A$ is in the second quadrant). Then $\\frac{|A F|}{|B F|}=$?",
    "fact_expressions": "M: Parabola;Expression(M) = (x^2 = 2*p*y);p: Number;p>0;F: Point;Focus(M) = F;PointOnCurve(F, l) = True;Slope(l) = 5/12;l: Line;Intersection(l, M) = {A, B};A: Point;B: Point;Quadrant(A) = 2",
    "query_expressions": "Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(B, F))",
    "answer_expressions": "4/9",
    "fact_spans": "[[[2, 28], [66, 72]], [[2, 28]], [[10, 28]], [[10, 28]], [[32, 35], [37, 41]], [[2, 35]], [[36, 65]], [[42, 65]], [[60, 65]], [[60, 83]], [[74, 77], [84, 88]], [[78, 81]], [[84, 93]]]",
    "query_spans": "[[[97, 120]]]",
    "process": "Let line CD be the directrix of the parabola, AC\\botCD, BD\\botCD, AE\\botBD, as shown in the figure: let |BE|=5x, then |AE|=12x, |AB||BE|=|BD|-|AC|=|BF|-|AF|=5x, |AB|=|AF|+|BF|=13x\\cdot|AF|=4x, |BF|=9x, \\therefore\\frac{|AF|}{|BF|}=\\frac{4x}{9x}=\\frac{4}{9}"
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=4x$, and a line passing through $F$ with slope $1$ intersects the parabola $C$ at points $A$ and $B$, then $|AB|=$?",
    "fact_expressions": "C: Parabola;G: Line;A: Point;B: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F, G);Slope(G) = 1;Intersection(G, C) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[6, 24], [43, 49]], [[40, 42]], [[50, 53]], [[56, 59]], [[2, 5], [29, 32]], [[6, 24]], [[2, 27]], [[28, 42]], [[33, 42]], [[40, 61]]]",
    "query_spans": "[[[63, 71]]]",
    "process": ""
  },
  {
    "text": "Given a moving chord $AB$ of length $2$ on the parabola $y = x^2$, what is the shortest distance from the midpoint $M$ of $AB$ to the $x$-axis?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;M: Point;Expression(G) = (y = x^2);IsChordOf(LineSegmentOf(A, B), G);Length(LineSegmentOf(A, B)) = 2;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Min(Distance(M, xAxis))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 14]], [[26, 31]], [[26, 31]], [[40, 43]], [[2, 14]], [[2, 31]], [[18, 31]], [[33, 43]]]",
    "query_spans": "[[[40, 55]]]",
    "process": ""
  },
  {
    "text": "Given a point $M$ on the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{20}=1$ such that the distance from $M$ to one of its foci is $6$, find the distance from point $M$ to the other focus?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/20 = 1);M: Point;PointOnCurve(M, G);F1: Point;F2: Point;OneOf(Focus(G)) = F1;OneOf(Focus(G)) = F2;Negation(F1=F2);Distance(M, F1) = 6",
    "query_expressions": "Distance(M, F2)",
    "answer_expressions": "14",
    "fact_spans": "[[[2, 42], [49, 50]], [[2, 42]], [[45, 48], [65, 69]], [[2, 48]], [], [], [[49, 55]], [[49, 75]], [[49, 75]], [[45, 63]]]",
    "query_spans": "[[[49, 79]]]",
    "process": ""
  },
  {
    "text": "If the line $y=k x+2$ has exactly one intersection point with the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1$, then the value of the slope $k$ is?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*x + 2);k: Number;G: Ellipse;Expression(G) = (x^2/3 + y^2/2 = 1);NumIntersection(H, G) = 1;Slope(H) = k",
    "query_expressions": "k",
    "answer_expressions": "pm*sqrt(6)/3",
    "fact_spans": "[[[1, 12]], [[1, 12]], [[62, 65]], [[13, 50]], [[13, 50]], [[1, 58]], [[1, 65]]]",
    "query_spans": "[[[62, 69]]]",
    "process": "From the system of equations, we obtain (2+3k^{2})x^{2}+12kx+6=0. According to the condition d=0, the answer can be obtained. Given that the line y=kx+2 intersects the ellipse \\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1 at exactly one point, by solving \\begin{cases}y=kx+2,\\\\\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1,\\end{cases} eliminating y and simplifying, we get (2+3k^{2})x^{2}+12kx+6=0. From the given condition, A=(12k)^{2}-4\\times6\\times(2+3k^{2})=0, solving gives k=\\pm\\frac{\\sqrt{6}}{3}."
  },
  {
    "text": "Given the coordinates of the three vertices of $\\triangle A B C$ are $(1,1)$, $(-1,-1)$, $(t, \\frac{1}{t})$, then what is the trajectory equation of the orthocenter of $\\triangle A B C$?",
    "fact_expressions": "A: Point;B: Point;C: Point;t: Number;D:Point;E:Point;F:Point;Coordinate(D) = (1, 1);Coordinate(E) = (-1, -1);Coordinate(F) = (t, 1/t);Vertex(TriangleOf(A, B, C)) = {D, E, F}",
    "query_expressions": "LocusEquation(Orthocenter(TriangleOf(A, B, C)))",
    "answer_expressions": "y=1/x",
    "fact_spans": "[[[2, 19]], [[2, 19]], [[2, 19]], [[46, 64]], [[28, 35]], [[36, 45]], [[46, 64]], [[28, 35]], [[36, 45]], [[46, 64]], [[2, 64]]]",
    "query_spans": "[[[66, 92]]]",
    "process": "Let P(x,y). To find the equation of the locus of the orthocenter P of \\triangle ABC, it suffices to find a relation between the coordinates x and y; this can be obtained by using the properties of the orthocenter to derive an equation in x and y and then simplifying. [Solution]: Let P(x,y), A(1,1), B(-1,-1), C(t,\\frac{1}{t}). \\overrightarrow{PA}=(1-x,1-y), \\overrightarrow{BC}=(t+1,\\frac{1}{t}+1), \\overrightarrow{PB}=(-1-x,-1-y), \\overrightarrow{AC}=(t-1,\\frac{1}{t}-1). By the definition of the orthocenter, we have: PA\\bot BC, PB\\bot AC. Therefore, \\begin{cases}(1-x)(t+1)+(1-y)(\\frac{1}{t}+1)=0\\\\(-1-x)(t-1)+(-1-y)(\\frac{1}{t}-1)=0\\end{cases}. Eliminating parameters and simplifying yields: y=\\frac{1}{x}-1."
  },
  {
    "text": "Given that the eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $\\frac{1}{3}$, then what are the equations of the two asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$?",
    "fact_expressions": "H: Ellipse;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);Eccentricity(H) = 1/3;b: Number;a: Number;a > b;b > 0;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(2*sqrt(2)/3)*x",
    "fact_spans": "[[[2, 54]], [[2, 54]], [[2, 72]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[74, 120]], [[74, 120]]]",
    "query_spans": "[[[74, 130]]]",
    "process": "Let the focal distance of the ellipse be 2c. According to the eccentricity formula of the ellipse, the relationship between a and c in the ellipse can be obtained. Then, using the relationship among a, b, c in the ellipse, the relationship between a and b is found. Finally, according to the asymptote equations of the hyperbola, the two asymptote equations of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 are obtained. Solution: Let the focal distance of the ellipse be 2c. From the given conditions: \\frac{c}{a}=\\frac{1}{3}\\Rightarrow a^{2}=9c^{2}. \\because c^{2}=a^{2}-b^{2}, \\therefore 8a^{2}=9b^{2}\\Rightarrow \\frac{b}{a}=\\pm\\frac{2\\sqrt{2}}{3}. Therefore, the two asymptote equations of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 are: y=\\pm\\frac{2\\sqrt{2}}{3}x"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus is $F$, and $A(-a , 0)$, $B(0 , b)$ are two vertices of the ellipse. If the distance from $F$ to $AB$ equals $\\frac{b}{\\sqrt{7}}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;B: Point;F: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(A)=(-a,0);Coordinate(B) = (0, b);LeftFocus(G)=F;In(A,Vertex(G));Distance(F, LineSegmentOf(A, B)) = b/sqrt(7);In(B,Vertex(G))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 54], [87, 89], [132, 134]], [[4, 54]], [[4, 54]], [[63, 74]], [[76, 86]], [[59, 62], [96, 99]], [[4, 54]], [[4, 54]], [[2, 54]], [[63, 74]], [[76, 86]], [[2, 62]], [[63, 94]], [[96, 130]], [[63, 94]]]",
    "query_spans": "[[[132, 140]]]",
    "process": "From the given conditions, the equation of line AB is: $ bx - ay + ab = 0 $. Using the distance formula from point $ F(-c, 0) $ to line AB, we can obtain $ d = \\frac{b}{\\sqrt{7}} $. Rearranging gives an equation in terms of $ e $, which can then be solved. [Detailed Solution] According to the conditions, the equation of AB is $ \\frac{x}{-a} + \\frac{y}{b} = 1 $, that is: $ bx - ay + ab = 0 $. Let $ d $ be the distance from point $ F(-c, 0) $ to line AB.  \n$ \\therefore d = \\frac{|-bc + ab|}{\\sqrt{a^{2} + b^{2}}} = \\frac{b}{\\sqrt{7}} $  \n$ \\therefore 5a^{2} \\cdot 14ac + 8c^{2} = 0 $,  \n$ \\therefore 8e^{2} \\cdot 14e + 5 = 0 $,  \n$ \\because e \\in (0, 1) $,  \n$ \\therefore e = \\frac{1}{2} $ or $ e = \\frac{5}{4} $ (discarded)."
  },
  {
    "text": "Given the equation of the parabola is $2 y = x^{2}$, then the equation of the directrix of this parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (2*y = x^2)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=-1/2",
    "fact_spans": "[[[2, 5], [23, 26]], [[2, 20]]]",
    "query_spans": "[[[23, 33]]]",
    "process": "2y=x^{2}\\Rightarrow x^{2}=2y\\Rightarrow \\frac{p}{2}=\\frac{1}{2}\\Rightarrow y=-\\frac{1}{2}"
  },
  {
    "text": "What is the length of the minor axis of the ellipse $x^{2}+\\frac{y^{2}}{16}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + y^2/16 = 1)",
    "query_expressions": "Length(MinorAxis(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 34]]]",
    "process": "From the given conditions, a=4, b=1, then the minor axis length of the ellipse x^{2}+\\frac{y^{2}}{16}=1 is 2b=2"
  },
  {
    "text": "Let point $P$ be a moving point on the curve $\\frac{x^{2}}{3}-y^{2}=1(x>0)$, point $Q$ be a moving point on the circle $x^{2}+(y-2)^{2}=1$, and point $A(-2,0)$. Then the minimum value of $|P A|+|P Q|$ is?",
    "fact_expressions": "G: Circle;H: Curve;A: Point;P: Point;Q: Point;Expression(G) = (x^2 + (y - 2)^2 = 1);Coordinate(A) = (-2, 0);Expression(H) = (x^2/3 - y^2 = 1)&(x>0);PointOnCurve(P, H);PointOnCurve(Q, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "2*sqrt(2)+2*sqrt(3)-1",
    "fact_spans": "[[[48, 68]], [[6, 38]], [[73, 83]], [[1, 5]], [[43, 47]], [[48, 68]], [[73, 83]], [[6, 38]], [[1, 42]], [[43, 72]]]",
    "query_spans": "[[[85, 104]]]",
    "process": "Let the right focus of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ be $F(2,0)$, and the center of the circle $x^{2}+(y-2)^{2}=1$ be $M(0,2)$, as shown in the figure. By the definition of the hyperbola, we have $|PA|-|PF|=2\\sqrt{3}$, so $|PA|=2\\sqrt{3}+|PF|$. Therefore, $|PA|+|PQ|=|PQ|+|PF|+2\\sqrt{3}\\geqslant|FQ|+2\\sqrt{3}\\geqslant|FM|-|MQ|+2\\sqrt{3}=2\\sqrt{2}+2\\sqrt{3}-1$, with equality if and only if $P$ and $Q$ are the intersection points of segment $FM$ with the right branch of the hyperbola and the circle, respectively. Hence, the minimum value of $|PA|+|PQ|$ is $2\\sqrt{2}+2\\sqrt{3}-1$."
  },
  {
    "text": "If the foci of the hyperbola lie on the $x$-axis, the focal distance is $4$, and it passes through the point $P(2,3)$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;P: Point;Coordinate(P) = (2, 3);PointOnCurve(Focus(G), xAxis);FocalLength(G) = 4;PointOnCurve(P, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/3 = 1",
    "fact_spans": "[[[1, 4], [34, 37]], [[23, 32]], [[23, 32]], [[1, 13]], [[1, 20]], [[1, 32]]]",
    "query_spans": "[[[34, 44]]]",
    "process": "Let the standard equation of the hyperbola be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0, b>0). According to the problem, the left and right foci of the hyperbola are F_{1}(-2,0) and F_{2}(2,0), respectively. By the definition of a hyperbola, we have 2a = ||PF_{1}|-|PF_{2}|| = |\\sqrt{(2+2)^{2}+3^{2}}-\\sqrt{(2-2)^{2}+3^{2}}| = 2. Therefore, a=1, then b=\\sqrt{2-a^{2}}=\\sqrt{3}. Thus, the standard equation of the hyperbola is x^{2}-\\frac{y^{2}}{3}=1."
  },
  {
    "text": "Given $M(-3,0)$, $N(3,0)$, $|P M|-|P N|=6$, then the locus of point $P$ is?",
    "fact_expressions": "M: Point;N: Point;P: Point;Coordinate(M) = (-3, 0);Coordinate(N) = (3, 0);Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)) = 6",
    "query_expressions": "Locus(P)",
    "answer_expressions": "A ray",
    "fact_spans": "[[[2, 11]], [[12, 21]], [[41, 44]], [[2, 11]], [[12, 21]], [[22, 37]]]",
    "query_spans": "[[[41, 49]]]",
    "process": "According to the definition of a hyperbola, it satisfies: ||PM| - |PN|| > |MN|; when ||PM| - |PN|| = |MN|, it represents two rays, thus the answer can be obtained. Since |PM| - |PN| = 6 = |MN|, the trajectory of the moving point P is a ray, with equation: y = 0, x \\geqslant 3."
  },
  {
    "text": "Given that point $P$ is the point on the parabola $y = x^{2}$ that is closest to the line $2x - y - 4 = 0$, then the coordinates of point $P$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2);H: Line;Expression(H) = (2*x - y - 4 = 0);P: Point;PointOnCurve(P,G) = True;WhenMin(Distance(P,H)) = True",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1,1)",
    "fact_spans": "[[[7, 19]], [[7, 19]], [[21, 34]], [[21, 34]], [[2, 6], [43, 47]], [[2, 20]], [[2, 41]]]",
    "query_spans": "[[[43, 52]]]",
    "process": "Let $ P(x_{0},y_{0}) $ be a point on the parabola $ y = x^{2} $. Then the distance from point $ P(x_{0},y_{0}) $ to the line $ 2x - y - 4 = 0 $ is:  \n$ d = \\frac{|2x_{0} - y_{0} - 4|}{\\sqrt{2^{2} + (-1)^{2}}} = \\frac{|2x_{0} - x_{0}^{2} - 4|}{\\sqrt{5}} = \\frac{-(x_{0} - 1)^{2} - 3}{\\sqrt{5}} \\leqslant \\frac{-3}{\\sqrt{5}} $, so $ \\frac{|2x_{0} - y_{0} - 4|}{\\sqrt{5}} \\geqslant \\frac{3}{\\sqrt{5}} $, with equality if and only if $ x_{0} = 1 $. At this time, $ y_{0} = (x_{0})^{2} = 1 $, so $ P(1,1) $.  \n【Analysis】This problem mainly examines the point-to-line distance formula, as well as the transformation idea and properties of quadratic functions. The computation is moderate, making it a medium-difficulty problem."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has an eccentricity of $\\frac{\\sqrt{3}}{2}$. A line passing through the right focus $F$ with slope $k$ $(k>0)$ intersects $C$ at points $A$ and $B$. If $\\overrightarrow{A F}=3\\overrightarrow{F B}$, then $k=$?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;A: Point;F: Point;B: Point;a > b;b > 0;k:Number;k>0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Eccentricity(C) = sqrt(3)/2;RightFocus(C)=F;PointOnCurve(F, G);Slope(G) = k;Intersection(G, C) = {A, B};VectorOf(A, F) = 3*VectorOf(F, B)",
    "query_expressions": "k",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 59], [108, 111]], [[9, 59]], [[9, 59]], [[105, 107]], [[114, 117]], [[89, 92]], [[118, 121]], [[9, 59]], [[9, 59]], [[172, 175]], [[96, 104]], [[2, 59]], [[2, 84]], [[2, 92]], [[85, 107]], [[93, 107]], [[105, 123]], [[126, 170]]]",
    "query_spans": "[[[172, 177]]]",
    "process": ""
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ has a focal distance of $2$, then $m=$?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/4 + y^2/m = 1);FocalLength(G) = 2",
    "query_expressions": "m",
    "answer_expressions": "{3, 5}",
    "fact_spans": "[[[0, 37]], [[46, 49]], [[0, 37]], [[0, 44]]]",
    "query_spans": "[[[46, 51]]]",
    "process": "Since the focal distance of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ is 2, we have $c=1$. If the foci are on the $x$-axis, then $4=m+c^{2}$, solving gives $m=3$; if the foci are on the $y$-axis, then $m=4+c^{2}$, solving gives $m=5$; in conclusion, $m=3$ or $5$."
  },
  {
    "text": "Given that the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ and the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=\\frac{1}{3}(a>0, b>0)$ have the same foci, what is the equation of the asymptotes of the hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Ellipse;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1/3);a1 > b1;b1 > 0;Expression(H) = (y^2/b1^2 + x^2/a1^2 = 1);Focus(G) = Focus(H);a1:Number;b1:Number",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(2)/2)*x",
    "fact_spans": "[[[55, 121], [128, 131]], [[58, 121]], [[58, 121]], [[2, 54]], [[58, 121]], [[58, 121]], [[55, 121]], [[58, 121]], [[58, 121]], [[2, 54]], [[2, 126]], [[4, 54]], [[4, 54]]]",
    "query_spans": "[[[128, 139]]]",
    "process": "Given that the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=\\frac{1}{3}$ $(a>0,b>0)$ have the same foci, solve $a^{2}-b^{2}=\\frac{a^{2}}{3}+\\frac{b^{2}}{3}$. [Detailed solution] Since the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=\\frac{1}{3}$ $(a>0,b>0)$ have the same foci, it follows that $a^{2}-b^{2}=\\frac{a^{2}}{3}+\\frac{b^{2}}{3}$, thus $a^{2}=2b^{2}$. Solving gives $a=\\sqrt{2}b$, so the asymptotes of the hyperbola are $y=\\pm\\frac{b}{a}x=\\pm\\frac{\\sqrt{2}}{2}x$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=6x$, a line $l$ passes through the point $P(2,2)$ and intersects the parabola $C$ at two points $M$ and $N$. If the midpoint of the segment $MN$ is exactly the point $P$, then what is the slope of the line $l$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 6*x);l: Line;PointOnCurve(P, l) = True;Coordinate(P) = (2, 2);P: Point;Intersection(l,C) = {M,N};N: Point;M: Point;MidPoint(LineSegmentOf(M,N)) = P",
    "query_expressions": "Slope(l)",
    "answer_expressions": "3/2",
    "fact_spans": "[[[40, 46], [2, 21]], [[2, 21]], [[22, 27], [78, 83]], [[22, 37]], [[28, 37]], [[28, 37], [72, 76]], [[22, 57]], [[52, 55]], [[48, 51]], [[59, 76]]]",
    "query_spans": "[[[78, 88]]]",
    "process": "Let the coordinates of points M and N be M(x_{1},y_{1}), N(x_{2},y_{2}). Since points M and N lie on the parabola, we have  \n\\begin{cases}y_{1}=6x_{1}\\\\y_{2}=6x_{2}\\end{cases}  \nSubtracting these two equations gives  \ny_{1}^{2}-y_{2}^{2}=6x_{1}-6x_{2}  \nThus, (y_{1}+y_{2})(y_{1}-y_{2})=6(x_{1}-x_{2})  \nSince x_{1}-x_{2}\\neq0 and y_{1}+y_{2}\\neq0,  \nk_{1}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{6}{y_{1}+y_{2}}  \nSince the midpoint of segment MN is exactly point P(2,2),  \ny_{1}+y_{2}=4  \nTherefore, k_{1}=\\frac{3}{2}"
  },
  {
    "text": "Given that the line $y=\\sqrt{2} x$ has no intersection points with the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, then the maximum value of the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Line;b: Number;b>0;Expression(G) = (x^2 - y^2/b^2 = 1);Expression(H) = (y = sqrt(2)*x);NumIntersection(G, H) = 0",
    "query_expressions": "Max(Eccentricity(G))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[19, 56], [62, 65]], [[2, 18]], [[22, 56]], [[22, 56]], [[19, 56]], [[2, 18]], [[2, 59]]]",
    "query_spans": "[[[62, 74]]]",
    "process": "The asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1$ ($b>0$) are: $y=\\pm bx$. Since the line $y=\\sqrt{2}x$ does not intersect the hyperbola, it follows that $b\\leqslant\\sqrt{2}$. The real semi-axis length of the hyperbola is $1$, so the eccentricity of the hyperbola is $e=\\sqrt{1+b^{2}}\\leqslant\\sqrt{3}$. Therefore, the maximum value of the eccentricity of this hyperbola is $\\sqrt{3}$."
  },
  {
    "text": "A point $A$ on the parabola $x^{2}=4 y$ has a vertical coordinate of $4$. What is the distance between point $A$ and the focus of the parabola?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);A: Point;PointOnCurve(A, G);YCoordinate(A) = 4",
    "query_expressions": "Distance(A, Focus(G))",
    "answer_expressions": "5",
    "fact_spans": "[[[0, 14], [35, 38]], [[0, 14]], [[17, 20], [30, 34]], [[0, 20]], [[17, 28]]]",
    "query_spans": "[[[30, 45]]]",
    "process": "From the given, the focus coordinates of the parabola \\( x^{2} = 4y \\) are \\( (0,1) \\), and the equation of the directrix is \\( y = -1 \\). According to the definition of a parabola, the distance from point A to the focus of the parabola is equal to the distance to the directrix, which is \\( 4 - (-1) = 5 \\)."
  },
  {
    "text": "Given that the length of the imaginary axis of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{m+6}=1$ $(m>0)$ is $2$ times the length of the real axis, what is the standard equation of the hyperbola?",
    "fact_expressions": "G: Hyperbola;m: Number;m>0;Expression(G) = (-y^2/(m + 6) + x^2/m = 1);Length(ImageinaryAxis(G)) = 2*Length(RealAxis(G))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/2 - y^2/8 = 1",
    "fact_spans": "[[[2, 47], [62, 65]], [[5, 47]], [[5, 47]], [[2, 47]], [[2, 60]]]",
    "query_spans": "[[[62, 72]]]",
    "process": "According to the given conditions, we obtain $2\\sqrt{m}\\times2=2\\sqrt{m+6}$, solve for $m$, and then find the hyperbola equation. From the problem, we have: $a^{2}=m$, $b^{2}=m+6$, so the length of the real axis is $2\\sqrt{m}$, and the length of the imaginary axis is $2\\sqrt{m+6}$. Since the length of the imaginary axis is twice the length of the real axis, we have $2\\sqrt{m}\\times2=2\\sqrt{m+6}$, solving gives $m=2$. Substituting into $\\frac{x^{2}}{m}-\\frac{y^{2}}{m+6}=1$, we get the hyperbola equation $\\frac{x^{2}}{2}-\\frac{y^{2}}{8}=1$."
  },
  {
    "text": "Given that the point $F(-c, 0)$ $(c>0)$ is the left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, a line passing through $F$ and parallel to the asymptotes of the hyperbola intersects the circle $x^{2}+y^{2}=c^{2}$ at another point $P$, and the point $P$ lies on the parabola $y^{2}=4 c x$. Then, the square of the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Circle;c: Number;C:Parabola;F: Point;P: Point;L:Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + y^2 = c^2);Coordinate(F) = (-c, 0);LeftFocus(G) = F;PointOnCurve(F,L);IsParallel(L,Asymptote(G));Intersection(L,H)=P;PointOnCurve(P,C);Expression(C)=(y^2 = 4*(c*x));c>0",
    "query_expressions": "(Eccentricity(G))^2",
    "answer_expressions": "(\\sqrt{5}+1)/2",
    "fact_spans": "[[[19, 75], [88, 91], [153, 156]], [[22, 75]], [[22, 75]], [[98, 118]], [[99, 118]], [[133, 149]], [[2, 18], [81, 84]], [[123, 126], [128, 132]], [[95, 97]], [[22, 75]], [[22, 75]], [[19, 75]], [[98, 118]], [[3, 18]], [[2, 79]], [[80, 97]], [[85, 97]], [[95, 126]], [[128, 150]], [[133, 149]], [[3, 18]]]",
    "query_spans": "[[[152, 165]]]",
    "process": "By using the equations of a circle, a parabola, and the positional relationships of lines, a system of equations is obtained, thereby deriving the relationship between $a$ and $b$; from the relations among $a$, $b$, and $c$ of the hyperbola, the expression for eccentricity is found. Let the directrix of the parabola be $l$, draw $PQ\\bot l$ at $Q$; let the right focus of the hyperbola be $F_{1}(c,0)$, and $P(x,y)$. From the given conditions, $FF_{1}$ is the diameter of the circle $x^{2}+y^{2}=c^{2}$, so $FF_{1}=2c$, thus $PF\\bot PF_{1}$, and $\\tan\\angle PFF_{1}=\\frac{b}{a}$. Therefore,\n$$\n\\begin{cases}\nx^{2}+y^{2}=c^{2} \\\\\n\\frac{y}{x+c}=\\frac{b}{a}\n\\end{cases}\n$$\nSimplifying yields $\\frac{b^{2}}{a^{2}}=\\frac{4\\sqrt{5}-8}{6-\\sqrt{5}}$. Also, in the hyperbola $c^{2}=a^{2}+b^{2}$, substituting gives $e^{2}=\\frac{\\sqrt{5}+1}{2}$."
  },
  {
    "text": "Given that the midpoint of chord $AB$ of the ellipse $x^{2}+\\frac{y^{2}}{4}=1$ is $M$, and $O$ is the coordinate origin, then the product of the slope of line $AB$ and the slope of line $OM$ equals?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + y^2/4 = 1);A: Point;B: Point;M: Point;IsChordOf(LineSegmentOf(A,B),G) = True;MidPoint(LineSegmentOf(A,B)) = M;O: Origin",
    "query_expressions": "Slope(LineOf(A,B)) * Slope(LineOf(O,M))",
    "answer_expressions": "-4",
    "fact_spans": "[[[2, 29]], [[2, 29]], [[56, 61]], [[56, 61]], [[40, 43]], [[2, 36]], [[31, 43]], [[44, 47]]]",
    "query_spans": "[[[54, 80]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, and $ x_{1} \\neq x_{2} $, then $ x_{1}^{2} + \\frac{y_{1}^{2}}{4} = 1 $, (1) $ x_{2}^{2} + \\frac{y_{2}^{2}}{4} = 1 $, (2) (1)-(2) yields: $ x_{1}^{2} - x_{2}^{2} = - $. Also $ k_{OM} = \\frac{y_{1}+y_{2}}{\\frac{x_{1}+x_{2}}{2}} = \\frac{y_{1}+y_{2}}{x_{1}+x_{2}} $, $ \\therefore k_{AB} = -4 \\cdot \\frac{1}{k_{OM}} $, $ \\therefore k_{AB} \\cdot k_{OM} = -4 $."
  },
  {
    "text": "The tangent line to the curve $y=x^{2}$ at the point $P(1,1)$ is parallel to the line $l$ and the distance between them is $\\sqrt{5}$. Then the equation of the line $l$ is?",
    "fact_expressions": "G: Curve;Expression(G) = (y = x^2);P: Point;Coordinate(P) = (1, 1);l: Line;IsParallel(TangentOnPoint(P, G), l);Distance(TangentOnPoint(P, G), l) = sqrt(5)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "{2*x - y + 4 = 0, 2*x - y - 6 = 0}",
    "fact_spans": "[[[0, 11]], [[0, 11]], [[12, 21]], [[12, 21]], [[26, 31], [50, 55]], [[0, 33]], [[0, 47]]]",
    "query_spans": "[[[50, 60]]]",
    "process": "The curve $ y = x^{2} $ has a tangent line at point $ P(1,1) $ given by $ 2x - y - 1 = 0 $. A line $ l $ parallel to it can be written as $ y = 2x + b \\Rightarrow 2x - y - b = 0 $. Using the distance formula between parallel lines, we obtain $ \\frac{|b - 1|}{\\sqrt{5}} = \\sqrt{5} \\Rightarrow b = -4 $ or $ 6 $. Substituting and simplifying gives the equations $ 2x - y + 4 = 0 $ or $ 2x - y - 6 = 0 $."
  },
  {
    "text": "A moving line passing through the focus $F$ of the parabola $C$: $y^{2}=4x$ intersects $C$ at points $A$ and $B$. The midpoint of segment $AB$ is $N$, and point $P(12,4)$. When the value of $|NA|+|NP|$ is minimized, what is the horizontal coordinate of point $N$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;G: Line;PointOnCurve(F, G);A: Point;B: Point;Intersection(G, C) = {A, B};N: Point;MidPoint(LineSegmentOf(A, B)) = N;P: Point;Coordinate(P) = (12, 4);WhenMin(Abs(LineSegmentOf(N, A))+Abs(LineSegmentOf(N, P)))",
    "query_expressions": "XCoordinate(N)",
    "answer_expressions": "9",
    "fact_spans": "[[[1, 20], [31, 34]], [[1, 20]], [[23, 26]], [[1, 26]], [[27, 30]], [[0, 30]], [[35, 38]], [[39, 42]], [[27, 44]], [[56, 59], [92, 96]], [[45, 59]], [[60, 70]], [[60, 70]], [[72, 91]]]",
    "query_spans": "[[[92, 102]]]",
    "process": "Let the directrix of parabola $ C $ be $ l $, draw $ BD\\bot l $, $ NG\\bot l $, $ AE\\bot l $, with feet at $ D $, $ G $, $ E $ respectively. Then $ |AB| = |BF| + |AF| = |BD| + |AE| = 2|NG| $. $ \\therefore |AN| = |NG| $, $ \\therefore |NA| + |NP| = |NG| + |NP| $. The distance from point $ P $ to line $ l $ is 13, $ \\therefore |NG| + |NP| \\geqslant 13 $. When points $ G $, $ N $, $ P $ are collinear and $ N $ lies between $ G $ and $ P $, $ |NG| + |NP| = |PG| = 13 $, at this time, the ordinate of point $ N $ is $ y_{N} = 4 $. Since $ AB $ passes through point $ F(1,0) $, let the equation of $ AB $ be $ x = my + 1 $, substitute into $ y^{2} = 4x $, we get $ y^{2} - 4my - 4 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ y_{1} + y_{2} = 4m $. When $ G $, $ N $, $ P $ are collinear, $ y_{1} + y_{2} = 2y_{N} = 8 $, $ \\therefore 4m = 8 $, $ m = 2 $. The equation of line $ AB $ is $ x = 2y + 1 $, $ N(9,4) $. Point $ N $ lies between $ G $ and $ P $, so $ |NG| + |NP| = |PG| = 13 $ holds. Therefore, when $ |NA| + |NP| $ is minimized, the abscissa of point $ N $ is 9."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the right focus is $F$, and point $P$ lies on the ellipse. A circle centered at $P$ is tangent to the $y$-axis and also tangent to the $x$-axis at the ellipse's right focus $F$. Then, the eccentricity of the ellipse $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Z: Ellipse;Expression(Z) = (y^2/a^2+x^2/b^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;RightFocus(G) = F;P: Point;PointOnCurve(P, G);H: Circle;Center(H) = P;IsTangent(H, yAxis);TangentPoint(H, xAxis) = F",
    "query_expressions": "Eccentricity(Z)",
    "answer_expressions": "(\\sqrt{5}-1)/2",
    "fact_spans": "[[[2, 54], [68, 70], [101, 103]], [[2, 54]], [[112, 157]], [[112, 157]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[59, 62], [107, 110]], [[2, 62]], [[63, 67], [73, 77]], [[63, 71]], [[81, 82]], [[72, 82]], [[81, 89]], [[81, 110]]]",
    "query_spans": "[[[112, 163]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the foci of the ellipse $4 x^{2}+y^{2}=16$ are?",
    "fact_expressions": "G: Ellipse;Expression(G) = (4*x^2 + y^2 = 16)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "{(0, -2*sqrt(3)), (0, 2*sqrt(3))}",
    "fact_spans": "[[[0, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 27]]]",
    "process": ""
  },
  {
    "text": "Points $A$ and $B$ move on the parabola $y = \\frac{1}{4}x^{2}$, and $|AB| = 5$. If the projection of the midpoint $M$ of segment $AB$ onto the $x$-axis is $M1$, then the minimum value of $|MM1|$ is?",
    "fact_expressions": "A: Point;B: Point;PointOnCurve(A, G) = True;PointOnCurve(B, G) = True;G: Parabola;Expression(G) = (y = x^2/4);Abs(LineSegmentOf(A, B)) = 5;MidPoint(LineSegmentOf(A,B)) = M;M1: Point;Projection(M,xAxis) = M1;M: Point",
    "query_expressions": "Min(Abs(LineSegmentOf(M, M1)))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[0, 3]], [[6, 9]], [[0, 38]], [[0, 38]], [[13, 37]], [[13, 37]], [[40, 49]], [[51, 64]], [[74, 78]], [[61, 78]], [[61, 64]]]",
    "query_spans": "[[[80, 94]]]",
    "process": "According to the problem, the standard equation of the parabola $ y = \\frac{1}{4}x^{2} $ is $ x^{2} = 4y $, and its directrix is $ y = -1 $; it opens upward. Let the focus of the parabola be $ F $, and let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, $ M(x_{0}, y_{0}) $. Then $ y_{1}, y_{2}, y_{0} $ are all greater than 0. The shortest distance from point $ M $ to the $ x $-axis is $ y_{0} $, so  \n$$\ny_{0} = \\frac{y_{1}+1 + y_{2}+1 - 2}{2} = \\frac{|AF| + |BF| - 2}{2} \\geqslant \\frac{|AB| - 2}{2} = \\frac{3}{2},\n$$\nwith equality if and only if $ F $, $ A $, $ B $ are collinear. Thus, the minimum value of $ |MM| $ is $ \\frac{3}{2} $."
  },
  {
    "text": "Write the standard equation of an ellipse with eccentricity $\\frac{1}{2}$ and foci on the $y$-axis.",
    "fact_expressions": "G: Ellipse;Eccentricity(G) = 1/2;PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/4 + x^2/3 = 1",
    "fact_spans": "[[[32, 34]], [[5, 34]], [[23, 34]]]",
    "query_spans": "[[[32, 40]]]",
    "process": "Since the eccentricity is $\\frac{1}{2}$, we can set $a=2$, $c=1$, $\\therefore b=\\sqrt{3}$, and the standard equation of the ellipse at this time is $\\frac{y^{2}}{4}+\\frac{x^{2}}{3}=1$."
  },
  {
    "text": "Let $A$ and $B$ be the left and right vertices of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$, respectively, and let $P$ be a point on the hyperbola different from $A$ and $B$. The slopes of lines $AP$ and $BP$ are $m$ and $n$, respectively. Then, when $\\frac{3 b}{a}+\\sqrt{\\frac{1}{m n}}$ attains its minimum value, what is the eccentricity of the hyperbola?",
    "fact_expressions": "A: Point;B: Point;LeftVertex(G) = A;RightVertex(G) = B;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;P: Point;PointOnCurve(P, G);Negation(P=A);Negation(P=B);m: Number;n: Number;Slope(LineOf(A, P)) = m;Slope(LineOf(B, P)) = n;WhenMin(sqrt(1/(m*n)) + (3*b)/a)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[1, 4], [90, 93]], [[6, 9], [95, 98]], [[1, 78]], [[1, 78]], [[12, 71], [83, 86], [175, 178]], [[12, 71]], [[15, 71]], [[15, 71]], [[15, 71]], [[15, 71]], [[79, 82]], [[79, 101]], [[79, 101]], [[79, 101]], [[122, 125]], [[127, 130]], [[102, 130]], [[102, 130]], [[132, 174]]]",
    "query_spans": "[[[175, 184]]]",
    "process": "First determine $ mn $ based on the relationship between points, then determine the minimum value using the basic inequality, and finally determine the eccentricity of the hyperbola according to the condition for achieving the minimum value. Let $ P(x_{1},y_{1}) $, then $ mn = \\frac{y_{1}}{x_{1}+a} \\cdot \\frac{y_{1}}{x_{1}-a} = \\frac{y_{1}}{x_{1}^{2}-a^{2}} = \\frac{b^{2}}{a^{2}} $. Therefore, $ \\frac{3b}{a} + \\sqrt{\\frac{1}{mn}} = \\frac{3b}{a} + \\frac{a}{b} \\geqslant 2\\sqrt{\\frac{3b}{a} \\cdot \\frac{a}{b}} = 2\\sqrt{3} $, with equality if and only if $ a = \\sqrt{3}b $. Thus, the eccentricity is $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\frac{2\\sqrt{3}}{3} $."
  },
  {
    "text": "The hyperbola $C$ passes through the point $(3,2 \\sqrt{2})$ and has asymptotes given by $y=\\pm \\frac{2}{3} x$. Then, the standard equation of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;G: Point;Coordinate(G) = (3, 2*sqrt(2));PointOnCurve(G, C);Expression(Asymptote(C)) = (y = pm*(2/3)*x)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/4 - x^2/9 = 1",
    "fact_spans": "[[[2, 8], [57, 60]], [[10, 27]], [[10, 27]], [[2, 27]], [[2, 55]]]",
    "query_spans": "[[[57, 67]]]",
    "process": "According to the family of hyperbolas sharing asymptotes with the given curve, the equation of this hyperbola can be written as $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=\\lambda$ ($\\lambda\\neq0$). Substituting the point $(3,2\\sqrt{2})$ yields $\\lambda=-1$, so the required hyperbola equation is $\\frac{y^{2}}{4}-\\frac{x^{2}}{9}=1$."
  },
  {
    "text": "Let point $M$ be a point on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. A circle centered at point $M$ is tangent to the $x$-axis at the focus $F$ of the ellipse, and intersects the $y$-axis at two distinct points $P$, $Q$, satisfying $|\\overrightarrow{M P}+\\overrightarrow{M Q}|=|\\overrightarrow{P Q}|$. Then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;M: Point;P: Point;Q: Point;F: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(M, G);Z: Circle;Center(Z) = M;Focus(G) = F;TangentPoint(Z, xAxis) = F;Intersection(Z, yAxis) = {P, Q};Negation(P = Q);Abs(VectorOf(M, P) + VectorOf(M, Q)) = Abs(VectorOf(P, Q))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(sqrt(6) - sqrt(2))/2",
    "fact_spans": "[[[6, 58], [80, 82], [186, 188]], [[8, 58]], [[8, 58]], [[63, 67], [1, 5]], [[106, 109]], [[110, 113]], [[85, 88]], [[8, 58]], [[8, 58]], [[6, 58]], [[1, 61]], [[71, 72], [89, 93]], [[62, 72]], [[80, 88]], [[71, 88]], [[89, 113]], [[101, 113]], [[117, 185]]]",
    "query_spans": "[[[186, 194]]]",
    "process": "By the parallelogram law of vector addition, since $|\\overrightarrow{MP}+\\overrightarrow{MQ}|=|\\overrightarrow{PQ}|$, the parallelogram with adjacent sides $MP$ and $MQ$ is a rectangle. Then, by symmetry, it is a square, thus the relationship among $a$, $b$, and $c$ can be easily obtained, and after rearrangement, the eccentricity $e$ can be found.  \n\nDetailed explanation:  \n$\\because |\\overrightarrow{MP}+\\overrightarrow{MQ}|=|\\overrightarrow{PQ}|$, by the parallelogram law of vector addition, the parallelogram with adjacent sides $MP$ and $MQ$ is a rectangle. Also, $|MP|=|MQ|$, therefore this quadrilateral is a square.  \n\nA circle centered at point $M$ is tangent to the $x$-axis at the ellipse's focus $F$, then $MF\\bot x$-axis,  \n$\\therefore |MF|=\\frac{b^{2}}{a}$, $x_{M}=c$,  \n$\\therefore \\frac{b^{2}}{a}=\\sqrt{2}c^{2}$  \n$\\therefore b^{2}=a^{2}-c^{2}=\\sqrt{2}ac$  \ni.e., $e^{2}+\\sqrt{2}e-1=0$,  \n$\\therefore e=\\frac{-\\sqrt{2}+\\sqrt{6}}{2}$ (the negative root $\\frac{-\\sqrt{2}-\\sqrt{6}}{2}$ is discarded),  \n$\\therefore e=\\frac{\\sqrt{6}-\\sqrt{2}}{2}$"
  },
  {
    "text": "Given the ellipse equation $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, when $a^{2}+\\frac{16}{b(a-b)}$ reaches its minimum value, what is the eccentricity $e$ of the ellipse?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>b;b>0;WhenMin(a^2 + 16/(b*(a - b)));e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[2, 4], [89, 91]], [[2, 56]], [[6, 56]], [[6, 56]], [[6, 56]], [[6, 56]], [[57, 88]], [[95, 98]], [[89, 98]]]",
    "query_spans": "[[[95, 100]]]",
    "process": "a^{2}+\\frac{16}{b(a-b)^{\\geqslant}a^{2}+\\frac{16}{(b+a-b)^{2}}}=a^{2}+\\frac{16\\times4}{a^{2}}\\geqslant2\\sqrt{a^{2}\\times\\frac{16\\times4}{a^{2}}}=16. Equality holds if and only if a-b=b, a=2b=2\\sqrt{2}. Equality holds when a=2\\sqrt{2}, b=\\sqrt{2}, so c=\\sqrt{a^{2}-b^{2}}=\\sqrt{6}, c=\\frac{c}{a}=\\frac{\\sqrt{3}}{}"
  },
  {
    "text": "Given the hyperbola equation $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0, b>0$), the distance from one focus of the hyperbola to an asymptote is $\\frac{\\sqrt{5} c}{3}$ ($c$ is the semi-focal length of the hyperbola). Then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a:Number;b:Number;a>0;b>0;Distance(OneOf(Focus(G)),OneOf(Asymptote(G)))=(sqrt(5)/3)*c;c:Number;HalfFocalLength(G)=c",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3/2",
    "fact_spans": "[[[2, 5], [119, 122], [63, 66], [108, 111]], [[2, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[63, 103]], [[104, 107]], [[104, 116]]]",
    "query_spans": "[[[119, 128]]]",
    "process": "From the given conditions: the asymptotes of the hyperbola are $ y = \\pm\\frac{b}{a}x $, so the distance from a focus of the hyperbola to an asymptote is $ \\frac{|bc|}{\\sqrt{a^{2}+b^{2}}} = \\frac{|bc|}{c} = b $, that is, $ b = \\frac{\\sqrt{5}c}{3} \\Rightarrow 4c^{2} = 9a^{2} \\Rightarrow e = \\frac{3}{2} $"
  },
  {
    "text": "The coordinates of the two foci of an ellipse are $(-2,0)$ and $(2,0)$, and it passes through the point $\\left(\\frac{5}{2}, -\\frac{3}{2}\\right)$. What is its standard equation?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Z: Point;Coordinate(F1) = (-2, 0);Coordinate(F2) = (2, 0);Coordinate(Z) = (5/2, -3/2);PointOnCurve(Z, G);Focus(G) = {F1, F2}",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/10 + y^2/6 = 1",
    "fact_spans": "[[[2, 4], [67, 68]], [[14, 22]], [[24, 31]], [[36, 65]], [[14, 22]], [[24, 31]], [[36, 65]], [[2, 65]], [[2, 31]]]",
    "query_spans": "[[[67, 75]]]",
    "process": "Since the foci of the ellipse lie on the x-axis, we set its standard equation as \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0). Thus, a=\\sqrt{10}. Also, since c=2, we have b^{2}=a^{2}-c^{2}=6. Therefore, the standard equation of the ellipse is \\frac{x^{2}}{10}+\\frac{y^{2}}{6}=1 \\text{guest head}+\\frac{y^{2}}{c}=1"
  },
  {
    "text": "Let the hyperbola be $\\frac{x^{2}}{4}-y^{2}=1$, $F_{1}$ its left focus, and line $l$ passes through its right focus $F_{2}$, intersecting the right branch of the hyperbola at points $A$ and $B$. Then the minimum value of $|\\overrightarrow{F_{1} A}| \\cdot|\\overrightarrow{F_{1} B}|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1);F1: Point;LeftFocus(G) = F1;l: Line;PointOnCurve(F2, l) = True;F2: Point;RightFocus(G) = F2;Intersection(l, RightPart(G)) = {A, B};A: Point;B: Point",
    "query_expressions": "Min(Abs(VectorOf(F1, A))*Abs(VectorOf(F1, B)))",
    "answer_expressions": "81/4",
    "fact_spans": "[[[1, 29], [40, 41], [53, 54], [68, 71]], [[1, 29]], [[32, 39]], [[32, 45]], [[46, 51]], [[46, 65]], [[58, 65]], [[53, 65]], [[46, 85]], [[76, 79]], [[80, 83]]]",
    "query_spans": "[[[87, 153]]]",
    "process": "According to the line passing through the right focus, when the slope exists, set the equation of the line as $ y = k(x - \\sqrt{5}) $, and substitute into the hyperbola equation. Eliminate $ y $ to obtain $ (1 - 4k^2)x^2 + 8\\sqrt{5}k^2x - 20k^2 - 4 = 0 $. By Vieta's formulas, we get $ x_1 + x_2 = \\frac{8\\sqrt{5}k^2}{4k^2 - 1} $, $ x_1x_2 = \\frac{20k^2 + 4}{4k^2 - 1} $. Then using the second definition, $ |\\overrightarrow{F_1A}| \\cdot |\\overrightarrow{F_1B}| = \\left( \\frac{\\sqrt{5}}{2}x_1 + 2 \\right) \\left( \\frac{\\sqrt{5}}{2}x_2 + 2 \\right) = \\frac{5}{4}x_1x_2 + \\sqrt{5}(x_1 + x_2) + 4 $. When the slope does not exist, $ |\\overrightarrow{F_2A}| = |\\overrightarrow{F_2B}| = \\frac{1}{2} $, by the first definition of the hyperbola, $ |\\overrightarrow{F_1A}| = |\\overrightarrow{F_1B}| = \\frac{1}{2} + 2a = \\frac{9}{2} $, so $ |\\overrightarrow{F_1A}| \\cdot |\\overrightarrow{F_1B}| = \\frac{81}{4} $. Solution: The right focus of the hyperbola is $ (\\sqrt{5}, 0) $. When the slope of the line exists, set the equation of the line as $ y = k(x - \\sqrt{5}) $. Substitute into the hyperbola equation, eliminate $ y $ to get $ (1 - 4k^2)x^2 + 8\\sqrt{5}k^2x - 20k^2 - 4 = 0 $. Let $ A(x_1, y_1) $, $ B(x_2, y_2) $. By Vieta's formulas, $ x_1 + x_2 = \\frac{8\\sqrt{5}k^2}{4k^2 - 1} $, $ x_1x_2 = \\frac{20k^2 + 4}{4k^2 - 1} $. According to the second definition of the hyperbola, $ |\\overrightarrow{F_1A}| \\cdot |\\overrightarrow{F_1B}| = \\left( \\frac{\\sqrt{5}}{2}x_1 + 2 \\right) \\left( \\frac{\\sqrt{5}}{2}x_2 + 2 \\right) = \\frac{5}{4}x_1x_2 + \\sqrt{5}(x_1 + x_2) + 4 $. When the slope does not exist, $ |\\overrightarrow{F_2A}| = |\\overrightarrow{F_2B}| = \\frac{1}{2} $. By the first definition of the hyperbola, $ |\\overrightarrow{F_1A}| = |\\overrightarrow{F_1B}| = \\frac{1}{2} + 2a = \\frac{9}{2} $, $ |\\overrightarrow{F_1A}| \\cdot |\\overrightarrow{F_1B}| = \\frac{81}{4} $. In conclusion: the minimum value of $ |\\overrightarrow{F_1A}| \\cdot |\\overrightarrow{F_1B}| $ is $ \\frac{81}{4} $."
  },
  {
    "text": "Let the left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ be $F_{1}$ and $F_{2}$, respectively. A line passing through focus $F_{1}$ intersects the ellipse at points $A$ and $B$. If the area of the incircle of $\\triangle A B F_{2}$ is $4 \\pi$, and the coordinates of points $A$ and $B$ are $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$, respectively, then what is the value of $|y_{1}-y_{2}|$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;Z: Line;PointOnCurve(F1, Z);A: Point;B: Point;Intersection(Z, G) = {A, B};Area(InscribedCircle(TriangleOf(A, B, F2))) = 4*pi;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2)",
    "query_expressions": "Abs(y1 - y2)",
    "answer_expressions": "5",
    "fact_spans": "[[[1, 39], [77, 79]], [[1, 39]], [[47, 54], [66, 73]], [[55, 62]], [[1, 62]], [[1, 62]], [[74, 76]], [[63, 76]], [[80, 83], [145, 162], [130, 133]], [[84, 87], [164, 181], [134, 137]], [[74, 89]], [[91, 127]], [[145, 162]], [[164, 181]], [[145, 162]], [[164, 181]], [[130, 181]], [[130, 181]]]",
    "query_spans": "[[[183, 201]]]",
    "process": "Because in the ellipse $\\frac{x^2}{25}+\\frac{y^{2}}{9}=1$, $c=\\sqrt{25-9}=4$, the foci are $F_{1}(-4,0)$, $F_{2}(4,0)$. Let the inradius of $\\triangle ABF_{2}$ be $r$, so $S=\\pi r^{2}=4\\pi$, yielding $r=2$. According to the definition of the ellipse, $|AB|+|AF_{2}|+|BF_{2}|=(|AF_{1}|+|AF_{2}|)+(|BF_{1}|+|BF_{2}|)=4a=20$, thus $S_{\\triangle ABF_{2}}=\\frac{1}{2}(|AB|+|AF_{2}|+|BF_{2}|)\\cdot r=\\frac{1}{2}\\times20\\times2=20$. Also, $S_{\\triangle ABF_{2}}=S_{\\triangle AF_{1}F_{2}}+S_{\\triangle BF_{1}F_{2}}=\\frac{1}{2}|y_{1}|\\cdot|F_{1}F_{2}|+\\frac{1}{2}|y_{2}|\\cdot|F_{1}F_{2}|=\\frac{1}{2}|y_{1}-y_{2}|\\cdot|F_{1}F_{2}|$, so $4|y_{1}-y_{2}|=20$, hence $|y_{1}-y_{2}|=5$."
  },
  {
    "text": "Given that the hyperbola $C$ passes through the point $P(\\sqrt{3}, 2\\sqrt{2})$, and one of its asymptotes has the equation $y = \\frac{2\\sqrt{3}}{3}x$, what is the standard equation of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;P: Point;Coordinate(P) = (sqrt(3), 2*sqrt(2));PointOnCurve(P, C);Expression(OneOf(Asymptote(C))) = (y = x*(2*sqrt(3)/3))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/4 - x^2/3 = 1",
    "fact_spans": "[[[2, 8], [74, 80]], [[9, 37]], [[9, 37]], [[2, 37]], [[2, 73]]]",
    "query_spans": "[[[74, 87]]]",
    "process": ""
  },
  {
    "text": "Given a point $M(1, m)$ $(m>0)$ on the parabola $y^{2}=2 p x$ $(p>0)$ such that the distance from $M$ to its focus is $5$, and the left vertex of the hyperbola $\\frac{x^{2}}{a}-y^{2}=1$ is $A$. If one asymptote of the hyperbola is parallel to the line $AM$, then the real number $a$ equals?",
    "fact_expressions": "A: Point;M: Point;G: Hyperbola;a: Real;H: Parabola;p: Number;Expression(G) = (-y^2 + x^2/a = 1);p>0;Expression(H) = (y^2 = 2*(p*x));Coordinate(M) = (1, m);PointOnCurve(M, H);Distance(M, Focus(H)) = 5;LeftVertex(G) = A;m:Number;m>0;IsParallel(OneOf(Asymptote(G)), LineOf(A, M))",
    "query_expressions": "a",
    "answer_expressions": "1/9",
    "fact_spans": "[[[87, 90]], [[26, 42]], [[54, 82], [93, 96]], [[113, 118]], [[2, 23], [43, 44]], [[5, 23]], [[54, 82]], [[5, 23]], [[2, 23]], [[26, 42]], [[2, 42]], [[26, 53]], [[54, 90]], [[26, 42]], [[26, 42]], [[93, 111]]]",
    "query_spans": "[[[113, 121]]]",
    "process": "From the given conditions: the directrix of the parabola $ y^{2} = 2px $ ($ p > 0 $) is $ x = 4 $, $ \\therefore p = 8 $, then point $ M(1, 4) $, the left vertex of the hyperbola $ \\frac{x^{2}}{a} - y^{2} = 1 $ is $ A(-\\sqrt{a}, 0) $, so the slope of line $ AM $ is $ \\frac{4}{1 + \\sqrt{a}} $, from the given conditions: $ \\frac{4}{1 + \\sqrt{a}} = \\frac{1}{\\sqrt{a}} $, $ \\therefore a = \\frac{1}{9} $."
  },
  {
    "text": "The standard equation of an ellipse with a major axis length of $4$ and one focus at $F(1,0)$ is?",
    "fact_expressions": "G: Ellipse;F: Point;Coordinate(F) = (1, 0);Length(MajorAxis(G)) = 4;OneOf(Focus(G)) = F",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 + y^2/3 = 1",
    "fact_spans": "[[[22, 24]], [[13, 21]], [[13, 21]], [[0, 24]], [[8, 24]]]",
    "query_spans": "[[[22, 31]]]",
    "process": "From the given information, the major axis length of the ellipse is 4 and one focus is at F(1,0). Therefore, 2a=4, c=1, and the foci lie on the x-axis. Thus, b^{2}=a^{2}-c^{2}=3, and the standard equation of the ellipse is: \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1"
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1  (a>0)$ is $y=\\frac{\\sqrt{3}}{3} x$, then $a$=?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (y = x*(sqrt(3)/3))",
    "query_expressions": "a",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 41]], [[76, 79]], [[5, 41]], [[2, 41]], [[2, 74]]]",
    "query_spans": "[[[76, 81]]]",
    "process": "\\because the hyperbola \\frac{x^{2}}{a^{2}}-y^{2}=1 (a>0), b=1, and one asymptote equation is y=\\frac{\\sqrt{3}}{3}x, \\therefore \\frac{1}{a}=\\frac{\\sqrt{3}}{3}, \\therefore a=\\sqrt{3}"
  },
  {
    "text": "Given that the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has an eccentricity of $2$, and its foci coincide with the foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, what is the equation of the asymptotes of the hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Ellipse;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2/25 + y^2/9 = 1);Eccentricity(G) = 2;Focus(G) = Focus(H)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[2, 48], [106, 109]], [[5, 48]], [[5, 48]], [[60, 98]], [[2, 48]], [[60, 98]], [[2, 56]], [[2, 103]]]",
    "query_spans": "[[[106, 117]]]",
    "process": "Since the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $2$, it follows that $1+\\frac{b^{2}}{a^{2}}=4$, $\\frac{b^{2}}{a^{2}}=3$, $\\frac{b}{a}=\\pm\\sqrt{3}$. Moreover, the foci of the hyperbola coincide with the foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, meaning the foci lie on the $x$-axis. Therefore, the asymptotes of the hyperbola are given by $y=\\pm\\sqrt{3}x$."
  },
  {
    "text": "In $Rt \\triangle A B C$, $A B = A C = 2$. If an ellipse passes through points $A$ and $B$, has one focus at point $C$, and the other focus on side $A B$, then what is the focal distance of this ellipse?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;C: Point;LineSegmentOf(A,B)=LineSegmentOf(A,C);LineSegmentOf(A,C)=2;PointOnCurve(A,G);PointOnCurve(B,G);OneOf(Focus(G))=C;F:Point;OneOf(Focus(G))=F;PointOnCurve(F, LineSegmentOf(A, B));Negation(C=F)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "sqrt(6)",
    "fact_spans": "[[[40, 42], [54, 55], [83, 85]], [[44, 47]], [[48, 51]], [[61, 65]], [[23, 34]], [[23, 34]], [[40, 53]], [[40, 53]], [[54, 65]], [], [[54, 71]], [[54, 79]], [[54, 71]]]",
    "query_spans": "[[[83, 90]]]",
    "process": "Let the other focus be F. In right triangle ABC, AB = AC = 2, so BC = 2\\sqrt{2}. Since AC + AF = BC + BF = 2a, it follows that AC + AF + BC + BF = 4 + 2\\sqrt{2} = 2a \\Rightarrow a = 2 + \\sqrt{2}. Given AC = 2, we have AF = \\sqrt{2}. Therefore, CF = \\sqrt{2^{2} + 2} = \\sqrt{6}, which is the focal distance of the ellipse."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{m}+\\frac{y^{2}}{2 m-1}=1$ represents an ellipse, then the condition that the real number $m$ must satisfy is?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G)=(x^2/m+y^2/(2*m-1)=1)",
    "query_expressions": "m",
    "answer_expressions": "{(1/2,+oo)&Negation(m=1)}",
    "fact_spans": "[[[44, 46]], [[48, 53]], [[1, 46]]]",
    "query_spans": "[[[48, 60]]]",
    "process": "The equation $\\frac{x^{2}}{m}+\\frac{y^{2}}{2m-1}=1$ represents an ellipse, then $\\begin{cases}m>0,\\\\2m-1>0,\\\\m\\neq2m-1,\\end{cases}$; solving the condition that the equation $\\frac{x^{2}}{m}+\\frac{y^{2}}{2m-1}=1$ represents an ellipse yields $\\begin{cases}m>0,\\\\2m-1>0,\\\\m\\neq2m-1,\\end{cases}$, resulting in $m>\\frac{1}{2}$ and $m\\neq1$."
  },
  {
    "text": "Given that the distance from a moving point $P$ to the fixed point $(2 , 0)$ is greater by $1$ than its distance to the fixed line $l$: $x = -1$, what is the equation of the trajectory of point $P$?",
    "fact_expressions": "P: Point;G: Point;Coordinate(G) = (2, 0);l: Line;Expression(l) = (x = -1);Distance(P, G) = Distance(P, l) + 1",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[4, 7], [23, 24], [48, 52]], [[10, 19]], [[10, 19]], [[28, 39]], [[28, 39]], [[4, 46]]]",
    "query_spans": "[[[48, 59]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{1}$ intersects the left branch of $C$ at points $A$ and $B$, such that $\\overrightarrow{A F_{1}}=3 \\overrightarrow{F_{1} B}$, $\\angle A B F_{2}=90^{\\circ}$. What is the eccentricity of $C$?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;A: Point;F1: Point;B: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, G);Intersection(G, LeftPart(C)) = {A, B};VectorOf(A, F1) = 3*VectorOf(F1, B);AngleOf(A, B, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(10)/2",
    "fact_spans": "[[[2, 63], [100, 103], [205, 208]], [[10, 63]], [[10, 63]], [[97, 99]], [[108, 111]], [[72, 79], [89, 96]], [[112, 115]], [[80, 87]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 87]], [[2, 87]], [[88, 99]], [[97, 117]], [[119, 172]], [[174, 203]]]",
    "query_spans": "[[[205, 214]]]",
    "process": "Draw the graph. According to the principle that scaling does not change shape, assume |F_{1}B|=1, and use the definition of hyperbola to express other related segments. Then, in Rt△ABF_{2} and Rt△F_{1}BF_{2}, apply the Pythagorean theorem to establish equation(s) and solve for the values of a and c, thereby obtaining the eccentricity. As shown in the figure, assume |F_{1}B|=1, then |AF_{1}|=3, |AB|=4, |F_{2}B|=2a+1, |F_{2}A|=2a+3. In Rt△ABF_{2}, by the Pythagorean theorem: 16+(2a+1)^{2}=(2a+3)^{2}, solving gives a=1. In Rt△F_{1}BF_{2}, |F_{1}B|=1, |F_{2}B|=2a+1=3, |F_{1}F_{2}|=2c."
  },
  {
    "text": "The coordinates of the focus of the parabola $y=\\frac{1}{8} x^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2/8)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,2)",
    "fact_spans": "[[[0, 24]], [[0, 24]]]",
    "query_spans": "[[[0, 31]]]",
    "process": ""
  },
  {
    "text": "If there exists a point $P$ on the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ such that $|P F_{1}|=8|P F_{2}|$, where $F_{1}$ and $F_{2}$ are the left and right foci of $C$ respectively, then the range of the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 8*Abs(LineSegmentOf(P, F2));LeftFocus(C) = F1;RightFocus(C) = F2",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "[7/9, 1)",
    "fact_spans": "[[[1, 58], [112, 115], [123, 126]], [[8, 58]], [[8, 58]], [[63, 66]], [[94, 101]], [[102, 109]], [[8, 58]], [[8, 58]], [[1, 58]], [[1, 66]], [[69, 91]], [[94, 121]], [[94, 121]]]",
    "query_spans": "[[[123, 137]]]",
    "process": "\\because|PF_{1}|+|PF_{2}|=2a=9|PF_{2}|,\\therefore|PF_{2}|=\\frac{2a}{9} and |PF_{2}|\\in[a-c,a+c],\\therefore a-c\\leqslant\\frac{2a}{9}\\leqslant a+c, solving gives e=\\frac{c}{a}\\geqslant\\frac{7}{9}, then e\\in[\\frac{7}{9},1)."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has an eccentricity of $\\frac{3}{2}$, and shares common foci with the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{3}=1$, then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Ellipse;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2/12 + y^2/3 = 1);Eccentricity(C) = 3/2;Focus(C) = Focus(G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4 - y^2/5 = 1",
    "fact_spans": "[[[2, 63], [129, 132]], [[10, 63]], [[10, 63]], [[84, 122]], [[10, 63]], [[10, 63]], [[2, 63]], [[84, 122]], [[2, 81]], [[2, 127]]]",
    "query_spans": "[[[129, 137]]]",
    "process": "【Solution】From the ellipse equation, find the coordinates of the foci to obtain the value of c. Then, using the eccentricity of the hyperbola, find a, and thus obtain the standard equation of the hyperbola. From the ellipse equation $\\frac{x^2}{12}+\\frac{y^{2}}{3}=1$, the foci are found to be $(3,0)$, $(-3,0)$. Let the semi-focal length of the hyperbola be $c$, then $c=3$. Since the eccentricity of the hyperbola is $\\frac{3}{2}$, we have $e=\\frac{c}{a}=\\frac{3}{a}=\\frac{3}{2}$. Hence, $a=2$, so $b=\\sqrt{c^{2}-a^{2}}=\\sqrt{5}$. Therefore, the standard equation of the hyperbola is: $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$"
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{64}+\\frac{y^{2}}{16}=1$, and let line $l$ passing through $F_{1}$ intersect the ellipse at points $A$, $B$. Then the maximum value of $|A F_{2}|+|B F_{2}|$ is?",
    "fact_expressions": "l: Line;G: Ellipse;A: Point;F2: Point;B: Point;F1: Point;Expression(G) = (x^2/64 + y^2/16 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, l);Intersection(l, G) = {A, B}",
    "query_expressions": "Max(Abs(LineSegmentOf(A, F2)) + Abs(LineSegmentOf(B, F2)))",
    "answer_expressions": "28",
    "fact_spans": "[[[71, 76]], [[17, 56], [77, 79]], [[80, 83]], [[9, 16]], [[84, 87]], [[1, 8], [63, 70]], [[17, 56]], [[1, 61]], [[1, 61]], [[62, 76]], [[71, 89]]]",
    "query_spans": "[[[91, 118]]]",
    "process": "By the given condition, the ellipse $\\frac{x^{2}}{64}+\\frac{y^{2}}{16}=1$ gives $a^{2}=64$, $b^{2}=16$, so $a=8$, $b=4$. According to the definition of an ellipse, $|AF_{1}|+|AF_{2}|=16$, $|BF_{1}|+|BF_{2}|=16$. Then $|AF_{2}|+|BF_{2}|+|AF_{1}|+|BF_{1}|=|AF_{2}|+|BF_{2}|+|AB|=32$, so $|AF_{2}|+|BF_{2}|=32-|AB|$. When $AB$ is perpendicular to the $x$-axis, $|AB|$ reaches its minimum value, and thus $|AF_{2}|+|BF_{2}|$ reaches its maximum value. At this time, $|AB|=\\frac{2b^{2}}{a}=\\frac{2\\times16}{8}=4$, so the maximum value of $|AF_{2}|+|BF_{2}|$ is $32-4=28$."
  },
  {
    "text": "Let $P$ be a point on the hyperbola $x^{2}-\\frac{y^{2}}{12}=1$, and let $F_{1}$, $F_{2}$ be the two foci of the hyperbola. If $|P F_{1}|=\\frac{3}{2}|P F_{2}|$, then what is $\\cos \\angle F_{1} PF_{2}$?",
    "fact_expressions": "P: Point;G: Hyperbola;Expression(G) = (x^2 - y^2/12 = 1);PointOnCurve(P, G) = True;F1: Point;F2: Point;Focus(G) = {F1, F2};Abs(LineSegmentOf(P, F1)) = (3/2)*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Cos(AngleOf(F1, P, F2))",
    "answer_expressions": "-13/4",
    "fact_spans": "[[[1, 4]], [[5, 34], [56, 59]], [[5, 34]], [[1, 38]], [[39, 46]], [[47, 54]], [[39, 64]], [[66, 98]]]",
    "query_spans": "[[[100, 128]]]",
    "process": ""
  },
  {
    "text": "The distance from the vertex of the hyperbola $x^{2}-y^{2}=1$ to its asymptote is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 1)",
    "query_expressions": "Distance(Vertex(G), Asymptote(G))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[0, 18], [22, 23]], [[0, 18]]]",
    "query_spans": "[[[0, 32]]]",
    "process": "The distance from the vertices $(\\pm1,0)$ of the hyperbola $x^{2}-y^{2}=1$ to its asymptotes $y=\\pm x$ is $\\frac{|\\pm1-0|}{\\sqrt{2}}=\\frac{\\sqrt{2}}{2}$."
  },
  {
    "text": "Given that two tangents are drawn from point $T(-1,2)$ to the parabola $C$: $y^{2}=2 p x$ ($p>0$), with points of tangency $A$ and $B$ respectively, and the line $AB$ passes through the focus $F$ of the parabola $C$, then $|T A|^2 +|T B|^2$=?",
    "fact_expressions": "C: Parabola;p: Number;A: Point;B: Point;T: Point;l1: Line;l2: Line;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(T) = (-1, 2);Focus(C)=F;TangentOfPoint(T, C) = {l1, l2};TangentPoint(l1,C)=A;TangentPoint(l2,C)=B;PointOnCurve(F, LineOf(A,B))",
    "query_expressions": "Abs(LineSegmentOf(T, A))^2 + Abs(LineSegmentOf(T, B))^2",
    "answer_expressions": "64",
    "fact_spans": "[[[14, 40], [69, 75]], [[21, 40]], [[51, 54]], [[55, 58]], [[3, 13]], [], [], [[78, 81]], [[21, 40]], [[14, 40]], [[3, 13]], [[69, 81]], [[2, 45]], [[2, 58]], [[2, 58]], [[60, 81]]]",
    "query_spans": "[[[83, 103]]]",
    "process": ""
  },
  {
    "text": "Given that the center of the circle $x^{2}-2 x+y^{2}-8=0$ is the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$), a line passing through point $F$ intersects the directrix of the parabola at point $A$, and intersects the parabola at point $B$, with $\\overrightarrow{F A}=-3 \\overrightarrow{F B}$. Then $|A B|$=?",
    "fact_expressions": "G: Parabola;p: Number;H: Circle;L: Line;F: Point;A: Point;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (y^2 + x^2 - 2*x - 8 = 0);Center(H)=F;Focus(G)=F;PointOnCurve(F, L);Intersection(L, Directrix(G)) = A;OneOf(Intersection(L,G))=B;VectorOf(F, A) = -3*VectorOf(F, B)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "32/3",
    "fact_spans": "[[[28, 49], [66, 69], [80, 83]], [[31, 49]], [[2, 24]], [[62, 64]], [[52, 55], [57, 61]], [[73, 77]], [[89, 92]], [[31, 49]], [[28, 49]], [[2, 24]], [[2, 55]], [[28, 55]], [[56, 64]], [[62, 77]], [[62, 92]], [[94, 140]]]",
    "query_spans": "[[[142, 151]]]",
    "process": "The circle $x^{2}-2x+y^{2}-8=0$, or $(x-1)^{2}+y^{2}=9$, has center coordinates $(1,0)$, then $\\frac{p}{2}=1$, and the parabola equation is $y^{2}=4x$, so $|DF|=2$. As shown in the figure, $\\overrightarrow{FA}=-3\\overrightarrow{FB}$, thus $|AF|:|FB|=3:1$. Also, $|DF|:|BC|=|AF|:|AB|$, so $2:|BC|=3:4$, yielding $|BC|=|BF|=\\frac{8}{3}$. Therefore, $|AB|=4|BF|=\\frac{32}{3}$."
  },
  {
    "text": "Let $F$ be the focus of the parabola $C$: $y^{2}=4x$. A line $l$ passing through point $P(-1,0)$ intersects the parabola $C$ at two points $A$, $B$, and point $Q$ is the midpoint of segment $AB$. If $|FQ|=2$, then the slope of line $l$ equals?",
    "fact_expressions": "F: Point;Focus(C) = F;C: Parabola;Expression(C) = (y^2 = 4*x);P: Point;Coordinate(P) = (-1, 0);PointOnCurve(P,l) = True;l: Line;Intersection(l, C) = {A, B};A: Point;B: Point;MidPoint(LineSegmentOf(A,B)) = Q;Q: Point;Abs(LineSegmentOf(F, Q)) = 2",
    "query_expressions": "Slope(l)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 4]], [[1, 26]], [[5, 23], [45, 51]], [[5, 23]], [[28, 38]], [[28, 38]], [[27, 44]], [[39, 44], [90, 95]], [[39, 61]], [[54, 57]], [[58, 61]], [[62, 76]], [[62, 66]], [[78, 88]]]",
    "query_spans": "[[[90, 101]]]",
    "process": "By the given condition, let the equation of the line be $my = x + 1$. Solving simultaneously  \n$$\n\\begin{cases}\nmy = x + 1 \\\\\ny^{2} = 4x\n\\end{cases}\n$$  \nwe obtain $y^{2} - 4my + 4 = 0$, $\\Delta = 16m^{2} - 16 = 16(m^{2} - 1) > 0$. Let $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$, $Q(x_{0}, y_{0})$. $\\therefore y_{1} + y_{2} = 4m$, $\\therefore y_{0} = \\frac{y_{1} + y_{2}}{2} = 2m$, $\\therefore x_{0} = my_{0} - 1 = 2m^{2} - 1$, $\\therefore Q(2m^{2} - 1, 2m)$. From the parabola $C: y^{2} = 4x$, we get the focus $F$. $\\because |QF| = 2$, $\\therefore \\sqrt{(2m^{2} - 2)^{2} + (2m)^{2}} = 2$, simplifying gives $m^{2} = 1$, solving yields $m = \\pm 1$, which does not satisfy $a > 0$. Hence, there is no line satisfying the condition."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{2-m}-\\frac{y^{2}}{m+1}=1$ represents a hyperbola, then what is the range of values for $m$?",
    "fact_expressions": "G: Hyperbola;Expression(G)=(x^2/(2-m)-y^2/(m+1)=1);m:Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-1,2)",
    "fact_spans": "[[[45, 48]], [[2, 48]], [[50, 53]]]",
    "query_spans": "[[[50, 60]]]",
    "process": "When the foci are on the x-axis, 2-m>0 and m+1>0, solving gives -1<m<2; when the foci are on the y-axis, 2-m<0 and m+1<0, which has no solution. In summary, when the equation represents a hyperbola, the range of values for m is -1<m<2. [Analysis] This problem mainly examines the standard equation of a hyperbola, and is a medium-difficulty question."
  },
  {
    "text": "Given that the length of the minor axis of an ellipse is $2$, and the distance between an endpoint of the major axis and an endpoint of the minor axis is $\\sqrt{5}$, then the standard equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;Length(MinorAxis(G)) = 2;Distance(Endpoint(MajorAxis(G)),Endpoint(MinorAxis(G)))=sqrt(5)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/4+y^2=1,x^2+y^2/4=1}",
    "fact_spans": "[[[2, 4], [43, 45]], [[2, 13]], [[2, 40]]]",
    "query_spans": "[[[43, 52]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has an asymptote with equation $y=\\sqrt{3} x$. If the distance from its right vertex to this asymptote is $\\sqrt{3}$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;J: Line;OneOf(Asymptote(G)) = J;Expression(J) = (y = sqrt(3)*x);Distance(RightVertex(G), J) = sqrt(3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/12 = 1",
    "fact_spans": "[[[2, 58], [83, 84], [109, 112]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [], [[2, 64]], [[2, 81]], [[2, 107]]]",
    "query_spans": "[[[109, 116]]]",
    "process": "The distance from the right vertex $(a,0)$ to the asymptote $y=\\sqrt{3}x$ is $d=\\frac{\\sqrt{3}a}{2}=\\sqrt{3}$, solving gives: $a=2$. From the hyperbola equation, its asymptotes are $y=\\pm\\frac{b}{a}x$, $\\therefore \\frac{b}{a}=\\sqrt{3}$, solving gives: $b=2\\sqrt{3}$, $\\therefore$ the hyperbola equation is $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$"
  },
  {
    "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$), and let $F_{1}$, $F_{2}$ be the left and right foci of the hyperbola $C$, respectively. If there exists a point $M$ on the hyperbola $C$ such that $\\frac{1}{3}|M F_{1}|=|M O|=|M F_{2}|$ ($O$ is the origin), then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;M: Point;F1: Point;O: Origin;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(M, C);(1/3)*Abs(LineSegmentOf(M,F1))=Abs(LineSegmentOf(M,O));Abs(LineSegmentOf(M,O))=Abs(LineSegmentOf(M,F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[1, 61], [82, 88], [97, 103], [160, 166]], [[8, 61]], [[8, 61]], [[105, 109]], [[62, 70]], [[151, 154]], [[72, 79]], [[8, 61]], [[8, 61]], [[1, 61]], [[62, 94]], [[62, 94]], [[97, 109]], [[112, 150]], [[112, 150]]]",
    "query_spans": "[[[160, 172]]]",
    "process": "\\because\\frac{1}{3}|MF_{1}|=|MO|=|MF_{2}|\\therefore M \\text{ lies on the right branch of the hyperbola, so } 2a=|MF_{1}|-|MF_{2}|=2|MF_{2}|. \\text{ In } \\triangle F_{1}OM, |F_{1}O|=c, |MF_{2}|=a, |MF_{1}|=3a, |MO|=a. \\therefore \\cos\\angle MF_{1}O = \\frac{9a^{2}+c^{2}-a^{2}}{2\\cdot3a\\cdot c} = \\frac{8a^{2}+c^{2}}{6ac}. \\text{ In } \\triangle F_{1}F_{2}M, |F_{1}F_{2}|=2c, |MF_{1}|=3a, |MF_{2}|=a. \\therefore \\cos\\angle MF_{1}F_{2} = \\frac{9a^{2}+4c^{2}-a^{2}}{2\\cdot3a\\cdot2c} = \\frac{8a^{2}+4c^{2}}{12ac}. \\because \\angle MF_{1}O = \\angle MF_{1}F_{2}, \\therefore \\frac{8a^{2}+c^{2}}{6ac} = \\frac{8a^{2}+4c^{2}}{12ac}, \\text{ thus } 4a^{2}=c^{2}. \\therefore \\frac{c}{a}=2"
  },
  {
    "text": "Given that point $P$ is an arbitrary point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, when the distance from point $P$ to the line $4x-5y+40=0$ is minimized, what are the coordinates of point $P$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);P: Point;PointOnCurve(P, G);H: Line;Expression(H) = (4*x - 5*y + 40 = 0);WhenMin(Distance(P,H))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(-4, 9/5)",
    "fact_spans": "[[[7, 45]], [[7, 45]], [[2, 6], [53, 57], [86, 90]], [[2, 50]], [[58, 74]], [[58, 74]], [[52, 83]]]",
    "query_spans": "[[[86, 95]]]",
    "process": "Let the line $ l_{1}: 4x - 5y + m = 0 $ ($ m \\in \\mathbb{R} $). When the line $ l_{1} $ is tangent to the ellipse, the distance from one of the tangent points to the line $ 4x - 5y + 40 = 0 $ is minimized. Thus, solving the system  \n\\[\n\\begin{cases}\n4x - 5y + m = 0 \\\\\n\\frac{x^{2}}{25} + \\frac{y^{2}}{9} = 1\n\\end{cases}\n\\]\nyields $ 25x^{2} + 8mx + m^{2} - 225 = 0 $. For tangency, $ \\Delta = b^{2} - 4ac = 0 \\Rightarrow m = \\pm 25 $. It is clear that when $ m = 25 $, the distance from the point to the line $ 4x - 5y + 40 = 0 $ is minimized. Substituting $ m = 25 $ into $ 25x^{2} + 8mx + m^{2} - 225 = 0 $ gives $ x = -4 $. Substituting $ x = -4 $ into $ 4x - 5y + 25 = 0 $ yields $ y = \\frac{9}{5} $. Hence, the coordinates of point $ P $ are $ (-4, \\frac{9}{5}) $."
  },
  {
    "text": "The hyperbola has the lines $x \\pm \\sqrt{2} y = 0$ as asymptotes and passes through the focus of the parabola $x^{2} - 4x + 4y + 8 = 0$. Then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;L: Line;Expression(L) = (x + pm*sqrt(2)*y = 0);Asymptote(G) = L;H: Parabola;Expression(H) = (4*y + x^2 - 4*x + 8 = 0);PointOnCurve(Focus(H), G) = True",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/2 - x^2/4 = 1",
    "fact_spans": "[[[0, 3], [63, 66]], [[4, 27]], [[4, 27]], [[0, 31]], [[35, 57]], [[35, 57]], [[0, 60]]]",
    "query_spans": "[[[63, 71]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, $P$ is a point on $C$ such that $P F_{2} \\perp x$-axis, the line $P F_{1}$ intersects $C$ at another point $Q$, if $|P F_{1}|=4|F_{1} Q|$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P, F2), xAxis);Q: Point;OneOf(Intersection(LineOf(P, F1), C)) = Q;Abs(LineSegmentOf(P, F1)) = 4*Abs(LineSegmentOf(F1, Q))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(21)/7",
    "fact_spans": "[[[2, 59], [88, 91], [127, 130], [166, 169]], [[9, 59]], [[9, 59]], [[84, 87]], [[68, 75]], [[76, 83]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 83]], [[2, 83]], [[84, 94]], [[96, 114]], [[137, 140]], [[115, 140]], [[142, 164]]]",
    "query_spans": "[[[166, 175]]]",
    "process": "According to the problem, we have $ P(c,\\frac{b^{2}}{a}) $. Draw $ QE\\bot x $-axis with foot at point $ E $, and let $ Q(x_{0},y_{0}) $. Using similarity of triangles, the coordinates of point $ Q $ can be found. Substituting the coordinates of $ Q $ into the ellipse equation and combining with $ b^{2}=a^{2}-c^{2} $, we obtain the solution.  \n\n**Detailed Solution:** Since $ PF_{2}\\bot x $-axis, the coordinates of point $ P $ are $ P(c,\\frac{b^{2}}{a}) $. As shown in the figure, draw $ QE\\bot x $-axis with foot at point $ E $, and let $ Q(x_{0},y_{0}) $.  \n\n$ \\because \\frac{|PF_{1}|}{|F_{1}Q|}=4 $,  \n$ \\therefore \\frac{|F_{1}F_{2}|}{|EF_{1}|}=\\frac{|PF_{2}|}{|QE|}=4 $,  \n$ \\therefore |EF|=\\frac{|F_{1}F_{2}|}{4}=\\frac{2c}{4}=\\frac{c}{2} $,  \n$ \\therefore x_{0}=-c-\\frac{c}{2}=-\\frac{3c}{2} $.  \n\nAlso, $ y_{0}=-|QE|=-\\frac{|PF_{2}|}{4}=-\\frac{b^{2}}{4a} $, so the coordinates of point $ Q $ are $ (-\\frac{3c}{2},-\\frac{b^{2}}{4a}) $.  \n\nSubstitute point $ Q $ into the ellipse equation:  \n$ \\frac{9c^{2}}{4a^{2}}+\\frac{b^{2}}{16a^{2}}=1 $, i.e.,  \n$ \\frac{9c^{2}}{4a^{2}}+\\frac{a^{2}-c^{2}}{16a^{2}}=1 $.  \n\nSolving gives:  \n$ \\frac{c^{2}}{a^{2}}=\\frac{3}{7} $, i.e., $ e=\\frac{c}{a}=\\frac{\\sqrt{21}}{7} $."
  },
  {
    "text": "If the point $M(-2 , 8)$ lies on the directrix of the parabola $y^{2}=2 px$, then the value of the real number $p$ is?",
    "fact_expressions": "G: Parabola;p: Real;M: Point;Expression(G) = (y^2 = 2*(p*x));Coordinate(M) = (-2, 8);PointOnCurve(M, Directrix(G))",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[14, 29]], [[35, 40]], [[1, 13]], [[14, 29]], [[1, 13]], [[1, 33]]]",
    "query_spans": "[[[35, 44]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the point where the directrix intersects the axis of symmetry for a parabola with vertex at the origin, axis of symmetry along the $y$-axis, and passing through the point $(4,1)$ are?",
    "fact_expressions": "G: Parabola;H: Point;O: Origin;Coordinate(H) = (4, 1);Vertex(G)=O;SymmetryAxis(G)=yAxis;PointOnCurve(H,G)",
    "query_expressions": "Coordinate(Intersection(Directrix(G),SymmetryAxis(G)))",
    "answer_expressions": "(0,-4)",
    "fact_spans": "[[[26, 29]], [[17, 25]], [[3, 5]], [[17, 25]], [[0, 29]], [[6, 29]], [[15, 29]]]",
    "query_spans": "[[[26, 43]]]",
    "process": "Let the equation of the parabola be $ x^{2} = 2py $ ($ p > 0 $), then we have $ 4^{2} = 2p \\cdot 1 $, that is, $ 2p = 16 $. Thus, the equation of the parabola is $ x^{2} = 16y $, and its directrix is $ y = -4 $. The coordinates of the intersection point of the directrix and the axis of symmetry are $ (0, -4) $. This question examines the standard equation, properties, and graph of a parabola, and belongs to a basic problem."
  },
  {
    "text": "Let the focus of the parabola $x^{2}=12 y$ be $F$, and let line $l$ passing through point $P(2 , 1)$ intersect the parabola at points $A$ and $B$. It is known that point $P$ is exactly the midpoint of $AB$. Then $|A F|+|B F|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 12*y);F: Point;Focus(G) = F;P: Point;Coordinate(P) = (2, 1);l: Line;PointOnCurve(P, l);A: Point;B: Point;Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 16], [44, 47]], [[1, 16]], [[20, 23]], [[1, 23]], [[26, 37], [64, 68]], [[26, 37]], [[38, 43]], [[25, 43]], [[50, 53]], [[56, 59]], [[38, 61]], [[64, 77]]]",
    "query_spans": "[[[79, 94]]]",
    "process": "Draw perpendiculars from points A, B, P to the directrix of the parabola $ y = .3 $, with feet at C, D, Q respectively. By the definition of a parabola, we obtain $ |AF| + |BF| = |AC| + |BD| = 2|PQ| = 8 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $P$ is a point on the hyperbola such that $S_{\\Delta P F_{1} F_{2}}=\\sqrt{3}$. Find $\\angle F_{1} P F_{2}$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/3 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G) = True;Area(TriangleOf(P,F1,F2))=sqrt(3)",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "2*pi/3",
    "fact_spans": "[[[2, 40], [70, 73]], [[2, 40]], [[50, 57]], [[58, 65]], [[2, 65]], [[2, 65]], [[66, 69]], [[66, 76]], [[78, 113]]]",
    "query_spans": "[[[115, 139]]]",
    "process": "According to the problem, $ a=2 $, $ b=\\sqrt{3} $, $ c=\\sqrt{7} $. Let $ |PF_{1}|=m $, $ |PF_{2}|=n $, without loss of generality assume $ m>n $, $ |F_{1}F_{2}|=2c=2\\sqrt{7} $. Let $ \\angle F_{1}PF_{2}=\\theta\\in(0,\\pi) $. Based on the definition of hyperbola, the law of cosines, and the triangle area formula, we have:\n$$\n\\begin{cases}\n(m-n)^{2}=16 \\\\\n28=m^{2}+n^{2}-2mn\\cos\\theta, \\\\\nmn\\sin\\theta=2\\sqrt{3}\n\\end{cases}\n.\n\\begin{cases}\nm^{2}+n^{2}-2mn=16 \\\\\n28=m^{2}+n^{2}-2mn\\cos\\theta \\\\\nmn\\sin\\theta=2\\sqrt{3}\n\\end{cases}\n\\begin{matrix}\nmn=\\frac{2\\sqrt{3}}{\\sin\\theta} \\\\\n12=2mn(1-\\cos\\theta) & m=\\frac{2\\sqrt{3}}{\\sin\\theta}\n\\end{matrix}\n=12=2\\cdot\\frac{2\\sqrt{3}}{\\sin\\theta}\\cdot(1-\\cos\\theta)' \\Rightarrow \\sqrt{3}\\sin\\theta+\\cos\\theta=1 \\Rightarrow 2\\sin(\\theta+\\frac{\\pi}{6})=1, \\sin(\\theta+\\frac{\\pi}{6})=\\frac{1}{2}\n$$\nSince $ 0<\\theta<\\pi $, $ \\frac{\\pi}{6}<\\theta+\\frac{\\pi}{6}<\\frac{7\\pi}{6} $, so $ \\theta+\\frac{\\pi}{6}=\\frac{5\\pi}{6} $, $ \\theta=\\frac{2\\pi}{3} $, therefore $ \\angle F_{1}PF_{2}=\\frac{2\\pi}{3} $."
  },
  {
    "text": "If the slope of an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $\\sqrt{2}$, then the eccentricity $e$=?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Slope(OneOf(Asymptote(G))) = sqrt(2);Eccentricity(G)=e;e:Number",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 77]], [[1, 85]], [[82, 85]]]",
    "query_spans": "[[[82, 87]]]",
    "process": "From the asymptote equation $ y = \\pm\\frac{b}{a}x' $, the slope of one asymptote is $ \\sqrt{2} $, so $ \\frac{b}{a} = \\sqrt{2} $. The eccentricity is found as follows: for the hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0 $, $ b > 0 $), the slope of one asymptote is $ \\sqrt{2} $, so $ \\frac{b}{a} = \\sqrt{2} $. The eccentricity of the hyperbola is $ e = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{3} $."
  },
  {
    "text": "Let the left and right foci of the ellipse $C$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ be $F_{1}$ and $F_{2}$ respectively, and let $A$ be any point on $C$. Then the perimeter of $\\Delta A F_{1} F_{2}$ is?",
    "fact_expressions": "C: Ellipse;A: Point;F1: Point;F2: Point;Expression(C) = (x^2/25 + y^2/9 = 1);LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(A,C)",
    "query_expressions": "Perimeter(TriangleOf(A, F1, F2))",
    "answer_expressions": "18",
    "fact_spans": "[[[1, 44], [73, 76]], [[69, 72]], [[53, 60]], [[61, 68]], [[1, 44]], [[1, 68]], [[1, 68]], [[69, 81]]]",
    "query_spans": "[[[83, 110]]]",
    "process": "According to the problem, the ellipse $ C: \\frac{x^{2}}{25} + \\frac{y^{2}}{9} = 1 $, where $ a = \\sqrt{25} = 5 $, $ b = \\sqrt{9} = 3 $, then $ c = \\sqrt{25 - 9} = 4 $. $ A $ is any point on $ C $, then the perimeter $ l $ of $ \\Delta AF_{1}F_{2} $ is $ |AF_{1}| + |AF_{2}| + |F_{1}F_{2}| = 2a + 2c = 10 + 8 = 18 $;"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$ with right focus $F$ and right directrix $l$, point $A \\in l$, segment $AF$ intersects $C$ at point $B$. If $\\overrightarrow{F A}=3 \\overrightarrow{F B}$, then $|\\overrightarrow{A F}|$=?",
    "fact_expressions": "C: Ellipse;A: Point;F: Point;B: Point;l: Line;Expression(C) = (x^2/2 + y^2 = 1);RightFocus(C) = F;RightDirectrix(C) = l;In(A, l);Intersection(LineSegmentOf(A, F), C) = B;VectorOf(F, A) = 3*VectorOf(F, B)",
    "query_expressions": "Abs(VectorOf(A, F))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 34], [70, 73]], [[51, 61]], [[39, 42]], [[74, 78]], [[47, 50]], [[2, 34]], [[2, 42]], [[2, 50]], [[51, 61]], [[62, 78]], [[80, 125]]]",
    "query_spans": "[[[128, 154]]]",
    "process": "Since the right directrix of the ellipse $ C: \\frac{x^2}{2} + y^2 = 1 $ is $ x = 2 $, the horizontal coordinate of point $ A $ is $ 2 $. Let $ B(x_1, y_1) $, and $ \\overrightarrow{FA} = 3\\overrightarrow{FB} $, then $ 2 - 1 = 3(x_1 - 1) $, hence $ x_1 = \\frac{4}{3} $. Therefore, $ y_1^2 = 1 - \\frac{16}{2 \\times 9} = \\frac{1}{9} $, so $ |y_1| = \\frac{1}{3} $, thus $ |y_A| = 3 \\times \\frac{1}{3} = 1 $, so $ |\\overrightarrow{AF}| = \\sqrt{(2 - 1)^2 + 1} = \\sqrt{2} $."
  },
  {
    "text": "A line with slope $1$ passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) intersects the parabola at points $A$ and $B$, and $|A F| \\cdot|B F|=8$. Find the value of $p$.",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);Slope(H) = 1;A: Point;B: Point;Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, F))*Abs(LineSegmentOf(B, F)) = 8",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 22], [39, 42]], [[1, 22]], [[75, 78]], [[4, 22]], [[25, 28]], [[1, 28]], [[36, 38]], [[0, 38]], [[29, 38]], [[43, 46]], [[47, 50]], [[36, 52]], [[53, 73]]]",
    "query_spans": "[[[75, 82]]]",
    "process": "The focus of the parabola $ y^{2}=2px $ is $ F(\\frac{p}{2},0) $, and the equation of the directrix is $ x=-\\frac{p}{2} $. Let $ A(x_{1},y_{2}) $, $ B(x_{2},y_{2}) $. Then the equation of line $ AB $ is $ y=x-\\frac{p}{2} $. Substituting into $ y^{2}=2px $ gives $ x^{2}-3px+\\frac{p^{2}}{4}=0 $. Therefore, $ x_{1}+x_{2}=3p $, $ x_{1}x_{2}=\\frac{p^{2}}{4} $. By the definition of the parabola, $ |AF|=x_{1}+\\frac{p}{2} $, $ |BF|=x_{2}+\\frac{p}{2} $, $ |_{AF}| \\cdot |_{BF}|=(x_{1}+\\frac{p}{2})(x_{2}+\\frac{p}{2})=x_{1}x_{2}+\\frac{p}{2}(x_{1}+x_{2})+\\frac{p^{2}}{4}=2p^{2}=8 $, solving gives $ p=2 $."
  },
  {
    "text": "Let the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ have eccentricity $\\sqrt{3}$, and suppose one of its directrices coincides with the directrix of the parabola $y^{2}=4 x$. Then the equation of this hyperbola is?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;a > 0;b > 0;Expression(C) = (x^2/a^2 - y^2/b^2 = 1);Eccentricity(C) = sqrt(3);E: Parabola;Expression(E) = (y^2 = 4*x);OneOf(Directrix(C)) = Directrix(E)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/3 - y^2/6 = 1",
    "fact_spans": "[[[1, 58], [75, 76], [104, 107]], [[4, 58]], [[4, 58]], [[4, 58]], [[4, 58]], [[1, 58]], [[1, 73]], [[82, 85]], [[82, 96]], [[75, 101]]]",
    "query_spans": "[[[104, 112]]]",
    "process": ""
  },
  {
    "text": "Let $M$ be a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and let $F_{1}$, $F_{2}$ be the foci. If $\\angle F_{1} M F_{2}=60^{\\circ}$, then the area of $\\triangle M F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);M: Point;PointOnCurve(M, G);F1: Point;F2: Point;Focus(G) = {F1, F2};AngleOf(F1, M, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(M, F1, F2))",
    "answer_expressions": "3*sqrt(3)",
    "fact_spans": "[[[5, 43]], [[5, 43]], [[1, 4]], [[1, 48]], [[49, 56]], [[57, 64]], [[5, 67]], [[68, 101]]]",
    "query_spans": "[[[103, 133]]]",
    "process": "M is a point on the ellipse \\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1, F_{1}, F_{2} are the two foci of the ellipse, \\angle F_{1}MF_{2}=60^{\\circ}. Let |MF_{1}|=x, |MF_{2}|=y. According to the cosine law: x^{2}+y^{2}-xy=64. Since x+y=10, we obtain: xy=12, thus S=\\frac{1}{2}xy\\sin60^{\\circ}=3\\sqrt{3}"
  },
  {
    "text": "If $a>2$, then the range of the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ is?",
    "fact_expressions": "G: Hyperbola;a: Number;Expression(G) = (-y^2 + x^2/a^2 = 1);a > 2",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1, \\sqrt{5}/2)",
    "fact_spans": "[[[8, 40]], [[1, 6]], [[8, 40]], [[1, 6]]]",
    "query_spans": "[[[8, 51]]]",
    "process": "a>2, then the eccentricity of the hyperbola \\frac{x^{2}}{a^{2}}-y^{2}=1 is \\frac{c}{a}=\\frac{\\sqrt{1+a^{2}}}{a}=\\sqrt{1+\\frac{1}{a^{2}}}\\in(1,\\frac{\\sqrt{5}}{2}),"
  },
  {
    "text": "Through the left focus $F_{1}$ of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, draw a straight line $AB$ with inclination angle $\\frac{\\pi}{6}$, where $A$ and $B$ are the intersection points of the line and the hyperbola. Then the length of $AB$ is?",
    "fact_expressions": "G: Hyperbola;A: Point;B: Point;F1: Point;Expression(G) = (x^2 - y^2/3 = 1);LeftFocus(G) = F1;PointOnCurve(F1,LineOf(A,B));Intersection(LineOf(A,B),G)={A,B};Inclination(LineOf(A,B))=pi/6",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 29], [85, 88]], [[72, 75]], [[76, 79]], [[33, 40]], [[1, 29]], [[1, 40]], [[0, 69]], [[72, 91]], [[42, 69]]]",
    "query_spans": "[[[93, 102]]]",
    "process": "Since the hyperbola equation is $x^{2}-\\frac{y^{2}}{3}=1$, the left focus is $F_{1}(-2,0)$. Since the inclination angle of line $AB$ is $\\frac{\\pi}{6}$, the slope of the line is $\\frac{\\sqrt{3}}{3}$. The equation of line $AB$ is $y=\\frac{\\sqrt{3}}{3}(x+2)$. Substituting into $x^{2}-\\frac{y^{2}}{3}=1$ gives $8x^{2}-4x-13=0$, $x_{1}+x_{2}=\\frac{1}{2}$, $x_{1}x_{2}=-\\frac{13}{8}$. Therefore, $\\frac{\\sqrt{1+3}|x_{1}-x_{2}|=-3}{2}-4(-\\frac{13}{8})=3'$"
  },
  {
    "text": "If the distance from point $P(2 , 0)$ to one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $\\sqrt{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (2, 0);Distance(P, OneOf(Asymptote(G))) = sqrt(2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[13, 59], [82, 85]], [[16, 59]], [[16, 59]], [[1, 12]], [[13, 59]], [[1, 12]], [[1, 80]]]",
    "query_spans": "[[[82, 91]]]",
    "process": ""
  },
  {
    "text": "What is the distance from the focus to the directrix of the parabola $x^{2}=y$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = y)",
    "query_expressions": "Distance(Focus(G),Directrix(G))",
    "answer_expressions": "1/2",
    "fact_spans": "[[[0, 12]], [[0, 12]]]",
    "query_spans": "[[[0, 23]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, $A$, $C$ are the upper and lower vertices of the ellipse, respectively, $B$ is the left vertex, $F$ is the left focus, the lines $AB$ and $FC$ intersect at point $D$, then $\\sin \\angle BDF$=?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;F: Point;C: Point;D: Point;Expression(G) = (x^2/4 + y^2/3 = 1);LeftFocus(G) = F;UpperVertex(G)=A;LowerVertex(G)=C;LeftVertex(G)=B;Intersection(LineOf(A,B), LineOf(F, C)) = D",
    "query_expressions": "Sin(AngleOf(B, D, F))",
    "answer_expressions": "3*sqrt(21)/14",
    "fact_spans": "[[[2, 39], [53, 55]], [[42, 45]], [[62, 65]], [[70, 73]], [[47, 50]], [[94, 98]], [[2, 39]], [[53, 77]], [[42, 61]], [[42, 61]], [[53, 69]], [[78, 98]]]",
    "query_spans": "[[[100, 121]]]",
    "process": "From $ a^{2}=4, b^{2}=3 $ we get: $ c^{2}=4-3=1 $, so $ F(-1,0) $, $ B(-2,0) $, $ A(0,\\sqrt{3}) $, $ C(0,-\\sqrt{3}) $, $ k_{AB}=\\frac{\\sqrt{3}}{2} $, $ k_{CF}=\\frac{-\\sqrt{3}-0}{0-(-1)}=-\\sqrt{3} $, using the angle formula we obtain: $ \\tan\\angle BDF = \\frac{-\\sqrt{3}-\\frac{\\sqrt{3}}{2}}{1+\\frac{\\sqrt{3}}{2}\\times(-\\sqrt{3})}=3\\sqrt{3} $, where $ \\angle BDF \\in (0,\\pi) $, so $ \\sin\\angle BDF = \\frac{3\\sqrt{21}}{14} $"
  },
  {
    "text": "The coordinates of the foci of the ellipse $\\frac{x^{2}}{100}+\\frac{y^{2}}{64}=1$ are?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/100 + y^2/64 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(pm*6,0)",
    "fact_spans": "[[[0, 40]], [[0, 40]]]",
    "query_spans": "[[[0, 47]]]",
    "process": ""
  },
  {
    "text": "If the distance from point $M$ on the parabola $y^{2}=4x$ to the focus is $10$, then what is the distance from $M$ to the $y$-axis?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(M, G);Distance(M, Focus(G)) = 10",
    "query_expressions": "Distance(M, yAxis)",
    "answer_expressions": "9",
    "fact_spans": "[[[1, 15]], [[17, 21], [34, 37]], [[1, 15]], [[1, 21]], [[1, 32]]]",
    "query_spans": "[[[34, 47]]]",
    "process": "x_{M}+1=10\\Rightarrow x_{M}=9. 【Topic】Definition of a parabola. 【Approach"
  },
  {
    "text": "Let the focus of the parabola $y^{2}=4x$ be $F$, and let $P$ be a point on the parabola (in the first quadrant). If the center of the circle with diameter $PF$ lies on the line $x+y=2$, then the radius of this circle is?",
    "fact_expressions": "G: Parabola;H: Circle;C:Line;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Expression(C) = (x + y = 2);Focus(G)=F;PointOnCurve(P,G);Quadrant(P)=1;IsDiameter(LineSegmentOf(P,F),H);PointOnCurve(Center(H),C)",
    "query_expressions": "Radius(H)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[1, 15], [27, 30]], [[53, 54], [71, 72]], [[58, 67]], [[23, 26]], [[19, 22]], [[1, 15]], [[58, 67]], [[1, 22]], [[23, 33]], [[23, 41]], [[43, 54]], [[53, 68]]]",
    "query_spans": "[[[71, 77]]]",
    "process": "Let P(x_{1},y_{1}), then the center of the circle is (\\frac{x_{1}+1}{2},\\frac{y_{1}}{2}). Substituting into x+y-2 and combining with y_{1}^{2}=4x_{1}, we can solve to get P(2,2). Hence, the radius of this circle is \\underline{\\sqrt{(2-1)^{2}+2^{2}}}=\\frac{\\sqrt{5}}{2}"
  },
  {
    "text": "Given $A(1,4)$, point $P$ is a moving point on the parabola $y^{2}=12x$, and the distance from point $P$ to the line $x=-2$ is $d$. Then the minimum value of $|PA|+d$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;P: Point;Expression(G) = (y^2 = 12*x);Expression(H) = (x = -2);Coordinate(A) = (1, 4);PointOnCurve(P, G);Distance(P, H) = d;d:Number",
    "query_expressions": "Min(d + Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "2*sqrt(5)-1",
    "fact_spans": "[[[16, 31]], [[41, 49]], [[2, 10]], [[36, 40], [11, 15]], [[16, 31]], [[41, 49]], [[2, 10]], [[11, 35]], [[36, 56]], [[53, 56]]]",
    "query_spans": "[[[58, 73]]]",
    "process": "As shown in the figure, for the parabola $ y^{2} = 12x $, the focus is at $ F(3,0) $, and the directrix is $ x = -3 $. Draw a perpendicular line from point $ P $ to the directrix, intersecting at point $ B $, and intersecting the line $ x = -2 $ at point $ C $. Then $ d = |PB| = |PC| - 1 $, so $ |PA| + d = |PA| + |PC| - 1 $. By the definition of the parabola, we have $ |PC| = |PF| $, thus $ |PA| + d = |PA| + |PF| - 1 $. From the figure, it follows that $ |PA| + |PF| $ reaches its minimum value when points $ A $, $ P $, and $ F $ are collinear. The minimum value is $ (|PA| + |PF|)_{\\min} = \\sqrt{(3-1)^{2} + (0-4)^{2}} = 2\\sqrt{5} $. Therefore, the minimum value of $ |PA| + d $ is $ 2\\sqrt{5} - 1 $."
  },
  {
    "text": "Given that line $l$ passes through the focus $F$ of the parabola $y^{2}=8x$, intersects the parabola at points $A$ and $B$, and intersects its directrix at point $C$. If point $F$ is the midpoint of $AC$, then the length of segment $BC$ is?",
    "fact_expressions": "l: Line;G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;PointOnCurve(F, l) = True;Intersection(l,G) = {A,B};B: Point;A: Point;Intersection(l,Directrix(G)) = C;C: Point;MidPoint(LineSegmentOf(A, C)) = F",
    "query_expressions": "Length(LineSegmentOf(B,C))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[2, 7]], [[8, 22], [30, 33], [46, 47]], [[8, 22]], [[25, 28], [58, 62]], [[8, 28]], [[2, 28]], [[2, 44]], [[39, 42]], [[35, 38]], [[2, 55]], [[51, 55]], [[58, 71]]]",
    "query_spans": "[[[73, 84]]]",
    "process": "From the equation of the parabola, the coordinates of the focus F and the equation of the directrix can be obtained. Since F is the midpoint of AC, the coordinates of A can be determined. Then find the equation of AB, obtain the coordinates of B, and then calculate the length of BC. According to the problem, the coordinates of the focus F of the parabola are (2,0), and the equation of the directrix is x = -2. Let the equation of line AB be x = my + 2. Draw AM, BN, FD perpendicular to the directrix, meeting it at M, N, D respectively. Since F is the midpoint of AC, AM = 2DF = 4, so x_{A} + 2 = 8, thus x_{A} = 6. Substituting into the parabola equation gives y = \\sqrt{48} = 4\\sqrt{3}. Therefore, k_{AB} = \\frac{4\\sqrt{3}}{6 - 2} = \\sqrt{3}, so the equation of line AB is y = \\sqrt{3}(x - 2). Solving the system of equations of the line and the parabola:  \n\\begin{cases} y = \\sqrt{3}(x - 2) \\\\ y^2 = 8x \\end{cases},  \nrearranging yields 3x^{2} - 20x + 12 = 0, so 6x_{B} = 4, giving x_{B} = \\frac{2}{3}. Then BF = BN = \\frac{2}{3} + 2 = \\frac{8}{3},  \nCB = CF - BF = AF - BF = 8 - \\frac{8}{3} = \\frac{16}{3}"
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=2 p x(p>0)$, $P(2, m)$ is a point on $C$, $d$ is the distance from $P$ to the origin, and $\\frac{d}{|P F|}=\\frac{8}{7}$, then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;P: Point;F: Point;p>0;m:Number;Expression(C) = (y^2 = 2*(p*x));Coordinate(P) = (2, m);Focus(C) = F;PointOnCurve(P, C);O:Origin;d:Number;Distance(P,O)=d;d/Abs(LineSegmentOf(P, F)) = 8/7",
    "query_expressions": "p",
    "answer_expressions": "{5/4, 3}",
    "fact_spans": "[[[6, 32], [46, 49]], [[99, 102]], [[36, 45], [57, 60]], [[2, 5]], [[14, 32]], [[36, 45]], [[6, 32]], [[36, 45]], [[2, 35]], [[36, 52]], [[61, 63]], [[53, 56]], [[53, 66]], [[68, 97]]]",
    "query_spans": "[[[99, 104]]]",
    "process": "Since P(2,m) is a point on C, we have m^{2}=2p\\times2, that is, m^{2}=4p. By the definition of the parabola, we get \\frac{d}{|PF|}=\\frac{\\sqrt{4+m^{2}}}{2+\\frac{p}{2}}=\\frac{\\sqrt{4+4p}}{2+\\frac{P}{2}}=\\frac{8}{7}. Simplifying yields (4p-5)(p-3)=0, hence p=\\frac{5}{4} or 3."
  },
  {
    "text": "Given the ellipse equation $\\frac{x^{2}}{6}+\\frac{y^{2}}{3}=1$, then its eccentricity is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/6 + y^2/3 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 4], [44, 45]], [[2, 42]]]",
    "query_spans": "[[[44, 50]]]",
    "process": "\\because the ellipse equation is \\frac{x^2}{6}+\\frac{y^{2}}{3}=1 \\therefore a^{2}=6, b^{2}=3 \\therefore a=\\sqrt{6}, c=\\sqrt{a^{2}-b^{2}}=\\sqrt{6-3}=\\sqrt{3} \\therefore eccentricity e=\\frac{c}{a}=\\frac{\\sqrt{3}}{\\sqrt{6}}=\\frac{\\sqrt{2}}{2}"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the ellipse $C$, and $\\angle P F_{1} F_{2}=60^{\\circ}$. Then the area of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/9 + y^2/5 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C);AngleOf(P, F1, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "(5*sqrt(3))/2",
    "fact_spans": "[[[2, 44], [74, 79]], [[2, 44]], [[53, 60]], [[61, 68]], [[2, 68]], [[2, 68]], [[69, 73]], [[69, 80]], [[82, 115]]]",
    "query_spans": "[[[117, 147]]]",
    "process": "By the law of cosines and combining with the definition of an ellipse, PF_{1} can be found, then use the area formula to solve. From the given conditions, we have a=3, c=\\sqrt{9-5}=2. Let |PF_{1}|=m, |PF_{2}|=n. By the definition of the ellipse and the law of cosines, we obtain: (m+n=6|n2=m^{2}+18m\\cos60^{\\circ}. Solving yields m=\\frac{5}{2}, n=\\frac{1}{2}. Therefore, the area of \\triangle PF_{1}F_{2} is \\frac{1}{2}m|F_{1}F_{2}|\\sin60^{\\circ}=\\frac{5\\sqrt{3}}{2}."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ has an eccentricity of $2$, and one focus coincides with the focus of the parabola $y^{2}=16 x$, then what are the coordinates of the foci of the hyperbola? What is the equation of its asymptotes?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 16*x);Eccentricity(G) = 2;OneOf(Focus(G)) = Focus(H)",
    "query_expressions": "Coordinate(Focus(G));Expression(Asymptote(G))",
    "answer_expressions": "(pm*4,0)\ny=pm*sqrt(3)*x",
    "fact_spans": "[[[2, 61], [97, 100]], [[5, 61]], [[5, 61]], [[75, 90]], [[5, 61]], [[5, 61]], [[2, 61]], [[75, 90]], [[2, 69]], [[2, 95]]]",
    "query_spans": "[[[97, 107]], [[97, 114]]]",
    "process": ""
  },
  {
    "text": "The minimum value of the sum of the distance from a point on the parabola $y^{2}=4 x$ to $(0,2)$ and the distance to its directrix is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);K: Point;PointOnCurve(K, G);H: Point;Coordinate(H) = (0, 2)",
    "query_expressions": "Min(Distance(K, H)+Distance(K, Directrix(G)))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[0, 14], [30, 31]], [[0, 14]], [[16, 17]], [[0, 17]], [[18, 25]], [[18, 25]]]",
    "query_spans": "[[[16, 43]]]",
    "process": "\\because the parabola y^{2}=4x, \\therefore F(1,0). As shown in the figure: let the projection of point P on the directrix be A'. According to the definition of the parabola, the distance from P to the directrix of this parabola is |PA'| = |PF|. Then the sum of the distance from point P to point A(0,2) and the distance from P to the directrix of this parabola is d = |PF| + |PA| \\geqslant |AF| = \\sqrt{5}"
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the left and right foci of the ellipse $\\frac{y^{2}}{16}+\\frac{y^{2}}{16}=1$, respectively, $P$ a point on the ellipse, $M$ the midpoint of $F_{1}P$, and $|OM|=3$. Then the distance from point $P$ to the left focus of the ellipse is?",
    "fact_expressions": "G: Ellipse;F1: Point;F2:Point;P: Point;O: Origin;M: Point;Expression(G) = (y^2/16 + y^2/16 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);MidPoint(LineSegmentOf(F1, P)) = M;Abs(LineSegmentOf(O, M)) = 3",
    "query_expressions": "Distance(P, LeftFocus(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[22, 61], [72, 74], [111, 113]], [[1, 8]], [[11, 19]], [[68, 71], [106, 110]], [[95, 103]], [[78, 81]], [[22, 61]], [[1, 67]], [[1, 67]], [[68, 77]], [[78, 94]], [[95, 103]]]",
    "query_spans": "[[[106, 120]]]",
    "process": ""
  },
  {
    "text": "Given a fixed point $Q(2,-1)$, $F$ is the focus of the parabola $y^{2}=4x$, and $P$ is an arbitrary point on the parabola. When $|PQ| + |PF|$ takes the minimum value, what are the coordinates of $P$?",
    "fact_expressions": "G: Parabola;P: Point;Q: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;Coordinate(Q) = (2, -1);PointOnCurve(P, G);WhenMin(Abs(LineSegmentOf(P, Q)) + Abs(LineSegmentOf(P, F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1/4, -1)",
    "fact_spans": "[[[19, 33], [43, 46]], [[39, 42], [73, 76]], [[4, 13]], [[15, 18]], [[19, 33]], [[15, 36]], [[4, 13]], [[39, 51]], [[53, 72]]]",
    "query_spans": "[[[73, 81]]]",
    "process": ""
  },
  {
    "text": "Let $P$ be a moving point on the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$. If the distances from $P$ to the two asymptotes are $d_{1}$ and $d_{2}$, respectively, then $d_{1} \\cdot d_{2}$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/2 = 1);P:Point;PointOnCurve(P,G);L1:Line;L2:Line;Asymptote(G)={L1,L2};Distance(P,L1)=d1;Distance(P,L2)=d2;d1:Number;d2:Number",
    "query_expressions": "d1*d2",
    "answer_expressions": "4/3",
    "fact_spans": "[[[5, 43]], [[5, 43]], [[1, 4], [48, 51]], [[1, 47]], [], [], [[5, 57]], [[5, 79]], [[5, 79]], [[63, 70]], [[72, 79]]]",
    "query_spans": "[[[81, 102]]]",
    "process": "From the given conditions: the two asymptotes are $x \\pm \\sqrt{2}y = 0$. Let $P(x, y)$ be a point on the hyperbola $C$, then the distances from point $P$ to the two asymptotes are $d_{1} = \\frac{|x + \\sqrt{2}y|}{\\sqrt{3}}$, $d_{2} = \\frac{|x - \\sqrt{2}y|}{\\sqrt{3}}$. Therefore, $d_{1} \\cdot d_{2} = \\frac{|x + \\sqrt{2}y|}{\\sqrt{3}} \\cdot \\frac{|x - \\sqrt{2}y|}{\\sqrt{3}} = \\frac{|x^{2} - 2y^{2}|}{3} = \\frac{4}{3}$."
  },
  {
    "text": "The equation of the locus of the midpoint $M$ of chord $AB$, where line $l$ passing through the focus $F$ of the parabola $x^{2}=4y$ intersects the parabola at points $A$ and $B$, is?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;Expression(G) = (x^2 = 4*y);Focus(G)=F;PointOnCurve(F,l);Intersection(l,G) = {A,B};IsChordOf(LineSegmentOf(A,B),G);MidPoint(LineSegmentOf(A,B))=M;M:Point;F:Point",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2=2*y-2",
    "fact_spans": "[[[22, 27]], [[1, 15], [28, 31]], [[32, 35]], [[38, 41]], [[1, 15]], [[1, 21]], [[0, 27]], [[22, 43]], [[28, 50]], [[46, 56]], [[53, 56]], [[18, 21]]]",
    "query_spans": "[[[53, 63]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the focus of the parabola $x=-\\frac{1}{4} y^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (x = -y^2/4)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(-1,0)",
    "fact_spans": "[[[0, 25]], [[0, 25]]]",
    "query_spans": "[[[0, 32]]]",
    "process": "The solution process is omitted"
  },
  {
    "text": "If a hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1(a>0)$ has an asymptote whose chord length intercepted by the circle $(x-2)^{2}+y^{2}=4$ is $2$, then the length of the real axis of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Circle;a>0;Expression(G) = (-y^2/3 + x^2/a^2 = 1);Expression(H) = (y^2 + (x - 2)^2 = 4);Length(InterceptChord(OneOf(Asymptote(G)),H))=2",
    "query_expressions": "Length(RealAxis(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 48], [88, 91]], [[4, 48]], [[55, 75]], [[4, 48]], [[1, 48]], [[55, 75]], [[1, 85]]]",
    "query_spans": "[[[88, 97]]]",
    "process": "One asymptote of $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1$ is $\\sqrt{3}x+ay=0$, and the chord length intercepted by the circle $(x-2)^{2}+y^{2}=4$ is 2, so the distance from the center of the circle to the line is $\\sqrt{3}$, that is, $\\frac{2\\sqrt{3}}{\\sqrt{3+a^{2}}}=\\sqrt{3}$, $a=1$. Therefore, the real axis length of the hyperbola is 2."
  },
  {
    "text": "The directrix of the parabola $y=ax^{2}$ is given by $y=-\\frac{1}{4}$. Then the value of the real number $a$ is?",
    "fact_expressions": "G: Parabola;a: Real;Expression(G) = (y = a*x^2);Expression(Directrix(G)) = (y = -1/4)",
    "query_expressions": "a",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 13]], [[37, 42]], [[0, 13]], [[0, 35]]]",
    "query_spans": "[[[37, 46]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has a focal distance of $2c$, $F$ is the right focus, $O$ is the coordinate origin, $P$ is a point on the hyperbola such that $|PO|=c$, and the area of $\\Delta POF$ is $\\frac{1}{2}ab$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;c: Number;FocalLength(G) = 2*c;F: Point;RightFocus(G) = F;O: Origin;P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, O)) = c;Area(TriangleOf(P, O, F)) = (1/2)*(a*b)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 59], [90, 93], [145, 148]], [[2, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[63, 68]], [[2, 68]], [[69, 72]], [[2, 76]], [[77, 80]], [[86, 89]], [[86, 96]], [[97, 106]], [[107, 142]]]",
    "query_spans": "[[[145, 154]]]",
    "process": "Let the left focus be $ F_{1} $, $ P(m,n) $. According to the given conditions, set up a system of equations, solve to obtain $ a=b $, then use the eccentricity formula to find the answer. Let the left focus be $ F_{1} $, $ P(m,n) $, then $ \\frac{m^{2}}{a^{2}}-\\frac{n^{2}}{b^{2}}=1 $\\textcircled{1}, since $ |PO|=c $, we have $ m^{2}+n^{2}=c^{2} $\\textcircled{2}. Solving \\textcircled{1} and \\textcircled{2} together yields $ n^{2}=\\frac{b^{4}}{c^{2}} $\\textcircled{3}. Since the area of $ \\triangle POF $ is $ \\frac{1}{2}ab $, we have $ \\frac{1}{2}cn=\\frac{1}{2}ab $, thus $ n=\\frac{ab}{c} $\\textcircled{4}. Solving \\textcircled{3} and \\textcircled{4} together and eliminating $ n $ gives $ \\frac{b^{4}}{c^{2}}=\\frac{a^{2}b^{2}}{c^{2}} $, simplifying yields $ a=b $, so the eccentricity $ e=\\frac{c}{a}=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{\\frac{2a^{2}}{a^{2}}}=\\sqrt{2} $"
  },
  {
    "text": "Given that the moving point $P$ lies on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, the point $M$ lies on the circle $C_{2}$: $(x-3)^{2}+y^{2}=1$, and the point $A(3 , 0)$ satisfies $P M \\perp A M$, then the minimum value of $|P M|$ is?",
    "fact_expressions": "G: Ellipse;C2: Circle;P: Point;M: Point;A: Point;Expression(G) = (x^2/25 + y^2/16 = 1);Expression(C2) = ((x-3)^2+y^2=1);Coordinate(A) = (3, 0);PointOnCurve(P, G);PointOnCurve(M, C2);IsPerpendicular(LineSegmentOf(P, M), LineSegmentOf(A, M))",
    "query_expressions": "Min(Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[8, 47]], [[54, 82]], [[4, 7]], [[49, 53]], [[84, 95]], [[8, 47]], [[54, 82]], [[84, 95]], [[4, 48]], [[49, 83]], [[97, 112]]]",
    "query_spans": "[[[114, 127]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of a hyperbola with foci on the $x$-axis are $y = \\pm 4x$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), xAxis);Expression(Asymptote(G)) = (y = pm*(4*x))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(17)",
    "fact_spans": "[[[11, 14], [35, 38]], [[2, 14]], [[11, 32]]]",
    "query_spans": "[[[35, 44]]]",
    "process": "Given the asymptote equations are y=\\pm4x, we obtain \\frac{b}{a}=4, e=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{17}."
  },
  {
    "text": "Let a point $P$ on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{4}=1$ $(a>0)$ with foci $F_{1}$, $F_{2}$ also lie on the parabola $y^{2}=x$, whose focus is $F_{3}$. If $|P F_{3}|=\\frac{9}{4}$, then the area of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "H: Ellipse;Expression(H) = (y^2/4 + x^2/a^2 = 1);F1: Point;F2: Point;Focus(H) = {F1, F2};a: Number;a>0;P: Point;PointOnCurve(P, H);G: Parabola;Expression(G) = (y^2 = x);PointOnCurve(P, G);F3: Point;Focus(G) = F3;Abs(LineSegmentOf(P, F3)) = 9/4",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[20, 66]], [[20, 66]], [[4, 11]], [[12, 19]], [[1, 66]], [[22, 66]], [[22, 66]], [[70, 73]], [[20, 73]], [[75, 87], [89, 92]], [[75, 87]], [[70, 88]], [[96, 103]], [[89, 103]], [[105, 128]]]",
    "query_spans": "[[[130, 157]]]",
    "process": "By symmetry, we may assume point P is above the x-axis. |PF_{3}| = x_{P} + \\frac{1}{4} = \\frac{9}{4}, so x_{P} = 2, hence P(2, \\sqrt{2}). Substituting into the ellipse equation yields a^{2} = 8, so c^{2} = a^{2} - b^{2} = 8 - 4 = 4, thus c = 2. Therefore, |F_{1}F_{2}| = 4, and S_{\\DeltaPF_{1}F_{2}} = \\frac{1}{2} \\times 4 \\times \\sqrt{2} = 2\\sqrt{2}."
  },
  {
    "text": "It is known that the line $l$ is perpendicular to the $x$-axis, and the segment length intercepted by the parabola $y^{2}=4 x$ on $l$ is $4 \\sqrt{3}$. Then the equation of the line $l$ is?",
    "fact_expressions": "l: Line;G: Parabola;Expression(G) = (y^2 = 4*x);IsPerpendicular(l,xAxis);Length(InterceptChord(l,G))=4*sqrt(3)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x=3",
    "fact_spans": "[[[2, 7], [16, 19], [55, 60]], [[20, 34]], [[20, 34]], [[2, 14]], [[16, 53]]]",
    "query_spans": "[[[55, 65]]]",
    "process": ""
  },
  {
    "text": "Given a parabola $ C $: $ y^{2} = 2 p x $ ($ p > 0 $) with a point $ P(2, m) $ ($ m > 0 $) on it. If the distance from $ P $ to the focus $ F $ is $ 4 $, then what is the standard equation of the circle centered at $ P $ that is tangent to the directrix of the parabola $ C $?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;P: Point;Coordinate(P) = (2, m);m: Number;m>0;PointOnCurve(P, C);F: Point;Focus(C)=F;Distance(P,F) = 4;G: Circle;Center(G) = P;IsTangent(G, Directrix(C))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "(x-2)^2+(y-4)^2=16",
    "fact_spans": "[[[2, 28], [74, 80]], [[2, 28]], [[10, 28]], [[10, 28]], [[31, 45], [47, 50], [66, 69]], [[31, 45]], [[31, 45]], [[31, 45]], [[2, 45]], [[53, 56]], [[2, 56]], [[47, 63]], [[86, 87]], [[65, 87]], [[73, 87]]]",
    "query_spans": "[[[86, 94]]]",
    "process": "By the definition of a parabola, the distance from point P to the focus of the parabola is equal to its distance to the directrix $ x = -\\frac{p}{2} $. Therefore, $ 2 + \\frac{p}{2} = 4 $, so $ p = 4 $, $ y^{2} = 8x $, $ m = 4 $. Hence, the center of the circle is $ P(2, 4) $, and the equation of the circle tangent to the parabola is $ (x - 2)^{2} + (y - 4)^{2} = 16 $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has eccentricity $\\frac{\\sqrt{2}}{2}$, and its right focus is $F(1,0)$. The three vertices of triangle $A B C$ lie on the ellipse. Let the midpoints of its three sides $A B$, $B C$, $A C$ be $D$, $E$, $F$, respectively, and the slopes of the lines containing the three sides be $k_{1}$, $k_{2}$, $k_{3}$ $(k_{1} k_{2} k_{3} \\neq 0)$. If the sum of the slopes of lines $O D$, $O E$, $O F$ is $-1$ ($O$ being the origin), then $\\frac{1}{k_{1}}+\\frac{1}{k_{2}}+\\frac{1}{k_{3}}=$?",
    "fact_expressions": "G: Ellipse;a: Number;b: Number;O: Origin;D: Point;A: Point;B: Point;C: Point;E: Point;F: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (1, 0);Eccentricity(G) = sqrt(2)/2;RightFocus(G) = F;PointOnCurve(Vertex(TriangleOf(A, B, C)), G);MidPoint(LineSegmentOf(A, B)) = D;MidPoint(LineSegmentOf(B, C)) = E;MidPoint(LineSegmentOf(A, C)) = F;k1: Number;k2: Number;k3: Number;Slope(OverlappingLine(LineSegmentOf(A, B))) = k1;Slope(OverlappingLine(LineSegmentOf(B, C))) = k2;Slope(OverlappingLine(LineSegmentOf(A, C))) = k3;Negation(k1*k2*k3 = 0);Slope(LineOf(O, D)) + Slope(LineOf(O, E)) + Slope(LineOf(O, F)) = -1",
    "query_expressions": "1/k2 + 1/k1 + 1/k3",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 54], [110, 112]], [[4, 54]], [[4, 54]], [[263, 266]], [[146, 149]], [[96, 103]], [[96, 103]], [[96, 103]], [[150, 153]], [[154, 157], [84, 92]], [[4, 54]], [[4, 54]], [[2, 54]], [[84, 92]], [[2, 79]], [[2, 92]], [[93, 113]], [[120, 157]], [[120, 157]], [[120, 157]], [[172, 179]], [[181, 188]], [[190, 197]], [[159, 197]], [[159, 197]], [[159, 197]], [[197, 225]], [[229, 262]]]",
    "query_spans": "[[[274, 325]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x$ ($p>0$) with focus $F$, a line $l$ passing through $F$ intersects the parabola $C$ at points $A$ and $B$, and $|A B|_{\\text{min}}=6$. Then the value of $p$ is?",
    "fact_expressions": "C: Parabola;p: Number;F: Point;l: Line;A: Point;B: Point;p > 0;Expression(C) = (y^2 = 2*p*x);Focus(C) = F;PointOnCurve(F,l);Intersection(l,C) = {A,B};Min(Abs(LineSegmentOf(A,B))) = 6",
    "query_expressions": "p",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 28], [47, 53]], [[91, 94]], [[32, 35], [37, 40]], [[41, 46]], [[54, 57]], [[58, 61]], [[10, 28]], [[2, 28]], [[2, 35]], [[36, 46]], [[41, 63]], [[65, 89]]]",
    "query_spans": "[[[91, 98]]]",
    "process": "Given $ F\\left(\\frac{P}{2},0\\right) $, let $ x = my + \\frac{p}{2} $, $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Then  \n$$\n\\begin{cases}\nx = my + \\frac{p}{2} \\\\\ny^{2} = 2px\n\\end{cases}\n\\Rightarrow y^{2} - 2pmy - p^{2} = 0.\n$$\nSince $ \\triangle > 0 $, we have $ y_{1} + y_{2} = 2pm $, $ y_{1}y_{2} = -p^{2} $.  \nTherefore,  \n$$\n|AB| = \\sqrt{1 + m^{2}} \\cdot |y_{1} - y_{2}| = \\sqrt{1 + m^{2}} \\cdot \\sqrt{(y_{1} + y_{2})^{2} - 4y_{1}y_{2}} = 2p(1 + m^{2}) \\geqslant 2p,\n$$\nwith equality if and only if $ m = 0 $.  \nHence, $ 2p = 6 \\Rightarrow p = 3 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Let $P$ be a point on the right branch of $C$, the line $P F_{1}$ is tangent to the circle $x^{2}+y^{2}=a^{2}$, and $|P F_{2}|=|F_{1} F_{2}|$. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = a^2);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, RightPart(C));IsTangent(LineOf(P, F1), G);Abs(LineSegmentOf(P, F2)) = Abs(LineSegmentOf(F1, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[2, 63], [92, 95], [165, 168]], [[10, 63]], [[10, 63]], [[114, 134]], [[88, 91]], [[72, 79]], [[80, 87]], [[10, 63]], [[10, 63]], [[2, 63]], [[114, 134]], [[2, 87]], [[2, 87]], [[88, 101]], [[102, 136]], [[138, 163]]]",
    "query_spans": "[[[165, 174]]]",
    "process": "Let PF₁ be tangent to the circle at point M. Since |PF₂| = |F₁F₂|, triangle PF₁F₂ is isosceles, and N is the midpoint of PF₁, so |F₁M| = ¹⁄₄|PF₁|. Also, in right triangle F₁MO, |F₁M|² = |F₁O|² − a² = c² − a², so |F₁M| = b = ¹⁄₄|PF₁| ①. Moreover, |PF₁| = |PF₂| + 2a = 2c + 2a ②, and c² = a² + b² ③. From ①②③ we obtain c² − a² = (⁽ᶜ⁺ᵃ⁾⁄₂)², which simplifies to 4(c − a) = c + a, so 3c = 5a, solving gives e = c⁄a = ⁵⁄₃."
  },
  {
    "text": "It is known that hyperbola $C$ shares common asymptotes with the curve $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ and passes through point $A(-3 , 4 \\sqrt{2})$. Then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Curve;A: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Coordinate(A) = (-3,4*sqrt(2));Asymptote(G)=Asymptote(C);PointOnCurve(A,C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/16 - x^2/9 = 1",
    "fact_spans": "[[[2, 8], [81, 84]], [[9, 47]], [[58, 79]], [[9, 47]], [[58, 79]], [[2, 54]], [[2, 79]]]",
    "query_spans": "[[[81, 89]]]",
    "process": ""
  },
  {
    "text": "The equation of the hyperbola passing through the point $(3, 2)$ and having the same asymptotes as the hyperbola $\\frac{y^{2}}{4}-\\frac{x^{2}}{3}=1$ is?",
    "fact_expressions": "H: Point;Coordinate(H) = (3, 2);G: Hyperbola;Expression(G) = (-x^2/3 + y^2/4 = 1);Z: Hyperbola;Asymptote(Z) = Asymptote(G);PointOnCurve(H, Z)",
    "query_expressions": "Expression(Z)",
    "answer_expressions": "x^2/8 - y^2/6 = 1",
    "fact_spans": "[[[2, 12]], [[2, 12]], [[14, 52]], [[14, 52]], [[59, 62]], [[13, 62]], [[0, 62]]]",
    "query_spans": "[[[59, 66]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$, passing through the point $(1,2)$, then its equation is?",
    "fact_expressions": "G: Parabola;p: Number;H: Point;p>0;Expression(G) = (y^2 = 2*p*x);Coordinate(H) = (1, 2);PointOnCurve(H,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[2, 23], [35, 36]], [[5, 23]], [[25, 33]], [[5, 23]], [[2, 23]], [[25, 33]], [[2, 33]]]",
    "query_spans": "[[[35, 41]]]",
    "process": "From the parabola, passing through the point, ∴ we have: p=2 ∴ y^{2}=4x"
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$ and the point $P(4,2)$, the line $l$ passes through point $P$ and intersects the ellipse at points $A$, $B$. When point $P$ is exactly the midpoint of segment $AB$, what is the slope of the line $l$ containing segment $AB$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/36 + y^2/9 = 1);P: Point;Coordinate(P) = (4, 2);l: Line;PointOnCurve(P, l);Intersection(l, G) = {A, B};A: Point;B: Point;MidPoint(LineSegmentOf(A, B)) = P;OverlappingLine(LineSegmentOf(A, B)) = l",
    "query_expressions": "Slope(l)",
    "answer_expressions": "-1/2",
    "fact_spans": "[[[0, 38], [62, 64]], [[0, 38]], [[39, 48], [56, 60], [78, 82]], [[39, 48]], [[49, 54], [106, 111]], [[49, 60]], [[49, 75]], [[66, 69]], [[70, 73]], [[78, 95]], [[97, 111]]]",
    "query_spans": "[[[106, 116]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then \n$$\n\\begin{cases}\n\\frac{x_{1}^{2}}{36} + \\frac{y_{1}}{9} = 1 \\\\\n\\frac{x_{2}^{2}}{36} + \\frac{y_{2}}{9} = 1\n\\end{cases}\n$$\nSubtracting the two equations gives \n$$\n\\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{36} + \\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{9} = 0.\n$$\nSince $ P(4,2) $ is the midpoint of segment $ AB $, we have $ \\frac{x_{1}+x_{2}}{2} = 4 $, $ \\frac{y_{1}+y_{2}}{2} = 2 $. Substituting into the above equation yields \n$$\n\\frac{8(x_{1}-x_{2})}{36} + \\frac{4(y_{1}-y_{2})}{9} = 0,\n$$\nwhich simplifies to \n$$\nk_{AB} = \\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = -\\frac{1}{2}.\n$$"
  },
  {
    "text": "The length of the real axis of the hyperbola $x^{2}-2 y^{2}=16$ is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - 2*y^2 = 16)",
    "query_expressions": "Length(RealAxis(G))",
    "answer_expressions": "8",
    "fact_spans": "[[[0, 21]], [[0, 21]]]",
    "query_spans": "[[[0, 28]]]",
    "process": "The hyperbola $x^{2}-2y^{2}=16$, converted into standard form, is $\\frac{x^{2}}{16}-\\frac{y^{2}}{8}=1$, thus the length of the real axis can be obtained. The hyperbola $x^{2}-2y^{2}=16$, converted into standard form, is $\\frac{x^{2}}{16}-\\frac{y^{2}}{8}=1$, $\\therefore a^{2}=16$, $\\therefore a=4$, $\\therefore 2a=8$, so the length of the real axis of the hyperbola $x^{2}-2y^{2}=16$ is $8$."
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{2}=1$ $(a>0)$ is $2$, then the value of $a$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/2 + x^2/a^2 = 1);a: Number;a>0;Eccentricity(G) = 2",
    "query_expressions": "a",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[2, 49]], [[2, 49]], [[59, 62]], [[5, 49]], [[2, 57]]]",
    "query_spans": "[[[59, 66]]]",
    "process": "By the given condition, for the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{2}=1$ $(a>0)$, we have $c=\\sqrt{a^{2}+2}$. Since the eccentricity of the hyperbola is $2$, we obtain $e=\\frac{c}{a}=\\frac{\\sqrt{a^{2}+2}}{a}=2^{n}$. Solving gives $a=\\frac{\\sqrt{6}}{3}$. Therefore, the value of the real number $a$ is $\\frac{\\sqrt{6}}{3}$."
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{4}=1$ are $F_{1}$ and $F_{2}$ respectively. A line $l$ passes through $F_{1}$ and intersects the ellipse at points $A$ and $B$. Then, the perimeter of $\\triangle A B F_{2}$ is?",
    "fact_expressions": "l: Line;G: Ellipse;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2/25 + y^2/4 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, l);Intersection(l, G) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "20",
    "fact_spans": "[[[63, 68]], [[0, 38], [78, 80]], [[81, 84]], [[85, 88]], [[55, 62]], [[47, 54], [70, 77]], [[0, 38]], [[0, 62]], [[0, 62]], [[63, 77]], [[63, 90]]]",
    "query_spans": "[[[92, 118]]]",
    "process": "Since the foci of the ellipse lie on the x-axis, a=5, b=2, ∴|AF₁|+|AF₂|=2a=10, |BF₁|+|BF₂|=2a=10, ∴ the perimeter of ΔABF₂ is |AB|+|AF₂|+|BF₂|=(|AF₁|+|AF₂|)+(|BF₁|+|BF₂|)=4a=20"
  },
  {
    "text": "Given the ellipse equation $\\frac{x^{2}}{4}+y^{2}=1$, a line $l$ passes through the point $(1,0)$ and intersects the ellipse at points $A$ and $B$, and $O$ is the origin. Then the maximum area of $\\triangle AOB$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);l: Line;H: Point;Coordinate(H) = (1, 0);PointOnCurve(H, l) = True;Intersection(l, G) = {A, B};A: Point;B: Point;O: Origin",
    "query_expressions": "Max(Area(TriangleOf(A, O, B)))",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[2, 4], [49, 51]], [[2, 32]], [[33, 38]], [[39, 47]], [[39, 47]], [[33, 47]], [[33, 62]], [[53, 56]], [[57, 60]], [[63, 66]]]",
    "query_spans": "[[[73, 99]]]",
    "process": "According to the problem, set the equation of line $ l $, combine it with the ellipse equation, transform it into a quadratic equation in terms of $ y $, use the relationship between roots and coefficients to find $ |y_{1}-y_{2}| $, substitute into the triangle area formula, and then find the value by substitution. Since line $ l $ passes through point $ E(1,0) $, we can set the equation of line $ l $ as $ x = my + 1 $. Then  \n$$\n\\begin{cases}\n\\frac{x^{2}}{4} + y^{2} = 1 \\\\\nx = my + 1,\n\\end{cases}\n$$  \neliminating $ x $ gives: $ (m^{2} + 4)y^{2} + 2my - 3 = 0 $. Since $ \\Delta = (2m)^{2} + 12(m^{2} + 4) > 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, define $ g(t) = t + \\frac{1}{t} $, where $ t = \\sqrt{m^{2} + 3} $, $ t \\geqslant \\sqrt{3} $. Then $ g(t) $ is increasing on the interval $ [\\sqrt{3}, +\\infty) $, so $ g(t) \\geqslant \\frac{4\\sqrt{3}}{3} $, with equality if and only if $ m = 0 $, i.e., $ (S_{\\Delta AOB})_{\\max} = \\frac{\\sqrt{3}}{2} $. Therefore, the maximum area of $ \\triangle AOB $ is $ \\frac{\\sqrt{3}}{2} $."
  },
  {
    "text": "The eccentricity $e$ of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{4}=1$ is?",
    "fact_expressions": "G: Hyperbola;e: Number;Expression(G) = (x^2/4 - y^2/4 = 1);Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 38]], [[42, 45]], [[0, 38]], [[0, 45]]]",
    "query_spans": "[[[42, 47]]]",
    "process": "From the given, we have $ a^{2} = b^{2} = 4 $, so $ a = 2 $, $ c = \\sqrt{a^{2} + b^{2}} = 2\\sqrt{2} $, therefore $ e = \\frac{c}{a} = \\sqrt{2} $, final answer is."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{a+7}+\\frac{y^{2}}{a^{2}+1}=1$ represents an ellipse with foci on the $x$-axis, then the range of real values for $a$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(a + 7) + y^2/(a^2 + 1) = 1);a: Real;PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Range(a)",
    "answer_expressions": "(-2,3)",
    "fact_spans": "[[[58, 60]], [[2, 60]], [[62, 67]], [[49, 60]]]",
    "query_spans": "[[[62, 74]]]",
    "process": "Since the foci of the ellipse lie on the x-axis, we have $ a+7 > a^{2}+1 $, from which the range of real number $ a $ can be determined. [Detailed solution] From the given condition: $ a+7 > a^{2}+1 $, rearranging yields $ a^{2}-a-6 < 0 $. Therefore, $ -2 < a < 3 $."
  },
  {
    "text": "Given an ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$ with eccentricity $\\frac{\\sqrt{6}}{3}$, the two foci are $F_{1}$ and $F_{2}$. Point $P$ lies on the ellipse such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$ and the area of $\\triangle P F_{1} F_{2}$ is $4$. Then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;a: Number;b: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2};Eccentricity(G) = sqrt(6)/3;DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Area(TriangleOf(P, F1, F2)) = 4",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/12 + y^2/4 = 1",
    "fact_spans": "[[[27, 81], [110, 112], [210, 212]], [[29, 81]], [[29, 81]], [[105, 109]], [[89, 96]], [[97, 104]], [[29, 81]], [[29, 81]], [[27, 81]], [[105, 113]], [[27, 104]], [[2, 81]], [[115, 174]], [[176, 208]]]",
    "query_spans": "[[[210, 217]]]",
    "process": "From the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with eccentricity $\\frac{\\sqrt{6}}{3}$, we get: $\\frac{c}{a}=\\frac{\\sqrt{6}}{3}$. Also, $a^{2}=b^{2}+c^{2}$, substituting into the above equation and simplifying gives: $a^{2}=3b^{2}$. From $\\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=0$, we obtain that $\\triangle PF_{1}F_{2}$ is a right triangle. Given the area of $\\triangle PF_{1}F_{2}$ is 4, let $|PF_{1}|=m$, $|PF_{2}|=n$, then \n$$\n\\begin{cases}\nm^{2}+n^{2}=(2c)^{2} \\\\\n\\frac{1}{2}mn=4\n\\end{cases}\n$$\nand $m+n=2a$. Solving yields: $a^{2}=12$, $b^{2}=4$. Therefore, the equation of the ellipse is: $\\frac{x^{2}}{12}+\\frac{y^{2}}{4}=1$."
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$ and the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ have common foci $F_{1}$, $F_{2}$, and $P$ is an intersection point of the two curves. Then the value of $\\cos \\angle F_{1} P F_{2}$ is?",
    "fact_expressions": "H: Ellipse;Expression(H) = (x^2/6 + y^2/2 = 1);G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1);F1: Point;F2: Point;Focus(H) = {F1, F2};Focus(G) = {F1, F2};P: Point;OneOf(Intersection(H, G)) = P",
    "query_expressions": "Cos(AngleOf(F1, P, F2))",
    "answer_expressions": "1/3",
    "fact_spans": "[[[0, 37]], [[0, 37]], [[38, 66]], [[38, 66]], [[72, 79]], [[80, 87]], [[0, 87]], [[0, 87]], [[88, 91]], [[88, 100]]]",
    "query_spans": "[[[103, 132]]]",
    "process": ""
  },
  {
    "text": "Given a point $P$ on the parabola $x^{2}=4 y$ such that the distance from $P$ to the focus $F$ is $5$, what is the horizontal coordinate of point $P$?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;Expression(G) = (x^2 = 4*y);PointOnCurve(P, G);Focus(G) = F;Distance(P, F) = 5",
    "query_expressions": "XCoordinate(P)",
    "answer_expressions": "pm*4",
    "fact_spans": "[[[2, 16]], [[19, 22], [37, 41]], [[25, 28]], [[2, 16]], [[2, 22]], [[2, 28]], [[19, 35]]]",
    "query_spans": "[[[37, 47]]]",
    "process": ""
  },
  {
    "text": "Given that the distance from a vertex of the hyperbola $\\frac{y^{2}}{9}-\\frac{x^{2}}{m^{2}}=1$ $(m>0)$ to one of its asymptotes is $1$, find $m=?$",
    "fact_expressions": "G: Hyperbola;Expression(G) = (y^2/9 - x^2/m^2 = 1);m: Number;m>0;Distance(OneOf(Vertex(G)), OneOf(Asymptote(G))) = 1",
    "query_expressions": "m",
    "answer_expressions": "3*sqrt(2)/4",
    "fact_spans": "[[[2, 49], [55, 56]], [[2, 49]], [[71, 74]], [[5, 49]], [[2, 69]]]",
    "query_spans": "[[[71, 76]]]",
    "process": ""
  },
  {
    "text": "The distance from the focus of the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$ to its asymptote is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/4 = 1)",
    "query_expressions": "Distance(Focus(G), Asymptote(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 41]]]",
    "process": ""
  },
  {
    "text": "The equations of the asymptotes of a hyperbola are $3 x \\pm 4 y=0$, then the eccentricity of the hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (pm*4*y + 3*x = 0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{5/3, 4/3}",
    "fact_spans": "[[[0, 3], [27, 30]], [[0, 25]]]",
    "query_spans": "[[[27, 37]]]",
    "process": ""
  },
  {
    "text": "The equation of the hyperbola passing through the points $(-\\sqrt{2}, \\sqrt{3})$, $(\\frac{\\sqrt{15}}{3}, \\sqrt{2})$ is?",
    "fact_expressions": "G: Hyperbola;H: Point;I: Point;Coordinate(H) = (-sqrt(2), sqrt(3));Coordinate(I) = (sqrt(15)/3, sqrt(2));PointOnCurve(H,G);PointOnCurve(I,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[61, 64]], [[2, 26]], [[27, 60]], [[2, 26]], [[27, 60]], [[0, 64]], [[0, 64]]]",
    "query_spans": "[[[61, 68]]]",
    "process": "Let the hyperbola equation be $mx^{2}+ny^{2}=1$. Since the hyperbola passes through the points $(-\\sqrt{2},\\sqrt{3})$ and $(\\frac{\\sqrt{15}}{3},\\sqrt{2})$, we have\n$$\n\\begin{cases}\n2m+3n=1 \\\\\n\\frac{5}{3}m+2n=1\n\\end{cases}\n\\Rightarrow\n\\begin{cases}\nm \\\\\nn\n\\end{cases}\n$$\nThen the hyperbola equation is $x^{2}-\\frac{y^{2}}{3}=1$. Hence, the answer is $x^{2}-\\frac{y^{2}}{3}=1$. This problem mainly examines the equation of a hyperbola, aiming to test the mastery of basic knowledge, and belongs to a basic-level question."
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $\\frac{x^{2}}{25}-\\frac{y^{2}}{16}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/25 - y^2/16 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(4/5)*x",
    "fact_spans": "[[[0, 40]], [[0, 40]]]",
    "query_spans": "[[[0, 48]]]",
    "process": ""
  },
  {
    "text": "Let point $P$ be an arbitrary point on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ other than the vertices, $F_{1}$, $F_{2}$ be the left and right foci respectively, $c$ be the semi-focal length, and the incircle of $\\triangle P F_{1} F_{2}$ touches the side $F_{1} F_{2}$ at point $M$. Find $|F_{1} M| \\cdot|F_{2} M|$=?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;F1: Point;F2: Point;P: Point;M: Point;c: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);Negation(P = Vertex(G));LeftFocus(G) = F1;RightFocus(G) = F2;HalfFocalLength(G) = c;TangentPoint(InscribedCircle(TriangleOf(P, F1, F2)), LineSegmentOf(F1, F2)) = M",
    "query_expressions": "Abs(LineSegmentOf(F1, M))*Abs(LineSegmentOf(F2, M))",
    "answer_expressions": "b^2",
    "fact_spans": "[[[6, 52]], [[9, 52]], [[9, 52]], [[63, 70]], [[71, 79]], [[1, 5]], [[142, 146]], [[88, 91]], [[6, 52]], [[1, 62]], [[1, 62]], [[6, 87]], [[6, 87]], [[6, 95]], [[96, 146]]]",
    "query_spans": "[[[148, 176]]]",
    "process": ""
  },
  {
    "text": "If the equation $\\frac{x^{2}}{1+m}-\\frac{y^{2}}{m-2}=1$ represents a hyperbola, then the range of values for $m$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) =  (x^2/(m + 1) - y^2/(m - 2) = 1);m: Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-\\infty,-1)+(2,+\\infty)",
    "fact_spans": "[[[44, 47]], [[1, 47]], [[49, 52]]]",
    "query_spans": "[[[49, 59]]]",
    "process": "The equation $\\frac{x^{2}}{1+m}-\\frac{y^{2}}{m-2}=1$ represents a hyperbola, from which we obtain $(1+m)(m-2)>0$, solving gives $m<-1$ or $m>2$."
  },
  {
    "text": "Let $F$ be the focus of the parabola $C$: $y^{2}=2 x$, and let $A$, $B$ be two distinct points on the parabola $C$. If the line $AB$ passes through the focus $F$, then the minimum value of $|AF|+4|BF|$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*x);F: Point;Focus(C) = F;B: Point;A: Point;PointOnCurve(A, C);PointOnCurve(B, C);Negation(A = B);PointOnCurve(F, LineOf(A, B))",
    "query_expressions": "Min(Abs(LineSegmentOf(A, F)) + 4*Abs(LineSegmentOf(B, F)))",
    "answer_expressions": "9/2",
    "fact_spans": "[[[5, 24], [36, 42]], [[5, 24]], [[1, 4], [64, 67]], [[1, 27]], [[32, 35]], [[28, 31]], [[28, 49]], [[28, 49]], [[28, 49]], [[51, 67]]]",
    "query_spans": "[[[69, 89]]]",
    "process": "If line AB coincides with the x-axis, then line AB intersects the parabola C at only one point, which does not satisfy the condition. It is easy to see that the focus of parabola C is $ F\\left(\\frac{1}{2},0\\right) $, and the equation of the directrix is $ x = -\\frac{1}{2} $. Let points $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, and let the equation of line AB be $ x = my + \\frac{1}{2} $. Combining\n$$\n\\begin{cases}\nx = my + \\frac{1}{2} \\\\\ny^{2} = 2x\n\\end{cases}\n$$\nand simplifying yields $ y^{2} - 2my - 1 = 0 $, $ \\Delta = 4m^{2} + 4 > 0 $. By Vieta's formulas, we have $ y_{1} + y_{2} = 2m $, $ y_{1}y_{2} = -1 $,\n$$\n\\frac{1}{|AF|} + \\frac{1}{|BF|} = \\frac{1}{my_{1} + 1} + \\frac{1}{my_{2} + 1} = \\frac{m(y_{1} + y_{2}) + 2}{(my_{1} + 1)(my_{2} + 1)} = \\frac{2m^{2} + 2}{-m^{2} + 2m^{2} + 1} = 2.\n$$\nThen,\n$$\n|AF| + 4|BF| = \\frac{1}{2}(|AF| + 4|BF|)\\left(\\frac{1}{|AF|} + \\frac{1}{|BF|}\\right) = \\frac{1}{2}\\left(5 + \\frac{|AF|}{|BF|} + \\frac{4|BF|}{|AF|}\\right) \\geqslant \\frac{1}{2}\\left(5 + 2\\sqrt{\\frac{|AF|}{|BF|} \\cdot \\frac{4|BF|}{|AF|}}\\right) = \\frac{9}{2}.\n$$\nThe equality holds if and only if $ |AF| = 2|BF| $. Therefore, the minimum value of $ |AF| + 4|BF| $ is $ \\frac{9}{2} $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ with foci $F_{1}$ and $F_{2}$. A line passing through the left focus $F_{1}$ intersects the left branch of the hyperbola at points $A$ and $B$. If $|A F_{2}|+|B F_{2}|=2|A B|$, then what is $|A B|$?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;A: Point;F2: Point;B: Point;F1: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Focus(G)={F1,F2};PointOnCurve(F1,H);Intersection(H,LeftPart(G)) = {A, B};Abs(LineSegmentOf(A, F2)) + Abs(LineSegmentOf(B, F2)) = 2*Abs(LineSegmentOf(A, B));LeftFocus(G)=F1",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "4*a",
    "fact_spans": "[[[2, 48], [86, 89]], [[5, 48]], [[5, 48]], [[83, 85]], [[92, 95]], [[62, 69]], [[96, 99]], [[52, 59], [75, 82]], [[2, 48]], [[2, 70]], [[71, 85]], [[83, 101]], [[103, 131]], [[2, 82]]]",
    "query_spans": "[[[133, 143]]]",
    "process": "According to the problem, draw the figure as follows: by the definition of hyperbola, we have: |AF_{2}|-|AF|=2a, |BF_{2}|-|BF_{1}|=2a. \\therefore |AF_{2}|+|BF_{2}|=4a+(|AF|+|BF_{1}|)=4a+|AB|. It is also given that |AF_{2}|+|BF_{2}|=2|AB|, \\therefore 2|AB|=4a+|AB|, thus |AB|=4a"
  },
  {
    "text": "A point $A(m, 2)$ on the parabola $y = a x^{2}$ ($a > 0$) is at a distance of $3$ from the focus. Then $a = $?",
    "fact_expressions": "G: Parabola;a: Number;A: Point;m:Number;a>0;Expression(G) = (y = a*x^2);Coordinate(A) = (m, 2);PointOnCurve(A, G);Distance(A, Focus(G)) = 3",
    "query_expressions": "a",
    "answer_expressions": "1/4",
    "fact_spans": "[[[0, 19]], [[43, 46]], [[22, 31]], [[22, 31]], [[3, 19]], [[0, 19]], [[22, 31]], [[0, 31]], [[0, 41]]]",
    "query_spans": "[[[43, 48]]]",
    "process": "First, find the equation of the directrix of the parabola. According to the definition of a parabola, the distance to the focus equals the distance to the directrix, leading to an equation which, when solved, gives the result; [Detailed solution] The parabola $ y = ax^{2} $, or $ x^{2} = \\frac{1}{a}y $, has the directrix equation $ y = -\\frac{1}{4a} $. The distance from point $ A(m, 2) $ to the focus is 3, so $ 2 - (-\\frac{1}{4a}) = 3 $, solving yields $ a = \\frac{1}{4} $."
  },
  {
    "text": "Let the parabola $C$: $y^{2}=4x$ have focus $F$, and let line $l$ passing through $F$ intersect the parabola at points $A$ and $B$. Let $M$ be the intersection point of the directrix of parabola $C$ and the $x$-axis. If $|AB|=8$, then $\\tan \\angle AMB = $?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, C) = {A, B};M: Point;Intersection(Directrix(C), xAxis) = M;Abs(LineSegmentOf(A, B)) = 8",
    "query_expressions": "Tan(AngleOf(A, M, B))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[1, 20], [58, 64], [39, 42]], [[1, 20]], [[24, 27], [29, 32]], [[1, 27]], [[33, 38]], [[28, 38]], [[44, 47]], [[48, 51]], [[33, 53]], [[54, 57]], [[54, 75]], [[77, 86]]]",
    "query_spans": "[[[88, 109]]]",
    "process": "Let x = 1 + ty, substitute into y^{2} = 4x to obtain y^{2} - 4ty - 4 = 0, then y_{1} + y_{2} = 4t, y_{1}y_{2} = -4, so x_{1} + x_{2} = 2 + t(y_{1} + y_{2}) = 2 + 4t^{2}. Since x_{1} + x_{2} + 2 = 8, that is, x_{1} + x_{2} = 6, which means 2 + 4t^{2} = 6, so t = \\pm1. Without loss of generality, let t = 1, then y^{2} - 4y - 4 = 0, solving gives y = 2 \\pm 2\\sqrt{2}. When y = 2 + 2\\sqrt{2}, x = 3 + 2\\sqrt{2}; then k_{1} = \\frac{2 + 2\\sqrt{2}}{3 + 2\\sqrt{2} + 1} = \\frac{1}{\\sqrt{2}}, when y = 2 - 2\\sqrt{2}, x = 3 - 2\\sqrt{2}, then k_{2} = \\frac{2 - 2\\sqrt{2}}{3 - 2\\sqrt{2} + 1} = -\\frac{1}{\\sqrt{2}}, so \\tan\\angle AMB = \\frac{\\frac{2}{\\sqrt{2}}}{1 - \\frac{1}{2}} = 2\\sqrt{2}."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $x^{2}=8 y$, $P$ is a point on $C$, and $M(-4,3)$, find the coordinates of point $P$ when the perimeter of $\\triangle P M F$ is minimized.",
    "fact_expressions": "C: Parabola;M: Point;P: Point;F: Point;Expression(C) = (x^2 = 8*y);Coordinate(M) = (-4, 3);Focus(C) = F;PointOnCurve(P, C);WhenMin(Perimeter(TriangleOf(P, M, F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(-4, 2)",
    "fact_spans": "[[[6, 24], [32, 35]], [[39, 48]], [[28, 31], [74, 78]], [[2, 5]], [[6, 24]], [[39, 48]], [[2, 27]], [[28, 38]], [[50, 73]]]",
    "query_spans": "[[[74, 82]]]",
    "process": "Draw a perpendicular line from P to the directrix, with foot S, connect MS, and draw a perpendicular line from M to the directrix, with foot T. Using the geometric properties of the parabola, we obtain |MP| + |PF| = |MP| + |PS|, thus the minimum value of |MP| + |PF| is 3, and the coordinates of P when the minimum is attained are obtained. The directrix of the parabola is y = -2. Draw a perpendicular line from P to the directrix, with foot S, and connect MS. Draw a perpendicular line from M to the directrix, with foot T. By the definition of the parabola, |PF| = |PS|. Also, |MP| + |PF| = |MP| + |PS| \\geqslant |MS| \\geqslant |MT| = 3, at which point M, P, T are collinear, so P(-4,2). Therefore, when the perimeter of \\triangle PMF is minimized, the coordinates of point P are (-4,2)."
  },
  {
    "text": "Given that point $F$ is the focus of the parabola $C$: $y^{2}=2 p x(p>0)$, the line $l$ passes through point $F$ and intersects the parabola $C$ at points $A$, $B$, and intersects the $y$-axis at point $M$. If $|A F|=|B M|$, $|A B|=2+\\sqrt{5}$, then $p=$?",
    "fact_expressions": "l: Line;C: Parabola;p: Number;A: Point;F: Point;B: Point;M: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(F,l);Intersection(l, C) = {A,B};Intersection(l, yAxis) = M;Abs(LineSegmentOf(A, F)) = Abs(LineSegmentOf(B, M));Abs(LineSegmentOf(A, B)) = 2 + sqrt(5)",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[36, 41]], [[7, 32], [49, 55]], [[112, 115]], [[56, 59]], [[2, 6], [43, 47]], [[60, 63]], [[72, 76]], [[14, 32]], [[7, 32]], [[2, 35]], [[36, 47]], [[36, 65]], [[36, 76]], [[78, 91]], [[92, 110]]]",
    "query_spans": "[[[112, 117]]]",
    "process": "From the given conditions, point B lies between A and M. Let the inclination angle of line / be \\theta, as shown in the figure, \\angle AFx = \\theta. Since |AF| = |BM|, then |MF| = |AB| = 2 + \\sqrt{5}. And |MF| = \\frac{|OF|}{\\cos\\angle OFM}, |AB| = \\frac{2p}{\\sin^{2}\\theta}, so \\frac{p}{2\\cos\\theta} = \\frac{2p}{\\sin^{2}\\theta} = 2 + \\sqrt{5}. From \\frac{2\\cos\\theta}{2\\cos\\theta} = \\frac{\\sin^{2}\\theta}{\\sin^{2}\\theta}, we obtain \\cos^{2}\\theta + 4\\cos\\theta - 1 = 0. Solving gives \\cos\\theta = \\sqrt{5} - 2 or \\cos\\theta = -\\sqrt{5} - 2 (discarded). Hence, p = 2\\cos\\theta \\cdot (2 + \\sqrt{5}) = 2"
  },
  {
    "text": "Given the parabola equation $y=-\\frac{1}{4} x^{2}$, then its focus coordinates are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = -x^2/4)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,-1)",
    "fact_spans": "[[[2, 5], [32, 33]], [[2, 30]]]",
    "query_spans": "[[[32, 39]]]",
    "process": "Since the parabola equation is $ y = -\\frac{1}{4}x^{2} $, which is equivalent to $ x^{2} = -4y $, it follows that $ 2p = -4 $, $ \\frac{p}{2} = -1 $, so the focus coordinates of the parabola are $ (0, -1) $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, a chord $AB$ of the ellipse passes through $F_{2}$. If the perimeter of $\\triangle A F_{1} B$ is $16$, and the eccentricity of the ellipse is $e=\\frac{\\sqrt{3}}{2}$, then what is the equation of the ellipse?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;B: Point;F1: Point;F2: Point;e:Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Focus(G) = {F1, F2};PointOnCurve(F2, LineSegmentOf(A, B));IsChordOf(LineSegmentOf(A, B), G);Perimeter(TriangleOf(A, F1, B)) = 16;Eccentricity(G) = e;e=sqrt(3)/2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16+y^2/4=1",
    "fact_spans": "[[[18, 70], [85, 87], [127, 129], [158, 160]], [[20, 70]], [[20, 70]], [[89, 94]], [[89, 94]], [[2, 9]], [[10, 17], [77, 84]], [[134, 156]], [[20, 70]], [[20, 70]], [[18, 70]], [[2, 75]], [[76, 94]], [[85, 94]], [[96, 125]], [[127, 156]], [[134, 156]]]",
    "query_spans": "[[[158, 165]]]",
    "process": ""
  },
  {
    "text": "Given that the line $l$ with slope $k$ ($k>0$) passes through the focus of the parabola $x^{2}=4y$ and intersects the parabola at points $A(x_{1}, y_{2})$, $B(x_{2}, y_{2})$ ($x_{1}<x_{2}$), and $|AB|=8$. Find the equation of line $l$?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;k: Number;x1:Number;x2:Number;y2:Number;Expression(G) = (x^2 = 4*y);Coordinate(A) = (x1,y1);Coordinate(B)=(x2,y2);PointOnCurve(Focus(G), l);Slope(l)=k;Intersection(l, G)={A,B};Abs(LineSegmentOf(A, B)) = 8;k>0;x1<x2;y1:Number",
    "query_expressions": "Expression(l)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[33, 38], [117, 122]], [[3, 17], [39, 42]], [[43, 61]], [[64, 82]], [[24, 32]], [[43, 61]], [[64, 82]], [[43, 61]], [[3, 17]], [[43, 61]], [[64, 82]], [[2, 38]], [[21, 38]], [[33, 82]], [[103, 114]], [[24, 32]], [[84, 99]], [[43, 61]]]",
    "query_spans": "[[[117, 126]]]",
    "process": ""
  },
  {
    "text": "In the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, what is the equation of the line containing the chord for which point $M(1, \\frac{1}{2})$ is the midpoint?",
    "fact_expressions": "G: Ellipse;H: LineSegment;M: Point;Expression(G) = (x^2/4 + y^2 = 1);Coordinate(M) = (1, 1/2);IsChordOf(H, G);MidPoint(H) = M",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "x + 2*y - 2 = 0",
    "fact_spans": "[[[0, 27]], [], [[30, 50]], [[0, 27]], [[30, 50]], [[0, 55]], [[0, 55]]]",
    "query_spans": "[[[0, 63]]]",
    "process": "Given: $\\frac{x^{2}}{4}+y^{2}=1$, let the coordinates of the two intersection points of the chord passing through the midpoint with the ellipse be $(x_{1},y_{1})$, $(x_{2},y_{2})$. Substituting into the ellipse equation gives: $\\frac{x_{1}^{2}}{4}+y_{1}^{2}=1$, $\\frac{x_{2}^{2}}{4}+y_{2}^{2}=1$. Subtracting these two equations yields: $\\frac{1}{4}(x_{1}+x_{2})(x_{1}-x_{2})=(y_{1}+y_{2})(y_{2}-y_{1})$, $k=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{1}{4}\\left(\\frac{x_{1}+x_{2}}{y_{1}+y_{2}}\\right)$. Also, from the midpoint coordinate formula: $x_{1}+x_{2}=2$, $y_{1}+y_{2}=1$, then: $k=-\\frac{1}{2}$. Hence, the equation of the line of the chord with this midpoint is: $y-\\frac{1}{2}=-\\frac{1}{2}(x-1)$, $x+2y-2=0$."
  },
  {
    "text": "Given the parabola $y^{2}=2 x$, where $A$ and $B$ are two points on the parabola, and the perpendicular bisector of segment $AB$ intersects the $x$-axis at point $P(x_{0},0)$, then what is the range of values for $x_{0}$? (Expressed in interval notation)",
    "fact_expressions": "G: Parabola;B: Point;A: Point;P: Point;Expression(G) = (y^2 = 2*x);x0: Number;Coordinate(P) = (x0, 0);PointOnCurve(A, G);PointOnCurve(B, G);Intersection(PerpendicularBisector(LineSegmentOf(A,B)), xAxis) = P",
    "query_expressions": "Range(x0)",
    "answer_expressions": "(1,+\\infty)",
    "fact_spans": "[[[2, 16], [26, 29]], [[22, 25]], [[18, 21]], [[55, 68]], [[2, 16]], [[70, 77]], [[55, 68]], [[17, 33]], [[17, 33]], [[34, 68]]]",
    "query_spans": "[[[70, 84]]]",
    "process": "Let the coordinates of points A and B be $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$, respectively. Since the perpendicular bisector of segment AB intersects the x-axis at point $P(x_{0}, 0)$, AB is not parallel to the y-axis, i.e., $x_{1} \\ne x_{2}$. Also, $|PA| = |PB|$, so $(x_{1}-x_{0})^{2} + y_{1}^{2} = (x_{2}-x_{0})^{2} + y_{2}^{2}$, which gives $(x_{1}-x_{2})(x_{1}+x_{2}-2x_{0}) = y_{2}^{2}-y_{1}^{2}$. Since A and B are two points on the parabola, $y_{1}^{2} = 2x_{1}$, $y_{2}^{2} = 2x_{2}$. Substituting into the above equation yields $x_{0} = 1 + \\frac{x_{1}+x_{2}}{2}$. Since $x_{1} \\geqslant 0$, $x_{2} \\geqslant 0$, $x_{1} \\ne x_{2}$, we have $x_{1}+x_{2} > 0$, thus $x_{0} > 1$."
  },
  {
    "text": "Given that the ellipse $C_{1}$: $\\frac{x^{2}}{m+2}-\\frac{y^{2}}{n}=1$ and the hyperbola $C_{2}$: $\\frac{x^{2}}{m}+\\frac{y^{2}}{n}=1$ have the same foci, what is the range of the eccentricity $e_{1}$ of the ellipse $C_{1}$?",
    "fact_expressions": "C1: Ellipse;Eccentricity(C1) = (x^2/(m + 2) - y^2/n = 1);m: Number;n: Number;C2: Hyperbola;Expression(C2) = (y^2/n + x^2/m = 1);Focus(C1) = Focus(C2);e1: Number;Eccentricity(C1) = e1",
    "query_expressions": "Range(e1)",
    "answer_expressions": "(\\sqrt{2}/2, 1)",
    "fact_spans": "[[[2, 50], [106, 115]], [[2, 50]], [[13, 50]], [[13, 50]], [[51, 98]], [[51, 98]], [[2, 104]], [[119, 126]], [[106, 126]]]",
    "query_spans": "[[[119, 133]]]",
    "process": "First, from the condition that ellipse $C_{1}$ and hyperbola $C_{2}$ have the same foci, we obtain $n = -1$, thereby obtaining ellipse $C_{1}: \\frac{x^{2}}{m+2} + y^{2} = 1$, $e^{2} = 1 - \\frac{1}{m+2}$, then find its range. Ellipse $C_{1}: \\frac{x^{2}}{m+2} + \\frac{y^{2}}{-n} = 1$, hyperbola $C_{2}: \\frac{x^{2}}{m} - \\frac{y^{2}}{-n} = 1$. Since ellipse $C_{1}$ and hyperbola $C_{2}$ have the same foci, $m+2 - (-n) = m - n$, solving gives $n = -1$. So ellipse $C_{1}: \\frac{x^{2}}{m+2} + y^{2} = 1$, $e^{2} = \\frac{m+}{m}$. Because $m > 0$, $m+2 > 2$, so $0 < \\frac{1}{m+2}$. Therefore, $\\frac{1}{2} < 1 - \\frac{1}{m+2} < 1$, i.e., $\\frac{1}{2} < e^{2} < 1$, $\\frac{\\sqrt{2}}{2} < e < 1$. Hence the answer is $(\\frac{\\sqrt{2}}{2}, 1)$"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$, and two arbitrary points $P_{1}(x_{1}, y_{1})$, $P_{2}(x_{2}, y_{2})$ on the right branch of the hyperbola satisfying $x_{1} x_{2}-y_{1} y_{2}>0$ always hold, then the range of values for $a$ is?",
    "fact_expressions": "G: Hyperbola;a: Number;P1: Point;P2: Point;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Coordinate(P1) = (x1, y1);Coordinate(P2) = (x2, y2);PointOnCurve(P1,RightPart(G));PointOnCurve(P2,RightPart(G));x1:Number;y1:Number;x2:Number;y2:Number;x1*x2-y1*y2>0",
    "query_expressions": "Range(a)",
    "answer_expressions": "[1,+\\infty)",
    "fact_spans": "[[[2, 39], [40, 43]], [[131, 134]], [[52, 74]], [[75, 96]], [[5, 39]], [[2, 39]], [[52, 74]], [[75, 96]], [[40, 96]], [[40, 96]], [[99, 126]], [[99, 126]], [[99, 126]], [[99, 126]], [[99, 126]]]",
    "query_spans": "[[[131, 141]]]",
    "process": "Let point $P_{3}(x_{2},-y_{2})$, then $\\overrightarrow{OP_{1}}\\cdot\\overrightarrow{OP_{3}}=x_{1}x_{2}-y_{1}y_{2}>0$, so $\\overrightarrow{OP_{1}}=\\overrightarrow{OP_{3}}$ or $\\angle P_{1}OP_{3}$ is an acute angle. Let point $M$ be a point on the hyperbola's asymptote $y=\\frac{1}{a}x$ in the first quadrant, and let point $N$ be a point on the hyperbola's asymptote $y=-\\frac{1}{a}x$ in the fourth quadrant. From the given condition, $0<\\angle MON\\leqslant\\frac{\\pi}{2}$, then $0<\\frac{1}{a}\\leqslant\\tan\\frac{\\pi}{4}=1$, solving yields $a\\geqslant1$."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) with an inclination angle of $60^{\\circ}$ intersects the parabola at points $A$ and $B$. If $|A F|>|B F|$ and $|A F|=2$, then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;A: Point;F: Point;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;Inclination(H)=ApplyUnit(60, degree) ;Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, F)) > Abs(LineSegmentOf(B, F));Abs(LineSegmentOf(A, F)) = 2;PointOnCurve(F,H)",
    "query_expressions": "p",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 22], [50, 53]], [[91, 94]], [[47, 49]], [[54, 57]], [[25, 28]], [[58, 61]], [[4, 22]], [[1, 22]], [[1, 28]], [[30, 49]], [[47, 63]], [[65, 78]], [[80, 89]], [[0, 49]]]",
    "query_spans": "[[[91, 96]]]",
    "process": "Using the properties of a parabola, we obtain |AF| = 3|BF|, then get |AB| = |AF| + |BF| = \\frac{8}{3}. Finally, combining the equation and |AB| = x_{1} + x_{2} + p = \\frac{8}{3}, using Vieta's formulas to eliminate parameters, and then solve for p. Draw perpendiculars from points A, B to the directrix of the parabola x = -\\frac{p}{2}, with feet C, D respectively. Draw a perpendicular from point B to AC, with foot E. \\because A, B lie on the parabola, \\therefore |AC| = |AF|, |BD| = |BF|. \\because BE \\bot AC, \\therefore |AE| = |AF| - |BF|. \\because the inclination angle of line AB is 60^{\\circ}, \\therefore in Rt\\triangle ABE, 2|AE| = |AB| = |AF| + |BF|, i.e., 2(|AF| - |BF|) = |AF| + |BF|, \\therefore |AF| = 3|BF|. \\because |AF| = 2, \\therefore |BF| = \\frac{2}{3}, \\therefore |AB| = |AF| + |BF| = \\frac{8}{3}. Let the equation of line AB be y = \\sqrt{3}(x - \\frac{p}{2}), substitute into y^{2} = 2px, we get 3x^{2} - 5px + \\frac{3p^{2}}{4} = 0. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), \\therefore x_{1} + x_{2} = \\frac{5p}{3}. \\because |AB| = x_{1} + x_{2} + p = \\frac{8}{3}, \\therefore p = 1"
  },
  {
    "text": "Given point $P$ is an arbitrary point on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse respectively, $I$ is the incenter of $\\triangle P F_{1} F_{2}$, if $S_{\\Delta P I F_{1}}+S_{\\Delta P I F_{2}}=\\lambda S_{\\Delta F_{1} I F_{2}}$ holds, then what is the value of $\\lambda$?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;I: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;Incenter(TriangleOf(P, F1, F2)) = I;Area(TriangleOf(P,I,F1))+Area(TriangleOf(P,I,F2))=lambda*Area(TriangleOf(F1,I,F2));lambda: Number",
    "query_expressions": "lambda",
    "answer_expressions": "a/sqrt(a^2-b^2)",
    "fact_spans": "[[[7, 59], [83, 85]], [[9, 59]], [[9, 59]], [[2, 6]], [[65, 72]], [[73, 80]], [[92, 95]], [[9, 59]], [[9, 59]], [[7, 59]], [[2, 64]], [[65, 91]], [[65, 91]], [[92, 124]], [[126, 202]], [[206, 215]]]",
    "query_spans": "[[[206, 219]]]",
    "process": "Let the inradius of $\\triangle PF_{1}F_{2}$ be $r$, $I$ be the incenter of $\\triangle PF_{1}F_{2}$, and $S_{\\triangle PIF_{1}} + S_{\\triangle PIF_{2}} = \\lambda S_{\\triangle F_{1}IF_{2}}$. Therefore, $\\frac{1}{2}r|PF_{1}| + \\frac{1}{2}r|PF_{2}| = \\frac{1}{2}\\lambda|F_{1}F_{2}|$, so $|PF_{1}| + |PF_{2}| = \\lambda|F_{1}F_{2}|$. Since $P$ is an arbitrary point on the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1$ $(a > b > 0)$, and $F_{1}, F_{2}$ are the left and right foci of the ellipse respectively, by the definition of an ellipse we have $|PF_{1}| + |PF_{2}| = 2a$, thus $2a = \\lambda \\times 2\\sqrt{a^{2} - b^{2}}$, therefore $\\lambda = \\frac{a}{\\sqrt{a^{2} - b^{2}}}$."
  },
  {
    "text": "Let there be two points $A$, $B$ on the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$, with the midpoint $M(1,2)$ of $AB$. Then the equation of line $AB$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/2 = 1);A: Point;B: Point;PointOnCurve(A, G) = True;PointOnCurve(B, G) = True;M: Point;Coordinate(M) = (1, 2);MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Expression(LineOf(A, B))",
    "answer_expressions": "y=x+1",
    "fact_spans": "[[[1, 29]], [[1, 29]], [[33, 36]], [[37, 41]], [[1, 41]], [[1, 41]], [[51, 59]], [[51, 59]], [[44, 59]]]",
    "query_spans": "[[[61, 73]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ x_{1}+x_{2}=2 $, $ y_{1}+y_{2}=4 $, then \n\\[\n\\begin{cases}\nx_{1}^{2}-\\frac{y_{1}^{2}}{2}=1 \\\\\nx_{2}^{2}-\\frac{y_{2}^{2}}{2}=1\n\\end{cases}\n\\]\nSubtracting the two equations gives $ (x_{1}-x_{2})(x_{1}+x_{2}) = \\frac{1}{2}(y_{1}-y_{2})(y_{1}+y_{2}) $, so $ k_{AB} = \\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = 2 \\times \\frac{x_{1}+x_{2}}{y_{1}+y_{2}} = 2 \\times \\frac{2}{4} = 2 \\times \\frac{1}{2} = 1 $. Therefore, the equation of line AB is $ y-2 = x-1 $, that is, $ y = x+1 $. Substituting into $ x^{2}-\\frac{y^{2}}{2}=1 $ satisfies $ 4>0 $, so the equation of line AB is $ y = x+1 $."
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 px$ has coordinates $(1,0)$, then what is $p$? What is the equation of the directrix?",
    "fact_expressions": "G: Parabola;p: Number;Expression(G) = (y^2 = 2*(p*x));Coordinate(Focus(G)) = (1, 0)",
    "query_expressions": "p;Expression(Directrix(G))",
    "answer_expressions": "2\nx=-1",
    "fact_spans": "[[[1, 16]], [[32, 35]], [[1, 16]], [[1, 29]]]",
    "query_spans": "[[[32, 37]], [[1, 43]]]",
    "process": ""
  },
  {
    "text": "Given two points $A$ and $B$ on the parabola $C$: $y^{2}=4x$ with focus $F$ such that $\\overrightarrow{A F}=3\\overrightarrow{F B}$, then $|A B|$=?",
    "fact_expressions": "C: Parabola;A: Point;F: Point;B: Point;Expression(C) = (y^2 = 4*x);Focus(C)=F;PointOnCurve(A,C);PointOnCurve(B,C);VectorOf(A,F)=3*VectorOf(F,B)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[10, 29]], [[33, 36]], [[3, 6]], [[37, 40]], [[10, 29]], [[2, 29]], [[10, 40]], [[10, 40]], [[42, 87]]]",
    "query_spans": "[[[89, 98]]]",
    "process": ""
  },
  {
    "text": "Given the line $l$: $4x - 3y + 8 = 0$, if $P$ is a moving point on the parabola $y^2 = 4x$, then the minimum value of the sum of the distance from point $P$ to the line $l$ and its distance to the $y$-axis is?",
    "fact_expressions": "l: Line;Expression(l) = (4*x - 3*y + 8 = 0);G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, l) + Distance(P, yAxis))",
    "answer_expressions": "7/5",
    "fact_spans": "[[[2, 22], [53, 58]], [[2, 22]], [[28, 42]], [[28, 42]], [[24, 27], [48, 52], [59, 60]], [[24, 46]]]",
    "query_spans": "[[[48, 76]]]",
    "process": "First, using the definition of a parabola, transform the distance from a point on the parabola to the y-axis into the distance from that point to the focus of the parabola minus 1. Thus, the problem is converted into finding the distance from the focus of the parabola to the line $4x - 3y + 8 = 0$ minus 1, leading to the result $|PA| + |PB| = (|PH| - 1) + |PB| = (|PF| + |PB|) - 1 \\geqslant d_{F \\rightarrow 1} - 1 = \\frac{12}{5} - 1 = \\frac{7}{5}$. Hence, the answer is: $\\frac{7}{5}$"
  },
  {
    "text": "If points ${O}$ and ${F}$ are the center and left focus of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, respectively, and point $P$ is any point on the right branch of the hyperbola, then the range of values of $\\overrightarrow{O P} \\cdot \\overrightarrow{F P}$ is?",
    "fact_expressions": "G: Hyperbola;O: Point;P: Point;F: Point;Expression(G) = (x^2/3 - y^2 = 1);PointOnCurve(P, RightPart(G));Center(G)=O;LeftFocus(G)=F",
    "query_expressions": "Range(DotProduct(VectorOf(O, P), VectorOf(F, P)))",
    "answer_expressions": "[3+2*\\sqrt{3},+\\infty)",
    "fact_spans": "[[[17, 45], [58, 61]], [[1, 7]], [[53, 57]], [[8, 14]], [[17, 45]], [[53, 69]], [[1, 52]], [[1, 52]]]",
    "query_spans": "[[[71, 127]]]",
    "process": "Let point $ P $ be given, substitute into the hyperbola equation to obtain the expression for $ y_{0} $. Express $ \\overrightarrow{OP} $ and $ \\overrightarrow{FP} $ using the coordinates of $ P $, $ F $, and $ O $, then find the expression for $ \\overrightarrow{OP} \\cdot \\overrightarrow{FP} $. Use the properties of quadratic functions to find its minimum value. Thus, the range of $ \\overrightarrow{OP} \\cdot \\overrightarrow{FP} $ can be obtained. The hyperbola equation is $ \\frac{x^{2}}{3} - y^{2} = 1 $. Since $ F(-2,0) $ is the left focus of the given hyperbola, let point $ P(x_{0}, y_{0}) $. Then $ \\frac{x_{0}^{2}}{3} - y_{0}^{2} = 1 $ ($ x_{0} \\geqslant \\sqrt{3} $), solving gives $ y_{0}^{2} = \\frac{x_{0}^{2}}{3} - 1 $ ($ x_{0} \\geqslant \\sqrt{3} $). Since $ \\overrightarrow{FP} = (x_{0} + 2, y_{0}) $, $ \\overrightarrow{OP} = (x_{0}, y_{0}) $, so $ \\overrightarrow{OP} \\cdot \\overrightarrow{FP} = x_{0}(x_{0} + 2) + y_{0}^{2} = x_{0}^{2}(x_{0} + 2) + \\frac{x_{0}^{2}}{3} - 1 = \\frac{4x_{0}^{2}}{3} + 2x_{0} - 1 $. The axis of symmetry of the parabola corresponding to this quadratic function is $ x_{0} = -\\frac{3}{4} $. Since $ x_{0} \\geqslant \\sqrt{3} $, when $ x_{0} = \\sqrt{3} $, $ \\overrightarrow{OP} \\cdot \\overrightarrow{FP} $ reaches its minimum value $ \\frac{4}{3} \\times 3 + 2\\sqrt{3} - 1 = 3 + 2\\sqrt{3} $. Hence, the range of $ \\overrightarrow{OP} \\cdot \\overrightarrow{FP} $ is $ [3 + 2\\sqrt{3}, +\\infty) $."
  },
  {
    "text": "If the center of circle $C$: $x^{2}+(y+1)^{2}=n$ is a focus of the ellipse $M$: $x^{2}+m y^{2}=1$, and circle $C$ passes through the other focus of $M$, then $\\frac{n}{m}=$?",
    "fact_expressions": "M: Ellipse;m: Number;C: Circle;n: Number;Expression(M) = (m*y^2 + x^2 = 1);Expression(C) = (x^2 + (y + 1)^2 = n);F1: Point;F2: Point;OneOf(Focus(C))=F1;OneOf(Focus(C))=F2;Negation(F1=F2);Center(C) = F1;PointOnCurve(F2, C)",
    "query_expressions": "n/m",
    "answer_expressions": "8",
    "fact_spans": "[[[30, 54], [67, 70]], [[78, 91]], [[0, 26], [61, 65]], [[78, 91]], [[30, 54]], [[0, 26]], [], [], [[30, 59]], [[67, 76]], [[30, 76]], [[1, 59]], [[61, 76]]]",
    "query_spans": "[[[78, 93]]]",
    "process": "\\because\\frac{1}{m}-1=1\\thereforem=\\frac{1}{2}\\because0+(1+1)^{2}=n\\thereforen=4,\\therefore\\frac{n}{m}=8"
  },
  {
    "text": "What is the equation of the directrix of the parabola $x=-2 y^{2}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x = -2*y^2)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=1/8",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 22]]]",
    "process": "From $ x = -2y^{2} $, we have $ y^{2} = -\\frac{1}{2}x $, hence the equation of the directrix is $ x = \\frac{1}{8} $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. If there exists a point $P$ on the ellipse such that $|P F_{1}|=e|P F_{2}|$, then the range of the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G) = True;Abs(LineSegmentOf(P, F1)) = e*Abs(LineSegmentOf(P, F2));e: Number;Eccentricity(G) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "[\\sqrt{2}-1,1)",
    "fact_spans": "[[[2, 54], [80, 82], [116, 118]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[63, 70]], [[71, 78]], [[2, 78]], [[2, 78]], [[87, 90]], [[80, 90]], [[91, 113]], [[122, 125]], [[116, 125]]]",
    "query_spans": "[[[122, 132]]]",
    "process": "Let the horizontal coordinate of point P be x. Since |PF₁| = e|PF₂|, then by the definition of the ellipse, we have e(x + \\frac{a^{2}}{c}) = e \\times e(\\frac{a^{2}}{c} - x). Therefore, x = \\frac{c - a}{e(e + 1)}. From the given condition, we get -a \\leqslant \\frac{c - a}{e(e + 1)} \\leqslant a, so -1 \\leqslant \\frac{e - 1}{e(e + 1)} \\leqslant 1. Hence, \\begin{cases} e - 1 \\geqslant -e^{2} \\\\ e - 1 \\leqslant e^{2} \\end{cases} and -e \\leqslant e < ] \\leqslant e^{2} + e. Then the eccentricity of the ellipse"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ with focus $F$, point $A(0, \\sqrt{3})$, and point $B$ on parabola $C$ satisfying $A B \\perp A F$ and $|B F|=4$, then $p=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;A: Point;Coordinate(A) = (0, sqrt(3));B: Point;PointOnCurve(B, C);IsPerpendicular(LineSegmentOf(A, B), LineSegmentOf(A, F));Abs(LineSegmentOf(B, F)) = 4",
    "query_expressions": "p",
    "answer_expressions": "{2, 6}",
    "fact_spans": "[[[2, 28], [54, 60]], [[2, 28]], [[96, 99]], [[10, 28]], [[32, 35]], [[2, 35]], [[37, 53]], [[37, 53]], [[62, 66]], [[54, 66]], [[68, 83]], [[85, 94]]]",
    "query_spans": "[[[96, 101]]]",
    "process": ""
  },
  {
    "text": "If the equation $\\frac{x^{2}}{5-m}+\\frac{y^{2}}{m-1}=1$ represents an ellipse with foci on the $y$-axis, then the range of real values for $m$ is?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (x^2/(5 - m) + y^2/(m - 1) = 1);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(3, 5)",
    "fact_spans": "[[[52, 54]], [[56, 61]], [[1, 54]], [[44, 54]]]",
    "query_spans": "[[[56, 68]]]",
    "process": "From the geometric properties of the ellipse, we obtain  \n\\begin{cases}5-m>0\\\\m-1>0\\\\m-1>5-m\\end{cases}.  \nSolving this system of inequalities gives the solution.  \nGiven that the equation $\\frac{x^2}{5-m}+\\frac{}{n}\\frac{\\sqrt{2}}{-1}=1$ represents an ellipse with foci on the $y$-axis, then  \n\\begin{cases}5-m>0\\\\m-1>0\\\\m-1>5-m\\end{cases},  \nsolving yields:  \n\\begin{cases}m<5\\\\m>1\\\\m>3\\end{cases},  \nthat is, $3<m<5$."
  },
  {
    "text": "Given an ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with one focus at $F(\\sqrt{3}, 0)$, $A$ is the right vertex of the ellipse $C$, and a circle $A1$ centered at $A$ intersects the line $y=\\frac{b}{a} x$ at points $P$ and $Q$, such that $\\overrightarrow{A P} \\cdot \\vec{A Q}=0$, $\\overrightarrow{O P}=3 \\overrightarrow{O Q}$, then the radius of circle $A1$ is?",
    "fact_expressions": "C: Ellipse;a: Number;b: Number;A1: Circle;G: Line;A: Point;P: Point;Q: Point;O: Origin;F: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y = x*(b/a));Coordinate(F) = (sqrt(3), 0);OneOf(Focus(C)) = F;RightVertex(C) = A;Center(A1) = A;Intersection(A1, G) = {P, Q};DotProduct(VectorOf(A, Q), VectorOf(A, P)) = 0;VectorOf(O, P) = 3*VectorOf(O, Q)",
    "query_expressions": "Radius(A1)",
    "answer_expressions": "2*sqrt(10)/5",
    "fact_spans": "[[[2, 59], [87, 92]], [[9, 59]], [[9, 59]], [[105, 110], [232, 237]], [[111, 130]], [[83, 86], [98, 101]], [[133, 136]], [[137, 140]], [[185, 230]], [[65, 81]], [[9, 59]], [[9, 59]], [[2, 59]], [[111, 130]], [[65, 81]], [[2, 81]], [[83, 96]], [[97, 110]], [[105, 142]], [[144, 184]], [[185, 230]]]",
    "query_spans": "[[[232, 242]]]",
    "process": "As shown in the figure, take the midpoint M of PQ, and connect AM, AP, AQ. From the given conditions: AP\\bot AQ, OP=3OQ, and combining the properties of the circle, we have: OQ=QM=MP, AM\\bot OP, AP=AQ. In the isosceles right triangle, let AP=AQ=2m (m>0), then: MA=MP=MQ=\\sqrt{2}m. In Rt\\Delta AMO, OA=\\sqrt{MO^{2}+MA^{2}}=\\sqrt{(2\\sqrt{2}m)^{2}+(\\sqrt{2}m)^{2}}=\\sqrt{10}m. k_{OP}=\\frac{AM}{OM}=\\frac{1}{2}, then \\frac{b}{a}=\\frac{1}{2}. Combining c=3, the equation of the ellipse is \\frac{x^{2}}{4}+y^{2}=1, then OA=\\sqrt{10}m=2=\\frac{2}{\\sqrt{10}}"
  },
  {
    "text": "Let points $A(-2, \\sqrt{3})$, $B(2,0)$, and point $M$ moves on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$. When $|M A|+|M B|$ is maximized, what are the coordinates of point $M$?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;M: Point;Expression(G) = (x^2/16 + y^2/12 = 1);Coordinate(A) = (-2,sqrt(3));Coordinate(B) = (2, 0);PointOnCurve(M, G);WhenMax(Abs(LineSegmentOf(M,A))+Abs(LineSegmentOf(M,B)))",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[36, 75]], [[1, 19]], [[21, 30]], [[31, 35], [31, 35]], [[36, 75]], [[1, 19]], [[21, 30]], [[31, 78]], [[79, 96]]]",
    "query_spans": "[[[97, 106]]]",
    "process": "B is the right focus of the ellipse. Let the left focus be F(-2,0). Then by the definition of the ellipse, |MB| + |MF| = 2a = 8. Thus, |MA| + |MB| = 8 + |MA| - |MF|. When M is not at the intersection points of line AF and the ellipse, points M, F, A form a triangle, so |MA| - |MF| < |AF|. When M is at the intersection points of line AF and the ellipse, at the intersection point in the second quadrant, |MA| - |MF| = -|AF|; at the intersection point in the third quadrant, |MA| - |MF| = |AF|. Clearly, when M is at the intersection point of line BF and the ellipse in the third quadrant, |MA| + |MB| reaches its maximum value. |MA| + |MB| = 8 + |MA| - |MF| = 8 + |AF| = 8 + \\sqrt{3}"
  },
  {
    "text": "The line $y=kx+1$ intersects the left branch of the hyperbola $x^{2}-y^{2}=1$ at points $A$ and $B$. Another line $l$ passes through the point $(-2, 0)$ and the midpoint of $AB$. Then the range of the intercept $b$ of line $l$ on the $y$-axis is?",
    "fact_expressions": "l: Line;G: Hyperbola;H: Line;k: Number;I: Point;A: Point;B: Point;Expression(G) = (x^2 - y^2 = 1);Expression(H) = (y = k*x + 1);Coordinate(I) = (-2, 0);Intersection(H, LeftPart(G)) = {A, B};PointOnCurve(I, l);PointOnCurve(MidPoint(LineSegmentOf(A, B)), l);b: Number;Intercept(l, yAxis) = b",
    "query_expressions": "Range(b)",
    "answer_expressions": "(-oo, -2-sqrt(2)) + (2, +oo)",
    "fact_spans": "[[[48, 53], [76, 81]], [[12, 30]], [[0, 11]], [[2, 11]], [[54, 65]], [[35, 38]], [[39, 42]], [[12, 30]], [[0, 11]], [[54, 65]], [[0, 44]], [[48, 65]], [[48, 74]], [[90, 93]], [[76, 93]]]",
    "query_spans": "[[[90, 100]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ satisfies $a^{2}+b^{2}-3 c^{2}=0$, where $c$ is the semi-focal distance, then $\\frac{a+c}{a-c}$=?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);-3*c^2 + a^2 + b^2 = 0;c: Number;HalfFocalLength(G) = c",
    "query_expressions": "(a + c)/(a - c)",
    "answer_expressions": "3 + 2*sqrt(2)",
    "fact_spans": "[[[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[2, 54]], [[56, 79]], [[82, 85]], [[2, 89]]]",
    "query_spans": "[[[91, 110]]]",
    "process": "From $a^{2}=b^{2}+c^{2}$ and $a^{2}+b^{2}\\cdot3c^{2}=0$ we get $a^{2}=2c^{2}\\Rightarrow a=\\sqrt{2}c$, then $\\frac{a+c}{a-c}=\\frac{\\sqrt{2}+1}{\\sqrt{2}-1}=3+2\\sqrt{2}$,"
  },
  {
    "text": "A line passing through a focus $F_1$ of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$ intersects the ellipse at points $A$ and $B$. What is the perimeter of triangle $\\triangle A B F_{2}$ formed by $A$, $B$, and the other focus of the ellipse?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2/4 + y^2/2 = 1);PointOnCurve(F1, H);Intersection(H, G) = {A, B};OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2)",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 38], [54, 56], [77, 79]], [[51, 53]], [[58, 61], [69, 72]], [[62, 65], [73, 76]], [[88, 109]], [[43, 50]], [[1, 38]], [[0, 53]], [[51, 67]], [1, 47], [75, 82], [1, 82]]",
    "query_spans": "[[[88, 114]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ and the parabola $y^{2}=8 x$ share a common focus $F$, and $P$ is an intersection point of the two curves. If $|F P|=5$, then the distance from point $F$ to the asymptotes of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;F: Point;P: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 8*x);OneOf(Focus(G)) = F;Focus(H) = F;OneOf(Intersection(G,H))=P;Abs(LineSegmentOf(F, P)) = 5",
    "query_expressions": "Distance(F,Asymptote(G))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 59], [116, 119]], [[5, 59]], [[5, 59]], [[60, 74]], [[82, 85], [111, 115]], [[95, 98]], [[5, 59]], [[5, 59]], [[2, 59]], [[60, 74]], [[2, 85]], [[2, 85]], [[86, 98]], [[100, 109]]]",
    "query_spans": "[[[111, 128]]]",
    "process": ""
  },
  {
    "text": "$P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, and $M$, $N$ are points on the circles $(x+5)^{2}+y^{2}=9$ and $(x-5)^{2}+y^{2}=4$, respectively. Then the maximum value of $|PM|-|PN|$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P, RightPart(G)) = True;G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1);C1: Circle;Expression(C1) = (y^2 + (x + 5)^2 = 9);PointOnCurve(M,C1) = True;M: Point;C2: Circle;Expression(C2) = (y^2 + (x - 5)^2 = 4);PointOnCurve(N,C2) = True;N: Point",
    "query_expressions": "Max(Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "13",
    "fact_spans": "[[[0, 3]], [[0, 49]], [[4, 43]], [[4, 43]], [[60, 80]], [[60, 80]], [[50, 104]], [[50, 53]], [[81, 100]], [[81, 100]], [[50, 104]], [[54, 57]]]",
    "query_spans": "[[[106, 123]]]",
    "process": ""
  },
  {
    "text": "Given that $c$ is the semi-focal distance of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, then the range of $\\frac{b+c}{a}$ is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;a>b;b>0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);HalfFocalLength(G) = c;c: Number",
    "query_expressions": "Range((b + c)/a)",
    "answer_expressions": "(1, \\sqrt{2}]",
    "fact_spans": "[[[6, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[6, 59]], [[2, 63]], [[2, 5]]]",
    "query_spans": "[[[65, 87]]]",
    "process": ""
  },
  {
    "text": "If the equation $(1-k) x^{2}+(3-k^{2}) y^{2}=4$ represents an ellipse, then the range of values for $k$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2*(1 - k) + y^2*(3 - k^2) = 4);k: Number",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-sqrt(3), 1)",
    "fact_spans": "[[[36, 38]], [[1, 38]], [[40, 43]]]",
    "query_spans": "[[[40, 50]]]",
    "process": ""
  },
  {
    "text": "The right focus of the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $F(c, 0)$, the intersection point of the line $x=\\frac{a^{2}}{c}$ and the $x$-axis is $A$. If there exists a point $P$ on the ellipse $E$ such that the perpendicular bisector of segment $AP$ passes through point $F$, then the range of the eccentricity of ellipse $E$ is?",
    "fact_expressions": "E: Ellipse;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;F: Point;c: Number;Coordinate(F) = (c, 0);RightFocus(E) = F;G: Line;Expression(G) = (x = a^2/c);A: Point;Intersection(G, xAxis) = A;P: Point;PointOnCurve(P, E);PointOnCurve(F, PerpendicularBisector(LineSegmentOf(A, P)))",
    "query_expressions": "Range(Eccentricity(E))",
    "answer_expressions": "[1/2, 1)",
    "fact_spans": "[[[0, 57], [107, 112], [141, 146]], [[0, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[62, 71], [135, 139]], [[62, 71]], [[62, 71]], [[0, 71]], [[72, 93]], [[72, 93]], [[102, 105]], [[72, 105]], [[115, 119]], [[106, 119]], [[121, 139]]]",
    "query_spans": "[[[141, 156]]]",
    "process": "From the property of the perpendicular bisector, we have: |PF| = |AF| = \\frac{a^{2}}{c} - c, and a - c \\leqslant |PF| \\leqslant a + c, thus a - c \\leqslant \\frac{a^{2}}{c} - c \\leqslant a + c,\\ ac - c^{2} \\leqslant a^{2} - c^{2} \\leqslant ac + c^{2},\\ e - e^{2} \\leqslant 1 - e^{2} \\leqslant e + e^{2}, and since e < 1, solving gives: \\frac{1}{2} \\leqslant e < 1."
  },
  {
    "text": "Given that a line with slope $1$ passes through the right focus of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ and intersects the ellipse at points $A$ and $B$, find the length of chord $AB$.",
    "fact_expressions": "H: Line;Slope(H) = 1;G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);PointOnCurve(RightFocus(G), H);A: Point;B: Point;Intersection(H, G) = {A, B};IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "8/5",
    "fact_spans": "[[[9, 11]], [[2, 11]], [[12, 39], [44, 46]], [[12, 39]], [[9, 43]], [[47, 50]], [[51, 54]], [[9, 56]], [[44, 64]]]",
    "query_spans": "[[[59, 68]]]",
    "process": "The right focus of the ellipse $\\frac{x^2}{4}+y^{2}=1$ is $F(\\sqrt{3},0)$, and the equation of line $AB$ is $y=x-\\sqrt{3}$. Solving the system  \n$$\n\\begin{cases}\ny=x-\\sqrt{3} \\\\\n\\frac{x^{2}}{4}+y^{2}=1\n\\end{cases}\n$$  \nyields $5x^{2}-8\\sqrt{3}x+8=0$, where $a=8^{2}\\times3-4\\times5\\times8=32>0$. Let points $A(x_{1},y_{1})$ and $B(x_{2},y_{2})$. By Vieta's formulas, we obtain $x_{1}+x_{2}=\\frac{8\\sqrt{3}}{5}$, $x_{1}x_{2}=\\frac{8}{5}$. Then  \n$$\n|AB|=\\sqrt{2}\\cdot\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\sqrt{2}\\times\\sqrt{\\left(\\frac{8\\sqrt{3}}{5}\\right)^{2}-4\\times\\frac{8}{5}}=\\frac{8}{5}\n$$"
  },
  {
    "text": "Given that point $P(1 , 2)$ lies on the parabola $C$: $y^{2}=2 p x$, what is the equation of the directrix of parabola $C$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;P: Point;Coordinate(P) = (1, 2);PointOnCurve(P, C)",
    "query_expressions": "Expression(Directrix(C))",
    "answer_expressions": "x=-1",
    "fact_spans": "[[[14, 35], [38, 44]], [[14, 35]], [[22, 35]], [[2, 13]], [[2, 13]], [[2, 36]]]",
    "query_spans": "[[[38, 51]]]",
    "process": "Substitute P(1,2) into the parabola equation to find p=2, then determine the directrix equation. P(1,2) lies on the parabola C: y^{2}=2px, so 2p=4, p=2, and the directrix equation is x=-\\frac{p}{2}=-1."
  },
  {
    "text": "It is known that $F$ is the focus of the parabola $C$: $y^{2}=4 x$. A line passing through $F$ with slope $\\sqrt{3}$ intersects $C$ at points $A$ and $B$. Given $|F A| > |F B|$, then the value of $\\frac{|F A|}{|F B|}$ is?",
    "fact_expressions": "C: Parabola;G: Line;F: Point;A: Point;B: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F, G);Slope(G) = sqrt(3);Intersection(G, C) = {A, B};Abs(LineSegmentOf(F, A)) > Abs(LineSegmentOf(F, B))",
    "query_expressions": "Abs(LineSegmentOf(F, A))/Abs(LineSegmentOf(F, B))",
    "answer_expressions": "-1/3",
    "fact_spans": "[[[6, 25], [51, 54]], [[48, 50]], [[2, 5], [30, 33]], [[55, 58]], [[59, 62]], [[6, 25]], [[2, 28]], [[29, 50]], [[34, 50]], [[48, 64]], [[67, 80]]]",
    "query_spans": "[[[82, 108]]]",
    "process": "F(1,0), let A(x_{1},y_{1}) B(x_{2},y_{2}) be given by \\begin{cases}y=\\sqrt{3}(x-1)\\\\y^{2}=4x\\end{cases}, rearranging yields 3x^{2}-10x+3=0, so x_{1}=3, x_{2}=\\frac{1}{3}, (x_{1}>x_{2}) \\therefore by the definition of the parabola, \\frac{|FA|}{|FB|}=\\frac{x_{1}+1}{x_{2}+1}=\\frac{3+1}{\\frac{1}{3}+1}"
  },
  {
    "text": "Given that the distance from the focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ to one of its asymptotes equals the length of the real axis, what is the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Distance(Focus(G),OneOf(Asymptote(G))) = Length(RealAxis(G))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 58], [79, 82]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 75]]]",
    "query_spans": "[[[79, 87]]]",
    "process": "From the given conditions, one focus of the hyperbola has coordinates (c,0), and one asymptote equation of the hyperbola is: $\\frac{x}{a}-\\frac{y}{b}=0$, that is, $bx-ay=0$. Based on this, we obtain: $\\frac{|bc-0|}{\\sqrt{a^{2}+b^{2}}}=\\frac{bc}{c}=b=2a$, then $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{5}a$. The eccentricity of the ellipse is $e=\\frac{c}{a}=\\sqrt{5}$. The eccentricity of a hyperbola is the most important geometric property of the hyperbola. To find the eccentricity of a hyperbola (or the range of eccentricity), common methods are: $\\textcircled{1}$ Find $a$ and $c$, then substitute into the formula $e=\\frac{c}{a}$; $\\textcircled{2}$ Only need to obtain a homogeneous equation in $a$, $b$, $c$ from one condition, combine with $b^{2}=c^{2}-a^{2}$ to transform it into a homogeneous equation in $a$ and $c$, then divide both sides of the equation (inequality) by $a$ or $a^{2}$ to convert it into an equation (inequality) in terms of $e$, solving which gives $e$ (or the range of $e$)."
  },
  {
    "text": "Given point $C(-1,3)$, line $CA$ intersects the $x$-axis at point $A$, and line $CB$, which passes through point $C$ and is perpendicular to line $CA$, intersects the $y$-axis at point $B$. Let $M$ be the midpoint of segment $AB$. Then the equation of the trajectory of point $M$ is?",
    "fact_expressions": "A: Point;C: Point;B: Point;M: Point;Coordinate(C) = (-1, 3);Intersection(LineOf(C,A),xAxis)=A;PointOnCurve(C,LineOf(C,B));IsPerpendicular(LineOf(C,A),LineOf(C,B));Intersection(LineOf(C,B),yAxis)=B;MidPoint(LineSegmentOf(A,B))=M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x-3*y+5=0",
    "fact_spans": "[[[27, 31]], [[2, 12], [33, 37]], [[63, 67]], [[84, 88], [68, 71]], [[2, 12]], [[13, 31]], [[32, 56]], [[38, 56]], [[49, 67]], [[68, 82]]]",
    "query_spans": "[[[84, 95]]]",
    "process": "In right triangle ACB, MC=\\frac{1}{2}AB; in right triangle AOB, MO=\\frac{1}{2}AB, thus MO=MC, so the locus of point M is the perpendicular bisector of segment OC, leading to the solution. As shown in the figure, CA\\bot CB, so in right triangle ACB, MC=\\frac{1}{2}AB; also in right triangle AOB, MO=\\frac{1}{4}AB, hence MO=MC. The locus of point M is the perpendicular bisector of segment OC. C(-1,3), O(0,0), k_{CO}=-3, the midpoint is (-\\frac{1}{2},\\frac{3}{7}), so the equation of the perpendicular bisector is: y-\\frac{3}{2}=\\frac{1}{3}(x+\\frac{1}{2}), simplifying to: x-3y+5=0. Answer: x-3y+5=0"
  },
  {
    "text": "The equation of the line on which the chord of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, passing through the point $M(4,1)$ and bisected by the point $M$, lies is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1);M: Point;Coordinate(M) = (4, 1);PointOnCurve(M, H);H: LineSegment;IsChordOf(H, G);MidPoint(H) = M",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "x-y-3=0",
    "fact_spans": "[[[19, 47]], [[19, 47]], [[1, 10], [12, 16]], [[1, 10]], [[0, 47]], [], [[19, 49]], [[0, 49]]]",
    "query_spans": "[[[19, 57]]]",
    "process": "Since the hyperbola is symmetric about the x-axis and point M does not lie on the x-axis, the required line is not parallel to the y-axis, meaning its slope is a real number. Let the slope of the required line be $ a $, and the coordinates of the two intersection points with the hyperbola be $ (4+t, 1+at) $ and $ (4-t, 1-at) $. Substituting these coordinates into the hyperbola equation gives:  \n$$\n\\frac{(4+t)^{2}}{4} - (1+at)^{2} = 1, \\quad \\frac{(4-t)^{2}}{4} - (1-at)^{2} = 1\n$$  \nSubtracting these two equations yields:  \n$$\n4t - 4at = 0 \\implies a = 1\n$$  \nThe required line equation is $ y - 1 = 1(x - 4) $, or $ x - y - 3 = 0 $."
  },
  {
    "text": "The ellipse $C$ with center at the origin passes through the point $(1, \\frac{2 \\sqrt{5}}{5})$, and one of its foci is $(2,0)$. Then the standard equation of $C$ is?",
    "fact_expressions": "C: Ellipse;G: Point;O: Origin;Coordinate(G) = (1, (2*sqrt(5))/5);Coordinate(OneOf(Focus(C))) = (2, 0);Center(C) = O;PointOnCurve(G, C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/5 + y^2 = 1",
    "fact_spans": "[[[10, 15], [46, 47], [62, 65]], [[16, 44]], [[5, 9]], [[16, 44]], [[46, 60]], [[2, 15]], [[10, 44]]]",
    "query_spans": "[[[62, 72]]]",
    "process": "Analysis: According to the given conditions, use the method of undetermined coefficients to find the values of $a$ and $b$, then obtain the standard equation of the ellipse. Since the foci of the ellipse lie on the $x$-axis, assume the equation of the ellipse is $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. The ellipse $C$ passes through the point $\\left(1,\\frac{\\sqrt[2]{5}}{5}\\right)$, then: $\\frac{1}{a^{2}}+\\frac{4}{5b^{2}}=1$, $\\textcircled{1}$ One of its foci is $(2,0)$, then $a^{2}-b^{2}=4$, $\\textcircled{2}$ Solving $\\textcircled{1}$ and $\\textcircled{2}$ together gives: $\\begin{cases}a^{2}=5\\\\b^{2}=1\\end{cases}$, thus the standard equation of $C$ is $\\frac{x^{2}}{5}+y^{2}=1$."
  },
  {
    "text": "Given two points $M(4,0)$, $N(1,0)$, point $P$ satisfies $\\overrightarrow{M N} \\cdot \\overrightarrow{M P}=6|\\overrightarrow{P N}|$, then the trajectory equation of point $P$ is?",
    "fact_expressions": "M: Point;N: Point;P: Point;Coordinate(M) = (4, 0);Coordinate(N) = (1, 0);DotProduct(VectorOf(M, N), VectorOf(M, P)) = 6*Abs(VectorOf(P, N))",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2/4 + y^2/3 = 1",
    "fact_spans": "[[[4, 12]], [[13, 21]], [[22, 26], [103, 107]], [[4, 12]], [[13, 21]], [[28, 101]]]",
    "query_spans": "[[[103, 114]]]",
    "process": ""
  },
  {
    "text": "The equation of the line containing the chord passing through the point $(2, 1)$ and bisected at this point within the parabola $y^{2}=16x$ is?",
    "fact_expressions": "G: Parabola;H: LineSegment;I: Point;Expression(G) = (y^2 = 16*x);Coordinate(I) = (2, 1);PointOnCurve(I,H);MidPoint(H)=I;IsChordOf(H,G)",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "8*x-y-15=0",
    "fact_spans": "[[[1, 16]], [], [[20, 30], [33, 34]], [[1, 16]], [[20, 30]], [[1, 39]], [[1, 39]], [[1, 39]]]",
    "query_spans": "[[[1, 48]]]",
    "process": ""
  },
  {
    "text": "Given that the focal distance of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $2$ times the length of the imaginary axis, find the equations of the asymptotes of the hyperbola.",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);FocalLength(G) = 2 * Length(ImageinaryAxis(G))",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(3)/3)*x",
    "fact_spans": "[[[2, 58], [72, 75]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 70]]]",
    "query_spans": "[[[72, 83]]]",
    "process": "According to the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$, the focal distance is twice the length of the imaginary axis, so $2c=4b$, then $\\frac{b}{a}=\\frac{b}{\\sqrt{c^{2}-b^{2}}}$ is solved. [Detailed solution] Since for the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$, the focal distance is twice the length of the imaginary axis, we have $2c=4b$, that is, $c=2b$, so $\\frac{b}{a}=\\frac{b}{\\sqrt{c^{2}-b^{2}}}=\\frac{b}{\\sqrt{3}b}=\\frac{\\sqrt{3}}{3}$. Therefore, the asymptotes of the hyperbola are $y=\\pm\\frac{\\sqrt{3}}{3}x^{x}$."
  },
  {
    "text": "Given an ellipse $\\frac{x^{2}}{10}+\\frac{y^{2}}{m}=1$ with foci on the $y$-axis and major axis length $8$, then $m=$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/10 + y^2/m = 1);m: Number;PointOnCurve(Focus(G), yAxis);Length(MajorAxis(G)) = 8",
    "query_expressions": "m",
    "answer_expressions": "16",
    "fact_spans": "[[[11, 49]], [[11, 49]], [[59, 62]], [[2, 49]], [[11, 57]]]",
    "query_spans": "[[[59, 64]]]",
    "process": "From the given condition we get: $2\\sqrt{m}=8$, that is, $m=16$"
  },
  {
    "text": "A line $l$ passing through the focus of the parabola $y^{2}=4x$ intersects the parabola at points $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$. If $x_{1}+x_{2}=6$, then $PQ$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);l: Line;PointOnCurve(Focus(G), l);P: Point;Q: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(P) = (x1, y1);Coordinate(Q) = (x2, y2);Intersection(l, G) = {P, Q};x1 + x2 = 6",
    "query_expressions": "LineSegmentOf(P, Q)",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 15], [25, 28]], [[1, 15]], [[19, 24]], [[0, 24]], [[29, 46]], [[48, 65]], [[29, 46]], [[48, 65]], [[29, 46]], [[48, 65]], [[29, 46]], [[48, 65]], [[19, 67]], [[70, 85]]]",
    "query_spans": "[[[88, 95]]]",
    "process": "Since line l passes through the focus F of the parabola, we have PQ = PF + QF. Then, according to the definition of the parabola, transform PF + QF into the distance from points on the parabola to the directrix, and the result can be obtained. The parabola y^{2} = 4x has focus F(1,0) and directrix equation x = -1. According to the given conditions, PQ = PF + QF = x_{1} + 1 + x + x_{2} + 2 = 8."
  },
  {
    "text": "If the right focus of the hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ $(m>0)$ coincides with the focus of the parabola $y^{2}=8x$, then the value of $m$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2 + x^2/m = 1);m: Number;m>0;H: Parabola;Expression(H) = (y^2 = 8*x);RightFocus(G) = Focus(H)",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 34]], [[1, 34]], [[60, 63]], [[4, 34]], [[39, 53]], [[39, 53]], [[1, 58]]]",
    "query_spans": "[[[60, 67]]]",
    "process": "Since the focus of the parabola \\( y^{2} = 8x \\) is \\( F(2,0) \\), it follows that \\( m+1 = 4 \\Rightarrow m = 3 \\). The answer to be filled in is 3."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{11}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $M$ is a point on the ellipse $C$, point $N$ is the midpoint of segment $M F_{1}$, and $O$ is the origin. If $|M N|=4$, then $|O N|=$?",
    "fact_expressions": "C: Ellipse;F1: Point;F2:Point;M: Point;N: Point;O: Origin;Expression(C)=(x^2/36 + y^2/11 = 1);LeftFocus(C) =F1;RightFocus(C)=F2;PointOnCurve(M, C);MidPoint(LineSegmentOf(M, F1)) = N;Abs(LineSegmentOf(M, N)) = 4",
    "query_expressions": "Abs(LineSegmentOf(O, N))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 41], [71, 76]], [[50, 57]], [[58, 65]], [[66, 70]], [[81, 85]], [[101, 104]], [[2, 41]], [[2, 65]], [[2, 65]], [[66, 80]], [[81, 100]], [[111, 120]]]",
    "query_spans": "[[[122, 131]]]",
    "process": "By the definition of an ellipse, |MF_{1}| + |MF_{2}| = 2a = 12. Since point N is the midpoint of segment MF_{1} and |MN| = 4, it follows that |MF| = 8, so |MF_{2}| = 4. Since point O is the midpoint of segment F_{1}F_{2}, ON is the midline of \\triangle MF_{1}F_{2}, so |ON| = \\frac{1}{3}|MF_{2}| = 2."
  },
  {
    "text": "Given that point $P$ is an arbitrary point on the right branch of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{3}=1$, and from point $P$ perpendiculars are drawn to the two asymptotes, with feet of the perpendiculars denoted as $M$ and $N$, then the area of $\\triangle P M N$ is?",
    "fact_expressions": "P: Point;G: Hyperbola;Expression(G) = (x^2/9 - y^2/3 = 1);PointOnCurve(P, RightPart(G)) = True;l1: Line;l2: Line;M: Point;N: Point;l3: Line;l4: Line;Asymptote(G) = {l3,l4};PointOnCurve(P,l1) =True;PointOnCurve(P,l2) = True;IsPerpendicular(l1,l3) = True;IsPerpendicular(l2,l4) = True;FootPoint(l1, l3) = M;FootPoint(l2, l4) = N",
    "query_expressions": "Area(TriangleOf(P, M, N))",
    "answer_expressions": "9/8",
    "fact_spans": "[[[2, 6], [54, 58]], [[7, 45]], [[7, 45]], [[2, 52]], [], [], [[73, 76]], [[77, 80]], [], [], [[2, 67]], [[2, 67]], [[2, 67]], [[2, 67]], [[2, 67]], [[2, 80]], [[2, 80]]]",
    "query_spans": "[[[82, 99]]]",
    "process": "The hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{3}=1$ has two asymptotes $x-\\sqrt{3}y=0$ and $x+\\sqrt{3}y=0$. Let $P(x_{0},y_{0})$, then the distance from $P$ to the asymptote $x-\\sqrt{3}y=0$ is $d_{1}=\\frac{|x_{0}-\\sqrt{3}y_{0}|}{2}$, the distance from $P$ to the asymptote $x+\\sqrt{3}y=0$ is $d_{2}=\\frac{|x_{0}+\\sqrt{3}y_{0}|}{2}$, $\\therefore$ the area of $\\triangle PMN = \\frac{1}{2}\\times\\frac{|x_{0}-\\sqrt{3}y_{0}|}{2}\\times\\frac{|x_{0}+\\sqrt{3}y_{0}|}{2}=\\frac{|x^{2}-3y_{0}^{2}|}{}$"
  },
  {
    "text": "If the length of the real axis of the hyperbola $\\frac{x^{2}}{4+k}+\\frac{y^{2}}{2-k}=1$ is $6$, then the value of $k$ is?",
    "fact_expressions": "G: Hyperbola;k: Number;Expression(G) = (x^2/(k + 4) + y^2/(2 - k) = 1);Length(RealAxis(G)) = 6",
    "query_expressions": "k",
    "answer_expressions": "{5,-7}",
    "fact_spans": "[[[1, 43]], [[53, 56]], [[1, 43]], [[1, 51]]]",
    "query_spans": "[[[53, 60]]]",
    "process": "If the hyperbola $\\frac{x^2}{4+k}+\\frac{y^{2}}{2-k}=1$ has foci on the $x$-axis, then $\\begin{cases}4+k>0\\\\2-k<0\\end{cases}$, solving gives $k>2$, so $2\\sqrt{4+k}=6$, solving gives $k=5$; if the hyperbola $\\frac{x^2}{4+k}+\\frac{y^{2}}{2-k}=1$ has foci on the $y$-axis, then $\\begin{cases}4+k<0\\\\2-k>0\\end{cases}$, solving gives $k<-4$, so $2\\sqrt{2-k}=6$, solving gives $k=-7$;"
  },
  {
    "text": "Given that the two foci of ellipse $C$ are $F_{1}(-1,0)$ and $F_{2}(1,0)$, a line passing through $F_{1}$ intersects ellipse $C$ at points $A$ and $B$. If $|B F_{1}|=3|A F_{1}|$ and $A B \\perp B F_{2}$, then the equation of $C$ is?",
    "fact_expressions": "C: Ellipse;F1: Point;F2: Point;Coordinate(F1) = (-1, 0);Coordinate(F2) = (1, 0);Focus(C) = {F1, F2};G: Line;PointOnCurve(F1, G);A: Point;B: Point;Intersection(G, C) = {A, B};Abs(LineSegmentOf(B, F1)) = 3*Abs(LineSegmentOf(A, F1));IsPerpendicular(LineSegmentOf(A, B), LineSegmentOf(B, F2))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/2+y^2=1",
    "fact_spans": "[[[2, 7], [54, 59], [117, 120]], [[13, 26], [43, 50]], [[28, 40]], [[13, 26]], [[28, 40]], [[2, 40]], [[51, 53]], [[42, 53]], [[61, 64]], [[65, 68]], [[51, 70]], [[72, 94]], [[96, 115]]]",
    "query_spans": "[[[117, 125]]]",
    "process": "As shown in the figure, let |AF₁| = x, ∴ |BF₁| = 3x, |BF₂| = 2a - 3x, |AF₂| = 2a - x. Since AB ⊥ BF₂, we have (4x)² + (2a - 3x)² = (2a - x)², (3x)² + (2a - 3x)² = (2c)² = 4. Given c = 1, solving yields a² = 2, and b² = a² - c² = 1, so the equation of C is \\frac{x^{2}}{2}+y^{2}=1."
  },
  {
    "text": "Write a hyperbola equation whose asymptotes are given by $y=\\pm \\sqrt{3} x$?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*sqrt(3)*x)",
    "query_expressions": "OneOf(Expression(G))",
    "answer_expressions": "x^2 - y^2/3 = 1",
    "fact_spans": "[[[30, 33]], [[3, 33]]]",
    "query_spans": "[[[0, 36]]]",
    "process": "If the foci are on the x-axis, according to the given conditions we have: a:b:c = 1:\\sqrt{3}:2. Without loss of generality, let a=1, b=\\sqrt{3}, c=2, then the hyperbola equation is x^2 - \\frac{y^{2}}{3} = 1."
  },
  {
    "text": "Let the parabola $C$: $y^{2}=2 x$ have focus $F$. If the point $P$ on the parabola $C$ has horizontal coordinate $2$, then $|P F|$=?",
    "fact_expressions": "C: Parabola;P: Point;F: Point;Expression(C) = (y^2 = 2*x);Focus(C) = F;PointOnCurve(P,C);XCoordinate(P)=2",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "5/2",
    "fact_spans": "[[[1, 20], [29, 35]], [[36, 40]], [[24, 27]], [[1, 20]], [[1, 27]], [[29, 40]], [[36, 48]]]",
    "query_spans": "[[[50, 59]]]",
    "process": "The distance from the point on the parabola with horizontal coordinate 2 to the focus is equal to the distance from this point to the directrix of the parabola. The equation of the directrix of the parabola is $x=-\\frac{1}{2}$, so $|PF|=2+\\frac{1}{2}=\\frac{5}{2}$."
  },
  {
    "text": "Given that line $l$ is a line passing through the focus of the parabola $x^{2}=4 y$, and $l$ intersects the parabola at points $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$. Then $x_{1} x_{2}=$?",
    "fact_expressions": "l: Line;G: Parabola;Expression(G) = (x^2 = 4*y);A: Point;B:Point;PointOnCurve(Focus(G),l);Coordinate(A) = (x1,y1);Coordinate(B)=(x2,y2);Intersection(l, G) = {A,B};x1:Number;x2:Number;y1:Number;y2:Number",
    "query_expressions": "x1*x2",
    "answer_expressions": "-4",
    "fact_spans": "[[[2, 7], [31, 34]], [[9, 23], [35, 38]], [[9, 23]], [[40, 57]], [[59, 76]], [[2, 30]], [[40, 57]], [[59, 76]], [[31, 78]], [[40, 57]], [[59, 76]], [[40, 57]], [[59, 76]]]",
    "query_spans": "[[[80, 95]]]",
    "process": "Since the focus of the parabola $ x^{2} = 4y $ is at $ (0,1) $, obviously the slope of the line exists. Let the equation of the line be $ y = kx + 1 $. Substituting $ y = kx + 1 $ into the parabola $ x^{2} = 4y $, we obtain $ x^{2} - 4kx - 4 = 0 $. $ \\therefore x_{1}x_{2} = -4 $"
  },
  {
    "text": "What is the standard equation of the parabola passing through the point $P(4,-2)$?",
    "fact_expressions": "G: Parabola;P: Point;Coordinate(P) = (4, -2);PointOnCurve(P, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2 = x, x^2 = -8*y}",
    "fact_spans": "[[[13, 16]], [[2, 12]], [[2, 12]], [[0, 16]]]",
    "query_spans": "[[[13, 23]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, with left and right foci $F_{1}$ and $F_{2}$ respectively. Point $M(\\sqrt{3}, 2)$ lies on the right branch of hyperbola $C$, and $F_{2}$ lies on the circumference of the circle with diameter $M F_{1}$. Then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C)=(x^2/a^2-y^2/b^2=1);G: Circle;a:Number;b:Number;a>0;b>0;F1: Point;F2:Point;M: Point;LeftFocus(C)=F1;RightFocus(C)=F2;Coordinate(M) = (sqrt(3), 2);PointOnCurve(M, RightPart(C));PointOnCurve(F2,G);IsDiameter(LineSegmentOf(M,F1),G)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 63], [106, 112], [150, 156]], [[2, 63]], [[143, 144]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[72, 79]], [[80, 87], [119, 126]], [[88, 105]], [[2, 87]], [[2, 87]], [[88, 105]], [[88, 117]], [[119, 148]], [[127, 144]]]",
    "query_spans": "[[[150, 162]]]",
    "process": "From the given condition, we know $\\angle F_{1}F_{2}M = 90^{\\circ}$, and since $M(\\sqrt{3}, 2)$, it follows that $c = \\sqrt{3}$. Also, the point $M(\\sqrt{3}, 2)$ lies on the hyperbola $C$, so $\\frac{3}{a^{2}} - \\frac{4}{b^{2}} = 1$. Solving this together yields $a = 1$, then the eccentricity of hyperbola $C$ is $\\sqrt{3}$."
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-y^{2}=1$, with points $F_{1}$ and $F_{2}$ being its two foci, and point $P$ a point on the hyperbola $C$ such that $P F_{1} \\perp P F_{2}$, then the value of $|P F_{1}|+|P F_{2}|$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2 = 1);F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))",
    "query_expressions": "Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[2, 25], [43, 44], [54, 60]], [[2, 25]], [[26, 34]], [[35, 42]], [[26, 48]], [[49, 53]], [[49, 63]], [[67, 90]]]",
    "query_spans": "[[[92, 118]]]",
    "process": "By the given condition, for the hyperbola $x^{2}-y^{2}=1$, we have $a=b=1$, $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{2}$. According to the definition of a hyperbola, $|PF_{1}|-|PF_{2}|=2a=2$. Since $PF_{1}\\bot PF_{2}$, it follows that $|PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}=(2c)^{2}=8$. From $(|PF_{1}|-|PF_{2}|)^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}||PF_{2}|=4$, we obtain $|PF_{1}||PF_{2}|=2$. Therefore, $(|PF_{1}|+|PF_{2}|)^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}+2|PF_{1}||PF_{2}|=8+4=12$, so $|PF_{1}|+|PF_{2}|=2\\sqrt{3}$."
  },
  {
    "text": "The distance from point $M(2,1)$ to the directrix of the parabola $y=ax^{2}$ is $2$. Then the value of $a$ is?",
    "fact_expressions": "G: Parabola;a: Number;M: Point;Expression(G)=(y = a*x^2);Coordinate(M)=(2,1);Distance(M,Directrix(G))=2",
    "query_expressions": "a",
    "answer_expressions": "1/4, -1/12",
    "fact_spans": "[[[10, 23]], [[34, 37]], [[0, 9]], [[10, 23]], [[0, 9]], [[0, 32]]]",
    "query_spans": "[[[34, 41]]]",
    "process": "The standard equation of the parabola $ y = ax^{2} $ is: $ x^{2} = \\frac{1}{a}y $, and the equation of the directrix is: $ y = -\\frac{1}{4a}|1 - (-\\frac{1}{43})| = 2 $. Solving gives $ a = \\frac{1}{4} $ or $ -\\frac{1}{12} $."
  },
  {
    "text": "The equation of the hyperbola with focus $F(2,0)$ and asymptotes $y=\\pm \\sqrt{3} x$ is?",
    "fact_expressions": "G: Hyperbola;F: Point;Coordinate(F) = (2, 0);OneOf(Focus(G))=F;Expression(Asymptote(G))=(y=pm*sqrt(3)*x)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[38, 41]], [[1, 9]], [[1, 9]], [[0, 41]], [[15, 41]]]",
    "query_spans": "[[[38, 45]]]",
    "process": "From the focus, we get $ c=2 $; from the asymptotes, we get $ \\frac{b}{a}=\\sqrt{3} $; together with the hyperbola condition $ c^{2}=a^{2}+b^{2} $, solving simultaneously gives: since $ F(2,0) $ is a focus, the foci lie on the x-axis and $ c=2 $. Since the asymptotes are $ y=\\pm\\sqrt{3}x $, we have $ \\frac{b}{a}=\\sqrt{3} $. Because the hyperbola satisfies $ c^{2}=a^{2}+b^{2} $, solving yields $ \\begin{cases}a=1\\\\b=\\sqrt{3}\\end{cases} $, so the hyperbola equation is $ x^{2}-\\frac{y^{2}}{3}=1 $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, point $P$ is the upper vertex of $C$, and $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$, $S_{\\Delta F_{1} P F_{2}}=\\sqrt{3}$. Then the equation of $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;F1: Point;P: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;UpperVertex(C)=P;AngleOf(F1, P, F2) = pi/3;Area(TriangleOf(F1,P,F2))=sqrt(3)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4+y^2/3=1",
    "fact_spans": "[[[20, 77], [89, 92], [175, 178]], [[26, 77]], [[26, 77]], [[2, 9]], [[84, 88]], [[10, 17]], [[26, 77]], [[26, 77]], [[20, 77]], [[2, 83]], [[2, 83]], [[84, 96]], [[98, 135]], [[137, 172]]]",
    "query_spans": "[[[175, 183]]]",
    "process": "\\angle F_{1}PF_{2}=\\frac{\\pi}{3} \\Rightarrow a=2c, S_{\\triangle F_{1}PF_{2}}=\\frac{1}{2}\\times 2c \\times b = \\sqrt{3}, hence b=\\frac{\\sqrt{3}}{c}. Also a^{2}=b^{2}+c^{2}, i.e., 4c^{2}=\\frac{3}{c^{2}}+c^{2}, solving gives: c^{2}=1, hence a^{2}=4, b^{2}=3, so the equation of C is \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, point $P$ is a point on the ellipse, $F_{1}$ and $F_{2}$ are the foci of the ellipse, and $\\angle P F_{1} F_{2}=120^{\\circ}$. Then the area of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);P: Point;PointOnCurve(P, G) = True;F1: Point;F2: Point;Focus(G) = {F1,F2};AngleOf(P, F1, F2) = ApplyUnit(120, degree)",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "3*sqrt(3)/5",
    "fact_spans": "[[[2, 39], [46, 48], [68, 70]], [[2, 39]], [[41, 45]], [[41, 51]], [[52, 59]], [[60, 67]], [[52, 73]], [[75, 109]]]",
    "query_spans": "[[[111, 141]]]",
    "process": "From $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, we know $a=2$, $b=\\sqrt{3}$, so $c=\\sqrt{a^{2}-b^{2}}=\\sqrt{4-3}=1$, thus $|F_{1}F_{2}|=2c=2$. In $\\triangle PF_{1}F_{2}$, by the cosine law we have $|PF_{2}|^{2}=|PF_{1}|^{2}+|F_{1}F_{2}|^{2}-2|PF_{1}||F_{1}F_{2}|\\cos\\angle PF_{1}F_{2}$, that is, $|PF_{2}|^{2}=|PF_{1}|^{2}+4-4|PF_{1}|\\times(-\\frac{1}{2})=|PF_{1}|^{2}+4+2|PF_{1}|$, $\\textcircled{1}$ From the ellipse definition, $|PF_{1}|+|PF_{2}|=2a=4$, $\\textcircled{2}$ Solving $\\textcircled{1}$ and $\\textcircled{2}$ together gives $|PF_{1}|=\\frac{6}{5}$. Therefore, $S_{\\Delta PF_{1}F_{2}}=\\frac{1}{2}|PF_{1}||F_{1}F_{2}|\\sin\\angle PF_{1}F_{2}=\\frac{1}{2}\\times\\frac{6}{5}\\times2\\times\\frac{\\sqrt{3}}{2}=\\frac{3\\sqrt{3}}{5}$"
  },
  {
    "text": "If the equation $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{a+12}=1$ represents an ellipse with foci on the $x$-axis, then the range of real values for $a$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/(a + 12) + x^2/a^2 = 1);a: Real;PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Range(a)",
    "answer_expressions": "(-12, -3) + (4, +\\infty)",
    "fact_spans": "[[[57, 59]], [[2, 59]], [[61, 66]], [[48, 59]]]",
    "query_spans": "[[[61, 73]]]",
    "process": "By the given condition, since the equation $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{a+12}=1$ represents an ellipse with foci on the x-axis, we have $a^{2}>a+12>0$. Solving this inequality yields $a>4$ or $-12<a<-3$. Therefore, the range of real values for $a$ is $(-12,-3)\\cup(4,+\\infty)$."
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{7}+\\frac{y^{2}}{3}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/7 + y^2/3 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(7)/7",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 43]]]",
    "process": "\\frac{x^{2}}{7}+\\frac{y^{2}}{3}=1\\Rightarrow a=\\sqrt{7},c=2\\Rightarrow e=\\frac{c}{a}=\\frac{2\\sqrt{7}}{7}"
  },
  {
    "text": "Through the right focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a tangent line $FM$ (with $M$ as the point of tangency) to the circle $x^{2}+y^{2}=a^{2}$, intersecting the $y$-axis at point $P$. If $M$ is the midpoint of segment $FP$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "F: Point;P: Point;G: Hyperbola;b: Number;a: Number;H: Circle;M: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + y^2 = a^2);RightFocus(G)=F;TangentOfPoint(F,H)=LineOf(F,M);TangentPoint(LineOf(F,M),H)=M;Intersection(LineOf(F,M), yAxis)=P;MidPoint(LineSegmentOf(F, P)) = M",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[61, 64]], [[107, 111]], [[1, 57], [129, 132]], [[4, 57]], [[4, 57]], [[65, 85]], [[96, 99], [114, 117]], [[4, 57]], [[4, 57]], [[1, 57]], [[65, 85]], [[1, 64]], [[0, 92]], [[65, 100]], [[88, 111]], [[114, 127]]]",
    "query_spans": "[[[129, 138]]]",
    "process": "From the given condition, we have $ F(c,0) $. Since the point of tangency $ M $ is the midpoint of segment $ FP $, it follows that triangle $ FPO $ is an isosceles right triangle. Therefore, point $ P(0,c) $. By the midpoint formula, we obtain $ M\\left(\\frac{c}{2},\\frac{c}{2}\\right) $. Substituting $ M\\left(\\frac{c}{2},\\frac{c}{2}\\right) $ into the equation of the circle gives $ \\frac{c^{2}}{4}+\\frac{c^{2}}{4}=a^{2} $, so $ \\frac{c}{a}=\\sqrt{2}=\\sqrt{2} $."
  },
  {
    "text": "The hyperbola $\\frac{y^{2}}{9}-\\frac{x^{2}}{4}=1$, what are the coordinates of the foci?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/4 + y^2/9 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, pm*sqrt(13))",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 45]]]",
    "process": "From the hyperbola equation $\\frac{y^{2}}{9}-\\frac{x^{2}}{4}=1$, we obtain $a=3$, $b=2$, $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{13}$. Since the foci of this hyperbola lie on the $y$-axis, the coordinates of the foci are $(0,\\pm\\sqrt{13})$."
  },
  {
    "text": "Given that $F$ is a focus of the ellipse $C$, $B$ is an endpoint of the minor axis, the extension of the line segment $BF$ intersects the ellipse $C$ at point $D$, and $\\overrightarrow{B F}+2 \\overrightarrow{D F}=\\overrightarrow{0}$, then the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;F: Point;B: Point;D: Point;OneOf(Focus(C)) = F;OneOf(Endpoint(MinorAxis(C)))=B;Intersection(OverlappingLine(LineSegmentOf(B,F)), C) = D;VectorOf(B, F) + 2*VectorOf(D, F) = 0",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[6, 11], [41, 46], [119, 124]], [[2, 5]], [[17, 20]], [[47, 51]], [[2, 16]], [[6, 28]], [[29, 51]], [[53, 117]]]",
    "query_spans": "[[[119, 130]]]",
    "process": "According to the problem, assume the foci of the ellipse lie on the x-axis, $ F(c,0) $, $ B(0,b) $, and let $ D(x,y) $. From $ \\overrightarrow{BF} + 2\\overrightarrow{DF} = \\overrightarrow{0} $, solve for the coordinates of point $ D $. Substitute into the ellipse equation and simplify to find the eccentricity. Assume the ellipse has its foci on the x-axis with equation $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $), $ F(c,0) $, $ B(0,b) $, and let $ D(x,y) $. From $ \\overrightarrow{BF} + 2\\overrightarrow{DF} = \\overrightarrow{0} $, and since $ \\overrightarrow{BF} = (c,-b) $, $ \\overrightarrow{DF} = (x-c,y) $, we obtain $ x = \\frac{3c}{2} $, $ y = -\\frac{1}{2}b $, so $ D\\left(\\frac{3c}{2}, -\\frac{1}{2}b\\right) $. Since point $ D $ lies on the ellipse, $ \\frac{\\left(\\frac{3c}{2}\\right)^{2}}{a^{2}} + \\frac{\\left(-\\frac{1}{2}b\\right)^{2}}{b^{2}} = 1 $, simplifying gives $ \\frac{c^{2}}{a^{2}} = \\frac{1}{3} $, hence the eccentricity $ e = \\frac{c}{a} = \\frac{\\sqrt{3}}{3} $."
  },
  {
    "text": "Given the parabola $y=\\frac{1}{4} x^{2}$ with focus $F$ and point $A(-1 , 7)$. Let $P$ be a point on the parabola. Then the minimum value of $|PA|+|PF|$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = (1/4)*x^2);F: Point;Focus(G) = F;A: Point;Coordinate(A) = (-1, 7);P: Point;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 26], [50, 53]], [[2, 26]], [[29, 32]], [[2, 32]], [[33, 45]], [[33, 45]], [[46, 49]], [[46, 57]]]",
    "query_spans": "[[[59, 76]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ and the ellipse $\\frac{x^{2}}{169}+\\frac{y^{2}}{144}=1$ have the same foci $F_{1}$, $F_{2}$, and point $P$ is an intersection point of the hyperbola and the ellipse, then $\\angle F_{1} P F_{2}$=?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/16 - y^2/9 = 1);Expression(H) = (x^2/169 + y^2/144 = 1);Focus(G) = {F1, F2};Focus(H) = {F1, F2};OneOf(Intersection(G,H))=P",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "arccos15/17",
    "fact_spans": "[[[2, 41], [110, 113]], [[42, 83], [114, 116]], [[89, 96]], [[105, 109]], [[97, 104]], [[2, 41]], [[42, 83]], [[2, 104]], [[2, 104]], [[105, 121]]]",
    "query_spans": "[[[123, 147]]]",
    "process": "The hyperbola $\\frac{x^2}{16}-\\frac{y^{2}}{9}=1$ and the ellipse $\\frac{x^2}{169}+\\frac{y^{2}}{144}=1$ have the same foci $F_{1}, F_{2}$. Let $P$ be their intersection point in the first quadrant. By the definition of the ellipse, we have $|PF_{1}|+|PF_{2}|=26$, and by the definition of the hyperbola, we have $|PF_{1}|-|PF_{2}|=8$. Therefore, $|PF_{1}|=17$, $|PF_{2}|=9$. Since $|F_{1}F_{2}|=2c=10$, by the cosine theorem, we obtain $\\cos\\angle F_{1}PF_{2}=\\frac{|PF_{1}|^{2}+|PF_{2}|^{2}-|F_{1}F_{2}|^{2}}{2\\cdot|PF_{1}|\\cdot|PF_{2}|}=\\frac{17^{2}+9^{2}-10^{2}}{2\\times17\\times9}=\\frac{15}{17}$. Hence, $\\angle F_{1}PF_{2}=\\arccos\\frac{15}{17}$."
  },
  {
    "text": "Ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has left and right foci $F_{1}$, $F_{2}$ respectively. Point $P$ is a point on ellipse $C$, $\\cos \\angle F_{1} P F_{2}=\\frac{11}{14}$, $S_{\\Delta F_{1} P F_{2}}=\\sqrt{3}$, then the length of the minor axis of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C) = True;Cos(AngleOf(F1, P, F2)) = 11/14;Area(TriangleOf(F1, P, F2)) = sqrt(3)",
    "query_expressions": "Length(MinorAxis(C))",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[0, 57], [87, 92], [177, 182]], [[0, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[66, 73]], [[74, 81]], [[0, 81]], [[0, 81]], [[82, 86]], [[82, 95]], [[96, 137]], [[140, 175]]]",
    "query_spans": "[[[177, 188]]]",
    "process": "The ellipse $ C: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $) has left and right foci $ F_{1} $, $ F_{2} $ respectively. Point $ P $ is on the ellipse $ C $, $ \\cos\\angle F_{1}PF_{2} = \\frac{11}{14} $, $ S_{\\Delta F_{1}PF_{2}} = \\sqrt{3} $, $ F_{1}P = m $, $ PF_{2} = n $. We obtain: $ 4c^{2} = m^{2} + n^{2} - 2mn\\cos\\angle F_{1}PF_{2} $, $ \\frac{1}{2}mn\\sin\\angle F_{1}PF_{2} = \\sqrt{3} $, $ m + n = 2a $, $ a^{2} = c^{2} + b^{2} $. Solving gives: $ b = \\sqrt{5} $. Thus, the minor axis length of ellipse $ C $ is $ 2\\sqrt{5} $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has eccentricity $\\frac{\\sqrt{5}}{5}$, $F$ is the right focus of the ellipse, $A$ is a moving point on the ellipse, line $l$: $b x+a y-a^{2}-b^{2}=0$, denote the distance from point $A$ to line $l$ as $d$, then what is the minimum value of $d-|A F|$? (express in terms of $a$ or $b$)",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Eccentricity(G) = sqrt(5)/5;F: Point;RightFocus(G) = F;A: Point;PointOnCurve(A, G);l: Line;Expression(l) = (b*x + a*y - a^2 - b^2 = 0);d: Number;Distance(A, l) = d",
    "query_expressions": "Min(d-Abs(LineSegmentOf(A, F)))",
    "answer_expressions": "{(11*sqrt(5)/15-2)*a, (11/6-sqrt(5))*b}",
    "fact_spans": "[[[2, 54], [85, 87], [96, 98]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[2, 79]], [[81, 84]], [[81, 91]], [[92, 95], [137, 141]], [[92, 104]], [[105, 135], [142, 147]], [[105, 135]], [[151, 154]], [[137, 154]]]",
    "query_spans": "[[[156, 171]]]",
    "process": "Based on the eccentricity of the ellipse, solve for the relationship among parameters a, b, c; then simplify d - |AF| according to the definition of the ellipse; finally, find the minimum value using plane geometric relationships. [Detailed Solution] According to the given conditions, the eccentricity of the ellipse is e = \\frac{c}{a} = \\frac{\\sqrt{5}}{5}, and a^{2} = b^{2} + c^{2}. \\therefore c = \\frac{\\sqrt{5}}{5}a, b = \\frac{2\\sqrt{5}}{5}a. Let the left focus of the ellipse be F_{1}(-c, 0), then its coordinates can be written as F_{1}\\left(-\\frac{\\sqrt{5}}{5}a, 0\\right). According to the definition of the ellipse, |AF_{1}| + |AF| = 2a. \\therefore |AF| = 2a - |AF_{1}|. \\therefore d - |AF| = d + |AF_{1}| - 2a. Based on plane geometry, the minimum value of d + |AF_{1}| is the distance from point F_{1} to line l. Therefore, the minimum value of d - |AF| is: \\frac{|-bc + a \\cdot 0 - a^{2} - b^{2}|}{\\sqrt{a^{2} + b^{2}}} - 2a. Simplifying the calculation yields \\frac{11\\sqrt{5}}{115} - 2|a|, which can be expressed in terms of b as \\left(\\frac{11}{6} - \\sqrt{5}\\right)b."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) with right focus $F$. A perpendicular is drawn from $F$ to an asymptote of the hyperbola $C$, and the foot of the perpendicular is $A$. If the area of $\\triangle O A F$ is $\\frac{a^{2}+b^{2}}{5}$ ($O$ being the origin), then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(C) = F;L: Line;PointOnCurve(F, L);IsPerpendicular(L, OneOf(Asymptote(C)));A: Point;FootPoint(L, OneOf(Asymptote(C))) = A;O: Origin;Area(TriangleOf(O, A, F)) = (a^2+b^2)/5",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "{sqrt(5), sqrt(5)/2}",
    "fact_spans": "[[[2, 64], [77, 83], [158, 164]], [[2, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[73, 76], [68, 71]], [[2, 71]], [], [[72, 92]], [[72, 92]], [[96, 99]], [[72, 99]], [[146, 149]], [[101, 145]]]",
    "query_spans": "[[[158, 170]]]",
    "process": "Let the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) have its right focus at $ F(c, 0) $. Let one asymptote of hyperbola $ C $ be $ bx + ay = 0 $. Then $ |AF| = \\frac{|bc|}{\\sqrt{b^{2} + a^{2}}} = b $, $ |OA| = \\sqrt{c^{2} - b^{2}} = a $. The area of quadrilateral $ AOAF $ is $ \\frac{a^{2} + b^{2}}{5} $, so $ \\frac{1}{2}ab = \\frac{a^{2} + b^{2}}{5} $. Simplifying yields $ b = 2a $ or $ b = \\frac{1}{2}a $. Since $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} $, it follows that $ e = \\sqrt{5} $ or $ \\frac{\\sqrt{5}}{2} $."
  },
  {
    "text": "A point $(2, t)$ on the parabola $y^{2}=2 p x$ is at a distance of $3$ from the point $(\\frac{p}{2}, 0)$. Then $t = $?",
    "fact_expressions": "G: Parabola;p: Number;H: Point;I: Point;t: Number;Expression(G) = (y^2 = 2*(p*x));Coordinate(H) = (2, t);Coordinate(I) = (p/2, 0);PointOnCurve(H, G);Distance(H, I) = 3",
    "query_expressions": "t",
    "answer_expressions": "pm*2*sqrt(2)",
    "fact_spans": "[[[0, 16]], [[29, 47]], [[19, 27]], [[28, 47]], [[57, 60]], [[0, 16]], [[19, 27]], [[28, 47]], [[0, 27]], [[19, 55]]]",
    "query_spans": "[[[57, 62]]]",
    "process": "From the definition of the parabola, we obtain $2+\\frac{p}{2}=3$, solving for $p$ gives $p=2$, hence $t^{2}=2\\times2\\times2$, solving for $t$ gives $t=\\pm2\\sqrt{2}$"
  },
  {
    "text": "Given the circle $C$: $x^{2}+(y-4)^{2}=4$ is tangent to the asymptotes of the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "E: Hyperbola;b: Number;a: Number;C: Circle;a>0;b>0;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);Expression(C) = (x^2 + (y - 4)^2 = 4);IsTangent(C, Asymptote(E))",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "2",
    "fact_spans": "[[[28, 89], [97, 100]], [[35, 89]], [[35, 89]], [[2, 27]], [[35, 89]], [[35, 89]], [[28, 89]], [[2, 27]], [[2, 95]]]",
    "query_spans": "[[[97, 106]]]",
    "process": "From the given condition, the asymptote equation of the hyperbola is $ bx - ay = 0 $, so $ \\frac{|-4a|}{\\sqrt{b^{2}+a^{2}}} = 2 $, $ \\therefore 4a = 2c $, $ \\therefore e = 2 $."
  },
  {
    "text": "Given that point $P$ lies on the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F_{1}$ is the left focus of the ellipse, and the midpoint of segment $P F_{1}$ lies on the circle $x^{2}+y^{2}=a^{2}-b^{2}$. Let $k$ denote the slope of line $P F_{1}$. If $k \\geq 1$, then the minimum value of the eccentricity of the ellipse is?",
    "fact_expressions": "C: Ellipse;a: Number;b: Number;G: Circle;P: Point;F1: Point;a > b;b > 0;k:Number;k>=1;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = a^2 - b^2);PointOnCurve(P, C);LeftFocus(C) = F1;PointOnCurve(MidPoint(LineSegmentOf(P,F1)), G);Slope(LineOf(P,F1)) = k",
    "query_expressions": "Min(Eccentricity(C))",
    "answer_expressions": "sqrt(2)-1",
    "fact_spans": "[[[7, 64], [74, 76], [157, 159]], [[14, 64]], [[14, 64]], [[96, 122]], [[2, 6]], [[66, 73]], [[14, 64]], [[14, 64]], [[140, 143]], [[145, 155]], [[7, 64]], [[96, 122]], [[2, 65]], [[66, 80]], [[81, 123]], [[125, 143]]]",
    "query_spans": "[[[157, 168]]]",
    "process": "Let the midpoint of segment $ PF_{1} $ be $ Q $, and connect $ OQ $. Since $ Q $ lies on the circle $ x^{2} + y^{2} = a^{2} - b^{2} = c^{2} $, we have $ OQ = c $. Since $ O $ is the midpoint of segment $ F_{1}F_{2} $, it follows that $ OQ \\parallel PF_{2} $, and $ PF_{2} = 2OQ = 2c $, so $ PF_{1} = 2a - 2c $. Let $ \\angle PF_{1}F_{2} = \\theta $, then $ k = \\tan\\theta \\geqslant 1 $, so $ \\theta \\in [\\frac{\\pi}{4}, \\frac{\\pi}{2}) $, $ \\cos\\theta \\in (0, \\frac{\\sqrt{2}}{2}] $. In triangle $ PF_{1}F_{2} $, by the law of cosines:\n\n$$\n\\frac{|PF_{1}|^{2} + |F_{1}F_{2}|^{2} - |PF_{2}|^{2}}{2|PF_{1}| \\cdot |F_{1}F_{2}|} = \\frac{(2a - 2c)^{2} + 4c^{2} - 4c^{2}}{2(2a - 2c) \\cdot 2c} = \\frac{(a - c)^{2}}{2(a - c) \\cdot c}\n$$\n\nSince $ a > c $, $ a - c > 0 $, so $ 0 < (a - c)^{2} \\leqslant \\sqrt{2}(a - c) \\cdot c $, thus $ 0 < (1 - \\frac{c}{a})^{2} \\leqslant \\sqrt{2}(1 - \\frac{c}{a}) \\cdot \\frac{c}{a} $, so $ 0 < (1 - e)^{2} \\leqslant \\sqrt{2}(1 - e) \\cdot e $. Since $ 0 < e < 1 $, the left side of the inequality holds; for the right side, $ (1 - e)^{2} \\leqslant \\sqrt{2}(1 - e) \\cdot e $, which can be rewritten as:\n\n$$\n(1 + \\sqrt{2})e^{2} - (2 + \\sqrt{2})e + 1 \\leqslant 0\n$$\n\n$$\n[(1 + \\sqrt{2})e - 1](e - 1) \\leqslant 0\n$$\n\nSolving gives $ \\sqrt{2} - 1 \\leqslant e \\leqslant 1 $, so the minimum value of $ e $ is $ \\sqrt{2} - 1 $."
  },
  {
    "text": "The left and right foci of the hyperbola $16 x^{2}-9 y^{2}=144$ are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the hyperbola, and $|P F_{1}| \\cdot |P F_{2}| = 64$. Find $\\angle F_{1} P F_{2}$=?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (16*x^2 - 9*y^2 = 144);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 64",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "ApplyUnit(60, degree)",
    "fact_spans": "[[[0, 25], [56, 59]], [[51, 55]], [[35, 42]], [[43, 50]], [[0, 25]], [[0, 50]], [[0, 50]], [[51, 60]], [[62, 91]]]",
    "query_spans": "[[[93, 117]]]",
    "process": ""
  },
  {
    "text": "Given point $P(2,-2)$ and parabola $C$: $y=\\frac{1}{4} x^{2}$, a line passing through the focus of parabola $C$ with slope $k$ intersects $C$ at points $A$ and $B$. If $\\overrightarrow{P A} \\cdot \\overrightarrow{P B}=25$, then $k=?$",
    "fact_expressions": "C: Parabola;G: Line;P: Point;A: Point;B: Point;k: Number;Expression(C) = (y = x^2/4);Coordinate(P) = (2, -2);PointOnCurve(Focus(C), G);Slope(G) = k;Intersection(G, C) = {A, B};DotProduct(VectorOf(P, A), VectorOf(P, B)) = 25",
    "query_expressions": "k",
    "answer_expressions": "{-1, 2}",
    "fact_spans": "[[[13, 42], [44, 50], [64, 67]], [[61, 63]], [[2, 12]], [[69, 72]], [[73, 76]], [[57, 60], [135, 138]], [[13, 42]], [[2, 12]], [[43, 63]], [[54, 63]], [[61, 78]], [[81, 133]]]",
    "query_spans": "[[[135, 140]]]",
    "process": "First, obtain the standard equation of parabola C and the coordinates of its focus. Assume the equation of the line, then solve it simultaneously with the parabola's equation to express Vieta's formulas, obtaining $x_{1}+x_{2}$, $x_{1}x_{2}$, $y_{1}+y_{2}$, $y_{1}y_{2}$. According to $\\overrightarrow{PA}\\cdot\\overrightarrow{PB}=25$, construct an equation in terms of $k$ using the dot product operation, then solve the equation to find the result. (Detailed solution) From the given information, the standard equation of parabola C is: $x^{2}=4y$, $\\therefore$ the focus coordinates are: $(0,1)$. Let the equation of line AB be: $y=kx+1$. From $\\begin{cases}y=kx+1\\\\x^{2}=4y\\end{cases}$, we get: $x^{2}-4kx-4=0$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $\\Delta=16k^{2}+16>0$, $x_{1}+x_{2}=4k$, $x_{1}x_{2}=-4$. $\\therefore y_{1}+y_{2}=k(x_{1}+x_{2})+2=4k^{2}+2$, $y_{1}y_{2}=\\frac{1}{4}x_{1}^{2}\\cdot\\frac{1}{4}x_{2}^{2}=\\frac{1}{16}(x_{1}x_{2})^{2}=1$. Also, $\\overrightarrow{PA}=(x_{1}-2,y_{1}+2)$, $\\overrightarrow{PB}=(x_{2}-2,y_{2}+2)$. $\\therefore \\overrightarrow{PA}\\cdot\\overrightarrow{PB}=(x_{1}-2)(x_{2}-2)+(y_{1}+2)(y_{2}+2)=25$, that is, $x_{1}x_{2}-2(x_{1}+x_{2})+4+y_{1}y_{2}+2(y_{1}+y_{2})+4=-8k+8k^{2}+9=25$. Solving gives: $k=-1$ or $2$. The correct result for this problem: $-1$ or $2$."
  },
  {
    "text": "What is the sum of the distances from any point on the ellipse $\\frac{x^{2}}{11}+\\frac{y^{2}}{6}=1$ to its two foci?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/11 + y^2/6 = 1);H: Point;PointOnCurve(H, G);F1: Point;F2: Point;Focus(G) = {F1, F2}",
    "query_expressions": "Distance(H, F1) + Distance(H, F2)",
    "answer_expressions": "2*sqrt(11)",
    "fact_spans": "[[[0, 38]], [[0, 38]], [], [[0, 43]], [], [], [[0, 47]]]",
    "query_spans": "[[[0, 54]]]",
    "process": "Since $a^{2}=11$, the sum of the distances from any point on the ellipse $\\frac{x^{2}}{11}+\\frac{y^{2}}{6}=1$ to the two foci is $2a=2\\sqrt{11}$."
  },
  {
    "text": "If the line $l$: $x - y + m = 0$ intersects the ellipse $x^{2} + \\frac{y^{2}}{2} = 1$ at points $A$ and $B$, and the midpoint of segment $AB$ lies on the circle $x^{2} + y^{2} = 1$, then $m = $?",
    "fact_expressions": "l: Line;G: Ellipse;H: Circle;B: Point;A: Point;Expression(G) = (x^2 + y^2/2 = 1);Expression(H) = (x^2 + y^2 = 1);Expression(l) = (m + x - y = 0);Intersection(l, G) = {A, B};PointOnCurve(MidPoint(LineSegmentOf(A,B)), H);m:Number",
    "query_expressions": "m",
    "answer_expressions": "pm*3*sqrt(5)/5",
    "fact_spans": "[[[1, 16]], [[17, 44]], [[68, 84]], [[50, 53]], [[46, 49]], [[17, 44]], [[68, 84]], [[1, 16]], [[1, 55]], [[57, 85]], [[87, 90]]]",
    "query_spans": "[[[87, 92]]]",
    "process": "Let points $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Solving the system \n\\[\n\\begin{cases}\ny = x + m \\\\\n2x^{2} + y^{2} = 2\n\\end{cases}\n\\]\nyields $ 3x^{2} + 2mx + m^{2} - 2 = 0 $. The discriminant is $ \\Delta = 4m^{2} - 12(m^{2} - 2) = 24 - 8m^{2} > 0 $, solving gives $ -\\sqrt{3} < m < \\sqrt{3} $. By Vieta's formulas, $ x_{1} + x_{2} = -\\frac{2m}{3} $, then $ y_{1} + y_{2} = x_{1} + x_{2} + 2m = \\frac{4m}{3} $. Therefore, the midpoint of segment $ AB $ is $ M(-\\frac{m}{3}, \\frac{2m}{3}) $. According to the problem, $ (-\\frac{m}{3})^{2} + (\\frac{2m}{3})^{2} = 1 $, solving gives $ m = \\pm \\frac{3\\sqrt{5}}{5} $."
  },
  {
    "text": "If a focus of the ellipse $m x^{2}+y^{2}=1$ coincides with the focus of the parabola $y^{2}=4 x$, then $m=$?",
    "fact_expressions": "G: Parabola;H: Ellipse;m: Number;Expression(G) = (y^2 = 4*x);Expression(H) = (m*x^2 + y^2 = 1);OneOf(Focus(H)) = Focus(G)",
    "query_expressions": "m",
    "answer_expressions": "1/2",
    "fact_spans": "[[[26, 40]], [[1, 20]], [[47, 50]], [[26, 40]], [[1, 20]], [[1, 45]]]",
    "query_spans": "[[[47, 52]]]",
    "process": ""
  },
  {
    "text": "The right focus of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$ is $F$, the right directrix is $l$, and from the upper vertex $A$ of the ellipse, draw $A M \\perp l$, with foot of perpendicular at $M$. Then the slope of the line $F M$ is?",
    "fact_expressions": "G: Ellipse;F: Point;M: Point;A: Point;l: Line;Expression(G) = (x^2/5 + y^2/4 = 1);RightFocus(G) = F;RightDirectrix(G) = l;UpperVertex(G)=A;PointOnCurve(A,LineSegmentOf(A,M));IsPerpendicular(LineSegmentOf(A,M),l);FootPoint(LineSegmentOf(A,M),l)=M",
    "query_expressions": "Slope(LineOf(F, M))",
    "answer_expressions": "1/2",
    "fact_spans": "[[[0, 37], [55, 57]], [[42, 45]], [[81, 84]], [[60, 63]], [[50, 53]], [[0, 37]], [[0, 45]], [[0, 53]], [[55, 63]], [[54, 77]], [[64, 77]], [[64, 84]]]",
    "query_spans": "[[[86, 98]]]",
    "process": "The right focus is $F(1,0)$, and $A(0,2)$, while $l:x=5$, so $M(5,2)$, hence $k_{FM}=\\frac{2-0}{5-1}=\\frac{1}{2}$, fill in $\\frac{1}{2}$"
  },
  {
    "text": "Given the hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has an eccentricity of $2$. If the distance from the focus of the parabola $C_{2}$: $x^{2}=2py$ $(p>0)$ to the asymptotes of the hyperbola $C_{1}$ is $2$, then what is the equation of the parabola $C_{2}$?",
    "fact_expressions": "C1: Hyperbola;Expression(C1) = ((x^2/a^2) - (y^2/b^2) = 1);a: Number;b: Number;a>0;Eccentricity(C1)=2;C2: Parabola;Expression(C2)=(x^2=2*p*y);p:Number;p>0;Distance(Focus(C2),Asymptote(C1))=2;b>0",
    "query_expressions": "Expression(C2)",
    "answer_expressions": "x^2=16*y",
    "fact_spans": "[[[2, 68], [112, 122]], [[2, 68]], [[13, 68]], [[13, 68]], [[13, 68]], [[2, 76]], [[79, 108], [135, 145]], [[79, 108]], [[91, 108]], [[91, 108]], [[79, 133]], [[13, 68]]]",
    "query_spans": "[[[135, 150]]]",
    "process": ""
  },
  {
    "text": "Given point $M(4,0)$, point $P$ moves on the curve $y^{2}=8x$, and point $Q$ moves on the curve $(x-2)^{2}+y^{2}=1$, then the minimum value of $\\frac{|PM|^{2}}{|PQ|}$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (4, 0);P: Point;PointOnCurve(P, G);G: Curve;Expression(G) = (y^2 = 8*x);Q: Point;G1:Curve;Expression(G1) = (y^2 + (x - 2)^2 = 1);PointOnCurve(Q, G1)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, M))^2/Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 11]], [[2, 11]], [[12, 16]], [[12, 33]], [[17, 30]], [[17, 30]], [[34, 38]], [[39, 60]], [[39, 60]], [[34, 63]]]",
    "query_spans": "[[[65, 96]]]",
    "process": ""
  },
  {
    "text": "Given that $P(x, y)$ is a point on the parabola $y^{2}=8 x$, then the maximum value of $\\sqrt{(x-4)^{2}+(y-1)^{2}}-x$ is?",
    "fact_expressions": "G: Parabola;P: Point;x1:Number;y1:Number;Expression(G) = (y^2 = 8*x);Coordinate(P) = (x1, y1);PointOnCurve(P, G)",
    "query_expressions": "Max(sqrt((x1-4)^2+(y1-1)^2)-x1)",
    "answer_expressions": "2+sqrt(5)",
    "fact_spans": "[[[12, 26]], [[2, 11]], [[2, 11]], [[2, 11]], [[12, 26]], [[2, 11]], [[2, 29]]]",
    "query_spans": "[[[31, 67]]]",
    "process": "According to the problem, draw the figure as shown; from the figure, we know $\\sqrt{(x-4)^{2}+(y-1)^{2}}$. $x=|PA|-x=|PA|-(|PM|-2)=|PA|-(|PF|-2)=|PA|-|PF|+2\\leqslant|AF|+2=\\sqrt{5}+2$; that is, the maximum value of $\\sqrt{(x-4)^{2}+(y-1)^{2}}-x$ is $\\sqrt{5}+2$."
  },
  {
    "text": "Given that the eccentricity of hyperbola $C$ is $2$, with left and right foci $F_{1}$ and $F_{2}$, and point $A$ lies on $C$. If $|F_{1} A|=2|F_{2} A|$, then $\\cos \\angle A F_{2} F_{1}$=?",
    "fact_expressions": "C: Hyperbola;F1: Point;A: Point;F2: Point;Eccentricity(C) = 2;LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(A, C);Abs(LineSegmentOf(F1, A)) = 2*Abs(LineSegmentOf(F2, A))",
    "query_expressions": "Cos(AngleOf(A, F2, F1))",
    "answer_expressions": "1/4",
    "fact_spans": "[[[2, 8], [44, 47]], [[23, 30]], [[39, 43]], [[31, 38]], [[2, 16]], [[2, 38]], [[2, 38]], [[39, 48]], [[50, 72]]]",
    "query_spans": "[[[74, 103]]]",
    "process": "By the definition of the hyperbola, |F_{1}A| - |F_{2}A| = |F_{2}A| = 2a, then |F_{1}A| = 4a. Since the eccentricity of the hyperbola is 2, |F_{1}F_{2}| = 2c = 4a. In \\triangle AF_{1}F_{2}, \\cos\\angle AF_{2}F_{1} = \\frac{16a^{2} + 4a^{2} - 16a^{2}}{2 \\times 4a \\times 2a} = \\frac{1}{4}; hence fill in \\frac{1}{4}."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$, respectively. If point $P$ lies on this hyperbola and $|P F_{1}|=5$, then $|P F_{2}|=$?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2/2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 5",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "{3, 7}",
    "fact_spans": "[[[19, 47], [61, 64]], [[55, 59]], [[1, 8]], [[9, 16]], [[19, 47]], [[1, 53]], [[1, 53]], [[55, 65]], [[67, 80]]]",
    "query_spans": "[[[82, 95]]]",
    "process": "From the hyperbola equation $ x^{2} - \\frac{y^{2}}{2} = 1 $, we obtain $ a = 1 $. Since point $ P $ lies on the hyperbola, by the definition of a hyperbola, $ ||PF_{1}| - |PF_{2}|| = 2 $. Given $ |PF_{1}| = 5 $, substituting yields $ |PF_{2}| = 3 $ or $ |PF_{2}| = 7 $."
  },
  {
    "text": "Given the curve $C$: $y=\\frac{1}{4} x^{2}$, with focus $F$, and $P$ being any point on the parabola, what is the minimum value of the sum of the distances from point $P$ to the focus $F$ and to the point $A(4,1)$?",
    "fact_expressions": "C: Parabola;A: Point;F: Point;P: Point;Expression(C) = (y = x^2/4);Coordinate(A) = (4, 1);Focus(C)=F;PointOnCurve(P, C)",
    "query_expressions": "Min(Distance(P,F)+Distance(P, A))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 30], [42, 45]], [[64, 73]], [[34, 37], [59, 62]], [[38, 41], [52, 56]], [[2, 30]], [[64, 73]], [[2, 37]], [[38, 50]]]",
    "query_spans": "[[[52, 84]]]",
    "process": "x^{2}=4y, the coordinates of the focus are F(0,1). From the given condition: |PF| + |PA| \\geqslant |AF| = 4. Therefore, the minimum value of the sum of distances from point P to the focus F and to point A(4,1) is 4."
  },
  {
    "text": "Given that one focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is the focus of the parabola $y^{2}=8x$, and the eccentricity of the hyperbola $C$ is $2$, then the equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;G: Parabola;Expression(G) = (y^2 = 8*x);OneOf(Focus(C)) = Focus(G);Eccentricity(C) = 2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 63], [88, 94], [105, 111]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[69, 83]], [[69, 83]], [[2, 86]], [[88, 102]]]",
    "query_spans": "[[[105, 116]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ is a moving point on the parabola $y^{2}=2x$, the projection of point $P$ onto the $y$-axis is $M$, and point $A\\left(\\frac{7}{2},-4\\right)$, then the minimum value of $|PA|+|PM|$ is?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;M: Point;Expression(G) = (y^2 = 2*x);Coordinate(A) = (7/2, -4);PointOnCurve(P, G);Projection(P, yAxis) = M",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "9/2",
    "fact_spans": "[[[7, 21]], [[44, 64]], [[2, 6], [26, 30]], [[40, 43]], [[7, 21]], [[44, 64]], [[2, 25]], [[26, 43]]]",
    "query_spans": "[[[66, 85]]]",
    "process": "The focus of the parabola $ y^{2}=2x $ is $ F(\\frac{1}{2},0) $, and the directrix is $ x=-\\frac{1}{2} $. By the definition of a parabola, we have $ |PM|=|PF|-\\frac{1}{2} $, so $ |PA|+|PM|=|PA|+|PF|-\\frac{1}{2}\\geqslant|AF|-\\frac{1}{2} $. Since $ A(\\frac{7}{2},-4) $, $ F(\\frac{1}{2},0) $, we have $ |AF|=\\sqrt{(\\frac{7}{2}-\\frac{1}{2})^{2}+(-4-0)^{2}}=5 $. Therefore, $ |PA|+|PM|=|PA|+|PF|-\\frac{1}{2}\\geqslant|AF|-\\frac{1}{2}=5-\\frac{1}{2}=\\frac{9}{2} $. The minimum value is attained if and only if points $ A $, $ P $, $ F $ are collinear and $ P $ lies on the segment $ AF $. Thus, the minimum value of $ |PA|+|PM| $ is $ \\frac{9}{2} $, which is: $ \\frac{9}{2} $"
  },
  {
    "text": "The foci of the hyperbola $9 y^{2}-4 x^{2}=36$ are? The eccentricity is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-4*x^2 + 9*y^2 = 36)",
    "query_expressions": "Coordinate(Focus(G));Eccentricity(G)",
    "answer_expressions": "(0,pm*sqrt(13))\nsqrt(13)/2",
    "fact_spans": "[[[0, 23]], [[0, 23]]]",
    "query_spans": "[[[0, 28]], [[0, 33]]]",
    "process": ""
  },
  {
    "text": "Let point $P$ be a point on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$, and let $F_{1}$, $F_{2}$ be the left and right foci of the ellipse respectively, and $I$ the incenter of $\\Delta PF_{1}F_{2}$. If $S_{\\Delta IPF_{1}}+S_{\\Delta IPF_{2}}=2 S_{\\Delta IF_{1}F_{2}}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;I: Point;a>0;b>0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;Incenter(TriangleOf(P,F1,F2))=I;Area(TriangleOf(I,P,F1))+Area(TriangleOf(I,P,F2))=2*Area(TriangleOf(I,F1,F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[6, 64], [86, 88], [86, 88]], [[8, 64]], [[8, 64]], [[1, 5]], [[68, 75]], [[76, 83]], [[95, 98]], [[8, 64]], [[8, 64]], [[6, 64]], [[1, 67]], [[68, 94]], [[68, 94]], [[95, 122]], [[124, 181]]]",
    "query_spans": "[[[184, 192]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the focus of the parabola $y^{2}=2 p x(p>0)$, chord $A B$ passes through $F$, and $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}=-3$, where $O$ is the origin. When the inclination angle of $A B$ is $60^{\\circ}$, $\\tan \\angle A O B$=?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;O: Origin;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;IsChordOf(LineSegmentOf(A,B),G);PointOnCurve(F, LineSegmentOf(A, B));DotProduct(VectorOf(O, A), VectorOf(O, B)) =-3;Inclination(LineSegmentOf(A,B))=ApplyUnit(60,degree)",
    "query_expressions": "Tan(AngleOf(A, O, B))",
    "answer_expressions": "-8*sqrt(3)/9",
    "fact_spans": "[[[6, 27]], [[9, 27]], [[32, 37]], [[32, 37]], [[99, 102]], [[2, 5], [39, 42]], [[9, 27]], [[6, 27]], [[2, 30]], [[6, 37]], [[32, 42]], [[44, 97]], [[109, 132]]]",
    "query_spans": "[[[134, 155]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then the equation of line $ AB $ is $ y=\\sqrt{3}(x-\\frac{p}{2}) $, i.e., $ x=\\frac{y}{\\sqrt{3}}+\\frac{p}{2} $. Substituting this into the parabola equation $ y^{2}=2px $, we obtain $ y^{2}-\\frac{2py}{\\sqrt{3}}-p^{2}=0 $, so $ y_{1}y_{2}=-p^{2} $, $ x_{1}x_{2}=\\frac{(y_{1}y_{2})^{2}}{4p^{2}}=\\frac{p^{2}}{4} $. From $ \\overrightarrow{OA}\\cdot\\overrightarrow{OB}=x_{1}x_{2}+y_{1}y_{2}=-3 $, we get $ -\\frac{3}{4}p^{2}=-3 $, solving gives $ p=2 $. At this time, the equation of line $ AB $ is $ y=\\sqrt{3}(x-1) $, and the parabola equation is $ y^{2}=4x $. Without loss of generality, assume point $ A $ lies in the first quadrant, thus we can solve to get $ A(3,2\\sqrt{3}) $, $ B(\\frac{1}{3},-\\frac{2\\sqrt{3}}{3}) $. Therefore, $ \\tan\\angle AOF=\\frac{2\\sqrt{3}}{3} $, $ \\tan\\angle BOF=2\\sqrt{3} $, hence $ \\tan\\angle AOB=\\tan(\\angle AOF+\\angle BOF)=\\frac{\\frac{2\\sqrt{3}}{3}+2\\sqrt{3}}{1-\\frac{2\\sqrt{3}}{3}\\times2\\sqrt{3}}=-\\frac{8\\sqrt{3}}{9} $."
  },
  {
    "text": "Let the hyperbola $ C $: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 $ ($ a>0, b>0 $) have foci $ F_{1} $ and $ F_{2} $. From a point $ P $ on the hyperbola $ C $, draw perpendiculars to the two asymptotes, with feet of the perpendiculars denoted as $ A $ and $ B $. If $ 3|F_{1} F_{2}|^{2}=64|P A| \\cdot|P B| $, then the eccentricity of the hyperbola $ C $ is?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;F1: Point;F2: Point;P: Point;A: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Focus(C) = {F1, F2};PointOnCurve(P, C);l: Line;l1: Line;l2: Line;Asymptote(C) = {l1, l2};PointOnCurve(P, l);IsPerpendicular(l, l1);IsPerpendicular(l, l2);FootPoint(l, l1) = A;FootPoint(l, l2) = B;3*Abs(LineSegmentOf(F1, F2))^2 = (64*Abs(LineSegmentOf(P, A)))*Abs(LineSegmentOf(P, B))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "{2*sqrt(3)/3, 2}",
    "fact_spans": "[[[1, 62], [84, 90], [160, 166]], [[9, 62]], [[9, 62]], [[67, 74]], [[75, 82]], [[93, 96]], [[110, 113]], [[114, 117]], [[9, 62]], [[9, 62]], [[1, 62]], [[1, 82]], [[84, 96]], [], [], [], [[84, 101]], [[83, 104]], [[83, 104]], [[83, 104]], [[83, 117]], [[83, 117]], [[119, 158]]]",
    "query_spans": "[[[160, 172]]]",
    "process": "From the hyperbola equation, its asymptotes are given by: $ y = \\pm\\frac{b}{a}x $, i.e., $ \\pmb{x} - ay = 0 $. Let $ P(x, y) $, then $ b^{2}x^{2} - a^{2}y^{2} = a^{2}b^{2} $. Therefore, $ |PA| \\cdot |PB| = \\frac{|bx - ay| \\cdot |-b}{a^{2} + b^{2}} \\frac{x - ay|}{c^{2}} = \\frac{|b^{2}x^{2} - a^{2}y^{2}|}{c^{2}} $. Also, $ |F_{1}F_{2}|^{2} = 4c^{2} $. Hence, $ 12c^{2} = \\frac{64a^{2}b^{2}}{c^{2}} = \\frac{64a^{2}(c^{2} - a^{2})}{c^{2}} $, i.e., $ 3c^{4} - 16a^{2}c^{2} + 16a^{4} = 0 $. Therefore, $ 3e^{4} - 16e^{2} + 16 = 0 $, solving gives: $ e^{2} = \\frac{4}{3} $ or $ e^{2} = 4 $. Since $ e > 1 $, $ \\therefore e = \\frac{2\\sqrt{3}}{3} $ or $ 2 $."
  },
  {
    "text": "Given the parabola $y = a x^{2}$ ($a > 0$) with directrix $l$, the chord length obtained by the intersection of $l$ and the circle $C$: $(x - 3)^{2} + y^{2} = 1$ is $\\sqrt{3}$. Then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;a: Number;C: Circle;l: Line;a>0;Expression(G) = (y = a*x^2);Expression(C) = (y^2 + (x - 3)^2 = 1);Directrix(G) = l;Length(InterceptChord(l,C))=sqrt(3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y=x^2/2",
    "fact_spans": "[[[2, 21], [77, 80]], [[5, 21]], [[34, 58]], [[25, 28], [30, 33]], [[5, 21]], [[2, 21]], [[34, 58]], [[2, 28]], [[30, 75]]]",
    "query_spans": "[[[77, 85]]]",
    "process": "Find the equation of the directrix $ l $. Given that the chord length formed by the intersection of $ l $ and the circle $ C: (x-3)^{2} + y^{2} = 1 $ is $ \\sqrt{3} $, find $ a $ and write the equation of the parabola. [Detailed solution] The directrix of the parabola $ y = ax^{2} $ ($ a > 0 $) is $ l: y = -\\frac{1}{4a} $. The circle $ C: (x-3)^{2} + y^{2} = 1 $ has center $ C(3, 0) $. The distance from $ C $ to $ l $ is $ \\therefore \\frac{1}{4a} = \\frac{1}{2} $, solving gives: $ a = \\frac{1}{2} $."
  },
  {
    "text": "Given that point $F$ is the focus of the parabola $y^{2}=-8x$, $O$ is the origin, point $P$ is a moving point on the directrix of the parabola, and point $A$ lies on the parabola such that $|AF|=4$, then the minimum value of $|PA|+|PO|$ is?",
    "fact_expressions": "G: Parabola;A: Point;F: Point;P: Point;O: Origin;Expression(G) = (y^2 = -8*x);Focus(G) = F;PointOnCurve(P,Directrix(G));PointOnCurve(A, G);Abs(LineSegmentOf(A, F)) = 4",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, O)))",
    "answer_expressions": "2*sqrt(13)",
    "fact_spans": "[[[7, 22], [38, 41], [38, 41]], [[48, 52]], [[2, 6]], [[33, 37]], [[26, 29]], [[7, 22]], [[2, 25]], [[33, 47]], [[48, 57]], [[59, 68]]]",
    "query_spans": "[[[70, 89]]]",
    "process": "\\because y^{2} = -8x, \\therefore F(-2,0), the directrix is x = 2. Let A(x_{A}, y_{A}), then -x_{A} + 2 = 4, so x_{A} = -2. Substituting into y^{2} = -8x gives y^{2} = 16. Without loss of generality, take y_{A} = 4, so A(-2,4). Let Q(x',y') be the symmetric point of A about the directrix x = 2, then Q(6,4). Hence |PA| + |PO| \\geqslant |OQ| = \\sqrt{6^{2} + 4^{2}} = 2\\sqrt{13}. Therefore, the minimum value of |PA| + |PO| is 2\\sqrt{13}."
  },
  {
    "text": "Given that there is a moving chord $AB$ of length $9$ on the parabola $y^2 = 8x$, then the shortest distance from the midpoint of $AB$ to the $y$-axis is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);IsChordOf(LineSegmentOf(A, B), G) = True;A: Point;B: Point;Length(LineSegmentOf(A, B)) = 9",
    "query_expressions": "Min(Distance(MidPoint(LineSegmentOf(A, B)), yAxis))",
    "answer_expressions": "5/2",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[2, 33]], [[28, 33]], [[28, 33]], [[20, 33]]]",
    "query_spans": "[[[35, 54]]]",
    "process": "It is easy to see that the directrix of the parabola $ y^{2} = 8x $ is $ l: x = -2 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, and let the midpoint of $ AB $ be $ C(x_{0}, y_{0}) $. Draw perpendiculars from points $ A $, $ B $, $ C $ to the line $ l: x = -2 $, with feet of the perpendiculars denoted as $ M $, $ N $, $ H $, respectively. Then $ |CH| = \\frac{|AM| + |BN|}{2} $. By the definition of a parabola, we obtain $ |MH| = \\frac{|AM| + |BN|}{2} = \\frac{|AF| + |BF|}{2} \\geqslant \\frac{|AB|}{2} = \\frac{9}{2} $ (equality holds if and only if points $ A $, $ B $, $ F $ are collinear)"
  },
  {
    "text": "If the foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ lie on the $x$-axis, and tangents are drawn from the point $(2,1)$ to the circle $x^{2}+y^{2}=4$, with points of tangency $A$ and $B$, respectively, such that the line $AB$ passes exactly through the right focus and the upper vertex of the ellipse, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;PointOnCurve(Focus(G), xAxis);I: Point;Coordinate(I) = (2, 1);L1: Line;L2: Line;TangentOfPoint(I, H) = {L1, L2};H: Circle;Expression(H) = (x^2 + y^2 = 4);TangentPoint(L1, H) = A;TangentPoint(L2, H) = B;A: Point;B: Point;PointOnCurve(RightFocus(G), LineOf(A, B));PointOnCurve(UpperVertex(G), LineOf(A, B))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/20+y^2/16=1",
    "fact_spans": "[[[1, 46], [111, 113], [123, 125]], [[1, 46]], [[3, 46]], [[3, 46]], [[1, 55]], [[57, 65]], [[57, 65]], [], [], [[56, 85]], [[66, 82]], [[66, 82]], [[56, 100]], [[56, 100]], [[91, 94]], [[97, 100]], [[101, 117]], [[101, 121]]]",
    "query_spans": "[[[123, 129]]]",
    "process": "Let M(2,1), and let the center of the circle x^{2}+y^{2}=4 be O. Then AB is the line containing the common chord of the circle x^{2}+y^{2}=4 and the circle with OM as diameter. The equation of the circle with OM as diameter is (x-1)^{2}+(y-\\frac{1}{2})^{2}=\\frac{5}{4}, or equivalently x^{2}+y^{2}-2x-y=0. Subtracting the two circle equations yields the equation of AB: 2x+y=4. The intersections of this line with the coordinate axes are (2,0) and (0,4). Since the foci lie on the x-axis, c=2, b=4, a^{2}=20. Therefore, the equation of the ellipse is \\frac{x^{2}}{20}+\\frac{y^{2}}{16}=1."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, if there exists a point $M$ on $C$ such that $|M F_{1}|+|M F_{2}|=3|M O|$ holds, where $O$ is the origin, then the range of the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;M: Point;F1: Point;F2: Point;O: Origin;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) =F1;RightFocus(C) = F2;PointOnCurve(M, C);Abs(LineSegmentOf(M, F1)) + Abs(LineSegmentOf(M, F2)) = 3*Abs(LineSegmentOf(M, O))",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "[3/2, +\\infty)",
    "fact_spans": "[[[18, 79], [87, 90], [144, 147]], [[26, 79]], [[26, 79]], [[95, 98]], [[2, 9]], [[10, 17]], [[134, 137]], [[26, 79]], [[26, 79]], [[18, 79]], [[2, 85]], [[2, 85]], [[87, 98]], [[101, 129]]]",
    "query_spans": "[[[144, 158]]]",
    "process": "Without loss of generality, assume point M lies on the right branch of the hyperbola. Let M(x,y), then x\\geqslant a. First, we find |MF_{1}|=\\frac{c}{a}x+a, |MF_{2}|=\\frac{c}{a}x-a. From the condition, we obtain |OM|=\\frac{2}{3}\\times\\frac{c}{a}x. Then, according to |OM|^{2}=\\frac{c^{2}}{a^{2}}x^{2}-b^{2}, and using x\\geqslant a, we establish an inequality to obtain the answer. [Solution] Without loss of generality, assume point M lies on the right branch =\\frac{1}{2}=\\sqrt{\\frac{c^{2}}{a^{2}}x^{2}+2cx+a^{2}}=\\sqrt{(\\frac{c}{a}x+a)^{2}}=\\frac{c}{a}x+a. Similarly, we get |MF_{2}|=\\frac{c}{a}x-a. From |MF_{1}|+|MF_{2}|=3|MO|, we have \\frac{c}{a}x+a+\\frac{c}{a}x-a=2\\times\\frac{c}{a}x=3|OM|, so |OM|=\\frac{2}{3}\\times\\frac{c}{a}x. Also, |OM|^{2}=x^{2}+y^{2}=x^{2}+\\frac{b^{2}}{a^{2}}x^{2}-b^{2}=\\frac{c^{2}}{a^{2}}x^{2}-b^{2}. Therefore, \\frac{c^{2}}{a^{2}}x^{2}-b^{2}=\\frac{4}{9}\\times\\frac{c^{2}}{a^{2}}x^{2}, i.e., \\frac{5}{9}\\times\\frac{c^{2}}{a^{2}}x^{2}=b^{2}, i.e., \\frac{5}{9}\\times x^{2}=b^{2}\\times\\frac{9a^{2}}{5c^{2}}\\geqslant a^{2}. Thus, 9b^{2}\\geqslant 5c^{2}, i.e., 9(c^{2}-a^{2})\\geqslant 5c^{2}, i.e., 4c^{2}\\geqslant 9a^{2}, i.e., \\frac{c^{2}}{a^{2}}\\geqslant\\frac{9}{4}. Hence, e^{2}\\geqslant\\frac{9}{4}, i.e., e\\geqslant\\frac{3}{2}."
  },
  {
    "text": "The asymptotes of the hyperbola are $y=\\pm 2 x$. Then, the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*(2*x))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{sqrt(3), sqrt(6)/2}",
    "fact_spans": "[[[0, 3], [22, 25]], [[0, 19]]]",
    "query_spans": "[[[22, 31]]]",
    "process": ""
  },
  {
    "text": "Given a hyperbola $ C $ centered at the origin with foci on the $ x $-axis, passing through the point $ P(2 , \\sqrt{3}) $ and having an eccentricity of $ 2 $, then the standard equation of the hyperbola $ C $ is?",
    "fact_expressions": "C: Hyperbola;O:Origin;Center(C)=O;PointOnCurve(Focus(C),xAxis);P: Point;Coordinate(P) = (2, sqrt(3));PointOnCurve(P,C);Eccentricity(C) = 2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/3 - y^2/9 = 1",
    "fact_spans": "[[[16, 22], [52, 58]], [[5, 7]], [[2, 22]], [[8, 22]], [[24, 42]], [[24, 42]], [[16, 42]], [[16, 50]]]",
    "query_spans": "[[[52, 65]]]",
    "process": ""
  },
  {
    "text": "$P$ is a point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, and $F_{1}$, $F_{2}$ are its two foci. Then the difference between the maximum and minimum values of $|P F_{1}| \\cdot |P F_{2}|$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2}",
    "query_expressions": "Max(Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)))-Min(Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)))",
    "answer_expressions": "5",
    "fact_spans": "[[[4, 41]], [[4, 41]], [[0, 3]], [[0, 44]], [[45, 52]], [[53, 60]], [[4, 65]]]",
    "query_spans": "[[[67, 105]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{4}-y^{2}=1$, $P$ is an arbitrary point on $C$, and the coordinates of point $A$ are $(3 , 0)$. Then the minimum value of $|P A|$ is?",
    "fact_expressions": "C: Hyperbola;P: Point;A: Point;Expression(C) = (x^2/4 - y^2 = 1);Coordinate(A) = (3, 0);PointOnCurve(P, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "2*sqrt(5)/5",
    "fact_spans": "[[[2, 35], [41, 44]], [[37, 40]], [[52, 56]], [[2, 35]], [[52, 69]], [[37, 50]]]",
    "query_spans": "[[[71, 84]]]",
    "process": "Let P(x,y), then \\frac{x^{2}}{4}-y^{2}=1, so |PA|=\\sqrt{(x-3)^{2}+y^{2}}=\\sqrt{(x-3)^{2}+\\frac{x^{2}}{4}-1}=\\sqrt{\\frac{5x^{2}}{4}-6x+8}=\\sqrt{\\frac{5}{4}(x-\\frac{12}{5})^{2}+\\frac{4}{5}}. Therefore, when x=\\frac{12}{5}, |PA| takes the minimum value of \\sqrt{\\frac{4}{5}}=\\frac{2\\sqrt{5}}{5}."
  },
  {
    "text": "If real numbers $x$, $y$ satisfy the equation $\\sqrt{x^{2}+(y+3)^{2}}+\\sqrt{x^{2}+(y-3)^{2}}=10$, then the range of values of $\\sqrt{(x-1)^{2}+y^{2}}+\\sqrt{x^{2}+(y-3)^{2}}$ is?",
    "fact_expressions": "x_: Real;y_: Real;sqrt(x_^2 + (y_ + 3)^2) + sqrt(x_^2 + (y_ - 3)^2) = 10",
    "query_expressions": "Range(sqrt(x_^2 + (y_ - 3)^2) + sqrt(y_^2 + (x_ - 1)^2))",
    "answer_expressions": "[10 - sqrt(10), 10 + sqrt(10)]",
    "fact_spans": "[[[1, 6]], [[9, 12]], [[1, 66]]]",
    "query_spans": "[[[68, 122]]]",
    "process": "From the given conditions, point P lies on an ellipse with foci at F_{1}(0,3) and F_{2}(0,-3), and major axis length 10. Then, by the distance formula between two points and the definition of an ellipse, the problem is transformed into finding the range of d = 10 + |PA| - |PF_{2}|. The range can be obtained using the distance formula. Let point P(x,y). According to the definition of the ellipse, point P lies on the ellipse with foci F_{1}(0,3), F_{2}(0,-3) and major axis length 10. The equation of this ellipse is: \\frac{x^{2}}{16}+\\frac{y^{2}}{25}=1. Meanwhile, \\sqrt{(x-1)^{2}+y^{2}}+\\sqrt{x^{2}+(y-3)^{2}} represents the sum of distances from point P(x,y) to points A(1,0) and F_{1}(0,3), i.e., d = |PA| + |PF_{1}|. By the definition of the ellipse, |PF_{1}| + |PF_{2}| = 2a = 10, so |PF_{1}| = 10 - |PF_{2}|. Therefore, d = |PA| + |PF_{1}| = |PA| + (10 - |PF_{2}|) = 10 + |PA| - |PF_{2}|. Since -|AF_{2}| \\leqslant |PA| - |PF_{2}| \\leqslant |AF_{2}|, and |AF_{2}| = \\sqrt{1^{2}+3^{2}} = \\sqrt{10}, it follows that \\sqrt{10} \\leqslant d = 10 + |PA| - |PF_{2}| \\leqslant 10 + \\sqrt{10}."
  },
  {
    "text": "The asymptotes of hyperbola $C$ passing through the point $(2 \\sqrt{2}, \\sqrt{3})$ are given by $y=\\pm \\frac{\\sqrt{3}}{2} x$. Let $P$ be a point on the right branch of hyperbola $C$, $F$ be the left focus of hyperbola $C$, and point $A(0,3)$. Then the minimum value of $|P A|+|P F|$ is?",
    "fact_expressions": "C: Hyperbola;G: Point;A: Point;P: Point;F: Point;Coordinate(G) = (2*sqrt(2), sqrt(3));Coordinate(A) = (0, 3);PointOnCurve(G,C);Expression(Asymptote(C)) = (y = pm*(x*(sqrt(3)/2)));PointOnCurve(P, RightPart(C));LeftFocus(C) = F",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "8",
    "fact_spans": "[[[27, 33], [90, 96], [74, 80]], [[1, 26]], [[101, 110]], [[70, 73]], [[86, 89]], [[1, 26]], [[101, 110]], [[0, 33]], [[27, 68]], [[69, 85]], [[86, 100]]]",
    "query_spans": "[[[113, 132]]]",
    "process": "From the given condition, the hyperbola equation can be set as: \\frac{x^{2}}{4}-\\frac{y^{2}}{3}=\\lambda. Substituting (2\\sqrt{2},\\sqrt{3}) gives \\lambda=1, so the hyperbola equation is \\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1. Let the right focus of the hyperbola be F_{1}. Since P lies on the right branch of the hyperbola and by the definition of hyperbola, |PF|-|PF_{1}|=2a=4, then |PA|+|PF|=|PA|+(4+|PF_{1}|)=|PA|+|PF_{1}|+4. When point P is the intersection point of segment AF_{1} and the hyperbola, |PA|+|PF|=|PA|+|PF_{1}|+4\\geqslant|AF_{1}|+4=8."
  },
  {
    "text": "Given a fixed point $M(-2,4)$ and the focus $F$ of the parabola $y=\\frac{1}{8} x^{2}$, find a point $P$ on the parabola such that the value of $|PM|+|PF|$ is minimized. Then the coordinates of point $P$ are?",
    "fact_expressions": "G: Parabola;M: Point;P: Point;F: Point;Expression(G) = (y = x^2/8);Coordinate(M) = (-2, 4);Focus(G)=F;PointOnCurve(P,G);WhenMin(Abs(LineSegmentOf(P,M))+Abs(LineSegmentOf(P,F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(-2,1/2)",
    "fact_spans": "[[[14, 38], [46, 49]], [[4, 13]], [[53, 56], [74, 78]], [[41, 44]], [[14, 38]], [[4, 13]], [[14, 44]], [[46, 56]], [[57, 72]]]",
    "query_spans": "[[[74, 83]]]",
    "process": ""
  },
  {
    "text": "It is known that a hyperbola shares foci with the ellipse $\\frac{y^{2}}{25}+\\frac{x^{2}}{9}=1$, and the sum of their eccentricities is $\\frac{24}{5}$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/9 + y^2/25 = 1);Focus(H) = Focus(G);Eccentricity(G) + Eccentricity(H) = 24/5",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 - x^2/15 = 1",
    "fact_spans": "[[[2, 5], [73, 76]], [[6, 44]], [[6, 44]], [[2, 47]], [[48, 71]]]",
    "query_spans": "[[[73, 80]]]",
    "process": "First, find the eccentricity and $ c $ of the ellipse from the ellipse equation, then combine with the eccentricity of the hyperbola to obtain the hyperbola equation. For the ellipse $\\frac{y^{2}}{25}+\\frac{x^{2}}{9}=1$, we have $a_{1}=5$, $b_{1}=3$, $c=\\sqrt{25-9}=4$, $e_{1}=\\frac{c}{a_{1}}=\\frac{4}{5}$. The eccentricity of the hyperbola is $e_{2}=\\frac{c}{a_{2}}=\\frac{4}{a_{2}}$. According to the problem, $\\frac{4}{5}+\\frac{4}{a_{2}}=\\frac{24}{5}$, solving gives $a_{2}=1$, $b_{2}^{2}=16-1=15$. Thus, the hyperbola equation is $y^{2}-\\frac{x^{2}}{15}=1$."
  },
  {
    "text": "If $F$ is the focus of the parabola $y^{2}=2 p x$, $M$ is a point on the parabola, $N$ is the intersection point of the directrix of the parabola and the coordinate axis, and $|M F|=\\frac{7}{2} p$, the area of $\\Delta M N F$ is $4 \\sqrt{6}$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;F: Point;Focus(G) = F;M: Point;PointOnCurve(M, G);N: Point;Intersection(Directrix(G),axis)=N;Abs(LineSegmentOf(M,F))=(7/2)*p;Area(TriangleOf(M,N,F))=4*sqrt(6)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=4*sqrt(2)*x",
    "fact_spans": "[[[5, 21], [29, 32], [40, 43], [109, 112]], [[5, 21]], [[8, 21]], [[1, 4]], [[1, 24]], [[25, 28]], [[25, 35]], [[36, 39]], [[36, 52]], [[54, 75]], [[77, 107]]]",
    "query_spans": "[[[109, 117]]]",
    "process": "From the given conditions: $ F(\\frac{p}{2},0) $, $ N(-\\frac{p}{2},0) $, the equation of the directrix is $ x=-\\frac{p}{2} $. Let $ M(\\frac{y_{0}^{2}}{2p}, y_{0}) $. Since $ |MF| = \\frac{7}{2}p $, we have $ \\frac{y_{0}^{2}}{2p} - (-\\frac{p}{2}) = \\frac{7}{2}p \\Rightarrow y_{0}^{2} = 6p^{2} \\Rightarrow |y_{0}| = \\sqrt{6}p $. Since the area of $ \\triangle MNF $ is $ 4\\sqrt{6} $ and $ FN = p $, we have $ \\frac{1}{2} \\cdot p \\cdot |y_{0}| = 4\\sqrt{6} \\Rightarrow p \\cdot \\sqrt{6}p = 8\\sqrt{6} \\Rightarrow p = 2\\sqrt{2} $. Therefore, the equation of the parabola is $ y^{2} = 4\\sqrt{2}x $."
  },
  {
    "text": "Given that point $E$ lies on the $y$-axis, point $F$ is the focus of the parabola $y^{2}=2 p x(p>0)$, the line $EF$ intersects the parabola at points $M$ and $N$, if point $M$ is the midpoint of segment $EF$ and $|N F|=12$, then $p=$?",
    "fact_expressions": "E: Point;PointOnCurve(E, yAxis);F: Point;Focus(G) = F;G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;M: Point;N: Point;Intersection(LineOf(E, F), G) = {M, N};MidPoint(LineSegmentOf(E, F)) = M;Abs(LineSegmentOf(N, F)) = 12",
    "query_expressions": "p",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 6]], [[2, 12]], [[13, 17]], [[13, 42]], [[18, 39], [51, 54]], [[18, 39]], [[96, 99]], [[21, 39]], [[56, 59], [67, 71]], [[60, 63]], [[43, 65]], [[67, 82]], [[84, 94]]]",
    "query_spans": "[[[96, 101]]]",
    "process": "Let E(0,b), and F(\\frac{p}{2},0). Since M is the midpoint of EF, the coordinates of point M are (\\frac{p}{4},y), then y^{2}=2p\\times\\frac{p}{4}=\\frac{p^{2}}{2}, so M(\\frac{p}{4},\\frac{\\sqrt{2}}{2}p). Also, from \\frac{0+b}{2}=\\frac{\\sqrt{2}}{2}p, it follows that b=\\sqrt{2}p, so E(0,\\sqrt{2}p). The equation of line EF is y=-2\\sqrt{2}x+\\sqrt{2}p. Substituting into y^{2}=2px, we get 4x^{2}-5px+p^{2}=0. Let N(x,y), then x+\\frac{p}{4}=\\frac{5}{4}p, solving gives x=p. By the definition of the parabola: |NF|=p+\\frac{p}{2}=12, solving gives: p=8."
  },
  {
    "text": "Given the hyperbola equation $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, $F_{1}$ and $F_{2}$ are the foci of the hyperbola, point $P$ lies on the hyperbola, and $|P F_{1}| \\cdot |P F_{2}| = 32$, then the measure of $\\angle F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Focus(G)={F1,F2};PointOnCurve(P, G);Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 32",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "pi/2",
    "fact_spans": "[[[2, 5], [64, 67], [76, 79]], [[71, 75]], [[48, 55]], [[56, 63]], [[2, 45]], [[48, 70]], [[71, 80]], [[82, 111]]]",
    "query_spans": "[[[113, 140]]]",
    "process": "By the given condition, the equation of the hyperbola is \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1, from which we obtain a=3, b=4, and then c=\\sqrt{a^{2}+b^{2}}=5. According to the definition of the hyperbola, ||PF_{1}|-|PF_{2}||=6, so |PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}|\\cdot|PF_{2}|=36. Given that |PF_{1}|\\cdot|PF_{2}|=32, it follows that |PF_{1}|^{2}+|PF_{2}|^{2}=100. In \\triangle F_{1}PF_{2}, by the law of cosines, \\cos\\angle F_{1}PF_{2}=\\frac{|PF_{1}|^{2}+|PF_{2}|^{2}-|F_{1}F_{2}|^{2}}{2|PF_{1}|\\cdot|PF_{2}|}=\\frac{100-100}{2\\times32}=0. Since \\angle F_{1}PF_{2}\\in(0,\\pi), it follows that \\angle F_{1}PF_{2}=\\frac{\\pi}{2}."
  },
  {
    "text": "The line passing through the left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ and perpendicular to the $x$-axis intersects the hyperbola at points $M$ and $N$. The circle with $MN$ as diameter passes exactly through the right vertex of the hyperbola. Then the eccentricity of the hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;PointOnCurve(LeftFocus(G), L) = True;IsPerpendicular(L, xAxis) = True;L: Line;Intersection(L, G) = {M, N};M: Point;N: Point;Q: Circle;IsDiameter(LineSegmentOf(M, N), Q) = True;PointOnCurve(RightVertex(G), Q) = True",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 57], [73, 76], [101, 104], [110, 113]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[0, 72]], [[62, 72]], [[70, 72]], [[70, 87]], [[78, 81]], [[82, 85]], [[97, 98]], [[88, 98]], [[97, 108]]]",
    "query_spans": "[[[110, 120]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the ellipse $C$: $\\frac{x^{2}}{4}+y^{2}=1$, and let $P$ be a point on the ellipse $C$ in the first quadrant with the abscissa of $P$ equal to $1$. Then the inradius of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/4 + y^2 = 1);Focus(C) = {F1, F2};PointOnCurve(P,C);Quadrant(P)=1;XCoordinate(P)=1",
    "query_expressions": "Radius(InscribedCircle(TriangleOf(P,F1,F2)))",
    "answer_expressions": "3-(3*sqrt(3)/2)",
    "fact_spans": "[[[17, 49], [59, 64]], [[55, 58], [74, 78]], [[1, 8]], [[9, 16]], [[17, 49]], [[1, 54]], [[55, 73]], [[55, 73]], [[74, 86]]]",
    "query_spans": "[[[88, 119]]]",
    "process": "Since the x-coordinate of point P is 1, the y-coordinate of point P is $ y_{p} = \\frac{\\sqrt{3}}{2} $, so the area of $ \\triangle PF_{1}F_{2} $ is $ S = \\frac{1}{2}|F_{1}F_{2}|\\cdot y_{P} = \\frac{3}{2} $. Let the inradius of $ \\triangle PF_{1}F_{2} $ be $ r $, then $ S = \\frac{1}{2}(|F_{1}F_{2}| + |PF_{1}| + |PF_{2}|) \\times r = (2 + \\sqrt{3})r $, that is, $ (2 + \\sqrt{3})r = \\frac{3}{2} $, so $ r = 3 - \\frac{3\\sqrt{3}}{2} $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$, the focus is $F$, the directrix is $l$, and point $M$ is a point on the parabola $C$. A perpendicular is drawn from point $M$ to the directrix $l$, intersecting $l$ at point $H$. If $|M H|=2$, $\\angle H F M=30^{\\circ}$, then the equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;p: Number;M: Point;H: Point;F: Point;l: Line;l1: Line;p>0;Expression(C) = (y^2 = 2*p*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(M, C);PointOnCurve(M, l1);IsPerpendicular(l,l1);Intersection(l,l1)=H;Abs(LineSegmentOf(M, H)) = 2;AngleOf(H, F, M) = ApplyUnit(30, degree)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2 = 6*x",
    "fact_spans": "[[[2, 28], [48, 54], [121, 127]], [[10, 28]], [[43, 47], [59, 63]], [[78, 82]], [[32, 35]], [[39, 42], [66, 69], [74, 77]], [], [[10, 28]], [[2, 28]], [[2, 35]], [[2, 42]], [[43, 57]], [[58, 72]], [[58, 72]], [[58, 82]], [[84, 93]], [[94, 119]]]",
    "query_spans": "[[[121, 132]]]",
    "process": "Since the distance from any point on a parabola to the focus is equal to the distance to the directrix, |MF| = |MH| = 2. Given ∠HFM = 30°, triangle MHF is an isosceles triangle with a vertex angle of 120°, so |HF| = 2√3. Let the directrix l intersect the x-axis at point Q, then ∠QHF = 60°, so p = |OF| = |HF| sin∠OHF = 2√3 sin60° = 3. Therefore, the equation of the parabola is y² = 6x. The answer is: y² = 6x."
  },
  {
    "text": "Given that $F_{1}$ is the left focus of the ellipse $9 x^{2}+25 y^{2}=225$, $P$ is a moving point on this ellipse, and $A(1,2)$ is a fixed point, then the maximum value of $|P A|+|P F_{1}|$ is?",
    "fact_expressions": "G: Ellipse;A: Point;P: Point;F1: Point;Expression(G) = (9*x^2 + 25*y^2 = 225);Coordinate(A) = (1, 2);LeftFocus(G) = F1;PointOnCurve(P, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F1)))",
    "answer_expressions": "10+sqrt(13)",
    "fact_spans": "[[[10, 34], [44, 46]], [[51, 59]], [[39, 42]], [[2, 9]], [[10, 34]], [[51, 59]], [[2, 38]], [[39, 50]]]",
    "query_spans": "[[[65, 88]]]",
    "process": "According to the problem, the equation of the ellipse is $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, so $a^{2}=25$, $b^{2}=9$, $a=5$, thus $c^{2}=25-9=16$, $c=4$, hence $F_{1}(-4,0)$, $F_{2}(4,0)$. As shown in the figure, according to the definition of the ellipse, we have: $|PA|+|PF_{1}|=2a-|PF_{2}|+|PA|=10+|PA|-|PF_{2}|$. When point $P$ moves to the intersection point $P$ of the extension line of $AF_{2}$ and the ellipse, $|PA|-|PF_{2}|$ reaches its maximum. At this time, $|PA|-|PF_{2}|=|AF_{2}|=\\sqrt{13}$, so the maximum value of $|PA|+|PF|$ is $10+\\sqrt{13}$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{3}=1$, and point $P$ lies on the ellipse. If the midpoint of segment $P F_{1}$ lies on the $y$-axis and $P F_{1}=t P F_{2}$, then what is the value of $t$?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/12 + y^2/3 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);PointOnCurve(MidPoint(LineSegmentOf(P,F1)), yAxis);LineSegmentOf(P, F1) = t*LineSegmentOf(P, F2);t:Number",
    "query_expressions": "t",
    "answer_expressions": "7",
    "fact_spans": "[[[18, 56], [67, 69]], [[2, 9]], [[62, 66]], [[10, 17]], [[18, 56]], [[2, 61]], [[62, 70]], [[73, 93]], [[95, 114]], [[116, 119]]]",
    "query_spans": "[[[116, 123]]]",
    "process": "\\because the origin O is the midpoint of F_{1}F_{2}, \\therefore PF_{2} is parallel to the y-axis, i.e., PF_{2} is perpendicular to the x-axis. \\because c=3, \\therefore |F_{1}F_{2}|=6. Let |PF_{1}|=x, then by the definition of ellipse, |PF_{2}|=4\\sqrt{3}-x. \\therefore (4\\sqrt{3}-x)^{2}+36=x^{2}, solving gives x=\\frac{7\\sqrt{3}}{2}. \\because PF_{2}=\\frac{\\sqrt{3}}{2}, {|PF_{1}|=|PF_{2}|, [Note] This problem mainly examines geometric properties of ellipses, equation thinking, and aims to assess students' transformation ability and computational solving skills"
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{8-m}+\\frac{y^{2}}{4-m}=1$ represents a hyperbola with foci on the $x$-axis, what is the range of values for $m$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(8 - m) + y^2/(4 - m) = 1);PointOnCurve(Focus(G), xAxis);m:Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(4,8)",
    "fact_spans": "[[[54, 57]], [[2, 57]], [[45, 57]], [[59, 62]]]",
    "query_spans": "[[[59, 69]]]",
    "process": "(8-m)(4-m)<0, then 4<m<8."
  },
  {
    "text": "Given the parabola equation $y^{2}=x$, point $M$ moves along this parabola. Then the minimum distance $|M A|$ between point $A(4,0)$ and point $M$ is?",
    "fact_expressions": "G: Parabola;A: Point;M: Point;Coordinate(A) = (4, 0);Expression(G) = (y^2 = x);PointOnCurve(M, G);Distance(A,M)=Abs(LineSegmentOf(M,A))",
    "query_expressions": "Min(Abs(LineSegmentOf(M, A)))",
    "answer_expressions": "sqrt(15)/2",
    "fact_spans": "[[[2, 5], [24, 27]], [[32, 41]], [[18, 22], [42, 46]], [[32, 41]], [[2, 17]], [[18, 28]], [[32, 58]]]",
    "query_spans": "[[[51, 64]]]",
    "process": ""
  },
  {
    "text": "A line $l$ is drawn through the focus of the parabola $y^{2}=-12x$. The line $l$ intersects the parabola at points $A$ and $B$. If the x-coordinate of the midpoint of segment $AB$ is $-9$, then $|AB|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -12*x);l: Line;PointOnCurve(Focus(G), l);A: Point;B: Point;Intersection(l, G) = {A, B};XCoordinate(MidPoint(LineSegmentOf(A, B))) = -9",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "24",
    "fact_spans": "[[[1, 17], [34, 37]], [[1, 17]], [[21, 26], [28, 33]], [[0, 26]], [[38, 41]], [[42, 45]], [[28, 47]], [[49, 66]]]",
    "query_spans": "[[[68, 77]]]",
    "process": "By the definition of the parabola given in the problem, |AB| = 3 - x_{A} + 3 - x_{B} = 6 - (x_{A} + x_{B}) = 6 - (-9) \\times 2 = 24"
  },
  {
    "text": "Given the ellipse $C_{1}$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1$ $(0<b<2)$ has eccentricity $\\frac{1}{2}$, $F_{1}$ and $F_{2}$ are the left and right foci of $C_{1}$, $M$ is a moving point on $C_{1}$, point $N$ lies on the extension of segment $F_{1} M$, $|M N|=| M F_{2} |$, and $P$ is the midpoint of segment $F_{2} N$. Then the maximum value of $|F_{1} P|$ is?",
    "fact_expressions": "C1: Ellipse;Expression(C1) = (x^2/4 + y^2/b^2 = 1);b: Number;0<b;b<2;Eccentricity(C1) = 1/2;F1: Point;F2: Point;LeftFocus(C1) = F1;RightFocus(C1) = F2;M: Point;PointOnCurve(M, C1);N: Point;PointOnCurve(N, OverlappingLine(LineSegmentOf(F1, M)));Abs(LineSegmentOf(M, N)) = Abs(LineSegmentOf(M, F2));P: Point;MidPoint(LineSegmentOf(F2, N)) = P",
    "query_expressions": "Max(Abs(LineSegmentOf(F1, P)))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 60], [96, 103], [113, 120]], [[2, 60]], [[13, 60]], [[13, 60]], [[13, 60]], [[2, 78]], [[80, 87]], [[88, 95]], [[80, 108]], [[80, 108]], [[109, 112]], [[109, 124]], [[125, 129]], [[125, 146]], [[147, 166]], [[182, 185]], [[167, 185]]]",
    "query_spans": "[[[187, 204]]]",
    "process": ""
  },
  {
    "text": "The length of the major axis of an ellipse is $2$ times the length of the minor axis, then its eccentricity?",
    "fact_expressions": "G: Ellipse;Length(MajorAxis(G)) = 2*Length(MinorAxis(G))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[0, 2], [17, 18]], [[0, 15]]]",
    "query_spans": "[[[17, 23]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=4x$, a line passing through the point $P(4,0)$ intersects the parabola at two points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. What is the minimum value of $y_{1}^{2}+y_{2}^{2}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Line;P: Point;Coordinate(P) = (4, 0);PointOnCurve(P, H);A: Point;B: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Intersection(H, G) = {A, B}",
    "query_expressions": "Min(y1^2 + y2^2)",
    "answer_expressions": "32",
    "fact_spans": "[[[2, 16], [32, 35]], [[2, 16]], [[29, 31]], [[19, 28]], [[19, 28]], [[18, 31]], [[38, 55]], [[57, 74]], [[38, 55]], [[57, 74]], [[38, 55]], [[57, 74]], [[38, 55]], [[57, 74]], [[29, 76]]]",
    "query_spans": "[[[78, 105]]]",
    "process": ""
  },
  {
    "text": "If one of the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{8-k}=1$ is given by the equation $x+2 y=0$, then $k$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/(8 - k) = 1);k: Number;Expression(OneOf(Asymptote(G))) = (x + 2*y = 0)",
    "query_expressions": "k",
    "answer_expressions": "7",
    "fact_spans": "[[[1, 41]], [[1, 41]], [[61, 64]], [[1, 59]]]",
    "query_spans": "[[[61, 66]]]",
    "process": "From the condition, we know that one asymptote of the hyperbola is $ y = -\\frac{1}{2}x $, so $ \\frac{8-k}{4} = \\frac{1}{4}k = 7 $."
  },
  {
    "text": "Through the focus $F$ of the parabola $x^{2}=8 y$, draw a line $l$ intersecting the parabola at points $A$ and $B$. If the ordinate of the midpoint $M$ of segment $A B$ is $4$, then $|A B|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 8*y);F: Point;Focus(G) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, G) = {A, B};M: Point;MidPoint(LineSegmentOf(A, B)) = M;YCoordinate(M) = 4",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "12",
    "fact_spans": "[[[1, 15], [27, 30]], [[1, 15]], [[17, 20]], [[1, 20]], [[21, 26]], [[0, 26]], [[31, 34]], [[35, 38]], [[21, 40]], [[51, 54]], [[42, 54]], [[51, 62]]]",
    "query_spans": "[[[64, 73]]]",
    "process": "The parabola $ x^{2} = 8y $ has focus $ F(0,2) $. A line $ l $ passing through the focus $ F $ of the parabola $ x^{2} = 8y $ intersects the parabola at points $ A $ and $ B $. If the vertical coordinate of the midpoint $ M $ of segment $ AB $ is 4, then $ y_{1} + y_{2} = 8 $. Thus, $ |AB| = y_{1} + y_{2} + p = 8 + 4 = 12 $."
  },
  {
    "text": "The equation of the hyperbola that shares the same asymptotes as $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ and passes through the point $(2 \\sqrt{3},-3)$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/16 - y^2/9 = 1);C1: Hyperbola;Asymptote(C) = Asymptote(C1);H: Point;Coordinate(H) = (2*sqrt(3), -3);PointOnCurve(H, C1)",
    "query_expressions": "Expression(C1)",
    "answer_expressions": "(y^{\\frac{9}{4}}) - \\frac{x^2}{4} = 1",
    "fact_spans": "[[[1, 40]], [[1, 40]], [[65, 68]], [[0, 68]], [[46, 64]], [[46, 64]], [[45, 68]]]",
    "query_spans": "[[[65, 72]]]",
    "process": "According to the problem, assume the required equation is $ 3\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=\\lambda $. Substituting $ (2\\sqrt{3},-3) $ yields $ \\lambda=-\\frac{1}{4} $. Thus, the required hyperbola equation is $ \\frac{\\frac{y^{2}}{9}}{4}-\\frac{x^{2}}{4}=1 $. Answer: $ \\frac{y^{2}}{\\frac{9}{4}}-\\frac{x^{2}}{4}=1 $"
  },
  {
    "text": "The standard equation of an ellipse passing through the point $N(1,2)$ and having the same eccentricity as the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{6}=1$ is?",
    "fact_expressions": "G: Ellipse;C:Ellipse;N: Point;Expression(G) = (x^2/12 + y^2/6 = 1);Coordinate(N) = (1, 2);PointOnCurve(N,C);Eccentricity(G)=Eccentricity(C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "{x^2/9+y^2/(9/2)=1,x^2/3+y^2/6=1}",
    "fact_spans": "[[[14, 52]], [[60, 62]], [[2, 11]], [[14, 52]], [[2, 11]], [[0, 62]], [[13, 62]]]",
    "query_spans": "[[[60, 69]]]",
    "process": "Let the equation of the desired ellipse be $\\frac{x^2}{12}+\\frac{y^{2}}{6}=m$ $(m>0)$ or $\\frac{y^{2}}{12}+\\frac{x^{2}}{6}=n$ $(n>0)$. Substituting the coordinates of point $N$, we obtain $\\frac{1}{12}+\\frac{2^{2}}{6}=m$ or $\\frac{2^{2}}{12}+\\frac{1}{6}=n$, that is, $m=\\frac{3}{4}$, $n=\\frac{1}{2}$. Hence, the standard equations of the desired ellipse are $\\frac{x^{2}}{12}+\\frac{y^{2}}{6}=\\frac{3}{4}$ or $\\frac{y^{2}}{12}+\\frac{x^{2}}{6}=\\frac{1}{2}$, which simplify to $\\frac{x^{2}}{9}+\\frac{y^{2}}{\\frac{9}{2}}=1$ or $\\frac{y^{2}}{6}+\\frac{x^{2}}{3}=1$."
  },
  {
    "text": "Find the equation of the conic section that shares foci with the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$ and passes through the point $(2 , 1)$.",
    "fact_expressions": "G: Hyperbola;H: Point;C: ConicSection;Expression(G) = (x^2/4 - y^2/2 = 1);Coordinate(H) = (2, 1);PointOnCurve(H,C);Focus(G)=Focus(C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 40]], [[46, 56]], [[57, 61]], [[2, 40]], [[46, 56]], [[45, 61]], [[1, 61]]]",
    "query_spans": "[[[57, 66]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus is $F$, the right vertex is $A$. If there exists a point $P$ on the ellipse $C$ such that $P F \\perp P A$, then the range of the eccentricity $e$ of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;F: Point;A: Point;e: Number;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;RightVertex(C)=A;PointOnCurve(P,C);IsPerpendicular(LineSegmentOf(P,F),LineSegmentOf(P,A));Eccentricity(C) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "(1/2,1)",
    "fact_spans": "[[[2, 59], [77, 82], [108, 113]], [[8, 59]], [[8, 59]], [[85, 89]], [[64, 67]], [[72, 75]], [[117, 120]], [[8, 59]], [[8, 59]], [[2, 59]], [[2, 67]], [[2, 75]], [[77, 89]], [[91, 106]], [[108, 120]]]",
    "query_spans": "[[[117, 127]]]",
    "process": "Let point $ P(x,y) $. Since $ \\angle APF = 90^{\\circ} $, point $ P $ lies on the circle with diameter $ FA $. The equation of the circle is $ (x - \\frac{a - c}{2})^{2} + y^{2} = (\\frac{a + c}{2})^{2} $, that is, $ y^{2} = (a - c)x - x^{2} + ac^{\\circ}\\textcircled{1} $. Also, point $ P $ lies on the ellipse, so $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1\\textcircled{2} $. Substituting \\textcircled{1} into \\textcircled{2}, we get $ (a^{2} - b^{2})x^{2} - a^{2}(a - c)x + a^{2}b^{2} - a^{3}c = 0 $. Hence, $ (x - a)[(a^{2} - b^{2})x - ab^{2} + a^{2}c] = 0 $. Since $ x \\ne a $, we have $ x = \\frac{ab^{2} - a^{2}c}{a^{2} - b^{2}} $. Also, $ -c < x < a $, so $ -c < \\frac{ab^{2} - a^{2}c}{a^{2} - b^{2}} < a $. Therefore, $ 2ab^{2} < a^{2}c + a^{3} $, that is, $ 2a(a^{2} - c^{2}) < a^{2}c + a^{3} $. Dividing both sides by $ c^{3} $ and simplifying yields: $ 2(\\frac{c}{a})^{2} + (\\frac{c}{a}) - 1 > 0 $, i.e., $ 2e^{2} + e - 1 > 0 $. Solving gives: $ e > \\frac{1}{2} $ or $ e < -1 $. But $ 0 < e < 1 $, therefore the required range of eccentricity of the ellipse is $ e \\in (\\frac{1}{2}, 1) $."
  },
  {
    "text": "Given that the ellipse $C$ with center at the origin has its left focus at $(-1,0)$ and eccentricity $\\frac{\\sqrt{3}}{2}$, what is the standard equation of the ellipse $C$?",
    "fact_expressions": "O: Origin;C: Ellipse;Center(C) = O;Coordinate(LeftFocus(C)) = (-1, 0);Eccentricity(C) = sqrt(3)/2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "(3*x^2)/4+3*y^2=1",
    "fact_spans": "[[[5, 9]], [[10, 15], [55, 60]], [[2, 15]], [[10, 28]], [[10, 53]]]",
    "query_spans": "[[[55, 67]]]",
    "process": "The left focus of ellipse C is (-1, 0), so c = 1. The eccentricity is \\frac{\\sqrt{3}}{2}, then \\frac{c}{a} = \\frac{\\sqrt{3}}{2} \\Rightarrow a = \\frac{2}{\\sqrt{3}}. Using the relationship among a, b, and c in an ellipse, we get b = \\frac{1}{3}. Thus, the equation of the ellipse is \\frac{3x^{2}}{4} + 3y^{2} = 1."
  },
  {
    "text": "The chord length intercepted by a circle centered at one of the foci of the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$ with radius $5$ on one of the asymptotes of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;G: Circle;Expression(C) = (x^2/4 - y^2/9 = 1);OneOf(Focus(C))=Center(G);Radius(G)=5",
    "query_expressions": "Length(InterceptChord(OneOf(Asymptote(C)),G))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 44], [65, 68]], [[61, 62]], [[1, 44]], [[0, 62]], [[53, 62]]]",
    "query_spans": "[[[61, 81]]]",
    "process": "a=2, b=3, then c=\\sqrt{13}, the asymptotes are y=\\pm\\frac{3}{2}x. Take a focus F(\\sqrt{13},0), and take one asymptote 3x-2y=0. The distance from F to the asymptote is d=\\frac{|3\\sqrt{13}|}{\\sqrt{3^{2}+(-2)^{2}}}=3, so the chord length is l=2\\sqrt{5^{2}-3^{2}}=8."
  },
  {
    "text": "Given that the curve $\\frac{y^{2}}{b}-\\frac{x^{2}}{a}=1(a \\cdot b \\neq 0, a \\neq b)$ intersects the line $x+y-2=0$ at two points $P$, $Q$, and $\\overrightarrow{O P} \\cdot \\overrightarrow{O Q}=0$ ($O$ is the origin), then the value of $\\frac{1}{b}-\\frac{1}{a}$ is?",
    "fact_expressions": "G: Line;H: Curve;b: Number;a: Number;Negation(a*b = 0);Negation(a = b);O: Origin;P: Point;Q: Point;Expression(G) = (x + y - 2 = 0);Expression(H) = (y^2/b - x^2/a = 1);Intersection(H, G) = {P, Q};DotProduct(VectorOf(O, P), VectorOf(O, Q)) = 0",
    "query_expressions": "1/b - 1/a",
    "answer_expressions": "1/2",
    "fact_spans": "[[[68, 79]], [[2, 67]], [[4, 67]], [[4, 67]], [[4, 67]], [[4, 67]], [[145, 148]], [[82, 85]], [[86, 89]], [[68, 79]], [[2, 67]], [[2, 91]], [[93, 144]]]",
    "query_spans": "[[[154, 183]]]",
    "process": "\\overrightarrow{OP}\\cdot\\overrightarrow{OQ}=0 implies \\overrightarrow{OP}\\bot\\overrightarrow{OQ}, let P(x_{1},y_{1}), Q(x_{2},y_{2}), then x_{1}x_{2}+y_{1}y_{2}=0. Combining the equation of the line and the hyperbola equation \\frac{y^{2}}{b}-\\frac{x^{2}}{a}=1, eliminating y gives (a-b)x^{2}-4ax+4a-ab=0, so x_{1}+x_{2}=\\frac{4a}{a-b}, x_{1}\\cdot x_{2}=\\frac{4a-ab}{a-b}, and \\frac{1}{2}\\frac{y_{1}y_{2}=(-x_{1}+2)(-x_{2}+2)=x_{1}x_{2}-2(x_{1}+x_{2})+4=4}{a-b}-\\frac{8a}{a-b}+4+\\frac{4a}{a-b}=0, simplifying yields \\frac{1}{b}-\\frac{1}{a}=\\frac{1}{2}"
  },
  {
    "text": "If point $P$ is a point on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ with foci $F_{1}$ and $F_{2}$, satisfying $P F_{2} \\perp F_{1} F_{2}$ and $|P F_{1}|=2|P F_{2}|$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;P: Point;F2: Point;F1: Point;PointOnCurve(P,G);Focus(G)={F1,F2};Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);IsPerpendicular(LineSegmentOf(P, F2), LineSegmentOf(F1, F2));Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[26, 72], [132, 135]], [[29, 72]], [[29, 72]], [[1, 5]], [[15, 22]], [[7, 14]], [[1, 75]], [[6, 72]], [[26, 72]], [[78, 105]], [[107, 129]]]",
    "query_spans": "[[[131, 141]]]",
    "process": ""
  },
  {
    "text": "Given that $P(x , y)$ is a moving point on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$, and $F_{1}$, $F_{2}$ are the foci, then the range of $|P F_{1}| \\cdot|P F_{2}|$ is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F2: Point;F1: Point;a > b;b > 0;x1:Number;y1:Number;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (x1, y1);PointOnCurve(P, G);Focus(G)={F1,F2}",
    "query_expressions": "Range(Abs(LineSegmentOf(P, F2))*Abs(LineSegmentOf(P, F1)))",
    "answer_expressions": "[b^2,a^2]",
    "fact_spans": "[[[13, 67]], [[15, 67]], [[15, 67]], [[2, 12]], [[80, 87]], [[72, 79]], [[15, 67]], [[15, 67]], [[2, 12]], [[2, 12]], [[13, 67]], [[2, 12]], [[2, 71]], [[13, 90]]]",
    "query_spans": "[[[92, 125]]]",
    "process": ""
  },
  {
    "text": "If the distance from the right vertex $A$ of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ to one of its asymptotes is $\\frac{2 \\sqrt{2}}{3} a$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;A: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightVertex(C)=A;Distance(A, OneOf(Asymptote(C))) = a*(2*sqrt(2))/3",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 62], [105, 108]], [[8, 62]], [[8, 62]], [[66, 69]], [[8, 62]], [[8, 62]], [[1, 62]], [[1, 69]], [[1, 103]]]",
    "query_spans": "[[[105, 114]]]",
    "process": "According to the distance from the right vertex $ A(a,0) $ of the hyperbola to the asymptote line $ y = \\frac{b}{a}x $ being $ \\frac{2\\sqrt{2}}{3}a $, using the point-to-line distance formula, we simplify to obtain $ b^{2} = \\frac{8}{9}c^{2} $. Combining this with $ c^{2} = a^{2} + b^{2} $ and the definition of eccentricity, we can solve the problem. For the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $, the right vertex is $ A(a,0) $, and one asymptote line is $ y = \\frac{b}{a}x $, i.e., $ bx - ay = 0 $. We get $ \\frac{ab}{\\sqrt{a^{2}+b^{2}}} = \\frac{ab}{c} = \\frac{2\\sqrt{2}}{3}a $, then $ b = \\frac{2\\sqrt{2}}{3}c $, i.e., $ b^{2} = \\frac{8}{9}c^{2} $. From $ c^{2} = a^{2} + b^{2} $, we obtain $ a = \\frac{1}{3}c $, so the eccentricity $ e = \\frac{c}{a} = 3 $."
  },
  {
    "text": "Given a point $P$ on the line $a x + y - 4 = 0$, $PA$ and $PB$ are two tangent lines from $P$ to the ellipse $C$: $\\frac{x^{2}}{a^{2}} + y^{2} = 1$ $(a > 0)$. If there exists exactly one point $P$ such that $PA \\perp PB$, then what is the eccentricity of the ellipse $C$?",
    "fact_expressions": "C: Ellipse;a: Number;G: Line;P: Point;A: Point;B: Point;a>0;Expression(C) = (y^2 + x^2/a^2 = 1);Expression(G) = (a*x + y - 4 = 0);PointOnCurve(P, G);IsTangent(LineOf(P, A), C);IsTangent(LineOf(P, B), C);IsPerpendicular(LineSegmentOf(P, A),LineSegmentOf(P, B))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[37, 78], [113, 118]], [[44, 78]], [[7, 20]], [[91, 94], [2, 6]], [[24, 29]], [[31, 36]], [[44, 78]], [[37, 78]], [[7, 20]], [[2, 23]], [[24, 83]], [[24, 83]], [[96, 111]]]",
    "query_spans": "[[[113, 124]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the left and right foci of the ellipse $E$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and let $P$ be a point on the ellipse $E$. Then the minimum value of $|P F_{1}| \\cdot|PF_{2}|$ is?",
    "fact_expressions": "E: Ellipse;Expression(E) = (x^2/25 + y^2/16 = 1);F1: Point;F2: Point;LeftFocus(E) = F1;RightFocus(E) = F2;P: Point;PointOnCurve(P, E)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)))",
    "answer_expressions": "16",
    "fact_spans": "[[[17, 61], [71, 76]], [[17, 61]], [[1, 8]], [[9, 16]], [[1, 66]], [[1, 66]], [[67, 70]], [[67, 79]]]",
    "query_spans": "[[[81, 112]]]",
    "process": "From the ellipse equation, we know that $ a=5 $, $ c=3 $. According to the definition of an ellipse, $ |PF_{2}| = 2a - |PF_{1}| = 10 - |PF_{1}| $, hence $ |PF_{1}| \\cdot |PF_{2}| = |PF_{1}| \\cdot (10 - |PF_{1}|) $. Since $ |PF_{1}| \\in [a-c, a+c] = [2,8] $, and noting that the quadratic function $ y = x(10 - x) $ has its axis of symmetry at $ x = 5 $, the minimum value of the function occurs at $ x = 2 $ and $ x = 8 $, which gives the minimum value as $ 2 \\times 8 = 16 $. Therefore, fill in $ 16 $."
  },
  {
    "text": "Given a line passing through the point $(0,1)$ intersects the parabola $x^{2}=4 y$ at two points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $y_{1}+y_{2}=\\frac{9}{4}$, then $|A B|$=?",
    "fact_expressions": "H: Line;I: Point;Coordinate(I) = (0, 1);PointOnCurve(I, H);G: Parabola;Expression(G) = (x^2 = 4*y);Intersection(H, G) = {A, B};A: Point;Coordinate(A) = (x1, y1);x1: Number;y1: Number;B: Point;Coordinate(B) = (x2, y2);x2: Number;y2: Number;y1 + y2 = 9/4",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "17/4",
    "fact_spans": "[[[12, 14]], [[3, 11]], [[3, 11]], [[2, 14]], [[15, 29]], [[15, 29]], [[12, 70]], [[31, 49]], [[31, 49]], [[31, 48]], [[31, 48]], [[51, 68]], [[51, 68]], [[51, 68]], [[51, 68]], [[72, 97]]]",
    "query_spans": "[[[99, 108]]]",
    "process": ""
  },
  {
    "text": "Given $F_{1}$, $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, a line $l$ passing through $F_{2}$ is perpendicular to one asymptote of the hyperbola $C$, intersecting the left and right branches of the hyperbola at points $P$ and $Q$ respectively, and point $P$ lies exactly on the perpendicular bisector of $Q F_{1}$. Then the equation of the asymptotes of hyperbola $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;Q: Point;F1: Point;F2: Point;P: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F2,l);IsPerpendicular(l,OneOf(Asymptote(C)));Intersection(l,LeftPart(C))=P;Intersection(l,RightPart(C))=Q;PointOnCurve(P, PerpendicularBisector(LineSegmentOf(Q, F1)))",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*(sqrt(3)+1)*x",
    "fact_spans": "[[[95, 100]], [[18, 79], [101, 107], [117, 120], [162, 168]], [[26, 79]], [[26, 79]], [[133, 136]], [[2, 9]], [[10, 17], [87, 94]], [[129, 132], [140, 144]], [[26, 79]], [[26, 79]], [[18, 79]], [[2, 85]], [[2, 85]], [[86, 100]], [[95, 115]], [[95, 138]], [[95, 138]], [[140, 160]]]",
    "query_spans": "[[[162, 176]]]",
    "process": "The line $ l $ passing through $ F_{2} $ is perpendicular to one of the asymptotes of hyperbola $ C $; let the foot of the perpendicular be $ A $. It is easy to obtain $ |F_{2}A| = b $, $ \\cos\\angle QF_{2}O = \\frac{b}{c} $. Since point $ P $ lies exactly on the perpendicular bisector of $ QF_{1} $, we have $ |PQ| = |PF_{1}| $. Thus, $ |PF_{2}| - |PF_{1}| = |PF_{2}| - |PQ| = |QF_{2}| = 2a $. Also, $ |QF_{1}| - |QF_{2}| = 2a $, hence $ |QF_{1}| = 4a $. In $ \\triangle QF_{1}F_{2} $, applying the law of cosines yields $ 16a^{2} = 4a^{2} + 4c^{2} - 2 \\times 2a \\times 2c \\times \\frac{b}{c} $, that is, $ 2a^{2} = b^{2} - 2ab $. Since $ \\frac{b}{a} > 0 $, we obtain $ \\frac{b}{a} = \\sqrt{3} + 1 $. Therefore, the asymptotic equations of hyperbola $ C $ are $ y = \\pm(\\sqrt{3}+1)x $."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ with focus $F$, point $M(-\\frac{p}{2}, 0)$, and a line passing through point $F$ intersecting the parabola at points $A$ and $B$. If $|A B|=24$ and $\\tan \\angle A M B=2 \\sqrt{2}$, then $p$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;H: Line;M: Point;A: Point;B: Point;F: Point;p>0;Coordinate(M) = (-p/2, 0);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 24;Tan(AngleOf(A, M, B)) = 2*sqrt(2)",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 23], [63, 66]], [[2, 23]], [[123, 126]], [[59, 61]], [[31, 52]], [[68, 71]], [[72, 75]], [[27, 30], [54, 58]], [[5, 23]], [[31, 52]], [[2, 30]], [[53, 61]], [[59, 77]], [[79, 89]], [[91, 121]]]",
    "query_spans": "[[[123, 128]]]",
    "process": "Let the equation of AB be $ x = my + \\frac{p}{2} $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Then from \n\\[\n\\begin{cases}\ny^{2} = 2px \\\\\nx = my + \\frac{p}{2}\n\\end{cases}\n\\]\nwe get $ y^{2} - 2pmy - p^{2} = 0 $, $ \\therefore y_{1} + y_{2} = 2pm $, $ y_{1}y_{2} = -p^{2} $.  \n$ \\therefore k_{MA} + k_{MB} = \\frac{y_{1}}{x_{1} + \\frac{p}{2}} + \\frac{y_{2}}{x_{2} + \\frac{p}{2}} = \\frac{y_{1}}{my_{1} + p} + \\frac{y_{2}}{my_{2} + p} = \\frac{y_{1}(my_{2} + p) + y_{2}(my_{1} + p)}{(my_{1} + p)(my_{2} + p)} = \\frac{2my_{1}y_{2} + p(y_{1} + y_{2})}{(my_{1} + p)(my_{2} + p)} = \\frac{2m(-p^{2}) + p(2pm)}{(my_{1} + p)(my_{2} + p)} = 0 $.  \n$ \\therefore \\angle AMF = \\angle BMF $,  \n$ \\because \\tan \\angle AMB = \\frac{2\\tan \\angle AMF}{1 - \\tan^{2} \\angle AMF} = 2\\sqrt{2} $, and $ \\angle AMF $ is acute,  \n$ \\therefore \\tan \\angle AMF = \\frac{\\sqrt{2}}{2} $.  \nWithout loss of generality, assume $ AF > BF $. As shown in the figure, draw $ AH \\perp x $-axis, with foot at $ H $, and draw from $ M $ a line $ l \\perp x $-axis, $ AA \\perp l $, with foot at $ A $. Then  \n$ \\because \\tan \\angle AMF = \\frac{AH}{MH} = \\frac{AH}{AA} = \\frac{AH}{AF} = \\sin \\angle AFH $,  \n$ \\therefore \\sin \\angle AFH = \\frac{\\sqrt{2}}{2} $,  \n$ \\therefore \\angle AFH = 45^{\\circ} $,  \n$ \\therefore m = 1 $.  \n$ \\therefore |AB| = \\sqrt{1 + m^{2}} |y_{1} - y_{2}| = \\sqrt{(1 + m^{2})[(y_{1} + y_{2})^{2} - 4y_{1}y_{2}]} = 4p = 24 $, hence $ p = 6 $."
  },
  {
    "text": "Given that point $P$ is any point on the ellipse $C$: $\\frac{x^{2}}{4}+y^{2}=1$, and the line $l$: $x-y+3=0$ intersects the two coordinate axes at points $A$ and $B$ respectively, then the maximum area of triangle $P A B$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P, C);C: Ellipse;Expression(C) = (x^2/4 + y^2 = 1);l: Line;Expression(l) = (x - y + 3 = 0);A: Point;B: Point;Intersection(l, xAxis) = A;Intersection(l, yAxis) = B",
    "query_expressions": "Max(Area(TriangleOf(P, A, B)))",
    "answer_expressions": "(3*sqrt(5)+9)/2",
    "fact_spans": "[[[2, 6]], [[2, 44]], [[7, 39]], [[7, 39]], [[45, 61]], [[45, 61]], [[70, 73]], [[74, 77]], [[45, 79]], [[45, 79]]]",
    "query_spans": "[[[81, 100]]]",
    "process": "Analysis: According to the problem, A(-3,0), B(0,3), then |AB|=3\\sqrt{2}. P is any point on the ellipse C: \\frac{x^{2}}{4}+y^{2}=1, so the coordinates of point P can be set as (2\\cos\\theta,\\sin\\theta). Then the distance from point P to line l is d=\\frac{|2\\cos\\theta-\\sin\\theta+3|}{\\sqrt{2}}\\leqslant\\frac{\\sqrt{5}+3}{\\sqrt{2}}, yielding S_{\\Delta PAB}=\\frac{\\sqrt{2}}{2}\\cdot|AB|\\cdot d\\leqslant\\frac{1}{2}\\cdot3\\sqrt{2}\\cdot\\frac{\\sqrt{5}+3}{\\sqrt{5}}+9"
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the foci of the curve $C_{1}$: $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$, and let $P$ be an intersection point of the curves $C_{2}$: $\\frac{x^{2}}{3}-y^{2}=1$ and $C_{1}$. Then the area of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "C1: Curve;C2: Curve;Expression(C1)=(x^2/6+y^2/2=1);Expression(C2)=(x^2/3-y^2=1);P: Point;F1: Point;F2: Point;Focus(C1)={F1,F2};OneOf(Intersection(C1,C2))=P",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[17, 63], [108, 115]], [[71, 107]], [[17, 63]], [[71, 107]], [[67, 70]], [[1, 8]], [[9, 16]], [[1, 66]], [[67, 120]]]",
    "query_spans": "[[[122, 152]]]",
    "process": "From the given conditions, |F_{1}F_{2}| = 2\\sqrt{6-2} = 4. Let the coordinates of point P be (x, y). From \\begin{cases}\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1\\\\\\frac{x^{2}}{3}-y^{2}=1\\end{cases}, we obtain \\begin{cases}x=\\pm\\frac{3\\sqrt{2}}{2}\\\\y=\\pm\\frac{\\sqrt{2}}{2}\\end{cases}. Then, S\\triangle PF_{1}F_{2} = \\frac{1}{2}|F_{1}F_{2}||y| = \\frac{1}{2}\\times4\\times\\frac{\\sqrt{2}}{2} = \\sqrt{2}. Answer: \\sqrt{2}"
  },
  {
    "text": "It is known that point $F$ is the right focus of the ellipse $\\frac{x^{2}}{a^{2}}+y^{2}=1$ $(a>1)$, and the maximum distance from point $P(0,3)$ to a moving point $Q$ on the ellipse does not exceed $2 \\sqrt{5}$. When the eccentricity of the ellipse reaches its maximum value, what is the maximum value of $|P Q|+|Q F|$?",
    "fact_expressions": "G: Ellipse;a: Number;P: Point;Q: Point;F: Point;a>1;Expression(G) = (y^2 + x^2/a^2 = 1);Coordinate(P) = (0, 3);RightFocus(G)=F;PointOnCurve(Q,G);Negation(Max(Distance(P,Q))>2*sqrt(5));WhenMax(Eccentricity(G))",
    "query_expressions": "Max(Abs(LineSegmentOf(P, Q)) + Abs(LineSegmentOf(Q, F)))",
    "answer_expressions": "3*sqrt(2)+2*sqrt(10)",
    "fact_spans": "[[[7, 43], [58, 60], [91, 93]], [[9, 43]], [[48, 57]], [[64, 67]], [[2, 6]], [[9, 43]], [[7, 43]], [[48, 57]], [[2, 47]], [[58, 67]], [[48, 89]], [[90, 103]]]",
    "query_spans": "[[[105, 125]]]",
    "process": "Let $ Q(x_{0},y_{0}) $, then $ \\frac{x_{0}^{2}}{ \\text{because} |PQ| } = \\sqrt{x_{0}^{2}+(y_{0}-3)^{2}} = \\sqrt{a^{2}-a^{2}y_{0}^{2}+y_{0}^{2}-6y_{0}+9} = \\sqrt{(1-a^{2})y_{0}^{2}-6y_{0}+9+a^{2}} $. Since $ a>1 $, we have $ 1-a^{2}<0 $, so when $ \\frac{3}{1-a^{2}} \\leqslant -1 $, i.e., $ 1 < a \\leqslant 2 $, $ |PQ| $ reaches its maximum when $ y_{0} = -1 $, and $ |PQ|_{\\max} = 4 \\leqslant 2\\sqrt{5} $. Let the left focus of the ellipse be $ F_{1} $, then $ F_{1}(-\\sqrt{3},0) $, thus $ |PQ| + |QF| = |PQ| + 2a - |QF_{1}| = |PQ| - |QF_{1}| + 4 $, so $ |PQ| - |QF_{1}| $ reaches its maximum when $ Q $ lies on the extension of $ PF_{1} $. $ (|PQ| - |QF_{1}|)_{\\max} = |PF_{1}| = \\sqrt{(\\sqrt{3})^{2}+3^{2}} = 2\\sqrt{3} $, therefore the maximum value of $ |PQ| + |QF| $ is $ 2\\sqrt{3} + 4 $. When $ \\frac{3}{1-a^{2}} > -1 $, i.e., $ a > 2 $, $ |PQ| $ reaches its maximum at $ y_{0} = \\frac{3}{1-a^{2}} $, and $ |PQ|_{\\max} = \\sqrt{-\\frac{9}{1-a^{2}}+a^{2}+9} $. $ \\frac{9}{2} + a^{2} + 9 \\leqslant 2\\sqrt{5} $ gives $ 2 \\leqslant a^{2} \\leqslant 10 $, i.e., $ 2 < a \\leqslant \\sqrt{10} $. Also, since the eccentricity of the ellipse is $ e = \\sqrt{\\frac{a^{2}}{a}}| = \\sqrt{1-\\frac{1}{2}} $, $ e $ is maximized when $ a = \\sqrt{10} $. Let the left focus of the ellipse be $ F_{1} $, then $ F_{1}(-3,0) $, thus $ |PQ| + |QF| = |PQ| + 2a - |QF_{1}| = |PQ| - |QF_{1}| + 2\\sqrt{10} $, so $ |PQ| - |QF_{1}| $ reaches its maximum when $ Q $ lies on the extension of $ PF $. $ (|PQ| - |QF_{1}|)_{\\max} = |PF_{1}| = \\sqrt{(-3)^{2}+3^{2}} = 3\\sqrt{2} $. Therefore, the maximum value of $ |PQ| + |QF| $ is $ 3\\sqrt{2} + 2\\sqrt{10} $. In conclusion, the maximum value of $ |PQ| + |QF| $ is $ 3\\sqrt{2} + 2\\sqrt{10} $."
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{16}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/16 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*4*x",
    "fact_spans": "[[[0, 29]], [[0, 29]]]",
    "query_spans": "[[[0, 37]]]",
    "process": "Find the values of a and b for the hyperbola to obtain the solution. From the standard equation of the hyperbola, we have a=1, b=4, and the foci of the hyperbola lie on the x-axis, so the asymptotes of the hyperbola are given by y=\\pm4x."
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=2x$, a line is drawn intersecting the parabola at points $A$ and $B$. If $|AB|=\\frac{25}{12}$ and $|AF|<|BF|$, then $|AF|=$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 2*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 25/12;Abs(LineSegmentOf(A, F)) < Abs(LineSegmentOf(B, F))",
    "query_expressions": "Abs(LineSegmentOf(A, F))",
    "answer_expressions": "5/6",
    "fact_spans": "[[[1, 15], [25, 28]], [[22, 24]], [[29, 32]], [[33, 36]], [[18, 21]], [[1, 15]], [[1, 21]], [[0, 24]], [[22, 38]], [[40, 60]], [[61, 72]]]",
    "query_spans": "[[[74, 82]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has a major axis length of $4$ and an eccentricity of $\\frac{\\sqrt{3}}{2}$, then the equation of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Length(MajorAxis(C))=4;Eccentricity(C)=sqrt(3)/2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4+y^2=1",
    "fact_spans": "[[[2, 59], [94, 99]], [[8, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[2, 59]], [[2, 67]], [[2, 92]]]",
    "query_spans": "[[[94, 104]]]",
    "process": "The foci of ellipse C lie on the x-axis, so a=2, \\frac{c}{a}=\\frac{\\sqrt{3}}{2}, then c=\\sqrt{3}, b=\\sqrt{a^2-c^{2}}=1. At this time, the equation of ellipse C is \\frac{x^{2}}{4}+y^{2}=1;"
  },
  {
    "text": "Given that one focus of the ellipse is $F(1,0)$ and the eccentricity is $\\frac{1}{2}$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;F: Point;Coordinate(F) = (1, 0);OneOf(Focus(G)) = F;Eccentricity(G) = 1/2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 + y^2/3 = 1",
    "fact_spans": "[[[2, 4], [38, 40]], [[10, 18]], [[10, 18]], [[2, 18]], [[2, 36]]]",
    "query_spans": "[[[38, 47]]]",
    "process": "Let the standard equation of the ellipse be: \\frac{x^2}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0). \\because one focus of the ellipse is F(1,0), and the eccentricity is e=\\frac{1}{2}, \\therefore \\begin{cases}c=1\\\\\\frac{c}{a}=\\frac{1}{2}\\\\a^{2}=b^{2}+c^2\\end{cases}, solving yields: \\begin{cases}a=2\\\\b^{2}=3\\end{cases}. The standard equation of the ellipse: \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1. The correct result for this problem: \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1"
  },
  {
    "text": "Line $l_{1}$: $4x - 3y + 6 = 0$ and line $l_{2}$: $x = -1$, a moving point $P$ on the parabola $y^{2} = 4x$ to the sum of the distances from line $l_{1}$ and line $l_{2}$ has a minimum value of?",
    "fact_expressions": "G: Parabola;l1: Line;l2: Line;Expression(G) = (y^2 = 4*x);P: Point;Expression(l1) = (4*x - 3*y + 6 = 0);Expression(l2) = (x = -1);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, l1) + Distance(P, l2))",
    "answer_expressions": "2",
    "fact_spans": "[[[43, 57]], [[0, 24], [65, 74]], [[25, 42], [75, 84]], [[43, 57]], [[61, 64]], [[0, 24]], [[25, 42]], [[43, 64]]]",
    "query_spans": "[[[61, 95]]]",
    "process": "Let the coordinates of a moving point P on the parabola \\( y^{2} = 4x \\) be \\( \\left( \\frac{t^{2}}{4}, t \\right) \\), and let \\( d \\) be the sum of its distances to the lines \\( l_{1} \\) and \\( l_{2} \\). Then \\( d = \\frac{|4 \\times \\frac{t^{2}}{4} - 3t + 6|}{5} + \\frac{|\\frac{t^{2}}{4} + 1|}{5} = \\frac{|t^{2} - 3t + 6|}{5} + \\frac{\\frac{t^{2}}{4} + 1}{5} = \\frac{t^{2} - 3t + 6}{5} + \\frac{t^{2}}{4} + 1 = \\frac{9}{20}t^{2} - \\frac{3}{5}t + \\frac{11}{5} \\). When \\( t = \\frac{2}{3} \\), \\( d_{\\min} = \\frac{9}{20} \\left( \\frac{2}{3} \\right)^{2} - \\frac{3}{5} \\times \\frac{2}{3} + \\frac{11}{5} \\)."
  },
  {
    "text": "Given that the right focus of the hyperbola is $(5 , 0)$ and one asymptote equation is $2 x - y = 0$, then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (5, 0);RightFocus(G) = H;Expression(OneOf(Asymptote(G))) = (2*x - y = 0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5 - y^2/20 = 1",
    "fact_spans": "[[[2, 5], [40, 43]], [[10, 19]], [[10, 19]], [[2, 19]], [[2, 37]]]",
    "query_spans": "[[[40, 50]]]",
    "process": "According to the problem, one asymptote of the hyperbola is $2x - y = 0$, so the equation of the hyperbola can be written as $x^{2} - \\frac{y^{2}}{4} = \\lambda$, $\\lambda \\neq 0$. Also, the right focus of the hyperbola is $(5, 0)$, meaning the foci lie on the x-axis and $c = 5$, thus $\\lambda > 0$. Then the equation of the hyperbola can be rewritten as $\\frac{x^{2}}{\\lambda} - \\frac{y^{2}}{4\\lambda} = 1$. Given $c = 5$, we have $5\\lambda = 25$, solving gives $\\lambda = 5$. Therefore, the standard equation of this hyperbola is $\\frac{x^{2}}{5} - \\frac{y^{2}}{20} = 1$."
  },
  {
    "text": "It is known that point $P$ is a moving point on the parabola $y^{2}=4x$, the projection of point $P$ on the $y$-axis is $M$, and the coordinates of point $A$ are $(4, a)$. Then, when $|a|>4$, the minimum value of $|PA|+|PM|$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P, G) = True;G: Parabola;Expression(G) = (y^2 = 4*x);Projection(P, yAxis) = M;M: Point;A: Point;Coordinate(A) = (4, a);a: Number;Abs(a)>4",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "sqrt(a^2+9)-1",
    "fact_spans": "[[[2, 6], [26, 30]], [[2, 25]], [[7, 21]], [[7, 21]], [[26, 43]], [[40, 43]], [[44, 48]], [[44, 60]], [[52, 60]], [[63, 70]]]",
    "query_spans": "[[[72, 91]]]",
    "process": "The focus of the parabola is $ F(1,0) $, and when $ |a| > 4 $, point $ A $ lies outside the parabola. According to the definition of the parabola, we have $ |PM| = |PF| - 1 $. Therefore, $ |PA| + |PM| = |PA| + |PF| - 1 $. Since $ |PA| + |PF| \\geqslant |AF| $, equality holds when points $ A $, $ P $, and $ F $ are collinear. Thus, the minimum value of $ |PA| + |PM| $ is $ |AF| - 1 = \\sqrt{a^2 + 9} - 1 $. When $ |a| > 4 $, the minimum value of $ |PA| + |PM| $ is $ \\sqrt{a^2 + 9} - 1 $."
  },
  {
    "text": "$P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and $M$, $N$ are points on the circles $(x+5)^{2}+y^{2}=4$ and $(x-5)^{2}+y^{2}=1$, respectively. Then the maximum value of $|P M|-|P N|$ is equal to?",
    "fact_expressions": "P: Point;PointOnCurve(P, RightPart(G)) = True;G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);M: Point;PointOnCurve(M,H1) = True;H1: Circle;Expression(H1) = (y^2 + (x + 5)^2 = 4);N: Point;PointOnCurve(N,H2) = True;H2: Circle;Expression(H2) = (y^2 + (x - 5)^2 = 1)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "9",
    "fact_spans": "[[[0, 3]], [[0, 49]], [[4, 43]], [[4, 43]], [[50, 53]], [[50, 103]], [[60, 80]], [[60, 80]], [[54, 57]], [[50, 103]], [[81, 100]], [[81, 100]]]",
    "query_spans": "[[[105, 125]]]",
    "process": ""
  },
  {
    "text": "The standard equation of a hyperbola with foci on the $y$-axis, a length of the imaginary axis of $8$, and a focal distance of $10$ is?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), yAxis);Length(ImageinaryAxis(G)) = 8;FocalLength(G) = 10",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/9 - x^2/16 = 1",
    "fact_spans": "[[[25, 28]], [[0, 28]], [[9, 28]], [[17, 28]]]",
    "query_spans": "[[[25, 35]]]",
    "process": "Since the length of the imaginary axis is 8, we have 2b=8, so b=4. Since the focal distance is 10, we have 2c=10, so c=5. Therefore, a^{2}=c^{2}-b^{2}=9. Thus, the standard equation of the hyperbola is \\frac{y^{2}}{9}-\\frac{x^{2}}{16}=1."
  },
  {
    "text": "Given a hyperbola $ D: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ $ (a > 0, b > 0) $ with asymptotes $ y = \\pm \\sqrt{3}x $, and left and right foci $ F_{1} $, $ F_{2} $ respectively. If $ P $ is any point on the right branch of hyperbola $ D $, then the range of $ \\frac{|P F_{1}| - |P F_{2}|}{|P F_{1}| + |P F_{2}|} $ is?",
    "fact_expressions": "Expression(Asymptote(D)) = (y = pm*sqrt(3)*x);D: Hyperbola;Expression(D) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(D) = F1;RightFocus(D) = F2;P: Point;PointOnCurve(P, RightPart(D)) = True",
    "query_expressions": "Range((Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)))/(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2))))",
    "answer_expressions": "(0, 1/2]",
    "fact_spans": "[[[2, 87]], [[26, 87], [116, 122]], [[26, 87]], [[34, 87]], [[34, 87]], [[34, 87]], [[34, 87]], [[95, 102]], [[103, 110]], [[26, 110]], [[26, 110]], [[112, 115]], [[112, 129]]]",
    "query_spans": "[[[131, 187]]]",
    "process": "Since the asymptotes of hyperbola D are $ y = \\pm\\sqrt{3}x $, it follows that $ b = \\sqrt{3}a $, thus $ c = \\sqrt{a^{2} + b^{2}} = 2a $. Since point $ P $ lies on the right branch of hyperbola D, we have $ |PF_{1}| - |PF_{2}| = 2a $. Combining with $ |PF_{1}| + |PF_{2}| \\geqslant |F_{1}F_{2}| = 2c $, substitute into the expression $ \\frac{|PF_{1}| - |PF_{2}|}{|PF_{1}| + |PF_{2}|} $ to compute. [Detailed Solution] $ \\because $ Hyperbola D: $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) has asymptotes $ y = \\pm\\sqrt{3}x $, $ \\therefore \\frac{b}{a} = \\sqrt{3} $, yielding $ b = \\sqrt{3}a $, $ c = \\sqrt{a^{2} + b^{2}} = 2a $. $ \\because P $ is a point on the right branch of hyperbola D, $ \\therefore |PF_{1}| - |PF_{2}| = 2a $. While $ |PF_{1}| + |PF_{2}| \\geqslant |F_{1}F_{2}| = 2c $, $ \\therefore 0 < \\frac{|PF_{1}| - |PF_{2}|}{|PF_{1}| + |PF_{2}|} \\leqslant \\frac{2a}{2c} = \\frac{a}{c} $. $ \\because c = 2a $, we get $ \\frac{a}{c} = \\frac{1}{2} $, the range of $ \\frac{|PF_{1}| - |PF_{2}|}{|PF_{1}| + |PF_{2}|} $ is $ (0, \\frac{1}{7}] $."
  },
  {
    "text": "If the equation $x^{2} \\cos \\alpha-y^{2} \\sin \\alpha+2=0$ represents an ellipse, then the center of the circle $(x+\\cos \\alpha)^{2}+(y+\\sin \\alpha)^{2}=1$ lies in which quadrant?",
    "fact_expressions": "G: Ellipse;H: Circle;alpha: Number;Expression(H) = ((x + Cos(alpha))^2 + (y + Sin(alpha))^2 = 1);Expression(G) = (x^2*Cos(alpha) - y^2*Sin(alpha) + 2 = 0)",
    "query_expressions": "Quadrant(Center(H))",
    "answer_expressions": "4",
    "fact_spans": "[[[48, 50]], [[52, 96]], [[3, 44]], [[52, 96]], [[1, 50]]]",
    "query_spans": "[[[52, 104]]]",
    "process": ""
  },
  {
    "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are given by $y=\\pm \\frac{1}{2} x$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(Asymptote(C)) = (y = pm*(x/2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[1, 57], [87, 90]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 85]]]",
    "query_spans": "[[[87, 96]]]",
    "process": "From the given condition: $\\frac{b}{a}=\\frac{1}{2}e=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{\\frac{5}{4}}=\\frac{\\sqrt{5}}{2}$"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus is $F$, $A(a, 0)$, $B(0, b)$, and point $M$ satisfies $\\overrightarrow{B M}=2 \\overrightarrow{M A}$. Then the range of the slope of line $F M$ is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;M: Point;F: Point;A: Point;B: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (a, 0);Coordinate(B) = (0, b);LeftFocus(G) = F;VectorOf(B, M) = 2*VectorOf(M, A)",
    "query_expressions": "Range(Slope(LineOf(F,M)))",
    "answer_expressions": "(0,1/2)",
    "fact_spans": "[[[2, 54]], [[4, 54]], [[4, 54]], [[85, 89]], [[59, 62]], [[64, 73]], [[75, 84]], [[4, 54]], [[4, 54]], [[2, 54]], [[64, 73]], [[75, 84]], [[2, 62]], [[91, 136]]]",
    "query_spans": "[[[138, 154]]]",
    "process": "Let M(x,y), from \\overrightarrow{BM}=2\\overrightarrow{MA}, then (x,y-b)=2(a-x,-y), so x=\\frac{2a}{3}, y=\\frac{b}{3}, thus M(\\frac{2a}{3},\\frac{b}{3}). Also F(-c,0), then the slope of line FM is k=\\frac{b}{2a+3c}=\\frac{-2}{\\frac{2}{\\triangle}}. From c^{2}=a^{2}-b^{2}, we have k=\\frac{1}{\\frac{2a}{b}+\\frac{3\\sqrt{a^{2}-b^{2}}}{b}}=\\frac{-}{2}. Let t=\\frac{a}{b}, t>1, then k=\\frac{1}{2t+3\\sqrt{t^{2}-1}}. Let y=2t+3\\sqrt{t^{2}-1}, y'=2+\\frac{3t}{\\sqrt{t^{2}-1}}. Thus, it can be seen that y=2t+3\\sqrt{t^{2}-1} is monotonically increasing on (1,+\\infty). Therefore, y=2t+3\\sqrt{t^{2}-1}>2, then 0<k=\\frac{1}{2t+3\\sqrt{t^{2}-1}}<\\frac{1}{2}. Hence, the range of the slope of line FM is (0,\\frac{1}{2})."
  },
  {
    "text": "What is the length of the real axis of the hyperbola $\\frac{y^{2}}{16}-\\frac{x^{2}}{9}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/9 + y^2/16 = 1)",
    "query_expressions": "Length(RealAxis(G))",
    "answer_expressions": "8",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 45]]]",
    "process": "Directly using the standard equation of the hyperbola, the length of the real axis can be found. The length of the real axis of the hyperbola $\\frac{y^{2}}{16}-\\frac{x^{2}}{9}=1$ is: $2a=2\\times4=8$."
  },
  {
    "text": "The standard equation of an ellipse with major axis length $8$ and one focus at the focus of the parabola $y^{2}=12x$ is?",
    "fact_expressions": "G: Parabola;H: Ellipse;Expression(G) = (y^2 = 12*x);Length(MajorAxis(H)) = 8;Focus(G)=OneOf(Focus(H))",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2/16+y^2/7=1",
    "fact_spans": "[[[9, 24]], [[33, 35]], [[9, 24]], [[0, 35]], [[8, 35]]]",
    "query_spans": "[[[33, 42]]]",
    "process": "Since the focus of the parabola $ y^{2} = 12x $ is $ (3,0) $, the semi-focal distance of the ellipse is $ c = 3 $. Given the major axis is 8, we have $ a = 4 $. Thus, $ b^{2} = a^{2} - c^{2} = 7 $. Therefore, the standard equation of the ellipse is $ \\frac{x^{2}}{16} + \\frac{y^{2}}{7} = 1 $."
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$ and the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{b^{2}}=1$ have common foci $F_{1}$, $F_{2}$, and $P$ is an intersection point of the two curves, then $\\cos \\angle F_{1} P F_{2}$=?",
    "fact_expressions": "H: Ellipse;Expression(H) = (x^2/6 + y^2/2 = 1);G: Hyperbola;Expression(G) = (x^2/3 - y^2/b^2 = 1);b: Number;F1: Point;F2: Point;Focus(H) = {F1, F2};Focus(G) = {F1, F2};P: Point;OneOf(Intersection(H, G)) = P",
    "query_expressions": "Cos(AngleOf(F1, P, F2))",
    "answer_expressions": "1/3",
    "fact_spans": "[[[0, 37]], [[0, 37]], [[38, 80]], [[38, 80]], [[41, 80]], [[86, 93]], [[94, 101]], [[0, 101]], [[0, 101]], [[102, 105]], [[102, 114]]]",
    "query_spans": "[[[116, 145]]]",
    "process": "Let the foci be $ F_{2}(2,0) $, $ F_{1}(-2,0) $. Therefore, $ 3 + b^{2} = 4 $, solving gives $ b^{2} = 1 $, the hyperbola $ \\frac{x^{2}}{3} - \\frac{y^{2}}{b^{2}} = 1 $, i.e., the hyperbola $ \\frac{x^{2}}{3} - y^{2} = 1 $. Without loss of generality, assume point $ P $ is in the first quadrant. According to the definitions and properties of the ellipse and hyperbola, we have $ |PF_{1}| + |PF_{2}| = 2\\sqrt{6} $, $ |PF_{1}| - |PF_{2}| = 2\\sqrt{3} $, solving gives $ |PF_{1}| = \\sqrt{6} + \\sqrt{3} $, $ |PF_{2}| = \\sqrt{6} - \\sqrt{3} $, and $ |F_{1}F_{2}| = 4 $. Then by the law of cosines, we obtain $ \\cos\\angle F_{1}PF_{2} = \\frac{PF_{1}^{2} + PF_{2}^{2} - F_{1}F_{2}^{2}}{2PF_{1} \\cdot PF_{2}} $, that is, $ \\frac{(\\sqrt{6}+\\sqrt{3})^{2}+(\\sqrt{6}-\\sqrt{3})^{2}-4^{2}}{2(\\sqrt{6}+\\sqrt{3})\\cdot(\\sqrt{6}-\\sqrt{3})} = \\frac{1}{3} $."
  },
  {
    "text": "Given points $A(-2 , 0)$, $B(2 , 0)$, if point $P(x, y)$ lies on the curve $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$, then $|PA|+|PB|$=?",
    "fact_expressions": "G: Curve;A: Point;B: Point;P: Point;x1: Number;y1: Number;Expression(G) = (x^2/16 + y^2/12 = 1);Coordinate(A) = (-2, 0);Coordinate(B) = (2, 0);Coordinate(P) = (x1, y1);PointOnCurve(P, G)",
    "query_expressions": "Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[40, 79]], [[2, 14]], [[16, 27]], [[29, 39]], [[30, 39]], [[30, 39]], [[40, 79]], [[2, 14]], [[16, 27]], [[29, 39]], [[29, 80]]]",
    "query_spans": "[[[82, 95]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $y=\\sqrt{3}x$, and one of its foci coincides with the focus of the parabola $y^{2}=16x$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x);H: Parabola;Expression(H) = (y^2 = 16*x);OneOf(Focus(G)) = Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/12 = 1",
    "fact_spans": "[[[2, 59], [83, 84], [112, 115]], [[2, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[2, 82]], [[90, 105]], [[90, 105]], [[83, 110]]]",
    "query_spans": "[[[112, 120]]]",
    "process": ""
  },
  {
    "text": "Given points $M(-3,0)$, $N(3,0)$, $B(1,0)$, a moving circle $C$ is tangent to the line $MN$ at point $B$. Two lines, each passing through points $M$ and $N$ respectively and tangent to the circle $C$, intersect at point $P$. Then the trajectory equation of point $P$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (-3, 0);N: Point;Coordinate(N) = (3, 0);B: Point;Coordinate(B) = (1, 0);C: Circle;TangentPoint(C,LineOf(M,N)) = B;L1: Line;L2: Line;PointOnCurve(M,L1);PointOnCurve(N,L2);IsTangent(L1,C);IsTangent(L2,C);Intersection(L1,L2) = P;P: Point",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(x^2-y^2/8=1)&(x>=1)",
    "fact_spans": "[[[2, 13], [58, 62]], [[2, 13]], [[15, 24], [63, 66]], [[15, 24]], [[26, 34], [50, 54]], [[26, 34]], [[37, 40], [68, 72]], [[37, 54]], [], [], [[55, 79]], [[55, 79]], [[55, 79]], [[55, 79]], [[55, 86]], [[82, 86], [88, 92]]]",
    "query_spans": "[[[88, 99]]]",
    "process": "As shown in the figure: Let PM and PN be tangent to circle C at points R and Q, respectively. By the tangent segment theorem of a circle, we have PQ = PR, MR = MB, NQ = NB. Therefore, PM - PN = RM - QN = MB - NB = 2 < MN. Thus, the locus of point P is the right branch of a hyperbola with foci M and N, where c = 3, a = 1. Hence, the equation of the locus of point P is $x^{2}-\\frac{y^{2}}{8}=1$ $(x\\geqslant1)$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ has its right focus at $F(3 , 0)$, and the point $(-3 ,\\frac{3 \\sqrt{2}}{2})$ lies on the ellipse $C$, then the standard equation of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Point;F: Point;a>0;b>0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(C)=F;Coordinate(F) = (3, 0);PointOnCurve(G, C);Coordinate(G)=(-3,3*sqrt(2)/2)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/18+y^2/9=1",
    "fact_spans": "[[[2, 63], [110, 115], [118, 123]], [[9, 63]], [[9, 63]], [[80, 109]], [[68, 78]], [[9, 63]], [[9, 63]], [[2, 63]], [[2, 78]], [[68, 78]], [[80, 116]], [[80, 109]]]",
    "query_spans": "[[[118, 130]]]",
    "process": ""
  },
  {
    "text": "The equation $\\frac{x^{2}}{2 m}+\\frac{y^{2}}{1-m}=1$ represents an ellipse with foci on the $y$-axis. Then, the range of values for $m$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/(1 - m) + x^2/((2*m)) = 1);PointOnCurve(Focus(G), yAxis);m: Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(0, 1/3)",
    "fact_spans": "[[[52, 54]], [[0, 54]], [[43, 54]], [[56, 59]]]",
    "query_spans": "[[[56, 66]]]",
    "process": ""
  },
  {
    "text": "The left focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $F$, and points $A(-a, 0)$, $B(0, b)$, $C(0,-b)$ are its three vertices. The line $CF$ intersects $AB$ at point $D$. If the eccentricity of the ellipse is $e=\\frac{1}{2}$, then $\\tan \\angle BDC$=?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;C: Point;F: Point;A: Point;B: Point;D: Point;a>0;b>0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (-a, 0);Coordinate(B) = (0, b);Coordinate(C) = (0, -b);LeftFocus(G) = F;Vertex(G)={A,B,C};Intersection(LineOf(C,F), LineOf(A, B)) = D;Eccentricity(G) = e ;e:Number;e=1/2",
    "query_expressions": "Tan(AngleOf(B, D, C))",
    "answer_expressions": "-3*sqrt(3)",
    "fact_spans": "[[[0, 55], [101, 102], [129, 131]], [[2, 55]], [[2, 55]], [[89, 98]], [[60, 63]], [[66, 76]], [[78, 87]], [[123, 127]], [[2, 55]], [[2, 55]], [[0, 55]], [[66, 76]], [[78, 87]], [[89, 98]], [[0, 63]], [[66, 106]], [[108, 127]], [[129, 150]], [[135, 150]], [[135, 150]]]",
    "query_spans": "[[[152, 173]]]",
    "process": "\\angleBDC=\\angleBAO+\\angleDFA=\\angleBAO+\\angleCFO so \\tan\\angleBDC=\\tan(\\angleBAO+\\angleCFO)=\\frac{\\tan\\angleBAO+\\tan\\angleCFO}{1-\\tan\\angleBAO\\tan\\angleCFO}=\\frac{\\frac{b}{a}+\\frac{b}{c}}{1-\\frac{b}{a}.\\frac{b}{c}} because the eccentricity e=\\frac{c}{a}=\\frac{1}{2}, so a=2m, c=m then b=\\sqrt{3}m, substitute into the above expression to get \\frac{\\frac{\\sqrt{3}m}{2m}+\\frac{\\sqrt{3}m}{m}}{1-\\frac{\\sqrt{3}m}{2m}.\\frac{\\sqrt{3}m}{m}}=-3\\sqrt{3}"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{m^{2}}-y^{2}=1$ $(m>0)$ has an asymptote $x-\\frac{y}{m}=0$, then the focal length of $C$ is?",
    "fact_expressions": "C: Hyperbola;m: Number;m>0;Expression(C) = (-y^2 + x^2/m^2 = 1);Expression(OneOf(Asymptote(C)))=(x-y/m=0)",
    "query_expressions": "FocalLength(C)",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 44], [70, 73]], [[9, 44]], [[9, 44]], [[2, 44]], [[2, 68]]]",
    "query_spans": "[[[70, 78]]]",
    "process": "The hyperbola $ C: \\frac{x^{2}}{m^{2}} - y^{2} = 1 $ ($ m > 0 $) has asymptotes $ y = \\pm \\frac{x}{m} $. From the given condition, we know $ m^{2} = 1 $, so $ c^{2} = 1 + 1 = 2 $, therefore the focal length of $ C $ is $ 2\\sqrt{2} $."
  },
  {
    "text": "The line passing through the focus of the parabola $y^{2}=4x$ intersects the parabola at two distinct points $A$ and $B$. What is the minimum value of $|AB|$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Line;PointOnCurve(Focus(G), H);A: Point;B: Point;Intersection(H, G) = {A, B};Negation(A=B)",
    "query_expressions": "Min(Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 15], [21, 24]], [[1, 15]], [[18, 20]], [[0, 20]], [[31, 34]], [[35, 38]], [[18, 38]], [[26, 38]]]",
    "query_spans": "[[[40, 53]]]",
    "process": ""
  },
  {
    "text": "If $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, and $A B$ is a chord passing through point $F_{1}$ on the left branch of the hyperbola with $|A B|=6$, then the perimeter of $\\triangle A B F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2/16 - y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(A, B)) = 6;IsChordOf(LineSegmentOf(A, B),LeftPart(G)) = True;PointOnCurve(F1,LineSegmentOf(A,B))",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "28",
    "fact_spans": "[[[20, 59], [72, 75]], [[66, 71]], [[66, 71]], [[10, 17]], [[2, 9], [79, 87]], [[20, 59]], [[2, 65]], [[2, 65]], [[91, 100]], [[66, 89]], [[66, 89]]]",
    "query_spans": "[[[102, 128]]]",
    "process": "From the given conditions, we have a=4, b=3, so c=5. By the definition of a hyperbola, |AF_{2}|-|AF_{1}|=8\\textcircled{1}, |BF_{2}|-|BF_{1}|=8\\textcircled{2}. Adding \\textcircled{1} and \\textcircled{2} gives: |AF_{2}|+|BF_{2}|-|AB|=16, so |AF_{2}|+|BF_{2}|=22. Therefore, the perimeter of \\triangle ABF_{2} is |AF_{2}|+|BF_{2}|+|AB|=28."
  },
  {
    "text": "The distance from a point $P$ on the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ to the point $(5,0)$ is $8.5$. What is the distance from point $P$ to the point $(-5,0)$?",
    "fact_expressions": "G: Hyperbola;H: Point;I: Point;P: Point;Expression(G) = (x^2/16 - y^2/9 = 1);Coordinate(H) = (5, 0);Coordinate(I) = (-5, 0);PointOnCurve(P, G);Distance(P, H) = 8.5",
    "query_expressions": "Distance(P, I)",
    "answer_expressions": "16.5",
    "fact_spans": "[[[0, 39]], [[46, 54]], [[70, 79]], [[41, 45], [65, 69]], [[0, 39]], [[46, 54]], [[70, 79]], [[0, 45]], [[41, 63]]]",
    "query_spans": "[[[65, 83]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, then the focal distance is?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[18, 45]], [[2, 9]], [[10, 17]], [[18, 45]], [[2, 51]], [[2, 51]]]",
    "query_spans": "[[[18, 57]]]",
    "process": ""
  },
  {
    "text": "The line $y=kx+2$ always has two common points with the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ whose foci lie on the $x$-axis. Then the range of real values for $b$ is?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*x + 2);k: Number;G: Ellipse;Expression(G) = (x^2/16 + y^2/b^2 = 1);b: Real;b>0;PointOnCurve(Focus(G), xAxis);NumIntersection(H, G) = 2",
    "query_expressions": "Range(b)",
    "answer_expressions": "(2, 4)",
    "fact_spans": "[[[0, 11]], [[0, 11]], [[2, 11]], [[21, 68]], [[21, 68]], [[77, 82]], [[23, 68]], [[12, 68]], [[0, 75]]]",
    "query_spans": "[[[77, 89]]]",
    "process": "Find the fixed point of the line; by combining algebra and geometry, transform the problem into the condition that the fixed point must lie inside the ellipse to obtain b>2. Since the foci of the ellipse lie on the x-axis, we have b<a=4; taking the intersection gives the solution. The line always passes through the fixed point (0,2). To ensure the line intersects the ellipse at two points, the fixed point must lie inside the ellipse, so \\frac{0}{16}+\\frac{4}{b^{2}}<1, solving yields b>2. Since the foci of the ellipse lie on the x-axis, we have b<a=4, thus b\\in(2,4)."
  },
  {
    "text": "A line with slope $1$ has only one common point with the parabola $y^{2}=x$. What is the equation of this line?",
    "fact_expressions": "G: Parabola;H: Line;Expression(G) = (y^2 = x);Slope(H)=1;NumIntersection(H,G)=1",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y=x+1/4",
    "fact_spans": "[[[10, 22]], [[7, 9], [32, 34]], [[10, 22]], [[0, 9]], [[7, 29]]]",
    "query_spans": "[[[32, 39]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the foci of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(pm*sqrt(5), 0)",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 35]]]",
    "process": "The hyperbola $\\frac{x^2}{4}-y^{2}=1$ has $a=2$, $b=1$, so $c=\\sqrt{2^{2}+1^{2}}=\\sqrt{5}$. Therefore, the foci are $(-\\sqrt{5},0)$, $(\\sqrt{5},0)$."
  },
  {
    "text": "Given that the line $y = -x - 4$ intersects the parabola $y^{2} = -12x$ at points $A$ and $B$, and the focus of the parabola is $F$, $\\overrightarrow{F A} \\cdot \\overrightarrow{F B}$ = ?",
    "fact_expressions": "H: Line;Expression(H) = (y = -x - 4);G: Parabola;Expression(G) = (y^2 = -12*x);A: Point;B: Point;Intersection(H, G) = {A, B};F: Point;Focus(G) = F",
    "query_expressions": "DotProduct(VectorOf(F, A), VectorOf(F, B))",
    "answer_expressions": "-83",
    "fact_spans": "[[[2, 12]], [[2, 12]], [[13, 29], [42, 45]], [[13, 29]], [[32, 35]], [[36, 39]], [[2, 41]], [[49, 52]], [[42, 52]]]",
    "query_spans": "[[[54, 105]]]",
    "process": ""
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$), draw two mutually perpendicular rays intersecting the parabola at points $M$ and $N$. From the midpoint $P$ of chord $MN$, draw a perpendicular $PQ$ to the directrix of the parabola, with foot $Q$. Then the maximum value of $|PQ|$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2=2*p*x);p: Number;p>0;F: Point;Focus(G) = F;L1: Ray;L2: Ray;PointOnCurve(F, L1);PointOnCurve(F, L2);IsPerpendicular(L1, L2);M: Point;N: Point;Intersection(L1, G) = M;Intersection(L2, G) = N;IsChordOf(LineSegmentOf(M, N), G);P: Point;MidPoint(LineSegmentOf(M, N)) = P;Q: Point;PointOnCurve(P, LineSegmentOf(P, Q));IsPerpendicular(LineSegmentOf(P, Q), Directrix(G));FootPoint(LineSegmentOf(P, Q), Directrix(G)) = Q",
    "query_expressions": "Max(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[1, 22], [42, 45], [71, 74]], [[1, 22]], [[4, 22]], [[4, 22]], [[25, 28]], [[1, 28]], [], [], [[0, 38]], [[0, 38]], [[0, 38]], [[48, 52]], [[53, 56]], [[0, 56]], [[0, 56]], [[42, 64]], [[67, 70]], [[59, 70]], [[88, 91]], [[57, 84]], [[71, 91]], [[71, 91]]]",
    "query_spans": "[[[93, 106]]]",
    "process": "Draw perpendiculars from points M and N to the directrix of the parabola, with feet of perpendiculars denoted as M' and N'. Then $|PQ|=\\frac{1}{2}(|MM'|+|NN'|)\\leqslant\\sqrt{\\frac{|MM'|^{2}+|NN'|^{2}}{2}}\\cdot\\frac{1}{|MN|}\\sqrt{\\frac{|MF|^{2}+|NF|^{2}}{2}}$, it follows that $\\frac{|PQ|}{|MN|}\\leqslant\\frac{\\sqrt{2}}{2}$, with equality if and only if $|MF|=|NF|$."
  },
  {
    "text": "The focal distance of the ellipse $\\frac{x^{2}}{a}+y^{2}=1$ $(a>1)$ is $2$, then $a$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2 + x^2/a = 1);FocalLength(G) = 2;a: Number;a>1",
    "query_expressions": "a",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 32]], [[0, 32]], [[0, 39]], [[41, 44]], [[2, 32]]]",
    "query_spans": "[[[41, 46]]]",
    "process": "The focal distance of the ellipse $\\frac{x^{2}}{a}+y^{2}=1$ $(a>1)$ is 2, that is, $c=\\sqrt{a-1}=1 \\Rightarrow a=2$."
  },
  {
    "text": "The equation of the ellipse that has the same foci as the ellipse $4 x^{2}+9 y^{2}=36$ and passes through the point $(-3 , 2)$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (4*x^2 + 9*y^2 = 36);G1: Ellipse;Focus(G) = Focus(G1);H: Point;Coordinate(H) = (-3, 2);PointOnCurve(H, G1) = True",
    "query_expressions": "Expression(G1)",
    "answer_expressions": "x^2/15+y^2/10=1",
    "fact_spans": "[[[1, 23]], [[1, 23]], [[44, 46]], [[0, 46]], [[32, 43]], [[32, 43]], [[31, 46]]]",
    "query_spans": "[[[44, 50]]]",
    "process": "[Analysis] Since the ellipse to be determined has the same foci as the given ellipse, we can assume the equation of the required ellipse. Then, using the fact that the ellipse passes through the point $(-3,2)$, we obtain a system of equations in parameters $a$ and $b$. Solving this system yields the values of $a$ and $b$, thereby giving the equation of the ellipse. From the given condition, the ellipse equation $4x^{2}+9y^{2}=36$ can be written as $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$. Thus, the foci of this ellipse are $(\\sqrt{5},0)$ and $(-\\sqrt{5},0)$. Let the equation of the required ellipse be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. Then  \n$$\n\\begin{cases}\na^{2}-b^{2}=5 \\\\\n\\frac{9}{a^{2}}+\\frac{4}{b^{2}}=1\n\\end{cases}\n$$\nSolving gives  \n$$\n\\begin{cases}\na^{2}=15 \\\\\nb^{2}=10\n\\end{cases}\n$$\nor  \n$$\n\\begin{cases}\na^{2}=3 \\\\\nb^{2}=-2\n\\end{cases}\n$$\n(the latter is discarded). Therefore, $a^{2}=15$, $b^{2}=10$. Hence, the equation of the required ellipse is $\\frac{x^{2}}{15}+\\frac{y^{2}}{10}=1$. First determine the position of the foci, then establish a system of equations in $a$ and $b$ based on the conditions. If the position of the foci is uncertain, consider whether there are two solutions. Sometimes, for convenience, the equation of the ellipse may also be assumed in the form $mx^{2}+ny^{2}=1$ $(m>0,n>0,m\\neq n)$."
  },
  {
    "text": "Draw a line through the right focus $F$ and the endpoint $B$ of the imaginary axis of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$). If the distance from the right vertex $A$ to the line $FB$ is equal to $\\frac{b}{\\sqrt{7}}$, then the eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F: Point;B: Point;A: Point;e: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G)=F;Endpoint(ImageinaryAxis(G))=B;RightVertex(G)=A;Distance(A, LineOf(F,B)) = b/sqrt(7);Eccentricity(G) = e;L:Line;PointOnCurve(F,L);PointOnCurve(B,L)",
    "query_expressions": "e",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 58], [121, 124]], [[4, 58]], [[4, 58]], [[62, 65]], [[70, 73]], [[83, 86]], [[128, 131]], [[4, 58]], [[4, 58]], [[1, 58]], [[1, 65]], [[1, 73]], [[1, 86]], [[83, 119]], [[121, 131]], [[76, 78]], [[0, 78]], [[0, 78]]]",
    "query_spans": "[[[128, 133]]]",
    "process": ""
  },
  {
    "text": "A line with slope $\\frac{\\sqrt{3}}{3}$ passes through the left focus $F_{1}$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ and intersects the right branch of the hyperbola at point $P$, and $P F_{2}$ is perpendicular to the $x$-axis ($F_{2}$ being the right focus). Then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;P: Point;F2: Point;F1:Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Slope(H)=sqrt(3)/3;LeftFocus(G)=F1;PointOnCurve(F1,H);Intersection(H,RightPart(G))=P;IsPerpendicular(LineSegmentOf(P,F2),xAxis);RightFocus(G)=F2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[27, 83], [95, 98], [142, 145]], [[30, 83]], [[30, 83]], [[24, 26]], [[103, 107]], [[127, 134]], [[87, 94]], [[30, 83]], [[30, 83]], [[27, 83]], [[0, 26]], [[27, 94]], [[24, 94]], [[24, 107]], [[109, 125]], [[95, 139]]]",
    "query_spans": "[[[142, 151]]]",
    "process": "Let the equation of the line be $ y = \\frac{\\sqrt{3}}{3}(x + c) $, find the y-coordinate of the intersection point of the line with the right branch. Since $ PF_2 $ is perpendicular to the x-axis, use the properties of the hyperbola to set up an equation and solve for the eccentricity. Given that a line with slope $ \\frac{\\sqrt{3}}{3} $ passes through the left focus $ F_1 $ of the hyperbola $ \\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1 $ ($ a > 0, b > 0 $), the equation of the line is $ y = \\frac{\\sqrt{3}}{3}(x + c) $. The y-coordinate of point $ P $ is $ \\frac{2\\sqrt{3}c}{3} $. Since $ PF_2 $ is perpendicular to the x-axis ($ F_2 $ being the right focus), we have $ \\frac{2\\sqrt{3}c}{3} = \\frac{b^2}{a} = \\frac{c^2 - a^2}{a} $, leading to $ e^2 - \\frac{2\\sqrt{3}}{3}e - 1 = 0 $, $ e > 1 $. Solving gives $ e = \\sqrt{3} $. Therefore, the eccentricity of the hyperbola is $ \\sqrt{3} $."
  },
  {
    "text": "Find the standard equation of a parabola whose focus lies on the line $3x - 4y - 12 = 0$?",
    "fact_expressions": "PointOnCurve(Focus(G), H) = True;H: Line;Expression(H) = (3*x - 4*y - 12 = 0);G: Parabola",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{(y^2=16*x),(x^2=-12*y)}",
    "fact_spans": "[[[1, 25]], [[4, 20]], [[4, 20]], [[22, 25]]]",
    "query_spans": "[[[22, 31]]]",
    "process": "Since 3x - 4y - 12 = 0, let x = 0, we get y = -3, so F(0, -3), thus the standard equation of the parabola is x^{2} = -12y. Let y = 0, we get x = 4, so F(4, 0), thus the standard equation of the parabola is y^{2} = 16x. In summary: the standard equations of the parabola are y^{2} = 16x, x^{2} = -12y."
  },
  {
    "text": "The circle $C$ passes through the point $(0,2)$, and its center $C$ lies on the parabola $y^{2}=x$ (not coinciding with the origin). If the circle $C$ intersects the $y$-axis at points $A$ and $B$, and $|AB|=4$, then the coordinates of the center $C$ are?",
    "fact_expressions": "G: Parabola;C: Circle;C1:Point;H: Point;A: Point;B: Point;O:Origin;Expression(G) = (y^2 = x);Coordinate(H) = (0, 2);Center(C)=C1;PointOnCurve(H, C);PointOnCurve(C1,G);Negation(C1=O);Intersection(C, yAxis) = {A, B};Abs(LineSegmentOf(A, B)) = 4",
    "query_expressions": "Coordinate(C1)",
    "answer_expressions": "(16,4)",
    "fact_spans": "[[[21, 33]], [[0, 4], [44, 48]], [[17, 20], [78, 81]], [[5, 13]], [[55, 59]], [[60, 63]], [[37, 39]], [[21, 33]], [[5, 13]], [[0, 20]], [[0, 13]], [[17, 34]], [[17, 42]], [[44, 63]], [[65, 74]]]",
    "query_spans": "[[[78, 86]]]",
    "process": "Let the center of the circle be $ C(m^{2},m) $. Use the distance formula between two points to find the radius, then write the standard equation of the circle, and finally determine the coordinates of the center based on the chord length. Let the center of the circle be $ C(m^{2},m) $, then the radius of the circle is $ r = \\sqrt{m^{4} + (m - 2)^{2}} $. The equation of circle $ C $ is $ (x - m^{2})^{2} + (y - m)^{2} = m^{4} + (m - 2)^{2} $. Letting $ x = 0 $, we obtain $ y^{2} - 2my + 4m - 4 = 0 $, then $ |AB| = |y_{1} - y_{2}| = \\sqrt{(y_{1} + y_{2})^{2} - 4y_{1}y_{2}} = \\sqrt{4m^{2} - 4(4m - 4)} = 4 $, and $ m \\neq 0 $, hence $ m = 4 $. Therefore, the coordinates of the center $ C $ are $ (16, 4) $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x$ has the directrix $x=-1$, if $M$ is a moving point on $C$, and the coordinates of point $N$ are $(3,0)$, then the minimum value of $|M N|$ is?",
    "fact_expressions": "C: Parabola;p: Number;M: Point;N: Point;Expression(C) = (y^2 = 2*(p*x));Coordinate(N) = (3, 0);Expression(Directrix(C)) = (x = -1);PointOnCurve(M, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(M, N)))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 23], [39, 42]], [[10, 23]], [[35, 38]], [[50, 54]], [[2, 23]], [[50, 65]], [[2, 33]], [[35, 48]]]",
    "query_spans": "[[[67, 80]]]",
    "process": "From the given conditions, p=2, therefore the parabola C: y^2=4x. Let M(x_{0},y_{0}) (x_{0}\\geqslant0), and from the conditions we have y_{0}^{2}=4x_{0}. Then |MN|^{2}=(x_{0}-3)^{2}+y_{0}^{2}=(x_{0}-3)^{2}+4x_{0}=(x_{0}-1)^{2}+8\\geqslant8. When x_{0}=1, |MN|^{2} attains the minimum value 8, therefore the minimum value of |MN| is 2,\\sqrt{5}."
  },
  {
    "text": "Given that the equation of the directrix of the parabola $x^{2}=a y$ is $y=-\\frac{1}{4}$, then $a$=?",
    "fact_expressions": "G: Parabola;a: Number;Expression(G) = (x^2 = a*y);Expression(Directrix(G)) = (y = -1/4)",
    "query_expressions": "a",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 16]], [[40, 43]], [[2, 16]], [[2, 38]]]",
    "query_spans": "[[[40, 45]]]",
    "process": "According to the given conditions, the parabola opens in the positive direction of the y-axis, and its standard equation is $x^{2}=2py$ ($p>0$). Since its directrix is $y=-\\frac{1}{4}$, it follows that $\\frac{p}{2}=\\frac{1}{4}$, so $p=\\frac{1}{2}$, and therefore $a=2p=1$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the common foci of the ellipse $C_{1}$ and the hyperbola $C_{2}$, $P$ is a common point of $C_{1}$ and $C_{2}$, and $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$. The eccentricities of the ellipse $C_{1}$ and the hyperbola $C_{2}$ are $e_{1}$ and $e_{2}$, respectively. Then the minimum value of $e_{1} \\cdot e_{2}$ is?",
    "fact_expressions": "C1: Ellipse;C2: Hyperbola;F1: Point;P: Point;F2: Point;Focus(C1)={F1,F2};Focus(C2)={F1,F2};OneOf(Intersection(C1,C2))=P;AngleOf(F1, P, F2) = pi/3;e1:Number;e2:Number;Eccentricity(C1)=e1;Eccentricity(C2)=e2",
    "query_expressions": "Min(e1*e2)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[18, 27], [48, 55], [108, 117]], [[28, 38], [56, 63], [118, 128]], [[2, 9]], [[44, 47]], [[10, 17]], [[2, 43]], [[2, 43]], [[44, 69]], [[71, 107]], [[135, 142]], [[143, 150]], [[108, 150]], [[108, 150]]]",
    "query_spans": "[[[152, 177]]]",
    "process": ""
  },
  {
    "text": "$M$ is a moving point on the ellipse $\\frac{x^{2}}{9}+y^{2}=1$, and $F_{1}$, $F_{2}$ are the two foci of the ellipse. Then the maximum value of $\\angle F_{1} M F_{2}$ is?",
    "fact_expressions": "M: Point;PointOnCurve(M, G);G: Ellipse;Expression(G) = (x^2/9 + y^2 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2}",
    "query_expressions": "Max(AngleOf(F1, M, F2))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 3]], [[0, 34]], [[4, 31], [51, 53]], [[4, 31]], [[35, 42]], [[43, 50]], [[35, 57]]]",
    "query_spans": "[[[59, 87]]]",
    "process": "From the standard equation of the ellipse $\\frac{x^{2}}{9}+y^{2}=1$, we know $a=3$, $b=1$, $c=\\sqrt{a^{2}-b^{2}}=2\\sqrt{2}$, so $|MF_{1}|+|MF_{2}|=2a=6$, $|F_{1}F_{2}|=2c=4\\sqrt{2}$. In the focal triangle $F_{1}MF_{2}$, by the cosine theorem, we have $|F_{2}|^{2}-|F_{1}F_{2}\\frac{2|MF_{1}|\\cdot|MF_{2}|-|F_{1}F_{2}|}{|\\times|MF_{2}|}$. By the basic inequality, we know $|MF_{1}|+|MF_{2}|\\geqslant2\\sqrt{|MF_{1}|\\cdot|MF_{2}|}$, so $|MF_{1}|\\cdot|MF_{2}|\\leqslant\\frac{1}{4}\\times6^{2}=9$, that is, $\\frac{2}{|MF|\\times|MF_{2}|}-1\\geqslant\\frac{2}{9}-1=-\\frac{7}{9}$. Combining the monotonicity of the cosine function, the maximum value of $\\angle F_{1}MF_{2}$ is $\\pi-\\arccos\\frac{7}{9}$."
  },
  {
    "text": "The equation of the directrix of the parabola $y^{2}=x$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = x)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x = -1/4",
    "fact_spans": "[[[0, 12]], [[0, 12]]]",
    "query_spans": "[[[0, 19]]]",
    "process": ""
  },
  {
    "text": "The foci of the ellipse $m x^{2}+y^{2}=1$ lie on the $y$-axis, and the length of the major axis is 3 times the length of the minor axis. Then $m=?$",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (m*x^2 + y^2 = 1);PointOnCurve(Focus(G), yAxis);Length(MajorAxis(G))=3*Length(MinorAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "9",
    "fact_spans": "[[[0, 19]], [[44, 47]], [[0, 19]], [[0, 28]], [[0, 42]]]",
    "query_spans": "[[[44, 49]]]",
    "process": ""
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{2+m}-\\frac{y^{2}}{m+1}=1$ represents a hyperbola with foci on the $y$-axis, what is the range of values for $m$?",
    "fact_expressions": "G: Hyperbola;m:Number;Expression(G) = (x^2/(m + 2) - y^2/(m + 1) = 1);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-oo, -2)",
    "fact_spans": "[[[54, 57]], [[59, 62]], [[2, 57]], [[45, 57]]]",
    "query_spans": "[[[59, 69]]]",
    "process": "According to the standard equation of a hyperbola with foci on the y-axis:  \n\\begin{cases}2+m<0\\\\m+1<0\\end{cases},  \nsolving gives $ m<-2 $, that is, the range of $ m $ is $ (-\\infty,-2) $."
  },
  {
    "text": "If $A$ and $B$ are the two vertices on the minor axis of the ellipse $E$: $x^{2}+\\frac{y^{2}}{m}=1 ,(m>1)$, and point $P$ is any point on the ellipse distinct from $A$ and $B$, and if the product of the slopes of lines $AP$ and $BP$ is $-\\frac{4}{m}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "E: Ellipse;m: Number;A: Point;P: Point;B: Point;m>1;Expression(E) = (x^2 + y^2/m = 1);Vertex(MinorAxis(E)) = {A, B};PointOnCurve(P, E);Negation(P = A);Negation(P = B);Slope(LineOf(A, P))*Slope(LineOf(B, P)) = -4/m",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[11, 50], [64, 66], [118, 120]], [[18, 50]], [[1, 4], [69, 72]], [[59, 63]], [[5, 8], [73, 76]], [[18, 50]], [[11, 50]], [[1, 58]], [[59, 81]], [[59, 81]], [[59, 81]], [[83, 116]]]",
    "query_spans": "[[[118, 126]]]",
    "process": "Let the equations of lines AP and BP be y = k_{AP}(x - 1) and y = k_{BP}(x + 1), respectively. Let point P(x_{0}, y_{0}), where k_{AP} = \\frac{y_{0}}{x_{0} - 1}, k_{BP} = \\frac{y_{0}}{x_{0} + 1}. Then k_{AP} \\cdot k_{BP} = \\frac{y_{0}^{2}}{x_{0}^{2} - 1} = -\\frac{4}{m}\\textcircled{1}. Also, since point P lies on the ellipse E: x^{2} + \\frac{y^{2}}{m} = 1, we have x_{0}^{2} - 1 = -\\frac{y_{0}^{2}}{m}\\textcircled{2}. From \\textcircled{1} and \\textcircled{2}, it follows that m^{2} = 4. Since m > 1, we obtain m = 2. Thus, the eccentricity is e = \\frac{c}{a} = \\frac{1}{\\sqrt{2}} = \\frac{\\sqrt{2}}{2}."
  },
  {
    "text": "Given that the line $y=k(x+2)$ intersects the parabola $C$: $y^{2}=8x$ at two points $A$ and $B$, and $F$ is the focus of the parabola $C$. If $|FA|=2|FB|$, then $k=$?",
    "fact_expressions": "C: Parabola;G: Line;F: Point;A: Point;B: Point;k: Number;Expression(C) = (y^2 = 8*x);Expression(G) = (y = k*(x + 2));Intersection(G, C) = {A, B};Focus(C) = F;Abs(LineSegmentOf(F, A)) = 2*Abs(LineSegmentOf(F, B))",
    "query_expressions": "k",
    "answer_expressions": "pm*2*sqrt(2)/3",
    "fact_spans": "[[[15, 34], [51, 57]], [[2, 14]], [[47, 50]], [[37, 40]], [[41, 44]], [[77, 80]], [[15, 34]], [[2, 14]], [[2, 46]], [[47, 60]], [[63, 75]]]",
    "query_spans": "[[[77, 82]]]",
    "process": "1. Solve the system of equations of the line and the parabola, then use the relationship between roots and coefficients, combined with the definition of the parabola and |FA| = 2|FB|, to set up a system of equations, from which the result can be obtained. According to the problem, let A(x_{1}, y_{1}), B(x_{2}, y_{2}). From the definition of the parabola, we have |FA| = x_{1} + 2, |FB| = x_{2} + 2. Since |FA| = 2|FB|, it follows that x_{1} + 2 = 2(x_{2} + 2), i.e., x_{1} = 2x_{2} + 2\\textcircled{1}; solving simultaneously \\begin{cases} y = k(x + 2) \\\\ y^{2} = 8x \\end{cases}, after rearranging we get k^{2}x^{2} + (4k^{2} - 8)x + 4k^{2} = 0, so \\Delta = (4k^{2} - 8)^{2} - 16k^{4} > 0, hence -1 < k < 1. Also, \\begin{cases} x_{1} + x_{2} = \\frac{8 - 4k^{2}}{k^{2}} \\textcircled{2} \\end{cases}, solving \\textcircled{1}\\textcircled{2}\\textcircled{3} gives k = \\pm\\frac{2\\sqrt{2}}{3}, which satisfies the conditions. x_{2} = 4\\textcircled{3} Note: This problem mainly examines the application of lines and parabolas, belonging to a medium-difficulty type."
  },
  {
    "text": "Given the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through $F_{1}$ intersects the left and right branches of the hyperbola at points $P$ and $Q$ respectively, and intersects the $y$-axis at point $C$. Let $M$ be the midpoint of segment $P Q$. If $C F_{2} \\perp O M$, then the eccentricity of hyperbola $E$ is?",
    "fact_expressions": "l: Line;E: Hyperbola;a: Number;b: Number;P: Point;Q: Point;C: Point;F2: Point;O: Origin;M: Point;F1: Point;a>0;b>0;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(E) = F2;LeftFocus(E) = F1;PointOnCurve(F1, l);Intersection(l, LeftPart(E)) = P;Intersection(l, RightPart(E)) = Q;Intersection(yAxis, l) = C;MidPoint(LineSegmentOf(P, Q)) = M;IsPerpendicular(LineSegmentOf(C, F2), LineSegmentOf(O, M))",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[99, 104]], [[2, 63], [105, 108], [178, 184]], [[10, 63]], [[10, 63]], [[118, 121]], [[122, 125]], [[135, 139]], [[81, 90]], [[157, 176]], [[140, 143]], [[72, 79], [91, 98]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 90]], [[2, 90]], [[90, 104]], [[99, 127]], [[99, 127]], [[99, 139]], [[140, 154]], [[157, 176]]]",
    "query_spans": "[[[178, 190]]]",
    "process": "From the given conditions, draw the graph. It is known that the slope of line PQ must exist; let the slope of line PQ be $k_{PQ}$ ($k_{PQ} \\neq 0$). By the symmetry of the hyperbola, the slope of line $CF_{2}$ is $k_{CF_{2}} = -k_{PQ}$. Since $CF_{2} \\perp OM$, the slope of line OM is $k_{OM} = \\frac{1}{k_{PQ}}$. Let $P(x_{1}, y_{1})$, $Q(x_{2}, y_{2})$, $M(x_{0}, y_{0})$, then $x_{1} + x_{2} = 2x_{0}$, $y_{1} + y_{2}$. Since $P$, $Q$ lie on the hyperbola, subtracting $\\textcircled{2}$ from $\\textcircled{1}$ gives $\\frac{(x_{1}+x_{2})(}{a^{2}}$, then $\\frac{b^{2}}{a^{2}} = 1$, hence the eccentricity of the hyperbola is $e = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{2}$, i.e., $\\frac{y_{1}+y_{2})(y_{1}-y_{2})}{2} = \\frac{y_{1}y_{2}y_{0}}{x_{1}-x_{2}} \\cdot \\frac{b^{2}}{x_{0}}$, i.e., $k_{PQ}^{2} k_{OM} = \\frac{b^{2}}{a^{2}}$. When the slope of line PQ is $k_{PQ} = 0$, $M(0,0)$, $C(0,0)$, which does not satisfy the conditions."
  },
  {
    "text": "In $\\triangle A B C$, $A B=4$, $B C=6 \\sqrt{2}$, $\\angle C B A=\\frac{\\pi}{4}$. If the hyperbola $\\Gamma$ has $A B$ as its transverse axis and passes through point $C$, then the focal length of $\\Gamma$ is?",
    "fact_expressions": "Gamma: Hyperbola;A: Point;B: Point;C: Point;LineSegmentOf(A, B) = 4;LineSegmentOf(B, C) = 6*sqrt(2);AngleOf(C, B, A) = pi/4;LineSegmentOf(A, B) = RealAxis(Gamma);PointOnCurve(C, Gamma)",
    "query_expressions": "FocalLength(Gamma)",
    "answer_expressions": "8",
    "fact_spans": "[[[79, 90], [108, 116]], [[1, 18]], [[1, 18]], [[102, 106]], [[20, 27]], [[30, 46]], [[49, 77]], [[79, 99]], [[79, 106]]]",
    "query_spans": "[[[108, 121]]]",
    "process": "As shown in the figure, let the equation of the hyperbola be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$, then according to the given conditions, $2a=4$, $a=2$. In $\\triangle ABC$, $AB=4$, $BC=6\\sqrt{2}$, $\\angle CBA=\\frac{\\pi}{4}$. $\\therefore$ The horizontal coordinate of $C$ is $-(BC\\cdot\\cos\\angle CBA-2)=-4$, and the vertical coordinate is $BC\\cdot\\sin\\angle CBA=6$. $\\because$ The hyperbola passes through point $C$, then $\\frac{16}{4}-\\frac{36}{b^{2}}=1$, solving gives: $b^{2}=12$, $\\therefore c^{2}=a^{2}+b^{2}=16$, $c=4$. Then the focal length of $T$ is $8$. Answer: $8$."
  },
  {
    "text": "If a point $M$ on the parabola $y^{2}=4x$ is at a distance $|MF|=5$ from the focus $F$ of the parabola, then what is the distance from point $M$ to the $x$-axis?",
    "fact_expressions": "G: Parabola;M: Point;F: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(M,G);Focus(G)=F;Distance(M,F)=Abs(LineSegmentOf(M,F));Abs(LineSegmentOf(M,F))=5",
    "query_expressions": "Distance(M, xAxis)",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 15], [23, 26]], [[18, 21], [46, 50]], [[29, 32]], [[1, 15]], [[1, 21]], [[23, 32]], [[18, 44]], [[35, 44]]]",
    "query_spans": "[[[46, 60]]]",
    "process": ""
  },
  {
    "text": "Given a point $M(4, y_{0})$ on the parabola $x^{2}=2 p y$ $(p>0)$, the distance from $M$ to the focus $F$ is $|M F|=\\frac{5}{4} y_{0}$. What are the coordinates of the focus $F$?",
    "fact_expressions": "G: Parabola;p: Number;M: Point;F: Point;p>0;Expression(G) = (x^2 = 2*(p*y));y0: Number;Coordinate(M) = (4, y0);PointOnCurve(M, G);Focus(G) = F;Distance(M, F) = Abs(LineSegmentOf(M, F)) ;Abs(LineSegmentOf(M, F)) = (5/4)*y0",
    "query_expressions": "Coordinate(F)",
    "answer_expressions": "(0, 1)",
    "fact_spans": "[[[2, 23]], [[5, 23]], [[26, 39]], [[42, 45], [77, 80]], [[5, 23]], [[2, 23]], [[26, 39]], [[26, 39]], [[2, 39]], [[2, 45]], [[26, 73]], [[48, 73]]]",
    "query_spans": "[[[77, 85]]]",
    "process": "From the given condition, |MF| = y_{0} + \\frac{p}{2} = \\frac{5}{4}y_{0} \\Rightarrow y_{0} = 2p, therefore 4^{2} = 2p(2p) \\Rightarrow p = 2, thus the coordinates of the focus F are (0,1)."
  },
  {
    "text": "Given the hyperbola $C$ with equation $\\frac{x^{2}}{a^{2}}-y^{2}=1(a>0)$, $A(-4,0)$, $B(-1,0)$, and there exists a point $P$ on hyperbola $C$ such that $|P A|=2|P B|$, then the maximum value of the real number $a$ is?",
    "fact_expressions": "C: Hyperbola;A: Point;B: Point;P: Point;a:Real;Coordinate(A) = (-4, 0);Coordinate(B) = (-1, 0);Expression(C) = (-y^2 + x^2/a^2 = 1);PointOnCurve(P, C);Abs(LineSegmentOf(P, A)) = 2*Abs(LineSegmentOf(P, B))",
    "query_expressions": "Max(a)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 8], [71, 77]], [[49, 58]], [[61, 70]], [[82, 85]], [[104, 109]], [[49, 58]], [[61, 70]], [[2, 46]], [[71, 85]], [[88, 102]]]",
    "query_spans": "[[[104, 115]]]",
    "process": "Let point $ P(x, y) $. From $ |PA| = 2|PB| $, we get: $ |PA|^2 = 4|PB|^2 $. Therefore, $ (x+4)^2 + y^2 = 4[(x+1)^2 + y^2] $, simplifying gives: $ x^2 + y^2 = 4 $. Thus, the point $ P(x, y) $ satisfying $ |PA| = 2|PB| $ moves on the circle $ x^2 + y^2 = 4 $. Also, point $ P(x, y) $ lies on $ \\frac{x^2}{a} - y^2 = 1 $ ($ a > 0 $). Hence, if the hyperbola and the circle have intersection points, then $ a < 2 $. That is, the maximum value of real number $ a $ is $ 2 $. The answer is: $ 2 $"
  },
  {
    "text": "5. In the rectangular coordinate system $x O y$, if the two asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ are perpendicular to each other, then what is the value of the positive real number $b$?",
    "fact_expressions": "E: Hyperbola;b: Real;b > 0;Expression(E) = (x**2 - y**2/b**2 = 1);l1: Line;l2: Line;Asymptote(E) = {l1, l2};IsPerpendicular(l1, l2)",
    "query_expressions": "b",
    "answer_expressions": "1",
    "fact_spans": "[[[23, 26]], [[75, 82]], [[27, 62]], [[23, 62]], [[64, 69]], [[64, 69]], [[23, 69]], [[23, 73]]]",
    "query_spans": "[[[75, 87]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=6x$ and a line $l$ intersecting at points $A$ and $B$, if the ordinate of the midpoint of segment $AB$ is $3$, then what is the inclination angle of $l$?",
    "fact_expressions": "l: Line;C: Parabola;B: Point;A: Point;Expression(C) = (y^2 = 6*x);Intersection(l,C) = {A, B};YCoordinate(MidPoint(LineSegmentOf(A,B))) = 3",
    "query_expressions": "Inclination(l)",
    "answer_expressions": "ApplyUnit(45,degree)",
    "fact_spans": "[[[22, 27], [59, 62]], [[2, 21]], [[33, 36]], [[29, 32]], [[2, 21]], [[2, 38]], [[40, 57]]]",
    "query_spans": "[[[59, 68]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ y_{1}^{2}=6x_{1} $, $ y_{2}^{2}=6x_{2} $. Subtracting these two equations gives $ (y_{1}-y_{2})(y_{1}+y_{2})=6(x_{1}-x_{2}) $, then $ \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\\cdot(y_{1}+y_{2})=k_{AB}\\cdot6=6 $. Hence the slope of $ l $ is 1, then the inclination angle of $ l $ is $ 45^{\\circ} $. The answer is $ \\cdot45c $"
  },
  {
    "text": "If a hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has an asymptote whose chord length intercepted by the circle $(x-2)^{2}+y^{2}=4$ is $2$, then the value of $\\frac{b}{a}$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y^2 + (x - 2)^2 = 4);Length(InterceptChord(OneOf(Asymptote(C)),G))=2",
    "query_expressions": "b/a",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[1, 62]], [[9, 62]], [[9, 62]], [[69, 89]], [[9, 62]], [[9, 62]], [[1, 62]], [[69, 89]], [[1, 99]]]",
    "query_spans": "[[[101, 118]]]",
    "process": "The hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) has an asymptote given by $ bx - ay = 0 $. The circle $ (x-2)^{2} + y^{2} = 4 $ has center $ (2, 0) $ and radius $ r = 2 $. The distance from the center to the asymptote is $ d = \\frac{|2b - 0|}{\\sqrt{a^{2} + b^{2}}} $. Then $ 2 = 2\\sqrt{4 - \\frac{4b^{2}}{a^{2} + b^{2}}} $, simplifying yields $ 3a^{2} = b^{2} $, that is, $ \\frac{b}{a} = \\sqrt{3} $."
  },
  {
    "text": "The parabola $C$: $y^{2}=4 x$ has focus $F$, point $P$ lies on the parabola, and $P F=3$. Then the distance from point $P$ to the line $x=-1$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;P: Point;PointOnCurve(P, C);LineSegmentOf(P, F) = 3;G: Line;Expression(G) = (x = -1)",
    "query_expressions": "Distance(P, G)",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 19], [32, 35]], [[0, 19]], [[23, 26]], [[0, 26]], [[27, 31], [47, 51]], [[27, 36]], [[38, 45]], [[52, 60]], [[52, 60]]]",
    "query_spans": "[[[48, 65]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the foci of the curve $C_{1}$: $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$, and let $P$ be an intersection point of the curves $C_{2}$: $\\frac{x^{2}}{3}-y^{2}=1$ and $C_{1}$. Then the value of $\\cos \\angle F_{1} P F_{2}$ is?",
    "fact_expressions": "C1: Curve;C2: Curve;F1: Point;P: Point;F2: Point;Expression(C1)= (x^2/6 + y^2/2 = 1);Expression(C2)=( x^2/3 - y^2 = 1);Focus(C1)={F1,F2};OneOf(Intersection(C1,C2))=P",
    "query_expressions": "Cos(AngleOf(F1, P, F2))",
    "answer_expressions": "1/3",
    "fact_spans": "[[[17, 63], [108, 115]], [[71, 107]], [[1, 8]], [[67, 70]], [[9, 16]], [[17, 63]], [[71, 107]], [[1, 66]], [[67, 120]]]",
    "query_spans": "[[[122, 153]]]",
    "process": "The curves $C_{1}: \\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$ and $C_{2}: \\frac{x^{2}}{3}-y^{2}=1$ have coincident foci, and the two curves have four intersection points. Let $P$ be the intersection point in the first quadrant. Then $PF_{1}+PF_{2}=2\\sqrt{6}$, $PF_{1}-PF_{2}=2\\sqrt{3}$, solving gives $PF_{1}=\\sqrt{6}+\\sqrt{3}$, $PF_{2}=\\sqrt{6}-\\sqrt{3}$. Also $F_{1}F_{2}=4$, in $\\triangle F_{1}PF_{2}$, by the law of cosines we obtain $\\cos\\angle F_{1}PF_{2}=\\frac{(\\sqrt{6}+\\sqrt{3})^{2}+(\\sqrt{6}-\\sqrt{3})^{2}-4^{2}}{2\\times(\\sqrt{6}+\\sqrt{3})(\\sqrt{6}-\\sqrt{3})}=\\frac{1}{3}$ Answer: $\\frac{1}{2}$"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has left and right foci $F_{1}$, $F_{2}$ respectively, the focus of the parabola $y^{2}=2 b x$ is $F$, and if $\\overrightarrow{F_{1} F}=\\frac{5}{3} \\overrightarrow{F F_{2}}$, then $\\frac{a}{b}=$?",
    "fact_expressions": "H: Ellipse;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(H) = F1;RightFocus(H) = F2;G: Parabola;Expression(G) = (y^2 = 2*(b*x));F: Point;Focus(G) = F;VectorOf(F1, F) = (5/3)*VectorOf(F, F2)",
    "query_expressions": "a/b",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 54]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[63, 70]], [[71, 78]], [[2, 78]], [[2, 78]], [[79, 95]], [[79, 95]], [[99, 102]], [[79, 102]], [[104, 167]]]",
    "query_spans": "[[[169, 184]]]",
    "process": "From the given conditions, we have: $\\frac{|F_{1}F|}{|FF_{2}|}=\\frac{5}{3}$. Without loss of generality, assume $|F_{1}F|=5m$ $(m>0)$, then $|F_{2}F|=3m$. Combining with the parabola equation, we get: $\\frac{b}{2}=m$, $\\therefore b=2m$. Combining with the ellipse equation, we obtain: $a=\\sqrt{b^{2}+c^{2}}=2\\sqrt{5}m$, $\\therefore \\frac{a}{b}=\\sqrt{5}$"
  },
  {
    "text": "It is known that an ellipse and a hyperbola share common foci $F_{1}$, $F_{2}$, and $P$ is one of their intersection points, $\\angle F_{1} P F_{2}=60^{\\circ}$. Denote the eccentricities of the ellipse and the hyperbola as $e_{1}$, $e_{2}$, respectively. Then the minimum value of $e_{1}^{2}+e_{2}^{2}$ is?",
    "fact_expressions": "H: Ellipse;G: Hyperbola;F1: Point;F2: Point;Focus(H) = {F1, F2};Focus(G) = {F1, F2};P: Point;OneOf(Intersection(H, G)) = P;AngleOf(F1, P, F2) = ApplyUnit(60, degree);e1: Number;e2: Number;Eccentricity(H) = e1;Eccentricity(G) = e2",
    "query_expressions": "Min(e1^2 + e2^2)",
    "answer_expressions": "1+(sqrt(3)/2)",
    "fact_spans": "[[[2, 4], [77, 79]], [[5, 8], [80, 83]], [[14, 21]], [[22, 29]], [[2, 29]], [[2, 29]], [[30, 33]], [[30, 41]], [[42, 75]], [[90, 97]], [[98, 105]], [[77, 105]], [[77, 105]]]",
    "query_spans": "[[[107, 134]]]",
    "process": "Set the standard equations of the ellipse and hyperbola, let |PF_{1}|=m, |PF_{2}|=n (m>n), use the definitions of the ellipse and hyperbola to solve for m and n, use the cosine law to obtain an equation involving e_{1} and e_{2}, then use the basic inequality to find the minimum value of e_{1}^{2}+e_{2}^{2}. Solution: Without loss of generality, assume the standard equations of the ellipse and hyperbola are \\frac{x^{2}}{a_{1}^{2}}+\\frac{y^{2}}{b_{1}^{2}}=1 (a_{1}>b_{1}>0), \\frac{x^{2}}{a_{2}^{2}}-\\frac{y^{2}}{b_{2}^{2}}=1 (a_{2}>0,b_{2}>0), respectively. Let the focal distance of both curves be 2c (c>0). Set |PF_{1}|=m, |PF_{2}|=n (m>n), then m+n=2a_{1}, m-n=2a_{2}, so \\begin{cases}m=a_{1}+a_{2}\\\\n=a_{1}-a_{2}\\end{cases} \\cos60^{\\circ}=\\frac{m^{2}+n^{2}-4c^{2}}{2mn}=\\frac{1}{2} becomes (a_{1}+a_{2})^{2}+(a_{1}-a_{2})^{2}-4c^{2}=(a_{1}+a_{2})(a_{1}-a_{2}), \\therefore a_{1}^{2}+3a_{2}^{2}-4c^{2}=0 \\therefore \\frac{1}{e_{1}^{2}}+\\frac{3}{e_{2}^{2}}=4, e_{1}^{2}+e_{2}^{2}=\\frac{1}{4}(e_{1}^{2}+e_{2}^{2})(\\frac{1}{e_{1}^{2}}+\\frac{3}{e_{2}^{2}})=\\frac{1}{4}(4+\\frac{e_{2}^{2}}{e_{1}^{2}}+\\frac{3e_{1}^{2}}{e_{2}^{2}})\\geqslant\\frac{1}{4}(4+2\\sqrt{3})=1+\\frac{\\sqrt{3}}{2}, equality holds if and only if e_{2}^{2}=\\sqrt{3}e_{1}^{2}, then the minimum value of e_{1}^{2}+e_{2}^{2} is 1+\\frac{\\sqrt{3}}{2}."
  },
  {
    "text": "Given that the line $y=x-1$ intersects the parabola $y^{2}=4 x$ at points $A$ and $B$, what is the length of chord $A B$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (y = x - 1);Intersection(H, G) = {A, B};IsChordOf(LineSegmentOf(A,B),G)",
    "query_expressions": "Length(LineSegmentOf(A,B))",
    "answer_expressions": "8",
    "fact_spans": "[[[12, 26]], [[2, 11]], [[28, 31]], [[32, 35]], [[12, 26]], [[2, 11]], [[2, 37]], [[12, 45]]]",
    "query_spans": "[[[40, 49]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $ C: y^{2} = 2 p x \\, (p > 0) $ with focus $ F $, point $ A(0, \\sqrt{3}) $, and point $ B $ on the parabola $ C $ satisfying $ AB \\perp AF $ and $ |BF| = 4 $, then $ p = $?",
    "fact_expressions": "C: Parabola;p: Number;A: Point;B: Point;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(A) = (0, sqrt(3));Focus(C) = F;PointOnCurve(B, C);IsPerpendicular(LineSegmentOf(A, B), LineSegmentOf(A, F));Abs(LineSegmentOf(B, F)) = 4",
    "query_expressions": "p",
    "answer_expressions": "{2, 6}",
    "fact_spans": "[[[2, 27], [55, 61]], [[97, 100]], [[37, 54]], [[63, 67]], [[31, 34]], [[9, 27]], [[2, 27]], [[37, 54]], [[2, 34]], [[55, 67]], [[69, 84]], [[86, 95]]]",
    "query_spans": "[[[97, 102]]]",
    "process": "Given $ F\\left(\\frac{P}{2},0\\right) $, $ k_{AF} = -\\frac{2\\sqrt{3}}{P} $. Since $ AB \\perp AF $, $ k_{AB} = \\frac{P}{2\\sqrt{3}} $. Thus, $ l_{AB}: y = \\frac{P}{2\\sqrt{3}}x + \\sqrt{3} $. From \n\\[\n\\begin{cases}\ny = \\frac{P}{2\\sqrt{3}}x + \\sqrt{3} \\\\\ny^2 = 2px\n\\end{cases}\n\\]\nwe get $ x_B = \\frac{6}{p} $. Since $ |BF| = 4 $, $ \\frac{6}{P} + \\frac{p}{2} = 4 $. Hence, $ p = 2 $ or $ p = 6 $."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=6 x$ is $F$, a line $l$ passing through point $F$ intersects the parabola at points $A$ and $B$, $O$ is the origin, if $\\overrightarrow{O M}=\\frac{1}{4} \\overrightarrow{O B}$, $\\overrightarrow{O A}=\\frac{1}{4} \\overrightarrow{O N}$, perpendiculars are drawn from points $M$ and $N$ to the directrix of the parabola, with feet of perpendiculars at points $C$ and $D$ respectively, then the minimum value of $|C D|$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 6*x);F: Point;Focus(G) = F;PointOnCurve(F, l) = True;l: Line;Intersection(l, G) = {A, B};B: Point;A: Point;O: Origin;VectorOf(O, M) = VectorOf(O, B)/4;VectorOf(O, A) = VectorOf(O, N)/4;L1: Line;L2: Line;PointOnCurve(M, L1) = True;PointOnCurve(N, L2) = True;M: Point;N: Point;IsPerpendicular(L1, Directrix(G)) = True;IsPerpendicular(L2, Directrix(G)) = True;FootPoint(L1, Directrix(G)) = C;FootPoint(L2, Directrix(G)) = D;C: Point;D: Point",
    "query_expressions": "Min(Abs(LineSegmentOf(C, D)))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 16], [36, 39], [187, 190]], [[2, 16]], [[20, 23], [25, 29]], [[2, 23]], [[24, 35]], [[30, 35]], [[30, 50]], [[45, 48]], [[41, 44]], [[51, 54]], [[61, 116]], [[119, 174]], [], [], [[175, 196]], [[175, 196]], [[176, 180]], [[181, 184]], [[175, 196]], [[175, 196]], [[175, 210]], [[175, 210]], [[202, 206]], [[207, 210]]]",
    "query_spans": "[[[212, 225]]]",
    "process": "Let the equation of line $ l $ be: $ x = ty + \\frac{3}{2} $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, $ M(x_{3}, y_{3}) $, $ N(x_{4}, y_{4}) $. From \n\\[\n\\begin{cases}\ny^2 = 6x \\\\\nx = ty + \\frac{3}{2}\n\\end{cases}\n\\Rightarrow y^{2} - 6ty - 9 = 0 \\Rightarrow y_{1}y_{2} = -9,\n\\]\nsince $ \\overrightarrow{OM} = \\frac{1}{4}\\overrightarrow{OB} $, $ \\overrightarrow{OA} = \\frac{1}{4}\\overrightarrow{ON} $, so $ (x_{3}, y_{3}) = \\frac{1}{4}(x_{2}, y_{2}) $, then $ |CD| = |y_{3}| + |y_{4}| = \\left|\\frac{1}{4}y_{2}\\right| + |4y_{1}| = -\\frac{1}{4}y_{2} + 4y_{1} \\geqslant 2\\sqrt{(x_{1}, y_{1}) = \\frac{1}{4}(x_{4}, y_{4})} $, that is, $ y_{3} = \\frac{1}{4}y_{2} $, $ y_{4} = 4y_{1} $, let $ y_{1} > 0 $, $ \\frac{y_{1}}{2} = 6 $, hence the minimum value of $ |CD| $ is 6 (equality holds if and only when $ y = \\frac{3}{4} $, $ y_{2} = -12 $)."
  },
  {
    "text": "The standard equation of a hyperbola with foci on the $x$-axis, focal length $10$, and the same asymptotes as the hyperbola $\\frac{y^{2}}{4}-x^{2}=1$ is?",
    "fact_expressions": "G: Hyperbola;C: Hyperbola;Expression(G) = (-x^2 + y^2/4 = 1);FocalLength(C) =10;PointOnCurve(Focus(C),xAxis);Asymptote(C)=Asymptote(G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/5 - y^2/20 = 1",
    "fact_spans": "[[[19, 47]], [[54, 57]], [[19, 47]], [[9, 57]], [[0, 57]], [[18, 57]]]",
    "query_spans": "[[[54, 64]]]",
    "process": "Since the two hyperbolas share asymptotes, assume the standard equation of the desired hyperbola is \\frac{y^{2}}{4}-x^{2}=-\\lambda(\\lambda>0). Combining with c^{2}=a^{2}+b^{2}, we can solve it. Let the standard equation of the desired hyperbola be \\frac{y^{2}}{4}-x^{2}=-\\lambda(\\lambda>0), that is, \\frac{x^{2}}{\\lambda}-\\frac{y^{2}}{4\\lambda}=1. Then 4\\lambda+\\lambda=25, solving gives \\lambda=5. Therefore, the standard equation of the desired hyperbola is \\frac{x^{2}}{5}-\\frac{y^{2}}{20}=1."
  },
  {
    "text": "Given the parabola $C$: $y=2 x^{2}$ and the line $y=k x+2$ intersect at points $A$ and $B$, $M$ is the midpoint of segment $AB$. A perpendicular is drawn from $M$ to the $x$-axis, with foot $N$. If $\\overrightarrow{N A} \\cdot \\overrightarrow{N B}=0$, then $k=?$",
    "fact_expressions": "C: Parabola;G: Line;k: Number;A: Point;B: Point;N: Point;M: Point;Expression(C) = (y = 2*x^2);Expression(G) = (y = k*x + 2);Intersection(C, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = M;l: Line;PointOnCurve(M, l);IsPerpendicular(l, xAxis);FootPoint(l, xAxis) = N;DotProduct(VectorOf(N, A), VectorOf(N, B)) = 0",
    "query_expressions": "k",
    "answer_expressions": "pm*4*sqrt(3)",
    "fact_spans": "[[[2, 21]], [[22, 33]], [[135, 138]], [[35, 38]], [[39, 42]], [[76, 79]], [[45, 48], [61, 64]], [[2, 21]], [[22, 33]], [[2, 44]], [[45, 59]], [], [[60, 72]], [[60, 72]], [[60, 79]], [[82, 133]]]",
    "query_spans": "[[[135, 140]]]",
    "process": "Let $ A(x_{1},2x_{1}), B(x_{2},2x_{2}) $. Solving the system of equations  \n$$\n\\begin{cases}\ny=2x^{2} \\\\\ny=kx+2\n\\end{cases}\n$$  \ngives $ 2x^{2}-kx-2=0 $.  \n$ x_{1}+x_{2}=\\frac{k}{2}, x_{1}x_{2}=-1 $, so the horizontal coordinate of $ M $ is $ \\frac{k}{4} $, that is, $ N(\\frac{k}{4},0) = x_{1}x_{2}-\\frac{k}{4}(x_{1}+x_{2})+\\frac{4}{16}+4(x_{1}x_{2})^{2} = 3-\\frac{k^{2}}{16}=0 $, so $ k=\\pm4\\sqrt{3} $."
  },
  {
    "text": "It is known that point $F$ is the focus of the parabola $y^{2}=2 p x(p>0)$, and points $A(2, y_{1})$, $B(\\frac{1}{2}, y_{2})$ are points on the parabola located in the first and fourth quadrants, respectively. If $|A F|=10$, then the area of $\\triangle A B F$ is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;F: Point;p>0;y1:Number;y2:Number;Expression(G) = (y^2 = 2*p*x);Coordinate(A) = (2, y1);Coordinate(B) = (1/2, y2);Focus(G) = F;Quadrant(A)=1;Quadrant(B)=4;Abs(LineSegmentOf(A, F)) = 10;PointOnCurve(A,G);PointOnCurve(B,G)",
    "query_expressions": "Area(TriangleOf(A, B, F))",
    "answer_expressions": "42",
    "fact_spans": "[[[7, 28], [74, 77]], [[10, 28]], [[32, 46]], [[48, 71]], [[2, 6]], [[10, 28]], [[33, 46]], [[48, 71]], [[7, 28]], [[32, 46]], [[48, 71]], [[2, 31]], [[32, 88]], [[32, 88]], [[90, 100]], [[32, 88]], [[32, 88]]]",
    "query_spans": "[[[102, 124]]]",
    "process": "Since |AF| = 2 + \\frac{p}{2} = 10, it follows that p = 16, and the equation of the parabola is y^{2} = 32x. Substituting x = \\frac{1}{2} into the equation gives y = -4 (y = 4 is discarded), so B(\\frac{1}{2}, -4). Similarly, A(2, 8). The equation of line AB is \\frac{y+4}{8+4} = \\frac{x-\\frac{1}{2}}{2-\\frac{1}{2}}, which simplifies to y = 8x - 8. Thus, line AB intersects the x-axis at point C(1, 0), and therefore S_{\\triangle ABF} = \\frac{1}{2} \\times (8 - 1) \\times |y_{1} - y_{2}| = 42."
  },
  {
    "text": "What is the equation of the hyperbola that shares the same foci as the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ and passes through the point $Q(2,1)$?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Q: Point;Expression(H) = (x^2/4 + y^2 = 1);Coordinate(Q) = (2, 1);Focus(G)=Focus(H);PointOnCurve(Q,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/2-y^2=1",
    "fact_spans": "[[[43, 46]], [[1, 28]], [[33, 42]], [[1, 28]], [[33, 42]], [[0, 46]], [[32, 46]]]",
    "query_spans": "[[[43, 50]]]",
    "process": "From the ellipse equation, the foci are at $(\\pm\\sqrt{3},0)$. Let the hyperbola equation be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, so we have\n\\begin{cases}\n\\frac{4}{a^{2}}-\\frac{1}{b^{2}}=1 \\\\\na^{2}+b^{2}=3\n\\end{cases},\nsolving gives: $a^{2}=2, b^{2}=1$. Therefore, the hyperbola equation is $\\frac{x^{2}}{2}-y^{2}=1$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of an ellipse, $P$ is a point on the ellipse, and $\\angle F_{1} P F_{2}=60^{\\circ}$, then the range of eccentricity $e$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Focus(G) = {F1, F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree);e: Number;Eccentricity(G) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "[1/2, 1)",
    "fact_spans": "[[[18, 20], [29, 31]], [[2, 9]], [[25, 28]], [[10, 17]], [[2, 24]], [[25, 34]], [[36, 69]], [[74, 77]], [[0, 77]]]",
    "query_spans": "[[[74, 82]]]",
    "process": "Let the equation of the ellipse be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $|PF_{1}|=m$, $|PF_{2}|=n$. In $\\triangle PF_{1}F_{2}$, by the cosine theorem, we have $4c^{2}=m^{2}+n^{2}-2mn\\cos60^{\\circ}$. Since $m+n=2a$, $\\therefore m^{2}+n^{2}=(m+n)^{2}-2mn=4a^{2}-2mn$. $\\therefore 4c^{2}=4a^{2}-3mn$, i.e., $3mn=4a^{2}-4c^{2}$. Also, $mn\\leqslant\\left(\\frac{m+n}{2}\\right)^{2}=a^{2}$ (equality holds if and only if $m=n$). $\\therefore 4a^{2}-4c^{2}\\leqslant3a^{2}$, $\\therefore \\frac{c^{2}}{a^{2}}\\geqslant\\frac{1}{4}$, i.e., $e\\geqslant\\frac{1}{2}$. $\\therefore$ The range of $e$ is $[\\frac{1}{2},1)$."
  },
  {
    "text": "When $a$ is an arbitrary real number, the line $(2 a+3) x+y-4 a+2=0$ always passes through a fixed point $P$. Then, the standard equation of the parabola passing through point $P$ is?",
    "fact_expressions": "a: Real;H: Line;Expression(H) = (-4*a + x*(2*a + 3) + y + 2 = 0);P: Point;PointOnCurve(P, H);PointOnCurve(P, G);G: Parabola",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{(y^2=32*x),(x^2=(-1/2)*y)}",
    "fact_spans": "[[[1, 4]], [[11, 34]], [[11, 34]], [[38, 41], [44, 48]], [[11, 41]], [[43, 52]], [[49, 52]]]",
    "query_spans": "[[[49, 59]]]",
    "process": "The linear equation is rearranged as $(2x-4)a+3x+y+2=0$. From $\\begin{cases}2x-4=0\\\\3x+y+2=0\\end{cases}$, we obtain $\\begin{cases}x=2\\\\y=-8\\end{cases}$, so the fixed point is $P(2,-8)$, located in the fourth quadrant. Assume the parabola equation is $y^{2}=ax$, then $(-8)^{2}=a\\times2$, $a=32$. Next, assume the parabola equation is $x^{2}=by$, then $2^{2}=b\\times(-8)$, $b=-\\frac{1}{2}$. Therefore, the parabola equations are $y^{2}=32x$ or $x^{2}=-\\frac{1}{2}y$."
  },
  {
    "text": "If point $P$ lies on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ with foci $F_{1}$, $F_{2}$, satisfying $P F_{1}\\perp P F_{2}$ and $|P F_{1}|=2|P F_{2}|$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[1, 5]], [[1, 76]], [[7, 14]], [[16, 23]], [[6, 73]], [[27, 73], [128, 131]], [[27, 73]], [[30, 73]], [[30, 73]], [[79, 101]], [[103, 125]]]",
    "query_spans": "[[[128, 137]]]",
    "process": ""
  },
  {
    "text": "The equation of the locus of a moving point in the plane whose sum of distances to two fixed points $(4,0)$ and $(-4,0)$ is $8$ is?",
    "fact_expressions": "G: Point;H: Point;P:Point;Coordinate(G) = (4, 0);Coordinate(H) = (-4, 0);Distance(P,G)+Distance(P,H)=8",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(y=0)&(-4<=x<=4)",
    "fact_spans": "[[[7, 14]], [[15, 23]], [[33, 35]], [[7, 14]], [[15, 23]], [[0, 35]]]",
    "query_spans": "[[[33, 42]]]",
    "process": "Let point A(4,0), B(-4,0), and let the desired point be P. From |PA| + |PB| = |AB|, the locus of point P can be determined, and thus the equation of the locus of point P can be obtained. Let point A(4,0), B(-4,0), and let the desired point be P, then |PA| + |PB| = |AB|. The locus of point P is the line segment AB, so the equation of the locus of the moving point is y=0 (-4\\leqslant x \\leqslant 4)."
  },
  {
    "text": "It is known that one focus of the ellipse $x^{2}+k y^{2}=3 k(k>0)$ coincides with the focus of the parabola $y^{2}=12 x$. What is the eccentricity of this ellipse?",
    "fact_expressions": "G: Parabola;H: Ellipse;k: Number;Expression(G) = (y^2 = 12*x);k>0;Expression(H) = (k*y^2 + x^2 = 3*k);OneOf(Focus(H))=Focus(G)",
    "query_expressions": "Eccentricity(H)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[34, 49]], [[2, 28], [57, 59]], [[4, 28]], [[34, 49]], [[4, 28]], [[2, 28]], [[2, 54]]]",
    "query_spans": "[[[57, 65]]]",
    "process": "The focus of $ y^{2}=12x $ is $ (3,0) $, so $ x^{2}+ky^{2}=3k $ is rewritten as $ \\frac{x^{2}}{3k}+\\frac{y^{2}}{3}=1 $. Therefore, $ 3k-3=9 $, so $ k=4 $, $ a^{2}=12 $, $ a=2\\sqrt{3} $, and $ e=\\frac{c}{a}=\\frac{3}{2\\sqrt{3}}=\\frac{\\sqrt{3}}{2} $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and $P$ is a moving point on this ellipse, then the maximum value of $|PF_{1}| \\cdot |PF_{2}|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P,F1))*Abs(LineSegmentOf(P,F2)))",
    "answer_expressions": "4",
    "fact_spans": "[[[18, 45], [56, 58]], [[51, 54]], [[2, 9]], [[10, 17]], [[18, 45]], [[2, 50]], [[51, 64]]]",
    "query_spans": "[[[66, 96]]]",
    "process": ""
  },
  {
    "text": "Let the parabola $C$: $y^{2}=4 x$ have focus $F$, and let line $l$ with slope $\\sqrt{3}$ pass through point $F$ and intersect the parabola $C$ at points $A$ and $B$. Then $|A B|$=?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F,l);Slope(l)=sqrt(3);Intersection(l, C) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[42, 47]], [[1, 20], [54, 60]], [[62, 65]], [[66, 69]], [[24, 27], [48, 52]], [[1, 20]], [[1, 27]], [[42, 52]], [[28, 47]], [[42, 71]]]",
    "query_spans": "[[[73, 82]]]",
    "process": "From the given conditions, the focus of the parabola is $ F(1,0) $, and the equation of line $ l $ is $ y = \\sqrt{3}(x - 1) $. Substituting into $ y^2 = 4x $ and simplifying yields $ 3x^2 - 10x + 3 = 0 $. Let $ A(x_1, y_1) $, $ B(x_2, y_2) $, then $ |AB| = \\sqrt{1 + 3}|x_1 - x_2| = 2\\sqrt{\\left(\\frac{10}{3}\\right)^2 - 4} = \\frac{16}{3} $."
  },
  {
    "text": "If the distance from the focus of the parabola $y^{2}=2 p x(p>0)$ to one focus of the hyperbola $2 y^{2}-x^{2}=2 p^{2}$ is $\\sqrt{13}$, then the value of $p$ is?",
    "fact_expressions": "G:Hyperbola;p: Number;H: Parabola;Expression(G) = (-x^2 + 2*y^2 = 2*p^2);p>0;Expression(H) = (y^2 = 2*p*x);Distance(Focus(H),OneOf(Focus(G))) = sqrt(13)",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[26, 52]], [[74, 77]], [[1, 22]], [[26, 52]], [[4, 22]], [[1, 22]], [[1, 72]]]",
    "query_spans": "[[[74, 81]]]",
    "process": "The focus of the parabola is $F\\left(\\frac{p}{2},0\\right)$, the equation of the hyperbola can be rewritten as $\\frac{y^{2}}{p^{2}}-\\frac{x^{2}}{2p^{2}}=1$, so $c^{2}=3p^{2}$,"
  },
  {
    "text": "If the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{a^{2}}=1$ and the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{2}=1$ have the same foci, then the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Ellipse;e: Number;Expression(G) = (-y^2/2 + x^2/a = 1);Expression(H) = (x^2/4 + y^2/a^2 = 1);Focus(H) = Focus(G);Eccentricity(H) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[43, 81]], [[46, 81]], [[1, 42], [88, 90]], [[94, 97]], [[43, 81]], [[1, 42]], [[1, 86]], [[88, 97]]]",
    "query_spans": "[[[94, 99]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. Let a line $l$ passing through $F_{2}$ intersect the right branch of $C$ at points $A$ and $B$, such that $|A F_{1}|=|F_{1} F_{2}|$, $|B F_{2}|=2|A F_{2}|$. Then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;l: Line;PointOnCurve(F2, l) ;Intersection(l, RightPart(C)) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(A, F1)) = Abs(LineSegmentOf(F1, F2));Abs(LineSegmentOf(B, F2)) = 2*Abs(LineSegmentOf(A, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[2, 63], [104, 107], [174, 180]], [[2, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[72, 79]], [[80, 87], [90, 97]], [[2, 87]], [[2, 87]], [[98, 103]], [[89, 103]], [[98, 122]], [[113, 116]], [[117, 120]], [[124, 149]], [[150, 172]]]",
    "query_spans": "[[[174, 186]]]",
    "process": "Let the midpoint of $AF_{2}$ be $M$, connect $F_{1}M$, $BF_{1}$. According to the problem, $|AF_{1}|=|F_{1}F_{2}|=2c$, $F_{1}M\\bot AF_{2}$. By the definition of the parabola, $|AF_{2}|=2c-2a$, $|MF_{2}|=c-a$, $|BF_{2}|=4c-4a$, $|BF_{1}|=4c-2a$, $\\angle BF_{2}F_{1}+\\angle MF_{2}F_{1}=\\pi$, $\\cos\\angle BF_{2}F_{1}+\\cos\\angle MF_{2}F_{1}=0$. Using the cosine law in $\\triangle MF_{1}F_{2}$ and $\\Delta BF_{1}F_{2}$ to express the cosine values of the two angles, the relationship between $a$ and $c$ can be found, and thus the eccentricity can be obtained. As shown in the figure: let the midpoint of $AF_{2}$ be $M$, connect $F_{1}M$, $BF_{1}$. Since $|AF_{1}|=|F_{1}F_{2}|=2c$, and $M$ is the midpoint of $AF_{2}$, then $F_{1}M\\bot AF_{2}$. From $|AF_{1}|-|AF_{2}|=2a$, we get $|AF_{2}|=2c-2a$, so $|MF_{2}|=\\frac{1}{2}|AF_{2}|=c-a$. In $\\Delta MF_{1}F_{2}$, $\\cos\\angle MF_{2}F_{1}=\\frac{|MF_{2}|}{|F_{1}F_{2}|}=\\frac{c-a}{2c}$. $|BF_{2}|=2|AF_{2}|=4c-4a$, so $|BF_{1}|=2a+|BF_{2}|=4c-2a$. In $\\triangle BF_{1}F_{2}$, $\\cos\\angle BF_{2}F_{1}=\\frac{|F_{1}F_{2}|^{2}+|BF_{2}|^{2}-|BF_{1}|^{2}}{2\\times|F_{1}F_{2}||BF_{2}|}=\\frac{4c^{2}+16(c-a)^{2}-4(2c-a)^{2}}{2\\times2c\\times4(c-a)}=\\frac{4c^{2}+12a^{2}-16ac}{16c(c-a)}$. Since $\\angle BF_{2}F_{1}+\\angle MF_{2}F_{1}=\\pi$, $\\cos\\angle BF_{2}F_{1}+\\cos\\angle MF_{2}F_{1}=0$, so $\\frac{c-a}{2c}+\\frac{4c^{2}+12a^{2}-16ac}{16c(c-a)}=0$. Simplifying gives: $16a^{2}-16ac+12c^{2}=0$, i.e., $5a^{2}-8ac+3c^{2}=0$, so $5a^{2}-8ac+3c^{2}=0$, i.e., $(5a-3c)(a-c)=0$. Thus $5a=3c$ or $a=c$ (discarded). Therefore, the eccentricity $e=\\frac{c}{a}=\\frac{5}{3}$."
  },
  {
    "text": "A line passing through the focus of the parabola $y^{2}=4 x$ intersects the parabola at two points. If the chord $A B=8$, then the equation of the line is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Abs(LineSegmentOf(A, B)) = 8;NumIntersection(H, G) = 2;IsChordOf(LineSegmentOf(A, B), G);PointOnCurve(Focus(G), H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y = pm*(x- 1)",
    "fact_spans": "[[[1, 15], [22, 25]], [[19, 21], [42, 44]], [[32, 39]], [[32, 39]], [[1, 15]], [[32, 39]], [[19, 29]], [[22, 39]], [[0, 21]]]",
    "query_spans": "[[[42, 49]]]",
    "process": "[Analysis] From the parabola equation, the focus is (1,0). Let the line equation be x=ty+1, solve it together with the parabola equation, and use Vieta's formulas and the chord length formula to find the value of t, thereby obtaining the answer. According to the problem, the focus of the parabola y^{2}=4x is (1,0). Let the line passing through (1,0) be x=ty+1, and let A(x_{1},y_{1}), B(x_{2},y_{2}). From \\begin{cases} y^{2}=4x \\\\ x=ty+1 \\end{cases}, we obtain: y^{2}-4ty-4=0, then \\begin{cases} y_{1}+y_{2}= \\\\ y_{1}y_{2}= \\end{cases} \\cdot \\sqrt{16t^{2}+16}=4(1+t^{2})=8. Therefore, t=\\pm1. Thus, the line equation is x=\\pm y+1, which means y=\\pm(x-1)."
  },
  {
    "text": "If the distance from one focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ to an asymptote is equal to $\\frac{1}{4}$ of the focal distance, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Distance(OneOf(Focus(G)),OneOf(Asymptote(G)))=1/4*FocalLength(G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(2*sqrt(3))/3",
    "fact_spans": "[[[1, 57], [92, 95]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 89]]]",
    "query_spans": "[[[92, 101]]]",
    "process": "A hyperbola has an asymptote equation $ bx - ay = 0 $, and a focus coordinate $ (c, 0) $. According to the problem:  \n$\\frac{|bc - a \\times 0|}{\\sqrt{b^{2} + a^{2}}} = \\frac{1}{4} \\times 2c$, so $ c = 2b $, $ a = \\sqrt{c^{2} - b^{2}} = \\sqrt{3}b $. Therefore, $ e = \\frac{c}{a} = \\frac{2}{\\sqrt{3}} = \\frac{2\\sqrt{3}}{3} $. So the answer should be: $\\frac{2\\sqrt{3}}{3}$"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, and its directrix intersects the $x$-axis at point $K$. A line passing through point $F$ intersects the parabola $C$ at points $A$ and $B$. If $|AF|-|BF|=\\frac{3}{2}$, then $\\left|\\frac{AK}{BK}\\right|$=?",
    "fact_expressions": "C: Parabola;H: Line;A: Point;F: Point;B: Point;K: Point;Expression(C)=(y^2=4*x);Focus(C)=F;Intersection(Directrix(C), xAxis) = K;PointOnCurve(F,H);Intersection(H, C) = {A, B};Abs(LineSegmentOf(A, F)) - Abs(LineSegmentOf(B, F)) = 3/2",
    "query_expressions": "Abs(LineSegmentOf(A, K)/LineSegmentOf(B, K))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 21], [54, 60], [29, 30]], [[51, 53]], [[63, 66]], [[25, 28], [46, 50]], [[67, 70]], [[41, 44]], [[2, 21]], [[2, 28]], [[29, 44]], [[45, 53]], [[51, 72]], [[74, 99]]]",
    "query_spans": "[[[101, 122]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has the equation $x-\\sqrt{3} y=0$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Expression(OneOf(Asymptote(C))) = (x - sqrt(3)*y = 0)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 63], [91, 97]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 89]]]",
    "query_spans": "[[[91, 103]]]",
    "process": "From the equation of an asymptote of hyperbola C being $x - \\sqrt{3}y = 0$, we obtain $\\frac{b}{a} = \\frac{\\sqrt{3}}{3}$. Therefore, the eccentricity of hyperbola C is $e = \\sqrt{1 + \\left(\\frac{b}{a}\\right)^2} = \\sqrt{1 + \\frac{1}{3}} = \\frac{2\\sqrt{3}}{3}$."
  },
  {
    "text": "Given that point $F$ is the focus of the parabola $y^{2}=x$, and $AB$ is a line passing through point $F$ intersecting the parabola at points $A$ and $B$, with $|AB|=3$, then the horizontal coordinate of the midpoint $M$ of segment $AB$ is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;F: Point;M: Point;Expression(G) = (y^2 = x);Focus(G) = F;PointOnCurve(F, LineOf(A, B));Intersection(LineOf(A, B), G) = {A, B};Abs(LineSegmentOf(A, B)) = 3;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "XCoordinate(M)",
    "answer_expressions": "5/4",
    "fact_spans": "[[[7, 19], [39, 42]], [[44, 47]], [[48, 51]], [[2, 6], [30, 34]], [[75, 78]], [[7, 19]], [[2, 22]], [[23, 37]], [[23, 53]], [[54, 63]], [[65, 78]]]",
    "query_spans": "[[[75, 84]]]",
    "process": "From the parabola equation, we know $ F\\left(\\frac{1}{4},0\\right) $. Assume the x-coordinates of $ A $ and $ B $ are $ x_{1} $, $ x_{2} $. By the property of the directrix of the parabola, $ |AB| = x_{1} + \\frac{1}{4} + x_{2} + \\frac{1}{4} = 3 \\Rightarrow x_{1} + x_{2} = \\frac{5}{2} $. The x-coordinate of the midpoint $ M $ of $ AB $ is $ \\frac{1}{2}(x_{1} + x_{2}) = \\frac{5}{4} $."
  },
  {
    "text": "Given that point $P$ lies on the parabola $y^{2}=4x$ and the distance from point $P$ to the origin is $2\\sqrt{3}$, what is the distance from point $P$ to the focus of the parabola?",
    "fact_expressions": "G: Parabola;P: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(P, G);O:Origin;Distance(P,O)=2*sqrt(3)",
    "query_expressions": "Distance(P, Focus(G))",
    "answer_expressions": "3",
    "fact_spans": "[[[7, 21], [57, 60]], [[2, 6], [26, 30], [51, 55]], [[7, 21]], [[2, 24]], [[31, 33]], [[26, 49]]]",
    "query_spans": "[[[51, 67]]]",
    "process": "Let P(x,y), then \\sqrt{x^{2}+y^{2}}=2\\sqrt{3}\\Rightarrow\\sqrt{x^{2}+4x}=2\\sqrt{3}\\Rightarrow x=2(-6 \\text{ discarded}), so the distance from point P to the focus of the parabola is 2+1=3"
  },
  {
    "text": "Let $M(x_{0}, y_{0})$ be a point on the parabola $C$: $x^{2}=8 y$, and let $F$ be the focus of the parabola $C$. The circle centered at $F$ with radius $|F M|$ intersects the directrix of the parabola $C$. Then the range of values for $y_{0}$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (x0, y0);x0: Number;y0: Number;C: Parabola;Expression(C) = (x^2 = 8*y);PointOnCurve(M, C) = True;Focus(C) = F;F:Point;G: Circle;Center(G) = F;Radius(G) = Abs(LineSegmentOf(F, M));IsIntersect(G , Directrix(C)) = True",
    "query_expressions": "Range(y0)",
    "answer_expressions": "(2,+oo)",
    "fact_spans": "[[[1, 18]], [[1, 18]], [[1, 18]], [[91, 98]], [[19, 38], [46, 52], [78, 84]], [[19, 38]], [[1, 41]], [[42, 55]], [[42, 45], [58, 61]], [[76, 77]], [[57, 77]], [[65, 77]], [[76, 89]]]",
    "query_spans": "[[[91, 105]]]",
    "process": "According to the definition of a parabola, the radius of the circle is $ y_{0}+2 $. To satisfy the condition that the circle intersects the directrix of the parabola, the distance from the center of the circle to the directrix must be less than the radius, i.e., $ 4 < y_{0}+2 $, solving this yields: $ y_{0} > 2 $. Therefore, the answer is: $ (2, +\\infty) $."
  },
  {
    "text": "The parabola $x^{2}=4(y-m)$ intersects the circle $x^{2} + y^{2}=1$ at point $P$ in the first quadrant, and the tangents to the two curves at $P$ are perpendicular to each other. Then $m=$?",
    "fact_expressions": "P: Point;m: Number;H: Circle;G: Parabola;Expression(G) = (x^2 = 4*(y-m));Expression(H) = (x^2 + y^2 = 1);Quadrant(P) = 1;Intersection(H, G)=P;IsPerpendicular(TangentOnPoint(P, G), TangentOnPoint(P, H))",
    "query_expressions": "m",
    "answer_expressions": "(\\sqrt{2}-1)/2",
    "fact_spans": "[[[44, 48], [51, 55]], [[68, 71]], [[18, 36]], [[0, 17]], [[0, 17]], [[18, 36]], [[39, 48]], [[0, 48]], [[50, 66]]]",
    "query_spans": "[[[68, 73]]]",
    "process": ""
  },
  {
    "text": "From any point $M$ on the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$, draw a line parallel to the transverse axis, intersecting its asymptotes at points $P$ and $Q$. Then the value of $|M P| \\cdot|M Q|$ is?",
    "fact_expressions": "G: Hyperbola;M: Point;P: Point;Q: Point;L:Line;Expression(G) = (x^2 - y^2/4 = 1);PointOnCurve(M,G);PointOnCurve(M,L);IsParallel(L,RealAxis(G));Intersection(L,Asymptote(G))={P,Q}",
    "query_expressions": "Abs(LineSegmentOf(M, P))*Abs(LineSegmentOf(M, Q))",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 29], [46, 47]], [[34, 37]], [[54, 57]], [[58, 61]], [], [[1, 29]], [[1, 37]], [[0, 44]], [[0, 44]], [[0, 63]]]",
    "query_spans": "[[[65, 87]]]",
    "process": "The asymptotes of the hyperbola are $ y = \\pm 2x $. Let $ M $ be the vertex of the real axis of the hyperbola at $ (1,0) $. Thus, points $ P $ and $ Q $ coincide at the origin, so $ |MP| \\cdot |MQ| = 1 $."
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=2 p x$ ($p<0$), draw two mutually perpendicular chords $AB$ and $CD$. If the minimum value of the sum of the areas of $\\triangle ACF$ and $\\triangle BDF$ is $32$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p<0;F: Point;Focus(G) = F;PointOnCurve(F, LineSegmentOf(A, B)) = True;PointOnCurve(F, LineSegmentOf(C, D)) = True;IsChordOf(LineSegmentOf(A, B), G)  = True;IsChordOf(LineSegmentOf(C, D), G) = True;IsPerpendicular(LineSegmentOf(A, B), LineSegmentOf(C, D)) = True;A: Point;B: Point;C: Point;D: Point;Min(Area(TriangleOf(A, C, F)) + Area(TriangleOf(B, D, F))) = 32",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = 8*x",
    "fact_spans": "[[[1, 22], [102, 105]], [[1, 22]], [[4, 22]], [[4, 22]], [[25, 28]], [[1, 28]], [[0, 50]], [[0, 50]], [[1, 50]], [[1, 50]], [[31, 50]], [[37, 42]], [[37, 42]], [[45, 50]], [[45, 50]], [[52, 100]]]",
    "query_spans": "[[[102, 110]]]",
    "process": "Let the angle between line AB and the x-axis be \\theta. From the focal radius formula, we obtain \\frac{p}{+\\sin\\theta}S_{\\triangle BDF}=\\frac{}{2(1+.)}S=\\frac{p^{2}}{2}\\left[\\frac{1}{(1-\\cos\\theta)(1+\\sin\\theta)}+\\frac{1}{(1+\\cos\\theta)(1-\\sin\\theta)}\\right]. \\therefore S=p^{2}\\times\\frac{1-\\sin\\theta\\cos\\theta}{\\sin^{2}\\theta\\cos2\\theta}=p^{2\\times}\\frac{1-\\frac{1}{2}\\sin2\\theta}{\\frac{1}{4}\\sin^{2}2\\theta}. Let t=\\sin2\\theta, \\theta\\in(0,\\frac{\\pi}{2})\\cup(\\frac{\\pi}{2},\\pi), t\\in[-1,0)\\cup(0,1]. The original expression simplifies to S=p^{2}\\times\\left(\\frac{4}{t^{2}}-\\frac{2}{t}\\right), \\frac{1}{t}\\in(-\\infty,-1]\\cup[1,+\\infty). According to the properties of quadratic functions, the minimum occurs when t=1, at which point S=2p^{2}=32, \\therefore p=4. The equation of the parabola is y^{2}=8x."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $x^{2}=12 y$, $P$ is a point on $C$, and the line $F P$ intersects the line $y=-3$ at point $Q$. If $\\overrightarrow{P Q}=2 \\overrightarrow{F Q}$, then $|P Q|$=?",
    "fact_expressions": "C: Parabola;G: Line;F: Point;P: Point;Q: Point;Expression(C) = (x^2 = 12*y);Expression(G) = (y=-3);Focus(C) = F;PointOnCurve(P, C);Intersection(LineOf(F,P),G) = Q;VectorOf(P, Q) = 2*VectorOf(F, Q)",
    "query_expressions": "Abs(LineSegmentOf(P, Q))",
    "answer_expressions": "8",
    "fact_spans": "[[[6, 26], [34, 37]], [[49, 57]], [[2, 5]], [[30, 33]], [[58, 62]], [[6, 26]], [[49, 57]], [[2, 29]], [[30, 40]], [[41, 62]], [[65, 110]]]",
    "query_spans": "[[[112, 121]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the focus of the parabola $y^{2}=4 ax(a<0)$ are?",
    "fact_expressions": "G: Parabola;a: Number;a<0;Expression(G) = (y^2 = 4*(a*x))",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(a, 0)",
    "fact_spans": "[[[0, 20]], [[3, 20]], [[3, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 27]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ is an intersection point of the ellipse $C_{1}$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the circle $C_{2}$: $x^{2}+y^{2}=a^{2}-b^{2}$, and $2 \\angle PF_{1}F_{2}=\\angle PF_{2} F_{1}$, where $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C_{1}$, respectively, then the eccentricity of the ellipse $C_{1}$ is?",
    "fact_expressions": "C1: Ellipse;Expression(C1) = (x^2/a^2+y^2/b^2 = 1);a: Number;b: Number;a>b;b>0;C2: Circle;Expression(C2) = (y^2+x^2 = a^2-b^2);P: Point;OneOf(Intersection(C1,C2)) = P;2*AngleOf(P,F1,F2) = AngleOf(P,F2,F1);F1: Point;F2: Point;LeftFocus(C1) = F1;RightFocus(C1) = F2",
    "query_expressions": "Eccentricity(C1)",
    "answer_expressions": "sqrt(3) - 1",
    "fact_spans": "[[[7, 68], [176, 185], [192, 201]], [[7, 68]], [[17, 68]], [[17, 68]], [[17, 68]], [[17, 68]], [[69, 104]], [[69, 104]], [[2, 6]], [[2, 109]], [[111, 153]], [[156, 163]], [[166, 173]], [[156, 190]], [[156, 190]]]",
    "query_spans": "[[[192, 207]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $C$ has an eccentricity of $2$. Write a standard equation for a hyperbola $C$ satisfying this condition.",
    "fact_expressions": "C: Hyperbola;Eccentricity(C) = 2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[0, 6], [0, 6]], [[0, 14]]]",
    "query_spans": "[[[24, 36]]]",
    "process": ""
  },
  {
    "text": "What is the length of the minor axis of the ellipse $4 x^{2}+6 y^{2}=24$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (4*x^2 + 6*y^2 = 24)",
    "query_expressions": "Length(MinorAxis(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 22]], [[0, 22]]]",
    "query_spans": "[[[0, 28]]]",
    "process": ""
  },
  {
    "text": "Given point $D(4,0)$, $F$ is the focus of the parabola $C$: $y^{2}=4x$. The line $l$ passing through point $F$ with slope $k$ intersects the parabola $C$ at points $A$ and $B$. If $\\overrightarrow{A D} \\cdot \\overrightarrow{B D} \\leqslant 1$, then what is the range of values for $k^{2}$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);D: Point;Coordinate(D) = (4, 0);F: Point;Focus(C) = F;PointOnCurve(F, l);Slope(l) = k;k: Number;l: Line;Intersection(l, C) = {A, B};A: Point;B: Point;DotProduct(VectorOf(A, D), VectorOf(B, D)) <= 1",
    "query_expressions": "Range(k^2)",
    "answer_expressions": "(0,4]",
    "fact_spans": "[[[16, 35], [58, 64]], [[16, 35]], [[2, 11]], [[2, 11]], [[12, 15], [40, 44]], [[12, 38]], [[39, 57]], [[45, 57]], [[48, 51]], [[52, 57]], [[52, 75]], [[66, 69]], [[70, 73]], [[77, 138]]]",
    "query_spans": "[[[140, 154]]]",
    "process": "From the given conditions, we know $ F(1,0) $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, and the equation of line $ l $ be $ x = my + 1 $. From \n\\[\n\\begin{cases}\nx = my + 1, \\\\\ny^2 = 4x,\n\\end{cases}\n\\]\nwe obtain $ y^{2} - 4my - 4 = 0 $, so $ y_{1} + y_{2} = 4m $, $ y_{1}y_{2} = -4 $. From $ \\overrightarrow{AD} \\cdot \\overrightarrow{BD} \\leqslant 1 $, we get\n\\[\n(4 - x_{1})(4 - x_{2}) + y_{1}y_{2} = (3 - my_{1})(3 - my_{2}) + y_{1}y_{2} = (m^{2} + 1)y_{1}y_{2} - 3m(y_{1} + y_{2}) + 9 \\leqslant 1.\n\\]\nSince $ y_{1} + y_{2} = 4m $, $ y_{1}y_{2} = -4 $, it follows that $ -16m^{2} + 5 \\leqslant 1 $, so $ \\frac{1}{4} \\leqslant m^{2} $. Since $ m = \\frac{1}{k} $, we have $ \\frac{1}{4} \\leqslant \\frac{1}{k^{2}} $, hence $ 0 < k^{2} \\leqslant 4 $."
  },
  {
    "text": "If $F_{1}$ and $F_{2}$ are the foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, and point $P$ lies on the hyperbola. If the distance from point $P$ to focus $F_{1}$ is equal to $9$, then what is the distance from point $P$ to focus $F_{2}$?",
    "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;P: Point;Expression(G) = (x^2/16 - y^2/9 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);Distance(P, F1) = 9",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "{1, 17}",
    "fact_spans": "[[[17, 56], [65, 68]], [[1, 8], [79, 86]], [[9, 16], [103, 110]], [[60, 64], [72, 76], [96, 100]], [[17, 56]], [[1, 59]], [[60, 69]], [[72, 94]]]",
    "query_spans": "[[[96, 115]]]",
    "process": "By the definition of a hyperbola, |PF₁| - |PF₂| = 2√16 = 8. Given |PF₁| = 9, it follows that |9 - |PF₂|| = 8, solving which yields |PF₂| = 1 or 17."
  },
  {
    "text": "Given a line segment $AB$ in the plane with length $6$, and a moving point $P$ satisfying $PA - PB = 4$. Let $O$ be the midpoint of $AB$. Then the minimum value of $PO$ is?",
    "fact_expressions": "A: Point;B: Point;Length(LineSegmentOf(A,B)) = 6;P: Point;LineSegmentOf(P,A) - LineSegmentOf(P,B) = 4;O: Origin;MidPoint(LineSegmentOf(A,B)) = O",
    "query_expressions": "Min(LineSegmentOf(P, O))",
    "answer_expressions": "2",
    "fact_spans": "[[[10, 15]], [[10, 15]], [[8, 22]], [[25, 28]], [[30, 41]], [[43, 47]], [[43, 56]]]",
    "query_spans": "[[[58, 69]]]",
    "process": ""
  },
  {
    "text": "Given that $A$ and $B$ are the two endpoints of the major axis of the ellipse $C$: $\\frac{x^{2}}{m+1}+\\frac{y^{2}}{m}=1$, and $P$ is a moving point on the ellipse $C$, and the maximum value of $\\angle A P B$ is $\\frac{2 \\pi}{3}$, then $m=$?",
    "fact_expressions": "C: Ellipse;m: Number;A: Point;P: Point;B: Point;Expression(C) = (x^2/(m + 1) + y^2/m = 1);Endpoint(MajorAxis(C))={A,B};PointOnCurve(P, C);Max(AngleOf(A,P,B))=2*pi/3",
    "query_expressions": "m",
    "answer_expressions": "1/2",
    "fact_spans": "[[[10, 53], [66, 71]], [[116, 119]], [[2, 5]], [[62, 65]], [[6, 9]], [[10, 53]], [[2, 61]], [[62, 75]], [[77, 113]]]",
    "query_spans": "[[[116, 121]]]",
    "process": ""
  },
  {
    "text": "If a point $M$ on the parabola $y^{2}=2 x$ is at a distance of $\\sqrt {3}$ from the coordinate origin $O$, then what is the distance from point $M$ to the focus of the parabola?",
    "fact_expressions": "G: Parabola;M: Point;O: Origin;Expression(G) = (y^2 = 2*x);PointOnCurve(M, G);Distance(M, O) = sqrt(3)",
    "query_expressions": "Distance(M, Focus(G))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[1, 15], [53, 56]], [[19, 22], [47, 51]], [[23, 30]], [[1, 15]], [[1, 22]], [[19, 45]]]",
    "query_spans": "[[[47, 63]]]",
    "process": ""
  },
  {
    "text": "Given that the distance between the directrix of the parabola $y^{2}=m x$ and the line $x=1$ is $3$, find the equation of the parabola?",
    "fact_expressions": "G: Parabola;m: Number;H: Line;Expression(G) = (y^2 = m*x);Expression(H) = (x = 1);Distance(Directrix(G), H) = 3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2 = 8*x, y^2 = -16*x}",
    "fact_spans": "[[[1, 15], [35, 38]], [[4, 15]], [[19, 26]], [[1, 15]], [[19, 26]], [[1, 33]]]",
    "query_spans": "[[[35, 42]]]",
    "process": "The directrix of the parabola is given by $ x = -\\frac{m}{4} $. From the condition, we have $ \\left|1 - \\left(-\\frac{m}{4}\\right)\\right| = 3 $, so $ m = 8 $ or $ m = -16 $. Thus, the equation of the parabola is $ y^{2} = 8x $ or $ y^{2} = -16x $."
  },
  {
    "text": "$P$ lies on the parabola $y^{2}=2x$. What is the minimum value of the sum of the distance from point $P$ to point $Q(0, 2)$ and the distance from $P$ to the directrix of the parabola?",
    "fact_expressions": "G: Parabola;Q: Point;P: Point;Expression(G) = (y^2 = 2*x);Coordinate(Q) = (0, 2);PointOnCurve(P,G)",
    "query_expressions": "Min(Distance(P,Q)+Distance(P,Directrix(G)))",
    "answer_expressions": "sqrt(17)/2",
    "fact_spans": "[[[4, 18], [46, 49]], [[27, 38]], [[22, 26], [42, 45], [0, 3]], [[4, 18]], [[27, 38]], [[0, 19]]]",
    "query_spans": "[[[22, 62]]]",
    "process": ""
  },
  {
    "text": "Point $P$ lies on the left branch of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, and its right focus is $F$. If $M$ is the midpoint of segment $FP$, and the distance from $M$ to the origin is $7$, then $|PF|=$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1);P: Point;PointOnCurve(P, LeftPart(G));F: Point;RightFocus(G) = F;M: Point;MidPoint(LineSegmentOf(F, P)) = M;O: Origin;Distance(M, O) = 7",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[5, 44], [51, 52]], [[5, 44]], [[0, 4]], [[0, 50]], [[56, 59]], [[51, 59]], [[61, 64], [78, 81]], [[61, 75]], [[82, 86]], [[78, 93]]]",
    "query_spans": "[[[95, 104]]]",
    "process": ""
  },
  {
    "text": "Given that the equation of the parabola $C$ is $y = -4x^{2}$, then the coordinates of the focus of $C$ are?",
    "fact_expressions": "C: Parabola;Expression(C) = (y = -4*x^2)",
    "query_expressions": "Coordinate(Focus(C))",
    "answer_expressions": "(0,-1/16)",
    "fact_spans": "[[[2, 8], [26, 29]], [[2, 24]]]",
    "query_spans": "[[[26, 36]]]",
    "process": "Given the equation of parabola C is $ y = -4x^{2} $, its standard form is $ x^{2} = -\\frac{1}{4}y $. Then the coordinates of the focus are $ \\left(0, -\\frac{1}{16}\\right) $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ with foci $F_{1}$, $F_{2}$, and a point $P$ in the first quadrant lying on $C$ such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=\\frac{9}{4}$, find the inradius of $\\Delta P F_{1} F_{2}$.",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/4 + y^2/3 = 1);Focus(C) = {F1, F2};Quadrant(P) = 1;PointOnCurve(P, C);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 9/4",
    "query_expressions": "Radius(InscribedCircle(TriangleOf(P, F1, F2)))",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 44], [73, 76]], [[69, 72]], [[48, 55]], [[56, 63]], [[2, 44]], [[2, 63]], [[64, 72]], [[69, 77]], [[79, 148]]]",
    "query_spans": "[[[150, 180]]]",
    "process": "From the given conditions, we have $ a^{2}=4 $, $ b^{2}=3^{2} $, $ c^{2}=a^{2}-b^{2}=1 $, then $ F_{1}(-1,0) $, $ F_{2}(1,0) $. Let the coordinates of point $ P $ be $ (x_{p},y_{p}) $, then $ \\overrightarrow{PF}_{1}=(-1-x_{p},-y_{p}) $, $ \\overrightarrow{PF_{2}}=(1-x_{p},-y_{p}) $. $ \\frac{13}{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=x_{p}^{2}+y_{p}^{2}-1=\\frac{9}{4} $, that is, $ x_{p}^{2}+y_{p}^{2}=\\frac{13}{4} $ ①. Since point $ P $ in the first quadrant lies on $ C $, $ \\therefore \\frac{x^{2}}{\\frac{p}{4}}+\\frac{y^{2}}{\\frac{p}{3}}=1 $, i.e., $ x_{P}^{2}=4-\\frac{4y_{P}^{2}}{3} $ ②. Solving ① and ② simultaneously gives $ y_{p}=\\frac{3}{2} $. From the definition of the ellipse, $ PF_{1}+|PF_{2}|=2a=4 $. Let the inradius of $ \\triangle PF_{1}F_{2} $ be $ r $, then $ S_{\\triangle PF_{1}F_{2}}=\\frac{1}{2}r(|PF_{1}|+|PF_{2}|+|F_{1}F_{2}|)=3r $. Also, $ \\because S_{\\triangle PF_{1}F_{2}}=\\frac{1}{2}\\cdot 2c\\cdot y_{p}=\\frac{3}{2} $, $ \\therefore 3r=\\frac{3}{2} $, i.e., $ r=\\frac{1}{2} $."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$, a line passing through its focus with slope $1$ intersects the parabola at points $A$ and $B$. If the ordinate of the midpoint of segment $AB$ is $2$, then the equation of the directrix of this parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;H: Line;Slope(H) = 1;PointOnCurve(Focus(G), H);A: Point;B: Point;Intersection(H, G) = {A, B};YCoordinate(MidPoint(LineSegmentOf(A, B))) = 2",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=-1",
    "fact_spans": "[[[2, 23], [40, 43], [76, 79], [26, 27]], [[2, 23]], [[5, 23]], [[5, 23]], [[37, 39]], [[30, 39]], [[25, 39]], [[44, 47]], [[48, 51]], [[37, 53]], [[55, 73]]]",
    "query_spans": "[[[76, 86]]]",
    "process": "From \\begin{cases}y^2=2px\\\\y=x-\\frac{p}{2}\\end{cases}, \\therefore y^2-2py-p^{2}=0, \\therefore y_{1}+y_{2}=2p=4, \\therefore p=2, directrix x=-1"
  },
  {
    "text": "Given that $P$ is a point on the ellipse $4 x^{2} + y^{2}=1$, and $F$ is one of its foci, then the minimum value of $P F$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F: Point;Expression(G) = (4*x^2 + y^2 = 1);PointOnCurve(P, G);OneOf(Focus(G))=F",
    "query_expressions": "Min(LineSegmentOf(P, F))",
    "answer_expressions": "1 - sqrt(3)/2",
    "fact_spans": "[[[6, 27]], [[2, 5]], [[31, 34]], [[6, 27]], [[2, 30]], [[6, 41]]]",
    "query_spans": "[[[43, 54]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $P$ is a point on the ellipse such that $\\overrightarrow{P F_{1}} \\cdot(\\overrightarrow{O F_{1}}+\\overrightarrow{O P})=0$ ($O$ is the origin). If $|P F_{1}|=\\sqrt{2}|P F_{2}|$, then the eccentricity of the ellipse is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;F1: Point;O: Origin;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(P,C);DotProduct(VectorOf(P,F1),(VectorOf(O,F1)+VectorOf(O,P)))=0;Abs(LineSegmentOf(P,F1))=sqrt(2)*Abs(LineSegmentOf(P,F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(6)-sqrt(3)",
    "fact_spans": "[[[2, 59], [88, 90], [220, 222]], [[8, 59]], [[8, 59]], [[84, 87]], [[68, 75]], [[179, 182]], [[76, 83]], [[8, 59]], [[8, 59]], [[2, 59]], [[2, 83]], [[2, 83]], [[84, 93]], [[97, 178]], [[189, 218]]]",
    "query_spans": "[[[220, 228]]]",
    "process": "As shown in the figure, take the midpoint A of $ PF_{1} $, connect $ OA $, $ \\overrightarrow{A} = \\overrightarrow{OF_{1}} + \\overrightarrow{OP} $, $ \\overrightarrow{OA} = \\frac{1}{2}\\overrightarrow{F_{2}P} \\therefore \\overrightarrow{OF_{1}} + \\overrightarrow{OP} = \\overrightarrow{F_{2}P} $, $ \\because \\overrightarrow{PF} \\cdot (\\overrightarrow{OF} + \\overrightarrow{OP}) = 0 $, $ \\therefore \\overrightarrow{PF_{1}} \\cdot \\overrightarrow{F_{2}P} = 0 $, $ \\therefore \\overrightarrow{PF_{1}} \\bot \\overrightarrow{F_{2}P} $, $ \\because |\\overrightarrow{PF_{1}}| = \\sqrt{2}|\\overrightarrow{PF_{2}}| $, without loss of generality, assume $ |PF_{2}| = m $, then $ |PF_{1}| = \\sqrt{2}m $, $ \\because |PF_{2}| + |PF_{1}| = 2a = m + \\sqrt{2}m $, $ \\therefore m = \\frac{2}{1+\\sqrt{2}}a = 2(\\sqrt{2}-1)a $, $ \\because |F_{1}F_{2}| = 2c $, $ \\therefore 4c^{2} = m^{2} + 2m^{2} = 3m^{2} = 3 \\times 4a^{2}(3-2\\sqrt{2}) $ $ \\therefore \\frac{c^{2}}{a^{2}} = 9 - 6\\sqrt{2} = (\\sqrt{6} - \\sqrt{3})^{2} $ $ \\therefore e = \\sqrt{6} - \\sqrt{3} $,"
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{m^{2}}=1$ $(0<m<2)$, and let $P(x_{0}, y_{0})$ be a point on $C$ satisfying that the area of $\\Delta P F_{1} F_{2}$ is $\\sqrt{3}$. Then the range of $|x_{0}|$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/m^2 = 1);m: Number;0 < m;m < 2;F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;x0: Number;y0: Number;Coordinate(P) = (x0, y0);PointOnCurve(P, C);Area(TriangleOf(P, F1, F2)) = sqrt(3)",
    "query_expressions": "Range(Abs(x0))",
    "answer_expressions": "[0, 1]",
    "fact_spans": "[[[17, 70], [94, 97]], [[17, 70]], [[24, 70]], [[24, 70]], [[24, 70]], [[1, 8]], [[9, 16]], [[1, 75]], [[76, 93]], [[76, 93]], [[76, 93]], [[76, 93]], [[76, 100]], [[104, 140]]]",
    "query_spans": "[[[143, 159]]]",
    "process": ""
  },
  {
    "text": "Given a line passing through the point $M(0,1)$ intersects the parabola $x^{2}=2 p y(p>0)$ at two distinct points $A$ and $B$, and the two tangents to the parabola at $A$ and $B$ intersect at point $N$. If $\\overrightarrow{N A} \\cdot \\overrightarrow{N B}=0$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;M: Point;N: Point;A: Point;B: Point;p>0;Expression(G) = (x^2 = 2*(p*y));Coordinate(M) = (0, 1);PointOnCurve(M, H);Intersection(H, G) = {A, B};Negation(A=B);Intersection(TangentOnPoint(A,G),TangentOnPoint(B,G))=N;DotProduct(VectorOf(N, A), VectorOf(N, B)) = 0",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[16, 37]], [[129, 132]], [[13, 15]], [[3, 12]], [[70, 74]], [[42, 45], [53, 56]], [[46, 49], [57, 60]], [[19, 37]], [[16, 37]], [[3, 12]], [[2, 15]], [[13, 51]], [[39, 51]], [[16, 74]], [[76, 127]]]",
    "query_spans": "[[[129, 136]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), and suppose the equation of line AB is known. Substituting into the equation of the parabola yields \\otimes. Then, from ▱ we obtain the center, so the slopes of the two tangents are respectively lower and region. From lower we know that the function implies the function, hence lower, i.e., p=2. Therefore, the answer is: 2"
  },
  {
    "text": "Given that point $P(1,-2)$ lies on the line $y=k x+2$, then the eccentricity of the conic section $C$: $\\frac{x^{2}}{k}+\\frac{5 y^{2}}{16}=1$ is?",
    "fact_expressions": "P: Point;Coordinate(P) = (1, -2);H: Line;Expression(H) = (y = k*x + 2);k: Number;PointOnCurve(P, H);C: ConicSection;Expression(C) = ((5*y^2)/16 + x^2/k = 1)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "3/2",
    "fact_spans": "[[[2, 12]], [[2, 12]], [[13, 24]], [[13, 24]], [[15, 24]], [[2, 25]], [[27, 74]], [[27, 74]]]",
    "query_spans": "[[[27, 80]]]",
    "process": "Analysis: Use the point on the line to find the value of $k$, then simplify the equation of the conic section and solve for the eccentricity. Since point $P(1,-2)$ lies on the line $y = kx + 2$, we obtain $k = -4$. Then the conic curve $C: \\frac{x^{2}}{-4} + \\frac{5y^{2}}{16} = 1$ is a hyperbola, from which we get $a = \\frac{4\\sqrt{5}}{5}$, $b = 2 \\Rightarrow c = \\sqrt{4 + \\frac{16}{5}} = \\frac{6\\sqrt{5}}{5}$. Therefore, the eccentricity of the hyperbola is $e = \\frac{c}{a} = \\frac{6\\sqrt{5}}{4\\sqrt{5}} = \\frac{3}{2}$."
  },
  {
    "text": "A focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0 )$ coincides with the focus of the parabola $y^{2}=20 x$, and the distance from this focus to an asymptote is $4$. What is the eccentricity of the hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;H: Parabola;Expression(H) = (y^2 = 20*x);OneOf(Focus(G)) = Focus(H);Distance(OneOf(Focus(G)), Asymptote(G)) = 4",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[0, 60], [105, 108]], [[0, 60]], [[3, 60]], [[3, 60]], [[3, 60]], [[3, 60]], [[66, 81]], [[66, 81]], [[0, 86]], [[0, 102]]]",
    "query_spans": "[[[105, 114]]]",
    "process": "The focus of the parabola $ y^{2} = 20x $ has coordinates $ (5, 0) $. One asymptote of the hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ $ (a > 0, b > 0) $ is given by $ bx + ay = 0 $. Since the distance from the focus of the parabola to the asymptote of the hyperbola is 4, we have $ \\frac{|5b|}{\\sqrt{b^{2} + a^{2}}} = 4 $, so $ b = 4 $. Given $ c = 5 $, we obtain $ a = 3 $. Therefore, the eccentricity of the hyperbola is $ e = \\frac{c}{a} = \\frac{5}{3} $."
  },
  {
    "text": "Given that point $E(2 , 2)$ lies on the parabola $C$: $y^{2}=2 px$, and a line $l$ passing through point $(2 , 0)$ intersects parabola $C$ at points $A$ and $B$ (distinct from point $E$). Lines $EA$ and $EB$ intersect the line $x=-2$ at points $M$ and $N$, respectively. Then, what is the measure of $\\angle MON$?",
    "fact_expressions": "E: Point;Coordinate(E) = (2, 2);C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;H: Point;Coordinate(H) = (2, 0);l: Line;PointOnCurve(H, l) = True;Intersection(l, C) = {A, B};A: Point;B: Point;Negation(A = E);Negation(B = E);G: Line;Expression(G) = (x = -2);Intersection(LineOf(E, A), G) = M;M: Point;Intersection(LineOf(E, B), G) = N;N: Point;O: Origin",
    "query_expressions": "AngleOf(M, O, N)",
    "answer_expressions": "pi/2",
    "fact_spans": "[[[2, 12], [77, 81]], [[2, 12]], [[13, 33], [56, 62]], [[13, 33]], [[21, 33]], [[39, 49]], [[39, 49]], [[50, 55]], [[37, 55]], [[50, 73]], [[64, 67]], [[68, 71]], [[64, 82]], [[64, 82]], [[97, 105]], [[97, 105]], [[83, 114]], [[106, 110]], [[83, 114]], [[111, 114]], [[116, 128]]]",
    "query_spans": "[[[116, 133]]]",
    "process": ""
  },
  {
    "text": "Let the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ be $\\sqrt{3}$, and suppose one of its foci coincides with the focus of the parabola $y^{2}=4 x$. Then the equation of this hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 4*x);Eccentricity(G) = sqrt(3);OneOf(Focus(G)) = Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/(1/3)-y^2/(2/3)=1",
    "fact_spans": "[[[1, 57], [103, 106], [74, 75]], [[4, 57]], [[4, 57]], [[81, 95]], [[4, 57]], [[4, 57]], [[1, 57]], [[81, 95]], [[1, 72]], [[74, 100]]]",
    "query_spans": "[[[103, 110]]]",
    "process": "From the given condition, \\frac{c}{a}=\\sqrt{3}. Since c=1, \\therefore a=\\frac{\\sqrt{3}}{3}, b=\\frac{\\sqrt{6}}{3}. \\therefore \\frac{x^{2}}{3}-\\frac{y^{2}}{3}=1"
  },
  {
    "text": "Given that a line $l$ with slope $2$ passes through the focus $F$ of the parabola $y^{2}=ax$ ($a>0$) and intersects the $y$-axis at point $A$, if the area of $\\triangle OAF$ ($O$ being the origin) is $4$, then the equation of the parabola is?",
    "fact_expressions": "l: Line;G: Parabola;a: Number;O: Origin;A: Point;F: Point;a>0;Expression(G) = (y^2 = a*x);Slope(l) = 2;Focus(G) = F;PointOnCurve(F, l);Intersection(l,yAxis) = A;Area(TriangleOf(O, A, F)) = 4",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = 8*x",
    "fact_spans": "[[[9, 14]], [[15, 33], [89, 92]], [[18, 33]], [[71, 74]], [[49, 53]], [[36, 39]], [[18, 33]], [[15, 33]], [[2, 14]], [[15, 39]], [[9, 39]], [[9, 53]], [[55, 87]]]",
    "query_spans": "[[[89, 96]]]",
    "process": ""
  },
  {
    "text": "The distance from the focus of the parabola $y^{2}=4 mx(m>0)$ to an asymptote of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ is $3$. Then the equation of this parabola is?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 4*(m*x));m: Number;m>0;G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1);Distance(Focus(H), OneOf(Asymptote(G))) = 3",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2=20*x",
    "fact_spans": "[[[0, 20], [79, 82]], [[0, 20]], [[3, 20]], [[3, 20]], [[24, 63]], [[24, 63]], [[0, 76]]]",
    "query_spans": "[[[79, 87]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the hyperbola $x^{2}+\\frac{y^{2}}{m}=1$ is $2$, then $m$=?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2 + y^2/m = 1);Eccentricity(G) = 2",
    "query_expressions": "m",
    "answer_expressions": "-3",
    "fact_spans": "[[[0, 28]], [[38, 41]], [[0, 28]], [[0, 36]]]",
    "query_spans": "[[[38, 43]]]",
    "process": "The hyperbola converted into standard form gives $x^{2}-\\frac{y^{2}}{m}=1$. The eccentricity is $e=\\frac{c}{a}=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{1-m}=2$, hence $m=-3$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, a circle with diameter $F_{1}F_{2}$ intersects the left branch of $C$ at point $A$, and line $AF_{2}$ intersects the right branch of $C$ at point $B$. If $\\cos \\angle F_{1} B F_{2}=-\\frac{3}{5}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;F1: Point;F2: Point;A: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;IsDiameter(LineSegmentOf(F1,F2),G);Intersection(G,LeftPart(C))=A;Intersection(LineSegmentOf(A,F2),RightPart(C))=B;Cos(AngleOf(F1, B, F2)) = -3/5",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(13)",
    "fact_spans": "[[[18, 80], [107, 110], [130, 133], [185, 188]], [[25, 80]], [[25, 80]], [[105, 106]], [[2, 9]], [[10, 17]], [[115, 119]], [[138, 142]], [[25, 80]], [[25, 80]], [[18, 80]], [[2, 86]], [[2, 86]], [[87, 106]], [[105, 119]], [[120, 142]], [[143, 183]]]",
    "query_spans": "[[[185, 194]]]",
    "process": "According to the problem, $\\angle F_{1}AF_{2}=90^{\\circ}$, $\\cos\\angle F_{1}BF_{2}=-\\frac{3}{5}$, further we can obtain $|AB|:|AF_{1}|:|BF_{1}|=3:4:5$. Then according to the definition of hyperbola, we can find $a$, $c$, and finally get the result using the eccentricity formula. [Detailed solution] From the given conditions, $\\angle F_{1}AF_{2}=90^{\\circ}$, $\\cos\\angle F_{1}BF_{2}=-\\frac{3}{5}$, so $\\cos\\angle ABF_{1}=\\frac{3}{5}$, i.e., $\\frac{|AB|}{|BF_{1}|}=\\frac{3}{5}$, it is easy to get $|AB|:|AF_{1}|:|BF_{1}|=3:4:5$. Let $|AB|=3$, $|AF_{1}|=4$, $|BF_{1}|=5$, $|BF_{2}|=x$. By the definition of hyperbola: $3+x-4=5-x$, solving gives $x=3$, so $|F_{1}F_{2}|=\\sqrt{4^{2}+6^{2}}=4\\sqrt{13}\\Rightarrow c=\\sqrt{13}$, since $2a=5-x=2\\Rightarrow a=1$, thus the eccentricity $e=\\sqrt{13}$."
  },
  {
    "text": "The coordinates of the foci of the hyperbola $x^{2}-2 y^{2}=2$ are? The eccentricity is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - 2*y^2 = 2)",
    "query_expressions": "Coordinate(Focus(G));Eccentricity(G)",
    "answer_expressions": "{(sqrt(3),0),(-sqrt(3),0)}\nsqrt(6)/2",
    "fact_spans": "[[[0, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 27]], [[0, 32]]]",
    "process": "By rearranging the given hyperbola equation into standard form and applying hyperbola properties, the coordinates of the foci and the eccentricity can be easily determined; the given hyperbola equation can be rewritten as \\frac{x^{2}}{2}-y^{2}=1, so the foci are at (\\sqrt{3},0), (-\\sqrt{3},0), and the eccentricity is \\frac{\\sqrt{6}}{2}."
  },
  {
    "text": "Given that point $P$ is an arbitrary point on the ellipse $\\frac{y^{2}}{4}+x^{2}=1$, and let the distances from point $P$ to the two lines $2 x \\pm y=0$ be $d_{1}$ and $d_{2}$, respectively. Then the maximum value of $d_{1}+d_{2}$ is?",
    "fact_expressions": "G: Ellipse;H1: Line;H2: Line;P: Point;d1: Number;d2: Number;Expression(G) = (x^2 + y^2/4 = 1);Expression(H1) = (2*x + y = 0);Expression(H2) = (2*x - y = 0);PointOnCurve(P, G);Distance(P, H1) = d1;Distance(P, H2) = d2",
    "query_expressions": "Max(d1 + d2)",
    "answer_expressions": "4*sqrt(5)/5",
    "fact_spans": "[[[7, 34]], [], [], [[2, 6], [41, 45]], [[68, 75]], [[77, 85]], [[7, 34]], [[46, 62]], [[46, 62]], [[2, 39]], [[41, 85]], [[41, 85]]]",
    "query_spans": "[[[87, 106]]]",
    "process": "According to the distance from a point to a line, $d_{1}^{2}+d_{2}^{2}$ is a constant. Using an inequality, the maximum and minimum values of $d_{1}+d_{2}$ can be found. Let $P(x_{0},y_{0})$ be any point on the ellipse, then $d_{1}=\\frac{|2x_{0}+y_{0}|}{\\sqrt{5}}$, $d_{2}=\\frac{|2x_{0}-y_{0}|}{\\sqrt{5}}$, so $d_{1}^{2}+d_{2}^{2}=\\frac{2(4x_{0}^{2}+y_{0}^{2})}{5}=\\frac{8}{5}$. Because $\\left(\\frac{d_{1}+d_{2}}{2}\\right)^{2}\\leqslant\\frac{d_{1}^{2}+d_{2}^{2}}{2}=\\frac{4}{5}$, it follows that $d_{1}+d_{2}\\leqslant\\frac{4\\sqrt{5}}{5}$. The equality holds if and only if $d_{1}=d_{2}=\\frac{2\\sqrt{5}}{5}$, which occurs when $P(\\pm1,0)$ or $P(0,\\pm2)$."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $C$: $y^{2}=2 p x$ ($p>0$) intersects the parabola $C$ at points $A$ and $B$, where point $A$ lies in the first quadrant. If $\\overrightarrow{A F}=5 \\overrightarrow{F B}$, then the slope of line $AB$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;L: Line;PointOnCurve(F,L) = True;Intersection(L,C) = {A,B};A: Point;B: Point;Quadrant(A) = 1;VectorOf(A, F) = 5*VectorOf(F, B)",
    "query_expressions": "Slope(LineOf(A,B))",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[1, 27], [37, 43]], [[1, 27]], [[9, 27]], [[9, 27]], [[30, 33]], [[1, 33]], [[34, 36]], [[0, 36]], [[34, 55]], [[46, 49], [58, 62]], [[50, 53]], [[58, 68]], [[70, 115]]]",
    "query_spans": "[[[117, 129]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $ ($ x_{1}>0 $, $ y_{1}>0 $). According to $ \\overrightarrow{AF}=5\\overrightarrow{FB} $, we get $ y_{1}=-5y_{2} $. Let the line equation be combined with the parabola; from the relationship between roots and coefficients, derive $ y_{2} $, then find point $ B $, and solve using the slope formula. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $ ($ x_{1}>0 $, $ y_{1}>0 $), $ \\overrightarrow{AF}=5\\overrightarrow{FB} $  \n$$\n\\begin{cases}\n\\therefore y_{1}=-5y_{2}' \\\\\ny^{2}=20x \\\\\ny=k(x-\\frac{1}{2})^{,}\n\\end{cases}\n$$\n$ y^{2}v=-p^{2} $"
  },
  {
    "text": "If a directrix of the hyperbola $\\frac{x^{2}}{8}-\\frac{y^{2}}{b^{2}}=1$ coincides with the directrix of the parabola $y^{2}=8 x$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;H: Parabola;Expression(G) = (x^2/8 - y^2/b^2 = 1);Expression(H) = (y^2 = 8*x);OneOf(Directrix(G))=Directrix(H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 43], [70, 73]], [[4, 43]], [[49, 63]], [[1, 43]], [[49, 63]], [[1, 68]]]",
    "query_spans": "[[[70, 79]]]",
    "process": "According to the problem and given conditions, we obtain the system of equations  \n\\begin{cases}\\frac{a2}{c}=2\\\\a^{2}=8\\end{cases}\\Rightarrow\\begin{cases}a=2\\sqrt{2}\\\\c=4\\end{cases}\\Rightarrow e=\\sqrt{2}  \nAnswer: \\sqrt{2}"
  },
  {
    "text": "Given that the focal length of the ellipse $x^{2}+\\frac{y^{2}}{\\cos ^{2} \\alpha}=1$ $(\\alpha \\in(0, \\frac{\\pi}{2}))$ equals the length of the minor axis of the ellipse. If the line $l$: $y=k x+1$ intersects the ellipse at points $A$ and $B$, and the midpoint of segment $AB$ lies on the line $y=x$, then what is the real number $k$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + y^2/Cos(alpha)^2 = 1);In(alpha,(0,pi/2));alpha: Number;FocalLength(G)=Length(MinorAxis(G));l: Line;Expression(l) = (y = k*x + 1);Intersection(l, G) = {A, B};A: Point;B: Point;l1: Line;Expression(l1) = (y=x);PointOnCurve(MidPoint(LineSegmentOf(A,B)),l1) = True;k: Real",
    "query_expressions": "k",
    "answer_expressions": "does not exist",
    "fact_spans": "[[[2, 76], [81, 83], [107, 109]], [[2, 76]], [[4, 76]], [[4, 76]], [[2, 87]], [[90, 106]], [[90, 106]], [[90, 120]], [[111, 114]], [[115, 118]], [[133, 140]], [[133, 140]], [[122, 141]], [[143, 148]]]",
    "query_spans": "[[[143, 149]]]",
    "process": "First, find the equation of the ellipse. Substitute $ l: y = kx + 1 $ into the ellipse equation $ x^{2} + 2y^{2} = 1 $, yielding $ (1+2k^{2})x^{2} + 4kx + 1 = 0 $. Use Vieta's formulas and the midpoint coordinate formula to find the value of $ k $, then verify.  \n[f} Solution] $ \\because 1 > \\cos^{2}\\alpha, \\therefore c^{2} = 1 - \\cos^{2}\\alpha = \\sin^{2}\\alpha = b^{2} = \\cos^{2}\\alpha $, obtain $ \\alpha = \\frac{\\pi}{4} $. Substitute $ l: y = kx + 1 $ into the ellipse equation $ x^{2} + 2y^{2} = 1 $, get $ (1+2k^{2})x^{2} + 4kx + 1 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = \\frac{-4k}{1+2k^{2}} $, $ y_{1} + y_{2} = k(x_{1} + x_{2}) + 2 = \\frac{2}{1+2k^{2}} $, from $ \\frac{x_{1}+x_{2}}{2} = \\frac{y_{1}+y_{2}}{2} $ solve to get $ k = -\\frac{1}{2} $. But substitute $ y = -\\frac{1}{2}x + 1 $ into $ x^{2} + 2y^{2} = 1 $, obtain: $ \\frac{3}{2}x^{2} - 2x + 1 = 0 $. At this time $ \\Delta = 4 - 4 \\cdot \\frac{3}{2} \\cdot 1 = -2 < 0 $, so $ k = -\\frac{1}{2} $ must be discarded, thus the real number $ k $ does not exist."
  },
  {
    "text": "On the hyperbola $\\frac{y^{2}}{64}-\\frac{x^{2}}{36}=1$, the distance from a point $P$ to one of its foci is equal to $3$. Then, what is the perimeter of the triangle formed by point $P$ and the two foci?",
    "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (-x^2/36 + y^2/64 = 1);PointOnCurve(P, G);Distance(P, OneOf(Focus(G))) = 3;F1:Point;F2:Point;Focus(G)={F1,F2}",
    "query_expressions": "Perimeter(TriangleOf(P,F1,F2))",
    "answer_expressions": "42",
    "fact_spans": "[[[0, 40], [47, 48]], [[43, 46], [64, 68]], [[0, 40]], [[0, 46]], [[43, 61]], [], [], [[47, 73]]]",
    "query_spans": "[[[47, 86]]]",
    "process": ""
  },
  {
    "text": "The foci of the hyperbola lie on the $x$-axis, the length of the real axis is $4$, and the eccentricity is $3$. Then, what is the standard equation of this hyperbola? What are the equations of its asymptotes?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), xAxis);Length(RealAxis(G)) = 4;Eccentricity(G) = 3",
    "query_expressions": "Expression(G);Expression(Asymptote(G))",
    "answer_expressions": "x^2/4 - y^2/32 = 1\ny= pm*2*sqrt(2)*x",
    "fact_spans": "[[[0, 3], [34, 37]], [[0, 12]], [[0, 21]], [[0, 30]]]",
    "query_spans": "[[[34, 44]], [[34, 51]]]",
    "process": ""
  },
  {
    "text": "What is the length of the real axis of the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/2 - y^2 = 1)",
    "query_expressions": "Length(RealAxis(G))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 34]]]",
    "process": "From $\\frac{x^{2}}{2}-y^{2}=1$ we get $a^{2}=2$, hence the length of the real axis is $2a=2\\sqrt{2}$."
  },
  {
    "text": "Given that points $C$ and $D$ are two moving points on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and point $M(0,2)$, if $\\overrightarrow{M D}=\\lambda \\overrightarrow{M C}$, then the range of real values for $\\lambda$ is?",
    "fact_expressions": "G: Ellipse;M: Point;D: Point;C: Point;lambda: Real;Expression(G) = (x^2/4 + y^2 = 1);Coordinate(M) = (0, 2);PointOnCurve(C, G);PointOnCurve(D, G);VectorOf(M, D) = lambda*VectorOf(M, C)",
    "query_expressions": "Range(lambda)",
    "answer_expressions": "[1/3, 3]",
    "fact_spans": "[[[11, 38]], [[46, 55]], [[7, 10]], [[2, 6]], [[110, 122]], [[11, 38]], [[46, 55]], [[2, 44]], [[2, 44]], [[57, 108]]]",
    "query_spans": "[[[110, 129]]]",
    "process": "\\textcircled{1} When the slope of the line exists, let the equation of the line passing through point $M(0,2)$ be $y = kx + 2$. Solving the system of equations\n\\[\n\\begin{cases}\ny = kx + 2 \\\\\n\\frac{x^2}{4} + y^2 = 1\n\\end{cases}\n\\]\nyields $(1 + 4k^2)x^2 + 16kx + 12 = 0$. Then $\\Delta = (16k)^2 - 4 \\times (1 + 4k^2) \\times 12 \\geqslant 0$, i.e., $k^2 \\geqslant \\frac{3}{4}$. Let $C(x_1, y_1)$, $D(x_2, y_2)$, then $x_1 + x_2 = -\\frac{16k}{1 + 4k^2}$, $x_1 \\cdot x_2 = \\frac{12}{1 + 4k^2}$. Since $\\overrightarrow{MD} = \\lambda \\overrightarrow{MC}$, we have $x_1 = \\lambda x_2$. Thus, $(1 + \\lambda)x_2 = -\\frac{16k}{1 + 4k^2}$, $(x_2)^2 \\cdot \\lambda = \\frac{12}{1 + 4k^2}$, i.e.,\n\\[\n\\frac{(1 + \\lambda)^2}{\\lambda} = \\left(\\frac{16k}{1 + 4k^2}\\right)^2 \\times \\frac{1 + 4k^2}{12} = \\frac{64k^2}{3(1 + 4k^2)} = \\frac{64}{3} \\times \\frac{1}{\\frac{1}{k^2} + 4}.\n\\]\nSince $k^2 \\geqslant \\frac{3}{4}$, we have $4 \\leqslant \\frac{(1 + \\lambda)^2}{\\lambda} < \\frac{16}{3}$, so $\\frac{1}{3} < \\lambda < 3$.\n\n\\textcircled{2} When the slope does not exist, the equation of the line passing through $M(0,2)$ is $x = 0$. In this case, $C(0,1), D(0,-1)$ or $C(0,-1), D(0,1)$. When $C(0,1), D(0,-1)$, $\\lambda = 3$; when $C(0,-1), D(0,1)$, $\\lambda = \\frac{1}{3}$. In conclusion, $\\frac{1}{3} \\leqslant \\lambda \\leqslant 3$."
  },
  {
    "text": "$P$ is the intersection point of the line $y=\\frac{b}{3 a} x$ and the left branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $F_{1}$ is the left focus, and $P F_{1}$ is perpendicular to the $x$-axis. Then the eccentricity $e$ of the hyperbola is $?$.",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;P: Point;F1: Point;e: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = x*(b/(3*a)));LeftFocus(G) = F1;Intersection(H,LeftPart(G))=P;IsPerpendicular(LineSegmentOf(P,F1),xAxis);Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "3*sqrt(2)/4",
    "fact_spans": "[[[26, 83], [119, 122]], [[29, 83]], [[29, 83]], [[4, 25]], [[0, 3]], [[89, 96]], [[126, 129]], [[29, 83]], [[29, 83]], [[26, 83]], [[4, 25]], [[26, 100]], [[0, 88]], [[101, 117]], [[119, 129]]]",
    "query_spans": "[[[126, 131]]]",
    "process": ""
  },
  {
    "text": "It is known that the foci of the ellipse $\\frac{x^{2}}{10-m}+\\frac{y^{2}}{m-2}=1$ lie on the $y$-axis. If the focal distance is $4$, then what is the value of $m$?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/(10 - m) + y^2/(m - 2) = 1);PointOnCurve(Focus(G),yAxis);FocalLength(G) = 4",
    "query_expressions": "m",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 44]], [[63, 66]], [[2, 44]], [[2, 53]], [[2, 61]]]",
    "query_spans": "[[[63, 69]]]",
    "process": "From the standard equation of the ellipse and the fact that the foci lie on the y-axis with 2c=4, m can be found using the relationship among ellipse parameters. From the given conditions:  \n\\begin{cases}10-m>0\\\\m-2>0\\end{cases}  \n(10-m>0 gives 6<m<10), and since 2c=4 and the foci are on the y-axis, m-2>10-m, therefore m-2=10-m+4, solving yields m=8."
  },
  {
    "text": "The coordinates of the focus of the parabola $x=\\frac{1}{4} y^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (x = y^2/4)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(1, 0)",
    "fact_spans": "[[[0, 24]], [[0, 24]]]",
    "query_spans": "[[[0, 31]]]",
    "process": "The standard equation of the parabola is: $y^{2}=4x$. Therefore, the coordinates of the focus are: $(1,0)$."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2x$ be $F$, and let a line $l$ passing through point $F$ intersect the parabola at points $A$ and $B$, such that $|AF|=4|BF|$. Then the chord length $AB=$?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 2*x);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, F)) = 4*Abs(LineSegmentOf(B, F));IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "LineSegmentOf(A, B)",
    "answer_expressions": "25/8",
    "fact_spans": "[[[29, 34]], [[1, 15], [35, 38]], [[40, 43]], [[19, 22], [24, 28]], [[44, 47]], [[1, 15]], [[1, 22]], [[23, 34]], [[29, 49]], [[51, 65]], [[35, 74]]]",
    "query_spans": "[[[69, 76]]]",
    "process": "Find the coordinates of the focus of the parabola, write the equation of line $ l $ using the point-slope form of a linear equation, and then solve the system with the parabola's equation to obtain the answer using the chord length formula. The focus of the parabola has coordinates $ F\\left(\\frac{1}{2}, 0\\right) $. Let point $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Let the equation of line $ l $ be $ x = my + \\frac{1}{2} $. By the definition of the parabola, $ |AF| = x_{1} + \\frac{p}{2} = x_{1} + \\frac{1}{2} $, $ |BF| = x_{2} + \\frac{p}{2} = x_{2} + \\frac{1}{2} $. From $ |AF| = 4|BF| $, we get $ x_{1} + \\frac{1}{2} = 4\\left(x_{2} + \\frac{1}{2}\\right) $, that is, $ my_{1} + 1 = 4(my_{2} + 1) $, so $ m(y_{1} - 4y_{2}) = 3 \\cdots\\cdots (1) $. Also, from $ \\begin{cases} x = my + \\frac{1}{2} \\\\ y^{2} = 2x \\end{cases} $, we obtain: $ y^{2} - 2my - 1 = 0 $, so $ y_{1} + y_{2} = 2m $, $ y_{1}y_{2} = -1 \\cdots\\cdots (2) $. Solving equations (1) and (2) simultaneously, we get $ m^{2} = \\frac{9}{16} $. Also, $ |AB| = |AF| + |BF| = x_{1} + x_{2} + 1 = my_{1} + my_{2} + 2 = m(y_{1} + y_{2}) + 2 = 2m^{2} + 2 = 2 \\times \\frac{9}{16} + 2 = \\frac{25}{8} $."
  },
  {
    "text": "Let $F$ be the focus of the parabola $C$: $y^{2}=12 x$. A line passing through point $P(1,0)$ intersects the parabola at points $A$ and $B$, and $2 \\overrightarrow{B P}=\\overrightarrow{P A}$. Then $|A F|+|B F|$=?",
    "fact_expressions": "C: Parabola;G: Line;P: Point;B: Point;A: Point;F: Point;Expression(C) = (y^2 = 12*x);Coordinate(P) = (1, 0);Focus(C) = F;PointOnCurve(P,G);Intersection(G,C) = {A, B};2*VectorOf(B, P) = VectorOf(P, A)",
    "query_expressions": "Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F))",
    "answer_expressions": "17/2",
    "fact_spans": "[[[5, 25], [44, 47]], [[41, 43]], [[31, 40]], [[53, 56]], [[49, 52]], [[1, 4]], [[5, 25]], [[31, 40]], [[1, 28]], [[29, 43]], [[41, 58]], [[60, 105]]]",
    "query_spans": "[[[107, 122]]]",
    "process": "From the given condition, a line passing through point P(1,0) must satisfy 2\\overrightarrow{BP}=\\overrightarrow{PA}, so the slope of this line must exist, and the line A is parallel to the x-axis, denoted as y=k(x-1), with k\\neq0. The focus of the parabola C: y^{2}=12x is F(3,0). Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since 2\\overrightarrow{BP}=\\overrightarrow{PA} and P(1,0), it follows that y_{1}=-2y_{2}. Solving the system of equations yields -1(x-1). Eliminating x gives y^{2}-\\frac{12}{k}y-12=0. Thus, y_{1}y_{2}=-12. From y_{1}=-2y_{2}, we obtain y_{1}^{2}=24, y_{2}^{2}=6. From the parabola equation, x_{1}=2, x_{2}=\\frac{1}{2}. Therefore, |AF|+|BF|=x_{1}+x_{2}+6=\\frac{17}{2}."
  },
  {
    "text": "Given that the line $l$: $y=2x-10$ is parallel to one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ and passes through one focus of the hyperbola, find the standard equation of the hyperbola.",
    "fact_expressions": "l: Line;G: Hyperbola;a: Number;b: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(l) = (y = 2*x - 10);IsParallel(l, OneOf(Asymptote(G)));PointOnCurve(OneOf(Focus(G)), l)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5 - y^2/20 = 1",
    "fact_spans": "[[[2, 19]], [[20, 76], [98, 101], [88, 91]], [[23, 76]], [[23, 76]], [[23, 76]], [[23, 76]], [[20, 76]], [[2, 19]], [[2, 84]], [[2, 96]]]",
    "query_spans": "[[[98, 108]]]",
    "process": "For the line $l: y = 2x - 10$, let $y = 0$, solving gives $x = 5$, so one focus of the hyperbola is $(5, 0)$. The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ ($a > 0, b > 0$) are $y = \\pm\\frac{b}{a}x$. According to the problem, $\\frac{b}{a} = 2$ and $c = 5$. Since $c^{2} = a^{2} + b^{2}$, solving gives $a^{2} = 5$, $b^{2} = 20$. Therefore, the equation of the hyperbola is $\\frac{x^{2}}{5} - \\frac{y^{2}}{20} = 1$."
  },
  {
    "text": "Given that a focus of the hyperbola $\\frac{x^{2}}{a}+y^{2}=1$ is $(0,-2)$, then the asymptotes of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Point;Expression(G) = (y^2 + x^2/a = 1);Coordinate(H) = (0,-2);OneOf(Focus(G))=H",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(3)/3)*x",
    "fact_spans": "[[[2, 30], [46, 49]], [[5, 30]], [[36, 44]], [[2, 30]], [[36, 44]], [[2, 44]]]",
    "query_spans": "[[[46, 57]]]",
    "process": "Because \\frac{x^{2}}{a}+y^{2}=1, so \\frac{x^{2}}{-a}+y^{2}=1, so _{4}=1-a, a=-3 \\therefore \\frac{x^{2}}{-3}+y^{2}=0 \\therefore y=\\pm\\frac{\\sqrt{3}}{3}x"
  },
  {
    "text": "The standard equation of the hyperbola with foci at $(\\pm 5,0)$ and asymptotes $y=\\pm \\frac{4}{3} x$ is?",
    "fact_expressions": "G: Hyperbola;Coordinate(Focus(G)) = (pm*5, 0);Expression(Asymptote(G)) = (y = pm*(4/3)*x)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/16 = 1",
    "fact_spans": "[[[46, 49]], [[0, 49]], [[18, 49]]]",
    "query_spans": "[[[46, 56]]]",
    "process": "From the given, assume the hyperbola equation is \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=\\lambda(\\lambda>0), that is, \\frac{x^{2}}{9\\lambda}-\\frac{y^{2}}{16\\lambda}=1, thus we obtain c^{2}=9\\lambda+16\\lambda=25, then solve for \\lambda to get the standard equation of the hyperbola. [Detailed solution] According to the problem, assume the hyperbola equation is \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=\\lambda(\\lambda>0), that is, \\frac{x^{2}}{9\\lambda}-\\frac{y^{2}}{16\\lambda}=1, then a^{2}=9\\lambda, b^{2}=16\\lambda. Since the coordinates of the foci are (\\pm5,0), it follows that c^{2}=9\\lambda+16\\lambda=25, solving gives \\lambda=1, so the standard equation of the hyperbola is \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1"
  },
  {
    "text": "The distance from the point $(2,0)$ to the asymptotes of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ is?",
    "fact_expressions": "G: Hyperbola;H: Point;Expression(G) = (x^2/4 - y^2 = 1);Coordinate(H) = (2, 0)",
    "query_expressions": "Distance(H, Asymptote(G))",
    "answer_expressions": "sqrt(5)*2/5",
    "fact_spans": "[[[9, 37]], [[0, 8]], [[9, 37]], [[0, 8]]]",
    "query_spans": "[[[0, 46]]]",
    "process": "From the equation of the hyperbola, it follows that an asymptote of the hyperbola has the equation $ y = \\frac{1}{2}x $, or $ x - 2y = 0 $. Thus, the distance from the point $ (2, 0) $ to the asymptote is $ d = \\frac{2}{\\sqrt{5}} = \\frac{2\\sqrt{5}}{5} $."
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x(p>0)$ coincides with the right focus of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, then the value of the real number $p$ is?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 2*(p*x));p: Real;p>0;G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1);Focus(H) = RightFocus(G)",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 22]], [[1, 22]], [[62, 67]], [[4, 22]], [[26, 54]], [[26, 54]], [[1, 60]]]",
    "query_spans": "[[[62, 71]]]",
    "process": "In the hyperbola: $c=\\sqrt{1+3}=2$, therefore, the focus of the parabola is $(2,0)$, $\\frac{p}{2}=2$, $p=4$"
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=4x$, and point $P(x, y)$ in the first quadrant is a moving point on the parabola $C$. If $|PF|=3$, then the coordinates of point $P$ are?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;P: Point;x1: Number;y1: Number;Coordinate(P) = (x1, y1);Quadrant(P) = 1;PointOnCurve(P, C);Abs(LineSegmentOf(P, F)) = 3",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2, 2*sqrt(2))",
    "fact_spans": "[[[6, 25], [44, 50]], [[6, 25]], [[2, 5]], [[2, 28]], [[33, 43], [66, 70]], [[34, 43]], [[34, 43]], [[33, 43]], [[29, 43]], [[33, 54]], [[56, 64]]]",
    "query_spans": "[[[66, 75]]]",
    "process": "From the definition of the parabola, we have $x_{P}+1=3$, $\\therefore x_{P}=2$, $\\therefore y_{p}^{2}=8$. $\\because y_{p}>0$, $\\therefore y_{P}=2\\sqrt{2}$, so the coordinates of point $P$ are $(2, 2\\sqrt{2})$."
  },
  {
    "text": "The equation of a hyperbola that shares common foci with the ellipse $\\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1$ and has eccentricity $e=\\frac{5}{3}$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;e:Number;Expression(H) = (x^2/49 + y^2/24 = 1);Focus(G)=Focus(H);Eccentricity(G)=e;e=5/3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/16 = 1",
    "fact_spans": "[[[66, 69]], [[1, 40]], [[50, 65]], [[1, 40]], [[0, 69]], [[47, 69]], [[50, 65]]]",
    "query_spans": "[[[66, 73]]]",
    "process": "The hyperbola sharing common foci with the ellipse $\\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1$ has foci at $(\\pm5,0)$, so $c=5$. The eccentricity of the hyperbola is $e=\\frac{5}{3}$, giving $a=3$, so $b=\\sqrt{c^{2}-a^{2}}=4$. Therefore, the equation of the hyperbola is: $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$."
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{4}=-1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/2 - y^2/4 = -1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(2)*x",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 47]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{m^{2}}=1$ $(0<m<4)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. If the maximum value of $|B F_{2}|+|A F_{2}|$ is $10$, then the value of $m$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/m^2 = 1);m: Number;0 < m;m < 4;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;l: Line;PointOnCurve(F1, l);A: Point;B: Point;Intersection(l, G) = {A, B};Max(Abs(LineSegmentOf(A, F2)) + Abs(LineSegmentOf(B, F2))) = 10",
    "query_expressions": "m",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[2, 51], [91, 93]], [[2, 51]], [[137, 140]], [[4, 51]], [[4, 51]], [[60, 67], [77, 84]], [[68, 75]], [[2, 75]], [[2, 75]], [[85, 90]], [[76, 90]], [[94, 97]], [[98, 101]], [[85, 103]], [[105, 135]]]",
    "query_spans": "[[[137, 144]]]",
    "process": "According to the definition of the ellipse, |BF_{2}| + |AF_{2}| + |AB| = 4a = 16, so |BF_{2}| + |AF_{2}| = 16 - |AB|. When AB \\bot x-axis, |AB| is minimized and |BF_{2}| + |AF_{2}| is maximized. At this time, |AB| = \\frac{2b^{2}}{a} = \\frac{2m^{2}}{4} = \\frac{m^{2}}{2}, so 16 - \\frac{m^{2}}{2} = 10, m^{2} = 12. Since 0 < m < 4, we have m = 2\\sqrt{3}."
  },
  {
    "text": "The minimum distance from a point on the ellipse $\\frac{x^{2}}{3}+y^{2}=1$ to the line $x+y=4$ is?",
    "fact_expressions": "G: Ellipse;H: Line;P:Point;Expression(G) = (x^2/3 + y^2 = 1);Expression(H) = (x + y = 4);PointOnCurve(P,G)",
    "query_expressions": "Min(Distance(P, H))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 27]], [[31, 40]], [[29, 30]], [[0, 27]], [[31, 40]], [[0, 30]]]",
    "query_spans": "[[[29, 47]]]",
    "process": ""
  },
  {
    "text": "The line $y = x + b$ intersects the curve $x = \\sqrt{1 - y^{2}}$ at exactly one point. What is the range of real values of $b$?",
    "fact_expressions": "G: Line;b: Real;H: Curve;Expression(G) = (y = b + x);Expression(H) = (x = sqrt(1 - y^2));NumIntersection(G, H)=1",
    "query_expressions": "Range(b)",
    "answer_expressions": "(-1,1]+{-\\sqrt{2}}",
    "fact_spans": "[[[0, 9]], [[41, 44]], [[10, 30]], [[0, 9]], [[10, 30]], [[0, 36]]]",
    "query_spans": "[[[41, 51]]]",
    "process": ""
  },
  {
    "text": "Let the eccentricity of ellipse $C_{1}$ be $\\frac{5}{13}$, with foci on the $x$-axis and major axis length $26$. If the absolute value of the difference of the distances from any point on curve $C_{2}$ to the two foci of ellipse $C_{1}$ is equal to $8$, then what is the standard equation of curve $C_{2}$?",
    "fact_expressions": "C1: Ellipse;C2: Curve;Eccentricity(C1) = 5/13;PointOnCurve(Focus(C1),xAxis);Length(MajorAxis(C1))=26;P:Point;F1:Point;F2:Point;Focus(C1)={F1,F2};PointOnCurve(P,C2);Abs(Distance(P,F1)-Distance(P,F2))=8",
    "query_expressions": "Expression(C2)",
    "answer_expressions": "x^2/16 - y^2/9 = 1",
    "fact_spans": "[[[1, 10], [63, 72]], [[50, 59], [93, 102]], [[1, 29]], [[1, 38]], [[1, 47]], [[61, 62]], [], [], [[63, 77]], [[50, 62]], [[61, 91]]]",
    "query_spans": "[[[93, 109]]]",
    "process": ""
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, the directrix is $l$, and if $l$ intersects the two asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0, b>0$) at points $A$ and $B$ respectively, and $|AB|=4|OF|$ ($O$ being the origin), then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;A: Point;B: Point;O: Origin;F: Point;l: Line;a>0;b>0;L1:Line;L2:Line;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 4*x);Focus(H) = F;Directrix(H) = l;Intersection(l,L1)=A;Intersection(l,L2)=B;Abs(LineSegmentOf(A, B)) = 4*Abs(LineSegmentOf(O, F));Asymptote(G)={L1,L2}",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[36, 92], [137, 140]], [[39, 92]], [[39, 92]], [[2, 16]], [[102, 106]], [[107, 111]], [[128, 131]], [[20, 23]], [[27, 30], [32, 35]], [[39, 92]], [[39, 92]], [], [], [[36, 92]], [[2, 16]], [[2, 23]], [[2, 30]], [[32, 111]], [[32, 111]], [[113, 127]], [[36, 98]]]",
    "query_spans": "[[[137, 146]]]",
    "process": "Calculate the directrix of the parabola and the asymptotes of the hyperbola, obtaining A(-1,\\frac{b}{a}), B(-1,-\\frac{b}{a}), simplifying yields b=2a, giving the eccentricity. [Detailed solution] The directrix $ l $ of the parabola $ y^{2}=4x $ has equation $ x=-1 $, the asymptotes of the hyperbola have equations $ y=\\pm\\frac{b}{a}x $. Thus, $ A(-1,\\frac{b}{a}) $, $ B(-1,-\\frac{b}{a}) $, $ \\therefore |AB|=\\frac{2b}{a} $, $ \\frac{2b}{a}=4 $, $ b=2a $. $ \\therefore e=\\frac{c}{a}=\\frac{\\sqrt{a^{2}+b^{2}}}{a}=\\sqrt{5} $"
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=2 p x$ ($p>0$), and the line $l$ intersects the parabola $C$ at points $M$ and $N$, satisfying $\\angle M F N=60^{\\circ}$. Let $d$ be the distance from the midpoint $P$ of segment $M N$ to the directrix of the parabola. If $d=\\frac{\\sqrt{3}}{2}|M N|$, then $\\frac{|M F|}{|N F|}$=?",
    "fact_expressions": "F: Point;Focus(C) = F;C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;p>0;l: Line;Intersection(l, C) = {M, N};M: Point;N: Point;AngleOf(M, F, N) = ApplyUnit(60, degree);MidPoint(LineSegmentOf(M, N)) = P;P: Point;Distance(P, Directrix(C)) = d;d: Number;d = Abs(LineSegmentOf(M, N))*sqrt(3)/2",
    "query_expressions": "Abs(LineSegmentOf(M, F))/Abs(LineSegmentOf(N, F))",
    "answer_expressions": "{2,1/2}",
    "fact_spans": "[[[2, 5]], [[2, 35]], [[6, 32], [42, 48], [104, 107]], [[6, 32]], [[14, 32]], [[14, 32]], [[36, 41]], [[36, 60]], [[51, 54]], [[55, 58]], [[63, 88]], [[90, 103]], [[100, 103]], [[100, 117]], [[114, 117]], [[119, 146]]]",
    "query_spans": "[[[148, 171]]]",
    "process": "According to the problem, construct the following figure: draw perpendiculars from points $M$ and $N$ to the directrix of the parabola, with feet of perpendiculars at $M$ and $N$, respectively. Let $|MF|=m$, $|NF|=n$. By the definition of a parabola, we have $|MM|=m$, $|NN|=n$, so $d=\\frac{m+n}{2}$. By the law of cosines, $|MN|=\\sqrt{m^{2}+n^{2}-mn}$. Thus, $\\frac{m+n}{2}=\\frac{\\sqrt{3}}{2}\\times\\sqrt{m^{2}+n^{2}-mn}$. Solving yields $m=2n$ or $n=2m$, so $\\frac{m}{n}=2$ or $\\frac{m}{n}=\\frac{1}{2}$."
  },
  {
    "text": "An equation of an ellipse with foci at the foci of the hyperbola $x^{2}-y^{2}=-2$ can be? (Write one equation that satisfies the condition) ?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(G) = (x^2 - y^2 =-2);Focus(G)=Focus(H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2/5 + x^2 = 1",
    "fact_spans": "[[[1, 20]], [[27, 29]], [[1, 20]], [[0, 29]]]",
    "query_spans": "[[[27, 36]]]",
    "process": ""
  },
  {
    "text": "Given that $F(\\sqrt{2}, 0)$ is the right focus of the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and that $E$ passes through the point $(\\sqrt{2}, 1)$, then the standard equation of the ellipse $E$ is?",
    "fact_expressions": "E: Ellipse;b: Number;a: Number;G: Point;F: Point;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(G) = (sqrt(2), 1);Coordinate(F) = (sqrt(2), 0);RightFocus(E) = F;PointOnCurve(G, E)",
    "query_expressions": "Expression(E)",
    "answer_expressions": "x^2/4 + y^2/2 = 1",
    "fact_spans": "[[[19, 76], [104, 109], [82, 85]], [[26, 76]], [[26, 76]], [[86, 102]], [[2, 18]], [[26, 76]], [[26, 76]], [[19, 76]], [[86, 102]], [[2, 18]], [[2, 80]], [[82, 102]]]",
    "query_spans": "[[[104, 116]]]",
    "process": "Since $F(\\sqrt{2},0)$ is the right focus of the ellipse $E:\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, we have $c=\\sqrt{2}$. Since the ellipse $E$ passes through the point $(\\sqrt{2},1)$, we have $\\begin{cases}\\frac{2}{a^{2}}+\\frac{1}{b^{2}}=1\\\\a^{2}=2+b^{2}\\end{cases}$, solving which yields $b^{2}=2$, $a^{2}=4$. Therefore, the standard equation of the ellipse is: $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$."
  },
  {
    "text": "Given the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has eccentricity $\\frac{\\sqrt{5}}{2}$. A line $l$ is drawn through the left focus $F(-5,0)$ of $E$. The line $l$ intersects the hyperbola $E$ at points $A$, $B$, and intersects the two asymptotes of $E$ at points $C$, $D$, respectively. If $\\overrightarrow{F A}=\\overrightarrow{A C}$, then $|\\overrightarrow{B D}|$=?",
    "fact_expressions": "l: Line;E: Hyperbola;b: Number;a: Number;F: Point;A: Point;C: Point;B: Point;D: Point;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (-5, 0);Eccentricity(E) = sqrt(5)/2;LeftFocus(E) = F;PointOnCurve(F,l);Intersection(l, E) = {A, B};L1:Line;L2:Line;Asymptote(E)={L1,L2};Intersection(l,L1)=C;Intersection(l,L2)=D;VectorOf(F, A) = VectorOf(A, C)",
    "query_expressions": "Abs(VectorOf(B, D))",
    "answer_expressions": "5*sqrt(5)/8",
    "fact_spans": "[[[97, 102], [103, 108]], [[2, 53], [80, 83], [109, 115], [129, 132]], [[10, 53]], [[10, 53]], [[87, 96]], [[119, 123]], [[141, 145]], [[124, 127]], [[146, 149]], [[2, 53]], [[87, 96]], [[2, 78]], [[80, 96]], [[79, 102]], [[103, 127]], [], [], [[129, 137]], [[103, 149]], [[103, 149]], [[151, 194]]]",
    "query_spans": "[[[196, 222]]]",
    "process": "According to the eccentricity of the hyperbola and the left focus $ F(-5,0) $, we obtain the hyperbola $ E: \\frac{x^{2}}{20}-\\frac{y^{2}}{5}=1 $. Then, from $ \\overrightarrow{FA} = \\overrightarrow{AC} $, it follows that $ A $ is the midpoint of $ F $ and $ C $. Let $ A(x_{A}, y_{A}) $, and from $ \\overrightarrow{FA} = \\overrightarrow{AC} $, we obtain the coordinates of $ C $. Substituting into the asymptote equation gives an expression for $ A(x_{A}, y_{A}) $ in terms of $ x_{A} $, which is then substituted into the hyperbola to find $ A(x_{A}, y_{A}) $, and thus the equation of line $ AF $ is obtained. Solving the hyperbola and its asymptotes simultaneously yields $ BD $. Since the hyperbola $ E: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ has eccentricity $ \\frac{\\sqrt{5}}{2} $ and left focus $ F(-5,0) $, we have $ c = 5 $, and $ \\frac{c}{a} = \\frac{\\sqrt{5}}{2} $, hence $ E: \\frac{x^{2}}{20} - \\frac{y^{2}}{5} = 1 $. Because $ \\overrightarrow{FA} = \\overrightarrow{AC} $, $ A $ is the midpoint of $ F $ and $ C $. Let $ A(x_{A}, y_{A}) $, $ C(x_{C}, y_{C}) $. Since $ \\overrightarrow{FA} = \\overrightarrow{AC} $, we have $ (x_{A} + 5, y_{A}) = (x_{C} - x_{A}, y_{C} - y_{A}) $, solving gives $ C(2x_{A} + 5, 2y_{A}) $. Assume $ C $ lies on the asymptote $ y = -\\frac{1}{2}x $, then $ x_{A} = -2y_{A} - \\frac{5}{2} $, so $ A(-2y_{A} - \\frac{5}{2}, y_{A}) $. Substituting into $ E: \\frac{x^{2}}{20} - \\frac{y^{2}}{5} = 1 $, we get $ \\frac{(-2y_{A} - \\frac{5}{2})^{2}}{20} - \\frac{y_{A}^{2}}{5} = 1 $, solving gives $ y_{A} = \\frac{11}{8} $, so $ A(-\\frac{21}{4}, \\frac{11}{8}) $. Thus, the slope of line $ l $ is $ k = \\frac{\\frac{11}{8} - 0}{-\\frac{21}{4} + 5} = -\\frac{11}{2} $, hence the equation of $ l $ is: $ y = -\\frac{11}{2}(x + 5) $. Solving the system:\n$$\n\\begin{cases}\n\\frac{x^{2}}{20} - \\frac{y^{2}}{5} = 1 \\\\\ny = -\\frac{11}{2}(x + 5)\n\\end{cases}\n\\Rightarrow 24x^{2} + 242x + 609 = 0\n\\Rightarrow (4x + 21)(6x + 29) = 0\n$$\nLet $ B(x_{B}, y_{B}) $, $ D(x_{D}, y_{D}) $, then $ x_{B} = -\\frac{29}{6} $. Solving with the asymptote $ y = \\frac{1}{2}x $:\n$$\n\\begin{cases}\ny = \\frac{1}{2}x \\\\\ny = -\\frac{11}{2}(x + 5)\n\\end{cases}\n\\Rightarrow x_{D} = -\\frac{55}{12}\n$$"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^2}=1$ $(a>0, b>0)$ with right focus $F$, if a line passing through point $F$ with an inclination angle of $\\frac{\\pi}{6}$ has exactly one common point with the right branch of the hyperbola, what is the range of values for the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;F: Point;a>0;b>0;Expression(G) = (-y^2/(b^2) + x^2/a^2 = 1);RightFocus(G) = F;PointOnCurve(F, H);Inclination(H) = pi/6;NumIntersection(H, RightPart(G)) = 1",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[2*\\sqrt{3}/3, +\\infty)",
    "fact_spans": "[[[2, 56], [95, 98], [113, 116]], [[5, 56]], [[5, 56]], [[92, 94]], [[61, 64], [67, 71]], [[5, 56]], [[5, 56]], [[2, 56]], [[2, 64]], [[66, 94]], [[72, 94]], [[92, 110]]]",
    "query_spans": "[[[113, 126]]]",
    "process": ""
  },
  {
    "text": "Let $M(-5,0)$, $N(5,0)$, and the perimeter of $\\triangle M N P$ be $36$. Then the trajectory equation of vertex $P$ of $\\triangle M N P$ is?",
    "fact_expressions": "M: Point;N: Point;P: Point;Coordinate(M) = (-5, 0);Coordinate(N) = (5, 0);Perimeter(TriangleOf(M, N, P)) = 36",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(x^2/169+y^2/144=1)&Negation(y=0)",
    "fact_spans": "[[[1, 10]], [[12, 20]], [[69, 72]], [[1, 10]], [[11, 20]], [[22, 47]]]",
    "query_spans": "[[[69, 79]]]",
    "process": ""
  },
  {
    "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(1/2)*x",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 36]]]",
    "process": ""
  },
  {
    "text": "Given that the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has an eccentricity of $2$, what are the asymptote equations of this hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = 2",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*sqrt(3)*x",
    "fact_spans": "[[[2, 59], [70, 73]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[2, 59]], [[2, 67]]]",
    "query_spans": "[[[70, 81]]]",
    "process": "Since the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $2$, then $2=\\sqrt{1+(\\frac{b}{a})^{2}}$, solving yields $\\frac{b}{a}=\\sqrt{3}$, hence the asymptotes of the hyperbola are $y=\\pm\\sqrt{3}x$."
  },
  {
    "text": "Let points $F_{1}(-c , 0)$ and $F_{2}(c, 0)$ be the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, respectively, and let $P$ be a point on the hyperbola such that $\\overrightarrow {P F_{1}} \\cdot \\overrightarrow {P F_{2}}=-\\frac{2 c^{2}}{3}$. Then the range of values for the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;c: Number;F1: Point;F2: Point;P: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = -2*c^2/3",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[\\sqrt{3}, +\\infty)",
    "fact_spans": "[[[36, 82], [92, 95], [182, 185]], [[39, 82]], [[39, 82]], [[2, 17]], [[1, 17]], [[20, 33]], [[88, 91]], [[36, 82]], [[2, 17]], [[20, 33]], [[1, 87]], [[1, 87]], [[88, 100]], [[102, 179]]]",
    "query_spans": "[[[182, 196]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{4}=1$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2/4 = 1)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*2*x",
    "fact_spans": "[[[0, 33]], [[0, 33]]]",
    "query_spans": "[[[0, 41]]]",
    "process": "Solve directly using the asymptote equation formula. From the hyperbola equation, we know that $ a^{2}=1 $, $ b^{2}=4 $, then $ a=1 $, $ b=2 $. The asymptote equations are $ y=\\pm\\frac{b}{a}x=\\pm2x $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has its right focus at $F$. A tangent line is drawn from point $F$ to the circle $(x-a)^{2}+y^{2}=\\frac{c^{2}}{16}$. If this tangent line is exactly perpendicular to one of the asymptotes of $C$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;c: Number;F: Point;a>0;b>0;L:Line;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y^2 + (-a + x)^2 = c^2/16);RightFocus(C) = F;TangentOfPoint(F,G)=L;IsPerpendicular(L,OneOf(Asymptote(C)))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 63], [137, 143], [124, 127]], [[10, 63]], [[10, 63]], [[78, 113]], [[79, 113]], [[68, 71], [73, 77]], [[10, 63]], [[10, 63]], [], [[2, 63]], [[78, 113]], [[2, 71]], [[72, 116]], [[72, 135]]]",
    "query_spans": "[[[137, 149]]]",
    "process": "From the given, the equation of the tangent line is $ y = -\\frac{a}{b}(x - c) $, that is, $ ax + by - ac = 0 \\Rightarrow $ the distance from the center $ C(a, 0) $ to this tangent line is $ d = \\frac{|a^{2} - ac|}{\\sqrt{a^{2} + b^{2}}} = \\frac{ac - a^{2}}{c} = \\frac{c}{4} \\Rightarrow e^{2} - 4e + 4 = 0 \\Rightarrow e = 2 $"
  },
  {
    "text": "It is known that $F$ is the focus of the parabola $C$: $y^{2}=2 p x$ ($p>0$). A line passing through $F$ intersects $C$ at points $P$ and $Q$, with $Q$ in the first quadrant. If $2 \\overrightarrow{P F}=\\overrightarrow{F Q}$, then what is the slope of the line $P Q$?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;Q: Point;P: Point;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(F,G);Intersection(G, C) = {P, Q};Quadrant(Q)=1;2*VectorOf(P, F) = VectorOf(F, Q)",
    "query_expressions": "Slope(LineOf(P,Q))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[6, 32], [44, 47]], [[14, 32]], [[41, 43]], [[54, 57], [61, 64]], [[50, 53]], [[2, 5], [37, 40]], [[14, 32]], [[6, 32]], [[2, 35]], [[36, 43]], [[41, 59]], [[61, 69]], [[71, 116]]]",
    "query_spans": "[[[118, 130]]]",
    "process": ""
  },
  {
    "text": "If the point $H(2,4)$ lies on the parabola $y^{2}=2 p x$, then what is the value of the real number $p$?",
    "fact_expressions": "G: Parabola;p: Real;H: Point;Expression(G) = (y^2 = 2*(p*x));Coordinate(H) = (2, 4);PointOnCurve(H, G)",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[11, 27]], [[30, 35]], [[1, 10]], [[11, 27]], [[1, 10]], [[1, 28]]]",
    "query_spans": "[[[30, 39]]]",
    "process": "Substituting the coordinates of point H into the equation of the parabola, we get $4^{2}=2p\\times2$, solving for $p$ gives $p=4$."
  },
  {
    "text": "Given point $A(0,5)$, from a point $P$ on the parabola $x^{2}=12 y$, draw a perpendicular to the line $y=-3$, with foot of perpendicular at $B$. If $|P B|=|P A|$, then $|P B|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 12*y);A: Point;P: Point;B: Point;Coordinate(A) = (0, 5);PointOnCurve(P,G);l1:Line;Expression(l1)=(y=-3);l2:Line;IsPerpendicular(l2,l1);PointOnCurve(P,l2);FootPoint(l1,l2)=B;Abs(LineSegmentOf(P, B)) = Abs(LineSegmentOf(P, A))",
    "query_expressions": "Abs(LineSegmentOf(P, B))",
    "answer_expressions": "7",
    "fact_spans": "[[[13, 28]], [[13, 28]], [[2, 11]], [[31, 34]], [[48, 51]], [[2, 11]], [[12, 34]], [[35, 41]], [[35, 41]], [], [[12, 44]], [[12, 44]], [[12, 51]], [[53, 66]]]",
    "query_spans": "[[[68, 77]]]",
    "process": "According to the problem, let P(x,y), |PB|=|PA|, we obtain y+3=\\sqrt{x^{2}+(y-5)^{2}}, combining with x^{2}=12y gives the solution. Let P(x,y), |PB|=|PA|, we obtain y+3=\\sqrt{x^{2}+(y-5)^{2}}, x^{2}-16y+16=0, substituting x^{2}=12y yields: y=4, so |PB|=y+3=7,"
  },
  {
    "text": "Let the semi-focal length of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ be $c$. It is known that the distance from the origin to the line $l$: $b x+a y=a b$ is equal to $\\frac{1}{4}  c+1$. Then the minimum value of $c$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;HalfFocalLength(G)=c;c:Number;O:Origin;l:Line;Expression(l) = (a*y + b*x = a*b);Distance(O,l)=(c/4+1)",
    "query_expressions": "Min(c)",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 58]], [[1, 58]], [[4, 58]], [[4, 58]], [[4, 58]], [[4, 58]], [[1, 66]], [[63, 66], [118, 121]], [[70, 72]], [[73, 93]], [[73, 93]], [[70, 116]]]",
    "query_spans": "[[[118, 127]]]",
    "process": ""
  },
  {
    "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{m}=1$ are given by $y=\\pm \\frac{\\sqrt{3}}{2} x$, then the focal distance of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/4 - y^2/m = 1);Expression(Asymptote(G)) = (y = pm*(x*(sqrt(3)/2)))",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(7)",
    "fact_spans": "[[[1, 39], [76, 79]], [[4, 39]], [[1, 39]], [[1, 74]]]",
    "query_spans": "[[[76, 84]]]",
    "process": ""
  },
  {
    "text": "The distance from the left focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ to its asymptote is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1)",
    "query_expressions": "Distance(LeftFocus(G), Asymptote(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 52]]]",
    "process": "(Analysis) First, according to the given hyperbola equation, find its left focus coordinates and asymptote equations, then use the point-to-line distance formula to obtain the result. According to the problem, the hyperbola equation is \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1, where a=3, b=4, so c=5, thus the coordinates of its left focus are (-5,0), and the asymptote equations are y=\\pm\\frac{4}{3}x, i.e., 4x\\pm3y=0. Then the distance from the left focus to its asymptote is d=\\frac{|-20\\pm0|}{\\sqrt{4^{2}+3^{2}}}=\\frac{20}{5}=4"
  },
  {
    "text": "Given the line $l$: $y = x + 1$ and the parabola $C$: $x^{2} = y$ intersect at points $A$ and $B$, point $P(0,1)$, $Q(-1,0)$, and $\\overrightarrow{P Q} = \\lambda \\overrightarrow{Q A} = \\mu \\overrightarrow{Q B}$ ($\\lambda, \\mu \\in \\mathbb{R}$), then $\\lambda + \\mu = $?",
    "fact_expressions": "l: Line;C: Parabola;P: Point;Q: Point;A: Point;B: Point;lambda:Real;mu:Real;Expression(C) = (x^2 = y);Expression(l) = (y = x + 1);Intersection(l, C) = {A, B};Coordinate(P) = (0, 1);Coordinate(Q) = (-1, 0);VectorOf(P, Q) = lambda*VectorOf(Q, A);lambda*VectorOf(Q, A) = mu*VectorOf(Q, B)",
    "query_expressions": "lambda + mu",
    "answer_expressions": "-3",
    "fact_spans": "[[[2, 16]], [[17, 34]], [[46, 55]], [[58, 67]], [[36, 39]], [[40, 43]], [[69, 159]], [[69, 159]], [[17, 34]], [[2, 16]], [[2, 45]], [[46, 55]], [[58, 67]], [[69, 159]], [[69, 159]]]",
    "query_spans": "[[[161, 176]]]",
    "process": "Let the coordinates of points A and B be given as per the problem, construct $\\overrightarrow{PQ}=(-1,-1)$, $\\lambda\\overrightarrow{QA}=\\lambda(x_{1}+1,y_{1})$, $\\mu\\overrightarrow{QB}=\\mu(x_{2}+1,y_{2})$. Using the vector parallel relationship, we obtain $x_{1}=\\frac{-1}{\\lambda}-1$, $y_{1}=\\frac{-1}{\\lambda}$, combined with $x^{2}=y$, thus $\\lambda^{2}+3\\lambda+1=0$, similarly $\\mu^{2}+3\\mu+1=0$, finally we get: $\\lambda+\\mu=-3$. [Solution] Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $\\overrightarrow{PQ}=(-1,-1)$, $\\lambda\\overrightarrow{QA}=\\lambda(x_{1}+1,y_{1})$, $\\mu\\overrightarrow{QB}=\\mu(x_{2}+1,y_{2})$. Then $x_{1}=\\frac{-1}{\\lambda}-1$, $y_{1}=\\frac{-1}{\\lambda}$, substitute into the equation $x^{2}=y$, hence $\\lambda^{2}+3\\lambda+1=0$, similarly $\\mu^{2}+3\\mu+1=0$, thus $\\lambda,\\mu$ can be viewed as the two roots of the equation $x^{2}+3x+1=0$, then $\\lambda+\\mu=-3$."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{25}-\\frac{y^{2}}{9}=1$, respectively. Point $P$ lies on the ellipse. If $|P F_{1}|=9|P F_{2}|$, then the coordinates of point $P$ are?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 - y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 9*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(5,0)",
    "fact_spans": "[[[19, 57], [69, 71]], [[64, 68], [98, 102]], [[1, 8]], [[9, 16]], [[19, 57]], [[1, 63]], [[1, 63]], [[64, 72]], [[74, 96]]]",
    "query_spans": "[[[98, 107]]]",
    "process": ""
  },
  {
    "text": "If an ellipse passes through the point $(2 , 3)$, and has foci $F_{1}(-2,0)$, $F_{2}(2,0)$, then the eccentricity of this ellipse is equal to?",
    "fact_expressions": "G: Ellipse;PointOnCurve(H, G) = True;H: Point;Coordinate(H) = (2, 3);F1: Point;F2: Point;Focus(G) = {F1,F2};Coordinate(F1) = (-2, 0);Coordinate(F2) = (2, 0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[1, 3], [51, 53]], [[1, 15]], [[5, 15]], [[5, 15]], [[20, 33]], [[35, 47]], [[1, 47]], [[20, 33]], [[35, 47]]]",
    "query_spans": "[[[51, 60]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ is a point on the parabola $y^{2} = -8x$, let $d_{1}$ be the distance from $P$ to the directrix of this parabola, and $d_{2}$ be the distance from $P$ to the line $x + y - 10 = 0$. Then the minimum value of $d_{1} + d_{2}$ is?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;d1: Number;d2: Number;Expression(G) = (y^2 = -8*x);Expression(H) = (x + y - 10 = 0);PointOnCurve(P, G);Distance(P, Directrix(G)) = d1;Distance(P, H) = d2",
    "query_expressions": "Min(d2 + d1)",
    "answer_expressions": "6*sqrt(2)",
    "fact_spans": "[[[7, 22], [32, 35]], [[50, 62]], [[2, 6], [27, 30]], [[41, 48]], [[66, 73]], [[7, 22]], [[50, 62]], [[2, 25]], [[27, 48]], [[27, 73]]]",
    "query_spans": "[[[75, 94]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus of the ellipse $C$ is $F$. A line passing through the origin intersects the ellipse $C$ at points $A$ and $B$. Connect $AF$, $BF$. If $AB=15$, $BF=12$, $\\sin \\angle ABF=\\frac{3}{5}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;A: Point;F: Point;B: Point;O: Origin;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;PointOnCurve(O, G);Intersection(C, G) = {A, B};LineSegmentOf(A, B) = 15;LineSegmentOf(B, F) = 12;Sin(AngleOf(A, B, F)) = 3/5",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "5/7",
    "fact_spans": "[[[2, 59], [68, 73], [161, 164]], [[9, 59]], [[9, 59]], [[78, 80]], [[83, 86]], [[64, 67]], [[87, 90]], [[75, 77]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 67]], [[74, 80]], [[68, 92]], [[110, 118]], [[119, 127]], [[128, 159]]]",
    "query_spans": "[[[161, 170]]]",
    "process": "According to the problem, draw the figure. In triangle AFB, given AB = 15, BF = 12, $\\sin\\angle ABF = \\frac{3}{5}$, using the law of cosines we obtain $AF^{2} = AB^{2} + BF^{2} - 2AB \\cdot BF \\cos\\angle ABF = 81$, $\\therefore AF = 9$. $\\therefore AF = \\frac{9}{1AF^{2} + |BF|} = |AB|^{2}$, then triangle AFB is a right triangle. Connect AF and BF, then quadrilateral AFBF is a rectangle. $\\therefore 2a = 9 + 12 = 21$, $2c = 15$, then $a = \\frac{21}{2}$, $c = \\frac{15}{2}$. $\\therefore$ the eccentricity of ellipse C is $e = \\frac{c}{a} = \\frac{15}{21} = \\frac{5}{7}$."
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x$ coincides with the left focus of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, then the real number $p$=?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Real;Expression(G) = (x^2/3 - y^2 = 1);Expression(H) = (y^2 = 2*(p*x));Focus(H) = LeftFocus(G)",
    "query_expressions": "p",
    "answer_expressions": "-4",
    "fact_spans": "[[[21, 49]], [[1, 17]], [[57, 62]], [[21, 49]], [[1, 17]], [[1, 55]]]",
    "query_spans": "[[[57, 64]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has a focal distance of $4$, and its asymptotes are tangent to the circle $C_{2}$: $(x-2)^{2}+y^{2}=1$. Then the standard equation of the hyperbola $C_{1}$ is?",
    "fact_expressions": "C1: Hyperbola;Expression(C1) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;FocalLength(C1) = 4;C2: Circle;Expression(C2) = ((x - 2)^2 + y^2 = 1);IsTangent(Asymptote(C1), C2)",
    "query_expressions": "Expression(C1)",
    "answer_expressions": "x^2/3-y^2=1",
    "fact_spans": "[[[0, 65], [74, 75], [111, 121]], [[0, 65]], [[12, 65]], [[12, 65]], [[12, 65]], [[12, 65]], [[0, 72]], [[79, 107]], [[79, 107]], [[74, 109]]]",
    "query_spans": "[[[111, 128]]]",
    "process": "Since the focal distance of the hyperbola $ C_{1}: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) is 4, we have $ c=2 $. Given that the two asymptotes of hyperbola $ C_{1} $, $ y = \\pm \\frac{b}{a}x $, are tangent to the circle $ C_{2}: (x-2)^{2} + y^{2} = 1 $, it follows that $ 1 = \\frac{2b}{\\sqrt{a^{2}+b^{2}}} $. Also, $ a^{2} + b^{2} = 4 $, so $ b=1 $, $ a = \\sqrt{3} $. Therefore, the standard equation of hyperbola $ C_{1} $ is $ \\frac{x^{2}}{3} - y^{2} = 1 $."
  },
  {
    "text": "Given two fixed points $A(-1,0)$, $B(1,0)$, and a point $P(x, y)$ moving on the line $y=2x+4$, find the maximum eccentricity of the ellipse with foci at $A$ and $B$ passing through point $P$.",
    "fact_expressions": "A: Point;Coordinate(A) = (-1, 0);B: Point;Coordinate(B) = (1, 0);x1: Number;y1: Number;P: Point;Coordinate(P) = (x1, y1);G: Ellipse;H: Line;Expression(H) = (y = 2*x + 4);PointOnCurve(P, H);Focus(G) = {A, B};PointOnCurve(P, G)",
    "query_expressions": "Max(Eccentricity(G))",
    "answer_expressions": "sqrt(85)/17",
    "fact_spans": "[[[5, 14], [56, 59]], [[5, 14]], [[16, 24], [60, 63]], [[16, 24]], [[26, 35]], [[26, 35]], [[25, 35], [68, 72]], [[25, 35]], [[73, 75]], [[36, 47]], [[36, 47]], [[25, 53]], [[55, 75]], [[67, 75]]]",
    "query_spans": "[[[73, 85]]]",
    "process": "From the given conditions: the ellipse has foci at A(-1,0) and B(1,0), so c=1, eccentricity e=\\frac{c}{a}. Since the ellipse passes through point P, |PA|+|PB|=2a. Let point C(m,n) be the symmetric point of point A about the line y=2x+4, then \\begin{cases}\\frac{n}{m+1}\\times2=-1\\\\\\frac{n}{2}=2\\times\\frac{m-1}{2}+4\\end{cases}, solving gives: \\begin{cases}m=-\\frac{13}{5}\\\\n=\\frac{4}{5}\\end{cases}. Thus, 2a=|PA|+|PB|=|PC|+|PB|\\geqslant|BC|, with equality if and only if points P, B, C are collinear. |BC|=\\sqrt{(1+\\frac{13}{5})^{2}+(0-\\frac{4}{5})^{2}}=\\frac{2\\sqrt{85}}{5}, so 2a\\geqslant\\frac{2\\sqrt{85}}{5}, i.e., a\\geqslant\\frac{\\sqrt{85}}{5}. Therefore, the eccentricity e=\\frac{c}{a}\\leqslant\\frac{1}{\\frac{\\sqrt{85}}{5}}=\\frac{5}{\\sqrt{85}}=\\frac{\\sqrt{85}}{17}. Hence, the maximum value of the ellipse's eccentricity is \\frac{\\sqrt{85}}{17}."
  },
  {
    "text": "Given that point $B$ is the intersection of the left directrix of the hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1(a>0, b>0)$ and the $x$-axis, point $A$ has coordinates $(0, b)$, and $\\overrightarrow{A P}=3 \\overrightarrow{A B}$. If point $P$ lies on the hyperbola, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;A: Point;P: Point;B: Point;a>0;b>0;Expression(G) = (-x^2/b^2 + y^2/a^2 = 1);Intersection(LeftDirectrix(G), xAxis) = B;Coordinate(A) = (0, b);VectorOf(A, P) = 3*VectorOf(A, B);PointOnCurve(P, G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3*sqrt(5)/5",
    "fact_spans": "[[[7, 63], [147, 150], [153, 156]], [[83, 91]], [[10, 63]], [[76, 80]], [[142, 146]], [[2, 6]], [[10, 63]], [[10, 63]], [[7, 63]], [[2, 75]], [[76, 91]], [[95, 140]], [[142, 151]]]",
    "query_spans": "[[[153, 162]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. If there exists a point $P$ on the ellipse such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}<0$, then what is the range of values for the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);DotProduct(VectorOf(P, F1), VectorOf(P, F2))<0",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(\\sqrt{2}/2, 1)",
    "fact_spans": "[[[17, 71], [79, 81], [152, 154]], [[17, 71]], [[19, 71]], [[19, 71]], [[19, 71]], [[19, 71]], [[1, 8]], [[9, 16]], [[1, 77]], [[1, 77]], [[84, 88]], [[79, 88]], [[91, 150]]]",
    "query_spans": "[[[152, 164]]]",
    "process": ""
  },
  {
    "text": "The right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $F$. An asymptote of the hyperbola $C$ intersects the circle with diameter $OF$ at point $M$ (distinct from point $O$), and intersects the line passing through $F$ and perpendicular to the $x$-axis at point $N$. If $S_{\\triangle O M F}=4 S_{\\triangle M N F}$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;O: Origin;F: Point;M: Point;N: Point;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;IsDiameter(LineSegmentOf(O,F),G);Intersection(OneOf(Asymptote(C)),G)=M;Negation(M=O);L:Line;PointOnCurve(F,L);IsPerpendicular(L,xAxis);Intersection(L,OneOf(Asymptote(C)))=N;Area(TriangleOf(O,M,F))=4*Area(TriangleOf(M,N,F))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[0, 51], [60, 66], [167, 173]], [[7, 51]], [[7, 51]], [[83, 84]], [[93, 97]], [[56, 59], [101, 104]], [[86, 90]], [[117, 120]], [[0, 51]], [[0, 59]], [[73, 84]], [[60, 90]], [[86, 98]], [[113, 115]], [[100, 115]], [[105, 115]], [[60, 120]], [[122, 165]]]",
    "query_spans": "[[[167, 179]]]",
    "process": "Without loss of generality, let the equation of one asymptote of the hyperbola be $ y = \\frac{b}{a}x $. From the given conditions, $ OM \\perp FM $, and $ |OF| = c $, $ c^{2} = a^{2} + b^{2} $, $ \\tan\\angle NOF = \\frac{b}{a} $, so $ |OM| = a $. If $ S_{\\triangle OMF} = 4S_{\\triangle MNF} $, then $ OM = 4MN $, i.e., $ |MN| = \\frac{a}{4} $. In right triangle $ \\triangle OFN $, by the Pythagorean theorem, we have $ |FN|^{2} = |ON|^{2} - |OF| = \\left(\\frac{5a}{4}\\right)^{2} - c^{2} $. Also, since $ \\tan\\angle NOF = \\frac{b}{a} $, we get $ |FN| = \\frac{b}{a} \\cdot c $. Therefore, $ \\left(\\frac{5a}{4}\\right)^{2} - c^{2} = \\left(\\frac{bc}{a}\\right)^{2} $. Simplifying yields $ (c^{2} - a^{2})c^{2} + a^{2}c^{2} = \\frac{25a^{4}}{16} $, i.e., $ c^{4} = \\frac{25a^{4}}{16} $, so $ e = \\frac{c}{a} = \\frac{\\sqrt{5}}{2} $."
  },
  {
    "text": "If the right focus $F$ of the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{m}=1$ lies on the circle $x^{2}+y^{2}-3 x-4 y+2=0$, then the equations of the asymptotes of this hyperbola are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/2 - y^2/m = 1);m: Number;H: Circle;Expression(H) = (-4*y - 3*x + x^2 + y^2 + 2 = 0);F: Point;RightFocus(G) = F;PointOnCurve(F, H)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*x",
    "fact_spans": "[[[1, 39], [77, 80]], [[1, 39]], [[4, 39]], [[47, 73]], [[47, 73]], [[43, 46]], [[1, 46]], [[43, 74]]]",
    "query_spans": "[[[77, 88]]]",
    "process": "Substitute y=0 into x^{2}+y^{2}-3x-4y+2=0, solving gives x=1 or x=2. Since \\sqrt{2+m}>1, it follows that \\sqrt{2+m}=2, solving gives m=2. Therefore, the asymptotes of the hyperbola \\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1 are y=\\pmx. The answer is: v=+x"
  },
  {
    "text": "The equation of the line $l$ passing through the focus of the parabola $y^{2}=4x$ and parallel to the line $3x-2y=0$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);l: Line;PointOnCurve(Focus(G), l);H: Line;Expression(H) = (3*x - 2*y = 0);IsParallel(H, l)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "3*x-2*y-3=0",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[37, 42]], [[0, 42]], [[23, 36]], [[23, 36]], [[20, 42]]]",
    "query_spans": "[[[37, 47]]]",
    "process": ""
  },
  {
    "text": "$P$ is an arbitrary point on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, $F_{1}$, $F_{2}$ are its two foci, $O$ is the coordinate origin, and there is a moving point $Q$ satisfying $\\overrightarrow{O Q}=\\overrightarrow{P F_{1}}+\\overrightarrow{P F_{2}}$. Then the trajectory equation of the moving point $Q$ is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;O: Origin;Q: Point;P: Point;F1: Point;F2:Point;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P,G);Focus(G)={F1,F2};VectorOf(O, Q) = VectorOf(P, F1) + VectorOf(P, F2)",
    "query_expressions": "LocusEquation(Q)",
    "answer_expressions": "x^2/(4*a^2)+y^2/(4*b^2)=1",
    "fact_spans": "[[[4, 49], [72, 73]], [[6, 49]], [[6, 49]], [[79, 82]], [[92, 95], [173, 176]], [[0, 3]], [[56, 63]], [[64, 71]], [[4, 49]], [[0, 55]], [[56, 78]], [[97, 169]]]",
    "query_spans": "[[[173, 183]]]",
    "process": ""
  },
  {
    "text": "Given points $F_{1}(-3,0)$, $F_{2}(3,0)$ are the left and right foci of the hyperbola $L$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, respectively. Let $P$ be a point on the hyperbola $L$ such that $4|P F_{1}|=5|P F_{2}|$. When $\\overrightarrow{P F_{2}} \\cdot \\overrightarrow{F_{1} F_{2}}=0$, what is the distance from the focus of the hyperbola $L$ to its asymptote?",
    "fact_expressions": "L: Hyperbola;b: Number;a: Number;F1: Point;F2: Point;P: Point;Expression(L) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-3, 0);Coordinate(F2) = (3, 0);LeftFocus(L) = F1;RightFocus(L) = F2;PointOnCurve(P, L);4*Abs(LineSegmentOf(P, F1)) = 5*Abs(LineSegmentOf(P, F2));DotProduct(VectorOf(P, F2), VectorOf(F1, F2)) = 0",
    "query_expressions": "Distance(Focus(L), Asymptote(L))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[34, 85], [96, 102], [197, 203]], [[42, 85]], [[42, 85]], [[2, 17]], [[19, 31]], [[92, 95]], [[34, 85]], [[2, 16]], [[19, 31]], [[2, 91]], [[2, 91]], [[92, 105]], [[107, 130]], [[132, 195]]]",
    "query_spans": "[[[197, 215]]]",
    "process": "From $\\overrightarrow{PF}_{2}\\cdot\\overrightarrow{F_{1}F_{2}}=0$ we get $\\angle PF_{2}F_{1}=90^{\\circ}$. Also, since $4|PF_{1}|=5|PF_{2}|$, $|F_{1}F_{2}|=6$, by the Pythagorean theorem we have $|PF_{1}|^{2}=|PF_{2}|^{2}+|F_{1}F_{2}|^{2}=|PF_{2}|^{2}+36$, solving gives $|PF_{1}|=10$, $|PF_{2}|=8$. From the definition of hyperbola we get $|PF_{1}|-|PF_{2}|=2a=2$, so $a=1$, thus $b=2\\sqrt{2}$. Therefore, the asymptotes of the hyperbola are $y=\\pm2\\sqrt{2}x$, so the distance from the focus to the asymptote is $d=\\frac{|2\\sqrt{2}\\times3|}{\\sqrt{1+8}}=2\\sqrt{2}$."
  },
  {
    "text": "Given the parabola $y^{2}=8x$, a line passing through its focus $F$ intersects the parabola at points $A$ and $B$. A perpendicular from the midpoint $M$ of $AB$ to the $Y$-axis meets the $Y$-axis at point $N$. If $|MN|=2$, then $|AB|=$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;M: Point;N: Point;F:Point;Expression(G) = (y^2 = 8*x);PointOnCurve(F,H);Focus(G)=F;Intersection(H,G)={A,B};MidPoint(LineSegmentOf(A,B))=M;L:Line;PointOnCurve(M,L);IsPerpendicular(L,yAxis);Intersection(L,yAxis)=N;Abs(LineSegmentOf(M, N)) = 2",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 16], [17, 18], [27, 30]], [[24, 26]], [[31, 34]], [[35, 38]], [[48, 51]], [[64, 68]], [[20, 23]], [[2, 16]], [[16, 26]], [[17, 23]], [[24, 40]], [[41, 51]], [], [[40, 58]], [[40, 58]], [[40, 68]], [[71, 80]]]",
    "query_spans": "[[[82, 91]]]",
    "process": ""
  },
  {
    "text": "Let the focus of the parabola $y^{2}=4x$ be $F$, the directrix be $L$, and let $P$ be a point on the parabola, with $PA \\perp L$ and $A$ the foot of the perpendicular. If the slope of the line $AF$ is $-\\sqrt{3}$, then what is the standard equation of the circle with $PF$ as diameter?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;L: Line;Directrix(G) = L;P: Point;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, A), L);A: Point;FootPoint(LineSegmentOf(P, A), L) = A;Slope(LineOf(A, F)) = -sqrt(3);H: Circle;IsDiameter(LineSegmentOf(P, F), H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x-2)^2+(y-sqrt(3))^2=4",
    "fact_spans": "[[[1, 15], [34, 37]], [[1, 15]], [[19, 22]], [[1, 22]], [[26, 29]], [[1, 29]], [[30, 33]], [[30, 40]], [[41, 54]], [[56, 59]], [[41, 62]], [[66, 87]], [[99, 100]], [[90, 100]]]",
    "query_spans": "[[[99, 107]]]",
    "process": "Using the definition of a parabola, |PF| = |PA|, let the projection of F on l be F', according to the problem, |FF'| and |AF| can be determined, thus the y-coordinate of point P can be found. Substituting into the parabola equation gives the x-coordinate of point P, allowing |PF| and |PA| to be determined. Then the equation of the circle can be found. [Detailed solution] ∵ The focus of the parabola y^{2} = 4x is F, the directrix is l, P is a point on the parabola, |PF| = |PA|, F(1,0), the equation of directrix l is: x = -1; let the projection of F on l be F', and PA ⊥ l, according to the problem ∠AFF' = 60^{\\circ}, |FF'| = 2, ∴ |AF| = 2\\sqrt{3}, PA // x-axis, ∴ the y-coordinate of point P is 2\\sqrt{3}, let the x-coordinate of point P be x_{0}, (2\\sqrt{3})^{2} = 4x_{0}, ∵ x_{0} = 3, ∴ |PF| = |PA| = x_{0} - (-1) = 3 - (-1) = 4. Hence, the center of the circle with PF as diameter is (2, \\sqrt{3}), radius is 2. The standard equation of the circle with PF as diameter is (x - 2)^{2} + (y - \\sqrt{3})^{2} = 4."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{m}=1$ $(m>0)$ with asymptotes given by $y=\\pm \\sqrt{2} x$, and let $F_{1}$, $F_{2}$ be the left and right foci of $C$, respectively. Let $P$ be a point on the right branch of $C$. If $|P F_{1}|=m-1$, then $|P F_{2}|=$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/4 - y^2/m = 1);m: Number;m>0;Expression(Asymptote(C)) = (y = pm*(sqrt(2)*x));F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, RightPart(C)) = True;Abs(LineSegmentOf(P, F1)) = m - 1",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 50], [98, 101], [112, 115]], [[2, 50]], [[10, 50]], [[10, 50]], [[2, 75]], [[78, 85]], [[88, 95]], [[78, 107]], [[78, 107]], [[108, 111]], [[108, 120]], [[123, 138]]]",
    "query_spans": "[[[140, 153]]]",
    "process": "From the given condition, we have $\\frac{\\sqrt{m}}{2}=\\sqrt{2}$, solving gives $m=8$, and $a=2$, so $|PF_{1}|=m-1=7$. Since $F_{1}, F_{2}$ are the left and right foci of $C$ respectively, and $P$ is a point on the right branch of $C$, $\\therefore |PF_{1}|-|PF_{2}|=2a=4$, hence $|PF_{2}|=3$."
  },
  {
    "text": "Given that point $F_{2}$ is the right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the line $y=k x$ intersects $C$ at points $A$ and $B$. If $\\angle A F_{2} B=\\frac{2 \\pi}{3}$, $S_{\\Delta A F_{2} B}=2 \\sqrt{3}$, then the length of the imaginary axis of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G:Line;k: Number;A: Point;F2: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y = k*x);RightFocus(C) = F2;Intersection(G, C) = {A, B};AngleOf(A, F2, B) = (2*pi)/3;Area(TriangleOf(A,F2,B))=2*sqrt(3)",
    "query_expressions": "Length(ImageinaryAxis(C))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[11, 72], [87, 90], [174, 177]], [[18, 72]], [[18, 72]], [[77, 86]], [[79, 86]], [[91, 94]], [[2, 10]], [[95, 98]], [[18, 72]], [[18, 72]], [[11, 72]], [[77, 86]], [[2, 76]], [[77, 100]], [[102, 137]], [[139, 172]]]",
    "query_spans": "[[[174, 183]]]",
    "process": "From the given conditions, point B is symmetric to point A with respect to the origin. Let the left focus of the hyperbola be $ F_{1} $. Connect $ AF_{1} $ and $ BF_{1} $. By symmetry, quadrilateral $ AF_{1}BF_{2} $ is a parallelogram, so $ \\angle F_{1}AF_{2} = \\frac{\\pi}{3} $, $ S_{AAF_{2}B} = 2\\sqrt{3} $. Let $ |AF_{2}| = m $, then $ |AF_{1}| = 2a + m $. In $ \\triangle AF_{1}F_{2} $, by the law of cosines:  \n$ 4c^{2} = m^{2} + (m + 2a)^{2} - m(m + 2a) $.  \nSimplifying yields: $ 4c^{2} - 4a^{2} = m^{2} + 2ma $, i.e., $ 4b^{2} = m(m + 2a) $.  \nAlso, $ S_{\\triangle ABF} = \\frac{1}{2} \\cdot m(m + 2a) \\cdot \\frac{\\sqrt{3}}{2} = 2\\sqrt{3} $, $ \\therefore b^{2} = 2 $."
  },
  {
    "text": "The equation of the circle with its center on the parabola $x^{2}=2 y$ and tangent to the line $2 x+2 y+3=0$, having the minimum area, is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 2*y);H: Circle;PointOnCurve(Center(H),G) = True;I: Line;Expression(I) = (2*x + 2*y + 3 = 0);IsTangent(I,H) = True;WhenMin(Area(H))",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x+1)^2+(y-1/2)^2=1/2",
    "fact_spans": "[[[3, 17]], [[3, 17]], [[38, 39], [46, 47]], [[0, 39]], [[20, 35]], [[20, 35]], [[19, 39]], [[41, 47]]]",
    "query_spans": "[[[46, 52]]]",
    "process": "Let the center of the circle be $(x_{0},\\frac{x_{0}^{2}}{2})$, the distance from the center to the line $2x+2y+3=0$ is $\\frac{2^{x}x_{+2}^{x_{0}}+3}{\\sqrt{\\frac{12+2^{2}}{2\\sqrt{++1}^{2}+2}}}$, $\\frac{|x_{0}^{2}+2x_{2}+3|}{\\sqrt[When $x_{0}-1$, the minimum distance is $\\frac{\\sqrt{2}}{2}$,}{the center is then $(-1,\\frac{1}{2})$}. Thus, the equation of the desired circle is $(x+1)^{2}+(y-\\frac{1}{2})^{2}=\\frac{1}{2}$."
  },
  {
    "text": "If the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=-1$ is $\\sqrt{3}$, then what are the equations of the two asymptotes?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = -1);Eccentricity(G) = sqrt(3)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*x*sqrt(2)/2",
    "fact_spans": "[[[1, 48]], [[4, 48]], [[4, 48]], [[1, 48]], [[1, 63]]]",
    "query_spans": "[[[1, 75]]]",
    "process": "Analysis: First, convert the hyperbola equation into standard form, then use the eccentricity to find the relationship between $a$ and $b$, and substitute into the asymptote equations. The hyperbola $\\frac{x^2}{a^{2}} - \\frac{y^{2}}{b^{2}} = -1$ can be rewritten as $\\frac{y^{2}}{b^{2}} - \\frac{x^{2}}{a^{2}} = 1$. From $e^{2} = 1 + \\frac{a^{2}}{b^{2}} = 3$, we obtain $a^{2} = 2b^{2}$. Therefore, the two asymptotes of the hyperbola are $y = \\pm\\frac{\\sqrt{2}}{2}x$."
  },
  {
    "text": "If $AB$ is a chord passing through the focus of the parabola $y^{2}=4x$, and $AB=4$, $O$ is the coordinate origin, then the area of $\\triangle OAB$ equals?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;O: Origin;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), LineSegmentOf(A, B));IsChordOf(LineSegmentOf(A, B), G);LineSegmentOf(A, B) = 4",
    "query_expressions": "Area(TriangleOf(O, A, B))",
    "answer_expressions": "2",
    "fact_spans": "[[[9, 23]], [[1, 6]], [[1, 6]], [[38, 41]], [[9, 23]], [[1, 27]], [[1, 27]], [[29, 36]]]",
    "query_spans": "[[[48, 71]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}) according to the problem. By the definition of the parabola, |AB| = x_{1}+1 + x_{2}+1 = x_{1}+x_{2}+2 = 4, solving gives: x_{1}+x_{2}=2 (1). Since the abscissas of chord AB passing through the focus also satisfy: x_{1}x_{2}=1 (2), solving equations (1) and (2) simultaneously yields: x_{1}=x_{2}=1. Therefore, A(1,2), B(1,-2), so S_{AAOB} = \\frac{1}{2}\\times4\\times1=2. The answer is: 2."
  },
  {
    "text": "If a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ is $F(5,0)$, and the slope of an asymptote is $\\frac{3}{4}$, then $a=$?",
    "fact_expressions": "C:Hyperbola;a:Number;b:Number;a>0;b>0;F:Point;Expression(C)=(x^2/a^2-y^2/b^2=1);OneOf(Focus(C))=F;Coordinate(F)=(5,0);Slope(OneOf(Asymptote(C)))=3/4",
    "query_expressions": "a",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 58]], [[96, 99]], [[4, 58]], [[4, 58]], [[4, 58]], [[63, 71]], [[1, 58]], [[1, 71]], [[63, 71]], [[1, 94]]]",
    "query_spans": "[[[96, 101]]]",
    "process": ""
  },
  {
    "text": "Given that the line $l$ intersects the parabola $y=x^{2}$ at points $A$ and $B$, with $|AB|=2$. Let $M$ be the midpoint of segment $AB$. As the line $l$ moves, the trajectory equation of point $M$ is?",
    "fact_expressions": "A: Point;B: Point;l: Line;G: Parabola;M: Point;Expression(G) = (y = x^2);Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, B)) = 2;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "(y-x^2)*(1+4*x^2)=1",
    "fact_spans": "[[[22, 25]], [[26, 29]], [[2, 7], [59, 64]], [[8, 20]], [[54, 57], [69, 73]], [[8, 20]], [[2, 31]], [[33, 41]], [[44, 57]]]",
    "query_spans": "[[[69, 80]]]",
    "process": "Let point M(x,y), A(x+m,y+n), B(x-m,y-n), then we have  \n\\begin{cases}y+n=(x+m)^{2}\\\\y-n=(x-m)^{2}\\end{cases}  \nSubtracting the two equations gives: 2n=4mx, \\Rightarrown=2mx  \nAdding the two equations gives: 2y=2x^{2}+2m^{2}\\Rightarrow y=x^{2}+m^{2}  \nSolving: m^{2}=y-x^{2}, n^{2}=4(y-x^{2})x^{2}  \nSince |AB|=2, we have |AB|=\\sqrt{(2m)^{2}+(2n)^{2}}=2\\sqrt{m^{2}+n^{2}}=2\\Rightarrow m^{2}+n^{2}=1  \nThus, y-x^{2}+4(y-x^{2})x^{2}=1, i.e., the trajectory equation of point M is (y-x^{2})(1+4x^{2})=1"
  },
  {
    "text": "If $(x+2)^{2}+\\frac{y^{2}}{4}=1$, then the range of $x^{2}+y^{2}$ is?",
    "fact_expressions": "x_: Number;y_: Number;y_^2/4 + (x_ + 2)^2 = 1",
    "query_expressions": "Range(x_^2 + y_^2)",
    "answer_expressions": "[1, 28/3]",
    "fact_spans": "[[[32, 45]], [[32, 45]], [[1, 30]]]",
    "query_spans": "[[[32, 52]]]",
    "process": "By the given condition: $(x+2)^{2}+\\frac{y^{2}}{4}=1$. Let $x=\\cos\\theta\\cdot2$, $y=2\\sin\\theta$, then:  \n$x^{2}+y^{2}=(\\cos\\theta\\cdot2)^{2}+4\\sin^{2}\\theta=\\cos^{2}\\theta\\cdot4+4\\sin^{2}\\theta=\\cos^{2}\\theta\\cdot4+4(1-\\cos^{2}\\theta)=4\\cos^{2}\\theta+4-4\\cos^{2}\\theta+4\\sin^{2}\\theta=4+4\\sin^{2}\\theta=4+4(1-\\cos^{2}\\theta)=8-4\\cos^{2}\\theta=-4\\cos^{2}\\theta+8$.  \nWhen $\\cos\\theta=-\\frac{2}{3}$, $x^{2}+y^{2}$ attains its maximum value $\\frac{28}{3}$.  \nWhen $\\cos\\theta=1$, $x^{2}+y^{2}$ attains its minimum value $1$.  \nThus, the range of $x^{2}+y^{2}$ is $[1,\\frac{28}{3}]$."
  },
  {
    "text": "Given that point $P$ is a moving point on the parabola $y^{2}=2x$, find the minimum value of the sum of the distance from point $P$ to $(0, 2)$ and the distance from $P$ to the directrix of this parabola.",
    "fact_expressions": "G: Parabola;H: Point;P: Point;Expression(G) = (y^2 = 2*x);Coordinate(H) = (0, 2);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P,H)+Distance(P,Directrix(G)))",
    "answer_expressions": "sqrt(17)/2",
    "fact_spans": "[[[7, 21], [53, 56]], [[34, 44]], [[29, 33], [2, 6], [48, 51]], [[7, 21]], [[34, 44]], [[2, 27]]]",
    "query_spans": "[[[29, 69]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the parabola with vertex at the center of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ and focus at the right focus of this ellipse?",
    "fact_expressions": "H: Ellipse;Expression(H) = (x^2/25 + y^2/16 = 1);Vertex(G) = Center(H);G: Parabola;RightFocus(H) = Focus(G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=12*x",
    "fact_spans": "[[[1, 40], [50, 52]], [[1, 40]], [[0, 63]], [[60, 63]], [[48, 63]]]",
    "query_spans": "[[[60, 67]]]",
    "process": "First calculate the ellipse \\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1 has center (0,0), right focus (3,0), then according to the parabola focus calculation obtain the detailed answer. The ellipse \\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1 has center (0,0), right focus (3,0). Let the parabola equation be y^{2}=2px, then \\frac{p}{2}=3 \\therefore p=6, so the parabola equation is y^{2}=12x"
  },
  {
    "text": "The asymptotes of a hyperbola centered at the origin with foci on the $y$-axis pass through the point $P(2, 1)$. Its eccentricity is?",
    "fact_expressions": "O: Origin;Center(G) = O;G: Hyperbola;PointOnCurve(Focus(G), yAxis);P: Point;Coordinate(P) = (2, 1);PointOnCurve(P, Asymptote(G))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[3, 7]], [[0, 20]], [[17, 20], [37, 38]], [[8, 20]], [[25, 36]], [[25, 36]], [[17, 36]]]",
    "query_spans": "[[[37, 43]]]",
    "process": ""
  },
  {
    "text": "In the Cartesian coordinate plane, a moving point $P$ and points $M(-2,0)$, $N(2,0)$ satisfy $|\\overline{M N}| \\cdot|\\overrightarrow{M P}|+\\overrightarrow{M N} \\cdot \\overrightarrow{N P}=0$. Then the trajectory equation of the moving point $P(x, y)$ is?",
    "fact_expressions": "M: Point;N: Point;P: Point;Coordinate(M) = (-2, 0);Coordinate(N) = (2, 0);Coordinate(P) = (x1, y1);Abs(VectorOf(N,P))*Abs(VectorOf(M,N))+DotProduct(VectorOf(M,P),VectorOf(M,N)) = 0;x1:Number;y1:Number",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "y^2=-8*x",
    "fact_spans": "[[[17, 26]], [[27, 35]], [[12, 15], [137, 146]], [[17, 26]], [[27, 35]], [[137, 146]], [[37, 133]], [[137, 146]], [[137, 146]]]",
    "query_spans": "[[[137, 153]]]",
    "process": "This problem can be solved using the basic method of finding trajectory equations—direct method. Express the given condition equation $|\\overrightarrow{MN}|\\overrightarrow{MP}|+\\overrightarrow{MN}\\cdot\\overrightarrow{NP}=0$ in coordinates: $4\\sqrt{(x+2)^{2}+y^{2}}+4(x-2)=0$, and simplifying and transforming it yields $y^{2}=-8x$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through the right focus $F_{2}$ intersects the ellipse at points $P$ and $Q$. Then the maximum area of $\\Delta F_{1} P Q$ is?",
    "fact_expressions": "l: Line;G: Ellipse;F1: Point;P: Point;Q: Point;F2: Point;Expression(G) = (x^2/4 + y^2/3 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, l);Intersection(l, G) = {P, Q}",
    "query_expressions": "Max(Area(TriangleOf(F1, P, Q)))",
    "answer_expressions": "3",
    "fact_spans": "[[[81, 86]], [[2, 39], [65, 67], [87, 89]], [[48, 55]], [[90, 94]], [[95, 98]], [[56, 63], [71, 78]], [[2, 39]], [[2, 63]], [[2, 63]], [[64, 86]], [[81, 98]]]",
    "query_spans": "[[[101, 127]]]",
    "process": "From the equation of the ellipse, we obtain $F_{2}(1,0)$. Let $P(x_{1},y_{1})$, $Q(x_{2},y_{2})$, and let the equation of line $l$ be $x=my+1$. Combining this with the equation of the ellipse yields $y_{1}+y_{2}$, $y_{1}y_{2}$, and thus $|y_{1}-y_{2}|$. Then, by using properties of functions, we can compute the minimum value of $S=\\frac{1}{2}|F_{1}F_{2}|\\cdot|y_{1}-y_{2}|$. Solution: From $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, we get $a^{2}=4$, $b^{2}=3$, so $c^{2}=a^{2}-b^{2}=4-3=1$, hence $a=2$, $b=\\sqrt{3}$, $c=1$, so $F_{2}(1,0)$. Let $P(x_{1},y_{1})$, $Q(x_{2},y_{2})$, and the equation of line $l$ be $x=my+1$. From\n$$\n\\begin{cases}\n\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1 \\\\\nx=my+1\n\\end{cases}\n$$\nwe obtain: $(3m^{2}+4)y^{2}+6my-9=0$, so\n$$\n|y_{1}-y_{2}| = \\sqrt{(y_{1}-y_{2})^{2}} = \\sqrt{(y_{1}+y_{2})^{2}-4y_{1}y_{2}} = \\sqrt{\\left(-\\frac{6m}{3m^{2}+4}\\right)^{2}-4\\times\\left(-\\frac{9}{3m^{2}+4}\\right)} = \\sqrt{\\frac{|y_{1}-y_{2}|=V(y_{1}-y_{2})}{(m^{2+1})}}{(3m^{2}+4)^{2}}} = \\frac{12\\sqrt{m^{2}+1}}{3m^{2}+4},\n$$\nso the area $S$ of $\\triangle F_{1}PQ$ is\n$$\nS = \\frac{1}{2}|F_{1}F_{2}|\\cdot|y_{1}-y_{2}| = \\frac{1}{2}\\times2\\cdot|y_{1}-y_{2}| = |y_{1}-y_{2}| = \\frac{12\\sqrt{m^{2}+1}}{3m^{2}+4}.\n$$\nLet $\\sqrt{m^{2}+1}=t \\geqslant 1$, then $m^{2}=t^{2}-1$, so\n$$\nS = \\frac{12t}{3(t^{2}-1)+}\\frac{12t}{4} = \\frac{12}{3t^{2}+1} = \\frac{12}{3t+\\frac{1}{t}}.\n$$\nFor $y=3t+\\frac{1}{t}$, $y=3-\\frac{1}{12}=\\frac{3t^{2}-1}{t^{2}}>0$, which holds for all $t \\geqslant 1$. Thus, $y=3t+\\frac{1}{t}$ is monotonically increasing on $[1,+\\infty)$. Therefore, when $t=1$, $y_{\\min}=3\\times1+\\frac{1}{1}=4$, so\n$$\nS_{\\max} = \\frac{12}{3t+\\frac{1}{t}} = \\frac{12}{4} = 3.\n$$"
  },
  {
    "text": "The line $y = kx + 2$ has exactly one common point with the parabola $y^2 = 8x$, then $k = $?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*x + 2);k: Number;G: Parabola;Expression(G) = (y^2 = 8*x);NumIntersection(H, G) = 1",
    "query_expressions": "k",
    "answer_expressions": "{0, 1}",
    "fact_spans": "[[[0, 11]], [[0, 11]], [[37, 40]], [[12, 26]], [[12, 26]], [[0, 35]]]",
    "query_spans": "[[[37, 42]]]",
    "process": "When $k=0$, the line is $y=2$, which intersects the parabola at only one point $\\left(\\frac{1}{2}, 2\\right)$. When $k \\neq 0$, substituting $y = kx + 2$ into the parabola $y^{2} = 8x$ gives: $k^{2}x^{2} + (4k - 8)x + 4 = 0$. Since the line $y = kx + 2$ and the parabola $y^{2} = 8x$ have exactly one common point, we have $\\Delta = (4k - 8)^{2} - 16k^{2} = 0$, solving which yields $k = 1$. In conclusion: $k = 0$ or $k = 1$."
  },
  {
    "text": "Let a point $P$ on the parabola $y^{2}=4x$ be at a distance of $5$ from the line $x+2=0$. Then, what is the distance from point $P$ to the focus $F$ of the parabola?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (x + 2 = 0);PointOnCurve(P, G);Distance(P, H) = 5;Focus(G)=F;F:Point",
    "query_expressions": "Distance(P, F)",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 15], [45, 48]], [[22, 31]], [[18, 21], [40, 44]], [[1, 15]], [[22, 31]], [[1, 21]], [[18, 38]], [[45, 53]], [[50, 53]]]",
    "query_spans": "[[[40, 58]]]",
    "process": "The directrix of the parabola $ y^{2} = 4x $ is $ x = -1 $. Since the distance from point $ P $ to the line $ x + 2 = 0 $ is 5, the distance from point $ P $ to the directrix $ x = -1 $ is $ 5 - 1 = 4 $. According to the definition of a parabola, the distance from point $ P $ to the focus of this parabola is 4."
  },
  {
    "text": "Given that the center of the ellipse is at the origin $O$, the axes of symmetry are the coordinate axes, the foci lie on the $x$-axis, the focal distance is $2 \\sqrt{6}$, and the ellipse passes through the point $(\\sqrt{3}, \\sqrt{2})$, what is the standard equation of the ellipse?",
    "fact_expressions": "G: Ellipse;H: Point;O: Origin;Coordinate(H) = (sqrt(3), sqrt(2));Center(G) = O;SymmetryAxis(G) = axis;PointOnCurve(Focus(G), xAxis);FocalLength(G) = 2*sqrt(6);PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 + y^2/3 = 1",
    "fact_spans": "[[[2, 4], [77, 79]], [[52, 75]], [[8, 15]], [[52, 75]], [[2, 15]], [[2, 23]], [[2, 32]], [[2, 48]], [[2, 75]]]",
    "query_spans": "[[[77, 86]]]",
    "process": "According to the problem, the focal distance of the ellipse is $2\\sqrt{6}$, and the foci lie on the x-axis; thus, the coordinates of the foci are $(-\\sqrt{6},0)$ and $(\\sqrt{6},0)$, where $c=\\sqrt{6}$. Since the ellipse passes through the point $(\\sqrt{3},\\sqrt{2})$, then $2a = \\sqrt{(\\sqrt{3}+\\sqrt{6})^{2}}\\sqrt{(\\sqrt{3}-\\sqrt{6})^{2}+(\\sqrt{2})^{2}} = \\sqrt{11+6\\sqrt{2}} + \\sqrt{11-6\\sqrt{2}} = \\sqrt{(3+\\sqrt{2})^{2}} + \\sqrt{(3-\\sqrt{2})^{2}} = 6$, so $a=3$, then $b^{2} = a^{2} - c^{2} = 9 - 6 = 3$, hence the standard equation of the ellipse is $\\frac{x^{2}}{9} + \\frac{y^{2}}{3} = 1$."
  },
  {
    "text": "The two endpoints $A$ and $B$ of a line segment $AB$ of length $a$ slide along the parabola $y^{2}=2 p x$ $(p>0, a>2 p)$. Then the shortest distance from the midpoint $M$ of segment $AB$ to the $y$-axis is?",
    "fact_expressions": "A: Point;B: Point;Length(LineSegmentOf(A, B)) = a;a: Number;Endpoint(LineSegmentOf(A, B)) = {A, B};G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;a>2*p;PointOnCurve(A,G) = True;PointOnCurve(B,G) = True;MidPoint(LineSegmentOf(A,B)) = M;M: Point",
    "query_expressions": "Min(Distance(M, yAxis))",
    "answer_expressions": "(a-p)/2",
    "fact_spans": "[[[18, 21]], [[24, 27]], [[0, 13]], [[3, 6]], [[7, 27]], [[29, 56]], [[29, 56]], [[32, 56]], [[32, 56]], [[32, 56]], [[7, 57]], [[7, 57]], [[61, 73]], [[70, 73]]]",
    "query_spans": "[[[70, 85]]]",
    "process": ""
  },
  {
    "text": "The eccentricity $e \\in (1,2)$ of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{k}=1$, then the range of values for $k$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/k = 1);k: Number;e: Number;In(e, (1, 2));Eccentricity(G) = e",
    "query_expressions": "Range(k)",
    "answer_expressions": "(0,12)",
    "fact_spans": "[[[0, 38]], [[0, 38]], [[56, 59]], [[42, 54]], [[42, 54]], [[0, 54]]]",
    "query_spans": "[[[56, 66]]]",
    "process": "Since the eccentricity of the curve $ e \\in (1,2) $, the curve is a hyperbola, so $ k > 0 $, and $ e = \\frac{c}{a} = \\sqrt{\\frac{a^2 + b^{2}}{a^{2}}} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{1 + \\frac{k}{4}} \\in (1,2) $. Solving gives $ 0 < k < 12 $, that is, the range of real number $ k $ is $ (0,12) $."
  },
  {
    "text": "It is known that the line $x-2 y+2=0$ passes through a vertex and a focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$. Then, what is the equation of this ellipse? What is its eccentricity?",
    "fact_expressions": "H: Line;Expression(H) = (x - 2*y + 2 = 0);G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;PointOnCurve(OneOf(Vertex(G)), H);PointOnCurve(OneOf(Focus(G)), H)",
    "query_expressions": "Expression(G);Eccentricity(G)",
    "answer_expressions": "x^2/5+y^2=1\n2*sqrt(5)/5",
    "fact_spans": "[[[2, 15]], [[2, 15]], [[17, 71], [86, 88]], [[17, 71]], [[19, 71]], [[19, 71]], [[19, 71]], [[19, 71]], [[2, 76]], [[2, 81]]]",
    "query_spans": "[[[86, 93]], [[86, 98]]]",
    "process": "One focus is $F(-2,0)$, and one vertex of the minor axis is $F(0,1)$, so $c=2$, $b=1$, hence $a=\\sqrt{5}$, leading to the ellipse equation $\\frac{x^{2}}{5}+y^{2}=1$. The line $x-2y+2=0$ intersects the $x$-axis at point $A(-2,0)$ and the $y$-axis at point $B(0,1)$. Therefore, one focus of the ellipse is $F(-2,0)$ and one vertex of the minor axis is $F(0,1)$. Thus, for the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $c=2$, $b=1$, $\\therefore a=\\sqrt{5}$. Hence, the equation of this ellipse is $\\frac{x^{2}}{5}+y^{2}=1$, and the eccentricity is $e=\\frac{2}{\\sqrt{5}}=\\frac{2\\sqrt{5}}{5}$."
  },
  {
    "text": "If an ellipse passes through the two points $(2,0)$ and $(0,1)$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;H: Point;I: Point;Coordinate(H) = (2, 0);Coordinate(I) = (0, 1);PointOnCurve(H,G);PointOnCurve(I,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4+y^2=1",
    "fact_spans": "[[[1, 3], [24, 26]], [[7, 14]], [[15, 22]], [[7, 14]], [[15, 22]], [[1, 22]], [[1, 22]]]",
    "query_spans": "[[[24, 33]]]",
    "process": "Discuss the cases where the foci of the ellipse lie on the x-axis and y-axis, substitute the coordinates of two points, and find the standard equation of the ellipse.  \nWhen the foci of the ellipse lie on the x-axis, let the equation of the ellipse be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. Since the ellipse passes through the points $(2,0)$ and $(0,1)$,  \n$\\therefore\\begin{cases}\\frac{4}{a^{2}}+\\frac{0}{b^{2}}=1\\\\\\frac{0}{a^{2}}+\\frac{1}{b^{2}}=1\\end{cases}$  \nSolving gives $\\begin{cases}a=2\\\\b=1\\end{cases}$  \nThus, the standard equation of the ellipse is $\\frac{x^{2}}{4}+y^{2}=1$;  \nWhen the foci of the ellipse lie on the y-axis, let the equation of the ellipse be $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$ $(a>b>0)$. Since the ellipse passes through the points $(2,0)$ and $(0,1)$,  \n$\\therefore\\begin{cases}\\frac{0}{a^{2}}+\\frac{4}{b^{2}}=1\\\\\\frac{1}{a^{2}}+\\frac{0}{b^{2}}=1\\end{cases}$,  \nSolving gives $\\begin{cases}a=1\\\\b=2\\end{cases}$, which contradicts $a>b$, so this case is discarded.  \nIn conclusion, the standard equation of the required ellipse is $\\frac{x^{2}}{4}+y^{2}=1$."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. Let $P$ be a point on its right directrix with ordinate $\\sqrt{3}c$ ($c$ is the semi-focal length), and suppose $F_{1}F_{2}=F_{2}P$. Then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;a: Number;b: Number;F1: Point;F2: Point;P: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P,RightDirectrix(G));YCoordinate(P)=sqrt(3)*c;c:Number;HalfFocalLength(G)=c;LineSegmentOf(F1, F2) = LineSegmentOf(F2, P)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[19, 73], [84, 85], [144, 146]], [[21, 73]], [[21, 73]], [[1, 8]], [[9, 16]], [[80, 83]], [[21, 73]], [[21, 73]], [[19, 73]], [[1, 79]], [[1, 79]], [[80, 119]], [[80, 119]], [[108, 111]], [[108, 115]], [[121, 142]]]",
    "query_spans": "[[[144, 152]]]",
    "process": ""
  },
  {
    "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ have its right focus at $F$, and let point $P$ lie on an asymptote of $C$ given by $x+\\sqrt{2} y=0$, with $O$ being the origin. If $|O F|=|P F|$ and the area of $\\Delta P O F$ is $2 \\sqrt{2}$, then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;O: Origin;F: Point;a > b;b > 0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;PointOnCurve(P,L);Expression(L)=(x+sqrt(2)*y=0);Abs(LineSegmentOf(O,F))=Abs(LineSegmentOf(P,F));Area(TriangleOf(P, O, F)) = 2*sqrt(2);L:Line;OneOf(Asymptote(C))=L",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4 - y^2/2 = 1",
    "fact_spans": "[[[1, 59], [73, 76], [157, 160]], [[9, 59]], [[9, 59]], [[68, 72]], [[100, 103]], [[64, 67]], [[9, 59]], [[9, 59]], [[1, 59]], [[1, 67]], [[68, 99]], [[82, 98]], [[110, 123]], [[125, 155]], [[82, 98]], [[73, 98]]]",
    "query_spans": "[[[157, 165]]]",
    "process": "The hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $) has its right focus at $ F $, and $ O $ is the origin. Point $ P $ lies on an asymptote of $ C $, $ x + \\sqrt{2}y = 0 $, and $ |OF| = |PF| $. Therefore, the asymptote in the fourth quadrant has slope $ -\\frac{\\sqrt{2}}{2} $, $ \\tan\\angle POF = \\frac{\\sqrt{2}}{2} $, so $ \\cos\\angle POF = \\frac{\\sqrt{6}}{3} $, $ \\sin\\angle POF = \\frac{\\sqrt{3}}{3} $. If $ |OF| = |PF| $ and the area of $ \\triangle POF $ is $ 2\\sqrt{2} $, then $ \\frac{1}{2}c^{2}\\sin(\\pi - 2\\angle POF) = 2\\sqrt{2} $, so $ c^{2}\\sin(2\\angle POF) = 4\\sqrt{2} $, $ \\therefore c^{2} \\times 2 \\times \\frac{\\sqrt{3}}{3} \\times \\frac{\\sqrt{6}}{3} = 4\\sqrt{2} $. Solving gives $ c = \\sqrt{6} $, $ \\frac{b}{a} = \\frac{\\sqrt{2}}{2} $, $ c^{2} = a^{2} + b^{2} $, solving yields $ b = \\sqrt{2} $, $ a = 2 $. Thus, the equation of the hyperbola is $ \\frac{x^{2}}{4} - \\frac{y^{2}}{2} = 1 $."
  },
  {
    "text": "Given circle $C_{1}$: $x^{2}+y^{2}=9$, circle $C_{2}$: $x^{2}+y^{2}=4$, fixed point $M(1,0)$, moving points $A$, $B$ lie on circles $C_{2}$ and $C_{1}$ respectively, satisfying $\\angle A M B=90^{\\circ}$, then what is the range of length of segment $A B$?",
    "fact_expressions": "C1: Circle;C2: Circle;B: Point;A: Point;M: Point;Expression(C1) = (x^2 + y^2 = 9);Expression(C2) = (x^2 + y^2 = 4);Coordinate(M) = (1, 0);PointOnCurve(A, C2);PointOnCurve(B, C1);AngleOf(A, M, B) = ApplyUnit(90, degree)",
    "query_expressions": "Range(LineSegmentOf(A,B))",
    "answer_expressions": "[2*sqrt(3)-1,2*sqrt(3)+1]",
    "fact_spans": "[[[2, 27], [87, 95]], [[28, 53], [78, 86]], [[72, 75]], [[68, 71]], [[57, 65]], [[2, 27]], [[28, 53]], [[57, 65]], [[68, 96]], [[68, 96]], [[99, 124]]]",
    "query_spans": "[[[127, 140]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a,b>0)$ is $\\frac{\\sqrt{6}}{2}$, what is the minimum value of $\\frac{a^{2}+4}{b}$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(G) = sqrt(6)/2",
    "query_expressions": "Min((a^2 + 4)/b)",
    "answer_expressions": "4*sqrt(2)",
    "fact_spans": "[[[2, 55]], [[2, 55]], [[5, 55]], [[5, 55]], [[5, 55]], [[5, 55]], [[2, 80]]]",
    "query_spans": "[[[83, 108]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}(-c, 0)$ and $F_{2}(c, 0)$ respectively. If there exists a point $P$ on the hyperbola such that $\\frac{\\sin \\angle P F_{1} F_{2}}{\\sin \\angle P F_{2} F_{1}}=\\frac{a}{c}$, then the range of values for the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F1: Point;F2: Point;P: Point;a>0;b>0;c:Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);LeftFocus(G) = F1;RightFocus(G)=F2;PointOnCurve(P,G);Sin(AngleOf(P,F1,F2))/Sin(AngleOf(P,F2,F1))=a/c",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1, \\sqrt{2}+1)",
    "fact_spans": "[[[2, 58], [99, 102], [187, 190]], [[5, 58]], [[5, 58]], [[67, 81]], [[84, 97]], [[107, 110]], [[5, 58]], [[5, 58]], [[84, 97]], [[2, 58]], [[67, 81]], [[84, 97]], [[2, 97]], [[2, 97]], [[99, 110]], [[111, 184]]]",
    "query_spans": "[[[187, 201]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y=\\frac{1}{4} x^{2}$, what is the equation of the line passing through its focus and perpendicular to its axis of symmetry?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2/4);l:Line;PointOnCurve(Focus(G),l);IsPerpendicular(l,SymmetryAxis(G))",
    "query_expressions": "Expression(l)",
    "answer_expressions": "y = 1",
    "fact_spans": "[[[2, 26], [29, 30], [35, 36]], [[2, 26]], [[40, 42]], [[28, 42]], [[32, 42]]]",
    "query_spans": "[[[40, 46]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $C$: $\\frac{x^{2}}{a^{2}}-y^{2}=1(a>0)$ intersects the line $l$: $x+y=1$ at two distinct points $A$, $B$. What is the range of values for the eccentricity $e$ of the hyperbola?",
    "fact_expressions": "l: Line;C: Hyperbola;a: Number;A: Point;B: Point;e: Number;a>0;Expression(C) = (-y^2 + x^2/a^2 = 1);Expression(l) = (x + y = 1);Intersection(l, C) = {A, B};Negation(A=B);Eccentricity(C) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "(\\sqrt{6}/2,\\sqrt{2})+(\\sqrt{2},+\\infty)",
    "fact_spans": "[[[43, 57]], [[0, 42], [75, 78]], [[7, 42]], [[65, 69]], [[70, 73]], [[82, 85]], [[7, 42]], [[0, 42]], [[43, 57]], [[0, 73]], [[60, 73]], [[75, 85]]]",
    "query_spans": "[[[82, 92]]]",
    "process": "From\n\\begin{cases}\n\\frac{x^{2}}{a}-y=1 \\\\\nx+y=1 \\\\\n\\text{According to the problem}, \\\\\n1-a^{2}\\neq0 \\\\\n\\triangle=4a+8a^{2}(1-a^{2})>0\n\\end{cases},\nwe obtain: $ a\\in(0,1) $.  \nSolving gives: $ a\\in(0,1)\\cup(1,\\sqrt{2}) $.  \nThus, $ e=\\frac{c}{a}=\\sqrt{\\frac{a^{2}+b^{2}}{a^{2}}+8a^{2}(1-a^{2})}=\\sqrt{1+\\frac{1}{a^{2}}} $, so $ e\\in\\left(\\frac{\\sqrt{6}}{2},\\sqrt{2}\\right)\\cup(\\sqrt{2},+\\infty) $."
  },
  {
    "text": "Given that $A$ and $B$ are any two distinct points on the parabola $y^{2}=2 p x(p>0)$, and the perpendicular bisector of segment $A B$ intersects the $x$-axis at point $P(x_{0} , 0)$, then what is the range of values of $x_{0}$? (Expressed in terms of $p$)",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;A: Point;B: Point;Negation(A=B);PointOnCurve(A, G);PointOnCurve(B, G);P: Point;x0: Number;Coordinate(P) = (x0, 0);Intersection(PerpendicularBisector(LineSegmentOf(A, B)), xAxis) = P",
    "query_expressions": "Range(x0)",
    "answer_expressions": "(p, +\\infty)",
    "fact_spans": "[[[10, 31]], [[10, 31]], [[13, 31]], [[13, 31]], [[2, 5]], [[6, 9]], [[2, 39]], [[2, 39]], [[2, 39]], [[61, 76]], [[78, 85]], [[61, 76]], [[40, 76]]]",
    "query_spans": "[[[78, 92]]]",
    "process": "Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$. Since the perpendicular bisector of segment $AB$ intersects the $x$-axis at point $P(x_{0},0)$, $AB$ is not parallel to the $y$-axis, i.e., $x_{1} \\ne x_{2}$. Then $|PA| = |PB|$, so $\\sqrt{(x_{1}-x_{0})^{2}+y_{1}^{2}} = \\sqrt{(x_{2}-x_{0})^{2}+y_{2}^{2}}$. Rearranging gives $(x_{1}-x_{2})(x_{1}+x_{2}-2x_{0}) = y_{2}^{2}-y_{1}^{2}$. Since $A$, $B$ are two points on the parabola, $y_{1}^{2} = 2px_{1}$, $y_{2}^{2} = 2px_{2}$. Substituting into the above equation yields $x_{0} = p + \\frac{x_{1}+x_{2}}{2}$. Since $x_{1} \\geqslant 0$, $x_{2} \\geqslant 0$, $x_{1} \\ne x_{2}$, we have $x_{1}+x_{2} > 0$, thus $x_{0} = p + \\frac{x_{1}+x_{2}}{2} > p$, i.e., the range of $x_{0}$ is $(p, +\\infty)$."
  },
  {
    "text": "Given that the foci of an ellipse are $F_{1}$, $F_{2}$, and $P$ is a moving point on the ellipse, if $M$ is the midpoint of the segment $F_{1} P$, then what is the trajectory of the moving point $M$?",
    "fact_expressions": "G: Ellipse;F1:Point;F2:Point;P: Point;M: Point;Focus(G)={F1,F2};PointOnCurve(P,G);MidPoint(LineSegmentOf(F1,P)) = M",
    "query_expressions": "Locus(M)",
    "answer_expressions": "Ellipse",
    "fact_spans": "[[[2, 4], [28, 30]], [[8, 15]], [[16, 23]], [[24, 27]], [[38, 41], [61, 64]], [[2, 23]], [[24, 35]], [[38, 56]]]",
    "query_spans": "[[[61, 69]]]",
    "process": "Connect MO. By the midline of a triangle, we have |F_{1}M| + |MO| = a (a > |F_{1}O|), which gives the locus of the moving point M. From the problem, |PF_{1}| + |PF_{2}| = 2a. Let the ellipse equation be: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 (where a > b > 0). Connect MO. By the midline of a triangle, we obtain |F_{1}M| + |MO| = a (a > |F_{1}O|). Thus, the locus of M is an ellipse with foci at F and O."
  },
  {
    "text": "If the coordinates of point $A$ are $(\\frac{5}{2}, 5)$, $F$ is the focus of the parabola $y^{2}=2 x$, and point $P$ moves along the parabola, then the minimum value of $|P A|+|P F|$ is?",
    "fact_expressions": "G: Parabola;P: Point;A: Point;F: Point;Expression(G) = (y^2 = 2*x);Coordinate(A) = (5/2, 5);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "sqrt(29)",
    "fact_spans": "[[[34, 48], [57, 60]], [[52, 56]], [[1, 5]], [[30, 33]], [[34, 48]], [[1, 27]], [[30, 51]], [[52, 61]]]",
    "query_spans": "[[[65, 84]]]",
    "process": "As shown in the figure, when $ x = \\frac{5}{2} $, $ y^2 = 5 < 25 $, so point $ A $ is outside the parabola. When points $ A $, $ P $, and $ F $ are collinear, the minimum value of $ |PA| + |PF| $ is $ |AF| $. Given $ F\\left(\\frac{1}{2}, 0\\right) $, $ |AF| = \\sqrt{\\left(\\frac{5}{2} - \\frac{1}{2}\\right)^2 + 5^2} = \\sqrt{29} $, so the minimum value of $ |PA| + |PF| $ is $ \\sqrt{29} $."
  },
  {
    "text": "The focus of the parabola $y^{2}=8 x$ is $F$, chord $A B$ passes through $F$, the origin is $O$, the directrix of the parabola intersects the $x$-axis at point $C$, $\\angle O F A=\\frac{2 \\pi}{3}$, then $\\tan \\angle A C B$=?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;O: Origin;F: Point;C: Point;Expression(G) = (y^2 = 8*x);Focus(G) = F;PointOnCurve(F, LineSegmentOf(A, B));IsChordOf(LineSegmentOf(A,B),G);Intersection(Directrix(G), xAxis) = C;AngleOf(O, F, A) = 2*pi/3",
    "query_expressions": "Tan(AngleOf(A, C, B))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[0, 14], [40, 43]], [[23, 28]], [[23, 28]], [[36, 39]], [[18, 21], [29, 32]], [[52, 56]], [[0, 14]], [[0, 21]], [[23, 32]], [[0, 28]], [[40, 57]], [[59, 89]]]",
    "query_spans": "[[[91, 112]]]",
    "process": "Analysis: Using the definition of the parabola and the tangent addition formula, we can solve the problem. The focus of the parabola is $ F(2,0) $. Since $ \\angle OFA = \\frac{2\\pi}{3} $, according to the property of the parabola, we have $ AF = \\frac{2\\sqrt{3}}{3}y_{A} = x_{A} + 2 $, $ BF = \\frac{2\\sqrt{3}}{3}y_{B} = x_{B} + 2 $. Hence, $ \\tan\\angle ACF = \\frac{y_{A}}{x_{A} + 2} = \\frac{\\sqrt{3}}{2} $, $ \\tan\\angle BCF = \\frac{|y_{B}|}{x_{B} + 2} = \\frac{\\sqrt{3}}{2} $. Therefore, by the tangent addition formula, $ \\tan\\angle ACB = \\frac{\\sqrt{3}}{1 - \\frac{3}{4}} = 4\\sqrt{3} $. The correct answer is $ 4\\sqrt{3} $."
  },
  {
    "text": "If the left and right foci of the hyperbola $\\frac{x^{2}}{9}-y^{2}=1$ are $F_{1}$, $F_{2}$ respectively, and $A$ is a point on the left branch of the hyperbola such that $|A F_{1}|=5$, then $|A F_{2}|=$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/9 - y^2 = 1);F1: Point;F2: Point;F1 = LeftFocus(C);F2 = RightFocus(C);A: Point;PointOnCurve(A, LeftPart(C));Abs(LineSegmentOf(A, F1)) = 5",
    "query_expressions": "Abs(LineSegmentOf(A, F2))",
    "answer_expressions": "11",
    "fact_spans": "[[[1, 29], [57, 60]], [[1, 29]], [[37, 44]], [[45, 52]], [[1, 52]], [[1, 52]], [[53, 56]], [[53, 66]], [[68, 81]]]",
    "query_spans": "[[[84, 97]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $P$ is a point on the ellipse $C$, $\\angle F_{1} P F_{2}=120^{\\circ}$, and the area of $\\triangle F_{1} P F_{2}$ is $4 \\sqrt{3}$, then $b=$?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;F1: Point;P: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Focus(C) = {F1, F2};PointOnCurve(P, C);AngleOf(F1, P, F2) = ApplyUnit(120, degree);Area(TriangleOf(F1, P, F2)) = 4*sqrt(3)",
    "query_expressions": "b",
    "answer_expressions": "2",
    "fact_spans": "[[[18, 75], [85, 90]], [[174, 177]], [[25, 75]], [[2, 9]], [[81, 84]], [[10, 17]], [[25, 75]], [[25, 75]], [[18, 75]], [[2, 80]], [[81, 94]], [[95, 129]], [[131, 172]]]",
    "query_spans": "[[[174, 179]]]",
    "process": "The area of \\triangleF_{1}PF_{2} = \\frac{1}{2}PF_{1}\\cdot PF_{2}\\sin120^{\\circ} = \\frac{\\sqrt{3}}{4}PF_{1}\\cdot PF_{2} = 4\\sqrt{3}, then PF_{1}\\cdot PF_{2} = 16, and according to the cosine law we have \\cos120^{\\circ} = \\frac{PF_{1}^{2} + PF_{2}^{2} - F_{1}F_{2}^{2}}{2PF_{1}\\cdot PF_{2}}, that is, 4c^{2} = PF_{1}^{2} + PF_{2}^{2} + 16 = (2a)^{2} \\cdot 32 + 16, so 4b^{2} = 16, solving gives b = 2,"
  },
  {
    "text": "Given that points $A$ and $B$ are the left focus and the upper vertex of the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1$, respectively, and $C$ is a moving point on the ellipse, then the maximum area of $\\triangle A B C$ is?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;C: Point;Expression(G) = (x^2/3 + y^2/2 = 1);LeftFocus(G) = A;UpperVertex(G)=B;PointOnCurve(C, G)",
    "query_expressions": "Max(Area(TriangleOf(A, B, C)))",
    "answer_expressions": "3*sqrt(2)/2",
    "fact_spans": "[[[14, 51], [64, 66]], [[2, 5]], [[6, 9]], [[60, 63]], [[14, 51]], [[2, 59]], [[2, 59]], [[60, 70]]]",
    "query_spans": "[[[72, 97]]]",
    "process": "From the equation of the ellipse, the coordinates of points A and B can be obtained, then the equation of line AB and the length |AB| can be found. When the area of triangle ABC is maximized, the line passing through point C is parallel to line AB and tangent to the ellipse. Let the line passing through C be combined with the ellipse equation; by setting the discriminant equal to zero, the parameter value can be obtained, thus solving the problem. From the ellipse equation, we get A(-1,0), B(0,\\sqrt{2}), \\therefore |AB|=\\sqrt{3}, so the equation of line AB is: \\sqrt{2}x-y+\\sqrt{2}=0. According to the problem, when the line passing through C is parallel to line AB and tangent to the ellipse, the distance between the two parallel lines is maximized, and thus the area of triangle ABC is maximized. Let the tangent line l passing through C and parallel to AB be: \\sqrt{2}x-y+m=0, the distance d between line l and line AB is d=\\frac{|\\sqrt{2}-m|}{\\sqrt{3}}. Combining the equations of line l and the ellipse gives: \\begin{cases}\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1\\\\\\sqrt{2}x-y+m=0\\end{cases}, simplifying yields: 8x^{2}+6\\sqrt{2}mx+3m^{2}-6=0. \\Delta=72m^{2}-96(m^{2}-2)=0, we get m^{2}=8, solving gives m=\\pm2\\sqrt{2}. Therefore, when m=-2\\sqrt{2}, d=\\frac{|\\sqrt{2}+2\\sqrt{2}|}{\\sqrt{3}}=\\frac{3\\sqrt{2}}{\\sqrt{3}} is maximized, then the maximum value of S_{\\triangle ABC} is: \\frac{1}{2}|AB|\\cdot d_{\\max}=\\frac{1}{2}\\times\\sqrt{3}\\times\\frac{3\\sqrt{2}}{\\sqrt{3}}=\\frac{3\\sqrt{2}}{2}"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, what are the coordinates of the foci of the ellipse?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(pm*3,0)",
    "fact_spans": "[[[2, 41], [43, 45]], [[2, 41]]]",
    "query_spans": "[[[43, 52]]]",
    "process": "From the given conditions: $ a^{2} = 25 $, $ b^{2} = 16 $. From $ a^{2} = b^{2} + c^{2} $, we get: $ c = \\sqrt{25 - 16} = 3 $. Therefore, the coordinates of the foci are $ (\\pm 3, 0) $. The correct result for this problem: $ (-3, 0) $, $ (3, 0) $."
  },
  {
    "text": "The coordinates of the midpoint of the line segment cut by the parabola $y^{2}=4x$ from the line $y=x-1$ are?",
    "fact_expressions": "G: Parabola;H: Line;Expression(G) = (y^2 = 4*x);Expression(H) = (y = x - 1)",
    "query_expressions": "Coordinate(MidPoint(InterceptChord(H,G)))",
    "answer_expressions": "(3,2)",
    "fact_spans": "[[[10, 24]], [[0, 9]], [[10, 24]], [[0, 9]]]",
    "query_spans": "[[[0, 35]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2} = -2px$ ($p > 0$), a line with slope $-2$ passes through the focus $F$ of $C$ and intersects the parabola $C$ at points $P$ and $Q$. If the y-coordinate of the midpoint $M$ of segment $PQ$ is $2$, then $p = $?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = -2*p*x);p: Number;p>0;Slope(G) = -2;G: Line;Focus(C) = F;F: Point;PointOnCurve(F, G) = True;Intersection(G, C) = {P, Q};P: Point;Q: Point;M: Point;MidPoint(LineSegmentOf(P, Q)) = M;YCoordinate(M) = 2",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 30], [42, 45], [53, 59]], [[2, 30]], [[95, 98]], [[9, 30]], [[31, 41]], [[39, 41]], [[42, 51]], [[48, 51]], [[39, 51]], [[39, 70]], [[61, 64]], [[65, 68]], [[82, 85]], [[72, 85]], [[82, 93]]]",
    "query_spans": "[[[95, 100]]]",
    "process": ""
  },
  {
    "text": "Let the parabola $C$: $y^{2}=8x$ have focus $F$. A line $l$ passing through point $A(-1,0)$ with slope $k$ intersects the parabola $C$ at points $M$ and $N$. If the area of $\\Delta AMF$ is twice the area of $\\Delta ANF$, then the value of $k$ is?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;M: Point;F: Point;N: Point;k:Number;Expression(C) = (y^2 = 8*x);Coordinate(A) = (-1, 0);Focus(C) = F;PointOnCurve(A, l);Slope(l) = k;Intersection(l, C) = {M, N};Area(TriangleOf(A,M,F))=2*Area(TriangleOf(A,N,F))",
    "query_expressions": "k",
    "answer_expressions": "pm*(4/3)",
    "fact_spans": "[[[48, 53]], [[1, 20], [54, 60]], [[30, 40]], [[62, 65]], [[24, 27]], [[66, 69]], [[114, 117], [44, 47]], [[1, 20]], [[30, 40]], [[1, 27]], [[28, 53]], [[41, 53]], [[48, 71]], [[73, 112]]]",
    "query_spans": "[[[114, 121]]]",
    "process": "Since the area of 4AMF is twice the area of 4ANF, N is the midpoint of segment AM. According to the problem, the equation of line l is y = k(x + 1). Let M(x_{1}, y_{1}), N(x_{2}, y_{2}). From \\begin{cases} y = k(x + 1) \\\\ y^{2} = 8x \\end{cases}, eliminating y gives k^{2}x^{2} + (2k^{2} - 8)x + k^{2} = 0. Thus, \\begin{cases} x_{1} + x_{2} = -\\frac{2k^{2} - 8}{k^{2}} \\\\ x_{1} \\cdot x_{2} = 1 \\end{cases} = -2 + \\frac{8}{k^{2}} \\textcircled{1}. Since N is the midpoint of segment AM and A(-1, 0), we have 2x_{2} = x_{1} - 1, so x_{1} = 2x_{2} + 1 \\textcircled{2}. Substituting \\textcircled{2} into \\textcircled{1} yields \\begin{cases} 3x_{2} + 1 = -2 + \\frac{8}{k^{2}}, \\\\ (2x_{2} + 1) \\cdot x_{2} = 1 \\end{cases}. Since x_{2} > 0, solving gives x_{2} = \\frac{1}{2}, k = \\pm \\frac{4}{3}."
  },
  {
    "text": "Given a point $A$ on the parabola $x^{2}=y$ such that the distance from $A$ to the directrix is $\\frac{5}{4}$, then the distance from $A$ to the vertex is equal to?",
    "fact_expressions": "G: Parabola;A: Point;Expression(G) = (x^2 = y);PointOnCurve(A, G);Distance(A, Directrix(G)) = 5/4",
    "query_expressions": "Distance(A, Vertex(G))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 14]], [[17, 20], [43, 46]], [[2, 14]], [[2, 20]], [[2, 40]]]",
    "query_spans": "[[[2, 55]]]",
    "process": ""
  },
  {
    "text": "The line $l$ intersects the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ at points $A$ and $B$, and the midpoint of segment $AB$ has coordinates $(-\\frac{4}{5}, \\frac{1}{5})$. Then the equation of line $l$ is?",
    "fact_expressions": "l: Line;G: Ellipse;B: Point;A: Point;Expression(G) = (x^2/4 + y^2 = 1);Intersection(l, G) = {A, B};Coordinate(MidPoint(LineSegmentOf(A,B)))=(-4/5,1/5)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "y=x+1",
    "fact_spans": "[[[0, 5], [87, 92]], [[6, 33]], [[38, 41]], [[34, 37]], [[6, 33]], [[0, 43]], [[44, 85]]]",
    "query_spans": "[[[87, 97]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), substituting into the ellipse equation yields \\begin{cases}\\frac{x^{2}}{4}+y^{2}=1\\\\-\\frac{1}{4}+\\frac{x+x}{4}+y^{2}=\\frac{y-y_{2}}{y_{1}}\\end{cases}. Subtracting the two equations and simplifying gives thus fill in: y=x+1."
  },
  {
    "text": "If points $A(1,1)$, $B(2, m)$ both lie on the curve represented by the equation $a x^{2}+xy-2=0$, then $m$=?",
    "fact_expressions": "G: Curve;A: Point;B: Point;Expression(G) = (a*x^2 + x*y - 2 = 0);Coordinate(A) = (1, 1);Coordinate(B) = (2, m);m: Number;PointOnCurve(A, G);PointOnCurve(B, G)",
    "query_expressions": "m",
    "answer_expressions": "-1",
    "fact_spans": "[[[43, 45]], [[1, 10]], [[11, 20]], [[22, 45]], [[1, 10]], [[11, 20]], [[49, 52]], [[1, 46]], [[1, 46]]]",
    "query_spans": "[[[49, 54]]]",
    "process": "From the given conditions, \\begin{cases}a+1-2=0\\\\4a+2m-2=0\\end{cases}, solving yields \\begin{cases}a=1\\\\m=-\\end{cases}"
  },
  {
    "text": "For a moving point $P$ on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$ with foci $F_{1}$, $F_{2}$, when $\\angle F_{1} P F_{2}$ is maximum, the tangent value of $\\angle P F_{1} F_{2}$ is $2$. Then the eccentricity of this ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;P: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P,G);Focus(G)={F1,F2};WhenMax(AngleOf(F1,P,F2));Tan(AngleOf(P, F1, F2)) = 2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/5",
    "fact_spans": "[[[20, 74], [139, 141]], [[22, 74]], [[22, 74]], [[1, 8]], [[78, 81]], [[9, 16]], [[22, 74]], [[22, 74]], [[20, 74]], [[20, 81]], [[0, 74]], [[82, 106]], [[106, 136]]]",
    "query_spans": "[[[139, 149]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the right focus is $F$, $O$ is the coordinate origin. A circle centered at $F$ with radius $OF$ intersects the $x$-axis at points $O$ and $A$, and intersects one asymptote of the hyperbola $C$ at points $O$ and $B$. If $AB=4a$, then the equation of one asymptote of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;O: Origin;F: Point;A: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;Center(G)=F;Radius(G)=LineSegmentOf(O,F);Intersection(G, xAxis) = {O, A};Intersection(G,OneOf(Asymptote(C)))={O,B};LineSegmentOf(A,B)=4*a",
    "query_expressions": "Expression(OneOf(Asymptote(C)))",
    "answer_expressions": "{y=2*x,y=-2*x}",
    "fact_spans": "[[[2, 63], [117, 123], [152, 158]], [[10, 63]], [[10, 63]], [[98, 99]], [[106, 109], [72, 75], [131, 134]], [[68, 71], [82, 85]], [[110, 113]], [[135, 138]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 71]], [[81, 99]], [[89, 99]], [[98, 115]], [[98, 140]], [[141, 150]]]",
    "query_spans": "[[[152, 168]]]",
    "process": "\\angleABO=90^{\\circ}, we have \\tan\\angleAOB=\\frac{AB}{OB}, that is, \\frac{b}{a}=\\frac{4a}{\\sqrt{4c^{2}-16a^{2}}}, simplifying yields \\frac{b}{a}=2^{\\circ}. Thus, the asymptote equation. From the problem, OA is the diameter of circle F, and B is a point on the circle, \\therefore\\angleABO=90^{\\circ}, \\because AB=4a, OA=2c, \\therefore OB=\\sqrt{(2c)^{2}-(4a)^{2}}=\\sqrt{4c^{2}-16a^{2}}. Assume B lies on the asymptote y=\\frac{b}{a}x, then in right triangle ABO, \\tan\\angleAOB=\\frac{AB}{OB}, i.e., \\frac{b}{a}=\\frac{4a}{\\sqrt{4c^{2}-16a^{2}}}, which implies \\frac{b^{2}}{a^{2}}=\\frac{16a^{2}}{4(a^{2}+b^{2})-16a^{2}}, solving gives b=2a, i.e., \\frac{b}{a}=2. Hence, one asymptote of hyperbola C is y=2x (or y=-2x)."
  },
  {
    "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=(pm*1/2)*x",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 34]]]",
    "process": ""
  },
  {
    "text": "Given a point $P$ on the parabola $y^{2}=4x$, the distance from $P$ to the directrix is $d_{1}$, and the distance from $P$ to the line $l$: $4x-3y+11=0$ is $d_{2}$. Then the minimum value of $d_{1}+d_{2}$ is?",
    "fact_expressions": "G: Parabola;l: Line;P: Point;Expression(G) = (y^2 = 4*x);Expression(l) = (4*x - 3*y + 11 = 0);PointOnCurve(P, G);Distance(P, Directrix(G)) = d1;Distance(P,l)=d2;d1:Number;d2:Number",
    "query_expressions": "Min(d1+d2)",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 16]], [[38, 59]], [[19, 22]], [[2, 16]], [[38, 59]], [[2, 22]], [[2, 36]], [[19, 70]], [[29, 36]], [[63, 70]]]",
    "query_spans": "[[[72, 89]]]",
    "process": "(Analysis) According to the definition of a parabola, the distance from point P to the directrix of the parabola is equal to the distance from point P to the focus F. Draw a perpendicular line from focus F to the line $ l: 4x - 3y + 11 = 0 $; at this moment, $ d_{1} + d_{2} $ attains its minimum value. Using the point-to-line distance formula, we obtain $ \\frac{|4 \\times 1 + 11|}{\\sqrt{4^{2} + (-3)^{2}}} = 3 $, that is, the minimum value of $ d_{1} + d_{2} $ is 3. This involves the standard equation of a parabola and the application of its simple geometric properties, as well as extremum problems related to parabolas. The key to solving the problem lies in recognizing, based on the definition of a parabola, that the distance from point P to the directrix equals the distance from point P to the focus F, and then applying the point-to-line distance formula. This problem mainly examines transformational thinking, as well as computational and problem-solving abilities, and is considered a medium-difficulty question."
  },
  {
    "text": "If the eccentricity of a hyperbola is $\\sqrt{2}$, and the coordinates of one vertex are $(0 , 3)$, then its standard equation is?",
    "fact_expressions": "G: Hyperbola;Eccentricity(G) = sqrt(2);Coordinate(OneOf(Vertex(G))) = (0, 3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 - x^2 = 9",
    "fact_spans": "[[[1, 4], [39, 40]], [[1, 19]], [[1, 37]]]",
    "query_spans": "[[[39, 47]]]",
    "process": ""
  },
  {
    "text": "The asymptotes of the hyperbola $C$: $x^{2}-y^{2}=1$ are given by? If the right vertex of hyperbola $C$ is $A$, and a line $l$ passing through $A$ intersects the two asymptotes of hyperbola $C$ at points $P$ and $Q$, such that $\\overrightarrow{P A}=2 \\overrightarrow{A Q}$, then the slope of line $l$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;P: Point;A: Point;Q: Point;Expression(C) = (x^2 - y^2 = 1);RightVertex(C) = A;PointOnCurve(A, l);l1: Line;l2: Line;Asymptote(C) = {l1, l2};Intersection(l, l1) = P;Intersection(l, l2) = Q;VectorOf(P, A) = 2*VectorOf(A, Q)",
    "query_expressions": "Expression(Asymptote(C));Slope(l)",
    "answer_expressions": "x + pm*y = 0",
    "fact_spans": "[[[53, 58], [131, 136]], [[0, 23], [33, 39], [59, 65]], [[73, 76]], [[44, 47], [49, 52]], [[77, 80]], [[0, 23]], [[33, 47]], [[48, 58]], [], [], [[59, 71]], [[53, 82]], [[53, 82]], [[84, 129]]]",
    "query_spans": "[[[0, 32]], [[131, 141]]]",
    "process": ""
  },
  {
    "text": "Find the standard equation of the hyperbola with foci at the endpoints of the minor axis of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ and passing through the point $A(4, -5)$.",
    "fact_expressions": "G: Hyperbola;H: Ellipse;A: Point;Expression(H) = (x^2/16 + y^2/9 = 1);Coordinate(A) = (4, -5);PointOnCurve(A, G);Focus(G)=Endpoint(MinorAxis(H))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/5 - x^2/4 = 1",
    "fact_spans": "[[[66, 69]], [[2, 40]], [[54, 65]], [[2, 40]], [[54, 65]], [[53, 69]], [[1, 69]]]",
    "query_spans": "[[[66, 75]]]",
    "process": "In \\frac{x^2}{16}+\\frac{y^{2}}{9}=1, the endpoints of the minor axis are (0,\\pm3), so in the hyperbola the foci are (0,\\pm3). The sum of distances from point A(4,-5) to the two foci is 2\\sqrt{5}, \\therefore 2a=2\\sqrt{5}, \\therefore a=\\sqrt{5}, \\therefore b^{2}=c^{2}-a^{2}=4, so the equation of the hyperbola is \\frac{y^{2}}{5}-\\frac{x^{2}}{4}=1"
  },
  {
    "text": "The semi-focal distance of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $c$, and the $x$-coordinate of an intersection point between the line $y=2x$ and the ellipse is exactly $c$. Then, the eccentricity of this ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;c: Number;HalfFocalLength(G) = c;H: Line;Expression(H) = (y = 2*x);XCoordinate(OneOf(Intersection(H, G))) = c",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2) - 1",
    "fact_spans": "[[[0, 52], [71, 73], [90, 92]], [[0, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[57, 60], [84, 87]], [[0, 60]], [[61, 70]], [[61, 70]], [[61, 87]]]",
    "query_spans": "[[[90, 98]]]",
    "process": "According to the problem, the y-coordinate of an intersection point between the line $ y = 2x $ and the ellipse is $ 2c $. Substituting into $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $, we get $ \\frac{c^{2}}{a^{2}} + \\frac{4c^{2}}{b^{2}} = 1 $. Since $ e^2 = \\frac{c^2}{a^2} $ and $ b^2 = a^2(1 - e^2) $, it follows that $ e^{2} + \\frac{4e^{2}}{1 - e^{2}} = 1 $. Solving gives $ e = \\sqrt{2} - 1 $; the other root does not satisfy the conditions and is discarded."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{b^{2}}=1$ $(3>b>0)$, if there exists a circle passing through the foci $F_{1}$ and $F_{2}$ that is tangent to the line $x+y+2=0$, then the maximum value of the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;H: Circle;I: Line;F1: Point;F2: Point;3 > b;b > 0;Expression(G) = (x^2/9 + y^2/b^2 = 1);Expression(I) = (x + y + 2 = 0);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1,H);PointOnCurve(F2,H);IsTangent(I,H)",
    "query_expressions": "Max(Eccentricity(G))",
    "answer_expressions": "2/3",
    "fact_spans": "[[[18, 68], [115, 117]], [[20, 68]], [[98, 99]], [[100, 111]], [[2, 9], [82, 89]], [[10, 17], [90, 97]], [[20, 68]], [[20, 68]], [[18, 68]], [[100, 111]], [[2, 75]], [[2, 75]], [[79, 99]], [[79, 99]], [[98, 113]]]",
    "query_spans": "[[[115, 126]]]",
    "process": "The equation of the circle passing through the foci $F_{1}, F_{2}$ is: $x^{2}+(y-m)^{2}=m^{2}+c^{2}$. Using the condition that this circle is tangent to the line $x+y+2=0$, properties of quadratic functions, and the eccentricity formula, the conclusion can be obtained. From the problem, the center of the circle passing through the foci $F_{1}, F_{2}$ lies on the $y$-axis; let its equation be: $x^{2}+(y-m)^{2}=m^{2}+c^{2}$. Since the circle passing through the foci $F_{1}, F_{2}$ is tangent to the line $x+y+2=0$, $\\therefore d=r$, i.e., $\\sqrt{m^{2}+c^{2}}=\\frac{|m+2|}{\\sqrt{1+1}}$. Solving gives: $c^{2}=-\\frac{m^{2}}{2}+2m+2$. Therefore, when $c$ is maximum, $e$ is maximum. While $-\\frac{m^{2}}{2}+2m+2=-\\frac{1}{2}(m-2)^{2}+4 \\leqslant 4$, the maximum value of $c$ is $2$, $\\therefore$ the maximum value of $e$ is $\\frac{2}{3}$."
  },
  {
    "text": "Given that point $A(1, \\sqrt{2})$ lies on the parabola $y^2=2 p x$. If the three vertices of $\\triangle A B C$ all lie on the parabola, and the slopes of the lines containing sides $A B$, $B C$, $C A$ are $k_{1}$, $k_{2}$, $k_{3}$ respectively, then $\\frac{1}{k_{1}}-\\frac{1}{k_{2}}+\\frac{1}{k_{3}}$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;A: Point;Coordinate(A) = (1, sqrt(2));PointOnCurve(A, G);B: Point;C: Point;PointOnCurve(Vertex(TriangleOf(A, B, C)), G);k1: Number;k2: Number;k3: Number;Slope(OverlappingLine(LineSegmentOf(A, B))) = k1;Slope(OverlappingLine(LineSegmentOf(B, C))) = k2;Slope(OverlappingLine(LineSegmentOf(C, A))) = k3",
    "query_expressions": "-1/k2 + 1/k1 + 1/k3",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[20, 34], [61, 64]], [[20, 34]], [[23, 34]], [[2, 19]], [[2, 19]], [[2, 35]], [[77, 82]], [[77, 82]], [[37, 65]], [[100, 107]], [[109, 116]], [[119, 126]], [[69, 127]], [[69, 127]], [[69, 127]]]",
    "query_spans": "[[[129, 180]]]",
    "process": "\\because point A(1,\\sqrt{2}) lies on the parabola r: y^{2}=2px. \\therefore 2=2p\\times1, solving gives p=1. \\therefore the equation of parabola r is: y^{2}=2x. Let B\\left(\\frac{y_{1}^{2}}{2},y_{1}\\right), C\\left(\\frac{y_{2}^{2}}{2},y_{2}\\right). k_{1}=\\frac{y_{1}-\\sqrt{2}}{\\frac{y_{1}^{2}}{2}-1}=\\frac{2}{y_{1}+\\sqrt{2}}, k_{2}=\\frac{y_{2}-\\sqrt{2}}{\\frac{y_{2}^{2}}{2}-1}=\\frac{2}{y_{2}+\\sqrt{2}}, k_{3}=\\frac{y_{2}-\\sqrt{2}}{\\frac{y_{2}^{2}}{2}-1}=\\frac{2}{y_{2}+\\sqrt{2}} + \\frac{1}{k_{3}} = \\frac{y_{1}+\\sqrt{2}}{2} - \\frac{y_{1}+y_{2}}{2} + \\frac{y_{2}+\\sqrt{2}}{2} = \\sqrt{2}k_{1}\\overline{k_{2}} + answer: \\sqrt{2}"
  },
  {
    "text": "Given that the eccentricity of ellipse $E$ is $e$, with foci $F_{1}$ and $F_{2}$, and a parabola $C$ has $F_{1}$ as its vertex and $F_{2}$ as its focus. $P$ is an intersection point of the two curves. If $\\frac{|P F_{1}|}{|P F_{2}|}=e$, then what is the value of $e$?",
    "fact_expressions": "C: Parabola;E: Ellipse;F1: Point;F2:Point;P: Point;e:Number;Eccentricity(E) = e;Focus(E)={F1,F2};Vertex(C)=F1;Focus(C)=F2;OneOf(Intersection(E,C))=P;Abs(LineSegmentOf(P,F1))/Abs(LineSegmentOf(P,F2)) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[38, 44]], [[2, 7]], [[20, 27], [45, 52]], [[30, 37], [56, 63]], [[67, 70]], [[12, 15], [114, 117]], [[2, 15]], [[2, 37]], [[38, 55]], [[38, 66]], [[67, 79]], [[81, 112]]]",
    "query_spans": "[[[114, 121]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively. Point $P$ lies on the hyperbola. If $\\overrightarrow{P F_{1}}\\cdot\\overrightarrow{P F_{2}}=0$, the area of $\\Delta P F_{1} F_{2}$ is $9$, and $a+b=7$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Area(TriangleOf(P, F1, F2)) = 9;a + b = 7",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/4",
    "fact_spans": "[[[19, 75], [87, 90], [194, 197]], [[22, 75]], [[22, 75]], [[82, 86]], [[1, 8]], [[9, 16]], [[22, 75]], [[22, 75]], [[19, 75]], [[1, 81]], [[1, 81]], [[82, 91]], [[93, 150]], [[153, 182]], [[184, 191]]]",
    "query_spans": "[[[194, 203]]]",
    "process": "Analysis: Let $|\\overrightarrow{PF_{1}}|=m$, $|\\overrightarrow{PF_{2}}|=n$. Since the area of $\\triangle PF_{1}F_{2}$ is 9, we obtain $mn=18$. Combining with the Pythagorean theorem, we derive $m^{2}+n^{2}=(m-n)^{2}+36=4c^{2}$. Using the definition of a hyperbola, we get $b^{2}=9$, then find $a$, use the square relationship to find $c$, and finally obtain the eccentricity of the hyperbola. Detailed solution: Let $|\\overrightarrow{PF_{1}}|=m$, $|\\overrightarrow{PF_{2}}|=n$. Since $\\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=0$ and the area of $\\triangle PF_{1}F_{2}$ is 9, we have $\\frac{1}{2}mn=9$, i.e., $mn=18$. In right triangle $\\triangle PF_{1}F_{2}$, by the Pythagorean theorem, $m^{2}+n^{2}=4c^{2}$. Thus, $(m-n)^{2}=m^{2}+n^{2}-2mn=4c^{2}-36$. By the definition of a hyperbola, $(m-n)^{2}=4a^{2}$. Therefore, $4c^{2}-36=4a^{2}$, simplifying to $c^{2}-a^{2}=9$, i.e., $b^{2}=9$, so $b=3$. Given $a+b=7$, we get $a=4$. Hence, $c=\\sqrt{a^{2}+b^{2}}=5$. Therefore, the eccentricity of the hyperbola is $e=\\frac{c}{a}=\\frac{5}{4}$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has a focal distance of $8$, the line $x-2 y=0$ intersects the hyperbola $C$ at points $A$ and $B$, and $M(-a, 2 b)$. If $|M A|=|M B|$, then the equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;M: Point;A: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x - 2*y = 0);Coordinate(M) = (-a, 2*b);FocalLength(C) = 8;Intersection(G, C) = {A, B};Abs(LineSegmentOf(M, A)) = Abs(LineSegmentOf(M, B))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/8 - y^2/8 = 1",
    "fact_spans": "[[[2, 63], [83, 89], [130, 136]], [[10, 63]], [[10, 63]], [[71, 82]], [[101, 113]], [[91, 94]], [[95, 98]], [[10, 63]], [[10, 63]], [[2, 63]], [[71, 82]], [[101, 113]], [[2, 70]], [[71, 100]], [[115, 128]]]",
    "query_spans": "[[[130, 141]]]",
    "process": "By the symmetry of the hyperbola, it is clear that points A and B are symmetric with respect to the origin O. Since |MA| = |MB|, we have MO \\bot AB, then -\\frac{2b}{a} \\times \\frac{1}{2} = -1, so a = b. Also, 2c = 8 implies c = 4, thus a^{2} + b^{2} = 16, so a^{2} = b^{2} = 8. Hence, the equation of hyperbola C is \\frac{x^{2}}{8} - \\frac{y^{2}}{8} = 1."
  },
  {
    "text": "Given $F_{1}(-c, 0)$, $F_{2}(c, 0)$ are the two foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, $P$ is a point on the ellipse, and $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=c^{2}$, then the range of the eccentricity of this ellipse is?",
    "fact_expressions": "F1: Point;F2: Point;Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);Focus(G) = {F1, F2};c: Number;G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;P: Point;PointOnCurve(P, G);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = c^2",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[sqrt(3)/3, sqrt(2)/2]",
    "fact_spans": "[[[2, 16]], [[19, 32]], [[2, 16]], [[19, 32]], [[2, 90]], [[2, 16]], [[33, 85], [95, 97], [168, 170]], [[33, 85]], [[35, 85]], [[35, 85]], [[35, 85]], [[35, 85]], [[91, 94]], [[91, 100]], [[102, 165]]]",
    "query_spans": "[[[168, 180]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a, b>0)$ has left focus at $F(-2,0)$. A moving line $l$ passes through point $A(2a, 0)$ and intersects the two asymptotes of $E$ at points $B$ and $C$, respectively. If $\\overrightarrow{A B}=\\overrightarrow{B C}$ and $CF$ is perpendicular to $l$, then the real axis length of $E$ is?",
    "fact_expressions": "E: Hyperbola;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;Coordinate(F) = (-2, 0);LeftFocus(E) = F;l: Line;A: Point;Coordinate(A) = (2*a, 0);PointOnCurve(A, l);B: Point;C: Point;L1: Line;L2: Line;Asymptote(E) = {L1, L2};Intersection(l, L1) = B;Intersection(l, L2) = C ;VectorOf(A, B) = VectorOf(B, C);IsPerpendicular(LineSegmentOf(C, F), l)",
    "query_expressions": "Length(RealAxis(E))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 61], [99, 102], [180, 183]], [[2, 61]], [[10, 61]], [[10, 61]], [[10, 61]], [[10, 61]], [[66, 75]], [[66, 75]], [[2, 75]], [[79, 82], [173, 176]], [[84, 96]], [[84, 96]], [[79, 96]], [[112, 115]], [[116, 119]], [], [], [[99, 108]], [[79, 121]], [[79, 121]], [[122, 165]], [[167, 178]]]",
    "query_spans": "[[[180, 189]]]",
    "process": "From the given conditions, it is easy to see that: c = 2 and B is the midpoint of AC, as shown in the figure below: if C(x, \\frac{b}{a}x), then B(\\frac{x+2a}{2}, \\frac{b}{2a}x) lies on y = -\\frac{b}{a}x, -b); also CF is perpendicular to l, i.e., CF \\bot AC, and \\overrightarrow{CF} = (a-2, b), \\overrightarrow{CA} = (3a, b), \\therefore 3a(a-2) + b^{2} = 0; since a^{2} + b^{2} = c^{2} = 4, then a^{2} - 3a + 2 = 0, solving gives a = 1 or a = 2; \\therefore from a < c we get: a = 1, hence the real axis length of E is 2a = 2."
  },
  {
    "text": "The hyperbola has its center at the origin and foci on the $x$-axis, with equal real and imaginary axes. The distance from one focus to an asymptote is $\\sqrt{2}$. Then the equation of the hyperbola is?",
    "fact_expressions": "Center(G) = O;O: Origin;PointOnCurve(Focus(G), xAxis) = True;G: Hyperbola;RealAxis(G) = ImageinaryAxis(G);Distance(OneOf(Focus(G)), OneOf(Asymptote(G))) = sqrt(2)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 = 2",
    "fact_spans": "[[[0, 18]], [[3, 5]], [[6, 18]], [[15, 18], [53, 56]], [[15, 26]], [[15, 51]]]",
    "query_spans": "[[[53, 60]]]",
    "process": ""
  },
  {
    "text": "Let point $P$ be a point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, and let $F_{1}$, $F_{2}$ be the two foci of the ellipse. If the area of $\\Delta P F_{1} F_{2}$ is $\\frac{5}{2}$, then $\\sin \\angle F_{1} P F_{2}$=?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 + y^2/5 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2};Area(TriangleOf(P, F1, F2)) = 5/2",
    "query_expressions": "Sin(AngleOf(F1, P, F2))",
    "answer_expressions": "4/5",
    "fact_spans": "[[[6, 43], [64, 66]], [[1, 5]], [[47, 54]], [[55, 62]], [[6, 43]], [[1, 46]], [[47, 71]], [[73, 112]]]",
    "query_spans": "[[[114, 143]]]",
    "process": "In the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, the semi-major axis $a=3$, the semi-focal distance $c=2$, by the definition of an ellipse we have $|PF_{1}|+|PF_{2}|=2a=6$. In $\\triangle PF_{1}F_{2}$, by the cosine theorem: $|F_{1}F_{2}|^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}|\\cdot|PF_{2}|\\cos\\angle F_{1}PF_{2}$, that is: $(2c)^{2}=(2a)^{2}-2|PF_{1}|\\cdot|PF_{2}|(1+\\cos\\angle F_{1}PF_{2})$, then $|PF_{1}|\\cdot|PF_{2}|(1+\\cos\\angle F_{1}PF_{2})=10$. Also, the area of $\\triangle PF_{1}F_{2}$ is $\\frac{5}{2}$, then $\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|\\sin\\angle F_{1}PF_{2}=\\frac{5}{2}$, i.e., $|PF_{1}|\\cdot|PF_{2}|\\sin\\angle F_{1}PF_{2}=5$. Thus we obtain $2\\sin\\angle F_{1}PF_{2}=1+\\cos\\angle F_{1}PF_{2}$, squaring both sides gives $(1+\\cos\\angle F_{1}PF_{2})^{2}=4\\sin^{2}\\angle F_{1}PF_{2}=4(1-\\cos\\angle F_{1}PF_{2})(1+\\cos\\angle F_{1}PF_{2})$, solving yields $\\cos\\angle F_{1}PF_{2}=\\frac{3}{5}$, then $\\sin\\angle F_{1}PF_{2}=\\frac{4}{5}$. Therefore, $\\sin\\angle F_{1}PF_{2}=\\frac{4}{5}$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, one of its asymptotes is given by $2 x-y=0$. Then, the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Expression(OneOf(Asymptote(C))) = (2*x - y = 0)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 64], [84, 87]], [[2, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[2, 82]]]",
    "query_spans": "[[[84, 93]]]",
    "process": "A asymptote equation of the hyperbola is: $2x - y = 0 \\Rightarrow y = 2x$, so we have $\\frac{b}{a} = 2 \\Rightarrow b^{2} = 4a^{2}$, and $c^{2} = a^{2} + b^{2}$, thus $c^{2} - a^{2} = 4a^{2} \\Rightarrow c^{2} = 5a^{2} \\Rightarrow c = \\sqrt{5}a \\Rightarrow e = \\sqrt{5}$"
  },
  {
    "text": "Given that the parabola $y^{2}=a x$ passes through the point $A(\\frac{1}{4}, 1)$, then the distance from point $A$ to the focus of this parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = a*x);a: Number;A: Point;Coordinate(A) = (1/4, 1);PointOnCurve(A, G) = True",
    "query_expressions": "Distance(A, Focus(G))",
    "answer_expressions": "5/4",
    "fact_spans": "[[[2, 16], [46, 49]], [[2, 16]], [[5, 16]], [[17, 37], [40, 44]], [[17, 37]], [[2, 37]]]",
    "query_spans": "[[[40, 57]]]",
    "process": "\\because the parabola y^{2}=ax passes through point A(\\frac{1}{4},1), \\therefore 1^{2}=a\\times\\frac{1}{4}, solving gives a=4, the equation of the parabola is y^{2}=4x. The directrix of the parabola is x=-1, the focus is F(1,0), by the definition of the parabola we obtain AF=\\frac{1}{4}+1=\\frac{5}{4}"
  },
  {
    "text": "Given that $F$ is the focus of the parabola $y^{2}=4x$, $P$ is a moving point on the parabola, and $A(3, 1)$, then the minimum perimeter of $\\triangle APF$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;P: Point;PointOnCurve(P, G);A: Point;Coordinate(A) = (3, 1)",
    "query_expressions": "Min(Perimeter(TriangleOf(A, P, F)))",
    "answer_expressions": "4+sqrt(5)",
    "fact_spans": "[[[6, 20], [28, 31]], [[6, 20]], [[2, 5]], [[2, 23]], [[24, 27]], [[24, 37]], [[38, 48]], [[38, 48]]]",
    "query_spans": "[[[50, 75]]]",
    "process": "The focus coordinates of $ y^2 = 4x $ are $ F(1,0) $. To find the minimum perimeter of $ \\triangle PAF $, we need to find the minimum value of $ |PA| + |PF| $. Let point $ D $ be the projection of point $ P $ on the directrix. According to the definition of a parabola, $ |PF| = |PD| $. Therefore, the minimum value of $ |PA| + |PF| $ is equivalent to the minimum value of $ |PA| + |PD| $. Based on plane geometry, $ |PA| + |PD| $ is minimized when points $ D $, $ P $, and $ A $ are collinear. Thus, the minimum value is $ x_A - (-1) = 3 + 1 = 4 $. Since $ |AF| = \\sqrt{5} $, the minimum perimeter of $ \\triangle PAF $ is $ 4 + \\sqrt{5} $, which corresponds to: $ A_1, F_5 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has eccentricity $3$, and $F_{1}$, $F_{2}$ are the left and right foci of the hyperbola. A line passing through the right focus $F_{2}$ and parallel to one asymptote intersects the hyperbola at point $A$. The inradius of $\\Delta A F_{1} F_{2}$ is $r$. Then $\\frac{b}{r}=$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(G) = 3;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Line;PointOnCurve(F2, H);IsParallel(H, OneOf(Asymptote(G)));A: Point;Intersection(H, G) = A;r: Number;Radius(InscribedCircle(TriangleOf(A, F1, F2))) = r",
    "query_expressions": "b/r",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 58], [85, 88], [118, 121]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 66]], [[68, 75]], [[77, 84], [98, 105]], [[68, 93]], [[68, 93]], [[115, 117]], [[94, 117]], [[85, 117]], [[122, 126]], [[115, 126]], [[157, 160]], [[128, 160]]]",
    "query_spans": "[[[162, 177]]]",
    "process": "By the given conditions: for the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$, the left and right foci are $F_{1}(-c,0)$, $F_{2}(c,0)$. Let one asymptote of the hyperbola be $y=\\frac{b}{a}x$, then the equation of line $AF_{2}$ is $y=\\frac{b}{a}(x-c)$. Solving the system of equations:\n\\[\n\\begin{cases}\ny=\\frac{b}{a}(x-c) \\\\\n\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1\n\\end{cases}\n\\Rightarrow\n\\begin{cases}\nx=\\frac{a^{2}+c^{2}}{2c} \\\\\ny=\\frac{b(c^{2}-a^{2})}{2ac}\n\\end{cases}\n\\]\nSince the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ is $3$, we have $e=\\frac{c}{a}=3$, $a^{2}+b^{2}=c^{2}$, so $c=3a$, $b=2\\sqrt{2}a$. Therefore, the coordinates of point $A$ are $(\\frac{5}{3}a,-\\frac{8}{3}\\sqrt{2}a)$. Thus, the height corresponding to base $F_{1}F_{2}$ in $\\triangle AF_{1}F_{2}$ is $h=\\frac{8}{3}\\sqrt{2}a$. Then,\n\\[\nS_{\\Delta AF_{1}F_{2}} = \\frac{1}{2}|F_{1}F_{2}|\\cdot h = \\frac{1}{2}\\times 2c \\times \\frac{8}{3}\\sqrt{2}a = \\frac{1}{2}\\times 2\\times 3a \\times \\frac{8}{3}\\sqrt{2}a = 8\\sqrt{2}a^{2}.\n\\]\nLet $|AF_{1}|=m$, $|AF_{2}|=n$, and the inradius of the triangle be $r$. In $\\triangle AF_{1}F_{2}$, let $\\angle F_{1}F_{2}A = \\theta$, where $\\theta$ is the inclination angle of line $AF_{2}$. Thus, $\\tan\\theta = \\frac{b}{a}$, and $\\sin^{2}\\theta + \\cos^{2}\\theta = 1$. Hence,\n\\[\n\\sin\\theta = \\frac{b}{\\sqrt{a^{2}+b^{2}}} = \\frac{b}{c} = \\frac{2}{3}\\sqrt{2}.\n\\]\nThen we obtain $n\\sin\\theta = \\frac{8}{3}\\sqrt{2}a \\Rightarrow n = 4a$. By the definition of the hyperbola: $m - n = 2a \\Rightarrow m = n + 2a = 6a$. Using the area equivalence method for triangles:\n\\[\n\\frac{1}{2}\\cdot r \\cdot (m + n + 2c) = S_{\\Delta AF_{1}F_{2}} = 8\\sqrt{2}a^{2} \\Rightarrow r = \\sqrt{2}a.\n\\]\nThus,\n\\[\n\\frac{b}{r} = \\frac{2\\sqrt{2}a}{\\sqrt{2}a} = 2.\n\\]\nFinal answer: $?$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the eccentricity is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 58]], [[2, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[8, 58]]]",
    "query_spans": "[[[2, 64]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has an eccentricity of $2$, then the distance from the point $(-2 \\sqrt{3}, 0)$ to the asymptote of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(G) = (-2*sqrt(3), 0);Eccentricity(C) = 2",
    "query_expressions": "Distance(G, Asymptote(C))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 63], [93, 96]], [[10, 63]], [[10, 63]], [[73, 92]], [[10, 63]], [[10, 63]], [[2, 63]], [[73, 92]], [[2, 71]]]",
    "query_spans": "[[[73, 105]]]",
    "process": "By the given condition, the eccentricity of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) is 2, that is, $ e^{2} = \\frac{c^{2}}{a^{2}} = \\frac{a^{2} + b^{2}}{a^{2}} = 1 + \\left( \\frac{b}{a} \\right)^{2} = 4 $, solving gives $ \\frac{b}{a} = \\sqrt{3} $. Therefore, an asymptote of the hyperbola is $ y = \\sqrt{3}x $, or $ \\sqrt{3}x - y = 0 $. Thus, the distance from the point $ (-2\\sqrt{3}, 0) $ to the asymptote of $ C $ is $ d = \\frac{\\sqrt{3} \\times (-2\\sqrt{3}) - 0}{\\sqrt{(\\sqrt{3})^{2} + (-1)^{2}}} = 3 $."
  },
  {
    "text": "Let the hyperbola $C$ be centered at the origin, with real axis length $4$ and eccentricity $\\frac{\\sqrt{5}}{2}$. Then the distance from the foci of $C$ to its asymptotes is?",
    "fact_expressions": "C: Hyperbola;O: Origin;Center(C) = O;Length(RealAxis(C)) = 4;Eccentricity(C) = sqrt(5)/2",
    "query_expressions": "Distance(Focus(C), Asymptote(C))",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 7], [48, 51], [55, 56]], [[11, 13]], [[1, 13]], [[1, 21]], [[1, 46]]]",
    "query_spans": "[[[48, 64]]]",
    "process": "Assume the hyperbola's foci lie on the x-axis. According to the given conditions, a=2, c=\\sqrt{5}, so the equation of the hyperbola is \\frac{x^{2}}{4}-y^{2}=1. One of its asymptotes is x+2y=0. The coordinates of the foci are (\\pm\\sqrt{5},0). Then the distance from a focus to the asymptote is d=\\frac{|\\sqrt{5}|}{\\sqrt{12}+2^{2}}=1"
  },
  {
    "text": "Given that the slope of an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $\\frac{1}{2}$, what is the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Slope(OneOf(Asymptote(G))) = 1/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 59], [85, 88]], [[2, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[2, 82]]]",
    "query_spans": "[[[85, 94]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ lies on the ellipse $C$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, $F_{1}$ and $F_{2}$ are its left and right foci, and $\\angle F_{1} P F_{2}=60^{\\circ}$, then the area of triangle $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "P: Point;C: Ellipse;Expression(C) = (x^2/25 + y^2/9 = 1);PointOnCurve(P, C);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "3*sqrt(3)",
    "fact_spans": "[[[2, 6]], [[7, 50], [70, 71]], [[7, 50]], [[2, 53]], [[54, 61]], [[62, 69]], [[54, 75]], [[54, 75]], [[77, 110]]]",
    "query_spans": "[[[112, 142]]]",
    "process": "From the ellipse equation, we know: $ a=5, b=3 $, then $ c^{2}=a^{2}-b^{2}=16 $. From the definition of the ellipse: $ |PF_{1}|+|PF_{2}|=2a=10 $. By the cosine law: $ |F_{1}F_{2}|^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}|\\cdot|PF_{2}|\\cos\\angle F_{1}PF_{2} $. Therefore, $ 4c^{2}=(|PF_{1}|+|PF_{2}|)^{2}-3|PF_{1}|\\cdot|PF_{2}|=100-3|PF_{1}|\\cdot|PF_{2}|=64 $, solving gives $ |PF_{1}|\\cdot|PF_{2}|=12 $. Thus, $ S_{\\Delta F_{1}PF_{2}}=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|\\sin\\angle F_{1}PF_{2}=6\\times\\frac{\\sqrt{3}}{2}=3\\sqrt{3} $."
  },
  {
    "text": "If a point $P$ on the ellipse $\\frac{x^{2}}{100}+\\frac{y^{2}}{36}=1$ has a distance of $6$ to focus $F_{1}$, then what is the distance from point $P$ to the other focus $F_{2}$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/100 + y^2/36 = 1);P: Point;PointOnCurve(P, G) = True;F1: Point;F2: Point;OneOf(Focus(G)) = F1;OneOf(Focus(G)) = F2;Negation(F1 = F2);Distance(P, F1) = 6",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "14",
    "fact_spans": "[[[2, 42]], [[2, 42]], [[45, 48], [68, 72]], [[2, 48]], [[51, 58]], [[78, 85]], [[2, 58]], [[2, 85]], [[2, 85]], [[45, 66]]]",
    "query_spans": "[[[68, 90]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, one of its asymptotes is perpendicular to the line $x+\\sqrt{3} y=0$. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x + sqrt(3)*y = 0);IsPerpendicular(OneOf(Asymptote(C)), G)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 63], [92, 95]], [[10, 63]], [[10, 63]], [[70, 88]], [[10, 63]], [[10, 63]], [[2, 63]], [[70, 88]], [[2, 90]]]",
    "query_spans": "[[[92, 101]]]",
    "process": "The slope of the line $x+\\sqrt{3}y=0$ is $-\\frac{\\sqrt{3}}{3}$, so the slope of the asymptote of the hyperbola perpendicular to the line $x+\\sqrt{3}y=0$ is $\\sqrt{3}$. Therefore, $\\frac{b}{a}=\\sqrt{3}$, so $e=\\frac{c}{a}=\\sqrt{\\frac{b^{2}+a^{2}}{a^{2}}}=\\sqrt{1+(\\frac{b}{a})^{2}}=2$"
  },
  {
    "text": "Given that the left focus of the hyperbola $\\frac{x^{2}}{3}-\\frac{16 y^{2}}{p^{2}}=1$ lies on the directrix of the parabola $y^{2}=2 p x$, then $p$=?",
    "fact_expressions": "G: Hyperbola;p: Number;H: Parabola;Expression(G) = (x^2/3 - 16*y^2/p^2 = 1);Expression(H) = (y^2 = 2*(p*x));PointOnCurve(LeftFocus(G), Directrix(H))",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 47]], [[74, 77]], [[52, 68]], [[2, 47]], [[52, 68]], [[2, 72]]]",
    "query_spans": "[[[74, 79]]]",
    "process": "In the hyperbola, $a^{2}=3$, $b^{2}=\\frac{p^{2}}{16}$, $\\therefore c^{2}=3+\\frac{p^{2}}{16}$, $\\therefore c=\\sqrt{3+\\frac{p^{2}}{16}}$. The directrix of the parabola is $x=-\\frac{p}{2}$, $\\therefore \\frac{p}{2}=\\sqrt{3+\\frac{p^{2}}{16}}$, $\\therefore p=4$. Key point: properties of hyperbola and parabola."
  },
  {
    "text": "Given that the distance from the point on the parabola $y^{2}=2 p x$ ($p>0$) with horizontal coordinate $3$ to its focus is $4$, then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));PointOnCurve(F, G);XCoordinate(F)=3;Distance(F, Focus(G)) = 4",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 23], [34, 35]], [[46, 49]], [[32, 33]], [[5, 23]], [[2, 23]], [[2, 33]], [[24, 33]], [[32, 44]]]",
    "query_spans": "[[[46, 51]]]",
    "process": "The directrix equation of the parabola $ y^{2}=2px $ ($ p>0 $) is: $ x=-\\frac{p}{2} $. Since the distance from the point on the parabola $ y^{2}=2px $ ($ p>0 $) with abscissa 3 to the focus equals 4, therefore according to the definition of the parabola, $ 3-(-\\frac{p}{2})=4 $, $ \\therefore p=2 $."
  },
  {
    "text": "Given that the angle of inclination of an asymptote of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ is $\\frac{\\pi}{3}$, then the value of $b$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/b^2 = 1);b: Number;b>0;Inclination(OneOf(Asymptote(G))) = pi/3",
    "query_expressions": "b",
    "answer_expressions": "3*sqrt(3)",
    "fact_spans": "[[[2, 49]], [[5, 49]], [[77, 80]], [[5, 49]], [[2, 75]]]",
    "query_spans": "[[[77, 84]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$, a line $l$ is drawn through the point $P(1, 1)$, intersecting the ellipse at points $A$ and $B$, such that point $P$ is the midpoint of segment $AB$. What is the slope of the line?",
    "fact_expressions": "l: Line;G: Ellipse;Expression(G) = (x^2/4 + y^2/2 = 1);A: Point;B: Point;P: Point;Coordinate(P) = (1, 1);PointOnCurve(P,l);Intersection(l,G)={A,B};MidPoint(LineSegmentOf(A,B)) = P",
    "query_expressions": "Slope(l)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[54, 59], [93, 95]], [[2, 39], [61, 63]], [[2, 39]], [[65, 68]], [[69, 72]], [[42, 53], [76, 80]], [[42, 53]], [[41, 59]], [[54, 74]], [[76, 91]]]",
    "query_spans": "[[[93, 100]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $x^{2}+4 y^{2}=16$, a line $l$ passes through its left focus $F_{1}$ and intersects the ellipse at points $A$ and $B$. If the slope of line $l$ is $1$, then the chord length $|A B|=$?",
    "fact_expressions": "l: Line;G: Ellipse;A: Point;B: Point;F1: Point;Expression(G) = (x^2 + 4*y^2 = 16);PointOnCurve(F1,l);Intersection(l,G)={A,B};LeftFocus(G) = F1;Slope(l) = 1;IsChordOf(LineSegmentOf(A, B),G)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16/5",
    "fact_spans": "[[[23, 28], [59, 64]], [[2, 22], [30, 31], [44, 46]], [[48, 51]], [[52, 55]], [[34, 41]], [[2, 22]], [[23, 41]], [[23, 57]], [[30, 41]], [[59, 71]], [[44, 82]]]",
    "query_spans": "[[[75, 84]]]",
    "process": "The standard equation of the ellipse is $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$, where $a=4$, $b=2$, then $c=\\sqrt{16-4}=2\\sqrt{3}$, $F_{1}(-2\\sqrt{3},0)$. Given that the slope of the line is 1, the equation of the line is $y=x+2\\sqrt{3}$. Solving this together with the ellipse's equation gives: $5x^{2}+16\\sqrt{3}x+32=0$, $x_{1}+x_{2}=-\\frac{16\\sqrt{3}}{5}$, $x_{1}x_{2}=\\frac{32}{5}$. The chord length $|AB|=\\sqrt{(1+k^{2})},-\\frac{16\\sqrt{3}}{5})-4\\times\\frac{32}{5}=\\frac{16}{5}$"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$, directrix $l$, and point $M$ is a point on the parabola $C$. $MH \\perp l$ at $H$. If $|MH|=4$, $\\angle HFM=60^{\\circ}$, then the equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;l: Line;Directrix(C) = l;M: Point;PointOnCurve(M, C);H: Point;IsPerpendicular(LineSegmentOf(M, H), l);FootPoint(LineSegmentOf(M, H), l) = H;Abs(LineSegmentOf(M, H)) = 4;AngleOf(H, F, M) = ApplyUnit(60, degree)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[2, 28], [48, 54], [115, 121]], [[2, 28]], [[10, 28]], [[10, 28]], [[32, 35]], [[2, 35]], [[39, 42]], [[2, 42]], [[43, 47]], [[43, 57]], [[72, 75]], [[58, 71]], [[58, 75]], [[77, 86]], [[88, 113]]]",
    "query_spans": "[[[115, 126]]]",
    "process": "Since the distance from any point on a parabola to the focus is equal to the distance to the directrix, |MF| = |MH| = 4. Also, \\angle HFM = 60^{\\circ}. Therefore, \\triangle MHF is an equilateral triangle, so |HF| = 4. Let the directrix l intersect the x-axis at point Q, then \\angle QHF = 30^{\\circ}. Thus, p = |QF| = |HF| \\sin \\angle QHF = 4 \\sin 30^{\\circ} = 2. Therefore, the equation of the parabola is: y^{2} = 4x."
  },
  {
    "text": "Given that the hyperbola $\\frac{x^{2}}{m} + \\frac{y^{2}}{n} = 1$ ($m, n \\in \\mathbb{R}$) with eccentricity $2$ has its right focus coinciding with the focus of the parabola $y^{2} = 4x$,  \nthen $\\frac{m}{n} = $?",
    "fact_expressions": "G: Hyperbola;n: Real;m: Real;H: Parabola;Expression(G) = (y^2/n + x^2/m = 1);Expression(H) = (y^2 = 4*x);Eccentricity(G)=2;RightFocus(G)=Focus(H)",
    "query_expressions": "m/n",
    "answer_expressions": "1/3",
    "fact_spans": "[[[10, 60]], [[13, 60]], [[13, 60]], [[65, 79]], [[10, 60]], [[65, 79]], [[2, 60]], [[10, 84]]]",
    "query_spans": "[[[87, 102]]]",
    "process": ""
  },
  {
    "text": "If the eccentricity of the hyperbola $x^{2}-\\frac{y^{2}}{m}=1$ is $2$, then what is the value of $m$?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2 - y^2/m = 1);Eccentricity(G) = 2",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 29]], [[39, 42]], [[1, 29]], [[1, 37]]]",
    "query_spans": "[[[39, 46]]]",
    "process": "According to the problem, this question examines the basic knowledge of hyperbolas; the key is to properly match the given equation with the standard equation."
  },
  {
    "text": "If the hyperbola $\\frac{x^{2}}{3 m}-\\frac{y^{2}}{m}=1$ has its foci on the $y$-axis and the focal distance is $8$, then the real number $m$=?",
    "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (x^2/(3*m) - y^2/m = 1);PointOnCurve(Focus(G), yAxis);FocalLength(G) = 8",
    "query_expressions": "m",
    "answer_expressions": "-4",
    "fact_spans": "[[[2, 42]], [[60, 65]], [[2, 42]], [[2, 51]], [[2, 58]]]",
    "query_spans": "[[[60, 67]]]",
    "process": "According to the problem, the hyperbola $\\frac{x^{2}}{3m}-\\frac{y^{2}}{m}=1$ has its foci on the $y$-axis, then $\\frac{y^{2}}{m-\\frac{x^2}{3m}}=1$, and the semi-focal distance is 4, so $-m-3m=16$, $\\therefore m=-4$."
  },
  {
    "text": "On the parabola $y^{2}=-12 x$, what are the coordinates of the points whose distance from the focus is equal to $9$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -12*x);P:Point;Distance(P,Focus(G))=9;PointOnCurve(P,G)",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(-6, \\pm 6\\sqrt{2})",
    "fact_spans": "[[[1, 17]], [[1, 17]], [[31, 32]], [[1, 32]], [[0, 32]]]",
    "query_spans": "[[[31, 37]]]",
    "process": "From the equation $ y^{2} = -12x $, we know the focus is $ F(-3,0) $, the directrix is $ l: x = 3 $. Let the desired point be $ P(x,y) $. Then by definition, $ |PF| = 3 - x $. Since $ |PF| = 9 $, we have $ 3 - x = 9 $, so $ x = -6 $. Substituting into $ y^{2} = -12x $, we get $ y = \\pm 6\\sqrt{2} $. Therefore, the coordinates of the desired point are $ (-6, 6\\sqrt{2}) $ or $ (-6, -6\\sqrt{2}) $."
  },
  {
    "text": "The line intersects the hyperbola $x^{2}-4 y^{2}=4$ at points $A$ and $B$. If the midpoint of segment $AB$ has coordinates $(8 , 1)$, then the equation of the line is?",
    "fact_expressions": "A: Point;B: Point;G: Hyperbola;H: Line;Expression(G) = (x^2 - 4*y^2 = 4);Intersection(H, G) = {A,B};Coordinate(MidPoint(LineSegmentOf(A,B))) =(8,1)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "2*x-y-15=0",
    "fact_spans": "[[[25, 28]], [[30, 33]], [[3, 23]], [[0, 2], [60, 62]], [[3, 23]], [[0, 35]], [[37, 58]]]",
    "query_spans": "[[[60, 67]]]",
    "process": ""
  },
  {
    "text": "Given that the line $l$ is the directrix of the parabola $y=4x^{2}$, then the distance from point $A(1,4)$ to $l$ is?",
    "fact_expressions": "l: Line;G: Parabola;Expression(G) = (y = 4*x^2);Directrix(G) = l;A: Point;Coordinate(A) = (1, 4)",
    "query_expressions": "Distance(A, l)",
    "answer_expressions": "65/16",
    "fact_spans": "[[[2, 7], [37, 40]], [[8, 22]], [[8, 22]], [[2, 25]], [[27, 36]], [[27, 36]]]",
    "query_spans": "[[[27, 45]]]",
    "process": "Analysis: First find the directrix, then calculate the distance from A to the directrix. Given $ y = 4x^{2} $, rewrite it into standard form: $ x^{2} = \\frac{1}{4}y $, so the equation of the directrix is $ y = -\\frac{1}{16} $, hence the distance from $ A(1,4) $ to $ l $ is $ \\frac{65}{16} $."
  },
  {
    "text": "Given that the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has an eccentricity of $2$, and its foci coincide with the foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, what are the coordinates of the foci of the hyperbola? What is the equation of its asymptotes?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Ellipse;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2/25 + y^2/9 = 1);Eccentricity(G) = 2;Focus(G) = Focus(H)",
    "query_expressions": "Coordinate(Focus(G));Expression(Asymptote(G))",
    "answer_expressions": "(pm*4, 0)\nsqrt(3)*x+pm*y=0",
    "fact_spans": "[[[2, 48], [106, 109]], [[5, 48]], [[5, 48]], [[60, 98]], [[2, 48]], [[60, 98]], [[2, 56]], [[2, 103]]]",
    "query_spans": "[[[106, 116]], [[106, 123]]]",
    "process": ""
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{a^{2}}=1$ and the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{2}=1$ have the same foci; then the value of $a$ is?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Ellipse;Expression(G) = (-y^2/2 + x^2/a = 1);Expression(H) = (x^2/4 + y^2/a^2 = 1);Focus(H) = Focus(G)",
    "query_expressions": "a",
    "answer_expressions": "1",
    "fact_spans": "[[[42, 80]], [[88, 91]], [[0, 41]], [[42, 80]], [[0, 41]], [[0, 86]]]",
    "query_spans": "[[[88, 95]]]",
    "process": ""
  },
  {
    "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ with focus on the $x$-axis is $\\frac{1}{2}$, then $m$=?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/4 + y^2/m = 1);PointOnCurve(Focus(G), xAxis);Eccentricity(G)=1/2",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[10, 47]], [[67, 70]], [[10, 47]], [[1, 47]], [[10, 65]]]",
    "query_spans": "[[[67, 72]]]",
    "process": "Given $ a^{2}=4, b^{2}=m $, then $ c^{2}=4-m $, so $ e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{4-m}{4}=\\frac{1}{4} $, solving gives $ m=3 $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through point $F_{1}$ intersects the ellipse at points $A$ and $B$, satisfying $|A F_{1}|=4|B F_{1}|$ and $\\angle A F_{2} F_{1}=90^{\\circ}$. Then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;G: Line;A: Point;F1: Point;B: Point;F2: Point;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, G);Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, F1)) = 4*Abs(LineSegmentOf(B, F1));AngleOf(A, F2, F1) = ApplyUnit(90, degree);b:Number;a:Number",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(21)/7",
    "fact_spans": "[[[2, 47], [85, 87], [158, 163]], [[82, 84]], [[88, 91]], [[56, 63], [73, 81]], [[92, 95]], [[64, 71]], [[2, 47]], [[2, 71]], [[2, 71]], [[72, 84]], [[82, 95]], [[98, 121]], [[123, 156]], [[2, 47]], [[2, 47]]]",
    "query_spans": "[[[158, 169]]]",
    "process": "Since $\\angle AF_{2}F_{1}=90^{\\circ}$, then $A(c,\\frac{b^{2}}{a})$ (without loss of generality, take $A$ in the first quadrant), and $F_{1}(-c,0)$, $|AF_{1}|=4|BF_{1}|$, $\\begin{cases}x_{A}=-4y_{B}\\end{cases}$ $\\begin{cases}x_{B}=-\\frac{3}{2}c\\\\y_{B}=-\\frac{b^{2}}{4a}\\end{cases}$, so $B(-\\frac{3}{2}c,-\\frac{b^{2}}{4a})$. Therefore, $\\frac{9c^{2}}{4a^{2}}+\\frac{b^{4}}{16a^{2}b^{2}}=1$, simplifying yields $3a^{2}=7c^{2}$, $e=\\frac{c}{a}=\\frac{\\sqrt{21}}{7}$."
  },
  {
    "text": "Draw a perpendicular line $PD$ from point $P$ on the ellipse $2x^{2} + y^{2} = 4$ to the $x$-axis, with $D$ as the foot of the perpendicular. Then, as point $P$ moves along the ellipse, the equation of the trajectory of the midpoint $M$ of segment $PD$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (2*x^2 + y^2 = 4);D: Point;P: Point;M: Point;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P,D),xAxis);FootPoint(LineSegmentOf(P,D),xAxis)=D;MidPoint(LineSegmentOf(P,D))=M;PointOnCurve(P,LineSegmentOf(P,D))",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2/2+y^2=1",
    "fact_spans": "[[[1, 20], [54, 56]], [[1, 20]], [[41, 44]], [[49, 53], [21, 25]], [[70, 73]], [[1, 25]], [[26, 38]], [[26, 47]], [[61, 73]], [[0, 38]]]",
    "query_spans": "[[[70, 80]]]",
    "process": ""
  },
  {
    "text": "The standard equation of an ellipse with foci on the $x$-axis, focal length equal to $4$, and passing through $P(3,-2 \\sqrt{6})$ is?",
    "fact_expressions": "G: Ellipse;PointOnCurve(Focus(G), xAxis);FocalLength(G) = 4;P: Point;Coordinate(P) = (3, -2*sqrt(6));PointOnCurve(P, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/36+y^2/32=1",
    "fact_spans": "[[[40, 42]], [[0, 42]], [[9, 42]], [[21, 39]], [[21, 39]], [[19, 42]]]",
    "query_spans": "[[[40, 49]]]",
    "process": "Since the foci of the ellipse lie on the x-axis, we set the standard equation of the ellipse as \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0). From the given conditions, we have {\\frac{2c=4}{a^{2}}+\\frac{(-2\\sqrt{6})^{2}}{b^{2}}=1, solving gives a^{2}=36, b^{2}=32, so the ellipse's \\overrightarrow{n}=b^{2}+c^{2}_{\\frac{x2}{36}}+\\frac{y^{2}}{32}=1,"
  },
  {
    "text": "Given that the line $l$ intersects the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1$ at points $A$ and $B$, and the midpoint of segment $AB$ is $M$. Let the slope of line $l$ be $k_{1}$ $(k_{1} \\neq 0)$, and the slope of line $OM$ be $k_{2}$. Then $k_{1} k_{2}=$?",
    "fact_expressions": "l: Line;G: Hyperbola;M: Point;O: Origin;A: Point;B: Point;k1: Number;k2: Number;Expression(G) = (x^2/9 - y^2/4 = 1);Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = M;Slope(l) = k1;Negation(k1 = 0);Slope(LineOf(O, M)) = k2",
    "query_expressions": "k1*k2",
    "answer_expressions": "4/9",
    "fact_spans": "[[[2, 7], [76, 81]], [[8, 46]], [[70, 73]], [[109, 114]], [[49, 52]], [[53, 56]], [[85, 106]], [[118, 125]], [[8, 46]], [[2, 58]], [[59, 73]], [[76, 106]], [[85, 106]], [[107, 125]]]",
    "query_spans": "[[[128, 143]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then the coordinates of point $ M $ are $ \\left( \\frac{x_{1}+x_{2}}{2}, \\frac{y_{1}+y_{2}}{2} \\right) $. Using the point difference method, the value of $ k_{1}k_{2} $ can be obtained. According to the problem, let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then the coordinates of point $ M $ are $ \\left( \\frac{x_{1}+x_{2}}{2}, \\frac{y_{1}+y_{2}}{2} \\right) $. Therefore, $ \\frac{x_{1}^{2}}{9} - \\frac{y_{1}^{2}}{4} = 1 $, $ \\frac{x_{2}^{2}}{9} - \\frac{y_{2}^{2}}{4} = 1 $. Taking the difference gives $ \\frac{(x_{1}-x_{2})(x_{1}+x_{2})}{9} - \\frac{(y_{1}-y_{2})(y_{1}+y_{2})}{4} = 0 $, that is, $ \\frac{x_{1}-x_{2}}{y_{1}-y_{2}} = \\frac{9}{4} \\cdot \\frac{y_{1}+y_{2}}{x_{1}+x_{2}} $, i.e., $ \\frac{1}{k_{1}} = \\frac{9}{4} k_{2} $, thus $ k_{1}k_{2} = \\frac{4}{9} $."
  },
  {
    "text": "Given that the circle $x^{2}+y^{2}-6 x-7=0$ is tangent to the directrix of the parabola $y^{2}=2 a x$, then the value of the real number $a$ is?",
    "fact_expressions": "G: Parabola;a: Real;H: Circle;Expression(G) = (y^2 = 2*(a*x));Expression(H) = (-6*x + x^2 + y^2 - 7 = 0);IsTangent(Directrix(G),H)",
    "query_expressions": "a",
    "answer_expressions": "{-14,2}",
    "fact_spans": "[[[25, 41]], [[48, 53]], [[2, 24]], [[25, 41]], [[2, 24]], [[2, 46]]]",
    "query_spans": "[[[48, 57]]]",
    "process": "The circle $ x^{2}+y^{2}-6x-7=0 $, or $ (x-3)^{2}+y^{2}=16 $, has center $ (3,0) $ and radius 4. The directrix of the parabola $ y^{2}=2ax $ is $ x=-\\frac{a}{2} $. Since the circle $ x^{2}+y^{2}-6x-7=0 $ is tangent to the directrix of the parabola $ y^{2}=2ax $, we have $ 3+\\frac{a}{2}=4 $ or $ -\\frac{a}{2}-3=4 $. Solving gives $ a=2 $ or $ a=-14 $."
  },
  {
    "text": "The line passing through the left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ and perpendicular to the $x$-axis intersects the hyperbola at points $M$ and $N$, $O$ is the center of the hyperbola, $\\overrightarrow{O M} \\cdot \\overrightarrow{O N}=0$. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;O: Origin;M: Point;N: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(LeftFocus(G), H);IsPerpendicular(H,xAxis);Intersection(H, G) = {M, N};Center(G) = O;DotProduct(VectorOf(O, M), VectorOf(O, N)) = 0",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(\\sqrt{5}+1)/2",
    "fact_spans": "[[[1, 57], [73, 76], [93, 96], [154, 157]], [[4, 57]], [[4, 57]], [[70, 72]], [[89, 92]], [[79, 82]], [[83, 86]], [[4, 57]], [[4, 57]], [[1, 57]], [[0, 72]], [[62, 72]], [[70, 88]], [[89, 99]], [[100, 151]]]",
    "query_spans": "[[[154, 163]]]",
    "process": ""
  },
  {
    "text": "Given points $A$ and $B$ lie on the parabola $\\Gamma$: $y^{2}=4x$, point $M$ lies on the directrix of $\\Gamma$, and the midpoints of segments $MA$ and $MB$ both lie on the parabola $\\Gamma$. Let line $AB$ intersect the $y$-axis at point $N(0, n)$. Then the minimum value of $|n|$ is?",
    "fact_expressions": "Gamma: Parabola;Expression(Gamma) = (y^2 = 4*x);A: Point;B: Point;PointOnCurve(A, Gamma);PointOnCurve(B, Gamma);M: Point;PointOnCurve(M, Directrix(Gamma));PointOnCurve(MidPoint(LineSegmentOf(M, A)), Gamma);PointOnCurve(MidPoint(LineSegmentOf(M, B)), Gamma);N: Point;n: Number;Coordinate(N) = (0, n);Intersection(LineOf(A, B), yAxis) = N",
    "query_expressions": "Min(Abs(n))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[11, 35], [74, 85], [42, 50]], [[11, 35]], [[2, 6]], [[7, 10]], [[2, 36]], [[7, 36]], [[37, 41]], [[37, 54]], [[55, 86]], [[55, 86]], [[102, 112]], [[114, 119]], [[102, 112]], [[88, 112]]]",
    "query_spans": "[[[114, 125]]]",
    "process": ""
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the common foci of the ellipse $C_{1}$ and the hyperbola $C_{2}$, $e_{1}$, $e_{2}$ are the eccentricities of the curves $C_{1}$, $C_{2}$ respectively, $P$ is a common point of the curves $C_{1}$, $C_{2}$, if $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$, and $e_{2} \\in[\\sqrt{3}, 2]$, then $e_{1} \\in$?",
    "fact_expressions": "C1:Ellipse;C2:Hyperbola;F1: Point;P: Point;F2: Point;e1:Number;e2:Number;Focus(C2)={F1,F2};Eccentricity(C1)=e1;Eccentricity(C2)=e2;OneOf(Intersection(C1,C2))=P;AngleOf(F1, P, F2) = pi/3;In(e2,[sqrt(3),2]);Focus(C1)={F1,F2}",
    "query_expressions": "Range(e1)",
    "answer_expressions": "[2*sqrt(13)/13, sqrt(3)/3]",
    "fact_spans": "[[[16, 25], [62, 71], [88, 97]], [[26, 36], [72, 79], [98, 105]], [[0, 7]], [[84, 87]], [[8, 15]], [[42, 49]], [[51, 59]], [[0, 41]], [[42, 83]], [[42, 83]], [[84, 111]], [[113, 149]], [[151, 175]], [[0, 41]]]",
    "query_spans": "[[[177, 189]]]",
    "process": "As shown in the figure, let the standard equation of hyperbola $ C_{2} $ be: $ \\frac{x^{2}}{a_{1}} - \\frac{y^{2}}{b_{1}^{2}} = 1 $ $ (a_{1}, b_{1} > 0) $, with semi-focal length $ c $. Let the ellipse $ C_{1}: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ $ (a > b > 0) $, with semi-focal length $ c $. Without loss of generality, assume point $ P $ lies in the first quadrant, and let $ |PF_{1}| = m $, $ |PF_{2}| = n $. $ \\therefore m + n = 2a $, $ m \\cdot n = 2a_{1} $. $ \\Rightarrow m = a + a_{1} $, $ n = a - a_{1} $. In $ \\triangle PF_{1}F_{2} $, by the law of cosines we obtain: $ 4c^{2} = m^{2} + n^{2} - 2mn\\cos\\frac{\\pi}{3} $. $ 4c^{2} = a^{2} + 3a_{1}^{2} $. Dividing both sides by $ c^{2} $, we get $ \\frac{1}{e_{1}^{2}} + \\frac{3}{e_{2}^{2}} = 4 $. $ \\because \\frac{1}{e_{2}} \\in \\left[\\frac{1}{4}, \\frac{1}{3}\\right] $, $ \\therefore \\frac{1}{e_{1}} \\in \\left[3, \\frac{13}{4}\\right] $."
  },
  {
    "text": "Let $F_{1}$ be a focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $A_{1}$, $A_{2}$ be the two vertices of $C$, and suppose there exists a point $M$ on $C$ such that $M F_{1}$ is tangent to the circle with diameter $A_{1} A_{2}$ at point $N$, and $N$ is the midpoint of segment $M F_{1}$. Then the asymptotes of $C$ have equations?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;OneOf(Focus(C)) = F1;A1: Point;A2: Point;Vertex(C) = {A1,A2};M: Point;PointOnCurve(M, C);G: Circle;IsDiameter(LineSegmentOf(A1,A2),G);TangentPoint(LineSegmentOf(M, F1), G) = N;MidPoint(LineSegmentOf(M, F1)) = N;N:Point",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y = pm*2*x",
    "fact_spans": "[[[9, 71], [93, 96], [102, 105], [173, 176]], [[9, 71]], [[16, 71]], [[16, 71]], [[16, 71]], [[16, 71]], [[1, 8]], [[1, 76]], [[77, 84]], [[85, 92]], [[77, 101]], [[110, 113]], [[102, 113]], [[144, 145]], [[126, 145]], [[116, 152]], [[154, 172]], [[148, 152], [154, 157]]]",
    "query_spans": "[[[173, 184]]]",
    "process": "By geometric conditions, we have $ON \\parallel MF_{2}$, $|ON| = \\frac{1}{2}|MF_{2}|$. Then, using the definition of the hyperbola combined with the Pythagorean theorem, the relationship between $a$ and $b$ can be obtained, thus giving the equation of the asymptotes. Let the other focus of the hyperbola be $F_{2}$. Then, by the midline of the triangle: $ON \\parallel MF_{2}$, $ON = \\frac{1}{2}|MF_{2}|$. Since $MF$ is tangent to the circle with diameter $A_{1}A_{2}$ at point $N$, it follows that $ON = a$, $|MF_{2}| = 2a$, $ON \\perp MF$, so $MF_{2} \\perp MF_{1}$. By the definition of the hyperbola, $MF_{1} - |MF_{2}| = 2a$, then $MF_{1} = 4a$. In the right triangle $\\triangle MF_{1}F_{2}$, $|F_{1}F_{2}|^{2} = |F_{1}M|^{2} + |F_{2}M|^{2}$, that is $(2c)^{2} = (2a)^{2} + (4a)^{2}$, so $4c^{2} = 20a^{2}$, i.e., $c^{2} = 5a^{2}$. Therefore, $c^{2} = a^{2} + b^{2} = 5a^{2}$, hence $b^{2} = 4a^{2}$, i.e., $b = 2a$. Thus, the asymptotes of the hyperbola are: $y = \\pm \\frac{b}{a}x = \\pm 2x$."
  },
  {
    "text": "If a focus of the hyperbola $x^{2}+k y^{2}=1$ is $(3 , 0)$, then the real number $k$=?",
    "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (x^2+k*y^2=1);Coordinate(OneOf(Focus(G))) = (3, 0)",
    "query_expressions": "k",
    "answer_expressions": "-1/8",
    "fact_spans": "[[[1, 21]], [[38, 43]], [[1, 21]], [[1, 36]]]",
    "query_spans": "[[[38, 45]]]",
    "process": ""
  },
  {
    "text": "Given that $P\\left(\\frac{2 \\sqrt{10}}{5}, y_{0}\\right)$ is a point on the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with eccentricity $e=\\frac{\\sqrt{3}}{2}$, and the line $y=-x$ intersects $C$ at points $A$ and $B$ ($A$ and $B$ do not coincide with $P$). If $|P A|=|P B|$, then the equation of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;P: Point;A: Point;B: Point;a > b;b > 0;y0:Number;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y = -x);Coordinate(P) = (2*sqrt(10)/5, y0);PointOnCurve(P,C);Eccentricity(C)=e;Intersection(G, C) = {B, A};Negation(A=P);Negation(B=P);Abs(LineSegmentOf(P, A)) = Abs(LineSegmentOf(P, B));e:Number;e=sqrt(3)/2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/8+y^2/2=1",
    "fact_spans": "[[[62, 119], [132, 135], [183, 188]], [[69, 119]], [[69, 119]], [[123, 131]], [[2, 35], [160, 163]], [[138, 141], [149, 152]], [[142, 145], [154, 157]], [[69, 119]], [[69, 119]], [[2, 35]], [[62, 119]], [[123, 131]], [[2, 35]], [[2, 122]], [[36, 119]], [[123, 148]], [[149, 165]], [[149, 165]], [[168, 181]], [[39, 61]], [[39, 61]]]",
    "query_spans": "[[[183, 193]]]",
    "process": "The midpoint of A and B is the origin O. Then, according to |PA| = |PB|, ∴ PO ⊥ AB. $\\frac{3}{2}$, ∴ a = 2b', ∴ let the equation of ellipse C be $\\frac{x^2}{4b^{2}} + \\frac{y^{2}}{b^{2}} = 1$. Substituting $P\\left(\\frac{2\\sqrt{10}}{5}, \\frac{2\\sqrt{10}}{5}\\right)$, solve for $b^{2} = 2$. Therefore, the equation of ellipse C is $\\frac{x^{2}}{8} + \\frac{y^{2}}{2} = 1$."
  },
  {
    "text": "If a point $M$ on the parabola $y^{2}=2 px(p>0)$ has distances of $10$ and $6$ to the line $x=-\\frac{p}{2}$ and to the axis of symmetry, respectively, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;p>0;M:Point;Expression(G) = (y^2 = 2*p*x);Expression(H) = (x = -p/2);PointOnCurve(M,G);Distance(M,H)=10;Distance(M,SymmetryAxis(G))=6",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=4*x,y^2=36*x}",
    "fact_spans": "[[[1, 21], [68, 71]], [[4, 21]], [[28, 46]], [[4, 21]], [[24, 27]], [[1, 21]], [[28, 46]], [[1, 27]], [[24, 65]], [[1, 65]]]",
    "query_spans": "[[[68, 76]]]",
    "process": ""
  },
  {
    "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ have left and right foci $F_{1}$ and $F_{2}$, respectively. A line passing through $F_{1}$ intersects the left branch of hyperbola $C$ at points $M$ and $N$. If $|M F_{2}|=|F_{1} F_{2}|$ and $2|M F_{1}|=|N F_{1}|$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;M: Point;F2: Point;F1: Point;N: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, G) = True;Intersection(G, LeftPart(C)) = {M, N};Abs(LineSegmentOf(M, F2)) = Abs(LineSegmentOf(F1, F2));2*Abs(LineSegmentOf(M, F1)) = Abs(LineSegmentOf(N, F1))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[1, 62], [99, 105], [171, 177]], [[9, 62]], [[9, 62]], [[96, 98]], [[109, 112]], [[79, 86]], [[71, 78], [88, 95]], [[113, 116]], [[8, 62]], [[8, 62]], [[1, 62]], [[1, 86]], [[1, 86]], [[87, 98]], [[87, 118]], [[120, 145]], [[147, 169]]]",
    "query_spans": "[[[171, 183]]]",
    "process": "Take the midpoint $ P $ of $ F_{1}M $, connect $ PF_{2} $, $ NF_{2} $. Using the definition of hyperbola and the given conditions, we obtain $ |NP|^{2}-|MP|^{2}=|NF_{2}|^{2}-|MF_{2}|^{2} $. Simplify and rearrange to get the result. Take the midpoint $ P $ of $ F_{1}M $, connect $ PF_{2} $, $ NF_{2} $. Since $ |MF_{2}|=|F_{1}F_{2}|=2c $, we have $ PF_{2}\\bot MN $. According to the definition of hyperbola: $ |MF_{1}|=|MF_{2}|-2a=2c-2a $, then $ |MP|=|F_{1}P|=c-a $. Also $ |NF_{1}|=2|MF_{1}| $, so $ |NF_{1}|=4(c-a) $, thus $ |NF_{2}|=4c-2a $. In right triangle $ \\triangle NPF_{2} $, $ |NP|^{2}+|PF_{2}|^{2}=|NF_{2}|^{2} $. In right triangle $ \\triangle MPF_{2} $, $ |MP|^{2}+|PF_{2}|^{2}=|MF_{2}|^{2} $. Therefore, $ |NP|^{2}-|MP|^{2}=|NF_{2}|^{2}-|MF_{2}|^{2} $, that is $ (|NF_{1}|+|F_{1}P|)^{2}-|MP|^{2}=|NF_{2}|^{2}-|MF_{2}|^{2} $, which is $ [5(c-a)]^{2}-(c-a)^{2}=(4c-2a)^{2}-(2c)^{2} $. Rearranging yields $ (c-a)(3c-5a)=0 $, solving gives $ e=\\frac{c}{a}=1 $ or $ \\frac{5}{2} $. Since $ e>1 $, we have $ e=\\frac{5}{3} $."
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ are $F_{1}$, $F_{2}$, and $P$ is a point on the ellipse. If $|P F_{1}|=5$, then $\\cos \\angle F_{1} P F_{2}$=?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 + y^2/4 = 1);Focus(G)={F1,F2};PointOnCurve(P,G);Abs(LineSegmentOf(P, F1)) = 5",
    "query_expressions": "Cos(AngleOf(F1, P, F2))",
    "answer_expressions": "3/5",
    "fact_spans": "[[[0, 37], [61, 63]], [[57, 60]], [[41, 48]], [[49, 56]], [[0, 37]], [[0, 56]], [[57, 66]], [[68, 81]]]",
    "query_spans": "[[[84, 113]]]",
    "process": "According to the definition of the ellipse: |PF₁| + |PF₂| = 2a, |F₁F₂| = 2c = 2√5. In triangle AF₁PF₂, applying the cosine law gives the answer. ∵ The ellipse x²/9 + y²/4 = 1 yields: a = 3, b = 2, c = √5. According to the definition of the ellipse: |PF₁| + |PF₂| = 2a = 6, |F₁F₂| = 2c = 2√5. It follows that 5 + |PF₂| = 2a, solving gives: |PF₂| = 2a - 5 = 6 - 5 = 1. In triangle AF₁PF₂, by the cosine law: cos∠F₁PF₂ = (|PF₁|² + |PF₂|² - |F₁F₂|²) / (2|PF₁|·|PF₂|) = (25 + 1 - 20) / (2×5×1) = 3/5,"
  },
  {
    "text": "Given the circle $C$: $(x-3)^{2}+y^{2}=1$, and point $M$ moving along the parabola $T$: $y^{2}=4x$. From point $M$, draw lines $l_{1}$, $l_{2}$ tangent to circle $C$, with points of tangency $A$ and $B$, respectively. Then the range of $|AB|$ is?",
    "fact_expressions": "C: Circle;Expression(C) = (y^2 + (x - 3)^2 = 1);M: Point;T: Parabola;Expression(T) = (y^2 = 4*x);PointOnCurve(M, T);l1: Line;l2: Line;TangentPoint(l1, C) = A;TangentPoint(l2, C) = B;A: Point;B: Point;TangentOfPoint(M,C)={l1,l2}",
    "query_expressions": "Range(Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "[\\sqrt{14}/2,2)",
    "fact_spans": "[[[2, 26], [81, 85]], [[2, 26]], [[27, 31], [56, 60]], [[32, 51]], [[32, 51]], [[27, 52]], [[61, 71]], [[73, 80]], [[61, 100]], [[61, 100]], [[93, 96]], [[97, 100]], [55, 86]]",
    "query_spans": "[[[102, 116]]]",
    "process": "Let $ M(x_{0},y_{0}) $, then $ y_{0}^{2}=4x_{0} $, $ C(3,0) $, $ |MC|=\\sqrt{(x_{0}-3)^{2}+y_{0}^{2}} $, the length of the tangent $ |MA|=\\sqrt{|MC|^{2}-1^{2}}=\\sqrt{(x_{0}-3)^{2}+y_{0}^{2}-1} $. Since $ AB\\perp MC $ and $ MC $ bisects segment $ AB $, it follows that $ |AB|=2\\times\\frac{|MA|}{|MC|}=\\frac{2\\sqrt{(x_{0}-3)^{2}+y_{0}^{2}-1}}{\\sqrt{(x_{0}-3)^{2}+y_{0}^{2}}}=2\\sqrt{1-\\frac{1}{(x_{0}-3)^{2}+8}} $. Because $ x_{0}\\geqslant0 $, we have $ (x_{0}-3)^{2}+8\\geqslant8 $, so $ \\frac{7}{8}\\leqslant1-\\frac{1}{(x_{0}-3)^{2}+8}<1 $, hence $ \\frac{\\sqrt{14}}{2}\\leqslant|AB|<2 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, one of its directrices coincides with the directrix of the parabola $y^{2}=4 x$. When $\\frac{a^{4}+4}{\\sqrt{a^{2}+b^{2}}}$ attains its minimum value, what is the eccentricity of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;G: Parabola;Expression(G) = (y^2 = 4*x);OneOf(Directrix(C)) = Directrix(G);WhenMin((a^4 + 4)/sqrt(a^2 + b^2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 63], [133, 139]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[69, 83]], [[69, 83]], [[2, 88]], [[89, 132]]]",
    "query_spans": "[[[133, 145]]]",
    "process": "The directrix equation of the parabola \\( y^{2} = 4x \\) is \\( x = -1 \\), and the directrix equations of the hyperbola \\( C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 \\) (\\( a > 0 \\), \\( b > 0 \\)) are \\( x = \\pm \\frac{a^{2}}{c} \\). Therefore, \\( \\frac{a^{2}}{c} = 1 \\), that is, \\( a^{2} = c \\), so \\( \\frac{a^{2} + 4}{\\sqrt{a^{2} + b^{2}}} = \\frac{c^{2} + 4}{c} = c + \\frac{4}{c} \\geqslant 4 \\), with equality holding if and only if \\( c = \\frac{4}{c} = 2 \\). Thus, \\( a^{2} = c = 2 \\), solving gives \\( a = \\sqrt{2} \\). Therefore, the eccentricity of the hyperbola is \\( e = \\frac{c}{a} = \\frac{2}{\\sqrt{2}} = \\sqrt{2} \\)."
  },
  {
    "text": "Given that the distance from point $A(2,-3)$ to the focus of the parabola $y^{2}=2 p x$ ($p>0$) is $5$, then $p=$?",
    "fact_expressions": "A: Point;Coordinate(A) = (2, -3);G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;Distance(A, Focus(G)) = 5",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 12]], [[2, 12]], [[13, 34]], [[13, 34]], [[46, 49]], [[16, 34]], [[2, 44]]]",
    "query_spans": "[[[46, 51]]]",
    "process": "The solution process is omitted"
  },
  {
    "text": "Through the focus $F$ of the parabola $x^{2}=4 y$, draw a line intersecting the parabola at points $P_{1}(x_{1}, y_{1})$ and $P_{2}(x_{2}, y_{2})$. If $y_{1}+y_{2}=6$, find the value of $|P_{1} P_{2}|$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);H: Line;F: Point;Focus(G) = F;PointOnCurve(F, H);P1: Point;P2: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(P1) = (x1, y1);Coordinate(P2) = (x2, y2);Intersection(H, G) = {P1, P2};y1 + y2 = 6",
    "query_expressions": "Abs(LineSegmentOf(P1, P2))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 15], [25, 28]], [[1, 15]], [[22, 24]], [[18, 21]], [[1, 21]], [[0, 24]], [[29, 50]], [[51, 70]], [[29, 50]], [[51, 70]], [[29, 50]], [[51, 70]], [[29, 50]], [[51, 70]], [[22, 72]], [[74, 89]]]",
    "query_spans": "[[[91, 109]]]",
    "process": ""
  },
  {
    "text": "Given that one focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{2}=1$ is $(2,0)$, what is the standard equation of the ellipse?",
    "fact_expressions": "G: Ellipse;a: Number;H: Point;Expression(G) = (y^2/2 + x^2/a^2 = 1);Coordinate(H) = (2, 0);OneOf(Focus(G))=H",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/6 + y^2/2 = 1",
    "fact_spans": "[[[2, 43], [58, 60]], [[4, 43]], [[49, 56]], [[2, 43]], [[49, 56]], [[2, 56]]]",
    "query_spans": "[[[58, 67]]]",
    "process": "From the coordinates of the foci of the ellipse, it is known that the foci lie on the x-axis. Then, using $a^{2}=b^{2}+c^{2}$, the value of $a$ can be found, thus obtaining the answer. Since one focus of $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{2}=1$ is $(2,0)$, the foci of the ellipse lie on the x-axis. Therefore, $a^{2}=b^{2}+c^{2}=2+2^{2}=6$. Hence, the standard equation of the ellipse is $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$."
  },
  {
    "text": "Given the parabola $ C $: $ y^{2} = 2 p x $ ($ p > 0 $) with focus $ F $, the line $ l $: $ 2x + 2y - p = 0 $ intersects the parabola $ C $ at points $ A $ and $ B $, and $ |BF| = 1 + |AF| $, then $ |AB| = $?",
    "fact_expressions": "l: Line;C: Parabola;p: Number;B: Point;F: Point;A: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;Expression(l)=(-p + 2*x + 2*y = 0);Intersection(l, C) = {A, B};Abs(LineSegmentOf(B, F)) = Abs(LineSegmentOf(A, F)) + 1",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[35, 54]], [[2, 27], [55, 61]], [[9, 27]], [[67, 70]], [[31, 34]], [[63, 66]], [[9, 27]], [[2, 27]], [[2, 34]], [[35, 54]], [[35, 72]], [[74, 89]]]",
    "query_spans": "[[[91, 100]]]",
    "process": "It is easy to see that line $ l $ passes through point $ F(\\frac{p}{2},0) $, and its angle of inclination is $ 135^{\\circ} $. As shown in the figure, let $ M, N $ be the projections of $ A, B $ on the directrix respectively. Draw $ AH\\bot BN $, with foot of perpendicular at $ H $, then $ \\angle ABH = 45^{\\circ} $, $ |BF| - |AF| = |BN| - |AM| = |BH| = 1 $, so $ |AB| = \\sqrt{2} $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a, b>0)$ with left and right foci $F_{1}$, $F_{2}$. Draw a line perpendicular to the $x$-axis through $F_{2}$ intersecting $C$ at points $A$, $B$. Line $F_{1} B$ intersects the $y$-axis at $D$. If $A D \\perp F_{1} B$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;A: Point;B: Point;L: Line;PointOnCurve(F2,L) = True;IsPerpendicular(L,xAxis) = True;Intersection(L,C) = {A,B};D: Point;Intersection(LineSegmentOf(F1,B),yAxis) = D;IsPerpendicular(LineSegmentOf(A,D),LineSegmentOf(F1,B)) = True",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 61], [102, 105], [162, 168]], [[2, 61]], [[10, 61]], [[10, 61]], [[10, 61]], [[10, 61]], [[67, 74]], [[76, 83], [86, 93]], [[2, 83]], [[2, 83]], [[108, 111]], [[112, 115]], [], [[85, 101]], [[85, 101]], [[85, 117]], [[135, 138]], [[118, 138]], [[141, 160]]]",
    "query_spans": "[[[162, 174]]]",
    "process": ""
  },
  {
    "text": "Given that $A$ and $B$ are two points on the parabola $y^{2}=8x$, if the distance from the midpoint of segment $AB$ to the directrix of the parabola is $5$, then a possible equation of line $AB$ is (this problem has multiple correct answers; any answer satisfying the conditions is acceptable)?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;Expression(G) = (y^2 = 8*x);PointOnCurve(A, G);PointOnCurve(B, G);Distance(MidPoint(LineSegmentOf(A,B)),Directrix(G))=5",
    "query_expressions": "Expression(LineOf(A,B))",
    "answer_expressions": "x=3",
    "fact_spans": "[[[10, 24], [40, 43]], [[2, 5]], [[6, 9]], [[10, 24]], [[2, 27]], [[2, 27]], [[29, 53]]]",
    "query_spans": "[[[55, 85]]]",
    "process": "From the given information, the directrix of the parabola $ y^{2} = 8x $ is $ l: x = -2 $. Since the distance from the midpoint of segment $ AB $ to the directrix of the parabola is 5, the midpoint of segment $ AB $ is $ M(3, y_{0}) $. When the slope of $ AB $ does not exist, $ x = 3 $ satisfies the condition. When the slope of line $ AB $ exists and is not zero, let the equation of line $ AB $ be $ x = ky + b $ ($ k \\neq 0 $). Substituting into $ y^{2} = 8x $, we obtain $ y^{2} - 8ky - 8b = 0 $. Thus, $ 8k = 2y_{0} = \\frac{2(3 - b)}{k} $, so $ b = 3 - 4k^{2} $. Therefore, the equation of line $ AB $ is $ x = ky + 3 - 4k^{2} $ ($ k \\neq 0 $). Letting $ k = 1 $, we get the equation of line $ AB $ as $ x = y - 1 $, or $ x - y + 1 = 0 $."
  },
  {
    "text": "The line $l$ intersects the parabola $y^{2}=2x$ at points $A$, $B$ and $\\angle AOB=90^{\\circ}$, then the minimum area of $\\triangle AOB$ is?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;O: Origin;B: Point;Expression(G) = (y^2 = 2*x);Intersection(l,G)={A,B};AngleOf(A, O, B) = ApplyUnit(90, degree)",
    "query_expressions": "Min(Area(TriangleOf(A, O, B)))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 5]], [[6, 20]], [[23, 27]], [[32, 57]], [[28, 31]], [[6, 20]], [[0, 31]], [[32, 57]]]",
    "query_spans": "[[[59, 84]]]",
    "process": "First, from $\\angle AOB = 90^{\\circ}$, assume the equations of lines $OA$ and $OB$, calculate that line $AB$ passes through a fixed point, then based on this fixed point, set up the equation of line $AB$ and express the area of the triangle, thus finding the minimum value of the area of $\\triangle AOB$. From the problem, we know: the slopes of lines $OA$ and $OB$ exist and are non-zero. Let the equation of line $OA$ be: $y = kx$, then the equation of line $OB$ is: $y = -\\frac{1}{k}x$. Solving the system:  \n$$\n\\begin{cases}\ny = kx \\\\\ny^{2} = 2x\n\\end{cases}\n$$  \nwe obtain: $A\\left(\\frac{2}{k^{2}}, \\frac{2}{k}\\right)$. Similarly, solving the system:  \n$$\n\\begin{cases}\ny = -\\frac{1}{k}x \\\\\ny^{2} = 2x\n\\end{cases}\n$$  \nwe obtain: $B(2k^{2}, -2k)$. Therefore, the equation of line $AB$ is:  \n$$\ny + 2k = \\frac{-2k - \\frac{2}{k}}{2k^{2} - \\frac{2}{k^{2}}}(x - 2k^{2})\n$$  \nwhich simplifies to $x - \\left(\\frac{1}{k} - k\\right)y - 2 = 0$, i.e., line $AB$ passes through the fixed point $P(2, 0)$. Thus, we can set the equation of line $AB$ as: $x = my + 2$. Solving the system:  \n$$\n\\begin{cases}\nx = my + 2 \\\\\ny^{2} = 2x\n\\end{cases}\n$$  \nwe get: $y^{2} - 2my - 4 = 0$, so $y_{1} + y_{2} = 2m$, $y_{1}y_{2} = -4$, hence  \n$$\n|y_{1} - y_{2}| = \\sqrt{(y_{1} + y_{2})^{2} - 4y_{1}y_{2}} = \\sqrt{4m^{2} + 16}\n$$  \nThen,  \n$$\nS_{\\Delta AOB} = \\frac{1}{2} \\times |OP| \\cdot |y_{1} - y_{2}| = \\frac{1}{2} \\times 2 \\times \\sqrt{4m^{2} + 16} = \\sqrt{4m^{2} + 16} = 2\\sqrt{m^{2} + 4}\n$$  \nTherefore, when $m = 0$, the minimum value of $S_{\\triangle AOB}$ is $4$."
  },
  {
    "text": "Given that the graph of $y=\\frac{1}{x}$ is a hyperbola, take points $P$ and $Q$ on the two branches of the hyperbola respectively. Then the minimum value of the segment $P Q$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;Q: Point;Expression(G) = (y=1/x);PointOnCurve(P,LeftPart(G));PointOnCurve(Q,RightPart(G))",
    "query_expressions": "Min(LineSegmentOf(P,Q))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[21, 24], [26, 29]], [[36, 40]], [[41, 44]], [[2, 24]], [[26, 44]], [[26, 44]]]",
    "query_spans": "[[[46, 59]]]",
    "process": ""
  },
  {
    "text": "A point $M$ on the parabola $y^{2}=4 x$ is at a distance of $5$ from the focus. What is the abscissa of point $M$?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(M, G);Distance(M, Focus(G)) = 5",
    "query_expressions": "XCoordinate(M)",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 14]], [[17, 20], [32, 36]], [[0, 14]], [[0, 20]], [[0, 30]]]",
    "query_spans": "[[[32, 42]]]",
    "process": ""
  },
  {
    "text": "Given two points $A$ and $B$ on the parabola $y^{2}=4x$ with focus $F$ such that $\\overrightarrow{A F}=3 \\overrightarrow{F B}$, what is the distance from the midpoint of chord $AB$ to the directrix of the parabola?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);VectorOf(A, F) = 3*VectorOf(F, B);IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A, B)), Directrix(G))",
    "answer_expressions": "8/3",
    "fact_spans": "[[[10, 24], [94, 97]], [[10, 24]], [[3, 6]], [[2, 24]], [[28, 31]], [[32, 35]], [[10, 35]], [[10, 35]], [[37, 82]], [[10, 90]]]",
    "query_spans": "[[[85, 105]]]",
    "process": "Let the coordinates of points A and B be $(x_{1},y_{1})$, $(x_{2},y_{2})$. It is known that the focus of the parabola $y^{2}=4x$ is $F(1,0)$, and the equation of the directrix is $x=-1$. From $\\overrightarrow{AF}=3\\overrightarrow{FB}$, we obtain $(1-x_{1},-y_{1})=3(x_{2}-1,y_{2})$, then \n$$\n\\begin{cases}\n3x_{2}+x_{1}=4\\\\\n-y_{1}=3y_{2}\n\\end{cases}\n$$\nSince A and B lie on the parabola $y^{2}=4x$, we have $y_{1}^{2}=4x_{1}$, $y_{2}^{2}=4x_{2}$, then $\\frac{3}{4}y_{2}^{2}+\\frac{1}{4}y_{1}^{2}=4$, i.e., $3y_{2}^{2}=4$, so $y_{2}^{2}=\\frac{4}{3}$, $y_{1}^{2}=12$, hence $x_{1}=\\frac{y_{1}^{2}}{4}=3$, $x_{2}=\\frac{y_{2}^{2}}{4}=\\frac{1}{3}$, therefore the distance from the midpoint of chord AB to the directrix is $d=\\frac{x_{1}+x_{2}}{2}+1=\\frac{8}{3}$"
  },
  {
    "text": "If the focus of the parabola ${y}^{2}=2 p x(p>0)$ coincides with the right focus of the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$, then $p$=?",
    "fact_expressions": "G: Parabola;p: Number;H: Ellipse;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (x^2/6 + y^2/2 = 1);Focus(G) = RightFocus(H)",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 24]], [[73, 76]], [[28, 65]], [[4, 24]], [[1, 24]], [[28, 65]], [[1, 71]]]",
    "query_spans": "[[[73, 78]]]",
    "process": ""
  },
  {
    "text": "Point $P$ is a moving point on the parabola $y^{2}=4x$. Then, the minimum value of the sum of the distance from point $P$ to point $A(0,1)$ and the distance from $P$ to the line $x=-1$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P:Point;PointOnCurve(P,G);A: Point;Coordinate(A) = (0, 1);l:Line;Expression(l) = (x = -1)",
    "query_expressions": "Min(Distance(P,A)+Distance(P,l))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[5, 19]], [[5, 19]], [[0, 4], [25, 29], [43, 46]], [[0, 23]], [[30, 39]], [[30, 39]], [[47, 55]], [[47, 55]]]",
    "query_spans": "[[[25, 65]]]",
    "process": ""
  },
  {
    "text": "Given that a focus of the conic section $x^{2}+a y^{2}=1$ has coordinates $F\\left(\\frac{2}{\\sqrt{|a|}}, 0\\right)$, what is the eccentricity of this conic section?",
    "fact_expressions": "G: ConicSection;Expression(G) = (a*y^2 + x^2 = 1);a: Number;OneOf(Focus(G)) = F;F: Point;Coordinate(F) = (2/sqrt(Abs(a)), 0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{2*sqrt(3)/3,2*sqrt(5)/5}",
    "fact_spans": "[[[2, 23], [62, 66]], [[2, 23]], [[6, 23]], [[2, 59]], [[31, 59]], [[31, 59]]]",
    "query_spans": "[[[62, 72]]]",
    "process": "When $ a>0 $ and $ a\\neq1 $, the curve is an ellipse with foci on the x-axis, and the standard equation is: $ x^{2}+\\frac{y^{2}}{\\frac{1}{a}}=1 $. Then $ 1-\\frac{1}{a}=\\frac{4}{a} $, solving gives $ a=5 $, so the eccentricity is $ e=\\frac{2}{5}\\sqrt{5} $. When $ a<0 $, the curve is a hyperbola with foci on the y-axis, and the equation is: $ x^{2}-\\frac{y^{2}}{-\\frac{1}{a}}=1 $. Then $ 1-\\frac{1}{a}=-\\frac{4}{a} $, solving gives $ a=-3 $, so the eccentricity is $ e=\\frac{2}{3}\\sqrt{3} $. Hence, fill in: $ e=\\frac{2}{5}\\sqrt{5} $ or $ e=\\frac{2}{3}\\sqrt{3} $."
  },
  {
    "text": "Given the ellipse $C$: $9x^{2}+y^{2}=1$, and a line $l$ that does not pass through the origin $O$ and is not parallel to the coordinate axes. The line $l$ intersects $C$ at two points $A$ and $B$, with $M$ being the midpoint of segment $AB$. Then, what is the product of the slope of line $OM$ and the slope of line $l$?",
    "fact_expressions": "l: Line;C: Ellipse;O: Origin;M:Point;A: Point;B: Point;Expression(C) = (9*x^2 + y^2 = 1);Negation(PointOnCurve(O,l));Negation(IsParallel(l,axis));Intersection(l, C) = {A, B};MidPoint(LineSegmentOf(A,B))=M",
    "query_expressions": "Slope(LineOf(O,M))*Slope(l)",
    "answer_expressions": "-9",
    "fact_spans": "[[[28, 33], [49, 52], [95, 98]], [[2, 26], [53, 56]], [[35, 40]], [[80, 83]], [[61, 64]], [[65, 68]], [[2, 26]], [[28, 40]], [[27, 48]], [[49, 68]], [[69, 83]]]",
    "query_spans": "[[[84, 106]]]",
    "process": "Let the equation of the line be y = kx + b, and let A(x_{1}, y_{1}), B(x_{2}, y_{2}). From \\begin{cases} y = kx + b \\\\ 9x^{2} + y^{2} = 1 \\end{cases}, we obtain (9 + k^{2})x^{2} + 2kbx + b^{2} - 1 = 0. \\therefore x_{1} + x_{2} = -\\frac{2kb}{9 + k^{2}}, \\therefore x_{M} = -\\frac{kb}{9 + k^{2}}, \\therefore y_{M} = kx_{M} + b = \\frac{9b}{9 + k^{2}}, \\therefore K_{OM} \\cdot K_{1} = \\frac{9b}{kb} \\cdot k = -9"
  },
  {
    "text": "Given the two foci of an ellipse are $F_{1}$ and $F_{2}$, if there exists a point $P$ on the ellipse such that $\\angle F_{1} P F_{2}=120^{\\circ}$, then the minimum value of the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Focus(G) = {F1,F2};PointOnCurve(P,G);AngleOf(F1,P,F2)=ApplyUnit(120,degree)",
    "query_expressions": "Min(Eccentricity(G))",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[2, 4], [27, 29], [75, 77]], [[10, 17]], [[34, 37]], [[18, 25]], [[2, 25]], [[27, 37]], [[39, 73]]]",
    "query_spans": "[[[75, 86]]]",
    "process": "Without loss of generality, assume the two foci of the ellipse lie on the x-axis; hence, when point P is at the upper or lower vertex of the ellipse, $\\angle F_{1}PF_{2}$ is maximized. Let the upper vertex of the ellipse be $P_{0}$, then $\\angle F_{1}P_{0}F_{2} \\geqslant 120^{\\circ}$. Combining with $\\tan\\angle OP_{0}F_{2} = \\frac{c}{b} \\geqslant \\sqrt{3}$, $e = \\frac{c}{a} = \\frac{c}{\\sqrt{b^{2}+c^{2}}}$, analysis yields the solution. Without loss of generality, assume the two foci of the ellipse lie on the x-axis; hence, when point P is at the upper or lower vertex of the ellipse, $\\angle F_{1}PF_{2}$ is maximized. Let the upper vertex of the ellipse be $P_{0}$. If there exists a point P on the ellipse such that $\\angle F_{1}PF_{2} = 120^{\\circ}$, then $\\angle F_{1}P_{0}F_{2} \\geqslant 120^{\\circ}$ and $\\tan\\angle OP_{0}F_{2} = \\frac{c}{b} \\geqslant \\tan 60^{\\circ} = \\sqrt{3}$, thus $b \\leqslant \\frac{c}{\\sqrt{3}}$. Hence, $e = \\frac{c}{a} = \\frac{c}{\\sqrt{b^{2}+c^{2}}} \\geqslant \\frac{\\sqrt{3}}{2}$. Then the minimum value of the ellipse's eccentricity is $\\frac{\\sqrt{3}}{2}$."
  },
  {
    "text": "The focus of the parabola $y^{2}=a x$ coincides exactly with the right focus of the hyperbola $x^{2}-y^{2}=2$. Then $a$=?",
    "fact_expressions": "G: Hyperbola;H: Parabola;a: Number;Expression(G) = (x^2 - y^2 = 2);Expression(H) = (y^2 = a*x);Focus(H) = RightFocus(G)",
    "query_expressions": "a",
    "answer_expressions": "8",
    "fact_spans": "[[[20, 38]], [[0, 14]], [[44, 47]], [[20, 38]], [[0, 14]], [[0, 42]]]",
    "query_spans": "[[[44, 49]]]",
    "process": "First, find the right focus of the hyperbola $x^{2}-y^{2}=2$, which gives the focus of the parabola $y^{2}=ax$. Then, determine its value based on the meaning of $p$. The right focus of the hyperbola $x^{2}-y^{2}=2$ is $(2,0)$, so the focus of the parabola $y^{2}=ax$ is $(2,0)$. $\\therefore\\frac{a}{4}=2,\\therefore a=8$."
  },
  {
    "text": "Suppose a hyperbola centered at the origin shares common foci with the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$, and their eccentricities are reciprocals of each other. Then the equations of the asymptotes of this hyperbola are?",
    "fact_expressions": "G: Hyperbola;O: Origin;Center(G) = O;H: Ellipse;Expression(H) = (x^2/16 + y^2/12 = 1);Focus(G) = Focus(H);InterReciprocal(Eccentricity(G), Eccentricity(H))",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[7, 10], [71, 74]], [[4, 6]], [[1, 10]], [[11, 50]], [[11, 50]], [[7, 56]], [[58, 68]]]",
    "query_spans": "[[[71, 83]]]",
    "process": "The foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$ are $(\\pm2,0)$, and the eccentricity is $\\frac{1}{2}$. Let the transverse axis length of the hyperbola be $2a$, the conjugate axis length be $2b$, and the focal distance be $2c$. Then $c=2$, and since $\\frac{c}{a}=2$, we obtain $a=1$, then $b=\\sqrt{c^{2}-a^{2}}=\\sqrt{3}$. The equations of the asymptotes of the hyperbola are $y=\\pm\\frac{b}{a}x=\\pm\\sqrt{3}x$."
  },
  {
    "text": "A line passing through the left focus $F_{1}$ of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{12}=1$ intersects the left branch of the hyperbola at points $M$ and $N$, and $F_{2}$ is its right focus. Then the value of $|M F_{2}|+|N F_{2}|-|M N|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/12 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Line;PointOnCurve(F1, H);M: Point;N: Point;Intersection(H, LeftPart(G)) = {M, N}",
    "query_expressions": "Abs(LineSegmentOf(M, F2)) + Abs(LineSegmentOf(N, F2)) - Abs(LineSegmentOf(M, N))",
    "answer_expressions": "16",
    "fact_spans": "[[[1, 41], [80, 81], [55, 58]], [[1, 41]], [[44, 51]], [[72, 79]], [[1, 51]], [[72, 84]], [[52, 54]], [[0, 54]], [[62, 65]], [[66, 69]], [[52, 71]]]",
    "query_spans": "[[[86, 117]]]",
    "process": "According to the definition of hyperbola, |MF_{2}|-|MF|=2a, |NF_{2}|-|NF|=2a; adding these two equations yields |MF_{2}|+|NF_{2}|-|MN|=4a=16"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of ellipse $C$, and $P$ is a point on $C$. If $P F_{1} \\perp P F_{2}$ and $\\angle P F_{2} F_{1}=60^{\\circ}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Focus(C) = {F1, F2};PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));AngleOf(P, F2, F1) = ApplyUnit(60, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[18, 23], [33, 36], [102, 105]], [[29, 32]], [[2, 9]], [[10, 17]], [[2, 28]], [[29, 40]], [[42, 65]], [[67, 100]]]",
    "query_spans": "[[[102, 111]]]",
    "process": "F_{1}, F_{2} are the two foci of ellipse C, and P is a point on C. If PF_{1} \\bot PF_{2} and \\angle PF_{2}F_{1} = 60^{\\circ}, then the coordinates of the focus can be obtained as F_{2}(c,0), so P(\\frac{1}{2}c,\\frac{\\sqrt{3}}{2}c). We obtain: \\frac{c^{2}}{4a^{2}}+\\frac{3c^{2}}{4b^{2}}=1, which leads to \\frac{1}{4}e^{2}+\\frac{3}{4(\\frac{1}{e^{2}}-1)}=1, resulting in e^{4}-8e^{2}+4=0, with e \\in (0,1). Solving gives e = \\sqrt{3}-1."
  },
  {
    "text": "$P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and $M$, $N$ are points on the circles $(x+5)^{2}+y^{2}=4$ and $(x-5)^{2}+y^{2}=1$, respectively. Then the maximum value of $|PM|-|PN|$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P, RightPart(G)) = True;G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);M: Point;C1: Circle;Expression(C1) = (y^2 + (x + 5)^2 = 4);PointOnCurve(M,C1) = True;N: Point;C2: Circle;Expression(C2) = (y^2 + (x - 5)^2 = 1);PointOnCurve(N,C2) = True",
    "query_expressions": "Max(Abs(LineSegmentOf(P,M)) - Abs(LineSegmentOf(P,N)))",
    "answer_expressions": "9",
    "fact_spans": "[[[0, 3]], [[0, 49]], [[4, 43]], [[4, 43]], [[50, 53]], [[61, 81]], [[61, 81]], [[50, 105]], [[54, 57]], [[82, 101]], [[82, 101]], [[50, 105]]]",
    "query_spans": "[[[107, 126]]]",
    "process": ""
  },
  {
    "text": "The equation of the circle with its center at the focus of the parabola $y^{2}=-6 x$ and tangent to the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;H: Circle;Expression(G) = (y^2 = -6*x);Focus(G) = Center(H);IsTangent(Directrix(G),H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x+3/2)^2+y^2=9",
    "fact_spans": "[[[1, 16], [25, 28]], [[34, 35]], [[1, 16]], [[0, 35]], [[24, 35]]]",
    "query_spans": "[[[34, 40]]]",
    "process": "First, find the focus coordinates and directrix equation of the parabola, then further determine the equation of the circle. The focus coordinates of the parabola $ y^{2} = -6x $ are $ (-\\frac{3}{2}, 0) $, and the equation of the directrix is $ x = \\frac{3}{2} $. Therefore, the distance from the focus to the directrix is 3. Hence, the equation of the circle centered at the focus and tangent to the directrix of the parabola is: $ (x + \\frac{3}{2})^{2} + y^{2} = 9 $."
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $\\sqrt{2}$, and it passes through the point $(3,1)$, then the focal distance of the hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(G) = sqrt(2);H: Point;Coordinate(H) = (3, 1);PointOnCurve(H, G)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 58], [86, 89]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 73]], [[76, 84]], [[76, 84]], [[2, 84]]]",
    "query_spans": "[[[86, 95]]]",
    "process": "From the given conditions, we have\n\\[\n\\begin{cases}\n\\frac{c}{a} = \\sqrt{2} \\\\\n\\frac{9}{a^{2}} - \\frac{1}{b^{2}} = 1\n\\end{cases},\n\\]\nand $ c^{2} = a^{2} + b^{2} $. Solving gives\n\\[\n\\begin{cases}\na = 2\\sqrt{2} \\\\\nb = 2\\sqrt{2} \\\\\nc = 4\n\\end{cases},\n\\]\nso the focal distance of the hyperbola is $ 2c = 8 $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$, a line $l$: $y=x-1$ passing through the right focus intersects the ellipse at points $A$ and $B$, and $O$ is the origin. Then the area of $\\triangle O A B$ is?",
    "fact_expressions": "C: Ellipse;l: Line;O: Origin;A: Point;B: Point;Expression(C) = (x^2/2 + y^2 = 1);Expression(l) = (y = x - 1);PointOnCurve(RightFocus(C),l);Intersection(l, C) = {A, B}",
    "query_expressions": "Area(TriangleOf(O, A, B))",
    "answer_expressions": "2/3",
    "fact_spans": "[[[2, 34], [55, 57]], [[40, 54]], [[69, 72]], [[59, 62]], [[63, 66]], [[2, 34]], [[40, 54]], [[2, 54]], [[40, 68]]]",
    "query_spans": "[[[79, 101]]]",
    "process": "According to the problem, find the right focus $ F $, substitute the line equation into the ellipse equation, eliminate $ x $ to obtain a quadratic equation in $ y $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, solve the equation to find the roots, and then find the area of $ \\triangle OAB $. From the problem, the ellipse is $ \\frac{x^{2}}{2} + y^{2} = 1 $, $ F(1,0) $, and the line $ l $ has equation $ y = x - 1 $. Substituting into the ellipse equation $ \\frac{x^{2}}{2} + y^{2} = 1 $, eliminating $ x $ and simplifying gives $ 3y^{2} + 2y - 1 = 0 $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ y_{1} = -1 $, $ y_{2} = \\frac{1}{3} $. Therefore, the area of $ \\triangle AOB $ is $ S_{\\triangle AOB} = \\frac{1}{2}|y_{1} - y_{2}| \\cdot |OF| = \\frac{1}{2} \\times \\frac{4}{3} \\times 1 = \\frac{2}{3} $."
  },
  {
    "text": "A line $l$ passing through the point $M(2,1)$ intersects the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$ at points $A$ and $B$, with $M$ being the midpoint of $AB$. Find the slope of line $l$.",
    "fact_expressions": "M: Point;Coordinate(M) = (2, 1);l: Line;G: Hyperbola;Expression(G) = (x^2 - y^2/2 = 1);PointOnCurve(M, l);A: Point;B: Point;Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Slope(l)",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 11], [59, 62]], [[2, 11]], [[12, 17], [73, 78]], [[19, 47]], [[19, 47]], [[1, 17]], [[48, 51]], [[52, 55]], [[12, 57]], [[59, 71]]]",
    "query_spans": "[[[73, 83]]]",
    "process": "Let point A(x_{1},y_{1}), point B(x_{2},y_{2}), M(x_{0},y_{0}). Then 2x_{1}^{2}-y^{2}=2\\textcircled{1}, 2x_{2}^{2}-y_{2}^{2}=2\\textcircled{2}. \\textcircled{1}-\\textcircled{2} gives 2(x_{1}+x_{2})(x_{1}-x_{2})-(y_{1}+y_{2})(y_{1}-y_{2})=0, 2\\times2x_{0}-2y_{0}\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=0, \\therefore8-2k=0, \\thereforek=4, \\thereforey-1=4(x-2), \\therefore the equation of line l is 4x-y-7=0"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1$ has a focal distance of $4 \\sqrt{3}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/4 + x^2/a^2 = 1);a: Number;FocalLength(C) = 4*sqrt(3)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[2, 48], [66, 69]], [[2, 48]], [[9, 48]], [[2, 64]]]",
    "query_spans": "[[[66, 75]]]",
    "process": "Given $2c=4\\sqrt{3}$, $c=2\\sqrt{3}$, and $b^{2}=4$, so $a^{2}=c^{2}-b^{2}=12-4=8$, $a=$, thus $e=\\frac{c}{a}=\\frac{2\\sqrt{3}}{2\\sqrt{6}}=\\frac{\\sqrt{6}}{2}$"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right vertices are denoted as $A$ and $B$, respectively. Let point $F$ be the left focus of hyperbola $C$. A line passing through point $F$ and perpendicular to the $x$-axis intersects hyperbola $C$ at points $P$ and $Q$, where point $P$ lies in the second quadrant. The line $PB$ intersects the $y$-axis at point $E$, and the line $AE$ intersects $QF$ at point $M$. If $\\overrightarrow{F M}=2 \\overrightarrow{M Q}$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;G: Line;P: Point;B: Point;A: Point;E: Point;Q: Point;F: Point;M: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftVertex(C) = A;RightVertex(C) = B ;LeftFocus(C) = F;PointOnCurve(F, G);IsPerpendicular(xAxis, G);Intersection(G, C) = {P, Q};Quadrant(P) = 2;Intersection(LineSegmentOf(P, B), yAxis) = E;Intersection(LineSegmentOf(A, E), LineSegmentOf(Q, F)) = M;VectorOf(F, M) = 2*VectorOf(M, Q)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "5",
    "fact_spans": "[[[2, 63], [85, 91], [113, 119], [228, 234]], [[10, 63]], [[10, 63]], [[110, 112]], [[121, 125], [133, 137]], [[76, 79]], [[72, 75]], [[156, 160]], [[126, 130]], [[80, 84], [97, 101]], [[175, 179]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 79]], [[2, 79]], [[80, 95]], [[96, 112]], [[102, 112]], [[110, 130]], [[133, 142]], [[145, 160]], [[163, 179]], [[181, 226]]]",
    "query_spans": "[[[228, 240]]]",
    "process": ""
  },
  {
    "text": "Given $\\frac{2}{m}+\\frac{1}{n}=1$ $(m>0, n>0)$, when $m n$ attains its minimum value, what is the equation of the asymptotes of the hyperbola $\\frac{x^{2}}{m^{2}}-\\frac{y^{2}}{n^{2}}=1$?",
    "fact_expressions": "C: Curve;Expression(C) = (2/m + 1/n = 1);m: Number;n: Number;m > 0;n > 0;WhenMin(m*n);G: Hyperbola;Expression(G) = (-y^2/n^2 + x^2/m^2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*x/2",
    "fact_spans": "[[[2, 39]], [[2, 39]], [[2, 39]], [[2, 39]], [[2, 39]], [[2, 39]], [[41, 53]], [[54, 100]], [[54, 100]]]",
    "query_spans": "[[[54, 108]]]",
    "process": "From the given condition, we have $\\frac{2}{m}+\\frac{1}{n}=1$, so $1=\\frac{2}{m}+\\frac{1}{n}\\geqslant2\\sqrt{\\frac{2}{m}\\cdot\\frac{1}{n}}=2\\sqrt{\\frac{2}{mn}}$, thus $mn\\geqslant8$. The equality holds if and only if $a=4$, $b=2$. Therefore, the equation of the hyperbola is $\\frac{x^{2}}{16}-\\frac{y^{2}}{4}=1$, and its asymptotes are given by $y=\\pm\\frac{1}{2}x$."
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $\\frac{\\sqrt{2}}{2}$. If the abscissa of an intersection point of the line $y=k x$ with the ellipse is $b$, then what is the value of $k$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;H: Line;k: Number;a > b;b > 0;Expression(H) = (y = k*x);Eccentricity(G) = sqrt(2)/2;XCoordinate(OneOf(Intersection(H,G)))=b",
    "query_expressions": "k",
    "answer_expressions": "pm*sqrt(2)/2",
    "fact_spans": "[[[0, 52], [89, 90]], [[0, 52]], [[99, 102]], [[2, 52]], [[79, 88]], [[104, 107]], [[2, 52]], [[2, 52]], [[79, 88]], [[0, 77]], [[79, 102]]]",
    "query_spans": "[[[104, 111]]]",
    "process": "Since the eccentricity is $\\frac{\\sqrt{2}}{2}$, it follows that $a^{2}=2c^{2}$, $b^{2}=c^{2}$, so the equation of the ellipse is $\\frac{x^{2}}{2b^{2}}+\\frac{y^{2}}{b^{2}}=1$. Since the point $(b,kb)$ lies on the ellipse, $\\frac{1}{2}+k^{2}=1$, thus $k=\\pm\\frac{\\sqrt{2}}{2}$."
  },
  {
    "text": "Let $F$ be the focus of the parabola $y^{2}=6 x$, and let $A$, $B$, $C$ be three points on this parabola. If $\\overrightarrow{F A}+\\overrightarrow{F B}=-\\overrightarrow{F C}$, then $|\\overrightarrow{F A}|+| \\overrightarrow{F B}|+| \\overrightarrow{F C} |$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 6*x);F: Point;Focus(G) = F;A: Point;B: Point;C: Point;PointOnCurve(A, G) = True;PointOnCurve(B, G) = True;PointOnCurve(C, G) = True;VectorOf(F, A) + VectorOf(F, B) = -VectorOf(F, C)",
    "query_expressions": "Abs(VectorOf(F,A))+Abs(VectorOf(F,B))+Abs(VectorOf(F,C))",
    "answer_expressions": "9",
    "fact_spans": "[[[5, 19], [36, 39]], [[5, 19]], [[1, 4]], [[1, 22]], [[23, 26]], [[27, 30]], [[31, 34]], [[23, 42]], [[23, 42]], [[23, 42]], [[44, 109]]]",
    "query_spans": "[[[111, 187]]]",
    "process": "Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, $C(x_{3},y_{3})$. The focus of the parabola $y^{2}=6x$ has coordinates $F(\\frac{3}{2},0)$, and the equation of the directrix is $x=-\\frac{3}{2}$. From the given condition, $\\overrightarrow{FA}+\\overrightarrow{FB}+\\overrightarrow{FC}=\\overrightarrow{0}$, so point $F$ is the centroid of $\\triangle ABC$, hence $x_{1}+x_{2}+x_{3}=3\\times\\frac{3}{2}=\\frac{9}{2}$. From the property that the distance from a point on the parabola to the focus equals the distance to the directrix (or directly from the focal radius formula), we obtain $|\\overrightarrow{FA}|+|\\overrightarrow{FB}|+|\\overrightarrow{FC}|=x_{1}+\\frac{3}{2}+x_{2}+\\frac{3}{2}+x_{3}+\\frac{3}{2}=x_{1}+x_{2}+x_{3}+\\frac{9}{2}=\\frac{9}{2}+\\frac{9}{2}=9$."
  },
  {
    "text": "Given that $F$ is the left focus of the ellipse $5 x^{2}+9 y^{2}=45$, $P$ is a moving point on this ellipse, and $A(1,1)$ is a fixed point, then the maximum value of $|P A|+|P F|$ is?",
    "fact_expressions": "G: Ellipse;A: Point;P: Point;F: Point;Expression(G) = (5*x^2 + 9*y^2 = 45);Coordinate(A) = (1, 1);LeftFocus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "6+sqrt(2)",
    "fact_spans": "[[[6, 28], [38, 40]], [[45, 53]], [[33, 36]], [[2, 5]], [[6, 28]], [[45, 53]], [[2, 32]], [[33, 44]]]",
    "query_spans": "[[[59, 78]]]",
    "process": "The standard equation of the ellipse $5x^{2}+9y^{2}=45$ is $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, $a=3$, $c=2$. Let the right focus of the ellipse be $F_{2}(2,0)$. According to the definition of the ellipse, we have $|PA|+|PF|=|PA|+2a-|PF_{2}|$. Therefore, $|PA|+|PF|$ is maximized when $|PA|-|PF_{2}|$ reaches its maximum value. As shown in the figure: since $|PA|+|PF|\\leqslant 2a+|AF_{2}|=6+\\sqrt{(2-1)^{2}+(0-1)^{2}}=6+\\sqrt{2}$, and equality holds if and only if points $P$, $A$, $F_{2}$ are collinear and $F_{2}$ lies on the segment $PA$, the maximum value of $|PA|+|PF|$ is $6+\\sqrt{2}$."
  },
  {
    "text": "The equation of the conic section that has the same foci as the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$ and passes through the point $Q(2,1)$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/2 = 1);Focus(G) = Focus(H);Q: Point;Coordinate(Q) = (2, 1);PointOnCurve(Q, H) = True;H: ConicSection",
    "query_expressions": "Expression(H)",
    "answer_expressions": "{(x^2/8+y^2/2=1),(x^2/3-y^2/3=1)}",
    "fact_spans": "[[[1, 39]], [[1, 39]], [[0, 62]], [[48, 57]], [[48, 57]], [[47, 62]], [[58, 62]]]",
    "query_spans": "[[[58, 66]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $P$ is a point on the ellipse satisfying $(\\overrightarrow{O F_{1}}-\\overrightarrow{O P}) \\cdot(\\overrightarrow{O F_{1}}+\\overrightarrow{O P})=0$ ($O$ being the coordinate origin). If $|P F_{1}|=2|P F_{2}|$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;O: Origin;F1: Point;P: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);DotProduct((VectorOf(O, F1) - VectorOf(O, P)), (VectorOf(O, F1) + VectorOf(O, P))) = 0;Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2));LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[2, 54], [83, 85], [232, 234]], [[4, 54]], [[4, 54]], [[196, 199]], [[63, 70]], [[79, 82]], [[71, 78]], [[4, 54]], [[4, 54]], [[2, 54]], [[91, 195]], [[208, 230]], [[2, 78]], [[2, 78]], [[79, 88]]]",
    "query_spans": "[[[232, 240]]]",
    "process": "From $(\\overrightarrow{OF_{1}}-\\overrightarrow{OP})\\cdot(\\overrightarrow{OF_{1}}+\\overrightarrow{OP})=0$ we get $|\\overrightarrow{OF_{1}}|=|\\overrightarrow{OP}|$, and combining with the properties of the ellipse, we obtain that $\\triangle PF_{1}F_{2}$ is a right triangle. By the given condition, let $|PF_{2}|=m$, then $|PF_{1}|=2m$. From the Pythagorean theorem we get $_{m}=\\frac{2\\sqrt{5}}{5}c$. Combining with the definition of the ellipse, the eccentricity can be found. Since $(\\overrightarrow{OF}_{1}-\\overrightarrow{OP})\\cdot(\\overrightarrow{OF_{1}}+\\overrightarrow{OP})=0$, it follows that $\\overrightarrow{OF}^{2}=\\overrightarrow{OP}^{2}$, so $|\\overrightarrow{OF_{1}}|=|\\overrightarrow{OP}|$. Because $|OF_{1}|=|OF_{2}|$, we have $|OF_{1}|=|OF_{2}|=|OP|$. Thus, $\\triangle PF_{1}F_{2}$ is a right triangle, i.e., $PF_{1}\\bot PF_{2}$. Therefore, $|PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}$. Let $|PF_{2}|=m$, then $|PF_{1}|=2m$, so $m^{2}+(2m)^{2}=(2c)^{2}$, yielding $_{m}=\\frac{2\\sqrt{5}}{5}c$. Since $|PF_{1}|+|PF_{2}|=3m=2a$, we have $\\frac{6\\sqrt{5}}{5}c=2a$, so $\\frac{c}{a}=\\frac{\\sqrt{5}}{3}$, i.e., the eccentricity is $\\frac{\\sqrt{5}}{3}$."
  },
  {
    "text": "Let points $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$, and let $P$ be a point on $C$. If the area of $\\triangle PF_{1} F_{2}$ is $6$, then $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$=?",
    "fact_expressions": "C: Hyperbola;P: Point;F1: Point;F2: Point;Expression(C) = (x^2 - y^2/3 = 1);LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(P, C);Area(TriangleOf(P, F1, F2)) = 6",
    "query_expressions": "DotProduct(VectorOf(P, F1), VectorOf(P, F2))",
    "answer_expressions": "9",
    "fact_spans": "[[[20, 53], [64, 67]], [[60, 63]], [[1, 10]], [[12, 19]], [[20, 53]], [[1, 59]], [[1, 59]], [[60, 70]], [[72, 103]]]",
    "query_spans": "[[[105, 164]]]",
    "process": ""
  },
  {
    "text": "There is a point $P$ on the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, and $F_{1}$, $F_{2}$ are the foci of the hyperbola, with $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$. Then the area of $\\triangle F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/16 - y^2/9 = 1);PointOnCurve(P,G);Focus(G)={F1,F2};AngleOf(F1, P, F2) = pi/3",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "9*sqrt(3)",
    "fact_spans": "[[[0, 39], [62, 65]], [[46, 53]], [[41, 45]], [[54, 61]], [[0, 39]], [[0, 45]], [[46, 68]], [[70, 106]]]",
    "query_spans": "[[[108, 138]]]",
    "process": "From the hyperbola, we obtain |PF_{1}-PF_{2}|=2a=8. Squaring both sides gives PF_{1}^{2}+PF_{2}^{2}-2PF_{1}\\cdot PF_{2}=64. In triangle PF_{1}F_{2}, by the law of cosines, we have PF_{1}^{2}+PF_{2}^{2}-2PF_{1}\\cdot PF_{2}\\cos60^{\\circ}=F_{1}F_{2}^{2}=100. Solving these two equations together yields PF_{1}\\cdot PF_{2}=36. Therefore, the area of \\triangle F_{1}PF_{2} is 9\\sqrt{3}."
  },
  {
    "text": "It is known that the center of the ellipse is at the origin and its foci lie on the $x$-axis. If the projections onto the $x$-axis of the intersection points between the line $l$: $3x - 2y = 0$ and the ellipse are exactly the foci of the ellipse, then the eccentricity of the ellipse equals?",
    "fact_expressions": "l: Line;G: Ellipse;O: Origin;Expression(l) = (3*x - 2*y=0);Center(G) = O;PointOnCurve(Focus(G), xAxis);Projection(Intersection(l, G), xAxis) = Focus(G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[22, 40]], [[2, 4], [41, 43], [57, 59], [64, 66]], [[8, 10]], [[22, 40]], [[2, 10]], [[2, 19]], [[22, 62]]]",
    "query_spans": "[[[64, 73]]]",
    "process": "Let the standard equation of the ellipse be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, with semi-focal length $c$. The $x$-coordinate of the intersection point between the line and the ellipse in the first quadrant is $c$. Substituting $x=c$ into the standard equation of the ellipse yields $y=\\frac{b^{2}}{a}$, so the coordinates of the intersection point are $(c,\\frac{b^{2}}{a})$. Since the intersection point lies on the line $3x-2y=0$, we have $3c-\\frac{2b^{2}}{a}=0$, which simplifies to $2c^{2}+3ac-2a^{2}=0$, or $2e^{2}+3e-2=0$. Solving gives $e=\\frac{1}{2}$."
  },
  {
    "text": "Given that the moving circle $M$ passes through a fixed point $A(-3,0)$ and is internally tangent to the fixed circle $B$: $(x-3)^{2}+y^{2}=64$, find the trajectory equation of the center $M$ of the moving circle.",
    "fact_expressions": "M: Circle;A: Point;B:Circle;Coordinate(A) = (-3, 0);PointOnCurve(A, M);IsInTangent(M,B);Expression(B)=((x-3)^2+y^2=64);M1:Point;Center(M)=M1",
    "query_expressions": "LocusEquation(M1)",
    "answer_expressions": "x^2/16+y^2/7=1",
    "fact_spans": "[[[4, 7], [53, 55]], [[10, 19]], [[27, 51]], [[10, 19]], [[4, 19]], [[4, 51]], [[27, 51]], [[57, 60]], [[53, 60]]]",
    "query_spans": "[[[57, 66]]]",
    "process": "From the standard equation of a circle, the center is at B(3,0) and the radius is 8. Since circle M is internally tangent to fixed circle B and passes through the fixed point A(-3,0), it follows that |AM| + |BM| = 8 and |AB| = 6, indicating that the locus of M is an ellipse. Thus, we can write the equation of this locus. From the circle equation: the center of circle B is B(3,0) and the radius is 8. Since circle M passes through the fixed point A(-3,0) and is internally tangent to circle B, letting the center of circle M be M(x,y), it follows from the conditions that |AM| + |BM| = 8, and since |AB| = 6, M lies on an ellipse with foci A and B. Therefore, a = 4, c = 3, b^{2} = a^{2} - c^{2} = 7, so the trajectory equation of center M is \\frac{x^{2}}{16} + \\frac{y^{2}}{7} = 1. Key point"
  },
  {
    "text": "Given that point $N$ is a moving point on the parabola $y^{2}=-4x$, and point $M$ is a moving point on the circle $O^{\\prime}$: $(x-1)^{2}+(y-2)^{2}=1$, denote the distance from the moving point $N$ to the $y$-axis as $m$. Then, the minimum value of $m+|MN|$ is?",
    "fact_expressions": "G: Parabola;O1: Circle;M: Point;N: Point;m: Number;Expression(G) = (y^2 = -4*x);PointOnCurve(N, G);Expression(O1) = ((x - 1)^2 + (y - 2)^2 = 1);PointOnCurve(M, O1);Distance(N, yAxis) = m",
    "query_expressions": "Min(m + Abs(LineSegmentOf(M, N)))",
    "answer_expressions": "2*sqrt(2) - 2",
    "fact_spans": "[[[7, 22]], [[32, 69]], [[27, 31]], [[2, 6], [77, 80]], [[88, 91]], [[7, 22]], [[2, 26]], [[32, 69]], [[27, 73]], [[77, 91]]]",
    "query_spans": "[[[93, 108]]]",
    "process": "As shown in the figure, connect O'B intersecting circle O at point M and the parabola at point N. From the parabola equation $ y^{2} = -4x $, we know the focus has coordinates $ B(-1,0) $ and the directrix equation is $ x = 1 $. According to the definition of a parabola, $ m = |BN| - 1 $. Therefore, $ m + |MN| = |MN| + |BN| - 1 $. When points B, N, and M are collinear, $ m + |MN| = |MN| + |BN| - 1 $ is minimized. Since point M lies on circle O, $ m + |MN| $ is minimized when points B, N, M, and O are collinear. As shown in the figure, the minimum value of $ |MN| + m $ is: $ |O'B| - |OM| - 1 = 2\\sqrt{2} - 2 $."
  },
  {
    "text": "It is known that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ is $y=\\sqrt{3} x$, and one of its foci coincides with the focus of the parabola $y^{2}=16 x$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x);H: Parabola;Expression(H) = (y^2 = 16*x);OneOf(Focus(G)) = Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/12 = 1",
    "fact_spans": "[[[2, 59], [83, 84], [112, 115]], [[2, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[2, 82]], [[90, 105]], [[90, 105]], [[83, 110]]]",
    "query_spans": "[[[112, 120]]]",
    "process": ""
  },
  {
    "text": "The right vertex of the ellipse $E$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ is $B$. A line $L$ with slope $1$ passing through the right focus of $E$ intersects $E$ at points $M$ and $N$. Then, the area of $\\triangle M B N$ is?",
    "fact_expressions": "E: Ellipse;Expression(E) = (x^2/4 + y^2/3 = 1);B: Point;RightVertex(E) = B;PointOnCurve(RightFocus(E), L) = True;Slope(L) = 1;L: Line;M: Point;N: Point;Intersection(L, E) = {M, N}",
    "query_expressions": "Area(TriangleOf(M,B,N))",
    "answer_expressions": "6*sqrt(2)/7",
    "fact_spans": "[[[0, 42], [73, 76], [52, 55]], [[0, 42]], [[47, 50]], [[0, 50]], [[51, 72]], [[60, 72]], [[67, 72]], [[78, 81]], [[82, 85]], [[67, 87]]]",
    "query_spans": "[[[89, 111]]]",
    "process": "From the right focus of ellipse $ E $ and the slope of line $ L $ being 1, we can assume the equation of line $ L $ is $ y = x - 1 $. Substituting into the ellipse equation, using Vieta's formulas and the chord length formula, we find $ |MN| $. Using the right vertex $ (2,0) $ and the point-to-line distance formula, we find the distance $ d $ from $ B $ to line $ L $. Combining with the area $ S = \\frac{1}{2} \\cdot |MN| \\cdot d $, the solution can be obtained. Solution: From the given conditions, ellipse $ E: \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $ has right focus $ (1,0) $, right vertex $ (2,0) $. Line $ L $ has slope 1 and passes through the right focus $ (1,0) $. Let the equation of line $ L $ be $ y = x - 1 $, $ M(x_{1}, y_{1}) $, $ N(x_{2}, y_{2}) $. From \n$$\n\\begin{cases}\ny = x - 1 \\\\\n\\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1\n\\end{cases}\n$$\nwe obtain after simplification: $ 7x^{2} - 8x - 8 = 0 $. By Vieta's formulas, $ x_{1} + x_{2} = \\frac{8}{7} $, $ x_{1}x_{2} = -\\frac{8}{7} $. By the chord length formula,\n$$\n|MN| = \\sqrt{1 + k^{2}} \\cdot \\sqrt{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}} = \\sqrt{2} \\cdot \\sqrt{\\left(\\frac{8}{7}\\right)^{2} - 4 \\times \\left(-\\frac{8}{7}\\right)} = \\frac{24}{7}\n$$\nBy the point-to-line distance formula, the distance $ d $ from $ B $ to line $ L $ is\n$$\nd = \\frac{|0 - 2 + 1|}{\\sqrt{1 + (-1)^{2}}} = \\frac{\\sqrt{2}}{2}\n$$\nThe area of quadrilateral $ AMBN $ is\n$$\nS = \\frac{1}{2} \\cdot |MN| \\cdot d = \\frac{1}{2} \\times \\frac{24}{7} \\times \\frac{\\sqrt{2}}{2} = \\frac{6\\sqrt{2}}{7}\n$$\nTherefore, the area of $ AMBN $ is $ \\frac{6\\sqrt{2}}{7} $."
  },
  {
    "text": "Given that point $P$ lies on the parabola $y^{2}=4$, let $d_{1}$ be the distance from point $P$ to the directrix of the parabola, and let $d_{2}$ be the distance from $P$ to a moving point $Q$ on the circle $(x+3)^{2}+(y-3)^{2}=1$. Then the minimum value of $d_{1}+d_{2}$ is?",
    "fact_expressions": "P: Point;G: Parabola;Expression(G) = (y^2 = 4);PointOnCurve(P, G);d1: Number;d2: Number;Distance(P, Directrix(G)) = d1;Distance(P, Q) = d2;H: Circle;Expression(H) = ((x + 3)^2 + (y - 3)^2 = 1);Q: Point;PointOnCurve(Q, H)",
    "query_expressions": "Min(d1+ d2)",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 6], [24, 28]], [[7, 19], [29, 32]], [[7, 19]], [[2, 22]], [[38, 45]], [[82, 89]], [[24, 45]], [[24, 89]], [[46, 70]], [[46, 70]], [[75, 78]], [[46, 78]]]",
    "query_spans": "[[[91, 110]]]",
    "process": ""
  },
  {
    "text": "Given that $A$ and $B$ are two points on the parabola $x^{2}=4 y$, and the midpoint of segment $AB$ is $M(2 , 2)$, then $|AB|$ equals?",
    "fact_expressions": "G: Parabola;B: Point;A: Point;M: Point;Expression(G) = (x^2 = 4*y);Coordinate(M) = (2, 2);PointOnCurve(A, G);PointOnCurve(B, G);MidPoint(LineSegmentOf(A,B)) = M",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "4*sqrt(2)",
    "fact_spans": "[[[10, 24]], [[6, 9]], [[2, 5]], [[40, 50]], [[10, 24]], [[40, 50]], [[2, 28]], [[2, 28]], [[29, 50]]]",
    "query_spans": "[[[52, 63]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1$ $(a>0)$, and that an asymptote of the hyperbola is given by $3x - y = 0$. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola, respectively. If $|P F_{2}| = 3$, then $|P F_{1}| = $?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/9 + x^2/a^2 = 1);a: Number;a>0;P: Point;PointOnCurve(P, RightPart(G));F1: Point;LeftFocus(G)=F1;F2: Point;RightFocus(G)=F2;Expression(OneOf(Asymptote(G)))= (3*x - y = 0);Abs(LineSegmentOf(P, F2)) = 3",
    "query_expressions": "Abs(LineSegmentOf(P, F1))",
    "answer_expressions": "5",
    "fact_spans": "[[[6, 53], [60, 63], [101, 104]], [[6, 53]], [[9, 53]], [[9, 53]], [[2, 5]], [[2, 59]], [[83, 90]], [[83, 110]], [[91, 98]], [[83, 110]], [[60, 81]], [[112, 125]]]",
    "query_spans": "[[[127, 140]]]",
    "process": "Since the asymptotes of the hyperbola are given by $3x - y = 0$, that is, $y = 3x = \\frac{b}{a}$, we have $\\frac{3}{a} = 3$, solving gives $a = 1$. According to the definition of a hyperbola, point $P$ lies on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{9} = 1$ ($a > 0$) satisfying $|PF_{1}| - |PF_{2}| = 2a = 2$, so $|PF_{1}| = |PF_{2}| + 2 = 5$."
  },
  {
    "text": "The curve $\\frac{x^{2}}{25-k}+\\frac{y^{2}}{k-9}=1$ is an ellipse with foci on the $x$-axis; then the range of $k$ is?",
    "fact_expressions": "H:Ellipse;k: Number;Expression(G) = (x^2/(25 - k) + y^2/(k - 9) = 1);PointOnCurve(Focus(H),xAxis);H = G;G: Curve",
    "query_expressions": "Range(k)",
    "answer_expressions": "(9,17)",
    "fact_spans": "[[[52, 54]], [[56, 59]], [[0, 54]], [[43, 54]], [0, 51], [0, 39]]",
    "query_spans": "[[[56, 64]]]",
    "process": "According to the curve representing an ellipse with foci on the x-axis, set up inequalities to solve. \\because the curve \\frac{x^2}{25-k}+\\frac{y^{2}}{k-9}=1 is an ellipse with foci on the x-axis, \\therefore 25-k>k-9>0, solving gives: 9<k<17."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and point $P$ is a moving point on this ellipse, then $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}} \\in$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G)",
    "query_expressions": "Range(DotProduct(VectorOf(P,F1),VectorOf(P,F2)))",
    "answer_expressions": "[-2,1]",
    "fact_spans": "[[[20, 47], [58, 60]], [[52, 56]], [[2, 9]], [[10, 17]], [[20, 47]], [[2, 51]], [[52, 64]]]",
    "query_spans": "[[[66, 128]]]",
    "process": "From the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, we know the foci are $F_{1}(-\\sqrt{3},0)$, $F_{2}(\\sqrt{3},0)$, let $P(x,y)$, $-2\\leqslant x \\leqslant 2$, then $\\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=(-\\sqrt{3}-x,-y)\\cdot(\\sqrt{3}-x,-y)=x^{2}+y^{2}-3=x^{2}+1-\\frac{x^{2}}{4}-3=\\frac{1}{4}(3x^{2}-8)$. Since $-2\\leqslant x \\leqslant 2$, $\\therefore 0\\leqslant x^{2} \\leqslant 4$, hence $\\overrightarrow{PF}\\cdot\\overrightarrow{PF} \\in [-2,1]$."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=x$ be $F$, and let point $M$ lie on the parabola. The extension of segment $MF$ intersects the line $x=-\\frac{1}{4}$ at point $N$. Then the value of $\\frac{1}{|MF|}+\\frac{1}{|NF|}$ is?",
    "fact_expressions": "G: Parabola;H: Line;F: Point;M: Point;N: Point;Expression(G) = (y^2 = x);Expression(H) = (x = -1/4);Focus(G) = F;PointOnCurve(M, G);Intersection(OverlappingLine(LineSegmentOf(M, F)), H) = N",
    "query_expressions": "1/Abs(LineSegmentOf(N, F)) + 1/Abs(LineSegmentOf(M, F))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 13], [26, 29]], [[43, 61]], [[17, 20]], [[21, 25]], [[63, 67]], [[1, 13]], [[43, 61]], [[1, 20]], [[21, 30]], [[31, 67]]]",
    "query_spans": "[[[69, 106]]]",
    "process": "From the given conditions, F is $(\\frac{1}{4},0)$ and the directrix equation is $x=-\\frac{1}{4}$. Draw MH perpendicular to the directrix from point M, with H as the foot of the perpendicular, and let K be the intersection point of the directrix and the x-axis. It follows that $\\triangle NFK \\sim \\triangle NMH$, thus obtaining $\\frac{|FK|}{|MH|}=\\frac{|NF|}{|NF|+|MF|}$, simplifying yields the required result. From the problem, F is $(\\frac{1}{4},0)$, the directrix equation is $x=-\\frac{1}{4}$, draw MH perpendicular to the directrix from point M, with H as the foot of the perpendicular, and let K be the intersection point of the directrix and the x-axis. Then by the definition of a parabola, $|MF|=|MH|$, $|FK|=\\frac{1}{2}$, and it is easy to obtain $\\frac{|NF|}{|NF|+|MF|}=\\frac{|FK|}{|MH|}$, i.e., $\\therefore\\frac{1}{2|MH|}=\\frac{|NF|}{|NF|+|MF|} \\therefore 2|MF|\\cdot|NF|=|NF|+|MF|$, dividing both sides by $|MF|\\cdot|NF|$ gives $\\therefore\\frac{1}{|MF|}+\\frac{1}{|NF|}=2$"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A point $P$ on the hyperbola satisfies $P F_{2} \\perp x$-axis. If $|F_{1} F_{2}|=12$, $|P F_{2}|=5$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F2: Point;F1: Point;a > b;b > 0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P,G);IsPerpendicular(LineSegmentOf(P,F2),xAxis);Abs(LineSegmentOf(F1,F2)) = 12;Abs(LineSegmentOf(P,F2)) = 5",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3/2",
    "fact_spans": "[[[2, 55], [79, 82], [146, 149]], [[5, 55]], [[5, 55]], [[85, 88]], [[71, 78]], [[63, 70]], [[5, 55]], [[5, 55]], [[2, 55]], [[2, 78]], [[2, 78]], [[79, 88]], [[90, 108]], [[111, 129]], [[130, 143]]]",
    "query_spans": "[[[146, 155]]]",
    "process": "In RtAPF_{1}F_{2}, DE, we obtain |PF_{1}|=\\sqrt{12^{2}+5^{2}}=13, then c=6, 2a=13-5=8, a=4, hence e=\\frac{6}{4}=\\frac{3}{2},"
  },
  {
    "text": "Find the equation of an ellipse that has the same eccentricity as $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$ and passes through the point $( \\sqrt{5}, 2)$.",
    "fact_expressions": "C: Curve;Expression(C) = (x^2/5 + y^2/4 = 1);G: Ellipse;Eccentricity(C) = Eccentricity(G);H: Point;Coordinate(H) = (sqrt(5), 2);PointOnCurve(H, G) = True",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{(x^2/10+y^2/8=1),(41*y^2/4+41*x^2/5=1)}",
    "fact_spans": "[[[2, 37]], [[2, 37]], [[64, 66]], [[1, 66]], [[46, 63]], [[46, 63]], [[45, 66]]]",
    "query_spans": "[[[64, 69]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $\\Gamma$: $x^{2}=2 y$, the line passing through points $A(0,-2)$ and $B(t, 0)$ has no common points with the parabola. Then the range of real values for $t$ is?",
    "fact_expressions": "Gamma:Parabola;H: Line;A: Point;B: Point;t: Real;Expression(Gamma) = (x^2=2*y);Coordinate(A) = (0,-2);Coordinate(B) = (t, 0);PointOnCurve(A,H);PointOnCurve(B,H);NumIntersection(H,Gamma)=0",
    "query_expressions": "Range(t)",
    "answer_expressions": "(-\\infty,-1)+(1,+\\infty)",
    "fact_spans": "[[[2, 26], [52, 55]], [[49, 51]], [[28, 38]], [[39, 48]], [[62, 67]], [[2, 26]], [[28, 38]], [[39, 48]], [[27, 51]], [[27, 51]], [[49, 60]]]",
    "query_spans": "[[[62, 74]]]",
    "process": "Clearly, $ t \\neq 0 $, the equation of line $ AB $ is $ \\frac{x}{t} + \\frac{y}{-2} = 1 $, that is, $ 2x - ty - 2t = 0 $. From \n\\[\n\\begin{cases}\n2x - ty - 2t = 0 \\\\\nx^{2} = 2y\n\\end{cases}\n\\],\neliminating $ y $ gives $ tx^{2} - 4x + 4t = 0 $. According to the problem, $ a = (-4)^{2} - 16t^{2} < 0 $, solving yields $ t < -1 $ or $ t > 1 $."
  },
  {
    "text": "It is known that the foci of hyperbola $C$ lie on the coordinate axes, the center is at the origin, and its asymptotes are given by $y = \\pm 2x$. Then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;O: Origin;Expression(Asymptote(C)) = (y = pm*2*x);PointOnCurve(Focus(C),axis);Center(C)=O",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "{sqrt(5),sqrt(5)/2}",
    "fact_spans": "[[[2, 8], [25, 26], [46, 52]], [[20, 24]], [[25, 43]], [[2, 16]], [[2, 24]]]",
    "query_spans": "[[[46, 58]]]",
    "process": "Let the real semi-axis length of the hyperbola be $ a $, and the imaginary semi-axis length be $ b $. When the foci are on the x-axis: $ \\frac{b}{a}=2 $, $ \\therefore e=\\sqrt{1+(\\frac{b}{a})^{2}}=\\sqrt{5} $; when the foci are on the y-axis: $ \\frac{a}{b}=2 $, $ e=\\frac{c}{a}=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{1+(\\frac{b}{a})^{2}}=\\frac{\\sqrt{5}}{2} $"
  },
  {
    "text": "Given the parabola $y^{2}=4x$, a chord passing through the focus $F$ intersects the parabola at points $A$ and $B$. Perpendiculars from $A$ and $B$ to the $y$-axis have feet at $C$ and $D$, respectively. Then the minimum value of $|AB|+|BD|$ is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;D: Point;C: Point;L1:Line;L2:Line;H:LineSegment;Expression(G) = (y^2 = 4*x);Focus(G)=F;PointOnCurve(F,H);Intersection(H,G)={A,B};PointOnCurve(A,L1);PointOnCurve(B,L2);IsPerpendicular(L1,yAxis);IsPerpendicular(L2,yAxis);FootPoint(L1,yAxis)=C;FootPoint(L2,yAxis)=D;IsChordOf(H,G);F:Point",
    "query_expressions": "Min(Abs(LineSegmentOf(A, B)) + Abs(LineSegmentOf(B, D)))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 16], [27, 30]], [[32, 35], [43, 46]], [[36, 39], [47, 50]], [[69, 72]], [[65, 68]], [], [], [], [[2, 16]], [[2, 24]], [[18, 26]], [[18, 41]], [[42, 59]], [[42, 59]], [[42, 59]], [[42, 59]], [[42, 72]], [[42, 72]], [[2, 26]], [[21, 24]]]",
    "query_spans": "[[[74, 91]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $2 x^{2}-y^{2}=2$, what is the equation of the line on which the chord of the hyperbola lies, with point $A(2,3)$ as its midpoint?",
    "fact_expressions": "G: Hyperbola;H: LineSegment;A: Point;Expression(G) = (2*x^2 - y^2 = 2);Coordinate(A) = (2, 3);IsChordOf(H,G);MidPoint(H)=A",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "4*x-3*y+1=0",
    "fact_spans": "[[[2, 22], [38, 41]], [], [[25, 34]], [[2, 22]], [[25, 34]], [[38, 43]], [[24, 43]]]",
    "query_spans": "[[[38, 52]]]",
    "process": "Let the endpoints of the chord with midpoint A(2,3) be P_{1}(x_{1},y_{1}), P_{2}(x_{2},y_{2}). Then x_{1}+x_{2}=4, y_{1}+y_{2}=6. Also, 2x_{1}^{2}-y_{1}^{2}=2,\\textcircled{1} 2x_{2}^{2}-y_{2}^{2}=2,\\textcircled{2} \\textcircled{1}-\\textcircled{2} gives: 2(x_{1}+x_{2})(x_{1}-x_{2})=(y_{1}+y_{2})(y_{1}-y_{2}). By symmetry, x_{1}\\neq x_{2}. Therefore, the line containing the chord with midpoint A(2,3) satisfies \\frac{y_{2}}{x_{2}}=\\frac{2(x_{1}+x}{y_{+}y_{0}}\\frac{x_{2}}{2}=\\frac{2\\times4}{6}=\\frac{4}{3}. Thus, the equation of the line containing the midpoint chord is y\\cdot3=\\frac{4}{3}), i.e., 4x-3v+1=0."
  },
  {
    "text": "Point $P$ lies on the ellipse $C_{1}$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and the right focus of $C_{1}$ is $F$. Point $Q$ lies on the circle $C_{2}$: $x^{2}+y^{2}+6 x-8 y+21=0$. Then the minimum value of $|P Q|-|P F|$ is?",
    "fact_expressions": "C1: Ellipse;Expression(C1) = (x^2/4 + y^2/3 = 1);P: Point;PointOnCurve(P, C1);F: Point;RightFocus(C1) = F;C2: Circle;Expression(C2) = (-8*y + 6*x + x^2 + y^2 + 21 = 0);Q: Point;PointOnCurve(Q, C2)",
    "query_expressions": "Min(-Abs(LineSegmentOf(P, F)) + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "2*sqrt(5)-6",
    "fact_spans": "[[[5, 51], [53, 60]], [[5, 51]], [[0, 4]], [[0, 52]], [[65, 68]], [[53, 68]], [[74, 110]], [[74, 110]], [[69, 73]], [[69, 111]]]",
    "query_spans": "[[[113, 132]]]",
    "process": "Let the left focus of the ellipse $ C_{1}: \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $ be $ E(-1,0) $. By the definition of an ellipse, we have $ |PE| + |PF| = 2a = 4 $. Then, from the equation of the circle, we obtain that the center of circle $ C_{2} $ is $ (-3,4) $ and the radius is $ r = 2 $. Drawing the graph and combining with the figure, we get $ |PQ| - |PF| = |PQ| + |PE| - 4 \\geqslant |PC_{2}| + |PE| - 6 \\geqslant |EC_{2}| - 6 $, thus the result can be obtained. Let the left focus of the ellipse $ C_{1}: \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $ be $ E(-1,0) $. By the definition of the ellipse, $ |PE| + |PF| = 2a = 4 $, so $ |PQ| - |PF| = |PQ| + |PE| - 4 $. From $ x^{2} + y^{2} + 6x - 8y + 21 = 0 $, we get $ (x+3)^{2} + (y-4)^{2} = 4 $, i.e., the center of circle $ C_{2} $ is $ (-3,4) $, and the radius is $ r = 2 $. The graph is shown below:  \n\\begin{matrix} \\text{By the triangle inequality,} & |PQ| \\geqslant |PC_{2}| - r = |PC_{2}| - 2, \\\\ |PQ| - |PF| = |PQ| + |PE| - 4 \\geqslant |PC_{2}| + |PE| - 6 \\geqslant |EC_{2}| - 6 = \\sqrt{(-3+1)^{2} + 4^{2}} - 6 = 2\\sqrt{5} - 6 \\end{matrix} (equality holds if and only if points $ C_{2}, Q, P, E $ are collinear.)"
  },
  {
    "text": "Let $P$ be a moving point on the ellipse $\\frac{x^{2}}{4^{2}}+\\frac{y^{2}}{3^{2}}=1$, then the sum of the distances from $P$ to the two foci of this ellipse is?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/4^2 + y^2/3^2 = 1);PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2}",
    "query_expressions": "Distance(P, F1) + Distance(P, F2)",
    "answer_expressions": "8",
    "fact_spans": "[[[5, 50], [61, 63]], [[1, 4], [56, 59]], [[5, 50]], [[1, 54]], [], [], [[61, 67]]]",
    "query_spans": "[[[56, 73]]]",
    "process": "From the ellipse equation, find a, then use the definition of the ellipse to obtain the result. From \\frac{x^{2}}{4^{2}}+\\frac{y^{2}}{3^{2}}=1, we get a=4. By the definition of the ellipse, the sum of the distances from point P to the two foci of this ellipse is 2a=8."
  },
  {
    "text": "A point $A$ on the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ is at a distance of $15$ from the point $(5,0)$. Then, the distance from point $A$ to the point $(-5,0)$ is?",
    "fact_expressions": "G: Hyperbola;H: Point;I: Point;A: Point;Expression(G) = (x^2/16 - y^2/9 = 1);Coordinate(H) = (5, 0);Coordinate(I) = (-5, 0);PointOnCurve(A, G);Distance(A, H) = 15",
    "query_expressions": "Distance(A, I)",
    "answer_expressions": "7, 23",
    "fact_spans": "[[[0, 39]], [[46, 54]], [[69, 78]], [[42, 45], [64, 68]], [[0, 39]], [[46, 54]], [[69, 78]], [[0, 45]], [[42, 62]]]",
    "query_spans": "[[[64, 83]]]",
    "process": "\\because the hyperbola \\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1, \\therefore 2a=8, (5,0) and (-5,0) are the two foci of the hyperbola, \\because point A lies on the hyperbola, \\therefore ||PF_{1}|-|PF_{2}||=8, \\because the distance from point A to (5,0) is 15, then the distance from point A to (-5,0) is 15+8=23 or 15-8=7,"
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $9 x^{2}-4 y^{2}=-36$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (9*x^2 - 4*y^2 = -36)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(3/2)*x",
    "fact_spans": "[[[0, 24]], [[0, 24]]]",
    "query_spans": "[[[0, 32]]]",
    "process": "Find the standard equation of the hyperbola, and solve by combining with the equation of the asymptotes of the hyperbola. The standard equation of the hyperbola is \\frac{y^{2}}{9}\\cdot\\frac{x^{2}}{4}=1, then the asymptotes of the hyperbola are y=\\pm\\frac{3}{2}x"
  },
  {
    "text": "It is known that the hyperbola $E$ passing through the point $(2, \\sqrt{3})$ has the same asymptotes as the hyperbola $C$: $\\frac{x^{2}}{4}-y^{2}=1$. Then the equation of the hyperbola $E$ is?",
    "fact_expressions": "H: Point;Coordinate(H) = (2, sqrt(3));E: Hyperbola;PointOnCurve(H, E);C: Hyperbola;Expression(C) = (x^2/4-y^2=1);Asymptote(E) = Asymptote(C)",
    "query_expressions": "Expression(E)",
    "answer_expressions": "y^2/2 - x^2/8 = 1",
    "fact_spans": "[[[3, 19]], [[3, 19]], [[20, 26], [68, 74]], [[2, 26]], [[27, 60]], [[27, 60]], [[20, 66]]]",
    "query_spans": "[[[68, 79]]]",
    "process": "Since hyperbola E has the same asymptotes as hyperbola C: \\frac{x^{2}}{4}-y^{2}=1, the equation of hyperbola E can be written as \\frac{x^{2}}{4}-y^{2}=\\lambda. Substituting the coordinates of the point (2,\\sqrt{3}) into \\frac{x^{2}}{4}-y^{2}=\\lambda gives: 1-3=\\lambda, \\therefore\\lambda=-2, \\therefore the standard equation of the required hyperbola is \\frac{x^{2}}{4}-y^{2}=-2, which is \\frac{y^{2}}{2}-\\frac{x^{2}}{8}=1."
  },
  {
    "text": "If a point $P$ on the right branch of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ is at a distance of $1$ from the right focus of the hyperbola, then what is the distance from point $P$ to the $y$-axis?",
    "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2/4 - y^2/5 = 1);PointOnCurve(P,RightPart(G));Distance(P,RightFocus(G)) = 1",
    "query_expressions": "Distance(P, yAxis)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 40], [49, 52]], [[45, 48], [65, 69]], [[2, 40]], [[2, 48]], [[45, 62]]]",
    "query_spans": "[[[65, 79]]]",
    "process": "In the hyperbola $\\frac{x^2}{4}-\\frac{y^{2}}{5}=1$, $a=2$, $b=\\sqrt{5}$, $c=\\sqrt{a^{2}+b^{2}}=3$. Therefore, the right focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ is $F(3,0)$, and the distance from the right vertex of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ to $F$ is $1$, so $P(2,0)$. Thus, the distance from point $P$ to the $y$-axis is $2$."
  },
  {
    "text": "Given that the line $l$ passing through the focus $F$ of the parabola $C$: $y^{2}=8x$ intersects the parabola $C$ at points $A$ and $B$, if $P$ is the midpoint of segment $AB$, $O$ is the origin, and the line $OP$ is extended to intersect the parabola $C$ at point $Q$, then the range of $\\frac{|OP|}{|OQ|}$ is?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;O: Origin;P: Point;Q: Point;F: Point;Expression(C) = (y^2 = 8*x);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};MidPoint(LineSegmentOf(A,B)) = P;Intersection(OverlappingLine(LineSegmentOf(O,P)),C)=Q",
    "query_expressions": "Range(Abs(LineSegmentOf(O, P))/Abs(LineSegmentOf(O, Q)))",
    "answer_expressions": "(0,1/2)",
    "fact_spans": "[[[28, 33]], [[3, 21], [34, 40], [88, 94]], [[41, 44]], [[45, 48]], [[67, 70]], [[52, 55]], [[95, 99]], [[24, 27]], [[3, 21]], [[3, 27]], [[2, 33]], [[28, 50]], [[52, 66]], [[76, 99]]]",
    "query_spans": "[[[101, 129]]]",
    "process": "The parabola $ C: y^2 = 8x $ has focus $ F(2,0) $. The slope of line $ l $ exists and is non-zero. Let the equation of line $ l $ be $ y = k(x - 2) $. Combining\n\\[\n\\begin{cases}\ny = k(x - 2) \\\\\ny^2 = 8x\n\\end{cases}\n\\]\nand eliminating $ y $, we obtain: $ k^{2}x^{2} - 4(k^{2} + 2)x + 4k^{2} = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, $ P(x_{0}, y_{0}) $, $ Q(x_{3}, y_{3}) $. Then $ x_{1} + x_{2} = \\frac{4(k^{2} + 2)}{k^{2}} $, so $ x_{0} = \\frac{x_{1} + x_{2}}{2} = \\frac{2(k^{2} + 2)}{k^{2}} $, $ y_{0} = k(x_{0} - 2) = \\frac{4}{k} $, thus $ k_{OQ} = \\frac{y_{0}}{x_{0}} = \\frac{2k}{k^{2} + 2} $. The equation of line $ OQ $ is $ y = \\frac{2k}{k^{2} + 2}x $. Solving simultaneously\n\\[\n\\begin{cases}\ny = \\frac{2k}{k^{2} + 2}x \\\\\ny^2 = 8x\n\\end{cases}\n\\]\ngives $ x_{3} = \\frac{2(k^{2} + 2)^{2}}{k^{2}} $. Since $ k^{2} > 0 $, then $ \\frac{|OP|}{|OQ|} = \\frac{x_{0}}{x_{3}} = \\frac{1}{k^{2} + 2} < \\frac{1}{2} $. Therefore, the range of $ \\frac{|OP|}{|OQ|} $ is $ \\left(0, \\frac{1}{2}\\right) $."
  },
  {
    "text": "Given a hyperbola $ C $ with foci on the $ x $-axis, the left focus is $ F $ and the right vertex is $ A $. If the perpendicular bisector of segment $ FA $ has no common point with the hyperbola $ C $, then the range of the eccentricity of hyperbola $ C $ is?",
    "fact_expressions": "C: Hyperbola;A: Point;F: Point;PointOnCurve(Focus(C), xAxis);LeftFocus(C) = F;RightVertex(C) = A;NumIntersection(PerpendicularBisector(LineSegmentOf(F,A)),C)=0",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(1, 3)",
    "fact_spans": "[[[11, 17], [49, 55], [62, 68]], [[30, 33]], [[22, 25]], [[2, 17]], [[11, 25]], [[11, 33]], [[35, 60]]]",
    "query_spans": "[[[62, 79]]]",
    "process": "\\because the left focus of hyperbola C with foci on the x-axis is F and the right vertex is A, \\therefore F(-c,0), A(a,0). \\because the perpendicular bisector of segment FA has no common point with hyperbola C, \\therefore \\frac{a-c}{2} > -a, \\therefore e = \\frac{c}{a} < 3. \\because e \\in (1,+\\infty), \\therefore 1 < e < 3"
  },
  {
    "text": "The line $l$ passing through the focus $F$ of the parabola $C: x^{2} = 2y$ intersects $C$ at two points $A$ and $B$. The tangent to the parabola at point $A$ intersects the $x$-axis and $y$-axis at points $P$ and $Q$, respectively. If the area of $\\triangle POQ$ ($O$ being the origin) is $1$, then $|AF| = $?",
    "fact_expressions": "l: Line;C: Parabola;F: Point;A: Point;B: Point;P:Point;Q:Point;Expression(C) = (x^2 = 2*y);Focus(C) = F;Intersection(l,C)={A,B};PointOnCurve(F, l);Intersection(TangentOnPoint(A, C),xAxis)=P;Intersection(TangentOnPoint(A, C),yAxis)=Q;Area(TriangleOf(P,O,Q))=1;O:Origin",
    "query_expressions": "Abs(LineSegmentOf(A,F))",
    "answer_expressions": "5/2",
    "fact_spans": "[[[26, 31]], [[1, 19], [32, 35]], [[22, 25]], [[38, 41], [46, 50]], [[42, 45]], [[70, 73]], [[74, 77]], [[1, 19]], [[1, 25]], [[26, 45]], [[0, 31]], [[32, 77]], [[32, 77]], [[79, 113]], [[97, 100]]]",
    "query_spans": "[[[115, 124]]]",
    "process": "Let $ A(t,\\frac{t^{2}}{2}) $ ($ t>0 $). From the parabola $ C: x^{2}=2y $, we get $ y=\\frac{x^{2}}{2} $, $ y'=x $, so $ k=t $. Then the tangent line at point $ A $ is $ y-\\frac{t^{2}}{2}=t(x-t) $, intersecting the $ x $-axis and $ y $-axis at points $ P(\\frac{t}{2},0) $ and $ Q(0,-\\frac{t^{2}}{2}) $, respectively. Since the area of $ \\triangle POQ $ is 1, we have $ \\frac{1}{2}\\times|\\frac{t}{2}|\\times|-\\frac{t^{2}}{2}| $. Solving gives $ t=2 $, so $ A(2,2) $, then $ |AF|=2+\\frac{1}{2}=\\frac{5}{2} $."
  },
  {
    "text": "Let point $P$ be a point on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$, and let $F_{1}$, $F_{2}$ be the left and right foci of the ellipse respectively, and $I$ the incenter of $\\Delta PF_{1} F_{2}$. If $S_{\\Delta IPF_{1}}+S_{\\Delta IPF_{2}}=2 S_{\\Delta IF_{1}F_{2}}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "P: Point;G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;PointOnCurve(P, G);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;I: Point;Incenter(TriangleOf(P,F1,F2))=I;Area(TriangleOf(I, P, F1)) + Abs(TriangleOf(I, P, F2)) = 2*Area(TriangleOf(I, F1, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[1, 5]], [[6, 64], [86, 88], [185, 187]], [[6, 64]], [[8, 64]], [[8, 64]], [[8, 64]], [[8, 64]], [[1, 67]], [[68, 75]], [[76, 83]], [[68, 94]], [[68, 94]], [[95, 98]], [[95, 123]], [[125, 182]]]",
    "query_spans": "[[[185, 193]]]",
    "process": ""
  },
  {
    "text": "Let the circle $C$: $(x-3)^{2}+y^{2}=4$ pass through the focus of the parabola $y^{2}=2 p x(p>0)$. Then the equation of the parabola is?",
    "fact_expressions": "C: Circle;Expression(C) = (y^2 + (x - 3)^2 = 4);G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;PointOnCurve(Focus(G), C)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=2*x, y^2=10*x}",
    "fact_spans": "[[[1, 26]], [[1, 26]], [[28, 49], [54, 57]], [[28, 49]], [[31, 49]], [[31, 49]], [[1, 52]]]",
    "query_spans": "[[[54, 62]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the curve $\\frac{x^{2}}{4}+\\frac{y^{2}}{1+m}=1$ is $e=\\frac{1}{2}$, then what is the value of $m$?",
    "fact_expressions": "E: Curve;m: Number;Expression(E) = (x**2/4 + y**2/(m + 1) = 1);Eccentricity(E) = e;e = 1/2;e: Number",
    "query_expressions": "m",
    "answer_expressions": "{2, 13/3}",
    "fact_spans": "[[[0, 39]], [[60, 63]], [[0, 39]], [[0, 58]], [[43, 58]], [[43, 58]]]",
    "query_spans": "[[[60, 67]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a moving point on the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, $F_{1}$ and $F_{2}$ are its left and right foci respectively, and $O$ is the origin, then the range of $\\frac{|P F_{1}|+|P F_{2}|}{| P O| }$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/12 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;LeftFocus(G)=F1;RightFocus(G)=F2;O: Origin",
    "query_expressions": "Range((Abs(LineSegmentOf(P,F1))+Abs(LineSegmentOf(P,F2)))/Abs(LineSegmentOf(P,O)))",
    "answer_expressions": "(2,4]",
    "fact_spans": "[[[6, 45], [68, 69]], [[6, 45]], [[2, 5]], [[2, 49]], [[50, 57]], [[58, 65]], [[50, 74]], [[50, 74]], [[75, 78]]]",
    "query_spans": "[[[85, 129]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of a hyperbola are $y=\\pm \\frac{x}{2}$ and the length of the conjugate axis is $4$, then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*(x/2));Length(ImageinaryAxis(G)) = 4",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/16-y^2/4=1,y^2-x^2/4=1}",
    "fact_spans": "[[[2, 5], [42, 45]], [[2, 31]], [[2, 39]]]",
    "query_spans": "[[[42, 52]]]",
    "process": ""
  },
  {
    "text": "Given that one of the asymptotes of the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{3}=1$ is $y=\\sqrt{3} x$, then what is the distance from a point $M(2, y_{0})$ on the parabola $y^{2}=4 a x$ to the focus $F$ of the parabola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/3 + x^2/a = 1);a: Number;Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x);H: Parabola;Expression(H) = (y^2 = 4*a*x);M: Point;Coordinate(M) = (2, y0);y0: Number;PointOnCurve(M, H);F: Point;Focus(H) = F",
    "query_expressions": "Distance(M, F)",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 40]], [[2, 40]], [[5, 40]], [[2, 64]], [[66, 82], [100, 103]], [[66, 82]], [[85, 98]], [[85, 98]], [[85, 98]], [[66, 98]], [[105, 108]], [[100, 108]]]",
    "query_spans": "[[[85, 113]]]",
    "process": "Since one asymptote of the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{3}=1$ is $y=\\sqrt{3}x$, it follows that $a>0$, $a=1$, so the equation of the parabola is $y^{2}=4x$. Therefore, the distance from $M(2,y_{0})$ to the focus $F(1,0)$ of the parabola equals $x_{M}+\\frac{p}{2}=2+1=3$. The answer is: $3$"
  },
  {
    "text": "The equation of an ellipse centered at the origin, with directrices $x = \\pm 4$ and eccentricity $\\frac{1}{2}$ is?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;Expression(Directrix(G)) = (x = pm*4);Eccentricity(G) = 1/2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[40, 42]], [[3, 5]], [[0, 42]], [[6, 42]], [[21, 42]]]",
    "query_spans": "[[[40, 46]]]",
    "process": "From the given conditions, we have \\frac{a2}{c}=4, e=\\frac{c}{a}=\\frac{1}{2} \\therefore a=2, c=1 \\therefore b^{2}=3, and the equation of the ellipse is \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1"
  },
  {
    "text": "If the hyperbola $C$ passes through the point $(2 , 2)$ and has the same asymptotes as the hyperbola $\\frac{y^{2}}{4}-x^{2}=1$, then the standard equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Hyperbola;P:Point;Coordinate(P) = (2, 2);PointOnCurve(P, C);Asymptote(C)=Asymptote(G);Expression(G)=(y^2/4-x^2=1)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/3 - y^2/12 = 1",
    "fact_spans": "[[[1, 7], [59, 65]], [[22, 50]], [[9, 19]], [[9, 19]], [[1, 19]], [[1, 57]], [[22, 50]]]",
    "query_spans": "[[[59, 72]]]",
    "process": "According to the problem, let the standard equation of hyperbola C be \\frac{y^{2}}{4}-x^{2}=\\lambda, and it passes through the point (2,2), so \\lambda=-3, \\frac{x^{2}}{3}-\\frac{y^{2}}{12}=1"
  },
  {
    "text": "Let the parabola $C$: $y^{2}=3x$ have focus $F$, and let point $A$ be a point on the parabola $C$. If $|FA|=3$, then the inclination angle of the line $FA$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 3*x);F: Point;Focus(C) = F;A: Point;PointOnCurve(A, C);Abs(LineSegmentOf(F, A)) = 3",
    "query_expressions": "Inclination(LineOf(F, A))",
    "answer_expressions": "{pi/3, 2*pi/3}",
    "fact_spans": "[[[1, 20], [33, 39]], [[1, 20]], [[24, 27]], [[1, 27]], [[28, 32]], [[28, 42]], [[44, 53]]]",
    "query_spans": "[[[55, 68]]]",
    "process": "Let the coordinates of point A be (x, y), and the focus F of the parabola C: y^{2} = 3x be F(\\frac{3}{4}, 0). According to the definition of a parabola, we have x + \\frac{3}{4} = 3, solving gives x = \\frac{9}{4}. Substituting into the parabola equation yields y = \\pm\\frac{3\\sqrt{3}}{2}. Thus, the coordinates of A are: \\frac{9}{4}, \\pm\\frac{3\\sqrt{3}}{2}. The slope of AF is: \\frac{3\\sqrt{3}}{\\frac{9}{4}-\\frac{3}{4}} = \\pm\\sqrt{3}, so the inclination angle of line FA is: \\frac{\\pi}{3} or \\frac{2\\pi}{3}."
  },
  {
    "text": "Given that the directrix of the parabola $x^{2}=2 p y(p>0)$ is the line $y=-1$, then the coordinates of the focus of the parabola are?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;p>0;Expression(G) = (x^2 = 2*(p*y));Expression(H) = (y = -1);Directrix(G) = H",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,1)",
    "fact_spans": "[[[2, 23], [37, 40]], [[5, 23]], [[27, 35]], [[5, 23]], [[2, 23]], [[27, 35]], [[2, 35]]]",
    "query_spans": "[[[37, 46]]]",
    "process": "\\because the directrix of the parabola \\(x^{2}=2py\\) (\\(p>0\\)) is the line \\(y=-1\\), \\(\\therefore -\\frac{p}{2}=-1\\), \\(p=2\\). Hence, the focus of the parabola has coordinates \\((0,1)\\)"
  },
  {
    "text": "A point $M$ on the parabola $y=4 x^{2}$ is at a distance of $1$ from the focus. What is the ordinate of point $M$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 4*x^2);M: Point;PointOnCurve(M, G);Distance(M, Focus(G)) = 1",
    "query_expressions": "YCoordinate(M)",
    "answer_expressions": "15/16",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[18, 21], [33, 37]], [[0, 21]], [[0, 31]]]",
    "query_spans": "[[[33, 43]]]",
    "process": "From $ y = 4x^{2} $, we obtain $ x^{2} = \\frac{1}{4}y $, so the focus of the parabola is $ F(0, \\frac{1}{16}) $, and the equation of the directrix is $ y = -\\frac{1}{16} $. Let $ M(x_{M}, y_{M}) $. By the definition of the parabola, we have $ MF = y_{M} + \\frac{1}{16} = 1 $, so $ y_{M} = \\frac{15}{16} $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{6}=1$, respectively, and $M$ is a point on the right branch of the hyperbola satisfying $\\overrightarrow{M F_{1}} \\cdot \\overrightarrow{M F_{2}}=0$. If the line $M F_{2}$ intersects the hyperbola again at point $N$, then the area of $\\Delta M F_{1} N$ is?",
    "fact_expressions": "G: Hyperbola;M: Point;F2: Point;F1: Point;N: Point;Expression(G) = (x^2/4 - y^2/6 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(M,RightPart(G));DotProduct(VectorOf(M,F1),VectorOf(M,F2))=0;OneOf(Intersection(LineOf(M,F2),G))=N",
    "query_expressions": "Area(TriangleOf(M, F1, N))",
    "answer_expressions": "24",
    "fact_spans": "[[[20, 58], [69, 72], [153, 156]], [[65, 68]], [[10, 17]], [[2, 9]], [[163, 166]], [[20, 58]], [[2, 64]], [[2, 64]], [[65, 77]], [[80, 139]], [[141, 166]]]",
    "query_spans": "[[[168, 191]]]",
    "process": ""
  },
  {
    "text": "The distance from point $M$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ to the focus $F_{1}$ is $2$, and $N$ is the midpoint of $M F_{1}$. Then the value of $|O N|$ (where $O$ is the coordinate origin) is?",
    "fact_expressions": "G: Ellipse;M: Point;F1: Point;O: Origin;N: Point;Expression(G) = (x^2/25 + y^2/9 = 1);OneOf(Focus(G))=F1;Distance(M,F1)=2;MidPoint(LineSegmentOf(M,F1))=N;PointOnCurve(M,G)",
    "query_expressions": "Abs(LineSegmentOf(O, N))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 38]], [[40, 44]], [[47, 54]], [[90, 93]], [[63, 67]], [[0, 38]], [[0, 54]], [[40, 61]], [[63, 80]], [[0, 44]]]",
    "query_spans": "[[[82, 103]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ is $y=\\frac{4}{3} x$, what is the eccentricity of the hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/b^2 + y^2/a^2 = 1);b: Number;a: Number;Expression(OneOf(Asymptote(G))) = (y = (4/3)*x)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/4",
    "fact_spans": "[[[2, 48], [76, 79]], [[2, 48]], [[5, 48]], [[5, 48]], [[2, 74]]]",
    "query_spans": "[[[76, 85]]]",
    "process": "Given the equation $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$, the asymptotes are $y=\\frac{4}{3}x$, so $\\frac{a}{b}=\\frac{4}{3}$. Also, $c^{2}=a^{2}+b^{2}$, $b=\\frac{3}{4}a$, $(\\frac{c}{a})^{2}=\\frac{25}{16}$, $e=\\frac{5}{4}$."
  },
  {
    "text": "Given circle $C_{1}$: $(x+1)^{2}+y^{2}=1$ and circle $C_{2}$: $(x-1)^{2}+y^{2}=25$, a moving circle $M$ is externally tangent to circle $C_{1}$ and internally tangent to circle $C_{2}$. Then the trajectory equation of the center $M$ of the moving circle is?",
    "fact_expressions": "C1: Circle;Expression(C1) = ((x+1)^2 + y^2 = 1);C2: Circle;Expression(C2) = ((x-1)^2 + y^2 = 25);M: Circle;IsOutTangent(M, C1) ;IsInTangent(M, C2);M1: Point;Center(M) = M1",
    "query_expressions": "LocusEquation(M1)",
    "answer_expressions": "x^2/9+y^2/8=1",
    "fact_spans": "[[[2, 30], [69, 77]], [[2, 30]], [[31, 60], [80, 88]], [[31, 60]], [[63, 66], [92, 94]], [[63, 90]], [[63, 90]], [[97, 100]], [[92, 100]]]",
    "query_spans": "[[[97, 107]]]",
    "process": "From circle $ C_{1}:(x+1)^{2}+y^{2}=1 $, we obtain center $ C_{1}(-1,0) $ and radius $ r_{1}=1 $. From circle $ C_{2}:(x-1)^{2}+y^{2}=25 $, we obtain center $ C_{2}(1,0) $ and radius $ r_{2}=5 $. Let the radius of moving circle $ M $ be $ r $. Since circle $ M $ is externally tangent to circle $ C_{1} $ and internally tangent to circle $ C_{2} $, it follows that $ MC_{1}=r+1 $, $ MC_{2}=5-r $. Therefore, $ MC_{1}+MC_{2}=r+1+5-r=6>C_{1}C_{2}=2 $. Thus, the locus of point $ M $ is an ellipse with foci at $ C_{1}(-1,0) $ and $ C_{2}(1,0) $, where $ 2a=6 $. Hence, $ a=3 $, $ c=1 $, $ b=\\sqrt{a^{2}-c^{2}}=2\\sqrt{2} $. The equation of the trajectory of the center $ M $ of the moving circle is: $ \\frac{x^{2}}{9}+\\frac{y^{2}}{8}=1 $."
  },
  {
    "text": "The coordinates of the focus of the parabola $y^{2}=2 x$ are? The equation of the directrix is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x)",
    "query_expressions": "Coordinate(Focus(G));Expression(Directrix(G))",
    "answer_expressions": "(1/2, 0)  \nx = -1/2",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]], [[0, 27]]]",
    "process": "According to the problem, the coordinates of the focus are $(\\frac{1}{2},0)$, and the equation of the directrix is $x=-\\frac{1}{2}$; thus, fill in: $(\\frac{1}{2},0)$, $x=-\\frac{1}{2}$."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $x^{2}-\\frac{y^{2}}{9}=1$, respectively. If point $P$ lies on the hyperbola such that $\\overrightarrow {P F_{1}} \\cdot \\overrightarrow {P F_{2}}=0$, then $|\\overrightarrow{P F_{1}}|+|\\overrightarrow{P F_{2}}|$=?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0",
    "query_expressions": "Abs(VectorOf(P, F1)) + Abs(VectorOf(P, F2))",
    "answer_expressions": "2*sqrt(10)",
    "fact_spans": "[[[19, 47], [60, 63]], [[55, 59]], [[1, 8]], [[9, 16]], [[19, 47]], [[1, 53]], [[1, 53]], [[55, 64]], [[66, 127]]]",
    "query_spans": "[[[129, 186]]]",
    "process": "From $\\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=0$ we get $\\overrightarrow{PF}\\cdot\\overrightarrow{PF_{2}}=0$. Since $|\\overrightarrow{PF_{1}}+\\overrightarrow{PF_{2}}|^{2}=(\\overrightarrow{PF}_{1}+\\overrightarrow{PF_{2}})^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}+2\\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=|F_{1}F_{2}|^{2}=(2c)^{2}=4(1+9)=40$, therefore $|\\overrightarrow{PF}_{1}+\\overrightarrow{PF}_{2}|=2\\sqrt{10}$."
  },
  {
    "text": "In the rectangular coordinate system $x O y$, if the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ is $\\frac{5}{4}$, then the equations of the asymptotes of this hyperbola are?",
    "fact_expressions": "E: Hyperbola;a: Number;b: Number;a>0;b>0;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(E) = 5/4",
    "query_expressions": "Expression(Asymptote(E))",
    "answer_expressions": "y=pm*3/4*x",
    "fact_spans": "[[[18, 77], [98, 101]], [[21, 77]], [[21, 77]], [[21, 77]], [[21, 77]], [[18, 77]], [[18, 95]]]",
    "query_spans": "[[[98, 109]]]",
    "process": ""
  },
  {
    "text": "The endpoints of the major axis of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ are $M$ and $N$. A point $P$, distinct from $M$ and $N$, lies on this ellipse. Then, what is the product of the slopes of $PM$ and $PN$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);M: Point;N: Point;Endpoint(MajorAxis(G)) = {M, N};P: Point;Negation(P=M);Negation(P=N);PointOnCurve(P, G)",
    "query_expressions": "Slope(LineSegmentOf(P, M))*Slope(LineSegmentOf(P, N))",
    "answer_expressions": "-3/4",
    "fact_spans": "[[[0, 37], [68, 70]], [[0, 37]], [[43, 46], [54, 57]], [[47, 50], [58, 61]], [[0, 50]], [[62, 66]], [[51, 66]], [[51, 66]], [[62, 71]]]",
    "query_spans": "[[[74, 93]]]",
    "process": "According to the problem, M(2,0), N(-2,0), and P is any point on the ellipse, with coordinates set as P(2\\cosw,\\sqrt{3}\\sinw). The slopes of PM and PN are k_{1}=\\frac{\\sqrt{3}\\sinv}{2(\\cos\\sqrt[b]{3})}, k_{2}=\\frac{\\sqrt{3}\\sinv}{2(\\cosw+1)}, respectively."
  },
  {
    "text": "If the line $y = kx + 1$ intersects the curve $x = \\sqrt{y^{2} + 1}$ at two distinct points, then what is the range of values for $k$?",
    "fact_expressions": "G: Line;k: Number;H: Curve;Expression(G) = (y = k*x + 1);Expression(H) = (x = sqrt(y^2 + 1));NumIntersection(G, H) = 2",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-sqrt(2),-1)",
    "fact_spans": "[[[1, 12]], [[43, 46]], [[13, 33]], [[1, 12]], [[13, 33]], [[1, 41]]]",
    "query_spans": "[[[43, 53]]]",
    "process": "The line y = kx + 1 passes through the point P(0,1), and the curve x = \\sqrt{y^{2}+1} is the right branch of the hyperbola x^{2}-y^{2}=1. As shown in the figure, its asymptotes are given by y = \\pm x. When the line PM is tangent to the right branch of the hyperbola, solving \\begin{cases} y = kx + 1 \\\\ x^{2} - y^{2} = 1 \\end{cases} yields (1 - k^{2})x^{2} - 2kx - 2 = 0. The discriminant is \\Delta = 4k^{2} + 8(1 - k^{2}) = 0, so k = -\\sqrt{2} (k = \\sqrt{2} is discarded). Therefore, when the line y = kx + 1 intersects the curve x = \\sqrt{y^{2}+1} at two distinct points, -\\sqrt{2} < k < -1."
  },
  {
    "text": "The distance from the focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$ to its asymptote is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/12 = 1)",
    "query_expressions": "Distance(Focus(G), Asymptote(G))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 51]]]",
    "process": ""
  },
  {
    "text": "It is known that the center of ellipse $C$ is at the origin of coordinates, its foci lie on the $x$-axis, and its eccentricity equals $\\frac{\\sqrt{2}}{2}$. One of its vertices coincides with the focus of the parabola $x^{2}=4 y$. Then the standard equation of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;O: Origin;Center(C) = O;PointOnCurve(Focus(C), xAxis);Eccentricity(C) = sqrt(2)/2;G: Parabola;Expression(G) = (x^2 = 4*y);OneOf(Vertex(C)) = Focus(G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/2+y^2=1",
    "fact_spans": "[[[2, 7], [79, 84], [51, 52]], [[11, 15]], [[2, 15]], [[2, 24]], [[2, 50]], [[60, 74]], [[60, 74]], [[51, 77]]]",
    "query_spans": "[[[79, 91]]]",
    "process": "\\because the ellipse C is centered at the origin, with foci on the x-axis and eccentricity equal to \\frac{\\sqrt{2}}{2}; one of its vertices coincides exactly with the focus of the parabola x^{2}=4y. According to the given conditions, assume the equation of the ellipse is \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0). Then we have \\begin{cases}\\frac{c}{a}=\\frac{\\sqrt{2}}{2}\\\\b=1\\end{cases}, solving gives a=\\sqrt{2}, b=c=1 \\therefore the equation of ellipse C: \\frac{x^{2}}{2}+y^{2}=1"
  },
  {
    "text": "The equation of the directrix of the parabola $x^{2}=y$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = y)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=-1/4",
    "fact_spans": "[[[0, 12]], [[0, 12]]]",
    "query_spans": "[[[0, 19]]]",
    "process": "Since the parabola equation is $x^{2}=y$, we have $p=\\frac{1}{2} \\Rightarrow \\frac{p}{2}=\\frac{1}{4}$. Also, because the focus of the parabola lies on the $y$-axis, the directrix equation of the parabola $x^{2}=y$ is $y=-\\frac{1}{4}$."
  },
  {
    "text": "It is known that the center of ellipse $G$ is at the origin of the coordinate system, the major axis lies on the $x$-axis, the eccentricity is $\\frac{\\sqrt{3}}{2}$, and the sum of the distances from a point on $G$ to the two foci of $G$ is $12$. Then, the equation of ellipse $G$ is?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;OverlappingLine(MajorAxis(G), xAxis);Eccentricity(G) = sqrt(3)/2;P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Distance(P, F1) + Distance(P, F2) = 12;Focus(G) ={F1, F2}",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/36+y^2/9 = 1",
    "fact_spans": "[[[2, 7], [52, 55], [59, 62], [80, 85]], [[11, 15]], [[2, 15]], [[2, 24]], [[2, 49]], [], [[52, 58]], [], [], [[52, 77]], [[59, 67]]]",
    "query_spans": "[[[80, 90]]]",
    "process": "Analysis: From the given conditions, we have 2a=12, so a=6. Also, since e=\\frac{c}{a}=\\frac{\\sqrt{3}}{2}, then c=3\\sqrt{3}. Thus, we can obtain b=3, from which the required ellipse equation follows. Specifically, from the given conditions, 2a=12 implies a=6. Since e=\\frac{c}{a}=\\frac{\\sqrt{3}}{2}, it follows that c=3\\sqrt{3}. Therefore, the required ellipse equation is \\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1."
  },
  {
    "text": "If the foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ lie on the $x$-axis, and tangents are drawn from the point $(1 , \\frac{1}{2})$ to the circle $x^{2}+y^{2}=1$, with points of tangency $A$ and $B$, respectively, such that the line $AB$ passes exactly through the right focus and the upper vertex of the ellipse, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;PointOnCurve(Focus(G), xAxis);H: Circle;Expression(H) = (x^2 + y^2 = 1);F: Point;Coordinate(F) = (1, 1/2);L1: Line;L2: Line;TangentOfPoint(F, H) = {L1, L2};A: Point;B: Point;TangentPoint(L1, H) = A;TangentPoint(L2, H) = B;PointOnCurve(RightFocus(G), LineOf(A, B));PointOnCurve(UpperVertex(G), LineOf(A, B))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5+y^2/4=1",
    "fact_spans": "[[[1, 46], [122, 124], [134, 136]], [[1, 46]], [[3, 46]], [[3, 46]], [[1, 55]], [[78, 94]], [[78, 94]], [[57, 77]], [[57, 77]], [], [], [[56, 97]], [[103, 106]], [[107, 110]], [[56, 110]], [[56, 110]], [[111, 128]], [[111, 132]]]",
    "query_spans": "[[[134, 141]]]",
    "process": ""
  },
  {
    "text": "Given that point $A$ is on the parabola $C$: $y^{2}=2 p x(p>0)$, the distance from point $A$ to the focus of $C$ is $12$, and the distance from point $A$ to the $y$-axis is $9$. Then $p=?$",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;A: Point;PointOnCurve(A, C);Distance(A, Focus(C)) = 12;Distance(A, yAxis) = 9",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[6, 32], [41, 44]], [[6, 32]], [[70, 73]], [[13, 32]], [[2, 5], [36, 40]], [[2, 35]], [[36, 55]], [[36, 68]]]",
    "query_spans": "[[[70, 75]]]",
    "process": "Let the focus of the parabola be F. Since the distance from point A to the focus C is 12, by the definition of a parabola, |AF| = x_{A} + \\frac{p}{2} = 12. Also, since the distance from point A to the y-axis is 9, we have x_{A} = 9. Therefore, \\frac{p}{2} = 3, solving gives p = 6."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the right focus is $F$, the right vertex is $A$, and $B$ is an endpoint of the imaginary axis. If the distance from point $F$ to the line $AB$ is $\\frac{b}{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;B: Point;A: Point;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;RightVertex(G)=A;OneOf(Endpoint(ImageinaryAxis(G)))=B;Distance(F,LineOf(A,B)) = b/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 58], [119, 122]], [[5, 58]], [[5, 58]], [[75, 78]], [[71, 74]], [[63, 66], [88, 92]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 66]], [[2, 74]], [[2, 86]], [[88, 117]]]",
    "query_spans": "[[[119, 128]]]",
    "process": "Let F(c,0), then A(a,0), without loss of generality take B(0,b). Therefore, the equation of line AB is \\frac{x}{a}+\\frac{y}{b}=1, i.e., bx+ay-ab=0. Thus, the distance from point F to AB is \\frac{|bc-ab|}{\\sqrt{a^{2}+b^{2}}}=\\frac{1}{2}b. Since a^{2}+b^{2}=c^{2}, we have c-a=\\frac{1}{2}c, so a=\\frac{1}{2}c, hence e=2"
  },
  {
    "text": "If the equation $\\frac{x^{2}}{2 \\lambda-1}+\\frac{y^{2}}{2-\\lambda}=1$ represents an ellipse, then the range of real values for $\\lambda$ is?",
    "fact_expressions": "G: Ellipse;lambda: Real;Expression(G)=(x^2/(2*lambda - 1) + y^2/(2 - lambda) = 1)",
    "query_expressions": "Range(lambda)",
    "answer_expressions": "{(1/2,1)+(1,2)}",
    "fact_spans": "[[[58, 60]], [[62, 73]], [[1, 60]]]",
    "query_spans": "[[[62, 80]]]",
    "process": "If the equation $\\frac{x^2}{2\\lambda-1}+\\frac{y^{2}}{2-\\lambda}=1$ represents an ellipse, then $\\begin{cases}2\\lambda-1>0\\\\2-\\lambda>0\\\\2\\lambda-1\\neq2-\\lambda\\end{cases}$ Solving this yields $\\frac{1}{2}<\\lambda<1$ or $1<\\lambda<2$."
  },
  {
    "text": "Given the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$, its left and right vertices coincide with the left and right foci of the ellipse $T$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and the left and right foci of $E$ coincide with the left and right vertices of $T$. Then the eccentricity of $E$ is?",
    "fact_expressions": "E: Hyperbola;b: Number;a: Number;T: Ellipse;a>0;b>0;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);Expression(T) = (x^2/4 + y^2/3 = 1);LeftVertex(E) = LeftFocus(T);RightVertex(E) = RightFocus(T);LeftVertex(T) = LeftFocus(E);RightVertex(T) = RightFocus(E)",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 62], [121, 124], [144, 147]], [[10, 62]], [[10, 62]], [[69, 111], [131, 134]], [[10, 62]], [[10, 62]], [[2, 62]], [[69, 111]], [[2, 119]], [[2, 119]], [[121, 142]], [[121, 142]]]",
    "query_spans": "[[[144, 153]]]",
    "process": "Let the focal distance of the hyperbola be $2c$. The foci of the ellipse $T: \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1$ are $(\\pm1,0)$, and the left and right vertices are $(-2,0)$, $(2,0)$. Since the left and right vertices of the hyperbola $E: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ ($a>0$, $b>0$) coincide with the left and right foci of the ellipse $T: \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1$, and the left and right foci of $E$ coincide with the left and right vertices of $T$, it follows that $a=1$, $c=2$. Therefore, the eccentricity of $E$ is $e = \\frac{c}{a} = 2$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, a line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. If $|F_{2} A|+| F_{2} B |=12$, then $|AB|$=?",
    "fact_expressions": "G: Ellipse;H: Line;F2: Point;A: Point;B: Point;F1: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(G) = {F1, F2};PointOnCurve(F1, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(F2, A)) + Abs(LineSegmentOf(F2, B)) = 12",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[18, 56], [74, 76]], [[71, 73]], [[10, 17]], [[77, 80]], [[81, 84]], [[2, 9], [63, 70]], [[18, 56]], [[2, 61]], [[62, 73]], [[71, 86]], [[88, 114]]]",
    "query_spans": "[[[116, 124]]]",
    "process": ""
  },
  {
    "text": "Given that $M(4, 2)$ is the midpoint of the segment $AB$ cut from the line $l$ by the ellipse $x^{2}+4 y^{2}=36$, then the equation of the line $l$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (4, 2);l: Line;G: Ellipse;Expression(G) = (x^2 + 4*y^2 = 36);A: Point;B: Point;MidPoint(LineSegmentOf(A, B)) = M;InterceptChord(l, G) = LineSegmentOf(A, B)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x+2*y-8=0",
    "fact_spans": "[[[2, 11]], [[2, 11]], [[12, 17], [54, 59]], [[18, 38]], [[18, 38]], [[44, 49]], [[44, 49]], [[2, 52]], [[12, 49]]]",
    "query_spans": "[[[54, 64]]]",
    "process": ""
  },
  {
    "text": "Pass a line $l$ with slope $1$ through the focus of the parabola $y^{2}=4x$, intersecting the parabola at points $A$ and $B$. Then the equation of the circle with segment $AB$ as diameter is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);l: Line;Slope(l) = 1;PointOnCurve(Focus(G), l);A: Point;B: Point;Intersection(l, G) = {A, B};H: Circle;IsDiameter(LineSegmentOf(A, B), H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x-3)^2+(y-2)^2=16",
    "fact_spans": "[[[1, 15], [32, 35]], [[1, 15]], [[26, 31]], [[19, 31]], [[0, 31]], [[36, 39]], [[40, 43]], [[26, 45]], [[59, 60]], [[47, 60]]]",
    "query_spans": "[[[59, 65]]]",
    "process": "The focus of the parabola $ y^2 = 4x $ is $ F(1,0) $. According to the problem, the equation of line $ l $ is $ y = x - 1 $. Let points $ A(x_1, y_1) $, $ B(x_2, y_2) $. Solving the system \n\\[\n\\begin{cases}\ny^2 = 4x \\\\\ny = x - 1\n\\end{cases}\n\\]\neliminating $ y $ gives $ x^2 - 6x + 1 = 0 $, $ \\Delta = 32 > 0 $. By Vieta's formulas, $ x_1 + x_2 = 6 $, then $ \\frac{x_1 + x_2}{2} = 3 $, $ \\frac{y_1 + y_2}{2} = \\frac{x_1 + x_2}{2} - 1 = 2 $. The midpoint of segment $ AB $ is $ M(3,2) $. Using the focal chord length formula for the parabola, $ |AB| = x_1 + x_2 + 2 = 8 $. Therefore, the equation of the circle with segment $ AB $ as diameter is $ (x - 3)^2 + (y - 2)^2 = 16 $."
  },
  {
    "text": "Given that $F_{1}$, $F_{2}$ are the left and right foci of the hyperbola $C$: $x^{2}-y^{2}=2$, point $P$ lies on $C$, and $|PF_{1}|=2|P F_{2}|$, then $\\cos \\angle F_{1} P F_{2}$?",
    "fact_expressions": "C: Hyperbola;P: Point;F1: Point;F2: Point;Expression(C) = (x^2 - y^2 = 2);LeftFocus(C) =F1;RightFocus(C)=F2;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Cos(AngleOf(F1, P, F2))",
    "answer_expressions": "3/4",
    "fact_spans": "[[[19, 42], [54, 57]], [[49, 53]], [[2, 9]], [[11, 18]], [[19, 42]], [[2, 48]], [[2, 48]], [[49, 58]], [[59, 80]]]",
    "query_spans": "[[[82, 110]]]",
    "process": ""
  },
  {
    "text": "Given a point $P$ on the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{4}=1$ such that its distance to one focus is $10$, then its distance to the other focus is?",
    "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2/16 - y^2/4 = 1);PointOnCurve(P, G);Distance(P, F1) = 10;F1: Point;F2: Point;OneOf(Focus(G)) = F1;OneOf(Focus(G)) = F2;Negation(F1 = F2)",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "{2, 18}",
    "fact_spans": "[[[2, 41]], [[44, 47], [62, 63]], [[2, 41]], [[2, 47]], [[2, 60]], [], [], [[2, 52]], [[2, 69]], [[2, 69]]]",
    "query_spans": "[[[2, 74]]]",
    "process": ""
  },
  {
    "text": "Let $F$ be the focus of the parabola $C_{1}$: $y^{2}=4 x$, and let point $A$ be a common point of the parabola and an asymptote of the hyperbola $C_{2}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, with $A F \\perp x$-axis. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "C1:Parabola;C2:Hyperbola;F: Point;A:Point;a:Number;b:Number;a>0;b>0;Expression(C1)=(y^2=4*x);Expression(C2)=(x^2/a^2-y^2/b^2=1);Focus(C1)=F;OneOf(Intersection(C1,OneOf(Asymptote(C2))))=A;IsPerpendicular(LineSegmentOf(A,F),xAxis)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[5, 28], [37, 40]], [[41, 106], [136, 139]], [[1, 4]], [[32, 36]], [[53, 106]], [[53, 106]], [[53, 106]], [[53, 106]], [[5, 28]], [[41, 106]], [[1, 31]], [[32, 118]], [[120, 134]]]",
    "query_spans": "[[[136, 145]]]",
    "process": "The focus of the parabola $ C_{1}: y^{2} = 4x $ is $ F(1,0) $. Without loss of generality, let $ A $ be the intersection point of $ y^{2} = 4x $ and $ y = \\frac{b}{a}x $. Since $ AF \\perp x $-axis, $ A(1,2) $. Substituting $ A(1,2) $ into $ y = \\frac{b}{a}x $ gives $ \\frac{b}{a} = 2 $, $ e = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{5} $."
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$ are $F_{1}$ and $F_{2}$ respectively. The line $l$ passes through $F_{2}$ and intersects the ellipse at points $A$ and $B$. $O$ is the origin. The circle with diameter $AB$ passes exactly through $O$. Find the equation of line $l$?",
    "fact_expressions": "l: Line;G: Ellipse;H: Circle;A: Point;B: Point;F1: Point;F2: Point;O: Origin;Expression(G) = (x^2/4 + y^2/2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2,l);Intersection(l, G) = {A,B};IsDiameter(LineSegmentOf(A, B),H);PointOnCurve(O, H)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "y=pm*sqrt(2)*(x-sqrt(2))",
    "fact_spans": "[[[62, 67], [119, 124]], [[0, 37], [76, 78]], [[110, 111]], [[81, 84]], [[85, 88]], [[46, 53]], [[54, 61], [68, 75]], [[91, 94], [114, 117]], [[0, 37]], [[0, 61]], [[0, 61]], [[62, 75]], [[62, 90]], [[100, 111]], [[110, 117]]]",
    "query_spans": "[[[119, 128]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has its right focus at $F$. A perpendicular is drawn from point $F$ to one asymptote of the hyperbola, with foot of perpendicular at $P$, intersecting the other asymptote at $Q$. If $5  \\overrightarrow{P F}=3 \\overrightarrow{F Q}$, then the eccentricity of hyperbola $E$ is?",
    "fact_expressions": "E: Hyperbola;b: Number;a: Number;P: Point;F: Point;Q: Point;a>0;b>0;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(E) = F;L1:Line;L2:Line;L:Line;OneOf(Asymptote(E))=L1;OneOf(Asymptote(E))=L2;Negation(L1=L2);PointOnCurve(F,L);IsPerpendicular(L1,L);FootPoint(L1,L)=P;Intersection(L2,L)=Q;5*VectorOf(P, F) = 3*VectorOf(F, Q)",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 62], [77, 80], [162, 168]], [[9, 62]], [[9, 62]], [[94, 97]], [[67, 70], [72, 76]], [[106, 109]], [[9, 62]], [[9, 62]], [[2, 62]], [[2, 70]], [], [], [], [[77, 86]], [[77, 105]], [[77, 105]], [[71, 89]], [[71, 89]], [[71, 97]], [[71, 109]], [[111, 159]]]",
    "query_spans": "[[[162, 174]]]",
    "process": ""
  },
  {
    "text": "Let the parabola $ C $: $ y^{2} = 2 p x $ ($ p > 0 $) have focus $ F $ and directrix $ l $. Let point $ A $ be a point on the parabola $ C $. A circle centered at $ F $ with radius $ FA $ intersects $ l $ at points $ B $ and $ D $. If $ \\angle BFD = 120^{\\circ} $, and the area of $ \\triangle ABD $ is $ 2\\sqrt{3} $, then $ p = $?",
    "fact_expressions": "C: Parabola;p: Number;G: Circle;F: Point;A: Point;B: Point;D: Point;l: Line;p>0;Expression(C) = (y^2 = 2*p*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(A, C);Center(G)=F;Radius(G)=LineSegmentOf(F,A);Intersection(G,l)={B,D};AngleOf(B, F, D) = ApplyUnit(120, degree);Area(TriangleOf(A,B,D))=2*sqrt(3)",
    "query_expressions": "p",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 26], [46, 52]], [[154, 157]], [[73, 74]], [[30, 33], [57, 60]], [[41, 45]], [[79, 82]], [[83, 86]], [[37, 40], [75, 78]], [[8, 26]], [[1, 26]], [[1, 33]], [[1, 40]], [[41, 55]], [[56, 74]], [[64, 74]], [[73, 88]], [[90, 117]], [[119, 152]]]",
    "query_spans": "[[[154, 159]]]",
    "process": "According to the problem, draw the figure and use the figure to find |BF| = |DF| = |AF| = 2p, |BD| = 2\\sqrt{3}p. Write the area of \\triangle ABD using the distance from point A to the directrix l, and thereby solve for p. \\because \\angle BFD = 120^{\\circ}, and \\because |FF| = p. \\therefore |BF| = |DF| = |AF| = 2p, |BD| = 2\\sqrt{3}p. The distance from A to the directrix l is d = |AF| = 2p, \\therefore S_{\\Delta ABD} = \\frac{1}{2} \\times d \\times |BD| = \\frac{1}{2} \\times 2p \\times 2\\sqrt{3}p = 2\\sqrt{3}p^{2} = 2\\sqrt{3}, solving gives p = 1"
  },
  {
    "text": "If the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1$ passes through the point $(-3 \\sqrt{2} , 2)$, then the focal length of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Point;Expression(G) = (-y^2/4 + x^2/a^2 = 1);Coordinate(H) = (-3*sqrt(2), 2);PointOnCurve(H, G)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(13)",
    "fact_spans": "[[[1, 43], [67, 70]], [[4, 43]], [[44, 64]], [[1, 43]], [[44, 64]], [[1, 64]]]",
    "query_spans": "[[[67, 75]]]",
    "process": ""
  },
  {
    "text": "The standard equation of a parabola with directrix $y=2$ is?",
    "fact_expressions": "G: Parabola;Expression(Directrix(G)) = (y = 2)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 = -8 \\cdot y",
    "fact_spans": "[[[11, 14]], [[0, 14]]]",
    "query_spans": "[[[11, 21]]]",
    "process": "From the directrix equation of the parabola, it can be seen that the parabola has its focus on the negative y-axis, and the value of p can be found, thus allowing the solution. Given the directrix equation of the parabola is y=2, it follows that the parabola has its focus on the negative y-axis. Let its equation be x^{2}=-2py (p>0), then its directrix equation is y=\\frac{p}{2}=2, yielding p=4. \\therefore The standard equation of this parabola is x^{2}=-8y."
  },
  {
    "text": "Given that a focus of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ is $(2,0)$, what is the equation of its asymptotes?",
    "fact_expressions": "G: Hyperbola;b: Number;b>0;Expression(G) = (x^2 - y^2/b^2 = 1);Coordinate(OneOf(Focus(G))) = (2, 0)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*sqrt(3)*x",
    "fact_spans": "[[[2, 39], [54, 55]], [[5, 39]], [[5, 39]], [[2, 39]], [[2, 52]]]",
    "query_spans": "[[[54, 63]]]",
    "process": "From the coordinates of the focus, we know that $ c = 2 \\therefore 1 + b^{2} = 4 \\therefore b = \\sqrt{3} $, and the asymptotes are $ y = \\pm\\sqrt{3}x $."
  },
  {
    "text": "The equation of the hyperbola passing through the point $M(10, \\frac{8}{3})$ with asymptotes given by ${y} = \\pm \\frac{1}{3} x$ is?",
    "fact_expressions": "G: Hyperbola;M: Point;Coordinate(M) = (10, 8/3);Expression(Asymptote(G)) = (y = pm*(x/3));PointOnCurve(M,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/36 - y^2/4 = 1",
    "fact_spans": "[[[54, 57]], [[2, 23]], [[2, 23]], [[24, 57]], [[0, 57]]]",
    "query_spans": "[[[54, 61]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $A$ is the left vertex, $B$ is the endpoint of the minor axis, $F$ is the right focus, and $AB \\perp BF$. Then the eccentricity of the ellipse is equal to?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;B: Point;F: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(G)=A;Endpoint(MinorAxis(G))=B;RightFocus(G)=F;IsPerpendicular(LineSegmentOf(A, B), LineSegmentOf(B, F))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(\\sqrt{5}-1)/2",
    "fact_spans": "[[[2, 54], [100, 102]], [[4, 55]], [[4, 54]], [[57, 60]], [[65, 68]], [[74, 77]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 64]], [[2, 73]], [[2, 81]], [[83, 98]]]",
    "query_spans": "[[[100, 109]]]",
    "process": "The left vertex of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $M(-a,0)$, the upper vertex is $N(0,b)$, and the right focus is $F(c,0)$. If $\\overrightarrow{NM}\\cdot\\overrightarrow{NF}=0$, then $NM\\bot NF$, so we obtain: $a^{2}+b^{2}+b^{2}+c^{2}=(a+c)^{2}$. Since $a^{2}=b^{2}+c^{2}$, it follows that $a^{2}\\cdot c^{2}=ac$, i.e., $e^{2}+e\\cdot1=0$, $e\\in(0,1)$, solving gives $c=\\frac{\\sqrt{5}-1}{2}$."
  },
  {
    "text": "The minor axis length of the ellipse $m x^{2}+y^{2}=1(m>1)$ is $\\frac{\\sqrt{2}}{2} m$, then $m$=?",
    "fact_expressions": "G: Ellipse;m: Number;m>1;Expression(G) = (m*x^2 + y^2 = 1);Length(MinorAxis(G)) = m*(sqrt(2)/2)",
    "query_expressions": "m",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 24]], [[53, 56]], [[2, 24]], [[0, 24]], [[0, 51]]]",
    "query_spans": "[[[53, 58]]]",
    "process": "From the ellipse equation $mx^{2}+y^{2}=1$ $(m>1)$, simplify to $\\frac{x^{2}}{m}+y^{2}=1$. Since the minor axis length of the ellipse $mx^{2}+y^{2}=1$ is $\\frac{\\sqrt{2}}{2}m$, we obtain $\\frac{1}{m}=\\left(\\frac{\\sqrt{2}}{4}m\\right)^{2}$, solving gives $m=2$."
  },
  {
    "text": "The standard equation of the hyperbola that shares the same asymptotes as $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ and passes through the point $A(2 \\sqrt {3},-3)$ is?",
    "fact_expressions": "G: Hyperbola;C:Hyperbola;A: Point;Expression(G) = (x^2/16 - y^2/9 = 1);Coordinate(A) = (2*sqrt(3), -3);Asymptote(G) = Asymptote(C);PointOnCurve(A, C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 40]], [[72, 75]], [[51, 71]], [[1, 40]], [[51, 71]], [[0, 75]], [[49, 75]]]",
    "query_spans": "[[[72, 81]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ is on the left branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola, and one asymptote of the hyperbola is exactly the perpendicular bisector of segment $P F_{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F2: Point;F1: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, LeftPart(G));LeftFocus(G) = F1;RightFocus(G) = F2;PerpendicularBisector(LineSegmentOf(P, F2)) = OneOf(Asymptote(G))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[7, 63], [85, 88], [95, 98], [124, 127]], [[10, 63]], [[10, 63]], [[2, 6]], [[77, 84]], [[69, 76]], [[10, 63]], [[10, 63]], [[7, 63]], [[2, 68]], [[69, 93]], [[69, 93]], [[95, 121]]]",
    "query_spans": "[[[124, 133]]]",
    "process": ""
  },
  {
    "text": "Let points $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$ be two points on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$. If the two lines passing through points $A$ and $B$ with slopes $\\frac{x_{1}}{4 y_{1}}$, $\\frac{x_{2}}{4 y_{2}}$ respectively intersect at point $P$, and the product of the slopes of lines $O A$ and $O B$ is $-\\frac{1}{4}$, $E(\\sqrt{6}, 0)$, then the minimum value of $|P E|$ is?",
    "fact_expressions": "G: Ellipse;O: Origin;A: Point;B: Point;E: Point;P: Point;Expression(G) = (x^2/4 + y^2 = 1);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);x1:Number;x2:Number;y1:Number;y2:Number;Coordinate(E) = (sqrt(6), 0);PointOnCurve(A, G);PointOnCurve(B, G);H:Line;C:Line;PointOnCurve(A,H);PointOnCurve(B,C);Slope(H)=x1/(4*y1);Slope(C)=x2/(4*y2);Intersection(H,C) = P;Slope(LineOf(O,A))*Slope(LineOf(O,B))=-1/4",
    "query_expressions": "Min(Abs(LineSegmentOf(P, E)))",
    "answer_expressions": "2*sqrt(2)-sqrt(6)",
    "fact_spans": "[[[39, 66]], [[148, 153]], [[1, 19], [72, 76]], [[21, 38], [77, 80]], [[182, 198]], [[140, 144]], [[39, 66]], [[1, 19]], [[21, 38]], [[86, 109]], [[111, 134]], [[86, 109]], [[111, 134]], [[182, 198]], [[1, 69]], [[1, 69]], [], [], [[71, 138]], [[71, 138]], [[81, 138]], [[81, 138]], [[71, 144]], [[146, 180]]]",
    "query_spans": "[[[200, 213]]]",
    "process": "From the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, let $A(2\\cos\\alpha,\\sin\\alpha)$, $B(2\\cos\\beta,\\sin\\beta)$, taking derivative with respect to $x$ on both sides of $\\frac{x^{2}}{4}+y^{2}=1$, we get $\\frac{x}{2}+2yy'=0$, thus the slope of the tangent line is $-\\frac{x}{4y}$. According to the problem, $AP$, $BP$ are both tangent lines to the ellipse, and $A$, $B$ are points of tangency. Then the equation of line $AP$ is $\\frac{xx_{1}}{4}+yy_{1}=1$, so $\\frac{x\\cos\\alpha}{2}+y\\sin\\alpha=1$. Similarly, the equation of line $BP$ is $\\frac{x\\cos\\beta}{2}+y\\sin\\beta=1$. The coordinates of intersection point $P$ are found to be $x=\\frac{2(\\sin\\beta-\\sin\\alpha)}{\\sin(\\beta-\\alpha)}$, $y=\\frac{\\cos\\beta-\\cos\\alpha}{\\sin(\\alpha-\\beta)}$, $x^{2}+y^{2}=\\frac{(\\sin\\beta-\\sin\\alpha)^{2}+(\\cos\\beta-\\cos\\alpha)^{2}}{\\sin^{2}(\\beta-\\alpha)}=\\frac{2-2\\cos(\\beta-\\alpha)}{\\sin^{2}(\\alpha-\\beta)}$. Since $k_{OA}\\cdot k_{OB}=$ $\\frac{\\sin\\alpha}{\\cos\\alpha}\\cdot\\frac{\\sin\\beta}{2\\cos\\beta}=-\\frac{1}{4}$, $\\therefore \\cos(\\beta-\\alpha)=0$, $\\sin(\\beta-\\alpha)=\\pm1$, $\\therefore \\sin^{2}(\\beta-\\alpha)=1$, $\\therefore |PE|=\\sqrt{(2\\sqrt{2}\\cos\\theta-\\sqrt{6})^{2}+(\\sqrt{2}\\sin\\theta)^{2}}=\\sqrt{6\\cos^{2}\\theta-8\\sqrt{3}\\cos\\theta+8}=\\sqrt{6}\\cos\\theta-2\\sqrt{2}$. Therefore, when $\\cos\\theta=1$, $|PE|_{\\min}=2\\sqrt{2}-\\sqrt{6}$."
  },
  {
    "text": "Given the parabola $C_{1}$: $y=a x^{2}  (a>0)$, the focus $F$ of which is also a focus of the ellipse $C_{2}$: $\\frac{y^{2}}{4}+\\frac{x^{2}}{b^{2}}=1  (b>0)$. Points $M$ and $P(\\frac{3}{2}, 1)$ are on curves $C_{1}$ and $C_{2}$ respectively. Then the minimum value of $|M P|+|M F|$ is?",
    "fact_expressions": "C1: Parabola;Expression(C1) = (y = a*x^2);a: Number;a>0;F: Point;Focus(C1) = F;C2: Ellipse;Expression(C2) = (y^2/4 + x^2/b^2 = 1);b: Number;b>0;OneOf(Focus(C2)) = F;M: Point;PointOnCurve(M, C1);P: Point;Coordinate(P) = (3/2, 1);PointOnCurve(P, C2)",
    "query_expressions": "Min(Abs(LineSegmentOf(M, F)) + Abs(LineSegmentOf(M, P)))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 32], [132, 141]], [[2, 32]], [[14, 32]], [[14, 32]], [[35, 38]], [[2, 38]], [[40, 97], [142, 149]], [[40, 97]], [[51, 97]], [[51, 97]], [[35, 102]], [[103, 107]], [[103, 152]], [[110, 129]], [[110, 129]], [[103, 152]]]",
    "query_spans": "[[[154, 173]]]",
    "process": "Since point $ P\\left(\\frac{3}{2},1\\right) $ lies on the ellipse $ C_{2} $, and $ b>0 $, we have $ \\frac{1}{4}+\\frac{\\left(\\frac{3}{2}\\right)^{2}}{b^{2}}=1 \\Rightarrow b=\\sqrt{3} $, so the coordinates of focus $ F $ are $ (0,1) $. Also, from the equation of parabola $ C_{1} $, we get $ F\\left(0,\\frac{1}{4a}\\right) $, so $ \\frac{1}{4a}=1 \\Rightarrow a=\\frac{1}{4} $, then $ c_{1}: y=\\frac{1}{4}x^{2} $. By the definition of the parabola, $ |MF| $ equals the distance $ d $ from point $ M $ to its directrix $ l: y=-1 $. Draw a perpendicular line $ l': x=\\frac{3}{2} $ from point $ P $ to the directrix $ l: y=-1 $. Then the intersection point of the perpendicular line $ l': x=\\frac{3}{2} $ and the parabola $ C_{1}: y=\\frac{1}{4}x^{2} $ is the desired point $ M $. Therefore, the minimum value of $ |MP|+|MF|=|MP|+d $ is $ 1-(-1)=2 $."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{k-3}+\\frac{y^{2}}{k+3}=1$ represents a hyperbola with foci on the $y$-axis, then the range of real values for $k$ is?",
    "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (x^2/(k - 3) + y^2/(k + 3) = 1);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-3, 3)",
    "fact_spans": "[[[53, 56]], [[58, 63]], [[1, 56]], [[44, 56]]]",
    "query_spans": "[[[58, 70]]]",
    "process": "Since the equation $\\frac{x^{2}}{k-3}+\\frac{y^{2}}{k+3}=1$ represents a hyperbola with foci on the $y$-axis, then $\\begin{cases}k+3>0\\\\k-3<0\\end{cases}$, solving gives $-3<k<3$."
  },
  {
    "text": "Given point $A(0 , b)$, and point $B$ is the intersection of the left directrix of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with the $x$-axis. If the midpoint $C$ of segment $AB$ lies on the ellipse, then the eccentricity of the ellipse is?",
    "fact_expressions": "A: Point;Coordinate(A) = (0, b);b: Number;B: Point;G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;a > b;b > 0;Intersection(LeftDirectrix(G), xAxis) = B;MidPoint(LineSegmentOf(A, B)) = C;C: Point;PointOnCurve(C, G) = True",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[2, 13]], [[2, 13]], [[3, 13]], [[16, 19]], [[20, 72], [101, 103], [107, 109]], [[20, 72]], [[22, 72]], [[22, 72]], [[22, 72]], [[16, 84]], [[87, 100]], [[97, 100]], [[97, 104]]]",
    "query_spans": "[[[107, 115]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left focus is $F$. A line $l$ passes through point $F$ and is perpendicular to one of the asymptotes of the hyperbola. Line $l$ intersects the left branch of the hyperbola at two distinct points $A$ and $B$. If $\\overrightarrow{A F}=2 \\overrightarrow{F B}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;LeftFocus(G) = F;l: Line;PointOnCurve(F, l);IsPerpendicular(OneOf(Asymptote(G)), l);A: Point;B: Point;Intersection(l, LeftPart(G)) = {A, B};VectorOf(A, F) = 2*VectorOf(F, B);Negation(A=B)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)/3",
    "fact_spans": "[[[2, 58], [79, 82], [97, 100], [166, 169]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[62, 65], [73, 77]], [[2, 65]], [[66, 71], [91, 96]], [[66, 77]], [[66, 90]], [[109, 112]], [[113, 116]], [[91, 116]], [[118, 163]], [103, 113]]",
    "query_spans": "[[[166, 175]]]",
    "process": "Assume that line $ l $ is perpendicular to the asymptote $ y = -\\frac{b}{a}x $, so the equation of line $ l $ is $ y = \\frac{a}{b}(x + c) $. From  \n$$\n\\begin{cases}\ny = \\frac{a}{b}(x + c) \\\\\n\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1\n\\end{cases},\n$$\nwe obtain  \n$$\nb^{2}\\left( \\frac{b^{2}y^{2}}{a^{2}} - \\frac{2bcy}{a} + c^{2} \\right) - a^{2}y^{2} = a^{2}b^{2},\n$$\ni.e.,  \n$$\nc^{2}(b^{2} - a^{2})y^{2} - 2ab^{3}cy + a^{2}b^{4} = 0.\n$$\nLet $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then  \n$$\ny_{1} + y_{2} = \\frac{}{c^{2}}\\frac{1}{2}\\frac{1}{3}, \\quad y_{1}y_{2} = \n$$\nAlso $ \\overrightarrow{AF} = 2\\overrightarrow{FB} $, $ F(-c, 0) $, so $ y_{1} = -2y_{2} $\\textcircled{3}. Substituting \\textcircled{3} into \\textcircled{1} gives  \n$$\ny_{2} = \\frac{2ab^{3}}{c(a^{2} - b^{2})}, \\quad \\text{so} \\quad y_{1} = -\\frac{4ab^{3}}{c(a^{2} - b^{2})}.\n$$\nSubstituting $ y_{1}, y_{2} $ into \\textcircled{2} yields  \n$$\n-\\frac{8a^{2}b^{6}}{c^{2}(a^{2} - b^{2})^{2}} = \\frac{a^{2}b^{4}}{c^{2}(b^{2} - a^{2})},\n$$\nsimplifying gives $ 9c^{2} = 10a^{2} $, so  \n$$\ne = \\frac{c}{a} = \\frac{\\sqrt{10}}{3}.\n$$"
  },
  {
    "text": "The focus of the parabola $C$: $x^{2}=4 a y$ has coordinates $(0,2)$, then the equation of the directrix of $C$ is?",
    "fact_expressions": "C: Parabola;a: Number;Expression(C) = (x^2 = 4*(a*y));Coordinate(Focus(C)) = (0, 2)",
    "query_expressions": "Expression(Directrix(C))",
    "answer_expressions": "y=-2",
    "fact_spans": "[[[0, 21], [36, 39]], [[7, 21]], [[0, 21]], [[0, 34]]]",
    "query_spans": "[[[36, 46]]]",
    "process": "Since the focus of the parabola \\( C: x^{2} = 4ay \\) is at \\( (0, 2) \\), the directrix of \\( C \\) is \\( y = -2 \\)."
  },
  {
    "text": "Given $B(-4,0)$, $C(4,0)$, and the perimeter of $\\triangle A B C$ equals $20$, find the trajectory equation of vertex $A$?",
    "fact_expressions": "B: Point;Coordinate(B) = (-4, 0);C: Point;Coordinate(C) = (4, 0);A: Point;Perimeter(TriangleOf(A, B, C)) = 20",
    "query_expressions": "LocusEquation(A)",
    "answer_expressions": "(x^2/36+y^2/20=1)&Negation(y=0)",
    "fact_spans": "[[[2, 11]], [[2, 11]], [[13, 21]], [[13, 21]], [[53, 56]], [[23, 49]]]",
    "query_spans": "[[[53, 62]]]",
    "process": "Since B(-4,0), C(4,0), we have |BC| = 8. Given that the perimeter of \\triangle ABC is 20, it follows that |AB| + |AC| = 12 > 8. Therefore, the locus of vertex A is an ellipse with foci at B and C (excluding the intersections with the x-axis). Thus, 2a = 12, c = 4, so a = 6, b^{2} = a^{2} - c^{2} = 20. Hence, the equation of the locus of vertex A is \\frac{x^{2}}{36} + \\frac{y^{2}}{20} = 1 (y \\neq 0)."
  },
  {
    "text": "If a hyperbola passes through the point $(3 , \\sqrt{2})$, and its asymptotes are given by $y=\\pm \\frac{1}{3} x$, then the equation of this hyperbola is?",
    "fact_expressions": "E: Hyperbola;F: Point;Coordinate(F) = (3, sqrt(2));PointOnCurve(F, E);Expression(Asymptote(E)) = (y = (pm*1/3)*x)",
    "query_expressions": "Expression(E)",
    "answer_expressions": "y^2 - x^2/9 = 1",
    "fact_spans": "[[[1, 4], [56, 59]], [[6, 23]], [[6, 23]], [[1, 23]], [[1, 52]]]",
    "query_spans": "[[[56, 64]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has an eccentricity of $2$, its two asymptotes intersect the directrix of the parabola $y^{2}=2 p x$ $(p>0)$ at points $A$ and $B$ respectively, $O$ is the origin, and $S_{\\triangle A O B} =\\frac{\\sqrt{3}}{4}$, then $p=$?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;p: Number;A: Point;O: Origin;B: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);p>0;Expression(H) = (y^2 = 2*(p*x));Eccentricity(G) = 2;L1: Line;L2: Line;Asymptote(G) = {L1, L2};Intersection(L1, Directrix(H)) = A;Intersection(L2, Directrix(H)) = B;Area(TriangleOf(A, O, B)) = sqrt(3)/4",
    "query_expressions": "p",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 58], [68, 69]], [[5, 58]], [[5, 58]], [[75, 96]], [[165, 168]], [[103, 106]], [[113, 116]], [[107, 110]], [[5, 58]], [[5, 58]], [[2, 58]], [[78, 96]], [[75, 96]], [[2, 67]], [], [], [[68, 74]], [[68, 112]], [[68, 112]], [[122, 163]]]",
    "query_spans": "[[[165, 170]]]",
    "process": "From the given conditions, we have: $ e = \\frac{c}{a} = 2 $, then: $ \\frac{b}{a} = \\sqrt{\\frac{c^{2}-a^{2}}{a^{2}}} = \\sqrt{3} $. The asymptotes of the hyperbola are: $ y = \\pm\\sqrt{3}x $. Letting $ x = -\\frac{p}{2} $, we get: $ y = \\pm\\frac{\\sqrt{3}}{2}p $. Based on this, we obtain $ S_{\\triangle OAB} = \\frac{1}{2} \\times \\sqrt{3}p \\times \\frac{p}{2} = \\frac{\\sqrt{3}}{4}p^{2} = \\frac{\\sqrt{3}}{4} $, solving yields: $ p = 1 $."
  },
  {
    "text": "Given the parabola $x^{2}=4 y$ with focus $F$ and point $A(-1,5)$, let point $P$ be a point on the parabola. Then the minimum value of $|P A|+|P F|$ is?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (x^2 = 4*y);Coordinate(A) = (-1, 5);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 16], [39, 42]], [[23, 33]], [[34, 38]], [[19, 22]], [[2, 16]], [[23, 33]], [[2, 22]], [[34, 45]]]",
    "query_spans": "[[[47, 66]]]",
    "process": "From the equation of the parabola, we know that $ p = 2 $, the focus coordinates are $ F(0,1) $, and the directrix equation is $ y = -1 $. Let the distance from point $ P $ to the directrix be $ PM $ (i.e., $ PM $ is perpendicular to the directrix, with $ M $ as the foot of the perpendicular). Then $ |PA| + |PF| = |PA| + |PM| \\geqslant |AM| = 6 $ (equality holds if and only if $ P $, $ A $, $ M $ are collinear)."
  },
  {
    "text": "If the directrix of the parabola $y^{2}=8x$ passes through one focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{5}=1$ $(a>\\sqrt{5})$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Parabola;H: Ellipse;a: Number;Expression(G) = (y^2 = 8*x);a>sqrt(5);Expression(H) = (y^2/5 + x^2/a^2 = 1);PointOnCurve(OneOf(Focus(H)), Directrix(G))",
    "query_expressions": "Eccentricity(H)",
    "answer_expressions": "2/3",
    "fact_spans": "[[[1, 15]], [[20, 73], [80, 82]], [[22, 73]], [[1, 15]], [[22, 73]], [[20, 73]], [[1, 78]]]",
    "query_spans": "[[[80, 88]]]",
    "process": "The directrix of the parabola \\( y^{2} = 8x \\) is \\( x = -2 \\), so for the ellipse, \\( c = \\sqrt{a^{2} - 5} = 2 \\), yielding \\( a = 3 \\); hence, the eccentricity of the ellipse is \\( \\frac{2}{3} \\)."
  },
  {
    "text": "The minimum distance from a point on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ to the line $x-y+3 \\sqrt{5}=0$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);P: Point;PointOnCurve(P, G);H: Line;Expression(H) = (x - y + 3*sqrt(5) = 0)",
    "query_expressions": "Min(Distance(P, H))",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[0, 27]], [[0, 27]], [[29, 30]], [[0, 30]], [[31, 51]], [[31, 51]]]",
    "query_spans": "[[[29, 60]]]",
    "process": "Let the equation of the line parallel to the line $x - y + 3\\sqrt{5} = 0$ be: $x - y + c = 0$. Combining it with the ellipse equation, eliminating variables, and setting $\\triangle = 0$, the value of $c$ can be obtained. Then, by finding the distance between the two parallel lines, the minimum distance from a point $P$ on the ellipse $\\frac{x^{2}}{4} + y^{2} = 1$ to the line $x - y + 3\\sqrt{5} = 0$ can be determined. Let the equation of the line parallel to $x - y + 3\\sqrt{5} = 0$ be: $x - y + c = 0$. Combining it with the ellipse equation and eliminating variables yields $5x^{2} + 8cx + 4c$. Let $\\triangle = 64c^{2} - 20(4c^{2} - 4) = 0$, we obtain $c = \\pm\\sqrt{5}$. Therefore, the distance between the two parallel lines is $\\frac{|\\pm\\sqrt{5} - 3\\sqrt{5}|}{\\sqrt{2}} = 2\\sqrt{10}$ or $\\sqrt{10}$. Hence, the minimum distance from a point on the ellipse $\\frac{x^{2}}{4} + y^{2} = 1$ to the line $x - y + 3\\sqrt{5} = 0$ is: $\\sqrt{10}$."
  },
  {
    "text": "The equation of the parabola with the center of the hyperbola $\\frac{x^{2}}{6}-\\frac{y^{2}}{10}=1$ as its vertex and the right focus as its focus is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (x^2/6 - y^2/10 = 1);Vertex(H) = Center(G);Focus(H) = RightFocus(G)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2=16*x",
    "fact_spans": "[[[1, 40]], [[55, 58]], [[1, 40]], [[0, 58]], [[0, 58]]]",
    "query_spans": "[[[55, 63]]]",
    "process": ""
  },
  {
    "text": "Let the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ have left focus $F$ and right directrix $l$. If there exists a point $P$ on $l$ such that the midpoint of segment $PF$ lies exactly on the ellipse $C$, then the minimum value of the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;F: Point;P: Point;l: Line;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;RightDirectrix(C) = l;PointOnCurve(P, l);PointOnCurve(MidPoint(LineSegmentOf(P,F)), C)",
    "query_expressions": "Min(Eccentricity(C))",
    "answer_expressions": "sqrt(2)-1",
    "fact_spans": "[[[1, 58], [102, 107], [110, 115]], [[8, 58]], [[8, 58]], [[63, 66]], [[82, 86]], [[71, 74], [76, 79]], [[8, 58]], [[8, 58]], [[1, 58]], [[1, 66]], [[1, 74]], [[76, 86]], [[89, 108]]]",
    "query_spans": "[[[110, 125]]]",
    "process": "Using the directrix equation of the ellipse, let point $ P\\left(\\frac{a^{2}}{c}, 2y\\right) $, obtain the midpoint coordinates, substitute into the ellipse equation, simplify to get $ y^{2} $, and since $ y^{2} \\geqslant 0 $, solve the inequality to obtain the minimum value of the eccentricity. Given $ C: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $), we have $ F(-c, 0) $, $ l: x = \\frac{a^{2}}{c} $. Let point $ P\\left(\\frac{a^{2}}{c}, 2y\\right) $, so the midpoint is $ \\left(\\frac{a^{2}-c^{2}}{2c}, y\\right) $. Since the midpoint lies on the ellipse, substituting into the ellipse equation gives $ \\frac{(a^{2}-c^{2})^{2}}{4a^{2}c^{2}} + \\frac{y^{2}}{b^{2}} = 1 $. Simplifying yields $ y^{2} = b^{2} \\cdot \\left[1 - \\frac{(a^{2}-c^{2})^{2}}{4a^{2}c^{2}}\\right] \\geqslant 0 $, hence $ 1 - \\frac{(a^{2}-c^{2})^{2}}{4a^{2}c^{2}} \\geqslant 0 $. Also, $ e = \\frac{c}{a} \\in (0,1) $, simplifying gives $ (e^{2}-3)^{2} \\leqslant 8 $, $ 3 - 2\\sqrt{2} \\leqslant e^{2} \\leqslant 3 + 2\\sqrt{2} $, thus $ e^{2} \\geqslant 3 - 2\\sqrt{2} = (\\sqrt{2} - 1)^{2} $, $ e \\geqslant \\sqrt{2} - 1 $."
  },
  {
    "text": "Given that the vertex of the parabola is at the origin, the coordinate axis is the axis of symmetry, and it passes through the point $P(1,1)$, then the equation of this parabola is?",
    "fact_expressions": "G: Parabola;P: Point;O: Origin;Coordinate(P) = (1, 1);Vertex(G) = O;SymmetryAxis(G) = axis;PointOnCurve(P,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=x,x^2=y}",
    "fact_spans": "[[[2, 5], [34, 37]], [[22, 31]], [[9, 11]], [[22, 31]], [[2, 11]], [[2, 19]], [[2, 31]]]",
    "query_spans": "[[[34, 42]]]",
    "process": "Since the vertex of the parabola is at the origin, the coordinate axes are the axes of symmetry, and it passes through the point P(1,1), the parabola opens either to the right or upward. We discuss the two cases separately: If the focus lies on the x-axis, let the equation be y^{2}=2px (p>0). Substituting point P(1,1) gives: 1^{2}=2p\\times1, so p=\\frac{1}{2}, hence the parabola's equation is y^{2}=x. If the focus lies on the y-axis, let the equation be x^{2}=2py (p>0). Substituting point P(1,1) gives: 1^{2}=2p\\times1, so p=\\frac{1}{2}, hence the parabola's equation is x^{2}=y."
  },
  {
    "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ are given by $y=\\pm \\frac{1}{2} x$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(Asymptote(G)) = (y = pm*(x/2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[1, 33], [63, 66]], [[4, 33]], [[1, 33]], [[1, 61]]]",
    "query_spans": "[[[63, 72]]]",
    "process": "\\because the asymptotes of the hyperbola \\frac{x^{2}}{a^{2}}-y^{2}=1 are given by y=\\pm\\frac{1}{2}x, \\therefore \\frac{b}{a}=\\frac{1}{2}, \\because b=1, \\therefore a=2, c=\\sqrt{5}, the eccentricity of the hyperbola is e=\\frac{c}{a}=\\frac{\\sqrt{5}}{2},"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $P$ is a point on the ellipse such that $\\overrightarrow{P F_{1}} \\cdot(\\overrightarrow{O F_{1}}+\\overrightarrow{O P})=0$ ($O$ is the origin). If $|P F_{1}|=2|P F_{2}|$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F1: Point;O: Origin;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P,G);DotProduct(VectorOf(P,F1),(VectorOf(O,F1)+VectorOf(O,P)))=0;Abs(LineSegmentOf(P,F1))=2*Abs(LineSegmentOf(P,F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[2, 54], [83, 85], [207, 209]], [[4, 54]], [[4, 54]], [[79, 82]], [[63, 70]], [[172, 175]], [[71, 78]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 78]], [[2, 78]], [[79, 88]], [[90, 171]], [[183, 205]]]",
    "query_spans": "[[[207, 215]]]",
    "process": "\\overrightarrow{PF}\\cdot(\\overrightarrow{OF}+\\overrightarrow{OP})=(\\overrightarrow{OF}-\\overrightarrow{OP})\\cdot(\\overrightarrow{OF}+\\overrightarrow{OP})=|\\overrightarrow{OF}|^{2}-|\\overrightarrow{OP}|^{2}=0 so |OF_{1}|=|OP|, and |OF_{1}|=|OF_{2}|, so \\triangle PF_{1}F_{2} is a right triangle, PF_{1}\\bot PF_{2}, |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}=4c^{2}, and |PF_{1}|+|PF_{2}|=2a, |PF_{1}|=2|PF_{2}|, so |PF_{2}|=\\frac{2}{3}a, |PF_{1}|=\\frac{4}{3}a. \\frac{4}{9}a^{2}+\\frac{16}{9}a^{2}=4c^{2}, so e=\\frac{c}{a}=\\frac{\\sqrt{5}}{3}. The answer is"
  },
  {
    "text": "The standard equation of an ellipse with foci on the $x$-axis, focal length $2$, and passing through $(\\sqrt{5}, 0)$ is?",
    "fact_expressions": "G: Ellipse;H: Point;Coordinate(H) = (sqrt(5), 0);PointOnCurve(Focus(G), xAxis);FocalLength(G) = 2;PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5 + y^2/4 = 1",
    "fact_spans": "[[[35, 37]], [[19, 34]], [[19, 34]], [[0, 37]], [[9, 37]], [[17, 37]]]",
    "query_spans": "[[[35, 44]]]",
    "process": "According to the problem: $2c=2 \\therefore c=1, a=\\sqrt{5} \\therefore b=2$, therefore the equation of the ellipse is: $\\frac{x^{2}}{5}+\\frac{y}{4}=1$"
  },
  {
    "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{2}=1$ is $\\frac{\\sqrt{2}}{2}$, then $m=$?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (y^2/2 + x^2/m = 1);Eccentricity(G) = sqrt(2)/2",
    "query_expressions": "m",
    "answer_expressions": "{4, 1}",
    "fact_spans": "[[[1, 38]], [[65, 68]], [[1, 38]], [[1, 63]]]",
    "query_spans": "[[[65, 70]]]",
    "process": "When the foci are on the horizontal axis, $m>2\\therefore c=\\sqrt{m-2}$, the eccentricity is $\\frac{\\sqrt{2}}{2}$, so $m>2\\therefore c=\\frac{\\sqrt{m-2}}{\\sqrt{m}}=\\frac{\\sqrt{2}}{2}\\Rightarrow m=4$; when the foci are on the vertical axis, $m<2\\therefore c=\\frac{\\sqrt{2-m}}{\\sqrt{2}}=\\frac{\\sqrt{2}}{2}\\Rightarrow m=1$, therefore the value of $m$ is $4$ or $1$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, respectively, if a point $P$ on the right branch of the hyperbola satisfies $|P F_{1}|=3|P F_{2}|$ and $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=\\frac{1}{2} a^{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;b: Number;a: Number;P: Point;PointOnCurve(P,RightPart(G));Abs(LineSegmentOf(P,F1))=3*Abs(LineSegmentOf(P,F2));DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = a^2/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3/2",
    "fact_spans": "[[[20, 66], [189, 192], [74, 77]], [[20, 66]], [[2, 9]], [[10, 17]], [[2, 72]], [[2, 72]], [[23, 66]], [[23, 66]], [[82, 85]], [[74, 85]], [[87, 109]], [[111, 186]]]",
    "query_spans": "[[[189, 198]]]",
    "process": "According to the definition of hyperbola, |PF_{1}|-|PF_{2}|=2a, and |PF_{1}|=3|PF_{2}|, so |PF_{1}|=3a, |PF_{2}|=a. Also, \\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=|\\overrightarrow{PF_{1}}|\\cdot|\\overrightarrow{PF_{2}}|\\cos\\angle F_{1}PF_{2}=\\frac{1}{2}a^{2}, thus \\cos\\angle F_{1}PF_{2}=\\frac{1}{6}. Moreover, |F_{1}F_{2}|=2c. In \\triangle F_{1}PF_{2}, by the law of cosines, |F_{1}F_{2}|^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}|\\cdot|PF_{2}|\\cos\\angle F_{1}PF_{2}, i.e., 4c^{2}=9a^{2}+a^{2}-2\\times3a\\times a\\times\\frac{1}{6}, which gives 2c=3a. Therefore, the eccentricity e=\\frac{c}{a}=\\frac{3}{2}."
  },
  {
    "text": "Let the right vertex of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ be $A$, the upper vertex be $B$, and the left focus be $F$. If $\\angle A B F=90^{\\circ}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;B: Point;F: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);RightVertex(G) = A;UpperVertex(G) = B;LeftFocus(G) = F;AngleOf(A, B, F) = ApplyUnit(90, degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(\\sqrt{5}-1)/2",
    "fact_spans": "[[[1, 53], [107, 109]], [[3, 53]], [[3, 53]], [[58, 61]], [[66, 69]], [[74, 77]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 61]], [[1, 69]], [[1, 77]], [[80, 105]]]",
    "query_spans": "[[[107, 115]]]",
    "process": "From the equation of the ellipse, we obtain A(a,0), B(0,b), F(-c,0). Since $\\angle ABF=90^{\\circ}$, it follows that $\\overrightarrow{BA} \\cdot \\overrightarrow{BF} = (a,-b) \\cdot (-c,-b) = 0$, which implies $b^{2} = ac$. Since $b^{2} = a^{2} - c^{2}$, we have $c^{2} + ac - a^{2} = 0$, then $e^{2} + e - 1 = 0$, where $e \\in (0,1)$. Solving gives $e = \\frac{\\sqrt{5}-1}{2}$."
  },
  {
    "text": "The line $l$ intersects the parabola $y^{2}=4x$ at two distinct points $A$ and $B$. If $M(x_{0}, 4)$ is the midpoint of $AB$, then the slope $k$ of the line $l$ is?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;M: Point;k: Number;x0: Number;Expression(G) = (y^2 = 4*x);Coordinate(M) = (x0, 4);Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = M;Slope(l) = k;Negation(A = B)",
    "query_expressions": "k",
    "answer_expressions": "1/2",
    "fact_spans": "[[[0, 5], [59, 64]], [[6, 20]], [[27, 30]], [[31, 34]], [[36, 49]], [[67, 70]], [[36, 49]], [[6, 20]], [[36, 49]], [[0, 34]], [[36, 57]], [[59, 70]], [[23, 34]]]",
    "query_spans": "[[[67, 72]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since line l intersects the parabola y^{2}=4x at two distinct points A and B, we have y_{1}^{2}=4x_{1}, y_{2}^{2}=4x_{2}. Subtracting these two equations gives (y_{1}+y_{2})(y_{1}-y_{2})=4(x_{1}-x_{2}). Since M(x_{0},4) is the midpoint of AB, it follows that 8(y_{1}-y_{2})=4(x_{1}-x_{2})."
  },
  {
    "text": "Let $P$ be the intersection point of the line $y=\\frac{b}{3 a} x$ and the left branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and let $F_{1}$ be the left focus. If $PF_{1}$ is perpendicular to the $x$-axis, then the eccentricity $e$ of the hyperbola is $?$.",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;H:Line;F1: Point;e: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y =(b/(3*a))*x);Intersection(H,LeftPart(G))=P;LeftFocus(G) = F1;IsPerpendicular(LineSegmentOf(P,F1),xAxis);Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "3*sqrt(2)/4",
    "fact_spans": "[[[27, 83], [119, 122]], [[30, 83]], [[30, 83]], [[1, 4]], [[5, 26]], [[89, 96]], [[126, 129]], [[30, 83]], [[30, 83]], [[27, 83]], [[5, 26]], [[1, 88]], [[27, 100]], [[101, 116]], [[119, 129]]]",
    "query_spans": "[[[126, 131]]]",
    "process": ""
  },
  {
    "text": "Given that the line $y = -x + 1$ intersects the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1\\ (a > b > 0)$ at points $A$ and $B$, and the midpoint of segment $AB$ lies on the line $x - 2y = 0$, then the eccentricity of this ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;L:Line;A: Point;B: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = 1 - x);Expression(L) = (x - 2*y = 0);Intersection(H, G) = {A, B};PointOnCurve(MidPoint(LineSegmentOf(A,B)), L)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[13, 65], [105, 107]], [[15, 65]], [[15, 65]], [[2, 12]], [[90, 101]], [[68, 71]], [[72, 75]], [[15, 65]], [[15, 65]], [[13, 65]], [[2, 12]], [[90, 101]], [[2, 77]], [[79, 102]]]",
    "query_spans": "[[[105, 113]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the focus $F$ of the parabola $y^{2}=4x$ are? If point $P$ is any point on this parabola, and point $A(6, 3)$, then what is the minimum value of $|PA| + |PF|$?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (6, 3);Focus(G)=F;PointOnCurve(P,G)",
    "query_expressions": "Coordinate(F);Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "(1,0)\n7",
    "fact_spans": "[[[2, 16], [34, 37]], [[42, 53]], [[28, 32]], [[19, 22]], [[2, 16]], [[42, 53]], [[2, 22]], [[28, 41]]]",
    "query_spans": "[[[19, 27]], [[55, 75]]]",
    "process": ""
  },
  {
    "text": "Let $F$ be the focus of the parabola $y^{2}=8 x$, and let $A$, $B$, $C$ be three points on this parabola. If $\\overrightarrow{F A}+\\overrightarrow{F B}+\\overrightarrow{F C}=0$, then what is $|\\overrightarrow{F A}|+|\\overrightarrow{F B}|+|\\overrightarrow{F C}|$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;A: Point;B: Point;C: Point;PointOnCurve(A, G);PointOnCurve(B, G);PointOnCurve(C, G);VectorOf(F, A) + VectorOf(F, B) + VectorOf(F, C) = 0",
    "query_expressions": "Abs(VectorOf(F, A)) + Abs(VectorOf(F, B)) + Abs(VectorOf(F, C))",
    "answer_expressions": "12",
    "fact_spans": "[[[5, 19], [36, 39]], [[5, 19]], [[1, 4]], [[1, 22]], [[23, 26]], [[27, 30]], [[31, 34]], [[23, 42]], [[23, 42]], [[23, 42]], [[44, 110]]]",
    "query_spans": "[[[112, 183]]]",
    "process": "Let the coordinates of points A, B, C be $(x_{1},y_{1})$, $(x_{2},y_{2})$, $(x_{3},y_{3})$, respectively. Transform $|\\overrightarrow{FA}|+|\\overrightarrow{FB}|+|\\overrightarrow{FC}|$ (the sum of three focal radii) into the sum of three point-to-line distances, and use the given conditions to solve. Given $p=4$, $F(2,0)$, then $\\overrightarrow{FA}=(x_{1}-2,y_{1})$, $\\overrightarrow{FB}=(x_{2}-2,y_{2})$, $\\overrightarrow{FC}=(x_{3}-2,y_{3})$. Since $\\overrightarrow{FA}+\\overrightarrow{FB}+\\overrightarrow{FC}=\\overrightarrow{0}$, we have $x_{1}-2+x_{2}-2+x_{3}-2=0$, so $x_{1}+x_{2}+x_{3}=6$. By the definition of the parabola, we get: $|\\overrightarrow{FA}|=x_{1}+\\frac{p}{2}$, $|\\overrightarrow{FB}|=x_{2}+\\frac{p}{2}$, $|\\overrightarrow{FC}|=x_{3}+\\frac{p}{2}$. Therefore, $|\\overrightarrow{FA}|+|\\overrightarrow{FB}|+|\\overrightarrow{FC}|=x_{1}+x_{2}+x_{3}+2=6+6=12$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through point $F_{1}$ and perpendicular to one asymptote of the hyperbola $C$ intersects the two asymptotes of $C$ at points $M$ and $N$ respectively. If $|N F_{1}|=2|M F_{1}|$, then the equation of the asymptotes of the hyperbola $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;N: Point;F1: Point;M: Point;F2: Point;a > b;b > 0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1,l);L1:Line;L2:Line;Asymptote(C)={L1,L2};IsPerpendicular(l,L1);Intersection(l,L1)=M;Intersection(l,L2)=N;Abs(LineSegmentOf(N, F1)) = 2*Abs(LineSegmentOf(M, F1))",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*(sqrt(3)/3)*x",
    "fact_spans": "[[[111, 116]], [[2, 60], [96, 102], [117, 120], [165, 171]], [[9, 60]], [[9, 60]], [[134, 137]], [[69, 76], [86, 94]], [[130, 133]], [[77, 84]], [[9, 60]], [[9, 60]], [[2, 60]], [[2, 84]], [[2, 84]], [[85, 116]], [], [], [[117, 126]], [[95, 116]], [[111, 139]], [[111, 139]], [[141, 163]]]",
    "query_spans": "[[[165, 179]]]",
    "process": "Solve the system of the line equation and the asymptote equations:  \n\\begin{cases} y = -\\frac{a}{b}( \\\\ y = \\frac{b}{a} \\end{cases}  \nSolving the system gives the coordinates of intersection point M: M  \nSolve the system of the line equation and the asymptote equations:  \nSolving the system gives the coordinates of intersection point N:  \nCombining |NF_{1}| = 2|MF_{1}| and the distance formula between two points yields:  \n\\frac{a^{2}}{b^{2}} = \\frac{1}{3}.  \nThen the asymptote equations of hyperbola C are y = \\pm\\frac{\\sqrt{3}}{3}x"
  },
  {
    "text": "The parabola $C$: $y^{2}=x$ has focus $F$, and the point $M$ in the first quadrant lies on $C$ such that $|M F|=\\frac{5}{4}$. Then the coordinates of $M$ are?",
    "fact_expressions": "C: Parabola;M: Point;F: Point;Expression(C) = (y^2 = x);Focus(C) = F;Quadrant(M)=1;PointOnCurve(M,C);Abs(LineSegmentOf(M, F)) = 5/4",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(1,1)",
    "fact_spans": "[[[0, 17], [35, 38]], [[30, 34], [62, 65]], [[21, 24]], [[0, 17]], [[0, 24]], [[25, 34]], [[30, 39]], [[41, 60]]]",
    "query_spans": "[[[62, 70]]]",
    "process": "According to the problem: let the coordinates of point M be (a, b), hence b^{2} = a. Since the coordinates of point F are (\\frac{1}{4}, 0), then |MF| = a + \\frac{p}{2} = a + \\frac{1}{4} = \\frac{5}{4}. Solving gives a = 1. Also, since M is in the first quadrant, a = 1, b = 1, so the coordinates of point M are (1, 1)."
  },
  {
    "text": "A point $(1, y_{0})$ on the parabola $C$: $y^{2}=2 p x$ is at a distance of $3$ from its focus. Then the equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;p: Number;G: Point;y0: Number;Expression(C) = (y^2 = 2*(p*x));Coordinate(G) = (1, y0);PointOnCurve(G, C);Distance(G, Focus(C)) = 3",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2 = 8*x",
    "fact_spans": "[[[0, 21], [37, 38], [49, 55]], [[8, 21]], [[24, 36]], [[24, 36]], [[0, 21]], [[24, 36]], [[0, 36]], [[24, 47]]]",
    "query_spans": "[[[49, 60]]]",
    "process": "The directrix equation of the parabola $ C: y^{2} = 2px $ ($ p > 0 $) is $ x = -\\frac{p}{2} $. According to the definition of the parabola, we have $ 1 + \\frac{p}{3} = 3 $. Solving for $ p $, we get $ p = 4 $. Therefore, the equation of $ C $ is $ y^{2} = 8x $."
  },
  {
    "text": "Given that $P$ is any point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, $F_{1}$, $F_{2}$ are the two foci of the ellipse, if $P F_{1} \\perp P F_{2}$, then $S_{\\triangle{PF_{1} F_{2}}}$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "9",
    "fact_spans": "[[[6, 44], [65, 67]], [[6, 44]], [[2, 5]], [[2, 48]], [[49, 56]], [[57, 64]], [[49, 71]], [[73, 96]]]",
    "query_spans": "[[[97, 128]]]",
    "process": ""
  },
  {
    "text": "It is known that one focus of a hyperbola, whose axes of symmetry are the coordinate axes and whose eccentricity equals $2$, coincides with the focus of the parabola $x = \\frac{1}{8} y^{2}$. Then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(H) = (x = y^2/8);SymmetryAxis(G)=axis;Eccentricity(G)=2;OneOf(Focus(G))=Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[20, 23], [61, 64]], [[29, 53]], [[29, 53]], [[2, 23]], [[11, 23]], [[20, 58]]]",
    "query_spans": "[[[61, 69]]]",
    "process": "According to the equation of the parabola, its focus is calculated as (2,0), thus the right focus of the hyperbola is F(2,0). Then, by setting up the equation of the hyperbola and using the formula for eccentricity along with the squared relationship among a, b, and c, a system of equations is established. Solving this system gives the values of a and b, thereby obtaining the equation of the hyperbola.  \nSince the parabola's equation is $ y^{2} = 8x $,  \n$ \\therefore 2p = 8 $, yielding the focus of the parabola as (2,0).  \nSince one focus of the hyperbola coincides with the focus of the parabola $ y^{2} = 8x $,  \n$ \\therefore $ the right focus of the hyperbola is $ F(2,0) $.  \nLet the equation of the hyperbola be $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $), then we have $ a^{2} + b^{2} = 4 \\cdots\\textcircled{1} $.  \nSince the eccentricity of the hyperbola is 2,  \n$ \\therefore \\frac{c}{a} = 2 $, i.e., $ \\frac{a^{2} + b^{2}}{a^{2}} = 4 \\cdots\\textcircled{2} $.  \nSolving equations \\textcircled{1} and \\textcircled{2} simultaneously, we get $ a^{2} = 1 $, $ b^{2} = 3 $. Therefore, the equation of the hyperbola is $ \\frac{x^{2}}{1} - \\frac{y^{2}}{3} = 1 $."
  },
  {
    "text": "The length of the major axis of the ellipse $(1-m) x^{2}-m y^{2}=1$ is?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (-m*y^2 + x^2*(1 - m) = 1)",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "2*sqrt(-1/m)",
    "fact_spans": "[[[0, 25]], [[2, 25]], [[0, 25]]]",
    "query_spans": "[[[0, 31]]]",
    "process": "According to the problem, $(1-m)x^{2}-my^{2}=1$ is the equation of an ellipse, i.e., $\\frac{x^{2}}{1-m}+\\frac{y^{2}}{\\frac{1}{m}}=1$, $\\frac{1}{1-m}>0$, $\\frac{1}{m}>0$, $m<0$, $a=\\frac{m+1-m}{m(1-m)}=\\frac{1}{m(1-m)}<0$, $\\therefore \\frac{1}{1-m}<\\frac{1}{x}$"
  },
  {
    "text": "Given $\\frac{1}{m}+\\frac{2}{n}=1$ $(m>0, n>0)$, when $m n$ attains its minimum value, the number of intersection points between the line $y=-\\sqrt{2} x+2$ and the curve $\\frac{x|x|}{m}+\\frac{y|y|}{n}=1$ is?",
    "fact_expressions": "G: Line;Expression(G) = (y = 2 - sqrt(2)*x);H: Curve;Expression(H) = ((y*Abs(y))/n + (x*Abs(x))/m = 1);m: Number;n: Number;m>0;n>0;2/n + 1/m = 1;WhenMin(m*n)",
    "query_expressions": "NumIntersection(G, H)",
    "answer_expressions": "2",
    "fact_spans": "[[[54, 73]], [[54, 73]], [[74, 109]], [[74, 109]], [[2, 39]], [[2, 39]], [[2, 39]], [[2, 39]], [[2, 39]], [[41, 53]]]",
    "query_spans": "[[[54, 116]]]",
    "process": ""
  },
  {
    "text": "From a point $P$ on the parabola $y=\\frac{1}{8} x^{2}$, draw a perpendicular to the directrix of the parabola, with foot of perpendicular at $M$, and $|P M|=6$. Let the focus of the parabola be $F$. Then the area of $\\triangle M P F$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2/8);P: Point;M: Point;FootPoint(l, Directrix(G)) = M;F: Point;Abs(LineSegmentOf(P, M)) = 6;Focus(G) = F;l:Line;PointOnCurve(P,l);IsPerpendicular(l,Directrix(G));PointOnCurve(P,G)",
    "query_expressions": "Area(TriangleOf(M, P, F))",
    "answer_expressions": "12*sqrt(2)",
    "fact_spans": "[[[1, 25], [32, 35], [60, 63]], [[1, 25]], [[28, 31]], [[44, 47]], [[0, 47]], [[67, 70]], [[49, 58]], [[60, 70]], [], [[0, 40]], [[0, 40]], [1, 29]]",
    "query_spans": "[[[72, 94]]]",
    "process": "As shown in the figure, let the directrix intersect the y-axis at point N, and let P(x_{0},y_{0}). Since the equation of the parabola is x^{2}=8y, we have F(0,2), N(0,-2). Also, |PM|=y_{0}+\\frac{p}{2}=y_{0}+2=6, so y_{0}=4. Thus, x_{0}^{2}=8y_{0}=32, so x_{0}=\\pm4\\sqrt{2}. By symmetry, take x_{0}=-4\\sqrt{2}. Therefore, the area of trapezoid MNFP is: \\frac{(4+6)\\cdot4\\sqrt{2}}{2}=20\\sqrt{2}. The area of \\triangle MNF is: \\frac{4\\sqrt{2}\\cdot4}{2}=8\\sqrt{2}. Hence, S_{\\triangle MPF}=20\\sqrt{2}-8\\sqrt{2}=12\\sqrt{2}."
  },
  {
    "text": "Given that the moving circle $M$ is externally tangent to the circle $C_{1}$: $(x+4)^{2}+y^{2}=4$ and internally tangent to the circle $C_{2}$: $(x-4)^{2}+y^{2}=4$, then the equation of the trajectory $C$ of the center of the moving circle $M$ is?",
    "fact_expressions": "M: Circle;C1: Circle;C2: Circle;M1: Point;Expression(C1)=((x+4)^2+y^2=4);Expression(C2)=((x-4)^2+y^2=4);IsOutTangent(M,C1);IsInTangent(M,C2);Center(M)=M1;Locus(M1)=C;C:Curve",
    "query_expressions": "Expression(C)",
    "answer_expressions": "(x^2/4-y^2/12=1)&(x>=2)",
    "fact_spans": "[[[4, 7], [71, 73]], [[8, 36]], [[39, 67]], [[75, 78]], [[8, 36]], [[39, 67]], [[4, 38]], [[4, 69]], [[71, 78]], [[75, 84]], [[81, 84]]]",
    "query_spans": "[[[81, 89]]]",
    "process": "Let the center of the moving circle be $ M(x,y) $ and its radius be $ r $. Since circle $ M $ is externally tangent to circle $ C_{1}:(x+4)^{2}+y^{2}=4 $ and internally tangent to circle $ C_{2}:(x-4)^{2}+y^{2}=4 $, with centers $ C_{1}(-4,0) $, $ C_{2}(4,0) $, and $ |C_{1}C_{2}|=8 $, we have\n$$\n\\begin{cases}\n|MC_{1}|=r+2 \\\\\n|MC_{2}|=r-2\n\\end{cases}\n$$\nThen $ |MC_{1}|-|MC_{2}|=4<8 $. Thus, the locus of point $ M $ is the right branch of a hyperbola with foci at $ C_{1} $ and $ C_{2} $. From the given conditions, $ 2a=4 $, $ 2c=8 \\Rightarrow a=2 $, $ c=4 $, $ b^{2}=12 $. Therefore, the equation of $ C $ is:\n$$\n\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1 \\quad (x\\geqslant2)\n$$"
  },
  {
    "text": "Given a point $A(1,1)$ on the curve $y=\\frac{1}{x}$, what is the equation of the tangent line to the curve at point $A$?",
    "fact_expressions": "G: Curve;Expression(G) = (y = 1/x);A: Point;Coordinate(A) = (1, 1);PointOnCurve(A, G) = True",
    "query_expressions": "Expression(TangentOnPoint(A, G))",
    "answer_expressions": "x+y-2=0",
    "fact_spans": "[[[2, 19], [33, 35]], [[2, 19]], [[22, 30], [36, 40]], [[22, 30]], [[2, 30]]]",
    "query_spans": "[[[33, 48]]]",
    "process": ""
  },
  {
    "text": "If the eccentricity $e\\in(1 , 2)$ of the hyperbola $\\frac{y^{2}}{5}-\\frac{x^{2}}{m}=1$, then what is the range of values for $m$?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (y^2/5 - x^2/m = 1);Eccentricity(G)=e;e:Number;In(e,(1,2))",
    "query_expressions": "Range(m)",
    "answer_expressions": "(0,15)",
    "fact_spans": "[[[1, 39]], [[58, 61]], [[1, 39]], [[1, 56]], [[43, 56]], [[43, 56]]]",
    "query_spans": "[[[58, 68]]]",
    "process": ""
  },
  {
    "text": "Given a point $P$ on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$ is at a distance of $6$ from one focus of the ellipse, then the distance from point $P$ to the other focus is?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/16 + y^2/4 = 1);PointOnCurve(P, G);F1:Point;F2:Point;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2);Distance(P, F1) = 6",
    "query_expressions": "Distance(P,F2)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 40], [48, 50]], [[44, 47], [63, 67]], [[2, 40]], [[2, 47]], [], [], [[48, 54]], [[48, 73]], [[48, 73]], [[44, 61]]]",
    "query_spans": "[[[48, 78]]]",
    "process": "By the definition of an ellipse, |PF_{1}| + |PF_{2}| = 2a. Using the ellipse definition |PF_{1}| + |PF_{2}| = 2a, with a = 4, we have 6 + |PF_{2}| = 8, thus |PF_{2}| = 2."
  },
  {
    "text": "Given that the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ and the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{5}=1$ have common foci, then $m$=?",
    "fact_expressions": "G: Hyperbola;m: Number;H: Ellipse;Expression(G) = (-y^2/5 + x^2/m = 1);Expression(H) = (x^2/25 + y^2/16 = 1);Focus(H) = Focus(G)",
    "query_expressions": "m",
    "answer_expressions": "4",
    "fact_spans": "[[[42, 80]], [[88, 91]], [[2, 41]], [[42, 80]], [[2, 41]], [[2, 86]]]",
    "query_spans": "[[[88, 93]]]",
    "process": "From the given conditions, the foci of the ellipse are (-3,0) and (3,0), so $3=\\sqrt{m+5}$, therefore $m=4$."
  },
  {
    "text": "Through the focus of the parabola $y^{2}=2 p x$ ($p>0$), draw a line intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$, then $\\frac{y_{1} y_{2}}{x_{1} x_{2}}=$?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;A: Point;B: Point;p>0;x1:Number;y1:Number;x2:Number;y2:Number;Expression(G) = (y^2 = 2*(p*x));Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);PointOnCurve(Focus(G), H);Intersection(H, G) = {A,B}",
    "query_expressions": "(y1*y2)/(x1*x2)",
    "answer_expressions": "-4",
    "fact_spans": "[[[1, 22], [31, 34]], [[4, 22]], [[28, 30]], [[35, 52]], [[55, 73]], [[4, 22]], [[35, 52]], [[35, 52]], [[55, 73]], [[55, 73]], [[1, 22]], [[35, 52]], [[55, 73]], [[0, 30]], [[28, 73]]]",
    "query_spans": "[[[75, 110]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $C$: $y^{2}=2 p x$ lies on the line $x+y-3=0$, then the real number $p$=? What is the equation of the directrix of the parabola $C$?",
    "fact_expressions": "C: Parabola;p: Real;G: Line;Expression(C) = (y^2 = 2*(p*x));Expression(G) = (x + y - 3 = 0);PointOnCurve(Focus(C), G)",
    "query_expressions": "p;Expression(Directrix(C))",
    "answer_expressions": "6\nx=-3",
    "fact_spans": "[[[1, 22], [47, 53]], [[40, 45]], [[26, 37]], [[1, 22]], [[26, 37]], [[1, 38]]]",
    "query_spans": "[[[40, 47]], [[47, 60]]]",
    "process": "The focus of the parabola $ C: y^{2} = 2px $ is $ \\left( \\frac{p}{2}, 0 \\right) $. According to the given condition, $ \\frac{p}{2} + 0 - 3 = 0 $, so $ p = 6 $, and the equation of the directrix is $ x = -3 $."
  },
  {
    "text": "Given the ellipse equation $\\frac{x^{2}}{5}+\\frac{y^{2}}{16}=1$ represents an ellipse with foci $F_{1}$, $F_{2}$, and a moving point $P$ on the ellipse, then $|P F_{1}|+|P F_{2}|$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/5 + y^2/16 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G)",
    "query_expressions": "Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 4], [44, 46], [65, 67]], [[2, 42]], [[49, 56]], [[57, 64]], [[44, 64]], [[72, 75]], [[65, 75]]]",
    "query_spans": "[[[77, 100]]]",
    "process": "First, obtain the length of the major axis from the ellipse equation, then use the definition of the ellipse to derive the result. Since the length of the major axis of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{16}=1$ is $2a=2\\sqrt{16}=8$, and $P$ is a point on the ellipse, $F_{1}$ and $F_{2}$ are the two foci of the ellipse, according to the definition of the ellipse, we have $|PF_{1}|+|PF_{2}|=2a=8$."
  },
  {
    "text": "Given point $Q(\\sqrt{2}, 0)$ and a moving point $P(x, y)$ on the parabola $x^{2}=2 y$, find the minimum value of $y+|P Q|$.",
    "fact_expressions": "G: Parabola;Q: Point;P: Point;x1: Number;y1: Number;Expression(G) = (x^2 = 2*y);Coordinate(Q) = (sqrt(2), 0);Coordinate(P) = (x1, y1);PointOnCurve(P, G)",
    "query_expressions": "Min(y1 + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "1",
    "fact_spans": "[[[20, 34]], [[2, 19]], [[38, 47]], [[38, 47]], [[38, 47]], [[20, 34]], [[2, 19]], [[38, 47]], [[20, 47]]]",
    "query_spans": "[[[49, 64]]]",
    "process": "The focus of the parabola $x^{2}=2y$ is $F(0,\\frac{1}{2})$. According to the definition of a parabola, we have $y+\\frac{1}{2}=|PF|$, so $y+|PQ|=|PF|+|PQ|-\\frac{1}{2}$. Therefore, when points $F$, $P$, and $Q$ are collinear, $|PF|+|PQ|-\\frac{1}{2}$ attains its minimum value $|FQ|-\\frac{1}{2}=\\sqrt{(\\sqrt{2})^{2}+(\\frac{1}{2})^{2}}-\\frac{1}{2}=\\frac{3}{2}-\\frac{1}{2}=1$."
  },
  {
    "text": "If the parabola $y^{2}=2 p x$ and the circle $(x-2)^{2}+y^{2}=3$ have exactly two common points, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;H: Circle;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (y^2 + (x - 2)^2 = 3);NumIntersection(G,H)=2",
    "query_expressions": "p",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 17]], [[46, 49]], [[18, 38]], [[1, 17]], [[18, 38]], [[1, 45]]]",
    "query_spans": "[[[46, 53]]]",
    "process": "From the condition, we have p>0, x^{2}+(2p-4)x+1=0. Since it is known that the parabola is tangent to the circle, \\therefore A=(2p-4)^{2}-4=0, solving gives: p=3 or p=1. When p=3, x^2+2x+1=0, solving gives: x=-1 (discarded). When p=1, x^{2}-2x+1=0, solving gives: x=1 (valid)."
  },
  {
    "text": "If the distance from the focus of the hyperbola $x^{2}-\\frac{y^{2}}{k}=1$ to its asymptote is $2 \\sqrt{2}$, then the value of the real number $k$ is?",
    "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (x^2 - y^2/k = 1);Distance(Focus(G),Asymptote(G))=2*sqrt(2)",
    "query_expressions": "k",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 29]], [[54, 59]], [[1, 29]], [[1, 52]]]",
    "query_spans": "[[[54, 63]]]",
    "process": "From the hyperbola $x^{2}-\\frac{y^{2}}{k}=1$, one of the foci is $(\\sqrt{k+1},0)$, and one of the asymptotes is $y=\\sqrt{k}x$. Therefore, the distance from the focus to the line is $\\frac{|\\sqrt{k}\\sqrt{k+1}|}{\\sqrt{k+1}}=2\\sqrt{2}$, so $k=8$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ and the circle $x^{2}+y^{2}=b^{2}$ intersect at two points $A$, $C$ in the second and fourth quadrants respectively, point $F$ is the right focus of the hyperbola, and $|A F|=2|C F|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;H: Circle;Expression(H) = (x^2 + y^2 = b^2);Intersection(H,G) = {A,C};A: Point;C: Point;Quadrant(A) = 2;Quadrant(C) = 4;RightFocus(G) = F;F: Point;Abs(LineSegmentOf(A, F)) = 2*Abs(LineSegmentOf(C, F))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(22)/2",
    "fact_spans": "[[[2, 58], [107, 110], [133, 136]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[59, 79]], [[59, 79]], [[2, 100]], [[93, 96]], [[97, 100]], [[79, 100]], [[79, 100]], [[101, 114]], [[101, 105]], [[116, 130]]]",
    "query_spans": "[[[133, 142]]]",
    "process": "The right focus of the hyperbola is F, the left focus is E. By symmetry, AFCE is a parallelogram, so |AF| = 2|CF| = 2|AE|. Since point A lies on the hyperbola, |AF| - |AE| = 2a. Because |AF| = 2|CF|, it follows that |AF| - |AE| = 2|CF| - |CF| = 2a. Thus |CF| = 2a. In triangle OFC, |FC| = 2a, |OC| = b, |OF| = c, |AF| = 4a, we obtain 16a^{2} = b^{2} + c^{2} - 2bc\\cos\\angleAOF and 4a^{2} = b^{2} + c^{2} - 2bc\\cos\\angleCOF. Adding these gives 20a^{2} = 2b^{2} + 2c^{2} = 4c^{2} - 2a^{2}. That is: 11a^{2} = 2c^{2}, so the eccentricity of the hyperbola is: e = \\frac{\\sqrt{22}}{2}"
  },
  {
    "text": "Through the point $P(2, 4)$, draw two perpendicular lines $l_{1}$ and $l_{2}$. If $l_{1}$ intersects the $x$-axis at point $A$, and $l_{2}$ intersects the $y$-axis at point $B$, and if point $M$ is a point on the segment $AB$ such that $|BM| = 2|AM|$, then the equation of the trajectory of point $M$ is?",
    "fact_expressions": "l1: Line;l2: Line;B: Point;A: Point;P: Point;M: Point;IsPerpendicular(l1, l2);PointOnCurve(P, l1);PointOnCurve(P, l2);Coordinate(P) = (2, 4);Intersection(l1, xAxis) = A;Intersection(l2, yAxis) = B;PointOnCurve(M, LineSegmentOf(A, B));Abs(LineSegmentOf(B, M)) = 2*Abs(LineSegmentOf(A, M))",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "3*x + 12*y - 20 = 0",
    "fact_spans": "[[[20, 29], [40, 47]], [[31, 38], [58, 65]], [[71, 75]], [[53, 57]], [[1, 12]], [[77, 81], [112, 116]], [[13, 38]], [[0, 38]], [[0, 38]], [[1, 12]], [[40, 57]], [[58, 75]], [[77, 92]], [[96, 110]]]",
    "query_spans": "[[[112, 123]]]",
    "process": "Let $ l_{1}: x = k(y - 4) + 2 $. When $ y = 0 $, we get $ x = -4k + 2 $. Let $ l_{2}: y = -k(x - 2) + 4 $. When $ x = 0 $, we get $ y = 2k + 4 $. Then $ A(-4k + 2, 0) $, $ B(0, 2k + 4) $. Let $ M(x, y) $. From the given $ \\overrightarrow{BM} = 2\\overrightarrow{MA} $, we have $ (x, y - 2k - 4) = 2(-4k + 2 - x, -y) $, which gives\n$$\n\\begin{cases}\ny = -8k + 4 - 2x \\\\\ny - 2k - 4 = -2y\n\\end{cases}\n$$\nEliminating $ k $ yields $ 3x + 12y - 20 = 0 $."
  },
  {
    "text": "Given the parabola $E$: $y^{2}=2px$ $(p>0)$ with focus $F$, a circle centered at $F$ with radius $3p$ intersects the parabola $E$ at points $P$ and $Q$. The circle with diameter $PF$ passes through the point $(0,-1)$. Then, the distance from point $F$ to the line $PQ$ is?",
    "fact_expressions": "E: Parabola;p: Number;G: Circle;Q: Point;P: Point;F: Point;p>0;Expression(E) = (y^2 = 2*(p*x));Focus(E) = F;Center(G)=F;Radius(G)=3*p;Intersection(G, E) = {P, Q};I:Point;Coordinate(I)=(0,-1);C:Circle;IsDiameter(LineSegmentOf(P,F),C);PointOnCurve(I,C)",
    "query_expressions": "Distance(F, LineOf(P,Q))",
    "answer_expressions": "4*sqrt(5)/5",
    "fact_spans": "[[[2, 30], [57, 63]], [[46, 51]], [[55, 56]], [[68, 71]], [[64, 67]], [[34, 37], [39, 42], [100, 104]], [[9, 30]], [[2, 30]], [[2, 37]], [[38, 56]], [[46, 56]], [[55, 73]], [[89, 98]], [[89, 98]], [[86, 87]], [[74, 87]], [[86, 98]]]",
    "query_spans": "[[[100, 117]]]",
    "process": ""
  },
  {
    "text": "Draw a line passing through the origin with inclination angle $\\theta$, intersecting the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ at points $A$ and $B$. Let $F$ be the left focus of the ellipse. If $AF \\perp BF$ and the eccentricity $e$ of the ellipse satisfies $e \\in[\\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{6}}{3}]$, then what is the range of values for $\\theta$?",
    "fact_expressions": "G: Ellipse;a: Number;b: Number;H: Line;A: Point;F: Point;B: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);theta: Number;Inclination(H) = theta;O: Origin;PointOnCurve(O, H);Intersection(H, G) = {A, B};LeftFocus(G) = F;IsPerpendicular(LineSegmentOf(A, F), LineSegmentOf(B, F));Eccentricity(G) = e;e: Number;In(e, [sqrt(2)/2, sqrt(6)/3])",
    "query_expressions": "Range(theta)",
    "answer_expressions": "[\\pi/6, 5\\pi/6]",
    "fact_spans": "[[[22, 74], [90, 92], [116, 118]], [[24, 74]], [[24, 74]], [[19, 21]], [[76, 79]], [[86, 89]], [[80, 83]], [[24, 74]], [[24, 74]], [[22, 74]], [[10, 18], [171, 179]], [[6, 21]], [[1, 3]], [[0, 21]], [[19, 85]], [[86, 96]], [[98, 113]], [[116, 169]], [[122, 169]], [[122, 169]]]",
    "query_spans": "[[[171, 186]]]",
    "process": "Let the right focus be F', connect AF', BF', then quadrilateral AFBF is a square. ∵ AF + AF' = 2a, AF + BF = 2a, OF = c, ∴ AB = 2c, ∵ ∠BAF = 1/2 θ, ∴ AF = 2c·cos(θ/2), BF = 2c·sin(θ/2), ∴ 2c sin(θ/2) + 2c cos(θ/2) = 2a, c/a = −/si. The eccentricity e of this ellipse ∈ [√2/2, 1/(2(θ/2 + π/4))] ∈ [√2/2, ,] ∵ θ ∈ [0, π) ∴ the range of θ is [π/6, 5π/6]."
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has left and right foci $F_{1}$, $F_{2}$, $b=4$, eccentricity $\\frac{3}{5}$. A line passing through $F_{1}$ intersects the ellipse at points $A$, $B$. Then the perimeter of $\\triangle A B F_{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F2: Point;F1: Point;LeftFocus(G) = F1;RightFocus(G) = F2;b = 4;Eccentricity(G) = 3/5;H: Line;PointOnCurve(F1,H) = True;Intersection(H,G) = {A,B};A: Point;B: Point",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "20",
    "fact_spans": "[[[0, 52], [113, 115]], [[0, 52]], [[76, 81]], [[2, 52]], [[2, 52]], [[2, 52]], [[67, 74]], [[58, 65], [102, 109]], [[0, 74]], [[0, 74]], [[76, 81]], [[0, 99]], [[110, 112]], [[101, 112]], [[110, 125]], [[116, 119]], [[120, 123]]]",
    "query_spans": "[[[127, 153]]]",
    "process": "The ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has foci $F_{1}, F_{2}$, $b=4$, eccentricity $\\frac{3}{5}$, $\\therefore a=5$, $\\because$ the perimeter of $\\triangle ABF_{2}$ is $(|AF_{1}|+|AF_{2}|)+(|BF_{1}|+|BF_{2}|)=2a+2a=4a=20$, hence choose D"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, point $P$ lies on the hyperbola and satisfies $\\angle F_{1} P F_{2}=90^{\\circ}$, then $|P F_{1}| \\cdot|P F_{2}|$=?",
    "fact_expressions": "G: Hyperbola;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 - y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "2",
    "fact_spans": "[[[18, 46], [57, 60]], [[2, 9]], [[52, 56]], [[10, 17]], [[18, 46]], [[2, 51]], [[52, 61]], [[65, 98]]]",
    "query_spans": "[[[100, 128]]]",
    "process": "The hyperbola $\\frac{x^2}{4}-y^{2}=1$ has $a=2$, $b=1$, $c=\\sqrt{5}$. Let $P$ be a point on the right branch of the hyperbola, $|PF_{1}|=m$, $|PF_{2}|=n$. Then $m-n=2a=4$, $4c^{2}=m^{2}+n^{2}-2mn\\cos90^{\\circ}=m^{2}+n^{2}=20$, so $mn=2$."
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and point $P$ lies on the ellipse such that $\\angle F_{1} P F_{2}=\\frac{\\pi}{2}$. Then the area of $\\Delta F_{1} P F_{2}$ is equal to?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = pi/2",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[17, 44], [55, 57]], [[1, 8]], [[50, 54]], [[9, 16]], [[17, 44]], [[1, 49]], [[50, 58]], [[62, 98]]]",
    "query_spans": "[[[100, 128]]]",
    "process": "Since $F_{1}F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and point $P$ lies on the ellipse satisfying $\\angle F_{1}PF_{2}=\\frac{\\pi}{2}$, we have  \n\\[\n\\begin{cases}\nPF_{1}+PF_{2}=2a=4 \\\\\nPF_{1}^{2}+PF_{2}^{2}=4c^{2}=12\n\\end{cases}\n\\Rightarrow 2PF_{1}\\cdot PF_{2} = (PF_{1}+PF_{2})^{2} - (PF_{1}^{2}+PF_{2}^{2}) = 16-12=4\n\\]  \nThus, $PF_{1}\\cdot PF_{2}=2$, then the area of $\\triangle F_{1}PF_{2}$ equals $\\frac{1}{2}PF_{1}\\cdot PF_{2}=1$."
  },
  {
    "text": "Given the parabola equation $y^{2}=2 x$, what is the distance from the point $P(\\frac{3}{2}, y_{0})$ on the parabola to the focus $F$?",
    "fact_expressions": "G: Parabola;P: Point;y0:Number;Coordinate(P) = (3/2, y0);Expression(G) = (y^2 = 2*x);PointOnCurve(P, G);Focus(G) = F;F:Point",
    "query_expressions": "Distance(P,F)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 5], [21, 24]], [[26, 50]], [[27, 50]], [[26, 50]], [[2, 19]], [[21, 50]], [[21, 56]], [[53, 56]]]",
    "query_spans": "[[[26, 61]]]",
    "process": "\\because the parabola equation is y^{2}=2x, \\therefore the directrix equation: x=-\\frac{1}{2} \\therefore the distance from point P(\\frac{3}{2},y_{0}) on the parabola to the focus F is d=\\frac{3}{2}-(-\\frac{1}{2})=2"
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x(p>0)$ is a focus of the ellipse $\\frac{x^{2}}{3 p}+\\frac{y^{2}}{p}=1$, then $p$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;H: Ellipse;Expression(H) = (x^2/((3*p)) + y^2/p = 1);Focus(G) = OneOf(Focus(H))",
    "query_expressions": "p",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 22]], [[1, 22]], [[72, 75]], [[4, 22]], [[26, 65]], [[26, 65]], [[1, 70]]]",
    "query_spans": "[[[72, 77]]]",
    "process": "Since the focus of the parabola $ y^{2} = 2px $ ($ p > 0 $) is $ \\left( \\frac{p}{2}, 0 \\right) $, one focus of the ellipse $ \\frac{x^{2}}{3p} + \\frac{y^{2}}{p} = 1 $ is also $ \\left( \\frac{p}{2}, 0 \\right) $."
  },
  {
    "text": "Given the ellipse: $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1$, with left and right foci $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. If the maximum value of $A F_{2}+B F_{2}$ is $5$, then what is the standard equation of the ellipse?",
    "fact_expressions": "l: Line;G: Ellipse;b: Number;A: Point;F2: Point;B: Point;F1: Point;Expression(G) = (x^2/4 + y^2/b^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(F1, l);Intersection(l, G) = {A, B};Max(LineSegmentOf(A, F2) + LineSegmentOf(B, F2)) = 5",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4+y^2/3=1",
    "fact_spans": "[[[77, 82]], [[2, 43], [82, 84], [123, 125]], [[5, 44]], [[86, 89]], [[60, 67]], [[90, 93]], [[52, 59], [69, 76]], [[2, 44]], [[2, 67]], [[2, 67]], [[68, 82]], [[77, 95]], [[97, 122]]]",
    "query_spans": "[[[124, 132]]]",
    "process": "From the ellipse equation, we know that $ a^{2} = 4 \\therefore a = 2 $. From the definition of the ellipse, the perimeter of $ 4ABF_{2} $ is a constant $ 4a = 8 $. Since the maximum value of $ AF_{2} + BF_{2} $ is 5, the length of the latus rectum is 3, i.e., $ \\therefore \\frac{2b^{2}}{a} = 3 \\therefore b^{2} = 3^{n} $. Therefore, the ellipse equation is $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $."
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{m}=1$ has eccentricity $\\sqrt{5}$, then $m$=?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/4 - y^2/m = 1);Eccentricity(G) = sqrt(5)",
    "query_expressions": "m",
    "answer_expressions": "16",
    "fact_spans": "[[[0, 38]], [[55, 58]], [[0, 38]], [[0, 53]]]",
    "query_spans": "[[[55, 60]]]",
    "process": "From the given conditions, we have $a^{2}=4$, $b^{2}=m$, $\\therefore c^{2}=a^{2}+b^{2}=4+m$, $\\therefore a=2$, $c=\\sqrt{4+m}$, $\\therefore$ eccentricity $e=\\frac{c}{a}=\\frac{\\sqrt{4+m}}{2}=\\sqrt{5}$, solving gives $m=16$."
  },
  {
    "text": "Given that $F_{1}$ is the left focus of the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$, and the line $l$: $y=x-1$ intersects the ellipse $C$ at points $A$ and $B$, what is the value of $|F_{1} A|+|F_{1} B|$?",
    "fact_expressions": "l: Line;C: Ellipse;F1: Point;A: Point;B: Point;Expression(C) = (x^2/2 + y^2 = 1);LeftFocus(C) = F1;Expression(l) = (y = x - 1);Intersection(l, C) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(F1, A)) + Abs(LineSegmentOf(F1, B))",
    "answer_expressions": "8*sqrt(2)/3",
    "fact_spans": "[[[47, 60]], [[10, 42], [61, 66]], [[2, 9]], [[68, 71]], [[72, 75]], [[10, 42]], [[2, 46]], [[47, 60]], [[47, 77]]]",
    "query_spans": "[[[80, 106]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the ellipse. When $P F_{1} \\cdot P F_{2}=0$, the area of $\\triangle F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G) = True;LineSegmentOf(P, F1)*LineSegmentOf(P, F2) = 0",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 29], [59, 61]], [[2, 29]], [[38, 45]], [[46, 53]], [[2, 53]], [[2, 53]], [[54, 58]], [[54, 62]], [[64, 89]]]",
    "query_spans": "[[[91, 121]]]",
    "process": ""
  },
  {
    "text": "Let $P$ and $Q$ be points on the circle $x^{2}+(y+1)^{2}=7$ and the ellipse $\\frac{x^{2}}{8}+y^{2}=1$, respectively. Then the maximum distance between $P$ and $Q$ is?",
    "fact_expressions": "P: Point;Q: Point;H: Circle;Expression(H) = (x^2 + (y + 1)^2 = 7);G: Ellipse;Expression(G) = (x^2/8 + y^2 = 1);PointOnCurve(P, H) = True;PointOnCurve(Q, G) = True",
    "query_expressions": "Max(Distance(P, Q))",
    "answer_expressions": "15*sqrt(7)/7",
    "fact_spans": "[[[1, 4], [64, 67]], [[5, 8], [68, 71]], [[11, 31]], [[11, 31]], [[32, 59]], [[32, 59]], [[1, 62]], [[1, 62]]]",
    "query_spans": "[[[64, 79]]]",
    "process": "Find the maximum distance from the center of the circle to a point on the ellipse, then add the radius of the circle. Let $ Q(x,y) $. The distance from the center $ (0,-1) $ to point $ Q $ can be expressed as a function of $ y $, and the properties of quadratic functions can be applied. [Detailed explanation] According to the properties of the circle, the maximum distance between points $ P $ and $ Q $ can be transformed into the maximum distance from the center of the circle to a point on the ellipse plus the radius $ \\sqrt{7} $. Let $ Q(x,y) $, then the distance $ d $ from the center $ (0,-1) $ to a point on the ellipse is $ d = \\sqrt{x^{2} + (y+1)^{2}} $, where $ -1 \\leqslant y \\leqslant 1 $. Therefore, when $ y = \\frac{1}{7} $, $ d $ takes the maximum value $ \\frac{8\\sqrt{7}}{7} $. Hence, the maximum distance between points $ P $ and $ Q $ is $ d_{\\max} + \\sqrt{7} = \\frac{15\\sqrt{7}}{7} $."
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ has two foci $F_{1}$ and $F_{2}$. A line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. Then, the perimeter of $\\triangle A B F_{2}$ is?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;F1: Point;F2: Point;Expression(G) = (x^2/16 + y^2/9 = 1);Focus(G)={F1,F2};PointOnCurve(F1, H);Intersection(H, G) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "16",
    "fact_spans": "[[[0, 38], [73, 75]], [[70, 72]], [[76, 79]], [[80, 83]], [[45, 52], [62, 69]], [[53, 60]], [[0, 38]], [[0, 60]], [[61, 72]], [[70, 85]]]",
    "query_spans": "[[[87, 112]]]",
    "process": "Since the ellipse is $\\frac{x^2}{16} + \\frac{y^2}{9} = 1$, $\\therefore a = 4$. By the definition of the ellipse, the perimeter of $ABF_{2}$ is $(|AF_{1}| + |AF_{2}|) + (|BF_{1}| + |BF_{2}|) = 2a + 2a = 4a = 16$. Hence, the answer is A."
  },
  {
    "text": "If there is a point on the ellipse $\\frac{x^{2}}{45}+\\frac{y^{2}}{b^{2}}=1$ with foci on the $x$-axis such that the lines connecting this point to the two foci are perpendicular to each other, then the range of values for $b$ is?",
    "fact_expressions": "G: Ellipse;b: Number;Expression(G) = (x^2/45 + y^2/b^2 = 1);PointOnCurve(Focus(G),xAxis);P:Point;PointOnCurve(P,G);F1:Point;F2:Point;IsPerpendicular(LineSegmentOf(P,F1),LineSegmentOf(P,F2));Focus(G)={F1,F2}",
    "query_expressions": "Range(b)",
    "answer_expressions": "[-3*sqrt(10)/2,3*sqrt(10)/2]&Negation(b=0)",
    "fact_spans": "[[[10, 52]], [[73, 76]], [[10, 52]], [[1, 52]], [[58, 59]], [[10, 56]], [], [], [[10, 71]], [[10, 64]]]",
    "query_spans": "[[[73, 83]]]",
    "process": "\\because the foci of the ellipse \\frac{x^{2}}{45}+\\frac{y^{2}}{b^{2}}=1 lie on the x-axis, thus b^{2}<45, i.e., b\\in(-3\\sqrt{5},3\\sqrt{5}) and b\\neq0 \\textcircled{1}. Let the focal distance of the ellipse be 2c, then the circle O with the origin as center and the segment joining the two foci as diameter has the equation x^{2}+y^{2}=c^{2}. To ensure that there exists a point on the ellipse \\frac{x^{2}}{45}+\\frac{y^{2}}{b^{2}}=1 such that the lines connecting it to the two foci are mutually perpendicular, it suffices that circle O and the ellipse intersect. By the geometric properties of the ellipse, this requires the radius c\\geqslant|b|, i.e., c^{2}\\geqslant b^{2}, which means 45-b^{2}\\geqslant b^{2}, so b^{2}\\leqslant\\frac{45}{2} \\textcircled{2}. From \\textcircled{1} and \\textcircled{2}, we obtain: -\\frac{3\\sqrt{10}}{2}\\leqslant b\\leqslant\\frac{3\\sqrt{10}}{2} and b\\neq0."
  },
  {
    "text": "Given the standard equation of an ellipse $\\frac{x^{2}}{a^{2}}+y^{2}=1$ $(a>1)$, the upper vertex is $A$, the left vertex is $B$, and let point $P$ be a point on the ellipse. The maximum area of $\\Delta P A B$ is $\\sqrt{2}+1$. If points $M(-\\sqrt{3}, 0)$, $N(\\sqrt{3}, 0)$ are given, and point $Q$ is any point on the ellipse, then the minimum value of $\\frac{1}{|Q N|}+\\frac{4}{|Q M|}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2 + x^2/a^2 = 1);a: Number;a>1;A: Point;UpperVertex(G) = A;B: Point;LeftVertex(G) = B;P: Point;PointOnCurve(P, G);Max(Area(TriangleOf(P, A, B))) = 1 + sqrt(2);M: Point;Coordinate(M) = (-sqrt(3), 0);N: Point;Coordinate(N) = (sqrt(3), 0);Q: Point;PointOnCurve(Q, G)",
    "query_expressions": "Min(1/Abs(LineSegmentOf(Q, N)) + 4/Abs(LineSegmentOf(Q, M)))",
    "answer_expressions": "9/4",
    "fact_spans": "[[[2, 4], [67, 69], [152, 154]], [[2, 44]], [[10, 44]], [[10, 44]], [[49, 52]], [[2, 52]], [[57, 60]], [[2, 60]], [[62, 66]], [[62, 72]], [[73, 107]], [[111, 129]], [[111, 129]], [[130, 146]], [[130, 146]], [[147, 151]], [[147, 159]]]",
    "query_spans": "[[[161, 200]]]",
    "process": "From the given conditions, we have A(0,1), B(-a,0), the slope of line AB is $ k_{AB} = \\frac{1}{a} $, and the equation of line AB is $ y = \\frac{1}{a}x + 1 $. When the area of $ \\triangle PAB $ is maximized, the line passing through point P is tangent to the ellipse and parallel to line AB. Thus, assume the equation of this line is $ y = \\frac{1}{a}x + m $ ($ m \\neq 1 $). Solving the system:\n$$\n\\begin{cases}\n\\frac{x^{2}}{a^{2}} + y^{2} = 1 \\\\\ny = \\frac{1}{a}x + m\n\\end{cases}\n$$\nRearranging gives $ 2x^{2} + 2amx + a^{2}m^{2} - a^{2} = 0 $. From $ \\Delta = 0 $, we get $ 4a^{2}m^{2} - 8(a^{2}m^{2} - a^{2}) = 0 $, solving yields $ m^{2} = 2 $. Analysis shows that when the area of $ \\triangle PAB $ is maximized, $ m = -\\sqrt{2} $, so the tangent line equation is $ y = \\frac{1}{a}x - \\sqrt{2} $. Then the distance from point P to line AB is $ d = \\frac{\\sqrt{2}+1}{\\sqrt{\\frac{1}{a^{2}}+1}} = \\frac{a(\\sqrt{2}+1)}{\\sqrt{a^{2}+1}} $. Also, $ |AB| = \\sqrt{a^{2}+1} $, so $ S_{\\triangle PAB} = \\frac{1}{2}|AB| \\cdot d = \\frac{a}{2}(\\sqrt{2}+1) = \\sqrt{2}+1 $, hence $ a = 2 $. Therefore, $ M(-\\sqrt{3},0) $, $ N(\\sqrt{3},0) $ are the left and right foci of the ellipse respectively. So $ |QM| + |QN| = 2a = 4 $, then $ \\frac{1}{|ON|} + \\frac{|QN| + |QM|}{4} = \\frac{1}{4}\\left(5 + \\frac{|QM|}{|QN|} + \\frac{4|QN|}{|QM|}\\right) \\geqslant \\frac{1}{4}\\left(5 + 2\\sqrt{\\frac{|QM|}{|QN|} \\cdot \\frac{4|QN|}{|QM|}}\\right) = \\frac{9}{4} $, with equality if and only if $ |QM| = 2|QN| $. Therefore, the minimum value of $ \\frac{1}{|ON|} + \\frac{4}{|OM|} $ is $ \\frac{9}{4} $."
  },
  {
    "text": "The standard equation of an ellipse with one focus at $(0 ,\\sqrt {5})$ and a minor axis length of $4 \\sqrt {5}$ is?",
    "fact_expressions": "G: Ellipse;H: Point;Coordinate(H) = (0, sqrt(5));OneOf(Focus(G)) = H;Length(MinorAxis(G))=4*sqrt(5)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/20 + y^2/25 = 1",
    "fact_spans": "[[[0, 2], [42, 44]], [[6, 22]], [[6, 22]], [[0, 22]], [[0, 41]]]",
    "query_spans": "[[[42, 50]]]",
    "process": ""
  },
  {
    "text": "$F$ is the right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$. From point $F$, a perpendicular is drawn to an asymptote of $C$, with foot at $A$, intersecting the other asymptote at point $B$. If $2 \\overrightarrow{A F}=\\overrightarrow{F B}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;A: Point;F: Point;B: Point;L1: Line;L2: Line;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;L:Line;OneOf(Asymptote(C))=L1;OneOf(Asymptote(C))=L2;Negation(L1=L2);PointOnCurve(F,L);IsPerpendicular(L,L1);FootPoint(L,L1)=A;Intersection(L,L2)=B;2*VectorOf(A, F) = VectorOf(F, B)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[4, 65], [76, 79], [159, 162]], [[11, 65]], [[11, 65]], [[92, 95]], [[0, 3], [71, 75]], [[104, 108]], [], [], [[11, 65]], [[11, 65]], [[4, 65]], [[0, 69]], [], [[76, 85]], [[76, 103]], [[76, 103]], [[70, 88]], [[70, 88]], [[70, 95]], [[70, 108]], [[111, 157]]]",
    "query_spans": "[[[159, 168]]]",
    "process": "The hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) has asymptotes $ y = \\pm \\frac{b}{a}x $. According to the problem, $ |AF| = b $, $ |BF| = 2b $. In right triangle $ \\triangle AOF $, $ |OF| = c $, then $ |OA| = \\sqrt{c^{2} - b^{2}} = a $. Let the inclination angle of $ l_{1} $ be $ \\theta $, i.e., $ \\angle AOF = \\theta $, then $ \\angle AOB = 2\\theta $, $ \\tan\\theta = \\frac{b}{a} $, $ \\tan2\\theta = \\frac{3b}{a} $, that is, $ \\frac{3b}{a} = \\frac{\\frac{2b}{a}}{1 - \\frac{b^{2}}{a^{2}}} $, thus $ a^{2} = 3b^{2} $. Then $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\frac{2\\sqrt{3}}{3} $."
  },
  {
    "text": "The two endpoints of a line segment $AB$ with length $6$ move on the parabola $y = x^2$. What is the minimum distance from the midpoint $M$ of segment $AB$ to the $x$-axis?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2);B: Point;A: Point;Length(LineSegmentOf(A,B)) = 6;PointOnCurve(Endpoint(LineSegmentOf(A,B)), G);MidPoint(LineSegmentOf(A,B)) = M;M: Point",
    "query_expressions": "Min(Distance(M, xAxis))",
    "answer_expressions": "11/4",
    "fact_spans": "[[[20, 32]], [[20, 32]], [[9, 14]], [[9, 14]], [[0, 14]], [[7, 33]], [[38, 51]], [[48, 51]]]",
    "query_spans": "[[[48, 64]]]",
    "process": "According to the problem, the parabola $ y = x^{2} $ has focus $ F(0, \\frac{1}{4}) $ and directrix $ l: y = -\\frac{1}{4} $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then the midpoint $ M $ of $ AB $ is $ (\\frac{x_{1}+x_{2}}{2}, \\frac{y_{1}+y_{2}}{2}) $. Draw perpendiculars from points $ A $, $ B $, $ M $ to line $ l $, with feet $ A_{1} $, $ B_{1} $, $ M_{1} $ respectively. Let $ MM_{1} $ intersect the $ x $-axis at point $ M_{0} $, as shown in the figure. By the definition of a parabola, we have: $ |AF| = |AA_{1}| = y_{1} + \\frac{1}{4} $, $ |BF| = |BB_{1}| = y_{2} + \\frac{1}{4} $. Thus, $ |AF| + |BF| = y_{1} + \\frac{1}{4} + y_{2} + \\frac{1}{4} = y_{1} + y_{2} + \\frac{1}{2} $. Since $ |AF| + |BF| \\geqslant |AB| = 6 $, with equality if and only if points $ A $, $ F $, $ B $ are collinear, it follows that $ y_{1} + y_{2} + \\frac{1}{2} \\geqslant 6 $, i.e., $ y_{1} + y_{2} \\geqslant \\frac{11}{2} $. Then $ |MM_{0}| = \\frac{y_{1} + y_{2}}{2} \\geqslant \\frac{11}{4} $. Therefore, the minimum distance from the midpoint $ M $ of segment $ AB $ to the $ x $-axis is $ \\frac{11}{4} $."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=4 x$, two mutually perpendicular lines $l_{1}$ and $l_{2}$ are drawn through $F$. The line $l_{1}$ intersects $C$ at points $A$ and $B$, and the line $l_{2}$ intersects $C$ at points $D$ and $E$. Then the minimum value of $|A B|+|D E|$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l1: Line;l2: Line;PointOnCurve(F, l1);PointOnCurve(F, l2);IsPerpendicular(l1, l2);A: Point;B: Point;Intersection(l1, C) = {A, B};D: Point;E: Point;Intersection(l2, C) = {D, E}",
    "query_expressions": "Min(Abs(LineSegmentOf(A, B)) + Abs(LineSegmentOf(D, E)))",
    "answer_expressions": "16",
    "fact_spans": "[[[6, 25], [70, 73], [95, 98]], [[6, 25]], [[2, 5], [30, 33]], [[2, 28]], [[41, 50], [60, 69]], [[52, 59], [85, 94]], [[29, 59]], [[29, 59]], [[34, 59]], [[75, 78]], [[79, 82]], [[60, 84]], [[100, 103]], [[104, 107]], [[85, 109]]]",
    "query_spans": "[[[111, 130]]]",
    "process": "From the given conditions, the parabola $ C: y^2 = 4x $ has focus $ F: (1, 0) $ and directrix $ x = -1 $. Let the equation of line $ l_1 $ be $ y = k(x - 1) $. Since lines $ l_1 $ and $ l_2 $ are perpendicular, the slope of $ l_2 $ is $ -\\frac{1}{k} $. Solving the system of equations with the parabola:\n$$\n\\begin{cases}\ny = k(x - 1)\n\\end{cases}\n$$\nEliminating $ y $ gives $ k^2x^2 - (2k^2 + 4)x + k^2 = 0 $. Let points $ A(x_1, y_1) $, $ B(x_2, y_2) $, $ C(x_3, y_3) $, $ D(x_4, y_4) $. By the relationship between roots and coefficients, $ x_1 + x_2 = \\frac{2k^2 + 4}{k^2} $, similarly $ x_3 + x_4 = \\frac{2\\frac{1}{k^2} + 4}{\\frac{1}{k^2}} $. According to the property of the parabola, the distance from a point on the parabola to the focus equals the distance to the directrix. Therefore, $ |AB| = x_1 + 1 + x_2 + 1 $, similarly $ |DE| = x_3 + 1 + x_4 + 1 $. Hence,\n$$\n|AB| + |DE| = \\frac{2k^2 + 4}{k^2} + \\frac{2\\frac{1}{k^2} + 4}{\\frac{1}{k^2}} + 4 = 8 + \\frac{4}{k^2} + 4k^2 \\geqslant 8 + 2\\sqrt{4 \\times 4} = 16,\n$$\nwith equality if and only if $ k^2 = 1 $."
  },
  {
    "text": "Given that a line passing through the focus $F$ of the parabola $y^2 = x$ intersects the parabola at points $A$ and $B$, and a line passing through the origin $O$ intersects the hyperbola $\\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1$ $(a > 0, b > 0)$ at points $M$ and $N$. Point $P$ is a point on the hyperbola, and the slopes of lines $PM$ and $PN$ are $k_1$ and $k_2$ respectively. If the inequality $(|k_1| + 4|k_2|)(|AF| \\cdot |BF|) \\geq |AF| + |BF|$ always holds, then what is the eccentricity of the hyperbola?",
    "fact_expressions": "P: Point;M: Point;N: Point;G: Hyperbola;H: Parabola;a: Number;b: Number;I: Line;Z: Line;A: Point;F: Point;B: Point;O: Origin;a>0;b>0;k1: Number;k2: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = x);Focus(H) = F;PointOnCurve(F, I);Intersection(I, H) = {A, B};PointOnCurve(O, Z);Intersection(Z, G) = {M, N};PointOnCurve(P, G);Slope(LineOf(P, M)) = k1;Slope(LineOf(P, N)) = k2;(Abs(k1) + 4*Abs(k2))*(Abs(LineSegmentOf(A, F))*Abs(LineSegmentOf(B, F))) >= Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[108, 112]], [[98, 101]], [[102, 105]], [[49, 96], [113, 116], [205, 208]], [[3, 13], [22, 25]], [[52, 96]], [[52, 96]], [[19, 21]], [[46, 48]], [[27, 30]], [[15, 18]], [[31, 34]], [[38, 45]], [[52, 96]], [[52, 96]], [[138, 143]], [[144, 149]], [[49, 96]], [[3, 13]], [[3, 18]], [[2, 21]], [[19, 36]], [[37, 48]], [[46, 107]], [[108, 119]], [[121, 149]], [[121, 149]], [[154, 200]]]",
    "query_spans": "[[[205, 214]]]",
    "process": ""
  },
  {
    "text": "The left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through $F_{1}$ intersects the hyperbola at exactly one point $P$. If $\\sin \\angle F_{1} P F_{2}=\\frac{4}{5}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "l: Line;G: Hyperbola;b: Number;a: Number;F1: Point;P: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(F1,l);Intersection(l,G)=P;Sin(AngleOf(F1, P, F2)) = 4/5",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{sqrt(2),sqrt(17)}",
    "fact_spans": "[[[92, 97]], [[0, 56], [98, 101], [153, 156]], [[3, 56]], [[3, 56]], [[64, 72], [84, 91]], [[106, 109]], [[73, 82]], [[3, 56]], [[3, 56]], [[0, 56]], [[0, 81]], [[0, 81]], [[83, 97]], [[92, 109]], [[111, 150]]]",
    "query_spans": "[[[153, 162]]]",
    "process": "First, set the equation of line $ l $, solve it together with the hyperbola equation to obtain the coordinates of point $ P $, use the chord length formula to find $ |PF_{1}| $, and express $ |PF_{2}| $ according to the definition. In $ \\triangle F_{1}PF_{2} $, by the cosine law, we express $ \\therefore 1-\\cos\\angle F_{1}PF_{2} = \\frac{8}{e^{2}+3} $, then find the eccentricity. Solution: As shown in the figure, when the line is parallel to the asymptote, $ l $ intersects the hyperbola at a unique point $ P $. Let $ l: y = \\frac{b}{a}(x+c) $, solving together with the hyperbola equation yields $ -2cx = a^{2}+c^{2} $, solving gives: $ x = -\\frac{a^{2}+c^{2}}{2c} $, $ |PF_{1}| = \\sqrt{1+\\frac{b^{2}}{a^{2}}} |x_{P}-(-c)| = \\sqrt{\\frac{c^{2}}{a^{2}}} \\left| -\\frac{a^{2}+c^{2}}{2c} + c \\right| = \\frac{b^{2}}{2a} $, $ |PF_{2}| = |PF_{1}| + 2a = \\frac{b^{2}+4a^{2}}{2a} $, $ |F_{1}F_{2}| = 2c $. In $ \\Delta F_{1}PF_{2} $, since $ \\sin\\angle F_{1}PF_{2} = \\frac{4}{5} $, $ \\therefore \\cos\\angle F_{1}PF_{2} = \\pm\\frac{3}{5} $. By the cosine law, $ |F_{1}F_{2}|^{2} = |PF_{1}|^{2} + |PF_{2}|^{2} - 2|PF_{1}||PF_{2}|\\cos\\angle F_{1}PF_{2} = (|PF_{1}| - |PF_{2}|)^{2} + 2|PF_{1}||PF_{2}|(1 - \\cos\\angle F_{1}PF_{2}) $,"
  },
  {
    "text": "The equations of the two asymptotes of the hyperbola are $y = \\pm \\sqrt{2}x$, and the hyperbola passes through the point $(3, -2\\sqrt{3})$. A line passing through the right focus $F$ of the hyperbola with an inclination angle of $60^{\\circ}$ intersects the hyperbola at points $A$ and $B$. Then the value of $|AB|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*(sqrt(2)*x));K: Point;Coordinate(K) = (3, -2*sqrt(3));PointOnCurve(K, G);H: Line;F: Point;RightFocus(G) = F;PointOnCurve(F, H);Inclination(H) = ApplyUnit(60, degree);A: Point;B: Point;Intersection(H, G) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16*sqrt(3)",
    "fact_spans": "[[[0, 3], [56, 59], [87, 90]], [[0, 31]], [[35, 53]], [[35, 53]], [[0, 53]], [[84, 86]], [[63, 66]], [[56, 66]], [[55, 86]], [[67, 86]], [[91, 94]], [[95, 98]], [[84, 100]]]",
    "query_spans": "[[[102, 113]]]",
    "process": "According to the problem, the equations of the two asymptotes of the hyperbola are $ y = \\pm\\sqrt{2}x $. Thus, we can assume the equation of the hyperbola is $ 2x^{2} - y^{2} = \\lambda $ ($ \\lambda \\neq 0 $). Since the hyperbola passes through the point $ (3, -2\\sqrt{3}) $, substituting into the equation gives $ \\lambda = 6 $, so the required hyperbola equation is $ \\frac{x^{2}}{3} - \\frac{y^{2}}{6} = 1 $, and the right focus is $ F(3, 0) $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, the line passing through focus $ F $ with an inclination angle of $ 60^{\\circ} $ has the equation $ y = \\sqrt{3}(x - 3) $. Solving the system of equations\n$$\n\\begin{cases}\ny = \\sqrt{3}(x - 3) \\\\\n\\frac{x^{2}}{3} - \\frac{y^{2}}{6} = 1\n\\end{cases}\n$$\nleads to the quadratic equation $ x^{2} - 18x + 33 = 0 $. Then $ x_{1} + x_{2} = 18 $, $ x_{1}x_{2} = 33 $. The chord length is\n$$\n|AB| = \\sqrt{1 + k^{2}} |x_{2} - x_{1}| = \\sqrt{1 + (\\sqrt{3})^{2}} \\cdot \\sqrt{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}} = 2\\sqrt{18^{2} - 4 \\times 33} = 16\\sqrt{3}\n$$\nThus, the value of the chord length $ |AB| $ is $ 16\\sqrt{3} $. This problem mainly examines the application of the positional relationship between a line and a hyperbola. The key to solving it lies in combining the system of equations, properly using the relationship between roots and coefficients, and the chord length formula, along with accurate computation. It emphasizes reasoning and computational ability and is a medium-difficulty problem."
  },
  {
    "text": "Given that point $F$ is the focus of the parabola $x^{2}=8 y$, $M(0,-2)$, and point $N$ is a moving point on the parabola. When $\\frac{|N F|}{|N M|}$ is minimized, point $N$ lies exactly on the hyperbola with foci at $M$ and $F$. Then, the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;M: Point;N: Point;F: Point;Expression(H) = (x^2 = 8*y);Coordinate(M) = (0, -2);Focus(H) = F;PointOnCurve(N, H);WhenMin(Abs(LineSegmentOf(N,F))/Abs(LineSegmentOf(N,M)));Focus(G)={M,F};PointOnCurve(N,G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3",
    "fact_spans": "[[[93, 96], [100, 103]], [[7, 21], [40, 43]], [[25, 34], [82, 85]], [[35, 39], [74, 78]], [[2, 6], [86, 89]], [[7, 21]], [[25, 34]], [[2, 24]], [[35, 47]], [[48, 73]], [[81, 96]], [[74, 97]]]",
    "query_spans": "[[[100, 109]]]",
    "process": "From the given conditions, F(0,2), M(0,-2), the directrix of the parabola is y = -2. Let point \\( N(x_{0},\\frac{x_{0}^{2}}{8}) \\). According to symmetry, assume without loss of generality that \\( x_{0} > 0 \\). By the definition of a parabola, \\( |NF| = \\frac{x_{0}^{2}}{8} + 2 \\), and \\( |NM| = \\sqrt{x_{0}^{2} + \\left(\\frac{x_{0}^{2}}{8} + 2\\right)^2} \\). Let the equation of the hyperbola with foci M and F be \\( \\frac{y^{2}}{a^{2}} - \\frac{x^{2}}{b^{2}} = 1 \\). Then \\( 2a = ||NF| - |NM|| = |4 - 4\\sqrt{2}| = 4(\\sqrt{2} - 1) \\), so \\( a = 2(\\sqrt{2} - 1) \\). Also, \\( 2c = |MF| = 4 \\), so \\( c = 2 \\), thus the eccentricity \\( e = \\frac{c}{a} = \\frac{2}{2(\\sqrt{2} - 1)} = \\sqrt{2} + 1 \\). Also, \\( 2c = |MF| = 4 \\), \\( c = \\)"
  },
  {
    "text": "A set of values of $m$, $n$ that illustrates \"if $m n \\neq 0$, then the equation $\\frac{x^{2}}{m}+\\frac{y^{2}}{n}=1$ represents a hyperbola with foci on the $y$-axis and asymptotes given by $y=2 x$\" is?",
    "fact_expressions": "m: Number;n: Number;Negation(m*n=0);H: Curve;Expression(H) = (x^2/m+y^2/n=1);G: Hyperbola;H=G;PointOnCurve(Focus(G), yAxis);Expression(OneOf(Asymptote(G))) = (y = 2*x)",
    "query_expressions": "m;n",
    "answer_expressions": "-1\n4",
    "fact_spans": "[[[92, 95]], [[96, 99]], [[5, 17]], [[59, 61]], [[19, 61]], [[85, 88]], [[59, 88]], [[62, 88]], [[71, 88]]]",
    "query_spans": "[[[92, 103]], [[96, 103]]]",
    "process": "Let the equation of a hyperbola with foci on the y-axis and asymptotes given by y = 2x be \\frac{y^{2}}{4} - x^{2} = \\lambda (\\lambda > 0), that is, \\frac{y^{2}}{4\\lambda} - \\frac{x^{2}}{\\lambda} = 1 (\\lambda > 0), so \\begin{cases} m = - \\\\ n = 4\\lambda \\end{cases}. 2(\\lambda > 0), without loss of generality let \\lambda = 1, so \\begin{cases} m = \\frac{7}{4} \\\\ n = 4 \\end{cases}"
  },
  {
    "text": "Given that $F$ is the left focus of the hyperbola $C$: $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, and $P$, $Q$ are points on $C$. If the length of $PQ$ is equal to twice the length of the imaginary axis, and point $A(5,0)$ lies on segment $PQ$, then the perimeter of $\\triangle PQF$ is?",
    "fact_expressions": "C: Hyperbola;P: Point;Q: Point;A: Point;F: Point;Expression(C) = (x^2/16 - y^2/9 = 1);Coordinate(A) = (5, 0);LeftFocus(C) = F;PointOnCurve(A, LineSegmentOf(P,Q));PointOnCurve(P,C);PointOnCurve(Q,C);Length(LineSegmentOf(P,Q))=2*Length(ImageinaryAxis(C))",
    "query_expressions": "Perimeter(TriangleOf(P, Q, F))",
    "answer_expressions": "40",
    "fact_spans": "[[[6, 50], [63, 66]], [[55, 58]], [[59, 62]], [[88, 97]], [[2, 5]], [[6, 50]], [[88, 97]], [[2, 54]], [[88, 106]], [[55, 69]], [[55, 69]], [[70, 87]]]",
    "query_spans": "[[[108, 130]]]",
    "process": "From the hyperbola equation, we have $ a=4, b=3, c=5 $, then the length of the imaginary axis is 6. The line segment $ PQ $ passes through point $ A(5,0) $, which is the right focus of the hyperbola. $ PF = PA + 2a $, $ OF = OA + 2a $, $ PF + OF = 4a + 12 = 28 $. The perimeter of quadrilateral $ APOF $ is $ 28 + 12 = 40 $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{m}+y^{2}=1$ $(m>1)$, if there exist two mutually perpendicular lines $l_{1}$, $l_{2}$ passing through point $A(1,2)$ such that both $l_{1}$ and $l_{2}$ have no common points with the ellipse $C$, then the range of the eccentricity of the ellipse is?",
    "fact_expressions": "C: Ellipse;m: Number;l1: Line;l2: Line;A: Point;m>1;Expression(C) = (y^2 + x^2/m = 1);Coordinate(A) = (1, 2);PointOnCurve(A, l1);PointOnCurve(A, l2);IsPerpendicular(l1, l2);NumIntersection(l1,C)=0;Negation(IsIntersect(l2, C))",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(0, \\sqrt{3}/2)",
    "fact_spans": "[[[2, 39], [97, 102], [110, 112]], [[9, 39]], [[61, 68], [79, 86]], [[70, 77], [89, 96]], [[44, 53]], [[9, 39]], [[2, 39]], [[44, 53]], [[43, 77]], [[43, 77]], [[54, 77]], [[79, 107]], [[79, 107]]]",
    "query_spans": "[[[110, 122]]]",
    "process": "Ellipse $ C: \\frac{x^{2}}{m} + y^{2} = 1 $ ($ m > 1 $)  \n① When one of the lines $ l_{1}, l_{2} $ has an undefined slope and the other has slope 0, the two lines intersect the ellipse.  \n② When the slopes of both lines $ l_{1}, l_{2} $ exist, let $ l_{1}: y - 2 = k(x - 1) $, i.e., $ y = kx + 2 - k $.  \nCombining with the ellipse equation gives:  \n$ (mk^{2} + 1)x^{2} + 2km(2 - k)x + m(2 - k)^{2} - m = 0 $.  \nSince line $ l_{1} $ and ellipse $ C $ have no intersection points, we have:  \n$ \\Delta = 4k^{2}m^{2}(2 - k)^{2} - 4(mk^{2} + 1)[m(2 - k)^{2} - m] < 0 $,  \nwhich simplifies to $ (m - 1)k^{2} + 4k - 3 < 0 $,  \nsolving gives $ \\frac{-2 - \\sqrt{3m + 1}}{m - 1} < k < \\frac{-2 + \\sqrt{3m + 1}}{m - 1} $.  \nUsing the condition that the two lines are perpendicular, replace $ k $ with $ -\\frac{1}{k} $, yielding:  \n$ \\frac{m - 1}{k^{2}} - \\frac{4}{k} - 3 < 0 $,  \nwhich becomes $ 3k^{2} + 4k - m + 1 > 0 $,  \n$ -2 + \\sqrt{3m + 1} $, $ m > 1 $,  \nleading to $ 8 - 2m < (4 - m)\\sqrt{3m + 1} $.  \nWhen $ m > 1 $, $ \\sqrt{3m + 1} > 2 $, so $ 4 - m > 0 $, giving $ m < 4 $. Thus $ 1 < m < 4 $.  \nSolving $ 3m + 1 $, we get $ 1 < m < 4 $, hence $ \\frac{1}{4} < \\frac{1}{m} < 1 $.  \nTherefore, the eccentricity of the ellipse is $ e = \\frac{c}{a} = \\sqrt{1 - \\frac{b^{2}}{a^{2}}} = \\sqrt{1 - \\frac{1}{m}} \\in \\left(0, \\frac{\\sqrt{3}}{2}\\right) $.  \nThus, the range of the ellipse's eccentricity is $ \\left(0, \\frac{\\sqrt{3}}{2}\\right) $."
  },
  {
    "text": "The coordinates of the two foci of an ellipse are $(-2 , 0)$ and $(2 , 0)$, and it passes through the point $\\left(\\frac{5}{2} , -\\frac{3}{2}\\right)$. Then its standard equation is?",
    "fact_expressions": "G: Ellipse;F1:Point;F2:Point;P:Point;Coordinate(F1)=(-2, 0);Coordinate(F2)=(2,0);Coordinate(P) = (5/2, -3/2);Focus(G)={F1,F2};PointOnCurve(P,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/10+y^2/6=1",
    "fact_spans": "[[[2, 4], [71, 72]], [[14, 24]], [[25, 34]], [[39, 69]], [[14, 24]], [[25, 34]], [[39, 69]], [[2, 34]], [[2, 69]]]",
    "query_spans": "[[[71, 79]]]",
    "process": ""
  },
  {
    "text": "A line passing through the focus of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. From point $B$, a perpendicular is drawn to the directrix $l$ of the parabola, with foot of the perpendicular at $C$. Given that point $A(4,4)$, find the equation of line $AC$.",
    "fact_expressions": "A: Point;C: Point;G: Parabola;H: Line;B: Point;l: Line;Z: Line;Expression(G) = (y^2 = 4*x);Coordinate(A) = (4, 4);PointOnCurve(Focus(G), H);Intersection(H, G) = {A, B};PointOnCurve(B, Z);IsPerpendicular(Z, l);Directrix(G) = l;FootPoint(Z, l) = C",
    "query_expressions": "Expression(LineOf(A, C))",
    "answer_expressions": "y = x",
    "fact_spans": "[[[25, 28], [63, 72]], [[57, 60]], [[1, 15], [21, 24], [41, 44]], [[18, 20]], [[29, 32], [36, 40]], [[47, 50]], [], [[1, 15]], [[63, 72]], [[0, 20]], [[18, 34]], [[35, 53]], [[35, 53]], [[41, 50]], [[35, 60]]]",
    "query_spans": "[[[74, 85]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the foci of the hyperbola $\\frac{y^{2}}{2}-x^{2}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2 + y^2/2 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, pm*sqrt(3))",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 35]]]",
    "process": "Since the hyperbola equation is \\frac{y^{2}}{2}-x^{2}=1, we have a^{2}=2, b^{2}=1, c^{2}=a^{2}+b^{2}=3. Since the foci lie on the y-axis, the coordinates of the foci are (0,-\\sqrt{3}), (0,\\sqrt{3}). [This question mainly examines the equation of a hyperbola and its simple geometric properties, and belongs to a medium-difficulty problem.]"
  },
  {
    "text": "For the equation $\\frac{x^{2}}{|m|-1}+\\frac{y^{2}}{2}=1$ to represent an ellipse with foci on the $y$-axis, what is the range of real values for $m$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(Abs(m) - 1) + y^2/2 = 1);m: Real;PointOnCurve(Focus(G), yAxis) = True",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-3,-1)+(1,3)",
    "fact_spans": "[[[52, 54]], [[0, 54]], [[56, 61]], [[43, 54]]]",
    "query_spans": "[[[56, 68]]]",
    "process": ""
  },
  {
    "text": "Given two fixed points $P_{1}(-1 , 0)$ and $P_{2}(3 , 0)$, what is the trajectory equation of points for which the sum of the squares of the distances to $P_{1}$ and $P_{2}$ equals $16$?",
    "fact_expressions": "P1: Point;P2: Point;Coordinate(P1) = (-1, 0);Coordinate(P2) = (3, 0);F: Point;(Distance(F, P1))^2 + (Distance(F, P2))^2 = 16",
    "query_expressions": "LocusEquation(F)",
    "answer_expressions": "x^2+y^2-2*x-3=0",
    "fact_spans": "[[[5, 20], [38, 46]], [[21, 35], [47, 54]], [[5, 20]], [[21, 35]], [[68, 69]], [[37, 69]]]",
    "query_spans": "[[[68, 76]]]",
    "process": "Let the sum of the squares of the distances from point P(x, y) to points $ P_{1} $ and $ P_{2} $ be equal to 16, then $ PP_{1}^{2} + PP_{2}^{2} = (x+1)^{2} + y^{2} + (x-3)^{2} + y^{2} = 16 $. Rearranging gives: $ x^{2} + y^{2} - 2x - 3 = 0 $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F_{1}$, $F_{2}$ are the left and right foci of $C$, $P$ is a point on the ellipse $C$, and the incenter of $\\Delta P F_{1} F_{2}$ is $I(s, 1)$. If the area of $\\Delta P F_{1} F_{2}$ is $2 b$, then the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;I: Point;P: Point;F1: Point;F2: Point;e: Number;s:Number;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(P, C);Incenter(TriangleOf(P, F1, F2)) = I;Coordinate(I)=(s,1);Area(TriangleOf(P, F1, F2)) = 2*b;Eccentricity(C) = e",
    "query_expressions": "e",
    "answer_expressions": "3/5",
    "fact_spans": "[[[2, 59], [78, 81], [92, 97], [171, 173]], [[8, 59]], [[8, 59]], [[127, 136]], [[88, 91]], [[61, 68]], [[70, 77]], [[177, 180]], [[127, 136]], [[8, 59]], [[8, 59]], [[2, 59]], [[61, 87]], [[61, 87]], [[88, 100]], [[102, 136]], [[127, 136]], [[138, 169]], [[171, 180]]]",
    "query_spans": "[[[177, 182]]]",
    "process": "As shown in the figure, let the extension of PI intersect the x-axis at point M, draw IH⊥x-axis at H. Without loss of generality, assume P is in the first quadrant, S_{\\DeltaPF_{1}F_{2}}=\\frac{1}{2}\\times2c\\timesy_{P}=2b, y_{P}=\\frac{2b}{c}. I is the incenter of \\DeltaPF_{1}F_{2}, then \\frac{|PI|}{|IM|}=\\frac{|PF_{1}|}{|F_{1}M|}=\\frac{|PF_{2}|}{|F_{2}M|}=\\frac{|PF_{1}|+|PF_{2}|}{|F_{1}M|+|F_{2}M|}=\\frac{2a}{2c}=\\frac{a}{c}. Therefore, \\frac{a+c}{c}=\\frac{|PI|+|IM|}{|IM|}=\\frac{y_{P}}{y_{I}}=\\frac{2b}{1}, a+c=2b=2\\sqrt{a^{2}-c^{2}}, a+c=4(a-c), e=\\frac{c}{a}=\\frac{3}{5}"
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2 p x  (p>0)$ be $F$, point $A(0 , 2)$, the segment $FA$ intersects the parabola at point $B$, and $\\overrightarrow{F B}=2 \\overrightarrow{B A}$, then $ |B F |$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(G) = F;A: Point;Coordinate(A) = (0, 2);Intersection(LineSegmentOf(F, A), G) = B;B: Point;VectorOf(F, B) = 2*VectorOf(B, A)",
    "query_expressions": "Abs(LineSegmentOf(B, F))",
    "answer_expressions": "8*sqrt(3)/9",
    "fact_spans": "[[[1, 24], [52, 55]], [[1, 24]], [[4, 24]], [[4, 24]], [[28, 31]], [[1, 31]], [[32, 43]], [[32, 43]], [[44, 61]], [[57, 61]], [[63, 108]]]",
    "query_spans": "[[[110, 121]]]",
    "process": "Let $ B(x, y) $. According to $ \\overrightarrow{FB} = 2\\overrightarrow{BA} $, the coordinates of point $ B $ expressed in terms of $ p $ can be obtained. Substituting into the parabola equation gives the value of $ p $, then the coordinates of points $ B $ and $ F $ can be found. Using the distance formula between two points yields the answer. From the problem, $ F\\left(\\frac{p}{2}, 0\\right) $ ($ p > 0 $). Let $ B(x, y) $, then $ \\overrightarrow{FB} = \\left(x - \\frac{p}{2}, y\\right) $, $ 2\\overrightarrow{BA} = 2(-x, 2 - y) = (-2x, 4 - 2y) $. From $ \\overrightarrow{FB} = 2\\overrightarrow{BA} $, we get\n$$\n\\begin{cases}\nx - \\frac{p}{2} = -2x \\\\\ny = 4 - 2y\n\\end{cases}\n$$\nSolving gives\n$$\n\\begin{cases}\nx = \\frac{p}{6} \\\\\ny = \\frac{4}{3}\n\\end{cases}\n$$\nSubstitute into the parabola equation: $ \\left(\\frac{4}{3}\\right)^2 = 2p \\times \\frac{p}{6} $, solving gives $ p = \\frac{4\\sqrt{3}}{3} $. Therefore, $ B\\left(\\frac{2\\sqrt{3}}{9}, \\frac{4}{3}\\right) $, $ F\\left(\\frac{2\\sqrt{3}}{3}, 0\\right) $."
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{8}=1$ is $\\sqrt{3}$, then the value of the real number $m$ is?",
    "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (-y^2/8 + x^2/m = 1);Eccentricity(G) = sqrt(3)",
    "query_expressions": "m",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 40]], [[57, 62]], [[2, 40]], [[2, 55]]]",
    "query_spans": "[[[57, 66]]]",
    "process": "From the given condition: $\\frac{\\sqrt{m+8}}{\\sqrt{m}}=\\sqrt{3}$, solving gives $m=4$. To solve such problems, clarify the correspondence: one is $a^{2}=m$, $b^{2}=8$, the other is that in hyperbola $c^{2}=a^{2}+b^{2}$."
  },
  {
    "text": "The coordinates of the foci of the hyperbola $x^{2}-2 y^{2}+8=0$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - 2*y^2 + 8 = 0)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,pm*2*sqrt(3))",
    "fact_spans": "[[[0, 22]], [[0, 22]]]",
    "query_spans": "[[[0, 29]]]",
    "process": ""
  },
  {
    "text": "Given that the line $y=x$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$, what are the coordinates of the midpoint of segment $AB$?",
    "fact_expressions": "G: Parabola;H: Line;B: Point;A: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (y = x);Intersection(H, G) = {A, B}",
    "query_expressions": "Coordinate(MidPoint(LineSegmentOf(A,B)))",
    "answer_expressions": "(2,2)",
    "fact_spans": "[[[10, 24]], [[2, 9]], [[31, 34]], [[27, 30]], [[10, 24]], [[2, 9]], [[2, 36]]]",
    "query_spans": "[[[41, 55]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-y^{2}=\\lambda$ ($\\lambda>0$) with a focus at $F$, and $O$ as the origin. Take a point $P$ on an asymptote of the hyperbola $C$ such that $|P F|=|P O|$, and the area of $\\Delta P O F$ is $1$. Then $\\lambda=$?",
    "fact_expressions": "C: Hyperbola;lambda: Number;P: Point;O: Origin;F: Point;lambda>0;Expression(C) = (x^2 - y^2 = lambda);OneOf(Focus(C)) = F;PointOnCurve(P,Asymptote(C));Abs(LineSegmentOf(P, F)) = Abs(LineSegmentOf(P, O));Area(TriangleOf(P, O, F)) = 1",
    "query_expressions": "lambda",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 41], [61, 67]], [[119, 128]], [[75, 78]], [[51, 54]], [[47, 50]], [[9, 41]], [[2, 41]], [[2, 50]], [[60, 78]], [[81, 94]], [[96, 117]]]",
    "query_spans": "[[[119, 130]]]",
    "process": "Let F be the right focus of the hyperbola and c be the semi-focal length of the hyperbola. According to the problem, the x-coordinate of point P is \\frac{c}{2}. The asymptotes of the hyperbola are given by y=\\pm x. Assume point P lies on the asymptote y=x; then the y-coordinate of P is \\frac{c}{2}. Thus, the area of \\triangle POF is \\frac{c^{2}}{4}=1, so c^{2}=4. From the problem, c^{2}=2\\lambda, hence 2\\lambda=4, solving gives \\lambda=2."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{2}-y^{2}=1$, the right focus is point $F$, point $B$ is an endpoint of the imaginary axis, and point $P$ is a moving point on the left branch of hyperbola $C$. Then the minimum perimeter of $\\Delta B P F$ is equal to?",
    "fact_expressions": "C: Hyperbola;B: Point;P: Point;F: Point;Expression(C) = (x^2/2 - y^2 = 1);RightFocus(C) = F;OneOf(Endpoint(ImageinaryAxis(C))) = B;PointOnCurve(P, LeftPart(C))",
    "query_expressions": "Min(Perimeter(TriangleOf(B, P, F)))",
    "answer_expressions": "4 + 2*sqrt(2)",
    "fact_spans": "[[[2, 35], [63, 69]], [[45, 49]], [[58, 62]], [[40, 44]], [[2, 35]], [[2, 44]], [[2, 57]], [[58, 77]]]",
    "query_spans": "[[[79, 102]]]",
    "process": "First, write the coordinates of B and F using the geometric properties of the hyperbola, and find the length of |BF|. Then let E be the left focus of the hyperbola. By the definition of the hyperbola, |PF| - |PE| = 2a. The perimeter of triangle BPF is |BF| + |PF| + |PB| = |BF| + 2a + (|PE| + |PB|). Thus, finding the minimum value of |PE| + |PB| gives the minimum perimeter of triangle BPF, which occurs if and only if points B, P, and E are collinear.  \n∵ The hyperbola C: \\frac{x^{2}}{2} - y^{2} = 1, ∴ F(\\sqrt{3},0). As shown in the figure, assume B is the endpoint of the imaginary axis above the x-axis, then B(0,1), |BF| = 2. Let E be the left focus of the hyperbola. By the definition of the hyperbola, |PF| - |PE| = 2a = 2\\sqrt{2}, that is, |PF| = |PE| + 2\\sqrt{2}.  \n∴ The perimeter of triangle BPF is |BF| + |PF| + |PB| = |BF| + (|PE| + 2\\sqrt{2}) + |PB| = 2 + 2\\sqrt{2} + |PE| + |PB| \\geqslant 2 + 2\\sqrt{2} + |BE| = 4 + 2\\sqrt{2}.  \nThe equality holds if and only if B, P, and E are collinear. Therefore, the minimum perimeter of triangle BPF equals 4 + 2\\sqrt{2}."
  },
  {
    "text": "Given that the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) coincides with the right focus of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, and if $A$ is a point on the parabola in the first quadrant such that $|A F|=3$, then the slope of the line $A F$ is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;A: Point;F: Point;Expression(G) = (x^2/3 - y^2 = 1);p>0;Expression(H) = (y^2 = 2*(p*x));Focus(H) = F ;RightFocus(G) = F;Quadrant(A) = 1;PointOnCurve(A, H);Abs(LineSegmentOf(A, F)) = 3",
    "query_expressions": "Slope(LineOf(A,F))",
    "answer_expressions": "-2*sqrt(2)",
    "fact_spans": "[[[30, 58]], [[2, 23], [70, 73]], [[5, 23]], [[66, 69]], [[26, 29]], [[30, 58]], [[5, 23]], [[2, 23]], [[2, 29]], [[26, 64]], [[66, 82]], [[66, 82]], [[84, 93]]]",
    "query_spans": "[[[95, 107]]]",
    "process": "Find the right focus of the hyperbola, determine the equation of the parabola, and use the definition of the parabola to find the coordinates of point A, then obtain the slope of line AF. Since the right focus of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ is $(2,0)$, the equation of the parabola is $y^{2}=8x$. Given $|AF|=3$, we have $x_{A}+2=3$, so $x_{A}=1$. Substituting into the parabola equation gives $y_{A}=\\pm2\\sqrt{2}$. Since point A lies in the first quadrant, $A(1,2\\sqrt{2})$. Therefore, the slope of line AF is $\\frac{2\\sqrt{2}}{1-2}=-2\\sqrt{2}$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus is $F$, the right vertex is $A$, and point $B$ lies on the ellipse such that $B F \\perp x$-axis. The line $AB$ intersects the $y$-axis at point $P$. If $|A P|= 2|P B|$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;LeftFocus(G) = F;A: Point;RightVertex(G) = A;B: Point;PointOnCurve(B, G);IsPerpendicular(LineSegmentOf(B, F), xAxis);P: Point;Intersection(LineOf(A, B), yAxis) = P;Abs(LineSegmentOf(A, P)) = 2*Abs(LineSegmentOf(P, B))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 54], [76, 78], [131, 133]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[59, 62]], [[2, 62]], [[67, 70]], [[2, 70]], [[71, 75]], [[71, 79]], [[81, 95]], [[108, 112]], [[96, 112]], [[114, 129]]]",
    "query_spans": "[[[131, 139]]]",
    "process": ""
  },
  {
    "text": "If the equation $\\frac{x^{2}}{m-2}+\\frac{y^{2}}{6-m}=1$ represents an ellipse, then the range of real values for $m$ is?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G)=(x^2/(m - 2) + y^2/(6 - m) = 1)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(2,4)+(4,6)",
    "fact_spans": "[[[46, 48]], [[50, 55]], [[1, 48]]]",
    "query_spans": "[[[50, 62]]]",
    "process": "From the ellipse equation, it is required that \\begin{cases}m-2>0\\\\6-m>0\\\\m-2+6-m\\end{cases}; solving the inequalities gives the range of real values of $m$ as $(2,4)\\cup(4,6)$."
  },
  {
    "text": "Let $P$ be a point on the hyperbola $x^{2}-\\frac{y^{2}}{12}=1$, and let $F_{1}$, $F_{2}$ be the left and right foci of this hyperbola. If the area of $\\Delta P F_{1} F_{2}$ is $12$, then $\\angle F_{1} PF_{2}$ equals?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/12 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;Area(TriangleOf(P, F1, F2)) = 12",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "pi/2",
    "fact_spans": "[[[5, 34], [56, 59]], [[5, 34]], [[1, 4]], [[1, 38]], [[39, 46]], [[47, 54]], [[39, 65]], [[39, 65]], [[67, 97]]]",
    "query_spans": "[[[99, 123]]]",
    "process": ""
  },
  {
    "text": "Point $M(20,40)$, the focus of the parabola $y^{2}=2 p x(p>0)$ is $F$. If for any point $P$ on the parabola, the minimum value of $|P M|+|P F|$ is $41$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;M: Point;P: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(M) = (20, 40);Focus(G) = F;PointOnCurve(P, G);Min(Abs(LineSegmentOf(P, F)) + Abs(LineSegmentOf(P, M))) = 41",
    "query_expressions": "p",
    "answer_expressions": "{42, 22}",
    "fact_spans": "[[[12, 33], [44, 47]], [[81, 84]], [[0, 11]], [[52, 55]], [[37, 40]], [[15, 33]], [[12, 33]], [[0, 11]], [[12, 40]], [[44, 56]], [[57, 79]]]",
    "query_spans": "[[[81, 89]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, respectively, a line $l$ passing through $F_{1}$ is tangent to the circle $x^{2}+y^{2}=a^{2}$ and intersects the two asymptotes of the hyperbola at points $A$ and $B$, respectively. If $(\\overrightarrow{F_{2} A}+\\overrightarrow{F_{2} B}) \\cdot \\overrightarrow{A B}=0$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;G: Circle;F2: Point;A: Point;B: Point;F1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = a^2);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1,l);IsTangent(l,G);L1:Line;L2:Line;Asymptote(C)={L1,L2};Intersection(L1,l)=A;Intersection(L2,l)=B;DotProduct((VectorOf(F2, A) + VectorOf(F2, B)),VectorOf(A, B)) = 0",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[98, 103]], [[20, 82], [129, 132], [236, 242]], [[28, 82]], [[28, 82]], [[104, 124]], [[10, 17]], [[141, 145]], [[146, 149]], [[2, 9], [90, 97]], [[28, 82]], [[28, 82]], [[20, 82]], [[104, 124]], [[2, 88]], [[2, 88]], [[89, 103]], [[98, 126]], [], [], [[129, 137]], [[98, 149]], [[98, 149]], [[151, 233]]]",
    "query_spans": "[[[236, 248]]]",
    "process": "Find the slope of the tangent line $ l $. There are two tangents; without loss of generality, consider one of them, as shown in the figure. Find the coordinates of point $ A $, confirming that $ A $ is the point of tangency. Let $ M $ be the midpoint of $ AB $. Using $ (\\overrightarrow{F_{2}A} + \\overrightarrow{F_{2}B}) \\cdot \\overrightarrow{AB} = 0 $, we obtain $ F_{2}M \\perp AB $, from which it follows that $ A $ and $ M $ trisect $ F_{1}B $, thus $ \\overrightarrow{F_{1}B} = 3\\overrightarrow{F_{1}A} $. From this, find the coordinates of point $ B $, substitute into the asymptote equation to obtain an equation in $ a, b, c $, and after manipulation, derive the eccentricity. [Detailed Solution] Since $ l $ is tangent to $ \\odot O: x^{2} + y^{2} = a^{2} $, $ \\sin\\angle BF_{1}O = \\frac{a}{c} $, $ \\angle BF_{1}O $ is acute, so $ \\cos\\angle BF_{1}O = \\sqrt{1 - \\frac{a^{2}}{c^{2}}} = \\frac{b}{c} $, $ \\tan\\angle GF_{2}O = \\frac{a}{b} $, hence the tangent slope $ k = \\pm\\frac{a}{b} $. By symmetry, consider without loss of generality the case $ k = \\frac{a}{b} $. The asymptotes of hyperbola $ C $ are $ y = \\pm\\frac{b}{a}x $, so $ l $ is perpendicular to one asymptote $ y = -\\frac{b}{a}x $. Therefore, the intersection point $ A $ of $ l $ with this asymptote is the intersection point of the asymptote and $ \\odot O $ in the second quadrant, and also the point of tangency. Solving\n$$\n\\begin{cases}\ny = -\\frac{b}{a}x \\\\\ny = \\frac{a}{b}(x + c)\n\\end{cases}\n$$\ngives $ A\\left(-\\frac{a^{2}}{c}, \\frac{ab}{c}\\right) $. Let $ M $ be the midpoint of $ AB $. From $ (\\overrightarrow{F_{2}A} + \\overrightarrow{F_{2}B}) \\cdot \\overrightarrow{AB} = 0 $, i.e., $ 2\\overrightarrow{F_{2}M} \\cdot \\overrightarrow{AB} = 0 $, we have $ F_{2}M \\perp l $. Also, $ OA \\perp l $, so $ OA \\parallel F_{2}M $, and since $ O $ is the midpoint of $ F_{1}F_{2} $, it follows that $ A $ is the midpoint of $ F_{1}M $, hence $ A $ and $ M $ trisect $ F_{1}B $. Thus $ \\overrightarrow{F_{1}B} = 3\\overrightarrow{F_{1}A} $,\n$$\n\\begin{cases}\nx_{B} + c = 3\\left(-\\frac{a^{2}}{c} + c\\right) \\\\\ny_{B} = 3 \\times \\frac{ab}{c}\n\\end{cases}\n$$\nsolving yields\n$$\n\\begin{cases}\nx_{B} = \\frac{3b^{2}}{c} - c \\\\\ny_{B} = \\frac{3ab}{c}\n\\end{cases}\n$$\ni.e., $ B\\left(\\frac{3b^{2}}{c} - c, \\frac{3ab}{c}\\right) B\\left(\\frac{3b}{c} - c, \\frac{3ab}{c}\\right)' $, therefore the eccentricity $ e = \\sqrt{3} $ ultimately holds. ,,,2"
  },
  {
    "text": "If a focus of the hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ is $F(2,0)$, then the real number $m$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2 + x^2/m = 1);m: Real;F: Point;Coordinate(F) = (2, 0);OneOf(Focus(G)) = F",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 29]], [[1, 29]], [[45, 50]], [[35, 43]], [[35, 43]], [[1, 43]]]",
    "query_spans": "[[[45, 52]]]",
    "process": "The hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ has a focus at $F(2,0)$. Therefore, $m>0$ and $m+1=4$, so $m=3$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. There exists a point $P$ on the right branch of $C$ such that $\\cos \\angle F_{1} P F_{2}=\\frac{3}{4}$, and $|P F_{2}|$ equals the length of the imaginary axis of the hyperbola $C$. Then the equation of the asymptotes of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, RightPart(C)) = True;Cos(AngleOf(F1, P, F2)) = 3/4;Abs(LineSegmentOf(P, F2)) = Length(ImageinaryAxis(C))",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*x",
    "fact_spans": "[[[2, 64], [89, 92], [160, 166], [172, 178]], [[2, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[73, 80]], [[81, 88]], [[2, 88]], [[2, 88]], [[100, 103]], [[89, 103]], [[106, 145]], [[147, 170]]]",
    "query_spans": "[[[172, 186]]]",
    "process": "Using the definition of a hyperbola and the cosine law in a triangle, simplifying and rearranging yields $a = b$, thus obtaining the asymptote equations of the required hyperbola. [Detailed solution] From the given conditions, $|PF_{2}| = 2b$. By the definition of a hyperbola, $|PF_{1}| = |PF_{2}| + 2a = 2b + 2a$, $|F_{1}F_{2}| = 2c$. In $\\triangle PF_{1}F_{2}$, $\\cos\\angle F_{1}PF_{2} = \\frac{(2b+2a)^{2}+(2b)^{2}-4c^{2}}{2\\cdot2b\\cdot(2b+2a)} = \\frac{3}{4}$. Using $c^{2} = a^{2} + b^{2}$, simplifying leads to $a = b$. Thus, the asymptote equations of the hyperbola are $y = \\pm\\frac{b}{a}x$, that is, $y = \\pm x$."
  },
  {
    "text": "The vertex of parabola $C$ is at the origin, and the focus is $F(1,0)$. A line $l$ passing through point $F$ intersects parabola $C$ at points $A$ and $B$. If the midpoint of $AB$ is $(2, m)$, then the chord length $|AB|$ is?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;F: Point;O: Origin;MidPoint(LineSegmentOf(A, B)) = Z;Z: Point;Coordinate(Z) = (2, m);Coordinate(F) = (1, 0);Vertex(C) = O;Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};IsChordOf(LineSegmentOf(A, B),C)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "6",
    "fact_spans": "[[[33, 38]], [[0, 6], [39, 45]], [[48, 51]], [[52, 55]], [[18, 26], [28, 32]], [[10, 14]], [[59, 76]], [[68, 76]], [[68, 76]], [[18, 26]], [[0, 14]], [[0, 26]], [[27, 38]], [[33, 57]], [[39, 87]]]",
    "query_spans": "[[[80, 89]]]",
    "process": "Since the line $ l $ passing through point $ F $ intersects the parabola $ C $ at points $ A $ and $ B $, and the midpoint of $ AB $ is $ (2, m) $, we have $ x_{A} + x_{B} = 4 $. Since the vertex of the parabola $ C $ is at the origin and the focus is $ F(1,0) $, the directrix of the parabola $ C $ is $ x = -1 $. By the definition of a parabola, we obtain $ |AB| = |AF| + |BF| = x_{A} + 1 + x_{B} + 1 = 4 + 2 = 6 $. The answer is $ 6 $."
  },
  {
    "text": "The point on the parabola $y^{2}=a x(a>0)$ with abscissa $6$ is at a distance of $10$ from the focus. Then $a=?$",
    "fact_expressions": "G: Parabola;a: Number;a>0;P:Point;Expression(G) = (y^2 = a*x);PointOnCurve(P,G);XCoordinate(P)=6;Distance(P, Focus(G)) = 10",
    "query_expressions": "a",
    "answer_expressions": "16",
    "fact_spans": "[[[0, 19]], [[42, 45]], [[3, 19]], [[28, 29]], [[0, 19]], [[0, 29]], [[20, 29]], [[0, 40]]]",
    "query_spans": "[[[42, 47]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=4x$ with focus $F$, a line passing through point $P(2, 0)$ intersects the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. Then: ($I$) $y_{1} y_{2}$=? ($II$) The minimum area of triangle $ABF$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;P: Point;Coordinate(P) = (2, 0);H: Line;PointOnCurve(P,H);A: Point;B: Point;Intersection(H, G) = {A, B};x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2)",
    "query_expressions": "y1*y2;Min(Area(TriangleOf(A,B,F)))",
    "answer_expressions": "-8; 2*sqrt(2)",
    "fact_spans": "[[[2, 16], [40, 43]], [[2, 16]], [[20, 23]], [[2, 23]], [[25, 36]], [[25, 36]], [[37, 39]], [[24, 39]], [[44, 62]], [[63, 81]], [[37, 83]], [[44, 62]], [[63, 81]], [[44, 62]], [[63, 81]], [[44, 62]], [[63, 81]]]",
    "query_spans": "[[[92, 107]], [[114, 130]]]",
    "process": ""
  },
  {
    "text": "If the line $x+2 y+m=0$ passes through the focus of the parabola $y=2 x^{2}$, then $m=$?",
    "fact_expressions": "G: Parabola;H:Line;m: Number;Expression(G) = (y = 2*x^2);Expression(H) = (m + x + 2*y = 0);PointOnCurve(Focus(G),H)",
    "query_expressions": "m",
    "answer_expressions": "-1/4",
    "fact_spans": "[[[16, 30]], [[1, 14]], [[35, 38]], [[16, 30]], [[1, 14]], [[1, 33]]]",
    "query_spans": "[[[35, 40]]]",
    "process": "y=2x^{2} can be rewritten as x^{2}=\\frac{1}{2}y, with the focus at (0,\\frac{1}{8}). Substituting into the line equation x+2y+m=0 gives 0+2\\times\\frac{1}{8}+m=0, solving for m yields m=-\\frac{1}{4}."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, the right focus is $F(3 \\sqrt{5} , 0)$, point $N$ has coordinates $(0 , 2)$, and point $M$ is a moving point on the left branch of hyperbola $C$. If the perimeter of $\\triangle M N F$ is not less than $20$, then the range of the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;Coordinate(F) = (3*sqrt(5), 0);RightFocus(C) = F;N: Point;Coordinate(N) = (0, 2);M: Point;PointOnCurve(M, LeftPart(C));Negation(Perimeter(TriangleOf(M, N, F)) < 20)",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(1, \\sqrt{5}]",
    "fact_spans": "[[[2, 64], [112, 118], [155, 161]], [[2, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[69, 88]], [[69, 88]], [[2, 88]], [[89, 93]], [[89, 106]], [[107, 111]], [[107, 124]], [[126, 153]]]",
    "query_spans": "[[[155, 172]]]",
    "process": "Let the left focus of the hyperbola be $ F $, then $ |MF|\\cdot|MF|=2a $, i.e., $ |MF|=|MF|+2a $, hence $ |MF|+|MN|=|MF|+|MN|+2a\\geqslant|FN|+2a $, with equality if and only if points $ N, M, F $ are collinear. From $ F(3\\sqrt{5},0), F(-3\\sqrt{5},0), N(0,2) $, we get $ |FN|=|FN|=\\sqrt{45+4}=7 $. Thus, the perimeter of $ \\triangle MNF $, $ |MN|+|MF|+|NF|\\geqslant|F|N|+2a+|NF|=14+2a $. According to the problem, the minimum perimeter of $ \\triangle MNF $ satisfies $ 14+2a\\geqslant20 $, solving gives $ a\\geqslant3 $. Therefore, the eccentricity of the hyperbola $ e=\\frac{c}{a}\\leqslant\\frac{3\\sqrt{5}}{3}=\\sqrt{5} $, and since $ e>1 $, we obtain $ 1<e\\leqslant\\sqrt{5} $. Hence, the range of the eccentricity of the hyperbola is $ (1,\\sqrt{5}] $."
  },
  {
    "text": "Given that $F$ is a focus of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$, point $P$ lies on $C$, and $O$ is the origin, if $|O P|=|O F|$, then the area of $\\triangle O P F$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2/3 = 1);F: Point;OneOf(Focus(C))=F;P: Point;PointOnCurve(P, C);O: Origin;Abs(LineSegmentOf(O, P)) = Abs(LineSegmentOf(O, F))",
    "query_expressions": "Area(TriangleOf(O, P, F))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[6, 39], [50, 53]], [[6, 39]], [[2, 5]], [[2, 44]], [[45, 49]], [[45, 54]], [[55, 58]], [[65, 78]]]",
    "query_spans": "[[[80, 102]]]",
    "process": "According to the problem, draw the figure. Without loss of generality, let F be the right focus of the hyperbola $ C: x^{2} - \\frac{y^{2}}{3} = 1 $, and P be a point in the first quadrant. Find the coordinates of point P, then use the triangle area formula to solve. As shown in the figure, without loss of generality, let F be the right focus of the hyperbola $ C: x^{2} - \\frac{y^{2}}{3} = 1 $, and P be a point in the first quadrant. From the hyperbola equation, we have $ a^{2} = 1 $, $ b^{2} = 3 $, then $ c = 2 $. Thus, the equation of the circle with center O and radius 2 is $ x^{2} + y^{2} = 4 $. Solving the system of equations  \n\\[\n\\begin{cases}\nx^{2} + y^{2} = 4 \\\\\nx^{2} - \\frac{y^{2}}{3} = 1\n\\end{cases}\n\\]\nyields $ P\\left( \\frac{\\sqrt{7}}{2}, \\frac{3}{2} \\right) $. Therefore, $ S_{\\triangle OPF} = \\frac{1}{2} \\times 2 \\times \\frac{3}{2} = \\frac{3}{2} $."
  },
  {
    "text": "Given that the moving point $P$ lies on the circle $(x+2)^{2}+(y-4)^{2}=4$, the right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $F(2,0)$. If there exists a point $Q$ on the asymptote of $C$ such that $\\overrightarrow{O P}+\\overrightarrow{O F}=2 \\overrightarrow{O Q}$, then the range of the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;F: Point;O: Origin;P: Point;Q: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = ((x + 2)^2 + (y - 4)^2 = 4);Coordinate(F) = (2, 0);PointOnCurve(P, G);RightFocus(C) = F;PointOnCurve(Q,Asymptote(C));VectorOf(O,P)+VectorOf(O,F)=2*VectorOf(O,Q)",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "[2,+oo)",
    "fact_spans": "[[[34, 95], [110, 113], [194, 197]], [[42, 95]], [[42, 95]], [[8, 32]], [[100, 108]], [[126, 192]], [[4, 7]], [[120, 124]], [[42, 95]], [[42, 95]], [[34, 95]], [[8, 32]], [[100, 108]], [[4, 33]], [[34, 108]], [[110, 124]], [[126, 192]]]",
    "query_spans": "[[[194, 208]]]",
    "process": "Let $ Q'(x,y) $, $ P(x_{0},y_{0}) $, satisfy $ \\overrightarrow{OP} + \\overrightarrow{OF} = 2\\overrightarrow{OQ} $. So $ (x_{0},y_{0}) + (2,0) = (2x,2y) $, thus $ x_{0} = 2x - 2 $, $ y_{0} = 2y $. Since $ P(x_{0},y_{0}) $ lies on the circle satisfying $ (x_{0}+2)^{2} + (y_{0}-4)^{2} = 4 $, we have $ (2x - 2 + 2)^{2} + (2y - 4)^{2} = 4 $, simplifying to $ x^{2} + (y - 2)^{2} = 1 $. Therefore, the trajectory of point $ Q' $ is a circle with center $ (0,2) $ and radius $ 1 $, as shown in the figure: When the asymptote intersects the circle, it indicates that there exists a point $ Q $ on the asymptote such that $ \\overrightarrow{OP} + \\overrightarrow{OF} = 2\\overrightarrow{OO} $. The critical case occurs when both asymptotes are tangent to the circle, then: the distance from the center $ (0,2) $ to the asymptote $ bx - ay = 0 $ is $ d = \\frac{|-2a|}{\\sqrt{b^{2} + a^{2}}} = 1 $. Since $ c = 2 $, i.e., $ a^{2} + b^{2} = 4 $, we get $ a = 1 $, at this time $ b = \\sqrt{3} $, $ \\frac{b}{a} = \\sqrt{3} $. When $ \\frac{b}{a} \\geqslant \\sqrt{3} $, the asymptote intersects the circle, then $ \\frac{c}{a} = \\sqrt{\\frac{a^{2} + b^{2}}{a^{2}}} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} \\geqslant \\sqrt{1 + (\\sqrt{3})^{2}} = 2 $."
  },
  {
    "text": "Given the parabola equation $y^{2}=x$, point $M$ moves along this parabola. Then the minimum value of the sum of the distances from point $M$ to point $A(4 , 1)$ and the focus $F$ is?",
    "fact_expressions": "G: Parabola;A: Point;M: Point;F:Point;Coordinate(A) = (4, 1);Expression(G) = (y^2 = x);PointOnCurve(M, G);Focus(G)=F",
    "query_expressions": "Min(Distance(M,A)+Distance(M,F))",
    "answer_expressions": "17/4",
    "fact_spans": "[[[2, 17], [24, 27]], [[37, 48]], [[18, 22], [32, 36]], [[51, 54]], [[37, 48]], [[2, 17]], [[18, 30]], [[24, 54]]]",
    "query_spans": "[[[32, 67]]]",
    "process": "As shown in the figure, the parabola $ y^{2} = x $ has focus $ F\\left(\\frac{1}{4}, 0\\right) $ and directrix $ x = -\\frac{1}{4} $. Draw a perpendicular line $ MM' $ from point $ M $ to the directrix. According to the definition of the parabola, $ |MF| + |MA| = |MM'| + |MA| $. From the figure, the minimum value of the sum of distances from point $ A(4,1) $ to the focus $ F $ is the distance from point $ A $ to the directrix: thus, the minimum sum of distances is $ 4 - \\left(-\\frac{1}{4}\\right) = \\frac{17}{4} $. The answer is: $ \\underline{17} $"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{4 x^{2}}{9}-\\frac{y^{2}}{b^{2}}=1(b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. The line $A B$ passes through the right focus $F_{2}$, intersects the right branch of the hyperbola $C$ at points $A$ and $B$, satisfying $|A F_{2}|=2|B F_{2}|$, $\\cos \\angle F_{1} A B=\\frac{3}{5}$. Then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = ((4*x^2)/9 - y^2/b^2 = 1);b: Number;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F2, LineOf(A, B)) = True;Intersection(LineOf(A, B), RightPart(C)) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(A, F2)) = 2*Abs(LineSegmentOf(B, F2));Cos(AngleOf(F1, A, B)) = 3/5",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(17)/3",
    "fact_spans": "[[[2, 55], [101, 107], [187, 193]], [[2, 55]], [[9, 55]], [[9, 55]], [[65, 72]], [[73, 80], [92, 99]], [[2, 80]], [[2, 80]], [[81, 99]], [[81, 121]], [[112, 115]], [[116, 119]], [[125, 147]], [[149, 185]]]",
    "query_spans": "[[[187, 199]]]",
    "process": "Let |BF₂| = m. In △ABF₁, by the cosine law, m = 1. In △AF₁F₂, by the cosine law, c = \\frac{\\sqrt{17}}{2}, and thus the solution is obtained. [Detailed solution] From the problem, we have \\frac{x^{2}}{4} - \\frac{y^{2}}{b^{2}} = 1, so a = \\frac{3}{2}, 2a = 3. Let |BF₂| = m, so |AF₂| = 2m. From the problem, |AF| = 2m + 3, |BF₁| = m + 3, |AB| = 3m. In △ABF₁, by the cosine law, \\cos\\angle F_{1}AB = \\frac{3}{5} = \\frac{(2m+3)^{2} + 9m^{2} - (m+3)^{2}}{6m(2m+3)}. Solving this gives m = 1. In △AF₁F₂, by the cosine law, \\cos\\angle F_{1}AF_{2} = \\frac{3}{5} = \\frac{5^{2} + 2^{2} - (2c)^{2}}{20}, so c = \\frac{\\sqrt{17}}{2}. Therefore, the eccentricity is e = \\frac{c}{a} = \\frac{\\sqrt{17}}{3}."
  },
  {
    "text": "Let the foci of the hyperbola $C$: $\\frac{y^{2}}{16}-\\frac{x^{2}}{64}=1$ be $F_{1}$ and $F_{2}$, and let point $P$ be a point on $C$ such that $|P F_{1}|=6$. Then what is $|P F_{2}|$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-x^2/64 + y^2/16 = 1);F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 6",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "14",
    "fact_spans": "[[[1, 46], [71, 74]], [[1, 46]], [[50, 57]], [[58, 65]], [[1, 65]], [[66, 70]], [[66, 77]], [[78, 91]]]",
    "query_spans": "[[[93, 106]]]",
    "process": "From \\frac{y^{2}}{16}-\\frac{x^{2}}{64}=1, we get a^{2}=16, then a=4. Since point P is a point on C, ||PF_{1}|-|PF_{2}||=2a=8. Given |PF_{1}|=6, we have |6-|PF_{2}||=2a=8, solving gives |PF_{2}|=14 or |PF_{2}|=-2 (discarded)."
  },
  {
    "text": "Given that the eccentricity of hyperbola $C$ is $\\sqrt{3}$, with foci $F_{1}$ and $F_{2}$, and point $A$ lies on curve $C$. If $|F_{1} A| = 3|F_{2} A|$, then $\\cos \\angle A F_{2} F_{1}$ = ?",
    "fact_expressions": "C: Hyperbola;F1: Point;A: Point;F2: Point;Eccentricity(C) = sqrt(3);Focus(C)={F1,F2};PointOnCurve(A, C);Abs(LineSegmentOf(F1, A)) = 3*Abs(LineSegmentOf(F2, A))",
    "query_expressions": "Cos(AngleOf(A, F2, F1))",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[2, 8], [48, 53]], [[27, 34]], [[43, 47]], [[35, 42]], [[2, 23]], [[2, 42]], [[43, 54]], [[56, 78]]]",
    "query_spans": "[[[80, 109]]]",
    "process": "Let the equation of the hyperbola be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a,b>0)$, take point $A$ on the right branch, and $|F_{2}A|=m$. Since $|F_{1}A|=3|F_{2}A|$, we have $|F_{1}A|=3m$. By the definition of the hyperbola, $|F_{1}A|-|F_{2}A|=2a$, solving gives $m=a$. Also, $e=\\frac{c}{a}=\\sqrt{3}$, we obtain $c=\\sqrt{3}a$. In $\\triangle AF_{1}F_{2}$, $|F_{1}A|=3a$, $|F_{2}A|=a$, $|F_{1}F_{2}|=2\\sqrt{3}a$. By the cosine law, $\\cos\\angle AF_{2}F_{1}=\\frac{a^{2}+12a^{2}-9a^{2}}{2a\\cdot2\\sqrt{3}a}=\\frac{\\sqrt{3}}{3}$."
  },
  {
    "text": "The two foci of an ellipse are $F_{1}$ and $F_{2}$. A line passing through $F_{1}$ intersects the ellipse at points $M$ and $N$, where $|M F_{1}| = \\frac{4}{3}|N F_{1}|$, and $|M F_{2}| = |F_{1} F_{2}|$. Find the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;H: Line;M: Point;F1: Point;N: Point;F2: Point;Focus(G) = {F1,F2};PointOnCurve(F1, H);Intersection(H, G) = {M, N};Abs(LineSegmentOf(M, F1)) = (4/3)*Abs(LineSegmentOf(N, F1));Abs(LineSegmentOf(M, F2)) = Abs(LineSegmentOf(F1, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/7",
    "fact_spans": "[[[0, 2], [36, 38], [110, 112]], [[33, 35]], [[39, 42]], [[8, 15], [25, 32]], [[43, 46]], [[16, 23]], [[0, 23]], [[24, 35]], [[33, 48]], [[49, 81]], [[83, 108]]]",
    "query_spans": "[[[110, 118]]]",
    "process": "Let the ellipse be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F_{1}(-c,0)$, $F_{2}(c,0)$, $|MF_{2}|=|F_{1}F_{2}|=2c$, as shown in the figure: let $|NF|=3t$ $(t>0)$, then $|MF_{1}|=\\frac{4}{3}|NF_{1}|=4t$. By the definition of the ellipse, we obtain $|NF_{2}|=2a-|NF|=2a-3t$, $|MF_{2}|+|MF|=2a$, thus $2c+4t=2a$, i.e., $a-c=2t$, $\\textcircled{1}$. Take point $A$ as the midpoint of $MF_{1}$, connect $AF_{2}$, then $|NA|=5t$. Since $|MF_{2}|=|F_{1}F_{2}|$, we have $AF_{2}\\bot MN$. By the Pythagorean theorem, $|MF_{2}|^{2}-|MA|^{2}=|NF_{2}|^{2}-|NA|^{2}$, that is, $4c^{2}-4t^{2}=(2a-3t)^{2}-25t^{2}$, $\\textcircled{2}$. Solving $\\textcircled{1}$ and $\\textcircled{2}$, we get $a=7t$, $c=5t$, then the eccentricity $e=\\frac{c}{a}=\\frac{5}{7}$."
  },
  {
    "text": "Given a fixed point $B(-1 , 0)$ and two moving points $P$, $Q$ on the parabola $y = x^{2} - 1$. When point $P$ moves along the parabola, $BP \\perp PQ$. Then the range of the horizontal coordinate of point $Q$ is?",
    "fact_expressions": "G: Parabola;B: Point;P: Point;Q: Point;Expression(G) = (y = x^2 - 1);Coordinate(B) = (-1, 0);PointOnCurve(B, G);PointOnCurve(P, G);PointOnCurve(Q, G);IsPerpendicular(LineSegmentOf(B, P), LineSegmentOf(P, Q))",
    "query_expressions": "Range(XCoordinate(Q))",
    "answer_expressions": "(-oo, -3] + [1, +oo)",
    "fact_spans": "[[[2, 16], [49, 52]], [[20, 31]], [[36, 39], [45, 48]], [[40, 43], [74, 78]], [[2, 16]], [[20, 31]], [[2, 31]], [[2, 43]], [[2, 43]], [[57, 72]]]",
    "query_spans": "[[[74, 89]]]",
    "process": ""
  },
  {
    "text": "The left focus of the ellipse $\\frac{x^{2}}{4 a^{2}}+\\frac{y^{2}}{3 b^{2}}=1(a>0)$ is $F$, and the line $x=m$ intersects the ellipse at points $A$ and $B$. When the perimeter of $\\Delta F A B$ is maximized, what is the area of $\\Delta F A B$?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;m: Number;F: Point;A: Point;B: Point;a>0;Expression(G) = (y^2/((3*b^2)) + x^2/((4*a^2)) = 1);Expression(H) = (x = m);LeftFocus(G) = F;Intersection(H, G) = {A, B};WhenMax(Perimeter(TriangleOf(F, A, B)))",
    "query_expressions": "Area(TriangleOf(F, A, B))",
    "answer_expressions": "S=3*(b^2)*c/a",
    "fact_spans": "[[[0, 54], [71, 73]], [[2, 54]], [[2, 54]], [[63, 70]], [[65, 70]], [[59, 62]], [[76, 80]], [[81, 84]], [[2, 54]], [[0, 54]], [[63, 70]], [[0, 62]], [[63, 84]], [[85, 106]]]",
    "query_spans": "[[[107, 126]]]",
    "process": "Let the right focus of the ellipse be F'. Since AF' = BF', it follows that AB ≤ 2AF', with equality if and only if points A, F', and B are collinear. At this point, the perimeter of triangle AFB is L = 2AF + AB ≤ 2AF + 2AF' = 8a, achieving its maximum value. In this case, AB = 2\\sqrt{1 - \\frac{c^{2}}{4a^{2}}} \\cdot \\sqrt{3}b = \\frac{3b^{2}}{a}, and the area of the triangle is S = \\frac{3b^{2}c}{a}."
  },
  {
    "text": "Given that $P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, not at a vertex, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola, respectively, and point $M$ is the incenter of $\\triangle P F_{1} F_{2}$. If $S_{\\triangle M P F_{1}}-S_{\\triangle M P F_{2}}=\\frac{\\sqrt{2}}{2} S_{\\triangle M F_{1} F_{2}}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "P: Point;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;PointOnCurve(P, RightPart(G)) = True;Negation(Vertex(G) = P);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;M: Point;Incenter(TriangleOf(P, F1, F2)) = M;Area(TriangleOf(M, P, F1)) - Area(TriangleOf(M, P, F2)) = sqrt(2)*Area(TriangleOf(M, F1, F2))/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 5]], [[6, 62], [92, 95], [235, 238]], [[6, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[2, 67]], [[2, 73]], [[74, 81]], [[82, 89]], [[74, 100]], [[74, 100]], [[101, 105]], [[101, 134]], [[136, 232]]]",
    "query_spans": "[[[235, 244]]]",
    "process": "Transform the condition into $\\frac{1}{2}|PF_{1}|r-\\frac{1}{2}|PF_{2}|r=\\frac{\\sqrt{2}}{2}\\frac{1}{2}|F_{1}F_{2}|r'$, then simplify using the definition of a hyperbola and compute the eccentricity. From the given, $M$ is the incenter of $\\triangle PF_{1}F_{2}$, let the inradius of $\\triangle PF_{1}F_{2}$ be $r$. Since $S_{\\Delta MPF_{1}}-S_{\\Delta MPF_{2}}=\\frac{\\sqrt{2}}{2}S_{\\Delta MF_{1}F_{2}}$, it follows that $\\frac{1}{2}|PF_{1}|r-\\frac{1}{2}|PF_{2}|r=\\frac{\\sqrt{2}}{2}\\frac{1}{2}|F_{1}F_{2}|r$, thus $|PF_{1}|-|PF_{2}|=2a=\\frac{\\sqrt{2}}{2}|F_{1}F_{2}|=\\sqrt{2}c$, so $e=\\sqrt{2}$."
  },
  {
    "text": "The coordinates of the focus of the parabola $y=-6 x^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = -6*x^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,-1/24)",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 22]]]",
    "process": "From y = -6x^{2}, we get x^{2} = -\\frac{1}{6}y, hence the focus of the parabola has coordinates (0, -\\frac{1}{24})."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with left and right foci $F_{1}$, $F_{2}$, and focal distance $2$. A line $l$ passing through $F_{2}$ intersects the ellipse $C$ at points $A$, $B$. If the perimeter of $\\triangle A F_{1} B$ is $4 \\sqrt{3}$, then the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "l: Line;C: Ellipse;b: Number;a: Number;A: Point;F1: Point;B: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;FocalLength(C) = 2;PointOnCurve(F2, l);Intersection(l, C) = {A, B};Perimeter(TriangleOf(A, F1, B)) = 4*sqrt(3)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[98, 103]], [[2, 59], [104, 109], [160, 165]], [[8, 59]], [[8, 59]], [[110, 113]], [[66, 73]], [[114, 117]], [[74, 81], [90, 97]], [[8, 59]], [[8, 59]], [[2, 59]], [[2, 81]], [[2, 81]], [[2, 88]], [[89, 103]], [[98, 119]], [[121, 158]]]",
    "query_spans": "[[[160, 171]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and let $P$ be a point on the right branch of $C$. If $|P F_{2}|=|F_{1} F_{2}|$ and the distance from point $F_{2}$ to the line $P F_{1}$ is $2 a$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, RightPart(C));Abs(LineSegmentOf(P, F2)) = Abs(LineSegmentOf(F1, F2));Distance(F2,LineOf(P,F1)) = 2*a",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[19, 80], [91, 94], [159, 162]], [[19, 80]], [[27, 80]], [[27, 80]], [[27, 80]], [[27, 80]], [[1, 8]], [[9, 16], [128, 136]], [[1, 86]], [[1, 86]], [[87, 90]], [[87, 99]], [[102, 127]], [[128, 157]]]",
    "query_spans": "[[[159, 168]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, $l_{1}$, $l_{2}$ are two asymptotes of $C$. Draw a perpendicular from the right focus $F$ of $C$ to $l_{1}$, with foot of perpendicular at $A$, and this perpendicular intersects $l_{2}$ at point $B$. If $\\overrightarrow{B A}=3 \\overrightarrow{A F}$, then the eccentricity $e$ of the curve $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;B: Point;A: Point;F: Point;l1: Line;l2: Line;e: Number;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Asymptote(C) = {l1, l2};RightFocus(C) = F;L: Line;PointOnCurve(F, L);IsPerpendicular(L, l1);FootPoint(L, l1) = A;Intersection(L, l2) = B;VectorOf(B, A) = 3*VectorOf(A, F);Eccentricity(C) = e",
    "query_expressions": "e",
    "answer_expressions": "2*sqrt(6)/3",
    "fact_spans": "[[[2, 53], [74, 77], [85, 88], [180, 185]], [[9, 53]], [[9, 53]], [[127, 131]], [[110, 113]], [[92, 95]], [[56, 63], [96, 103]], [[66, 73], [119, 126]], [[189, 192]], [[2, 53]], [[55, 83]], [[85, 95]], [], [[84, 106]], [[84, 106]], [[84, 113]], [[84, 131]], [[133, 178]], [[180, 192]]]",
    "query_spans": "[[[189, 194]]]",
    "process": ""
  },
  {
    "text": "It is known that $F_{1}$ and $F_{2}$ are the left and right foci of an ellipse, respectively. A circle centered at $F_{2}$ passes exactly through the center of the ellipse and intersects the ellipse at points $M$ and $N$. If the line $M F_{1}$ passing through $F_{1}$ is tangent to the circle $F_{2}$, then what is the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;H: Circle;F2: Point;F1: Point;M:Point;N:Point;LeftFocus(G) = F1;RightFocus(G) = F2;Center(H)=F2;PointOnCurve(Center(G),H);Intersection(H,G)={M,N};PointOnCurve(F1,LineOf(M,F1));IsTangent(LineOf(M,F1),H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[20, 22], [20, 22], [20, 22], [20, 22]], [[44, 45], [91, 99]], [[10, 17], [31, 38]], [[2, 9], [70, 77]], [[59, 63]], [[64, 67]], [[2, 28]], [[2, 28]], [[31, 45]], [[44, 53]], [[44, 67]], [[69, 90]], [[78, 102]]]",
    "query_spans": "[[[104, 112]]]",
    "process": "\\because F_{1}F_{2} are the left and right foci of the ellipse, respectively. Now, taking F_{2} as the center, draw a circle passing exactly through the center of the ellipse and intersecting the ellipse at points M, N. The line MF_{1} passing through F_{1} is tangent to the circle F_{2}. \\therefore |MF_{2}|=c, |F_{1}F_{2}|=2c, \\angle F_{1}MF_{2}=90^{\\circ}. \\therefore |MF_{1}|=\\sqrt{4c^{2}-c^{2}}=\\sqrt{3}c, \\therefore 2a=\\sqrt{3}c+c=(\\sqrt{3}+1)c, e=\\sqrt{3}-1. The answer is \\sqrt{3}-1"
  },
  {
    "text": "The number of intersection points between the line $y=2x+5$ and the curve $\\frac{x |x|}{9}+\\frac{y^{2}}{25}=1$ is?",
    "fact_expressions": "G: Line;H: Curve;Expression(G) = (y = 2*x + 5);Expression(H) = (y^2/25 + (x*Abs(x))/9 = 1)",
    "query_expressions": "NumIntersection(G, H)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 11]], [[12, 50]], [[0, 11]], [[12, 50]]]",
    "query_spans": "[[[0, 57]]]",
    "process": ""
  },
  {
    "text": "The line $l$ intersects a hyperbola centered at the origin with foci on the $x$-axis, real axis length $2$, and eccentricity $\\sqrt{3}$ at points $A$ and $B$. If the midpoint of $AB$ is $(2,1)$, then the equation of line $l$ is?",
    "fact_expressions": "l: Line;G: Hyperbola;H: Point;A: Point;B: Point;O:Origin;Coordinate(H) = (2, 1);Center(G)=O;PointOnCurve(Focus(G), xAxis);Length(RealAxis(G)) = 2;Eccentricity(G)=sqrt(3);Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = H",
    "query_expressions": "Expression(l)",
    "answer_expressions": "4*x-y-7=0",
    "fact_spans": "[[[0, 5], [78, 83]], [[44, 47]], [[69, 76]], [[49, 52]], [[53, 56]], [[9, 11]], [[69, 76]], [[6, 47]], [[12, 47]], [[21, 47]], [[29, 47]], [[0, 58]], [[60, 76]]]",
    "query_spans": "[[[78, 88]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^2=4x$, with focus $F$, and the directrix of $C$ intersects the $x$-axis at point $F'$. Points $A$ and $B$ are moving points on the parabola $C$ such that the line $AB$ passes through point $F'$. Through $F$, draw lines $l_1$ and $l_2$ parallel to $OA$ and $OB$ respectively ($O$ is the origin). The lines $l_1$ and $l_2$ intersect at point $P$. Let the trajectory of point $P$ be curve $E$. The line $y=kx$ ($k\\ge 0$) has no intersection with curve $E$. Then the range of values for $k$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;F1:Point;Intersection(Directrix(C),xAxis)=F1;A: Point;B: Point;PointOnCurve(A,C);PointOnCurve(B,C);PointOnCurve(F1,LineOf(A,B));l1:Line;l2:Line;PointOnCurve(F,l1);PointOnCurve(F,l2);IsParallel(l1,LineOf(O,A));IsParallel(l2,LineOf(O,B));O: Origin;P:Point;Intersection(l1,l2)=P;E: Curve;Locus(P)=E;H:Line;k:Number;k>=0;Expression(H)=(y=k*x);NumIntersection(H,E)=0",
    "query_expressions": "Range(k)",
    "answer_expressions": "[2,+oo)+{0}",
    "fact_spans": "[[[2, 17], [18, 19], [26, 29], [52, 58]], [[2, 17]], [[22, 25], [77, 80]], [[18, 25]], [[38, 43], [70, 75]], [[26, 43]], [[44, 47]], [[48, 51]], [[44, 61]], [[48, 61]], [[63, 75]], [[96, 101], [118, 125]], [[102, 107], [126, 131]], [[76, 107]], [[76, 107]], [[83, 107]], [[83, 107]], [[108, 111]], [[134, 138], [140, 144]], [[118, 138]], [[150, 155], [173, 178]], [[140, 155]], [[156, 172]], [[183, 186]], [[158, 172]], [[156, 172]], [[156, 181]]]",
    "query_spans": "[[[183, 193]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, and $P$ is a point on the hyperbola such that $\\angle F_{1} P F_{2}=60^{\\circ}$, then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 - y^2/12 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "12*sqrt(3)",
    "fact_spans": "[[[18, 57], [66, 69]], [[2, 9]], [[62, 65]], [[10, 17]], [[18, 57]], [[2, 61]], [[62, 73]], [[75, 108]]]",
    "query_spans": "[[[110, 137]]]",
    "process": ""
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2x$ be $F$, and let a line $l$ passing through point $F$ intersect the parabola at points $A$ and $B$ such that $|AF|=4|BF|$. Point $O$ is the origin. Then the area of $\\triangle AOB$ is?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;O: Origin;B: Point;F: Point;Expression(G) = (y^2 = 2*x);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, F)) = 4*Abs(LineSegmentOf(B, F))",
    "query_expressions": "Area(TriangleOf(A, O, B))",
    "answer_expressions": "5/8",
    "fact_spans": "[[[29, 34]], [[1, 15], [35, 38]], [[40, 43]], [[66, 70]], [[44, 47]], [[19, 22], [24, 28]], [[1, 15]], [[1, 22]], [[23, 34]], [[29, 49]], [[51, 65]]]",
    "query_spans": "[[[77, 99]]]",
    "process": "Without loss of generality, assume the equation of line AB is $x=ty+\\frac{1}{2}$. By simultaneously solving the line and parabola equations, and combining with $|AF|=4|BF|$, we obtain $\\overrightarrow{AF}=4\\overrightarrow{FB}$. Using the relationship between roots and coefficients of the equation and the coordinate representation of vectors, we can solve for $t$, then compute the area according to $S_{\\triangle AOB}=\\frac{1}{2}\\cdot\\frac{1}{2}|y_{1}-y_{2}|$. Without loss of generality, assume the equation of line AB is $x=ty+\\frac{1}{2}$. Solving the system of equations\n$$\n\\begin{cases}\nx=ty+\\frac{1}{2}\\\\\ny^2=2x\n\\end{cases}\n$$\nwe get $y^{2}-2ty-1=0$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$. Since $|AF|=4|BF|$, it follows that $\\overrightarrow{AF}=4\\overrightarrow{FB}$, so $y_{1}=-4y_{2}$. Then $y_{1}y_{2}=-4y_{2}^{2}$, hence $y_{2}^{2}=\\frac{1}{4}$, i.e., $|y_{2}|=\\frac{1}{2}$. Therefore, $S_{\\Delta AOB}=\\frac{1}{2}\\cdot\\frac{1}{2}|y_{1}-y_{2}|=\\frac{1}{4}\\times|5y_{2}|=\\frac{5}{4}\\times\\frac{1}{2}=\\frac{5}{8}$."
  },
  {
    "text": "Given that the directrix of the parabola $y^{2}=-16 x$ is tangent to the circle $x^{2}-2 a x+y^{2}=0$, and the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ passes through the point $P(4,3)$, then the asymptotes of the hyperbola $C$ have equations?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Parabola;H: Circle;P: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y^2 = -16*x);Expression(H) = (y^2 - 2*a*x + x^2 = 0);Coordinate(P) = (4, 3);IsTangent(Directrix(G),H);PointOnCurve(P, C)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*(sqrt(3)/2)*x",
    "fact_spans": "[[[47, 108], [120, 126]], [[55, 108]], [[55, 108]], [[2, 18]], [[22, 44]], [[109, 118]], [[55, 108]], [[55, 108]], [[47, 108]], [[2, 18]], [[22, 44]], [[109, 118]], [[2, 46]], [[47, 118]]]",
    "query_spans": "[[[120, 134]]]",
    "process": "From the given conditions, the directrix of the parabola $ y^{2} = -16x $ is $ x = 4 $; the center of the circle $ x^{2} - 2ax + y^{2} = 0 $ is $ (a, 0) $, and the radius is $ a $ ($ a > 0 $); since the directrix of the parabola $ y^{2} = -16x $ is tangent to the circle $ x^{2} - 2ax + y^{2} = 0 $, we have $ 2a = 4 $, so $ a = 2 $; also, the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) passes through point $ P(4, 3) $, so $ \\frac{16}{4} - \\frac{9}{b^{2}} = 1 $, thus $ b = \\sqrt{3} $; therefore, the asymptotes of the hyperbola $ C $ are $ y = \\pm \\frac{\\sqrt{3}}{2}x $."
  },
  {
    "text": "Given that $F$ is the right focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and a line passing through point $A(0, \\frac{b}{2})$ perpendicular to the $y$-axis intersects the ellipse at points $B$ and $C$. If $\\angle B F C=90^{\\circ}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;A: Point;B: Point;F: Point;C: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (0, b/2);RightFocus(G)=F;PointOnCurve(A,H);IsPerpendicular(H,yAxis);Intersection(H,G)={B,C};AngleOf(B, F, C) = ApplyUnit(90, degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[6, 58], [96, 98], [140, 142]], [[8, 58]], [[8, 58]], [[93, 95]], [[64, 84]], [[100, 103]], [[2, 5]], [[104, 107]], [[8, 58]], [[8, 58]], [[6, 58]], [[64, 84]], [[2, 62]], [[63, 95]], [[85, 95]], [[93, 109]], [[112, 137]]]",
    "query_spans": "[[[140, 148]]]",
    "process": "According to the problem, construct the following figure: It is easy to obtain $ B\\left(-\\frac{\\sqrt{3}}{2}a,\\frac{b}{2}\\right) $, $ c\\left(\\frac{\\sqrt{3}}{2}a,\\frac{b}{2}\\right) $, $ F(c,0) $, then $ \\overrightarrow{FB}=\\left(\\frac{-\\sqrt{3}a}{2}-c,\\frac{b}{2}\\right) $, $ \\overrightarrow{FC}=\\left(\\frac{1}{1}\\left(\\frac{\\sqrt{3}}{2}\\right)0\\right) $, that is $ \\left(-\\frac{\\sqrt{3}}{3}a-c\\right)\\left(\\frac{\\sqrt{3}a}{2}-c\\right)+\\frac{b^{2}}{4}=0 $, $ c^{2}-\\frac{3}{4}a^{2}+\\frac{b^{2}}{4}=0 $, $ b^{2}=a^{2}-c^{2} $, solving yields $ \\frac{c}{a}=\\frac{\\sqrt{6}}{3}^{2} $, $ e=\\frac{\\sqrt{6}}{3} $"
  },
  {
    "text": "An equation of an ellipse tangent to the line $l$: $x+y=1$ could be?",
    "fact_expressions": "l: Line;Expression(l)=(x+y=1);G: Ellipse;IsTangent(l,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "(x^2/a^2+y^2/b^2=1)&(a^2+b^2=1)",
    "fact_spans": "[[[1, 14]], [[1, 14]], [[19, 21]], [[0, 21]]]",
    "query_spans": "[[[19, 28]]]",
    "process": "Let the point of tangency be $ P(x_{0},y_{0}) $, then the tangent line to the ellipse at $ P $ is $ \\frac{x_{0}x}{a^{2}}+\\frac{y_{0}y}{b^{2}}=1 $. Since $ x+y=1 $ is a tangent line, $ \\frac{x_{0}}{a^{2}}=1 $, $ \\frac{y_{0}}{b^{2}}=1 $, and $ y_{0}+x_{0}=1 $, therefore $ a^{2}+b^{2}=1 $."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2x$ be $F$, and let a line $l$ passing through $F$ intersect the parabola at points $A$ and $B$. Draw a perpendicular from the midpoint $M$ of $AB$ to the $y$-axis, intersecting the parabola at point $P$. If $|PF|=\\frac{3}{2}$, then the equation of line $l$ is?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;P: Point;F: Point;Expression(G) = (y^2 = 2*x);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A,B))=M;L:Line;PointOnCurve(M,L);IsPerpendicular(L,yAxis);Intersection(L,G)=P;Abs(LineSegmentOf(P, F)) = 3/2;M:Point",
    "query_expressions": "Expression(l)",
    "answer_expressions": "2*x+pm*sqrt(2)*y-1=0",
    "fact_spans": "[[[28, 33], [101, 106]], [[1, 15], [34, 37], [69, 72]], [[38, 41]], [[42, 45]], [[74, 78]], [[19, 22], [24, 27]], [[1, 15]], [[1, 22]], [[23, 33]], [[28, 47]], [[49, 60]], [], [[48, 68]], [[48, 68]], [[49, 78]], [[80, 99]], [[57, 60]]]",
    "query_spans": "[[[101, 111]]]",
    "process": "Since the parabola equation is $ y^{2} = 2x $, the focus is $ F\\left(\\frac{1}{2}, 0\\right) $, and the directrix is $ l: x = -\\frac{1}{2} $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, and the line $ AB $ be $ y = k\\left(x - \\frac{1}{2}\\right) $. Substituting into the parabola equation and eliminating $ y $, we get $ k^{2}x^{2} - (k^{2} + 2)x + \\frac{k^{2}}{4} = 0 $. Thus, $ x_{1} + x_{2} = \\frac{k^{2} + 2}{k^{2}} $, $ x_{1}x_{2} = \\frac{1}{4} $. Also, drawing a perpendicular from the midpoint $ M $ of $ AB $ to the directrix, intersecting the parabola at point $ P $, let $ P(x_{0}, y_{0}) $, we obtain $ y_{0} = \\frac{1}{2}(y_{1} + y_{2}) $. Since $ y_{1} = k\\left(x_{1} - \\frac{1}{2}\\right) $, $ y_{2} = k\\left(x_{2} - \\frac{1}{2}\\right) $, so $ y_{1} + y_{2} = k(x_{1} + x_{2}) - k = k \\cdot \\frac{k^{2} + 2}{k^{2}} - k = \\frac{2}{k} $, yielding $ y_{0} = \\frac{2}{k} $, $ x_{0} = \\frac{1}{2k^{2}} $, hence $ P\\left(\\frac{1}{2k^{2}}, \\frac{2}{k}\\right) $. Since $ |PF| = \\frac{3}{2} $, we have $ \\sqrt{\\left(\\frac{1}{2} - \\frac{1}{2k^{2}}\\right)^{2} + \\left(0 - \\frac{2}{k}\\right)^{2}} = \\frac{3}{2} $, solving gives $ k^{2} = \\frac{1}{2} $. Therefore, $ k = \\pm \\frac{\\sqrt{2}}{2} $, and the line equations are $ y = \\pm \\frac{\\sqrt{2}}{2}\\left(x - \\frac{1}{2}\\right) $, i.e., $ 2x \\pm 2\\sqrt{2}y - 1 = 0 $."
  },
  {
    "text": "The standard equation of a parabola with focus $(0,-\\frac{1}{2})$ is?",
    "fact_expressions": "G: Parabola;Coordinate(Focus(G)) = (0,-1/2)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 = -2*y",
    "fact_spans": "[[[22, 25]], [[0, 25]]]",
    "query_spans": "[[[22, 32]]]",
    "process": ""
  },
  {
    "text": "Let the line $l$ with slope $2$ pass through the focus $F$ of the parabola $y^2 = ax$ ($a > 0$), and intersect the $y$-axis at point $A$. If the area of $\\triangle OAF$ ($O$ being the origin) is $4$, then the value of $a$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = a*x);a: Number;a>0;F:Point;Focus(G)=F;l: Line;PointOnCurve(F, l);Slope(l)=2;A:Point;Intersection(l,yAxis)=A;O: Origin;Area(TriangleOf(O, A, F))=4",
    "query_expressions": "a",
    "answer_expressions": "8",
    "fact_spans": "[[[14, 36]], [[14, 36]], [[94, 97]], [[17, 36]], [[39, 42]], [[14, 42]], [[8, 13]], [[8, 42]], [[1, 13]], [[51, 55]], [[8, 55]], [[75, 78]], [[57, 92]]]",
    "query_spans": "[[[94, 101]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-\\frac{y^{2}}{b^{2}}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. The line $l$ passes through $F_{2}$ and intersects the left and right branches of the hyperbola $C$ at points $A$ and $B$ respectively. It is known that $\\angle F_{1} A F_{2}=90^{\\circ}$, and the inradius of $\\triangle A B F_{1}$ is $1$. Then $|A B|$=?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2/b^2 = 1);b: Number;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;l: Line;PointOnCurve(F2, l);A: Point;Intersection(l, LeftPart(C)) = A;B: Point;Intersection(l, RightPart(C)) = B;AngleOf(F1, A, F2) = ApplyUnit(90, degree);Radius(InscribedCircle(TriangleOf(A, B, F1))) = 1",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 39], [79, 85]], [[2, 39]], [[10, 39]], [[49, 56]], [[57, 64], [71, 78]], [[2, 64]], [[2, 64]], [[65, 70]], [[65, 78]], [[96, 99]], [[65, 105]], [[100, 103]], [[65, 105]], [[108, 141]], [[143, 173]]]",
    "query_spans": "[[[175, 184]]]",
    "process": "The hyperbola $ C: x^{2} - \\frac{y^{2}}{b^{2}} = 1 $ has $ a = 1 $. Let $ |AF_{1}| = m $, $ |BF_{1}| = n $. By the definition of the hyperbola, we obtain $ |AF_{2}| = |AF_{1}| + 2a = m + 2 $, $ |BF_{2}| = |BF_{1}| - 2a = n - 2 $, and $ |AB| = |AF_{2}| - |BF_{2}| = m - n + 4 $. By the tangent segment theorem, the inradius of a right triangle is half the difference between the sum of the two legs and the hypotenuse. Thus, in the right triangle $ \\triangle ABF_{1} $, $ \\frac{1}{2}(|AB| + |AF_{1}| - |BF_{1}|) = \\frac{1}{2}(m - n + 4 + m - n) = 1 $, which gives $ m - n = -1 $. Therefore, $ |AB| = -1 + 4 = 3 $."
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ are $F_{1}$ and $F_{2}$, respectively. A line passing through focus $F_{1}$ intersects the ellipse at points $A$ and $B$. If the area of the incircle of $\\triangle ABF_{2}$ is $\\pi$, and the coordinates of points $A$ and $B$ are $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$, respectively, then the value of $|y_{2}-y_{1}|$ is?",
    "fact_expressions": "G: Ellipse;I: Line;A: Point;B: Point;F1: Point;F2: Point;Expression(G) = (x^2/16 + y^2/9 = 1);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, I);Intersection(I, G) = {A, B};Area(InscribedCircle(TriangleOf(A,B,F2)))=pi;x1:Number;x2:Number;y1:Number;y2:Number",
    "query_expressions": "Abs(y2-y1)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 38], [77, 79]], [[74, 76]], [[80, 83], [127, 130]], [[84, 87], [132, 135]], [[47, 54], [66, 73]], [[55, 62]], [[0, 38]], [[127, 178]], [[127, 178]], [[0, 62]], [[0, 62]], [[63, 76]], [[74, 89]], [[91, 124]], [[143, 160]], [[161, 178]], [[143, 160]], [[161, 178]]]",
    "query_spans": "[[[180, 199]]]",
    "process": ""
  },
  {
    "text": "If the equation $\\frac{x^{2}}{1-k}+\\frac{y^{2}}{2+k}=1$ represents an ellipse, then what is the range of values for $k$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(1 - k) + y^2/(k + 2) = 1);k: Number",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-2,1)&Negation(k=-1/2)",
    "fact_spans": "[[[44, 46]], [[1, 46]], [[48, 51]]]",
    "query_spans": "[[[48, 58]]]",
    "process": "From the ellipse equation, we obtain  \n\\begin{cases}1-k>0\\\\2+k>0\\\\1-k+2+k\\end{cases},  \nsolving the inequalities gives the range of $k$ as $-2<k<1$ and $k\\neq-\\frac{1}{2}$."
  },
  {
    "text": "Given that the parabola $y^{2}=2 x$ intersects the line $y-x+1=0$ at points $A$ and $B$, find the length of the chord $|A B|$.",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);H: Line;Expression(H) = (-x + y + 1 = 0);A: Point;B: Point;Intersection(G, H) = {A, B};IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "2*sqrt(6)",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[17, 28]], [[17, 28]], [[30, 33]], [[34, 37]], [[2, 39]], [[2, 50]]]",
    "query_spans": "[[[43, 52]]]",
    "process": "Substitute the line $ x = y + 1 $ into the parabola equation to obtain: $ y^{2} - 2y - 2 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ y_{1} + y_{2} = 2 $, $ y_{1}y_{2} = -2 $. Therefore, $ |AB| = \\sqrt{2} \\cdot \\sqrt{(y_{1} + y_{2})^{2}} - 4y_{1}y_{2} = \\sqrt{2} \\times \\sqrt{4 + 8} = 2\\sqrt{6} $."
  },
  {
    "text": "If the eccentricity $e \\in (1, 2)$ of the hyperbola $\\frac{x^{2}}{4}+\\frac{y^{2}}{k}=1$, then what is the range of values for $k$?",
    "fact_expressions": "G: Hyperbola;k: Number;Expression(G) = (x^2/4 + y^2/k = 1);e: Number;Eccentricity(G) = e ;In(e, (1, 2))",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-12, 0)",
    "fact_spans": "[[[1, 39]], [[59, 62]], [[1, 39]], [[43, 57]], [[1, 57]], [[43, 57]]]",
    "query_spans": "[[[59, 69]]]",
    "process": ""
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, point $A$ is a point on this parabola, and $\\angle OFA=120^{\\circ}$ (where $O$ is the origin), then the distance from the midpoint $M$ of segment $AF$ to the $y$-axis is?",
    "fact_expressions": "A: Point;F: Point;G: Parabola;O: Origin;M: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(A, G);AngleOf(O, F, A) = ApplyUnit(120, degree);MidPoint(LineSegmentOf(A, F)) = M",
    "query_expressions": "Distance(M, yAxis)",
    "answer_expressions": "2",
    "fact_spans": "[[[24, 28]], [[20, 23]], [[2, 16], [30, 33]], [[67, 70]], [[87, 90]], [[2, 16]], [[2, 23]], [[24, 36]], [[38, 64]], [[78, 90]]]",
    "query_spans": "[[[87, 100]]]",
    "process": "The focus of the parabola $ y^{2} = 4x $ is $ F(1,0) $, the directrix is $ x = -1 $, and $ \\angle OFA = 120^{\\circ} $ (where $ O $ is the origin), so $ k_{AF} = \\sqrt{3} $. The line $ AF: y = \\sqrt{3}(x - 1) $, when solved simultaneously with the parabola equation, gives $ 3x^{2} - 10x + 3 = 0 \\Rightarrow x = 3, x = \\frac{1}{3} $. Since $ \\angle OFA = 120^{\\circ} $, $ x = 3 $, $ A(3, \\pm 2\\sqrt{2}) $. The distance from the midpoint $ M $ of segment $ AF $ to the $ y $-axis is $ \\frac{3 + 1}{2} = 2 $."
  },
  {
    "text": "The hyperbola $C$: $x^{2}-y^{2}=a^{2}$ is centered at the origin, with foci on the $x$-axis. $C$ intersects the directrix of the parabola $y^{2}=16x$ at points $A$ and $B$, and $|AB|=4\\sqrt{3}$. Then the equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2 = a^2);a: Number;G: Parabola;Expression(G) = (y^2 = 16*x);A: Point;B: Point;O: Origin;Center(C) = O;PointOnCurve(Focus(C), xAxis);Intersection(C, Directrix(G)) = {A,B};Abs(LineSegmentOf(A, B)) = 4*sqrt(3)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4 - y^2/4 = 1",
    "fact_spans": "[[[0, 27], [43, 46], [97, 103]], [[0, 27]], [[8, 27]], [[47, 62]], [[47, 62]], [[67, 70]], [[71, 74]], [[31, 33]], [[0, 33]], [[0, 42]], [[43, 76]], [[77, 95]]]",
    "query_spans": "[[[97, 108]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, a line passing through the point $(4,0)$ intersects the ellipse $E$ at points $A$ and $B$. If the midpoint of $AB$ has coordinates $(2,-1)$, then the eccentricity of the ellipse $E$ is?",
    "fact_expressions": "E: Ellipse;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;H: Point;Coordinate(H) = (4, 0);L: Line;PointOnCurve(H, L) = True;Intersection(L, E) = {A, B};A: Point;B: Point;Coordinate(MidPoint(LineSegmentOf(A,B))) = (2,-1)",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[2, 59], [73, 78], [111, 116]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[61, 69]], [[61, 69]], [[70, 72]], [[60, 72]], [[70, 88]], [[79, 82]], [[83, 86]], [[90, 109]]]",
    "query_spans": "[[[111, 122]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), substitute into the ellipse equation, subtract the two equations, and use the eccentricity formula to solve. Let A(x_{1},y_{1}), B(x_{2},y_{2}), then \\frac{x_{1}^{2}}{a^{2}}+\\frac{y_{1}^{2}}{b^{2}}=1,\\textcircled{1} \\frac{y_{2}^{2}}{b^{2}}=1,\\textcircled{2} \\textcircled{1}-\\textcircled{2} yields \\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{a^{2}}+\\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{b^{2}}=0. Since the midpoint coordinates of AB are (2,-1), then x_{1}+x_{2}=4, y_{1}+y_{2}=-2. Therefore, \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{2b^{2}}{a^{2}}=\\frac{0-(-1)}{4-2}=\\frac{1}{2}. So a^{2}=4b^{2}. Since b^{2}=a^{2}-c^{2}, we have 3a^{2}=4c^{2}, thus e=\\frac{c}{a}=\\frac{\\sqrt{3}}{2}"
  },
  {
    "text": "$P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and $M$, $N$ are points on the circles $(x+5)^{2}+y^{2}=4$ and $(x-5)^{2}+y^{2}=1$, respectively. Then the maximum value of $P M-P N$ is?",
    "fact_expressions": "G: Hyperbola;H: Circle;Z: Circle;P: Point;M: Point;N: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Expression(H) = (y^2 + (x + 5)^2 = 4);Expression(Z) = (y^2 + (x - 5)^2 = 1);PointOnCurve(P, RightPart(G));PointOnCurve(M, H);PointOnCurve(N, Z)",
    "query_expressions": "Max(LineSegmentOf(P, M) - LineSegmentOf(P, N))",
    "answer_expressions": "9",
    "fact_spans": "[[[4, 43]], [[60, 80]], [[81, 100]], [[0, 3]], [[50, 53]], [[54, 57]], [[4, 43]], [[60, 80]], [[81, 100]], [[0, 49]], [[50, 103]], [[50, 103]]]",
    "query_spans": "[[[105, 120]]]",
    "process": ""
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and point $P$ lies on the hyperbola such that $|PF_{1}| \\cdot |PF_{2}|=32$. Then $\\angle F_{1}PF_{2}$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 32",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "ApplyUnit(90, degree)",
    "fact_spans": "[[[17, 56], [66, 69]], [[17, 56]], [[0, 7]], [[9, 16]], [[0, 61]], [[62, 65]], [[62, 70]], [[73, 100]]]",
    "query_spans": "[[[103, 126]]]",
    "process": ""
  },
  {
    "text": "A chord $AB$ of the parabola $y^{2}=2 px(p>0)$ passes through the focus $F$, and $|AF|=1$, $|BF|=\\frac{1}{3}$. Then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G) = True;F: Point;Focus(G) = F;PointOnCurve(F, LineSegmentOf(A, B)) = True;Abs(LineSegmentOf(A, F)) = 1;Abs(LineSegmentOf(B, F)) = 1/3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=x",
    "fact_spans": "[[[0, 20], [66, 69]], [[0, 20]], [[3, 20]], [[3, 20]], [[24, 28]], [[24, 28]], [[0, 28]], [[31, 34]], [[0, 34]], [[24, 34]], [[36, 44]], [[46, 64]]]",
    "query_spans": "[[[66, 73]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y=\\frac{1}{8} x^2$ with focus $F$, and point $M$ a point on its directrix $l$, the segment $MF$ intersects the parabola $C$ at point $N$. When $\\overrightarrow{M N}=\\frac{2}{3} \\overrightarrow{M F}$, the area of $\\triangle N O F$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y = x^2/8);F: Point;Focus(C) = F;M: Point;l: Line;Directrix(C) = l;PointOnCurve(M, l);Intersection(LineSegmentOf(M,F), C) = N;N: Point;VectorOf(M,N) = (2/3)*VectorOf(M,F);O: Origin",
    "query_expressions": "Area(TriangleOf(N, O, F))",
    "answer_expressions": "4*sqrt(3)/3",
    "fact_spans": "[[[2, 29], [42, 43], [60, 66]], [[2, 29]], [[33, 36]], [[2, 36]], [[37, 41]], [[45, 48]], [[42, 48]], [[37, 51]], [[52, 71]], [[67, 71]], [[73, 128]], [[130, 147]]]",
    "query_spans": "[[[130, 152]]]",
    "process": "From the equation of the parabola, we can obtain the coordinates of the focus $ F $ and the equation of the directrix. Since $ \\overrightarrow{MN} = \\frac{2}{3}\\overrightarrow{MF} $, it follows that $ N $ lies between $ M $ and $ F $. Let $ NN' $ be perpendicular to the directrix, intersecting it at $ N' $. By the property of the parabola, we have $ NN' = NF $. Then $ \\tan\\angle FMN' = \\frac{\\sqrt{3}}{3} $. Find the equation of line $ MF $, substitute it into the parabola's equation to find the x-coordinate of $ N $, and then calculate the area of quadrilateral $ ANOF $. Solution: According to the problem, the standard equation of the parabola is $ x^2 = 8y $, so the focus is $ F(0,2) $, and the directrix equation is $ y = -2 $. Let $ NN' $ be perpendicular to the directrix, intersecting it at $ N' $, as shown in the figure. By the property of the parabola, $ NN' = NF $. Since $ \\overrightarrow{MN} = \\frac{2}{3}\\overrightarrow{MF} $, we know $ N $ lies between $ M $ and $ F $, so $ MN = 2NF = 2NN' $. Therefore, $ \\sin\\angle FMN' = \\frac{NN'}{MN} = \\frac{1}{2} $, so $ \\tan\\angle FMN' = \\frac{\\sqrt{3}}{3} $, i.e., the slope of line $ MF $ is $ \\frac{\\sqrt{3}}{3} $. Thus, the equation of line $ MF $ is $ y = \\frac{\\sqrt{3}}{3}x + 2 $. Substituting the equation of line $ MF $ into the parabola's equation gives: $ x^2 - \\frac{8\\sqrt{3}}{3}x - 16 = 0 $. Solving yields $ x = -\\frac{4}{\\sqrt{3}} $ or $ x = 4\\sqrt{3} $ (discarded). Therefore, $ S_{ANOF} = \\frac{1}{2}|OF| \\cdot |x| = \\frac{1}{2} \\times 2 \\times \\frac{4\\sqrt{3}}{3} = \\frac{4\\sqrt{3}}{3} $."
  },
  {
    "text": "Given that the ordinate of the midpoint of chord $AB$ of the parabola $x^{2}=8 y$ is $4$, then the maximum value of $|A B|$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 8*y);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);YCoordinate(MidPoint(LineSegmentOf(A, B))) = 4",
    "query_expressions": "Max(Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "12",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[18, 23]], [[18, 23]], [[2, 23]], [[18, 34]]]",
    "query_spans": "[[[36, 49]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since the ordinate of the midpoint of AB is 4, we have y_{1}+y_{2}=8. By the triangle inequality, |AB|\\leqslant|AF|+|BF|=y+y+4=8+4=12."
  },
  {
    "text": "Given points $A(-1,0)$, $B(1,0)$ and the parabola $y^{2}=2x$, if point $P$ on the parabola satisfies $|PA|=m|PB|$, then the range of values for $m$ is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;P: Point;Expression(G) = (y^2 = 2*x);Coordinate(A) = (-1, 0);Coordinate(B) = (1, 0);PointOnCurve(P, G);m: Number;Abs(LineSegmentOf(P, A)) = m*Abs(LineSegmentOf(P, B))",
    "query_expressions": "Range(m)",
    "answer_expressions": "[1, sqrt(3)]",
    "fact_spans": "[[[24, 38], [40, 43]], [[2, 12]], [[15, 23]], [[44, 48]], [[24, 38]], [[2, 12]], [[15, 23]], [[40, 48]], [[66, 69]], [[50, 64]]]",
    "query_spans": "[[[66, 76]]]",
    "process": "Let $ P\\left(\\frac{y^{2}}{2}, y\\right) $. From the given condition, we have  \n$ m^{2} = \\frac{|PA|^{2}}{|PB|^{2}} = \\frac{(y^{2}+1)+y^{2}}{\\left(\\frac{y^{2}}{2}-1\\right)^{2}+y^{2}} = \\frac{y^{4}+4+8y^{2}}{y^{4}+4} = 1 + \\frac{8y^{2}}{y^{4}+4} $.  \n$ \\therefore m^{2} \\leqslant 1 + \\frac{8y^{2}}{2\\sqrt{4y^{4}}} = 3 $, with equality if and only if $ y^{2} = 2 $.  \nAlso, $ m^{2} = 1 + \\frac{8y^{2}}{y^{4}+4} \\geqslant 1 $, with equality when $ y = 0 $.  \n$ \\therefore 1 \\leqslant m < \\sqrt{3} $. The maximum value of $ m $ is attained when $ y^{2} = 2 $, and the minimum value when $ y = 0 $. Hence, the range of $ m $ is $ [1, \\sqrt{3}] $."
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1$ is $y=\\frac{\\sqrt{3}}{3} x$, then the length of the real axis of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/4 + x^2/a^2 = 1);a: Number;Expression(OneOf(Asymptote(G))) = (y = x*(sqrt(3)/3))",
    "query_expressions": "Length(RealAxis(G))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[2, 44], [80, 83]], [[2, 44]], [[5, 44]], [[2, 77]]]",
    "query_spans": "[[[80, 89]]]",
    "process": "From the given conditions, we have\n\\begin{cases}\\frac{b}{a}=\\frac{\\sqrt{3}}{3}\\\\b^{2}=4\\end{cases},\nsolving gives $ a=2\\sqrt{3} $, then the real axis length of this hyperbola is $ 2a=4\\sqrt{3} $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, with left and right foci $F_{1}$, $F_{2}$ respectively, and a point $P(3, \\frac{\\sqrt{10}}{2})$ on the right branch of $C$, such that $|P F_{1}|-|P F_{2}|=4$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (3, sqrt(10)/2);LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(P, RightPart(C));Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)) = 4",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[2, 63], [120, 123], [156, 159]], [[10, 63]], [[10, 63]], [[92, 119]], [[72, 79]], [[81, 89]], [[10, 63]], [[10, 63]], [[2, 63]], [[92, 119]], [[2, 89]], [[2, 89]], [[92, 129]], [[131, 154]]]",
    "query_spans": "[[[156, 165]]]",
    "process": "Given that 2a=4, so a=2, and the point P(3,\\frac{\\sqrt{10}}{2}) lies on the hyperbola, therefore \\frac{9}{4}-\\frac{10}{b^{2}}=1, solving gives b^{2}=2, thus e=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{1+\\frac{2}{4}}=\\frac{\\sqrt{6}}{2}."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and the line $l$: $ax+by-4a+2b=0$, how many common points do the line $l$ and the ellipse $C$ have?",
    "fact_expressions": "l: Line;C: Ellipse;a: Number;b: Number;Expression(C) = (x^2/25+y^2/16 = 1);Expression(l) = (a*x + b*y - 4*a + 2*b = 0)",
    "query_expressions": "NumIntersection(l, C)",
    "answer_expressions": "2",
    "fact_spans": "[[[47, 70], [72, 77]], [[2, 46], [78, 83]], [[9, 46]], [[9, 46]], [[2, 46]], [[47, 70]]]",
    "query_spans": "[[[72, 90]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola $y^{2}=4 m x(m>0)$ is $F$, point $P$ is a moving point on the parabola, and point $A(-m, 0)$. Then the minimum value of $\\frac{|P F|}{|P A|}$ is?",
    "fact_expressions": "G: Parabola;m: Number;A: Point;P: Point;F: Point;m>0;Expression(G) = (y^2 = 4*(m*x));Coordinate(A) = (-m, 0);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F))/Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[0, 21], [35, 38]], [[3, 21]], [[45, 56]], [[30, 34]], [[25, 28]], [[3, 21]], [[0, 21]], [[45, 56]], [[0, 28]], [[30, 42]]]",
    "query_spans": "[[[59, 86]]]",
    "process": ""
  },
  {
    "text": "Given a curve $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{7}=1$ with eccentricity $e$, its right focus coincides with the focus of the parabola $y^{2}=16x$. Find the value of $e$.",
    "fact_expressions": "G: Parabola;H: Curve;a: Number;Expression(G) = (y^2 = 16*x);Expression(H) = (-y^2/7 + x^2/a^2 = 1);e:Number;Eccentricity(H)=e;RightFocus(H)=Focus(G)",
    "query_expressions": "e",
    "answer_expressions": "4/3",
    "fact_spans": "[[[57, 72]], [[10, 51], [52, 53]], [[12, 51]], [[57, 72]], [[10, 51]], [[79, 82], [6, 9]], [[2, 51]], [[52, 77]]]",
    "query_spans": "[[[79, 86]]]",
    "process": ""
  },
  {
    "text": "The equation of the hyperbola that shares the same asymptotes as $\\frac{x^{2}}{3}-\\frac{y^{2}}{2}=1$ and passes through the point $A(\\sqrt{3}, 2 \\sqrt{5})$ is?",
    "fact_expressions": "G: Hyperbola;C:Hyperbola;A: Point;Expression(G) = (x^2/3 - y^2/2 = 1);Coordinate(A) = (sqrt(3), 2*sqrt(5));Asymptote(C) = Asymptote(G);PointOnCurve(A, C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/18 - x^2/27 = 1",
    "fact_spans": "[[[1, 39]], [[77, 80]], [[50, 76]], [[1, 39]], [[50, 76]], [[0, 80]], [[48, 80]]]",
    "query_spans": "[[[77, 85]]]",
    "process": "Set up the equation of a hyperbola sharing the same asymptotes with $\\frac{x^{2}}{3}-\\frac{y^{2}}{2}=1$, and substitute point $A(\\sqrt{3},2\\sqrt{5})$ to solve. Substituting point $A(\\sqrt{3},2\\sqrt{5})$ into the hyperbola equation yields $\\underline{(\\sqrt{3})^{2}}-\\frac{(2\\sqrt{5})^{2}}{2}=2$, solving gives $\\lambda=-9$, then the hyperbola's equation is $\\frac{y^{2}}{18}-\\frac{x^{2}}{27}=1$."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. Suppose line $l$ passes through $F_{1}$, intersects ellipse $C$ at points $A$ and $B$, and intersects the $y$-axis at point $P$. If $\\overrightarrow{F_{1} P}=\\overrightarrow{A F_{1}}$ and $\\overrightarrow{F_{2} A} \\cdot \\overrightarrow{F_{2} P}=0$, then the eccentricity of the ellipse is?",
    "fact_expressions": "F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;l: Line;PointOnCurve(F1, l);Intersection(l, C) = {A, B};A: Point;B: Point;Intersection(l, yAxis) = P;P: Point;VectorOf(F1, P) = VectorOf(A, F1);DotProduct(VectorOf(F2, A), VectorOf(F2, P)) = 0",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(\\sqrt{3} - 1)/2",
    "fact_spans": "[[[1, 8], [89, 96]], [[9, 16]], [[1, 82]], [[1, 82]], [[19, 76], [97, 102], [238, 240]], [[19, 76]], [[26, 76]], [[26, 76]], [[26, 76]], [[26, 76]], [[83, 88]], [[83, 96]], [[83, 112]], [[103, 106]], [[107, 110]], [[83, 123]], [[119, 123]], [[125, 176]], [[177, 236]]]",
    "query_spans": "[[[238, 246]]]",
    "process": "Let $ P(0,t) $, since $ \\overrightarrow{F_{1}P} = \\overline{AI} $, therefore $ F_{1} $ is the midpoint of $ AP $, so $ A(-2c,-t) $, $ \\overrightarrow{F_{2}A} \\cdot \\overrightarrow{F_{2}P} = 0 $, thus $ (-3c_{1}-t)(-c,t) = 0 $, $ t^{2} = 3c^{2} $. Since $ A $ lies on the ellipse, $ \\frac{4c^{2}}{a^{2}} + \\frac{3c^{2}}{b^{2}} = 1 $, i.e., $ 4b^{2}c^{2} + 3a^{2}c^{2} = a^{2}b^{2} $. Rearranging yields: $ a^{4} - 8a^{2}c^{2} + 4c^{4} = 0 $, so $ 1 - 8e^{2} + 4e^{4} = 0 $, hence $ e^{2} = \\frac{8 \\pm 4\\sqrt{3}}{8} = \\frac{4 \\pm 2\\sqrt{3}}{4} $, $ e \\in (0,1) $, therefore $ e = \\frac{4 - 2\\sqrt{3}}{4} = \\left( \\frac{\\sqrt{3} - 1}{2} \\right) \\overset{3}{\\cdot} e = \\frac{\\sqrt{3} - 1}{} $"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $F(5,0)$ is the right focus of the hyperbola. A line passing through $F$ intersects the hyperbola at points $A$ and $B$, and the midpoint of $AB$ is $M\\left(-\\frac{45}{7},-\\frac{80}{7}\\right)$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;Coordinate(F) = (5, 0);RightFocus(G) = F;H: Line;PointOnCurve(F, H);A: Point;B: Point;Intersection(H, G) = {A, B};M: Point;Coordinate(M) = (-45/7, -80/7);MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/16 = 1",
    "fact_spans": "[[[2, 58], [70, 73], [87, 90], [146, 149]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[60, 68], [79, 82]], [[60, 68]], [[60, 77]], [[83, 85]], [[78, 85]], [[91, 94]], [[95, 98]], [[83, 100]], [[111, 143]], [[111, 143]], [[102, 143]]]",
    "query_spans": "[[[146, 154]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then \n\\[\n\\begin{cases}\n\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 \\\\\n\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{\\frac{2}{2}}=1\n\\end{cases}\n\\]\nSubtracting the two equations gives: \n\\[\n\\frac{(x_{1}+x_{2})(x_{1}-}{a^{2}}\n\\]\nSince the midpoint of $ AB $ is $ M(-\\frac{45}{7},-\\frac{80}{7}(-\\frac{1}{2},-\\frac{1}{7})^{2}=-\\frac{160}{7} $, \n\\[\n\\therefore k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{b^{2}(x_{1}+x_{2})}{a^{2}(y_{1}+y_{2})}=\\frac{9b^{2}}{16a^{2}}\n\\]\nAlso $ k_{AB}=k_{FM}=\\frac{-\\frac{80}{7}}{-\\frac{45}{7}-5}=1 $, so $ \\frac{9b^{2}}{16a^{2}}=1 $, i.e., $ 16a^{2}=9b^{2} $.  \nMoreover, $ a^{2}+b^{2}=c^{2}=25 $. From \n\\[\n\\begin{cases}\n9b^{2}=16a^{2} \\\\\na2+b^{2}=25\n\\end{cases}\n\\]\nsolving gives \n\\[\n\\begin{cases}\na=3 \\\\\nb=4\n\\end{cases}\n\\]\nThe hyperbola equation is $ \\frac{x^{2}}{O}-\\frac{y^{2}}{1c}= $"
  },
  {
    "text": "Given the parabola $C$: $x^{2}=2 p y(p>0)$ with focus $F$, point $M$ lies on $C$, and the distance from point $M$ to point $F$ is $13$, and the distance from $M$ to the $x$-axis is $9$, then $p=?$",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 2*(p*y));p: Number;p>0;F: Point;Focus(C) = F;M: Point;PointOnCurve(M, C);Distance(M, F) = 13;Distance(M, xAxis) = 9",
    "query_expressions": "p",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 28], [41, 44]], [[2, 28]], [[79, 82]], [[10, 28]], [[32, 35], [52, 56]], [[2, 35]], [[36, 40], [47, 51]], [[36, 45]], [[47, 64]], [[47, 77]]]",
    "query_spans": "[[[79, 84]]]",
    "process": "According to the definition of a parabola, the distance from point M to point F is equal to the distance from point M to the directrix $ y = -\\frac{p}{2} $, which is 13. Also, the distance to the x-axis is 9. Thus, $ 13 - 9 = \\frac{p}{2} $, solving gives $ p = 8 $."
  },
  {
    "text": "If a point $A\\left(\\frac{p}{4}, y_{1}\\right)$ on the parabola $C$: $y^{2}=2 p x$ $(p>0)$ is at a distance of $6$ from its focus, then $p=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;A: Point;y1: Number;Coordinate(A) = (p/4, y1);PointOnCurve(A, C);Distance(A, Focus(C)) = 6",
    "query_expressions": "p",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 27], [55, 56]], [[1, 27]], [[68, 71]], [[9, 27]], [[31, 54]], [[31, 54]], [[31, 54]], [[1, 54]], [[31, 66]]]",
    "query_spans": "[[[68, 73]]]",
    "process": "According to the definition of a parabola, $\\frac{p}{4}+\\frac{p}{2}=6$, so $p=8$."
  },
  {
    "text": "What is the standard equation of a parabola with focus at $(0,1)$?",
    "fact_expressions": "G: Parabola;Coordinate(Focus(G)) = (0, 1)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 = 4*y",
    "fact_spans": "[[[13, 16]], [[0, 16]]]",
    "query_spans": "[[[13, 23]]]",
    "process": "Given that the focus of the parabola has coordinates (0,1), the parabola opens upward, so \\frac{p}{2}=1, thus p=2, therefore the standard equation of the parabola is x^2=2pv=4v. Hence the answer is: x^2=4v"
  },
  {
    "text": "Given that the eccentricity of the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{m}=1$ is $\\frac{1}{2}$, then $m$=?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/3 + y^2/m = 1);Eccentricity(G) = 1/2",
    "query_expressions": "m",
    "answer_expressions": "{4,9/4}",
    "fact_spans": "[[[2, 39]], [[59, 62]], [[2, 39]], [[2, 57]]]",
    "query_spans": "[[[59, 64]]]",
    "process": "From the standard equation of the ellipse: (1) When $0 < m < 3$, we obtain $a = \\sqrt{3}$, $b = \\sqrt{m}$, then $c = \\sqrt{3 - m}$, so the eccentricity of the ellipse is $e = \\frac{c}{a} = \\frac{\\sqrt{3 - m}}{\\sqrt{3}} = \\frac{1}{2}$, which gives $m = \\frac{9}{4}$; (2) When $m > 3$, we obtain $b = \\sqrt{3}$, $a = \\sqrt{m}$, then $c = \\sqrt{-3 + m}$, so the eccentricity of the ellipse is $e = \\frac{c}{a} = \\frac{\\sqrt{-3 + m}}{\\sqrt{m}} = \\frac{1}{2}$, which gives $m = 4$; in conclusion, $m = 4$ or $\\frac{9}{4}$."
  },
  {
    "text": "Let $F$ be the focus of the parabola $y=-\\frac{1}{4} x^{2}$, and let line $l$, tangent to the parabola at point $P(-4,-4)$, intersect the $x$-axis at point $Q$. Then $\\angle P Q F =$?",
    "fact_expressions": "l: Line;G: Parabola;P: Point;Q: Point;F: Point;Expression(G) = (y = -x^2/4);Coordinate(P) = (-4,-4);Focus(G) = F;Intersection(l,xAxis)=Q;TangentPoint(l,G)=P",
    "query_expressions": "AngleOf(P, Q, F)",
    "answer_expressions": "pi/2",
    "fact_spans": "[[[54, 59]], [[5, 30], [35, 38]], [[41, 53]], [[68, 71]], [[1, 4]], [[5, 30]], [[41, 53]], [[1, 33]], [[54, 71]], [[34, 59]]]",
    "query_spans": "[[[73, 89]]]",
    "process": ""
  },
  {
    "text": "If the focal distance of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{m}=1$ is $6$, then the length of the imaginary axis of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/4 - y^2/m = 1);FocalLength(G) = 6",
    "query_expressions": "Length(ImageinaryAxis(G))",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[1, 39], [49, 52]], [[4, 39]], [[1, 39]], [[1, 46]]]",
    "query_spans": "[[[49, 58]]]",
    "process": "Given that $2c=6$, so $c=3$, and $a=2$, $b^{2}=m$, therefore $4+m=3^{2}$, so $m=5$. Thus, the length of the imaginary axis of the hyperbola is $2\\sqrt{5}$."
  },
  {
    "text": "Given the parabola equation $y^{2}=4x$, a line $l$ passing through its focus $F$ intersects the parabola at points $A$ and $B$. If $S_{\\triangle AOF} = 3S_{\\triangle BOF}$ ($O$ being the origin), then $|AB|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, G) = {A, B};O: Origin;Area(TriangleOf(A, O, F)) = 3*Area(TriangleOf(B, O ,F))",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[2, 5], [22, 23], [35, 38]], [[2, 20]], [[25, 28]], [[22, 28]], [[29, 34]], [[21, 34]], [[40, 43]], [[44, 47]], [[29, 49]], [[96, 99]], [[51, 94]]]",
    "query_spans": "[[[107, 116]]]",
    "process": "Analysis: First find the coordinates of intersection point F, then use $ S_{AAOF} = 3S_{ABOF} $ to obtain the coordinate relationship between A and B, solve the system of equations to find the x-coordinates of A and B, and finally use the definition of the parabola to find $ |AB| $. Detailed solution: Since $ y^{2} = 4x $, we have $ F(1,0) $. Because $ S_{AAOF} = 3S_{ABOF} $, it follows that  \n$$\n\\begin{cases}\n3(1 - x_{B}) = x_{A} - 1 \\\\\ny = -3y_{R}\n\\end{cases}\n$$\nSince  \n$$\n\\begin{cases}\ny_{B}^{2} = 4x_{B} \\\\\ny_{A}^{2} = 4x_{A}\n\\end{cases}\n$$\nwe obtain  \n$$\n\\begin{cases}\nx_{A} = 3 \\\\\nx_{B} = \\frac{1}{3}\n\\end{cases}\n$$\nThus, $ |AB| = x_{A} + x_{B} + 2 = 3 + \\frac{1}{3} + 2 = \\frac{16}{3} $."
  },
  {
    "text": "Given that the line $x - y + m = 0$ intersects the hyperbola $x^{2} - \\frac{y^{2}}{2} = 1$ at two distinct points $A$, $B$, and the midpoint of segment $AB$ lies on the circle $x^{2} + y^{2} = 5$, then the value of $m$ is?",
    "fact_expressions": "G: Hyperbola;H: Circle;l:Line;m: Number;B: Point;A: Point;Expression(G) = (x^2 - y^2/2 = 1);Expression(H) = (x^2 + y^2 = 5);Expression(l)=(x-y+m=0);Intersection(l,G)={A,B};Negation(A=B);PointOnCurve(MidPoint(LineSegmentOf(A,B)),H)",
    "query_expressions": "m",
    "answer_expressions": "pm*1",
    "fact_spans": "[[[14, 42]], [[69, 85]], [[2, 13]], [[88, 91]], [[53, 56]], [[49, 52]], [[14, 42]], [[69, 85]], [[2, 13]], [[2, 56]], [[44, 56]], [[58, 86]]]",
    "query_spans": "[[[88, 95]]]",
    "process": "Solve the system \\begin{cases}x-y+m=0\\\\x^{2}-\\frac{y^{2}}{2}=1\\end{cases}, eliminate $ y $ to obtain $ x^{2}-2mx-m^{2}-2=0 $, $ \\Delta=4m^{2}+4(m^{2}+2)=8(m^{2}+1)>0 $, since $ x_{1}+x_{2}=2m $, the horizontal coordinate of the midpoint of $ AB $ is $ \\frac{2m}{2}=m $, substitute into $ x-y+m=0 $, the vertical coordinate of the midpoint of $ AB $ is $ 2m $. The midpoint of $ AB $ is $ (m,2m) $, substitute into the circle equation $ x^{2}+y^{2}=5 $, we get $ m^{2}+4m^{2}=5 $, thus $ m=\\pm1 $."
  },
  {
    "text": "The asymptotes of the hyperbola $x^{2}-2 y^{2}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - 2*y^2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(2)/2)*x",
    "fact_spans": "[[[0, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 28]]]",
    "process": "From the equation of the hyperbola, we know that $ a=1 $, $ b=\\frac{\\sqrt{2}}{2} $, so the asymptotes of the hyperbola are given by $ y=\\pm\\frac{b}{a}x=\\pm\\frac{\\sqrt{2}}{2}x $."
  },
  {
    "text": "Let $AB$ be a moving chord on the parabola $y^{2}=x$, with $|AB|=2$. Then the minimum distance from the midpoint $M$ of chord $AB$ to the $y$-axis is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = x);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);Abs(LineSegmentOf(A, B)) = 2;M: Point;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Min(Distance(M, yAxis))",
    "answer_expressions": "3/4",
    "fact_spans": "[[[7, 19]], [[7, 19]], [[1, 6]], [[1, 6]], [[1, 23]], [[25, 34]], [[45, 48]], [[37, 48]]]",
    "query_spans": "[[[45, 60]]]",
    "process": "According to the problem, the focus of the parabola $ y^{2} = x $ has coordinates $ \\left( \\frac{1}{4}, 0 \\right) $, and the equation of the directrix is $ x = -\\frac{1}{4} $. By the definition of a parabola, since $ |AB| = 2 $, the sum of the distances from points $ A $ and $ B $ to the directrix is at least 2 (with equality if and only if points $ A $, $ B $, and $ F $ are collinear). Therefore, the minimum distance from the midpoint of chord $ AB $ to the directrix is 1. Hence, the minimum distance from the midpoint of chord $ AB $ to the $ y $-axis is $ 1 - \\frac{1}{4} = \\frac{3}{4} $. [Note] This problem examines the definition of a parabola and tests students' computational ability; correctly applying the definition of a parabola is key."
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with foci $F_{1}$ and $F_{2}$, if $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$ and $\\tan \\angle P F_{1} F_{2}=\\frac{1}{2}$, then the eccentricity of this ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Tan(AngleOf(P, F1, F2)) = 1/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[26, 78], [187, 189]], [[26, 78]], [[28, 78]], [[28, 78]], [[28, 78]], [[28, 78]], [[2, 5]], [[2, 82]], [[7, 14]], [[15, 22]], [[6, 78]], [[84, 143]], [[145, 184]]]",
    "query_spans": "[[[187, 195]]]",
    "process": "From $\\overrightarrow{PF}_{1}\\cdot\\overrightarrow{PF_{2}}=0$, we get $\\angle F_{1}PF_{2}=90^{\\circ}$. Also, $\\tan \\angle PF_{1}F_{2}=\\frac{1}{2}$, $\\therefore \\frac{|PF_{2}|}{|PF_{1}|}=\\frac{1}{2}$, i.e., $|PF_{1}|=2|PF_{2}|$. By the definition of ellipse, $|PF_{1}|+|PF_{2}|=2a$, so $|PF_{2}|=\\frac{2}{3}a$. Thus, $|F_{1}F_{2}|^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}$, i.e., $4c^{2}=(\\frac{4}{3}a)^{2}+(\\frac{2}{3}a)^{2}$. Simplifying gives $\\frac{c^{2}}{a^{2}}=\\frac{5}{9}$, i.e., $e=\\frac{c}{a}=\\frac{\\sqrt{5}}{3}$."
  },
  {
    "text": "Given that the midpoint $M$ of chord $AB$ passing through the focus $F(-1, 0)$ of the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has coordinates $\\left(-\\frac{2}{3},\\frac{1}{3}\\right)$, then the equation of the ellipse $E$ is?",
    "fact_expressions": "E: Ellipse;b: Number;a: Number;A: Point;B: Point;F: Point;M:Point;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (-1, 0);OneOf(Focus(E))=F;PointOnCurve(F,LineSegmentOf(A,B));MidPoint(LineSegmentOf(A,B))=M;Coordinate(M)=(-2/3,1/3);IsChordOf(LineSegmentOf(A,B),E)",
    "query_expressions": "Expression(E)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[3, 61], [122, 127]], [[9, 61]], [[9, 61]], [[77, 82]], [[77, 82]], [[64, 75]], [[85, 88]], [[9, 61]], [[9, 61]], [[3, 61]], [[64, 75]], [[3, 75]], [[2, 82]], [[77, 88]], [[85, 120]], [[3, 82]]]",
    "query_spans": "[[[122, 132]]]",
    "process": ""
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2 p x(p>0)$ be $F$, and let point $A(0,2)$. If the midpoint $B$ of segment $F A$ lies on the parabola, then the distance from $B$ to the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;F: Point;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(A) = (0, 2);Focus(G) = F;MidPoint(LineSegmentOf(F, A)) = B;PointOnCurve(B, G)",
    "query_expressions": "Distance(B, Directrix(G))",
    "answer_expressions": "sqrt(2)*3/4",
    "fact_spans": "[[[1, 22], [56, 59], [67, 70]], [[4, 22]], [[30, 39]], [[26, 29]], [[52, 55], [62, 65]], [[4, 22]], [[1, 22]], [[30, 39]], [[1, 29]], [[42, 55]], [[52, 60]]]",
    "query_spans": "[[[62, 77]]]",
    "process": "According to the parabolic equation, the coordinates of point B can be obtained and substituted into the parabolic equation to solve for $ p $. Then the coordinates of point B and the equation of the directrix of the parabola can be determined, thereby finding the distance from point B to the directrix of the parabola. As given, the coordinates of point F are $ \\left( \\frac{p}{2}, 0 \\right) $, so the coordinates of point B are $ \\left( \\frac{p}{4}, 1 \\right) $. Substituting into the parabolic equation gives $ \\frac{p^{2}}{2} = 1 $, solving yields $ p = \\sqrt{2} $. Therefore, the equation of the directrix of the parabola is $ x = -\\frac{\\sqrt{2}}{2} $. Hence, the distance from point B to the directrix of the parabola is $ \\frac{\\sqrt{2}}{4} + \\frac{\\sqrt{2}}{2} = \\frac{3}{4}\\sqrt{2} $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the common foci of an ellipse and a hyperbola, $P$ is one of their common points, and $\\angle F_{1} P F_{2} = \\frac{\\pi}{3}$, with the eccentricities of the ellipse and hyperbola being $e_{1}$ and $e_{2}$ respectively, find the minimum value of $e_{1}^{2} + e_{2}^{2}$?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;F1: Point;P: Point;F2: Point;e1: Number;e2: Number;Focus(G) = {F1, F2};Focus(H) = {F1, F2};OneOf(Intersection(G, H)) = P;AngleOf(F1, P, F2) = pi/3;Eccentricity(H) = e1;Eccentricity(G) = e2",
    "query_expressions": "Min(e1^2 + e2^2)",
    "answer_expressions": "1 + \\sqrt{3}/2",
    "fact_spans": "[[[21, 24], [83, 86]], [[18, 20], [80, 82]], [[2, 9]], [[30, 33]], [[10, 17]], [[93, 100]], [[102, 109]], [[2, 29]], [[2, 29]], [[30, 41]], [[43, 79]], [[80, 109]], [[80, 109]]]",
    "query_spans": "[[[111, 137]]]",
    "process": "According to the problem, let the semi-major axis of the ellipse be $a_{1}$ and the semi-transverse axis of the hyperbola be $a_{2}$. By the definitions of the ellipse and hyperbola, we have $PF_{1}+PF_{2}=2a_{1}$, $PF_{1}-PF_{2}=2a_{2}$, so $PF_{1}=a_{1}+a_{2}$, $PF_{2}=a_{1}-a_{2}$. Also, $\\angle F_{1}PF_{2}=60^{\\circ}$. By the law of cosines, we obtain $(2c)^{2}=(a_{1}+a_{2})^{2}+(a_{1}-a_{2})^{2}-2(a_{1}+a_{2})(a_{1}-a_{2})\\cos60^{\\circ}$, simplifying yields $4c^{2}=a_{1}^{2}+3a_{2}$, i.e., $\\frac{1}{e_{1}^{2}}+\\frac{3}{e_{2}^{2}}=4$, then $\\frac{1}{4e_{1}^{2}}+\\frac{3}{4e_{2}}=1$. Thus, $e_{1}^{2}+e_{2}^{2}=(\\frac{1}{4e_{1}}+\\frac{3}{4e_{2}})(e_{1}^{2}+e_{2})\\geqslant\\frac{1}{2e_{1}}\\cdot e_{1}+\\frac{\\sqrt{3}}{2e_{2}}\\cdot e_{2}^{2}=1+\\frac{\\sqrt{3}}{2}$"
  },
  {
    "text": "Let the coordinates of point $A$ be $(a, 0)$, $(a \\in {R})$. Then the minimum distance from a point on the curve $y^{2}=2 x$ to point $A$ is?",
    "fact_expressions": "A: Point;a: Real;Coordinate(A) = (a, 0);G: Curve;Expression(G) = (y^2 = 2*x);P: Point;PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, A))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 5], [49, 53]], [[9, 30]], [[1, 30]], [[32, 45]], [[32, 45]], [[47, 48]], [[32, 48]]]",
    "query_spans": "[[[47, 62]]]",
    "process": "Let the distance from a point on the parabola to point A be $ d $, and let the coordinates of an arbitrary point on the parabola be $ P(x, y) $. Then  \n$ |PA|^2 = (x - a)^2 + y^2 = x^2 - (2a - 2)x + a^2 = [x - (a - 1)]^2 + (2a - 1) $.  \nSince $ x \\in [0, +\\infty) $,  \nwhen $ a - 1 \\geqslant 0 $, i.e., $ a \\geqslant 1 $, $ d_{\\min}^2 = 2a - 1 $, $ d_{\\min} = \\sqrt{2a - 1} $;  \nwhen $ a - 1 < 0 $, i.e., $ a < 1 $, at $ x = 0 $, $ d_{\\min}^2 = a^2 $, $ d_{\\min} = |a| $.  \nIn conclusion, the minimum distance from a point on the curve $ y^2 = 2x $ to point A is  \n$$\n\\begin{cases}\n\\sqrt{2a - 1}, & a \\geqslant 1 \\\\\n|a|, & a < 1\n\\end{cases}\n$$"
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ is $x-2 y=0$, find the eccentricity of this hyperbola.",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (x - 2*y = 0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 59], [80, 83]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[2, 59]], [[2, 77]]]",
    "query_spans": "[[[80, 89]]]",
    "process": "According to the problem, the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ are given by $y=\\pm\\frac{b}{a}x$. Since one asymptote of this hyperbola is $x-2y=0$, or $y=\\frac{1}{2}x$, it follows that $\\frac{b}{a}=\\frac{1}{2}$, so $a=2b$. Then $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{5}b$, and the eccentricity of the hyperbola is $e=\\frac{c}{a}=\\frac{\\sqrt{5}b}{2b}=\\frac{\\sqrt{5}}{2}$;"
  },
  {
    "text": "A chord $PQ$ passes through the left focus $F_1$ of the hyperbola $x^{2}-y^{2}=8$, intersecting the left branch at points $P$ and $Q$. If $|PQ|=7$ and $F_2$ is the right focus of the hyperbola, then the perimeter of $\\Delta P F_{2} Q$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;Q: Point;F1:Point;F2: Point;Expression(G) = (x^2 - y^2 = 8);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(F1,LineSegmentOf(P,Q));IsChordOf(LineSegmentOf(P,Q),G);Intersection(LineSegmentOf(P,Q),LeftPart(G))={P,Q};Abs(LineSegmentOf(P,Q))=7",
    "query_expressions": "Perimeter(TriangleOf(P, F2, Q))",
    "answer_expressions": "14+8*sqrt(2)",
    "fact_spans": "[[[1, 19], [72, 75]], [[43, 46]], [[47, 50]], [[23, 30]], [[64, 71]], [[1, 19]], [[1, 30]], [[64, 79]], [[0, 39]], [[1, 39]], [[1, 52]], [[54, 63]]]",
    "query_spans": "[[[81, 104]]]",
    "process": "From the hyperbola equation, we obtain $ a = 2\\sqrt{2} $. Using the definition of the hyperbola, express $ |PF_{2}| $ and $ |QF_{2}| $ in terms of $ |PF_{1}| $ and $ |QF_{1}| $ respectively to get the answer. From the given information, $ a = b = 2\\sqrt{2} $, $ c = 4 $. According to the definition of the hyperbola, $ |PF_{2}| - |PF| = |QF_{2}| - |QF| = 4\\sqrt{2} $. Therefore, $ |PQ| + |PF_{2}| + |QF_{2}| = |PQ| + (4\\sqrt{2} + |PF|) + (4\\sqrt{2} + |QF|) = 2|PQ| + 8\\sqrt{2} = 14 + 8\\sqrt{2} $."
  },
  {
    "text": "The standard equation of an ellipse with foci at $(-4,0)$, $(4, 0)$ and passing through the point $(-4, \\frac{9}{5})$ is?",
    "fact_expressions": "G: Ellipse;A:Point;B:Point;C:Point;Coordinate(A)=(-4, 0);Coordinate(B)=(4, 0);Focus(G)={A,B};Coordinate(C)=(-4,9/5);PointOnCurve(C,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/25+y^2/9=1",
    "fact_spans": "[[[0, 2]], [[8, 16]], [[18, 27]], [[31, 51]], [[8, 16]], [[18, 27]], [[0, 27]], [[31, 51]], [[0, 51]]]",
    "query_spans": "[[[0, 58]]]",
    "process": ""
  },
  {
    "text": "$P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and $M$, $N$ are points on the circles $x^{2}+y^{2}+10 x+21=0$ and $x^{2}+y^{2}-10 x+24=0$, respectively. Then the maximum value of $|P M|-|P N|$ is?",
    "fact_expressions": "G: Hyperbola;H: Circle;C:Circle;P: Point;M: Point;N: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Expression(H) = (10*x + x^2 + y^2 + 21 = 0);Expression(C)=(x^2+y^2-10*x+24=0);PointOnCurve(P, RightPart(G));PointOnCurve(M, H);PointOnCurve(N,C)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "9",
    "fact_spans": "[[[4, 43]], [[60, 84]], [[85, 108]], [[0, 3]], [[50, 53]], [[54, 57]], [[4, 43]], [[60, 84]], [[85, 108]], [[0, 49]], [[50, 111]], [[50, 111]]]",
    "query_spans": "[[[113, 132]]]",
    "process": "According to the problem: the foci of the ellipse are the centers of the two circles. The maximum value of |PM| - |PN| is the maximum value of |PM| minus the minimum value of |PN|. The maximum value of |PM| is |PF_{1}| + 2, and the minimum value of |PN| is |PF_{2}| - 1. Subtracting these gives |PF_{1}| - |PF_{2}| + 3 = 2a + 3 = 6 + 3 = 9."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2 p x(p>0)$ be $F$, and let point $A(0 , 2)$. If the midpoint $B$ of segment $FA$ lies on the parabola, then the distance from $B$ to the directrix of the parabola is?",
    "fact_expressions": "F: Point;A: Point;G: Parabola;p: Number;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(A) = (0, 2);Focus(G) = F;MidPoint(LineSegmentOf(F, A)) = B;PointOnCurve(B, G)",
    "query_expressions": "Distance(B, Directrix(G))",
    "answer_expressions": "3*sqrt(2)/4",
    "fact_spans": "[[[26, 29]], [[30, 41]], [[1, 22], [68, 71], [57, 60]], [[4, 22]], [[63, 66], [53, 56]], [[4, 22]], [[1, 22]], [[30, 41]], [[1, 29]], [[44, 56]], [[53, 61]]]",
    "query_spans": "[[[63, 78]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a point on the right branch of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, then the minimum value of the sum of the distance from $P$ to the line $y=2x$ and the distance from $P$ to the point $F(-2,0)$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P, RightPart(G)) = True;G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1);H: Line;Expression(H) = (y = 2*x);F: Point;Coordinate(F) = (-2, 0)",
    "query_expressions": "Min(Distance(P,H)+Distance(P,F))",
    "answer_expressions": "2+4*sqrt(5)/5",
    "fact_spans": "[[[2, 5], [41, 44], [58, 61]], [[2, 39]], [[6, 34]], [[6, 34]], [[45, 54]], [[45, 54]], [[62, 72]], [[62, 72]]]",
    "query_spans": "[[[41, 83]]]",
    "process": "From $ x^2 - \\frac{y^{2}}{3} = 1 $, we know $ c^{2} = 1 + 3 = 4 $, so the coordinates of the hyperbola's foci are $ (\\pm2,0) $. Since $ |PF| - |PF| = 2 $, $ |PF| + |PD| = 2 + |PF| + |PD| $, clearly points P, D, F' are collinear, and when PF is perpendicular to the line $ y = 2x $, the minimum value of the sum of the distance from P to the line $ y = 2x $ and the distance from P to point $ F(-2,0) $ is: $ 2 + \\frac{|4|}{\\sqrt{1+2^{2}}} = 2 + \\frac{4\\sqrt{5}}{5} $."
  },
  {
    "text": "Given that the hyperbola $ D: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) has asymptotes $ y = \\pm \\sqrt{3}x $, and the left and right foci are $ F_{1} $ and $ F_{2} $ respectively, if $ P $ is any point on the right branch of the hyperbola $ D $, then the range of $ \\frac{|P F_{1}| - |P F_{2}|}{|P F_{1}| + |P F_{2}|} $ is?",
    "fact_expressions": "D: Hyperbola;Expression(D) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Expression(Asymptote(D)) = (y = pm*sqrt(3)*x);F1: Point;F2: Point;LeftFocus(D) = F1;RightFocus(D) = F2;P: Point;PointOnCurve(P, RightPart(D))",
    "query_expressions": "Range((Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)))/(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2))))",
    "answer_expressions": "(0, 1/2]",
    "fact_spans": "[[[26, 88], [118, 124]], [[26, 88]], [[34, 88]], [[34, 88]], [[34, 88]], [[34, 88]], [[2, 88]], [[97, 104]], [[105, 112]], [[26, 112]], [[26, 112]], [[114, 117]], [[114, 131]]]",
    "query_spans": "[[[133, 189]]]",
    "process": ""
  },
  {
    "text": "Point $P$ is an arbitrary point on the curve $x^{2}-y-2 \\ln \\sqrt{x}=0$. Then, the minimum distance from point $P$ to the line $4 x+4 y+1=0$ is?",
    "fact_expressions": "C: Line;Expression(C) = ( 4*x + 4*y + 1 = 0 );D: Curve;Expression(D) = (x^2-y-2*ln(sqrt(x))=0);P: Point;PointOnCurve(P, D) = True",
    "query_expressions": "Min(Distance(P, C))",
    "answer_expressions": "(sqrt(2)/2)*(1+ln(2))",
    "fact_spans": "[[[45, 60]], [[45, 60]], [[5, 33]], [[5, 33]], [[0, 4], [0, 4]], [[0, 38]]]",
    "query_spans": "[[[40, 67]]]",
    "process": ""
  },
  {
    "text": "If for any point on the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the absolute value of the difference of the distances to the two foci is $6$, and the eccentricity is $2$, then what is the standard equation of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;D: Point;PointOnCurve(D, C) = True;P1: Point;P2: Point;Focus(C) = {P1, P2};Abs(Distance(D, P1) - Distance(D, P2)) = 6;Eccentricity(C) = 2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/9 - y^2/27 = 1",
    "fact_spans": "[[[1, 62], [95, 101]], [[1, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [], [[1, 67]], [], [], [[1, 71]], [[1, 84]], [[1, 93]]]",
    "query_spans": "[[[95, 108]]]",
    "process": "Since the absolute value of the difference of distances from any point on the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) to the two foci is 6, we have $ 2a = 6 $, so $ a = 3 $. Also, since $ e = \\frac{c}{a} = 2 $, it follows that $ c = 6 $, and thus $ b^{2} = c^{2} - a^{2} = 27 $. Therefore, the standard equation of the hyperbola $ C $ is $ \\frac{x^{2}}{9} - \\frac{y^{2}}{27} = 1 $."
  },
  {
    "text": "Let the right focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ coincide with the focus of the parabola $y^{2}=16x$, and the eccentricity is $\\frac{\\sqrt{6}}{3}$. Then the equation of this ellipse is?",
    "fact_expressions": "H: Ellipse;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;G: Parabola;Expression(G) = (y^2 = 16*x);RightFocus(H) = Focus(G);Eccentricity(H) = sqrt(6)/3",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2/24 + y^2/8 = 1",
    "fact_spans": "[[[1, 53], [106, 108]], [[1, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[58, 73]], [[58, 73]], [[1, 78]], [[1, 103]]]",
    "query_spans": "[[[106, 113]]]",
    "process": "From the given condition, the focus of the parabola $ y^{2} = 16x $ is $ (4,0) $, so $ c = 4 $. Since $ e = \\frac{c}{a} = \\frac{4}{a} = \\frac{\\sqrt{6}}{3} $, we have $ a = 2\\sqrt{6} $, thus $ b^{2} = a^{2} - c^{2} = 8 $. Therefore, the equation of the ellipse is $ \\frac{x^{2}}{24} + \\frac{y^{2}}{8} = 1 $. Answer: $ \\frac{x^{2}}{24} + \\frac{y^{2}}{4} = 1 $"
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ with focus $F$, and a point $M(x_{0}, 2 \\sqrt{2})$ on the parabola. A circle centered at $M$ passes through the origin $O$ and is tangent to the directrix of the parabola, with point of tangency $H$. The line segment $H F$ intersects the parabola at point $B$. Then $\\frac{|\\overrightarrow{B F}|}{|\\overrightarrow{H B}|}$=?",
    "fact_expressions": "G: Parabola;X: Circle;p: Number;H: Point;F: Point;O: Origin;M: Point;B: Point;x0: Number;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(M) = (x0, 2*sqrt(2));Focus(G) = F;PointOnCurve(M, G);Center(X) = M;PointOnCurve(O, X);IsTangent(Directrix(G), X);TangentPoint(Directrix(G), X)=H;Intersection(LineSegmentOf(H, F), G) = B",
    "query_expressions": "Abs(VectorOf(B, F))/Abs(VectorOf(H, B))",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[2, 23], [55, 58], [81, 84], [105, 108]], [[70, 71]], [[5, 23]], [[93, 96]], [[27, 30]], [[73, 78]], [[31, 54], [63, 66]], [[109, 113]], [[32, 54]], [[5, 23]], [[2, 23]], [[31, 54]], [[2, 30]], [[31, 61]], [[62, 71]], [[70, 78]], [[70, 89]], [[70, 96]], [[97, 113]]]",
    "query_spans": "[[[115, 172]]]",
    "process": "Analysis shows that $\\triangle MFO$ is an isosceles triangle, yielding $M\\left(\\frac{p}{4},2\\sqrt{2}\\right)$. Substituting the coordinates of point $M$ into the parabola's equation allows solving for $p$, thus obtaining the parabola's equation and the coordinates of point $P$. The coordinates of point $H$ are found. Let point $B\\left(\\frac{t^{2}}{8},t\\right)$, where $t>0$. Analysis shows $\\overrightarrow{FH} \\parallel \\overrightarrow{FB}$. Using the coordinate representation of collinear planar vectors, solve for $t$, thereby obtaining the result. From the definition of the parabola combined with the given conditions, $|MF|=|MO|$, so $\\triangle MFO$ is an isosceles triangle. The focus of the parabola $y^{2}=2px$ ($p>0$) is $F\\left(\\frac{p}{2},0\\right)$, hence $x_{0}=\\frac{p}{4}$, i.e., point $M\\left(\\frac{p}{4},2\\sqrt{2}\\right)$. Since point $P$ lies on the parabola $y^{2}=2px$ ($p>0$), then $2p \\cdot \\frac{p}{4}=8$, solving gives $p=4$. Therefore, the parabola's equation is $y^{2}=8x$, so point $M(1,2\\sqrt{2})$, $F(2,0)$. Since the circle centered at $M$ with radius $|MF|$ is tangent to the line $x=-2$ at point $H$, then $H(-2,2\\sqrt{2})$. Let point $B\\left(\\frac{t^{2}}{8},t\\right)$, where $t>0$, $\\overrightarrow{FB}=\\left(\\frac{t^{2}}{8}-2,t\\right)$, $\\overrightarrow{FH}=(-4,2\\sqrt{2})$. From the condition $\\overrightarrow{FH} \\parallel \\overrightarrow{FB}$, we have $2\\sqrt{2}\\left(\\frac{t^{2}}{8}-2\\right)=-4t$, simplifying yields $t^{2}+8\\sqrt{2}t-16=0$, solving gives $t=4(\\sqrt{3}-\\sqrt{2})$. Therefore, $\\frac{|\\overrightarrow{BF}|}{|\\overrightarrow{HB}|}=\\frac{t}{|t-2\\sqrt{2}|}=\\frac{4(\\sqrt{3}-\\sqrt{2})}{6\\sqrt{2}-4\\sqrt{3}}=\\frac{4(\\sqrt{3}-\\sqrt{2})}{2\\sqrt{6}(\\sqrt{3}-\\sqrt{2})}=\\frac{\\sqrt{6}}{3}$."
  },
  {
    "text": "Given that the hyperbola shares a common focus with the ellipse $\\frac{x^{2}}{40}+\\frac{y^{2}}{15}=1$ and has an eccentricity of $\\frac{5}{3}$, then its standard equation is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/40 + y^2/15 = 1);Focus(G)=Focus(H);Eccentricity(G)=5/3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/16 = 1",
    "fact_spans": "[[[2, 5], [71, 72]], [[6, 45]], [[6, 45]], [[2, 50]], [[2, 68]]]",
    "query_spans": "[[[71, 78]]]",
    "process": "The hyperbola shares common foci with the ellipse \\frac{x^{2}}{40}+\\frac{y^{2}}{15}=1, giving c=5. The eccentricity of the hyperbola is \\frac{5}{2}, yielding a=3, then b=4. Thus, the equation of the hyperbola is: \\frac{x^{2}}{0}-\\frac{y^{2}}{1}=1"
  },
  {
    "text": "A line passing through the point $(0,-2)$ intersects the parabola $y^{2}=16x$ at two points $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$, and $y_{1}^{2}-y_{2}^{2}=1$. Then the area of $\\triangle OAB$ ($O$ being the origin) is?",
    "fact_expressions": "I: Point;Coordinate(I) = (0, -2);H: Line;PointOnCurve(I, H);G: Parabola;Expression(G) = (y^2 = 16*x);Intersection(H, G) = {A, B};A: Point;B: Point;Coordinate(A) = (x1, y1);x1: Number;y1: Number;Coordinate(B) = (x2, y2);x2: Number;y2: Number;y1^2 - y2^2 = 1;O: Origin",
    "query_expressions": "Area(TriangleOf(O, A, B))",
    "answer_expressions": "1/16",
    "fact_spans": "[[[1, 10]], [[1, 10]], [[11, 13]], [[0, 13]], [[14, 29]], [[14, 29]], [[11, 66]], [[30, 47]], [[49, 66]], [[30, 47]], [[30, 47]], [[30, 47]], [[49, 66]], [[49, 66]], [[49, 66]], [[70, 93]], [[113, 116]]]",
    "query_spans": "[[[95, 127]]]",
    "process": "Let the equation of line AB be x = m(y + 2), (m > 0), A(x_{1}, y_{1}), B(x_{2}, y_{2}). Substituting the equation of line AB into y^{2} = 16x, we obtain y^{2} - 16my - 32m = 0, \\therefore y_{1} + y_{2} = 16m, y_{1}y_{2} = -32m. \\because y_{1} - y_{2} = 1, \\therefore y_{1} - y_{2} = \\frac{y_{1} - y_{2}}{y_{1} + y_{2}} = \\frac{1}{16m}. \\therefore The area of \\Delta OAB = \\frac{1}{2} \\times 2m \\times (y_{1} - y_{2}) = \\frac{1}{16}. The numerical answer is: 1"
  },
  {
    "text": "Given that the line $y = kx - 1$ always has common points with the ellipse $\\frac{x^{2}}{2} + \\frac{y^{2}}{b} = 1$ whose foci lie on the $x$-axis, then the range of values for $b$ is?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*x - 1);k: Number;PointOnCurve(Focus(G), xAxis) = True;G: Ellipse;Expression(G) = (x^2/2 + y^2/b = 1);IsIntersect(H, G) = True;b: Number",
    "query_expressions": "Range(b)",
    "answer_expressions": "[1,2)",
    "fact_spans": "[[[2, 13]], [[2, 13]], [[4, 13]], [[14, 60]], [[23, 60]], [[23, 60]], [[2, 65]], [[67, 70]]]",
    "query_spans": "[[[67, 77]]]",
    "process": "The line $ y = kx - 1 $ always passes through the fixed point $ N(0, -1) $. To ensure that the line $ y = kx - 1 $ always has common points with the ellipse $ \\frac{x^{2}}{2} + \\frac{y^{2}}{b} = 1 $ whose foci lie on the x-axis, it is sufficient that the point $ N(0, -1) $ lies on or inside the ellipse, so $ b \\geqslant 1 $. Since the foci lie on the x-axis, $ b < 2 $. Therefore, $ 1 < b < 2 $. Answer: $ 1 < b < 2 $"
  },
  {
    "text": "Given that $P$ is a point on the parabola $C$: $y^{2}=4 \\sqrt{2} x$, point $M(\\sqrt{2}, 0)$, and $|P M|=4 \\sqrt{2}$, then the area of $\\triangle P O M$ ($O$ is the origin) is?",
    "fact_expressions": "P: Point;C: Parabola;Expression(C) = (y^2 = 4*sqrt(2)*x);PointOnCurve(P, C);M: Point;Coordinate(M) = (sqrt(2), 0);Abs(LineSegmentOf(P, M)) = 4*sqrt(2);O: Origin",
    "query_expressions": "Area(TriangleOf(P, O, M))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[2, 5]], [[6, 34]], [[6, 34]], [[2, 37]], [[38, 55]], [[38, 55]], [[57, 75]], [[95, 98]]]",
    "query_spans": "[[[77, 110]]]",
    "process": "\\because the equation of parabola C is y^{2}=4\\sqrt{2}x, \\therefore M(\\sqrt{2},0) is the focus of the parabola. Let P(m,n). According to the definition of the parabola, |PM|=m + \\frac{p}{2}=4\\sqrt{2}, i.e., m+\\sqrt{2}=4\\sqrt{2}. Solving gives m=3\\sqrt{2}. \\because point P lies on parabola C, we have n^{2}=4\\sqrt{2}\\times3\\sqrt{2}=24, \\therefore n=\\pm2\\sqrt{6}. \\because |OM|=\\sqrt{2}, \\therefore the area of \\triangle POF is S=\\frac{1}{2}|OM|\\times|n|=2\\sqrt{3}."
  },
  {
    "text": "Let point $A(-2 , \\sqrt{3})$, $F$ be the right focus of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$, and point $M$ be a moving point on the ellipse. When $|M A|+2|M F|$ takes the minimum value, what are the coordinates of point $M$?",
    "fact_expressions": "G: Ellipse;A: Point;M: Point;F: Point;Expression(G) = (x^2/16 + y^2/12 = 1);Coordinate(A) = (-2, sqrt(3));RightFocus(G) = F;PointOnCurve(M, G);WhenMin(Abs(LineSegmentOf(M, A)) + 2*Abs(LineSegmentOf(M, F)))",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(2*sqrt(3),sqrt(3))",
    "fact_spans": "[[[27, 66], [76, 78]], [[1, 21]], [[71, 75], [103, 107]], [[23, 26]], [[27, 66]], [[1, 21]], [[23, 70]], [[71, 81]], [[82, 102]]]",
    "query_spans": "[[[103, 112]]]",
    "process": ""
  },
  {
    "text": "The circle centered at the left focus $F(-c, 0)$ of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with radius $c$ intersects the left directrix of the ellipse at two distinct points. Find the range of values for the eccentricity of the ellipse.",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;c: Number;Coordinate(F) = (-c, 0);LeftFocus(G) = F;H: Circle;Center(H) = F;Radius(H) = c;NumIntersection(H, LeftDirectrix(G)) = 2",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(\\sqrt{2}/2, 1)",
    "fact_spans": "[[[1, 53], [82, 84], [98, 100]], [[1, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[57, 69]], [[73, 76]], [[57, 69]], [[1, 69]], [[80, 81]], [[0, 81]], [[73, 81]], [[80, 95]]]",
    "query_spans": "[[[98, 111]]]",
    "process": ""
  },
  {
    "text": "From a point $P$ on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, draw $P M \\perp x$-axis at $M$ ($M$, $P$ not coincident), and let $A_{1} A_{2}$ be the major axis of the ellipse. Then the value of $\\frac{|M A_{1}| \\cdot|M A_{2}|}{|M P|^{2}}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;P: Point;PointOnCurve(P, G);M: Point;IsPerpendicular(LineSegmentOf(P, M), xAxis);FootPoint(LineSegmentOf(P, M), xAxis) = M;Negation(M=P);A1: Point;A2: Point;MajorAxis(G) = LineSegmentOf(A1, A2);PointOnCurve(P, LineSegmentOf(P, M))",
    "query_expressions": "(Abs(LineSegmentOf(M, A1))*Abs(LineSegmentOf(M, A2)))/(Abs(LineSegmentOf(M, P)))^2",
    "answer_expressions": "a^2/b^2",
    "fact_spans": "[[[1, 53], [105, 107]], [[1, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[55, 59], [83, 86]], [[1, 59]], [[75, 78], [79, 82]], [[60, 74]], [[60, 78]], [[79, 89]], [[91, 104]], [[91, 104]], [[91, 110]], [[0, 73]]]",
    "query_spans": "[[[112, 160]]]",
    "process": "Let P(x,y), then M(x,0), |MA_{1}|\\cdot|MA_{2}|=(x+a)(a-x)=a^{2}-x^{2}, |MP|^{2}=y^{2}, \\frac{|MA_{1}|\\cdot|MA_{2}|}{|MP|^{2}}=\\frac{a^{2}-x^{2}}{y^{2}}=\\frac{a^{2}-x^{2}}{b^{2}-\\frac{b^{2}}{a^{2}}x^{2}}=\\frac{a^{2}}{b^{2}}. The correct result of this problem: \\frac{a^{2}}{b^{2}}"
  },
  {
    "text": "Given the circle $(x+2)^{2}+y^{2}=64$ with center $M$, let $A$ be any point on the circle, and let $N(2,0)$. The perpendicular bisector of segment $AN$ intersects $MA$ at point $P$. What is the equation of the locus of the moving point $P$?",
    "fact_expressions": "G: Circle;Expression(G) = (y^2 + (x + 2)^2 = 64);M: Point;Center(G) = M;A: Point;PointOnCurve(A, G);N: Point;Coordinate(N) = (2, 0);P: Point;Intersection(PerpendicularBisector(LineSegmentOf(A, N)), LineSegmentOf(M, A)) = P",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2/16+y^2/12=1",
    "fact_spans": "[[[2, 23], [36, 37]], [[2, 23]], [[27, 30]], [[2, 30]], [[32, 35]], [[32, 41]], [[43, 52]], [[43, 52]], [[73, 77], [81, 84]], [[53, 77]]]",
    "query_spans": "[[[81, 91]]]",
    "process": "According to the problem, the circle $(x+2)^{2}+y^{2}=64$ has center $M(-2,0)$, and point $N(2,0)$. The perpendicular bisector of segment $AN$ intersects $MA$ at point $P$. Therefore, $P$ lies on the perpendicular bisector of $AN$, so $PA=PN$. Also, since $|AM|=8$, the point $P$ satisfies $|PM|+|PN|=8>4$. According to the definition of an ellipse, the locus of point $P$ is an ellipse with foci $M$ and $N$, where $2a=8$, $2c=4$, giving $a=4$, $c=2$, so $b=\\sqrt{a^{2}-c^{2}}=\\sqrt{12}$. Thus, the equation of the ellipse is $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$."
  },
  {
    "text": "The sum of the distances from two points $A$ and $B$ on the parabola $y^{2}=2 x$ to the focus $F$ is $5$. Then, what is the horizontal coordinate of the midpoint $M$ of segment $A B$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);F: Point;Focus(G) = F;Distance(A, F) + Distance(B, F) = 5;M: Point;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "XCoordinate(M)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[18, 21]], [[22, 25]], [[0, 25]], [[0, 25]], [[28, 31]], [[0, 31]], [[18, 40]], [[52, 55]], [[42, 55]]]",
    "query_spans": "[[[52, 61]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4+k}=1$ with foci on the $x$-axis is $\\frac{2}{3}$. Find the value of the real number $k$.",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (x^2/9 + y^2/(k + 4) = 1);PointOnCurve(Focus(G), xAxis);Eccentricity(G) = 2/3",
    "query_expressions": "k",
    "answer_expressions": "1",
    "fact_spans": "[[[9, 48]], [[68, 73]], [[9, 48]], [[0, 48]], [[9, 66]]]",
    "query_spans": "[[[68, 77]]]",
    "process": "Since the foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4+k}=1$ lie on the $x$-axis, its semi-major axis length is $a$, semi-minor axis length is $b$, semi-focal distance is $c$, and eccentricity is $e$. Thus, we have $a^{2}=9$, $b^{2}=4+k$, $e=\\frac{2}{3}$. Therefore: $e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{a^{2}-b^{2}}{a^{2}}=\\frac{5-k}{9}=\\frac{4}{9}$. Solving gives $k=1$. Hence, the value of the real number $k$ is $1$."
  },
  {
    "text": "If the focus of the parabola $x^{2}=2 p y(p>0)$ coincides with a vertex of the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1$, then the distance from the focus to the directrix of this parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 2*(p*y));p: Number;p>0;H: Ellipse;Expression(H) = (x^2/3 + y^2/4 = 1);Focus(G) = OneOf(Vertex(H))",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 22], [73, 76]], [[1, 22]], [[4, 22]], [[4, 22]], [[26, 63]], [[26, 63]], [[1, 70]]]",
    "query_spans": "[[[73, 87]]]",
    "process": "First consider which vertex of the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1$ coincides with the focus of the parabola $x^{2}=2py$ $(p>0)$, then calculate the distance from the focus to the directrix of the parabola. The focus of the parabola $x^{2}=2py$ $(p>0)$ has coordinates $(0,\\frac{p}{2})$, and it coincides with a vertex of the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1$. Therefore, the coincident point is the upper vertex of the ellipse, with coordinates $(0,2)$. Hence, $\\frac{p}{2}=2$, solving gives $p=4$. Thus, the directrix of the parabola is $y=-\\frac{p}{2}=-2$, so the distance from the focus to the directrix of the parabola is $2+2=4$."
  },
  {
    "text": "The center $C$ of circle $C$ lies on the parabola $y^{2}=2 x$, and the circle $C$ is tangent to the $y$-axis at point $A$, intersecting the $x$-axis at points $P$ and $Q$. If $\\overrightarrow{O C} \\cdot \\overrightarrow{O A}=9$ ($O$ is the origin), then $|P Q|=$?",
    "fact_expressions": "C1: Circle;C: Point;Center(C1) = C;G: Parabola;Expression(G) = (y^2 = 2*x);PointOnCurve(C, G) = True;TangentPoint(C1, yAxis) = A;A: Point;Intersection(C1, xAxis) = {P, Q};P: Point;Q: Point;O: Origin;DotProduct(VectorOf(O, C), VectorOf(O, A)) = 9",
    "query_expressions": "Abs(LineSegmentOf(P, Q))",
    "answer_expressions": "3*sqrt(5)",
    "fact_spans": "[[[0, 4], [28, 32]], [[7, 10]], [[0, 10]], [[11, 25]], [[11, 25]], [[7, 26]], [[28, 44]], [[40, 44]], [[28, 62]], [[53, 56]], [[57, 60]], [[117, 120]], [[64, 116]]]",
    "query_spans": "[[[128, 137]]]",
    "process": "Without loss of generality, assume point C is in the first quadrant, let C(\\frac{y_{0}^{2}}{2}, y_{0}), then A(0, y_{0}), hence \\overrightarrow{OC} \\cdot \\overrightarrow{OA} = (\\frac{y_{0}^{2}}{2}, y_{0}) \\cdot (0, y_{0}) = y_{0}^{2} = 9, solving gives y_{0} = 3, thus the center C(\\frac{9}{2}, 3). Therefore, the radius of circle C equals \\frac{9}{2}. So the equation of circle C is (x - \\frac{9}{2})^{2} + (y - 3)^{2} = \\frac{81}{4}. When y = 0, x = \\frac{3\\sqrt{5} + 9}{2}, so |PQ| = |\\frac{-3\\sqrt{5} + 9}{2} - \\frac{-3\\sqrt{5} + 9}{2}| = 3\\sqrt{5}"
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the hyperbola $x^{2}-\\frac{y^{2}}{24}=1$, and let $P$ be a common point of the hyperbola and the ellipse $\\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1$. Then the area of $\\triangle P F_{1} F_{2}$ is equal to?",
    "fact_expressions": "G: Hyperbola;P: Point;H: Ellipse;F1: Point;F2: Point;Expression(G) = (x^2 - y^2/24 = 1);Expression(H) = (x^2/49 + y^2/24 = 1);Focus(G) = {F1, F2};OneOf(Intersection(G, H)) = P",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "24",
    "fact_spans": "[[[17, 46], [56, 59]], [[52, 55]], [[60, 99]], [[1, 8]], [[9, 16]], [[17, 46]], [[60, 99]], [[1, 51]], [[52, 105]]]",
    "query_spans": "[[[107, 138]]]",
    "process": "Since the hyperbola equation is $x^{2}-\\frac{y^{2}}{24}=1$, we have $c^{2}=a^{2}+b^{2}=25$, $c=5$, so the focal distance $|F_{1}F_{2}|=2c=10$. Solving $\\begin{cases}x^{2}-\\frac{y^{2}}{24}=1\\\\\\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1\\end{cases}$ yields $y=\\frac{24}{5}$ (without loss of generality, take the positive value), so the area of $APF_{1}F_{2}$ equals"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$, and $P$ is a point on the right branch of the hyperbola such that $|P F_{1}|=2|P F_{2}|$, then the area of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2/4 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "4",
    "fact_spans": "[[[18, 46], [57, 60]], [[53, 56]], [[2, 9]], [[10, 17]], [[18, 46]], [[2, 52]], [[2, 52]], [[53, 65]], [[67, 89]]]",
    "query_spans": "[[[91, 118]]]",
    "process": "By the given condition, |PF| = 2|PF₂|. Also, |PF₁| - |PF₂| = 2, so |PF| = 4, |PF₂| = 2. Moreover, |F₁F₂| = 2√5, so |PF₁|² + |PF₂|² = |F₁F₂|². Hence, ∠F₁PF₂ = π/2. Therefore, S△PF₁F₂ = (1/2)·|PF₁|·|PF₂| = 4. [Note: This problem mainly examines the standard equation of a hyperbola, the definition of a hyperbola, and the area of a triangle; it is a medium-difficulty problem.]"
  },
  {
    "text": "From the left focus $F$ of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, draw a chord $AB$ with an inclination angle of $\\frac{\\pi}{6}$. Then $|AB|=$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1);F: Point;LeftFocus(G) = F;A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);Inclination(LineSegmentOf(A, B)) = pi/6;PointOnCurve(F, LineSegmentOf(A, B))",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 29]], [[1, 29]], [[33, 36]], [[1, 36]], [[58, 63]], [[58, 63]], [[1, 63]], [[37, 63]], [[0, 63]]]",
    "query_spans": "[[[65, 75]]]",
    "process": "\\because F_{1}(-2,0), let the equation of AB be: y=\\frac{\\sqrt{3}}{3}(x+2), substituting into x^{2}-\\frac{y^{2}}{3}=1 yields: 8x^{2}-4x-13=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), then x_{1}+x_{2}=\\frac{1}{2}, x_{1}x_{2}=-\\frac{13}{8}. \\therefore |AB|=\\sqrt{1+k^{2}}\\cdot|x_{1}-x_{2}|=\\sqrt{1+\\frac{1}{3}}\\cdot\\frac{\\sqrt{16+4\\times8\\times13}}{8}=3."
  },
  {
    "text": "If the focal distance of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ is $1$, then $m$=?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/4 + y^2/m = 1);FocalLength(G) = 1",
    "query_expressions": "m",
    "answer_expressions": "{15/4, 17/4}",
    "fact_spans": "[[[1, 38]], [[47, 50]], [[1, 38]], [[1, 45]]]",
    "query_spans": "[[[47, 52]]]",
    "process": "\\because the ellipse \\frac{x^2}{4} + \\frac{y^2}{m} = 1 has a focal distance of 1, when the foci are on the x-axis, a^2 = 4, b^2 = m, \\therefore 2c = 2\\sqrt{a^2 - b^2} = 2\\sqrt{4 - m} = 1, solving gives: m = \\frac{15}{4}; when the foci are on the y-axis, a^2 = m, b^2 = 4, \\therefore 2c = 2\\sqrt{a^2 - b^2} = 2\\sqrt{m - 4} = 1, solving gives: m = \\frac{17}{4}"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x$ ($p>0$) with focus $F$ and directrix $l$, a line passing through $F$ intersects the parabola $C$ at points $P$ and $Q$, and intersects $l$ at point $A$. If $\\overrightarrow{P F}=3 \\overrightarrow{F Q}$, then $\\frac{|A Q|}{|Q F|}$=?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;P: Point;F: Point;Q: Point;A: Point;l: Line;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;Directrix(C) = l;PointOnCurve(F, G);Intersection(G, C) = {P, Q};Intersection(G, l) = A;VectorOf(P, F) = 3*VectorOf(F, Q)",
    "query_expressions": "Abs(LineSegmentOf(A, Q))/Abs(LineSegmentOf(Q, F))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 29], [52, 58]], [[10, 29]], [[49, 51]], [[59, 62]], [[33, 36], [45, 48]], [[63, 66]], [[74, 78]], [[40, 43], [70, 73]], [[10, 29]], [[2, 29]], [[2, 36]], [[2, 43]], [[44, 51]], [[49, 68]], [[49, 78]], [[80, 125]]]",
    "query_spans": "[[[127, 150]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively. A line passing through point $F_{2}$ intersects the right branch of hyperbola $C$ at points $P$ and $Q$, and $(\\overrightarrow{F_{1} P}+\\overrightarrow{F_{1} Q}) \\cdot \\overrightarrow{P Q}=0$. A line $l$ is drawn through the right vertex of hyperbola $C$ parallel to one asymptote of hyperbola $C$. If line $l$ intersects segment $P Q$ at point $M$ and $|Q M|=3|P M|$, then the eccentricity $e$ of hyperbola $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;a: Number;b: Number;G: Line;Q: Point;P: Point;F1: Point;M: Point;F2: Point;e: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F2, G);Intersection(G, RightPart(C)) = {P, Q};DotProduct((VectorOf(F1, P) + VectorOf(F1, Q)), VectorOf(P, Q)) = 0;PointOnCurve(RightVertex(C), l);IsParallel(l, OneOf(Asymptote(C)));Intersection(l, LineSegmentOf(P, Q)) = M;Abs(LineSegmentOf(Q, M)) = 3*Abs(LineSegmentOf(P, M));Eccentricity(C) = e",
    "query_expressions": "e",
    "answer_expressions": "5/3",
    "fact_spans": "[[[235, 240], [242, 247]], [[20, 82], [102, 108], [208, 214], [222, 228], [278, 284]], [[28, 82]], [[28, 82]], [[99, 101]], [[116, 119]], [[112, 115]], [[2, 9]], [[256, 260]], [[10, 17], [90, 98]], [[288, 291]], [[28, 82]], [[28, 82]], [[20, 82]], [[2, 88]], [[2, 88]], [[89, 101]], [[99, 121]], [[123, 205]], [[207, 240]], [[219, 240]], [[242, 260]], [[262, 276]], [[278, 291]]]",
    "query_spans": "[[[288, 293]]]",
    "process": ""
  },
  {
    "text": "Given foci $F_{1}(5,0)$, $F_{2}(-5,0)$, and a point $P$ on the hyperbola such that the absolute difference of distances from $P$ to $F_{1}$ and $F_{2}$ is $6$, what is the standard equation of the hyperbola?",
    "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;Coordinate(F1) = (5, 0);Coordinate(F2) = (-5, 0);Focus(G)={F1,F2};PointOnCurve(P,G);Abs(Distance(P,F1)-Distance(P,F2))=6;P:Point",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/16 = 1",
    "fact_spans": "[[[32, 35], [72, 75]], [[4, 16], [43, 50]], [[17, 31], [51, 58]], [[4, 16]], [[17, 31]], [[2, 35]], [[32, 42]], [[39, 71]], [[39, 42]]]",
    "query_spans": "[[[72, 82]]]",
    "process": "Since the foci of the hyperbola are $F_{1}(5,0)$, $F_{2}(-5,0)$, its equation can be written as $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, and $a^{2}+b^{2}=25$. According to the definition of a hyperbola, from the given condition we have: $2a=6$, so $a=3$, hence $a^{2}=9$, $b^{2}=16$. Therefore, the required equation of the hyperbola is: $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ with focus $F$, directrix $l$: $x=-\\frac{3}{2}$, point $M$ lies on the parabola $C$, point $A$ lies on the directrix $l$. If $M A \\perp l$ and the slope of line $A F$ is $k_{A F}=-\\sqrt{3}$, then the area of $\\Delta A F M$ is?",
    "fact_expressions": "C: Parabola;p: Number;A: Point;F: Point;M: Point;l:Line;p>0;Expression(C) = (y^2=2*(p*x));Focus(C) = F;Directrix(C)=l;Expression(l)=(x=-3/2);PointOnCurve(M, C);PointOnCurve(A, l);IsPerpendicular(LineSegmentOf(M,A),l);Slope(LineOf(A,F))=k1;k1=-sqrt(3);k1:Number",
    "query_expressions": "Area(TriangleOf(A, F, M))",
    "answer_expressions": "9*sqrt(3)",
    "fact_spans": "[[[2, 28], [65, 71]], [[10, 28]], [[73, 77]], [[32, 35]], [[60, 64]], [[38, 59], [80, 83]], [[10, 28]], [[2, 28]], [[2, 35]], [[2, 59]], [[38, 59]], [[60, 72]], [[73, 84]], [[86, 99]], [[101, 130]], [[111, 130]], [[111, 130]]]",
    "query_spans": "[[[132, 151]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{n}=1$ has an eccentricity of $2$, and one of its foci coincides with the focus of the parabola $y^{2}=4 x$. What is the value of $m n$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/n + x^2/m = 1);Eccentricity(G) = 2;m: Number;n: Number;H: Parabola;Expression(H) = (y^2 = 4*x);OneOf(Focus(G)) = Focus(H)",
    "query_expressions": "m*n",
    "answer_expressions": "3/16",
    "fact_spans": "[[[0, 38]], [[0, 38]], [[0, 46]], [[74, 79]], [[74, 79]], [[53, 67]], [[53, 67]], [[0, 72]]]",
    "query_spans": "[[[74, 83]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{9}=1$, and the line $y=x-4$ passes through one focus of the ellipse $C$, then the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/9 + x^2/a^2 = 1);a: Number;G: Line;Expression(G) = (y = x - 4);PointOnCurve(OneOf(Focus(C)),G) = True",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "4/5",
    "fact_spans": "[[[2, 48], [60, 65], [72, 77]], [[2, 48]], [[9, 48]], [[49, 58]], [[49, 58]], [[49, 70]]]",
    "query_spans": "[[[72, 83]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the right focus of the hyperbola $x^{2}-\\frac{y^{2}}{8}=1$, $P$ lies on the left branch of the hyperbola, and $A(0,6 \\sqrt{6})$, when the perimeter of $\\triangle A P F$ is minimized, what is the area of this triangle?",
    "fact_expressions": "G: Hyperbola;A: Point;P: Point;F: Point;Expression(G) = (x^2 - y^2/8 = 1);Coordinate(A) = (0, 6*sqrt(6));RightFocus(G) = F;PointOnCurve(P, LeftPart(G));WhenMin(Perimeter(TriangleOf(A, P, F)))",
    "query_expressions": "Area(TriangleOf(A,P,F))",
    "answer_expressions": "12*sqrt(6)",
    "fact_spans": "[[[5, 33], [42, 45]], [[50, 67]], [[38, 41]], [[2, 5]], [[5, 33]], [[50, 67]], [[2, 37]], [[38, 49]], [[68, 93]]]",
    "query_spans": "[[[68, 103]]]",
    "process": "By the given condition, let F' be the left focus, then the perimeter of △APF is |AF| + |AP| + |PF| = |AF| + |AP| + |PF'| + 2 ≥ |AF| + |AF'| + 2 (equality holds when points A, P, F are collinear). The equation of line AF is $\\frac{x}{-3}+\\frac{y}{6\\sqrt{6}}=1$. Solving this together with $x^{2}-\\frac{y^{2}}{8}=1$ yields $y^{2}+6\\sqrt{6}y-96=0$, thus the y-coordinate of point P is $2\\sqrt{6}$. Therefore, when the perimeter of △APF is minimized, the area of the triangle is $\\frac{1}{2}\\times6\\times6\\sqrt{6}-\\frac{1}{2}\\times6\\times2\\sqrt{6}=12\\sqrt{6}$."
  },
  {
    "text": "Let the parabola $C$: $y^{2}=4x$ have focus $F$, and let a line $l$ passing through $F$ intersect $C$ at points $A$ and $B$, such that $|AF|-|BF|=\\frac{3}{2}$. Then $\\frac{|AF|}{|BF|}=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, F)) - Abs(LineSegmentOf(B, F)) = 3/2",
    "query_expressions": "Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(B, F))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 20], [40, 43]], [[1, 20]], [[24, 27], [29, 33]], [[1, 27]], [[34, 39]], [[28, 39]], [[46, 49]], [[50, 53]], [[34, 53]], [[55, 80]]]",
    "query_spans": "[[[82, 105]]]",
    "process": "Let the equation of line AB be $ y = k(x - 1) $. Combining it with the parabola equation yields Vieta's formulas. From the definition of the parabola, we have: $ |AF| = x_{1} + 1 $, $ |BF| = x_{2} + 1 $, leading to \n\\[\n\\begin{cases}\nx_{1}x_{2} = 1 \\\\\nx_{1} - x_{2} = \\frac{3}{2}\n\\end{cases}\n\\]\nSolving for $ x_{1}x_{2} $ gives the answer. The parabola $ C: y^{2} = 4x $ has focus $ F(1,0) $. Let the equation of line AB be $ y = k(x - 1) $. Substituting into $ y^{2} = 4x $, we get $ k^{2}x^{2} - (2k^{2} + 4)x + k^{2} = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = 2 + \\frac{4}{k^{2}} $, $ x_{1}x_{2} = 1 $. From the definition of the parabola, we obtain: $ |AF| = x_{1} + 1 $, $ |BF| = x_{2} + 1 $. Given $ |AF| - |BF| = \\frac{3}{2} $, we have $ (x_{1} + 1) - (x_{2} + 1) = \\frac{3}{2} $, i.e., $ x_{1} - x_{2} = \\frac{3}{2} $. From \n\\[\n\\begin{cases}\nx_{1}x_{2} = 1 \\\\\nx_{1} - x_{2} = \\frac{3}{2}\n\\end{cases}\n\\]\nwe get $ \\frac{1}{x_{2}} - x_{2} = \\frac{3}{2} $, solving which yields $ x_{2} = \\frac{1}{2} $ or $ x_{2} = -2 $ (discarded). Thus, $ x_{1} = 2 $. Therefore, $ \\frac{|AF|}{|BF|} = \\frac{x_{1} + 1}{x_{2} + 1} = \\frac{3}{\\frac{1}{2} + 1} = 2 $."
  },
  {
    "text": "The maximum distance from a point on an ellipse to its focus is $5$, and the minimum distance is $3$. What is the eccentricity of this ellipse?",
    "fact_expressions": "G: Ellipse;P: Point;PointOnCurve(P, G);Max(Distance(P, Focus(G))) = 5;Min(Distance(P, Focus(G))) = 3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/4",
    "fact_spans": "[[[0, 2], [5, 6], [27, 29]], [[3, 4]], [[0, 4]], [[3, 17]], [[3, 24]]]",
    "query_spans": "[[[27, 35]]]",
    "process": ""
  },
  {
    "text": "Let point $P$ move on the circle $x^{2}+(y-2)^{2}=1$, and point $Q$ move on the ellipse $\\frac{x^{2}}{9}+y^{2}=1$. Then the maximum value of $|P Q|$ is?",
    "fact_expressions": "P: Point;H: Circle;Expression(H) = (x^2 + (y - 2)^2 = 1);PointOnCurve(P, H);G: Ellipse;Expression(G) = (x^2/9 + y^2 = 1);Q: Point;PointOnCurve(Q, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "1+3*sqrt(6)/2",
    "fact_spans": "[[[1, 5]], [[6, 26]], [[6, 26]], [[1, 29]], [[35, 62]], [[35, 62]], [[30, 34]], [[30, 65]]]",
    "query_spans": "[[[67, 80]]]",
    "process": "Let the coordinates of a point on the ellipse be $ Q(3\\cos\\theta,\\sin\\theta) $. Use the distance formula between two points to obtain the expression for $ |AQ| $ in terms of $ \\theta $. Then, by substitution and using properties of quadratic functions, find the maximum value of $ |AQ| $, and subsequently determine the maximum value of $ |PQ| $. Let the coordinates of a point on the ellipse be $ Q(3\\cos\\theta,\\sin\\theta) $, then  \n$$\n|AQ| = \\sqrt{(3\\cos\\theta)^2 + (\\sin\\theta - 2)^2} = \\sqrt{5 + 8\\cos 2\\theta - 4\\sin\\theta} = \\sqrt{-8\\sin^2\\theta - 4\\sin\\theta + 13}.\n$$\nLet $ t = \\sin\\theta \\in [-1,1] $, $ u = -8t^2 - 4t + 13 $, so $ |AQ| = \\sqrt{u} $. Since $ u = -8t^2 - 4t + 13 $ is a downward-opening parabola with axis of symmetry at $ t = -\\frac{1}{4} $, substituting gives $ |AQ| \\leqslant \\frac{3\\sqrt{6}}{2} $. Hence, the maximum value of $ |PQ| $ is $ 1 + \\frac{3\\sqrt{6}}{2} $.  \n\n【Insight】Setting the coordinates of a point on the ellipse as $ Q(3\\cos\\theta,\\sin\\theta) $ is a parametric approach. For problems involving the distance from a point on an ellipse to a point on another curve, it is common to represent the ellipse using parametric equations, then apply the distance formula between two points or the distance from a point to a line."
  },
  {
    "text": "If an ellipse passes through the point $(2 , 3)$, and its foci are $F_{1}(-2 , 0)$, $F_{2}(2 , 0)$, then the eccentricity of this ellipse is equal to?",
    "fact_expressions": "G: Ellipse;H: Point;F1: Point;F2: Point;Coordinate(H) = (2, 3);Coordinate(F1) = (-2, 0);Coordinate(F2) = (2, 0);PointOnCurve(H, G);Focus(G) = {F1,F2}",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[1, 3], [55, 57]], [[5, 15]], [[20, 36]], [[37, 51]], [[5, 15]], [[20, 36]], [[37, 51]], [[1, 15]], [[1, 51]]]",
    "query_spans": "[[[55, 64]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$, with focus $F$, and let $K$ be the intersection point of the directrix and the $x$-axis. Let $P$ be a point on the parabola $C$, located in the first quadrant. When $\\frac{|PF|}{|PK|}$ attains its minimum value, what are the coordinates of point $P$?",
    "fact_expressions": "C: Parabola;P: Point;F: Point;K: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;Intersection(Directrix(C), xAxis) = K;PointOnCurve(P, C);Quadrant(P) = 1;WhenMin(Abs(LineSegmentOf(P, F))/Abs(LineSegmentOf(P, K)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1, 2)",
    "fact_spans": "[[[2, 21], [48, 54]], [[44, 47], [59, 62], [97, 101]], [[25, 28]], [[40, 43]], [[2, 21]], [[2, 28]], [[2, 43]], [[44, 57]], [[59, 67]], [[68, 96]]]",
    "query_spans": "[[[97, 106]]]",
    "process": "First, find the expression for the ratio of the distance from point P to point K and the focus F, then solve using geometric relationships to find the coordinates of point P when $\\frac{|PF|}{|PK|}$ reaches its minimum value. From the problem, the focus is $F(1,0)$, the directrix equation is $x=-1$. Draw $PM$ perpendicular to the directrix from point $P$, with $M$ as the foot of the perpendicular. $P_{1}KF<\\frac{\\pi}{2}$. Finding the minimum value of $\\cos\\angle PKF$ is equivalent to finding the maximum value of $\\tan\\angle PKF$, that is, $\\tan\\angle PKF = \\frac{y}{x+1} = \\frac{y}{2} \\cdot \\frac{1}{y} \\leqslant 1$. Hence, $\\cos\\angle PKF \\geqslant \\frac{\\sqrt{2}}{2}$, with equality holding if and only if $\\begin{cases}x=1\\\\v=2\\end{cases}$, i.e., $P(1,2)$."
  },
  {
    "text": "If the line $y=2$ and the curve $y=x^{2}-|x|+a$ have two intersection points, then the range of values for $a$ is?",
    "fact_expressions": "G: Line;Expression(G) = (y = 2);H: Curve;Expression(H) = (y = a + x^2 - Abs(x));a: Number;NumIntersection(G, H) = 2",
    "query_expressions": "Range(a)",
    "answer_expressions": "{(-oo, 2), 9/4}",
    "fact_spans": "[[[1, 8]], [[1, 8]], [[9, 26]], [[9, 26]], [[33, 36]], [[1, 31]]]",
    "query_spans": "[[[33, 43]]]",
    "process": ""
  },
  {
    "text": "Given that the hyperbola $C$ has its center at the origin and foci on the coordinate axes, and its eccentricity is $\\sqrt{2}$, what is the equation of the asymptotes of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;O: Origin;Center(C)=O;PointOnCurve(Focus(C),axis);Eccentricity(C)=sqrt(2)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*x",
    "fact_spans": "[[[16, 22], [39, 45]], [[5, 7]], [[2, 22]], [[8, 22]], [[16, 37]]]",
    "query_spans": "[[[39, 53]]]",
    "process": "Since the center of the hyperbola is at the origin and the foci are on the coordinate axes, the equation of the hyperbola can be written as \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a,b>0). Since the eccentricity of hyperbola C is \\sqrt{2}, \\therefore c=\\sqrt{2}a, \\because c^{2}=2a^{2}=a^{2}+b^{2}, \\therefore b=a, \\frac{b}{a}=1, \\therefore the equation of the hyperbola is \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a,b>0), and the asymptotes are y=\\pm\\frac{b}{a}x=\\pm x."
  },
  {
    "text": "Given the curve $C$: $\\frac{x|x|}{4}+y|y|=1$, let point $P(m, n)$ be an arbitrary point on curve $C$. If points $A(-2,1)$ and $B(4,-2)$, then the maximum area of $\\triangle P A B$ is?",
    "fact_expressions": "C: Curve;Expression(C) = (y*Abs(y) + (x*Abs(x))/4 = 1);P: Point;Coordinate(P) = (m, n);PointOnCurve(P, C) = True;A: Point;Coordinate(A) = (-2, 1);B: Point;Coordinate(B) = (4, -2);m: Number;n: Number",
    "query_expressions": "Max(Area(TriangleOf(P, A, B)))",
    "answer_expressions": "3*sqrt(2)",
    "fact_spans": "[[[2, 32], [44, 49]], [[2, 32]], [[33, 43]], [[33, 43]], [[33, 54]], [[56, 66]], [[56, 66]], [[69, 78]], [[69, 78]], [[34, 43]], [[34, 43]]]",
    "query_spans": "[[[80, 105]]]",
    "process": ""
  },
  {
    "text": "Let the right vertex of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ be $A$, and the right focus be $F$. A line passing through point $F$ and parallel to one asymptote of the hyperbola intersects the hyperbola at point $B$. Then the area of $\\triangle A F B$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);A: Point;RightVertex(G)=A;F: Point;RightFocus(G)=F;H: Line;PointOnCurve(F,H);IsParallel(H,OneOf(Asymptote(G)));B: Point;Intersection(H,G)=B",
    "query_expressions": "Area(TriangleOf(A, F, B))",
    "answer_expressions": "32/15",
    "fact_spans": "[[[1, 40], [64, 67], [77, 80]], [[1, 40]], [[45, 48]], [[1, 48]], [[53, 56], [58, 62]], [[1, 56]], [[74, 76]], [[57, 76]], [[62, 76]], [[82, 86]], [[74, 86]]]",
    "query_spans": "[[[88, 110]]]",
    "process": "It is easy to obtain: $\\alpha=3$, $b=4$, then $c=\\sqrt{a^{2}+b^{2}}=5$, $A(3,0)$, $F(5,0)$. The asymptotes of the hyperbola are $y=\\pm\\frac{4}{3}x$, so the line passing through $F(5,0)$ and parallel to the asymptote $y=\\frac{4}{3}x$ has the equation $y=\\frac{4}{3}(x-5)$. Solving the system of equations\n\\[\n\\begin{cases}\n\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1, \\\\\ny=\\frac{4}{3}(x-5),\n\\end{cases}\n\\]\nwe get $B\\left(\\frac{1}{5},-\\frac{32}{15}\\right)$. $S_{\\triangle ABC}=\\frac{1}{2}|AB|\\cdot|y_{B}|=\\frac{1}{2} \\cdot 2 \\cdot \\frac{32}{15}=\\frac{32}{15}$"
  },
  {
    "text": "If the equation $\\frac{x^{2}}{9-k}+\\frac{y^{2}}{k-1}=1$ represents an ellipse, then the range of values for $k$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(9 - k) + y^2/(k - 1) = 1);k: Number",
    "query_expressions": "Range(k)",
    "answer_expressions": "(1,5) + (5,9)",
    "fact_spans": "[[[44, 46]], [[1, 46]], [[48, 51]]]",
    "query_spans": "[[[48, 58]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line perpendicular to the $x$-axis is drawn through $F_{1}$, intersecting the ellipse at point $P$. If $\\angle F_{1} P F_{2}=60^{\\circ}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;P: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;L:Line;PointOnCurve(F1, L);IsPerpendicular(L,xAxis);Intersection(L, G) = P;AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[2, 54], [96, 98], [141, 143]], [[4, 54]], [[4, 54]], [[63, 70], [81, 88]], [[99, 102]], [[71, 78]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 78]], [[2, 78]], [], [[80, 95]], [[80, 95]], [[80, 102]], [[104, 137]]]",
    "query_spans": "[[[141, 149]]]",
    "process": "According to the given condition: $\\triangle F_{1}PF_{2}$ is a right triangle, $\\angle F_{1}PF_{2}=60^{\\circ}$, $\\angle PF_{2}F_{1}=30^{\\circ}$, so $|PF_{1}|:|PF_{2}|:|F_{1}F_{2}|=1:2:\\sqrt{3}$, therefore $e=\\frac{c}{a}=\\frac{2c}{2a}=\\frac{|F_{1}F_{2}|}{|PF_{1}|+|PF_{2}|}=\\frac{\\sqrt{3}}{1+2}=\\frac{\\sqrt{3}}{3}$"
  },
  {
    "text": "The standard equation of a parabola with focus on $x-y-1=0$ is?",
    "fact_expressions": "G: Parabola;H: Line;Expression(H) = (x - y - 1 = 0);PointOnCurve(Focus(G), H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=4*x, x^2=-4*y}",
    "fact_spans": "[[[14, 17]], [[3, 12]], [[3, 12]], [[0, 17]]]",
    "query_spans": "[[[14, 24]]]",
    "process": "Since the parabola has a standard equation, the focus can be determined accordingly. Then, by finding the intersection points of the line with the coordinate axes, the coordinates of the focus can be obtained, leading to the standard equation.  \nBecause the focus coordinates of the parabola are the intersection points of the line $ x - y - 1 = 0 $ with the coordinate axes, the focus coordinates are $ (1, 0) $ and $ (0, -1) $.  \nWhen the focus is $ (1, 0) $, assume the standard form of the parabola is $ y^2 = 2px $, we find $ p = 2 $, so the equation is $ y^{2} = 4x $.  \nWhen the focus is $ (0, -1) $, assume the standard form of the parabola is $ x^{2} = -2py $, we find $ p = 2 $, so the equation is $ x^{2} = -4y $."
  },
  {
    "text": "Draw a line through the focus of the parabola $y^{2}=8x$ intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $|AB|=16$, then $x_{1}+x_{2}$=?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;x1:Number;x2:Number;y2:Number;Expression(G) = (y^2 = 8*x);Coordinate(A) = (x1,x2);Coordinate(B) = (x2,y2);PointOnCurve(Focus(G),H);Intersection(H, G) = {A,B};Abs(LineSegmentOf(A,B))=16",
    "query_expressions": "x1 + x2",
    "answer_expressions": "12",
    "fact_spans": "[[[1, 15], [22, 25]], [[19, 21]], [[26, 44]], [[47, 65]], [[26, 44]], [[47, 65]], [[47, 65]], [[1, 15]], [[26, 44]], [[47, 65]], [[0, 21]], [[19, 67]], [[69, 79]]]",
    "query_spans": "[[[82, 97]]]",
    "process": "A line passing through the focus of the parabola $ y^{2} = 8x $ intersects the parabola at points $ A(x_{1}, y_{1}) $ and $ B(x_{2}, y_{2}) $, yielding $ |AB| = x_{1} + x_{2} + 2 = 16 $. From this, the answer is easily obtained. Given $ p = 4 $, the directrix of the parabola is $ x = -2 $. Since a line passing through the focus of the parabola $ y^{2} = 8x $ intersects the parabola at points $ A(x_{1}, y_{1}) $ and $ B(x_{2}, y_{2}) $, it follows that $ |AB| = x_{1} + x_{2} + 4 = 16 $, solving which gives $ x_{1} + x_{2} = 12 $."
  },
  {
    "text": "A moving point $P(x, y)$ on the plane satisfies $\\sqrt{(x+1)^{2}+y^{2}}+\\sqrt{(x-1)^{2}+y^{2}}=4$, then the trajectory equation of $P$ is?",
    "fact_expressions": "P: Point;x_: Number;y_: Number;Coordinate(P) = (x_, y_);sqrt(y_^2 + (x_ - 1)^2) + sqrt(y_^2 + (x_ + 1)^2) = 4",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2/4+y^2/3=1",
    "fact_spans": "[[[6, 15], [68, 71]], [[6, 15]], [[6, 15]], [[6, 15]], [[17, 66]]]",
    "query_spans": "[[[68, 78]]]",
    "process": "The coordinates of the moving point $ P(x,y) $ satisfy $ \\sqrt{(x+1)^{2}+y^{2}}+\\sqrt{(x-1)^{2}+y^{2}}=4 $. Therefore, the sum of the distances from the moving point $ P(x,y) $ to $ A(-1,0) $ and $ B(1,0) $ is equal to $ 4 > |AB| = 2 $. Hence, the trajectory of the moving point $ P $ is an ellipse with foci at points $ A $ and $ B $. Let its equation be $ \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 $ ($ a>b>0 $). From the problem, we have $ c=1 $, $ 2a=4 $, so $ a=2 $, $ b^{2}=4-1=3 $. Therefore, the trajectory equation of the moving point $ P $ is $ \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{2}$ intersects the right branch of the hyperbola at points $P$ and $Q$. If $|P F_{1}|=|F_{1} F_{2}|$ and $3|P F_{2}|=2|Q F_{2}|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;P: Point;F1: Point;F2: Point;Q: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, H);Intersection(H, RightPart(G)) = {P, Q};Abs(LineSegmentOf(P, F1)) = Abs(LineSegmentOf(F1, F2));3*Abs(LineSegmentOf(P, F2)) = 2*Abs(LineSegmentOf(Q, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "7/5",
    "fact_spans": "[[[2, 58], [95, 98], [167, 170]], [[5, 58]], [[5, 58]], [[92, 94]], [[102, 105]], [[67, 74]], [[75, 82], [84, 91]], [[106, 109]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 82]], [[2, 82]], [[83, 94]], [[92, 111]], [[113, 138]], [[141, 164]]]",
    "query_spans": "[[[167, 176]]]",
    "process": "From the properties of the hyperbola, |PF|=2c, |PF_{2}|=2c-2a, \\therefore aF_{2}=3c-3a, |FQ|=3c-a, (c-a)(5c-7a)=0 \\Rightarrow e=\\frac{c}{a}=\\frac{7}{5}, hence fill in: \\frac{7}{5}"
  },
  {
    "text": "If the ellipse $a x^{2}+b y^{2}=1$ intersects the line $x+y=1$ at points $A$ and $B$, and point $M$ is the midpoint of $AB$, the slope of the line $OM$ ($O$ being the origin) is $\\frac{\\sqrt{2}}{2}$, what is the value of $\\frac{b}{a}$?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;O: Origin;M: Point;A: Point;B: Point;Expression(G) = (a*x^2 + b*y^2 = 1);Expression(H) = (x + y = 1);Intersection(G, H) = {A, B};MidPoint(LineSegmentOf(A, B)) = M;Slope(LineOf(O,M))=sqrt(2)/2",
    "query_expressions": "b/a",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 22]], [[102, 115]], [[102, 115]], [[23, 32]], [[66, 69]], [[44, 48]], [[34, 37]], [[38, 41]], [[1, 22]], [[23, 32]], [[1, 43]], [[44, 57]], [[58, 100]]]",
    "query_spans": "[[[102, 119]]]",
    "process": "Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $x_{M}=\\frac{x_{1}+x_{2}}{2}$, $y_{M}=\\frac{y_{1}+y_{2}}{2}$, $\\frac{y_{M}}{x_{M}}=\\frac{\\sqrt{2}}{2}$, $\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=-1$. From $\\begin{cases}ax_{1}^{2}+by_{1}^{2}=1\\\\ax_{2}^{2}+by_{2}^{2}=1\\end{cases}$, subtracting gives $a(x_{2}^{2}-x_{1}^{2})+b(y_{2}^{2}-y_{1}^{2})=0$, i.e., $a+b\\frac{y_{2}+y_{1}}{x_{2}+x_{1}}\\cdot\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=0$. Therefore, $a+b\\times\\frac{\\sqrt{2}}{2}\\times(-1)=0$, solving yields $\\frac{b}{a}=\\sqrt{2}$."
  },
  {
    "text": "Given that point $P$ lies on the hyperbola $C$: $x^{2}-y^{2}=1$ with foci $F_{1}$ and $F_{2}$, and $|P F_{1}|=3|P F_{2}|$, then the perimeter of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2 = 1);F1: Point;F2: Point;Focus(C)={F1,F2};P: Point;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Perimeter(TriangleOf(P, F1, F2))",
    "answer_expressions": "4+2*sqrt(2)",
    "fact_spans": "[[[27, 50]], [[27, 50]], [[8, 15]], [[16, 23]], [[7, 50]], [[2, 6]], [[2, 54]], [[56, 78]]]",
    "query_spans": "[[[80, 110]]]",
    "process": "Analysis: According to the given conditions, from the standard equation of the hyperbola, we can obtain the values of $ a $ and $ b $. By the definition of the hyperbola, we have $ |PF_{1}| - |PF_{2}| = 2a = 2 $. Given that $ |PF_{1}| = 3|PF_{2}| $, we can calculate $ |PF_{1}| = 3 $, $ |PF_{2}| = 1 $. Also, $ |F_{1}F_{2}| = 2c = 2\\sqrt{2} $. Using the perimeter formula for a triangle, we can obtain the answer. Detailed solution: According to the given conditions, the equation of hyperbola $ C $ is $ x^{2} - y^{2} = 1 $, so $ a = 1 $, $ b = 1 $, then $ c = \\sqrt{2} $. Thus, $ |PF_{1}| - |PF_{2}| = 2a = 2 $. Since $ |PF_{1}| = 3|PF_{2}| $, we get $ |PF_{1}| = 3 $, $ |PF_{2}| = 1 $. Also, since $ c = \\sqrt{2} $, then $ F_{1}F_{2} = 2c = 2\\sqrt{2} $. Therefore, the perimeter of $ \\triangle PF_{1}F_{2} $ is $ |PF_{1}| + |PF_{2}| + |F_{1}F_{2}| = 4 + 2\\sqrt{2} $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ with focus $F$, a line $l$ passing through $F$ intersects the parabola $C$ at points $A$ and $B$. If $|A F|=\\frac{2}{3}$, $|B F|=2$, then $p=$?",
    "fact_expressions": "l: Line;C: Parabola;p: Number;A: Point;F: Point;B: Point;p>0;Expression(C) = (y^2 = 2*p*x);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, F)) = 2/3;Abs(LineSegmentOf(B, F)) = 2",
    "query_expressions": "p",
    "answer_expressions": "1",
    "fact_spans": "[[[41, 46]], [[2, 28], [47, 53]], [[97, 100]], [[54, 57]], [[32, 35], [37, 40]], [[58, 61]], [[10, 28]], [[2, 28]], [[2, 35]], [[36, 46]], [[41, 63]], [[65, 84]], [[86, 95]]]",
    "query_spans": "[[[97, 102]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), then \\because |AF| = \\frac{2}{3}, |BF| = 2 \\therefore according to the definition of parabola, we obtain x_{1} = \\frac{2}{3} - \\frac{p}{2}, x_{2} = 2 - \\frac{p}{2} \\therefore \\frac{y_{2}}{y_{2}} = \\frac{x_{1}}{x_{2}} = \\frac{\\frac{2}{3}}{2} = \\frac{1}{9}, \\therefore 9(\\frac{2}{3} - \\frac{p}{2}) = 2 - \\frac{p}{2} \\therefore p = 1."
  },
  {
    "text": "Draw a straight line through the point $M(1 , 1)$ intersecting the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ at points $A$ and $B$. If point $M$ is exactly the midpoint of chord $AB$, then what is the equation of the line on which $AB$ lies?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;M: Point;H:Line;Expression(G) = (x^2/9 + y^2/4 = 1);Coordinate(M) = (1, 1);PointOnCurve(M,H);Intersection(H,G)={A,B};IsChordOf(LineSegmentOf(A,B),G);MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Expression(OverlappingLine(LineSegmentOf(A,B)))",
    "answer_expressions": "4*x+9*y-13=0",
    "fact_spans": "[[[17, 54]], [[57, 60]], [[61, 64]], [[1, 12], [68, 72]], [[14, 16]], [[17, 54]], [[1, 12]], [[0, 16]], [[14, 66]], [[17, 81]], [[68, 84]]]",
    "query_spans": "[[[87, 100]]]",
    "process": ""
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ has two foci $F_{1}$, $F_{2}$. A line passes through $F_{1}$ and intersects the ellipse at points $P$, $Q$. Then, what is the perimeter of $\\triangle PQF_{2}$?",
    "fact_expressions": "G: Ellipse;H: Line;P: Point;Q: Point;F2: Point;F1: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(G)={F1,F2};PointOnCurve(F1,H);Intersection(H,G)={P,Q}",
    "query_expressions": "Perimeter(TriangleOf(P, Q, F2))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 38], [74, 76]], [[63, 65]], [[77, 80]], [[81, 84]], [[53, 60]], [[43, 50], [66, 73]], [[0, 38]], [[0, 60]], [[63, 73]], [[63, 84]]]",
    "query_spans": "[[[86, 110]]]",
    "process": ""
  },
  {
    "text": "A point $P$ on the hyperbola $\\frac{y^{2}}{16}-\\frac{x^{2}}{9}=1$ is at a distance of $10$ from one focus. What is the distance from point $P$ to the other focus?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/9 + y^2/16 = 1);P: Point;PointOnCurve(P, G) = True;P1: Point;P2: Point;OneOf(Focus(G))=P1;OneOf(Focus(G))=P2;Negation(P1=P2);Distance(P, P1) = 10",
    "query_expressions": "Distance(P, P2)",
    "answer_expressions": "{2,18}",
    "fact_spans": "[[[0, 39]], [[0, 39]], [[42, 45], [61, 65]], [[0, 45]], [], [], [[0, 50]], [[0, 71]], [[0, 71]], [[0, 58]]]",
    "query_spans": "[[[0, 76]]]",
    "process": "In the hyperbola $\\frac{y^{2}}{16}-\\frac{x^{2}}{9}=1$, $2a=8$. Let the distance from point $P$ to the other focus be $d$, then $|10-d|=2a=8$, solving gives: $d=2$ or $18$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, the right vertex is $A$, the right focus is $F$. A circle centered at $A$ with radius $R$ intersects the ellipse at points $B$ and $C$. If the line $BC$ passes through point $F$, then the value of $R$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);A: Point;RightVertex(G) = A;F: Point;RightFocus(G) = F;H: Circle;Center(H) = A;R: Number;Radius(H) = R;B: Point;C: Point;Intersection(H, G) = {B, C};PointOnCurve(F,LineOf(B,C)) = True",
    "query_expressions": "R",
    "answer_expressions": "sqrt(13)/2",
    "fact_spans": "[[[2, 39], [75, 77]], [[2, 39]], [[44, 47], [59, 62]], [[2, 47]], [[53, 56], [99, 103]], [[2, 56]], [[73, 74]], [[58, 74]], [[66, 69], [106, 109]], [[66, 74]], [[80, 83]], [[84, 87]], [[73, 89]], [[91, 103]]]",
    "query_spans": "[[[106, 113]]]",
    "process": "Given A(2,0), F(1,0), since BC passes through the focus F, by symmetry BC\\botx-axis, so |BC|=\\frac{2b^{2}}{a}=\\frac{2\\times3}{2}=3, |FA|=1, thus _{R}=\\sqrt{1^{2}+(\\frac{3}{2})^{2}}=\\frac{\\sqrt{13}}{2}"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ with left and right foci $F_{1}$ and $F_{2}$ respectively, point $P$ is a point on the ellipse, and point $A(-4,4)$. Then the minimum value of $|P A|-|P F_{2}|$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);F2: Point;F1: Point;LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G) = True;P: Point;A: Point;Coordinate(A) = ((-1)*4, 4)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) - Abs(LineSegmentOf(P, F2)))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 39], [69, 71]], [[2, 39]], [[56, 63]], [[48, 55]], [[2, 63]], [[2, 63]], [[64, 74]], [[64, 68]], [[75, 85]], [[75, 85]]]",
    "query_spans": "[[[87, 110]]]",
    "process": "According to the problem, the left focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ is $F_{1}(-1,0)$, the right focus is $F_{2}(1,0)$, point $P$ is a point on the ellipse, and point $A$ lies outside the ellipse. By the definition of an ellipse, we have $|PF_{2}|=4-|PF_{1}|$. Therefore, $|PA|-|PF_{2}|=|PA|+|PF_{1}|-4\\geqslant|AF_{1}|-4=\\sqrt{(-1-(-1))^{2}+(4-0)^{2}}-4=1$, with equality holding if and only if point $P$ is the intersection point of segment $AF_{1}$ and the ellipse. Hence, the minimum value of $|PA|-|PF_{2}|$ is $1$."
  },
  {
    "text": "A point $(2, t)$ on the parabola $y^{2}=2 p x$ is at a distance of $3$ from the point $(\\frac{p}{2}, 0)$. Then $p=?$",
    "fact_expressions": "G: Parabola;p: Number;t:Number;H: Point;I: Point;Expression(G) = (y^2 = 2*(p*x));Coordinate(H) = (2, t);Coordinate(I) = (p/2, 0);PointOnCurve(H, G);Distance(H, I) = 3",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 16]], [[57, 60]], [[19, 27]], [[19, 27]], [[28, 47]], [[0, 16]], [[19, 27]], [[28, 47]], [[0, 27]], [[19, 55]]]",
    "query_spans": "[[[57, 62]]]",
    "process": "From the definition of the parabola, we obtain $2+\\frac{p}{2}=3$, from which the answer can be derived. $(\\frac{p}{2},0)$"
  },
  {
    "text": "Given that point $M$ lies on the parabola $y^{2}=4x$, $F$ is the focus of the parabola, and point $A$ lies on the circle $C$: $(x-3)^{2}+(y-1)^{2}=1$, find the minimum value of $|MA|+|MF|$.",
    "fact_expressions": "G: Parabola;C: Circle;M: Point;A: Point;F: Point;Expression(G) = (y^2 = 4*x);Expression(C) = ((x - 3)^2 + (y - 1)^2 = 1);PointOnCurve(M, G);Focus(G) = F;PointOnCurve(A, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F)))",
    "answer_expressions": "3",
    "fact_spans": "[[[6, 20], [29, 32]], [[41, 69]], [[2, 6]], [[36, 40]], [[25, 28]], [[6, 20]], [[41, 69]], [[2, 24]], [[25, 35]], [[36, 70]]]",
    "query_spans": "[[[72, 90]]]",
    "process": ""
  },
  {
    "text": "Given that point $A(0 , 1)$ lies on the ellipse $x^{2}+4 y^{2}=4$, and $P$ is a moving point on the ellipse, when the length of chord $AP$ is maximized, what are the coordinates of point $P$?",
    "fact_expressions": "G: Ellipse;A: Point;P: Point;Expression(G) = (x^2 + 4*y^2 = 4);Coordinate(A) = (0, 1);PointOnCurve(A, G);PointOnCurve(P, G);IsChordOf(LineSegmentOf(A,P),G);WhenMax(Length(LineSegmentOf(A,P)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(pm*(4*sqrt(3))/3,-1/3)",
    "fact_spans": "[[[14, 33], [42, 44]], [[2, 13]], [[38, 41], [63, 67]], [[14, 33]], [[2, 13]], [[2, 37]], [[38, 48]], [[42, 55]], [[49, 61]]]",
    "query_spans": "[[[63, 72]]]",
    "process": ""
  },
  {
    "text": "Given that the focal distance of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $2 \\sqrt{5}$, and one asymptote of the hyperbola is perpendicular to the line $2x+y=0$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (2*x + y = 0);FocalLength(G)=2*sqrt(5) ;IsPerpendicular(OneOf(Asymptote(G)), H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4-y^2=1",
    "fact_spans": "[[[2, 58], [76, 79], [102, 105]], [[5, 58]], [[5, 58]], [[86, 97]], [[5, 58]], [[5, 58]], [[2, 58]], [[86, 97]], [[2, 74]], [[76, 99]]]",
    "query_spans": "[[[102, 110]]]",
    "process": "Since the focal distance of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ is $2\\sqrt{5}$, we have $2c=2\\sqrt{5}$, thus $c=\\sqrt{5}$. Since one asymptote of the hyperbola is perpendicular to the line $2x+y=0$, it follows that $\\frac{b}{a}\\cdot(-2)=-1$, i.e., $a=2b$. Because $c^{2}=a^{2}+b^{2}$, we get $5b^{2}=5$, so $b=1$, $a=2$. Therefore, the equation of the hyperbola is $\\frac{x^{2}}{4}-y^{2}=1$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$, a line $l$ passing through point $F$ intersects the parabola $C$ at points $A$ and $B$. Point $D$ is the midpoint of $OA$, and the projections of $B$ and $D$ onto the $y$-axis are $P$ and $Q$, respectively. Then the minimum value of $|PQ|$ is?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;P: Point;Q: Point;F: Point;B: Point;D: Point;Expression(C) = (y^2 = 8*x);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};O: Origin;MidPoint(LineSegmentOf(O, A)) = D;Projection(B, yAxis) = P;Projection(D, yAxis) = Q",
    "query_expressions": "Min(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "4*sqrt(2)",
    "fact_spans": "[[[35, 40]], [[2, 21], [41, 47]], [[49, 52]], [[92, 95]], [[96, 99]], [[25, 28], [30, 34]], [[53, 56], [73, 76]], [[59, 63], [77, 80]], [[2, 21]], [[2, 28]], [[29, 40]], [[35, 58]], [[64, 69]], [[59, 72]], [[73, 99]], [[73, 99]]]",
    "query_spans": "[[[101, 114]]]",
    "process": "As shown in the figure, let the equation of line $ l $ be $ x = my + 2 $, and let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Solving the system \n\\[\n\\begin{cases}\nx = my + 2 \\\\\ny^{2} = 8x\n\\end{cases}\n\\]\nand simplifying yields $ y^{2} - 8my - 16 = 0 $. Then $ y_{1} + y_{2} = 8m $, $ y_{1}y_{2} = -16 $. Since $ D $ is the midpoint of $ OA $, $ D\\left(\\frac{x_{1}}{2}, \\frac{y_{1}}{2}\\right) $, so $ Q\\left(0, \\frac{y_{1}}{2}\\right) $, $ P(0, y_{2}) $. Thus, $ |PQ| = |OP| + |OQ| = |y_{2}| + \\left|\\frac{y_{1}}{2}\\right| \\geqslant 2\\sqrt{\\frac{|y_{1}y_{2}|}{2}} = 4\\sqrt{2} $, with equality if and only if $ |y_{2}| = \\left|\\frac{y_{1}}{2}\\right| $, that is, when $ y_{1} = 4\\sqrt{2}, y_{2} = -2\\sqrt{2} $ or $ y_{1} = -4\\sqrt{2}, y_{2} = 2\\sqrt{2} $."
  },
  {
    "text": "Given that point $P$ is a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and points $M$ and $N$ are points on the circles $(x+3)^{2}+y^{2}=4$ and $(x-3)^{2}+y^{2}=1$ respectively, then the maximum value of $|P M|+|P N|$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);P: Point;PointOnCurve(P, G);H: Circle;Expression(H) = (y^2 + (x + 3)^2 = 4);M: Point;PointOnCurve(M, H);H1: Circle;Expression(H1) = (y^2 + (x - 3)^2 = 1);N: Point;PointOnCurve(N, H1)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, M)) + Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "13",
    "fact_spans": "[[[7, 46]], [[7, 46]], [[2, 6]], [[2, 49]], [[61, 81]], [[61, 81]], [[50, 54]], [[50, 105]], [[82, 102]], [[82, 102]], [[55, 58]], [[50, 105]]]",
    "query_spans": "[[[107, 126]]]",
    "process": "Given the problem, let the centers of the circles $(x+3)^{2}+y^{2}=4$ and $(x-3)^{2}+y^{2}=1$ be $A$ and $B$, respectively, with radii $r_{1}$, $r_{2}$. Then the foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ are $A(-3,0)$, $B(3,0)$. Also, $|PA|+r_{1} \\geqslant |PM|$, $|PB|+r_{2} \\geqslant |PN|$, hence $|PM|+|PN| \\leqslant |PA|+|PB|+r_{1}+r_{2}$, with equality if and only if $M$, $N$ lie on the extensions of $PA$, $PB$, respectively. At this time, the maximum value is $|PA|+|PB|+r_{1}+r_{2}=2\\sqrt{25}+2+1=13$."
  },
  {
    "text": "The ellipse $a x^{2}+b^{2}=1$ intersects the line $y=-x+1$ at points $A$ and $B$. The slope of the line passing through the origin and the midpoint of segment $AB$ is $\\frac{\\sqrt{2}}{2}$. Then $\\frac{a}{b}=$?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;A: Point;B: Point;Expression(G) = (a*x^2 + b^2 = 1);Expression(H) = (y = 1 - x);Intersection(G, H) = {A, B};L:Line;PointOnCurve(O,L);O:Origin;PointOnCurve(MidPoint(LineSegmentOf(A,B)),L);Slope(L)=sqrt(2)/2",
    "query_expressions": "a/b",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[0, 19]], [[2, 19]], [[2, 19]], [[20, 30]], [[32, 35]], [[36, 39]], [[0, 19]], [[20, 30]], [[0, 41]], [[56, 58]], [[42, 58]], [[43, 45]], [[42, 58]], [[56, 82]]]",
    "query_spans": "[[[84, 99]]]",
    "process": ""
  },
  {
    "text": "Given a point $A(1, m)$ on the parabola $y^{2}=2 p x  (p>0)$ such that the distance from $A$ to its focus is $3$, then $p$=?",
    "fact_expressions": "G: Parabola;p: Number;m: Number;A: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(A) = (1, m);PointOnCurve(A, G);Distance(A, Focus(G)) = 3",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 25], [38, 39]], [[50, 53]], [[28, 37]], [[28, 37]], [[5, 25]], [[2, 25]], [[28, 37]], [[2, 37]], [[28, 48]]]",
    "query_spans": "[[[50, 55]]]",
    "process": "First, use the equation of the parabola to find the equation of the directrix. Given that the distance from a point to the focus of the parabola is 3, by the definition of the parabola, the distance from the point to the directrix is also 3. Using $1+\\frac{p}{2}=3$, solve for $p$. From the equation of the parabola, it is known that the directrix equation is $x=-\\frac{p}{2}$. $\\because$ For the point $A(1,m)$ on the parabola $y^{2}=2px$ $(p>0)$, the distance to the focus is 3, $\\therefore$ according to the definition of the parabola, its distance to the directrix is 3, $\\therefore$ $1+\\frac{p}{2}=3$, .p"
  },
  {
    "text": "The equation of an asymptote of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1$ ($b>0$) is $y=\\sqrt{3} x$, then $b$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/b^2 = 1);b: Number;b>0;Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x)",
    "query_expressions": "b",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[0, 37]], [[0, 37]], [[62, 65]], [[3, 37]], [[0, 60]]]",
    "query_spans": "[[[62, 67]]]",
    "process": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) are given by $y=\\pm\\frac{b}{a}x$, hence $\\frac{b}{a}=\\sqrt{3}$, where $a=1$, therefore $b=\\sqrt{3}$."
  },
  {
    "text": "The distance $d$ from a point $(4, 1)$ on the parabola $x^{2}=2 p y(p>0)$ to its focus is $?$.",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 2*p*y);p: Number;p>0;H: Point;Coordinate(H) = (4, 1);PointOnCurve(H, G);Distance(H, Focus(G)) = d;d: Number",
    "query_expressions": "d",
    "answer_expressions": "5",
    "fact_spans": "[[[0, 21], [34, 35]], [[0, 21]], [[3, 21]], [[3, 21]], [[24, 33]], [[24, 33]], [[0, 33]], [[24, 43]], [[40, 43]]]",
    "query_spans": "[[[40, 45]]]",
    "process": "According to the problem, the parabola $ x^{2} = 2py $ ($ p > 0 $) passes through the point $ (4,1) $, then we have $ 16 = 2p $. Solving gives $ p = 8 $, so the standard equation of the parabola is: $ x^{2} = 16y $. Its focus is at $ (0,4) $. The distance from the point $ (4,1) $ to the focus is $ d = \\sqrt{4^{2} + 3^{2}} = 5 $."
  },
  {
    "text": "A line passing through the focus of the parabola $y^{2}=2 p x$ ($p>0$) intersects the parabola at points $A$ and $B$, $|A B|=3$, and the ordinate of the midpoint of $A B$ is $\\frac{1}{2}$. Find the value of $p$.",
    "fact_expressions": "G: Parabola;p: Number;H: Line;A: Point;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));PointOnCurve(Focus(G), H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 3;YCoordinate(MidPoint(LineSegmentOf(A, B))) = 1/2",
    "query_expressions": "p",
    "answer_expressions": "(3+pm*sqrt(5))/4",
    "fact_spans": "[[[1, 22], [28, 31]], [[81, 84]], [[25, 27]], [[33, 36]], [[37, 40]], [[4, 22]], [[1, 22]], [[0, 27]], [[25, 42]], [[43, 52]], [[54, 79]]]",
    "query_spans": "[[[81, 88]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), using the midpoint coordinate formula we obtain y_{1}+y_{2}=1. Let the equation of AB be: x=ky+\\frac{p}{2}, substitute into the parabola equation, and using Vieta's formulas we can obtain y_{1}^{2}+y_{2}^{2}-2p^{2}=1, that is, 2px_{1}+2px_{2}-2p^{2}=1, then using the focal radius formula we can solve it. [Detailed solution] Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since the ordinate of the midpoint of AB is \\frac{1}{2}, \\therefore y_{1}+y_{2}=1. Let the equation of AB be: x=ky+\\frac{p}{2}. Substituting into the parabola equation gives y^{2}=2p(ky+\\frac{p}{2}), i.e., y^{2}-2pky-p^{2}=0, \\therefore y_{1}y_{2}=-p^{2}, \\therefore y_{1}^{2}+y_{2}^{2}-2p^{2}=1, \\therefore 2px_{1}+2px_{2}-2p^{2}=1, \\therefore 2p(x_{1}+x_{2}-p)=1. Also, \\because |AB|=x_{1}+x_{2}+p=3, so x_{1}+x_{2}=3-p, \\therefore 2p(3-2p)=1, \\therefore 4p^{2}-6p+1=0, solving gives p=\\frac{3\\pm\\sqrt{5}}{4}"
  },
  {
    "text": "If the equations of the two asymptotes of hyperbola $C$ are $y = \\pm \\frac{3}{4} x$, then an equation of the hyperbola can be? (only write one hyperbola equation satisfying the condition)",
    "fact_expressions": "C: Hyperbola;Expression(Asymptote(C))=(y = pm*((3/4)*x))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/16 - y^2/9 = 1",
    "fact_spans": "[[[1, 7], [43, 46]], [[1, 40]]]",
    "query_spans": "[[[43, 52]]]",
    "process": ""
  },
  {
    "text": "Draw a line through the focus of the parabola $y^{2}=4x$ intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $x_{1}+x_{2}=8$, then $|AB|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Line;PointOnCurve(Focus(G), H);A: Point;B: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Intersection(H, G) = {A, B};x1+x2=8",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "10",
    "fact_spans": "[[[1, 15], [22, 25]], [[1, 15]], [[19, 21]], [[0, 21]], [[26, 44]], [[46, 64]], [[26, 44]], [[46, 64]], [[26, 44]], [[46, 64]], [[26, 44]], [[46, 64]], [[19, 64]], [[67, 82]]]",
    "query_spans": "[[[84, 93]]]",
    "process": ""
  },
  {
    "text": "The line passing through the focus $F$ of the parabola $y^{2}=2x$ with an inclination angle of $45^{\\circ}$ intersects the parabola at points $A$ and $B$. What is the length of segment $AB$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);F: Point;Focus(G) = F;PointOnCurve(F, H) = True;Inclination(H) = ApplyUnit(45, degree);H: Line;Intersection(H, G) = {A, B};B: Point;A: Point",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 15], [42, 45]], [[1, 15]], [[18, 21]], [[1, 21]], [[0, 41]], [[22, 41]], [[39, 41]], [[39, 53]], [[50, 53]], [[46, 49]]]",
    "query_spans": "[[[55, 66]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola $y^{2}=4 x$ is $F$, point $A(2 , 1)$, and $M$ is a point on the parabola such that $M$ does not lie on the line $A F$. Then the minimum perimeter of $\\triangle M A F$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;A: Point;M: Point;Coordinate(A) = (2, 1);Focus(G) = F;PointOnCurve(M, G);Negation(PointOnCurve(M,LineOf(A,F)))",
    "query_expressions": "Min(Perimeter(TriangleOf(M,A,F)))",
    "answer_expressions": "3+sqrt(2)",
    "fact_spans": "[[[0, 14], [40, 43]], [[0, 14]], [[18, 21]], [[22, 33]], [[36, 39], [48, 51]], [[22, 33]], [[0, 21]], [[37, 46]], [[48, 61]]]",
    "query_spans": "[[[63, 88]]]",
    "process": "As shown in the figure, draw MN perpendicular to the directrix $ l $ of the parabola from point M, with foot of perpendicular N. It is easy to know that $ F(1,0) $. Since the perimeter of $ \\triangle MAF $ is $ |AF| + |MF| + |AM| $, $ |AF| = \\sqrt{(2-1)^{2}+1} = \\sqrt{2} $, and $ |MF| + |AM| = |AM| + |MN| $, the perimeter of $ \\triangle MAF $ reaches its minimum when points A, M, N are collinear, and the minimum value is $ 2 + 1 + \\sqrt{2} = 3 + \\sqrt{2} $."
  },
  {
    "text": "If the distance from point $P(2,0)$ to one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $\\sqrt{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (2, 0);Distance(P,OneOf(Asymptote(G)))=sqrt(2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[11, 57], [79, 82]], [[14, 57]], [[14, 57]], [[1, 10]], [[11, 57]], [[1, 10]], [[1, 77]]]",
    "query_spans": "[[[79, 88]]]",
    "process": "Since one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) has the equation $bx+ay=0$, the distance from point $P(2,0)$ to the asymptote is $d=\\frac{|2b|}{\\sqrt{a^{2}+b^{2}}}=\\frac{2b}{c}=\\sqrt{2}$, so $c^{2}=2b^{2}$, hence $a=b$, and thus the eccentricity is $e=\\frac{c}{a}=\\frac{\\sqrt{a^{2}+b^{2}}}{a}=\\sqrt{2}$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $M$ lies on the hyperbola, and $M F_{1} \\perp x$-axis. Find the distance from $F_{1}$ to the line $F_{2} M$.",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/6 - y^2/3 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;M: Point;PointOnCurve(M, G);IsPerpendicular(LineSegmentOf(M, F1), xAxis)",
    "query_expressions": "Distance(F1, LineOf(F2, M))",
    "answer_expressions": "6/5",
    "fact_spans": "[[[2, 40], [70, 73]], [[2, 40]], [[49, 56], [96, 103]], [[57, 64]], [[2, 64]], [[2, 64]], [[65, 69]], [[65, 74]], [[76, 94]]]",
    "query_spans": "[[[96, 120]]]",
    "process": "Given the hyperbola $\\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1$ with foci $F_{1}$ and $F_{2}$, assume point $M$ lies above the $x$-axis on the hyperbola, and $MF_{1} \\perp x$-axis. Then $M(-3,\\frac{\\sqrt{6}}{2})$, so $MF_{1}=\\frac{\\sqrt{6}}{2}$, hence $MF_{2}=2\\sqrt{6}+\\frac{\\sqrt{6}}{2}=\\frac{5\\sqrt{6}}{2}$. Therefore, the distance from $F_{1}$ to line $F_{2}M$ is $\\frac{F_{1}F_{2}\\cdot MF_{1}}{MF_{2}}=\\frac{6\\times\\frac{\\sqrt{6}}{2}}{\\frac{5\\sqrt{6}}{5}}$."
  },
  {
    "text": "From a point $Q$ on the hyperbola $x^{2}-y^{2}=1$, draw a perpendicular to the line $x+y=2$, with foot of perpendicular at $N$. Then the equation of the locus of the midpoint $P$ of segment $QN$ is?",
    "fact_expressions": "Q: Point;N: Point;G: Hyperbola;H: Line;L:Line;Expression(G) = (x^2 - y^2 = 1);PointOnCurve(Q,G);PointOnCurve(Q,L);IsPerpendicular(L,H);FootPoint(L,H)=N;Expression(H) = (x + y = 2);MidPoint(LineSegmentOf(Q, N)) = P;P:Point",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "2*x**2-2*y**2-2*x-2*y-1=0",
    "fact_spans": "[[[22, 25]], [[42, 45]], [[1, 19]], [[26, 35]], [], [[1, 19]], [[1, 25]], [[0, 38]], [[0, 38]], [[0, 45]], [[26, 35]], [[47, 59]], [[56, 59]]]",
    "query_spans": "[[[56, 66]]]",
    "process": "Let $ P(x,y) $, $ Q(x_{1},y_{1}) $, then $ N(2x-x_{1},2y-y_{1}) $. Since $ N $ lies on the line $ x+y=2 $, we have $ 2x-x_{1}+2y-y_{1}=2 $ $\\textcircled{1}$. Also, since $ PQ $ is perpendicular to the line $ x+y=2 $, we have $ \\frac{y-y_{1}}{x-x_{1}}=1 $, i.e., $ x-y+y_{1}-x_{1}=0 $ $\\textcircled{2}$. From $\\textcircled{1}$ and $\\textcircled{2}$, we obtain  \n$$\n\\begin{cases}\nx_{1}=\\frac{3}{2}x+\\frac{1}{2}y-1\\\\\ny_{1}=\\frac{1}{2}x+\\frac{3}{2}y-1\n\\end{cases}\n$$\nAlso, since $ Q $ lies on the hyperbola $ x^{2}-y^{2}=1 $, we have $ x_{1}^{2}-y_{1}^{2}=1 $. Therefore,  \n$$\n\\left(\\frac{3}{2}x+\\frac{1}{2}y-1\\right)^{2}-\\left(\\frac{1}{2}x+\\frac{3}{2}y-1\\right)^{2}=1.\n$$\nSimplifying, we get $ 2x^{2}-2y^{2}-2x+2y-1=0 $, which is the trajectory equation of the midpoint $ P $."
  },
  {
    "text": "Given that a point of intersection between an asymptote and a directrix of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has coordinates $(1, \\sqrt{3})$, then the focal distance of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(OneOf(Intersection(Asymptote(G), Directrix(G)))) = (1, sqrt(3))",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 58], [90, 93]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 88]]]",
    "query_spans": "[[[90, 98]]]",
    "process": ""
  },
  {
    "text": "If the equation $\\frac{x^{2}}{4-k}+\\frac{y^{2}}{k-1}=1$ represents an ellipse, then the range of real values for $k$ is?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G)=(x^2/(4 - k) + y^2/(k - 1) = 1)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(1,5/2)+(5/2,4)",
    "fact_spans": "[[[44, 46]], [[48, 53]], [[1, 46]]]",
    "query_spans": "[[[48, 60]]]",
    "process": "From the equation representing the ellipse, we obtain the system of inequalities \\begin{cases}k-1>0\\\\4-k>0\\\\k-1\\neq4-k\\end{cases}, which can be solved to get the answer. According to the problem, the equation \\frac{x^2}{4-k}+7 satisfies \\begin{cases} \\text{equation } x^2 \\\\ k-1-\\frac{4-k>0}{k-1\\neq4-k} \\end{cases}, solving yields 1<k<4 and k\\neq\\frac{5}{2}, so the range of real values for k is (1,\\frac{5}{2})\\cup(\\frac{5}{2},4)."
  },
  {
    "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ are tangent to the circle given by $(x-2)^{2}+y^{2}=3$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;H: Circle;Expression(H)=((x - 2)^2 + y^2 = 3);IsTangent(Asymptote(G), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 47], [81, 84]], [[1, 47]], [[4, 47]], [[4, 47]], [[75, 76]], [[52, 76]], [[1, 78]]]",
    "query_spans": "[[[81, 90]]]",
    "process": "First, according to the hyperbola equation, obtain the asymptotes of the hyperbola $ bx \\pm ay = 0 \\therefore \\frac{2b}{c} = \\sqrt{3} \\therefore c = \\frac{1}{2}a, e = 2 $,"
  },
  {
    "text": "If one focus of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$ has coordinates $(0,1)$, then the value of the real number $m$ is?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (x^2/5 + y^2/m = 1);Coordinate(OneOf(Focus(G))) = (0, 1)",
    "query_expressions": "m",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 38]], [[55, 60]], [[1, 38]], [[1, 53]]]",
    "query_spans": "[[[55, 65]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $\\frac{\\sqrt {3}}{3}$. If the abscissa of an intersection point of the line $y=k x$ with the ellipse is $b$, then what is the value of $k$?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;k: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = k*x);Eccentricity(G) = sqrt(3)/3;XCoordinate(OneOf(Intersection(H,G))) = b",
    "query_expressions": "k",
    "answer_expressions": "pm*sqrt(3)/3",
    "fact_spans": "[[[0, 52], [90, 91]], [[100, 103]], [[2, 52]], [[80, 89]], [[105, 108]], [[2, 52]], [[2, 52]], [[0, 52]], [[80, 89]], [[0, 78]], [[80, 103]]]",
    "query_spans": "[[[105, 112]]]",
    "process": ""
  },
  {
    "text": "If the line $y=kx-2$ always has common points with the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$ whose foci lie on the $x$-axis, then the range of real values for $m$ is?",
    "fact_expressions": "G: Ellipse;m: Real;H: Line;k: Number;Expression(G) = (x^2/5 + y^2/m = 1);Expression(H) = (y = k*x - 2);IsIntersect(H, G);PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Range(m)",
    "answer_expressions": "[4, 5)",
    "fact_spans": "[[[22, 59]], [[66, 71]], [[1, 12]], [[3, 12]], [[22, 59]], [[1, 12]], [[1, 64]], [[13, 59]]]",
    "query_spans": "[[[66, 78]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. If point $P$ lies on ellipse $C$, then $|P F_{1}|+|P F_{2}|$=?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/16 + y^2/7 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C)",
    "query_expressions": "Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 45], [75, 80]], [[70, 74]], [[54, 61]], [[62, 69]], [[2, 45]], [[2, 69]], [[2, 69]], [[70, 81]]]",
    "query_spans": "[[[83, 106]]]",
    "process": "Since the equation of ellipse C is $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$, we have $a^{2}=16$, $a=4$. Since point P lies on ellipse C, $|PF|+|PF_{2}|=2a=8$."
  },
  {
    "text": "The equation of the locus of the midpoint of a chord passing through the focus of the parabola $y^{2}=4 x$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: LineSegment;PointOnCurve(Focus(G), H);IsChordOf(H, G)",
    "query_expressions": "LocusEquation(MidPoint(H))",
    "answer_expressions": "y^2=2*x-2",
    "fact_spans": "[[[2, 16]], [[2, 16]], [], [[0, 20]], [[2, 20]]]",
    "query_spans": "[[[0, 30]]]",
    "process": "The focus of the parabola $ y^{2} = 4x $ has coordinates $ F(1,0) $. Then the equation of a line passing through the focus is $ y = kx - k $. Let the midpoint of a focal chord have coordinates $ (x, y) $. Solving the system \n\\[\n\\begin{cases}\ny = kx - k \\\\\ny^{2} = 4x\n\\end{cases}\n\\]\nwe obtain $ k^{2}x^{2} - (2k^{2} + 4)x + k^{2} = 0 $. Then \n\\[\n\\begin{cases}\nx_{1} + x_{2} = \\frac{2k^{2} + 4}{k^{2}} = 2 + \\frac{4}{k^{2}} = 2x \\\\\ny_{1} + y_{2} = k(x_{1} + x_{2}) - 2k = 2kx - 2k = 2y,\n\\end{cases}\n\\]\nwhich simplifies to $ y^{2} = 2x - 2 $."
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$ are $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{1}$ with an inclination angle of $45^{\\circ}$ intersects the ellipse at a point $M$. If $M F_{2}$ is perpendicular to the $x$-axis, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;M: Point;F2: Point;F1: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(F1,H);Inclination(H)=ApplyUnit(45,degree);OneOf(Intersection(H,G))=M;IsPerpendicular(LineSegmentOf(M,F2),xAxis)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)-1",
    "fact_spans": "[[[0, 54], [111, 113], [142, 144]], [[2, 54]], [[2, 54]], [[108, 110]], [[119, 122]], [[72, 80]], [[83, 90], [63, 71]], [[2, 54]], [[2, 54]], [[0, 54]], [[0, 81]], [[0, 81]], [[82, 110]], [[91, 110]], [[108, 122]], [[124, 140]]]",
    "query_spans": "[[[142, 150]]]",
    "process": ""
  },
  {
    "text": "The maximum distance from a point $P$ on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ to the upper vertex $B$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);P: Point;PointOnCurve(P, G);B: Point;UpperVertex(G) = B",
    "query_expressions": "Max(Distance(P, B))",
    "answer_expressions": "4*sqrt(3)/3",
    "fact_spans": "[[[0, 27]], [[0, 27]], [[31, 34]], [[0, 34]], [[38, 41]], [[0, 41]]]",
    "query_spans": "[[[31, 50]]]",
    "process": "The upper vertex of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ is $B(0,1)$. Let the coordinates of point $P$ be $(m,n)$, then $\\frac{m^{2}}{4}+n^{2}=1$. Since $-1\\leqslant n\\leqslant 1$, when $n=\\frac{1}{3}$, $|PB|$ attains the maximum value $\\frac{4\\sqrt{3}}{3}$."
  },
  {
    "text": "The length of the major axis of the ellipse $(1-m) x^{2}-m y^{2}=1$ is?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (-m*y^2 + x^2*(1 - m) = 1)",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "2*sqrt(-1/m)",
    "fact_spans": "[[[0, 25]], [[2, 25]], [[0, 25]]]",
    "query_spans": "[[[0, 31]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, $P$ is a point on the ellipse, $Q$ is the midpoint of $P F_{1}$, if $|OQ|=1$, then $|PF_{1}|=$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;O: Origin;Q: Point;F2: Point;Expression(G) = (x^2/16 + y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);MidPoint(LineSegmentOf(P, F1)) = Q;Abs(LineSegmentOf(O, Q)) = 1",
    "query_expressions": "Abs(LineSegmentOf(P, F1))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 40], [69, 71]], [[65, 68]], [[49, 56]], [[94, 102]], [[76, 79]], [[57, 64]], [[2, 40]], [[2, 64]], [[2, 64]], [[65, 75]], [[76, 92]], [[94, 102]]]",
    "query_spans": "[[[104, 116]]]",
    "process": ""
  },
  {
    "text": "Given that $c$ is the semi-focal distance of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, then the range of $\\frac{b+c}{a}$ is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;c: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);HalfFocalLength(G) = c",
    "query_expressions": "Range((b + c)/a)",
    "answer_expressions": "(1, \\sqrt{2}]",
    "fact_spans": "[[[6, 60]], [[66, 81]], [[66, 81]], [[2, 5]], [[8, 60]], [[8, 60]], [[6, 60]], [[2, 64]]]",
    "query_spans": "[[[66, 88]]]",
    "process": ""
  },
  {
    "text": "If the chord of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ is bisected by the point $(2,1)$, then the slope $k$ of the line containing the chord is $k=$?",
    "fact_expressions": "G: Ellipse;H: LineSegment;I: Point;Expression(G) = (x^2/9 + y^2/4 = 1);Coordinate(I) = (2, 1);IsChordOf(H,G);MidPoint(H)=I;Slope(OverlappingLine(H)) = k;k:Number",
    "query_expressions": "k",
    "answer_expressions": "-8/9",
    "fact_spans": "[[[1, 38]], [], [[41, 49]], [[1, 38]], [[41, 49]], [[1, 40]], [[1, 51]], [[1, 65]], [[62, 65]]]",
    "query_spans": "[[[62, 67]]]",
    "process": "Let the coordinates of the two intersection points of the line and the ellipse be $(x_{1},y_{1})$, $(x_{2},y_{2})$. Then $\\frac{x_{1}^2}{9}+\\frac{y_{1}^2}{4}=1$, $\\frac{x_{2}^2}{9}+\\frac{y_{2}^2}{4}=1$. Subtracting these two equations gives $\\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{9}+\\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{4}=0$. Since the chord is bisected by the point $(2,1)$, we have $x_{1}+x_{2}=4$, $y_{1}+y_{2}=2$. Therefore, $\\frac{4(x_{1}-x_{2})}{9}+\\frac{2(y_{1}-y_{2})}{4}=0$, which simplifies to $\\frac{4(x_{1}-x_{2})}{9}+\\frac{y_{1}-y_{2}}{2}=0$, i.e., $\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{8}{9}$. Hence, the slope of the line is $-\\frac{8}{9}$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A point $P$ on $C$ satisfies $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$. Then, the area of the incircle of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/49 + y^2/24 = 1);LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(P,C);DotProduct(VectorOf(P,F1),VectorOf(P,F2))=0",
    "query_expressions": "Area(InscribedCircle(TriangleOf(P,F1,F2)))",
    "answer_expressions": "4*pi",
    "fact_spans": "[[[2, 46], [70, 73]], [[76, 79]], [[54, 61]], [[62, 69]], [[2, 46]], [[2, 69]], [[2, 69]], [[70, 79]], [[81, 140]]]",
    "query_spans": "[[[142, 172]]]",
    "process": "From the ellipse equation, we have $a^{2}=49$, $c^{2}=49-24=25$, $\\therefore a=7$, $c=5$, so $|F_{1}F_{2}|=2c=10$. By the definition of an ellipse, $|PF_{1}|+|PF_{2}|=2a=14$. Since $\\overrightarrow{PF}_{1}\\cdot\\overrightarrow{PF_{2}}=0$, $\\therefore \\overrightarrow{PF}_{1}\\bot\\overrightarrow{PF_{2}}$. Thus, $|PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}=100$.  \n$$\n\\begin{cases}\n|PF_{1}|+|PF_{2}|=14 \\\\\n|PF_{1}|^{2}+|PF_{2}|^{2}=100\n\\end{cases}\n\\Rightarrow |PF_{1}|\\cdot|PF_{2}|=48,\n\\therefore S_{\\triangle F_{1}PF_{2}}=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|=24.\n$$\nLet $r$ be the inradius of $\\triangle PF_{1}F_{2}$, then $\\frac{1}{2}(|PF_{1}|+|PF_{2}|+|F_{1}F_{2}|)r=24$, that is, $\\frac{1}{2}(14+10)r=24$, solving gives $r=2$. Therefore, the area of the incircle of $\\triangle PF_{1}F_{2}$ is $S=\\pi r^{2}=4\\pi$."
  },
  {
    "text": "The asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1  (a>0)$ is given by $y=\\frac{3}{5} x$. Then $a$=?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2/9 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (y = (3/5)*x)",
    "query_expressions": "a",
    "answer_expressions": "5",
    "fact_spans": "[[[0, 49]], [[77, 80]], [[3, 49]], [[0, 49]], [[0, 75]]]",
    "query_spans": "[[[77, 82]]]",
    "process": "From the standard equation of the hyperbola, the asymptotes are given by $ y = \\pm\\frac{3}{a}x $. Combining with the given condition, we obtain $ a = 5 $. [Master Teacher"
  },
  {
    "text": "Let the hyperbola $ C $: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 $ ($ a>0, b>0 $). Taking the right focus $ F $ of $ C $ as the center, a circle with radius $ \\sqrt{3} $ intersects one asymptote of $ C $ at points $ P $ and $ Q $. If $ \\overrightarrow{PQ} = 2 \\overrightarrow{QO} $ ($ O $ being the origin), and $ PF $ is perpendicular to the $ x $-axis, then the standard equation of the hyperbola $ C $ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F: Point;Q: Point;O: Origin;G: Circle;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;Center(G) = F;Radius(G) = sqrt(3);Intersection(G, OneOf(Asymptote(C))) = {P, Q};VectorOf(P, Q) = 2*VectorOf(Q, O);IsPerpendicular(LineSegmentOf(P, F), xAxis)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4 - y^2/2 = 1",
    "fact_spans": "[[[1, 62], [89, 92], [183, 189]], [[9, 62]], [[9, 62]], [[101, 104]], [[66, 69]], [[105, 108]], [[158, 161]], [[87, 88]], [[9, 62]], [[9, 62]], [[1, 62]], [[1, 69]], [[66, 88]], [[73, 88]], [[87, 110]], [[112, 157]], [[169, 181]]]",
    "query_spans": "[[[183, 196]]]",
    "process": "The right focus of the hyperbola is $(c,0)$, and the asymptotes are $y=\\frac{b}{a}x$. Since $PF$ is perpendicular to the $x$-axis, $P(c,\\frac{bc}{a})$, i.e., $\\frac{bc}{a}=\\sqrt{3}$\\textcircled{1}. Let $Q(n_1,n)$, from $\\overrightarrow{PQ}=2\\overrightarrow{QO}$ we get $(m-c,n-\\frac{bc}{a})=(-2m,-2n)$, solving gives $Q(\\frac{c}{3},\\frac{bc}{3a})$. From $|QF|=\\sqrt{\\frac{4c^{2}}{9}+\\frac{b^{2}c^{2}}{9a^{2}}}=\\sqrt{3}$\\textcircled{2}, using \\textcircled{1}\\textcircled{2} and $c^{2}=a^{2}+b^{2}$, we solve $c^{2}=6$, $a^{2}=4$, $b^{2}=2$. Hence, the equation of the hyperbola is a routine problem."
  },
  {
    "text": "The minimum distance from a point on the parabola $y = -x^2$ to the line $4x + 3y - 8 = 0$ is?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Expression(G) = (y = -x^2);Expression(H) = (4*x + 3*y - 8 = 0);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, H))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[0, 13]], [[17, 32]], [[15, 16]], [[0, 13]], [[17, 32]], [[0, 16]]]",
    "query_spans": "[[[15, 41]]]",
    "process": "Method 1: As shown in the figure below, the line is $34x+3y+b=0$ and combining with the equation gives $\\begin{cases}\\\\4\\end{cases}4x+3y+b=0$. Eliminating $y$, we obtain $3x^{2}-4x-b=0$. Then $A=16+12b=0$, solving gives $b=-\\frac{4}{3}$. Therefore, the tangent line equation is $4x+3y-\\frac{4}{3}=0$. The minimum distance from a point on the parabola $y=-x^{2}$ to the line $4x+3y-8=0$ is the distance between these two parallel lines, $d=\\frac{|8-\\frac{4}{3}|}{5}=\\frac{4}{3}$."
  },
  {
    "text": "$P$ is any point on the line $x - y + 3 = 0$, and an ellipse has two foci $F_{1}(-1 , 0)$, $F_{2}(1 , 0)$. Find the equation of the ellipse passing through point $P$ with the shortest possible major axis.",
    "fact_expressions": "H: Line;Expression(H) = (x - y + 3 = 0);P: Point;PointOnCurve(P, H);G: Ellipse;F1: Point;F2: Point;Coordinate(F1) = (-1, 0);Coordinate(F2) = (1, 0);Focus(G) = {F1, F2};PointOnCurve(P, G);WhenMin(MajorAxis(G))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5+y^2/4=1",
    "fact_spans": "[[[4, 15]], [[4, 15]], [[0, 3], [65, 69]], [[0, 19]], [[21, 23], [62, 64]], [[28, 43]], [[45, 60]], [[28, 43]], [[45, 60]], [[21, 60]], [[62, 69]], [[62, 75]]]",
    "query_spans": "[[[62, 80]]]",
    "process": ""
  },
  {
    "text": "The upper vertex of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $B$, the left focus is $F$, the line $BF$ is perpendicular to the line $x+y-3 \\sqrt{2}=0$, with foot of perpendicular at $M$, and point $B$ is the midpoint of segment $MF$. The equation of the ellipse is?",
    "fact_expressions": "B: Point;F: Point;M: Point;G: Ellipse;b: Number;a: Number;H: Line;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x + y - 3*sqrt(2) = 0);UpperVertex(G) = B;LeftFocus(G) = F;IsPerpendicular(LineOf(B,F),H);FootPoint(LineOf(B,F),H)=M;MidPoint(LineSegmentOf(M, F)) = B",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 + y^2/2 = 1",
    "fact_spans": "[[[57, 60], [107, 111]], [[65, 68]], [[102, 105]], [[0, 52], [123, 125]], [[2, 52]], [[2, 52]], [[76, 96]], [[2, 52]], [[2, 52]], [[0, 52]], [[76, 96]], [[0, 60]], [[0, 68]], [[69, 98]], [[69, 105]], [[107, 121]]]",
    "query_spans": "[[[123, 129]]]",
    "process": "(Analysis) Since the line BF is perpendicular to the line $x + y - 3\\sqrt{2} = 0$, the slope of line BF can be obtained. Find the coordinates of point M, substitute into the directrix equation to obtain $b$ and $c$, then solve for $a$ to get the ellipse equation. Let $F(-c, 0)$, $B(0, b)$. Since line BF is perpendicular to $x + y - 3\\sqrt{2} = 0$, we have $k_{BF} = \\frac{b}{c} = 1$, so $b = c$. Also, point B is the midpoint of segment MF. Using the midpoint formula, we get $M(b, 2b)$. Substituting into the line $x + y - 3\\sqrt{2} = 0$, we obtain $b = c = \\sqrt{2}$. Since $a^{2} = b^{2} + c^{2}$, it follows that $a = 2$. Therefore, the equation of the ellipse is: $\\frac{x^{2}}{4} + \\frac{y^{2}}{2} = 1$."
  },
  {
    "text": "Given that point $A$ is an arbitrary point on the line $x+y-3=0$, and for any point $P$ on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, it always holds that $P A>m$, then the range of real number $m$ is?",
    "fact_expressions": "G: Ellipse;H: Line;P: Point;A: Point;m: Real;Expression(G) = (x^2/4 + y^2 = 1);Expression(H) = (x + y - 3 = 0);PointOnCurve(A, H);PointOnCurve(P,G);LineSegmentOf(P,A)>m",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-oo,(3*sqrt(2)-sqrt(10))/2)",
    "fact_spans": "[[[26, 53]], [[7, 18]], [[58, 61]], [[2, 6]], [[73, 78]], [[26, 53]], [[7, 18]], [[2, 24]], [[26, 61]], [[64, 71]]]",
    "query_spans": "[[[73, 85]]]",
    "process": "Let the line $ l: y = -x + b $ be tangent to the ellipse. It is clear that line $ l $ is parallel to the line $ x + y - 3 = 0 $. Then  \n\\[\n\\begin{cases}\ny = -x + b \\\\\n\\frac{x^{2}}{4} + y^{2} = 1\n\\end{cases}\n\\]\nyields $ 5x^{2} - 8bx + 4(b^{2} - 1) = 0 $. Let $ \\triangle = 80 - 16b^{2} = 0 $, then $ b = \\pm\\sqrt{5} $, so the equation of line $ l $ is $ y = -x + \\sqrt{5} $ or $ y = -x - \\sqrt{5} $. The line closer to $ x + y - 3 = 0 $ is $ y = -x + \\sqrt{5} $, and the distance between these two lines is:  \n$ d = \\frac{|3 - \\sqrt{5}|}{\\sqrt{2}} = \\frac{3\\sqrt{2} - \\sqrt{10}}{2} $,  \n$ \\therefore |PA|_{\\min} = \\frac{3\\sqrt{2} - \\sqrt{10}}{2} $. From $ |PA| > m $ always holding, we get $ m < \\frac{3\\sqrt{2} - \\sqrt{10}}{2} $."
  },
  {
    "text": "The standard equation of a parabola with directrix $x=-2$ is?",
    "fact_expressions": "G: Parabola;Expression(Directrix(G)) = (x = -2)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = 8*x",
    "fact_spans": "[[[11, 14]], [[0, 14]]]",
    "query_spans": "[[[11, 21]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, a line $l$ passes through point $F$ and intersects the parabola $C$ at points $A$ and $B$, such that $|AB|=6$. If the perpendicular bisector of $AB$ intersects the $x$-axis at point $P$, then the coordinates of point $P$ are?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;F: Point;P: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, B)) = 6;Intersection(PerpendicularBisector(LineSegmentOf(A, B)), xAxis) = P",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(4, 0)",
    "fact_spans": "[[[29, 34]], [[2, 21], [40, 46]], [[48, 51]], [[54, 57]], [[25, 28], [35, 39]], [[87, 91], [93, 97]], [[2, 21]], [[2, 28]], [[29, 39]], [[29, 59]], [[61, 69]], [[71, 91]]]",
    "query_spans": "[[[93, 102]]]",
    "process": "From the parabola $ y^{2} = 4x $, we get $ p = 2 $. It is clear that the slope of line $ l $ exists. Let the line $ l $ passing through point $ F $ be $ y = k(x - 1) $, and let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Substituting $ y = k(x - 1) $ into $ y^{2} = 4x $ yields $ k^{2}x^{2} - (2k^{2} + 4)x + k^{2} = 0 $. Therefore, $ x_{1} + x_{2} = 2 + \\frac{4}{k^{2}} $. Using the definition of the parabola, $ x_{1} + x_{2} = |AB| - p = 6 - 2 = 4 $, so $ 2 + \\frac{4}{k^{2}} = 4 $, thus $ k = \\pm\\sqrt{2} $. Since the midpoint of $ AB $ has coordinates $ (2, k) $, the perpendicular bisector of $ AB $ has equation $ y - k = -\\frac{1}{k}(x - 2) $. Setting $ y = 0 $, we get $ x = 4 $, so the coordinates of point $ P $ are $ (4, 0) $."
  },
  {
    "text": "Given the parabola $C$: $y = m x^{2}$ ($m \\in \\mathbb{R}, m \\neq 0$) passes through the point $P(-1, 4)$, then the equation of the directrix of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y = m*x^2);m: Real;Negation(m=0);P: Point;Coordinate(P) = (-1, 4);PointOnCurve(P, C)",
    "query_expressions": "Expression(Directrix(C))",
    "answer_expressions": "y = -1/16",
    "fact_spans": "[[[2, 40], [53, 59]], [[2, 40]], [[10, 40]], [[10, 40]], [[41, 51]], [[41, 51]], [[2, 51]]]",
    "query_spans": "[[[53, 66]]]",
    "process": "Substitute P(-1,4) into the parabola $ C: y = mx^{2} $ ($ m \\in \\mathbb{R}, m \\neq 0 $), then simplify to standard form to find the equation of the directrix. [Detailed solution] From the problem, $ 4 = m \\cdot (-1)^{2} \\Rightarrow m = 4 $, thus $ C: y = 4x^{2} \\Rightarrow x^{2} = \\frac{1}{4}y $. Therefore, the directrix of parabola $ C $ is $ y = -\\frac{1}{16} $."
  },
  {
    "text": "Given that $M$ and $N$ are the intersection points of line $l$ passing through the focus $F$ of the parabola $C$: $y^{2}=2 p x$ ($p>0$) with the parabola $C$, $O$ is the origin, and satisfying $\\overrightarrow{M F}=3 \\overrightarrow{F N}$, $S_{\\triangle O M N}=\\sqrt{3}|M N|$, then the value of $p$ is?",
    "fact_expressions": "l: Line;C: Parabola;p: Number;M: Point;F: Point;N: Point;O: Origin;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {M, N};VectorOf(M, F) = 3*VectorOf(F, N);Area(TriangleOf(O, M, N)) = sqrt(3)*Abs(LineSegmentOf(M, N))",
    "query_expressions": "p",
    "answer_expressions": "8",
    "fact_spans": "[[[44, 49]], [[11, 37], [50, 56]], [[157, 160]], [[2, 5]], [[40, 43]], [[6, 9]], [[60, 63]], [[19, 37]], [[11, 37]], [[11, 43]], [[10, 49]], [[2, 59]], [[72, 117]], [[120, 155]]]",
    "query_spans": "[[[157, 164]]]",
    "process": "According to the problem, express the area of $\\triangle OMN$ in terms of the areas of $\\triangle OMF$ and $\\triangle ONF$ to obtain $\\frac{\\sqrt{3}}{8}p|MN|$; combining with $S_{\\triangle OMN} = \\sqrt{3}|MN|$ allows solving. Without loss of generality, assume the slope $k > 0$ of line $MN$, draw perpendiculars from $M, N$ to the directrix of the parabola, with feet $G, H$ respectively, and draw $NK \\bot MG$ at $K$. From $\\overrightarrow{MF} = 3\\overrightarrow{FN}$, we get $|MF| = 3|FN|$, $\\therefore |MG| = 3|NH|$. $\\therefore |MK| = 2|NH| = 2|NF| = \\frac{1}{2}|MN|$, $\\therefore |NK| = \\sqrt{|MN|^{2} - |MK|^{2}} = \\frac{\\sqrt{3}}{2}|MN|$. By $S_{\\Delta OMN} = S_{\\Delta OMF} + S_{\\Delta ONF} = \\frac{1}{2}|OF| \\cdot |NK| = \\frac{\\sqrt{3}}{8}p|MN|$, and $S_{\\Delta OMN} = \\sqrt{3}|MN|$, so $\\frac{\\sqrt{3}}{8}p|MN| = \\sqrt{3}|MN|$, $\\therefore p = 8$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through the right focus $F_{2}$ intersects the ellipse at points $P$ and $Q$. Then, the maximum area of the incircle of $\\Delta F_{1} P Q$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;l: Line;PointOnCurve(F2, l);P: Point;Q: Point;Intersection(l, G) = {P, Q}",
    "query_expressions": "Max(Area(InscribedCircle(TriangleOf(F1, P, Q))))",
    "answer_expressions": "9*pi/16",
    "fact_spans": "[[[2, 39], [65, 67], [87, 89]], [[2, 39]], [[48, 55]], [[56, 63], [71, 78]], [[2, 63]], [[65, 78]], [[81, 86]], [[64, 86]], [[90, 94]], [[95, 98]], [[81, 98]]]",
    "query_spans": "[[[101, 130]]]",
    "process": "Let the line $ l: x = my + 1 $, and combine it with the ellipse equation, eliminating $ x $ to obtain $ (3m^2 + 4)y^2 + 6my - 9 = 0 $. Let $ Q(x_2, y_2) $, then $ y_1 + y_2 = \\frac{6m}{y_2^2 - 4y_1y_2} = 12\\sqrt{\\frac{3m^2 + 4}{(3m^2 + 4)^2}} $, also $ = \\frac{1}{9(m^2 + 1) + \\frac{1}{m^2 + }} - \\frac{1}{16} $, hence $ S_{\\triangle F_1PQ} \\leqslant 3 $. The sum of the triangle's perimeter and the radius of its incircle is twice the triangle's area, then the inradius $ r = \\frac{2S_{\\Delta FPQ}}{8} \\leqslant \\frac{3}{4} $, and its maximum area is $ \\frac{9\\pi}{16} $. Therefore, the answer should be $ \\frac{9\\pi}{16} $."
  },
  {
    "text": "Given that point $P(1,1)$ lies on the parabola $C$: $y^{2}=2px$ ($p>0$), and $F$ is the focus of parabola $C$, then the value of $|PF|$ is?",
    "fact_expressions": "C: Parabola;p: Number;P: Point;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(P) = (1, 1);PointOnCurve(P, C);Focus(C) = F",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "5/4",
    "fact_spans": "[[[12, 38], [44, 50]], [[19, 38]], [[2, 11]], [[40, 43]], [[19, 38]], [[12, 38]], [[2, 11]], [[2, 39]], [[40, 53]]]",
    "query_spans": "[[[55, 66]]]",
    "process": "Since point P(1,1) lies on the parabola C: y^{2}=2px (p>0), we have 1=2p, so p=\\frac{1}{2}, hence |PF|=1+\\frac{1}{2}\\times\\frac{1}{2}=\\frac{5}{4}."
  },
  {
    "text": "Given that $A$, $B$, $C$ are three points on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$, the line $AB$ passes through the origin $O$, the line $AC$ passes through the right focus $F$ of the hyperbola, if $BF \\perp AC$ and $|BF|=|CF|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "A: Point;B: Point;C: Point;PointOnCurve(A, G) = True;PointOnCurve(B, G) = True;PointOnCurve(C, G) = True;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;O: Origin;PointOnCurve(O,LineOf(A,B)) = True;F: Point;RightFocus(G) = F;PointOnCurve(F,LineOf(A,C)) = True;IsPerpendicular(LineSegmentOf(B,F),LineSegmentOf(A,C));Abs(LineSegmentOf(B, F)) = Abs(LineSegmentOf(C, F))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)/2",
    "fact_spans": "[[[2, 5]], [[6, 9]], [[10, 13]], [[2, 78]], [[2, 78]], [[2, 78]], [[14, 73], [103, 106], [147, 150]], [[14, 73]], [[17, 73]], [[17, 73]], [[17, 73]], [[17, 73]], [[88, 93]], [[79, 93]], [[110, 113]], [[103, 113]], [[94, 113]], [[115, 130]], [[132, 145]]]",
    "query_spans": "[[[147, 156]]]",
    "process": "In the right triangle ABF, OF is the median on the hypotenuse AB, so |AB| = 2|OA| = 2|OF| = 2c. Let A(m, n), then m^{2} + n^{2} = c^{2}, and \\frac{m^{2}}{a^{2}} - \\frac{n^{2}}{b^{2}} = 1. Solving gives m = \\frac{a\\sqrt{c^{2}+b^{2}}}{c}, so A\\left(\\frac{a\\sqrt{c^{2}+b^{2}}}{c}, \\frac{b^{2}}{c}\\right). Also F(c, 0). Since BF \\perp AC, let C(x, y), we obtain x = \\frac{b^{2}+c^{2}}{c}, y = \\frac{a\\sqrt{c^{2}+b^{2}} + c^{2}}{c}. Substituting C\\left(\\frac{b^{2}+c^{2}}{c}, \\frac{a\\sqrt{c^{2}+b^{2}} + c^{2}}{c}\\right) into the hyperbola equation yields \\frac{(b^{2}+c^{2})^{2}}{a^{2}c^{2}} - \\frac{(a\\sqrt{c^{2}+b^{2}} + c^{2})^{2}}{b^{2}c^{2}} = 1, which leads to (2e^{2}-1)(e^{2}-2)^{2} = 1, solving gives e = \\frac{\\sqrt{10}}{2}."
  },
  {
    "text": "Write the standard equation of a hyperbola whose eccentricity is 4 times the eccentricity of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$ and whose foci lie on the $x$-axis?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/16 + y^2/12 = 1);Eccentricity(G)=4*Eccentricity(H);PointOnCurve(Focus(G),xAxis)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[64, 67]], [[7, 46]], [[7, 46]], [[3, 67]], [[55, 67]]]",
    "query_spans": "[[[64, 72]]]",
    "process": "From the ellipse equation, we know $a^{2}=16$, $b^{2}=12$, then $c^{2}=16-12=4$, so the eccentricity of the ellipse is $e=\\frac{c}{a}=\\frac{2}{4}=\\frac{1}{2}$. Then the eccentricity of the hyperbola is $e=2$, so in the hyperbola $\\frac{c}{a}=2 \\Rightarrow c=2a$, that is, $c^{2}=4a^{2}=a^{2}+b^{2}$, we get $b^{2}=3a^{2}$. Let $a^{2}=1$, then $b^{2}=3$, so one hyperbola equation satisfying the condition is $x^{2}-\\frac{y^{2}}{3}=1$."
  },
  {
    "text": "Given that $P$ is a point on the parabola $y=x^{2}$ distinct from the origin $O$, $P Q \\perp x$-axis with foot $Q$, and a line parallel to the $x$-axis is drawn through the midpoint of $P Q$, intersecting the parabola at point $M$. The line $Q M$ intersects the $y$-axis at point $N$. Then $\\frac{|P Q|}{|N O|}$=?",
    "fact_expressions": "P: Point;G: Parabola;Expression(G) = (y = x^2);PointOnCurve(P, G) ;Negation(P = O);O: Origin;IsPerpendicular(LineSegmentOf(P, Q), xAxis);FootPoint(LineSegmentOf(P, Q), xAxis) = Q;Q: Point;H: Line;PointOnCurve(MidPoint(LineSegmentOf(P, Q)), H);IsParallel(H, xAxis) ;Intersection(H, G) = M;M: Point;Intersection(LineOf(Q, M), yAxis) = N;N: Point",
    "query_expressions": "Abs(LineSegmentOf(P, Q))/Abs(LineSegmentOf(N, O))",
    "answer_expressions": "2+pm*sqrt(2)",
    "fact_spans": "[[[2, 5]], [[6, 18], [70, 73]], [[6, 18]], [[2, 28]], [[2, 28]], [[21, 26]], [[29, 43]], [[29, 50]], [[47, 50]], [], [[51, 69]], [[51, 69]], [[51, 78]], [[74, 78]], [[79, 96]], [[92, 96]]]",
    "query_spans": "[[[98, 122]]]",
    "process": "Let point $ P(x_{0},x_{0}^{2}) $ be a point in the first quadrant. Classify and discuss point $ M $ as a point in the first or second quadrant, find the coordinates of point $ M $, obtain the equation of line $ MQ $, and then determine the coordinates of point $ N $. Thus, the value of $ \\frac{|PQ|}{|NO|} $ can be found.\n\n① Let point $ P(x_{0},x_{0}^{2}) $ be a point in the first quadrant, then point $ Q(x_{0},0) $, and the midpoint of segment $ PQ $ is $ A(x_{0},\\frac{x_{0}^{2}}{2}) $. If point $ M $ is in the first quadrant, then $ M(\\frac{\\sqrt{2}x_{0}}{2},\\frac{x_{0}^{2}}{2}) $,  \n$ k_{MQ} = \\frac{x_{0}^{2}}{\\frac{\\sqrt{2}}{2}x_{0}-x_{0}} = -\\frac{(2+\\sqrt{2})x_{0}}{2} $,  \nthe equation of line $ MQ $ is $ y = -\\frac{(2+\\sqrt{2})x_{0}}{2}(x-x_{0}) $.  \nIn the equation of line $ MQ $: $ y = -\\frac{2+\\sqrt{2}}{2}x_{0}(x-x_{0}) $,  \nso, $ \\frac{|PQ|}{|NO|} = \\frac{x_{0}^{2}}{\\frac{(2+\\sqrt{2})x_{0}^{2}}{2}} = 2-\\sqrt{2} $.  \n\nIf point $ M $ is in the second quadrant, then point $ M(-\\frac{\\sqrt{2}}{2}x_{0},\\frac{x_{0}^{2}}{2}) $,  \n$ k_{MQ} = \\frac{\\frac{x_{0}^{2}}{2}}{-\\frac{\\sqrt{2}}{2}x_{0}-x_{0}} = \\frac{(\\sqrt{2}-2)x_{0}}{2} $,  \nthe equation of line $ MQ $ is $ y = \\frac{(\\sqrt{2}-2)x_{0}}{2}(x-x_{0}) $.  \nIn line $ MQ $, let $ x = 0 $, we get $ y = \\frac{2-\\sqrt{2}}{2}x_{0}^{2} $, so point $ N(0,\\frac{(2-\\sqrt{2})x_{0}^{2}}{2}) $. At this time,  \n$ \\frac{|PQ|}{|NO|} = \\frac{x_{0}^{2}}{\\frac{(2-\\sqrt{2})x_{0}^{2}}{2}} = 2+\\sqrt{2} $.  \n\nIn summary, $ \\frac{|PQ|}{|NO|} = 2\\pm\\sqrt{2} $."
  },
  {
    "text": "The hyperbola $C$ has its foci on the $x$-axis, eccentricity $e=2$, and passes through the point $P$ $(\\sqrt{2}, \\sqrt{3})$. Then the standard equation of hyperbola $C$ is_?",
    "fact_expressions": "C: Hyperbola;PointOnCurve(Focus(C), xAxis);e: Number;Eccentricity(C) = e;e = 2;P: Point;Coordinate(P) = (sqrt(2), sqrt(3));PointOnCurve(P, C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2/3 = 1",
    "fact_spans": "[[[0, 3]], [[0, 19]], [[24, 29]], [[0, 29]], [[24, 29]], [[36, 58]], [[34, 58]], [[0, 58]]]",
    "query_spans": "[[[61, 77]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the directrix of the parabola $y=\\frac{1}{4} x^{2}+2 x$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2/4 + 2*x)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=-5",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 35]]]",
    "process": "By the given condition, $ y = \\frac{1}{4}x^{2} + 2x = \\frac{1}{4}(x+4)^{2} - 4 $. Since the directrix of $ y = \\frac{1}{4}x^2 $ is $ y = -1 $, the parabola $ y = \\frac{1}{4}x^{2} + 2x $ is obtained by shifting the graph of $ y = \\frac{1}{4}x^{2} $ left by 4 units and down by 4 units. Therefore, the directrix of the parabola $ y = \\frac{1}{4}x^{2} + 2x $ is $ y = -5 $."
  },
  {
    "text": "If the circle $C$: $x^{2}+(y-2)^{2}=5$ intersects with lines passing through the fixed point $P(0,1)$ at points $A$ and $B$, then the trajectory equation of the midpoint $M$ of chord $AB$ is?",
    "fact_expressions": "C: Circle;Expression(C) = (x^2 + (y - 2)^2 = 5);P: Point;Coordinate(P) = (0, 1);PointOnCurve(P, G);G: Line;Intersection(C, G) = {A, B};A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), C) ;M: Point;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2+y^2-3*y+2=0",
    "fact_spans": "[[[1, 26]], [[1, 26]], [[29, 38]], [[29, 38]], [[28, 41]], [[39, 41]], [[1, 52]], [[43, 46]], [[47, 50]], [[1, 60]], [[63, 66]], [[55, 66]]]",
    "query_spans": "[[[63, 73]]]",
    "process": "Given that C(0,2), let the moving point be M(x,y). When x=0, M(0,1); when x≠0, by the perpendicular chord theorem, we have MN⊥MC. Therefore, \\frac{y-2}{x} \\cdot \\frac{y-1}{x} = -1. Rearranging gives x^{2} + y^{2} - 3y + 2 = 0. Since (0,1) satisfies this equation, the trajectory equation of the midpoint M of chord AB is x^{2} + y^{2} - 3y + 2 = 0."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively. If there exists a point $P$ on the hyperbola such that $\\angle F_{1} P F_{2}=60^{\\circ}$ and $|P F_{1}|=2|P F_{2}|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree);Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[19, 75], [83, 86], [154, 157]], [[19, 75]], [[22, 75]], [[22, 75]], [[22, 75]], [[22, 75]], [[1, 8]], [[9, 16]], [[1, 81]], [[1, 81]], [[89, 93]], [[83, 93]], [[95, 128]], [[130, 152]]]",
    "query_spans": "[[[154, 163]]]",
    "process": "Let |PF_{2}| = m, then |PF_{1}| = 2m. Obviously, point P lies on the right branch of the hyperbola, so |PF_{1}| - |PF_{2}| = 2a. Therefore, m = 2a, \\therefore |PF_{1}| = 4a, |PF_{2}| = 2a, and |F_{1}F_{2}| = 2c, \\angle F_{1}PF_{2} = 60^{\\circ}. Hence, by the law of cosines: |F_{1}F_{2}|^{2} = |PF_{1}|^{2} + |PF_{2}|^{2} - 2|PF_{1}| \\cdot |PF_{2}| \\cdot \\cos\\angle F_{1}PF_{2}, that is, 4c^{2} = 16a^{2} + 4a^{2} - 2 \\cdot 4a \\cdot 2a \\cdot \\frac{1}{2}. Simplifying yields: c = \\sqrt{3}a \\Rightarrow e = \\frac{c}{a} = \\sqrt{3}"
  },
  {
    "text": "A line $l$ passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) with an inclination angle of $120^{\\circ}$ intersects the parabola in the first and fourth quadrants at points $A$ and $B$, respectively. Then $\\frac{|A F|}{|B F|}$=?",
    "fact_expressions": "l: Line;G: Parabola;p: Number;A: Point;F: Point;B: Point;p>0;Expression(G) = (y^2 = 2*p*x);Focus(G) = F;PointOnCurve(F, l);Inclination(l) = ApplyUnit(120, degree);Intersection(l, G) = {A, B};Quadrant(A)=1;Quadrant(B)=4",
    "query_expressions": "Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(B, F))",
    "answer_expressions": "1/3",
    "fact_spans": "[[[47, 52]], [[1, 22], [53, 56]], [[4, 22]], [[67, 70]], [[25, 28]], [[71, 74]], [[4, 22]], [[1, 22]], [[1, 28]], [[0, 52]], [[29, 52]], [[47, 76]], [[56, 76]], [[56, 76]]]",
    "query_spans": "[[[78, 101]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), the equation of line AB is y=-\\sqrt{3}(x-\\frac{p}{2}). From the focal chord formula of the parabola, |AB|=x_{1}+x_{2}+p=\\frac{2p}{\\sin^{2}\\theta}=\\frac{8p}{3}, \\therefore x_{1}+x_{2}=\\frac{5p}{3}. Solving simultaneously the equations of the line and the parabola:  \n\\begin{cases}y=-\\sqrt{3}(x-\\frac{p}{2})\\\\y^{2}=2px\\end{cases}  \nEliminating y gives 3x^{2}-5px+\\frac{3p^{2}}{4}=0, hence x_{1}x_{2}=\\frac{p^{2}}{4}. Solving the system of equations (1) yields x_{2}=\\frac{3}{2}p_{1}, then \\frac{|AF|}{|BF|}=\\frac{x_{1}+\\frac{p}{2}}{x_{2}+\\frac{p}{2}}=\\frac{\\frac{p}{6}+\\frac{p}{2}}{\\frac{p}{2}+\\frac{p}{2}}=\\frac{1}{3}"
  },
  {
    "text": "If the curve $2|x|-y-4=0$ and the curve $x^{2}+\\lambda y^{2}=4$ ($\\lambda<0$) have exactly two distinct common points, then the range of real values for $\\lambda$ is?",
    "fact_expressions": "G: Curve;Expression(G) = (-y + 2*Abs(x) - 4 = 0);G1:Curve;Expression(G1) = (lambda*y^2 + x^2 = 4);lambda: Real;lambda < 0;NumIntersection(G, G1) = 2",
    "query_expressions": "Range(lambda)",
    "answer_expressions": "[1/4,0)",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[17, 53]], [[17, 53]], [[66, 77]], [[19, 53]], [[2, 64]]]",
    "query_spans": "[[[66, 84]]]",
    "process": "Since the curve 2|x|-y-4=0 and the curve x^{2}+\\lambda y^{2}=4 (\\lambda<0) both pass through the points (\\pm2,0), the slope of the hyperbola's asymptote y=\\sqrt{-\\frac{1}{\\lambda}}x is not less than the slope of the line y=2x-4, i.e., \\sqrt{-\\frac{1}{\\lambda}}\\geqslant2 \\Rightarrow -\\frac{1}{4}\\leqslant\\lambda<0"
  },
  {
    "text": "A line with slope $1$ passes through a focus of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ and intersects the hyperbola at points $A$ and $B$. Then $|A B|$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1);H: Line;Slope(H) = 1;PointOnCurve(OneOf(Focus(G)), H);A: Point;B: Point;Intersection(H, G) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "6",
    "fact_spans": "[[[11, 39], [46, 49]], [[11, 39]], [[7, 9]], [[0, 9]], [[7, 44]], [[51, 54]], [[55, 58]], [[7, 60]]]",
    "query_spans": "[[[62, 71]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. From $ x^{2}-\\frac{y^{2}}{3}=1 $, we get $ a^{2}=1 $, $ b^{2}=3 $, $ c^{2}=a^{2}+b^{2}=4 $, $ k=1 $. One of the foci is $ (2,0) $. Substituting into the point-slope equation of line $ l $ gives $ y=x-2 $. From \n\\[\n\\begin{cases}\nx^{2}-\\frac{y^{2}}{3}=1 \\\\\ny=x-2\n\\end{cases}\n\\]\nwe obtain $ 3x^{2}-(x-2)^{2}-3=0 $, that is, $ 2x^{2}+4x-7=0 $. Therefore, $ x_{1}+x_{2}=-2 $, $ x_{1}x_{2}=-\\frac{7}{2} $, $ A=16+56=72 $, $ |x_{2}-x_{1}|=\\sqrt{(x_{2}-x_{1})^{2}}=\\sqrt{(x_{2}+x_{1})^{2}-4x_{2}x_{1}}=\\sqrt{4+14}=3\\sqrt{2} $, $ AB=\\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y)^{2}}=\\sqrt{(1+k^{2})(x_{2}-x_{1}^{2}}}=\\sqrt{1+k^{2}}|x_{2}-x_{1}|=6 $"
  },
  {
    "text": "Given that $A$ and $B$ are the points of intersection of a line passing through the focus $F$ of the parabola $C$: $y^{2}=2px$ ($p>0$) with the parabola, $O$ is the origin, and it satisfies $\\overrightarrow{AB}=3\\overrightarrow{FB}$, $S_{\\triangle OAB}=\\frac{\\sqrt{2}}{3}|AB|$, then the value of $|AB|$ is?",
    "fact_expressions": "C: Parabola;p: Number;A: Point;B: Point;F: Point;O: Origin;L:Line;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(F, L)=True;Intersection(C, L) = {A, B};VectorOf(A, B) = 3*VectorOf(F, B);Area(TriangleOf(O, A, B)) = (sqrt(2)/3)*Abs(LineSegmentOf(A, B))",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "9/2",
    "fact_spans": "[[[11, 37], [47, 50]], [[18, 37]], [[2, 5]], [[6, 9]], [[40, 43]], [[54, 57]], [[44, 46]], [[18, 37]], [[11, 37]], [[11, 43]], [[10, 46]], [[2, 53]], [[66, 111]], [[113, 160]]]",
    "query_spans": "[[[162, 173]]]",
    "process": "By the given condition, it is clear that the slope of line $ AB $ exists. Due to the symmetry of the parabola, without loss of generality, assume the slope $ k > 0 $. Draw perpendiculars from points $ A $ and $ B $ to the directrix of the parabola, with feet at $ C $ and $ D $, respectively. Draw $ BE \\perp AC $ meeting at point $ E $. Then from $ \\overrightarrow{AB} = 3\\overrightarrow{FB} $, we obtain $ \\overrightarrow{AF} = 2\\overrightarrow{FB} $, i.e., $ |\\overrightarrow{AF}| = 2|\\overrightarrow{FB}| $, so $ |AC| = 2|BD| $. Therefore, point $ E $ is the midpoint of $ AC $, so $ |AE| = \\frac{1}{3}|AB| $. Hence, $ |BE| = \\sqrt{|AB|^2 - |AE|^2} = \\frac{2\\sqrt{2}}{3}|AB| $. Then $ S_{\\Delta OAB} = S_{\\Delta OAF} + S_{\\Delta OBF} = \\frac{1}{2}|BE| \\cdot |OF| = \\frac{\\sqrt{2}}{6}p \\cdot |AB| = \\frac{\\sqrt{2}}{3}|AB| $. Solving gives $ p = 2 $, so the equation of line $ AB $ is $ y = k(x - 1) $. From \n\\[\n\\begin{cases}\ny = k(x - 1) \\\\\ny^2 = 4x\n\\end{cases}\n\\]\nwe get $ k^2x^2 - 2(k^2 + 2)x + k^2 = 0 $, then \n\\[\n\\begin{cases}\ny^2 = 4x \\\\\nx_A + x_B = \\frac{2k^2 + 4}{k^2} \\\\\nx_A x_B = 1\n\\end{cases}\n\\]\nFrom $ \\overrightarrow{AF} = 2\\overrightarrow{FB} $, we have $ x_A - 1 = 2(1 - x_B) $, i.e., $ x_A = 3 - 2x_B $. Combining with $ k > 0 $, solving yields \n\\[\n\\begin{cases}\nx_A = 2 \\\\\nx_B = \\frac{1}{2} \\\\\nk = 2\\sqrt{2}\n\\end{cases}\n\\]\nthen $ |AB| = x_A + x_B + 2 = \\frac{9}{2} $. This involves memorizing properties of parabolas and the relationship between lines and parabolas, which are commonly tested."
  },
  {
    "text": "Let point $F$ be the focus of the parabola $y^{2}=2 p x(p>0)$, and $E$ a point on its directrix such that $E F=\\frac{2 \\sqrt{3}}{3}$. If a line passing through the focus $F$ and perpendicular to $E F$ intersects the parabola at points $A$ and $B$, and $\\overrightarrow{A F}=3 \\overrightarrow{F B}$, then $p=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;E: Point;PointOnCurve(E, Directrix(G));LineSegmentOf(E, F) = (2*sqrt(3))/3;H: Line;PointOnCurve(F, H);IsPerpendicular(H, LineSegmentOf(E, F));A: Point;B: Point;Intersection(H, G) = {A, B};VectorOf(A, F) = 3*VectorOf(F, B)",
    "query_expressions": "p",
    "answer_expressions": "1",
    "fact_spans": "[[[5, 26], [34, 35], [90, 93]], [[5, 26]], [[152, 155]], [[8, 26]], [[0, 4], [74, 77]], [[0, 29]], [[30, 33]], [[30, 40]], [[42, 68]], [[87, 89]], [[71, 89]], [[78, 89]], [[94, 97]], [[98, 101]], [[87, 103]], [[105, 150]]]",
    "query_spans": "[[[152, 157]]]",
    "process": "From the focal radius formula we get: \\frac{p}{1-\\cos\\theta}=3\\times\\frac{p}{1+\\cos\\theta}, then \\tan\\theta=\\sqrt{3}, based on which the equation of AB is y=\\sqrt{3}(x-\\frac{p}{2}), and the equation of EF is y=-\\frac{\\sqrt{3}}{3}(x-\\frac{p}{2})', combining with the given conditions, an equation in terms of p is obtained from the length of EF, solving this equation yields the value of the real number p. Solution: According to the problem and the focal radius formula we get: \\frac{p}{1-\\cos\\theta}=3\\times\\frac{p}{1+\\cos\\theta}, rearranging gives: \\cos\\theta=\\frac{1}{2}, \\therefore \\tan\\theta=\\sqrt{3}, based on this, the equation of line AB is: y=\\sqrt{3}(x-\\frac{P}{2}), the equation of line EF is y=-\\frac{\\sqrt{3}}{3}(x-\\frac{P}{2})^{x}, solving gives: p=1."
  },
  {
    "text": "Parabola $E$: $y^{2}=2 p x(p>0)$ has focus $F$, point $A$ is the intersection of the directrix of $E$ and the coordinate axis, point $P$ lies on $E$. If $\\angle P A F=30^{\\circ}$, then $\\sin \\angle P F A$=?",
    "fact_expressions": "E: Parabola;p: Number;P: Point;A: Point;F: Point;p>0;Expression(E) = (y^2 = 2*(p*x));Focus(E) = F;Intersection(Directrix(E),axis)=A;PointOnCurve(P, E);AngleOf(P, A, F) = ApplyUnit(30, degree)",
    "query_expressions": "Sin(AngleOf(P, F, A))",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[0, 26], [39, 42], [58, 61]], [[7, 26]], [[53, 57]], [[34, 38]], [[30, 33]], [[7, 26]], [[0, 26]], [[0, 33]], [[34, 52]], [[53, 62]], [[64, 89]]]",
    "query_spans": "[[[91, 112]]]",
    "process": ""
  },
  {
    "text": "Given that the right focus $F$ of a hyperbola is the center of the circle $x^{2}+y^{2}-4 x+3=0$, and its asymptotes are tangent to the circle, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Circle;Expression(H) = (-4*x + x^2 + y^2 + 3 = 0);Center(H)=F;RightFocus(G)=F;IsTangent(Asymptote(G),H);F: Point",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/3-y^2=1",
    "fact_spans": "[[[2, 5], [40, 41], [51, 54]], [[13, 35], [46, 47]], [[13, 35]], [[9, 38]], [[2, 12]], [[40, 49]], [9, 11]]",
    "query_spans": "[[[51, 61]]]",
    "process": "The circle $x^{2}+y^{2}-4x+3=0$ has center $(2,0)$ and radius $1$, so $F(2,0)$, thus $c=2$, hence $a^{2}+b^{2}=4$. The asymptotes of the hyperbola are given by $y=\\pm\\frac{b}{a}x$. From the condition that the line is tangent to the circle, we obtain: $\\frac{|2b|}{\\sqrt{a^{2}+b^{2}}}=1 \\Rightarrow b=1, a=\\sqrt{3}$. Thus, the standard equation of the hyperbola is $\\frac{x^{2}}{3}-y^{2}=1$."
  },
  {
    "text": "The line $y=2x-1$ passes through a vertex and a focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;H: Line;Expression(H) = (y = 2*x - 1);PointOnCurve(OneOf(Vertex(G)),H) = True;PointOnCurve(OneOf(Focus(G)),H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/5",
    "fact_spans": "[[[12, 64], [74, 76]], [[12, 64]], [[14, 64]], [[14, 64]], [[14, 64]], [[14, 64]], [[0, 11]], [[0, 11]], [[0, 69]], [[0, 72]]]",
    "query_spans": "[[[74, 82]]]",
    "process": "From the given conditions: the line $ y = 2x - 1 $ intersects the $ y $-axis and $ x $-axis at points $ (0, -1) $ and $ \\left(\\frac{1}{2}, 0\\right) $. Since the line $ y = 2x - l $ passes through a vertex and a focus of the ellipse $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $), and the foci of the ellipse lie on the $ x $-axis, $ \\therefore b = 1 $, $ c = \\frac{1}{2} $, $ a^{2} = b^{2} + c^{2} = 1 + \\frac{1}{4} = \\frac{5}{4} $, so $ a = \\frac{\\sqrt{5}}{2} $. $ \\therefore e = \\frac{c}{a} = \\frac{\\frac{1}{2}}{\\frac{\\sqrt{5}}{2}} = \\frac{\\sqrt{5}}{5} $."
  },
  {
    "text": "Let the eccentricity of ellipse $C_{1}$ be $\\frac{5}{13}$, with foci on the $x$-axis and major axis length $26$. If the absolute value of the difference of the distances from any point on curve $C_{2}$ to the two foci of ellipse $C_{1}$ is equal to $8$, then what is the standard equation of curve $C_{2}$?",
    "fact_expressions": "C1:Ellipse;C2:Curve;Eccentricity(C1) = 5/13;PointOnCurve(Focus(C1),xAxis);Length(MajorAxis(C1))=26;P:Point;PointOnCurve(P,C2);F1:Point;F2:Point;Focus(C1)={F1,F2};Abs(Distance(P,F1)-Distance(P,F2))=8",
    "query_expressions": "Expression(C2)",
    "answer_expressions": "x^2/16 - y^2/9 = 1",
    "fact_spans": "[[[1, 10], [62, 71]], [[49, 58], [92, 101]], [[1, 29]], [[1, 38]], [[1, 47]], [[60, 61]], [[49, 61]], [], [], [[62, 76]], [[60, 90]]]",
    "query_spans": "[[[92, 108]]]",
    "process": "According to the problem, in the ellipse equation, $ a = 13 $. Since $ \\frac{c}{a} = \\frac{5}{13} $, $ \\therefore c = 5 $. According to the definition of hyperbola, the curve $ C_{2} $ is a hyperbola with semi-focal length 5 and real axis length 8. $ \\therefore $ the imaginary axis length is 6. $ \\therefore $ the hyperbola equation is $ \\frac{x^2}{16} - \\frac{y^{2}}{9} = 1 $."
  },
  {
    "text": "The equation of the locus of the centers of circles passing through the point $(0,-2)$ and tangent to the line $y=2$ is?",
    "fact_expressions": "G: Circle;I: Point;Coordinate(I) = (0, -2);PointOnCurve(I,G)  =True;H: Line;Expression(H) = (y = 2);IsTangent(G,H) = True",
    "query_expressions": "LocusEquation(Center(G))",
    "answer_expressions": "x^2 = -8y",
    "fact_spans": "[[[22, 23]], [[1, 10]], [[1, 10]], [[0, 23]], [[12, 19]], [[12, 19]], [[11, 23]]]",
    "query_spans": "[[[22, 33]]]",
    "process": "Let the coordinates of the circle's center be $ M(x, y) $. Then the distance from $ M $ to the point $ (0, -2) $ is equal to the distance from $ M $ to the line $ y = 2 $. According to the definition of a parabola, the locus of the center $ M(x, y) $ is a parabola. Thus, $ -\\frac{p}{2} = -2 $, solving gives $ p = 4 $. Since the directrix equation is $ y = 2 $, the standard equation of the parabola is $ x^{2} = -8y $."
  },
  {
    "text": "The point $(2,3)$ lies on the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0, b>0)$, and the focal length of $C$ is $4$. What is its eccentricity?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;G: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(G) = (2, 3);PointOnCurve(G, C);FocalLength(C) = 4",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[11, 74], [76, 79], [88, 89]], [[19, 74]], [[19, 74]], [[2, 10]], [[19, 74]], [[19, 74]], [[11, 74]], [[2, 10]], [[2, 75]], [[76, 86]]]",
    "query_spans": "[[[88, 95]]]",
    "process": "Since the focal length of C is 4, we have 2c = 4, that is, c = 2. Because the foci of the hyperbola lie on the x-axis, the coordinates of the foci are (-2, 0), (2, 0). Thus, 2a = |\\sqrt{(2+2)^{2}+3^{2}} - \\sqrt{(2-2)^{2}+3^{2}}| = 2, so a = 1, e = 2."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{100}+\\frac{y^{2}}{64}=1$, and $P$ is a point on the ellipse, then the maximum value of $|P F_{1}| \\cdot|P F_{2}|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/100 + y^2/64 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)))",
    "answer_expressions": "100",
    "fact_spans": "[[[18, 58], [68, 70]], [[64, 67]], [[2, 9]], [[10, 17]], [[18, 58]], [[2, 63]], [[64, 73]]]",
    "query_spans": "[[[75, 107]]]",
    "process": "According to the definition of an ellipse and combining with the basic inequality, find the maximum value of |PF_{1}|\\cdot|PF_{2}|. \\because F_{1}, F_{2} are the two foci of the ellipse \\frac{x^{2}}{100}+\\frac{y^{2}}{64}=1, let |PF_{1}|=m, |PF_{2}|=n, then by the definition of the ellipse we obtain m+n=20, \\because m+n=20 \\geqslant 2\\sqrt{mn}, \\therefore mn \\leqslant \\left(\\frac{m+n}{2}\\right)^{2}=100, equality holds if and only if m=n=10; \\therefore the maximum value of |PF_{1}|\\cdot|PF_{2}| is 100."
  },
  {
    "text": "Given the parabola $y^{2}=4 x$, line $l$ passes through the fixed point $(-1,0)$. When line $l$ has only one common point with the parabola, what is the slope of line $l$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);l: Line;H: Point;Coordinate(H) = (-1, 0);PointOnCurve(H, l) = True;NumIntersection(l, G) = 1",
    "query_expressions": "Slope(l)",
    "answer_expressions": "{0,1,-1}",
    "fact_spans": "[[[2, 16], [40, 43]], [[2, 16]], [[17, 22], [34, 39], [52, 57]], [[25, 33]], [[25, 33]], [[17, 33]], [[34, 50]]]",
    "query_spans": "[[[52, 62]]]",
    "process": "According to the problem, the equation of the line can be set as: $ y = k(x+1) $. By combining equations, we have  \n\\[\n\\begin{cases}\ny = k(x+1) \\\\\ny^{2} = 4x\n\\end{cases}\n\\]  \nRearranging gives  \n\\[\nk^{2}x^{2} + (2k^{2} \\cdot 4)x + k^{2} = 0 \\quad (*)\n\\]  \nThe line and the parabola have only one common point, so equation $(*)$ has only one root.  \n① When $ k = 0 $, $ y = 0 $ satisfies the condition.  \n② When $ k \\neq 0 $, $ \\Delta = (2k^{2} \\cdot 4)^{2} - 4 \\cdot k^{2} \\cdot k^{2} = 0 $. Rearranging gives $ k^{2} = 1 $, solving yields $ k = 1 $ or $ k = -1 $.  \nIn conclusion, $ k = 1 $ or $ k = -1 $ or $ k = 0 $."
  },
  {
    "text": "Given that point $P(x, y)$ satisfies the ellipse equation $2x^{2} + y^{2} = 1$, then the maximum value of $\\frac{y}{x-1}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;Coordinate(P) = (x1, y1);x1:Number;y1:Number;Expression(G)=(2*x^2+y^2=1)",
    "query_expressions": "Max(y1/(x1 - 1))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[15, 17]], [[2, 13]], [[2, 13]], [[3, 13]], [[3, 13]], [[15, 36]]]",
    "query_spans": "[[[38, 59]]]",
    "process": ""
  },
  {
    "text": "The equation of the parabola with vertex at the origin and the right directrix of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ as its directrix is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (x^2/3 - y^2 = 1);O: Origin;Vertex(H) = O;Directrix(H) = RightDirectrix(G)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2 = -6*x",
    "fact_spans": "[[[7, 35]], [[43, 46]], [[7, 35]], [[3, 5]], [[0, 46]], [[6, 46]]]",
    "query_spans": "[[[43, 50]]]",
    "process": ""
  },
  {
    "text": "Given that the focal distance of the hyperbola $\\frac{x^{2}}{m+1}-\\frac{y^{2}}{2 m-1}=1$ is $6$, then $m$=?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/(m + 1) - y^2/(2*m - 1) = 1);FocalLength(G)=6",
    "query_expressions": "m",
    "answer_expressions": "pm*3",
    "fact_spans": "[[[2, 46]], [[55, 58]], [[2, 46]], [[2, 53]]]",
    "query_spans": "[[[55, 60]]]",
    "process": "When a focus is (3,0), $ m > \\frac{1}{2} $, $ m+1+2m-1=9 $, $ \\therefore m=3 $; when a focus is (0,3), $ m < -1 $, $ -(m+1)-(2m-1)=9 $, $ \\therefore m=-3 $"
  },
  {
    "text": "Given that the line $y=k(x+2)$ intersects the parabola $C$: $y^{2}=8x$ at points $A$ and $B$, and $F$ is the focus of the parabola $C$. If $|\\overrightarrow{F A}|=2|\\overrightarrow{F B}|$, then the real number $k$=?",
    "fact_expressions": "G: Line;Expression(G) = (y = k*(x + 2));k: Real;C: Parabola;Expression(C) = (y^2 = 8*x);A: Point;B: Point;Intersection(G, C) = {A, B};F: Point;Focus(C) = F;Abs(VectorOf(F, A)) = 2*Abs(VectorOf(F, B))",
    "query_expressions": "k",
    "answer_expressions": "pm*((2*sqrt(2))/3)",
    "fact_spans": "[[[2, 14]], [[2, 14]], [[113, 118]], [[15, 34], [51, 57]], [[15, 34]], [[37, 40]], [[41, 44]], [[2, 46]], [[47, 50]], [[47, 60]], [[63, 111]]]",
    "query_spans": "[[[113, 120]]]",
    "process": ""
  },
  {
    "text": "A point $M$ on the parabola $y^{2}=x$ is at a distance of $2$ from the focus. Then the horizontal coordinate of point $M$ is?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = x);PointOnCurve(M, G);Distance(M, Focus(G)) = 2",
    "query_expressions": "XCoordinate(M)",
    "answer_expressions": "7/4",
    "fact_spans": "[[[0, 12]], [[15, 18], [30, 34]], [[0, 12]], [[0, 18]], [[0, 28]]]",
    "query_spans": "[[[30, 40]]]",
    "process": "It is easy to obtain that the directrix equation of the parabola $ y^{2} = x $ is $ x = -\\frac{1}{4} $. Given that the distance from a point $ M $ on the parabola $ y^{2} = x $ to the focus is 2, and since the distance from any point on the parabola to the focus equals the distance from that point to the directrix, the horizontal coordinate of point $ M $ is: $ 2 - \\frac{1}{4} = \\frac{7}{4} $."
  },
  {
    "text": "Let a circle pass through a vertex and a focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, with its center on this hyperbola. Then the distance from the center of the circle to the center of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);H: Circle;PointOnCurve(OneOf(Vertex(G)),H) = True;PointOnCurve(OneOf(Focus(G)),H) = True;PointOnCurve(Center(H),G) = True",
    "query_expressions": "Distance(Center(H),Center(G))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[3, 42], [56, 59], [64, 67]], [[3, 42]], [[1, 2]], [[1, 47]], [[1, 52]], [[1, 60]]]",
    "query_spans": "[[[61, 74]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0, b>0)$, a line with slope $2$ is drawn from the lower focus $F$ intersecting one of the asymptotes of the hyperbola at point $A$, where $A$ lies in the first quadrant. If $|O A|=|O F|$ ($O$ is the origin), then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;G: Line;O: Origin;A: Point;F: Point;a>0;b>0;Expression(C) = (-x^2/b^2 + y^2/a^2 = 1);LowerFocus(C)=F;PointOnCurve(F,G);Slope(G)=2;Intersection(G,OneOf(Asymptote(C)))=A;Quadrant(A)=1;Abs(LineSegmentOf(O, A)) = Abs(LineSegmentOf(O, F))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[2, 63], [82, 85], [136, 142]], [[10, 63]], [[10, 63]], [[79, 81]], [[125, 128]], [[94, 98], [100, 103]], [[68, 71]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 71]], [[64, 81]], [[72, 81]], [[79, 98]], [[100, 108]], [[110, 124]]]",
    "query_spans": "[[[136, 148]]]",
    "process": "Let the equation of line AF be y = 2x - c, and the asymptotes of hyperbola C be y = \\pm\\frac{a}{b}x. From \\begin{cases} y = 2x - c \\\\ y = \\frac{a}{b}x \\end{cases}, solving gives \\begin{cases} x = \\frac{bc}{2b - a} \\\\ y = \\frac{ac}{2b - a} \\end{cases}. Thus, A\\left(\\frac{bc}{2b - a}, \\frac{ac}{2b - a}\\right). Given |OA| = |OF| = c, we have \\left(\\frac{bc}{2b - a}\\right)^{2} + \\left(\\frac{ac}{2b - a}\\right)^{2} = c^{2}. Simplifying yields b^{2} + a^{2} = (2b - a)^{2}, which rearranges to 4a = 3b. Hence, 16a^{2} = 9c^{2} - 9a^{2}, i.e., 25a^{2} = 9c^{2}, and e^{2} = \\frac{25}{9}. Therefore, the eccentricity is e = \\frac{5}{3}."
  },
  {
    "text": "Given points $A$ and $B$ lie on the ellipse $G$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and the segment $AB$ is exactly a diameter of the circle $x^{2}+y^{2}=R^{2}$ $(R>0)$. Let $M$ be a point on ellipse $G$ distinct from $A$ and $B$, such that the product of the slopes of lines $MA$ and $MB$ is $-\\frac{1}{4}$. Find the eccentricity of ellipse $G$.",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);Z: Circle;Expression(Z) = (x^2 + y^2 = R^2);IsDiameter(LineSegmentOf(A, B),Z);R: Number;R>0;M: Point;PointOnCurve(M, G);Negation(M=A);Negation(M=B);Slope(LineOf(M, A))*Slope(LineOf(M, B)) = -1/4",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[11, 68], [119, 124], [177, 182]], [[11, 68]], [[18, 68]], [[18, 68]], [[18, 68]], [[18, 68]], [[2, 6], [126, 129]], [[7, 10], [130, 133]], [[2, 72]], [[2, 72]], [[84, 109]], [[84, 109]], [[74, 114]], [[85, 109]], [[85, 109]], [[115, 118]], [[115, 139]], [[115, 139]], [[115, 139]], [[141, 175]]]",
    "query_spans": "[[[177, 188]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ M(x_{0},y_{0}) $. According to the problem,\n$$\n\\begin{cases}\n\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 \\\\\n\\frac{x_{0}^{2}}{a^{2}}+\\frac{y_{0}}{\\frac{0}{2}}=1\n\\end{cases}\n$$\nSubtracting the two equations gives\n$$\n\\frac{x^{2}-x_{0}^{2}}{a^{2}} = \\frac{y_{\\frac{1}{b^{2}}-y_{0}^{2}}\n$$\nSince segment $ AB $ is exactly a diameter of the circle $ x^{2}+y^{2}=R^{2} $ ($ R>0 $), the product of the slopes of lines $ MA $ and $ MB $ is\n$$\n\\frac{y_{1}-y_{0}}{x_{1}}-x_{0} \\cdot \\frac{x}{x_{1}}-x_{0} = \\frac{y_{2}-y_{0}^{2}}{x_{1}^{2}-x_{0}^{2}} = -\\frac{b^{2}}{a^{2}} = -\\frac{1}{4},\n$$\nsolving gives $ \\frac{b^{2}}{a^{2}} = \\frac{1}{4} $. Then $ B(-x_{1} $, so the eccentricity of ellipse $ G $ is\n$$\ne = \\frac{\\sqrt{a2-b^{2}}}{a^{2}} = \\sqrt{1-\\frac{b^{2}}{a^{2}}} = \\frac{\\sqrt{3}}{2}.\n$$"
  },
  {
    "text": "Let $e_{1}$, $e_{2}$ be the eccentricities of an ellipse and a hyperbola having common foci $F_{1}$ and $F_{2}$, and let $P$ be a common point of the two curves such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$. Then the minimum value of $4 e_{1}+e_{2}^2$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;P: Point;F1: Point;F2: Point;e1:Number;e2:Number;Focus(G)={F1,F2};Focus(H)={F1,F2};Eccentricity(H)=e1;Eccentricity(G)=e2;OneOf(Intersection(H,G))=P;DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0",
    "query_expressions": "Min(4*e1+e2^2)",
    "answer_expressions": "9/2",
    "fact_spans": "[[[46, 49]], [[43, 45]], [[54, 57]], [[27, 34]], [[35, 42]], [[1, 8]], [[10, 18]], [[21, 49]], [[21, 49]], [[1, 53]], [[1, 53]], [[54, 67]], [[71, 130]]]",
    "query_spans": "[[[132, 155]]]",
    "process": "According to the definitions of the ellipse and hyperbola, find the relationship between |PF_{1}| and |PF_{2}|. Then, from \\overrightarrow{PF}_{1}\\cdot\\overrightarrow{PF_{2}}=0, we get PF_{1}\\bot PF_{2}. Using the Pythagorean theorem, we obtain \\frac{1}{e_{1}^{2}}+\\frac{1}{e_{2}^{2}}=2. Combining with the basic inequality, the solution can be obtained. [Solution] Let the semi-major axis of the ellipse be a_{1}, the semi-transverse axis of the hyperbola be a_{2} (a_{1}>a_{2}), and their semi-focal length be c. Let P be a common point of the two curves, and assume |PF_{1}|>|PF_{2}|. Thus, |PF_{1}|+|PF_{2}|=2a_{1}, |PF_{1}|-|PF_{2}|=2a_{2}. Therefore, |PF_{1}|=a_{1}+a_{2}, |PF_{2}|=a_{1}-a_{2}. Since \\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=0, we have PF_{1}\\bot PF_{2}, and |PF_{1}|^{2}+|PF_{2}|^{2}=(2c)^{2}. Hence, (a_{1}+a_{2})^{2}+(a_{1}-a_{2})^{2}=(2c)^{2}, which implies 2c^{2}=a_{1}^{2}+a_{2}^{2}, so 2=\\frac{a_{1}^{2}}{c^{2}}+\\frac{a_{2}^{2}}{c^{2}}, i.e., \\frac{1}{e_{1}^{2}}+\\frac{1}{e_{2}^{2}}=2. Thus, 4e_{1}^{2}+e_{2}^{2}=\\frac{1}{2}(4e_{1}^{2}+e_{2}^{2})\\left(\\frac{1}{e_{1}^{2}}+\\frac{1}{e_{2}^{2}}\\right)=\\frac{1}{2}\\left(5+\\frac{e_{2}^{2}}{e_{1}^{2}}+\\frac{4e_{1}^{2}}{e_{2}^{2}}\\right)\\geqslant\\frac{1}{2}\\left(5+2\\sqrt{\\frac{e_{2}^{2}}{e_{1}^{2}}\\cdot\\frac{4e_{1}^{2}}{e_{2}^{2}}}\\right)=\\frac{9}{2}, where equality holds if and only if e_{2}^{2}=2e_{1}^{2}=\\frac{3}{2}. Therefore, the minimum value of 4e_{1}^{2}+e_{2}^{2} is \\frac{9}{2}."
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ are $F_{1}$ and $F_{2}$ respectively, and the coordinates of the upper vertex $A$ are $(0, \\sqrt{3})$. If the area of the incircle of $\\Delta A F_{1} F_{2}$ is $\\frac{1}{3} \\pi$, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;A: Point;UpperVertex(G) = A;Coordinate(A) = (0, sqrt(3));Area(InscribedCircle(TriangleOf(A, F1, F2))) = (1/3)*pi",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4+y^2/3=1",
    "fact_spans": "[[[0, 52], [153, 155]], [[0, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[61, 68]], [[69, 76]], [[0, 76]], [[0, 76]], [[80, 83]], [[0, 83]], [[80, 102]], [[104, 151]]]",
    "query_spans": "[[[153, 159]]]",
    "process": "From the given conditions, $AAF_{1}F_{2}$ is an isosceles triangle. According to equal areas, determine $a+c=\\sqrt{3}b$, and $b=\\sqrt{3}$, $a^{2}=b^{2}+c^{2}$, then solve. Let the inradius of $AAF_{1}F_{2}$ be $r$, then its incircle area is $S=\\pi r^{2}=\\frac{1}{3}\\pi$, solving gives $r=\\frac{\\sqrt{3}}{3}$. From the given conditions, $AAF_{1}F_{2}$ is an isosceles triangle, and $|AF_{1}|=|AF_{2}|=a$, $|F_{1}F_{2}|=2c$. $\\because OA\\bot F_{1}F_{2}$, $|OA|=b$ $\\therefore S_{\\Delta AF_{1}F_{2}}=\\frac{1}{2}(|AF_{1}|+|AF_{2}|+|F_{1}F_{2}|)\\times r=\\frac{1}{2}\\times |F_{1}F_{2}|\\times |OA|$, i.e., $S_{AAF_{1}F_{2}}=\\frac{1}{2}(2a+2c)\\times r=\\frac{1}{2}\\times 2c\\times b$, yielding $(a+c)r=bc$ $\\therefore a+c=\\sqrt{3}bc$. Also $\\because b=\\sqrt{3}$, $a^{2}=b^{2}+c^{2}$ $\\therefore c=1$, $a=2$ $\\therefore$ the ellipse equation is $\\frac{x^{2}}{4}+\\frac{3}{2}$"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a, b>0)$ has its right focus at $F$. If there exists a point $P$ on the line $x=-a$ such that $\\angle O P F=30^{\\circ}$, where $O$ is the origin, then the minimum value of the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;O: Origin;P: Point;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x = -a);RightFocus(G) = F;PointOnCurve(P, H);AngleOf(O, P, F) = ApplyUnit(30, degree)",
    "query_expressions": "Min(Eccentricity(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 56], [122, 125]], [[5, 56]], [[5, 56]], [[66, 74]], [[112, 115]], [[77, 81]], [[61, 64]], [[5, 56]], [[5, 56]], [[2, 56]], [[66, 74]], [[2, 64]], [[66, 81]], [[84, 109]]]",
    "query_spans": "[[[122, 135]]]",
    "process": "Let the line $ x = -a $ intersect the x-axis at point $ H $. Let $ P(-a, t) $ ($ t > 0 $), then  \n$ \\tan 30^{\\circ} = \\tan \\angle OPF = \\tan(\\angle HPF - \\angle HPO) = t + \\frac{a^{2} + ac}{t} \\geqslant 2\\sqrt{t \\cdot \\frac{a^{2} + ac}{t}} = 2\\sqrt{a^{2} + ac} $,  \nso $ \\frac{\\sqrt{3}}{3} \\leqslant \\frac{c}{2\\sqrt{a^{2} + ac}} $.  \nInequality (1) simplifies to $ 3c^{2} - 4a^{2} - 4ac \\geqslant 0 $,  \n$ 3e^{2} - 4e - 4 \\geqslant 0 $, solving gives $ e \\geqslant 2 $,  \nthus the minimum eccentricity of the hyperbola is $ 2 $."
  },
  {
    "text": "The line $l$ intersects the parabola $y^{2}=4x$ at two points $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$, and $O$ is the origin. If $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}=-4$, then $x_{1} x_{2}$=?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;O: Origin;Expression(G) = (y^2 = 4*x);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Intersection(l, G) = {A,B};DotProduct(VectorOf(O, A), VectorOf(O, B)) = -4;x1:Number;y1:Number;x2:Number;y2:Number",
    "query_expressions": "x1*x2",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 5]], [[6, 20]], [[24, 41]], [[43, 60]], [[63, 66]], [[6, 20]], [[24, 41]], [[43, 60]], [[0, 60]], [[73, 125]], [[24, 41]], [[24, 41]], [[43, 60]], [[43, 60]]]",
    "query_spans": "[[[127, 142]]]",
    "process": "Since the line $ l $ intersects the parabola $ y^{2} = 4x $ at two points $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, it follows from $ \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = -4 \\Rightarrow x_{1}x_{2} + y_{1}y_{2} = -4 $. Since $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $ lie on the parabola $ y^{2} = 4x $, we have $ y_{1}^{2}y_{2}^{2} = 16x_{1}x_{2} $, that is, $ x_{1}x_{2} = \\frac{y_{1}^{2}y_{2}^{2}}{16} $. Therefore, $ \\frac{y_{1}^{2}y_{2}^{2}}{16} + y_{1}y_{2} = -4 $. Solving gives: $ y_{1}y_{2} = -8 $, so $ x_{1}x_{2} $"
  },
  {
    "text": "The equation of the locus of points at a distance of $2$ from the line $3x - 4y - 2 = 0$ is?",
    "fact_expressions": "G: Line;P: Point;Expression(G) = (3*x - 4*y - 2 = 0);Distance(P, G) = 2",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "{3*x-4*y-12=0, 3*x-4*y+8=0}",
    "fact_spans": "[[[1, 16]], [[24, 25]], [[1, 16]], [[0, 25]]]",
    "query_spans": "[[[24, 32]]]",
    "process": "Let the coordinates of points whose distance to the line $3x - 4y - 2 = 0$ is equal to 2 be $(x, y)$. According to the problem: $\\frac{|3x - 4y - 2|}{\\sqrt{3^{2} + (-4)^{2}}} = 2 \\Rightarrow |3x - 4y - 2| = 10 \\Rightarrow 3x - 4y - 2 = \\pm 10 \\Rightarrow 3x - 4y - 12 = 0$ or $3x - 4y + 8 = 0$,"
  },
  {
    "text": "The center is at the origin, and the foci are on the $y$-axis. If the major axis has length $10$ and the focal distance is $8$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;PointOnCurve(Focus(G), yAxis);Length(MajorAxis(G)) = 10;FocalLength(G) = 8",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/25 + x^2/9 = 1",
    "fact_spans": "[[[32, 34]], [[3, 5]], [[0, 34]], [[6, 34]], [[16, 34]], [[24, 34]]]",
    "query_spans": "[[[32, 41]]]",
    "process": "The center is at the origin, and the foci are on the y-axis. Let the standard equation of the ellipse be $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$. Since the major axis length is 10 and the focal distance is 8, $2a=10$, $2c=8$, so $a=5$, $c=4$. Since $a^{2}=b^{2}+c^{2}$, then $b^{2}=25-16=9$. Therefore, the standard equation of the ellipse is $\\frac{y^{2}}{25}+\\frac{x^{2}}{9}=1$."
  },
  {
    "text": "Given the parabola $y^{2}=2 x$, then its focus coordinates are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(1/2,0)",
    "fact_spans": "[[[2, 16], [18, 19]], [[2, 16]]]",
    "query_spans": "[[[18, 26]]]",
    "process": "From the parabola equation $ y^{2} = 2x $, we can obtain $ 2p = 2 $, that is, $ p = 1 $, and the focus lies on the positive x-axis. Then the coordinates of the focus are $ \\left( \\frac{1}{7}, 0 \\right) $."
  },
  {
    "text": "Given that point $P$ lies on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ with foci $F_{1}$ and $F_{2}$, then $|P F_{1}|+|P F_{2}|$=?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/16 + y^2/9 = 1);Focus(G)={F1,F2};PointOnCurve(P,G)",
    "query_expressions": "Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "8",
    "fact_spans": "[[[26, 64]], [[2, 6]], [[10, 17]], [[18, 25]], [[26, 64]], [[7, 64]], [[2, 65]]]",
    "query_spans": "[[[67, 90]]]",
    "process": "Since point P lies on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ with foci $F_{1}$ and $F_{2}$, we have $a^{2}=16$, so $a=4$. Therefore, $|PF_{1}|+|PF_{2}|=2a=8$,"
  },
  {
    "text": "Given that the right vertex of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ is $A$, and the upper vertex is $B$, then $|A B|$ equals?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;Expression(G) = (x^2/9 + y^2/4 = 1);RightVertex(G)=A;UpperVertex(G)=B",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "sqrt(13)",
    "fact_spans": "[[[2, 39]], [[44, 47]], [[52, 55]], [[2, 39]], [[2, 47]], [[2, 55]]]",
    "query_spans": "[[[57, 67]]]",
    "process": "\\because the standard equation of the ellipse is \\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1, \\therefore the right vertex is A(3,0), the upper vertex is B(0,2), \\therefore |AB|=\\sqrt{3^{2}+2^{2}}=\\sqrt{13}"
  },
  {
    "text": "What is the focal length of the hyperbola $9 x^{2}-16 y^{2}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (9*x^2 - 16*y^2 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "5/6",
    "fact_spans": "[[[0, 23]], [[0, 23]]]",
    "query_spans": "[[[0, 28]]]",
    "process": "The standard form of the hyperbola $9x^{2}-16y^{2}=1$ is $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, then $a^{2}=\\frac{1}{9}$, $b^{2}=\\frac{1}{16}$, so $c^{2}=a^{2}+b^{2}=\\frac{25}{16\\times9}$, then $c=\\frac{5}{12}$, so the focal distance is $2c=\\frac{5}{6}$."
  },
  {
    "text": "Given the ellipse $ C $: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 $ ($ a>b>0 $), the right focus is $ F $, the left and right vertices are $ A $, $ B $, $|F A|=3$, $|F B|=1$. Then the length of the chord intercepted by the line $ y=x+\\frac{1}{2} $ on the ellipse $ C $ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;F: Point;RightFocus(C) = F;A: Point;B: Point;LeftVertex(C) = A;RightVertex(C) = B;Abs(LineSegmentOf(F,A)) = 3;Abs(LineSegmentOf(F,B)) = 1;G: Line;Expression(G) = (y = x + 1/2)",
    "query_expressions": "Length(InterceptChord(G,C))",
    "answer_expressions": "18*sqrt(2)/7",
    "fact_spans": "[[[2, 59], [128, 133]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[64, 67]], [[2, 67]], [[74, 77]], [[79, 83]], [[2, 83]], [[2, 83]], [[85, 94]], [[96, 105]], [[108, 127]], [[108, 127]]]",
    "query_spans": "[[[108, 140]]]",
    "process": "Let the semi-focal distance of the ellipse be $ c $. From $ |FA| = 3 $, $ |FB| = 1 $, we obtain $ a + c = 3 $, $ a - c = 1 $. Solving gives $ a = 2 $, $ c = 1 $, then $ b = \\sqrt{a^{2} - c^{2}} = \\sqrt{4 - 1} = \\sqrt{3} $. Thus, the equation of the ellipse is $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $. Combining the line $ y = x + \\frac{1}{2} $ and the ellipse $ 3x^{2} + 4y^{2} = 12 $, we get $ 7x^{2} + 4x - 11 = 0 $. Let the horizontal coordinates of the endpoints of the chord intercepted by the ellipse $ C $ be $ x_{1} $, $ x_{2} $, then $ x_{1} + x_{2} = -\\frac{4}{7} $, $ x_{1}x_{2} = -\\frac{11}{7} $. The chord length is $ \\sqrt{1 + k^{2}} \\cdot \\sqrt{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}} = \\sqrt{1 + 1} \\cdot \\sqrt{(-\\frac{4}{7})^{2} - 4 \\times (-\\frac{11}{7})} = \\frac{18\\sqrt{2}}{7} $."
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{t}+\\frac{y^{2}}{t-2}=1$ represents a hyperbola, what is the range of real values for $t$?",
    "fact_expressions": "G: Hyperbola;t: Real;Expression(G)=(x^2/t+y^2/(t-2)=1)",
    "query_expressions": "Range(t)",
    "answer_expressions": "(0,2)",
    "fact_spans": "[[[43, 46]], [[48, 53]], [[2, 46]]]",
    "query_spans": "[[[48, 60]]]",
    "process": "According to the problem, for $\\frac{x^{2}}{t}+\\frac{y^{2}}{t-2}=1$ to represent a hyperbola, it is sufficient that $t(t-2)<0$, which gives $0<t<2$. Therefore, the range of real values for $t$ is $(0,2)$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$ and directrix $l$, let $P$ be a point on $l$, and $Q$ be an intersection point of line $PF$ and the parabola $C$. If $\\overrightarrow{PF}=4\\overrightarrow{FQ}$, then $|QF|$=?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l: Line;Directrix(C) = l;P: Point;PointOnCurve(P, l);Q: Point;OneOf(Intersection(LineOf(P, F), C)) = Q;VectorOf(P, F) = 4*VectorOf(F, Q)",
    "query_expressions": "Abs(LineSegmentOf(Q, F))",
    "answer_expressions": "5/2",
    "fact_spans": "[[[2, 21], [60, 66]], [[2, 21]], [[25, 28]], [[2, 28]], [[32, 35], [41, 44]], [[2, 35]], [[37, 40]], [[37, 47]], [[48, 51]], [[48, 71]], [[74, 119]]]",
    "query_spans": "[[[121, 130]]]",
    "process": "Let P(-1,t), Q(x_{0},y_{0}). Since F(1,0), we have \\overrightarrow{PF}=(2,-t), \\overrightarrow{FQ}=(x_{0}-1,y_{0}). From \\overrightarrow{PF}=4\\overrightarrow{FQ}, we get 2=4(x_{0}-1), so x_{0}=\\frac{3}{2}. According to the definition of the parabola, |OF|=x_{0}+1=\\frac{5}{2}. The final answer is \\frac{5}{2}."
  },
  {
    "text": "Given that the equation of the directrix of a parabola is $x=1$, then the standard equation of this parabola is?",
    "fact_expressions": "G: Parabola;Expression(Directrix(G)) = (x = 1)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=-4*x",
    "fact_spans": "[[[3, 6], [20, 23]], [[3, 17]]]",
    "query_spans": "[[[20, 30]]]",
    "process": ""
  },
  {
    "text": "If the ellipse $C$: $\\frac{x^{2}}{m^{2}+1}+\\frac{y^{2}}{m}=1$ $(m>0)$ has a focal distance of $2 \\sqrt{3}$, then what is the length of the major axis of ellipse $C$?",
    "fact_expressions": "C: Ellipse;m: Number;m>0;Expression(C) = (x^2/(m^2 + 1) + y^2/m = 1);FocalLength(C) = 2*sqrt(3)",
    "query_expressions": "Length(MajorAxis(C))",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[1, 54], [72, 77]], [[8, 54]], [[8, 54]], [[1, 54]], [[1, 70]]]",
    "query_spans": "[[[72, 83]]]",
    "process": "According to the problem, the ellipse $ C: \\frac{x^2}{m^{2}+1} + \\frac{y^{2}}{m} = 1 $ ($ m > 0 $) has a focal distance of $ 2\\sqrt{3} $. Then $ m^{2}+1 - m = (\\sqrt{3})^{2} $, solving gives $ m = 2 $, so $ m^{2}+1 = 5 $, therefore the major axis length of ellipse $ C $ is $ 2\\sqrt{m^{2}+1} = 2\\sqrt{5} $."
  },
  {
    "text": "Given that the line $2x - y + 2 = 0$ passes through a vertex and a focus of the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1$ $(a > b > 0)$, then the equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (2*x - y + 2 = 0);PointOnCurve(OneOf(Vertex(G)),H);PointOnCurve(OneOf(Focus(G)),H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5 + y^2/4 = 1",
    "fact_spans": "[[[17, 69], [84, 86]], [[19, 69]], [[19, 69]], [[2, 15]], [[19, 69]], [[19, 69]], [[17, 69]], [[2, 15]], [[2, 74]], [[2, 79]]]",
    "query_spans": "[[[84, 91]]]",
    "process": "The line $x - 2y + 2 = 0$ intersects the coordinate axes at points $(-2, 0)$ and $(0, 1)$. According to the problem, $c = 2$, $b = 1$ $\\Rightarrow a = \\sqrt{5}$ $\\Rightarrow \\frac{x^{2}}{5} + \\frac{y^{2}}{4} = 1$"
  },
  {
    "text": "Given that a line with slope $1$ passing through the right focus $F$ of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ intersects the ellipse at points $A$ and $B$, and if $AF=2FB$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G:Ellipse;Expression(G)=(x^2/a^2+y^2/b^2=1);A:Point;B:Point;F:Point;a:Number;b:Number;a>b;b>0;H:Line;RightFocus(G)=F;PointOnCurve(F,H);Slope(H)=1;Intersection(H,G)={A,B};LineSegmentOf(A,F)=2*LineSegmentOf(F,B)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/3",
    "fact_spans": "[[[3, 55], [72, 74], [99, 101]], [[3, 55]], [[75, 78]], [[81, 84]], [[59, 62]], [[5, 55]], [[5, 55]], [[5, 55]], [[5, 55]], [[69, 71]], [[3, 62]], [[2, 71]], [[62, 71]], [[69, 86]], [[88, 97]]]",
    "query_spans": "[[[99, 107]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the right branch of the hyperbola such that $|P F_{1}|=4|P F_{2}|$. Then, the maximum value of the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F1)) = 4*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Max(Eccentricity(G))",
    "answer_expressions": "5/3",
    "fact_spans": "[[[2, 58], [88, 91], [122, 125]], [[5, 58]], [[5, 58]], [[83, 87]], [[67, 74]], [[75, 82]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 82]], [[2, 82]], [[83, 95]], [[97, 119]]]",
    "query_spans": "[[[121, 135]]]",
    "process": "Solution 1: Since  \n\\[\n\\begin{cases}\n|PF_{1}|-|PF_{2}|=2a, \\\\\n|PF_{1}|=4|PF_{2}|,\n\\end{cases}\n\\]\nwe have  \n\\[\n\\begin{cases}\n|PF_{1}|=\\frac{8a}{3}, \\\\\n|PF_{2}|=\\frac{2a}{3}.\n\\end{cases}\n\\]  \nIn $\\triangle PF_{1}F_{2}$, by the cosine law,  \n\\[\n4c^{2} = \\left(\\frac{8a}{3}\\right)^{2} + \\left(\\frac{2a}{3}\\right)^{2} - 2 \\times \\frac{8a}{3} \\times \\frac{2a}{3} \\cos\\theta = \\frac{68}{9}a^{2} - \\frac{32}{9}a^{2}\\cos\\theta.\n\\]  \nDividing both sides by $a^{2}$, we get  \n\\[\n4e^{2} = \\frac{68}{9} - \\frac{32}{9}\\cos\\theta.\n\\]  \nSince $\\cos\\theta \\in (-1,1)$, it follows that  \n\\[\n4 < 4e^{2} < \\frac{100}{9}, \\quad 1 < e < \\frac{5}{3}.\n\\]  \nWhen points $P$, $F_{1}$, $F_{2}$ are collinear, $\\theta = 180^{\\circ}$, $e = \\frac{5}{3}$, thus $1 < e \\leqslant \\frac{5}{3}$, and the maximum value of $e$ is $\\frac{5}{3}$.\n\nSolution 2: From  \n\\[\n\\begin{cases}\n|PF_{1}|-|PF_{2}|=2a, \\\\\n|PF_{1}|=4|PF_{2}|,\n\\end{cases}\n\\]\nwe obtain $|PF_{1}| = \\frac{8a}{3}$, $|PF_{2}| = \\frac{2a}{3}$. Let $|PP'|$ be the distance from point $P$ to the directrix, then  \n\\[\nPP' = \\frac{3}{e} \\geqslant a - \\frac{a^{2}}{c} \\Leftrightarrow \\frac{5}{3e} \\geqslant 1 \\Leftrightarrow e \\leqslant \\frac{5}{3}.\n\\]"
  },
  {
    "text": "Given that point $P$ moves on the parabola $y^{2}=8x$, $F$ is the focus of the parabola, and point $A$ has coordinates $(5,2)$, then the minimum value of $PA + PF$ is?",
    "fact_expressions": "G: Parabola;P: Point;A: Point;F: Point;Expression(G) = (y^2 = 8*x);Coordinate(A) = (5, 2);PointOnCurve(P, G);Focus(G) = F",
    "query_expressions": "Min(LineSegmentOf(P, A) + LineSegmentOf(P, F))",
    "answer_expressions": "7",
    "fact_spans": "[[[7, 21], [29, 32]], [[2, 6]], [[36, 40]], [[25, 28]], [[7, 21]], [[36, 51]], [[2, 22]], [[25, 35]]]",
    "query_spans": "[[[53, 68]]]",
    "process": "PA+PF\\geqslantd_{A-L}=5+\\frac{P}{2}=5+2=7"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, one of its asymptotes is perpendicular to the line $l$: $x+\\sqrt{3} y=0$, and the distance from a focus of the hyperbola $C$ to the line $l$ is $1$. Then the equation of the hyperbola $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(l)=(x+sqrt(3)*y=0);IsPerpendicular(OneOf(Asymptote(C)),l);Distance(OneOf(Focus(C)), l) = 1",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[69, 91], [106, 111]], [[2, 62], [94, 100], [120, 126]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[2, 62]], [[69, 91]], [[2, 93]], [[94, 118]]]",
    "query_spans": "[[[120, 131]]]",
    "process": "Since one asymptote of the hyperbola is perpendicular to the line $ l $, use the relationship between the slopes of two perpendicular lines to obtain $ \\frac{b}{a} = \\sqrt{3} $. Combine this with the point-to-line distance formula to find $ c = 2 $, then solve the system of equations involving $ a $, $ b $, and $ c $ to obtain the equation. Because one asymptote of the hyperbola is perpendicular to the line $ l: x + \\sqrt{3}y = 0 $, the slope of the hyperbola's asymptotes is $ \\sqrt{3} $, so $ \\frac{b}{a} = \\sqrt{3} $.\\textcircled{1} From the given conditions, the foci of the hyperbola lie on the $ x $-axis; thus, one focus can be set as $ (c, 0) $. Using the point-to-line distance formula, we get $ \\frac{|c|}{2} = 1 $, so $ c = 2 $, which implies $ a^{2} + b^{2} = 4 $.\\textcircled{2} Solving equations \\textcircled{1} and \\textcircled{2} simultaneously gives $ a^{2} = 1 $, $ b^{2} = 3 $. Therefore, the standard equation of the hyperbola is $ x^{2} - \\frac{y^{2}}{3} = 1 $."
  },
  {
    "text": "The coordinates of the focus of the parabola $y=-\\frac{1}{4} x^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = -x^2/4)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,-1)",
    "fact_spans": "[[[0, 25]], [[0, 25]]]",
    "query_spans": "[[[0, 32]]]",
    "process": "The parabola equation is transformed into $x^{2}=-4y\\therefore2p=4\\therefore\\frac{p}{2}=1$, the focus is $(0,-1)$"
  },
  {
    "text": "If the foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ lie on the $x$-axis, and tangents are drawn from the point $(1 , \\frac{1}{2})$ to the circle $x^{2}+y^{2}=1$, with points of tangency $A$ and $B$, respectively, such that the line $AB$ passes exactly through the right focus and the upper vertex of the ellipse, then the equation of the ellipse is?",
    "fact_expressions": "A: Point;B: Point;G: Ellipse;b: Number;a: Number;H: Circle;P: Point;l1: Line;l2:Line;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + y^2 = 1);Coordinate(P) = (1, 1/2);PointOnCurve(Focus(G), xAxis);TangentOfPoint(P,H)={l1,l2};TangentPoint(l1,H)=A;TangentPoint(l2,H)=B;PointOnCurve(RightFocus(G),LineOf(A,B));PointOnCurve(UpperVertex(G),LineOf(A,B))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5+y^2/4=1",
    "fact_spans": "[[[103, 106]], [[107, 110]], [[1, 46], [121, 123], [121, 123]], [[3, 46]], [[3, 46]], [[78, 94]], [[57, 77]], [], [], [[1, 46]], [[78, 94]], [[57, 77]], [[1, 55]], [[56, 97]], [[56, 110]], [[56, 110]], [[111, 127]], [[111, 131]]]",
    "query_spans": "[[[133, 139]]]",
    "process": ""
  },
  {
    "text": "Let the focus of the parabola $y^{2}=4 x$ be $F$, the directrix be $l$, and let $P$ be a point on the parabola. Let $P Q \\perp x$-axis at $Q$. If $P F=3$, then what is the length of the segment $P Q$?",
    "fact_expressions": "G: Parabola;P: Point;Q: Point;F: Point;l: Line;Expression(G) = (y^2 = 4*x);Focus(G) = F;Directrix(G) = l;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P,Q),xAxis);FootPoint(LineSegmentOf(P,Q),xAxis)=Q;LineSegmentOf(P,F)=3",
    "query_expressions": "Length(LineSegmentOf(P,Q))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[1, 15], [35, 38]], [[31, 34]], [[58, 61]], [[19, 22]], [[26, 29]], [[1, 15]], [[1, 22]], [[1, 29]], [[31, 42]], [[43, 57]], [[43, 61]], [[63, 70]]]",
    "query_spans": "[[[72, 83]]]",
    "process": "Given PF=3, using the definition of the parabola to find the x-coordinate of P, substitute the x-coordinate into the parabola equation to obtain the absolute value of the y-coordinate. The directrix of the parabola y^{2}=4x is x=-1. Since PF=3, by the definition of the parabola, the distance from P to the directrix equals 3, that is, x_{p}+1=3\\Rightarrow x_{p}=2. Substituting x_{p}=2 into the parabola equation gives y_{p}^{2}=8\\Rightarrow |y_{p}|=2\\sqrt{2}. Therefore, the length of segment PQ is 2\\sqrt{2}."
  },
  {
    "text": "Given that one focus of the ellipse $k x^{2}+5 y^{2}=5$ is $(2,0)$, then $k=?$",
    "fact_expressions": "G: Ellipse;Expression(G) = (k*x^2 + 5*y^2 = 5);k: Number;Coordinate(OneOf(Focus(G))) = (2, 0)",
    "query_expressions": "k",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 23]], [[2, 23]], [[40, 43]], [[2, 38]]]",
    "query_spans": "[[[40, 45]]]",
    "process": "According to the problem, the equation of the ellipse can be written as \\frac{x^{2}}{k}+\\frac{y^{2}}{1}=1. Since the coordinates of the focus are (2,0), it follows that \\frac{5}{k}-1=2^{2}\\Rightarrow k=1."
  },
  {
    "text": "The foci of the ellipse $x^{2}+\\frac{y^{2}}{m}=1$ lie on the $y$-axis, and the length of the major axis is twice the length of the minor axis. Then $m=$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + y^2/m = 1);m: Number;PointOnCurve(Focus(G), yAxis);Length(MajorAxis(G)) = 2*Length(MinorAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 27]], [[0, 27]], [[51, 54]], [[0, 36]], [[0, 49]]]",
    "query_spans": "[[[51, 56]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ $(m>0)$ is $\\frac{2 \\sqrt{3}}{3}$, then the value of $m$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2 + x^2/m = 1);m: Number;m>0;Eccentricity(G) = 2*sqrt(3)/3",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 35]], [[2, 35]], [[64, 67]], [[5, 35]], [[2, 62]]]",
    "query_spans": "[[[64, 71]]]",
    "process": "e=\\frac{c}{a}=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{\\frac{a^{2}+b^{2}}{a^{2}}}=\\sqrt{\\frac{m+1}{m}}=\\frac{2\\sqrt{3}}{3}, solving gives m=3"
  },
  {
    "text": "The line $2x + y - 4 = 0$ passes through the focus of the parabola $y^2 = 2px$. What is the equation of the directrix of the parabola?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (2*x + y - 4 = 0);PointOnCurve(Focus(G), H)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x = -2",
    "fact_spans": "[[[15, 31], [36, 39]], [[18, 31]], [[0, 13]], [[15, 31]], [[0, 13]], [[0, 34]]]",
    "query_spans": "[[[36, 46]]]",
    "process": "According to the problem, the focus of the parabola $ y^{2} = 2px $ is $ F\\left(\\frac{p}{2}, 0\\right) $. Since the focus lies on the line $ 2x + y - 4 = 0 $, we have $ p - 4 = 0 $, so $ p = 4 $. Therefore, the equation of the directrix of the parabola is $ x = -\\frac{p}{2} = -2 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $A(a, 0)$, $B(2 a, 0)$, and point $P$ is a point on the right branch of hyperbola $C$ (distinct from point $A$), satisfying $\\overrightarrow{P A}^{2}+\\overrightarrow{P B}^{2}=a^{2}$, then the range of the eccentricity $e$ of this hyperbola is?",
    "fact_expressions": "C: Hyperbola;P: Point;A: Point;B: Point;a: Number;b: Number;a>0;b>0;e: Number;Expression(C) = (x^2/a^2 - y^2/b^2 = 1);Coordinate(A) = (a, 0);Coordinate(B) = (2*a, 0);PointOnCurve(P,RightPart(C));Negation(P=A);VectorOf(P, A)^2 + VectorOf(P, B)^2 = a^2;Eccentricity(C) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "(1, \\sqrt{6}/2)",
    "fact_spans": "[[[2, 63], [93, 99], [176, 179]], [[88, 92]], [[65, 74], [108, 112]], [[76, 87]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[182, 185]], [[2, 63]], [[65, 74]], [[76, 87]], [[88, 104]], [[88, 113]], [[116, 173]], [[176, 185]]]",
    "query_spans": "[[[182, 192]]]",
    "process": ""
  },
  {
    "text": "Given a point $Q(m, n)$ on the parabola $y^{2}=5x$ such that its distance to the focus is $\\frac{25}{4}$, then $m+|n|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 5*x);Q: Point;Coordinate(Q) = (m, n);m: Number;n: Number;PointOnCurve(Q, G);Distance(Q, Focus(G)) = 25/4",
    "query_expressions": "m + Abs(n)",
    "answer_expressions": "10",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[19, 28]], [[19, 28]], [[51, 58]], [[51, 58]], [[2, 28]], [[2, 49]]]",
    "query_spans": "[[[51, 60]]]",
    "process": "A point Q(m,n) on the parabola y^{2}=5x satisfies: n^{2}=5m. The equation of the directrix is x=-\\frac{5}{4}, and the focus is F(\\frac{5}{4},0). Since the distance from the point to the focus is \\frac{25}{4}, by the definition of a parabola, |QF|=m+\\frac{5}{4}=\\frac{25}{4}, so m=5, thus |n|=5, and m+|n|=10."
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{3-k}+\\frac{y^{2}}{5+k}=1$ represents an ellipse, what is the range of real values for $k$?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G)=(x^2/(3 - k) + y^2/(k + 5) = 1)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-5,-1)+(-1,3)",
    "fact_spans": "[[[45, 47]], [[49, 54]], [[2, 47]]]",
    "query_spans": "[[[49, 61]]]",
    "process": "From the characteristics of the standard equation of an ellipse, we have:  \n\\begin{cases}3-k>0\\\\5+k>0\\\\3-k\\neq5+k\\end{cases},  \nsolving yields: -5<k<3 and k\\neq-1,  \ntherefore, the range of values for k is (-5,-1)\\cup(-1,3)"
  },
  {
    "text": "If the line passing through the two points $A(a, 0)$ and $B(0, a)$ does not intersect the parabola $y=x^{2}-2 x-3$, then what is the range of real values for $a$?",
    "fact_expressions": "a: Real;A: Point;Coordinate(A) = (a, 0);B: Point;Coordinate(B) = (0, a);H: Line;PointOnCurve(A, H);PointOnCurve(B, H);G: Parabola;Expression(G) = (y = x^2 - 2*x - 3);NumIntersection(H, G) = 0",
    "query_expressions": "Range(a)",
    "answer_expressions": "(-oo, -13/4)",
    "fact_spans": "[[[54, 59]], [[5, 14]], [[5, 14]], [[15, 24]], [[15, 24]], [[25, 27]], [[2, 27]], [[2, 27]], [[28, 46]], [[28, 46]], [[25, 50]]]",
    "query_spans": "[[[54, 66]]]",
    "process": "Solve the system of the line AB equation and the parabola equation, eliminate y, obtain a quadratic equation in x: $x^{2}-x-a-3=0$ with no real roots, thus $\\triangle=1+4(a+3)<0\\Rightarrow a<-\\frac{13}{4}$"
  },
  {
    "text": "A point $P$ on the hyperbola $\\frac{x^{2}}{64}-\\frac{y^{2}}{36}=1$ is at a distance of $4$ from the right focus of the hyperbola. What is the distance from $P$ to the left directrix?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/64 - y^2/36 = 1);P: Point;PointOnCurve(P, G);Distance(P, RightFocus(G)) = 4",
    "query_expressions": "Distance(P, LeftDirectrix(G))",
    "answer_expressions": "16",
    "fact_spans": "[[[0, 40], [47, 50]], [[0, 40]], [[43, 46], [63, 66]], [[0, 46]], [[43, 60]]]",
    "query_spans": "[[[47, 75]]]",
    "process": "First, determine the basic quantities and eccentricity from the equation of the hyperbola. By considering the range of distances from points on the left and right branches to the right focus, determine that point P lies on the right branch of the hyperbola. Then, using the definition of a hyperbola, find the distance from P to the left focus, and subsequently use the property that the ratio of the distance from a point on the hyperbola to a focus and to the corresponding directrix equals the eccentricity to find the distance from P to the left directrix. [Detailed Solution] The hyperbola $\\frac{x^2}{64}-\\frac{y^{2}}{36}=1$ has its center at the origin, foci on the x-axis, $a=8$, $b=6$, $c=10$, and eccentricity $e=\\frac{c}{a}=\\frac{5}{4}$. Since the distance from point P to the right focus of the hyperbola is 4, which is less than $a+c=18$, point P lies on the right branch of the hyperbola. Therefore, the distance from point P to the left focus is $4+2a=4+16=20$, so the distance from point P to the left directrix is equal to $\\frac{|PF_{1}|}{e}=\\frac{20}{5}=16$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and $P$ is a point on the ellipse, what is the perimeter of $\\Delta P F_{1} F_{2}$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G) = True",
    "query_expressions": "Perimeter(TriangleOf(P, F1, F2))",
    "answer_expressions": "16",
    "fact_spans": "[[[18, 57], [67, 69]], [[18, 57]], [[2, 9]], [[10, 17]], [[2, 62]], [[63, 66]], [[63, 72]]]",
    "query_spans": "[[[74, 100]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}(-6,0)$ and $F_{2}(6,0)$ respectively. Point $M$ lies on the right branch of hyperbola $C$, and point $N(0,4)$. If the minimum perimeter of $\\Delta M N F_{1}$ is $4 \\sqrt{13}+4$, then the asymptotes of hyperbola $C$ are?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;F1: Point;F2: Point;N: Point;M: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-6, 0);Coordinate(F2) = (6, 0);Coordinate(N) = (0, 4);LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(M, RightPart(C));Min(Perimeter(TriangleOf(M, N, F1))) = 4 + 4*sqrt(13)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*2*sqrt(2)*x",
    "fact_spans": "[[[2, 62], [105, 111], [170, 176]], [[9, 62]], [[9, 62]], [[71, 84]], [[87, 99]], [[116, 125]], [[100, 104]], [[9, 62]], [[9, 62]], [[2, 62]], [[71, 84]], [[87, 99]], [[116, 125]], [[2, 99]], [[2, 99]], [[100, 115]], [[128, 168]]]",
    "query_spans": "[[[170, 184]]]",
    "process": "According to the definition of hyperbola and the triangle inequality that the sum of any two sides is greater than the third side, the minimum perimeter of $\\triangle MNF_{1}$ is $4a$, thus $a=2$. Then, using the properties of the hyperbola, the result can be obtained. [Detailed solution] The perimeter of $\\triangle MNF_{1}$ is $|MN| + |MF_{1}| + |NF_{1}| = |MN| + 2a + |MF_{2}| + |NF_{1}| \\geqslant |NF_{1}| + 2a + |NF_{2}| = 2a + 4\\sqrt{13} = 4\\sqrt{13} + 4$, hence $a=2$, and since $c=6$, we have $b = \\sqrt{c^{2} - a^{2}} = 4\\sqrt{2}$. Therefore, the asymptotes of hyperbola $C$ are $v = \\pm 2\\sqrt{2}x$."
  },
  {
    "text": "The parabola $C$: $y^{2}=2 x$ and the line $l$: $y=x-\\frac{1}{2}$ intersect at points $A$ and $B$, then $|A B|=$?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;Expression(C) = (y^2 = 2*x);Expression(l) = (y = x - 1/2);Intersection(C, l) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "4",
    "fact_spans": "[[[20, 44]], [[0, 19]], [[46, 49]], [[50, 53]], [[0, 19]], [[20, 44]], [[0, 55]]]",
    "query_spans": "[[[57, 66]]]",
    "process": "The equation of parabola C is $ y^{2} = 2px $, then \n$$\n\\begin{cases}\ny^{2} = 2px \\\\\ny = x - \\frac{p}{2}\n\\end{cases}\n$$\nyields $ \\left(x - \\frac{p}{2}\\right)^{2} = 2px $, that is, $ x^{2} - 3px + \\frac{p^{2}}{4} = 0 $, $ x_{1} + x_{2} = 3p $. By the focal chord length formula, $ |AB| = x_{1} + x_{2} + p = 4 $."
  },
  {
    "text": "Let $F$ be the focus of the parabola $C$: $y^{2}=4x$, and let $A$, $B$ be two distinct points on the parabola $C$. If the line $AB$ passes through the focus $F$, then the minimum value of $|AF|+4|BF|$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;A: Point;B: Point;PointOnCurve(A, C);PointOnCurve(B, C);Negation(A=B);PointOnCurve(F, LineOf(A, B))",
    "query_expressions": "Min(Abs(LineSegmentOf(A, F)) + 4*Abs(LineSegmentOf(B, F)))",
    "answer_expressions": "9",
    "fact_spans": "[[[5, 24], [36, 42]], [[5, 24]], [[1, 4], [64, 67]], [[1, 27]], [[28, 31]], [[32, 35]], [[28, 49]], [[28, 49]], [[28, 49]], [[51, 67]]]",
    "query_spans": "[[[69, 89]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x$ coincides with the right vertex of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, then $p$=?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;Expression(G) = (x^2/4 - y^2 = 1);Expression(H) = (y^2 = 2*(p*x));Focus(H) = RightVertex(G)",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[21, 49]], [[1, 17]], [[57, 60]], [[21, 49]], [[1, 17]], [[1, 55]]]",
    "query_spans": "[[[57, 62]]]",
    "process": "The coordinates of the right vertex of the hyperbola are (2,0), that is: \\frac{p}{2}=2,\\therefore p=4"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$, the line $y=4$ intersects $C$ at point $P$, intersects the $y$-axis at point $Q$, and $|P F|=\\frac{3}{2}|P Q|$, then the equation of parabola $C$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;L: Line;Expression(L) = (y = 4);Intersection(L,C) = P;P: Point;Intersection(L,yAxis) = Q;Q: Point;Abs(LineSegmentOf(P, F)) = (3/2)*Abs(LineSegmentOf(P, Q))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=4*sqrt(2)*x",
    "fact_spans": "[[[2, 27], [43, 46], [94, 100]], [[2, 27]], [[9, 27]], [[9, 27]], [[31, 34]], [[2, 34]], [[35, 42]], [[35, 42]], [[35, 53]], [[50, 53]], [[35, 66]], [[63, 66]], [[68, 92]]]",
    "query_spans": "[[[94, 105]]]",
    "process": "Let P(x_{0},4), substituting into the parabola equation gives x_{0}=\\frac{8}{p}, and |PF|=x_{0}+\\frac{p}{2}, |PQ|=x_{0}, from which p can be determined based on the given conditions, yielding the parabola equation. [Detailed solution] Let P(x_{0},4), substitute the coordinates of point P into y^{2}=2px (p>0), we get x_{0}=\\frac{8}{p}, so |PQ|=\\frac{8}{p}, |PF|=\\frac{p}{2}+\\frac{8}{p}. According to the problem, \\frac{p}{2}+\\frac{8}{p}=\\frac{3}{2}\\times\\frac{8}{p}, and since p>0, solving gives p=2\\sqrt{2}, thus the equation of parabola C is y^{2}=4\\sqrt{2}x."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{9-t}+\\frac{y^{2}}{t-3}=1$ represents an ellipse, then the range of real values for $t$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(9 - t) + y^2/(t - 3) = 1);t: Real",
    "query_expressions": "Range(t)",
    "answer_expressions": "(3, 6)+(6, 9)",
    "fact_spans": "[[[44, 46]], [[1, 46]], [[48, 53]]]",
    "query_spans": "[[[48, 60]]]",
    "process": ""
  },
  {
    "text": "Given point $M(\\frac{3}{2},-1)$, line $l$ passes through the focus of the parabola $C$: $x^{2}=4 y$, intersects parabola $C$ at points $A$ and $B$, and $AM$ is tangent to parabola $C$. Then, what is the slope of line $l$?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;M: Point;B: Point;Expression(C) = (x^2 = 4*y);Coordinate(M) = (3/2, -1);PointOnCurve(Focus(C), l);Intersection(l, C) = {A, B};IsTangent(LineSegmentOf(A, M), C)",
    "query_expressions": "Slope(l)",
    "answer_expressions": "3/4",
    "fact_spans": "[[[23, 28], [88, 93]], [[29, 48], [52, 58], [77, 83]], [[59, 62]], [[2, 22]], [[63, 66]], [[29, 48]], [[2, 22]], [[23, 51]], [[23, 68]], [[70, 85]]]",
    "query_spans": "[[[88, 98]]]",
    "process": "Let M(x_{0},y_{0}), then x_{0}^{2}=4y_{0}' \\because y=\\frac{x}{2} \\Rightarrow \\frac{x_{0}}{2}=\\frac{y_{0}+1}{x_{0}-\\frac{3}{2}}. Solving the two equations simultaneously gives: x_{0}=4 or x_{0}=-1, \\therefore y_{0}=4 or y_{0}=\\frac{1}{4}. \\therefore A(4,4) or A(-1,\\frac{1}{4}). The focus of the parabola is (0,1). \\therefore k=\\frac{4-1}{4-0}=\\frac{3}{4}, or k=\\frac{\\frac{1}{4}-1}{1-0}=\\frac{3}{4}"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with two foci $F_{1}$ and $F_{2}$, and its two asymptotes are perpendicular to each other. If a point $P$ on $C$ satisfies $|P F_{1}|=3|P F_{2}|$, then the cosine value of $\\angle F_{1} P F_{2}$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;Focus(C) = {F1, F2};L1: Line;L2: Line;Asymptote(C) = {L1, L2};IsPerpendicular(L1, L2);P: Point;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Cos(AngleOf(F1, P, F2))",
    "answer_expressions": "1/3",
    "fact_spans": "[[[2, 63], [99, 102]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[71, 78]], [[79, 86]], [[2, 86]], [], [], [[2, 93]], [[2, 97]], [[105, 108]], [[99, 108]], [[110, 132]]]",
    "query_spans": "[[[134, 162]]]",
    "process": "Since the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $), the asymptotes are given by $ y = \\pm \\frac{b}{a}x $. Because the two asymptotes are perpendicular, $ -\\left( \\frac{b}{a} \\right)^{2} = -1^{\\circ} $, so $ \\frac{b}{a} = 1 $, that is, $ b = a $, hence $ c = \\sqrt{2}a $. Therefore, $ |PF_{1}| = 3|PF_{2}| $, and from the definition of the hyperbola, $ |PF_{1}| - |PF_{2}| = 2a $, then $ |PF_{1}| = 3a $, $ |PF_{2}| = a $. In $ \\triangle F_{1}PF_{2} $, by the cosine law, we have $ \\cos \\angle F_{1} = \\frac{|PF_{1}|^{2} + |PF_{2}|^{2} - |F_{2}F_{1}|^{2}}{2|PF_{1}||PF_{2}|} = \\frac{(3a)^{2} + a^{2} - (2\\sqrt{2}a)^{2}}{2 \\cdot 3a \\cdot a} = \\frac{1}{3} $,"
  },
  {
    "text": "Given that line $l$ passes through the focus of parabola $C$ and is perpendicular to the axis of symmetry of $C$, $l$ intersects $C$ at points $A$ and $B$, with $|AB| = 18$. Point $P$ lies on the directrix of $C$. Then, the area of $\\triangle ABP$ is?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;P: Point;PointOnCurve(Focus(C),l);IsPerpendicular(l,SymmetryAxis(C));Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, B)) = 18;PointOnCurve(P, Directrix(C))",
    "query_expressions": "Area(TriangleOf(A, B, P))",
    "answer_expressions": "81",
    "fact_spans": "[[[2, 7], [30, 33]], [[8, 14], [20, 23], [34, 37], [66, 69]], [[39, 42]], [[43, 46]], [[61, 65]], [[2, 17]], [[2, 29]], [[30, 48]], [[49, 59]], [[61, 75]]]",
    "query_spans": "[[[77, 99]]]",
    "process": "(Analysis) First, assume the parabola equation, write down the equation of the directrix and the coordinates of the focus, use |AB| = 18 to obtain the parabola equation, then apply the triangle area formula to solve. Let the equation of parabola C be y^{2} = 2px (p > 0); then the focus is F(\\frac{p}{2}, 0), the directrix equation is m: x = -\\frac{p}{2}. From the given conditions, we have P(-\\frac{p}{2}, y_{0}), l: x = \\frac{p}{2}, A(\\frac{p}{2}, \\pm9). Thus, 81 = 2p \\times \\frac{p}{2}, solving gives p = 9, so S_{\\Delta PAB} = \\frac{1}{2}|AB| \\cdot p = \\frac{1}{2} \\times 18 \\times 9 = 81"
  },
  {
    "text": "Given points $O(0,0)$, $A(1,1)$, and point $P$ lies on the right branch of the hyperbola $x^{2}-y^{2}=1$, then the range of $\\overrightarrow{O A} \\cdot \\overrightarrow{O P}$ is?",
    "fact_expressions": "G: Hyperbola;O: Point;A: Point;P: Point;Expression(G) = (x^2 - y^2 = 1);Coordinate(O) = (0, 0);Coordinate(A) = (1, 1);PointOnCurve(P,RightPart(G))",
    "query_expressions": "Range(DotProduct(VectorOf(O, A), VectorOf(O, P)))",
    "answer_expressions": "(0, +\\infty)",
    "fact_spans": "[[[27, 45]], [[2, 11]], [[13, 21]], [[22, 26]], [[27, 45]], [[2, 11]], [[13, 21]], [[22, 49]]]",
    "query_spans": "[[[51, 107]]]",
    "process": "Let point $ P(x,y) $, $ (x>1) $, so $ \\overrightarrow{OA}=(1,1) $, $ \\overrightarrow{OP}=(x,y) $, $ \\therefore \\overrightarrow{OA}\\cdot\\overrightarrow{OP}=x+y $. Because $ x^{2}-y^{2}=1 $, when $ y>0 $, $ y=\\sqrt{x^{2}-1} $, so $ \\overrightarrow{OA}\\cdot\\overrightarrow{OP}=x+\\sqrt{x^{2}-1} $. Since the functions $ g(x)=x $ and $ h(x)=\\sqrt{x^{2}-1} $ are both increasing on $ [1,+\\infty) $, the function $ f(x)=x+\\sqrt{x^{2}-1} $ is increasing on $ [1,+\\infty) $. Therefore, when $ y>0 $, the minimum value of $ f(x) $ is $ f(1)=1 $, i.e., $ f(x)\\geqslant1 $. When $ y\\leqslant0 $, $ y=-\\sqrt{x^{2}-1} $, so $ \\overrightarrow{OA}\\cdot\\overrightarrow{OP}=x-\\sqrt{x^{2}-1}=\\frac{1}{x+\\sqrt{x^{2}-1}} $. Since the functions $ g(x)=x $ and $ h(x)=\\sqrt{x^{2}-1} $ are both increasing on $ (1,+\\infty) $, the function $ k(x)=\\frac{x+1x-1}{x} $ on $ (1,+\\infty) $ is decreasing, so when $ y\\leqslant0 $, $ k(x)>0 $. In summary, the range of $ \\overrightarrow{OA}\\cdot\\overrightarrow{OP} $ is $ (0,+\\infty) $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$ and directrix $l$, $P$ is a point on $l$, and $Q$ is an intersection point of line $PF$ and $C$. If $\\overrightarrow{FP}=3\\overrightarrow{FQ}$, then $|QF|$=?",
    "fact_expressions": "C: Parabola;F: Point;P: Point;Q: Point;l: Line;Expression(C) = (y^2 = 8*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(P, l);OneOf(Intersection(LineOf(P,F), C))=Q;VectorOf(F, P) = 3*VectorOf(F, Q)",
    "query_expressions": "Abs(LineSegmentOf(Q, F))",
    "answer_expressions": "8/3",
    "fact_spans": "[[[2, 21], [61, 64]], [[25, 28]], [[38, 41]], [[49, 52]], [[32, 36], [42, 45]], [[2, 21]], [[2, 28]], [[2, 35]], [[37, 48]], [[49, 69]], [[71, 116]]]",
    "query_spans": "[[[118, 127]]]",
    "process": "First, using similarity, find the length of segment |MQ|, then use the definition of the parabola to convert |QF| into |MQ|. Let |MQ| be the perpendicular segment from Q to the directrix of the parabola, then |MQ| = |QF|. The distance from the focus of the parabola to the directrix is 4. As shown in the figure, by the definition of the parabola and \\overrightarrow{FP} = 3\\overrightarrow{FQ}, we obtain \\frac{|MQ|}{4} = \\frac{2}{3}, |MQ| = \\frac{8}{3}. \\therefore |QF| = |MQ| = \\frac{8}{3}."
  },
  {
    "text": "The line $y = kx + 1$ and the parabola $y^2 = 4x$ have at most one common point; then the range of values for $k$ is?",
    "fact_expressions": "G: Parabola;H: Line;k: Number;Expression(G) = (y^2 = 4*x);Expression(H) = (y = k*x + 1);NumIntersection(H, G) <= 1",
    "query_expressions": "Range(k)",
    "answer_expressions": "{0}+[1,+oo)",
    "fact_spans": "[[[12, 26]], [[0, 11]], [[36, 39]], [[12, 26]], [[0, 11]], [[0, 34]]]",
    "query_spans": "[[[36, 46]]]",
    "process": "When $k=0$, solving the system $\\begin{cases}y=1\\\\y^{2}=4x\\end{cases}$ yields $\\begin{cases}x=\\frac{1}{4}\\\\y=1\\end{cases}$; in this case, the line $y=1$ and the parabola $y^{2}=4x$ have one common point, which satisfies the condition. When $k\\neq0$, solving the system $\\begin{cases}y=kx+1\\\\y^{2}=4x\\end{cases}$ gives $k^{2}x^{2}+(2k-4)x+1=0$. According to the problem, we require $\\triangle=(2k-4)^{2}-4k^{2}=16-16k\\leqslant0$, solving which yields $k\\geqslant1$. In conclusion, the range of values for $k$ is $\\{0\\}\\cup[1,+\\infty)$."
  },
  {
    "text": "It is known that the center of the hyperbola is at the origin, one focus is $F_{1}(-\\sqrt{5}, 0)$, point $P$ lies on the hyperbola, and the midpoint coordinates of segment $PF_{1}$ are $(0,2)$. Then the eccentricity of this hyperbola is?",
    "fact_expressions": "P: Point;F1: Point;G: Hyperbola;O:Origin;Center(G)=O;Coordinate(MidPoint(LineSegmentOf(P,F1)))=(0, 2);Coordinate(F1) = (-sqrt(5), 0);PointOnCurve(P, G);OneOf(Focus(G))=F1",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[38, 42]], [[16, 37]], [[2, 5], [2, 5], [2, 5]], [[8, 10]], [[2, 10]], [[49, 72]], [[16, 37]], [[38, 47]], [[2, 37]]]",
    "query_spans": "[[[75, 84]]]",
    "process": "From the given focus coordinates, it is determined that the focus lies on the x-axis. Let the equation of the hyperbola be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$. One focus is $(-\\sqrt{5},0)$, so $a^{2}+b^{2}=5$ $\\textcircled{1}$. Since the midpoint coordinates of segment $PF_{1}$ are $(0,2)$, the coordinates of point $P$ are $(\\sqrt{5},4)$. Substituting into the hyperbola equation gives $\\frac{5}{a^{2}}-\\frac{16}{b^{2}}=1$ $\\textcircled{2}$. Solving $\\textcircled{1}$ and $\\textcircled{2}$ yields $a^{2}=1$, $b^{2}=4$. Therefore, the equation of the hyperbola is $x^{2}-\\frac{y^{2}}{4}=1$. The eccentricity of the hyperbola is: $e=\\frac{c}{a}=\\sqrt{5}$."
  },
  {
    "text": "Given the hyperbola $C_{1}$: $x^{2}-\\frac{y^{2}}{3}=1$, if the distance from the focus of the parabola $C_{2}$: $x^{2}=2 p y$ $(p>0)$ to the asymptotes of the hyperbola $C_{1}$ is $2$, then what is the equation of the parabola $C_{2}$?",
    "fact_expressions": "C1: Hyperbola;Expression(C1) = (x^2 - y^2/3 = 1);C2: Parabola;Expression(C2) = (x^2 = 2*p*y) ;p: Number;p>0;Distance(Focus(C2),Asymptote(C1)) = 2",
    "query_expressions": "Expression(C2)",
    "answer_expressions": "x^2=16*y",
    "fact_spans": "[[[2, 39], [75, 85]], [[2, 39]], [[41, 71], [98, 108]], [[41, 71]], [[53, 71]], [[53, 71]], [[41, 96]]]",
    "query_spans": "[[[98, 113]]]",
    "process": "\\because hyperbola C_{1}: x^{2}-\\frac{y^{2}}{3}=1, \\therefore the asymptotes of hyperbola C_{1} are given by y=\\pm\\sqrt{3}x, i.e., \\pm\\sqrt{3}x+y=0. The distance from the focus F(0,\\frac{p}{2}) of parabola C_{2}: x^{2}=2py (p>0) to the asymptotes of hyperbola C_{1} is 2. \\therefore 2=\\frac{p}{2}, solving gives p=8. \\therefore the equation of parabola C_{2} is x^{2}=16y"
  },
  {
    "text": "Given that the focal distance of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $4 \\sqrt{2}$, and the two asymptotes are perpendicular to each other, then the length of the imaginary axis of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;FocalLength(G) = 4*sqrt(2);l1: Line;l2: Line;Asymptote(G) = {l1, l2};IsPerpendicular(l1, l2) = True",
    "query_expressions": "Length(ImageinaryAxis(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 58], [87, 90]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 74]], [], [], [[2, 85]], [[2, 85]]]",
    "query_spans": "[[[87, 96]]]",
    "process": "From the given conditions, the asymptotes of the hyperbola are $ y = \\pm\\frac{b}{a}x $. Since the two asymptotes are perpendicular to each other, $ -\\frac{b}{a} \\times \\frac{b}{a} = -1 $, we obtain $ a^{2} = b^{2} $. Given the focal distance is $ 4\\sqrt{2} $, that is $ 2c = 4\\sqrt{2} $, so $ c = 2\\sqrt{2} $. Since $ c^{2} = a^{2} + b^{2} $, we have $ a = b = 2 $. Therefore, the length of the imaginary axis of the hyperbola is $ 4 $."
  },
  {
    "text": "If $F_{1}$, $F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and $P$ is a point on the hyperbola such that $|P F_{1}| \\cdot|P F_{2}|=64$, then $\\angle F_{1} P F_{2}$=?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 64",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "pi/3",
    "fact_spans": "[[[21, 60], [70, 73]], [[66, 69]], [[1, 8]], [[13, 20]], [[21, 60]], [[1, 65]], [[66, 77]], [[79, 108]]]",
    "query_spans": "[[[110, 134]]]",
    "process": "By the definition of a hyperbola, we know $|PF_{1}|-|PF_{2}|=\\pm6$. In $\\triangle PF_{1}F_{2}$, by the law of cosines:  \n$\\cos\\angle F_{1}PF_{2}=\\frac{|PF_{1}|^{2}+|PF_{2}|^{2}-100}{2|PF_{1}||PF_{2}|}=\\frac{(|PF_{1}|-|PF_{2}|)^{2}+2|PF_{1}||PF_{2}|-100}{2|PF_{1}||PF_{2}|}=\\frac{36+128-100}{128}=\\frac{1}{2}$,  \nso $\\angle F_{1}PF_{2}=\\frac{\\pi}{3}$. Therefore, the answer should be: $\\frac{\\pi}{3}$."
  },
  {
    "text": "Given a hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with eccentricity $e=\\frac{\\sqrt{5}}{2}$, the right focus of the hyperbola is $F$, and $O$ is the origin. The circle with diameter $OF$ intersects an asymptote of the hyperbola $C$ at points $O$ and $A$. If the area of $\\triangle AOF$ is $1$, then the value of the real number $a$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Real;G: Circle;O: Origin;F: Point;A: Point;e: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(C) = e;e = sqrt(5)/2;RightFocus(C) = F;IsDiameter(LineSegmentOf(O, F), G);Intersection(G, OneOf(Asymptote(C))) = {O, A};Area(TriangleOf(A, O, F)) = 1",
    "query_expressions": "a",
    "answer_expressions": "2",
    "fact_spans": "[[[28, 89], [119, 125]], [[35, 89]], [[172, 177]], [[117, 118]], [[98, 101], [134, 137]], [[94, 97]], [[138, 141]], [[5, 27]], [[35, 89]], [[35, 89]], [[28, 89]], [[2, 89]], [[5, 27]], [[28, 97]], [[107, 118]], [[117, 143]], [[146, 170]]]",
    "query_spans": "[[[172, 181]]]",
    "process": "The angle subtended by a diameter on the circumference is a right angle, hence OA\\botAF. The distance from a hyperbola's focus to its asymptote is b, so |OA|=\\sqrt{c^{2}-b^{2}}=a. Thus, the area of right triangle AOF is \\frac{1}{2}ab=1, ab=2. Combining the hyperbola's eccentricity and c^{2}=a^{2}+b^{2}, the value of a can be found. [Detailed explanation] The angle subtended by a diameter on the circumference is a right angle, hence OA\\botAF. The distance from a hyperbola's focus to its asymptote is b, so |OA|=\\sqrt{c^{2}-b^{2}}=a. Thus, the area of right triangle AOF is \\frac{1}{2}ab=1, ab=2. Solving the system of equations \\begin{cases}ab=2\\\\\\frac{c}{a}=\\frac{\\sqrt{5}}{2}\\\\c^{2}=a^{2}+b^{2}\\end{cases} yields a=2. [Comments] This question mainly examines geometric properties of a circle, the distance from a hyperbola's focus to its asymptote, and the method for finding the standard equation of a hyperbola. In a circle, the angle subtended by a diameter on the circumference is a right angle. The distance from a hyperbola's focus to its asymptote is b; this should be remembered as a conclusion. The standard equation of a hyperbola has two parameters, a and b, requiring two equations for solution. In this problem, the system of equations is established using the conditions of area and eccentricity."
  },
  {
    "text": "The line passing through the point $M(1,1)$ intersects the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ at points $A$ and $B$, and the point $M$ bisects the chord $AB$. Then the equation of the line $AB$ is?",
    "fact_expressions": "G: Ellipse;H: Line;B: Point;A: Point;M: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Coordinate(M) = (1, 1);PointOnCurve(M, H);Intersection(H, G) = {A, B};IsChordOf(LineSegmentOf(A,B),G);MidPoint(LineSegmentOf(A,B))=M",
    "query_expressions": "Expression(LineOf(A,B))",
    "answer_expressions": "3*x+4*y-7=0",
    "fact_spans": "[[[14, 51]], [[11, 13]], [[57, 60]], [[53, 56]], [[64, 68], [1, 10]], [[14, 51]], [[1, 10]], [[0, 13]], [[11, 62]], [[14, 76]], [[64, 76]]]",
    "query_spans": "[[[78, 90]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). According to the midpoint coordinate formula, x_{1}+x_{2}=2, y_{1}+y_{2}=2, and \\frac{x_{1}}{4}+\\frac{y_{1}}{3}=1, \\frac{x_{2}}{4}+\\frac{y_{2}}{3}=1. Subtracting these two equations and simplifying yields \\frac{(y_{1}-y_{2})(y_{1}+y_{2})}{(x_{1}-x_{2})(x_{1}+x_{2})}=-\\frac{3}{4}, so \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{3}{4}, meaning the slope of the line is -\\frac{3}{4}. Using the point-slope form, the equation of line AB is 3x+4y-7=0."
  },
  {
    "text": "If the distance from the focus $F(1,0)$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ to an asymptote is $\\frac{\\sqrt{5}}{5}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Focus(G) = F;F: Point;Coordinate(F) = (1, 0);Distance(F, Asymptote(G)) = sqrt(5)/5",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[1, 57], [98, 101]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 68]], [[60, 68]], [[60, 68]], [[1, 96]]]",
    "query_spans": "[[[98, 107]]]",
    "process": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) has an asymptote $bx-ay=0$. According to the given conditions, $\\frac{b}{\\sqrt{a^{2}+b^{2}}}=\\frac{\\sqrt{5}}{5}$. Since $c=\\sqrt{a^{2}+b^{2}}=1$, it follows that $a=\\frac{2\\sqrt{5}}{5}$, $b=\\frac{\\sqrt{5}}{5}$. Therefore, the eccentricity of the hyperbola is $e=\\frac{c}{a}=\\frac{1}{\\frac{2\\sqrt{5}}{5}}=\\frac{\\sqrt{5}}{2}$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, with $A$ and $B$ being the endpoints of the minor axis, and point $P$ being any point on the ellipse other than $A$ and $B$, what is the product of the slopes of lines $PA$ and $PB$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;A: Point;B: Point;Endpoint(MinorAxis(G)) = {A, B};P: Point;PointOnCurve(P, G);Negation(P = A);Negation(P = B)",
    "query_expressions": "Slope(LineOf(P, A))*Slope(LineOf(P, B))",
    "answer_expressions": "-(b^2/a^2)",
    "fact_spans": "[[[2, 54], [72, 74]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[59, 62], [76, 79]], [[63, 66], [80, 83]], [[2, 66]], [[67, 71]], [[67, 88]], [[67, 88]], [[67, 88]]]",
    "query_spans": "[[[90, 112]]]",
    "process": ""
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. Perpendiculars are drawn from $A$ and $B$ to the directrix $l$, with feet $C$ and $D$, respectively. If $|A F|=4|B F|$, then $|C D|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;PointOnCurve(F, H);H: Line;Intersection(H, G) = {A, B};A: Point;B: Point;l1: Line;l2: Line;PointOnCurve(A, l1) ;PointOnCurve(B, l2) ;Directrix(G) = l;l: Line;IsPerpendicular(l1, l) ;IsPerpendicular(l2, l) ;FootPoint(l1, l) = C;FootPoint(l2, l) = D;C: Point;D: Point;Abs(LineSegmentOf(A, F)) = 4*Abs(LineSegmentOf(B, F))",
    "query_expressions": "Abs(LineSegmentOf(C, D))",
    "answer_expressions": "5",
    "fact_spans": "[[[1, 15], [24, 27]], [[1, 15]], [[17, 20]], [[1, 20]], [[0, 23]], [[21, 23]], [[21, 37]], [[28, 31], [41, 44]], [[32, 35], [45, 48]], [], [], [[38, 57]], [[38, 57]], [[24, 54]], [[51, 54]], [[38, 57]], [[38, 57]], [[38, 70]], [[38, 70]], [[63, 66]], [[67, 70]], [[73, 87]]]",
    "query_spans": "[[[89, 98]]]",
    "process": "Let the angle of inclination of line AB be $\\theta$, and assume $\\theta$ is acute. Given $|AF|=4|BF|$, we have $\\frac{2}{1-\\frac{1}{4}\\cos\\theta}=\\frac{1}{1}2$. Solving yields $\\cos\\theta=\\frac{3}{5}$, then $\\sin\\theta=\\frac{4}{5}$. Using the focal chord length formula of the parabola, we get $|AB|=\\frac{4}{\\sin^{2}\\theta}=\\frac{4}{(\\frac{4}{5})^{2}}=\\frac{25}{4}$. Therefore, $|CD|=|AB|\\sin\\theta=\\frac{25}{4}\\times\\frac{4}{5}=5$."
  },
  {
    "text": "In triangle $ABC$ with perimeter $18$, $A(-4,0)$, $B(4,0)$, $O$ is the origin, $D$ is the midpoint of $AC$, when $AC=4$, what is the length of $OD$?",
    "fact_expressions": "A: Point;Coordinate(A) = (-4, 0);B: Point;Coordinate(B) = (4, 0);C: Point;Perimeter(TriangleOf(A, B, C)) = 18;O: Origin;D: Point;MidPoint(LineSegmentOf(A, C)) = D;LineSegmentOf(A, C) = 4",
    "query_expressions": "Length(LineSegmentOf(O, D))",
    "answer_expressions": "3",
    "fact_spans": "[[[20, 29]], [[20, 29]], [[32, 40]], [[32, 40]], [[56, 61]], [[0, 18]], [[43, 46]], [[52, 55]], [[52, 63]], [[65, 72]]]",
    "query_spans": "[[[74, 83]]]",
    "process": "From the given conditions, we have: AC + BC = 10, so point C lies on an ellipse with foci A and B. If AC = 4, then BC = 6. Noticing that AD = CD and AO = BO, by the midpoint theorem of a triangle, we obtain OD = \\frac{1}{2}BC = 3"
  },
  {
    "text": "Given fixed points $A(-2,0)$, $B(2,0)$, and a moving point $P(a, b)$ satisfying: $|\\overrightarrow{P A}|-|\\overrightarrow{P B}|=2$, then the trajectory equation of the moving point $P$ is?",
    "fact_expressions": "A: Point;B: Point;P: Point;a:Number;b:Number;Coordinate(A) = (-2, 0);Coordinate(B) = (2, 0);Coordinate(P) = (a, b);Abs(VectorOf(P, A)) - Abs(VectorOf(P, B)) = 2",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(x^2-y^2/3=1)&(x>=sqrt(3))",
    "fact_spans": "[[[3, 12]], [[15, 23]], [[26, 35], [91, 94]], [[26, 35]], [[26, 35]], [[3, 12]], [[15, 23]], [[26, 35]], [[38, 87]]]",
    "query_spans": "[[[91, 101]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the right focus of the ellipse $\\frac{x^{2}}{5}+y^{2}=1$, $M$ is a point on the ellipse in the first quadrant, and $M F \\perp x$-axis. The line $M N$ is tangent to the circle $x^{2}+y^{2}=1$ at point $N$. Then $|M N|$ equals?",
    "fact_expressions": "G: Ellipse;H: Circle;M: Point;N: Point;F: Point;Expression(G) = (x^2/5 + y^2 = 1);Expression(H) = (x^2 + y^2 = 1);RightFocus(G) = F;Quadrant(M) = 1;PointOnCurve(M, G);IsPerpendicular(LineSegmentOf(M,F),xAxis);TangentPoint(LineOf(M,N),H)=N",
    "query_expressions": "Abs(LineSegmentOf(M, N))",
    "answer_expressions": "4*sqrt(5)/5",
    "fact_spans": "[[[6, 33], [46, 48]], [[76, 92]], [[38, 41]], [[95, 99]], [[2, 5]], [[6, 33]], [[76, 92]], [[2, 37]], [[38, 51]], [[38, 51]], [[53, 67]], [[68, 99]]]",
    "query_spans": "[[[101, 111]]]",
    "process": "From the latus rectum formula, we have: MF = \\frac{b^{2}}{a} = \\frac{1}{\\sqrt{5}}, then M(2, \\frac{1}{\\sqrt{5}}). From the distance formula between two points, we obtain: OM = . Combining with the Pythagorean theorem, we get: MN = \\sqrt{OM^{2}} \\frac{4\\sqrt{5}}{5}"
  },
  {
    "text": "The length of the real axis of the hyperbola $x^{2}-m y^{2}=1$ is twice the length of the imaginary axis, then the value of $m$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-m*y^2 + x^2 = 1);m: Number;Length(RealAxis(G)) = 2*Length(ImageinaryAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 20]], [[0, 20]], [[35, 38]], [[0, 33]]]",
    "query_spans": "[[[35, 42]]]",
    "process": "Using the standard equation of the hyperbola, the relationship between the lengths of the real axis and the imaginary axis can be obtained. Then solve for the value of $ m $ to obtain the answer. [Detailed solution] From the given condition, the hyperbola $ x^{2} - my^{2} = 1 $ can be rewritten as $ x^{2} - \\frac{y^{2}}{\\frac{1}{m}} = 1 $, so $ a^{2} = 1 $, $ b^{2} = \\frac{1}{m} $. Since the length of the real axis is twice that of the imaginary axis, we have $ 2a = 2 \\times 2b $, i.e., $ a^{2} = 4b^{2} $. Therefore, $ 1 = \\frac{4}{m} $, solving gives $ m = 4 $."
  },
  {
    "text": "Let the focus of the parabola $x = \\frac{1}{6} y^2$ be $F$, the directrix be $l$, and let $P$ be a point on the parabola such that $PA \\perp l$, with foot of perpendicular at $A$. If $\\angle APF = 60^\\circ$, then $|PF|$ equals?",
    "fact_expressions": "G: Parabola;P: Point;A: Point;F: Point;l: Line;Expression(G) = (x = y^2/6);Focus(G) = F;Directrix(G)=l;PointOnCurve(P, G);IsPerpendicular(l,LineSegmentOf(P, A));FootPoint(l,LineSegmentOf(P, A))=A;AngleOf(A, P, F) = ApplyUnit(60, degree)",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 25], [44, 47]], [[40, 43]], [[69, 72]], [[29, 32]], [[36, 39]], [[1, 25]], [[1, 32]], [[1, 39]], [[40, 50]], [[52, 65]], [[52, 72]], [[75, 100]]]",
    "query_spans": "[[[102, 112]]]",
    "process": "\\because the parabola equation is y^{2}=6x, \\therefore the focus F is (1.5,0), the directrix l has equation x=-1.5, \\because \\angle APF=60^{\\circ}, \\therefore triangle APF is equilateral, \\therefore the slope of line AF is -\\sqrt{3}, \\therefore the equation of line AF is y=-\\sqrt{3}(x-1.5); solving together with x=-1.5, we obtain point A at (-1.5,3\\sqrt{3}); \\because PA \\perp l and A is the foot of the perpendicular, \\therefore the y-coordinate of point P is 3\\sqrt{3}; substituting into the parabola equation, we find point P at (4.5,3\\sqrt{3}); \\therefore |PF|=|PA|=4.5-(-1.5)=6; therefore, the answer is: 6"
  },
  {
    "text": "Given point $A(0,2)$, if the line $l$: $y=x+\\frac{3}{2}$ has only one common point with the parabola $C$: $y^{2}=2 p x(p>0)$, then the minimum value of the sum of the distance from a moving point $P$ on the parabola $C$ to the directrix of $C$ and to point $A$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (0, 2);l: Line;Expression(l) = (y = x + 3/2);C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;NumIntersection(l, C) = 1;P: Point;PointOnCurve(P, C)",
    "query_expressions": "Min(Distance(P, Directrix(C)) + Distance(P, A))",
    "answer_expressions": "5/2",
    "fact_spans": "[[[2, 11], [99, 103]], [[2, 11]], [[13, 37]], [[13, 37]], [[38, 64], [75, 81], [89, 95]], [[38, 64]], [[46, 64]], [[46, 64]], [[13, 73]], [[85, 88]], [[75, 88]]]",
    "query_spans": "[[[85, 114]]]",
    "process": "Solving the system \\begin{cases}y=x+\\frac{3}{2}\\\\y^2=2px\\end{cases}, we obtain y^{2}-2py+3p=0. Since p>0, from \\Delta=4p^{2}-12p=0, we get p=3. Therefore, the equation of parabola C is y^{2}=6x. The focus of parabola C is F(\\frac{3}{2},0), and the directrix equation is x=-\\frac{3}{2}, as shown in the figure below: Let E be the point on the directrix. By the definition of a parabola, |PE|=|PF|. Thus, |AP|+|PE|=|AP|+|PF|\\geqslant|AF|=\\sqrt{(0-\\frac{3}{2})^{2}+(2-0)^{2}}=\\frac{5}{2}. The minimum value \\frac{5}{2} of |AP|+|PE| is attained if and only if points A, P, and F are collinear and P is the intersection point of line segment AF and parabola C."
  },
  {
    "text": "If the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1$ $(a>0)$ has eccentricity $\\sqrt{3}$, then what is the focal length of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/4 + x^2/a^2 = 1);a: Number;a>0;Eccentricity(C) = sqrt(3)",
    "query_expressions": "FocalLength(C)",
    "answer_expressions": "2*sqrt(6)",
    "fact_spans": "[[[1, 53], [70, 76]], [[1, 53]], [[9, 53]], [[9, 53]], [[1, 68]]]",
    "query_spans": "[[[70, 81]]]",
    "process": "From the given condition, we have \\sqrt{\\frac{a^{2}+4}{a^{2}}}=\\sqrt{3}, solving gives a^{2}=2, then c=\\sqrt{a^{2}+b^{2}}=\\sqrt{6}, hence the focal length of hyperbola C is 2\\sqrt{6}."
  },
  {
    "text": "The line $y = x - 1$ passes through the focus $F$ of the parabola $C$: $y^{2} = 2 p x$ ($p > 0$), and intersects $C$ at points $A$ and $B$. Then $|A B|$ = ?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;A: Point;B: Point;F: Point;p>0;Expression(C) = (y^2 = 2*p*x);Expression(G) = (y = x - 1);Focus(C) = F;PointOnCurve(F,G);Intersection(G,C)={A,B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[10, 35], [44, 47]], [[17, 35]], [[0, 9]], [[49, 52]], [[53, 56]], [[38, 41]], [[17, 35]], [[10, 35]], [[0, 9]], [[10, 41]], [[0, 41]], [[0, 58]]]",
    "query_spans": "[[[60, 69]]]",
    "process": "Since the focus of the parabola $ C: y^{2} = 2px $ ($ p > 0 $) has coordinates $ F\\left(\\frac{p}{2}, 0\\right) $, and the line $ y = x - l $ passes through the focus $ F $ of the parabola $ C: y^{2} = 2px $ ($ p > 0 $), it follows that $ p = 2 $. The equation of the parabola $ C $ is $ y^{2} = 4x $. From \n\\[\n\\begin{cases}\ny = x - 1 \\\\\ny^{2} = 4x\n\\end{cases}\n\\]\nwe obtain $ x^{2} - 6x + 1 = 0 $, so $ x_{A} + x_{B} = 6 $. Therefore, $ |AB| = x_{A} + x_{B} + p = 6 + 2 = 8 $."
  },
  {
    "text": "The equation of the line that has exactly one common point with the parabola $y^{2}=x$ and passes through the point $(1,1)$ is?",
    "fact_expressions": "G: Parabola;H: Line;P:Point;Expression(G) = (y^2 = x);Coordinate(P) = (1, 1);NumIntersection(H,G)=1;PointOnCurve(P, H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "{x-2*y+1=0,y=1}",
    "fact_spans": "[[[1, 13]], [[35, 37]], [[26, 34]], [[1, 13]], [[26, 34]], [[1, 37]], [[25, 37]]]",
    "query_spans": "[[[35, 41]]]",
    "process": ""
  },
  {
    "text": "Given that a line $ l $ with slope $ k(k>0) $ passes through the focus $ F $ of the parabola $ C: y^{2}=6x $, intersecting the parabola $ C $ at points $ A $ and $ B $. Perpendiculars are drawn from $ A $ and $ B $ to the $ x $-axis, with feet at $ A_{1} $, $ B_{1} $ respectively. If $ \\frac{S_{A A B B}}{S_{\\triangle A B A_{1}}}=2 $, then the value of $ k $ is?",
    "fact_expressions": "l: Line;C: Parabola;k: Number;k > 0;F: Point;A: Point;B: Point;A1: Point;B1: Point;Expression(C) = (y^2 = 6*x);Slope(l) = k;Focus(C) = F;PointOnCurve(F ,l);Intersection(l, C) = {A, B};l1: Line;l2: Line;PointOnCurve(A, l1);PointOnCurve(B, l2);IsPerpendicular(l1, xAxis);IsPerpendicular(l2, xAxis);FootPoint(l1, xAxis) = A1;FootPoint(l2, xAxis) = B1;Area(TriangleOf(A, B, B1))/Area(TriangleOf(A, B, A1)) = 2",
    "query_expressions": "k",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[14, 19]], [[20, 39], [47, 53]], [[154, 157], [5, 13]], [[5, 13]], [[42, 45]], [[55, 58], [66, 69]], [[59, 62], [70, 73]], [[87, 94]], [[97, 104]], [[20, 39]], [[2, 19]], [[20, 45]], [[14, 45]], [[14, 64]], [], [], [[65, 81]], [[65, 81]], [[65, 81]], [[65, 81]], [[65, 104]], [[65, 104]], [[105, 152]]]",
    "query_spans": "[[[154, 161]]]",
    "process": "A(x_{1},y_{1}),B(x_{2},y_{2}), solve the system of the line and the parabola, by Vieta's formulas we get x_{1}+x_{2}=\\frac{3k^{2}+6}{k^{2}}=3+\\frac{6}{k^{2}},x_{1}x_{2}=\\frac{9}{4}. Let the distances from A_{1},B_{1} to the line l be d_{1},d_{2} respectively, according to \\frac{S_{AABB}}{S_{1ABA_{1}}}=\\frac{d_{2}}{d_{1}}=2 we obtain x_{2}+2x_{1}=\\frac{9}{2}, solving the system of equations yields k=2\\sqrt{2}. By the given conditions, F(\\frac{3}{2},0), the line l: y=k(x-\\frac{3}{2}). Let A(x_{1},y_{1}),B(x_{2},y_{2}), then A_{1}(x_{1},0),B_{1}(x_{2},0). Solve the system \\begin{cases}y=k(x-\\frac{3}{2})\\\\v2=6x\\end{cases}, eliminate y and rearrange to get k^{2}x^{2}-(3k^{2}+6)x+\\frac{9}{4}k^{2}=0. Thus x_{1}+x_{2}=\\frac{3k^{2}+6}{k^{2}}=3+\\frac{6}{k^{2}},x_{1}x_{2}=\\frac{9}{4}. Let the distances from A_{1},B_{1} to the line l be d_{1},d_{2} respectively, then d_{1}=\\frac{kx_{1}-\\frac{3}{2}k}{\\sqrt{k^{2}+1}}=\\frac{k|x_{1}-\\frac{3}{2}|}{\\sqrt{k^{2}+1}},d_{2}=\\frac{|kx_{2}-\\frac{3}{2}k|}{\\sqrt{k^{2}+1}}=\\frac{k|x_{2}-\\frac{3}{2}|}{\\sqrt{k^{2}+1}}, so \\frac{S_{\\DeltaABB_{1}}}{S_{\\triangleABA_{1}}}=\\frac{d_{2}}{d_{1}}=\\frac{|x_{2}-\\frac{3}{2}|}{|x_{1}-\\frac{3}{2}|}=2. Since (x_{1}-\\frac{3}{2})(x_{2}-\\frac{3}{2})<0, we have x_{2}+2x_{1}=\\frac{9}{2}, solve to get x_{1}=\\frac{3}{4},x_{2}=3,"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, let the left vertex of $C$ be $A$. A circle $G$ is drawn with center $A$ and radius $b$. The circle $G$ intersects an asymptote of the hyperbola $C$ at points $M$ and $N$. If $\\angle M A N=120^{\\circ}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;G:Circle;a: Number;b: Number;A: Point;M: Point;N: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftVertex(C) = A;Center(G)=A;Radius(G)=b;Intersection(OneOf(Asymptote(C)), G) = {M, N};AngleOf(M, A, N) = ApplyUnit(120, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 63], [98, 104], [152, 155]], [[88, 92], [93, 97]], [[10, 63]], [[81, 84]], [[68, 72], [74, 77]], [[112, 115]], [[116, 120]], [[9, 63]], [[9, 63]], [[2, 63]], [[2, 72]], [[73, 92]], [[81, 92]], [[93, 122]], [[124, 150]]]",
    "query_spans": "[[[152, 161]]]",
    "process": "Take the asymptote as $ bx - ay = 0 $, and sketch the diagram according to the given conditions. From the figure, the distance from point $ A $ to the asymptote $ bx - ay = 0 $ is $ b\\sin30^{\\circ} = \\frac{1}{2}b $, so $ \\frac{1}{2}b = \\frac{ab}{c} $, therefore the eccentricity $ e = \\frac{c}{a} = 2 $."
  },
  {
    "text": "If the equation of the hyperbola $C$ is $\\frac{x^{2}}{k^{2}-1}-\\frac{y^{2}}{4-k^{2}}=1$, then what is the range of values for $k$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/(k^2 - 1) - y^2/(4 - k^2) = 1);k: Number",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-2, -1) + (1, 2)",
    "fact_spans": "[[[1, 7]], [[1, 58]], [[60, 63]]]",
    "query_spans": "[[[60, 70]]]",
    "process": "From the given condition: $(k^{2}-1)(4-k^{2})>0$, we have $\\begin{cases}k^{2}-1>0\\\\4-k^{2}>0\\end{cases}$ or $\\begin{cases}k^{2}-1<0\\\\4-k^{2}<0\\end{cases}$, solving gives $k\\in(-2,-1)\\cup(1,2)$"
  },
  {
    "text": "The foci of ellipse $C$ are $F_{1}(-2 \\sqrt{2}, 0)$ and $F_{2}(2 \\sqrt{2}, 0)$, and the length of the major axis is $6$. The line $y = x + 2$ intersects ellipse $C$ at points $A$ and $B$. Find the coordinates of the midpoint of segment $AB$.",
    "fact_expressions": "C: Ellipse;E: Line;A: Point;B: Point;F1: Point;F2: Point;Expression(E) = (y = x + 2);Coordinate(F1) = (-2*sqrt(2), 0);Coordinate(F2) = (2*sqrt(2), 0);Focus(C) = {F1, F2};Intersection(E, C) = {A, B};Length(MajorAxis(C)) = 6",
    "query_expressions": "Coordinate(MidPoint(LineSegmentOf(A, B)))",
    "answer_expressions": "(-9/5,1/5)",
    "fact_spans": "[[[2, 7], [79, 84]], [[69, 78]], [[85, 88]], [[89, 92]], [[13, 36]], [[37, 59]], [[69, 78]], [[13, 36]], [[37, 59]], [[2, 59]], [[69, 94]], [[2, 67]]]",
    "query_spans": "[[[96, 110]]]",
    "process": ""
  },
  {
    "text": "Given that $O$ is the coordinate origin, and for a point $A$ on the parabola $C$: $y^{2}=2 p x$, the distance from $A$ to the focus is $4$. If point $M$ is a moving point on the directrix of parabola $C$, and the minimum value of $|O M|+|M A|$ is $2 \\sqrt{13}$, then what is the value of $p$?",
    "fact_expressions": "O: Origin;C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;A: Point;PointOnCurve(A, C);Distance(A, Focus(C)) = 4;M: Point;PointOnCurve(M, Directrix(C));Min(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(O, M))) = 2*sqrt(13)",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 5]], [[11, 32], [55, 61]], [[11, 32]], [[101, 104]], [[35, 38]], [[11, 38]], [[11, 48]], [[50, 54]], [[50, 67]], [[69, 99]]]",
    "query_spans": "[[[101, 107]]]",
    "process": "Let point B be the symmetric point of the origin with respect to the directrix, then |MB| = |MO|. When point M is the intersection of AB and the directrix, |OM| + |MA| reaches its minimum value. At this time, |AB| = 2\\sqrt{13}. Without loss of generality, let the focus of the parabola be F. According to the problem, the distance from A to the directrix is |AF| = 4. Let A(x_{0}, y_{0}), then x_{0} + \\frac{p}{2} = 4, so x_{0} = 4 - \\frac{p}{2}. Since A lies on the parabola, y_{0}^{2} = 2px_{0} = 2p(4 - \\frac{p}{2}) = 8p - p^{2}. Draw a perpendicular line from A to the x-axis, with foot point C, then |BC| = \\frac{p}{2} + 4. In triangle ABC, by the Pythagorean theorem, |AC|^{2} + |BC|^{2} = |AB|^{2}, that is, y_{0}^{2} + (\\frac{p}{2} + 4)^{2} = (8p - p^{2}) + (\\frac{p}{2} + 4)^{2} = (2\\sqrt{13})^{2}. Simplifying yields 3p^{2} - 48p + 144 = 0, solving gives p = 4 or 12. However, when p = 12, x_{0} = 4 - \\frac{p}{2} = 4 - 6 = -2 < 0, which does not satisfy the condition. Therefore, p = 4."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with left and right foci $F_{1}(-c, 0)$, $F_{2}(c, 0)$, if there exists a moving point $Q$ satisfying $|\\overrightarrow {F_{1} Q}|=2 a$ and the area of $\\Delta F_{1} Q F_{2}$ equals $b^{2}$, then the range of values for the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;c:Number;F1: Point;F2: Point;Q: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);LeftFocus(G) =F1;RightFocus(G)=F2;Abs(VectorOf(F1, Q)) = 2*a;Area(TriangleOf(F1, Q, F2)) = b^2",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[\\sqrt{2}-1,1)",
    "fact_spans": "[[[2, 54], [172, 174]], [[163, 170]], [[4, 54]], [[60, 74]], [[60, 74]], [[76, 89]], [[95, 98]], [[4, 54]], [[4, 54]], [[2, 54]], [[60, 74]], [[76, 89]], [[2, 89]], [[2, 89]], [[101, 134]], [[136, 170]]]",
    "query_spans": "[[[172, 184]]]",
    "process": ""
  },
  {
    "text": "If the length of the real axis of the hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ is $2$ times its eccentricity, then $m$=?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (-y^2 + x^2/m = 1);Length(RealAxis(G)) = 2*Eccentricity(G)",
    "query_expressions": "m",
    "answer_expressions": "(1+sqrt(5))/2",
    "fact_spans": "[[[1, 29]], [[44, 47]], [[1, 29]], [[1, 42]]]",
    "query_spans": "[[[44, 49]]]",
    "process": "\\because2e=2\\sqrt{\\frac{c^{2}}{a^{2}}}=2\\sqrt{\\frac{a^{2}+b^{2}}{a^{2}}}=2\\sqrt{\\frac{m+1}{m}}=2\\sqrt{m}, and m>0, \\therefore m^{2}-1=m, i.e., m^{2}-m-1=0, solving yields m=\\frac{1+\\sqrt{5}}{2} or m=\\frac{1-\\sqrt{5}}{2} (discarded)"
  },
  {
    "text": "In triangle $A B C$, $A B=8$, $A C=4$, $\\angle B A C=60^{\\circ}$, a hyperbola has foci at $A$ and $B$ and passes through point $C$. What is the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;A: Point;B: Point;C: Point;LineSegmentOf(A, B) = 8;LineSegmentOf(A,C) = 4;AngleOf(B,A,C)=ApplyUnit(60,degree);Focus(G)={A,B};PointOnCurve(C,G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)+1",
    "fact_spans": "[[[55, 58], [80, 83]], [[59, 62]], [[63, 66]], [[73, 77]], [[13, 20]], [[21, 28]], [[29, 54]], [[55, 69]], [[55, 77]]]",
    "query_spans": "[[[80, 89]]]",
    "process": "Because in triangle ABC, AB=8, AC=4, \\angle BAC=60^{\\circ}, thus CB=4\\sqrt{3}; in the hyperbola, 2c=|AB|=8 \\Rightarrow c=4, 2a=|CB|-|CA|=4\\sqrt{3}-4 \\Rightarrow a=2\\sqrt{3}-2, so the eccentricity e=\\frac{c}{a}=\\frac{2}{\\sqrt{3}-1}=\\sqrt{3}+"
  },
  {
    "text": "The coordinates of the focus of the parabola $y=\\frac{1}{12} x^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2/12)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, 3)",
    "fact_spans": "[[[0, 25]], [[0, 25]]]",
    "query_spans": "[[[0, 32]]]",
    "process": "Transforming gives $x^{2}=12y$, calculating the focus yields the answer: the standard equation of the parabola $y=\\frac{1}{12}x^{2}$ is $x^{2}=12y$, $p=6$, so the focus coordinates are $(0,3)$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. If there exists a point $P$ on the asymptote of the hyperbola such that $|P F_{1}|=2|P F_{2}|$, then the range of the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P,Asymptote(C));Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(1,5/3]",
    "fact_spans": "[[[2, 63], [89, 92], [130, 136]], [[10, 63]], [[10, 63]], [[99, 103]], [[72, 79]], [[80, 87]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 87]], [[2, 87]], [[89, 103]], [[106, 128]]]",
    "query_spans": "[[[130, 147]]]",
    "process": "Let P(x,y), then (x+c)^{2}+y^{2}=4[(x-c)^{2}+y^{2}], simplifying yields (x-\\frac{5}{3}c)^{2}+y^{2}=\\frac{16}{9}c^{2}, so point P lies on a circle with center M(\\frac{5c}{3},0) and radius \\frac{4}{3}c. Since point P also lies on the asymptotes bx\\pm ay=0 of the hyperbola, the asymptotes intersect the circle M, thus \\frac{\\frac{5}{3}bc}{\\sqrt{b^{2}+a^{2}}}\\leqslant\\frac{4}{3}c, solving gives 5b\\leqslant4c, i.e., \\frac{c}{a}\\leqslant\\frac{5}{3}, therefore the range of eccentricity of the hyperbola is (1,\\frac{5}{3}]."
  },
  {
    "text": "Given that lines $l_{1}$, $l_{2}$ are the two asymptotes of the hyperbola $C$: $\\frac{x^{2}}{4}-y^{2}=1$, and point $P$ is a point on the hyperbola $C$. If the distance from point $P$ to asymptote $l_{1}$ lies in the range $[\\frac{1}{2}, 1]$, then the range of the distance from point $P$ to asymptote $l_{2}$ is?",
    "fact_expressions": "l1: Line;l2: Line;C: Hyperbola;Expression(C) = (x^2/4 - y^2 = 1);Asymptote(C) = {l1, l2};P: Point;PointOnCurve(P, C);Range(Distance(P, l1)) = [1/2,1]",
    "query_expressions": "Range(Distance(P,l2))",
    "answer_expressions": "[4/5, 8/5]",
    "fact_spans": "[[[2, 11], [85, 92]], [[13, 20], [129, 136]], [[21, 54], [66, 72]], [[21, 54]], [[2, 60]], [[61, 65], [77, 81], [121, 125]], [[61, 75]], [[77, 119]]]",
    "query_spans": "[[[121, 146]]]",
    "process": "Let point $ P(x_{0},y_{0}) $. According to the problem, the asymptotes are set as $ l_{1}: x-2y=0 $ and $ l_{2}: x+2y=0 $. The distance from point $ P $ to line $ l_{1} $ is $ d_{1}=\\frac{|x_{0}-2y_{0}|}{\\sqrt{5}} $, and the distance from point $ P $ to line $ l_{2} $ is $ d_{2}=\\frac{|x_{0}+2y_{0}|}{\\sqrt{5}} $. Then, $ d_{1}d_{2}=\\frac{|x_{0}-2y_{0}|}{\\sqrt{5}} \\cdot \\frac{|x_{0}+2y_{0}|}{\\sqrt{5}}=\\frac{|x_{0}^{2}-4y_{0}^{2}|}{5} $. Also, $ \\frac{x_{0}^{2}}{4}-y_{0}^{2}=1 $, i.e., $ x_{0}^{2}-4y_{0}^{2}=4 $, so $ d_{1}d_{2}=\\frac{4}{5} $, hence $ d_{2}=\\frac{4}{5d_{1}} $. Since $ d_{2} $ is inversely proportional to $ d_{1} $, and $ d_{1}\\in\\left[\\frac{1}{2},1\\right] $, it follows that $ d_{2}\\in\\left[\\frac{4}{5},\\frac{8}{5}\\right] $."
  },
  {
    "text": "Given that the parabola $y^{2}=2 p x$ passes through the point $A(2 , 2)$, then the distance from point $A$ to the directrix is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;Expression(G) = (y^2 = 2*p*x);Coordinate(A) = (2, 2);PointOnCurve(A, G)",
    "query_expressions": "Distance(A, Directrix(G))",
    "answer_expressions": "5/2",
    "fact_spans": "[[[2, 18]], [[5, 18]], [[32, 36], [19, 30]], [[2, 18]], [[19, 30]], [[2, 30]]]",
    "query_spans": "[[[2, 44]]]",
    "process": "From the given condition, $2^{2}=2p\\times2$, we get $p=1$, so the parabola equation is $y^{2}=2x$, and the directrix equation is $x=-\\frac{1}{2}$. Then the distance from point $A(2,2)$ to the directrix $x=-\\frac{1}{2}$ is $d=2-(-\\frac{1}{2})=\\frac{5}{2}$, which is the answer."
  },
  {
    "text": "Given the hyperbola $x^{2}-y^{2}=4$, $F_{1}$ is the left focus, $P_{1}$ and $P_{2}$ are two moving points on the right branch. Then the minimum value of $| F_{1} P_{1}|+| F_{1} P_{2}|-| P_{1} P_{2}|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 4);F1: Point;LeftFocus(G) = F1;P1: Point;P2: Point;PointOnCurve(P1, RightPart(G));PointOnCurve(P2, RightPart(G))",
    "query_expressions": "Min(Abs(LineSegmentOf(F1, P1)) + Abs(LineSegmentOf(F1, P2)) - Abs(LineSegmentOf(P1, P2)))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 20]], [[2, 20]], [[22, 29]], [[2, 33]], [[34, 41]], [[42, 49]], [[2, 58]], [[2, 58]]]",
    "query_spans": "[[[60, 112]]]",
    "process": "Directly use the definition of hyperbola to solve. Let the right focus of the hyperbola be $F_{2}$, $\\because|F_{1}P_{1}|=2a+|F_{2}P_{1}|$, $|F_{1}P_{2}|=2a+|F_{2}P_{2}|$, $\\therefore|F_{1}P_{1}|+|F_{1}P_{2}|-|P_{1}P_{2}|=2a+|F_{2}P_{1}|+2a+|F_{2}P_{2}|-|P_{1}P_{2}|$, $=8+(|F_{2}P_{1}|+|F_{2}P_{2}|-|P_{1}P_{2}|)\\geqslant8$, (equality holds if and only if $P_{1}$, $P_{2}$, $F_{2}$ are collinear) $\\therefore|F_{1}P_{1}|+|F_{1}P_{2}|-|P_{1}P_{2}|$ has minimum value $8$. Hence the answer is: $8$"
  },
  {
    "text": "Given that $P$ and $Q$ are two points on the parabola $x^{2}=2 y$, with the horizontal coordinates of points $P$ and $Q$ being $4$ and $-2$ respectively, tangents to the parabola are drawn at $P$ and $Q$, and these two tangents intersect at point $A$. Then, what is the vertical coordinate of point $A$?",
    "fact_expressions": "G: Parabola;P: Point;Q: Point;A: Point;l1:Line;l2:Line;Expression(G) = (x^2 = 2*y);PointOnCurve(P,G);PointOnCurve(Q,G);XCoordinate(P)=4;XCoordinate(Q)=-2;TangentOfPoint(P,G)=l1;TangentOfPoint(Q,G)=l2;Intersection(l1,l2)=A",
    "query_expressions": "YCoordinate(A)",
    "answer_expressions": "-4",
    "fact_spans": "[[[10, 24], [64, 67]], [[2, 5], [28, 32], [54, 57]], [[33, 36], [58, 61], [58, 61]], [[76, 80], [76, 80]], [], [], [[10, 24]], [[2, 27]], [[2, 27]], [[28, 52]], [[28, 52]], [[53, 70]], [[53, 70]], [[53, 80]]]",
    "query_spans": "[[[82, 92]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and $M$, $N$ are points on the circles $(x+3)^{2}+y^{2}=1$ and $(x-3)^{2}+y^{2}=4$ respectively, find the minimum value of $|P M|+|P N|$.",
    "fact_expressions": "G: Ellipse;H1: Circle;H2: Circle;P: Point;M: Point;N: Point;Expression(G) = (x^2/25 + y^2/16 = 1);Expression(H1) = (y^2 + (x + 3)^2 = 1);Expression(H2) = (y^2 + (x - 3)^2 = 4);PointOnCurve(P, G);PointOnCurve(M, H1);PointOnCurve(N, H2)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, M)) + Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "7",
    "fact_spans": "[[[6, 45]], [[60, 80]], [[81, 101]], [[2, 5]], [[50, 53]], [[54, 57]], [[6, 45]], [[60, 80]], [[81, 101]], [[2, 49]], [[50, 104]], [[50, 104]]]",
    "query_spans": "[[[106, 125]]]",
    "process": "First, determine a, b, c from the ellipse equation. It follows that the centers of the two circles are the left and right foci $F_{1}, F_{2}$ of the ellipse, and then the solution can be obtained according to the definition of the ellipse. From the ellipse equation, we have $a=5$, $b=4$, $c=3$. The centers of the two circles are respectively the left and right foci $F_{1}, F_{2}$ of the ellipse. Let the radii of the two circles be $r_{1}, r_{2}$, then $r_{1}=1$, $r_{2}=2$. Therefore, $|PM|_{\\min}=|PF_{1}|-r_{1}=|PF_{1}|-1$, $|PN|_{\\min}=|PF_{2}|-r_{2}=|PF_{2}|-2$. Hence, the minimum value of $|PM|+|PN|$ is $|PF_{1}|+|PF_{2}|-3=2a-3=7$. Answer: This problem mainly involves the definition of an ellipse; one needs to memorize the definition of an ellipse, and it is a basic problem."
  },
  {
    "text": "If the line $a x - y + 1 = 0$ passes through the right focus of the ellipse $\\frac{x^{2}}{25} + \\frac{y^{2}}{16} = 1$, then the real number $a = $?",
    "fact_expressions": "G: Ellipse;H: Line;a: Real;Expression(G) = (x^2/25 + y^2/16 = 1);Expression(H) = (a*x - y + 1 = 0);PointOnCurve(RightFocus(G),H)",
    "query_expressions": "a",
    "answer_expressions": "-1/3",
    "fact_spans": "[[[16, 55]], [[1, 14]], [[61, 66]], [[16, 55]], [[1, 14]], [[1, 59]]]",
    "query_spans": "[[[61, 68]]]",
    "process": "The right focus of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ is $(3,0)$, so $3a-0+1=0$, $a=-\\frac{1}{3}$"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+y^{2}=1$ with right focus $F$, point $P$ lies on ellipse $C$, and $O$ is the origin. If $|OP|=|OF|$, then the area of $\\triangle OPF$ is?",
    "fact_expressions": "C: Ellipse;O: Origin;P: Point;F: Point;Expression(C) = (x^2/4 + y^2 = 1);RightFocus(C) = F;PointOnCurve(P, C);Abs(LineSegmentOf(O, P)) = Abs(LineSegmentOf(O, F))",
    "query_expressions": "Area(TriangleOf(O, P, F))",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 33], [47, 52]], [[54, 57]], [[42, 46]], [[38, 41]], [[2, 33]], [[2, 41]], [[42, 53]], [[64, 77]]]",
    "query_spans": "[[[79, 101]]]",
    "process": "Solve for the upper right, set up the system of equations, find the absolute value of the y-coordinate of point P, and calculate the area of the triangle. From the equation of ellipse C, $\\frac{x^{2}}{4}+y^{2}=1$, we obtain: $c^{2}=a^{2}-b^{2}=4-1=3$, $F(\\sqrt{3},0)$. As shown in the figure, let $P(x,y)$. Since $P$ lies on ellipse C and $|OP|=|OF|$, the coordinates of point P satisfy\n\\[\n\\begin{cases}\n\\frac{x^{2}}{4}+y^{2}=1 \\\\\nx^{2}+y^{2}=3\n\\end{cases}\n\\]\nEliminating $x$, we get $y=\\frac{1}{3}$, so $y=\\frac{\\sqrt{3}}{3}$. Therefore, the area $S$ of $\\triangle OPF$ is $S=\\frac{1}{2}|OF||y|=\\frac{1}{2}\\times\\sqrt{3}\\times\\frac{\\sqrt{3}}{3}=\\frac{1}{2}$."
  },
  {
    "text": "Let $F$ be the focus of the parabola $C$: $y=\\frac{1}{4} x^{2}$, and the curve $y=\\frac{k}{x}$ $(k>0)$ intersects $C$ at point $P$. If $P F \\perp y$-axis, then $k=$?",
    "fact_expressions": "C: Parabola;G: Curve;k: Number;P: Point;F: Point;Expression(C) = (y = x^2/4);k>0;Expression(G) = (y = k/x);Focus(C) = F;Intersection(G, C) = P;IsPerpendicular(LineSegmentOf(P, F), yAxis)",
    "query_expressions": "k",
    "answer_expressions": "2",
    "fact_spans": "[[[5, 34], [61, 64]], [[38, 60]], [[89, 92]], [[66, 70]], [[1, 4]], [[5, 34]], [[40, 60]], [[38, 60]], [[1, 37]], [[38, 70]], [[73, 87]]]",
    "query_spans": "[[[89, 94]]]",
    "process": ""
  },
  {
    "text": "If a point $M$ on the parabola $y^{2}=4x$ is at a distance of $2$ from its focus, then what is the distance from $M$ to its vertex?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);M: Point;PointOnCurve(M, G) = True;Distance(M, Focus(G)) = 2",
    "query_expressions": "Distance(M, Vertex(G))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[1, 15], [22, 23], [39, 40]], [[1, 15]], [[18, 21], [35, 38]], [[1, 21]], [[18, 33]]]",
    "query_spans": "[[[35, 48]]]",
    "process": "\\because the parabola equation is y^{2}=4x, \\therefore the focus is F(1,0), the directrix is l:x=-1. \\because a point P on the parabola y^{2}=4x has a distance of 2 to the focus, \\therefore according to the definition of a parabola, the distance from P to the directrix is 2, i.e., x+1=2, solving gives x=1, substituting into the parabola equation yields y=\\pm2, \\therefore the coordinates of point P are (1,\\pm2), hence its distance to the vertex is \\sqrt{1^{2}+(\\pm2)^{2}}=\\sqrt{5}"
  },
  {
    "text": "Given the parabola $x^{2}=4 y$ with focus $F$ and directrix $l$, let $P$ be a point on the parabola, and let $PA \\perp l$ intersecting $l$ at point $A$. When $\\angle AFO=30^{\\circ}$ ($O$ is the origin), $|PF|=$?",
    "fact_expressions": "G: Parabola;P: Point;A: Point;F: Point;O: Origin;l: Line;Expression(G) = (x^2 = 4*y);Focus(G) = F;Directrix(G) = l;PointOnCurve(P, G);PointOnCurve(P, LineSegmentOf(P,A));IsPerpendicular(LineSegmentOf(P,A),l);FootPoint(LineSegmentOf(P,A),l)=A;AngleOf(A, F, O) = ApplyUnit(30, degree)",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[2, 16], [37, 40]], [[32, 36], [45, 48]], [[63, 67]], [[20, 23]], [[95, 98]], [[27, 30]], [[2, 16]], [[2, 23]], [[2, 30]], [[33, 43]], [[44, 62]], [[49, 62]], [[49, 67]], [[69, 94]]]",
    "query_spans": "[[[107, 116]]]",
    "process": "Problem Analysis: Let the intersection point of line $ l $ with the $ y $-axis be $ B $. In right triangle $ \\triangle ABF $, $ \\angle AFB = 30^{\\circ} $, $ BF = 2 $, so $ AB = \\frac{2\\sqrt{3}}{3} $. If $ P(x_0, y_0) $, then $ x_0 = \\frac{2\\sqrt{3}}{3} $. Substituting into $ x^2 = 4y $, we get $ y_0 = \\frac{1}{3} $, and $ |PF| = |PA| = y_0 + 1 = \\frac{4}{3} $."
  },
  {
    "text": "The equation of the directrix of the parabola $x^{2}=y$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = y)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "2*x+1=0",
    "fact_spans": "[[[0, 12]], [[0, 12]]]",
    "query_spans": "[[[0, 19]]]",
    "process": ""
  },
  {
    "text": "If a point $P(a, b)$ on the left branch of the hyperbola $x^{2}-y^{2}=1$ is at a distance of $\\sqrt{2}$ from the asymptote $y=x$, then the value of $a+b$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2 - y^2 = 1);Coordinate(P) = (a, b);PointOnCurve(P, LeftPart(G));L:Line;OneOf(Asymptote(G))=L;Expression(L)=(y = x);Distance(P, L) = sqrt(2);a:Number;b:Number",
    "query_expressions": "a + b",
    "answer_expressions": "-1/2",
    "fact_spans": "[[[1, 19]], [[25, 34]], [[1, 19]], [[25, 34]], [[1, 34]], [[38, 43]], [[1, 43]], [[38, 43]], [[25, 57]], [[59, 64]], [[59, 64]]]",
    "query_spans": "[[[59, 68]]]",
    "process": ""
  },
  {
    "text": "Draw a line $l$ through the focus of the parabola $y^{2}=4x$, intersecting the parabola at points $A$ and $B$. If the x-coordinate of the midpoint of segment $AB$ is $3$, then $|AB|$ equals?",
    "fact_expressions": "A: Point;B: Point;l: Line;G: Parabola;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), l);Intersection(l, G) = {A,B};XCoordinate(MidPoint(LineSegmentOf(A, B))) = 3",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[29, 32]], [[35, 38]], [[19, 24]], [[1, 15], [25, 28]], [[1, 15]], [[0, 24]], [[19, 40]], [[42, 58]]]",
    "query_spans": "[[[60, 70]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$, a line passing through point $F_{1}$ intersects the ellipse at points $A$ and $B$. If $|F_{2} A|+|A B|=10$, then $|F_{2} B|$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/4 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};H: Line;PointOnCurve(F1, H);Intersection(H, G) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(A, B)) + Abs(LineSegmentOf(F2, A)) = 10",
    "query_expressions": "Abs(LineSegmentOf(F2, B))",
    "answer_expressions": "6",
    "fact_spans": "[[[18, 56], [75, 77]], [[18, 56]], [[2, 9], [63, 71]], [[10, 17]], [[2, 61]], [[72, 74]], [[62, 74]], [[72, 87]], [[78, 81]], [[82, 85]], [[89, 109]]]",
    "query_spans": "[[[111, 124]]]",
    "process": "Using the definition of an ellipse, we can directly solve the problem. Let the semi-major axis of the ellipse be $ a $. By the definition of an ellipse, $ |F_{2}A| + |AB| + |F_{2}B| = 4a = 16 $, hence $ |F_{2}B| = 16 - 10 = 6 $."
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{2}=1$ is $\\sqrt{3}$, then the equations of the asymptotes of this hyperbola are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/2 + x^2/a^2 = 1);a: Number;Eccentricity(G) = sqrt(3)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(2)*x",
    "fact_spans": "[[[2, 44], [62, 65]], [[2, 44]], [[5, 44]], [[2, 59]]]",
    "query_spans": "[[[62, 73]]]",
    "process": "Using the eccentricity of the hyperbola to find $ a $, then solve for the asymptotes of the hyperbola. The hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{2} = 1 $ ($ a > 0 $) has an eccentricity of $ \\sqrt{3} $, so we have: $ \\frac{\\sqrt{a^{2} + 2}}{a} = \\sqrt{3} $, solving gives $ a = 1 $, thus the equation of the hyperbola is: $ \\frac{x^{2}}{1} - \\frac{y^{2}}{2} = 1 $, so the asymptotes of this hyperbola are $ y = \\pm\\sqrt{2}x $."
  },
  {
    "text": "Given the equation of the ellipse is $\\frac{x^{2}}{16}+\\frac{y^{2}}{m^{2}}=1$ $(m>0)$. If a point of intersection $M$ between the line $y=\\frac{\\sqrt{2}}{2} x$ and the ellipse has its projection on the $x$-axis exactly at the right focus $F$ of the ellipse, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/m^2 = 1);m: Number;m > 0;H: Line;Expression(H) = (y = x*sqrt(2)/2);OneOf(Intersection(H, G)) = M;M: Point;Projection(M, xAxis) = F;F: Point;RightFocus(G) = F",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 4], [83, 85], [103, 105], [114, 116]], [[2, 53]], [[8, 53]], [[8, 53]], [[56, 82]], [[56, 82]], [[56, 93]], [[90, 93]], [[90, 112]], [[109, 112]], [[103, 112]]]",
    "query_spans": "[[[114, 122]]]",
    "process": ""
  },
  {
    "text": "Ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ has left and right foci $F_{1}$, $F_{2}$ respectively, with focal distance $2 \\sqrt{3}$. Point $E$ lies on $C$ such that $E F_{1} \\perp E F_{2}$, and the slope of line $E F_{1}$ is $\\frac{b}{c}$ ($c$ being the semi-focal length). Then the equation of $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;F1: Point;E: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;FocalLength(C) = 2*sqrt(3);PointOnCurve(E, C);IsPerpendicular(LineSegmentOf(E, F1), LineSegmentOf(E, F2));Slope(LineOf(E,F1))=b/c;c:Number;HalfFocalLength(C)=c",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/6 + y^2/3 = 1",
    "fact_spans": "[[[0, 57], [103, 106], [171, 174]], [[7, 57]], [[7, 57]], [[66, 73]], [[98, 102]], [[74, 81]], [[7, 57]], [[7, 57]], [[0, 57]], [[0, 81]], [[0, 81]], [[0, 97]], [[98, 107]], [[108, 131]], [[132, 160]], [[161, 164]], [[103, 169]]]",
    "query_spans": "[[[171, 179]]]",
    "process": "Since $EF_{1}\\bot EF_{2}$, and the slope of line $EF_{1}$ is $\\frac{b}{c}$, by the definition of slope, the tangent of the inclination angle $\\frac{EF_{2}}{EF_{1}} = \\frac{b}{c}$. Thus, according to the proportional relationship, $EF_{2}:EF_{1}:F_{1}F_{2} = b:c:\\sqrt{b^{2}+c^{2}} = b:c:a$. Hence, the eccentricity $\\frac{c}{a} = \\frac{2c}{2a} = \\frac{F_{1}F_{2}}{EF_{2}+EF_{1}} = \\frac{a}{b+c}$, that is, $\\frac{c}{a} = \\frac{a}{b+c}$. Therefore, $a^{2} = bc + c^{2} \\Rightarrow a^{2} - c^{2} = bc \\Rightarrow b^{2} = bc$, so $b = c$. Also, $2c = 2\\sqrt{3}$, hence $b = c = \\sqrt{3}$. Thus, $a = \\sqrt{3+3} = \\sqrt{6}$. Therefore, the equation of $C$ is $\\frac{x^{2}}{6} + \\frac{y^{2}}{3} = 1$."
  },
  {
    "text": "Given that the circle $x^{2}+y^{2}-6 x-7=0$ is tangent to the directrix of the parabola $y^{2}=-2 p x$ ($p>0$), then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;H: Circle;p>0;Expression(G) = (y^2 = -2*p*x);Expression(H) = (-6*x + x^2 + y^2 - 7 = 0);IsTangent(Directrix(G),H)",
    "query_expressions": "p",
    "answer_expressions": "14",
    "fact_spans": "[[[25, 47]], [[54, 57]], [[2, 24]], [[28, 47]], [[25, 47]], [[2, 24]], [[2, 52]]]",
    "query_spans": "[[[54, 59]]]",
    "process": ""
  },
  {
    "text": "From point $M(2,-2p)$, draw two tangents to the parabola $x^{2}=2py$ $(p>0)$, with points of tangency $A$ and $B$. If the ordinate of the midpoint of segment $AB$ is $6$, then the equation of the parabola is?",
    "fact_expressions": "M: Point;Coordinate(M) = (2, -2*p);p: Number;G: Parabola;Expression(G) = (x^2 = 2*p*y);p>0;l1: Line;l2: Line;TangentOfPoint(M, G) = {l1, l2};TangentPoint(l1, G) = A;TangentPoint(l2, G) = B;B: Point;A: Point;YCoordinate(MidPoint(LineSegmentOf(A, B))) = 6",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{(x^2=2*y),(x^2=4*y)}",
    "fact_spans": "[[[1, 13]], [[1, 13]], [[17, 35]], [[14, 35], [75, 78]], [[14, 35]], [[17, 35]], [], [], [[0, 40]], [[0, 53]], [[0, 53]], [[50, 53]], [[46, 49]], [[56, 73]]]",
    "query_spans": "[[[75, 83]]]",
    "process": ""
  },
  {
    "text": "The equation of the circle centered at the focus of the parabola $y^{2}=20 x$ and tangent to both asymptotes of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;C:Circle;Expression(G) = (x^2/16 - y^2/9 = 1);Expression(H) = (y^2 = 20*x);Center(C)=Focus(H);IsTangent(Asymptote(G),C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "(x-5)^2+y^2=9",
    "fact_spans": "[[[25, 64]], [[1, 16]], [[74, 75]], [[25, 64]], [[1, 16]], [[0, 75]], [[24, 75]]]",
    "query_spans": "[[[74, 80]]]",
    "process": ""
  },
  {
    "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ are tangent to the circle $x^{2}+(y-m)^{2}=4$, then $m=$?",
    "fact_expressions": "G: Hyperbola;H: Circle;m: Number;Expression(G) = (x^2/16 - y^2/9 = 1);Expression(H) = (x^2 + (-m + y)^2 = 4);IsTangent(Asymptote(G),H)",
    "query_expressions": "m",
    "answer_expressions": "pm*(5/2)",
    "fact_spans": "[[[1, 40]], [[45, 65]], [[69, 72]], [[1, 40]], [[45, 65]], [[1, 67]]]",
    "query_spans": "[[[69, 74]]]",
    "process": "From $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, the asymptotes are given by $y=\\frac{3}{4}x$, i.e., $3x-4y=0$. Since the distance from the center $(0,m)$ to the asymptote equals the radius, $\\therefore d=\\frac{|4m|}{5}=2$, $m=\\pm\\frac{5}{2}$,"
  },
  {
    "text": "If the curve $y=\\sqrt{|x^{2}-4|}$ and the line $y=x+m$ have exactly two intersection points, then the range of real values for $m$ is?",
    "fact_expressions": "H: Curve;Expression(H) = (y = sqrt(Abs(x^2 - 4)));G: Line;Expression(G) = (y = m + x);m: Real;NumIntersection(H, G) = 2",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-2, 0)+{2}+{2*sqrt(2)}",
    "fact_spans": "[[[1, 23]], [[1, 23]], [[24, 33]], [[24, 33]], [[42, 47]], [[1, 40]]]",
    "query_spans": "[[[42, 54]]]",
    "process": "From $ y = \\sqrt{|x^2 - 4|} $, we get $ y^2 = |x^2 \\cdot 4| $. When $ x^2 \\cdot 4 \\geqslant 0 $, it becomes $ x^2 \\cdot y^2 = 4 $ ($ y \\geqslant 0 $); when $ x^2 \\cdot 4 < 0 $, it becomes $ x^2 + y^2 = 4 $ ($ y \\geqslant 0 $). The graph is as shown. When the line is tangent to the semicircle, $ m = 2\\sqrt{2} $. The asymptotes of the hyperbola are $ y = \\pm x $. Therefore, the range of real number $ m $ is $ (-2, 0) \\cup \\{2\\} \\cup \\{2\\sqrt{2}\\} $."
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the two foci of ellipse $C$, $P$ is a point on ellipse $C$ distinct from the vertices, $I$ is the incenter of $\\Delta P F_{1} F_{2}$, and if the area of $\\Delta P F_{1} F_{2}$ is 3 times the area of $\\Delta I F_{1} F_{2}$, then what is the eccentricity of ellipse $C$?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;I: Point;Focus(C) = {F1, F2};PointOnCurve(P, C);Negation(P=Vertex(C));Center(InscribedCircle(TriangleOf(P,F1,F2)))=I;Area(TriangleOf(P,F1,F2))=3*Area(TriangleOf(I,F1,F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[16, 21], [31, 36], [138, 143]], [[27, 30]], [[0, 7]], [[8, 15]], [[45, 48]], [[0, 26]], [[27, 44]], [[27, 44]], [[45, 77]], [[79, 136]]]",
    "query_spans": "[[[138, 149]]]",
    "process": ""
  },
  {
    "text": "It is known that point $M$ lies on the parabola $C$: $y^{2}=2 p x$ ($p>0$), $F$ is the focus of $C$, and the coordinates of the midpoint of $MF$ are $(2,2)$. Then the value of $p$ is?",
    "fact_expressions": "C: Parabola;p: Number;M: Point;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(MidPoint(LineSegmentOf(M,F))) = (2, 2);PointOnCurve(M, C);Focus(C) = F",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[7, 33], [41, 44]], [[68, 71]], [[2, 6]], [[37, 40]], [[15, 33]], [[7, 33]], [[48, 66]], [[2, 36]], [[37, 47]]]",
    "query_spans": "[[[68, 75]]]",
    "process": "By analysis, according to the given conditions, $ F\\left(\\frac{p}{2},0\\right) $, let $ M\\left(\\frac{y_{0}^{2}}{2p},y_{0}\\right) $, then $ \\frac{p}{2} + \\frac{y_{0}^{2}}{2p} = 2 \\times 2 $, $ 0 + y_{0} = 2 \\times 2^{-P} $, so $ p = 4 $."
  },
  {
    "text": "Given the two foci of a hyperbola $F_{1}(-\\sqrt{10} , 0)$, $F_{2}(\\sqrt{10} , 0)$, and $M$ is a point on this hyperbola satisfying $\\overrightarrow{M F_{1}} \\cdot \\overrightarrow{M F_{2}}=0$, $|\\overrightarrow{M F_{1}}| \\cdot |\\overrightarrow{M F_{2}}|=2$, then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;M: Point;Coordinate(F1) = (-sqrt(10), 0);Coordinate(F2) = (sqrt(10), 0);Focus(G) = {F1,F2};PointOnCurve(M, G);DotProduct(VectorOf(M, F1),VectorOf(M,F2))=0;Abs(VectorOf(M, F1))*Abs(VectorOf(M, F2)) = 2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9-y^2=1",
    "fact_spans": "[[[2, 5], [63, 66], [63, 66]], [[11, 34]], [[35, 57]], [[58, 61]], [[11, 34]], [[35, 57]], [[2, 57]], [[58, 70]], [[75, 134]], [[136, 198]]]",
    "query_spans": "[[[201, 209]]]",
    "process": ""
  },
  {
    "text": "The line $l$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$, $O$ is the origin, and the product of the slopes of lines $OA$ and $OB$ is $-1$. A circle with radius $\\sqrt{2}$ centered at the midpoint of segment $AB$ intersects line $l$ at points $P$ and $Q$. Then the minimum value of $|OP|^{2}+|OQ|^{2}$ is?",
    "fact_expressions": "l: Line;G: Parabola;Expression(G) = (y^2 = 4*x);Intersection(l, G) = {A, B};A: Point;B: Point;O: Origin;Slope(LineOf(O,A)) * Slope(LineOf(O,B)) = -1;C: Circle;Center(C) = MidPoint(LineSegmentOf(A,B));Radius(C) = sqrt(2);Intersection(l,C) = {P,Q};P: Point;Q: Point",
    "query_expressions": "Min(Abs(LineSegmentOf(O, P))^2 + Abs(LineSegmentOf(O, Q))^2)",
    "answer_expressions": "36",
    "fact_spans": "[[[0, 5], [98, 103]], [[6, 20]], [[6, 20]], [[0, 31]], [[22, 25]], [[26, 29]], [[32, 35]], [[41, 66]], [[96, 97]], [[67, 97]], [[82, 97]], [[96, 114]], [[105, 108]], [[109, 112]]]",
    "query_spans": "[[[116, 143]]]",
    "process": ""
  },
  {
    "text": "If the directrix of the parabola $y^{2}=2 p x(p>0)$ passes through a focus of the hyperbola $x^{2}-y^{2}=1$, then $p$=?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;Expression(G) = (x^2 - y^2 = 1);p>0;Expression(H) = (y^2 = 2*p*x);PointOnCurve(OneOf(Focus(G)), Directrix(H))",
    "query_expressions": "p",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[27, 45]], [[1, 22]], [[52, 55]], [[27, 45]], [[4, 22]], [[1, 22]], [[1, 50]]]",
    "query_spans": "[[[52, 57]]]",
    "process": "From the given conditions, one focus of the hyperbola is $ F(\\sqrt{2},0) $, hence $ \\frac{p}{2} = \\sqrt{2} \\Rightarrow p = 2\\sqrt{2} $, so the answer is $ 2\\sqrt{2} $."
  },
  {
    "text": "Given that point $P$ is a moving point on the parabola $y^{2}=2x$, $F$ is the focus of the parabola, and point $A(3,2)$, then the minimum value of $|PA|+|PF|$ is?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 2*x);Coordinate(A) = (3, 2);PointOnCurve(P, G);Focus(G) = F",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "7/2",
    "fact_spans": "[[[7, 21], [30, 33]], [[38, 47]], [[2, 6]], [[26, 29]], [[7, 21]], [[38, 47]], [[2, 25]], [[26, 36]]]",
    "query_spans": "[[[49, 68]]]",
    "process": ""
  },
  {
    "text": "Let points $A(1,0)$, $B(-1,0)$, and $M$ be a moving point. It is known that the product of the slopes of lines $AM$ and $BM$ is a constant $m$ $(m \\neq 0)$. If the trajectory of point $M$ is a hyperbola with eccentricity $2$ (excluding points $A$ and $B$), then the value of $m$ is?",
    "fact_expressions": "M:Point;A:Point;B:Point;G:Hyperbola;m:Number;Negation(m=0);Coordinate(A)=(1,0);Coordinate(B)=(-1,0);Slope(LineOf(A,M))*Slope(LineOf(B,M))=m;Locus(M)=G;Eccentricity(G)=2;Negation(PointOnCurve(A,G));Negation(PointOnCurve(B,G))",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[24, 28], [72, 76]], [[1, 10], [94, 98]], [[13, 22], [101, 104]], [[88, 91]], [[107, 110], [57, 70]], [[57, 70]], [[1, 10]], [[13, 22]], [[34, 70]], [[72, 91]], [[80, 91]], [[88, 105]], [[88, 105]]]",
    "query_spans": "[[[107, 114]]]",
    "process": "Let point M(x,y), then \\frac{y}{x-1}\\cdot\\frac{y}{x+1}=m, that is, y^{2}=m(x^{2}-1), or x^{2}-\\frac{y^{2}}{m}=1 (x\\neq\\pm1). Since the trajectory of point M is a hyperbola with eccentricity 2, we have m>0 and a^{2}=1, b^{2}=m, thus c^{2}=1+m. Since e=\\frac{c}{a}=\\sqrt{1+m}=2, solving gives: m=3."
  },
  {
    "text": "Given that $F$ and $G$ are the two foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, a line passing through point $G$ intersects the ellipse at points $A$ and $B$, and $|A B|=5$, then the value of $|A F|+|B F|$ is?",
    "fact_expressions": "C: Ellipse;H: Line;A: Point;B: Point;F: Point;G: Point;Expression(C) = (x^2/16 + y^2/9 = 1);Focus(C) = {F, G};PointOnCurve(G, H);Intersection(H, C) = {A, B};Abs(LineSegmentOf(A, B)) = 5",
    "query_expressions": "Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F))",
    "answer_expressions": "11",
    "fact_spans": "[[[10, 48], [63, 65]], [[60, 62]], [[66, 69]], [[70, 73]], [[2, 5]], [[6, 9], [55, 59]], [[10, 48]], [[2, 53]], [[54, 62]], [[60, 75]], [[77, 86]]]",
    "query_spans": "[[[88, 105]]]",
    "process": "From the equation $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, we obtain: $a=4$, $b=3$. Hence, by the definition of an ellipse, $|AF|+|BF|+|AB|=4a=16$. Since $|AB|=5$, it follows that $|AF|+|BF|=11$. Therefore, the answer is: $11$."
  },
  {
    "text": "Given that $O$ is the coordinate origin, $A(0,3)$, and a moving point $N$ in the plane satisfies $|NO| = \\frac{1}{2}|NA|$, the trajectory of point $N$ is curve $C$. Let circle $M$ have radius $1$ and center $M$ lying on the line $2x - y - 4 = 0$. If circle $M$ and curve $C$ have exactly one common point, then the value of the horizontal coordinate of center $M$ is?",
    "fact_expressions": "M: Circle;G: Line;C: Curve;A: Point;N: Point;O: Origin;M1: Point;Expression(G) = (2*x - y - 4 = 0);Coordinate(A) = (0, 3);Abs(LineSegmentOf(N, O)) = Abs(LineSegmentOf(N, A))/2;Locus(N)=C;Radius(M) = 1;Center(M)=M1;PointOnCurve(M1, G);NumIntersection(M, C)=1",
    "query_expressions": "XCoordinate(M1)",
    "answer_expressions": "{0, 12/5}",
    "fact_spans": "[[[71, 75], [105, 109]], [[89, 102]], [[64, 69], [110, 115]], [[11, 19]], [[25, 28], [57, 60]], [[2, 5]], [[85, 88], [128, 131]], [[90, 103]], [[11, 19]], [[30, 54]], [[55, 69]], [[71, 82]], [[71, 88]], [[85, 103]], [[105, 124]]]",
    "query_spans": "[[[128, 138]]]",
    "process": "First, according to the problem, denote the coordinates of the moving point and write down the equation that the coordinates satisfy, then simplify to obtain curve C. Let N(x, y), from |NO| = \\frac{1}{2}|NA|, we get 4(x^{2}+y^{2}) = x^{2}+(y-3)^{2}, simplifying yields x^{2}+(y+1)^{2} = 4; thus, curve C represents a circle with center C(0, -1) and radius 2. According to the conditions, circle C and circle M can only be externally tangent. Let M(a, 2a-4), then the distance between the centers equals the sum of the radii: a^{2}+(2a-4+1)^{2} = (2+1)^{2}, solving gives a = 0 or \\frac{12}{5}."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, $(a>0, b>0)$, the focal distance is $10$, and the slope of one asymptote is $\\frac{3}{4}$. Then, what is the standard equation of this hyperbola? What is the distance from a focus to an asymptote?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);FocalLength(G)=10;Slope(OneOf(Asymptote(G)))=3/4",
    "query_expressions": "Expression(G);Distance(Focus(G), Asymptote(G))",
    "answer_expressions": "x^2/16-y^2/9=1;3",
    "fact_spans": "[[[2, 59], [94, 97]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[2, 59]], [[2, 67]], [[2, 91]]]",
    "query_spans": "[[[94, 104]], [[94, 115]]]",
    "process": ""
  },
  {
    "text": "Given a point $P$ on the hyperbola $\\frac{x^{2}}{36}-\\frac{y^{2}}{64}=1$ such that the distance from $P$ to focus $F_{1}$ is $15$, find the distance from $P$ to focus $F_{2}$.",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/36 - y^2/64 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2};Distance(P, F1) = 15",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "27",
    "fact_spans": "[[[2, 42]], [[45, 48], [68, 71]], [[51, 58]], [[74, 81]], [[2, 42]], [[2, 48]], [[2, 81]], [[45, 66]]]",
    "query_spans": "[[[68, 86]]]",
    "process": "In the hyperbola $\\frac{x^{2}}{36}-\\frac{y^{2}}{64}=1$, $a=6$, $b=8$, $c=10$, $\\therefore|PF_{2}|\\geqslant c-a=$ by the definition of the hyperbola, $||PF_{1}|-|PF_{2}||=2a=12$, and $|PF_{1}|=15$, so $|PF_{2}|=27$ or $|PF_{2}|=3$ (discarded)"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{25}-\\frac{y^{2}}{9}=1$, the two foci are $F_{1}$ and $F_{2}$. If a point $P$ on the hyperbola is at a distance of $12$ from $F_{1}$, then what is the distance from point $P$ to $F_{2}$?",
    "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;P:Point;Expression(G) = (x^2/25 - y^2/9 = 1);Focus(G)={F1,F2};PointOnCurve(P,G);Distance(P,F1)=12",
    "query_expressions": "Distance(P,F2)",
    "answer_expressions": "{22,2}",
    "fact_spans": "[[[2, 41], [66, 69]], [[49, 56], [76, 84]], [[57, 64], [98, 106]], [[71, 75], [71, 75]], [[2, 41]], [[2, 64]], [[66, 75]], [[71, 92]]]",
    "query_spans": "[[[93, 111]]]",
    "process": "From the given conditions, $ a=5 $, $ b=3 $, $ c=\\sqrt{34} $ and $ c-a=\\sqrt{34}-5<1 $. Let $ F_{1} $ be the left focus and $ F_{2} $ be the right focus. When point $ P $ is on the left branch of the hyperbola, $ |PF_{2}|-|PF|=10 $, hence $ |PF_{2}|=22 $; when point $ P $ is on the right branch of the hyperbola, $ |PF|-|PF_{2}|=10 $, hence $ |PF_{2}|=2 $. In conclusion, the distance from point $ P $ to point $ F_{2} $ is $ 22 $ or $ 2 $. Answer: $ 22 $ or $ 2 $."
  },
  {
    "text": "Given that the equation $(m-1) x^{2}+(3-m) y^{2}= (m-1) (3-m)$ represents an ellipse, find the range of values for $m$.",
    "fact_expressions": "G: Ellipse;m:Number;Expression(G)=(x^2*(m - 1) + y^2*(3 - m) = (3 - m)*(m - 1))",
    "query_expressions": "Range(m)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[44, 46]], [[48, 51]], [[2, 46]]]",
    "query_spans": "[[[48, 58]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is parallel to the line $l$: $y=2x+10$, and one focus of the hyperbola lies on the line $l$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;l: Line;Expression(l) = (y = 2*x + 10);IsParallel(OneOf(Asymptote(G)),l) = True;PointOnCurve(OneOf(Focus(G)),l) = True",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5 - y^2/20 = 1",
    "fact_spans": "[[[2, 58], [86, 89], [103, 106]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[67, 84], [95, 100]], [[67, 84]], [[2, 84]], [[86, 101]]]",
    "query_spans": "[[[103, 111]]]",
    "process": "Determine the relationship between a and b based on the asymptotes' parallelism with line l, then determine the value of c based on the focus lying on l, and solve for a^{2} and b^{2} using a^{2}+b^{2}=c^{2} to obtain the hyperbola's equation. Since one asymptote is parallel to y=2x+10, \\frac{b}{a}=2. Also, since the foci of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0) are (\\pm c,0), and line l passes through the point (-5,0), it follows that c=5. Therefore, \\begin{cases}b=2a\\\\a^{2}+b^{2}=c^{2}=25\\end{cases}, so \\begin{cases}a^{2}=5\\\\b^{2}=20\\end{cases}. Thus, the equation of the hyperbola is: \\frac{x^{2}}{5}-\\frac{y^{2}}{20}=1"
  },
  {
    "text": "Given that $O$ is the coordinate origin, a line $l$ not passing through point $O$ intersects the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ at points $A$ and $B$, $M$ is the midpoint of segment $AB$, and the intersection point of the perpendicular bisector of segment $AB$ with the $x$-axis is $N$. Then the maximum value of the tangent of $\\angle O M N$ is?",
    "fact_expressions": "l: Line;G: Ellipse;B: Point;A: Point;O: Origin;M: Point;N: Point;Expression(G) = (x^2/2 + y^2 = 1);Negation(PointOnCurve(O, l));Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A,B)) = M;Intersection(PerpendicularBisector(LineSegmentOf(A,B)), xAxis) = N",
    "query_expressions": "Max(Tan(AngleOf(O, M, N)))",
    "answer_expressions": "sqrt(2)/4",
    "fact_spans": "[[[19, 24]], [[25, 52]], [[58, 61]], [[54, 57]], [[2, 5], [14, 18]], [[64, 67]], [[99, 102]], [[25, 52]], [[11, 24]], [[19, 63]], [[64, 78]], [[79, 102]]]",
    "query_spans": "[[[104, 128]]]",
    "process": "From the given, the slope of line $ l $ exists and is non-zero. Let the equation of line $ l $ be $ y = kx + b $. Then  \n\\[\n\\begin{cases}\n\\frac{x^{2}}{2} + y^{2} = 1 \\\\\ny = kx + b\n\\end{cases}\n\\]\nyields  \n$$\n(1 + 2k^{2})x^{2} + 4bkx + 2b^{2} - 2 = 0.\n$$\nLet $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then  \n$$\nM\\left(\\frac{x_{1} + x_{2}}{2}, \\frac{y_{1} + y_{2}}{2}\\right),\n$$\nso  \n$$\nx_{1} + x_{2} = \\frac{-4bk}{1 + 2k^{2}}, \\quad\nx_{1}x_{2} = \\frac{2b^{2} - 2}{1 + 2k^{2}},\n$$\n$$\ny_{1} + y_{2} = k(x_{1} + x_{2}) + 2b = \\frac{-4bk^{2}}{1 + 2k^{2}} + 2b = \\frac{2b}{1 + 2k^{2}}.\n$$\nThen  \n$$\nk_{OM} = \\frac{y_{1} + y_{2}}{x_{1} + x_{2}} = \\frac{2b}{-4bk} = \\frac{1}{-2k}, \\quad\nk_{MN} = -\\frac{1}{k}.\n$$\nBy  \n$$\n\\tan\\angle OMN = \\frac{k_{OM} - k_{NM}}{1 + k_{OM}k_{NM}} = \\frac{\\frac{1}{-2k} + \\frac{1}{k}}{1 + \\frac{1}{2k} \\times \\frac{1}{k}} = \\frac{1}{2k + \\frac{1}{k}}.\n$$\nWhen $ k > 0 $,  \n$$\n\\tan\\angle OMN = \\frac{k_{OM} - k_{NM}}{1 + k_{OM}k_{NM}} = \\frac{\\frac{1}{-2k} + \\frac{1}{k}}{1 + \\frac{1}{2k} \\times \\frac{1}{k}} = \\frac{1}{2k + \\frac{1}{k}} \\leqslant \\frac{1}{2\\sqrt{2k \\times \\frac{1}{k}}} = \\frac{\\sqrt{2}}{4},\n$$\nwith equality if and only if $ 2k = \\frac{1}{k} $, i.e., $ k = \\frac{\\sqrt{2}}{2} $.  \nWhen $ k < 0 $,  \n$$\n\\tan\\angle OMN = \\frac{1}{2k + \\frac{1}{k}} < 0,\n$$\nwhich does not satisfy the condition."
  },
  {
    "text": "Given the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus is $F$. A line $l$ passing through the origin $O$ intersects the ellipse $E$ at points $P$ and $Q$. If $|P F|=3|Q F|$ and $\\angle P F Q=120^{\\circ}$, then the eccentricity of the ellipse $E$ is?",
    "fact_expressions": "l: Line;E: Ellipse;b: Number;a: Number;P: Point;F: Point;Q: Point;O:Origin;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(E) = F;PointOnCurve(O,l);Intersection(l, E) = {P, Q};Abs(LineSegmentOf(P, F)) = 3*Abs(LineSegmentOf(Q, F));AngleOf(P, F, Q) = ApplyUnit(120, degree)",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(7)/4",
    "fact_spans": "[[[76, 81]], [[2, 59], [82, 87], [144, 149]], [[8, 59]], [[8, 59]], [[89, 92]], [[64, 67]], [[93, 96]], [[70, 75]], [[8, 59]], [[8, 59]], [[2, 59]], [[2, 67]], [[68, 81]], [[76, 98]], [[100, 114]], [[116, 142]]]",
    "query_spans": "[[[144, 155]]]",
    "process": "Take the right focus $ F $ of the ellipse. Since line $ l $ passes through the origin and by the symmetry of the ellipse, the quadrilateral $ PFQF' $ is a parallelogram. From $ |PF| = 3|QF| $ and the properties of the ellipse, we obtain $ |PF'| = \\frac{a}{2} $, $ |PF| = \\frac{3a}{2} $, $ \\angle PFQ = 120^{\\circ} $. Applying the cosine law gives the value of the eccentricity. [Solution] Take the right focus $ F $ of the ellipse, connect $ QF $, $ PF' $. By the symmetry of the ellipse, the quadrilateral $ PFQF' $ is a parallelogram, then $ |PF| = |QF'| $, $ \\angle FPF' = 180^{\\circ} - \\angle PFQ = 180^{\\circ} - 120^{\\circ} = 60^{\\circ} $, $ |PF| = 3|QF| = 3|PF'| $, and $ |PF| + |PF'| = 2a $, so $ |PF'| = \\frac{a}{2} $, hence $ |PF| = \\frac{3a}{2} $. In $ \\triangle PFF' $, $ \\cos \\angle FPF' = \\frac{|PF|^2 + |PF'|^{2} - |FF'|^{2}}{2|PF||PF'|} = \\frac{\\frac{9}{4}a^{2} + \\frac{1}{4}a^{2} - 4c^{2}}{2 \\times \\frac{3}{2}a \\times \\frac{a}{2}} = \\frac{5}{3} - \\frac{8}{3}e^{2} = \\frac{1}{2} $, solving gives: $ e = \\frac{\\sqrt{7}}{2} $."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2 p x  (p>0)$ be $F$, and the directrix be $l$. A line passing through the focus intersects the parabola at points $A$ and $B$. Perpendiculars from $A$ and $B$ to $l$ have feet $C$ and $D$, respectively. If $|A F|=2|B F|$ and the area of triangle $C D F$ is $\\sqrt{2}$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;l: Line;Directrix(G) = l;H: Line;PointOnCurve(F,H);A: Point;B: Point;Intersection(H, G) = {A, B};L1: Line;L2: Line;C: Point;D: Point;PointOnCurve(A, L1);PointOnCurve(B, L2);IsPerpendicular(L1, l);IsPerpendicular(L2, l);FootPoint(L1, l) = C;FootPoint(L2, l) = D;Abs(LineSegmentOf(A, F)) = 2*Abs(LineSegmentOf(B, F));Area(TriangleOf(C, D, F)) = sqrt(2)",
    "query_expressions": "p",
    "answer_expressions": "(2*sqrt(3))/3",
    "fact_spans": "[[[1, 24], [49, 52]], [[1, 24]], [[135, 138]], [[4, 24]], [[28, 31]], [[1, 31]], [[35, 38], [74, 77]], [[1, 38]], [[44, 46]], [[1, 46]], [[53, 56], [66, 69]], [[57, 60], [70, 73]], [[44, 62]], [], [], [[83, 86]], [[87, 90]], [[63, 80]], [[63, 80]], [[63, 80]], [[63, 80]], [[63, 90]], [[63, 90]], [[93, 107]], [[109, 133]]]",
    "query_spans": "[[[135, 142]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since the line AB passes through the focus F, we have y_{1}y_{2} = -p^{2} (without loss of generality, assume C is in the first quadrant). Also, from |AF| = 2|BF|, it follows that |y_{1}| = 2|y_{2}|, i.e., y_{1} = -2y_{2}. Therefore, -2y_{2}^{2} = -p^{2}, y_{2} = -\\frac{\\sqrt{2}}{2}p, y_{1} = -2y_{2} = \\sqrt{2}p. Thus, S_{\\triangle CDF} = \\frac{1}{2}|y_{1}-y_{2}| \\times p = \\frac{1}{2} \\times \\frac{3\\sqrt{2}}{2}p^{2} = \\sqrt{2}. Solving gives p = \\frac{2\\sqrt{3}}{3}."
  },
  {
    "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are given by $y=\\pm x$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(Asymptote(G)) = (y = pm*x)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 57], [75, 78]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 73]]]",
    "query_spans": "[[[75, 84]]]",
    "process": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ are given by $y=\\pm\\frac{b}{a}x$. According to the problem, $\\pm\\frac{b}{a}=\\pm1$, so $\\frac{b}{a}=1$. The eccentricity of the hyperbola is $e=\\frac{c}{a}=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{\\frac{a^{2}+b^{2}}{a^{2}}}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{2}$."
  },
  {
    "text": "A vertex of the hyperbola is $(0,2)$, and the focal length is $6$. What is its standard equation?",
    "fact_expressions": "G: Hyperbola;Coordinate(OneOf(Vertex(G))) = (0, 2);FocalLength(G) = 6",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/4 - x^2/5 = 1",
    "fact_spans": "[[[0, 3], [24, 25]], [[0, 16]], [[0, 23]]]",
    "query_spans": "[[[24, 31]]]",
    "process": "One vertex of the hyperbola is (0,2), so a=2. The focal distance is 6, thus c=3, hence b^{2}=c^{2}-a^{2}=5. Therefore, the standard equation of the hyperbola is: \\frac{y^{2}}{4}-\\frac{x^{2}}{5}=1"
  },
  {
    "text": "The directrix of the parabola $C$: $y^{2}=2 p x$ ($p>0$) intersects the $x$-axis at point $M$. From point $M$, two tangents are drawn to $C$, with points of tangency $P$ and $Q$, respectively. Then $\\angle P M Q$=?",
    "fact_expressions": "C: Parabola;p: Number;P: Point;M: Point;Q:Point;L1: Line;L2: Line;p>0;Expression(C) = (y^2 = 2*(p*x));Intersection(Directrix(C), xAxis) = M;TangentOfPoint(M, C) = {L1, L2};TangentPoint(L1,C)=P;TangentPoint(L2,C)=Q",
    "query_expressions": "AngleOf(P,M,Q)",
    "answer_expressions": "pi/2",
    "fact_spans": "[[[0, 26], [48, 51]], [[7, 26]], [[62, 65]], [[38, 41], [43, 47]], [[66, 69]], [], [], [[7, 26]], [[0, 26]], [[0, 41]], [[42, 56]], [[42, 69]], [[42, 69]]]",
    "query_spans": "[[[70, 86]]]",
    "process": "Find the equation of the directrix of parabola C, yielding M(-\\frac{p}{2},0). Let the tangent line passing through M(-\\frac{p}{2},0) be x=my-\\frac{p}{2}; combine this with the parabola equation and eliminate variables. Using A=0, the slope of the tangent can be found, thus solving the problem. [Detailed Solution] Since the parabola C: y^{2}=2px (p>0), the directrix equation is: x=-\\frac{p}{2}. Hence, M(-\\frac{p}{2},0). Let the tangent line through point M be x=my-\\frac{p}{2}. Substituting into y^{2}=2px gives y^{2}-2pmy+p^{2}=0. Thus, A=4p^{2}m^{2}-4p^{2}=0. Since p\\neq0, we have m^{2}=1, solving gives: m=\\pm1. Therefore, the tangent slope k=\\pm1, so MQ\\bot MP. Hence \\angle PMQ=\\frac{\\pi}{2}. It follows that \\angle PMQ=\\frac{1}{2}. Using A=0 to find the tangent slope, \\angle PMO can then be determined."
  },
  {
    "text": "What is the length of the minor axis of the ellipse $9 x^{2}+8 y^{2}=72$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (9*x^2 + 8*y^2 = 72)",
    "query_expressions": "Length(MinorAxis(G))",
    "answer_expressions": "4*sqrt(2)",
    "fact_spans": "[[[0, 22]], [[0, 22]]]",
    "query_spans": "[[[0, 28]]]",
    "process": "From the ellipse equation $9x^{2}+8y^{2}=72$, it can be rewritten as $\\frac{x^{2}}{8}+\\frac{y^{2}}{9}=1$, $\\therefore a^{2}=9$, $b^{2}=8$, $\\therefore b=2\\sqrt{2}$, $\\therefore$ the minor axis length is $2b=4\\sqrt{2}$."
  },
  {
    "text": "Let the parabola $y^{2}=8x$ intersect the line passing through its focus with slope $1$ at points $A$ and $B$, and let $O$ be the origin. Then $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}$=?",
    "fact_expressions": "G: Parabola;H: Line;O: Origin;A: Point;B: Point;Expression(G) = (y^2 = 8*x);PointOnCurve(Focus(G), H);Intersection(G, H) = {A, B};Slope(H) = 1",
    "query_expressions": "DotProduct(VectorOf(O, A), VectorOf(O, B))",
    "answer_expressions": "-12",
    "fact_spans": "[[[1, 15], [17, 18]], [[28, 30]], [[42, 45]], [[32, 35]], [[36, 39]], [[1, 15]], [[16, 30]], [[1, 41]], [[21, 30]]]",
    "query_spans": "[[[52, 103]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with eccentricity $e$, and the line $l$: $y=x$ intersecting the hyperbola $C$ at points $M$ and $N$. If $|M N|=\\sqrt{2} b$, then the value of $e$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;M: Point;N: Point;a>0;b>0;e:Number;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(C) = e;Expression(l)=(y=x);Intersection(l, C) = {M, N};Abs(LineSegmentOf(M, N)) = sqrt(2)*b",
    "query_expressions": "e",
    "answer_expressions": "sqrt(6)",
    "fact_spans": "[[[72, 84]], [[2, 63], [85, 91]], [[10, 63]], [[10, 63]], [[93, 96]], [[97, 100]], [[10, 63]], [[10, 63]], [[124, 127], [68, 71]], [[2, 63]], [[2, 71]], [[72, 84]], [[72, 102]], [[104, 122]]]",
    "query_spans": "[[[124, 131]]]",
    "process": "Solve the system of equations to find the coordinates of point M, use |MN| = \\sqrt{2}b, and simplify to obtain b^{2} = 5a^{2}, then find the eccentricity. Without loss of generality, assume point M(x,y) lies in the first quadrant. Solve the system \\begin{cases}\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1\\\\y=x\\end{cases} to get x^{2} = y^{2} = \\frac{a^{2}b^{2}}{b^{2}-a^{2}}, and since |MN| = \\sqrt{2}b, we have x^{2} + y^{2} = \\frac{b^{2}}{2}, thus \\frac{2a^{2}b^{2}}{b^{2}-a^{2}} = \\frac{b^{2}}{2}. Simplify to obtain b^{2} = 5a^{2}, so e = \\sqrt{1+\\frac{b^{2}}{a^{2}}} = \\sqrt{6}."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ shares the same foci as the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, and the asymptotes of hyperbola $C$ are given by $y=\\pm 2 x$, then the equation of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Ellipse;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2/9 + y^2/4 = 1);Focus(C) = Focus(G);Expression(Asymptote(C)) = (y = pm*(2*x))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2/4 = 1",
    "fact_spans": "[[[2, 53], [99, 105], [125, 131]], [[10, 53]], [[10, 53]], [[54, 91]], [[2, 53]], [[54, 91]], [[2, 97]], [[99, 123]]]",
    "query_spans": "[[[125, 136]]]",
    "process": "In the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, $\\therefore a^{2}+b^{2}=5$ $\\therefore a=1, b=2$, the asymptotes of hyperbola $C$ are $y=\\pm2x$ $\\therefore \\frac{b}{a}=2$ $\\therefore a=1, b=2$, the equation of the hyperbola is $x^{2}-\\frac{y^{2}}{4}=1$"
  },
  {
    "text": "The standard equation of a hyperbola passing through the point $P(3,2)$ and having the same asymptotes as the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$ is?",
    "fact_expressions": "P: Point;Coordinate(P) = (3, 2);PointOnCurve(P, Z);G: Hyperbola;Expression(G) = (x^2/4 - y^2/2 = 1);Z: Hyperbola;Expression(Asymptote(Z))=Expression(Asymptote(G))",
    "query_expressions": "Expression(Z)",
    "answer_expressions": "x^2 - y^2/(1/2) = 1",
    "fact_spans": "[[[1, 10]], [[1, 10]], [[0, 62]], [[12, 50]], [[12, 50]], [[59, 62]], [[11, 62]]]",
    "query_spans": "[[[59, 69]]]",
    "process": "Let the required hyperbola equation be \\frac{x^{2}}{4}-\\frac{y^{2}}{2}=\\lambda. Since it passes through the point P(3,2), we have \\frac{3^{2}}{4}-\\frac{2^{2}}{2}=\\lambda, so \\lambda=\\frac{1}{4}. Therefore, the required hyperbola equation is x^{2}-\\frac{y^{2}}{\\frac{1}{2}}=1."
  },
  {
    "text": "Given the equation of ellipse $C$ is $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, $B(-2,0)$, $A(4,2)$, and $M$ is any point on $C$, then the minimum value of $|M A|-|M B|$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/9 + y^2/5 = 1);B: Point;Coordinate(B) = (-2, 0);A: Point;Coordinate(A) = (4, 2);M: Point;PointOnCurve(M, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(M, A)) - Abs(LineSegmentOf(M, B)))",
    "answer_expressions": "2*sqrt(2)-6",
    "fact_spans": "[[[2, 7], [74, 77]], [[2, 46]], [[48, 57]], [[48, 57]], [[59, 67]], [[59, 67]], [[70, 73]], [[70, 82]]]",
    "query_spans": "[[[84, 103]]]",
    "process": "By the given conditions, $ a=3 $, $ b=\\sqrt{5} $, $ c=2 $, so $ B(-2,0) $ is the left focus, $ D(2,0) $ is the right focus. Thus, $ |MA| - |MB| = |MA| - (2a - |MD|) = |MA| + |MD| - 2a \\geqslant |AD| - 2a = 2\\sqrt{2} - 6 $. The equality holds if and only if $ M $, $ D $, $ A $ are collinear."
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ are $F_{1}$ and $F_{2}$, and $P$ is a point on the ellipse such that $P F_{1} \\perp P F_{2}$. Find the area of $\\triangle F_{1} P F_{2}$.",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "9",
    "fact_spans": "[[[0, 38], [61, 63]], [[0, 38]], [[41, 48]], [[49, 56]], [[0, 56]], [[57, 60]], [[57, 67]], [[70, 92]]]",
    "query_spans": "[[[95, 125]]]",
    "process": ""
  },
  {
    "text": "It is known that a hyperbola has the same focal distance as the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{25}=1$, and the sum of their eccentricities is $\\frac{14}{5}$. Then, the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/9 + y^2/25 = 1);FocalLength(G) = FocalLength(H);Eccentricity(G) + Eccentricity(H) = 14/5",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/4 - x^2/12 = 1",
    "fact_spans": "[[[2, 5], [76, 79]], [[6, 44]], [[6, 44]], [[2, 50]], [[51, 73]]]",
    "query_spans": "[[[76, 86]]]",
    "process": ""
  },
  {
    "text": "If the parabola $y^{2}=mx$ and the ellipse $\\frac{x^{2}}{9} + \\frac{y^{2}}{5}=1$ have a common focus, then $m$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = m*x);m: Number;H: Ellipse;Expression(H) = (x^2/9 + y^2/5 = 1);Focus(G) = OneOf(Focus(H))",
    "query_expressions": "m",
    "answer_expressions": "pm*8",
    "fact_spans": "[[[1, 14]], [[1, 14]], [[64, 67]], [[15, 54]], [[15, 54]], [[1, 62]]]",
    "query_spans": "[[[64, 69]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>b, b>0$) are $y=\\pm x$, and it passes through the point $(\\sqrt{2}, 1)$, then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>b;b>0;Expression(Asymptote(G)) = (y = pm*x);H: Point;Coordinate(H) = (sqrt(2), 1);PointOnCurve(H,G) = True",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 = 1",
    "fact_spans": "[[[2, 58], [97, 100]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 74]], [[78, 94]], [[78, 94]], [[2, 94]]]",
    "query_spans": "[[[97, 105]]]",
    "process": "The asymptotes of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>b, b>0) are given by y=\\pm\\frac{b}{a}x=\\pm x, so a=b; substituting the point (\\sqrt{2},1) into the hyperbola equation yields a=b=1, therefore the equation of this hyperbola is x^{2}-y^{2}=1"
  },
  {
    "text": "If point $M$ is a moving point on the line $l$: $y = -2$, and two tangent lines are drawn from point $M$ to the parabola $C$: $y = \\frac{1}{4} x^{2}$, with points of tangency $A$ and $B$ respectively, then $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}$ = ?",
    "fact_expressions": "l: Line;C: Parabola;O: Origin;A: Point;B: Point;M: Point;l1: Line;l2: Line;Expression(C) = (y = x^2/4);Expression(l) = (y=-2);PointOnCurve(M, l);TangentOfPoint(M, C) = {l1, l2};TangentPoint(l1,C)=A;TangentPoint(l2,C)=B",
    "query_expressions": "DotProduct(VectorOf(O, A), VectorOf(O, B))",
    "answer_expressions": "-4",
    "fact_spans": "[[[6, 19]], [[30, 59]], [[79, 128]], [[70, 73]], [[74, 77]], [[1, 5], [25, 29]], [], [], [[30, 59]], [[6, 19]], [[1, 23]], [[30, 64]], [[24, 77]], [[24, 77]]]",
    "query_spans": "[[[79, 130]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = x_{1}x_{2} + y_{1}y_{2} $. Since $ A $, $ B $ lie on $ y = \\frac{1}{4}x^{2} $, $ \\therefore \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = x_{1}x_{2} + y_{1}y_{2} = x_{1}x_{2} + \\frac{1}{16}(x_{1}x_{2})^{2} $. Since the derivative of $ y = \\frac{1}{4}x^{2} $ is $ y = \\frac{1}{2}x $, the slope of tangent $ MA $ is $ \\frac{1}{2}x_{1} $, so the equation of tangent $ MA $ is $ y - y_{1} = \\frac{1}{2}x_{1}(x - x_{1}) $, i.e., $ 2y = x_{1}x - 2y_{1} $. Similarly, the equation of tangent $ MB $ is $ 2y = x_{2}x - 2y_{2} $ $ \\textcircled{2} $. Let $ M(m, -2) $, substituting into $ \\textcircled{1} $ and $ \\textcircled{2} $ gives $ -4 = x_{1}m - 2y_{1} $ and $ -4 = x_{2}m - 2y_{2} $. $ \\therefore $ the equation of line $ AB $ is $ mx - 2y + 4 = 0 $. Solving the system: \n\\[\n\\begin{cases}\nmx - 2y + 4 = 0 \\\\\ny = \\frac{1}{4}x^{2}\n\\end{cases}\n\\]\ngives $ x^{2} - 2mx - 8 = 0 $. $ \\therefore x_{1}x_{2} = -8 $. $ \\therefore \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = x_{1}x_{2} + \\frac{1}{16}(x_{1}x_{2})^{2} = -4 $."
  },
  {
    "text": "The standard equation of a hyperbola passing through the point $(3 , 0)$ with eccentricity $\\frac{5}{3}$ is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (3, 0);Eccentricity(G)=5/3;PointOnCurve(H,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/16 = 1",
    "fact_spans": "[[[31, 34]], [[2, 12]], [[2, 12]], [[13, 34]], [[0, 34]]]",
    "query_spans": "[[[31, 41]]]",
    "process": "Since the hyperbola passes through the point (3,0) and has an eccentricity of $\\frac{5}{3}$, the standard equation of the hyperbola is $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$. Thus, we know $e=\\frac{5}{3}$ and $a=3$, so $c=5$. Using the relationship among $a$, $b$, and $c$, the equation is obtained as $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$."
  },
  {
    "text": "Given that $F_{1}$, $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $P$ is a point on the ellipse $C$, and $\\overrightarrow{P F_{1}} \\perp \\overrightarrow{P F_{2}}$. If the area of $\\triangle P F_{1} F_{2}$ is $9$, then $b=?$",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Focus(C)={F1,F2};PointOnCurve(P, C);IsPerpendicular(VectorOf(P, F1),VectorOf(P,F2));Area(TriangleOf(P, F1, F2)) = 9",
    "query_expressions": "b",
    "answer_expressions": "3",
    "fact_spans": "[[[20, 76], [86, 91]], [[190, 193]], [[26, 76]], [[82, 85]], [[2, 10]], [[12, 19]], [[26, 76]], [[26, 76]], [[20, 76]], [[2, 81]], [[82, 94]], [[96, 153]], [[156, 188]]]",
    "query_spans": "[[[190, 195]]]",
    "process": ""
  },
  {
    "text": "Hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has left and right foci $F_{1}$, $F_{2}$ respectively, $|F_{1} F_{2}|=10$, $P$ is a point on the right branch of $E$, $P F_{1}$ intersects the $y$-axis at point $A$, the incircle of $\\Delta P A F_{2}$ touches side $A F_{2}$ at point $Q$. If $|A Q|=\\sqrt{3}$, then the eccentricity of $E$ is?",
    "fact_expressions": "E: Hyperbola;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(E) = F1;RightFocus(E) = F2;Abs(LineSegmentOf(F1, F2)) = 10;P: Point;PointOnCurve(P, RightPart(E));A: Point;Intersection(LineSegmentOf(P, F1), yAxis) = A;Q: Point;TangentPoint(LineSegmentOf(A, F2), InscribedCircle(TriangleOf(P, A, F2))) = Q;Abs(LineSegmentOf(A, Q)) = sqrt(3)",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "5*sqrt(3)/3",
    "fact_spans": "[[[0, 61], [114, 117], [208, 211]], [[0, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[70, 78]], [[80, 87]], [[0, 87]], [[0, 87]], [[89, 108]], [[110, 113]], [[110, 123]], [[140, 144]], [[124, 144]], [[184, 187]], [[146, 187]], [[190, 206]]]",
    "query_spans": "[[[208, 217]]]",
    "process": "Let the incircle of triangle $APAF_{2}$ touch side $PF_{2}$ at point $M$ and touch $AP$ at point $N$. Then $|PM| = |PN|$, $|AQ| = |AN| = \\sqrt{3}$, $|QF_{2}| = |MF_{2}|$. By the symmetry of the hyperbola, we have $|AF_{1}| = |AF_{2}| = |AQ| + |QF_{2}| = \\sqrt{3} + |QF_{2}|$. From the definition of the hyperbola, $|PF_{1}| - |PF_{2}| = |PA| + |AF_{1}| - |PM| - |MF_{2}| = \\sqrt{3} + |QF_{2}| + |AN| + |NP| - |PM| - |MF_{2}| = 2\\sqrt{3} = 2a$, solving gives $a = \\sqrt{3}$. Also $|F_{1}F_{2}| = 10$, so $c = 5$, and the eccentricity $e = \\frac{c}{a} = \\frac{5}{\\sqrt{3}} = \\frac{5\\sqrt{3}}{3}$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$ and directrix $l$, a line passing through point $F$ intersects $l$ at point $A$ and intersects the parabola at a point $B$, such that $\\overrightarrow{F A}=3 \\overrightarrow{F B}$. Then $|A B|$=?",
    "fact_expressions": "C: Parabola;F: Point;A: Point;B: Point;l: Line;H: Line;Expression(C) = (y^2 = 4*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(F, H);Intersection(H, l) = A;OneOf(Intersection(H, C)) = B;VectorOf(F, A) = 3*VectorOf(F, B)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8/3",
    "fact_spans": "[[[2, 21], [53, 56]], [[24, 27], [35, 39]], [[47, 51]], [[62, 65]], [[30, 33], [43, 46]], [[40, 42]], [[2, 21]], [[2, 27]], [[2, 33]], [[34, 42]], [[40, 51]], [[40, 65]], [[67, 112]]]",
    "query_spans": "[[[115, 124]]]",
    "process": "Find the focus and directrix of the parabola. Let A(-1, a), B(m, n). From the given conditions and the definition of a parabola, we have m + 1 = \\frac{4}{3}, then from |AB| = 2|BF|, we can solve the problem. The parabola C: y^{2} = 4x has focus F(1, 0) and directrix l: x = -1. Let A(-1, a), B(x, y). Since \\overrightarrow{FA} = 3\\overrightarrow{FB}, it follows that \\frac{x+1}{2} = \\frac{2}{3}, thus |BF| = x + 1 = \\frac{4}{3}, |AB| = 2|BF| = \\frac{8}{3}."
  },
  {
    "text": "The distance from the right focus of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ to one of its asymptotes is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1)",
    "query_expressions": "Distance(RightFocus(G), OneOf(Asymptote(G)))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 28], [33, 34]], [[0, 28]]]",
    "query_spans": "[[[0, 44]]]",
    "process": "Find the coordinates of the right focus and the equations of the asymptotes, then use the point-to-line distance formula to solve. From $\\frac{x^{2}}{4}-y^{2}=1$, we have: $a^{2}=4$, $b^{2}=1$, so $c^{2}=a^{2}+b^{2}=5$, thus $c=\\sqrt{5}$. The right focus is $(\\sqrt{5},0)$, and one of the asymptotes is $x-2y=0$. Therefore, the distance from the right focus $(\\sqrt{5},0)$ to the asymptote $x-2y=0$ is $d=\\frac{\\sqrt{5}-0}{\\sqrt{1+2^{2}}}=1$."
  },
  {
    "text": "Write the standard equation of a hyperbola that shares asymptotes with the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$ (different from the original hyperbola)?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/4 = 1);H: Hyperbola;Asymptote(G) = Asymptote(H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2/2 - y^2/8 = 1",
    "fact_spans": "[[[5, 33]], [[5, 33]], [[38, 41]], [[4, 41]]]",
    "query_spans": "[[[38, 57]]]",
    "process": "The hyperbola sharing asymptotes with the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$ is $x^{2}-\\frac{y^{2}}{4}=2(\\lambda\\neq0)$, that is, $\\frac{x^{2}}{2}-\\frac{y^{2}}{42}=1(\\lambda\\neq0)$, so one can fill in $\\frac{x^{2}}{2}-\\frac{y^{2}}{8}=1$."
  },
  {
    "text": "Hyperbola $C$: $\\frac{x^{2}}{2}-\\frac{y^{2}}{3}=1$ has left and right vertices $A_{1}$, $A_{2}$ respectively. Point $P$ lies on $C$ and the slope of line $P A_{2}$ ranges in $[1,2]$. What is the range of the slope of line $P A_{1}$?",
    "fact_expressions": "C:Hyperbola;P: Point;A2: Point;A1: Point;Expression(C) = (x^2/2 - y^2/3 = 1);LeftVertex(C)=A1;RightVertex(C)=A2;PointOnCurve(P,C);Range(Slope(LineOf(P,A2)))=[1,2]",
    "query_expressions": "Range(Slope(LineOf(P,A1)))",
    "answer_expressions": "[3/4,3/2]",
    "fact_spans": "[[[0, 43], [76, 79]], [[71, 75]], [[62, 69]], [[52, 59]], [[0, 43]], [[0, 69]], [[0, 69]], [[71, 80]], [[81, 107]]]",
    "query_spans": "[[[110, 130]]]",
    "process": "From the hyperbola $ C: \\frac{x^2}{2} - \\frac{y^{2}}{3} = 1 $, we get $ A_1(-\\sqrt{2}, 0) $, $ A_2(\\sqrt{2}, 0) $. Let point $ P(x_0, y_0) $, then $ \\frac{x_0^2}{2} - \\frac{y_0^2}{3} = 1 $, i.e., $ y_0^2 = \\frac{3}{2}(x_0^2 - 2) $. Also, $ k_{PA_1} = \\frac{y_0}{x_0 + \\sqrt{2}} $, $ k_{PA_2} = \\frac{y_0}{x_0 - \\sqrt{2}} $. Therefore, $ k_{PA_1} k_{PA_2} = \\frac{y_0^2}{x_0^2 - 2} = \\frac{\\frac{3}{2}(x_0^2 - 2)}{x_0^2 - 2} = \\frac{3}{2} $. Since $ k_{PA_2} \\in [1, 2] $, it follows that $ k_{PA_1} \\in \\left[\\frac{3}{4}, \\frac{3}{2}\\right] $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, $P$ is on the ellipse, and $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$, then the distance from point $P$ to the $x$-axis is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/9 + y^2/5 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = pi/3",
    "query_expressions": "Distance(P, xAxis)",
    "answer_expressions": "5*sqrt(3)/6",
    "fact_spans": "[[[18, 55], [63, 65]], [[2, 9]], [[59, 62], [106, 110]], [[10, 17]], [[18, 55]], [[2, 58]], [[59, 66]], [[68, 104]]]",
    "query_spans": "[[[106, 120]]]",
    "process": "From the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, we obtain: $a=3$, $b=\\sqrt{5}$, $c=\\sqrt{a^{2}-b^{2}}=2$. Let $|PF_{1}|=m$, $|PF_{2}|=n$. Then $m+n=2a=6$, $(2\\times2)^{2}=m^{2}+n^{2}-2mn\\cos\\frac{\\pi}{3}$, which gives: $mn=\\frac{20}{3}$. $\\therefore S_{\\triangle F_{1}F_{2}P}=\\frac{1}{2}\\times2c\\cdot|y_{P}|=\\frac{1}{2}mn\\sin\\frac{\\pi}{3}$, $\\therefore 2\\times2|y_{P}|=\\frac{20}{3}\\times\\frac{\\sqrt{3}}{2}$, solving yields $|y_{P}|=\\frac{5\\sqrt{3}}{6}$."
  },
  {
    "text": "What are the coordinates of the focus $F$ of the parabola $x=8 a y^{2}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x = 8*(a*y^2));a: Number;F: Point;Focus(G) = F",
    "query_expressions": "Coordinate(F)",
    "answer_expressions": "(1/(32*a), 0)",
    "fact_spans": "[[[0, 16]], [[0, 16]], [[3, 16]], [[19, 22]], [[0, 22]]]",
    "query_spans": "[[[19, 27]]]",
    "process": "Since the equation of the parabola is $x=8ay^{2}$, it follows that $y^{2}=\\frac{1}{8a}x$, so the focus of the parabola lies on the $x$-axis, and $2p=\\frac{1}{8a}$, that is, $p=\\frac{1}{16a}$, so the coordinates of its focus $F$ are $(\\frac{1}{32a},0)$. Therefore, the answer is $(\\frac{1}{32a},0)$."
  },
  {
    "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1$ $(a>0)$ are tangent to the circle $x^{2}+(y-2)^{2}=1$, then the length of the real axis of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Circle;a>0;Expression(G) = (-y^2/4 + x^2/a^2 = 1);Expression(H) = (x^2 + (y - 2)^2 = 1);IsTangent(Asymptote(G),H)",
    "query_expressions": "Length(RealAxis(G))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[1, 48], [78, 81]], [[4, 48]], [[53, 73]], [[4, 48]], [[1, 48]], [[53, 73]], [[1, 75]]]",
    "query_spans": "[[[78, 87]]]",
    "process": "From the given conditions, the asymptote equation is y=\\frac{2}{2}x, and the center of the circle is (0,2) with radius 1, so the length of the transverse axis of the hyperbola is 2a=\\sqrt{3}"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ has its right vertex at $A$ and upper vertex at $B$, and a line $l$ intersects the ellipse at points $C$ and $D$. If line $l$ is parallel to line $AB$, and the slopes of lines $AC$ and $BD$ are $k_{1}$ and $k_{2}$ respectively, then the value of $k_{1} k_{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);A: Point;RightVertex(G) = A;B: Point;UpperVertex(G) = B;l: Line;C: Point;D: Point;Intersection(l, G) = {C, D};IsParallel(l, LineOf(A, B));k1: Number;k2: Number;Slope(LineOf(A, C)) = k1;Slope(LineOf(B, D)) = k2",
    "query_expressions": "k1*k2",
    "answer_expressions": "1/4",
    "fact_spans": "[[[2, 29], [53, 55]], [[2, 29]], [[34, 37]], [[2, 37]], [[42, 45]], [[2, 45]], [[47, 52], [68, 73]], [[57, 60]], [[61, 64]], [[47, 66]], [[68, 83]], [[104, 111]], [[112, 119]], [[85, 119]], [[85, 119]]]",
    "query_spans": "[[[121, 138]]]",
    "process": "According to the problem, point A(2,0), B(0,1), the slope of line AB is $-\\frac{1}{2}$. Since line $l$ is parallel to line AB, let the equation of line $l$ be:  \n\\[\n\\begin{cases}\n-\\frac{1}{2}x + t, & t \\neq 1, \\\\\ny = -\\frac{1}{2}x + t \\\\\nx^{2} + 4y^{2} = 4\n\\end{cases}\n\\]  \nEliminating $y$ and simplifying yields: $x^{2} - 2tx + 2t^{2} - 2 = 0$,  \n$\\triangle = 4t^{2} - 4(2t^{2} - 2) = -4(t^{2} - 2) > 0$, solving gives $-\\sqrt{2} < t < \\sqrt{2}$, thus $-\\sqrt{2} < t < 1$ or $1 < t < \\sqrt{2}$.  \nLet $C(x_{1}, y_{1})$, $D(x_{2}, y_{2})$, then $x_{1} + x_{2} = 2t$, $x_{1}x_{2} = 2t^{2} - 2$,  \nand $k_{1} = \\frac{y_{1}}{x_{1} - 2}$, $k_{2} = \\frac{y_{2} - 1}{x_{2}}$.  \nTherefore,  \n\\[\nk_{1}k_{2} = \\frac{y_{1}y_{2} - 1}{(x_{1} - 2)x_{2}} = \\frac{(-\\frac{1}{2}x_{1} + t)(-\\frac{1}{2}x_{2} + t - 1)}{x_{1}x_{2} + 4t^{2} + 2(x_{1} - 2t)} = \\frac{1}{4} \\cdot \\frac{x_{1}x_{2} - 2t \\cdot 2t + 4t^{2} - 12}{x_{2}x_{2}} = \\frac{1}{4} - 2x_{2}^{\\frac{1}{4}}\n\\]  \nSo the value of $k_{1}k_{2}$ is $\\frac{1}{4}$."
  },
  {
    "text": "If one of the asymptotes of the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{4}=1$ is given by the equation $x-2 y=0$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;Expression(G) = (-y^2/4 + x^2/a = 1);Expression(OneOf(Asymptote(G))) = (x - 2*y = 0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[1, 39], [60, 63]], [[4, 39]], [[1, 39]], [[1, 57]]]",
    "query_spans": "[[[60, 69]]]",
    "process": "From the hyperbola equation, its asymptotes are given by $ y = \\pm\\frac{2}{\\sqrt{a}}x $. Given that $ x - 2y = 0 $ is an asymptote, we have $ \\frac{\\sqrt{a}}{\\sqrt{a}} = \\frac{1}{2} $, solving gives $ a = 16 $, and $ b = 2 $, $ c = \\sqrt{a + 4} = 2\\sqrt{5} $, then $ e = \\frac{\\sqrt{5}}{2} $."
  },
  {
    "text": "The eccentricity of an ellipse is $e=\\frac{1}{2}$, and its foci coincide with the foci of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$. If $P$ is any point on the ellipse, then the sum of the distances from $P$ to the two foci of the ellipse is?",
    "fact_expressions": "H: Ellipse;e: Number;Eccentricity(H) = e;e = 1/2;G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1);Focus(H) = Focus(G);P: Point;PointOnCurve(P, H);F1: Point;F2: Point;Focus(H) = {F1, F2}",
    "query_expressions": "Distance(P, F1)+Distance(P, F2)",
    "answer_expressions": "8",
    "fact_spans": "[[[19, 21], [22, 23], [65, 67], [78, 80]], [[3, 18]], [[0, 21]], [[3, 18]], [[27, 55]], [[27, 55]], [[22, 60]], [[61, 64], [74, 77]], [[61, 72]], [], [], [[78, 83]]]",
    "query_spans": "[[[74, 89]]]",
    "process": ""
  },
  {
    "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ have its right focus at $F(c, 0)$. A circle centered at $F(c, 0)$ with radius $a$ intersects an asymptote of $C$ at points $A$ and $B$. If $|AB|=\\sqrt{2}c$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;F: Point;A: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (c, 0);RightFocus(C)=F;Center(G)=F;Radius(G)=a;Intersection(G,OneOf(Asymptote(C)))={A, B};Abs(LineSegmentOf(A, B)) = sqrt(2)*c;c:Number",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[1, 62], [88, 91], [130, 136]], [[9, 62]], [[79, 82]], [[86, 87]], [[66, 75]], [[99, 102]], [[103, 106]], [[9, 62]], [[9, 62]], [[1, 62]], [[66, 75]], [[1, 75]], [[0, 87]], [[79, 87]], [[86, 108]], [[110, 128]], [[66, 75]]]",
    "query_spans": "[[[130, 142]]]",
    "process": "Since the distance from a focus of the hyperbola to one of its asymptotes is $ b $, it follows from the given condition that $ b^{2} + \\left(\\frac{\\sqrt{2}c}{2}\\right)^{2} = a^{2} $, that is, $ c^{2} - a^{2} + \\frac{c^{2}}{2} = a^{2} $, then $ \\frac{c^{2}}{a^{2}} = \\frac{4}{3} $, so the eccentricity $ e = \\frac{c}{a} = \\frac{2\\sqrt{3}}{3} $."
  },
  {
    "text": "If the chord $AB$ formed by the intersection of line $l$ and the parabola $y^{2}=16x$ is bisected by the point $P(3,2)$, then the equation of line $l$ is?",
    "fact_expressions": "l: Line;G: Parabola;Expression(G) = (y^2 = 16*x);A: Point;B: Point;Intersection(l, G) = {A, B};IsChordOf(LineSegmentOf(A, B), G);P: Point;Coordinate(P) = (3, 2);MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Expression(l)",
    "answer_expressions": "4*x - y -10 =0",
    "fact_spans": "[[[1, 6], [47, 52]], [[7, 22]], [[7, 22]], [[28, 33]], [[28, 33]], [[1, 33]], [[7, 33]], [[34, 43]], [[34, 43]], [[28, 45]]]",
    "query_spans": "[[[47, 57]]]",
    "process": "Let points $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. If $ l\\bot x $ axis, then the midpoint of segment $ AB $ lies on the $ x $-axis, which does not satisfy the condition. Therefore, the slope of line $ l $ exists. From the given, we have\n$$\n\\begin{cases}\nx_{1}+x_{2}=6 \\\\\ny_{1}+y_{2}=4\n\\end{cases}\n$$\nSince points $ A $ and $ B $ lie on the parabola, then\n$$\n\\begin{cases}\ny_{1}=16x_{1} \\\\\ny_{2}=16x_{2}\n\\end{cases}\n$$\nSubtracting the two equations gives $ (y_{1}-y_{2})(y_{1}+y_{2})=16(x_{1}-x_{2}) $. Thus, the slope of line $ l $ is\n$$\n\\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = \\frac{16}{y_{1}+y_{2}} = 4\n$$\nTherefore, the equation of line $ l $ is $ y-2=4(x-3) $, i.e., $ 4x-y-10=0 $. The answer is $ 4x-y-10=0 $."
  },
  {
    "text": "Let the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ always pass through the fixed point $A(1,2)$. What is the minimum distance from the center of the ellipse to its directrix?",
    "fact_expressions": "C: Ellipse;a: Number;b: Number;A: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (1, 2);PointOnCurve(A, C)",
    "query_expressions": "Min(Distance(Center(C), Directrix(C)))",
    "answer_expressions": "sqrt(5) + 2",
    "fact_spans": "[[[1, 58], [72, 74]], [[8, 58]], [[8, 58]], [[62, 70]], [[8, 58]], [[8, 58]], [[1, 58]], [[62, 70]], [[1, 70]]]",
    "query_spans": "[[[72, 88]]]",
    "process": "Let the focal distance of the ellipse be $2c$, $\\frac{a^{2}}{c}=\\frac{1}{t}$, $\\therefore c=ta^{2}$. The ellipse passes through the fixed point $A(1,2)$, so $\\frac{1}{a^{2}}+\\frac{4}{b^{2}}=1 \\Rightarrow b^{2}+4a^{2}=a^{2}b^{2} \\Rightarrow 5a^{2}-c^{2}=a^{2}(a^{2}-c^{2})$, $\\Rightarrow 5a^{2}-(ta^{2})^{2}=a^{2}(a^{2}-(ta^{2})^{2}] \\Rightarrow t^{2}a^{4}-(t^{2}+1)a^{2}+5=0$. $\\Delta=(t^{2}+1)^{2}-20t^{2}\\geqslant0 \\Leftrightarrow t^{2}-2\\sqrt{5}t+1\\geqslant0$. $\\therefore t\\geqslant\\sqrt{5}+2$ or $0<t\\leqslant\\sqrt{5}-2$. $\\therefore 0<\\frac{1}{t}\\leqslant\\sqrt{5}-2$ or $\\frac{1}{t}\\geqslant\\sqrt{5}+2$. The ellipse passes through the fixed point $A(1,2)$, $\\therefore \\frac{a^{2}}{c}=\\frac{1}{t}>1$. Therefore, the minimum distance from the center of the ellipse to the directrix is: $\\sqrt{5}+2$."
  },
  {
    "text": "Given the parabola: $x^{2}=2 p y(p>0)$ with focus $F$, directrix $l$, point $P$ on $C$, the perpendicular from $P$ to $l$ intersects $l$ at point $E$, and $\\angle P F E=60^{\\circ}$, $|P F|=6$, then the equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 2*p*y);p: Number;p>0;F: Point;Focus(C) = F;l: Line;Directrix(C) = l;P: Point;PointOnCurve(P,C) = True;E: Point;l1: Line;PointOnCurve(P,l1) = True;IsPerpendicular(l,l1) = True;Intersection(l,l1) = E;AngleOf(P, F, E) = ApplyUnit(60, degree);Abs(LineSegmentOf(P, F)) = 6",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2=6*y",
    "fact_spans": "[[[2, 24], [44, 47], [109, 115]], [[2, 24]], [[6, 24]], [[6, 24]], [[28, 31]], [[2, 31]], [[35, 38], [55, 58], [62, 65]], [[2, 38]], [[39, 43], [50, 54]], [[39, 48]], [[66, 70]], [], [[49, 61]], [[49, 61]], [[49, 70]], [[72, 97]], [[98, 107]]]",
    "query_spans": "[[[109, 120]]]",
    "process": "According to the definition of a parabola, $\\triangle PFE$ is an equilateral triangle, thus the length $|EF|$ is obtained. Then in right triangle $\\triangle FEH$, the length $|HF|$ is found, and consequently the value of $p$ is determined, leading to the result. Let $H$ be the intersection point of the directrix and the $x$-axis, the directrix is $x = -\\frac{p}{2}$, and the focus is $(\\frac{p}{2}, 0)$. By the definition of the parabola, $|PE| = |PF|$. Since $\\angle PFE = 60^{\\circ}$, $\\triangle PFE$ is an equilateral triangle and $\\angle FEH = 30^{\\circ}$. Therefore, $|EF| = |PF| = 6$, so $|HF| = \\frac{1}{2}|EF| = 3$. Since $|HF| = p$, it follows that $p = 3$. Hence, the equation of parabola $C$ is $x^2 = 6y$."
  },
  {
    "text": "Ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{9}=1$, point $A(0, \\frac{1}{2})$, point $P$ is a moving point on the ellipse, then the maximum value of $|P A|$ is?",
    "fact_expressions": "G: Ellipse;A: Point;P: Point;Expression(G) = (x^2/12 + y^2/9 = 1);Coordinate(A) = (0, 1/2);PointOnCurve(P, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "sqrt(13)",
    "fact_spans": "[[[0, 38], [65, 67]], [[39, 59]], [[60, 64]], [[0, 38]], [[39, 59]], [[60, 71]]]",
    "query_spans": "[[[73, 86]]]",
    "process": "From the ellipse $\\frac{x^2}{12}+\\frac{y^{2}}{9}=1$, let point $P(2\\sqrt{3}\\cos\\theta,3\\sin\\theta)$, so equality holds if and only if: $\\sin\\theta=-\\frac{1}{2}$, therefore the maximum value is $\\sqrt{13}$."
  },
  {
    "text": "Given $F_{1}(-2,0)$, $F_{2}(2,0)$, let $P$ be one of the intersection points of the ellipse $x^{2}+2 y^{2}=8$ and the hyperbola $x^{2}-y^{2}=2$. Then $|P F_{1}| \\cdot|P F_{2}|$=?",
    "fact_expressions": "F1: Point;F2: Point;Coordinate(F1) = (-2, 0);Coordinate(F2) = (2, 0);H: Ellipse;Expression(H) = (x^2 + 2*y^2 = 8);G: Hyperbola;Expression(G) = (x^2 - y^2 = 2);P: Point;OneOf(Intersection(H, G)) = P",
    "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 16]], [[17, 30]], [[2, 16]], [[17, 30]], [[36, 55]], [[36, 55]], [[56, 74]], [[56, 74]], [[32, 35]], [[32, 79]]]",
    "query_spans": "[[[81, 109]]]",
    "process": "The ellipse and hyperbola are reduced to the standard equations $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$, $\\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1$, from which it is known that the two curves share foci.  \n\\begin{cases}Let |PF_{1}|=r_{1}, |PF_{2}|=r_{2},\\\\r_{1}+r_{2}=4\\sqrt{2}\\\\|r_{1}-r_{2}|=2\\sqrt{2}\\end{cases}  \n\\begin{cases}r_{1}=3\\sqrt{2}\\\\r_{2}=\\sqrt{2}\\end{cases} or \\begin{cases}r_{1}=\\sqrt{2}\\\\r_{2}=3\\sqrt{2}\\end{cases} \\Rightarrow r_{1} \\times r_{2}=6"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively, and that a line $l$ passing through the origin $O$ with an inclination angle of $60^{\\circ}$ intersects the ellipse $C$ at a point $M$, such that $|\\overrightarrow{M F_{1}}+\\overrightarrow{M F_{2}}|=|\\overrightarrow{M F_{1}}-\\overrightarrow{M F_{2}}|$, find the eccentricity of the ellipse $C$.",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;O: Origin;l: Line;PointOnCurve(O, l);Inclination(l) = ApplyUnit(60, degree);M: Point;OneOf(Intersection(l, C)) = M;Abs(VectorOf(M, F1) + VectorOf(M, F2)) = Abs(VectorOf(M, F1) - VectorOf(M, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[20, 79], [116, 121], [238, 243]], [[20, 79]], [[27, 79]], [[27, 79]], [[27, 79]], [[27, 79]], [[2, 9]], [[10, 17]], [[2, 85]], [[2, 85]], [[87, 92]], [[110, 115]], [[86, 115]], [[93, 115]], [[127, 130]], [[110, 130]], [[132, 237]]]",
    "query_spans": "[[[238, 249]]]",
    "process": "Without loss of generality, assume point $ M $ is in the first quadrant. Squaring both sides of $ |\\overrightarrow{MF_{1}} + \\overrightarrow{MF_{2}}| = |\\overrightarrow{MF_{1}} - \\overrightarrow{MF_{2}}| $ and simplifying yields: $ \\overrightarrow{MF} \\cdot \\overrightarrow{MF_{2}} = 0 $. In right triangle $ \\triangle MF_{1}F_{2} $, $ \\angle MF_{2}F_{1} = 60^{\\circ} $, $ MF_{1} = 2c\\sin60^{\\circ} = \\sqrt{3}c $, $ MF_{2} = 2c\\sin30^{\\circ} = c $. From $ MF_{1} + MF_{2} = 2a $, we have $ (\\sqrt{3} + 1)c = 2a $, so $ e = \\frac{c}{a} = \\frac{2}{\\sqrt{3} + 1} = \\sqrt{3} - 1 $."
  },
  {
    "text": "The focal distance of the ellipse $\\frac{x^{2}}{20}+\\frac{y^{2}}{k}=1$ is $6$, then the value of $k$ is?",
    "fact_expressions": "G: Ellipse;k: Number;Expression(G) = (x^2/20 + y^2/k = 1);FocalLength(G) = 6",
    "query_expressions": "k",
    "answer_expressions": "{11, 29}",
    "fact_spans": "[[[0, 38]], [[47, 50]], [[0, 38]], [[0, 45]]]",
    "query_spans": "[[[47, 54]]]",
    "process": ""
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=8x$ intersects the parabola at points $A$ and $B$, $O$ is the origin. If $|AF|=6$, then the area of $\\triangle BOF$ is?",
    "fact_expressions": "G: Parabola;H: Line;B: Point;O: Origin;F: Point;A: Point;Expression(G) = (y^2 = 8*x);Focus(G)=F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, F)) = 6",
    "query_expressions": "Area(TriangleOf(B, O, F))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[1, 15], [26, 29]], [[22, 24]], [[36, 39]], [[42, 45]], [[18, 21]], [[32, 35]], [[1, 15]], [[1, 21]], [[0, 24]], [[22, 41]], [[52, 61]]]",
    "query_spans": "[[[63, 85]]]",
    "process": "From the equation of the parabola, the directrix equation can be obtained. Using the properties of the parabola and the value of |AF|, the x-coordinate of A can be found. Substituting into the parabola's equation gives the y-coordinate of A, then the equation of line AB is determined. Solving this equation together with the parabola's equation yields the y-coordinate of B. Substituting into the area formula gives the area of triangle OBF. [Detailed Solution] According to the problem, the directrix equation of the parabola is: $x = -2$, focus $F(2,0)$. Let A be above the x-axis, with x-coordinate $x_{1}$. Since $|AF| = 6$, we have $x_{1} + 2 = 6$, so $x_{1} = 4$. Substituting into the parabola's equation, we get $y_{1}^{2} = 8 \\times 4$, so $y_{1} = 4\\sqrt{2}$, i.e., $A(4, 4\\sqrt{2})$. Thus, $k_{AB} = \\frac{4\\sqrt{2}}{4 - 2} = 2\\sqrt{2}$. Therefore, the equation of line AB is: $y = 2\\sqrt{2}(x - 2)$. Solving simultaneously \n$$\n\\begin{cases}\ny = 2\\sqrt{2}(x - 2) \\\\\ny^2 = 8x\n\\end{cases}\n$$\nwe obtain $x^{2} - 5x + 4 = 0$, solving gives $x_{B} = 1$, $x_{A} = 4$. Substituting the x-coordinate of B, 1, into the parabola's equation, $y_{B} = -\\sqrt{8 \\times 1} = -2\\sqrt{2}$. Therefore, $S_{\\triangle BOF} = \\frac{1}{2}|OF| \\cdot |y_{B}| = \\frac{1}{2} \\times 2 \\times 2\\sqrt{2} = 2\\sqrt{2}$."
  },
  {
    "text": "Given that the curve $\\frac{x^{2}}{{a}}-\\frac{y^{2}}{b}=1$ intersects the line $x+y-1=0$ at points $P$ and $Q$, and $\\overrightarrow{O P} \\cdot \\overrightarrow{O Q}=0$ ($O$ is the origin), then the value of $\\frac{1}{a}-  \\frac{1}{b}$ is?",
    "fact_expressions": "G: Line;H: Curve;b: Number;a: Number;O: Origin;P: Point;Q: Point;Expression(G) = (x + y - 1 = 0);Expression(H) = (-y^2/b + x^2/a = 1);Intersection(H, G) = {P, Q};DotProduct(VectorOf(O, P), VectorOf(O, Q)) = 0",
    "query_expressions": "1/a - 1/b",
    "answer_expressions": "2",
    "fact_spans": "[[[42, 53]], [[2, 41]], [[4, 41]], [[4, 41]], [[119, 122]], [[56, 59]], [[60, 63]], [[42, 53]], [[2, 41]], [[2, 65]], [[67, 118]]]",
    "query_spans": "[[[128, 159]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of ellipse $C$, and $P$ is a point on $C$. If $|P F_{1}|+|P F_{2}|=2|F_{1} F_{2}|$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Focus(C) = {F1, F2};PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 2*Abs(LineSegmentOf(F1, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[18, 23], [33, 36], [79, 82]], [[29, 32]], [[2, 9]], [[10, 17]], [[2, 28]], [[29, 39]], [[41, 77]]]",
    "query_spans": "[[[79, 88]]]",
    "process": "P is a point on the ellipse C. By the definition of an ellipse, |PF_{1}| + |PF_{2}| = 2a. Since |PF_{1}| + |PF_{2}| = 2|F_{1}F_{2}| = 4c, it follows that 2a = 4c, so e = \\frac{c}{a} = \\frac{1}{2}."
  },
  {
    "text": "If the asymptotes of a hyperbola are given by $y = \\pm 3x$, and it passes through the point $A(1, \\sqrt{10})$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;A: Point;Coordinate(A) = (1, sqrt(10));Expression(Asymptote(G)) = (y = pm*(3*x));PointOnCurve(A,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2-9*x^2=1",
    "fact_spans": "[[[1, 4], [45, 48]], [[25, 43]], [[25, 43]], [[1, 22]], [[1, 43]]]",
    "query_spans": "[[[45, 53]]]",
    "process": "According to the asymptote equations and the coordinates of the given point, analyze the position of the hyperbola's foci, set up the equation, and determine the coefficients. Since the asymptotes of the hyperbola are y = \\pm3x, and it passes through point A(1,\\sqrt{10}), it is not difficult to determine that point A(1,\\sqrt{10}) lies above the lines y = \\pm3x; hence, the foci of this hyperbola lie on the y-axis. Let the hyperbola equation be \\frac{y^{2}}{a^{2}} - \\frac{x^{2}}{b^{2}} = 1, then \\frac{a}{b} = 3, \\frac{10}{a^{2}} - \\frac{1}{b^{2}} = 1. Solving gives b = \\frac{1}{2}, a = \\frac{3}{2}, then the hyperbola's equation is y^{2} - 9x^{2} = 1. The answer is: y^{2} - 9x^{2} = 1. This is a basic problem involving solving equations, and the key point is to determine the focus position based on the point through which the hyperbola passes."
  },
  {
    "text": "Write the standard equation of a hyperbola with eccentricity $2$ and foci on the $y$-axis.",
    "fact_expressions": "G: Hyperbola;Eccentricity(G) = 2;PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 - x^2/3 = 1",
    "fact_spans": "[[[21, 24]], [[4, 24]], [[12, 24]]]",
    "query_spans": "[[[21, 31]]]",
    "process": "Let the required hyperbola equation be $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0,b>0)$, and let $2c$ be the focal distance. Since $e=\\frac{c}{a}=2$, it follows that $c=2a$. Assume $a=1$, then $c=2$, so $c^{2}=4a^{2}$, hence $b^{2}=3a^{2}$. Therefore, the hyperbola equation can be $y^{2}-\\frac{x^{2}}{3}=1$."
  },
  {
    "text": "The tangent line to the circle $x^{2}+y^{2}=1$ intersects the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ at two points $A$, $B$. The tangent lines to the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ at points $A$ and $B$, respectively, intersect at point $P$. Then the equation of the trajectory of point $P$ is?",
    "fact_expressions": "H: Circle;Expression(H) = (x^2 + y^2 = 1);L1: Line;IsTangent(L1, H) = True;G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);A: Point;B: Point;Intersection(L1, G) = {A, B};Intersection(TangentOnPoint(A, G), TangentOnPoint(B, G)) = P;P: Point",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2/16+y^2/9=1",
    "fact_spans": "[[[0, 16]], [[0, 16]], [], [[0, 19]], [[20, 57], [82, 117]], [[20, 57]], [[61, 64], [71, 74]], [[65, 68], [75, 78]], [[0, 68]], [[68, 126]], [[122, 126], [128, 132]]]",
    "query_spans": "[[[128, 139]]]",
    "process": "Let the points of tangency be $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Then the tangent line passing through point $ A(x_{1},y_{1}) $ has equation $ y - y_{1} = k(x - x_{1}) $, i.e., $ y = kx + y_{1} - kx_{1} $. Substituting into $ 3x^{2} + 4y^{2} - 12 = 0 $, after rearranging and simplifying we obtain $ (3 + 4k^{2})x^{2} + 8k(y_{1} - kx_{1})x + 4(y_{1} - kx_{1})^{2} - 12 = 0 $. From the given condition we get $ 64k^{2}(y_{1} - kx_{1})^{2} - 4 \\times 4[(y_{1} - kx_{1})^{2} - 3](3 + 4k^{2}) = 0 $, that is, $ 3 + 4k^{2} = (y_{1} - kx_{1})^{2} $. Combining with $ 3x_{1}^{2} + 4y_{1}^{2} - 12 = 0 $, we obtain $ k = -\\frac{3x_{1}}{4y_{1}} $, then the tangent equation is $ 3x_{1}x + 4y_{1}y - 12 = 0 $. Similarly, the tangent line passing through point $ B(x_{2},y_{2}) $ is $ 3x_{2}x + 4y_{2}y - 12 = 0 $. Let the intersection point be $ P(x_{0},y_{0}) $, thus from the given condition we have $ 3x_{1}x_{0} + 4y_{1}y_{0} - 12 = 0 $ and $ 3x_{2}x_{0} + 4y_{2}y_{0} - 12 = 0 $. Observing these two equations, it can be seen that the line passing through points $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $ is $ 3x_{0}x + 4y_{0}y - 12 = 0 $. Since this line is tangent to $ x^{2} + y^{2} = 1 $, then $ \\frac{12}{\\sqrt{(3x_{0})^{2} + (4y_{0})^{2}}} = 1 $, i.e., $ 9x_{0}^{2} + 16y_{0}^{2} = 144 $, so the intersection point $ P $ moves along the curve $ \\frac{x^{2}}{16} + \\frac{y^{2}}{9} = 1 $. The answer to be filled is $ \\frac{x^{2}}{16} + \\frac{y^{2}}{9} = 1 $."
  },
  {
    "text": "If the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ intersects the directrix of the parabola $y^{2}=4 x$ to form a line segment of length $b$, then $a$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;H: Parabola;Expression(H) = (y^2 = 4*x);Length(InterceptChord(Directrix(H), G)) = b",
    "query_expressions": "a",
    "answer_expressions": "2*sqrt(5)/5",
    "fact_spans": "[[[1, 57]], [[1, 57]], [[81, 84]], [[86, 89]], [[4, 57]], [[4, 57]], [[58, 72]], [[58, 72]], [[1, 84]]]",
    "query_spans": "[[[86, 91]]]",
    "process": "The directrix of the parabola is given by the equation $x = -1$. Since the line $x = -1$ intersects the hyperbola and the length of the chord is $b$, we have $\\frac{1}{a^{2}} - \\frac{b^{2}}{b^{2}} = 1$, solving which gives: $a = \\frac{2\\sqrt{5}}{5}$."
  },
  {
    "text": "The vertex of the parabola $C$ is at the origin, its axis of symmetry is the $y$-axis, and its focus lies on the line $2x - 3y - 5 = 0$. Then the equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;G: Line;O:Origin;Expression(G) = (2*x - 3*y - 5 = 0);Vertex(C)=O;SymmetryAxis(C)=yAxis;PointOnCurve(Focus(C),G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2=-(20/3)*y",
    "fact_spans": "[[[0, 6], [47, 53]], [[28, 43]], [[10, 14]], [[28, 43]], [[0, 14]], [[0, 23]], [[0, 44]]]",
    "query_spans": "[[[47, 58]]]",
    "process": "\\because the vertex of the parabola is at the origin, and the axis of symmetry is the y-axis, \\therefore let the standard equation of the parabola be x^{2}=2my. The focus lies on the line 2x-3y-5=0, \\therefore letting x=0 gives y=-\\frac{5}{3}, \\therefore the focus F(0,-\\frac{5}{3}). \\frac{2m}{4}=-\\frac{5}{3}, solving for m yields m=-\\frac{10}{3}, \\therefore the standard equation of the parabola is x^{2}=-\\frac{20}{3}y"
  },
  {
    "text": "Given that vertices $B$ and $C$ of $\\triangle A B C$ lie on the ellipse $\\frac{x^{2}}{3}+y^{2}=1$, vertex $A$ coincides with the focus $F_{1}$ of the ellipse, and the other focus $F_{2}$ of the ellipse lies on side $BC$, then what is the perimeter of $\\triangle A B C$?",
    "fact_expressions": "G: Ellipse;B: Point;C: Point;A: Point;Expression(G) = (x^2/3 + y^2 = 1);PointOnCurve(F2, LineSegmentOf(B, C));PointOnCurve(B, G);PointOnCurve(C, G);F1: Point;F2: Point;OneOf(Focus(G)) = F1;OneOf(Focus(G)) = F2;Negation(F1 = F2);A = F1",
    "query_expressions": "Perimeter(TriangleOf(A, B, C))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[30, 57], [81, 83], [65, 67]], [[22, 25]], [[26, 29]], [[61, 64]], [[30, 57]], [[90, 105]], [[22, 58]], [[22, 58]], [[70, 77]], [[90, 97]], [[65, 77]], [[81, 97]], [[65, 97]], [[61, 79]]]",
    "query_spans": "[[[107, 129]]]",
    "process": ""
  },
  {
    "text": "The standard equation of an ellipse with focal length $2$ passing through the point $P(-\\sqrt{5}, 0)$ is?",
    "fact_expressions": "G: Ellipse;P: Point;Coordinate(P) = (-sqrt(5), 0);FocalLength(G) = 2;PointOnCurve(P, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/5+y^2/4=1,x^2/5+y^2/6=1}",
    "fact_spans": "[[[29, 31]], [[9, 28]], [[9, 28]], [[0, 31]], [[8, 31]]]",
    "query_spans": "[[[29, 38]]]",
    "process": "By the given condition, 2c=2, so c=1. Also, the ellipse passes through point P(-\\sqrt{5},0). If the foci are on the x-axis, then a=\\sqrt{5}, and b^{2}=a^{2}-c^{2}=4, so the ellipse equation is \\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1; if the foci are on the y-axis, then b=\\sqrt{5}, and a^{2}=b^{2}+c^{2}=6, so the ellipse equation is \\frac{y^{2}}{6}+\\frac{x^{2}}{5}=1. Therefore, the standard equations of the ellipse are: \\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1 or \\frac{y^{2}}{6}+\\frac{x^{2}}{5}=1."
  },
  {
    "text": "The focus of the parabola $y^{2}=4 x$ is $F$, the line $y=x$ intersects the parabola at points $O$ and $A$ ($O$ being the origin), and intersects the directrix of the parabola at point $B$. The line $A F$ intersects the parabola again at point $C$. Then $\\cos \\angle A B C$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;H: Line;Expression(H) = (y = x);O: Origin;A: Point;Intersection(H, G) = {O, A};B: Point;Intersection(H, Directrix(G)) = B;C: Point;OneOf(Intersection(LineOf(A,F),G))=C",
    "query_expressions": "Cos(AngleOf(A, B, C))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[0, 14], [31, 34], [57, 60], [78, 81]], [[0, 14]], [[18, 21]], [[0, 21]], [[22, 29]], [[22, 29]], [[36, 39], [46, 49]], [[40, 43]], [[22, 45]], [[65, 69]], [[22, 69]], [[87, 90]], [[40, 90]]]",
    "query_spans": "[[[92, 113]]]",
    "process": "\\begin{cases}y=x\\\\y^{2}=4x\\end{cases}\\Rightarrow A(4,4),\\begin{cases}y=x\\\\x=-1\\end{cases}\\Rightarrow B(-1,-1),AF:y=\\frac{4-0}{4-1}(x-1),\\begin{cases}y=\\frac{4}{3}(x-1)\\\\y^{2}=4x\\end{cases}\\Rightarrow C\\left(\\frac{1}{4},-1\\right)\\therefore\\angle ABC=\\frac{\\pi}{4},\\cos\\angle ABC=\\frac{\\sqrt{2}}{2}"
  },
  {
    "text": "The maximum eccentricity of the hyperbola $C$: $\\frac{x^{2}}{2}-\\frac{y^{2}}{\\sin ^{2} \\theta}=1(0<\\theta \\leq \\frac{\\pi}{2})$ is?",
    "fact_expressions": "C: Hyperbola;theta: Number;0<theta ;theta <= pi/2;Expression(C) = (x^2/2 - y^2/Sin(theta)^2 = 1)",
    "query_expressions": "Max(Eccentricity(C))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 87]], [[8, 87]], [[8, 87]], [[8, 87]], [[0, 87]]]",
    "query_spans": "[[[0, 97]]]",
    "process": "According to the problem, the eccentricity of hyperbola $ C $ is $ e = \\frac{2 + \\sin^{2}\\theta}{2} \\in \\left(1, \\frac{\\sqrt{6}}{2}\\right] $."
  },
  {
    "text": "The parabola $C$: $y^{2}=4 x$ has focus $F$. From a point $P$ on $C$, draw a perpendicular to the directrix $l$ of $C$, with foot of perpendicular at $A$. If the slope of line $A F$ is $-2$, then the area of $\\triangle P A F$ is?",
    "fact_expressions": "C: Parabola;G: Line;A: Point;F: Point;P: Point;l:Line;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(P,C);Directrix(C)=l;PointOnCurve(P,G);IsPerpendicular(l,G);FootPoint(l,G)=A;Slope(LineOf(A,F))=-2",
    "query_expressions": "Area(TriangleOf(P, A, F))",
    "answer_expressions": "10",
    "fact_spans": "[[[0, 19], [28, 31], [28, 31]], [], [[54, 57]], [[23, 26]], [[34, 37]], [[44, 47]], [[0, 19]], [[0, 26]], [[28, 37]], [[38, 47]], [[27, 50]], [[27, 50]], [[27, 57]], [[59, 74]]]",
    "query_spans": "[[[76, 98]]]",
    "process": "From the equation of the parabola, we get F(1,0), and the directrix equation is x = -1. Let P(m,n). From the given conditions, we have A(-1,n), then k_{AF} = \\frac{n}{-1-1} = -2, solving gives n = 4. Substituting P(m,4) into the parabola equation yields 4^{2} = 4m, solving gives m = 4, so P(4,4). Then |PA| = 4 + 1 = 5, thus the area S of \\triangle PAF is \\frac{1}{2} \\times 5 \\times 4 = 10."
  },
  {
    "text": "Given that the two foci of ellipse $E$ are $F_{1}$ and $F_{2}$ respectively, point $P$ lies on the ellipse, and $\\tan \\angle P F_{1} F_{2}=\\frac{1}{3}$, $\\tan \\angle P F_{2} F_{1}=-3$, then the eccentricity of ellipse $E$ is?",
    "fact_expressions": "E: Ellipse;P: Point;F1: Point;F2: Point;Focus(E)={F1,F2};PointOnCurve(P, E);Tan(AngleOf(P, F1, F2)) = 1/3;Tan(AngleOf(P, F2, F1)) = -3",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(10)/4",
    "fact_spans": "[[[2, 7], [36, 38], [116, 121]], [[31, 35]], [[15, 22]], [[23, 30]], [[2, 30]], [[31, 41]], [[43, 82]], [[84, 114]]]",
    "query_spans": "[[[116, 127]]]",
    "process": "From the given conditions, since $\\tan\\angle PF_{1}F_{2} = \\frac{1}{3}$, it follows that $\\sin\\angle PF_{1}F_{2} = \\frac{\\sqrt{10}}{10}$; and since $\\tan\\angle PF_{2}F_{1} = -3$, it follows that $\\sin\\angle PF_{2}F_{1} = \\frac{3\\sqrt{10}}{10}$. Then, $\\tan\\angle F_{2}PF_{1} = \\tan(\\pi - \\angle PF_{1}F_{2} - \\angle PF_{2}F_{1}) \\frac{PF_{F}_{2} + \\angle PF_{2}F_{1})}{\\angle PF_{1}F_{2} + \\tan\\angle PF_{2}F_{1}} = \\frac{4}{3}$. Solving gives $\\sin\\angle F_{2}PF_{1} = \\frac{4}{5}$. Then, in $\\triangle PF_{1}F_{2}$, by the law of sines, we obtain $\\frac{\\sin\\angle PF_{1}F_{2} + \\sin\\angle PF_{2}F}{F}$"
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. Then $\\frac{1}{|A F|}+\\frac{1}{|B F|}$=?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B}",
    "query_expressions": "1/Abs(LineSegmentOf(B, F)) + 1/Abs(LineSegmentOf(A, F))",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 15], [25, 28]], [[22, 24]], [[29, 32]], [[18, 21]], [[33, 36]], [[1, 15]], [[1, 21]], [[0, 24]], [[22, 38]]]",
    "query_spans": "[[[40, 75]]]",
    "process": "From $ y^{2} = 4x $, the focus $ F $ has coordinates $ (1,0) $, and the directrix equation is $ x = -1 $. Let the line passing through $ F $ be $ y = k(x - 1) $. Substituting into the parabola equation gives $ k^{2}(x - 1)^{2} = 4x $. Simplifying yields: $ k^{2}x^{2} - (2k^{2} + 4)x + k^{2} = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1}x_{2} = 1 $. According to the definition of a parabola, $ |AF| = x_{1} + 1 $, $ |BF| = x13.(1000 - 10) = 1 $, so the answer is $ 31 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has eccentricity $2$, point $P$ is a moving point on the hyperbola, $F_{1}$ and $F_{2}$ are its left and right foci respectively, and $O$ is the origin. Then the range of $\\frac{|P F_{1}|+|P F_{2}|}{|O P|}$ is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;O: Origin;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = 2;PointOnCurve(P, G);LeftFocus(G)=F1;RightFocus(G)=F2",
    "query_expressions": "Range((Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)))/Abs(LineSegmentOf(O, P)))",
    "answer_expressions": "(2,4]",
    "fact_spans": "[[[2, 48], [61, 64], [87, 88]], [[5, 48]], [[5, 48]], [[56, 60]], [[69, 76]], [[77, 84]], [[94, 97]], [[2, 48]], [[2, 55]], [[56, 68]], [[69, 93]], [[69, 93]]]",
    "query_spans": "[[[104, 146]]]",
    "process": "According to the definition of hyperbola, express $\\frac{|PF|+|PF_{2}|}{|OP|}$ in terms of the horizontal coordinate of point $P$, and use the monotonicity of the function to find the range of $\\frac{|PF|+|PF_{2}|}{|OP|}$. [Detailed solution] $\\because$ the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $2$, $\\therefore b^{2}=3a^{2}$, hence the equation of the hyperbola is $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3a^{2}}=1$. Without loss of generality, assume $P$ is a point on the right branch, $P(x,y)$ where $x\\geqslant a$, $(x+c)^{2}+y^{2}=\\sqrt{(x+c)^{2}+3x^{2}-3a^{2}}=2x+a=|PF_{1}|-2a=2x-a$, that is, $|PF_{1}|+|PF_{2}|=4x$. $|x^{2}+y^{2}|=\\sqrt{4x^{2}-3a^{2}}$, $\\frac{|PF_{1}|+|PF_{2}|}{|OD|}=\\frac{\\sqrt{1}}{5}$, at this time $0$, $\\therefore$ when $0<\\frac{a}{x}\\leqslant1$, $\\therefore 1\\leqslant4-3(\\frac{a}{x})^{2}$, $\\frac{}{\\sqrt{4-3(\\frac{a}{x})^{2}}}\\in(2,4]$"
  },
  {
    "text": "The ellipse $3 x^{2}+k y^{2}=3$ has a focal distance of $2 \\sqrt{2}$. Then $k=$?",
    "fact_expressions": "G: Ellipse;k: Number;Expression(G) = (k*y^2 + 3*x^2=3);FocalLength(G) = 2*sqrt(2)",
    "query_expressions": "k",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 21]], [[39, 42]], [[0, 21]], [[0, 37]]]",
    "query_spans": "[[[39, 44]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a point on the curve $C$: $x=3 \\sqrt{y}$, $T(0, \\frac{9}{4})$, $A(3,3)$, then the minimum value of $|P T|+|P A|$ is?",
    "fact_expressions": "C: Curve;Expression(C) = (x = 3*sqrt(y));P: Point;PointOnCurve(P, C);T: Point;Coordinate(T) = (0, 9/4);A: Point;Coordinate(A) = (3, 3)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, T)))",
    "answer_expressions": "21/4",
    "fact_spans": "[[[6, 27]], [[6, 27]], [[2, 5]], [[2, 30]], [[31, 50]], [[31, 50]], [[52, 61]], [[52, 61]]]",
    "query_spans": "[[[63, 82]]]",
    "process": "It is easy to see that curve C is the right half of the parabola $ x^{2} = 9y $. As shown in the figure, since the directrix of the parabola $ x^{2} = 9y $ is $ y = -\\frac{9}{4} $ and T is the focus of the parabola, $ |PT| $ equals the distance $ |PM| $ from P to the line $ l: y = -\\frac{9}{4} $. Draw a perpendicular from A to this line, with foot H. Then the minimum value of $ |PT| + |PA| = |PA| + |PM| $ is $ |AH| = 3 + \\frac{9}{4} = \\frac{21}{4} $."
  },
  {
    "text": "It is known that the hyperbola $C$ passes through the point $P(6, \\sqrt{3})$, and the equations of its asymptotes are $y = \\pm \\frac{x}{3}$. Then the equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;P: Point;Coordinate(P) = (6, sqrt(3));PointOnCurve(P, C);Expression(Asymptote(C)) = (y = pm*(x/3))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/9-y^2=1",
    "fact_spans": "[[[2, 8], [56, 62]], [[10, 27]], [[10, 27]], [[2, 27]], [[2, 54]]]",
    "query_spans": "[[[56, 66]]]",
    "process": "Combining with the asymptote equations, set the hyperbola equation as \\frac{x^{2}}{9}-y^{2}=\\lambda, substitute the coordinates of point P to obtain \\lambda=1, hence the hyperbola equation is \\frac{x^{2}}{9}-y^{2}=1. [Note] This problem examines the method of calculating hyperbola equations; the key is to combine the asymptote equations and set up the hyperbola equation accordingly. Moderate difficulty."
  },
  {
    "text": "If the standard equation of a hyperbola is $x^{2}-\\frac{y^{2}}{4}=1$, then the equation of the directrix of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/4 = 1)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=pm*(\\sqrt{5}/5)",
    "fact_spans": "[[[1, 4], [38, 41]], [[1, 35]]]",
    "query_spans": "[[[38, 48]]]",
    "process": ""
  },
  {
    "text": "If the line $y=2 x+b$ has no common points with the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, then the range of values for $b$ is?",
    "fact_expressions": "G: Ellipse;H: Line;b: Number;Expression(G) = (x^2/4 + y^2 = 1);Expression(H) = (y = b + 2*x);NumIntersection(H,G)=0",
    "query_expressions": "Range(b)",
    "answer_expressions": "{(-oo,-sqrt(17))+(sqrt(17),+oo)}",
    "fact_spans": "[[[13, 40]], [[1, 12]], [[46, 49]], [[13, 40]], [[1, 12]], [[1, 44]]]",
    "query_spans": "[[[46, 56]]]",
    "process": "Analysis: From \\begin{cases}y=2x+b\\\\\\frac{x^{2}}{4}+y^{2}=1\\end{cases}, we obtain \\frac{x^{2}}{4}+(2x+b)^{2}=1, simplifying to 17x^{2}+16bx+4b^{2}-4=0. Therefore, A=(16b)^{2}-4\\times17\\times(4b^{2}-4)<0, solving gives b>\\sqrt{17} or b<-\\sqrt{17}, i.e., b\\in(-\\infty,-\\sqrt{17})\\cup(\\sqrt{17},+\\infty)"
  },
  {
    "text": "It is known that the hyperbola $C$ is centered at the origin and its axes of symmetry are the coordinate axes. One of its foci coincides with the focus of the parabola $y^{2}=8x$, and one of its asymptotes has the equation $x+y=0$. Then the equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Parabola;O: Origin;Expression(G) = (y^2 = 8*x);Center(C) = O;SymmetryAxis(C)=axis;OneOf(Focus(C) )= Focus(G);Expression(OneOf(Asymptote(C))) = (x + y = 0)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/2 - y^2/2 = 1",
    "fact_spans": "[[[2, 8], [23, 24], [67, 73]], [[30, 44]], [[12, 14]], [[30, 44]], [[2, 14]], [[2, 22]], [[23, 49]], [[23, 65]]]",
    "query_spans": "[[[67, 78]]]",
    "process": "The focus coordinates of the parabola $ y^2 = 8x $ are $ (2, 0) $, so the right focus coordinates of the hyperbola $ C $ are $ (2, 0) $. Since one asymptote equation of the hyperbola is $ x + y = 0 $, we have $ a = b $, so $ a^2 + a^2 = 4 $, hence $ a^2 = 2 $. Therefore, the hyperbola equation is $ \\frac{x^2}{2} - \\frac{y^2}{2} = 1 $."
  },
  {
    "text": "A line $ l $ passing through the focus of the parabola $ C: y^{2} = 6x $ intersects $ C $ at points $ A $ and $ B $. If $ |AB| = 9 $, then what is the x-coordinate of the midpoint of segment $ AB $?",
    "fact_expressions": "l: Line;C: Parabola;B: Point;A: Point;Expression(C) = (y^2 = 6*x);PointOnCurve(Focus(C), l);Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, B)) = 9",
    "query_expressions": "XCoordinate(MidPoint(LineSegmentOf(A, B)))",
    "answer_expressions": "3",
    "fact_spans": "[[[24, 29]], [[1, 20], [30, 33]], [[38, 41]], [[34, 37]], [[1, 20]], [[0, 29]], [[24, 43]], [[45, 54]]]",
    "query_spans": "[[[56, 71]]]",
    "process": "As shown in the figure, the parabola $ y^{2} = 6x $ has focus $ F\\left(\\frac{3}{2}, 0\\right) $ and directrix $ x = -\\frac{3}{2} $. From points $ A $ and $ B $, perpendiculars are drawn to the directrix, with feet at $ A' $ and $ B' $, respectively. Then $ |AB| = |AF| + |BF| = |AA'| + |BB'| = 9 $. From the midpoint $ M $ of $ AB $, a perpendicular is drawn to the directrix, with foot at $ M' $. Then $ M'M' $ is the median line of the right trapezoid $ ABB'A' $, so $ |M'M'| = \\frac{1}{2}(|AA'| + |BB'|) = \\frac{9}{2} $, that is, $ x_{M} + \\frac{3}{2} = \\frac{9}{2} $, solving gives $ x_{M} = 3 $. Therefore, the horizontal coordinate of $ M $ is 3."
  },
  {
    "text": "The line passing through the right focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ and perpendicular to the $x$-axis intersects the ellipse at points $M$ and $N$. The circle with $MN$ as diameter passes exactly through the left focus. Then, the eccentricity of the ellipse is equal to?",
    "fact_expressions": "G: Ellipse;a: Number;b: Number;H: Circle;I: Line;M: Point;N: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(RightFocus(G), I);IsPerpendicular(I, xAxis);Intersection(I, G) = {M, N};IsDiameter(LineSegmentOf(M, N), H);PointOnCurve(LeftFocus(G), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2) - 1",
    "fact_spans": "[[[1, 53], [69, 71], [102, 104]], [[3, 53]], [[3, 53]], [[93, 94]], [[66, 68]], [[73, 76]], [[77, 80]], [[3, 53]], [[3, 53]], [[1, 53]], [[0, 68]], [[58, 68]], [[66, 82]], [[83, 94]], [[0, 100]]]",
    "query_spans": "[[[102, 111]]]",
    "process": "\\because the line passing through the right focus of the ellipse and perpendicular to the x-axis intersects the ellipse at points M, N, let x=c, substitute into \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0), we get y=\\pm\\frac{b^{2}}{a}. Since the circle with MN as diameter passes exactly through the left focus, \\therefore\\frac{b^{2}}{a}=2c, which becomes a^{2}-c^{2}=2ac, \\therefore e^{2}+2e-1=0, e>0, solving gives e=\\frac{-2+2\\sqrt{2}}{2}=\\sqrt{2}"
  },
  {
    "text": "The parabola $y^{2}=2 p x(p>0)$ passes through the point $P(4,4)$. Then the distance from point $P$ to the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;P: Point;Coordinate(P) = (4, 4);PointOnCurve(P, G)",
    "query_expressions": "Distance(P, Directrix(G))",
    "answer_expressions": "5",
    "fact_spans": "[[[0, 21], [38, 41]], [[0, 21]], [[3, 21]], [[3, 21]], [[22, 31], [33, 37]], [[22, 31]], [[0, 31]]]",
    "query_spans": "[[[33, 48]]]",
    "process": "\\because point P(4,4) is a point on the parabola y^{2}=2px, \\therefore 4^{2}=2p\\times4, solving gives p=2, \\therefore the equation of the parabola is y^{2}=4x, \\therefore the equation of the directrix is x=-1, \\therefore the distance from point P to the directrix of the parabola is 4+1=5."
  },
  {
    "text": "The two foci of the ellipse are $F_{1}(-1,0)$, $F_{2}(1,0)$, and the length of the major axis is $10$. Then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Coordinate(F1) = (-1, 0);Coordinate(F2) = (1, 0);Focus(G) = {F1,F2};Length(MajorAxis(G)) = 10",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/25+y^2/24=1",
    "fact_spans": "[[[0, 2], [47, 49]], [[8, 21]], [[22, 35]], [[8, 21]], [[22, 35]], [[0, 35]], [[0, 45]]]",
    "query_spans": "[[[47, 54]]]",
    "process": "From the focus of the ellipse, we know that $ c = 1 $, and the major axis lies on the horizontal axis. The length of the major axis is 10, so $ 2a = 10 $, $ a = 5 $. Then the semi-minor axis is $ b = \\sqrt{a^{2} - c^{2}} = 2\\sqrt{6} $. Substituting $ a = 5 $, $ b = 2\\sqrt{6} $ into the ellipse equation $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $, the equation of the ellipse is $ \\frac{x^{2}}{25} + \\frac{y^{2}}{24} = 1 $."
  },
  {
    "text": "If the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{n}=1$ has an eccentricity of $2$, and one focus of the hyperbola coincides exactly with the focus of the parabola $y^{2}=8x$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/n + x^2/m = 1);m: Number;n: Number;Eccentricity(G) = 2;H: Parabola;Expression(H) = (y^2 = 8*x);OneOf(Focus(G)) = Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[1, 39], [49, 52], [79, 82]], [[1, 39]], [[4, 39]], [[4, 39]], [[1, 47]], [[60, 74]], [[60, 74]], [[49, 77]]]",
    "query_spans": "[[[79, 89]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{11}+\\frac{y^{2}}{7}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/11 + y^2/7 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(11)/11",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 44]]]",
    "process": "From the given conditions, since $a^{2}=11$, $b^{2}=7$, $\\therefore c^{2}=a^{2}-b^{2}=11-7=4$. Thus $a=\\sqrt{11}$, $c=2$, so the eccentricity of the ellipse is $e=\\frac{c}{a}=\\frac{2}{\\sqrt{11}}=\\frac{2\\sqrt{11}}{11}$."
  },
  {
    "text": "It is known that the foci of ellipse $C$ lie on the $x$-axis, and the eccentricity is $\\frac{1}{2}$. Then an equation of $C$ could be?",
    "fact_expressions": "C: Ellipse;PointOnCurve(Focus(C),xAxis) = True;Eccentricity(C) = 1/2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4+y^2/3=1",
    "fact_spans": "[[[2, 7], [37, 40]], [[2, 16]], [[2, 35]]]",
    "query_spans": "[[[37, 47]]]",
    "process": "Since the foci are on the x-axis, let the equation of the ellipse be \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1, a>b>0. Since the eccentricity is \\frac{1}{2}, we have \\frac{c}{a}=\\frac{1}{2}, so \\frac{c^{2}}{a^{2}}=\\frac{a^{2}-b^{2}}{a^{2}}=\\frac{1}{4}, then \\frac{b^{2}}{a^{2}}=\\frac{3}{4}."
  },
  {
    "text": "The point on the parabola $y^{2}=a x  (a>0)$ with abscissa $6$ is at a distance of $10$ from the focus. Then $a=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = a*x);a: Number;a>0;V: Point;PointOnCurve(V, G) = True;XCoordinate(V) = 6;Distance(V, Focus(G)) = 10",
    "query_expressions": "a",
    "answer_expressions": "16",
    "fact_spans": "[[[0, 21]], [[0, 21]], [[44, 47]], [[3, 21]], [[30, 31]], [[0, 31]], [[22, 31]], [[0, 42]]]",
    "query_spans": "[[[44, 49]]]",
    "process": "From the parabola $ y^{2} = ax $ ($ a > 0 $), its directrix equation is $ x = -\\frac{a}{4} $. Given that the distance from the point on the parabola with horizontal coordinate 6 to the focus is 10, according to the definition of a parabola, the distance from this point to the directrix is also 10, i.e., $ 6 + \\frac{a}{4} = 10 $. Solving gives $ a = 16 $. This problem mainly examines the definition and application of the standard equation of a parabola. The key to solving is using the definition of the parabola to convert the distance to the focus into the distance to the directrix and setting up the equation. It emphasizes transformation thinking, as well as reasoning and computational ability, and is a basic-level problem."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{m-1}+\\frac{y^{2}}{2-m}=1$ represents a hyperbola, then what is the range of values for $m$?",
    "fact_expressions": "G: Hyperbola;m:Number;Expression(G)=(x^2/(m-1)+y^2/(2-m)=1)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-oo,1)+(2+oo)",
    "fact_spans": "[[[44, 47]], [[49, 52]], [[1, 47]]]",
    "query_spans": "[[[49, 59]]]",
    "process": "Since $\\frac{x^{2}}{m-1}+\\frac{y^{2}}{2-m}=1$ represents a hyperbola, it follows that $(m-1)(2-m)<0$, solving which gives $m<1$ or $m>2$, i.e., the range of $m$ is $(-\\infty,1)\\cup(2,+\\infty)$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$, the focus is $F$, and $P$ is a point on $C$. If $|P F|=4$ and the distance from point $P$ to the $y$-axis is equal to $3$, then what are the coordinates of point $F$?",
    "fact_expressions": "C: Parabola;p: Number;P: Point;F: Point;p>0;Expression(C) = (y^2 = 2*p*x);Focus(C)=F;PointOnCurve(P, C);Abs(LineSegmentOf(P, F)) = 4;Distance(P, yAxis) = 3",
    "query_expressions": "Coordinate(F)",
    "answer_expressions": "(1,0)",
    "fact_spans": "[[[2, 28], [40, 43]], [[10, 28]], [[36, 39], [58, 62]], [[32, 35], [77, 81]], [[10, 28]], [[2, 28]], [[2, 35]], [[36, 46]], [[48, 57]], [[58, 75]]]",
    "query_spans": "[[[77, 86]]]",
    "process": "Solve using the parabola equation and definition. From the given condition, we have \\frac{p}{2}=|PF|-3=1, so p=2, therefore the coordinates of point F are (1,0)."
  },
  {
    "text": "Given that $F$ is the right focus of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{8}=1$, and a tangent line drawn at a point $M$ (where $M$ lies in the first quadrant) on the circle $x^{2}+y^{2}=8$ intersects the ellipse at points $P$ and $Q$, then the set of possible values for the perimeter of $\\Delta P F Q$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/8 = 1);F: Point;RightFocus(G) = F;H: Circle;Expression(H) = (x^2 + y^2 = 8);M: Point;Quadrant(M) = 1;Z: Line;TangentOnPoint(M, H) = Z;P: Point;Q: Point;Intersection(Z, G) = {P, Q}",
    "query_expressions": "Range(Perimeter(TriangleOf(P, F, Q)))",
    "answer_expressions": "{6}",
    "fact_spans": "[[[6, 43], [87, 89]], [[6, 43]], [[2, 5]], [[2, 47]], [[49, 65], [82, 83]], [[49, 65]], [[68, 71], [72, 75]], [[72, 80]], [], [[48, 86]], [[90, 93]], [[94, 97]], [[48, 99]]]",
    "query_spans": "[[[101, 125]]]",
    "process": "Let P(x_{1},y_{1}),Q(x_{2},y_{2}),|PF_{2}|=\\sqrt{(x_{1}-1)^{2}+y_{1}^{2}}=\\sqrt{(x_{1}-1)^{2}+8(1-\\frac{x_{1}^{2}}{9})}=\\sqrt{(\\frac{x_{1}}{3}-3)^{2}},0<x_{1}<3, so |PF_{2}|=3-\\frac{1}{3}x_{1}, similarly |QF_{2}|=3-\\frac{1}{3}x_{2}, M is the tangent point of circle x^{2}+y^{2}=8, connect OP,OM, |PM|=\\sqrt{|OP|^{2}-|OM|^{2}}=\\sqrt{x_{1}^{2}+y_{1}^{2}-8}=\\sqrt{x_{1}^{2}+8(1-\\frac{x_{1}^{2}}{9})-8}=\\frac{1}{3}x_{1}, so |PF_{2}|+|PM|=3-\\frac{1}{3}x_{1}+\\frac{1}{3}x_{1}=3,|QF_{2}|+|QM|=3, so the perimeter of quadrilateral PF_{2}QM is |PF_{2}|+|QF_{2}|+|PQ|=3+3=6."
  },
  {
    "text": "The semi-focal distance of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$ is $c$. If the abscissa of an intersection point of the line $y=2 x$ and the ellipse is exactly $c$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = 2*x);HalfFocalLength(G) = c;XCoordinate(OneOf(Intersection(H,G)))=c;c:Number",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)-1",
    "fact_spans": "[[[0, 54], [74, 76], [92, 94]], [[2, 54]], [[2, 54]], [[64, 73]], [[2, 54]], [[2, 54]], [[0, 54]], [[64, 73]], [[0, 62]], [[64, 90]], [[87, 90], [59, 62]]]",
    "query_spans": "[[[92, 100]]]",
    "process": ""
  },
  {
    "text": "Given line $l_{1}$: $3x - 4y - 9 = 0$ and line $l_{2}$: $y = \\frac{1}{4}$, the minimum value of the sum of distances from a moving point $P$ on the parabola $y = x^{2}$ to line $l_{1}$ and line $l_{2}$ is?",
    "fact_expressions": "l1: Line;Expression(l1) = (3*x - 4*y - 9 = 0);l2: Line;Expression(l2) = (y = 1/4);G: Parabola;Expression(G) = (y = x^2);P: Point;PointOnCurve(P,G) = True",
    "query_expressions": "Min(Distance(P,l1) + Distance(P,l2))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 26], [74, 83]], [[2, 26]], [[27, 53], [84, 93]], [[27, 53]], [[54, 66]], [[54, 66]], [[70, 73]], [[54, 73]]]",
    "query_spans": "[[[70, 104]]]",
    "process": ""
  },
  {
    "text": "Given the line $l$: $y = kx + 2$ passes through the upper vertex $B$ and the left focus $F$ of the ellipse $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$ $(a > b > 0)$, and the chord length intercepted by the circle $x^2 + y^2 = 4$ is $L$. If $L \\geq \\frac{4\\sqrt{5}}{5}$, then the range of the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "l: Line;k:Number;Expression(l) = (y = k*x + 2);G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;B: Point;UpperVertex(G) = B;F: Point;LeftFocus(G) = F;PointOnCurve(B, l);PointOnCurve(F, l);H: Circle;Expression(H) = (x^2 + y^2 = 4);L: Number;Length(InterceptChord(l, H)) = L;L >= 4*sqrt(5)/5;e: Number;Eccentricity(G) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "(0,2*sqrt(5)/5)",
    "fact_spans": "[[[2, 18]], [[9, 18]], [[2, 18]], [[19, 71], [146, 148]], [[19, 71]], [[21, 71]], [[21, 71]], [[21, 71]], [[21, 71]], [[75, 78]], [[19, 78]], [[82, 85]], [[19, 85]], [[19, 78]], [[19, 85]], [[88, 104]], [[88, 104]], [[110, 113]], [[2, 113]], [[115, 144]], [[151, 154]], [[146, 154]]]",
    "query_spans": "[[[151, 161]]]",
    "process": "According to the problem, we have $ b=2 $, $ kc=2 $. Using the chord length formula for the intersection of a line and a circle, and given $ L \\geqslant \\frac{4\\sqrt{5}}{5} $, we obtain $ d^{2} \\leqslant \\frac{16}{5} $. Then, using the point-to-line distance formula, we simplify to get $ k^{2} \\geqslant \\frac{1}{4} $. Combining this with the eccentricity formula, we find the range of the ellipse's eccentricity. From the problem, we know $ b=2 $, $ kc=2 $. Let $ d $ be the distance from the center of the circle to the line $ l $, then $ L = 2\\sqrt{4 - d^{2}} \\geqslant \\frac{4\\sqrt{5}}{5} $, solving gives $ d^{2} \\leqslant \\frac{16}{5} $. Also, since $ d = \\frac{2}{\\sqrt{1 + k^{2}}} $, it follows that $ \\frac{1}{1 + k^{2}} \\leqslant \\frac{4}{5} $, solving gives $ k^{2} \\geqslant \\frac{1}{4} $. Thus, $ e^{2} = \\frac{4}{a^{2}} = \\frac{c^{2}}{b^{2} + c^{2}} = \\frac{1}{1 + k^{2}} $, so $ 0 < e^{2} \\leqslant \\frac{4}{5} $, solving gives $ 0 < e \\leqslant \\frac{2\\sqrt{5}}{5} $."
  },
  {
    "text": "Given that the parabola $y^{2}=4x$ is intersected by the line $y=2x+b$ to form a chord of length $3\\sqrt{5}$, then $b=$?",
    "fact_expressions": "G: Parabola;H: Line;b: Number;Expression(G) = (y^2 = 4*x);Expression(H) = (y = b + 2*x);Length(InterceptChord(H,G))=3*sqrt(5)",
    "query_expressions": "b",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 15]], [[16, 27]], [[48, 51]], [[1, 15]], [[16, 27]], [[1, 46]]]",
    "query_spans": "[[[48, 53]]]",
    "process": ""
  },
  {
    "text": "$F$ is the focus of the parabola $y^{2}=2x$, and $A$, $B$ are two points on the parabola such that $|AF| + |BF| = 6$. Then, what is the distance from the midpoint of segment $AB$ to the $y$-axis?",
    "fact_expressions": "A: Point;B: Point;G: Parabola;F: Point;Expression(G) = (y^2 = 2*x);Focus(G) = F;PointOnCurve(A, G);PointOnCurve(B, G);Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F)) = 6",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A, B)), yAxis)",
    "answer_expressions": "5/2",
    "fact_spans": "[[[22, 25]], [[26, 29]], [[4, 18], [30, 33]], [[0, 3]], [[4, 18]], [[0, 21]], [[22, 37]], [[22, 37]], [[38, 53]]]",
    "query_spans": "[[[55, 74]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, a perpendicular is drawn from point $F_{2}$ to one asymptote of this hyperbola, with the foot of the perpendicular at $M$, and it satisfies $| \\overrightarrow{M F_{1}}|=3| \\overrightarrow{M F_{2}} |$. Then the equation of the asymptotes of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;M: Point;F1: Point;F2: Point;l: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, l);IsPerpendicular(l, OneOf(Asymptote(G)));FootPoint(l, OneOf(Asymptote(G))) = M;Abs(VectorOf(M, F1)) = 3*Abs(VectorOf(M, F2))",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*(sqrt(2)/2)*x",
    "fact_spans": "[[[20, 77], [95, 98], [179, 182]], [[23, 77]], [[23, 77]], [[110, 113]], [[2, 9]], [[12, 19], [85, 93]], [], [[23, 77]], [[23, 77]], [[20, 77]], [[2, 83]], [[2, 83]], [[84, 106]], [[84, 106]], [[84, 113]], [[117, 176]]]",
    "query_spans": "[[[179, 190]]]",
    "process": ""
  },
  {
    "text": "Given that the moving circle $C$ is internally tangent to both the circle $(x+1)^{2}+y^{2}=1$ and the circle $(x-1)^{2}+y^{2}=25$, find the equation of the trajectory of the center of the moving circle $C$.",
    "fact_expressions": "C: Circle;H:Circle;G:Circle;C1:Point;Expression(H)=((x+1)^2+y^2=1);Expression(G)=((x-1)^2+y^2=25);IsIntersect(C,H);IsTangent(C,G);Center(C)=C1",
    "query_expressions": "LocusEquation(C1)",
    "answer_expressions": "x^2/4 + y^2/3 = 1",
    "fact_spans": "[[[4, 7]], [[8, 28]], [[29, 50]], [[59, 62]], [[8, 28]], [[29, 50]], [[4, 53]], [[4, 53]], [[55, 62]]]",
    "query_spans": "[[[59, 69]]]",
    "process": "Let the radius of the moving circle be $ r $, let the center of the circle $ (x+1)^{2} + y^{2} = 1 $ be $ A(-1,0) $, and its radius be 1. Let the center of the circle $ (x-1)^{2} + y^{2} = 25 $ be $ B(1,0) $, and its radius be 5, with $ |AB| = 2 $. Since the moving circle $ C $ is internally tangent to both circles $ (x+1)^{2} + y^{2} = 1 $ and $ (x-1)^{2} + y^{2} = 25 $, we have $ |CB| = 5 - r $, $ |CA| = r - 1 $. Adding these two equations gives: $ |CB| + |CA| = 4 > |AB| $. Therefore, the locus of the center $ C $ of the moving circle is an ellipse with foci at points $ A $ and $ B $. Thus, we have $ 2a = 4 $, $ 2c = 2 \\Rightarrow a = 2 $, $ c = 1 \\Rightarrow b^{2} = a^{2} - c^{2} = 4 - 1 = 3 $. Therefore, the standard equation of this ellipse is: $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $."
  },
  {
    "text": "It is known that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $y=\\sqrt{3} x$, and one of its foci lies on the directrix of the parabola $y^{2}=8 x$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 8*x);Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x);PointOnCurve(OneOf(Focus(G)), Directrix(H))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[2, 58], [82, 83], [109, 112]], [[5, 58]], [[5, 58]], [[89, 103]], [[5, 58]], [[5, 58]], [[2, 58]], [[89, 103]], [[2, 81]], [[82, 107]]]",
    "query_spans": "[[[109, 117]]]",
    "process": "The directrix of the parabola \\( y^{2} = 8x \\) is \\( x = -2 \\), so for the hyperbola, \\( a^{2} + b^{2} = 4 \\). Since the asymptotes of the hyperbola are \\( y = \\sqrt{3}x \\), it follows that \\( \\frac{b}{a} = \\sqrt{3} \\). Solving these two equations simultaneously gives \\( a = 1 \\), \\( b = \\sqrt{3} \\). Therefore, the equation of the hyperbola is \\( x^{2} - \\frac{y^{2}}{3} = 1 \\)."
  },
  {
    "text": "Given that $F_{1}$, $F_{2}$ are the common foci of the ellipse $C_{1}$: $\\frac{x^{2}}{3}+y^{2}=1$ and the hyperbola $C_{2}$, and $P$ is a common point of $C_{1}$ and $C_{2}$, if $O P=O F_{1}$, then the asymptotes of $C_{2}$ are?",
    "fact_expressions": "C1: Ellipse;Expression(C1) = (x^2/3 + y^2 = 1);C2: Hyperbola;F1: Point;F2: Point;Focus(C1) = {F1, F2};Focus(C2) = {F1, F2};P: Point;Intersection(C1, C2) = P;O: Origin;LineSegmentOf(O, P) = LineSegmentOf(O, F1)",
    "query_expressions": "Expression(Asymptote(C2))",
    "answer_expressions": "y=pm*x",
    "fact_spans": "[[[18, 54], [75, 82]], [[18, 54]], [[55, 65], [83, 90], [111, 118]], [[2, 9]], [[10, 17]], [[2, 70]], [[2, 70]], [[71, 74]], [[71, 94]], [[96, 109]], [[96, 109]]]",
    "query_spans": "[[[111, 126]]]",
    "process": "Since $F_{1}, F_{2}$ are the common foci of the ellipse $C_{1}: \\frac{x^{2}}{3} + y^{2} = 1$ and the hyperbola $C_{2}$, it follows that $F_{1}(-2,0)$. Let point $P(\\sqrt{3}\\cos\\theta, \\sin\\theta)$ satisfy $|OP| = |OF_{1}| \\rightarrow 3\\cos^{2}\\theta + \\sin^{2}\\theta = c = 2 \\Rightarrow \\cos\\theta = \\pm\\frac{\\sqrt{2}}{2}$. Without loss of generality, take the positive value, so $P\\left(\\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{6}}{2}\\right)$. Substituting into the hyperbola equation gives: $\\frac{6}{4a^{2}} - \\frac{2}{4b^{2}} = 1$, and since $a^{2} + b^{2} = 4$, we get $a = b = 1$; thus, the asymptotes of $C_{2}$ are $y = \\pm x$."
  },
  {
    "text": "Given that a line passing through the right focus $F_{2}$ of the ellipse $4 x^{2}+8 y^{2}=1$ intersects the ellipse at two points $A$, $B$, and $F_{1}$ is the left focus of the ellipse, then the perimeter of $\\triangle F_{1} AB$ is?",
    "fact_expressions": "G: Ellipse;H: Line;F1:Point;F2:Point;A: Point;B: Point;Expression(G) = (4*x^2 + 8*y^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(F2,H);Intersection(H,G)={A,B}",
    "query_expressions": "Perimeter(TriangleOf(F1, A, B))",
    "answer_expressions": "2",
    "fact_spans": "[[[4, 25], [39, 41], [62, 64]], [[36, 38]], [[54, 61]], [[28, 35]], [[46, 49]], [[50, 53]], [[4, 25]], [[54, 68]], [[4, 35]], [[2, 38]], [[36, 53]]]",
    "query_spans": "[[[70, 95]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of a hyperbola whose real axis length is equal to its imaginary axis length is?",
    "fact_expressions": "G: Hyperbola;Length(RealAxis(G)) = Length(ImageinaryAxis(G))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[10, 13]], [[0, 13]]]",
    "query_spans": "[[[10, 19]]]",
    "process": "From the given condition, we know that a = b. Squaring both sides simultaneously, we get a^{2} = b^{2}, that is, a^{2} = c^{2} - a^{2}, so 2a^{2} = c^{2}. Therefore, the eccentricity e = \\frac{c}{a} = \\sqrt{2}. Final answer is."
  },
  {
    "text": "Given the parabola $y^{2}=4 x$ and the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ intersect at a point $M$, $F$ is the focus of the parabola. If $M F=3$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;M: Point;F: Point;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 4*x);OneOf(Intersection(H, G)) = M;Focus(H) = F;LineSegmentOf(M, F) = 3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(13)/2",
    "fact_spans": "[[[17, 49], [81, 84]], [[20, 49]], [[2, 16], [63, 66]], [[55, 58]], [[59, 62]], [[17, 49]], [[2, 16]], [[2, 58]], [[59, 69]], [[71, 78]]]",
    "query_spans": "[[[81, 90]]]",
    "process": "From the definition of the parabola: MF=3\\Rightarrow x_{M}+1=3\\Rightarrow x_{M}=2\\Rightarrow y_{M}^{2}=8, therefore \\frac{4}{a^{2}}-8=1\\Rightarrow a^{2}=\\frac{4}{9}\\Rightarrow c^{2}=\\frac{4}{9}+1=\\frac{13}{9}\\Rightarrow e=\\frac{c}{a}=\\frac{\\sqrt{13}}{2}"
  },
  {
    "text": "What are the equations of the two asymptotes of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1);L1:Line;L2:Line;Asymptote(G)={L1,L2}",
    "query_expressions": "Expression(L1);Expression(L2)",
    "answer_expressions": "y=(-1/2)*x\ny=(1/2)*x",
    "fact_spans": "[[[0, 28]], [[0, 28]], [], [], [[0, 34]]]",
    "query_spans": "[[[0, 40]], [[0, 40]]]",
    "process": "Analysis: Using the formula for the asymptotes of a hyperbola, the asymptotes of this hyperbola can be obtained. Specifically, from the given hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, we have $a=2$, $b=1$, so the asymptotes of the hyperbola are $y=-\\frac{1}{2}x$, $y=\\frac{1}{2}x$. Thus, the answer is $y=-\\frac{1}{2}x$, $y=\\frac{1}{2}x\\cdots$"
  },
  {
    "text": "Given that point $P$ is a moving point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$, and $EF$ is any diameter of the circle $N$: $x^{2}+(y-1)^{2}=1$, then the maximum value of $\\overrightarrow{P E} \\cdot \\overrightarrow{P F}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/12 = 1);P: Point;PointOnCurve(P, G);N: Circle;Expression(N) = (x^2 + (y - 1)^2 = 1);E: Point;F: Point;IsDiameter(LineSegmentOf(E, F), N) = True",
    "query_expressions": "Max(DotProduct(VectorOf(P, E), VectorOf(P, F)))",
    "answer_expressions": "19",
    "fact_spans": "[[[7, 46]], [[7, 46]], [[2, 6]], [[2, 50]], [[57, 82]], [[57, 82]], [[51, 56]], [[51, 56]], [[51, 89]]]",
    "query_spans": "[[[91, 146]]]",
    "process": "The circle $ N: x^{2} + (y - 1)^{2} = 1 $ has center $ N(0,1) $ and radius length 1. Let point $ P(x,y) $. Since point $ P $ lies on the ellipse $ \\frac{x^{2}}{16} + \\frac{y^{2}}{12} = 1 $, we have: $ x^{2} = 16 - \\frac{4}{3}y^{2} $ and $ -2\\sqrt{3} \\leqslant y \\leqslant 2\\sqrt{3} $. Since $ EF $ is any diameter of the circle $ N: x^{2} + (y - 1)^{2} = 1 $, we obtain: $ \\overrightarrow{PE} = \\overrightarrow{PN} + \\overrightarrow{NE} $, $ \\overrightarrow{PF} = \\overrightarrow{PN} + \\overrightarrow{NF} = \\overrightarrow{PN} - \\overrightarrow{NE} $, $ \\therefore \\overrightarrow{PE} \\cdot \\overrightarrow{PF} = (\\overrightarrow{PN} + \\overrightarrow{NE}) \\cdot (\\overrightarrow{PN} - \\overrightarrow{NE}) = \\overrightarrow{PN}^{2} - \\overrightarrow{NE}^{2} = x^{2} + (y - 1)^{2} - 1 = 16 - \\frac{4}{3}y^{2} + y^{2} - 2y + 1 - 1 = -\\frac{1}{3}y^{2} - 2y + 16 = -\\frac{1}{3}(y + 3)^{2} + 19 $, $ (-2\\sqrt{3} \\leqslant y \\leqslant 2\\sqrt{3}) $. $ \\therefore $ When $ y = -3 $, $ \\overrightarrow{PE} \\cdot \\overrightarrow{PF} $ attains its maximum value, i.e., $ (\\overrightarrow{PE} \\cdot \\overrightarrow{PF})_{\\text{max}} = 19 $."
  },
  {
    "text": "Given that point $B(8,8)$ lies on the parabola $C$: $y^{2}=2 p x$, the tangent to $C$ at point $B$ intersects the directrix of $C$ at point $A$, and $F$ is the focus of $C$, then the slope of line $A F$ is?",
    "fact_expressions": "B: Point;Coordinate(B) = (8, 8);C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;PointOnCurve(B, C) = True;Intersection(TangentOnPoint(B,C),Directrix(C))=A;A: Point;F: Point;Focus(C) = F",
    "query_expressions": "Slope(LineOf(A, F))",
    "answer_expressions": "-3/4",
    "fact_spans": "[[[2, 11], [39, 43]], [[2, 11]], [[12, 33], [35, 38], [48, 51], [65, 68]], [[12, 33]], [[20, 33]], [[2, 34]], [[35, 60]], [[56, 60]], [[61, 64]], [[61, 71]]]",
    "query_spans": "[[[73, 85]]]",
    "process": "By the given condition, $8^{2}=2p\\times8$, $p=4$, the equation of the parabola is $y^{2}=8x$. The equation of the parabola at point $B$ can be expressed as $y=\\sqrt{8x}=2\\sqrt{2}x^{1}$, so $y=2\\sqrt{2}\\times\\frac{1}{2}x^{-1}=\\frac{\\sqrt{2}}{\\sqrt{x}}$. When $x=8$, $y=\\frac{\\sqrt{2}}{\\sqrt{8}}=\\frac{1}{2}$. The tangent line equation is $y-8=\\frac{1}{2}(x-8)$, i.e., $y=\\frac{1}{2}x+4$. The directrix equation is $x=-2$. From $\\begin{cases}y=\\frac{1}{2}x+4\\\\x=-2\\end{cases}$, we get $\\begin{cases}x=-2\\\\y=3\\end{cases}$, so $A(-2,3)$. Also, $F(2,0)$, thus $k_{AF}=\\frac{3-0}{-2-2}=-\\frac{3}{4}$."
  },
  {
    "text": "It is known that the focus of the parabola $y^{2}=-8 x$ coincides with the left focus of the hyperbola $\\frac{x^{2}}{m}-y^{2}=1$. Find the value of the real number $m$.",
    "fact_expressions": "G: Hyperbola;m: Real;H: Parabola;Expression(G) = (-y^2 + x^2/m = 1);Expression(H) = (y^2 = -8*x);Focus(H) = LeftFocus(G)",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[21, 49]], [[57, 62]], [[2, 17]], [[21, 49]], [[2, 17]], [[2, 55]]]",
    "query_spans": "[[[57, 66]]]",
    "process": "Since the focus of the parabola \\( y^{2} = -8x \\) is \\( (-2, 0) \\), and the focus of the parabola \\( y^{2} = -8x \\) coincides with the left focus of the hyperbola \\( \\frac{x^{2}}{m} - y^{2} = 1 \\), it follows that \\( m + 1 = (-2)^{2} \\), so \\( m = 3 \\);"
  },
  {
    "text": "Given that the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) has eccentricity $\\sqrt{3}$, and the directrix of the parabola $y^{2}=8x$ is the left directrix of the hyperbola, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 8*x);Eccentricity(G) = sqrt(3);LeftDirectrix(G)=Directrix(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/12 - y^2/24 = 1",
    "fact_spans": "[[[2, 59], [94, 97], [103, 106]], [[5, 59]], [[5, 59]], [[76, 90]], [[5, 59]], [[5, 59]], [[2, 59]], [[76, 90]], [[2, 75]], [[76, 101]]]",
    "query_spans": "[[[103, 111]]]",
    "process": ""
  },
  {
    "text": "The line passing through the point $(2, 0)$ with an inclination angle of $60^{\\circ}$ intersects the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{3}=1$ at points $A$ and $B$. Then the coordinates of the midpoint of $AB$ are?",
    "fact_expressions": "P: Point;Coordinate(P) = (2, 0);H: Line;Inclination(H) = ApplyUnit(60, degree);PointOnCurve(P, H);G: Ellipse;Expression(G) = (x^2/5 + y^2/3 = 1);A: Point;B: Point;Intersection(H, G) = {A, B}",
    "query_expressions": "Coordinate(MidPoint(LineSegmentOf(A, B)))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 11]], [[1, 11]], [[29, 31]], [[12, 31]], [[0, 31]], [[32, 69]], [[32, 69]], [[72, 75]], [[76, 79]], [[29, 81]]]",
    "query_spans": "[[[83, 95]]]",
    "process": ""
  },
  {
    "text": "If a point $M(x, y)$ moves in such a way that it always satisfies the equation $\\sqrt{(x-1)^{2}+y^{2}}+\\sqrt{(x+1)^{2}+y^{2}}=4$, denote the locus of all such points $M$ as $C$. The line $x=m$ intersects $C$ at points $D$ and $E$, and given $A(-1,0)$, then the maximum perimeter of $\\triangle A D E$ is?",
    "fact_expressions": "G: Line;m: Number;M: Point;A: Point;D: Point;E: Point;C:Curve;x1:Number;y1:Number;Expression(G) = (x = m);Coordinate(M) = (x1, y1);Coordinate(A) = (-1, 0);Locus(M)=C;Intersection(G, C) = {D, E};sqrt((x1-1)^2+y1^2)+sqrt((x1+1)^2+y1^2)=4",
    "query_expressions": "Max(Perimeter(TriangleOf(A, D, E)))",
    "answer_expressions": "8",
    "fact_spans": "[[[94, 101]], [[96, 101]], [[2, 12], [82, 86]], [[117, 126]], [[107, 110]], [[111, 114]], [[90, 93], [102, 105]], [[3, 12]], [[3, 12]], [[94, 101]], [[2, 12]], [[117, 126]], [[82, 93]], [[94, 114]], [[25, 74]]]",
    "query_spans": "[[[128, 153]]]",
    "process": "\\because\\sqrt{(x-1)^{2}+y^{2}}+\\sqrt{(x+1)^{2}+y^{2}}=4.\\thereforeM(x,y) to the fixed points (1,0), (-1,0) has distance sum equal to constant 4>2\\therefore the locus C of point M is an ellipse, with a=2,c=1, hence its equation is \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1. Then A(-1,0) is the left focus. Since the line x=m intersects C at D,E, then -2<m<2, assume without loss of generality that D is above the x-axis and E is below the x-axis. Let the right focus of the ellipse be A', connect DA',EA'. Since DA'+EA'\\geqslant DE, it follows that DA+EA+DA'+EA'\\geqslant DA+EA+DE, i.e., 4a\\geqslant DA+EA+DE. Therefore, the perimeter of \\triangle ADE \\leqslant 4a=8, which achieves maximum value 8 when m=1."
  },
  {
    "text": "Given the parabola $y^{2}=4x$ with focus $F$, let $P$ be a moving point on the parabola and $A(1,1)$ a fixed point. When the perimeter of $\\Delta PAF$ is minimized, what is the slope of the line containing $PF$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;P: Point;PointOnCurve(P, G) = True;A: Point;Coordinate(A) = (1, 1);WhenMin(Perimeter(TriangleOf(P,A,F))) = True",
    "query_expressions": "Slope(OverlappingLine(LineSegmentOf(P, F)))",
    "answer_expressions": "-4/3",
    "fact_spans": "[[[2, 16], [28, 31]], [[2, 16]], [[20, 23]], [[2, 23]], [[24, 27]], [[24, 35]], [[36, 46]], [[36, 46]], [[47, 67]]]",
    "query_spans": "[[[68, 82]]]",
    "process": "The focus of the parabola $ y^{2}=4x $ is $ F(1,0) $, the directrix is $ x=-1 $. Draw a perpendicular from point $ P $ to the directrix, with foot $ B $, and draw a perpendicular from point $ A $ to the directrix, with foot $ A_{1} $. Then the perimeter of $ \\triangle PAF $ is $ |PF|+|PA|+|AF|=|PB|+|PA|+1\\geqslant|AA_{1}|+1=3 $, with equality if and only if points $ A $, $ P $, $ A_{1} $ are collinear. At this time, $ P\\left(\\frac{1}{4},1\\right) $, $ k_{PF}=-\\frac{4}{3} $."
  },
  {
    "text": "$P$ is any point on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, excluding the vertex, $A_{1}$ and $A_{2}$ are the left and right vertices of the hyperbola, the slopes of lines $A_{1} P$ and $A_{2} P$ are $k_{1}$ and $k_{2}$ respectively, then $k_{1} \\cdot k_{2}$=?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;A1: Point;P: Point;A2: Point;a>0;b>0;k1:Number;k2:Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, RightPart(G));Negation(P=Vertex(G));LeftVertex(G)=A1;RightVertex(G)=A2;Slope(LineOf(A1,P))=k1;Slope(LineOf(A2,P))=k2",
    "query_expressions": "k1*k2",
    "answer_expressions": "b^2/a^2",
    "fact_spans": "[[[4, 60], [87, 90]], [[7, 60]], [[7, 60]], [[71, 78]], [[0, 3]], [[79, 86]], [[7, 60]], [[7, 60]], [[125, 132]], [[133, 140]], [[4, 60]], [[0, 70]], [[0, 70]], [[71, 95]], [[71, 95]], [[96, 140]], [[96, 140]]]",
    "query_spans": "[[[142, 163]]]",
    "process": "From the given conditions, we have: $A_{1}(-a,0)$, $A_{2}(a,0)$, let $P(x_{0},y_{0})$, then $\\frac{x_{0}^{2}}{a^{2}}-\\frac{y_{0}^{2}}{b^{2}}=1$, so $y_{0}^{2}=b^{2}\\left(\\frac{x_{0}^{2}}{a^{2}}-1\\right)=b^{2}\\left(\\frac{x_{0}^{2}-a^{2}}{a^{2}}\\right)$. Since $k_{1}=\\frac{y_{0}}{x_{0}+a}$, $k_{2}=\\frac{y_{0}}{x_{0}-a}$, therefore $k_{1}k_{2}=\\frac{y_{0}}{x_{0}+a}\\cdot\\frac{y_{0}}{x_{0}-a}=\\frac{y_{0}^{2}}{x_{0}^{2}-a^{2}}=\\frac{b^{2}\\left(\\frac{x_{0}^{2}-a^{2}}{a^{2}}\\right)}{x_{0}^{2}-a^{2}}=\\frac{b^{2}}{a^{2}}$"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. If there exists a point $P$ on the hyperbola such that $\\angle P F_{2} F_{1}=120^{\\circ}$, then the range of the eccentricity is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F2: Point;F1: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G) = True;AngleOf(P, F2, F1) = ApplyUnit(120, degree)",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1,2)",
    "fact_spans": "[[[2, 58], [84, 87]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[75, 82]], [[67, 74]], [[2, 82]], [[2, 82]], [[90, 94]], [[84, 94]], [[95, 129]]]",
    "query_spans": "[[[84, 141]]]",
    "process": "According to the properties of hyperbolas, if there exists a point $P$ on the hyperbola such that $\\angle PF_{2}F_{1}=120^{\\circ}$, then the slope of the asymptote $-\\frac{b}{a}>\\tan120^{\\circ}=-\\sqrt{3}$, that is, $\\frac{b}{a}<\\sqrt{3}$. Since the eccentricity $e=\\frac{c}{a}$ and $c^{2}=a^{2}+b^{2}$, it follows that $e^{2}=1+\\frac{b^{2}}{a^{2}}<4$. Because $e>1$, the range of eccentricity $e$ is $(1,2)$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, one of its asymptotes has the equation $y=x$, and the point $P(\\sqrt{3}, y_{0})$ lies on the hyperbola. Then $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/2 - y^2/b^2 = 1);b: Number;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;Expression(OneOf(Asymptote(G))) = (y = x);P: Point;Coordinate(P) = (sqrt(3), y0);y0: Number;PointOnCurve(P, G)",
    "query_expressions": "DotProduct(VectorOf(P, F1), VectorOf(P, F2))",
    "answer_expressions": "0",
    "fact_spans": "[[[2, 49], [75, 76], [113, 116]], [[2, 49]], [[5, 49]], [[5, 49]], [[58, 65]], [[67, 74]], [[2, 74]], [[2, 74]], [[75, 89]], [[90, 111]], [[90, 111]], [[91, 111]], [[90, 117]]]",
    "query_spans": "[[[119, 178]]]",
    "process": ""
  },
  {
    "text": "Given real numbers $x$, $y$ satisfying: $x|x| + y|y| = 1$, then the range of $|x + y + \\sqrt{2}|$ is?",
    "fact_expressions": "x_:Real;y_:Real;(x_*Abs(x_))+(y_*Abs(y_)) = 1",
    "query_expressions": "Range(Abs(x_ + y_ + sqrt(2)))",
    "answer_expressions": "(\\sqrt{2}, 2\\sqrt{2}]",
    "fact_spans": "[[[2, 7]], [[9, 12]], [[15, 28]]]",
    "query_spans": "[[[30, 53]]]",
    "process": "When $ x \\geqslant 0, y \\geqslant 0 $, $ x^{2} + y^{2} = 1 $, its graph is the part of the unit circle in the first quadrant. When $ x < 0, y \\geqslant 0 $, $ -x^{2} + y^{2} = 1 $, its graph is the part of the hyperbola $ -x^{2} + y^{2} = 1 $ in the second quadrant. When $ x \\geqslant 0, y < 0 $, $ x^{2} - y^{2} = 1 $, its graph is the part of the hyperbola $ x^{2} - y^{2} = 1 $ in the fourth quadrant. When $ x < 0, y < 0 $, $ -x^{2} - y^{2} = 1 $, the graph does not exist. Therefore, the graph corresponding to the equation $ x|x| + y|y| = 1 $ is as follows: $ |x + y + \\sqrt{2}| = \\frac{|x + y + \\sqrt{2}|}{\\sqrt{2}} \\cdot \\sqrt{2} $, which represents $ \\sqrt{2} $ times the distance from the point $ (x, y) $ to the line $ x + y + \\sqrt{2} = 0 $. Thus, the maximum value of $ |x + y + \\sqrt{2}| $ is $ \\sqrt{2} $ times the maximum distance from a point on the unit circle to the line $ x + y + \\sqrt{2} = 0 $, which is $ \\left( \\frac{\\sqrt{2}}{\\sqrt{2}} + 1 \\right) \\sqrt{2} = 2\\sqrt{2} $. One asymptote of the hyperbola has the equation $ y = -x $. The distance between the line $ y = -x $ and the line $ x + y + \\sqrt{2} = 0 $ is 1. Hence, $ |x + y + \\sqrt{2}| > \\sqrt{2} $. In summary, the range of $ |x + y + \\sqrt{2}| $ is $ (\\sqrt{2}, 2\\sqrt{2}] $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$, with right focus $F$. A perpendicular is drawn from $F$ to an asymptote of the hyperbola $C$, intersecting the hyperbola at point $M$, and the foot of the perpendicular is $N$. If $M$ is the midpoint of segment $F N$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(C) = F;Z: Line;PointOnCurve(F, Z);IsPerpendicular(Z, OneOf(Asymptote(C)));M: Point;Intersection(Z, C) = M;N: Point;FootPoint(Z, OneOf(Asymptote(C))) = N;MidPoint(LineSegmentOf(F, N)) = M",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 65], [79, 85], [96, 99], [129, 135]], [[2, 65]], [[9, 65]], [[9, 65]], [[9, 65]], [[9, 65]], [[70, 73], [75, 78]], [[2, 73]], [], [[74, 94]], [[74, 94]], [[101, 104], [113, 116]], [[74, 104]], [[108, 111]], [[74, 111]], [[113, 127]]]",
    "query_spans": "[[[129, 141]]]",
    "process": ""
  },
  {
    "text": "A point $M$ on the parabola $y^{2}=4x$ is at a distance of $3$ from the focus. What is the horizontal coordinate of point $M$?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(M, G);Distance(M, Focus(G)) = 3",
    "query_expressions": "XCoordinate(M)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 14]], [[17, 20], [32, 36]], [[0, 14]], [[0, 20]], [[0, 30]]]",
    "query_spans": "[[[32, 42]]]",
    "process": ""
  },
  {
    "text": "Given that the x-coordinate of a point $M$ on the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ is $5$, what is the distance from point $M$ to the left focus?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);M:Point;PointOnCurve(M,G);XCoordinate(M)=5",
    "query_expressions": "Distance(M, LeftFocus(G))",
    "answer_expressions": "34/3",
    "fact_spans": "[[[2, 41]], [[2, 41]], [[44, 47], [57, 61]], [[2, 47]], [[44, 55]]]",
    "query_spans": "[[[2, 70]]]",
    "process": "Analysis: Since the right focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ is $F(5,0)$, substituting $x_{M}=5$ into the hyperbola equation yields $|y_{M}|=\\frac{16}{3}$, which is the distance from point $M$ to the right focus. By the definition of the hyperbola, the distance from $M$ to the left focus is $\\frac{16}{3}+2\\times3=\\frac{34}{3}$."
  },
  {
    "text": "Given that the distance from a point $M$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ to one of its foci $F$ is $2$, $N$ is the midpoint of $MF$, and $O$ is the coordinate origin, what is the length of the segment $ON$?",
    "fact_expressions": "G: Ellipse;O: Origin;N: Point;M: Point;F: Point;Expression(G) = (x^2/25 + y^2/9 = 1);PointOnCurve(M,G);OneOf(Focus(G))=F;Distance(M,F)=2;MidPoint(LineSegmentOf(M,F))=N",
    "query_expressions": "Length(LineSegmentOf(O,N))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 40], [48, 50]], [[78, 81]], [[65, 68]], [[42, 46]], [[54, 57]], [[2, 40]], [[2, 46]], [[48, 57]], [[42, 64]], [[65, 77]]]",
    "query_spans": "[[[89, 100]]]",
    "process": "Let the other focus of the ellipse be E, then |MF| + |ME| = 10. Since |MF| = 2, ∴ |ME| = 8. Also, ON is the midline of triangle MEF, ∴ ON = \\frac{1}{2}|ME| = 4. b female safety as."
  },
  {
    "text": "The standard equation of a parabola with vertex at the origin, symmetric about the $x$-axis, and passing through the point $M(-2,3)$ is?",
    "fact_expressions": "G: Parabola;M: Point;O: Origin;Coordinate(M) = (-2, 3);Vertex(G) = O;SymmetryAxis(G) = xAxis;PointOnCurve(M, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = (-9/2)*x",
    "fact_spans": "[[[29, 32]], [[18, 28]], [[3, 5]], [[18, 28]], [[0, 32]], [[6, 32]], [[16, 32]]]",
    "query_spans": "[[[29, 39]]]",
    "process": "According to the problem, we can assume the standard equation of the parabola is $ y^{2} = -2px $. Therefore, $ 3^{2} = -2p \\cdot (-2) $, solving gives $ p = \\frac{9}{4} $. Thus, the standard equation of the parabola is $ y^{2} = -\\frac{9}{2}x $. This problem mainly examines finding the standard equation of a parabola and is a basic question."
  },
  {
    "text": "Given that the center of the hyperbola is at the origin, one focus is at $F(10,0)$, and the equations of the two asymptotes are $y = \\pm \\frac{4}{3}x$, then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Center(G) = O;O: Origin;F: Point;OneOf(Focus(G)) = F;Coordinate(F) = (10, 0);Expression(Asymptote(G)) = (y = pm*(4/3)*x)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/36 - y^2/64 = 1",
    "fact_spans": "[[[2, 5], [62, 65]], [[2, 13]], [[9, 13]], [[19, 28]], [[2, 28]], [[19, 28]], [[2, 59]]]",
    "query_spans": "[[[62, 72]]]",
    "process": "From the given conditions, $ c=10 $, $ \\frac{b}{a}=\\frac{4}{3} $, $ 100=a^{2}+b^{2} $, $ \\therefore a=6 $, $ b=8 $. Hence, the standard equation of the hyperbola is $ \\frac{x^{2}}{36}-\\frac{y^{2}}{64}=1 $."
  },
  {
    "text": "Given the line $y = kx + m$ ($k > 0$) intersects the parabola $C$: $y^2 = 4x$ and its directrix at points $M$ and $N$ respectively, $F$ is the focus of the parabola. If $2 \\overrightarrow{FM} = \\overrightarrow{MN}$, then what is the value of $k$?",
    "fact_expressions": "C: Parabola;G: Line;m: Number;k: Number;F: Point;M: Point;N: Point;Expression(C) = (y^2 = 4*x);k>0;Expression(G) = (y = k*x + m);Intersection(G,C)=M;Intersection(G, Directrix(C))=N;Focus(C) = F;2*VectorOf(F, M) = VectorOf(M, N)",
    "query_expressions": "k",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[19, 38], [39, 40], [60, 63]], [[2, 18]], [[4, 18]], [[115, 118]], [[56, 59]], [[46, 49]], [[50, 53]], [[19, 38]], [[4, 18]], [[2, 18]], [[2, 55]], [[2, 55]], [[56, 66]], [[68, 113]]]",
    "query_spans": "[[[115, 121]]]",
    "process": "As shown in the figure, since $k>0$, let the inclination angle of line $l$ be $\\alpha$. By the definition of the parabola, the distance from point $M$ to the directrix $|MQ|=|MF|$, hence $\\sin(\\frac{\\pi}{2}-\\alpha)=\\frac{|MQ|}{|MN|}=\\frac{|MF|}{|MN|}=\\frac{1}{2}$, therefore $\\cos\\alpha=\\frac{1}{2}$, then $\\alpha=\\frac{\\pi}{3}$, thus $k=\\tan\\frac{\\pi}{3}=\\sqrt{3}$."
  },
  {
    "text": "The standard equation of the hyperbola that shares the same two foci as the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ and passes through the point $(\\sqrt{2}, \\sqrt{3})$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;I: Point;Expression(H) = (x^2/9 + y^2/5 = 1);Focus(G) = Focus(H);Coordinate(I) = (sqrt(2), sqrt(3));PointOnCurve(I, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/3 = 1",
    "fact_spans": "[[[74, 77]], [[1, 38]], [[50, 73]], [[1, 38]], [[0, 77]], [[50, 73]], [[48, 77]]]",
    "query_spans": "[[[74, 84]]]",
    "process": "The foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ are $(2,0)$, $(-2,0)$. Let the standard equation of the hyperbola be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$. According to the problem, the foci of the hyperbola are $(2,0)$, $(-2,0)$. By the definition of a hyperbola: $2a=|\\sqrt{(\\sqrt{2}+2)^{2}+(\\sqrt{3}-0)^{2}}-\\sqrt{(\\sqrt{2}-2)^{2}+(\\sqrt{3}-0)^{2}}|=\\sqrt{8}+1-\\sqrt{8}+1 \\Rightarrow a=1$, and $c=2$, $\\therefore b=\\sqrt{c^{2}-a^{2}}=\\sqrt{3}$, so the equation of the hyperbola is: $x^{2}-\\frac{y^{2}}{3}=1$"
  },
  {
    "text": "If the distance from the coordinate origin to the directrix of the parabola $y=m x^{2}$ is $2$, then $m$=?",
    "fact_expressions": "G: Parabola;m: Number;Expression(G) = (y = m*x^2);O:Origin;Distance(O,Directrix(G))=2",
    "query_expressions": "m",
    "answer_expressions": "pm*(1/8)",
    "fact_spans": "[[[6, 20]], [[31, 34]], [[6, 20]], [[1, 5]], [[1, 29]]]",
    "query_spans": "[[[31, 36]]]",
    "process": "According to the properties of a parabola, transforming $ y = mx^2 $ into the standard form gives $ x^2 = \\frac{1}{m}y $, and the equation of the directrix is $ y = -\\frac{1}{4m} $. Thus, from the given condition $ \\left| -\\frac{1}{4m} \\right| = 2 $, we obtain $ m = \\pm \\frac{1}{8} $."
  },
  {
    "text": "Draw a line $l$ through the focus $F$ of the parabola $C$: $y^{2}=4x$, intersecting the parabola $C$ at points $A$ and $B$. If $|AF|=3|BF|$, then what is the slope of line $l$?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;F: Point;B: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, F)) = 3*Abs(LineSegmentOf(B, F))",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*sqrt(3)",
    "fact_spans": "[[[27, 32], [65, 70]], [[1, 20], [33, 39]], [[40, 43]], [[23, 26]], [[44, 47]], [[1, 20]], [[1, 26]], [[0, 32]], [[27, 47]], [[49, 63]]]",
    "query_spans": "[[[65, 75]]]",
    "process": "\\because the equation of parabola C is y^{2}=4x, its focus is F(1,0), \\therefore let the equation of line l be y=k(x-1). From \\begin{cases}y=k(x-1)\\\\y^{2}=4x\\end{cases}, eliminating x gives \\frac{k}{4}y^{2}-y-k=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), then y_{1}+y_{2}=\\frac{4}{k}, y_{1}y_{2}=-4\\textcircled{1}. \\because |AF|=3|BF|, \\therefore y_{1}+3y_{2}=0, so y_{1}=-3y_{2}. Substituting into \\textcircled{1} gives -2y_{2}=\\frac{4}{k} and -3y_{2}^{2}=-4. Eliminating y_{2} yields k^{2}=3, solving gives k=\\pm\\sqrt{3}"
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, a line $l$ passing through point $F$ intersects the parabola at points $M$ and $N$, and $O$ is the origin. If the area of $\\triangle MON$ is $\\sqrt{5}$, then $|MN|=$?",
    "fact_expressions": "l: Line;G: Parabola;M: Point;O: Origin;N: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {M, N};Area(TriangleOf(M, O, N)) = sqrt(5)",
    "query_expressions": "Abs(LineSegmentOf(M, N))",
    "answer_expressions": "5",
    "fact_spans": "[[[29, 34]], [[2, 16], [35, 38]], [[40, 43]], [[50, 53]], [[44, 47]], [[20, 23], [24, 28]], [[2, 16]], [[2, 23]], [[23, 34]], [[29, 49]], [[60, 91]]]",
    "query_spans": "[[[94, 103]]]",
    "process": "Let $ l: x = my + 1 $,  \n$$\n\\begin{cases}\nx = my + 1 \\\\\ny^{2} = 4x\n\\end{cases}\n\\Rightarrow y^{2} - 4my - 4 = 0\n$$\nSince the area of $ AMON $ is $ \\sqrt{5} $,  \n$$\n\\frac{1}{2}|y_{1} - y_{2}| \\times 1 = \\sqrt{5}\n\\quad\\therefore\\quad |y_{1} - y_{2}| = 2\\sqrt{5}\n\\quad\\therefore\\quad m^{2} = \\frac{1}{4}\n$$\nThus,  \n$$\n|MN| = \\sqrt{1 + m^{2}}|y_{1} - y_{2}| = \\sqrt{1 + \\frac{1}{4}} \\cdot 2\\sqrt{5} = 5\n$$"
  },
  {
    "text": "Let $P$ be a moving point on the curve $y^{2}=4(x-1)$. Then, the minimum value of the sum of the distance from point $P$ to the point $(0 , 1)$ and the distance from point $P$ to the $y$-axis is?",
    "fact_expressions": "G: Curve;H: Point;P: Point;Expression(G) = (y^2 = 4*(x - 1));Coordinate(H) = (0, 1);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P,H)+Distance(P,yAxis))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[5, 21]], [[34, 44]], [[1, 4], [29, 33], [48, 52]], [[5, 21]], [[34, 44]], [[1, 27]]]",
    "query_spans": "[[[29, 68]]]",
    "process": ""
  },
  {
    "text": "In the plane, $A(-1,0)$, $B(1,0)$, $C$ is a moving point. If $\\overrightarrow{A C} \\cdot \\overrightarrow{B C}=2$, then the trajectory equation of point $C$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (-1, 0);B: Point;Coordinate(B) = (1, 0);C: Point;DotProduct(VectorOf(A, C), VectorOf(B, C)) = 2",
    "query_expressions": "LocusEquation(C)",
    "answer_expressions": "x^2+y^2=3",
    "fact_spans": "[[[5, 14]], [[5, 14]], [[17, 25]], [[17, 25]], [[28, 31], [89, 93]], [[36, 87]]]",
    "query_spans": "[[[89, 100]]]",
    "process": "Let the moving point be C(x,y). Given A(-1,0), B(1,0), we have \\overrightarrow{AC}=(x+1,y), \\overrightarrow{BC}=(x-1,y). From the dot product of planar vectors, \\overrightarrow{AC}\\cdot\\overrightarrow{BC}=(x+1)(x-1)+y^{2}=x^{2}-1+y^{2}=2, that is, x^{2}+y^{2}=3. Therefore, the trajectory equation of point C is x^{2}+y^{2}=3"
  },
  {
    "text": "The distance from the focus of the hyperbola $\\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1$ to its asymptote is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/5 - y^2/4 = 1)",
    "query_expressions": "Distance(Focus(G), Asymptote(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 51]]]",
    "process": "By the given condition, $ c = \\sqrt{5 + 4} = 3 $, the equations of the asymptotes are: $ 2x \\pm \\sqrt{5}y = 0 $, and the distance from the foci $ (\\pm 3, 0) $ to the asymptote $ 2x \\pm \\sqrt{5}y = 0 $ is: $ \\frac{|2 \\times (-3) + 0|}{\\sqrt{4 + 5}} = 2 $."
  },
  {
    "text": "Let $P$ be the intersection point of the line $y=\\frac{b}{3 a} x$ and the left branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$, and let $F_{1}$ be the left focus. If $P F_{1}$ is perpendicular to the $x$-axis, then the eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Line;P:Point;e: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = x*(b/(3*a)));Eccentricity(G) = e;Intersection(H,LeftPart(G))=P;LeftFocus(G)=F1;IsPerpendicular(LineSegmentOf(P,F1),xAxis);F1:Point",
    "query_expressions": "e",
    "answer_expressions": "3*sqrt(2)/4",
    "fact_spans": "[[[27, 86], [124, 127]], [[30, 86]], [[30, 86]], [[5, 26]], [[1, 4]], [[131, 134]], [[30, 86]], [[30, 86]], [[27, 86]], [[5, 26]], [[124, 134]], [[1, 91]], [[27, 103]], [[104, 122]], [89, 95]]",
    "query_spans": "[[[131, 136]]]",
    "process": ""
  },
  {
    "text": "The line intersects the hyperbola $x^{2}-4 y^{2}=4$ at points $A$ and $B$. If point $P(4,1)$ is the midpoint of segment $AB$, then the equation of the line is?",
    "fact_expressions": "G: Hyperbola;H: Line;B: Point;A: Point;P: Point;Expression(G) = (x^2 - 4*y^2 = 4);Coordinate(P) = (4, 1);Intersection(H, G) = {A, B};MidPoint(LineSegmentOf(A,B))=P",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x-y-3=0",
    "fact_spans": "[[[3, 23]], [[0, 2], [59, 61]], [[30, 33]], [[26, 29]], [[37, 46]], [[3, 23]], [[37, 46]], [[0, 35]], [[37, 57]]]",
    "query_spans": "[[[59, 66]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since P(4,1) is the midpoint of AB, we have x_{1}+x_{2}=8, y_{1}+y_{2}=2. From \\begin{cases}x_{1}^{2}-4y_{1}^{2}=4\\\\x_{2}^{2}-4y_{2}^{2}=4\\end{cases}, subtracting the two equations yields: (x_{1}+x_{2})(x_{1}-x_{2})=4(y_{1}+y_{2})(y_{1}-y_{2}). Therefore, the slope of the line k=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{x_{1}+x_{2}}{4(y_{1}+y_{2})}=1. Hence, the equation of the line is: y-1=1\\times(x-4), i.e., x-y-3=0."
  },
  {
    "text": "What is the length of the real axis of the hyperbola $2 x^{2}-y^{2}=8$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (2*x^2 - y^2 = 8)",
    "query_expressions": "Length(RealAxis(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 26]]]",
    "process": "Standardize the hyperbola equation to obtain [detailed explanation] from the given, we have \\frac{x^{2}}{4}-\\frac{y^{2}}{8}=1, hence a=2, and the length of the real axis is 2a=4"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $|F_{1} F_{2}|=2 c$. Point $A$ lies on the ellipse, $\\overrightarrow{A F_{1}} \\cdot \\overrightarrow{F_{1} F_{2}}=0$, $\\overrightarrow{A F_{1}} \\cdot \\overrightarrow{A F_{2}}=c^{2}$. Then the eccentricity $e$ of the ellipse equals?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;F2: Point;A: Point;e: Number;c:Number;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(F1, F2)) = 2*c;PointOnCurve(A, G);DotProduct(VectorOf(A, F1), VectorOf(F1, F2)) = 0;DotProduct(VectorOf(A, F1), VectorOf(A, F2)) = c^2;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "(\\sqrt{5}-1)/2",
    "fact_spans": "[[[2, 47], [98, 100], [232, 234]], [[4, 47]], [[4, 47]], [[56, 63]], [[64, 71]], [[93, 97]], [[238, 241]], [[73, 92]], [[2, 47]], [[2, 71]], [[2, 71]], [[73, 92]], [[93, 101]], [[102, 165]], [[167, 230]], [[232, 241]]]",
    "query_spans": "[[[238, 244]]]",
    "process": "\\because\\overrightarrow{AF_{1}}\\cdot\\overrightarrow{F_{1}F_{2}}=0\\therefore AF_{1}\\bot F_{1}F_{2}, i.e., the x-coordinate of point A is the same as the left focus. Also \\because A is on the ellipse,\\therefore A(-c,\\pm\\frac{b^{2}}{a}), also \\overrightarrow{AF}_{1}\\cdot\\overrightarrow{AF}_{2}=c^{2}\\therefore\\overrightarrow{AF_{1}}\\cdot(\\overrightarrow{AF_{1}}+\\overrightarrow{F_{1}F_{2}})=c^{2},|\\overrightarrow{AF_{1}}|^{2}=c^{2}, i.e., |\\overrightarrow{AF_{1}}|=c, then 2a=c+\\sqrt{5}c\\therefore e=\\frac{\\sqrt{5}-1}{2}. So the answer is \\frac{\\sqrt{5}-1}{2}."
  },
  {
    "text": "It is known that hyperbola $C$ shares the same asymptotes with the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and $C$ passes through the point $M(-3, 2 \\sqrt{3})$. Then the length of the real axis of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Hyperbola;M: Point;Expression(G) = (x^2/9-y^2/16=1);Asymptote(C) = Asymptote(G);Coordinate(M) = (-3, 2*sqrt(3));PointOnCurve(M,C)",
    "query_expressions": "Length(RealAxis(C))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 8], [83, 89], [57, 60]], [[9, 48]], [[62, 81]], [[9, 48]], [[2, 55]], [[62, 81]], [[57, 81]]]",
    "query_spans": "[[[83, 95]]]",
    "process": "Since hyperbola C shares the same asymptotes with the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, assume the equation, substitute the point $(-3,2\\sqrt{3})$, solve for $\\lambda$, and then simplify. According to the problem, hyperbola C shares the same asymptotes with the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, so let the desired hyperbola's equation be $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=\\lambda$ ($\\lambda\\neq0$). Since C passes through the point $(-3,2\\sqrt{3})$, we have $1-\\frac{3}{4}\\lambda$, thus $\\lambda=\\frac{1}{4}$. Substituting into the equation and simplifying yields $\\frac{x^{2}}{\\frac{9}{4}}-\\frac{y^{2}}{4}=1$. The length of the real axis of hyperbola C is: $3$."
  },
  {
    "text": "The standard equation of a parabola symmetric about the $x$-axis and passing through the point $P(-2, -4)$ is?",
    "fact_expressions": "G: Parabola;P: Point;Coordinate(P) = (-2, -4);PointOnCurve(P, G);SymmetryAxis(G) = xAxis",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = -8*x",
    "fact_spans": "[[[25, 28]], [[12, 24]], [[12, 24]], [[11, 28]], [[0, 28]]]",
    "query_spans": "[[[25, 35]]]",
    "process": ""
  },
  {
    "text": "The moving circle $P$ is tangent to both circle $M$: $(x-2)^{2}+y^{2}=25$ and circle $N$: $(x+2)^{2}+y^{2}=1$. What is the trajectory equation of the center of the moving circle $P$?",
    "fact_expressions": "P: Circle;M:Circle;N:Circle;Expression(M)=((x-2)^2+y^2=25);Expression(N)=((x+2)^2+y^2=1);IsTangent(P,M);IsTangent(P,N)",
    "query_expressions": "LocusEquation(Center(P))",
    "answer_expressions": "{(x^2/9+y^2/5=1)&Negation(x=-3),(y=0)&Negation(x=-3)&Negation(x=-2)&Negation(x=2)}",
    "fact_spans": "[[[2, 5], [64, 67]], [[6, 31]], [[32, 56]], [[6, 31]], [[32, 56]], [[2, 60]], [[2, 60]]]",
    "query_spans": "[[[64, 77]]]",
    "process": "The center of circle M is M(2,0) with radius 5; the center of circle N is N(-2,0) with radius 1. When the center of moving circle P lies on the x-axis, the three circles are tangent at (-3,0). Note that the center of moving circle P does not coincide with M or N, so the trajectory equation of the center of moving circle P is y=0 (x≠-3 and x≠-2 and x≠2). When the center of moving circle P does not lie on the x-axis, |PM|+|PN|=5+1=6>|MN|, which satisfies the definition of an ellipse, where a=3, c=2, b^{2}=a^{2}-c^{2}=5. Thus, the trajectory equation of the center of moving circle P is \\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1 (x≠-3). In summary, the trajectory equation of the center of moving circle P is \\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1 (x≠-3) or y=0 (x≠-3 and x≠-2 and x≠2)."
  },
  {
    "text": "If the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ is $\\frac{\\sqrt{10}}{3}$, then what is the equation of the asymptotes of this hyperbola?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Eccentricity(G) = sqrt(10)/3",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "pm*3*y+x=0",
    "fact_spans": "[[[1, 38], [67, 70]], [[4, 38]], [[4, 38]], [[1, 38]], [[1, 64]]]",
    "query_spans": "[[[67, 78]]]",
    "process": "The hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ has eccentricity $\\frac{\\sqrt{10}}{3}$, then $\\begin{cases}\\frac{c}{a}=\\frac{\\sqrt{10}}{3}\\\\c^{2}-a^{2}=1\\end{cases}$ so $a=3$, $b=1$, then the asymptotes are $v=\\pm\\frac{1}{3}x$. Hence the answer is: $x+3v=0$."
  },
  {
    "text": "If the asymptotes of the hyperbola $C$: $\\frac{y^{2}}{5}-\\frac{x^{2}}{4}=1$ are tangent to the circle $(x-3)^{2}+y^{2}=r^{2}$ $(r>0)$, then $r=$?",
    "fact_expressions": "C: Hyperbola;G: Circle;r: Number;Expression(C) = (-x^2/4 + y^2/5 = 1);r>0;Expression(G) = (y^2 + (x - 3)^2 = r^2);IsTangent(Asymptote(C),G)",
    "query_expressions": "r",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[1, 44]], [[49, 78]], [[82, 85]], [[1, 44]], [[50, 78]], [[49, 78]], [[1, 80]]]",
    "query_spans": "[[[82, 87]]]",
    "process": "First, find the asymptotes of the hyperbola, then use the condition that the asymptotes are tangent to the circle, transforming it into the distance from the center of the circle to the asymptote being equal to the radius; thus, the value of r can be obtained. Solution: The asymptotes of hyperbola C are given by y=\\pm\\frac{\\sqrt{5}}{2}x, i.e., \\sqrt{5}x\\pm2y=0. The circle is (x-3)^{2}+y^{2}=r^{2}, with center at (3,0) and radius r. Since the asymptotes of hyperbola C are tangent to the circle, then r=\\frac{|3\\sqrt{5}|}{\\sqrt{(\\sqrt{5})^{2}+(\\pm2)^{2}}}=\\sqrt{5}."
  },
  {
    "text": "What is the focal distance of the ellipse $4 x^{2}+3 y^{2}=12$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (4*x^2 + 3*y^2 = 12)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 22]], [[0, 22]]]",
    "query_spans": "[[[0, 27]]]",
    "process": "4x^{2}+3y^{2}=12\\therefore\\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1\\thereforea^{2}=4,b^{2}=3\\thereforec^{2}=1,c=1\\therefore2c=2, so the focal distance is 2"
  },
  {
    "text": "If a point $P$ on the parabola $C$: $x^{2}=4 y$ is at a distance of $2$ from the fixed point $A(0,1)$, then what is the distance from point $P$ to the $x$-axis?",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 4*y);P: Point;PointOnCurve(P, C);A: Point;Coordinate(A) = (0, 1);Distance(P, A) = 2",
    "query_expressions": "Distance(P, xAxis)",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 20]], [[1, 20]], [[23, 26], [46, 50]], [[1, 26]], [[29, 37]], [[29, 37]], [[23, 44]]]",
    "query_spans": "[[[46, 60]]]",
    "process": "A(0,1) is the focus of the parabola. According to the definition of a parabola, the distance from any point on the parabola to the focus is equal to the distance to the directrix. Therefore, the distance from point P to the directrix of the parabola is 2. Since the equation of the directrix is $x = -1$, the distance from P to the x-axis is $2 - 1 = 1$."
  },
  {
    "text": "The line segment $P Q$ is a moving chord of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ passing through $M(1,0)$, and the line $P Q$ intersects the line $x=4$ at point $S$. Then $\\frac{|S M|}{|S P|}+\\frac{|S M|}{|S Q|}$=?",
    "fact_expressions": "G: Ellipse;P: Point;Q: Point;H:Line;M: Point;S: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Expression(H) = (x = 4);Coordinate(M) = (1, 0);PointOnCurve(M,LineSegmentOf(P,Q));IsChordOf(LineSegmentOf(P,Q),G);Intersection(LineOf(P,Q), H) = S",
    "query_expressions": "Abs(LineSegmentOf(S, M))/Abs(LineSegmentOf(S, Q)) + Abs(LineSegmentOf(S, M))/Abs(LineSegmentOf(S, P))",
    "answer_expressions": "2",
    "fact_spans": "[[[8, 45]], [[2, 7]], [[2, 7]], [[68, 75]], [[46, 54]], [[77, 81]], [[8, 45]], [[68, 75]], [[46, 54]], [[0, 58]], [[0, 58]], [[60, 81]]]",
    "query_spans": "[[[83, 126]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a moving point on the parabola $y^{2}=4x$, and $A(0,2\\sqrt{2})$ is a fixed point. From point $P$, draw $PQ \\perp y$-axis, meeting it at point $Q$. Then the minimum value of $|PA|+|PQ|$ is?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;Q: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (0, 2*sqrt(2));PointOnCurve(P, G);PointOnCurve(P, LineSegmentOf(P, Q));IsPerpendicular(LineSegmentOf(P, Q), yAxis);FootPoint(LineSegmentOf(P, Q), yAxis) = Q",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "2",
    "fact_spans": "[[[6, 20]], [[27, 44]], [[2, 5], [46, 50]], [[66, 70]], [[6, 20]], [[27, 44]], [[2, 24]], [[45, 64]], [[51, 65]], [[51, 70]]]",
    "query_spans": "[[[73, 92]]]",
    "process": "From the parabola $ y^{2} = 4x $, we know its focus is at $ F(1,0) $ and its directrix is $ x = -1 $. Let $ d $ be the distance from point $ P $ to the directrix. According to the definition of a parabola, $ d = |PF| $. Then the distance from point $ P $ to the $ y $-axis is $ |PQ| = |PF| - 1 $, and $ |FA| = \\sqrt{1^{2} + (2\\sqrt{2})^{2}} = 3 $. Thus, $ |PA| + |PQ| = |PA| + |PF| - 1 \\geqslant |FA| - 1 = 2 $ (equality holds if and only if points $ A $, $ P $, and $ F $ are collinear). Therefore, the minimum value of $ |PA| + |PQ| $ is 2. (Analysis) This problem mainly examines the application of the definition of a parabola. The key to solving it lies in using the definition to transform $ |PQ| = |PF| - 1 $, and then using geometric insight to obtain $ |PA| + |PQ| = |PA| + |PF| - 1 \\geqslant |FA| - 1 $. This problem emphasizes transformation thinking, integration of numerical and geometric methods, as well as computational and problem-solving abilities, and is considered fundamental."
  },
  {
    "text": "Given that point $P$ is any point on the ellipse $\\frac{x^{2}}{3}+y^{2}=1$, and point $Q$ is any point on the directrix of the parabola $x^{2}=2 \\sqrt{6} y$, the circle with diameter $PQ$ passes through the origin $O$. Determine $\\frac{1}{|O P|^{2}}+\\frac{1}{|O Q|^{2}}$=?",
    "fact_expressions": "G: Parabola;H: Ellipse;I: Circle;P: Point;Q: Point;O: Origin;Expression(G) = (x^2 = 2*(sqrt(6)*y));Expression(H) = (x^2/3 + y^2 = 1);PointOnCurve(P, H);PointOnCurve(Q, Directrix(G));IsDiameter(LineSegmentOf(P,Q),I);PointOnCurve(O, I)",
    "query_expressions": "1/Abs(LineSegmentOf(O,P))^2+1/Abs(LineSegmentOf(O,Q))^2",
    "answer_expressions": "1",
    "fact_spans": "[[[44, 67]], [[7, 34]], [[87, 88]], [[2, 6]], [[39, 43]], [[89, 94]], [[44, 67]], [[7, 34]], [[3, 38]], [[39, 76]], [[77, 88]], [[87, 94]]]",
    "query_spans": "[[[98, 141]]]",
    "process": "The standard equation of the parabola C is $x^{2}=2\\sqrt{6}y$, and its directrix equation is: $y=-\\frac{\\sqrt{6}}{2}$. Let $P(x_{P},y_{P})$, $Q(x_{Q},-\\frac{\\sqrt{6}}{2})$. Since the circle with diameter $PQ$ passes through the origin, $OP\\bot OQ$, so $x_{P}\\neq0$. Therefore, $x_{P}x_{Q}-\\frac{\\sqrt{6}}{2}=0$, i.e., $x_{Q}=\\frac{\\sqrt{6}y_{P}}{2x_{P}}$, so $|\\frac{1}{|OP}$"
  },
  {
    "text": "What is the length of the major axis of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1)",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "6",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 43]]]",
    "process": "From the given, $a^{2}=9$, that is, $a=3$, so the major axis length is $2a=6$."
  },
  {
    "text": "Let the origin of the coordinate system be $O$, and let the parabola $y^{2}=4x$ intersect a line passing through the focus at points $A$ and $B$. Then $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}$=?",
    "fact_expressions": "G: Parabola;H: Line;O: Origin;A: Point;B: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G),H);Intersection(G, H) = {A, B}",
    "query_expressions": "DotProduct(VectorOf(O, A), VectorOf(O, B))",
    "answer_expressions": "-3",
    "fact_spans": "[[[10, 24]], [[29, 31]], [[6, 9]], [[33, 36]], [[37, 40]], [[10, 24]], [[10, 31]], [[10, 42]]]",
    "query_spans": "[[[44, 95]]]",
    "process": "The focus of the parabola $ y^{2}=4x $ is $ F(1,0) $. Let the equation of line $ AB $ be $ x=my+1 $, and let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $.  \n$$\n\\begin{cases}\nx=my+1 \\\\\ny^{2}=4x \\\\\n\\overrightarrow{OA}\\cdot\\overrightarrow{OB}=(x_{1},y_{1})(x_{2},y_{2})=(my_{1}+1)(my_{2}+1)+y_{1}y_{2}=(1+m^{2})y_{1}y_{2}+m(y_{1}+y_{2})+1\n\\end{cases}\n=-4(1+m^{2})-4m^{2}+1=-3."
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ are $F_{1}$ and $F_{2}$, respectively. Point $M$ is a point on the ellipse such that $\\angle F_{1} M F_{2}=90^{\\circ}$. The line $M F_{1}$ intersects the ellipse again at point $N$, and $4|N F_{2}|=5|M F_{2}|$. What is the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;M: Point;PointOnCurve(M, G);AngleOf(F1, M, F2) = ApplyUnit(90, degree);N: Point;Intersection(LineOf(M, F1), G) = N;4*Abs(LineSegmentOf(N, F2)) = 5*Abs(LineSegmentOf(M, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[0, 52], [82, 84], [134, 136], [170, 172]], [[0, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[61, 68]], [[69, 76]], [[0, 76]], [[0, 76]], [[77, 81]], [[77, 87]], [[88, 121]], [[140, 143]], [[77, 143]], [[145, 168]]]",
    "query_spans": "[[[170, 178]]]",
    "process": "Let $|MF_{2}|=m$ $(m>0)$, from $4|NF_{2}|=5|MF_{2}|$, we get $|NF_{2}|=\\frac{5}{4}m$. From $\\angle F_{1}MF_{2}=90^{\\circ}$, we get $|MN|^{2}=|NF_{2}|^{2}-|MF_{2}|^{2}=\\frac{25}{16}m^{2}-m^{2}=\\frac{9}{16}m^{2}$, so $|MN|=\\frac{3}{4}m$. Also, $|MN|+|MF_{2}|+|NF_{2}|=|MF_{1}|+|NF_{1}|+|MF_{2}|+|NF_{2}|=4a$, that is, $\\frac{3}{4}m+m+\\frac{5}{4}m=4a$, simplifying gives $m=\\frac{4}{3}a$, so $|MF_{2}|=\\frac{4}{3}a$. According to $|MF_{1}|+|MF_{2}|=2a$, we get $|MF_{1}|=\\frac{2}{3}a$. Also, $|F_{1}F_{2}|^{2}=|MF_{1}|^{2}+|MF_{2}|^{2}$, so $4c^{2}=\\frac{4}{9}a^{2}+\\frac{16}{9}a^{2}$, therefore the ellipse's eccentricity is $e=\\frac{c}{a}=\\frac{\\sqrt{5}}{3}$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the two foci are $F_{1}$ and $F_{2}$, the eccentricity is $e=\\frac{\\sqrt{2}}{2}$, point $P$ lies on the ellipse, $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, and the area of $\\Delta P F_{1} F_{2}$ is $1$. Then the coordinates of the right focus $F_{2}$ are?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;Focus(G) = {F1, F2};e: Number;Eccentricity(G) = e;e = sqrt(2)/2;P: Point;PointOnCurve(P, G) = True;DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Area(TriangleOf(P, F1, F2)) = 1;RightFocus(G) = F2",
    "query_expressions": "Coordinate(F2)",
    "answer_expressions": "(1, 0)",
    "fact_spans": "[[[2, 54], [109, 111]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[62, 69]], [[70, 77], [208, 215]], [[2, 77]], [[81, 103]], [[2, 103]], [[81, 103]], [[104, 108]], [[104, 112]], [[113, 172]], [[174, 203]], [[109, 215]]]",
    "query_spans": "[[[208, 220]]]",
    "process": "According to the given conditions, find c, and thus obtain the coordinates of the right focus. [f solution]  \n\\begin{matrix}|PF_{1}|+|PF_{2}|=2a\\\\\\frac{|PF|+|PF|}{|PF_{1}|+|PF_{2}|^{2}}&\\frac{c}{4}c^{2}\\end{matrix}  \nB yields a=\\sqrt{2}, c=1, so the coordinates of the right focus are (1,0)"
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{3-m}=1$ has a focus at $(0 , 1)$, then what is the value of $m$?",
    "fact_expressions": "G: Ellipse;m: Number;H: Point;Expression(G) = (y^2/(3 - m) + x^2/m^2 = 1);Coordinate(H) = (0, 1);OneOf(Focus(G)) = H",
    "query_expressions": "m",
    "answer_expressions": "-2, 1",
    "fact_spans": "[[[0, 43]], [[60, 63]], [[49, 58]], [[0, 43]], [[49, 58]], [[0, 58]]]",
    "query_spans": "[[[60, 66]]]",
    "process": ""
  },
  {
    "text": "Let $P$ be a point on the curve $2 x=\\sqrt{4+y^{2}}$, $A(-\\sqrt{5}, 0)$, $B(\\sqrt{5}, 0)$. If $|P B|=2$, then $|P A|$=?",
    "fact_expressions": "G: Curve;A: Point;B: Point;P: Point;Expression(G) = (2*x = sqrt(y^2 + 4));Coordinate(A) = (-sqrt(5), 0);Coordinate(B) = (sqrt(5), 0);PointOnCurve(P, G);Abs(LineSegmentOf(P, B)) = 2",
    "query_expressions": "Abs(LineSegmentOf(P, A))",
    "answer_expressions": "4",
    "fact_spans": "[[[5, 27]], [[31, 48]], [[50, 67]], [[1, 4]], [[5, 27]], [[31, 48]], [[50, 67]], [[1, 30]], [[69, 78]]]",
    "query_spans": "[[[80, 89]]]",
    "process": "From $ 2x = \\sqrt{4 + y^{2}} $, we get $ 4x^{2} = 4 + y^{2} $ ($ x > 0 $), that is, $ x^{2} - \\frac{y^{2}}{4} = 1 $ ($ x > 0 $). Hence, $ P $ is a point on the right branch of the hyperbola $ x^{2} - \\frac{y^{2}}{4} = 1 $ ($ x > 0 $), and $ A $, $ B $ are the left and right foci of this hyperbola, respectively. Then $ |PA| - |PB| = 2a = 2 $, so $ |PA| = 2 + 2 = 4 $."
  },
  {
    "text": "The equation $\\frac{x^{2}}{22-m}+\\frac{y^{2}}{6+m}=1$ represents an ellipse with foci on the $y$-axis. Then the range of values for $m$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(22 - m) + y^2/(m + 6) = 1);PointOnCurve(Focus(G), yAxis);m:Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(8,22)",
    "fact_spans": "[[[53, 55]], [[0, 55]], [[44, 55]], [[57, 60]]]",
    "query_spans": "[[[57, 67]]]",
    "process": "According to the problem, we have  \n\\begin{cases}22-m>0\\\\6+m>0\\\\2m>16\\end{cases}\\Rightarrow8<m<22."
  },
  {
    "text": "The line passing through the left focus $F(-3,0)$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ intersects the hyperbola at points $M$ and $N$, and the midpoint of segment $MN$ has coordinates $(3,6)$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;N: Point;M: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(MidPoint(LineSegmentOf(M,N))) = (3, 6);Coordinate(F) = (-3, 0);LeftFocus(G)=F;PointOnCurve(F, H);Intersection(H, G) = {M, N};F:Point",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/3 - y^2/6 = 1",
    "fact_spans": "[[[1, 57], [74, 77], [111, 114]], [[4, 57]], [[4, 57]], [[71, 73]], [[82, 85]], [[78, 81]], [[4, 57]], [[4, 57]], [[1, 57]], [[89, 109]], [[61, 70]], [[1, 70]], [[0, 73]], [[71, 87]], [[61, 70]]]",
    "query_spans": "[[[111, 118]]]",
    "process": "Let M(x_{1},y_{1}), N(x_{2},y_{2}), we have x_{1}+x_{2}=6, y_{1}+y_{2}=12. Substitute the coordinates of points M and N into the hyperbola equation. Subtracting the two equations and simplifying yields k_{MN}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{x_{1}+x_{2}}{y_{1}+y_{2}}\\times\\frac{b^{2}}{a^{2}}. Using the given point coordinates to find the slope of line MN, the relationship between a^{2} and b^{2} can be obtained. Combining with c^{2}=a^{2}+b^{2}=9, the values of a^{2} and b^{2} can be determined, thus obtaining the hyperbola equation. [Solution] Let M(x_{1},y_{1}), N(x_{2},y_{2}), we have: ^{\\textcircled{1}} Therefore \\xrightarrow{(x_{1}-x_{2})(x_{1}} Since point (3,6) is the midpoint of segment MN, we have x_{1}+x_{2}=6, y_{1}+y_{2}=12. Thus k_{MN}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{x_{1}+x_{2}}{y_{1}+y_{2}}\\times\\frac{b^{2}}{a^{2}}=\\frac{6}{12}\\times\\frac{b^{2}}{a^{2}}=\\frac{b^{2}}{2a^{2}}. Since k_{MN}=\\frac{6-}{3-(}\\frac{-0}{-3)}=1, we get \\frac{b^{2}}{2a^{2}}=1, i.e., b^{2}=2a^{2}. Because c^{2}=a^{2}+b^{2}=3a^{2}=9, so a^{2}=3, b^{2}=6. Therefore, the hyperbola equation is \\frac{x^{2}}{3}-\\frac{y^{2}}{6}=1."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4 x$ is $F$, the directrix is $l$, $P$ is a point on $l$, and $Q$ is an intersection point of the line $PF$ with the parabola, if $\\overrightarrow{F P}=4 \\overrightarrow{F Q}$, then $|Q F|$=?",
    "fact_expressions": "G: Parabola;F: Point;P: Point;Q: Point;l: Line;Expression(G) = (y^2 = 4*x);Focus(G) = F;Directrix(G)=l;PointOnCurve(P, l);OneOf(Intersection(LineOf(P,F),G))=Q;VectorOf(F, P) = 4*VectorOf(F, Q)",
    "query_expressions": "Abs(LineSegmentOf(Q, F))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[2, 16], [56, 59]], [[20, 23]], [[33, 36]], [[44, 47]], [[27, 31], [37, 40]], [[2, 16]], [[2, 23]], [[2, 31]], [[33, 43]], [[44, 64]], [[66, 111]]]",
    "query_spans": "[[[113, 122]]]",
    "process": "The focus of the parabola $ C: y^{2} = 4x $ is $ F(1,0) $. Let $ P(-1,t) $, $ Q(x,y) $. Since $ \\overrightarrow{FP} = 4\\overrightarrow{FQ} $, then $ (-2,t) = 4(x-1,y) $, so $ x = \\frac{1}{2} $, $ y = \\frac{t}{4} $. Substituting into $ y^{2} = 4x $, we get $ \\left(\\frac{t}{4}\\right)^{2} = 4 \\cdot \\frac{1}{2} $, which gives $ t^{2} = 32 $."
  },
  {
    "text": "The line passing through the left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ and perpendicular to the $x$-axis intersects the hyperbola at points $M$ and $N$. The circle with diameter $MN$ passes exactly through the right vertex of the hyperbola. Then the eccentricity of the hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;M: Point;N: Point;H:Line;C:Circle;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(LeftFocus(G),H);IsPerpendicular(H,xAxis);Intersection(H,G) = {M, N};IsDiameter(LineSegmentOf(M,N),C);PointOnCurve(RightVertex(G),C)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 57], [73, 76], [102, 105], [111, 114]], [[4, 57]], [[4, 57]], [[79, 82]], [[83, 86]], [[70, 72]], [[98, 99]], [[4, 57]], [[4, 57]], [[1, 57]], [[0, 72]], [[62, 72]], [[70, 88]], [[89, 99]], [[98, 109]]]",
    "query_spans": "[[[111, 121]]]",
    "process": "Let the left focus of the hyperbola be $ F(-c,0) $, $ M(-c,y_{1}) $, $ N(-c,y_{1}) $, $ (y_{1}>0) $. Since point $ M(-c,y_{1}) $ lies on the hyperbola, $ \\frac{(-c)^{2}}{a^{2}} - \\frac{y_{1}^{2}}{b^{2}} = 1 $. Therefore, $ y_{1} = \\frac{b^{2}}{a} $, that is, $ FM = \\frac{b^{2}}{a} + a $. Hence, $ c^{2} - a^{2} = c + c $. Dividing both sides by $ a $, we get $ \\frac{c}{a} = 2 $, i.e., the eccentricity $ e = 2 $."
  },
  {
    "text": "Point $P$ is a point on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and $F_{1}$, $F_{2}$ are the left and right foci of the ellipse, respectively. If $|P F_{1}| \\cdot|P F_{2}|=\\frac{4}{3}$, then the measure of $\\angle F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 4/3",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "ApplyUnit(60,degree)",
    "fact_spans": "[[[5, 32], [55, 57]], [[0, 4]], [[36, 43]], [[45, 52]], [[5, 32]], [[0, 35]], [[36, 63]], [[36, 63]], [[66, 104]]]",
    "query_spans": "[[[107, 134]]]",
    "process": "From the ellipse $\\frac{x^2}{4}+y^{2}=1$, we get $a=2$, $b=1$, $c=\\sqrt{3}$. In $\\triangle PF_{1}F_{2}$, by the definition of the ellipse, $|PF_{1}|+|PF_{2}|=2a=4$, so $|PF_{1}|^{2}+|PF_{2}|^{2}+2|PF_{1}|\\cdot|PF_{2}|=16$. Since $|PF_{1}|\\cdot|PF_{2}|=\\frac{4}{3}$, it follows that $|PF_{1}|^{2}+|PF_{2}|^{2}=\\frac{40}{3}$. Therefore, $\\cos\\angle F_{1}PF_{2}=\\frac{|PF_{1}|^{2}+|PF_{2}|^{2}-4c^{2}}{2|PF_{1}||PF_{2}|}=\\frac{\\frac{40}{3}-12}{2\\times\\frac{4}{3}}=\\frac{1}{2}$, so the measure of $\\angle F_{1}PF_{2}$ is $60^{\\circ}$."
  },
  {
    "text": "Given an ellipse $ r $: $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ $ (a > b > 0) $ with eccentricity $ \\frac{\\sqrt{3}}{2} $, and a triangle $ ABC $ whose three vertices lie on the ellipse $ r $. Let the midpoints of its three sides $ AB $, $ BC $, $ AC $ be $ D $, $ E $, $ M $, respectively, and let the slopes of the lines containing these three sides be $ k_{1} $, $ k_{2} $, $ k_{3} $, all nonzero. Let $ O $ be the origin. If the sum of the slopes of the lines $ OD $, $ OE $, $ OM $ is $ 2 $, then $ \\frac{1}{k_{1}} + \\frac{1}{k_{2}} + \\frac{1}{k_{3}} = $?",
    "fact_expressions": "r: Ellipse;Expression(r) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;Eccentricity(r) = sqrt(3)/2;A: Point;B: Point;C: Point;PointOnCurve(Vertex(TriangleOf(A, B, C)), r);D: Point;E: Point;M: Point;MidPoint(LineSegmentOf(A, B)) = D;MidPoint(LineSegmentOf(B, C)) = E;MidPoint(LineSegmentOf(A, C)) = M;k1: Number;k2: Number;k3: Number;Slope(OverlappingLine(LineSegmentOf(A, B))) = k1;Slope(OverlappingLine(LineSegmentOf(B, C))) = k2;Slope(OverlappingLine(LineSegmentOf(A, C))) = k3;Negation(k1=0);Negation(k2=0);Negation(k3=0);O: Origin;Slope(LineOf(O, D)) + Slope(LineOf(O, E)) + Slope(LineOf(O, M)) = 2",
    "query_expressions": "1/k1 + 1/k2 + 1/k3",
    "answer_expressions": "-8",
    "fact_spans": "[[[2, 59], [102, 107]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[2, 84]], [[88, 95]], [[88, 95]], [[88, 95]], [[85, 108]], [[142, 145]], [[146, 149]], [[150, 153]], [[115, 153]], [[115, 153]], [[115, 153]], [[167, 174]], [[177, 184]], [[187, 194]], [[115, 194]], [[115, 194]], [[115, 194]], [[167, 203]], [[167, 203]], [[167, 202]], [[205, 208]], [[215, 247]]]",
    "query_spans": "[[[249, 300]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has its right focus at $F(3 , 0)$, and a line passing through point $F$ intersects the ellipse at points $A$ and $B$. If the midpoint of segment $AB$ has coordinates $(1 ,-1)$, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;B: Point;A: Point;F: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (3, 0);RightFocus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Coordinate(MidPoint(LineSegmentOf(A,B)))=(1,-1)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/18+y^2/9=1",
    "fact_spans": "[[[2, 54], [79, 81], [118, 120]], [[4, 54]], [[4, 54]], [[76, 78]], [[86, 89]], [[82, 85]], [[71, 75], [59, 69]], [[4, 54]], [[4, 54]], [[2, 54]], [[59, 69]], [[2, 69]], [[70, 78]], [[76, 91]], [[94, 116]]]",
    "query_spans": "[[[118, 125]]]",
    "process": ""
  },
  {
    "text": "Given that $P(x, y)$ is an arbitrary point on the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, $F_{1}$ is the left focus of the hyperbola, and $O$ is the coordinate origin, then the minimum value of $\\overrightarrow{P O} \\cdot \\overrightarrow{P F_{1}}$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;O: Origin;F1: Point;x1:Number;y1:Number;Expression(G) = (x^2/4 - y^2 = 1);Coordinate(P) = (x1, y1);PointOnCurve(P, G);LeftFocus(G) = F1",
    "query_expressions": "Min(DotProduct(VectorOf(P, O), VectorOf(P, F1)))",
    "answer_expressions": "4-2*sqrt(5)",
    "fact_spans": "[[[12, 40], [54, 57]], [[2, 11]], [[62, 65]], [[46, 53]], [[2, 11]], [[2, 11]], [[12, 40]], [[2, 11]], [[2, 45]], [[46, 61]]]",
    "query_spans": "[[[72, 131]]]",
    "process": "First calculate the expression of $\\overrightarrow{PO}\\cdot\\overrightarrow{PF_{1}}$, and based on the range of $x$, find the extreme value of $\\overrightarrow{PO}\\cdot\\overrightarrow{PF_{1}}$. From the given, the coordinates of $F_{1}$ are $(\\sqrt{5},0)$. Let $P(x,y)$, then $\\overrightarrow{PO}\\cdot\\overrightarrow{PF_{1}}=(-x,-y)\\cdot(\\sqrt{5}-x,-y)=x^{2}-\\sqrt{5}x+y^{2}=x^{2}-\\sqrt{5}x+\\frac{x^{2}}{4}-1=\\frac{5}{4}x^{2}-\\sqrt{5}x-1=(\\frac{\\sqrt{5}}{2}x-1)^{2}-2$, $x\\in(-\\infty,-2]\\cup[2,+\\infty)$. Therefore, when $x=-2$, the minimum value of $\\overrightarrow{PO}\\cdot\\overrightarrow{PF_{1}}$ is $4-2\\sqrt{5}$."
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$, $O$ be the coordinate origin, and point $P$ lies on $C$ such that $|O P|=2$. Then the area of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2/3 = 1);F1: Point;F2: Point;Focus(C) = {F1, F2};O: Origin;P: Point;PointOnCurve(P, C);Abs(LineSegmentOf(O, P)) = 2",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "3",
    "fact_spans": "[[[17, 50], [70, 73]], [[17, 50]], [[1, 8]], [[9, 16]], [[1, 55]], [[56, 59]], [[65, 69]], [[65, 74]], [[75, 84]]]",
    "query_spans": "[[[86, 116]]]",
    "process": "From the given, without loss of generality, let $F_{1}(-2,0)$, $F_{2}(2,0)$, then $a=1$, $c=2$. Since $|OP|=2=\\frac{1}{2}|F_{1}F_{2}|$, point $P$ lies on the circle with $F_{1}F_{2}$ as diameter, i.e., $\\triangle F_{1}F_{2}P$ is a right triangle with the right angle at $P$. Hence, $|PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}$, so $|PF_{1}|^{2}+|PF_{2}|^{2}=16$. Also, $||PF_{1}|-|PF_{2}||=2a=2$, thus $4=||PF_{1}|-|PF_{2}||^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}||PF_{2}|=16-2|PF_{1}||PF_{2}|$. Solving gives $|PF_{1}||PF_{2}|=6$, therefore $S_{\\triangle F_{1}F_{2}P}=\\frac{1}{2}|PF_{1}||PF_{2}|=3$."
  },
  {
    "text": "Let the parabola $ C $: $ y^{2} = 2 p x $ ($ p > 0 $) have focus $ F $ and directrix $ l $. Let $ A $ be a point on $ C $, and a circle centered at $ F $ with radius $ |FA| $ intersects $ l $ at points $ B $ and $ D $. If $ \\angle ABD = 90^{\\circ} $, and the area of $ \\Delta ABF $ is $ 9 \\sqrt{3} $, then the equation of this parabola is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;l: Line;Directrix(C) = l;A: Point;PointOnCurve(A, C);G: Circle;Center(G) = F;Radius(G) = Abs(LineSegmentOf(F, A));B: Point;D: Point;Intersection(G, l) = {B, D};AngleOf(A, B, D) = ApplyUnit(90, degree);Area(TriangleOf(A, B, F)) = 9*sqrt(3)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2 = 6*x",
    "fact_spans": "[[[1, 27], [48, 51], [151, 154]], [[1, 27]], [[9, 27]], [[9, 27]], [[56, 59], [31, 34]], [[1, 34]], [[38, 41], [76, 79]], [[1, 41]], [[44, 47]], [[44, 54]], [[74, 75]], [[55, 75]], [[63, 75]], [[80, 83]], [[84, 87]], [[74, 89]], [[91, 116]], [[118, 148]]]",
    "query_spans": "[[[151, 159]]]",
    "process": "Since the circle with center F and radius |FA| intersects line l at points B and D, ∠ABD = 90°. By the definition of the parabola, we have |AB| = |AF| = |BF|. ∴ △ABF is an equilateral triangle. ∴ ∠FBD = 30°. ∵ The area of △ABF is 9√3 = (√3/4)|BF|², ∴ |BF| = 6. The distance from F to the directrix is |BF| sin30° = 3 = p. The equation of this parabola is y² = 6x."
  },
  {
    "text": "Given that the equation of ellipse $C$ is $\\frac{x^{2}}{4}+y^{2}=1$, hyperbola $D$ shares the same foci $F_{1}$, $F_{2}$ with the ellipse, and $P$ is one of their intersection points such that $P F_{1} \\perp P F_{2}$, then the eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "D: Hyperbola;C: Ellipse;P: Point;F1: Point;F2: Point;e: Number;Expression(C) = (x^2/4 + y^2 = 1);Focus(D) = {F1,F2};Focus(C) = {F1,F2};OneOf(Intersection(C,D))=P;IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));Eccentricity(D) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[37, 43], [109, 112]], [[2, 7], [44, 46]], [[72, 75]], [[52, 59]], [[62, 69]], [[116, 119]], [[2, 36]], [[37, 69]], [[37, 69]], [[71, 83]], [[84, 107]], [[109, 119]]]",
    "query_spans": "[[[116, 121]]]",
    "process": ""
  },
  {
    "text": "Given that a focus of the hyperbola $x^{2}-\\frac{y^{2}}{b}=1$ is $(2,0)$. If point $M(4,0)$ is known and point $N(x, y)$ is an arbitrary point on the hyperbola, then the minimum value of $ |M N|$ is?",
    "fact_expressions": "G: Hyperbola;b: Number;H: Point;M: Point;N: Point;x1:Number;y1:Number;Expression(G) = (x^2 - y^2/b = 1);Coordinate(H) = (2, 0);OneOf(Focus(G))=H;Coordinate(M) = (4, 0);Coordinate(N) = (x1, y1);PointOnCurve(N, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(M, N)))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 30], [69, 72]], [[5, 30]], [[36, 43]], [[48, 57]], [[58, 68]], [[59, 68]], [[59, 68]], [[2, 30]], [[36, 43]], [[2, 43]], [[48, 57]], [[58, 68]], [[58, 78]]]",
    "query_spans": "[[[80, 94]]]",
    "process": "From the given condition, we have 1 + b = 4, so b = 3, thus obtaining y^{2} = 3x^{2} - 3. Using the distance formula between two points, |MN| = \\sqrt{(x - 4)^{2} + y^{2}} = \\sqrt{4x^{2} - 8x + 13} = \\sqrt{4(x^{2} - 2x + 1) + 9} = \\sqrt{4(x - 1)^{2} + 9}. Since x \\leqslant -1 or x \\geqslant 1, when x = 1, |MN| attains its minimum value of 3."
  },
  {
    "text": "Given that the directrix of the parabola $y=a x^{2}$ is tangent to the circle $x^{2}+y^{2}-6 y-7=0$, then the value of $a$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = a*x^2);a: Number;H: Circle;Expression(H) = (-6*y + x^2 + y^2 - 7 = 0);IsTangent(Directrix(G), H) = True",
    "query_expressions": "a",
    "answer_expressions": "1/4, -1/28",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[46, 49]], [[20, 42]], [[20, 42]], [[2, 44]]]",
    "query_spans": "[[[46, 53]]]",
    "process": "First express the equation of the directrix, then since the directrix of the parabola $ y = ax^2 $ is tangent to the circle $ x^{2} + y^{2} - 6y - 7 = 0 $, the distance from the center of the circle to the directrix equals the radius, thus obtaining the value of $ p $. [Detailed solution] The parabola $ y = ax^2 $, i.e., $ x^2 = \\frac{1}{a}y $, has the directrix equation $ y = -\\frac{1}{4a} $. Since the directrix of the parabola $ x^2 = \\frac{1}{a}y $ is tangent to the circle $ x^2 + (y - 3)^2 = 16 $, when $ a > 0 $, $ 3 + \\frac{1}{4a} = 4 $, solving gives $ a = \\frac{1}{4} $; when $ a < 0 $, $ -3 + \\frac{1}{4a} = -4 $, solving gives $ a = -\\frac{1}{28} $."
  },
  {
    "text": "It is known that one focus of the ellipse $x^{2}+k^{2}=3 k(k>0)$ coincides with the focus of the parabola $y^{2}=12 x$. Then, what is the eccentricity of this ellipse?",
    "fact_expressions": "H: Ellipse;Expression(H) = (k^2 + x^2 = 3*k);k: Number;k>0;G: Parabola;Expression(G) = (y^2 = 12*x);OneOf(Focus(H)) = Focus(G)",
    "query_expressions": "Eccentricity(H)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[2, 26], [55, 57]], [[2, 26]], [[4, 26]], [[4, 26]], [[32, 47]], [[32, 47]], [[2, 52]]]",
    "query_spans": "[[[55, 63]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, point $A$ is a point on the parabola $C$. If $|AF|=3$, then the horizontal coordinate of point $A$ is?",
    "fact_expressions": "C: Parabola;A: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(A, C);Abs(LineSegmentOf(A, F)) = 3",
    "query_expressions": "XCoordinate(A)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 21], [34, 40]], [[29, 33], [56, 60]], [[25, 28]], [[2, 21]], [[2, 28]], [[29, 43]], [[45, 54]]]",
    "query_spans": "[[[56, 66]]]",
    "process": "According to the definition of a parabola, |AF| = x_{1} + \\frac{p}{2} = 3, thus the horizontal coordinate of point A can be obtained. Let point A(x_{1}, y_{1}). Since |AF| = 3, by the definition of the parabola, we have |AF| = x_{1} + \\frac{p}{2} = x_{1} + 1 = 3. Solving gives x_{1} = 2, so the horizontal coordinate of point A is 2. The answer is."
  },
  {
    "text": "Given that the hyperbola $G$ has its center at the origin $O$, passes through the point $(\\sqrt{5}, 4)$, and has the focus of the parabola $C$: $y^{2}=4x$ as its right vertex, what is the equation of the hyperbola $G$?",
    "fact_expressions": "G: Hyperbola;C: Parabola;P: Point;O: Origin;Expression(C) = (y^2 = 4*x);Coordinate(P) = (sqrt(5), 4);Center(G) = O;PointOnCurve(P, G);RightVertex(G) = Focus(C)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/4 = 1",
    "fact_spans": "[[[2, 8], [67, 73]], [[38, 57]], [[19, 35]], [[9, 14]], [[38, 57]], [[19, 35]], [[2, 17]], [[2, 35]], [[2, 64]]]",
    "query_spans": "[[[67, 78]]]",
    "process": "The focus of the parabola $ y^{2}=4x $ is $ (1,0) $, so the equation of the hyperbola can be written as $ x^{2}-\\frac{y^{2}}{b^{2}}=1 $ $ (b>0) $. Substituting the point $ (\\sqrt{5},4) $ into the equation gives $ b^{2}=4 $. Therefore, the equation of the hyperbola is $ x^{2}-\\frac{y^{2}}{4}=1 $."
  },
  {
    "text": "What is the standard equation of a parabola with focus $F(2,0)$?",
    "fact_expressions": "G: Parabola;F: Point;Coordinate(F) = (2, 0);Focus(G) = F",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = 8*x",
    "fact_spans": "[[[12, 15]], [[3, 11]], [[3, 11]], [[0, 15]]]",
    "query_spans": "[[[12, 22]]]",
    "process": "Since the focus of the parabola is $ F(2,0) $, $ \\frac{p}{2} = 2 $, $ \\therefore 2p = 8 $, it follows that the parabola opens to the right, $ \\therefore $ the standard equation of the parabola is $ y^2 = 8x $."
  },
  {
    "text": "The line passing through point $A(4,0)$ intersects the circle $x^{2}+y^{2}=4$ at point $B$. What is the trajectory equation of the midpoint $P$ of $AB$?",
    "fact_expressions": "H: Line;A: Point;Coordinate(A) = (4, 0);PointOnCurve(A, H);G: Circle;Expression(G) = (x^2 + y^2 = 4);B: Point;Intersection(H, G) = B;P: Point;MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(x-2)^2+y^2=1",
    "fact_spans": "[[[10, 12]], [[1, 10]], [[1, 10]], [[0, 12]], [[13, 29]], [[13, 29]], [[31, 34]], [[10, 34]], [[43, 46]], [[36, 46]]]",
    "query_spans": "[[[43, 52]]]",
    "process": "Analysis: Since point B moves along the circle, expressing the coordinates of B in terms of the coordinates of M and then substituting into the equation of the circle gives the trajectory equation of M. Let M(x, y), then B(2x-4, 2y). Therefore, (2x-4)^{2}+4y^{2}=4. Simplifying yields the trajectory equation of M as (x-2)^{2}+y^{2}=1."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, a line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. If $|F_{2} A|+|F_{2} B|=12$, then $|A B|$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};H: Line;PointOnCurve(F1, H);A: Point;B: Point;Intersection(H, G) = {A, B};Abs(LineSegmentOf(F2, A)) + Abs(LineSegmentOf(F2, B)) = 12",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[18, 56], [74, 76]], [[18, 56]], [[2, 9], [63, 70]], [[10, 17]], [[2, 61]], [[71, 73]], [[62, 73]], [[77, 80]], [[81, 84]], [[71, 86]], [[88, 112]]]",
    "query_spans": "[[[114, 123]]]",
    "process": "The ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ has $a=5$. By the definition given in the problem, $|AF_{1}|+|AF_{2}|=|BF_{1}|+|BF_{2}|=2a$. Then the perimeter of triangle $ABF_{2}$ is $4a=20$. If $|F_{2}A|+|F_{2}B|=12$, then $|AB|=20-12=8$."
  },
  {
    "text": "The equation of the directrix of the parabola $x^{2}=-a y$ is $y=2$. Then the value of the real number $a$ is?",
    "fact_expressions": "G: Parabola;a: Real;Expression(G) = (x^2 = -a*y);Expression(Directrix(G)) = (y = 2)",
    "query_expressions": "a",
    "answer_expressions": "8",
    "fact_spans": "[[[0, 15]], [[28, 33]], [[0, 15]], [[0, 26]]]",
    "query_spans": "[[[28, 37]]]",
    "process": "From the given condition, a > 0, and \\frac{a}{4} = 2, \\therefore a = 8."
  },
  {
    "text": "A line passing through the focus of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$, with $|AB|=8$. Then, the horizontal coordinate of the midpoint of segment $AB$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Line;PointOnCurve(Focus(G), H) = True;Intersection(H, G) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(A, B)) = 8",
    "query_expressions": "XCoordinate(MidPoint(LineSegmentOf(A, B)))",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 15], [21, 24]], [[1, 15]], [[18, 20]], [[0, 20]], [[18, 35]], [[26, 29]], [[30, 33]], [[36, 45]]]",
    "query_spans": "[[[47, 62]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ with two foci $F_{1}$, $F_{2}$, a line passing through $F_{1}$ intersects the ellipse at points $A$, $B$. Then the range of the inradius of triangle $F_{2} A B$ is?",
    "fact_expressions": "G: Ellipse;Z: Line;F2: Point;A: Point;F1: Point;B: Point;Expression(G) = (x^2/4 + y^2 = 1);Focus(G) = {F1,F2};PointOnCurve(F1, Z);Intersection(Z, G) = {A, B}",
    "query_expressions": "Range(Radius(InscribedCircle(TriangleOf(F2, A, B))))",
    "answer_expressions": "(0, 1/2]",
    "fact_spans": "[[[2, 29], [63, 65]], [[60, 62]], [[43, 50]], [[66, 69]], [[35, 42], [52, 59]], [[70, 73]], [[2, 29]], [[2, 50]], [[51, 62]], [[60, 75]]]",
    "query_spans": "[[[77, 104]]]",
    "process": "Let $ AB: x = my - \\sqrt{3} $, and let the inradius be $ r $. Using the area equivalence method, we obtain $ 4r = \\sqrt{3}|y_{1} - y_{2}| $. By solving the system of the line equation and the ellipse equation, we get $ r = \\sqrt{3} \\frac{\\sqrt{m^{2} + 1}}{m^{2} + 4} $. The range of $ r $ can be found using the basic inequality. [Detailed Solution] As shown in the figure, from the ellipse $ \\frac{x^{2}}{4} + y^{2} = 1 $, we have $ a^{2} = 4 $, $ b^{2} = 1 $. When the line $ AB $ approaches the $ x $-axis infinitely, $ \\angle AF_{2}B $ approaches $ 0^{\\circ} $, then the inradius of $ \\triangle ABC $ approaches 0. Let $ AB: x = my - \\sqrt{3} $, solve the system:\n$$\n\\begin{cases}\nx = my - \\sqrt{3} \\\\\n\\frac{x^{2}}{4} + y^{2} = 1\n\\end{cases}\n$$\nwe obtain $ (m^{2} + 4)y^{2} - 2\\sqrt{3}my - 1 = 0 $,\n$ y_{1} + y_{2} = \\frac{2\\sqrt{3}m}{m^{2} + 4} $, $ y_{1}y_{2} = \\frac{-1}{m^{2} + 4} $,\n$ y_{1} + y_{2} = \\frac{2\\sqrt{5}m}{m^{2} + 4} $, $ y_{1}y_{2} = \\frac{1}{2} \\times 2\\sqrt{3} \\times |y_{1} - y_{2}| $,\nthus $ 4r = \\sqrt{3}|y_{1} - y_{2}| $,\n$ \\left( \\frac{2\\sqrt{3}m}{m^{2} + 4} \\right)^{2} + \\frac{4}{m^{2} + 4} = \\sqrt{3} \\frac{\\sqrt{m^{2} + 1}}{m^{2} + 4} $,\nlet $ t = m^{2} + 1 $ ($ t \\geqslant 1 $), we get $ r = \\sqrt{3} \\times \\frac{1}{t + \\frac{9}{t} + 6} \\leqslant \\frac{1}{2} $, equality holds if and only if $ t = 3 $. Therefore, the range of the inradius of triangle $ F_{2}AB $ is $ \\left(0, \\frac{1}{2}\\right] $."
  },
  {
    "text": "$F$ is the focus of the parabola $y^{2}=4x$, and point $A(2,2)$ is fixed. If point $P$ moves along the parabola, then the minimum value of $|AP|+|PF|$ is?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (2, 2);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(A, P)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "3",
    "fact_spans": "[[[4, 18], [39, 42]], [[24, 32]], [[34, 38]], [[0, 3]], [[4, 18]], [[24, 32]], [[0, 21]], [[34, 43]]]",
    "query_spans": "[[[48, 67]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ intersects the $y$-axis at points $M$, $N$, the line $y=x$ intersects the ellipse at points $A_{1}$, $A_{2}$, and $P$ is a point on the ellipse distinct from $A_{1}$, $A_{2}$. The point $Q$ satisfies $Q A_{1} \\perp P A_{1}$, $Q A_{2} \\perp P A_{2}$. Then $|Q M|+|Q N|$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);Intersection(G, yAxis) = {M, N};M: Point;N: Point;H: Line;Expression(H) = (y = x);A1: Point;A2: Point;Intersection(H, G) = {A1, A2};P: Point;PointOnCurve(P, G) = True;Negation(P = A1);Negation(P = A2);Q: Point;IsPerpendicular(LineSegmentOf(Q, A1), LineSegmentOf(P, A1)) = True;IsPerpendicular(LineSegmentOf(Q, A2), LineSegmentOf(P, A2)) = True",
    "query_expressions": "Abs(LineSegmentOf(Q, M)) + Abs(LineSegmentOf(Q, N))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 29], [53, 55], [78, 80]], [[2, 29]], [[2, 44]], [[36, 40]], [[41, 44]], [[45, 52]], [[45, 52]], [[56, 63], [83, 90]], [[64, 71], [91, 98]], [[45, 73]], [[74, 77]], [[74, 100]], [[74, 100]], [[74, 100]], [[101, 105]], [[107, 130]], [[132, 155]]]",
    "query_spans": "[[[157, 172]]]",
    "process": "Solve: From the given conditions, solve simultaneously $\\frac{x^{2}}{2}+y^{2}=1$ and $y=x$  \n\\[\n\\begin{cases}\n\\frac{x^{2}}{2}+y=1 \\\\\n\\frac{y=x}{3},\\frac{\\sqrt{6}}{3},\n\\end{cases}\n\\]\nSince point $Q$ satisfies $QA_{1}\\bot PA_{1}$, $QA_{2}\\bot PA_{2}$, by the symmetry of the ellipse, $Q(-\\frac{\\sqrt{6}}{3},\\frac{\\sqrt{6}}{3})$. Without loss of generality, let $M(0,1)$, $N(0,-1)$, $-1^{2}$, $|52N|=\\sqrt{(\\frac{y}{3}}(|QM|+|QN|)^{2}=|QM|^{2}+|_{2N|}^{2}+2|QM|\\cdot|QN|=(\\frac{\\sqrt{6}}{3})^{2}+(\\frac{\\sqrt{6}}{3}-1)^{2}+(\\frac{\\sqrt{6}}{3})^{2}+\\frac{\\sqrt{6}}{3}+1^{2}=8$, so $|QM|+|QN|=2\\sqrt{2}$."
  },
  {
    "text": "Given that the line $y=x-1$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$, $O$ is the origin, and the slopes of $OA$ and $OB$ are $k_{1}$, $k_{2}$ respectively, then $\\frac{1}{k_{1}}+\\frac{1}{k_{2}}$=?",
    "fact_expressions": "G: Parabola;H: Line;O: Origin;A: Point;B: Point;k1:Number;k2:Number;Expression(G) = (y^2 = 4*x);Expression(H) = (y = x - 1);Intersection(H, G) = {A, B};Slope(LineSegmentOf(O, A)) = k1;Slope(LineSegmentOf(O,B))=k2",
    "query_expressions": "1/k1 + 1/k2",
    "answer_expressions": "1",
    "fact_spans": "[[[12, 26]], [[2, 11]], [[38, 41]], [[28, 31]], [[32, 35]], [[65, 72]], [[74, 81]], [[12, 26]], [[2, 11]], [[2, 37]], [[47, 81]], [[47, 81]]]",
    "query_spans": "[[[83, 118]]]",
    "process": "Let $ A\\left(\\frac{y_{1}^{2}}{4}, y_{1}\\right), B\\left(\\frac{y_{2}^{2}}{4}, y_{2}\\right) $, then $ k_{1} = \\frac{4}{y_{1}}, k_{2} = \\frac{4}{y_{2}}, \\frac{1}{k_{1}} + \\frac{1}{k_{2}} = \\frac{y_{1} + y_{2}}{4} $\\textcircled{1}. From $ \\begin{cases} y = x - 1 \\\\ y^{2} = 4x \\end{cases} $, eliminating $ x $ gives $ y^{2} - 4y - 4 = 0 $, so $ y_{1} + y_{2} = 4 $. Hence, from \\textcircled{1}, $ \\frac{1}{k_{1}} + \\frac{1}{k_{2}} = \\frac{4}{4} = 1 $."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, point $M(a, b)$, $\\angle M F_{1} F_{2}=30^{\\circ}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;M: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(M) = (a, b);LeftFocus(C) = F1;RightFocus(C) = F2;AngleOf(M, F1, F2) = ApplyUnit(30, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[19, 80], [134, 137]], [[26, 80]], [[26, 80]], [[86, 96]], [[1, 8]], [[9, 16]], [[27, 80]], [[27, 80]], [[19, 80]], [[86, 96]], [[1, 85]], [[1, 85]], [[99, 132]]]",
    "query_spans": "[[[134, 143]]]",
    "process": "From the given conditions, we have F_{1}(-c,0), M(a,b), and the slope of line MF_{1} is \\tan30^{\\circ}=\\frac{\\sqrt{3}}{3}, so \\frac{b}{a+c}=\\frac{\\sqrt{3}}{3}. Therefore, a+c=\\sqrt{3}b. Squaring and simplifying yields a+c=3(c-a), so c=2a, thus e=2"
  },
  {
    "text": "The distance from the left focus of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ to its asymptote is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1)",
    "query_expressions": "Distance(LeftFocus(G), Asymptote(G))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[0, 28], [33, 34]], [[0, 28]]]",
    "query_spans": "[[[0, 42]]]",
    "process": "Find the left focus and asymptotes of the hyperbola, use the point-to-line distance formula to obtain the answer. The left focus of the hyperbola is $(-2,0)$, the asymptotes are $y=\\pm\\sqrt{3}x$. Then the distance from the left focus to its asymptote is $\\frac{|2\\sqrt{3}|}{\\sqrt{(\\sqrt{3})^{2}+1^{2}}}=\\sqrt{3}$"
  },
  {
    "text": "Given the parabola $y^{2}=2 x$, with focus $F$. A line $l$ passing through $F$ intersects the parabola at points $A$ and $B$. The minimum value of $|A F|+2|B F|$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);F: Point;Focus(G) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, G) = {A, B}",
    "query_expressions": "Min(Abs(LineSegmentOf(A, F)) + 2*Abs(LineSegmentOf(B, F)))",
    "answer_expressions": "3/2+sqrt(2)",
    "fact_spans": "[[[2, 16], [36, 39]], [[2, 16]], [[20, 23], [25, 29]], [[2, 23]], [[30, 35]], [[24, 35]], [[40, 43]], [[44, 47]], [[30, 49]]]",
    "query_spans": "[[[51, 71]]]",
    "process": "Analysis: Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. When the slope of line $ AB $ exists, let the equation of line $ AB $ be $ y=k(x-\\frac{1}{2}) $, $ (k\\neq0) $. By solving simultaneously with the parabola equation, we obtain the relationship between roots and coefficients. Using $ |AF|+4|BF|=x_{1}+\\frac{1}{2}+2(x_{2}+\\frac{1}{2}) $ and the properties of the basic inequality, the result can be derived. When the slope of line $ AB $ does not exist, compute directly. Let $ F(\\frac{1}{2},0) $, $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. When the slope of line $ AB $ exists, let the equation of line $ AB $ be $ y=k(x-\\frac{1}{2}) $, $ (k\\neq0) $. Solving simultaneously:\n$$\n\\begin{cases}\ny^{2}=2x \\\\\ny=k(x-\\frac{1}{2})\n\\end{cases}\n$$\nwe obtain $ k^{2}x^{2}-(k^{2}+2)x+\\frac{1}{4}k^{2}=0 $. Thus, $ x_{1}x_{2}=\\frac{1}{4} $. Since\n$$\n|AF|+2|BF|=x_{1}+\\frac{1}{2}+2(x_{2}+\\frac{1}{2})=x_{1}+2x_{2}+\\frac{3}{2}\\geqslant2\\sqrt{2x_{1}x_{2}}+\\frac{3}{2}=\\frac{3}{2}+\\sqrt{2},\n$$\nequality holds if and only if $ x_{1}=2x_{2}=\\frac{\\sqrt{2}}{2} $. When the slope of line $ AB $ does not exist, $ |AF|+2|BF|=3p=3 $. Therefore, the minimum value of $ |AF|+2|BF| $ is $ \\frac{3}{2}+\\sqrt{2} $."
  },
  {
    "text": "Given that a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $F(2,0)$, and the asymptotes of the hyperbola are tangent to the circle $(x-2)^{2}+y^{2}=3$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;Coordinate(F) = (2, 0);OneOf(Focus(G)) = F;H: Circle;Expression(H) = (y^2 + (x - 2)^2 = 3);IsTangent(Asymptote(G), H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[2, 58], [74, 77], [106, 109]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[64, 72]], [[64, 72]], [[2, 72]], [[82, 102]], [[82, 102]], [[74, 104]]]",
    "query_spans": "[[[106, 114]]]",
    "process": "The asymptotes of the hyperbola are given by $ y = \\pm\\frac{b}{a}x $, or $ bx \\pm ay = 0 $. These lines are tangent to the circle $ (x-2)^{2} + y^{2} = 3 $, so $ \\frac{2b}{\\sqrt{a^{2}+b^{2}}} = \\sqrt{3} $. Also, $ c = \\sqrt{a^{2}+b^{2}} = 2 $. Solving these equations simultaneously gives $ \\begin{cases} a = 1 \\\\ b = \\sqrt{3} \\end{cases} $. Therefore, the equation of the hyperbola is $ x^{2} - \\frac{y^{2}}{3} = 1 $."
  },
  {
    "text": "The equation of the hyperbola that shares the same asymptotes as $\\frac{x^{2}}{3}-\\frac{y^{2}}{4}=1$ and passes through the point $(3 , 2)$ is?",
    "fact_expressions": "G: Hyperbola;H: Point;C:Hyperbola;Expression(G) = (x^2/3 - y^2/4 = 1);Coordinate(H) = (3, 2);Asymptote(C) = Asymptote(G);PointOnCurve(H,C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/6 - y^2/8 = 1",
    "fact_spans": "[[[1, 39]], [[49, 59]], [[60, 63]], [[1, 39]], [[49, 59]], [[0, 63]], [[48, 63]]]",
    "query_spans": "[[[60, 67]]]",
    "process": "Let the hyperbola sharing the same asymptotes with $\\frac{x^{2}}{3}-\\frac{y^{2}}{4}=1$ be: $\\frac{x^{2}}{3}-\\frac{y^{2}}{4}=m$, $m\\neq0$, and $m\\neq1$. Then, according to the given conditions, $3-1=m$. This problem examines the application of hyperbola properties and the method of finding hyperbola equations, belonging to basic problems."
  },
  {
    "text": "A point $M$ on the parabola $y^{2}=-4 x$ is at a distance of $3$ from the focus. Then the coordinates of point $M$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -4*x);M: Point;PointOnCurve(M, G);Distance(M, Focus(G)) = 3",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(-2, \\pm 2\\sqrt{2})",
    "fact_spans": "[[[0, 15]], [[0, 15]], [[18, 21], [33, 37]], [[0, 21]], [[0, 31]]]",
    "query_spans": "[[[33, 42]]]",
    "process": "The solution process is omitted"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the right vertex is $A$, the left and right foci are $F_{1}$, $F_{2}$, respectively. A line passing through point $F_{2}$ and perpendicular to the $x$-axis intersects the ellipse at a point $B$. If $|F_{1} F_{2}|=2$, $|F_{2} B|=\\frac{3}{2}$, then the distance from point $F_{1}$ to the line $A B$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;A: Point;RightVertex(G) = A;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Line;PointOnCurve(F2, H);IsPerpendicular(H, xAxis);B: Point;OneOf(Intersection(H, G)) = B;Abs(LineSegmentOf(F1, F2)) = 2;Abs(LineSegmentOf(F2, B)) = 3/2",
    "query_expressions": "Distance(F1, LineOf(A, B))",
    "answer_expressions": "9*sqrt(13)/13",
    "fact_spans": "[[[2, 54], [105, 107]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[59, 62]], [[2, 62]], [[69, 76], [163, 171]], [[77, 84], [86, 94]], [[2, 84]], [[2, 84]], [[102, 104]], [[85, 104]], [[94, 104]], [[113, 116]], [[102, 116]], [[119, 136]], [[138, 161]]]",
    "query_spans": "[[[163, 184]]]",
    "process": "Based on the given information, the equation of the ellipse can be determined, and then the distance from a point to a line formula can be used to obtain the result. According to the problem, the diagram is drawn as follows: substitute $ x = c $ into the ellipse equation to get $ \\because |F_{1}F_{2}| = 2, |F_{2}B| = \\frac{3}{2}, (2c = 2) $, solving gives $ a = 2, b = \\sqrt{3} $. Without loss of generality, assume $ B $ lies in the first quadrant, then $ A(2,0), B(1,\\frac{3}{2}), F_{1}(-1,0) $. The equation of line $ AB $ is $ y = -\\frac{3}{2}x + 3 $, or $ 3x + 2y - 6 = 0 $. $ \\therefore $ the distance from point $ F_{1} $ to line $ AB $ is $ d = \\frac{9}{\\sqrt{9+4}} = \\frac{9\\sqrt{13}}{13} $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ and the hyperbola $x^{2}-\\frac{y^{2}}{n}=1$ share common foci $F_{1}$, $F_{2}$, if $P$ is an intersection point of the two curves, then $|P F_{1}| \\cdot|P F_{2}|$=?",
    "fact_expressions": "H: Ellipse;Expression(H) = (x^2/4 + y^2/m = 1);m: Number;G: Hyperbola;Expression(G) = (x^2 - y^2/n = 1);n: Number;F1: Point;F2: Point;Focus(H) = {F1, F2};Focus(G) = {F1, F2};P: Point;OneOf(Intersection(H, G)) = P",
    "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 39]], [[2, 39]], [[4, 39]], [[40, 68]], [[40, 68]], [[43, 68]], [[74, 81]], [[82, 89]], [[2, 89]], [[2, 89]], [[91, 94]], [[91, 103]]]",
    "query_spans": "[[[105, 133]]]",
    "process": "Since the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ and the hyperbola $x^{2}-\\frac{y^{2}}{n}=1$ have common foci $F_{1}, F_{2}$, and $P$ is an intersection point of the two curves, we have  \n$$\n\\begin{cases}\n|PF_{1}|+|PF_{2}|=4 \\\\\n||PF_{1}|-|PF_{2}||=2\n\\end{cases}\n$$\nThus  \n$$\n\\begin{matrix}\n|P \\\\\n|P\n\\end{matrix}\n\\overset{\\rightarrow}{F_{1}}|^{2}+2|\\overset{\\rightarrow}{PF_{1}}\\cdot|PF_{2}|+|PF_{2}|^{2}=16\n\\quad\n\\begin{cases}\n|PF_{1}|^{2}-2|PF_{1}|\\cdot|PF_{2}|+|PF_{2}|^{2}=4\n\\end{cases}\n$$\nSubtracting the two equations gives $4|PF_{1}|\\cdot|PF_{2}|=12$, so $|PF_{1}|\\cdot|PF_{2}|=3$.  \nHence the answer is: $3$"
  },
  {
    "text": "It is known that an ellipse and a hyperbola share the same foci $F_{1}$ and $F_{2}$. Let the eccentricities of the ellipse and the hyperbola be $e_{1}$ and $e_{2}$, respectively, and let $P$ be a common point on both curves such that $|\\overrightarrow{P F_{1}}-\\overrightarrow{P F_{2}}|=2|\\overrightarrow{P O}|$, where $O$ is the origin. If $e_{1} \\in\\left(\\frac{\\sqrt{2}}{2},\\frac{\\sqrt{3}}{2}\\right]$, then the range of values for $e_{2}$ is?",
    "fact_expressions": "H: Ellipse;G: Hyperbola;F1: Point;F2: Point;Focus(H) = {F1, F2};Focus(G) = {F1, F2};e1: Number;e2: Number;Eccentricity(H) = e1;Eccentricity(G) = e2;OneOf(Intersection(H, G)) = P;P: Point;Abs(VectorOf(P, F1) - VectorOf(P, F2)) = 2*Abs(VectorOf(P, O));O: Origin;In(e1, (sqrt(2)/2, sqrt(3)/2])",
    "query_expressions": "Range(e2)",
    "answer_expressions": "[\\sqrt{6}/2,+\\infty)",
    "fact_spans": "[[[2, 4], [31, 33]], [[5, 8], [34, 37]], [[14, 21]], [[22, 29]], [[2, 29]], [[2, 29]], [[44, 51]], [[54, 61], [223, 230]], [[31, 61]], [[31, 61]], [[63, 77]], [[64, 67]], [[79, 156]], [[158, 161]], [[171, 221]]]",
    "query_spans": "[[[223, 237]]]",
    "process": "Let the semi-major axis of the ellipse be $ a $, the semi-transverse axis of the hyperbola be $ a $, and their semi-focal distance be $ c $. Thus we obtain $ a = \\frac{c}{e_{1}} $, $ a = \\frac{c}{e_{2}} $. By the symmetry of the ellipse and hyperbola, assume without loss of generality that foci $ F_{1} $ and $ F_{2} $ lie on the x-axis, and point $ P $ lies to the right of the y-axis. From the definitions of the ellipse and hyperbola, we have:\n\\[\n\\begin{cases}\n|PF_{1}| + |PF_{2}| = 2a \\\\\n|PF_{1}| - |PF_{2}| = 2a\n\\end{cases}\n\\]\nSolving gives $ |PF_{1}| = a + a $, $ |PF_{2}| = a - a $. Since $ |\\overrightarrow{PF_{1}} - \\overrightarrow{PF_{2}}| = 2|\\overrightarrow{PO}| $, i.e., $ |\\overrightarrow{F_{1}F_{2}}| = 2|\\overrightarrow{PO}| $, and $ O $ is the midpoint of segment $ F_{1}F_{2} $, it follows that $ \\angle F_{1}PF_{2} = 90^{\\circ} $. Then $ |PF_{1}|^{2} + |PF_{2}|^{2} = |F_{1}F_{2}|^{2} $, i.e., $ (a + a)^{2} + (a - a)^{2} = 4c^{2} $. Simplifying yields: $ a^{2} + a^{2} = 2c^{2} $. Hence $ \\left(\\frac{c}{e_{1}}\\right)^{2} + \\left(\\frac{c}{e_{2}}\\right)^{2} = 2c^{2} $, which implies $ \\frac{1}{e_{1}^{2}} + \\frac{1}{e_{2}^{2}} = 2 $. Given $ \\frac{\\sqrt{2}}{2} < e_{1} \\leqslant \\frac{\\sqrt{3}}{2} $, we have $ 0 < 2 - \\frac{1}{e_{1}^{2}} \\leqslant \\frac{2}{3} $, i.e., $ e_{2}^{2} \\geqslant \\frac{3}{2} $, solving gives $ e_{2} \\geqslant \\frac{\\sqrt{6}}{2} $. Therefore, the range of $ e_{2} $ is $ \\frac{\\sqrt{6}}{2} \\leqslant e_{2} < 1 $."
  },
  {
    "text": "The distance from the right focus $F$ of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ to one of its asymptotes is?",
    "fact_expressions": "G: Hyperbola;F:Point;Expression(G) = (x^2/9 - y^2/16 = 1);RightFocus(G)=F",
    "query_expressions": "Distance(F,OneOf(Asymptote(G)))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 39]], [[43, 46]], [[0, 39]], [[0, 46]]]",
    "query_spans": "[[[0, 57]]]",
    "process": "From the given conditions, we have: $a^{2}=9$, $b^{2}=16 \\Rightarrow c=\\sqrt{a^{2}+b^{2}}=5$, so the coordinates of the right focus $F$ are $(5,0)$. An asymptote of this hyperbola has the equation: $y=\\frac{4}{3}x \\Rightarrow 4x-3y=0$. Therefore, the distance from $F$ to one asymptote is: $\\frac{|4\\times5-3\\times0|}{\\sqrt{4^{2}+(-3)^{2}}}=4$"
  },
  {
    "text": "Given the ellipse $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$ $(a>b>0)$ has its right vertex at $A(1,0)$, and the length of the chord passing through its focus and perpendicular to the major axis is $1$, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;a > b;b > 0;Expression(G) = (x^2/b^2 + y^2/a^2 = 1);Coordinate(A) = (1, 0);RightVertex(G) = A;H:LineSegment;IsChordOf(H,G);PointOnCurve(Focus(G),H);IsPerpendicular(H,MajorAxis(G));Length(H)=1",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/4 + x^2 = 1",
    "fact_spans": "[[[2, 54], [70, 71], [87, 89]], [[4, 54]], [[4, 54]], [[59, 67]], [[4, 54]], [[4, 54]], [[2, 54]], [[59, 67]], [[2, 67]], [], [[70, 80]], [[68, 80]], [[68, 80]], [[69, 85]]]",
    "query_spans": "[[[87, 93]]]",
    "process": "\\because the right vertex of the ellipse \\frac{x^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1 is A(1,0), \\therefore b=1, the focus coordinates are (0,c), and the chord length passing through the focus and perpendicular to the major axis is 1, i.e., 1=2|x|=2b\\sqrt{1-\\frac{c^{2}}{a^{2}}}=\\frac{2b^{2}}{a}=\\frac{2}{a}, a=2, then the equation of the ellipse is \\frac{z}{4}+x^{2}=1."
  },
  {
    "text": "A line $l$ passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) intersects the parabola at points $A$ and $B$, where point $A(3, y_{1})$ and $|A F|=4$. Then $|A B|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, G) = {A, B};y1: Number;Coordinate(A) = (3, y1);Abs(LineSegmentOf(A, F)) = 4",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[1, 22], [34, 37]], [[1, 22]], [[4, 22]], [[4, 22]], [[24, 27]], [[1, 27]], [[28, 33]], [[0, 33]], [[38, 41], [50, 64]], [[42, 45]], [[28, 47]], [[51, 64]], [[50, 64]], [[66, 75]]]",
    "query_spans": "[[[77, 86]]]",
    "process": "Since the directrix of the parabola $ y^{2} = 2px $ is $ x = -\\frac{p}{2} $, and point $ A(3, y_{1}) $ lies on the parabola, we have $ |AF| = 3 + \\frac{p}{2} = 4 $, solving gives $ p = 2 $, so the equation of the parabola is $ y^{2} = 4x $. Let $ B(x_{2}, y_{2}) $, since point $ A(3, y_{1}) $ lies on the parabola, we get $ y_{1} = \\pm 2\\sqrt{3} $. By symmetry of the parabola, without loss of generality, let $ A(3, 2\\sqrt{3}) $, and $ F(1, 0) $, so the slope of line $ AF $ is $ k = \\frac{2\\sqrt{3}}{3-1} = \\sqrt{3} $. Thus, the equation of line $ AF $ is $ y = \\sqrt{3}(x - 1) $. Substituting into the parabola equation $ y^{2} = 4x $ yields $ 3x^{2} - 10x + 3 = 0 $, so $ x_{1} + x_{2} = \\frac{10}{3} $. Therefore, $ |AB| = x_{1} + x_{2} + p = \\frac{16}{3} $."
  },
  {
    "text": "Let $F$ be the focus of the parabola $y^{2}=8x$, and let $A$, $B$ be two points on the parabola such that $\\overrightarrow{A F}=2 \\overrightarrow{F B}$. Then $|\\overrightarrow{F A}|+2|\\overrightarrow{F B}|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);VectorOf(A, F) = 2*VectorOf(F, B)",
    "query_expressions": "Abs(VectorOf(F,A))+2*Abs(VectorOf(F,B))",
    "answer_expressions": "12",
    "fact_spans": "[[[5, 19], [31, 34]], [[5, 19]], [[1, 4]], [[1, 22]], [[23, 26]], [[27, 30]], [[23, 37]], [[23, 37]], [[39, 84]]]",
    "query_spans": "[[[87, 137]]]",
    "process": "Analysis: Draw perpendicular lines from points A and B to the directrix, and draw a perpendicular line from point B to AC with foot E. In right triangle ABE, we obtain $\\cos\\angle BAE = \\frac{AE}{AB} = \\frac{1}{3}$, then deduce that the slope of line AB is $k = 2\\sqrt{2}$. Thus, the equation of line AB can be found. By solving the system of equations, the coordinates of points A and B are obtained, allowing us to find the answer. Specifically, draw perpendicular lines from points A and B to the directrix, and draw a perpendicular from point B to AC with foot E. Let $BF = m$, then $BD = m$. Since $\\overrightarrow{AF} = 2\\overrightarrow{FB}$, it follows that $AC = AF = 2m$. In right triangle ABE, $AE = AC - BD = 2m - m = m$, $AB = 3m$, so $\\cos\\angle BAE = \\frac{AE}{AB} = \\frac{1}{3}$. Therefore, the slope of line AB is $k = \\tan\\angle BAE = 2\\sqrt{2}$. Hence, the equation of line AB is $y = 2\\sqrt{2}(x - 2)$. Substituting into the parabola's equation yields $x^{2} - 5x + 4 = 0$, solving gives $x_{1} = 1$, $x_{2} = 4$. Thus, point $A(4, 4\\sqrt{2})$, $B(1, -2\\sqrt{2})$. Given $F(2, 0)$, we have $|BF| = \\sqrt{(1 - 2)^{2} + (-2\\sqrt{2})^{2}} = 3$. Therefore, $|\\overrightarrow{FA}| + 2|\\overrightarrow{FB}| = 4|\\overrightarrow{FB}| = 12$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. If there exists a point $P$ on the ellipse such that $\\angle F_{1} P F_{2}=\\frac{2}{3} \\pi$, then the range of values for the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G) ;AngleOf(F1, P, F2) = 2*pi/3",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[\\sqrt{3}/2, 1)",
    "fact_spans": "[[[2, 54], [80, 82], [133, 135]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[63, 70]], [[71, 78]], [[2, 78]], [[2, 78]], [[87, 90]], [[80, 90]], [[92, 130]]]",
    "query_spans": "[[[133, 145]]]",
    "process": "From the definition of the ellipse, we know: $ PF_{1} + PF_{2} = 2a $. In $ \\triangle PF_{1}F_{2} $, by the law of cosines:  \n$ \\cos\\angle F_{1}PF_{2} = \\frac{F_{1}P^{2} + F_{2}P^{2} - F_{1}F_{2}^{2}}{2F_{1}P \\cdot F_{2}P} = \\frac{(F_{1}P + F_{2}P)^{2} - 2F_{1}P \\cdot F_{2}P - F_{1}F_{2}^{2}}{2F_{1}P \\cdot F_{2}P} = \\frac{4b^{2} - 2F_{1}P \\cdot F_{2}P}{2F_{1}P \\cdot F_{2}P} = -\\frac{1}{2} $,  \nso $ F_{1}P \\cdot F_{2}P = 4b^{2} $. Also, $ F_{1}P \\cdot F_{2}P \\leqslant \\frac{(F_{1}P + F_{2}P)^{2}}{4} = a^{2} $, that is, $ 4b^{2} \\leqslant a^{2} $, with equality if and only if $ F_{1}P = F_{2}P $. Hence, $ 4a^{2} - 4c^{2} \\leqslant a^{2} $, so $ 3a^{2} \\leqslant 4c^{2} $, $ e^{2} \\geqslant \\frac{3}{4} $, solving gives: $ e \\in \\left[ \\frac{\\sqrt{3}}{2}, 1 \\right) $."
  },
  {
    "text": "It is known that the vertex of the parabola is at the coordinate origin, and the focus $F$ coincides with the left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$. If the two curves intersect at points $M$ and $N$, and the midpoint of the segment $MN$ is the point $F$, then the eccentricity of the hyperbola equals?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;N: Point;M: Point;O: Origin;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Vertex(H) = O;Focus(H) = F;LeftFocus(G)=F;Intersection(H,G)={M,N};MidPoint(LineSegmentOf(M,N)) = F",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)+1",
    "fact_spans": "[[[20, 76], [119, 122]], [[23, 76]], [[23, 76]], [[2, 5]], [[94, 97]], [[90, 93]], [[9, 13]], [[16, 19], [112, 116]], [[23, 76]], [[23, 76]], [[20, 76]], [[2, 13]], [[2, 19]], [[16, 82]], [[84, 99]], [[101, 116]]]",
    "query_spans": "[[[119, 129]]]",
    "process": "From the given conditions: $-\\frac{p}{2}=-c$, $\\therefore p=2c$, $\\therefore$ the equation of the parabola is: $y^{2}=-2px=-4cx$. Since $M$ lies on the parabola, $M(-c,2c)$. Since $M$ lies on the hyperbola, $\\therefore \\frac{c^{2}}{a^{2}}-\\frac{4c^{2}}{k^{2}}=1$. $\\because b^{2}=c^{2}-a^{2}$, $\\therefore c^{4}-6a^{2}c^{2}+a^{4}=0$. $\\therefore e^{2}=3\\pm2\\sqrt{2}$. Also $e\\in(1,+\\infty)$, $\\therefore e=\\sqrt{2}+1$."
  },
  {
    "text": "Given that $F$ is the right focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, and point $P$ lies on the left branch of the hyperbola, $A(0,6 \\sqrt{6})$, then the minimum perimeter of $\\triangle A P F$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/5 = 1);F: Point;RightFocus(G) = F;P: Point;PointOnCurve(P, LeftPart(G));A: Point;Coordinate(A) = (0, 6*sqrt(6))",
    "query_expressions": "Min(Perimeter(TriangleOf(A, P, F)))",
    "answer_expressions": "34",
    "fact_spans": "[[[6, 44], [55, 58]], [[6, 44]], [[2, 5]], [[2, 48]], [[50, 54]], [[50, 64]], [[65, 82]], [[65, 82]]]",
    "query_spans": "[[[84, 109]]]",
    "process": "Transforming the distance from P to the right focus F into the distance from P to the left focus, the minimum value is easily obtained. For the hyperbola $\\frac{x^2}{4}-\\frac{y^{2}}{5}=1$, $a=2$, $b=\\sqrt{5}$, $c=\\sqrt{4+5}=3$, so $F(3,0)$. Let $F'(-3,0)$ be the left focus of the hyperbola, then $|AF'|=|AF|=\\sqrt{3^{2}+(-6\\sqrt{6})^{2}}=15$. Since $P$ lies on the left branch of the hyperbola, $|PF'|-|PF|=2a=4$, i.e., $|PF'|=|PF|+4$. Therefore, the perimeter of $\\Delta APF$ is $l=|PF|+|PA|+|AF|=4+|PF'|+|PA|+15=|PA|+|PF'|+19$. Clearly, $|PA|+|PF'|\\geqslant|AF'|=\\sqrt{3^{2}+(6\\sqrt{6})^{2}}=15$. The equality holds if and only if $P$ is the intersection point of segment $AF'$ and the hyperbola. Therefore, the minimum perimeter $l$ of $\\Delta APF$ is $15+19=34$."
  },
  {
    "text": "A point $P$ on the parabola $x^{2}=2 y$ has a vertical coordinate of $3$. What is the distance from point $P$ to the focus of the parabola?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 2*y);P: Point;PointOnCurve(P, G) = True;YCoordinate(P) = 3",
    "query_expressions": "Distance(P, Focus(G))",
    "answer_expressions": "7/2",
    "fact_spans": "[[[0, 14], [35, 38]], [[0, 14]], [[17, 20], [30, 34]], [[0, 20]], [[17, 28]]]",
    "query_spans": "[[[30, 45]]]",
    "process": "By the given condition, the directrix of the parabola is $ y = -\\frac{1}{2} $. Therefore, the distance from point $ P $ to the parabola is $ 3 + \\frac{1}{2} = \\frac{7}{2} $. By the definition of the parabola, the distance from point $ P $ to the focus of the parabola is $ \\frac{7}{2} $. Answer: $ 7 $"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, let $A$, $B$ be two points on the ellipse, and the perpendicular bisector of segment $AB$ intersects the $x$-axis at point $M(\\frac{a}{5}, 0)$. Then the range of the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;B: Point;A: Point;M: Point;e: Number;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(A,G);PointOnCurve(B,G);a > b;b > 0;Coordinate(M) = (a/5, 0);Intersection(PerpendicularBisector(LineSegmentOf(A,B)), xAxis) = M;Eccentricity(G) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "(\\sqrt{5}/5,1)",
    "fact_spans": "[[[2, 55], [67, 69], [115, 117]], [[4, 54]], [[4, 54]], [[63, 66]], [[57, 60]], [[93, 113]], [[121, 124]], [[2, 55]], [[57, 73]], [[57, 73]], [[4, 54]], [[4, 54]], [[93, 113]], [[74, 113]], [[115, 124]]]",
    "query_spans": "[[[121, 131]]]",
    "process": "Let $ A(x_{1},y_{1}), B(x_{2},y_{2}), x_{1} \\ne x_{2} $, then \n$$\n\\begin{cases}\n(x_{1}-\\frac{a}{5})^{2}+y_{1}=\\frac{x_{1}}{a_{2}}+\\frac{y_{2}}{b^{2}}=1 \\\\\n\\frac{x_{2}}{y_{2}}=\n\\end{cases}\ny_{1}^{2}=(x_{2}-\\frac{a}{5})^{2}+y_{2}2\n$$\ni.e.\n$$\n\\begin{cases}\n\\frac{2a}{5}(x_{1}-x_{2})=x_{1}2- \\text{ i.e. } y_{1}=b^{2}-\\frac{b^{2}}{a^{2}1}2 \\\\\ny_{2}=b^{2}-\\frac{b^{2}}{x_{2}}\n\\end{cases}_{1}^{2}-x_{2}2+y_{1}^{2}-y_{2}2\n$$\nso $ \\frac{a^{2}}{5}(x_{1}-x_{2})=\\frac{a^{2}-b^{2}}{a^{2}}(x_{1}^{2}-x_{2})^{,} $ so $ \\frac{2^{-}}{5(a^{2}-b^{2})}=x_{1}+x_{2} $. Also $ -a \\leqslant x_{1} \\leqslant a, -a \\leqslant x_{2} \\leqslant a, x_{1} \\ne x_{2} $, so $ -2a < x_{1}+x_{2} < 2a $, then $ \\frac{2a3}{5(a2-b^{2})} < 2a $ i.e. $ \\frac{b^{2}}{a^{2}} < \\frac{4}{5} $, so $ e^{2}=1-\\frac{b^{2}}{a^{2}} > \\frac{1}{5} $. Also $ 0 < e < 1 $, so $ \\frac{\\sqrt{5}}{5} < e < 1 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, a line passing through the origin with an inclination angle of $\\frac{\\pi}{3}$ intersects the left and right branches of the hyperbola at points $P$ and $Q$, respectively. The circle with diameter $PQ$ passes through the right focus $F$. Then, the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;O: Origin;L: Line;PointOnCurve(O, L);Inclination(L) = pi/3;P: Point;Q: Point;Intersection(L, LeftPart(G)) = P;Intersection(L, RightPart(G)) = Q;H: Circle;IsDiameter(LineSegmentOf(P, Q), H);PointOnCurve(F, H);RightFocus(G) = F;F:Point",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3) + 1",
    "fact_spans": "[[[2, 58], [89, 92], [130, 133]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[60, 62]], [[84, 86]], [[59, 86]], [[65, 86]], [[97, 100]], [[101, 104]], [[84, 106]], [[84, 106]], [[119, 120]], [[107, 120]], [[119, 127]], [[89, 127]], [121, 123]]",
    "query_spans": "[[[130, 138]]]",
    "process": "A line passing through the origin with an inclination angle of $\\frac{\\pi}{3}$ has the equation $y=\\sqrt{3}x$. Solve the system of equations:  \n\\[\n\\begin{cases}\ny=\\sqrt{3}x \\\\\n\\frac{a}{a}-b=1\n\\end{cases}\n\\quad \\frac{\\sqrt{3}ab}{b^{2}-3a^{2}}),\\ F(c,0)\n\\]\nSince the segment $PQ$ is a diameter passing through the right focus $F$, we have $\\overrightarrow{FP}\\cdot\\overrightarrow{FQ}=0$, thus  \n\\[\n+\\frac{\\sqrt{3}ab}{\\sqrt{b^{2}-3a^{2}}}\\left(-\\frac{\\sqrt{3}ab}{\\sqrt{b^{2}-3a^{2}}}\\right)=0,\\ c^{2}=a^{2}+b^{2}\n\\]\nSimplifying yields $c^{4}-8c^{2}a^{2}+4a^{4}=0$, so $e^{4}-8e^{2}+4=0$. Solving gives $e^{2}=4+2\\sqrt{3} \\Rightarrow e=\\sqrt{3}+1$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $P$ is a point on the ellipse $C$, and $\\overrightarrow{P F_{1}} \\perp \\overrightarrow{P F_{2}}$. If the area of $\\triangle P F_{1} F_{2}$ is $9$, then $b=$?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Focus(C) = {F1, F2};PointOnCurve(P, C);IsPerpendicular(VectorOf(P, F2),VectorOf(P, F1));Area(TriangleOf(P, F1, F2)) = 9",
    "query_expressions": "b",
    "answer_expressions": "3",
    "fact_spans": "[[[18, 75], [85, 90]], [[188, 191]], [[24, 75]], [[81, 84]], [[2, 9]], [[10, 17]], [[24, 75]], [[24, 75]], [[18, 75]], [[2, 80]], [[81, 93]], [[95, 152]], [[154, 186]]]",
    "query_spans": "[[[188, 193]]]",
    "process": ""
  },
  {
    "text": "A focus of the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1$ is $F$. A perpendicular is drawn from point $F$ to an asymptote, with foot of perpendicular at $E$, and $O$ is the origin. Then the area of $\\Delta E O F$ is?",
    "fact_expressions": "G: Hyperbola;E: Point;O: Origin;F: Point;Expression(G) = (x^2/2 - y^2/2 = 1);OneOf(Focus(G)) = F;L:Line;PointOnCurve(F,L);IsPerpendicular(OneOf(Asymptote(G)),L);FootPoint(OneOf(Asymptote(G)),L)=E",
    "query_expressions": "Area(TriangleOf(E, O, F))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 38]], [[66, 69]], [[70, 73]], [[44, 47], [49, 53]], [[0, 38]], [[0, 47]], [], [[48, 62]], [[48, 62]], [[48, 69]]]",
    "query_spans": "[[[80, 99]]]",
    "process": "According to the problem, draw the figure as follows: $a^{2}=b^{2}=2$, $c^{2}=4$, thus $F(2,0)$, $OF=2$, the equation of asymptote $OE$ is $y=x$, $\\angle EOF=\\frac{\\pi}{4}$, $\\triangle EOF$ is an isosceles right triangle with right angle at $\\angle E$, $OE=EF=\\sqrt{2}$, its area is: $\\frac{1}{2}\\times\\sqrt{2}\\times\\sqrt{2}=1$"
  },
  {
    "text": "It is known that a line passing through the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ with an inclination angle of $45^{\\circ}$ intersects the right branch of the hyperbola at two points. Then, the range of the eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;H: Line;PointOnCurve(RightFocus(G), H) = True;Inclination(H) = ApplyUnit(45, degree);NumIntersection(H, RightPart(G)) = 2;e: Number;Eccentricity(G) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "(1,\\sqrt{2})",
    "fact_spans": "[[[3, 49], [73, 76], [85, 88]], [[3, 49]], [[6, 49]], [[6, 49]], [[70, 72]], [[2, 72]], [[53, 72]], [[70, 83]], [[93, 96]], [[85, 96]]]",
    "query_spans": "[[[93, 103]]]",
    "process": ""
  },
  {
    "text": "$P$ is any point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$. When the distance from $P$ to the line $x-2y-12=0$ is minimized, what are the coordinates of point $P$?",
    "fact_expressions": "G: Ellipse;H: Line;P: Point;Expression(G) = (x^2/16 + y^2/12 = 1);Expression(H) = (x - 2*y - 12 = 0);PointOnCurve(P, G);WhenMin(Distance(P,H))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2, -3)",
    "fact_spans": "[[[4, 43]], [[53, 67]], [[0, 3], [49, 52], [75, 79]], [[4, 43]], [[53, 67]], [[0, 47]], [[48, 74]]]",
    "query_spans": "[[[75, 84]]]",
    "process": ""
  },
  {
    "text": "Let $P(x_{0}, y_{0})$ be a point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, and let $F_{1}$, $F_{2}$ be the two foci of the ellipse. Then the maximum value of $\\sqrt{|P F_{1}|} \\cdot \\sqrt{|P F_{2}|}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;x0:Number;y0:Number;Expression(G) = (x^2/16 + y^2/9 = 1);Coordinate(P) = (x0, y0);PointOnCurve(P, G);Focus(G) = {F1, F2}",
    "query_expressions": "Max(sqrt(Abs(LineSegmentOf(P, F1)))*sqrt(Abs(LineSegmentOf(P, F2))))",
    "answer_expressions": "4",
    "fact_spans": "[[[19, 57], [78, 80]], [[1, 18]], [[62, 69]], [[70, 77]], [[1, 18]], [[1, 18]], [[19, 57]], [[1, 18]], [[1, 61]], [[62, 85]]]",
    "query_spans": "[[[87, 134]]]",
    "process": ""
  },
  {
    "text": "Given that the ellipse $\\frac{x^{2}}{3m^{2}}+\\frac{y^{2}}{5 n^{2}}=1$ and the hyperbola $\\frac{x^{2}}{2 m^{2}}-\\frac{y^{2}}{3 n^{2}}=1$ have common foci, what is the eccentricity of the hyperbola?",
    "fact_expressions": "G: Hyperbola;n: Number;m: Number;H: Ellipse;Expression(G) = (x^2/(2*m^2)-y^2/(3*n^2)= 1);Expression(H) = (y^2/(5*n^2) + x^2/(3*m^2) = 1);Focus(H) = Focus(G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(19)/4",
    "fact_spans": "[[[51, 101], [110, 113]], [[4, 50]], [[4, 50]], [[2, 50]], [[51, 101]], [[2, 50]], [[2, 107]]]",
    "query_spans": "[[[110, 119]]]",
    "process": ""
  },
  {
    "text": "If the eccentricity of the hyperbola $x^{2}-\\frac{y^{2}}{a}=1$ is $\\sqrt{2}$, then the value of the real number $a$ is?",
    "fact_expressions": "G: Hyperbola;a: Real;Expression(G) = (x^2 - y^2/a = 1);Eccentricity(G) = sqrt(2)",
    "query_expressions": "a",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 29]], [[46, 51]], [[1, 29]], [[1, 44]]]",
    "query_spans": "[[[46, 55]]]",
    "process": "The eccentricity of the hyperbola $x^{2}-\\frac{y^{2}}{a}=1$ is $\\sqrt{2}$, so we have $e=\\frac{\\sqrt{a+1}}{1}=\\sqrt{2}$, solving gives $a=1$,"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $x^{2}-y^{2}=2$, point $P$ lies on $C$, and $|P F_{1}|=2|P F_{2}|$, then $\\cos \\angle F_{1} P F_{2}$=?",
    "fact_expressions": "F1: Point;F2: Point;C: Hyperbola;Expression(C) = (x^2 - y^2 = 2);LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C) = True;Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Cos(AngleOf(F1, P, F2))",
    "answer_expressions": "3/4",
    "fact_spans": "[[[2, 9]], [[10, 17]], [[18, 41], [53, 56]], [[18, 41]], [[2, 47]], [[2, 47]], [[48, 52]], [[48, 57]], [[58, 80]]]",
    "query_spans": "[[[82, 111]]]",
    "process": "Hyperbola $ C: x^{2} - y^{2} = 2 $, that is $ \\frac{x^{2}}{2} - \\frac{y^{2}}{2} = 1 $, its semi-major axis length is $ a = \\sqrt{2} $, the semi-focal distance $ c $ satisfies $ c = \\sqrt{2 + 2} = 2 $. By the definition of hyperbola, $ |PF_{1}| - |PF_{2}| = 2a = 2\\sqrt{2} $, and $ |PF_{1}| = 2|PF_{2}| $, then $ |PF_{1}| = 4\\sqrt{2} $, $ |PF_{2}| = 2\\sqrt{2} $, and $ |F_{1}F_{2}| = 4 $. Thus, $ \\cos\\angle F_{1}PF_{2} = \\frac{|PF_{1}|^{2} + |PF_{2}|^{2}}{2|PF_{1}|\\cdot|PF_{2}|} = \\frac{(4\\sqrt{2})^{2} + (2\\sqrt{2})^{2}}{2 \\cdot 4\\sqrt{2} \\cdot 2\\sqrt{2}} = \\frac{3}{4} $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A perpendicular is drawn from $F_{2}$ to an asymptote, with foot of the perpendicular at $P$, and $O$ is the origin. If $\\tan \\angle P F_{2} O = \\frac{1}{3}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1:Point;F2: Point;O: Origin;a>0;b>0;L:Line;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(F2,L);IsPerpendicular(L,Asymptote(G));FootPoint(L,Asymptote(G))=P;LeftFocus(G)=F1;RightFocus(G)=F2;Tan(AngleOf(P, F2, O)) = 1/3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[2, 59], [156, 159]], [[5, 59]], [[5, 59]], [[105, 108]], [[67, 74]], [[76, 84], [87, 94]], [[109, 112]], [[5, 59]], [[5, 59]], [], [[2, 59]], [[86, 101]], [[2, 101]], [[2, 108]], [[2, 84]], [[2, 84]], [[119, 154]]]",
    "query_spans": "[[[156, 165]]]",
    "process": "Given that $ F_{2}(c,0) $, the asymptotes of the hyperbola are $ y = \\pm\\frac{b}{a}x $. As shown in the figure, the distance from the focus to the asymptote is:  \n$ PF_{2} = \\frac{|bc|}{\\sqrt{a^{2}+b^{2}}} = b $.  \nSince $ OF_{2} = c $, in right triangle $ \\triangle PF_{2}O $, $ OP = a $.  \nThus, $ \\tan\\angle PF_{2}O = \\frac{OP}{PF_{2}} = \\frac{a}{b} = \\frac{1}{3} $, so $ b = 3a $.  \nTherefore, the eccentricity is:  \n$ e = \\sqrt{1+\\left(\\frac{b}{a}\\right)^{2}} = \\sqrt{10} $.  \nThis problem examines computational and problem-solving skills, combining numerical and geometric reasoning. It is a medium-difficulty question. The key to solving this problem lies in understanding that the distance from the focus of a hyperbola to its asymptote is $ b $."
  },
  {
    "text": "If the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1(a>0)$ has an eccentricity of $2$, then what is the value of $a$?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2/3 + x^2/a^2 = 1);Eccentricity(G) = 2",
    "query_expressions": "a",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 48]], [[58, 61]], [[4, 48]], [[1, 48]], [[1, 56]]]",
    "query_spans": "[[[58, 64]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the right focus is $F$. The circle with diameter $OF$ ($O$ being the origin) intersects one asymptote of the hyperbola $C$ at another point $A$, and $\\sqrt{3}|OF|=2|OA|$. The distance from point $F$ to this asymptote is $1$. Then the standard equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;O: Origin;F: Point;A: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;IsDiameter(LineSegmentOf(O,F),G);Intersection(G,OneOf(Asymptote(C)))=A;sqrt(3)*Abs(LineSegmentOf(O, F)) = 2*Abs(LineSegmentOf(O, A));Distance(F,l) = 1;l: Line;OneOf(Asymptote(C)) = l",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x**2/3-y**2=1",
    "fact_spans": "[[[2, 63], [95, 101], [158, 164]], [[9, 63]], [[9, 63]], [[93, 94]], [[79, 82]], [[68, 71], [140, 144]], [[111, 114]], [[9, 63]], [[9, 63]], [[2, 63]], [[2, 71]], [[72, 94]], [[93, 114]], [[116, 138]], [[95, 156]], [-1, -1], [93, 104]]",
    "query_spans": "[[[158, 171]]]",
    "process": "According to the problem, connect FA and construct the figure as follows: |OA| can be obtained: |OA| = \\frac{\\sqrt{3}}{2}c', and |OF| = c, thus \\angle AOF = 30^{\\circ}, |AF| = \\frac{1}{2}c, then \\frac{b}{a} = \\tan\\angle AOF = \\frac{\\sqrt{3}}{3}, i.e., 3b^{2} = a^{2}. Since |AF| is the distance from point F to this asymptote, it follows that |AF| = 1, solving gives c = 2; also a^{2} + b^{2} = c^{2}, hence a^{2} = 3, b^{2} = 1, c^{2} = 4, then the hyperbola equation is: \\frac{x^{2}}{2} - y^{2} = 1."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ with right focus $F$, $P$ a moving point on the ellipse $C$, and fixed point $A(2 , 4)$, then the minimum value of $|P A|-|P F|$ is?",
    "fact_expressions": "C: Ellipse;A: Point;P: Point;F: Point;Expression(C) = (x^2/4 + y^2/3 = 1);RightFocus(C)=F;Coordinate(A) = (2, 4);PointOnCurve(P, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) - Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 44], [57, 62]], [[69, 79]], [[53, 56]], [[49, 52]], [[2, 44]], [[2, 52]], [[69, 79]], [[53, 66]]]",
    "query_spans": "[[[81, 100]]]",
    "process": "Let the left focus of the ellipse be F, then |PF| + |PF| = 4, it follows that |PA| - |PF| = |PA| + |PF| - 4. The expression |PA| + |PF| reaches its minimum value |AF| if and only if points P, A, and F are collinear, from which the conclusion can be drawn. As shown in the figure, let the left focus of the ellipse be F, then |PF| + |PF| = 4, so |PF| = 4 - |PF|, thus |PA| - |PF| = |PA| + |PF| - 4. If and only if P, A, F are collinear, |PA| + |PF| attains the minimum value |AF| = \\sqrt{(2+1)^{2}+16} = 5, so the minimum value of |PA| - |PF| is 1."
  },
  {
    "text": "The asymptotes of the hyperbola $C$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ are given by? The focus of a certain parabola coincides with the right focus of the hyperbola $C$, then the standard equation of this parabola is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/9 - y^2/16 = 1);G: Parabola;Focus(G) = RightFocus(C)",
    "query_expressions": "Expression(Asymptote(C));Expression(G)",
    "answer_expressions": "y=pm*4*x/3\ny^2=20*x",
    "fact_spans": "[[[0, 43], [60, 66]], [[0, 43]], [[53, 56], [75, 78]], [[53, 72]]]",
    "query_spans": "[[[0, 52]], [[75, 85]]]",
    "process": "By using the simple properties of hyperbolas and parabolas under the given conditions, we draw the conclusion. The asymptotes of the hyperbola $ C: \\frac{x^{2}}{9} - \\frac{y^{2}}{16} = 1 $ are $ y = \\pm\\frac{4}{3}x $. Since the right focus of the hyperbola $ C: \\frac{x^{2}}{9} - \\frac{y^{2}}{16} = 1 $ is $ (5,0) $, let the standard equation of this parabola be $ y^{2} = 2px $. Then $ \\frac{p}{2} = 5 $, so $ p = 10 $, hence the equation of the parabola is $ y^{2} = 20x $."
  },
  {
    "text": "The range of real values of $a$ such that the directrix of the parabola $y^{2}=4x$ intersects the curve $\\frac{x^{2}}{a}+\\frac{y^{2}}{4}=1$ $(y \\geq 0)$ at exactly one point is?",
    "fact_expressions": "G: Parabola;H: Curve;a: Real;Expression(G) = (y^2 = 4*x);Expression(H) = And(x^2/a + y^2/4 = 1, y>=0);NumIntersection(Directrix(G), H) = 1",
    "query_expressions": "Range(a)",
    "answer_expressions": "(-oo, 0) + [1, +oo)",
    "fact_spans": "[[[1, 15]], [[19, 66]], [[74, 79]], [[1, 15]], [[19, 66]], [[1, 72]]]",
    "query_spans": "[[[74, 86]]]",
    "process": "From the parabola, we know its directrix is $x = -1$. When $a > 0$, to make the directrix intersect $\\frac{x^{2}}{a} + \\frac{y^{2}}{4} = 1$ ($y \\geqslant 0$) at only one point, it suffices that $a \\geqslant 1$. When $a < 0$, the directrix always intersects $\\frac{x^{2}}{a} + \\frac{y^{2}}{4} = 1$ ($y \\geqslant 0$) at exactly one point. Thus, we can write the range of $a$. When $a > 0$, the semi-ellipse $\\frac{x^{2}}{a} + \\frac{y^{2}}{4} = 1$ ($y \\geqslant 0$), the parabola $y^{2} = 4x$, and its directrix are as follows: $\\therefore$ at this time, only when $a \\geqslant 1$ do they have exactly one intersection point; together with the case of the parabola $y^{2} = 4x$ and its directrix as shown, therefore, in conclusion, $a \\in (-\\infty, 0) \\cup [1, +\\infty)$."
  },
  {
    "text": "The distance from the point $(1,0)$ to the asymptotes of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ is?",
    "fact_expressions": "G: Hyperbola;H: Point;Expression(G) = (x^2/4 - y^2 = 1);Coordinate(H) = (1, 0)",
    "query_expressions": "Distance(H, Asymptote(G))",
    "answer_expressions": "sqrt(5)/5",
    "fact_spans": "[[[9, 37]], [[0, 8]], [[9, 37]], [[0, 8]]]",
    "query_spans": "[[[0, 46]]]",
    "process": "The asymptote equation is $ y = \\frac{b}{a}x $, that is, $ bx - ay = 0 $; thus the distance from point $ (1,0) $ to the asymptote is $ d = \\frac{|b|}{\\sqrt{b^{2} + a^{2}}} = \\frac{|}{\\sqrt{12}} $."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{k-1}+\\frac{y^{2}}{k-3}=1$ represents an ellipse, then the range of real values for $k$ is?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (x^2/(k - 1) + y^2/(k - 3) = 1)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(3, +oo)",
    "fact_spans": "[[[44, 46]], [[48, 53]], [[1, 46]]]",
    "query_spans": "[[[48, 60]]]",
    "process": "According to the problem, the equation $\\frac{x^2}{k-1} + \\frac{y^{2}}{k-3} = 1$ represents an ellipse, so it satisfies $\\begin{cases} k-1 > 0 \\\\ k-3 > 0 \\\\ k-1 \\neq k-3 \\end{cases}$, solving which yields $k > 3$, that is, the range of real number $k$ is $(3, +\\infty)$."
  },
  {
    "text": "Given that $P$ is a moving point on the parabola $x^{2}=4 y$, the distance from point $P$ to the line $y=-1$ is $d$, and the fixed point $A(2,0)$, then the minimum value of $d+|P A|$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;P: Point;d: Number;Expression(G) = (x^2 = 4*y);Expression(H) = (y = -1);Coordinate(A) = (2, 0);PointOnCurve(P, G);Distance(P, H) = d",
    "query_expressions": "Min(d + Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[6, 20]], [[30, 38]], [[48, 56]], [[2, 5], [25, 29]], [[42, 45]], [[6, 20]], [[30, 38]], [[48, 56]], [[2, 24]], [[25, 45]]]",
    "query_spans": "[[[58, 73]]]",
    "process": "From the definition of the parabola, we have $ d + |PA| = |PF| + |PA| \\geqslant |AF| = \\sqrt{5} $."
  },
  {
    "text": "Given that point $P$ lies on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola respectively, and $|F_{1} F_{2}|=\\frac{b^{2}}{a}$, $I$ is the incenter of $\\triangle P F_{1} F_{2}$. If $\\lambda S_{\\Delta I P F_{1}}= \\lambda S_{\\Delta I P F_{2}}+S_{\\Delta I  F_{1} F_{2}}$ holds, then the value of $\\lambda$ is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;I: Point;lambda:Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, RightPart(G));LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(F1, F2)) = b^2/a;Incenter(TriangleOf(P,F1,F2))=I;lambda*Area(TriangleOf(I,P,F1))=lambda*Area(TriangleOf(I,P,F2))+Area(TriangleOf(I,F1,F2))",
    "query_expressions": "lambda",
    "answer_expressions": "sqrt(2)+1",
    "fact_spans": "[[[7, 63], [87, 90]], [[10, 63]], [[10, 63]], [[2, 6]], [[69, 76]], [[77, 84]], [[129, 132]], [[253, 262]], [[10, 63]], [[10, 63]], [[7, 63]], [[2, 68]], [[69, 95]], [[69, 95]], [[97, 128]], [[129, 161]], [[163, 249]]]",
    "query_spans": "[[[253, 266]]]",
    "process": "Since $|F_{1}F_{2}|=\\frac{b^{2}}{a}$, it follows that $\\frac{b^{2}}{a}=2c$, $b^{2}=c^{2}-a^{2}=2ac$, solving this yields $e=\\sqrt{2}+1$. Let the inradius of triangle $PF_{1}F_{2}$ be $r$. From $\\lambda S_{\\triangle APF_{1}}=\\lambda S_{\\triangle AIPF_{2}}+S_{\\triangle AIF_{1}F_{2}}$, we obtain $\\lambda=\\frac{S_{\\Delta IF_{1}F_{2}}}{S_{AIPF_{1}}-S_{AIPF_{2}}}=\\frac{1}{\\frac{1}{2}}\\frac{\\frac{1}{2}\\times2c\\times r}{\\frac{1}{2}\\times|PF_{1}|\\times r-\\frac{1}{2}\\times|PF_{2}|\\times r}=\\frac{2c}{|PF_{1}|-|PF_{2}|}=\\frac{2c}{2a}=e=\\sqrt{2}+1$"
  },
  {
    "text": "Given that the difference between the distance from a moving point $M(x, y)$ to the fixed point $F(1,0)$ and to the fixed line $x=0$ is $1$. Then the trajectory equation of the moving point $M$ is?",
    "fact_expressions": "M: Point;x0: Number;y0: Number;Coordinate(M) = (x0, y0);F: Point;Coordinate(F) = (1, 0);G: Line;Expression(G) = (x = 0);Distance(M, F) - Distance(M, G) = 1",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[4, 13], [47, 50]], [[4, 13]], [[4, 13]], [[4, 13]], [[16, 24]], [[16, 24]], [[28, 33]], [[28, 33]], [[4, 42]]]",
    "query_spans": "[[[47, 57]]]",
    "process": "By the given condition, if $ x \\geqslant 0 $, the problem is equivalent to $ |MF| = |x + 1| $, then $ \\sqrt{(x - 1)^2 + y^2} = (x + 1)^2 $, simplifying yields $ y^2 = 4x $ ($ x \\geqslant 0 $). If $ x < 0 $, $ y = 0 $ also satisfies the condition. Therefore, the trajectory equation of the moving point $ M $ is $ y^2 = 4x $ ($ x \\geqslant 0 $), $ y = 0 $ ($ x < 0 $). Alternatively, according to the condition, $ |MF| - |x| = 1 $, then $ \\sqrt{(x - 1)^2 + y^2} - |x| = 1 $, simplifying and rearranging gives: $ y^2 = 2|x| + 2x $. Thus, the trajectory equation of the moving point $ M $ is $ y^2 = 2|x| + 2x $."
  },
  {
    "text": "Let a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ be $F$, and let an endpoint of the imaginary axis be $B$. The line segment $BF$ intersects an asymptote of the hyperbola at point $A$. If $\\overrightarrow{F A}=2 \\overrightarrow{A B}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;B: Point;F: Point;A: Point;a>0;b>0;OneOf(Focus(G))=F;OneOf(Endpoint(ImageinaryAxis(G)))=B;Intersection(LineSegmentOf(B,F),OneOf(Asymptote(G))) = A;VectorOf(F, A) = 2*VectorOf(A, B)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 57], [87, 90], [151, 154]], [[1, 57]], [[4, 57]], [[4, 57]], [[75, 78]], [[63, 66]], [[98, 102]], [[4, 57]], [[4, 57]], [[1, 66]], [[1, 78]], [[79, 102]], [[104, 149]]]",
    "query_spans": "[[[151, 160]]]",
    "process": ""
  },
  {
    "text": "Let $ A $ be any point on the ellipse $ \\frac{x^{2}}{25} + \\frac{y^{2}}{9} = 1 $, and $ B $ be any point on the circle $ (x-1)^{2} + y^{2} = 1 $. Then the maximum value of $ |AB| $ is? The minimum value is?",
    "fact_expressions": "G: Ellipse;H: Circle;A: Point;B: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Expression(H) = (y^2 + (x - 1)^2 = 1);PointOnCurve(A, G);PointOnCurve(B, H)",
    "query_expressions": "Max(Abs(LineSegmentOf(A,B)));Min(Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "7;3*sqrt(15)/4-1",
    "fact_spans": "[[[4, 42]], [[52, 72]], [[0, 3]], [[48, 51]], [[4, 42]], [[52, 72]], [[0, 47]], [[48, 77]]]",
    "query_spans": "[[[79, 91]], [[79, 96]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the ellipse. When $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, what is the area of $\\Delta F_{1} P F_{2}$?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 29], [59, 61]], [[38, 45]], [[54, 58]], [[46, 53]], [[2, 29]], [[2, 53]], [[2, 53]], [[54, 62]], [[64, 123]]]",
    "query_spans": "[[[125, 152]]]",
    "process": "From the given conditions, $ F_{1}F_{2}=2\\sqrt{3} $, $ \\angle F_{1}PF_{2}=90^{\\circ} $, $ |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}=12 $. Also, by the definition of an ellipse, $ |PF_{1}|+|PF_{2}|=4 $, thus $ |PF_{1}|^{2}+2|PF_{1}||PF_{2}|+|PF_{2}|^{2}=16 $. Hence, $ |PF_{1}||PF_{2}|=2 $. Therefore, the area of $ \\triangle F_{1}PF_{2} $ is $ \\frac{1}{2}|PF_{1}||PF_{2}|=1 $."
  },
  {
    "text": "Given that a hyperbola has one focus $F_{1}(0 , 5)$ and passes through the point $(0 , 4)$, then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;F1: Point;Coordinate(H) = (0, 4);Coordinate(F1) = (0, 5);OneOf(Focus(G)) = F1;PointOnCurve(H,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/16 - x^2/9 = 1",
    "fact_spans": "[[[2, 5], [40, 43]], [[27, 37]], [[10, 24]], [[27, 37]], [[10, 24]], [[2, 24]], [[2, 37]]]",
    "query_spans": "[[[39, 50]]]",
    "process": ""
  },
  {
    "text": "If the distance from a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ to an asymptote is $2 a$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Distance(OneOf(Focus(G)),OneOf(Asymptote(G)))=2*a",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[1, 47], [69, 72]], [[4, 47]], [[4, 47]], [[1, 47]], [[1, 67]]]",
    "query_spans": "[[[69, 78]]]",
    "process": "The distance from the focus of the hyperbola to the asymptote is $ b $, $\\therefore b=2a$, $\\therefore b^{2}=4a^{2}$, $\\therefore c^{2}-a^{2}=4a^{2}$, $\\therefore 5a^{2}=c^{2}$, $\\therefore e=\\sqrt{5}$"
  },
  {
    "text": "Given that $A$ is the right vertex of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{7}=1$, and $F$ is the right focus of the hyperbola, then $|A F|=$?",
    "fact_expressions": "G: Hyperbola;A: Point;F: Point;Expression(G) = (x^2/9 - y^2/7 = 1);RightVertex(G) = A;RightFocus(G) = F",
    "query_expressions": "Abs(LineSegmentOf(A, F))",
    "answer_expressions": "1",
    "fact_spans": "[[[6, 44], [53, 56]], [[2, 5]], [[49, 52]], [[6, 44]], [[2, 48]], [[49, 60]]]",
    "query_spans": "[[[62, 71]]]",
    "process": ""
  },
  {
    "text": "If the curve represented by the equation $(k-1) x^{2}+(5-2 k) y^{2}=1$ is a hyperbola, then the range of real values for $k$ is?",
    "fact_expressions": "G: Hyperbola;H: Curve;k: Real;Expression(H) = (x^2*(k - 1) + y^2*(5 - 2*k) = 1);H = G",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-oo, 1) + (5/2, +oo)",
    "fact_spans": "[[[38, 41]], [[35, 37]], [[43, 48]], [[1, 37]], [[35, 41]]]",
    "query_spans": "[[[43, 55]]]",
    "process": "According to the characteristics of the hyperbola equation, set up the inequality and solve. Since the equation $(k-1)x^{2}+(5-2k)y^{2}=1$ represents a hyperbola, it follows that $(k-1)(5-2k)<0$, which is equivalent to $(k-1)(2k-5)>0$. Solving yields $k\\in(-\\infty,1)\\cup(\\frac{5}{2},+\\infty)$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b<0)$, the right vertex and right focus are denoted as $A$ and $F$, respectively. The intersection point of its left directrix with the $x$-axis is $B$. If $A$ is the midpoint of segment $B F$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b<0;A: Point;RightVertex(C) = A;F: Point;RightFocus(C) = F;Intersection(LeftDirectrix(C), xAxis) = B;B: Point;MidPoint(LineSegmentOf(B, F)) = A",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)+1",
    "fact_spans": "[[[2, 65], [84, 85], [119, 125]], [[2, 65]], [[9, 65]], [[9, 65]], [[9, 65]], [[9, 65]], [[76, 79], [103, 106]], [[2, 83]], [[80, 83]], [[2, 83]], [[84, 101]], [[98, 101]], [[103, 117]]]",
    "query_spans": "[[[119, 131]]]",
    "process": ""
  },
  {
    "text": "Given a parabola $C$: $x^{2}=4 y$, draw two tangent lines $PA$ and $PB$ from an arbitrary point $P$ on the directrix of the parabola, with points of tangency $A$ and $B$ respectively. Then, the minimum value of the sum of the distances from point $A$ to the directrix and from point $B$ to the directrix is?",
    "fact_expressions": "C: Parabola;P: Point;A: Point;B: Point;Expression(C) = (x^2 = 4*y);PointOnCurve(P,Directrix(C));TangentOfPoint(P,C)={LineOf(P,A),LineOf(P,B)};TangentPoint(LineOf(P,A),C)=A;TangentPoint(LineOf(P,B),C)=B",
    "query_expressions": "Min(Distance(A,Directrix(C))+Distance(B,Directrix(C)))",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 20], [32, 35]], [[28, 31]], [[57, 60], [66, 70]], [[61, 64], [77, 81]], [[1, 20]], [[1, 31]], [[0, 51]], [[32, 64]], [[32, 64]]]",
    "query_spans": "[[[32, 95]]]",
    "process": "First, find the equations of lines PA and PB. Solving them simultaneously gives $x_{P}=\\frac{x_{1}+x_{2}}{2}$. From the fact that point P is the common point of the two tangents, the equation of AB is obtained as $y=\\frac{x\\cdot x_{P}}{2}+1$. Expressing the sum of the distances from points A and B to the directrix and simplifying yields $\\underline{(x_{1}+x_{2})^{2}}+4$, thus finding the minimum value. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then the equations of lines PA and PB are respectively $y=\\frac{x_{1}}{2}x-\\frac{x_{1}^{2}}{4}$, $y=\\frac{x_{2}}{2}x-\\frac{x_{2}^{2}}{4}$. Solving them simultaneously gives $x_{P}=\\frac{x_{1}+x_{2}}{2}$, $y_{P}=\\frac{x_{1}\\cdot x_{2}}{4}$. Also, the equations of lines PA and PB can be written as $y=\\frac{x_{1}}{2}x-y_{1}$, $y=\\frac{x_{2}}{2}x-y_{2}$. Substituting the coordinates of point P into the two equations gives $\\begin{cases}y_{P}=\\frac{x_{1}}{2}.\\end{cases}\\frac{1\\cdot x_{P}}{2}-y_{1}$, $\\begin{matrix}x_{2}\\cdot x_{P}&y_{2},\\\\y_{P}&2\\end{matrix}-y_{2}$. Therefore, the equation of line AB is $\\frac{x\\cdot x_{P}}{2}-y=-1$, i.e., $y=-\\frac{x\\cdot x_{P}^{2}}{}+1$. Hence, the sum of the distances from point A and point B to the directrix is $y+y,+2=(\\frac{x_{P}}{2}x_{1}+1)+(\\frac{x_{P}}{2}x_{2}+1)+2=\\frac{x_{P}}{2}(x_{1}+x_{2})+4=\\frac{(x_{1}+x_{2})^{2}}{4}+4\\geqslant4$"
  },
  {
    "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are perpendicular to each other, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;L1: Line;L2: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Asymptote(G) = {L1, L2};IsPerpendicular(L1, L2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 57], [68, 71]], [[4, 57]], [[4, 57]], [], [], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 61]], [[1, 65]]]",
    "query_spans": "[[[68, 77]]]",
    "process": "From the given conditions, it can be deduced that the hyperbola is equilateral, thus $ a = b $, and the eccentricity can be further determined. Since the asymptotes of the hyperbola are perpendicular to each other, it is an equilateral hyperbola, i.e., $ a = b $, so $ c = \\sqrt{2}a $. Therefore, the eccentricity $ e = \\frac{c}{a} = \\sqrt{2} $."
  },
  {
    "text": "The distance from the focus of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ to its asymptote is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1)",
    "query_expressions": "Distance(Focus(G), Asymptote(G))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 42]]]",
    "process": ""
  },
  {
    "text": "What is the eccentricity of the hyperbola $C$: $\\frac{x^{2}}{2}-y^{2}=1$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/2 - y^2 = 1)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[0, 33]], [[0, 33]]]",
    "query_spans": "[[[0, 39]]]",
    "process": ""
  },
  {
    "text": "Hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has left and right foci $F_{1}$, $F_{2}$ respectively, with focal length $2c$. A circle centered at the right vertex $A$ with radius $\\frac{a+c}{2}$ is tangent to a line $l$ passing through $F_{1}$ at point $N$. Let the intersection points of $l$ and $C$ be $P$, $Q$. If $\\overrightarrow{P Q}=2 \\overrightarrow{P N}$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;c: Number;G: Circle;P: Point;Q: Point;N: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;FocalLength(C) = 2*c;A: Point;RightVertex(C) = A;Center(G) = A;Radius(G) = (a + c)/2;PointOnCurve(F1, l);TangentPoint(G, l) = N;Intersection(l, C) = {P, Q};VectorOf(P, Q) = 2*VectorOf(P, N)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[136, 141], [150, 153]], [[0, 61], [154, 157], [217, 223]], [[8, 61]], [[8, 61]], [[89, 94]], [[125, 126]], [[161, 164]], [[165, 168]], [[144, 148]], [[70, 77], [128, 135]], [[78, 85]], [[8, 61]], [[8, 61]], [[0, 61]], [[0, 85]], [[0, 85]], [[0, 94]], [[99, 102]], [[0, 102]], [[99, 126]], [[106, 126]], [[125, 143]], [[125, 148]], [[150, 168]], [[170, 215]]]",
    "query_spans": "[[[217, 229]]]",
    "process": "Since the circle centered at the right vertex A with radius $\\frac{a+c}{2}$ is tangent to the line $l$ passing through $F_{1}$ at point N, and $AF_{1}=a+c$, it follows that the inclination angle of the line is $30^{0}$. Let the equations be $\\begin{cases}x=\\sqrt{3}y-c\\\\\\frac{x^{2}}{a}-\\frac{y^{2}}{b}=1\\end{cases}\\Rightarrow(3b^{2}-a^{2})y^{2}-2\\sqrt{3}b^{2}cy+b^{4}=0$. Let points $P(x_{1},y_{1})$, $Q(x_{2},y_{2})$. According to the condition $\\overrightarrow{PQ}=2\\overrightarrow{PN}$, N is the midpoint of PQ, thus $N(\\frac{a^{2}c}{3b^{2}-a^{2}},\\frac{\\sqrt{3}b^{2}c}{3b^{2}-a^{2}})$. Since $NA\\perp l \\Rightarrow \\frac{\\frac{\\sqrt{3}b^{2}c}{3b^{2}-a^{2}}}{\\frac{a^{2}c}{3b^{2}-a^{2}}-a}=-\\sqrt{3}$, we obtain $e^{3}-3e^{2}+4=0 \\Rightarrow e=2$."
  },
  {
    "text": "Given the parabola $y = a x^{2}$ ($a > 0$) with directrix $l$, if the chord length obtained by the intersection of $l$ and the circle $C$: $(x - 3)^{2} + y^{2} = 1$ is $\\sqrt{3}$, then $a = $?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = a*x^2);a: Number;a>0;l: Line;Directrix(G) = l;C: Circle;Expression(C) = (y^2 + (x - 3)^2 = 1);Length(InterceptChord(l,C)) = sqrt(3)",
    "query_expressions": "a",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 21]], [[2, 21]], [[78, 81]], [[5, 21]], [[25, 28], [31, 34]], [[2, 28]], [[35, 59]], [[35, 59]], [[31, 76]]]",
    "query_spans": "[[[78, 83]]]",
    "process": "The directrix of the parabola $ y = ax^{2} $ ($ a > 0 $) is $ l: y = -\\frac{1}{4a} $. Therefore, the distance from the center of the circle $ (3, 0) $ to it is $ d = \\sqrt{1 - \\left( \\frac{\\sqrt{3}}{2} \\right)^{2}} = \\frac{1}{2} $. Hence, $ \\frac{1}{4a} = \\frac{1}{2} $, so $ a = \\frac{1}{2} $."
  },
  {
    "text": "What is the standard equation of a parabola passing through the point $(1 , 2)$ with its focus on the $x$-axis?",
    "fact_expressions": "G: Parabola;H: Point;Coordinate(H) = (1, 2);PointOnCurve(H, G);PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = 4*x",
    "fact_spans": "[[[22, 25]], [[2, 12]], [[2, 12]], [[0, 25]], [[13, 25]]]",
    "query_spans": "[[[22, 32]]]",
    "process": ""
  },
  {
    "text": "The tangent to the parabola $y=x^{2}$ at the point ? is parallel to the line $y=4 x-5$",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Expression(G) = (y = x^2);Expression(H) = (y = 4*x - 5);IsParallel(TangentOnPoint(P, G), H)",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2,4)",
    "fact_spans": "[[[0, 12]], [[22, 33]], [[13, 14]], [[0, 12]], [[22, 33]], [[0, 33]]]",
    "query_spans": "[[[13, 15]]]",
    "process": ""
  },
  {
    "text": "Given that $O$ is the coordinate origin, and point $P$ is a moving point on the curve $D$: $2 x^{2}+x y-3 y^{2}=5$, then the equation of the trajectory of the midpoint $M$ of segment $O P$ is?",
    "fact_expressions": "D: Curve;O: Origin;P: Point;Expression(D) = (-3*y^2 + 2*x^2 + x*y = 5);PointOnCurve(P, D);MidPoint(LineSegmentOf(O,P))=M;M:Point",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "8*x^2+4*x*y-12*y^2=5",
    "fact_spans": "[[[16, 46]], [[2, 5]], [[11, 15]], [[16, 46]], [[11, 50]], [[52, 64]], [[61, 64]]]",
    "query_spans": "[[[61, 71]]]",
    "process": "Let M(x, y), P(x_{0}, y_{0}). Since M is the midpoint of segment OP, we have \\begin{cases}x=\\frac{x_{0}}{2}\\\\y=\\frac{y_{0}}{2}\\Rightarrow\\\\x_{0}=2x\\end{cases}. Since point P lies on the curve D: 2x^{2}+xy-3y^{2}=5, it follows that 2(2x)^{2}+4xy-3(2y)^{2}=5, i.e., 8x^{2}+4xy-12y^{2}=5."
  },
  {
    "text": "The line $l$ passes through the focus $F$ of the parabola $y^{2}=2 p x(p>0)$, intersects the parabola at points $A$ and $B$, and intersects the line $x=-\\frac{p}{2}$ at point $M$. If $\\overrightarrow{A F}=\\overrightarrow{F M}$ and $|A B|=\\frac{16}{3}$, then the equation of the parabola is?",
    "fact_expressions": "l: Line;G: Parabola;p: Number;H: Line;A: Point;F: Point;M: Point;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (x = -p/2);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};Intersection(l, H) = M;VectorOf(A, F) = VectorOf(F, M) ;Abs(LineSegmentOf(A, B)) = 16/3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = 4*x",
    "fact_spans": "[[[0, 5]], [[7, 28], [36, 39], [146, 149]], [[10, 28]], [[52, 70]], [[41, 44]], [[31, 34]], [[72, 76]], [[45, 48]], [[10, 28]], [[7, 28]], [[52, 70]], [[7, 34]], [[0, 34]], [[0, 50]], [[0, 76]], [[78, 122]], [[124, 144]]]",
    "query_spans": "[[[146, 154]]]",
    "process": "From the equation of the parabola, the coordinates of the focus F can be obtained. From the vector relationship, it follows that F is the midpoint of AM, so the x-coordinate of A can be found. Substituting into the parabola's equation gives the y-coordinate of A. Then, the equation of line AB is derived and solved together with the parabola to find the sum of the roots. Then, using the properties of the parabola, the value of AB is obtained. According to the given conditions, the value of p is determined, and thus the equation of the parabola is found. As shown in the figure, since $\\overrightarrow{AF}=\\overrightarrow{FM}$, F is the midpoint of AM, so $AF=AA=|NF|=2p$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$. Therefore, $2p=x_{1}+\\frac{p}{2}$, so $x_{1}=\\frac{3p}{2}$. Substituting into the parabola's equation yields $y_{1}=\\sqrt{3}p$, i.e., $A(\\frac{3p}{2},\\sqrt{3}p)$. Thus, $k_{AB}=\\frac{\\sqrt{3}p}{\\frac{3p}{2}-\\frac{p}{2}}=\\sqrt{3}$, so the equation of line AB is: $y=\\sqrt{3}(x-\\frac{p}{2})$. Solving this line equation together with the parabola's equation gives: after simplifying, $3x^{2}-5px+\\frac{3p^{2}}{4}=0$, so $x_{1}+x_{2}=\\frac{5p}{3}$. From the property of the parabola, $AB=x_{1}+x_{2}+p=\\frac{5p}{3}+p=\\frac{16}{3}$, solving gives $p=2$. Therefore, the equation of the parabola is: $y^2=4x$. Answer: $y^2=4x$. This problem involves the relationship between vectors and point positions, as well as the properties of a parabola, and is of medium difficulty."
  },
  {
    "text": "Given the line $y=b$ intersects an asymptote of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ at point $P$, the left and right vertices of hyperbola $C$ are $A_{1}$ and $A_{2}$, respectively, and the left focus is $F$. If $|P F|=|A_{1} A_{2}|$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Line;Expression(G) = (y = b);C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;Intersection(G,OneOf(Asymptote(C))) = P;P: Point;LeftVertex(C) = A1;RightVertex(C) = A2;A1: Point;A2: Point;F: Point;LeftFocus(C) = F;Abs(LineSegmentOf(P, F)) = Abs(LineSegmentOf(A1, A2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 9]], [[2, 9]], [[10, 71], [84, 90], [149, 152]], [[10, 71]], [[18, 71]], [[18, 71]], [[18, 71]], [[18, 71]], [[2, 83]], [[79, 83]], [[84, 114]], [[84, 114]], [[99, 106]], [[107, 114]], [[119, 122]], [[84, 122]], [[125, 146]]]",
    "query_spans": "[[[149, 158]]]",
    "process": "First, find the asymptotes given by $ y = \\frac{b}{a}x $ or $ y = -\\frac{b}{a}x $, then the coordinates of point $ P $ are $ P_{1}(a,b) $ or $ P_{2}(-a,b) $. Since the x-coordinate of point $ P $ equals the x-coordinates of the left and right vertices $ A_{1}, A_{2} $, it follows that $ P_{1}A_{2} \\perp FA_{2} $, $ P_{2}A_{1} \\perp FA_{2} $. Applying the Pythagorean theorem in right triangles $ \\triangle PFA_{2} $ and $ \\triangle P_{2}A_{1}F $, and using $ b^{2} = c^{2} - a^{2} $, we can solve as follows. When point $ P $ lies on $ y = \\frac{b}{a}x $, from \n$$\n\\begin{cases}\ny = b \\\\\ny = \\frac{b}{a}x\n\\end{cases}\n$$\nwe get $ P_{1}(a,b) $, and $ P_{1}A_{2} \\perp FA_{2} $. In right triangle $ \\triangle PFA_{2} $, $ |PF| = |A_{1}A_{2}| = 2a $, $ |P_{1}A_{2}| = b $, $ |P_{1}A_{2}|^{2} + |A_{2}F|^{2} = |PF|^{2} $, so $ b^{2} + (a+c)^{2} = 4a^{2} $, i.e., $ b^{2} + a^{2} + c^{2} + 2ac = 4a^{2} $. Since $ b^{2} = c^{2} - a^{2} $, we have $ c^{2} + ac - 2a^{2} = 0 $, i.e., $ e^{2} + e - 2 = 0 $, solving gives $ e = 1 $ or $ e = -2 $, which does not satisfy the conditions. When point $ P $ lies on $ y = -\\frac{b}{a}x $, from \n$$\n\\begin{cases}\ny = b \\\\\ny = -\\frac{b}{a}x\n\\end{cases}\n$$\nwe obtain $ P_{2}(-a,b) $, and $ P_{2}A_{1} \\perp FA_{2} $. In right triangle $ \\triangle P_{2}A_{1}F $, $ |P_{2}F| = |A_{1}A_{2}| = 2a $, $ |P_{2}A_{1}| = b $, $ |A_{1}F| = c - a $, $ |P_{2}A_{1}|^{2} + |A_{1}F|^{2} = |P_{2}F|^{2} $, so $ b^{2} + (c - a)^{2} = 4a^{2} $, i.e., $ c^{2} - a^{2} + c^{2} + a^{2} - 2ac = 4a^{2} $, thus $ c^{2} - ac - 2a^{2} = 0 $, so $ e^{2} - e - 2 = 0 $, solving gives $ e = 2 $ or $ e = -1 $ (discarded)."
  },
  {
    "text": "The left focus of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ is $F_{1}$, $P$ is a moving point on the ellipse, and $M$ is a moving point on the circle $x^{2}+(y-2 \\sqrt{5})^{2}=1$. Then the maximum value of $|P M|+|P F_{1}|$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);F1: Point;LeftFocus(G) = F1;P: Point;PointOnCurve(P, G) = True;H: Circle;Expression(H) = (x^2 + (y - 2*sqrt(5))^2 = 1);M: Point;PointOnCurve(M, H) = True",
    "query_expressions": "Max(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "17",
    "fact_spans": "[[[0, 38], [55, 57]], [[0, 38]], [[43, 50]], [[0, 50]], [[51, 54]], [[51, 61]], [[66, 95]], [[66, 95]], [[62, 65]], [[62, 99]]]",
    "query_spans": "[[[101, 124]]]",
    "process": "The circle $x^{2}+(y-2\\sqrt{5})^{2}=1$ has center $C(0,2\\sqrt{5})$ and radius $1$. From the ellipse equation $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}-1$, we know $a^{2}=25$, $b^{2}=9$, so $a=5$, left focus $F(-4,0)$. $|PC|+|PF_{1}|=|PC|+2a-|PF_{2}|=10+|PC|-|PF_{2}|\\leqslant10+|CF_{2}|=10+\\sqrt{4^{2}+(2\\sqrt{5})^{2}}=16$. $(|PM|+|PF_{1})_{\\max}=(|PC|+|PF_{1}|)_{\\max}+1=17$"
  },
  {
    "text": "Let the right vertex of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ be $A$, and the right focus be $F$. A line passing through point $F$ and parallel to one asymptote of the hyperbola intersects the other asymptote at point $B$. Then the area of $\\triangle A F B$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;A: Point;F: Point;B: Point;L1:Line;L2:Line;Expression(G) = (x^2/9 - y^2/16 = 1);RightVertex(G)=A;RightFocus(G)=F;PointOnCurve(F, H);OneOf(Asymptote(G)) = L1;OneOf(Asymptote(G)) = L2;Negation(L1 = L2);IsParallel(L1,H);Intersection(H,L2) = B",
    "query_expressions": "Area(TriangleOf(A, F, B))",
    "answer_expressions": "10/3",
    "fact_spans": "[[[1, 40], [64, 67]], [[76, 78]], [[45, 48]], [[53, 56], [58, 62]], [[87, 91]], [], [], [[1, 40]], [[1, 48]], [[1, 56]], [[57, 78]], [[64, 73]], [[64, 85]], [[64, 85]], [[63, 78]], [[64, 91]]]",
    "query_spans": "[[[93, 115]]]",
    "process": ""
  },
  {
    "text": "Given point $A(0 , 2)$, the focus of the parabola $y^{2}=2 px(p>0)$ is $F$, the directrix is $l$, the line segment $FA$ intersects the parabola at point $B$, and a perpendicular is drawn from $B$ to $l$ with foot $M$. If $A M \\perp M F$, then $p=$?",
    "fact_expressions": "F: Point;A: Point;G: Parabola;p: Number;M: Point;l: Line;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(A) = (0, 2);Directrix(G)=l;Focus(G) = F;Intersection(LineSegmentOf(F, A), G) = B;L:Line;IsPerpendicular(l,L);PointOnCurve(B,L);FootPoint(l,L)=M;IsPerpendicular(LineSegmentOf(A, M), LineSegmentOf(M, F))",
    "query_expressions": "p",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[37, 40]], [[2, 13]], [[13, 33], [55, 58]], [[101, 104]], [[79, 82]], [[44, 47]], [[59, 63], [65, 68]], [[16, 33]], [[13, 33]], [[2, 13]], [[13, 47]], [[13, 40]], [[48, 63]], [], [[64, 75]], [[64, 75]], [[64, 82]], [[84, 99]]]",
    "query_spans": "[[[101, 106]]]",
    "process": ""
  },
  {
    "text": "The standard equation of a hyperbola with focal length $10$, conjugate axis length $8$, and passing through the point $(3 \\sqrt{2}, 4)$ is?",
    "fact_expressions": "E: Hyperbola;F: Point;Coordinate(F) = (3*sqrt(2), 4);FocalLength(E) = 10;Length(ImageinaryAxis(E)) = 8;PointOnCurve(F, E)",
    "query_expressions": "Expression(E)",
    "answer_expressions": "x**2/9-y**2/16=1",
    "fact_spans": "[[[37, 40]], [[17, 36]], [[17, 36]], [[0, 40]], [[8, 40]], [[16, 40]]]",
    "query_spans": "[[[37, 47]]]",
    "process": ""
  },
  {
    "text": "The standard equation of a parabola with vertex at the origin and passing through the point $P(-2 , 3)$ is?",
    "fact_expressions": "G: Parabola;P: Point;Coordinate(P) = (-2, 3);PointOnCurve(P, G);O: Origin;Vertex(G) = O",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2 = (-9/2)*x, x^2 = (4/3)*y}",
    "fact_spans": "[[[21, 24]], [[8, 20]], [[8, 20]], [[7, 24]], [[3, 5]], [[0, 24]]]",
    "query_spans": "[[[21, 31]]]",
    "process": "Let the standard equation of the parabola be $ y^{2} = 2mx $ or $ x^{2} = 2my $. Since point $ P(-2,3) $ lies on the parabola, then $ 3^{2} = 2m \\cdot (-2) \\Rightarrow m = -\\frac{9}{4} $ or $ (-2)^{2} = 2m \\cdot 3 \\Rightarrow m = \\frac{2}{3} $. Therefore, the standard equation of the parabola is $ y^{2} = -\\frac{9}{2}x $ or $ x^{2} = \\frac{4}{3}y $."
  },
  {
    "text": "Through the right focus $F$ of the hyperbola $x^{2}-y^{2}=4$, draw a line with an inclination angle of $105^{\\circ}$, intersecting the hyperbola at points $P$ and $Q$. Then the value of $|FP| \\cdot |FQ|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 4);F: Point;RightFocus(G) = F;H: Line;PointOnCurve(F, H);Inclination(H) = ApplyUnit(105, degree);P: Point;Q: Point;Intersection(H, G) = {P, Q}",
    "query_expressions": "Abs(LineSegmentOf(F, P))*Abs(LineSegmentOf(F, Q))",
    "answer_expressions": "8*sqrt(3)/3",
    "fact_spans": "[[[1, 19], [49, 52]], [[1, 19]], [[23, 26]], [[1, 26]], [[45, 47]], [[0, 47]], [[27, 47]], [[53, 56]], [[57, 60]], [[45, 62]]]",
    "query_spans": "[[[64, 84]]]",
    "process": ""
  },
  {
    "text": "A line passing through the focus of the parabola $y^{2}=4 x$ intersects the parabola at points $A$ and $B$, and the x-coordinate of point $A$ is $4$. The line passing through point $A$ and the vertex of the parabola intersects the directrix of the parabola at point $C$. Then, the area of $\\triangle A B C$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;C: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), H);Intersection(H, G) = {A, B};XCoordinate(A) = 4;L:Line;PointOnCurve(A,L);PointOnCurve(Vertex(G),L);Intersection(L,Directrix(G))=C",
    "query_expressions": "Area(TriangleOf(A, B, C))",
    "answer_expressions": "25/8",
    "fact_spans": "[[[1, 15], [22, 25], [55, 58], [64, 67]], [[19, 21]], [[26, 30], [36, 40], [50, 54]], [[31, 34]], [[71, 75]], [[1, 15]], [[0, 21]], [[19, 34]], [[36, 48]], [[61, 63]], [[49, 63]], [[49, 63]], [[61, 76]]]",
    "query_spans": "[[[77, 99]]]",
    "process": "Assume point A is in the first quadrant, let point A be (4,n), where n>0, then n^{2}=4\\times4=16, so n=4. Thus, point A is (4,4). The focus of the parabola y^{2}=4x is F(1,0), k_{AF}=\\frac{4}{4-1}=\\frac{4}{3}. Therefore, the equation of line AB is y=\\frac{4}{3}(x-1). Solving the system \\begin{cases}y=\\frac{4}{3}(x-1)\\\\y^{2}=4x\\end{cases}, we obtain \\begin{cases}x=4\\\\y=4\\end{cases} or \\begin{cases}x=\\frac{1}{4}\\\\y=-1\\end{cases}, so point B is (\\frac{1}{4},-1). Hence, |AB|=\\sqrt{(4-\\frac{1}{4})^{2}+(4+1)^{2}}=\\frac{25}{4}. The equation of line AC is y=x. The directrix of the parabola is x=-1. Solving the system \\begin{cases}y=x\\\\x=-1\\end{cases}, we get point C(-1,-1). The distance from point C to line AB: 4x-3y-4=0 is d=\\frac{|-4+3-4|}{5}=1. Therefore, S_{\\triangle ABC}=\\frac{1}{2}|AB|\\cdot d=\\frac{1}{2}\\times\\frac{25}{4}\\times1=\\frac{25}{8}."
  },
  {
    "text": "If a line passing through the point $P(8,1)$ intersects the hyperbola $x^{2}-4 y^{2}=4$ at points $A$ and $B$, and $P$ is the midpoint of segment $AB$, then the equation of line $AB$ is?",
    "fact_expressions": "P: Point;G: Hyperbola;A: Point;B: Point;H: Line;Expression(G) = (x^2 - 4*y^2 = 4);Coordinate(P) = (8, 1);PointOnCurve(P, H) = True;Intersection(H, G) = {A , B};MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Expression(LineOf(A, B))",
    "answer_expressions": "2*x-y-15=0",
    "fact_spans": "[[[2, 11], [50, 53]], [[15, 35]], [[38, 41]], [[43, 46]], [[12, 14]], [[15, 35]], [[2, 11]], [[1, 14]], [[12, 48]], [[50, 63]]]",
    "query_spans": "[[[65, 76]]]",
    "process": ""
  },
  {
    "text": "There is a moving point $A$ on the parabola with vertex at the origin and focus $F(0 , 1)$, and a fixed point $M(-1 , 4)$. Then the minimum value of $|AM| + |AF|$ is?",
    "fact_expressions": "G: Parabola;F: Point;M: Point;A: Point;O:Origin;Coordinate(F) = (0, 1);Coordinate(M) = (-1, 4);Vertex(G)=O;Focus(G)=F;PointOnCurve(A,G)",
    "query_expressions": "Min(Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(A, M)))",
    "answer_expressions": "5",
    "fact_spans": "[[[22, 25]], [[11, 21]], [[36, 47]], [[30, 33]], [[3, 7]], [[11, 21]], [[36, 47]], [[0, 25]], [[8, 25]], [[22, 33]]]",
    "query_spans": "[[[49, 68]]]",
    "process": "Let the distance from point A to the directrix be |AE|; by definition, |AF| = |AE|, so |AM| + |AF| = |AE| + |AM| \\geqslant |ME| > |MN| = 4 + 1 (let N be the foot of the perpendicular from M to the directrix). The equality holds when points M, A, E are collinear; therefore, the minimum value of |AM| + |AF| equals 5."
  },
  {
    "text": "The line $l$ passes through the focus $F$ of the parabola $C$: $y^{2}=2 p x$ ($p>0$), intersects the parabola at points $A$ and $B$, and intersects its directrix at point $D$. If $|A F|=6$ and $\\overrightarrow{D B}=2 \\overrightarrow{B F}$, then $p=$?",
    "fact_expressions": "l: Line;C: Parabola;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;p: Number;A: Point;F: Point;D: Point;B: Point;p>0;Intersection(l,C)={A,B};Intersection(l,Directrix(C))=D;PointOnCurve(F, l);Abs(LineSegmentOf(A, F)) = 6;VectorOf(D, B) = 2*VectorOf(B, F)",
    "query_expressions": "p",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 5]], [[6, 32], [40, 43], [56, 57]], [[6, 32]], [[6, 38]], [[126, 129]], [[45, 48]], [[35, 38]], [[61, 65]], [[49, 52]], [[13, 32]], [[0, 54]], [[0, 65]], [[0, 38]], [[67, 76]], [[79, 124]]]",
    "query_spans": "[[[126, 131]]]",
    "process": ""
  },
  {
    "text": "Given that point $P(2, t)$ lies on the parabola $x^{2} = 4y$, and $M$, $N$ are two points on the parabola distinct from $P$. If the sum of the slopes of lines $PM$ and $PN$ is $\\frac{3}{2}$, and $Q$ is the midpoint of segment $MN$, then the range of the distance $d$ from point $Q$ to the origin is?",
    "fact_expressions": "t: Number;P: Point;Coordinate(P) = (2, t);G: Parabola;Expression(G) = (x^2 = 4*y);PointOnCurve(P, G);M: Point;N: Point;PointOnCurve(M, G);PointOnCurve(N, G);Negation(M=P);Negation(N=P);Slope(LineOf(P, M))+Slope(LineOf(P, N)) = 3/2;Q: Point;MidPoint(LineSegmentOf(M, N)) = Q;d: Number;O: Origin;Distance(Q, O) = d",
    "query_expressions": "Range(d)",
    "answer_expressions": "(\\sqrt{17}/4, +\\infty)",
    "fact_spans": "[[[2, 12]], [[2, 12], [45, 48]], [[2, 12]], [[13, 27], [39, 42]], [[13, 27]], [[2, 30]], [[31, 34]], [[35, 38]], [[31, 51]], [[31, 51]], [[31, 51]], [[31, 51]], [[53, 87]], [[99, 102], [104, 108]], [[88, 102]], [[116, 119]], [[109, 113]], [[104, 119]]]",
    "query_spans": "[[[116, 126]]]",
    "process": "Given that the sum of the slopes of line PM and line PN is \\frac{3}{2}, we obtain x_{1}+x_{2}=2, and determine the range of x_{1}^{2}+x_{2}^{2} to arrive at the detailed solution. From the problem, we have P(2,1), let M(x_{1},\\frac{1}{4}x_{1}), N(x_{2},\\frac{1}{4}x_{2}^{2}), Q(\\frac{x_{1}+x_{2}}{2},\\frac{1}{2}\\frac{x_{1}^{2}}{2}\\frac{1}{2})^{2}. Since M and N are two points on the parabola P, and the sum of the slopes of line PM and line PN is \\frac{3}{2}, it follows that \\frac{1}{x_{1}-2}+\\frac{1}{x_{2}-2}=\\frac{3}{2}\\frac{+(\\frac{1}{4}x_{2}^{2}-1)(x_{1}-2)}{(x_{2}-2)}=\\frac{3}{2}. Simplifying yields x_{1}+x_{2}=2. Since 2=4-2x_{1}x_{2}\\geqslant^{2}4-2(\\frac{x_{1}+x_{2}}{2})^{2}=2, thus \\frac{1}{2}.\\frac{2}{\\sqrt{17}+[\\frac{1}{8}(x^{2}+x^{2})}]^{2}}\\geqslant\\sqrt{1+\\frac{1}{16}}=\\frac{\\sqrt{17}}{4}. This problem mainly examines the comprehensive topic of parabolas and lines, involving using the AM-GM inequality to find extrema."
  },
  {
    "text": "The standard equation of a hyperbola that shares the same asymptotes as the hyperbola $\\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1$ and has a focal length of $12$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/5 - y^2/4 = 1);C: Hyperbola;Asymptote(C) = Asymptote(G);FocalLength(C) = 12",
    "query_expressions": "Expression(C)",
    "answer_expressions": "{x^2/20-y^2/16=1, y^2/16-x^2/20=1}",
    "fact_spans": "[[[1, 39]], [[1, 39]], [[54, 57]], [[0, 57]], [[46, 57]]]",
    "query_spans": "[[[54, 64]]]",
    "process": ""
  },
  {
    "text": "It is known that the line $y=1-x$ intersects the asymptotes of the hyperbola $a x^{2}+b y^{2}=1$ $(a>0, b<0)$ at points $A$ and $B$, and the slope of the line passing through the origin and the midpoint of segment $AB$ is $-\\frac{\\sqrt{3}}{2}$. Then the value of $\\frac{a}{b}$ is?",
    "fact_expressions": "H: Line;Expression(H) = (y = 1 - x);G: Hyperbola;Expression(G) = (a*x^2 + b*y^2 = 1);b: Number;a: Number;a>0;b<0;Intersection(H, Asymptote(G)) = {A, B};A: Point;B: Point;O: Origin;Z: Line;PointOnCurve(O, Z) ;PointOnCurve(MidPoint(LineSegmentOf(A, B)), Z) ;Slope(Z) = -sqrt(3)/2",
    "query_expressions": "a/b",
    "answer_expressions": "-sqrt(3)/2",
    "fact_spans": "[[[2, 11]], [[2, 11]], [[12, 47]], [[12, 47]], [[15, 47]], [[15, 47]], [[15, 47]], [[15, 47]], [[2, 62]], [[53, 56]], [[57, 60]], [[65, 67]], [[78, 80]], [[64, 80]], [[64, 80]], [[78, 105]]]",
    "query_spans": "[[[107, 124]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, and the midpoint of $ AB $ be $ M(x_{0},y_{0}) $. Thus,\n$$\n\\begin{cases}\nax_{1}^{2}+by_{1}^{2}=1 \\\\\nax_{2}^{2}+by_{2}^{2}=1\n\\end{cases}\n$$\nSo $ a(x_{1}-y_{1})(x_{1}+y_{1})+b(x_{2}-y_{2})(x_{2}+y_{2})=0 $, that is, $ a(x_{1}-x_{2})x_{0}+b(y_{1}-y_{2})y_{0}=0 $, so\n$$\na+b\\times\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\\times\\frac{y_{0}}{x_{0}}=0\n$$\nSince the slope of the line passing through the origin and the midpoint of segment $ AB $ is $ -\\frac{\\sqrt{3}}{2} $, we have $ \\frac{y_{0}}{x_{0}}=-\\frac{\\sqrt{3}}{2} $. From $ AB: y=-x+1 $, we get $ \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-1 $, so\n$$\na+b\\times(-1)\\times\\left(-\\frac{\\sqrt{3}}{2}\\right)=0\n$$\nThus, $ \\frac{a}{b}=-\\frac{\\sqrt{3}}{2} $."
  },
  {
    "text": "Given that an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is tangent to the circle $(x-4)^{2}+(y-3)^{2}=5$, find the eccentricity of this hyperbola.",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = ((x - 4)^2 + (y - 3)^2 = 5);IsTangent(OneOf(Asymptote(G)),H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{5*sqrt(5)/11,sqrt(5)}",
    "fact_spans": "[[[2, 58], [94, 97]], [[5, 58]], [[5, 58]], [[65, 89]], [[5, 58]], [[5, 58]], [[2, 58]], [[65, 89]], [[2, 91]]]",
    "query_spans": "[[[94, 103]]]",
    "process": "Assume the equation of the asymptote tangent to the circle is $ y = \\frac{b}{a}x' $, i.e., $ bx - ay = 0 $, then:  \n$ d = \\frac{|4b - 3a|}{\\sqrt{a^{2} + b^{2}}} = \\sqrt{5} $.  \nRearranging yields: $ (11b - 2a)(b - 2a) = 0 $, so $ \\frac{b}{a} = \\frac{2}{11} $ or $ \\frac{b}{a} = 2 $.  \nThus, the eccentricity of the ellipse is:  \n$ e = \\sqrt{1 + \\left( \\frac{b}{a} \\right)^{2}} = \\frac{5\\sqrt{5}}{11} $ or $ e = \\sqrt{1 + \\left( \\frac{b}{a} \\right)^{2}} = \\sqrt{5} $."
  },
  {
    "text": "The coordinates of the foci of the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/2 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(pm*sqrt(3), 0)",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 35]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ and the hyperbola $C_{2}$: $\\frac{y^{2}}{m^{2}}-\\frac{x^{2}}{n^{2}}=1(m>0, n>0)$ have the same asymptotes, and the eccentricity of $C_{1}$ is $\\frac{\\sqrt{5}}{2}$, then the eccentricity of $C_{2}$ is?",
    "fact_expressions": "C1: Hyperbola;Expression(C1) = (x^2/a^2 - y^2/b^2 = 1);a: Number;b: Number;a>0;b>0;C2: Hyperbola;Expression(C2) = (y^2/m^2 - x^2/n^2 = 1);m: Number;n: Number;m>0;n>0;Asymptote(C1) = Asymptote(C2);Eccentricity(C1) = sqrt(5)/2",
    "query_expressions": "Eccentricity(C2)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 67], [142, 149]], [[2, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[68, 133], [176, 183]], [[68, 133]], [[80, 133]], [[80, 133]], [[80, 133]], [[80, 133]], [[2, 140]], [[142, 174]]]",
    "query_spans": "[[[176, 189]]]",
    "process": "Since the eccentricity of $C_{1}$ is $\\frac{\\sqrt{5}}{2}$, we have $\\frac{c}{m}=\\sqrt{1+(\\frac{b}{a})^{2}}=\\frac{\\sqrt{5}}{2}\\Rightarrow\\frac{b}{n^{2}}=\\frac{1}{2}$, so the slopes of the asymptotes of $C_{1}$ are $\\pm\\frac{1}{2}$, then $\\frac{m^{2}}{n^{2}}=(\\frac{1}{2})^{2}=\\frac{1}{4}, \\frac{n^{2}}{m^{2}}=4$, thus the eccentricity of $C_{2}$ is $e=\\sqrt{\\sqrt{1+\\frac{n^{2}}{3}}}=\\sqrt{5}$."
  },
  {
    "text": "The product of the slopes of the lines joining a moving point $P(x, y)$ in the plane to two fixed points $A(-2,0)$, $B(2,0)$ is equal to $-\\frac{1}{3}$. Then the equation of the trajectory of point $P$ is?",
    "fact_expressions": "P: Point;A: Point;B: Point;x1:Number;y1:Number;Coordinate(P) = (x1, y1);Coordinate(A) = (-2, 0);Coordinate(B) = (2, 0);Slope(LineSegmentOf(P,A))*Slope(LineSegmentOf(P,B))=-1/3",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(x^2/4+3*y^2/4=1)&Negation(x=pm*2)",
    "fact_spans": "[[[5, 14], [62, 66]], [[18, 27]], [[29, 37]], [[5, 14]], [[5, 14]], [[5, 14]], [[18, 27]], [[29, 37]], [[5, 60]]]",
    "query_spans": "[[[62, 73]]]",
    "process": "K_{AP}=\\frac{y}{x+2},K_{BP}=\\frac{y}{x-2}K_{AP}\\cdot K_{BP}=\\frac{y^{2}}{x^{2}-4}=-\\frac{1}{3} \\text{ that is } x^{2}+3y^{2}=4 \\frac{x^{2}}{4}+\\frac{3y^{2}}{4}=1 |x\\neq\\pm2"
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ with focus $F$, and points $A$, $B$ on this parabola distinct from the origin $O$, satisfying $|\\overrightarrow{F A}|+|\\overrightarrow{F B}|+|\\overrightarrow{F O}|=12$ and $\\overrightarrow{F A}+\\overrightarrow{F B}+\\overrightarrow{F O}=\\overrightarrow{0}$, then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;F: Point;A: Point;B: Point;O: Origin;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(A, G);PointOnCurve(B, G);Negation(A = O);Negation(B = O);Negation(A = B);Abs(VectorOf(F, A)) + Abs(VectorOf(F, B)) + Abs(VectorOf(F, O)) = 12;VectorOf(F, A) + VectorOf(F, B) + VectorOf(F, O) = 0",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 23], [41, 44]], [[225, 228]], [[27, 30]], [[32, 35]], [[36, 39]], [[48, 55]], [[5, 23]], [[2, 23]], [[2, 30]], [[32, 62]], [[32, 62]], [[32, 62]], [[32, 62]], [[32, 62]], [[65, 138]], [[140, 223]]]",
    "query_spans": "[[[225, 230]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has eccentricity $\\sqrt{3}$. If the curve $y(y-k x)=0$ intersects the hyperbola $C$ at exactly $2$ points, what is the range of real values for $k$?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Curve;k: Real;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y*(-k*x + y) = 0);Eccentricity(C) = sqrt(3);NumIntersection(G,C)=2",
    "query_expressions": "Range(k)",
    "answer_expressions": "{(-∞,-√(2)], [√(2),+∞)}",
    "fact_spans": "[[[2, 63], [95, 101]], [[10, 63]], [[10, 63]], [[80, 94]], [[113, 118]], [[10, 63]], [[10, 63]], [[2, 63]], [[80, 94]], [[2, 78]], [[80, 111]]]",
    "query_spans": "[[[113, 124]]]",
    "process": "From the given, $\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{3}$, $\\frac{b}{a}=\\sqrt{2}$. $y(y-kx)=0$, i.e., $y=0$ or $y=kx$. Since $y=0$ intersects the hyperbola at two points, $y=kx$ does not intersect the hyperbola. Therefore, $y=kx$ lies within the region bounded by $y=\\pm\\sqrt{2}x$ containing the vertical axis, hence $k\\leqslant-\\sqrt{2}$ or $k\\geqslant\\sqrt{2}$."
  },
  {
    "text": "There exists a point $M$ on an ellipse centered at $O$ with foci $F_{1}$ and $F_{2}$, satisfying $2|\\overrightarrow{M F_{1}}|=3|\\overrightarrow{M O}|=3|\\overrightarrow{M F_{2}}|$. What is the eccentricity of this ellipse?",
    "fact_expressions": "G: Ellipse;M: Point;F1: Point;O: Origin;F2: Point;Center(G)=O;Focus(G)={F1,F2};PointOnCurve(M,G);2*Abs(VectorOf(M,F1))=3*Abs(VectorOf(M,O));3*Abs(VectorOf(M,O))=3*Abs(VectorOf(M,F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)/5",
    "fact_spans": "[[[27, 29], [124, 126]], [[34, 37]], [[8, 15]], [[1, 4]], [[16, 23]], [[0, 29]], [[8, 29]], [[27, 37]], [[40, 121]], [[40, 121]]]",
    "query_spans": "[[[124, 132]]]",
    "process": "Since |MF_{1}|+|MF_{2}|=\\frac{3}{2}|MF_{2}|+|MF_{2}|=\\frac{5}{2}|MF_{2}|=2a, because |MF_{2}|=\\frac{4}{5}a=|MO|, |MF_{1}|=\\frac{6}{5}a, |F_{1}F_{2}|=2c, since \\cos\\angle MOF_{1}=-\\cos\\angle MOF_{2}, so \\frac{(\\frac{4}{5}a)^{2}+c^{2}-(\\frac{6}{5}a)^{2}}{2\\times\\frac{4}{5}a\\times c} = \\frac{(\\frac{4}{5}a)^{2}+c^{2}-(\\frac{4}{5}a)^{2}}{2\\times\\frac{4}{5}a\\times c}, so \\frac{c^{2}}{a^{2}}=\\frac{2\\times\\frac{10}{25}}{25}, so e=\\sqrt{\\frac{10}{25}}=\\frac{\\sqrt{10}}{5}"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$, a line $l$ passes through the focus of the parabola $C$ and intersects the parabola at points $A$ and $B$. The circle with diameter $AB$ intersects the directrix of the parabola at the common point $M(-1,-1)$. Then the slope $k$ of line $l$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;l: Line;PointOnCurve(Focus(C), l);A: Point;B: Point;Intersection(l, C) = {A, B};H: Circle;IsDiameter(LineSegmentOf(A, B), H);M: Point;Coordinate(M) = (-1, -1);Intersection(H, Directrix(C)) = M;k: Number;Slope(l) = k",
    "query_expressions": "k",
    "answer_expressions": "-2",
    "fact_spans": "[[[2, 28], [35, 41], [45, 48], [73, 76]], [[2, 28]], [[10, 28]], [[10, 28]], [[29, 34], [96, 101]], [[29, 44]], [[50, 53]], [[54, 57]], [[29, 59]], [[71, 72]], [[61, 72]], [[84, 94]], [[84, 94]], [[71, 94]], [[104, 107]], [[96, 107]]]",
    "query_spans": "[[[104, 109]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), using the point difference method, the slope of the line can be obtained; let A(x_{1},y_{1}), B(x_{2},y_{2}), since -\\frac{p}{2} = -1 \\Rightarrow p = 2, the common point M(-1,-1) of the circle with AB as diameter and the directrix of the parabola, so \\frac{y_{1}+y_{2}}{2} = -1, x_{1} = \\frac{1}{2}\\frac{y_{2}^{2}px_{2}}{x_{2}} = \\frac{2p}{y_{1}+y_{3}} = \\frac{4}{-2} = -2,"
  },
  {
    "text": "Given that $F$ is a focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the line $l$ passing through point $F$ and perpendicular to the $x$-axis intersects the ellipse $C$ at points $A$ and $B$. If the origin $O$ lies on the circle with diameter $AB$, then the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "l: Line;C: Ellipse;G: Circle;A: Point;B: Point;F: Point;O:Origin;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);OneOf(Focus(C)) = F;PointOnCurve(F, l);IsPerpendicular(l,xAxis);Intersection(l, C) = {A, B};IsDiameter(LineSegmentOf(A, B), G);PointOnCurve(O, G);a:Number;b:Number",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(\\sqrt{5}-1)/2",
    "fact_spans": "[[[83, 88]], [[6, 63], [89, 94], [124, 129]], [[120, 121]], [[95, 99]], [[100, 103]], [[2, 5], [70, 74]], [[105, 110]], [[13, 63]], [[13, 63]], [[6, 63]], [[2, 68]], [[69, 88]], [[75, 88]], [[83, 103]], [[111, 121]], [[105, 122]], [[13, 63]], [[13, 63]]]",
    "query_spans": "[[[124, 135]]]",
    "process": "Let F(c,0), substitute x=c into the ellipse equation to obtain y=\\pm\\frac{b^{2}}{a}, hence |FA|=\\frac{b^{2}}{a}. According to the problem, \\frac{b^{2}}{a}=c, that is, b^{2}=ac=a^{2}-c^{2}, so e^{2}+e-1=0, solving yields e=\\frac{-1+\\sqrt{5}}{2},"
  },
  {
    "text": "The asymptote of a hyperbola centered at the origin with foci on the $x$-axis is given by $4x + 3y = 0$. What is the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;O: Origin;PointOnCurve(Focus(G), xAxis);Expression(OneOf(Asymptote(G))) = (4*x + 3*y = 0);Center(G) = O",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[17, 20], [43, 46]], [[3, 7]], [[8, 20]], [[17, 40]], [[0, 20]]]",
    "query_spans": "[[[43, 52]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=6x$, a line passing through the focus $F$ with slope $\\sqrt{3}$ intersects $C$ at points $P$ and $Q$. The projections of points $P$ and $Q$ onto the directrix are $M$ and $N$, respectively. Then $S_{\\Delta MFN}$=?",
    "fact_expressions": "C: Parabola;G: Line;M: Point;N:Point;F: Point;P: Point;Q: Point;Expression(C) = (y^2 = 6*x);Focus(C)=F;PointOnCurve(F, G);Slope(G) = sqrt(3);Intersection(G, C) = {P, Q};Projection(P,Directrix(C))=M;Projection(Q,Directrix(C))=N",
    "query_expressions": "Area(TriangleOf(M,F,N))",
    "answer_expressions": "6*sqrt(3)",
    "fact_spans": "[[[2, 21], [46, 49]], [[43, 45]], [[82, 85]], [[86, 89]], [[25, 28]], [[52, 55], [52, 55]], [[67, 70], [67, 70]], [[2, 21]], [[2, 28]], [[22, 45]], [[29, 45]], [[43, 61]], [[46, 91]], [[46, 91]]]",
    "query_spans": "[[[93, 113]]]",
    "process": "According to the parabolic equation, the coordinates of the focus and the equation of the directrix are obtained. From the slope of the line, the equation of the line is derived. By solving the system of equations consisting of the line equation and the parabola equation, and combining with Vieta's theorem, |y_{1}-y_{2}| is found, thus obtaining S_{AMFN}. The parabola C: y^{2}=6x has focus coordinates F(\\frac{3}{2},0) and directrix equation x=-\\frac{3}{2}. The line passing through the focus F with slope \\sqrt{3} has equation y=\\sqrt{3}(x-\\frac{3}{2}), which simplifies to x=\\frac{\\sqrt{3}}{3}y+\\frac{3}{2}. The parabola C intersects the line at points P and Q, where P(x_{1},y_{1}), Q(x_{2},y_{2}), and the projections of P and Q on the directrix are M and N respectively. Then  \n\\begin{cases}x=\\frac{\\sqrt{3}}{3}y+\\frac{3}{2}, simplified to y^{2}-2\\sqrt{3}y-9=0\\\\y^{2}=6x\\end{cases}  \nThus, y_{1}+y_{2}=2\\sqrt{3}, y_{1}\\cdot y_{2}=-9. Then |MN|=|y_{1}-y_{2}|=\\sqrt{(y_{1}+y_{2})^{2}-4y_{1}\\cdot y_{2}}. Therefore, S_{AMFN}=\\frac{1}{2}\\times4\\sqrt{3}\\times3=6\\sqrt{3}"
  },
  {
    "text": "Given that $A$ and $B$ are moving points on the parabola $y^{2}=4x$ satisfying $|AB|=15$, what is the minimum value of the horizontal coordinate $x_{0}$ of the midpoint $M$ of $AB$?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;M: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(A, G);PointOnCurve(B, G);Abs(LineSegmentOf(A, B)) = 15;MidPoint(LineSegmentOf(A, B)) = M;XCoordinate(M)=x0;x0:Number",
    "query_expressions": "Min(x0)",
    "answer_expressions": "13/2",
    "fact_spans": "[[[10, 24]], [[2, 5]], [[6, 9]], [[51, 54]], [[10, 24]], [[2, 28]], [[6, 28]], [[32, 42]], [[44, 54]], [[51, 65]], [[58, 65]]]",
    "query_spans": "[[[58, 71]]]",
    "process": "Let the directrix of the parabola $ y^{2} = 4x $ be $ l $, and the focus be $ F $. Then $ l: x = -1 $. Draw $ AA_{1} \\perp l $ from point $ A $ to $ A_{1} $, and $ BB_{1} \\perp l $ from point $ B $ to $ B_{1} $. Connect $ FA $ and $ FB $. Then $ 2(x_{0} + 1) = |AA_{1}| + |BB_{1}| = |FA| + |FB| \\geqslant |AB| = 15 $, so $ x_{0} + 1 \\geqslant \\frac{15}{2} $, solving gives $ x_{0} \\geqslant \\frac{13}{2} $. The equality holds if and only if points $ A $, $ B $, $ F $ are collinear. Therefore, the minimum value of $ x_{0} $ is $ \\frac{13}{2} $."
  },
  {
    "text": "The line passing through point $P(2,0)$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$. If the focus of the parabola is $F$, then the minimum area of $\\triangle ABF$ is?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(P) = (2, 0);PointOnCurve(P, H);Intersection(H, G) = {A, B};Focus(G) = F",
    "query_expressions": "Min(Area(TriangleOf(A, B, F)))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[14, 28], [40, 43]], [[11, 13]], [[1, 10]], [[29, 32]], [[33, 36]], [[47, 50]], [[14, 28]], [[1, 10]], [[0, 13]], [[11, 38]], [[40, 50]]]",
    "query_spans": "[[[52, 77]]]",
    "process": "Let points $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $ where $ y_{1}>0 $, $ y_{2}<0 $.  \n① When the slope of the line does not exist, it is easy to see that the equation of line $ AB $ is $ x=2 $. Substituting $ x=2 $ into the parabola's equation $ y^{2}=4x $, we get $ y^{2}=8 $, solving gives $ y=\\pm2\\sqrt{2} $. Thus, the coordinates of points $ A $, $ B $ are $ (2,2\\sqrt{2}) $, $ (2,-2\\sqrt{2}) $ respectively. Therefore, the area of $ \\triangle ABF $ is $ S=\\frac{1}{2}\\times|PF|\\times|y_{1}-y_{2}|=\\frac{1}{2}|2\\sqrt{2}-(-2\\sqrt{2})|=2\\sqrt{2} $;  \n② When the slope of line $ AB $ exists, let the slope be $ k $, clearly $ k\\neq0 $, so the equation of the line is $ y=k(x-2) $. Solving simultaneously  \n$$\n\\begin{cases}\ny= \\\\\n\\end{cases}\nk(x-2,\n$$\neliminating $ y $, we obtain $ k^{2}x^{2}-(4k^{2}+4)x+4k^{2}=0 $, and $ \\triangle=32k^{2}+16>0=4x $. By the relationship between roots and coefficients, we have $ x_{1}+x_{2}=\\frac{4k^{2}+4}{l^{2}} $, $ x_{1}x_{2}=4 $. $ \\therefore y_{1}y_{2}=(2\\sqrt{x_{1}})-(-2\\sqrt{x_{2}})=-4\\sqrt{x_{1}x_{2}}=-8 $, thus the area of $ \\triangle ABF $. In conclusion: the minimum area is $ 2\\sqrt{2} $."
  },
  {
    "text": "From a point $M$ on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, draw two tangents to the circle $x^{2}+y^{2}=2$, with points $A$ and $B$ as the points of tangency. The line $l$ passing through $A$ and $B$ intersects the $x$-axis and $y$-axis at points $P$ and $Q$, respectively. Then the minimum area of $\\Delta P O Q$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/9 = 1);M: Point;PointOnCurve(M, G);H: Circle;Expression(H) = (x^2 + y^2 = 2);L1: Line;L2: Line;TangentOfPoint(M, H) = {L1, L2};A: Point;B: Point;TangentPoint(L1, H) = A;TangentPoint(L2, H) = B;l: Line;PointOnCurve(A, l);PointOnCurve(B, l);P: Point;Intersection(l, xAxis) = P;Q: Point;Intersection(l, yAxis) = Q;O: Origin",
    "query_expressions": "Min(Area(TriangleOf(P, O, Q)))",
    "answer_expressions": "1/3",
    "fact_spans": "[[[1, 39]], [[1, 39]], [[42, 45]], [[1, 45]], [[46, 62]], [[46, 62]], [], [], [[0, 67]], [[80, 83], [68, 72]], [[73, 76], [84, 87]], [[0, 79]], [[0, 79]], [[88, 93]], [[79, 93]], [[79, 93]], [[107, 111]], [[88, 117]], [[112, 115]], [[88, 117]], [[119, 133]]]",
    "query_spans": "[[[119, 141]]]",
    "process": "Analysis: Let $ M(4\\cos\\theta,3\\sin\\theta) $, then the equation of line $ l $ is $ (4\\cos\\theta)x+(3\\sin\\theta)y=2 $, $ x_{P}=\\frac{1}{2\\cos\\theta} $, $ y_{Q}=\\frac{2}{3\\sin\\theta} $, $ S=\\frac{1}{2}|x_{P}|\\cdot|y_{Q}|=\\frac{1}{3|\\sin2\\theta|}\\geqslant\\frac{1}{3} $, with equality holding if and only if $ \\theta=\\frac{\\pi}{4} $."
  },
  {
    "text": "A point $A$ on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ is at a distance of $2$ from one of its foci. What is the distance from point $A$ to the other focus?",
    "fact_expressions": "G: Ellipse;A: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2/3 = 1);PointOnCurve(A, G);OneOf(Focus(G)) = F1;OneOf(Focus(G)) = F2;Negation(F1 = F2);Distance(A, F1) = 2",
    "query_expressions": "Distance(A, F2)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 37], [44, 45]], [[40, 43], [61, 65]], [], [], [[0, 37]], [[0, 43]], [[44, 50]], [[44, 71]], [[44, 71]], [[40, 58]]]",
    "query_spans": "[[[44, 77]]]",
    "process": "From the equation of the ellipse, we know that $ a^{2} = 4 $, $ 2a = 4 $. By the definition of the ellipse, the distance from point A to the other focus is equal to $ 2a - 2 = 4 - 2 = 2 $."
  },
  {
    "text": "Given a moving point $M$ on the ellipse $\\frac{x^{2}}{3}+y^{2}=1$, then the minimum distance from point $M$ to the line $l$: $x-y-8=0$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/3 + y^2 = 1);M: Point;PointOnCurve(M, G) = True;l: Line;Expression(l) = (x - y - 8 = 0)",
    "query_expressions": "Min(Distance(M, l))",
    "answer_expressions": "3*sqrt(2)",
    "fact_spans": "[[[2, 29]], [[2, 29]], [[33, 36], [38, 42]], [[2, 36]], [[43, 59]], [[43, 59]]]",
    "query_spans": "[[[38, 68]]]",
    "process": "Let $ P(\\sqrt{3}\\cos\\theta,\\sin\\theta) $, $ \\theta \\in [0,2\\pi] $. Find the distance from point $ P $ to the line $ l: x - y - 8 = 0 $, and use the properties of trigonometric functions to find the minimum value. Let $ P(\\sqrt{3}\\cos\\theta,\\sin\\theta) $, $ \\theta \\in [0,2\\pi] $. Then the distance from point $ P $ to the line $ l: x - y - 8 = 0 $ is so when $ \\sin(\\theta - \\frac{\\pi}{3}) = -1 $, $ \\frac{|-\\sqrt{3}\\cos\\theta + 8|}{\\sin = \\frac{\\sqrt{2}}{2} + 8|} = \\frac{|2\\sin(\\theta - \\frac{\\pi}{3}) + 8|}{\\sqrt{2}} = 3\\sqrt{2} $,"
  },
  {
    "text": "If the asymptotes of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ have no intersection points with the circle $(x-2)^{2}+y^{2}=1$, then the range of the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y^2 + (x - 2)^2 = 1);NumIntersection(Asymptote(C), G)=0",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(2*sqrt(3)/3, +oo)",
    "fact_spans": "[[[1, 62], [92, 95]], [[9, 62]], [[9, 62]], [[67, 87]], [[9, 62]], [[9, 62]], [[1, 62]], [[67, 87]], [[1, 90]]]",
    "query_spans": "[[[92, 106]]]",
    "process": "Analysis: Since the distance from the center of the circle to the line is greater than the radius, set up an inequality; combining with $c^{2}=b^{2}+a^{2}$, we obtain the range of the eccentricity.  \n$\\because$ The asymptotes of curve $C:\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ do not intersect the circle $(x-2)^{2}+y^{2}=1$,  \n$\\therefore$ the distance from the center $(2,0)$ to the line $y=\\frac{b}{a}x$ is greater than the radius $1$, i.e., $\\frac{2b}{\\sqrt{a^{2}+b^{2}}}>1$,  \n$4b^{2}>a^{2}+b^{2}$,  \n$3b^{2}=3(c^{2}-a^{2})>a^{2}$,  \n$3c^{2}>4a^{2}$,  \n$\\frac{c^{2}}{a^{2}}>\\frac{4}{3}$,  \n$e>\\frac{2\\sqrt{3}}{3}$,  \nthus the range of eccentricity of $C$ is $(\\frac{2\\sqrt{3}}{3},+\\infty)$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$, the focus is $F$, the intersection point of the directrix and the $x$-axis is $K$, point $A$ lies on the parabola, and $|AK|=\\sqrt{2}|AF|$, $O$ is the origin. Then $|OA|=$?",
    "fact_expressions": "C: Parabola;A: Point;K: Point;F: Point;O: Origin;Expression(C) = (y^2 = 8*x);Focus(C) = F;Intersection(Directrix(C), xAxis) = K;PointOnCurve(A, C);Abs(LineSegmentOf(A, K)) = sqrt(2)*Abs(LineSegmentOf(A, F))",
    "query_expressions": "Abs(LineSegmentOf(O, A))",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[2, 21], [49, 52]], [[44, 48]], [[40, 43]], [[25, 28]], [[79, 82]], [[2, 21]], [[2, 28]], [[2, 43]], [[44, 53]], [[55, 76]]]",
    "query_spans": "[[[89, 98]]]",
    "process": "Let the distance from point A to the directrix be equal to AM. By the definition of a parabola, we have |AF| = |AM|. From |AK| = \\sqrt{2}|AF|, it follows that triangle AMK is an isosceles right triangle. Let point A(\\frac{s^{2}}{8}, s). Since the equation of the directrix is x = -2 and |AM| = |MK|, we have \\frac{s^{2}}{8} + 2 = |s|, hence s = \\pm4, therefore A(2, \\pm4), thus |AO| = \\sqrt{4+16} = 2\\sqrt{5}"
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1$ $(a>0)$ is $\\sqrt{3}$, then the coordinates of the right focus of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2/4 + x^2/a^2 = 1);Eccentricity(G) = sqrt(3)",
    "query_expressions": "Coordinate(RightFocus(G))",
    "answer_expressions": "(\\sqrt{6}, 0)",
    "fact_spans": "[[[2, 49], [66, 69]], [[5, 49]], [[5, 49]], [[2, 49]], [[2, 64]]]",
    "query_spans": "[[[66, 77]]]",
    "process": "\\because the hyperbola's eccentricity is \\sqrt{3}, \\therefore e^{2} = \\frac{a^{2}+4}{a^{2}} = 3 \\Rightarrow a^{2} = 2 \\Rightarrow c^{2} = 2 + 4 = 6, \\therefore c = \\sqrt{6}, \\therefore the coordinates of the right focus of the hyperbola are (\\sqrt{6}, 0)."
  },
  {
    "text": "The hyperbola shares the same asymptotes as $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and its foci lie on the $y$-axis. What is the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);Asymptote(G) = Asymptote(H);PointOnCurve(Focus(H), yAxis);H: Hyperbola",
    "query_expressions": "Eccentricity(H)",
    "answer_expressions": "5/4",
    "fact_spans": "[[[1, 40]], [[1, 40]], [[0, 61]], [[49, 61]], [[58, 61]]]",
    "query_spans": "[[[58, 67]]]",
    "process": ""
  },
  {
    "text": "Given that the point $P$ on the parabola $y^{2}=4x$ is at a distance $d_{1}$ from the directrix of the parabola and at a distance $d_{2}$ from the line $3x-4y+9=0$, then the minimum value of $d_{1}+d_{2}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G);d1: Number;Distance(P, Directrix(G)) = d1;H: Line;Expression(H) = (3*x - 4*y + 9 = 0);d2: Number;Distance(P, H) = d2",
    "query_expressions": "Min(d1 + d2)",
    "answer_expressions": "12/5",
    "fact_spans": "[[[2, 16], [23, 26]], [[2, 16]], [[18, 22]], [[2, 22]], [[32, 39]], [[18, 39]], [[41, 56]], [[41, 56]], [[60, 67]], [[18, 67]]]",
    "query_spans": "[[[69, 88]]]",
    "process": ""
  },
  {
    "text": "If the equation $\\frac{x^{2}}{2+m}+\\frac{y^{2}}{m+1}=1$ represents a hyperbola, then the range of real values for $m$ is?",
    "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (x^2/(m + 2) + y^2/(m + 1) = 1)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-2, -1)",
    "fact_spans": "[[[44, 47]], [[49, 54]], [[1, 47]]]",
    "query_spans": "[[[49, 61]]]",
    "process": "Analysis: Using the characteristics of the hyperbola equation, we obtain (2+m)(m+1)<0. Solving this inequality gives the range of real values for m. Since the equation \\frac{x^{2}}{2+m}+\\frac{y^{2}}{m+1}=1 represents a hyperbola, it follows that (2+m)(m+1)<0. Solving this yields -2<m<-1. Therefore, the range of real values for m is (-2,-1)."
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 px(p>0)$ is also a focus of the hyperbola $x^{2}-y^{2}=8$, then $p$=?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;Expression(G) = (x^2 - y^2 = 8);p>0;Expression(H) = (y^2 = 2*(p*x));Focus(H) = OneOf(Focus(G))",
    "query_expressions": "p",
    "answer_expressions": "8",
    "fact_spans": "[[[26, 44]], [[1, 21]], [[51, 54]], [[26, 44]], [[4, 21]], [[1, 21]], [[1, 49]]]",
    "query_spans": "[[[51, 56]]]",
    "process": ""
  },
  {
    "text": "Given that a line passing through the focus $F$ of the parabola $x=4 y^{2}$ intersects the parabola at points $M$ and $N$, if $|M F|=\\frac{1}{8}$, then $|M N|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x = 4*y^2);F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);M: Point;N: Point;Intersection(H, G) = {M, N};Abs(LineSegmentOf(M, F)) = 1/8",
    "query_expressions": "Abs(LineSegmentOf(M, N))",
    "answer_expressions": "1/4",
    "fact_spans": "[[[3, 17], [28, 31]], [[3, 17]], [[20, 23]], [[3, 23]], [[24, 26]], [[2, 26]], [[32, 35]], [[36, 39]], [[24, 41]], [[43, 62]]]",
    "query_spans": "[[[64, 73]]]",
    "process": "The parabola $x=4y^{2}$ can be rewritten as $y^{2}=\\frac{1}{4}x$, with focus $F(\\frac{1}{16},0)$ and directrix equation $x=-\\frac{1}{16}$. Since $|MF|=\\frac{1}{8}$, the distance from point $M$ to the directrix of the parabola is $\\frac{1}{8}$, so the x-coordinate of point $M$ is $\\frac{1}{16}$. Hence, line $MF$ is perpendicular to the x-axis, so $|NF|=|MF|=\\frac{1}{8}$, and therefore $|MN|=\\frac{1}{4}$."
  },
  {
    "text": "Draw a line $l$ with slope $1$ through the right focus $F$ of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$, intersecting the ellipse $C$ at points $A$ and $B$. Let $P$ be the intersection point of the right directrix and the $x$-axis. Denote the slope of line $PA$ as $k_{1}$ and the slope of line $PB$ as $k_{2}$. Then the value of $\\frac{1}{k_{1}^{2}}+\\frac{1}{k_{2}^{2}}$ is?",
    "fact_expressions": "P: Point;A: Point;B: Point;l: Line;F:Point;C: Ellipse;Expression(C) = (x^2/4 + y^2/2 = 1);RightFocus(C)=F;PointOnCurve(F,l);Slope(l)=1;Intersection(l,C)={A,B};Intersection(RightDirectrix(C),xAxis)=P;Slope(LineOf(P,A))=k1;Slope(LineOf(P,B))=k2;k1:Number;k2:Number",
    "query_expressions": "1/k2^2 + 1/k1^2",
    "answer_expressions": "8",
    "fact_spans": "[[[83, 86]], [[71, 74]], [[77, 80]], [[58, 63]], [[47, 50]], [[1, 43], [64, 69]], [[1, 43]], [[1, 50]], [[0, 63]], [[51, 63]], [[58, 82]], [[64, 98]], [[100, 117]], [[118, 135]], [[110, 117]], [[128, 135]]]",
    "query_spans": "[[[137, 182]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ P(2\\sqrt{2},0) $, $ k_{1}=\\frac{y_{1}}{x_{1}-2\\sqrt{2}} $, $ k_{2}=\\frac{y_{2}}{x_{2}-2\\sqrt{2}} $, $ l:y=x-\\sqrt{2} $. From $ y=x-\\sqrt{2} $, $ x^{2}+2y^{2}=4 $, we obtain $ 3y^{2}+2\\sqrt{2}y-2=0 $. $ \\therefore y_{1}=\\frac{\\sqrt{2}}{3} $, $ y_{1}=-\\sqrt{2} $. So $ \\frac{1}{k_{1}^{2}}+\\frac{1}{k_{2}^{2}}=\\frac{\\sqrt{2}-\\sqrt{2}}{\\frac{2}{0}}+\\frac{(-\\sqrt{2}-\\sqrt{2})^{2}}{2}=8 $."
  },
  {
    "text": "The hyperbola $\\frac{y^{2}}{16}-\\frac{x^{2}}{m}=1$ has eccentricity $e=2$. Then the asymptotes of the hyperbola are given by?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (y^2/16 - x^2/m = 1);Eccentricity(G) = e;e = 2;e:Number",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(3)/3)*x",
    "fact_spans": "[[[0, 39], [50, 53]], [[3, 39]], [[0, 39]], [[0, 48]], [[43, 48]], [[43, 48]]]",
    "query_spans": "[[[50, 61]]]",
    "process": "e=\\frac{c}{a}=\\frac{\\sqrt{16+m}}{4}=2, solving gives: m=48, so the hyperbola equation is \\frac{y^{2}}{16}-\\frac{x^{2}}{48}=0. a=4, b=4\\sqrt{3}, so the asymptote equations are y=\\pm\\frac{a}{b}x=\\pm\\frac{\\sqrt{3}}{3}x"
  },
  {
    "text": "If $a^{2}+2 b^{2}=3$ $(a>0 , b>0)$, then the maximum value of $a+2 b$ is?",
    "fact_expressions": "a^2 + 2*b^2 = 3;a: Number;b: Number;a>0;b>0",
    "query_expressions": "Max(a + 2*b)",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 29]], [[1, 29]], [[1, 29]], [[1, 29]], [[1, 29]]]",
    "query_spans": "[[[31, 44]]]",
    "process": ""
  },
  {
    "text": "A asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is perpendicular to the line $x+2 y-1=0$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;Expression(H) = (x + 2*y - 1 = 0);IsPerpendicular(OneOf(Asymptote(G)), H);H: Line",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[0, 46], [70, 73]], [[0, 46]], [[3, 46]], [[3, 46]], [[53, 66]], [[0, 68]], [[53, 66]]]",
    "query_spans": "[[[70, 79]]]",
    "process": "The asymptotes of the hyperbola are $ y = \\pm\\frac{b}{a} $. The slope of the line $ x + 2y - 1 = 0 $ is $ y = -\\frac{1}{2} $. Since $ y = \\frac{b}{a}x $ is perpendicular to the line $ x + 2y - 1 = 0 $, it follows that $ \\frac{b}{a} \\cdot \\left(-\\frac{1}{2}\\right) = -1 $, so $ b = 2a $. Therefore, $ c^{2} = a^{2} + b^{2} = 5a^{2} $, which implies $ e^{2} = 5 $, $ e = \\sqrt{5} $."
  },
  {
    "text": "Let the directrix of the parabola $y^{2}=2 p x$ ($p$ is a constant) intersect the $X$-axis at point $K$. A line $l$ passing through $K$ intersects the parabola at points $A$ and $B$. Then $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;K: Point;Intersection(Directrix(G), xAxis) = K;l: Line;PointOnCurve(K, l);A: Point;B: Point;Intersection(l, G) = {A, B};O: Origin",
    "query_expressions": "DotProduct(VectorOf(O, A), VectorOf(O, B))",
    "answer_expressions": "(5/4)*p^2",
    "fact_spans": "[[[1, 17], [51, 54]], [[1, 17]], [[18, 21]], [[35, 39], [41, 44]], [[1, 39]], [[45, 50]], [[40, 50]], [[56, 59]], [[61, 64]], [[45, 66]], [[68, 117]]]",
    "query_spans": "[[[68, 119]]]",
    "process": ""
  },
  {
    "text": "The parabola $C$: $y^{2}=4 x$ has focus $F$, directrix $l$, and $P$ is a point on the parabola $C$ in the first quadrant. $P M \\perp l$ at point $M$, and the segment $M F$ intersects the parabola $C$ at point $N$. If the slope of $P F$ is $\\frac{3}{4}$, then $\\frac{|M N|}{|N F|}=$?",
    "fact_expressions": "C: Parabola;M: Point;F: Point;P: Point;N: Point;l: Line;Expression(C) = (y^2 = 4*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(P, C);Quadrant(P)=1;IsPerpendicular(LineSegmentOf(P,M),l);FootPoint(LineSegmentOf(P,M),l)=M;Intersection(LineSegmentOf(M,F), C) = N;Slope(LineSegmentOf(P, F)) = 3/4",
    "query_expressions": "Abs(LineSegmentOf(M, N))/Abs(LineSegmentOf(N, F))",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[0, 19], [40, 46], [40, 46]], [[74, 78]], [[23, 26]], [[51, 54], [51, 54]], [[95, 99]], [[30, 33]], [[0, 19]], [[0, 26]], [[0, 33]], [[36, 49]], [[51, 59]], [[60, 73]], [[60, 78]], [[79, 99]], [[101, 123]]]",
    "query_spans": "[[[125, 148]]]",
    "process": "Draw a perpendicular line from N, with foot of perpendicular at Q, then |NF| = |NQ|, |PF| = |PM|, thus ∠PFM = ∠PMF = ∠MFO = ∠MNQ, let = λ, then cos∠MNQ = 1/λ, use the double angle formula to find cos∠PFx, set up an equation and solve for λ. ∠A = 1010 draw a perpendicular line, with foot of perpendicular at Q, then NF = NQ, let |MN| / |NF| = λ, then |MN| / |NQ| = λ, ∴ cos∠MNQ = 1/2 ⋅ cos∠MFO = 1/λ. ∵ |PM| = |PF|, ∴ ∠PMF = ∠PFM, ∴ ∠PFM = ∠MFO. ∴ cos∠PFx = −cos2∠MFO = 1 − 2cos²∠MFO = 1 − 2/2². ∵ tan∠PFx = 3/4, ∴ cos∠PFx = 4/5, ∴ 1 − 2/2 = 4/5, solving gives λ² = 10, i.e., λ = √10. The answer is √10."
  },
  {
    "text": "Given the parabola $C$: $x^{2}=8 y$ with focus $F$, and the line $y=k x+m$ ($k \\in \\mathbb{R}$) intersects the parabola $C$ at points $A$ and $B$. If $\\angle A F B>90^{\\circ}$ always holds, then the range of values for $m$ is?",
    "fact_expressions": "C: Parabola;G: Line;k: Real;m:Number;A: Point;F: Point;B: Point;Expression(C) = (x^2 = 8*y);Expression(G) = (y = k*x + m);Focus(C) = F;Intersection(G, C) = {A, B};AngleOf(A,F,B)>ApplyUnit(90,degree)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(6-4*sqrt(2),6+4*sqrt(2))",
    "fact_spans": "[[[2, 20], [49, 55]], [[28, 48]], [[28, 48]], [[99, 102]], [[57, 60]], [[24, 27]], [[61, 64]], [[2, 20]], [[28, 48]], [[2, 27]], [[28, 66]], [[69, 94]]]",
    "query_spans": "[[[99, 109]]]",
    "process": ""
  },
  {
    "text": "If the coordinate of a vertex of a hyperbola is $(3 , 0)$ and the focal length is $10$, then its standard equation is?",
    "fact_expressions": "G: Hyperbola;Coordinate(OneOf(Vertex(G))) = (3, 0);FocalLength(G) = 10",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/16 = 1",
    "fact_spans": "[[[1, 4], [31, 32]], [[1, 21]], [[1, 29]]]",
    "query_spans": "[[[31, 39]]]",
    "process": ""
  },
  {
    "text": "If the asymptotes of a hyperbola are given by $y = \\pm 3x$, and its focal distance is $2\\sqrt{10}$, then what is the standard equation of this hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*(3*x));FocalLength(G) = 2*sqrt(10)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/9 = \\pm 1",
    "fact_spans": "[[[1, 4], [44, 47], [23, 24]], [[1, 22]], [[23, 41]]]",
    "query_spans": "[[[44, 54]]]",
    "process": "According to the position of the hyperbola's foci, classify and solve for the standard equation of the hyperbola. The focal distance of the hyperbola is $2\\sqrt{10}$, so $c=\\sqrt{10}$. When the foci of the hyperbola are on the horizontal axis, since the asymptotes are given by $y=\\pm3x$, it follows that $\\frac{b}{a}=3 \\Rightarrow b=3a$. Also, because $c^{2}=a^{2}+b^{2}$, solving yields $a^{2}=1$, $b^{2}=9$, so the hyperbola equation is: $x^{2}-\\frac{y^{2}}{9}=1$. When the foci of the hyperbola are on the vertical axis, since the asymptotes are given by $y=\\pm3x$, it follows that $\\frac{a}{b}=3 \\Rightarrow a=3b$. Also, because $c^{2}=a^{2}+b^{2}$, solving yields $a^{2}=9$, $b^{2}=1$, so the hyperbola equation is: $\\frac{y^{2}}{9}-x^{2}=1$. Therefore, the standard equation of the hyperbola is $x^{2}-\\frac{y^{2}}{9}=\\pm1$."
  },
  {
    "text": "Given that the ellipse $\\frac{x^{2}}{3 m}+\\frac{y^{2}}{5 n}=1$ and the hyperbola $\\frac{x^{2}}{2 m}-\\frac{y^{2}}{3 n}=1$ have common foci, what is the equation of the asymptotes of the hyperbola?",
    "fact_expressions": "G: Hyperbola;n: Number;m: Number;H: Ellipse;Expression(G) = (-y^2/(3*n) + x^2/(2*m) = 1);Expression(H) = (y^2/(5*n) + x^2/(3*m) = 1);Focus(H) = Focus(G)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[44, 86], [95, 98]], [[4, 43]], [[4, 43]], [[2, 43]], [[44, 86]], [[2, 43]], [[2, 92]]]",
    "query_spans": "[[[95, 106]]]",
    "process": ""
  },
  {
    "text": "Given $M(2,0)$, $N(3,0)$, and $P$ is a point on the parabola $C$: $y^{2}=3x$, then the minimum value of $\\overrightarrow{P M} \\cdot \\overrightarrow{P N}$ is?",
    "fact_expressions": "C: Parabola;M: Point;P: Point;N: Point;Expression(C) = (y^2 = 3*x);Coordinate(M) = (2, 0);Coordinate(N)=(3, 0);PointOnCurve(P, C)",
    "query_expressions": "Min(DotProduct(VectorOf(P, M), VectorOf(P, N)))",
    "answer_expressions": "5",
    "fact_spans": "[[[28, 47]], [[2, 11]], [[24, 27]], [[12, 21]], [[28, 47]], [[2, 11]], [[12, 21]], [[24, 50]]]",
    "query_spans": "[[[52, 107]]]",
    "process": "Let $ P(x, y) $. Using the coordinate representation of the dot product of vectors, we obtain $ \\overrightarrow{PM} \\cdot \\overrightarrow{PN} = x^{2} - 5x + 6 + y^{2} $. According to $ y^{2} = 3x $, eliminate $ y $ to obtain a quadratic function in $ x $, which can then be solved. Let $ P(x, y) $, then $ \\overrightarrow{PM} = (2 - x, -y) $, $ \\overrightarrow{PN} = (3 - x, -y) $. Thus, $ \\overrightarrow{PM} \\cdot \\overrightarrow{PN} = (3 - x)(2 - x) + y^{2} = x^{2} - 5x + 6 + y^{2} $. Since point $ P $ lies on the parabola $ C $, $ y^{2} = 3x $, so $ \\overrightarrow{PM} \\cdot \\overrightarrow{PN} = (3 - x)(2 - x) + y^{2} = x^{2} - 5x + 6 + 3x = x^{2} - 2x + 6 \\geqslant 5 $, with equality if and only if $ x = 1 $."
  },
  {
    "text": "If the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has eccentricity $4$, then a standard equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(C) = 4",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2/15 = 1",
    "fact_spans": "[[[1, 52], [62, 65]], [[8, 52]], [[8, 52]], [[1, 52]], [[1, 60]]]",
    "query_spans": "[[[62, 74]]]",
    "process": "Since the eccentricity of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ is 4, let the semi-focal distance of $ C $ be $ c $, then $ e^{2} = \\frac{c^{2}}{a^{2}} = 1 + \\frac{b^{2}}{a^{2}} = 16 $, that is, $ \\frac{b^{2}}{a^{2}} = 15 $. Taking $ a^{2} = 1 $, we get $ b^{2} = 15 $. Therefore, a standard equation of $ C $ is $ x^{2} - \\frac{y^{2}}{15} = 1 $."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, point $P$ lies on the parabola, and point $Q(9,0)$. If $|QF|=2|PF|$, then the area of $\\Delta PQF$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;P: Point;PointOnCurve(P, G) = True;Q: Point;Coordinate(Q) = (9, 0);Abs(LineSegmentOf(Q, F)) = 2*Abs(LineSegmentOf(P, F))",
    "query_expressions": "Area(TriangleOf(P, Q, F))",
    "answer_expressions": "8*sqrt(3)",
    "fact_spans": "[[[2, 16], [29, 32]], [[2, 16]], [[20, 23]], [[2, 23]], [[24, 28]], [[24, 33]], [[34, 43]], [[34, 43]], [[45, 59]]]",
    "query_spans": "[[[61, 80]]]",
    "process": "Let point $ P\\left(\\frac{m^{2}}{4}, m\\right) $. It is easy to know that the focus of the parabola $ y^{2} = 4x $ is $ F(1, 0) $. From $ |QF| = 2|PF| $, we get $ 2\\left(\\frac{m^{2}}{4} + 1\\right) = 8 $, solving yields $ m = \\pm 2\\sqrt{3} $. Therefore, $ S_{\\triangle POF} = \\frac{1}{2}|QF| \\cdot |m| = \\frac{1}{2} \\times 8 \\times 2\\sqrt{3} = 8\\sqrt{3} $. Hence, the answer is: $ 8.5 $"
  },
  {
    "text": "Given that the parabola $y^{2}=2 p x$ passes through the point $M(2 , 2)$, then the distance from point $M$ to the focus of the parabola is?",
    "fact_expressions": "G: Parabola;p: Number;M: Point;Expression(G) = (y^2 = 2*(p*x));Coordinate(M) = (2, 2);PointOnCurve(M, G)",
    "query_expressions": "Distance(M, Focus(G))",
    "answer_expressions": "5/2",
    "fact_spans": "[[[2, 18], [37, 40]], [[5, 18]], [[19, 30], [32, 36]], [[2, 18]], [[19, 30]], [[2, 30]]]",
    "query_spans": "[[[32, 47]]]",
    "process": ""
  },
  {
    "text": "What is the standard equation of a parabola with focus $(0,2)$?",
    "fact_expressions": "G: Parabola;H: Point;Coordinate(H) = (0, 2);Focus(G) = H",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2=8*y",
    "fact_spans": "[[[11, 14]], [[3, 10]], [[3, 10]], [[0, 14]]]",
    "query_spans": "[[[11, 20]]]",
    "process": "The focus is (0,2), so p=4, and the equation is x^{2}=2py=8y"
  },
  {
    "text": "Given that the parabola $y^{2}=2 p x(p>0)$ and the hyperbola $\\frac{x^{2}}{2}-y^{2}=\\frac{1}{3}$ share a common focus $F$, point $P$ is an arbitrary point on the parabola, $M(-1,0)$, then the minimum value of $\\frac{|P F|}{|P M|}$ is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;M: Point;P: Point;F: Point;Expression(G) = (x^2/2 - y^2 = 1/3);p>0;Expression(H) = (y^2 = 2*(p*x));Coordinate(M) = (-1, 0);Focus(H) = F;OneOf(Focus(G)) = F;Focus(H) = OneOf(Focus(G));PointOnCurve(P, H)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F))/Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[24, 62]], [[2, 23], [79, 82]], [[5, 23]], [[88, 97]], [[74, 78]], [[70, 73]], [[24, 62]], [[5, 23]], [[2, 23]], [[88, 97]], [[2, 73]], [[2, 73]], [[2, 73]], [[74, 87]]]",
    "query_spans": "[[[99, 126]]]",
    "process": "\\because\\frac{x^{2}}{2}-y^{2}=\\frac{1}{3},\\therefore\\frac{x^{2}}{3}-\\frac{y^{2}}{3}=1, therefore the foci of the hyperbola are (\\pm1,0). Since the parabola y^{2}=2px (p>0) and the hyperbola \\frac{x^{2}}{2}-y^{2}=\\frac{1}{3} share a common focus F, then \\frac{p}{2}=1,\\therefore p=2,y^{2}=4x,F(1,0). Let P(x,y), then y^{2}=4x,x\\geqslant0. When x>0, \\frac{|PF|}{|PM|}=\\frac{\\sqrt{(x-1)^{2}+y^{2}}}{\\sqrt{(x+1)^{2}+y^{2}}}=\\sqrt{\\frac{x^{2}+2x+1}{x^{2}+6x+1}}=\\sqrt{\\frac{1}{1+\\frac{4x}{x^{2}+2x+1}}}\\geqslant\\frac{\\sqrt{1+2}}{1+2}, with equality if and only if x=1; when x=0, \\frac{|PF|}{|PM|}=1"
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{n}-\\frac{y^{2}}{12-n}=1$ is $\\sqrt{3}$. Then $n$=?",
    "fact_expressions": "G: Hyperbola;n: Number;Expression(G) = (-y^2/(12 - n) + x^2/n = 1);Eccentricity(G) = sqrt(3)",
    "query_expressions": "n",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 43]], [[61, 64]], [[2, 43]], [[2, 58]]]",
    "query_spans": "[[[61, 66]]]",
    "process": "Since the equation $\\frac{x^{2}}{n}-\\frac{y^{2}}{12-n}=1$ is a hyperbola, it follows that $n(12-n)>0$, solving gives $0<n<12$. Thus, the foci of this hyperbola lie on the $x$-axis. Let the eccentricity of the hyperbola be $e$, then $e^{2}=\\frac{n+(12-n)}{n}=\\frac{12}{n}=3$, solving gives $n=4$. Therefore, $n=4$."
  },
  {
    "text": "The center of ellipse $E$ is at the origin, and $F(3,0)$ is a focus of $E$. A line $l$ passing through $F$ intersects $E$ at points $A$ and $B$, and the midpoint of $AB$ is $(2,-1)$. Then the eccentricity $e$ of $E$ is?",
    "fact_expressions": "l: Line;E: Ellipse;O:Origin;A: Point;B: Point;F: Point;e: Number;Coordinate(MidPoint(LineSegmentOf(A,B))) = (2, -1);Coordinate(F) = (3, 0);Center(E)=O;OneOf(Focus(E)) = F;PointOnCurve(F, l);Intersection(l, E) = {A, B};Eccentricity(E) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[35, 40]], [[2, 7], [23, 26], [41, 44], [76, 79]], [[11, 13]], [[47, 50]], [[51, 54]], [[31, 34], [14, 22]], [[83, 86]], [[58, 74]], [[14, 22]], [[2, 13]], [[14, 29]], [[30, 40]], [[35, 56]], [[76, 86]]]",
    "query_spans": "[[[83, 88]]]",
    "process": "Let the equation of the ellipse be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$, $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $x_{1}+x_{2}=4$, $y_{1}+y_{2}=-2$, $k=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{0+1}{3-2}=1$, from $\\frac{x_{1}^{2}}{a^{2}}+\\frac{y_{1}^{2}}{b^{2}}=1$, $\\frac{x_{2}^{2}}{a^{2}}+\\frac{y_{2}^{2}}{b^{2}}=1$, subtracting gives $\\frac{x_{1}+x_{2}}{a^{2}}+\\frac{y_{1}+y_{2}}{b^{2}}\\cdot\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=0$, $\\frac{4}{a^{2}}-\\frac{2}{b^{2}}=0$, $\\frac{b^{2}}{a^{2}}=\\frac{1}{2}$, $e=\\frac{c}{a}=\\sqrt{1-\\frac{b^{2}}{a^{2}}}=\\frac{\\sqrt{2}}{2}$,"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$, and $P$ a point on $C$. If $A(-2,0)$, then the maximum value of $\\frac{|PA|}{|PF|}$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 8*x);F: Point;Focus(C) = F;P: Point;PointOnCurve(P, C);A: Point;Coordinate(A) = (-2, 0)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, A))/Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 21], [33, 36]], [[2, 21]], [[25, 28]], [[2, 28]], [[29, 32]], [[29, 39]], [[41, 50]], [[41, 50]]]",
    "query_spans": "[[[52, 79]]]",
    "process": "From the problem, we obtain that A(-2,0) is the intersection point of the directrix x = -2 and the x-axis. Draw PB perpendicular from P to the directrix, with foot at B. By the definition of a parabola, |PB| = |PF|. Then, \\frac{|PA|}{|PF|} = \\frac{|PA|}{|PB|} = \\frac{1}{\\sin\\angle PAB} = \\frac{1}{\\cos\\angle PAF}. Thus, when \\cos\\angle PAF is minimized, i.e., when \\angle PAF is maximized, \\frac{|PA|}{|PF|} reaches its maximum value. From the figure, it can be seen that \\angle PAF is maximized when line AP is tangent to the parabola. Let the equation of line AP be x = my - 2. Substituting into the parabola yields y^{2} - 8my + 16 = 0. Then, from \\triangle = (-8m)^{2} - 4 \\times 16 = 0, solving gives m = \\pm1. Due to the symmetry of the parabola, taking m = 1 gives \\angle PAF = 45^{\\circ} at this time. Therefore, the maximum value of \\frac{|PA|}{|PF|} is \\sqrt{2}."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$, if the distance from point $A(-2,3)$ to its focus is $5$, then $p$=?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(A) = ((-1)*2, 3);Distance(A, Focus(G))=5",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 23], [36, 37]], [[48, 51]], [[25, 35]], [[5, 23]], [[2, 23]], [[25, 35]], [[25, 46]]]",
    "query_spans": "[[[48, 53]]]",
    "process": "According to the problem, the focus of the parabola has coordinates $(\\frac{p}{2},0)$, so the distance from point $A(-2,3)$ to the focus is $\\sqrt{(-2-\\frac{p}{2})^{2}+(3-0)^{2}}=5$. Solving gives $p=4$ or $p=-12$ (discarded)."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$, the distance from the right focus $F$ to an asymptote is $2$, and the distance from point $F$ to the origin is $3$. Then the eccentricity $e$ of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;e: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);F:Point;RightFocus(C)=F;Distance(F,Asymptote(C))=2;Eccentricity(C) = e;O:Origin;Distance(F,O)=3",
    "query_expressions": "e",
    "answer_expressions": "3*sqrt(5)/5",
    "fact_spans": "[[[2, 64], [99, 105]], [[10, 64]], [[10, 64]], [[109, 112]], [[10, 64]], [[10, 64]], [[2, 64]], [[68, 71], [83, 87]], [[2, 71]], [[2, 82]], [[99, 112]], [[88, 90]], [[83, 97]]]",
    "query_spans": "[[[109, 114]]]",
    "process": "The distance from point F to the origin is 3, so c=3. Write the asymptote equations and find the distance from the right focus to the asymptote. This distance being 2 allows us to find b and a, thus obtaining the eccentricity. \\because the distance from the right focus F to the asymptote is 2, \\therefore the distance from F(c,0) to y=\\pm\\frac{b}{a}x is 2. The asymptote equation is bx+ay=0, so \\frac{|bc|}{\\sqrt{a^{2}+b^{2}}}=2. Also, b>0, c>0, a^{2}+b^{2}=c^{2}, \\therefore \\frac{bc}{c}=b=2. Also, \\because the distance from point F to the origin is 3, \\therefore c=3, a=\\sqrt{c^{2}-b^{2}}=\\sqrt{5}, \\therefore the eccentricity e=\\frac{c}{a}=\\frac{3}{\\sqrt{5}}=\\frac{3\\sqrt{5}}{5}"
  },
  {
    "text": "A point $M$ on the parabola $x^{2}=y$ is at a distance of $1$ from the focus. Then the horizontal coordinate of point $M$ is?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (x^2 = y);PointOnCurve(M, G);Distance(M, Focus(G)) = 1",
    "query_expressions": "XCoordinate(M)",
    "answer_expressions": "pm*sqrt(3)/2",
    "fact_spans": "[[[0, 12]], [[15, 18], [30, 34]], [[0, 12]], [[0, 18]], [[0, 28]]]",
    "query_spans": "[[[30, 40]]]",
    "process": "Analysis: According to the problem, let the coordinates of point M be M(m, n), (n > 0). Find the equation of the directrix of the parabola x^{2} = y. By the definition of a parabola, the distance from M to the directrix is also 1. Thus, we have n - (-\\frac{1}{4}) = 1. Solving this gives n = \\frac{3}{4}. Substituting the coordinates of M into the equation of the parabola, we calculate m. Detailed solution: According to the problem, let the coordinates of point M be M(m, n), (n > 0). The parabola y = 4x^{2}, its standard form is x^{2} = y, and its directrix is y = -\\frac{1}{4}. If the distance from M to the focus is 1, then the distance from M to the directrix is also 1. Hence, n - (-\\frac{1}{4}) = 1. Solving gives n = \\frac{3}{4}. Since M lies on the parabola, m^{2} = \\frac{3}{4}, solving yields m = \\pm\\frac{\\sqrt{3}}{2}."
  },
  {
    "text": "Given the line $l$: $y = 2x + b$ is intersected by the parabola $C$: $y^2 = 2px$ $(p > 0)$ with a chord length of $5$, and the line $l$ passes through the focus of $C$. Let $M$ be a moving point on $C$, and let the coordinates of point $N$ be $(3, 0)$. Then the minimum value of $|MN|$ is?",
    "fact_expressions": "l: Line;Expression(l) = (y = 2*x + b);b: Number;C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;Length(InterceptChord(l, C)) = 5;PointOnCurve(Focus(C), l);M: Point;PointOnCurve(M, C);N: Point;Coordinate(N) = (3, 0)",
    "query_expressions": "Min(Abs(LineSegmentOf(M, N)))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 18], [57, 62]], [[2, 18]], [[9, 18]], [[19, 47], [64, 67], [75, 78]], [[19, 47]], [[27, 47]], [[27, 47]], [[2, 56]], [[57, 70]], [[71, 74]], [[71, 84]], [[86, 90]], [[86, 102]]]",
    "query_spans": "[[[104, 117]]]",
    "process": "Solve the system of the line equation and the parabola equation to obtain the relationship between roots and coefficients, use the chord length formula to derive the relationship between $ p $ and $ b $; then, based on the condition that the line passes through the focus of $ C $, obtain another relationship between $ p $ and $ b $, solve for the equation of the parabola, and then find the minimum value of $ |MN| $. Solution: (1) $\\because \\begin{cases} y = 2x + b \\\\ y^{2} = 2px \\end{cases} \\Rightarrow 4x^{2} + (4b - 2p)x + b^{2} = 0 $, then $ 5^{2} = (1 + 2^{2})\\left[\\left(\\frac{2b - p}{2}\\right)^{2} - 4 \\times \\frac{b^{2}}{4}\\right] $, and since the line $ l $ passes through the focus of $ C $, we have $ -\\frac{b}{2} = \\frac{p}{2} $, $ \\therefore b = -p $. Solving this gives $ p = 2 $, so the equation of the parabola is $ y^{2} = 4x $. Let $ M(x_{0}, y_{0}) $, $ \\therefore y_{0}^{2} = 4x_{0} $, then $ |MN|^{2} = (x_{0} - 3)^{2} + y_{0}^{2} = (x_{0} - 3)^{2} + 4x_{0} = (x_{0} - 1)^{2} + 8 $. Hence, when $ x_{0} = 1 $, $ |MN| = 2\\sqrt{2} $, so the answer is $ 2\\sqrt{2} $."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. Let $M$ be a point on $C$ such that $M F_{2}$ is perpendicular to the $x$-axis, and let $N$ be the other intersection point of line $M F_{1}$ with the ellipse $C$. If the slope of line $M N$ is $\\frac{3}{4}$, then the eccentricity of $C$ equals?",
    "fact_expressions": "C: Ellipse;a:Number;b:Number;a>b;b>0;M: Point;F1: Point;N: Point;F2: Point;Expression(C)=(x^2/a^2 +y^2/b^2=1);LeftFocus(C) =F1;RightFocus(C)=F2;PointOnCurve(M,C);IsPerpendicular(LineSegmentOf(M,F2),xAxis);Intersection(LineOf(M,F1),C)={M,N};Slope(LineOf(M,N)) = 3/4",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[19, 78], [89, 92], [125, 130], [168, 171]], [[26, 78]], [[26, 78]], [[26, 78]], [[26, 78]], [[85, 88]], [[1, 8]], [[137, 140]], [[9, 16]], [[19, 78]], [[1, 84]], [[1, 84]], [[85, 95]], [[96, 112]], [[85, 140]], [[142, 166]]]",
    "query_spans": "[[[168, 178]]]",
    "process": "Let the semi-focal length of ellipse C be c, then F_{1}(-c,0), F_{2}(c,0), line MF_{2}: x=c, from \\begin{cases}x=c\\\\\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}\\end{cases}=1 we obtain \\begin{cases}x=c\\\\y=\\pm\\frac{b^{2}}{a}\\end{cases}. Since the slope of line MF_{1} is positive, then M(c,\\frac{b^{2}}{a}), thus we get \\frac{b^{2}}{c-(-c)}=\\frac{3}{4}, i.e., a^{2}-c^{2}=\\frac{3}{2}ac, then e^{2}+\\frac{3}{2}e-1=0. Since 0<e<1, solving gives e=\\frac{1}{2}. Therefore, the eccentricity of C equals \\frac{1}{2}."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively, if the distance from point $F_{2}$ to an asymptote of this hyperbola is $2$, and point $P$ lies on the hyperbola such that $\\angle F_{1} P F_{2}=60^{\\circ}$, then the area of triangle $F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F1: Point;P: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;Distance(F2, Asymptote(G)) = 2;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[20, 76], [94, 97], [114, 117]], [[23, 76]], [[23, 76]], [[2, 9]], [[109, 113]], [[10, 17], [84, 92]], [[23, 76]], [[23, 76]], [[20, 76]], [[2, 82]], [[2, 82]], [[84, 108]], [[109, 118]], [[120, 153]]]",
    "query_spans": "[[[155, 178]]]",
    "process": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ are given by $y=\\pm\\frac{b}{a}x$. The distance from the right focus $F_{2}(c,0)$ to an asymptote of this hyperbola is 2, so we have $\\frac{\\frac{bc}{a}}{\\sqrt{1+(\\frac{b}{a})^{2}}}=2$, which gives $b=2$. From $\\begin{cases}||PF_{1}|-|PF_{2}|=2a\\\\(2c)^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}||PF_{2}|\\cos60^{\\circ}\\end{cases}$, the area of triangle $F_{1}PF_{2}$ is $\\frac{1}{2}|PF_{1}||PF_{2}|\\sin60^{\\circ}=\\frac{1}{2}\\times16\\times\\frac{\\sqrt{3}}{2}=4\\sqrt{3}$, and $|PF_{1}||PF_{2}|=4b^{2}=16$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the right focus is $F$. A line $l$ passing through $F$ intersects the asymptotes of the hyperbola at points $A$ and $B$, and is perpendicular to one of the asymptotes. If $\\overrightarrow{A F}=3 \\overrightarrow{F B}$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "l: Line;G: Hyperbola;b: Number;a: Number;A: Point;F: Point;B: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;PointOnCurve(F, l);Intersection(l, Asymptote(G)) = {A, B};IsPerpendicular(l,OneOf(Asymptote(G)));VectorOf(A, F) = 3*VectorOf(F, B)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[72, 77]], [[2, 58], [78, 81], [158, 161]], [[5, 58]], [[5, 58]], [[87, 90]], [[63, 66], [68, 71]], [[91, 94]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 66]], [[67, 77]], [[72, 96]], [[72, 108]], [[110, 155]]]",
    "query_spans": "[[[158, 167]]]",
    "process": "By the given condition, the right focus is $ F(c,0) $. Let the equation of one asymptote $ OA $ (where $ O $ is the origin) be $ y = \\frac{b}{a}x $, then the equation of the other asymptote $ OB $ is $ y = -\\frac{b}{a}x $. Let $ A(m, \\frac{bm}{a}) $, $ B(n, -\\frac{bn}{a}) $. Since $ \\overrightarrow{AF} = 3\\overrightarrow{FB} $, we have $ (c - m, -\\frac{bm}{a}) = 3(n - c, -\\frac{bn}{a}) $. Therefore, $ c - m = 3(n - c) $, $ -\\frac{bm}{a} = -\\frac{3bn}{a} $. Solving gives $ m = 2c $, $ n = \\frac{2}{3}c $. Hence, $ B(\\frac{2}{3}c, -\\frac{2bc}{3a}) $. By the given condition, $ FB \\perp OB $, then $ \\frac{2b}{a} \\cdot (-\\frac{b}{a}) = -1 $. Simplifying yields $ 2b^{2} = a^{2} $, i.e., $ 2(c^{2} - a^{2}) = a^{2} $. Solving gives $ 2c^{2} = 3a^{2} $, so $ e = \\frac{c}{a} = \\frac{\\sqrt{6}}{2} $."
  },
  {
    "text": "Given that $F$ is a focus of the hyperbola $C$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, what is the distance from point $F$ to an asymptote of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;F: Point;Expression(C) = (x^2/9 - y^2/16 = 1);OneOf(Focus(C)) = F",
    "query_expressions": "Distance(F, OneOf(Asymptote(C)))",
    "answer_expressions": "4",
    "fact_spans": "[[[6, 50], [62, 68]], [[2, 5], [57, 61]], [[6, 50]], [[2, 55]]]",
    "query_spans": "[[[57, 79]]]",
    "process": "The hyperbola $ C: \\frac{x^{2}}{9} - \\frac{y^{2}}{16} = 1 $ has foci at $ (-5,0) $, $ (5,0) $. By the symmetry of hyperbola $ C $, without loss of generality, take focus $ F(5,0) $, and asymptote $ y = \\frac{4}{3}x $. Then the distance from point $ F $ to the asymptote is $ d = \\frac{|\\frac{4}{3} \\times 5 - 0|}{\\sqrt{1 + (\\frac{4}{3})^{2}}} = 4 $."
  },
  {
    "text": "A focus of the hyperbola $C$ is $F(3,0)$, and its center is at the origin. A line $l$ passing through $F$ intersects $C$ at points $A$ and $B$. If the midpoint of $AB$ is $E(-12,-15)$, then the equations of the asymptotes of this hyperbola are?",
    "fact_expressions": "C: Hyperbola;F: Point;Coordinate(F) = (3, 0);OneOf(Focus(C)) = F;O: Origin;Center(C) = O;l: Line;PointOnCurve(F,l) = True;Intersection(l, C) = {A, B};A: Point;B: Point;E: Point;MidPoint(LineSegmentOf(A,B)) = E;Coordinate(E) = (-12, -15)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y = pm*sqrt(5)*x/2",
    "fact_spans": "[[[0, 6], [38, 41], [78, 81]], [[28, 31], [12, 20]], [[12, 20]], [[0, 20]], [[24, 26]], [[0, 26]], [[32, 37]], [[27, 37]], [[32, 52]], [[43, 46]], [[47, 50]], [[63, 75]], [[54, 75]], [[63, 75]]]",
    "query_spans": "[[[78, 89]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. According to the given conditions, assume the hyperbola equation is $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ $ (a>0, b>0) $. Substitute the coordinates of the two points into the hyperbola equation, subtract the two equations and simplify, obtaining $ k_{AB} = \\frac{b^{2}}{a^{2}} \\cdot \\frac{x_{1}+x_{2}}{y_{1}+y_{2}} $. Find $ \\frac{b^{2}}{a^{2}} $, then the asymptotes of the hyperbola can be determined. From the problem, assume the hyperbola equation is $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ $ (a>0, b>0) $, $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Since line $ l $ passing through $ F $ intersects $ C $ at points $ A $, $ B $, and the midpoint of $ AB $ is $ E(-12,-15) $, so $ x_{1} \\neq x_{2} $, $ y_{1}<0 $, $ y_{2}<0 $, \n$$\n\\begin{cases}\nx_{1}+x_{2}=-24 \\\\\ny_{1}+y_{2}=-30\n\\end{cases}\n$$\nSubtracting the two equations gives $ \\frac{x_{1}^{2}-x_{2}^{2}}{a^{2}} - \\frac{y_{1}^{2}-y_{2}^{2}}{b^{2}} = 0 $, that is, $ \\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{a^{2}} = \\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{b^{2}} $, then $ \\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = \\frac{b^{2}}{a^{2}} \\cdot \\frac{x_{1}+x_{2}}{y_{1}+y_{2}} = \\frac{4b^{2}}{5a^{2}} $, i.e., $ k_{AB} = \\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = \\frac{4b^{2}}{5a^{2}} $. Also $ k_{AB} = k_{EF} = \\frac{0+15}{3+12} = 1 $, so $ \\frac{4b^{2}}{5a^{2}} = 1 $, therefore $ \\frac{b^{2}}{a^{2}} = \\frac{5}{4} $, then $ \\frac{b}{a} = \\frac{\\sqrt{5}}{2} $. Thus, the asymptotes of this hyperbola are $ y = \\pm \\frac{b}{a}x = \\pm \\frac{\\sqrt{5}}{2}x $."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$, respectively, $E$ be any point on the ellipse, and the coordinates of point $N$ be $(5,1)$. Then the maximum value of $|E N|+|E F_{1}|$ is?",
    "fact_expressions": "F1: Point;F2: Point;G: Ellipse;Expression(G) = (x^2/16 + y^2/7 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;E: Point;PointOnCurve(E, G);N: Point;Coordinate(N) = (5, 1)",
    "query_expressions": "Max(Abs(LineSegmentOf(E, F1)) + Abs(LineSegmentOf(E, N)))",
    "answer_expressions": "8+sqrt(5)",
    "fact_spans": "[[[1, 8]], [[9, 16]], [[19, 57], [68, 70]], [[19, 57]], [[1, 63]], [[1, 63]], [[64, 67]], [[64, 74]], [[75, 79]], [[75, 92]]]",
    "query_spans": "[[[94, 117]]]",
    "process": "First, using the definition of the ellipse, transform |EN|+|EF_{1}|=8+(|EN|-|EF_{2}|), and use |EN|-|EF_{2}|\\leqslant|NF_{2}| combined with numerical-geometric analysis to determine the maximum value of the sum of distances. \\because|EF_{1}|+|EF_{2}|=8 \\therefore|EN|+|EF_{1}|=8+(|EN|-|EF_{2}|) \\because|EN|-|EF_{2}|\\leqslant|NF_{2}|, as shown in the figure, equality holds when points E, F_{2}, N are collinear. Therefore, the maximum value of |EN|-|EF_{2}| is |F_{2}N|=\\sqrt{(5-3)^{2}+(1-0)^{2}}=\\sqrt{5}, so the maximum value of |EN|+|EF_{1}| is 8+\\sqrt{5}."
  },
  {
    "text": "If the center of an ellipse is at the origin, the length of the major axis is $4$, and one directrix is given by $x = -4$, then what is the length of the chord intercepted by the line $y = x + 1$ on this ellipse?",
    "fact_expressions": "G: Ellipse;H: Line;O: Origin;Expression(OneOf(Directrix(G))) = (x = -4);Center(G) = O;Length(MajorAxis(G)) = 4;Expression(H) = (y = x + 1)",
    "query_expressions": "Length(InterceptChord(H,G))",
    "answer_expressions": "24/7",
    "fact_spans": "[[[1, 3], [36, 38]], [[39, 48]], [[7, 11]], [[1, 33]], [[1, 11]], [[1, 19]], [[39, 48]]]",
    "query_spans": "[[[35, 55]]]",
    "process": "Let the equation of the ellipse be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. According to the problem, the foci of the ellipse lie on the $x$-axis, and $2a=4$, $\\frac{a^{2}}{c}=4$, solving gives $a=2$, $c=1$, $\\therefore b^{2}=3$, $\\therefore$ the equation of the ellipse is $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$. Combining with $y=x+1$ yields $7x^{2}+8x-8=0$. Let the line $y=x+1$ intersect the ellipse at points $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $x_{1}+x_{2}=-\\frac{8}{7}$, $x_{1}x_{2}=-\\frac{8}{7}$. $\\therefore$ the chord length cut by the line $y=x+1$ on the ellipse is: $|AB|=\\sqrt{2}\\left[(-\\frac{8}{7})^{2}+4\\times\\frac{8}{7}\\right]=\\frac{24}{7}$"
  },
  {
    "text": "Given that $F$ is the focus of the parabola $x^{2}=8 y$, $O$ is the origin, point $P$ is a moving point on the directrix of the parabola, point $A$ is on the parabola and $|A F|=4$, then the minimum value of $|P A|+|P O|$ is?",
    "fact_expressions": "G: Parabola;A: Point;F: Point;P: Point;O: Origin;Expression(G) = (x^2 = 8*y);Focus(G) = F;PointOnCurve(P, Directrix(G));PointOnCurve(A, G);Abs(LineSegmentOf(A, F)) = 4",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, O)))",
    "answer_expressions": "2*sqrt(13)",
    "fact_spans": "[[[6, 20], [36, 39], [51, 54]], [[46, 50]], [[2, 5]], [[31, 35]], [[24, 27]], [[6, 20]], [[2, 23]], [[31, 45]], [[46, 55]], [[57, 66]]]",
    "query_spans": "[[[68, 87]]]",
    "process": "\\because|AF|=4, by the definition of the parabola, \\therefore the distance from A to the directrix is 4, that is, the ordinate of point A is 2; since point A lies on the parabola, \\therefore without loss of generality, take the coordinates of point A as A(4,2). The coordinates of the point symmetric to the origin with respect to the directrix are B(0,4), then the minimum value of |PA|+|PO| is: |AB|=\\sqrt{(4-0)^{2}+(2+4)^{2}}=2\\sqrt{13}, hence the answer is 2\\sqrt{13}"
  },
  {
    "text": "What is the eccentricity of the hyperbola $\\frac{y^{2}}{4}-\\frac{x^{2}}{12}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/12 + y^2/4 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 45]]]",
    "process": "The solution process is omitted"
  },
  {
    "text": "Given that on the parabola $y^{2}=2 p x(p>0)$, the point with horizontal coordinate $1$ has equal distance to the vertex and to the directrix, then the equation of this parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;A:Point;XCoordinate(A) = 1;PointOnCurve(A,G) = True;Distance(A,Vertex(G)) = Distance(A,Directrix(G))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=8*x",
    "fact_spans": "[[[2, 23], [51, 54]], [[2, 23]], [[5, 23]], [[5, 23]], [[32, 33]], [[24, 33]], [[2, 33]], [[2, 48]]]",
    "query_spans": "[[[51, 59]]]",
    "process": "Let point P be the point on the parabola with horizontal coordinate 1, and let F be the focus of the parabola. By the definition of a parabola, |PO| = |PF|. Since $ x_{F} = 2x_{P} = 2 $, it follows that $ \\frac{p}{2} = 2 $. $ \\therefore p = 4 $, and the equation of the parabola is $ y^{2} = 8x $."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=2 x$ is $F$, point $P$ is a moving point on the parabola, and there is a point $A(0,2)$, find the minimum value of $P A+P F$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);F: Point;Focus(G) = F;P: Point;PointOnCurve(P, G);A: Point;Coordinate(A) = (0, 2)",
    "query_expressions": "Min(LineSegmentOf(P, A) + LineSegmentOf(P, F))",
    "answer_expressions": "sqrt(17)/2",
    "fact_spans": "[[[2, 16], [29, 32]], [[2, 16]], [[20, 23]], [[2, 23]], [[24, 28]], [[24, 36]], [[39, 48]], [[39, 48]]]",
    "query_spans": "[[[50, 64]]]",
    "process": "The focus of the parabola \\( y^{2} = 2x \\) is \\( F\\left(\\frac{1}{2}, 0\\right) \\), hence \\( PA + PF \\geqslant AF = \\sqrt{2^{2} + \\left(\\frac{1}{2}\\right)^{2}} = \\frac{\\sqrt{17}}{2} \\), and equality holds when \\( A \\), \\( P \\), and \\( F \\) are collinear."
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ has eccentricity $\\sqrt{2}$, then the distance from the right vertex of the hyperbola $C$ to the asymptote of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2/b^2 = 1);b: Number;b>0;Eccentricity(C) = sqrt(2)",
    "query_expressions": "Distance(RightVertex(C), Asymptote(C))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 44], [61, 67], [72, 75]], [[2, 44]], [[10, 44]], [[10, 44]], [[2, 59]]]",
    "query_spans": "[[[61, 84]]]",
    "process": ""
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$. If a point $M$ on the hyperbola is at a distance of $7$ from one of its foci, find the distance from point $M$ to the other focus.",
    "fact_expressions": "F1: Point;F2: Point;Focus(G) = {F1, F2};G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);M: Point;PointOnCurve(M, G);Distance(M, F1) = 7",
    "query_expressions": "Distance(M, F2)",
    "answer_expressions": "13",
    "fact_spans": "[[[0, 7]], [[8, 15]], [[0, 60]], [[16, 55], [62, 65], [72, 73]], [[16, 55]], [[68, 71], [88, 92]], [[62, 71]], [[68, 86]]]",
    "query_spans": "[[[72, 103]]]",
    "process": "First, derive $a$, $b$, $c$ from the equation of the hyperbola. Determine on which branch the point $M$ lies based on the properties of the hyperbola, and combine with the definition of the hyperbola to obtain the answer. [Detailed solution] From the hyperbola $\\frac{x^2}{9}-\\frac{y^{2}}{16}=1$, we get $a=3$, $b=4$, $c=5$. By the definition of the hyperbola, $||MF_1|-|MF_{2}||=2a=6$. Without loss of generality, assume point $M$ lies on the right branch of the hyperbola, and let $F_1$, $F_2$ be the left and right foci of the hyperbola, respectively. Then $|MF_{1}|\\geq a+c=8$. Given that the distance from point $M$ to one of its foci is 7, it must be that $|MF_{2}|=7$. By the definition of the hyperbola, $||MF_1|-|MF_{2}||=2a=6$, that is, $|MF_{1}|-|MF_{2}|=2a=6$. Thus $|MF_1|-7=2a=6$, so $|MF_1|=13$. Therefore, the distance from point $M$ to the other focus is 13."
  },
  {
    "text": "Let $P$ be a moving point on the parabola $y^{2}=4x$. Then the minimum value of the sum of the distance from point $P$ to the $y$-axis and the distance from $P$ to the point $A(2,3)$ is equal to?",
    "fact_expressions": "G: Parabola;A: Point;P:Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (2, 3);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P,yAxis)+Distance(P,A))",
    "answer_expressions": "sqrt(10)-1",
    "fact_spans": "[[[4, 18]], [[37, 46]], [[0, 3], [24, 28]], [[4, 18]], [[37, 46]], [[0, 22]]]",
    "query_spans": "[[[24, 57]]]",
    "process": ""
  },
  {
    "text": "The equation of an ellipse that has the same foci as the hyperbola $\\frac{x^{2}}{5}-\\frac{y^{2}}{4}=-1$ and has eccentricity $0.6$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/5 - y^2/4 = -1);H: Ellipse;Focus(G) = Focus(H);Eccentricity(H) = 0.6",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2/16+y^2/25=1",
    "fact_spans": "[[[1, 40]], [[1, 40]], [[57, 59]], [[0, 59]], [[47, 59]]]",
    "query_spans": "[[[57, 63]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x$ coincides with the right focus of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, then the equation of the directrix of this parabola is?",
    "fact_expressions": "G: Parabola;p: Number;H: Ellipse;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (x^2/9 + y^2/5 = 1);Focus(G) = RightFocus(H)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=-2",
    "fact_spans": "[[[1, 17], [67, 70]], [[4, 17]], [[21, 58]], [[1, 17]], [[21, 58]], [[1, 64]]]",
    "query_spans": "[[[66, 77]]]",
    "process": "The right focus of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ is $(2,0)$, which is also the focus of the parabola $y^{2}=2px$. Thus, $2=\\frac{p}{2}$, so $p=4$. Therefore, the directrix of the parabola is $x=-\\frac{p}{2}=-2$. Hence, the directrix equation of the parabola is $x=-2$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the distance from its right focus to one of its asymptotes is $4$, and the focal length is $10$. What is the equation of the asymptotes of this hyperbola?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Distance(RightFocus(C),OneOf(Asymptote(C)))=4;FocalLength(C) = 10",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*(4/3)*x",
    "fact_spans": "[[[2, 63], [68, 69], [94, 97]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 82]], [[68, 91]]]",
    "query_spans": "[[[94, 105]]]",
    "process": "From the standard equation, the asymptote equations can be obtained. By using the distance from a point to a line to construct an equation, the value of \\frac{b}{a} is found, thus obtaining the asymptote equation. [Detailed solution] \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 \\Rightarrow the asymptote equations are: y=\\pm\\frac{b}{a}x. Due to the symmetry of the hyperbola, the distances from both foci to the two asymptotes are equal. \\therefore take the asymptote y=\\frac{b}{a}x^{\\circ}. Also, since the focal distance is 10, the right focus is (5,0). \\therefore \\frac{|5b|}{\\sqrt{a^{2}+b^{2}}}=4 \\Rightarrow \\frac{b}{a}=\\frac{4}{3}, \\therefore the asymptote equations are: y=\\pm\\frac{4}{3}x"
  },
  {
    "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are $y=\\pm 2 x$, then the value of its eccentricity is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(Asymptote(G)) = (y = pm*(2*x))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[1, 57], [75, 76]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 73]]]",
    "query_spans": "[[[75, 83]]]",
    "process": ""
  },
  {
    "text": "Given that point $P(x, y)$ is an arbitrary point on the parabola $y^{2}=4x$, and point $Q$ is an arbitrary point on the circle $(x+2)^{2}+(y-4)^{2}=1$, then the minimum value of $|PQ|+x$ is?",
    "fact_expressions": "P: Point;x0: Number;y0: Number;Coordinate(P) = (x0, y0);PointOnCurve(P, G);G: Parabola;Expression(G) = (y^2 = 4*x);H: Circle;Expression(H) = ((x + 2)^2 + (y - 4)^2 = 1);Q: Point;PointOnCurve(Q, H)",
    "query_expressions": "Min(x0 + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 13]], [[3, 13]], [[3, 13]], [[2, 13]], [[2, 33]], [[14, 28]], [[14, 28]], [[38, 62]], [[38, 62]], [[34, 37]], [[34, 67]]]",
    "query_spans": "[[[69, 84]]]",
    "process": "Draw the graph, let the focus be F(1,0). By the definition of a parabola, PF = x + 1, so x = PF - 1. Also, PQ + QC \\geqslant CP, with equality if and only if C, Q, P are collinear and Q is the intersection point of CP with circle C, then |PQ| takes the minimum value of |PC| - |QC| = |PC| - 1. Hence, the minimum value of |PQ| + x is |PC| - 1 + |PF| - 1 = |PC| + |PF| - 2. Moreover, when P is the intersection point of segment CF and the parabola, |PC| + |PF| takes the minimum value. At this time, |PQ| + x = |PC| + |PF| - 2 = |CF| - 2 = \\sqrt{[1-(-2)]^{2}+(0-4)^{2}}-2="
  },
  {
    "text": "Let $P$ be a moving point on the curve $2 x^{2}-y^{2}=1$, $O$ be the coordinate origin, and $M$ be the midpoint of the segment $O P$. Then the equation of the trajectory of point $M$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P, G);G: Curve;Expression(G) = (2*x^2 - y^2 = 1);O: Origin;M: Point;MidPoint(LineSegmentOf(O, P)) = M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "8*x^2-4*y^2=1",
    "fact_spans": "[[[1, 4]], [[1, 29]], [[5, 24]], [[5, 24]], [[30, 33]], [[39, 42], [55, 59]], [[39, 53]]]",
    "query_spans": "[[[55, 66]]]",
    "process": "Let $ P(x,y) $, $ M(x_{0},y_{0}) $. Since $ M $ is the midpoint of segment $ OP $, we have \n\\[\n\\begin{cases}\nx_{0} = \\frac{x}{2} \\\\\ny_{0} = \\frac{y}{2}\n\\end{cases}\n\\quad\n(x = 2x_{0}\n\\begin{cases}\ny = 2y \\\\\n\\end{cases}\n\\text{so} \\quad\n2 \\times (2x_{0})^{2} - (y_{0}^{2} = 1, \\quad\n\\text{i.e.,} \\quad\n8x_{0}^{2} - y_{0}^{2}.\n\\]"
  },
  {
    "text": "The locus of the center of a moving circle that is externally tangent to circle $C_{1}$: $(x+3)^{2}+y^{2}=9$ and internally tangent to circle $C_{2}$: $(x-3)^{2}+y^{2}=1$ is?",
    "fact_expressions": "Z: Circle;C1: Circle;Expression(C1) = ((x + 3)^2 + y^2 = 9);IsOutTangent(Z, C1);C2: Circle;Expression(C2) = ((x - 3)^2 + y^2 = 1);IsInTangent(Z, C2)",
    "query_expressions": "LocusEquation(Center(Z))",
    "answer_expressions": "(x^2/4-y^2/5=1)&(x>=2)",
    "fact_spans": "[[[64, 66]], [[1, 29]], [[1, 29]], [[0, 66]], [[33, 61]], [[33, 61]], [[32, 66]]]",
    "query_spans": "[[[64, 72]]]",
    "process": ""
  },
  {
    "text": "Draw a line $l$ with slope $\\frac{1}{2}$ passing through the left focus $F$ of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, intersecting $C$ at points $A$ and $B$. If $|O F|=|O A|$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;LeftFocus(C) = F;l: Line;PointOnCurve(F, l);Slope(l) = 1/2;A: Point;B: Point;Intersection(l, C) = {A, B};O: Origin;Abs(LineSegmentOf(O, F)) = Abs(LineSegmentOf(O, A))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[1, 58], [120, 125], [89, 92]], [[1, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[62, 65]], [[1, 65]], [[83, 88]], [[0, 88]], [[66, 88]], [[94, 97]], [[98, 101]], [[83, 103]], [[105, 118]], [[105, 118]]]",
    "query_spans": "[[[120, 131]]]",
    "process": "Draw a schematic diagram, denote the right focus as $ F_{2} $. Calculate the lengths of $ AF $ and $ AF_{2} $ based on length and position relationships. Then, set up the corresponding equation according to the shape of $ \\triangle AFF_{2} $, and solve for the eccentricity $ e $. As shown in the figure, let $ M $ be the midpoint of $ AF $, the right focus be $ F_{2} $, connect $ MO $, $ AF_{2} $, so $ MO \\parallel AF_{2} $. Since $ |OA| = |OF| $, it follows that $ OM \\perp AF $, hence $ AF \\perp AF_{2} $. Given $ k_{AF} = \\frac{1}{2} $, we have $ \\frac{AF_{2}}{AF} = \\frac{1}{2} $ and $ AF_{2} + AF = 2a $, thus $ AF = \\frac{4a}{3} $, $ AF_{2} = \\frac{2a}{3} $. Also, since $ AF^{2} + AF_{2}^{2} = FF_{2}^{2} $, it follows that $ \\frac{16a^{2}}{9} + \\frac{4a^{2}}{9} = 4c^{2} $, so $ \\frac{c^{2}}{a^{2}} = \\frac{5}{9} $, therefore $ e = \\frac{\\sqrt{5}}{3} $."
  },
  {
    "text": "Given a line passing through the focus $F$ of the parabola $y^{2}=8x$ with an inclination angle of $60^{\\circ}$, intersecting the parabola at points $A$ and $B$, then the distance from the midpoint of segment $AB$ to the $y$-axis is?",
    "fact_expressions": "G: Parabola;H: Line;B: Point;A: Point;F: Point;Expression(G) = (y^2 = 8*x);Focus(G) = F;PointOnCurve(F, H);Inclination(H) = ApplyUnit(60, degree);Intersection(H, G) = {A, B}",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A,B)),yAxis)",
    "answer_expressions": "10/3",
    "fact_spans": "[[[3, 17], [44, 47]], [[41, 43]], [[52, 55]], [[48, 51]], [[20, 23]], [[3, 17]], [[3, 23]], [[2, 43]], [[24, 43]], [[41, 57]]]",
    "query_spans": "[[[59, 79]]]",
    "process": "The focus of the parabola $ y^{2} = 8x $ is $ F(2,0) $. Let points $ A(x_{1}, y_{1}) $ and $ B(x_{2}, y_{2}) $. According to the given conditions, the equation of line $ AB $ is $ y = \\sqrt{3}(x - 2) $. Solving the system \n\\[\n\\begin{cases}\ny^{2} = 8x \\\\\ny = \\sqrt{3}(x - 2)\n\\end{cases}\n\\]\nby eliminating $ y $ and simplifying, we obtain $ 3x^{2} - 20x + 12 = 0 $. By Vieta's formulas, we have $ x_{1} + x_{2} = \\frac{20}{3} $, so the x-coordinate of the midpoint of segment $ AB $ is $ \\frac{x_{1} + x_{2}}{2} = \\frac{10}{3} $. Therefore, the distance from the midpoint of segment $ AB $ to the y-axis is $ \\frac{10}{3} $."
  },
  {
    "text": "Given point $P$ is a moving point on the hyperbola $C$: $\\frac{x^{2}}{36}-\\frac{y^{2}}{64}=1$, point $A(-10,0)$, point $B(10,0)$. If $|P A|=15$, then $|P B|$=?",
    "fact_expressions": "P: Point;C: Hyperbola;Expression(C) = (x^2/36 - y^2/64 = 1);PointOnCurve(P, C) = True;A: Point;Coordinate(A) = (-10, 0);B: Point;Coordinate(B) = (10, 0);Abs(LineSegmentOf(P, A)) = 15",
    "query_expressions": "Abs(LineSegmentOf(P, B))",
    "answer_expressions": "27",
    "fact_spans": "[[[2, 6]], [[7, 52]], [[7, 52]], [[2, 56]], [[57, 68]], [[57, 68]], [[69, 79]], [[69, 79]], [[82, 92]]]",
    "query_spans": "[[[94, 103]]]",
    "process": "According to the problem, for the hyperbola, $ a=6 $, $ b=8 $, $ c=\\sqrt{36+64}=10 $. Thus, $ A $ and $ B $ are the left and right foci of the hyperbola. By the definition of a hyperbola, we have $ ||PA|-|PB|| = |15-|PB|| = 2a = 12 $, so $ |PB| = 27 $ or $ |PB| = 3 $. Since $ c-a = 10-6 = 4 $, it follows that $ |PB| \\geqslant 4 $. Therefore, $ |PB| = 3 $ is invalid, and hence $ |PB| = 27 $."
  },
  {
    "text": "Given the equation of the parabola is $y^{2}=8x$, the right focus of the hyperbola coincides with the focus of the parabola, and the eccentricity is $2$. Then, what is the standard equation of the hyperbola? What are its asymptote equations?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 8*x);G: Hyperbola;RightFocus(G) = Focus(H);Eccentricity(G) = 2",
    "query_expressions": "Expression(G);Expression(Asymptote(G))",
    "answer_expressions": "x^2 - y^2 / 3 = 1  \ny = pm * sqrt(3) * x",
    "fact_spans": "[[[2, 5], [29, 32]], [[2, 20]], [[21, 24], [45, 48], [55, 56]], [[21, 35]], [[21, 43]]]",
    "query_spans": "[[[45, 55]], [[55, 63]]]",
    "process": ""
  },
  {
    "text": "The distance from the focus to the directrix of the parabola $y^{2}=10 x$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 10*x)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "5",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 26]]]",
    "process": "Since 2p=10, we have p=5, that is, the distance from the focus to the directrix is 5"
  },
  {
    "text": "The distance from the focus of the parabola $y^{2}=-x$ to its directrix is equal to?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -x)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "1/2",
    "fact_spans": "[[[0, 13], [17, 18]], [[0, 13]]]",
    "query_spans": "[[[0, 27]]]",
    "process": "According to the distance from the focus of the parabola to the directrix being equal to p, the desired result is \\frac{1}{2}."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{36}-\\frac{y^{2}}{9}=1$, what are the equations of its asymptotes?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/36 - y^2/9 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*x/2",
    "fact_spans": "[[[2, 41], [42, 43]], [[2, 41]]]",
    "query_spans": "[[[42, 51]]]",
    "process": "From the given: \\frac{x^{2}}{36}-\\frac{y^{2}}{9}=0,\\frac{x^{2}}{36}=\\frac{y^{2}}{9},y=\\pm\\frac{1}{2}x"
  },
  {
    "text": "Let a line passing through the origin intersect the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) at two distinct points $P$ and $Q$, and let $F$ be a focus of $C$. If $\\tan \\angle P F Q = \\frac{4}{3}$ and $|Q F| = 3|P F|$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;P: Point;F: Point;Q: Point;O:Origin;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(O,G);Intersection(G, C) = {P, Q};Negation(P=Q);OneOf(Focus(C)) = F;Tan(AngleOf(P, F, Q)) = 4/3;Abs(LineSegmentOf(Q, F)) = 3*Abs(LineSegmentOf(P, F))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(85)/5",
    "fact_spans": "[[[8, 69], [88, 91], [147, 153]], [[16, 69]], [[16, 69]], [[5, 7]], [[71, 74]], [[84, 87]], [[75, 78]], [[2, 4]], [[16, 69]], [[16, 69]], [[8, 69]], [[1, 7]], [[5, 83]], [[71, 83]], [[84, 96]], [[98, 129]], [[131, 145]]]",
    "query_spans": "[[[147, 159]]]",
    "process": ""
  },
  {
    "text": "Let the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ always pass through the fixed point $A(1,2)$. What is the minimum distance from the center of the ellipse to its directrix?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;A: Point;Coordinate(A) = (1, 2);PointOnCurve(A,C) = True",
    "query_expressions": "Min(Distance(Center(C),Directrix(C)))",
    "answer_expressions": "sqrt(5)+2",
    "fact_spans": "[[[1, 58], [72, 74]], [[1, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[62, 70]], [[62, 70]], [[1, 70]]]",
    "query_spans": "[[[72, 88]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$ and directrix $l$, $P$ is a point on $l$, $Q$ is an intersection point of line $PF$ and $C$, if $\\overrightarrow{FP}=4\\overrightarrow{FQ}$, then $|QF|$=?",
    "fact_expressions": "C: Parabola;F: Point;P: Point;Q: Point;l: Line;Expression(C) = (y^2 = 8*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(P, l);OneOf(Intersection(LineOf(P,F),C))=Q;VectorOf(F, P) = 4*VectorOf(F, Q)",
    "query_expressions": "Abs(LineSegmentOf(Q, F))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 21], [60, 63]], [[25, 28]], [[37, 40]], [[48, 51]], [[32, 35], [41, 44]], [[2, 21]], [[2, 28]], [[2, 35]], [[36, 47]], [[48, 68]], [[70, 115]]]",
    "query_spans": "[[[117, 126]]]",
    "process": "As shown in the figure: draw $ QQ \\bot l $ passing through point $ Q $, intersecting $ l $ at point $ Q $. Using the definition of a parabola, we obtain $ |QF| = |QQ| $. Let the directrix $ l $ intersect the $ x $-axis at point $ M $. Since $ \\overrightarrow{FP} = 4\\overrightarrow{FO} $, it follows that $ |QQ| : |FM| = |PQ| : |PF| = 3 : 4 $. Given that the distance from the focus $ F $ to the directrix $ l $ is $ 4 $, we have $ |QQ| = 3 $. Therefore, $ |OF| = |OO| = 3 $. The answer is: $ 3 $"
  },
  {
    "text": "A point $P$ on the parabola $y^{2}=x$ is at a distance of $2$ from the focus. Then the coordinates of point $P$ are?",
    "fact_expressions": "G: Parabola;P: Point;Expression(G) = (y^2 = x);PointOnCurve(P, G);Distance(P, Focus(G)) = 2",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(7/4, pm*sqrt(7)/2)",
    "fact_spans": "[[[0, 12]], [[15, 18], [30, 34]], [[0, 12]], [[0, 18]], [[0, 28]]]",
    "query_spans": "[[[30, 38]]]",
    "process": "From the given conditions, the equation of the directrix of the parabola is $x = -\\frac{1}{4}$. Let the coordinates of point $P$ be $(x_0, y_0)$. Then, by the definition of a parabola, $x_0 + \\frac{1}{4} = 2$, solving gives $x_0 = \\frac{7}{4}$. At this time, $y_0^2 = \\frac{7}{4}$, solving gives $y_0 = \\pm\\frac{\\sqrt{7}}{2}$. Therefore, the coordinates of point $P$ are $\\frac{7}{4}, \\pm\\frac{\\sqrt{7}}{2}$. Answer: $\\frac{7}{4}, \\pm\\frac{\\sqrt{7}}{2}$"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the right branch of the hyperbola, and $|P F_{1}| \\cdot |P F_{2}| = 32$. Then $\\angle F_{1} P F_{2} = $?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 32",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "pi/2",
    "fact_spans": "[[[2, 41], [71, 74]], [[2, 41]], [[50, 57]], [[58, 65]], [[2, 65]], [[2, 65]], [[66, 70]], [[66, 78]], [[80, 109]]]",
    "query_spans": "[[[111, 135]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ has a focal distance of $10 \\sqrt{5}$, and the point $P(1,2)$ lies on the asymptote of the hyperbola $C$, then the equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-x^2/b^2 + y^2/a^2 = 1);b: Number;a: Number;FocalLength(C) = 10*sqrt(5);P: Point;Coordinate(P) = (1, 2);PointOnCurve(P, Asymptote(C))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/100 - x^2/25 = 1",
    "fact_spans": "[[[2, 53], [81, 87], [94, 100]], [[2, 53]], [[10, 53]], [[10, 53]], [[2, 70]], [[71, 80]], [[71, 80]], [[71, 92]]]",
    "query_spans": "[[[94, 105]]]",
    "process": "The hyperbola $ C: \\frac{y^{2}}{a^{2}} - \\frac{x^{2}}{b^{2}} = 1 $ has a focal distance of $ 10\\sqrt{5} $, so $ 2c = 10\\sqrt{5} $, that is, $ c = 5\\sqrt{5} $, and thus $ a^{2} + b^{2} = 125 $. The asymptotes of the hyperbola are given by $ y = \\pm \\frac{a}{b}x $. Since point $ P(1,2) $ lies on an asymptote of hyperbola $ C $, we have $ a = 2b $. Solving gives $ a = 10 $, $ b = 5 $, yielding the hyperbola equation $ \\frac{y^{2}}{100} - \\frac{x^{2}}{25} = 1 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with right focus $F$, if a line passing through point $F$ with an inclination angle of $60^{\\circ}$ intersects the right branch of the hyperbola at exactly one point, then what is the range of values for the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(G) = F;PointOnCurve(F, H) = True;Inclination(H) = ApplyUnit(60, degree);H: Line;NumIntersection(H, RightPart(G)) = 1",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[2,+oo)",
    "fact_spans": "[[[2, 61], [97, 100], [114, 117]], [[2, 61]], [[5, 61]], [[5, 61]], [[5, 61]], [[5, 61]], [[66, 69], [72, 76]], [[2, 69]], [[71, 96]], [[77, 96]], [[94, 96]], [[94, 111]]]",
    "query_spans": "[[[114, 127]]]",
    "process": ""
  },
  {
    "text": "Suppose the vertex of the parabola is at the origin, and its focus $F$ lies on the $x$-axis. If the point $P(2 , k)$ on the parabola is at a distance of $3$ from the point $F$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;Vertex(G) = O;O: Origin;F: Point;Focus(G) = F;PointOnCurve(F, xAxis);P: Point;Coordinate(P) = (2, k);k: Number;PointOnCurve(P, G) = True;Distance(P, F) = 3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[1, 4], [11, 12], [24, 27], [54, 57]], [[1, 10]], [[8, 10]], [[14, 17], [41, 45]], [[11, 17]], [[14, 23]], [[29, 40]], [[29, 40]], [[30, 40]], [[24, 40]], [[29, 52]]]",
    "query_spans": "[[[54, 61]]]",
    "process": "From the given conditions, the parabola opens to the right. Let the equation of the parabola be $ y^{2} = 2px $, $ (p > 0) $, with focus $ F\\left(\\frac{p}{2}, 0\\right) $ and directrix $ l: x = -\\frac{p}{2} $. According to the definition of the parabola, $ 2 - \\left(-\\frac{p}{2}\\right) = 3 $, solving gives $ p = 2 $. Therefore, the equation of this parabola is $ y^{2} = 4x $. Key point: definition and equation of the parabola"
  },
  {
    "text": "Given that the distance from the left focus to the right directrix of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $\\frac{7 \\sqrt{3}}{3}$, and the distance from the center to the directrix is $\\frac{4 \\sqrt{3}}{3}$, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;a: Number;b: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Distance(LeftFocus(G), RightDirectrix(G)) = 7*sqrt(3)/3;Distance(Center(G), Directrix(G)) = 4*sqrt(3)/3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 + y^2 = 1",
    "fact_spans": "[[[2, 55], [123, 125]], [[4, 55]], [[4, 55]], [[4, 55]], [[4, 55]], [[2, 55]], [[2, 89]], [[2, 121]]]",
    "query_spans": "[[[123, 130]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}(-c, 0)$ and $F_{2}(c, 0)$ respectively. If there exists a point $P$ on the hyperbola such that $\\frac{\\sin \\angle P F_{1} F_{2}}{\\sin \\angle P F_{2} F_{1}}=\\frac{a}{c}$, then the range of values for the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F1: Point;F2: Point;P: Point;a>0;b>0;c:Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(P,G);Sin(AngleOf(P,F1,F2))/Sin(AngleOf(P,F2,F1))=a/c",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1, \\sqrt{2}+1)",
    "fact_spans": "[[[2, 58], [98, 101], [186, 189]], [[5, 58]], [[5, 58]], [[67, 81]], [[83, 96]], [[106, 109]], [[5, 58]], [[5, 58]], [[67, 81]], [[2, 58]], [[67, 81]], [[83, 96]], [[2, 96]], [[2, 96]], [[98, 109]], [[110, 183]]]",
    "query_spans": "[[[186, 200]]]",
    "process": ""
  },
  {
    "text": "Given that $O$ is the coordinate origin, $A$ and $B$ are the left and right vertices of the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$, respectively, and $M$ is a moving point on the hyperbola $C$ distinct from $A$ and $B$. The lines $AM$ and $BM$ intersect the $y$-axis at points $P$ and $Q$, respectively. Then $|OP| \\cdot |OQ|$ = ?",
    "fact_expressions": "C: Hyperbola;M: Point;A: Point;B: Point;O: Origin;P: Point;Q: Point;Expression(C) = (x^2/4 - y^2/3 = 1);LeftVertex(C) = A;RightVertex(C) = B;PointOnCurve(M, C);Negation(M = A);Negation(M = B);Intersection(LineOf(A, M),yAxis)=P;Intersection(LineOf(B, M), yAxis) = Q",
    "query_expressions": "Abs(LineSegmentOf(O, P))*Abs(LineSegmentOf(O, Q))",
    "answer_expressions": "3",
    "fact_spans": "[[[21, 64], [75, 81]], [[71, 74]], [[11, 14], [85, 88]], [[15, 18], [89, 92]], [[2, 5]], [[120, 124]], [[125, 128]], [[21, 64]], [[11, 70]], [[11, 70]], [[71, 95]], [[71, 95]], [[71, 95]], [[96, 130]], [[96, 130]]]",
    "query_spans": "[[[132, 152]]]",
    "process": "According to the problem, A(-2,0), B(2,0). Let M(x_{0},y_{0}), then \\frac{x_{0}^{2}}{4}-\\frac{y_{0}^{2}}{3}=1, x_{0}^{2}-4=\\frac{4}{3}y_{0}^{2}. The equation of line AM is y=\\frac{y_{0}}{x_{0}+2}(x+2), so y_{P}=\\frac{2y_{0}}{x_{0}+2}. The equation of line BM is y=\\frac{y_{0}}{x_{0}-2}(x-2), so y_{Q}=\\frac{-2y_{0}}{x_{0}-2}. Therefore, |OP|\\cdot|OQ|=|\\frac{4y_{0}^{2}}{x_{0}^{2}-4}|=|\\frac{4y_{0}^{2}}{\\frac{4}{3}y_{0}^{2}}|=3."
  },
  {
    "text": "Given that there are three points $A$, $B$, $C$ on the hyperbola $x^{2}-\\frac{y^{2}}{8}=1$, and the midpoints of $AB$, $BC$, $AC$ are $D$, $E$, $F$ respectively. Let $k$ denote the slope. If $k_{OD}+k_{OE}+k_{OF}=-8$ (where point $O$ is the origin, and $k_{OD}$, $k_{OE}$, $k_{OF}$ are all non-zero), then $\\frac{1}{k_{AB}}+\\frac{1}{k_{BC}}+\\frac{1}{k_{AC}}$=?",
    "fact_expressions": "G: Hyperbola;A: Point;B: Point;C: Point;O: Origin;D: Point;E: Point;F: Point;Expression(G) = (x^2 - y^2/8 = 1);PointOnCurve(A, G);PointOnCurve(B, G);PointOnCurve(C, G);MidPoint(LineSegmentOf(A, B)) = D;MidPoint(LineSegmentOf(B, C)) = E;MidPoint(LineSegmentOf(A, C)) = F;k1: Number;k2: Number;k3: Number;Slope(LineSegmentOf(O, F)) = k3;Slope(LineSegmentOf(O, D)) = k1;Slope(LineSegmentOf(O, E)) = k2;k1 + k2 + k3 = -8;Negation(k1 = 0);Negation(k2 = 0);Negation(k3 = 0);k4: Number;k5: Number;k6: Number;Slope(LineSegmentOf(A, B)) = k4;Slope(LineSegmentOf(B, C)) = k5;Slope(LineSegmentOf(A, C)) = k6",
    "query_expressions": "1/k4 + 1/k5 + 1/k6",
    "answer_expressions": "-1",
    "fact_spans": "[[[2, 30]], [[35, 38]], [[39, 42]], [[43, 46]], [[123, 127]], [[70, 73]], [[74, 77]], [[78, 81]], [[2, 30]], [[2, 46]], [[2, 46]], [[2, 46]], [[47, 81]], [[47, 81]], [[47, 81]], [[134, 143]], [[146, 156]], [[158, 167]], [[158, 167]], [[134, 143]], [[146, 156]], [[94, 122]], [[134, 171]], [[134, 171]], [[134, 171]], [[174, 229]], [[174, 229]], [[174, 229]], [[174, 229]], [[174, 229]], [[174, 229]]]",
    "query_spans": "[[[174, 231]]]",
    "process": ""
  },
  {
    "text": "What is the sum of the distances from a point on the ellipse $7 x^{2}+3 y^{2}=21$ to its two foci?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Expression(G) = (7*x^2 + 3*y^2 = 21);Focus(G) = {F1, F2};P:Point;PointOnCurve(P,G)",
    "query_expressions": "Distance(P, F1)+Distance(P, F2)",
    "answer_expressions": "2*sqrt(7)",
    "fact_spans": "[[[0, 22]], [], [], [[0, 22]], [[0, 30]], [], [[0, 25]]]",
    "query_spans": "[[[0, 37]]]",
    "process": "According to the problem, the standard equation of the ellipse is $\\frac{x^{2}}{3}+\\frac{y^{2}}{7}=1$, so $a=\\sqrt{7}$, $2a=2\\sqrt{7}$."
  },
  {
    "text": "If the asymptotes of a hyperbola with foci on the $x$-axis are given by $y = \\pm \\frac{3}{2} x$, and the length of the imaginary axis is $6$, then the length of the real axis is?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), xAxis);Expression(Asymptote(G)) = (y = pm*(3/2)*x);Length(ImageinaryAxis(G)) = 6",
    "query_expressions": "Length(RealAxis(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[10, 13]], [[1, 13]], [[10, 41]], [[10, 49]]]",
    "query_spans": "[[[10, 56]]]",
    "process": "The equation of a hyperbola with foci on the x-axis is given by $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$. From the asymptotes $y=\\pm\\frac{3}{2}x$, it follows that $\\frac{b}{a}=\\frac{3}{2}$, i.e., $2b=3a$. The length of the imaginary axis is 6, i.e., $2b=6$, so $b=3$, and thus $a=2$. Therefore, the length of the real axis is 4."
  },
  {
    "text": "Given that lines $l_{1}$ and $l_{2}$ pass through the origin and both have positive slopes, the inclination angle of $l_{2}$ is twice that of $l_{1}$, and $l_{2}$ is an asymptote of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$. A perpendicular is drawn from the right focus $F$ of $C$ to $l_{1}$, with foot $M$. If $M$ lies exactly on $C$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);l1:Line;l2:Line;O:Origin;PointOnCurve(O,l1);PointOnCurve(O,l2);Slope(l1)>0;Slope(l2)>0;Inclination(l2)=2*Inclination(l1);OneOf(Asymptote(C))=l2;RightFocus(C)=F;L:Line;PointOnCurve(F,L);IsPerpendicular(l1,L);FootPoint(l1,L)=M;PointOnCurve(M,C);M:Point;F:Point",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)-1",
    "fact_spans": "[[[68, 129], [137, 140], [172, 175], [178, 181]], [[75, 129]], [[75, 129]], [[75, 129]], [[75, 129]], [[68, 129]], [[2, 11], [49, 56], [148, 155]], [[13, 20], [36, 44], [60, 67]], [[21, 25]], [[2, 25]], [[2, 25]], [[2, 34]], [[2, 34]], [[37, 59]], [[60, 135]], [[137, 147]], [], [[136, 158]], [[136, 158]], [[136, 165]], [[166, 176]], [[162, 165], [166, 169]], [[144, 147]]]",
    "query_spans": "[[[178, 187]]]",
    "process": "Extend FM to intersect $ l_{2} $ at point N. Since the inclination angle of $ l_{2} $ is twice that of $ l_{1} $, $ l_{1} $ is the angle bisector of the angle between $ l_{2} $ and the x-axis, so FM = MN, hence OF = ON = c. Since $ l_{2} $ is an asymptote of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $), the equation of $ l_{2} $ is $ y = \\frac{b}{a}x $. Thus, N has coordinates $ (a, b) $. Given F($ c, 0 $), M has coordinates $ \\left( \\frac{a+c}{2}, \\frac{b}{2} \\right) $. Since M lies exactly on C, we have $ \\frac{\\left( \\frac{a+c}{2} \\right)^{2}}{a^{2}} - \\frac{\\left( \\frac{b}{2} \\right)^{2}}{b^{2}} = 1 $. Simplifying yields $ c^{2} + 2ac - 4a^{2} = 0 $, thus $ e^{2} + 2e - 4 = 0 $. Solving gives $ e = -1 \\pm \\sqrt{5} $. Since $ e > 1 $, $ e = \\sqrt{5} - 1 $. Therefore, the answer is $ \\sqrt{5} - 1 $."
  },
  {
    "text": "If the vertex of a parabola is at the origin, the axis of symmetry is a coordinate axis, and the focus lies on the line $3x - 4y - 12 = 0$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;Vertex(G) = O;O: Origin;SymmetryAxis(G) = axis;H: Line;Expression(H) = (3*x - 4*y - 12 = 0);PointOnCurve(Focus(G), H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{(y^2=16*x), (x^2=-12*y)}",
    "fact_spans": "[[[1, 4], [41, 44]], [[1, 10]], [[8, 10]], [[1, 18]], [[22, 38]], [[22, 38]], [[1, 39]]]",
    "query_spans": "[[[41, 48]]]",
    "process": "The line $3x-4y-12=0$ intersects the coordinate axes at points $(4,0)$ and $(0,-3)$, which are the foci of the parabola; when the focus is $(4,0)$, the equation of the parabola is $y^{2}=16x$; when the focus is $(0,-3)$, the equation of the parabola is $x^{2}=-12y$; therefore, the equation of the parabola is $y^{2}=16x$ or $x^{2}=-12y$."
  },
  {
    "text": "What is the distance from the focus to the directrix of the parabola $x^{2}=-y$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = -y)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "1/2",
    "fact_spans": "[[[0, 13]], [[0, 13]]]",
    "query_spans": "[[[0, 24]]]",
    "process": "According to the problem, the focus of the parabola is $(0, -\\frac{1}{4})$, and the directrix is $y = \\frac{1}{4}$. Therefore, the distance from the focus to the directrix is $\\frac{1}{2}$."
  },
  {
    "text": "The ellipse $C$ has its center at the origin of the coordinate system, with the left and right foci $F_{1}$, $F_{2}$ on the $x$-axis. Given that $A$ and $B$ are respectively the upper vertex and the right vertex of the ellipse, and $P$ is a point on the ellipse such that $PF_{1} \\perp x$-axis and $PF_{2} \\| AB$, then the eccentricity of this ellipse is?",
    "fact_expressions": "C: Ellipse;O: Origin;Center(C) = O;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, xAxis) = True;PointOnCurve(F2, xAxis) = True;A: Point;B: Point;UpperVertex(C) = A;RightVertex(C) = B;P: Point;PointOnCurve(P, C) = True;IsPerpendicular(LineSegmentOf(P,F1),xAxis) = True;IsParallel(LineSegmentOf(P,F2),LineSegmentOf(A,B)) = True",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/5",
    "fact_spans": "[[[0, 5], [55, 57], [70, 72], [112, 114]], [[9, 13]], [[0, 13]], [[19, 26]], [[28, 35]], [[0, 35]], [[0, 35]], [[19, 41]], [[19, 41]], [[45, 48]], [[49, 52]], [[45, 65]], [[45, 65]], [[66, 69]], [[66, 75]], [[77, 94]], [[95, 109]]]",
    "query_spans": "[[[112, 120]]]",
    "process": "By the given conditions, let A(0,b), B(a,0), P(-c,t), then |PF_{1}| = |t| = \\sqrt{(1-\\frac{c^{2}}{a^{2}})}b = \\frac{b^{2}}{a}, so k_{AB} = -\\frac{b}{a} = -\\frac{|PF_{1}|}{|F_{F}|} = -\\frac{b^{2}}{2ac}, which implies b = 2c, that is, a^{2} - c^{2} = 4c^{2} \\Rightarrow a = \\sqrt{5}c, thus e = \\frac{\\sqrt{5}}{5}. The answer to be filled is \\frac{\\sqrt{5}}{5}."
  },
  {
    "text": "The ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ has left and right vertices $A_{1}$ and $A_{2}$ respectively. Point $P$ lies on $C$, and the slope of line $P A_{2}$ ranges in $[-2,-1]$. Then, what is the range of the slope of line $P A_{1}$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);A1: Point;A2: Point;LeftVertex(C) = A1;RightVertex(C) = A2;P: Point;PointOnCurve(P, C);Range(Slope(LineOf(P, A2))) = [-2, -1]",
    "query_expressions": "Range(Slope(LineOf(P, A1)))",
    "answer_expressions": "[3/8, 3/4]",
    "fact_spans": "[[[0, 42], [72, 75]], [[0, 42]], [[51, 58]], [[59, 66]], [[0, 66]], [[0, 66]], [[67, 71]], [[67, 76]], [[77, 105]]]",
    "query_spans": "[[[108, 128]]]",
    "process": "According to the problem, the diagram is drawn as follows: $A_{1}(-2,0)$, $A_{2}(2,0)$. When the slope of $PA_{2}$ is $-2$, point $P$ coincides with point $D$. The equation of line $PA_{2}$ is $y=-2(x-2)$. Substituting into the ellipse equation, eliminating $y$ and simplifying yields $19x^{2}-64x+52=0$. Solving gives $x=2$ or $x=\\frac{26}{19}$, so point $D(\\frac{26}{19},\\frac{24}{19})$. At this time, the slope of line $PA_{1}$ is $k=\\frac{\\frac{24}{19}}{\\frac{26}{19}+2}=\\frac{3}{8}$. Similarly, when the slope of line $PA_{2}$ is $-1$, point $P$ coincides with point $C$. The equation of line $PA_{2}$ is $y=-(x-2)$, substituting into the ellipse equation, eliminating $y$ and simplifying yields $7x^{2}-16x+4=0$. Solving gives $x=2$ or $x=\\frac{2}{7}$, so point $C(\\frac{2}{7},\\frac{12}{7})$. At this time, the slope of line $PA_{1}$ is $k=\\frac{\\frac{12}{7}}{\\frac{2}{7}+2}=\\frac{3}{4}$. Combining numerical and graphical analysis, the range of the slope of line $PA_{1}$ is $[\\frac{3}{8},\\frac{3}{4}]$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $P$ is a point on the ellipse $C$, and $P F_{1} \\perp P F_{2}$. If the area of $\\triangle P F_{1} F_{2}$ is $9$, then $b=$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));Area(TriangleOf(P, F1, F2)) = 9",
    "query_expressions": "b",
    "answer_expressions": "3",
    "fact_spans": "[[[18, 81], [91, 96]], [[18, 81]], [[160, 163]], [[25, 81]], [[25, 81]], [[25, 81]], [[2, 9]], [[10, 17]], [[2, 86]], [[87, 90]], [[87, 99]], [[101, 124]], [[126, 158]]]",
    "query_spans": "[[[160, 165]]]",
    "process": "Let |PF_{1}| = m, |PF_{2}| = n. Since PF_{1} \\perp PF_{2}, we have \\begin{cases} m^{2} + n^{2} = 4c^{2} \\\\ m + n = 2a \\end{cases}, so 2mn = (m + n)^{2} - (m^{2} + n^{2}) = 4a^{2} - 4c^{2} = 4b^{2}, mn = 2b^{2}. Also, S_{\\triangle PF_{1}F_{2}} = \\frac{1}{2}mn = 9, so 2b^{2} = 18, b^{2} = 9, b = 3."
  },
  {
    "text": "Given that the coordinates of the focus of a parabola are $(1 , 0)$, then the standard equation of this parabola is?",
    "fact_expressions": "G: Parabola;Coordinate(Focus(G)) = (1,0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[1, 4], [22, 25]], [[1, 19]]]",
    "query_spans": "[[[22, 32]]]",
    "process": "Since the focus is (1,0), let the standard equation of the parabola be: $y^{2}=2px$ $(p>0)$. From the focus coordinates, we have $\\frac{p}{2}=1$, so $p=2$. Therefore, the standard equation of the parabola is: $y^{2}=4x$."
  },
  {
    "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1$ is $\\frac{\\sqrt{6}}{2}$. Its two asymptotes are tangent to the circle $(x-6)^{2}+y^{2}=r^{2}  (r>0)$. Then $r=$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/3 + x^2/a^2 = 1);a: Number;Eccentricity(G) = sqrt(6)/2;H: Circle;Expression(H) = (y^2 + (x - 6)^2 = r^2);r: Number;r>0;IsTangent(Asymptote(G), H)",
    "query_expressions": "r",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[0, 42], [68, 69]], [[0, 42]], [[3, 42]], [[0, 67]], [[76, 107]], [[76, 107]], [[112, 115]], [[77, 107]], [[68, 110]]]",
    "query_spans": "[[[112, 117]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola coincides with the right focus of the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$, then the standard equation of the parabola is?",
    "fact_expressions": "G: Parabola;H: Ellipse;Expression(H) = (x^2/6 + y^2/2 = 1);Focus(G) = RightFocus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = 8*x",
    "fact_spans": "[[[0, 3], [52, 55]], [[9, 46]], [[9, 46]], [[0, 50]]]",
    "query_spans": "[[[52, 62]]]",
    "process": "Find the values of a, b, c of the ellipse to obtain the right focus. Set up the equation of the parabola to find the focus coordinates. Solve the equation to obtain p and then the required equation. For the ellipse \\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1, a=\\sqrt{6}, b=\\sqrt{2}, c=\\sqrt{a^{2}-b^{2}}=2, thus the right focus is (2,0). Let the equation of the parabola be y^{2}=2px, p>0, with focus (\\frac{p}{2},0). Then \\frac{p}{2}=2, solving gives p=4. Hence, the standard equation of the parabola is y^{2}=8x"
  },
  {
    "text": "Let circle $C$: $(x-1)^{2}+y^{2}=1$. Draw any chord of the circle passing through the origin $O$. Find the equation of the locus of the midpoint of such chords.",
    "fact_expressions": "C: Circle;Expression(C) = (y^2 + (x - 1)^2 = 1);O: Origin;H: LineSegment;PointOnCurve(O,H);IsChordOf(H,C)",
    "query_expressions": "LocusEquation(MidPoint(H))",
    "answer_expressions": "((x - 1/2)^2 + y^2 = 1/4)&(0 < x <= 1)",
    "fact_spans": "[[[1, 25], [33, 34]], [[1, 25]], [[27, 32]], [], [[26, 38]], [[26, 38]]]",
    "query_spans": "[[[26, 53]]]",
    "process": "\\because\\angleOPC=90^{\\circ}, the moving point P lies on a circle with center M(\\frac{1}{2},0) and OC as diameter, \\therefore the trajectory equation of the required point is (x-\\frac{1}{2})^{2}+y^{2}=\\frac{1}{4} (0<x\\leqslant1."
  },
  {
    "text": "The line passing through the origin intersects the circle $x^{2}+y^{2}+6 y+5=0$ at two points $A$, $B$. What is the equation of the locus of the midpoint $M$ of chord $AB$?",
    "fact_expressions": "G: Circle;H: Line;A: Point;B: Point;O:Origin;Expression(G) = (6*y + x^2 + y^2 + 5 = 0);PointOnCurve(O,H);Intersection(H,G)={A,B};IsChordOf(LineSegmentOf(A,B),G);MidPoint(LineSegmentOf(A,B))=M;M:Point",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "{(y^2+x^2+3*y=0)&(5/3<x<=3)}",
    "fact_spans": "[[[7, 29]], [[4, 6]], [[32, 36]], [[38, 41]], [[1, 3]], [[7, 29]], [[0, 6]], [[4, 43]], [[7, 50]], [[46, 55]], [[52, 55]]]",
    "query_spans": "[[[52, 62]]]",
    "process": ""
  },
  {
    "text": "Let the hyperbola $C$ pass through the point $(2,2)$ and have the same asymptotes as $\\frac{y^{2}}{4}-x^{2}=1$. Then the equation of $C$ is? The equations of the asymptotes are?",
    "fact_expressions": "C: Hyperbola;G: Point;Coordinate(G) = (2, 2);PointOnCurve(G, C);Q:Curve;Expression(Q)=(-x^2 + y^2/4 = 1);Asymptote(C) = Asymptote(Q)",
    "query_expressions": "Expression(C);Expression(Asymptote(C))",
    "answer_expressions": "x^2/3 - y^2/12 = 1  \ny = pm*2*x",
    "fact_spans": "[[[1, 7], [54, 57]], [[9, 17]], [[9, 17]], [[1, 17]], [[20, 45]], [[20, 45]], [[1, 52]]]",
    "query_spans": "[[[54, 63]], [[54, 70]]]",
    "process": ""
  },
  {
    "text": "The right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1(a>0)$ is $F$, and the left and right vertices are $A$ and $B$, respectively. A line passing through $F$ and perpendicular to $AB$ intersects the hyperbola at points $M$ and $N$. If $AM \\perp BN$, then the asymptotes of this hyperbola are?",
    "fact_expressions": "G: Hyperbola;a: Number;A: Point;B: Point;M: Point;N: Point;F: Point;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);RightFocus(G) = F;LeftVertex(G)=A;RightVertex(G)=B;L:Line;PointOnCurve(F,L);IsPerpendicular(LineSegmentOf(A, B),L);Intersection(L, G) = {M, N};IsPerpendicular(LineSegmentOf(A, M), LineSegmentOf(B, N))",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*x",
    "fact_spans": "[[[0, 37], [77, 80], [111, 114]], [[3, 37]], [[52, 56]], [[57, 62]], [[82, 85]], [[86, 89]], [[42, 45], [64, 67]], [[3, 37]], [[0, 37]], [[0, 45]], [[0, 61]], [[0, 61]], [], [[63, 76]], [[63, 76]], [[63, 91]], [[93, 108]]]",
    "query_spans": "[[[111, 120]]]",
    "process": "From $\\frac{x^{2}}{a^{2}}-y^{2}=1$ we obtain $c=\\sqrt{a^{2}+1}$, $A(-a,0)$, $B(a,0)$. Substituting $x=\\sqrt{a^{2}+1}$ into $\\frac{x^{2}}{a^{2}}-y^{2}=1$ gives $y=\\pm\\frac{1}{a}$, so $M(\\sqrt{a^{2}+1},\\frac{1}{a})$, $N(\\sqrt{a^{2}+1},-\\frac{1}{a})$. Since $AM\\perp BN$, it follows that $k_{AM}k_{BN}=\\frac{\\frac{1}{a}}{\\sqrt{a^{2}+1}+a}$. Moreover, since $a>0$, $\\therefore a=1$, hence the asymptotes of the hyperbola are $y=\\pm\\frac{b}{a}x$, i.e., $y=\\pm x$."
  },
  {
    "text": "$P$ is a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and $F_{1}$, $F_{2}$ are its left and right foci, respectively. Then the perimeter of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2",
    "query_expressions": "Perimeter(TriangleOf(P, F1, F2))",
    "answer_expressions": "16",
    "fact_spans": "[[[4, 43], [65, 66]], [[0, 3]], [[47, 54]], [[55, 62]], [[4, 43]], [[0, 46]], [[47, 71]], [[47, 71]]]",
    "query_spans": "[[[73, 99]]]",
    "process": ""
  },
  {
    "text": "If the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ passes through the point $P(0, \\sqrt{3})$, and the length of the major axis of the ellipse is twice the focal distance, then $a$=?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;P: Point;Coordinate(P) = (0, sqrt(3));PointOnCurve(P,C);Length(MajorAxis(C)) = 2 * FocalLength(C)",
    "query_expressions": "a",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 58], [79, 81]], [[1, 58]], [[93, 96]], [[8, 58]], [[8, 58]], [[8, 58]], [[60, 77]], [[60, 77]], [[1, 77]], [[79, 91]]]",
    "query_spans": "[[[93, 98]]]",
    "process": "From the ellipse $ C: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ $(a > b > 0)$ passing through the point $ P(0, \\sqrt{3}) $, we have $ b = \\sqrt{3} $. Also, the length of the major axis of the ellipse is twice the focal distance, i.e., $ 2a = 2 \\cdot 2c \\Rightarrow a = 2c $. Therefore, $ a^{2} = b^{2} + c^{2} \\Rightarrow a^{2} = 4 \\Rightarrow a = 2 $."
  },
  {
    "text": "Given circle $C_{1}$: $x^{2}+(y+3)^{2}=9$ and circle $C_{2}$: $x^{2}+(y-3)^{2}=1$, a moving circle $M$ is externally tangent to both circle $C_{1}$ and circle $C_{2}$. Then the trajectory equation of the center $M$ of the moving circle is?",
    "fact_expressions": "C1: Circle;Expression(C1) = (x^2+(y+3)^2=9);C2: Circle;Expression(C2) = (x^2+(y-3)^2=1);M: Circle;IsOutTangent(M, C1)\t;IsOutTangent(M, C2);M1: Point;Center(M) = M1",
    "query_expressions": "LocusEquation(M1)",
    "answer_expressions": "(y^2-x^2/8=1)&(y>=1)",
    "fact_spans": "[[[2, 31], [70, 78]], [[2, 31]], [[32, 61], [79, 87]], [[32, 61]], [[64, 67], [91, 93]], [[64, 89]], [[64, 89]], [[96, 99]], [[91, 99]]]",
    "query_spans": "[[[96, 106]]]",
    "process": "According to the condition that moving circle M is externally tangent to both circle $ C_{1} $ and circle $ C_{2} $, the geometric relationship can be obtained. Combining this with the definition of a hyperbola gives the trajectory equation of the moving point. [Detailed solution] From the problem, let the radius of moving circle M be $ r $, the radius of circle $ C_{1} $ be $ r_{1} = 3 $, and the radius of circle $ C_{2} $ be $ r_{2} = 1 $. When moving circle M is externally tangent to circles $ C_{1} $ and $ C_{2} $, $ |MC_{1}| = 3 + r $, $ |MC_{2}| = 1 + r $. Therefore, $ |MC_{1}| - |MC_{2}| = (3 + r) - (1 + r) = 2 $. Since the centers are $ C_{1}(0, -3) $, $ C_{2}(0, 3) $, we have $ |C_{1}C_{2}| = 6 $, and $ 2 < |C_{1}C_{2}| $. According to the definition of a hyperbola, the trajectory of moving point M is the upper branch of a hyperbola, where $ a = 1 $, $ c = 3 $. Thus, $ b^{2} = c^{2} - a^{2} = 8 $. Then, the trajectory equation of the center M of the moving circle is $ y^{2} - \\frac{x^{2}}{8} = 1 $ ($ y \\geqslant 1 $);"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, with left and right foci $F_{1}$ and $F_{2}$ respectively. Point $M$ lies on the hyperbola $C$. If $|M O|=|O F_{1}|$ and the area of $\\triangle M O F_{1}$ is $2 a b$, then the asymptotes of hyperbola $C$ have equations?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;M: Point;O: Origin;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(M, C);Abs(LineSegmentOf(M, O)) = Abs(LineSegmentOf(O, F1));Area(TriangleOf(M, O, F1)) = 2*a*b",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*4*x",
    "fact_spans": "[[[2, 63], [93, 99], [155, 161]], [[10, 63]], [[10, 63]], [[88, 92]], [[102, 119]], [[72, 79]], [[80, 87]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 87]], [[2, 87]], [[88, 100]], [[102, 119]], [[121, 153]]]",
    "query_spans": "[[[155, 169]]]",
    "process": "Assume point M lies on the right branch of the hyperbola. According to the given conditions, |MO| = |OF₁| = |OF₂|. Hence, ∠F₁MF₂ = 90°. Since S△MOF₁ = 2ab, then S△MF₁F₂ = 4ab. Let |MF₁| = m, |MF₂| = n, so (1/2)mn = 4ab, ∴ mn = 8ab. Thus, m − n = 2a, m² + n² = 4c². Therefore, (m − n)² + 2mn = 4(a² + b²). So, 4a² + 16ab = 4(a² + b²), then b/a = 4. Hence, the asymptotes of hyperbola C are y = ±4x."
  },
  {
    "text": "The left focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ is $F$, and the line $x=m$ intersects the ellipse at points $A$ and $B$. When the perimeter of $\\triangle F A B$ is maximized, what is the area of $\\triangle F A B$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);F: Point;LeftFocus(G) = F;H: Line;Expression(H) = (x = m);m: Number;Intersection(H, G) = {A, B};A: Point;B: Point;WhenMax(Perimeter(TriangleOf(F, A, B)))",
    "query_expressions": "Area(TriangleOf(F, A, B))",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 37], [54, 56]], [[0, 37]], [[42, 45]], [[0, 45]], [[46, 53]], [[46, 53]], [[48, 53]], [[46, 67]], [[59, 63]], [[64, 67]], [[68, 92]]]",
    "query_spans": "[[[93, 115]]]",
    "process": "Let the right focus of the ellipse be E. As shown in the figure: by the definition of the ellipse, the perimeter of AFAB is: AB + AF + BF = AB + (2a - AE) + (2a - BE) = 4a + AB - AE - BE; \\because AE + BE \\geqslant AB; \\therefore AB - AE - BE \\leqslant 0, with equality when AB passes through point E; \\therefore AB + AF + BF = 4a + AB - AE - BE \\leqslant 4a; that is, the perimeter of AFAB is maximized when the line x = m passes through the right focus E of the ellipse, at which time the height of AFAB is: EF = 2. At this moment, the line x = m = c = 1; substitute x = 1 into the equation of the ellipse \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1, we get: y = \\pm\\frac{3}{2}, \\therefore AB = 3. Therefore: the area of AFAB equals: S_{AFAB} = \\frac{1}{2} \\times 3 \\times EF = \\frac{1}{2} \\times 3 \\times 2 = 3."
  },
  {
    "text": "A asymptote of the hyperbola $4 x^{2}-y^{2}=1$ is perpendicular to the line $t x+y+1=0$, then $t=?$",
    "fact_expressions": "G: Hyperbola;H: Line;t: Number;Expression(G) = (4*x^2 - y^2 = 1);Expression(H) = (t*x + y + 1 = 0);IsPerpendicular(OneOf(Asymptote(G)), H)",
    "query_expressions": "t",
    "answer_expressions": "pm*1/2",
    "fact_spans": "[[[0, 20]], [[27, 40]], [[44, 47]], [[0, 20]], [[27, 40]], [[0, 42]]]",
    "query_spans": "[[[44, 49]]]",
    "process": ""
  },
  {
    "text": "Given that the vertex of the parabola is at the origin, the axis of symmetry coincides with the line containing the minor axis of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{16}=1$, and the distance from the focus of the parabola to the vertex is $5$, then the standard equation of this parabola is?",
    "fact_expressions": "G: Parabola;H: Ellipse;O: Origin;Expression(H) = (x^2/9 + y^2/16 = 1);Vertex(G) = O;SymmetryAxis(G)=OverlappingLine(MinorAxis(H));Distance(Focus(G),Vertex(G))=5",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = pm*20*x",
    "fact_spans": "[[[2, 5], [64, 67], [83, 86]], [[18, 56]], [[9, 11]], [[18, 56]], [[2, 11]], [[2, 63]], [[64, 80]]]",
    "query_spans": "[[[83, 93]]]",
    "process": "Let the equation of the parabola be y^{2}=2px. Since the distance from the focus to the vertex of the parabola is 5, we have |\\frac{p}{2}|=5, so p=\\pm10. Therefore, the standard equation of the parabola is y^{2}=20x or y^{2}=-20x. Fill in y^{2}=20x or y^{2}=-20x. [Green] This question tests the method for finding the standard equation of a parabola and belongs to basic problems."
  },
  {
    "text": "A point $M$ on the parabola $y^{2}=4x$ is at a distance of $3$ from the focus. Find the abscissa $x$ of point $M$.",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(M, G);Distance(M, Focus(G)) = 3;x1: Number;XCoordinate(M) = x1",
    "query_expressions": "x1",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 14]], [[17, 20], [32, 36]], [[0, 14]], [[0, 20]], [[0, 30]], [[40, 43]], [[32, 43]]]",
    "query_spans": "[[[40, 45]]]",
    "process": "From the given conditions, the parabola $ y^{2} = 4x $ has focus $ F(1,0) $ and directrix equation $ x = -1 $. By the definition of a parabola, $ MF = x + 1 = 3 $, solving gives $ x = 2 $."
  },
  {
    "text": "Draw a perpendicular line from a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ to one of its asymptotes, such that the foot of the perpendicular lies exactly on the curve $\\frac{x^{2}}{b^{2}}+\\frac{y^{2}}{a^{2}}=1$. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Curve;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2/b^2 + y^2/a^2 = 1);L:Line;PointOnCurve(OneOf(Focus(G)),L);IsPerpendicular(OneOf(Asymptote(G)),L);PointOnCurve(FootPoint(OneOf(Asymptote(G)),L),H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 58], [127, 130]], [[4, 58]], [[4, 58]], [[79, 124]], [[4, 58]], [[4, 58]], [[1, 58]], [[79, 124]], [], [[0, 72]], [[0, 72]], [[0, 125]]]",
    "query_spans": "[[[127, 136]]]",
    "process": "Let a focus of the hyperbola be $(c,0)$, $(c>0)$, and an asymptote be $y=\\frac{b}{a}x$. From\n\\[\n\\begin{cases}\ny-0=-\\frac{a}{b}(x-c)\\\\\ny=\\frac{b}{a}x\n\\end{cases}\n\\]\nthe coordinates of the foot of the perpendicular are $\\left(\\frac{a^2}{c},\\frac{ab}{c}\\right)$. Substituting this point into the equation $\\frac{x^{2}}{b^{2}}+\\frac{y^{2}}{a^{2}}=1$, we get $\\frac{a^{4}}{b^{2}c^{2}}+\\frac{a}{a}\\frac{a}{a2c}=1$. Simplifying, and using $c^{2}=a^{2}+b^{2}$, we obtain $a=b$, $c=\\sqrt{2}a$, $\\therefore e=\\frac{c}{a}=\\sqrt{2}$."
  },
  {
    "text": "The chord $AB$ of the parabola $y^{2}=-2 p x(p>0)$ passes through its focus and is perpendicular to its axis of symmetry, $O$ is the origin. If the area of $\\triangle A O B$ is $3$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;O: Origin;p>0;Expression(G) = (y^2 = -2*p*x);PointOnCurve(Focus(G),LineSegmentOf(A,B));IsChordOf(LineSegmentOf(A,B),G);IsPerpendicular(LineSegmentOf(A,B),SymmetryAxis(G));Area(TriangleOf(A, O, B)) = 3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=-2*sqrt(6)*x",
    "fact_spans": "[[[0, 22], [38, 39], [30, 31], [78, 81]], [[3, 22]], [[24, 29]], [[24, 29]], [[44, 47]], [[3, 22]], [[0, 22]], [[24, 33]], [[0, 29]], [[24, 43]], [[52, 76]]]",
    "query_spans": "[[[78, 85]]]",
    "process": "The focus of the parabola $ y^{2} = -2px $ ($ p > 0 $) has coordinates $ \\left( -\\frac{p}{2}, 0 \\right) $. According to the problem, assume $ A\\left( -\\frac{p}{2}, p \\right) $, $ B\\left( -\\frac{p}{2}, -p \\right) $, then $ |AB| = 2p $. Thus, $ S_{\\triangle AOB} = \\frac{1}{2}|AB| \\cdot |OF| = \\frac{1}{2} \\cdot 2p \\cdot \\frac{p}{2} = 3 $, solving gives $ p = \\sqrt{6} $, so the equation of the parabola is $ y^{2} = -2\\sqrt{6}x $."
  },
  {
    "text": "Given that point $F$ is the focus of the parabola $y=2 x^{2}$, and $M$, $N$ are two points on this parabola such that $|M F|+|N F|=\\frac{17}{4}$, then what is the vertical coordinate of the midpoint of segment $M N$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 2*x^2);F: Point;Focus(G) = F;M: Point;N: Point;PointOnCurve(M, G);PointOnCurve(N, G);Abs(LineSegmentOf(M, F)) + Abs(LineSegmentOf(N, F)) = 17/4",
    "query_expressions": "YCoordinate(MidPoint(LineSegmentOf(M, N)))",
    "answer_expressions": "2",
    "fact_spans": "[[[7, 21], [34, 37]], [[7, 21]], [[2, 6]], [[2, 24]], [[25, 28]], [[29, 32]], [[25, 41]], [[25, 41]], [[43, 69]]]",
    "query_spans": "[[[71, 86]]]",
    "process": ""
  },
  {
    "text": "If an ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{3}=1$ $(m>3)$ has a focus $F$, and the maximum distance from a point $P$ on the ellipse to the focus $F$ is $3$, then what is the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;m: Number;F: Point;P: Point;m>3;Expression(G) = (y^2/3 + x^2/m = 1);OneOf(Focus(G)) = F;PointOnCurve(P, G);Max(Distance(P, F)) = 3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[1, 43], [78, 80], [53, 55]], [[3, 43]], [[49, 52], [64, 67]], [[58, 61]], [[3, 43]], [[1, 43]], [[1, 52]], [[53, 61]], [[58, 76]]]",
    "query_spans": "[[[78, 86]]]",
    "process": "From $\\frac{x^{2}}{m}+\\frac{y^{2}}{3}=1$ $(m>3)$, we obtain $a^{2}=m$, $b^{2}=3$, $c^{2}=m-3$, so that $a=\\sqrt{m}$, $b=\\sqrt{3}$, $c=\\sqrt{m-3}$. Since the maximum distance from a point $P$ on the ellipse to the focus $F$ is 3, then $a+c=\\sqrt{m}+\\sqrt{m-3}=3$, solving gives $m=4$. Therefore, the eccentricity of the ellipse is $\\frac{c}{a}=\\frac{\\sqrt{m-3}}{\\sqrt{m}}=\\frac{1}{2}$."
  },
  {
    "text": "If real numbers $x$, $y$ satisfy the equation $\\frac{x^{2}}{16}+\\frac{y^{2}}{25}=1$, then the range of $\\sqrt{(x-1)^{2}+y^{2}}+\\sqrt{x^{2}+(y-3)^{2}}$ is?",
    "fact_expressions": "x_: Real;y_: Real;x_^2/16 + y_^2/25 = 1",
    "query_expressions": "Range(sqrt(x_^2 + (y_ - 3)^2) + sqrt(y_^2 + (x_ - 1)^2))",
    "answer_expressions": "[10-sqrt(10),10+sqrt(10)]",
    "fact_spans": "[[[1, 6]], [[9, 12]], [[1, 53]]]",
    "query_spans": "[[[55, 109]]]",
    "process": "From the problem, $\\sqrt{(x-1)^{2}+y^{2}}+\\overline{x}$ represents the sum of the distances from a point $P(x,y)$ on the ellipse to the point $A(1,0)$ and the upper focus $F_{2}(0,3)$. Let the lower focus of the ellipse be $F_{1}(0,-3)$. Then, by the definition of an ellipse, the problem is transformed into finding the range of $10+|PA|-|PF_{1}|$. $\\sqrt{(x-1)^{2}+y^{2}}+\\sqrt{x^{2}+(y-3)^{2}}$ can represent the sum of the distances from a point $P(x,y)$ on the ellipse to the point $A(1,0)$ and the upper focus $F_{2}(0,3)$, that is, $\\sqrt{(x-1)^{2}+y^{2}}+\\sqrt{x^{2}+(y-3)^{2}}=|PA|+|PF_{2}|$. Let the lower focus of the ellipse be $F_{1}(0,-3)$. By the definition of the ellipse, $|PF_{1}|+|PF_{2}|=10$, so $|PA|+|PF_{2}|=10+|PA|-|PF_{1}|$. Since $||PA|-|PF_{1}||\\leqslant|AF_{1}|=\\sqrt{10}$, it follows that $-\\sqrt{10}\\leqslant|PA|-|PF_{1}|\\leqslant\\sqrt{10}$. Hence, $10-\\sqrt{10}\\leqslant|PA|+|PF_{2}|\\leqslant10+\\sqrt{10}$."
  },
  {
    "text": "A line $l$ passing through the point $M(0,1)$ intersects the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ at points $A$ and $B$, and $F_{1}$ is the left focus of the ellipse. When the perimeter of $\\triangle A B F_{1}$ is maximized, what is the equation of line $l$?",
    "fact_expressions": "l: Line;C: Ellipse;M: Point;A: Point;B: Point;F1: Point;Expression(C) = (x^2/4 + y^2/3 = 1);Coordinate(M) = (0, 1);PointOnCurve(M, l);Intersection(l, C) = {A, B};LeftFocus(C)=F1;WhenMax(Perimeter(TriangleOf(A,B,F1)))",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x+y-1=0",
    "fact_spans": "[[[11, 16], [113, 118]], [[17, 59], [78, 80]], [[1, 10]], [[60, 63]], [[64, 67]], [[70, 77]], [[17, 59]], [[1, 10]], [[0, 16]], [[11, 69]], [[70, 84]], [[85, 112]]]",
    "query_spans": "[[[113, 123]]]",
    "process": "Let the right focus be $ F_{2}(1,0) $, then $ |AF_{1}| = 4 - |AF_{2}| $, $ |BF_{1}| = 4 - |BF_{2}| $, so $ |AF_{1}| + |BF_{1}| + |AB| = 8 + |AB| - (|AF_{2}| + |BF_{2}|) $. Clearly, $ |AF_{2}| + |BF_{2}| \\geqslant |AB| $, with equality if and only if $ A $, $ B $, $ F_{2} $ are collinear. Therefore, when line $ l $ passes through point $ F_{2} $, the perimeter of $ \\triangle ABF_{1} $ attains the maximum value of 8, and at this time the equation of the line is $ y = -x + 1 $, i.e., $ x + y - 1 = 0 $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the ellipse, and let $M$ be the midpoint of segment $P F_{1}$. If $|O F_{2}|=|O M|$, then the area of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;O: Origin;M: Point;Expression(G) = (x^2/9 + y^2/5 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);MidPoint(LineSegmentOf(P,F1))=M;Abs(LineSegmentOf(O,F2))=Abs(LineSegmentOf(O,M))",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "sqrt(15)",
    "fact_spans": "[[[2, 39], [68, 70]], [[47, 54]], [[63, 67]], [[55, 62]], [[94, 112]], [[88, 92]], [[2, 39]], [[2, 62]], [[2, 62]], [[63, 71]], [[73, 91]], [[94, 112]]]",
    "query_spans": "[[[114, 144]]]",
    "process": "By the law of cosines and combining with the definition of an ellipse, we can find |PF₂||PF₁||F₁F₂|, then solve using the law of cosines and the area formula. From the given conditions, we have a=3, b=\\sqrt{5}, c=\\sqrt{9-5}=2. Since O and M are the midpoints of F₁F₂ and F₁P respectively, |PF₂|=2|OM|=2|OF₂|=2c=4. According to the definition of the ellipse, |PF₁|=2a-2c=2. Also, F₁F₂=2c=4, so \\cos\\angle PF₂F₁=\\frac{|PF₂|^{2}+|F₁F₂|^{2}-|PF₁|^{2}}{2|PF₂|\\cdot|F₁F₂|}=\\frac{16+16-4}{2\\times4\\times4}=\\frac{7}{8}. Therefore, \\sin\\angle PF₂F₁=\\sqrt{1-\\cos^{2}\\angle PF₂F₁}=\\frac{\\sqrt{15}}{8}. Hence, the area of \\triangle PF₁F₂ is \\frac{1}{2}|PF₂|\\cdot|F₁F₂|\\cdot\\sin\\angle PF₂F₁=\\sqrt{15}."
  },
  {
    "text": "If point $P$ is a moving point on the parabola $x^{2}=8y$, then the minimum value of the sum of the distance from point $P$ to point $A(4,0)$ and the distance from point $P$ to the line $y=-2$ is?",
    "fact_expressions": "P: Point;G: Parabola;Expression(G) = (x^2 = 8*y);PointOnCurve(P, G) = True;A: Point;Coordinate(A) = (4, 0);H: Line;Expression(H) = (y = -2)",
    "query_expressions": "Min(Distance(P, A) + Distance(P, H))",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[1, 5], [26, 30]], [[6, 20]], [[6, 20]], [[1, 24]], [[31, 40]], [[31, 40]], [[45, 53]], [[45, 53]]]",
    "query_spans": "[[[26, 64]]]",
    "process": "The focus of the parabola $ x^{2} = 8y $ is $ F(0,2) $, and the equation of the directrix is $ y = -2 $. For a moving point $ P $ on the parabola $ x^{2} = 8y $, the distance from $ P $ to the line $ y = -2 $ is equal to the distance from $ P $ to the focus $ F(0,2) $. Therefore, the minimum value of the sum of the distance from point $ P $ to point $ A(4,0) $ and the distance from $ P $ to the line $ y = -2 $ is $ |FA| = 2\\sqrt{5} $."
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{10-k}+\\frac{y^{2}}{k-4}=1$ represents an ellipse, what is the range of real values for $k$?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G)=(x^2/(10 - k) + y^2/(k - 4) = 1)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(4,7)+(7,10)",
    "fact_spans": "[[[46, 48]], [[50, 55]], [[2, 48]]]",
    "query_spans": "[[[50, 62]]]",
    "process": "The equation $\\frac{x^2}{10-k}+\\frac{y^{2}}{k-4}=1$ represents an ellipse. According to the standard form of an ellipse; since $\\frac{x^{2}}{10-k}+\\frac{y^{2}}{k-4}=1$ represents an ellipse, then $\\begin{cases}10-k>0\\\\k-4>0\\\\10-k+k-4\\end{cases}$ yields $k\\in(4,7)\\cup(7,10)$"
  },
  {
    "text": "A line $l$ passing through the focus $F$ of the parabola $C$: $x^{2}=4 y$ intersects $C$ at points $A$ and $B$. The tangent to $C$ at point $A$ intersects the $x$-axis and $y$-axis at points $M$ and $N$, respectively. If the area of $\\Delta M O N$ is $\\frac{1}{2}$, then $|A F|=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 4*y);F: Point;Focus(C) = F;l: Line;PointOnCurve(F, l);Intersection(l, C) = {A, B};A: Point;B: Point;Intersection(TangentOnPoint(A,C),xAxis) = M;Intersection(TangentOnPoint(A,C),yAxis) = N;M: Point;O: Origin;N: Point;Area(TriangleOf(M, O, N)) = 1/2",
    "query_expressions": "Abs(LineSegmentOf(A, F))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 20], [33, 36]], [[1, 20]], [[23, 26]], [[1, 26]], [[27, 32]], [[0, 32]], [[27, 46]], [[37, 40], [48, 52]], [[41, 44]], [[47, 78]], [[47, 78]], [[70, 74]], [[80, 94]], [[75, 78]], [[80, 111]]]",
    "query_spans": "[[[113, 122]]]",
    "process": "By the given condition, the focus is $ F(0,1) $. Let the line be $ y = kx + 1 $, and assume $ A $ is the left intersection point, $ A(x_{0},y_{0}) $. Then the tangent line at $ A $ is $ x_{0}x = 2y_{0} + 2y $, so $ M\\left(\\frac{x_{0}}{2},0\\right) $, $ N(0,-y_{0}) $. Therefore, $ S = \\frac{1}{2} \\cdot \\frac{x_{0}}{2} \\cdot (-y_{0}) = \\frac{1}{2} $, solving gives $ x_{0} = -2 $, then $ A(-2,1) $. According to the definition of the parabola, we obtain $ |AF| = 1 + \\frac{p}{2} = 1 + 1 = 2 $."
  },
  {
    "text": "Given the circle $C$: $x^{2}+(y-2)^{2}=26$ and a fixed point $A(0,-2)$, let points $P$, $Q$ be two moving points on circle $C$, and point $M$ be the midpoint of chord $P Q$. If $\\overrightarrow{A P} \\cdot \\overrightarrow{A Q}=0$, then the maximum distance from point $M$ to point $N(3,4)$ is?",
    "fact_expressions": "C: Circle;Expression(C) = (x^2 + (y - 2)^2 = 26);A: Point;Coordinate(A) = (0, -2);P: Point;Q: Point;PointOnCurve(P, C);PointOnCurve(Q, C);M: Point;MidPoint(LineSegmentOf(P, Q)) = M;IsChordOf(LineSegmentOf(P, Q), C);DotProduct(VectorOf(A, P), VectorOf(A, Q)) = 0;N: Point;Coordinate(N) = (3, 4)",
    "query_expressions": "Max(Distance(M, N))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 28], [50, 54]], [[2, 28]], [[31, 40]], [[31, 40]], [[41, 45]], [[46, 49]], [[41, 58]], [[41, 58]], [[59, 63], [128, 132]], [[59, 73]], [[50, 70]], [[75, 126]], [[133, 142]], [[133, 142]]]",
    "query_spans": "[[[128, 151]]]",
    "process": "From the given, circle $ C: x^{2}+(y-2)^{2}=26 $ has center $ C(0,2) $ and radius $ R=\\sqrt{26} $. Let $ M(x_{0},y_{0}) $. Since $ \\overrightarrow{AP} \\cdot \\overrightarrow{AQ}=0 $, we have $ AP \\perp AQ $. Also, point $ M $ is the midpoint of chord $ PQ $, so in right triangle $ \\Delta APQ $, $ PM=AM $. Moreover, $ CM \\perp PQ $, thus $ CM^{2}+PM^{2}=CM^{2}+AM^{2}=CP^{2}=R^{2}=26 $. Therefore, $ (x_{0}-0)^{2}+(y_{0}-2)^{2}+(x_{0}-0)^{2}+(y_{0}+2)^{2}=26 $, which simplifies to $ x_{0}^{2}+y_{0}^{2}=9 $. Hence, point $ M $ lies on a circle centered at $ O(0,0) $ with radius 3. The maximum distance from point $ M $ to point $ N(3,4) $ is $ |ON|+3=\\sqrt{3^{2}+4^{2}}+3=8 $."
  },
  {
    "text": "The equation $\\frac{x^{2}}{k}-\\frac{y^{2}}{2-k^{2}}=1$ represents an ellipse; then the range of real values for $k$ is?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G)=(x^2/k-y^2/(2-k^2)=1)",
    "query_expressions": "Range(k)",
    "answer_expressions": "{(sqrt(2),2)+(2,+oo)}",
    "fact_spans": "[[[45, 47]], [[49, 54]], [[0, 47]]]",
    "query_spans": "[[[49, 61]]]",
    "process": "The original equation can be rewritten as \\frac{x^{2}}{k}+\\frac{y^{2}}{k^{2}-2}=1. According to the problem, we have \\begin{cases}k>0\\\\k^{2}-2>0\\\\k^{2}-2\\neq k\\end{cases}. Solving gives k\\in(\\sqrt{2},2)\\cup(2,+\\infty)"
  },
  {
    "text": "The distance between the foci of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{4}=1$ with foci on the $x$-axis is $2$, then what is the value of $m$?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (y^2/4 + x^2/m = 1);PointOnCurve(Focus(G), xAxis);FocalLength(G) = 2",
    "query_expressions": "m",
    "answer_expressions": "5",
    "fact_spans": "[[[9, 46]], [[55, 58]], [[9, 46]], [[0, 46]], [[9, 53]]]",
    "query_spans": "[[[55, 62]]]",
    "process": "From the given conditions, 2c=2, so c=1. From the properties of the ellipse: m=b^{2}+c^{2}. Thus, m=4+1=5,"
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=4x$, draw a chord $AB$ with an inclination angle of $45^{\\circ}$. Then the length of chord $AB$ is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;IsChordOf(LineSegmentOf(A, B), G);Inclination(LineSegmentOf(A, B)) = ApplyUnit(45, degree);PointOnCurve(F, LineSegmentOf(A, B))",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 15]], [[40, 45]], [[40, 45]], [[1, 15]], [[18, 21]], [[1, 21]], [[0, 45]], [[22, 45]], [[0, 45]]]",
    "query_spans": "[[[47, 57]]]",
    "process": "This is a problem of finding the length of a focal chord of a parabola. One can first determine the equation of the line containing the chord passing through the focus of the parabola, then solve the system of equations formed by the line and the parabola, and apply Vieta's formulas to obtain the result. Since the focus of the parabola is $ F(1,0) $, the equation of the line is $ y = x - 1 $. Substituting and eliminating $ y $ yields $ x^{2} - 6x + 1 = 0 $. Therefore, $ |AB| = x_{1} + x_{2} + p = 6 + 2 = 8 $. Hence, the answer should be $ 8 $."
  },
  {
    "text": "$P$ is an arbitrary point on the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{81}=1$, and $F_{1}$, $F_{2}$ are its left and right foci respectively. Given that $|P F_{1}|=9$, then $|P F_{2}|$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/81 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(P, F1)) = 9",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "17",
    "fact_spans": "[[[4, 44], [68, 69]], [[4, 44]], [[0, 3]], [[0, 49]], [[50, 57]], [[58, 65]], [[50, 75]], [[50, 75]], [[77, 90]]]",
    "query_spans": "[[[92, 105]]]",
    "process": "According to the problem, the hyperbola is $\\frac{x^2}{16}-\\frac{y^{2}}{81}=1$, where $a=4$, $c=\\sqrt{97}$. Since $P$ is a point on the hyperbola, we have $|PF_{1}|-|PF_{2}|=2a=8$. Given $|PF_{1}|=9$, then $|PF_{2}|=1<c-a=\\sqrt{97}-4$ (discarded) or $17$."
  },
  {
    "text": "Given that the distance from point $P(1,0)$ to one of the asymptotes of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ is $\\frac{1}{2}$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (1, 0);Distance(P, OneOf(Asymptote(C))) = 1/2",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[12, 74], [99, 105]], [[19, 74]], [[19, 74]], [[2, 11]], [[19, 74]], [[19, 74]], [[12, 74]], [[2, 11]], [[2, 97]]]",
    "query_spans": "[[[99, 111]]]",
    "process": ""
  },
  {
    "text": "Given that line $l_{1}$ passes through point $P(1,4)$ and intersects the $x$-axis at point $A$, line $l_{2}$ passes through point $Q(3,-1)$ and intersects the $y$-axis at point $B$. If $l_{1} \\perp l_{2}$ and $\\overrightarrow{A M}=2 \\overrightarrow{M B}$, then the trajectory equation of point $M$ is?",
    "fact_expressions": "l1:Line;l2:Line;P: Point;Q: Point;A: Point;M: Point;B: Point;PointOnCurve(P,l1);Intersection(l1,xAxis)=A;Coordinate(P) = (1, 4);Coordinate(Q) = (3, -1);PointOnCurve(Q,l2);Intersection(l2,yAxis)=B;IsPerpendicular(l1,l2);VectorOf(A, M) = 2*VectorOf(M, B)",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "9*x+6*y+1=0",
    "fact_spans": "[[[2, 11]], [[34, 43]], [[12, 21]], [[44, 54]], [[29, 33]], [[136, 140]], [[62, 66]], [[2, 21]], [[2, 33]], [[12, 21]], [[44, 54]], [[34, 54]], [[34, 66]], [[68, 87]], [[89, 134]]]",
    "query_spans": "[[[136, 147]]]",
    "process": "Let M(x,y). (1) If the slope does not exist, then: l_{1} is perpendicular to the x-axis, l_{2} is perpendicular to the y-axis; \\because A(1,0), B(0,-1); (1)=2(-x,-1-y); -2x, thus M(\\frac{1}{3},\\frac{2}{3}). (2) If the slope of l_{1} is k, and the slope of l_{2} is -\\frac{1}{k}, then: l_{1}: y-4=k(x-1), set y=0, x=-\\frac{4}{k}+1. \\therefore A(-\\frac{4}{k}+1,0); l_{2}: y+1=-\\frac{1}{k}(x-3), set x=0, y=\\frac{3}{k}-1; \\therefore B(0,\\frac{3}{k}-1); \\frac{\\sqrt{b}}{1}=\\frac{k}{AM}=2\\overrightarrow{MB} gives, (x+\\frac{4}{k}-1,y)=2(-x,\\frac{3}{k}-1-y) \\Rightarrow x+\\frac{4}{k}-1=-2x \\begin{cases} \\\\ y=\\frac{6}{k}-2-2y \\end{cases} \\therefore eliminating k and simplifying yields: 9x+6y+1=0; the point M(\\frac{1}{3},-\\frac{2}{3}) satisfies the equation 9x+6y+1=0; combining (1) and (2), the trajectory equation of point M is 9x+6y+1=0."
  },
  {
    "text": "Given the circle $C$: $(x-3)^{2}+y^{2}=4$, and point $M$ moving along the parabola $T$: $y^{2}=4x$. From point $M$, draw lines $l_{1}$, $l_{2}$ tangent to circle $C$, with points of tangency $P$ and $Q$, respectively. Then the range of $|P Q|$ is?",
    "fact_expressions": "T: Parabola;C: Circle;l1: Line;l2: Line;P: Point;Q: Point;M: Point;Expression(T) = (y^2 = 4*x);Expression(C) = (y^2 + (x - 3)^2 = 4);PointOnCurve(M, T);TangentOfPoint(M, C) = {l1, l2};TangentPoint(l1, C) = P;TangentPoint(l2, C) = Q",
    "query_expressions": "Range(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "[2*sqrt(2), 4)",
    "fact_spans": "[[[32, 51]], [[2, 26], [81, 85]], [[61, 70]], [[72, 80]], [[93, 96]], [[97, 100]], [[27, 31], [56, 60]], [[32, 51]], [[2, 26]], [[27, 52]], [[55, 87]], [[61, 100]], [[61, 100]]]",
    "query_spans": "[[[102, 116]]]",
    "process": "As shown in the figure, connect CP, CQ, CM. According to the problem, CP\\bot MP, CQ\\bot MQ, and |CP|=|CQ|=2, |MP|=|MQ|. Then CM is the perpendicular bisector of segment PQ. Thus, the area of quadrilateral MPCQ is twice the area of right triangle CPIM, so we obtain \\frac{1}{2}|PQ|\\cdot|CM|=2\\cdot\\frac{1}{2}|CP|\\cdot|MP|, that is, |PQ|=\\frac{2|CP|\\cdot|MP|}{|CM|}=\\frac{4\\sqrt{|CM|^{2}-|CP|^2}}{|CM|}=4\\sqrt{1-\\frac{4}{|CM|^2}}. Let point M(t,s), and C(3,0), s^{2}=4t (t\\geqslant0), then |CM|^{2}=(t-3)^{2}+s^{2}=t^{2}-2t+9=(t-1)^{2}+8\\geqslant8. Therefore, 0<\\frac{1}{|CM|^{2}}\\leqslant\\frac{1}{8}, i.e., \\frac{7}{8}\\leqslant1-\\frac{4}{|CM|^{2}}<1, so 2\\sqrt{2}\\leqslant|PQ|<4. Hence, the range of |PQ| is [2\\sqrt{2},4)."
  },
  {
    "text": "Given that the line $y=x-1$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$, and point $O$ is the origin, what is the area of $\\triangle OAB$?",
    "fact_expressions": "H: Line;Expression(H) = (y = x - 1);G: Parabola;Expression(G) = (y^2 = 4*x);Intersection(H,G) = {A,B};A: Point;B: Point;O: Origin",
    "query_expressions": "Area(TriangleOf(O, A, B))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 11]], [[2, 11]], [[12, 26]], [[12, 26]], [[2, 36]], [[27, 30]], [[31, 34]], [[37, 41]]]",
    "query_spans": "[[[49, 71]]]",
    "process": "The focus of the parabola $ y^{2}=4x $ is $ F(1,0) $, which lies on the line $ y=x-1 $. Solving the system of equations:  \n$$\n\\begin{cases}\ny^{2}=4x \\\\\ny=x-1\n\\end{cases}\n$$\nwe obtain $ y^{2}-4y-4=0 $, with solutions $ y_{1}=2+2\\sqrt{2} $, $ y_{2}=2-2\\sqrt{2} $.  \n$ S_{\\Delta AOB}=S_{\\Delta OFA}+S_{\\Delta OFB}=\\frac{1}{2}|OF|\\cdot|y_{1}-y_{2}|=\\frac{1}{2}\\times1\\times4\\sqrt{2}=2\\sqrt{2} $"
  },
  {
    "text": "Given that the vertices of ellipse $E$ are $A(-1,0)$ and $B(1,0)$, if a line $l$ passing through its focus $F(0,1)$ intersects the ellipse at points $C$ and $D$, and intersects the $x$-axis at point $P$ (where $P$ is distinct from points $A$ and $B$), and lines $AC$ and $BD$ intersect at point $Q$, then $\\overrightarrow{O P} \\cdot \\overrightarrow{O Q}$=?",
    "fact_expressions": "l: Line;E: Ellipse;A: Point;C: Point;B: Point;D: Point;F: Point;O: Origin;P: Point;Q: Point;Coordinate(A) = (-1, 0);Coordinate(B) = (1, 0);Coordinate(F) = (0, 1);Vertex(E)={A,B};OneOf(Focus(E)) = F;PointOnCurve(F, l);Intersection(l, E) = {C, D};Intersection(l, xAxis) = P;Negation(P=A);Negation(P=B);Intersection(LineOf(A,C), LineOf(B, D)) = Q",
    "query_expressions": "DotProduct(VectorOf(O, P), VectorOf(O, Q))",
    "answer_expressions": "1",
    "fact_spans": "[[[46, 51]], [[2, 7], [34, 35], [52, 54]], [[11, 20], [84, 88]], [[56, 59]], [[22, 31], [89, 92]], [[60, 63]], [[37, 45]], [[115, 164]], [[74, 78], [79, 82]], [[109, 113]], [[11, 20]], [[22, 31]], [[37, 45]], [[2, 31]], [[34, 45]], [[33, 51]], [[46, 65]], [[46, 78]], [[79, 88]], [[79, 92]], [[94, 113]]]",
    "query_spans": "[[[115, 166]]]",
    "process": "First, find the ellipse equation from the given conditions, set up the line equation, solve the system of the line and the ellipse to obtain Vieta's formulas, express the relationship between coordinates, then substitute into \\overrightarrow{OP}\\cdot\\overrightarrow{OQ} for calculation. From the problem, the foci of the ellipse lie on the y-axis, with c=1, b=1, \\therefore a^{2}=b^{2}+c^{2}=2, \\therefore the ellipse equation is x^{2}+\\frac{y^{2}}{2}=1. It can be seen that when the slope of line l does not exist, it does not satisfy the condition. Let the equation of line l be y=kx+1. Since P is distinct from points A and B, \\therefore k\\neq\\pm1, then P(-\\frac{1}{k},0). Let C(x_{1},y_{1}), D(x_{2},y_{2}). Solving the system of the line and the ellipse: \\begin{cases}y=kx+1\\\\x^{2}+\\frac{y^{2}}{2}=1\\end{cases}, we get (2+k^{2})x^{2}+2kx-1=0, \\therefore x_{1}+x_{2}=-\\frac{2k}{2+k^{2}}, x_{1}x_{2}=-\\frac{1}{2+k^{2}}. The equation of line AC is y=\\frac{y_{1}}{x_{1}+1}(x+1), the equation of line BD is y=\\frac{y_{2}}{x_{2}-1}(x-1). Solving the system of lines AC and BD gives \\frac{x+1}{x-1}=\\frac{y_{2}(x_{1}+1)}{y_{1}(x_{2}-1)}. Since -1<x_{1},x_{2}<1, \\therefore \\frac{x+1}{x-1} and \\frac{y_{2}}{y_{1}} have opposite signs. (\\frac{x+1}{x-1})^{2}=\\frac{y_{2}^{2}(x_{1}+1)^{2}}{y_{1}^{2}(x_{2}-1)^{2}}=\\frac{2-2x_{2}^{2}}{2-2x_{1}^{2}}\\cdot\\frac{(x_{1}+1)^{2}}{(x_{2}-1)^{2}}=\\frac{(1+x_{1})(1+x_{2})}{(1-x_{1})(1-x_{2})}=\\frac{1+\\frac{2k}{k^{2}+2}}{1-\\frac{-2k}{k^{2}+2}}\\cdot\\frac{2}{-1}=(\\frac{k-1}{k+1}). Also, y_{1}y_{2}=k^{2}x_{1}x_{2}+k(x_{1}+x_{2})+1=\\frac{2(1-k^{2})}{k^{2}+2}=-\\frac{2(1+k)^{2}}{k^{2}+2}\\cdot\\frac{k-1}{k+1}. Therefore, \\frac{k-1}{k+1} and y_{1}y_{2} have opposite signs, so \\frac{x+1}{x-1} and \\frac{k-1}{k+1} have the same sign, thus \\frac{x+1}{x-1}=\\frac{k-1}{k+1}, solving gives x=-k, hence Q(-k,y_{0}). Then \\overrightarrow{OP}\\cdot\\overrightarrow{OQ}=(-\\frac{1}{k},0)\\cdot(-k,y_{0})=1."
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ are $F_{1}$ and $F_{2}$ respectively, and $P$ is a point on the ellipse such that $\\angle F_{1} P F_{2}=60^{\\circ}$. Then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/25 + y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "3*sqrt(3)",
    "fact_spans": "[[[0, 38], [66, 68]], [[46, 53]], [[62, 65]], [[54, 61]], [[0, 38]], [[0, 61]], [[0, 61]], [[62, 71]], [[73, 106]]]",
    "query_spans": "[[[108, 135]]]",
    "process": "Since the left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ are $F_{1}$ and $F_{2}$ respectively, and $P$ is a point on the ellipse, we have $|PF_{1}|+|PF_{2}|=10$, $|F_{1}F_{2}|=8$. Given $\\angle F_{1}PF_{2}=60^{\\circ}$, in $\\triangle F_{1}PF_{2}$, by the cosine law:  \n$|F_{1}F_{2}|^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}|\\cdot|PF_{2}|\\cos60^{\\circ}=(|PF_{1}|+|PF_{2}|)^{2}-2|PF_{1}||PF_{2}|-2|PF_{1}|\\cdot|PF_{2}|\\cos60^{\\circ}=(|PF_{1}|+|PF_{2}|)^{2}-3|PF_{1}|\\cdot|PF_{2}|=100-3|PF_{1}|\\cdot|PF_{2}|=64$,  \nso $|PF_{1}|\\cdot|PF_{2}|=12$, thus $S_{\\triangle PF_{1}F_{2}}=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|\\sin60^{\\circ}=3\\sqrt{3}$."
  },
  {
    "text": "A line with slope $-2$ is drawn through the focus of the parabola $y^{2}=4x$, intersecting the parabline at points $A$ and $B$. Then $|AB|=$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), H);Slope(H)=-2;Intersection(H, G) = {A,B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "5",
    "fact_spans": "[[[1, 15], [29, 32]], [[26, 28]], [[33, 36]], [[38, 42]], [[1, 15]], [[0, 28]], [[18, 28]], [[26, 44]]]",
    "query_spans": "[[[46, 54]]]",
    "process": "First, find the coordinates of the focus of the parabola from the parabolic equation, then use the point-slope form to obtain the equation of the line. Combine the line equation with the parabolic equation, eliminate $ y $, and use Vieta's formulas to find the value of $ x_{1}+x_{2} $. Then, according to the definition of the parabola, $ |AB| = x_{1}+\\frac{p}{2}+x_{2}+\\frac{p}{2} = x_{1}+x_{2}+p $, yielding the answer. The focus of the parabola is $ (1,0) $; thus, the equation of the line is $ y = -2x+2 $. Substituting into the parabolic equation gives $ x^{2} \\cdot 3x+1 = 0 $. Therefore, $ x_{1}+x_{2} = 3 $. According to the definition of the parabola, $ |AB| = x_{1}+\\frac{p}{2}+x_{2}+\\frac{p}{2} = x_{1}+x_{2}+p = 3+2 = 5 $."
  },
  {
    "text": "$AB$ is a chord passing through the focus $F$ of the parabola $y^{2}=2x$, and $|AB|=4$. Then, the distance from the midpoint $C$ of $AB$ to the line $x+\\frac{1}{2}=0$ is?",
    "fact_expressions": "A: Point;B: Point;G: Parabola;Expression(G) = (y^2 = 2*x);Focus(G) = F;F: Point;IsChordOf(LineSegmentOf(A, B), G) = True;PointOnCurve(F, LineSegmentOf(A, B)) = True;Abs(LineSegmentOf(A, B)) = 4;C: Point;MidPoint(LineSegmentOf(A, B)) = C;H: Line;Expression(H) = (x + 1/2 = 0)",
    "query_expressions": "Distance(C,H)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 4]], [[0, 4]], [[6, 20]], [[6, 20]], [[6, 26]], [[23, 26]], [[0, 28]], [[0, 28]], [[30, 38]], [[47, 50]], [[40, 50]], [[51, 70]], [[51, 70]]]",
    "query_spans": "[[[47, 75]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ is a point on the parabola $y^{2}=8x$, let $d_{1}$ be the distance from $P$ to the directrix of this parabola, and $d_{2}$ be the distance from $P$ to the line $4x+3y+8=0$. Then the minimum value of $d_{1}+d_{2}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);P: Point;PointOnCurve(P, G);d1: Number;d2: Number;H: Line;Expression(H) = (4*x + 3*y + 8 = 0);Distance(P, Directrix(G)) = d1;Distance(P, H) = d2",
    "query_expressions": "Min(d1 + d2)",
    "answer_expressions": "16/5",
    "fact_spans": "[[[7, 21], [31, 34]], [[7, 21]], [[2, 6], [26, 29]], [[2, 24]], [[41, 48]], [[69, 76]], [[50, 65]], [[50, 65]], [[26, 48]], [[26, 76]]]",
    "query_spans": "[[[78, 97]]]",
    "process": "The focus of $ y^{2}=8x $ is $ F(2,0) $. By the definition of a parabola, the distance from point $ P $ to the directrix of this parabola is $ d_{1}=PF $. Therefore, the minimum value of $ d_{1}+d_{2} $ is the distance from $ F $ to the line $ 4x+3y+8=0 $. Hence, $ d=\\frac{|8+8|}{\\sqrt{4^{2}+3^{2}}}=\\frac{16}{5} $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has an eccentricity of $2$, and its right focus coincides with the focus of the parabola $y^{2}=16 x$. Then, what is the distance from the vertex of the hyperbola $C$ to its asymptote?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(C) = 2;G: Parabola;Expression(G) = (y^2 = 16*x);RightFocus(C) = Focus(G)",
    "query_expressions": "Distance(Vertex(C), Asymptote(C))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 63], [98, 104]], [[2, 63]], [[9, 63]], [[10, 63]], [[9, 63]], [[9, 63]], [[2, 71]], [[76, 91]], [[76, 91]], [[2, 96]]]",
    "query_spans": "[[[98, 116]]]",
    "process": "Find the focus of the parabola to obtain $ c $ for the hyperbola, use the eccentricity formula to find $ a $, then use the relationship among $ a, b, c $ to find $ b $. Calculate the distance from the vertex to the asymptote to obtain the desired value. The focus of the parabola $ y^2 = 16x $ is $ (4, 0) $, so $ c = 4 $ for the hyperbola. The eccentricity of the hyperbola is 2, thus $ \\frac{c}{a} = 2 $, giving $ a = 2 $, and $ b = \\sqrt{c^2 - a^2} = 2\\sqrt{3} $. Therefore, the asymptotes of the hyperbola are $ v = \\pm\\sqrt{3}x $, and the vertices are at $ (\\pm2, 0) $. The distance $ l $ from the vertex of the hyperbola to its asymptote is $ \\frac{2\\sqrt{3}}{\\sqrt{1+3}} = \\sqrt{3} $."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ has directrix $l$, and $l$ intersects the asymptotes of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ at points $A$ and $B$, respectively. If $|AB|=4$, then $p=$?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 2*(p*x));p: Number;p>0;l: Line;Directrix(H) = l;G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1);Z1: Line;Z2: Line;Asymptote(G) = {Z1, Z2};A: Point;Intersection(l, Z1) = A;B: Point;Intersection(l, Z2) = B;Abs(LineSegmentOf(A, B)) = 4",
    "query_expressions": "p",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 23]], [[2, 23]], [[96, 99]], [[5, 23]], [[27, 30], [33, 36]], [[2, 31]], [[37, 65]], [[37, 65]], [], [], [[37, 69]], [[73, 76]], [[33, 83]], [[78, 81]], [[33, 83]], [[86, 94]]]",
    "query_spans": "[[[96, 101]]]",
    "process": "The directrix of the parabola $ y^{2} = 2px $ ($ p > 0 $) is $ l: x = -\\frac{p}{2} $. The equations of the two asymptotes of the hyperbola $ \\frac{x^{2}}{4} - y^{2} = 1 $ are $ y = \\pm\\frac{1}{2}x $. We obtain $ A(-\\frac{p}{2}, -\\frac{p}{4}) $, $ B(-\\frac{p}{2}, \\frac{p}{4}) $. From $ |AB| = \\frac{p}{2} = 4 $, we get $ p = 8 $."
  },
  {
    "text": "Given the parabola $C$: $x^{2}=8 y$, the directrix intersects the $y$-axis at point $A$, and the focus is $F$. Point $P$ is an arbitrary point on the parabola $C$. Let $t=\\frac{|P A|}{|P F|}$. When $t$ attains its maximum value, what is the slope of the line $P A$?",
    "fact_expressions": "C: Parabola;P: Point;A: Point;F: Point;t:Number;Expression(C) = (x^2 = 8*y);Intersection(Directrix(C), yAxis) = A;Focus(C) = F;PointOnCurve(P, C);t=Abs(LineSegmentOf(P,A))/Abs(LineSegmentOf(P,F));WhenMax(t)",
    "query_expressions": "Slope(LineOf(P,A))",
    "answer_expressions": "pm*1",
    "fact_spans": "[[[2, 21], [48, 54]], [[43, 47]], [[31, 35]], [[39, 42]], [[86, 89]], [[2, 21]], [[2, 35]], [[2, 42]], [[43, 59]], [[61, 84]], [[85, 95]]]",
    "query_spans": "[[[96, 108]]]",
    "process": "From the given conditions, we know A(0,-2), F(0,2). Draw PB\\bot l (where l is the directrix of the parabola), with foot B. By the definition of a parabola, |PF| = |PB|. Let \\angle PAB = \\alpha, then t = \\frac{|PA|}{|PF|} = \\frac{|PA|}{|PB|} = \\frac{1}{\\sin\\alpha}. When \\sin\\alpha is minimum, t is maximum. When line PA is tangent to the parabola x^{2} = 8y, \\sin\\alpha is minimum, i.e., t is maximum. Let P(x_{0}, \\frac{x_{0}^{2}}{8}). Since y' = \\frac{x}{4}, the slope of the tangent at point P is k = \\frac{x_{0}}{4}. Thus, the equation of the tangent at P is y - \\frac{x_{0}^{2}}{8} = \\frac{x_{0}}{4}(x - x_{0}). Since the tangent passes through A(0,-2), we have -2 - \\frac{x_{0}^{2}}{8} = -\\frac{x_{0}^{2}}{4}. Solving gives x_{0} = \\pm4. Therefore, when t reaches its maximum value, the slope of line PA is \\pm1."
  },
  {
    "text": "Given that the distance from a point $P$ on the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ to the point $(5,0)$ is $8.5$, then the distance from point $P$ to the point $(-5,0)$ is?",
    "fact_expressions": "G: Hyperbola;H: Point;P: Point;D:Point;Expression(G) = (x^2/16 - y^2/9 = 1);Coordinate(H) = (5, 0);Coordinate(D)=(-5,0);PointOnCurve(P, G);Distance(P, H) = 8.5",
    "query_expressions": "Distance(P, D)",
    "answer_expressions": "16.5",
    "fact_spans": "[[[2, 41]], [[48, 56]], [[43, 47], [67, 71]], [[72, 81]], [[2, 41]], [[48, 56]], [[72, 81]], [[2, 47]], [[43, 65]]]",
    "query_spans": "[[[67, 86]]]",
    "process": "Let the two foci of the hyperbola be $ F_{1}(-5,0) $ and $ F_{2}(5,0) $. By the definition of a hyperbola, $ ||PF_{1}|-|PF_{2}||=8 $, and $ |PF_{2}|=8.5 $. Therefore, $ |PF_{1}|=16.5 $ or $ |PF_{1}|=0.5 $. According to the problem, the shortest distance from a point on the left branch of the hyperbola to the left focus is 1. Hence, $ |PF_{1}|=0.5 $ does not satisfy the condition, so only $ |PF_{1}|=16.5 $ is valid."
  },
  {
    "text": "The line $y=k(x-1)$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$. If $|AB|=\\frac{16}{3}$, then $k=$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;k:Number;Expression(G) = (y^2 = 4*x);Expression(H) = (y = k*(x - 1));Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 16/3",
    "query_expressions": "k",
    "answer_expressions": "pm*sqrt(3)",
    "fact_spans": "[[[13, 27]], [[0, 12]], [[29, 32]], [[33, 36]], [[63, 66]], [[13, 27]], [[0, 12]], [[0, 38]], [[40, 60]]]",
    "query_spans": "[[[63, 68]]]",
    "process": "From \\begin{cases}y=k(x-1)\\\\y^{2}=4x\\end{cases}, eliminating $y$, we obtain $k^{2}x^{2}-(2k^{2}+4)x+k^{2}=0$. Since the line intersects the parabola at two points $A$ and $B$, we have $\\begin{cases}\\\\\\end{cases}=(2k^{2}+4_{1}^{2}-\\angle4k^{4}>0$, solving gives $k\\neq0$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $x_{1}+x_{2}=\\frac{2k^{2}+4}{k^{2}}$. Since $|AB|=x_{1}+x_{2}+2=\\frac{16}{3}$, we have $x_{1}+x_{2}=\\frac{2k^{2}+4}{k^{2}}=\\frac{10}{3}$, thus $k^{2}=3$, $k=\\pm\\sqrt{3}$. Verification shows that $k=+\\sqrt{3}$ satisfies the condition."
  },
  {
    "text": "The directrix of the parabola $y^{2}=4 x$ intersects the $x$-axis at point $P$. A line with slope $k$ ($k>0$) passing through point $P$ intersects the parabola at points $A$ and $B$. $F$ is the focus of the parabola. If $|F A|=3|F B|$, then the slope $k$ of line $A B$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);Intersection(Directrix(G), xAxis) = P;P: Point;PointOnCurve(P, H);Slope(H) = k;k: Number;k>0;H: Line;Intersection(H, G) = {A, B};B: Point;A: Point;Focus(G) = F;F: Point;Abs(LineSegmentOf(F, A)) = 3*Abs(LineSegmentOf(F, B));Slope(LineOf(A, B)) = k",
    "query_expressions": "k",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[0, 14], [50, 53], [68, 71]], [[0, 14]], [[0, 29]], [[25, 29], [31, 35]], [[30, 49]], [[36, 49]], [[38, 46], [102, 105]], [[38, 46]], [[47, 49]], [[47, 63]], [[58, 61]], [[54, 57]], [[64, 74]], [[64, 67]], [[76, 90]], [[92, 105]]]",
    "query_spans": "[[[102, 107]]]",
    "process": "Solve the system of the line AB equation and the parabola equation; according to the definition of the parabola and the focal radius formula, the coordinates of A or B can be obtained, and k can be calculated using the slope formula between two points. [Detailed solution] From the problem, we know P(-1,0), let A(x_{1},y_{1}), B(x_{2},y_{2}). From the given |FA|=3|FB|, we have x_{1}+1=3(x_{2}+1), that is, x_{1}=3x_{2}+2\\textcircled{1}. The equation of AB: y=kx+k, combined with y^{2}=4x gives: k^{2}x^{2}+(2k^{2}-4)x+k^{2}=0, then x_{1}x_{2}=1\\textcircled{2}. Solving \\textcircled{1}\\textcircled{2} yields x_{2}=\\frac{1}{3}, x_{1}=3. Substituting x_{1}=3 into y^{2}=4x, and knowing k>0 implies y_{1}>0, we obtain A(3,2\\sqrt{3}). Therefore, k=\\frac{2\\sqrt{3}-0}{3-(-1)}=\\frac{\\sqrt{3}}{2}"
  },
  {
    "text": "Let the focus of the parabola $y^{2}=8 x$ coincide with the right focus of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1(b>0)$, then $b$=?",
    "fact_expressions": "G: Hyperbola;b: Number;H: Parabola;b>0;Expression(G) = (x^2 - y^2/b^2 = 1);Expression(H) = (y^2 = 8*x);Focus(H) = RightFocus(G)",
    "query_expressions": "b",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[19, 56]], [[64, 67]], [[1, 15]], [[22, 56]], [[19, 56]], [[1, 15]], [[1, 62]]]",
    "query_spans": "[[[64, 69]]]",
    "process": "From the given conditions, it is known that the focus of the parabola has coordinates (2,0). For the hyperbola: $a^{2}=1$, $c^{2}=2^{2}=4$, $b^{2}=c^{2}-a^{2}=3 \\Rightarrow b=\\sqrt{3}$"
  },
  {
    "text": "What is the equation of the parabola with the center of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ as its vertex and the left focus as its focus?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (x^2/16 - y^2/9 = 1);Center(G) = Vertex(H);LeftFocus(G) = Focus(H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2 = - 20*x",
    "fact_spans": "[[[1, 40]], [[54, 57]], [[1, 40]], [[0, 57]], [[0, 57]]]",
    "query_spans": "[[[54, 61]]]",
    "process": ""
  },
  {
    "text": "The right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $F_{2}$, and the line $l$ passes through the point $F_{2}$ and intersects the right branch of the hyperbola $C$ at points $A$ and $B$. If the chord length $|A B|=5 a$ and $2|B F_{2}|=3|A F_{2}|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;A: Point;B: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F2;PointOnCurve(F2, l);Intersection(l, RightPart(C)) = {A, B};IsChordOf(LineSegmentOf(A, B), C);Abs(LineSegmentOf(A, B)) = 5*a;2*Abs(LineSegmentOf(B, F2)) = 3*Abs(LineSegmentOf(A, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(85)/5",
    "fact_spans": "[[[74, 79]], [[0, 61], [90, 96], [152, 155]], [[8, 61]], [[8, 61]], [[100, 103]], [[104, 107]], [[66, 73], [80, 88]], [[8, 61]], [[8, 61]], [[0, 61]], [[0, 73]], [[74, 88]], [[74, 109]], [[90, 124]], [[113, 124]], [[126, 149]]]",
    "query_spans": "[[[152, 161]]]",
    "process": "According to the problem, |AF₁| = 4a, |BF₁| = 5a, then cos∠F₁AB = 2/5. In △AF₁F₂, using the law of cosines, the relationship between a and c can be obtained, thus yielding the eccentricity. As shown in the figure: therefore |AF₂| = 2a. Let the left focus be F₁, by the definition of hyperbola, |AF₁| = 4a, |BF₁| = 5a, then △ABF is an isosceles triangle, cos∠F₁AB = 2/5. In △AF₁F₂, |F₁F₂| = 2c, by the law of cosines we have 4c² = (2a)² + (4a)² - 2×2a×4a×cos∠F₁AB. Simplifying yields c/a = √85/5. The eccentricity of this hyperbola is √85."
  },
  {
    "text": "Given that $P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{8}=1$, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola, respectively, $O$ is the origin, and points $M$, $N$ satisfy $\\overrightarrow{F_{1} P}=\\lambda \\overrightarrow{P M}(\\lambda>0)$, $\\overrightarrow{P N}=\\mu\\left(\\frac{\\overrightarrow{P M}}{| \\overrightarrow{P M}|}+\\frac{\\overrightarrow{P F_{2}}}{| \\overrightarrow{P F_{2}|}}\\right)$, $\\overrightarrow{P N} \\cdot \\overrightarrow{F_{2} N}=0$. If $|\\overrightarrow{P F_{2}}|=4$, then what is the area of the circle with center $O$ and radius $ON$?",
    "fact_expressions": "G: Hyperbola;H: Circle;O: Origin;N: Point;F1: Point;P: Point;M: Point;F2: Point;lambda:Number;lambda>0;mu:Number;Expression(G) = (x^2/16 - y^2/8 = 1);PointOnCurve(P, RightPart(G));LeftFocus(G) = F1;RightFocus(G) = F2;VectorOf(F1, P) = lambda*VectorOf(P, M);VectorOf(P,N)=mu*(VectorOf(P,M)/Abs(VectorOf(P,M))+VectorOf(P,F2)/Abs(VectorOf(P,F2)));Abs(VectorOf(P, F2)) = 4;DotProduct(VectorOf(P,N),VectorOf(F2,N))=0;Center(H)=O;Radius(H)=LineSegmentOf(O,N)",
    "query_expressions": "Area(H)",
    "answer_expressions": "64*pi",
    "fact_spans": "[[[6, 45], [69, 72]], [[416, 417]], [[79, 82], [400, 403]], [[93, 96]], [[51, 58]], [[2, 5]], [[88, 92]], [[59, 66]], [[98, 164]], [[98, 164]], [[166, 307]], [[6, 45]], [[2, 50]], [[51, 78]], [[51, 78]], [[98, 164]], [[166, 307]], [[366, 396]], [[309, 364]], [[399, 417]], [[407, 417]]]",
    "query_spans": "[[[416, 422]]]",
    "process": ""
  },
  {
    "text": "If the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has a focal distance of $2 \\sqrt{6}$, an asymptote $l$, and the distance from the point $(1,0)$ to $l$ is $\\frac{\\sqrt{6}}{3}$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Point;a>0;b>0;l:Line;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(H) = (1, 0);FocalLength(G) = 2*sqrt(6);OneOf(Asymptote(G)) = l;Distance(H, l) = sqrt(6)/3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/2 - y^2/4 = 1",
    "fact_spans": "[[[1, 57], [123, 126]], [[4, 57]], [[4, 57]], [[85, 93]], [[4, 57]], [[4, 57]], [[80, 83], [94, 97]], [[1, 57]], [[85, 93]], [[1, 73]], [[1, 83]], [[85, 121]]]",
    "query_spans": "[[[123, 131]]]",
    "process": "Analysis: According to the given data, the semi-focal distance $ c $ can be first obtained. Since the point $ (1,0) $ lies on the symmetry axis of the hyperbola, selecting any asymptote equation of the hyperbola and combining it with the point-to-line distance formula yields the relationship between $ a $ and $ b $, thereby solving for the equation of the hyperbola. From the problem, the semi-focal distance of the hyperbola is $ c = \\sqrt{6} $, and the distances from the point $ (1,0) $ to the two asymptotes of the hyperbola are equal. Therefore, the equation of line $ l $ can be chosen as $ bx - ay = 0 $, then $ \\frac{b}{\\sqrt{b^{2} + a^{2}}} = \\frac{b}{c} = \\frac{b}{\\sqrt{6}} = \\frac{\\sqrt{6}}{3} $, yielding $ b = 2 $, so $ a^{2} = c^{2} - b^{2} = 2 $; thus, the standard equation of the hyperbola is $ \\frac{x^{2}}{2} - \\frac{y^{2}}{4} = 1 $."
  },
  {
    "text": "Given fixed points $M(2,0)$, $N(-2,0)$, and $P$ is a moving point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, then the minimum value of $\\frac{9}{|P M|}+\\frac{1}{|P N|}$ is?",
    "fact_expressions": "G: Ellipse;M: Point;N: Point;P: Point;Expression(G) = (x^2/9 + y^2/5 = 1);Coordinate(M) = (2, 0);Coordinate(N) = (-2, 0);PointOnCurve(P, G)",
    "query_expressions": "Min(1/Abs(LineSegmentOf(P, N)) + 9/Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "8/3",
    "fact_spans": "[[[31, 68]], [[4, 12]], [[15, 25]], [[27, 30]], [[31, 68]], [[4, 12]], [[15, 25]], [[26, 72]]]",
    "query_spans": "[[[74, 114]]]",
    "process": "kg); according to the definition of an ellipse, |PM| + |PN| = 6, simplify (\\frac{9}{|PM|} + \\frac{1}{|PN|}) \\cdot \\frac{|PM| + |PN|}{6} and combine with the basic inequality to find the minimum value of \\frac{9}{|PM|} + \\frac{1}{|PN|}. From the problem: points M(-2,0) and N(2,0) are the foci of the ellipse \\frac{x^{2}}{9} + \\frac{y^{2}}{5} = 1, so |PM| + |PN| = 2a = 6, thus \\frac{9}{|PM|} + \\frac{1}{|PN|} \\overset\\rightarrow M| + |PN| + \\frac{| + |PN|}{|PN|} = \\frac{10}{6} + \\frac{3|PN|}{2|PM|} + \\frac{|PM|}{6|PN|} \\geqslant \\frac{5}{3} + 2\\sqrt{\\frac{3|PN|}{2|PM|} \\cdot \\frac{|PM|}{6|PN|} = \\frac{8}{3}, equality holds if and only if \\frac{DM|}{PN|}, i.e., when 3|PN| = |PM|, the minimum value of \\frac{9}{|PM|} + \\frac{1}{|PN|} is \\frac{8}{2},"
  },
  {
    "text": "In the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, what is the equation of the line containing the chord for which point $M(-1 , 2)$ is the midpoint?",
    "fact_expressions": "G: Ellipse;H: LineSegment;M: Point;Expression(G) = (x^2/16 + y^2/9 = 1);Coordinate(M) = (-1, 2);IsChordOf(H, G);MidPoint(H) = M",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "9*x - 32*y + 73 = 0",
    "fact_spans": "[[[0, 38]], [], [[41, 53]], [[0, 38]], [[41, 53]], [[0, 58]], [[0, 58]]]",
    "query_spans": "[[[0, 66]]]",
    "process": ""
  },
  {
    "text": "The line $l$ passes through the focus $F$ of the parabola $C$: $y^{2}=12x$, and intersects the parabola $C$ at points $A$ and $B$. The length of chord $AB$ is $16$. Then, what is the inclination angle of the line $l$?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;F: Point;Expression(C) = (y^2 = 12*x);Focus(C) = F;PointOnCurve(F, l);Intersection(l,C)={A,B};IsChordOf(LineSegmentOf(A,B),C);Length(LineSegmentOf(A, B)) = 16",
    "query_expressions": "Inclination(l)",
    "answer_expressions": "pi/3, 2*pi/3",
    "fact_spans": "[[[0, 5], [69, 74]], [[7, 27], [36, 42]], [[44, 47]], [[48, 51]], [[30, 33]], [[7, 27]], [[7, 33]], [[0, 33]], [[0, 53]], [[36, 60]], [[55, 67]]]",
    "query_spans": "[[[69, 81]]]",
    "process": "Find the coordinates of the focus of the parabola, write the equation of the line, solve the system of equations with the parabola, use the chord length formula to transform and solve for the slope of the line, and then obtain the inclination angle. The line $ l $ passes through the focus $ F(3,0) $ of the parabola $ C: y^{2} = 12x $. If the slope of the line does not exist, then the chord length is 12, which does not meet the condition; hence, the slope of line $ l $ exists, denoted as $ k $, and the equation of line $ l $ is $ y = k(x - 3) $, intersecting the parabola $ C $ at points $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. We obtain $ k^{2}(x - 3)^{2} = 12x $, that is, $ k^{2}x^{2} - (6k^{2} + 12)x + 9k^{2} = 0 $, yielding $ x_{1} + x_{2} = \\frac{6k^{2} + 12}{k^{2}} $. The length of chord $ AB $ is 16, so $ \\frac{6k^{2} + 12}{k^{2}} + 6 = 16 $, solving gives $ k = \\pm\\sqrt{3} $. Therefore, the inclination angle of the line is: $ \\frac{\\pi}{3} $ or $ \\frac{2\\pi}{3} $."
  },
  {
    "text": "Given that point $A$ lies on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F$ is one focus of the ellipse, $AF \\perp x$-axis, and $|AF|=c$ ($c$ being the semi-focal length of the ellipse), then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;F: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(A, G);OneOf(Focus(G)) = F;IsPerpendicular(LineSegmentOf(A,F), xAxis);c: Number;Abs(LineSegmentOf(A, F)) = c;HalfFocalLength(G) = c",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(\\sqrt{5} - 1)/2",
    "fact_spans": "[[[7, 59], [67, 69], [107, 109], [116, 118]], [[9, 59]], [[9, 59]], [[2, 6]], [[63, 66]], [[9, 59]], [[9, 59]], [[7, 59]], [[2, 62]], [[63, 74]], [[76, 90]], [[103, 106]], [[91, 100]], [[103, 113]]]",
    "query_spans": "[[[116, 124]]]",
    "process": "Without loss of generality, let $ A(c,c) $. Substituting into the ellipse equation yields: $\\frac{c^{2}}{a^{2}}+\\frac{c^{2}}{b^{2}}=1$, solving gives $ e^{2}=\\frac{3-\\sqrt{5}}{2} $, thus $ e=\\frac{\\sqrt{5}-1}{2} $. Therefore, the answer should be filled in as: $\\frac{\\sqrt{5}-1}{2}$."
  },
  {
    "text": "The midpoint of a chord of the parabola $y^{2}=-12 x$ is $M(-2,-3)$. Then the equation of the line containing this chord is?",
    "fact_expressions": "G: Parabola;H: LineSegment;M: Point;Expression(G) = (y^2 = -12*x);Coordinate(M) = (-2, -3);MidPoint(H)=M;IsChordOf(H,G)",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "y=2*x+1",
    "fact_spans": "[[[0, 16]], [], [[23, 33]], [[0, 16]], [[23, 33]], [[0, 33]], [[0, 20]]]",
    "query_spans": "[[[0, 46]]]",
    "process": ""
  },
  {
    "text": "The point $M(-4, m)$ on the parabola $y^{2}=-2 p x(p>0)$ is at a distance of $5$ from the focus. Find the value of $m$.",
    "fact_expressions": "G: Parabola;p: Number;M: Point;p>0;Expression(G) = (y^2 = -2*p*x);Coordinate(M) = (-4, m);PointOnCurve(M, G);Distance(M, Focus(G)) = 5;m:Number",
    "query_expressions": "m",
    "answer_expressions": "pm*4",
    "fact_spans": "[[[0, 22]], [[3, 22]], [[24, 35]], [[3, 22]], [[0, 22]], [[24, 35]], [[0, 35]], [[0, 45]], [[47, 50]]]",
    "query_spans": "[[[47, 54]]]",
    "process": "According to the definition of a parabola, we have $4+\\frac{p}{2}=5$, so $p=2$, $2p=4$. Therefore, the equation of the parabola is $y^{2}=-4x$. When $x=-4$, $y^{2}=16$, $y=\\pm4$. [Note: This question mainly examines the definition of a parabola, solving the equation of a parabola, and finding coordinates of points on a parabola, belonging to a basic-level problem."
  },
  {
    "text": "The left vertex of hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ is $A$, the right focus is $F$, and a moving point $B$ lies on hyperbola $C$. When $B F \\perp A F$, $|A F|=|B F|$, then the asymptote equations of hyperbola $C$ are?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;B: Point;F: Point;A: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftVertex(C) = A;RightFocus(C) = F;PointOnCurve(B, C);IsPerpendicular(LineSegmentOf(B, F), LineSegmentOf(A, F));Abs(LineSegmentOf(A, F)) = Abs(LineSegmentOf(B, F))",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y = pm*sqrt(3)*x",
    "fact_spans": "[[[0, 64], [87, 93], [129, 135]], [[7, 64]], [[7, 64]], [[83, 86]], [[77, 80]], [[69, 72]], [[7, 64]], [[7, 64]], [[0, 64]], [[0, 72]], [[0, 80]], [[83, 94]], [[97, 112]], [[114, 127]]]",
    "query_spans": "[[[129, 143]]]",
    "process": "According to the given conditions, we obtain $ c^{2} - ac - 2a^{2} = 0 $. From this homogeneous equation, we get $ e = 2 $. Then, using the relationship among $ a $, $ b $, and $ c $, we can find $ \\frac{b}{a} $ to obtain the solution. [Solution] Let the semi-focal length of the hyperbola be $ c $, then $ F(c,0) $, $ B\\left(c,\\pm\\frac{b^{2}}{a}\\right) $. Since $ |AF| = |BF| $, we have $ \\frac{b^{2}}{a} = a + c $. Therefore, $ c^{2} - ac - 2a^{2} = 0 $, which implies $ e^{2} - e - 2 = 0 $. Hence, $ e = 2 $. Thus, the asymptotic equations of the hyperbola are $ y = \\pm\\sqrt{3}x $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively, $P$ is a point on $C$, and the incenter of $\\triangle P F_{1} F_{2}$ is point $I$. A line passing through $I$ and parallel to the $x$-axis intersects $P F_{1}$ and $P F_{2}$ at points $A$ and $B$, respectively. If the eccentricity of ellipse $C$ is $e=\\frac{1}{2}$, then $\\frac {S_ {\\Delta{P A B}}} {S_ \\Delta{P F_{1} F_{2}}}$=?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;P: Point;F1: Point;F2: Point;A: Point;I: Point;B: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);Incenter(TriangleOf(P,F1,F2))=I;PointOnCurve(I,G);IsParallel(G,xAxis);Intersection(G,LineSegmentOf(P,F1))=A;Intersection(G,LineSegmentOf(P,F2))=B;Eccentricity(C)=e;e:Number;e=1/2",
    "query_expressions": "Area(TriangleOf(P,A,B))/Area(TriangleOf(P,F1,F2))",
    "answer_expressions": "4/9",
    "fact_spans": "[[[20, 77], [87, 90], [177, 182]], [[27, 77]], [[27, 77]], [[141, 143]], [[83, 86]], [[2, 9]], [[10, 17]], [[167, 171]], [[123, 127], [129, 132]], [[172, 175]], [[27, 77]], [[27, 77]], [[20, 77]], [[2, 82]], [[2, 82]], [[83, 93]], [[94, 127]], [[128, 143]], [[133, 143]], [[141, 175]], [[141, 175]], [[177, 201]], [[186, 201]], [[186, 201]]]",
    "query_spans": "[[[203, 260]]]",
    "process": "According to the eccentricity of the ellipse, we have $ a = 2c $. According to the definition of the ellipse, the perimeter of $ \\triangle PF_{1}F_{2} $ is $ 2(a + c) = 6c $. Let the inradius of $ \\triangle PF_{1}F_{2} $ be $ r $, and let point $ P(x, y) $. Using $ S_{\\triangle PF_{1}F_{2}} = \\frac{1}{2}|F_{1}F_{2}||y| = c|y| = pr $ (where $ p $ is half the perimeter of $ \\triangle PF_{1}F_{2} $), we obtain $ |y| = 3r $. Then, according to $ \\frac{S_{\\triangle PAB}}{S_{\\triangle PF_{1}F_{2}}} = \\left( \\frac{y - r}{y} \\right)^{2} $, the result for 4 can be found. [Solution] Let the ellipse $ C: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $) have focal length $ 2c $. From the given condition $ \\frac{c}{a} = \\frac{1}{2} $, we get $ a = 2c $. From the definition of the ellipse, $ |PF_{1}| + |PF_{2}| = 2a $, $ |F_{1}F_{2}| = 2c $. The perimeter of $ \\triangle PF_{1}F_{2} $ is $ 2(a + c) = 6c $. Let the inradius of $ \\triangle PF_{1}F_{2} $ be $ r $, and point $ P(x, y) $. Also, $ S_{\\triangle PF_{1}F_{2}} = \\frac{1}{2}|F_{1}F_{2}||y| = c|y| $. Let $ p $ be half the perimeter of $ \\triangle PF_{1}F_{2} $, then $ S_{\\triangle PF_{1}F_{2}} = pr = 3cr $. Thus, $ 3cr = c|y| $, so $ |y| = 3r $. From the given condition, $ \\triangle PAB \\sim \\triangle PF_{1}F_{2} $, we get $ \\frac{S_{\\triangle PAB}}{S_{\\triangle PF_{1}F_{2}}} = \\left( \\frac{y - r}{y} \\right)^{2} $. Therefore, $ \\frac{S_{\\triangle PAB}}{S_{\\triangle PF_{1}F_{2}}} = \\left( \\frac{3r - r}{3r} \\right)^{2} = \\frac{4}{9} $."
  },
  {
    "text": "Ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, $A(a , 0)$, $B(0 , b)$, the distance from the origin to line $AB$ is $c$ ($c$ is the semi-focal length), then the eccentricity of the ellipse $e$=?",
    "fact_expressions": "A: Point;B: Point;G: Ellipse;b: Number;a: Number;e: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (a, 0);Coordinate(B) = (0, b);Eccentricity(G) = e;O: Origin;Distance(O, LineOf(A, B)) = c ;c: Number;HalfFocalLength(G) = c",
    "query_expressions": "e",
    "answer_expressions": "(\\sqrt{5} - 1)/2",
    "fact_spans": "[[[55, 65]], [[68, 78]], [[0, 52], [107, 109]], [[2, 52]], [[2, 52]], [[112, 115]], [[2, 52]], [[2, 52]], [[0, 52]], [[55, 65]], [[68, 78]], [[107, 115]], [[79, 81]], [[79, 95]], [[92, 95], [96, 99]], [[0, 104]]]",
    "query_spans": "[[[112, 117]]]",
    "process": ""
  },
  {
    "text": "Given that the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ and the hyperbola $\\frac{x^{2}}{a1^{2}}-\\frac{y^{2}}{b1^{2}}=\\frac{1}{2}(a1>0, b1>0)$ have the same foci, what is the equation of the asymptotes of the hyperbola?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;E: Ellipse;a1: Number;b1: Number;a1 > 0;b1 > 0;Expression(G) = (x^2/a1^2 - y^2/b1^2 = 1/2);a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);Focus(G) = Focus(E)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = (pm*sqrt(3)/3)*x",
    "fact_spans": "[[[55, 125], [132, 135]], [[4, 54]], [[4, 54]], [[2, 54]], [[58, 125]], [[58, 125]], [[58, 125]], [[58, 125]], [[55, 125]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 130]]]",
    "query_spans": "[[[132, 142]]]",
    "process": "From the same foci, we have $a^{2}-b^{2}=\\frac{1}{2}a^{2}+\\frac{1}{2}b^{2}$, that is, $a^{2}=3b^{2}$. Thus, the asymptotes can be found. According to the problem, the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=\\frac{1}{2}$ $(a>0,b>0)$ have the same foci, which means they share the same foci with $\\frac{x^{2}}{2}-\\frac{y^{2}}{\\frac{b^{2}}{2}}=1$ $(a>0,b>0)$. Therefore, we obtain: $a^{2}-b^{2}=\\frac{1}{2}a^{2}+\\frac{1}{2}b^{2}$, that is, $a^{2}=3b^{2}$. Hence, $\\frac{b}{a}=\\frac{\\sqrt{3}}{3}$, and we get $\\frac{\\frac{b}{\\sqrt{2}}}{\\frac{a}{\\sqrt{2}}}=\\frac{\\sqrt{3}}{3}$. Therefore, the asymptotes of the hyperbola are: $y=\\pm\\frac{\\frac{b}{\\sqrt{2}}}{\\sqrt{2}}x$, $y=\\pm\\frac{\\sqrt{3}}{3}x$."
  },
  {
    "text": "The vertex of a parabola is at the origin, and its focus is one of the foci of the hyperbola $x^{2}-2 y^{2}=8$. Then, the distance from the focus of this parabola to its directrix is equal to?",
    "fact_expressions": "G: Hyperbola;H: Parabola;O: Origin;Expression(G) = (x^2 - 2*y^2 = 8);Vertex(H) = O;Focus(H) = OneOf(Focus(G))",
    "query_expressions": "Distance(Focus(H), Directrix(H))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[15, 35]], [[0, 3], [43, 46], [50, 51]], [[7, 11]], [[15, 35]], [[0, 11]], [[0, 40]]]",
    "query_spans": "[[[43, 60]]]",
    "process": ""
  },
  {
    "text": "Draw a line $l$ with slope $1$ passing through the point $M(2,0)$, intersecting the parabola $y^{2}=4x$ at points $A$ and $B$. Then $|AB|=$?",
    "fact_expressions": "l: Line;M: Point;Coordinate(M) = (2, 0);PointOnCurve(M, l);Slope(l) = 1;G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;B: Point;Intersection(l, G) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "4*sqrt(6)",
    "fact_spans": "[[[18, 23]], [[1, 10]], [[1, 10]], [[0, 23]], [[11, 23]], [[25, 39]], [[25, 39]], [[40, 43]], [[45, 48]], [[18, 50]]]",
    "query_spans": "[[[52, 60]]]",
    "process": ""
  },
  {
    "text": "A focus of the hyperbola $8 k x^{2}-k y^{2}=8$ is $(0,3)$, then what is the value of $k$? What is the equation of the asymptotes of the hyperbola?",
    "fact_expressions": "G: Hyperbola;k: Number;H: Point;Expression(G) = (8*(k*x^2) - k*y^2 = 8);Coordinate(H) = (0, 3);OneOf(Focus(G)) = H",
    "query_expressions": "k;Expression(Asymptote(G))",
    "answer_expressions": "-1\ny=pm*2*sqrt(2)*x",
    "fact_spans": "[[[0, 24], [46, 49]], [[39, 42]], [[30, 37]], [[0, 24]], [[30, 37]], [[0, 37]]]",
    "query_spans": "[[[39, 46]], [[46, 57]]]",
    "process": ""
  },
  {
    "text": "The right focus of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ is $F$, and point $P$ is a point on the asymptote such that $|O P|=2$. Then $|P F|=$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1);F: Point;RightFocus(G) = F;P: Point;PointOnCurve(P, Asymptote(G));O: Origin;Abs(LineSegmentOf(O, P)) = 2",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "{2, 2*sqrt(3)}",
    "fact_spans": "[[[0, 28]], [[0, 28]], [[33, 36]], [[0, 36]], [[37, 41]], [[0, 48]], [[50, 59]], [[50, 59]]]",
    "query_spans": "[[[61, 70]]]",
    "process": "According to the hyperbola equation, the asymptotes and the angle between them are obtained. Combining with the six-chord theorem, we know |OF| = c = 2, and the asymptote equations are y = \\pm\\sqrt{3}x, so the inclination angles of the two asymptotes are 60^{\\circ}, 120^{\\circ}. There are two possible positions for point P as shown in the figure. When P is at position P_{1}, \\angle P_{1}OF = 60^{\\circ}, then \\triangle OP_{1}F is an equilateral triangle, so |PF| = 2; when P is at position P_{2}, \\angle P_{2}OF = 120^{\\circ}, by the cosine law, PF^{2} = OF^{2} + OP^{2} - 2 \\times OF \\times OP \\times \\cos120^{\\circ}, then |PF| = 2\\sqrt{3}."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$, the left and right vertices are $A$ and $B$ respectively. Point $P$ is a point on the ellipse, and the slopes of lines $PA$ and $PB$ are $k_{1}$ and $k_{2}$ respectively. If $k_{1}=2$, then $k_{2}=$?",
    "fact_expressions": "G: Ellipse;A: Point;P: Point;B:Point;k1:Number;k2:Number;Expression(G) = (x^2/8 + y^2/4 = 1);LeftVertex(G)=A;RightVertex(G)=B;PointOnCurve(P,G);Slope(LineOf(P,A))=k1;Slope(LineOf(P,B))=k2;k1=2",
    "query_expressions": "k2",
    "answer_expressions": "-1/4",
    "fact_spans": "[[[2, 39], [62, 64]], [[48, 51]], [[57, 61]], [[52, 55]], [[88, 95]], [[97, 104], [117, 124]], [[2, 39]], [[2, 55]], [[2, 55]], [[57, 67]], [[68, 104]], [[68, 104]], [[106, 115]]]",
    "query_spans": "[[[117, 126]]]",
    "process": "From the given conditions, we know A(-2\\sqrt{2},0), B(2\\sqrt{2},0). Let P(x,y), then k_{1}=\\frac{y}{x+2\\sqrt{2}}, k_{2}=\\frac{y}{x-2\\sqrt{2}}. k_{1}k_{2}=\\frac{y^{2}}{x^{2}-8}=\\frac{4(1-\\frac{x^{2}}{8})}{x^{2}-8}=-\\frac{1}{2}. Also, k_{1}=2, so k_{2}=-\\frac{1}{4}."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=4 x$ be $F$, and let a moving line passing through the point $(\\frac{1}{2}, 0)$ intersect the parabola at two distinct points $P$ and $Q$. Let $M$ be the midpoint of segment $P Q$. Then the equation of the trajectory of point $M$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;J: Point;Coordinate(J) = (1/2, 0);PointOnCurve(J, H) ;H: Line;Intersection(H, G) = {P, Q};Q: Point;P: Point;MidPoint(LineSegmentOf(P, Q)) = M;M: Point;Negation(P=Q)",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "y^2=2*x-1",
    "fact_spans": "[[[1, 15], [48, 51]], [[1, 15]], [[19, 22]], [[1, 22]], [[24, 43]], [[24, 43]], [[23, 47]], [[44, 47]], [[44, 64]], [[61, 64]], [[57, 60]], [[65, 79]], [[76, 79], [81, 85]], [51, 62]]",
    "query_spans": "[[[81, 92]]]",
    "process": "Let the equation of line PQ be $x=ty+\\frac{1}{2}$, let $P(x_{1},y_{1})$, $Q(x_{2},y_{2})$, then solve simultaneously the line equation and the parabola equation, apply Vieta's formulas to obtain $y_{1}+y_{2}$, then derive $x_{1}+x_{2}$, and further obtain the coordinates of point M using the midpoint formula; eliminate parameter $t$ to get the trajectory equation. Let the equation of line PQ be $x=ty+\\frac{1}{2}$, let $P(x_{1},y_{1})$, $Q(x_{2},y_{2})$, $M(x,y)$. Solve the system:  \n$$\n\\begin{cases}\ny^2=4x \\\\\nx=ty+\\frac{1}{2}\n\\end{cases}\n$$\nobtain $y^{2}-4ty-2=0$, then $\\Delta=16t^{2}+8>0$, so $y_{1}+y_{2}=4t$, $x_{1}+x_{2}=4t^{2}+1$. By the midpoint formula: $M(2t^{2}+\\frac{1}{2}, 2t)$. Then from  \n$$\n\\begin{cases}\nx=2t^{2}+\\frac{1}{2} \\\\\ny=2t\n\\end{cases}\n$$\neliminate $t$ to obtain the trajectory equation of point M: $y^{2}=2x-1$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{a^{2}-4}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. If there exists a point $P$ on the ellipse such that $P F_{1} \\perp P F_{2}$, and the area of $\\Delta P F_{1} F_{2}$ is $2$, then what is the length of the major axis of the ellipse?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/(a^2 - 4) + x^2/a^2 = 1);a: Number;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G) ;IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));Area(TriangleOf(P, F1, F2)) = 2",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "2*sqrt(6)",
    "fact_spans": "[[[2, 49], [76, 78], [144, 146]], [[2, 49]], [[4, 49]], [[58, 65]], [[66, 73]], [[2, 73]], [[2, 73]], [[81, 85]], [[76, 85]], [[87, 110]], [[112, 141]]]",
    "query_spans": "[[[144, 152]]]",
    "process": "According to the definition of an ellipse, |PF_{1}| + |PF_{2}| = 2a. Since PF_{1} \\perp PF_{2}, triangle PF_{1}F_{2} is a right triangle, so (|PF_{1}|)^{2} + (|PF_{2}|)^{2} = (2c)^{2}. Also, since the area of triangle PF_{1}F_{2} is 2, \\frac{1}{2} \\cdot |PF_{1}| \\cdot |PF_{2}| = 2, then |PF_{1}| \\cdot |PF_{2}| = 4. (2a)^{2} = (|PF_{1}| + |PF_{2}|)^{2} = (|PF_{1}|)^{2} + (|PF_{2}|)^{2} + 2|PF_{1}| \\cdot |PF_{2}| = 4c^{2} + 8, we obtain a^{2} - c^{2} = 2 = b^{2}. From \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{a^{2}-4} = 1, we get b^{2} = a^{2} - 4, so a^{2} - 4 = 2, a = \\sqrt{6}, thus 2a = 2\\sqrt{6}."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$, and point $P$ lies on $C$ such that $|P F_{1}|=2|P F_{2}|$, then $\\cos \\angle F_{1} P F_{2}$=?",
    "fact_expressions": "C: Hyperbola;P: Point;F1: Point;F2: Point;Expression(C) = (x^2 - y^2/3 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Cos(AngleOf(F1, P, F2))",
    "answer_expressions": "1/4",
    "fact_spans": "[[[18, 51], [63, 66]], [[58, 62]], [[2, 9]], [[10, 17]], [[18, 51]], [[2, 57]], [[2, 57]], [[58, 67]], [[68, 90]]]",
    "query_spans": "[[[92, 121]]]",
    "process": "Given $ a=1, b=\\sqrt{3}, c=2 $, and $ \\begin{cases} |PF_1|+|PF_2|=2 \\\\ |PF_1|=2|PF_2| \\end{cases} $, solving yields $ \\begin{cases} |PF_1|=4 \\\\ |PF_2|=2 \\end{cases} $. Then in $ \\triangle PF_1F_2 $, $ |F_1F_2|=2c=4 $, so $ \\cos\\angle F_1PF_2 = \\frac{16+4-16}{2\\times4\\times2} = \\frac{1}{4} $."
  },
  {
    "text": "Given the parabola $E$: $y^{2}=2 p x$ ($p>0$), a line $l$ passing through its focus $F$ intersects the parabola $E$ at points $A$ and $B$ (point $A$ is in the first quadrant). If $S_{\\triangle O A B}=-\\frac{3}{2} \\tan \\angle A O B$, then the value of $p$ is?",
    "fact_expressions": "l: Line;E: Parabola;p: Number;A: Point;O: Origin;B: Point;p>0;Expression(E)=(y^2=2*p*x);Focus(E)=F;PointOnCurve(F,l);Intersection(l,E)={A,B};Quadrant(A)=1;Area(TriangleOf(O,A,B))=(-3/2)*Tan(AngleOf(A,O,B));F:Point",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[37, 42]], [[2, 28], [30, 31], [43, 49]], [[126, 129]], [[50, 53], [60, 64]], [[72, 124]], [[54, 57]], [[10, 28]], [[2, 28]], [[30, 36]], [[29, 42]], [[37, 59]], [[60, 69]], [[72, 124]], [33, 35]]",
    "query_spans": "[[[126, 133]]]",
    "process": "S_{AOAB}=-\\frac{3}{2}\\tan\\angleAOB=\\frac{1}{2}|\\overrightarrow{OA}|\\overrightarrow{|OB|}\\sin\\angleAOB\\therefore\\overrightarrow{OA}\\cdot\\overrightarrow{C}\\overrightarrow{2B}=-3  \nLet A(x_{1},y_{1}), B(x_{2},y_{2}),  \ni.e., x_{1}x_{2}+y_{1}y_{2}=-3, \\frac{(y_{1}y_{2})^{2}}{4p^{2}}+y_{1}y_{2}=-3.  \nLet the line equation be x=my+\\frac{p}{2}, hence  \n\\begin{cases}x=my+\\frac{p}{2}\\\\y^{2}=2px\\end{cases}  \nobtaining y^{2}-2pmy-p^{2}=0, y_{1}y_{2}=-p^{2}.  \nSubstituting into calculation yields: p=2"
  },
  {
    "text": "Given that $A$ and $B$ are the intersection points of a line passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) with the parabola, $O$ is the origin, satisfying $\\overrightarrow{A F}=2 \\overrightarrow{F B}$, $S_{\\Delta O A B }=\\sqrt{3}|A B|$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;A: Point;F: Point;B: Point;O: Origin;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G)=F;PointOnCurve(F,H);Intersection(H, G) = {A, B};VectorOf(A, F) = 2*VectorOf(F, B);Area(TriangleOf(O,A,B))= sqrt(3)*Abs(LineSegmentOf(A, B))",
    "query_expressions": "p",
    "answer_expressions": "3*sqrt(6)",
    "fact_spans": "[[[13, 34], [43, 46]], [[145, 148]], [[40, 42]], [[2, 5]], [[36, 39]], [[8, 11]], [[50, 53]], [[16, 34]], [[13, 34]], [[13, 39]], [[12, 42]], [[2, 49]], [[61, 106]], [[109, 142]]]",
    "query_spans": "[[[145, 152]]]",
    "process": ""
  },
  {
    "text": "If the asymptotes of a hyperbola are given by $y=\\pm \\frac{4}{3} x$, and the foci lie on the circle $x^{2}+y^{2}=100$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Circle;Expression(Asymptote(G)) = (y = pm*((4/3)*x));Expression(H) = (x^2 + y^2 = 100);PointOnCurve(Focus(G), H) = True",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/36-y^2/64=1,y^2/64-x^2/36=1}",
    "fact_spans": "[[[1, 4], [58, 61]], [[37, 55]], [[1, 31]], [[37, 55]], [[1, 56]]]",
    "query_spans": "[[[58, 65]]]",
    "process": "If the foci of the hyperbola lie on the x-axis, then $\\frac{b}{a}=\\frac{4}{3}$, and the coordinates of the foci are $(10,0)$. From $a^{2}+b^{2}=100$, we obtain $a^{2}=36$, $b^{2}=64$, so the equation of the hyperbola is $\\frac{x^{2}}{36}-\\frac{y^{2}}{64}=1$; if the foci of the hyperbola lie on the y-axis, then $\\frac{a}{b}=\\frac{4}{3}$, and the coordinates of the foci are $(0,10)$. From $a^{2}+b^{2}=100$, we obtain $a^{2}=64$, $b^{2}=36$, so the equation of the hyperbola is $\\frac{y^{2}}{64}-x^{2}-2x+1=0$."
  },
  {
    "text": "If the center of the circle $(x-a)^{2}+(y-b)^{2}=1$ coincides with the right focus of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, then the coordinates of the center of the circle are?",
    "fact_expressions": "G: Ellipse;H: Circle;b: Number;a: Number;Expression(G) = (x^2/9 + y^2/5 = 1);Expression(H) = ((-a + x)^2 + (-b + y)^2 = 1);Center(H)=RightFocus(G)",
    "query_expressions": "Coordinate(Center(H))",
    "answer_expressions": "(2,0)",
    "fact_spans": "[[[29, 66]], [[1, 25], [75, 76]], [[2, 25]], [[2, 25]], [[29, 66]], [[1, 25]], [[1, 72]]]",
    "query_spans": "[[[75, 83]]]",
    "process": "From the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, we obtain $a^{2}=9$, $b^{2}=5$, then $c^{2}=a^{2}-b^{2}=4$, so $c=2$, thus the right focus of the ellipse is $(2,0)$, and therefore the center of the circle is $(2,0)$."
  },
  {
    "text": "The line $y = x - 2$ intersects the parabola $y^2 = 2x$ at points $A$ and $B$. Then $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}$ = ?",
    "fact_expressions": "G: Parabola;H: Line;O: Origin;A: Point;B: Point;Expression(G) = (y^2 = 2*x);Expression(H) = (y = x - 2);Intersection(H, G) = {A, B}",
    "query_expressions": "DotProduct(VectorOf(O, A), VectorOf(O, B))",
    "answer_expressions": "0",
    "fact_spans": "[[[10, 24]], [[0, 9]], [[38, 87]], [[27, 30]], [[31, 34]], [[10, 24]], [[0, 9]], [[0, 36]]]",
    "query_spans": "[[[38, 89]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = (x_{1},y_{1}) \\cdot (x_{2},y_{2}) = x_{1}x_{2} + y_{1}y_{2} $. From $ \\begin{cases} y = x - \\\\ y^{2} = 2x \\end{cases} 2 $, we obtain $ y^{2} - 2y - 4 = 0 $ or $ x^{2} - 6x + 4 = 0 $, so $ x_{1}x_{2} = 4 $, $ y_{1}y_{2} = -4 $, thus $ \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = x_{1}x_{2} + y_{1}y_{2} = -4 + 4 = 0 $."
  },
  {
    "text": "If an ellipse with center at the origin and axes of symmetry along the coordinate axes passes through the point $(4,0)$ and has eccentricity $\\frac{\\sqrt{3}}{2}$, then what is the standard equation of the ellipse?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;SymmetryAxis(G) = axis;H: Point;Coordinate(H) = (4, 0);PointOnCurve(H, G);Eccentricity(G) = sqrt(3)/2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/16+y^2/64=1, x^2/16+y^2/4=1}",
    "fact_spans": "[[[17, 19], [56, 58]], [[4, 8]], [[1, 19]], [[9, 19]], [[21, 29]], [[21, 29]], [[17, 29]], [[17, 54]]]",
    "query_spans": "[[[56, 65]]]",
    "process": "Since the ellipse is centered at the origin and its axes of symmetry are the coordinate axes, and the ellipse passes through the point (4,0) with eccentricity $\\frac{\\sqrt{3}}{2}$, when the foci are on the x-axis, let the equation be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, then\n$$\n\\begin{matrix}\na=4 \\\\\ne=\\frac{c}{a}=\\frac{\\sqrt{3}}{2}, \\text{ solving gives } b=2\n\\end{matrix}\n$$\n$a=4$, $a^{2}=b^{2}+c^{2}$, so the ellipse equation is $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$. When the foci are on the y-axis, $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$, then\n$$\n\\begin{matrix}\nb=4 & \\\\\ne=\\frac{c}{a}=\\frac{\\sqrt{3}}{2}, \\text{ solving gives } \n\\begin{cases}\na=8 \\\\\nb=4 \\\\\nc=4\n\\end{cases} \\\\\nc=4\\sqrt{3}\n\\end{matrix}\n$$\nso the ellipse equation is $\\frac{x^{2}}{16}+\\frac{y^{2}}{64}=1$; $a^{2}=b^{2}+c^{2}$, so the ellipse equation is $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$ or $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$."
  },
  {
    "text": "What is the distance from the point $(2,2 \\sqrt{2})$ on the parabola $y^{2}=4 x$ to the focus of this parabola?",
    "fact_expressions": "G: Parabola;H: Point;Expression(G) = (y^2 = 4*x);Coordinate(H) = (2, 2*sqrt(2));PointOnCurve(H, G)",
    "query_expressions": "Distance(H, Focus(G))",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 14], [35, 38]], [[17, 33]], [[0, 14]], [[17, 33]], [[0, 33]]]",
    "query_spans": "[[[17, 45]]]",
    "process": "The distance from the point (2,2\\sqrt{2}) to the focus of this parabola is 2+\\frac{p}{2}=2+1=3"
  },
  {
    "text": "Given a point $P(6, m)$ on the parabola $C$: $y^{2}=2px$ ($p>0$) such that the distance from $P$ to the focus $F$ is 7, find the equation of the line containing the chord $AB$ of the parabola $C$ with midpoint $M(2, 1)$.",
    "fact_expressions": "C: Parabola;p: Number;A: Point;B: Point;P: Point;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(P) = (6, m);Coordinate(M) = (2, 1);PointOnCurve(P, C);Focus(C) = F;Distance(P, F) = 7;IsChordOf(LineSegmentOf(A,B),C);MidPoint(LineSegmentOf(A,B))=M;m:Number;M:Point",
    "query_expressions": "Expression(OverlappingLine(LineSegmentOf(A,B)))",
    "answer_expressions": "48*x-y-95=0",
    "fact_spans": "[[[2, 27], [55, 61], [40, 41]], [[10, 27]], [[79, 84]], [[79, 84]], [[30, 39]], [[43, 46]], [[10, 27]], [[2, 27]], [[30, 39]], [[63, 74]], [[2, 39]], [[40, 46]], [[30, 53]], [[55, 84]], [[62, 84]], [[30, 39]], [[63, 74]]]",
    "query_spans": "[[[78, 93]]]",
    "process": ""
  },
  {
    "text": "The standard equation of an ellipse passing through the point $(2,-3)$ and having the same foci as the ellipse $9x^{2}+4y^{2}=36$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (9*x^2 + 4*y^2 = 36);H: Point;Coordinate(H) = (2, -3);S: Ellipse;PointOnCurve(H, S);Focus(G) = Focus(S)",
    "query_expressions": "Expression(S)",
    "answer_expressions": "y^2/15 + x^2/10 = 1",
    "fact_spans": "[[[12, 34]], [[12, 34]], [[1, 10]], [[1, 10]], [[41, 43]], [[0, 43]], [[11, 43]]]",
    "query_spans": "[[[41, 50]]]",
    "process": "From $9x^{2}+4y^{2}=36$ we get: $\\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$, so $c^{2}=5$. Let the desired ellipse be $\\frac{x^{2}}{b^{2}}+\\frac{y^{2}}{a^{2}}=1$ $(a>b>0)$. Substituting the point $(2,-3)$ gives $\\frac{4}{b^{2}}+\\frac{9}{a^{2}}=1$. Solving simultaneously with $c^{2}=5=a^{2}-b^{2}$ yields $a^{2}=15$, $b^{2}=10$. Therefore, the required equation is $\\frac{y^{2}}{15}+\\frac{x^{2}}{10}=1$."
  },
  {
    "text": "Given that the center of the hyperbola is at the origin, one focus is $F_{1}(-\\sqrt{5}, 0)$, point $P$ lies on the hyperbola, and the midpoint of segment $P F_{1}$ has coordinates $(0 , 2)$, then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;O: Origin;Center(G) = O;F1: Point;Coordinate(F1) = (-sqrt(5), 0);OneOf(Focus(G)) = F1;P: Point;PointOnCurve(P, G);Coordinate(MidPoint(LineSegmentOf(P, F1))) = (0, 2)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/4 = 1",
    "fact_spans": "[[[2, 5], [43, 46], [78, 81]], [[8, 10]], [[2, 10]], [[16, 37]], [[16, 37]], [[2, 37]], [[38, 42]], [[38, 47]], [[49, 75]]]",
    "query_spans": "[[[78, 86]]]",
    "process": "Let the hyperbola equation be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1. From the problem, we have P(\\sqrt{5},4). Substituting P(\\sqrt{5},4) into the standard hyperbola equation, \\therefore\\frac{5}{a^{2}}-\\frac{16}{b^{2}}=1. \\because c^{2}=a^{2}+b^{2}=5, solving gives a^{2}=1, b^{2}=4. \\because the equation of this hyperbola is x^{2}-\\frac{y^{2}}{4}=1"
  },
  {
    "text": "The standard equation of an ellipse centered at the origin, with coordinate axes as symmetry axes, eccentricity $\\frac{1}{2}$, and major axis length $8$ is?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;SymmetryAxis(G) = axis;Eccentricity(G) = 1/2;MajorAxis(G) = 8",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/16+y^2/12=1, y^2/16+x^2/12=1}",
    "fact_spans": "[[[39, 41]], [[3, 5]], [[0, 41]], [[6, 41]], [[14, 41]], [[32, 41]]]",
    "query_spans": "[[[39, 48]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a moving point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and $EF$ is any diameter of the circle $N$: $(x-1)^{2}+y^{2}=1$, then the maximum value of $\\overrightarrow{P E} \\cdot \\overrightarrow{P F}$ is?",
    "fact_expressions": "G: Ellipse;N: Circle;E: Point;F: Point;P: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Expression(N) = (y^2 + (x - 1)^2 = 1);PointOnCurve(P, G);IsDiameter(LineSegmentOf(E, F), N)",
    "query_expressions": "Max(DotProduct(VectorOf(P, E), VectorOf(P, F)))",
    "answer_expressions": "8",
    "fact_spans": "[[[6, 43]], [[54, 78]], [[48, 53]], [[48, 53]], [[2, 5]], [[6, 43]], [[54, 78]], [[2, 47]], [[48, 85]]]",
    "query_spans": "[[[88, 143]]]",
    "process": "Because: $\\overrightarrow{PE}\\cdot\\overrightarrow{PF}=(\\overrightarrow{NE}-\\overrightarrow{NP})\\cdot(\\overrightarrow{NF}-\\overrightarrow{NP})=\\overrightarrow{NE}\\cdot\\overrightarrow{NF}-\\overrightarrow{NP}\\cdot(\\overrightarrow{NE}+\\overrightarrow{NF})+\\overrightarrow{NP}^{2}=-|NE|\\cdot|NF|\\cdot\\cos\\pi-0+|NP|^{2}=-1+|NP|^{2}$. Also, since $N$ is the right focus of the ellipse, $\\therefore|NP|\\in[a-c,a+c]=[1,3]$, $\\therefore\\overrightarrow{PE}\\cdot\\overrightarrow{PF}\\in[0,8]$"
  },
  {
    "text": "Given point $E(2,-2)$ and parabola $C$: $x^{2}=8 y$, a line passing through the focus of $C$ with slope $k$ intersects $C$ at points $P$ and $Q$. If $\\angle P E Q=90^{\\circ}$, then $k=?$",
    "fact_expressions": "C: Parabola;G: Line;E: Point;P: Point;Q: Point;k: Number;Expression(C) = (x^2 = 8*y);Coordinate(E) = (2, -2);PointOnCurve(Focus(C), G);Slope(G) = k;Intersection(G, C) = {P, Q};AngleOf(P, E, Q) = ApplyUnit(90, degree)",
    "query_expressions": "k",
    "answer_expressions": "1/2",
    "fact_spans": "[[[13, 32], [34, 37], [51, 54]], [[48, 50]], [[2, 12]], [[56, 59]], [[60, 63]], [[44, 47], [94, 97]], [[13, 32]], [[2, 12]], [[33, 50]], [[41, 50]], [[48, 65]], [[67, 92]]]",
    "query_spans": "[[[94, 99]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the hyperbola $\\frac{x^{2}}{25}-\\frac{y^{2}}{9}=1$, and let $P$ be a point on the hyperbola such that $\\angle F_{1} P F_{2}=120^{\\circ}$. Then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/25 - y^2/9 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(120, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "3*sqrt(3)",
    "fact_spans": "[[[17, 56], [65, 68]], [[17, 56]], [[1, 8]], [[9, 16]], [[1, 60]], [[61, 64]], [[61, 72]], [[74, 108]]]",
    "query_spans": "[[[110, 137]]]",
    "process": "From the given conditions, we obtain the hyperbola $\\frac{x^2}{25}-\\frac{y^{2}}{9}=1$, $a=5$, $b=3$, $c=\\sqrt{34}$, so $F_{1}(-\\sqrt{34},0)$, $F_{2}(\\sqrt{34},0)$, and $|F_{1}F_{2}|^{2}=34\\times4$, $|PF_{1}|-|PF_{2}|=10$. By the cosine law:  \n$|F_{1}F_{2}|^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}|\\cdot|PF_{2}|\\cos120^{\\circ}=(|PF_{1}|-|PF_{2}|)^{2}+3|PF_{1}|\\cdot|PF_{2}|=100+3|PF_{1}|\\cdot|PF_{2}|=136$,  \n$\\therefore |PF_{1}|\\cdot|PF_{2}|=12$,  \n$\\therefore$ the area of $\\triangle F_{1}PF_{2}$ is $S=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|\\sin120^{\\circ}=\\frac{1}{2}\\times12\\times\\frac{\\sqrt{3}}{2}=3\\sqrt{3}$."
  },
  {
    "text": "Let a point $P$ on the parabola $y^{2}=4x$ be at a distance of $5$ from the line $x=-2$. Then, the distance from point $P$ to the focus of the parabola is?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (x = -2);PointOnCurve(P, G);Distance(P, H) = 5",
    "query_expressions": "Distance(P, Focus(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 15], [45, 48]], [[22, 30]], [[18, 21], [39, 43]], [[1, 15]], [[22, 30]], [[1, 21]], [[18, 37]]]",
    "query_spans": "[[[39, 55]]]",
    "process": "By the definition of a parabola: the distance from point P to the focus of the parabola is equal to the distance from point P to the directrix x=1, \\overrightarrow{m}, ----------- the distance to the parabola's focus is 5-1=4."
  },
  {
    "text": "Given the parabola equation $y^{2}=4 x$, then its directrix equation is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=-1",
    "fact_spans": "[[[2, 5], [20, 21]], [[2, 18]]]",
    "query_spans": "[[[20, 27]]]",
    "process": ""
  },
  {
    "text": "Given that a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with eccentricity $\\sqrt{5}$ coincides with the focus of the parabola $y^{2}=20x$, then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 20*x);Eccentricity(G)=sqrt(5);OneOf(Focus(G))=Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5 - y^2/20 = 1",
    "fact_spans": "[[[17, 73], [102, 105]], [[20, 73]], [[20, 73]], [[79, 94]], [[20, 73]], [[20, 73]], [[17, 73]], [[79, 94]], [[2, 73]], [[17, 99]]]",
    "query_spans": "[[[102, 112]]]",
    "process": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ has eccentricity $\\sqrt{5}$, and one of its foci coincides with the focus of the parabola $y^{2}=20x$, leading to $\\begin{cases}c=5\\\\\\frac{c}{a}=\\sqrt{5}\\end{cases}$, yielding $a=\\sqrt{5}$; then $b^{2}=c^{2}-a^{2}=20$. The required equation of the hyperbola is: $\\frac{x^{2}}{5}-\\frac{y^{2}}{20}=1$"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$, a line $l$ passes through the left focus $F$ of the ellipse $C$ and intersects the ellipse at points $A$ and $B$. The perpendicular bisector of $AB$ intersects the $x$-axis at point $M$. Then the range of $\\frac{|F M|}{|A B|^{2}}$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/2 + y^2 = 1);l: Line;F: Point;LeftFocus(C) = F;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, C) = {A, B};M: Point;Intersection(PerpendicularBisector(LineSegmentOf(A, B)), xAxis) = M",
    "query_expressions": "Range(Abs(LineSegmentOf(F, M))/Abs(LineSegmentOf(A, B))^2)",
    "answer_expressions": "[1/8,1/4)",
    "fact_spans": "[[[2, 34], [41, 46], [55, 57]], [[2, 34]], [[35, 40]], [[50, 53]], [[41, 53]], [[35, 53]], [[58, 61]], [[62, 65]], [[35, 67]], [[83, 87]], [[68, 87]]]",
    "query_spans": "[[[89, 121]]]",
    "process": "From the ellipse equation: $\\frac{x^{2}}{2}+y^{2}=1$, we obtain the left focus $F(-1,0)$. (i) When the slope of line $l$ is $0$, then line $l$ is the $x$-axis, and the perpendicular bisector of $AB$ is the $y$-axis. At this time, $M$ coincides with the origin $O$, $|AB|=2a=2\\sqrt{2}$, $|FM|=c=1$, so $\\frac{|FM|}{|AB|^{2}}=\\frac{1}{8}$. (ii) When the slope of line $l$ does not exist, the perpendicular bisector of $AB$ is the $x$-axis, which is discarded. (iii) When the slope of the line is not $0$, let the equation of line $l$ be $x=my-1$. Since the slope of line $l$ exists, $m\\neq0$. Let the coordinates of $A,B$ be $(x_{1},y_{1})$, $(x_{2},y_{2})$. Solving the system of equations of the line and the ellipse:\n$$\n\\begin{cases}\nx=my-1 \\\\\n\\frac{x^{2}}{2}+y^{2}=1\n\\end{cases}\n$$\n$\\Delta=(-2m)^{2}+4(2+m^{2})=8(m^{2}+1)>0$, $\\therefore y_{1}+y_{2}=\\frac{2m}{2+m^{2}}$, $y_{1}y_{2}=-\\frac{1}{2+m^{2}}$, then the chord length $|AB|=\\sqrt{1+m^{2}}|y_{1}-y_{2}|=\\frac{\\sqrt{1+m^{2}} \\cdot \\sqrt{8(m^{2}+1)}}{2+m^{2}}=\\frac{2\\sqrt{2}(1+m^{2})}{2+m^{2}}$, $x_{1}+x_{2}=m(y_{1}+y_{2})-2=\\frac{2m^{2}}{2+m^{2}}-2=-\\frac{4}{2+m^{2}}$. So the midpoint coordinates of $AB$ are $\\left(-\\frac{2}{2+m^{2}},\\frac{m}{2+m^{2}}\\right)$, thus the equation of the perpendicular bisector of line $AB$ is: $y-\\frac{m}{2+m^{2}}=-m\\left(x+\\frac{2}{2+m^{2}}\\right)$. Let $y=0$, we get $x=-\\frac{1}{2}$, then $M\\left(-\\frac{1}{2},0\\right)$. So $|FM|=\\left|-1+\\frac{1}{2}\\right|=\\frac{1}{2}$, actually $|FM|=\\left|-1+\\frac{1+m^{2}}{2+m^{2}}\\right|=\\frac{1+m^{2}}{2+m^{2}}$, so $\\frac{|FM|}{|AB|^{2}}=\\frac{1}{8}\\times\\frac{2+m^{2}}{1+m^{2}}=\\frac{1}{8}\\left(1+\\frac{1}{1+m^{2}}\\right)\\in\\left(\\frac{1}{8},\\frac{1}{4}\\right)$. In summary, the range of $\\frac{|FM|}{|AB|^{2}}$ is $\\left(\\frac{1}{8},\\frac{1}{4}\\right)$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, let $F_{1}$ and $F_{2}$ be its left and right foci respectively, and $P$ be a point on the hyperbola. If $|P F_{1}|=7$, then the value of $|P F_{2}|$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 - y^2/16 = 1);PointOnCurve(P, G);LeftFocus(G) =F1;RightFocus(G)=F2;Abs(LineSegmentOf(P, F1)) = 7",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "13",
    "fact_spans": "[[[2, 41], [72, 75], [60, 61]], [[68, 71]], [[42, 49]], [[50, 57]], [[2, 41]], [[68, 78]], [[42, 67]], [[42, 67]], [[80, 93]]]",
    "query_spans": "[[[95, 110]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $4 x^{2} + y^{2} = 1$ and the line $y = x + m$, what is the equation of the line when the chord intercepted by the ellipse is the longest?",
    "fact_expressions": "G: Ellipse;Expression(G) = (4*x^2 + y^2 = 1);H: Line;Expression(H) = (y = m + x);m: Number;WhenMax(Length(InterceptChord(H, G)))",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y=x",
    "fact_spans": "[[[2, 22], [38, 40]], [[2, 22]], [[23, 32], [35, 37], [48, 50]], [[23, 32]], [[25, 32]], [[34, 47]]]",
    "query_spans": "[[[48, 54]]]",
    "process": ""
  },
  {
    "text": "Let point $A(3 , 2)$ and let $S$ be the sum of the distances from the focus $F$ of the parabola $y^{2}=2 x$ and a moving point $M$ on the parabola to point $A$, that is, $S = |MA| + |MF|$. When $S$ takes its minimum value, what are the coordinates of point $M$?",
    "fact_expressions": "G: Parabola;A: Point;M: Point;F: Point;S:Number;Expression(G) = (y^2 = 2*x);Coordinate(A) = (3, 2);Focus(G)=F;PointOnCurve(M,G);Distance(A,F)+Distance(A,M)=Abs(LineSegmentOf(M,A))+Abs(LineSegmentOf(M,F));Abs(LineSegmentOf(M,A))+Abs(LineSegmentOf(M,F))=S;WhenMin(S)",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(2,2)",
    "fact_spans": "[[[14, 28], [35, 38]], [[1, 12]], [[42, 45], [77, 81]], [[31, 34]], [[62, 65], [67, 70]], [[14, 28]], [[1, 12]], [[14, 34]], [[35, 45]], [[1, 61]], [[50, 65]], [[66, 75]]]",
    "query_spans": "[[[77, 86]]]",
    "process": ""
  },
  {
    "text": "If at a point $P$ on the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{16}=1$, the lines connecting $P$ to the two foci $F_{1}$ and $F_{2}$ of the ellipse are perpendicular to each other, then the area of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/36 + y^2/16 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "16",
    "fact_spans": "[[[1, 40], [47, 49]], [[1, 40]], [[43, 46]], [[1, 46]], [[54, 61]], [[62, 69]], [[47, 69]], [[43, 76]]]",
    "query_spans": "[[[78, 105]]]",
    "process": "c^{2}=36-16=20,a^{2}=36\\begin{cases}|PF_{1}|+|PF_{2}|=2a=12\\\\|PF_{1}|^{2}+|PF_{2}|^{2}=4c^{2}=80\\end{cases}|PF_{1}||PF_{2}|=\\frac{(|PF_{1}|+|PF_{2}|)^{2}-(|PF_{2}|^{2}+|PF_{2}|^{2})}{2}=32S_{PF_{1}F_{2}}=\\frac{1}{2}|PF_{1}||PF_{2}|=16."
  },
  {
    "text": "It is known that the foci of the ellipse lie on the $y$-axis, the sum of the distances from any point on it to the two foci is $8$, and the focal distance is $2 \\sqrt{15}$. Then the standard equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;PointOnCurve(Focus(G),yAxis);D: Point;PointOnCurve(D, G) ;F1: Point;F2: Point;Focus(G) = {F1, F2};Distance(D, F1) + Distance(D, F2) = 8;FocalLength(G) = 2*sqrt(15)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/16 + x^2 = 1",
    "fact_spans": "[[[2, 4], [14, 15], [52, 54]], [[2, 13]], [], [[14, 20]], [], [], [[14, 24]], [[14, 32]], [[14, 49]]]",
    "query_spans": "[[[52, 61]]]",
    "process": "According to the definition of an ellipse, we have $ a=4 $, $ c=\\sqrt{15} $, and then by $ b^{2}=a^{2}-c^{2} $, the solution can be obtained. From the given, $ 2a=8 $, $ 2c=2\\sqrt{15} $, so $ a=4 $, $ c=\\sqrt{15} $, thus $ b^{2}=a^{2}-c^{2}=16-15=1 $. Since the foci of the ellipse lie on the y-axis, the standard equation of the ellipse is $ \\frac{y^{2}}{16}+x^{2}=1 $."
  },
  {
    "text": "The equation of the hyperbola with focus $(3, 0)$ and having the same asymptotes as the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/2 - y^2 = 1);Coordinate(Focus(Z)) = (3, 0);Z: Hyperbola;Asymptote(G) = Asymptote(Z)",
    "query_expressions": "Expression(Z)",
    "answer_expressions": "x^2/6 - y^2/3 = 1",
    "fact_spans": "[[[15, 43]], [[15, 43]], [[0, 54]], [[51, 54]], [[14, 54]]]",
    "query_spans": "[[[51, 58]]]",
    "process": ""
  },
  {
    "text": "Point $P$ is a moving point on the parabola $y^{2}=4x$. Then, the minimum value of the sum of the distance from point $P$ to the point $(0,-1)$ and the distance from point $P$ to the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;H: Point;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(H) = (0, -1);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P,H)+Distance(P,Directrix(G)))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[5, 19], [44, 47]], [[30, 39]], [[0, 4], [25, 29]], [[5, 19]], [[30, 39]], [[0, 23]]]",
    "query_spans": "[[[25, 60]]]",
    "process": "The focus of the parabola $ y^{2} = 4x $ is $ F(1,0) $. The distance from point $ P $ to the directrix is equal to the distance from point $ P $ to point $ F $. This problem is to find the minimum value of the sum of the distance from point $ P $ to point $ (0,-1) $ and the distance from point $ P $ to point $ F(1,0) $. From the graph, it can be seen that the minimum value is the distance between point $ (0,-1) $ and point $ F(1,0) $, and the minimum value is $ \\sqrt{2} $."
  },
  {
    "text": "If the standard equation of a hyperbola is $x^{2}-\\frac{y^{2}}{4}=1$, then what is the equation of the directrix of this hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/4 = 1)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x = pm*sqrt(5)/5",
    "fact_spans": "[[[1, 4], [39, 42]], [[1, 35]]]",
    "query_spans": "[[[39, 49]]]",
    "process": ""
  },
  {
    "text": "If $a>1$, then the range of eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1$ is?",
    "fact_expressions": "G: Hyperbola;a: Number;Expression(G) = (-y^2/3 + x^2/a^2 = 1);a > 1",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1, 2)",
    "fact_spans": "[[[8, 50]], [[1, 6]], [[8, 50]], [[1, 6]]]",
    "query_spans": "[[[8, 61]]]",
    "process": "By the given condition, for the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1$, the eccentricity of the hyperbola is $e=\\frac{c}{a}=\\frac{\\sqrt{a^{2}+3}}{a}=\\sqrt{1+\\frac{3}{a^{2}}}$. Since $a>1$, it follows that $e=\\sqrt{1+\\frac{3}{a^{2}}}\\in(1,2)$, that is, the range of the eccentricity of the hyperbola is $(1,2)$."
  },
  {
    "text": "A line $ l $ with slope $ 2 $ intersects the parabola $ y^{2} = 2 p x $ ($ p > 0 $) at points $ A $ and $ B $. Point $ P(1,1) $ is the midpoint of chord $ AB $. Then the equation of the directrix of the parabola is?",
    "fact_expressions": "l: Line;G: Parabola;p: Number;A: Point;B: Point;P: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(P) = (1, 1);Slope(l) = 2;Intersection(l, G) = {A, B};IsChordOf(LineSegmentOf(A, B), G);MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x = -1",
    "fact_spans": "[[[7, 12]], [[13, 34], [67, 70]], [[16, 34]], [[36, 39]], [[40, 43]], [[47, 55]], [[16, 34]], [[13, 34]], [[47, 55]], [[0, 12]], [[7, 45]], [[13, 62]], [[46, 65]]]",
    "query_spans": "[[[67, 77]]]",
    "process": "Let the equation of line $ l $ be: $ y - 1 = 2(x - 1) $, that is, $ y = 2x - 1 $. Together with $ y^{2} = 2px $, we obtain: $ 4x^{2} - (4 + 2p)x + 1 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, $ \\triangle = (4 + 2p)^{2} - 16 > 0 $, so $ x_{1} + x_{2} = \\frac{4 + 2p}{4} = 2 \\Rightarrow p = 2 $, satisfying $ \\triangle > 0 $. The equation of the directrix of the parabola is: $ x = -\\frac{p}{2} = -1 $."
  },
  {
    "text": "Let the parabola $E$: $y^{2}=4x$ have focus $F$, and let the line $l$: $y=k(x-1)$ intersect $E$ at points $A$, $B$, and intersect the $y$-axis at point $C$. If $|AF|=|BC|$, then $|AB|=$?",
    "fact_expressions": "l: Line;E: Parabola;A: Point;F: Point;B: Point;C: Point;Expression(E) = (y^2 = 4*x);Focus(E) = F;Expression(l)=(y = k*(x - 1));Intersection(l, E) = {A, B};Intersection(l,yAxis)=C;Abs(LineSegmentOf(A, F)) = Abs(LineSegmentOf(B, C));k:Number",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "sqrt(5)+2",
    "fact_spans": "[[[28, 45]], [[1, 20], [46, 49]], [[51, 54]], [[24, 27]], [[55, 58]], [[66, 69]], [[1, 20]], [[1, 27]], [[28, 45]], [[28, 58]], [[28, 69]], [[71, 84]], [[34, 45]]]",
    "query_spans": "[[[86, 95]]]",
    "process": "From the given conditions, line $ l $ must pass through point $ F $, and $ B $ lies between $ A $ and $ C $, $ |AB| = |FC| = \\sqrt{1 + k^{2}} $. By solving the parabola and line equations simultaneously, rearranging, and applying Vieta's formulas, we obtain $ x_{1} + x_{2} = 2 + \\frac{2}{k^{2}} $. Furthermore, from the definition of the parabola, $ |AB| = x_{1} + x_{2} + p $, allowing us to set up an equation to solve for $ k^{2} $, and then find $ |AB| $. Solution: From the given conditions, $ F(1, 0) $, and line $ l: y = k(x - 1) $ passes through $ F $. Since $ |AF| = |BC| $, $ B $ lies between $ A $ and $ C $, and $ |AB| = |FC| $, $ C(0, -k) $, so $ |FC| = \\sqrt{1 + k^{2}} $. Solving the parabola and line equations simultaneously,\n$$\n\\begin{cases}\ny^2 = 4x \\\\\ny = k(x - 1)\n\\end{cases}\n$$\nRearranging yields $ k^{2}x^{2} - 2(k^{2} + 2)x + k^{2} = 0 $ with discriminant $ \\Delta > 0 $. If $ A(x_{1}, y_{1}) $, $ C(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = \\frac{2(k^{2} + 2)}{k^{2}} = 2 + \\frac{2}{k^{2}} $, and $ |AB| = x_{1} + x_{2} + p = x_{1} + x_{2} + 2 $. Therefore, $ 2 + \\frac{2}{k^{2}} + 2 = \\sqrt{1 + k^{2}} $, i.e., $ 4 + \\frac{2}{k^{2}} = \\sqrt{1 + k^{2}} $. Solving gives $ k^{2} = 8 + 4\\sqrt{5} $, thus $ |AB| = |FC| = \\sqrt{1 + k^{2}} = \\sqrt{9 + 4\\sqrt{5}} = \\sqrt{5} + 2 $."
  },
  {
    "text": "Let the line $ l $: $ y = \\frac{1}{2}x + 1 $ intersect the two asymptotes of the hyperbola $ C $: $ \\frac{x^2}{a^2} - y^2 = 1 $ $ (a > 0) $ at points $ P $ and $ Q $, respectively. If the midpoint of segment $ PQ $ lies on the line $ x = 2 $, then the eccentricity of the hyperbola $ C $ is?",
    "fact_expressions": "l: Line;Expression(l) = (y = x/2 + 1);C: Hyperbola;Expression(C) = (-y^2 + x^2/a^2 = 1);a: Number;a>0;L1: Line;L2: Line;Asymptote(C) = {L1, L2};P: Point;Q: Point;Intersection(l, L1) = P;Intersection(l, L2) = Q;G: Line;Expression(G) = (x = 2);PointOnCurve(MidPoint(LineSegmentOf(P, Q)), G)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[1, 27]], [[1, 27]], [[28, 72], [114, 120]], [[28, 72]], [[36, 72]], [[36, 72]], [], [], [[28, 78]], [[82, 85]], [[86, 89]], [[1, 91]], [[1, 91]], [[104, 111]], [[104, 111]], [[93, 112]]]",
    "query_spans": "[[[114, 126]]]",
    "process": "The hyperbola $ C: \\frac{x^{2}}{a^{2}} - y^{2} = 1 $ ($ a > 0 $) has two asymptotes $ y = \\pm \\frac{1}{a}x $,  \n$ \\begin{cases} y = \\frac{1}{2}x + 1 \\\\ y = \\frac{1}{1}x \\\\ y = \\frac{x}{2}x + 1 \\\\ y = -\\frac{1}{1}x \\end{cases} $  \nSolving gives $ x_{P} = \\frac{2a}{2 - a} $, so $ \\frac{x_{P} + x_{Q}}{2} = \\frac{ \\frac{2a}{2 - a} + \\frac{-2a}{2 + a} }{2} = 2 $. Solving gives $ a^{2} = 2 $, and since $ b^{2} = 1 $, we have $ c^{2} = a^{2} + b^{2} = 3^{n} $, so $ e = \\frac{c}{a} = \\sqrt{ \\frac{c^{2}}{a^{2}} } = \\sqrt{ \\frac{3}{2} } = \\frac{ \\sqrt{6} }{2} $."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $E$: $x^{2}+\\frac{y^{2}}{b^{2}}=1$ $(0<b<1)$, respectively. A line passing through point $F_{1}$ intersects the ellipse $E$ at points $A$ and $B$. If $|A F_{1}|=3|B F_{1}|$ and $A F_{2} \\perp x$-axis, then the equation of the ellipse $E$ is?",
    "fact_expressions": "E: Ellipse;b: Number;G: Line;A: Point;F1: Point;B: Point;F2: Point;0 < b;b < 1;Expression(E) = (x^2 + y^2/b^2 = 1);LeftFocus(E) = F1;RightFocus(E) = F2;PointOnCurve(F1, G);Intersection(G, E) = {A, B};Abs(LineSegmentOf(A, F1)) = 3*Abs(LineSegmentOf(B, F1));IsPerpendicular(LineSegmentOf(A,F2),xAxis)",
    "query_expressions": "Expression(E)",
    "answer_expressions": "x^2+(3/2)*y^2=1",
    "fact_spans": "[[[19, 62], [82, 87], [143, 148]], [[26, 62]], [[79, 81]], [[88, 91]], [[1, 8], [70, 78]], [[92, 95]], [[9, 16]], [[26, 62]], [[26, 62]], [[19, 62]], [[1, 68]], [[1, 68]], [[69, 81]], [[79, 97]], [[99, 121]], [[123, 141]]]",
    "query_spans": "[[[143, 153]]]",
    "process": "As shown in the figure, since $AF_{2}\\bot x$-axis, $\\therefore AF_{2} = \\frac{b^{2}}{a} = b^{2}$, let $A(c, b^{2})$, and since $|AF_{1}| = 3|BF|$, then the coordinates of point $B$ are $(-\\frac{5}{3}c, -\\frac{1}{3}b^{2})$. Substituting into the ellipse equation: $(-\\frac{5}{3}c)^{2} + \\frac{(-\\frac{1}{3}b^{2})}{b^{2}} = 1$, solving yields $c^{2} = \\frac{1}{3}$, $b^{2} = \\frac{2}{3}$, so the equation of the ellipse is $x^{2} + \\frac{3}{2}y^{2} = 1$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the hyperbola such that $P F_{2} \\perp x$-axis. Find the distance from $F_{2}$ to the line $PF_{1}$.",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F2), xAxis)",
    "query_expressions": "Distance(F2, LineOf(P, F1))",
    "answer_expressions": "4/7",
    "fact_spans": "[[[2, 30], [61, 64]], [[2, 30]], [[39, 46]], [[47, 54], [87, 94]], [[2, 54]], [[2, 54]], [[56, 60]], [[56, 65]], [[67, 85]]]",
    "query_spans": "[[[87, 110]]]",
    "process": ""
  },
  {
    "text": "Given two distinct points $A$ and $B$ on the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{4}=1$ satisfying $OA \\perp OB$, where $O$ is the coordinate origin, then the minimum value of $|OA| \\cdot |OB|$ is?",
    "fact_expressions": "G: Hyperbola;O: Origin;A: Point;B: Point;Expression(G) = (x^2/3 - y^2/4 = 1);PointOnCurve(A,G);PointOnCurve(B,G);Negation(A=B);IsPerpendicular(LineSegmentOf(O,A),LineSegmentOf(O,B))",
    "query_expressions": "Min(Abs(LineSegmentOf(O, A))*Abs(LineSegmentOf(O, B)))",
    "answer_expressions": "24",
    "fact_spans": "[[[2, 40]], [[73, 76]], [[46, 49]], [[50, 53]], [[2, 40]], [[2, 53]], [[2, 53]], [[41, 53]], [[55, 70]]]",
    "query_spans": "[[[83, 107]]]",
    "process": "From the given conditions, the slopes of lines OA and OB both exist and are non-zero and not equal to \\underline{2\\sqrt{3}}. Without loss of generality, let the equation of line OA be $ y = kx $ ($ k \\neq 0 $), $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Since $ OA \\perp OB $, the equation of line OB is $ y = -\\frac{1}{k}x $. From\n$$\n\\begin{cases}\n\\frac{x^{2}}{3} - \\frac{y^{2}}{4} = 1, \\\\\ny = kx\n\\end{cases}\n$$\nwe obtain $ x^{2} = \\frac{12}{4 - 3k^{2}} $. Therefore, $ |OA| \\cdot |OB| \\geqslant 24 $, with equality if and only if $ |OA| = |OB| = 2\\sqrt{6} $. Hence, the minimum value of $ |OA| \\cdot |OB| $ is 24."
  },
  {
    "text": "It is known that the center of the ellipse is at the origin of coordinates, the foci lie on the $x$-axis, and the sum of the distances from point $P(3 \\sqrt{2}, 4)$ on the ellipse to the two foci is $12$. Then, what is the standard equation of the ellipse?",
    "fact_expressions": "G: Ellipse;P: Point;O:Origin;F1:Point;F2:Point;Coordinate(P) = (3*sqrt(2), 4);Center(G)=O;PointOnCurve(Focus(G), xAxis);PointOnCurve(P, G);Focus(G)={F1,F2};Distance(P,F1)+Distance(P,F2)=12",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/36+y^2/32=1",
    "fact_spans": "[[[2, 4], [22, 24], [61, 63]], [[25, 45]], [[8, 12]], [], [], [[25, 45]], [[2, 12]], [[2, 21]], [[22, 45]], [[22, 49]], [[22, 59]]]",
    "query_spans": "[[[61, 70]]]",
    "process": ""
  },
  {
    "text": "Let $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$ be two distinct points on the curve $y=\\sqrt{x}$, and $F(\\frac{1}{4}, 0)$. If $|A F|=2|B F|$ and $x_{1}=p x_{2}+q$, then $\\frac{p}{q}=?$",
    "fact_expressions": "A: Point;B: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Negation(A=B);PointOnCurve(A, G);PointOnCurve(B, G);G: Curve;Expression(G) = (y = sqrt(x));F: Point;Coordinate(F) = (1/4, 0);Abs(LineSegmentOf(A, F)) = 2*Abs(LineSegmentOf(B, F));p: Number;q: Number;x1 = p*x2 + q",
    "query_expressions": "p/q",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 18]], [[21, 38]], [[1, 18]], [[21, 38]], [[1, 18]], [[21, 38]], [[1, 18]], [[21, 38]], [[1, 61]], [[1, 61]], [[1, 61]], [[41, 55]], [[41, 55]], [[62, 81]], [[62, 81]], [[83, 97]], [[99, 116]], [[99, 116]], [[99, 116]]]",
    "query_spans": "[[[118, 133]]]",
    "process": "The curve $ y = \\sqrt{x} $, simplified to $ y^2 = x $, $ |AF| = 2|BF| $. According to the definition of a parabola, we obtain $ x_{1} + \\frac{1}{4} = 2(x_{2} + \\frac{1}{4}) \\Rightarrow x_{1} = 2x_{2} + \\frac{1}{4} $. Since $ x_{1} = px_{2} + q $, it follows that $ p = 2 $, $ q = \\frac{1}{4} $, $ \\frac{p}{q} = 8 $."
  },
  {
    "text": "The directrix $l$ of the parabola $y^{2}=8x$ is tangent to the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-y^{2}=1$, then the eccentricity $e$ of $C$ is $?$.",
    "fact_expressions": "C: Hyperbola;a: Number;G: Parabola;l: Line;e: Number;Expression(C) = (-y^2 + x^2/a^2 = 1);Expression(G) = (y^2 = 8*x);Directrix(G) = l;IsTangent(l, C);Eccentricity(C) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[21, 58], [62, 65]], [[29, 58]], [[0, 14]], [[17, 20]], [[69, 72]], [[21, 58]], [[0, 14]], [[0, 20]], [[17, 60]], [[62, 72]]]",
    "query_spans": "[[[69, 74]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}(-1 , 0)$, $F_{2}(1 , 0)$ are the two foci of an ellipse, a line $l$ passing through $F_{1}$ intersects the ellipse at points $M$ and $N$. If the perimeter of $\\triangle M F_{2} N$ is $8$, then what is the standard equation of the ellipse?",
    "fact_expressions": "l: Line;G: Ellipse;F1: Point;M: Point;F2: Point;N: Point;Coordinate(F1) = (-1, 0);Coordinate(F2)=(1,0);Focus(G)={F1,F2};PointOnCurve(F1, l);Intersection(l, G) = {M, N};Perimeter(TriangleOf(M, F2, N)) = 8",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4+y^2/3=1",
    "fact_spans": "[[[52, 57]], [[35, 37], [58, 60], [102, 104]], [[2, 17], [44, 51]], [[61, 64]], [[19, 34]], [[65, 68]], [[2, 17]], [[19, 34]], [[2, 42]], [[43, 57]], [[52, 70]], [[72, 100]]]",
    "query_spans": "[[[102, 111]]]",
    "process": "From the given condition, the perimeter of $\\triangle MF_{2}N$ is $4a$, thus $a$ can be determined, and further $b$ can be found, so the equation of the ellipse can be obtained. According to the condition, $4a=8$, $\\therefore a=2$, since $F_{1}(-1,0)$, $F_{2}(1,0)$ are the foci of the ellipse, $\\therefore b^{2}=2^{2}-1=3$, $\\therefore$ the equation of the ellipse is: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$."
  },
  {
    "text": "Let the right focus of the hyperbola $\\frac{x^{2}}{m^{2}}-\\frac{y^{2}}{n^{2}}=1$ $(m n \\neq 0)$ coincide with the focus of the parabola $y^{2}=8 x$, and the eccentricity is $2$. Then the equation of this hyperbola is?",
    "fact_expressions": "C: Hyperbola;m: Number;n: Number;Negation(m*n=0);Expression(C) = (x^2/m^2 - y^2/n^2 = 1);E: Parabola;Expression(E) = (y^2 = 8*x);RightFocus(C) = Focus(E);Eccentricity(C) = 2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2/3 = 1",
    "fact_spans": "[[[1, 59], [94, 97]], [[4, 59]], [[4, 59]], [[4, 59]], [[1, 59]], [[64, 78]], [[64, 78]], [[1, 83]], [[1, 91]]]",
    "query_spans": "[[[94, 102]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, find the equation of the parabola that has the center of the hyperbola as its focus and the left focus of the hyperbola as its vertex.",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (x^2/4 - y^2/5 = 1);Focus(H)= Center(G);LeftFocus(G)=Vertex(H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2=12*(x+3)",
    "fact_spans": "[[[2, 40], [43, 46], [53, 56]], [[63, 66]], [[2, 40]], [[42, 66]], [[52, 66]]]",
    "query_spans": "[[[63, 70]]]",
    "process": "The hyperbola \\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1 has center at O(0,0), and the left focus of this hyperbola is F(-3,0). Then the vertex of the parabola is (3,0), and the focus is (0,0), so p=6, hence the equation of the parabola is y^{2}=12(x+3)"
  },
  {
    "text": "If the left and right foci of the hyperbola $\\frac{x^{2}}{9}-y^{2}=1$ are $F_{1}$ and $F_{2}$ respectively, and $A$ is a point on the left branch of the hyperbola such that $|AF_{1}|=5$, then $|AF_{2}|=$?",
    "fact_expressions": "G: Hyperbola;A: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 - y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(A, LeftPart(G));Abs(LineSegmentOf(A, F1)) = 5",
    "query_expressions": "Abs(LineSegmentOf(A, F2))",
    "answer_expressions": "11",
    "fact_spans": "[[[1, 29], [61, 64]], [[57, 60]], [[37, 44]], [[47, 54]], [[1, 29]], [[1, 54]], [[1, 54]], [[57, 70]], [[72, 84]]]",
    "query_spans": "[[[87, 99]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$, the focus is $F$, the directrix $l$ intersects the $x$-axis at point $A$, and $M$ is a point on the parabola $C$ such that $M F \\perp x$-axis. If the chord length intercepted on line $A M$ by the circle with diameter $A F$ is $2$, then $p=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;l: Line;Directrix(C) = l;A: Point;Intersection(l, xAxis) = A;M: Point;PointOnCurve(M, C);IsPerpendicular(LineSegmentOf(M, F), xAxis);G: Circle;IsDiameter(LineSegmentOf(A, F), G);Length(InterceptChord(LineOf(A, M), G)) = 2",
    "query_expressions": "p",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 28], [58, 64]], [[2, 28]], [[116, 119]], [[10, 28]], [[32, 35]], [[2, 35]], [[38, 41]], [[2, 41]], [[50, 53]], [[38, 53]], [[54, 57]], [[54, 67]], [[69, 83]], [[96, 97]], [[86, 97]], [[96, 114]]]",
    "query_spans": "[[[116, 121]]]",
    "process": "From the given conditions, we can obtain the following rough graph: From $ y^{2} = 2px $, we get: $ A(-\\frac{p}{2}, 0) $, $ F(\\frac{p}{2}, 0) $, $ M(\\frac{p}{2}, \\pm p) $. By the symmetry of the parabola, taking $ M(\\frac{p}{2}, p) $ or $ M(\\frac{p}{2}, -p) $ yields the same result; without loss of generality, let $ M(\\frac{p}{2}, p) $. Therefore, the equation of the circle with diameter $ AF $ is: $ x^{2} + y^{2} = \\frac{p^{2}}{4} $; the equation of line $ AM $ is: $ x - y + \\frac{p}{2} = 0 $. Let $ d $ be the distance from the center of the circle to the line, then $ d = \\frac{\\frac{p}{2}}{\\sqrt{1^{2} + 1^{2}}} = \\frac{\\sqrt{2}p}{4} $. Therefore, the chord length intercepted by line $ AM $ on the circle is: $ 2\\sqrt{\\frac{p^{2}}{4} - \\frac{p^{2}}{8}} = 2 \\Rightarrow p = 2\\sqrt{2} $. The correct answer to this problem: $ 2\\sqrt{2} $."
  },
  {
    "text": "Given that $A(0,1)$ is a fixed point on the ellipse $x^{2}+4 y^{2}=4$, and point $P$ is a moving point on the ellipse distinct from $A$, then the maximum value of $|A P|$ is?",
    "fact_expressions": "G: Ellipse;A: Point;P: Point;Expression(G) = (x^2 + 4*y^2 = 4);Coordinate(A) = (0, 1);PointOnCurve(A, G);PointOnCurve(P, G);Negation(P = A)",
    "query_expressions": "Max(Abs(LineSegmentOf(A, P)))",
    "answer_expressions": "4*sqrt(3)/3",
    "fact_spans": "[[[11, 30], [40, 42]], [[2, 10], [45, 48]], [[35, 39]], [[11, 30]], [[2, 10]], [[2, 34]], [[35, 52]], [[35, 52]]]",
    "query_spans": "[[[54, 67]]]",
    "process": "The standard equation of the ellipse is $\\frac{x^{2}}{4}+y^{2}=1$. Let point $P(x,y)$, then $-1\\leqslant y<1$. From $x^{2}+4y^{2}=4$ we get $x^{2}=4-4y^{2}$, $|AP|=\\sqrt{3}\\frac{x^{2}}{x}$. Therefore, when $y=-\\frac{1}{3}$, $|AP|$ takes the maximum value $\\frac{4\\sqrt{3}}{3}$."
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$), draw a line intersecting the parabola at points $A$ and $B$. Let $O$ be the origin, and denote the slopes of lines $OA$ and $OB$ as $k_{1}$ and $k_{2}$, respectively. Then $k_{1} \\cdot k_{2} = ?$",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);A: Point;B: Point;Intersection(H, G) = {A, B};O: Origin;k1: Number;k2: Number;Slope(LineOf(O, A)) = k1;Slope(LineOf(O, B)) = k2",
    "query_expressions": "k1*k2",
    "answer_expressions": "-4",
    "fact_spans": "[[[1, 22], [32, 35]], [[1, 22]], [[4, 22]], [[4, 22]], [[25, 28]], [[1, 28]], [[29, 31]], [[0, 31]], [[36, 39]], [[40, 43]], [[29, 45]], [[46, 49]], [[76, 83]], [[85, 92]], [[56, 92]], [[56, 92]]]",
    "query_spans": "[[[94, 115]]]",
    "process": "The focus of the parabola $ y^{2}=2px $ ($ p>0 $) is $ F(\\frac{p}{2},0) $. When the slope of a line passing through focus $ F $ does not exist, the line equation can be set as $ x=\\frac{p}{2} $. Without loss of generality, let $ A(\\frac{p}{2},p) $, $ B(\\frac{p}{2},-p) $, then $ k_{1}=\\frac{p}{\\frac{p}{2}}=2 $, $ k_{2}=\\frac{-p}{\\frac{p}{2}}=-2 $, hence $ k_{1}\\cdot k_{2}=2\\times(-2)=-4 $. When the slope of the line passing through focus $ F $ exists, the line equation can be set as $ y=k(x-\\frac{p}{2}) $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. From $ \\begin{cases} y=k(x-\\frac{p}{2}) \\\\ y^{2}=2px \\end{cases} $, rearranging yields $ 4k^{2}x^{2}-4p(k^{2}+2)x+k^{2}p^{2}=0 $. Then $ x_{1}+x_{2}=\\frac{p(k^{2}+2)}{k^{2}} $, $ x_{1}x_{2}=\\frac{p^{2}}{4} $, $ y_{1}y_{2}=k^{2}(x_{1}-\\frac{p}{2})(x_{2}-\\frac{p}{2})=k^{2}x_{1}x_{2}-\\frac{pk^{2}}{2}(x_{1}+x_{2})+\\frac{p^{2}k^{2}}{4} $. $ k_{1}\\cdot k_{2}=\\frac{y_{1}}{x_{1}}\\times\\frac{y_{2}}{x_{2}}=\\frac{y_{1}y_{2}}{x_{1}x_{2}}=k^{2}+\\frac{-\\frac{pk^{2}}{2}(x_{1}+x_{2})+\\frac{p^{2}k^{2}}{4}}{x_{1}x_{2}}=k^{2}+\\frac{-\\frac{pk^{2}}{2}\\times\\frac{p(k^{2}+2)}{k^{2}}+\\frac{p^{2}k^{2}}{4}}{\\frac{p^{2}}{4}}=-4 $. In conclusion, $ k_{1}\\cdot k_{2}=-4 $."
  },
  {
    "text": "Given $\\odot C_{1}$: $x^{2}+(y-2)^{2}=1$, $\\odot C_{2}$: $x^{2}+(y+2)^{2}=9$, a moving circle is externally tangent to both $\\odot C_{1}$ and $\\odot C_{2}$. Then the trajectory equation of the center $M$ of the moving circle is?",
    "fact_expressions": "G: Circle;M:Point;C1:Circle;C2:Circle;Expression(C1) = (x^2 + (y - 2)^2 = 1);Expression(C2) = (x^2 + (y + 2)^2 = 9);IsOutTangent(G,C1);IsOutTangent(G,C2);Center(G)=M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "(y^2-x^2/3=1)&(y>0)",
    "fact_spans": "[[[74, 76], [111, 113]], [[115, 118]], [[2, 36], [77, 90]], [[38, 73], [93, 106]], [[2, 36]], [[38, 73]], [[74, 109]], [[74, 109]], [[111, 118]]]",
    "query_spans": "[[[115, 125]]]",
    "process": "Find the centers and radii of the two circles, and set up equations based on the positional relationship between the moving circle and the given circles.  \nGiven circle $ C_{1}: x^{2} + (y - 2)^{2} = 1 $ and circle $ x^{2} + (y + 2)^{2} = 9 $, we obtain $ C_{1}(0, 2) $, $ r_{1} = 1 $, $ C_{2}(0, -2) $, $ r_{2} = 3 $.  \nLet the center of the moving circle be $ M(x, y) $. Since it is tangent to both $ C_{1} $ and $ C_{2} $, we have $ MC_{1} = r + 1 $, $ MC_{2} = 3 + r $.  \nThus, $ MC_{2} - MC_{1} = (r + 3) - (1 + r) = 2 < 4 = r_{1} + r_{2} $, so the trajectory of point $ M $ is the upper branch of a hyperbola with foci $ C_{1} $ and $ C_{2} $, where $ a = 1 $, $ c = 2 $. Therefore, the trajectory equation of point $ M $ is $ y^{2} - \\frac{x^{2}}{3} = 1 $ ($ y > 0 $)."
  },
  {
    "text": "Given that the focus of the parabola $y = a x^{2} - 1$ is at the origin, the area of the triangle with vertices at the three intersection points of the parabola and the two coordinate axes is?",
    "fact_expressions": "G: Parabola;a: Number;Expression(G) = (y = a*x^2 - 1);O: Origin;Focus(G) = O;A: Point;B: Point;C: Point;Intersection(G, axis) = {A,B,C}",
    "query_expressions": "Area(TriangleOf(A, B, C))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 18], [29, 32]], [[5, 18]], [[2, 18]], [[22, 26]], [[2, 26]], [], [], [], [[29, 42]]]",
    "query_spans": "[[[28, 54]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, $A$ is a point on the hyperbola located in the first quadrant, $\\overrightarrow{O A} \\cdot \\overrightarrow{O F}=|\\overrightarrow{O F}|^{2}$, and the equation of line $O A$ is $y=\\frac{2 \\sqrt{3}}{3} x$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;O: Origin;A: Point;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;Quadrant(A) = 1;PointOnCurve(A, G);DotProduct(VectorOf(O, A), VectorOf(O, F)) = Abs(VectorOf(O, F))^2;Expression(LineOf(O,A)) = (y = x*(2*sqrt(3)/3))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[6, 63], [72, 75], [203, 206]], [[9, 63]], [[9, 63]], [[166, 171]], [[68, 71]], [[2, 5]], [[9, 63]], [[9, 63]], [[6, 63]], [[2, 67]], [[68, 86]], [[68, 86]], [[87, 163]], [[164, 201]]]",
    "query_spans": "[[[203, 212]]]",
    "process": "Analysis: From $\\overrightarrow{OA}\\cdot\\overrightarrow{OF}=|\\overrightarrow{OF}|^{2}$, it follows that $AF\\bot x$-axis, thus obtaining $A(c,\\frac{b^{2}}{a})$. Substituting into the equation of line $OA$, $y=\\frac{2\\sqrt{3}}{3}x$, gives the result. Details: $\\because \\overrightarrow{OA}\\cdot\\overrightarrow{OF}=|\\overrightarrow{OA}|\\cdot|\\overrightarrow{OF}|\\cos<AOF=|\\overrightarrow{OF}|^{2}$, $\\therefore |\\overrightarrow{OA}|\\cos<A\\overrightarrow{OF}=|\\overrightarrow{OF}|$, $\\therefore AF\\bot x$-axis. Let $x=c$, then $y_{A}=\\frac{b^{2}}{a}$, $\\therefore A(c,\\frac{b^{2}}{a})$. Also, $\\because$ the equation of $OA$ is $y=\\frac{2\\sqrt{3}}{3}x$, $\\therefore \\frac{b^{2}}{c}=\\frac{2\\sqrt{3}}{3}$, $\\therefore \\frac{b^{2}}{ac}=\\frac{a^{2}-c^{2}}{ac}=\\frac{2\\sqrt{3}}{3}$, i.e., $e-\\frac{1}{e}=\\frac{2\\sqrt{3}}{3}$, $e^{2}-\\frac{2\\sqrt{3}}{3}e-1=0$, $e=\\sqrt{3}$."
  },
  {
    "text": "Draw two tangents from the left focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ to the circle $\\odot O$: $x^{2}+y^{2}=a^{2}$, and denote the points of tangency as $A$, $B$. Let $C$ be the left vertex of the hyperbola. If $\\angle A C B=120^{\\circ}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;F: Point;LeftFocus(G) = F;O: Circle;Expression(O) = (x^2 + y^2 = a^2);L1: Line;L2: Line;TangentOfPoint(F, O) = {L1, L2};A: Point;C: Point;B: Point;TangentPoint(L1, O) = A;TangentPoint(L2, O) = B;LeftVertex(G) = C;AngleOf(A, C, B) = ApplyUnit(120, degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 47], [104, 107], [144, 147]], [[1, 47]], [[4, 47]], [[4, 47]], [[51, 54]], [[1, 54]], [[55, 85]], [[55, 85]], [], [], [[0, 90]], [[95, 98]], [[111, 114]], [[99, 102]], [[0, 102]], [[0, 102]], [[104, 114]], [[116, 142]]]",
    "query_spans": "[[[144, 153]]]",
    "process": ""
  },
  {
    "text": "The line $y = x + 2$ intersects the ellipse $\\frac{x^{2}}{m} + \\frac{y^{2}}{3} = 1$ at two points; then the range of values for $m$ is?",
    "fact_expressions": "G: Ellipse;m: Number;H: Line;Expression(G) = (y^2/3 + x^2/m = 1);Expression(H) = (y = x + 2);NumIntersection(H, G) = 2",
    "query_expressions": "Range(m)",
    "answer_expressions": "(1,3)+(3,+oo)",
    "fact_spans": "[[[10, 47]], [[55, 58]], [[0, 9]], [[10, 47]], [[0, 9]], [[0, 53]]]",
    "query_spans": "[[[55, 65]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the midpoint of the segment cut by the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ from the line passing through the point $(3,0)$ with slope $\\frac{4}{5}$ are?",
    "fact_expressions": "J: Point;Coordinate(J) = (3, 0);PointOnCurve(J,L) = True;L: Line;Slope(L) = 4/5;G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1)",
    "query_expressions": "Coordinate(MidPoint(InterceptChord(L,G)))",
    "answer_expressions": "(3/2,-6/5)",
    "fact_spans": "[[[1, 9]], [[1, 9]], [[0, 29]], [[27, 29]], [[10, 29]], [[30, 69]], [[30, 69]]]",
    "query_spans": "[[[27, 80]]]",
    "process": "The equation of the line passing through the point (3,0) with slope \\frac{4}{5} is y-0=\\frac{4}{5}(x-3), that is, y=\\frac{4}{5}(x-3). Combining with the ellipse equation \\begin{cases}y=\\frac{4}{5}(x-3)\\\\\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1\\end{cases}, we obtain x^{2}-3x-8=0. Let the intersection points of the line and the ellipse be (x_{1},y_{1}), (x_{2},y_{2}). Then x_{1}+x_{2}=3, y_{1}+y_{2}=\\frac{4}{5}(x_{1}+x_{2}-3-3)=-. Therefore, the coordinates of the midpoint of the intercepted segment are (\\frac{3}{2},-\\frac{6}{5}). Hence, fill in: (\\frac{3}{2},-\\frac{6}{5})"
  },
  {
    "text": "Given real numbers $x$, $y$ satisfying the equation $x^{2}+2 y^{2}-2=0$, what is the maximum value of $x+y$?",
    "fact_expressions": "x: Real;y: Real;x^2 + 2*y^2 - 2 = 0",
    "query_expressions": "Max(x + y)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 8]], [[10, 13]], [[17, 36]]]",
    "query_spans": "[[[38, 49]]]",
    "process": ""
  },
  {
    "text": "A chord of the parabola $y^{2}=2 x$ is bisected by the point $A(4,2)$. What is the equation of the line containing this chord?",
    "fact_expressions": "G: Parabola;H: LineSegment;A: Point;Expression(G) = (y^2 = 2*x);Coordinate(A) = (4, 2);MidPoint(H) = A;IsChordOf(H, G)",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "x - 2*y = 0",
    "fact_spans": "[[[0, 14]], [], [[19, 27]], [[0, 14]], [[19, 27]], [[0, 29]], [[0, 18]]]",
    "query_spans": "[[[0, 44]]]",
    "process": "Let the coordinates of the two endpoints of the chord be found using the point difference method: substitute into the parabola equation, subtract, substitute the coordinates of point A to obtain the slope of the line containing the chord, and use the point-slope form to find the equation of the line. Let the two endpoints of the chord be M(x_{1},y_{1}), N(x_{2},y_{2}). Substituting into the parabola equation gives: y_{1}^{2}=2x\\textcircled{1}\\textcircled{1}\\textcircled{1}-\\textcircled{2} yields: y_{1}^{2}-y_{2}^{2}=2(x_{1}-x_{2}), i.e., \\frac{y_{1}^{2}}{x_{1}-x_{2}}= . Moreover, since MN is bisected by point A(4,2), it follows that y_{1}+y_{2}. Thus, the slope k of the chord MN is k=\\frac{1}{2}. Therefore, the equation of the line is: y-2=\\frac{1}{2}(x-4), i.e., x-2y=0."
  },
  {
    "text": "The two foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ are $F_{1}$ and $F_{2}$. If $P$ is a point on it such that $|P F_{1}|=2 |P F_{2}|$, then the range of the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;Focus(G)={F1,F2};P: Point;PointOnCurve(P,G);Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1,3]",
    "fact_spans": "[[[0, 57], [84, 85], [115, 118]], [[0, 57]], [[3, 57]], [[3, 57]], [[3, 57]], [[3, 57]], [[63, 70]], [[71, 78]], [[0, 78]], [[80, 83]], [[80, 88]], [[90, 113]]]",
    "query_spans": "[[[115, 128]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ is $\\frac{\\sqrt{5}}{2}$, then the real number $m$=?",
    "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (-y^2 + x^2/m = 1);Eccentricity(G) = sqrt(5)/2",
    "query_expressions": "m",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 30]], [[57, 62]], [[2, 30]], [[2, 55]]]",
    "query_spans": "[[[57, 64]]]",
    "process": "From the given conditions, we have $ a = \\sqrt{m} $, $ b = 1 $, $ c = \\sqrt{1 + m} $, then $ e = \\frac{c}{a} = \\frac{\\sqrt{1 + m}}{\\sqrt{m}} = \\frac{\\sqrt{5}}{2} $, solving yields $ m = 4 $."
  },
  {
    "text": "Given that the parabola $y^{2}=2 p x(p>0)$ intersects the circle $x^{2}+(y-1)^{2}=1$ at two points, and the distance between these two points is $\\frac{2 \\sqrt{3}}{3}$, then the distance from the focus to the directrix of this parabola is?",
    "fact_expressions": "G: Parabola;p: Number;H: Circle;p>0;Z: Point;X: Point;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (x^2 + (y - 1)^2 = 1);Intersection(G, H) = {Z, X};Distance(Z, X) = 2*sqrt(3)/3",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "sqrt(2)/6",
    "fact_spans": "[[[2, 23], [84, 87]], [[5, 23]], [[24, 44]], [[5, 23]], [], [], [[2, 23]], [[24, 44]], [[2, 49]], [[2, 81]]]",
    "query_spans": "[[[84, 98]]]",
    "process": "Analysis: First, determine that O(0,0) is a common point of the two curves. Use the distance between two points and the condition that the point lies on the circle to find the coordinates of the other intersection point, then substitute the coordinates into the parabola equation to solve. Clearly, O(0,0) is an intersection point of the parabola $ y^{2}=2px $ ($ p>0 $) and the circle $ x^{2}+(y-1)^{2}=1 $. Let the other intersection point be $ A(x,y) $, $ x>0 $. Since $ |OA|=\\frac{2\\sqrt{3}}{3} $, we have $ x^{2}+y^{2}=\\frac{4}{3} $. Solving the system \n\\[\n\\begin{cases}\nx^{2}+y^{2}=\\frac{4}{3} \\\\\nx^{2}+(y-1)^{2}=1\n\\end{cases}\n\\]\nyields $ A\\left(\\frac{2\\sqrt{2}}{3},\\frac{2}{3}\\right) $. Since $ A\\left(\\frac{2\\sqrt{2}}{3},\\frac{2}{3}\\right) $ lies on the parabola $ y^{2}=2px $, we have $ \\frac{4}{9}=2p\\times\\frac{2\\sqrt{2}}{3} $, solving which gives $ p=\\frac{1}{3\\sqrt{2}}=\\frac{\\sqrt{2}}{6} $. Thus, the distance from the focus to the directrix of the parabola is $ \\frac{\\sqrt{2}}{6} $."
  },
  {
    "text": "The eccentricity $e$ of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$ is $?$.",
    "fact_expressions": "G: Hyperbola;e: Number;Expression(G) = (x^2/4 - y^2/3 = 1);Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(7)/2",
    "fact_spans": "[[[0, 38]], [[42, 45]], [[0, 38]], [[0, 45]]]",
    "query_spans": "[[[42, 47]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus of which is $F$. A line passing through the origin intersects $C$ at points $A$ and $B$, and it always holds that $\\angle A F B \\geq 120^{\\circ}$. Then the range of eccentricity of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;A: Point;F: Point;B: Point;O:Origin;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;PointOnCurve(O,G);Intersection(G,C) = {A, B};AngleOf(A,F,B)>=ApplyUnit(120,degree)",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(0, 1/2]",
    "fact_spans": "[[[2, 59], [76, 79], [127, 132]], [[9, 59]], [[9, 59]], [[73, 75]], [[81, 84]], [[64, 67]], [[85, 88]], [[70, 72]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 67]], [[68, 75]], [[73, 90]], [[93, 124]]]",
    "query_spans": "[[[127, 142]]]",
    "process": "As shown in the figure, let the right focus of the ellipse be $F_{2}$. By symmetry, $AFBF_{2}$ is a parallelogram, $\\angle AF_{2}F = \\angle BFF_{2}$. Since $\\angle FB \\geqslant 120^{\\circ}$, it follows that $\\angle FAF_{2} \\leqslant 60^{\\circ}$. Let $|AF| = m$, $|AF_{2}| = n$. From the definition of the ellipse, $m + n = 2a$, then $mn \\leqslant \\frac{(m+n)^{2}}{4} = a^{2}$, with equality if and only if $m = n$. In $\\triangle AFF_{2}$, by the law of cosines, $\\cos\\angle FAF_{2} = \\frac{m^{2} + n^{2} - |FF_{2}|^{2}}{2mn} = \\frac{}{}\\frac{1 - 4c^{2}}{= \\frac{4a^{2} - 4c^{2}}{2mn} - 1 \\geqslant \\frac{4a^{2} - 4c^{2}}{2a^{2}} - 1 = 1 - 2e^{2}}$. Also, since $\\angle FAF_{2} \\leqslant 60^{\\circ}$, $\\cos\\angle FAF_{2} \\geqslant \\frac{1}{2}$, thus $1 - 2e^{2} \\geqslant \\frac{1}{2}$, solving gives $0 < e \\leqslant \\frac{1}{2}$."
  },
  {
    "text": "What is the focal distance of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "6",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 44]]]",
    "process": "According to the problem, in the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, $a=5$, $b=4$, then $c=\\sqrt{a^{2}-b^{2}}=3$, so the focal length $2c=6$;"
  },
  {
    "text": "If a hyperbola centered at the origin has a focus on the $x$-axis and one of its asymptotes given by the equation $2x - y = 0$, then what is the eccentricity of the hyperbola?",
    "fact_expressions": "G: Hyperbola;O: Origin;Center(G)=O;PointOnCurve(Focus(G), xAxis);Expression(OneOf(Asymptote(G))) = (2*x - y = 0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[16, 19], [39, 42]], [[4, 6]], [[1, 19]], [[7, 19]], [[16, 37]]]",
    "query_spans": "[[[39, 47]]]",
    "process": ""
  },
  {
    "text": "Given that a line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$, $O$ is the coordinate origin, $|AF|=2$, then the area of $\\Delta OAB$ is?",
    "fact_expressions": "G: Parabola;H: Line;O: Origin;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, F)) = 2",
    "query_expressions": "Area(TriangleOf(O, A, B))",
    "answer_expressions": "2",
    "fact_spans": "[[[3, 17], [27, 30]], [[24, 26]], [[41, 44]], [[31, 34]], [[35, 38]], [[20, 23]], [[3, 17]], [[3, 23]], [[2, 26]], [[24, 40]], [[50, 58]]]",
    "query_spans": "[[[59, 76]]]",
    "process": "Analysis: According to the given conditions and the properties of the parabola, find the length of BF; combining the geometric properties of the figure and performing calculations yields the final result. The focus of the parabola $ y^{2}=4x $ is $ F(1,0) $, $ p=2 $. From $ \\frac{1}{|AF|}+\\frac{1}{|BF|}=\\frac{2}{p} $, we have $ \\frac{1}{2}+\\frac{1}{|BF|}=\\frac{2}{2} $, thus $ |BF|=2 $. Since $ |AF|=2 $, $ |BF|=2 $, and in the parabola equation, when $ x=1 $, $ y=\\pm2 $, therefore $ AB=4 $, meaning AB is the latus rectum of the parabola, hence $ S_{\\DeltaABO}=\\frac{1}{2}\\times OF\\times AB=\\frac{1}{2}\\times 1\\times 4=2 $."
  },
  {
    "text": "It is known that the focus of the parabola $y^{2}=20 x$ coincides with the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1(a>0)$. Then, what is the equation of the asymptotes of this hyperbola?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;a>0;Expression(G) = (-y^2/9 + x^2/a^2 = 1);Expression(H) = (y^2 = 20*x);Focus(H) = RightFocus(G)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(4/3)*x",
    "fact_spans": "[[[21, 68], [77, 80]], [[24, 68]], [[2, 17]], [[24, 68]], [[21, 68]], [[2, 17]], [[2, 74]]]",
    "query_spans": "[[[77, 88]]]",
    "process": ""
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{10-k}+\\frac{y^{2}}{k-4}=1$ represents a hyperbola, what is the range of real values for $k$?",
    "fact_expressions": "G: Hyperbola;k: Real;Expression(G)=(x^2/(10 - k) + y^2/(k - 4) = 1)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-oo,4)+(10,+oo)",
    "fact_spans": "[[[46, 49]], [[51, 56]], [[2, 49]]]",
    "query_spans": "[[[51, 63]]]",
    "process": "Since the equation $\\frac{x^{2}}{10-k}+\\frac{y^{2}}{k-4}=1$ represents a hyperbola, we have $(10-k)(k-4)<0$, solving which yields $k>10$ or $k<4$. Therefore, the range of real values for $k$ is $(-\\infty,4)\\cup(10,+\\infty)$."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $C$: $y^{2}=2 p x$ ($p>0$) intersects the parabola at points $A$ and $B$, with $|A F|=2|B F|$. $O$ is the origin, and the area of $\\triangle A O B$ is $6 \\sqrt{2}$. Then the standard equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;p: Number;A: Point;O: Origin;B: Point;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;G: Line;PointOnCurve(F, G);Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, F)) = 2*Abs(LineSegmentOf(B, F));Area(TriangleOf(A, O, B)) = 6*sqrt(2)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 27], [36, 39], [114, 120]], [[9, 27]], [[42, 45]], [[69, 72]], [[46, 49]], [[29, 32]], [[9, 27]], [[1, 27]], [[1, 32]], [[33, 35]], [[0, 35]], [[33, 51]], [[52, 66]], [[79, 112]]]",
    "query_spans": "[[[114, 127]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $x^{2}=2py$ has coordinates $(0 , 1)$, then the equation of the directrix is?",
    "fact_expressions": "G: Parabola;p: Number;Expression(G) = (x^2 = 2*p*y);Coordinate(Focus(G)) = (0, 1)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y = -1",
    "fact_spans": "[[[1, 15]], [[4, 15]], [[1, 15]], [[1, 30]]]",
    "query_spans": "[[[1, 38]]]",
    "process": ""
  },
  {
    "text": "Given point $A(0,4)$, the parabola $C$: $x^{2}=2 p y$ $(0<p<4)$ has directrix $l$, point $P$ lies on $C$, draw $P H \\perp l$ at $H$, such that $|P H|=|P A|$ and $\\angle A P H=120^{\\circ}$. Then the equation of the parabola is?",
    "fact_expressions": "C: Parabola;p: Number;A: Point;P: Point;H: Point;0 < p;p < 4;Expression(C) = (x^2 = 2*(p*y));Coordinate(A) = (0, 4);Directrix(C)=l;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P,H),l);FootPoint(LineSegmentOf(P,H),l)=H;Abs(LineSegmentOf(P, H)) = Abs(LineSegmentOf(P, A));AngleOf(A, P, H) = ApplyUnit(120, degree);l:Line",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2=(16/5)*y",
    "fact_spans": "[[[12, 40], [120, 123], [53, 56]], [[19, 40]], [[2, 11]], [[48, 52]], [[73, 76]], [[19, 40]], [[19, 40]], [[12, 40]], [[2, 11]], [[12, 47]], [[48, 57]], [[59, 72]], [[59, 76]], [[78, 91]], [[92, 118]], [[44, 47]]]",
    "query_spans": "[[[120, 127]]]",
    "process": "Let the focus of the parabola be $ F(0,\\frac{p}{2}) $, $ |AF|=4-\\frac{p}{2} $. By the definition of a parabola, $ |PH|=|PF| $. Since $ |PH|=|PA| $, $ \\therefore |PA|=|PF| $. Without loss of generality, assume point $ P $ lies in the first quadrant. Draw $ PQ\\bot y $-axis at point $ Q $, then $ Q $ is the midpoint of $ AF $, $ |AQ|=|FQ|=\\frac{1}{2}|AF|=\\frac{1}{2}(4-\\frac{p}{2}) $. $ \\because \\frac{1}{2}(4-\\frac{\\frac{x}{2}}{2}) $, $ \\because \\angle APH=120^{\\circ} $, $ \\therefore \\angle APQ=120^{\\circ}-90^{\\circ}=30^{\\circ} $, $ \\therefore |PQ|=\\sqrt{3}|AQ|=\\frac{\\sqrt{3}}{2}(4-\\frac{p}{2}) $, $ |OQ|=|FQ|+|OF|=\\frac{1}{2}(4-\\frac{p}{2})+\\frac{p}{2}=2+\\frac{p}{4} $, $ \\therefore $ the coordinates of point $ P $ are $ (\\frac{\\sqrt{3}}{2}(4-\\frac{p}{2}), 2+\\frac{p}{4}) $. Since point $ P $ lies on parabola $ C $, $ \\therefore \\sqrt{3}(4-\\frac{p}{2})]=2p\\times(2+\\frac{p}{4}) $, simplifying yields $ 5p^{2}+112p-192=0 $, solving gives $ p=\\frac{8}{5} $ or $ -24 $ (discard negative), $ \\therefore $ the equation of the parabola is $ x^{2}=\\frac{16}{5}y $."
  },
  {
    "text": "The two endpoints $A$ and $B$ of a line segment $AB$ of length $a$ both slide along the parabola $y^{2}=2 p x$ $(p>0 , a>2 p)$. Then, the shortest distance from the midpoint $M$ of segment $AB$ to the $y$-axis is?",
    "fact_expressions": "A: Point;B: Point;a: Number;Length(LineSegmentOf(A, B)) = a;Endpoint(LineSegmentOf(A, B)) = {A, B};G: Parabola;Expression(G) = (y^2=2*p*x);p>0;a>2*p;PointOnCurve(A, G) = True;PointOnCurve(B, G) = True;M: Point;MidPoint(LineSegmentOf(A, B)) = M;p: Number",
    "query_expressions": "Min(Distance(M, yAxis))",
    "answer_expressions": "(1/2)*(a-p)",
    "fact_spans": "[[[19, 22]], [[23, 26]], [[3, 6]], [[0, 14]], [[7, 26]], [[28, 58]], [[28, 58]], [[31, 58]], [[31, 58]], [[19, 59]], [[19, 59]], [[73, 76]], [[63, 76]], [[31, 58]]]",
    "query_spans": "[[[73, 88]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a point on the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F_{1}$, $F_{2}$ are the left and right foci, and $\\angle F_{1} PF_{2}=90^{\\circ}$, if $S_{\\triangle P F_{1} F_{2}}=9$, then $b=$?",
    "fact_expressions": "C: Ellipse;a: Number;b: Number;F1: Point;P: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, C);LeftFocus(C) =F1;RightFocus(C)=F2;AngleOf(F1, P, F2) = ApplyUnit(90, degree);Area(TriangleOf(P,F1,F2)) = 9",
    "query_expressions": "b",
    "answer_expressions": "3",
    "fact_spans": "[[[6, 63]], [[12, 63]], [[158, 161]], [[67, 74]], [[2, 5]], [[75, 83]], [[12, 63]], [[12, 63]], [[6, 63]], [[2, 66]], [[6, 88]], [[6, 88]], [[90, 122]], [[125, 156]]]",
    "query_spans": "[[[158, 163]]]",
    "process": "Let $|F_{1}P|=m$, $|PF_{2}|=n$. By the definition of the ellipse, we have $m+n=2a$ ①. In right triangle $\\triangle F_{1}PF_{2}$, by the Pythagorean theorem, we obtain $(2c)^{2}=m^{2}+n^{2}$ ②. From ① and ②, we get $mn=2b^{2}$. Therefore, the area of $\\triangle F_{1}PF_{2}$ is $S=\\frac{1}{2}mn=\\frac{1}{2}\\times2b^{2}=b^{2}=9$, so $b=3$."
  },
  {
    "text": "Given that the center of the ellipse is at the origin, the eccentricity is $e=\\frac{\\sqrt{3}}{2}$, and one of its foci coincides with the focus of the parabola $x^{2}=-4\\sqrt{3}y$, then the equation of this ellipse is?",
    "fact_expressions": "G: Parabola;H: Ellipse;O:Origin;Expression(G) = (x^2 = -4*sqrt(3)*y);Center(H)=O;Eccentricity(H)=e;e:Number;e=sqrt(3)/2;OneOf(Focus(H)) = Focus(G)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2 + y^2/4 = 1",
    "fact_spans": "[[[46, 71]], [[2, 4], [39, 40], [79, 81]], [[8, 10]], [[46, 71]], [[2, 10]], [[2, 37]], [[14, 37]], [[14, 37]], [[39, 76]]]",
    "query_spans": "[[[79, 85]]]",
    "process": ""
  },
  {
    "text": "The directrix of the parabola $x^{2}=2 p y(p>0)$ intersects the circle $x^{2}+y^{2}+6 y-16=0$ at points $A$, $B$. If $AB=8$, then the focus of the parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 2*(p*y));p: Number;p>0;H: Circle;Expression(H) = (6*y + x^2 + y^2 - 16 = 0);A: Point;B: Point;Intersection(Directrix(G), H) = {A, B};LineSegmentOf(A, B) = 8",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, 6)",
    "fact_spans": "[[[0, 21], [67, 70]], [[0, 21]], [[3, 21]], [[3, 21]], [[25, 48]], [[25, 48]], [[49, 53]], [[54, 57]], [[0, 57]], [[59, 65]]]",
    "query_spans": "[[[67, 75]]]",
    "process": "The directrix equation of the parabola is: $ y = -\\frac{P}{2} $. The circle $ x^{2} + y^{2} + 6y \\cdot 16 = 0 $ has center $ (0, \\cdot 3) $ and radius: $ 5 $. The directrix of the parabola $ x^{2} = 2py $ ($ p > 0 $) intersects the circle $ x^{2} + y^{2} + 6y - 16 = 0 $ at points $ A $, $ B $. If $ AB = 8 $, then: $ \\left| \\frac{p}{2} - 3 \\right| = \\sqrt{25 - 16} $. Solving gives: $ p = 12 $. The parabola is $ x^{2} = 24y $. The focus coordinates of the parabola: $ (0, 6) $."
  },
  {
    "text": "Given the standard equation of ellipse $C$ is $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, and $c=\\sqrt{a^{2}-b^{2}}$, point $A$ has coordinates $(0 , b)$, point $B$ has coordinates $(0,-b)$, point $F$ has coordinates $(c, 0)$, point $T$ has coordinates $(3 c, 0)$. If the intersection point of line $AT$ and line $BF$ lies on the ellipse, then the eccentricity of the ellipse is?",
    "fact_expressions": "A: Point;T: Point;B: Point;F: Point;C: Ellipse;Coordinate(A) = (0, b);Coordinate(B) = (0, -b);Coordinate(F) = (c, 0);Coordinate(T) = (3*c, 0);Expression(C) = (y^2/b^2 + x^2/a^2 = 1);a:Number;b:Number;c:Number;a>b;b>0;c = sqrt(a^2 - b^2);PointOnCurve(Intersection(LineOf(A,T),LineOf(B,F)), C)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[90, 94]], [[143, 147]], [[108, 112]], [[125, 130]], [[2, 7], [179, 181], [184, 186]], [[90, 105]], [[108, 123]], [[125, 141]], [[143, 159]], [[2, 63]], [[65, 87]], [[65, 87]], [[65, 87]], [[13, 63]], [[13, 63]], [[65, 87]], [[161, 182]]]",
    "query_spans": "[[[184, 192]]]",
    "process": ""
  },
  {
    "text": "Given that the distance from the right focus $F$ of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$ to the asymptotes of the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is less than $\\sqrt{3}$, find the range of the eccentricity of hyperbola $E$.",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/12 = 1);F: Point;RightFocus(G) = F;E: Hyperbola;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;Distance(F, Asymptote(E))<sqrt(3)",
    "query_expressions": "Range(Eccentricity(E))",
    "answer_expressions": "(1, 2)",
    "fact_spans": "[[[2, 41]], [[2, 41]], [[45, 48]], [[2, 48]], [[49, 110], [131, 137]], [[49, 110]], [[57, 110]], [[57, 110]], [[57, 110]], [[57, 110]], [[45, 129]]]",
    "query_spans": "[[[131, 148]]]",
    "process": "The right focus of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$ is $F(2,0)$. From the condition, we obtain $\\frac{2b}{\\sqrt{a^{2}+b^{2}}}<\\sqrt{3}$, that is, $4b^{2}<3c^{2}$, so $4(c^{2}-a^{2})<3c^{2}$, thereby yielding $e^{2}<4$, and further solving gives the range of eccentricity as $(1,2)$."
  },
  {
    "text": "If the left and right foci of the hyperbola $\\frac{x^{2}}{9}-y^{2}=1$ are $F_{1}$, $F_{2}$ respectively, and $A$ is a point on the left branch of the hyperbola such that $|A F_{1}|=5$, then $|A F_{2}|$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;A: Point;PointOnCurve(A, LeftPart(G));Abs(LineSegmentOf(A, F1)) = 5",
    "query_expressions": "Abs(LineSegmentOf(A, F2))",
    "answer_expressions": "11",
    "fact_spans": "[[[1, 29], [57, 60]], [[1, 29]], [[37, 44]], [[45, 52]], [[1, 52]], [[1, 52]], [[53, 56]], [[53, 66]], [[68, 81]]]",
    "query_spans": "[[[85, 98]]]",
    "process": "Because the equation of the hyperbola is $\\frac{x^{2}}{9}-y^{2}=1$, $\\therefore a=3$, $b=1$. By the definition of the hyperbola, and since point A is on the left branch of the hyperbola, it follows: $|AF_{2}|-|AF_{1}|=2a=6$, $\\therefore |AF_{2}|=6+|AF_{1}|=1$"
  },
  {
    "text": "Given the two foci of a hyperbola $F_{1}(-\\sqrt{5}, 0)$, $F_{2}(\\sqrt{5}, 0)$, and a point $P$ on the hyperbola such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, $|P F_{1}| \\cdot |P F_{2}|=2$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;P: Point;Coordinate(F1) = (-sqrt(5), 0);Coordinate(F2) = (sqrt(5), 0);Focus(G) = {F1,F2};PointOnCurve(P, G);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4-y^2=1",
    "fact_spans": "[[[2, 5], [61, 64], [159, 162]], [[10, 31]], [[34, 55]], [[57, 60]], [[10, 31]], [[34, 55]], [[2, 54]], [[56, 67]], [[68, 128]], [[129, 157]]]",
    "query_spans": "[[[159, 169]]]",
    "process": "This problem first assumes the coordinates of point P. Combining $\\overrightarrow{PF}_{1}\\cdot\\overrightarrow{PF_{2}}=0$, $|PF_{1}|\\cdot|PF_{2}|=2$, calculate the coordinates of point P. Using the properties of the hyperbola, compute the values of $a$ and $b$ respectively to obtain the answer. [Detailed solution] Let point $P(x,y)$, $\\overrightarrow{PF_{1}}=(-\\sqrt{5}-x,-y)$, $\\overrightarrow{PF_{2}}=(\\sqrt{5}-x,-y)$. Combining $\\overrightarrow{PF}_{1}\\cdot\\overrightarrow{PF_{2}}=0$, substitute the point coordinates to obtain $x^{2}+y^{2}=5$. Using the vector magnitude formula, then $\\overrightarrow{PF}\\cdot\\overrightarrow{PF}=100-20x^{2}=4$, solving gives $P\\left(\\frac{2\\sqrt{30}}{5},\\frac{\\sqrt{5}}{5}\\right)$. Compute $|PF_{1}|-|PF_{2}|=2a=4$, so $a=2$. Given $a^{2}+b^{2}=c^{2}$, $c=\\sqrt{5}$, thus $b=1$. Therefore, the hyperbola equation is $\\frac{x^{2}}{4}-y^{2}=1$."
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $y=2 x$, what is the eccentricity of the hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G)))=(y=2*x)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 48], [64, 67]], [[5, 48]], [[5, 48]], [[2, 48]], [[2, 62]]]",
    "query_spans": "[[[64, 73]]]",
    "process": "From the equation of the asymptotes of the hyperbola, we know that $\\frac{b}{a}=2$, so $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{a^{2}+(2a)^{2}}=\\sqrt{5}a$, therefore the eccentricity of this hyperbola is $e=\\frac{c}{a}=\\sqrt{5}$."
  },
  {
    "text": "Given the ellipse $x^{2}+\\frac{y^{2}}{2}=1$ with foci $F_{1}$ and $F_{2}$, point $M$ lies on the ellipse and satisfies $\\overrightarrow{M F_{1}} \\cdot \\overrightarrow{M F_{2}}=0$. Then the distance from point $M$ to the $x$-axis is?",
    "fact_expressions": "G: Ellipse;M: Point;F1: Point;F2: Point;Expression(G) = (x^2 + y^2/2 = 1);Focus(G) = {F1, F2};PointOnCurve(M, G);DotProduct(VectorOf(M, F1), VectorOf(M, F2)) = 0",
    "query_expressions": "Distance(M, xAxis)",
    "answer_expressions": "0",
    "fact_spans": "[[[2, 29], [54, 56]], [[49, 53], [119, 123]], [[33, 40]], [[41, 48]], [[2, 29]], [[2, 48]], [[49, 57]], [[58, 117]]]",
    "query_spans": "[[[119, 133]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the right focus of the hyperbola $M$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, point $P$ lies on $M$, $O$ is the coordinate origin, and $|O P|=2 b$, $\\angle P O F=\\frac{\\pi}{6}$, then the eccentricity of $M$ is?",
    "fact_expressions": "F: Point;M: Hyperbola;Expression(M) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;RightFocus(M) = F;P: Point;PointOnCurve(P, M);O: Origin;Abs(LineSegmentOf(O, P)) = 2*b;AngleOf(P, O, F) = pi/6",
    "query_expressions": "Eccentricity(M)",
    "answer_expressions": "sqrt(15)/3",
    "fact_spans": "[[[2, 5]], [[6, 67], [77, 80], [136, 139]], [[6, 67]], [[13, 67]], [[13, 67]], [[13, 67]], [[13, 67]], [[2, 71]], [[72, 76]], [[72, 81]], [[82, 85]], [[92, 104]], [[106, 134]]]",
    "query_spans": "[[[136, 145]]]",
    "process": "Without loss of generality, assume point $ P $ lies in the first quadrant. Given $ |OP| = 2b $ and $ \\angle POF = \\frac{\\pi}{6} $, we obtain $ P(\\sqrt{3}b, b) $. Substituting into the hyperbola equation yields $ \\frac{3b^{2}}{a^{2}} - \\frac{b^{2}}{b^{2}} = 1 $, then $ a^{2} = \\frac{3}{2}b^{2} $. Therefore, $ c^{2} = a^{2} + b^{2} = \\frac{5}{2}b^{2} $, so the eccentricity is $ e = \\sqrt{\\frac{c^{2}}{a^{2}}} = \\sqrt{\\frac{5}{3}} = \\frac{\\sqrt{15}}{3} $."
  },
  {
    "text": "It is known that the line $x - 2y + 2 = 0$ passes through a vertex and a focus of the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1$ $(a > b > 0)$. Then, what is the equation of this ellipse? What is its eccentricity?",
    "fact_expressions": "H: Line;Expression(H) = (x - 2*y + 2 = 0);G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;PointOnCurve(OneOf(Vertex(G)), H) = True;PointOnCurve(OneOf(Focus(G)), H) = True",
    "query_expressions": "Expression(G);Eccentricity(G)",
    "answer_expressions": "x^2/5+y^2=1\n2*sqrt(5)/5",
    "fact_spans": "[[[2, 15]], [[2, 15]], [[17, 69], [84, 86]], [[17, 69]], [[19, 69]], [[19, 69]], [[19, 69]], [[19, 69]], [[2, 79]], [[2, 79]]]",
    "query_spans": "[[[84, 91]], [[84, 96]]]",
    "process": ""
  },
  {
    "text": "Let the lower and upper vertices of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$ be $B_{1}$ and $B_{2}$, respectively. If point $P$ is a point on the ellipse such that the slopes of lines $PB_{1}$ and $PB_{2}$ are $\\frac{1}{4}$ and $-1$, respectively, then the eccentricity of the ellipse is?",
    "fact_expressions": "P: Point;B1: Point;G: Ellipse;a: Number;b: Number;B2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);UpperVertex(G) = B2;LowerVertex(G) = B1;PointOnCurve(P, G);Slope(LineOf(P, B1)) = 1/4;Slope(LineOf(P, B2)) = -1",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[81, 85]], [[64, 71]], [[1, 55], [86, 88], [139, 141]], [[3, 55]], [[3, 55]], [[72, 79]], [[3, 55]], [[3, 55]], [[1, 55]], [[1, 79]], [[1, 79]], [[81, 92]], [[94, 137]], [[94, 137]]]",
    "query_spans": "[[[139, 147]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ is a moving point on the parabola $y^{2}=4x$, the projection of point $P$ on the $y$-axis is $M$, and point $A(4,6)$, then the minimum value of $|PA|+|PM|$ is?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;M: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (4, 6);PointOnCurve(P, G);Projection(P, yAxis) = M",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "3*sqrt(5) - 1",
    "fact_spans": "[[[7, 21]], [[43, 52]], [[2, 6], [26, 30]], [[39, 42]], [[7, 21]], [[43, 52]], [[2, 25]], [[26, 42]]]",
    "query_spans": "[[[54, 73]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=8 x$, with focus $F$ and directrix $l$, let $P$ be a point on the parabola, $P A \\perp l$, where $A$ is the foot of the perpendicular. If the slope of the line $A F$ is $-\\sqrt{3}$, then $|P F|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;l: Line;Directrix(G) = l;P: Point;PointOnCurve(P,G) = True;IsPerpendicular(LineSegmentOf(P,A),l) = True;A: Point;FootPoint(LineSegmentOf(P,A),l) = A;Slope(LineOf(A,F)) = -sqrt(3)",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 16], [36, 39]], [[2, 16]], [[21, 24]], [[2, 24]], [[27, 30]], [[2, 30]], [[32, 35]], [[32, 42]], [[43, 56]], [[58, 61]], [[43, 64]], [[66, 88]]]",
    "query_spans": "[[[92, 101]]]",
    "process": "From the parametric equations of the parabola $\\begin{cases}x=8t^{2}\\\\y=8t\\end{cases}$, we obtain its standard equation $y^{2}=8x$, $\\therefore$ directrix: $x=-2$. The focus of the parabola is $F(2,0)$, the equation of line $AF$ is $y=-\\sqrt{3}(x-2)$, so point $A(-2,4\\sqrt{3})$, $P(6,4\\sqrt{3})$, thus $|PF|=6+2=8$, hence the answer is B."
  },
  {
    "text": "Given that one focus of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$ is $(0,1)$, then $m$=?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/5 + y^2/m = 1);Coordinate(OneOf(Focus(G))) = (0, 1)",
    "query_expressions": "m",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 39]], [[54, 57]], [[2, 39]], [[2, 52]]]",
    "query_spans": "[[[54, 59]]]",
    "process": "\\because the ellipse \\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1 has a focus at (0,1) \\therefore the focus is on the y-axis, and c=1, b^{2}=5, \\therefore m=b^{2}+c^{2}=6, the answer is: 6"
  },
  {
    "text": "The point $M(m, 1)$ on the parabola $x^{2}=a y(a>0)$ is at a distance of $2$ from its directrix $l$. Then $a=?$",
    "fact_expressions": "G: Parabola;a: Number;M: Point;l: Line;a>0;m:Number;Expression(G) = (x^2 = a*y);Coordinate(M) = (m, 1);PointOnCurve(M, G);Directrix(G) = l;Distance(M, l) = 2",
    "query_expressions": "a",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 19], [32, 33]], [[47, 50]], [[21, 31]], [[35, 38]], [[3, 19]], [[22, 31]], [[0, 19]], [[21, 31]], [[0, 31]], [[32, 38]], [[21, 45]]]",
    "query_spans": "[[[47, 52]]]",
    "process": "The equation of the directrix $ l $ of the parabola $ x^{2} = ay $ ($ a > 0 $) is $ y = -\\frac{a}{4} $. Since the distance from point $ M(m, 1) $ to the directrix $ l $ is 2, we obtain $ 1 + \\frac{a}{4} = 2 $. Solving this gives $ a = 4 $. Therefore, $ a = 4 $. Hence, the answer is: 4"
  },
  {
    "text": "Given the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with right focus $F$, a perpendicular is drawn from point $F$ to one asymptote of the hyperbola, with foot at $P$, intersecting the other asymptote at $Q$. If $5 \\overrightarrow{P F}=3 \\overrightarrow{F Q}$, then the eccentricity of hyperbola $E$ is?",
    "fact_expressions": "E: Hyperbola;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F: Point;RightFocus(E) = F;l1: Line;l2: Line;OneOf(Asymptote(E)) = l1;OneOf(Asymptote(E)) = l2;Negation(l1 = l2);H: Line;PointOnCurve(F, H);IsPerpendicular(l1,H);FootPoint(l1,H) = P;P: Point;Intersection(H,l2) = Q;Q: Point;5*VectorOf(P, F) = 3*VectorOf(F, Q)",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 63], [78, 81], [161, 167]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 71], [73, 77]], [[2, 71]], [], [], [[78, 87]], [[78, 105]], [[78, 105]], [], [[72, 90]], [[72, 90]], [[72, 97]], [[94, 97]], [[72, 109]], [[106, 109]], [[111, 158]]]",
    "query_spans": "[[[161, 173]]]",
    "process": "From the hyperbola equation, the asymptotes are given by $ y = \\pm\\frac{b}{a}x $. Let $ P $ lie on $ y = \\frac{b}{a}x $, $ Q $ lie on $ y = -\\frac{b}{a}x $, and $ F(c,0) $. Therefore, $ |PF| = \\frac{|bc|}{\\sqrt{a^{2}+b^{2}}} = b $. Also, $ 5\\overrightarrow{PF} = 3\\overrightarrow{FQ} $, so $ |FQ| = \\frac{5b}{3} $. Since $ \\tan\\angle POF = \\frac{b}{a} $, it follows that $ \\tan\\angle POQ = \\frac{2\\tan\\angle POF}{1-\\tan^{2}\\angle POF} = \\frac{\\frac{2b}{a}}{1-\\frac{b^{2}}{a^{2}}} $. Moreover, $ OP \\perp PF $, so $ \\tan\\angle POQ = \\frac{|PQ|}{|OP|} = \\frac{8b}{a} $. Hence, $ \\frac{8b}{a} = \\frac{2b}{1-\\frac{b^{2}}{a^{2}}} $, solving gives $ \\frac{b^{2}}{a^{2}} = \\frac{1}{4} $, therefore $ e = \\sqrt{\\frac{c^{2}}{a^{2}}} = \\sqrt{1+\\frac{b^{2}}{a^{2}}} = \\frac{\\sqrt{5}}{2} $."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2 p x(p>0)$ be $F$, and let its directrix intersect the $x$-axis at point $C$. A line passing through point $F$ intersects the parabola at points $A$ and $B$. If $\\overrightarrow{B C} \\cdot \\overrightarrow{B F}=0$, then $\\overrightarrow{B A} \\cdot \\overrightarrow{C F}$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;C: Point;Intersection(Directrix(G), xAxis) = C;H: Line;PointOnCurve(F, H);A: Point;B: Point;Intersection(H, G) = {A, B};DotProduct(VectorOf(B, C), VectorOf(B, F)) = 0",
    "query_expressions": "DotProduct(VectorOf(B, A), VectorOf(C, F))",
    "answer_expressions": "2*p^2",
    "fact_spans": "[[[1, 22], [54, 57], [30, 31]], [[1, 22]], [[4, 22]], [[4, 22]], [[26, 29], [46, 50]], [[1, 29]], [[40, 44]], [[30, 44]], [[51, 53]], [[45, 53]], [[59, 62]], [[63, 66]], [[51, 68]], [[70, 121]]]",
    "query_spans": "[[[123, 174]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. From $ \\overrightarrow{BC} \\cdot \\overrightarrow{BF} = 0 $, we obtain that point $ B(x_{2},y_{2}) $ is the intersection point of the circle with $ CF $ as diameter and the parabola. Solving the system of equations, we get $ x_{2} = \\frac{\\sqrt{5}-2}{2}p $. Let the equation of line $ AB $ be $ x = my + \\frac{p}{2} $. Solving the system of equations, we find $ x_{1} = \\frac{\\sqrt{5}+2}{2}p $. Then using the dot product formula of vectors, we can solve it. As shown in the figure, let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. From $ \\overrightarrow{BC} \\cdot \\overrightarrow{BF} = 0 $, we have $ x_{1} > x_{2} > 0 $, and point $ B(x_{2},y_{2}) $ is the intersection point of the circle with $ CF $ as diameter and the parabola $ y^{2} = 2px $. Solving the system of equations\n$$\n\\begin{cases}\nx_{2}^{2} + y_{2}^{2} = \\frac{p^{2}}{4} \\\\\ny_{2}^{2} = 2px_{2}\n\\end{cases}\n$$\nwe obtain $ x_{2} = \\frac{\\sqrt{5}-2}{2}p $. Let the equation of line $ AB $ be $ x = my + \\frac{p}{2} $. Solving the system of equations\n$$\n\\begin{cases}\nx = my + \\frac{p}{2} \\\\\ny^{2} = 2px\n\\end{cases}\n$$\nand simplifying, we get $ x^{2} - (1+2m^{2})px + \\frac{p^{2}}{4} = 0 $, from which $ x_{1}x_{2} = \\frac{p^{2}}{4} $, so $ x_{1} = \\frac{\\sqrt{5}+2}{2}p $. Then $ \\overrightarrow{BA} \\cdot \\overrightarrow{CF} = (x_{1}-x_{2}, y_{1}-y_{2}) \\cdot (p, 0) = p(x_{1}-x_{2}) = 2p^{2} $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the hyperbola and satisfies $|P F_{1}|+|P F_{2}|=2 \\sqrt{5}$. Find the area of $\\Delta P F_{1} F_{2}$.",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/3 - y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 2*sqrt(5)",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 30], [60, 63]], [[55, 59]], [[39, 46]], [[47, 54]], [[2, 30]], [[2, 54]], [[2, 54]], [[55, 64]], [[68, 100]]]",
    "query_spans": "[[[103, 130]]]",
    "process": "Assume point $ P $ lies on the right branch of the hyperbola. By the definition of a hyperbola, we have $ |PF_{1}| - |PF_{2}| = 2\\sqrt{3} $. Also, $ |PF| + |PF_{2}| = 2\\sqrt{5} $. Solving these two equations simultaneously gives $ |PF| = \\sqrt{5} + \\sqrt{3} $, $ |PF_{2}| = \\sqrt{5} - \\sqrt{3} $. Moreover, $ |F_{1}F_{2}| = 4 $, so $ |PF_{1}|^{2} + |PF_{2}|^{2} = |F_{1}F_{2}|^{2} $, which implies that $ \\triangle PF_{1}F_{2} $ is a right triangle. Therefore, $ S_{\\triangle PF_{1}F_{2}} = \\frac{1}{2}|PF_{1}||PF_{2}| = 1 $."
  },
  {
    "text": "Given that the distance from a focus $F$ of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{m-3}=1$ to one of its asymptotes is $3$, then the value of the real number $m$ is?",
    "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (-y^2/(m - 3) + x^2/m = 1);OneOf(Focus(G))=F;Distance(F,OneOf(Asymptote(G)))=3;F: Point",
    "query_expressions": "m",
    "answer_expressions": "12, -9",
    "fact_spans": "[[[2, 42], [51, 52]], [[66, 71]], [[2, 42]], [[2, 50]], [[47, 64]], [[47, 50]]]",
    "query_spans": "[[[66, 75]]]",
    "process": "From the given condition, we have $ b = 3 $, therefore  \n$$\n\\begin{cases}\nm > 3, \\\\\nm - 3 = 3^{2}\n\\end{cases}\n\\quad \\text{or} \\quad\n\\begin{cases}\nm < 0, \\\\\n-m = 3^{3},\n\\end{cases}\n$$\nthen the real value of $ m $ is $ 12 $ or $ -9 $."
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(2)*x/2",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 46]]]",
    "process": "\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1,a=2,b=\\sqrt{2}, hence the asymptotes are: y=\\pm\\frac{\\sqrt{2}}{2}x"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, with left and right foci $F_{1}$, $F_{2}$ respectively. The line $l$ passes through point $F_{2}$ and intersects the right branch of the hyperbola at points $P$, $Q$. If $|P F_{1}|=3|P F_{2}|$, $|P Q|=3|P F_{2}|$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;l: Line;PointOnCurve(F2, l) = True;Intersection(l, RightPart(C)) = {P, Q};P: Point;Q: Point;Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, F2));Abs(LineSegmentOf(P, Q)) = 3*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(21)/3",
    "fact_spans": "[[[2, 63], [102, 105], [162, 168]], [[2, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[71, 78]], [[79, 86], [93, 101]], [[2, 86]], [[2, 86]], [[87, 92]], [[87, 101]], [[87, 117]], [[108, 111]], [[112, 115]], [[119, 141]], [[142, 160]]]",
    "query_spans": "[[[162, 174]]]",
    "process": "Let |PF_{2}| = m, then |PF_{1}| = 3m, |PQ| = 3m, implying |QF_{2}| = 2m. By the definition of hyperbola, we have \\begin{cases} |QF_{1}| = 4a \\\\ m = a \\end{cases}. Then applying the law of cosines in \\triangle PQF_{1} and \\triangle QF_{1}F_{2}, we get \\frac{5a^{2}-c^{2}}{4a^{2}} = \\frac{2}{3}, leading to the answer. Let |PF_{2}| = m, then |PF_{1}| = 3m, |PQ| = 3m, \\therefore |QF_{2}| = 2m. By the definition of hyperbola, we obtain \\begin{cases} |PF_{1}| - |PF_{2}| = 2m = 2a \\\\ |QF_{1}| - |QF_{2}| = |QF_{1}| - 2m = 2a \\end{cases} \\Rightarrow \\begin{cases} |QF_{1}| = 4a \\\\ m = a \\end{cases}. At this point, applying the law of cosines in \\triangle PQF_{1} and \\triangle QF_{1}F_{2}, we get \\cos\\angle F_{1}QF_{2} = \\frac{|QF_{1}|^{2} + |PQ|^{2} - |PF_{1}|^{2}}{2|QF_{1}||PQ|} = \\frac{16a^{2}}{2} \\frac{+9a^{2}-6}{\\times4a\\times3a} \\frac{-9a^{2}}{3a} = \\frac{2}{3}, \\cos\\angle F_{1}QF_{2} = \\frac{|QF_{1}|^{2} + |QF_{2}|^{2} - ||}{2|QF_{1}||QF_{2}|} \\frac{2}{3F_{2}} |F_{1}F_{2}|^{2} = \\frac{16a^{2} + 4a^{2} - 4c^{2}}{2 \\times 4a \\times 2a} = \\frac{5a^{2} - c^{2}}{4a^{2}}; thus \\frac{5a^{2} - c^{2}}{4a^{2}} = \\frac{2}{3}, i.e., 3c^{2} = 7a^{2}, hence \\frac{c^{2}}{a^{2}} = \\frac{7}{3}, so e = \\frac{c}{a} = \\frac{\\sqrt{21}}{3}"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$, $F_{2}$ respectively, and $O$ is the coordinate origin. A perpendicular is drawn from point $F_{2}$ to an asymptote of $C$, with foot of the perpendicular at $P$. The line $P F_{1}$ intersects the other asymptote of the hyperbola at point $Q$, and satisfies $3\\overrightarrow{F_{1} Q}=2 \\overrightarrow{F_{1} P}$. Then, what is the slope of the asymptotes of the hyperbola?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;O: Origin;L1: Line;L2: Line;OneOf(Asymptote(C)) = L1;OneOf(Asymptote(C)) = L2;Negation(L1=L2);L: Line;PointOnCurve(F2, L);IsPerpendicular(L, L1);P: Point;FootPoint(L, L1) = P;Q: Point;Intersection(LineSegmentOf(P, F1), L2) = Q;3*VectorOf(F1, Q) = 2*VectorOf(F1, P)",
    "query_expressions": "Slope(Asymptote(C))",
    "answer_expressions": "pm*sqrt(3)",
    "fact_spans": "[[[2, 63], [137, 140], [213, 216], [106, 109]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[71, 78]], [[79, 86], [97, 105]], [[2, 86]], [[2, 86]], [[87, 90]], [], [], [[106, 115]], [[137, 147]], [[106, 147]], [], [[96, 118]], [[96, 118]], [[122, 125]], [[96, 125]], [[148, 152]], [[127, 152]], [[156, 210]]]",
    "query_spans": "[[[213, 225]]]",
    "process": "Without loss of generality, assume the line $ PF_{2} $ is perpendicular to the asymptote $ y = \\frac{b}{a}x $. Solving $ \\begin{cases} \\\\ \\end{cases} y = \\frac{b}{a}x $, we obtain point $ P\\left(\\frac{a^{2}}{c}, \\frac{ab}{c}\\right) $, $ = -\\frac{a}{b}(x - c) $. Also, $ \\overrightarrow{F_{1}Q} = \\frac{2}{3}\\overrightarrow{F_{1}P} $, and $ F_{1}(-c, 0) $, then $ Q\\left(\\frac{2a^{2} - c^{2}}{3c}, \\frac{2ab}{3c}\\right) $. Since point $ Q $ lies on the line $ y = -\\frac{b}{a}x $, it follows that $ \\frac{2ab}{3c} = -\\frac{b}{a}\\left(\\frac{2a^{2} - c^{2}}{3c}\\right) $, $ \\therefore b^{2} = 3a^{2} $. Hence, the slopes of the hyperbola's asymptotes are $ \\pm\\sqrt{3} $."
  },
  {
    "text": "$P$, $Q$ are moving points on the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, then the maximum value of $|P Q|$ is?",
    "fact_expressions": "C: Ellipse;P: Point;Q: Point;Expression(C) = (x^2/4 + y^2/3 = 1);PointOnCurve(P, C);PointOnCurve(Q, C)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "4",
    "fact_spans": "[[[8, 50]], [[0, 3]], [[4, 7]], [[8, 50]], [[0, 53]], [[0, 53]]]",
    "query_spans": "[[[55, 68]]]",
    "process": "Since the major axis is the longest chord in an ellipse, |PQ|_{\\max}=4"
  },
  {
    "text": "Let point $P$ be a point on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and let $F_{1}$, $F_{2}$ be the left and right foci of the ellipse, respectively. Let $I$ be the incenter of $\\Delta P F_{1} F_{2}$. If $S_{\\Delta {P I F_{1}}}+S_{\\Delta P I F_{2}}=2 S_{\\Delta I F_{1} F_{2}}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;I: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;Incenter(TriangleOf(P, F1, F2)) = I;Area(TriangleOf(P, I, F1)) + Area(TriangleOf(P, I, F2)) = 2*Area(TriangleOf(I, F1, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[6, 58], [80, 82], [195, 197]], [[8, 58]], [[8, 58]], [[1, 5]], [[62, 69]], [[70, 77]], [[89, 92]], [[8, 58]], [[8, 58]], [[6, 58]], [[1, 61]], [[62, 88]], [[62, 88]], [[89, 118]], [[120, 192]]]",
    "query_spans": "[[[195, 203]]]",
    "process": "Let the inradius of $\\triangle PF_{1}F_{2}$ be $r$. Then $S_{\\Delta PIF_{1}} = \\frac{1}{2}|PF_{1}|\\cdot r$, $S_{\\Delta PIF_{2}} = \\frac{1}{2}|PF_{2}|\\cdot r$, $S_{\\Delta IF_{1}F_{2}} = \\frac{1}{2}|F_{1}F_{2}|\\cdot r$. Since $S_{\\triangle PIF_{1}} + S_{\\triangle PIF_{2}} = 2S_{\\Delta IF_{1}F_{2}}$, we have $\\frac{1}{2}|PF_{1}|\\cdot r + \\frac{1}{2}|PF_{2}|\\cdot r = 2 \\times \\frac{1}{2} \\times |F_{1}F_{2}|\\cdot r$, which gives $|PF_{1}| + |PF_{2}| = 2|F_{1}F_{2}|$. Therefore, the eccentricity of the ellipse is $e = \\frac{c}{a} = \\frac{2c}{2a} = \\frac{|F_{1}F_{2}|}{|PF_{1}| + |PF_{2}|} = \\frac{1}{2}$."
  },
  {
    "text": "Given that hyperbola $C$ and ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$ share the same foci $F_{1}$, $F_{2}$, and their eccentricities are reciprocals of each other. If a point $P$ on the right branch of the hyperbola is at a distance of $4$ from the right focus $F_{2}$, then the distance from the midpoint $M$ of $P F_{2}$ to the origin $O$ equals?",
    "fact_expressions": "C: Hyperbola;G: Ellipse;P: Point;F2: Point;M: Point;Expression(G) = (x^2/16 + y^2/12 = 1);InterReciprocal(Eccentricity(C), Eccentricity(G));PointOnCurve(P, RightPart(C));Focus(C) = {F1,F2};Distance(P, F2) = 4;MidPoint(LineSegmentOf(P, F2)) = M;O:Origin;F1:Point;Focus(G)={F1,F2};RightFocus(C)=F2",
    "query_expressions": "Min(Distance(M, O))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 8], [81, 84]], [[9, 48]], [[89, 92]], [[62, 69], [96, 103]], [[124, 127]], [[9, 48]], [[2, 78]], [[81, 92]], [[2, 69]], [[81, 110]], [[112, 127]], [[128, 135]], [[54, 61]], [[2, 69]], [[81, 103]]]",
    "query_spans": "[[[124, 141]]]",
    "process": ""
  },
  {
    "text": "If the parabola $C$: $y=a x^{2}$ passes through the point $(4,2)$, then the coordinates of the focus of the parabola are?",
    "fact_expressions": "C: Parabola;Expression(C) = (y = a*x^2);a: Number;G: Point;Coordinate(G) = (4, 2);PointOnCurve(G, C)",
    "query_expressions": "Coordinate(Focus(C))",
    "answer_expressions": "(0, 2)",
    "fact_spans": "[[[1, 20], [32, 35]], [[1, 20]], [[9, 20]], [[22, 30]], [[22, 30]], [[1, 30]]]",
    "query_spans": "[[[32, 42]]]",
    "process": "From the given conditions, we have $2 = a \\cdot 4^{2}$, so $a = \\frac{1}{8}$. Therefore, the equation of the parabola is $x^{2} = 8y$, and its focus has coordinates $(0, 2)$."
  },
  {
    "text": "If the line $y=k x-1$ intersects the hyperbola $x^{2}-y^{2}=4$ at exactly one point, then the set of real numbers $k$ satisfying this condition is?",
    "fact_expressions": "G: Hyperbola;H: Line;k: Real;Expression(G) = (x^2 - y^2 = 4);Expression(H) = (y = k*x - 1);NumIntersection(H, G)=1",
    "query_expressions": "Range(k)",
    "answer_expressions": "{-sqrt(5)/2, -1, 1, sqrt(5)/2}",
    "fact_spans": "[[[13, 31]], [[1, 12]], [[45, 50]], [[13, 31]], [[1, 12]], [[1, 38]]]",
    "query_spans": "[[[45, 57]]]",
    "process": ""
  },
  {
    "text": "Given that point $A$ is a point on the parabola $C$: $x^{2}=4 y$, if the distance from point $A$ to the focus of parabola $C$ is $2$, then the vertical coordinate $y_{A}$ of point $A$ is?",
    "fact_expressions": "C: Parabola;A: Point;Expression(C) = (x^2 = 4*y);PointOnCurve(A, C);Distance(A, Focus(C)) = 2;yA: Number;YCoordinate(A) = yA",
    "query_expressions": "yA",
    "answer_expressions": "1",
    "fact_spans": "[[[7, 26], [36, 42]], [[2, 6], [31, 35], [53, 57]], [[7, 26]], [[2, 29]], [[31, 51]], [[61, 68]], [[53, 68]]]",
    "query_spans": "[[[61, 70]]]",
    "process": "From the parabola $ C: x^{2} = 4y $, it can be seen that the directrix of the parabola is $ y = -1 $. Let $ F $ be the focus of the parabola $ C $, and $ AH \\perp $ the directrix of the parabola. By the definition of a parabola, $ |AH| = |AF| = 2 $. From the graph, it follows that the distance from point $ A $ to the directrix is 1, so the ordinate of point $ A $ is $ y_{A} = 1 $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$, a line passing through point $M(-2,0)$ intersects $C$ at points $A$ and $B$. If $OA \\parallel BF$ ($O$ is the origin), then $|AB|=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 8*x);F: Point;Focus(C) = F;G: Line;M: Point;Coordinate(M) = (-2, 0);PointOnCurve(M, G);A: Point;B: Point;Intersection(G, C) = {A, B};O: Origin;IsParallel(LineSegmentOf(O, A), LineSegmentOf(B, F))",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "sqrt(17)",
    "fact_spans": "[[[2, 21], [45, 48]], [[2, 21]], [[25, 28]], [[2, 28]], [[42, 44]], [[31, 41]], [[31, 41]], [[30, 44]], [[49, 52]], [[53, 56]], [[42, 58]], [[75, 78]], [[60, 72]]]",
    "query_spans": "[[[86, 95]]]",
    "process": "The focus of the parabola $ C: y^{2} = 8x $ is $ F $. A line passing through point $ M(-2,0) $ intersects $ C $ at points $ A $ and $ B $. Given that $ OA \\parallel BF $ (where $ O $ is the origin), it follows that $ A $ is the midpoint of $ BM $. Let $ B(m, 2\\sqrt{2m}) $, then $ A\\left( \\frac{m-2}{2}, \\sqrt{2m} \\right) $."
  },
  {
    "text": "A point $P$ on the left branch of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ has a distance of $13$ to the right focus. What is the distance from point $P$ to the left directrix?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1);P: Point;PointOnCurve(P, LeftPart(G)) = True;Distance(P, RightFocus(G)) = 13",
    "query_expressions": "Distance(P, LeftDirectrix(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 39]], [[0, 39]], [[45, 48], [62, 66]], [[0, 48]], [[0, 60]]]",
    "query_spans": "[[[0, 75]]]",
    "process": "Let point P(x,y), then x\\leqslant-4, we can derive y^{2}=\\frac{9x^{2}}{16}-9. From the distance from point P to the right focus being 13, the value of x can be found, and then the distance from point P to the left directrix of this hyperbola can be determined. Let point P(x,y), then x\\leqslant-4. For the hyperbola \\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1, a=4, b=3, then c=5. The right focus of this hyperbola is (5,0). According to the condition, 13=\\sqrt{(x-5)^{2}+y^{2}}=\\sqrt{(x-5)^{2}+y^{2}}=\\sqrt{x^{2}-10x+25+\\frac{9x^{2}}{16}-9}, so x^{2}-10x+16=|\\frac{5x}{4}-4|=4-\\frac{5x}{4}=13, solving gives x=-\\frac{36}{5}. The equation of the left directrix of this hyperbola is x=-\\frac{a^{2}}{c}=-\\frac{16}{5}. Therefore, the distance from point P to the left directrix is d=-\\frac{36}{5}+\\frac{16}{5}=4"
  },
  {
    "text": "Given that the coordinates of the two foci of an ellipse are $F_{1}(4,0)$ and $F_{2}(-4,0)$, and the sum of the distances from a point $M$ on the ellipse to points $F_{1}$ and $F_{2}$ is equal to $10$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Coordinate(F1) = (4, 0);Coordinate(F2) = (-4, 0);Focus(G)={F1,F2};M:Point;PointOnCurve(M,G);Distance(M,F1)+Distance(M,F2)=10",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/25 + y^2/9 = 1",
    "fact_spans": "[[[2, 4], [2, 4], [2, 4]], [[15, 27], [56, 64]], [[30, 43], [65, 72]], [[15, 27]], [[30, 43]], [[2, 43]], [[52, 55]], [[47, 55]], [[52, 83]]]",
    "query_spans": "[[[86, 95]]]",
    "process": "From |MF₁| + |MF₂| = 2a = 10, we can find a. From the coordinates of the foci, we can obtain the value of c. Then, using b² = a² - c², we can find b², thus obtaining the equation of the ellipse. Let the equation of the ellipse be \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1. Since |MF₁| + |MF₂| = 10, it follows that 2a = 10, so a = 5. Given c = 4, we have b² = a² - c² = 5^{2} - 4^{2} = 9. Therefore, the equation of the ellipse is \\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1."
  },
  {
    "text": "Let a line pass through one focus of hyperbola $C$ and be perpendicular to one of the symmetry axes of $C$, intersecting $C$ at points $A$ and $B$, such that $|AB|$ is twice the length of the real axis of $C$. Then the eccentricity of $C$ is?",
    "fact_expressions": "G: Line;C: Hyperbola;PointOnCurve(OneOf(Focus(C)), G);IsPerpendicular(G, OneOf(SymmetryAxis(C)));A: Point;B: Point;Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, B)) = 2*Length(RealAxis(C))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[1, 3]], [[4, 10], [18, 21], [31, 34], [54, 57], [68, 71]], [[1, 15]], [[1, 29]], [[36, 39]], [[40, 43]], [[1, 45]], [[46, 66]]]",
    "query_spans": "[[[68, 77]]]",
    "process": "Let the standard equation of the hyperbola be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$). From the given conditions, we have $\\frac{2b^{2}}{a}=4a'$, that is, $b^{2}=2a^{2}$. $c^{2}=3a^{2}$, so the eccentricity of the hyperbola is $e=\\sqrt{3}$."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{k-5}+\\frac{y^{2}}{10-k}=1$ represents an ellipse with foci on the $y$-axis, then the range of real values for $k$ is?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (x^2/(k - 5) + y^2/(10 - k) = 1);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(5, 15/2)",
    "fact_spans": "[[[54, 56]], [[58, 63]], [[1, 56]], [[45, 56]]]",
    "query_spans": "[[[58, 70]]]",
    "process": "Since the equation $\\frac{x^{2}}{k-5}+\\frac{y^{2}}{10-k}=1$ represents an ellipse with foci on the $y$-axis, it follows that $10-k>k-5>0$, solving which yields $5<k<\\frac{15}{2}$. Therefore, the range of $k$ is $(5,\\frac{15}{2})$."
  },
  {
    "text": "The coordinates of the focus of the parabola $2 x^{2}+y=0$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (2*x^2 + y = 0)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,-1/8)",
    "fact_spans": "[[[0, 16]], [[0, 16]]]",
    "query_spans": "[[[0, 23]]]",
    "process": "From $2x^{2}+y=0$, we get $x^{2}=-\\frac{1}{2}y$, so the parabola opens downward and $2p=\\frac{1}{2}$, thus $\\frac{p}{2}=\\frac{1}{8}$, so the focus coordinates are $(0,-\\frac{1}{8})$. This question examines finding the focus from the standard equation of a parabola, and is a simple problem."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with left focus $F$. A line passing through the origin intersects the hyperbola at points $A$ and $B$. If $|AB|=\\sqrt{10}$, $|AF|=\\sqrt{2}$, $\\cos \\angle ABF = \\frac{2\\sqrt{5}}{5}$, then the length of the real axis of the hyperbola $C$, $2a = $?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;O:Origin;A: Point;B: Point;F: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;PointOnCurve(O,G);Intersection(G,C)={A,B};Abs(LineSegmentOf(A, B)) = sqrt(10);Abs(LineSegmentOf(A, F)) = sqrt(2);Cos(AngleOf(A, B, F)) = (2*sqrt(5))/5;Length(RealAxis(C)) = 2*a",
    "query_expressions": "2*a",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 63], [79, 82], [175, 181]], [[10, 63]], [[10, 63]], [[76, 78]], [[73, 75]], [[85, 88]], [[89, 92]], [[68, 71]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 71]], [[72, 78]], [[76, 94]], [[95, 112]], [[114, 130]], [[133, 173]], [[175, 190]]]",
    "query_spans": "[[[185, 192]]]",
    "process": "In triangle AFB, by the law of cosines, |AF|^{2} = |AB|^{2} + |BF|^{2} - 2|AB||BF|\\cos\\angle ABF, from which |BF| can be obtained. Let F be the right focus of the hyperbola, connect BF and AF. By symmetry, quadrilateral AFBF is a rectangle, then using the definition of the hyperbola, the solution can be found. In triangle AFB, |AB| = \\sqrt{10}, |AF| = \\sqrt{2}, \\cos\\angle ABF = \\frac{2\\sqrt{5}}{5}. By the law of cosines, |AF|^{2} = |AB|^{2} + |BF|^{2} - 2|AB||BF|\\cos\\angle ABF, thus (|BF| - 2\\sqrt{2})^{2} = 0, solving gives |BF| = 2\\sqrt{2}, so triangle AFB is a right triangle. Let F be the right focus of the hyperbola, connect BF and AF. By symmetry, quadrilateral AFBF is a rectangle, so |BF| = |AF| = \\sqrt{2}, hence 2a = |BF| - |BF| = \\sqrt{2}."
  },
  {
    "text": "Draw a line through the focus of the parabola $y^{2}=8x$ intersecting the parabola at two points $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$. If $x_{1}+x_{2}=16$, then $|AB|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);H: Line;PointOnCurve(Focus(G), H);A: Point;B: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Intersection(H, G) = {A, B};x1 + x2 = 16",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "20",
    "fact_spans": "[[[1, 15], [22, 25]], [[1, 15]], [[19, 21]], [[0, 21]], [[26, 43]], [[45, 62]], [[26, 43]], [[45, 62]], [[26, 43]], [[45, 62]], [[26, 43]], [[45, 62]], [[19, 64]], [[66, 82]]]",
    "query_spans": "[[[84, 93]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the foci of the ellipse $\\frac{x^{2}}{3}+y^{2}=1$, and let points $A$ and $B$ lie on the ellipse such that $\\overrightarrow{F_{1} A}=5 \\overrightarrow{F_{2} B}$. Then the coordinates of point $A$ are?",
    "fact_expressions": "G: Ellipse;F1: Point;A: Point;F2: Point;B: Point;Expression(G) = (x^2/3 + y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(A, G);PointOnCurve(B, G);VectorOf(F1, A) = 5*VectorOf(F2, B)",
    "query_expressions": "Coordinate(A)",
    "answer_expressions": "(0,pm*1)",
    "fact_spans": "[[[19, 46], [59, 61]], [[1, 8]], [[50, 54], [120, 124]], [[9, 16]], [[55, 58]], [[19, 46]], [[1, 49]], [[50, 62]], [[50, 62]], [[64, 117]]]",
    "query_spans": "[[[120, 129]]]",
    "process": "Method 1: The extension of line $F_{1}A$ in the opposite direction intersects the ellipse at point $B$. Since $\\overrightarrow{F_{1}A}=5\\overrightarrow{F_{2}B}$, by the symmetry of the ellipse, we have $\\overrightarrow{F_{1}A}=5\\overrightarrow{B'}F_{1}$. Let $A(x_{1},y_{1})$, $B'(x_{2},y_{2})$. For the ellipse $\\frac{x^{2}}{3}+y^{2}=1$, we have $a=\\sqrt{3}$, $b=1$, $c=\\sqrt{2}$, so $c=\\frac{c}{a}=\\frac{\\sqrt{2}}{\\sqrt{3}}=\\frac{\\sqrt{6}}{3}$, $F_{1}(-\\sqrt{2},0)$. Since $|F_{1}A|=\\frac{\\sqrt{6}}{3}|x_{1}+\\frac{3\\sqrt{2}}{2}|$, $|F_{1}B'|=\\frac{\\sqrt{6}}{3}|x_{2}+\\frac{3\\sqrt{2}}{2}|$, it follows that:  \n$\\frac{\\sqrt{6}}{3}|x_{1}+\\frac{3\\sqrt{2}}{2}|=5\\times\\frac{\\sqrt{6}}{3}|x_{2}+\\frac{3\\sqrt{2}}{2}|$.  \nSince $-\\sqrt{3}\\leqslant x_{1},x_{2}<\\sqrt{3}$, and $\\frac{3\\sqrt{2}}{2}>0$, $x_{2}+\\frac{3\\sqrt{2}}{2}>0$, we have:  \n$\\frac{\\sqrt{6}}{3}(\\frac{3\\sqrt{2}}{2}+x_{1})=5\\times\\frac{\\sqrt{6}}{3}(x_{2}+\\frac{3\\sqrt{2}}{2})$,  \nthus $\\frac{3\\sqrt{2}}{2}+x_{1}=5(x_{2}+\\frac{3\\sqrt{2}}{2})$. ①  \nAlso, since points $A$, $F_{1}$, $B'$ are collinear and $\\overrightarrow{F_{1}A}=5\\overrightarrow{B'}F_{1}$,  \nwe have $(x_{1}-(-\\sqrt{2}), y_{1}-0)=5(-\\sqrt{2}-x_{2}, 0-y_{2})$,  \nso $x_{1}+\\sqrt{2}=5(-\\sqrt{2}-x_{2})$. ②  \nAdding ① and ② gives: $x_{1}=0$. Substituting into the ellipse equation yields: $y_{1}=\\pm1$.  \nTherefore, the coordinates of point $A$ are $(0,1)$ or $(0,-1)$.  \n\nMethod 2: Since $F_{1}$, $F_{2}$ are the foci of the ellipse $\\frac{x^{2}}{3}+y^{2}=1$, we have $F_{1}(-\\sqrt{2},0)$, $F_{2}(\\sqrt{2},0)$. Let the coordinates of $A$, $B$ be $A(x_{A},y_{A})$, $B(x_{B},y_{B})$. If $\\overrightarrow{F_{1}A}=5\\overrightarrow{F_{2}B}$, then  \n$$\n\\begin{cases}\nx_{A}+\\sqrt{2}=5(x_{B}-\\sqrt{2}) \\\\\ny_{A}=5y_{B}\n\\end{cases}\n$$\nThus,\n$$\n\\begin{cases}\nx_{B}=\\frac{x_{A}+6\\sqrt{2}}{5} \\\\\ny_{A}=5y_{B}\n\\end{cases}\n$$\nSince $A$, $B$ lie on the ellipse,  \n$$\n\\begin{cases}\n\\frac{x_{A}^{2}}{3}+y_{A}^{2}=1 \\\\\n\\frac{x_{B}^{2}}{3}+y_{B}^{2}=1\n\\end{cases}\n$$\nSubstituting and solving yields  \n$$\n\\begin{cases}\nx_{A}=0 \\\\\ny_{A}=\\pm1\n\\end{cases}\n$$\nHence $A(0,\\pm1)$."
  },
  {
    "text": "The equation of the parabola with vertex at the origin, symmetric about the $y$-axis, and passing through the point $P(-4,-2)$ is?",
    "fact_expressions": "G: Parabola;P: Point;O: Origin;Coordinate(P) = (-4, -2);PointOnCurve(P, G);Vertex(G)=O;SymmetryAxis(G)=yAxis",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 = -8y",
    "fact_spans": "[[[31, 34]], [[19, 30]], [[3, 5]], [[19, 30]], [[17, 34]], [[0, 34]], [[6, 34]]]",
    "query_spans": "[[[31, 38]]]",
    "process": "According to the problem, assume the equation of the parabola is $x^{2}=ay$. Substituting the point $P(-4,-2)$ into the equation gives $16=-2a$, so $a=-8$. Therefore, the equation of the parabola is $x^{2}=-8y$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, the right focus is $F$. A perpendicular is drawn from point $F$ to one asymptote of the hyperbola, with foot of perpendicular at $M$, intersecting the other asymptote at $N$. If $2 \\overrightarrow{M F}=\\overrightarrow{F N}$, then what is the eccentricity of the hyperbola?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;M: Point;F: Point;N: Point;l1: Line;l2: Line;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;OneOf(Asymptote(G))=l1;OneOf(Asymptote(G))=l2;Negation(l1=l2);L:Line;PointOnCurve(F, L);IsPerpendicular(l1, L);FootPoint(l1,L)=M;Intersection(l2,L)=N;2*VectorOf(M, F) = VectorOf(F, N)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 53], [68, 71], [148, 151]], [[10, 53]], [[10, 53]], [[84, 87]], [[58, 61], [63, 67]], [[96, 99]], [], [], [[2, 53]], [[2, 61]], [[68, 77]], [[68, 95]], [[68, 95]], [], [[62, 80]], [[62, 80]], [[62, 87]], [[62, 99]], [[101, 146]]]",
    "query_spans": "[[[148, 156]]]",
    "process": "As shown in the figure, the equation of asymptote OM is $bx+ay=0$, and the right focus is $F(c,0)$, so $|FM|=\\frac{bc}{\\sqrt{a^{2}+b^{2}}}=b$. Draw a perpendicular line from point $F$ to $ON$, with foot $P$, then $|FP|=|FM|=b$. Since $2\\overrightarrow{MF}=\\overrightarrow{FN}$, it follows that $|FN|=2b$. In the right triangle $FPN$, $\\sin\\angle FNP=\\frac{|PF|}{|FN|}=\\frac{b}{2b}=\\frac{1}{2}$, so $\\angle FNP=\\frac{\\pi}{6}$. Therefore, in triangle $OMN$, $\\angle MON=\\frac{\\pi}{3}$, hence $\\angle FON=\\frac{\\pi}{6}$, so $\\frac{b}{a}=\\frac{\\sqrt{3}}{3}$, i.e., $a=\\sqrt{3}b$, $c=\\sqrt{a^{2}+b^{2}}=2b$. Thus, the eccentricity of the hyperbola is $e=\\frac{c}{a}=\\frac{2b}{\\sqrt{3}b}=\\frac{2\\sqrt{3}}{3}$."
  },
  {
    "text": "What is the distance from the point $(m, 2)$ on the parabola $y^{2}=4 x$ to its focus?",
    "fact_expressions": "G: Parabola;H: Point;Expression(G) = (y^2 = 4*x);Coordinate(H) = (m, 2);PointOnCurve(H,G);m:Number",
    "query_expressions": "Distance(H, Focus(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[26, 27], [0, 14]], [[16, 25]], [[0, 14]], [[16, 25]], [[0, 25]], [[17, 25]]]",
    "query_spans": "[[[16, 34]]]",
    "process": "Substituting the point $(m,2)$ into the equation of the parabola gives $2^{2}=4m$, solving yields $m=1$, so the coordinates of the point are $(1,2)$. Since $\\frac{p}{2}=1$, according to the definition of the parabola, the distance from the point to the focus is $1+\\frac{p}{2}=1+1=2$. Qing] This question mainly examines the method of finding the coordinates of a point on a parabola and the definition of a parabola, belonging to basic problems."
  },
  {
    "text": "Given that the directrix of the parabola $y^{2}=2 ax$ is $x=-\\frac{1}{4}$, then its focus coordinates are?",
    "fact_expressions": "G: Parabola;a: Number;Expression(G) = (y^2 = 2*a*x);Expression(Directrix(G)) = (x = -1/4)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(1/4,0)",
    "fact_spans": "[[[2, 17], [39, 40]], [[5, 17]], [[2, 17]], [[2, 37]]]",
    "query_spans": "[[[39, 46]]]",
    "process": ""
  },
  {
    "text": "Given that $A$, $B$, $C$ are three points on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, point $F(3 , 0)$, and if $\\overrightarrow{F A}+\\overrightarrow{F B}+\\overrightarrow{F C}={0}$, then $|F A|+|F B|+|F C|$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);A: Point;B: Point;C: Point;F: Point;Coordinate(F) = (3, 0);PointOnCurve(A, G);PointOnCurve(B, G);PointOnCurve(C, G);VectorOf(F, A) + VectorOf(F, B) + VectorOf(F, C) = 0",
    "query_expressions": "Abs(LineSegmentOf(F, A)) + Abs(LineSegmentOf(F, B)) + Abs(LineSegmentOf(F, C))",
    "answer_expressions": "48/5",
    "fact_spans": "[[[16, 55]], [[16, 55]], [[2, 5]], [[7, 10]], [[12, 15]], [[60, 71]], [[60, 71]], [[2, 59]], [[2, 59]], [[2, 59]], [[73, 141]]]",
    "query_spans": "[[[143, 164]]]",
    "process": ""
  },
  {
    "text": "The equation of the directrix of the parabola $x^{2}=20 y$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 20*y)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=-5",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 22]]]",
    "process": "From the parabola equation, we know: the focus of the parabola lies on the y-axis, and p=10, ∴ the directrix equation is y=-\\frac{p}{2}=-5"
  },
  {
    "text": "The standard equation of a parabola with directrix $y+1=0$ is?",
    "fact_expressions": "G: Parabola;Expression(Directrix(G)) = (y + 1 = 0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2=4*y",
    "fact_spans": "[[[13, 16]], [[0, 16]]]",
    "query_spans": "[[[13, 22]]]",
    "process": "From the given condition, the directrix of the parabola is $ y = -1 $, so the parabola opens upwards. Let the equation of the parabola be $ x^{2} = 2py $. Therefore, $ \\frac{p}{2} = 1 $, so $ p = 2 $. Thus, the standard equation of the parabola is $ x^{2} = 4y $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, its upper vertex coincides with the focus $F$ of the parabola $C^{\\prime}$: $x^{2}=2 p y$ $(p>0)$, and $P$ is a common point of $C$ and $C^{\\prime}$. If the eccentricity of $C$ is $\\frac{\\sqrt{3}}{3}$, and $|P F|=2$, then $p=$?",
    "fact_expressions": "C: Ellipse;a: Number;b: Number;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);E: Parabola;p: Number;p>0;Expression(E) = (x^2 = 2*(p*y));F: Point;Focus(E) = F;UpperVertex(C) = F;P: Point;OneOf(Intersection(C, E)) = P;Eccentricity(C) = sqrt(3)/3;Abs(LineSegmentOf(P, F)) = 2",
    "query_expressions": "p",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 59], [112, 115], [136, 139]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[2, 59]], [[64, 99], [116, 128]], [[177, 180]], [[81, 99]], [[64, 99]], [[102, 105]], [[64, 105]], [[3, 107]], [[108, 111]], [[108, 134]], [[136, 164]], [[166, 175]]]",
    "query_spans": "[[[177, 182]]]",
    "process": "Since the upper vertex of the ellipse coincides with the focus of the parabola, and using the eccentricity of the ellipse, express a and b in terms of p. Use the focal radius formula to find the coordinates of point P, substitute into the ellipse equation to solve for p. Solution: From the given condition, b = \\frac{p}{2}, and e = \\frac{c}{a} = \\frac{\\sqrt{a^{2}-b^{2}}}{a} = \\frac{\\sqrt{3}}{3}, so a = \\frac{\\sqrt{6}}{4}p. Also |PF| = 2, so y_{p} = 2 - \\frac{p}{2}, then x_{P}^{2} = 2py_{P} = 2p(2 - \\frac{p}{2}) = 4p - p^{2}. Therefore, \\frac{4p - p^{2}}{\\frac{6}{16}p^{2}} + \\frac{(2 - \\frac{p}{2})^{2}}{\\frac{1}{4}p^{2}} = 1, simplifying yields p^{2} - p - 6 = 0, solving gives p = 3 (negative value discarded)."
  },
  {
    "text": "The equation of the circle centered at the right focus of the ellipse $\\frac{x^{2}}{169}+\\frac{y^{2}}{144}=1$ and tangent to the asymptotes of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ is?",
    "fact_expressions": "H: Ellipse;Expression(H) = (x^2/169 + y^2/144 = 1);Z: Circle;Center(Z) = RightFocus(H);G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);IsTangent(Asymptote(G), Z)",
    "query_expressions": "Expression(Z)",
    "answer_expressions": "(x-5)^2+y^2=16",
    "fact_spans": "[[[1, 42]], [[1, 42]], [[98, 99]], [[0, 99]], [[52, 91]], [[52, 91]], [[51, 99]]]",
    "query_spans": "[[[98, 104]]]",
    "process": ""
  },
  {
    "text": "The length of the imaginary axis of the hyperbola $mx^{2}-y^{2}=1$ is $2$ times the length of the real axis, then $m$=?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (m*x^2 - y^2 = 1);Length(ImageinaryAxis(G)) = 2*Length(RealAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 19]], [[34, 37]], [[0, 19]], [[0, 32]]]",
    "query_spans": "[[[34, 39]]]",
    "process": ""
  },
  {
    "text": "Draw two tangents from a focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ to the circle $x^{2}+y^{2}=a^{2}$, with points of tangency $A$ and $B$, respectively. If $\\angle AOB = 90^{\\circ}$ ($O$ is the origin), then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;A: Point;O: Origin;B: Point;l1: Line;l2: Line;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = a^2);TangentOfPoint(OneOf(Focus(C)), G) = {l1, l2};TangentPoint(l1, G) = A;TangentPoint(l2, G) = B;AngleOf(A, O, B) = ApplyUnit(90, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 62], [147, 153]], [[9, 62]], [[9, 62]], [[68, 88]], [[99, 102]], [[136, 139]], [[104, 107]], [], [], [[9, 62]], [[9, 62]], [[1, 62]], [[68, 88]], [[0, 93]], [[0, 107]], [[0, 107]], [[110, 135]]]",
    "query_spans": "[[[147, 159]]]",
    "process": ""
  },
  {
    "text": "What is the focal distance of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{5}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/5 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(14)",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 43]]]",
    "process": ""
  },
  {
    "text": "An equation of a hyperbola that shares the same asymptotes with the hyperbola $\\frac{x^{2}}{2}-y=1$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/2 - y = 1);H: Hyperbola;Asymptote(H) = Asymptote(G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4 - y^2/2 = 1",
    "fact_spans": "[[[1, 25]], [[1, 25]], [[34, 37]], [[0, 37]]]",
    "query_spans": "[[[34, 41]]]",
    "process": "The hyperbola equation sharing the same asymptotes with $\\frac{x^{2}}{2}-y=1$ is $\\frac{x^{2}}{2}-y=k$, $k\\neq0$. For example, taking $k=2$, the equation becomes $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$."
  },
  {
    "text": "The equation of the ellipse that has the same foci as the ellipse $4 x^{2}+9 y^{2}=36$ and passes through the point $(-3 , 2)$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (4*x^2 + 9*y^2 = 36);Z: Ellipse;Focus(G) = Focus(Z);H: Point;Coordinate(H) = (-3, 2);PointOnCurve(H, Z)",
    "query_expressions": "Expression(Z)",
    "answer_expressions": "x^2/15+y^2/10=1",
    "fact_spans": "[[[1, 23]], [[1, 23]], [[45, 47]], [[0, 47]], [[33, 44]], [[33, 44]], [[32, 47]]]",
    "query_spans": "[[[45, 51]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $P$ is a point on the ellipse $C$, and $\\overrightarrow{P F_{1}} \\perp \\overrightarrow{P F_{2}}$. If the area of $\\triangle P F_{1} F_{2}$ is $9$, then $b=$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C);IsPerpendicular(VectorOf(P, F1), VectorOf(P, F2));Area(TriangleOf(P, F1, F2)) = 9",
    "query_expressions": "b",
    "answer_expressions": "3",
    "fact_spans": "[[[18, 77], [87, 92]], [[18, 77]], [[191, 194]], [[25, 77]], [[25, 77]], [[25, 77]], [[2, 9]], [[10, 17]], [[2, 82]], [[83, 86]], [[83, 95]], [[97, 154]], [[157, 189]]]",
    "query_spans": "[[[191, 196]]]",
    "process": ""
  },
  {
    "text": "The foci of ellipse $C$ lie on the $x$-axis, and the focal distance is $2$. The line $n$: $x-y-1=0$ intersects the ellipse $C$ at points $A$ and $B$, $F_{1}$ is the left focus, and $F_{1} A \\perp F_{1} B$. Then, what is the standard equation of ellipse $C$?",
    "fact_expressions": "C: Ellipse;PointOnCurve(Focus(C), xAxis) = True;FocalLength(C) = 2;n: Line;Expression(n) = (x - y - 1 = 0);Intersection(n, C) = {A, B};A: Point;B: Point;F1: Point;LeftFocus(C) = F1;IsPerpendicular(LineSegmentOf(F1, A), LineSegmentOf(F1, B)) = True",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/(2+sqrt(3))+y^2/(1+sqrt(3))=1",
    "fact_spans": "[[[0, 5], [39, 44], [94, 99]], [[0, 14]], [[0, 21]], [[22, 38]], [[22, 38]], [[22, 55]], [[46, 49]], [[50, 53]], [[56, 63]], [[39, 67]], [[69, 92]]]",
    "query_spans": "[[[94, 106]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 x$ is $F$, point $P$ is a moving point on the parabola, and there is a point $A(3,2)$, then when $|P A|+|P F|$ attains its minimum value, what are the coordinates of point $P$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);F: Point;Focus(G) = F;P: Point;PointOnCurve(P, G) = True;A: Point;Coordinate(A) = (3, 2);WhenMin(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2,2)",
    "fact_spans": "[[[1, 15], [28, 31]], [[1, 15]], [[19, 22]], [[1, 22]], [[23, 27], [70, 74]], [[23, 35]], [[38, 47]], [[38, 47]], [[49, 69]]]",
    "query_spans": "[[[70, 79]]]",
    "process": "Substitute $x=3$ into the parabola equation $y^{2}=2x$, we get $y=\\pm\\sqrt{6}$. Since $\\sqrt{6}>2$, point $A$ lies inside the parabola. As shown in the figure, let the distance from point $P$ on the parabola to the directrix $l: x=-\\frac{1}{2}$ be $d$. By definition, $|PA|+|PF|=|PA|+d$. When $PA\\perp l$, $|PA|+d$ is minimized, and the minimum value is $\\frac{7}{2}$. At this time, the ordinate of point $P$ is $2$. Substituting into $y^{2}=2x$, we get $x=2$. Therefore, the coordinates of point $P$ are $(2,2)$."
  },
  {
    "text": "It is known that $F$ is the focus of the parabola $C$: $y^{2}=4x$. A line $l$ passing through point $F$ intersects the parabola $C$ at points $A$ and $B$, and intersects the directrix of the parabola $C$ at point $D$. If $F$ is the midpoint of $AD$, then $|FB|$=?",
    "fact_expressions": "F: Point;C: Parabola;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F, l) = True;l: Line;Intersection(l, C) = {A, B};A: Point;B: Point;Intersection(l, Directrix(C)) = D;D: Point;MidPoint(LineSegmentOf(A, D)) = F",
    "query_expressions": "Abs(LineSegmentOf(F, B))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[2, 5], [30, 34], [77, 80]], [[6, 25], [41, 47], [60, 66]], [[6, 25]], [[2, 28]], [[29, 40]], [[35, 40]], [[35, 58]], [[49, 52]], [[53, 56]], [[35, 75]], [[71, 75]], [[77, 89]]]",
    "query_spans": "[[[91, 100]]]",
    "process": ""
  },
  {
    "text": "If the line $y=k x+2$ intersects the right branch of the hyperbola $x^{2}-y^{2}=6$ at two distinct points, then the range of values for $k$ is?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*x + 2);k: Number;G: Hyperbola;Expression(G) = (x^2 - y^2 = 6);NumIntersection(H, RightPart(G)) = 2",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-sqrt(15)/3,-1)",
    "fact_spans": "[[[1, 12]], [[1, 12]], [[43, 46]], [[13, 31]], [[13, 31]], [[1, 41]]]",
    "query_spans": "[[[43, 53]]]",
    "process": "From \\begin{cases}y=kx+2\\\\x^{2}-y^{2}=6\\end{cases} we obtain: (1-k^{2})x^{2}-4kx-10=0. (1) The line y=kx+2 intersects the right branch of the hyperbola x^{2}-y^{2}=6 at two distinct points, meaning equation (1) has two distinct solutions, so \\frac{4>0}{1-k^{2}}>0, solving gives -\\frac{\\sqrt{15}}{3}<k<-1"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F_{1}$, $F_{2}$ are its left and right foci, $|F_{1} F_{2}|=2 \\sqrt{2}$, $B$ is an endpoint of the minor axis, the area of triangle $B F_{1} O$ ($O$ is the origin) is $\\sqrt{7}$, then the length of the major axis of the ellipse is?",
    "fact_expressions": "C: Ellipse;Expression(C)=(x^2/a^2+y^2/b^2=1);a: Number;b: Number;a > b;b > 0;F1:Point;F2:Point;LeftFocus(C)=F1;RightFocus(C)=F2;Abs(LineSegmentOf(F1,F2))=2*sqrt(2);O:Origin;B:Point;OneOf(Endpoint(MinorAxis(C)))=B;Area(TriangleOf(B,F1,O))=sqrt(7)",
    "query_expressions": "Length(MajorAxis(C))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 59], [78, 79], [167, 169]], [[2, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[61, 68]], [[70, 77]], [[61, 83]], [[61, 83]], [[84, 110]], [[140, 143]], [[112, 115]], [[78, 123]], [[125, 164]]]",
    "query_spans": "[[[167, 175]]]",
    "process": "Given the problem, let $ c^{2} = a^{2} - b^{2} $, $ c > 0 $. Since $ |F_{1}F_{2}| = 2\\sqrt{2} $, it follows that $ c = \\sqrt{2} $. Also, the area of triangle $ BF_{1}O $ is $ \\frac{1}{2}bc = \\sqrt{7} \\Rightarrow b = \\sqrt{14} $. Hence, $ a = \\sqrt{14 + 2} = 4 $. Therefore, the major axis length of the ellipse is $ 2a = 8 $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8 x$, then the coordinates of the focus of the parabola are?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 8*x)",
    "query_expressions": "Coordinate(Focus(C))",
    "answer_expressions": "(2,0)",
    "fact_spans": "[[[2, 21], [23, 26]], [[2, 21]]]",
    "query_spans": "[[[23, 33]]]",
    "process": "From the parabola equation $ y^{2} = 8x $, we obtain $ 2p = 8 $, that is, $ p = 4 $, and the focus lies on the positive x-axis; thus, the coordinates of the focus are $ (2, 0) $."
  },
  {
    "text": "Given that $P$ is a moving point on the parabola $y^{2}=8x$, fixed point $A(3,1)$, and $F$ is the focus of the parabola, then the minimum value of $PF + PA$ is?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;A: Point;Expression(G) = (y^2 = 8*x);PointOnCurve(P, G);Coordinate(A)=(3, 1);Focus(G) = F",
    "query_expressions": "Min(LineSegmentOf(P, A) + LineSegmentOf(P, F))",
    "answer_expressions": "5",
    "fact_spans": "[[[6, 20], [41, 44]], [[2, 5]], [[36, 39]], [[26, 34]], [[6, 20]], [[2, 23]], [[26, 34]], [[35, 47]]]",
    "query_spans": "[[[49, 64]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, a line $l$ passing through point $P(1, \\frac{1}{2})$ intersects the ellipse $C$ at points $A$ and $B$, and point $P$ is the midpoint of $AB$. Then the equation of line $l$ is?",
    "fact_expressions": "l: Line;C: Ellipse;A: Point;B: Point;P: Point;Coordinate(P) = (1, 1/2);Expression(C) = (x^2/4 + y^2 = 1);PointOnCurve(P, l);Intersection(l, C) = {A, B};MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x+2*y-2=0",
    "fact_spans": "[[[53, 58], [91, 96]], [[2, 30], [59, 64]], [[65, 68]], [[69, 72]], [[76, 80], [32, 52]], [[32, 52]], [[2, 30]], [[31, 58]], [[53, 74]], [[76, 89]]]",
    "query_spans": "[[[91, 101]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ x_{1}^{2} + 4y_{1}^{2} = 4 $, $ x_{2}^{2} + 4y_{2}^{2} = 4 $, $ \\therefore (x_{1}+x_{2})(x_{1}-x_{2}) + 4(y_{1}+y_{2})(y_{1}-y_{2}) = 0 $. $ \\because P(1,\\frac{1}{2}) $ is exactly the midpoint of segment $ AB $, so $ x_{1}+x_{2} = 2 $, $ y_{1}+y_{2} = 1 $, $ \\therefore (x_{1}-x_{2}) + 2(y_{1}-y_{2}) = 0 $, $ \\therefore $ the slope of line $ AB $ is $ k = \\frac{y_{1}-y_{2}}{x_{1}} = -\\frac{1}{2} $. $ \\therefore $ the equation of line $ AB $ is $ y - \\frac{1}{2} = -\\frac{1}{2}(x-1) $, i.e., $ x + 2v - 2 = 0 $. Since $ P $ is inside the ellipse, it holds that the answer is actually $ x + 2y - 2 = 0 $."
  },
  {
    "text": "The equation of the hyperbola passing through the point $(2, -2)$ and having the same asymptotes as $\\frac{x^{2}}{2}-y^{2}=1$ is?",
    "fact_expressions": "G: Hyperbola;H:Point;C:Curve;Expression(C)=(x^2/2-y^2=1);Coordinate(H) = (2,-2);PointOnCurve(H, G);Asymptote(G)=Asymptote(C)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/2 - x^2/4 = 1",
    "fact_spans": "[[[47, 50]], [[1, 11]], [[13, 38]], [[13, 38]], [[1, 11]], [[0, 50]], [[12, 50]]]",
    "query_spans": "[[[47, 54]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, a line passing through point $F_{1}$ intersects the ellipse at points $A$ and $B$. If $|A B|=5$, then the area of $\\triangle A B F_{2}$ is?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2/9 + y^2/5 = 1);Focus(G) = {F1, F2};PointOnCurve(F1, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 5",
    "query_expressions": "Area(TriangleOf(A, B, F2))",
    "answer_expressions": "5",
    "fact_spans": "[[[18, 55], [74, 76]], [[71, 73]], [[77, 80]], [[81, 84]], [[10, 17]], [[2, 9], [62, 70]], [[18, 55]], [[2, 60]], [[61, 73]], [[71, 86]], [[88, 97]]]",
    "query_spans": "[[[99, 125]]]",
    "process": "Let the equation of line AB be $ y = k(x + 2) $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. By solving the system of equations and using the chord length formula and the definition of the ellipse, we find $ k = \\pm\\frac{\\sqrt{3}}{3} $, obtain the equation of line AB, then combine with the point-to-line distance formula and the triangle area formula to solve. From the ellipse $ \\frac{x^{2}}{9} + \\frac{y^{2}}{5} = 1 $, we get $ a^{2} = 9 $, $ b^{2} = 5 $, then $ c = \\sqrt{a^{2} - b^{2}} = 2 $, so the foci of the ellipse are $ F_{1}(-2, 0) $, $ F_{2}(2, 0) $. Let the equation of line AB be $ y = k(x + 2) $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Solve the system:\n$$\n\\begin{cases}\ny = k(x + 2) \\\\\n\\frac{x^{2}}{9} + \\frac{y^{2}}{5} = 1\n\\end{cases}\n$$\nWe obtain $ (5 + 9k^{2})x^{2} + 36k^{2}x + 36k^{2} - 45 = 0 $. Then,\n$ x_{1} + x_{2} = -\\frac{36k^{2}}{5 + 9k^{2}} $, $ x_{1}x_{2} = \\frac{36k^{2} - 45}{5 + 9k^{2}} $. According to the definition of the ellipse, we derive $ |AB| = e(x_{1} + x_{2}) + 2a = \\frac{2}{3} \\times \\left(-\\frac{36k^{2}}{5 + 9k^{2}}\\right) + 6 = 5 $. Solving gives $ k^{2} = \\frac{1}{3} $, i.e., $ k = \\pm\\frac{\\sqrt{3}}{3} $. At this time, the line equations are $ y = \\pm\\frac{\\sqrt{3}}{3}(x + 2) $, that is, $ \\sqrt{3}x - 3y + 2\\sqrt{3} = 0 $ or $ -\\sqrt{3}x + 3y - 2\\sqrt{3} = 0 $. Also, the distances from point $ F_{2} $ to line AB are $ d_{1} = \\frac{|2\\sqrt{3} + 2\\sqrt{3}|}{\\sqrt{3 + 9}} = 2 $ or $ d_{2} = \\frac{|-2\\sqrt{3} - 2\\sqrt{3}|}{\\sqrt{3 + 9}} = 2 $. Therefore, the area of $ \\triangle ABF_{2} $ is $ S = \\frac{1}{2}|AB| \\cdot d = \\frac{1}{2} \\times 5 \\times 2 = 5 $."
  },
  {
    "text": "Given that the right focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{a}=1$ is $(\\sqrt{13} , 0)$, then the asymptotes of this hyperbola are?",
    "fact_expressions": "G: Hyperbola;a: Number;Expression(G) = (x^2/9 - y^2/a = 1);Coordinate(RightFocus(G)) = (sqrt(13), 0)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*(2/3)*x",
    "fact_spans": "[[[2, 40], [65, 68]], [[5, 40]], [[2, 40]], [[2, 62]]]",
    "query_spans": "[[[65, 76]]]",
    "process": ""
  },
  {
    "text": "Given that the equation of hyperbola $C$ is $\\frac{x^{2}}{8}-\\frac{y^{2}}{4}=1$, then the equations of the asymptotes of $C$ are?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/8 - y^2/4 = 1)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y = pm*(sqrt(2)/2)*x",
    "fact_spans": "[[[2, 8], [49, 52]], [[2, 47]]]",
    "query_spans": "[[[49, 60]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=8x$, a line passing through its focus $F$ intersects the parabola at points $A$ and $B$. A perpendicular from the midpoint $M$ of $AB$ to the $y$-axis meets the $y$-axis at point $N$. If $|MN|=2$, then $|AB|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);A: Point;B: Point;Intersection(H, G) = {A, B};M: Point;MidPoint(LineSegmentOf(A, B)) = M;Z: Line;PointOnCurve(M, Z);IsPerpendicular(Z, yAxis);N: Point;Intersection(Z, yAxis) = N;Abs(LineSegmentOf(M, N)) = 2",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 16], [17, 18], [27, 30]], [[2, 16]], [[20, 23]], [[17, 23]], [[24, 26]], [[16, 26]], [[31, 34]], [[35, 38]], [[24, 40]], [[49, 52]], [[42, 52]], [], [[41, 52]], [[41, 59]], [[65, 69]], [[41, 69]], [[72, 81]]]",
    "query_spans": "[[[84, 93]]]",
    "process": "As shown in the figure: $BB_{1}\\bot l$ at $B_{1}$, extend $MN$ to intersect $l$ at $D$, then $MD\\bot l$ at $D$. Since $MN\\bot y$-axis, $MD\\bot l$. In trapezoid $AA_{1}B_{1}B$, $M$ is the midpoint of $AB$, so $MD$ is the median line of trapezoid $AA_{1}B_{1}B$, $|AA_{1}|+|BB_{1}|=2|MD|=2(2+2)=8$. By the definition of parabola, $|AB|=|AA_{1}|+|BB_{1}|=8$."
  },
  {
    "text": "Given that the vertex of the parabola $C$ is at the origin and the equation of the directrix is $y=-2$, then the standard equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;O: Origin;Vertex(C) = O;Expression(Directrix(C)) = (y = -2)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2=8*y",
    "fact_spans": "[[[2, 8], [28, 34]], [[12, 14]], [[2, 14]], [[2, 26]]]",
    "query_spans": "[[[28, 41]]]",
    "process": "Let the standard equation of parabola C be: $x^{2}=2py$ ($p>0$), with focus at $(0,\\frac{p}{2})$ and directrix equation $y=-\\frac{p}{2}=-2$. It follows that: $p=4$, so the standard equation of parabola C is: $x^{2}=8y$"
  },
  {
    "text": "Given that $A$ and $F$ are the right vertex and the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ with eccentricity $2$, and denoting the distances from $A$ and $F$ to the line $b x - a y = 0$ as $d_{1}$ and $d_{2}$ respectively, then $\\frac{d_{1}}{d_{2}}$=?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;F: Point;A:Point;d1:Number;d2:Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (-a*y + b*x = 0);Eccentricity(G)=2;RightVertex(G)=A;RightFocus(G)=F;Distance(A, H) = d1;Distance(F,H)=d2",
    "query_expressions": "d1/d2",
    "answer_expressions": "1/2",
    "fact_spans": "[[[18, 64]], [[21, 64]], [[21, 64]], [[82, 95]], [[6, 9], [78, 81]], [[2, 5], [74, 77]], [[101, 108]], [[110, 118]], [[18, 64]], [[82, 95]], [[10, 64]], [[2, 72]], [[2, 72]], [[74, 118]], [[74, 118]]]",
    "query_spans": "[[[120, 143]]]",
    "process": "Let $ a>0, b>0 $, then the eccentricity of the hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ is $ e = \\frac{c}{a} = 2 $, so $ c = 2a $, $ b = \\sqrt{c^{2} - a^{2}} = \\sqrt{3}a $. From the given conditions, $ A(a,0) $, $ F(c,0) $, then $ d_{1} = \\frac{ab}{\\sqrt{a^{2} + b^{2}}} = \\frac{ab}{c} = \\frac{1}{2}b $, $ d_{2} = \\frac{bc}{\\sqrt{a^{2} + b^{2}}} = b $, therefore, $ \\frac{d_{1}}{d_{2}} = \\frac{1}{2} $."
  },
  {
    "text": "A line passing through the focus of the parabola $y^{2}=6x$ intersects the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $x_{1}+x_{2}=8$, then the value of $|AB|$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;Expression(G) = (y^2 = 6*x);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);PointOnCurve(Focus(G), H);Intersection(H, G) = {A,B};x1+x2=8;x1:Number;y1:Number;x2:Number;y2:Number",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "11",
    "fact_spans": "[[[1, 15], [22, 25]], [[19, 21]], [[26, 43]], [[46, 63]], [[1, 15]], [[26, 43]], [[46, 63]], [[0, 21]], [[19, 65]], [[68, 83]], [[26, 43]], [[26, 43]], [[46, 63]], [[45, 63]]]",
    "query_spans": "[[[86, 97]]]",
    "process": "The focus of the parabola $ y^{2} = 6x $ is $ \\left( \\frac{3}{2}, 0 \\right) $, then according to the problem, $ |AB| = x_{1} + x_{2} + p $. Since $ x_{1} + x_{2} = 8 $, $ p = 3 $, so $ |AB| = x_{1} + x_{2} + p = 8 + 3 = 11 $. Hence, the answer is: $ 11 $"
  },
  {
    "text": "Through the center of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, draw a line intersecting the ellipse at points $P$ and $Q$, and let $F$ be a focus of the ellipse. Then the minimum value of the perimeter of $\\triangle P Q F$ is?",
    "fact_expressions": "G: Ellipse;H: Line;P: Point;Q: Point;F: Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(Center(G), H);Intersection(H, G) = {P, Q};OneOf(Focus(G)) = F",
    "query_expressions": "Min(Perimeter(TriangleOf(P, Q, F)))",
    "answer_expressions": "18",
    "fact_spans": "[[[1, 40], [48, 50], [65, 67]], [[45, 47]], [[51, 54]], [[55, 58]], [[61, 64]], [[1, 40]], [[0, 47]], [[45, 60]], [[61, 72]]]",
    "query_spans": "[[[74, 99]]]",
    "process": "As shown in the figure, denote the other focus of the ellipse as $ F_{1} $. According to the symmetry of the ellipse, we know: $ |QF| = |PF_{1}| $, $ |PQ| = 2|PO| $. Let $ P(a\\cos\\theta, b\\sin\\theta) $, then $ |PO|^{2} = a^{2}\\cos^{2}\\theta + b^{2}\\sin^{2}\\theta = (a^{2} - b^{2})\\cos^{2}\\theta + b^{2} $. Since $ a^{2} - b^{2} > 0 $, $ \\cos2\\theta \\geqslant 0 $, it follows that $ |PO|^{2} \\geqslant b^{2} $, i.e., $ |PO| \\geqslant b $, and $ |PQ| = 2|PO| \\geqslant 2b $. Therefore, the perimeter of quadrilateral $ APQF $ is $ |QF| + |PF| + |PQ| = |PF_{1}| + |PF| + |PQ| = 2a + |PQ| \\geqslant 2a + 2b = 10 + 8 = 18 $. Hence, fill in $ 18 $."
  },
  {
    "text": "If the line $y=k x-2$ intersects the parabola $y^{2}=8 x$ at points $A$ and $B$, and the horizontal coordinate of the midpoint of segment $AB$ is $3$, then $|AB|$=?",
    "fact_expressions": "G: Parabola;H: Line;k: Number;B: Point;A: Point;Expression(G) = (y^2 = 8*x);Expression(H) = (y = k*x - 2);Intersection(H, G) = {A, B};XCoordinate(MidPoint(LineSegmentOf(A, B))) = 3",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "6",
    "fact_spans": "[[[13, 27]], [[1, 12]], [[3, 12]], [[33, 36]], [[29, 32]], [[13, 27]], [[1, 12]], [[1, 38]], [[40, 58]]]",
    "query_spans": "[[[60, 69]]]",
    "process": ""
  },
  {
    "text": "On the circle $x^{2}+y^{2}=8$, take any point $P$. Draw a perpendicular segment $P D$ from point $P$ to the $x$-axis, where $D$ is the foot of the perpendicular. As point $P$ moves along the circle, what is the equation of the trajectory of the midpoint $M$ of segment $P D$?",
    "fact_expressions": "G: Circle;P: Point;D: Point;M: Point;Expression(G) = (x^2 + y^2 = 8);PointOnCurve(P, G);PointOnCurve(P, LineSegmentOf(P, D));IsPerpendicular(LineSegmentOf(P, D), xAxis);FootPoint(LineSegmentOf(P, D), xAxis) = D;MidPoint(LineSegmentOf(P, D)) = M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2/8 + y^2/2 = 1",
    "fact_spans": "[[[1, 17], [61, 62]], [[22, 25], [27, 31], [56, 60]], [[48, 51]], [[77, 80]], [[1, 17]], [[1, 25]], [[26, 45]], [[32, 45]], [[32, 54]], [[67, 80]]]",
    "query_spans": "[[[77, 87]]]",
    "process": "Let M(x, y), then P(x, 2y); since point P moves on the circle x^{2}+y^{2}=8, we have x^{2}+(2y)^{2}=8, so the trajectory equation of the midpoint M of segment PD is \\frac{x^{2}}{8}+\\frac{y^{2}}{2}=1"
  },
  {
    "text": "If the line $y=kx$ ($k>0$) is an asymptote of the hyperbola $x^{2}-y^{2}=1$, then $k=$?",
    "fact_expressions": "G: Hyperbola;H: Line;k: Number;Expression(G) = (x^2 - y^2 = 1);k > 0;Expression(H) = (y = k*x);OneOf(Asymptote(G)) = H",
    "query_expressions": "k",
    "answer_expressions": "1",
    "fact_spans": "[[[16, 34]], [[1, 15]], [[42, 45]], [[16, 34]], [[3, 15]], [[1, 15]], [[1, 40]]]",
    "query_spans": "[[[42, 47]]]",
    "process": "\\because hyperbola x^{2}-y^{2}=1 \\therefore a=1, b=1 \\therefore asymptotes are y=\\pm\\frac{b}{a}x=\\pm x \\because line y=kx (k>0) is an asymptote of the hyperbola x^{2}-y^{2}=1 \\therefore k=1"
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4 x$ is $F$, and points $A$, $B$ are two points on the parabola with point $A$ located in the first quadrant. If $\\overrightarrow{A F}=4 \\overrightarrow{F B}$, then what is the general form of the equation of line $A B$?",
    "fact_expressions": "G: Parabola;B: Point;A: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(A, G);PointOnCurve(B, G);VectorOf(A, F) = 4*VectorOf(F, B);Quadrant(A)=1",
    "query_expressions": "Expression(LineOf(A,B))",
    "answer_expressions": "4*x-3*y-4=0",
    "fact_spans": "[[[2, 16], [33, 36]], [[29, 32]], [[24, 28], [41, 45]], [[20, 23]], [[2, 16]], [[2, 23]], [[24, 39]], [[24, 39]], [[54, 99]], [[41, 51]]]",
    "query_spans": "[[[101, 118]]]",
    "process": "Given that F(1,0), the slope of line AB is not 0. Let A(x_{1},y_{1}), B(x_{2},y_{2}) (y_{1}>0, y_{2}<0). The equation of line AB is x=my+1. Combining with the parabola equation yields y^{2}-4my-4=0, so y_{1}+y_{2}=4m, y_{1}y_{2}=-4. Also, \\overrightarrow{AF}=4\\overrightarrow{FB}, thus (1-x_{1},-y_{1})=4(x_{2}-1,y_{2}). Therefore, y_{1}=-4y_{2}, y_{1}y_{2}=-4y_{2}^{2}=-4, so y_{2}=-1, y_{1}=4, and hence y_{1}+y_{2}=3=4m, solving gives m=\\frac{3}{4}. Thus, the equation of AB is x=\\frac{3}{4}y+1, or 4x-3y-4=0."
  },
  {
    "text": "A line with slope $1$ intersects the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ at two points $A$ and $B$. The midpoint $M(m, 1)$ of $AB$, then $m=$?",
    "fact_expressions": "G: Ellipse;H:Line;A: Point;B: Point;M: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Coordinate(M) = (m, 1);Slope(H)=1;Intersection(H, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = M;m:Number",
    "query_expressions": "m",
    "answer_expressions": "-4/3",
    "fact_spans": "[[[10, 47]], [[7, 9]], [[50, 53]], [[54, 57]], [[68, 77]], [[10, 47]], [[68, 77]], [[0, 9]], [[7, 59]], [[60, 77]], [[79, 82]]]",
    "query_spans": "[[[79, 84]]]",
    "process": ""
  },
  {
    "text": "Let points $M$ and $N$ move on the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$, and let $F_{1}$, $F_{2}$ be the left and right foci of hyperbola $C$, respectively. Then the minimum value of $|\\overrightarrow{M F_{1}}+\\overrightarrow{M F_{2}}-2 \\overrightarrow{M N}|$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/4 - y^2/3 = 1);M: Point;N: Point;PointOnCurve(M, C);PointOnCurve(N, C);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2",
    "query_expressions": "Min(Abs(VectorOf(M, F1) + VectorOf(M, F2) - 2*VectorOf(M, N)))",
    "answer_expressions": "4",
    "fact_spans": "[[[11, 54], [74, 80]], [[11, 54]], [[1, 5]], [[6, 9]], [[1, 57]], [[1, 57]], [[58, 65]], [[66, 73]], [[58, 86]], [[58, 86]]]",
    "query_spans": "[[[88, 170]]]",
    "process": "\\because O is the midpoint of F_{1}F_{2}, \\therefore |\\overrightarrow{MF}_{1}+\\overrightarrow{MF_{2}}-2\\overrightarrow{MN}|=|2\\overrightarrow{MO}-2\\overrightarrow{MN}|=2|\\overrightarrow{NO}|\\geqslant2a=4"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has a focal distance of $4$, and point $A(2,3)$ lies on $C$, then the equations of the asymptotes of $C$ are?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;FocalLength(C) = 4;A: Point;Coordinate(A) = (2, 3);PointOnCurve(A, C)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*(\\sqrt{3})*x",
    "fact_spans": "[[[2, 63], [82, 85], [90, 93]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 70]], [[73, 81]], [[73, 81]], [[73, 88]]]",
    "query_spans": "[[[90, 101]]]",
    "process": "According to the problem, the coordinates of the two foci of hyperbola C can be determined. Using the definition, the value of $ a $ can be found. Combining with the focal distance of the hyperbola, the value of $ b $ can be obtained, thus yielding the asymptotic equations of hyperbola C. From the given conditions, the left focus of hyperbola C is $ F_{1}(-2,0) $, and the right focus is $ F_{2}(2,0) $. Since $ |AF_{1}| = \\sqrt{(2+2)^{2}+3^{2}} = 5 $, $ |AF_{2}| = \\sqrt{(2-2)^{2}+3^{2}} = 3 $, by the definition of a hyperbola, we have $ 2a = ||AF_{1}| - |AF_{2}|| = 2 $, so $ a = 1 $, $ b = \\sqrt{2^{2}-a^{2}} = \\sqrt{3} $. Therefore, the asymptotic equations of hyperbola C are $ y = \\pm\\sqrt{3}x $."
  },
  {
    "text": "Given the parabola $y^{2}=-4 x$ with focus $F$, point $P$ lies on the parabola, and point $A(-2,1)$. Find the coordinates of point $P$ that minimize the value of $|P F|+|P A|$.",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = -4*x);Coordinate(A) = (-2,1);Focus(G) = F;PointOnCurve(P, G);WhenMin(Abs(LineSegmentOf(P,F))+Abs(LineSegmentOf(P,A)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(-1/4, 1)",
    "fact_spans": "[[[2, 17], [30, 33]], [[35, 45]], [[25, 29], [67, 71]], [[21, 24]], [[2, 17]], [[35, 45]], [[2, 24]], [[25, 34]], [[49, 66]]]",
    "query_spans": "[[[67, 76]]]",
    "process": "Using the definition of a parabola, transform the distance |PF| from point P(m,n) to the focus F into the distance from P to the directrix l: x=1, and use the inequality to obtain the answer. Since the focus F of the parabola y^{2}=-4x is F(-1,0), its directrix equation is l: x=1. Since point P lies on the parabola and point A(-2,1) is given, let P' be the projection of point P onto the directrix l: x=1. Then |PF|=|PP'|, and |PF|+|PA|=|PA|+|PP'|\\geqslant|AP'|=3 (equality holds when A, P, P' are collinear). At this time, the ordinate of point P is n=1, so 1^{2}=-4m, solving gives m=-\\frac{1}{4}. Therefore, the coordinates of point P are (-\\frac{1}{4},1)."
  },
  {
    "text": "The equation of the circle with its center at the focus of the parabola $x^{2}=16 y$ and tangent to the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 16*y);H: Circle;Center(H) = Focus(G);IsTangent(Directrix(G), H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2+(y-4)^2=64",
    "fact_spans": "[[[1, 16], [25, 28]], [[1, 16]], [[34, 35]], [[1, 35]], [[24, 35]]]",
    "query_spans": "[[[34, 40]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, $M$ is a point on the ellipse, $N$ is the midpoint of $M F_{1}$, if $O N=1$, then $|M F_{1}|$=?",
    "fact_expressions": "G: Ellipse;M: Point;F1: Point;F2: Point;O: Origin;N: Point;Expression(G) = (x^2/16 + y^2/12 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(M, G);MidPoint(LineSegmentOf(M, F1)) = N;LineSegmentOf(O, N) = 1",
    "query_expressions": "Abs(LineSegmentOf(M, F1))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 41], [70, 72]], [[66, 69]], [[50, 57]], [[58, 65]], [[94, 101]], [[76, 79]], [[2, 41]], [[2, 65]], [[2, 65]], [[66, 75]], [[76, 92]], [[94, 101]]]",
    "query_spans": "[[[103, 116]]]",
    "process": "As shown in the figure, since N is the midpoint of MF, ON is the midline of \\triangleMF_{1}F_{2}, so MF_{2}=2. According to the definition of the ellipse, |MF|+|MF_{2}|=2a=8, so |MF|=8-2=6. Hence, fill in 6."
  },
  {
    "text": "Given the hyperbola $\\frac{y^{2}}{4}-x^{2}=1$, then its asymptotes are? The eccentricity $e$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2 + y^2/4 = 1);e: Number;Eccentricity(G) = e",
    "query_expressions": "Expression(Asymptote(G));e",
    "answer_expressions": "y=pm*2*x;sqrt(5)/2",
    "fact_spans": "[[[2, 30], [32, 33]], [[2, 30]], [[43, 46]], [[32, 46]]]",
    "query_spans": "[[[32, 40]], [[43, 48]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=3x$ with focus $F$, a line $l$ with slope $\\frac{3}{2}$ intersects $C$ at points $A$ and $B$, and intersects the $x$-axis at point $P$. If $|AF|+|BF|=4$, then the equation of $l$ is?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;F: Point;B: Point;Expression(C) = (y^2 = 3*x);Focus(C) = F;Slope(l)=3/2;Intersection(l, C) = {A, B};Intersection(l,xAxis)=P;Abs(LineSegmentOf(A,F))+Abs(LineSegmentOf(B,F))=4;P:Point",
    "query_expressions": "Expression(l)",
    "answer_expressions": "12*x-8*y-7=0",
    "fact_spans": "[[[46, 51], [98, 101]], [[2, 21], [52, 55]], [[59, 62]], [[25, 28]], [[63, 66]], [[2, 21]], [[2, 28]], [[29, 51]], [[46, 66]], [[46, 79]], [[81, 96]], [75, 77]]",
    "query_spans": "[[[98, 106]]]",
    "process": "By the given condition, the focus of the parabola $ C: y^{2} = 3x $ is $ F\\left(\\frac{3}{4}, 0\\right) $, and let the equation of line $ l $ be $ y = \\frac{3}{2}x + m $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. By the focal radius formula of the parabola, we have $ |AF| + |BF| = x_{1} + x_{2} + \\frac{3}{2} = 4 $, which gives $ x_{1} + x_{2} = \\frac{5}{2} $. Solving the system of equations\n$$\n\\begin{cases}\ny = \\frac{3}{2}x + m \\\\\ny^{2} = 3x\n\\end{cases}\n$$\nwe obtain $ 9x^{2} - (12m - 12)x + 4m^{2} = 0 $. Moreover, from $ \\Delta = (12m - 12)^{2} - 144m^{2} > 0 $, solving yields $ m < \\frac{1}{2} $, and $ x_{1} + x_{2} = -\\frac{12m - 12}{9} $. Therefore, $ -\\frac{12m - 12}{9} = \\frac{5}{2} $, solving gives $ m = -\\frac{7}{8} $. Thus, the equation of line $ l $ is $ y = \\frac{3}{2}x - \\frac{7}{8} $, or $ 12x - 8y - 7 = 0 $."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{k-1}+\\frac{y^{2}}{k-3}=1$ represents a hyperbola, then the range of real values for $k$ is?",
    "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (x^2/(k - 1) + y^2/(k - 3) = 1)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(1,3)",
    "fact_spans": "[[[44, 47]], [[49, 54]], [[1, 47]]]",
    "query_spans": "[[[49, 61]]]",
    "process": "The equation $\\frac{x2}{k-1}+\\frac{y^{2}}{k-3}=1$ represents a hyperbola if $(k-1)(k-3)<0$, therefore $1<k<3$."
  },
  {
    "text": "From the left focus $F$ of the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{5}=1$, draw a tangent line $FP$ to the circle $x^{2}+y^{2}=3$, intersecting the right branch of the hyperbola at point $P$, $T$ being the point of tangency, $M$ the midpoint of segment $FP$, and $O$ the origin. Then $| MO |-| MT |$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2/5 = 1);F: Point;LeftFocus(G) = F;H: Circle;Expression(H) = (x^2 + y^2 = 3);P: Point;T: Point;TangentOfPoint(F, H) = LineOf(F,P);Intersection(LineSegmentOf(F, P), RightPart(G)) = P;TangentPoint(LineOf(F, P), H) = T;M: Point;MidPoint(LineSegmentOf(F, P)) = M;O: Origin",
    "query_expressions": "Abs(LineSegmentOf(M, O)) - Abs(LineSegmentOf(M, T))",
    "answer_expressions": "sqrt(5)-sqrt(3)",
    "fact_spans": "[[[1, 39], [71, 74]], [[1, 39]], [[43, 46]], [[1, 46]], [[47, 63]], [[47, 63]], [[77, 81]], [[83, 86]], [[0, 70]], [[66, 81]], [[47, 89]], [[90, 93]], [[90, 103]], [[104, 107]]]",
    "query_spans": "[[[114, 131]]]",
    "process": ""
  },
  {
    "text": "If the equation $\\frac{x^{2}}{10-k}+\\frac{y^{2}}{k-4}=1$ represents an ellipse with foci on the $x$-axis, then the range of values for $k$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(10 - k) + y^2/(k - 4) = 1);PointOnCurve(Focus(G), xAxis);k: Number",
    "query_expressions": "Range(k)",
    "answer_expressions": "(4, 7)",
    "fact_spans": "[[[54, 56]], [[1, 56]], [[45, 56]], [[58, 61]]]",
    "query_spans": "[[[58, 68]]]",
    "process": "\\because the equation \\frac{x^2}{10-k}+\\frac{y^{2}}{k-4}=1 represents an ellipse with foci on the x-axis. \\therefore 10-k>k-4>0, solving gives 4<k<7."
  },
  {
    "text": "The line $y=\\sqrt{3} x$ intersects the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ at points $A$ and $B$, $|A B|=4 \\sqrt{3}$, and $F$ is the right focus of the ellipse. If $A F \\perp B F$, then $a=$?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;A: Point;F: Point;B: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y = sqrt(3)*x);Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, B)) = 4*sqrt(3);RightFocus(C) = F;IsPerpendicular(LineSegmentOf(A, F), LineSegmentOf(B, F))",
    "query_expressions": "a",
    "answer_expressions": "3+sqrt(3)",
    "fact_spans": "[[[17, 74], [108, 110]], [[24, 74]], [[133, 136]], [[0, 16]], [[75, 78]], [[104, 107]], [[79, 82]], [[24, 74]], [[24, 74]], [[17, 74]], [[0, 16]], [[0, 84]], [[85, 103]], [[104, 114]], [[116, 131]]]",
    "query_spans": "[[[133, 138]]]",
    "process": "As shown in the figure, connect $AF_{1}$, $BF_{1}$. Since $|OA|=|OB|$, $|OF_{1}|=|OF|$, the quadrilateral $AF_{1}BF$ is a parallelogram. Also, $AF\\perp BF$, so the quadrilateral $AF_{1}BF$ is a rectangle. Therefore, $\\angle F_{1}AF = \\frac{\\pi}{2}$, then $|OF_{1}| = |OF| = |OA| = 2\\sqrt{3}$. From the line $y = \\sqrt{3}x$, it follows that $\\angle AOF = \\frac{\\pi}{3}$, so $|AF| = |OF| = |OA| = 2\\sqrt{3}$. By the Pythagorean theorem, $|AF_{1}| = 6$. From the definition of the ellipse, $|AF| + |AF_{1}| = 2\\sqrt{3} + 6 = 2a$, so $a = 3 + \\sqrt{3}$."
  },
  {
    "text": "The equation of the directrix of the parabola $y=a x^{2}$ is $y=-1$. What is the value of $a$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = a*x^2);a: Number;Expression(Directrix(G)) = (y = -1)",
    "query_expressions": "a",
    "answer_expressions": "1/4",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[29, 32]], [[0, 26]]]",
    "query_spans": "[[[29, 36]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the focus of the parabola $y^{2}=2 p x$ ($p>0$), a line passing through point $F$ with slope $1$ intersects the parabola at points $A$ and $B$. If $|A F|-|B F|=\\sqrt{6}$, then the length of segment $A B$ is?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;B: Point;A: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(F, H);Slope(H) = 1;Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, F)) - Abs(LineSegmentOf(B, F)) = sqrt(6)",
    "query_expressions": "Length(LineSegmentOf(A,B))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[6, 27], [47, 50]], [[9, 27]], [[44, 46]], [[57, 60]], [[53, 56]], [[2, 5], [32, 36]], [[9, 27]], [[6, 27]], [[2, 30]], [[31, 46]], [[37, 46]], [[44, 62]], [[65, 87]]]",
    "query_spans": "[[[89, 100]]]",
    "process": "The focus of the parabola $ y^{2}=2px $ ($ p>0 $) is $ F(\\frac{p}{2},0) $, and the equation of the directrix is $ x=-\\frac{p}{2} $. Thus, the equation of line $ AB $ is: $ y=x- $—. Solving together with the parabola equation yields: $ \\frac{1}{2}x_{1}+x_{2}=3p $, $ x_{1}x_{2}=\\frac{p^{2}}{4} $. Since $ |AF|-|BF|=\\sqrt{6} $, we have $ x_{1}-(-\\frac{p}{2})-[x_{2}-(-\\frac{p}{2})]=\\sqrt{6} \\Rightarrow x_{1}-x_{2}=\\sqrt{6} \\Rightarrow \\sqrt{(x_{1}+x_{2})}\\frac{1}{2}-_{2}=\\sqrt{6} $, that is, $ 9p^{2}-4\\cdot\\frac{p^{2}}{4}=6 \\Rightarrow p=\\frac{\\sqrt{3}}{2} $. Therefore, $ AB=x_{1}-(-\\frac{p}{2})+x_{2}-(-\\frac{p}{2})=x_{1}+x_{2}+p=4p=2\\sqrt{3} $."
  },
  {
    "text": "The equation of the directrix of the parabola $y=a x^{2}$ is $y-2=0$, then the value of $a$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = a*x^2);a: Number;Expression(Directrix(G)) = (y - 2 = 0)",
    "query_expressions": "a",
    "answer_expressions": "-1/8",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[29, 32]], [[0, 27]]]",
    "query_spans": "[[[29, 36]]]",
    "process": "The parabolic equation $ y = ax^{2} $ can be rewritten as $ x^{2} = \\frac{1}{a}y $. Since the equation of the directrix of the parabola is $ x - 2 = 0 $, it follows that $ -\\frac{1}{4a} = 2 $. Solving gives $ a = -\\frac{1}{8} $."
  },
  {
    "text": "If the distance from a moving point $P$ to the fixed point $F(1,1)$ is equal to the distance from the moving point $P$ to the line $l$: $3x + y - 4 = 0$, then what is the equation of the trajectory of the moving point $P$?",
    "fact_expressions": "l: Line;F: Point;P: Point;Expression(l) = (3*x + y - 4 = 0);Coordinate(F) = (1, 1);Distance(P, F) = Distance(P, l)",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x-3*y+2=0",
    "fact_spans": "[[[27, 45]], [[9, 17]], [[3, 6], [23, 26], [54, 57]], [[27, 45]], [[9, 17]], [[3, 50]]]",
    "query_spans": "[[[54, 64]]]",
    "process": "This problem examines the definition and equation of a parabola, primarily focusing on accurately applying the definition of a parabola. The key is to first determine that the point lies on a line, then conclude that $ P $ lies on the line passing through the fixed point $ F $ and perpendicular to the given line. Thus, the trajectory equation of $ P $ can be obtained using the perpendicular relationship between lines. Since the fixed point $ F(1,1) $ lies on the line $ l: 3x + y - 4 = 0 $, the locus of points equidistant from the fixed point $ F $ and the fixed line $ l $ is a line—specifically, the line passing through $ F $ and perpendicular to $ l: 3x + y - 4 = 0 $. Therefore, the trajectory equation of the moving point $ P $ is $ y - 1 = \\frac{1}{3}(x - 1) $, or $ x - 3y + 2 = 0 $."
  },
  {
    "text": "Given that a line passing through the focus $F$ of the parabola $x^{2}=8y$ intersects the parabola at points $A$ and $B$, and $|AF|=4$, then $|BF|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 8*y);F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);A: Point;B: Point;Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, F)) = 4",
    "query_expressions": "Abs(LineSegmentOf(B, F))",
    "answer_expressions": "4",
    "fact_spans": "[[[3, 17], [28, 31]], [[3, 17]], [[20, 23]], [[3, 23]], [[24, 26]], [[2, 26]], [[32, 35]], [[36, 39]], [[24, 41]], [[42, 51]]]",
    "query_spans": "[[[53, 62]]]",
    "process": ""
  },
  {
    "text": "The line $l$ passing through the point $M(-1,1)$ intersects the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1$ at points $A$ and $B$. If $M$ is exactly the midpoint of segment $AB$, then the equation of line $l$ is?",
    "fact_expressions": "l: Line;G: Ellipse;B: Point;A: Point;M: Point;Expression(G) = (x^2/3 + y^2/2 = 1);Coordinate(M) = (-1, 1);PointOnCurve(M, l);Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A,B)) = M",
    "query_expressions": "Expression(l)",
    "answer_expressions": "2*x - 3*y + 5 = 0",
    "fact_spans": "[[[12, 17], [84, 89]], [[18, 55]], [[60, 63]], [[56, 59]], [[1, 11], [67, 70]], [[18, 55]], [[1, 11]], [[0, 17]], [[12, 65]], [[67, 82]]]",
    "query_spans": "[[[84, 94]]]",
    "process": ""
  },
  {
    "text": "A line $l$ with slope $\\frac{1}{3}$ passes through the left focus $F_{1}$ of hyperbola $C$ and intersects the left and right branches of hyperbola $C$ at points $A$ and $B$, respectively. If the perpendicular bisector of segment $AB$ passes through the right focus $F_{2}$ of hyperbola $C$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;F1: Point;LeftFocus(C) = F1;PointOnCurve(F1, l);Slope(l) = 1/3;B: Point;A: Point;Intersection(l, LeftPart(C)) = A;Intersection(l, RightPart(C)) = B;F2: Point;RightFocus(C) = F2;PointOnCurve(F2, PerpendicularBisector(LineSegmentOf(A, B)))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[36, 41]], [[1, 7], [42, 48], [78, 84], [98, 104]], [[11, 18]], [[1, 18]], [[0, 41]], [[19, 41]], [[58, 61]], [[54, 57]], [[36, 63]], [[36, 63]], [[88, 95]], [[78, 95]], [[65, 95]]]",
    "query_spans": "[[[98, 110]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, the left and right vertices are $A$ and $B$ respectively. Point $P$ lies on the ellipse and is distinct from $A$ and $B$. Lines $AP$ and $BP$ intersect the line $x=10$ at points $M$ and $N$ respectively. Then the minimum length of segment $MN$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);A: Point;B: Point;LeftVertex(G) = A;RightVertex(G) = B;P: Point;PointOnCurve(P, G);Negation(P = A);Negation(P = B);H: Line;Expression(H) = (x = 10);M: Point;Intersection(LineOf(A,P),H)=M;N: Point;Intersection(LineOf(B,P),H)=N",
    "query_expressions": "Min(LineSegmentOf(M,N))",
    "answer_expressions": "6*sqrt(3)",
    "fact_spans": "[[[2, 40], [62, 64]], [[2, 40]], [[49, 52], [67, 71]], [[53, 56], [72, 75]], [[2, 56]], [[2, 56]], [[57, 61]], [[57, 65]], [[57, 71]], [[57, 75]], [[91, 99]], [[91, 99]], [[103, 106]], [[76, 110]], [[107, 110]], [[76, 110]]]",
    "query_spans": "[[[113, 126]]]",
    "process": "From the given conditions, we have: A(-5,0), B(5,0). Since point P lies on the ellipse and is distinct from points A and B, let P(x_{0},y_{0}) (x_{0}\\neq\\pm5), so \\frac{x_{0}^{2}}{25}+\\frac{y_{0}^{2}}{9}=1. Thus, we obtain: k_{PA}\\cdot k_{PB} = \\frac{y_{0}}{x_{0}+5} \\cdot \\frac{y_{0}}{x_{0}-5} = \\frac{\\frac{9}{25}(25-x_{0}^{2})}{x_{0}^{2}-25} = -\\frac{9}{25}. The equation of line AP is: y = k_{PA}(x+5). Letting x=10, the coordinates of M are: (10, 15k_{PA}). The equation of line BP is: y = k_{PB}(x-5). Letting x=10, the coordinates of N are: (10, 5k_{PB}). Then, MN = |15k_{PA} - 5k_{PB}| = 5\\left|3k_{PA} + \\frac{9}{25k_{PA}}\\right| = 5\\left(3|k_{PA}| + \\left|\\frac{9}{25k_{PA}}\\right|\\right) \\geqslant 5 \\cdot 2\\sqrt{3 \\cdot |k_{PA}| \\cdot \\left|\\frac{9}{25k_{PA}}\\right|}, with equality holding if and only if 3|k_{PA}| = \\left|\\frac{9}{25k_{PA}}\\right|, that is, when k_{PA} = \\frac{\\sqrt{3}}{5}, k_{PB} = -\\frac{3\\sqrt{3}}{5} or k_{PA} = -\\frac{\\sqrt{3}}{5}, k_{PB} = \\frac{3\\sqrt{3}}{5}."
  },
  {
    "text": "Given that the midpoint of the chord cut by line $l$ on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{16}=1$ is $(\\frac{1}{2}, 1)$, find the equation of line $l$?",
    "fact_expressions": "l: Line;G: Ellipse;Expression(G) = (x^2/4 + y^2/16 = 1);Coordinate(MidPoint(InterceptChord(l,G))) = (1/2, 1)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "2*x+y-2=0",
    "fact_spans": "[[[41, 46], [75, 80]], [[2, 40]], [[2, 40]], [[2, 73]]]",
    "query_spans": "[[[75, 84]]]",
    "process": "Substituting the coordinates of the midpoint into the ellipse equation, we obtain $\\frac{1}{16}+\\frac{1}{16}<1$, indicating that line $l$ intersects the ellipse. Let the intersection points of the line and the ellipse be $A(x_{1},y_{1})$, $B(x_{2},y_{2})$. Then we have $\\frac{x_{1}^2}{4}+\\frac{y_{1}^2}{16}=1$, $\\frac{x_{2}^2}{4}+\\frac{y_{2}^2}{16}=1$. Subtracting these two equations gives $\\frac{(x_{1}-x_{2})(x_{1}+x_{2})}{4}+\\frac{(y_{1}-y_{2})(y_{1}+y_{2})}{16}=0$. From the midpoint formula, we get $x_{1}+x_{2}=1$, $y_{1}+y_{2}=2$. Substituting into the above equation, the slope of line $l$ is $k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{4(x_{1}+x_{2})}{y_{1}+y_{2}}=-2$. Thus, the equation of line $l$ is $y-1=-2(x-\\frac{1}{2})$, which simplifies to $2x+y-2=0$."
  },
  {
    "text": "Given that the distance from a point $A(3, b)$ in the first quadrant on the parabola $y^{2}=2 p x$ $(p>0)$ to the focus $F$ of the parabola is $4$, if $P$ is an arbitrary point on the directrix of the parabola, then when the perimeter of $\\Delta P A F$ is minimized, what is the distance from point $P$ to the line $A F$?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;F: Point;P: Point;p>0;Expression(G) = (y^2 = 2*p*x);Coordinate(A) = (3, b);Quadrant(A)=1;Focus(G)=F;Distance(A,F)=4;PointOnCurve(P, Directrix(G));WhenMin(Perimeter(TriangleOf(P,A,F)));b:Number",
    "query_expressions": "Distance(P,LineOf(A,F))",
    "answer_expressions": "4*sqrt(3)/3",
    "fact_spans": "[[[2, 23], [42, 45], [63, 66]], [[5, 23]], [[32, 41]], [[47, 50]], [[59, 62], [97, 101]], [[5, 23]], [[2, 23]], [[32, 41]], [[24, 41]], [[42, 50]], [[32, 57]], [[59, 73]], [[75, 96]], [[32, 41]]]",
    "query_spans": "[[[97, 114]]]",
    "process": "From the given information and the definition of a parabola, the distance from point A to the directrix is 4, so we have $3+\\frac{p}{2}=4$, solving gives $p=2$. Thus, the equation of the parabola is $y^{2}=4x$, and hence $A(3,2\\sqrt{3})$. When the perimeter of $\\triangle PAF$ is minimized, i.e., the value of $|PA|+|PF|$ is minimized, let $F_1$ be the symmetric point of focus $F$ with respect to the directrix, then $F_1(-3,0)$. Connect $AF_1$, the intersection point of line $AF_1$ and the directrix is the point $P$ that minimizes $|PA|+|PF|$, at this time $P\\left(-1,\\frac{2\\sqrt{3}}{2}\\right)$. Since $k_{AF}=\\frac{2\\sqrt{3}-0}{3-1}=\\sqrt{3}$, the equation of line $AF$ is $y-0=\\sqrt{3}(x-1)$, i.e., $\\sqrt{3}x-y-\\sqrt{3}=0$. Therefore, the distance $d$ from point $P$ to line $AF$ is $d=\\frac{|-\\sqrt{3}-\\frac{2\\sqrt{3}}{3}-\\sqrt{3}|}{\\sqrt{(\\sqrt{3})^{2}+1}}=\\frac{4\\sqrt{3}}{3}$."
  },
  {
    "text": "Let $A$ and $B$ be two moving points on the parabola $x^{2}=4 y$, with the length of segment $AB$ being $6$, and let $M$ be the midpoint of segment $AB$. Then the shortest distance from point $M$ to the $x$-axis is?",
    "fact_expressions": "A: Point;B: Point;G: Parabola;Expression(G) = (x^2 = 4*y);PointOnCurve(A, G);PointOnCurve(B, G);Length(LineSegmentOf(A,B)) = 6;MidPoint(LineSegmentOf(A,B)) = M;M: Point",
    "query_expressions": "Min(Distance(M, xAxis))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 4]], [[5, 8]], [[9, 23]], [[9, 23]], [[1, 28]], [[1, 28]], [[30, 43]], [[46, 60]], [[46, 49], [62, 66]]]",
    "query_spans": "[[[62, 78]]]",
    "process": "AA'\\bot x\\text{-axis}, BB'\\bot x\\text{-axis}. When line AB does not pass through focus F, points A, B, F form triangle ABF. At this time, the distance from point M to the x-axis is d=\\frac{|AA'|+|BB'|}{2}, and |AA'|=|AF|-1, |BB'|=|BF|-1, while |AF|+|BF|>|AB|=6, so d=\\frac{|AF|+|BF|-2}{2}>2. When line AB passes through focus F, points A, B, F are collinear, and the distance from point M to the x-axis is d=\\frac{|AA'|+|BB'|}{2}=\\frac{|AF|+|BF|-2}{2}=2. Therefore, the shortest distance from point M to the x-axis is 2. Key point: geometric properties of parabola"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has eccentricity $\\sqrt{5}$, and the distance from the focus of $C$ to its asymptote is $5$, then $a=?$",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(C) = sqrt(5);Distance(Focus(C), Asymptote(C)) = 5",
    "query_expressions": "a",
    "answer_expressions": "5/2",
    "fact_spans": "[[[2, 63], [80, 83], [87, 88]], [[2, 63]], [[10, 63]], [[100, 103]], [[10, 63]], [[10, 63]], [[2, 78]], [[80, 98]]]",
    "query_spans": "[[[100, 105]]]",
    "process": "From the given information, $ e = \\frac{c}{a} = \\sqrt{5} $. Let one asymptote of the hyperbola be $ y = \\frac{b}{a}x $, that is, $ bx - ay = 0 $. Combining with the point-to-line distance formula, we obtain $ d = \\frac{bc}{\\sqrt{a^{2} + b^{2}}} = \\frac{bc}{c} = b $, so $ b = 5 $. Since $ c^{2} = a^{2} + b^{2} $, solving simultaneously gives $ a = \\frac{5}{2} $."
  },
  {
    "text": "If the line $2x + 4y + m = 0$ passes through the focus of the parabola $y = 2x^2$, then $m =$?",
    "fact_expressions": "L: Line;Expression(L) = (m + 2*x + 4*y = 0);m: Number;G: Parabola;Expression(G) = (y = 2*x^2);PointOnCurve(Focus(G), L) = True",
    "query_expressions": "m",
    "answer_expressions": "-1/2",
    "fact_spans": "[[[1, 16]], [[1, 16]], [[37, 40]], [[18, 32]], [[18, 32]], [[1, 35]]]",
    "query_spans": "[[[37, 42]]]",
    "process": "From the equation of the parabola, the coordinates of the focus can be obtained. Substituting into the line equation gives the value of $ m $. $ y = 2x^{2} $ can be rewritten as $ x^{2} = \\frac{1}{2}y $, and the focus coordinates are $ \\left(0, \\frac{1}{8}\\right) $. According to the problem: $ 2 \\times 0 + 4 \\times \\frac{1}{8} + m = 0 $, thus $ m = -\\frac{1}{2} $."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. A line $l$ passing through the right focus intersects the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ at points $A$ and $B$, and intersects the $y$-axis at point $M$, satisfying $\\overrightarrow{A F_{2}}=3 \\overrightarrow{F_{2} B}$ and $\\overrightarrow{M A}=\\overrightarrow{A F_{2}}$. Then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;l: Line;PointOnCurve(RightFocus(G), l);A: Point;B: Point;Intersection(l, G) = {A, B};M: Point;Intersection(l, yAxis) = M;VectorOf(A, F2) = 3*VectorOf(F2, B);VectorOf(M, A) = VectorOf(A, F2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[17, 69], [87, 139], [269, 271]], [[17, 69]], [[19, 69]], [[19, 69]], [[19, 69]], [[19, 69]], [[1, 8]], [[9, 16]], [[1, 75]], [[1, 75]], [[81, 86]], [[17, 86]], [[141, 144]], [[145, 148]], [[81, 150]], [[158, 162]], [[81, 162]], [[166, 219]], [[220, 267]]]",
    "query_spans": "[[[269, 277]]]",
    "process": "From $\\overrightarrow{MA}=\\overrightarrow{AF_{2}}$ we know that $A$ is the midpoint of $MF_{2}$, therefore $x_{A}=\\frac{c}{2}$, then $y_{A}=\\frac{b\\sqrt{4a^{2}-c^{2}}}{2a}$, further from $\\overrightarrow{AF}_{2}=3\\overrightarrow{F_{2}B}$ we get $x_{B}=\\frac{7c}{6}$, $y_{B}=-\\frac{b\\sqrt{4a^{2}-c^{2}}}{6a}$, so $\\frac{49c^{2}}{36a^{2}}+\\frac{4a^{2}-c^{2}}{36a^{2}}=1$, $\\frac{c^{2}}{a^{2}}=\\frac{2}{3}$, $e=\\frac{c}{a}=\\frac{\\sqrt{6}}{3}$."
  },
  {
    "text": "The equation of the locus of the center of a moving circle that is externally tangent to circle $C_{1}$: $(x+3)^{2}+y^{2}=9$ and internally tangent to circle $C_{2}$: $(x-3)^{2}+y^{2}=1$ is?",
    "fact_expressions": "C: Circle;C1: Circle;Expression(C1) = ((x + 3)^2 + y^2 = 9);IsOutTangent(C,C1) = True;C2: Circle;Expression(C2) = ((x - 3)^2 + y^2 = 1);IsInTangent(C,C2) = True",
    "query_expressions": "LocusEquation(Center(C))",
    "answer_expressions": "x^2/4 - y^2/5 = 1 & (x >= 2)",
    "fact_spans": "[[[64, 66]], [[1, 29]], [[1, 29]], [[0, 66]], [[33, 61]], [[33, 61]], [[32, 66]]]",
    "query_spans": "[[[64, 74]]]",
    "process": "Let the center of the moving circle be $ P(x, y) $, with radius $ r $. According to the problem, $ C_{1}(-3, 0) $, $ r_{1} = 3 $; $ C_{2}(3, 0) $, $ r_{2} = 1 $. The moving circle is externally tangent to circle $ C_{1} $ and internally tangent to circle $ C_{2} $, so we have $ |PC_{1}| = r + 3 $, $ |PC_{2}| = r - 1 $. Subtracting these two equations gives $ |PC_{1}| - |PC_{2}| = 4 $, thus the center of the moving circle lies on the right branch of a hyperbola with foci $ C_{1} $ and $ C_{2} $. Here, $ a = 2 $, $ c = 3 $, $ b^{2} = c^{2} - a^{2} = 5 $, hence the trajectory equation is $ \\frac{x^{2}}{4} - \\frac{y^{2}}{5} = 1 $ ($ x \\geqslant 2 $)."
  },
  {
    "text": "The parabola $C$: $y = a x^{2}$ ($a > 0$) passes through the point $(4 , 2)$, then the coordinates of the focus of parabola $C$ are?",
    "fact_expressions": "C: Parabola;Expression(C) = (y=a*x^2);a: Number;a>0;P: Point;Coordinate(P) = (4,2);PointOnCurve(P,C) = True",
    "query_expressions": "Coordinate(Focus(C))",
    "answer_expressions": "(0,2)",
    "fact_spans": "[[[0, 24], [37, 43]], [[0, 24]], [[7, 24]], [[7, 24]], [[25, 35]], [[25, 35]], [[0, 35]]]",
    "query_spans": "[[[37, 50]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{1}$ and perpendicular to the $x$-axis intersects the ellipse at points $A$ and $B$. The line $A F_{2}$ intersects the ellipse again at point $C$. If $S_{\\triangle A B C}=4 S_{\\triangle B C F_{2}}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;a: Number;b: Number;H: Line;F2: Point;A: Point;B: Point;C: Point;F1: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, H);IsPerpendicular(H, xAxis);Intersection(H, G) = {A, B};Intersection(LineOf(A,F2),G)={A,C};Area(TriangleOf(A, B, C)) = 4*Area(TriangleOf(B, C, F2));Negation(A=C)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(1/3)*sqrt(3)",
    "fact_spans": "[[[2, 54], [98, 100], [124, 126], [187, 189]], [[4, 54]], [[4, 54]], [[95, 97]], [[70, 77]], [[102, 105]], [[106, 109]], [[133, 136]], [[62, 69], [79, 86]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 77]], [[2, 77]], [[78, 97]], [[87, 97]], [[95, 111]], [[112, 137]], [[138, 185]], [110, 133]]",
    "query_spans": "[[[187, 195]]]",
    "process": "Let the left and right foci of the ellipse be $ F_{1}(-c,0) $, $ F_{2}(c,0) $. Substituting $ x = -c $ into the ellipse equation yields $ y = \\pm\\frac{b^{2}}{a} $, so we can set $ A(-c,\\frac{b^{2}}{a}) $, $ C(x,y) $. From $ S_{\\triangle ABC} = 4S_{\\triangle BCF_{2}} $, we obtain $ \\overrightarrow{AF_{2}} = 3\\overrightarrow{F_{2}C} $, that is, $ (2c,-\\frac{b^{2}}{a}) = 3(x-c,y) $. Thus, $ 2c = 3x - 3c $, $ -\\frac{b^{2}}{a} = 3y $. Solving gives $ x = \\frac{5c}{3} $, $ y = -\\frac{b^{2}}{3a} $. Substituting into the ellipse equation yields $ \\frac{25c^{2}}{9a^{2}} + \\frac{b^{2}}{9a^{2}} = 1 $, i.e., $ 25c^{2} + b^{2} = 9a^{2} $. Since $ b^{2} = a^{2} - c^{2} $, it follows that $ a^{2} = 3c^{2} $, so $ a = \\sqrt{3}c $. Therefore, $ e = \\frac{c}{a} = \\frac{\\sqrt{3}}{3} $. Solving gives $ e = \\frac{\\sqrt{3}}{3} $."
  },
  {
    "text": "Let $m$ be a constant. If $F(0,5)$ is a focus of the hyperbola $\\frac{y^{2}}{m}-\\frac{x^{2}}{9}=1$, then $m=$?",
    "fact_expressions": "G: Hyperbola;m: Number;F: Point;Expression(G) = (-x^2/9 + y^2/m = 1);Coordinate(F) = (0, 5);OneOf(Focus(G)) = F",
    "query_expressions": "m",
    "answer_expressions": "16",
    "fact_spans": "[[[18, 56]], [[1, 4], [63, 66]], [[9, 17]], [[18, 56]], [[9, 17]], [[9, 61]]]",
    "query_spans": "[[[63, 68]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x$ coincides with the right focus of the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{3}=1$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;H: Ellipse;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (x^2/12 + y^2/3 = 1);Focus(G) = RightFocus(H)",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 17]], [[67, 70]], [[21, 59]], [[1, 17]], [[21, 59]], [[1, 65]]]",
    "query_spans": "[[[67, 74]]]",
    "process": ""
  },
  {
    "text": "Draw a line $l$ through the focus $F$ of the parabola $C$: $y^{2}=4x$, intersecting the parabola $C$ at points $A$ and $B$. If the distance from $A$ to the directrix of the parabola is $4$, then $|AB|=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, C) = {A, B};Distance(A, Directrix(C)) = 4",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[1, 20], [33, 39], [55, 58]], [[1, 20]], [[23, 26]], [[1, 26]], [[27, 32]], [[0, 32]], [[40, 43], [51, 54]], [[44, 47]], [[27, 49]], [[51, 68]]]",
    "query_spans": "[[[70, 79]]]",
    "process": "\\because y^2=4x, \\therefore the directrix of the parabola is x=-1, F(1,0), and the distance from A to the directrix of the parabola is 4, x_{A}+1=4, \\therefore x_{A}=3, \\because x_{A}x_{B}=\\frac{p^{2}}{4}=1, \\therefore x_{B}=\\frac{1}{3}, \\therefore |AB|=x_{A}+x_{B}+p=3+\\frac{1}{3}+2=\\frac{16}{3}."
  },
  {
    "text": "Let the right focus of the ellipse $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{n^{2}}=1$ $(m>0, n>0)$ be $F(2,0)$, and the eccentricity be $\\frac{1}{2}$. Then the equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;m: Number;n: Number;F: Point;m>0;n>0;Expression(G) = (y^2/n^2 + x^2/m^2 = 1);Coordinate(F) = (2, 0);RightFocus(G) = F;Eccentricity(G) = 1/2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16 + y^2/12 = 1",
    "fact_spans": "[[[1, 56], [90, 92]], [[3, 56]], [[3, 56]], [[61, 69]], [[3, 56]], [[3, 56]], [[1, 56]], [[61, 69]], [[1, 69]], [[1, 87]]]",
    "query_spans": "[[[90, 97]]]",
    "process": "Since the right focus of the ellipse $\\frac{x^{2}}{m^2}+\\frac{y^{2}}{n^2}=1$ $(m>0,n>0)$ is $F(2,0)$, the foci of the ellipse lie on the x-axis and $c=2$. Because the eccentricity is $\\frac{1}{2}$, we have $\\frac{c}{m}=\\frac{1}{2}$, solving gives $m=4$, so $n^{2}=m^{2}-c^{2}=16-4=12$. Therefore, the equation of the ellipse is $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, and a line passing through the focus intersects the parabola at points $A$ and $B$, for what slope of the line does $|AF|+4|BF|$ attain its minimum value?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Line;A: Point;F: Point;B: Point;Focus(G) = F;PointOnCurve(F,H);Intersection(H, G) = {A, B};WhenMin(Abs(LineSegmentOf(A, F)) + 4*Abs(LineSegmentOf(B, F)))",
    "query_expressions": "Slope(H)",
    "answer_expressions": "pm*2*sqrt(2)",
    "fact_spans": "[[[2, 16], [31, 34]], [[2, 16]], [[47, 49], [28, 30]], [[36, 39]], [[20, 23]], [[40, 43]], [[2, 23]], [[2, 30]], [[28, 45]], [[56, 75]]]",
    "query_spans": "[[[47, 54]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=16 x$, $M$ is a point on $C$, and the extension of $F M$ intersects the $y$-axis at point $N$. If $M$ is the midpoint of $F N$, then the value of $|F N|$ is?",
    "fact_expressions": "C: Parabola;F: Point;M: Point;N: Point;Expression(C) = (y^2 = 16*x);Focus(C) = F;PointOnCurve(M, C);Intersection(OverlappingLine(LineSegmentOf(F,M)),yAxis)=N;MidPoint(LineSegmentOf(F, N)) = M",
    "query_expressions": "Abs(LineSegmentOf(F, N))",
    "answer_expressions": "12",
    "fact_spans": "[[[6, 26], [34, 37]], [[2, 5]], [[30, 33], [62, 65]], [[56, 60]], [[6, 26]], [[2, 29]], [[30, 40]], [[41, 60]], [[62, 74]]]",
    "query_spans": "[[[76, 87]]]",
    "process": "From the parabolic equation, obtain the coordinates of the focus. Let the coordinates of point N be (0,t). Using the midpoint formula, find the coordinates of point M, substitute into the parabolic equation and simplify, then compute the value of |FN|. [Detailed Solution] According to the given conditions, the focus of the parabola is F(4,0). Let N be (0,t). By the midpoint formula, M has coordinates (2,\\frac{t}{2}). Substituting into the parabolic equation yields (\\frac{t}{2})^{2}=32, that is, t^{2}=128. Therefore, |FN|=\\sqrt{16+t^{2}}=\\sqrt{16+128}=12. [Comment] This question mainly examines the relationship between a line and a parabola, as well as the midpoint coordinate formula and the distance formula between two points, belonging to basic problems."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, if there exists a line $AB$ passing through point $F_{1}$ intersecting the ellipse at points $A$ and $B$ such that $2|A F_{1}|=3|F_{1} B|$ and $|B F_{2}|=2|B F_{1}|$, then the eccentricity of this ellipse is?",
    "fact_expressions": "F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;PointOnCurve(F1,LineOf(A,B)) = True;Intersection(LineOf(A,B),G) = {A,B};A: Point;B: Point;2*Abs(LineSegmentOf(A, F1)) = 3*Abs(LineSegmentOf(F1, B));Abs(LineSegmentOf(B, F2)) = 2*Abs(LineSegmentOf(B, F1))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "Answer missing",
    "fact_spans": "[[[2, 9], [80, 88]], [[10, 17]], [[2, 75]], [[2, 75]], [[18, 70], [97, 99], [161, 163]], [[18, 70]], [[20, 70]], [[20, 70]], [[20, 70]], [[20, 70]], [[79, 96]], [[89, 109]], [[100, 103]], [[104, 107]], [[112, 135]], [[136, 158]]]",
    "query_spans": "[[[161, 169]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $E$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, with left and right foci $F_{1}$, $F_{2}$ respectively, $P$ a moving point on the ellipse, and $Q$ a moving point on the circle $M$: $x^{2}+y^{2}-10 x-8 y+40=0$, then the maximum value of $|P F_{1}|+|P Q|$ is?",
    "fact_expressions": "E: Ellipse;M: Circle;P: Point;F1: Point;F2: Point;Q: Point;Expression(E) = (x^2/9 + y^2/5 = 1);Expression(M) = (-8*y - 10*x + x^2 + y^2 + 40 = 0);PointOnCurve(Q, M);LeftFocus(E) = F1;RightFocus(E) = F2;PointOnCurve(P, E)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "12",
    "fact_spans": "[[[2, 44], [73, 75]], [[85, 118]], [[69, 72]], [[53, 60]], [[61, 68]], [[81, 84]], [[2, 44]], [[85, 118]], [[81, 123]], [[2, 68]], [[2, 68]], [[69, 80]]]",
    "query_spans": "[[[125, 148]]]",
    "process": "From the given conditions: $ F_{1}(-2,0), F_{2}(2,0) $, by the definition of an ellipse, $ |PF_{1}| + |PF_{2}| = 2a = 6 $, so $ |PF_{1}| = 6 - |PF_{2}| $. The circle $ M: x^{2} + y^{2} - 10x - 8y + 40 = 0 $ can be rewritten as $ (x-5)^{2} + (y-4)^{2} = 1 $, which means the center is $ M(5,4) $ and the radius $ r = 1 $. To maximize $ |PF_{1}| + |PQ| $, we need to maximize $ |PF_{1}| + |PM| + r $. Since $ |PF_{1}| + |PM| = 6 - |PF_{2}| + |PM| $, it suffices to maximize $ |PM| - |PF_{2}| $. As shown in the figure, when $ P, F_{2}, M $ are collinear, $ |PM| - |PF_{2}| $ reaches its maximum value of $ |F_{2}M| = \\sqrt{(5-2)^{2} + 4^{2}} = 5 $. Therefore, the maximum value of $ |PF_{1}| + |PM| = 6 - |PF_{2}| + |PM| $ is $ 5 + 6 = 11 $. Hence, the maximum value of $ |PF_{1}| + |PQ| $, i.e., the maximum value of $ |PF_{1}| + |PM| + r $, is $ 11 + 1 = 12 $."
  },
  {
    "text": "Given $\\frac{1}{m}+\\frac{2}{n}=1$ $(m>0, n>0)$, when $m n$ attains its minimum value, what is the eccentricity of the ellipse $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{n^{2}}=1$?",
    "fact_expressions": "G: Ellipse;m: Number;n: Number;m > 0;n > 0;Expression(G) = (y^2/n^2 + x^2/m^2 = 1);2/n + 1/m = 1;WhenMin(m*n)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[54, 99]], [[42, 47]], [[42, 47]], [[2, 39]], [[2, 39]], [[54, 99]], [[2, 39]], [[41, 53]]]",
    "query_spans": "[[[54, 105]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ is $y=-2 x$, then the eccentricity of this hyperbola equals?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (y = -2*x)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 59], [79, 82]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[2, 59]], [[2, 76]]]",
    "query_spans": "[[[79, 88]]]",
    "process": "From the given condition: $-\\frac{b}{a} = -2$, that is, $b = 2a$, $\\therefore_{e = \\frac{c}{a}} = \\frac{\\sqrt{a^{2 + b^{2}}}}{a} = \\sqrt{5}$"
  },
  {
    "text": "If the line $l$: $y=\\sqrt{2}(x+2 \\sqrt{3})$ passes through the left focus $F$ of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and has only one common point with the hyperbola $C$, then the equation of the hyperbola $C$ is?",
    "fact_expressions": "l: Line;Expression(l) = (y = sqrt(2)*(x + 2*sqrt(3)));C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F: Point;LeftFocus(C) = F;PointOnCurve(F, l) = True;NumIntersection(l, C)=1",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4 - y^2/8 = 1",
    "fact_spans": "[[[1, 33]], [[1, 33]], [[34, 95], [105, 111], [120, 126]], [[34, 95]], [[42, 95]], [[42, 95]], [[42, 95]], [[42, 95]], [[99, 102]], [[34, 102]], [[1, 102]], [[1, 118]]]",
    "query_spans": "[[[120, 131]]]",
    "process": "According to the problem, line $ l $ is parallel to the asymptote and passes through the left focus $ F $, so three equations involving $ a $, $ b $, $ c $ can be established. Solving for $ a $, $ b $, $ c $ yields the equation of the hyperbola. [Detailed solution] The asymptotes of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0 $, $ b > 0 $) are $ y = \\pm \\frac{b}{a}x $. Since the line $ l: y = \\sqrt{2}(x + 2\\sqrt{3}) $ passing through the left focus $ F $ of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0 $, $ b > 0 $) has only one common point with hyperbola $ C $, line $ l $ is parallel to an asymptote, thus $ \\frac{b}{a} = \\sqrt{2} $, $ 0 = \\sqrt{2}(-c + 2\\sqrt{3}) $. Also, $ a^{2} + b^{2} = c^{2} $. Solving gives $ c = 2\\sqrt{3} $, $ a = 2 $, $ b = 2\\sqrt{2} $. Therefore, the equation of hyperbola $ C $ is $ \\frac{x^{2}}{4} - \\frac{y^{2}}{8} = 1 $."
  },
  {
    "text": "If a focus of the hyperbola $3 m x^{2}-m y^{2}=3$ has coordinates $(0,-2)$, then $m=?$",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (3*(m*x^2) - m*y^2 = 3);Coordinate(OneOf(Focus(G))) = (0, -2)",
    "query_expressions": "m",
    "answer_expressions": "-1",
    "fact_spans": "[[[1, 25]], [[43, 46]], [[1, 25]], [[1, 41]]]",
    "query_spans": "[[[43, 48]]]",
    "process": "The hyperbola $3mx^{2}-my^{2}=3$ has a focus at $(0,-2)$, so $m<0$. This hyperbola can be rewritten as: $\\frac{y^{2}}{-\\frac{3}{m}}-\\frac{x^{2}}{-\\frac{1}{m}}=1$, hence $\\frac{1}{m}+\\frac{3}{m}=4$, solving gives $m=-1$."
  },
  {
    "text": "Let $k \\in R$. If $\\frac{y^{2}}{k}-\\frac{x^{2}}{k-2}=1$ represents a hyperbola with foci on the $y$-axis, then the range of values for the semi-focal length is?",
    "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (-x^2/(k - 2) + y^2/k = 1);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(HalfFocalLength(G))",
    "answer_expressions": "(\\sqrt{2}, +\\infty)",
    "fact_spans": "[[[60, 63]], [[1, 10]], [[12, 63]], [[51, 63]]]",
    "query_spans": "[[[60, 75]]]",
    "process": "Since the foci are on the y-axis, we have \\begin{cases}k>0\\\\k-2>0\\end{cases}, which leads to k>2. Then, solving for the range of the semi-focal distance c gives: If \\frac{y^{2}}{k}-\\frac{x^{2}}{k-2}=1 represents a hyperbola with foci on the y-axis, we obtain \\begin{cases}k>0\\\\k-2>0\\end{cases}, which implies k>2. The semi-focal distance is c=\\sqrt{k+k-2}=\\sqrt{2k-2}>\\sqrt{2}. Thus, the range of the semi-focal distance is: (\\sqrt{2},+\\infty)."
  },
  {
    "text": "Given that the line $MN$ intersects the left and right branches of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ at points $M$ and $N$ respectively, and intersects the right directrix of the hyperbola $C$ at point $P$, with $F$ being the right focus. If $|\\overrightarrow{F M}|=2|\\overrightarrow{F N}|$, and $\\overrightarrow{N P}=\\lambda \\overrightarrow{P M}$ $(0<\\lambda<1)$, then the value of the real number $\\lambda$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;M: Point;N: Point;Intersection(LineOf(M, N), LeftPart(C)) = M;Intersection(LineOf(M, N), RightPart(C)) = N;P: Point;Intersection(LineOf(M, N), RightDirectrix(C)) = P;F: Point;RightFocus(C) = F;lambda: Real;0<lambda;lambda<1;Abs(VectorOf(F, M)) = 2*Abs(VectorOf(F, N));VectorOf(N, P) = lambda*VectorOf(P, M)",
    "query_expressions": "lambda",
    "answer_expressions": "1/2",
    "fact_spans": "[[[10, 61], [81, 87]], [[10, 61]], [[18, 61]], [[18, 61]], [[70, 73]], [[74, 77]], [[2, 79]], [[2, 79]], [[94, 98]], [[2, 98]], [[99, 103]], [[81, 107]], [[225, 236]], [[159, 223]], [[159, 223]], [[109, 157]], [[159, 223]]]",
    "query_spans": "[[[225, 240]]]",
    "process": "Denote the projections of points M and N on the right directrix as points M_{1} and N_{1}, respectively. According to the second definition of the hyperbola, we obtain |MM_{1}|=2|NN_{1}|. Then, from \\triangle MM_{1}P \\sim \\triangle NN_{1}P, it follows that |MP|=2|NP|, from which we determine the value of the real number \\lambda. [Solution] Denote the projections of points M and N on the right directrix as points M_{1} and N_{1}, respectively. From |\\overrightarrow{FM}|=2|\\overrightarrow{FN}| and the second definition of the hyperbola, we have |MM_{1}|=2|NN_{1}|, and since \\triangle MM_{1}P \\sim \\triangle NN_{1}P, it follows that |MP|=2|NP|. Thus, \\overrightarrow{NP}=\\frac{1}{2}\\overrightarrow{PM}, so \\lambda=\\frac{1}{2}."
  },
  {
    "text": "The equation of the directrix of the parabola $y^{2}=6 x$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 6*x)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x = -3/2",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": "Since the focus of the parabola lies on the x-axis, 2p=6, then its directrix equation is x=-\\frac{3}{2}"
  },
  {
    "text": "Given that the focus $F$ of the parabola $x^{2}=2 p y(p>0)$ is a focus of the ellipse $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1(a>b>0)$, and that $P$, $Q$ are common points of the ellipse and the parabola, and the line $P Q$ passes through the focus $F$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Parabola;p: Number;H: Ellipse;a: Number;b: Number;P: Point;Q: Point;F: Point;p>0;Expression(G) = (x^2 = 2*(p*y));a > b;b > 0;Expression(H)=(x^2/b^2+y^2/a^2=1);Focus(G) = F;OneOf(Focus(H)) = F;Intersection(G,H)={P,Q};PointOnCurve(F,LineOf(P,Q))",
    "query_expressions": "Eccentricity(H)",
    "answer_expressions": "sqrt(2)-1",
    "fact_spans": "[[[2, 23], [100, 103]], [[5, 23]], [[30, 82], [97, 99], [126, 128]], [[32, 82]], [[32, 82]], [[89, 92]], [[93, 96]], [[26, 29], [120, 123]], [[5, 23]], [[2, 23]], [[32, 82]], [[32, 82]], [[30, 82]], [[2, 29]], [[26, 87]], [[89, 107]], [[109, 123]]]",
    "query_spans": "[[[126, 134]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $\\Gamma$: $y^{2}=2 p x  (p>0)$ with focus $F$, and a line $l$ of slope $1$ intersecting the parabola $\\Gamma$ at points $A$ and $B$. If $|A F|=3$, $|B F|=5$, then $|A B|$=?",
    "fact_expressions": "Gamma: Parabola;Expression(Gamma) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(Gamma) = F;l: Line;Slope(l) = 1;A: Point;B: Point;Intersection(l, Gamma) = {A, B};Abs(LineSegmentOf(A, F)) = 3;Abs(LineSegmentOf(B, F)) = 5",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 35], [56, 67]], [[2, 35]], [[15, 35]], [[15, 35]], [[39, 42]], [[2, 42]], [[50, 55]], [[43, 55]], [[70, 73]], [[74, 77]], [[50, 79]], [[81, 90]], [[92, 101]]]",
    "query_spans": "[[[103, 112]]]",
    "process": "As shown in the figure: since |AF| = 3, |BF| = 5, then |AA| = 3, |BB| = 5, so |BD| = 2. Because the slope of line l is 1, |AB| = \\sqrt{2}|BD| = 2\\sqrt{2}"
  },
  {
    "text": "Given a point $P$ on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ such that the distance from $P$ to its right focus $F_{2}$ is $5$, what is the distance from point $P$ to its left directrix?",
    "fact_expressions": "G: Ellipse;P: Point;F2: Point;Expression(G) = (x^2/16 + y^2/9 = 1);PointOnCurve(P, G);RightFocus(G) = F2;Distance(P, F2) = 5",
    "query_expressions": "Distance(P, LeftDirectrix(P))",
    "answer_expressions": "12*sqrt(7)/7",
    "fact_spans": "[[[2, 40], [47, 48], [72, 73]], [[43, 46], [67, 71]], [[51, 58]], [[2, 40]], [[2, 46]], [[47, 58]], [[43, 65]]]",
    "query_spans": "[[[67, 81]]]",
    "process": "Combining the equation of the ellipse, we obtain: $a^{2}=16$, $b^{2}=9$, $c$, and its eccentricity $e=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\frac{\\sqrt{7}}{4}$. From the first definition of the ellipse, it follows that: $PF_{1}=2a-PF_{2}=8-5=3$. Let $d$ be the distance from point $P$ to its left directrix. By the second definition of the ellipse, we have: $\\frac{PF_{1}}{d}=\\frac{3}{d}=\\frac{\\sqrt{7}}{4}$, solving for $d$ gives: $d=\\frac{12\\sqrt{7}}{7}$. Thus, the distance from point $P$ to its left directrix is $\\underline{12\\sqrt{7}}$."
  },
  {
    "text": "What is the distance from the focus to the directrix of the parabola $y=10 x^{2}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 10*x^2)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "1/20",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 26]]]",
    "process": "The standard equation of the parabola $ y = 10x^{2} $ is $ x^{2} = \\frac{1}{10}y $, then $ 2p = \\frac{1}{10} $, $ \\therefore p = \\frac{1}{20} $, that is, the distance from the focus to the directrix of the parabola $ y = 10x^{2} $ is $ \\frac{1}{20} $."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$, respectively. If $P$ is a moving point on this ellipse, then the maximum value of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/5 + y^2/4 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G)",
    "query_expressions": "Max(DotProduct(VectorOf(P, F1),VectorOf(P,F2)))",
    "answer_expressions": "4",
    "fact_spans": "[[[19, 56], [70, 72]], [[65, 68]], [[1, 8]], [[9, 16]], [[19, 56]], [[1, 62]], [[1, 62]], [[65, 78]]]",
    "query_spans": "[[[80, 143]]]",
    "process": ""
  },
  {
    "text": "If $F_{1}$ is the left focus of the ellipse, $A$ and $B$ are the right vertex and upper vertex of the ellipse respectively, and $P$ is a point on the ellipse such that ${P F_{1}} \\perp {F_{1} A}$ and $P O \\parallel A B$ ($O$ being the center of the ellipse), then what is the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;A: Point;O: Origin;B: Point;LeftFocus(G) = F1;UpperVertex(G)=A;UpperVertex(G) = B;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P,F1),LineSegmentOf(F1,A));IsParallel(LineSegmentOf(P,O),LineSegmentOf(A,B));Center(G)=O",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[10, 12], [27, 29], [42, 44], [94, 96], [102, 104]], [[38, 41]], [[2, 9]], [[17, 20]], [[90, 93]], [[21, 24]], [[2, 16]], [[17, 37]], [[17, 37]], [[38, 47]], [[49, 76]], [[77, 89]], [[90, 99]]]",
    "query_spans": "[[[102, 110]]]",
    "process": ""
  },
  {
    "text": "What is the eccentricity of the hyperbola $\\frac{y^{2}}{9}-\\frac{x^{2}}{16}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/16 + y^2/9 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 45]]]",
    "process": "From the hyperbola $\\frac{y^{2}}{9}-\\frac{x^{2}}{16}=1$, we obtain: $a=3$, $b=4$, $c=\\sqrt{3^{2}+4^{2}}=5$, hence $e=\\frac{c}{a}=\\frac{5}{3}$"
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$), draw a straight line inclined at an angle of $90^{\\circ}$ intersecting the parabola at points $A$ and $B$. If the length of segment $AB$ is $8$, then the equation of the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);Inclination(H) = ApplyUnit(90,degree);A: Point;B: Point;Intersection(H, G) = {A, B};Length(LineSegmentOf(A, B)) = 8",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=-2",
    "fact_spans": "[[[1, 22], [49, 52], [79, 82]], [[1, 22]], [[4, 22]], [[4, 22]], [[25, 28]], [[1, 28]], [[46, 48]], [[0, 48]], [[29, 48]], [[53, 56]], [[57, 60]], [[46, 62]], [[64, 77]]]",
    "query_spans": "[[[79, 89]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left focus is $F$, $O$ is the coordinate origin, and $P$ is a point on the right branch of the hyperbola $C$ such that $|P F|-|P O|=2 a$. Then the range of the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F: Point;O: Origin;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;PointOnCurve(P, RightPart(C));Abs(LineSegmentOf(P, F)) - Abs(LineSegmentOf(P, O)) = 2*a",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "[2,+oo)",
    "fact_spans": "[[[2, 63], [85, 91], [116, 122]], [[10, 63]], [[10, 63]], [[81, 84]], [[68, 71]], [[72, 75]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 71]], [[81, 96]], [[97, 114]]]",
    "query_spans": "[[[116, 133]]]",
    "process": "Let the right focus of hyperbola C be $ F_{1} $. By the definition of a hyperbola, $ |PF| - |PF_{1}| = 2a $, and also $ |PF| - |PO| = 2a $. Therefore, $ |PO| = |PF_{1}| $. Let $ c^{2} = a^{2} + b^{2} $, then the x-coordinate of point $ P $ is $ \\frac{c}{2} $. Since point $ P $ lies on the hyperbola, clearly $ \\frac{c}{2} \\geqslant a $, that is, $ e = \\frac{c}{a} \\geqslant 2 $. Therefore, the range of eccentricity $ e $ is $ [2, +\\infty) $."
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-\\frac{y^{2}}{n}=1$ $(n>0)$, the distance from its focus to its asymptote is $\\frac{1}{2}$. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;n: Number;n>0;Expression(C) = (x^2 - y^2/n = 1);Distance(Focus(C),Asymptote(C))=1/2",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 39], [43, 44], [67, 70]], [[9, 39]], [[9, 39]], [[2, 39]], [[2, 65]]]",
    "query_spans": "[[[67, 76]]]",
    "process": "From the given conditions, we have $a^{2}=1$, $b^{2}=n$, then $c=\\sqrt{1+n}$, the coordinates of the foci are $(\\pm\\sqrt{1+n},0)$, and the equations of the asymptotes are $y=\\pm\\frac{b}{a}x \\Leftrightarrow bx\\pm ab$'s value. The value. Hence, $c=\\sqrt{1+\\frac{1}{4}}=\\frac{\\sqrt{5}}{2}$. Therefore, the eccentricity is $e=\\frac{\\sqrt{5}}{1}=\\frac{\\sqrt{5}}{2}$. Thus, the answer is $5$."
  },
  {
    "text": "From a point $P$ on the parabola $y^{2}=4 x$, draw a perpendicular to its directrix, with foot of perpendicular at $M$. Let the focus of the parabola be $F$, and $|P F|=5$. Then the area of $\\triangle M P F$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P,G) = True;L: Line;PointOnCurve(P,L) = True;IsPerpendicular(L,Directrix(G)) = True;FootPoint(L,Directrix(G)) = M;M: Point;Focus(G) = F;F: Point;Abs(LineSegmentOf(P, F)) = 5",
    "query_expressions": "Area(TriangleOf(M, P, F))",
    "answer_expressions": "10",
    "fact_spans": "[[[1, 15], [22, 23], [37, 40]], [[1, 15]], [[18, 21]], [[1, 21]], [], [[0, 28]], [[0, 28]], [[0, 35]], [[32, 35]], [[37, 47]], [[44, 47]], [[49, 58]]]",
    "query_spans": "[[[60, 82]]]",
    "process": "From the definition of the parabola, we have |PF| = |PM| = 5, and the distance from point P to the directrix is $ x_{p} + 1 = 5 $, $ \\therefore x_{p} = 4, y_{p} = \\pm 4 $, $ \\therefore S = \\frac{1}{2} \\times 5 \\times 4 = 10 $."
  },
  {
    "text": "It is known that the hyperbola $C$ passes through the point $(1, \\sqrt{5})$, and its asymptotes have equations $y = \\pm 2x$. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Point;Coordinate(G) = (1, sqrt(5));PointOnCurve(G, C);Expression(Asymptote(C)) = (y = pm*(2*x))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 8], [46, 49]], [[9, 25]], [[9, 25]], [[2, 25]], [[2, 44]]]",
    "query_spans": "[[[46, 55]]]",
    "process": "Let the equation of C be $4x^{2}-y^{2}=\\lambda$ $(\\lambda\\neq0)$. Since C passes through the point $(1,\\sqrt{5})$, substituting into $4x^{2}-y^{2}=\\lambda$ gives: $\\lambda=-1$. Therefore, the equation of C is $y^{2}-4x^{2}=1$, i.e., $y^{2}-\\frac{x^{2}}{4}=1$. Thus, $a=1$, $b=\\frac{1}{2}$, hence: $c=\\sqrt{a^{2}+b^{2}}=\\frac{\\sqrt{5}}{2}$. Therefore, the eccentricity $e=\\frac{c}{a}=\\frac{\\sqrt{5}}{2}$."
  },
  {
    "text": "Given $M(2,0)$, $P$ is a moving point on the circle $N$: $x^{2}+4x+y^{2}-32=0$, and the perpendicular bisector of segment $MP$ intersects $NP$ at point $Q$. Then the trajectory equation of the moving point $Q$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (2, 0);N: Circle;Expression(N) = (y^2 + x^2 + 4*x - 32 = 0);P: Point;PointOnCurve(P, N);Q: Point;Intersection(PerpendicularBisector(LineSegmentOf(M, P)), LineSegmentOf(N, P)) = Q",
    "query_expressions": "LocusEquation(Q)",
    "answer_expressions": "x^2/9+y^2/5=1",
    "fact_spans": "[[[2, 10]], [[2, 10]], [[16, 44]], [[16, 44]], [[12, 15]], [[12, 48]], [[69, 73], [77, 80]], [[49, 73]]]",
    "query_spans": "[[[77, 87]]]",
    "process": "The circle $ N: x^{2} + 4x + y^{2} - 32 = 0 $ can be rewritten as $ (x+2)^{2} + y^{2} = 6^{2} $, so $ N(-2,0) $, radius $ r = 6 $, $ |MN| = 4 $. According to the condition, $ |QP| = |QM| $, so $ |QN| + |QM| = |QN| + |QP| = |NP| = 6 > |MN| $. Therefore, the trajectory of $ Q $ is an ellipse, $ 2a = 6 $, $ 2c = 4 $, $ a = 3 $, $ c = 2 $, $ b = \\sqrt{a^{2} - c^{2}} = \\sqrt{5} $. Thus, the equation of the trajectory of the moving point $ Q $ is $ \\frac{x^{2}}{9} + \\frac{y^{2}}{5} = 1 $."
  },
  {
    "text": "Given that $A$, $B$, $P$ are three distinct points on the ellipse $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{n^{2}}=1$ $(m, n>0)$, and the line connecting $A$ and $B$ passes through the origin, if the product of the slopes of lines $PA$ and $PB$ is $k_{PA} \\cdot k_{PB} = -2$, then what is the eccentricity of the ellipse?",
    "fact_expressions": "P: Point;A: Point;B: Point;G: Ellipse;Expression(G) = (y^2/n^2 + x^2/m^2 = 1);m: Number;n: Number;m>0;n>0;PointOnCurve(A, G);PointOnCurve(B, G);PointOnCurve(P, G);Negation(A = B);Negation(A = P);Negation(B = P);O: Origin;PointOnCurve(O, LineSegmentOf(A, B)) ;k1: Number;k2: Number;Slope(LineOf(P, A)) = k1;Slope(LineOf(P, B)) = k2;k1*k2 = -2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[10, 13]], [[2, 5], [75, 78]], [[6, 9], [79, 82]], [[14, 67], [141, 143]], [[14, 67]], [[16, 67]], [[16, 67]], [[16, 67]], [[16, 67]], [[2, 73]], [[2, 73]], [[2, 73]], [[2, 73]], [[2, 73]], [[2, 73]], [[86, 90]], [[75, 90]], [[112, 138]], [[112, 138]], [[93, 138]], [[93, 138]], [[112, 138]]]",
    "query_spans": "[[[141, 149]]]",
    "process": ""
  },
  {
    "text": "A moving circle $M$ is externally tangent to the circle $O_{1}$: $x^{2}+(y+3)^{2}=1$, and internally tangent to the circle $O_{2}$: $x^{2}+(y-3)^{2}=81$. Then, the trajectory equation of the center of the moving circle $M$ is?",
    "fact_expressions": "M: Circle;O1: Circle;Expression(O1) = (x^2 + (y + 3)^2 = 1);IsOutTangent(M, O1);O2: Circle;Expression(O2) = (x^2 + (y - 3)^2 = 81);IsInTangent(M, O2);M1: Point;Center(M) = M1",
    "query_expressions": "LocusEquation(M1)",
    "answer_expressions": "y^2/25 + x^2/16 = 1",
    "fact_spans": "[[[3, 6]], [[7, 36]], [[7, 36]], [[3, 38]], [[40, 70]], [[40, 70]], [[3, 72]], [[78, 81]], [[74, 81]]]",
    "query_spans": "[[[78, 88]]]",
    "process": "Let the radius of the moving circle be $ r $. According to the problem, $ MO_{1} = 1 + r $, $ MO_{2} = 9 - r $, hence $ MO_{1} + MO_{2} = 10 > 6 $, so the trajectory is an ellipse. $ 2a = 10 $, $ 2c = 6 $, thus $ a = 5 $, $ b = 4 $, therefore the trajectory equation is: $ \\frac{y^{2}}{25} + \\frac{x^{2}}{16} = 1 $."
  },
  {
    "text": "If $A(3,2)$, $F$ is the focus of the parabola $y^{2}=2x$, and $P$ is any point on the parabola, then the minimum value of $|PF|+|PA|$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (3, 2);F: Point;Focus(G) = F;G: Parabola;Expression(G) = (y^2 = 2*x);P: Point;PointOnCurve(P, G) = True",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "7/2",
    "fact_spans": "[[[1, 9]], [[1, 9]], [[12, 15]], [[12, 33]], [[16, 30], [38, 41]], [[16, 30]], [[34, 37]], [[34, 46]]]",
    "query_spans": "[[[48, 67]]]",
    "process": "Let the projection of point P on the directrix be D. Then, according to the definition of the parabola, |PF| = |PD|. Therefore, to minimize |PA| + |PF|, we need to minimize |PA| + |PD|. When points D, P, and A are collinear, |PA| + |PD| is minimized, giving 3 + \\frac{1}{2} = \\frac{7}{2}."
  },
  {
    "text": "The equation of the ellipse sharing foci with $\\frac{x^{2}}{6}+\\frac{y^{2}}{10}=1$ and passing through the point $P(-3,2)$ is?",
    "fact_expressions": "G: Ellipse;Z: Ellipse;P: Point;Expression(G) = (x^2/6 + y^2/10 = 1);Coordinate(P) = (-3, 2);PointOnCurve(P, Z);Focus(G) = Focus(Z)",
    "query_expressions": "Expression(Z)",
    "answer_expressions": "x^2/12 + y^2/16 = 1",
    "fact_spans": "[[[1, 39]], [[57, 59]], [[46, 56]], [[1, 39]], [[46, 56]], [[44, 59]], [[0, 59]]]",
    "query_spans": "[[[57, 63]]]",
    "process": ""
  },
  {
    "text": "Given a point $R$ on the directrix of the parabola $x^{2}=4 y$, two tangent lines are drawn from $R$ to the parabola, touching it at points $M$ and $N$. If $O$ is the origin, then $\\overrightarrow{OM} \\cdot \\overrightarrow{ON}$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);R: Point;PointOnCurve(R, Directrix(G)) ;L1: Line;L2: Line;TangentOfPoint(R, G) = {L1, L2};TangentPoint(L1, G) = M;TangentPoint(L2, G) = N;M: Point;N: Point;O: Origin",
    "query_expressions": "DotProduct(VectorOf(O, M), VectorOf(O, N))",
    "answer_expressions": "-3",
    "fact_spans": "[[[0, 14], [24, 27]], [[0, 14]], [[20, 23]], [[0, 23]], [], [], [[0, 32]], [[0, 45]], [[0, 45]], [[38, 41]], [[42, 45]], [[47, 50]]]",
    "query_spans": "[[[57, 106]]]",
    "process": ""
  },
  {
    "text": "What is the eccentricity of $x^{2}+4 y^{2}=16$? What is the equation of the hyperbola that shares the same foci with this ellipse and has an asymptote $x+\\sqrt{3} y=0$?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H)=(x^2+4*y^2=16);Focus(G)=Focus(H);Expression(OneOf(Asymptote(G)))=(x + sqrt(3)*y = 0)",
    "query_expressions": "Eccentricity(H);Expression(G)",
    "answer_expressions": "sqrt(3)/2\nx^2/9-y^2/3=1",
    "fact_spans": "[[[59, 62]], [[0, 18], [27, 29]], [[0, 18]], [[25, 62]], [[36, 62]]]",
    "query_spans": "[[[0, 25]], [[59, 66]]]",
    "process": "The standard equation of the ellipse $x^{2}+4y^{2}=16$ is $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$, where $a=4$, $b=2$, $c=2\\sqrt{3}$, $e=\\frac{c}{a}=\\frac{\\sqrt{3}}{2}$. Since one asymptote of the hyperbola is $x+\\sqrt{3}y=0$, the equation of the hyperbola can be assumed as $\\frac{x^{2}}{\\lambda}+\\frac{y^{2}}{\\frac{2}{3}}=1$ ($\\lambda>0$). Since the coordinates of the foci of the ellipse are $(\\pm2\\sqrt{3},0)$, the coordinates of the foci of the hyperbola are $(\\pm2\\sqrt{3},0)$. Therefore, $\\lambda+\\frac{\\lambda}{3}=12$, $\\lambda=9$, so the equation of the hyperbola is $\\frac{x^{2}}{9}-\\frac{y^{2}}{3}=1$. Answer: $\\frac{\\sqrt{3}}{2}$, $\\frac{x^{2}}{9}-\\frac{y^{2}}{3}=1$"
  },
  {
    "text": "Given that point $P(x_1, y_1)$ is an arbitrary point on the curve $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, then the maximum value of $x_1 - 2 y_1$ is?",
    "fact_expressions": "G: Curve;P: Point;x1: Number;y1: Number;Expression(G) = (x^2/16 + y^2/9 = 1);Coordinate(P) = (x1, y1);PointOnCurve(P, G)",
    "query_expressions": "Max(x1 - 2*y1)",
    "answer_expressions": "sqrt(52)",
    "fact_spans": "[[[16, 54]], [[2, 15]], [[3, 15]], [[3, 15]], [[16, 54]], [[2, 15]], [[2, 59]]]",
    "query_spans": "[[[61, 76]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ is an arbitrary point on the circle $C$: $x^{2}+y^{2}-8x-8y+28=0$, the curve $N$: $x^{2}+4y^2=4$ intersects the $x$-axis at points $A$ and $B$. The line $OP$ intersects curve $N$ at point $M$. Denote the slopes of lines $MA$, $MB$, and $OP$ as $k_{1}$, $k_{2}$, $k_{3}$, respectively. Then the range of values of $k_{1} \\cdot k_{2} \\cdot k_{3}$ is?",
    "fact_expressions": "C: Circle;Expression(C) = (-8*y - 8*x + x^2 + y^2 + 28 = 0);P: Point;PointOnCurve(P, C);N: Curve;Expression(N) = (x^2 + 4*y^2 = 4);A: Point;B: Point;Intersection(N, xAxis) = {A, B};O: Origin;M: Point;Intersection(LineOf(O, P), N) = M;k1: Number;k2: Number;k3: Number;Slope(LineOf(M, A)) = k1;Slope(LineOf(M, B)) = k2;Slope(LineOf(O, P)) = k3",
    "query_expressions": "Range(k1*k2*k3)",
    "answer_expressions": "[(-4-sqrt(7))/12, (-4+sqrt(7))/12]",
    "fact_spans": "[[[7, 39]], [[7, 39]], [[2, 6]], [[2, 44]], [[45, 66], [90, 95]], [[45, 66]], [[73, 76]], [[77, 80]], [[45, 82]], [[85, 89]], [[97, 101]], [[83, 101]], [[125, 132]], [[133, 140]], [[141, 148]], [[103, 148]], [[103, 148]], [[103, 148]]]",
    "query_spans": "[[[150, 188]]]",
    "process": ""
  },
  {
    "text": "It is known that $F$ is the focus of the parabola $C$: $y^{2}=4x$. A line passing through $F$ with slope $\\sqrt{3}$ intersects $C$ at points $A$ and $B$. Given $|FA| > |FB|$, then the value of $\\frac{|FA|}{|FB|}$ is equal to?",
    "fact_expressions": "F: Point;C: Parabola;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F, G) = True;Slope(G) = sqrt(3);G: Line;Intersection(G, C) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(F, A)) > Abs(LineSegmentOf(F, B))",
    "query_expressions": "Abs(LineSegmentOf(F, A))/Abs(LineSegmentOf(F, B))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 5], [29, 32]], [[6, 24], [50, 53]], [[6, 24]], [[2, 27]], [[28, 49]], [[33, 49]], [[47, 49]], [[47, 63]], [[54, 57]], [[58, 61]], [[66, 77]]]",
    "query_spans": "[[[79, 101]]]",
    "process": ""
  },
  {
    "text": "There exists a point $M(x_{0}, y_{0})$ in the first quadrant on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, such that the line passing through point $M$ and perpendicular to the tangent line of the ellipse at this point, $\\frac{x_{0} x}{a^{2}}+\\frac{y_{0} y}{b^{2}}=1$, passes through the point $(\\frac{c}{2}, 0)$ ($c$ is the semi-focal distance of the ellipse). Then, the range of values for the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;I: Point;M: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(I) = (c/2, 0);Coordinate(M) = (x0, y0);Quadrant(M)=1;PointOnCurve(M,G);Expression(TangentOnPoint(M,G))=(x0*x/a^2+y0*y/b^2=1);PointOnCurve(M,H);IsPerpendicular(TangentOnPoint(M,G),H);PointOnCurve(I,H);HalfFocalLength(G)=c;c:Number;x0:Number;y0:Number",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1/2, 1)",
    "fact_spans": "[[[26, 78], [89, 91], [175, 177], [184, 186]], [[28, 78]], [[28, 78]], [[147, 149]], [[151, 170]], [[7, 25], [83, 87], [93, 94]], [[28, 78]], [[28, 78]], [[26, 78]], [[151, 170]], [[7, 25]], [[0, 25]], [[7, 79]], [[89, 144]], [[82, 149]], [[88, 149]], [[147, 170]], [[171, 180]], [[171, 174]], [[8, 25]], [[8, 25]]]",
    "query_spans": "[[[184, 196]]]",
    "process": "Since the tangent line to the ellipse at a point is $\\frac{x_{0}x}{a^{2}}+\\frac{y_{0}y}{b^{2}}=1$, the slope of the tangent line is $-\\frac{b^{2}x_{0}}{a^{2}y_{0}}$. From $\\frac{y_{0}}{x_{0}-\\frac{c}{2}}\\times(-\\frac{b^{2}x_{0}}{a^{2}y_{0}})=-1$, solving gives $x_{0}=\\frac{a^{2}}{2}<a$, leading to $\\frac{c}{a}>\\frac{1}{2}$. Therefore, the range of the ellipse's eccentricity is $(\\frac{1}{2},1)$."
  },
  {
    "text": "Given that the point $F(1,0)$ is the focus of the parabola $C$: $y^{2}=2 p x$, then $p$=?",
    "fact_expressions": "C: Parabola;p: Number;F: Point;Expression(C) = (y^2 = 2*(p*x));Coordinate(F) = (1, 0);Focus(C) = F",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[12, 33]], [[38, 41]], [[2, 11]], [[12, 33]], [[2, 11]], [[2, 36]]]",
    "query_spans": "[[[38, 43]]]",
    "process": ""
  },
  {
    "text": "Given point $N(\\frac{5}{2}, 1)$, the focus of the parabola $y^{2}=6 x$ is $F$, and point $M$ is any point on the parabola. Then the minimum perimeter of $\\triangle M N F$ is?",
    "fact_expressions": "G: Parabola;N: Point;M: Point;F: Point;Expression(G) = (y^2 = 6*x);Coordinate(N) = (5/2, 1);Focus(G) = F;PointOnCurve(M, G)",
    "query_expressions": "Min(Perimeter(TriangleOf(M, N, F)))",
    "answer_expressions": "4+sqrt(2)",
    "fact_spans": "[[[23, 37], [50, 53]], [[2, 22]], [[45, 49]], [[41, 44]], [[23, 37]], [[2, 22]], [[23, 44]], [[45, 58]]]",
    "query_spans": "[[[60, 85]]]",
    "process": "The focus of the parabola $ y^{2}=6x $ is $ F(\\frac{3}{2},0) $, and the equation of the directrix $ l $ is $ x=-\\frac{3}{2} $. Draw $ MM \\perp l $ from point $ M $, with foot of perpendicular $ M $. Then $ |MM|=|MF| $. Therefore, the perimeter of $ \\triangle MNF $ is $ |MF|+|MN|+|FN|=|MM|+|MN|+\\sqrt{2} \\geqslant |MN|+\\sqrt{2} = \\frac{5}{2}-(-\\frac{3}{2})+\\sqrt{2}=4+\\sqrt{2} $. The equality holds if and only if points $ M $, $ M $, and $ N $ are collinear."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, the intersection point of the directrix and the $x$-axis is $K$, and point $P$ is a point on the parabola, then the minimum value of $\\frac{|PF|}{|PK|}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;Intersection(Directrix(G), xAxis) = K;K: Point;P: Point;PointOnCurve(P, G) = True",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F))/Abs(LineSegmentOf(P, K)))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 16], [44, 47]], [[2, 16]], [[20, 23]], [[2, 23]], [[2, 38]], [[35, 38]], [[39, 43]], [[39, 50]]]",
    "query_spans": "[[[52, 79]]]",
    "process": "Draw PA perpendicular to the directrix, intersecting the directrix at point A, then $\\frac{|PF|}{|PK|}=\\frac{|PA|}{|PK|}=\\sin\\angle AKP$. When the line KP is tangent to the parabola, $\\frac{|PF|}{|PK|}$ reaches its minimum value. Suppose the equation of line KP at this moment is $x=my-1$. Solving simultaneously with the parabola equation and eliminating variables, an equation is obtained based on $A=0$, which can then be solved. [Detailed explanation] Draw PA perpendicular to the directrix, intersecting the directrix at point A, then $|PF|=|PA|$, so $\\frac{|PF|}{|PK|}=\\frac{|PA|}{|PK|}=\\sin\\angle AKP$. When the line KP is tangent to the parabola, $\\frac{|PF|}{|PK|}$ reaches its minimum value. Let the equation of KP at this time be $x=my-1$, combining with $y^{2}=4x$, we obtain $y^{2}-4my+4=0$, $A=16m^{2}-16=0$, thus $m=\\pm1$, then $\\sin\\angle AKP=\\frac{\\sqrt{2}}{2}$, hence the minimum value of $\\frac{|PF|}{|PK|}$ is $\\frac{\\sqrt{2}}{2}$."
  },
  {
    "text": "A line $l$ passing through the focus $F$ of the parabola $C$: $y^{2}=2 p x(p>0)$, with an inclination angle of $30^{\\circ}$, intersects $C$ at points $A$ and $B$. If the horizontal coordinate of the midpoint $M$ of segment $A B$ is $7$, then $p=$?",
    "fact_expressions": "l: Line;C: Parabola;p: Number;B: Point;A: Point;F: Point;M:Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;Inclination(l)=ApplyUnit(30,degree);PointOnCurve(F, l);Intersection(l, C) = {A, B};MidPoint(LineSegmentOf(A,B))=M;XCoordinate(M)=7",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[52, 57]], [[2, 28], [58, 61]], [[98, 101]], [[67, 70]], [[63, 66]], [[31, 34]], [[84, 87]], [[10, 28]], [[2, 28]], [[2, 34]], [[35, 57]], [[0, 57]], [[52, 72]], [[74, 87]], [[84, 95]]]",
    "query_spans": "[[[98, 103]]]",
    "process": "According to the problem, the equation of the line passing through the focus with an inclination angle of $30^{\\circ}$ is $y=\\frac{\\sqrt{3}}{3}(x-\\frac{p}{2})$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$. Solving the system of equations:  \n$$\n\\begin{cases}\ny^{2}=2px \\\\\ny=\\frac{\\sqrt{3}}{3}(x-\\frac{p}{2})\n\\end{cases}\n$$  \nyields: $\\Rightarrow x^{2}-7px+\\frac{p^{2}}{4}=0$.  \nSince the x-coordinate of the midpoint $M$ of $AB$ is 7,  \n$\\therefore x_{1}+x_{2}=7p=14$, solving gives: $p=2$,"
  },
  {
    "text": "Let the right focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$ be $F$, then the distance from point $F$ to the asymptotes of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;F: Point;Expression(G) = (x^2/4 - y^2/9 = 1);RightFocus(G) = F",
    "query_expressions": "Distance(F, Asymptote(G))",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 39], [55, 58]], [[44, 47], [49, 53]], [[1, 39]], [[1, 47]]]",
    "query_spans": "[[[49, 67]]]",
    "process": ""
  },
  {
    "text": "If the maximum distance from any point on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ to its upper vertex is exactly equal to the distance from the center of the ellipse to its directrix, then the range of values for the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;a: Number;b: Number;a > b;b > 0;P:Point;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P,G);Max(Distance(P,UpperVertex(G)))=Distance(Center(G),Directrix(G))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[\\sqrt{2}/2,1)",
    "fact_spans": "[[[1, 53], [58, 59], [72, 74], [78, 79], [87, 89]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [], [[1, 53]], [[1, 57]], [[1, 84]]]",
    "query_spans": "[[[87, 100]]]",
    "process": ""
  },
  {
    "text": "If the coordinate of a vertex of a hyperbola is $(3,0)$ and the focal length is $10$, then its standard equation is?",
    "fact_expressions": "G: Hyperbola;Coordinate(OneOf(Vertex(G)))=(3,0);FocalLength(G) = 10",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/16 = 1",
    "fact_spans": "[[[1, 4], [29, 30]], [[1, 19]], [[1, 27]]]",
    "query_spans": "[[[29, 37]]]",
    "process": "Given the vertex coordinates, we obtain a; given the focal distance, we obtain c; then, using b^{2}=c^{2}-a^{2}, we obtain b, and thus derive the standard equation of the hyperbola. According to the problem, a=3, c=5, \\therefore b=\\sqrt{25-9}=4. From the vertex coordinates, we know the foci lie on the x-axis, \\therefore the equation of the hyperbola is \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1."
  },
  {
    "text": "It is known that the center of ellipse $G$ is at the origin of coordinates, the major axis lies on the $x$-axis, the eccentricity is $\\frac{\\sqrt{3}}{2}$, and the sum of the distances from a point on $G$ to the two foci of $G$ is $12$. Then the equation of ellipse $G$ is?",
    "fact_expressions": "G: Ellipse;O:Origin;Center(G)=O;OverlappingLine(MajorAxis(G),xAxis);F1:Point;F2:Point;P:Point;Focus(G)={F1,F2};Distance(P, F1)+Distance(P,F2) = 12;Eccentricity(G)=sqrt(3)/2;PointOnCurve(P,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/36+y^2/9=1",
    "fact_spans": "[[[2, 7], [52, 55], [59, 62], [79, 84]], [[11, 15]], [[2, 15]], [[2, 24]], [], [], [], [[59, 67]], [[52, 77]], [[2, 50]], [[52, 58]]]",
    "query_spans": "[[[79, 89]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=8 x$, and a line $l$ passing through $F$ intersects $C$ at points $A$ and $B$, with the ordinate of the midpoint of segment $A B$ being $4$, then $|A B|$=?",
    "fact_expressions": "F: Point;Focus(C) = F;C: Parabola;Expression(C) = (y^2 = 8*x);l: Line;PointOnCurve(F, l) = True;Intersection(l, C) = {A, B};A: Point;B: Point;YCoordinate(MidPoint(LineSegmentOf(A,B))) = 4",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16",
    "fact_spans": "[[[2, 5], [30, 33]], [[2, 28]], [[6, 25], [40, 43]], [[6, 25]], [[34, 39]], [[29, 39]], [[34, 54]], [[45, 48]], [[49, 52]], [[55, 72]]]",
    "query_spans": "[[[74, 83]]]",
    "process": "By the given condition, the focus is $ F(2,0) $. Let the midpoint of $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $ be $ M(x_{0},4) $, and let the slope of line $ l $ be $ k $. Thus, $ \\frac{y_{1}+y_{2}}{2}=4 $. Also, $ y_{1}^{2}=8x_{1} $, $ y_{2}^{2}=8x_{2} $. Subtracting these two equations gives: $ (y_{1}-y_{2})(y_{1}+y_{2})=8(x_{1}-x_{2}) $, hence $ \\frac{y_{1}-y_{2}}{x_{1}-x_{2}} \\cdot \\frac{y_{1}+y_{2}}{2}=4 $, so $ k=1 $. The line $ l $ is: $ y=x-2 $. Therefore, $ x_{0}=6 $, $ |AB|=x_{1}+x_{2}+p=2x_{0}+4=16 $."
  },
  {
    "text": "Given that a line intersects the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ at points $A$ and $B$, let $P$ be the midpoint of segment $AB$. If the slope of the line is $k_{1}$ and the slope of line $OP$ is $k_{2}$, then what is $k_{1} k_{2}$ equal to?",
    "fact_expressions": "G: Ellipse;H:Line;B: Point;A: Point;O: Origin;P: Point;Expression(G) = (x^2/9 + y^2/4 = 1);Intersection(H, G) = {A, B};MidPoint(LineSegmentOf(A,B)) = P;Slope(H) = k1;Slope(LineOf(O, P)) = k2;k1:Number;k2:Number",
    "query_expressions": "k1*k2",
    "answer_expressions": "-4/9",
    "fact_spans": "[[[5, 42]], [[2, 4], [71, 73]], [[48, 51]], [[44, 47]], [[87, 92]], [[66, 69]], [[5, 42]], [[2, 53]], [[55, 69]], [[71, 84]], [[85, 103]], [[77, 84]], [[96, 103]]]",
    "query_spans": "[[[105, 121]]]",
    "process": ""
  },
  {
    "text": "It is known that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ passes through the point $(1,2)$. What is the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(H) = (1, 2);PointOnCurve(H,OneOf(Asymptote(G)))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 59], [78, 81]], [[5, 59]], [[5, 59]], [[67, 75]], [[5, 59]], [[5, 59]], [[2, 59]], [[67, 75]], [[2, 75]]]",
    "query_spans": "[[[78, 89]]]",
    "process": ""
  },
  {
    "text": "Given the equation of the hyperbola is $\\frac{x^{2}}{3}-y^{2}=1$, then the eccentricity of this hyperbola is? The distance from its focus to the asymptote is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1)",
    "query_expressions": "Eccentricity(G);Distance(Focus(G), Asymptote(G))",
    "answer_expressions": "2*sqrt(3)/3\n1",
    "fact_spans": "[[[2, 5], [37, 40], [46, 47]], [[2, 34]]]",
    "query_spans": "[[[37, 46]], [[46, 58]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ with foci $F_{1}$ and $F_{2}$, let point $M$ lie on the major axis $A_{1} A_{2}$. A line passing through point $M$ and perpendicular to $A_{1} A_{2}$ intersects the ellipse at point $P$. Find the range of the horizontal coordinate of point $M$ such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}<0$.",
    "fact_expressions": "G: Ellipse;H: Line;A1: Point;A2: Point;P: Point;F1: Point;F2: Point;M: Point;Expression(G) = (x^2/4 + y^2 = 1);Focus(G) = {F1,F2};PointOnCurve(M, LineSegmentOf(A1, A2));IsPerpendicular(LineSegmentOf(A1, A2), H);Intersection(H, G) = P;DotProduct(VectorOf(P,F1),VectorOf(P,F2))<0",
    "query_expressions": "Range(XCoordinate(M))",
    "answer_expressions": "(-2*sqrt(6)/3,2*sqrt(6)/3)",
    "fact_spans": "[[[2, 29], [98, 100]], [[95, 97]], [[57, 70]], [[57, 70]], [[101, 104]], [[33, 40]], [[41, 48]], [[50, 54], [73, 77], [168, 172]], [[2, 29]], [[2, 48]], [[50, 71]], [[78, 97]], [[95, 104]], [[108, 167]]]",
    "query_spans": "[[[168, 182]]]",
    "process": "Since point $ P $ satisfies $ \\overrightarrow{PF_1} \\cdot \\overrightarrow{PF_2} < 0 $, point $ P $ lies inside the circle with $ F_1F_2 $ as diameter, and the equation of the circle is $ x^2 + y^2 = 3 $. Solving the system \n\\[\n\\begin{cases}\nx^2 + y^2 = 3 \\\\\nx^2 + 4y^2 = 4\n\\end{cases}\n\\]\neliminating $ y $ gives: $ x^2 = \\frac{8}{3} $, $ x = \\pm \\frac{2\\sqrt{6}}{3} $. The range of the horizontal coordinate of point $ M $ is $ \\left( -\\frac{2\\sqrt{6}}{3}, \\frac{2\\sqrt{6}}{3} \\right) $."
  },
  {
    "text": "The line $ l $ with slope $ -\\frac{1}{3} $ intersects the ellipse $ C: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $) such that the chord is bisected by the point $ M(1,1) $. Then the eccentricity of $ C $ is?",
    "fact_expressions": "l: Line;C: Ellipse;a: Number;b: Number;M: Point;a > b;b > 0;Slope(l) = -1/3;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);MidPoint(InterceptChord(l, C)) = M;Coordinate(M) = (1, 1)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[18, 23]], [[24, 81], [100, 103]], [[31, 81]], [[31, 81]], [[87, 96]], [[31, 81]], [[31, 81]], [[0, 23]], [[24, 81]], [[18, 98]], [[87, 96]]]",
    "query_spans": "[[[100, 109]]]",
    "process": "Let the points of intersection of line $ l $ and the ellipse be $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Since the chord is bisected by point $ M(1,1) $, we have $ x_{1}+x_{2}=2 $, $ y_{1}+y_{2}=2 $. From $ \\frac{x_{1}^{2}}{a^{2}}+\\frac{y_{1}^{2}}{b^{2}}=1 $, $ \\frac{x_{2}^{2}}{a^{2}}+\\frac{y_{2}^{2}}{b^{2}}=1 $, subtracting these two equations yields: $ \\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{a^{2}}+\\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{b^{2}}=0 $. Simplifying gives: $ \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{b^{2}}{a^{2}} $. Since the slope of line $ l $ is $ -\\frac{1}{3} $, we have $ \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{1}{3}=-\\frac{b^{2}}{a^{2}} $, thus $ \\frac{b^{2}}{a^{2}}=\\frac{1}{3} $. Therefore, the eccentricity $ e=\\sqrt{1-\\frac{b^{2}}{a^{2}}}=\\frac{\\sqrt{6}}{3} $."
  },
  {
    "text": "Let the parabola $C$: $y^{2}=2 p x$ ($p>0$) have focus $F$, and let point $M$ lie on $C$ such that $|M F|=2$. If the circle with diameter $M F$ passes through the point $(0,1)$, then the distance from the focus of $C$ to its directrix is?",
    "fact_expressions": "C: Parabola;p: Number;G: Circle;H: Point;M: Point;F: Point;p>0;Expression(C) = (y^2 = 2*p*x);Coordinate(H) = (0, 1);Focus(C) = F;PointOnCurve(M, C);Abs(LineSegmentOf(M, F)) = 2;IsDiameter(LineSegmentOf(M,F),G);PointOnCurve(H,G)",
    "query_expressions": "Distance(Focus(C), Directrix(C))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 26], [39, 42], [77, 80], [84, 85]], [[8, 26]], [[65, 66]], [[67, 75]], [[34, 38]], [[30, 33]], [[8, 26]], [[1, 26]], [[67, 75]], [[1, 33]], [[34, 43]], [[44, 53]], [[55, 66]], [[65, 75]]]",
    "query_spans": "[[[77, 92]]]",
    "process": "\\because the parabola C has equation y^{2}=2px (p>0), \\therefore the focus is F(\\frac{p}{2},0), and the directrix has equation x=-\\frac{p}{2}. Let M(x,y). By the property of the parabola, |MF|=x+\\frac{p}{2}=2, so x=2-\\frac{p}{2}. Since the center of the circle is the midpoint of MF, by the midpoint formula, the x-coordinate of the center is \\frac{2-\\frac{p}{2}+\\frac{p}{2}}{2}=1. Given that the radius of the circle is also 1, it follows that the circle is tangent to the y-axis at point (0,1). Hence, the y-coordinate of the center is 1, so the y-coordinate of M is 2, i.e., M(2-\\frac{p}{2},2). Substituting into the parabola equation gives 4=2p(2-\\frac{p}{2}), so p=2. Therefore, the distance from the focus of C to the directrix is 2."
  },
  {
    "text": "If the product of the distances from a point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{8}=1$ to its two foci is $M$, then what is the maximum value of $M$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/8 = 1);K: Point;PointOnCurve(K, G);F1: Point;F2: Point;Focus(G) = {F1, F2};M: Number;Distance(K, F1)*Distance(K, F2) = M",
    "query_expressions": "Max(M)",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 38]], [[1, 38]], [], [[1, 42]], [], [], [[1, 46]], [[53, 56], [58, 61]], [[1, 56]]]",
    "query_spans": "[[[58, 67]]]",
    "process": "Let the two foci of the ellipse be denoted as $ F_{1} $ and $ F_{2} $, and let $ P $ be an arbitrary point on the ellipse such that $ |PF_{1}| + |PF_{2}| = 2a = 4\\sqrt{2} $. Then $ M = |PF_{1}| \\cdot |PF_{2}| \\leqslant \\left( \\frac{|PF_{1}| + |PF_{2}|}{2} \\right)^{2} = 8 $. The equality holds if and only if $ |PF_{1}| = |PF_{2}| = 2\\sqrt{2} $, that is, when point $ P $ is located at a vertex on the minor axis of the ellipse. Thus, the maximum value of $ M $ is 8."
  },
  {
    "text": "Given a parabola $C$: $y^{2}=4x$ with a point $M(4,-4)$, points $A$ and $B$ are two moving points on the parabola $C$ such that $\\overrightarrow{MA} \\cdot \\overrightarrow{MB}=0$. Then, the maximum value of the distance from point $M$ to the line $AB$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);M: Point;Coordinate(M) = (4, -4);PointOnCurve(A, C) = True;PointOnCurve(B, C) = True;A: Point;B: Point;DotProduct(VectorOf(M, A), VectorOf(M, B)) = 0;PointOnCurve(M, C)",
    "query_expressions": "Max(Distance(M, LineOf(A, B)))",
    "answer_expressions": "4*sqrt(5)",
    "fact_spans": "[[[2, 21], [43, 49]], [[2, 21]], [[24, 33], [109, 113]], [[24, 33]], [[34, 54]], [[34, 54]], [[34, 38]], [[39, 42]], [[56, 107]], [[2, 33]]]",
    "query_spans": "[[[109, 130]]]",
    "process": "Let $ A(x_{1},y_{1}), B(x_{2},y_{2}) $, $\\therefore\\begin{cases}y_{1}^{2}=4x_{1},\\\\y_{2}^{2}=4x_{2},\\end{cases}\\therefore k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{4}{y_{1}+y_{2}},\\therefore l_{AB}: y-y_{1}=\\frac{4}{y_{1}+y_{2}}(x-x_{1}) $. $\\because$ point $ M $ lies on the parabola, $\\therefore k_{MA}=\\frac{4}{y_{1}-4}, k_{MB}=\\frac{y}{y_{2}}\\frac{4}{-4} \\because \\overrightarrow{MA}\\cdot\\overrightarrow{MB}=(-1,\\therefore y_{1}y_{2}-4(y_{1}+y_{2})+32=0,\\therefore l_{AB}: x-\\frac{y^{2}}{y_{1}+y_{2}}+y_{1}=\\frac{4}{y_{1}+y_{2}}x+\\frac{y_{1}y_{2}}{y_{1}+y_{2}}-(x-8)+4,\\therefore$ line $ AB $ always passes through point $ N(8,4) $, then the maximum distance from point $ M $ to line $ AB $ is $ |MN|=\\sqrt{(8-4)^{2}+(4+4)^{2}}=4\\sqrt{5} $"
  },
  {
    "text": "Given that a line $l$ with slope $1$ intersects the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{8}=1$ at points $A$ and $B$, and $|A B|=4 \\sqrt{5}$, find the equation of line $l$.",
    "fact_expressions": "l: Line;Slope(l) = 1;G: Hyperbola;Expression(G) = (x^2/4 - y^2/8 = 1);Intersection(l,G) = {A,B};A: Point;B: Point;Abs(LineSegmentOf(A, B)) = 4*sqrt(5)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "y=x+pm*1",
    "fact_spans": "[[[9, 14], [87, 92]], [[2, 14]], [[15, 53]], [[15, 53]], [[9, 64]], [[55, 58]], [[59, 62]], [[66, 84]]]",
    "query_spans": "[[[87, 97]]]",
    "process": "Let the equation of line $ l $ be $ y = x + m $. Substituting into the hyperbola equation gives $ x^{2} - 2mx - m^{2} - 8 = 0 $. Since $ |AB| = \\sqrt{2} \\cdot \\sqrt{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}} = 4\\sqrt{5} $, it follows that $ \\sqrt{2} \\cdot \\sqrt{4m^{2} - 4 \\times (-m^{2} - 8)} = 4\\sqrt{m^{2} + 4} = 4\\sqrt{5} $. Solving yields $ m = \\pm 1 $. Therefore, the equation of line $ l $ is $ y = x \\pm 1 $."
  },
  {
    "text": "An ellipse $\\frac{x^{2}}{4}+y^{2}=1$ has a focus $F$, and a moving point $P$ lies on the ellipse. The circle with diameter $PF$ is always tangent to a fixed circle. Then, what is the equation of the fixed circle?",
    "fact_expressions": "G: Ellipse;H: Circle;C:Circle;F: Point;P: Point;Expression(G) = (x^2/4 + y^2 = 1);OneOf(Focus(G)) = F;PointOnCurve(P, G);IsDiameter(LineSegmentOf(P,F),H);IsTangent(C, H)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2+y^2=4",
    "fact_spans": "[[[0, 27], [43, 45]], [[61, 62]], [[66, 68], [72, 74]], [[33, 36]], [[39, 42]], [[0, 27]], [[0, 36]], [[39, 48]], [[49, 62]], [[61, 70]]]",
    "query_spans": "[[[72, 79]]]",
    "process": "Let $ F_{1} $ be the other focus of the ellipse, and $ M $ be the midpoint of segment $ PF $. According to the midline theorem of a triangle and the definition of an ellipse, we have $ |MO| = \\frac{1}{2}|PF_{1}| = \\frac{1}{2}(2a - |PF|) = a - \\frac{1}{2}|PF| $. Then, based on the positional relationship between two circles, we obtain the conclusion. [Detailed explanation] Let $ F_{1} $ be the other focus of the ellipse, and $ M $ be the midpoint of segment $ PF $. According to the given conditions, $ |MO| = \\frac{1}{2}|PF_{1}| = \\frac{1}{2}(2a - |PF|) = a - \\frac{1}{2}|PF| $, which implies that the circle with the major axis as diameter is internally tangent to the circle with segment $ PF $ as diameter. Therefore, the fixed circle has center $ O(0,0) $ and radius $ r = a = 2 $. Hence, the equation of the fixed circle is $ x^{2} + y^{2} = 4 $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ has two foci $F_{1}$, $F_{2}$, and there exists a point $P$ on the ellipse $C$ such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}<0$, then the range of real values for $m$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/m = 1);m: Real;F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C) = True;DotProduct(VectorOf(P, F1), VectorOf(P, F2)) < 0",
    "query_expressions": "Range(m)",
    "answer_expressions": "(0,2)+(8,+oo)",
    "fact_spans": "[[[2, 44], [67, 72]], [[2, 44]], [[142, 147]], [[50, 57]], [[58, 65]], [[2, 65]], [[75, 79]], [[67, 79]], [[81, 140]]]",
    "query_spans": "[[[142, 154]]]",
    "process": "By the given condition, there exists a point $ P $ on the ellipse $ C $ such that $ \\overrightarrow{PF_1} \\cdot \\overrightarrow{PF_2} < 0 $, i.e., there exists a point $ P $ for which $ \\angle F_1PF_2 $ is obtuse. Thus, it suffices that the maximum value of $ \\angle F_1PF_2 $ can be obtuse. When the foci are on the $ x $-axis, by the properties of the ellipse, $ \\angle F_1PF_2 $ reaches its maximum when $ P $ is at the upper or lower vertex. At this time, if $ \\angle F_1PF_2 > \\frac{\\pi}{2} $, then $ \\angle F_1PO > \\frac{\\pi}{4} $, hence $ \\sqrt{4 - m} > \\sqrt{m} $, solving gives $ m < 2 $. When the foci are on the $ y $-axis, by the properties of the ellipse, $ \\angle F_1PF_2 $ reaches its maximum when $ P $ is at the left or right vertex. At this time, if $ \\angle F_1PF_2 > \\frac{\\pi}{2} $, then $ \\angle F_1PO > \\frac{\\pi}{4} $, hence $ \\sqrt{m - 4} > 2 $, solving gives $ m > 8 $. Therefore, the range of $ m $ is $ (0, 2) \\cup (8, +\\infty) $."
  },
  {
    "text": "Let $b \\in R$. If the curve $y^{2}=-|x|+1$ and the line $y=-x+b$ have common points, then the range of values for $b$ is?",
    "fact_expressions": "G: Line;b: Real;H: Curve;Expression(G) = (y = b - x);Expression(H) = (y^2 = 1 - Abs(x));IsIntersect(H, G)",
    "query_expressions": "Range(b)",
    "answer_expressions": "[-5/4, 5/4]",
    "fact_spans": "[[[29, 39]], [[1, 10], [45, 48]], [[12, 28]], [[29, 39]], [[12, 28]], [[12, 43]]]",
    "query_spans": "[[[45, 55]]]",
    "process": "y^{2}=-|x|+1=\\begin{cases}-x+1,x\\geqslant0\\\\x+1,x\\leqslant0\\end{cases} as shown: when the line y=-x+b is tangent to the curve y^{2}=-x+1, \\begin{cases}y2=-x+1\\\\v=-x+b\\end{cases}\\Rightarrow x^{2}+(1-2b)x+b^{2}-1=0,\\Delta=(1-2b)^{2}-4(b^{2}-1)=0, solving gives b=\\frac{5}{4}. When the line y=-x+b is tangent to the curve y^{2}=x+1, \\begin{cases}y^{2}=x+1\\\\v=-x+b\\end{cases}\\Rightarrow x^{2}-(2b+1)x+b^{2}-1=0,\\triangle=(2b+1)^{2}-4(b^{2}-1)=0, solving gives b=-\\frac{5}{4}. Since the curve y^{2}=-|x|+1 and the line y=-x+b have common points, the range of values for b is [-\\frac{5}{4},\\frac{5}{4}]."
  },
  {
    "text": "The line $y = x + 1$ intersects the hyperbola $\\frac{x^{2}}{2} - \\frac{y^{2}}{3} = 1$ at points $A$ and $B$, $|AB| =$?",
    "fact_expressions": "H: Line;Expression(H) = (y = x + 1);G: Hyperbola;Expression(G) = (x^2/2 - y^2/3 = 1);A: Point;B: Point;Intersection(H, G) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "4*sqrt(6)",
    "fact_spans": "[[[0, 9]], [[0, 9]], [[10, 48]], [[10, 48]], [[51, 54]], [[55, 58]], [[0, 60]]]",
    "query_spans": "[[[61, 70]]]",
    "process": "Let points A(x_{1},y_{1}) and B(x_{2},y_{2}). Solving the system \\begin{cases}y=x+1\\\\3x^{2}-2y^{2}=6\\end{cases}, eliminate y and simplify to obtain x^{2}-4x-8=0. By Vieta's formulas, x_{1}+x_{2}=4, x_{1}x_{2}=-8. Using the chord length formula, |AB|=\\sqrt{1+1^{2}}\\cdot|x_{1}-x_{2}|=\\sqrt{2}\\cdot\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\sqrt{2}\\times\\sqrt{4^{2}-4\\times(-8)}=4\\sqrt{6}"
  },
  {
    "text": "Given that the equation of the parabola is $y^{2}=4 x$, then the distance from its focus to the directrix is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x)",
    "query_expressions": "Distance(Focus(G),Directrix(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 5], [22, 23]], [[2, 20]]]",
    "query_spans": "[[[22, 33]]]",
    "process": "The focus of the parabola \\( y^{2} = 4x \\) is \\( F(1,0) \\), and the equation of the directrix is \\( x = -1 \\); the distance from its focus to the directrix is 2."
  },
  {
    "text": "Given that $M$ is a moving point on the curve $y=4 x^{2}+1$, and point $M$ is the midpoint of line segment $O P$, then the equation of the trajectory of moving point $P$ is?",
    "fact_expressions": "G: Curve;P: Point;O: Origin;M: Point;Expression(G) = (y = 4*x^2 + 1);PointOnCurve(M, G);MidPoint(LineSegmentOf(O,P)) = M",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "y=2*x^2+2",
    "fact_spans": "[[[6, 21]], [[48, 51]], [[36, 41]], [[2, 5], [29, 33]], [[6, 21]], [[2, 27]], [[29, 44]]]",
    "query_spans": "[[[48, 58]]]",
    "process": "Let P(x,y), M(x_{0},y_{0}). Since M is the midpoint of OP, and M lies on the curve y=4x^{2}+1, then \\begin{cases}x_{0}=\\frac{x}{2}\\\\y_{0}=\\frac{y}{2}\\end{cases}, i.e., M(\\frac{x}{2},\\frac{y}{2}). Therefore, the trajectory equation of point P is: y=2x^{2}+2"
  },
  {
    "text": "If $x$, $y$ satisfy the equation $\\frac{x^{2}}{4}+y^{2}=1$, then the maximum value of $y-x$ is?",
    "fact_expressions": "x_:Number;y_: Number;x_^2/4 + y_^2 = 1",
    "query_expressions": "Max(-x_ + y_)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[1, 4]], [[7, 10]], [[14, 39]]]",
    "query_spans": "[[[41, 52]]]",
    "process": "Since the equation of the ellipse is $\\frac{x^{2}}{4}+y^{2}=1$, the parametric equations of the ellipse can be set as: $\\begin{cases}x=2\\cos\\theta\\\\v=\\sin\\theta\\end{cases}$, then $y-x=\\sin\\theta-2\\cos\\theta=\\sqrt{5}\\sin(\\theta+\\varphi)$, $\\sqrt{5}\\sin(\\theta+\\varphi)\\in[-\\sqrt{5},\\sqrt{5}]$, so the maximum value of $y-x$ is $\\sqrt{5}$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, draw a line $l$ through the point $(1,0)$ with an inclination angle of $45^{\\circ}$, intersecting the ellipse at points $A$ and $B$. Let $O$ be the origin. Then the area of $\\triangle A O B$ is?",
    "fact_expressions": "l: Line;G: Ellipse;H: Point;A: Point;O: Origin;B: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Coordinate(H) = (1, 0);PointOnCurve(H, l);Inclination(l) = ApplyUnit(45, degree);Intersection(l, G) = {A, B}",
    "query_expressions": "Area(TriangleOf(A, O, B))",
    "answer_expressions": "6*sqrt(2)/7",
    "fact_spans": "[[[68, 73]], [[2, 39], [74, 76]], [[42, 50]], [[77, 80]], [[87, 90]], [[81, 84]], [[2, 39]], [[42, 50]], [[41, 73]], [[51, 73]], [[68, 86]]]",
    "query_spans": "[[[97, 119]]]",
    "process": ""
  },
  {
    "text": "If a point $(5, t)$ on the parabola $C$: $y^{2}=2 p x(p>0)$ is at a distance of $6$ from the focus, and $P$, $Q$ are moving points on the parabola and the circle $(x-6)^{2}+y^{2}=1$ respectively, then the minimum value of $|P Q|$ is?",
    "fact_expressions": "C: Parabola;p: Number;G: Circle;H: Point;P: Point;Q: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Expression(G) = (y^2 + (x - 6)^2 = 1);Coordinate(H) = (5, t);PointOnCurve(H, C);PointOnCurve(P,C);PointOnCurve(Q,G);Distance(H, Focus(C)) = 6;t:Number",
    "query_expressions": "Min(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "2*sqrt(5)-1",
    "fact_spans": "[[[1, 27], [62, 65]], [[9, 27]], [[66, 86]], [[30, 38]], [[50, 53]], [[56, 59]], [[9, 27]], [[1, 27]], [[66, 86]], [[30, 38]], [[1, 38]], [[50, 90]], [[50, 90]], [[1, 48]], [[30, 38]]]",
    "query_spans": "[[[92, 105]]]",
    "process": "According to the definition of a parabola, $ 5 + \\frac{p}{2} = 6 $, the parameter $ p $ can be determined. Then the problem is transformed into finding the minimum distance from the center of the circle $ A(6,0) $ to a point on the parabola. By using the distance formula between two points and the properties of quadratic functions, the minimum value of $ |PQ| $ can be obtained. [Detailed Solution] From the given condition and the definition of the parabola: $ 5 + \\frac{p}{2} = 6 $, we get $ p = 2 $, hence $ C: y^{2} = 4x $. The circle $ (x-6)^{2} + y^{2} = 1 $ has center $ A(6,0) $ and radius $ 1 $. Therefore, $ |PQ| $ is minimized when $ A $, $ Q $, and $ P $ are collinear and $ |PQ| = |AP| - 1 $. Thus, it suffices to minimize $ |AP| $. Let $ P(x,y) $, then $ |AP| = \\sqrt{(x-6)^{2} + y^{2}} = \\sqrt{(x-4)^{2} + 20} $, with $ x \\geqslant 0 $. When $ x = 4 $, $ |AP|_{\\min} = 2\\sqrt{5} $, so the minimum value of $ |PQ| $ is $ 2\\sqrt{5} - 1 $."
  },
  {
    "text": "The standard equation of a hyperbola with a focus at $(-6,0)$ and passing through the point $(5,-2)$ is?",
    "fact_expressions": "G: Hyperbola;Coordinate(OneOf(Focus(G))) = (-6, 0);I: Point;Coordinate(I) = (5,-2);PointOnCurve(I , G) = True",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/20 - y^2/16 = 1",
    "fact_spans": "[[[26, 29]], [[0, 29]], [[15, 25]], [[15, 25]], [[14, 29]]]",
    "query_spans": "[[[26, 35]]]",
    "process": ""
  },
  {
    "text": "If points $O$ and $F$ are the center and the left focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, respectively, and point $P$ is any point on the ellipse, then the maximum value of $\\overrightarrow{O P} \\cdot \\overrightarrow{F P}$ is?",
    "fact_expressions": "G: Ellipse;O: Point;P: Point;F: Point;Expression(G) = (x^2/4 + y^2/3 = 1);PointOnCurve(P, G);Center(G)=O;LeftFocus(G)=F",
    "query_expressions": "Max(DotProduct(VectorOf(O, P), VectorOf(F, P)))",
    "answer_expressions": "6",
    "fact_spans": "[[[13, 50], [63, 65]], [[1, 5]], [[58, 62]], [[6, 10]], [[13, 50]], [[58, 71]], [[1, 57]], [[1, 57]]]",
    "query_spans": "[[[73, 128]]]",
    "process": "From the ellipse equation, obtain the coordinates of F and O. Let P(x, y) (-2 \\leqslant x \\leqslant 2). Use coordinate operations of the dot product to transform \\overrightarrow{OP} \\cdot \\overrightarrow{FP} into a quadratic function and find its maximum value. From the ellipse \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1, we get F(-1, 0), point O(0, 0). Let P(x, y) (-2 \\leqslant x \\leqslant 2). Then \\overrightarrow{OP} \\cdot \\overrightarrow{FP} = x^{2} + x + y^{2} = x^{2} + x + 3\\left(1 - \\frac{x^{2}}{4}\\right) = \\frac{1}{4}x^{2} + x + 3 = \\frac{1}{4}(x + 2)^{2} + 2, -2 \\leqslant x \\leqslant 2. When x = 2, \\overrightarrow{OP} \\cdot \\overrightarrow{FP} reaches the maximum value 6."
  },
  {
    "text": "The focal distance of the hyperbola $\\frac{x^{2}}{10}-\\frac{y^{2}}{2}=1$ is (fill in the blank with a number)?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/10 - y^2/2 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 52]]]",
    "process": "In the hyperbola $\\frac{x^2}{10}-\\frac{y^{2}}{2}=1$, we have: $a^{2}=10$, $b^{2}=2$, so $c^{2}=a^{2}+b^{2}=12$, $c=2\\sqrt{3}$, therefore the focal distance is $2c=4\\sqrt{3}$."
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $4 x^{2}-3 y^{2}=-12$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (4*x^2 - 3*y^2 = -12)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*2*sqrt(3)*x/3",
    "fact_spans": "[[[0, 24]], [[0, 24]]]",
    "query_spans": "[[[0, 32]]]",
    "process": "The standard equation of this hyperbola is \\frac{y^{2}}{4}-\\frac{x^{2}}{3}=1, its foci lie on the y-axis, where a=2, b=\\sqrt{3}, so the asymptote equations are y=\\pm\\frac{a}{b}x=\\pm\\frac{2}{\\frac{2}{5}}x=\\pm\\frac{2\\sqrt{3}}{3}x"
  },
  {
    "text": "Given the hyperbola $x^{2}-y^{2}=1$, points $F_{1}$ and $F_{2}$ are its two foci, and point $P$ is a point on the hyperbola. If $PF_{1} \\perp PF_{2}$, then the value of $|PF_{1}|+|PF_{2}|$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))",
    "query_expressions": "Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[2, 20], [38, 39], [49, 52]], [[44, 48]], [[21, 29]], [[30, 37]], [[2, 20]], [[21, 43]], [[44, 55]], [[57, 78]]]",
    "query_spans": "[[[80, 103]]]",
    "process": ""
  },
  {
    "text": "The foci of the hyperbola $\\frac{x^{2}}{n}-y^{2}=1(n>1)$ are $F_{1}$, $F_{2}$, and point $P$ lies on the hyperbola such that $|P F_{1}|+|P F_{2}|=2 \\sqrt{n+2}$. Then the area of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2 + x^2/n = 1);n: Number;n>1;F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 2*sqrt(n + 2)",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 33], [59, 62]], [[0, 33]], [[3, 33]], [[3, 33]], [[37, 44]], [[45, 54]], [[0, 54]], [[55, 58]], [[55, 63]], [[68, 102]]]",
    "query_spans": "[[[104, 131]]]",
    "process": "Given: Let $ F_{1} $, $ F_{2} $ be the left and right foci of the hyperbola, and $ P $ a point on the right branch, $ |PF_{1}| - |PF_{2}| = 2\\sqrt{n} $ ①, $ |PF_{1}| + |PF_{2}| = 2\\sqrt{n+2} $ ②. Solving ① and ② yields: $ |PF_{1}| = \\sqrt{n} + \\sqrt{n+2} $, $ |PF_{2}| = \\sqrt{n+2} - \\sqrt{n} $. Hence, $ |PF_{1}|^{2} + |PF_{2}|^{2} = |F_{1}F_{2}|^{2} $, therefore $ PF_{1} \\bot PF_{2} $. Also, by squaring ① and ② separately and subtracting, we obtain: $ PF_{1} \\cdot PF_{2} = 2 $, then the area of $ \\triangle PF_{1}F_{2} $ is: 1"
  },
  {
    "text": "Draw a perpendicular line from a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ to the real axis, intersecting the hyperbola at points $A$ and $B$. If the length of segment $AB$ is exactly equal to the focal distance, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;A: Point;B: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);L:Line;PointOnCurve(OneOf(Focus(G)),L);IsPerpendicular(L,RealAxis(G));Intersection(L,G)={A,B};Length(LineSegmentOf(A,B))=FocalLength(G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(\\sqrt{5}+1)/2",
    "fact_spans": "[[[1, 57], [70, 73], [102, 105]], [[4, 57]], [[4, 57]], [[74, 77]], [[78, 81]], [[4, 57]], [[4, 57]], [[1, 57]], [], [[0, 68]], [[0, 68]], [[0, 83]], [[85, 100]]]",
    "query_spans": "[[[102, 111]]]",
    "process": ""
  },
  {
    "text": "The left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are $F_{1}$ and $F_{2}$ respectively, and $P$ is a point on the right branch of the hyperbola. $P F_{2}$ is tangent to the circle $x^{2}+y^{2}=b^{2}$ at point $G$, and $G$ is the midpoint of $P F_{2}$. Then the eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;H: Circle;P: Point;F1:Point;F2: Point;G: Point;e: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + y^2 = b^2);LeftFocus(C) =F1;RightFocus(C)=F2;PointOnCurve(P,RightPart(C));TangentPoint(LineSegmentOf(P,F2),H)=G;MidPoint(LineSegmentOf(P, F2)) = G;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[0, 56], [85, 88], [152, 155]], [[3, 56]], [[3, 56]], [[104, 124]], [[81, 84]], [[65, 72]], [[73, 80]], [[126, 130], [133, 136]], [[159, 162]], [[3, 56]], [[3, 56]], [[0, 56]], [[104, 124]], [[0, 80]], [[0, 80]], [[81, 93]], [[94, 130]], [[133, 149]], [[152, 162]]]",
    "query_spans": "[[[159, 164]]]",
    "process": "According to the problem, |PF₂| = 2a, |PF₁| = 4a, ∠F₁PF₂ = 90°. By the Pythagorean theorem, we obtain the relationship between a and c, and then find the eccentricity. Connect PF₁. Since O and G are the midpoints of F₁F₂ and PF₂ respectively, OG ∥ PF₁. Because OG ⊥ PF₂, it follows that PF₁ ∥ OG and PF₁ ⊥ PF₂. Since OG = b, PF₁ = 2b, so PF₂ = 2b − 2a. By the Pythagorean theorem, (2b)² + (2b − 2a)² = (2c)² = 4a² + 4b². Thus, b = 2a, therefore e = c/a = √5."
  },
  {
    "text": "The coordinates of the point on the right branch of the hyperbola $x^{2}-y^{2}=1$ whose distance to the line $y=x$ is $\\sqrt{2}$ are?",
    "fact_expressions": "G: Hyperbola;H: Line;Expression(G) = (x^2 - y^2 = 1);Expression(H) = (y = x);P:Point;Distance(P,H)=sqrt(2);PointOnCurve(P,RightPart(G))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(5/4,-3/4)",
    "fact_spans": "[[[0, 18]], [[23, 30]], [[0, 18]], [[23, 30]], [[45, 46]], [[0, 46]], [[0, 46]]]",
    "query_spans": "[[[45, 51]]]",
    "process": "Let the coordinates of the desired point be $(a,b)$ $(a>0)$, then this point is the solution of the system of equations  \n$$\n\\begin{cases}\na2 - b^{2} = 1 \\\\\n\\frac{|a - b|}{\\sqrt{1^{2} + 1^{2}}} = \\sqrt{2}\n\\end{cases}\n$$\nCombining with $a > 0$, the coordinates of the point can be obtained. Let the coordinates of the desired point be $(a,b)$ $(a>0)$, then  \n$$\n\\begin{cases}\na2 - b^{2} = 1 \\textcircled{1} \\\\\n\\frac{|a - b|}{\\sqrt{1^{2} + 1^{2}}} = \\sqrt{2} \\textcircled{2}\n\\end{cases}\n$$\nFrom \\textcircled{1} we get $a^{2} = 1 + b^{2}$, that is, $a = \\sqrt{1 + b^{2}} > b$, so equation \\textcircled{2} becomes: $a - b = 2 \\textcircled{3}$. Substituting $a = b + 2$ into $a^{2} - b^{2} = 1$ gives $b = -\\frac{3}{4}$, so $a = \\frac{5}{4}$. The coordinates of the desired point are: $\\left(\\frac{5}{4}, -\\frac{3}{4}\\right)$."
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and let $P$ lie on the ellipse such that $\\angle F_{1} P F_{2}=60^{\\circ}$. Then the area of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[17, 44], [54, 56]], [[50, 53]], [[1, 8]], [[9, 16]], [[17, 44]], [[1, 49]], [[50, 57]], [[61, 94]]]",
    "query_spans": "[[[96, 126]]]",
    "process": "From the given conditions, we have  \n\\begin{cases}  \n|PF_{1}|+|PF_{2}|=4 \\\\  \n|PF_{1}|^{2}+|PF_{2}|^{2}-|PF_{1}|\\cdot|PF_{2}|=12  \n\\end{cases},  \nthen $3|PF_{1}|\\cdot|PF_{2}|=4^{2}-12$, that is, $|PF_{1}|\\cdot|PF_{2}|=\\frac{4}{3}$, so the area of $\\triangle PF_{1}F_{2}$ is $S=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|\\sin60^{\\circ}=\\frac{\\sqrt{3}}{3}$."
  },
  {
    "text": "The focal length of the ellipse $\\frac{x^{2}}{15}+\\frac{y^{2}}{2+m}=1$ is $4$, then $m$=?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/15 + y^2/(m + 2) = 1);FocalLength(G) = 4",
    "query_expressions": "m",
    "answer_expressions": "{9, 17}",
    "fact_spans": "[[[0, 40]], [[49, 52]], [[0, 40]], [[0, 47]]]",
    "query_spans": "[[[49, 54]]]",
    "process": "Because $\\frac{x2}{15}+\\frac{y^{2}}{y+m}=1$ represents an ellipse, $m>-2$ and $m\\neq13$. When the foci of the ellipse are on the $x$-axis, $a^{2}=15$, $b^{2}=2+m$, so $15=2+m+2^{2}$, i.e., $m=9$; when the foci of the ellipse are on the $y$-axis, $a2=2+m$, $b^{2}=15$, so $2+m=15+2^{2}$, i.e., $m=17$; hence the answer is: $9$ or $17$."
  },
  {
    "text": "The major axis length of the ellipse $x^{2}+9 y^{2}=25$ is? The minor axis length is? The coordinates of the foci are? The coordinates of the vertices are?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 9*y^2 = 25)",
    "query_expressions": "Length(MajorAxis(G));Length(MinorAxis(G));Coordinate(Focus(G));Coordinate(Vertex(G))",
    "answer_expressions": "10\n10/3\n(pm*(10/3)*sqrt(2),0)\n{(pm*5,0),(0,pm*5/3)}",
    "fact_spans": "[[[0, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 26]], [[0, 31]], [[0, 37]], [[0, 43]]]",
    "process": "From the given conditions: the standard equation of the ellipse is $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, $\\therefore a=5$, $b=\\frac{5}{3}$, $c=\\frac{10\\sqrt{2}}{3}$, i.e., the major axis length is $10$, the minor axis length is $\\frac{10}{3}$, the foci coordinates are $(\\pm\\frac{10}{3}\\sqrt{2},0)$, and the vertices coordinates are $(\\pm5,0)$, $(0,\\pm\\frac{5}{3})$."
  },
  {
    "text": "The equation of the ellipse with the two foci and the two endpoints of the minor axis of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ as its four vertices is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);H: Ellipse;Vertex(H) = {Focus(G), Endpoint(MinorAxis(G))}",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2/16 + x^2/9 = 1",
    "fact_spans": "[[[1, 40]], [[1, 40]], [[59, 61]], [[0, 61]]]",
    "query_spans": "[[[59, 66]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has an asymptote with equation $y=\\frac{3}{4}x$, $P$ is a point on this hyperbola, $F_{1}$, $F_{2}$ are its left and right foci, respectively, such that $P F_{1} \\perp P F_{2}$ and $|P F_{1}| \\cdot |P F_{2}|=18$. Then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G)))=(y=(3/4)*x);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 18",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16 - y^2/9 = 1",
    "fact_spans": "[[[2, 58], [90, 93], [113, 114], [177, 180]], [[5, 58]], [[5, 58]], [[85, 88]], [[97, 104]], [[105, 112]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 84]], [[85, 96]], [[97, 119]], [[97, 119]], [[121, 144]], [[145, 174]]]",
    "query_spans": "[[[177, 185]]]",
    "process": "From the asymptote equation, the slope gives $\\frac{b}{a}=\\frac{3}{4}$. According to the definition of the hyperbola and the Pythagorean theorem, $b=3$, then $c=5$, $a=4$, thus the equation of the hyperbola can be obtained. Let $c=\\sqrt{a^{2}+b^{2}}$, then from the asymptote equation $y=\\frac{3}{4}x$, $\\frac{b}{a}=\\frac{3}{4}$, and  \n\\begin{cases}\n||PF_{1}|-|PF_{2}||=2a,\\\\\n|PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2},\n\\end{cases}  \n$\\left|\\frac{|F_{1}F_{2}|}{2}\\right||PF_{1}|\\cdot|PF_{2}|=4a^{2}$, so  \n\\begin{cases}\n|PF_{1}|^{2}+|PF_{2}|^{2}-2|\\overrightarrow{P}\\\\\n|PF_{1}|^{2}+|PF_{2}|^{2}=4c^{2}.\n\\end{cases}  \nSubtracting the two equations yields $2|PF_{1}|\\cdot|PF_{2}|=4b^{2}$, and since $|PF_{1}|\\cdot|PF_{2}|=18$, it follows that $b^{2}=9$, so $b=3$, hence $c=5$, $a=4$. Therefore, the equation of the hyperbola is $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$."
  },
  {
    "text": "The focus of the parabola $C$: $y^{2}=2 p x(p>0)$ is $F$, and $M$ is a point on the parabola $C$. If the circumcircle of $\\triangle O F M$ is tangent to the directrix of the parabola $C$, and the area of this circle is $36 \\pi$, then the value of $p$ is?",
    "fact_expressions": "C: Parabola;p: Number;O: Origin;F: Point;M: Point;p>0;Expression(C) = (y^2 = 2*p*x);Focus(C)=F;PointOnCurve(M,C);Area(CircumCircle(TriangleOf(O,F,M)))=36*pi;IsTangent(CircumCircle(TriangleOf(O,F,M)),Directrix(C))",
    "query_expressions": "p",
    "answer_expressions": "8",
    "fact_spans": "[[[0, 26], [38, 44], [38, 44]], [[99, 102]], [[48, 65]], [[30, 33]], [[34, 37]], [[8, 26]], [[0, 26]], [[0, 33]], [[34, 47]], [[48, 97]], [[48, 81]]]",
    "query_spans": "[[[99, 106]]]",
    "process": "Let the center of the circumcircle of $\\triangle OFM$ be $O_{1}$, then $|O_{1}O|=|O_{1}F|=|O_{1}M|$, so $O_{1}$ lies on the perpendicular bisector of segment $OF$. Since circle $O_{1}$ is tangent to the directrix of the parabola, point $O_{1}$ lies on the parabola, so $O_{1}\\left(\\frac{p}{4},\\frac{\\sqrt{2}}{2}\\right)$. Since the area of the circle is $36\\pi$, the radius of the circle is 6, so $\\frac{p^{2}}{16}+\\frac{1}{2}p^{2}=36$, solving gives $p=8$."
  },
  {
    "text": "Given that the eccentricity of the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{4}=1$ $(a>2)$ is $\\frac{\\sqrt{5}}{5}$, then $a$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/4 + x^2/a^2 = 1);a: Number;a>2;Eccentricity(G) = sqrt(5)/5",
    "query_expressions": "a",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 50]], [[2, 50]], [[77, 80]], [[4, 50]], [[2, 75]]]",
    "query_spans": "[[[77, 82]]]",
    "process": ""
  },
  {
    "text": "Given $a > b > 0$, $e_{1}$, $e_{2}$ are the eccentricities of the conic sections $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1$ and $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$, respectively. Let $m = \\lg e_{1} + \\lg e_{2}$, then the range of values of $m$ is?",
    "fact_expressions": "G: ConicSection;H: ConicSection;b: Number;a: Number;a>b;b>0;e1:Number;e2:Number;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(H)=e1;Eccentricity(G)=e2;m:Number;m=lg(e1)+lg(e2)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-oo,0)",
    "fact_spans": "[[[28, 75]], [[76, 119]], [[2, 9]], [[2, 9]], [[2, 9]], [[2, 9]], [[10, 17]], [[18, 25]], [[28, 75]], [[76, 119]], [[10, 123]], [[10, 123]], [[150, 153]], [[125, 148]]]",
    "query_spans": "[[[150, 160]]]",
    "process": ""
  },
  {
    "text": "Given that the distance from a moving point $P$ to the fixed point $(2,0)$ is equal to its distance to the fixed line $l$: $x=-2$, then the equation of the trajectory of point $P$ is?",
    "fact_expressions": "P: Point;G: Point;Coordinate(G) = (2, 0);l: Line;Expression(l) = (x=-2);Distance(P, G) = Distance(P, l)",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "y^2=8*x",
    "fact_spans": "[[[4, 7], [44, 48], [21, 22]], [[10, 17]], [[10, 17]], [[26, 37]], [[26, 37]], [[4, 42]]]",
    "query_spans": "[[[44, 55]]]",
    "process": ""
  },
  {
    "text": "The center is at the origin, and one asymptote of a hyperbola with foci on the $x$-axis passes through the point $(2,-1)$. What is its eccentricity?",
    "fact_expressions": "Center(G) = O;O: Origin;PointOnCurve(Focus(G), xAxis);G: Hyperbola;PointOnCurve(H, OneOf(Asymptote(G)));H: Point;Coordinate(H) = (2, -1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[0, 18]], [[3, 5]], [[6, 18]], [[15, 18], [37, 38]], [[15, 35]], [[26, 35]], [[26, 35]]]",
    "query_spans": "[[[37, 44]]]",
    "process": "From the given condition, $\\frac{b}{a}=\\frac{1}{2}$, therefore the eccentricity $e=\\sqrt{1+(\\frac{b}{a})^{2}}=\\frac{\\sqrt{5}}{2}$, hence fill in: $\\frac{\\sqrt{5}}{2}$."
  },
  {
    "text": "For the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{t}=1$ with foci on the $x$-axis, the distance from any point on the ellipse to one of its foci is always greater than $1$. Then the range of values for $t$ is?",
    "fact_expressions": "G: Ellipse;t: Number;Expression(G) = (x^2/4 + y^2/t = 1);PointOnCurve(Focus(G), xAxis);P: Point;PointOnCurve(P, G);Distance(P, OneOf(Focus(G))) > 1",
    "query_expressions": "Range(t)",
    "answer_expressions": "(3, 4)",
    "fact_spans": "[[[9, 46]], [[69, 72]], [[9, 46]], [[0, 46]], [], [[9, 51]], [[9, 67]]]",
    "query_spans": "[[[69, 79]]]",
    "process": "From the given condition we have $ a - c > 1 $, that is, $ 2 - \\sqrt{4 - t} > 1 $, solving yields $ 3 < t < 4 $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and point $M$ lies on $C$, then the maximum value of $|M F_{1}|\\cdot|M F_{2}|$ is $25$. Find $a=?$",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;M: Point;F1: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Focus(C) = {F1, F2};PointOnCurve(M, C);Max(Abs(LineSegmentOf(M, F1))*Abs(LineSegmentOf(M, F2))) = 25",
    "query_expressions": "a",
    "answer_expressions": "5",
    "fact_spans": "[[[18, 75], [86, 89]], [[25, 75]], [[128, 131]], [[81, 85]], [[2, 9]], [[10, 17]], [[25, 75]], [[25, 75]], [[18, 75]], [[2, 80]], [[81, 90]], [[92, 126]]]",
    "query_spans": "[[[128, 133]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $E$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1$ $(0<b<2)$ with right focus $F$. Let $M$ be one endpoint of the minor axis, and the line $l$: $3x-4y=0$. If the distance from point $M$ to line $l$ is at least $\\frac{4}{5}$, then the range of values for the eccentricity of ellipse $E$ is?",
    "fact_expressions": "l: Line;E: Ellipse;b: Number;F: Point;M: Point;0 < b;b < 2;Expression(E) = (x^2/4 + y^2/b^2 = 1);Expression(l)=(3*x-4*y=0);RightFocus(E) = F;OneOf(Endpoint(MinorAxis(E)))=M;Negation(Distance(M, l) < 4/5)",
    "query_expressions": "Range(Eccentricity(E))",
    "answer_expressions": "(0, \\sqrt{3}/2]",
    "fact_spans": "[[[77, 95], [102, 107]], [[2, 55], [130, 135]], [[9, 55]], [[60, 63]], [[73, 76], [97, 101]], [[9, 55]], [[9, 55]], [[2, 55]], [[77, 95]], [[2, 63]], [[2, 76]], [[97, 127]]]",
    "query_spans": "[[[130, 146]]]",
    "process": "By the symmetry of the ellipse, we may take M(0,b). The distance from point M to line l is d = \\frac{|b \\times 0 - 4 \\times b|}{\\sqrt{3^{2} + (-4)^{2}}} = \\frac{4b}{5} \\geqslant \\frac{4}{5}, \\therefore b \\geqslant 1. Also, since 0 < b < 2, \\therefore 1 \\leqslant b < 2. \\therefore e^{2} = \\frac{4 - b^{2}}{4} = 1 - \\frac{b^{2}}{4} \\in (0, \\frac{3}{4}], \\therefore e \\in (0, \\frac{\\sqrt{3}}{2}]"
  },
  {
    "text": "Let the line $y=k x$ intersect the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ at points $A$ and $B$. Perpendiculars are drawn from $A$ and $B$ to the $x$-axis. If the feet of the perpendiculars coincide with the two foci of the ellipse, then what is the value of $k$?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*x);k: Number;G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);A: Point;B: Point;Intersection(H, G) = {A, B};L1: Line;L2: Line;PointOnCurve(A, L1);PointOnCurve(B, L2);IsPerpendicular(L1, xAxis);IsPerpendicular(L2, xAxis);F1: Point;F2: Point;Focus(G) = {F1, F2};FootPoint(L1, xAxis) = F1;FootPoint(L2, xAxis) = F2",
    "query_expressions": "k",
    "answer_expressions": "pm*(3/2)",
    "fact_spans": "[[[1, 10]], [[1, 10]], [[94, 97]], [[11, 48], [85, 87]], [[11, 48]], [[51, 54], [64, 67]], [[55, 58], [68, 71]], [[1, 60]], [], [], [[63, 79]], [[63, 79]], [[63, 79]], [[63, 79]], [], [], [[85, 92]], [[63, 92]], [[63, 92]]]",
    "query_spans": "[[[94, 100]]]",
    "process": ""
  },
  {
    "text": "The standard equation of an ellipse with foci on the $y$-axis, focal length $2 \\sqrt{6}$, and passing through the point $(\\sqrt{2}, \\sqrt{3})$ is?",
    "fact_expressions": "PointOnCurve(Focus(G), yAxis);FocalLength(G) = 2*sqrt(6);G: Ellipse;H: Point;Coordinate(H) = (sqrt(2), sqrt(3));PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/9 + x^2/3 = 1",
    "fact_spans": "[[[0, 52]], [[9, 52]], [[50, 52]], [[26, 49]], [[26, 49]], [[25, 52]]]",
    "query_spans": "[[[50, 59]]]",
    "process": "By the given conditions, the equation of the ellipse can be written as: $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$ $(a>b>0)$. Since the focal distance is $2\\sqrt{6}$, i.e., $2c=2\\sqrt{6}$, we have $c=\\sqrt{6}$, thus $a^{2}=b^{2}+6$, and the equation becomes $\\frac{y^{2}}{b^{2}+6}+\\frac{x^{2}}{b^{2}}=1$. Substituting the point $(\\sqrt{2},\\sqrt{3})$ gives: $\\frac{3}{b^{2}+6}+\\frac{2}{b^{2}}=1$, solving yields $b^{2}=3$, $a^{2}=9$. Therefore, the standard equation of the ellipse is: $\\frac{y^{2}}{9}+\\frac{x^{2}}{3}=1$."
  },
  {
    "text": "If the directrix of a parabola is given by the equation $2x + 3y - 1 = 0$ and the focus is $(-2, 1)$, then what is the equation of the axis of symmetry of the parabola?",
    "fact_expressions": "G: Parabola;H: Point;Coordinate(H) = (-2, 1);Expression(Directrix(G)) = (2*x + 3*y - 1 = 0);Focus(G) = H",
    "query_expressions": "Expression(SymmetryAxis(G))",
    "answer_expressions": "3*x-2*y+8=0",
    "fact_spans": "[[[1, 4], [39, 42]], [[28, 36]], [[28, 36]], [[1, 23]], [[1, 36]]]",
    "query_spans": "[[[39, 50]]]",
    "process": ""
  },
  {
    "text": "The eccentricity $e$ of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$ is?",
    "fact_expressions": "G: Hyperbola;e: Number;Expression(G) = (x^2/4 - y^2/9 = 1);Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(13)/2",
    "fact_spans": "[[[0, 38]], [[42, 45]], [[0, 38]], [[0, 45]]]",
    "query_spans": "[[[42, 47]]]",
    "process": "From $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$, we get $a=2$, $b=3$, hence $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{13}$, and the eccentricity $e=\\frac{c}{a}=\\frac{\\sqrt{13}}{2}$"
  },
  {
    "text": "The equation of the right directrix of the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/8 + y^2/4 = 1)",
    "query_expressions": "Expression(RightDirectrix(G))",
    "answer_expressions": "x=4",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 45]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ with right focus $F$, and let $A$ be a point on the ellipse in the first quadrant. Connect $AF$ and extend it to intersect the ellipse again at point $B$. Connect $AO$ ($O$ being the origin) and extend it to intersect the ellipse again at point $C$. If $S_{\\triangle ABC}=3$, then the coordinates of point $C$ are?",
    "fact_expressions": "G: Ellipse;A: Point;F: Point;O: Origin;B: Point;C: Point;Expression(G) = (x^2/4 + y^2/3 = 1);RightFocus(G) = F;PointOnCurve(A,G);Quadrant(A) = 1;Intersection(OverlappingLine(LineSegmentOf(A, F)), G) = B;Intersection(OverlappingLine(LineSegmentOf(A, O)), G) = C;Area(TriangleOf(A, B, C)) = 3",
    "query_expressions": "Coordinate(C)",
    "answer_expressions": "(-1,-3/2)",
    "fact_spans": "[[[2, 39], [52, 54], [74, 76], [104, 106]], [[48, 51]], [[44, 47]], [[90, 93]], [[77, 81]], [[107, 111], [138, 142]], [[2, 39]], [[2, 47]], [[48, 62]], [[48, 62]], [[63, 81]], [[82, 111]], [[113, 136]]]",
    "query_spans": "[[[138, 147]]]",
    "process": "Obtain F(1,0), let the equation of AB be x=my+1, combine with the ellipse equation, apply Vieta's formulas and the perfect square formula, and according to the problem condition we have S_{\\triangleABO}=S_{\\triangleAOF}+S_{\\triangleBOF}=\\frac{1}{2}\\cdot|OF|\\cdot|y_{1}-y_{2}|=\\frac{3}{2}, thus |y_{1}-y_{2}|=3. Squaring both sides and using Vieta's formulas, we can find the coordinates of C. [Solution] Draw the graph according to the problem: from the given conditions we get F(1,0). Let the equation of AB be x=my+1. Combining with the ellipse equation yields (4+3m^{2})y^{2}+6my-9=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), then y_{1}+y_{2}=-\\frac{6m}{4+3m^{2}}, y_{1}y_{2}=-\\frac{9}{4+3m^{2}}. Since O is the midpoint of AC and the area of \\triangleABC is 3, the area of \\triangleABO is \\frac{3}{2}. S_{\\DeltaABO}=S_{\\DeltaAOF}+S_{\\DeltaBOF}=\\frac{1}{2}\\cdot|OF|\\cdot|y_{1}-y_{2}|=\\frac{3}{2}, hence |y_{1}-y_{2}|=3. We obtain \\frac{36m^{2}}{(4+3m^{2})^{2}}+\\frac{36}{4+3m^{2}}=9, which simplifies to 9m^{4}+8m^{2}=0, so m=0. Then AB\\botx-axis, giving A(1,\\frac{3}{2}). Since point A and point C are symmetric about the origin, the coordinates of point C are (-1,-\\frac{3}{2})."
  },
  {
    "text": "If the line $y=kx-2$ intersects the parabola $y^{2}=8x$ at two distinct points $A$ and $B$, and the ordinate of the midpoint of $AB$ is $2$, then $k=$?",
    "fact_expressions": "G: Parabola;H: Line;k: Number;A: Point;B: Point;Expression(G) = (y^2 = 8*x);Expression(H) = (y = k*x - 2);Intersection(H, G) = {A, B};Negation(A=B);YCoordinate(MidPoint(LineSegmentOf(A, B))) = 2",
    "query_expressions": "k",
    "answer_expressions": "2",
    "fact_spans": "[[[13, 27]], [[1, 12]], [[60, 63]], [[35, 38]], [[39, 42]], [[13, 27]], [[1, 12]], [[1, 42]], [[30, 42]], [[44, 58]]]",
    "query_spans": "[[[60, 65]]]",
    "process": "Substitute the line $ y = kx - 2 $ into the parabola $ y^2 = 8x $, eliminate $ y $, and we obtain a quadratic equation. Using the fact that the ordinate of the midpoint of segment $ AB $ is 2, combined with Vieta's formulas, we can find the value of $ k $. Substituting $ y = kx - 2 $ into $ y^2 = 8x $, eliminating $ y $ gives $ k^2x^2 + (-4k - 8)x + 4 = 0 $. Let $ A(x_1, y_1) $, $ B(x_2, y_2) $, then $ x_1 + x_2 = \\frac{4k + 8}{k^2} $. Since the ordinate of the midpoint of segment $ AB $ is 2, $ y_1 + y_2 = 4 $, so $ k(x_1 + x_2) - 4 = 4 $, thus $ k \\cdot \\frac{4k + 8}{k^2} - 4 = 4 $, therefore $ k = 2 $."
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-y^{2}=1$ with right focus $F$, what is the distance from point $F$ to an asymptote of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;F: Point;Expression(C) = (x^2 - y^2 = 1);RightFocus(C) = F",
    "query_expressions": "Distance(F, OneOf(Asymptote(C)))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 25], [40, 46]], [[30, 33], [35, 39]], [[2, 25]], [[2, 33]]]",
    "query_spans": "[[[35, 57]]]",
    "process": ""
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2x$ be $F$, the directrix be $l$, and let chord $AB$ pass through point $F$ with midpoint $M$. Draw perpendiculars from points $F$ and $M$ to $AB$, intersecting $l$ at points $P$ and $Q$, respectively. If $|AF|=3|BF|$, then $|FP|\\cdot|MQ|=$?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;F: Point;M: Point;Q: Point;P: Point;l: Line;Expression(G) = (y^2 = 2*x);Focus(G) = F;Directrix(G) = l;IsChordOf(LineSegmentOf(A,B),G);PointOnCurve(F,LineSegmentOf(A,B));MidPoint(LineSegmentOf(A,B))=M;L1:Line;L2:Line;PointOnCurve(F,L1);PointOnCurve(M,L2);IsPerpendicular(L1,LineSegmentOf(A,B));IsPerpendicular(L2,LineSegmentOf(A,B));Intersection(L1,l)=P;Intersection(L2,l)=Q;Abs(LineSegmentOf(A,F))=3*Abs(LineSegmentOf(B,F))",
    "query_expressions": "Abs(LineSegmentOf(F, P))*Abs(LineSegmentOf(M, Q))",
    "answer_expressions": "16/9",
    "fact_spans": "[[[1, 15]], [[31, 36]], [[31, 36]], [[37, 41], [19, 22], [50, 54]], [[45, 48], [55, 58]], [[80, 84]], [[74, 78]], [[26, 29], [70, 73]], [[1, 15]], [[1, 22]], [[1, 29]], [[1, 36]], [[31, 41]], [[31, 48]], [], [], [[49, 69]], [[49, 69]], [[49, 69]], [[49, 69]], [[49, 85]], [[49, 85]], [[87, 101]]]",
    "query_spans": "[[[103, 124]]]",
    "process": ""
  },
  {
    "text": "Given that $l$ is an asymptote of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and $l$ intersects the circle $(x-c)^{2}+y^{2}=a^{2}$ (where $c^{2}=a^{2}+b^{2}$) at points $A$ and $B$, if $|A B|=a$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;l: Line;OneOf(Asymptote(C)) = l;G: Circle;Expression(G) = (y^2 + (-c + x)^2 = a^2);c: Number;A: Point;B: Point;Intersection(l, G) = {A, B};c^2 = a^2 + b^2;Abs(LineSegmentOf(A, B)) = a",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(7)/2",
    "fact_spans": "[[[6, 67], [151, 154]], [[6, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[2, 5], [74, 77]], [[2, 73]], [[78, 102]], [[78, 102]], [[105, 124]], [[128, 131]], [[132, 135]], [[74, 137]], [[105, 124]], [[139, 148]]]",
    "query_spans": "[[[151, 160]]]",
    "process": "From the given conditions, one asymptote of the hyperbola is: $ bx + ay = 0 $. The circle $ (x - c)^2 + y^2 = a^2 $ has center $ (c, 0) $ and radius $ a $. Let $ l $ be an asymptote of the hyperbola $ C: \\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1 $ ($ a > 0 $, $ b > 0 $). The line $ l $ intersects the circle $ (x - c)^2 + y^2 = a^2 $ (where $ c^2 = a^2 + b^2 $) at points $ A $ and $ B $. If $ |AB| = a $, then $ \\frac{bc}{\\sqrt{a^2} + b^2} + \\left( \\frac{a}{2} \\right)^2 = a^2 $, leading to $ 4b^2 = 3a^2 $, which gives $ 4(c^2 - a^2) = 3a^2 $. Solving yields $ e = \\frac{c}{a} = \\frac{\\sqrt{7}}{2} $."
  },
  {
    "text": "Let $F$ be the focus of the parabola $C$: $y^{2}=6 x$. A line passing through $F$ with an inclination angle of $30^{\\circ}$ intersects $C$ at points $A$ and $B$. Then, the distance from the midpoint of segment $A B$ to the $x$-axis is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 6*x);F: Point;Focus(C) = F;G: Line;A: Point;B: Point;PointOnCurve(F, G);Inclination(G) = ApplyUnit(30, degree);Intersection(G, C) = {A, B}",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A, B)), xAxis)",
    "answer_expressions": "3*sqrt(3)",
    "fact_spans": "[[[5, 24], [53, 56]], [[5, 24]], [[1, 4], [29, 32]], [[1, 27]], [[50, 52]], [[57, 60]], [[61, 64]], [[28, 52]], [[33, 52]], [[50, 66]]]",
    "query_spans": "[[[68, 88]]]",
    "process": "(Analysis) First, find the coordinates of the focus from the parabola equation, then obtain the equation of the line using its slope. Substitute into the parabola equation and eliminate $ y $. Use Vieta's formulas to find the value of $ x_{A}+x_{B} $, which allows finding the midpoint coordinates, thus obtaining the answer. According to the problem, the parabola $ C: y^{2}=6x $ has focus $ \\left(\\frac{3}{2}, 0\\right) $. The equation of line $ AB $ is $ y=\\frac{\\sqrt{3}}{3}\\left(x-\\frac{3}{2}\\right) $. Substituting into the parabola equation yields $ x^{2}-21x+\\frac{9}{4}=0 $, giving $ x_{A}+x_{B}=21 $. Let the midpoint of segment $ AB $ be $ (x_{0}, y_{0}) $. $ \\therefore x_{0}=\\frac{21}{2} $, $ \\therefore y_{0}=\\frac{\\sqrt{3}}{3}\\left(\\frac{21}{2}-\\frac{3}{2}\\right)=3\\sqrt{3} $. Note that during solving, pay attention to the flexible application of the definition of a parabola, i.e., connecting the chord length formula with the midpoint coordinates of the chord."
  },
  {
    "text": "Given that the ratio of the distance from a point $M$ to a fixed point $F$ $(1,0)$ and the distance from $M$ to a fixed line $l$: $x=4$ is a constant $\\frac{1}{2}$, let the locus of point $M$ be curve $C$. Then the equation of the trajectory of curve $C$ is?",
    "fact_expressions": "M: Point;F: Point;Coordinate(F) = (1, 0);l: Line;Expression(l) = (x = 4);Distance(M, F)/Distance(M, l) = 1/2;C: Curve;Locus(M)=C",
    "query_expressions": "LocusEquation(C)",
    "answer_expressions": "x^2/4 + y^2/3 = 1",
    "fact_spans": "[[[2, 6], [24, 25], [62, 66]], [[9, 20]], [[9, 20]], [[26, 39]], [[26, 39]], [[2, 60]], [[70, 75], [77, 82]], [[62, 75]]]",
    "query_spans": "[[[77, 89]]]",
    "process": "Let point $ M(x,y) $. Using the given condition, we obtain $ \\frac{\\sqrt{(x-1)^{2}+y^{2}}}{|x-4|} = \\frac{1}{2} $, which simplifies to the trajectory equation of curve $ C $, leading to the conclusion. Let point $ M(x,y) $, then according to the problem, $ \\frac{\\sqrt{(x-1)^{2}+y^{2}}}{|x-4|} = \\frac{1}{2} $. Then $ 4[(x-1)^{2}+y^{2}] = (x-4)^{2} $, that is, $ 3x^{2}+4y^{2} = 12 $, $ \\therefore \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $. Hence, the trajectory equation of curve $ C $ is $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $."
  },
  {
    "text": "The coordinates of the focus of the parabola $y^{2}=4 x$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(1,0)",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": ""
  },
  {
    "text": "The equation $\\frac{x^{2}}{24-k}+\\frac{y^{2}}{16+k}=1$ represents an ellipse; then the range of real values for $k$ is?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (x^2/(24 - k) + y^2/(k + 16) = 1)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-16,4)+(4,24)",
    "fact_spans": "[[[45, 47]], [[49, 54]], [[0, 47]]]",
    "query_spans": "[[[49, 61]]]",
    "process": "\\because the equation \\frac{x^{2}}{24-k}+\\frac{y^{2}}{16+k}=1 represents an ellipse, \\begin{cases}24-k>0\\\\16+k>0\\end{cases} (24-k\\neq16+k), solving gives -16<k<4 or 4<k<24. \\therefore the range of k is: (-16,4)\\cup(4,24)"
  },
  {
    "text": "Let $P$ be the intersection point of the line $y=\\frac{b}{3 a} x$ and the left branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, and let $F_{1}$ be the left focus. If $P F_{1}$ is perpendicular to the $x$-axis, then the eccentricity $e$ of the hyperbola is $?$.",
    "fact_expressions": "H: Line;Expression(H) = (y = x*b/(3*a));G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Intersection(H,LeftPart(G)) = P;P: Point;F1: Point;LeftFocus(G) = F1;IsPerpendicular(LineSegmentOf(P,F1),xAxis) = True;e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "3*sqrt(2)/4",
    "fact_spans": "[[[5, 26]], [[5, 26]], [[27, 83], [119, 122]], [[27, 83]], [[7, 26]], [[7, 26]], [[30, 83]], [[30, 83]], [[1, 88]], [[1, 4]], [[89, 96]], [[27, 100]], [[101, 117]], [[126, 129]], [[119, 129]]]",
    "query_spans": "[[[126, 131]]]",
    "process": "Let $ P\\left(x, \\frac{b}{3a}x\\right) $, then from the given condition, $ c = |x| $. Since $ PF_1 $ is perpendicular to the x-axis, by the latus rectum formula of the hyperbola, $ \\left|\\frac{b}{3a}\\right| = \\frac{b^{2}}{a} $, that is, $ \\frac{b}{3a}c = \\frac{b^{2}}{a} $, so $ b = \\frac{c}{3} $. Also, from $ a^{2} = c^{2} - b^{2} $, we get $ a^{2} = \\frac{8}{9}c^{2} $, hence $ e = \\frac{c}{a} = \\frac{3\\sqrt{2}}{4} $."
  },
  {
    "text": "If the slope of an asymptote of the hyperbola $x^{2}-\\frac{y^{2}}{k}=1$ is $-2$, then the value of the real number $k$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/k = 1);k: Real;Slope(OneOf(Asymptote(G))) = -2",
    "query_expressions": "k",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 29]], [[1, 29]], [[45, 50]], [[1, 43]]]",
    "query_spans": "[[[45, 54]]]",
    "process": "Find a, b. From the given condition, we have -\\frac{b}{a} = -2. Furthermore, in the hyperbola x^{2} - \\frac{y^{2}}{k} = 1, k > 0, a = 1, b = \\sqrt{k}. Since the slope of one asymptote of this hyperbola is -2, then -\\frac{b}{a} = -\\frac{\\sqrt{k}}{1} = -2, solving gives k = 4."
  },
  {
    "text": "The standard equation of the hyperbola that has the same asymptotes as the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$ and passes through the point $P(2,1)$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/2 = 1);Z: Hyperbola;Asymptote(Z) = Asymptote(G);P: Point;Coordinate(P) = (2, 1);PointOnCurve(P, Z)",
    "query_expressions": "Expression(Z)",
    "answer_expressions": "x^2/2-y^2=1",
    "fact_spans": "[[[1, 39]], [[1, 39]], [[59, 62]], [[0, 62]], [[49, 58]], [[49, 58]], [[48, 62]]]",
    "query_spans": "[[[59, 68]]]",
    "process": "Let the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$ have the same asymptotes as the hyperbola given by $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=\\lambda$, $(\\lambda\\neq0)$. Substituting the point $P(2,1)$ yields: $\\lambda=\\frac{1}{2}$. Therefore, the required hyperbola equation is $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=\\frac{1}{2} \\Rightarrow \\frac{x^{2}}{2}-y^{2}=1$."
  },
  {
    "text": "A line passing through the point $P(2,2)$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$, and $\\overrightarrow{P A}+\\overrightarrow{P B}=\\overrightarrow{0}$. Then the equation of this line is?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Coordinate(P) = (2, 2);PointOnCurve(P,H);Intersection(H, G) = {A, B};VectorOf(P, A) + VectorOf(P, B) = 0",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y = x",
    "fact_spans": "[[[14, 28]], [[11, 13], [106, 108]], [[1, 10]], [[30, 33]], [[34, 37]], [[14, 28]], [[1, 10]], [[0, 13]], [[11, 39]], [[41, 103]]]",
    "query_spans": "[[[106, 113]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ y_{1}^{2}=4x_{1} $, $ y_{2}^{2}=4x_{2} $. Hence $ y_{1}^{2}-y_{2}^{2}=4x_{1}-4x_{2} $, rearranging gives $ (y_{1}-y_{2})(y_{1}+y_{2})=4(x_{1}-x_{2}) $, $ k_{AB}=\\frac{4}{y_{1}+y_{2}} $. Since $ \\overrightarrow{PA}+\\overrightarrow{PB}=\\overrightarrow{0} $, $ P $ is the midpoint of $ AB $, so $ y_{1}+y_{2}=4 $. Therefore $ k_{AB}=1 $, hence the equation of the line is $ y=1\\times(x-2)+2=x $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{2^{t}}-\\frac{y^{2}}{4^{t}+4}=1$ $(t \\in \\mathbb{R})$, then the minimum value of the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/(4^t + 4) + x^2/2^t = 1);t: Real",
    "query_expressions": "Min(Eccentricity(C))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 64], [66, 69]], [[2, 64]], [[10, 64]]]",
    "query_spans": "[[[66, 79]]]",
    "process": "From the hyperbola equation, we obtain $ a^{2} = 2^{t} $, $ b^{2} = 4^{t} + 4 $, then $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{1 + \\frac{4^{t} + 4}{2^{t}}} = \\sqrt{2^{t} + \\frac{4}{2^{t}} + 1} $. Then, using the basic inequality, we can solve it. Since the hyperbola $ C: \\frac{x^{2}}{2^{t}} $, so $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{1 + \\frac{4^{t} + 4}{2^{t}}} = \\sqrt{2^{t} + \\frac{4}{2^{t}} + 1} \\geqslant \\sqrt{2\\sqrt{4} + 1} = \\sqrt{5} $, with equality if and only if $ 2^{t} = \\frac{4}{2^{t}} $, that is, when $ t = 1 $."
  },
  {
    "text": "The point $P(x, y)$ is a moving point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and $F_{1}$, $F_{2}$ are its left and right foci, respectively. Then the range of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);x1: Number;y1: Number;P: Point;Coordinate(P) = (x1, y1);PointOnCurve(P, G);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2",
    "query_expressions": "Range(DotProduct(VectorOf(P, F1), VectorOf(P, F2)))",
    "answer_expressions": "[2, 3]",
    "fact_spans": "[[[11, 48], [69, 70]], [[11, 48]], [[1, 10]], [[1, 10]], [[0, 10]], [[0, 10]], [[0, 52]], [[53, 60]], [[61, 68]], [[53, 75]], [[53, 75]]]",
    "query_spans": "[[[77, 141]]]",
    "process": "Since $F_{1}, F_{2}$ are the left and right foci of the ellipse $\\frac{x^2}{4} + \\frac{y^{2}}{3} = 1$, we have $F_{1}(-1,0)$, $F_{2}(1,0)$. Also, point $P(x,y)$ is a moving point on the ellipse $\\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1$, so we can set $P(2\\cos\\theta, \\sqrt{3}\\sin\\theta)$. Thus, $\\overrightarrow{PF}_{1} = (-1 - 2\\cos\\theta, -\\sqrt{3}\\sin\\theta)$, $\\overrightarrow{PF_{2}} = (1 - 2\\cos\\theta, -\\sqrt{3}\\sin\\theta)$. Therefore, $\\overrightarrow{PF}_{1} \\cdot \\overrightarrow{PF_{2}} = (-1 - 2\\cos\\theta)(1 - 2\\cos\\theta) + 3\\sin^{2}\\theta = 4\\cos^{2}\\theta + 3\\sin^{2}\\theta - 1 = 2 + \\cos 2\\theta$. Since $0 \\leqslant \\cos^{2}\\theta \\leqslant 1$, it follows that $2 \\leqslant 2 + \\cos 2\\theta \\leqslant 3$, so the range of $\\overrightarrow{PF}_{1} \\cdot \\overrightarrow{PF}_{2}$ is $[2,3]$."
  },
  {
    "text": "The standard equation of the circle centered at the right focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ and tangent to its asymptotes is?",
    "fact_expressions": "G: Hyperbola;H: Circle;Expression(G) = (x^2/9 - y^2/16 = 1);RightFocus(G) = Center(H);IsTangent(Asymptote(G),H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x-5)^2+y^2=16",
    "fact_spans": "[[[1, 40], [50, 51]], [[57, 58]], [[1, 40]], [[0, 58]], [[49, 58]]]",
    "query_spans": "[[[57, 65]]]",
    "process": ""
  },
  {
    "text": "The line $l$ intersects the parabola $y^{2}=2x$ at points $A$ and $B$. The circle $M$, with diameter $AB$, is tangent to the directrix of the parabola. If the area of circle $M$ is $16\\pi$, then what is the slope of line $l$?",
    "fact_expressions": "l: Line;G: Parabola;M: Circle;B:Point;A:Point;Expression(G) = (y^2 = 2*x);Intersection(l,G)={A,B};IsDiameter(LineSegmentOf(A,B),M);IsTangent(Directrix(G),M);Area(M)=16*pi",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*(sqrt(3)/3)",
    "fact_spans": "[[[0, 5], [79, 84]], [[6, 20], [51, 54]], [[45, 49], [61, 65]], [[29, 32]], [[23, 27]], [[6, 20]], [[0, 32]], [[33, 49]], [[45, 59]], [[61, 77]]]",
    "query_spans": "[[[79, 89]]]",
    "process": ""
  },
  {
    "text": "Given that the vertex of parabola $C$ is at the origin and its focus lies on the $x$-axis, the line $y = x$ intersects parabola $C$ at points $A$ and $B$. If $P(2,2)$ is the midpoint of segment $AB$, then the equation of parabola $C$ is?",
    "fact_expressions": "C: Parabola;G: Line;B: Point;A: Point;P: Point;O: Origin;Expression(G) = (y = x);Coordinate(P) = (2, 2);Vertex(C) = O;PointOnCurve(Focus(C), xAxis);Intersection(G, C) = {A, B};MidPoint(LineSegmentOf(A,B)) = P",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[2, 8], [34, 40], [75, 81]], [[26, 33]], [[46, 49]], [[42, 45]], [[54, 62]], [[14, 16]], [[26, 33]], [[54, 62]], [[2, 16]], [[2, 25]], [[26, 51]], [[54, 73]]]",
    "query_spans": "[[[75, 86]]]",
    "process": "Let the equation of the parabola be $ y^{2} = 2px $. By eliminating $ y $ from the system of the parabola equation and the line $ y = x $, we obtain an equation in $ x $, and solve for the x-coordinates of points $ A $ and $ B $. Then, combining with the midpoint formula and the fact that point $ P(2,2) $ is the midpoint of $ AB $, we get $ p = 2 $, thereby obtaining the equation of the parabola. [Detailed solution] Let the equation of the parabola be $ y^{2} = 2px $, ($ p > 0 $). From\n$$\n\\begin{cases}\ny^{2} = 2px \\\\\ny = x\n\\end{cases}\n$$\neliminating $ y $, we get $ x^{2} - 2px = 0 $, yielding $ x_{1} = 0 $, $ x_{2} = 2p $. The line is intercepted by the parabola forming chord $ AB $, and point $ P(2,2) $ is the midpoint of $ AB $, so $ \\frac{0 + 2p}{2} = 2 $, giving $ p = 2 $. Therefore, the equation of the parabola is $ y^{2} = 4x $."
  },
  {
    "text": "If the distance from a moving point $P$ to point $A(0 , 1)$ is less than its distance to the line $l$: $y=-2$ by $1$, then what is the equation of the trajectory of point $P$?",
    "fact_expressions": "P: Point;A: Point;Coordinate(A) = (0, 1);l: Line;Expression(l) = (y=-2);Distance(P, A) = Distance(P, l) - 1",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2=4*y",
    "fact_spans": "[[[3, 6], [47, 50]], [[7, 18]], [[7, 18]], [[23, 36]], [[23, 36]], [[3, 43]]]",
    "query_spans": "[[[47, 57]]]",
    "process": ""
  },
  {
    "text": "The minor axis length is $\\sqrt{5}$, and the eccentricity is $e = \\frac{2}{3}$. The foci of the ellipse are $F_{1}$ and $F_{2}$. A line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. Then, the perimeter of $\\triangle A B F_{2}$ is?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;F2: Point;F1: Point;e:Number;Length(MinorAxis(G))=sqrt(5);Eccentricity(G)=e;e=2/3;Focus(G) = {F1, F2};PointOnCurve(F1, H);Intersection(H, G) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "6",
    "fact_spans": "[[[34, 36], [68, 70]], [[65, 67]], [[71, 74]], [[75, 78]], [[48, 55]], [[40, 47], [57, 64]], [[18, 33]], [[0, 36]], [[15, 36]], [[18, 33]], [[34, 55]], [[56, 67]], [[65, 80]]]",
    "query_spans": "[[[82, 108]]]",
    "process": "First, solve for the semi-major axis length $ a $ of the ellipse according to the given conditions, then analyze that the perimeter of $ \\triangle ABF_{2} $ is $ 4a $, from which the result can be obtained. Let the major axis length of the ellipse be $ 2a $, the minor axis length be $ 2b $, and the focal distance be $ 2c $. Since the minor axis length is $ \\sqrt{5} $ and the eccentricity is $ e = \\frac{2}{3} $, we have  \n$$\n\\begin{cases}\n\\frac{c}{a} = \\frac{2}{3} \\\\\n2b = \\sqrt{5} \\\\\na^{2} = b^{2} + c^{2}\n\\end{cases}\n$$  \nThus, $ a^{2} = \\frac{9}{4} $, so $ a = \\frac{3}{2} $. Moreover, the perimeter of $ \\triangle ABF_{2} $ is $ |AF_{1}| + |AF_{2}| + |BF_{1}| + |BF_{2}| $. By the definition of an ellipse, the perimeter of $ \\triangle ABF_{2} $ is $ 4a = 6 $."
  },
  {
    "text": "What is the eccentricity of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 34]]]",
    "process": "From $x^{2}-\\frac{y^{2}}{3}=1$, we have $a=1$, $b=\\sqrt{3} \\Rightarrow c=2$, so the eccentricity of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ is $e=\\frac{c}{a}=2$."
  },
  {
    "text": "If an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is perpendicular to the line $y=2 x$, then its eccentricity is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = 2*x);IsPerpendicular(OneOf(Asymptote(G)),H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[1, 57], [77, 78]], [[4, 57]], [[4, 57]], [[64, 73]], [[4, 57]], [[4, 57]], [[1, 57]], [[64, 73]], [[1, 75]]]",
    "query_spans": "[[[77, 83]]]",
    "process": "According to the asymptote equation $ y = \\pm\\frac{b}{a}x' $, we have $ a = 2b $. From $ c = \\sqrt{a^{2} + b^{2}} = \\sqrt{5b^{2}} = \\sqrt{5}b $ and the eccentricity formula, the answer can be obtained. Since the asymptote equation is $ y = \\pm\\frac{b}{a}x' $, it follows that $ \\frac{b}{a} = \\frac{1}{2} $, then $ a = 2b $, $ c = \\sqrt{a^{2} + b^{2}} = \\sqrt{5b^{2}} = \\sqrt{5}b $. Hence, the eccentricity is $ \\frac{c}{a} = \\frac{\\sqrt{5}b}{2b} = \\frac{\\sqrt{5}}{2} $."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{m}+\\frac{y^{2}}{2-m^{2}}=1$ represents an ellipse, then the range of real values for $m$ is?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G)=(y^2/(2 - m^2) + x^2/m = 1)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(0,1) + (1,\\sqrt{2})",
    "fact_spans": "[[[46, 48]], [[50, 55]], [[3, 44]]]",
    "query_spans": "[[[50, 62]]]",
    "process": "The equation $\\frac{x^{2}}{m}+\\frac{y^{2}}{2-m^{2}}=1$ represents an ellipse, then $\\begin{cases}m>0\\\\2-m^{2}>0\\\\m\\neq2-m^{2}\\end{cases}$ solving gives $m\\in(0,1)\\cup(1,\\sqrt{2})$."
  },
  {
    "text": "The distance from the focus of the parabola $y^{2}=4 x$ to the asymptote of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (x^2/16 - y^2/9 = 1);Expression(H) = (y^2 = 4*x)",
    "query_expressions": "Distance(Focus(H), Asymptote(G))",
    "answer_expressions": "3/5",
    "fact_spans": "[[[18, 57]], [[0, 14]], [[18, 57]], [[0, 14]]]",
    "query_spans": "[[[0, 65]]]",
    "process": "The focus of the parabola $ y^{2}=4x $ is $ (1,0) $, the asymptotes of the hyperbola $ \\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1 $ are $ y=\\pm\\frac{3}{4}x $, $ 3x\\pm4y=0 $, the required distance is $ d=\\frac{3}{\\sqrt{3^{2}+4^{2}}}=\\frac{3}{5} $."
  },
  {
    "text": "Given that $F$ is the upper focus of the hyperbola $C$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1(a>0, b>0)$, $A$ is the upper vertex of $C$, and $B$ is a point on $C$ such that $BF$ is perpendicular to the $y$-axis. If the slope of $AB$ is $\\frac{1}{3}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;B: Point;F: Point;A: Point;a>0;b>0;Expression(C) = (-x^2/b^2 + y^2/a^2 = 1);UpperFocus(C) = F;UpperVertex(C) = A ;PointOnCurve(B, C);IsPerpendicular(LineSegmentOf(B, F), yAxis);Slope(LineSegmentOf(A, B)) = 1/3",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[6, 67], [76, 79], [88, 91], [135, 138]], [[14, 67]], [[14, 67]], [[84, 87]], [[2, 5]], [[72, 75]], [[14, 67]], [[14, 67]], [[6, 67]], [[2, 71]], [[72, 83]], [[84, 94]], [[96, 108]], [[111, 133]]]",
    "query_spans": "[[[135, 144]]]",
    "process": "Since A is the upper vertex of C, A(0,a). Since F is the upper focus of the hyperbola, F(0,c). Since B is a point on C and BF is perpendicular to the y-axis, B(\\frac{b^{2}}{a},c). Thus, k_{AB} = \\frac{c-a}{\\frac{b^{2}}{a}-0} = \\frac{1}{3}, i.e., c^{2}-3ac+2a^{2}=0, i.e., e^{2}-3e+2=0. Solving gives e=2 or e=1 (discarded)."
  },
  {
    "text": "Given that point $M(1,0)$ lies on the axis of symmetry of the parabola $y^{2}=2 p x$ $(p>0)$, and a line passing through this point intersects the parabola at points $A$ and $B$, then the product of the slopes of lines $OA$ and $OB$ is?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;O: Origin;A: Point;M: Point;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(M) = (1, 0);PointOnCurve(M,SymmetryAxis(G));PointOnCurve(M,H);Intersection(H, G) = {A, B}",
    "query_expressions": "Slope(LineOf(O,A))*Slope(LineOf(O,B))",
    "answer_expressions": "-2*p",
    "fact_spans": "[[[11, 32], [47, 50]], [[14, 32]], [[44, 46]], [[65, 70]], [[52, 55]], [[2, 10], [42, 43]], [[56, 59]], [[14, 32]], [[11, 32]], [[2, 10]], [[2, 38]], [[40, 46]], [[44, 61]]]",
    "query_spans": "[[[63, 84]]]",
    "process": "Let points $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ k_{OA} = \\frac{y_{1}}{x_{1}} = \\frac{y_{1}}{\\frac{y_{1}^{2}}{2p}} = \\frac{2p}{y_{1}} $, similarly we get $ k_{OB} = \\frac{2p}{y_{2}} $. If line $ AB $ coincides with the $ x $-axis, then line $ AB $ intersects the parabola $ y^{2} = 2px $ ($ p > 0 $) at only one point, which does not satisfy the condition. Let the equation of line $ AB $ be $ x = my + 1 $, solving simultaneously \n\\[\n\\begin{cases}\ny^{2} = 2px \\\\\nx = my + 1\n\\end{cases}\n\\]\nwe obtain $ y^{2} - 2mpy - 2p = 0 $. By Vieta's formulas, we have $ y_{1}y_{2} = -2p $. Therefore, the product of the slopes of lines $ OA $ and $ OB $ is $ k_{OA} \\cdot k_{OB} = \\frac{4p^{2}}{y_{1}y_{2}} = \\frac{4p^{2}}{-2p} = -2p $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the foci of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and point $P$ is a moving point on $C$, then the range of $P F_{1}$ is?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/4 + y^2/3 = 1);Focus(C) = {F1, F2};PointOnCurve(P, C)",
    "query_expressions": "Range(LineSegmentOf(P, F1))",
    "answer_expressions": "[1, 3]",
    "fact_spans": "[[[18, 60], [69, 72]], [[64, 68]], [[2, 9]], [[10, 17]], [[18, 60]], [[2, 63]], [[64, 76]]]",
    "query_spans": "[[[78, 94]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{y^{2}}{6}-\\frac{x^{2}}{3}=1$, find the equation of the parabola that has the center of the hyperbola as its vertex and the directrix of the hyperbola as its directrix.",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (-x^2/3 + y^2/6 = 1);Center(G) = Vertex(H);Directrix(H) = Directrix(G)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2 = pm*8*y",
    "fact_spans": "[[[2, 40], [43, 46], [53, 56]], [[62, 65]], [[2, 40]], [[42, 65]], [[52, 65]]]",
    "query_spans": "[[[62, 69]]]",
    "process": "The directrix equations of the hyperbola $\\frac{y^{2}}{6}-\\frac{x^{2}}{3}=1$ are: $y=\\pm\\frac{a^{2}}{c}\\Rightarrow y=\\pm2$. According to the problem, the directrix equations of the parabola are $y=\\pm2$, so the standard equation is $x^{2}=\\pm8y$."
  },
  {
    "text": "Let the line $y=2x+t$ ($t \\ne 0$) intersect the two asymptotes of the hyperbola $\\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1$ ($a>0$, $b>0$) at points $A$ and $B$, respectively. If the point $P(4t, 0)$ satisfies $|PA| = |PB|$, then the equation of the asymptotes of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;t: Number;Negation(t=0);P: Point;A: Point;B: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = t + 2*x);Coordinate(P) = (4*t, 0);L1:Line;L2:Line;Asymptote(G)={L1,L2};Intersection(H, L1) = A;Intersection(H,L2)=B;Abs(LineSegmentOf(P, A)) = Abs(LineSegmentOf(P, B))",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*3*x",
    "fact_spans": "[[[23, 79], [128, 131]], [[26, 79]], [[26, 79]], [[1, 22]], [[3, 22]], [[3, 22]], [[98, 110]], [[88, 92]], [[93, 96]], [[26, 79]], [[26, 79]], [[23, 79]], [[1, 22]], [[98, 110]], [], [], [[23, 84]], [[1, 96]], [[1, 96]], [[112, 125]]]",
    "query_spans": "[[[128, 139]]]",
    "process": ""
  },
  {
    "text": "The line $l$: $2x - y + 2 = 0$ passes through the left focus $F_1$ and a vertex $B$ of an ellipse. Then the eccentricity of this ellipse is?",
    "fact_expressions": "E: Ellipse;l: Line;Expression(l) = (2*x - y + 2 = 0);F1: Point;F1 = LeftFocus(E);B: Point;B = OneOf(Vertex(E));PointOnCurve(F1, l);PointOnCurve(B, l)",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(5)/5",
    "fact_spans": "[[[18, 20], [41, 43]], [[0, 17]], [[0, 17]], [[23, 30]], [[18, 30]], [[35, 38]], [[18, 38]], [[0, 30]], [[0, 38]]]",
    "query_spans": "[[[41, 49]]]",
    "process": ""
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{2}=1$ are $F_{1}$, $F_{2}$, and point $P$ lies on the ellipse. If $|P F_{1}|=4$, then what is $|P F_{2}|$? What is the measure of $\\angle F_{1} P F_{2}$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 + y^2/2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 4",
    "query_expressions": "Abs(LineSegmentOf(P, F2));AngleOf(F1, P, F2)",
    "answer_expressions": "2\n2*pi/3",
    "fact_spans": "[[[1, 38], [64, 66]], [[59, 63]], [[42, 49]], [[50, 58]], [[1, 38]], [[1, 58]], [[59, 67]], [[69, 82]]]",
    "query_spans": "[[[84, 97]], [[98, 125]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, $F_{1}$, $F_{2}$ are the left and right foci of this ellipse, point $A(4,1)$, $P$ is a moving point on the ellipse. When the perimeter of $\\triangle A P F_{1}$ is maximized, what is the area of $\\triangle A P F_{1}$?",
    "fact_expressions": "C: Ellipse;A: Point;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/25 + y^2/9 = 1);Coordinate(A) = (4, 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);WhenMax(Perimeter(TriangleOf(A, P, F1)))",
    "query_expressions": "Area(TriangleOf(A, P, F1))",
    "answer_expressions": "56/5",
    "fact_spans": "[[[2, 45], [65, 67], [90, 92]], [[74, 84]], [[86, 89]], [[46, 54]], [[56, 63]], [[2, 45]], [[74, 84]], [[47, 73]], [[47, 73]], [[86, 98]], [[99, 129]]]",
    "query_spans": "[[[130, 156]]]",
    "process": "Analysis: First, use the definition of the ellipse to convert the distance from a point on the ellipse to the left focus into the distance to the right focus, then solve using plane geometry knowledge. Detailed solution: connect AF_{2}, PF_{2}. From the ellipse equation \\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1, we get a=5, F_{1}(-4,0), F_{2}(4,0). Then the perimeter of AAPF_{1} is |PF_{1}|+|PA|+|AF_{1}|=10+|PA|-|PF_{2}|+|AF_{1}|\\leqslant10+|AF_{2}|+|AF_{1}| (equality holds if and only if P lies on the ray AF_{2}). In the ellipse equation \\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1, let x=4, we obtain y=-\\frac{9}{5}, then |AP|=\\frac{14}{5}, S_{AAPF_{1}}=\\frac{1}{2}\\times8\\times\\frac{14}{5}=\\frac{56}{5}"
  },
  {
    "text": "Given the hyperbola equation $\\frac{y^{2}}{16}-\\frac{x^{2}}{9}=1$, then the coordinates of the upper focus of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/9 + y^2/16 = 1)",
    "query_expressions": "Coordinate(UpperFocus(G))",
    "answer_expressions": "(0, 5)",
    "fact_spans": "[[[2, 44], [46, 49]], [[2, 44]]]",
    "query_spans": "[[[46, 58]]]",
    "process": "The foci of the hyperbola \\frac{y^{2}}{16}-\\frac{x^{2}}{9}=1 lie on the y-axis, and the focal distance is 2\\sqrt{16+9}=10. Therefore, the coordinates of the upper focus of this hyperbola are (0,5)."
  },
  {
    "text": "Let $F$ be the focus of the parabola $C$: $x^{2}=2 p y(p>0)$, the line $y-x+1=0$ intersects the parabola $C$ at a common point $M$, and $MF$ is perpendicular to the axis of symmetry of the parabola $C$, then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;M: Point;F: Point;p>0;Expression(C) = (x^2 = 2*(p*y));Expression(G) = (-x + y + 1 = 0);Focus(C) = F;Intersection(G, C) = M;IsPerpendicular(LineSegmentOf(M,F),SymmetryAxis(C))",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[5, 31], [47, 53], [68, 74]], [[82, 85]], [[35, 46]], [[57, 60]], [[1, 4]], [[12, 31]], [[5, 31]], [[35, 46]], [[1, 34]], [[35, 60]], [[62, 80]]]",
    "query_spans": "[[[82, 87]]]",
    "process": "The parabola $ C: x^{2} = 2py $ ($ p > 0 $) has the $ y $-axis as its axis of symmetry, and $ F(0, \\frac{p}{2}) $. Since $ MF $ is perpendicular to the axis of symmetry of the parabola $ C $, let $ M(x_{0}, y_{0}) $, then $ y_{0} = \\frac{p}{2} $, $ x_{0} = y_{0} + 1 = \\frac{p}{2} + 1 $. Since point $ M(x_{0}, y_{0}) $ lies on the parabola, we have $ (\\frac{p}{2} + 1)^{2} = 2p \\cdot \\frac{p}{2} $. Solving gives $ p = 2 $ or $ p = -\\frac{2}{3} $ (discarded). Therefore, $ p = 2 $."
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and $F_{1}$, $F_{2}$ are the two foci of the ellipse, then the perimeter of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2/3 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2}",
    "query_expressions": "Perimeter(TriangleOf(P, F1, F2))",
    "answer_expressions": "6",
    "fact_spans": "[[[6, 43], [63, 65]], [[2, 5]], [[47, 54]], [[55, 62]], [[6, 43]], [[2, 46]], [[47, 69]]]",
    "query_spans": "[[[71, 98]]]",
    "process": "In the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, $a=2$, $b=\\sqrt{3}$, $c=1$. According to the definition of the ellipse, the perimeter of $APF_{1}F_{2}$ is $|PF_{1}|+|PF_{2}|+|F_{1}F_{2}|=2a+2c$. Therefore, the perimeter of $APF_{1}F_{2}$ is $4+2=6$."
  },
  {
    "text": "Draw a tangent from the left focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ to the circle $x^{2}+y^{2}=\\frac{a^{2}}{4}$, with point of tangency $E$. Extend $FE$ to intersect the right branch of the hyperbola at point $P$. If $E$ is the midpoint of $PF$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;LeftFocus(G) = F;H: Circle;Expression(H) = (x^2 + y^2 = a^2/4);L: Line;TangentOfPoint(F, H) = L;TangentPoint(L, H) = E;E: Point;Intersection(OverlappingLine(LineSegmentOf(F, E)), RightPart(G)) = P;P: Point;MidPoint(LineSegmentOf(P, F)) = E",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)/2",
    "fact_spans": "[[[1, 58], [115, 118], [141, 144]], [[1, 58]], [[4, 58]], [[4, 58]], [[4, 58]], [[4, 58]], [[62, 65]], [[1, 65]], [[66, 96]], [[66, 96]], [], [[0, 99]], [[0, 106]], [[103, 106], [127, 130]], [[107, 125]], [[121, 125]], [[127, 139]]]",
    "query_spans": "[[[141, 150]]]",
    "process": "Let the right focus be $ F_{2} $. From the given conditions, $ E $ is the midpoint of $ FP $, $ O $ is the midpoint of $ FF_{2} $, therefore $ OE \\parallel PF_{2} $ and $ |PF_{2}| = 2|OE| = a $. Also, $ E $ is the point of tangency, i.e., $ OE \\perp PF $, so $ PF_{2} \\perp PF $. By the definition of hyperbola, $ |PF| = 2a + |PF_{2}| = 3a $, thus $ (3a)^{2} + a^{2} = (2c)^{2} $, solving gives $ e = \\frac{c}{a} = \\frac{\\sqrt{10}}{2} $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has eccentricity $e=\\frac{\\sqrt{5}}{5}$, and one of its vertex coordinates is $(0,2)$, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;e: Number;Eccentricity(G) = e ;e = sqrt(5)/5;Coordinate(OneOf(Vertex(G))) = (0, 2)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5+y^2/4=1",
    "fact_spans": "[[[2, 54], [99, 101]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[58, 80]], [[2, 80]], [[58, 80]], [[2, 97]]]",
    "query_spans": "[[[99, 106]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the foci of the hyperbola $\\frac{x^{2}}{25}+\\frac{y^{2}}{m}=1$ are $(-6,0)$ and $(6,0)$. What is the value of $m$?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/25 + y^2/m = 1);F1: Point;F2: Point;Coordinate(F1) = (-6, 0);Coordinate(F2) = (6, 0);Focus(G) = {F1, F2}",
    "query_expressions": "m",
    "answer_expressions": "-11",
    "fact_spans": "[[[0, 39]], [[63, 66]], [[0, 39]], [[45, 53]], [[54, 61]], [[45, 53]], [[54, 61]], [[0, 61]]]",
    "query_spans": "[[[63, 70]]]",
    "process": "\\because the standard equation of a hyperbola with foci on the x-axis is \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1, (a>0,b>0) \\therefore a^{2}=25, b^{2}=-m^{,} c^{2}=6^{2}=36 and \\because c^{2}=a^{2}+b^{2} \\therefore 36=25-m, i.e. m=-11"
  },
  {
    "text": "A hyperbola has an asymptote given by $x+\\sqrt{3} y=0$, and one of its foci coincides with the focus of the parabola given by $y^{2}=16 x$. What is the standard equation of this hyperbola?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(OneOf(Asymptote(G))) = (x + sqrt(3)*y = 0);OneOf(Focus(G)) = Focus(H);Expression(H) = (y^2 = 16*x)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/12 - y^2/4 = 1",
    "fact_spans": "[[[25, 28], [29, 30], [62, 65]], [[52, 55]], [[0, 28]], [[29, 60]], [[36, 55]]]",
    "query_spans": "[[[62, 72]]]",
    "process": "Since the parabola is $ y^{2} = 16x $, its focus is $ (4, 0) $. Because one asymptote of the hyperbola is $ x + \\sqrt{3}y = 0 $, we set the hyperbola equation as $ \\frac{x^{2}}{3\\lambda} - \\frac{y^{2}}{\\lambda} = 1 $. Therefore, $ 3\\lambda + \\lambda = 16 $, $ \\lambda = 4 $. Hence, the required hyperbola equation is $ \\frac{x^{2}}{12} - \\frac{y^{2}}{4} = 1 $. This problem examines finding the standard equation of a hyperbola using its asymptotes and focus, and is a simple question."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $y^{2}=2 x$, $A$ is a moving point on the parabola, and point $B(-\\frac{1}{2}, 0)$, then when $\\frac{|A B|}{|A F|}$ takes its maximum value, the value of $|A B|$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);F: Point;Focus(G) = F;A: Point;PointOnCurve(A, G);B: Point;Coordinate(B) = (-1/2, 0);WhenMax(Abs(LineSegmentOf(A, B))/Abs(LineSegmentOf(A, F))) = True",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[6, 20], [28, 31]], [[6, 20]], [[2, 5]], [[2, 23]], [[24, 27]], [[24, 35]], [[36, 57]], [[36, 57]], [[59, 86]]]",
    "query_spans": "[[[87, 98]]]",
    "process": "Let $ A(x,y) $, then $ AB = \\sqrt{(x+\\frac{1}{2})^{2}+y^{2}} = \\sqrt{x^{2}+3x+\\frac{1}{4}} $, and $ |AF| = x+\\frac{1}{2} $, so $ \\frac{|AB|^{2}}{|AF|^{2}} = \\frac{x^{2}+3x+\\frac{1}{4}}{x^{2}+x+\\frac{1}{4}} = 1 + \\frac{2}{x+\\frac{1}{4x}+1} \\leqslant 1 + \\frac{2}{2\\sqrt{x \\cdot \\frac{1}{4x}}+1} = 2 $, with equality if and only if $ x = \\frac{1}{2} $. Therefore, when $ \\frac{|AB|}{|AF|} $ reaches its maximum value, $ x = \\frac{1}{2} $, and at this time $ |AB| = \\sqrt{\\frac{1}{4}+\\frac{3}{2}+\\frac{1}{4}} = \\sqrt{2} $."
  },
  {
    "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$ equals? The equations of the asymptotes are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/12 = 1)",
    "query_expressions": "Eccentricity(G);Expression(Asymptote(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 46]], [[0, 53]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y=2 x^{2}$ with focus $F$ and directrix $l$, a circle centered at $F$ and tangent to $l$ intersects the parabola at points $A$ and $B$. Then $|AB|=$?",
    "fact_expressions": "C: Parabola;G:Circle;F: Point;l: Line;A: Point;B: Point;Expression(C) = (y = 2*x^2);Focus(C) = F;Directrix(C) = l;Center(G)=F;IsTangent(l,G);Intersection(G, C) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A,B))",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 20], [54, 57]], [[51, 52]], [[24, 27], [37, 40]], [[31, 34], [45, 48]], [[60, 63]], [[64, 67]], [[2, 20]], [[2, 27]], [[2, 34]], [[36, 52]], [[44, 52]], [[51, 69]]]",
    "query_spans": "[[[71, 79]]]",
    "process": ""
  },
  {
    "text": "The left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1(a>0)$ are $F_{1}$ and $F_{2}$ respectively, and $P$ is a point on the right branch of the hyperbola. If $|P F_{1}|=6$ and $\\cos \\angle F_{1} P F_{2}=\\frac{5}{6}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;P: Point;F1: Point;F2: Point;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(P,RightPart(G));Abs(LineSegmentOf(P, F1)) = 6;Cos(AngleOf(F1, P, F2)) = 5/6",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[0, 37], [65, 68], [131, 134]], [[3, 37]], [[61, 64]], [[45, 52]], [[53, 60]], [[3, 37]], [[0, 37]], [[0, 60]], [[0, 60]], [[61, 73]], [[75, 88]], [[90, 129]]]",
    "query_spans": "[[[131, 140]]]",
    "process": "Since |PF_{1}| = 6, then |PF_{2}| = 6 - 2a, and |F_{1}F_{2}| = 2c = 2\\sqrt{a^{2} + 1}. Also, \\cos\\angle F_{1}PF_{2} = \\frac{5}{6}, by the law of cosines: \\cos\\angle F_{1}PF_{2} = \\frac{|PF_{1}|^{2} + |PF_{2}|^{2} - |F_{1}F_{2}|^{2}}{2|PF_{1}||PF_{2}|} = \\frac{5}{6}, that is, \\frac{5}{6} = \\frac{36 + 36 - 24a + 4a^{2} - 4a^{2} - 4}{2 \\times 6 \\times (6 - 2a)}, solving gives a = 2. Therefore, c = \\sqrt{5}, so the eccentricity of the hyperbola e = \\frac{c}{a} = \\frac{\\sqrt{5}}{2}."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{m-1}-\\frac{y^{2}}{m}=1$ ($m \\in \\mathbb{R}$) represents a hyperbola, then the range of real values for $m$ is?",
    "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (x^2/(m - 1) - y^2/m = 1)",
    "query_expressions": "Range(m)",
    "answer_expressions": "{(1, +oo), (-oo, 0)}",
    "fact_spans": "[[[51, 54]], [[56, 61]], [[1, 54]]]",
    "query_spans": "[[[56, 68]]]",
    "process": "Using the characteristics of the hyperbola equation, we obtain $ m(m-1) > 0 $. Solving this inequality gives the range of real values for $ m $. [Detailed solution] Since the equation $ \\frac{x^2}{m-1} - \\frac{y^{2}}{m} = 1 $ ($ m \\in \\mathbb{R} $) represents a hyperbola, it follows that $ m(m-1) > 0 $. Solving this yields $ m > 1 $ or $ m < 0 $. Therefore, the range of real values for $ m $ is $ m > 1 $ or $ m < 0 $."
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $2$, what is the angle of inclination of the line $b x-3 a y=0$?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (-3*a*y + b*x = 0);Eccentricity(G) = 2",
    "query_expressions": "Inclination(H)",
    "answer_expressions": "pi/6",
    "fact_spans": "[[[2, 58]], [[5, 58]], [[5, 58]], [[68, 83]], [[5, 58]], [[5, 58]], [[2, 58]], [[68, 83]], [[2, 66]]]",
    "query_spans": "[[[68, 89]]]",
    "process": "\\because e=\\frac{c}{a}=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{\\frac{a^{2}+b^{2}}{a^{2}}}=\\sqrt{1+(\\frac{b}{a})^{2}}=2, then \\frac{b}{a}=\\sqrt{3}, so the slope of the line bx-3ay=0 is k=\\frac{b}{3a}=\\frac{\\sqrt{3}}{3}. Therefore, the inclination angle of the line bx-3ay=0 is \\frac{\\pi}{6}."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ with focus $F$, $O$ as the coordinate origin, and $M$ a point on the parabola such that $|M F|=4|O F|$ and the area of $\\Delta MFO$ is $4 \\sqrt{3}$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;p: Number;M: Point;F: Point;O: Origin;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(M, G);Abs(LineSegmentOf(M, F)) = 4*Abs(LineSegmentOf(O, F));Area(TriangleOf(M, F, O)) = 4*sqrt(3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = 8*x",
    "fact_spans": "[[[2, 23], [44, 47], [98, 101]], [[5, 23]], [[40, 43]], [[27, 30]], [[31, 34]], [[5, 23]], [[2, 23]], [[2, 30]], [[40, 50]], [[52, 66]], [[67, 95]]]",
    "query_spans": "[[[98, 106]]]",
    "process": ""
  },
  {
    "text": "Let the hyperbola $C$: $\\frac{x^{2}}{2}+\\frac{y^{2}}{m}=1$ have eccentricity $e$, and its asymptotes are tangent to the circle $M$: $(x-2)^{2}+y^{2}=e^{2}$. Then $m$=?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/2 + y^2/m = 1);m: Number;e: Number;Eccentricity(C) = e;M: Circle;Expression(M) = (y^2 + (x - 2)^2 = e^2);IsTangent(Asymptote(C), M) = True",
    "query_expressions": "m",
    "answer_expressions": "-2",
    "fact_spans": "[[[1, 44], [53, 54]], [[1, 44]], [[90, 93]], [[49, 52]], [[1, 52]], [[58, 86]], [[58, 86]], [[53, 88]]]",
    "query_spans": "[[[90, 95]]]",
    "process": "From the given conditions, we know that $ m < 0 $. The asymptotes of the hyperbola are $ \\frac{x}{\\sqrt{2}} \\pm \\frac{y}{\\sqrt{-m}} = 0 $, that is, $ \\sqrt{-m}x \\pm \\sqrt{2}y = 0 $, and $ e^{2} = 1 + \\frac{-m}{2} $. The distance from the center of the circle to the asymptote is $ \\frac{|2\\sqrt{-m}|}{\\sqrt{2 - m}} = e = \\sqrt{1 - \\frac{m}{2}} $. Simplifying yields $ (m + 2)^{2} = 0 $, solving gives $ m = -2 $."
  },
  {
    "text": "If there are two points $P$, $Q$ (not the endpoints of the major axis) on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, $O$ is the origin, and the slopes of lines $OP$, $OQ$ are $K_{1}$, $K_{2}$ respectively, satisfying $K_{1} K_{2}=-\\frac{3}{4}$, then $\\overrightarrow{O P}^{2}+\\overrightarrow{O Q}^{2}$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);P: Point;Q: Point;PointOnCurve(P, G);PointOnCurve(Q, G);Negation(P=Endpoint(MajorAxis(G)));Negation(Q=Endpoint(MajorAxis(G)));O: Origin;K1: Number;K2: Number;Slope(LineOf(O, P)) = K1;Slope(LineOf(O, Q)) = K2;K1*K2 = -3/4",
    "query_expressions": "VectorOf(O, P)^2+VectorOf(O, Q)^2",
    "answer_expressions": "7",
    "fact_spans": "[[[1, 38]], [[1, 38]], [[42, 45]], [[46, 49]], [[1, 49]], [[1, 49]], [[1, 58]], [[1, 58]], [[59, 62]], [[86, 93]], [[94, 101]], [[67, 101]], [[67, 101]], [[105, 131]]]",
    "query_spans": "[[[133, 186]]]",
    "process": "Let the coordinates of P and Q be $(2\\cos\\alpha,\\sqrt{3}\\sin\\alpha)$, $(2\\cos\\beta,\\sqrt{3}\\sin\\beta)$. Since $K_{1}K_{2}=-\\frac{3}{4}$, we have $\\frac{\\sqrt{3}\\sin\\alpha}{2\\cos\\alpha} \\cdot \\frac{\\sqrt{3}\\sin\\beta}{2\\cos\\beta}=-\\frac{3}{4}$, i.e., $\\cos(\\alpha-\\beta)=0$, so $\\alpha-\\beta=\\frac{\\pi}{2}+k\\pi$, $k\\in\\mathbb{Z}$. Without loss of generality, take $\\beta=\\alpha+\\frac{\\pi}{2}$. Therefore, $(\\overrightarrow{OP})^{2}+(\\overrightarrow{OQ})^{2}=4\\cos^{2}\\alpha+3\\sin^{2}\\alpha+4\\sqrt{3}\\sin\\alpha\\cos\\alpha+4\\cos^{2}\\beta+3\\sin^{2}\\beta+4\\sqrt{3}\\sin\\beta\\cos\\beta=4\\cos^{2}\\alpha+3\\sin^{2}\\alpha+4\\sqrt{3}\\sin\\alpha\\cos\\alpha+4\\sin^{2}\\alpha+3\\cos^{2}\\alpha-4\\sqrt{3}\\sin\\alpha\\cos\\alpha=4+3=7$."
  },
  {
    "text": "The equation of an ellipse centered at the origin, with directrices $x = \\pm 4$ and eccentricity $\\frac{1}{2}$ is?",
    "fact_expressions": "O:Origin;Center(G)=O;G: Ellipse;Eccentricity(G)=1/2;Expression(Directrix(G))=(x=pm*4)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 + y^2/3 = 1",
    "fact_spans": "[[[3, 5]], [[0, 44]], [[42, 44]], [[23, 44]], [[6, 44]]]",
    "query_spans": "[[[42, 48]]]",
    "process": "From the given conditions, we have \\frac{a2}{c}=4, e=\\frac{c}{a}=\\frac{1}{2} \\therefore a=2, c=1 \\therefore b^{2}=3, and the equation of the ellipse is \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1"
  },
  {
    "text": "The equation of a circle whose center lies on the parabola $y=\\frac{1}{2} x^{2}(x<0)$ and is tangent to both the directrix of the parabola and the $y$-axis is?",
    "fact_expressions": "G: Parabola;Expression(G) = ((y = x^2/2)&(x<0));H: Circle;PointOnCurve(Center(H), G);IsTangent(Directrix(G), H);IsTangent(yAxis, H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x+1)^2+(y-1/2)^2=1",
    "fact_spans": "[[[3, 32], [37, 40]], [[3, 32]], [[52, 53]], [[0, 53]], [[36, 53]], [[36, 53]]]",
    "query_spans": "[[[52, 58]]]",
    "process": ""
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ are $F_{1}$ and $F_{2}$ respectively, the eccentricity is $\\frac{1}{2}$, a line passing through $F_{2}$ intersects the ellipse at points $A$ and $B$, and the perimeter of $\\triangle A B F_{1}$ is $8$. Then the length of the minor axis of this ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;Eccentricity(G) = 1/2;H: Line;PointOnCurve(F2, H);Intersection(H, G) = {A, B};A: Point;B: Point;Perimeter(TriangleOf(A, B, F1)) = 8",
    "query_expressions": "Length(MinorAxis(G))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[0, 52], [107, 109], [151, 153]], [[0, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[61, 68]], [[69, 76], [96, 103]], [[0, 76]], [[0, 76]], [[0, 94]], [[104, 106]], [[95, 106]], [[104, 119]], [[110, 113]], [[114, 117]], [[120, 148]]]",
    "query_spans": "[[[151, 159]]]",
    "process": "Given that the perimeter of AABF is 8, using the definition of an ellipse, we can obtain the value of a. Then, based on the eccentricity being \\frac{1}{2}, we find the value of c, and thus determine the value of b, which leads to the result. [Detailed solution] Since the perimeter of 4ABF is 8, it follows that F_{1}A + F_{1}B + F_{2}A + F_{2}B = 4a = 8, so a = 2. Because the eccentricity is \\frac{1}{2}, we have \\frac{c}{a} = \\frac{1}{2}, hence c = \\frac{1}{2}a = 1. From a^{2} = b^{2} + c^{2}, solving gives b = \\sqrt{3}. Therefore, the minor axis length of the ellipse is 2\\sqrt{3}."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $y^{2}=4 \\sqrt{3} x$, a line passing through point $F$ intersects the parabola at points $A$ and $B$. If $\\overrightarrow{A F}=3 \\overrightarrow{F B}$, then $|A B|$=?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 4*sqrt(3)*x);Focus(G)=F;PointOnCurve(F,H);Intersection(H,G)={A,B};VectorOf(A,F)=3*VectorOf(F,B)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16*sqrt(3)/3",
    "fact_spans": "[[[6, 29], [42, 45]], [[39, 41]], [[46, 49]], [[2, 5], [34, 38]], [[50, 53]], [[6, 29]], [[2, 32]], [[33, 41]], [[39, 55]], [[57, 102]]]",
    "query_spans": "[[[105, 114]]]",
    "process": "The line passing through the focus $ F $ of the parabola $ y^{2} = 4\\sqrt{3}x $ intersects the parabola at points $ A $ and $ B $, and $ \\overrightarrow{AF} = 3\\overrightarrow{FB} $. The slope of the line exists; let the line $ AB $ be $ y = k(x - \\sqrt{3}) $, with $ k > 0 $. Thus,\n$$\n\\begin{cases}\ny = k(x - \\sqrt{3}) \\\\\ny^{2} = 4\\sqrt{3}x\n\\end{cases}\n$$\nSimplifying yields $ k^{2}x^{2} - (2k^{2} + 4)x + k^{2} = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1}x_{2} = 3 $, and $ x_{1} > \\sqrt{3} > x_{2} $ (1). From $ \\overrightarrow{AF} = 3\\overrightarrow{FB} $, we have $ \\sqrt{3} - x_{1} = 3(x_{2} - \\sqrt{3}) $ (2). Solving (1) and (2) together gives\n$$\n\\begin{cases}\nx_{1} = 3\\sqrt{3} \\\\\nx_{2} = \\frac{\\sqrt{3}}{3}\n\\end{cases}\n\\quad \\text{or} \\quad\n\\begin{cases}\nx_{1} = \\sqrt{3} \\\\\nx_{2} = \\sqrt{3}\n\\end{cases}\n\\quad \\text{(discarded)}.\n$$\nTherefore, $ |AB| = x_{1} + x_{2} + p = 3\\sqrt{3} + \\frac{\\sqrt{3}}{3} + 2\\sqrt{3} = \\frac{16\\sqrt{3}}{3} $."
  },
  {
    "text": "It is known that the distance from the focus of the parabola $y^{2}=-4 \\sqrt{2} x$ to one of the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $\\frac{\\sqrt{5}}{5}$. Then, the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = -4*sqrt(2)*x);Distance(Focus(H), OneOf(Asymptote(G))) = sqrt(5)/5",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)/3",
    "fact_spans": "[[[30, 86], [119, 122]], [[33, 86]], [[33, 86]], [[2, 26]], [[33, 86]], [[33, 86]], [[30, 86]], [[2, 26]], [[2, 116]]]",
    "query_spans": "[[[119, 128]]]",
    "process": "Since the focus of the parabola has coordinates $(-\\sqrt{2},0)$, and one asymptote of the hyperbola has the equation $bx-ay=0$, it follows that $\\frac{|-\\sqrt{2}b-0|}{\\sqrt{a^{2}+b^{2}}}=\\frac{\\sqrt{5}}{5}$, i.e., $c=\\sqrt{10}b$, $c^{2}=10(c^{2}-a^{2})$, so $e^{2}=\\frac{10}{9}$ and thus $e=\\frac{\\sqrt{10}}{3}$."
  },
  {
    "text": "The equation of the trajectory of the center of a moving circle that is externally tangent to the circle $x^{2} + y^{2}-4 x=0$ and tangent to the $y$-axis is?",
    "fact_expressions": "G: Circle;Expression(G) = (-4*x + x^2 + y^2 = 0);Z:Circle;IsOutTangent(Z, G);IsTangent(Z, yAxis)",
    "query_expressions": "LocusEquation(Center(Z))",
    "answer_expressions": "{(y^2 = 8*x)&(x > 0), (y = 0)&(x < 0)}",
    "fact_spans": "[[[1, 23]], [[1, 23]], [[35, 37]], [[0, 37]], [[27, 37]]]",
    "query_spans": "[[[35, 46]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{3}=1$ $(a>\\sqrt{3})$, and a line passing through the origin $O$ with an inclination angle of $60^{\\circ}$ intersects the ellipse $C$ at a point $M$. If $|\\overrightarrow{M F_{1}}+\\overrightarrow{M F_{2}}|=|\\overrightarrow{M F_{1}}-\\overrightarrow{M F_{2}}|$, then what is the length of the major axis of ellipse $C$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/3 + x^2/a^2 = 1);a: Number;a>sqrt(3);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;O: Origin;G: Line;PointOnCurve(O, G);Inclination(G) = ApplyUnit(60, degree);M: Point;OneOf(Intersection(G, C)) = M;Abs(VectorOf(M, F1) + VectorOf(M, F2)) = Abs(VectorOf(M, F1) - VectorOf(M, F2))",
    "query_expressions": "Length(MajorAxis(C))",
    "answer_expressions": "2*sqrt(3+2*sqrt(3))",
    "fact_spans": "[[[18, 76], [110, 115], [233, 238]], [[18, 76]], [[25, 76]], [[25, 76]], [[2, 9]], [[10, 17]], [[2, 82]], [[2, 82]], [[84, 89]], [[107, 109]], [[83, 109]], [[90, 109]], [[121, 124]], [[107, 124]], [[126, 231]]]",
    "query_spans": "[[[233, 244]]]",
    "process": "Let the line be $ y = \\sqrt{3}x $, substitute into $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{3} = 1 $, solving gives $ x^{2} = \\frac{a^{2}}{1+a^{2}} $, $ y^{2} = \\frac{3a^{2}}{1+a^{2}} $. Since $ |\\overrightarrow{MF}_{1} + \\overrightarrow{MF_{2}}| = |\\overrightarrow{MF_{1}} - \\overrightarrow{MF_{2}}| $, it follows that $ 2|\\overrightarrow{OM}| = |\\overrightarrow{F_{1}F_{2}}| $, so $ 4\\left( \\frac{a^{2}}{1+a^{2}} + \\frac{3a^{2}}{1+a^{2}} \\right) = 4c^{2} $. Also, since $ a^{2} = b^{2} + c^{2} $, $ b^{2} = 3 $, solving gives $ a^{2} = 3 + 2\\sqrt{3} $. Therefore, the major axis length of ellipse $ C $ is $ 2\\sqrt{3 + 2\\sqrt{3}} $."
  },
  {
    "text": "The left focus of the ellipse $\\frac{x^{2}}{4 a^{2}}+\\frac{y^{2}}{3 a^{2}}=1(a>0)$ is $F$, and the line $x=m$ intersects the ellipse at points $A$ and $B$. When the perimeter of $\\triangle F A B$ is maximized, what is the area of $\\triangle F A B$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(4*a^2) + y^2/(3*a^2) = 1);a: Number;a>0;F: Point;LeftFocus(G) = F;L: Line;Expression(L) = (x = m);m: Number;Intersection(G,L) = {A,B};A: Point;B: Point;WhenMax(Perimeter(TriangleOf(F,A,B))) = True",
    "query_expressions": "Area(TriangleOf(F, A, B))",
    "answer_expressions": "3*a^2",
    "fact_spans": "[[[0, 54], [71, 73]], [[0, 54]], [[2, 54]], [[2, 54]], [[59, 62]], [[0, 62]], [[63, 70]], [[63, 70]], [[65, 70]], [[63, 84]], [[76, 80]], [[81, 84]], [[85, 109]]]",
    "query_spans": "[[[110, 132]]]",
    "process": ""
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{20}=1$, and point $P$ lies on the hyperbola. If the distance from point $P$ to focus $F_{1}$ is equal to $9$, then the distance from point $P$ to focus $F_{2}$ is equal to?",
    "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;P: Point;Expression(G) = (x^2/16 - y^2/20 = 1);Focus(G)={F1,F2};PointOnCurve(P, G);Distance(P,F1) = 9",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "17",
    "fact_spans": "[[[16, 56], [65, 68]], [[0, 7], [78, 85]], [[8, 15], [102, 109]], [[60, 64], [71, 75], [95, 99]], [[16, 56]], [[0, 59]], [[60, 69]], [[71, 93]]]",
    "query_spans": "[[[95, 115]]]",
    "process": ""
  },
  {
    "text": "Given that a moving circle passes through a fixed point $A(-4,0)$ and is tangent to the circle $x^{2}+y^{2}-8 x-84=0$, then the trajectory equation of the center $P$ of the moving circle is?",
    "fact_expressions": "G: Circle;C:Circle;A: Point;Expression(G) = (-8*x + x^2 + y^2 - 84 = 0);Coordinate(A) = (-4, 0);PointOnCurve(A, C);IsTangent(G,C);P:Point;Center(C)=P",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2/25+y^2/9=1",
    "fact_spans": "[[[19, 42]], [[2, 4], [46, 48]], [[7, 16]], [[19, 42]], [[7, 16]], [[2, 16]], [[2, 44]], [[51, 54]], [[46, 54]]]",
    "query_spans": "[[[51, 61]]]",
    "process": "According to the sum of the distance from the center P of a circle to the fixed point A(-4,0) and the center of the circle $ x^{2}+y^{2}-8x-84=0 $ being constant, we can determine that the circle $ x^{2}+y^{2}-8x-84=0 $, which is equivalent to $ (x-4)^{2}+y^{2}=100 $, has center B(4,0) and radius 10. Since A(-4,0) lies inside the circle $ (x-4)^{2}+y^{2}=100 $, the moving circle is internally tangent to $ (x-4)^{2}+y^{2}=100 $. Let the radius of the moving circle be $ r $, then the sum of distances from center P to A(-4,0) and B(4,0) is $ d = r + 10 - r = 10 $. Therefore, the center P of the moving circle traces an ellipse with foci at A(-4,0) and B(4,0), where $ 2a = 10 $, so $ a = 5 $, $ c = 4 $, $ b = \\sqrt{5^{2}-4^{2}} = 3 $. Hence, the trajectory equation of the center P of the moving circle is $ \\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1 $."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$, through the fixed point $T(p , 0)$ draw two mutually perpendicular lines $l_{1}$, $l_{2}$. If $l_{1}$ intersects the parabola at points $P$, $Q$, and $l_{2}$ intersects the parabola at points $M$, $N$, with the slope of $l_{1}$ being $k$. A student has correctly found that the midpoint of chord $P Q$ is $(\\frac{p}{k^2}+p , \\frac{p}{k})$. Please write the midpoint of chord $M N$?",
    "fact_expressions": "G: Parabola;p: Number;l1: Line;l2: Line;P: Point;Q: Point;M: Point;N: Point;T: Point;k: Number;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(X) = (p + p/k^2, p/k);Coordinate(T) = (p, 0);PointOnCurve(T, l1);PointOnCurve(T, l2);IsPerpendicular(l1, l2);Intersection(l1, G) = {P, Q};Intersection(l2, G) = {M, N};Slope(l1) = k;IsChordOf(LineSegmentOf(P, Q), G);IsChordOf(LineSegmentOf(M, N), G);X: Point;MidPoint(LineSegmentOf(P, Q)) = X",
    "query_expressions": "Coordinate(MidPoint(LineSegmentOf(M, N)))",
    "answer_expressions": "(p*k^2 + p, -p*k)",
    "fact_spans": "[[[2, 23], [72, 75], [94, 97]], [[5, 23]], [[45, 54], [64, 71], [110, 117]], [[55, 62], [86, 93]], [[77, 80]], [[81, 84]], [[99, 102]], [[105, 108]], [[27, 37]], [[121, 124]], [[5, 23]], [[2, 23]], [[146, 179]], [[27, 37]], [[24, 62]], [[24, 62]], [[38, 62]], [[64, 84]], [[86, 108]], [[110, 124]], [[94, 142]], [[94, 189]], [[146, 179]], [[137, 179]]]",
    "query_spans": "[[[184, 193]]]",
    "process": ""
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=a x$ is $F(1 , 0)$, a line $l$ passing through the focus $F$ intersects the parabola at points $A$ and $B$. If $AB=8$, then the equation of the line is?",
    "fact_expressions": "l: Line;G: Parabola;a: Number;F: Point;A: Point;B: Point;Expression(G) = (y^2 = a*x);Coordinate(F) = (1, 0);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};LineSegmentOf(A, B) = 8",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x \\pm y - 1 = 0",
    "fact_spans": "[[[38, 43], [68, 70]], [[2, 16], [44, 47]], [[5, 16]], [[34, 37], [20, 30]], [[48, 51]], [[52, 55]], [[2, 16]], [[20, 30]], [[2, 30]], [[31, 43]], [[38, 57]], [[59, 66]]]",
    "query_spans": "[[[68, 75]]]",
    "process": ""
  },
  {
    "text": "Given a point $P$ on the left branch of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ such that the distance from $P$ to the left focus is $10$, then the distance from point $P$ to the right focus is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1);P: Point;PointOnCurve(P, LeftPart(G));Distance(P, LeftFocus(G)) = 10",
    "query_expressions": "Distance(P, RightFocus(G))",
    "answer_expressions": "18",
    "fact_spans": "[[[2, 41]], [[2, 41]], [[47, 50], [64, 68]], [[2, 50]], [[2, 62]]]",
    "query_spans": "[[[2, 77]]]",
    "process": "From the equation of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, we obtain $a=4$. By the definition of a hyperbola, the distance from point $P$ to the right focus is equal to $2a$ plus the distance from point $P$ to the left focus. Hence, the distance from point $P$ to the right focus is $8+10=18$."
  },
  {
    "text": "Draw a line through the focus of the parabola $y^{2}=4x$ intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $x_{1}+x_{2}=6$, then $|AB|$ equals?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;x1:Number;y1:Number;x2:Number;y2:Number;Expression(G) = (y^2 = 4*x);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);PointOnCurve(Focus(G), H);Intersection(H, G) = {A,B};x1 + x2 = 6",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 15], [22, 25]], [[19, 21]], [[26, 43]], [[44, 62]], [[26, 43]], [[26, 43]], [[44, 62]], [[44, 62]], [[1, 15]], [[26, 43]], [[44, 62]], [[0, 21]], [[19, 64]], [[66, 81]]]",
    "query_spans": "[[[84, 93]]]",
    "process": ""
  },
  {
    "text": "It is known that $F$ is the right focus of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$, $P$ is a point on $C$, and $PF$ is perpendicular to the $x$-axis. The coordinates of point $A$ are $(1,\\ 3)$. Then the area of $\\triangle APF$ is?",
    "fact_expressions": "C: Hyperbola;P: Point;F: Point;A: Point;Expression(C) = (x^2 - y^2/3 = 1);Coordinate(A) = (1, 3);RightFocus(C)=F;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P,F),xAxis)",
    "query_expressions": "Area(TriangleOf(A,P,F))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[6, 39], [48, 51]], [[44, 47]], [[2, 5]], [[69, 73]], [[6, 39]], [[69, 87]], [[2, 43]], [[44, 54]], [[56, 68]]]",
    "query_spans": "[[[89, 111]]]",
    "process": "From the hyperbola equation, find the coordinates of focus F; from the perpendicular condition, find the coordinates of point P; then find the distance from A to line PF; and use the triangle area formula to obtain the area. Since F is the right focus of the hyperbola $ C: x^{2} - \\frac{y^{2}}{3} = 1 $, it follows that $ F(2,0) $. Since $ PF \\perp x $-axis, we can let the coordinates of P be $ (2, y_{P}) $. Since P lies on C, $ 4 - \\frac{y_{P}^{2}}{3} = 1 $, solving gives $ y_{P} = \\pm 3 $, so $ P(2, \\pm 3) $, $ |PF| = 3 $. Also, since $ A(1,3) $, the distance from point A to line PF is 1, so $ S_{\\triangle APF} = \\frac{1}{2} \\times |PF| \\times 1 = \\frac{1}{2} \\times 3 \\times 1 = \\frac{3}{2} $."
  },
  {
    "text": "The line $y = kx + 1$ always has common points with the ellipse $\\frac{x^{2}}{5} + \\frac{y^{2}}{m} = 1$, then what is the value of $m$?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*x + 1);k: Number;G: Ellipse;Expression(G) = (x^2/5 + y^2/m = 1);m: Number;IsIntersect(H, G)",
    "query_expressions": "m",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 11]], [[0, 11]], [[2, 11]], [[12, 49]], [[12, 49]], [[56, 59]], [[0, 54]]]",
    "query_spans": "[[[56, 63]]]",
    "process": ""
  },
  {
    "text": "The left and right vertices of the ellipse $x^{2}+m y^{2}=1(m>1)$ are $A$ and $B$, respectively. Draw a line $l$ perpendicular to the $x$-axis through point $B$. Point $P$ is a point on line $l$. Connect $PA$ intersecting the ellipse at point $C$. The origin is $O$, and $OP \\perp BC$. Then $m=$?",
    "fact_expressions": "l: Line;G: Ellipse;m: Number;P: Point;A: Point;O: Origin;B: Point;C: Point;m>1;Expression(G) = (m*y^2 + x^2 = 1);LeftVertex(G) = A;RightVertex(G) = B;PointOnCurve(B, l);IsPerpendicular(l, xAxis);PointOnCurve(P, l);Intersection(LineSegmentOf(P,A),G)=C;IsPerpendicular(LineSegmentOf(O, P), LineSegmentOf(B, C))",
    "query_expressions": "m",
    "answer_expressions": "2",
    "fact_spans": "[[[52, 55], [61, 66]], [[0, 24], [79, 81]], [[114, 117]], [[56, 60]], [[31, 34]], [[92, 95]], [[35, 38], [40, 44]], [[82, 86]], [[2, 24]], [[0, 24]], [[0, 38]], [[0, 38]], [[39, 55]], [[45, 55]], [[56, 70]], [[71, 86]], [[97, 112]]]",
    "query_spans": "[[[114, 119]]]",
    "process": "The left and right vertices of the ellipse $x^{2}+my^{2}=1$ $(m>1)$ are $A$ and $B$, respectively. Draw a perpendicular line $l$ to the $x$-axis through point $B$. Point $P$ is a point on line $l$, which can be set as $P(1,t)$. Then $k_{PO}=t$, $k_{BC}=-\\frac{1}{t}$, and the equation of $BC$ is: $y=-\\frac{1}{t}(x-1)$. $k_{AP}=\\frac{t}{2}$, and the equation of $AP$ is: $y=\\frac{t}{2}(x+1)$. Therefore, $\\begin{cases}y=-\\frac{1}{t}(x-1)\\\\y=\\frac{t}{2}(x+1)\\end{cases}$, solving gives: $C\\left(\\frac{2-t^{2}}{2+t^{2}},\\frac{2t}{2+t^{2}}\\right)$. Thus, $\\left(\\frac{2-t^{2}}{2+t^{2}}\\right)^{2}+m\\left(\\frac{2t}{2+t^{2}}\\right)^{2}=1$, leading to $4-4t^{2}+4t^{4}+4mt^{2}=4+4t^{2}+4t^{4}$, solving gives $m=2$."
  },
  {
    "text": "Given that the hyperbola passes through the point $(4, \\sqrt{3})$ and has asymptotes with equations $y = \\pm \\frac{1}{2} x$, then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (4, sqrt(3));PointOnCurve(H,G) = True;Expression(Asymptote(G)) = (y=pm*(1/2)*x)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4-y^2=1",
    "fact_spans": "[[[2, 5], [56, 59]], [[6, 22]], [[7, 22]], [[2, 22]], [[2, 52]]]",
    "query_spans": "[[[56, 66]]]",
    "process": "When the foci are on the x-axis, set the hyperbola equation as \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1,(a>0,b>0). According to the problem, we have \\begin{cases}42(\\sqrt{3})^{2}\\\\\\frac{4}{a}-\\frac{\\sqrt{a}}{b^{2}}=1\\end{cases}\\Rightarrow\\begin{cases}a=4\\\\a=\\frac{1}{2}\\end{cases}, so the hyperbola equation is \\frac{x2}{4}-y^{2}=1; when the foci are on the y-axis, set the hyperbola equation as \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1,(a>0,b>0). According to the problem, we have \\begin{cases}\\sqrt{3})^{2}\\\\\\frac{a^{2}}{a^{2}}-\\frac{4^{2}}{b^{2}}=1,a\\end{cases}. In conclusion, the hyperbola equation is \\frac{x^{2}}{4}-y^{2}=1"
  },
  {
    "text": "It is known that the asymptotes of a hyperbola are given by $y=\\pm \\frac{1}{2} x$, and that the hyperbola shares common foci with the ellipse $4 x^{2}+9 y^{2}=36$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (4*x^2 + 9*y^2 = 36);Expression(Asymptote(G)) = (y = pm*(x/2));Focus(G) = Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2 = 1",
    "fact_spans": "[[[2, 5], [35, 38], [68, 71]], [[39, 61]], [[39, 61]], [[2, 33]], [[35, 66]]]",
    "query_spans": "[[[68, 76]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has center $O$, right focus $F$, right vertex $A$, and the intersection point of the right directrix with the $x$-axis is $H$, then the maximum value of $\\frac{|FA|}{|OH|}$ is?",
    "fact_expressions": "G: Ellipse;a: Number;b: Number;F: Point;A: Point;O:Origin;H: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Center(G) = O;RightFocus(G)=F;RightVertex(G)=A;Intersection(RightDirectrix(G),xAxis)=H",
    "query_expressions": "Max(Abs(LineSegmentOf(F, A))/Abs(LineSegmentOf(O, H)))",
    "answer_expressions": "1/4",
    "fact_spans": "[[[2, 56]], [[4, 56]], [[4, 56]], [[68, 71]], [[76, 79]], [[60, 63]], [[92, 95]], [[4, 56]], [[4, 56]], [[2, 56]], [[2, 63]], [[2, 71]], [[2, 79]], [[2, 95]]]",
    "query_spans": "[[[97, 123]]]",
    "process": ""
  },
  {
    "text": "Draw a tangent line $l$ to the parabola $y^{2}=2x$ at point $P(2,2)$. What is the intercept of tangent line $l$ on the $y$-axis?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);P: Point;Coordinate(P) = (2, 2);l: Line;TangentOfPoint(P, G) = l",
    "query_expressions": "Intercept(l, yAxis)",
    "answer_expressions": "1",
    "fact_spans": "[[[11, 25]], [[11, 25]], [[1, 10]], [[1, 10]], [[28, 31], [34, 37]], [[0, 31]]]",
    "query_spans": "[[[34, 48]]]",
    "process": "Let the slope of the tangent line be $ k $, then the equation of the tangent line is $ y - 2 = k(x - 2) $. Solving the system of equations  \n\\[\n\\begin{cases}\ny - 2 = k(x - 2) \\\\\ny^2 = 2x\n\\end{cases}\n\\]  \nwe obtain $ ky^{2} - 2y - 4k + 4 = 0 $. Then $ \\Delta = 4 - 4k(-4k + 4) = 0 $, solving gives $ k = \\frac{1}{2} $, so the tangent line equation is $ y - 2 = \\frac{1}{2}(x - 2) $. Let $ x = 0 $, we get $ y = 1 $. $ \\therefore $ the intercept of tangent line $ l $ on the $ y $-axis is $ 1 $+-----------------"
  },
  {
    "text": "If the asymptotes of a hyperbola are given by $y = \\pm 3x$, and one of its foci is $(\\sqrt{10}, 0)$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*(3*x));Coordinate(OneOf(Focus(G))) = (sqrt(10), 0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/9 = 1",
    "fact_spans": "[[[1, 4], [23, 24], [48, 51]], [[1, 22]], [[23, 46]]]",
    "query_spans": "[[[48, 58]]]",
    "process": ""
  },
  {
    "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1  (b>0)$ are given by $y=\\pm \\frac{1}{2} x$, then what is the value of $b$?",
    "fact_expressions": "G: Hyperbola;b: Number;b>0;Expression(G) = (x^2/4 - y^2/b^2 = 1);Expression(Asymptote(G)) = (y = pm*(x/2))",
    "query_expressions": "b",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 50]], [[81, 84]], [[4, 50]], [[1, 50]], [[1, 79]]]",
    "query_spans": "[[[81, 87]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$, $F$ is the focus of $C$. The line $l$ passing through the point $(-2,0)$ with slope $\\frac{2}{3}$ intersects the parabola at points $A$ and $B$. Then $|AF| + |BF|$ = ?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;G: Point;Coordinate(G) = (-2, 0);l: Line;Slope(l) = 2/3;PointOnCurve(G, l);A: Point;B: Point;Intersection(l, C) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F))",
    "answer_expressions": "7",
    "fact_spans": "[[[2, 21], [69, 72], [28, 31]], [[2, 21]], [[24, 27]], [[24, 34]], [[36, 45]], [[36, 45]], [[63, 68]], [[46, 68]], [[35, 68]], [[73, 76]], [[77, 80]], [[63, 82]]]",
    "query_spans": "[[[84, 99]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. From the given conditions, the equation of the line is $ y = \\frac{2}{3}(x+2) $. Solving simultaneously with the parabola equation:\n\\[\n\\begin{cases}\ny = \\frac{2}{3}(x+2) \\\\\ny^{2} = 4x\n\\end{cases}\n\\]\nwe obtain $ x^{2} - 5x + 4 = 0 $, so $ x_{1} + x_{2} = 5 $. Using the focal radius formula for the parabola, we get $ |AF| + |BF| = x_{1} + \\frac{p}{3} + x_{2} + \\frac{p}{2} = x_{1} + x_{2} + p = 5 + 2 = 7 $. Hence, the answer is: $ 7 $."
  },
  {
    "text": "Let $P$ be a moving point on the parabola $y^{2}=4x$, $F$ be the focus, and $B(4,2)$. Then the minimum value of $|PB| + |PF|$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G);F: Point;Focus(G) = F;B: Point;Coordinate(B) = (4, 2)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, B)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "5",
    "fact_spans": "[[[5, 19]], [[5, 19]], [[1, 4]], [[1, 25]], [[26, 29]], [[5, 32]], [[34, 42]], [[34, 42]]]",
    "query_spans": "[[[44, 63]]]",
    "process": "Find the equation of the directrix of the parabola, transform the distance from point P to the focus F into the distance from P to the directrix, then use the collinearity of three points to obtain the minimum value. As shown in the figure, draw PM perpendicular to the directrix $ x = -1 $, with foot at M, then $ |PF| = |PM| $, so $ |PF| + |PB| = |PM| + |PB| $. It is clear that $ |PM| + |PB| $ is minimized when points B, P, and M are collinear, and the minimum value is $ 4 - (-1) = 5 $. Therefore, the minimum value of $ |PB| + |PF| $ is 5."
  },
  {
    "text": "If the hyperbola passes through the point $P(6 , \\sqrt{3})$ and has asymptotes given by $y=\\pm \\frac{x}{3}$, then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;P: Point;Coordinate(P) = (6, sqrt(3));PointOnCurve(P, G);Expression(Asymptote(G)) = (y = pm*x/3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2 = 1",
    "fact_spans": "[[[2, 5], [53, 56]], [[6, 24]], [[6, 24]], [[2, 24]], [[2, 50]]]",
    "query_spans": "[[[53, 61]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{k+2}+\\frac{y^{2}}{k+1}=1$, a line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. If the perimeter of $\\triangle A B F_{2}$ is $8$, then the value of $k$ is?",
    "fact_expressions": "G: Ellipse;k: Number;H: Line;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2/(k + 2) + y^2/(k + 1) = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, H);Intersection(H, G) = {A, B};Perimeter(TriangleOf(A, B, F2)) = 8",
    "query_expressions": "k",
    "answer_expressions": "2",
    "fact_spans": "[[[18, 59], [78, 80]], [[122, 125]], [[75, 77]], [[81, 84]], [[85, 88]], [[10, 17]], [[2, 9], [67, 74]], [[18, 59]], [[2, 65]], [[2, 65]], [[66, 77]], [[75, 90]], [[92, 120]]]",
    "query_spans": "[[[122, 129]]]",
    "process": ""
  },
  {
    "text": "The equation of the parabola with vertex at the center of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ and focus at the right focus of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (x^2/4 - y^2/5 = 1);Vertex(H) = Center(G);RightFocus(G)=Focus(H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2=12*x",
    "fact_spans": "[[[1, 39], [49, 52]], [[60, 63]], [[1, 39]], [[0, 63]], [[47, 63]]]",
    "query_spans": "[[[60, 67]]]",
    "process": ""
  },
  {
    "text": "It is known that the hyperbola $C$ shares common asymptotes with $\\frac{x^{2}}{5}-\\frac{y^{2}}{3}=1$, and one focus is at $(4,0)$. Then, the standard equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Curve;Coordinate(OneOf(Focus(C))) = (4, 0);Expression(G) = (x^2/5 - y^2/3 = 1);Asymptote(C) = Asymptote(G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/10 - y^2/6 = 1",
    "fact_spans": "[[[2, 8], [66, 72]], [[9, 44]], [[2, 64]], [[9, 44]], [[2, 50]]]",
    "query_spans": "[[[66, 79]]]",
    "process": "Let the hyperbola $ C: \\frac{x^2}{5} - \\frac{y^{2}}{3} = t $, then $ 5t + 3t = 4^{2} $, $ t = 2 \\therefore \\frac{x^{2}}{10} - \\frac{y^{2}}{6} = 1 $"
  },
  {
    "text": "If one asymptote of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is intercepted by the circle $(x-2)^{2}+y^{2}=4$ to form a chord of length $2 \\sqrt{3}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y^2 + (x - 2)^2 = 4);Length(InterceptChord(OneOf(Asymptote(C)),G))=2*sqrt(3)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[1, 62], [110, 113]], [[9, 62]], [[9, 62]], [[69, 89]], [[9, 62]], [[9, 62]], [[1, 62]], [[69, 89]], [[1, 108]]]",
    "query_spans": "[[[110, 119]]]",
    "process": "Without loss of generality, let one asymptote of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) be $ bx + ay = 0 $. Since the chord length intercepted by the circle $ (x-2)^{2} + y^{2} = 4 $ on the asymptote is $ 2\\sqrt{3} $, the distance from the center of the circle to the asymptote is $ d = \\sqrt{2^{2} - (\\sqrt{3})^{2}} = 1 $, that is, $ \\frac{|2b + 0|}{\\sqrt{a^{2} + b^{2}}} = 1 $. Therefore, $ \\frac{b^{2}}{a^{2}} = \\frac{1}{3} $, so the eccentricity of the hyperbola $ C $ is $ e = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\frac{2\\sqrt{3}}{3} $."
  },
  {
    "text": "Given the equation of ellipse $C$ is $\\frac{x^{2}}{25}+\\frac{y^{2}}{b^{2}}=1$ $(0<b<5)$, $A(-4,0)$, $B(-1,0)$, there exists a point $P$ on ellipse $C$ such that $|PA|=2|PB|$, then the maximum value of the real number $b$ is?",
    "fact_expressions": "C: Ellipse;A: Point;B: Point;P: Point;b:Real;b>0;b<5;Coordinate(A) = (-4, 0);Coordinate(B) = (-1, 0);Expression(C) = (x^2/25 + y^2/b^2 = 1);PointOnCurve(P,C);Abs(LineSegmentOf(P,A))=2*Abs(LineSegmentOf(P,B))",
    "query_expressions": "Max(b)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 7], [83, 88]], [[60, 70]], [[73, 82]], [[93, 96]], [[114, 119]], [[11, 58]], [[11, 58]], [[60, 70]], [[73, 82]], [[2, 58]], [[83, 96]], [[98, 112]]]",
    "query_spans": "[[[114, 125]]]",
    "process": "Let P(x, y), since A(-4, 0), B(-1, 0), |PA| = 2|PB|, ∴ (x + 4)^{2} + y^{2} = 4[(x + 1)^{2} + y^{2}], ∴ x^{2} + y^{2} = 4, i.e., the locus of point P is a circle centered at the origin with radius 2. Also, point P lies on ellipse C, ∴ the circle x^{2} + y^{2} = 4 and ellipse C have intersection points, ∴ 0 < b ≤ 2. The maximum value of real number b is 2."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=x$ with focus $F$, and a point $A(x_{0}, y_{0})$ on $C$ such that $|A F|=\\frac{5}{4} x_{0}$, find $x_{0}$.",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = x);F: Point;Focus(C) = F;A: Point;x0: Number;y0: Number;Coordinate(A) = (x0, y0);PointOnCurve(A, C);Abs(LineSegmentOf(A, F)) = (5/4)*x0",
    "query_expressions": "x0",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 18], [44, 47]], [[2, 18]], [[22, 25]], [[2, 25]], [[26, 43]], [[78, 85]], [[26, 43]], [[26, 43]], [[26, 50]], [[51, 76]]]",
    "query_spans": "[[[78, 87]]]",
    "process": ""
  },
  {
    "text": "The equation of the directrix of the parabola $y=-\\frac{x^{2}}{8}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = -x^2/8)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=2",
    "fact_spans": "[[[0, 23]], [[0, 23]]]",
    "query_spans": "[[[0, 30]]]",
    "process": ""
  },
  {
    "text": "Given point $P(\\frac{\\sqrt{2}}{2}, 1)$, $M$, $N$ are two moving points on the ellipse $x^{2}+\\frac{y^{2}}{2}=1$. Let the slopes of lines $PM$, $PN$, $MN$ be $k_{1}$, $k_{2}$, $k$, respectively. If $k_{1}+k_{2}=0$, then $k=$?",
    "fact_expressions": "P: Point;Coordinate(P) = (sqrt(2)/2, 1);M: Point;N: Point;PointOnCurve(M, G);PointOnCurve(N, G);G: Ellipse;Expression(G) = (x^2 + y^2/2 = 1);k1: Number;k2: Number;k: Number;Slope(LineOf(P, M)) = k1;Slope(LineOf(P, N)) = k2;Slope(LineOf(M, N)) = k;k1 + k2 = 0",
    "query_expressions": "k",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 29]], [[2, 29]], [[31, 34]], [[36, 39]], [[31, 73]], [[31, 73]], [[40, 67]], [[40, 67]], [[102, 109]], [[111, 118]], [[120, 123], [142, 145]], [[75, 123]], [[75, 123]], [[75, 123]], [[125, 140]]]",
    "query_spans": "[[[142, 147]]]",
    "process": "Let $ M(x_{1},y_{1}) $, $ N(x_{2},y_{2}) $, the line $ MN: y = kx + m $. Solving the system \n\\[\n\\begin{cases}\ny = kx + m \\\\\nx^{2} + \\frac{y^{2}}{2} = 1\n\\end{cases}\n\\]\nyields $ (2 + k^{2})x^{2} + 2kmx + m^{2} - 2 = 0 $, then we have $ x_{1} + x_{2} = -\\frac{2km}{2 + k^{2}} $, $ x_{1}x_{2} = \\frac{m^{2} - 2}{2 + k^{2}} $; since $ k_{1} + k_{2} = 0 $, $ m_{n} $, $ (y_{1} - 1)(x_{2} - \\frac{\\sqrt{2}}{2}) + (y_{2} - 1)(x_{1} - \\frac{\\sqrt{2}}{2}) = 0 $, simplifying gives $ 2kx_{1}x_{2} + (m - 1 - \\frac{\\sqrt{2}}{2}k)(x_{1} + x_{2}) - \\sqrt{2}(m - 1) = 0^{x} $, so $ k = \\sqrt{2} $."
  },
  {
    "text": "As shown in the figure, in the rectangular coordinate system $x O y$, $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively, and $B$, $C$ are the upper and lower vertices of the ellipse, respectively. The line $B F_{2}$ intersects the ellipse at another point $D$. If $\\cos \\angle F_{1} B F_{2} = \\frac{7}{25}$, then the slope of line $CD$ is?",
    "fact_expressions": "F1: Point;F2: Point;E: Ellipse;LeftFocus(E) = F1;RightFocus(E) = F2;a: Number;b: Number;a > b;b > 0;Expression(E) = (x^2/a^2 + y^2/b^2 = 1);B: Point;C: Point;UpperVertex(E) = B;LowerVertex(E) = C;D: Point;Intersection(LineOf(B,F2),E)={B,D};Cos(AngleOf(F1, B, F2)) = 7/25;Negation(B=D)",
    "query_expressions": "Slope(LineOf(C, D))",
    "answer_expressions": "12/25",
    "fact_spans": "[[[20, 27]], [[28, 35]], [[38, 90], [108, 110], [129, 131]], [[20, 97]], [[20, 97]], [[40, 90]], [[40, 90]], [[40, 90]], [[40, 90]], [[38, 90]], [[98, 101]], [[102, 105]], [[98, 116]], [[98, 116]], [[138, 141]], [[117, 141]], [[143, 185]], [114, 137]]",
    "query_spans": "[[[187, 198]]]",
    "process": ""
  },
  {
    "text": "For the line $y=k x+1  (k \\in R)$ to always have common points with the ellipse $\\frac{x^{2}}{7}+\\frac{y^{2}}{a}=1$ whose foci lie on the $x$-axis, what is the range of real values of $a$?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*x + 1);k: Real;G: Ellipse;Expression(G) = (x^2/7 + y^2/a = 1);a: Real;PointOnCurve(Focus(G), xAxis);IsIntersect(H, G)",
    "query_expressions": "Range(a)",
    "answer_expressions": "[1, 7)",
    "fact_spans": "[[[2, 24]], [[2, 24]], [[4, 24]], [[34, 71]], [[34, 71]], [[77, 82]], [[25, 71]], [[2, 76]]]",
    "query_spans": "[[[77, 89]]]",
    "process": ""
  },
  {
    "text": "The standard equation of an ellipse with eccentricity $e=\\frac{2}{3}$ and minor axis length $8 \\sqrt{5}$ is?",
    "fact_expressions": "G: Ellipse;Length(MinorAxis(G)) = 8*sqrt(5);e: Number;Eccentricity(G) = e;e = 2/3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/144 + y^2/80 = 1, y^2/144 + x^2/80 = 1}",
    "fact_spans": "[[[36, 38]], [[19, 38]], [[3, 18]], [[0, 38]], [[3, 18]]]",
    "query_spans": "[[[36, 44]]]",
    "process": ""
  },
  {
    "text": "Let the focus of the parabola $y^{2}=4x$ be $F$. Draw a line $l$ passing through point $F$, intersecting the parabola at two points $A$ and $B$. Point $M$ satisfies $\\overrightarrow{O M}=\\frac{1}{2}(\\overrightarrow{O A}+\\overrightarrow{O B})$. Draw a perpendicular from $M$ to the $y$-axis, intersecting the parabola at point $P$. If $|P F|=2$, then the horizontal coordinate of point $M$ is?",
    "fact_expressions": "l: Line;G: Parabola;O: Origin;M: Point;A: Point;B: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};VectorOf(O, M) = (VectorOf(O, A) + VectorOf(O, B))/2;Z: Line;PointOnCurve(M, Z);IsPerpendicular(yAxis, Z);Intersection(Z, G) = P;Abs(LineSegmentOf(P, F)) = 2",
    "query_expressions": "XCoordinate(M)",
    "answer_expressions": "3",
    "fact_spans": "[[[29, 34]], [[1, 15], [35, 38], [150, 153]], [[59, 136]], [[53, 57], [138, 141], [172, 176]], [[44, 47]], [[48, 51]], [[155, 159]], [[19, 22], [24, 28]], [[1, 15]], [[1, 22]], [[23, 34]], [[29, 51]], [[59, 136]], [], [[137, 149]], [[137, 149]], [[137, 159]], [[161, 170]]]",
    "query_spans": "[[[172, 182]]]",
    "process": "The focus of the parabola $ y^{2}=4x $ is $ F(1,0) $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, $ M(x_{0},y_{0}) $, $ P(x,y_{0}) $, and the equation of line $ AB $ be $ y=k(x-1) $. Therefore,\n$$\n\\begin{cases}\ny^{2}=4x \\\\\ny=k(x-1)\n\\end{cases}\n\\Rightarrow\nky^{2}-4y+4=0\n\\Rightarrow\ny_{1}+y_{2}=\\frac{4}{k}\n\\Rightarrow\ny_{0}=\\frac{y_{1}+y_{2}}{2}=\\frac{2}{k}\n$$\nSince $ y_{0}^{2}=4x $, we have $ x=\\frac{1}{k^{2}} $. Given $ |PF|=2 $, we get $ x=1 $, so $ k^{2}=1 $. Hence, $ x_{0}=\\frac{x_{1}+x_{2}}{2}+1=3 $."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=2 p x(p>0)$ is one of the foci of the hyperbola $\\frac{x^{2}}{8}-\\frac{y^{2}}{p}=1$, then the asymptotes of the hyperbola are?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 2*(p*x));p: Number;p>0;G: Hyperbola;Expression(G) = (x^2/8 - y^2/p = 1);Focus(H) = OneOf(Focus(G))",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*x",
    "fact_spans": "[[[2, 23]], [[2, 23]], [[5, 23]], [[5, 23]], [[27, 65], [72, 75]], [[27, 65]], [[2, 70]]]",
    "query_spans": "[[[72, 83]]]",
    "process": "According to the problem, an equation in terms of $ p $ can be obtained. Solving for the positive value of $ p $, the standard equation of the hyperbola can be determined, and then the asymptotes of the hyperbola can be found. The focus of the parabola $ y^{2} = 2px $ ($ p > 0 $) is at $ \\left( \\frac{p}{2}, 0 \\right) $. According to the problem, the point $ \\left( \\frac{p}{2}, 0 \\right) $ is a focus of the hyperbola $ \\frac{x^{2}}{8} - \\frac{y^{2}}{p} = 1 $, so $ \\sqrt{8 + p} = \\frac{p}{2} $. Rearranging gives $ p^{2} - 4p - 32 = 0 $. Since $ p > 0 $, solving yields $ p = 8 $. Therefore, the standard equation of the hyperbola is $ \\frac{x^{2}}{8} - \\frac{y^{2}}{8} = 1 $. Thus, the asymptotes of the hyperbola are $ y = \\pm x $."
  },
  {
    "text": "Given that hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ and hyperbola $C_{2}$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{16}=1$ have the same asymptotes, and the right focus of $C_{1}$ is $F(\\sqrt{5} , 0)$, then $a$=? $b$=?",
    "fact_expressions": "C1: Hyperbola;C2: Hyperbola;a:Number;b:Number;F: Point;Expression(C1) = (x^2/a^2-y^2/b^2=1);Expression(C2) = (x^2/4-y^2/16=1);Coordinate(F) = (sqrt(5), 0);RightFocus(C1) = F;a>0;b>0;Asymptote(C1)=Asymptote(C2)",
    "query_expressions": "a;b",
    "answer_expressions": "1\n2",
    "fact_spans": "[[[2, 68], [126, 133]], [[69, 117]], [[157, 160]], [[163, 166]], [[138, 155]], [[2, 68]], [[69, 117]], [[138, 155]], [[126, 155]], [[13, 68]], [[13, 68]], [2, 119]]",
    "query_spans": "[[[157, 162]], [[163, 168]]]",
    "process": ""
  },
  {
    "text": "Let $P$ be a point on the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{32}=1$, and let $M$, $N$ be points on the circles $(x-2)^{2}+y^{2}=1$ and $(x+2)^{2}+y^{2}=\\frac{1}{4}$, respectively. Then the range of values of $|P M|+|P N|$ is?",
    "fact_expressions": "G: Ellipse;H: Circle;C:Circle;P: Point;M: Point;N: Point;Expression(G) = (x^2/36 + y^2/32 = 1);Expression(H) = (y^2 + (x - 2)^2 = 1);Expression(C) = (y^2 + (x + 2)^2 = 1/4);PointOnCurve(P, G);PointOnCurve(M, H);PointOnCurve(N, C)",
    "query_expressions": "Range(Abs(LineSegmentOf(P, M)) + Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "[21/2, 27/2]",
    "fact_spans": "[[[5, 44]], [[60, 79]], [[80, 109]], [[1, 4]], [[48, 51]], [[52, 55]], [[5, 44]], [[60, 79]], [[80, 109]], [[1, 47]], [[48, 112]], [[48, 112]]]",
    "query_spans": "[[[114, 134]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ is any point on the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$, the minimum distance from point $P$ to the line $l$: $x-y+2 \\sqrt{3}=0$ is?",
    "fact_expressions": "l: Line;C: Ellipse;P: Point;Expression(C) = (x^2/2 + y^2 = 1);PointOnCurve(P, C);Expression(l)=(x - y + 2*sqrt(3)=0)",
    "query_expressions": "Min(Distance(P,l))",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[51, 76]], [[7, 39]], [[2, 6], [46, 50]], [[7, 39]], [[2, 44]], [[51, 76]]]",
    "query_spans": "[[[46, 84]]]",
    "process": "Let the line tangent to the ellipse C and parallel to l be y = x + m. Combining with the ellipse and simplifying yields: 3x^{2}+4mx+2m^{2}-2=0, then \\triangle=16m^{2}-24(m^{2}-1)=0 \\therefore m=\\pm\\sqrt{3}. Since the distance between two parallel lines is d=\\frac{|m-2\\sqrt{3}|}{\\sqrt{2}}, the minimum distance from P to the line l: x-y+2\\sqrt{3}=0 is d=\\frac{|\\sqrt{3}-2\\sqrt{3}|}{\\sqrt{6}}=\\frac{\\sqrt{6}}{2}"
  },
  {
    "text": "The coordinates of the vertices of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{7}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/7 = 1)",
    "query_expressions": "Coordinate(Vertex(G))",
    "answer_expressions": "(pm*3, 0)",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 45]]]",
    "process": "From the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{7}=1$, we get $a=3$, so the vertex coordinates are $(3,0)$, $(-3,0)$."
  },
  {
    "text": "$P$ is a point on the right branch of the hyperbola $x^{2}-\\frac{y^{2}}{15}=1$, and $M$, $N$ are points on the circles $(x+4)^{2}+y^{2}=4$ and $(x-4)^{2}+y^{2}=4$, respectively. Then the maximum value of $|P M|-|P N|$ is?",
    "fact_expressions": "P: Point;G: Hyperbola;Expression(G) = (x^2 - y^2/15 = 1);PointOnCurve(P, RightPart(G)) = True;M: Point;N: Point;H1: Circle;Expression(H1) = (y^2 + (x + 4)^2 = 4);H2: Circle;Expression(H2) = (y^2 + (x - 4)^2 = 4);PointOnCurve(M, H1) = True;PointOnCurve(N, H2) = True",
    "query_expressions": "Max(Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "6",
    "fact_spans": "[[[0, 3]], [[4, 33]], [[4, 33]], [[0, 38]], [[39, 42]], [[43, 46]], [[49, 69]], [[49, 69]], [[70, 89]], [[70, 89]], [[39, 92]], [[39, 92]]]",
    "query_spans": "[[[94, 113]]]",
    "process": "Note that the centers of the two circles are the two foci of the hyperbola. Using the definition of a hyperbola, connect P to the left focus $ F_{1} $ and extend it to intersect the lower semicircle at point M, and let $ PF_{2} $ intersect the upper semicircle at point N. Clearly, $ |PM| - |PN| = (|PF_{1}| + 2) - (|PF_{2}| - 2) = 2a + 4 $ is the maximum value. [Solution] The circle $ (x+4)^{2} + y^{2} = 4 $ has center $ (-4, 0) $ and radius 2; the circle $ (x-4)^{2} + y^{2} = 4 $ has center $ (4, 0) $ and radius 2. Connect P to the left focus $ F_{1} $, extend it to intersect the lower semicircle at M, and let $ PF_{2} $ intersect the upper semicircle at N. Clearly, $ |PM| - |PN| = (|PF_{1}| + 2) - (|PF_{2}| - 2) = 2a + 4 = 6 $ is the maximum value. Hence, the answer is: 6."
  },
  {
    "text": "What is the length of the minor axis of the ellipse $x^{2}+3 y^{2}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 3*y^2 = 1)",
    "query_expressions": "Length(MinorAxis(G))",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[0, 19]], [[0, 19]]]",
    "query_spans": "[[[0, 25]]]",
    "process": "From the given problem, the standard equation of the ellipse is $x^{2}+\\frac{y^{2}}{3}=1$, then $a^{2}=1$ so $a=1$, $b^{2}=\\frac{1}{3}$ so $b=\\frac{\\sqrt{3}}{3}$. Therefore, the length of the minor axis is $\\frac{2\\sqrt{3}}{3}$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $A(2,0)$ is an endpoint of the major axis, chord $BC$ passes through the center $O$ of the ellipse, and $\\overrightarrow{A C} \\cdot \\overrightarrow{B C}=0$, $|\\overrightarrow{O C}-\\overrightarrow{O B}|=2|\\overrightarrow{B C}-\\overrightarrow{B A}|$. Then the focal distance of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;A: Point;Coordinate(A) = (2, 0);OneOf(Endpoint(MajorAxis(G))) = A;B: Point;C: Point;IsChordOf(LineSegmentOf(B, C), G);O: Point;Center(G) = O;PointOnCurve(O, LineSegmentOf(B, C));DotProduct(VectorOf(A, C), VectorOf(B, C)) = 0;Abs(-VectorOf(O, B) + VectorOf(O, C)) = 2*Abs(-VectorOf(B, A) + VectorOf(B, C))",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "(4/3)*sqrt(6)",
    "fact_spans": "[[[2, 54], [79, 81], [234, 236]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[56, 64]], [[56, 64]], [[2, 71]], [[73, 78]], [[73, 78]], [[2, 78]], [[84, 87]], [[79, 87]], [[73, 87]], [[89, 140]], [[142, 232]]]",
    "query_spans": "[[[234, 241]]]",
    "process": "As shown in the figure: since $\\overrightarrow{AC}\\cdot\\overrightarrow{BC}=0$, then $\\overrightarrow{AC}\\bot\\overrightarrow{BC}$. Also, because $|\\overrightarrow{OC}-\\overrightarrow{OB}|=2|\\overrightarrow{BC}-\\overrightarrow{BA}|$, so $|\\overrightarrow{BC}|=2|\\overrightarrow{AC}|$, thus $\\triangle AOC$ is an isosceles right triangle. Since $A(2,0)$, then $C(1,1)$. Also, since $C(1,1)$ lies on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, we have $\\frac{1}{a^{2}}+\\frac{1}{b^{2}}=1$. Given $a=2$, solving gives $b^{2}=\\frac{4}{3}$. Therefore, $c=\\sqrt{4-\\frac{4}{3}}=\\frac{\\sqrt[2]{6}}{3}$, and the focal distance is $\\frac{4}{3}\\sqrt{6}$."
  },
  {
    "text": "Given that one of the asymptotes of the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{3}=1$ is $y=\\sqrt{3} x$, then what is the distance from a point $M(2, y_0)$ on the parabola $y^{2}=4 a x$ to the focus $F$ of the parabola?",
    "fact_expressions": "G: Hyperbola;H:Parabola;a: Number;M: Point;y0:Number;Expression(G) = (-y^2/3 + x^2/a = 1);Expression(H) = (y^2 = 4*(a*x));Coordinate(M) = (2, y0);Expression(OneOf(Asymptote(G)))= (y = sqrt(3)*x);PointOnCurve(M, H);Focus(H) = F;F:Point",
    "query_expressions": "Distance(M,F)",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 40]], [[66, 82], [98, 101]], [[5, 40]], [[85, 96]], [[85, 96]], [[2, 40]], [[66, 82]], [[85, 96]], [[2, 64]], [[66, 96]], [[98, 106]], [[103, 106]]]",
    "query_spans": "[[[85, 111]]]",
    "process": ""
  },
  {
    "text": "The left and right foci of the ellipse $M$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ are $F_{1}$ and $F_{2}$, respectively. Let $P$ be any point on the ellipse $M$, and the maximum value of $|P F_{1}| \\cdot |P F_{2}|$ lies in the range $[2 c^{2}, 3 c^{2}]$, where $c=\\sqrt{a^{2}-b^{2}}$. Then, what is the range of the eccentricity of the ellipse $M$?",
    "fact_expressions": "M: Ellipse;a: Number;b: Number;c: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(M) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(M) = F1;RightFocus(M) = F2;PointOnCurve(P, M);c = sqrt(a^2 - b^2);Range(Max(Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)))) = [2*c^2, 3*c^2]",
    "query_expressions": "Range(Eccentricity(M))",
    "answer_expressions": "[sqrt(3)/3, sqrt(2)/2]",
    "fact_spans": "[[[0, 56], [85, 90], [179, 184]], [[6, 56]], [[6, 56]], [[155, 177]], [[81, 84]], [[65, 72]], [[73, 80]], [[6, 56]], [[6, 56]], [[0, 56]], [[0, 80]], [[0, 80]], [[81, 94]], [[155, 177]], [[96, 152]]]",
    "query_spans": "[[[179, 195]]]",
    "process": "\\because the maximum value of |PF_{1}||PF_{2}| is a^{2},\\therefore by the problem 2c^{2}\\leqslanta^{2}\\leqslant3c^{2},\\therefore\\sqrt{2}c\\leqslanta\\leqslant\\sqrt{3}c,\\therefore\\frac{\\sqrt{3}}{3}\\leqslante\\leqslant\\frac{\\sqrt{2}}{2}\\therefore the range of ellipse eccentricity e is [\\frac{\\sqrt{3}}{3},\\frac{\\sqrt{2}}{2}]"
  },
  {
    "text": "The moving point $P$ travels along the curve $y=2x^{2}+1$. What is the equation of the locus of the midpoint of the line segment joining point $P$ and the fixed point $A(0,-1)$?",
    "fact_expressions": "G: Curve;A: Point;P: Point;Expression(G) = (y = 2*x^2 + 1);Coordinate(A) = (0, -1);PointOnCurve(P, G)",
    "query_expressions": "LocusEquation(MidPoint(LineSegmentOf(P,A)))",
    "answer_expressions": "y=4*x^2",
    "fact_spans": "[[[6, 21]], [[33, 42]], [[2, 5], [26, 30]], [[6, 21]], [[33, 42]], [[2, 24]]]",
    "query_spans": "[[[26, 54]]]",
    "process": "Let $ P(x_{0},y_{0}) $, and let the midpoint of the line segment joining point $ P $ and the fixed point $ A(0,-1) $ be $ (x,y) $. Then  \n$$\n\\begin{cases}\n\\frac{x_{0}}{2}=x \\\\\n\\frac{y_{0}-1}{2}=y\n\\end{cases},\n$$\nand $ y_{0} = 2x_{0}^{2} + 1 $. Therefore,\n$$\n\\begin{cases}\nx_{0} = 2x \\\\\ny_{0} = 2y + 1\n\\end{cases}.\n$$\nHence, $ 2y + 1 = 2 \\times (2x)^{2} + 1 $, which simplifies to $ y = 4x^{2} $. Thus, the trajectory equation of the midpoint of the line segment joining point $ P $ and the fixed point $ A(0,-1) $ is $ y = 4x^{2} $."
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$, point $P$ moves on the ellipse, then the maximum value of $|P F_{1}| \\cdot|P F_{2}|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/5 + y^2/4 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)))",
    "answer_expressions": "5",
    "fact_spans": "[[[16, 53], [65, 67]], [[60, 64]], [[0, 7]], [[8, 15]], [[16, 53]], [[0, 59]], [[0, 59]], [[60, 70]]]",
    "query_spans": "[[[72, 104]]]",
    "process": "Since point $P$ lies on the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$, by the definition of an ellipse, we have $|PF_{1}|+|PF_{2}|=2a=2\\sqrt{5}$. Also, by the inequality $|PF_{1}||PF_{2}|\\leqslant\\left(\\frac{|PF_{1}|+|PF_{2}|}{2}\\right)^{2}=(\\sqrt{5})^{2}=5$, equality holds if and only if $|PF_{1}|=|PF_{2}|$. Therefore, the maximum value of $|PF_{1}||PF_{2}|$ is 5."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=2 p x(p>0)$ is $F(2,0)$, then $p=$? Draw a perpendicular from point $A(3,2)$ to its directrix, and denote the intersection point with the parabola as $E$, then $|E F|=$?",
    "fact_expressions": "G: Parabola;p: Number;F: Point;A: Point;E: Point;p>0;Expression(G) = (y^2 = 2*p*x);Coordinate(F) = (2, 0);Coordinate(A) = (3, 2);Focus(G) = F;l:Line;PointOnCurve(A,l);IsPerpendicular(l,Directrix(G));Intersection(l,G)=E",
    "query_expressions": "p;Abs(LineSegmentOf(E, F))",
    "answer_expressions": "4\n5/2",
    "fact_spans": "[[[2, 23], [53, 54], [62, 65]], [[37, 40]], [[27, 35]], [[43, 52]], [[69, 72]], [[5, 23]], [[2, 23]], [[27, 35]], [[43, 52]], [[2, 35]], [], [[42, 59]], [[42, 59]], [[42, 72]]]",
    "query_spans": "[[[37, 42]], [[74, 83]]]",
    "process": "Since the focus of the parabola is $ F\\left(\\frac{p}{2}, 0\\right) $, we have $ \\frac{p}{2} = 2 $, so $ p = 4 $. Thus, the equation of the parabola is $ y^2 = 8x $. It can be seen that point $ A(3, 2) $ lies inside the parabola. Drawing a perpendicular line from point $ A(3, 2) $ to the directrix $ x = -2 $, this perpendicular line has the equation $ y = 2 $. Substituting $ y = 2 $ into the parabola equation $ y^2 = 8x $ gives $ x = \\frac{1}{2} $, so point $ E\\left(\\frac{1}{2}, 2\\right) $. Therefore, $ |EF| = \\sqrt{\\left(2 - \\frac{1}{2}\\right)^2 + 2^2} = \\frac{5}{2} $."
  },
  {
    "text": "Let the constant $m \\in R$. The equation of the directrix of the parabola $y = m x^{2}$ is $y = -1$. Then $m =$?",
    "fact_expressions": "G: Parabola;m: Real;Expression(G) = (y = m*x^2);Expression(Directrix(G)) = (y = -1)",
    "query_expressions": "m",
    "answer_expressions": "1/4",
    "fact_spans": "[[[13, 27]], [[41, 44]], [[13, 27]], [[13, 39]]]",
    "query_spans": "[[[41, 46]]]",
    "process": "The standard equation of the parabola $ y = mx^{2} $ is $ x^{2} = \\frac{1}{m}y^{\\circ} $, then the equation of the directrix of the parabola is: $ y = -\\frac{1}{4m} $. Given that the directrix of the parabola $ y = mx^{2} $ is $ y = -1 $, so $ -\\frac{1}{4m} = -1 $, we get $ m = \\frac{1}{4} $,"
  },
  {
    "text": "It is known that the foci and the endpoints of the real axis of hyperbola $C$ are exactly the endpoints of the major axis and the foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$, respectively. Then, the equations of the asymptotes of hyperbola $C$ are?",
    "fact_expressions": "G: Ellipse;Expression(G)=(x^2/16+y^2/7=1);C:Hyperbola;Focus(C)=Endpoint(MajorAxis(G));Endpoint(RealAxis(C))=Focus(G)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*(sqrt(7)/3)*x",
    "fact_spans": "[[[21, 59]], [[21, 59]], [[2, 8], [69, 75]], [[2, 67]], [[2, 67]]]",
    "query_spans": "[[[69, 83]]]",
    "process": "\\because the equation of the ellipse is \\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1, \\therefore the endpoints of the major axis of the ellipse are (-4,0), (4,0), and the foci are (-3,0), (3,0). \\because the foci and real axis endpoints of hyperbola C are exactly the major axis endpoints and foci of the ellipse \\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1, \\therefore for the hyperbola, c=4, a=3, then b=\\sqrt{7}. \\therefore the equations of the asymptotes of the hyperbola are y=\\pm\\frac{\\sqrt{7}}{3}x"
  },
  {
    "text": "The distance from the focus to the directrix of the parabola $y=12 x^{2}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 12*x^2)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "1/24",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 26]]]",
    "process": ""
  },
  {
    "text": "If the equation $\\frac{x^{2}}{10-m}+\\frac{y^{2}}{m-2}=1$ represents an ellipse with foci on the $x$-axis, then the range of values for $m$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(10 - m) + y^2/(m - 2) = 1);PointOnCurve(Focus(G), xAxis);m:Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(2,6)",
    "fact_spans": "[[[54, 56]], [[1, 56]], [[45, 56]], [[58, 61]]]",
    "query_spans": "[[[58, 68]]]",
    "process": ""
  },
  {
    "text": "It is known that the ellipse $C$ has its center at the origin of the coordinate system, with foci on the $y$-axis. Let $F_{1}$ and $F_{2}$ be the two foci of $C$, the minor axis of $C$ has length $4$, and there exists a point $P$ on $C$ such that $|P F_{1}| = 6|P F_{2}|$. Write a standard equation of $C$?",
    "fact_expressions": "O: Origin;Center(C) = O;C: Ellipse;PointOnCurve(Focus(C),yAxis);P: Point;F1: Point;F2: Point;Focus(C) = {F1, F2};Length(MinorAxis(C)) = 4;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 6*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4 + y^2/9 = 1",
    "fact_spans": "[[[11, 15]], [[2, 15]], [[2, 7], [41, 44], [50, 53], [63, 66], [102, 105]], [[2, 24]], [[71, 74]], [[25, 32]], [[33, 40]], [[25, 49]], [[50, 61]], [[63, 74]], [[77, 99]]]",
    "query_spans": "[[[102, 113]]]",
    "process": "Since |PF_{1}| = 6|PF_{2}|, it follows that |PF_{1}| + |PF_{2}| = 7|PF_{2}| = 2a, then |PF_{2}| = \\frac{2a}{7}. Also, since a - c \\leqslant |PF_{2}| \\leqslant a + c, we have \\frac{2a}{7} \\geqslant a - c, which implies \\frac{c}{a} \\geqslant \\frac{5}{7}. According to the problem, suppose the equation of C is \\frac{x^{2}}{b^{2}} + \\frac{y^{2}}{a^{2}} = 1 (a > b > 0). Since the minor axis length of ellipse C is 4, then 2b = 4 gives b = 2. From \\frac{c}{a} \\geqslant \\frac{5}{7}, we obtain \\sqrt{1-}\\frac{2}{2} = \\sqrt{1 - \\frac{4}{a^{2}}} \\geqslant \\frac{5}{7}, solving yields a^{2} \\geqslant \\frac{49}{6}. Thus, one standard equation of the ellipse C is \\frac{x^{2}}{4} + \\frac{y^{2}}{9} = 1."
  },
  {
    "text": "The standard equation of a hyperbola passing through the point $P(3 , 2)$ and having the same asymptotes as the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$ is?",
    "fact_expressions": "E: Hyperbola;P: Point;Expression(E) = (x^2/4 - y^2/2 = 1);Coordinate(P) = (3, 2);C: Hyperbola;PointOnCurve(P, C);Expression(Asymptote(C)) = Expression(Asymptote(E))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2/(1/2) = 1",
    "fact_spans": "[[[14, 52]], [[1, 12]], [[14, 52]], [[1, 12]], [[61, 64]], [[0, 64]], [[13, 64]]]",
    "query_spans": "[[[61, 71]]]",
    "process": ""
  },
  {
    "text": "If a point $M$ on the parabola $y^{2}=16 x$ is at a distance of $12$ from the $x$-axis, then what is the distance from point $M$ to the focus of this parabola?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = 16*x);PointOnCurve(M, G);Distance(M, xAxis) = 12",
    "query_expressions": "Distance(M, Focus(G))",
    "answer_expressions": "13",
    "fact_spans": "[[[1, 16], [44, 47]], [[19, 22], [38, 42]], [[1, 16]], [[1, 22]], [[19, 36]]]",
    "query_spans": "[[[38, 55]]]",
    "process": "According to the problem, the vertical coordinate of point M satisfies |y_{M}|=12. Substituting into the parabolic equation gives x_{M}=9. The directrix of the parabola is x=-\\frac{p}{2}=-4. By the definition of a parabola, the distance between point M and the focus F is MF=x_{M}+\\frac{p}{2}=9+4=13. The answer is 13."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Let point $P(0, y_{0})(y_{0}>0)$, and let the circle with diameter $OP$ intersect the line $y=\\frac{b}{a}x$ at points $O$ and $M$, where point $M$ lies on segment $PF_{2}$. If $\\frac{S_{\\Delta M F_{2} O} }{S_{\\Delta P M O}}=7$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;H: Line;P: Point;F2: Point;O: Origin;M: Point;Area(TriangleOf(M,F2,O))/Area(TriangleOf(P,M,O))=7;F1: Point;y0:Number;y0>0;a > b;b > 0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = x*(b/a));Coordinate(P) = (0, y0);LeftFocus(C) = F1;RightFocus(C) = F2;IsDiameter(LineSegmentOf(O, P), G);Intersection(G, H) = {O, M};PointOnCurve(M, LineSegmentOf(P,F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 60], [224, 230]], [[10, 60]], [[10, 60]], [[119, 120]], [[121, 140]], [[85, 108]], [[77, 84]], [[144, 147]], [[148, 151], [153, 157]], [[172, 222]], [[69, 76]], [[85, 108]], [[85, 108]], [[10, 60]], [[10, 60]], [[2, 60]], [[121, 140]], [[85, 108]], [[2, 84]], [[2, 84]], [[109, 120]], [[119, 151]], [[153, 170]]]",
    "query_spans": "[[[224, 236]]]",
    "process": "Since $\\frac{S_{AMF_{2}O}}{S_{APMO}}=7$, i.e., $\\frac{|MF_{2}|}{|MP|}=7$, and $OM\\bot PF_{2}$, hence $|MF_{2}|=\\frac{bc}{\\sqrt{a^{2}+b^{2}}}=\\frac{bc}{c}=b$, then $|PM|=\\frac{b}{7}$. In right triangle $POF_{2}$, we obtain $|OF_{2}^{2}|=|MF_{2}||F_{2}P|$, that is, $c^{2}=b\\cdot\\frac{8}{7}b=\\frac{8}{7}(c^{2}-a^{2})$, thus $c^{2}=8a^{2}$, hence $e=\\frac{c}{a}=2\\sqrt{2}$. Answer: $2\\sqrt{2}$"
  },
  {
    "text": "Let $D$ be an arbitrary point on the ellipse $x^{2}+\\frac{y^{2}}{5}=1$, $A(0,-2)$, $B(0,2)$. Extend $AD$ to a point $P$ such that $|PD|=|BD|$. Then the equation of the trajectory of point $P$ is?",
    "fact_expressions": "G: Ellipse;A: Point;D: Point;B: Point;P: Point;Expression(G) = (x^2 + y^2/5 = 1);Coordinate(A) = (0, -2);Coordinate(B) = (0, 2);PointOnCurve(D, G);Abs(LineSegmentOf(P, D)) = Abs(LineSegmentOf(B, D));PointOnCurve(P,OverlappingLine(LineSegmentOf(A,D)))",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2+(y+2)^2=20",
    "fact_spans": "[[[5, 32]], [[38, 47]], [[1, 4]], [[50, 58]], [[67, 71], [89, 93]], [[5, 32]], [[38, 47]], [[50, 58]], [[1, 37]], [[74, 87]], [[59, 71]]]",
    "query_spans": "[[[89, 100]]]",
    "process": "As shown in the figure, from the ellipse equation $ x^{2} + \\frac{y^{2}}{5} = 1 $, we obtain $ a^{2} = 5 $, $ b^{2} = 1 $, so $ c = \\sqrt{a^{2} - b^{2}} = 2 $, then $ A(0, -2) $, $ B(0, 2) $ are the two foci of the ellipse, thus $ |DA| + |DB| = 2a = 2\\sqrt{5} $. Since $ |PD| = |BD| $, then $ |PA| = |PD| + |DA| = |BD| + |DA| = 2\\sqrt{5} $. Therefore, the locus of point $ P $ is a circle with center $ A $ and radius $ 2\\sqrt{5} $, its equation is $ x^{2} + (y + 2)^{2} = 20 $."
  },
  {
    "text": "Let $A$ and $B$ be two fixed points, $|AB|=2$, and a moving point $P$ satisfies $|PA|+|PB|=t$. If the trajectory of point $P$ is an ellipse, then the range of values for $t$ is?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;P: Point;t: Number;Abs(LineSegmentOf(A, B)) = 2;Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B)) = t;Locus(P) = G",
    "query_expressions": "Range(t)",
    "answer_expressions": "(2, +oo)",
    "fact_spans": "[[[57, 59]], [[1, 4]], [[6, 10]], [[27, 30], [49, 53]], [[61, 64]], [[16, 24]], [[32, 47]], [[49, 59]]]",
    "query_spans": "[[[61, 71]]]",
    "process": ""
  },
  {
    "text": "The asymptotes of the hyperbola $C$: $x^{2}-y^{2}=1$ are given by? If the right vertex of hyperbola $C$ is $A$, and a line $l$ passing through $A$ intersects the two asymptotes of hyperbola $C$ at points $P$ and $Q$, such that $PA = 2AQ$, then the slope of line $l$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2 = 1);A: Point;RightVertex(C) = A;l: Line;PointOnCurve(A, l) = True;J1: Line;J2: Line;Asymptote(C) = {J1, J2};Intersection(l, J1) = P;Intersection(l, J2) = Q;P: Point;Q: Point;LineSegmentOf(P, A) = 2*LineSegmentOf(A, Q)",
    "query_expressions": "Expression(Asymptote(C));Slope(l)",
    "answer_expressions": "x+pm*y=0;pm*3",
    "fact_spans": "[[[0, 23], [32, 38], [58, 64]], [[0, 23]], [[43, 46], [48, 51]], [[32, 46]], [[52, 57], [96, 101]], [[47, 57]], [], [], [[58, 70]], [[52, 81]], [[52, 81]], [[72, 75]], [[76, 79]], [[83, 94]]]",
    "query_spans": "[[[0, 31]], [[96, 106]]]",
    "process": ""
  },
  {
    "text": "Given that the line $l$ passes through the focus of the parabola $C$: $y^{2}=4x$, and intersects $C$ at points $A$ and $B$. The tangents to $C$ at points $A$ and $B$ are drawn respectively, intersecting at point $P$. Then, the equation of the locus of point $P$ is?",
    "fact_expressions": "l: Line;C: Parabola;Expression(C) = (y^2 = 4*x);PointOnCurve(Focus(C), l);A: Point;B: Point;Intersection(l, C) = {A, B};Z1: Line;Z2: Line;TangentOfPoint(A, C) = Z1;TangentOfPoint(B, C) = Z2;P: Point;Intersection(Z1, Z2) = P",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x=-1",
    "fact_spans": "[[[2, 7], [31, 34]], [[8, 27], [35, 38], [62, 65]], [[8, 27]], [[2, 30]], [[40, 43], [51, 55]], [[44, 47], [56, 59]], [[31, 49]], [], [], [[50, 68]], [[50, 68]], [[72, 76], [78, 82]], [[50, 76]]]",
    "query_spans": "[[[78, 89]]]",
    "process": "Without loss of generality, flip the parabola to $x^{2}=4y$, and let the equation of the flipped line $l$ be $y=kx+1$. Let the coordinates of the flipped points $A$ and $B$ be $(x_{1},y_{1})$ and $(x_{2},y_{2})$, respectively. Then, by solving the system\n\\[\n\\begin{cases}\nx^{2}=4y \\\\\ny=kx+1\n\\end{cases}\n\\]\nwe obtain $x^{2}-4kx-4=0$ \\textcircled{1}. It is easy to find that the tangent line to the parabola $x^{2}=4y$ at point $A$ is given by $y-\\frac{1}{4}x_{1}^{2}=\\frac{1}{2}x_{1}(x-x_{1})$. Similarly, the tangent line to the parabola $x^{2}=4y$ at point $B$ is $y-\\frac{1}{4}x_{2}^{2}=\\frac{1}{2}x_{2}(x-x_{2})$. Solving the system\n\\[\n\\begin{cases}\ny-\\frac{1}{4}x_{1}^{2}=\\frac{1}{2}x_{1}(x-x_{1}) \\\\\ny-\\frac{1}{4}x_{2}^{2}=\\frac{1}{2}x_{2}(x-x_{2})\n\\end{cases}\n\\]\nyields $y=\\frac{1}{4}x_{1}x_{2}$. From \\textcircled{1}, we have $x_{1}x_{2}=-4$, so $y=-1$. Hence, the trajectory equation of the corresponding point $P$ on the original parabola $C$ is $x=-1$."
  },
  {
    "text": "Given that the point $(4 , 2)$ is the midpoint of the segment cut by the line $l$ on the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$, then the slope of the line $l$ is?",
    "fact_expressions": "l: Line;G: Ellipse;I: Point;Expression(G) = (x^2/36 + y^2/9 = 1);Coordinate(I) = (4, 2);MidPoint(InterceptChord(l, G)) = I",
    "query_expressions": "Slope(l)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[13, 18], [68, 73]], [[19, 57]], [[2, 12]], [[19, 57]], [[2, 12]], [[2, 66]]]",
    "query_spans": "[[[68, 78]]]",
    "process": ""
  },
  {
    "text": "If the two asymptotes of a hyperbola are perpendicular to each other, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;l1: Line;l2: Line;Asymptote(G) = {l1, l2};IsPerpendicular(l1, l2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 4], [17, 20]], [], [], [[1, 10]], [[1, 14]]]",
    "query_spans": "[[[17, 26]]]",
    "process": "The solution process is omitted"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1  (a>0)$ has an asymptote $y=k x  (k>0)$ and eccentricity $e=\\sqrt{5} k$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2 + x^2/a^2 = 1);a: Number;a>0;k: Number;k>0;Expression(OneOf(Asymptote(G))) = (y = k*x);e: Number;Eccentricity(G) = e;e = sqrt(5)*k",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4-y^2=1",
    "fact_spans": "[[[2, 41], [81, 84]], [[2, 41]], [[5, 41]], [[5, 41]], [[48, 61]], [[48, 61]], [[2, 61]], [[65, 79]], [[2, 79]], [[65, 79]]]",
    "query_spans": "[[[81, 88]]]",
    "process": ""
  },
  {
    "text": "The curve $C$: $|x^{2}-y^{2}-1|=1$ and the line $l$: $y=k x-1$ have 4 intersection points; then the range of values for $k$ is?",
    "fact_expressions": "C: Curve;Expression(C) = (Abs(x^2 - y^2 - 1) = 1);l: Line;Expression(l) = (y = k*x - 1);k: Number;NumIntersection(C, l) = 4",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-sqrt(6)/2, -1) + (-1, 1) + (1, sqrt(6)/2)",
    "fact_spans": "[[[0, 25]], [[0, 25]], [[26, 42]], [[26, 42]], [[51, 54]], [[0, 49]]]",
    "query_spans": "[[[51, 61]]]",
    "process": "The curve $ C: |x^{2}-y^{2}-1|=1 $ simplifies to $ x^{2}-y^{2}=0 $, i.e., $ y=\\pm x $, or $ x^{2}-y^{2}=2 $. Therefore, the curve $ C $ represents two intersecting lines and a hyperbola. The curve $ C $ and the line $ l $ have 4 intersection points, so there are two intersection points with the lines $ y=\\pm x $. It suffices that $ k\\neq\\pm1 $, and the line $ l $ intersects the hyperbola $ x^{2}-y^{2}=2 $ at two points. Solving the system \n$$\n\\begin{cases}\ny=kx-1 \\\\\nx^{2}-y^{2}=2\n\\end{cases}\n$$\nby eliminating $ y $, we obtain $ (1-k^{2})x^{2}+2kx-3=0 $. For this equation to have two solutions, we require\n$$\n\\begin{cases}\nk^{2}-1\\neq0 \\\\\n4k^{2}-12(1-k^{2})>0\n\\end{cases}\n$$\nSolving gives $ -\\frac{\\sqrt{6}}{2}<k<\\frac{\\sqrt{6}}{2} $ and $ k\\neq\\pm1 $. Thus, the desired range of $ k $ is $ \\left(-\\frac{\\sqrt{6}}{2},-1\\right)\\cup( $"
  },
  {
    "text": "A line with slope $\\frac{1}{2}$ intersects the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ at two points $A$ and $B$. The midpoint $M(m, \\frac{1}{2})$ of $AB$ is given. Find $m=?$",
    "fact_expressions": "L: Line;Slope(L) = 1/2;G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);Intersection(L,G) = {A,B};A: Point;B: Point;M: Point;MidPoint(LineSegmentOf(A, B)) = M;Coordinate(M) = (m, 1/2);m: Number",
    "query_expressions": "m",
    "answer_expressions": "-1/3",
    "fact_spans": "[[[17, 19]], [[0, 19]], [[20, 57]], [[20, 57]], [[17, 69]], [[60, 63]], [[64, 67]], [[78, 97]], [[70, 97]], [[78, 97]], [[99, 102]]]",
    "query_spans": "[[[99, 104]]]",
    "process": "First, assume the line AB is $ y = \\frac{1}{2}x + b $, then substitute it into the ellipse equation to eliminate $ x $, obtaining a quadratic equation in $ y $. Then, the sum of the $ y $-coordinates of points A and B can be expressed, and thus the coordinates of M can be determined. Combining this with the midpoint $ M(m, \\frac{1}{2}) $ of AB, the answer can be found. Solution: Let the line AB be $ y = \\frac{1}{2}x + b $. Substituting into the ellipse equation $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $, we get: $ 4y^{2} - 6by + 3b^{2} - 3 = 0 $, $ y_{A} + y_{B} = \\frac{3}{2}b $, $ y_{M} = \\frac{1}{2}(y_{A} + y_{B}) = \\frac{3}{4}b = \\frac{1}{2} $, so $ b = \\frac{2}{3} $. The line AB is $ y = \\frac{1}{2}x + \\frac{2}{3} $. Given the midpoint $ M(m, \\frac{1}{2}) $ of AB, we have $ \\frac{1}{2} = \\frac{1}{2}m + \\frac{2}{3} $, $ \\therefore m = -\\frac{1}{3} $."
  },
  {
    "text": "The coordinates of the left focus of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ are?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1)",
    "query_expressions": "Coordinate(LeftFocus(G))",
    "answer_expressions": "(-1, 0)",
    "fact_spans": "[[[0, 27]], [[0, 27]]]",
    "query_spans": "[[[0, 36]]]",
    "process": "From the standard equation of the ellipse, we can find $ c $, and thus obtain the coordinates of the left focus of the ellipse. From the standard equation of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, we have $ a^{2}=2 $, $ b^{2}=1 $, $\\therefore c^{2}=1$, $\\therefore c=1$, so the coordinates of the left focus are $(-1,0)$."
  },
  {
    "text": "A line $l$ passing through the focus $F$ of the parabola $C$: $y^{2}=8x$ intersects the parabola at points $P$ and $Q$, and intersects the directrix of the parabola at point $M$, with $\\overrightarrow{F M}=3 \\overrightarrow{F P}$. Then $|\\overrightarrow{F P}|$=?",
    "fact_expressions": "l: Line;C: Parabola;F: Point;M: Point;P: Point;Q: Point;Expression(C) = (y^2 = 8*x);Focus(C)=F;PointOnCurve(F,l);Intersection(l, C) = {P, Q};Intersection(l,Directrix(C))=M;VectorOf(F, M) = 3*VectorOf(F, P)",
    "query_expressions": "Abs(VectorOf(F, P))",
    "answer_expressions": "8/3",
    "fact_spans": "[[[27, 32]], [[1, 20], [33, 36], [33, 36]], [[23, 26]], [[57, 61]], [[38, 41]], [[42, 45]], [[1, 20]], [[1, 26]], [[0, 32]], [[27, 47]], [[27, 61]], [[63, 108]]]",
    "query_spans": "[[[110, 136]]]",
    "process": "As shown in the figure: draw PH perpendicular to the directrix from point P, then |PH| = |PF|. From $\\overrightarrow{FM} = 3\\overrightarrow{FP}$, we have $\\frac{|MP|}{|MF|} = \\frac{2}{3}$. By similar triangles, $\\frac{|PH|}{P} = \\frac{|PH|}{4} = \\frac{|MP|}{|MF|} = \\frac{2}{3}$, so $|PH| = \\frac{8}{3}$, hence $|\\overrightarrow{FP}| = \\frac{8}{3}$."
  },
  {
    "text": "Through the left focus $F_{1}$ of the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$, draw a line $l$ intersecting the left branch of the hyperbola at points $P$ and $Q$. If $|PQ|=4$ and $F_{2}$ is the right focus of the hyperbola, then the perimeter of $\\triangle PF_{2}Q$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1:Point;F2: Point;Q: Point;Expression(G) = (x^2 - y^2/4 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(F1, l);Intersection(l, LeftPart(G)) = {P, Q};Abs(LineSegmentOf(P, Q)) = 4;l:Line",
    "query_expressions": "Perimeter(TriangleOf(P, F2, Q))",
    "answer_expressions": "12",
    "fact_spans": "[[[1, 29], [47, 50], [83, 86]], [[53, 56]], [[33, 40]], [[75, 82]], [[57, 60]], [[1, 29]], [[1, 40]], [[75, 90]], [[0, 46]], [[43, 62]], [[64, 72]], [[43, 46]]]",
    "query_spans": "[[[92, 117]]]",
    "process": "By the given condition, |PF_{2}|-|PF_{1}|=2, |QF_{2}|-|QF_{1}|=2. \\because |PF_{1}|+|QF_{1}|=|PQ|=4, \\therefore |PF_{2}|+|QF_{2}|-4=4, \\therefore |PF_{2}|+|QF_{2}|=8. \\therefore The perimeter of \\triangle PF_{2}Q is |PF_{2}|+|QF_{2}|+|PQ|=8+4=12."
  },
  {
    "text": "If the eccentricity of the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{m}=1$ is $2$, then the value of the real number $m$ is?",
    "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (x^2/2 - y^2/m = 1);Eccentricity(G) = 2",
    "query_expressions": "m",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 39]], [[49, 54]], [[1, 39]], [[1, 47]]]",
    "query_spans": "[[[49, 58]]]",
    "process": "Since the eccentricity of the hyperbola is 2, using the simple properties of hyperbolas, the value of the real number m can be determined from the given conditions. Because the eccentricity of the hyperbola \\frac{x^{2}}{2}-\\frac{y^{2}}{m}=1 is 2, \\frac{\\sqrt{2+m}}{\\sqrt{2}}=2, solving gives m=6."
  },
  {
    "text": "The coordinates of the foci of the ellipse $\\frac{x^{2}}{4+k}+\\frac{y^{2}}{3+k}=1$ are?",
    "fact_expressions": "G: Ellipse;k: Number;Expression(G) = (x^2/(k + 4) + y^2/(k + 3) = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(pm*1,0)",
    "fact_spans": "[[[0, 41]], [[2, 41]], [[0, 41]]]",
    "query_spans": "[[[0, 48]]]",
    "process": ""
  },
  {
    "text": "Given that the focus $F$ of the parabola $y^{2}=2 p x$ coincides with the right focus of the hyperbola $\\frac{x^{2}}{7}-\\frac{y^{2}}{9}=1$, the intersection point of the directrix of the parabola and the $x$-axis is $K$, and point $A$ lies on the parabola such that $|A K|=\\sqrt{2}|A F|$, then the area of $\\triangle A F K$ is?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 2*(p*x));p: Number;F: Point;Focus(H) = F;G: Hyperbola;Expression(G) = (x^2/7 - y^2/9 = 1);RightFocus(G) = F;Intersection(Directrix(H), xAxis) = K;K: Point;A: Point;PointOnCurve(A, H) = True;Abs(LineSegmentOf(A, K)) = sqrt(2)*Abs(LineSegmentOf(A, F))",
    "query_expressions": "Area(TriangleOf(A, F, K))",
    "answer_expressions": "32",
    "fact_spans": "[[[2, 18], [70, 73], [94, 97]], [[2, 18]], [[5, 18]], [[21, 24]], [[2, 24]], [[25, 63]], [[25, 63]], [[2, 69]], [[70, 88]], [[85, 88]], [[89, 93]], [[89, 98]], [[99, 120]]]",
    "query_spans": "[[[122, 144]]]",
    "process": "From the hyperbola $\\frac{x^{2}}{7}-\\frac{y^{2}}{9}=1$, the right focus is $(4,0)$, which is also the focus of the parabola $y^{2}=2px$. Therefore, $\\frac{p}{2}=4$, solving gives $p=8$. Hence, the equation of the parabola is $y^{2}=16x$. Its directrix is $x=-4$, so $K(-4,0)$. Draw $AM\\bot$ directrix from point $A$, with foot at $M$, then $|AM|=|AF|$. Therefore, $|AK|=\\sqrt{2}|AM|$, so $\\angle MAK=45^{\\circ}$. Thus, $|KF|=|AF|$. $S_{AAKF}=\\frac{1}{2}|KF|^{2}=\\frac{1}{3}\\times8^{2}=32$."
  },
  {
    "text": "If the line $y=k x-1$ always has common points with the hyperbola $x^{2}-y^{2}=4$, then the range of values for $k$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;k: Number;Expression(G) = (x^2 - y^2 = 4);Expression(H) = (y = k*x - 1);IsIntersect(H, G)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(pm*1)+[-sqrt(5)/2,]",
    "fact_spans": "[[[13, 31]], [[1, 12]], [[39, 42]], [[13, 31]], [[1, 12]], [[1, 37]]]",
    "query_spans": "[[[39, 48]]]",
    "process": "\\begin{cases}x^{2}-y^{2}=4\\\\y=kx-1\\end{cases},x^{2}-(kx-1)^{2}=4,(1-k^{2})x+2kx-5=0 when 1-k^{2}=0,k=\\pm1, obviously satisfies the condition; when 1-k^{2}\\neq0, then_{A}=20-16k^{2}=0,k=\\pm\\frac{\\sqrt{5}}{2}"
  },
  {
    "text": "The coordinates of the foci of the curve $x^{2}+\\frac{y^{2}}{5}=1$ are?",
    "fact_expressions": "G: Curve;Expression(G) = (x^2 + y^2/5 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, pm*2)",
    "fact_spans": "[[[0, 27]], [[0, 27]]]",
    "query_spans": "[[[0, 34]]]",
    "process": "c=\\sqrt{5-1}=2, and the foci are on the y-axis, so the coordinates of the foci are (0,\\pm2)"
  },
  {
    "text": "If the coordinates of point $A$ are $(3 , 2)$, $F$ is the focus of the parabola $y^{2}=2 x$, and point $P$ moves along this parabola, then what are the coordinates of point $P$ such that $PA+PF$ is minimized?",
    "fact_expressions": "G: Parabola;P: Point;A: Point;F: Point;Expression(G) = (y^2 = 2*x);Coordinate(A) = (3, 2);Focus(G) = F;PointOnCurve(P, G);WhenMin(LineSegmentOf(P, A) + LineSegmentOf(P, F))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2, 2)",
    "fact_spans": "[[[25, 39], [49, 52]], [[43, 47], [73, 77]], [[1, 5]], [[21, 24]], [[25, 39]], [[1, 18]], [[21, 42]], [[43, 53]], [[59, 71]]]",
    "query_spans": "[[[73, 82]]]",
    "process": ""
  },
  {
    "text": "The minimum distance between the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1(x \\geq 0, y \\geq 0)$ and the line $x-y-5=0$ is?",
    "fact_expressions": "G: Ellipse;H: Line;Expression(G) = And((x^2/9 + y^2/4 = 1), (x >= 0), (y >= 0));Expression(H) = (x - y - 5 = 0)",
    "query_expressions": "Min(Distance(G, H))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 57]], [[58, 69]], [[0, 57]], [[58, 69]]]",
    "query_spans": "[[[0, 78]]]",
    "process": "Draw the graph of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ $(x\\geqslant0,y\\geqslant0)$ and the line $x-y-5=0$ in the coordinate system as shown: it can be seen that the minimum distance from $(3,0)$ to the line $x-y+5=0$ is $d=\\frac{|3-5|}{\\sqrt{2}}=\\sqrt{2}$"
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ is $y= \\sqrt{3} x$, and one of its foci coincides with the focus of the parabola $y^{2}=32 x$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x);H: Parabola;Expression(H) = (y^2 = 32*x);OneOf(Focus(G)) = Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16 - y^2/48 = 1",
    "fact_spans": "[[[2, 59], [84, 85], [114, 117]], [[2, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[2, 83]], [[91, 106]], [[91, 106]], [[84, 111]]]",
    "query_spans": "[[[114, 122]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/4",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 45]]]",
    "process": "From the given information: a=4, b=3, and c=\\sqrt{a^{2}+b^{2}}=5, so the eccentricity e=\\frac{c}{a}=\\frac{5}{4}"
  },
  {
    "text": "The equation of the trajectory of the center of the moving circle $x^{2}+y^{2}-(4 m+2) x-2 m y+4 m^{2}+4 m+1=0$ is?",
    "fact_expressions": "G: Circle;Expression(G) = (4*m + 4*m^2 - 2*m*y - x*(4*m + 2) + x^2 + y^2 + 1 = 0);m: Number",
    "query_expressions": "LocusEquation(Center(G))",
    "answer_expressions": "(x-2*y-1=0)&(Negation(x=1))",
    "fact_spans": "[[[0, 47]], [[0, 47]], [[2, 47]]]",
    "query_spans": "[[[0, 57]]]",
    "process": "Rewriting the equation of the circle into standard form gives: $[x-(2m+1)]^{2}+(y-m)^{2}=m^{2}$ $(m\\neq0)$, then the coordinates of the center are $\\begin{cases}x=2m+1\\\\y=m\\end{cases}$, since $m\\neq2m+1$, and because $m\\neq0$, we obtain $x\\neq1$, eliminating $m$ yields $x-2y-1=0$."
  },
  {
    "text": "Given the parabola $y^{2}=4x$, a line passing through the point $P(4,0)$ intersects the parabola at two points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. What is the minimum value of $y_{1}^{2}+y_{2}^{2}$?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;A: Point;B: Point;x1:Number;x2:Number;y1:Number;y2:Number;Expression(G) = (y^2 = 4*x);Coordinate(P) = (4, 0);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);PointOnCurve(P, H);Intersection(H, G) = {A,B}",
    "query_expressions": "Min(y1^2 + y2^2)",
    "answer_expressions": "32",
    "fact_spans": "[[[2, 16], [32, 35]], [[29, 31]], [[19, 28]], [[38, 55]], [[58, 75]], [[38, 55]], [[58, 75]], [[38, 55]], [[58, 75]], [[2, 16]], [[19, 28]], [[38, 55]], [[58, 75]], [[18, 31]], [[29, 77]]]",
    "query_spans": "[[[80, 107]]]",
    "process": "Let the equation of line AB be x = my + 4. Substituting into the parabola equation and simplifying yields y^{2} - 4my - 16 = 0. Then y_{1} + y_{2} = 4m, y_{1}y_{2} = -16, and y_{1}^{2} + y_{2}^{2} = (y_{1} + y_{2})^{2} - 2y_{1}y_{2} = 16m^{2} + 32. Therefore, when m = 0, y_{1}^{2} + y_{2}^{2} attains its minimum value of 32."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) intersects the parabola at points $A$ and $B$, and intersects its directrix $l$ at point $C$. If $\\overrightarrow{C F}=2 \\overrightarrow{F A}$ and $A F=8$, then what is the length of segment $A B$?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;l: Line;A: Point;B: Point;C: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Directrix(G) = l;Intersection(H, l) = C;VectorOf(C, F) = 2*VectorOf(F, A);LineSegmentOf(A, F) = 8",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "12",
    "fact_spans": "[[[1, 22], [32, 35], [47, 48]], [[4, 22]], [[29, 31]], [[50, 53]], [[36, 39]], [[40, 43]], [[54, 58]], [[25, 28]], [[4, 22]], [[1, 22]], [[1, 28]], [[0, 31]], [[29, 45]], [[47, 53]], [[29, 58]], [[61, 106]], [[108, 115]]]",
    "query_spans": "[[[118, 129]]]",
    "process": "Draw AD\\bot l and BE\\bot l as shown. By the definition of a parabola, AD=AF=8, BE=BF. From \\overrightarrow{CF}=2\\overrightarrow{FA}, we have CF=16, CA=24. Let BF=x, then BE=x, CB=16-x. Therefore, \\frac{CB}{CA}=\\frac{BE}{AD}, that is, \\frac{16-x}{24}=\\frac{x}{8}. Solving gives x=4. Hence AB=AF+FB=12. The answer is: 12"
  },
  {
    "text": "The focal length of the ellipse $x^{2}+3 y^{2}=6$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 3*y^2 = 6)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 19]], [[0, 19]]]",
    "query_spans": "[[[0, 24]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y=ax^{2}$ ($a<0$) with focus $F$, a line $L$ passing through $F$ intersects the parabola at points $A$ and $B$. Then $\\frac{1}{|AF|}+\\frac{1}{|BF|}$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = a*x^2);a: Number;a<0;F: Point;Focus(G) = F;L: Line;PointOnCurve(F,L) = True;Intersection(L,G) = {A,B};A: Point;B: Point",
    "query_expressions": "1/Abs(LineSegmentOf(B, F)) + 1/Abs(LineSegmentOf(A, F))",
    "answer_expressions": "-4*a",
    "fact_spans": "[[[2, 20], [38, 41]], [[2, 20]], [[5, 20]], [[5, 20]], [[23, 26], [28, 31]], [[2, 26]], [[32, 37]], [[27, 37]], [[32, 53]], [[42, 45]], [[48, 51]]]",
    "query_spans": "[[[55, 88]]]",
    "process": ""
  },
  {
    "text": "Given that the distance from the focus to the directrix of the parabola $y^{2}=2 p x$ is $1$, then the shortest chord length among all chords passing through the focus of this parabola is?",
    "fact_expressions": "G: Parabola;p: Number;Expression(G) = (y^2 = 2*(p*x));Distance(Focus(G),Directrix(G))=1;H:LineSegment;PointOnCurve(Focus(G),H);IsChordOf(H,G)",
    "query_expressions": "Min(Length(H))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 18], [34, 37]], [[5, 18]], [[2, 18]], [[2, 31]], [], [[34, 46]], [[34, 46]]]",
    "query_spans": "[[[34, 54]]]",
    "process": "\\because the distance from the focus to the directrix of the parabola \\( y^{2} = 2px \\) is 1, \\( \\therefore p = 1 \\). Let the intersection points of the line and the parabola be \\( A(x_{1}, y_{1}) \\), \\( B(x_{2}, y_{2}) \\). When the slope of the line does not exist, the equation of the line is \\( x = \\frac{1}{2} \\), and the intersection points are \\( A\\left(\\frac{1}{2}, 1\\right) \\), \\( B\\left(\\frac{1}{2}, -1\\right) \\), so the chord length is \\( AB = 2 \\). When the slope exists, it can be set as \\( y = k\\left(x - \\frac{1}{2}\\right) \\) (\\( k \\neq 0 \\)). Solving simultaneously \n\\[\n\\begin{cases}\ny^{2} = 2x \\\\\ny = k\\left(x - \\frac{1}{2}\\right)\n\\end{cases}\n\\]\nsimplifies to \\( k^{2}x^{2} - (k^{2} + 2)x + \\frac{k^{2}}{4} = 0 \\), \\( \\therefore |AB| = x_{1} + x_{2} + 1 = 2 + \\frac{2}{k^{2}} > 2 \\). Hence, among all chords of this parabola passing through the focus, the shortest chord length is 2."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{a^{2}+5}+\\frac{y^{2}}{2 a^{2}+1}=1$ represents an ellipse with foci on the $x$-axis, then the range of real values for $a$ is?",
    "fact_expressions": "G: Ellipse;a: Real;Expression(G) = (x^2/(a^2 + 5) + y^2/(2*a^2 + 1) = 1);PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Range(a)",
    "answer_expressions": "(-2, 2)",
    "fact_spans": "[[[63, 65]], [[67, 72]], [[1, 65]], [[54, 65]]]",
    "query_spans": "[[[67, 79]]]",
    "process": "From the given condition, we have $a^{2}+5>2a^{2}+1$, which implies $a^{2}<4$, so $-2<a<2$, meaning the range of $a$ is $(-2,2)$."
  },
  {
    "text": "What is the length of the real axis of the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{4}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2/4 = 1)",
    "query_expressions": "Length(RealAxis(G))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 44]]]",
    "process": "Since in the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{4}=1$, $a^{2}=3$, so $a=\\sqrt{3}$, therefore the length of the real axis is $2a=2\\sqrt{3}$."
  },
  {
    "text": "The right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ is $F(3,0)$, and the distance from point $F$ to an asymptote of the hyperbola $C$ is $1$. Then the standard equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;RightFocus(C) = F;F: Point;Coordinate(F) = (3, 0);Distance(F,OneOf(Asymptote(C))) = 1",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/8-y^2=1",
    "fact_spans": "[[[0, 64], [84, 90], [105, 111]], [[0, 64]], [[8, 64]], [[8, 64]], [[8, 64]], [[8, 64]], [[0, 77]], [[69, 77], [79, 83]], [[69, 77]], [[79, 103]]]",
    "query_spans": "[[[105, 118]]]",
    "process": "Since the right focus of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) is $ F(3,0) $, we have $ c = 3 $. Also, since the distance from point $ F $ to one asymptote $ bx - ay = 0 $ of the hyperbola $ C $ is 1, we have $ \\frac{|bc|}{\\sqrt{b^{2} + a^{2}}} = b = 1 $. Thus, $ a^{2} = c^{2} - b^{2} = 8 $. Therefore, the standard equation of the hyperbola $ C $ is $ \\frac{x^{2}}{8} - y^{2} = 1 $."
  },
  {
    "text": "A point $P$ on the ellipse $\\frac{x^{2}}{100}+\\frac{y^{2}}{64}=1$ has a distance of $7$ to the left focus of the ellipse. Then, what is the distance from point $P$ to the right focus?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/100 + y^2/64 = 1);PointOnCurve(P, G);Distance(P, LeftFocus(G)) = 7",
    "query_expressions": "Distance(P,RightFocus(G))",
    "answer_expressions": "13",
    "fact_spans": "[[[0, 40], [47, 49]], [[43, 46], [61, 65]], [[0, 40]], [[0, 46]], [[43, 59]]]",
    "query_spans": "[[[47, 74]]]",
    "process": "According to the definition of an ellipse, |PF_{1}| + |PF_{2}| = 2a \\Leftrightarrow 7 + |PF_{2}| = 20, solving gives: |PF_{2}| = 20 - 7 = 13, therefore fill in: 13."
  },
  {
    "text": "Given that the focal distance of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is equal to the length of the chord passing through a focus and perpendicular to the major axis, then the eccentricity of this ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: LineSegment;PointOnCurve(Focus(G),H);a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);IsPerpendicular(MajorAxis(G), H);IsChordOf(H, G) ;FocalLength(G) = Length(H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(\\sqrt{5} - 1)/2",
    "fact_spans": "[[[2, 54], [59, 60], [75, 77]], [[4, 54]], [[4, 54]], [], [[59, 71]], [[4, 54]], [[4, 54]], [[2, 54]], [[59, 71]], [[59, 71]], [[2, 72]]]",
    "query_spans": "[[[75, 83]]]",
    "process": "Draw the graph. Let AB be the chord passing through the right focus $ F_{2} $ of the ellipse and perpendicular to the major axis. Calculate $ |AF_{1}| $, then use the definition of the ellipse to obtain an equation in terms of $ a $ and $ c $, from which the eccentricity of the ellipse can be found. As shown in the figure below, let the ellipse $ \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 $ ($ a>b>0 $) have left and right foci $ F_{1} $ and $ F_{2} $, respectively. Let AB be the chord passing through the right focus $ F_{2} $ of the ellipse and perpendicular to the major axis. Then $ |AB|=2c $, $ |AF_{2}|=\\frac{1}{2}|AB|=c $. By the Pythagorean theorem, we get $ |AF_{1}|=\\sqrt{|AF_{2}|^{2}+|F_{1}F_{2}|^{2}}=\\sqrt{5}c $. From the definition of the ellipse, $ |AF_{1}|+|AF_{2}|=2a $, that is, $ \\sqrt{5}c+c=2a $. Therefore, the eccentricity of the ellipse is $ e=\\frac{c}{a}=\\frac{2}{\\sqrt{5}+1}=\\frac{2(\\sqrt{5}-1)}{(\\sqrt{5}+1)(\\sqrt{5}} $"
  },
  {
    "text": "Given the circle $C$: $x^{2}+y^{2}-16 y+48=0$ is tangent to the asymptotes of the hyperbola $E$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1(a>0, b>0)$, then the eccentricity of $E$ is?",
    "fact_expressions": "E: Hyperbola;a: Number;b: Number;C: Circle;a>0;b>0;Expression(E) = (-x^2/b^2 + y^2/a^2 = 1);Expression(C) = (-16*y + x^2 + y^2 + 48 = 0);IsTangent(C, Asymptote(E))",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[32, 93], [101, 104]], [[40, 93]], [[40, 93]], [[2, 31]], [[40, 93]], [[40, 93]], [[32, 93]], [[2, 31]], [[2, 99]]]",
    "query_spans": "[[[101, 110]]]",
    "process": "First, find the asymptotes of the hyperbola, then use the distance from the center of the circle to the asymptote equal to the radius to obtain the relationship among a, b, and c, where c^{2}=a^{2}+b^{2}, and thus find the eccentricity. From x^{2}+y^{2}-16y+48=0 we get x^{2}+(y-8)^{2}=4^{2}, so the center of the circle is C(0,8), and the radius r=4. One asymptote of the hyperbola E:\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1(a>0,b>0) is ax-by=0. According to the problem, the distance from the center to the asymptote is d=\\frac{|-8b|}{\\sqrt{a^{2}+b^{2}}}=\\frac{8b}{c}=4, so b=\\frac{1}{2}c, hence a=\\sqrt{c^{2}-b^{2}}=\\frac{\\sqrt{3}}{2}c, so e=\\frac{c}{a}=\\frac{2\\sqrt{3}}{3}."
  },
  {
    "text": "If an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is parallel to the line $\\sqrt{3} x-y+\\sqrt{3}=0$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (sqrt(3)*x - y + sqrt(3) = 0);IsParallel(OneOf(Asymptote(G)),H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 58], [96, 99]], [[5, 58]], [[5, 58]], [[65, 92]], [[5, 58]], [[5, 58]], [[2, 58]], [[65, 92]], [[2, 94]]]",
    "query_spans": "[[[96, 105]]]",
    "process": "\\because the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0,b>0) has an asymptote parallel to the line \\sqrt{3}x-y+\\sqrt{3}=0, \\frac{b}{a}=\\sqrt{3}, \\therefore the eccentricity e=\\frac{c}{a}=2."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and $P$ is a point on the asymptote. If $|P F_{1}|=2|P F_{2}|$ and $\\angle F_{1} P F_{2}=60^{\\circ}$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, Asymptote(G));Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2));AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(21)/3",
    "fact_spans": "[[[20, 76], [155, 158]], [[23, 76]], [[23, 76]], [[83, 86]], [[2, 9]], [[10, 17]], [[23, 76]], [[23, 76]], [[20, 76]], [[2, 82]], [[2, 82]], [[20, 93]], [[95, 117]], [[119, 152]]]",
    "query_spans": "[[[155, 164]]]",
    "process": "First, use the law of cosines in $\\triangle PF_{1}F_{2}$ to find $|F_{1}F_{2}|$, then use the Pythagorean theorem to determine that $\\triangle PF_{1}F_{2}$ is a right triangle, then use the right triangle to find the value of $\\frac{b}{a}$, and finally use $\\frac{c}{a}=\\sqrt{1+(\\frac{b}{a})^{2}}$ to solve. [Detailed solution] The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ are given by $y=\\pm\\frac{b}{a}x$. Let $|PF_{1}|=2|PF_{2}|=2x>0$. In $\\triangle PF_{1}F_{2}$, since $\\angle F_{1}PF_{2}=60^{\\circ}$, we have $|F_{1}F_{2}|^{2}=x^{2}+4x^{2}-2x\\times2x\\cos60^{\\circ}=3x^{2}$, so $|F_{1}F_{2}|=\\sqrt{3}x$, and $\\triangle PF_{1}F_{2}$ is a right triangle. Therefore, in $\\triangle OPF_{2}$, $PF_{2}\\bot OF_{2}$, $|PF_{2}|=x$, $|OF_{2}|=\\frac{\\sqrt{3}}{2}x$, so $\\frac{b}{a}=\\frac{x}{\\frac{\\sqrt{3}}{3}}=\\frac{2\\sqrt{3}}{3}$. Then the eccentricity of the hyperbola is $e=\\frac{c}{a}=\\sqrt{1+(\\frac{b}{a})^{2}}=\\sqrt{1+\\frac{4}{3}}=\\frac{\\sqrt{21}}{3}$"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Let $P$ be a point on the ellipse in the second quadrant. Connect $P F_{2}$ intersecting the $y$-axis at point $N$. If $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, $|F_{1} F_{2}|=4|O N|$, where $O$ is the origin, then the eccentricity of the ellipse is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;F2: Point;F1: Point;O: Origin;N: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) =F1;RightFocus(C)=F2;PointOnCurve(P,C);Quadrant(P)=2;Intersection(LineSegmentOf(P, F2), yAxis) = N;DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Abs(LineSegmentOf(F1, F2)) = 4*Abs(LineSegmentOf(O, N))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[2, 59], [93, 95], [220, 222]], [[8, 59]], [[8, 59]], [[84, 87]], [[76, 83]], [[68, 75]], [[209, 212]], [[117, 121]], [[8, 59]], [[8, 59]], [[2, 59]], [[2, 83]], [[2, 83]], [[84, 99]], [[84, 99]], [[102, 121]], [[123, 182]], [[184, 206]]]",
    "query_spans": "[[[220, 228]]]",
    "process": "Since $\\overrightarrow{PF}_{1}\\cdot\\overrightarrow{PF}_{2}=0$, it follows that $\\angle F_{1}PF_{2}=90^{\\circ}$. From the given condition, $\\triangle F_{2}ON\\sim\\triangle F_{2}PF_{1}$, then $\\frac{|PF_{1}|}{|PF_{2}|}=\\frac{|ON|}{|OF_{2}|}$. Since $|F_{1}F_{2}|=4|ON|$, we have $\\frac{|ON|}{|OF_{2}|}=\\frac{1}{2}$, so $\\frac{|PF_{1}|}{|PF_{2}|}=\\frac{|ON|}{|OF_{2}|}=\\frac{1}{2}$. Since $|PF_{1}|+|PF_{2}|=2a$, it follows that $|PF_{1}|=\\frac{2a}{3}$, $|PF_{2}|=\\frac{4a}{3}$. Therefore, $|PF_{1}|^{2}+|PF_{2}|^{2}=4c^{2}$, which gives $\\frac{4a^{2}}{9}+\\frac{16a^{2}}{9}=4c^{2}$, solving yields $e=\\frac{c}{a}=\\frac{\\sqrt{5}}{3}$."
  },
  {
    "text": "Let a point $P(2, m)$ on the parabola $y^{2}=2 p x(p>0)$ be such that its distance to the $y$-axis is half of its distance to the focus. Then the standard equation of the parabola is?",
    "fact_expressions": "G: Parabola;p: Number;P: Point;m:Number;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(P) = (2, m);PointOnCurve(P,G);Distance(P,yAxis)=Distance(P,Focus(G))/2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=8*x",
    "fact_spans": "[[[1, 22], [53, 56]], [[4, 22]], [[25, 34]], [[25, 34]], [[4, 22]], [[1, 22]], [[25, 34]], [[1, 34]], [[1, 51]]]",
    "query_spans": "[[[53, 63]]]",
    "process": "From the given condition, we have $2 + \\frac{p}{2} = 2 \\times 2$, solving for $p$ gives $p = 4$, hence the standard equation of the parabola is $y^2 = 8x$."
  },
  {
    "text": "Given that $F$ is the right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $O$ is the coordinate origin, and the line passing through $F$ intersects the two asymptotes of $C$ at points $M$ and $N$, respectively. If $\\overrightarrow{O M} \\cdot \\overrightarrow{M F}=0$ and $|M N|=b$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;O: Origin;M: Point;F: Point;N: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;PointOnCurve(F,G);L1:Line;L2:Line;Asymptote(C)={L1,L2};Intersection(G,L1)=M;Intersection(G,L2)=N;DotProduct(VectorOf(O, M), VectorOf(M, F)) = 0;Abs(LineSegmentOf(M, N)) = b",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[6, 67], [89, 92], [176, 179]], [[14, 67]], [[14, 67]], [[86, 88]], [[72, 75]], [[104, 107]], [[82, 85], [2, 5]], [[108, 111]], [[14, 67]], [[14, 67]], [[6, 67]], [[2, 71]], [[81, 88]], [], [], [[89, 98]], [[86, 111]], [[86, 111]], [[113, 164]], [[165, 174]]]",
    "query_spans": "[[[176, 185]]]",
    "process": "First, from $\\overrightarrow{OM}\\cdot\\overrightarrow{MF}=0$, it follows that $OM\\bot MF$, and $|MF|=b$ can be calculated. Combining with $|MN|=b$, it follows that $\\triangle OFN$ is an isosceles triangle, and $|ON|=c$. Then, using the slope of the asymptote, the coordinates of point $N$ can be calculated, thus obtaining the coordinates of point $M$. Using $|OM|=a$ and combining with $b^{2}=c^{2}-a^{2}$, the relationship between $a$ and $c$ can be obtained, allowing the solution to be found. Since $\\overrightarrow{OM}\\cdot\\overrightarrow{MF}=0$, we have $\\overrightarrow{OM}\\bot\\overrightarrow{MF}$, i.e., $OM\\bot MF$. Therefore, $|MF|$ is the distance from point $F(c,0)$ to the asymptote $bx-ay=0$:  \n$$\n|MF|=\\frac{bc}{\\sqrt{b^{2}+a^{2}}}=\\frac{bc}{c}=b\n$$\nThus, $|MF|=|MN|=b$, so point $M$ is the midpoint of $NF$. Also, since $OM\\bot MF$, we have $|ON|=|OF|=c$, so $|OM|^{2}=c^{2}-b^{2}=a^{2}$. Let the left focus of the hyperbola be $F_{1}$, $\\angle F_{1}ON=\\theta$, $N(x,y)$. Then  \n$$\n\\tan\\theta=\\tan(\\pi-\\angle FON)=-\\tan\\angle FON=\\frac{b}{a}\n$$\nSince $c^{2}=a^{2}+b^{2}$, we have  \n$$\n\\cos\\theta=\\frac{a}{c},\\quad \\sin\\theta=\\frac{b}{c}\n$$\nThus,  \n$$\nx=-|ON|\\cos\\theta=-c\\cdot\\frac{a}{c}=-a,\\quad y=|ON|\\sin\\theta=c\\cdot\\frac{b}{c}=b\n$$\nSo $N(-a,b)$. Since $M$ is the midpoint of $NF$,  \n$$\nM\\left(\\frac{c-a}{2},\\frac{b}{2}\\right)\n$$\nThen  \n$$\n|OM|^{2}=\\left(\\frac{c-a}{2}\\right)^{2}+\\left(\\frac{b}{2}\\right)^{2}=a^{2}\n$$\nSubstituting $b^{2}=c^{2}-a^{2}$ and simplifying gives:  \n$$\n(c-a)^{2}+c^{2}-a^{2}=4a^{2}\n$$\ni.e.,  \n$$\n2c^{2}-2ac-4a^{2}=0\n$$\nThus,  \n$$\ne^{2}-e-2=0\n$$\nwhich factors as $(e-2)(e+1)=0$, solving gives $e=2$ or $e=-1$ (discarded),"
  },
  {
    "text": "Given that $F$ is the right focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the line $y=\\frac{b}{a} x$ intersects the ellipse at points $A$ and $B$, and if $\\cos \\angle A F B = \\frac{1}{3}$, then the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "F: Point;C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;RightFocus(C) = F;G: Line;Expression(G) = (y = x*b/a);Intersection(G, C) = {A, B};A: Point;B: Point;Cos(AngleOf(A, F, B)) = 1/3",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(5)/5",
    "fact_spans": "[[[2, 5]], [[6, 58], [83, 85], [130, 135]], [[6, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[2, 62]], [[63, 82]], [[63, 82]], [[63, 95]], [[86, 89]], [[90, 93]], [[97, 128]]]",
    "query_spans": "[[[130, 141]]]",
    "process": ""
  },
  {
    "text": "The equation of the circle centered at the right focus of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ and tangent to its asymptotes is?",
    "fact_expressions": "G: Hyperbola;H:Circle;Expression(G) = (x^2/4 - y^2 = 1);RightFocus(G)=Center(H);IsTangent(Asymptote(G),H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x-\\sqrt{5})^2+y^2=1",
    "fact_spans": "[[[1, 29], [39, 40]], [[46, 47]], [[1, 29]], [[0, 47]], [[38, 47]]]",
    "query_spans": "[[[46, 52]]]",
    "process": "According to the hyperbola equation, the right focus and one of the asymptotes are obtained. Using the point-to-line distance formula, the radius of the circle is found, and then the equation of the circle is determined. From the given condition, the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, we have $a=2$, $b=1$, then $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{5}$. Thus, the right focus of the hyperbola is $F(\\sqrt{5},0)$, and one of its asymptotes is $y=-\\frac{1}{2}x$, or $x+2y=0$. Therefore, the distance from point $F$ to the asymptote $x+2y=0$ is $d=\\frac{|\\sqrt{5}\\times1-2\\times0|}{\\sqrt{1^{2}+2^{2}}}=1$. Hence, the equation of the circle with the right focus of the hyperbola as its center and tangent to its asymptote is $(x-\\sqrt{5})^{2}+y^{2}=1$."
  },
  {
    "text": "The line with slope $1$ passing through the focus of the parabola $y=\\frac{1}{4} x^{2}$ is intercepted by the parabola, forming a chord of length?",
    "fact_expressions": "G: Parabola;H: Line;Expression(G) = (y = x^2/4);Slope(H)=1;PointOnCurve(Focus(G),H)",
    "query_expressions": "Length(InterceptChord(H,G))",
    "answer_expressions": "8",
    "fact_spans": "[[[9, 33], [40, 43]], [[37, 39]], [[9, 33]], [[0, 39]], [[8, 39]]]",
    "query_spans": "[[[37, 50]]]",
    "process": "First, transform the parabola $ y = \\frac{1}{4}x^{2} $ into its standard form and find the focus coordinates. Given the slope is 1, obtain the line equation $ y = x + 1 $, then solve it together with the parabola equation using the chord length formula. [Detailed solution] From the parabola $ y = \\frac{1}{4}x^{2} $, we get $ x^{2} = 4y $, so $ p = 2 $, thus the focus coordinates are $ (0, 1) $. Since the slope is 1, the line passing through the focus has the equation $ y = x + 1 $. From \n$$\n\\begin{cases}\ny = x + 1, \\\\\nx^{2} = 4y,\n\\end{cases}\n$$\neliminating $ x $, we obtain $ y^{2} - 6y + 1 = 0 $. Let the intersection points of this line with the parabola be $ A $ and $ B $, with coordinates $ (x_{1}, y_{1}) $ and $ (x_{2}, y_{2}) $, respectively. Then $ y_{1} + y_{2} = 6 $, so the chord length intercepted by the parabola is $ y_{1} + \\frac{p}{2} + y_{2} + \\frac{p}{2} = y_{1} + y_{2} + p = 6 + 2 = 8 $."
  },
  {
    "text": "Given that point $F$ is the left focus of the ellipse $\\Gamma$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, point $P$ is any point on the ellipse $\\Gamma$, and point $O$ is the origin, then the maximum value of $\\overrightarrow{O P} \\cdot \\overrightarrow{F P}$ is?",
    "fact_expressions": "Gamma: Ellipse;O: Origin;P: Point;F: Point;Expression(Gamma) = (x^2/4 + y^2/3 = 1);LeftFocus(Gamma) = F;PointOnCurve(P, Gamma)",
    "query_expressions": "Max(DotProduct(VectorOf(O, P), VectorOf(F, P)))",
    "answer_expressions": "6",
    "fact_spans": "[[[7, 54], [64, 74]], [[80, 84]], [[59, 63]], [[2, 6]], [[7, 54]], [[2, 58]], [[59, 79]]]",
    "query_spans": "[[[91, 146]]]",
    "process": "Let the coordinates of point P be (x,y), then -2\\leqslant x\\leqslant 2, it follows that y^{2}=3-\\frac{3}{4}x^{2}. Using the coordinate operation of the dot product of planar vectors combined with the basic properties of quadratic functions, the maximum value of \\overrightarrow{OP}\\cdot\\overrightarrow{FP} can be found. Let the coordinates of point P be (x,y), then -2\\leqslant x\\leqslant 2, and \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1, yielding y^{2}=3-\\frac{3}{4}x^{2}. The left focus of ellipse \\Gamma is F(-1,0), \\overrightarrow{OP}=(x,y), \\overrightarrow{FP}=(x+1,y), then \\overrightarrow{OP}\\cdot\\overrightarrow{FP}=x(x+1)+y^{2}=x^{2}+x+3-\\frac{3}{4}x^{2}=\\frac{1}{4}x^{2}+x+3=\\frac{1}{4}(x+2)^{2}+2. The quadratic function f(x)=\\frac{1}{4}(x+2)^{2}+2 is monotonically increasing on the interval [-2,2]. Therefore, f(x)_{\\max}=f(2)=\\frac{1}{4}\\times 4^{2}+2=6. Hence, the maximum value of \\overrightarrow{OP}\\cdot\\overrightarrow{FP} is 6. The answer is: 6. This problem examines the solution of the maximum value of the dot product of vectors in an ellipse, involving the boundedness of the ellipse and the application of basic properties of quadratic functions, testing computational ability."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ with focus $F$ and directrix $l$, draw a line through point $Q(-\\frac{p}{2}, 0)$ intersecting $C$ at points $A$ and $B$. From $A$ and $B$, draw perpendiculars to $l$, intersecting $l$ at points $D$ and $E$ respectively. Let the slopes of $A E$ and $B D$ be $k_{1}$ and $k_{2}$, then the minimum value of $|\\frac{1}{k_{1}}-\\frac{1}{k_{2}}|$ is?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;A: Point;E: Point;B: Point;D: Point;Q: Point;F: Point;l: Line;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(Q) = (-p/2, 0);Focus(C) = F;Directrix(C) = l;PointOnCurve(Q, G);Intersection(G, C) = {A, B};L1:Line;L2:Line;PointOnCurve(A, L1);PointOnCurve(B,L2);Intersection(l,L1)=D;Intersection(l, L2)=E;Slope(LineSegmentOf(A, E))=k1;Slope(LineSegmentOf(B, D)) = k2;k1:Number;k2:Number;IsPerpendicular(l,L1);IsPerpendicular(l,L2)",
    "query_expressions": "Min(Abs(-1/k2 + 1/k1))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 28], [69, 72]], [[10, 28]], [[66, 68]], [[73, 76], [84, 87]], [[109, 112]], [[77, 80], [88, 91]], [[105, 108]], [[44, 65]], [[32, 35]], [[39, 42], [94, 97], [101, 104]], [[10, 28]], [[2, 28]], [[44, 65]], [[2, 35]], [[2, 42]], [[43, 68]], [[66, 82]], [], [], [[83, 100]], [[83, 100]], [[83, 114]], [[83, 114]], [[116, 152]], [[116, 152]], [[135, 143]], [[145, 152]], [[83, 100]], [[83, 100]]]",
    "query_spans": "[[[154, 195]]]",
    "process": "From the given, let $ l: x = ky - \\frac{p}{2} $. Substituting into $ y^{2} = 2px $, we get: $ y^{2} - 2pky + p^{2} = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ y_{1} + y_{2} = 2pk $, $ y_{1}y_{2} = p^{2} $. From $ \\Delta = 4p^{2}k^{2} - 4p^{2} > 0 $, we obtain $ k^{2} > 1 $. $ \\left| \\frac{1}{k_{1}} - \\frac{1}{k_{2}} \\right| = \\left| \\frac{x_{1} + x_{2} + p}{y_{2} - y_{1}} \\right| = \\frac{k^{2}}{\\sqrt{k^{2} - 1}} $. Let $ \\sqrt{k^{2} - 1} = t > 0 $, then $ \\left| \\frac{1}{k_{1}} - \\frac{1}{k_{2}} \\right| = \\frac{k^{2}}{\\sqrt{k^{2} - 1}} = \\frac{t^{2} + 1}{t} = t + \\frac{1}{t} \\geqslant 2 $. The minimum value $ 2 $ is attained if and only if $ t = 1 $, i.e., $ k = \\pm\\sqrt{2} $."
  },
  {
    "text": "Let $P$ be a moving point on the parabola $y^{2}=4 x$, and let $F$ be the focus of the parabola $y^{2}=4 x$. If $B(3 , 2)$, then the minimum value of $| P B|+| P F|$ is?",
    "fact_expressions": "G: Parabola;B: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(B) = (3, 2);PointOnCurve(P, G);Focus(G) = F",
    "query_expressions": "Min(Abs(LineSegmentOf(P, B)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "4",
    "fact_spans": "[[[5, 19], [30, 44]], [[49, 59]], [[1, 4]], [[26, 29]], [[5, 19], [30, 44]], [[49, 59]], [[1, 25]], [[26, 47]]]",
    "query_spans": "[[[61, 82]]]",
    "process": "Draw BQ perpendicular to the directrix from point B, intersecting the parabola at point $ P_{1} $, as shown in the figure. Then $ |P_{1}Q| = |P_{1}F| $. Thus, $ |PB| + |PF| \\geqslant |P_{1}B| + |P_{1}Q| = |BQ| = 4 $, that is, the minimum value of $ |PB| + |PF| $ is 4."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has eccentricity $\\frac{2 \\sqrt{3}}{3}$, focal length $2 c$, and $2 a^{2}=3 c$. A point $P$ on the hyperbola satisfies $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=2$ ($F_{1}$, $F_{2}$ are the left and right foci), then $|\\overrightarrow{PF_{1}}| \\cdot |\\overrightarrow{P F_{2}}|$=?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;c: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = (2*sqrt(3))/3;FocalLength(G) = 2*c;2*a^2 = 3*c;PointOnCurve(P, G);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 2;LeftFocus(G) = F1;RightFocus(G) = F2",
    "query_expressions": "Abs(VectorOf(P, F1))*Abs(VectorOf(P, F2))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 58], [110, 113]], [[5, 58]], [[5, 58]], [[89, 94]], [[116, 119]], [[181, 188]], [[191, 198]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 85]], [[2, 94]], [[96, 109]], [[110, 119]], [[121, 180]], [[110, 204]], [[110, 204]]]",
    "query_spans": "[[[207, 269]]]",
    "process": ""
  },
  {
    "text": "Given that the focus of a parabola is the intersection point of the line $l$: $y = -x - t$ ($t > 0$) and the $x$-axis, if the parabola intersects the line $l$ at two points $A$ and $B$, and $|AB| = 2\\sqrt{6}$, then $t = $?",
    "fact_expressions": "l: Line;G: Parabola;t: Number;A: Point;B: Point;t>0;Expression(l) = (y = -t - x);Intersection(l, xAxis) =Focus(G);Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, B)) = 2*sqrt(6)",
    "query_expressions": "t",
    "answer_expressions": "sqrt(6)/4",
    "fact_spans": "[[[11, 31], [45, 50]], [[4, 7], [41, 44]], [[82, 85]], [[53, 56]], [[57, 60]], [[17, 31]], [[11, 31]], [[4, 39]], [[41, 60]], [[62, 80]]]",
    "query_spans": "[[[82, 87]]]",
    "process": "The intersection point of the line and the x-axis is (-t, 0). Let the equation of the parabola be y^{2} = -4tx, and the equation of the line be x = t. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}). Solving the system of equations of the line and the parabola yields: x^{2} + 6tx + t^{2} = 0. Then: |AB| = 2t - (x_{1} + x_{2}) = 8t = 2\\sqrt{6}, \\sqrt{6}"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, point $A$ lies on the parabola $C$ and satisfies $|AF|=3$. What is the length of the chord intercepted on the $y$-axis by the circle centered at point $A$ with radius $AF$?",
    "fact_expressions": "C: Parabola;G: Circle;A: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(A, C);Abs(LineSegmentOf(A, F)) = 3;Center(G)=A;Radius(G)=LineSegmentOf(A,F)",
    "query_expressions": "Length(InterceptChord(yAxis, G))",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[2, 20], [33, 39]], [[73, 74]], [[28, 32], [56, 60]], [[24, 27]], [[2, 20]], [[2, 27]], [[28, 40]], [[44, 53]], [[55, 74]], [[64, 74]]]",
    "query_spans": "[[[73, 85]]]",
    "process": "By the given condition, the parabola $ y^{2} = 4x $ has focus $ F(1,0) $. Let $ A(x_{0},y_{0}) $. According to the definition of a parabola, $ |AF| = x_{0} + 1 = 3 $, solving gives $ x_{0} = 2 $, so the distance from $ A $ to the $ y $-axis is $ d = 2 $. Therefore, the chord length intercepted by the circle on the $ y $-axis is $ 2\\sqrt{R^{2}-d^{2}} = 2\\sqrt{3^{2}-2^{2}} = 2\\sqrt{5} $."
  },
  {
    "text": "It is known that one focus of a hyperbola coincides with the focus $F$ of the parabola $y^{2}=8x$, and the directrix of the parabola intersects the hyperbola at points $A$ and $B$. Given that the area of $\\triangle OAB$ is $6$ ($O$ being the origin), find the standard equation of the hyperbola.",
    "fact_expressions": "G: Hyperbola;H: Parabola;O: Origin;A: Point;B: Point;Expression(H) = (y^2 = 8*x);Focus(H)=F;OneOf(Focus(G))=F;Intersection(Directrix(H), G) = {A, B};Area(TriangleOf(O,A,B))=6;F:Point",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[2, 5], [41, 44], [91, 94]], [[11, 25], [34, 37]], [[82, 85]], [[46, 49]], [[50, 53]], [[11, 25]], [[11, 31]], [[2, 33]], [[34, 55]], [[57, 81]], [[28, 31]]]",
    "query_spans": "[[[91, 101]]]",
    "process": "\\because y^{2}=8x,\\therefore\\frac{p}{2}=2, i.e., the focus of y2=8x is (2,0), i.e., the focus of \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 is (2,0), \\therefore a^{2}+b^{2}=4,\\textcircled{1} also \\because the area of \\triangle OAB is 6, when x=-c, y=\\pm\\frac{b^{2}}{a}, \\therefore A(-c,\\frac{b^{2}}{a}), B(-c, S_{\\Delta AOB}=\\frac{1}{2}\\times2\\times\\frac{2b^{2}}{a}=6, we get b^{2}=3a,\\textcircled{2} from \\textcircled{1}\\textcircled{2}, \\begin{cases}a2=1\\\\b^{2}=3\\end{cases}, the equation of the hyperbola is x^{2}-\\frac{y^{2}}{2}=1"
  },
  {
    "text": "Given the parabola $x^{2}=4 y$ with focus $F$ and directrix $l$, let $P$ be a point on the parabola, and draw $P A \\perp l$ intersecting at point $A$. When $\\angle A F O=30^{\\circ}$ ($O$ is the origin), find $P F=$?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;A: Point;O: Origin;l: Line;Expression(G) = (x^2 = 4*y);Focus(G) = F;Directrix(G)=l;PointOnCurve(P,G);PointOnCurve(P,LineSegmentOf(P,A));IsPerpendicular(LineSegmentOf(P,A),l);FootPoint(LineSegmentOf(P,A),l) = A;AngleOf(A, F, O) = ApplyUnit(30, degree)",
    "query_expressions": "LineSegmentOf(P, F)",
    "answer_expressions": "4/3",
    "fact_spans": "[[[2, 16], [37, 40]], [[33, 36], [45, 48]], [[20, 23]], [[63, 67]], [[95, 98]], [[27, 31]], [[2, 16]], [[2, 23]], [[2, 30]], [[33, 43]], [[44, 62]], [[49, 62]], [[49, 67]], [[69, 94]]]",
    "query_spans": "[[[106, 113]]]",
    "process": "From the parabola equation $x^{2}=4y$, we can obtain the distance from the focus to the directrix. Then, given $\\angle AFO=30^{\\circ}$, we can find the x-coordinate of point $P$. Substituting into the parabola equation gives the y-coordinate of point $P$, and then by the definition of the parabola, $PF$ can be found. As shown in the figure, let $l$ intersect the y-axis at point $B$. In right triangle $\\triangle ABF$, $\\angle AFB=30^{\\circ}$, $BF=2$, so $AB=\\frac{2\\sqrt{3}}{3}$. If $P(x_{0},y_{0})$, then $x_{0}=\\frac{2\\sqrt{3}}{3}$. Substituting into $x^{2}=4y$, we get $y_{0}=\\frac{1}{3}$, so $PF=PA=y_{0}+1=\\frac{4}{3}$."
  },
  {
    "text": "A hyperbola has an asymptote with equation $x+\\sqrt{3} y=0$, and one of its foci coincides with the focus of the parabola given by $y^{2}=16 x$. What is the standard equation of this hyperbola?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(OneOf(Asymptote(G)))=(x + sqrt(3)*y = 0);Expression(H)=(y^2=16*x);OneOf(Focus(G))=Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/12 - y^2/4 = 1",
    "fact_spans": "[[[26, 29], [63, 66], [30, 31]], [[53, 56]], [[0, 29]], [[37, 56]], [[30, 61]]]",
    "query_spans": "[[[63, 73]]]",
    "process": ""
  },
  {
    "text": "A point $M$ on the parabola $y^{2}=12 x$ is at a distance of $9$ from the focus of the parabola. Then, what is the distance from point $M$ to the $x$-axis?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = 12*x);PointOnCurve(M, G);Distance(M, Focus(G)) = 9",
    "query_expressions": "Distance(M, xAxis)",
    "answer_expressions": "6*sqrt(2)",
    "fact_spans": "[[[0, 15], [22, 25]], [[18, 21], [36, 40]], [[0, 15]], [[0, 21]], [[18, 34]]]",
    "query_spans": "[[[36, 50]]]",
    "process": "The directrix of the parabola $ y^{2} = 12x $ is $ x = -3 $. The distance from point $ M $ to the focus of the parabola equals the distance from point $ M $ to the directrix. Therefore, the horizontal coordinate of point $ M $ is $ 6 $. Substituting into the equation of the parabola, the vertical coordinate is found to be $ \\pm 6\\sqrt{2} $. Thus, the distance from point $ M $ to the $ x $-axis is $ 6\\sqrt{2} $."
  },
  {
    "text": "Ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$ intersects the $x$-axis at points $A$ and $B$. Point $M$ is any point on ellipse $C$ distinct from $A$ and $B$. Let the slopes of lines $MA$ and $MB$ be $k_{MA}$ and $k_{MB}$, respectively. Then $k_{MA} \\cdot k_{MB}$ = ?",
    "fact_expressions": "C: Ellipse;A: Point;M: Point;B: Point;Expression(C) = (x^2/4 + y^2/2 = 1);Intersection(C, xAxis) = {A, B};PointOnCurve(M, C);Negation(M = A);Negation(M = B);Slope(LineOf(M,A))=k1;Slope(LineOf(M,B))=k2;k1:Number;k2:Number",
    "query_expressions": "k1*k2",
    "answer_expressions": "-1/2",
    "fact_spans": "[[[0, 42], [64, 69]], [[49, 52], [72, 75]], [[59, 63]], [[53, 56], [76, 79]], [[0, 42]], [[0, 58]], [[59, 84]], [[59, 84]], [[59, 84]], [[86, 126]], [[86, 126]], [[107, 116]], [[117, 126]]]",
    "query_spans": "[[[128, 153]]]",
    "process": "Using the properties of the ellipse, we obtain A(-2,0), B(2,0). Let M(x_{0},y_{0}). Using the slope formula for two points and simplifying k_{MA}\\cdot k_{MB} with the equation of the ellipse, we can derive the answer. From the given conditions, A(-2,0), B(2,0). Let M(x_{0},y_{0}), then k_{MA}\\cdot k_{MB} = \\frac{y_{0}}{x_{0}+2} \\cdot \\frac{y_{0}}{x_{0}-2} = \\frac{y_{0}^{2}}{x_{0}^{2}-4} = \\frac{2-\\frac{1}{2}x_{0}^{2}}{x_{0}^{2}-4} = -\\frac{1}{2}"
  },
  {
    "text": "Given the parabola $H$: $4x^{2}=y$, the directrix $l$ intersects the asymptotes of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) at points $A$ and $B$. If $|AB|=\\frac{1}{8}$, then the eccentricity $e$ of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;H: Parabola;A: Point;B: Point;e: Number;l:Line;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (4*x^2=y);Directrix(H)=l;Intersection(l,Asymptote(C))={A, B};Abs(LineSegmentOf(A, B)) = 1/8;Eccentricity(C)=e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[28, 89], [128, 134]], [[36, 89]], [[36, 89]], [[2, 21]], [[95, 98]], [[99, 102]], [[138, 141]], [[24, 27]], [[36, 89]], [[36, 89]], [[28, 89]], [[2, 21]], [[2, 27]], [[24, 104]], [[106, 125]], [[128, 141]]]",
    "query_spans": "[[[138, 143]]]",
    "process": "Given the parabola $ H: 4x^{2} = y $, i.e., $ x^{2} = \\frac{1}{4}y $, the directrix $ l $ is $ y = \\frac{1}{16} $, and the asymptotes of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) are $ y = \\pm\\frac{b}{a}x $. When $ y = \\frac{1}{16} $, $ x = \\pm\\frac{a}{16b} $. Also, since $ |AB| = \\frac{1}{8} $, we have $ \\frac{a}{8b} = \\frac{1}{8} $, $ \\therefore a = b $, $ c^{2} = a^{2} + b^{2} = 2a^{2} $, so $ c = \\sqrt{2}a $. Then the eccentricity of hyperbola $ C $ is $ e = \\frac{c}{a} = \\sqrt{2} $."
  },
  {
    "text": "Given that a line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$, and $|A F|=2$, then $|B F|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);A: Point;B: Point;Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, F)) = 2",
    "query_expressions": "Abs(LineSegmentOf(B, F))",
    "answer_expressions": "2",
    "fact_spans": "[[[3, 17], [28, 31]], [[3, 17]], [[20, 23]], [[3, 23]], [[24, 26]], [[2, 26]], [[32, 35]], [[36, 39]], [[24, 41]], [[42, 51]]]",
    "query_spans": "[[[53, 62]]]",
    "process": "The focus $ F $ of the parabola $ y^{2}=4x $ is $ (1,0) $, and the directrix is $ x=-1 $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. By the definition of the parabola, we have $ |AF|=x_{1}+1=2 $, solving gives $ x_{1}=1 $, $ y_{1}=\\pm2 $. Thus, $ AB\\bot x $-axis, and we obtain $ BF=|AF|=2 $."
  },
  {
    "text": "Find a point $P$ on the parabola $y^{2}=-4 x$ such that the sum of the distance from $P$ to the focus $F$ and the distance from $P$ to $A(-2 , 1)$ is minimized. Then what are the coordinates of this point?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;Expression(G) = (y^2 = -4*x);Coordinate(A) = (-2, 1);PointOnCurve(P, G);WhenMin(Distance(P,F)+Distance(P,A));Focus(G)=F;F:Point",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(-1/4,1)",
    "fact_spans": "[[[1, 16]], [[37, 48]], [[20, 23], [25, 26], [58, 59]], [[1, 16]], [[37, 48]], [[1, 23]], [[25, 55]], [[1, 32]], [[29, 32]]]",
    "query_spans": "[[[58, 64]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, and point $P$ lies on the hyperbola such that $\\angle F_{1} P F_{2}=90°$. Then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "F1: Point;F2: Point;G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1);Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G) = True;AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 8]], [[9, 16]], [[17, 45], [56, 59]], [[17, 45]], [[1, 50]], [[51, 55]], [[51, 60]], [[63, 89]]]",
    "query_spans": "[[[91, 118]]]",
    "process": ""
  },
  {
    "text": "Given that the equations of the two asymptotes of a hyperbola are $2x \\pm y = 0$, and the distance from the focus to the asymptote is $2$, then what is the standard equation of this hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (2*x + pm*y = 0);Distance(Focus(G), Asymptote(G)) = 2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2-y^2/4=1, y^2/16-x^2/4=1}",
    "fact_spans": "[[[2, 5], [42, 45]], [[2, 25]], [[2, 39]]]",
    "query_spans": "[[[42, 51]]]",
    "process": "Since the asymptotes are given by $2x \\pm y = 0$, we can assume the hyperbola equation is $4x^{2} - y^{2} = \\lambda$. When $\\lambda > 0$, the equation becomes $\\frac{x^{2}}{\\frac{\\lambda}{4}} - \\frac{y^{2}}{\\lambda} = 1$, which simplifies to $\\frac{x^{2}}{4} - \\frac{y^{2}}{2} = 1$. The foci are at $(\\pm\\frac{\\sqrt{5\\lambda}}{2}, 0)$, and the distance from a focus to an asymptote is $\\frac{\\sqrt{5\\lambda}}{\\sqrt{5}} = 2$, so $\\lambda = 4$. The standard equation is $x^{2} - \\frac{y^{2}}{4} = 1$. When $\\lambda < 0$, the equation becomes $\\frac{y^{2}}{-\\lambda} - \\frac{x^{2}}{-\\frac{\\lambda}{4}} = 1$, or $\\frac{y^{2}}{-\\lambda} - \\frac{x^{2}}{-\\frac{2}{4}} = 1$. The foci are at $(0, \\pm\\frac{\\sqrt{-5\\lambda}}{2})$, and the distance from a focus to an asymptote is $\\frac{\\sqrt{-5\\lambda}}{\\sqrt{5}} = 2$, solving gives $\\lambda = -16$. The standard equation is $\\frac{y^{2}}{16} - \\frac{x^{2}}{4} = 1$. Therefore, the standard equations of the hyperbola are $x^{2} - \\frac{y^{2}}{4} = 1$ or $\\frac{y^{2}}{16} - \\frac{x^{2}}{4} = 1$."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively. Point $P$ lies on the right branch of the hyperbola and satisfies $|P F_{2}|=|F_{1} F_{2}|$. The equations of the asymptotes of the hyperbola are $4x \\pm 3y = 0$. Then $\\cos \\angle P F_{1} F_{2} = ?$",
    "fact_expressions": "F2: Point;F1: Point;LeftFocus(G) = F1;RightFocus(G) = F2;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;P: Point;PointOnCurve(P,RightPart(G)) = True;Abs(LineSegmentOf(P, F2)) = Abs(LineSegmentOf(F1, F2));Expression(Asymptote(G)) = (4*x+pm*3*y = 0)",
    "query_expressions": "Cos(AngleOf(P, F1, F2))",
    "answer_expressions": "4/5",
    "fact_spans": "[[[9, 16]], [[1, 8]], [[1, 82]], [[1, 82]], [[19, 76], [88, 91], [123, 126]], [[19, 76]], [[22, 76]], [[22, 76]], [[22, 76]], [[22, 76]], [[83, 87]], [[83, 94]], [[97, 122]], [[123, 148]]]",
    "query_spans": "[[[150, 179]]]",
    "process": "Let the semi-focal length of the hyperbola be $ c $. From the asymptotes of the hyperbola, the relationship among $ a $, $ b $, and $ c $ can be obtained. Find the three sides of $ \\triangle PF_{1}F_{2} $, and apply the cosine law to find the value of $ \\cos\\angle PF_{1}F_{2} $. Let the semi-focal length of the hyperbola be $ c $. From the asymptote equations of the hyperbola, we have $ \\frac{b}{a} = \\frac{4}{3} $, then $ c = \\sqrt{a^{2} + b^{2}} = \\sqrt{a^{2} + \\frac{16}{9}a^{2}} = \\frac{5}{3}a $. In $ \\triangle PF_{1}F_{2} $, $ |PF_{2}| = |F_{1}F_{2}| = 2c $, $ |PF_{1}| = 2c + 2a $. By the cosine law, $ \\cos\\angle PF_{1}F_{2} = \\frac{(2c)^{2} + (2c+2a)^{2} - (2c)^{2}}{2 \\times 2c(2c+2a)} = \\frac{a+c}{2c} = \\frac{a + \\frac{5}{3}a}{\\frac{10}{3}a} = \\frac{4}{5} $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, a line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. If $|F_{2} A|+|F_{2} B|=14$, then $|A B|$=?",
    "fact_expressions": "G: Ellipse;H: Line;F2: Point;A: Point;B: Point;F1: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(G) = {F1, F2};PointOnCurve(F1, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(F2, A)) + Abs(LineSegmentOf(F2, B)) = 14",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "6",
    "fact_spans": "[[[18, 56], [74, 76]], [[71, 73]], [[10, 17]], [[77, 80]], [[81, 84]], [[2, 9], [63, 70]], [[18, 56]], [[2, 61]], [[62, 73]], [[71, 86]], [[88, 112]]]",
    "query_spans": "[[[114, 123]]]",
    "process": ""
  },
  {
    "text": "The circle $x^{2}+y^{2}+4 y=0$ intersects the directrix of the parabola $x^{2}=4 y$, what is the length of the chord?",
    "fact_expressions": "H: Circle;Expression(H) = (4*y + x^2 + y^2 = 0);G: Parabola;Expression(G) = (x^2 = 4*y)",
    "query_expressions": "Length(InterceptChord(Directrix(G), H))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[0, 20]], [[0, 20]], [[21, 35]], [[21, 35]]]",
    "query_spans": "[[[0, 44]]]",
    "process": "From $ x^{2} + y^{2} + 4y = 0 $, we obtain $ x^{2} + (y + 2)^{2} = 4 $, so the center of the circle is $ (0, -2) $ and the radius $ r = 2 $. The directrix of the parabola $ x^{2} = 4y $ is $ y = -1 $. Then, the distance from the center $ (0, -2) $ to the directrix $ y = -1 $ is $ d = 1 $. Therefore, the chord length intercepted by the circle $ x^{2} + y^{2} + 4y = 0 $ on the directrix of the parabola $ x^{2} = 4y $ is $ 2\\sqrt{r^{2} - d^{2}} = 2\\sqrt{2^{2} - 1^{2}} = 2\\sqrt{3} $."
  },
  {
    "text": "Let $P$ be a point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, and let $F_{1}$, $F_{2}$ be the left and right foci of the ellipse, respectively. If $|P F_{1}|\\cdot | P F_{2} |=12$, then what is the measure of $\\angle F_{1} P F_{2}$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/16 + y^2/9 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 12",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "ApplyUnit(60,degree)",
    "fact_spans": "[[[5, 43], [65, 67]], [[1, 4]], [[47, 54]], [[55, 62]], [[5, 43]], [[1, 46]], [[47, 73]], [[47, 73]], [[75, 106]]]",
    "query_spans": "[[[108, 134]]]",
    "process": "The ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ gives $2a=8$. Let $|PF_{1}|=m$, $|PF_{2}|=n$, then we have $\\begin{cases}m+n=2a=8\\\\mn=12\\end{cases}$. Simplifying yields: $\\cos\\angle F_{1}PF_{2}=\\frac{1}{2}$, $\\therefore\\angle F_{1}PF_{2}=60^{\\circ}$."
  },
  {
    "text": "Given that the line $l$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$, if the midpoint of segment $AB$ is $(3,2)$, then the length of segment $AB$ is?",
    "fact_expressions": "l: Line;G: Parabola;B: Point;A: Point;Expression(G) = (y^2 = 4*x);Intersection(l, G) = {A, B};Coordinate(MidPoint(LineSegmentOf(A,B)))=(3,2)",
    "query_expressions": "Length(LineSegmentOf(A,B))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 7]], [[8, 22]], [[28, 31]], [[24, 27]], [[8, 22]], [[2, 33]], [[35, 53]]]",
    "query_spans": "[[[55, 67]]]",
    "process": "According to the problem, the slope of the line clearly exists. Let the line be $ y - 2 = k(x - 3) $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. From \n\\[\n\\begin{cases}\ny - 2 = k(x - 3) \\\\\ny^{2} = 4x\n\\end{cases}\n\\]\neliminating $ y $ and simplifying gives \n\\[\nk^{2}x^{2} - (6k^{2} + 4 - 4k)x + (2 - 3k)^{2} = 0.\n\\]\nWhen $ k = 0 $, it clearly does not hold. When $ k \\neq 0 $, \n\\[\nx_{1} + x_{2} = \\frac{6k^{2} + 4 - 4k}{k^{2}}.\n\\]\nAlso, since $ \\frac{x_{1} + x_{2}}{2} = 3 $, we have \n\\[\n\\frac{6k^{2} + 4 - 4k}{k^{2}} = 6,\n\\]\nsolving gives $ k = 1 $. When $ k = 1 $, the line is $ x - y - 1 = 0 $, and the focus $ F(1, 0) $ satisfies the line $ x - y - 1 = 0 $. Thus, \n\\[\n|AB| = |FA| + |FB| = (x_{1} + 1) + (x_{2} + 1) = x_{1} + x_{2} + 2,\n\\]\nand since $ x_{1} + x_{2} = 6 $, $ \\therefore |AB| = 8 $. The final answer is: $ 8 $"
  },
  {
    "text": "Given that point $F$ is the right focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, point $B$ is an endpoint of the minor axis, and the extension of line segment $BF$ intersects the ellipse $C$ at point $D$, with $\\overrightarrow{B F}=2 \\overrightarrow{F D}$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;B: Point;F: Point;D: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;OneOf(Endpoint(MinorAxis(C)))=B;Intersection(OverlappingLine(LineSegmentOf(B,F)), C) = D;VectorOf(B, F) = 2*VectorOf(F, D)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[7, 64], [94, 99], [153, 158]], [[14, 64]], [[14, 64]], [[69, 73]], [[2, 6]], [[100, 104]], [[14, 64]], [[14, 64]], [[7, 64]], [[2, 68]], [[7, 81]], [[82, 104]], [[106, 151]]]",
    "query_spans": "[[[153, 164]]]",
    "process": "Find the value of BF using the properties of the ellipse, and use the known vector relationships and similar triangles to find the x-coordinate of D. Then, use the second definition of the ellipse to find the value of \\overrightarrow{FD}. Given that \\overrightarrow{BF}=2\\overrightarrow{FD}, establish an equation in terms of a and c, and solve the equation to find the value of \\frac{c}{a}. As shown in the figure, BF=\\sqrt{b^{2}+c^{2}}=a. Draw DD_{1}\\bot y-axis at point D_{1}, then from \\overrightarrow{BF}=2\\overrightarrow{FD}, we have: \\frac{|\\overrightarrow{OF}|}{|DD_{1}|}=\\frac{|\\overrightarrow{BF}|}{|BD|}=\\frac{2}{3}. Therefore, |\\overrightarrow{DD_{1}}|=\\frac{3}{2}|\\overrightarrow{OF}|=\\frac{3}{2}c, i.e., x_{D}=\\frac{3}{2}c. From the second definition of the ellipse, \\overrightarrow{FD}=\\left(\\frac{a^{2}}{c}-\\frac{3}{2}c\\right)=a-\\frac{3c^{2}}{2a}. Again, from \\overrightarrow{BF}=2|\\overrightarrow{FD}|, we get a=2\\left(a-\\frac{3c^{2}}{2a}\\right), a^{2}=3c^{2}, solving gives e=\\frac{c}{a}=\\frac{\\sqrt{3}}{3}."
  },
  {
    "text": "What are the coordinates of the points on the parabola $y^{2}=12 x$ whose distance from the focus is equal to $6$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 12*x);P:Point;Distance(P,Focus(G))=6;PointOnCurve(P,G)",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "{(3,6),(3,-6)}",
    "fact_spans": "[[[0, 15]], [[0, 15]], [[28, 29]], [[0, 29]], [[0, 29]]]",
    "query_spans": "[[[28, 34]]]",
    "process": "From the definition of a parabola, the distance from a point $(x, y)$ on the parabola $y^{2} = 12x$ to the focus is $x + 3$. Given that $x = 3$, substituting into the parabola equation yields $y = \\pm 6$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{25}=1$ has its left focus at $F$, point $P$ lies on the right branch of the hyperbola, and $PF$ is tangent to the circle $x^{2}+y^{2}=16$ at point $N$. Let $M$ be the midpoint of segment $PF$, and $O$ be the origin. Then $|MN|-|MO|=$?",
    "fact_expressions": "G: Hyperbola;H: Circle;P: Point;F: Point;M: Point;N: Point;O: Origin;Expression(G) = (x^2/16 - y^2/25 = 1);Expression(H) = (x^2 + y^2 = 16);LeftFocus(G) = F;PointOnCurve(P, RightPart(G));TangentPoint(LineSegmentOf(P, F), H) = N;MidPoint(LineSegmentOf(P, F)) = M",
    "query_expressions": "Abs(LineSegmentOf(M, N)) - Abs(LineSegmentOf(M, O))",
    "answer_expressions": "-1",
    "fact_spans": "[[[2, 42], [56, 59]], [[72, 89]], [[51, 55]], [[47, 50]], [[97, 100]], [[92, 96]], [[112, 115]], [[2, 42]], [[72, 89]], [[2, 50]], [[51, 64]], [[66, 96]], [[97, 111]]]",
    "query_spans": "[[[122, 137]]]",
    "process": ""
  },
  {
    "text": "Let the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, with its left focus at $F(-2,0)$. Draw a perpendicular line from $F$ to one asymptote of the hyperbola, with foot of perpendicular at point $N$, and intersecting the other asymptote at point $M$. If $\\overrightarrow {M N}=\\overrightarrow{N F}$, then the equations of the asymptotes of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F: Point;N: Point;M: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (-2, 0);LeftFocus(G)=F;L:Line;PointOnCurve(F,L);IsPerpendicular(L,L1);FootPoint(L,L1)=N;VectorOf(M, N) = VectorOf(N, F);L1:Line;L2:Line;Intersection(L,L2)=M;OneOf(Asymptote(G))=L1;OneOf(Asymptote(G))=L2;Negation(L1=L2)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[1, 57], [58, 59], [78, 81], [160, 163]], [[4, 57]], [[4, 57]], [[63, 72], [74, 77]], [[93, 97]], [[108, 112]], [[4, 57]], [[4, 57]], [[1, 57]], [[63, 72]], [[58, 72]], [], [[73, 89]], [[73, 89]], [[73, 97]], [[114, 158]], [], [], [[73, 112]], [76, 83], [76, 103], [76, 103]]",
    "query_spans": "[[[160, 171]]]",
    "process": "Draw the figure. From $\\overrightarrow{MN}=\\overrightarrow{NF}$, it follows that $N$ is the midpoint of $FM$. Combining this with the given conditions, $ON$ perpendicularly bisects $FM$. Since the two asymptotes of the hyperbola are symmetric about the $y$-axis, we have $\\angle FON = \\angle MOE = \\angle NOM = 60^{\\circ}$, and thus the equations of the asymptotes of the hyperbola can be found. Because $\\overrightarrow{MN}=\\overrightarrow{NF}$, $N$ is the midpoint of $FM$. Since $FM\\perp ON$, $ON$ perpendicularly bisects $FM$. Therefore, $\\angle FON = \\angle NOM$. Because the two asymptotes of the hyperbola are symmetric about the $y$-axis, $\\angle FON = \\angle MOE$. Since $\\angle FON + \\angle MOE + \\angle NOM = 180^{\\circ}$, it follows that $\\angle FON = \\angle MOE = \\angle NOM = 60^{\\circ}$. Hence, the equations of the asymptotes of the hyperbola are $y = \\pm \\tan 60^{\\circ}x = \\pm \\sqrt{3}x$."
  },
  {
    "text": "The length of the segment cut by the line $y=k x+1$ from the ellipse $\\frac{x^{2}}{a^{2}}+y^{2}=1(a>1)$ is (expressed in terms of $a$, $k$)?",
    "fact_expressions": "G: Ellipse;a: Number;H: Line;k: Number;a>1;Expression(G) = (y^2 + x^2/a^2 = 1);Expression(H) = (y = k*x + 1)",
    "query_expressions": "Length(InterceptChord(H, G))",
    "answer_expressions": "(2*a^2*Abs(k)/(1 + a^2*k^2))*sqrt(1 + k^2)",
    "fact_spans": "[[[12, 48]], [[58, 61]], [[0, 11]], [[63, 67]], [[14, 48]], [[12, 48]], [[0, 11]]]",
    "query_spans": "[[[0, 71]]]",
    "process": "Solve the system of equations by classifying into cases $ k \\neq 0 $ and $ k = 0 $, and use Vieta's formulas and the chord length formula to obtain the solution. From the given conditions  \n\\[\n\\begin{cases}\n\\frac{x^{2}}{a^{2}} + y^{2} = 1 \\\\\ny = kx + 1\n\\end{cases}\n\\]\nsimplifying yields $ (1 + a^{2}k^{2})x^{2} + 2ka^{2}x = 0 $. When $ k \\neq 0 $, $ A > 0 $, let the intersection points be $ (x_{1}, y_{1}) $, $ (x_{2}, y_{2}) $, then  \n$ x_{1} + x_{2} = -\\frac{2ka^{2}}{1 + a^{2}k^{2}} $, $ x_{1}x_{2} = 0 $,  \nthe length of the intercepted segment is  \n$ \\sqrt{1 + k^{2}}|x_{1} - x_{2}| = \\sqrt{1 + k^{2}}\\sqrt{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}} = \\frac{2a^{2}|k|}{1 + a^{2}k^{2}}\\sqrt{1 + k^{2}} $.  \nWhen $ k = 0 $, the length of the intercepted segment is 0, which satisfies the above expression. In conclusion, the length of the segment intercepted by the line $ y = kx + 1 $ on the ellipse $ \\frac{x^{2}}{a^{2}} + y^{2} = 1 $ ($ a > 1 $) is $ \\frac{2a^{2}|k|}{1 + a^{2}k^{2}}\\sqrt{1 + k^{2}} $."
  },
  {
    "text": "A moving circle $M$ is internally tangent to the circle $C_{1}$: $(x+1)^{2}+y^{2}=25$, and externally tangent to the circle $C_{2}$: $(x-1)^{2}+y^{2}=1$. What is the trajectory equation of the center of the moving circle $M$?",
    "fact_expressions": "M: Circle;C1:Circle;Expression(C1)=((x+1)^2+y^2=25);C2:Circle;Expression(C2)=((x-1)^2+y^2=1);IsInTangent(M,C1);IsOutTangent(M,C2);M1:Point;Center(M)=M1",
    "query_expressions": "LocusEquation(M1)",
    "answer_expressions": "x^2/9+y^2/8=1",
    "fact_spans": "[[[3, 6], [73, 75]], [[7, 36]], [[7, 36]], [[41, 69]], [[41, 69]], [[3, 38]], [[3, 71]], [[77, 80]], [[73, 80]]]",
    "query_spans": "[[[77, 87]]]",
    "process": "From the positional relationship between circles, we have |MC_{1}| + |MC_{2}| = 6 > |C_{1}C_{2}|, and then by the definition of an ellipse, the solution can be obtained. According to the problem, circle C_{1}: (x+1)^{2} + y^{2} = 25 has center (-1, 0) and radius 5; circle C_{2}: (x-1)^{2} + y^{2} = 1 has center (1, 0) and radius 1. Let M(x, y) be the center of the moving circle with radius R. The moving circle is internally tangent to circle C_{1}: (x+1)^{2} + y^{2} = 25 and externally tangent to circle C_{2}: (x-1)^{2} + y^{2} = 1. Thus, |MC_{1}| = 5 - R, |MC_{2}| = 1 + R. Therefore, |MC_{1}| + |MC_{2}| = 5 - R + 1 + R = 6 > |C_{1}C_{2}|. Hence, the locus of M is an ellipse centered at the origin with foci on the x-axis, where 2a = 6, c = 1. Thus, b^{2} = a^{2} - c^{2} = 8, \\therefore the equation of the ellipse is \\frac{x^{2}}{9} + \\frac{y^{2}}{8} = 1"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, with left and right foci $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through $F_{1}$ is tangent to the circle $x^{2}+y^{2}=a^{2}$ at point $T$, and intersects the right branch of the hyperbola $C$ at point $P$. If $\\overrightarrow{T P}=3 \\overrightarrow{F_{1} T}$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;G: Circle;T: Point;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = a^2);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, l);TangentPoint(l, G) = T;Intersection(l, RightPart(C)) = P;VectorOf(T, P) = 3*VectorOf(F1, T)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[96, 101], [131, 136]], [[2, 63], [137, 143], [205, 211]], [[10, 63]], [[10, 63]], [[102, 122]], [[125, 129]], [[148, 152]], [[71, 78], [88, 95]], [[79, 86]], [[10, 63]], [[10, 63]], [[2, 63]], [[102, 122]], [[2, 86]], [[2, 86]], [[87, 101]], [[96, 129]], [[131, 152]], [[154, 203]]]",
    "query_spans": "[[[205, 217]]]",
    "process": "As shown in the figure, from the given conditions we have |OF| = |OF₂| = c, |OT| = a, then |F₁T| = b. Also, since $\\overrightarrow{TP} = 3\\overrightarrow{F_{1}T}$, we have |TP| = 3b, so |F₁P| = 4b. Moreover, since |PF₁| - |PF₂| = 2a, it follows that |PF₂| = 4b - 2a. Construct F₂M // OT, we obtain |F₂M| = 2a, |TM| = b, then |PM| = 2b. In $\\triangle MPF_{2}$, $|PM|^{2} + |MF_{2}|^{2} = |PF_{2}|^{2}$, thus $c^{2} = (2b - a)^{2}$, which gives 2b = a + c. Also, since $c^{2} = a^{2} + b^{2}$, simplifying yields $3c^{2} - 2ac - 5a^{2} = 0$. Dividing through by $a^{2}$ gives $3e^{2} - 2e - 5 = 0$. Solving yields $e = \\frac{5}{2}$ or $e = -1$ (discarded). Therefore, the eccentricity of the hyperbola is $\\frac{5}{2}$."
  },
  {
    "text": "If there are two points $A$ and $B$ on the parabola $y^{2}=2x$, and $AB$ is perpendicular to the $x$-axis, and $|AB|=2\\sqrt{2}$, then what is the distance from point $A$ to the directrix of the parabola?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);IsPerpendicular(LineSegmentOf(A, B), xAxis);Abs(LineSegmentOf(A, B)) = 2*sqrt(2)",
    "query_expressions": "Distance(A, Directrix(G))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[1, 15], [67, 70]], [[1, 15]], [[19, 22], [62, 66]], [[23, 26]], [[1, 26]], [[1, 26]], [[28, 40]], [[42, 60]]]",
    "query_spans": "[[[62, 78]]]",
    "process": "Find the directrix of the parabola, then use the chord length to determine the y-coordinate of point A, substitute into the parabola to find point A, and solve using the definition of the parabola. Given the parabola $ y^{2} = 2x $, its directrix is $ x = -\\frac{1}{2} $. Since $ AB $ is perpendicular to the x-axis, $ |AB| = 2\\sqrt{2} $, and the distance from A to the x-axis is $ \\sqrt{2} $. Assume A lies above the x-axis, so $ y = \\sqrt{2} $; substituting into the parabola $ y^{2} = 2x $ gives $ x = 1 $. The distance from point A to the directrix of the parabola is $ d = 1 + \\frac{1}{2} = \\frac{3}{2} $."
  },
  {
    "text": "The directrix of the parabola $y^{2}=a x$ is given by $x=\\frac{1}{2}$, then $a$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = a*x);a: Number;Expression(Directrix(G)) = (x = 1/2)",
    "query_expressions": "a",
    "answer_expressions": "-2",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[37, 40]], [[0, 35]]]",
    "query_spans": "[[[37, 42]]]",
    "process": "\\because the directrix equation of the parabola y^{2}=ax is x=\\frac{1}{2}, \\therefore x=-\\frac{a}{4}=\\frac{1}{2}, solving gives: a=-2,"
  },
  {
    "text": "It is known that the hyperbola shares the same foci as the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{6}=1$, and the asymptotes of the hyperbola are given by $y=\\pm \\frac{1}{3} x$. Then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/16 + y^2/6 = 1);Focus(G) = Focus(H);Expression(Asymptote(G)) = (y = pm*(x/3))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9-y^2=1",
    "fact_spans": "[[[2, 5], [86, 89], [52, 55]], [[6, 44]], [[6, 44]], [[2, 50]], [[52, 83]]]",
    "query_spans": "[[[86, 93]]]",
    "process": "Find the coordinates of the ellipse's foci, which are also the foci of the hyperbola, giving the value of $ c $. From the asymptote equation, we have $ \\frac{b}{a} = \\frac{1}{3} $. Using $ a^{2} + b^{2} = c^{2} $, we can solve for $ a $ and $ b $ to obtain the hyperbola equation. According to the problem, the ellipse's foci are $ (\\pm\\sqrt{10}, 0) $, so $ c = \\sqrt{10} $. Let the hyperbola equation be $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0 $, $ b > 0 $), then $ \\frac{b}{a} = \\frac{1}{3} $. Solving the system \n\\[\n\\begin{cases}\n\\frac{b}{a} = \\frac{1}{3} \\\\\na^{2} + b^{2} = c^{2} = 10\n\\end{cases}\n\\]\nyields $ a = $. Therefore, the hyperbola equation is $ \\frac{x^{2}}{9} - y^{2} = 1 $."
  },
  {
    "text": "Let the parabola $C$: $y^{2}=2 m x$ have focus $F$. If for a point $M$ on the parabola $C$, the difference between the distance from $M$ to the focus $F$ and the distance from $M$ to the $y$-axis is $2$, then $m=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*m*x);m: Number;F: Point;Focus(C) = F;M: Point;PointOnCurve(M, C);Distance(M, F) - Distance(M, yAxis) = 2",
    "query_expressions": "m",
    "answer_expressions": "pm*4",
    "fact_spans": "[[[1, 22], [31, 37]], [[1, 22]], [[69, 72]], [[26, 29], [46, 49]], [[1, 29]], [[39, 43]], [[31, 43]], [[39, 67]]]",
    "query_spans": "[[[69, 74]]]",
    "process": "The parabola $ C: y^2 = 2mx $ has focus $ F\\left(\\frac{m}{2}, 0\\right) $, and directrix equation $ x = -\\frac{m}{2} $. Let $ M(x_0, y_0) $. By the definition of a parabola, we have $ \\left|\\left(x_0 + \\frac{m}{2}\\right) - x_0\\right| = 2 $, solving gives $ m = \\pm 4 $."
  },
  {
    "text": "Let the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ be $F_{1}$ and $F_{2}$, respectively, and let $P$ be a point on the ellipse such that $|P F_{1}|=\\lambda|P F_{2}|$ $\\left(\\frac{1}{2} \\leq \\lambda \\leq 2\\right)$, $\\angle F_{1} P F_{2}=\\frac{\\pi}{2}$. Then the range of values for the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G) = True;lambda: Number;Abs(LineSegmentOf(P, F1)) = lambda*Abs(LineSegmentOf(P, F2));1/2 <= lambda;lambda <= 2;AngleOf(F1, P, F2) = pi/2",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[sqrt(2)/2,sqrt(5)/3]",
    "fact_spans": "[[[1, 53], [82, 84], [194, 196]], [[1, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[62, 69]], [[70, 77]], [[1, 77]], [[1, 77]], [[78, 81]], [[78, 87]], [[88, 116]], [[88, 116]], [[118, 153]], [[118, 153]], [[156, 192]]]",
    "query_spans": "[[[194, 206]]]",
    "process": "Let $|PF_{2}| = t$, then $|PF_{1}| = \\lambda t$. By the definition of an ellipse, we have $(\\lambda + 1)t = 2a$, that is, $(\\lambda + 1)^{2}t^{2} = 4a^{2}$. By the Pythagorean theorem, we have $(\\lambda^{2} + 1)t^{2} = 4c^{2}$. Dividing the two equations yields $e^{2} = \\frac{\\lambda^{2} + 1}{(\\lambda + 1)^{2}}$. Let $m = \\lambda + 1 \\in \\left[\\frac{3}{2}, 3\\right]$, then by the properties of the function, the range of $e^{2}$ can be obtained, and thus the range of the ellipse's eccentricity can be determined. Solution: Let $F_{1}(-c, 0)$, $F_{2}(c, 0)$. By the definition of the ellipse, $|PF_{1}| + |PF_{2}| = 2a$. Let $|PF_{2}| = t$, then $|PF_{1}| = \\lambda t$, so $(\\lambda + 1)t = 2a$, i.e., $(\\lambda + 1)^{2}t^{2} = 4a^{2}$, $\\textcircled{1}$. Since $\\angle F_{1}PF_{2} = \\frac{\\pi}{2}$, we have $(\\lambda^{2} + 1)t^{2} = 4c^{2}$, $\\textcircled{2}$. Dividing the two equations gives $\\frac{\\lambda^{2} + 1}{(\\lambda + 1)^{2}} = \\frac{c^{2}}{a^{2}} = e^{2}$. Let $m = \\lambda + 1 \\in \\left[\\frac{3}{2}, 3\\right]$, then $\\lambda = m - 1$, so $e^{2} = \\frac{\\lambda^{2} + 1}{(\\lambda + 1)^{2}} = \\frac{(m - 1)^{2} + 1}{m^{2}} = \\frac{m^{2} - 2m + 2}{m^{2}} = 2\\cdot\\left(\\frac{1}{m}\\right)^{2} - 2\\cdot\\frac{1}{m} + 1 = 2\\cdot\\left(\\frac{1}{m} - \\frac{1}{2}\\right)^{2} + \\frac{1}{2}$. Since $\\frac{3}{2} \\leqslant m \\leqslant 3$, we have $\\frac{1}{3} \\leqslant \\frac{1}{m} \\leqslant \\frac{2}{3}$. Thus, when $\\frac{1}{m} = \\frac{1}{2}$, i.e., $m = 2$, $\\lambda = 1$, $e^{2}$ attains its minimum value $\\frac{1}{2}$, and $e$ reaches its minimum $\\frac{\\sqrt{2}}{2}$. When $\\frac{1}{m} = \\frac{1}{3}$ or $\\frac{2}{3}$, i.e., $m = 3$, $\\lambda = 2$, $e^{2}$ attains its maximum value $\\frac{5}{9}$, and $e$ reaches its maximum $\\frac{\\sqrt{5}}{3}$. Therefore, the range of the ellipse's eccentricity is $\\frac{\\sqrt{2}}{2} \\leqslant e \\leqslant \\frac{\\sqrt{5}}{3}$."
  },
  {
    "text": "Suppose the foci of the hyperbola lie on the $x$-axis, and the two asymptotes are $y=\\pm \\frac{1}{2} x$. Then the eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), xAxis);Expression(Asymptote(G)) = (y = pm*(x/2));e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[1, 4], [43, 46]], [[1, 13]], [[1, 41]], [[50, 53]], [[43, 53]]]",
    "query_spans": "[[[50, 55]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola is $y=2x$, and it passes through the focus of the parabola $y^{2}=4x$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(OneOf(Asymptote(G))) = (y = 2*x);H: Parabola;Expression(H) = (y^2 = 4*x);PointOnCurve(Focus(H), G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/4 = 1",
    "fact_spans": "[[[2, 5], [42, 45]], [[2, 19]], [[23, 37]], [[23, 37]], [[2, 40]]]",
    "query_spans": "[[[42, 52]]]",
    "process": ""
  },
  {
    "text": "A line segment $MN$ of fixed length $4$ has its endpoints moving on the parabola $y^2 = x$. Let point $P$ be the midpoint of segment $MN$. Then, the minimum distance from point $P$ to the $y$-axis is?",
    "fact_expressions": "M: Point;N: Point;Length(LineSegmentOf(M, N)) = 4;Endpoint(LineSegmentOf(M, N)) = {M, N};G: Parabola;Expression(G) = (y^2 = x);PointOnCurve(M, G);PointOnCurve(N, G);P: Point;MidPoint(LineSegmentOf(M, N)) = P",
    "query_expressions": "Min(Distance(P, yAxis))",
    "answer_expressions": "7/4",
    "fact_spans": "[[[9, 14]], [[9, 14]], [[1, 14]], [[7, 17]], [[18, 30]], [[18, 30]], [[7, 31]], [[7, 31]], [[35, 39], [52, 56]], [[35, 50]]]",
    "query_spans": "[[[52, 69]]]",
    "process": ""
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$ has left and right foci $F_{1}$ and $F_{2}$. A line passes through $F_{1}$ and intersects the ellipse at points $A$ and $B$. Then, what is the perimeter of $\\triangle A B F_{2}$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/7 = 1);F2: Point;F1: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Line;A: Point;B: Point;PointOnCurve(F1, H);Intersection(H, G) = {A,B}",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "16",
    "fact_spans": "[[[0, 38], [75, 77]], [[0, 38]], [[54, 61]], [[44, 51], [67, 74]], [[0, 61]], [[0, 61]], [[64, 66]], [[78, 81]], [[82, 85]], [[64, 74]], [[64, 87]]]",
    "query_spans": "[[[89, 115]]]",
    "process": "Since $F_{1}, F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$, and a line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$, we have: $|AF_{1}|+|AF_{2}|=2a=8$, $|BF_{1}|+|BF_{2}|=2a=8$. Therefore, the perimeter of $ABF_{2}$ is $|AB|+|AF_{2}|+|BF_{2}|=|AF_{1}|+|AF_{2}|+|BF_{1}|+|BF_{2}|=16$."
  },
  {
    "text": "The left and right foci of the ellipse $C$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ are $F_{1}$, $F_{2}$, and $M$ is a moving point on the ellipse $C$. Then the minimum value of $\\frac{1}{M F_{1}}+\\frac{1}{M F_{2}}$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/25 + y^2/16 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;M: Point;PointOnCurve(M, C)",
    "query_expressions": "Min(1/LineSegmentOf(M, F2) + 1/LineSegmentOf(M, F1))",
    "answer_expressions": "2/5",
    "fact_spans": "[[[0, 43], [69, 74]], [[0, 43]], [[49, 56]], [[57, 64]], [[0, 64]], [[0, 64]], [[65, 68]], [[65, 78]]]",
    "query_spans": "[[[80, 123]]]",
    "process": "By $\\frac{1}{IF_{1}}+\\frac{1}{IF_{2}}=\\frac{MF_{1}+MF_{2}}{MF_{1}\\cdot MF_{2}}=\\frac{10}{MF_{1}\\cdot MF_{2}}$, the maximum value of $MF_{1}\\cdot MF_{2}$ is $a^{2}=25$, the minimum value of $\\frac{1}{MF_{1}}+\\frac{1}{MF_{2}}$ can be found. $\\because$ ellipse $C: \\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ has left and right foci $F_{1}, F_{2}$, and $M$ is a moving point on ellipse $C$, $\\therefore \\frac{1}{MF_{1}}+\\frac{1}{MF_{2}}=\\frac{MF_{1}+MF_{2}}{MF_{1}\\cdot MF_{2}}=\\frac{10}{MF_{1}\\cdot MF_{2}}$. $\\because$ the maximum value of $MF_{1}\\cdot MF_{2}$ is $a^{2}=25$, the minimum value of $\\frac{1}{MF_{1}}+\\frac{1}{MF_{2}}$ is $d_{\\min}=\\frac{10}{25}=\\frac{2}{5}$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ has left and right foci $F_{1}$ and $F_{2}$ respectively, and point $P$ lies on the line $x-\\sqrt{3} y+4+\\sqrt{3}=0$. When $\\angle F_{1} P F_{2}$ reaches its maximum value, the value of $\\frac{|P F_{1}|}{|P F_{2}|}$ is?",
    "fact_expressions": "G: Ellipse;H: Line;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);Expression(H) = (x - sqrt(3)*y + 4 + sqrt(3) = 0);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, H);WhenMax(AngleOf(F1,P,F2))",
    "query_expressions": "Abs(LineSegmentOf(P, F1))/Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[2, 29]], [[59, 88]], [[38, 45]], [[54, 58]], [[46, 53]], [[2, 29]], [[59, 88]], [[2, 53]], [[2, 53]], [[54, 89]], [[91, 119]]]",
    "query_spans": "[[[120, 153]]]",
    "process": "From $\\frac{x^{2}}{4}+y^{2}=1$, we have $a=2$, $b=1$, so $c=\\sqrt{3}$, thus the foci are $F_{1}(-\\sqrt{3},0)$, $F_{2}(\\sqrt{3},0)$. To maximize $\\angle F_{1}PF_{2}$, the circle passing through points $F_{1}$, $F_{2}$, and $P$ must be tangent to the line $x-\\sqrt{3}y+4+\\sqrt{3}=0$ at point $P$. The line $x-\\sqrt{3}y+4+\\sqrt{3}=0$ intersects the $x$-axis at point $A(-4-\\sqrt{3},0)$. Since $\\angle APF_{1}=\\angle AF_{2}P$, $\\angle PAF_{1}=\\angle F_{2}AP$, it follows that $\\triangle APF_{1} \\sim \\triangle AF_{2}P$, hence $\\frac{|PF_{1}|}{|PF_{2}|}=\\frac{|AP|}{|AF_{2}|}$. Therefore, $\\frac{|PF_{1}|}{|PF_{2}|}=\\sqrt{\\frac{|AP|^{2}}{|AF_{2}|^{2}}}=\\sqrt{\\frac{|AF_{1}|=4,|AF_{2}|=4+2\\sqrt{3}}{|AF_{2}|^{2}}}=\\sqrt{\\frac{|AF_{1}|}{|AF_{2}|}}=\\sqrt{\\frac{4}{4+2\\sqrt{3}}}=\\sqrt{4-2\\sqrt{3}}=\\sqrt{3}-1$."
  },
  {
    "text": "Given the ellipse $M$: $\\frac{x^{2}}{4}+y^{2}=1$, a line $l$ intersects the ellipse $M$ at points $A$ and $B$, and point $D(1, \\frac{1}{2})$ is the midpoint of chord $AB$. Then the equation of line $l$ is?",
    "fact_expressions": "M: Ellipse;Expression(M) = (x^2/4 + y^2 = 1);l: Line;Intersection(l, M) = {A, B};A: Point;B: Point;D: Point;Coordinate(D) = (1, 1/2);IsChordOf(LineSegmentOf(A,B),M) ;MidPoint(LineSegmentOf(A, B)) = D",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x + 2*y-2 = 0",
    "fact_spans": "[[[2, 34], [41, 46]], [[2, 34]], [[35, 40], [91, 96]], [[35, 58]], [[49, 52]], [[53, 56]], [[59, 79]], [[59, 79]], [[41, 86]], [[59, 89]]]",
    "query_spans": "[[[91, 101]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Since the line $ l $ intersects the ellipse $ M $ at points $ A $ and $ B $, we have \n$$\n\\begin{cases}\n\\frac{x_{1}^{2}}{4}+y_{1}^{2}=1 \\\\\n\\frac{x_{2}^{2}}{4}+y_{2}^{2}=1\n\\end{cases}\n$$\nSubtracting the two equations gives: \n$ y_{1}^{2}-y_{2}^{2}=\\frac{x_{2}^{2}}{4}-\\frac{x_{1}^{2}}{4} $, which simplifies to \n$ k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{1}{4}\\times\\frac{x_{1}+x_{2}}{y_{1}+y_{2}} $.  \nSince point $ D(1,\\frac{1}{2}) $ is the midpoint of chord $ AB $, we have $ x_{1}+x_{2}=2 $, $ y_{1}+y_{2}=1 $, so $ k_{AB}=-\\frac{1}{2} $.  \nTherefore, the equation of line $ l $ is $ y-\\frac{1}{2}=-\\frac{1}{2}(x-1) $, which simplifies to $ x+2y-2=0 $."
  },
  {
    "text": "Let point $P$ be a point on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ with $F_{1}$ and $F_{2}$ as the left and right foci, respectively, such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$. The line $P F_{1}$ has exactly one common point with the circle $x^{2}+y^{2}=\\frac{a^{2}}{4}$. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;P: Point;F1: Point;F2: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + y^2 = a^2/4);PointOnCurve(P,RightPart(G));LeftFocus(G)=F1;RightFocus(G)=F2;DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;NumIntersection(LineOf(P,F1), H) = 1",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)/2",
    "fact_spans": "[[[28, 74], [196, 199]], [[31, 74]], [[31, 74]], [[155, 185]], [[1, 5]], [[7, 14]], [[15, 22]], [[28, 74]], [[155, 185]], [[1, 79]], [[6, 74]], [[6, 74]], [[83, 142]], [[143, 194]]]",
    "query_spans": "[[[196, 205]]]",
    "process": "By the given condition, $ PF_{2} = a $, $ PF_{1} = 3a $, so $ a^{2} + (3a)^{2} = (2c)^{2} $, therefore the eccentricity $ e = \\frac{c}{a} = \\frac{\\sqrt{10}}{2} $."
  },
  {
    "text": "The ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has a focal distance of $4$, and the maximum distance from a point on the ellipse to one focus is $5$. Then, the minimum distance from a point on the ellipse to one focus is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);FocalLength(C) = 4;P:Point;PointOnCurve(P,C);Max(Distance(P,OneOf(Focus(C))))=5",
    "query_expressions": "Min(Distance(P,OneOf(Focus(C))))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 56], [64, 66], [85, 87]], [[6, 56]], [[6, 56]], [[6, 56]], [[6, 56]], [[0, 56]], [[0, 63]], [[68, 69], [89, 90]], [[64, 69]], [[64, 83]]]",
    "query_spans": "[[[85, 102]]]",
    "process": "According to the geometric properties of an ellipse, the maximum distance from a point on the ellipse to one focus is $a + c$. Given that the focal distance is 4 and the maximum distance from a point on the ellipse to one focus is 5, it follows that $c = 2$ and $a = 3$. Therefore, the minimum distance from a point on the ellipse to one focus is $a - c = 1$."
  },
  {
    "text": "Let $P$ be an arbitrary point on the parabola $y^{2}=4x$, and let $Q$ be the projection of $P$ onto the $y$-axis. Given point $M(4, 5)$, find the minimum value of the sum of the lengths $PQ$ and $PM$.",
    "fact_expressions": "P: Point;G: Parabola;Expression(G) = (y^2 = 4*x);PointOnCurve(P, G) = True;Projection(P, yAxis) = Q;Q: Point;M: Point;Coordinate(M) = (4, 5)",
    "query_expressions": "Min(Length(LineSegmentOf(P,Q))+Length(LineSegmentOf(P,M)))",
    "answer_expressions": "sqrt(34)-1",
    "fact_spans": "[[[0, 3], [24, 27]], [[4, 18]], [[4, 18]], [[0, 23]], [[24, 40]], [[37, 40]], [[41, 52]], [[41, 52]]]",
    "query_spans": "[[[54, 75]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a moving point on the parabola $y^{2}=16 x$, let $d$ be the distance from $P$ to the directrix of the parabola, and let point $Q(0,-3)$. Then the minimum value of $|P Q|+d$ is?",
    "fact_expressions": "P: Point;G: Parabola;Expression(G) = (y^2 = 16*x);PointOnCurve(P, G);Distance(P, Directrix(G)) = d;d: Number;Q: Point;Coordinate(Q) = (0, -3)",
    "query_expressions": "Min(d + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "5",
    "fact_spans": "[[[2, 5], [29, 32]], [[6, 21], [33, 36]], [[6, 21]], [[2, 27]], [[29, 45]], [[42, 45]], [[46, 56]], [[46, 56]]]",
    "query_spans": "[[[59, 74]]]",
    "process": "The focus of the parabola $ y^{2}=16x $ is $ F(4,0) $, and the directrix is $ x=-4 $. As shown in the figure, draw $ PM $ perpendicular to the directrix at point $ M $, then $ |PF|=|PM|=d $. Therefore, $ |PQ|+d=|PQ|+|PF| $. From the figure, it can be seen that when points $ P $, $ Q $, and $ F $ are collinear, $ |PQ|+|PF| $ reaches its minimum value, which is $ |FQ| $. $ |FQ|=\\sqrt{4^{2}+3^{2}}=5 $. Hence, the minimum value of $ |PQ|+d $ is 5."
  },
  {
    "text": "Given that circle $C$ passes through a vertex and a focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and the center of the circle lies on this hyperbola, then the distance from the center of the circle to the center of the hyperbola is?",
    "fact_expressions": "C: Circle;G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);PointOnCurve(OneOf(Vertex(G)), C);PointOnCurve(OneOf(Focus(G)), C);PointOnCurve(Center(C), G)",
    "query_expressions": "Distance(Center(C), Center(G))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[2, 6]], [[7, 46], [62, 65], [71, 74]], [[7, 46]], [[2, 51]], [[2, 56]], [[2, 66]]]",
    "query_spans": "[[[2, 81]]]",
    "process": "By the geometric properties of the hyperbola, it is clear that circle C passes through the vertex and focus on the same branch of the hyperbola, so the x-coordinate of the center of circle C is 4. Hence, the center coordinates are (4,\\pm\\frac{4\\sqrt{7}}{3}). Therefore, its distance to the center (0,0) is d=\\sqrt{16+\\frac{112}{9}}=\\frac{16}{3}"
  },
  {
    "text": "Given two points $A(-3,0)$, $B(3,0)$, if $|P A|-|P B|=\\pm 4$, then what is the trajectory equation of point $P$?",
    "fact_expressions": "A: Point;Coordinate(A) = (-3, 0);B: Point;Coordinate(B) = (3, 0);P: Point;Abs(LineSegmentOf(P, A)) - Abs(LineSegmentOf(P, B)) = pm*4",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2/4 - y^2/5 = 1",
    "fact_spans": "[[[4, 13]], [[4, 13]], [[15, 23]], [[15, 23]], [[47, 51]], [[25, 44]]]",
    "query_spans": "[[[47, 58]]]",
    "process": "Let the coordinates of point P be (x, y). Since |PA| - |PB| = ±4 ⇒ ||PA| - |PB|| = 4, the locus of point P is a hyperbola with foci on the x-axis, where c = 3, 2a = 4 ⇒ a = 2, so b = \\sqrt{c^{2} - a^{2}} = \\sqrt{5}. Therefore, the equation of the locus of point P is: \\frac{x^{2}}{4} - \\frac{y^{2}}{5} = 1"
  },
  {
    "text": "A point $P$ on the hyperbola $\\frac{x^{2}}{36}-\\frac{y^{2}}{45}=1$ has a distance of $13$ to the left focus $F_{1}$. Then, what is the distance from point $P$ to the right focus $F_{2}$?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2:Point;Expression(G) = (x^2/36 - y^2/45 = 1);PointOnCurve(P, G);LeftFocus(G) =F1;RightFocus(G)=F2;Distance(P, F1) = 13",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "25",
    "fact_spans": "[[[0, 40]], [[43, 46], [67, 71]], [[50, 57]], [[75, 82]], [[0, 40]], [[0, 46]], [[0, 57]], [[0, 82]], [[43, 65]]]",
    "query_spans": "[[[67, 87]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the ellipse such that $P F_{2}$ is perpendicular to the $x$-axis. If the slope of line $P F_{1}$ is $\\frac{\\sqrt{3}}{3}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;P: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P,F2),xAxis);Slope(LineOf(P,F1)) = sqrt(3)/3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[2, 54], [84, 86], [144, 146]], [[4, 54]], [[4, 54]], [[63, 70]], [[79, 83]], [[71, 78]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 78]], [[2, 78]], [[79, 87]], [[89, 104]], [[106, 141]]]",
    "query_spans": "[[[144, 152]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ are?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(pm*sqrt(5),0)",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 44]]]",
    "process": "From the given, $a^{2}=9$, $b^{2}=4$, $c^{2}=a^{2}-b^{2}=5$, so $c=\\pm\\sqrt{5}$, hence the coordinates of the foci are $(\\pm\\sqrt{5},0)$"
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{3+k}+\\frac{y^{2}}{2-k}=1$ represents a hyperbola, what is the range of values for $k$?",
    "fact_expressions": "G: Hyperbola;Expression(G)=(x^2/(k + 3) + y^2/(2 - k) = 1);k:Number",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-\\infty,-3)+(2,+\\infty)",
    "fact_spans": "[[[45, 48]], [[2, 48]], [[50, 53]]]",
    "query_spans": "[[[50, 60]]]",
    "process": ""
  },
  {
    "text": "The equation of a circle whose center lies on the line $y = x$, in the first quadrant, passing through the point $(-1, 2)$, and with a chord length of $4 \\sqrt{2}$ intercepted by the $x$-axis is?",
    "fact_expressions": "G: Circle;H: Line;Expression(H) = (y = x);PointOnCurve(Center(G), H);Quadrant(Center(G)) = 1;P: Point;Coordinate(P) = (-1, 2);PointOnCurve(P, G);Length(InterceptChord(xAxis, G)) = 4*sqrt(2)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "(x-3)^2+(y-3)^2=17",
    "fact_spans": "[[[58, 59]], [[3, 10]], [[3, 10]], [[0, 59]], [[0, 59]], [[23, 32]], [[23, 32]], [[21, 59]], [[34, 59]]]",
    "query_spans": "[[[58, 64]]]",
    "process": "Let the center of the circle be (a, a) and the radius be r. From the given conditions, a > 0, and the circle passes through the point (-1, 2), then (a+1)^{2}+(a-2)^{2}=r^{2}. The chord length intercepted by the x-axis is 4\\sqrt{2}, so a_{2}+(\\frac{4\\sqrt{2}}{2})^{2}=r^{2}. From \\begin{cases}(a+1)\\\\\\end{cases}^{2}+(a-2), solving gives \\begin{cases}a=3\\\\r=\\sqrt{17}\\end{cases}. Hence, the equation of the circle is (x-3)^{2}+(y-3)^{2}=17"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$, $F_{2}$ respectively, the right vertex is $A$, and point $P$ lies on the line $l$ passing through $A$ and perpendicular to the $x$-axis. When the area of the circumcircle of $\\Delta F_{1} F_{2} P$ is minimized, the length $PA$ equals the semi-focal distance. Then the eccentricity of the ellipse is?",
    "fact_expressions": "l: Line;G: Ellipse;b: Number;a: Number;P: Point;A: Point;F1: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;RightVertex(G)=A;PointOnCurve(P,l);PointOnCurve(A,l);IsPerpendicular(l,xAxis);WhenMin(Area(CircumCircle(TriangleOf(F1,F2,P))));Length(LineSegmentOf(P,A))=HalfFocalLength(G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[104, 109]], [[2, 54], [159, 161]], [[4, 54]], [[4, 54]], [[86, 90]], [[92, 95], [82, 85]], [[62, 69]], [[70, 77]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 77]], [[2, 77]], [[2, 85]], [[86, 110]], [[91, 109]], [[96, 109]], [[111, 144]], [[2, 156]]]",
    "query_spans": "[[[159, 167]]]",
    "process": "Let the circumcenter of $\\triangle F_{1}F_{2}P$ be $O$, then $O'$ lies on the perpendicular bisector of $F_{1}F_{2}$. Since $P$ lies on $x=a$ and $O$ lies on the $y$-axis, $|OP|\\geqslant a$, i.e., the area is minimized when the circumradius of $\\triangle F_{1}F_{2}P$ is $|OP|=a$. From the given conditions, $|PA|=|OO'|=c$, $|OP|=|OF_{2}|=a$. In the right triangle $O'OF_{2}$, $c^{2}+c^{2}=a^{2}$, so $e=\\frac{c}{a}=\\frac{\\sqrt{2}}{2}$."
  },
  {
    "text": "What is the focal distance of the ellipse $x^{2}+\\frac{y^{2}}{4}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + y^2/4 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[0, 27]], [[0, 27]]]",
    "query_spans": "[[[0, 32]]]",
    "process": "Ellipse $x^{2}+\\frac{y^{2}}{4}=1$, $a=2$, $b=1$, then $c=\\sqrt{4-1}=\\sqrt{3}$. The focal distance of the ellipse $x^{2}+\\frac{y^{2}}{4}=1$ is: $2\\sqrt{3}$"
  },
  {
    "text": "The focus of the parabola $M$: $y^{2}=8x$ is $F$, and an asymptote of the hyperbola $x^{2}-y^{2}=1$ intersects the parabola $M$ at points $A$ and $B$. Then, the area of $\\triangle ABF$ is?",
    "fact_expressions": "G: Hyperbola;M: Parabola;A: Point;B: Point;F: Point;Expression(G) = (x^2 - y^2 = 1);Expression(M) = (y^2 = 8*x);Focus(M) = F;Intersection(OneOf(Asymptote(G)), M) = {A, B}",
    "query_expressions": "Area(TriangleOf(A, B, F))",
    "answer_expressions": "8",
    "fact_spans": "[[[26, 44]], [[0, 18], [51, 57]], [[59, 62]], [[63, 66]], [[22, 25]], [[26, 44]], [[0, 18]], [[0, 25]], [[26, 68]]]",
    "query_spans": "[[[70, 92]]]",
    "process": "Given F(2,0), one asymptote of the hyperbola is y = x. From \\begin{cases}y=x\\\\y^2=8x\\end{cases}, we obtain \\begin{cases}x=0\\\\y=0\\end{cases} or \\begin{cases}x=8\\\\y=8\\end{cases}, that is, A(0,0), B(8,8). Therefore, S_{\\triangle ABF} = \\frac{1}{2} \\times 2 \\times 8 = 8."
  },
  {
    "text": "Given the curve $Q$: $\\frac{x^{2}}{2 a^{2}}-\\frac{y^{2}}{a^{2}}=1(x>0)$, points $A$ and $B$ lie on curve $Q$, and the equation of the circle with $AB$ as diameter is $(x-2)^{2}+(y-1)^{2}=1$. Then $a$=?",
    "fact_expressions": "G: Circle;Q: Curve;a: Number;A: Point;B: Point;Expression(Q) = (x^2/(2*a^2) - y^2/a^2 = 1)&(x>0);Expression(G)=((x-2)^2+(y-1)^2=1);PointOnCurve(A, Q);PointOnCurve(B, Q);IsDiameter(LineSegmentOf(A,B),G)",
    "query_expressions": "a",
    "answer_expressions": "pm*(sqrt(3)/2)",
    "fact_spans": "[[[87, 88]], [[2, 59], [69, 74]], [[118, 121]], [[60, 64]], [[65, 68]], [[2, 59]], [[87, 115]], [[60, 75]], [[65, 75]], [[77, 88]]]",
    "query_spans": "[[[118, 123]]]",
    "process": "Since AB is the diameter of the circle, it must pass through the center of the circle at point (2,1). Let the equation of line AB be $ l_{AB}: y - 1 = k(x - 2) $. Let the coordinates of points A and B be $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, both lying on Q, hence  \n$$\n\\begin{cases}\n\\frac{x^{2}}{2a^{2}} - \\frac{y^{2}}{a^{2}} = 1 \\\\\n\\frac{x^{2}}{2a} - \\frac{y^{2}}{a^{2}} = 1\n\\end{cases}\n$$\n(since (2,1) is the midpoint of AB), i.e., $ k = $ Solving the system of equations of line AB and Q:  \n$$\n\\begin{cases}\ny = x - 1 \\\\\n\\frac{x^{2}}{2a^{2}} - \\frac{y^{2}}{a^{2}} = 1\n\\end{cases}\n\\Rightarrow x^{2} - 4x + 2 + 2a^{2} = 0\n$$\nAlso $ |AB| = 2 $, i.e., $ |AB|^{2} = 4 $, i.e., $ (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} = 4 $. Since $ y_{1} - y_{2} = x_{1} - x_{2} $, we have  \n$ 4 = 2(x_{1} - x_{2})^{2} = 2[(x_{1} + x_{2})^{2} - 4x_{1}x_{2}] = 2[4^{2} - 4(2 + 2a^{2})] $,  \ni.e., $ 8 - 8a^{2} = 2 $, $ \\therefore a = \\pm \\frac{\\sqrt{3}}{2} $"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, a line $l$ passing through $F_{1}$ is perpendicular to one asymptote of the hyperbola $C$, intersecting the left and right branches of the hyperbola at points $Q$ and $P$ respectively, and $|P Q|-|P F_{2}|=a$. What is the equation of the asymptotes of the hyperbola $C$?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;P: Point;Q: Point;F2: Point;F1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1,l);IsPerpendicular(l,OneOf(Asymptote(C)));Intersection(l,LeftPart(C))=Q;Intersection(l,RightPart(C))=P;-Abs(LineSegmentOf(P, F2)) + Abs(LineSegmentOf(P, Q)) = a",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*((sqrt(5)+1)/2)*x",
    "fact_spans": "[[[95, 100]], [[18, 79], [101, 107], [117, 120], [159, 165]], [[26, 79]], [[26, 79]], [[132, 135]], [[128, 131]], [[10, 17]], [[2, 9], [87, 94]], [[26, 79]], [[26, 79]], [[18, 79]], [[2, 85]], [[2, 85]], [[86, 100]], [[95, 115]], [[95, 137]], [[95, 137]], [[139, 158]]]",
    "query_spans": "[[[159, 173]]]",
    "process": "The line $ l $ passing through $ F_{1} $ is perpendicular to one asymptote of the hyperbola $ C $, with foot of perpendicular at $ A $. It follows that $ |F_{1}A| = b $, $ \\cos\\angle QF_{1}O = \\frac{b}{c} $. Also, $ |PQ| - |PF_{2}| = a $, so $ |PF_{1}| - |QF_{1}| - |PF_{2}| = a $. Since $ |PF_{1}| - |PF_{2}| = 2a $, it follows that $ |QF_{1}| = a $, $ |QF_{2}| = 3a $. In $ \\triangle QF_{1}O $, applying the law of cosines gives: $ 9a^{2} = a^{2} + 4c^{2} - 2a \\times 2c \\times \\frac{b}{c} $, which simplifies to $ 2a^{2} = c^{2} - ab $, or $ a^{2} + ab - b^{2} = 0 $. Solving yields $ \\frac{b}{a} = \\frac{\\sqrt{5}+1}{2} $. Hence, the equations of the asymptotes are: $ y = \\pm\\frac{\\sqrt{5}+1}{2}x $."
  },
  {
    "text": "Given that the right focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b}=1$ is $(3,0)$, what is the equation of the asymptotes of this hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;H: Point;Expression(G) = (x^2/4 - y^2/b = 1);Coordinate(H) = (3, 0);RightFocus(G) = H",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*x*sqrt(5)/2",
    "fact_spans": "[[[2, 40], [55, 58]], [[5, 40]], [[45, 52]], [[2, 40]], [[45, 52]], [[2, 52]]]",
    "query_spans": "[[[55, 66]]]",
    "process": ""
  },
  {
    "text": "A moving circle is externally tangent to both circles $(x+2)^{2}+y^{2}=1$ and $(x-2)^{2}+y^{2}=4$. What is the trajectory equation of the center of the moving circle?",
    "fact_expressions": "G: Circle;Expression(G) = (y^2 + (x + 2)^2 = 1);C:Circle;Expression(C)=((x-2)^2+y^2=4);H:Circle;IsOutTangent(H,G);IsOutTangent(H,C)",
    "query_expressions": "LocusEquation(Center(H))",
    "answer_expressions": "(x^2/(1/4)-y^2/(15/4)=1)&(x<0)",
    "fact_spans": "[[[6, 25]], [[6, 25]], [[26, 45]], [[26, 45]], [[1, 3], [50, 52]], [[0, 48]], [[0, 48]]]",
    "query_spans": "[[[50, 61]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$ $(a>b>0)$, the upper and lower foci are $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{1}$ and perpendicular to the $y$-axis intersects the ellipse at points $A$ and $B$. The line $A F_{2}$ intersects the ellipse again at point $C$. If $S_{\\triangle A B C}=3 S_{\\triangle B C F_{2}}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;a: Number;b: Number;H: Line;A: Point;F2: Point;B: Point;C: Point;F1: Point;a > b;b > 0;Expression(G) = (x^2/b^2 + y^2/a^2 = 1);UpperFocus(G) = F1;LowerFocus(G) = F2;PointOnCurve(F1, H);IsPerpendicular(H, yAxis);Intersection(H, G) = {A, B};Intersection(LineOf(A,F2),G)={A,C};Area(TriangleOf(A, B, C)) = 3*Area(TriangleOf(B, C, F2));Negation(A=C)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/5",
    "fact_spans": "[[[2, 54], [102, 104], [127, 129], [191, 193]], [[4, 54]], [[4, 54]], [[99, 101]], [[105, 108]], [[73, 80]], [[109, 112]], [[136, 139]], [[63, 70], [83, 90]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 80]], [[2, 80]], [[82, 101]], [[91, 101]], [[99, 114]], [[115, 139]], [[141, 189]], [113, 136]]",
    "query_spans": "[[[191, 199]]]",
    "process": "From the given conditions, we have $F_{1}(0,c)$, $F_{2}(0,-c)$. Since $AB$ passes through point $F_{1}$ and is perpendicular to the $y$-axis, the coordinates of point $A$ are $(-\\frac{b^{2}}{a},c)$. Also, $S_{\\Delta ABC}=3S_{ABCF_{2}}$, so $|AF_{2}|=2|F_{2}C|$, i.e., $\\overrightarrow{AF}_{2}=2\\overrightarrow{F_{2}C}$. Therefore, $\\overrightarrow{F_{2}C}=\\frac{1}{2}\\overrightarrow{AF_{2}}=\\frac{1}{2}(\\frac{b^{2}}{a},-2c)=(\\frac{b^{2}}{2a},-c)$. Hence, point $C$ is $(\\frac{b^{2}}{2a},-2c)$. Since point $C$ lies on the ellipse, $\\frac{4c^{2}}{a^{2}}+\\frac{b^{2}}{4a^{2}}=1$. Using $b^{2}=a^{2}-c^{2}$, we get $\\frac{4c^{2}}{a^{2}}+\\frac{a^{2}-c^{2}}{4a^{2}}=1$, which simplifies to $4e^{2}-\\frac{1}{4}e^{2}+\\frac{1}{4}=1$. Solving gives $e=\\frac{\\sqrt{5}}{5}$."
  },
  {
    "text": "Given that the distance from the focus $F$ of the parabola $C$: $y^{2}=2 p x$ ($p>0$) to its directrix is equal to the length of the major axis of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, then the standard equation of the parabola is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1);Distance(F, Directrix(C)) = Length(MajorAxis(G))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=12*x",
    "fact_spans": "[[[2, 28], [86, 89]], [[2, 28]], [[10, 28]], [[10, 28]], [[31, 34]], [[2, 34]], [[41, 78]], [[41, 78]], [[2, 84]]]",
    "query_spans": "[[[86, 96]]]",
    "process": "In the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, $a=3$, $b=2$, so the major axis length is 6. Since the distance from the focus $F$ of the parabola to the directrix is equal to the major axis length of the ellipse, we have $p=6$. Hence, the standard equation of the parabola is $y^{2}=12x$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$, point $M(-1,1)$, a line passing through the focus of $C$ with slope $k$ intersects $C$ at points $A$ and $B$. If $\\overrightarrow{M A} \\cdot \\overrightarrow{M B}=0$, then the real value of $k$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);M: Point;Coordinate(M) = (-1, 1);L: Line;PointOnCurve(Focus(C), L) = True;Slope(L) = k;k: Real;Intersection(L, C) = {A, B};A: Point;B: Point;DotProduct(VectorOf(M, A), VectorOf(M, B)) = 0",
    "query_expressions": "k",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 21], [34, 37], [51, 54]], [[2, 21]], [[22, 32]], [[22, 32]], [[48, 50]], [[33, 50]], [[41, 50]], [[44, 47], [120, 125]], [[48, 65]], [[56, 59]], [[60, 63]], [[67, 118]]]",
    "query_spans": "[[[120, 129]]]",
    "process": "Since the focus of the parabola $ C: y^{2} = 4x $ is $ F(1,0) $, the line passing through the focus of $ C $ with slope $ k $ is $ y = k(x - 1) $. Solving the system of equations  \n\\[\n\\begin{cases}\ny^{2} = 4x \\\\\ny = k(x - 1)\n\\end{cases}\n\\]\nand eliminating $ x $, we obtain $ ky^{2} - 4y - 4k = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ y_{1} + y_{2} = \\frac{4}{k} $, $ y_{1}y_{2} = -4 $. Solving the system of equations  \n\\[\n\\begin{cases}\ny^{2} = 4x \\\\\ny = k(x - 1)\n\\end{cases}\n\\]\nand eliminating $ y $, we get $ k^{2}x^{2} - (2k^{2} + 4)x + k^{2} = 0 $. From the relationship between roots and coefficients of a quadratic equation, we have $ x_{1} + x_{2} = \\frac{2k^{2} + 4}{k^{2}} $, $ x_{1}x_{2} = 1 $. Since $ \\overrightarrow{MA} \\cdot \\overrightarrow{MB} = 0 $, it follows that $ (x_{1} + 1, y_{1} + 1) \\cdot (x_{2} + 1, y_{2} + 1) = 0 $, so  \n\\[\nx_{1}x_{2} + x_{1} + x_{2} + 1 + y_{1}y_{2} - (y_{1} + y_{2}) + 1 = 0\n\\]\nThus,  \n\\[\n1 + \\frac{2k^{2} + 4}{k^{2}} + 1 - 4 - \\frac{4}{k} + 1 = 0\n\\]\nSolving this equation gives $ k = 2 $."
  },
  {
    "text": "The standard equation of a parabola with vertex at the origin, symmetric axis being the $y$-axis, and focus lying on the line $3x - 4y - 24 = 0$ is?",
    "fact_expressions": "G: Parabola;O: Origin;Vertex(G) = O;SymmetryAxis(G) = yAxis;H: Line;Expression(H) = (3*x - 4*y - 24 = 0);PointOnCurve(Focus(G), H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2=-24*y",
    "fact_spans": "[[[37, 40]], [[3, 5]], [[0, 40]], [[6, 40]], [[19, 35]], [[19, 35]], [[16, 40]]]",
    "query_spans": "[[[37, 47]]]",
    "process": "According to the problem, the standard equation of the parabola is $x^{2}=2my$. In the line $3x-4y-24=0$, setting $x=0$ allows us to find the focus coordinates of the parabola, thereby obtaining the answer. [Detailed solution] $\\because$ the vertex of the parabola is at the origin and the axis of symmetry is the $y$-axis, $\\therefore$ the standard equation of the parabola is $x^{2}=2my$, $\\because$ its focus lies on the line $3x-4y-24=0$, $\\therefore$ setting $x=0$ gives $y=-6$, $\\therefore$ the focus $F(0,-6)$, $\\therefore m=-12$. $\\therefore$ the standard equation of the parabola is $x^{2}=$.A_{1},"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{m^{2}-1}-y^{2}=1(m>1)$, the distance from its right focus to the line $x+y=0$ is $\\sqrt{2}$. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;m: Number;G: Line;m>1;Expression(C) = (x^2/(m^2 - 1) - y^2 = 1);Expression(G) = (x + y = 0);Distance(RightFocus(C), G) = sqrt(2)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 46], [76, 79]], [[10, 46]], [[51, 60]], [[10, 46]], [[2, 46]], [[51, 60]], [[2, 74]]]",
    "query_spans": "[[[76, 85]]]",
    "process": ""
  },
  {
    "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ are given by $x \\pm \\sqrt{3} y=0$, then the eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;e: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(Asymptote(G)) = (x + pm*sqrt(3)*y = 0);Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[0, 57], [86, 89]], [[3, 57]], [[3, 57]], [[92, 95]], [[3, 57]], [[3, 57]], [[0, 57]], [[0, 84]], [[86, 95]]]",
    "query_spans": "[[[92, 97]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the focus of the parabola $(y-1)^{2}=4(x-1)$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = ((y - 1)^2 = 4*(x - 1))",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,1)",
    "fact_spans": "[[[0, 21]], [[0, 21]]]",
    "query_spans": "[[[0, 28]]]",
    "process": ""
  },
  {
    "text": "Let $O$ be the origin, $P$ an arbitrary point on the parabola $y^{2}=2 p x  (p>0)$ with focus $F$, and $Q$ a point on the segment $P F$ such that $2|\\overrightarrow{P Q}|=|\\overrightarrow{Q F}|$. Then the maximum value of the slope of the line $O Q$ is?",
    "fact_expressions": "G: Parabola;p: Number;Q: Point;O: Origin;F: Point;P: Point;p>0;Expression(G) = (y^2 = 2*(p*x));PointOnCurve(P, G);Focus(G)=F;PointOnCurve(Q, LineSegmentOf(P,F));2*Abs(VectorOf(P, Q)) = Abs(VectorOf(Q, F))",
    "query_expressions": "Max(Slope(LineOf(O,Q)))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[22, 45]], [[25, 45]], [[51, 54]], [[1, 4]], [[15, 18]], [[10, 13]], [[25, 45]], [[22, 45]], [[10, 50]], [[14, 45]], [[51, 65]], [[67, 115]]]",
    "query_spans": "[[[117, 133]]]",
    "process": ""
  },
  {
    "text": "The equation of the directrix of the parabola $y^{2}=4 a x$ is $x=-2$, then $a$=?",
    "fact_expressions": "G: Parabola;a: Number;Expression(G) = (y^2 = 4*(a*x));Expression(Directrix(G)) = (x = -2)",
    "query_expressions": "a",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 16]], [[30, 33]], [[0, 16]], [[0, 28]]]",
    "query_spans": "[[[30, 35]]]",
    "process": "Using the equation of the directrix of the parabola, the value of the real number $ a $ can be found. [Detailed solution] The directrix equation of the parabola $ y^{2} = 4ax $ is $ x = -a $, so $ -a = -2 $, therefore, $ a = 2 $."
  },
  {
    "text": "If the distance from the point $(m, 3)$ on the parabola $x^{2}=2 p y(p>0)$ to the focus is $5$, then $m^{2}=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 2*(p*y));p: Number;p>0;H: Point;Coordinate(H) = (m, 3);m: Number;PointOnCurve(H, G);Distance(H, Focus(G)) = 5",
    "query_expressions": "m^2",
    "answer_expressions": "24",
    "fact_spans": "[[[1, 22]], [[1, 22]], [[4, 22]], [[4, 22]], [[24, 33]], [[24, 33]], [[25, 33]], [[1, 33]], [[1, 43]]]",
    "query_spans": "[[[46, 55]]]",
    "process": "Since the point $(m,3)$ lies on the parabola $x^{2}=2py$ $(p>0)$, the distance from the point $(m,3)$ to the focus is $3 - (-\\frac{p}{2}) = 5$. Solving gives $p=4$, so the equation of the parabola is $x^{2}=8y$. Substituting the point $(m,3)$ into $x^{2}=8y$ yields $m^{2}=24$."
  },
  {
    "text": "Given that the line $y = kx - 2$ ($k > 0$) intersects the parabola $C$: $x^2 = 8y$ at points $A$ and $B$, and $F$ is the focus of $C$, if $|FA| = 2|FB|$, then $k = $?",
    "fact_expressions": "G: Line;Expression(G) = (y = k*x - 2);k: Number;k>0;C: Parabola;Expression(C) = (x^2 = 8*y);A: Point;B: Point;Intersection(G, C) = {A, B};F: Point;Focus(C) = F;Abs(LineSegmentOf(F, A)) = 2*Abs(LineSegmentOf(F, B))",
    "query_expressions": "k",
    "answer_expressions": "3*sqrt(2)/4",
    "fact_spans": "[[[2, 18]], [[2, 18]], [[79, 82]], [[4, 18]], [[19, 38], [55, 58]], [[19, 38]], [[41, 44]], [[45, 48]], [[2, 50]], [[51, 54]], [[51, 61]], [[63, 77]]]",
    "query_spans": "[[[79, 84]]]",
    "process": "From |FA|=2|FB| we get: |AM|=2|BN|, so point B is the midpoint of AP, |OB|=\\frac{1}{2}|AF|, |OB|=|BF|, yielding B(2\\sqrt{2},1), thus k_{AB}=k_{PB}=\\frac{1-(-2)}{2\\sqrt{2}-0}=\\frac{3}{2\\sqrt{2}}=\\frac{3\\sqrt{2}}{4}. It is known that the line y=kx-2(k>0) passes through point P(0,-2), which is exactly the intersection point of the directrix l:y=-2 of the parabola C:x^{2}=8y and the y-axis, as shown in the figure below. Draw AM\\botl at M and BN\\botl at N from points A and B respectively. From |FA|=2|FB|, it follows that |AM|=2|BN|. Therefore, point B is the midpoint of AP. Connect OB, then |OB|=\\frac{1}{2}|AF|, so OB=BF, hence the y-coordinate of point B is 1, B(2\\sqrt{2},1). Thus, k_{AB}=k_{PB}=\\frac{1-(-2)}{2\\sqrt{2}-0}=\\frac{3}{2\\sqrt{2}}=\\frac{3\\sqrt{2}}{4}"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-y^{2}=1(a>0)$, one of its asymptotes is tangent to the circle $C$: $(x-1)^{2}+(y+2)^{2}=4$, and the point of tangency lies in the third quadrant. Then $a$=?",
    "fact_expressions": "C1: Circle;C:Hyperbola;a: Number;a>0;Expression(C) = (-y^2 + x^2/a^2 = 1);Expression(C1) = ((x - 1)^2 + (y + 2)^2 = 4);IsTangent(OneOf(Asymptote(C)),C1);Quadrant(TangentPoint(OneOf(Asymptote(C)),C1))=3",
    "query_expressions": "a",
    "answer_expressions": "3/4",
    "fact_spans": "[[[51, 79]], [[2, 44]], [[91, 94]], [[10, 44]], [[2, 44]], [[51, 81]], [[2, 81]], [[2, 89]]]",
    "query_spans": "[[[91, 96]]]",
    "process": "Given that $ y = \\frac{1}{a}x $ is tangent to the circle $ (x-1)^2 + (y+2)^2 = 4 $, the distance from the center to the asymptote is $ d = \\frac{|1 + 2a|}{\\sqrt{1 + a^2}} = 2 $. Solving gives $ a = \\frac{3}{4} $."
  },
  {
    "text": "The equation of the hyperbola passing through the point $A(-2,2)$ and having common asymptotes with the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$ is?",
    "fact_expressions": "G: Hyperbola;C: Hyperbola;A: Point;Expression(G) = (x^2/2 - y^2 = 1);Coordinate(A) = (-2, 2);PointOnCurve(A, C);Asymptote(G) = Asymptote(C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/2 - x^2/4 = 1",
    "fact_spans": "[[[12, 40]], [[47, 50]], [[1, 11]], [[12, 40]], [[1, 11]], [[0, 50]], [[11, 50]]]",
    "query_spans": "[[[47, 54]]]",
    "process": "Using the method of undetermined coefficients, assume the required hyperbola standard equation is \\frac{x^{2}}{2}-y^{2}=\\lambda(\\lambda\\neq0), then substitute point A(-2,2) to solve for the result. [Detailed solution] Assume the standard equation of a hyperbola having common asymptotes with \\frac{x^{2}}{2}-y^{2}=1 is \\frac{x^{2}}{2}-y^{2}=\\lambda(\\lambda\\neq0). Since the hyperbola \\frac{x^{2}}{2}-y^{2}=1 passes through A(-2,2), we have \\frac{4}{2}-4=\\lambda, thus \\lambda=-2. Therefore, the standard equation of the required hyperbola is \\frac{y^{2}}{2}-\\frac{x^{2}}{4}=1"
  },
  {
    "text": "What is the focal distance of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/5 + y^2/4 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 43]]]",
    "process": "According to the ellipse equation, find a, b to solve. Let the focal distance of the ellipse be 2c, the ellipse equation is \\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1 \\therefore a^{2}=5, b^{2}=4, \\therefore c=1"
  },
  {
    "text": "Given two points $A$, $B$ on the parabola $y^{2}=4x$ with focus $F$ such that $\\overrightarrow{A F}=3 \\overrightarrow{F B}$, then the distance from the midpoint of chord $AB$ to the directrix is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);VectorOf(A, F) = 3*VectorOf(F, B);IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A, B)), Directrix(G))",
    "answer_expressions": "8/3",
    "fact_spans": "[[[10, 24]], [[10, 24]], [[3, 6]], [[2, 24]], [[28, 31]], [[33, 36]], [[10, 36]], [[10, 36]], [[38, 83]], [[10, 90]]]",
    "query_spans": "[[[10, 101]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola $y^{2}=-4 \\sqrt{2} x$ is $F$, and the directrix is $l$. The equation of the circle with center $F$ that is tangent to $l$ is?",
    "fact_expressions": "G: Parabola;H: Circle;F: Point;l: Line;Expression(G) = (y^2 = -4*sqrt(2)*x);Focus(G) = F;Directrix(G) = l;Center(H)=F;IsTangent(l,H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x+sqrt(2))^2+y^2=8",
    "fact_spans": "[[[0, 24]], [[54, 55]], [[28, 31], [40, 43]], [[35, 38], [48, 51]], [[0, 24]], [[0, 31]], [[0, 38]], [[39, 55]], [[47, 55]]]",
    "query_spans": "[[[54, 60]]]",
    "process": "Since the parabola equation is $ y^{2} = -4\\sqrt{2}x $, it follows that $ F(-\\sqrt{2}, 0) $, and the directrix equation is $ x = \\sqrt{2} $. Then, the radius of the circle centered at $ F $ and tangent to $ l $ is $ 2\\sqrt{2} $. Hence, the equation of the circle satisfying the condition is $ (x + \\sqrt{2})^{2} + y^{2} = 8 $."
  },
  {
    "text": "If the curve $x^{2}-x y+2 y+k=0$ passes through the point $P(-t, t)$ $(t \\in R)$, then the range of values for $k$ is?",
    "fact_expressions": "G: Curve;k: Number;t: Real;P: Point;Expression(G) = (k + 2*y + x^2 - x*y = 0);Coordinate(P) = (-t, t);PointOnCurve(P, G)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-oo, 1/2]",
    "fact_spans": "[[[1, 22]], [[46, 49]], [[25, 44]], [[24, 44]], [[1, 22]], [[25, 44]], [[1, 44]]]",
    "query_spans": "[[[46, 56]]]",
    "process": "The curve $x^{2}-xy+2y+k=0$ passes through the point $P(-t,t)$, then $t^{2}+t^{2}+2t+k=0$, that is, $k=-2t^{2}-2t=-2(t+\\frac{1}{2})^{2}+\\frac{1}{2}$, hence $k\\in(-\\infty,\\frac{1}{2}]$"
  },
  {
    "text": "The focus of the parabola $y^{2}=2 p x(p>0)$ is $F$, and the chord $A B$ passes through $F$. The midpoint of $A B$ is $M(x_{m}, y_{m})$, where $x_{m}=\\frac{3}{4} p$. Then $y_{m}$=?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;M: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(M) = (xm, ym);xm:Number;ym:Number;Focus(G) = F;IsChordOf(LineSegmentOf(A, B), G);PointOnCurve(F, LineSegmentOf(A, B));MidPoint(LineSegmentOf(A, B)) = M;xm = (3/4)*p",
    "query_expressions": "ym",
    "answer_expressions": "pm*p/2",
    "fact_spans": "[[[0, 21]], [[3, 21]], [[41, 46]], [[41, 46]], [[48, 65]], [[24, 27], [35, 38]], [[3, 21]], [[0, 21]], [[48, 65]], [[68, 89]], [[92, 99]], [[0, 27]], [[0, 34]], [[29, 38]], [[41, 65]], [[68, 89]]]",
    "query_spans": "[[[92, 101]]]",
    "process": "Let A(x₁,y₁), B(x₂,y₂), and the focal chord AB have the equation: x = my + \\frac{p}{2} \\textcircled{1}. Combining this with the parabola equation yields y² - 2pmy - p² = 0, then y₁ + y₂ = 2pm ⇒ m = \\frac{y₁ + y₂}{2p}. Substituting into \\textcircled{1}, the equation of line AB becomes 2px - (y₁ + y₂)y - p² = 0. Since point M(\\frac{3}{4}p, yₘ) is the midpoint of segment AB, \\frac{1}{2}p² = (y₁ + y₂)yₘ = 2yₘ², ∴ yₘ = ±\\frac{p}{2}"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and the two asymptotes are $l_{1}$ and $l_{2}$. A line passing through point $F_{2}$ and perpendicular to $l_{1}$ intersects $l_{1}$ and $l_{2}$ at points $P$ and $Q$ respectively. $O$ is the coordinate origin. If $\\overrightarrow{O F_{2}}+\\overrightarrow{O Q}=2 \\overrightarrow{O P}$ is satisfied, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;Asymptote(G) = {l1, l2};l1: Line;l2: Line;PointOnCurve(F2, V) = True;IsPerpendicular(V, l1) = True;V: Line;Intersection(V, l1) = P;Intersection(V, l2) = Q;P: Point;Q: Point;O: Origin;VectorOf(O, F2) + VectorOf(O, Q) = 2*VectorOf(O, P)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 58], [252, 255]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[67, 75]], [[77, 84], [114, 122]], [[2, 84]], [[2, 84]], [[2, 111]], [[94, 101], [124, 131], [139, 146]], [[104, 111], [149, 156]], [[113, 136]], [[123, 136]], [[134, 136]], [[134, 166]], [[134, 166]], [[157, 160]], [[161, 164]], [[167, 170]], [[179, 249]]]",
    "query_spans": "[[[252, 261]]]",
    "process": "As shown in the figure, without loss of generality, assume that the slope of asymptote $ l_{1} $ is greater than 0. From $ \\overrightarrow{OF_{2}} + \\overrightarrow{OQ} = 2\\overrightarrow{OP} $, it follows that $ P $ is the midpoint of segment $ F_{2}Q $. Since $ OP \\perp F_{2}Q $, we have $ \\angle QOP = \\angle F_{2}OP = \\angle F_{1}OQ $. Also, $ \\angle F_{1}OQ + \\angle F_{2}OP + \\angle QOP = \\pi $, so $ \\angle F_{2}OP = \\frac{\\pi}{3} $. Hence, the slope of line $ l_{1} $ is $ \\sqrt{3} $, i.e., $ \\frac{b}{a} = \\sqrt{3} $, and therefore $ e = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = 2 $."
  },
  {
    "text": "The ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$ has left and right foci $F_{1}$ and $F_{2}$, respectively. A line passing through $F_{2}$ intersects the ellipse at points $A$ and $B$. If the area of $\\triangle A B F_{1}$ is $\\frac{4}{3}$, then $\\frac{|F_{2} A|}{|F_{2} B|}=$?",
    "fact_expressions": "C: Ellipse;G: Line;A: Point;B: Point;F1: Point;F2: Point;Expression(C) = (x^2/2 + y^2 = 1);LeftFocus(C) =F1;RightFocus(C)=F2;PointOnCurve(F2, G);Intersection(G, C) = {A, B};Area(TriangleOf(A, B, F1)) = 4/3",
    "query_expressions": "Abs(LineSegmentOf(F2, A))/Abs(LineSegmentOf(F2, B))",
    "answer_expressions": "{1/3, 3}",
    "fact_spans": "[[[0, 32], [72, 74]], [[69, 71]], [[76, 79]], [[80, 83]], [[41, 48]], [[51, 58], [61, 68]], [[0, 32]], [[0, 59]], [[0, 59]], [[60, 71]], [[69, 85]], [[87, 125]]]",
    "query_spans": "[[[127, 158]]]",
    "process": "As shown in the figure: when the slope of line AB is zero, it clearly does not satisfy the condition. According to the problem, $ F_{1}(-1,0) $, $ F_{2}(1,0) $. Let line AB: $ x = my + 1 $, $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, without loss of generality, assume point A is above the x-axis. From \n\\[\n\\begin{cases}\n\\frac{x^{2}}{2} + y^{2} = 1 \\\\\nx = my + 1\n\\end{cases}\n\\]\nsimplifying yields $ (m^{2} + 2)y^{2} + 2my - 1 = 0 $. Thus, $ y_{1} + y_{2} = -\\frac{2m}{m^{2} + 2} $, $ y_{1}y_{2} = \\frac{-}{m^{2}}\\frac{-1}{+2} $, that is, the area of $ \\triangle ABF_{1} $ is $ \\frac{1}{2} \\times |F_{1}F_{2}| \\times |y_{1} - y_{2}| = \\frac{4}{3} $, so $ |y_{1} - y_{2}| = \\frac{4}{3} $. Then, when $ m = 1 $, $ y_{1} + y_{2} = -\\frac{2}{3} $, so $ y_{1} = +\\frac{4}{m^{2} + 2} = \\frac{16}{9} $, simplifying gives $ 2m^{4} - m^{2} - 1 = 0 $, solving yields $ m^{2} = 1 $, so $ y_{1}y_{2} = -\\frac{1}{3}\\frac{1}{3} $, $ y_{2}y_{2} = -1 $, thus $ \\frac{|F_{2}A|}{|F_{2}B|} = \\left| \\frac{y_{1}}{y_{2}} \\right| = \\frac{1}{3} $. When $ m = -1 $, $ y_{1} + y_{2} = \\frac{2}{3} $, $ y_{1} = 1 $, $ y_{2} = -\\frac{1}{3} $, so $ \\frac{|F_{2}A|}{|F_{2}B|} = \\left| \\frac{y_{1}}{y_{2}} \\right| = 3 $. Similarly, when point B is above the x-axis, $ \\frac{|F_{2}A|}{|F_{2}B|} = \\frac{1}{3} $ or $ 3 $."
  },
  {
    "text": "Given that point $P$ lies on the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$, and $F_{1}$, $F_{2}$ are the two foci of the hyperbola. If $|P F_{1}|+|P F_{2}|=4 \\sqrt{2}$, then the area of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;P: Point;Expression(G) = (x^2/2 - y^2 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2};Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 4*sqrt(2)",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[7, 35], [56, 59]], [[40, 47]], [[48, 55]], [[2, 6]], [[7, 35]], [[2, 39]], [[40, 64]], [[66, 98]]]",
    "query_spans": "[[[101, 131]]]",
    "process": "Assume point P lies on the right branch of the hyperbola. By the definition of a hyperbola, we have |PF₁|⋅|PF₂|=2√2, and |PF₁|+|PF₂|=4√2, thus |PF₁|=3√2, |PF₂|=√2. Also, |F₁F₂|=2c=2√3, therefore cos∠F₁PF₂=(PF₁²+PF₂²−F₁F₂²)/(2⋅PF₁⋅PF₂)=2/3, sin∠F₁PF₂=√(1−cos²∠F₁PF₂)=√5/3. Hence, the area of triangle PF₁F₂ is (1/2)×3√2×√2×(√5/3)=√5."
  },
  {
    "text": "Given that $A$, $B$, $C$ are three points on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $AB$ passes through the origin $O$, $AC$ passes through the right focus $F$, if $BF \\perp AC$ and $3|AF|=|AC|$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;A: Point;B: Point;F: Point;C: Point;O:Origin;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(A, G);PointOnCurve(B, G);PointOnCurve(C, G);PointOnCurve(O,LineSegmentOf(A,B));RightFocus(G)=F;PointOnCurve(F,LineSegmentOf(A,C));IsPerpendicular(LineSegmentOf(B, F), LineSegmentOf(A, C));3*Abs(LineSegmentOf(A, F)) = Abs(LineSegmentOf(A, C))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(17)/3",
    "fact_spans": "[[[14, 70], [139, 142]], [[17, 70]], [[17, 70]], [[2, 5]], [[6, 9]], [[101, 104]], [[10, 13]], [[83, 88]], [[17, 70]], [[17, 70]], [[14, 70]], [[2, 75]], [[2, 75]], [[2, 75]], [[76, 89]], [[14, 104]], [[91, 104]], [[106, 121]], [[122, 136]]]",
    "query_spans": "[[[139, 148]]]",
    "process": "From the given conditions: as shown in the figure, let the left focus be F, |AF| = m, connect AF', BF, CF. Since 3|AF| = |AC|, then |FC| = 2m, |AF| = 2a + m, |CF| = 2a + 2m, |FF'| = 2c. Therefore, quadrilateral FAF'B is a rectangle. In right triangle \\triangle AFC, |AF|^{2} + |AC|^{2} = |CF|^{2}, that is, (2a + m)^{2} + (3m)^{2} = (2a + 2m)^{2}. Simplifying yields m = \\frac{2a}{3}. Also, in right triangle \\triangle AFF', |AF|^{2} + |AF'|^{2} = |FF'|^{2}, so (2a + \\frac{2a}{3})^{2} + (\\frac{2a}{3})^{2} = (2c)^{2}, simplifying gives \\frac{c^{2}}{a^{2}} = \\frac{17}{9}, thus e = \\frac{\\sqrt{17}}{3}."
  },
  {
    "text": "If the hyperbola $C$: $x^{2}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ has eccentricity $2$, then $b=$?",
    "fact_expressions": "C: Hyperbola;b: Number;b>0;Expression(C) = (x^2 - y^2/b^2 = 1);Eccentricity(C) = 2",
    "query_expressions": "b",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[1, 43]], [[53, 56]], [[9, 43]], [[1, 43]], [[1, 51]]]",
    "query_spans": "[[[53, 58]]]",
    "process": "\\because the hyperbola C: x^{2}-\\frac{y^{2}}{b^{2}}=1 (b>0) has eccentricity 2, \\therefore \\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{1}}=2, \\therefore b=\\sqrt{3}"
  },
  {
    "text": "The distance from the focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{5}=1(a>0)$ to its asymptote is?",
    "fact_expressions": "C: Hyperbola;a: Number;a>0;Expression(C) = (-y^2/5 + x^2/a^2 = 1)",
    "query_expressions": "Distance(Focus(C), Asymptote(C))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[0, 52]], [[8, 52]], [[8, 52]], [[0, 52]]]",
    "query_spans": "[[[0, 64]]]",
    "process": "Let the right focus of the hyperbola be $ F(c,0) $, and one of its asymptotes be $ y = \\frac{b}{a}x $, i.e., $ bx - ay = 0 $. Then the distance $ d $ from the point $ (c,0) $ to the line $ bx - ay = 0 $ is $ d = \\frac{|bc|}{\\sqrt{a^{2}+b^{2}}} = \\frac{bc}{c} = b $, i.e., the distance from the focus of the hyperbola to the asymptote is $ d = b $. From the hyperbola equation, we know $ C: \\frac{x^2}{a^2} - \\frac{y^{2}}{5} = 1 $ $ (a > 0) $, $ b = \\sqrt{5} $, i.e., the distance from the focus to the asymptote is $ \\sqrt{5} $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ is a moving point on the line $x=-\\frac{a^{2}}{c}$ (not on the coordinate axes). When the maximum value of $\\angle F_{1} P F_{2}$ is $\\frac{\\pi}{6}$, what is the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;H: Line;Expression(H) = (x = -a^2/c);c: Number;PointOnCurve(P, H);Negation(PointOnCurve(P, axis));Max(AngleOf(F1, P, F2)) = pi/6",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 54], [168, 170]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[63, 70]], [[71, 78]], [[2, 78]], [[2, 78]], [[79, 83]], [[84, 106]], [[84, 106]], [[86, 106]], [[79, 112]], [[79, 121]], [[124, 166]]]",
    "query_spans": "[[[168, 176]]]",
    "process": "By the given condition, point P is a moving point on the line $ x = -\\frac{a^2}{c} $. Let point $ P\\left(-\\frac{a^{2}}{c}, m\\right) $. Since $ F_{1}(-c, 0) $, $ F_{2}(c, 0) $, we obtain $ k_{PF_{1}} = -\\frac{b^{2}}{m} $, $ k_{PF_{2}} = -\\frac{c^{2}+a^{2}}{m} $, then $ \\tan\\angle F_{1}PF_{2} \\cdot \\frac{c^2}{2} = \\frac{c}{\\sqrt{(c^{2}+a^{2})b^{2}}} = \\tan\\frac{\\pi}{6} = \\frac{\\sqrt{3}}{3} $. Also, from $ c^{2} = a^{2} - $"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $x^{2}-y^{2}=2$, point $P$ lies on $C$, and $\\angle F_{1} P F_{2}=60^{\\circ}$, then the area of $\\triangle F_{1} P F_{2}$ is?",
    "fact_expressions": "C: Hyperbola;F1: Point;P: Point;F2: Point;Expression(C) = (x^2 - y^2 = 2);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[18, 41], [53, 56]], [[2, 9]], [[48, 52]], [[10, 17]], [[18, 41]], [[2, 47]], [[2, 47]], [[48, 57]], [[58, 91]]]",
    "query_spans": "[[[93, 123]]]",
    "process": "Assume point $ P $ lies on the right branch of the hyperbola, then $ |PF_{1}| - |PF_{2}| = 2a = 2\\sqrt{2} $, $ |F_{1}F_{2}| = 2c = 4 $. In $ \\triangle F_{1}PF_{2} $, by the cosine law,  \n$ \\cos\\angle F_{1}PF_{2} = \\frac{|PF_{1}|^{2} + |PF_{2}|^{2} - |F_{1}F_{2}|^{2}}{2|PF_{1}|\\cdot|PF_{2}|} = \\frac{(|PF_{1}| - |PF_{2}|)^{2} + 2|PF_{1}|\\cdot|PF_{2}| - |F_{1}F_{2}|^{2}}{2|PF_{1}|\\cdot|PF_{2}|} = \\frac{1}{2} $,  \n$ \\therefore |PF_{1}|\\cdot|PF_{2}| = 8 $,  \n$ \\therefore S = \\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|\\sin60^{\\circ} = 2\\sqrt{3} $"
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{m-1}+\\frac{y^{2}}{3-m}=1$ lie on the $x$-axis. What is the range of real values for $m$?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (x^2/(m - 1) + y^2/(3 - m) = 1);PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(2, 3)",
    "fact_spans": "[[[0, 41]], [[52, 57]], [[0, 41]], [[0, 50]]]",
    "query_spans": "[[[52, 64]]]",
    "process": "The foci of the ellipse $\\frac{x^{2}}{m-1}+\\frac{y^{2}}{3-m}=1$ lie on the $x$-axis, $\\therefore m-1>3-m>0$, solving gives: $2<m<3$,"
  },
  {
    "text": "The length of the real axis of the hyperbola $m x^{2}+y^{2}=1$ is twice the length of the imaginary axis. Then $m=?$",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (m*x^2 + y^2 = 1);Length(RealAxis(G)) = 2*Length(ImageinaryAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "-4",
    "fact_spans": "[[[0, 20]], [[35, 38]], [[0, 20]], [[0, 33]]]",
    "query_spans": "[[[35, 40]]]",
    "process": "From the given conditions, the standard equation of the hyperbola $ mx^{2} + y^{2} = 1 $ is $ y^{2} - \\frac{x^{2}}{-\\frac{1}{m}} = 1 $, then $ a^{2} = 1 $, $ b^{2} = -\\frac{1}{m} $, the length of the real axis is $ 2 $, and the length of the imaginary axis is $ 2\\sqrt{-\\frac{1}{m}} $."
  },
  {
    "text": "Given the parabola $y^{2}=8x$ with focus $F$, its directrix intersects the $x$-axis at point $C$. A line passing through point $F$ intersects the parabola at points $A$ and $B$. If $\\angle CBF=90^{\\circ}$, then $|AF|-|BF|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;C: Point;Intersection(Directrix(G), xAxis) = C;H: Line;PointOnCurve(F, H);A: Point;B: Point;Intersection(H, G) = {A, B};AngleOf(C, B, F) = ApplyUnit(90, degree)",
    "query_expressions": "Abs(LineSegmentOf(A, F)) - Abs(LineSegmentOf(B, F))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 16], [24, 25], [48, 51]], [[2, 16]], [[20, 23], [39, 43]], [[2, 23]], [[33, 37]], [[24, 37]], [[44, 46]], [[38, 46]], [[52, 55]], [[56, 59]], [[44, 61]], [[63, 88]]]",
    "query_spans": "[[[90, 105]]]",
    "process": "Set the equation of the line and solve it together with the parabola to find the product of the two roots. If $\\angle CBF=90^{\\circ}$, then $\\overrightarrow{CB}\\cdot\\overrightarrow{FB}=0$, from which the coordinates of $B$ can be obtained, and then the coordinates of $A$ can be found. Based on the property of the parabola: the distance to the focus equals the distance to the directrix, we obtain the conclusion. From the equation of the parabola, we get: focus $F(2,0)$, directrix equation $x=-2$. $\\therefore$ the coordinates of $C$ are $(-2,0)$. By the symmetry of the parabola, assume the slope of line $AB$ is greater than 0, let the equation of line $AB$ be: $y=k(x-2)$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, above the x-axis, i.e., $x_{1}>x_{2}$. Solving the line and parabola equations together gives:\n\\[\n\\begin{cases}\ny=k(x-\\\\\ny^{2}=8x\n\\end{cases}-2)\n\\]\nEliminating $y$ yields: $k^{2}x^{2}-(4k^{2}+8)x+4k^{2}=0$, $x_{1}+x_{2}=\\frac{8+4k^{2}}{k^{2}}$, $x_{1}x_{2}=4$--$\\textcircled{1}$. If $\\angle CBF=90^{\\circ}$, then $\\overrightarrow{CB}\\cdot\\overrightarrow{FB}=0$. That is $(x_{2}+2,y_{2})(x_{2}-2,y_{2})=0$, i.e., $x_{2}^{2}-4+y_{2}^{2}=0$, thus $x_{2}^{2}+8x_{2}-4=0$, giving $x_{2}=-4+2\\sqrt{5}$. Substituting into $\\textcircled{1}$ gives $x_{1}=2\\sqrt{5}+4$. By the property of the parabola: the distance to the focus equals the distance to the directrix, we get $|AF|-|BF|=(x_{1}+2)-(x_{2}+2)=x_{1}-x_{2}=(2\\sqrt{5}+4)-(2\\sqrt{5}-4)=8$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has an eccentricity of $\\frac{\\sqrt{3}}{2}$. A line passing through the right focus $F$ with slope $k$ $(k>0)$ intersects the ellipse $C$ at points $A$ and $B$. If $\\overrightarrow{A F}=3 \\overrightarrow{F B}$, then $k=$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/a^2+y^2/b^2=1);a: Number;b: Number;a>b;b>0;Eccentricity(C) = sqrt(3)/2;F: Point;RightFocus(C) = F;G: Line;PointOnCurve(F, G);k: Number;k>0;Slope(G) = k;A: Point;B: Point;Intersection(G, C) = {A, B};VectorOf(A, F) = 3*VectorOf(F, B)",
    "query_expressions": "k",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 59], [108, 113]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[2, 84]], [[89, 92]], [[2, 92]], [[105, 107]], [[85, 108]], [[175, 178]], [[96, 104], [175, 178]], [[93, 107]], [[116, 119]], [[120, 123]], [[105, 125]], [[128, 173]]]",
    "query_spans": "[[[175, 180]]]",
    "process": "From the given $ e = \\frac{c}{a} = \\sqrt{1 - \\frac{b^{2}}{a^{2}}} = \\frac{\\sqrt{3}}{2} $, we have $ a = 2b $, so $ a = \\frac{2}{\\sqrt{3}}c $, $ b = \\frac{c}{\\sqrt{3}} $. Then the ellipse equation $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ becomes $ \\frac{3}{4}x^{2} + 3y^{2} = c^{2} $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, and $ \\overrightarrow{AF} = 3\\overrightarrow{FB} $, so $ (c - x_{1}, -y_{1}) = 3(x_{2} - c, y_{2}) $, thus  \n$$\n\\begin{cases}\nc - x_{1} = 3(x_{2} - c) \\\\\n-y_{1} = 3y_{2}\n\\end{cases}\n$$\nso  \n$$\n\\begin{cases}\nx_{1} + 3x_{2} = 4c \\\\\ny_{1} + 3y_{2} = 0\n\\end{cases}\n$$\n$$\n\\frac{3}{4}x_{1}^{2} + 3y_{1}^{2} = c^{2} \\quad \\textcircled{1}\n$$\n$$\n\\frac{3}{4}x_{2}^{2} + 3y_{2}^{2} = c^{2} \\quad \\textcircled{2}\n$$\n$ \\textcircled{1} - 9 \\times \\textcircled{2} $ gives  \n$$\n\\frac{3}{4}(x_{1} + 3x_{2})(x_{1} - 3x_{2}) + 3(y_{1} + 3y_{2})(y_{1} - 3y_{2}) = -8c^{2}\n$$\nso  \n$$\n\\frac{3}{4} \\times 4c(x_{1} - 3x_{2}) = -8c^{2}\n$$\nthus  \n$$\nx_{1} - 3x_{2} = -\\frac{8}{3}c\n$$\nso  \n$$\nx_{1} = \\frac{2}{3}c, \\quad x_{2} = \\frac{10}{9}c\n$$\nand  \n$$\ny_{1} = -\\frac{\\sqrt{2}}{3}c, \\quad y_{2} = \\frac{\\sqrt{2}}{9}c\n$$\ntherefore  \n$$\nA\\left( \\frac{2}{3}c, -\\frac{\\sqrt{2}}{3}c \\right), \\quad B\\left( \\frac{10}{9}c, \\frac{\\sqrt{2}}{9}c \\right)\n$$\nhence $ k = \\sqrt{2} $"
  },
  {
    "text": "Let the focus of the parabola $x^{2}=4 y$ be $F$, and let $A$ be a point on the parabola in the first quadrant such that $|A F|=2$. Given that $P$ is any point on the directrix of the parabola, when $|P A|+|P F|$ attains its minimum value, what is the circumradius of $\\Delta P A F$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);F: Point;Focus(G) = F;A: Point;Quadrant(A) = 1;PointOnCurve(A, G);Abs(LineSegmentOf(A, F)) = 2;P: Point;PointOnCurve(P, Directrix(G));WhenMin(Abs(LineSegmentOf(P, A))+ Abs(LineSegmentOf(P, F)))",
    "query_expressions": "Radius(CircumCircle(TriangleOf(P, A, F)))",
    "answer_expressions": "5/4",
    "fact_spans": "[[[1, 15], [27, 30], [57, 60]], [[1, 15]], [[19, 22]], [[1, 22]], [[23, 26]], [[23, 38]], [[23, 38]], [[41, 50]], [[53, 56]], [[53, 66]], [[67, 87]]]",
    "query_spans": "[[[88, 110]]]",
    "process": "Analysis: According to the definition of the parabola, we have |AF| = y_{0} + \\frac{p}{2} = y_{0} + 1 = 2. Solving gives y_{0} = 1, so A(2,1). Let F(0,1) be the focus of the parabola. The symmetric point of F(0,1) with respect to the directrix y = -1 is F_{1}(0,-3). Connect AF_{1}, intersecting the directrix y = -1 at point P, such that |PA| + |PF| reaches its minimum value. At this time, the coordinates of point P are (1,-1). In triangle APF, applying the law of cosines and the law of sines yields the solution. Detailed: From the equation of the parabola x^{2} = 4y, we know F(0,1). Let A(x_{0},y_{0}). Given |AF| = 2, by the definition of the parabola, |AF| = y_{0} + \\frac{p}{2} = y_{0} + 1 = 2. Solving gives y_{0} = 1. Substituting into the parabola's equation yields x_{0} = 2, thus A(2,1). Taking the focus F(0,1), the symmetric point with respect to the directrix y = -1 is F_{1}(0,-3). Connecting AF_{1} intersects the directrix y = -1 at point P, where |PA| + |PF| reaches its minimum. At this time, the coordinates of point P are (1,-1). By the law of cosines, \\cos\\angleAPF = \\frac{(\\sqrt{5})+(\\sqrt{5})}{2}\\times\\sqrt{5}\\times\\sqrt{5}-\\frac{3}{5}, then \\sin\\angleAPF = \\frac{4}{5}. By the law of sines, 2R = \\frac{|AF|}{\\sin\\angleAPF} = \\frac{2\\times5}{4} = \\frac{5}{2}, so R = \\frac{5}{4}, i.e., the circumradius of the triangle is R = \\frac{5}{4}."
  },
  {
    "text": "Given that the center of the ellipse is at the origin, a focus on the $x$-axis is connected to the two endpoints of the minor axis by lines that are perpendicular to each other, and the shortest distance from this focus to the ellipse is $4(\\sqrt{2}-1)$. Find the equation of the ellipse?",
    "fact_expressions": "G: Ellipse;O:Origin;Center(G)=O;F:Point;OneOf(Focus(G))=F;PointOnCurve(F,xAxis);A:Point;B:Point;Endpoint(MinorAxis(G))={A,B};IsPerpendicular(LineSegmentOf(F,A),LineSegmentOf(F,B));Min(Distance(F,G))=4*(sqrt(2)-1)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 4], [10, 11], [43, 45], [69, 71]], [[7, 9]], [[2, 9]], [], [[10, 22]], [[10, 22]], [], [], [[10, 28]], [[10, 35]], [[10, 67]]]",
    "query_spans": "[[[69, 76]]]",
    "process": ""
  },
  {
    "text": "It is known that the length of the major axis of an ellipse is $2$ times the length of the minor axis. What is the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;Length(MajorAxis(G)) = 2 * Length(MinorAxis(G))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[2, 4], [2, 4]], [[2, 17]]]",
    "query_spans": "[[[19, 28]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$ and directrix $l$, let point $A \\in l$, and segment $AF$ intersect the parabola $C$ at point $B$. If $\\overrightarrow{FA}=3\\overrightarrow{FB}$, then $|\\overrightarrow{AF}|=$?",
    "fact_expressions": "C: Parabola;A: Point;F: Point;B: Point;l: Line;Expression(C) = (y^2 = 4*x);Focus(C) = F;Directrix(C) = l;In(A, l);Intersection(LineSegmentOf(A,F), C) = B;VectorOf(F, A) = 3*VectorOf(F, B)",
    "query_expressions": "Abs(VectorOf(A, F))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 20], [54, 60]], [[35, 45]], [[24, 27]], [[61, 65]], [[31, 34]], [[2, 20]], [[2, 27]], [[2, 34]], [[35, 45]], [[46, 65]], [[67, 112]]]",
    "query_spans": "[[[113, 139]]]",
    "process": "Draw a perpendicular from B to the directrix l, with foot of perpendicular at B_{1}, and let the directrix l intersect the x-axis at point M. Since AABB_{1} is similar to AAFM, we have: \\frac{|AB|}{|AF|}=\\frac{|B_{1}B|}{|MF|}. Let |FB|=m. By the definition of the parabola, |B_{1}B|=m, |AB|=2m, |AF|=3m. Substituting into \\frac{|AB|}{|AF|}=\\frac{|B_{1}B|}{|MF|} gives: \\frac{2m}{3m}=\\frac{m}{2}, so m=\\frac{4}{3}. Therefore, |\\overrightarrow{AF}|=3m=4."
  },
  {
    "text": "What is the distance from the focus to the directrix of the parabola $x=\\frac{1}{10} y^{2}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x = y^2/10)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "5",
    "fact_spans": "[[[0, 25]], [[0, 25]]]",
    "query_spans": "[[[0, 36]]]",
    "process": "Convert the equation of the parabola into standard form to obtain the result. The standard equation of the parabola is y^{2}=10x, then 2p=10, so p=5. Therefore, the distance from the focus to the directrix of the parabola x=\\frac{1}{10}y^{2} is 5"
  },
  {
    "text": "Given point $A(1,0)$, line $l$: $y=2x-4$, and point $R$ is a point on line $l$. If $\\overrightarrow{RA}=\\overrightarrow{AP}$, then the equation of the trajectory of point $P$ is?",
    "fact_expressions": "l: Line;A: Point;R: Point;P: Point;Expression(l) = (y = 2*x - 4);Coordinate(A) = (1, 0);PointOnCurve(R, l);VectorOf(R, A) = VectorOf(A, P)",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "y=2*x",
    "fact_spans": "[[[12, 28], [34, 39]], [[2, 11]], [[29, 33]], [[91, 95]], [[12, 28]], [[2, 11]], [[29, 43]], [[46, 89]]]",
    "query_spans": "[[[91, 102]]]",
    "process": "Let the coordinates of point P be (x, y) and point R be (m, n), then n = 2m - 4 ①. From \\overrightarrow{RA} = \\overrightarrow{AP}, we get (1 - m, -n) = (x - 1, y). Therefore, 1 - m = x - 1, -n = y, that is, m = 2 - x, n = -y. Substituting into ① gives y = 2(2 - x) - 4, simplifying yields y = 2x."
  },
  {
    "text": "Through a point $P$ on the right branch of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$), draw two lines parallel to the asymptotes, intersecting the other asymptote at points $M$ and $N$, respectively. Let $O$ be the origin. Denote the area of $\\Delta O M N$ by $S$. If $S \\geq \\frac{b^{2}}{2}$, then the range of the eccentricity of hyperbola $C$ is (answer in interval notation)?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;P:Point;PointOnCurve(P, RightPart(C)) = True;L1: Line;L2: Line;PointOnCurve(P, L1) = True;PointOnCurve(P, L2) = True;J1: Line;J2: Line;Asymptote(C) = {J1, J2};IsParallel(L1, J1) = True;IsParallel(L2, J2) = True;Intersection(L1, J2) = M;Intersection(L2, J1) = N;M: Point;N: Point;O: Origin;Area(TriangleOf(O, M, N)) = S;S: Number;S >= b^2/2",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(1,\\sqrt{5}/2]",
    "fact_spans": "[[[1, 63], [159, 165]], [[1, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[68, 71]], [[1, 71]], [], [], [[0, 81]], [[0, 81]], [], [], [[1, 77]], [[0, 81]], [[0, 81]], [[0, 99]], [[0, 99]], [[91, 95]], [[96, 99]], [[100, 103]], [[110, 131]], [[128, 131]], [[133, 157]]]",
    "query_spans": "[[[159, 183]]]",
    "process": "Let $ P(m,n) $, and $ y = -\\frac{b}{a}x + d $ be the line passing through $ P $ parallel to the asymptote $ y = -\\frac{b}{a}x $, intersecting the $ y $-axis at point $ D(0,d) $, and intersecting the asymptote $ y = \\frac{b}{a}x $ at point $ M(x_{1},y_{1}) $. Solving the system \n\\[\n\\begin{cases}\ny = -\\frac{b}{a}x + d \\\\\ny = \\frac{b}{a}x\n\\end{cases}\n\\]\ngives $ x_{1} = \\frac{bm + an}{2b} $. Then $ S_{\\triangle DOM} = \\frac{1}{2}|x_{1} \\cdot d| $. From the problem, quadrilateral $ OMPN $ is a parallelogram. Since $ P(m,n) $ lies on the hyperbola, it satisfies $ \\frac{m^{2}}{a^{2}} - \\frac{n^{2}}{b^{2}} = 1 $, i.e., $ b^{2}m^{2} - a^{2}n^{2} = a^{2}b^{2} $. Then \n\\[\nS_{OMPN} = 2S_{\\Delta OMP} = 2(S_{\\Delta OPD} - S_{\\Delta DMO}) = |md| - |x_{1} \\cdot d| = |(m - x_{1})d| = \\left| \\frac{(bm - an)(bm + an)}{2ab} \\right| = \\frac{a^{2}b^{2}}{2ab} = \\frac{ab}{2}\n\\]\nThus $ S = \\frac{1}{2}S_{OMPN} = \\frac{ab}{4} \\geqslant \\frac{b^{2}}{2} $, solving gives $ \\frac{b}{a} \\leqslant \\frac{1}{2} $, so the eccentricity $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} \\leqslant \\frac{\\sqrt{5}}{2} $. Therefore, the range of eccentricity is $ \\left(1, \\frac{\\sqrt{5}}{2}\\right] $."
  },
  {
    "text": "Given that an asymptote of the hyperbola $k x^{2}-y^{2}=1$ is perpendicular to the line $2 x+y+1=0$, then what is the eccentricity of the hyperbola? What are the equations of the asymptotes?",
    "fact_expressions": "G: Hyperbola;k: Number;H: Line;Expression(G) = (k*x^2 - y^2 = 1);Expression(H) = (2*x + y + 1 = 0);IsPerpendicular(OneOf(Asymptote(G)),H)",
    "query_expressions": "Eccentricity(G);Expression(Asymptote(G))",
    "answer_expressions": "sqrt(5)/2\nx/2 + pm*y=0",
    "fact_spans": "[[[2, 22], [47, 50]], [[5, 22]], [[29, 42]], [[2, 22]], [[29, 42]], [[2, 44]]]",
    "query_spans": "[[[47, 56]], [[47, 63]]]",
    "process": ""
  },
  {
    "text": "The line $l$ passes through the point $(a, 0)$ and is perpendicular to the $x$-axis. If the line $l$ is intersected by the parabola $y^{2}=4 a x$ to form a line segment of length $4$, then the coordinates of the focus of the parabola are?",
    "fact_expressions": "l: Line;P: Point;Coordinate(P) = (a, 0);a: Number;PointOnCurve(P,l);IsPerpendicular(l,xAxis);Length(InterceptChord(L, C)) = 4;C: Parabola;Expression(C) = (y^2 = 4*(a*x))",
    "query_expressions": "Coordinate(Focus(C))",
    "answer_expressions": "{(1,0),(-1,0)}",
    "fact_spans": "[[[0, 5], [25, 30]], [[6, 15]], [[6, 15]], [[7, 15]], [[0, 15]], [[0, 23]], [[25, 56]], [[31, 47], [58, 61]], [[31, 47]]]",
    "query_spans": "[[[58, 68]]]",
    "process": "From the given conditions, we know that $ a \\neq 0 $. When $ a > 0 $,  \n$$\n\\begin{cases}\nx = a \\\\\ny^2 = 4a^2\n\\end{cases}\n\\Rightarrow y = \\pm 2a,\n$$\nso $ 4a = 4 $, $ a = 1 $, and the focus of the parabola is $ (1, 0) $. When $ a < 0 $, $ x = a $, $ y^2 = 4a^2 \\Rightarrow y = \\pm 2a $, so $ 4a = 4 $, $ a = 1 $, and the focus of the parabola is $ (1, 0) $."
  },
  {
    "text": "Given that $F_{1}(0,-2)$, $F_{2}(0,2)$ are the two foci of an ellipse, and a chord $AB$ of the ellipse passes through $F_{2}$. If the perimeter of $\\triangle A F_{1} B$ is $16$, then what is the standard equation of the ellipse?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;F1: Point;F2: Point;Coordinate(F1) = (0, -2);Coordinate(F2) = (0 , 2);Focus(G)={F1,F2};PointOnCurve(F2, LineSegmentOf(A, B));IsChordOf(LineSegmentOf(A, B), G);Perimeter(TriangleOf(A, F1, B)) = 16",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/12 + y^2/16 = 1",
    "fact_spans": "[[[31, 33], [48, 50], [91, 93]], [[52, 57]], [[52, 57]], [[2, 15]], [[18, 30], [40, 47]], [[2, 15]], [[18, 30]], [[2, 38]], [[39, 57]], [[48, 57]], [[59, 88]]]",
    "query_spans": "[[[91, 100]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ with left vertex $A$. A line $MN$ passing through the point $(-\\frac{6}{5}, 0)$ intersects the ellipse at points $M$ and $N$ (both $M$ and $N$ are distinct from $A$). If the slopes of lines $AM$ and $AN$ are $k_{1}$ and $k_{2}$ respectively, then the minimum value of $|4 k_{1}-k_{2}|$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);A: Point;LeftVertex(G) = A;I: Point;Coordinate(I) = (-6/5, 0);M: Point;N: Point;PointOnCurve(I, LineOf(M, N));Intersection(LineOf(M, N), G) = {M, N};Negation(A=M);Negation(A=N);k1: Number;Slope(LineOf(A, M)) = k1;k2: Number;Slope(LineOf(A, N)) = k2",
    "query_expressions": "Min(Abs(4*k1 - k2))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 29], [68, 70]], [[2, 29]], [[34, 37], [95, 99]], [[2, 37]], [[39, 59]], [[39, 59]], [[72, 76], [84, 87]], [[77, 80], [89, 92]], [[38, 67]], [[60, 82]], [[84, 99]], [[84, 99]], [[122, 129]], [[102, 138]], [[130, 138]], [[102, 138]]]",
    "query_spans": "[[[140, 163]]]",
    "process": "When the slope of MN does not exist, we can obtain M(-\\frac{6}{5},\\frac{4}{5}), N(-\\frac{6}{5},-\\frac{4}{5}), and then find k_{1}=1, k_{2}=-1, yielding |4k_{1}-k_{2}|=5; when the slope exists, let the line be y=k(x+\\frac{6}{5}), combine with the ellipse, use Vieta's formulas to obtain k_{1}k_{2}=-1, then use the basic inequality to find the minimum value of |4k_{1}-k_{2}|. Solution: When the slope of line MN does not exist, the line equation is x=-\\frac{6}{5}, obtaining M(-\\frac{6}{5},\\frac{4}{5}), N(-\\frac{6}{5},-\\frac{4}{5}), and A(-2,0), \\therefore k_{1}=\\frac{\\frac{4}{5}-0}{-\\frac{6}{5}+2}=1, k_{2}=\\frac{-\\frac{4}{5}-0}{-\\frac{6}{5}+2}=-1, thus |4k_{1}-k_{2}|=5; when the slope of line MN exists, let the line equation be y=k(x+\\frac{6}{5}), combining with the ellipse equation yields (100k^{2}+25)x^{2}+240k^{2}x+144k^{2}-100=0, \\because (-\\frac{6}{5},0) lies inside the ellipse, then clearly \\Delta>0, let M(x_{1},y_{1}), N(x_{2},y_{2}), then x_{1}+x_{2}, \\therefore y_{1}y_{2}=k^{2}(x_{1}+\\frac{6}{5})(x_{2}+\\frac{6}{5})=k^{2}[x_{1}x_{2}+ \\frac{6}{5}(x_{1}+x_{2})+\\frac{36}{25}], \\therefore k_{1}k_{2}=\\frac{y_{1}}{x_{1}+2}\\cdot\\frac{y_{2}}{x_{2}+2}=\\frac{y_{1}y_{2}}{x_{1}x_{2}+2(x_{1}+x_{2})+4}=-1, then |4k_{1}-k_{2}|=|4k_{1}+\\frac{1}{k_{1}}|=|4k_{1}|+\\frac{1}{|k_{1}|}\\geqslant2\\sqrt{|4k_{1}|\\cdot\\frac{1}{|k_{1}|}}=4, equality holds if and only if |4k_{1}|=\\frac{1}{|k_{1}|}, i.e., k_{1}=\\pm\\frac{1}{2}, in conclusion, the minimum value of |4k_{1}-k_{2}| is 4."
  },
  {
    "text": "It is known that the hyperbola $C$ has its center at the origin of the coordinate system, and the point $F(2,0)$ is one of the foci of the hyperbola $C$. A line $l$ perpendicular to an asymptote is drawn through point $F$, with foot of perpendicular at $M$. The line $l$ intersects the $y$-axis at point $E$. If $|FM| = 3|ME|$, then the equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;O: Origin;Center(C) = O;F: Point;OneOf(Focus(C)) = F;Coordinate(F) = (2, 0);l: Line;PointOnCurve(F, l);IsPerpendicular(l, Asymptote(C));M: Point;FootPoint(l, Asymptote(C)) = M;E: Point;Intersection(l, yAxis) = E;Abs(LineSegmentOf(F, M)) = 3*Abs(LineSegmentOf(M, E))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[2, 8], [27, 33], [95, 101]], [[12, 16]], [[2, 16]], [[17, 26], [40, 44]], [[17, 38]], [[17, 26]], [[51, 54], [62, 67]], [[39, 54]], [[27, 54]], [[58, 61]], [[27, 61]], [[73, 77]], [[62, 77]], [[79, 93]]]",
    "query_spans": "[[[95, 106]]]",
    "process": "Let the equation of hyperbola C be: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1. From the given conditions, using the point-to-line distance formula, we obtain |FM|=b. From |FM|=3|ME| and the Pythagorean theorem, we get OE=\\sqrt{(\\frac{4b}{3})^{2}-4}. Since FE is perpendicular to the asymptote, \\therefore\\frac{\\sqrt{(\\frac{4b}{3})^{2}-4}{2}}=\\frac{a}{b}. Combining with a^{2}=4-b^{2}, we obtain b^{2}=3, a^{2}=1. \\therefore The equation of hyperbola C is: x^{2}-\\frac{y^{2}}{3}=1. Hence, the answer"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2x$ with focus $F$, a line $l$ passes through $F$ and intersects $C$ at points $A$ and $B$. If $|AF|=|BF|$, then the length of the chord intercepted on the $y$-axis by the circle with diameter $AB$ is?",
    "fact_expressions": "l: Line;C: Parabola;G: Circle;B: Point;A: Point;F: Point;Expression(C) = (y^2 = 2*x);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, F)) = Abs(LineSegmentOf(B, F));IsDiameter(LineSegmentOf(A,B),G)",
    "query_expressions": "Length(InterceptChord(yAxis,G))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[29, 34]], [[2, 21], [39, 42]], [[87, 88]], [[48, 51]], [[44, 47]], [[25, 28], [35, 38]], [[2, 21]], [[2, 28]], [[29, 38]], [[29, 53]], [[55, 68]], [[75, 88]]]",
    "query_spans": "[[[70, 95]]]",
    "process": "As shown in the figure: Since |AF| = |BF|, it follows that l \\bot x-axis, so the center of the circle is at F(\\frac{1}{2},0), the radius is r = 1, and the chord length is 2\\sqrt{1-(\\frac{1}{2})^{2}} = \\sqrt{3}"
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{m+4}+\\frac{y^{2}}{9}=1$ is $\\frac{1}{2}$; then the value of the real number $m$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(m + 4) + y^2/9 = 1);m: Real;Eccentricity(G) = 1/2",
    "query_expressions": "m",
    "answer_expressions": "{8, 11/4}",
    "fact_spans": "[[[0, 39]], [[0, 39]], [[59, 64]], [[0, 57]]]",
    "query_spans": "[[[59, 68]]]",
    "process": ""
  },
  {
    "text": "The parabola with vertex at the origin and focus on the $x$-axis intersects the line $2x - y + 1 = 0$ to form a chord of length $\\sqrt{15}$. Then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;O: Origin;Vertex(G) = O;PointOnCurve(Focus(G), xAxis);H: Line;Expression(H) = (2*x - y + 1 = 0);Length(InterceptChord(H, G)) = sqrt(15)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=12*x, y^2=-4*x}",
    "fact_spans": "[[[15, 18], [51, 54]], [[3, 5]], [[0, 18]], [[6, 18]], [[20, 33]], [[20, 33]], [[15, 49]]]",
    "query_spans": "[[[51, 58]]]",
    "process": "1 Let the equation of the required parabola be expressed in the form [express the chord length in terms of $ a $, then solve for $ a $, and thus obtain the parabola's equation. Assume the required parabola equation is $ y^2 = ax $ ($ a \\neq 0 $). The given line rewritten is $ y = 2x + 1 $. Let the line intersect the parabola at points $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Solving the system:  \n\\[\n\\begin{cases}\ny^{2} = ax, \\\\\ny = 2x + 1,\n\\end{cases}\n\\]  \neliminate $ y $ to get $ (2x + 1)^2 = ax $, which simplifies to $ 4x^{2} + (4 - a)x + 1 = 0 $.  \nThen $ x_{1} + x_{2} = -\\frac{a - 4}{4} $, $ x_{1}x_{2} = \\frac{1}{4} = \\sqrt{15} $. Solving gives $ a = 12 $ or $ a = -4 $, both satisfying the conditions. Therefore, the required parabola equations are $ y^2 = 12x $ or $ y^2 = -4x $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, a line $l$ passing through point $P(2,1)$ intersects the ellipse $C$ at points $A$ and $B$, and point $P$ is the midpoint of $AB$. Then the equation of line $l$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/9 + y^2/4 = 1);P: Point;Coordinate(P) = (2, 1);PointOnCurve(P, l) = True;l: Line;Intersection(l, C) = {A, B};A: Point;B: Point;MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Expression(l)",
    "answer_expressions": "8*x+9*y-25=0",
    "fact_spans": "[[[2, 39], [58, 63]], [[2, 39]], [[42, 51], [75, 79]], [[42, 51]], [[41, 57]], [[52, 57], [90, 95]], [[52, 73]], [[64, 67]], [[68, 71]], [[75, 88]]]",
    "query_spans": "[[[90, 100]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since point P(2,1) is the midpoint of AB, we have x_{1}+x_{2}=4, y_{1}+y_{2}=2. From \\begin{cases}\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1\\\\\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1\\end{cases}, subtracting the two equations gives \\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=-\\frac{4(x_{1}+x_{2})}{9(y_{1}+y_{2})}=-\\frac{4\\times4}{9\\times2}=-\\frac{8}{9}, i.e., k_{AB}=-\\frac{8}{9}. Therefore, the equation of line AB is y-1=-\\frac{8}{9}(x-2), or 8x+9y-25=0. The answer is: 8x+9y-25=0."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, point $M$ lies on the parabola, and $MN$ is perpendicular to the $x$-axis at point $N$. If $|MF|=6$, then the horizontal coordinate of $M$ is?",
    "fact_expressions": "G: Parabola;M: Point;N: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(M, G);IsPerpendicular(LineSegmentOf(M, N), xAxis);Intersection(LineSegmentOf(M, N), xAxis) = N;Abs(LineSegmentOf(M, F)) = 6",
    "query_expressions": "XCoordinate(M)",
    "answer_expressions": "5",
    "fact_spans": "[[[2, 16], [29, 32]], [[24, 28], [64, 67]], [[46, 50]], [[20, 23]], [[2, 16]], [[2, 23]], [[24, 33]], [[34, 45]], [[34, 50]], [[53, 62]]]",
    "query_spans": "[[[64, 73]]]",
    "process": "Let M(x_{0},y_{0}), then |MF| = x_{0} + 1 = 6, so x_{0} = 5. Therefore, the answer is: 5"
  },
  {
    "text": "Given the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ with left vertex $A_{1}$ and right focus $F_{2}$, and a point $P$ on the right branch of the hyperbola, find the minimum value of $\\overrightarrow{P A_{1}} \\cdot \\overrightarrow{P F_{2}}$.",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1);A1: Point;LeftVertex(G) = A1;F2: Point;RightFocus(G) = F2;P: Point;PointOnCurve(P, RightPart(G))",
    "query_expressions": "Min(DotProduct(VectorOf(P, A1), VectorOf(P, F2)))",
    "answer_expressions": "-2",
    "fact_spans": "[[[2, 30], [59, 62]], [[2, 30]], [[35, 42]], [[2, 42]], [[47, 54]], [[2, 54]], [[55, 58]], [[55, 67]]]",
    "query_spans": "[[[69, 132]]]",
    "process": ""
  },
  {
    "text": "Through the point $M(2, -2p)$, draw two tangent lines to the parabola $x^{2} = 2py$ $(p > 0)$, with points of tangency $A$ and $B$. If the vertical coordinate of the midpoint of segment $AB$ is $6$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;M: Point;p>0;Expression(G) = (x^2 = 2*(p*y));Coordinate(M) = (2, -2*p);l1: Line;l2: Line;TangentOfPoint(M, G) = {l1, l2};TangentPoint(l1, G) = A;TangentPoint(l2, G) = B;YCoordinate(MidPoint(LineSegmentOf(A, B))) = 6",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2 = 2*y, x^2 = 4*y}",
    "fact_spans": "[[[15, 36], [75, 78]], [[18, 36]], [[47, 50]], [[51, 54]], [[1, 14]], [[18, 36]], [[15, 36]], [[1, 14]], [], [], [[0, 41]], [[0, 54]], [[0, 54]], [[56, 73]]]",
    "query_spans": "[[[75, 83]]]",
    "process": ""
  },
  {
    "text": "The equation of a parabola with vertex at the origin, the $x$-axis as the axis of symmetry, and focus on the line $2 x-4 y+3=0$ is?",
    "fact_expressions": "G: Parabola;C:Curve;O:Origin;Expression(C)=(2*x-4*y+3=0);SymmetryAxis(G)=xAxis;Vertex(G)=O;PointOnCurve(Focus(G),C)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = -6x",
    "fact_spans": "[[[34, 37]], [[19, 32]], [[1, 3]], [[19, 32]], [[7, 37]], [[1, 37]], [[16, 37]]]",
    "query_spans": "[[[34, 41]]]",
    "process": ""
  },
  {
    "text": "If the circle $(x-\\sqrt{3})^{2}+(y-1)^{2}=9$ intersects the asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ passing through the second and fourth quadrants at points $A$ and $B$, and $|A B|=2 \\sqrt{6}$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "H: Circle;Expression(H) = ((x - sqrt(3))^2 + (y - 1)^2 = 9);G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Intersection(H,Asymptote(G)) = {A,B};Quadrant(A) = 2;Quadrant(B) = 4;A: Point;B: Point;Abs(LineSegmentOf(A, B)) = 2*sqrt(6)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[1, 32]], [[1, 32]], [[33, 89], [135, 138]], [[33, 89]], [[36, 89]], [[36, 89]], [[36, 89]], [[36, 89]], [[1, 112]], [[91, 112]], [[91, 112]], [[103, 106]], [[107, 110]], [[114, 132]]]",
    "query_spans": "[[[135, 144]]]",
    "process": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ has an asymptote passing through the second and fourth quadrants with equation $bx+ay=0$. Since $|AB|=2\\sqrt{6}$, and the circle $(x-\\sqrt{3})^{2}+(y-1)^{2}=9$ has center $(\\sqrt{3},1)$ and radius $3$, the distance from the center to the asymptote is $d=\\sqrt{9-6}=\\sqrt{3}$, that is, $\\frac{|\\sqrt{3}b+a|}{\\sqrt{a^{2}+b^{2}}}=\\sqrt{3}$. Solving gives $a=\\sqrt{3}b$, so $c=\\sqrt{a^{2}+b^{2}}=2b$, thus the eccentricity of the hyperbola is $e=\\frac{c}{a}=\\frac{2b}{\\sqrt{3}b}=\\frac{2\\sqrt{3}}{3}$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and $P$ is a moving point on the ellipse, then the maximum value of $|P F_{1}| \\cdot|P F_{2}|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)))",
    "answer_expressions": "25",
    "fact_spans": "[[[18, 56], [66, 68]], [[62, 65]], [[2, 9]], [[10, 17]], [[18, 56]], [[2, 61]], [[62, 74]]]",
    "query_spans": "[[[76, 108]]]",
    "process": "|PF_{1}|\\cdot|PF_{2}|\\leqslant\\left(\\frac{|PF_{1}|+|PF_{2}|}{2}\\right)^{2}=\\left(\\frac{2a}{2}\\right)^{2}=25 \\quad \\text{(equality holds if and only if } |PF_{1}|=|PF_{2}|)"
  },
  {
    "text": "Draw a line $l$ through the left focus $F_{1}$ of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ with an inclination angle of $60^{\\circ}$. The line $l$ intersects the ellipse at points $A$ and $B$. What is the length of segment $AB$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);F1: Point;LeftFocus(G) = F1;l: Line;PointOnCurve(F1, l) = True;Inclination(l) = ApplyUnit(60, degree);Intersection(l, G) = {A, B};A: Point;B: Point",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "8*sqrt(2)/7",
    "fact_spans": "[[[2, 29], [70, 72]], [[2, 29]], [[33, 40]], [[2, 40]], [[58, 63], [64, 69]], [[0, 63]], [[41, 63]], [[64, 84]], [[75, 78]], [[79, 82]]]",
    "query_spans": "[[[86, 97]]]",
    "process": "In $\\frac{x^{2}}{2}+y^{2}=1$, $a^{2}=2$, $b^{2}=1$, so $c^{2}=a^{2}-b^{2}=1$, that is, $c=1$, hence the left focus is $F_{1}(-1,0)$, and $\\tan60^{\\circ}=\\sqrt{3}$, thus the line $l$ is: $y=\\sqrt{3}(x+1)$, solving simultaneously with $\\frac{x^{2}}{2}+y^{2}=1$, then by the chord length formula: $AB=\\sqrt{1+(\\sqrt{3})^{2}}\\cdot\\sqrt{(-\\frac{12}{7})^{2}-4\\times\\frac{4}{7}}=\\frac{8\\sqrt{2}}{7}$, by Vieta's theorem: $x_{1}+x_{2}=-\\frac{12}{7}$, $x_{1}x_{2}=\\frac{4}{7}$"
  },
  {
    "text": "Given that $F$ is the right focus of the ellipse $C$: $\\frac{x^{2}}{20}+\\frac{y^{2}}{4}=1$, $P$ is a point on $C$, $A(-2,1)$, when the perimeter of $\\triangle A P F$ is minimized, its area is?",
    "fact_expressions": "F: Point;C: Ellipse;Expression(C) = (x^2/20 + y^2/4 = 1);RightFocus(C) = F;P: Point;PointOnCurve(P, C) = True;A: Point;Coordinate(A) = (-2, 1);WhenMin(Perimeter(TriangleOf(A, P, F))) = True",
    "query_expressions": "Area(TriangleOf(A, P, F))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 5]], [[6, 49], [58, 61]], [[6, 49]], [[2, 53]], [[54, 57]], [[54, 64]], [[65, 74]], [[65, 74]], [[76, 99]]]",
    "query_spans": "[[[77, 105]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{k+8}+\\frac{y^{2}}{9}=1$ is $\\frac{1}{2}$, then the value of $k$ is?",
    "fact_expressions": "G: Ellipse;k: Number;Expression(G) = (x^2/(k + 8) + y^2/9 = 1);Eccentricity(G) = 1/2",
    "query_expressions": "k",
    "answer_expressions": "{4,-5/4}",
    "fact_spans": "[[[0, 39]], [[59, 62]], [[0, 39]], [[0, 57]]]",
    "query_spans": "[[[59, 66]]]",
    "process": "The eccentricity of the ellipse satisfies $ e = \\frac{c}{a} = \\sqrt{1 - \\left( \\frac{b}{a} \\right)^2} $. When the foci of the ellipse are on the x-axis, $ \\sqrt{1 - \\frac{9}{k+8}} = \\frac{1}{2} $, solving gives $ k = 4 $. When the foci of the ellipse are on the y-axis, $ \\sqrt{1 - \\frac{k+8}{9}} = \\frac{1}{2} $, solving gives $ k = -\\frac{5}{4} $. Therefore, fill in $ 4 $ or $ -\\frac{5}{4} $."
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{2+m}-\\frac{y^{2}}{m+1}=1$ represents an ellipse, what is the range of real values for $m$?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G)=(x^2/(m + 2) - y^2/(m + 1) = 1)",
    "query_expressions": "Range(m)",
    "answer_expressions": "{(-2,3/2)+(-3/2,-1)}",
    "fact_spans": "[[[45, 47]], [[49, 54]], [[2, 47]]]",
    "query_spans": "[[[49, 61]]]",
    "process": ""
  },
  {
    "text": "Given a point $(x_{0}, y_{0})$ on the parabola $y^{2}=8x$ such that its distance to the focus is $x_{0}^{2}$, then $x_{0}=$?",
    "fact_expressions": "G: Parabola;H: Point;x0:Number;y0:Number;Expression(G) = (y^2 = 8*x);Coordinate(H) = (x0, y0);PointOnCurve(H, G);Distance(H, Focus(G)) = x0^2",
    "query_expressions": "x0",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 16]], [[19, 35]], [[56, 63]], [[19, 35]], [[2, 16]], [[19, 35]], [[2, 35]], [[2, 54]]]",
    "query_spans": "[[[56, 65]]]",
    "process": ""
  },
  {
    "text": "If $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and $M$ is a moving point on the ellipse, then the minimum value of $\\frac{1}{|M F_{1}|}+\\frac{1}{|M F_{2}|}$ is?",
    "fact_expressions": "F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);M: Point;PointOnCurve(M, G) = True",
    "query_expressions": "Min(1/Abs(LineSegmentOf(M, F2)) + 1/Abs(LineSegmentOf(M, F1)))",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 8]], [[9, 16]], [[1, 52]], [[1, 52]], [[17, 44], [57, 59]], [[17, 44]], [[53, 56]], [[53, 63]]]",
    "query_spans": "[[[65, 112]]]",
    "process": "(Analysis) By the definition of an ellipse, the given expression can be transformed into \\frac{4}{|MF_{1}|}|\\overrightarrow{MF_{2}}|. Using the basic inequality, the value of (|MF_{1}|\\cdot|MF_{2}|)^{\\frac{1}{2}} can be found, and substituting it yields the minimum value of the required expression. From the definition of an ellipse: |MF_{1}|+|MF_{2}|=2a=4, \\frac{1}{|MF_{1}|}+\\frac{1}{|MF_{2}|}=\\frac{|MF_{1}|+|MF_{2}|}{|MF_{1}|\\cdot|MF_{2}|}=\\frac{4}{|MF_{1}|\\cdot|MF_{2}|}, |MF_{1}|\\cdot|MF_{2}|\\leqslant4 (equality holds if and only if |MF_{1}|=|MF_{2}|), \\frac{1}{|MF|}+\\frac{1}{|MF_{2}|}>1, i.e., \\frac{1}{|MF|}+\\frac{1}{|MF|} has a minimum value of 1."
  },
  {
    "text": "It is known that a hyperbola whose axis of symmetry is a coordinate axis has an asymptote parallel to the line $x+2 y-3=0$. What is the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;H: Line;Expression(H) = (x + 2*y - 3 = 0);SymmetryAxis(G) = axis;IsParallel(OneOf(Asymptote(G)),H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{sqrt(5), sqrt(5)/2}",
    "fact_spans": "[[[10, 13], [38, 41]], [[22, 35]], [[22, 35]], [[2, 13]], [[10, 35]]]",
    "query_spans": "[[[38, 47]]]",
    "process": ""
  },
  {
    "text": "Given that the moving point $M(x, y)$ satisfies $5 \\sqrt{(x-1)^{2}+(y-2)^{2}}=|3 x+4 y+12|$, what is the trajectory curve of point $M$?",
    "fact_expressions": "M: Point;x_:Number;y_:Number;Coordinate(M) = (x_, y_);5*sqrt((x_ - 1)^2 + (y_ - 2)^2) = Abs(3*x_ + 4*y_ + 12)",
    "query_expressions": "Locus(M)",
    "answer_expressions": "Parabola",
    "fact_spans": "[[[4, 13], [60, 64]], [[4, 13]], [[4, 13]], [[4, 13]], [[15, 58]]]",
    "query_spans": "[[[60, 71]]]",
    "process": "From $5\\sqrt{(x-1)^{2}+(y-2)^{2}}=|3x+4y+12|$ we get $\\sqrt{(x-1)^{2}+(y-2)^{2}}=\\frac{|3x+4y+12|}{5}$, where the left side represents the distance between the point $(x,y)$ and the point $(1,2)$, and the right side of the equation represents the distance from the point $(x,y)$ to the line $3x+4y+12=0$. The entire equation indicates that the distance from the point $(x,y)$ to the point $(1,2)$ is equal to the distance from the point $(x,y)$ to the line $3x+4y+12=0$, hence its trajectory is a parabola."
  },
  {
    "text": "Given the parabola $C$: $x^{2}=4 y$ with focus $F$, and point $P$ moving on the line $x-y-2=0$. From point $P$, draw two tangents $P A$ and $P B$ to the parabola $C$, where $A$ and $B$ are the points of tangency. Then, what is the maximum distance from the origin $O$ to the line $A B$?",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 4*y);F: Point;Focus(C) = F;P: Point;G: Line;Expression(G) = (x - y - 2 = 0);PointOnCurve(P, G);TangentOfPoint(P, C) = {LineOf(P, A), LineOf(P, B)};TangentPoint(LineOf(P, A), C) = A;TangentPoint(LineOf(P, B), C) = B;B: Point;A: Point;O: Origin",
    "query_expressions": "Max(Distance(O, LineOf(A, B)))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 21], [56, 62]], [[2, 21]], [[25, 28]], [[2, 28]], [[29, 33], [51, 55]], [[34, 45]], [[34, 45]], [[29, 49]], [[50, 79]], [[50, 92]], [[50, 92]], [[86, 89]], [[82, 85]], [[94, 99]]]",
    "query_spans": "[[[94, 115]]]",
    "process": ""
  },
  {
    "text": "The vertex of the parabola $y=8 x^{2} -(m+1) x+m-7$ lies on the $x$-axis, then $m=$?",
    "fact_expressions": "G: Parabola;m: Number;Expression(G) = (y = m + 8*x^2 - x*(m + 1) - 7);PointOnCurve(Vertex(G), xAxis)",
    "query_expressions": "m",
    "answer_expressions": "15",
    "fact_spans": "[[[0, 26]], [[37, 40]], [[0, 26]], [[0, 35]]]",
    "query_spans": "[[[37, 42]]]",
    "process": ""
  },
  {
    "text": "The standard equation of a parabola with directrix $x=2$ is?",
    "fact_expressions": "G: Parabola;Expression(Directrix(G)) = (x = 2)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = -8*x",
    "fact_spans": "[[[11, 14]], [[0, 14]]]",
    "query_spans": "[[[11, 21]]]",
    "process": "The directrix of the parabola is given by the equation x=2, indicating that the parabola opens to the left and p=2\\times2=4, so the standard equation of the parabola is v^2=-8x."
  },
  {
    "text": "A hyperbola that shares the same asymptotes as the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ and passes through the point $(-3 , 2 \\sqrt{3})$, what is the distance from one focus of this hyperbola to one of its asymptotes?",
    "fact_expressions": "G: Hyperbola;C:Hyperbola;H: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Coordinate(H) = (-3, 2*sqrt(3));Asymptote(G) = Asymptote(C);PointOnCurve(H,C)",
    "query_expressions": "Distance(OneOf(Focus(C)),OneOf(Asymptote(C)))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 40]], [[72, 75]], [[51, 71]], [[1, 40]], [[51, 71]], [[0, 75]], [[49, 75]]]",
    "query_spans": "[[[72, 91]]]",
    "process": ""
  },
  {
    "text": "The chord of the parabola $y^{2}=4 x$ is bisected by the point $A(4 , 2)$, then what is the equation of the line on which this chord lies?",
    "fact_expressions": "G: Parabola;H: LineSegment;A: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (4, 2);MidPoint(H)=A;IsChordOf(H,G)",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "x-y-2=0",
    "fact_spans": "[[[0, 14]], [], [[19, 30]], [[0, 14]], [[19, 30]], [[0, 32]], [[0, 18]]]",
    "query_spans": "[[[0, 48]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{n}=1$ $(m>0, n>0)$ has an eccentricity of $2$, and one of its foci coincides with the focus of the parabola $y^{2}=4 m x$. Find the value of $n$.",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/n + x^2/m = 1);m: Number;n: Number;m>0;n>0;Eccentricity(G) = 2;H: Parabola;Expression(H) = (y^2 = 4*(m*x));OneOf(Focus(G)) = Focus(H)",
    "query_expressions": "n",
    "answer_expressions": "12",
    "fact_spans": "[[[0, 48]], [[0, 48]], [[3, 48]], [[86, 89]], [[3, 48]], [[3, 48]], [[0, 56]], [[63, 79]], [[63, 79]], [[0, 84]]]",
    "query_spans": "[[[86, 93]]]",
    "process": "The focus of the parabola $ y^{2} = 4mx $ has coordinates $ (m, 0) $. Since the hyperbola $ \\frac{x^{2}}{m} - \\frac{y^{2}}{n} = 1 $ $ (m > 0, n > 0) $ has eccentricity 2, and one of its foci coincides with the focus of the parabola $ y^{2} = 4mx $, we have $ c = m $. Since the eccentricity of the hyperbola $ \\frac{x^{2}}{m} - \\frac{y^{2}}{n} = 1 $ $ (m > 0, n > 0) $ is 2, we have $ a = \\frac{1}{2}m $. Therefore, $ a^{2} = m = \\frac{1}{4}m^{2} $, so $ m = 4 $. Since $ c^{2} = a^{2} + b^{2} $, we have $ 4 + n = 16 $, thus $ n = 12 $."
  },
  {
    "text": "If the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a, b>0)$ is $2$, then $\\frac{b}{a}$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;Eccentricity(G) = 2",
    "query_expressions": "b/a",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[1, 55]], [[1, 55]], [[4, 55]], [[4, 55]], [[4, 55]], [[4, 55]], [[1, 63]]]",
    "query_spans": "[[[65, 80]]]",
    "process": ""
  },
  {
    "text": "The tangent line $M T$ to the circle $x^{2}+y^{2}=9$ passes through the left focus $F$ of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{12}=1$, where $T$ is the point of tangency, $M$ is the intersection point of the tangent line with the right branch of the hyperbola, and $P$ is the midpoint of $M F$. Then $|P O|-|P T|=$?",
    "fact_expressions": "G: Hyperbola;H: Circle;M: Point;T: Point;F: Point;P: Point;O: Origin;Expression(G) = (x^2/9 - y^2/12 = 1);Expression(H) = (x^2 + y^2 = 9);LeftFocus(G)=F;IsTangent(LineOf(M, T), H);PointOnCurve(F, LineOf(M, T));TangentPoint(LineOf(M,T),H)=T;Intersection(LineOf(M,T),RightPart(G))=M;MidPoint(LineSegmentOf(M, F)) = P",
    "query_expressions": "Abs(LineSegmentOf(P, O)) - Abs(LineSegmentOf(P, T))",
    "answer_expressions": "2*sqrt(3) - 3",
    "fact_spans": "[[[25, 64], [88, 91]], [[0, 16]], [[81, 84]], [[74, 77]], [[68, 71]], [[97, 100]], [[111, 124]], [[25, 64]], [[0, 16]], [[25, 71]], [[0, 24]], [[19, 71]], [[0, 80]], [[19, 96]], [[97, 109]]]",
    "query_spans": "[[[111, 126]]]",
    "process": "Let the right focus be F, |FF| = \\sqrt{OF^{2} - |or|} = b \\Rightarrow |P| = |PF| - |TF| = \\frac{1}{2}|MF| - b, |PO| = \\frac{1}{2}|PF| \\Rightarrow |PO| - |PF| = b - \\frac{1}{2}|MF| - |MF| = b - a = 2\\sqrt{3} - 3"
  },
  {
    "text": "Given that the center of the ellipse is at the origin, the foci are on the $x$-axis, the eccentricity is $\\frac{\\sqrt{5}}{5}$, and it passes through the point $P(-5,4)$, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;PointOnCurve(Focus(G), xAxis);Eccentricity(G) = sqrt(5)/5;P: Point;Coordinate(P) = (-5, 4);PointOnCurve(P, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/45 + y^2/36 = 1",
    "fact_spans": "[[[2, 4], [59, 61]], [[8, 10]], [[2, 10]], [[2, 19]], [[2, 44]], [[47, 57]], [[47, 57]], [[2, 57]]]",
    "query_spans": "[[[59, 66]]]",
    "process": "Let the equation of the ellipse be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. From the given conditions, $e=\\frac{c}{a}=\\frac{\\sqrt{5}}{5}$, $\\therefore c=\\frac{\\sqrt{5}}{5}a$, $\\therefore b^{2}=a^{2}-c^{2}=\\frac{4a^{2}}{5}$. Since point $P(-5,4)$ lies on the ellipse, $\\therefore \\frac{25}{a^{2}}+\\frac{80}{4a^{2}}=\\frac{45}{a^{2}}=1$, solving gives $a^{2}=45$, $\\therefore b^{2}=36$, $\\therefore$ the equation of the ellipse is $\\frac{x^{2}}{45}+\\frac{y^{2}}{36}=1$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, its two asymptotes are the lines $l_{1}$ and $l_{2}$. The line $l$, passing through the right focus $F$ and perpendicular to $l_{1}$, intersects $l_{1}$ and $l_{2}$ at points $A$ and $B$, respectively, and $\\overrightarrow{F B}=3 \\overrightarrow{A F}$. Then the eccentricity of this hyperbola is?",
    "fact_expressions": "l: Line;G: Hyperbola;b: Number;a: Number;l1:Line;F: Point;B: Point;A: Point;l2: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G)=F;Asymptote(G)={l1,l2};PointOnCurve(F,l);IsPerpendicular(l,l1);Intersection(l,l1)=A;Intersection(l,l2)=B;VectorOf(F, B) = 3*VectorOf(A, F)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[106, 111]], [[2, 58], [191, 194]], [[5, 58]], [[5, 58]], [[67, 76], [114, 121], [98, 105]], [[91, 94]], [[136, 139]], [[132, 135]], [[79, 86], [79, 86]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 94]], [[2, 86]], [[86, 111]], [[95, 111]], [[106, 141]], [[106, 141]], [[143, 188]]]",
    "query_spans": "[[[191, 200]]]",
    "process": "From the given conditions, $|\\overrightarrow{FA}|=b$, $|\\overrightarrow{FB}|=3b$, $|\\overrightarrow{OA}|=a$, $\\tan\\angle BOF=\\tan\\angle AOF=\\frac{b}{a}$, $\\tan\\angle BOA=\\tan 2\\angle BOF=\\frac{4b}{a}$, solve the equation to obtain the solution. [Solution] From the given conditions, $|\\overrightarrow{FA}|=b$, $|\\overrightarrow{FB}|=3b$, $|\\overrightarrow{OA}|=a$, and $\\tan\\angle BOF=\\tan\\angle AOF=\\frac{b}{a}$. Therefore, $\\tan\\angle BOA=\\frac{4b}{a}=\\tan 2\\angle BOF=\\frac{\\frac{b}{a}+\\frac{b}{a}}{1-(\\frac{b}{a})^{2}}$. Simplifying yields $a^{2}=2b^{2}$, i.e., $a^{2}=2(c^{2}-a^{2})$, hence $3a^{2}=2c^{2}$, $e^{2}=\\frac{3}{2}$, so $e=\\frac{\\sqrt{6}}{2}$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $A$ lies on the hyperbola, and $A F_{2} \\perp x$-axis. If $\\frac{|A F_{1}|}{|A F_{2}|}=\\frac{5}{3}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;A: Point;F2: Point;F1: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(A, G);IsPerpendicular(LineSegmentOf(A, F2), xAxis);Abs(LineSegmentOf(A, F1))/Abs(LineSegmentOf(A, F2)) = 5/3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 61], [90, 93], [159, 162]], [[5, 61]], [[5, 61]], [[85, 89]], [[77, 84]], [[69, 76]], [[5, 61]], [[5, 61]], [[2, 61]], [[2, 84]], [[2, 84]], [[85, 94]], [[96, 114]], [[116, 157]]]",
    "query_spans": "[[[159, 169]]]",
    "process": ""
  },
  {
    "text": "A line $l$ with slope $1$ passes through the focus $F$ of the parabola $y^2 = 2px$ ($p > 0$). If $l$ is tangent to the circle $(x - 5)^2 + y^2 = 8$, then $p$ equals?",
    "fact_expressions": "l: Line;G: Parabola;p: Number;H: Circle;p>0;Expression(G) = (y^2 = 2*p*x);Slope(l)=1;Focus(G)=F;Expression(H) = (y^2 + (x - 5)^2 = 8);IsTangent(l,H);F:Point;PointOnCurve(F,l)",
    "query_expressions": "p",
    "answer_expressions": "{2,18}",
    "fact_spans": "[[[7, 12], [42, 45]], [[13, 34]], [[70, 73]], [[46, 66]], [[16, 34]], [[13, 34]], [[0, 12]], [[13, 40]], [[46, 66]], [[42, 68]], [37, 39], [7, 39]]",
    "query_spans": "[[[70, 76]]]",
    "process": "First find the coordinates of the focus, then determine the value of p using the distance from the center of the circle to the line l equal to the radius. The focus of the parabola y^{2}=2px (p>0) is F(\\frac{p}{2},0), so the line l: y=x-\\frac{p}{2}. Since l is tangent to the circle (x-5)^{2}+y^{2}=8, we have \\frac{|5-\\frac{p}{2}|}{\\sqrt{2}}=2\\sqrt{2}. \\therefore p=2^{or 18}"
  },
  {
    "text": "A line $l$ passing through the focus $F$ of the parabola $C$: $y^{2}=2 p x$ ($p>0$) intersects the directrix of the parabola $C$ at point $A$, and intersects the parabola $C$ at a point $B$, such that $\\overrightarrow{A B}=k \\overrightarrow{B F}$ ($k \\geq \\sqrt{2}$). If $l$ is perpendicular to an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$), then the range of the eccentricity of this hyperbola is?",
    "fact_expressions": "l: Line;G: Hyperbola;b: Number;a: Number;C: Parabola;p: Number;A: Point;B: Point;F: Point;k:Number;k>=sqrt(2);a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);p>0;Expression(C) = (y^2 = 2*p*x);Focus(C)=F;PointOnCurve(F,l);Intersection(l,Directrix(C))=A;OneOf(Intersection(l,C))=B;VectorOf(A, B) = k*VectorOf(B, F);IsPerpendicular(l,OneOf(Asymptote(G)))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1,\\sqrt{2}]",
    "fact_spans": "[[[34, 39], [139, 142]], [[143, 199], [210, 213]], [[146, 199]], [[146, 199]], [[1, 27], [41, 47], [57, 63]], [[8, 27]], [[51, 55]], [[69, 72]], [[30, 33]], [[74, 136]], [[74, 136]], [[146, 199]], [[146, 199]], [[143, 199]], [[8, 27]], [[1, 27]], [[1, 33]], [[0, 39]], [[34, 55]], [[34, 72]], [[74, 136]], [[139, 207]]]",
    "query_spans": "[[[210, 223]]]",
    "process": "According to the problem, the slope of line $ l $ exists and is non-zero. Draw $ BC $ from $ B $ perpendicular to the directrix of the parabola, with foot $ C $. By the definition of a parabola, $ BF = BC $. Since $ \\overrightarrow{AB} = k\\overrightarrow{BF} $ ($ k \\geqslant \\sqrt{2} $), it follows that $ |AB| = k|BF| = k|BC| $, so $ \\frac{1}{k} = \\frac{|BC|}{|AB|} = \\cos\\angle ABC $. Because $ k \\geqslant \\sqrt{2} $, $ \\frac{1}{k} \\in (0, \\frac{\\sqrt{2}}{2}] $, so $ \\cos\\angle ABC \\in (0, \\frac{\\sqrt{2}}{2}] $, hence $ \\angle ABC \\in [\\frac{\\pi}{4}, \\frac{\\pi}{2}) $. Therefore, $ \\tan\\angle ABC \\in [1, +\\infty) $, meaning the range of the slope of line $ l $ is $ [1, +\\infty) $. Since $ l $ is perpendicular to one asymptote $ y = -\\frac{b}{a}x $ of the hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $), we have $ \\frac{a}{b} \\geqslant 1 $. Thus, the eccentricity of the hyperbola $ e = \\frac{c}{a} = \\sqrt{\\frac{a^{2}+b^{2}}{a^{2}}} = \\sqrt{1+(\\frac{b}{a})^{2}} \\leqslant \\sqrt{1+1} = \\sqrt{2} $, and since $ e > 1 $, we get $ 1 < e \\leqslant \\sqrt{2} $, i.e., the range of the eccentricity of the hyperbola is $ 1 < e \\leqslant \\sqrt{2} $."
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=x$, draw a line intersecting the parabola at points $M(x_{1}, y_{1})$ and $N(x_{2}, y_{2})$. If $x_{1}+x_{2}=1$, then the distance from the midpoint of segment $MN$ to the directrix is equal to?",
    "fact_expressions": "G: Parabola;H: Line;M: Point;N: Point;F:Point;x1:Number;y1:Number;x2:Number;y2:Number;Expression(G)=(y^2 = x);Coordinate(M)=(x1,y1);Coordinate(N)=(x2,y2);Focus(G)=F;PointOnCurve(F,H);Intersection(H,G)={M,N};x1+x2=1",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(M,N)),Directrix(G))",
    "answer_expressions": "3/4",
    "fact_spans": "[[[1, 13], [24, 27]], [[21, 23]], [[28, 45]], [[46, 63]], [[16, 19]], [[28, 45]], [[28, 45]], [[46, 63]], [[46, 63]], [[1, 13]], [[28, 45]], [[46, 63]], [[1, 19]], [[0, 23]], [[21, 65]], [[68, 83]]]",
    "query_spans": "[[[24, 104]]]",
    "process": "From the parabola equation $ y^{2} = x $, we get $ p = \\frac{1}{2} $. Since $ x_{1} + x_{2} + p = \\frac{3}{2} $, the distance from the midpoint of $ MN $ to the directrix of the parabola is $ \\frac{x_{1} + x_{2}}{2} + \\frac{p}{2} = \\frac{3}{4} $."
  },
  {
    "text": "On the parabola $y^{2}=2 p x(p>0)$, the distance from the point with abscissa $4$ to the focus is $5$. Then the value of $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;p>0;Expression(G) = (y^2 = 2*(p*x));P: Point;XCoordinate(P) = 4;PointOnCurve(P, G);Distance(P, Focus(G)) = 5",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 22]], [[45, 48]], [[4, 22]], [[1, 22]], [[32, 33]], [[25, 33]], [[0, 33]], [[1, 43]]]",
    "query_spans": "[[[45, 52]]]",
    "process": "The directrix of the parabola is $x = -\\frac{p}{2}$. The distance from the point with abscissa 4 to the focus is 5, so $\\frac{p}{2} = 1 \\therefore p = 2$"
  },
  {
    "text": "Given that a vertex of the ellipse is $(0 , \\sqrt{3})$ and the eccentricity is $e=\\frac{\\sqrt{3}}{2}$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;Coordinate(OneOf(Vertex(G))) = (0, sqrt(3));Eccentricity(G)=e;e:Number;e=sqrt(3)/2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/12+y^2/3=1,x^2/(3/4)+y^2/3=1}",
    "fact_spans": "[[[2, 4], [55, 57]], [[2, 26]], [[2, 53]], [[31, 53]], [[31, 53]]]",
    "query_spans": "[[[55, 64]]]",
    "process": "a is the semi-major axis length, b is the semi-minor axis length. Since $ b = \\sqrt{b^{2}} = \\sqrt{\\frac{\\sqrt{2-c}}{a^{2}}} = \\sqrt{1-e^{2}} = \\sqrt{1-(\\frac{\\sqrt{3}}{2}} = \\frac{1}{2} $, $ \\therefore a = 2b $. If the foci of the ellipse lie on the x-axis, then $ b = \\sqrt{3} $, $ a = 2\\sqrt{3} $; if the foci of the ellipse lie on the y-axis, then $ a = \\sqrt{3} $, $ b = \\frac{\\sqrt{3}}{2} $. Therefore, the standard equation of the ellipse is $ \\frac{x^{2}}{12} + \\frac{y^{2}}{3} = 1 $ or $ \\frac{y^{2}}{3} + \\frac{x^{2}}{\\frac{3}{4}} = 1 $."
  },
  {
    "text": "The vertices and foci of a hyperbola centered at the origin are respectively the foci and vertices of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$. If the eccentricity of the hyperbola is $2$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;b: Number;a: Number;a > b;b > 0;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = 2;Center(G)=O;O:Origin;Vertex(G)=Focus(H);Focus(G)=Vertex(H)",
    "query_expressions": "Eccentricity(H)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[8, 11], [80, 83]], [[20, 72], [93, 95]], [[22, 72]], [[22, 72]], [[22, 72]], [[22, 72]], [[20, 72]], [[80, 91]], [[2, 11]], [[5, 7]], [[8, 78]], [[8, 78]]]",
    "query_spans": "[[[93, 100]]]",
    "process": ""
  },
  {
    "text": "The standard equation of the hyperbola that has the same asymptotes as $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ and passes through the point $(3 \\sqrt{3}, -3)$ is?",
    "fact_expressions": "G: Hyperbola;C:Hyperbola;H: Point;Expression(G) = (x^2/16 - y^2/9 = 1);Coordinate(H) = (3*sqrt(3), -3);Asymptote(C) = Asymptote(G);PointOnCurve(H, C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/11 - y^2/(99/16) = 1",
    "fact_spans": "[[[1, 40]], [[70, 73]], [[50, 69]], [[1, 40]], [[50, 69]], [[0, 73]], [[48, 73]]]",
    "query_spans": "[[[70, 80]]]",
    "process": "Analysis: Let the equation of the desired hyperbola be $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=\\lambda$ ($\\lambda\\neq0$). Since the hyperbola passes through the point $(3\\sqrt{3},-3)$, we have $\\frac{27}{16}-\\frac{9}{9}=\\lambda \\Rightarrow \\lambda=\\frac{11}{16}$. Therefore, the standard equation of the desired hyperbola is $\\frac{x^{2}}{11}-\\frac{y^{2}}{\\frac{99}{16}}=$"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $M$ lies on the asymptote of $C$, with $M F_{1} \\perp M F_{2}$ and $|M F_{1}|=2 b+|M F_{2}|$. Then $\\frac{b^{2}}{a^{2}}$=?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;M: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(M, Asymptote(C));IsPerpendicular(LineSegmentOf(M, F1), LineSegmentOf(M, F2));Abs(LineSegmentOf(M, F1)) = 2*b + Abs(LineSegmentOf(M, F2))",
    "query_expressions": "b^2/a^2",
    "answer_expressions": "(\\sqrt{5}-1)/2",
    "fact_spans": "[[[2, 63], [93, 96]], [[10, 63]], [[10, 63]], [[88, 92]], [[72, 79]], [[80, 87]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 87]], [[2, 87]], [[88, 101]], [[103, 127]], [[128, 153]]]",
    "query_spans": "[[[155, 178]]]",
    "process": "Without loss of generality, assume point $ M $ is in the first quadrant. Let $ |MF_{1}| = m $, $ |MF_{2}| = n $, then $ m = 2b + n $. Since $ MF_{1} \\perp MF_{2} $, it follows that $ m^{2} + n^{2} = 4c^{2} $. Solving these two equations simultaneously gives $ mn = 2c^{2} - 2b^{2} $. Solving the system \n\\[\n\\begin{cases}\ny = \\frac{b}{a}x \\\\\nx^{2} + y^{2} = c^{2}\n\\end{cases}\n\\]\nyields $ M(a, b) $. Using the triangle area formula, we obtain $ \\frac{1}{2}mn = \\frac{1}{2} \\cdot 2cb $, i.e., $ c^{2} - b^{2} = cb $, hence $ a^{2} = bc $, which implies $ a^{4} = b^{2}c^{2} $. Thus, $ a^{4} = b^{2}(a^{2} + b^{2}) $, leading to $ b^{4} + a^{2}b^{2} - a^{4} = 0 $. Then $ \\left(\\frac{b}{a}\\right)^{4} + \\left(\\frac{b}{a}\\right)^{2} - 1 = 0 $, solving this yields $ \\frac{b^{2}}{a^{2}} = \\frac{\\sqrt{5}-1}{2} $. Hence, fill in: $ \\frac{\\sqrt{5}-1}{2} $"
  },
  {
    "text": "The asymptotes of a hyperbola are given by $y=\\pm \\frac{1}{2} x$, and the distance between the two vertices is $4$. What is the standard equation of the hyperbola?",
    "fact_expressions": "G: Hyperbola;F:Point;H:Point;Expression(Asymptote(G)) = (y = pm*(x*1/2));Distance(F,H)=4;Vertex(G)={F,H}",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/4-y^2=1，y^2/4-x^2/16=1}",
    "fact_spans": "[[[2, 5], [47, 50]], [], [], [[2, 33]], [[2, 46]], [[2, 37]]]",
    "query_spans": "[[[47, 57]]]",
    "process": ""
  },
  {
    "text": "Let the line $l$ passing through the focus $F$ of the parabola $C$: $y^{2}=2 p x$ ($p>0$) intersect $C$ at points $A$ and $B$. The projections of points $A$ and $B$ onto the directrix are $M$ and $N$, respectively. Given $\\frac{S_{\\Delta M F N}}{S_{\\triangle A F M}}=\\lambda$, $\\frac{S_{\\triangle B F N}}{S_{\\triangle M F N}}=\\mu$, then $\\frac{\\lambda}{\\mu}=?$",
    "fact_expressions": "l: Line;C: Parabola;p: Number;M: Point;F: Point;A: Point;B: Point;N: Point;p>0;Expression(C) = (y^2 = 2*p*x);Focus(C)=F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Projection(A,Directrix(C))=M;Projection(B, Directrix(C)) = N;Area(TriangleOf(M,F,N))/Area(TriangleOf(A,F,M))=lambda;Area(TriangleOf(B,F,N))/Area(TriangleOf(M,F,N))=mu;lambda:Number;mu:Number",
    "query_expressions": "lambda/mu",
    "answer_expressions": "4",
    "fact_spans": "[[[34, 39]], [[1, 27], [40, 43]], [[9, 27]], [[76, 79]], [[30, 33]], [[46, 49], [57, 60]], [[50, 53], [61, 64]], [[80, 84]], [[9, 27]], [[1, 27]], [[1, 33]], [[0, 39]], [[34, 55]], [[40, 84]], [[40, 84]], [[86, 141]], [[143, 196]], [[198, 219]], [[198, 219]]]",
    "query_spans": "[[[198, 221]]]",
    "process": "As shown in the figure: let $\\angle MAF = \\theta$, $AF = a$, $BF = b$, by the definition of parabola, we get: $AM = a$, $BN = b$, $\\angle MFO + \\angle NFO = \\angle MFA + \\angle NFB = \\frac{\\pi}{2}$. In $\\triangle MAF$, by the law of cosines, we obtain: $MF^{2} = 2a^{2}(1 - \\cos\\theta)$, similarly: $MF^{2} = 2b^{2}(1 + \\cos\\theta)$. Hence, $S_{\\Delta MAF} = \\frac{1}{2}a^{2}\\sin\\theta$, $S_{\\Delta NBF} = \\frac{1}{2}b^{2}\\sin\\theta$, $(S_{AMNF})^{2} = \\frac{1}{4}MF^{2} \\cdot NF^{2} = a^{2}b^{2}\\sin^{2}\\theta$, therefore $\\frac{\\lambda}{\\mu} = \\frac{(S_{AMNF})^{2}}{S_{\\Delta MAF} \\cdot S_{\\Delta NBF}} = 4$."
  },
  {
    "text": "The focus of the parabola $y^{2}=4 x$ is $F$, and point $P(x, y)$ is a moving point on this parabola. Given point $A(-1,0)$, then the range of $\\frac{|P F|}{| P A|}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;P: Point;Coordinate(P) = (x1, y1);x1: Number;y1: Number;PointOnCurve(P, G) = True;A: Point;Coordinate(A) = (-1, 0)",
    "query_expressions": "Range(Abs(LineSegmentOf(P, F))/Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "[sqrt(2)/2, 1]",
    "fact_spans": "[[[0, 14], [34, 37]], [[0, 14]], [[18, 21]], [[0, 21]], [[22, 32]], [[22, 32]], [[23, 32]], [[23, 32]], [[22, 41]], [[43, 53]], [[43, 53]]]",
    "query_spans": "[[[55, 84]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of the ellipse $C$: $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$ and the eccentricity of the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1(b>0)$ are reciprocals of each other, then $b$=?",
    "fact_expressions": "C: Ellipse;C1:Hyperbola;b: Number;b>0;Expression(C1) = (x^2/4 - y^2/b^2 = 1);Expression(C) = (x^2/16 + y^2/12 = 1);InterReciprocal(Eccentricity(C1), Eccentricity(C))",
    "query_expressions": "b",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[2, 46]], [[51, 103]], [[115, 118]], [[59, 103]], [[51, 103]], [[2, 46]], [[2, 113]]]",
    "query_spans": "[[[115, 120]]]",
    "process": "From the equations of the ellipse and hyperbola, combined with the fact that their eccentricities are reciprocals of each other, the eccentricity of the hyperbola is found to be 2. The value of b can then be determined. From the ellipse equation: e = \\frac{1}{2}, so the eccentricity of the hyperbola is 2, \\therefore \\frac{b^{2}+4}{4} = 4, solving gives b = -2\\sqrt{3} (discarded) or b = 2\\sqrt{3}."
  },
  {
    "text": "Point $P$ is an intersection point of the ellipse $C_{1}$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ and the circle $x^{2}+y^{2}=a^{2}-b^{2}$, and $2 \\angle P F_{2} F_{1}=\\angle P F_{1} F_{2}$, where $F_{1}$, $F_{2}$ are the left and right foci of the ellipse $C_{1}$, respectively. Then the eccentricity of the ellipse $C_{1}$ is?",
    "fact_expressions": "C1: Ellipse;Expression(C1) = (x^2/a^2 + y^2/b^2 = 1);b: Number;a: Number;a > b;b > 0;H: Circle;Expression(H) = (x^2 + y^2 = a^2 - b^2);P: Point;OneOf(Intersection(C1,H))=P;F2: Point;F1: Point;LeftFocus(C1) = F1;RightFocus(C1) = F2;2*AngleOf(P, F2, F1) = AngleOf(P, F1, F2)",
    "query_expressions": "Eccentricity(C1)",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[5, 66], [183, 192], [166, 175]], [[5, 66]], [[16, 66]], [[16, 66]], [[16, 66]], [[16, 66]], [[67, 93]], [[67, 93]], [[0, 4]], [[0, 98]], [[156, 163]], [[148, 155]], [[148, 181]], [[148, 181]], [[100, 145]]]",
    "query_spans": "[[[183, 198]]]",
    "process": "As shown in the figure: since the ellipse $ C_{1}: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ has foci $ F_{1}(-\\sqrt{a^{2}-b^{2}},0) $, $ F_{2}(\\sqrt{a^{2}-b^{2}},0) $, and the circle $ x^{2} + y^{2} = a^{2} - b^{2} $ has radius $ r = \\sqrt{a^{2} - b^{2}} $, therefore $ \\triangle PF_{1}F_{2} $ is a right triangle. Given $ 2\\angle PF_{2}F_{1} = \\angle PF_{1}F_{2} $, so $ \\angle PF_{2}F_{1} = 30^{\\circ} $, $ \\angle PF_{1}F_{2} = 60^{\\circ} $, $ |F_{1}F_{2}| = 2r = 2c $, $ |PF_{2}| = \\sqrt{3}c $, $ |PF_{1}| = c $. By the definition of the ellipse, $ |PF_{2}| + |PF_{1}| = \\sqrt{3}c + c = 2a $, the eccentricity of the ellipse $ e = \\frac{2c}{2a} = \\frac{2c}{\\sqrt{3}c + c} = \\sqrt{3} - 1 $."
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x  (p>0)$ coincides with the right focus of the hyperbola $\\frac{x^{2}}{6}-\\frac{y^{2}}{10}=1$, then the value of the real number $p$ is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Real;Expression(G) = (x^2/6 - y^2/10 = 1);p>0;Expression(H) = (y^2 = 2*(p*x));Focus(H) = RightFocus(G)",
    "query_expressions": "p",
    "answer_expressions": "8",
    "fact_spans": "[[[28, 67]], [[1, 24]], [[75, 80]], [[28, 67]], [[4, 24]], [[1, 24]], [[1, 73]]]",
    "query_spans": "[[[75, 84]]]",
    "process": "The right focus of \\frac{x^{2}}{6}-\\frac{y^{2}}{10}=1 is (4,0), so \\frac{p}{2}=4, \\therefore p=8."
  },
  {
    "text": "The chord $AB$ passing through the focus of the parabola $y^2 = 8x$ has midpoint $(4, a)$, then $|AB| =$?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;Expression(G) = (y^2 = 8*x);IsChordOf(LineSegmentOf(A, B), G);a: Number;Coordinate(MidPoint(LineSegmentOf(A, B))) = (4, a);PointOnCurve(Focus(G), LineSegmentOf(A, B))",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "12",
    "fact_spans": "[[[1, 15]], [[20, 25]], [[20, 25]], [[1, 15]], [[0, 25]], [[26, 35]], [[20, 38]], [[0, 25]]]",
    "query_spans": "[[[40, 49]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the hyperbola $\\frac{y^{2}}{2}-\\frac{x^{2}}{4}=1$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/4 + y^2/2 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 44]]]",
    "process": "From the standard equation of the hyperbola, we know $ a^{2}=2 $, $ b^{2}=4 $, $ c^{2}=a^{2}+b^{2}=6 $, then $ e=\\frac{c}{a}=\\frac{\\sqrt{6}}{\\sqrt{2}}=\\sqrt{3} $. Therefore, the answer should be $ \\sqrt{3} $"
  },
  {
    "text": "Given point $M(0,5)$, point $P$ moves along the curve $x^{2}=8 y$, and point $Q$ moves along the circle $x^{2}+(y-2)^{2}=1$. Then the minimum value of $\\frac{|P M|^{2}}{|P Q|}$ is?",
    "fact_expressions": "G: Circle;H:Curve;M: Point;P: Point;Q: Point;Expression(G) = (x^2 + (y - 2)^2 = 1);Expression(H) = (x^2 = 8*y);Coordinate(M) = (0, 5);PointOnCurve(M,H);PointOnCurve(Q, G);PointOnCurve(P,H)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, M))^2/Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "4*sqrt(10)-8",
    "fact_spans": "[[[38, 58]], [[16, 29]], [[2, 11]], [[11, 15]], [[33, 37]], [[38, 58]], [[16, 29]], [[2, 11]], [[2, 32]], [[33, 61]], [[11, 32]]]",
    "query_spans": "[[[63, 94]]]",
    "process": "The directrix of the parabola $ x^{2} = 8y $ is $ l: y = -2 $, and the focus is $ F(0,2) $. Draw $ PB \\perp l $ from point $ P $, with foot $ B $. By the definition of a parabola, we have $ |PF| = |PB| $. The circle $ x^{2} + (y - 2)^{2} = 1 $ has center $ F(0,2) $ and radius $ r = 1 $. The maximum value of $ |PQ| $ is $ |PF| + r = |PF| + 1 $. From $ \\frac{|PM|^{2}}{|PQ|} \\geqslant \\frac{|PM|^{2}}{|PF| + 1} $, let $ |PF| + 1 = t $ ($ t > 1 $), then $ |PF| = t - 1 = |PB| = y_{P} + 2 $, so $ y_{P} = t - 3 $, $ x_{P}^{2} = 8(t - 3) $. We obtain  \n$$\n\\frac{|PM|^{2}}{|PF| + 1} = \\frac{(t - 3 - 5)^{2} + 8(t - 3)}{t} = t + \\frac{40}{t} - 8 \\geqslant 2\\sqrt{t \\times \\frac{40}{t}} - 8 = 4\\sqrt{10} - 8,\n$$\nwith equality if and only if $ t = \\frac{40}{t} $, i.e., $ t = 2\\sqrt{10} $. Thus, the minimum value of $ \\frac{|PM|^{2}}{|PQ|} $ is $ 4\\sqrt{10} - 8 $."
  },
  {
    "text": "The standard equation of the ellipse $C$ whose center is the center of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, and whose foci are the foci of the hyperbola, and which passes through the point $(5 , 0)$, is?",
    "fact_expressions": "G: Hyperbola;C: Ellipse;H: Point;Expression(G) = (x^2/4 - y^2/5 = 1);Coordinate(H) = (5, 0);Center(G) = Center(C);Focus(G) = Focus(C);PointOnCurve(H,C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/25 + y^2/16 = 1",
    "fact_spans": "[[[1, 39], [54, 57]], [[43, 48], [61, 66], [82, 87]], [[71, 81]], [[1, 39]], [[71, 81]], [[0, 51]], [[53, 69]], [[70, 87]]]",
    "query_spans": "[[[82, 94]]]",
    "process": "The foci of the hyperbola \\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1 are F_{1}(-3,0), F_{2}(3,0). Since the foci of the hyperbola are the foci of ellipse C, the semi-focal length of the ellipse is c=3. Also, since ellipse C passes through the point (5,0), the semi-major axis is a=5. Therefore, b^{2}=16, so the standard equation of ellipse C is \\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1"
  },
  {
    "text": "If the hyperbola $\\frac{x^{2}}{9 k^{2}}-\\frac{y^{2}}{4 k^{2}}=1$ and the circle $x^{2}+y^{2}=1$ have no common points, find the range of real values for $k$.",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/(4*k^2) + x^2/(9*k^2) = 1);k: Real;H: Circle;Expression(H) = (x^2 + y^2 = 1);NumIntersection(G,H)=0",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-\\infty,-1/3)+(1/3,+\\infty)",
    "fact_spans": "[[[1, 51]], [[1, 51]], [[75, 80]], [[52, 68]], [[52, 68]], [[1, 73]]]",
    "query_spans": "[[[75, 87]]]",
    "process": "Since the hyperbola $\\frac{x^{2}}{9k^{2}}-\\frac{y^{2}}{4k^{2}}=1$ and the circle $x^{2}+y^{2}=1$ have no common points, the radius of the circle is less than the length of the real semi-axis of the hyperbola. Thus, we can find the range of values for $k$. The hyperbola $\\frac{x^{2}}{9k^{2}}-\\frac{y^{2}}{4k^{2}}=1$ has its foci on the $x$-axis, so $a^{2}=9k^{2}$, giving the semi-major axis length $a=3|k|$. From $x^{2}+y^{2}=1$, the center of the circle is $(0,0)$ and the radius is $1$. If the hyperbola $\\frac{x^{2}}{9k^{2}}-\\frac{y^{2}}{4k^{2}}=1$ and the circle $x^{2}+y^{2}=1$ have no common points, then $3|k|>1$, that is, $|k|>\\frac{1}{3}$, so $k>\\frac{1}{3}$ or $k<-\\frac{1}{3}$. Therefore, the range of real values of $k$ is $(-\\infty,-\\frac{1}{3})\\cup(\\frac{1}{3},+\\infty)$."
  },
  {
    "text": "The coordinates of the focus of the parabola $y^{2}=-8 x$ are? The equation of the directrix is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -8*x)",
    "query_expressions": "Coordinate(Focus(G));Expression(Directrix(G))",
    "answer_expressions": "(-2, 0)  \nx = 2",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 22]], [[0, 28]]]",
    "process": ""
  },
  {
    "text": "As shown in the figure, through the focus $F$ of the parabola $C$: $y^{2}=2 p x$ ($p>0$), draw a line intersecting $C$ at points $A$ and $B$. From $A$ and $B$, drop perpendiculars to the directrix $l$ of $C$, with feet of perpendiculars $A_{1}$ and $B_{1}$, respectively. Given that the areas of $\\triangle A A_{1} F$ and $\\triangle B B_{1} F$ are $9$ and $1$, respectively, then the area of $\\triangle A_{1} B_{1} F$ is?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;A: Point;A1: Point;F: Point;B: Point;B1: Point;l1: Line;l: Line;l2: Line;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(F, G);Intersection(G, C) = {A, B};PointOnCurve(A, l1);PointOnCurve(B, l2);IsPerpendicular(l1, l);IsPerpendicular(l2, l);FootPoint(l1, l) = A1;FootPoint(l2, l) = B1;Area(TriangleOf(A, A1, F)) = 9;Area(TriangleOf(B, B1, F)) = 1;Directrix(C) = l",
    "query_expressions": "Area(TriangleOf(A1, B1, F))",
    "answer_expressions": "6",
    "fact_spans": "[[[4, 30], [40, 43], [65, 68]], [[12, 30]], [[37, 39]], [[44, 47], [55, 58]], [[81, 88]], [[33, 36]], [[48, 51], [59, 62]], [[89, 96]], [], [[71, 74]], [], [[12, 30]], [[4, 30]], [[4, 36]], [[3, 39]], [[37, 53]], [[54, 77]], [[54, 77]], [[54, 77]], [[54, 77]], [[54, 96]], [[54, 96]], [[99, 155]], [[99, 155]], [[65, 74]]]",
    "query_spans": "[[[157, 187]]]",
    "process": "Let the line $ AB: x = my + \\frac{p}{2} $ be substituted into the parabola equation, eliminating variables yields $ y^{2} - 2pmy - p^{2} = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ y_{1}y_{2} = -p^{2} $, $ y_{1} + y_{2} = 2pm $.  \n$ S_{\\Delta AA_{1}F} = \\frac{1}{2}|AA_{1}| \\times |y_{1}| = \\frac{1}{2}|x_{1} + \\frac{p}{2}| \\times |y_{1}| = \\frac{1}{2}\\left(\\frac{y_{1}^{2}}{2p} + \\frac{p}{2}\\right) \\times |y_{1}| = 9 $  \n$ S_{\\Delta BB_{1}F} = \\frac{1}{2}|BB_{1}| \\times |y_{2}| = \\frac{1}{2}|x_{2} + \\frac{p}{2}| \\times |y_{2}| = \\frac{1}{2}\\left(\\frac{y_{2}^{2}}{2p} + \\frac{p}{2}\\right) \\times |y_{2}| = 1 $  \n$ \\therefore S_{\\Delta AA_{1}F} \\cdot S_{\\Delta BB_{1}F} = \\left[\\frac{(y_{1}y_{2})^{2}}{4p^{2}} + \\frac{p^{2}}{4} + \\frac{1}{4}(y_{1}^{2} + y_{2}^{2})\\right] \\times |y_{1}y_{2}| = \\left[\\frac{p^{4}}{4p^{2}} + \\frac{p^{2}}{4} + \\frac{1}{4}(4p^{2}m^{2} + 2p^{2})\\right] \\times p^{2} = \\frac{p^{2}}{4}(m^{2} + 1) = 9 $,  \n$ m^{2} + 1 = \\frac{36}{p^{4}} $  \n$ \\therefore s_{\\Delta A_{1}B_{1}F} = \\frac{p}{2}|y_{1} - y_{2}| = \\frac{p}{2}\\sqrt{(y_{1} + y_{2})^{2} - 4y_{1}y_{2}} = p\\sqrt{m^{2} + 1} = 6 $"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ with foci $F_{1}$, $F_{2}$, and a point $P$ on the ellipse such that $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$, find the inradius of $\\Delta F_{1} P F_{2}$.",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);AngleOf(F1, P, F2) = pi/3",
    "query_expressions": "Radius(InscribedCircle(TriangleOf(F1, P, F2)))",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 41], [64, 66]], [[2, 41]], [[44, 51]], [[52, 59]], [[2, 59]], [[60, 63]], [[60, 69]], [[71, 107]]]",
    "query_spans": "[[[109, 139]]]",
    "process": "From the given equation, we have $a=5$, $b=4$, $\\therefore c=\\sqrt{a^{2}-b^{2}}=3$, i.e., $|F_{1}F_{2}|=6$. Let $|PF_{1}|=t_{1}$, $|PF_{2}|=t_{2}$, then according to the definition of ellipse, we get: $t_{1}+t_{2}=10$, $\\textcircled{1}$ In $\\triangle F_{1}PF_{2}$, since $\\angle F_{1}PF_{2}=\\frac{\\pi}{3}$, $\\therefore$ by the law of cosines: $t_{1}^{2}+t_{2}^{2}-2t_{1}t_{2}\\cdot\\cos\\frac{\\pi}{3}=6^{2}$, $\\textcircled{2}$ Solving $\\textcircled{1}$ and $\\textcircled{2}$ together gives $t_{1}\\cdot t_{2}=\\frac{64}{3}$, $\\therefore S_{\\Delta F_{1}PF_{2}}=\\frac{1}{2}t_{1}t_{2}\\cdot\\sin\\frac{\\pi}{3}=\\frac{1}{2}\\times\\frac{64}{3}\\times\\frac{\\sqrt{3}}{2}=\\frac{16\\sqrt{3}}{3}$. Let the inradius of $\\triangle F_{1}PF_{2}$ be $r$, since the perimeter of $\\triangle F_{1}PF_{2}$ is $L=10+6=16$, area is $S=\\frac{16\\sqrt{3}}{3}$, then $S_{\\triangle F_{1}PF_{2}}=\\frac{1}{2}Lr$, $\\therefore r=\\frac{2S}{L}=\\frac{2\\sqrt{3}}{3}$."
  },
  {
    "text": "The parabola $C$: $y^{2}=4 x$ has focus $F$, and the line $l$ intersects $C$ at points $A$ and $B$. The perpendicular bisector of segment $A B$ intersects the $x$-axis at point $P(4,0)$. Then $|A F|+|B F|=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);Focus(C) = F;F: Point;l: Line;Intersection(l,C) = {A,B};A: Point;B: Point;Intersection(PerpendicularBisector(LineSegmentOf(A,B)),xAxis) = P;P: Point;Coordinate(P) = (4, 0)",
    "query_expressions": "Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F))",
    "answer_expressions": "6",
    "fact_spans": "[[[0, 19], [33, 36]], [[0, 19]], [[0, 26]], [[23, 26]], [[27, 32]], [[27, 47]], [[38, 41]], [[42, 45]], [[48, 76]], [[67, 76]], [[67, 76]]]",
    "query_spans": "[[[78, 93]]]",
    "process": "From the given conditions, F(1,0). Let M(x_{0},y_{0}) be the midpoint of segment AB. Then |AF|+|BF|=x_{A}+1+x_{B}+1=2x_{0}+2. Let the slope of line l be k. Then the equation of the perpendicular bisector of segment AB is y-y_{0}=-\\frac{1}{k}(x-x_{0}). Setting y=0, we get x=ky_{0}+x_{0}, so 4=ky_{0}+x_{0}. Also, \\begin{cases}y_{A}^{2}=4x_{A}\\\\y_{B}^{2}=4x_{B}\\end{cases}, subtracting gives \\frac{y_{A}-y_{B}}{x_{A}-x_{B}}=\\frac{4}{y_{A}+y_{B}}, simplifying yields ky_{0}=2. Thus x_{0}=2, |AF|+|BF|=6"
  },
  {
    "text": "Given the hyperbola $x^{2}-y^{2}=1$, $A_{1}$, $A_{2}$ are its two vertices, point $P$ is a point on the hyperbola, and the slope of line $P A_{1}$ is $\\frac{1}{2}$, then the slope of line $P A_{2}$ is?",
    "fact_expressions": "G: Hyperbola;A1: Point;P: Point;A2: Point;Expression(G) = (x^2 - y^2 = 1);Vertex(G)={A1,A2};PointOnCurve(P, G);Slope(LineOf(P,A1)) = 1/2",
    "query_expressions": "Slope(LineOf(P,A2))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 20], [37, 38], [49, 52]], [[21, 28]], [[44, 48]], [[29, 36]], [[2, 20]], [[21, 43]], [[44, 55]], [[57, 85]]]",
    "query_spans": "[[[87, 103]]]",
    "process": "Let $ P(x_{0},y_{0}) $, then $ x_{0}^{2}-y_{0}^{2}=1 $, $ \\frac{y_{0}^{2}}{x_{0}^{2}-1}=1 $. Since $ A_{1}(-1,0) $, $ A_{2}(1,0) $, let the slope of line $ PA_{1} $ be $ k_{1} $, and the slope of line $ PA_{2} $ be $ k_{2} $. Therefore, $ k_{1}k_{2}=\\frac{y_{0}}{x_{0}+1}\\cdot\\frac{y_{0}}{x_{0}-1}=\\frac{y_{0}^{2}}{x_{0}^{2}-1}=1 $. Since $ k_{1}=\\frac{1}{2} $, $ \\therefore k_{2}=2 $."
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ is $3x+2y=0$, then $b$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/b^2 = 1);b: Number;b>0;Expression(OneOf(Asymptote(G))) = (3*x + 2*y = 0)",
    "query_expressions": "b",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 50]], [[2, 50]], [[72, 75]], [[5, 50]], [[2, 70]]]",
    "query_spans": "[[[72, 77]]]",
    "process": "The asymptotes of $\\frac{x^{2}}{16}-\\frac{y^{2}}{b^{2}}=1$ are given by $y=\\pm\\frac{b}{4}x$. Since the slope of $3x+2y=0$ is $-\\frac{3}{2}$, it follows that $\\frac{b}{4}=\\frac{3}{2} \\Rightarrow b=6$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$ with foci $F_{1}$ and $F_{2}$, and a point $P(x_{0}, y_{0})$ satisfying $0<\\frac{x_{0}^{2}}{2}+y_{0}^{2}<1$, then the range of values for $|P F_{1}|+|P F_{2}|$ is?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/2 + y^2 = 1);Coordinate(P) = (x0, y0);Focus(C) = {F1,F2};0<x0^2/2+y0^2;x0^2/2+y0^2<1;x0:Number;y0:Number",
    "query_expressions": "Range(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)))",
    "answer_expressions": "[2,2*sqrt(2)]",
    "fact_spans": "[[[2, 34]], [[55, 73]], [[39, 46]], [[47, 54]], [[2, 34]], [[55, 73]], [[2, 54]], [[75, 110]], [[75, 110]], [[75, 110]], [[75, 110]]]",
    "query_spans": "[[[112, 140]]]",
    "process": ""
  },
  {
    "text": "If $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{4}-y^{2}=1$, and point $P$ lies on the hyperbola $C$ such that $\\angle F_{1} P F_{2}=60^{\\circ}$, then what is the distance from $P$ to the $x$-axis?",
    "fact_expressions": "C: Hyperbola;F1: Point;P: Point;F2: Point;Expression(C) = (x^2/4 - y^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Distance(P, xAxis)",
    "answer_expressions": "sqrt(15)/5",
    "fact_spans": "[[[17, 50], [62, 68]], [[1, 8]], [[57, 61], [105, 108]], [[9, 16]], [[17, 50]], [[1, 56]], [[1, 56]], [[57, 69]], [[70, 103]]]",
    "query_spans": "[[[105, 118]]]",
    "process": "According to the hyperbola equation, |F_{1}F_{2}| can be obtained; according to the area formula of the hyperbola focal triangle, S_{AF}PF_{2} can be obtained. Without using the area bridge, the detailed solution is as follows: From the hyperbola equation, we know: F_{1}(-\\sqrt{5},0), F_{2}(\\sqrt{5},0), |F_{1}F_{2}|=2\\sqrt{5}. From the hyperbola focal triangle area formula: S_{\\triangle F_{1}PF_{2}}=b^{2}\\cot\\frac{\\angle F_{1}PF_{2}}{2}=\\cot30^{\\circ}=\\sqrt{3}. Also, S_{AF_{1}PF_{2}}=\\frac{1}{2}|F_{1}F_{2}|\\cdot|y_{P}|, |y_{P}|=\\frac{2\\sqrt{3}}{2\\sqrt{5}}=\\frac{\\sqrt{15}}{5}, so the distance from P to the x-axis is \\frac{\\sqrt{15}}{5}."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ with focus $F$, points $A$ and $B$ are two moving points on the parabola such that $\\angle A F B=60^{\\circ}$. Draw a perpendicular line $M N$ from the midpoint $M$ of chord $A B$ to the directrix of the parabola, with foot $N$. Then the minimum value of $\\frac{|A B|}{|M N|}$ is?",
    "fact_expressions": "C: Parabola;p: Number;A: Point;B: Point;M: Point;N: Point;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(A,C);PointOnCurve(B,C);AngleOf(A, F, B) = ApplyUnit(60, degree);IsChordOf(LineSegmentOf(A, B), C);MidPoint(LineSegmentOf(A, B)) = M;PointOnCurve(M,LineSegmentOf(M,N));IsPerpendicular(Directrix(C),LineSegmentOf(M,N));FootPoint(Directrix(C),LineSegmentOf(M,N))=N",
    "query_expressions": "Min(Abs(LineSegmentOf(A, B))/Abs(LineSegmentOf(M, N)))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 28], [45, 48], [96, 99]], [[10, 28]], [[36, 40]], [[41, 44]], [[92, 95]], [[113, 116]], [[32, 35]], [[10, 28]], [[2, 28]], [[2, 35]], [[36, 54]], [[36, 54]], [[56, 81]], [[45, 89]], [[84, 95]], [[82, 109]], [[96, 109]], [[96, 116]]]",
    "query_spans": "[[[118, 145]]]",
    "process": "As shown in the figure: draw perpendiculars from A and C to the directrix, with feet of perpendiculars at C and D respectively. Let |AF| = m, |BF| = n, then |AC| = m, |BD| = n. M is the midpoint of AB, and AC, MN, BD are all perpendicular to the directrix l, hence they are parallel, so |MN| = \\frac{1}{2}(|AC| + |BD|) = \\frac{m+n}{2}. By the cosine law: \\times(\\frac{m+n)^{2}}{2} = \\frac{m+n}{2}, equality holds if and only if m = n. Therefore, \\frac{|AB|}{|MN|} \\geqslant \\frac{\\frac{1}{2}(m+n)}{2}(m+n) = 1, that is, the minimum value of \\frac{|AB|}{|MN|} is 1."
  },
  {
    "text": "If the line $y=kx+2$ and the parabola $y^{2}=4x$ have only one common point, then the real number $k$=?",
    "fact_expressions": "G: Parabola;H: Line;k: Real;Expression(G) = (y^2 = 4*x);Expression(H) = (y = k*x + 2);NumIntersection(H, G)=1",
    "query_expressions": "k",
    "answer_expressions": "{0,1/2}",
    "fact_spans": "[[[12, 26]], [[1, 11]], [[35, 40]], [[12, 26]], [[1, 11]], [[1, 33]]]",
    "query_spans": "[[[35, 42]]]",
    "process": ""
  },
  {
    "text": "If point $A(3 , 1)$, $F$ is the focus of the parabola $y^{2}=2 x$, and point $M$ moves on the parabola, then the coordinates of point $M$ when $|MA|+|MF|$ takes the minimum value are?",
    "fact_expressions": "G: Parabola;A: Point;M: Point;F: Point;Expression(G) = (y^2 = 2*x);Coordinate(A) = (3, 1);Focus(G) = F;PointOnCurve(M, G);WhenMin(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F)))",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(1/2,1)",
    "fact_spans": "[[[17, 31], [40, 43]], [[1, 12]], [[35, 39], [66, 70]], [[13, 16]], [[17, 31]], [[1, 12]], [[13, 34]], [[35, 46]], [[49, 65]]]",
    "query_spans": "[[[66, 75]]]",
    "process": ""
  },
  {
    "text": "If the distance from a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ to an asymptote is equal to $\\frac{1}{4}$ of the focal distance, then what is the equation of the asymptotes of this hyperbola?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Distance(OneOf(Focus(G)),OneOf(Asymptote(G)))=(1/4)*FocalLength(G)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "x \\pm \\sqrt{3} y = 0",
    "fact_spans": "[[[1, 58], [93, 96]], [[4, 58]], [[4, 58]], [[4, 58]], [[4, 58]], [[1, 58]], [[1, 90]]]",
    "query_spans": "[[[93, 104]]]",
    "process": ""
  },
  {
    "text": "The asymptotes of the hyperbola $C$: $\\frac{x^{2}}{3}-y^{2}=1$ are? If the focus of the parabola $y^{2}=2 p x(p>0)$ coincides with one focus of the hyperbola $C$, then $p$=?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/3 - y^2 = 1);G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;Focus(G) = OneOf(Focus(C))",
    "query_expressions": "Expression(Asymptote(C));p",
    "answer_expressions": "y=pm*sqrt(3)*x/3\n4",
    "fact_spans": "[[[0, 33], [67, 73]], [[0, 33]], [[42, 63]], [[42, 63]], [[82, 85]], [[45, 63]], [[42, 80]]]",
    "query_spans": "[[[0, 41]], [[82, 87]]]",
    "process": "From the given conditions, $ a = \\sqrt{3} $, $ b = 1 $, $ c = \\sqrt{a^{2} + b^{2}} = 2 $, $ \\therefore $ the asymptotes are $ y = \\pm\\frac{b}{a}x = \\pm\\frac{\\sqrt{3}}{3}x $, $ \\frac{p}{2} = 2 \\Rightarrow p = 4 $, hence fill in: $ y = \\pm\\frac{\\sqrt{3}}{3}x'4 $."
  },
  {
    "text": "Let the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ be $F_{1}$, $F_{2}$ respectively, the right vertex be $A$, and the upper vertex be $B$. Given that $A B=\\frac{\\sqrt{3}}{2} F_{1} F_{2}$, what is the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;A: Point;B: Point;RightVertex(G) = A;UpperVertex(G) = B;LineSegmentOf(A, B) = (sqrt(3)/2)*LineSegmentOf(F1, F2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[1, 53], [135, 137]], [[1, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[61, 68]], [[69, 76]], [[1, 76]], [[1, 76]], [[82, 85]], [[90, 93]], [[1, 85]], [[1, 93]], [[97, 133]]]",
    "query_spans": "[[[135, 143]]]",
    "process": "From the given conditions, we have |AB|^{2}=a^{2}+b^{2}, |F_{1}F_{2}|^{2}=4c^{2}, and combining with AB=\\frac{\\sqrt{3}}{2}|F_{1}F_{2}|, simplifying yields a^{2}=2c^{2}, thus solving. Solution: From the given, |AB|^{2}=a^{2}+b^{2}, |F_{1}F_{2}|^{2}=4c^{2}, AB=\\frac{\\sqrt{3}}{2}|F_{1}F_{2}|. \\cdota^{2}+b^{2}=\\frac{3}{4}\\cdot4c^{2}, i.e., a^{2}+a^{2}-c^{2}=3c^{2}, so a^{2}=2c^{2}, hence e=\\frac{c}{a}=\\frac{\\sqrt{2}}{2}"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the right focus is $F$. A line with slope $-1$ passing through $F$ intersects the asymptotes of the hyperbola at point $P$, where $P$ lies in the first quadrant. $O$ is the origin. If the area of $\\triangle O F P$ is $\\frac{a^{2}+b^{2}}{8}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;O: Origin;F: Point;P: Point;a>0;b>0;l:Line;RightFocus(G) = F;PointOnCurve(F,l);Slope(l)=-1;Intersection(l,Asymptote(G))=P;Quadrant(P)=1;Area(TriangleOf(O, F, P)) = (a^2 + b^2)/8",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)/3",
    "fact_spans": "[[[2, 58], [83, 86], [163, 166]], [[2, 58]], [[5, 58]], [[5, 58]], [[106, 109]], [[63, 66], [68, 71]], [[91, 95], [96, 100]], [[5, 58]], [[5, 58]], [[80, 82]], [[2, 66]], [[67, 82]], [[72, 82]], [[80, 95]], [[96, 105]], [[116, 160]]]",
    "query_spans": "[[[163, 172]]]",
    "process": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) has an asymptote with equation $y=\\frac{b}{a}x$. The line passing through the focus with slope $-1$ has equation $y=-(x-c)$. Solving the system\n\\[\n\\begin{cases}\ny=\\frac{b}{a}x \\\\\ny=-x+c\n\\end{cases}\n\\]\nyields $ay=b(c-y)$, so $y=\\frac{bc}{a+b}$; then $S_{\\triangle OPF}=\\frac{1}{2}\\times\\frac{bc}{a+b}\\times c=\\frac{a^{2}+b^{2}}{8}$, solving gives $a=3b$, thus $c=\\sqrt{10}b$, and the eccentricity of the hyperbola is $e=\\frac{c}{a}=\\frac{\\sqrt{10}}{3}$."
  },
  {
    "text": "Given that the directrix of the parabola $y^{2}=2 p x(p>0)$ is tangent to the circle $(x-3)^{2}+y^{2}=225$, and one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $y=\\sqrt{3} x$, with one of its foci being the focus of the parabola, find the length of the real axis of the hyperbola?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;I: Circle;Expression(I) = (y^2 + (x - 3)^2 = 225);IsTangent(Directrix(G), I);C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;Expression(OneOf(Asymptote(C))) = (y = sqrt(3)*x);OneOf(Focus(C)) = Focus(G)",
    "query_expressions": "Length(RealAxis(C))",
    "answer_expressions": "12",
    "fact_spans": "[[[2, 23], [140, 143]], [[2, 23]], [[5, 23]], [[5, 23]], [[27, 49]], [[27, 49]], [[2, 51]], [[52, 108], [132, 133], [148, 151]], [[52, 108]], [[55, 108]], [[55, 108]], [[55, 108]], [[55, 108]], [[52, 131]], [[132, 146]]]",
    "query_spans": "[[[148, 155]]]",
    "process": ""
  },
  {
    "text": "If the equation of the hyperbola is $x^{2}-2 y^{2}=1$, then the coordinates of its left focus are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - 2*y^2 = 1)",
    "query_expressions": "Coordinate(LeftFocus(G))",
    "answer_expressions": "(-sqrt(6)/2,0)",
    "fact_spans": "[[[1, 4], [26, 27]], [[1, 24]]]",
    "query_spans": "[[[26, 36]]]",
    "process": ""
  },
  {
    "text": "It is known that an ellipse and a hyperbola share common foci $F_{1}$, $F_{2}$, and $P$ is one of their intersection points, with $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$. Let the eccentricities of the ellipse and the hyperbola be $e_{1}$, $e_{2}$ respectively. Then the maximum value of $\\frac{1}{e_{1} e_{2}}$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;F1: Point;P: Point;F2: Point;e1:Number;e2:Number;Focus(G)={F1,F2};Focus(H)={F1,F2};OneOf(Intersection(G,H))=P;AngleOf(F1, P, F2) = pi/3;Eccentricity(H)=e1;Eccentricity(G)=e2",
    "query_expressions": "Max(1/(e1*e2))",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[5, 8], [83, 86]], [[2, 4], [80, 82]], [[13, 20]], [[29, 32]], [[21, 28]], [[93, 100]], [[102, 109]], [[2, 28]], [[2, 28]], [[29, 40]], [[42, 78]], [[80, 109]], [[80, 109]]]",
    "query_spans": "[[[111, 140]]]",
    "process": "Let the semi-major axis of the ellipse be $a_{1}$, and the semi-transverse axis of the hyperbola be $a_{2}$. According to the definitions of the ellipse and hyperbola: $|PF_{1}|+|PF_{2}|=2a_{1}$, $|PF_{1}|-|PF_{2}|=2a_{2}$, solving gives $|PF_{1}|=a_{1}+a_{2}$, $|PF_{2}|=a_{1}-a_{2}$. Let $|F_{1}F_{2}|=2c$, $\\angle F_{1}PF_{2}=\\frac{\\pi}{3}$, then in $\\triangle F_{1}PF_{2}$, by the cosine law we obtain: $4c^{2}=(a_{1}+a_{2})^{2}+(a_{1}-a_{2})^{2}-2(a_{1}+a_{2})(a_{1}-a_{2})\\cos\\frac{\\pi}{3}$, simplifying yields $a_{1}^{2}+3a_{2}^{2}=4c^{2}$, i.e., $\\frac{1}{e_{1}^{2}}+\\frac{3}{e_{2}^{2}}=4\\geqslant\\frac{2\\sqrt{3}}{e_{1}e_{2}}$, $\\therefore \\frac{1}{e_{1}e_{2}}\\leqslant\\frac{2}{3}\\sqrt{3}$, hence fill in $\\frac{2}{3}\\sqrt{3}$."
  },
  {
    "text": "If a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1$ lies on the directrix of the parabola $y^{2}=8 x$, then the equations of the asymptotes of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;Expression(G) = (-y^2/3 + x^2/a^2 = 1);Expression(H) = (y^2 = 8*x);PointOnCurve(OneOf(Focus(G)), Directrix(H))",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[1, 43], [69, 72]], [[4, 43]], [[49, 63]], [[1, 43]], [[49, 63]], [[1, 67]]]",
    "query_spans": "[[[69, 80]]]",
    "process": "The directrix of the parabola \\( y^{2} = 8x \\) is \\( x = -2 \\). According to the problem, one focus of the hyperbola is \\( (-2, 0) \\). From the hyperbola \\( \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{3} = 1 \\), we obtain \\( a^{2} + 3 = 4 \\), solving gives \\( a^{2} = 1 \\), so the equation of the hyperbola is \\( x^{2} - \\frac{y^{2}}{3} = 1 \\). Thus, the asymptotes are \\( y = \\pm\\sqrt{3}x \\)."
  },
  {
    "text": "Given $E(1,-2)$, $F(-3,4)$, and $M$ is a moving point on the plane satisfying $|M E| = \\frac{3}{2}|M F|$, then the equation of the trajectory of $M$ is?",
    "fact_expressions": "E: Point;F: Point;M: Point;Coordinate(E) = (1, -2);Coordinate(F) = (-3, 4);Abs(LineSegmentOf(M,E))=(3/2)*Abs(LineSegmentOf(M,F))",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2+y^2+(62/5)*x-(88/5)*y+41=0",
    "fact_spans": "[[[2, 11]], [[14, 24]], [[25, 29], [65, 68]], [[2, 11]], [[14, 24]], [[39, 63]]]",
    "query_spans": "[[[65, 75]]]",
    "process": "Let M(x, y). From the condition |ME| = \\frac{3}{2}|MF|, we obtain \\sqrt{(x+1)^{2}+(y+2)^{2}} = \\frac{3}{2}\\sqrt{(x+3)^{2}+(y-4)}. Squaring both sides and simplifying yields x^{2}+y^{2}+\\frac{62}{5}x-\\frac{88}{5}y+41=0"
  },
  {
    "text": "The equation of the directrix of the parabola $y=a x^{2}$ is $y=1$. Then the real number $a$=?",
    "fact_expressions": "G: Parabola;a: Real;Expression(G) = (y = a*x^2);Expression(Directrix(G)) = (y = 1)",
    "query_expressions": "a",
    "answer_expressions": "-1/4",
    "fact_spans": "[[[0, 14]], [[27, 32]], [[0, 14]], [[0, 25]]]",
    "query_spans": "[[[27, 34]]]",
    "process": "The parabola $ y = ax^{2} $ is transformed into the standard form: $ x^{2} = \\frac{1}{a}y' $. Its directrix equation is $ y = 1 $, and $ x^{2} = -2 \\times \\frac{1}{(-2a)}y $, $ a < 0 $, so $ \\frac{1}{4a} = 1 $, thus $ a = -\\frac{1}{4} $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4 x$, what is the equation of the directrix of the parabola $C$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x)",
    "query_expressions": "Expression(Directrix(C))",
    "answer_expressions": "x=-1",
    "fact_spans": "[[[2, 21], [23, 29]], [[2, 21]]]",
    "query_spans": "[[[23, 36]]]",
    "process": "From the parabola equation, solve for p=2, and find the directrix equation. Parabola C: y^{2}=4x, so 2p=4, p=2, the directrix equation is x=-\\frac{p}{2}=-1,"
  },
  {
    "text": "The standard equation of an ellipse that has the same foci as the hyperbola $\\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1$ and has eccentricity $\\frac{3}{5}$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/5 - y^2/4 = 1);Focus(G) = Focus(H);H: Ellipse;Eccentricity(H) = 3/5",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2/25 + y^2/16 = 1",
    "fact_spans": "[[[1, 39]], [[1, 39]], [[0, 67]], [[65, 67]], [[47, 67]]]",
    "query_spans": "[[[65, 74]]]",
    "process": ""
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ are $F_{1}$ and $F_{2}$ respectively, and the chord $AB$ passes through $F_{1}$. If the circumference of the incircle of $\\triangle ABF_{2}$ is $\\pi$, and the coordinates of points $A$ and $B$ are $(x_{1}, y_{1})$, $(x_{2}, y_{2})$, then what is the value of $|y_{1}-y_{2}|$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);PointOnCurve(F1, LineSegmentOf(A, B));Perimeter(InscribedCircle(TriangleOf(A, B, F2))) = pi;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2)",
    "query_expressions": "Abs(y1 - y2)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[0, 39]], [[0, 39]], [[47, 54], [70, 77]], [[55, 62]], [[0, 62]], [[0, 62]], [[114, 117]], [[119, 122]], [[0, 69]], [[64, 77]], [[79, 112]], [[130, 146]], [[148, 164]], [[130, 146]], [[148, 164]], [[114, 164]], [[114, 164]]]",
    "query_spans": "[[[167, 185]]]",
    "process": "By the given conditions, we have $F_{1}(-3,0)$, $F_{2}(3,0)$. Since chord $AB$ passes through point $F_{1}$, the perimeter $l$ of $\\Delta ABF_{2}$ is $|AB|+|AF_{2}|+|BF_{2}|=|AF_{1}|+|AF_{2}|+|BF_{1}|+|BF_{2}|=4a=20$. Because the circumference of the incircle of $\\triangle ABF_{2}$ is $\\pi$, the radius of its incircle is $r=\\frac{1}{2}$. Then $S_{\\triangle ABF_{2}}=\\frac{1}{2}rl=5$, while $S_{\\Delta ABF_{2}}=S_{\\Delta AF_{1}F_{2}}+S_{\\Delta BF_{1}F_{2}}=\\frac{1}{2}|y_{1}-y_{2}|\\cdot|F_{1}F_{2}|$, so $|y_{1}-y_{2}|=\\frac{2S_{\\triangle ABF}}{|F_{1}F_{2}|}=\\frac{2\\cdot5}{6}=\\frac{5}{3}$."
  },
  {
    "text": "Given that point $P(m , 4)$ lies on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F_{1}$ and $F_{2}$ are the two foci of the ellipse, and the radius of the incircle of $\\triangle P F_{1} F_{2}$ is $\\frac{3}{2}$, then what is the eccentricity of this ellipse?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;m:Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (m, 4);PointOnCurve(P, G);Focus(G) = {F1, F2};Radius(InscribedCircle(TriangleOf(P,F1,F2)))=3/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3/5",
    "fact_spans": "[[[14, 66], [87, 89], [145, 147]], [[16, 66]], [[16, 66]], [[2, 13]], [[71, 78]], [[79, 86]], [[3, 13]], [[16, 66]], [[16, 66]], [[14, 66]], [[2, 13]], [[2, 70]], [[71, 94]], [[96, 142]]]",
    "query_spans": "[[[145, 153]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the directrix of the parabola $y=a x^{2}  (a \\neq 0)$?",
    "fact_expressions": "G: Parabola;a: Number;Expression(G) = (y = a*x^2);Negation(a=0)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=-1/(4*a)",
    "fact_spans": "[[[0, 26]], [[3, 26]], [[0, 26]], [[3, 26]]]",
    "query_spans": "[[[0, 33]]]",
    "process": ""
  },
  {
    "text": "Given that the hyperbola passes through the point $(2, \\sqrt{2})$ and has two asymptotes $y = \\pm \\frac{1}{2}x$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (2, sqrt(2));PointOnCurve(H, G);Expression(Asymptote(G))=(y=pm*(x/2))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 - x^2/4 = 1",
    "fact_spans": "[[[2, 5], [2, 5]], [[6, 22]], [[6, 22]], [[2, 22]], [[2, 51]]]",
    "query_spans": "[[[54, 61]]]",
    "process": "Since the point (2,\\sqrt{2}) is above and to the left of y=\\frac{1}{2}x, the hyperbola's foci lie on the y-axis. Let the equation be \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1, (a>0,b>0). Then \\begin{cases}b=2a\\\\\\frac{2}{a^{2}}-\\frac{4}{b^{2}}=1\\end{cases}. Solving gives \\begin{cases}a=1\\\\b=2\\end{cases}. Thus, the equation is y^{2}-\\frac{x^{2}}{4}=1."
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has an asymptote with equation $y=\\frac{4}{5} x$. What is its eccentricity?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (y = (4/5)*x)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(41)/5",
    "fact_spans": "[[[0, 46], [73, 74]], [[3, 46]], [[3, 46]], [[0, 46]], [[0, 71]]]",
    "query_spans": "[[[73, 79]]]",
    "process": ""
  },
  {
    "text": "Given that point $M$ moves on the ellipse $\\frac{x^{2}}{18}+\\frac{y^{2}}{9}=1$, and point $N$ moves on the circle $x^{2}+(y-1)^{2}=1$, then the maximum value of $|M N|$ is?",
    "fact_expressions": "G: Ellipse;H: Circle;M: Point;N: Point;Expression(G) = (x^2/18 + y^2/9 = 1);Expression(H) = (x^2 + (y - 1)^2 = 1);PointOnCurve(M, G);PointOnCurve(N, H)",
    "query_expressions": "Max(Abs(LineSegmentOf(M, N)))",
    "answer_expressions": "2*sqrt(5) + 1",
    "fact_spans": "[[[7, 45]], [[54, 74]], [[2, 6]], [[49, 53]], [[7, 45]], [[54, 74]], [[2, 46]], [[49, 75]]]",
    "query_spans": "[[[79, 92]]]",
    "process": "Let point M be $(x_{0},y_{0})$, $y_{0}\\in[-3,3]$, then $\\frac{x_{0}^{2}}{18}+\\frac{y_{0}^{2}}{9}=1^{2}$. Then $x_{0}^{2}=18-2y_{0}^{2}$. Let the center of the circle $x^{2}+(y-1)^{2}=1$ be P, then the coordinates of P are $(0,1)$. The maximum value of $|MN|$ is the sum of the maximum value of $|MP|$ and the radius $1$ of the circle $x^{2}+(y-1)^{2}=1$. When $y_{0}\\in[-3,3]$, $|MP|\\leqslant2\\sqrt{5}$, with equality if and only if $y_{0}=-1$. Hence, $|MN|\\leqslant2\\sqrt{5}+1$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+y^{2}=1$ $(a>0)$, if the distance from its left focus to the right vertex is $2$, then the value of $a$ is?",
    "fact_expressions": "G: Ellipse;a: Number;a>0;Expression(G) = (y^2 + x^2/a^2 = 1);Distance(LeftFocus(G),RightVertex(G))=2",
    "query_expressions": "a",
    "answer_expressions": "5/4",
    "fact_spans": "[[[2, 38], [40, 41]], [[57, 60]], [[4, 38]], [[2, 38]], [[40, 55]]]",
    "query_spans": "[[[57, 64]]]",
    "process": "From the given information, the left focus is $(-\\sqrt{a^{2}-1},0)$, and the right vertex is $(a,0)$. The distance from the left focus to the right vertex is $\\sqrt{a^{2}-1}+a=2$, leading to the answer. [Detailed solution] From the given, $a^{2}>b^{2}=1^{2}=a^{2}-b^{2}=a^{2}-1$ $(a>0)$, so the left focus is $(-\\sqrt{a^{2}-1},0)$, the right vertex is $(a,0)$, and the distance from the left focus to the right vertex is $\\sqrt{a^{2}-1}+a=2$. Solving gives $a=\\frac{5}{4}$."
  },
  {
    "text": "If the circle $(x-4)^{2}+y^{2}=4$ is tangent to the asymptotes of the hyperbola $C$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0, b>0)$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "G: Circle;Expression(G) = (y^2 + (x - 4)^2 = 4);C: Hyperbola;Expression(C) = (-x^2/b^2 + y^2/a^2 = 1);b: Number;a: Number;a>0;b>0;IsTangent(G, Asymptote(C))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 21]], [[1, 21]], [[22, 83], [91, 97]], [[22, 83]], [[30, 83]], [[30, 83]], [[30, 83]], [[30, 83]], [[1, 89]]]",
    "query_spans": "[[[91, 103]]]",
    "process": "The asymptotes of the hyperbola are given by $ ax-by=0 $. The distance from the center of the circle to the line equals the radius, leading to $ \\frac{b^{2}}{a^{2}}=3 $. Finally, the eccentricity is obtained using its definition. Let one asymptote of the hyperbola be $ y=\\frac{a}{b}x $, that is, $ ax-by=0 $. Since it is tangent to the circle $ (x-4)^{2}+y^{2}=4 $, we have $ \\frac{|4a|}{\\sqrt{a^{2}+b^{2}}}=2 $. Rearranging yields $ \\frac{b^{2}}{a^{2}}=3 $, hence the eccentricity is $ e=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=2 $."
  },
  {
    "text": "A line $l$ passing through the point $M(2,1)$ intersects the ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ at points $A$ and $B$, such that $M$ is the midpoint of $AB$. What is the equation of line $l$? (Write in general form)",
    "fact_expressions": "l: Line;C: Ellipse;A: Point;B: Point;M: Point;Expression(C) = (x^2/9 + y^2/5 = 1);Coordinate(M) = (2, 1);PointOnCurve(M, l);Intersection(l, C) = {A, B};MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Expression(l)",
    "answer_expressions": "10*x+9*y-29=0",
    "fact_spans": "[[[12, 17], [86, 91]], [[18, 60]], [[61, 64]], [[65, 68]], [[2, 11], [72, 75]], [[18, 60]], [[2, 11]], [[0, 17]], [[12, 70]], [[72, 84]]]",
    "query_spans": "[[[86, 96]]]",
    "process": "Given the problem, let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Then  \n$$\n\\begin{cases}\n\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1 \\\\\n\\frac{x^{2}}{9}+\\frac{y_{2}}{5}=1\n\\end{cases}\n$$  \nSubtracting gives  \n$$\n\\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = -\\frac{5(x_{1}+x_{2})}{9(y_{1}+y_{2})}\n$$  \nSince $ M $ is the midpoint of $ AB $, we have  \n$$\n\\begin{cases}\nx_{1}+x_{2}=4 \\\\\ny_{1}+y_{2}=2\n\\end{cases}\n$$  \nThen  \n$$\n\\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = -\\frac{5\\times4}{9\\times2} = -\\frac{10}{9},\n$$  \nso $ k_{1} = -\\frac{10}{9} $. Thus, the line $ l $ is  \n$$\ny-1 = -\\frac{10}{9}(x-2).\n$$  \nRearranging gives  \n$$\n10x + 9y - 29 = 0.\n$$"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ has eccentricity $\\sqrt{3}$, then the distance from one of its foci to one of its asymptotes is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/b^2 = 1);b: Number;b>0;Eccentricity(G) = sqrt(3)",
    "query_expressions": "Distance(OneOf(Focus(G)), OneOf(Asymptote(G)))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 49], [66, 67]], [[2, 49]], [[5, 49]], [[5, 49]], [[2, 64]]]",
    "query_spans": "[[[66, 85]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{b^{2}}=1$, a perpendicular is drawn from $F_{1}$ to an asymptote of $C$, with foot of the perpendicular at $P$. If the area of $\\Delta F_{1} P F_{2}$ is $\\sqrt{3}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;F1: Point;P: Point;F2: Point;L:Line;Expression(C) = (x^2 - y^2/b^2 = 1);Focus(C) = {F1, F2};PointOnCurve(F1,L);IsPerpendicular(L,Asymptote(C));FootPoint(L,Asymptote(C))=P;Area(TriangleOf(F1, P, F2)) = sqrt(3)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[18, 54], [69, 72], [127, 130]], [[25, 54]], [[2, 9], [61, 68]], [[83, 86]], [[10, 17]], [], [[18, 54]], [[2, 59]], [[60, 79]], [[60, 79]], [[60, 86]], [[89, 125]]]",
    "query_spans": "[[[127, 136]]]",
    "process": "Given that $ a=1 $, the focus is $ F_{1}(-c,0) $, and the asymptote equation is $ y = -bx $. Using the point-to-line distance formula, we get $ PF_{1} = \\frac{bc}{\\sqrt{b^{2}+1}} = b $. By the Pythagorean theorem, $ PO = a $. In right triangle $ \\triangle F_{1}PO $, using the equal area method, the distance from $ P $ to the $ x $-axis is $ h = \\frac{b}{c} $. Therefore, $ S_{\\triangle F_{1}PF_{2}} = \\frac{1}{2} \\times 2c \\times \\frac{b}{c} = b = \\sqrt{3} $, and the eccentricity is $ e = \\frac{c}{a} = \\sqrt{1 + \\left( \\frac{b}{a} \\right)^{2}} = \\sqrt{1 + 3} = 2 $."
  },
  {
    "text": "Given that $P$ is any point on the parabola $y^{2}=4x$, and $Q$ is any point on the circle $(x-4)^{2}+y^{2}=1$, then the minimum value of $|PQ|$ is?",
    "fact_expressions": "G: Parabola;H: Circle;P: Point;Q: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (y^2 + (x - 4)^2 = 1);PointOnCurve(P, G);PointOnCurve(Q, H)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "2*sqrt(3)-1",
    "fact_spans": "[[[6, 20]], [[30, 50]], [[2, 5]], [[26, 29]], [[6, 20]], [[30, 50]], [[2, 25]], [[26, 55]]]",
    "query_spans": "[[[57, 70]]]",
    "process": "Let the coordinates of point P be $(\\frac{1}{4}m^{2}, m)$, the center A of the circle $(x \\cdot 4)^{2} + y^{2} = 1$ has coordinates $(4, 0)$. $\\therefore |PA|^{2} = (\\frac{1}{4}m^{2} - 4)^{2} + m^{2} = \\frac{1}{16}(m^{2} - 8)^{2} + 12 \\geqslant 12$, $\\therefore |PA| \\geqslant 2\\sqrt{3}$. Since Q is any point on the circle $(x \\cdot 4)^{2} + y^{2} = 1$, $\\therefore$ the minimum value of PQ is $2\\sqrt{3}$.1"
  },
  {
    "text": "Let $AB$ be the major axis of ellipse $\\Gamma$, point $C$ lies on $\\Gamma$, and $\\angle CBA = \\frac{\\pi}{4}$. If $AB = 4$, $BC = \\sqrt{2}$, then the distance between the two foci of $\\Gamma$ is?",
    "fact_expressions": "Gamma: Ellipse;A: Point;B: Point;MajorAxis(Gamma) = LineSegmentOf(A, B);C: Point;PointOnCurve(C, Gamma);AngleOf(C, B, A) = pi/4;LineSegmentOf(A, B) = 4;LineSegmentOf(B, C) = sqrt(2);F1: Point;F2: Point;Focus(Gamma) = {F1, F2}",
    "query_expressions": "Distance(F1, F2)",
    "answer_expressions": "(4*sqrt(6))/3",
    "fact_spans": "[[[7, 17], [26, 34], [90, 98]], [[1, 6]], [[1, 6]], [[1, 20]], [[21, 25]], [[21, 35]], [[37, 63]], [[65, 71]], [[74, 88]], [], [], [[90, 103]]]",
    "query_spans": "[[[90, 110]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$, and $Q$ is a point on the parabola $C$ located in the first quadrant. If $|QF|=5$, then what are the coordinates of point $Q$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 8*x);F: Point;Focus(C) = F;Q: Point;PointOnCurve(Q, C);Quadrant(Q) = 1;Abs(LineSegmentOf(Q, F)) = 5",
    "query_expressions": "Coordinate(Q)",
    "answer_expressions": "(3, 2*sqrt(6))",
    "fact_spans": "[[[2, 21], [33, 39]], [[2, 21]], [[25, 28]], [[2, 28]], [[29, 32], [43, 47], [65, 69]], [[29, 42]], [[43, 52]], [[54, 63]]]",
    "query_spans": "[[[65, 74]]]",
    "process": "Let the coordinates of the point be (x, y). According to the definition of the parabola, we have x + 2 = 5, solving gives x = 3. Substituting into the parabola equation yields y = \\pm2\\sqrt{6}. Since P is in the first quadrant, \\therefore P(3, 2\\sqrt{6})"
  },
  {
    "text": "The standard equation of the hyperbola with foci at the two vertices of the minor axis of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ and passing through the point $A(4, -5)$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;A: Point;Expression(H) = (x^2/16 + y^2/9 = 1);Coordinate(A) = (4, -5);PointOnCurve(A, G);Vertex(MinorAxis(H))=Focus(G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[64, 67]], [[1, 39]], [[52, 63]], [[1, 39]], [[52, 63]], [[51, 67]], [0, 64]]",
    "query_spans": "[[[64, 74]]]",
    "process": ""
  },
  {
    "text": "It is known that the center of the hyperbola is at the origin, the axes of symmetry are the coordinate axes, and it passes through the points $(2 , \\sqrt{2})$ and $(\\sqrt{2} , 0)$. What are the coordinates of the foci of the hyperbola?",
    "fact_expressions": "G: Hyperbola;H: Point;I: Point;O: Origin;Coordinate(H) = (2, sqrt(2));Coordinate(I) = (sqrt(2), 0);Center(G) = O;SymmetryAxis(G)=axis;PointOnCurve(H,G);PointOnCurve(I,G)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(pm*2,0)",
    "fact_spans": "[[[2, 5], [61, 64]], [[23, 41]], [[42, 59]], [[9, 11]], [[23, 41]], [[42, 59]], [[2, 11]], [[2, 19]], [[2, 41]], [[2, 59]]]",
    "query_spans": "[[[61, 71]]]",
    "process": ""
  },
  {
    "text": "Let the center of circle $C$ coincide with the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{2}=1(a>0)$, and suppose that the circle is tangent to the asymptotes of this hyperbola. If the chord length intercepted by the line $l$: $x-\\sqrt{3} y=0$ on the circle $C$ equals $2$, then the value of $a$ is?",
    "fact_expressions": "l: Line;G: Hyperbola;a: Number;C: Circle;a>0;Expression(G) = (-y^2/2 + x^2/a^2 = 1);Center(C)=RightFocus(G);IsTangent(C, Asymptote(G));Expression(l)=(x-sqrt(3)*y=0);Length(InterceptChord(l,C))=2",
    "query_expressions": "a",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[79, 102]], [[9, 56], [68, 71]], [[119, 122]], [[1, 5], [103, 107], [65, 66]], [[12, 56]], [[9, 56]], [[1, 62]], [[65, 77]], [[79, 102]], [[79, 117]]]",
    "query_spans": "[[[119, 126]]]",
    "process": "The right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{2}=1$ has coordinates $(\\sqrt{a^{2}+2},0)$. From the problem, the center of the circle is $C(\\sqrt{a^{2}+2},0)$, and the asymptotes of the hyperbola are $\\sqrt{2}x\\pm ay=0$. The distance from the center $C$ to the asymptote is $d=\\frac{\\sqrt{2}\\sqrt{a^{2}+2}}{\\sqrt{a^{2}+2}}=\\sqrt{2}$, which is the radius of circle $C$. Given that the chord length intercepted by line $l$ on circle $C$ is 2 and the radius of circle $C$ is $\\sqrt{2}$, the distance from center $C$ to line $l$ is $d=\\sqrt{(\\sqrt{2})^{2}-1^{2}}=1$, that is, $d=\\frac{|\\sqrt{a^{2}+2}-0\\times\\sqrt{3}|}{\\sqrt{1+(-\\sqrt{3})^{2}}}=\\frac{|\\sqrt{a^{2}+2}|}{2}=1$. Solving gives $a=\\pm\\sqrt{2}$, and since $a>0$, we have $a=\\sqrt{2}$."
  },
  {
    "text": "Given two fixed points $A(-1,0)$, $B(1,0)$, if a moving point $C$ in the plane satisfies the condition $|CA| = \\sqrt{3}|CB|$, then the maximum value of $S_\\triangle{ABC}$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (-1, 0);B: Point;Coordinate(B) = (1, 0);C: Point;Abs(LineSegmentOf(C, A)) = sqrt(3)*Abs(LineSegmentOf(C, B))",
    "query_expressions": "Max(Area(TriangleOf(A, B, C)))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[5, 14]], [[5, 14]], [[16, 24]], [[16, 24]], [[32, 35]], [[39, 60]]]",
    "query_spans": "[[[62, 88]]]",
    "process": "Let the coordinates of moving point C be C(x,y). From |CA| = \\sqrt{3}|CB|, we have \\sqrt{(x+1)^{2}+(y-0)^{2}} = \\sqrt{3}\\sqrt{(x-1)^{2}+(y-0)^{2}}. Simplifying yields: x^{2}+y^{2}-4x+1=0, or (x-2)^{2}+y^{2}=3. Therefore, S_{AABC} = \\frac{1}{2} \\times |AB| \\times h_{AB} (where h_{AB} denotes the height from C to side AB in \\triangle ABC). Clearly, (h_{AB})_{\\max} = \\sqrt{3}, so the maximum value of S_{AABC} is \\sqrt{3}."
  },
  {
    "text": "Given the parabola $x^{2}=4 y$, a line with slope $-\\frac{1}{2}$ intersects the parabola at points $A$ and $B$. If the circle with diameter $AB$ is tangent to the directrix of the parabola at point $P$, then the distance from point $P$ to the line $AB$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);L: Line;Slope(L) = -1/2;A: Point;B: Point;Intersection(L, G) = {A, B};H: Circle;IsDiameter(LineSegmentOf(A, B), H);P: Point;TangentPoint(H, Directrix(G)) = P",
    "query_expressions": "Distance(P, LineOf(A, B))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 16], [38, 41], [66, 69]], [[2, 16]], [[35, 37]], [[17, 37]], [[42, 45]], [[46, 49]], [[35, 51]], [[64, 65]], [[52, 65]], [[74, 78], [80, 84]], [[64, 78]]]",
    "query_spans": "[[[80, 97]]]",
    "process": "Let the equation of the line be $ y = -\\frac{1}{2}x + b $. Combining with the parabola's equation $ x^2 = 4y $, eliminating $ y $ gives $ x^2 + 2x - 4b = 0 $, so $ A = 4 + 16b > 0 \\frac{a}{a} + \\frac{4 + 100 > 0}{1} $, $ y_1 $, $ B(x_2, y_2) $, thus $ x_1 + x_2 = -2 $, $ x_1 \\cdot x_2 = -4b $. Since $ P\\left( \\frac{x_1 + x_2}{2}, -1 \\right) $, $ \\therefore P(-1, -1) $. Therefore, $ \\overrightarrow{PA} = (x_1 + 1, y_1 + 1) $, $ \\overrightarrow{PB} = (x_2 + 1, y_2 + 1) $. From the condition, $ \\overrightarrow{PA} \\cdot \\overrightarrow{PB} = (x_1 + 1)(x_2 + 1) + (y_1 + 1)(y_2 + 1) = 0 $, $ \\therefore b = $. So the equation of the line is $ y = -\\frac{1}{2}x + 1 $, i.e., $ x + 2y - 2 = 0 $. Thus, the distance from point $ P $ to line $ AB $ is $ \\frac{|-1 - 2 - 2|}{\\sqrt{1^2 + 2^2}} = \\sqrt{5} $. Hence, fill in $ \\sqrt{5} $."
  },
  {
    "text": "Given the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$ has its right focus at $F$, then the distance from $F$ to one of its asymptotes is?",
    "fact_expressions": "G: Hyperbola;F: Point;Expression(G) = (x^2 - y^2/4 = 1);RightFocus(G) = F",
    "query_expressions": "Distance(F,OneOf(Asymptote(G)))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 30]], [[35, 38], [40, 43]], [[2, 30]], [[2, 38]]]",
    "query_spans": "[[[2, 56]]]",
    "process": "For any hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, the distance from a focus $F(\\pm c,0)$ to an asymptote $y=\\pm\\frac{b}{a}x$ (i.e., $bx\\pm ay=0$) is $d=\\frac{|bc\\pm a\\times0|}{\\sqrt{b^{2}+(\\pm a)^{2}}}=\\frac{|bc|}{c}=b$. Since $b^{2}=4 \\Rightarrow b=2$, the distance from focus $F$ to one of the asymptotes is $2$. Hence, fill in: $2$. (This question mainly examines the distance from a hyperbola's focus to its asymptote and the point-to-line distance formula, and is a basic problem.)"
  },
  {
    "text": "Given that the eccentricity of hyperbola $C$ is $\\frac{5}{2}$, with left and right foci $F_{1}$ and $F_{2}$, and point $A$ lies on $C$ such that $|F_{1} A|=2|F_{2} A|$, then $\\cos \\angle A F_{2} F_{1}$=?",
    "fact_expressions": "C: Hyperbola;F1: Point;A: Point;F2: Point;Eccentricity(C) = 5/2;LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(A, C);Abs(LineSegmentOf(F1, A)) = 2*Abs(LineSegmentOf(F2, A))",
    "query_expressions": "Cos(AngleOf(A, F2, F1))",
    "answer_expressions": "13/16",
    "fact_spans": "[[[2, 8], [54, 57]], [[33, 40]], [[49, 53]], [[41, 48]], [[2, 26]], [[2, 48]], [[2, 48]], [[49, 58]], [[60, 82]]]",
    "query_spans": "[[[84, 113]]]",
    "process": "By the definition of a hyperbola, |F_{1}A| - |F_{2}A| = |F_{2}A| = 2a, then |F_{1}A| = 4a. Since the eccentricity of the hyperbola is \\frac{5}{2}, then |F_{1}F_{2}| = 2c = 5a. In \\triangle AF_{1}F_{2}, \\cos\\angle AF_{2}F_{1} = \\frac{25a^{2}+4a^{2}-16a^{2}}{2\\times5a\\times2a} = \\frac{13}{20}; hence fill in \\frac{13}{20}."
  },
  {
    "text": "If the left and right foci of the hyperbola $x^{2}-4 y^{2}=4$ are $F_{1}$, $F_{2}$, and a line passing through $F_{1}$ intersects the left branch at points $A$, $B$, and if $|A B|=5$, then the perimeter of $\\triangle A F_{2} B$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - 4*y^2 = 4);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;L: Line;PointOnCurve(F1,L) = True;A: Point;B: Point;Intersection(L, LeftPart(G)) = {A, B};Abs(LineSegmentOf(A, B)) = 5",
    "query_expressions": "Perimeter(TriangleOf(A, F2, B))",
    "answer_expressions": "18",
    "fact_spans": "[[[1, 21]], [[1, 21]], [[28, 35], [45, 52]], [[36, 43]], [[1, 43]], [[1, 43]], [[53, 55]], [[44, 55]], [[59, 62]], [[63, 66]], [[1, 68]], [[70, 79]]]",
    "query_spans": "[[[81, 107]]]",
    "process": "\\because If the left and right foci of the hyperbola \\( x^{2} - 4y^{2} = 4 \\) are \\( F_{1}, F_{2} \\), and a line passing through \\( F_{1} \\) intersects the left branch at points \\( A, B \\), then \\( | |AF_{2}| - |AF_{1}| | = 2a = 4 \\), \\( | |BF_{2}| - |BF_{1}| | = 2a = 4 \\), so \\( |AF_{2}| + |BF_{2}| = 8 + |AF_{1}| + |BF_{1}| = 8 + |AB| \\). \\because \\( |AB| = 5 \\), \\therefore \\( |AF_{2}| + |BF_{2}| = 13 \\), \\therefore the perimeter of \\( \\triangle AF_{2}B \\) is 18."
  },
  {
    "text": "Given a point $P$ on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{16}=1$ such that the distance from $P$ to one focus of the ellipse is $5$, then the distance from point $P$ to the other focus is?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/4 + y^2/16 = 1);PointOnCurve(P, G);F1:Point;F2:Point;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2);Distance(P, F1) = 5",
    "query_expressions": "Distance(P,F2)",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 40], [48, 50]], [[44, 47], [63, 67]], [[2, 40]], [[2, 47]], [], [], [[48, 54]], [[48, 73]], [[48, 73]], [[44, 61]]]",
    "query_spans": "[[[48, 78]]]",
    "process": "Let the required distance be $ d $. From the given conditions, we have $ a = 4 $. According to the definition of an ellipse, $ 2a = 5 + d \\Rightarrow d = 2a - 5 = 3 $. Therefore, the distance from point $ P $ to the other focus is $ 3 $. [Analysis] This problem examines the definition of an ellipse and is a basic question."
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, $F_{1}$ and $F_{2}$ are the two foci of the ellipse, and $\\angle F_{1} P F_{2}=60^{\\circ}$, then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2};AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[6, 33], [54, 56]], [[38, 45]], [[2, 5]], [[46, 53]], [[6, 33]], [[2, 37]], [[38, 61]], [[63, 96]]]",
    "query_spans": "[[[98, 125]]]",
    "process": "In the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, $a=2$, $b=1$, $c=\\sqrt{3}$. By the definition of an ellipse, we have $|PF_{1}|+|PF_{2}|=2a=4$, $|F_{1}F_{2}|=2\\sqrt{3}$. In $\\triangle F_{1}PF_{2}$, $\\angle F_{1}PF_{2}=60^{\\circ}$. By the law of cosines, we get $|F_{1}F_{2}|^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}|\\cdot|PF_{2}|\\cos60^{\\circ}=(|PF_{1}|+|PF_{2}|)^{2}-3|PF_{1}|\\cdot|PF_{2}|=16-3|PF_{1}|\\cdot|PF_{2}|$. Solving gives $|PF_{1}|\\cdot|PF_{2}|=\\frac{4}{3}$. Therefore, $S_{\\triangle PF_{1}F_{2}}=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|\\sin60^{\\circ}=\\frac{\\sqrt{3}}{3}$."
  },
  {
    "text": "Given $A(-2,0)$, $B(2,0)$, if there exists a point $P$ on the curve $\\left(\\frac{x}{a}+\\frac{y}{b}\\right)\\left(\\frac{x}{a}-\\frac{y}{b}\\right)=0$ $(a>0, b>0)$ such that $|PA|-|PB|=2$, then the range of values for $\\frac{b}{a}$ is?",
    "fact_expressions": "G: Curve;b: Number;a: Number;A: Point;P: Point;B: Point;a>0;b>0;Expression(G) = ((-y/b + x/a)*(y/b + x/a) = 0);Coordinate(A) = (-2, 0);Coordinate(B) = (2, 0);PointOnCurve(P, G);Abs(LineSegmentOf(P, A)) - Abs(LineSegmentOf(P, B)) = 2",
    "query_expressions": "Range(b/a)",
    "answer_expressions": "(0, \\sqrt{3})",
    "fact_spans": "[[[24, 90]], [[116, 129]], [[116, 129]], [[2, 11]], [[93, 97]], [[14, 22]], [[26, 90]], [[26, 90]], [[24, 90]], [[2, 11]], [[14, 22]], [[24, 97]], [[99, 114]]]",
    "query_spans": "[[[116, 136]]]",
    "process": "If A(-2,0), B(2,0), and |PA| - |PB| = 2 < |AB| = 4, then point P lies on the right branch of the hyperbola with foci at A and B, and 2a = 2, c = 2, ∴ a = 1, b = \\sqrt{4 - 1} = \\sqrt{3}, ∴ the equation of the hyperbola is x^{2} - \\frac{y^{2}}{4} = 1 (x > 0); its asymptotes are y = \\pm\\sqrt{3}x. Then there exists a point P on the curve (\\frac{x}{a} + \\frac{y}{b})(\\frac{x}{a} - \\frac{y}{b}) = 0 (a > 0, b > 0) satisfying |PA| - |PB| = 2. This is equivalent to y = \\pm\\frac{b}{a}x intersecting the hyperbola x^{2} - \\frac{y^{2}}{4} = 1 (x > 0), ∴ 0 < \\frac{b}{a} < \\sqrt{3}"
  },
  {
    "text": "Given that point $P$ is any point on the curve $x = \\frac{1}{4} y^{2}$, a perpendicular is drawn from point $P$ to the $y$-axis with foot $H$, and point $Q$ is any point on the curve $y = e^{x}$. Then the minimum value of $|PH| + |PQ|$ is?",
    "fact_expressions": "G: Curve;C:Curve;P: Point;H: Point;Q: Point;Expression(G) = (x = y^2/4);Expression(C) = (y = e^x);PointOnCurve(P, G);PointOnCurve(Q, C);L:Line;PointOnCurve(P,L);IsPerpendicular(L,yAxis);FootPoint(L,yAxis)=H",
    "query_expressions": "Min(Abs(LineSegmentOf(P, H)) + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "sqrt(2)-1",
    "fact_spans": "[[[7, 30]], [[62, 73]], [[2, 6], [37, 41]], [[53, 56]], [[57, 61]], [[7, 30]], [[62, 73]], [[2, 35]], [[57, 78]], [], [[36, 49]], [[36, 49]], [[36, 56]]]",
    "query_spans": "[[[80, 97]]]",
    "process": "As shown in the figure: |PH| = |PF| - 1, |PH| + |PQ| = |PQ| + |PF| - 1 \\geqslant |QF| - 1, i.e., find the minimum value of |QF|. Let Q(x, e^{x}), then |QF|^{2} = (x - 1)^{2} + (e^{x})^{2} = e^{2x} + x^{2} - 2x + 1. Define function f(x) = e^{2x} + x^{2} - 2x + 1, f'(x) = 2e^{2x} + 2x - 2, f'(0) = 0, f''(x) = 4e^{2x} + 2 > 0, f'(x) is monotonically increasing. Hence f(x) is monotonically decreasing on (-\\infty, 0) and monotonically increasing on (0, +\\infty). f(x)_{\\min} = f(0) = 2, the minimum value of |QF| is \\sqrt{2}. When Q(0, 1), P, Q, F are collinear, |PH| + |PQ| has the minimum value of \\sqrt{2} - 1."
  },
  {
    "text": "Given that the line $x - y - 4 = 0$ intersects the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1$ $(a > b > 0)$ at points $A$ and $B$, and the abscissa of the midpoint of $AB$ is $3$, then the eccentricity of the ellipse is?",
    "fact_expressions": "H: Line;Expression(H) = (x - y - 4 = 0);G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Intersection(H, G) = {A, B};A: Point;B: Point;XCoordinate(MidPoint(LineSegmentOf(A, B))) = 3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[2, 13]], [[2, 13]], [[14, 66], [95, 97]], [[14, 66]], [[16, 66]], [[16, 66]], [[16, 66]], [[16, 66]], [[2, 77]], [[68, 71]], [[72, 75]], [[79, 93]]]",
    "query_spans": "[[[95, 103]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), then x_{1}+x_{2}=6. From \\begin{cases}x-y-4=0\\\\b^{2}x^{2}+a^{2}y^{2}=a^{2}b^{2}\\end{cases}, we obtain (a^{2}+b^{2})x^{2}-8a^{2}x+16a^{2}-a^{2}b^{2}=0, so x_{1}+x_{2}=\\frac{8a^{2}}{a^{2}+b^{2}}=6, a^{2}=3b^{2}, that is, 2a^{2}=3c^{2}, the eccentricity of the ellipse e=\\sqrt{\\frac{2}{3}}=\\frac{\\sqrt{6}}{3}."
  },
  {
    "text": "Given $A(-\\frac{1}{2}, 0)$, $B$ is a moving point on the circle $F$: $(x-\\frac{1}{2})^{2}+y^{2}=4$ ($F$ is the center of the circle), the perpendicular bisector of segment $AB$ intersects $BF$ at $P$. Then the trajectory equation of the moving point $P$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (-1/2, 0);F: Circle;Expression(F) = (y^2 + (x - 1/2)^2 = 4);B: Point;PointOnCurve(B, F);F1: Point;Center(F) = F1;P: Point;Intersection(PerpendicularBisector(LineSegmentOf(A, B)), LineSegmentOf(B, F)) = P",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2+(4/3)*y^2=1",
    "fact_spans": "[[[2, 22]], [[2, 22]], [[28, 62]], [[28, 62]], [[24, 27]], [[24, 75]], [[63, 66]], [[28, 70]], [[96, 99], [103, 106]], [[76, 99]]]",
    "query_spans": "[[[103, 113]]]",
    "process": "According to the given condition, |BP| + |PF| = 2 and |PB| = |PA|, thus |AP| + |PF| = 2. By the definition of an ellipse, the locus of point P is an ellipse with foci at A and F. Given A and F, we find a = 1, c = \\frac{1}{2}, then b = \\frac{\\sqrt{3}}{2}. Therefore, the equation of the locus of point P is x^{2} + \\frac{4}{3}y^{2} = 1."
  },
  {
    "text": "Given that point $P$ is a moving point on the parabola $y^{2} = -2x$, find the minimum value of the sum of the distance from point $P$ to point $M(0,2)$ and the distance from point $P$ to the directrix of this parabola.",
    "fact_expressions": "P: Point;G: Parabola;Expression(G) = (y^2 = -2*x);PointOnCurve(P, G) = True;M: Point;Coordinate(M) = (0, 2)",
    "query_expressions": "Min(Distance(P, M) + Distance(P, Directrix(G)))",
    "answer_expressions": "sqrt(17)/2",
    "fact_spans": "[[[2, 6], [30, 34], [48, 52]], [[7, 22], [54, 57]], [[7, 22]], [[2, 28]], [[35, 44]], [[35, 44]]]",
    "query_spans": "[[[30, 70]]]",
    "process": "As shown in the figure, let P' be the projection of point P onto the directrix of the parabola, and let F be the focus of the parabola. Then F(-\\frac{1}{2},0). By the definition of the parabola, the distance from P to the directrix of the parabola is |PP'| = |PF|."
  },
  {
    "text": "If the left focus of the ellipse $\\frac{x^{2}}{5}-\\frac{y^{2}}{p^{2}}=1$ lies on the directrix of the parabola $y^{2}=2 p x$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;H: Ellipse;Expression(G) = (y^2 = 2*p*x);Expression(H) = (x^2/5 - y^2/p^2 = 1);PointOnCurve(LeftFocus(H),Directrix(G))",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[47, 63]], [[69, 72]], [[1, 42]], [[47, 63]], [[1, 42]], [[1, 67]]]",
    "query_spans": "[[[69, 76]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola $x^{2}=2 p y(p>0)$ is $F$. Points $A$ and $B$ are two moving points on the parabola such that $\\angle A F B=60^{\\circ}$. From the midpoint $C$ of chord $A B$, a perpendicular $C D$ is drawn to the directrix of the parabola, with $D$ being the foot of the perpendicular. Then the minimum value of $\\frac{|A B|}{|C D|}$ is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;C: Point;D: Point;F: Point;p>0;Expression(G) = (x^2 = 2*(p*y));Focus(G) = F;PointOnCurve(A,G);PointOnCurve(B,G);AngleOf(A, F, B) = ApplyUnit(60, degree);IsChordOf(LineSegmentOf(A,B),G);MidPoint(LineSegmentOf(A,B))=C;PointOnCurve(C,LineOf(C,D));IsPerpendicular(Directrix(G),LineOf(C,D));FootPoint(Directrix(G),LineOf(C,D))=D",
    "query_expressions": "Min(Abs(LineSegmentOf(A, B))/Abs(LineSegmentOf(C, D)))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 21], [40, 43], [94, 97]], [[3, 21]], [[31, 35]], [[36, 39]], [[89, 92]], [[111, 114]], [[25, 28]], [[3, 21]], [[0, 21]], [[0, 28]], [[31, 49]], [[31, 49]], [[53, 78]], [[40, 86]], [[81, 92]], [[79, 107]], [[94, 107]], [[94, 114]]]",
    "query_spans": "[[[116, 143]]]",
    "process": "Let |AF| = a, |BF| = b. Draw AQ perpendicular to the directrix at Q, and BP perpendicular to the directrix at P. By the definition of the parabola, we have |AF| = |AQ|, |BF| = |BP|. In \\triangle AFB, by the law of cosines, |AB|^{2} = (a + b)^{2} - 3ab. Then by the AM-GM inequality, we obtain |AB| \\geqslant \\frac{1}{2}(a + b) = |CD|, thus solving it. Let |AF| = a, |BF| = b. Draw AQ perpendicular to the directrix at Q, BP perpendicular to the directrix at P. By the definition of the parabola, we have |AF| = |AQ|, |BF| = |BP|. In trapezoid ABPQ, we have 2|CD| = |AQ| + |BP| = a + b. By the law of cosines, |AB|^{2} = a^{2} + b^{2} - 2ab\\cos60^{\\circ} = a^{2} + b^{2} - ab. Completing the square gives |AB|^{2} = (a + b)^{2} - 3ab. Again, since ab \\leqslant \\left(\\frac{a + b}{2}\\right)^{2}, therefore (a + b)^{2} - 3ab \\geqslant (a + b)^{2} - \\frac{3}{4}(a + b)^{2} = \\frac{1}{4}(a + b)^{2}. Thus we get |AB| \\geqslant \\frac{1}{2}(a + b) = |CD|, with equality if and only if a = b. Hence \\frac{|AB|}{|CD|} \\geqslant 1, that is, the minimum value of \\frac{|AB|}{|CD|} is 1. Answer is"
  },
  {
    "text": "Given that point $A(\\sqrt{2},1)$ lies on the parabola $x^{2}=2 p y$ $(p>0)$, then the distance from $A$ to its focus $F$ is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;p>0;Expression(G) = (x^2 = 2*(p*y));Coordinate(A) = (sqrt(2), 1);PointOnCurve(A, G);F: Point;Focus(G) = F",
    "query_expressions": "Distance(A, F)",
    "answer_expressions": "3/2",
    "fact_spans": "[[[20, 41], [50, 51]], [[23, 41]], [[2, 19], [46, 49]], [[23, 41]], [[20, 41]], [[2, 19]], [[2, 44]], [[53, 56]], [[50, 56]]]",
    "query_spans": "[[[46, 61]]]",
    "process": "Since point A(\\sqrt{2},1) lies on the parabola, we have 2=2p, so p=1. The equation of the parabola is x^{2}=2y, the focus coordinates are F(0,\\frac{1}{2}), and the directrix equation is y=-\\frac{1}{2}. Therefore, the distance from point A(\\sqrt{2},1) to the directrix is 1+\\frac{1}{2}=\\frac{3}{2}. Hence, AF=\\frac{3}{2}."
  },
  {
    "text": "If $a>2$, then the range of the eccentricity $e$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{(a+1)^{2}}=1$ is?",
    "fact_expressions": "a: Number;a > 2;G: Hyperbola;Expression(G) = (-y^2/(a + 1)^2 + x^2/a^2 = 1);e: Number;Eccentricity(G) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "(\\sqrt{2},\\sqrt{13}/2)",
    "fact_spans": "[[[1, 6]], [[1, 6]], [[8, 58]], [[8, 58]], [[62, 65]], [[8, 65]]]",
    "query_spans": "[[[62, 72]]]",
    "process": "Using the eccentricity formula of a hyperbola and the monotonicity of a quadratic function, the range of values for $ e $ can be calculated. [Detailed solution] For the hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{(a+1)^{2}} = 1 $, we obtain $ e^{2} = \\frac{a^{2} + (a+1)^{2}}{a^{2}} = 2 + \\frac{2}{a} + \\frac{1}{a^{2}} = \\left( \\frac{1}{a} + 1 \\right)^{2} + 1 $. Given $ a > 2 $, it follows that $ 0 < \\frac{1}{a} < \\frac{1}{2} $, then $ e^{2} \\in \\left( 2, \\frac{13}{4} \\right) $, hence $ e \\in \\left( \\sqrt{2}, \\frac{\\sqrt{13}}{2} \\right) $."
  },
  {
    "text": "Given that the directrix of the parabola $y^{2}=2 p x$ is $x=-2$, and point $P$ is a point on the parabola, then the minimum distance from point $P$ to the line $y=x+3$ is?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;P: Point;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (y = x + 3);Expression(Directrix(G)) = (x = -2);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, H))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 18], [36, 39]], [[5, 18]], [[50, 59]], [[31, 35], [45, 49]], [[2, 18]], [[50, 59]], [[2, 30]], [[31, 43]]]",
    "query_spans": "[[[45, 68]]]",
    "process": "From the given conditions, the equation of the parabola is $ y^{2} = 8x $. Let the coordinates of point $ P $ be $ P(x, y) $. Then the distance from point $ P $ to the line $ y = x + 3 $ is:  \n$ d = \\frac{|x - y + 3|}{\\sqrt{2}} = \\frac{|8x - 8y + 24|}{8\\sqrt{2}} = \\frac{|y^{2} - 8y + 24|}{8\\sqrt{2}} = \\frac{|(y - 4)^{2} + 8|}{8\\sqrt{2}} \\geqslant \\frac{\\sqrt{2}}{2} $,  \nwith equality if and only when $ y = 4 $, the minimum value is $ \\underline{\\sqrt{2}} $."
  },
  {
    "text": "Given that the eccentricity of the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $\\frac{\\sqrt{3}}{2}$, and the sum of the distances from a point on the ellipse to the two foci is $12$, find $a=?$ $b=?$",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/a^2+y^2/b^2 = 1);a: Number;b: Number;a > b;b > 0;Eccentricity(G) = sqrt(3)/2;P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};Distance(P, F1) + Distance(P, F2) = 12",
    "query_expressions": "a;b",
    "answer_expressions": "6;3",
    "fact_spans": "[[[2, 56], [83, 85]], [[2, 56]], [[104, 107]], [[110, 113]], [[4, 56]], [[4, 56]], [[2, 81]], [], [[83, 88]], [], [], [[83, 92]], [[83, 102]]]",
    "query_spans": "[[[104, 109]], [[110, 115]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, one of its asymptotes is perpendicular to the line $l$: $x+\\sqrt{3} y=0$, and the distance from one focus of $C$ to $l$ is $2$. Then the standard equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;l: Line;Expression(l) = (x + sqrt(3)*y = 0);IsPerpendicular(OneOf(Asymptote(C)), l);Distance(OneOf(Focus(C)), l) = 2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4 - y^2/12 = 1",
    "fact_spans": "[[[2, 63], [97, 100], [118, 121]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[70, 93], [106, 109]], [[70, 93]], [[2, 95]], [[97, 116]]]",
    "query_spans": "[[[118, 128]]]",
    "process": "From the given conditions: $\\frac{a}{b}=\\frac{1}{\\sqrt{3}}$, and the distance from a focus of $C$ to the center is $\\frac{c}{2}=2$, so $c=4$. Therefore, $a=2$, $b=2\\sqrt{3}$, and the standard equation of $c$ is $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$."
  },
  {
    "text": "Given that $A$ and $F$ are the lower vertex and the left focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. The line $l$ passing through $A$ with an inclination angle of $60^{\\circ}$ intersects the $x$-axis and the ellipse $C$ at points $M$ and $N$, respectively, and the ordinate of point $N$ is $\\frac{3}{5} b$. If the perimeter of $\\Delta F M N$ is $6$, then the area of $\\Delta F A N$ is?",
    "fact_expressions": "A: Point;F: Point;LowerVertex(C) = A;LeftFocus(C) = F;C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;l: Line;PointOnCurve(A,l) = True;Inclination(l) = ApplyUnit(60,degree);Intersection(l,xAxis) = M;Intersection(l,C) = N;M: Point;N: Point;YCoordinate(N) = 3*b/5;Perimeter(TriangleOf(F, M, N)) = 6",
    "query_expressions": "Area(TriangleOf(F, A, N))",
    "answer_expressions": "8*sqrt(3)/5",
    "fact_spans": "[[[2, 5], [79, 82]], [[6, 9]], [[2, 77]], [[2, 77]], [[12, 69], [113, 118]], [[12, 69]], [[19, 69]], [[19, 69]], [[19, 69]], [[19, 69]], [[100, 105]], [[78, 105]], [[83, 105]], [[100, 128]], [[100, 128]], [[119, 122]], [[123, 126], [130, 134]], [[130, 154]], [[156, 177]]]",
    "query_spans": "[[[179, 198]]]",
    "process": "Draw the graph. From the conditions, we obtain $\\frac{b}{a}=\\frac{\\sqrt{3}}{2}$, $b=\\sqrt{3}c$, then deduce that $M$ is the right focus of the ellipse. Then, by the definition of the ellipse, $2a+2c=6$, thus the values of $a$, $b$, and $c$ can be calculated. Then use $S_{\\triangle FAN}=\\frac{1}{2}\\cdot|FM|\\cdot[\\frac{3}{5}b-(-b)]$ to compute the answer. [Solution] As shown in the figure, from the given conditions, $A(0,-b)$, $F(-c,0)$, the equation of line $MN$ is $y=\\sqrt{3}x-b$. Substitute $y=\\frac{3}{5}b$ into the ellipse equation to get $x=\\frac{4}{5}a$, so $N(\\frac{4}{5}a,\\frac{3}{5}b)$. Since $N$ lies on line $MN$, $\\frac{3}{5}b=\\sqrt{3}\\times\\frac{4}{5}a-b$, solving gives $\\frac{b}{a}=\\frac{\\sqrt{3}}{2}$. Also $a^{2}=b^{2}+c^{2}$, so $(\\frac{2}{\\sqrt{3}}b)^{2}=b^{2}+c^{2}$, solving gives $b=\\sqrt{3}c$. Let $y=\\sqrt{3}x-b=0$, then $M(\\frac{b}{\\sqrt{3}},0)$, i.e., $M(c,0)$, so $M$ is the right focus of the ellipse, thus $|FM|=2c$. By the definition of the ellipse, $|NF|+|NM|=2a$. Since the perimeter of $\\triangle FMN$ is 6, $2a+2c=6$. Because $\\frac{b}{a}=\\frac{\\sqrt{3}}{2}$, $a=2c$, so $c=1$, $a=2$, $b=\\sqrt{3}$. Therefore, $S_{\\Delta FAN}=\\frac{1}{2}\\cdot|FM|\\cdot[\\frac{3}{5}b-(-b)]=c\\cdot\\frac{8}{5}b=\\frac{8\\sqrt{3}}{5}$."
  },
  {
    "text": "What is the length of the minor axis of the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{9}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/6 + y^2/9 = 1)",
    "query_expressions": "Length(MinorAxis(G))",
    "answer_expressions": "2*sqrt(6)",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 43]]]",
    "process": "Determine the position of the foci based on the ellipse equation. Since 6 < 9, the foci of the ellipse lie on the y-axis, so b^{2} = 6, and thus the minor axis length of the ellipse \\frac{x^{2}}{6} + \\frac{y^{2}}{9} = 1 is 2b = 2\\sqrt{6}."
  },
  {
    "text": "The length of the chord intercepted by the line $y=x-1$ on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ is?",
    "fact_expressions": "H: Line;Expression(H) = (y = x - 1);G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1)",
    "query_expressions": "Length(InterceptChord(H,G))",
    "answer_expressions": "8*sqrt(2)/5",
    "fact_spans": "[[[0, 9]], [[0, 9]], [[10, 37]], [[10, 37]]]",
    "query_spans": "[[[0, 44]]]",
    "process": "By the given condition,  \n\\begin{cases}\\frac{x^{2}}{4}+y^{2}=1\\\\y=x-1\\end{cases}  \neliminating $ y $ and simplifying yields $ x(5x-8)=0 $. Let the intersection points of the line $ y=x-1 $ and the ellipse $ \\frac{x^{2}}{4}+y^{2}=1 $ be $ (m,m-1) $ and $ (n,n-1) $, hence $ |m-n|=\\frac{8}{5} $. Therefore, the chord length intercepted by the line $ y=x-1 $ on the ellipse $ \\frac{x^{2}}{4}+y^{2}=1 $ is  \n$ d = \\sqrt{1+k^{2}}|x_{1}-x_{2}| = \\sqrt{2} \\times \\frac{8}{5} = \\frac{8\\sqrt{2}}{5} $."
  },
  {
    "text": "If the chord of the parabola $y^{2}=4x$ passing through the focus is the diameter of a circle, and the chord length intercepted by the $y$-axis is $4$, then what is the equation of the circle?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);L: LineSegment;PointOnCurve(Focus(G),L) = True;IsChordOf(L,G) = True;IsDiameter(L,H) = True;Length(InterceptChord(yAxis,H)) = 4;H: Circle",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x-3/2)^2+(y+pm*1)^2=25/4",
    "fact_spans": "[[[3, 17]], [[3, 17]], [], [[3, 22]], [[3, 22]], [[2, 27]], [[26, 41]], [[26, 27], [44, 45]]]",
    "query_spans": "[[[44, 50]]]",
    "process": "Let the line passing through the focus intersect the parabola at points A and B with coordinates (x_{1},y_{1}) and (x_{2},y_{2}), respectively. From the definition of the parabola, the relationship between the radius and the center of the circle can be obtained. Then, using the fact that the chord length intercepted by the circle on the y-axis is 4 and applying the Pythagorean theorem, the corresponding relations are derived. Set up the equation of the line, solve it simultaneously with the parabola equation, and use Vieta's formulas to obtain the result. [Detailed solution] Let the intersection points A and B of the line passing through the focus and the parabola have coordinates (x_{1},y_{1}) and (x_{2},y_{2}), respectively. The center C of the circle, which is the midpoint of AB, has coordinates (x_{0},y_{0}). From the definition of the parabola: |AB| = x_{1}+x_{2}+p = 2x_{0}+2, so r = x_{0}+1. Since the chord length intercepted by the circle on the y-axis is 4, by the Pythagorean theorem we have r^{2} = 4 + x_{0}^{2}, that is,\n\\begin{cases}\nr = x_{0}+1 \\\\\nr^{2} = 4 + x_{0}^{2}\n\\end{cases}\nSolving gives x_{0} = \\frac{3}{2}, so r = \\frac{5}{2}. Let the equation of the line passing through the focus be x = ay + 1, then\n\\begin{cases}\nx = ay + 1 \\\\\ny^{2} = 4x\n\\end{cases}\nSubstituting and simplifying yields y^{2} - 4ay - 4 = 0. Thus, y_{1} + y_{2} = 4a, so y_{0} = 2a. Eliminating y gives x^{2} - (2 + 4a^{2})x + 1 = 0. Therefore, x_{1} + x_{2} = 2 + 4a^{2}, so x_{0} = 1 + 2a^{2} = \\frac{3}{2}, solving gives a = \\pm\\frac{1}{2}, y_{0} = 2a = \\pm1. Hence, the equation of the circle is (x - \\frac{3}{2})^{2} + (y \\pm 1)^{2} = \\frac{25}{4}."
  },
  {
    "text": "Given the curve $\\frac{x^{2}}{2}+\\frac{y^{2}}{k^{2}-k}=1$. When the curve represents a circle, what is the value of $k$?",
    "fact_expressions": "H: Curve;Expression(H) = (x^2/2 + y^2/(k^2 - k) = 1);k: Number;G: Circle;H = G",
    "query_expressions": "k",
    "answer_expressions": "{2, -1}",
    "fact_spans": "[[[2, 45], [48, 50]], [[2, 45]], [[54, 57]], [[52, 53]], [[48, 53]]]",
    "query_spans": "[[[54, 62]]]",
    "process": "Because the curve is $\\frac{x^{2}}{2}+\\frac{y^{2}}{k^{2}-k}=1$, when the curve represents a circle, the condition $k^{2}-k=2$ must be satisfied, solving which gives $2$ or $-1$; when the curve represents an ellipse with foci on the $y$-axis, the condition $k^{2}-k>2$, i.e., $k>2$ or $k<-1$, must be satisfied; when the curve represents a hyperbola, the condition $k^{2}-k<0$, i.e., $0<k<1$, must be satisfied. Therefore, the answers are $2$ or $-1$; $k>2$ or $k<-1$; $0<k<1$."
  },
  {
    "text": "Let the common foci of the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$ and the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1$ be $F_{1}$, $F_{2}$, and let $P$ be an intersection point of the two curves. Then $\\angle {F}_{1} P {F}_{2}$=?",
    "fact_expressions": "H: Ellipse;Expression(H) = (x^2/6 + y^2/2 = 1);G: Hyperbola;Expression(G) = (x^2/2 - y^2/2 = 1);F1: Point;F2: Point;Focus(H) = {F1, F2};P: Point;OneOf(Intersection(H, G)) = P;Focus(G) = {F1, F2}",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "Answer missing",
    "fact_spans": "[[[1, 38]], [[1, 38]], [[39, 77]], [[39, 77]], [[83, 90]], [[91, 98]], [[1, 98]], [[99, 102]], [[99, 111]], [[1, 98]]]",
    "query_spans": "[[[113, 141]]]",
    "process": ""
  },
  {
    "text": "Given that the distance from the focus of a hyperbola to its asymptote equals the length of the hyperbola's real axis, then the equation of the asymptotes of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Distance(Focus(G), Asymptote(G)) = Length(RealAxis(G))",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "{y=pm*2*x, y=pm*(1/2)*x}",
    "fact_spans": "[[[2, 5], [28, 31], [18, 21], [9, 10]], [[2, 25]]]",
    "query_spans": "[[[28, 39]]]",
    "process": "Discuss the hyperbola with foci on the x-axis or y-axis, find the coordinates of the foci and the equations of the asymptotes. From the given conditions, the relationship between $a$ and $b$ can be obtained, thereby determining the asymptote equations of the required hyperbola. When the foci are on the x-axis, the hyperbola equation is set as $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a,b>0)$, the asymptote equations are $bx\\pm ay=0$, and the foci are $(\\pm c,0)$. The distance from a focus to an asymptote is $\\frac{bc}{\\sqrt{b^{2}+a^{2}}}=b=2a$, so the required asymptote equations are $y=\\pm 2x$. When the foci are on the y-axis, the hyperbola equation is set as $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a,b>0)$, the asymptote equations are $by\\pm ax=0$, and the foci are $(0,\\pm c)$. The distance from a focus to an asymptote is $\\frac{bc}{\\sqrt{b^{2}+a^{2}}}=b=2a$, so the required asymptote equations are $y=\\pm \\frac{1}{2}x$. In summary, the asymptote equations are $y=\\pm 2x$ or $y=\\pm \\frac{1}{2}x$."
  },
  {
    "text": "If the parabola $y^{2}=2 px(p>0)$ passes through the point $(1 , 2)$, then what is the value of $p$? What are the coordinates of the focus $F$ of this parabola?",
    "fact_expressions": "G: Parabola;p: Number;H:Point;F:Point;p>0;Focus(G)=F;Expression(G) = (y^2 = 2*(p*x));Coordinate(H) = (1, 2);PointOnCurve(H, G)",
    "query_expressions": "p;Coordinate(F)",
    "answer_expressions": "2\n(1,0)",
    "fact_spans": "[[[1, 21], [41, 44]], [[34, 37]], [[23, 32]], [[47, 50]], [[4, 21]], [[41, 50]], [[1, 21]], [[22, 32]], [[1, 32]]]",
    "query_spans": "[[[34, 40]], [[47, 55]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $E$: $x^{2}+\\frac{y^{2}}{b^{2}}=1$ $(0<b<1)$, respectively. A line passing through point $F_{1}$ intersects the ellipse $E$ at points $A$ and $B$. If $|A F_{1}|=3|B F_{1}|$ and $A F_{2} \\perp x$-axis, then the value of $b$ is?",
    "fact_expressions": "E: Ellipse;Expression(E) = (x^2 + y^2/b^2 = 1);b: Number;0 < b;b < 1;F1: Point;F2: Point;LeftFocus(E) = F1;RightFocus(E) = F2;G: Line;PointOnCurve(F1, G);A: Point;B: Point;Intersection(G, E) = {A, B};Abs(LineSegmentOf(A, F1)) = 3*Abs(LineSegmentOf(B, F1));IsPerpendicular(LineSegmentOf(A, F2), xAxis)",
    "query_expressions": "b",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[19, 62], [82, 87]], [[19, 62]], [[144, 147]], [[26, 62]], [[26, 62]], [[1, 8], [70, 78]], [[9, 16]], [[1, 68]], [[1, 68]], [[79, 81]], [[69, 81]], [[88, 91]], [[92, 95]], [[79, 97]], [[99, 121]], [[123, 142]]]",
    "query_spans": "[[[144, 151]]]",
    "process": "As shown in the figure: let F_{2}(c,0), by the focal radius formula we have |AF_{1}| = a + ex_{0} = 1 + c^{2}. Since |AF_{1}| = 3|BF_{1}|, it follows that 2c = 3(-c - x_{B}), so x_{B} = -\\frac{5c}{3}. Therefore, |BF| = 1 - \\frac{5c^{2}}{3}. From |AF_{1}| = 3|BF_{1}|, we get 1 + c^{2} = 3(1 - \\frac{5c^{2}}{3}), which yields c^{2} = \\frac{1}{3}, so b^{2} = \\frac{2}{3}. The answer should be x^{2} + \\frac{3}{2}y^{2} = 1."
  },
  {
    "text": "The line $y=k x-3$ and the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$ have exactly one common point, but the line is not tangent to the hyperbola. Then the value of the real number $k$ is?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*x - 3);k: Real;G: Hyperbola;Expression(G) = (x^2/2 - y^2 = 1);NumIntersection(H, G) = 1;Negation(IsTangent(H, G))",
    "query_expressions": "k",
    "answer_expressions": "pm*sqrt(2)/2",
    "fact_spans": "[[[0, 11], [51, 53]], [[0, 11]], [[62, 67]], [[12, 40], [54, 57]], [[12, 40]], [[0, 49]], [[51, 60]]]",
    "query_spans": "[[[62, 71]]]",
    "process": "\\because the line y=kx-3 and the hyperbola \\frac{x^{2}}{2}-y^{2}=1 have exactly one common point, but the line is not tangent to the hyperbola, \\therefore the line y=kx-3 is parallel to the asymptote of the hyperbola \\frac{x^{2}}{2}-y^{2}=1. \\because the asymptotes of the hyperbola \\frac{x^{2}}{2}-y^{2}=1 are y=\\pm\\frac{\\sqrt{2}}{2}x, \\therefore k=\\pm\\frac{\\sqrt{2}}{2}"
  },
  {
    "text": "Given that the distance from one endpoint of the imaginary axis of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1(b>0)$ to one of its asymptotes is $\\frac{\\sqrt{3}}{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;b>0;Expression(G) = (x^2 - y^2/b^2 = 1);Distance(OneOf(Endpoint(ImageinaryAxis(G))),OneOf(Asymptote(G)))=sqrt(3)/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 39], [47, 48], [80, 83]], [[5, 39]], [[5, 39]], [[2, 39]], [[2, 78]]]",
    "query_spans": "[[[80, 89]]]",
    "process": "Let the distance from an endpoint (0,b) of the imaginary axis of the hyperbola $ x^{2}-\\frac{y^{2}}{b^{2}}=1 $ to one of its asymptotes $ y=bx $ ($ b>0 $) be $ \\frac{\\sqrt{3}}{2} $. Then we have $ \\frac{b}{\\sqrt{1+b^{2}}}=\\frac{\\sqrt{3}}{2} $. Solving gives $ b=\\sqrt{3} $. Thus, the eccentricity of the hyperbola is $ e=\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{1+3}=2 $."
  },
  {
    "text": "If the foci of the curve $\\frac{x^{2}}{a-4}+\\frac{y^{2}}{a+5}=1$ are fixed points, then the coordinates of the foci are?",
    "fact_expressions": "G: Curve;a: Number;Expression(G) = (x^2/(a - 4) + y^2/(a + 5) = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,pm*3)",
    "fact_spans": "[[[1, 42]], [[3, 42]], [[1, 42]]]",
    "query_spans": "[[[1, 56]]]",
    "process": "When $a-4>0$, $a+5>0$, the curve represents an ellipse with foci on the $y$-axis, so $c=\\sqrt{(a+5)-(a-4)}=3$, thus the foci are $(0,\\pm3)$; when $a-4<0$, $a+5>0$, the curve represents a hyperbola with foci on the $y$-axis, i.e., $\\frac{y^{2}}{a+5}-\\frac{x^{2}}{4-a}=1$, so $c=\\sqrt{(a+5)+(4-a)}=3$, thus the foci are $(0,\\pm3)$. Therefore, the coordinates of the foci are $(0,\\pm3)$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{100}+\\frac{y^{2}}{64}=1$, and $P$ is a point on the ellipse such that $P F_{1} \\perp P F_{2}$, then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/100 + y^2/64 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "64",
    "fact_spans": "[[[18, 58], [69, 71]], [[65, 68]], [[2, 9]], [[10, 17]], [[18, 58]], [[2, 64]], [[65, 74]], [[76, 99]]]",
    "query_spans": "[[[101, 128]]]",
    "process": "Let $|PF_{1}|=r_{1}$, $|PF_{2}|=r_{2}$. According to the definition of the ellipse and the Pythagorean theorem, we find $r_{1}r_{2}=128$, then the result can be obtained using the triangle area formula. From $\\frac{x^{2}}{100}+\\frac{y^{2}}{64}=1$, we get $a^{2}=100$, $b^{2}=64$, so $a=10$, $c=\\sqrt{a^{2}-b^{2}}=\\sqrt{100-64}=6$, hence $|F_{1}F_{2}|=2c=12$. Let $|PF_{1}|=r_{1}$, $|PF_{2}|=r_{2}$, so $r_{1}+r_{2}=2a=20$, thus $r_{1}^{2}+r_{2}^{2}+2r_{1}r_{2}=400$. Since $PF_{1}\\bot PF_{2}$, we have $r_{1}^{2}+r_{2}^{2}=4c^{2}=144$, so $144+2r_{1}r_{2}=400$, hence $r_{1}r_{2}=128$. Therefore, the area of $\\triangle F_{1}PF_{2}$ is $\\frac{1}{2}r_{1}r_{2}=\\frac{1}{2}\\times128=64$."
  },
  {
    "text": "Given that the line $y=x-1$ intersects the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ at points $A$ and $B$, what is the length of segment $AB$?",
    "fact_expressions": "G: Ellipse;H: Line;B: Point;A: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Expression(H) = (y = x - 1);Intersection(H, G) = {A, B}",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "24/7",
    "fact_spans": "[[[12, 49]], [[2, 11]], [[55, 58]], [[51, 54]], [[12, 49]], [[2, 11]], [[2, 60]]]",
    "query_spans": "[[[62, 73]]]",
    "process": "Let $ A(x_{1},y_{1}), B(x_{2},y_{2}) $. Solving the system \n\\[\n\\begin{cases}\ny = x - 1 \\\\\n\\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1\n\\end{cases}\n\\Rightarrow 7x^{2} - 8x - 8 = 0, \\quad \\Delta = 64 + 4 \\times 7 \\times 8 = 288 > 0, \\quad x_{1} + x_{2} = \\frac{8}{7}, \\quad x_{1}x_{2} = -\\frac{8}{7},\n\\]"
  },
  {
    "text": "Let the line $ l $ passing through the focus $ F $ of the parabola $ C: y^2 = 2px $ ($ p > 0 $) intersect the parabola at points $ M $ and $ N $. If $ \\overrightarrow{MF} = 4 \\overrightarrow{FN} $, then what is the slope of line $ l $?",
    "fact_expressions": "l: Line;C: Parabola;p: Number;M: Point;F: Point;N: Point;p>0;Expression(C) = (y^2 = 2*p*x);Focus(C)=F;PointOnCurve(F, l);Intersection(l, C) = {M, N};VectorOf(M, F) = 4*VectorOf(F, N)",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*4/3",
    "fact_spans": "[[[34, 39], [103, 108]], [[1, 27], [40, 43]], [[9, 27]], [[45, 48]], [[30, 33]], [[49, 52]], [[9, 27]], [[1, 27]], [[1, 33]], [[0, 39]], [[34, 54]], [[56, 101]]]",
    "query_spans": "[[[103, 113]]]",
    "process": "As shown in the figure, draw MB perpendicular to the directrix at B, and draw NC perpendicular to the directrix at C. According to the definition of a parabola, we have MB = MF and NC = NF. Draw NA perpendicular to MB at A. Let FN = m, then MN = 5m, MA = MF - NF = 3m. In the right triangle AMN, \\tan\\angle NMA = \\frac{AN}{AM} = \\frac{4}{3}, \\therefore the slope of line l is \\pm\\frac{4}{3}."
  },
  {
    "text": "The trajectory equation of the center $M$ of a moving circle passing through the fixed point $F(0,1)$ and tangent to the line $y=-1$ is?",
    "fact_expressions": "F: Point;Coordinate(F) = (0, 1);PointOnCurve(F, G) = True;H: Line;Expression(H) = (y = -1);IsTangent(H, G) = True;G: Circle;M: Point;Center(G) = M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2=4*y",
    "fact_spans": "[[[3, 11]], [[3, 11]], [[0, 26]], [[13, 21]], [[13, 21]], [[12, 26]], [[24, 26]], [[28, 31]], [[24, 31]]]",
    "query_spans": "[[[28, 38]]]",
    "process": "From the given condition, the distance from point M to point F is equal to its distance to the line y = -1. Therefore, the trajectory of point M is a parabola with focus F(0,1) and directrix y = -1. Hence, the required trajectory equation is x^{2} = 4y."
  },
  {
    "text": "The hyperbola $x^{2}-y^{2}=2019$ has left and right vertices denoted by $A_{1}$, $A_{2}$ respectively. Let $P$ be a point on the right branch of the hyperbola such that $\\angle A_{1} P A_{2}=10^{\\circ}$. Then $\\angle P A_{1} A_{2}$=?",
    "fact_expressions": "G: Hyperbola;A1: Point;P: Point;A2: Point;Expression(G) = (x^2 - y^2 = 2019);LeftVertex(G)=A1;RightVertex(G)=A2;PointOnCurve(P,RightPart(G));AngleOf(A1, P, A2) = ApplyUnit(10, degree)",
    "query_expressions": "AngleOf(P, A1, A2)",
    "answer_expressions": "ApplyUnit(40,degree)",
    "fact_spans": "[[[0, 21], [49, 52]], [[29, 36]], [[45, 48]], [[37, 44]], [[0, 21]], [[0, 44]], [[0, 44]], [[45, 57]], [[59, 92]]]",
    "query_spans": "[[[94, 118]]]",
    "process": "Let the angles of inclination of lines $A_{1}P$ and $A_{2}P$ be $\\alpha$ and $\\beta$, respectively. Then $\\tan\\alpha\\tan\\beta = \\frac{2019}{2019} = 1$. Hence, $\\alpha + \\beta = 90^{\\circ}$, and $\\beta - \\alpha = 10^{\\circ}$, so $\\alpha = 40^{\\circ}$."
  },
  {
    "text": "Given that the parabola $C$: $4 x + a y^{2} = 0$ passes exactly through the center of the circle $M$: $(x-1)^{2} + (y-2)^{2} = 1$, then the coordinates of the focus of the parabola $C$ are?",
    "fact_expressions": "C: Parabola;a: Number;M: Circle;Expression(C) = (a*y^2 + 4*x = 0);Expression(M) = ((x - 1)^2 + (y - 2)^2 = 1);PointOnCurve(Center(M),C)",
    "query_expressions": "Coordinate(Focus(C))",
    "answer_expressions": "(1,0)",
    "fact_spans": "[[[2, 24], [61, 67]], [[9, 24]], [[28, 56]], [[2, 24]], [[28, 56]], [[2, 59]]]",
    "query_spans": "[[[61, 74]]]",
    "process": "The center of circle M is (1,2). Substituting into 4x + ay^2 = 0 gives a = -1. Rewriting the equation of parabola C into standard form yields y^{2} = 4x, so the focus coordinates are (1,0)."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+y^{2}=1$, its two foci are $F_{1}$ and $F_{2}$ respectively. A line passing through point $F_{1}$ and not parallel to the coordinate axes intersects the ellipse at points $M$ and $N$. Then, what is the perimeter of $\\Delta M N F_{2}$?",
    "fact_expressions": "C: Ellipse;G: Line;M: Point;N: Point;F2: Point;F1: Point;Expression(C) = (x^2/4 + y^2 = 1);Focus(C) = {F1, F2};PointOnCurve(F1, G);Negation(IsParallel(G, axis));Intersection(G, C) = {M, N}",
    "query_expressions": "Perimeter(TriangleOf(M, N, F2))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 34], [79, 81]], [[76, 78]], [[83, 87]], [[88, 91]], [[50, 57]], [[42, 49], [59, 67]], [[2, 34]], [[2, 57]], [[58, 78]], [[68, 78]], [[76, 91]]]",
    "query_spans": "[[[93, 116]]]",
    "process": "According to the definition of an ellipse, the perimeter of $\\triangle MNF_{2}$ can be found. The perimeter of $\\triangle MNF_{2}$ is $|MN| + |MF_{2}| + |NF_{2}| = |MF_{1}| + |NF| + |MF_{2}| + |NF_{2}| = 2 \\times 2 + 2 \\times 2 = 8$."
  },
  {
    "text": "Given that the vertex of the parabola $C$ is at the origin, and the focus is $F(1,0)$. A line $l$ passing through the focus $F$ intersects the parabola $C$ at two points $A$ and $B$. If the inclination angle of the line $l$ is $45^{\\circ}$, then what are the coordinates of the midpoint of chord $AB$?",
    "fact_expressions": "C: Parabola;O: Origin;Vertex(C) = O;F: Point;Coordinate(F) = (1, 0);Focus(C) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, C) = {A, B};Inclination(l) = ApplyUnit(45, degree);IsChordOf(LineSegmentOf(A, B), C)",
    "query_expressions": "Coordinate(MidPoint(LineSegmentOf(A, B)))",
    "answer_expressions": "(3, 2)",
    "fact_spans": "[[[2, 8], [42, 48]], [[12, 16]], [[2, 16]], [[20, 28], [32, 35]], [[20, 28]], [[2, 28]], [[36, 41], [63, 68]], [[29, 41]], [[51, 54]], [[55, 59]], [[36, 61]], [[63, 85]], [[42, 92]]]",
    "query_spans": "[[[88, 99]]]",
    "process": "Given the parabola equation $ y^{2} = 4x $ and the line $ l $ equation $ y = x - 1 $, substituting into the parabola equation yields $ (x - 1)^{2} = 4x $, i.e., $ x^{2} - 6x + 1 = 0 $. Then $ x_{1} + x_{2} = 6 $, the horizontal coordinate of the midpoint of $ AB $ is $ x_{0} = \\frac{6}{2} = 3 $, so $ y_{0} = x_{0} - 1 = 2 $. Hence, the midpoint is $ (3, 2) $."
  },
  {
    "text": "The coordinates of the intersection points of the line $y=x+4$ and the hyperbola $x^{2}-y^{2}=1$ are?",
    "fact_expressions": "G: Hyperbola;H: Line;Expression(G) = (x^2 - y^2 = 1);Expression(H) = (y = x + 4)",
    "query_expressions": "Coordinate(Intersection(H, G))",
    "answer_expressions": "(-17/8, 15/8)",
    "fact_spans": "[[[10, 28]], [[0, 9]], [[10, 28]], [[0, 9]]]",
    "query_spans": "[[[0, 35]]]",
    "process": "From \\begin{cases}y=x+4\\\\x^2-y^2=1\\end{cases}, solve to get: x=-\\frac{17}{8}, y=\\frac{15}{8}. Therefore, the intersection point of the line y=x+4 and the hyperbola x^{2}-y^{2}=1 is (-\\frac{17}{8},\\frac{15}{8}). [Key point] This problem examines the intersection of a line and a curve; solving the system of equations gives the solution."
  },
  {
    "text": "Let the parabola $C$: $y^{2}=4x$ have focus $F$, and let point $A$ have coordinates $(-2,0)$. The line $x+2=ky$ $(k>0)$ intersects $C$ at points $M$ and $N$, with $\\overrightarrow{AN}=2\\overrightarrow{AM}$. Then $\\overrightarrow{FM} \\cdot \\overrightarrow{FN}$ = ?",
    "fact_expressions": "C: Parabola;G: Line;k: Number;A: Point;N: Point;M: Point;F: Point;Expression(C) = (y^2 = 4*x);k>0;Expression(G) = (x + 2 = k*y)&(k>0);Coordinate(A) = (-2, 0);Focus(C) = F;Intersection(G, C) = {M, N};VectorOf(A, N) = 2*VectorOf(A, M)",
    "query_expressions": "DotProduct(VectorOf(F, M), VectorOf(F, N))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 20], [62, 65]], [[45, 61]], [[47, 61]], [[28, 32]], [[71, 74]], [[67, 70]], [[24, 27]], [[1, 20]], [[47, 61]], [[45, 61]], [[28, 44]], [[1, 27]], [[45, 76]], [[77, 122]]]",
    "query_spans": "[[[124, 176]]]",
    "process": "】Solving the system \\begin{cases}x+2=ky\\\\y^2=4x\\end{cases}, we obtain y^{2}-4ky+8=0, yielding y_{1}+y_{2}=4k, y_{1}y_{2}=8. Then, given \\overrightarrow{AN}=2\\overrightarrow{AM}, we find y_{1}=2, y_{2}=4, and further determine the coordinates of M and N. Using the dot product of vectors, we can solve the problem. According to the problem, the parabola C: y^{2}=4x has focus F(1,0). Solving the system \\begin{cases}x+2=ky\\\\y^2=4x\\end{cases}, we rearrange to get y^{2}-4ky+8=0. Let M(x_{1},y_{1}), N(x_{2},y_{2}), then y_{1}+y_{2}=4k, y_{1}y_{2}=8. Since \\overrightarrow{AN}=2\\overrightarrow{AM}, we have (x_{2}+2,y_{2})=2(x_{1}+2,y_{1}), so y_{2}=2y_{1}. Solving gives y_{1}=2, y_{2}=4, thus k=\\frac{3}{2}, and the line equation is 2x-3y+4=0. We obtain M(1,2), N(4,4), so \\overrightarrow{FM}\\cdot\\overrightarrow{FN}=(0,2)\\cdot(3,4)=8."
  },
  {
    "text": "Given that point $P(2, -3)$ lies on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and the distance between the two foci of the hyperbola is $4$, what is the equation of this hyperbola?",
    "fact_expressions": "P: Point;Coordinate(P) = (2, -3);PointOnCurve(P, G);G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F1: Point;F2: Point;Focus(G) = {F1, F2};Distance(F1, F2) = 4",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^{-y^2/3}=1",
    "fact_spans": "[[[2, 13]], [[2, 13]], [[2, 74]], [[14, 71], [75, 78], [94, 97]], [[14, 71]], [[17, 71]], [[17, 71]], [[17, 71]], [[17, 71]], [], [], [[75, 82]], [[75, 91]]]",
    "query_spans": "[[[94, 101]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{n}-\\frac{y^{2}}{12-n}=1$ is $\\sqrt{3}$, then $n=$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/(12 - n) + x^2/n = 1);n: Number;Eccentricity(G) = sqrt(3)",
    "query_expressions": "n",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 43]], [[2, 43]], [[60, 63]], [[2, 58]]]",
    "query_spans": "[[[60, 65]]]",
    "process": ""
  },
  {
    "text": "Let $A$ and $B$ be two moving points on the parabola $y = -\\frac{1}{4} x^{2}$, with $|AB| = 6$. Then the minimum distance from the midpoint $M$ of $AB$ to the $x$-axis is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;M: Point;Expression(G) = (y = -x^2/4);PointOnCurve(A,G);PointOnCurve(B, G);Abs(LineSegmentOf(A, B)) = 6;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Min(Distance(M, xAxis))",
    "answer_expressions": "2",
    "fact_spans": "[[[11, 36]], [[1, 4]], [[7, 10]], [[63, 66]], [[11, 36]], [[1, 42]], [[1, 42]], [[44, 53]], [[56, 66]]]",
    "query_spans": "[[[63, 80]]]",
    "process": ""
  },
  {
    "text": "A moving circle is externally tangent to the circle $x^{2}+y^{2}+4 x+3=0$ and internally tangent to the circle $x^{2}+y^{2}-4 x-45=0$. Let $M$ denote the locus of the center of the moving circle. Then, the maximum distance from points on the curve $M$ to the line $x-2 y-1=0$ is?",
    "fact_expressions": "G: Circle;C:Circle;T:Circle;H: Line;M: Curve;P0: Point;Expression(G) = (4*x + x^2 + y^2 + 3 = 0);Expression(T) = (-4*x + x^2 + y^2 - 45 = 0);Expression(H) = (x - 2*y - 1 = 0);IsOutTangent(C,G);IsInTangent(C,T);C1:Point;Center(C)=C1;Locus(C1)=M;PointOnCurve(P0, M)",
    "query_expressions": "Max(Distance(P0, H))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[4, 26]], [[1, 3], [60, 62]], [[32, 55]], [[82, 95]], [[68, 71], [73, 78]], [[80, 81]], [[4, 26]], [[32, 55]], [[82, 95]], [[1, 28]], [[1, 57]], [], [[60, 64]], [[60, 71]], [[73, 81]]]",
    "query_spans": "[[[80, 104]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{9}=1$ $(a>0)$, the right focus is $F$, point $M$ lies on $C$, point $N$ is the midpoint of segment $MF$, point $O$ is the origin, and $|MF|=2|ON|=4$. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/9 + x^2/a^2 = 1);a: Number;a>0;F: Point;RightFocus(C) = F;M: Point;PointOnCurve(M, C) = True;N: Point;MidPoint(LineSegmentOf(M, F)) = N;O: Origin;Abs(LineSegmentOf(M, F)) = 2*Abs(LineSegmentOf(O, N));2*Abs(LineSegmentOf(O, N)) = 4",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(7)/4",
    "fact_spans": "[[[2, 53], [67, 70], [117, 120]], [[2, 53]], [[9, 53]], [[9, 53]], [[58, 61]], [[2, 61]], [[62, 66]], [[62, 71]], [[72, 76]], [[72, 87]], [[88, 92]], [[99, 115]], [[99, 115]]]",
    "query_spans": "[[[117, 126]]]",
    "process": "Let the left focus of ellipse C be F. By the definition of an ellipse, |MF| + |MF| = 2a. That is, 4 + |MF| = 2a (*). Since O is the midpoint of segment FF and N is the midpoint of segment MF, by the property of the midline, |MF| = 2|ON| = 4. Substituting into (*) yields a = 4. Hence, the eccentricity e = \\frac{\\sqrt{a^{2}-9}}{a} = \\frac{\\sqrt{7}}{4}."
  },
  {
    "text": "The parabola $C$: $y^{2}=4 x$ has focus $F$, and its directrix $l$ intersects the $x$-axis at point $A$. Point $M$ lies on the parabola $C$. When $\\frac{|M A|}{|M F|}=\\sqrt{2}$, the area of $\\Delta A M F$ is?",
    "fact_expressions": "C: Parabola;A: Point;M: Point;F: Point;l: Line;Expression(C) = (y^2 = 4*x);Focus(C) = F;Directrix(C) = l;Intersection(l, xAxis) = A;PointOnCurve(M, C);Abs(LineSegmentOf(M, A))/Abs(LineSegmentOf(M, F)) = sqrt(2)",
    "query_expressions": "Area(TriangleOf(A, M, F))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 19], [50, 56], [27, 28]], [[40, 44]], [[45, 49]], [[23, 26]], [[30, 33]], [[0, 19]], [[0, 26]], [[27, 33]], [[30, 44]], [[45, 57]], [[59, 89]]]",
    "query_spans": "[[[91, 110]]]",
    "process": "Without loss of generality, let point M be a point in the first quadrant. As shown in the figure below, draw MP perpendicular from point M to the directrix of the parabola, with foot of perpendicular at P. Then |MF| = |MP|. When $\\frac{|MA|}{|MF|} = \\sqrt{2} = \\frac{|MA|}{|MP|} = \\frac{1}{\\cos\\angle A}$, then $\\cos\\angle AMP = \\frac{\\sqrt{2}}{2}$, so $\\angle AMP = \\frac{\\pi}{4}$. It is clear that $MP \\parallel AF$, hence $\\angle MAF = \\angle AMP = \\frac{\\pi}{4}$. Therefore, the slope of line AM is 1, and the equation of line AM is $y = x + 1$. Solving the system \n$$\n\\begin{cases}\ny = x + 1 \\\\\ny^2 = 4x\n\\end{cases}\n$$\nyields \n$$\n\\begin{cases}\nx = 1 \\\\\ny = 2\n\\end{cases}\n$$\ni.e., point $M(1,2)$. It is clear that $|AF| = 2$, thus, $S_{\\triangle AMF} = \\frac{1}{2}|AF| \\times 2 = \\frac{1}{2} \\times 2 \\times 2 = 2$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has an eccentricity of $\\sqrt{5}$, and its imaginary axis length is greater than $1$, then a standard equation of the hyperbola $C$ could be?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(C) = sqrt(5);Length(ImageinaryAxis(C))>1",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2/4 = 1",
    "fact_spans": "[[[2, 63], [80, 81], [91, 97]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 78]], [[80, 89]]]",
    "query_spans": "[[[91, 108]]]",
    "process": "According to the problem, $\\frac{c}{a}=\\sqrt{5}$, then $\\frac{b}{a}=\\sqrt{(\\frac{c}{a})^{2}-1}=2^{n}$. Moreover, since the length of its imaginary axis is greater than 1, $2b>1$. Let $b=2$, we get $a=1$, thus obtaining a standard equation of hyperbola $C$ as $x^{2}-\\frac{y^{2}}{4}=1$."
  },
  {
    "text": "The distance from $M(1,1)$ to the directrix of the parabola $y=a x^{2}$ is $2$, then $a$=?",
    "fact_expressions": "G: Parabola;a: Number;M: Point;Expression(G) = (y = a*x^2);Coordinate(M) = (1, 1);Distance(M, Directrix(G)) = 2",
    "query_expressions": "a",
    "answer_expressions": "{1/4,-1/12}",
    "fact_spans": "[[[9, 23]], [[34, 37]], [[0, 8]], [[9, 23]], [[0, 8]], [[0, 32]]]",
    "query_spans": "[[[34, 39]]]",
    "process": "The parabola $ y = ax^{2} $ can be rewritten as: $ x^{2} = \\frac{1}{a}y $, its directrix equation is: $ y = -\\frac{1}{4a} $. The distance from point $ M(1,1) $ to the directrix of the parabola $ y = ax^{2} $ is 2, which gives $ \\left|1 + \\frac{1}{4a}\\right| = 2 $. Solving this yields $ a = \\frac{1}{4} $ or $ a = -\\frac{1}{12} $."
  },
  {
    "text": "Given that the center of the ellipse is at the origin, one focus is $F(0,-2 \\sqrt{3})$, and the length of the major axis is twice the length of the minor axis, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;F: Point;OneOf(Focus(G)) = F;Coordinate(F) = (0, -2*sqrt(3));Length(MajorAxis(G)) = 2*Length(MinorAxis(G))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/16 + x^2/4 = 1",
    "fact_spans": "[[[2, 4], [50, 52]], [[7, 9]], [[2, 9]], [[15, 33]], [[2, 33]], [[15, 33]], [[2, 47]]]",
    "query_spans": "[[[50, 59]]]",
    "process": "According to the coordinates of the foci, the relationship between the major and minor axes, and the relation $ a^{2} = b^{2} + c^{2} $ in an ellipse, solve the system of equations to obtain $ a^{2} $, $ b^{2} $, thus finding the equation of the ellipse. Given that the focus of the ellipse is $ F(0, -2\\sqrt{3}) $, the foci lie on the y-axis, so assume the equation is $ \\frac{y^{2}}{a^{2}} + \\frac{x^{2}}{b^{2}} = 1 $ ($ a > b > 0 $). From the given conditions, $ c = 2\\sqrt{3} $, $ a = 2b $. Since $ a^{2} = b^{2} + c^{2} $, we have $ 4b^{2} = b^{2} + 12 $, solving gives $ a^{2} = 16 $, $ b^{2} = 4 $. Therefore, the equation of the ellipse is: $ \\frac{y^{2}}{16} + \\frac{x^{2}}{4} = 1 $."
  },
  {
    "text": "Given that the focal distance of a hyperbola is $10$, and the absolute difference of distances from any point on the hyperbola to the two foci is $8$, what are the coordinates of the foci in the standard equation of this hyperbola?",
    "fact_expressions": "G: Hyperbola;FocalLength(G)=10;P:Point;PointOnCurve(P,G);F1:Point;F2:Point;Focus(G) = {F1,F2};Abs(Distance(P,F1)-Distance(P,F2))=8",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "{(pm*5,0),(0,pm*5)}",
    "fact_spans": "[[[2, 5], [14, 17], [43, 46]], [[2, 13]], [[19, 20]], [[14, 20]], [], [], [[14, 25]], [[14, 40]]]",
    "query_spans": "[[[43, 57]]]",
    "process": "According to the problem: when the foci are on the x-axis, the coordinates are (5,0), (-5,0); when the foci are on the y-axis, the coordinates are (0,5), (0,-5). Therefore, the foci are: (5,0), (-5,0) or (0,5), (0,-5)"
  },
  {
    "text": "Given that $F$ is the right focus of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{8}=1$, $P$ is a moving point on the left branch of $C$, and $A(0,6 \\sqrt{6})$, then the minimum perimeter of $\\triangle A P F$ is?",
    "fact_expressions": "C: Hyperbola;A: Point;P: Point;F: Point;Expression(C) = (x^2 - y^2/8 = 1);Coordinate(A) = (0, 6*sqrt(6));RightFocus(C) = F;PointOnCurve(P, LeftPart(C))",
    "query_expressions": "Min(Perimeter(TriangleOf(A,P,F)))",
    "answer_expressions": "32",
    "fact_spans": "[[[6, 39], [48, 51]], [[59, 76]], [[44, 47]], [[2, 5]], [[6, 39]], [[59, 76]], [[2, 43]], [[44, 58]]]",
    "query_spans": "[[[78, 103]]]",
    "process": "From the hyperbola equation $ x^{2}-\\frac{y^{2}}{8}=1 $, we know $ a=1 $, $ c=3 $, hence $ F(3,0) $, $ F_{1}(-3,0) $. When point $ P $ moves on the left branch of the hyperbola, by the definition of the hyperbola, $ |PF|-|PF_{1}|=2 $, so $ |PF|=|PF_{1}|+2 $. Thus, the perimeter of $ \\triangle APF $ is $ |AP|+|PF|+|AF|=|AP|+|PF_{1}|+2+|AF| $. Since $ |AF|=\\sqrt{3^{2}+(6\\sqrt{6})^{2}}=15 $ is a constant, the perimeter of $ \\triangle APF $ is minimized when $ (|AP|+|PF_{1}|) $ is minimized. From the graph, this occurs when point $ P $ lies at the intersection of segment $ AF_{1} $ and the hyperbola (as shown in the figure). At this time, $ |AF_{1}|=\\sqrt{3^{2}+(6\\sqrt{6})^{2}}=15 $, so the minimum perimeter of $ \\triangle APF $ is 32."
  },
  {
    "text": "Given that the distance from a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ to one of its asymptotes is $3a$, then the equations of the asymptotes of this hyperbola are?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Distance(OneOf(Focus(G)),OneOf(Asymptote(G)))=3*a",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*3*x",
    "fact_spans": "[[[2, 58], [81, 84]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 78]]]",
    "query_spans": "[[[81, 92]]]",
    "process": "From the standard equation, the asymptote equations can be obtained. By using the distance from a point to a line to construct an equation, the value of $\\frac{b}{a}$ is found, thus obtaining the asymptote equations. [Detailed solution] $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 \\Rightarrow$ the asymptote equations are: $y=\\pm\\frac{b}{a}x$. Due to the symmetry of the hyperbola, the distances from the two foci to the two asymptotes are equal. $\\therefore$ take the asymptote $y=\\frac{b}{a}x$ and focus $(c,0)$. $\\therefore 3a = \\frac{|bc|}{\\sqrt{a^{2}+b^{2}}} = \\frac{bc}{c} = b \\Rightarrow \\frac{b}{a} = 3$. $\\therefore$ the asymptote equations are: $y = \\pm 3x$. The correct result for this problem: $y = \\pm 3x$"
  },
  {
    "text": "Given that $P$ is a point on the hyperbola $\\frac{y^{2}}{3}-x^{2}=1$, if the circle with diameter $OP$ ($O$ being the origin) intersects the two asymptotes of the hyperbola at points $A$ and $B$ respectively, then the minimum value of $|A B|$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P, G);G: Hyperbola;Expression(G) = (-x^2 + y^2/3 = 1);H: Circle;IsDiameter(LineSegmentOf(O, P), H) = True;O: Origin;L1: Line;L2: Line;Asymptote(G) = {L1, L2};Intersection(H, L1) = A;Intersection(H, L2) = B;A: Point;B: Point",
    "query_expressions": "Min(Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[2, 5]], [[2, 37]], [[6, 34], [62, 65]], [[6, 34]], [[60, 61]], [[39, 61]], [[46, 49]], [], [], [[62, 71]], [[60, 85]], [[60, 85]], [[76, 79]], [[80, 83]]]",
    "query_spans": "[[[87, 100]]]",
    "process": "Write the asymptotic equations of the hyperbola. It is clear that \\angleAOB=\\frac{\\pi}{3}. Combining that points O, P, A, B are concyclic, let the radius of this circle be R. By the law of sines, we have \\frac{|AB|}{\\sin\\frac{\\pi}{3}}=2R, thus |AB|=\\sqrt{3}R. Therefore, to find the minimum value of |AB|, it suffices to find the minimum value of R. Clearly, when point P is at the vertex of the hyperbola, the diameter |OP| is minimized, i.e., R is minimized. Finding the minimum value of R gives the solution. According to the problem, the hyperbola \\frac{y^{2}}{3}-x^{2}=1 has asymptotes given by y=\\pm\\sqrt{3}x. It is clear that points O, P, A, B are concyclic. Let the radius of this circle be R. It is clear that \\angleAOB=\\frac{\\pi}{3}. By the law of sines, we obtain \\frac{|AB|}{\\sin\\frac{\\pi}{2}}=2R, hence |AB|=\\sqrt{3}R. Therefore, to find the minimum value of |AB|, it suffices to find the minimum value of R. Clearly, when point P is at the vertex of the hyperbola, |OP| is minimized, i.e., R is minimized, and R_{\\min}=\\frac{|OP|}{2}=\\frac{\\sqrt{3}}{2}. Thus |AB|_{\\min}=\\sqrt{3}R_{\\min}=\\frac{3}{2}."
  },
  {
    "text": "Let point $P(x_{1}, y_{1})$ lie on the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{2}=1$, and point $Q(x_{2}, y_{2})$ lie on the line $x+2 y-8=0$. Then the minimum value of $3|x_{2}-x_{1}|+6|y_{2}-y_{1}|$ is?",
    "fact_expressions": "G: Ellipse;H: Line;P: Point;Q: Point;x1:Number;x2:Number;y1:Number;y2:Number;Expression(G) = (x^2/8 + y^2/2 = 1);Expression(H) = (x + 2*y - 8 = 0);Coordinate(P) = (x1,y1);Coordinate(Q) = (x2,y2);PointOnCurve(P, G);PointOnCurve(Q, H)",
    "query_expressions": "Min(3*Abs(x2-x1) + 6*Abs(y2-y1))",
    "answer_expressions": "12",
    "fact_spans": "[[[20, 57]], [[78, 91]], [[1, 19]], [[59, 77]], [[2, 19]], [[60, 77]], [[2, 19]], [[60, 77]], [[20, 57]], [[78, 91]], [[1, 19]], [[59, 77]], [[1, 58]], [[59, 92]]]",
    "query_spans": "[[[94, 131]]]",
    "process": "According to the problem, let \\begin{cases}x_{1}=2\\sqrt{2}\\cos\\alpha\\\\y=\\sqrt{2}\\sin\\alpha\\end{cases},\\alpha\\in[0,2\\pi), then 3|x_{2}-x_{1}|+6|y_{2}-y_{1}|=3|x_{2}-2\\sqrt{2}\\cos\\alpha|+6|y_{2}-\\sqrt{2}\\sin\\alpha|=3(|x_{2}-2\\sqrt{2}\\cos\\alpha|+2|y_{2}-\\sqrt{2}\\sin\\alpha|)3(|x|2\\sqrt{2}\\cos\\alpha|+2|,-\\sqrt{2}\\sin\\alpha)\\geqslant3|x_{2}+2y.-2\\sqrt{2}(\\cos\\alpha+\\sin\\alpha)=3|8-4\\sin(\\alpha+\\frac{\\pi}{4})|\\geqslant3|8-4|=12, equality holds if and only if \\alpha=\\frac{\\pi}{4}."
  },
  {
    "text": "The focal distance of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[0, 27]], [[0, 27]]]",
    "query_spans": "[[[0, 32]]]",
    "process": "From the given conditions, we have $a^{2}=4$, $b^{2}=1$, then $c^{2}=a^{2}-b^{2}=3$, $2c=2\\sqrt{3}$, that is, the focal distance is $2\\sqrt{3}$."
  },
  {
    "text": "Given a point $P$ on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ such that its distance to the left focus is $\\frac{5}{2}$, then what is its distance to the right directrix?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/4 + y^2/3 = 1);PointOnCurve(P, G);Distance(P, LeftFocus(G)) = 5/2",
    "query_expressions": "Distance(P,RightDirectrix(G))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 39]], [[42, 45], [68, 69]], [[2, 39]], [[2, 45]], [[2, 66]]]",
    "query_spans": "[[[2, 78]]]",
    "process": ""
  },
  {
    "text": "The line $ l: y - x + 2 = 0 $ intersects the parabola $ y^2 = 2px $ ($ p > 0 $) at points $ A $ and $ B $. A perpendicular from the midpoint $ M $ of $ AB $ to the $ y $-axis intersects the parabola at point $ N $. If $ \\angle ANB = \\frac{\\pi}{2} $, then $ p = $?",
    "fact_expressions": "l: Line;G: Parabola;p: Number;A: Point;B: Point;N: Point;M:Point;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(l) = (-x + y + 2 = 0);Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, B))=M;L:Line;PointOnCurve(M, L);IsPerpendicular(L, yAxis);Intersection(L, G) = N;AngleOf(A, N, B) = pi/2",
    "query_expressions": "p",
    "answer_expressions": "4/7",
    "fact_spans": "[[[0, 16]], [[17, 38], [71, 74]], [[113, 116]], [[40, 43]], [[44, 47]], [[75, 79]], [[59, 62]], [[20, 38]], [[17, 38]], [[0, 16]], [[0, 49]], [[51, 62]], [], [[50, 70]], [[50, 70]], [[50, 79]], [[83, 111]]]",
    "query_spans": "[[[113, 118]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Solving the system of equations \\begin{cases}y-x+2=0\\\\y^{2}=2px\\end{cases} yields y^{2}-2py-4p=0. Clearly, A>0. By Vieta's formulas, we obtain: y_{1}+y_{2}=2p, y_{1}y_{2}=-4p. Therefore, |AB|=\\sqrt{1+(\\frac{1}{k})^{2}}\\frac{1}{4})^{2}|y_{1}-y_{2}|=\\sqrt{2}\\times\\sqrt{4p^{2}+16p}. Let y_{M}=p, then x_{M}=p+2, x_{N}=\\frac{p}{2}, so |MN|=|x_{M}-x_{1}|=\\frac{p}{2}+2. Since M is the midpoint of AB and \\angle ANB=\\frac{\\pi}{2}, it follows that |AB|=2|MN|. Hence, \\sqrt{2}\\times\\sqrt{4p^{2}+16p}=p+4. Solving gives p=\\frac{4}{7}."
  },
  {
    "text": "The focus of the parabola $y=a x^{2}(a>0)$ coincides with one focus of the ellipse $\\frac{y^{2}}{10}+x^{2}=1$. Then, the equation of the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = a*x^2);a: Number;a>0;H: Ellipse;Expression(H) = (x^2 + y^2/10 = 1);Focus(G)=OneOf(Focus(H))",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=-3",
    "fact_spans": "[[[0, 19], [60, 63]], [[0, 17]], [[3, 19]], [[3, 19]], [[23, 51]], [[23, 51]], [[0, 58]]]",
    "query_spans": "[[[60, 70]]]",
    "process": "\\because the foci of the ellipse \\frac{y^{2}}{10}+x^{2}=1 are (0,\\pm3), and the focus of the parabola y=ax^{2} (a>0) is at (0,3), \\therefore \\frac{1}{4a}=3, solving gives a=\\frac{1}{12}, so the standard equation of the parabola is x^{2}=12y. Therefore, the directrix of the parabola is y=-3."
  },
  {
    "text": "The standard equation of the hyperbola with foci on the $x$-axis passing through points $P(4, -2)$ and $Q(2\\sqrt{6}, 2\\sqrt{2})$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;Q: Point;Coordinate(P) = (4, -2);Coordinate(Q) = (2*sqrt(6), 2*sqrt(2));PointOnCurve(Focus(G), xAxis);PointOnCurve(P, G);PointOnCurve(Q, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/8 - y^2/4 = 1",
    "fact_spans": "[[[53, 56]], [[11, 22]], [[23, 52]], [[11, 22]], [[23, 52]], [[0, 56]], [[9, 56]], [[9, 56]]]",
    "query_spans": "[[[53, 63]]]",
    "process": "Let the hyperbola equation be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$. Substituting the points $(4,-2)$ and $(2\\sqrt{6},2\\sqrt{2})$ into the equation yields\n\\[\n\\begin{cases}\n\\frac{16}{a^{2}}-\\frac{4}{b^{2}}=1, \\\\\n\\frac{24}{a^{2}}-\\frac{8}{b^{2}}=1,\n\\end{cases}\n\\]\nsolving gives $a^{2}=8$, $b^{2}=4$, so the standard equation of the hyperbola is $\\frac{x^{2}}{8}-\\frac{y^{2}}{4}=1$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{k+2}+\\frac{y^{2}}{k+1}=1$, and chord $A B$ passes through $F_{1}$. If the perimeter of $\\triangle A B F_{2}$ is $8$, then what is the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;k: Number;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2/(k + 2) + y^2/(k + 1) = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, LineSegmentOf(A, B));IsChordOf(LineSegmentOf(A, B), G);Perimeter(TriangleOf(A, B, F2)) = 8",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[18, 59], [111, 113]], [[20, 59]], [[66, 71]], [[66, 71]], [[10, 17]], [[2, 9], [72, 79]], [[18, 59]], [[2, 64]], [[2, 64]], [[66, 79]], [[2, 71]], [[81, 109]]]",
    "query_spans": "[[[111, 119]]]",
    "process": ""
  },
  {
    "text": "Given that a focus of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{3 m}=1$ is $F(0 , 2)$, then $m$=?",
    "fact_expressions": "G: Hyperbola;m: Number;F: Point;Expression(G) = (-y^2/(3*m) + x^2/m = 1);Coordinate(F) = (0, 2);OneOf(Focus(G))=F",
    "query_expressions": "m",
    "answer_expressions": "-1",
    "fact_spans": "[[[2, 42]], [[60, 63]], [[48, 58]], [[2, 42]], [[48, 58]], [[2, 58]]]",
    "query_spans": "[[[60, 65]]]",
    "process": ""
  },
  {
    "text": "It is known that the line $y=2x$ is an asymptote of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y = 2*x);OneOf(Asymptote(C))=G",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[12, 63], [71, 77]], [[19, 63]], [[19, 63]], [[2, 11]], [[12, 63]], [[2, 11]], [[2, 69]]]",
    "query_spans": "[[[71, 83]]]",
    "process": "\\because the line y=2x is an asymptote of the hyperbola C: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0, b>0), \\therefore \\frac{b}{a}=2, \\therefore b=2a, \\therefore b^{2}=4a^{2}, \\therefore c^{2}-a^{2}=4a^{2}, \\therefore c^{2}=5a^{2}, \\therefore e=\\frac{c}{a}=\\sqrt{5}"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{8}=1$ with left and right foci $F_{1}$ and $F_{2}$, a point $P$ on the ellipse satisfies $|P F_{1}|=\\frac{5}{2}$, then $|P F_{2}|=$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 + y^2/8 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P,G);Abs(LineSegmentOf(P,F1))=5/2",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "7/2",
    "fact_spans": "[[[2, 39], [62, 64]], [[67, 70]], [[46, 53]], [[54, 61]], [[2, 39]], [[2, 61]], [[2, 61]], [[62, 70]], [[72, 95]]]",
    "query_spans": "[[[97, 110]]]",
    "process": "In the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{8}=1$, $a^{2}=9$, so $a=3$, $|PF_{1}|+|PF_{2}|=2a=6$. Since $|PF_{1}|=\\frac{5}{2}$, it follows that $|PF_{2}|=6-\\frac{5}{2}=\\frac{7}{2}$."
  },
  {
    "text": "Given that the left and right foci of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ are $F_{1}$ and $F_{2}$ respectively, and $P$ is an arbitrary point on the circle with the minor axis of the ellipse as its diameter, then $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$=?",
    "fact_expressions": "G: Ellipse;H: Circle;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2/3 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P,H);IsDiameter(MinorAxis(G),H)",
    "query_expressions": "DotProduct(VectorOf(P, F1), VectorOf(P, F2))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 39], [69, 71]], [[77, 78]], [[64, 67]], [[48, 55]], [[56, 63]], [[2, 39]], [[2, 63]], [[2, 63]], [[64, 83]], [[68, 78]]]",
    "query_spans": "[[[85, 144]]]",
    "process": "By the given condition, the left and right foci of $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ are $F_1(-1,0)$, $F_{2}(1,0)$ respectively. Let $P(\\sqrt{3}\\cos\\theta,\\sqrt{3}\\sin\\theta)$ be an arbitrary point on the circle with diameter equal to the minor axis of the ellipse. Then $\\overrightarrow{PF}_{1}=(-1-\\sqrt{3}\\cos\\theta,-\\sqrt{3}\\sin\\theta)$, $\\overrightarrow{PF_{2}}=(1-\\sqrt{3}\\cos\\theta,-\\sqrt{3}\\sin\\theta)$, and $\\overrightarrow{PF}_{1}\\cdot\\overrightarrow{PF_{2}}=(1-\\sqrt{3}\\cos\\theta)(-1-\\sqrt{3}\\cos\\theta)+3\\sin^{2}\\theta=3\\sin^{2}\\theta+3\\cos^{2}\\theta-1=2$; hence, fill in $2$. [Technique"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $A$ lies on one of the asymptotes of $C$, such that $O A \\perp F_{1} A $ and $|A F_{2}|=2|A F_{1}|$. Then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;O: Origin;A: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) =F1;RightFocus(C)=F2;PointOnCurve(A,OneOf(Asymptote(C)));IsPerpendicular(LineSegmentOf(O, A), LineSegmentOf(F1, A));Abs(LineSegmentOf(A, F2)) = 2*Abs(LineSegmentOf(A, F1))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(21)/3",
    "fact_spans": "[[[2, 63], [93, 96], [153, 159]], [[10, 63]], [[10, 63]], [[108, 128]], [[89, 92]], [[73, 80]], [[81, 88]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 88]], [[2, 88]], [[89, 106]], [[108, 128]], [[129, 151]]]",
    "query_spans": "[[[153, 165]]]",
    "process": "\\because the hyperbola C has equation: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0, b>0), \\therefore the asymptotes of hyperbola C are given by: bx\\pm ay=0. Assume point A lies on the asymptote bx+ay=0 of hyperbola C, the left focus of the hyperbola is F_{1}(-c,0) and the right focus is F_{2}(c,0). As shown in the figure, the distance is: |F_{1}A|=\\frac{bc}{\\sqrt{a^{2}+b^{2}}}=b, \\therefore |AO|=\\sqrt{|OF_{1}|^{2}-|AF_{1}|^{2}}=a. In right triangle \\triangle AOF_{1}, \\cos\\angle AOF_{1}=\\frac{|OA|}{|OF_{1}|}=\\frac{a}{c}. Also \\because |AF_{2}|=2|AF_{1}|, \\therefore |AF_{2}|=2|AF_{1}|=2b. In \\triangle AOF_{2}, by the law of cosines: \\cos\\angle AOF_{2}=\\frac{|OA|^{2}+|OF_{2}|^{2}-|AF_{2}|^{2}}{2|OA|\\times|OF_{2}|}=\\frac{5a^{2}-3c^{2}}{2ac}. Also \\because \\angle AOF_{1} + \\angle AOF_{2}=180^{\\circ}, \\therefore \\cos\\angle AOF_{1} + \\cos\\angle AOF_{2}=0. \\therefore \\frac{a}{c}+\\frac{5a^{2}-3c^{2}}{2ac}=0, \\therefore 7a^{2}=3c^{2}, \\therefore \\frac{c^{2}}{a^{2}}=\\frac{7}{3}, \\therefore e=\\frac{\\sqrt{21}}{3}."
  },
  {
    "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ intersect the circle $x^{2}-4x+y^{2}+2=0$, then the range of the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 + x^2 - 4*x + 2 = 0);IsIntersect(Asymptote(G),H)",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1,\\sqrt{2})",
    "fact_spans": "[[[2, 59], [90, 93]], [[5, 59]], [[5, 59]], [[64, 86]], [[5, 59]], [[5, 59]], [[2, 59]], [[64, 86]], [[2, 88]]]",
    "query_spans": "[[[90, 104]]]",
    "process": "From the hyperbola equation, its asymptotes are given by: $ y = \\pm\\frac{b}{a}x $, i.e., $ bx \\pm ay = 0 $. The standard equation of the circle is: $ (x-2)^{2} + y^{2} = 2 $. Without loss of generality, consider the asymptote $ bx + ay = 0 $ intersecting the circle, then: $ \\frac{|2b+0|}{\\sqrt{a^{2}+b^{2}}} < \\sqrt{2} $. Simplifying yields: $ \\frac{2b}{c} < \\sqrt{2} $, i.e., $ 2(c^{2}-a^{2}) < c^{2} $. Then $ e^{2} = \\frac{c^{2}}{a^{2}} < 2 $, so $ e < \\sqrt{2} $. From the properties of a hyperbola, the eccentricity satisfies $ e > 1 $. Combining these results, the range of the hyperbola's eccentricity is $ (1,\\sqrt{2}) $."
  },
  {
    "text": "If the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1$ $(a>0)$ has eccentricity $2$, then $a$=?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2/3 + x^2/a^2 = 1);Eccentricity(G) = 2",
    "query_expressions": "a",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 48]], [[58, 61]], [[4, 48]], [[1, 48]], [[1, 56]]]",
    "query_spans": "[[[58, 63]]]",
    "process": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1$ ($a>0$) has eccentricity 2, $|1+\\frac{3}{a^{2}}|=2$ gives $a=1$"
  },
  {
    "text": "Given that the vertex of the parabola $C$ is at the origin and the focus is $F(1 , 0)$, the line $l$ intersects the parabola $C$ at points $A$ and $B$. If the midpoint of $AB$ is $(2 , 2)$, then what is the equation of the line $l$?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;F: Point;O: Origin;Focus(C)=F;Coordinate(F) = (1, 0);Vertex(C) = O;Intersection(l, C) = {A, B};Coordinate(MidPoint(LineSegmentOf(A, B))) = (2,2)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "y=x",
    "fact_spans": "[[[32, 37], [79, 84]], [[3, 9], [38, 44]], [[47, 50]], [[51, 54]], [[21, 31]], [[13, 17]], [[3, 31]], [[21, 31]], [[3, 17]], [[32, 56]], [[58, 77]]]",
    "query_spans": "[[[79, 89]]]",
    "process": "The equation of the parabola is $ y^{2} = 4x $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} \\neq x_{2} $, and\n\\[\n\\begin{cases}\ny_{1}^{2} = 4x_{1} \\\\\ny_{2}^{2} = 4x_{2}\n\\end{cases}\n\\]\nSubtracting these two equations gives $ y_{1}^{2} - y_{2}^{2} = 4(x_{1} - x_{2}) $, $ \\therefore \\frac{y_{1} - y_{2}}{x_{1} - x_{2}} = \\frac{4}{y_{1} + y_{2}} $. $ \\therefore $ The equation of line $ l $ is $ y - 2 = x - 2 $, that is, $ y = x $."
  },
  {
    "text": "Given the hyperbola $C$ with equation $x^{2}-\\frac{y^{2}}{8}=1$, and its right focus at $F$. If point $N(0,6)$, and $M$ is a point on the left branch of hyperbola $C$, then the minimum value of the perimeter of $\\Delta F M N$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2/8 = 1);F: Point;RightFocus(C) = F;N: Point;Coordinate(N) = (0, 6);M: Point;PointOnCurve(M, LeftPart(C))",
    "query_expressions": "Min(Perimeter(TriangleOf(F, M, N)))",
    "answer_expressions": "6*sqrt(5)+2",
    "fact_spans": "[[[2, 8], [62, 68]], [[2, 37]], [[42, 45]], [[2, 45]], [[47, 56]], [[47, 56]], [[58, 61]], [[58, 74]]]",
    "query_spans": "[[[76, 98]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ and the circle $x^{2}+y^{2}-8 x+15=0$, then the distance from the center $C$ of the circle to the asymptotes of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/5 = 1);H: Circle;Expression(H) = (x^2 + y^2 - 8*x + 15 = 0);C: Point;Center(H) = C",
    "query_expressions": "Distance(C, Asymptote(G))",
    "answer_expressions": "(4*sqrt(5))/3",
    "fact_spans": "[[[2, 40], [72, 75]], [[2, 40]], [[41, 64]], [[41, 64]], [[68, 71]], [[41, 71]]]",
    "query_spans": "[[[68, 83]]]",
    "process": "The circle $x^{2}+y^{2}-8x+15=0$ has center $(4,0)$. The asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ are given by $y=\\frac{\\sqrt{5}}{2}x$, or equivalently $\\sqrt{5}x-2y=0$. Therefore, the distance from the center $C$ to the asymptote of the hyperbola is $\\frac{|4\\sqrt{5}|}{\\sqrt{5+4}}=\\frac{4}{3}\\sqrt{5}$."
  },
  {
    "text": "Let a line passing through the focus $F$ of the parabola $C$: $y^{2}=2 p x$ ($p>0$) intersect the parabola $C$ at points $A$ and $B$. If $|A F|=p+2$, $|B F|=p-1$, then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;A: Point;F: Point;B: Point;p>0;Expression(C) = (y^2 = 2*p*x);PointOnCurve(F, G);Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, F)) = p + 2;Abs(LineSegmentOf(B, F)) = p - 1;Focus(C)=F",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 27], [37, 43]], [[82, 85]], [[34, 36]], [[45, 48]], [[30, 33]], [[49, 52]], [[9, 27]], [[1, 27]], [[0, 36]], [[34, 54]], [[56, 67]], [[69, 80]], [[1, 33]]]",
    "query_spans": "[[[82, 87]]]",
    "process": "Let the coordinates of points A and B be (x_{1},y_{1}) and (x_{2},y_{2}), respectively. Clearly, the slope of line AB exists; let the equation of line AB be y=k(x-\\frac{p}{2}) (k\\neq0). Solving the system of equations \\begin{cases}y^{2}=2px\\\\y=k(x-\\frac{p}{2})\\end{cases}. Given |AF|=p+2, |BF|=p-1, we have \\begin{cases}\\frac{p^{2k2}}{4}=0,\\\\x_{1}+\\frac{p}{2}=p+2,\\\\x_{2}+\\frac{p}{2}=p-1,\\end{cases}, which yields \\begin{cases}x_{1}=p+\\frac{p}{2},\\\\x_{2}=\\frac{p}{2}-1,\\end{cases}. Then (p+\\frac{p}{2})(\\frac{p}{2}-1)=\\frac{p^{2}}{4}, solving gives p=4."
  },
  {
    "text": "Let $P$ be a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$. If $F_{1}$ and $F_{2}$ are the two foci of the ellipse, then $|P F_{1}|+|P F_{2}|$ equals?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(P, G);Focus(G)={F1,F2}",
    "query_expressions": "Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "10",
    "fact_spans": "[[[5, 44], [66, 68]], [[1, 4]], [[49, 56]], [[58, 65]], [[5, 44]], [[1, 47]], [[49, 73]]]",
    "query_spans": "[[[75, 99]]]",
    "process": ""
  },
  {
    "text": "For any value of $a$, the line $(a-1) x - y + 2a + 1 = 0$ always passes through a fixed point $P$. Then, what is the standard equation of the parabola passing through point $P$?",
    "fact_expressions": "G: Parabola;H: Line;a: Number;P: Point;Expression(H) = (2*a + x*(a - 1) - y + 1 = 0);PointOnCurve(P,G);PointOnCurve(P,H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=-(9/2)*x,x^2=(4/3)*y}",
    "fact_spans": "[[[46, 49]], [[10, 31]], [[2, 5]], [[35, 38], [41, 45]], [[10, 31]], [[10, 38]], [[40, 49]]]",
    "query_spans": "[[[46, 56]]]",
    "process": "Since the line $(a-1)x - y + 2a + 1 = 0$ always passes through a fixed point $P$ regardless of the value of $a$, then passing through the point $(x+2)a + (-x - y + 1) = 0$, hence $x = -2$, $y = 3$, therefore the equations of the parabolas passing through point $P$ are $x^{2} = \\frac{4}{3}y$ or $y^{2} = -\\frac{9}{2}x$."
  },
  {
    "text": "The equation of an asymptote of the hyperbola $x^{2}-\\frac{y^{2}}{m}=1$ $(m>0)$ is $y= \\sqrt{2} x$, then $m=$?",
    "fact_expressions": "G: Hyperbola;m: Number;m>0;Expression(G) = (x^2 - y^2/m = 1);Expression(OneOf(Asymptote(G))) = (y = (sqrt(2)*x))",
    "query_expressions": "m",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 33]], [[60, 63]], [[3, 33]], [[0, 33]], [[0, 58]]]",
    "query_spans": "[[[60, 65]]]",
    "process": ""
  },
  {
    "text": "Given point $A(3,4)$, $F$ is the focus of the parabola $y^{2}=8x$, and $M$ is a moving point on the parabola. When $|AM|+|MF|$ is minimized, the coordinates of point $M$ are?",
    "fact_expressions": "G: Parabola;A: Point;M: Point;F: Point;Expression(G) = (y^2 = 8*x);Coordinate(A) = (3, 4);Focus(G)=F;PointOnCurve(M, G);WhenMin(Abs(LineSegmentOf(A,M))+Abs(LineSegmentOf(M,F)))",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(2,4)",
    "fact_spans": "[[[16, 30], [38, 41]], [[2, 11]], [[34, 37], [64, 68]], [[12, 15]], [[16, 30]], [[2, 11]], [[12, 33]], [[34, 45]], [[46, 63]]]",
    "query_spans": "[[[64, 72]]]",
    "process": "From the given information, point A lies inside the parabola. Let the distance from M to the directrix be |MK|, then |MA| + |MF| = |MA| + |MK|. When |MA| + |MK| is minimized, the coordinates of point M are (2,4)."
  },
  {
    "text": "Given that $F$ is the left focus of the ellipse $C$: $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$, a perpendicular line to the $x$-axis is drawn through $F$, intersecting $C$ at points $A$ and $B$. Then $|A B|=$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/16 + y^2/4 = 1);F: Point;LeftFocus(C) = F;L: Line;PointOnCurve(F, L);IsPerpendicular(L, xAxis);A: Point;B: Point;Intersection(L, C) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "2",
    "fact_spans": "[[[6, 49], [67, 70]], [[6, 49]], [[2, 5], [55, 58]], [[2, 53]], [], [[54, 66]], [[54, 66]], [[71, 74]], [[75, 78]], [[54, 80]]]",
    "query_spans": "[[[82, 91]]]",
    "process": "Since F is the left focus of the ellipse $ C: \\frac{x^2}{16} + \\frac{y^2}{4} = 1 $, we have $ F(-2\\sqrt{3}, 0) $. Drawing a vertical line through F intersecting C at points A and B, with A above B, we obtain $ A(-2\\sqrt{3}, 1) $, $ B(-2\\sqrt{3}, -1) $, so $ |AB| = 2 $. Hence, the answer is: $ 2 $. This problem mainly examines properties of ellipses and computational ability, and is a basic-level question."
  },
  {
    "text": "Let $O$ be the origin, and let the line $x=a$ intersect the two asymptotes of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ at points $D$ and $E$, respectively. If the area of $\\triangle O D E$ is $1$, then the minimum value of the focal length of hyperbola $C$ is?",
    "fact_expressions": "O: Origin;G: Line;Expression(G) = (x = a);C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;L1: Line;L2: Line;Asymptote(C) = {L1, L2};Intersection(G, L1) = D;Intersection(G, L2) = E;D: Point;E: Point;Area(TriangleOf(O, D, E)) = 1",
    "query_expressions": "Min(FocalLength(C))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[1, 4]], [[10, 17]], [[10, 17]], [[18, 82], [129, 135]], [[18, 82]], [[26, 82]], [[26, 82]], [[26, 82]], [[26, 82]], [], [], [[18, 88]], [[10, 101]], [[10, 101]], [[92, 95]], [[96, 99]], [[103, 127]]]",
    "query_spans": "[[[129, 144]]]",
    "process": "From the given conditions, we have D(a,b), E(a,-b). Using the area of \\triangle ODE, we obtain ab=1. According to c^{2}=a^{2}+b^{2}, the basic inequality can be applied. The asymptotes of the hyperbola are y=\\pm\\frac{b}{a}x', so D(a,b), E(a,-b). Since the area of \\triangle ODE is 1, we have a\\cdot2b\\cdot\\frac{1}{2}=1, that is, ab=1. Because c^{2}=a^{2}+b^{2}, it follows that 2c=2\\sqrt{a^{2}+b^{2}}\\geqslant2\\sqrt{2ab}=2\\sqrt{2}, with equality holding if and only if a=b=1. Thus, the minimum value of the focal length of the hyperbola is 2\\sqrt{2}."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$, satisfying $|A F_{1}|=2|B F_{1}|=|F_{1} F_{2}|$. Then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Line;PointOnCurve(F1, H);A: Point;B: Point;Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, F1)) = 2*Abs(LineSegmentOf(B, F1));2*Abs(LineSegmentOf(B, F1)) = Abs(LineSegmentOf(F1, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(\\sqrt{13} - 2)/3",
    "fact_spans": "[[[2, 54], [93, 95], [147, 149]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[62, 69], [82, 89]], [[72, 79]], [[2, 79]], [[2, 79]], [[90, 92]], [[81, 92]], [[96, 99]], [[100, 103]], [[90, 105]], [[109, 145]], [[109, 145]]]",
    "query_spans": "[[[147, 155]]]",
    "process": "Let |AF_{1}|=2|BF_{1}|=|F_{1}F_{2}|=2c, then |AF_{2}|=2a-2c, |BF_{2}|=2a-c. In triangles AF_{1}F_{2} and BF_{1}F_{2}, applying the cosine law respectively gives \\cos\\angle AF_{1}F_{2}+\\cos\\angle BF_{1}F_{2}=0, \\frac{2-(2a-c)^{2}}{4c^{2}}=0"
  },
  {
    "text": "Given that the eccentricity of the ellipse $x^{2}+m y^{2}=1$ $(m>1)$ is $\\frac{\\sqrt{3}}{2}$, then $m=$?",
    "fact_expressions": "G: Ellipse;m: Number;m>1;Expression(G) = (m*y^2 + x^2 = 1);Eccentricity(G) = sqrt(3)/2",
    "query_expressions": "m",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 26]], [[53, 56]], [[4, 26]], [[2, 26]], [[2, 51]]]",
    "query_spans": "[[[53, 58]]]",
    "process": "From the given conditions: the ellipse equation is transformed into standard form as $x^{2}+\\frac{y^{2}}{\\frac{1}{m}}=1$. Since the eccentricity of the ellipse is $\\frac{\\sqrt{3}}{2}$, we have $e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{a^{2}-b^{2}}{a^{2}}=\\frac{3}{4}$, so $a^{2}=4b^{2}$. Given $m>1$, it follows that $\\frac{4}{m}=1$, hence $m=4$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has the directrix equation $x=-2$, focus $F$, and point $A$ is the intersection of the directrix and the $x$-axis. Let $B$ be a point on the parabola $C$ satisfying $\\sqrt{5}|B F|=2|A B|$. Then the distance from point $F$ to line $A B$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*p*x);Expression(Directrix(C)) = (x = -2);p: Number;p>0;F: Point;Focus(C) = F;A: Point;Intersection(Directrix(C), xAxis) = A;B: Point;PointOnCurve(B, C);sqrt(5)*Abs(LineSegmentOf(B, F)) = 2*Abs(LineSegmentOf(A, B))",
    "query_expressions": "Distance(F, LineSegmentOf(A, B))",
    "answer_expressions": "4*sqrt(5)/5",
    "fact_spans": "[[[2, 28], [67, 73]], [[2, 28]], [[2, 40]], [[10, 28]], [[10, 28]], [[44, 47], [104, 108]], [[2, 47]], [[59, 62]], [[2, 62]], [[63, 66]], [[63, 76]], [[80, 102]]]",
    "query_spans": "[[[104, 119]]]",
    "process": ""
  },
  {
    "text": "If the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has left focus $F$, right vertex $A$, and $P$ is a point on the left branch of $E$ such that $\\angle P A F=60^{\\circ}$, $|P A|=|A F|$, then the eccentricity of $E$ is?",
    "fact_expressions": "E: Hyperbola;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;P: Point;A: Point;F: Point;a>0;b>0;LeftFocus(E) = F;RightVertex(E)=A;PointOnCurve(P,LeftPart(E));AngleOf(P, A, F) = ApplyUnit(60, degree);Abs(LineSegmentOf(P,A))=Abs(LineSegmentOf(A,F))",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 62], [83, 86], [136, 139]], [[1, 62]], [[9, 62]], [[9, 62]], [[79, 82]], [[75, 78]], [[67, 70]], [[9, 62]], [[9, 62]], [[1, 70]], [[1, 78]], [[79, 92]], [[94, 119]], [[121, 134]]]",
    "query_spans": "[[[136, 145]]]",
    "process": "From the given, |PA| = |AF| = a + c, ∴ |PF| = a + c. Let the right focus of E be F'. By the law of cosines, |PF'|² = (a + c)² + (2c)² - 2(a + c)(2c) cos60°, ∴ |PF'| = √(3c² + a²). By the hyperbola definition, |PF'| - |PF| = 2a, i.e., √(3c² + a²) - (a + c) = 2a, ∴ c² - 3ac - 4a² = 0, i.e., e² - 3e - 4 = 0, ∴ e = 4 or -1 (discarded)."
  },
  {
    "text": "The asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{4}=2$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/4 = 2)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*2*x",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 36]]]",
    "process": "The asymptotes of \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 are y=\\pm\\frac{b}{a}x. [Detailed explanation] x^{2}-\\frac{y^{2}}{4}=2, i.e., \\frac{x^{2}}{2}-\\frac{y^{2}}{8}=1, has asymptotes given by y=\\pm\\sqrt{\\frac{8}{2}}x, i.e., y=\\pm2x."
  },
  {
    "text": "Given point $M(\\sqrt{3}, 0)$, the line $y=k(x+\\sqrt{3})$ intersects the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ at points $A$, $B$. Then the perimeter of $\\triangle A B M$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (sqrt(3), 0);H: Line;Expression(H) = (y = k*(x + sqrt(3)));G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);A: Point;B: Point;Intersection(H, G) = {A, B};k: Number",
    "query_expressions": "Perimeter(TriangleOf(A, B, M))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 19]], [[2, 19]], [[20, 39]], [[20, 39]], [[40, 67]], [[40, 67]], [[70, 73]], [[76, 79]], [[20, 81]], [[20, 39]]]",
    "query_spans": "[[[83, 105]]]",
    "process": "The point $M(\\sqrt{3},0)$ is the right focus of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and the line $y=k(x+\\sqrt{3})$ passes through the left focus of the ellipse. By the definition of an ellipse, the perimeter of $\\triangle ABM$ can be obtained. According to the problem, in the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, $a=2$, $b=1$, $c=\\sqrt{3}$. Therefore, the point $M(\\sqrt{3},0)$ is the right focus of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and the line $y=k(x+\\sqrt{3})$ passes through the left focus of the ellipse. Hence, by the definition of the ellipse, the perimeter of $\\triangle ABM$ is $4a=4\\times2=8$."
  },
  {
    "text": "A point $P$ on the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{64}=1$ is at a distance of $7$ from one of its foci. Then, the distance from point $P$ to the other focus is equal to?",
    "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2/16 - y^2/64 = 1);PointOnCurve(P, G);F1:Point;F2:Point;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2);Distance(P, F1) = 7",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "15",
    "fact_spans": "[[[0, 40], [47, 48]], [[43, 46], [62, 66]], [[0, 40]], [[0, 46]], [], [], [[47, 53]], [[47, 72]], [[47, 72]], [[43, 60]]]",
    "query_spans": "[[[47, 78]]]",
    "process": "The hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{64}=1$ has $a=4$, $b=8$, $c=\\sqrt{16+64}=4\\sqrt{5}$. Let the left and right foci be $F_{1}$, $F_{2}$, then by the definition of the hyperbola, $|PF|-|PF_{2}|=2a=8$. Suppose $|PF|=7$, then $|PF_{2}|=15$ or $|PF_{2}|=-1$ (discarded)."
  },
  {
    "text": "Given that the distance from the focus to the directrix of the parabola $y^{2}=2 p x(p>0)$ is $\\frac{1}{2}$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;Distance(Focus(G),Directrix(G))=1/2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=x",
    "fact_spans": "[[[2, 23], [27, 28], [50, 53]], [[2, 23]], [[5, 23]], [[5, 23]], [[2, 47]]]",
    "query_spans": "[[[50, 58]]]",
    "process": "\\because given that the distance from the focus to the directrix of the parabola y^{2}=2px(p>0) is p=\\frac{1}{2}, \\therefore the equation of the parabola is y2=x,"
  },
  {
    "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ are $y=\\pm \\frac{3}{4} x$, then its eccentricity is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(Asymptote(G)) = (y = pm*((3/4)*x))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/4",
    "fact_spans": "[[[2, 48], [79, 80]], [[5, 48]], [[5, 48]], [[2, 48]], [[2, 76]]]",
    "query_spans": "[[[79, 86]]]",
    "process": "\\because the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 has its foci on the x-axis, \\therefore the asymptotes are given by y=\\pm\\frac{b}{a}x. Again, \\because the asymptotes are given by y=\\frac{3}{4}x, \\therefore \\frac{b}{a}=\\frac{3}{4}, \\therefore \\frac{b^{2}}{a^{2}}=\\frac{9}{16}. \\because b^{2}=c^{2}-a^{2}, \\therefore \\frac{c^{2}-a^{2}}{a^{2}}=\\frac{9}{16}. Simplifying gives \\frac{c^{2}}{a^{2}}-1=\\frac{9}{16}, i.e., e^{2}=\\frac{25}{16}, e=\\frac{5}{4}."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=4 x$ be $F$, the directrix be $l$, and let $P$ be a point on the parabola, with $PA \\perp l$ and $A$ the foot of the perpendicular. If the angle of inclination of $AF$ is $\\frac{2 \\pi}{3}$, then $|PF|=$?",
    "fact_expressions": "G: Parabola;A: Point;F: Point;P: Point;l: Line;Expression(G) = (y^2 = 4*x);Focus(G) = F;Directrix(G)=l;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P,A),l);FootPoint(LineSegmentOf(P,A),l)=A;Inclination(LineSegmentOf(A, F)) = (2*pi)/3",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 15], [36, 39]], [[57, 60]], [[19, 22]], [[32, 35]], [[26, 29]], [[1, 15]], [[1, 22]], [[1, 29]], [[32, 42]], [[43, 55]], [[43, 63]], [[66, 92]]]",
    "query_spans": "[[[94, 102]]]",
    "process": "From the definition of a parabola, we know: |PF| = |PA|, ∴ △APF is an isosceles triangle. Also, ∠AFx = \\frac{2\\pi}{3}, ∴ △APF is an equilateral triangle."
  },
  {
    "text": "Given that an asymptote of the hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ is tangent to the circle $x^{2}+y^{2}-4 x+3=0$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2 + x^2/m = 1);m: Number;H: Circle;Expression(H) = (-4*x + x^2 + y^2 + 3 = 0);IsTangent(OneOf(Asymptote(G)), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 30], [64, 67]], [[2, 30]], [[5, 30]], [[37, 59]], [[37, 59]], [[2, 61]]]",
    "query_spans": "[[[64, 73]]]",
    "process": ""
  },
  {
    "text": "The equation of a hyperbola that shares the same foci with the ellipse $x^{2} + 4 y^{2}=16$ and has an asymptote given by $x+\\sqrt{3} y=0$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2 + 4*y^2 = 16);Focus(G)=Focus(H);Expression(OneOf(Asymptote(G))) = (x + sqrt(3)*y = 0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/3 = 1",
    "fact_spans": "[[[55, 58]], [[1, 23]], [[1, 23]], [[0, 58]], [[30, 58]]]",
    "query_spans": "[[[55, 63]]]",
    "process": ""
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$ and the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ have common foci $F_{1}$, $F_{2}$, and $P$ is an intersection point of the two curves. What is the value of $\\cos \\angle F_{1} PF_{2}$?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;P:Point;F1:Point;F2:Point;Expression(G) = (x^2/3 - y^2 = 1);Expression(H) = (x^2/6 + y^2/2 = 1);Focus(G)={F1,F2};Focus(H)={F1,F2};OneOf(Intersection(G,H))=P",
    "query_expressions": "Cos(AngleOf(F1,P,F2))",
    "answer_expressions": "1/3",
    "fact_spans": "[[[38, 66]], [[0, 37]], [[88, 91]], [[72, 79]], [[80, 87]], [[38, 66]], [[0, 37]], [[0, 87]], [[0, 87]], [[88, 100]]]",
    "query_spans": "[[[103, 132]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has left and right foci $F_{1}$, $F_{2}$, respectively, and asymptotes $l_{1}$, $l_{2}$. Point $P$ lies in the first quadrant and on $l_{1}$. If $l_{2} \\parallel P F_{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F2: Point;F1: Point;l1: Line;l2: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;Asymptote(G)={l1,l2};Quadrant(P) = 1;PointOnCurve(P, l1);IsParallel(l2,LineSegmentOf(P,F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 56], [148, 151]], [[3, 56]], [[3, 56]], [[104, 108]], [[73, 80]], [[65, 72]], [[87, 94], [116, 123]], [[96, 103]], [[3, 56]], [[3, 56]], [[0, 56]], [[0, 80]], [[0, 80]], [[0, 103]], [[104, 114]], [[104, 124]], [[126, 146]]]",
    "query_spans": "[[[148, 157]]]",
    "process": ""
  },
  {
    "text": "If the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{2}=1$ and $\\frac{x^{2}}{6}+\\frac{y^{2}}{3}=1$ have the same eccentricity, then $m$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/2 + x^2/m = 1);m: Number;G1: Ellipse;Expression(G1) = (x^2/6 + y^2/3 = 1);Eccentricity(G) = Eccentricity(G1)",
    "query_expressions": "m",
    "answer_expressions": "{1,4}",
    "fact_spans": "[[[1, 38]], [[1, 38]], [[83, 86]], [[39, 74]], [[39, 74]], [[1, 81]]]",
    "query_spans": "[[[83, 88]]]",
    "process": ""
  },
  {
    "text": "Given that the line $l$ intersects the parabola $C$: $y^{2}=4x$ at two points $A$ and $B$, and $|AB|=6$, then when the x-coordinate of the midpoint of $AB$ is minimized, what is the slope of the line $l$?",
    "fact_expressions": "l: Line;C: Parabola;Expression(C) = (y^2 = 4*x);A: Point;B: Point;Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, B)) = 6;WhenMin(XCoordinate(MidPoint(LineSegmentOf(A, B)))) = True",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*sqrt(2)",
    "fact_spans": "[[[2, 7], [68, 73]], [[8, 27]], [[8, 27]], [[30, 33]], [[34, 37]], [[2, 39]], [[41, 50]], [[52, 67]]]",
    "query_spans": "[[[68, 78]]]",
    "process": "Let the focus of the parabola be $ F(1,0) $. Connect $ AF $ and $ BF $, then $ |AF| + |BF| \\geqslant |AB| $, so $ (x_{A}+1) + (x_{B}+1) \\geqslant 6 $, then $ \\frac{x_{A}+x_{B}}{2} \\geqslant 2 $, that is, the minimum value of the horizontal coordinate of the midpoint of $ AB $ is 2, and the minimum value is attained if and only if $ AB $ passes through the focus $ F $. Let the line $ l: y = k(x-1) $, substitute into $ C: y^{2} = 4x $, we obtain $ k^{2}x^{2} - (2k^{2}+4)x + k^{2} = 0 $, so $ x_{A} + x_{B} = \\frac{2k^{2}+4}{k^{2}} = 4 $, which gives $ k = \\pm\\sqrt{2} $."
  },
  {
    "text": "Given that $F$ is the left focus of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and $A$, $B$ are two distinct moving points on the ellipse $C$, if the horizontal coordinate of the midpoint of $AB$ is $1$, then the minimum distance from $F$ to the line $AB$ is?",
    "fact_expressions": "C: Ellipse;B: Point;A: Point;F: Point;Expression(C) = (x^2/4 + y^2/3 = 1);LeftFocus(C) = F;PointOnCurve(A, C);PointOnCurve(B, C);Negation(A=B);XCoordinate(MidPoint(LineSegmentOf(A, B))) = 1",
    "query_expressions": "Min(Distance(F,LineOf(A,B)))",
    "answer_expressions": "sqrt(15)/2",
    "fact_spans": "[[[6, 48], [61, 66]], [[57, 60]], [[53, 56]], [[2, 5], [93, 96]], [[6, 48]], [[2, 52]], [[53, 74]], [[53, 74]], [[53, 74]], [[76, 91]]]",
    "query_spans": "[[[93, 112]]]",
    "process": ""
  },
  {
    "text": "A point $M$ on the parabola $y=2 x^{2}$ is at a distance of $1$ from the focus. What is the vertical coordinate of point $M$?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y = 2*x^2);PointOnCurve(M, G);Distance(M, Focus(G)) = 1",
    "query_expressions": "YCoordinate(M)",
    "answer_expressions": "7/8",
    "fact_spans": "[[[0, 14]], [[18, 21], [33, 37]], [[0, 14]], [[0, 21]], [[0, 31]]]",
    "query_spans": "[[[33, 43]]]",
    "process": "According to the problem, the standard equation of the parabola is $x^{2}=\\frac{1}{2}y$, and the equation of the directrix is $y=-\\frac{1}{8}$. Let point $M(-6,6)$. Based on the definition of a parabola, the distance from point $M$ to the focus is equal to the distance from point $M$ to the directrix. Therefore, $y_{0}+\\frac{1}{8}=1$, solving gives $y_{0}=\\frac{7}{8}$."
  },
  {
    "text": "The standard equation of a parabola with vertex at the origin, coordinate axes as symmetry axes, and passing through the point $P(-2, -4)$ is?",
    "fact_expressions": "G: Parabola;P: Point;Coordinate(P) = (-2, -4);PointOnCurve(P, G);O:Origin;Vertex(G)=O;SymmetryAxis(G)=axis",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=-8*x ,x^2=-y}",
    "fact_spans": "[[[31, 34]], [[18, 30]], [[18, 30]], [[17, 34]], [[1, 3]], [[0, 34]], [[7, 34]]]",
    "query_spans": "[[[31, 40]]]",
    "process": ""
  },
  {
    "text": "Given that $B$ is the intersection point of the left directrix of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ and the $x$-axis, and point $A(0 , b)$, if point $P$ satisfying $A P=2 A B$ lies on the hyperbola, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;B: Point;Intersection(LeftDirectrix(G), xAxis) = B;A: Point;Coordinate(A) = (0, b);P: Point;LineSegmentOf(A, P) = 2*LineSegmentOf(A, B);PointOnCurve(P, G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[6, 65], [110, 113], [117, 120]], [[6, 65]], [[9, 65]], [[9, 65]], [[9, 65]], [[9, 65]], [[2, 5]], [[2, 77]], [[78, 89]], [[78, 89]], [[105, 109]], [[93, 104]], [[105, 114]]]",
    "query_spans": "[[[117, 126]]]",
    "process": ""
  },
  {
    "text": "Let the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ be $F_{1}$ and $F_{2}$, respectively. The point $P(a , b)$ satisfies $|P F_{2}|=|F_{1} F_{2}|$. Then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P:Point;F2: Point;F1: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (a, b);LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(P, F2)) = Abs(LineSegmentOf(F1, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[1, 53], [118, 120]], [[3, 53]], [[3, 53]], [[78, 89]], [[70, 77]], [[62, 69]], [[3, 53]], [[3, 53]], [[1, 53]], [[78, 89]], [[1, 77]], [[1, 77]], [[91, 116]]]",
    "query_spans": "[[[118, 126]]]",
    "process": "Let $F_{1}(-c,0)$, $F_{2}(c,0)$ $(c>0)$, since $|PF_{2}|=|F_{1}F_{2}|$, then $\\sqrt{(a-c)^{2}+b^{2}}=2c$, simplifying yields $2c^{2}+ac-a^{2}=0$. Dividing both sides by $a^{2}$ gives $2(\\frac{c}{a})^{2}+\\frac{c}{a}-1=0$, so $\\frac{c}{a}=-1$ (discarded) or $\\frac{c}{a}=\\frac{1}{2}$. Therefore, the eccentricity of the ellipse is $e=\\frac{1}{2}$."
  },
  {
    "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{m}=1$ $(m>0)$ is $\\frac{5}{4}$. Then $m$ equals?",
    "fact_expressions": "G: Hyperbola;m: Number;m>0;Expression(G) = (x^2/16 - y^2/m = 1);Eccentricity(G) = 5/4",
    "query_expressions": "m",
    "answer_expressions": "9",
    "fact_spans": "[[[0, 44]], [[64, 67]], [[3, 44]], [[0, 44]], [[0, 62]]]",
    "query_spans": "[[[64, 70]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola equation $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, what is the equation of the parabola with the left vertex of the hyperbola as its vertex and the right focus as its focus?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (x^2/16 - y^2/9 = 1);LeftVertex(G) = Vertex(H);Focus(H) = RightFocus(G)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2 = -36*(x - 4)",
    "fact_spans": "[[[2, 5], [47, 50]], [[64, 67]], [[2, 44]], [[46, 67]], [[46, 67]]]",
    "query_spans": "[[[64, 71]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse equation $\\frac{x^{2}}{m+3}+\\frac{y^{2}}{m-6}=1(m>6)$, then its focal distance is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(m + 3) + y^2/(m - 6) = 1);m: Number;m > 6",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 4], [53, 54]], [[2, 51]], [[7, 51]], [[7, 51]]]",
    "query_spans": "[[[53, 58]]]",
    "process": "Find c from the ellipse equation to obtain the focal distance. \\because\\frac{x^{2}}{m+3}+\\frac{y^{2}}{m-6}=1(m>6)\\thereforec^{2}=m+3-(m-6)=9\\thereforec=3,2c=6."
  },
  {
    "text": "There is a point $P$ on the parabola $y^{2}=8x$, and its distance to the focus is $20$. Then the coordinates of point $P$ are?",
    "fact_expressions": "G: Parabola;P: Point;Expression(G) = (y^2=8*x);PointOnCurve(P,G);Distance(P,Focus(G))=20",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(18,pm*12)",
    "fact_spans": "[[[1, 15]], [[19, 22], [23, 24], [37, 41]], [[1, 15]], [[1, 22]], [[1, 35]]]",
    "query_spans": "[[[37, 46]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the focus of the parabola $y=3 x^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 3*x^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, 1/12)",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$, and let point $P$ lie on the hyperbola. If the midpoint of segment $P F_{1}$ lies on the $y$-axis, then the value of $\\frac{|P F_{1} |}{|P F_{2}|}$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/4 - y^2/3 = 1);F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C);PointOnCurve(MidPoint(LineSegmentOf(P, F1)), yAxis)",
    "query_expressions": "Abs(LineSegmentOf(P, F1))/Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "11/3",
    "fact_spans": "[[[17, 60], [71, 74]], [[17, 60]], [[1, 8]], [[9, 16]], [[1, 65]], [[66, 70]], [[66, 75]], [[77, 97]]]",
    "query_spans": "[[[99, 133]]]",
    "process": "According to the equation, find the coordinates of the foci. Since the midpoint of segment $ PF_{1} $ lies on the y-axis, solve for $ x_{p} = \\sqrt{7} $. Substituting into the hyperbola equation gives $ P(\\sqrt{7}, \\frac{3}{2}) $. Find $ |PF_{1}| $ and $ |PF_{2}| $ respectively to obtain the solution. From the hyperbola equation $ \\frac{x^{2}}{4} - \\frac{y^{2}}{3} = 1 $, we know $ a = 2 $, $ b = \\sqrt{3} $, $ c = \\sqrt{4 + 3} = \\sqrt{7} $. Let $ F_{1}(-\\sqrt{7}, 0) $, $ F_{2}(\\sqrt{7}, 0) $. Since the midpoint of segment $ PF_{1} $ lies on the y-axis, then $ \\frac{x_{p} + (-\\sqrt{7})}{2} = 0 $, solving gives $ x_{p} = \\sqrt{7} $. Since point $ P $ lies on the hyperbola and satisfies the hyperbola equation, $ \\frac{7}{4} - \\frac{y_{p}^{2}}{3} = 1 $, so $ y_{p} = \\pm\\frac{3}{2} $. Therefore, $ \\frac{|PF_{1}|}{|PF_{2}|} = \\frac{11}{3} $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively, if there exists a point $P$ on the left branch of the hyperbola such that $\\frac{|PF_{2}|^{2}}{|PF_{1}|}=8a$, then the range of the eccentricity of the hyperbola is?",
    "fact_expressions": "F1: Point;F2: Point;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, LeftPart(G)) = True;Abs(LineSegmentOf(P, F2))^2/Abs(LineSegmentOf(P, F1)) = 8*a",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1, 3]",
    "fact_spans": "[[[2, 9]], [[10, 17]], [[20, 81], [89, 92], [142, 145]], [[20, 81]], [[23, 81]], [[23, 81]], [[23, 81]], [[23, 81]], [[2, 87]], [[2, 87]], [[100, 103]], [[89, 103]], [[105, 140]]]",
    "query_spans": "[[[142, 156]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the right branch of hyperbola $C$, and $|O P|=|O F_{2}|$ ($O$ is the coordinate origin). If the line $P F_{2}$ intersects the left branch of $C$, then the range of the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;F2: Point;P: Point;O: Origin;F1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P,RightPart(C));Abs(LineSegmentOf(O, P)) = Abs(LineSegmentOf(O, F2));IsIntersect(LineOf(P,F2),LeftPart(C))",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(\\sqrt{5}, +\\infty)",
    "fact_spans": "[[[2, 63], [93, 99], [156, 159], [145, 148]], [[10, 63]], [[10, 63]], [[80, 87]], [[88, 92]], [[122, 125]], [[72, 79]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 87]], [[2, 87]], [[88, 103]], [[104, 121]], [[133, 154]]]",
    "query_spans": "[[[156, 170]]]",
    "process": "Let $ P $ be in the fourth quadrant, then $ 0 < k_{PF_{2}} < \\frac{b}{a} $. Let $ P(x, y) $. From $ |OP| = |OF_{2}| $ and $ P $ lying on the hyperbola, a system of equations can be constructed to solve for the coordinates of $ P $, thereby expressing $ k_{PF_{2}} $. Using $ 0 < k_{PF_{2}} < \\frac{b}{a} $, simplification yields $ \\frac{b}{a} > 2 $, and from $ e = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} $, the result can be obtained. [Detailed solution] From the hyperbola equation, its asymptotes are: $ y = \\pm\\frac{b}{a}x $. Without loss of generality, assume $ P $ is in the fourth quadrant; then if line $ PF_{2} $ intersects the left branch of $ C $, we have $ 0 < k_{PF_{2}} < \\frac{b}{a} $. Let $ P(x, y) $. From $ |OP| = |OF_{2}| $, we get $ x^{2} + y^{2} = c^{2} $, and $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $, $ \\frac{-\\frac{b^{2}}{c} - 0}{b^{2} + c^{2}} = \\frac{}{a\\sqrt{l}} $. Rearranging gives $ a^{2} < c^{2} - 2ab $, i.e., $ c^{2} - a^{2} = b^{2} > 2ab' \\therefore \\frac{b}{a} > 2 \\therefore e = \\sqrt{\\frac{c^{2}}{a^{2}}} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} > \\sqrt{5} $, so the range of eccentricity of $ C $ is $ (\\sqrt{5}, +\\infty) $."
  },
  {
    "text": "The distance from the vertex of the hyperbola $C$: $\\frac{x^{2}}{4}-y^{2}=1$ to its asymptote is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/4 - y^2 = 1)",
    "query_expressions": "Distance(Vertex(C), Asymptote(C))",
    "answer_expressions": "2*sqrt(5)/5",
    "fact_spans": "[[[0, 33], [37, 38]], [[0, 33]]]",
    "query_spans": "[[[0, 46]]]",
    "process": "According to the problem, the vertices of the hyperbola are $(\\pm2,0)$, and the asymptotes are $y=\\pm\\frac{1}{2}x$. By symmetry, without loss of generality, take the vertex $(2,0)$ and the asymptote $y=\\frac{1}{2}x$, so the distance from the vertex $(2,0)$ to the asymptote $y=\\frac{1}{2}x$ of the hyperbola $C: \\frac{x^{2}}{4}-y^{2}=1$ is $\\underline{2\\sqrt{5}}$."
  },
  {
    "text": "Given that the distance from the focus of the parabola $y^{2}=2 p x(p>0)$ to the origin is $5$, then the value of the real number $p$ is?",
    "fact_expressions": "G: Parabola;p: Real;O:Origin;p>0;Expression(G) = (y^2 = 2*(p*x));Distance(Focus(G),O)=5",
    "query_expressions": "p",
    "answer_expressions": "10",
    "fact_spans": "[[[2, 23]], [[38, 43]], [[27, 29]], [[5, 23]], [[2, 23]], [[2, 36]]]",
    "query_spans": "[[[38, 47]]]",
    "process": "The distance from the focus of the parabola $ y^{2} = 2px $ ($ p > 0 $) to the origin is 5, then $ \\frac{p}{2} = 5 $, solving gives $ p = 10 $."
  },
  {
    "text": "The equation of the circle centered at the right focus of the hyperbola $x^{2}-y^{2}=2$ and tangent to its asymptotes is?",
    "fact_expressions": "G: Hyperbola;E: Circle;Expression(G) = (x^2 - y^2 = 2);Center(E) = RightFocus(G);IsTangent(Asymptote(G), E) = True",
    "query_expressions": "Expression(E)",
    "answer_expressions": "(x-2)^2+y^2=2",
    "fact_spans": "[[[1, 19], [29, 30]], [[36, 37]], [[1, 19]], [[0, 37]], [[28, 37]]]",
    "query_spans": "[[[36, 42]]]",
    "process": ""
  },
  {
    "text": "Given an ellipse $ C $ with foci on the $ x $-axis: $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ $ (a > b > 0) $, the left and right foci are $ F_{1} $ and $ F_{2} $ respectively. The line $ l $ passes through $ F_{1} $ and intersects the ellipse $ C $ at points $ A $ and $ B $. It is known that $ \\overrightarrow{A F_{1}} = 3 \\overrightarrow{F_{1} B} $, and $ 3|\\overrightarrow{B F_{2}}| = 5|\\overrightarrow{A F_{2}}| $. Find the eccentricity of the ellipse $ C $.",
    "fact_expressions": "l: Line;C: Ellipse;b: Number;a: Number;A: Point;F1: Point;B: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(Focus(C),xAxis);LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(F1, l);Intersection(l,C)={A,B};VectorOf(A, F1) = 3*VectorOf(F1, B);3*Abs(VectorOf(B, F2)) = 5*Abs(VectorOf(A, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[93, 98]], [[11, 68], [109, 114], [241, 246]], [[18, 68]], [[18, 68]], [[116, 119]], [[77, 84], [99, 106]], [[120, 123]], [[85, 92]], [[18, 68]], [[18, 68]], [[11, 68]], [[2, 68]], [[11, 92]], [[11, 92]], [[93, 106]], [[93, 125]], [[126, 179]], [[182, 239]]]",
    "query_spans": "[[[241, 252]]]",
    "process": "Since |AF₁| + |AF₂| + |BF₁| + |BF₂| = 4a, and \\overrightarrow{AF} = 3\\overrightarrow{F₁B}, 3|\\overrightarrow{BF₂}| = 5|\\overrightarrow{AF₂}|, it follows that \\frac{4}{3}|AF₁| + \\frac{8}{3}|AF₂| = 4a, i.e., |AF₁| + 2|AF₂| = 3a. Since |AF₁| + |AF₂| = 2a, we have |AF₁| = |AF₂| = a. Thus, A lies on the minor axis of the ellipse. Without loss of generality, assume A is the upper endpoint. Let B(x₀, y₀), then A(0, b). Since F₁(-c, 0), we have \\overrightarrow{AF₁} = (-c, -b), \\overrightarrow{F₁B} = (x₀ + c, y₀). Therefore, -b = 3y₀ and x₀ + c = -3c, so B(-\\frac{4}{3}c, -\\frac{b}{3})."
  },
  {
    "text": "Given that the hyperbola $x^{2}-y^{2}=m$ and the ellipse $2 x^{2}+3 y^{2}=m+1$ have the same foci, then the real number $m$=?",
    "fact_expressions": "G: Hyperbola;m: Real;H: Ellipse;Expression(G) = (x^2 - y^2 = m);Expression(H) = (2*x^2 + 3*y^2 = m + 1);Focus(G) = Focus(H)",
    "query_expressions": "m",
    "answer_expressions": "1/11",
    "fact_spans": "[[[2, 20]], [[52, 57]], [[21, 44]], [[2, 20]], [[21, 44]], [[2, 50]]]",
    "query_spans": "[[[52, 59]]]",
    "process": "Since the foci are the same, the focal lengths are equal; establish an equation to solve. [Detailed solution] From $2x^{2}+3y^{2}=m+1$, we obtain $\\frac{x^{2}}{\\frac{m+1}{2}}+\\frac{y^{2}}{\\frac{m+1}{3}}=1$. From $x^{2}-y^{2}=m$, we obtain $\\frac{x^{2}}{m}-\\frac{y^{2}}{m}=1$. Thus, the foci lie on the $x$-axis, and $\\frac{m+1}{2}-\\frac{m+1}{3}=m+m$. Solving gives $m=\\frac{1}{11}$."
  },
  {
    "text": "Given that the equation $2$$(\\lambda+4) x^{2}+(\\lambda^{2}-3 \\lambda+2) y^{2}=1$ represents an ellipse, find the range of values for $\\lambda$.",
    "fact_expressions": "G: Ellipse;lambda: Number;Expression(G) = (x^2*(lambda + 4) + y^2*(lambda^2 - 3*lambda + 2) = 1)",
    "query_expressions": "Range(lambda)",
    "answer_expressions": "{(-4, 1), (2, +oo)&Negation(In(lambda, {-1, 6}))}",
    "fact_spans": "[[[62, 64]], [[66, 75]], [[2, 64]]]",
    "query_spans": "[[[66, 82]]]",
    "process": ""
  },
  {
    "text": "If the hyperbola $C$: $2 x^{2}-y^{2}=m(m>0)$ intersects the directrix of the parabola $y^{2}=16 x$ at points $A$ and $B$, and $|A B|=4 \\sqrt{3}$, then the value of $m$ is?",
    "fact_expressions": "C: Hyperbola;m: Number;G: Parabola;A: Point;B: Point;m>0;Expression(C) = (2*x^2 - y^2 = m);Expression(G) = (y^2 = 16*x);Intersection(C, Directrix(G)) = {A,B};Abs(LineSegmentOf(A, B)) = 4*sqrt(3)",
    "query_expressions": "m",
    "answer_expressions": "20",
    "fact_spans": "[[[1, 31]], [[83, 86]], [[32, 47]], [[52, 55]], [[57, 60]], [[9, 31]], [[1, 31]], [[32, 47]], [[1, 62]], [[64, 82]]]",
    "query_spans": "[[[83, 90]]]",
    "process": "The directrix of the parabola $ y^{2}=16x $ is $ l: x=-4 $. Since hyperbola $ C $ intersects the directrix $ l: x=-4 $ of the parabola $ y^{2}=16x $ at points $ A $ and $ B $, and $ |AB|=4\\sqrt{3} $, it follows that $ A(-4,2\\sqrt{3}) $, $ B(-4,-2\\sqrt{3}) $. Substituting the coordinates of point $ A $ into the hyperbola equation gives $ 2(-4)^{2}-(2\\sqrt{3})^{2}=m $, so $ m=20 $."
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $4 x^{2}-y^{2}=16$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (4*x^2 - y^2 = 16)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*2*x",
    "fact_spans": "[[[0, 21]], [[0, 21]]]",
    "query_spans": "[[[0, 29]]]",
    "process": "Letting the right-hand side of the hyperbola be 0, we obtain $4x^{2}-y^{2}=0$, which simplifies to the asymptotes of the hyperbola $y=\\underline{+}2x$."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $x^{2}=4 y$, $P$ is a moving point on the parabola, and the coordinates of $A$ are $(\\frac{3}{2},-1)$, then the minimum value of $\\frac{|P F|}{|P A|}$ is?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;A: Point;Expression(G) = (x^2 = 4*y);Coordinate(A) = (3/2, -1);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F))/Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "2*sqrt(5)/5",
    "fact_spans": "[[[6, 20], [28, 31]], [[24, 27]], [[2, 5]], [[37, 40]], [[6, 20]], [[37, 62]], [[2, 23]], [[24, 35]]]",
    "query_spans": "[[[64, 91]]]",
    "process": "Through point P, draw PM perpendicular to the directrix, with M as the foot of the perpendicular. Then by the definition of the parabola, |PF| = |PM|, so \\frac{|PF|}{|PA|} = \\frac{|PM|}{|PA|} = \\sin\\angle PAM. Therefore, when PA is tangent to the parabola, \\frac{|PF|}{|PA|} is minimized. Then, using the slope formula and the geometric meaning of derivatives, determine the coordinates of the tangent point P to solve the problem. [Detailed solution] The parabola x^{2} = 4y has focus F(0,1) and directrix equation y = -1. Draw PM perpendicular to the directrix through point P, with M as the foot of the perpendicular. Then by the definition of the parabola, |PF| = |PM|, so \\frac{|PF|}{|PA|} = \\frac{|PM|}{|PA|} = \\sin\\angle PAM, where \\angle PAM is an acute angle. Hence, when \\angle PAM is minimized, \\frac{|PF|}{|PA|} is minimized; thus, when PA is tangent to the parabola, \\frac{|PF|}{|PA|} is minimized. Let the tangent point be P(2\\sqrt{a}, a). Since the derivative of y = \\frac{1}{4}x^{2} is y' = \\frac{1}{2}x, the slope of PA is \\frac{1}{2} \\cdot 2\\sqrt{a} = \\sqrt{a} = \\frac{a+1}{2\\sqrt{a}-\\frac{3}{3}}. Solving gives a = 4, yielding P(4,4). Thus |PM| = 5, |PA| = \\frac{5\\sqrt{5}}{2}, \\frac{|PF|}{|PA|} = \\frac{|PM|}{|PA|}, so the minimum value of \\frac{|PF|}{|DA|} is \\frac{2\\sqrt{5}}{5}. (Insight) This problem mainly examines the definition of the parabola, the slope of a line, and the geometric meaning of derivatives, and is a medium-difficulty problem."
  },
  {
    "text": "The moving point $M$ lies on the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$. A perpendicular is drawn from $M$ to the $x$-axis, with foot $N$. The point $P$ satisfies $\\overrightarrow{N P}=\\sqrt{2} \\overrightarrow{N M}$. What is the trajectory equation of point $P$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/2 + y^2 = 1);M: Point;PointOnCurve(M, C);l: Line;PointOnCurve(M, l);IsPerpendicular(l,xAxis);N: Point;FootPoint(l,xAxis) = N;P: Point;VectorOf(N, P) = sqrt(2)*VectorOf(N, M)",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2 + y^2 = 2",
    "fact_spans": "[[[6, 38]], [[6, 38]], [[2, 5], [41, 44]], [[2, 39]], [], [[40, 52]], [[40, 52]], [[56, 59]], [[40, 59]], [[60, 64], [122, 126]], [[66, 119]]]",
    "query_spans": "[[[122, 132]]]",
    "process": "Let $ M(x_{0},y_{0}) $, $ N(x_{0},0) $, $ P(x,y) $. Then $ \\overrightarrow{NP}=(x-x_{0},y) $, $ \\overrightarrow{NM}=(0,y_{0}) $. Since $ \\overrightarrow{NP}=\\sqrt{2}\\overrightarrow{NM} $, therefore $ (x-x_{0},y)=\\sqrt{2}(0,y_{0}) $, so $ \\begin{cases} x-x_{0}=0 \\\\ y=\\sqrt{2}y_{0} \\end{cases} $, i.e., $ \\therefore \\begin{cases} x_{0}=x \\\\ y_{0}=\\frac{\\sqrt{2}}{2}y \\end{cases} $. Substituting into $ \\frac{x^{2}}{2}+y^{2}=1 $ gives $ \\frac{x^{2}}{2}+\\frac{y^{2}}{2}=1 $, i.e., $ x^{2}+y^{2}=2 $."
  },
  {
    "text": "Given that $A$ and $B$ are two points on the parabola $E$: $y^{2}=4x$, find the standard equation of the circle passing through the origin $O$ with $AB$ as diameter and having the minimum area.",
    "fact_expressions": "E: Parabola;G: Circle;A: Point;B: Point;O: Origin;Expression(E) = (y^2 = 4*x);PointOnCurve(A, E);PointOnCurve(B, E);PointOnCurve(O, G);IsDiameter(LineSegmentOf(A, B), G);WhenMin(Area(G))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "(x - 4)^2 + y^2 = 16",
    "fact_spans": "[[[10, 29]], [[57, 58]], [[2, 5]], [[6, 9]], [[35, 42]], [[10, 29]], [[2, 32]], [[2, 32]], [[33, 58]], [[42, 58]], [[52, 58]]]",
    "query_spans": "[[[57, 65]]]",
    "process": "From the given conditions, it follows that the slope of line AB is certainly not zero; thus, let the line $ l_{AB}: x = my + n $. Combining this with the parabola equation and eliminating $ x $ yields $ y^2 - 4my - 4n = 0 $. Let points $ A(x_1, y_1) $, $ B(x_2, y_2) $, then $ y_1 + y_2 = 4m $, $ y_1 y_2 = -4n $. Since point $ O $ lies on the circle with $ AB $ as diameter, $ \\therefore \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = x_1 x_2 + y_1 y_2 = 0 $, solving gives $ n = 0 $ (discarded) or $ n = 4 $, i.e., line $ AB $ always passes through the fixed point $ (4, 0) $. Using geometric properties, to minimize the area of the circle, the diameter $ AB $ must be shortest. $ \\therefore |AB| = \\sqrt{1 + m^2} \\cdot \\sqrt{(y_1 + y_2)^2 - 4y_1 y_2} = 4\\sqrt{m^4 + 5m^2 + 4} > 8 $ (achieved when $ m = 0 $, i.e., line $ AB $ is perpendicular to the $ x $-axis). $ \\therefore $ The standard equation of the circle is $ (x - 4)^2 + y^2 = 16 $."
  },
  {
    "text": "A line $l$ passing through the point $P(3 \\sqrt{3}, 4)$ intersects the parabola $C$: $x^{2}=18 y$ at points $A$ and $B$. If the midpoint of chord $AB$ is exactly $P$, then what is the inclination angle of line $l$?",
    "fact_expressions": "P: Point;Coordinate(P) = (3*sqrt(3), 4);l: Line;PointOnCurve(P, l);C: Parabola;Expression(C) = (x^2 = 18*y);A: Point;B: Point;Intersection(l, C) = {A, B};IsChordOf(LineSegmentOf(A, B), C);MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Inclination(l)",
    "answer_expressions": "pi/6",
    "fact_spans": "[[[1, 20], [71, 74]], [[1, 20]], [[21, 26], [76, 81]], [[0, 26]], [[27, 47]], [[27, 47]], [[48, 51]], [[52, 55]], [[21, 57]], [[27, 65]], [[60, 74]]]",
    "query_spans": "[[[76, 87]]]",
    "process": "Obviously, the slope of the line exists. Let the slope of the line be k, intersecting the parabola at points A(x_{1},y_{1}) and B(x_{2},y_{2}). Substituting into the parabola equation and taking the difference yields: x_{1}^{2}-x_{2}^{2}=18(y_{1}-y_{2}), rearranging gives: \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{x_{1}+x_{2}}{18}, so k=\\frac{\\sqrt{3}}{3}, hence the inclination angle is \\frac{\\pi}{6}."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ has its right focus at $F_{2}$, and point $M$ lies on the circle $O$: $x^{2}+y^{2}=3$, with $M$ in the first quadrant. The tangent to circle $O$ at point $M$ intersects the ellipse at points $P$ and $Q$. Then, the perimeter of $\\Delta P F_{2} Q$ is?",
    "fact_expressions": "G: Ellipse;O: Circle;P: Point;F2: Point;Q: Point;M: Point;Expression(G) = (x^2/4 + y^2/3 = 1);RightFocus(G) = F2;Expression(O) = (x^2 + y^2 = 3);PointOnCurve(M, O) = True;Quadrant(M) = 1;Intersection(TangentOfPoint(M,O), G) = {P, Q}",
    "query_expressions": "Perimeter(TriangleOf(P, F2, Q))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 39], [104, 106]], [[57, 78], [96, 100]], [[107, 110]], [[44, 51]], [[111, 114]], [[52, 56], [81, 84], [91, 95]], [[2, 39]], [[2, 51]], [[57, 78]], [[52, 79]], [[81, 89]], [[90, 116]]]",
    "query_spans": "[[[118, 141]]]",
    "process": "Let $ P(x_{1},y_{1}) $, $ Q(x_{2},y_{2}) $. Using the focal radius formula, $ |PF_{2}| $ and $ |QF_{2}| $ can be found. Then, using the Pythagorean theorem, $ |PM| $ and $ |QM| $ can be determined. Note that by simplifying $ |PM| $ and $ |QM| $ using the condition that the points lie on the ellipse, the perimeter of $ PF_{2}Q $ can be obtained. The radius of circle $ O $ is $ \\sqrt{3} $, and the semi-minor axis of the ellipse is $ \\sqrt{3} $. Since $ M $ lies in the first quadrant, $ P $ and $ Q $ are on the right side of the y-axis. Let $ P(x_{1},y_{1}) $, $ Q(x_{2},y_{2}) $, then $ |PF_{2}| = 2 - \\frac{1}{2}x_{1} $, $ |QF_{2}| = 2 - \\frac{1}{2}x_{2} $, and $ x_{1} > 0 $, $ x_{2} > 0 $. Also, $ |PM| = \\sqrt{x_{1}^{2} + y_{1}^{2} - 3} = \\sqrt{x_{1}^{2} + 3(1 - \\frac{x^{2}}{4}) - 3} = \\frac{1}{2}x_{1}' $. Similarly, $ |QM| = \\frac{1}{2}x_{2}' $. Therefore, the perimeter of $ PF_{2}Q $ is $ 2 - \\frac{1}{2}x_{1} + 2 - \\frac{1}{2}x_{2} + \\frac{1}{2}x_{1} + \\frac{1}{2}x_{2} = 4 $."
  },
  {
    "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>b>0)$ have its right focus at $F$, and left and right vertices at $A_{1}$, $A_{2}$ respectively. A line passing through $F$ and parallel to one asymptote of the hyperbola $C$ intersects the other asymptote at point $P$. If $P$ lies exactly on the circle with diameter $A_{1} A_{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/a^2 - y^2/b^2 = 1);a: Number;b: Number;a>b;b>0;F: Point;RightFocus(C) = F;A1: Point;A2: Point;LeftVertex(C) = A1;RightVertex(C) = A2;L1: Line;L2: Line;OneOf(Asymptote(C)) = L1;OneOf(Asymptote(C)) = L2;Negation(L1 = L2);H: Line;PointOnCurve(F,H);IsParallel(L1,H);P: Point;Intersection(H,L2) = P;G: Circle;IsDiameter(LineSegmentOf(A1,A2),G);PointOnCurve(P,G)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 61], [101, 107], [162, 165]], [[1, 61]], [[9, 61]], [[9, 61]], [[9, 61]], [[9, 61]], [[66, 69], [96, 99]], [[1, 69]], [[77, 84]], [[87, 94]], [[1, 94]], [[1, 94]], [], [], [[101, 113]], [[101, 125]], [[101, 125]], [[116, 118]], [[95, 118]], [[100, 118]], [[128, 131], [133, 136]], [[101, 131]], [[157, 158]], [[139, 158]], [[133, 160]]]",
    "query_spans": "[[[162, 171]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the right focus $F$ of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ are? If the focus of a parabola $C$ with vertex at the origin is also $F$, then its standard equation is?",
    "fact_expressions": "C: Parabola;G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);RightFocus(G)=F;Vertex(C)=O;O:Origin;Focus(C)=F;F:Point",
    "query_expressions": "Coordinate(F);Expression(C)",
    "answer_expressions": "(3,0)\ny^2=12*x",
    "fact_spans": "[[[58, 64], [74, 75]], [[0, 39]], [[0, 39]], [[0, 46]], [[52, 64]], [[55, 57]], [[58, 72]], [[43, 46], [69, 72]]]",
    "query_spans": "[[[43, 51]], [[74, 81]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=8x$, $M$ is a point on $C$, and the extension of $FM$ intersects the $y$-axis at point $N$. If $M$ is the midpoint of $FN$, then $|FN|=$?",
    "fact_expressions": "C: Parabola;F: Point;M: Point;N: Point;Expression(C) = (y^2 = 8*x);Focus(C) = F;PointOnCurve(M, C);Intersection(OverlappingLine(LineSegmentOf(F, M)), yAxis) = N;MidPoint(LineSegmentOf(F, N)) = M",
    "query_expressions": "Abs(LineSegmentOf(F, N))",
    "answer_expressions": "6",
    "fact_spans": "[[[6, 25], [33, 36]], [[2, 5]], [[29, 32], [62, 65]], [[55, 59]], [[6, 25]], [[2, 28]], [[29, 39]], [[40, 59]], [[62, 74]]]",
    "query_spans": "[[[76, 86]]]",
    "process": ""
  },
  {
    "text": "What is the focal distance of the ellipse $3 x^{2}+4 y^{2}=12$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (3*x^2 + 4*y^2 = 12)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 22]], [[0, 22]]]",
    "query_spans": "[[[0, 27]]]",
    "process": "According to the given condition, the standard equation of the ellipse is $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, from which its focal distance can be obtained as $2c=2\\sqrt{4-1}=2$."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $C$: $x^{2}=2 p y$ ($p>0$) intersects the parabola at points $A$ and $B$. If $3|A F|=|B F|$, and $O$ is the origin, then $\\frac{|O F|}{|A F|}=$?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;A: Point;F: Point;B: Point;O: Origin;p>0;Expression(C) = (x^2 = 2*p*y);Focus(C) = F;PointOnCurve(F, G);Intersection(G, C) = {A, B};3*Abs(LineSegmentOf(A, F)) = Abs(LineSegmentOf(B, F))",
    "query_expressions": "Abs(LineSegmentOf(O, F))/Abs(LineSegmentOf(A, F))",
    "answer_expressions": "3/4",
    "fact_spans": "[[[1, 27], [38, 41]], [[9, 27]], [[34, 36]], [[42, 45]], [[30, 33]], [[46, 49]], [[69, 73]], [[9, 27]], [[1, 27]], [[1, 33]], [[0, 36]], [[34, 51]], [[53, 67]]]",
    "query_spans": "[[[80, 103]]]",
    "process": "Draw AE\\bot directrix from A, draw BG\\bot directrix from B, draw AD\\bot BG intersecting BG at D and the y-axis at C. Let |AF|=x, then |BF|=3x, F(0,\\frac{p}{2}), directrix: y=-\\frac{p}{2}. According to the property of parabola: |AE|=|AF|=x, |BG|=|BF|=3x, |AB|=x+3x=4x, |BD|=3x-x=2x, |FC|=p-x. From the figure we have: \\frac{AF}{AB}=\\frac{FC}{BD}, i.e., \\frac{x}{4x}=\\frac{p-x}{2x}, solving gives x=\\frac{2}{3}p, then \\frac{|OF|}{|AF|}=\\frac{\\frac{p}{2}}{\\frac{2}{3}p}=\\frac{3}{4}"
  },
  {
    "text": "The line $ l $ with slope $ 1 $ intersects the ellipse $ \\frac{x^{2}}{4} + y^{2} = 1 $ at points $ A $ and $ B $. Then the maximum value of $ |AB| $ is?",
    "fact_expressions": "l: Line;Slope(l) = 1;G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);Intersection(l, G) = {A, B};A: Point;B: Point",
    "query_expressions": "Max(Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "4*sqrt(10)/5",
    "fact_spans": "[[[7, 12]], [[0, 12]], [[13, 40]], [[13, 40]], [[7, 52]], [[43, 46]], [[47, 50]]]",
    "query_spans": "[[[54, 67]]]",
    "process": "Let the equation of the line be y = x + b. Substituting into the ellipse equation and simplifying yields 5x^{2} + 8bx + 4b^{2} - 4 = 0 - \\frac{8b}{5}, x_{1} \\cdot x_{2} = \\frac{4b^{2} - 4}{5}, 4 = 64b^{2} - 20(4b^{2} - 4) = -16b^{2} + 1 + 1 \\cdot \\sqrt{\\frac{64b^{2}}{25} - \\frac{16b^{2} - 16}{5}} = \\sqrt{2} \\cdot \\sqrt{\\frac{-16b^{2} + 80}{25}}, when b = 0, b^{2} - 20(4b^{2} - 4) = -16b^{2} + 80 > 0, -\\sqrt{5} < b < \\sqrt{5}, |AB|_{\\max} = \\sqrt{2} \\cdot \\sqrt{\\frac{80}{25}} = \\frac{4}{}\\sqrt{10}. This question mainly examines the relationship between a line and an ellipse, and the method for finding the maximum chord length formed by the intersection of a line and an ellipse, belonging to a medium-difficulty problem."
  },
  {
    "text": "Given a moving point $P(x, y)$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$. If $A(3,0)$, and point $M$ satisfies $|A M|=1$ and $\\overrightarrow{P M} \\cdot \\overrightarrow{A M}=0$, then the minimum value of $|P M|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;A: Point;M: Point;x1:Number;y1:Number;Expression(G) = (x^2/25 + y^2/16 = 1);Coordinate(P) = (x1, y1);Coordinate(A) = (3, 0);PointOnCurve(P, G);Abs(LineSegmentOf(A, M)) = 1;DotProduct(VectorOf(P, M), VectorOf(A, M)) = 0",
    "query_expressions": "Min(Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[14, 53]], [[4, 13]], [[56, 64]], [[65, 69]], [[4, 13]], [[4, 13]], [[14, 53]], [[4, 13]], [[56, 64]], [[4, 54]], [[71, 80]], [[82, 133]]]",
    "query_spans": "[[[135, 148]]]",
    "process": "\\because\\overrightarrow{PM}\\cdot\\overrightarrow{AM}=0,\\therefore\\overrightarrow{PM}\\bot\\overrightarrow{AM},\\therefore|\\overrightarrow{PM}|^{2}=|\\overrightarrow{AP}|^{2}-|\\overrightarrow{AM}|^{2}=|\\overrightarrow{AP}|^{2}-1,\\therefore the trajectory of point M is a circle centered at point A with radius 1,\\because|\\overrightarrow{PM}|^{2}=|\\overrightarrow{AP}|^{2}-1, the smaller |\\overrightarrow{AP}| is, the smaller |\\overrightarrow{PM}| is. Combined with the graph, when point P is the right vertex of the ellipse, |\\overrightarrow{AP}| takes the minimum value a-c=5-3=2,\\therefore the minimum value of |\\overrightarrow{PM}| is \\sqrt{4-1}=\\sqrt{3}. So choose B."
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ are $F_{1}$ and $F_{2}$, respectively. A line passing through $F_{2}$ with an inclination angle of $120^{\\circ}$ intersects the ellipse at a point $M$. If $M F_{1}$ is perpendicular to $M F_{2}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;Z: Line;PointOnCurve(F2,Z);Inclination(Z) = ApplyUnit(120,degree);OneOf(Intersection(Z,G)) = M;M: Point;IsPerpendicular(LineSegmentOf(M,F1),LineSegmentOf(M,F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3) - 1",
    "fact_spans": "[[[0, 53], [108, 110], [144, 146]], [[0, 53]], [[2, 53]], [[2, 53]], [[2, 53]], [[2, 53]], [[62, 69]], [[70, 77], [79, 86]], [[0, 77]], [[0, 77]], [[105, 107]], [[78, 107]], [[87, 107]], [[105, 119]], [[116, 119]], [[121, 142]]]",
    "query_spans": "[[[144, 152]]]",
    "process": ""
  },
  {
    "text": "A ray with endpoint at the focus $F$ of the parabola $y^{2}=2 p x(p>0)$ intersects the parabola at point $A$ and the $y$-axis at point $B$. If $|A F|=2$, $|B F|=3$, then $p=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;Endpoint(Z) = F;A: Point;Intersection(Z, G) = A;B: Point;Intersection(Z, yAxis) = B;Abs(LineSegmentOf(A, F)) = 2;Abs(LineSegmentOf(B, F)) = 3;Z: Ray",
    "query_expressions": "p",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 22], [36, 39]], [[1, 22]], [[79, 82]], [[4, 22]], [[24, 27]], [[1, 27]], [[0, 35]], [[40, 44]], [[33, 44]], [[51, 55]], [[33, 55]], [[57, 66]], [[68, 77]], [[33, 35]]]",
    "query_spans": "[[[79, 84]]]",
    "process": "According to the problem, we have $ F\\left(\\frac{p}{2},0\\right) $. Let $ A(x_{1},y_{1}) $, then $ |AF| = x_{1} + \\frac{p}{2} = 2 $. Since $ |AF| = 2 $, $ |BF| = 3 $, it follows that $ |AB| = |BF| - |AF| = 3 - 2 = 1 $, so $ \\frac{|AB|}{|BF|} = \\frac{1}{3} $. Also, $ \\frac{|AB|}{|BF|} = \\frac{x_{1}}{\\frac{P}{2}} = \\frac{2x_{1}}{P} $, hence $ \\frac{2x_{1}}{p} = \\frac{1}{3} $, so $ p = 6x_{1} $. Therefore, $ x_{1} + \\frac{6x_{1}}{2} = 2 $, solving gives $ x_{1} = \\frac{1}{2} $, so $ p = 6x_{1} = 6 \\times \\frac{1}{2} = 3 $."
  },
  {
    "text": "Given the parabola $C$: $x^{2}=8y$ with focus $F$, a line $l$ passing through point $P(0,-2)$ intersects the parabola at points $M$ and $N$, and $|MF|+|NF|=32$. If $Q$ is a moving point on line $l$ and $B(0,3)$, then the minimum value of $|QF|+|QB|$ is?",
    "fact_expressions": "l: Line;C: Parabola;P: Point;B: Point;M: Point;F: Point;N: Point;Q: Point;Expression(C) = (x^2 = 8*y);Coordinate(P) = (0, -2);Coordinate(B) = (0, 3);Focus(C) = F;PointOnCurve(P, l);Intersection(l, C) = {M, N};Abs(LineSegmentOf(M, F)) + Abs(LineSegmentOf(N, F)) = 32;PointOnCurve(Q, l)",
    "query_expressions": "Min(Abs(LineSegmentOf(Q, B)) + Abs(LineSegmentOf(Q, F)))",
    "answer_expressions": "sqrt(17)",
    "fact_spans": "[[[40, 45], [85, 90]], [[2, 20], [46, 49]], [[29, 39]], [[97, 105]], [[52, 55]], [[24, 27]], [[56, 59]], [[81, 84]], [[2, 20]], [[29, 39]], [[97, 105]], [[2, 27]], [[28, 45]], [[40, 61]], [[63, 79]], [[81, 96]]]",
    "query_spans": "[[[107, 126]]]",
    "process": "Since the line $ l $ passes through point $ P(0,-2) $, let the equation of line $ l $ be $ y = kx - 2 $. Let $ M(x_{1},y_{1}) $, $ N(x_{2},y_{2}) $. Solving the system of equations \n\\[\n\\begin{cases}\nx = \\frac{y+2}{k}, \\\\\nx^{2} = 8y\n\\end{cases}\n\\]\nyields $ y^{2} + (4 - 8k^{2})y + 4 = 0 $, then $ y_{1} + y_{2} = 8k^{2} - 4 $. According to the definition of a parabola, $ |MF| + |NF| = y_{1} + y_{2} + 4 = 32 $. Solving gives $ k = \\pm 2 $. Take $ k = 2 $ (the result is the same when $ k = -2 $), then the equation of line $ l $ is $ y = 2x - 2 $. Let point $ B(0,3) $ have symmetric point $ B'(x_{0},y_{0}) $ about line $ l $. Based on perpendicular bisector properties, we can set up the system of equations \n\\[\n\\begin{cases}\n\\frac{y_{0}+3}{2} = x_{0} - 2 \\\\\n\\frac{y_{0}-3}{x_{0}} = -\\frac{1}{2}\n\\end{cases}\n\\Rightarrow\n\\begin{cases}\nx_{0} = 4 \\\\\ny_{0} = 1\n\\end{cases}\n\\]\nThus, $ B'(4,1) $. The intersection point of segment $ FB' $ and line $ l $ is the point where $ |QF| + |QB| $ reaches its minimum value. Since $ F(0,2) $, the minimum distance is $ |FB'| = \\sqrt{(4-0)^{2} + (1-2)^{2}} = \\sqrt{17} $."
  },
  {
    "text": "Point $P$ is a point on the parabola $y^{2}=4x$, and the distance from $P$ to the focus of this parabola is $4$. Then the horizontal coordinate of point $P$ is?",
    "fact_expressions": "G: Parabola;P: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(P, G);Distance(P, Focus(G)) = 4",
    "query_expressions": "XCoordinate(P)",
    "answer_expressions": "3",
    "fact_spans": "[[[5, 19], [28, 31]], [[0, 4], [23, 26], [42, 46]], [[5, 19]], [[0, 22]], [[23, 40]]]",
    "query_spans": "[[[42, 52]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $x^{2}-\\frac{y^{2}}{a^{2}}=1$ $(a>0)$ with two foci $F_{1}(-2,0)$, $F_{2}(2,0)$, and a point $P$ on the hyperbola such that $\\angle F_{1} P F_{2}=60^{\\circ}$, then $|P F_{1}| \\cdot|P F_{2}|$ equals?",
    "fact_expressions": "G: Hyperbola;a: Number;F1: Point;F2: Point;P: Point;a>0;Expression(G) = (x^2 - y^2/a^2 = 1);Coordinate(F1) = (-2, 0);Coordinate(F2) = (2, 0);Focus(G) = {F1,F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "12",
    "fact_spans": "[[[2, 39], [80, 83]], [[5, 39]], [[47, 60]], [[62, 74]], [[76, 79]], [[5, 39]], [[2, 39]], [[47, 60]], [[62, 74]], [[2, 74]], [[76, 86]], [[88, 121]]]",
    "query_spans": "[[[123, 152]]]",
    "process": "From the coordinates of the foci of the hyperbola, find $ c $, then derive the equation of the hyperbola. Using the definition of the hyperbola, $ ||PF_{1}|-|PF_{2}||=2 $, combined with the law of cosines, the result can be obtained. Since the two foci of the hyperbola $ x^{2}-\\frac{y^{2}}{a^{2}}=1 $ ($ a>0 $) are $ F_{1}(-2,0) $, $ F_{2}(2,0) $, respectively, then $ c=2 $, $ a=1 $. From the definition of the hyperbola, we get $ ||PF_{1}|-|PF_{2}||=2 $, $ 4=|PF_{1}|^{2}-2|PF_{1}||PF_{2}|+|PF_{2}|^{2} $. By the law of cosines, $ 16=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}||PF_{2}|\\cos60^{\\circ}=|PF_{1}|^{2}+|PF_{2}|^{2}-|PF_{1}||PF_{2}| $, so $ |PF_{1}|\\cdot|PF_{2}|=12 $."
  },
  {
    "text": "Given the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the right vertex is $A(2,0)$. If the distance from $A$ to one of the asymptotes of $E$ is $1$, then the eccentricity of $E$ is?",
    "fact_expressions": "E: Hyperbola;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;A: Point;Coordinate(A) = (2, 0);RightVertex(E) = A;Distance(A,OneOf(Asymptote(E))) = 1",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 63], [83, 86], [101, 104]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 76], [79, 82]], [[68, 76]], [[2, 76]], [[79, 99]]]",
    "query_spans": "[[[101, 110]]]",
    "process": "From the given conditions, the hyperbola $ E: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) has asymptotes $ y = \\pm \\frac{b}{a}x $. Since the right vertex of hyperbola $ E $ is $ A(2,0) $, we have $ a = 2 $. Also, the distance from $ A $ to one asymptote of $ E $ is 1, so $ \\frac{|b|}{\\sqrt{1 + \\frac{b^{2}}{4}}} = 1 $, thus $ b = \\frac{2\\sqrt{3}}{3} $, and therefore $ \\frac{b}{a} = \\frac{\\sqrt{3}}{3} $. Hence, the eccentricity of $ E $ is $ \\sqrt{1 + \\frac{b^{2}}{2}} = \\frac{\\sqrt[2]{3}}{3} $."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, a line passing through point $F$ intersects the parabola at points $A$ and $B$, and $|FA| \\cdot |FB|=6$, then $|AB|=$?",
    "fact_expressions": "G: Parabola;H: Line;F: Point;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A,B};Abs(LineSegmentOf(F, A))*Abs(LineSegmentOf(F, B)) = 6",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 16], [33, 36]], [[30, 32]], [[20, 23], [25, 29]], [[37, 40]], [[42, 45]], [[2, 16]], [[2, 23]], [[24, 32]], [[30, 47]], [[49, 69]]]",
    "query_spans": "[[[71, 80]]]",
    "process": "From the parabola $ y^{2} = 4x $, we get $ F(1,0) $. When the line $ AB $ is perpendicular to the $ x $-axis, $ |FA| = |FB| = 2 $, which does not satisfy the condition. Thus, we can set the line $ AB: y = k(x - 1) $. Combining with the parabola gives $ k^{2}x^{2} - 2(k^{2} + 2)x + k^{2} = 0 $, so $ x_{1}x_{2} = 1 $. From the focal radius of the parabola, we have $ |FA| = x_{1} + 1 $, $ |FB| = x_{2} + 1 $. Therefore, $ |FA| \\cdot |FB| = (x_{1} + 1)(x_{2} + 1) = x_{1}x_{2} + x_{1} + x_{2} + 1 = 2 + x_{1} + x_{2} = 6 $, so $ x_{1} + x_{2} = 4 $, and $ |AB| = |FA| + |FB| = x_{1} + x_{2} + 2 = 6 $."
  },
  {
    "text": "The line $l$ passes through the focus $F$ of the parabola $y^{2}=8x$ and intersects the parabola at points $A$ and $B$. If the distance from the midpoint of segment $AB$ to the $y$-axis is $2$, then $|AB|=$?",
    "fact_expressions": "A: Point;B: Point;l: Line;G: Parabola;F: Point;Expression(G) = (y^2 = 8*x);Focus(G) = F;PointOnCurve(F, l);Intersection(l,G)={A,B};Distance(MidPoint(LineSegmentOf(A,B)), yAxis) = 2",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[34, 37]], [[38, 41]], [[0, 5]], [[6, 20], [29, 32]], [[23, 26]], [[6, 20]], [[6, 26]], [[0, 26]], [[0, 43]], [[45, 66]]]",
    "query_spans": "[[[68, 76]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$. If there exists a point $P$ on the right branch of the hyperbola such that $(\\overrightarrow{O P}+\\overrightarrow{O F_{2}}) \\cdot \\overrightarrow{P F_{2}}=0$ ($O$ being the origin), and $3|P F_{1}|=4|P F_{2}|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;O: Origin;P: Point;F2: Point;F1: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));DotProduct((VectorOf(O,P)+VectorOf(O,F2)),VectorOf(P,F2))=0;3*Abs(LineSegmentOf(P, F1)) = 4*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5",
    "fact_spans": "[[[17, 73], [81, 84], [217, 220]], [[20, 73]], [[20, 73]], [[180, 184]], [[91, 94]], [[9, 16]], [[1, 8]], [[20, 73]], [[20, 73]], [[17, 73]], [[1, 79]], [[1, 79]], [[81, 94]], [[96, 179]], [[192, 215]]]",
    "query_spans": "[[[217, 226]]]",
    "process": "Let M be the midpoint of PF₂, then from (→OP + →OF₂) · →PF₂ = 0, we get 2→OM · →PF₂ = 0, i.e., →PF₁ · →PF₂ = 0, so PF₁ ⊥ PF₂. Given 3|→PF₁| = 4|→PF₂|, let |PF₁| = 4x, |PF₂| = 3x, |F₁F₂| = 2c, then (4x)² + (3x)² = 4c², so x = (2/5)c. By the definition of ellipse, 2a = |PF₁| + |PF₂| = 7x = (14/5)c, a = (7/5)c, then the eccentricity of the ellipse e = c/a = 5/7"
  },
  {
    "text": "The hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has left and right foci $F_{1}$ and $F_{2}$, respectively. A line passing through $F_{2}$ intersects the right branch of curve $C$ at points $P$ and $Q$, such that $P Q \\perp P F_{1}$ and $3|P Q|=4|P F_{1}|$. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;LeftFocus(C) = F1;RightFocus(C) = F2;F1: Point;F2: Point;PointOnCurve(F2,L) = True;L: Line;Intersection(L,RightPart(C)) = {P,Q};P: Point;Q: Point;IsPerpendicular(LineSegmentOf(P, Q), LineSegmentOf(P, F1)) = True;3*Abs(LineSegmentOf(P, Q)) = 4*Abs(LineSegmentOf(P, F1))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(10)/2",
    "fact_spans": "[[[0, 61], [98, 103], [159, 162]], [[0, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[0, 85]], [[0, 85]], [[70, 77]], [[78, 85], [87, 94]], [[86, 97]], [[95, 97]], [[95, 115]], [[106, 109]], [[110, 113]], [[117, 136]], [[138, 157]]]",
    "query_spans": "[[[159, 169]]]",
    "process": "As shown in the figure, let |PQ| = 4t (t > 0). From 3|PQ| = 4|PF_{1}|, we get |PF_{1}| = 3t. By the definition of hyperbola, |PF_{1}| - |PF_{2}| = 2a, so |PF_{2}| = 3t - 2a, |QF_{2}| = |PQ| - |PF_{1}| = t + 2a. Also, |QF_{1}| - |QF_{2}| = 2a, so |QF_{1}| = t + 4a. Since PQ \\bot PF_{1}, we have |PF_{1}|^{2} + |PF_{2}|^{2} = 4c^{2}, |PF_{1}|^{2} + |PQ|^{2} = |QF_{1}|^{2}, that is, (3t)^{2} + (3t - 2a)^{2} = 4c^{2\\textcircled{1}}, (3t)^{2} + (4t)^{2} = (t + 4a)^{2}\\textcircled{2}. Solving \\textcircled{2} gives t = a. Substituting into \\textcircled{1}, we get (3a)^{2} + (3a - 2a)^{2} = 4c^{2}, i.e., 10a^{2} = 4c^{2}. Therefore, e = \\frac{c}{a} = \\sqrt{\\frac{10}{4}} = \\frac{\\sqrt{10}}{2}"
  },
  {
    "text": "A point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ has a distance of $6$ to the left focus $F$. Then, what is the distance from point $P$ to the left directrix?",
    "fact_expressions": "G: Ellipse;P: Point;F: Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(P, G);LeftFocus(G) = F;Distance(P, F) = 6",
    "query_expressions": "Distance(P,LeftDirectrix(G))",
    "answer_expressions": "10",
    "fact_spans": "[[[0, 39]], [[42, 45], [61, 65]], [[49, 52]], [[0, 39]], [[0, 45]], [[0, 52]], [[42, 59]]]",
    "query_spans": "[[[0, 74]]]",
    "process": ""
  },
  {
    "text": "If $A$ and $B$ are two distinct points on the curve $x = \\sqrt{y^{2} + 2}$, and $O$ is the coordinate origin, then the range of $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}$ is?",
    "fact_expressions": "A: Point;B: Point;G: Curve;Expression(G) = (x = sqrt(y^2 + 2));PointOnCurve(A, G) = True;PointOnCurve(B, G) = True;Negation(A = B);O: Origin",
    "query_expressions": "Range(DotProduct(VectorOf(O, A), VectorOf(O, B)))",
    "answer_expressions": "[2, +oo)",
    "fact_spans": "[[[1, 4]], [[5, 8]], [[9, 29]], [[9, 29]], [[1, 35]], [[1, 35]], [[1, 35]], [[36, 39]]]",
    "query_spans": "[[[46, 102]]]",
    "process": "First simplify to obtain $\\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1(x\\geqslant0)$, let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then obtain $\\overrightarrow{OA}\\cdot\\overrightarrow{OB}=x_{1}x_{2}+y_{1}y_{2}$, discuss two cases: $AB\\bot x$-axis and $AB$ not perpendicular to the $x$-axis. When $AB$ is not perpendicular to the $x$-axis, let $l_{AB}:y=kx+m$, combine the two equations and eliminate $y$, obtaining a quadratic equation in $x$. Then use Vieta's formulas, substitute into $\\overrightarrow{OA}\\cdot\\overrightarrow{OB}=x_{1}x_{2}+y_{1}y_{2}$, simplify and organize to obtain the result. Solution: $\\because x=\\sqrt{y^{2}+2}$ can be rewritten as $\\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1(x\\geqslant\\sqrt{2})$, let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $x_{1}\\cdot x_{2}>0$, so $\\overrightarrow{OA}=(x_{1},y_{1})$, $\\overrightarrow{OB}=(x_{2},y_{2})$, $\\therefore \\overrightarrow{OA}\\cdot\\overrightarrow{OB}=x_{1}x_{2}+y_{1}y_{2}$. If $AB\\bot x$-axis, at this time $x_{1}=x_{2}$, $y_{1}=-y_{2}$. $\\therefore \\overrightarrow{OA}\\cdot\\overrightarrow{OB}=x^{2}-y^{2}=2$. If $AB$ is not perpendicular to the $x$-axis, let $l_{AB}:y=kx+m$, $(y=kx+m \\begin{cases} x^2-y^2=2 \\end{cases}$ $\\therefore (1-k^{2})x^{2}-2kmx-m^{2}-2=$ $\\therefore x_{1}+x_{2}=\\frac{2km}{1-k^{2}}$, $x_{1}\\cdot x_{2}==\\frac{-m^{2}-2}{1-k^{2}}=\\frac{m^{2}+2}{k^{2}-1}>0$, then $k^{2}>1$, $\\overrightarrow{OA}\\cdot\\overrightarrow{OB}=x_{1}x_{2}+y_{1}y_{2}=x_{1}x_{2}+(kx_{1}+m)(kx_{2}+m)=(1+k)^{2}\\frac{-m^{2}-2}{1-k^{2}}+km\\frac{2km}{1-b^{2}}+m^{2}=\\frac{2k^{2}+2}{k^{2}-1}=2+\\frac{4}{k^{2}-1}$. Also $\\because k^{2}>1$, $k^{2}-1>0$, $\\therefore \\overrightarrow{OA}\\cdot\\overrightarrow{OB}>2$, $\\therefore \\overrightarrow{OA}\\cdot\\overrightarrow{OB}\\in[2,+\\infty)$"
  },
  {
    "text": "Given that $F$ is the right focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, point $P$ lies on the ellipse, and the distance from $P$ to the origin $O$ equals the semi-focal length, the area of $\\triangle P O F$ is $6$, then $b=$?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;O: Origin;F: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;PointOnCurve(P, G);Distance(P,O)=HalfFocalLength(G);Area(TriangleOf(P, O, F)) = 6",
    "query_expressions": "b",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[6, 58], [68, 70]], [[117, 120]], [[8, 58]], [[63, 67], [73, 76]], [[77, 82]], [[2, 5]], [[8, 58]], [[8, 58]], [[6, 58]], [[2, 62]], [[63, 71]], [[68, 90]], [[91, 115]]]",
    "query_spans": "[[[117, 122]]]",
    "process": "Let $ P(x_{0},y_{0}) $. Since point $ P $ lies on the ellipse, $ \\frac{x_{0}^{2}}{a^{2}} + \\frac{y_{0}^{2}}{b^{2}} = 1 $ ①. Also, since the distance from point $ P $ to the origin $ O $ equals the semi-focal distance, $ \\sqrt{x_{0}^{2} + y_{0}^{2}} = c $, i.e., $ x_{0}^{2} + y_{0}^{2} = c^{2} $ ②. Given that the area of $ \\triangle POF $ is 6, $ \\frac{1}{2} \\times c \\times |y_{0}| = 6 $, we obtain $ |y_{0}| = \\frac{12}{c} $ ③. Substituting ③ into ② gives $ x_{0}^{2} = c^{2} - \\frac{144}{c^{2}} $. Substituting ③ into ① gives $ x_{0}^{2} = a^{2} - \\frac{144a^{2}}{b^{2}c^{2}} $. Hence, $ c^{2} - \\frac{144}{c^{2}} = a^{2} - \\frac{144a^{2}}{b^{2}c^{2}} $, leading to $ b = 2\\sqrt{:} $"
  },
  {
    "text": "Let the focus of the parabola $x^{2}=12 y$ be $F$, and let line $l$ passing through point $P(2 , 1)$ intersect the parabola at points $A$ and $B$. It is known that point $P$ is exactly the midpoint of $AB$. Then $|AF|+|BF|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 12*y);F: Point;Focus(G) = F;P: Point;Coordinate(P) = (2, 1);l: Line;PointOnCurve(P, l)  ;Intersection(l, G) = {A, B};A: Point;B: Point;MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 16], [44, 47]], [[1, 16]], [[20, 23]], [[1, 23]], [[26, 37], [62, 66]], [[26, 37]], [[38, 43]], [[24, 43]], [[38, 59]], [[50, 53]], [[54, 57]], [[62, 75]]]",
    "query_spans": "[[[77, 90]]]",
    "process": ""
  },
  {
    "text": "If the two distinct intersection points of the line $m x - y + m = 0$ and the parabola $y = x^{2} - 4 x + 3$ both lie in the first quadrant, then the range of real values for $m$ is?",
    "fact_expressions": "G: Parabola;H: Line;m: Real;Expression(G) = (y = x^2 - 4*x + 3);Expression(H) = (m + m*x - y = 0);P:Point;P1:Point;Negation(P=P1);Quadrant(P)=1;Quadrant(P1)=1;Intersection(H,G)={P,P1}",
    "query_expressions": "Range(m)",
    "answer_expressions": "(0,3)",
    "fact_spans": "[[[15, 33]], [[1, 14]], [[48, 53]], [[15, 33]], [[1, 14]], [], [], [[1, 40]], [[1, 46]], [[1, 46]], [[1, 40]]]",
    "query_spans": "[[[48, 60]]]",
    "process": "\\because y=x^{2}-4x+3=(x-2)^{2}-1, from mx-y+m=0, we get y=m(x+1), so the line mx-y+m=0 passes through the fixed point (-1,0). In the same plane rectangular coordinate system, draw the approximate graphs of the line and the parabola as shown in the figure. \\because if the two distinct intersection points of the line and the parabola are both in the first quadrant, \\therefore from the figure, we can see that 0<m<\\frac{3-0}{0-(-1)}, that is, 0<m<3,"
  },
  {
    "text": "If the standard equation of hyperbola $C$ is $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, point $P(\\sqrt{2}, 1)$ lies on $C$, and the distance from a focus to an asymptote is $2$, then the standard equation of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;P: Point;Coordinate(P) = (sqrt(2), 1);PointOnCurve(P, C);Distance(OneOf(Focus(C)), OneOf(Asymptote(C))) = 2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/(8/5)-y^2/4=1",
    "fact_spans": "[[[1, 7], [101, 104], [75, 78]], [[1, 56]], [[13, 56]], [[13, 56]], [[58, 74]], [[58, 74]], [[58, 81]], [[75, 99]]]",
    "query_spans": "[[[101, 111]]]",
    "process": "Let the right focus of the hyperbola be $(c,0)$, and one asymptote be $bx-ay=0$. According to the point-to-line distance formula, $\\frac{|bc|}{\\sqrt{a^{2}+b^{2}}}=b=2$, $\\left(\\frac{2}{2}\\right)^{2}-\\frac{1^{2}}{b^{2}}=1$, $\\therefore a^{2}=\\frac{8}{5}, b=2$, so the standard equation of the hyperbola is $\\frac{x^{2}}{8}-\\frac{y^{2}}{4}=1$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, with upper vertex $B$, right focus $F(2,0)$, and point $M\\left(-\\frac{\\sqrt{2} a}{2}, 0\\right)$, satisfying $B F \\perp B M$, then the standard equation of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;B: Point;F: Point;M: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (2, 0);Coordinate(M) = (-sqrt(2)*a/2, 0);UpperVertex(C) = B;RightFocus(C) = F;IsPerpendicular(LineSegmentOf(B, F), LineSegmentOf(B, M))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/8+y^2/4=1",
    "fact_spans": "[[[2, 59], [133, 138]], [[8, 59]], [[8, 59]], [[64, 67]], [[72, 80]], [[83, 112]], [[8, 59]], [[8, 59]], [[2, 59]], [[72, 80]], [[83, 112]], [[2, 67]], [[2, 80]], [[116, 131]]]",
    "query_spans": "[[[133, 145]]]",
    "process": "According to the problem, the upper vertex of the ellipse $ C: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $) is $ B $, so $ B(0, b) $. Then $ \\overrightarrow{BF} = (2, -b) $, $ \\overrightarrow{BM} = \\left(-\\frac{\\sqrt{2}a}{2}, -b\\right) $. If $ BF \\perp BM $, then $ \\overrightarrow{BF} \\cdot \\overrightarrow{BM} = -\\sqrt{2}a + b^{2} = 0 $, thus $ b^{2} = \\sqrt{2}a $. Also, since the right focus of the ellipse is $ F(2, 0) $, i.e., $ c = 2 $, we have $ a^{2} - b^{2} = c^{2} = 4 $. Solving gives $ a = 2\\sqrt{2} $, then $ b = 2 $. Therefore, the equation of ellipse $ C $ is $ \\frac{x^{2}}{8} + \\frac{y^{2}}{4} = 1 $."
  },
  {
    "text": "What is the length of the chord intercepted by the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ from a line passing through its left focus with slope $1$?",
    "fact_expressions": "C: Ellipse;G: Line;Expression(C) = (x^2/4 + y^2/3 = 1);PointOnCurve(LeftFocus(C), G);Slope(G)=1",
    "query_expressions": "Length(InterceptChord(G,C))",
    "answer_expressions": "24/7",
    "fact_spans": "[[[1, 43], [58, 63]], [[55, 57]], [[1, 43]], [[0, 57]], [[48, 57]]]",
    "query_spans": "[[[55, 70]]]",
    "process": "Let the coordinates of the two intersection points between the line and the ellipse be $(x_{1},y_{1}),(x_{2},y_{2})$. The ellipse $C: \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ has its left focus at $(-1,0)$, so the equation of the line is $y=x+1$. Then  \n\\[\n\\begin{cases}\ny=x+1 \\\\\n\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1\n\\end{cases}\n\\Rightarrow 7x^{2}+8x-8=0\n\\]  \nThus, $x_{1}+x_{2}=-\\frac{8}{7}, x_{1}x_{2}=-\\frac{8}{7}$. Therefore, the length of the chord intercepted by the ellipse on this line is  \n$\\sqrt{1+1^{2}}\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\frac{24}{7}$.  \nThe answer is: 24"
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$, the right focus is $F$, $P$ is a point on the left branch of hyperbola $C$, and $M(0,2)$. Then the minimum perimeter of $\\triangle P F M$ is?",
    "fact_expressions": "C: Hyperbola;M: Point;P: Point;F: Point;Expression(C) = (x^2 - y^2/3 = 1);Coordinate(M) = (0, 2);RightFocus(C)=F;PointOnCurve(P,LeftPart(C))",
    "query_expressions": "Min(Perimeter(TriangleOf(P, F, M)))",
    "answer_expressions": "2+4*sqrt(2)",
    "fact_spans": "[[[2, 35], [48, 54]], [[61, 69]], [[44, 47]], [[40, 43]], [[2, 35]], [[61, 69]], [[2, 43]], [[44, 60]]]",
    "query_spans": "[[[71, 95]]]",
    "process": "\\because F(2,0), M(0,2) \\therefore |FM| = 2\\sqrt{2}, the perimeter of \\triangle PFM is minimized, i.e., PM + PF is minimized. Let the left focus be F_{1}, by the definition of hyperbola we obtain PM + PF = PM + 2 + PF_{1}, and the minimum value of PM + PF_{1} is MF_{1} = 2\\sqrt{2}, \\therefore the minimum value of PM + 2 + PF_{1} is 2 + 4\\sqrt{2}, hence fill in 2 + 4\\sqrt{2}."
  },
  {
    "text": "Given points $P(x_{1}, y_{1})$, $Q(x_{2}, y_{2})$ on the parabola $y^{2}=4 x$, and $x_{1}+x_{2}+2=\\frac{2}{3} \\sqrt{3}|P Q|$, where $F$ is the focus, then the maximum value of $\\angle P F Q$ is?",
    "fact_expressions": "P: Point;Coordinate(P) = (x1, y1);x1: Number;y1: Number;Q: Point;Coordinate(Q) = (x2, y2);x2: Number;y2: Number;G: Parabola;Expression(G) = (y^2 = 4*x);PointOnCurve(P, G) = True;PointOnCurve(Q, G) = True;x1 + x2 + 2 = (2/3)*sqrt(3)*Abs(LineSegmentOf(P, Q));F: Point;Focus(G) = F",
    "query_expressions": "Max(AngleOf(P, F, Q))",
    "answer_expressions": "2*pi/3",
    "fact_spans": "[[[2, 20]], [[2, 20]], [[2, 19]], [[2, 19]], [[21, 39]], [[21, 39]], [[21, 39]], [[21, 39]], [[40, 54]], [[40, 54]], [[2, 57]], [[2, 57]], [[59, 101]], [[103, 106]], [[40, 109]]]",
    "query_spans": "[[[111, 130]]]",
    "process": "From the given condition, $x_{1}+x_{2}+2=x_{1}+1+x_{2}+1=|PF|+|QF|=\\frac{2}{3}\\sqrt{3}|PQ|$, that is, $|PQ|=\\frac{\\sqrt{3}}{2}(|PF|+|QF|)$. Hence, $\\cos\\angle PFQ = \\frac{|PF|^{2}+|OF|^{2}-|PO|^{2}}{2|PF||OF|} = \\frac{|PF|^{2}+|QF|^{2}-\\frac{3}{4}(|PF|+|QF|)^{2}}{2|PF||QF|}$, i.e., $\\cos\\angle PFQ \\geqslant -\\frac{1}{2}$. Since $\\angle PFQ \\in (0,\\pi)$ and the cosine function is monotonically decreasing in $(0,\\pi)$, it follows that $\\angle PFQ \\leqslant \\frac{2\\pi}{3}$. Equality holds if and only if $|PF|=|QF|$."
  },
  {
    "text": "The coordinates of the point $P$ on the parabola $y^{2}=2 x$ that is closest to the line $x-y+3=0$ are?",
    "fact_expressions": "G: Parabola;H: Line;Expression(G) = (y^2 = 2*x);Expression(H) = (x - y + 3 = 0);P:Point;PointOnCurve(P,G);Q:Point;PointOnCurve(Q,H);WhenMin(Distance(P,Q))",
    "query_expressions": "Coordinate(Q)",
    "answer_expressions": "(1/2,-1)",
    "fact_spans": "[[[0, 14]], [[20, 31]], [[0, 14]], [[20, 31]], [[15, 19]], [[0, 19]], [[37, 38]], [[20, 38]], [[15, 38]]]",
    "query_spans": "[[[37, 43]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a point on the hyperbola $3 x^{2}-5 y^{2}=15$, $F_{1}$, $F_{2}$ are its two foci, and the area of $\\triangle F_{1} PF_{2}$ is $3 \\sqrt{3}$, find the measure of $\\angle F_{1} PF_{2}$.",
    "fact_expressions": "G: Hyperbola;F1: Point;P: Point;F2: Point;Expression(G) = (3*x^2 - 5*y^2 = 15);PointOnCurve(P, G);Focus(G)={F1,F2};Area(TriangleOf(F1, P, F2)) = 3*sqrt(3)",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "ApplyUnit(60,degree)",
    "fact_spans": "[[[6, 29], [50, 51]], [[33, 40]], [[2, 5]], [[42, 49]], [[6, 29]], [[2, 32]], [[33, 55]], [[57, 97]]]",
    "query_spans": "[[[100, 126]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the right branch of the hyperbola, and $|P F_{1}|=4|P F_{2}|$. What is the maximum value of the eccentricity $e$ of this hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;e: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F1)) = 4*Abs(LineSegmentOf(P, F2));Eccentricity(G) = e",
    "query_expressions": "Max(e)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[2, 61], [91, 94], [126, 129]], [[5, 61]], [[5, 61]], [[86, 90]], [[70, 77]], [[78, 85]], [[133, 136]], [[5, 61]], [[5, 61]], [[2, 61]], [[2, 85]], [[2, 85]], [[86, 98]], [[100, 122]], [[126, 136]]]",
    "query_spans": "[[[133, 142]]]",
    "process": ""
  },
  {
    "text": "Given that points $A$ and $B$ on the parabola $y^{2}=8x$ satisfy $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}=0$, a perpendicular is drawn from the origin $O$ to the line $AB$, with foot of the perpendicular at $P$. Then the maximum area of $\\triangle O F P$ ($F$ being the focus of the parabola) is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);A: Point;B: Point;PointOnCurve(A,G) = True;PointOnCurve(B,G) = True;O: Origin;DotProduct(VectorOf(O, A), VectorOf(O, B)) = 0;L: Line;PointOnCurve(O,L) = True;IsPerpendicular(L,LineOf(A,B)) = True;FootPoint(L,LineOf(A,B)) = P;P: Point;F: Point;Focus(G) = F",
    "query_expressions": "Max(Area(TriangleOf(O,F,P)))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 16], [130, 133]], [[2, 16]], [[17, 20]], [[21, 24]], [[2, 26]], [[2, 26]], [[81, 88]], [[28, 79]], [], [[80, 99]], [[80, 99]], [[80, 106]], [[103, 106]], [[126, 129]], [[126, 136]]]",
    "query_spans": "[[[108, 146]]]",
    "process": "According to the problem, let $ A\\left(\\frac{y_1^{2}}{8}, y_{1}\\right), B\\left(\\frac{y_{2}^{2}}{8}, y_{2}\\right) $ ($ y_{1}y_{2} \\neq 0 $). From $ \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = 0 $, we get: $ \\frac{y_1^{2}y_{2}^{2}}{64} + y_{1}y_{2} = 0 $, solving yields $ y_{1}y_{2} = -64 $. Let $ M(x, y) $ be an arbitrary point on line $ AB $, then $ \\overrightarrow{AM} \\parallel \\overrightarrow{AB} $, where $ \\overrightarrow{AM} = \\left(x - \\frac{y_1^{2}}{8}, y - y_{1}\\right) $, $ \\overrightarrow{AB} = \\left(\\frac{y_{2}^{2} - y_{1}^{2}}{8}, y_{2} - y_{1}\\right) $. Thus we obtain: $ \\left(x - \\frac{y_1^{2}}{8}\\right)(y_{2} - y_{1}) - (y - y_{1})\\left(\\frac{y_{2}^{2} - y_{1}^{2}}{8}\\right) = 0 $, and since $ y_{1} \\neq y_{2} $, simplifying gives: $ x - 8 - \\frac{y_{1} + y_{2}}{8}y = 0 $. Therefore, the equation of line $ AB $ is: $ x - 8 - \\frac{y_{1} + y_{2}}{8}y = 0 $, which always passes through the fixed point $ Q(8, 0) $. Since $ OP \\perp AB $ at point $ P $, the trajectory of point $ P $ is a circle with diameter $ OQ $ (excluding the origin $ O $). Hence, the maximum distance from point $ P $ to the $ x $-axis is the radius of the circle, which is 4. Given focus $ F(2, 0) $, the maximum area of $ \\triangle OFP $ is $ S = \\frac{1}{2} \\times 2 \\times 4 = 4 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, one of its asymptotes is parallel to $3 x+y-4=0$. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Z: Curve;Expression(Z) = (3*x + y - 4 = 0);IsParallel(OneOf(Asymptote(C)), Z)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[2, 63], [85, 88]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[70, 81]], [[70, 81]], [[2, 83]]]",
    "query_spans": "[[[85, 94]]]",
    "process": "From the given condition: $-\\frac{b}{a} = -3$, then the eccentricity of $C$ is $e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{10}$"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, and $P Q$ is a line segment passing through focus $F_{1}$, then the value of $|P F_{2}|+|Q F_{2}|-| P Q|$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;Q: Point;F2: Point;F1: Point;Expression(G) = (x^2/16 - y^2/9 = 1);Focus(G) = {F1, F2};PointOnCurve(F1,LineSegmentOf(P,Q))",
    "query_expressions": "Abs(LineSegmentOf(P,F2))+Abs(LineSegmentOf(Q,F2))-Abs(LineSegmentOf(P,Q))",
    "answer_expressions": "16",
    "fact_spans": "[[[18, 57]], [[61, 66]], [[61, 66]], [[10, 17]], [[2, 9], [70, 77]], [[18, 57]], [[2, 60]], [[61, 76]]]",
    "query_spans": "[[[79, 111]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of hyperbola $C$ with foci on the $x$-axis are $y = \\pm x$, what is the eccentricity of this hyperbola?",
    "fact_expressions": "C: Hyperbola;PointOnCurve(Focus(C), xAxis);Expression(Asymptote(C)) = (y = pm*x)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[11, 17], [36, 39]], [[2, 17]], [[11, 33]]]",
    "query_spans": "[[[35, 45]]]",
    "process": "Given that the foci of the hyperbola lie on the x-axis, let the equation of the hyperbola be: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0). Then the asymptotes of hyperbola C are given by: y=\\pm\\frac{b}{a}x. Since the asymptotes of hyperbola C are also given by y=\\pm x, it follows that \\frac{b}{a}=1. The eccentricity of the hyperbola is e=\\sqrt{1+(\\frac{b}{a})^{2}}=\\sqrt{2}."
  },
  {
    "text": "If the distance from point $M$ to point $F(4,0)$ is less than its distance to the fixed line $x+5=0$ by $1$, then the equation satisfied by point $M$ is?",
    "fact_expressions": "G: Line;F: Point;M: Point;Expression(G) = (x + 5 = 0);Coordinate(F) = (4, 0);Distance(M, F) = Distance(M, G) - 1",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "y^2=16*x",
    "fact_spans": "[[[24, 31]], [[6, 15]], [[1, 5], [19, 20], [40, 44]], [[24, 31]], [[6, 15]], [[1, 38]]]",
    "query_spans": "[[[40, 51]]]",
    "process": "The distance from point P to point F(4,0) is 1 less than its distance to the line x+5=0. Therefore, the distance from point P to point F(4,0) is equal to its distance to the line x+4=0. Thus, its locus is a parabola with focus F(4,0) and directrix x+4=0, so the equation is $y^2=16x$."
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$ are $F_{1}$ and $F_{2}$, respectively. Chord $AB$ passes through point $F_{1}$. If the circumference of the incircle of triangle $ABF_{2}$ is $\\pi$, and the coordinates of points $A$ and $B$ are $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$, respectively, then $|y_{1}-y_{2}|=$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/7 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);PointOnCurve(F1, LineSegmentOf(A, B));Perimeter(InscribedCircle(TriangleOf(A, B, F2))) = pi;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2)",
    "query_expressions": "Abs(y1 - y2)",
    "answer_expressions": "4/3",
    "fact_spans": "[[[0, 38]], [[0, 38]], [[47, 54], [72, 80]], [[56, 63]], [[0, 63]], [[0, 63]], [[118, 121]], [[123, 126]], [[0, 71]], [[66, 80]], [[83, 116]], [[134, 151]], [[152, 169]], [[134, 151]], [[152, 169]], [[118, 169]], [[118, 169]]]",
    "query_spans": "[[[171, 188]]]",
    "process": "By drawing a diagram according to the given conditions, it is easy to see that the inradius $ r $ of $\\triangle ABF_{2} $ is $ \\frac{1}{2} $. Thus, using the area of the triangle and the equal-area method, we can solve the problem. Draw the diagram as follows: $ \\therefore $ the inradius $ r $ of $ \\triangle ABF_{2} $ is $ \\frac{1}{2} $. Also, since the perimeter of $ \\triangle ABF_{2} $ is $ 1 = 4a = 16 $, then $ S_{\\triangle ABF} = \\frac{1}{2} \\times 16 \\times \\frac{1}{2} = 4 $, and $ S_{\\triangle ABF} = \\frac{1}{2} \\times |F_{1}F_{2}| \\times |y_{1} \\cdot y_{2}| = 3|y_{1} \\cdot y_{2}| $, hence $ |y_{1} \\cdot y_{2}| = \\frac{4}{3} $."
  },
  {
    "text": "The right focus of the ellipse $C$: $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$ is $F$. A line passing through the origin intersects the ellipse $C$ at two points $A$ and $B$. What is the maximum area of $\\triangle A B F$?",
    "fact_expressions": "C: Ellipse;G: Line;A: Point;B: Point;F: Point;O: Origin;Expression(C) = (x^2/8 + y^2/4 = 1);RightFocus(C) = F;PointOnCurve(O, G);Intersection(G, C) = {A, B}",
    "query_expressions": "Max(Area(TriangleOf(A, B, F)))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 42], [58, 63]], [[55, 57]], [[67, 70]], [[71, 74]], [[47, 50]], [[52, 54]], [[0, 42]], [[0, 50]], [[51, 57]], [[55, 74]]]",
    "query_spans": "[[[76, 102]]]",
    "process": "In ellipse C, a=2\\sqrt{2}, b=2, then c=\\sqrt{a^{2}-b^{2}}=2, so F(2,0). From the problem, A and B are symmetric about the origin. When A and B are the endpoints of the minor axis of ellipse C, the area of \\triangle ABF reaches its maximum value, and the maximum value is \\frac{1}{3}\\times c\\times 2b=4. The final answer is 4."
  },
  {
    "text": "Point $F$ is the focus of the parabola $C$: $y^{2}=4 x$, and point $P$ lies on the parabola $C$. If $|PF|=5$, then the coordinates of point $P$ are?",
    "fact_expressions": "C: Parabola;P: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(P, C);Abs(LineSegmentOf(P, F)) = 5",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "{(4,4),(4,-4)}",
    "fact_spans": "[[[5, 24], [33, 39]], [[28, 32], [52, 56]], [[0, 4]], [[5, 24]], [[0, 27]], [[28, 40]], [[42, 50]]]",
    "query_spans": "[[[52, 61]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $P$ is a point on the ellipse $C$, and $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$. If the area of $\\Delta P F_{1} F_{2}$ is $9$, then $b=$?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Focus(C) = {F1, F2};PointOnCurve(P, C);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Area(TriangleOf(P, F1, F2)) = 9",
    "query_expressions": "b",
    "answer_expressions": "3",
    "fact_spans": "[[[18, 77], [87, 92]], [[190, 193]], [[25, 77]], [[83, 86]], [[2, 9]], [[10, 17]], [[25, 77]], [[25, 77]], [[18, 77]], [[2, 82]], [[83, 96]], [[98, 157]], [[159, 188]]]",
    "query_spans": "[[[190, 195]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with left focus $F_{1}(-2 \\sqrt{5}, 0)$, right focus $F_{2}(2 \\sqrt{5}, 0)$, and eccentricity $e=\\frac{\\sqrt{5}}{2}$. If point $P$ is a point on the right branch of hyperbola $C$, then $|P F_{1}|-|P F_{2}|$=?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F1: Point;Coordinate(F1) = (-2*sqrt(5), 0);LeftFocus(C) = F1;F2: Point;Coordinate(F2) = (2*sqrt(5), 0);RightFocus(C) = F2;e: Number;Eccentricity(C)=e;e = sqrt(5)/2;P: Point;PointOnCurve(P, RightPart(C))",
    "query_expressions": "Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 58], [145, 151]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[62, 85]], [[62, 85]], [[2, 85]], [[89, 111]], [[89, 111]], [[2, 111]], [[115, 137]], [[2, 137]], [[115, 137]], [[140, 144]], [[140, 156]]]",
    "query_spans": "[[[158, 181]]]",
    "process": "Given that $c=2\\sqrt{5}$, $e=\\frac{c}{a}=\\frac{\\sqrt{5}}{2}$, $\\therefore a=4$, $|PF_{1}|-|PF_{2}|=2a=8$."
  },
  {
    "text": "Given that point $P$ lies on the line $x + y + 5 = 0$, and point $Q$ lies on the parabola $y^{2} = 2x$, then the minimum value of $PQ$ is equal to?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Q: Point;Expression(G) = (y^2 = 2*x);Expression(H) = (x + y + 5 = 0);PointOnCurve(P, H);PointOnCurve(Q, G)",
    "query_expressions": "Min(LineSegmentOf(P, Q))",
    "answer_expressions": "9*sqrt(2)/4",
    "fact_spans": "[[[25, 39]], [[7, 18]], [[2, 6]], [[20, 24]], [[25, 39]], [[7, 18]], [[2, 19]], [[20, 40]]]",
    "query_spans": "[[[42, 54]]]",
    "process": ""
  },
  {
    "text": "A point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ has a distance of $7$ to one focus; then the distance from $P$ to the other focus is?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(P, G);F1:Point;F2:Point;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2);Distance(P, F1) = 7",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 39]], [[42, 45], [59, 62]], [[0, 39]], [[0, 45]], [], [], [[0, 50]], [[0, 68]], [[0, 68]], [[0, 57]]]",
    "query_spans": "[[[0, 73]]]",
    "process": "From the ellipse $\\frac{x^2}{25} + \\frac{y^2}{16} = 1$, we know $a = 5$. By the definition of an ellipse, the sum of the distances from any point on the ellipse to the two foci is $2a = 10$. If the distance from point $P$ to one focus is $7$, then the distance from $P$ to the other focus is $10 - 7 = 3$."
  },
  {
    "text": "Given that the curve represented by the equation $x^{2} +\\frac {y^{2}}{m}=1$ is an ellipse with foci on the $y$-axis and eccentricity $\\frac{1}{2}$, then $m$=?",
    "fact_expressions": "G: Ellipse;C:Curve;Expression(C) = (x^2+y^2/m=1);C=G;PointOnCurve(Focus(G), yAxis);Eccentricity(G) = 1/2;m: Number",
    "query_expressions": "m",
    "answer_expressions": "4/3",
    "fact_spans": "[[[64, 66]], [[34, 36]], [[2, 36]], [[34, 66]], [[37, 66]], [[46, 66]], [[68, 71]]]",
    "query_spans": "[[[68, 73]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus is $F$. If there exists a line passing through the origin intersecting the ellipse at points $A$ and $B$ such that $A F \\perp B F$, then the range of the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;LeftFocus(G) = F;O: Origin;H: Line;PointOnCurve(O,H);A: Point;B: Point;Intersection(H, G) = {A, B};IsPerpendicular(LineSegmentOf(A, F), LineSegmentOf(B, F))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[\\sqrt{2}/2,1)",
    "fact_spans": "[[[2, 54], [73, 75], [104, 106]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[59, 62]], [[2, 62]], [[67, 69]], [[70, 72]], [[66, 72]], [[76, 79]], [[80, 83]], [[70, 85]], [[87, 102]]]",
    "query_spans": "[[[104, 117]]]",
    "process": ""
  },
  {
    "text": "Given $2 x^{2}+3 y^{2}=6$, then the maximum value $M$ of $|x|+\\sqrt{3} y$ is?",
    "fact_expressions": "2*x^2 + 3*y^2 = 6;Max(sqrt(3)*y + Abs(x)) = M;M:Number",
    "query_expressions": "M",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 21]], [[23, 47]], [[44, 47]]]",
    "query_spans": "[[[44, 49]]]",
    "process": ""
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{13-m}+\\frac{y^{2}}{m-2}=1$ has a focal distance of $6$, then $m=$?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/(13 - m) + y^2/(m - 2) = 1);FocalLength(G) = 6",
    "query_expressions": "m",
    "answer_expressions": "{3, 12}",
    "fact_spans": "[[[0, 42]], [[51, 54]], [[0, 42]], [[0, 49]]]",
    "query_spans": "[[[51, 56]]]",
    "process": "The ellipse $\\frac{x^{2}}{13-m}+\\frac{y^{2}}{m-2}=1$ has a focal distance of 6, that is, $c=3$. The foci of the ellipse $\\frac{x^{2}}{13-m}+\\frac{y^{2}}{m-2}=1$ may lie on the $x$-axis or on the $y$-axis. When the foci are on the $x$-axis, $c^{2}=13-m-(m-2)=9$, solving gives $m=3$; when the foci are on the $y$-axis, $c^{2}=(m-2)-(13-m)=9$, solving gives $m=12$. In conclusion, $m=3$ or $12$."
  },
  {
    "text": "If the chord of the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$ is bisected by the point $(4 , 2)$, then what is the equation of the line on which this chord lies?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/36 + y^2/9 = 1);H: LineSegment;IsChordOf(H,G) = True;Coordinate(MidPoint(H)) = (4,2)",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "y = -0.5*x+4",
    "fact_spans": "[[[2, 40]], [[2, 40]], [], [[2, 42]], [[2, 55]]]",
    "query_spans": "[[[2, 69]]]",
    "process": "Let the chord be AB, with A(x_{1},y_{1}), B(x_{2},y_{2}). Substituting into the ellipse equation gives \\frac{x_{1}^{2}}{36}+\\frac{y_{1}^{2}}{9}=1, \\frac{x^{2}}{36}+\\frac{y_{2}^{2}}{9}=1. Subtracting these two equations and simplifying yields \\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=-\\frac{x_{1}+x_{2}}{y_{1}+y}=-\\frac{1}{2}, so the slope of the chord is -\\frac{1}{2}. Using the point-slope form, y-2=-\\frac{1}{2}(x-4), which simplifies to v=-0.5x+4."
  },
  {
    "text": "What is the standard equation of the parabola passing through the point $P(-2,-4)$?",
    "fact_expressions": "G: Parabola;P: Point;Coordinate(P) = (-2,-4);PointOnCurve(P,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=-8*x,x^2=-y}",
    "fact_spans": "[[[13, 16]], [[1, 12]], [[1, 12]], [[0, 16]]]",
    "query_spans": "[[[13, 23]]]",
    "process": "Since P(-2,-4) is in the third quadrant, the parabola opens to the left or downward. When the parabola opens downward, let the equation of the parabola be y^{2}=-2px; substituting the coordinates of point P gives 16=-2p\\times(-2), so p=4, and the equation of the parabola is y^{2}=-8x. When the parabola opens downward, let the equation of the parabola be x^{2}=-2py; substituting the coordinates of point P gives 4=-2p\\times(-4), so p=\\frac{1}{2}, thus the equation of the parabola is x^{2}=-y. In conclusion, the equations of the parabola are y^{2}=-8x or x^{2}=-y."
  },
  {
    "text": "On the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, the distance from a point $M$ to the left focus $F_{1}$ is $2$. $N$ is the midpoint of $M F_{1}$. Then $|O N|$ equals?",
    "fact_expressions": "G: Ellipse;M: Point;F1: Point;O: Origin;N: Point;Expression(G) = (x^2/25 + y^2/9 = 1);PointOnCurve(M, G);LeftFocus(G)=F1;Distance(M, F1) = 2;MidPoint(LineSegmentOf(M, F1)) = N",
    "query_expressions": "Abs(LineSegmentOf(O, N))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 38]], [[41, 44]], [[48, 55]], [[83, 90]], [[65, 68]], [[0, 38]], [[0, 44]], [[0, 55]], [[41, 62]], [[65, 81]]]",
    "query_spans": "[[[83, 93]]]",
    "process": "According to the definition of an ellipse: $|MF_{1}|+|MF_{2}|=10$, so $|MF_{2}|=8$. $N$ is the midpoint of $MF_{1}$, $O$ is the midpoint of $F_{1}F_{2}$, so $ON=\\frac{1}{2}|MF_{2}|=4$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $E$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$, and $P$ is any point on the ellipse $E$, then the range of values of $\\overrightarrow{F_{1} P} \\cdot \\overrightarrow{F_{2} P}$ is?",
    "fact_expressions": "E: Ellipse;Expression(E) = (x^2/4 + y^2/2 = 1);F1: Point;F2: Point;Focus(E) = {F1, F2};P: Point;PointOnCurve(P, E)",
    "query_expressions": "Range(DotProduct(VectorOf(F1, P), VectorOf(F2, P)))",
    "answer_expressions": "[0, 2]",
    "fact_spans": "[[[18, 60], [69, 75]], [[18, 60]], [[2, 9]], [[10, 17]], [[2, 65]], [[66, 69]], [[66, 79]]]",
    "query_spans": "[[[81, 145]]]",
    "process": "From $a^{2}=4$, $b^{2}=2$, solving gives: $c^{2}=a^{2}-b^{2}=2$, so $c=\\sqrt{2}$. Let $F_{1}(-\\sqrt{2},0)$, $F_{2}(\\sqrt{2},0)$. Since $P$ is an arbitrary point on the ellipse $E$, let $P(m,n)$ ($-\\sqrt{2}\\leqslant n \\leqslant \\sqrt{2}$), then $\\frac{m^{2}}{4}+\\frac{n^{2}}{2}=1$, that is, $m^{2}=4-2n^{2}$. Here, $\\overrightarrow{F_{1}P}\\cdot\\overrightarrow{F_{2}P}=(m+\\sqrt{2},n)\\cdot(m-\\sqrt{2},n)=m^{2}-2+n^{2}=2-n^{2}$. Since $-\\sqrt{2}\\leqslant n \\leqslant \\sqrt{2}$, we have $0\\leqslant n^{2}\\leqslant 2$, $0\\leqslant 2-n^{2}\\leqslant 2$. Therefore, the range of $\\overrightarrow{F_{1}P}\\cdot\\overrightarrow{F_{2}P}$ is $[0,2]$."
  },
  {
    "text": "The line $y=\\sqrt{3} x$ is an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$. What is the eccentricity of the hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = sqrt(3)*x);OneOf(Asymptote(G)) = H",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[17, 63], [70, 73]], [[20, 63]], [[20, 63]], [[0, 16]], [[17, 63]], [[0, 16]], [[0, 69]]]",
    "query_spans": "[[[70, 79]]]",
    "process": "Analysis: Using the asymptote equation of the hyperbola, derive the relationship between $a$ and $b$, then solve for the eccentricity of the hyperbola. Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $y=\\sqrt{3}x$, we obtain $\\frac{b}{a}=\\sqrt{3}$, that is, $\\frac{c^{2}-a^{2}}{a^{2}}=3$, solving yields $e=2$."
  },
  {
    "text": "The line $y=2x$ and the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ have no intersection points; then the maximum value of the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;H: Line;b>0;Expression(G) = (x^2 - y^2/b^2 = 1);Expression(H) = (y = 2*x);NumIntersection(H, G)=0",
    "query_expressions": "Max(Eccentricity(G))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[10, 47], [53, 56]], [[13, 47]], [[0, 9]], [[13, 47]], [[10, 47]], [[0, 9]], [[0, 50]]]",
    "query_spans": "[[[53, 65]]]",
    "process": ""
  },
  {
    "text": "The equation of the hyperbola passing through the point $M(-6,3)$ and having the same asymptotes as the hyperbola $x^{2}-2 y^{2}=2$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - 2*y^2 = 2);M: Point;Coordinate(M) = (-6, 3);Z: Hyperbola;PointOnCurve(M, Z);Asymptote(G) = Asymptote(Z)",
    "query_expressions": "Expression(Z)",
    "answer_expressions": "x^2/18 - y^2/9 = 1",
    "fact_spans": "[[[13, 33]], [[13, 33]], [[1, 11]], [[1, 11]], [[41, 44]], [[0, 44]], [[12, 44]]]",
    "query_spans": "[[[41, 48]]]",
    "process": "Let the hyperbola equation be: $x^{2}-2y^{2}=\\lambda$. Since the hyperbola passes through point $M(-6,3)$, then: $\\lambda=x^{2}-2y^{2}=36-2\\times9=18$. Thus, the hyperbola equation is: $x^{2}-2y^{2}=18$, or $\\frac{x^{2}}{18}-\\frac{y^{2}}{9}=1$. ] The basic method for finding the standard equation of a hyperbola is the method of undetermined coefficients. The specific process is first to determine the form, then the quantities; that is, first determine the form of the standard hyperbola equation, and then use the relationships among $a,b,c,e$ and the asymptotes to find the values of $a$ and $b$. If the equations of the asymptotes of the hyperbola are known, to find the standard equation of the hyperbola, one can use the form of hyperbola equations sharing common asymptotes: $\\frac{x^{2}}{2}+\\frac{y^{2}}{2}=\\lambda(\\lambda\\neq0)$, and then determine the value of $2$ from the given conditions."
  },
  {
    "text": "Let the line $x - 3y + m = 0$ ($m \\neq 0$) intersect the two asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ ($a > 0$, $b > 0$) at points $A$ and $B$, respectively. If point $P(m, 0)$ satisfies $|PA| = |PB|$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;m: Number;Negation(m=0);P: Point;A: Point;B: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (m + x - 3*y = 0);Coordinate(P) = (m, 0);L1:Line;L2:Line;Asymptote(G)={L1,L2};Intersection(H,L1)=A;Intersection(H,L2)=B;Abs(LineSegmentOf(P, A)) = Abs(LineSegmentOf(P, B))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[25, 81], [129, 132]], [[28, 81]], [[28, 81]], [[1, 24]], [[3, 24]], [[3, 24]], [[101, 111]], [[91, 95]], [[96, 99]], [[28, 81]], [[28, 81]], [[25, 81]], [[1, 24]], [[101, 111]], [], [], [[25, 87]], [[1, 99]], [[1, 99]], [[113, 126]]]",
    "query_spans": "[[[129, 138]]]",
    "process": "From the equation of the hyperbola, the asymptotes are given by $ y = \\pm\\frac{b}{a}x $. Setting these equal to $ x - 3y + m = 0 $ ($ m \\neq 0 $) and solving, since point $ P(m, 0) $ satisfies $ |PA| = |PB| $, it follows that $ \\frac{3mb}{\\frac{3b^{2}-a}{a^{2}}-0} = -3 $, so $ a = 2b $, hence $ c = \\sqrt{5}b $, and therefore $ e = \\frac{c}{a} = \\frac{\\sqrt{5}}{2} $."
  },
  {
    "text": "The equation of the hyperbola that shares the same asymptotes as $\\frac{x^{2}}{4}-y^{2}=1$ and passes through the point $(2, \\sqrt{5})$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1);Asymptote(G) = Asymptote(G1);H: Point;Coordinate(H) = (2, sqrt(5));PointOnCurve(H, G1) = True;G1: Hyperbola",
    "query_expressions": "Expression(G1)",
    "answer_expressions": "y^2/4 - x^2/16 = 1",
    "fact_spans": "[[[1, 29]], [[1, 29]], [[0, 61]], [[41, 57]], [[41, 57]], [[39, 61]], [[58, 61]]]",
    "query_spans": "[[[58, 65]]]",
    "process": "Let the equation of the hyperbola sharing the same asymptotes with $\\frac{x^{2}}{4}-y^{2}=1$ be $x^{2}-4y^{2}=\\lambda$. Since this hyperbola passes through the point $(2,\\sqrt{5})$, we have $\\lambda=4-4\\times5=-16$. Therefore, the required hyperbola equation is $x^{2}-4y^{2}=-16$, which simplifies to $\\frac{y^{2}}{4}-\\frac{x^{2}}{16}=1$."
  },
  {
    "text": "Point $M$ is any point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{16}=1$. Then, the maximum distance from point $M$ to the line $x+y-7=0$ is?",
    "fact_expressions": "M: Point;PointOnCurve(M, G);G: Ellipse;Expression(G) = (x^2/9 + y^2/16 = 1);H: Line;Expression(H) = (x + y - 7 = 0)",
    "query_expressions": "Max(Distance(M, H))",
    "answer_expressions": "6*sqrt(2)",
    "fact_spans": "[[[0, 4], [49, 53]], [[0, 47]], [[5, 43]], [[5, 43]], [[54, 65]], [[54, 65]]]",
    "query_spans": "[[[49, 74]]]",
    "process": "Let the line $x + y = m$, parallel to the line $x + y - 7 = 0$, be tangent to the ellipse $\\frac{x^{2}}{9} + \\frac{y^{2}}{16} = 1$. Solving the system \n\\[\n\\begin{cases}\n\\frac{x^{2}}{9} + \\frac{y^{2}}{16} = 1 \\\\\nx + y = m\n\\end{cases}\n\\]\nyields $25x^{2} - 18mx + 9m^{2} - 144 = 0$. Then $A = (18k)^{2} - 4 \\times 25 \\times (9k^{2} - 144) = 0$, solving gives $k = 5$ or $k = -5$. According to the position relationship between the ellipse and the line $x + y - 7 = 0$, take the tangent line $x + y = 5$ farther from the line $x + y - 7 = 0$. At this time, the point of tangency $M$ is the point on the ellipse $\\frac{x^{2}}{9} + \\frac{y^{2}}{16} = 1$ farthest from the line $x + y - 7 = 0$, and the distance equals the distance between the two parallel lines $d = \\frac{|5 - 7|}{\\sqrt{2^{2} + 1^{2}}} = 6\\sqrt{2}$."
  },
  {
    "text": "Given that point $P$ is a point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, and the area of the triangle with vertices at the two foci and $P$ is $2 \\sqrt{5}$, find the coordinates of point $P$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};Area(TriangleOf(P, F1, F2)) = 2*sqrt(5)",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(0,pm*2)",
    "fact_spans": "[[[7, 44]], [[7, 44]], [[2, 6], [51, 54], [82, 86]], [[2, 48]], [], [], [[7, 57]], [[7, 80]]]",
    "query_spans": "[[[82, 90]]]",
    "process": ""
  },
  {
    "text": "The distance from any point on the parabola $y^{2}=4 x$ to the fixed line $l$: $x=-1$ is equal to its distance to the fixed point $F$. Then, the coordinates of the fixed point $F$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);D: Point;PointOnCurve(D, G) = True;l: Line;Expression(l) = (x = -1);Distance(D, l) = Distance(D, F);F: Point",
    "query_expressions": "Coordinate(F)",
    "answer_expressions": "(1, 0)",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[37, 38]], [[0, 18]], [[20, 33]], [[20, 33]], [[0, 49]], [[41, 44], [54, 57]]]",
    "query_spans": "[[[54, 62]]]",
    "process": "Since 2p=4, we have p=2, thus \\frac{p}{2}=1, so the focus coordinates are (1,0), that is, the fixed point coordinates are (1,0)"
  },
  {
    "text": "If a focus of the hyperbola $x^{2}-\\frac{y^{2}}{m}=1$ coincides with the focus of the parabola $y^{2}=8 x$, then the value of $m$ is?",
    "fact_expressions": "G: Hyperbola;m: Number;H: Parabola;Expression(G) = (x^2 - y^2/m = 1);Expression(H) = (y^2 = 8*x);OneOf(Focus(G)) = Focus(H)",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 29]], [[56, 59]], [[35, 49]], [[1, 29]], [[35, 49]], [[1, 54]]]",
    "query_spans": "[[[56, 63]]]",
    "process": ""
  },
  {
    "text": "If the equation $\\frac{x^{2}}{16-m}+\\frac{y^{2}}{m+4}=1$ represents an ellipse, then the range of real values for $m$ is?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (x^2/(16 - m) + y^2/(m + 4) = 1)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-4, 6) + (6, 16)",
    "fact_spans": "[[[45, 47]], [[49, 54]], [[1, 47]]]",
    "query_spans": "[[[49, 61]]]",
    "process": "Since the equation $\\frac{x^{2}}{16-m}+\\frac{y^{2}}{m+4}=1$ represents an ellipse, we have $\\begin{cases}16-m>0\\\\m+4>0\\end{cases}$ and $16-m\\neq m+4$. Solving gives $-4<m<16$ and $m\\neq6$. Therefore, the range of real number $m$ is $(-4,6)\\cup(6,16)$."
  },
  {
    "text": "Let point $P$ be an intersection point of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the circle $x^{2}+y^{2}=3 b^{2}$, and let $F_{1}$, $F_{2}$ be the left and right foci of the ellipse, respectively. If $|P F_{1}|=3|P F_{2}|$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Circle;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + y^2 = 3*b^2);OneOf(Intersection(G, H))=P;LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(14)/4",
    "fact_spans": "[[[6, 58], [105, 107], [139, 141]], [[8, 58]], [[8, 58]], [[59, 81]], [[1, 5]], [[87, 94]], [[95, 102]], [[8, 58]], [[8, 58]], [[6, 58]], [[59, 81]], [[1, 86]], [[87, 113]], [[87, 113]], [[115, 137]]]",
    "query_spans": "[[[139, 147]]]",
    "process": "Let PO = x, ∠POF₁ = θ. Applying the cosine law in △POF₁ and △POF₂ respectively, we obtain cosθ = (c² + x² - (3a/2)²)/(2×c×x) ①, -cosθ = (c² + x² - (a/2)²)/(2×c×x) ②. Solving ① and ② yields x² = (5a²)/4 - c² = 3b², so (7a²)/4 - 2c² = 0, hence e = √(7/8) = √14/4"
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the two foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, and let point $P$ lie on the ellipse. If the midpoint of segment $P F_{1}$ lies on the $y$-axis, then $|P F_{1}|$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/9 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);PointOnCurve(MidPoint(LineSegmentOf(P, F1)), yAxis) = True",
    "query_expressions": "Abs(LineSegmentOf(P, F1))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[19, 57], [68, 70]], [[19, 57]], [[1, 8]], [[9, 16]], [[1, 62]], [[63, 67]], [[63, 71]], [[73, 93]]]",
    "query_spans": "[[[95, 108]]]",
    "process": "Let $ P(x_{p}, y_{p}) $, midpoint $ m(0, n) $. From the given conditions, $ F_{1}(\\sqrt{7}, 0) $, $ F_{2}(-\\sqrt{7}, 0) $, $ a = 4 $, $ e = \\frac{\\sqrt{7}}{4} $. Since the midpoint of segment $ PF $ lies on the y-axis, we have $ \\frac{x_{p} + \\sqrt{7}}{2} = 0 $, so $ x_{p} = -\\sqrt{7} $. Substituting into $ \\frac{x^{2}}{16} + \\frac{y^{2}}{9} = 1 $, the coordinates of point $ P $ are $ (-\\sqrt{7}, \\frac{9}{4}) $ or $ (-\\sqrt{7}, -\\frac{9}{4}) $. According to the focal radius formula, $ |PF_{1}| = \\frac{23}{4} $, $ |PF_{2}| = \\frac{9}{4} $, $ \\frac{|PF_{1}|}{|PF_{2}|} = \\frac{23}{9} $."
  },
  {
    "text": "The standard equation of an ellipse with foci at $(-4 , 0)$ and $(4 , 0)$, and the sum of distances from any point $P$ on the ellipse to the two foci equal to $10$, is?",
    "fact_expressions": "G: Ellipse;H: Point;P: Point;D: Point;Coordinate(H) = (-4, 0);Coordinate(D) = (4, 0);PointOnCurve(P, G);Focus(G) = {H, D};Distance(P, H) + Distance(P, D) = 10",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/25 + y^2/9 = 1",
    "fact_spans": "[[[32, 34], [56, 58]], [], [[37, 40]], [], [[0, 58]], [[0, 58]], [[32, 40]], [[0, 58]], [[37, 58]]]",
    "query_spans": "[[[56, 64]]]",
    "process": ""
  },
  {
    "text": "Given two fixed points $P_{1}(1,0)$, $P_{2}(4,0)$, find the trajectory equation of a moving point $Q$ such that its distance to point $P_{2}$ is twice its distance to point $P_{1}$.",
    "fact_expressions": "P1: Point;P2: Point;Coordinate(P1) = (1, 0);Coordinate(P2) = (4, 0);Q: Point;Distance(Q, P2) = Distance(Q, P1)*2",
    "query_expressions": "LocusEquation(Q)",
    "answer_expressions": "x^2 + y^2 = 4",
    "fact_spans": "[[[5, 17], [47, 55]], [[19, 31], [34, 42]], [[5, 17]], [[19, 31]], [[66, 69]], [[33, 69]]]",
    "query_spans": "[[[66, 76]]]",
    "process": "Let the coordinates of the moving point Q be (x, y). Therefore, $ P_{2}Q = 2P_{1}Q $, that is, $ 2\\sqrt{(x-1)^{2}+y^{2}} = \\sqrt{(x-4)^{2}+y^{2}} $. Rearranging gives: $ x^{2}+y^{2} = 4 $, so the trajectory equation of the moving point Q is: $ x^{2}+y^{2} = 4 $."
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{m}=1$ with foci on the $y$-axis is $\\frac{1}{2}$, then $m$=?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/3 + y^2/m = 1);PointOnCurve(Focus(G), yAxis);Eccentricity(G) = (1/2)",
    "query_expressions": "m",
    "answer_expressions": "4",
    "fact_spans": "[[[9, 46]], [[66, 69]], [[9, 46]], [[0, 46]], [[9, 64]]]",
    "query_spans": "[[[66, 71]]]",
    "process": "From the given, we know that $a^{2}=m$, $b^{2}=3$, $\\therefore c^{2}=a^{2}-b^{2}=m-3$. $\\because e=\\frac{c}{a}=\\frac{1}{2}$, $\\therefore e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{m-3}{m}=1-\\frac{3}{m}=\\frac{1}{4}$, hence $m=4$."
  },
  {
    "text": "If the length of the imaginary axis of the hyperbola $3 x^{2}-y^{2}=m$ is $2$, then what is the value of the real number $m$?",
    "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (3*x^2 - y^2 = m);Length(ImageinaryAxis(G)) = 2",
    "query_expressions": "m",
    "answer_expressions": "{-3,1}",
    "fact_spans": "[[[1, 21]], [[31, 36]], [[1, 21]], [[1, 29]]]",
    "query_spans": "[[[31, 40]]]",
    "process": "Discuss the two cases $ m>0 $ and $ m<0 $ separately; based on the length of the imaginary axis of the hyperbola, the result can be obtained. Since the length of the imaginary axis of the hyperbola $ 3x^{2}-y^{2}=m $ is $ 2 $,  \n\\textcircled{1} when $ m>0 $, the hyperbola equation can be rewritten as $ \\frac{x^{2}}{3}-\\frac{y^{2}}{m}=1 $, from which $ \\sqrt{m}=1 $, yielding $ m=1 $;  \n\\textcircled{2} when $ m<0 $, the hyperbola equation can be rewritten as $ \\frac{y^{2}}{m}-\\frac{x^{2}}{-\\frac{m}{3}}=1 $, giving $ m=-3 $.  \nHence, the real values of $ m $ are $ -3 $ or $ 1 $."
  },
  {
    "text": "It is known that a line passing through the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ with an inclination angle of $45^{\\circ}$ intersects the right branch of the hyperbola at two points. Then, the range of the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(RightFocus(G), H);Inclination(H) = ApplyUnit(45, degree);NumIntersection(H, RightPart(G)) = 2",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1,\\sqrt{2})",
    "fact_spans": "[[[3, 62], [86, 89], [98, 101]], [[6, 62]], [[6, 62]], [[83, 85]], [[6, 62]], [[6, 62]], [[3, 62]], [[2, 85]], [[66, 85]], [[83, 96]]]",
    "query_spans": "[[[98, 112]]]",
    "process": "Since the line passing through the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ with an inclination angle of $45^\\circ$ intersects the right branch of the hyperbola at two points, the asymptote of the hyperbola $y=\\frac{b}{a}x$ has an inclination angle less than $45^{\\circ}$, so $\\frac{b}{a}<1$, that is, $b^{2}<a^{2}$, $c^{2}-a^{2}<a^{2}$, solving yields $1<e<\\sqrt{2}$."
  },
  {
    "text": "Draw a line through the focus of the parabola $y=\\frac{1}{4} x^{2}$ intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $y_{1}+y_{2}=2 \\sqrt {2}$, then the length of the chord $|AB|$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;Expression(G) = (y = x^2/4);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);x1:Number;x2:Number;y1:Number;y2:Number;PointOnCurve(Focus(G), H);Intersection(H, G) = {A,B};y1+y2=2*sqrt(2);IsChordOf(LineSegmentOf(A,B),G)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "2+2*sqrt(2)",
    "fact_spans": "[[[1, 25], [32, 35]], [[29, 31]], [[36, 54]], [[56, 74]], [[1, 25]], [[36, 54]], [[56, 74]], [[36, 54]], [[56, 74]], [[36, 54]], [[56, 74]], [[0, 31]], [[29, 76]], [[78, 103]], [[32, 113]]]",
    "query_spans": "[[[107, 117]]]",
    "process": ""
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{n+1}+\\frac{y^{2}}{n+5}=1$ represents an ellipse, what are the coordinates of the foci of this ellipse?",
    "fact_expressions": "G: Ellipse;Expression(G)=(x^2/(n + 1) + y^2/(n + 5) = 1) ;n:Number",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,pm*2)",
    "fact_spans": "[[[45, 47], [50, 52]], [[2, 47]], [[4, 43]]]",
    "query_spans": "[[[50, 59]]]",
    "process": "Since $n+5 > n+1$ by the given condition, the foci lie on the y-axis. Because $n+5 - n - 1 = 4$, the coordinates of the foci of the ellipse are $(0, \\pm 2)$."
  },
  {
    "text": "Given that the standard equation of the ellipse $C$ is $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, if a line $l$ passing through the point $P(2,1)$ is tangent to the ellipse $C$ in the first quadrant at point $M$, then what are the coordinates of point $M$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);P: Point;Coordinate(P) = (2, 1);l: Line;PointOnCurve(P, l);M: Point;TangentPoint(l, C) = M;Quadrant(M) = 1",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(1, 3/2)",
    "fact_spans": "[[[2, 7], [67, 72]], [[2, 48]], [[51, 60]], [[51, 60]], [[61, 66]], [[50, 66]], [[80, 84], [86, 90]], [[61, 84]], [[72, 84]]]",
    "query_spans": "[[[86, 95]]]",
    "process": "When the tangent point is in the first quadrant, the slope exists and is not zero. Let the equation of the tangent line be: $ y = kx + m $, where $ k < 0 $. Since it passes through point $ P $, we have: $ 1 = 2k + m $, ①  \nCombining the equations of line $ l $ and the ellipse:  \n\\[\n\\begin{cases}\ny = kx + m \\\\\n\\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1\n\\end{cases}\n\\]  \nRearranging gives: $ (3 + 4k^{2})x^{2} + 8kmx + 4m^{2} - 12 = 0 $.  \nThen, $ \\Delta = (8mk)^{2} - 4(3 + 4k^{2})(4m^{2} - 12) = 0 $, which yields $ m^{2} = 3 + 4k^{2} $ ②  \nFrom ① and ②, we obtain: $ k = -\\frac{1}{2} $, $ m = 2 $. Thus, the tangent line equation is: $ y = -\\frac{1}{2}x + 2 $.  \nThe rearranged equation becomes: $ x^{2} - 2x + 1 = 0 $, solving gives $ x = 1 $. Substituting into the tangent equation yields $ y = \\frac{3}{2} $, so the tangent point is $ M(1, \\frac{3}{2}) $. Therefore, the equation of line $ l $ is $ y = -\\frac{1}{2}x + 2 $, and the coordinates of tangent point $ M $ are $ (1, \\frac{3}{2}) $."
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{3+k}+\\frac{y^{2}}{2-k}=1$ represents an ellipse, what is the range of real values for $k$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(k + 3) + y^2/(2 - k) = 1);k: Real",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-3, -1/2)+(-1/2, 2)",
    "fact_spans": "[[[45, 47]], [[2, 47]], [[49, 54]]]",
    "query_spans": "[[[49, 61]]]",
    "process": ""
  },
  {
    "text": "If a point $P(a, b)$ on the right branch of the hyperbola $x^{2}-y^{2}=1$ is at a distance of $\\sqrt{2}$ from the line $y=x$, then the value of $a+b$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;P: Point;a: Number;b: Number;Expression(G) = (x^2 - y^2 = 1);Expression(H) = (y = x);Coordinate(P) = (a, b);PointOnCurve(P, RightPart(G));Distance(P, H) = sqrt(2)",
    "query_expressions": "a + b",
    "answer_expressions": "1/2",
    "fact_spans": "[[[1, 19]], [[36, 43]], [[26, 35]], [[59, 64]], [[59, 64]], [[1, 19]], [[36, 43]], [[26, 35]], [[1, 35]], [[26, 57]]]",
    "query_spans": "[[[59, 68]]]",
    "process": "Since point P(a,b) lies on the hyperbola, then a^{2}-b^{2}=1, that is, (a+b)(a-b)=1. Also, the distance from point P to the line y=x is d=\\frac{|a-b|}{\\sqrt{2}}=\\sqrt{2}, so |a-b|=2. Since point P is on the right branch, then a>b, so a-b=2. Therefore, (a+b)\\times2=1, so a+b=\\frac{1}{2}"
  },
  {
    "text": "The minimum distance from a moving point $P$ on the parabola $y^{2}=2x$ to the line $y=2x+1$ is?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Expression(G) = (y^2 = 2*x);Expression(H) = (y = 2*x + 1);PointOnCurve(P,G)",
    "query_expressions": "Min(Distance(P, H))",
    "answer_expressions": "3*sqrt(5)/20",
    "fact_spans": "[[[0, 14]], [[22, 33]], [[18, 21]], [[0, 14]], [[22, 33]], [[0, 21]]]",
    "query_spans": "[[[18, 41]]]",
    "process": "Since point P lies on the parabola $ y^{2} = 2x $, it can be expressed as $ P\\left(\\frac{y_{0}^{2}}{2}, y_0\\right) $. The distance from this point to the line $ y = 2x + 1 $ is $ \\frac{|y_{0}^{2} - y_{0} + 1|}{\\sqrt{5}} $. The minimum value occurs when $ y_{0} = \\frac{1}{2} $, and the minimum value is $ \\frac{3\\sqrt{5}}{20} $."
  },
  {
    "text": "Given that $F$ is a focus of the ellipse $C$, $B$ is an endpoint of the minor axis, the extension of line segment $BF$ intersects $C$ at point $D$, and $\\overrightarrow{B F}=2 \\overrightarrow{F D}$, then the eccentricity of $C$ is?",
    "fact_expressions": "B: Point;F: Point;C: Ellipse;D: Point;OneOf(Focus(C)) = F;OneOf(Endpoint(MinorAxis(C)))=B;Intersection(OverlappingLine(LineSegmentOf(B, F)), C) = D;VectorOf(B, F) = 2*VectorOf(F, D)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[17, 20]], [[2, 5]], [[6, 11], [40, 43], [97, 100]], [[44, 48]], [[2, 16]], [[6, 28]], [[29, 48]], [[50, 95]]]",
    "query_spans": "[[[97, 106]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1$, and point $P$ lies on $C$ such that $|P F_{1}|=2|P F_{2}|$, then $\\cos \\angle F_{1} P F_{2}$=?",
    "fact_expressions": "C: Hyperbola;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/2 - y^2/2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Cos(AngleOf(F1, P, F2))",
    "answer_expressions": "3/4",
    "fact_spans": "[[[18, 61], [73, 76]], [[68, 72]], [[2, 9]], [[10, 17]], [[18, 61]], [[2, 67]], [[2, 67]], [[68, 77]], [[78, 100]]]",
    "query_spans": "[[[102, 131]]]",
    "process": "According to the definition of hyperbola, |PF_{1}|-|PF_{2}|=2\\sqrt{2}; combining with the given conditions, we obtain |PF_{1}| and |PF_{2}|; then by the cosine law, \\cos\\angle F_{1}PF_{2} can be derived. From the problem, we have \\begin{cases} |PF_{1}|-|PF_{2}|=2\\sqrt{2}, \\\\ |PF_{1}|=2|PF_{2}|, \\end{cases} solving gives: |PF_{1}|=4\\sqrt{2}, |PF_{2}|=2\\sqrt{2}. Since |F_{1}F_{2}|=2\\sqrt{2+2}=4, therefore \\cos\\angle F_{1}PF_{2}=\\frac{|PF_{1}|^{2}+|PF_{2}|^{2}-|F_{1}F_{2}|^{2}}{2|PF_{1}|\\cdot|PF_{2}|}=\\frac{32+8-16}{2\\times4\\sqrt{2}\\times2\\sqrt{2}}=\\frac{3}{4}"
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{b^{2}}=1(b>0)$ and the hyperbola $\\frac{x^{2}}{8}-y^{2}=1$ have common foci, then $b$=?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;b: Number;Expression(G) = (x^2/8 - y^2 = 1);b>0;Expression(H) = (x^2/25 + y^2/b^2 = 1);Focus(H) = Focus(G)",
    "query_expressions": "b",
    "answer_expressions": "4",
    "fact_spans": "[[[48, 76]], [[0, 47]], [[84, 87]], [[48, 76]], [[2, 47]], [[0, 47]], [[0, 82]]]",
    "query_spans": "[[[84, 89]]]",
    "process": "From the given condition, the $ c^{2} $ values of the two curves are equal, $\\therefore 25 - b^{2} = 8 + 1$, solving yields $ b^{2} = 16 $, and since $ b > 0 $, then $ b = 4 $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ with focus $F$, point $M(-\\frac{p}{2}, 0)$, a line passing through point $F$ intersects the parabola at points $A$ and $B$. If $|A B|=12$ and $\\tan \\angle A M B=2 \\sqrt{2}$, then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;M: Point;A: Point;B: Point;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(M) = (-p/2, 0);Focus(C) = F;PointOnCurve(F, G);Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, B)) = 12 ;Tan(AngleOf(A, M, B)) = 2*sqrt(2)",
    "query_expressions": "p",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 28], [68, 71]], [[129, 132]], [[64, 66]], [[36, 57]], [[73, 76]], [[77, 80]], [[32, 35], [59, 63]], [[10, 28]], [[2, 28]], [[36, 57]], [[2, 35]], [[58, 66]], [[64, 82]], [[84, 94]], [[97, 127]]]",
    "query_spans": "[[[129, 134]]]",
    "process": "Let the line $ AB: x = my + \\frac{p}{2} $, and let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Solving the system \n\\[\n\\begin{cases}\nx = my + \\frac{p}{2} \\\\\ny^{2} = 2px\n\\end{cases}\n\\]\nand simplifying yields: $ y^{2} - 2mpy - p^{2} = 0 $, from which we get $ y_{1} + y_{2} = 2mp $, $ y_{1}y_{2} = -p^{2} $. Thus,\n\\[\nk_{AM} + k_{BM} = \\frac{y_{1}(my_{2} + p) + y_{2}(my_{1} + p)}{(my_{1} + p)(my_{2} + p)} = \\frac{2m \\cdot (-p^{2}) + p \\cdot (2mp)}{(my_{1} + p)(my_{2} + p)} = 0,\n\\]\nso we obtain $ \\angle AMF = \\angle BMF $, hence\n\\[\n\\tan \\angle AMB = \\frac{2 \\tan \\angle AMF}{1 - \\tan^{2} \\angle AMF} = 2\\sqrt{2}.\n\\]\nSince $ \\angle AMF $ is acute, solving gives $ \\tan \\angle AMF = \\frac{\\sqrt{2}}{2} $. Assume $ AF > BF $. As shown in the figure, draw $ AH \\perp x $-axis intersecting at $ H $. By the given condition, $ M $ lies on the directrix of the parabola. Draw the directrix $ l $, and draw $ AA' \\perp l $ with foot $ A' $. Then\n\\[\n\\tan \\angle AMF = \\frac{AH}{MH} = \\frac{AH}{AA'} = \\frac{AH}{AF} = \\sin \\angle AFH = \\frac{\\sqrt{2}}{2},\n\\]\nso $ \\angle AFH = \\frac{\\pi}{4} $, thus $ m = 1 $. Therefore,\n\\[\n|AB| = \\sqrt{1 + m^{2}} |y_{1} - y_{2}| = \\sqrt{(1 + m^{2})[(y_{1} + y_{2})^{2} - 4y_{1}y_{2}]} = 4p = 12,\n\\]\nso $ p = 3 $."
  },
  {
    "text": "Given the parabola $C$: $x^{2}=2 p y(p>0)$ with focus $F$, point $P(x_{0}, \\frac{1}{2})$ lies on $C$, and $|P F|=\\frac{3}{4}$, then $p=?$",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 2*(p*y));p: Number;p>0;F: Point;Focus(C) = F;x0: Number;P: Point;Coordinate(P) = (x0, 1/2);PointOnCurve(P, C);Abs(LineSegmentOf(P, F)) = 3/4",
    "query_expressions": "p",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 28], [61, 64]], [[2, 28]], [[88, 91]], [[10, 28]], [[32, 35]], [[2, 35]], [[37, 60]], [[36, 60]], [[36, 60]], [[36, 65]], [[67, 86]]]",
    "query_spans": "[[[88, 93]]]",
    "process": "By the focal radius formula, |PF| = y_{0} + \\frac{p}{2} = \\frac{1}{2} + \\frac{p}{2} = \\frac{3}{4}, solving gives p = \\frac{1}{2}"
  },
  {
    "text": "The focus of the parabola $y^{2}=2 p x(p>0)$ is $F$. A line passing through $F$ and perpendicular to the $x$-axis intersects the parabola at points $A$ and $B$. If $|A B|=3$, then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;A: Point;B: Point;F: Point;p>0;Expression(G) = (y^2 = 2*p*x);Focus(G) = F;PointOnCurve(F, H);IsPerpendicular(H,xAxis);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 3",
    "query_expressions": "p",
    "answer_expressions": "3/2",
    "fact_spans": "[[[0, 21], [45, 48]], [[71, 74]], [[42, 44]], [[49, 52]], [[53, 56]], [[25, 28], [30, 33]], [[3, 21]], [[0, 21]], [[0, 28]], [[29, 44]], [[34, 44]], [[42, 58]], [[60, 69]]]",
    "query_spans": "[[[71, 76]]]",
    "process": "Since the line passing through the focus F and perpendicular to the x-axis intersects the parabola at points A and B, AB is the latus rectum, so 2p=3, solving gives p=\\frac{3}{7}."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, respectively, $P$ a point on the ellipse, $M$ the midpoint of $F_{1}P$, and $|OM|=3$. Then the distance from point $P$ to the left focus of the ellipse is?",
    "fact_expressions": "F1: Point;F2: Point;G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G) = True;O: Origin;M: Point;MidPoint(LineSegmentOf(F1, P)) = M;Abs(LineSegmentOf(O, M)) = 3",
    "query_expressions": "Distance(P, LeftFocus(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 8]], [[9, 16]], [[19, 58], [69, 71], [108, 110]], [[19, 58]], [[1, 64]], [[1, 64]], [[65, 68], [103, 107]], [[65, 74]], [[92, 101]], [[75, 78]], [[75, 91]], [[92, 101]]]",
    "query_spans": "[[[103, 118]]]",
    "process": "According to the problem, OM is the midline of $\\triangle PF_{1}F_{2}$, $\\because OM=3$, $\\therefore |PF_{2}|=6$, $\\because PF\\perp|PF_{2}|=2a=10$, $\\therefore |PF|=4$."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $x^{2}+\\frac{y^{2}}{b^{2}}=1$ $(0<b<1)$, respectively. A line with slope $1$ passing through $F_{1}$ intersects the ellipse at points $A$ and $B$, and $|A B|=\\frac{4}{3}$. Then the value of $b$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + y^2/b^2 = 1);b: Number;0 < b;b < 1;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Line;Slope(H) = 1;PointOnCurve(F1, H);A: Point;B: Point;Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 4/3",
    "query_expressions": "b",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[19, 59], [84, 86]], [[19, 59]], [[121, 124]], [[21, 59]], [[21, 59]], [[1, 8], [67, 74]], [[9, 16]], [[1, 65]], [[1, 65]], [[81, 83]], [[74, 83]], [[66, 83]], [[89, 92]], [[93, 96]], [[81, 98]], [[100, 119]]]",
    "query_spans": "[[[121, 128]]]",
    "process": "The coordinates of point $F_{1}$ can be obtained from the ellipse equation. Since the slope of line $l$ passing through $F_{1}$ is 1, the equation of line $l$ can be determined. By solving the system of equations of line $l$ and the ellipse, and using the distance formula between two points, the value of $b$ can be obtained. [Detailed solution] According to the given conditions, $F_{1}(-\\sqrt{1-b^{2}},0)$, $F_{2}(\\sqrt{1-b^{2}},0)$. Since the slope of line $l$ passing through $F_{1}$ is 1, the equation of line $l$ is: $y = x + \\sqrt{1-b^{2}}$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$. Solving the system of equations of line $l$ and the ellipse, eliminating $y$ and simplifying yields: $(1+b^{2})x^{2}+2\\sqrt{1-b^{2}}x+1-2b^{2}=0$, $\\frac{\\sqrt{3}}{2}$. Solving gives: $b=\\sqrt{2}$"
  },
  {
    "text": "It is known that the focus $F$ of the parabola $y^{2}=2 p x(p>0)$ is exactly the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, and the hyperbola passes through the point $(\\frac{3 a^{2}}{p}, \\frac{2 b^{2}}{p})$. Then the equations of the asymptotes of this hyperbola are?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;p: Number;F: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);p>0;Expression(H) = (y^2 = 2*p*x);Focus(H)=F;RightFocus(G)=F;P:Point;Coordinate(P)=(3*a^2/p,2*b^2/p);PointOnCurve(P,G)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(10)/4)*x",
    "fact_spans": "[[[31, 77], [83, 86], [131, 134]], [[34, 77]], [[34, 77]], [[2, 23]], [[5, 23]], [[25, 28]], [[31, 77]], [[5, 23]], [[2, 23]], [[2, 28]], [[25, 81]], [[87, 128]], [[87, 128]], [[83, 128]]]",
    "query_spans": "[[[131, 142]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, point $P$ lies on the hyperbola. If $P F_{2} \\perp F_{1} F_{2}$ and $\\angle P F_{1} F_{2}=30^{\\circ}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(F1, F2),LineSegmentOf(P, F2));AngleOf(P,F1,F2)=ApplyUnit(30,degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[20, 76], [89, 92], [160, 163]], [[23, 76]], [[23, 76]], [[84, 88]], [[2, 9]], [[10, 17]], [[23, 76]], [[23, 76]], [[20, 76]], [[2, 83]], [[2, 83]], [[84, 93]], [[95, 123]], [[125, 158]]]",
    "query_spans": "[[[160, 169]]]",
    "process": "Assume without loss of generality that point P lies on the right branch of the hyperbola, then |PF_{1}|-|PF_{2}|=2a30^{\\circ}, hence |PF_{1}|=2|PF_{2}|. Thus |PF_{1}|=4a, |PF_{2}|=2a, and \\tan\\angle PF_{1}F_{2}=\\frac{|PF_{2}|}{|F_{1}F_{2}|}=\\frac{2a}{2c}=\\frac{\\sqrt{3}}{3}, hence e=\\frac{c}{a}=\\sqrt{3}"
  },
  {
    "text": "Given point $P(-3,3)$, draw a line through point $M(3,0)$ intersecting the parabola $y^{2}=4x$ at points $A$ and $B$. Let the slopes of lines $PA$ and $PB$ be $k_{1}$ and $k_{2}$ respectively, then $k_{1}+k_{2}=$?",
    "fact_expressions": "P: Point;Coordinate(P) = (-3, 3);M: Point;Coordinate(M) = (3, 0);L: Line;PointOnCurve(M,L) = True;G: Parabola;Expression(G) = (y^2 = 4*x);Intersection(L,G) = {A,B};A: Point;B: Point;Slope(LineOf(P,A)) = k1;k1: Number;Slope(LineOf(P,B)) = k2;k2: Number",
    "query_expressions": "k1 + k2",
    "answer_expressions": "-1",
    "fact_spans": "[[[2, 12]], [[2, 12]], [[14, 23]], [[14, 23]], [[24, 26]], [[13, 26]], [[28, 42]], [[28, 42]], [[24, 54]], [[45, 48]], [[49, 52]], [[56, 94]], [[77, 84]], [[56, 94]], [[87, 94]]]",
    "query_spans": "[[[96, 111]]]",
    "process": ""
  },
  {
    "text": "Given $a > b > 0$, the equation of ellipse $C_{1}$ is $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1$, the equation of hyperbola $C_{2}$ is $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$, and the product of the eccentricities of $C_{1}$ and $C_{2}$ is $\\frac{\\sqrt{3}}{2}$. Then, what is the equation of the asymptotes of $C_{2}$?",
    "fact_expressions": "a: Number;b: Number;a > b;b > 0;C1: Ellipse;Expression(C1) = (y^2/b^2 + x^2/a^2 = 1);C2: Hyperbola;Expression(C2) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(C1)*Eccentricity(C2) = sqrt(3)/2",
    "query_expressions": "Expression(Asymptote(C2))",
    "answer_expressions": "x \\pm \\sqrt{2} \\cdot y = 0",
    "fact_spans": "[[[2, 9]], [[2, 9]], [[2, 9]], [[2, 9]], [[10, 19], [127, 134]], [[10, 66]], [[67, 77], [135, 142], [171, 178]], [[67, 124]], [[127, 169]]]",
    "query_spans": "[[[171, 186]]]",
    "process": "The eccentricity of ellipse $C_{1}$ is $\\frac{\\sqrt{a^{2}-b^{2}}}{a}$, and the eccentricity of hyperbola $C_{2}$ is $\\frac{\\sqrt{a^{2}+b^{2}}}{a}$. Therefore, $\\frac{\\sqrt{a^{2}-b^{2}}}{a} \\cdot \\frac{\\sqrt{a^{2}+b^{2}}}{a} = \\frac{\\sqrt{3}}{2}$, which implies $a^{4} = 4b^{4}$, so $a = \\sqrt{2}b$. Thus, the asymptotes of hyperbola $C_{2}$ are $y = \\pm \\frac{1}{\\sqrt{5}}x$, or $x + \\sqrt{2}y = 0$."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=2 p x$ ($p>0$) is $F(1,0)$, and the line $l$: $y=x+m$ intersects the parabola at two distinct points $A$, $B$. If $0 \\leq m<1$, then the maximum area of $\\Delta F A B$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;Coordinate(F) = (1, 0);l: Line;Expression(l) = (y = m + x);A: Point;B: Point;Intersection(l, G) = {A, B};m: Number;m >= 0;m < 1;Negation(A=B)",
    "query_expressions": "Max(Area(TriangleOf(F, A, B)))",
    "answer_expressions": "8*sqrt(6)/9",
    "fact_spans": "[[[2, 23], [51, 54]], [[2, 23]], [[5, 23]], [[5, 23]], [[27, 35]], [[2, 35]], [[27, 35]], [[36, 50]], [[36, 50]], [[61, 64]], [[65, 68]], [[36, 68]], [[71, 83]], [[71, 83]], [[71, 83]], [[36, 68]]]",
    "query_spans": "[[[85, 108]]]",
    "process": "Since the focus of the parabola is (1,0), we have p=2, and the equation of the parabola is y^{2}=4x. Solving the system \\begin{cases}y=x+m\\\\y^{2}=4x\\end{cases}, we obtain x^{2}+(2m-4)x+m^{2}=0, with x_{1}+x_{2}=4-2m, x_{1}\\cdot x_{2}=m^{2}. Since the line and the parabola have two intersection points, the discriminant A=(2m-4)^{2}-4m^{2}>0, solving gives m<1. By the chord length formula, |AB|=\\sqrt{1+1}\\sqrt{(4-2m)^{2}-4m^{2}}=4\\sqrt{2}\\cdot\\sqrt{1-m}. The distance from the focus (1,0) to the line x-y+m=0 is d=\\frac{|1+m|}{\\sqrt{2}}. Thus, the area of triangle FAB is \\frac{1}{2}\\cdot4\\sqrt{2}\\cdot\\sqrt{1-m}\\cdot\\frac{|1+m|}{\\sqrt{2}}=2\\sqrt{(1-m)}\\cdot|1+m|. Since 0\\leqslant m<1, the expression simplifies to 2\\sqrt{(1-m)(1+m)^{2}}=2\\sqrt{-m^{3}-m^{2}+m+1}. Let f(m)=-m^{3}-m^{2}+m+1 (0\\leqslant m<1), then f^{'}(m)=-3m^{2}-2m+1=-(3m-1)(m+1). Therefore, when m\\in(0,\\frac{1}{3}), the function increases, and when m\\in(\\frac{1}{3},1), the function decreases. Hence, the maximum occurs at m=\\frac{1}{3}, at which time"
  },
  {
    "text": "Let the standard equation of the ellipse be $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, then the eccentricity of this ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3/5",
    "fact_spans": "[[[1, 3], [48, 50]], [[1, 45]]]",
    "query_spans": "[[[48, 56]]]",
    "process": "In the ellipse $\\frac{x^2}{25}+\\frac{y^{2}}{16}=1$, $a=5$, $b=4$, then $c=\\sqrt{a^{2}-b^{2}}=3$. Therefore, the eccentricity of this ellipse is $e=\\frac{c}{a}=\\frac{3}{5}$."
  },
  {
    "text": "The slope of an asymptote of the hyperbola $x^{2}+ky^{2}=1$ is $2$, then the value of $k$ is?",
    "fact_expressions": "G: Hyperbola;k: Number;Expression(G) = (k*y^2 + x^2 = 1);Slope(OneOf(Asymptote(G)))=2",
    "query_expressions": "k",
    "answer_expressions": "-1/4",
    "fact_spans": "[[[0, 19]], [[34, 37]], [[0, 19]], [[0, 32]]]",
    "query_spans": "[[[34, 41]]]",
    "process": "Since $ k < 0 $, the asymptotes are $ x \\pm \\sqrt{-k}y = 0 $. Therefore, $ \\frac{1}{\\sqrt{-k}} = 2 $, solving gives $ k = -\\frac{1}{4} $."
  },
  {
    "text": "For the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ satisfying $a \\leq \\sqrt{3}$, if the eccentricity is $e$, then the minimum value of $e^{2}+\\frac{1}{e^{2}}$ is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;a > b;b > 0;e:Number;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a <= sqrt(3);Eccentricity(G)=e",
    "query_expressions": "Min(e^2 + 1/(e^2))",
    "answer_expressions": "13/6",
    "fact_spans": "[[[0, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[78, 81]], [[0, 52]], [[55, 72]], [[0, 81]]]",
    "query_spans": "[[[83, 112]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b<0)$, the right vertex and right focus are denoted as $A$ and $F$, respectively. The intersection point of its left directrix with the $x$-axis is $B$. If $A$ is the midpoint of segment $B F$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;F: Point;B: Point;A: Point;a>0;b<0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightVertex(C) = A;RightFocus(C) = F;Intersection(LeftDirectrix(C), xAxis) = B;MidPoint(LineSegmentOf(B,F)) = A",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)+1",
    "fact_spans": "[[[2, 64], [118, 124], [83, 84]], [[10, 64]], [[10, 64]], [[79, 82]], [[97, 100]], [[75, 78], [102, 105]], [[10, 64]], [[10, 64]], [[2, 64]], [[2, 82]], [[2, 82]], [[83, 100]], [[102, 116]]]",
    "query_spans": "[[[118, 130]]]",
    "process": ""
  },
  {
    "text": "Point $P(a , b)$ lies on the right branch of the hyperbola $x^{2}-y^{2}=1$, and the distance from $P$ to the asymptotes is $\\sqrt{2}$. Then $a+b=$?",
    "fact_expressions": "P: Point;a: Number;b: Number;G: Hyperbola;Coordinate(P) = (a, b);Expression(G) = (x^2 - y^2 = 1);PointOnCurve(P, RightPart(G));Distance(P, Asymptote(G)) = sqrt(2)",
    "query_expressions": "a + b",
    "answer_expressions": "{2, 1/2}",
    "fact_spans": "[[[0, 11], [37, 40]], [[1, 11]], [[1, 11]], [[12, 30]], [[0, 11]], [[12, 30]], [[0, 35]], [[12, 57]]]",
    "query_spans": "[[[59, 66]]]",
    "process": ""
  },
  {
    "text": "A line segment $AB$ of length $3$ has its endpoints $A$ and $B$ moving on the $x$-axis and $y$-axis respectively. A moving point $C(x, y)$ satisfies $\\overrightarrow{AC} = 2 \\overrightarrow{CB}$. Then, what is the trajectory equation of the moving point $C$?",
    "fact_expressions": "A: Point;B: Point;C: Point;x1: Number;y1: Number;Coordinate(C) = (x1, y1);Length(LineSegmentOf(A, B)) = 3;Endpoint(LineSegmentOf(A, B)) = {A, B};PointOnCurve(A, xAxis);PointOnCurve(B, yAxis);VectorOf(A, C) = 2*VectorOf(C, B)",
    "query_expressions": "LocusEquation(C)",
    "answer_expressions": "x^2 + y^2/4 = 1",
    "fact_spans": "[[[16, 20]], [[22, 25]], [[44, 54], [105, 108]], [[44, 54]], [[44, 54]], [[44, 54]], [[0, 13]], [[6, 25]], [[16, 39]], [[16, 39]], [[56, 101]]]",
    "query_spans": "[[[105, 115]]]",
    "process": ""
  },
  {
    "text": "Given that the line $y = x + m$ is intersected by the ellipse $4x^{2} + y^{2} = 1$ to form a chord of length $\\frac{2\\sqrt{2}}{5}$, find the value of $m$.",
    "fact_expressions": "G: Ellipse;H: Line;m: Number;Expression(G) = (4*x^2 + y^2 = 1);Expression(H) = (y = m + x);Length(InterceptChord(H,G))=2*sqrt(2)/5",
    "query_expressions": "m",
    "answer_expressions": "pm*1",
    "fact_spans": "[[[12, 31]], [[2, 11]], [[61, 64]], [[12, 31]], [[2, 11]], [[2, 59]]]",
    "query_spans": "[[[61, 68]]]",
    "process": "Substituting the line $ y = x + m $ into the ellipse equation gives: $ 4x^{2} + y^{2} = 1 $, that is: $ 5x^{2} + 2mx + m^{2} - 1 = 0 $. Let the line intersect the ellipse at two points $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1}, x_{2} $ are the two roots of the equation $ 5x^{2} + 2mx + m^{2} - 1 = 0 $. By Vieta's formulas, $ \\frac{2}{2} = \\frac{2m}{5} $, $ x_{1}x_{2} = \\frac{m^{2} - 1}{5} $, $ \\therefore |AB| = \\sqrt{2} \\sqrt{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}} = \\sqrt{\\frac{4m^{2} - 4}{5}} = \\frac{2\\sqrt{2}}{5} \\therefore m = \\pm \\frac{\\sqrt{5}}{2} $"
  },
  {
    "text": "Two fixed points $A(-2,0)$, $B(2,0)$ and a fixed line $l$: $x=\\frac{10}{3}$. Point $P$ is a moving point on $l$. A perpendicular to $BP$ is drawn through $B$, intersecting $AP$ at point $Q$. Then the equation of the trajectory of point $Q$ is?",
    "fact_expressions": "l: Line;B: Point;P: Point;A: Point;Q:Point;Coordinate(A) = (-2, 0);Coordinate(B) = (2, 0);Expression(l)=(x=10/3);PointOnCurve(P, l);L:Line;PointOnCurve(B,L);IsPerpendicular(LineSegmentOf(B,P),L);Intersection(L,LineSegmentOf(A,P))=Q",
    "query_expressions": "LocusEquation(Q)",
    "answer_expressions": "(x^2/4+y^2=1)&Negation(x=-2)",
    "fact_spans": "[[[27, 48], [54, 57]], [[15, 23], [64, 67]], [[49, 53]], [[3, 13]], [[84, 88], [90, 94]], [[3, 13]], [[15, 23]], [[27, 48]], [[49, 62]], [], [[63, 76]], [[63, 76]], [[63, 88]]]",
    "query_spans": "[[[90, 101]]]",
    "process": "Let $ P\\left(\\frac{10}{3}, y_{1}\\right) $. If $ y_{1} = 0 $, then $ BP: x = 0 $, the slope of $ BP $ is 0; at this time, drawing a perpendicular line to $ BP $ through $ B $ intersects $ AP $ at point $ Q $, so the coordinates of point $ Q $ are $ (2, 0) $; if $ y_{1} \\neq 0 $, then $ k_{BP} = \\frac{y_{1}}{\\frac{10}{3} - 2} = \\frac{3y_{1}}{4} \\neq 0 $, so the slope of the perpendicular line to $ BP $ through $ B $ is $ \\frac{-1}{\\frac{3y_{1}}{4}} = -\\frac{4}{3y_{1}} $, while $ k_{AP} = \\frac{y_{1}}{\\frac{10}{3} + 2} = \\frac{3y_{1}}{16} \\neq 0 $. Let $ Q(x, y) $ ($ x \\neq \\pm 2 $), $ k_{BQ} = \\frac{y}{x - 2} $, $ k_{AQ} = \\frac{y}{x + 2} $. According to points $ A $, $ Q $, $ P $ being collinear, and $ AQ \\perp BQ $, we obtain $ \\frac{3y}{16} = \\frac{y}{x + 2} $, $ \\frac{3y}{4} \\cdot \\frac{y}{x - 2} = -1 $, so $ 4y^{2} = 4 - x^{2} \\Rightarrow \\frac{x^{2}}{4} + y^{2} = 1 $. In summary, the trajectory equation of point $ Q $ is $ \\frac{x^{2}}{4} + y^{2} = 1 $ ($ x \\neq -2 $)."
  },
  {
    "text": "Given point $Q(-2 \\sqrt{2} , 0)$ and a moving point $P(x , y)$ on the parabola $x^{2}=-4 y$, find the minimum value of $|y|+| P Q |$.",
    "fact_expressions": "G: Parabola;Q: Point;P: Point;x0: Number;y0: Number;Expression(G) = (x^2 = -4*y);Coordinate(Q) = (-2*sqrt(2), 0);Coordinate(P) = (x0, y0);PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, Q)) + Abs(y0))",
    "answer_expressions": "2",
    "fact_spans": "[[[24, 39]], [[2, 23]], [[43, 53]], [[43, 53]], [[43, 53]], [[24, 39]], [[2, 23]], [[43, 53]], [[24, 53]]]",
    "query_spans": "[[[55, 74]]]",
    "process": "The parabola's focus is $ F(0,-1) $, and the directrix of the parabola is $ y=1 $. Then $ |y| + |PQ| = d - 1 + |PQ| = |PF| + |PQ| - 1 \\geqslant |QF| - 1 $, and the answer is obtained by calculation. As shown in the figure, the parabola's focus is $ F(0,-1) $, and the directrix of the parabola is $ y=1 $. Let $ d $ be the distance from point $ P $ to the directrix, then $ |y| + |PQ| = d - 1 + |PQ| = |PF| + |PQ| - 1 \\geqslant |QF| - 1 = \\sqrt{8} $. The equality holds when points $ P $, $ Q $, and $ F $ are collinear; thus, the minimum value of $ |y| + |PQ| $ is $ 2 $."
  },
  {
    "text": "Let the line $l$ with slope $1$ pass through the focus $F$ of the parabola $y^{2}=a x$ ($a>0$), and intersect the $y$-axis at point $A$. If the area of $\\triangle O A F$ (where $O$ is the origin) is $8$, then the value of $a$ is?",
    "fact_expressions": "l: Line;G: Parabola;a: Number;O: Origin;A: Point;F: Point;a>0;Expression(G) = (y^2 = a*x);Slope(l) = 1;Focus(G) = F;PointOnCurve(F, l);Intersection(l, yAxis) = A;Area(TriangleOf(O, A, F)) = 8",
    "query_expressions": "a",
    "answer_expressions": "16",
    "fact_spans": "[[[8, 13]], [[14, 33]], [[90, 93]], [[72, 75]], [[48, 52]], [[36, 39]], [[17, 33]], [[14, 33]], [[1, 13]], [[14, 39]], [[8, 39]], [[8, 52]], [[54, 88]]]",
    "query_spans": "[[[90, 97]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with left and right foci $F_{1}$, $F_{2}$. A line $l$ passing through point $F_{1}$ intersects the left branch of hyperbola $C$ at points $A$, $B$. The area of $\\triangle B F_{1} F_{2}$ is three times the area of $\\triangle A F_{1} F_{2}$, and $\\angle F_{1} A F_{2}=90^{\\circ}$. Then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;B: Point;F1: Point;F2: Point;A: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, l);Intersection(l, LeftPart(C)) = {A, B};Area(TriangleOf(B, F1, F2)) = 3*Area(TriangleOf(A, F1, F2));AngleOf(F1, A, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(10)/2",
    "fact_spans": "[[[96, 101]], [[2, 63], [102, 108], [218, 224]], [[10, 63]], [[10, 63]], [[117, 120]], [[70, 77], [87, 95]], [[78, 85]], [[113, 116]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 85]], [[2, 85]], [[86, 101]], [[96, 122]], [[123, 182]], [[183, 216]]]",
    "query_spans": "[[[218, 230]]]",
    "process": "Since the area of $\\triangle BF_{1}F_{2}$ is three times the area of $\\triangle AF_{1}F_{2}$, and $\\angle F_{1}AF_{2}=90^{\\circ}$, we obtain $\\frac{|AF_{1}|}{|AB|}=\\frac{1}{4}$. Let $|AF_{1}|=m$, $|BF_{1}|=3m$, then $|AF_{2}|=m+2a$, $|BF_{2}|=3m+2a$. From $|AB|^{2}+|AF_{2}|^{2}=|BF_{2}|^{2}$, solving gives $m=a$, so $|AF_{1}|=a$, $|AF_{2}|=3a$. Then from $|AF_{1}|^{2}+|AF_{2}|^{2}=|F_{1}F_{2}|^{2}$, we get $10a^{2}=4c^{2}$. Therefore, the eccentricity of hyperbola $C$ is $e=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\frac{\\sqrt{10}}{2}$."
  },
  {
    "text": "The distance from point $M(1,1)$ to the directrix of the parabola $y=a x^{2}$ is $2$, then $a$=?",
    "fact_expressions": "G: Parabola;a: Number;M: Point;Expression(G) = (y = a*x^2);Coordinate(M) = (1, 1);Distance(M,Directrix(G))=2",
    "query_expressions": "a",
    "answer_expressions": "{-1/12, 1/4}",
    "fact_spans": "[[[10, 24]], [[37, 40]], [[0, 9]], [[10, 24]], [[0, 9]], [[0, 34]]]",
    "query_spans": "[[[37, 42]]]",
    "process": "From the standard equation of the parabola: $x^{2}=\\frac{1}{a}y$, the focus coordinates of the parabola are $(0,\\frac{1}{4a})$ and the directrix equation is: $y=-\\frac{1}{4a}$. Given that the distance from $M(1,1)$ to the directrix of the parabola $y=ax^{2}$ is 2, then $\\left|1-\\left(-\\frac{1}{4a}\\right)\\right|=2$, solving gives: $a=\\frac{1}{4}$ or $a=-\\frac{1}{12}$, $a=\\frac{1}{4}$ or $a=-\\frac{1}{1}$."
  },
  {
    "text": "The standard equation of a circle whose center lies on the parabola $y=\\frac{1}{2} x^{2}$ and is tangent to both the directrix of the parabola and the $y$-axis is?",
    "fact_expressions": "H: Circle;G: Parabola;Expression(G) = (y = x^2/2);PointOnCurve(Center(H), G);IsTangent(Directrix(G), H);IsTangent(yAxis, H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x+pm*1)^2+(y-(1/2))^2=1",
    "fact_spans": "[[[49, 50]], [[3, 27], [34, 37]], [[3, 27]], [[0, 50]], [[32, 50]], [[32, 50]]]",
    "query_spans": "[[[49, 57]]]",
    "process": "\\because the center lies on the parabola y=\\frac{1}{2}x^{2}, \\therefore let P(t,\\frac{1}{2}t^{2}) be the center, and the directrix is y=-\\frac{1}{2}. \\because it is tangent to the directrix of the parabola and the y-axis, \\therefore |t|=\\frac{1}{2}t^{2}+\\frac{1}{2}. \\therefore t=\\pm1. \\therefore the standard equation of the circle is (x\\pm1)^{2}+(y-\\frac{1}{2})^{2}=1"
  },
  {
    "text": "What is the equation of the circle centered at the focus of the parabola $y=\\frac{1}{4} x^{2}$ and tangent to an asymptote of the hyperbola $\\frac{y^{2}}{4}-x^{2}=1$?",
    "fact_expressions": "G: Hyperbola;H: Parabola;C:Circle;Expression(G) = (-x^2 + y^2/4 = 1);Expression(H) = (y = x^2/4);Center(C)=Focus(H);IsTangent(OneOf(Asymptote(G)),C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2+(y-1)^2=1/5",
    "fact_spans": "[[[34, 62]], [[1, 25]], [[71, 72]], [[34, 62]], [[1, 25]], [[0, 72]], [[33, 72]]]",
    "query_spans": "[[[71, 76]]]",
    "process": "From the given conditions, the parabola is $x^{2}=4y$, so the focus coordinates are $(0,1)$. The asymptotes of the hyperbola $\\frac{y^{2}}{4}-x^{2}=1$ are $y=\\pm2x$. Without loss of generality, take $y=2x$, i.e., $2x-y=0$. Let the radius of the circle be $r$. From the conditions, we have $r=\\frac{|2\\times0-1|}{\\sqrt{2^{2}+1^{2}}}=\\frac{\\sqrt{5}}{5}$. Therefore, the equation of the circle is: $x^{2}+(y-1)^{2}=\\frac{1}{5}$."
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{k}=1$ is $\\frac{2}{3}$, then the real number $k$=?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (x^2/4 + y^2/k = 1);Eccentricity(G) = 2/3",
    "query_expressions": "k",
    "answer_expressions": "{20/9, 36/5}",
    "fact_spans": "[[[0, 37]], [[57, 62]], [[0, 37]], [[0, 55]]]",
    "query_spans": "[[[57, 64]]]",
    "process": "From the given condition, we have $\\frac{b^{2}}{a^{2}}=\\frac{5}{9}$. By classifying the position of the ellipse's foci, we can derive an equation involving the real number $k$, from which the range of values for the real number $k$ can be determined. Solution: Since $e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{a^{2}-b^{2}}{a^{2}}=1-\\frac{b^{2}}{a^{2}}=\\frac{4}{9}$, we obtain $\\frac{b^{2}}{a^{2}}=\\frac{5}{9}$. If the foci of the ellipse lie on the $x$-axis, then $\\frac{b^{2}}{a^{2}}=\\frac{k}{4}=\\frac{5}{9}$, solving gives $k=\\frac{20}{9}$. If the foci of the ellipse lie on the $y$-axis, then $\\frac{b^{2}}{a^{2}}=\\frac{4}{k}=\\frac{5}{9}$, solving gives $k=\\frac{36}{5}$. In conclusion, $k=\\frac{20}{9}$ or $\\frac{36}{5}$."
  },
  {
    "text": "Let the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ be $F_{1}$ and $F_{2}$, respectively. Given point $M\\left(0, \\frac{\\sqrt{15}}{2} b\\right)$, the segment $M F_{2}$ intersects the ellipse at point $P$, and $O$ is the origin. If $|P O|+|P F_{1}|=2 a$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;a: Number;b: Number;F2: Point;M: Point;P: Point;O: Origin;F1: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(M) = (0, b*(sqrt(15)/2));LeftFocus(G) = F1;RightFocus(G) = F2;Intersection(LineSegmentOf(M,F2),G)=P;Abs(LineSegmentOf(P,O))+Abs(LineSegmentOf(P,F1))=2*a",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[1, 53], [124, 126], [166, 168]], [[3, 53]], [[3, 53]], [[71, 78]], [[81, 111]], [[127, 131]], [[132, 135]], [[63, 70]], [[3, 53]], [[3, 53]], [[1, 53]], [[81, 111]], [[1, 78]], [[1, 78]], [[112, 131]], [[141, 162]]]",
    "query_spans": "[[[166, 174]]]",
    "process": "According to the definition of an ellipse, |PF₂| + |PF₁| = 2a. Also, since |PO| + |PF₁| = 2a, it follows that |PF₂| = |PO|. Since triangle MOF₂ is a right triangle, point P is the midpoint of MF₂. Given F₂(c,0), then P( c/2, (√15 b)/4 ). Substituting point P into the ellipse equation gives (c²)/(a²) + ((√15 b)²)/(b²) = 1 ⇒ c²/a² = 1/4 ⇒ e = 1/2"
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the two foci of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, and let point $P$ lie on the hyperbola such that $\\angle F_{1} P F_{2}=90^{\\circ}$. Then the area of $\\triangle F_{1} P F_{2}$ is?",
    "fact_expressions": "F1: Point;F2: Point;Focus(G) = {F1, F2};G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1);P: Point;PointOnCurve(P, G);AngleOf(F1,P,F2) = ApplyUnit(90,degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 8]], [[9, 16]], [[1, 50]], [[17, 45], [56, 59]], [[17, 45]], [[51, 55]], [[51, 60]], [[64, 97]]]",
    "query_spans": "[[[99, 129]]]",
    "process": ""
  },
  {
    "text": "Given that $AB$ is the diameter of the circle $O$: $x^{2}+y^{2}=1$, and point $P$ is a moving point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, then the minimum value of $\\overrightarrow{PA} \\cdot \\overrightarrow{PB}$ is?",
    "fact_expressions": "G: Ellipse;O: Circle;A: Point;B: Point;P: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Expression(O) = (x^2 + y^2 = 1);IsDiameter(LineSegmentOf(A, B), O);PointOnCurve(P, G)",
    "query_expressions": "Min(DotProduct(VectorOf(P, A), VectorOf(P, B)))",
    "answer_expressions": "2",
    "fact_spans": "[[[38, 75]], [[8, 29]], [[2, 7]], [[2, 7]], [[33, 37]], [[38, 75]], [[8, 29]], [[2, 32]], [[33, 79]]]",
    "query_spans": "[[[81, 136]]]",
    "process": "Method 1: Based on symmetry, without loss of generality, assume diameter AB lies on the x-axis, P(2\\cosx,\\sqrt{3}\\sinx), A(-1,0), B(1,0). Thus, \\overrightarrow{PA}\\cdot\\overrightarrow{PB}=(2\\cosx-1)(2\\cosx+1)+3\\sin^{2}x=2+\\cos^{2}x\\geqslant2"
  },
  {
    "text": "Given the parabola $y^{2}=8x$, a line passing through the focus $F$ intersects the parabola at points $A$ and $B$. The x-coordinate of the midpoint of segment $AB$ is $2$. Then $|AB|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);Focus(G) = F;F: Point;H: Line;PointOnCurve(F, H) = True;Intersection(H, G) = {A, B};A: Point;B: Point;XCoordinate(MidPoint(LineSegmentOf(A,B))) = 2",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 15], [26, 29]], [[1, 15]], [[1, 22]], [[19, 22]], [[23, 25]], [[16, 25]], [[23, 39]], [[30, 33]], [[34, 37]], [[40, 58]]]",
    "query_spans": "[[[60, 69]]]",
    "process": "From the parabola equation, its directrix is $x = -2$. Let $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$, then $\\frac{x_{1}+x_{2}}{2} = 2$, that is, $x_{1} + x_{2} = 4$. By the definition of the parabola, $|AF| = x_{1} + 2$, $|BF| = x_{2} + 2$, so $|AB| = |AF| + |BF| = (x_{1} + 2) + (x_{2} + 2) = (x_{1} + x_{2}) + 4 = 8$."
  },
  {
    "text": "Let the right focus $F$ of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ be the center of a circle with radius equal to the semi-focal distance. This circle intersects the asymptotes of the hyperbola at points $O$, $A$, and $B$. If the perimeter of $\\triangle AOB$ is $7a$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G:Circle;b: Number;a: Number;A: Point;O: Origin;B: Point;F: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;Center(G)=F;Radius(G)=HalfFocalLength(C);Intersection(G,Asymptote(C))={O,A,B};Perimeter(TriangleOf(A, O, B)) = 7*a",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "4*sqrt(7)/7",
    "fact_spans": "[[[1, 62], [83, 86], [136, 142]], [[80, 81]], [[9, 62]], [[9, 62]], [[96, 99]], [[92, 95]], [[100, 103]], [[66, 69]], [[9, 62]], [[9, 62]], [[1, 62]], [[1, 69]], [[66, 81]], [[1, 81]], [[80, 105]], [[108, 134]]]",
    "query_spans": "[[[136, 148]]]",
    "process": "\\because one focus of the hyperbola is F(c,0), and one asymptote of the hyperbola is y=\\frac{b}{a}x, i.e., bx-ay=0, \\therefore the distance from the focus to the asymptote is d=\\frac{|bc|}{\\sqrt{a^{2}+b^{2}}}=\\frac{bc}{c}=b. Thus, the distance from the center of the circle to the asymptote is b. Since the radius of the circle is c, |OA|=2\\sqrt{c^{2}-b^{2}}=2a. Similarly, |OB|=2a. Because \\tan\\angle AOx=\\frac{b}{a}, it follows that \\sin\\angle AOx=\\frac{b}{c}. Therefore, |AB|=2|OA|\\sin\\angle AOx=2\\times2a\\times\\frac{b}{c}=\\frac{4ab}{c}. Hence, 2a+2a+\\frac{4ab}{c}=7a, yielding 4b=3c. Thus, 16(c^{2}-a^{2})=9c^{2}, solving gives e=\\frac{c}{a}=\\frac{4\\sqrt{7}}{7}"
  },
  {
    "text": "The coordinates of the focus of the parabola $y^{2}+8 x=0$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (8*x + y^2 = 0)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(-2, 0)",
    "fact_spans": "[[[0, 16]], [[0, 16]]]",
    "query_spans": "[[[0, 23]]]",
    "process": "\\because the parabola equation is y^{2}=-8x, \\therefore the focus lies on the x-axis, p=4, \\therefore the focus coordinates are (-2,0)"
  },
  {
    "text": "The chord $AB$ of the parabola $x^{2}=4 y$ passes through the focus $F$, and the length of $AB$ is $6$. What is the ordinate of the midpoint $M$ of $AB$?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;F: Point;Expression(G) = (x^2 = 4*y);Focus(G) = F;PointOnCurve(F, LineSegmentOf(A, B));IsChordOf(LineSegmentOf(A, B), G);Length(LineSegmentOf(A, B)) = 6;MidPoint(LineSegmentOf(A, B)) = M;M: Point",
    "query_expressions": "YCoordinate(M)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 14]], [[16, 21]], [[16, 21]], [[24, 27]], [[0, 14]], [[0, 27]], [[16, 27]], [[0, 21]], [[29, 40]], [[42, 53]], [[50, 53]]]",
    "query_spans": "[[[50, 58]]]",
    "process": "Given p=2, let A(x_{1},y_{1}), B(x_{2},y_{2}), then |AB|=y_{1}+y_{2}+p=6, y_{1}+y_{2}=4, so \\frac{y_{1}+y_{2}}{2}=2"
  },
  {
    "text": "Given the parabola $C$: $x^{2}=4 y$ with focus $F$, and a line $AB$ intersecting the parabola $C$ at points $A$ and $B$. If $2 \\overrightarrow{O A}+\\overrightarrow{O B}-3 \\overrightarrow{O F}=\\overrightarrow{0}$, then the distance from the midpoint of chord $AB$ to the directrix of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;B: Point;A: Point;O: Origin;F: Point;Expression(C) = (x^2 = 4*y);Focus(C) = F;Intersection(LineOf(A, B), C) = {A, B};2*VectorOf(O, A) + VectorOf(O, B) - 3*VectorOf(O, F) = 0;IsChordOf(LineSegmentOf(A, B), C)",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A, B)), Directrix(C))",
    "answer_expressions": "9/4",
    "fact_spans": "[[[2, 21], [37, 43], [156, 162]], [[50, 53]], [[46, 49]], [[57, 144]], [[25, 28]], [[2, 21]], [[2, 28]], [[29, 55]], [[57, 144]], [[29, 152]]]",
    "query_spans": "[[[146, 170]]]",
    "process": "According to the problem, the parabola $ x^{2} = 4y $ has focus $ F(0,1) $ and directrix $ y = -1 $. Since $ 2\\overrightarrow{OA} + \\overrightarrow{OB} - 3\\overrightarrow{OF} = \\overrightarrow{0} $, i.e., $ 2(\\overrightarrow{OA} - \\overrightarrow{OF}) + (\\overrightarrow{OB} - \\overrightarrow{OF}) = \\overrightarrow{0} $, it follows that $ 2\\overrightarrow{FA} + \\overrightarrow{FB} = \\overrightarrow{0} $, so points $ F, A, B $ are collinear. From the conditions, the slope of line $ AB $ exists and is non-zero; let the equation of line $ AB $ be $ y = kx + 1 $ ($ k \\neq 0 $). From $ \\begin{cases} y = kx + 1 \\\\ x^{2} = 4y \\end{cases} $, eliminating $ y $ gives $ x^{2} - 4kx - 4 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1}x_{2} = -4 $, $\\textcircled{1}$ and from $ 2\\overrightarrow{FA} + \\overrightarrow{FB} = \\overrightarrow{0} $ we get $ 2x_{1} + x_{2} = 0 $, $\\textcircled{2}$ thus $ x^{2}_{1} = 2 $. Let $ M $ be the midpoint of chord $ AB $, and draw perpendiculars from $ A, B, M $ to the directrix with feet $ A_{1}, B_{1}, M_{1} $, respectively. Then $ |MM_{1}| = \\frac{1}{2}(|AA_{1}| + |BB_{1}|) = \\frac{1}{2}[(y_{1}+1)+(y_{2}+1)] = \\frac{1}{2}(y_{1}+y_{2}) + 1 = \\frac{1}{8}(x_{1}^{2} + x_{2}^{2}) + 1 = \\frac{1}{8}(x_{1}^{2} + \\frac{16}{x_{1}^{2}}) + 1 = \\frac{1}{8}(2 + 8) + 1 = \\frac{9}{4} $. The answer is: $ \\frac{9}{4} $"
  },
  {
    "text": "Given that $P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1$, and an asymptote of the hyperbola is $3x - y = 0$. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola, respectively. If $|P F_{2}|=3$, then $|P F_{1}|=$?",
    "fact_expressions": "G:Hyperbola;Expression(G) = (-y^2/9 + x^2/a^2 = 1);a: Number;P: Point;PointOnCurve(P, RightPart(G));Expression(OneOf(Asymptote(G))) = (3*x - y = 0);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(P, F2)) = 3",
    "query_expressions": "Abs(LineSegmentOf(P, F1))",
    "answer_expressions": "5",
    "fact_spans": "[[[6, 48], [55, 58], [97, 100]], [[6, 48]], [[9, 48]], [[2, 5]], [[2, 54]], [[55, 76]], [[78, 85]], [[87, 94]], [[78, 106]], [[78, 106]], [[108, 121]]]",
    "query_spans": "[[[123, 137]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+\\frac{y^{2}}{8}=1$, and points $A(a, 2a)$, $B(a-2,2a-4)$, if the line segment $AB$ has a common point with the ellipse $C$, then the range of real values for $a$ is?",
    "fact_expressions": "C: Ellipse;A: Point;B: Point;a: Real;Expression(C) = (x^2/2 + y^2/8 = 1);Coordinate(A) = (a, 2*a);Coordinate(B) = (a - 2, 2*a - 4);IsIntersect(LineSegmentOf(A,B), C)",
    "query_expressions": "Range(a)",
    "answer_expressions": "[-1, 3]",
    "fact_spans": "[[[2, 44], [83, 88]], [[45, 57]], [[58, 73]], [[94, 99]], [[2, 44]], [[45, 57]], [[58, 73]], [[75, 92]]]",
    "query_spans": "[[[94, 106]]]",
    "process": "Since points A and B both lie on the line $ l: y = 2x $, solving the system of the line and the ellipse \n\\[\n\\begin{cases}\n\\frac{x^{2}}{2} + \\frac{y^{2}}{8} = \\\\\ny = 2x\n\\end{cases}\n\\]\ngives the intersection points of $ l $ and the ellipse $ C $ as $ M(1,2) $, $ N(-1,-2) $, and $ |MN| = |AB| = 2\\sqrt{5} $. Thus, as long as point A or point B lies on the segment $ MN $, the segment $ AB $ will intersect the ellipse. Therefore, $ -1 \\leqslant a \\leqslant 1 $ or $ -1 \\leqslant a - 2 \\leqslant 1 $, solving yields $ -1 \\leqslant a \\leqslant 3 $."
  },
  {
    "text": "If the length of the major axis of an ellipse is twice the length of the minor axis, then the eccentricity of this ellipse is?",
    "fact_expressions": "G: Ellipse;Length(MajorAxis(G))=2*Length(MinorAxis(G))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[4, 6], [22, 24]], [[4, 17]]]",
    "query_spans": "[[[22, 30]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ lies on an ellipse $C$ centered at the origin, the right focus of the ellipse is $F_{2}(\\sqrt{5}, 0)$, and the perpendicular bisector of segment $P F_{2}$ is $y=2 x$, then the equation of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;O: Origin;Center(C) = O;P: Point;PointOnCurve(P, C);F2: Point;Coordinate(F2) = (sqrt(5), 0);RightFocus(C) = F2;Expression(PerpendicularBisector(LineSegmentOf(P, F2))) = (y = 2*x)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/9+y^2/4=1",
    "fact_spans": "[[[15, 20], [26, 28], [81, 86]], [[10, 14]], [[7, 20]], [[2, 6]], [[2, 24]], [[33, 53]], [[33, 53]], [[26, 53]], [[54, 79]]]",
    "query_spans": "[[[81, 91]]]",
    "process": "Point P is on the ellipse C centered at the origin, and the right focus of the ellipse is F_{2}(\\sqrt{5},0), so c=\\sqrt{5}. The equation of the line perpendicular to PF_{2} passing through F_{2} is: y=-\\frac{1}{2}(x-\\sqrt{5}), or x+2y-\\sqrt{5}=0. The distance from F_{2} to the perpendicular bisector y=2x is: \\frac{2\\sqrt{5}}{\\sqrt{5}}=2. The distance from the origin to the line x+2y-\\sqrt{5}=0 is: a=2+1=3, so b=2. Then the equation of the ellipse C is \\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1"
  },
  {
    "text": "Given three distinct points $A$, $B$, $C$ on the parabola $y^{2}=2 p x(p>0)$, with focus $F$, satisfying $\\overrightarrow{F A}+\\overrightarrow{F B}+\\overrightarrow{F C}=0$. If the equation of the line containing side $B C$ is $4 x+y-20=0$, then $p=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;A: Point;B: Point;C: Point;Negation(A=B);Negation(A=C);Negation(B=C);PointOnCurve(A, G);PointOnCurve(B, G);PointOnCurve(C, G);F: Point;Focus(G) = F;VectorOf(F, A) + VectorOf(F, B) + VectorOf(F, C) = 0;Expression(OverlappingLine(LineSegmentOf(B, C))) = (4*x + y - 20 = 0)",
    "query_expressions": "p",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 23], [43, 46]], [[2, 23]], [[153, 156]], [[5, 23]], [[30, 34]], [[35, 38]], [[39, 42]], [[25, 42]], [[25, 42]], [[25, 42]], [[2, 42]], [[2, 42]], [[2, 42]], [[50, 53]], [[43, 53]], [[57, 123]], [[126, 151]]]",
    "query_spans": "[[[153, 158]]]",
    "process": "From \\begin{cases}4x+y-2\\\\y^2=2px\\end{cases}=0 we obtain $2y^{2}+py-20p=0$. From $\\triangle>0$, we have $p>0$, or $p<-160$. Let $B(x_{1},y_{1})$, $C(x_{2},y_{2})$, then $y_{1}+y_{2}=-\\frac{p}{2}$, $\\therefore x_{1}+x_{2}=(5-\\frac{y_{1}}{4})+(5-\\frac{y_{2}}{4})=10+\\frac{p}{8}$. Let $A(x_{3},y_{3})$, the focus of the parabola be $F$, and satisfy $\\overrightarrow{FA}+\\overrightarrow{FB}+\\overrightarrow{FC}=\\overrightarrow{0}$, $\\therefore (x_{1}-\\frac{p}{2},y_{1})+(x_{2}-\\frac{p}{2},y_{2})+(x_{3}-\\frac{p}{2},y_{3})=0$, $\\therefore x_{1}+x_{2}+x_{3}=\\frac{3p}{2}$, $y_{1}+y_{2}+y_{3}=0$, $\\therefore x_{3}=\\frac{11}{8}p-10$, $y_{3}=$. Since point $A$ lies on the parabola, $\\therefore (\\frac{k}{4})^{2}=2p(\\frac{11}{8}p-10)$, $\\therefore p=8$."
  },
  {
    "text": "What are the coordinates of the focus of the parabola $y=\\frac{1}{a} x^{2}(a \\neq 0)$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = (1/a)*x^2);a: Number;Negation(a=0)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, a/4)",
    "fact_spans": "[[[0, 34]], [[0, 34]], [[3, 34]], [[3, 34]]]",
    "query_spans": "[[[0, 41]]]",
    "process": ""
  },
  {
    "text": "Given $A(3 , 1)$, $B(-4 , 0)$, and $P$ is a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, then the maximum value of $P A+P B$ is?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;P: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Coordinate(A) = (3, 1);Coordinate(B) = (-4, 0);PointOnCurve(P, G)",
    "query_expressions": "Max(LineSegmentOf(P, A) + LineSegmentOf(P, B))",
    "answer_expressions": "10 + sqrt(2)",
    "fact_spans": "[[[33, 71]], [[2, 12]], [[15, 26]], [[29, 32]], [[33, 71]], [[2, 12]], [[15, 26]], [[29, 75]]]",
    "query_spans": "[[[77, 92]]]",
    "process": "According to the problem, for the ellipse $ a=5 $, $ b=3 $, $ c=4 $, thus $ B $ is the left focus of the ellipse. Let the right focus of the ellipse be $ F(4,0) $. By the definition of an ellipse, $ |PA| + |PB| = |PA| + 2a - |PF| = 2a + |PA| - |PF| $. From the triangle inequality that the difference of two sides is less than the third side, we have $ |PA| - |PF| \\leqslant |AF| $. Hence, the maximum value of $ |PA| + |PB| $ is $ 2a + |AF| = 10 + \\sqrt{(3-4)^{2}+}\\frac{1}{2} $"
  },
  {
    "text": "What is the distance from the focus to the directrix of the parabola $y=4 x^{2}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 4*x^2)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "1/8",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 25]]]",
    "process": ""
  },
  {
    "text": "Given that $M$ and $N$ are two points on the parabola $y^{2}=8x$, $O$ is the origin, and $\\angle MON=90^{\\circ}$, then the minimum value of $|MN|$ is?",
    "fact_expressions": "G: Parabola;M: Point;O: Origin;N: Point;Expression(G) = (y^2 = 8*x);PointOnCurve(M,G);PointOnCurve(N,G);AngleOf(M, O, N) = ApplyUnit(90, degree)",
    "query_expressions": "Min(Abs(LineSegmentOf(M, N)))",
    "answer_expressions": "16",
    "fact_spans": "[[[10, 24]], [[2, 5]], [[28, 31]], [[6, 9]], [[10, 24]], [[2, 27]], [[2, 27]], [[38, 63]]]",
    "query_spans": "[[[65, 78]]]",
    "process": ""
  },
  {
    "text": "The two foci of ellipse $C$ are $F_{1}(-2,0)$ and $F_{2}(2,0)$, respectively, the eccentricity is $e=\\frac{1}{2}$, point $P$ lies on ellipse $C$, and $\\angle F_{1} P F_{2}=30^{\\circ}$. Then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "C: Ellipse;F1: Point;F2: Point;Focus(C)={F1,F2};Coordinate(F1) = (-2, 0);Coordinate(F2) = (2, 0);Eccentricity(C)=e;e:Number;e=1/2;PointOnCurve(P, C);AngleOf(F1, P, F2) = ApplyUnit(30, degree);P:Point",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "24-12*sqrt(3)",
    "fact_spans": "[[[2, 7], [69, 74]], [[15, 28]], [[31, 43]], [[2, 43]], [[15, 28]], [[31, 43]], [[2, 63]], [[48, 63]], [[48, 63]], [[64, 75]], [[77, 110]], [[64, 68]]]",
    "query_spans": "[[[112, 139]]]",
    "process": "From the definition of the ellipse, |F_{1}P|+|PF_{2}|=8. By the cosine law, \\cos\\angle F_{1}PF_{2}=\\frac{|F_{1}P|^{2}+|PF_{2}|^{2}-|F_{1}F_{2}|^{2}}{2|F_{1}P|\\times|PF_{2}|}. Combining these gives the value of |F_{1}P|\\times|PF_{2}|, leading to the answer. Given c=2, e=\\frac{1}{2}, so a=4. From the ellipse definition, |F_{1}P|+|PF_{2}|=2\\times4=8. By the cosine law, \\cos\\angle F_{1}PF_{2}=\\frac{|F_{1}P|^{2}+|PF_{2}|^{2}-|F_{1}F_{2}|^{2}}{2|F_{1}P|\\times|PF_{2}|}=\\cos30^{\\circ}=\\frac{\\sqrt{3}}{2}, that is, (|F_{1}P|+|PF_{2}|)^{2}-2|F_{1}P|\\times|PF_{2}|-16=\\sqrt{3}|F_{1}P|\\times|PF_{2}|, so |F_{1}P|\\times|PF_{2}|=\\frac{48}{2+\\sqrt{3}}. Then the area of \\triangle F_{1}PF_{2} is S=\\frac{1}{2}\\times|F_{1}P|\\times|PF_{2}|\\sin30^{\\circ}=\\frac{1}{2}\\times\\frac{48}{2+\\sqrt{3}}\\times\\frac{1}{2}=24-12\\sqrt{3}."
  },
  {
    "text": "The standard equation of a hyperbola passing through $(3,-6)$ and having the same asymptotes as the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/2 = 1);Z: Hyperbola;H: Point;Coordinate(H) = (3, -6);PointOnCurve(H, Z);Asymptote(G) = Asymptote(Z)",
    "query_expressions": "Expression(Z)",
    "answer_expressions": "y^2/18 - x^2/9 = 1",
    "fact_spans": "[[[11, 39]], [[11, 39]], [[46, 49]], [[1, 9]], [[1, 9]], [[0, 49]], [[10, 49]]]",
    "query_spans": "[[[46, 56]]]",
    "process": "\\because with the hyperbola \\(x^{2}-\\frac{y^{2}}{2}=1\\) having the same asymptotes, \\therefore let the hyperbola equation be \\(x^{2}-\\frac{y^{2}}{2}=\\lambda\\ (\\lambda\\neq0)\\). Substituting \\((3,-6)\\), we get \\(3^{2}-\\frac{(-6)^{2}}{2}=\\lambda\\). \\therefore \\(\\lambda=-9\\), \\therefore the required standard equation of the hyperbola is \\(\\frac{y^{2}}{18}-\\frac{x^{2}}{9}=1\\)."
  },
  {
    "text": "If a hyperbola centered at the origin has foci on the $y$-axis and an eccentricity of $\\sqrt{3}$, then what is the equation of the asymptotes of this hyperbola?",
    "fact_expressions": "G: Hyperbola;O: Origin;Center(G) = O;PointOnCurve(Focus(G), yAxis) = True;Eccentricity(G) = sqrt(3)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(2)/2)*x",
    "fact_spans": "[[[16, 19], [36, 39]], [[4, 6]], [[1, 19]], [[7, 19]], [[16, 33]]]",
    "query_spans": "[[[36, 47]]]",
    "process": "Let the hyperbola equation be \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1, then its asymptotes are y=\\pm\\frac{a}{b}x. \\because e=\\sqrt{3}, \\therefore e^{2}-\\frac{b^{2}}{a^{2}}=3-\\frac{b^{2}}{a^{2}}=1, solving gives \\frac{b}{a}=\\sqrt{2}, \\therefore \\frac{a}{b}=\\frac{\\sqrt{2}}{2}. \\therefore The asymptotes of the hyperbola are y=\\pm\\frac{\\sqrt{2}}{2}x"
  },
  {
    "text": "Given that $O$ is the coordinate origin, $A(-5,0)$, $B(5,0)$, point $P$ satisfies $|P A|+|P B|=14$, and point $P(x_0, y_0)$ also satisfies $\\sqrt{(x+5)^{2}+y^{2}}-\\sqrt{(x-5)^{2}+y^{2}}=2$, then the coordinates of point $P$ are?",
    "fact_expressions": "A: Point;B: Point;P: Point;O: Origin;G: Curve;x0: Number;y0: Number;Coordinate(A) = (-5, 0);Coordinate(B) = (5, 0);Coordinate(P) = (x0, y0);Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B)) = 14;PointOnCurve(P, G);Expression(G) = (sqrt((x + 5)^2 + y^2) - sqrt((x - 5)^2 + y^2) = 2)",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(7/5, pm*24/5)",
    "fact_spans": "[[[11, 20]], [[23, 31]], [[32, 36], [55, 67], [121, 125]], [[2, 5]], [[70, 119]], [[56, 67]], [[56, 67]], [[11, 20]], [[23, 31]], [[55, 67]], [[38, 54]], [[55, 119]], [[70, 119]]]",
    "query_spans": "[[[121, 130]]]",
    "process": "Given points A(-5,0), B(5,0), point P satisfies |PA| + |PB| = 14 > |AB|, 2a = 14, c = 5 ⇒ b² = a² - c² = 24. By the definition of an ellipse, point P lies on the ellipse \\frac{x^{2}}{49} + \\frac{y^{2}}{24} = 1. Also, point P(x,y) satisfies \\sqrt{(x+5)^{2}+y^{2}} - \\sqrt{(x-5)^{2}+y^{2}} = 2 ⇒ |PA| - |PB| = 2 < |AB|. By the definition of a hyperbola, point P lies on x^{2} - \\frac{y^{2}}{24} = 1 (x \\geqslant 1). Solving the ellipse and hyperbola equations simultaneously gives y = \\pm\\frac{24}{5}, x = \\frac{7}{5}. Therefore, the coordinates of point P are (\\frac{7}{5}, \\pm\\frac{24}{5})."
  },
  {
    "text": "Given $A(-1,0)$, $B(2,0)$, there exists a point $M$ on the line $l$: $x+2 y+a=0$ such that $M A^{2}+2 M B^{2}=10$. Then the range of real values for $a$ is?",
    "fact_expressions": "l: Line;A: Point;B: Point;M: Point;a: Real;Expression(l)=(x+2*y+a=0);Coordinate(A) = (-1, 0);Coordinate(B) = (2, 0);PointOnCurve(M,l);LineSegmentOf(M, A)^2 + 2*LineSegmentOf(M, B)^2 = 10",
    "query_expressions": "Range(a)",
    "answer_expressions": "[-1-(2*sqrt(15)/3),(2*sqrt(15)/3)-1]",
    "fact_spans": "[[[23, 41]], [[2, 11]], [[12, 21]], [[44, 48]], [[77, 82]], [[23, 41]], [[2, 11]], [[12, 21]], [[23, 48]], [[52, 74]]]",
    "query_spans": "[[[77, 89]]]",
    "process": "Let M(x,y). From MA^{2}+2MB^{2}=10, we get (x+1)^{2}+y^{2}+2[(x-2)^{2}+y^{2}]=10. Simplifying yields 3x^{2}-6x+3y^{2}=1. According to the problem, the line l: x+2y+a=0 intersects with 3x^{2}-6x+3y^{2}=1. Combining them gives 15x^{2}-(24-6a)x+3a^{2}-4=0. Therefore, \\Delta=(24-6a)^{2}-60(3a^{2}-4)\\geqslant0. Simplifying yields 3a^{2}+6a-17\\leqslant0. Solving gives -1-\\frac{\\sqrt[2]{15}}{3}\\leqslant\\frac{a}{5}\\leqslant-1+\\frac{2\\sqrt{15}}{3}."
  },
  {
    "text": "The chord length intercepted by the directrix of the parabola $y^{2}=2 p x(p>0)$ on the circle $x^{2}+y^{2}-2 y-1=0$ is $2$, then the coordinates of the focus of the parabola are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;H: Circle;Expression(H) = (-2*y + x^2 + y^2 - 1 = 0);Length(InterceptChord(Directrix(G), H)) = 2",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(1, 0)",
    "fact_spans": "[[[0, 21], [57, 60]], [[0, 21]], [[3, 21]], [[3, 21]], [[25, 47]], [[25, 47]], [[0, 55]]]",
    "query_spans": "[[[57, 67]]]",
    "process": "The directrix of the parabola $ y^{2} = 2px $ ($ p > 0 $) is $ x = -\\frac{p}{2} $. Converting the circle into standard form gives $ x^{2} + (y - 1)^{2} = 2 $, yielding the center $ M(0,1) $ and radius $ r = \\sqrt{2} $. The distance from the center to the directrix is $ \\frac{p}{2} $, so $ \\left( \\frac{p}{2} \\right)^{2} + \\left( \\frac{2}{2} \\right)^{2} = (\\sqrt{2})^{2} $, which gives $ p = 2 $. Therefore, the coordinates of the focus are $ (1,0) $. This problem examines finding parameters in the standard equation of a parabola, then determining the focus coordinates, involving the directrix of a parabola and chord length of a circle. Difficulty: easy"
  },
  {
    "text": "The foci of the ellipse $x^{2}+m y^{2}=1$ lie on the $y$-axis, and the length of the major axis is twice the length of the minor axis. What is the value of $m$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (m*y^2 + x^2 = 1);m: Number;PointOnCurve(Focus(G), yAxis);Length(MajorAxis(G)) = 2*Length(MinorAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "1/4",
    "fact_spans": "[[[0, 19]], [[0, 19]], [[41, 44]], [[0, 28]], [[0, 39]]]",
    "query_spans": "[[[41, 48]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ is a point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, and the area of the triangle with vertices at $P$ and the two foci is $2 \\sqrt{5}$, find the coordinates of point $P$?",
    "fact_expressions": "P: Point;F1: Point;F2: Point;Focus(G) = {F1, F2};G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1);PointOnCurve(P, G);Vertex(TriangleOf(P, F1, F2)) = {P, F1, F2};Area(TriangleOf(P, F1, F2)) = 2*sqrt(5)",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(0, pm*2)",
    "fact_spans": "[[[2, 6], [52, 55], [85, 89]], [], [], [[7, 59]], [[7, 44]], [[7, 44]], [[2, 48]], [[7, 66]], [[7, 82]]]",
    "query_spans": "[[[85, 93]]]",
    "process": "Let $ F_{1}F_{2} $ be the left and right foci of the ellipse $ \\frac{x^{2}}{4} + \\frac{y^{2}}{4} = 1 $, then $ F_{1}(-\\sqrt{5},0), F_{2}(\\sqrt{5},0) $. Let $ P(x,y) $ be a point on the ellipse, then $ \\frac{1}{2} \\times 2\\sqrt{5} \\times |y| = 2\\sqrt{5} $, $ \\therefore y = \\pm 2 $. Substituting $ y = \\pm 2 $ into the ellipse equation gives: $ \\frac{x^{2}}{0} + \\frac{2^{2}}{4} = 1 $, $ \\therefore x = 0 $, then the coordinates of point $ P $ are $ (0, \\pm 2) $. The answer is: $ (0+2) $"
  },
  {
    "text": "If one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1$ $(a>0)$ is given by $3 x-2 y=0$, then the value of $a$ is?",
    "fact_expressions": "G: Hyperbola;a: Number;Expression(G) = (-y^2/9 + x^2/a^2 = 1);a>0;Expression(OneOf(Asymptote(G))) = (3*x - 2*y = 0)",
    "query_expressions": "a",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 48]], [[70, 73]], [[1, 48]], [[4, 48]], [[1, 68]]]",
    "query_spans": "[[[70, 77]]]",
    "process": ""
  },
  {
    "text": "The equation of the hyperbola that shares the same asymptotes as the hyperbola $x^{2}-4 y^{2}=4$ and passes through the point $(2 , \\sqrt {5})$ is?",
    "fact_expressions": "G: Hyperbola;C: Hyperbola;H: Point;Expression(G) = (x^2 - 4*y^2 = 4);Coordinate(H) = (2, sqrt(5));Asymptote(G) = Asymptote(C);PointOnCurve(H, C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/4 - x^2/16 = 1",
    "fact_spans": "[[[1, 21]], [[52, 55]], [[33, 51]], [[1, 21]], [[33, 51]], [[0, 55]], [[31, 55]]]",
    "query_spans": "[[[52, 59]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of a hyperbola are $y=\\pm \\frac{5}{12} x$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*(5/12)*x)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{13/12, 13/5}",
    "fact_spans": "[[[2, 5], [36, 39]], [[2, 34]]]",
    "query_spans": "[[[36, 45]]]",
    "process": ""
  },
  {
    "text": "The asymptotes of the hyperbola $\\frac{y^{2}}{a^{2}}-x^{2}=1$ are $y=\\pm \\sqrt{2} x$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;Expression(G) = (-x^2 + y^2/a^2 = 1);Expression(Asymptote(G)) = (y = pm*sqrt(2)*x)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[0, 32], [58, 61]], [[3, 32]], [[0, 32]], [[0, 55]]]",
    "query_spans": "[[[58, 67]]]",
    "process": "Let the focal distance of the hyperbola be $2c$. For the hyperbola $\\frac{y^{2}}{a^{2}}-x^{2}=1$, we have $b=1$, and the slope of the asymptote equation is $\\frac{a}{b}=a=\\sqrt{2}c=\\sqrt{a^{2}+b^{2}}=\\sqrt{3}$, $e=\\frac{\\sqrt{3}}{\\sqrt{2}}=\\frac{\\sqrt{6}}{2}$"
  },
  {
    "text": "Given that $F_{1}(-1,0)$, $F_{2}(1,0)$ are the two foci of ellipse $C$, and a line passing through $F_{2}$ and perpendicular to the $x$-axis intersects $C$ at points $A$ and $B$ with $|A B|=3$, then the equation of $C$ is?",
    "fact_expressions": "C: Ellipse;G: Line;F1: Point;F2: Point;A: Point;B: Point;Coordinate(F1) = (-1, 0);Coordinate(F2) = (1, 0);Focus(C)={F1,F2};PointOnCurve(F2, G);IsPerpendicular(G, xAxis);Intersection(G,C)={A,B};Abs(LineSegmentOf(A, B)) = 3",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4+y^2/3=1",
    "fact_spans": "[[[30, 35], [61, 64], [87, 90]], [[58, 60]], [[2, 15]], [[17, 29], [42, 49]], [[65, 68]], [[69, 72]], [[2, 15]], [[17, 29]], [[2, 40]], [[41, 60]], [[50, 60]], [[58, 74]], [[75, 84]]]",
    "query_spans": "[[[87, 95]]]",
    "process": "Let the equation of ellipse $ C $ be $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $). From the given conditions, we have $ A(1, \\frac{b^{2}}{a}) $, $ B(1, -\\frac{b^{2}}{a}) $. Since $ |AB| = \\frac{b^{2}}{a} - (-\\frac{b^{2}}{a}) = \\frac{2b^{2}}{a} = 3 $, it follows that $ 2b^{2} = 3a $. Thus,\n$$\n\\begin{cases}\n2b^{2} = 3a, \\\\\na^{2} - b^{2} = c^{2} = 1,\n\\end{cases}\n$$\nSolving gives\n$$\n\\begin{cases}\na = 2, \\\\\nb = \\sqrt{3},\n\\end{cases}\n$$\nTherefore, the equation of ellipse $ C $ is $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $. Hence, choose option C."
  },
  {
    "text": "What is the standard equation of a parabola whose focus lies on the line $3x - 4y - 12 = 0$?",
    "fact_expressions": "G: Parabola;H: Line;Expression(H) = (3*x - 4*y - 12 = 0);PointOnCurve(Focus(G), H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2 = 16*x, x^2 = -12*y}",
    "fact_spans": "[[[21, 24]], [[3, 19]], [[3, 19]], [[0, 24]]]",
    "query_spans": "[[[21, 31]]]",
    "process": ""
  },
  {
    "text": "Given the equation of the parabola $y=2 x^{2}$, what is its directrix equation?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 2*x^2)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y = -1/8",
    "fact_spans": "[[[2, 5], [20, 21]], [[2, 19]]]",
    "query_spans": "[[[20, 27]]]",
    "process": "Since the parabola equation is $ y = 2x^{2} $, converting it to standard form gives $ x^{2} = \\frac{1}{2}y $, so $ 2p = \\frac{1}{2} $. Therefore, its directrix equation is $ y = -\\frac{p}{2} = -\\frac{1}{8} $, that is, $ y = -\\frac{1}{8} $."
  },
  {
    "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1  (m<4)$ is $\\frac{1}{2}$, then $m$=?",
    "fact_expressions": "G: Ellipse;m: Number;m<4;Expression(G) = (x^2/4 + y^2/m = 1);Eccentricity(G) = 1/2",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 45]], [[65, 68]], [[3, 45]], [[1, 45]], [[1, 63]]]",
    "query_spans": "[[[65, 70]]]",
    "process": ""
  },
  {
    "text": "It is known that the hyperbola $C$ is centered at the origin of the coordinate system, with left and right foci $F_{1}$, $F_{2}$, and asymptotes $l_{1}$, $l_{2}$. A line passing through point $F_{2}$ and perpendicular to $l_{1}$ intersects $l_{1}$ and $l_{2}$ at points $P$ and $Q$, respectively, and $\\overrightarrow{O F_{2}}+\\overrightarrow{O Q}=2 \\overrightarrow{O P}$. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;l1:Line;l2:Line;O: Origin;F2: Point;Q: Point;P: Point;F1: Point;Center(C)=O;LeftFocus(C) = F1;RightFocus(C) = F2;Asymptote(C) = {l1,l2};L:Line;PointOnCurve(F2, L);IsPerpendicular(L,l1);Intersection(L, l1) = P;Intersection(L,l2)=Q;VectorOf(O, F2) + VectorOf(O, Q) = 2*VectorOf(O, P)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 8], [188, 191]], [[46, 53], [74, 81], [89, 96]], [[55, 62], [97, 104]], [[12, 16]], [[32, 39], [64, 72]], [[109, 112]], [[105, 108]], [[24, 31]], [[2, 16]], [[2, 39]], [[2, 39]], [[2, 62]], [[84, 86]], [[63, 86]], [[73, 86]], [[84, 114]], [[84, 114]], [[116, 186]]]",
    "query_spans": "[[[188, 197]]]",
    "process": "Let the hyperbola equation be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0,b>0)$, the asymptotes of the hyperbola are given by $y=\\pm\\frac{b}{a}x$, and the right focus is $F_{2}(c,0)$. Without loss of generality, let $l_{1}:y=\\frac{b}{a}x$, $l_{2}:y=-\\frac{b}{a}x$. Since $\\overrightarrow{OF_{2}}+\\overrightarrow{OQ}=2\\overrightarrow{OP}$, point $P$ is the midpoint of segment $QF_{2}$. Since $QF_{2}\\bot l_{1}$, $l_{1}$ is the perpendicular bisector of segment $QF_{2}$. Therefore, triangle $OF_{2}Q$ is isosceles, so $|OQ|=|OF_{2}|=c$. The slope of line $l_{1}$ is $\\frac{b}{a}$, thus the slope of line $QF_{2}$ is $-\\frac{a}{b}$. Hence, the equation of line $QF_{2}$ is $y=-\\frac{a}{b}(x-c)$. Solving the system:\n$$\n\\begin{cases}\ny=-\\frac{a}{b}(x-c)\\\\\ny=-\\frac{b}{a}x\n\\end{cases}\n$$\nwe obtain $x_{Q}=\\frac{a^{2}c}{a^{2}-b^{2}}$, $y_{Q}=\\frac{abc}{b^{2}-a^{2}}$. Then $x_{Q}^{2}+y_{Q}^{2}=c^{2}$, that is,\n$$\n\\left(\\frac{a^{2}c}{a^{2}-b^{2}}\\right)^{2}+\\left(\\frac{abc}{b^{2}-a^{2}}\\right)^{2}=c^{2},\n$$\nsimplifying yields $\\frac{b^{2}}{a^{2}}=3$. Therefore, the eccentricity of the hyperbola is $e=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=2$."
  },
  {
    "text": "Given point $A(0,2)$ and any point $P$ on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, what is the maximum value of $|P A|$?",
    "fact_expressions": "G: Ellipse;A: Point;P: Point;Expression(G) = (x^2/4 + y^2 = 1);Coordinate(A) = (0, 2);PointOnCurve(P, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "(2/3)*sqrt(21)",
    "fact_spans": "[[[12, 39]], [[2, 11]], [[44, 47]], [[12, 39]], [[2, 11]], [[12, 47]]]",
    "query_spans": "[[[49, 61]]]",
    "process": "Let P(x_{0},y_{0}), then -2\\leqslant x_{0}\\leqslant 2, -1\\leqslant y_{0}\\leqslant 1. \\because |PA|^{2}=x_{0}^{2}+(y_{0}-2)^{2}=1 \\begin{matrix} 4 & 0 \\\\ |PA| & =4(1-y_{0}^{2})+(y_{0}^{-2})^{2}=-3y_{0}^{2}-4y_{0}+8=-3(y_{0}+\\frac{2}{3})^{2}+\\frac{28}{3} \\because -1\\leqslant y_{0}\\leqslant 1, and -1\\leqslant -\\frac{2}{3}\\leqslant 1, \\therefore when y_{0}=-\\frac{2}{3}, |PA|^{2}=\\frac{28}{3}, i.e., |PA|=\\frac{2\\sqrt{21}}{3}"
  },
  {
    "text": "Let a focus of the hyperbola be $F$; let an endpoint of the imaginary axis be $B$. If the line $F B$ is perpendicular to an asymptote of the hyperbola, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;B: Point;F: Point;OneOf(Focus(G))=F;OneOf(Endpoint(ImageinaryAxis(G)))=B;IsPerpendicular(LineOf(F,B),OneOf(Asymptote(G)))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 4], [38, 41], [53, 56]], [[23, 26]], [[10, 13]], [[1, 13]], [[1, 26]], [[29, 49]]]",
    "query_spans": "[[[53, 62]]]",
    "process": ""
  },
  {
    "text": "A line passing through the focus of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$, with $|AB|=8$. Then the x-coordinate of the midpoint of segment $AB$ is?",
    "fact_expressions": "A: Point;B: Point;G: Parabola;H: Line;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 8",
    "query_expressions": "XCoordinate(MidPoint(LineSegmentOf(A, B)))",
    "answer_expressions": "3",
    "fact_spans": "[[[26, 29]], [[30, 33]], [[1, 15], [21, 24]], [[18, 20]], [[1, 15]], [[0, 20]], [[18, 35]], [[36, 45]]]",
    "query_spans": "[[[47, 61]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively, $|F_{1} F_{2}|=2$, a line passing through the left focus of the ellipse with slope $2$ intersects the ellipse at points $A$ and $B$, and $S_{\\triangle A B F_{2}}=4$, then the chord length $|A B|$=?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;F1: Point;F2: Point;A: Point;B: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(F1, F2)) = 2;PointOnCurve(LeftFocus(G), H);Slope(H) = 2;Intersection(H, G) = {A, B};Area(TriangleOf(A,B,F2))= 4;IsChordOf(LineSegmentOf(A,B),G)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[20, 72], [98, 100], [114, 116]], [[22, 72]], [[22, 72]], [[111, 113]], [[2, 9]], [[10, 17]], [[117, 120]], [[121, 124]], [[22, 72]], [[22, 72]], [[20, 72]], [[2, 78]], [[2, 78]], [[79, 96]], [[97, 113]], [[104, 113]], [[111, 126]], [[128, 155]], [[114, 166]]]",
    "query_spans": "[[[159, 168]]]",
    "process": "\\because S_{\\triangle ABF}=4, \\therefore \\frac{1}{2} \\times 2c \\times |y_{A}-y_{B}|=4, also \\because |F_{1}F_{2}|=2, \\therefore |y_{A}-y_{B}|=4, \\because the line passes through the left focus of the ellipse and has slope \\frac{1}{2}, \\therefore |AB|=\\sqrt{1+(\\frac{1}{2})^{2}} \\times 4=2\\sqrt{5}."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, what is the equation of its right directrix?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1)",
    "query_expressions": "Expression(RightDirectrix(G))",
    "answer_expressions": "x = 4",
    "fact_spans": "[[[2, 39], [41, 42]], [[2, 39]]]",
    "query_spans": "[[[41, 51]]]",
    "process": "The equation of the right directrix is $x=\\frac{a^{2}}{c}$, which is $x=\\frac{4}{\\sqrt{4-3}}$, $x=1$"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has a focal distance of $2c$, and the circle $M$: $x^{2}+y^{2}-2cy=0$ intersects the ellipse $C$ at points $A$ and $B$. If $OA \\perp OB$ ($O$ being the origin), then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;FocalLength(C) = 2*c;c: Number;M: Circle;Expression(M) = (-2*c*y + x^2 + y^2 = 0);Intersection(M, C) = {A, B};A: Point;B: Point;IsPerpendicular(LineSegmentOf(O, A), LineSegmentOf(O, B)) = True;O: Origin",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(\\sqrt{5}-1)/2",
    "fact_spans": "[[[2, 59], [97, 102], [142, 147]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[2, 68]], [[75, 96]], [[69, 96]], [[69, 96]], [[69, 113]], [[104, 107]], [[108, 111]], [[115, 130]], [[131, 134]]]",
    "query_spans": "[[[142, 153]]]",
    "process": "The equation of circle M is $x^{2}+(y-c)^{2}=c^{2}$, representing a circle with center $(0,c)$ and radius $c$. Since $OA\\bot OB$, $AB$ is the diameter of circle M, and $|AB|=2c$. Thus, the coordinates of points $A$ and $B$ are $(-c,c)$ and $(c,c)$, respectively. Since points $A$ and $B$ lie on the ellipse $C$, it follows that $\\frac{c^{2}}{a^{2}}+\\frac{c^{2}}{b^{2}}=1$, so $\\frac{c^{2}}{b^{2}}=1-\\frac{c^{2}}{a^{2}}=\\frac{a^{2}-c^{2}}{a^{2}}=\\frac{b^{2}}{a^{2}}$. Rearranging gives $b^{2}=ac$, hence $a^{2}-c^{2}=ac$, which implies $e^{2}+e-1=0$. Solving yields $e=\\frac{\\sqrt{5}-1}{2}$ (discarding the negative value). Answer: $\\frac{\\sqrt{5}-1}{2}$"
  },
  {
    "text": "A hyperbola with an asymptote given by $y=x$ and passing through the point $(2,4)$ has the equation?",
    "fact_expressions": "E: Hyperbola;Expression(OneOf(Asymptote(E))) = (y=x);F: Point;Coordinate(F) = (2, 4);PointOnCurve(F, E)",
    "query_expressions": "Expression(E)",
    "answer_expressions": "y^2 - x^2 = 12",
    "fact_spans": "[[[25, 28]], [[0, 28]], [[16, 24]], [[16, 24]], [[15, 28]]]",
    "query_spans": "[[[25, 32]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $C$ has the same foci as the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{16}=1$, and the asymptotes of $C$ are $x \\pm \\sqrt{3} y=0$. Then the equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Ellipse;Expression(G) = (x^2/36 + y^2/16 = 1);Focus(C) = Focus(G);Expression(Asymptote(C))=(pm*sqrt(3)*y+x=0)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/15 - y^2/5 = 1",
    "fact_spans": "[[[0, 6], [54, 57], [86, 92]], [[7, 46]], [[7, 46]], [[0, 52]], [[54, 84]]]",
    "query_spans": "[[[86, 97]]]",
    "process": ""
  },
  {
    "text": "If the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ intersects the line $y=2 x$, then the range of eccentricity $e$ is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = 2*x);IsIntersect(G, H);Eccentricity(G)=e;e:Number",
    "query_expressions": "Range(e)",
    "answer_expressions": "(\\sqrt{5}, +\\infty)",
    "fact_spans": "[[[1, 58]], [[4, 58]], [[4, 58]], [[59, 68]], [[4, 58]], [[4, 58]], [[1, 58]], [[59, 68]], [[1, 71]], [[1, 79]], [[76, 79]]]",
    "query_spans": "[[[76, 86]]]",
    "process": "As shown in the figure, since the asymptotes of the hyperbola are given by $ y = \\pm\\frac{b}{a}x $, and the hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0 $, $ b>0 $) intersects the line $ y = 2x $, then: $ \\frac{b}{a} > 2 $. Therefore, $ e = \\sqrt{1 + \\left(\\frac{b}{a}\\right)^{2}} > \\sqrt{5} $, that is, the range of the eccentricity $ e $ is $ (\\sqrt{5}, +\\infty) $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, its asymptotes are tangent to the circle $F$: $(x-2)^{2}+y^{2}=3$, and one focus of the hyperbola $C$ coincides with the center of the circle $F$. Then the equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;F: Circle;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(F) = (y^2 + (x - 2)^2 = 3);IsTangent(Asymptote(C),F);OneOf(Focus(C))=Center(F)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[2, 63], [96, 102], [119, 125]], [[10, 63]], [[10, 63]], [[68, 92], [108, 112]], [[10, 63]], [[10, 63]], [[2, 63]], [[68, 92]], [[2, 94]], [[96, 117]]]",
    "query_spans": "[[[119, 130]]]",
    "process": "The asymptotes of the hyperbola are given by $ y = \\pm\\frac{b}{a}x^{x} $. Since the asymptotes of the hyperbola are tangent to the circle $ F: (x-2)^{2} + y^{2} = 3 $, we have:  \n$ \\frac{2b}{\\sqrt{a^{2}+b^{2}}} = \\frac{2b}{c} = \\sqrt{3} $.  \nAlso, one focus of the hyperbola $ C $ coincides with the center of the circle $ F $, so $ c = 2 $, hence $ b = \\sqrt{3} $, $ a = 1 $. Therefore, the equation of the hyperbola is $ x^{2} - \\frac{y^{2}}{2} = $"
  },
  {
    "text": "Given that the asymptotes of the hyperbola $C$ are $y = \\pm \\frac{5}{2}x$, and it passes through the point $(3,8)$, then the equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Point;Coordinate(G) = (3, 8);Expression(Asymptote(C)) = (y = pm*(5/2)*x);PointOnCurve(G,C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "4*y^2/31 - 25*x^2/31 = 1",
    "fact_spans": "[[[2, 8], [49, 55]], [[39, 47]], [[39, 47]], [[2, 36]], [2, 45]]",
    "query_spans": "[[[49, 60]]]",
    "process": "Let the equation of hyperbola C be \\frac{25}{4}x^{2}-y^{2}=\\lambda. Substituting the coordinates of the point (3,8) into the equation of hyperbola C, we find the value of \\lambda, thus obtaining the equation of hyperbola C. Since the asymptotes of hyperbola C are given by y=\\pm\\frac{5}{2}x, the equation of hyperbola C can be written as \\frac{25}{4}x^{2}-y^{2}=\\lambda. Since hyperbola C passes through the point (3,8), it follows that \\lambda=\\frac{25}{4}\\times3^{2}-8^{2}=-\\frac{31}{4}. Therefore, the equation of hyperbola C is \\frac{25}{4}x^{2}-y^{2}=-\\frac{31}{4}, or equivalently \\frac{4y^{2}}{31}-\\frac{25x^{2}}{31}=1."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the right vertex of $C$ is point $A$. A circle $A$ is drawn with center $A$ and radius $b$. The circle $A$ intersects an asymptote of the hyperbola $C$ at points $M$ and $N$. If $|MN|=b$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;A: Circle;M: Point;N: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightVertex(C) = A1;A1:Point;Center(A)=A1;Radius(A)=b;Intersection(A, OneOf(Asymptote(C))) = {M, N};Abs(LineSegmentOf(M, N)) = b",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 63], [97, 103], [133, 136]], [[80, 83]], [[10, 63]], [[92, 96], [92, 96]], [[111, 114]], [[115, 118]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 71]], [[68, 71], [73, 76]], [[72, 91]], [[80, 91]], [[92, 120]], [[122, 131]]]",
    "query_spans": "[[[133, 142]]]",
    "process": "From the problem, the asymptote equation of the hyperbola is $ bx - ay = 0 $. It is given that $ \\triangle AMN $ is an equilateral triangle with side length $ b $. Therefore, the distance from point $ A $ to the asymptote is $ \\frac{\\sqrt{3}}{2}b = \\frac{|ab|}{\\sqrt{a^{2}+b^{2}}} = \\frac{ab}{c} $, $ \\therefore e = \\frac{c}{a} = \\frac{2}{3}\\sqrt{3} $"
  },
  {
    "text": "$P$ is a point on the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$, one asymptote of the hyperbola is $3 x-2 y=0$, and $F_{1}$, $F_{2}$ are the left and right foci respectively. If $|P F_{1}|=5$, then what is the distance from $P$ to the right directrix of the hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/b^2 = 1);b: Number;P: Point;PointOnCurve(P, G);Expression(OneOf(Asymptote(G))) = (3*x - 2*y = 0);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(P, F1)) = 5",
    "query_expressions": "Distance(P, RightDirectrix(G))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[4, 46], [50, 53], [118, 121]], [[4, 46]], [[7, 46]], [[0, 3], [114, 117]], [[0, 49]], [[50, 71]], [[73, 80]], [[82, 89]], [[50, 97]], [[50, 97]], [[99, 112]]]",
    "query_spans": "[[[114, 129]]]",
    "process": ""
  },
  {
    "text": "If $A$ and $B$ are two distinct points on the parabola $y^{2}=4x$, and the perpendicular bisector of chord $AB$ (not parallel to the $y$-axis) intersects the $x$-axis at point $P(4,0)$, then what is the horizontal coordinate of the midpoint of chord $AB$?",
    "fact_expressions": "A: Point;B: Point;PointOnCurve(A, G) = True;PointOnCurve(B, G) = True;Negation(A = B);G: Parabola;Expression(G) = (y^2 = 4*x);IsChordOf(LineSegmentOf(A, B), G) = True;Negation(IsParallel(LineSegmentOf(A, B), yAxis));Intersection(PerpendicularBisector(LineSegmentOf(A, B)), xAxis) = P;P: Point;Coordinate(P) = (4, 0)",
    "query_expressions": "XCoordinate(MidPoint(LineSegmentOf(A, B)))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 4]], [[5, 8]], [[1, 29]], [[1, 29]], [[1, 29]], [[9, 23]], [[9, 23]], [[9, 36]], [[31, 47]], [[31, 70]], [[61, 70]], [[61, 70]]]",
    "query_spans": "[[[73, 86]]]",
    "process": "Let the coordinates of points A and B be (x_{1},y_{1}) and (x_{2},y_{2}) respectively, where x_{1} \\neq x_{2}. Then y_{1}^{2} = 4x_{1}, y_{2}^{2} = 4x_{2}. Subtracting these two equations gives (y_{1}+y_{2})(y_{1}-y_{2}) = 4(x_{1}-x_{2}). Since x_{1} \\neq x_{2}, it follows that y_{1}+y_{2} \\neq 0. Let k be the slope of line AB, and M(x_{M},y_{M}) be the midpoint of chord AB. Then k = \\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = \\frac{4}{y_{1}+y_{2}} = \\frac{2}{y_{M}}. Thus, the equation of the perpendicular bisector l of AB is y - y_{M} = -\\frac{y_{M}}{2}(x - x_{M}). Since point P(4,0) lies on line l, we have -y_{M} = -\\frac{y_{M}}{2}(4 - x_{M}). Given y_{M} \\neq 0, solving yields x_{M} = 2. The x-coordinate of the midpoint of chord AB is 2."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$, what is the equation of the line containing the chord for which point $M(-1,2)$ is the midpoint?",
    "fact_expressions": "G: Ellipse;H: LineSegment;M: Point;Expression(G) = (x^2/16 + y^2/12 = 1);Coordinate(M) = (-1, 2);IsChordOf(H, G);MidPoint(H) = M",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "3*x-8*y+19=0",
    "fact_spans": "[[[2, 41]], [], [[44, 54]], [[2, 41]], [[44, 54]], [[2, 59]], [[2, 59]]]",
    "query_spans": "[[[2, 67]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of an ellipse, and there exists a point $P$ on the ellipse such that $\\angle F_{1} PF_{2}=60^{\\circ}$, then the range of the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Focus(G) = {F1, F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[1/2, 1)",
    "fact_spans": "[[[18, 20], [26, 28], [72, 74]], [[2, 9]], [[33, 36]], [[10, 17]], [[2, 25]], [[26, 36]], [[38, 70]]]",
    "query_spans": "[[[72, 84]]]",
    "process": ""
  },
  {
    "text": "Given that $O$ is the coordinate origin, two lines are drawn through the point $P(a,-2)$ tangent to the parabola $C$: $x^{2}=4 y$ at points $A$ and $B$, then the minimum area of $\\triangle A O B$ is?",
    "fact_expressions": "C: Parabola;P: Point;A: Point;O: Origin;B: Point;a:Number;Expression(C) = (x^2 = 4*y);Coordinate(P) = (a,-2);L1:Line;L2:Line;PointOnCurve(P,L1);PointOnCurve(P,L2);TangentPoint(L1,C)=A;TangentPoint(L2,C)=B",
    "query_expressions": "Min(Area(TriangleOf(A, O, B)))",
    "answer_expressions": "4*sqrt(2)",
    "fact_spans": "[[[28, 47]], [[12, 22]], [[50, 53]], [[2, 5]], [[54, 57]], [[13, 22]], [[28, 47]], [[12, 22]], [], [], [[11, 27]], [[11, 27]], [[11, 59]], [[11, 59]]]",
    "query_spans": "[[[61, 86]]]",
    "process": "Analysis: The tangent line equation at point A is $ y - y_{1} = \\frac{1}{2}x_{1}(x - x_{1}) $, and the tangent line equation at point B is $ y = \\frac{1}{2}x_{2}x - \\frac{1}{4}x_{2}^{2} $. Substituting point $ P(a, -2) $, we get $ -2 = \\frac{a}{2}x_{2} - y_{2} $, $ -2 = \\frac{a}{2}x_{1} - y_{1} $. Therefore, the line equation passing through AB is $ \\frac{a}{2}x - y + 2 = 0 $. Using Vieta's formulas, chord length formula, and point-to-line distance formula, we obtain $ S = 2\\sqrt{a^{2} + 8} \\geqslant 2\\sqrt{8} = 4\\sqrt{2} $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Since $ y = \\frac{1}{4}x^{2} $, $ y' = \\frac{1}{2}x $, the tangent line equation at point A is $ y - y_{1} = \\frac{1}{2}x_{1}(x - x_{1}) $, i.e., $ y - \\frac{1}{4}x_{1}^{2} = \\frac{1}{2}x_{1}x - \\frac{1}{2}x_{1}^{2} $, thus $ y = \\frac{1}{2}x_{1}x - \\frac{1}{4}x_{1}^{2} $. Similarly, the tangent line equation at point B is $ y = \\frac{1}{2}x_{2}x - \\frac{1}{4}x_{2}^{2} $. Substituting $ P(a, -2) $ yields $ -2 = \\frac{a}{2}x_{2} - y_{2} $, $ -2 = \\frac{a}{2}x_{1} - y_{1} $. Therefore, the line equation passing through AB is $ \\frac{a}{2}x - y + 2 = 0 $. Solving the system $ \\begin{cases} x^{2} = 4y \\\\ \\frac{a}{2}x - y + 2 = 0 \\end{cases} $, $ 8 = 0 $, $ .4a^{2} + 32 > 0 $, $ x_{1} \\frac{a^{2}}{4}|x_{1} - x_{2}| = $. Also, the distance from O to line AB is $ ^{d} \\frac{a^{2} + 4}{} $. $ S = \\frac{1}{2} \\sqrt{a^{2} + 4} \\cdot \\sqrt{a^{2} + 8} $. $ \\frac{4}{1} $ When $ a = 0 $, equality holds."
  },
  {
    "text": "If the asymptotes of a hyperbola are given by $y=\\pm \\frac{3}{4} x$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*((3/4)*x))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{5/3,5/4}",
    "fact_spans": "[[[1, 4], [34, 37]], [[1, 32]]]",
    "query_spans": "[[[34, 43]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. There exists a point $P$ on the ellipse such that $|P F_{1}|-|P F_{2}|=2 b$, $|P F_{1}| \\cdot |P F_{2}|=\\frac{3}{2} a b$. Then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)) = 2*b;Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = (3/2)*(a*b)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[19, 71], [78, 80], [162, 164]], [[19, 71]], [[21, 71]], [[21, 71]], [[21, 71]], [[21, 71]], [[1, 8]], [[9, 16]], [[1, 77]], [[1, 77]], [[85, 88]], [[78, 88]], [[91, 116]], [[117, 159]]]",
    "query_spans": "[[[162, 170]]]",
    "process": "By the definition of hyperbola, |PF_{1}|+|PF_{2}|=2a, and given |PF|-|PF_{2}|=2b, |PF|\\cdot|PF_{2}|=\\frac{3}{2}ab, then |PF|=a+b, |PF_{2}|=a-b, \\therefore |PF|\\cdot|PF_{2}|=a^{2}-b^{2}=\\frac{3}{2}ab, thus (3b-4a)(3b+a)=0, so a=2b, therefore c^{2}=a^{2}-b^{2}=3b^{2}, \\therefore e=\\frac{c}{a}=\\frac{\\sqrt{3}b}{2b}=\\frac{\\sqrt{3}}{2}"
  },
  {
    "text": "If one of the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1(a>0)$ is given by $y=\\frac{1}{2} x$, then $a$=?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (y = x/2)",
    "query_expressions": "a",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 38]], [[66, 69]], [[4, 38]], [[1, 38]], [[1, 64]]]",
    "query_spans": "[[[66, 71]]]",
    "process": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ ($a>0$) are given by $y=\\pm\\frac{1}{a}x$, so $\\frac{1}{a}=\\frac{1}{2}$, solving gives $a=2$."
  },
  {
    "text": "Through the left vertex $A$ of the hyperbola $M$: $x^{2}-\\frac{y^{2}}{b^{2}}=1$, draw a line $l$ with slope $1$. If $l$ intersects the two asymptotes of hyperbola $M$ at points $B$ and $C$ respectively, and $|A B|=|B C|$, then the eccentricity of hyperbola $M$ is?",
    "fact_expressions": "l: Line;M: Hyperbola;b: Number;A: Point;B: Point;C: Point;Expression(M) = (x^2 - y^2/b^2 = 1);LeftVertex(M)=A;PointOnCurve(A,l);Slope(l)=1;L1:Line;L2:Line;Asymptote(M)={L1,L2};Intersection(l,L1) = B;Intersection(l,L2) = C;Abs(LineSegmentOf(A, B)) = Abs(LineSegmentOf(B, C))",
    "query_expressions": "Eccentricity(M)",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[53, 58], [60, 63]], [[1, 38], [64, 70], [105, 111]], [[8, 38]], [[42, 45]], [[81, 84]], [[85, 88]], [[1, 38]], [[1, 45]], [[0, 58]], [[46, 58]], [], [], [[64, 76]], [[60, 88]], [[60, 88]], [[90, 103]]]",
    "query_spans": "[[[105, 117]]]",
    "process": "Through the left vertex A(-1,0) of the hyperbola M: $ x^{2}-\\frac{y^{2}}{b^{2}}=1 $, draw a line $ l: y=x+1 $ with slope 1. If $ l $ intersects the two asymptotes of hyperbola M, $ x^{2}-\\frac{y^{2}}{b^{2}}=0 $, at points $ B(x_{1},y_{1}) $, $ C(x_{2},y_{2}) $, then solve the system of equations:\n$$\n\\begin{cases}\nx^{2}-\\frac{y^{2}}{b^{2}}=0 \\\\\ny=x+1\n\\end{cases}\n$$\nSubstituting and eliminating variables yields $ (b^{2}-1)x^{2}-2x-1=0 $,\n$$\n\\begin{cases}\nx_{1}+x_{2}=\\frac{2}{b^{2}-1}\n\\end{cases}_{2}=-2x_{1}x. |x_{1}x_{2}=\\frac{1}{1-b^{2}}|=|BC|,\n$$\nthen B is the midpoint of AC, $ 2x_{1}=-1+ $, substituting gives\n$$\n\\begin{cases}\nx_{1}=-\\frac{1}{4} \\\\\nx_{2}=\\frac{1}{3}\n\\end{cases}\n\\therefore b^{2}=9,\n$$\nthe eccentricity of hyperbola M is $ e=\\frac{c}{a}=\\sqrt{10} $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the right focus is $F$, the left vertex is $A$, and points $P$, $Q$ lie on the ellipse such that $P F \\perp A F$. If $\\tan \\angle P A F = \\frac{1}{2}$, then the eccentricity $e$ of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F: Point;A: Point;Q: Point;e: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(G)=F;LeftVertex(G)=A;PointOnCurve(P, G);PointOnCurve(Q, G);IsPerpendicular(LineSegmentOf(P, F), LineSegmentOf(A, F));Tan(AngleOf(P, A, F)) = 1/2;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 54], [135, 187], [80, 82]], [[4, 54]], [[4, 54]], [[71, 75]], [[59, 62]], [[67, 70]], [[76, 79]], [[191, 194]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 62]], [[2, 70]], [[71, 83]], [[76, 83]], [[85, 100]], [[102, 133]], [[135, 194]]]",
    "query_spans": "[[[191, 196]]]",
    "process": "[Analysis] From \\( x_{P} = c \\), we obtain \\( y_{P} = \\frac{b^{2}}{a} \\). Then, using \\( \\tan\\angle PAF = \\frac{PF}{AF} = \\frac{\\frac{b^{2}}{a}}{a + c} = \\frac{1}{2} \\), the solution can be obtained. Since \\( PF \\perp AF \\), we have \\( x_{P} = c \\). Substituting into the ellipse equation \\( \\frac{x_{P}^{2}}{a^{2}} + \\frac{y_{P}^{2}}{b^{2}} = 1 \\) (\\( a > b > 0 \\)), we get \\( y_{P} = \\pm \\frac{b^{2}}{a} \\). From \\( \\tan\\angle PAF = \\frac{PF}{AF} = \\frac{\\frac{b^{2}}{a}}{a + c} = \\frac{1}{2} \\), it follows that \\( 2b^{2} = a(a + c) \\). Since \\( b^{2} = a^{2} - c^{2} \\), we obtain \\( 2(a^{2} - c^{2}) = a^{2} + ac \\Rightarrow 2c^{2} + ac - a^{2} = 0 \\Rightarrow 2e^{2} + e - 1 = 0 \\). Solving gives: \\( e = \\frac{1}{2} \\) or \\( -1 \\) (discarded)."
  },
  {
    "text": "If the directrix of the parabola $y^{2}=2 p x(p>0)$ passes through the left focus of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, then the real number $p$=?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Real;Expression(G) = (x^2 - y^2/3 = 1);p>0;Expression(H) = (y^2 = 2*(p*x));PointOnCurve(LeftFocus(G), Directrix(H))",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[27, 55]], [[1, 22]], [[61, 66]], [[27, 55]], [[4, 22]], [[1, 22]], [[1, 59]]]",
    "query_spans": "[[[61, 68]]]",
    "process": "Since the directrix of the parabola \\( y^{2} = 2px \\) is \\( x = -\\frac{p}{2} \\), according to the given condition, the line \\( x = -\\frac{p}{2} \\) passes through the left focus \\( (-2, 0) \\) of the hyperbola, so \\( p = 4 \\)."
  },
  {
    "text": "Let the line $l$: $y = x + 1$ intersect the ellipse $C$: $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1$ $(a > b > 0)$ at points $A$ and $B$, and intersect the $x$-axis at the left focus $F$, with $\\overrightarrow{A F} = 3 \\overrightarrow{F B}$. Then the eccentricity $e$ of the ellipse is $?$.",
    "fact_expressions": "l: Line;C: Ellipse;b: Number;a: Number;A: Point;F: Point;B: Point;e: Number;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(l)=(y=x+1);Intersection(l,C)={A,B};Intersection(l,xAxis)=F;LeftFocus(C)=F;VectorOf(A, F) = 3*VectorOf(F, B);Eccentricity(C) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[1, 15]], [[16, 73], [149, 151]], [[23, 73]], [[23, 73]], [[76, 79]], [[97, 100]], [[80, 83]], [[155, 158]], [[23, 73]], [[23, 73]], [[16, 73]], [[1, 15]], [[1, 85]], [[1, 100]], [[16, 100]], [[102, 147]], [[149, 158]]]",
    "query_spans": "[[[155, 160]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Solving the system of equations yields $ y_{1}+y_{2} $, $ y_{1}y_{2} $. From $ \\overrightarrow{AF}=3\\overrightarrow{FB} $ we get $ y_{1}=-3y_{2} $, then $ a^{2}(a^{2}+b^{2})-(a^{2}+4b^{2})=0 $. Using the focus coordinates of the ellipse we obtain $ a $, thus the solution. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Substitute line $ l: y=x+1 $ into the ellipse equation, eliminate $ x $ and simplify to get $ (a^{2}+b^{2})y^{2}-2b^{2}y+b^{2}(1-a^{2})=0 $. Therefore, $ y_{1}+y_{2}=\\frac{2b^{2}}{a^{2}+b^{2}} $, $ y_{1}y_{2}=\\frac{b^{2}(1-a^{2})}{a^{2}+b^{2}} $. Since $ \\overrightarrow{AF}=3\\overrightarrow{FB} $, we have $ y_{1}=-3y_{2} $, so $ -2y_{2}=\\frac{2b^{2}}{a_{2}+b^{2}} $, $ -3y_{2}=\\frac{b^{2}(1-a2)}{a^{2}+b^{2}} $. Also, line $ l: y=x+l $ passes through the left focus $ F $ of ellipse $ C $, so $ F(-1,0) $, hence $ a2-b^{2}=c2=1 $, so $ a2=2 $ or $ a2=1 $ (discarded). Thus $ a=\\sqrt{2} $, the eccentricity of the ellipse is $ e=\\frac{c}{a}=\\frac{\\sqrt{2}}{2} $."
  },
  {
    "text": "If the focal length of an ellipse is equal to its minor axis length, then the eccentricity of the ellipse equals?",
    "fact_expressions": "G: Ellipse;Length(FocalLength(G)) = Length(MinorAxis(G))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 3], [9, 10], [16, 18]], [[1, 14]]]",
    "query_spans": "[[[16, 25]]]",
    "process": ""
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. If $\\angle AOB=120^{\\circ}$, then $|AB|=$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;O: Origin;B: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};AngleOf(A, O, B) = ApplyUnit(120, degree)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "27/4",
    "fact_spans": "[[[1, 15], [24, 27]], [[21, 23]], [[28, 31]], [[37, 63]], [[32, 35]], [[17, 20]], [[1, 15]], [[1, 20]], [[0, 23]], [[21, 35]], [[37, 63]]]",
    "query_spans": "[[[65, 74]]]",
    "process": "As shown in the figure, the focus of the parabola $ y^{2} = 4x $ is $ F(1,0) $. Let the slope of line $ AB $ be $ k $, then the equation of line $ AB $ is $ y = k(x - 1) $. Substituting into the parabola equation and eliminating $ x $, we obtain $ y^{2} - \\frac{4}{k}y - 4 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $; by the relationship between roots and coefficients, we get: $ y_{1} + y_{2} = \\frac{4}{k} $, $ y_{1}y_{2} = -4 $. Let $ |OA| = m $, $ |OB| = n $, since $ \\angle AOB = 120^{\\circ} $, $ \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = mn\\cos120^{\\circ} = -\\frac{1}{2}m $. Since $ \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = x_{1}x_{2} + y_{1}y_{2} = -3 $, thus $ mn = 6 $, then the area of $ \\triangle AOB $ is $ \\frac{1}{2}mn\\sin120^{\\circ} = \\frac{3\\sqrt{3}}{2} $, solving gives $ k = \\pm\\frac{4\\sqrt{11}}{11} $. Without loss of generality, take $ k = \\frac{4\\sqrt{11}}{11} $, then $ AB = x_{1} + x_{2} + 2 = \\frac{y_{1} + y_{2}}{k} + 4 = \\frac{4}{k^{2}} + 4 = \\frac{27}{4} $."
  },
  {
    "text": "The coordinates of the left focus of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ are? The equation of the right directrix is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/5 = 1)",
    "query_expressions": "Coordinate(LeftFocus(G));Expression(RightDirectrix(G))",
    "answer_expressions": "(-2,0)  \nx=9/2",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 45]], [[0, 52]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, point $M$ lies on the ellipse $C$, the segment $M F_{2}$ is tangent to the circle $x^{2}+y^{2}=b^{2}$ at point $E$, and point $E$ is the midpoint of segment $M F_{2}$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Circle;F2: Point;M: Point;F1: Point;E: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = b^2);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(M, C);TangentPoint(LineSegmentOf(M,F2), G) = E;MidPoint(LineSegmentOf(M,F2)) = E",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[18, 75], [87, 92], [156, 161]], [[25, 75]], [[25, 75]], [[106, 126]], [[10, 17]], [[82, 86]], [[2, 9]], [[135, 139], [129, 133]], [[25, 75]], [[25, 75]], [[18, 75]], [[106, 126]], [[2, 81]], [[2, 81]], [[82, 93]], [[94, 133]], [[135, 154]]]",
    "query_spans": "[[[156, 167]]]",
    "process": "Let the circle with the minor axis of the ellipse as its diameter be tangent to the segment $ PF_2 $ at point $ M $. Connect $ OM $, $ PF_{2} $. Since $ M $ and $ O $ are the midpoints of $ PF_{2} $ and $ F_{1}F_{2} $ respectively, $ \\therefore MO \\parallel PF_{1} $, and $ |PF_{1}| = 2|MO| = 2b $, $ OM \\perp PF_{2} $, $ \\therefore PF_{1} \\perp PF_{2} $, $ |F_{1}F_{2}| = 2c $, $ \\therefore |PF_{2}| = 2\\sqrt{c^{2}-b^{2}} $. According to the definition of the ellipse, $ |PF_{1}| + |PF_{2}| = 2a $, $ \\therefore 2b + 2\\sqrt{c^{2}-b^{2}} = 2a \\Rightarrow a - b = \\sqrt{c^{2}-b^{2}} $. Squaring both sides yields: $ a^{2} - 2ab + b^{2} = c^{2} - b^{2} $. Substituting $ c^{2} = a^{2} - b^{2} $ and simplifying gives: $ 2a = 3b \\Rightarrow b = \\frac{2}{3} $, $ a = 1 $, $ c = \\frac{\\sqrt{5}}{3} $, $ e = \\frac{c}{a} = \\frac{\\sqrt{5}}{3} $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$ and directrix $l$, $P$ is a point on $l$, and $Q$ is an intersection point of line $PF$ and $C$. If $\\overrightarrow{PF}=3\\overrightarrow{QF}$, then $|QF|=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 8*x);F: Point;Focus(C) = F;l: Line;Directrix(C) = l;P: Point;PointOnCurve(P, l);Q: Point;OneOf(Intersection(LineOf(P, F), C)) = Q;VectorOf(P, F) = 3*VectorOf(Q, F)",
    "query_expressions": "Abs(LineSegmentOf(Q, F))",
    "answer_expressions": "8/3",
    "fact_spans": "[[[2, 21], [61, 64]], [[2, 21]], [[25, 28]], [[2, 28]], [[32, 35], [42, 45]], [[2, 35]], [[38, 41]], [[38, 48]], [[49, 52]], [[49, 69]], [[71, 116]]]",
    "query_spans": "[[[118, 127]]]",
    "process": "Find the equation of line PF, and solve it together with $ y^{2} = 8x $ to obtain $ x = 1 $, then use $ |QF| = d $ to determine. Let the distance from Q to 1 be $ d $, then $ |QF| = d $, $ \\because\\overrightarrow{PF} = 3\\overrightarrow{QF} $, $ \\therefore|PQ| = 2d $, $ \\therefore $ without loss of generality, assume the slope of line PF is $ \\sqrt{3} $, $ \\because F(2,0) $, $ \\therefore $ the equation of line PF is $ y = \\sqrt{3}(x - 2) $, solving together with $ y^{2} = 8x $ yields $ x = \\frac{2}{3} $, $ |F| = d = \\frac{8}{a} $,"
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2 p x(p>0)$ be $F$, and let its directrix intersect the $x$-axis at point $C$. Draw a chord $AB$ through point $F$. If $\\angle C B F=90^{\\circ}$, then $|AF|-|BF|$=?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;C: Point;F: Point;p>0;Expression(G) = (y^2 = 2*p*x);Focus(G)=F;Intersection(Directrix(G), xAxis) = C;IsChordOf(LineSegmentOf(A,B),G);PointOnCurve(F,LineSegmentOf(A,B));AngleOf(C, B, F) = ApplyUnit(90, degree)",
    "query_expressions": "Abs(LineSegmentOf(A, F)) - Abs(LineSegmentOf(B, F))",
    "answer_expressions": "2*p",
    "fact_spans": "[[[1, 22], [30, 31], [51, 52]], [[4, 22]], [[54, 59]], [[54, 59]], [[40, 44]], [[26, 29], [46, 50]], [[4, 22]], [[1, 22]], [[1, 29]], [[30, 44]], [[50, 59]], [[45, 59]], [[61, 86]]]",
    "query_spans": "[[[88, 101]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the coordinates of the right focus $F_{2}$ are $(2,0)$, $F_{1}$ is the left focus of the ellipse $C$, and $P$ is a point on the ellipse. If $\\tan \\angle F_{1} P F_{2}=\\frac{4}{3}$, $S_{\\triangle P F_{1} F_{2}}=6$, then the equation of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;F1: Point;P: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F2) = (2, 0);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);Tan(AngleOf(F1, P, F2)) = 4/3;Area(TriangleOf(P,F1,F2)) = 6",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/16+y^2/12=1",
    "fact_spans": "[[[2, 59], [91, 96], [105, 107], [186, 191]], [[9, 59]], [[9, 59]], [[83, 90]], [[101, 104]], [[63, 70]], [[9, 59]], [[9, 59]], [[2, 59]], [[63, 81]], [[83, 100]], [[2, 70]], [[101, 110]], [[112, 151]], [[152, 184]]]",
    "query_spans": "[[[186, 195]]]",
    "process": "From the area formula and the law of cosines, we obtain the values of |PF_{1}||PF_{2}| and |PF_{1}|^{2}+|PF_{2}|^{2}, and then find the value of |PF_{1}|+|PF_{2}| to solve the problem. [Detailed solution] From the given information, c=2, and \\tan\\angle F_{1}PF_{2}=\\frac{4}{3}, then \\sin\\angle F_{1}PF_{2}=\\frac{4}{5}, \\cos\\angle F_{1}PF_{2}=\\frac{3}{5}. Hence, S_{\\Delta PF_{1}F_{2}}=\\frac{1}{2}\\times|PF_{1}||PF_{2}|\\times\\frac{4}{5}=6, \\therefore |PF_{1}||PF_{2}|=15. Then \\cos\\angle F_{1}PF_{2}=\\frac{3}{5}=\\frac{|PF_{1}|^{2}+|PF_{2}|^{2}-16}{2\\times15}, \\therefore |PF_{1}|^{2}+|PF_{2}|^{2}=34. Then (|PF_{1}|+|PF_{2}|)^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}+2|PF_{1}||PF_{2}|=64, \\therefore 2a=8, a=4, then b^{2}=16-4=12. Therefore, the equation of ellipse C is \\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1."
  },
  {
    "text": "Let the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ be tangent to the parabola $y=\\frac{1}{2} x^{2}+2$. Then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;H: Parabola;Expression(H) = (y = x^2/2 + 2);IsTangent(Asymptote(G), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[1, 57], [93, 96]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[62, 88]], [[62, 88]], [[1, 90]]]",
    "query_spans": "[[[93, 102]]]",
    "process": "The asymptotes of the hyperbola are $ y = \\pm\\frac{b}{a}x $. Without loss of generality, take $ y = \\frac{b}{a}x $. Combining the asymptote with the parabola equation gives $ x^{2} - \\frac{2b}{a}x + 4 = 0 $. Since the asymptote is tangent to the parabola, $ \\left(-\\frac{2b}{a}\\right)^{2} - 4 \\times 1 \\times 4 = 0 $, so $ \\frac{4b^{2}}{a^{2}} = 16 $, hence $ b^{2} = 4a^{2} $. Therefore, $ c^{2} = a^{2} + b^{2} = 5a^{2} $, and $ e = \\sqrt{\\frac{c^{2}}{a^{2}}} = \\sqrt{5} $."
  },
  {
    "text": "If a hyperbola passes through the point $(3, \\sqrt{2})$ and has asymptotes given by $y = \\pm \\frac{1}{3} x$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (3, sqrt(2));PointOnCurve(H, G);Expression(Asymptote(G)) = (y = pm*(x/3))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 - x^2/9 = 1",
    "fact_spans": "[[[1, 4], [53, 56]], [[6, 22]], [[6, 22]], [[1, 22]], [[1, 51]]]",
    "query_spans": "[[[53, 61]]]",
    "process": "Using the asymptotes equation $ y = \\pm\\frac{1}{3}x $, let the hyperbola equation be $ y^{2} - \\frac{x^{2}}{9} = \\lambda $. Substituting the point $ (3, \\sqrt{2}) $ allows solving. According to the asymptotes equation $ y = \\pm\\frac{1}{3}x $, let the hyperbola equation be $ y^{2} - \\frac{x^{2}}{9} = \\lambda $. Since the hyperbola passes through the point $ (3, \\sqrt{2}) $, we have $ \\lambda = 2 - \\frac{9}{9} = 1 $. Therefore, the hyperbola equation is $ y^{2} - \\frac{x^{2}}{9} = 1 $."
  },
  {
    "text": "The equation of a hyperbola that shares the same foci as the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ and for which the distance from each focus to the asymptotes is $1$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/9 + y^2/5 = 1);Focus(G)=Focus(H);Distance(Focus(G),Asymptote(G))=1",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/3 - y^2 = 1",
    "fact_spans": "[[[59, 62]], [[1, 38]], [[1, 38]], [[0, 62]], [[45, 62]]]",
    "query_spans": "[[[59, 66]]]",
    "process": "\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1 has foci at (\\pm\\sqrt{9-5},0)=(\\pm2,0). Since the distance from the focus of a hyperbola to its asymptote equals the semi-imaginary axis length, b=1, \\because c=2 \\therefore a=\\sqrt{3} \\therefore \\frac{x^{2}}{3}-y^{2}=1"
  },
  {
    "text": "Given that $F_{1}$ is the left focus of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, and the line $y=x-1$ intersects the ellipse at points $A$ and $B$, then $|F_{1} A|+|F_{1} B|$=?",
    "fact_expressions": "G: Ellipse;H: Line;F1: Point;A: Point;B: Point;Expression(G) = (x^2/2 + y^2 = 1);Expression(H) = (y = x - 1);LeftFocus(G) = F1;Intersection(H, G) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(F1, A)) + Abs(LineSegmentOf(F1, B))",
    "answer_expressions": "8*sqrt(2)/3",
    "fact_spans": "[[[10, 37], [52, 54]], [[42, 51]], [[2, 9]], [[56, 59]], [[60, 63]], [[10, 37]], [[42, 51]], [[2, 41]], [[42, 65]]]",
    "query_spans": "[[[68, 91]]]",
    "process": "Solving the system of equations of the line and the ellipse yields $3x^{2}-4x=0$. $\\therefore A(0,-1), B(\\frac{4}{3},\\frac{1}{3}), F_{1}=(-1,0), |F_{1}A|=\\sqrt{2}, F_{1}B=\\frac{5}{3}\\sqrt{2}, F_{1}A+F_{1}B=\\frac{8\\sqrt{2}}{3}$"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{20}=1$, $M$ is a point on the ellipse such that $M F_{1} \\perp M F_{2}$, then the area of $\\Delta F_{1} M F_{2}$ is?",
    "fact_expressions": "F1: Point;F2: Point;Focus(G) = {F1, F2};G: Ellipse;Expression(G) = (x^2/36 + y^2/20 = 1);M: Point;PointOnCurve(M, G) = True;IsPerpendicular(LineSegmentOf(M, F1), LineSegmentOf(M, F2)) = True",
    "query_expressions": "Area(TriangleOf(F1, M, F2))",
    "answer_expressions": "20",
    "fact_spans": "[[[2, 9]], [[10, 17]], [[2, 62]], [[18, 57], [67, 69]], [[18, 57]], [[63, 66]], [[63, 72]], [[74, 97]]]",
    "query_spans": "[[[99, 126]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola equation $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, what is the equation of its asymptotes?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/12 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[2, 5]], [[2, 45]]]",
    "query_spans": "[[[2, 54]]]",
    "process": "According to the geometric meaning of the hyperbola, the asymptotes are given by $ y = \\pm\\frac{b}{a}x $. Given the hyperbola equation $ \\frac{x^{2}}{4} - \\frac{y^{2}}{12} = 1 $, substituting the basic quantities yields $ y = \\pm\\sqrt{3}x $. Hence, the result is $ v = +\\sqrt{3}x $."
  },
  {
    "text": "Given a point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ such that the distance from $P$ to its left directrix is $10$, and $F$ is the left focus of the ellipse. If point $M$ satisfies $\\overrightarrow{O M}=\\frac{1}{2}(\\overrightarrow{O P}+\\overrightarrow{O F})$, where $O$ is the origin, then $|O M|$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);P: Point;PointOnCurve(P, G);Distance(P, LeftDirectrix(G)) = 10;F: Point;LeftFocus(G) = F;M: Point;O: Origin;VectorOf(O, M) = (1/2)*(VectorOf(O, P) + VectorOf(O, F))",
    "query_expressions": "Abs(LineSegmentOf(O, M))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 41], [67, 69], [48, 49]], [[2, 41]], [[44, 47]], [[2, 47]], [[44, 60]], [[62, 65]], [[62, 73]], [[75, 79]], [[161, 164]], [[81, 158]]]",
    "query_spans": "[[[172, 181]]]",
    "process": "First, obtain the distance from P to F using the unified definition of conic sections, then obtain the distance from P to the right focus $F_{2}$ according to the definition of the ellipse. The magnitude $|\\overrightarrow{OM}|$ is the midline parallel to $PF_{2}$ in quadrilateral $APFF_{2}$, thus yielding the length of $|\\overrightarrow{OM}|$. [Detailed Explanation] Let the right focus of the ellipse be $F_{2}$ and the semi-focal length be $c$. Then the eccentricity of the ellipse is $\\frac{c}{a} = \\frac{3}{5}$. From $\\frac{|PF|}{10} = e = \\frac{3}{5}$, we get $|PF| = 6$, hence $|PF_{2}| = 10 - 6 = 4$. Since $\\overrightarrow{OM} = \\frac{1}{2}(\\overrightarrow{OP} + \\overrightarrow{OF})$, point M is the midpoint of PF. And since $OF = OF_{2}$, it follows that $|\\overrightarrow{OM}| = \\frac{1}{2}|PF_{2}| = 2$."
  },
  {
    "text": "If the distance from one focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ to an asymptote is equal to $\\frac{1}{4}$ of the focal length, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Distance(OneOf(Focus(G)),OneOf(Asymptote(G)))=(1/4)*FocalLength(G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[1, 60], [95, 98]], [[4, 60]], [[4, 60]], [[4, 60]], [[4, 60]], [[1, 60]], [[1, 92]]]",
    "query_spans": "[[[95, 104]]]",
    "process": "The distance from the point $(c,0)$ to the line $y=\\frac{b}{a}x$ is: $d=\\frac{|bc|}{\\sqrt{a^{2}+b^{2}}}=\\frac{bc}{c}=b=\\frac{1}{4}\\times2c=\\frac{c}{2}$, so $4b^{2}=c^{2}$, $4(c^{2}-a^{2})=c^{2}$, $\\frac{c^{2}}{a^{2}}=\\frac{4}{3}$, therefore $\\frac{c}{a}=\\frac{2\\sqrt{3}}{3}$."
  },
  {
    "text": "It is known that a hyperbola passes through the point $(3,-2)$ and has the same foci as the ellipse $4 x^{2}+9 y^{2}=36$. Then, what is the standard equation of the hyperbola?",
    "fact_expressions": "G: Hyperbola;I: Point;Coordinate(I) = (3, -2);PointOnCurve(I, G);H: Ellipse;Expression(H) = (4*x^2 + 9*y^2 = 36);Focus(G) = Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 5], [47, 50]], [[6, 15]], [[6, 15]], [[2, 15]], [[18, 40]], [[18, 40]], [[2, 45]]]",
    "query_spans": "[[[47, 57]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with right focus $F$, a line $l$: $2 kx-2 y-3 ka=0$ passing through point $F$ intersects the hyperbola $C$ at points $A$ and $B$. If $\\overrightarrow{AF}=7\\overrightarrow{FB}$, then the real number $k$=?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;A: Point;F: Point;B: Point;k: Real;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(l) = (-3*a*k + 2*(k*x) - 2*y = 0);RightFocus(C) = F;PointOnCurve(F,l);Intersection(l, C) = {A, B};VectorOf(A, F) = 7*VectorOf(F, B)",
    "query_expressions": "k",
    "answer_expressions": "pm*sqrt(3)",
    "fact_spans": "[[[78, 102]], [[2, 63], [103, 109]], [[10, 63]], [[10, 63]], [[111, 114]], [[68, 71], [73, 77]], [[115, 118]], [[167, 172]], [[10, 63]], [[10, 63]], [[2, 63]], [[78, 102]], [[2, 71]], [[72, 102]], [[78, 120]], [[123, 165]]]",
    "query_spans": "[[[167, 174]]]",
    "process": "In $2kx-2y-3ka=0$, let $y=0$, then $x=\\frac{3a}{2}$, so $c=\\frac{3a}{2}$, then $b^{2}=c^{2}-a^{2}=\\frac{5a^{2}}{4}$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, from $\\begin{cases}\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1\\\\2kx-2y-3ka=0\\end{cases}$, eliminating $x$ yields $(\\frac{b^{2}}{k^{2}}-a^{2})y^{2}+\\frac{3ab^{2}}{k}y+\\frac{5a2b^{2}}{4}=0$. $y_{1}+y_{2}=\\frac{3kab^{2}}{a^{2k}^{2}-b^{2}}$, $y_{1}y_{2}=\\frac{5k^{2}a2b^{2}}{4(b^{2}-a^{2k^{2})}$. From $\\overrightarrow{AF}=\\overrightarrow{7FB}$, we get $y_{1}=-7y_{2}'$, $y_{1}+y_{2}=-6y_{2}$, so $y_{1}y_{2}=-7y_{2}^{2}=-7\\times\\frac{k^{2}a_{2}b^{4}}{4(a2k^{2}-b}$, $k=\\pm\\sqrt{3}$"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively, if there exists a point $P$ on the right directrix of the ellipse such that the perpendicular bisector of the segment $PF_{1}$ passes through the point $F_{2}$, then what is the range of the eccentricity?",
    "fact_expressions": "P: Point;F1: Point;F2:Point;G: Ellipse;b: Number;a: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(P, RightDirectrix(G));PointOnCurve(F2,PerpendicularBisector(LineSegmentOf(P,F1)))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[\\sqrt{3}/3,1)",
    "fact_spans": "[[[92, 95]], [[2, 9]], [[10, 17], [115, 123]], [[20, 72], [80, 82]], [[22, 72]], [[22, 72]], [[22, 72]], [[22, 72]], [[20, 72]], [[2, 78]], [[2, 78]], [[80, 95]], [[98, 123]]]",
    "query_spans": "[[[80, 133]]]",
    "process": ""
  },
  {
    "text": "A line with inclination angle $\\frac{\\pi}{4}$ passes through the focus $F$ of the parabola $y^{2}=2 x$, intersecting the parabola at points $A$ and $B$. Then $|A B|=$?",
    "fact_expressions": "H: Line;Inclination(H) = pi/4;G: Parabola;Expression(G) = (y^2 = 2*x);F: Point;Focus(G) = F;A: Point;B: Point;Intersection(H, G) = {A, B};PointOnCurve(F, H)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "4",
    "fact_spans": "[[[20, 22]], [[0, 22]], [[23, 37], [45, 48]], [[23, 37]], [[40, 43]], [[23, 43]], [[49, 52]], [[53, 56]], [[20, 58]], [[20, 43]]]",
    "query_spans": "[[[60, 69]]]",
    "process": "From the parabola $ y^{2} = 2x $, the focus is $ F\\left(\\frac{1}{2}, 0\\right) $. Then, the equation of the line is found. By solving the line equation and the parabola equation simultaneously, the relationship of the intersection points' coordinates is obtained: $ x_{1} + x_{2} = 3 $. Then, by the definition of the parabola, the length of the segment can be found. From the parabola $ y^{2} = 2x $, the focus is $ F\\left(\\frac{1}{2}, 0\\right) $. $ \\therefore $ the equation of the line with inclination angle $ \\frac{\\pi}{4} $ passing through focus $ F $ is: $ y = x - \\frac{1}{2} $. Solving this together with the parabola $ y^{2} = 2x $ gives $ x^{2} - 3x + \\frac{1}{4} = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = 3 $. By the definition of the parabola, $ |AF| = x_{1} + \\frac{1}{2} $, $ |BF| = x_{2} + \\frac{1}{2} $, $ \\therefore |AB| = x_{1} + \\frac{1}{2} + x_{2} + \\frac{1}{2} = x_{1} + x_{2} + 1 = 4 $."
  },
  {
    "text": "If a moving point $M$ satisfies $\\sqrt{(x+5)^{2}+y^{2}}-\\sqrt{(x-5)^{2}+y^{2}}=6$, then what is the trajectory equation of point $M$?",
    "fact_expressions": "-sqrt(y^2 + (x - 5)^2) + sqrt(y^2 + (x + 5)^2) = 6;M:Point",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "(x^2/9-y^2/16=1)&(x>0)",
    "fact_spans": "[[[8, 57]], [[3, 6], [59, 63]]]",
    "query_spans": "[[[59, 70]]]",
    "process": "Let A(-5,0), B(5,0). Since the trajectory equation of the moving point M(x,y) is \\sqrt{(x+5)^{2}+y^{2}}-\\sqrt{(x-5)^{2}+y^{2}}=6, then |MA|-|MB|=6<10. Hence, the difference of the distances from point M to the fixed points A(-5,0) and B(5,0) is 6. Thus, the trajectory of the moving point M(x,y) is the right branch of a hyperbola with foci at (\\pm5,0) and a real axis length of 6. Since 2a=6, c=5, then b^{2}=c^{2}-a^{2}=25-9=16. Therefore, the standard equation of the trajectory of M is: \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1 (x>0)."
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{2}=1$ are $F_{1}$, $F_{2}$, point $P$ lies on the ellipse. If $|P F_{1}|=4$, then what is $|P F_{2}|$? What is the measure of $\\angle F_{1} P F_{2}$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 + y^2/2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 4",
    "query_expressions": "Abs(LineSegmentOf(P, F2));AngleOf(F1, P, F2)",
    "answer_expressions": "Answer is incorrect\nApplyUnit(120, degree)",
    "fact_spans": "[[[0, 37], [62, 64]], [[57, 61]], [[41, 48]], [[49, 56]], [[0, 37]], [[0, 56]], [[57, 65]], [[67, 80]]]",
    "query_spans": "[[[82, 95]], [[95, 122]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, with left and right foci $F_{1}$, $F_{2}$ respectively, and $P(\\sqrt{2}, 1)$ a point on $C$, eccentricity $e=\\sqrt{2}$. Then the standard equation of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;e: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (sqrt(2), 1);RightFocus(C) = F2;LeftFocus(C) = F1;PointOnCurve(P, C);Eccentricity(C) = e;e = sqrt(2)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2 = 1",
    "fact_spans": "[[[2, 63], [109, 112], [133, 136]], [[9, 63]], [[9, 63]], [[92, 108]], [[72, 79]], [[82, 89]], [[119, 131]], [[9, 63]], [[9, 63]], [[2, 63]], [[91, 108]], [[2, 89]], [[2, 89]], [[92, 115]], [[109, 131]], [[119, 131]]]",
    "query_spans": "[[[133, 143]]]",
    "process": "According to the problem, we have \\begin{cases}\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{2}\\\\\\frac{(\\sqrt{2})^{2}}{a}-\\frac{1^{2}}{2}=1\\end{cases}, so $ a = b = 1 $, therefore the standard equation of the hyperbola is $ x^{2} - y^{2} = 1 $.  \nAnswer: $ x^{2} - y^{2} = 1 $"
  },
  {
    "text": "Real numbers $x$, $y$ satisfy $x^{2}-y^{2}=4$. If $\\frac{y+2}{x}+m>0$ always holds, then the range of values for the real number $m$ is?",
    "fact_expressions": "m: Real;x: Real;y: Real;x^2 - y^2 = 4;(y+2)/x + m > 0",
    "query_expressions": "Range(m)",
    "answer_expressions": "[\\sqrt{2}, +\\infty)",
    "fact_spans": "[[[52, 57]], [[0, 5]], [[6, 9]], [[11, 26]], [[28, 47]]]",
    "query_spans": "[[[52, 64]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{m+2}=1(m>0)$ has an asymptote with equation $y=2x$, then $m=$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/(m + 2) + x^2/m = 1);m: Number;m>0;Expression(OneOf(Asymptote(G))) = (y = 2*x)",
    "query_expressions": "m",
    "answer_expressions": "2/3",
    "fact_spans": "[[[0, 45]], [[0, 45]], [[63, 66]], [[3, 45]], [[0, 61]]]",
    "query_spans": "[[[63, 68]]]",
    "process": "From the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{m+2}=1$ $(m>0)$, the asymptotes are given by $y=\\pm\\sqrt{\\frac{m+2}{m}}x$. Since one asymptote of the given hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{m+2}=1$ $(m>0)$ is $y=2x$, we have $\\sqrt{\\frac{m+2}{m}}=2$, thus $m=\\frac{2}{3}$."
  },
  {
    "text": "In ellipse $C$, $F$ is a focus, and $A$, $B$ are two vertices. If $|F A|=3$, $|F B|=2$, then all possible values of $|A B|$ are?",
    "fact_expressions": "C: Ellipse;F: Point;OneOf(Focus(C)) = F;A: Point;B: Point;Vertex(C) = {A, B};Abs(LineSegmentOf(F, A)) = 3;Abs(LineSegmentOf(F, B)) = 2",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "{5, sqrt(7), sqrt(17)}",
    "fact_spans": "[[[1, 6]], [[7, 10]], [[0, 15]], [[16, 19]], [[20, 23]], [[0, 28]], [[30, 39]], [[41, 50]]]",
    "query_spans": "[[[52, 67]]]",
    "process": "Discuss the positions of points F, A, B by cases and use the geometric properties of the ellipse to solve. According to the problem, |FA| = 3, |FB| = 2; points A and B cannot both be the two fixed endpoints of the minor axis simultaneously.  \nWhen F, A, B are all on the major axis, then  \n\\begin{cases}a+c=3\\\\a-c=2\\end{cases},  \nsolving gives a = \\frac{5}{2}, so |AB| = 2a = 5;  \nWhen F and A are on the major axis and lie on opposite sides of the y-axis, and B is on the minor axis, then  \n\\begin{cases}|FB|=a=2\\\\|FA|=a+c=3\\end{cases},  \nsolving gives c = 1, b = \\sqrt{3}, so |AB| = \\sqrt{2^{2}+3} = \\sqrt{7};  \nWhen F and B are both on the major axis and lie on the same side of the y-axis, then  \n\\begin{cases}|FB|=a-c=2\\\\|FA|=a=3\\end{cases},  \nsolving gives a = 3, c = 1, b^{2} = a^{2}-c^{2} = 2\\sqrt{2}, so |AB| = \\sqrt{3^{2}+8} = \\sqrt{17}."
  },
  {
    "text": "Given that point $P(2,4)$ lies on the parabola $C$: $y^{2}=2 p x$, a line $l$ passing through its focus $F$ with an inclination angle of $45^{\\circ}$ intersects $C$ at points $M$ and $N$. Then the area of $\\triangle P M N$ is?",
    "fact_expressions": "P: Point;Coordinate(P) = (2, 4);PointOnCurve(P, C);C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;F: Point;Focus(C) = F;l: Line;Inclination(l) = ApplyUnit(45, degree);PointOnCurve(F, l);M: Point;N: Point;Intersection(l, C) = {M, N}",
    "query_expressions": "Area(TriangleOf(P, M, N))",
    "answer_expressions": "16*sqrt(2)",
    "fact_spans": "[[[2, 11]], [[2, 11]], [[2, 34]], [[12, 33], [36, 37], [66, 69]], [[12, 33]], [[20, 33]], [[39, 42]], [[36, 42]], [[60, 65]], [[43, 65]], [[35, 65]], [[71, 74]], [[75, 78]], [[60, 80]]]",
    "query_spans": "[[[82, 104]]]",
    "process": "According to the problem, find the parabolic equation, then determine the coordinates of the focus, and thus obtain the equation of line l. Solve the system of equations of the line and the parabola, use the parabolic focal chord formula to find chord MN, then use the point-to-line distance formula to find the distance from point P(2,4) to line l, and thus find the area of the triangle. Since P(2,4) lies on the parabola C: y^{2}=2px, we have 4^{2}=2p\\times2, so p=4, hence the parabolic equation is y^{2}=8x, with focus F(2,0). The inclination angle of line l is 45^{\\circ}, so the equation of line l is y=x-2. Solving the system of equations:\n\\begin{cases}y^{2}=8x\\\\y=x-2\\end{cases}\nEliminating y gives x^{2}-12x+4=0. Let M(x_{1},y_{1}), N(x_{2},y_{2}), so x_{1}+x_{2}=12, thus MN=x_{1}+x_{2}+p=16. The distance d from point P(2,4) to line l is d=\\frac{4}{\\sqrt{1^{2}+(-1)^{2}}}=2\\sqrt{2}. Therefore, S_{\\DeltaPMN}=\\frac{1}{2}MN\\cdot d=\\frac{1}{2}\\times16\\times2\\sqrt{2}=16\\sqrt{2}"
  },
  {
    "text": "Given that $O$ is the coordinate origin, $A$ and $B$ are two moving points on the parabola $y^{2}=2 p x(p \\geq 0)$ distinct from point $P$, and $\\angle A O B=90^{\\circ}$. If the maximum distance from point $O$ to the line $A B$ is $8$, then the value of $p$ is?",
    "fact_expressions": "O: Origin;G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>=0;A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);P: Point;Negation(P=A);Negation(P=B);AngleOf(A, O, B) = ApplyUnit(90, degree);Max(Distance(O, LineOf(A, B))) = 8",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 5], [86, 90]], [[19, 45]], [[19, 45]], [[111, 114]], [[22, 45]], [[11, 14]], [[15, 18]], [[11, 57]], [[11, 57]], [[48, 52]], [[11, 57]], [[11, 57]], [[59, 84]], [[86, 109]]]",
    "query_spans": "[[[111, 118]]]",
    "process": "Let the equation of line OA be y = kx. Solving the system of equations  \n\\begin{cases} y = kx \\\\ y^2 = 2px \\end{cases},  \nwe obtain point A\\left(\\frac{2p}{k^2}, \\frac{2p}{k}\\right).  \nSince the equation of line OB is y = -\\frac{1}{k}x,  \nthen B(2pk^2, -2pk).  \nThus, the equation of line AB is  \ny + 2pk = \\frac{\\frac{2p}{k} + 2pk}{\\frac{2p}{k^2} - 2pk^2}(x - 2pk^2),  \nthat is,  \ny + 2pk = \\frac{k}{1 - k^2}(x - 2pk^2).  \nSetting y = 0, we get x = 2p.  \nTherefore, line AB always passes through the fixed point (2p, 0).  \nHence, when line AB is perpendicular to the x-axis, the distance from point O to line AB is maximized.  \nThus, 2p = 8, so p = 4. The answer is: 4"
  },
  {
    "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ are $y=\\pm \\sqrt{3} x$, then its eccentricity is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;Expression(Asymptote(G)) = (y = pm*(sqrt(3)*x))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 48], [75, 76]], [[2, 48]], [[5, 48]], [[5, 48]], [[2, 73]]]",
    "query_spans": "[[[75, 82]]]",
    "process": "Using the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ given by $y=\\pm\\sqrt{3}x$, we obtain $\\frac{b}{a}=\\sqrt{3}$; combining with the eccentricity formula, the conclusion can be found. From the problem, $\\because$ the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ are $y=\\pm\\sqrt{3}x^{2}$, $\\therefore \\frac{b}{a}=\\sqrt{3}$, $\\therefore e^{2}=\\frac{c^{2}}{a^{2}}=1+(\\frac{b}{a})^{2}=4$."
  },
  {
    "text": "Given that $F$ is the right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, point $M$ lies on the right branch of $C$, the coordinate origin is $O$, if $|F M|=2|O F|$ and $\\angle O F M=120^{\\circ}$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "F: Point;RightFocus(C) = F;C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;M: Point;PointOnCurve(M,RightPart(C)) = True;O: Origin;Abs(LineSegmentOf(F, M)) = 2*Abs(LineSegmentOf(O, F));AngleOf(O, F, M) = ApplyUnit(120, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(\\sqrt{3}+1)/2",
    "fact_spans": "[[[2, 5]], [[2, 71]], [[6, 67], [77, 80], [139, 145]], [[6, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[72, 76]], [[72, 84]], [[90, 93]], [[95, 109]], [[111, 137]]]",
    "query_spans": "[[[139, 151]]]",
    "process": "Let the left focus of hyperbola $ C $ be $ F_{1} $. From the given conditions, we have $ |MF| = |F_{1}F| = 2c $, $ \\angle MFF_{1} = 120^{\\circ} $. By the law of cosines,  \n$ |MF_{1}|^{2} = |MF|^{2} + |F_{1}F|^{2} - 2|MF| \\cdot |F_{1}F| \\cos\\angle MFF_{1} = 4c^{2} + 4c^{2} - 2 \\cdot 4c^{2} \\cdot \\left(-\\frac{1}{2}\\right) = 12c^{2} $,  \nso $ |MF_{1}| = 2\\sqrt{3}c $. By the definition of a hyperbola, $ |MF_{1}| - |MF| = 2a $, which gives $ 2\\sqrt{3}c - 2c = 2a $, thus $ c = \\frac{\\sqrt{3}+1}{2}a $. Therefore, the eccentricity of hyperbola $ C $ is $ e = \\frac{c}{a} = \\frac{\\sqrt{3}+1}{2} $."
  },
  {
    "text": "Given that the parabola $y^{2}=ax$ passes through the point $A(\\frac{1}{4}, 1)$, what is the distance from point $A$ to the focus of this parabola?",
    "fact_expressions": "G: Parabola;a: Number;A: Point;Expression(G) = (y^2 = a*x);Coordinate(A) = (1/4, 1);PointOnCurve(A, G)",
    "query_expressions": "Distance(A, Focus(G))",
    "answer_expressions": "5/4",
    "fact_spans": "[[[2, 15], [45, 48]], [[5, 15]], [[16, 36], [39, 43]], [[2, 15]], [[16, 36]], [[2, 36]]]",
    "query_spans": "[[[39, 56]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, the two endpoints $A$ and $B$ of segment $AB$ move on the ellipse, and $|AB|=2$. Let $M(x_{0}, y_{0})$ be the midpoint of $AB$. Then the maximum value of $x_{0}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);A: Point;B: Point;Endpoint(LineSegmentOf(A, B)) = {A, B};PointOnCurve(A, G);PointOnCurve(B, G);Abs(LineSegmentOf(A, B)) = 2;x0: Number;y0: Number;M: Point;Coordinate(M) = (x0, y0);MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Max(x0)",
    "answer_expressions": "(2*sqrt(3))/3",
    "fact_spans": "[[[2, 29], [50, 52]], [[2, 29]], [[42, 45]], [[46, 49]], [[30, 49]], [[42, 55]], [[46, 55]], [[57, 66]], [[96, 103]], [[68, 85]], [[68, 85]], [[68, 85]], [[68, 94]]]",
    "query_spans": "[[[96, 109]]]",
    "process": "Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, and $F$ be the right focus of the ellipse. According to the problem, the semi-major axis length of the ellipse is $a=2$, the semi-minor axis length is $b=1$, and the semi-focal distance is $c=\\sqrt{3}$. Therefore, $|AF| = \\sqrt{(x_{1}-\\sqrt{3})^{2}+y_{1}^{2}} = \\sqrt{(x_{1}-\\sqrt{3})^{2}+1-\\frac{x_{1}^{2}}{4}} = \\sqrt{\\frac{(4-\\sqrt{3}x_{1})^{2}}{4}} = 2-\\frac{\\sqrt{3}}{2}x_{1}$. Similarly, $|BF| = 2-\\frac{\\sqrt{3}}{2}x_{2}$. Since $2 = |AB| \\leqslant |AF| + |BF| = (2-\\frac{\\sqrt{3}}{2}x_{1}) + (2-\\frac{\\sqrt{3}}{2}x_{2}) = 4-\\frac{\\sqrt{3}}{2}(x_{1}+x_{2}) = 4-\\sqrt{3}x_{0}$, we have $2 \\leqslant 4-\\sqrt{3}x_{0}$, solving gives $x_{0} \\leqslant \\frac{2\\sqrt{3}}{3}$."
  },
  {
    "text": "It is known that an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is parallel to $y=\\sqrt{3} x-1$, and one of its foci lies on the directrix of the parabola $y^{2}=8 \\sqrt{2} x$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;a>0;b>0;l:Line;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 8*sqrt(2)*x);Expression(l)=(y=sqrt(3)*x-1);IsParallel(l,OneOf(Asymptote(G)));PointOnCurve(OneOf(Focus(G)),Directrix(H))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/2 - y^2/6 = 1",
    "fact_spans": "[[[2, 58], [85, 86], [121, 124]], [[5, 58]], [[5, 58]], [[92, 115]], [[5, 58]], [[5, 58]], [[65, 81]], [[2, 58]], [[92, 115]], [[65, 81]], [[2, 83]], [[85, 119]]]",
    "query_spans": "[[[121, 129]]]",
    "process": "The directrix of the parabola $ y^{2}=8\\sqrt{2}x $ is $ x=-2\\sqrt{2} $. From the given conditions, we have $ c=2\\sqrt{2} $. Let one asymptote of the hyperbola be parallel to $ y=\\sqrt{3}x-1 $. Then, from the given conditions, $ \\frac{b}{a}=\\sqrt{3} $, so $ b^{2}=3a^{2} $. Solving gives $ a^{2}=2 $, $ b^{2}=6 $. Therefore, the standard equation of the hyperbola is $ \\frac{x^{2}}{2}-\\frac{y^{2}}{6}=1 $. The answer should be: $ \\frac{x^{2}}{2}-\\frac{y^{2}}{6}=1 $"
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=4x$, draw a line $l$ intersecting the parabola at points $A$ and $B$, and intersecting the directrix at point $C$. If $\\overrightarrow{F C}=4 \\overrightarrow{F B}$, then $|\\overrightarrow{A B}|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, G) = {A, B};C: Point;Intersection(l, Directrix(G)) = C;VectorOf(F, C) = 4*VectorOf(F, B)",
    "query_expressions": "Abs(VectorOf(A, B))",
    "answer_expressions": "9/2",
    "fact_spans": "[[[1, 15], [29, 32]], [[1, 15]], [[18, 21]], [[1, 21]], [[22, 27]], [[0, 27]], [[34, 37]], [[38, 41]], [[22, 43]], [[49, 53]], [[22, 53]], [[55, 100]]]",
    "query_spans": "[[[102, 128]]]",
    "process": "According to the parabolic equation $ y^{2} = 4x $, we obtain the focus coordinate $ F(1,0) $, directrix $ l: x = -1 $. Draw $ BB_{1} \\perp l $ with foot of perpendicular $ B_{1} $, then $ |BB_{1}| = |BF| $. Since $ \\overrightarrow{FC} = 4\\overrightarrow{FB} $, it follows that $ |\\overrightarrow{FC}| = 4|\\overrightarrow{FB}| $, so $ |BC| = 3|BF| = 3|BB_{1}| $. In right triangle $ \\triangle BCB_{1} $, since $ |BC| = 3|BB_{1}| $, we have $ \\tan\\angle B_{1}BC = \\frac{B_{1}C}{BC} = 2\\sqrt{2} $, thus the slope of line $ AB $ is $ k = 2\\sqrt{2} $. Therefore, the equation of line $ AB $ is $ y = 2\\sqrt{2}(x - 1) $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, solve the system of equations:\n$$\n\\begin{cases}\ny = 2\\sqrt{2}(x - 1) \\\\\ny^{2} = 4x\n\\end{cases}\n$$\nSimplifying yields $ 2x^{2} - 5x + 2 = 0 $. Thus, $ x_{1} + x_{2} = \\frac{5}{2} $, so $ |AB| = |AF| + |BF| = x_{1} + x_{2} + p = \\frac{5}{2} + 2 = \\frac{9}{2} $."
  },
  {
    "text": "The distance from the vertex of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ to its asymptote is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/5 = 1)",
    "query_expressions": "Distance(Vertex(G), Asymptote(G))",
    "answer_expressions": "2*sqrt(5)/3",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 50]]]",
    "process": ""
  },
  {
    "text": "Given that one focus of the ellipse $5 x^{2}-k y^{2}=5$ is $(0 , 2)$, then $k=$?",
    "fact_expressions": "G: Ellipse;k: Number;H: Point;Expression(G) = (-k*y^2 + 5*x^2 = 5);Coordinate(H) = (0, 2);OneOf(Focus(G)) = H",
    "query_expressions": "k",
    "answer_expressions": "-1",
    "fact_spans": "[[[2, 23]], [[40, 43]], [[29, 38]], [[2, 23]], [[29, 38]], [[2, 38]]]",
    "query_spans": "[[[40, 45]]]",
    "process": "\\because5x^{2}-ky^{2}=5\\therefore\\frac{y^{2}}{-k}+\\frac{x^{2}}{1}=1 Because one focus of the ellipse 5x^{2}-ky^{2}=5 is (0,2), so \\frac{5}{-1}-1=2^{2}\\therefore k=-1"
  },
  {
    "text": "Given $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, $M$, $N$ are the left and right vertices of the ellipse, $P$ is an arbitrary point on the ellipse, and the slopes of lines $PM$, $PN$ are $k_{1}$, $k_{2}$ respectively, $(k_{1} k_{2} \\neq 0)$. If the minimum value of $|k_{1}|+|k_{2}|$ is $1$, then what is the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;M: Point;P: Point;N: Point;a:Number;b:Number;a>b;b>0;k1:Number;k2:Number;Expression(G)=(x^2/a^2+y^2/b^2=1);LeftVertex(G)=M;RightVertex(G)=N;PointOnCurve(P, G);Slope(LineOf(P, M)) = k1;Slope(LineOf(P,N))=k2;Negation(k1*k2=0);Min(Abs(k1) + Abs(k2)) = 1",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[2, 52], [62, 64], [75, 77], [174, 176]], [[53, 56]], [[71, 74]], [[58, 61]], [[2, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[105, 112]], [[114, 121]], [[2, 52]], [[53, 70]], [[53, 70]], [[71, 82]], [[84, 121]], [[84, 121]], [[123, 145]], [[147, 172]]]",
    "query_spans": "[[[174, 182]]]",
    "process": "According to the problem, the coordinates of M and N are (-a,0) and (a,0), respectively. Let P(x,y), then k_{1}=\\frac{y}{x+a}, k_{2}=\\frac{y}{x-a}, then |k_{1}|+|k_{2}|=|\\frac{y}{x+a}|+|\\frac{y}{x-a}|=\\frac{2|y|}{a^{2}-x^{2}}=\\frac{2b^{2}}{a|y|}\\geqslant\\frac{2b^{2}}{ab}=\\frac{2b}{a}, so \\frac{2b}{a}=1, that is, b=\\frac{1}{2}a, thus c=\\sqrt{a^{2}-b^{2}}=\\frac{\\sqrt{3}}{2}a, then e=\\frac{c}{a}=\\frac{\\sqrt{3}}{2}"
  },
  {
    "text": "If the focus of a parabola lies on the line $x-2 y+2=0$, then the standard equation of the parabola is?",
    "fact_expressions": "G: Parabola;H: Line;Expression(H) = (x - 2*y + 2 = 0);PointOnCurve(Focus(G), H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{(x^2=4*y),(y^2=-8*x)}",
    "fact_spans": "[[[1, 4], [24, 27]], [[8, 21]], [[8, 21]], [[1, 22]]]",
    "query_spans": "[[[24, 34]]]",
    "process": "Since the line $x-2y+2=0$ intersects the $x$-axis at $(-2,0)$ and the $y$-axis at $(0,1)$, when the focus of the parabola is $(-2,0)$, the equation of the parabola is $y^{2}=-8x$; when the focus of the parabola is $(0,1)$, the equation of the parabola is $x^{2}=4y$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the right focus is $F$. The line $y=\\frac{b}{3}$ intersects $C$ at points $A$ and $B$. If the circle with diameter $AB$ passes through point $F$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;F: Point;RightFocus(C)=F;G: Circle;H: Line;Expression(H) = (y = b/3);A: Point;B: Point;Intersection(H, C) = {A, B};IsDiameter(LineSegmentOf(A,B),G);PointOnCurve(F, G)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(14)/4",
    "fact_spans": "[[[2, 59], [86, 89], [121, 124]], [[2, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[64, 67]], [[2, 67]], [[112, 113]], [[68, 85]], [[68, 85]], [[91, 94]], [[95, 98]], [[68, 100]], [[102, 113]], [[112, 119]]]",
    "query_spans": "[[[121, 130]]]",
    "process": "Let $ F(c,0) $ ($ c>0 $), substitute $ y=\\frac{b}{3} $ into the ellipse equation to obtain $ x=\\pm\\frac{2\\sqrt{2}}{3}a $. Without loss of generality, let $ A(-\\frac{2\\sqrt{2}a}{3},\\frac{b}{3}) $, $ B(\\frac{2\\sqrt{2}a}{3},\\frac{b}{3}) $, $ \\overrightarrow{AF}=(c+\\frac{2\\sqrt{2}a}{3},-\\frac{b}{3}) $, $ \\overrightarrow{BF}=(c-\\frac{2\\sqrt{2}a}{3},-\\frac{b}{3}) $. Since the circle with diameter $ AB $ passes through point $ F $, we have $ \\overrightarrow{AF}\\cdot\\overrightarrow{BF}=0 $, that is, $ (c+\\frac{2\\sqrt{2}a}{3})(c-\\frac{2\\sqrt{2}a}{3})+\\frac{b^{2}}{9}=0 $. Rearranging gives $ c^{2}-\\frac{8}{9}a^{2}+\\frac{b^{2}}{9}=0 $."
  },
  {
    "text": "Given that the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{4}=1$ $(m>0)$ has eccentricity $\\sqrt{3}$, what is the equation of its asymptotes?",
    "fact_expressions": "G: Hyperbola;m: Number;m>0;Expression(G) = (-y^2/4 + x^2/m = 1);Eccentricity(G) = sqrt(3)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(2)*x)",
    "fact_spans": "[[[2, 45], [62, 63]], [[5, 45]], [[5, 45]], [[2, 45]], [[2, 60]]]",
    "query_spans": "[[[62, 70]]]",
    "process": "Analysis: Using the eccentricity formula, we can calculate $ m $. Given that the hyperbola $ \\frac{x^{2}}{m} - \\frac{y^{2}}{4} = 1 $ ($ m > 0 $) has an eccentricity of $ \\sqrt{3} $, we have $ b = 2 $, $ c = \\sqrt{m + 4} $. Therefore, from the given condition, $ e = \\frac{c}{a} = \\frac{\\sqrt{m + 4}}{\\sqrt{m}} = \\sqrt{3} $, solving this gives $ m = 2 $. Therefore, the equation of the hyperbola is $ \\frac{x^{2}}{2} - \\frac{y^{2}}{4} = 1 $. Hence, the asymptotes are $ y = \\pm\\sqrt{2}x $."
  },
  {
    "text": "Through the right vertex $A$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a line with slope $-1$. The points of intersection of this line with the two asymptotes of the hyperbola are $B$ and $C$, respectively. If $\\overrightarrow{A B}=\\frac{1}{2} \\overrightarrow{B C}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;A: Point;B: Point;C: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightVertex(G)=A;PointOnCurve(A,H);Slope(H)=-1;L1:Line;L2:Line;Asymptote(G)={L1,L2};Intersection(H,L1)=B;Intersection(H,L2)=C;VectorOf(A, B) = VectorOf(B, C)/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[1, 57], [80, 83], [162, 165]], [[4, 57]], [[4, 57]], [[73, 75], [77, 79]], [[61, 64]], [[95, 98]], [[99, 102]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 64]], [[0, 75]], [[65, 75]], [], [], [[80, 89]], [[77, 102]], [[77, 102]], [[105, 160]]]",
    "query_spans": "[[[162, 171]]]",
    "process": "The line: $ y = -x + a $ intersects the asymptote $ l_{1}: bx \\cdot ay = 0 $ at $ B\\left(\\frac{a^{2}}{a+b}, \\frac{ab}{a+b}\\right) $, and intersects the asymptote $ l_{2}: bx + ay = 0 $ at $ C\\left(\\frac{a2}{a-b}, \\frac{-ab}{a-b}\\right) $. $ \\because A(a, 0) $, $ \\therefore \\overrightarrow{AB} = \\left(\\frac{ab}{a+b}, \\frac{ab}{a+b}\\right) $, $ \\overrightarrow{BC} = \\left(\\frac{2a2b}{a^{2}-b^{2}}, \\frac{2a2b}{a^{2}-b^{2}}\\right) \\cdot \\frac{ab}{a+b} = \\frac{a^{2}b}{a^{2}-b^{2}}b = 2a \\cdot c^{2} \\cdot a^{2} = 4a^{2}2 = \\frac{c^{2}}{a^{2}} = 5 $, $ \\therefore e = \\sqrt{ $"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$, the left focus is $F$, the left directrix is $l$, point $A \\in l$, the segment $AF$ intersects the ellipse $C$ at point $B$. If $\\overrightarrow{FA} = 3\\overrightarrow{FB}$, then $|\\overrightarrow{AF}| = $?",
    "fact_expressions": "C: Ellipse;A: Point;F: Point;B: Point;l: Line;Expression(C) = (x^2/2 + y^2 = 1);LeftFocus(C) = F;LeftDirectrix(C) = l;PointOnCurve(A,l);VectorOf(F, A) = 3*VectorOf(F, B);Intersection(LineSegmentOf(A,F), C) = B",
    "query_expressions": "Abs(VectorOf(A, F))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 34], [66, 71]], [[51, 57]], [[39, 42]], [[72, 76]], [[47, 50]], [[2, 34]], [[2, 42]], [[2, 50]], [[51, 57]], [[77, 125]], [[58, 76]]]",
    "query_spans": "[[[127, 153]]]",
    "process": ""
  },
  {
    "text": "The equation of the directrix of the parabola $y^{2}=-4 x$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -4*x)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x = 1",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 22]]]",
    "process": "The focus of the parabola \\( y^{2} = -4x \\) lies on the x-axis, and it opens to the left. \\( 2p = 4 \\), \\( \\frac{p}{2} = 1 \\). Therefore, the directrix equation of the parabola \\( y^{2} = -4x \\) is \\( x = 1 \\)."
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and let $P$ be a point on the right branch of the hyperbola $C$. If $|P F_{1}|+|P F_{2}|=4 a $ and $\\angle F_{1} P F_{2}=60^{\\circ}$, then what is the equation of the asymptotes of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, RightPart(C));Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 4*a;AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "sqrt(3)*x+pm*2*y=0",
    "fact_spans": "[[[17, 78], [89, 95], [165, 171]], [[24, 78]], [[24, 78]], [[85, 88]], [[1, 8]], [[9, 16]], [[24, 78]], [[24, 78]], [[17, 78]], [[1, 84]], [[1, 84]], [[85, 100]], [[102, 128]], [[130, 163]]]",
    "query_spans": "[[[165, 179]]]",
    "process": "From the definition of hyperbola, we have |PF_{1}|-|PF_{2}|=2a, and |PF_{1}|+|PF_{2}|=4a. Then we can find |PF_{1}|=3a, |PF_{2}|=a. In \\triangle PF_{1}F_{2}, by the cosine law, we get \\cos60^{\\circ}=\\frac{|PF_{1}|^{2}+|PF_{2}|^{2}-|F_{1}F_{2}|^{2}}{2|PF_{1}|\\cdot|PF_{2}|}, that is, \\frac{1}{2}=\\frac{(3a)^{2}+a^{2}-4c^{2}}{2\\times3a\\times a}, which implies 3a^{2}=10a^{2}-4c^{2}, 4c^{2}=7a^{2}. Since a^{2}+b^{2}=c^{2}, then \\frac{b^{2}}{a^{2}}=\\frac{3}{4}. Hence, the asymptotes are y=\\pm\\frac{\\sqrt{3}}{2}x, or \\sqrt{3}x\\pm2y=0."
  },
  {
    "text": "Let the parabola $E$: $y^{2}=8x$ have focus $F$. A line passing through point $F$ intersects the parabola at points $A$ and $B$. From the midpoint $M$ of chord $AB$, a perpendicular is drawn to the directrix of $E$, intersecting the parabola $E$ at point $P$. If $|PF|=\\frac{7}{2}$, then $|AB|=$?",
    "fact_expressions": "E: Parabola;Expression(E) = (y^2 = 8*x);F: Point;Focus(E) = F;PointOnCurve(F, G) = True;G: Line;Intersection(G, E) = {A, B};A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), E) = True;MidPoint(LineSegmentOf(A, B)) = M;M: Point;L: Line;PointOnCurve(M, L) = True;IsPerpendicular(L, Directrix(E)) = True;Intersection(L, E) = P;P: Point;Abs(LineSegmentOf(P, F)) = 7/2",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "14",
    "fact_spans": "[[[1, 20], [37, 40], [66, 69], [77, 83]], [[1, 20]], [[24, 27], [29, 33]], [[1, 27]], [[28, 36]], [[34, 36]], [[34, 51]], [[42, 45]], [[46, 49]], [[37, 59]], [[54, 65]], [[62, 65]], [], [[52, 75]], [[52, 75]], [[52, 89]], [[85, 89]], [[91, 110]]]",
    "query_spans": "[[[112, 121]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$ and directrix $l$, $P$ is a point on $l$, $Q$ is an intersection point of line $PF$ and $C$, and if $\\overrightarrow{FP}=4\\overrightarrow{FQ}$, then $|QO|$=?",
    "fact_expressions": "C: Parabola;l: Line;P: Point;F: Point;Q: Point;O: Origin;Expression(C) = (y^2 = 8*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(P, l);OneOf(Intersection(LineOf(P, F), C)) = Q;VectorOf(F, P) = 4*VectorOf(F, Q)",
    "query_expressions": "Abs(LineSegmentOf(Q, O))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 20], [60, 63]], [[31, 34], [41, 44]], [[37, 40]], [[24, 27]], [[48, 51]], [[117, 124]], [[2, 20]], [[2, 27]], [[2, 34]], [[37, 47]], [[48, 68]], [[70, 115]]]",
    "query_spans": "[[[117, 126]]]",
    "process": "Let $ Q\\left(\\frac{t^{2}}{8}, t\\right) $, $ P(-2, m) $. Since $ \\overrightarrow{FP} = 4\\overrightarrow{FQ} $, we have $ (4, -m) = 4\\left(\\frac{t^{2}}{8} - 2, t\\right) $, so $ t^{2} = 8 $. Hence, $ |QO| = \\sqrt{\\frac{t^{4}}{64} + t^{2}} = \\sqrt{1 + 8} = 3 $."
  },
  {
    "text": "Given that the distance from the right focus of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{m^{2}+2}=1$ to its asymptote is $\\sqrt{3}$, find the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (-y^2/(m^2 + 2) + x^2/m = 1);Distance(RightFocus(G),Asymptote(G))=sqrt(3)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 46], [51, 52], [72, 75]], [[5, 46]], [[2, 46]], [[2, 69]]]",
    "query_spans": "[[[72, 81]]]",
    "process": "Let the angle of inclination be _{\\theta}, \\tan^{2}\\theta = \\frac{m2+2}{m} = \\frac{(\\sqrt{3})^{2}}{m^{2}+m+2-3} \\Rightarrow (m+1)(m-1)(m^{2}+2)=0 \\Rightarrow m=1 \\Rightarrow e=\\sqrt{\\frac{4}{1}}=2."
  },
  {
    "text": "The coordinates of the foci of the ellipse $8 x^{2}+3 y^{2}=24$ are?",
    "fact_expressions": "G: Ellipse;Expression(G) = (8*x^2 + 3*y^2 = 24)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, pm*sqrt(5))",
    "fact_spans": "[[[0, 22]], [[0, 22]]]",
    "query_spans": "[[[0, 29]]]",
    "process": "The ellipse equation simplifies to the standard form: \\frac{x2}{3}+\\frac{y^{2}}{8}=1. From this, the coordinates of the foci of the ellipse are (0,-\\sqrt{5}), (0,\\sqrt{5})."
  },
  {
    "text": "It is known that the focus of the parabola $y^{2}=4 x$ coincides with the center of the circle $x^{2}+y^{2}+m x-4=0$. Then the value of $m$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Circle;m: Number;Expression(H) = (m*x + x^2 + y^2 - 4 = 0);Focus(G) = Center(H)",
    "query_expressions": "m",
    "answer_expressions": "-2",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[20, 42]], [[49, 52]], [[20, 42]], [[2, 47]]]",
    "query_spans": "[[[49, 56]]]",
    "process": "The focus of the parabola has coordinates (1,0), and the center of the circle has coordinates $(-\\frac{m}{2},0)$, so $-\\frac{m}{2}=1$, which gives $m=-2$. Fill in $-2$."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $y^{2}=3x$, $P$ is any point on the parabola, and $A(3,\\ 2)$ is a fixed point on the plane, then the minimum value of $|PF|+|PA|$ is?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 3*x);Coordinate(A) = (3, 2);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "15/4",
    "fact_spans": "[[[6, 20], [28, 31]], [[36, 46]], [[24, 27]], [[2, 5]], [[6, 20]], [[36, 46]], [[2, 23]], [[24, 35]]]",
    "query_spans": "[[[55, 72]]]",
    "process": ""
  },
  {
    "text": "Let point $P$ be a moving point on the curve $y=x^{2}$, and let $l$ be the tangent line to the curve $y=x^{2}$ at point $P$. The line passing through point $P$ and perpendicular to line $l$ intersects the curve $y=x^{2}$ at another point $Q$. Then the minimum value of $PQ$ is?",
    "fact_expressions": "P: Point;G: Curve;Expression(G) = (y = x^2);PointOnCurve(P, G);l: Line;TangentOnPoint(P, G) = l;H: Line;PointOnCurve(P, H);IsPerpendicular(H, l);Q: Point;Intersection(H, G) = {P, Q}",
    "query_expressions": "Min(LineSegmentOf(P, Q))",
    "answer_expressions": "(3*sqrt(3))/2",
    "fact_spans": "[[[1, 5], [36, 40], [50, 54]], [[6, 17], [24, 35], [67, 78]], [[6, 17]], [[1, 23]], [[45, 48], [56, 61]], [[24, 48]], [[64, 66]], [[49, 66]], [[55, 66]], [[84, 87]], [[50, 87]]]",
    "query_spans": "[[[89, 99]]]",
    "process": ""
  },
  {
    "text": "Point $A(1,1)$, $F_{1}$ is the left focus of the ellipse $5 x^{2} + 9 y^{2} = 45$, and point $P$ is a moving point on the ellipse. Then the maximum value of $|P A| + |P F_{1}|$ is?",
    "fact_expressions": "G: Ellipse;A: Point;P: Point;F1: Point;Expression(G) = (5*x^2 + 9*y^2 = 45);Coordinate(A) = (1, 1);LeftFocus(G) = F1;PointOnCurve(P, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F1)))",
    "answer_expressions": "6+sqrt(2)",
    "fact_spans": "[[[19, 41], [51, 53]], [[0, 9]], [[46, 50]], [[11, 18]], [[19, 41]], [[0, 9]], [[11, 45]], [[46, 57]]]",
    "query_spans": "[[[59, 82]]]",
    "process": "Transform $5x^{2}+9y^{2}=45$ into $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$. Let $F_{2}$ be the right focus of the ellipse, then $F_{2}(2,0)$. By the definition of the ellipse, $|PA|+|PF_{1}|=|PA|+6-|PF_{2}|\\leqslant6+|AF_{2}|=6+\\sqrt{2}$, with equality if and only if $P$ is the intersection point of the extension of $AF_{2}$ and the ellipse."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, a line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. Then, what is the perimeter of $\\triangle A B F_{2}$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);F2: Point;F1: Point;Focus(G) = {F1, F2};H: Line;PointOnCurve(F1, H) = True;Intersection(H, G) = {A, B};A: Point;B: Point",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "20",
    "fact_spans": "[[[18, 56], [74, 76]], [[18, 56]], [[10, 17]], [[2, 9]], [[2, 61]], [[71, 73]], [[62, 73]], [[71, 86]], [[77, 80]], [[81, 84]]]",
    "query_spans": "[[[88, 114]]]",
    "process": "According to the definition of an ellipse, compute directly. [Detailed solution] The perimeter $ l $ of $ \\Delta ABF_{2} $ is $ |AB| + |AF_{2}| + |BF_{2}| = |AF_{1}| + |BF_{1}| + |AF_{2}| + |BF_{2}| = (|AF_{1}| + |AF_{2}|) + (|BF_{1}| + |BF_{2}|) = 4a $. From the ellipse equation, $ a^{2} = 25 \\Rightarrow a = 5 $, so the perimeter $ l $ of $ \\triangle ABF_{2} $ is $ 20 $."
  },
  {
    "text": "Given that the line $y = kx + 1$ ($k \\neq 0$) intersects the parabola $x^2 = 4y$ at two points $E$ and $F$, and the chord length intercepted on the $x$-axis by the circle with diameter $EF$ is $2\\sqrt{7}$, then $k = $?",
    "fact_expressions": "Z: Line;Expression(Z) = (y = k*x + 1);k: Number;Negation(k=0);G: Parabola;Expression(G) = (x^2 = 4*y);E: Point;F: Point;Intersection(Z, G) = {E, F};H: Circle;IsDiameter(LineSegmentOf(E, F), H);Length(InterceptChord(xAxis, H)) = 2*sqrt(7)",
    "query_expressions": "k",
    "answer_expressions": "pm*1",
    "fact_spans": "[[[2, 23]], [[2, 23]], [[85, 88]], [[4, 23]], [[24, 38]], [[24, 38]], [[39, 42]], [[43, 46]], [[2, 48]], [[59, 60]], [[49, 60]], [[59, 83]]]",
    "query_spans": "[[[85, 90]]]",
    "process": "From \\begin{cases}y=kx+1\\\\x^{2}=4y\\end{cases}, eliminating $y$ and simplifying gives $x^{2}-4kx-4=0$. Let $E(x_{1},y_{1})$, $F(x_{2},y_{2})$, then $x_{1}+x_{2}=4k$, $x_{1}x_{2}=-4$, $\\therefore y_{1}+y_{2}=k(x_{1}+x_{2})+2=4k^{2}+2$. By the definition of the parabola, $|EF|=y_{1}+y_{2}+2=4k^{2}+4$. $\\therefore$ the radius of the circle with $EF$ as diameter is $\\frac{1}{2}|EF|=2k^{2}+2$, and the distance from the center of the circle to the $x$-axis is $\\frac{1}{2}(y_{1}+y_{2})=2k^{2}+1$. According to the problem, $(2k^{2}+2)^{2}=(\\sqrt{7})^{2}+(2k^{2}+1)^{2}$, solving yields $k=+1$."
  },
  {
    "text": "Given the ellipse equation $9 x^{2}+16 y^{2}=144$, what is its major axis length? minor axis length? focal distance? eccentricity?",
    "fact_expressions": "G: Ellipse;Expression(G) = (9*x^2 + 16*y^2 = 144)",
    "query_expressions": "Length(MajorAxis(G));Length(MinorAxis(G));FocalLength(G);Eccentricity(G)",
    "answer_expressions": "8\n6\n2*sqrt(7)\nsqrt(7)/4",
    "fact_spans": "[[[2, 4], [31, 32]], [[2, 29]]]",
    "query_spans": "[[[31, 38]], [[31, 43]], [[31, 47]], [[31, 52]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\Gamma$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the right focus is $F(3,0)$, and the upper and lower vertices are $A$ and $B$, respectively. The line $AF$ intersects $\\Gamma$ at another point $M$. If the line $BM$ intersects the $x$-axis at point $N(12,0)$, then the eccentricity of $\\Gamma$ is?",
    "fact_expressions": "Gamma: Ellipse;A: Point;F: Point;B: Point;M: Point;N: Point;Expression(Gamma) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (3, 0);Coordinate(N) = (12, 0);RightFocus(Gamma) = F;UpperVertex(Gamma) = A;LowerVertex(Gamma) = B;Intersection(LineOf(A,F), Gamma) = {A,M};Intersection(LineOf(B,M), xAxis) = N;a:Number;b:Number;a>b;b>0",
    "query_expressions": "Eccentricity(Gamma)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 64], [102, 110], [144, 152]], [[86, 89]], [[69, 77]], [[90, 93]], [[114, 117]], [[132, 142]], [[2, 64]], [[69, 77]], [[132, 142]], [[2, 77]], [[2, 93]], [[2, 93]], [[86, 117]], [[119, 142]], [[2, 64]], [[2, 64]], [[2, 64]], [[2, 64]]]",
    "query_spans": "[[[144, 158]]]",
    "process": "From the given conditions, we have A(0,b), B(0,-b). Then the equations of lines AM and BN are respectively $\\frac{x}{3}+\\frac{y}{b}=1$, $\\frac{x}{12}-\\frac{y}{b}=1$. Solving these two line equations simultaneously, we obtain $M(\\frac{24}{5},-\\frac{3b}{5})$. Then $\\frac{24^{2}}{25a^2}+\\frac{9}{25}=1$, solving gives $a=6$. Thus, the eccentricity of the ellipse is $e=\\frac{3}{6}=\\frac{1}{2}$."
  },
  {
    "text": "Given that the length of the major axis of an ellipse is $\\sqrt{3}$ times the length of the minor axis, what is the eccentricity of this ellipse?",
    "fact_expressions": "G: Ellipse;Length(MajorAxis(G)) = sqrt(3) * Length(MinorAxis(G))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[2, 4], [28, 30]], [[2, 24]]]",
    "query_spans": "[[[28, 36]]]",
    "process": "From the given condition, $ a = \\sqrt{3}b $, combining with $ a^{2} = b^{2} + c^{2} $, derive $ \\frac{c}{a} $. From the condition $ 2a = \\sqrt{3} \\times 2b $, $ \\therefore a = \\sqrt{3}b $, so $ \\therefore e = \\frac{c}{a} = \\frac{\\sqrt{6}}{3} $. [Note] This problem examines finding the eccentricity of an ellipse; mastering the relation $ a^{2} = b^{2} + c^{2} $ is key to solving the problem."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively, and $P$ is an intersection point of the circle with diameter $F_{1}F_{2}$ and the hyperbola such that $\\angle P F_{1} F_{2}=2 \\angle P F_{2} F_{1}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "F1: Point;F2: Point;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;LeftFocus(G) = F1;RightFocus(G) = F2;IsDiameter(LineSegmentOf(F1, F2), H) = True;H: Circle;OneOf(Intersection(G, H)) = P;P: Point;AngleOf(P, F1, F2) = 2*AngleOf(P, F2, F1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)+1",
    "fact_spans": "[[[2, 9]], [[10, 17]], [[20, 76], [108, 111], [166, 169]], [[20, 76]], [[23, 76]], [[23, 76]], [[23, 76]], [[23, 76]], [[2, 82]], [[2, 82]], [[87, 106]], [[105, 106]], [[83, 116]], [[83, 86]], [[118, 164]]]",
    "query_spans": "[[[166, 175]]]",
    "process": "Let |F_{1}F_{2}|=2c. Since P is an intersection point of a circle with diameter F_{1}F_{2} and the hyperbola, \\triangle F_{1}F_{2}P is a right triangle, \\angle F_{1}PF_{2}=90^{\\circ}. Given \\angle PF_{1}F_{2}=2\\angle PF_{2}F_{1}, then \\angle PF_{1}F_{2}=60^{\\circ}, \\therefore |PF_{2}|=\\sqrt{3}c, |PF_{1}|=c, \\therefore |PF_{2}|-|PF_{1}|=\\sqrt{3}c-c=2a, \\therefore e=\\frac{c}{a}=\\frac{2}{\\sqrt{3}-1}=\\sqrt{3}+1."
  },
  {
    "text": "7. Given that points $F_{1}$, $F_{2}$ are the common foci of the ellipse $C_{1}$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the hyperbola $C_{2}$: $\\frac{x^{2}}{a'^{2}}-\\frac{y^{2}}{b'^{2}}=1$ $(a'>0, b'>0)$, and point $P$ is an intersection point of the two curves satisfying $\\angle F_{1} P F_{2}=90°$. Let the eccentricities of the ellipse and the hyperbola be $e_{1}$, $e_{2}$ respectively, then $\\frac{1}{e_{1}^{2}}+\\frac{1}{e_{2}^{2}}=$?",
    "fact_expressions": "C1: Ellipse;C2: Hyperbola;a, b: Number;a_, b_: Number;a > b;b > 0;Expression(C1) = (x^2/a^2 + y^2/b^2 = 1);a_ > 0;b_ > 0;Expression(C2) = (x^2/a_^2 - y^2/b_^2 = 1);F1: Point;F2: Point;Focus(C1) = {F1, F2};Focus(C2) = {F1, F2};P: Point;OneOf(Intersection(C1, C2)) = P;AngleOf(F1, P, F2) = ApplyUnit(90, degree);e1: Number;e2: Number;Eccentricity(C1) = e1;Eccentricity(C2) = e2",
    "query_expressions": "1/e1^2 + 1/e2^2",
    "answer_expressions": "2",
    "fact_spans": "[[[26, 36], [218, 220]], [[91, 102], [221, 224]], [[38, 89]], [[102, 162]], [[38, 89]], [[38, 89]], [[26, 89]], [[102, 162]], [[102, 162]], [[95, 162]], [[5, 15]], [[17, 24]], [[5, 168]], [[5, 168]], [[169, 174]], [[169, 184]], [[189, 215]], [[232, 240]], [[241, 249]], [[219, 249]], [[219, 249]]]",
    "query_spans": "[[[253, 296]]]",
    "process": ""
  },
  {
    "text": "If an ellipse with center at the origin and axes of symmetry along the coordinate axes passes through the points $(4 , 0)$ and $(0 , 2)$, then the eccentricity of this ellipse is equal to?",
    "fact_expressions": "G: Ellipse;H: Point;I: Point;O: Origin;Coordinate(H) = (4, 0);Coordinate(I) = (0, 2);Center(G) = O;SymmetryAxis(G)=axis;PointOnCurve(H,G);PointOnCurve(I,G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[17, 19], [45, 47]], [[23, 32]], [[33, 42]], [[4, 8]], [[23, 32]], [[33, 42]], [[1, 19]], [[9, 19]], [[17, 42]], [[17, 42]]]",
    "query_spans": "[[[45, 54]]]",
    "process": ""
  },
  {
    "text": "Given circle $C_{1}$: $(x+2)^{2}+y^{2}=1$ and circle $C_{2}$: $(x-2)^{2}+y^{2}=81$, a moving circle $M$ is tangent to both circle $C_{1}$ and circle $C_{2}$. Then the trajectory equation of the center of the moving circle $M$ is?",
    "fact_expressions": "M: Circle;C1:Circle;C2:Circle;M1:Point;Expression(C1)=((x+2)^2+y^2=1);Expression(C2)=((x-2)^2+y^2=81);IsTangent(M,C1);IsTangent(M,C2);Center(M)=M1",
    "query_expressions": "LocusEquation(M1)",
    "answer_expressions": "{x^2/25+y^2/21=1,x^2/16+y^2/12=1}",
    "fact_spans": "[[[63, 66], [90, 92]], [[2, 30], [69, 77]], [[31, 60], [78, 86]], [[94, 97]], [[2, 30]], [[31, 60]], [[63, 88]], [[63, 88]], [[90, 97]]]",
    "query_spans": "[[[94, 104]]]",
    "process": "By the problem, let the radius of the moving circle $M$ be $r$, the radius of circle $C_{1}$ be $r_{1}=1$, and the radius of circle $C_{2}$ be $r_{2}=9$. Then, when the moving circle $M$ is externally tangent to circle $C_{1}$ and internally tangent to circle $C_{2}$, $|MC_{1}|=r_{1}+r$, $|MC_{2}|=r_{2}-r$, so $|MC_{1}|+|MC_{2}|=(r_{1}+r)+(r_{2}-r)=r_{1}+r_{2}=10$. Since the centers are $(-2,0)$ and $(2,0)$, according to the definition of an ellipse, $2a=10$, then $a=5$, $c=2$, so $b^{2}=a^{2}-c^{2}=25-4=21$. Thus, the trajectory equation of the center $M$ of the moving circle is $\\frac{x^{2}}{25}+\\frac{y^{2}}{21}=1$. When the moving circle $M$ is internally tangent to both circles $C_{1}$ and $C_{2}$, $|MC_{1}|=r-r_{1}$, $|MC_{2}|=r_{2}-r$, so $|MC_{1}|+|MC_{2}|=(r-r_{1})+(r_{2}-r)=r_{2}-r_{1}=8$, then $2a=8$, i.e., $a=4$, so $b^{2}=a^{2}-c^{2}=16-4=12$, thus the trajectory equation of the center $M$ of the moving circle is $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$."
  },
  {
    "text": "A line $ l $ passing through the focus of the parabola $ x^{2} = -4y $ intersects the parabola at points $ P(x_{1}, y_{1}) $ and $ Q(x_{2}, y_{2}) $. If $ y_{1} + y_{2} = -6 $, then $ |PQ| = $?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = -4*y);l: Line;PointOnCurve(Focus(G), l);P: Point;Q: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(P) = (x1, y1);Coordinate(Q) = (x2, y2);Intersection(l, G) = {P, Q};y1 + y2 = -6",
    "query_expressions": "Abs(LineSegmentOf(P, Q))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 16], [26, 29]], [[1, 16]], [[20, 25]], [[0, 25]], [[30, 47]], [[49, 66]], [[30, 47]], [[49, 66]], [[30, 47]], [[49, 66]], [[30, 47]], [[49, 66]], [[20, 68]], [[70, 86]]]",
    "query_spans": "[[[88, 97]]]",
    "process": "According to the problem, the focus of $x^{2}=-4y$ is $(0,-1)$. Let the line passing through the focus $(0,-1)$ be $y=kx-1$. Then set $kx-1=-\\frac{x^{2}}{4}$, which gives $x^{2}+4kx-4=0$. Since the line $l$ intersects the parabola at points $P(x_{1},y_{1})$ and $Q(x_{2},y_{2})$, by Vieta's formulas we have $x_{1}+x_{2}=-4k$, $x_{1}x_{2}=-4$. Since $y_{1}+y_{2}=-6$, we have $y_{1}+y_{2}=kx_{1}-1+kx_{2}-1=k(x_{1}+x_{2})-2=-4k^{2}-2=-6$. Thus $k^{2}=1$. Therefore, $|PQ|=\\sqrt{1+k^{2}}|x_{1}-x_{2}|=\\sqrt{1+k^{2}}\\cdot\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\sqrt{2}\\cdot\\sqrt{16k^{2}+16}=8$."
  },
  {
    "text": "$P$ is a moving point on the right branch of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, $F$ is the right focus of the hyperbola, and $A(3,1)$ is given. Then the minimum value of $|P A|+|P F|$ is?",
    "fact_expressions": "G: Hyperbola;A: Point;P: Point;F: Point;Expression(G) = (x^2/3 - y^2 = 1);Coordinate(A) = (3, 1);PointOnCurve(P, RightPart(G));RightFocus(G) = F",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[4, 32], [44, 47]], [[54, 62]], [[0, 3]], [[40, 43]], [[4, 32]], [[54, 62]], [[0, 39]], [[40, 51]]]",
    "query_spans": "[[[65, 84]]]",
    "process": ""
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{3}=1$ are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the ellipse. If the midpoint of segment $P F_{1}$ lies on the $y$-axis, then $\\cos \\angle F_{1} P F_{2}$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/12 + y^2/3 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G) = True;PointOnCurve(MidPoint(LineSegmentOf(P, F1)), yAxis) = True",
    "query_expressions": "Cos(AngleOf(F1, P, F2))",
    "answer_expressions": "1/7",
    "fact_spans": "[[[0, 38], [65, 67]], [[0, 38]], [[44, 51]], [[52, 59]], [[0, 59]], [[60, 64]], [[60, 68]], [[71, 91]]]",
    "query_spans": "[[[94, 123]]]",
    "process": "According to the problem, we can obtain $ a=2\\sqrt{3} $, $ b=\\sqrt{3} $, $ c=3 $. Let the coordinates of point $ P $ be $ (x, y) $. Since the midpoint of segment $ PF_1 $ lies on the $ y $-axis, we can find $ P\\left(3, \\pm\\frac{\\sqrt{3}}{3}\\right) $. Then, $ |PF_1| $ and $ |PF_2| $ can be determined. Using the cosine theorem, the answer can be obtained."
  },
  {
    "text": "Given that the line $y=2x-3$ intersects the parabola $y^2=4x$ at points $A$ and $B$, $O$ is the origin, and the slopes of $OA$ and $OB$ are $k_1$ and $k_2$ respectively, then $\\frac{1}{k_1}+\\frac{1}{k_2}$=?",
    "fact_expressions": "G: Parabola;H: Line;O: Origin;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (y = 2*x - 3);Intersection(H, G) = {A, B};Slope(LineOf(O, A)) = k1;Slope(LineOf(O, B)) = k2;k1:Number;k2:Number",
    "query_expressions": "1/k2 + 1/k1",
    "answer_expressions": "1/2",
    "fact_spans": "[[[14, 28]], [[2, 13]], [[40, 43]], [[30, 33]], [[34, 37]], [[14, 28]], [[2, 13]], [[2, 39]], [[49, 85]], [[49, 85]], [[68, 76]], [[77, 85]]]",
    "query_spans": "[[[87, 122]]]",
    "process": "The line $ y = 2x - 3 $ and the parabola $ y^2 = 4x $ are solved simultaneously to find the coordinates of points $ A $ and $ B $, then $ \\frac{1}{k_{1}} + \\frac{1}{k_{2}} $ can be calculated. Solving $ y = 2x - 3 $ and $ y^2 = 4x $ simultaneously yields $ y^2 - 2y - 6 = 0 $, $ \\therefore y = 1 \\pm \\sqrt{7} $. $ \\therefore A\\left(2 + \\frac{\\sqrt{7}}{2}, 1 + \\sqrt{7}\\right) $, $ B\\left(2 - \\frac{\\sqrt{7}}{2}, 1 - \\sqrt{7}\\right) $. $ \\therefore \\frac{1}{k_{1}} + \\frac{1}{k_{2}} = \\frac{2 + \\frac{\\sqrt{7}}{2}}{1 + \\sqrt{7}} + \\frac{2 - \\frac{\\sqrt{7}}{2}}{1 - \\sqrt{7}} = \\frac{1}{2} $."
  },
  {
    "text": "$P$ is a point on the parabola $x^{2}=4 y$, the focus of the parabola is $F$, and $|P F|=5$. Then the ordinate of point $P$ is?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;Expression(G) = (x^2 = 4*y);PointOnCurve(P, G);Focus(G) = F;Abs(LineSegmentOf(P, F)) = 5",
    "query_expressions": "YCoordinate(P)",
    "answer_expressions": "4",
    "fact_spans": "[[[4, 18], [22, 25]], [[0, 3], [45, 49]], [[29, 32]], [[4, 18]], [[0, 21]], [[22, 32]], [[34, 43]]]",
    "query_spans": "[[[45, 55]]]",
    "process": "The directrix equation of the parabola \\( x^{2} = 4y \\) is \\( y = -1 \\), and the coordinates of the focus are \\( (0, 1) \\). Let the ordinate of point \\( P \\) be \\( (x, y) \\), then  \n\\[\n\\begin{cases}\nx^{2} = 4y \\\\\ny + 1 = 5\n\\end{cases}\n\\quad\n\\begin{cases}\nx = \\pm 4 \\\\\ny = 4\n\\end{cases}\n\\]  \nTherefore, the answer should be filled in as 4."
  },
  {
    "text": "The left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ are $F_{1}$ and $F_{2}$, respectively. A line passing through $F_{1}$ with an inclination angle of $30^{\\circ}$ intersects the right branch of the hyperbola at point $M$. If $M F_{2}$ is perpendicular to the $x$-axis, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;M: Point;F2: Point;F1: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, H);Inclination(H) = ApplyUnit(30, degree);Intersection(H, RightPart(G)) = M;IsPerpendicular(LineSegmentOf(M, F2), xAxis)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[0, 58], [112, 115], [142, 145]], [[3, 58]], [[3, 58]], [[109, 111]], [[118, 122]], [[75, 82]], [[67, 74], [84, 91]], [[3, 58]], [[3, 58]], [[0, 58]], [[0, 82]], [[0, 82]], [[83, 111]], [[92, 111]], [[109, 122]], [[124, 140]]]",
    "query_spans": "[[[142, 151]]]",
    "process": "In \\triangle MF_{1}F_{2}, since \\angle MF_{1}F_{2}=30^{0}, F_{1}F_{2}=2c, then MF_{1}=\\frac{4\\sqrt{3}}{3}c, MF_{2}=\\frac{2\\sqrt{3}}{3}c. By the definition of hyperbola, 2a=|MF_{1}|-|MF_{2}|=\\frac{2\\sqrt{3}}{3}c, so e=\\sqrt{3}."
  },
  {
    "text": "The standard equation of the circle with center at the focus of the parabola $y^{2}=4x$ and passing through the point $P(5,-2\\sqrt{5})$ is?",
    "fact_expressions": "G: Parabola;H: Circle;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(P) = (5, -2*sqrt(5));Focus(G)=Center(H);PointOnCurve(P,H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x-1)^2+y^2=36",
    "fact_spans": "[[[1, 15]], [[43, 44]], [[23, 42]], [[1, 15]], [[23, 42]], [[0, 44]], [[22, 44]]]",
    "query_spans": "[[[43, 51]]]",
    "process": "It is easy to see that point P lies on the parabola, and the coordinates of F are (1,0); then |PF| = 5 + \\frac{p}{2} = 5 + 1 = 6. Therefore, the standard equation of the required circle is (x-1)^{2} + y^{2} = 36."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(b>a>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, with eccentricity $e$. Point $P$ lies on the right branch of hyperbola $C$ such that $P F_{2} \\perp F_{1} F_{2}$. Point $B$ lies on the circle $x^{2}+y^{2}=b^{2}$ such that $\\overrightarrow{O B}=\\frac{2}{\\sqrt{21}} \\overrightarrow{O P}$. Then the minimum value of $\\frac{a^{2}+e^{2}}{b}$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;P: Point;F2: Point;F1: Point;O:Origin;B: Point;e:Number;b > a;a > 0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = b^2);LeftFocus(C) = F1;RightFocus(C) = F2;Eccentricity(C)=e;PointOnCurve(P, RightPart(C));IsPerpendicular(LineSegmentOf(P, F2), LineSegmentOf(F1, F2));PointOnCurve(B, G);VectorOf(O, B) = (2/sqrt(21))*VectorOf(O, P)",
    "query_expressions": "Min((a^2 + e^2)/b)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 60], [98, 104]], [[10, 60]], [[10, 60]], [[146, 166]], [[94, 97]], [[77, 84]], [[69, 76]], [[172, 235]], [[142, 145]], [[89, 93]], [[10, 60]], [[10, 60]], [[2, 60]], [[146, 166]], [[2, 84]], [[2, 84]], [[2, 92]], [[94, 110]], [[112, 139]], [[142, 170]], [[172, 235]]]",
    "query_spans": "[[[237, 266]]]",
    "process": "From the given conditions, we have |\\overrightarrow{OB}|=b, so |\\overrightarrow{OP}|=\\frac{\\sqrt{21}}{2}|\\overrightarrow{OB}|=\\frac{\\sqrt{21}}{2}b, thus \\frac{\\sqrt{21}}{2}b=\\sqrt{(\\frac{b^{2}}{a})^{2}+c^{2}}, so \\frac{21}{4}b^{2}=\\frac{b^{4}}{a^{2}}+a^{2}+b^{2}, hence 4a^{4}+4b^{4}-17a^{2}b^{2}=0, so (4a^{2}-b^{2})(a^{2}-4b^{2})=0, therefore 4a^{2}=b^{2} or a^{2}=4b^{2}, so 2a=b or a=2b. Since b>a>0, we have b=2a. Thus c=\\sqrt{a^{2}+b^{2}}=\\sqrt{a^{2}+4a^{2}}=\\sqrt{5}a, so e=\\frac{c}{a}=\\sqrt{5}, hence \\frac{a^{2}+e^{2}}{b}=\\frac{a^{2}+5}{2a}=\\frac{1}{2}(a+\\frac{5}{a})\\geqslant\\frac{1}{2}\\times2\\sqrt{a\\cdot\\frac{5}{a}}=\\sqrt{5}, with equality holding if and only if a=\\sqrt{5}."
  },
  {
    "text": "Given that points $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1$ $(a>0)$, respectively, and $P$ is a point on this hyperbola such that $|P F_{1}|=2|P F_{2}|=16$, then the perimeter of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;a: Number;P: Point;F1: Point;F2: Point;a>0;Expression(G) = (-y^2/9 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);Abs(LineSegmentOf(P,F1))=2*Abs(LineSegmentOf(P,F2));Abs(LineSegmentOf(P,F1))=16",
    "query_expressions": "Perimeter(TriangleOf(P, F1, F2))",
    "answer_expressions": "34",
    "fact_spans": "[[[21, 68], [80, 83]], [[24, 68]], [[75, 78]], [[2, 10]], [[11, 18]], [[24, 68]], [[21, 68]], [[2, 74]], [[2, 74]], [[75, 87]], [[89, 114]], [[89, 114]]]",
    "query_spans": "[[[116, 143]]]",
    "process": "\\because|PF_{1}|=2|PF_{2}|=16,\\therefore|PF_{1}|-|PF_{2}|=16-8=8=2a,\\therefore a=4. Also b^{2}=9,\\therefore c^{2}=25,\\therefore 2c=10.\\therefore the perimeter of \\triangle PF_{1}F_{2} is |PF_{1}|+|PF_{2}|+|F_{1}F_{2}|=16+8+10=34"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the common foci of the ellipse $C_{1}: \\frac{x^{2}}{4}+y^{2}=1$ and the hyperbola $C_{2}$, and $A$ is a common point of $C_{1}$ and $C_{2}$ in the second quadrant, if $\\angle F_{1} A F_{2}=\\frac{\\pi}{3}$, then the eccentricity of $C_{2}$ is?",
    "fact_expressions": "C2: Hyperbola;C1: Ellipse;F1: Point;A: Point;F2: Point;Expression(C1) = (x^2/4 + y^2 = 1);Focus(C1) = {F1, F2};Focus(C2) = {F1, F2};Intersection(C1, C2) = A;Quadrant(A) = 2;AngleOf(F1, A, F2) = pi/3",
    "query_expressions": "Eccentricity(C2)",
    "answer_expressions": "3*sqrt(2)/4",
    "fact_spans": "[[[55, 65], [83, 90], [140, 147]], [[18, 54], [75, 82]], [[2, 9]], [[71, 74]], [[10, 17]], [[18, 54]], [[2, 70]], [[2, 70]], [[71, 100]], [[71, 100]], [[102, 138]]]",
    "query_spans": "[[[140, 153]]]",
    "process": "According to the problem, first find the value of $ c $ in the hyperbola, then combine the definition of the ellipse and the cosine law to find the value of $ a $, and finally use the definition of eccentricity to find the eccentricity of the hyperbola. [Solution] Let $ C_2: \\frac{x^2}{a_1^2} - \\frac{y^2}{b_1^2} = 1 $ ($ a_1b_1 > 0 $), from the given condition we have $ c = c_1 = \\sqrt{3} $. By the definition of the ellipse, $ AF_1 + AF_2 = 2a = 4 $. In $ \\triangle F_1AF_2 $, by the cosine law:  \n$$\n(2c)^2 = 12 = AF_1^2 + AF_2^2 - 2AF_1 \\cdot AF_2 \\cos\\frac{\\pi}{3} = (AF_1 + AF_2)^2 - 3AF_1 \\cdot AF_2 = 16 - 3AF_1 \\cdot AF_2\n$$  \nSolving gives $ AF_1 \\cdot AF_2 = \\frac{4}{3} $,  \n$ \\therefore (AF_1 - AF_2)^2 = (AF_1 + AF_2)^2 - 4AF_1 \\cdot AF_2 = \\frac{32}{3} $.  \nAssume $ F_1 $, $ F_2 $ are the left and right foci respectively, and $ AF_2 > AF_1 $, then $ AF_2 - AF_1 = \\frac{4}{3}\\sqrt{6} = 2a_1 $, solving gives $ a_1 = \\frac{2}{3}\\sqrt{6} $.  \nTherefore, the eccentricity of $ C_2 $ is $ e = \\frac{c_1}{a_1} = \\frac{3\\sqrt{2}}{4} $.  \n[Note] The eccentricity of a hyperbola is its most important geometric property. To find the eccentricity (or the range of eccentricity) of a hyperbola, there are two common methods:  \n\\textcircled{1} Find $ a $ and $ c $, then substitute into the formula $ e = \\frac{c}{a} $;  \n\\textcircled{2} Just obtain a homogeneous equation in terms of $ a $, $ b $, $ c $ from one condition, combine with $ b^2 = c^2 - a^2 $ to transform it into a homogeneous equation in $ a $ and $ c $, then divide both sides of the equation (inequality) by $ a $ or $ a^2 $ to convert it into an equation (inequality) in terms of $ e $, and solve the equation (inequality) to obtain $ e $ (or the range of $ e $)."
  },
  {
    "text": "If the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has an eccentricity of $2$, then what is the equation of the asymptotes of this hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;Eccentricity(G) = 2",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*sqrt(3)*x",
    "fact_spans": "[[[1, 47], [58, 61]], [[1, 47]], [[4, 47]], [[4, 47]], [[1, 55]]]",
    "query_spans": "[[[58, 68]]]",
    "process": "From the given information, the eccentricity $ e = \\frac{c}{a} = 2 $, so $ c = 2a $. Also, $ a^2 + b^{2} = c^{2} = 4a^{2} $, thus $ b^{2} = 3a^{2} $, then $ \\frac{b}{a} = \\sqrt{3} $, hence the asymptotes of this hyperbola are $ v = +\\sqrt{3}x $."
  },
  {
    "text": "The coordinates of the foci of the hyperbola $2 x^{2}-y^{2}=2$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (2*x^2 - y^2 = 2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(pm*sqrt(3), 0)",
    "fact_spans": "[[[0, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 27]]]",
    "process": "The equation of the hyperbola is: $x^{2}-\\frac{y^{2}}{2}=1$, where $a^{2}=1$, $b^{2}=2$, hence $c^{2}=3$, $c=\\sqrt{3}$. From the equation of the hyperbola, it can be seen that the foci lie on the x-axis, so the coordinates of the foci are $(\\pm\\sqrt{3},0)$."
  },
  {
    "text": "A line passing through the point $(3,0)$ intersects the parabola $y^{2}=6x$ at points $A$ and $B$, and intersects the $y$-axis at point $C$. If $\\overrightarrow{AB}=3\\overrightarrow{BC}$, then $|AB|=$?",
    "fact_expressions": "G: Parabola;H: Line;I: Point;A: Point;B: Point;C: Point;Expression(G) = (y^2 = 6*x);Coordinate(I) = (3, 0);PointOnCurve(I, H);Intersection(H, G) = {A, B};Intersection(H,yAxis)=C;VectorOf(A, B) = 3*VectorOf(B, C)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "9*sqrt(5)/2",
    "fact_spans": "[[[13, 27]], [[10, 12]], [[1, 9]], [[32, 35]], [[36, 39]], [[49, 52]], [[13, 27]], [[1, 9]], [[0, 12]], [[10, 39]], [[10, 52]], [[54, 99]]]",
    "query_spans": "[[[101, 110]]]",
    "process": "Let the equation of AB be $ y = k(x - 3) $. Solving the system yields $ k^{2}x^{2} - (6k^{2} + 6)x + 9k^{2} = 0 $. Using Vieta's formulas, $ x_{1} + x_{2} = \\frac{6k^{2} + 6}{k^{2}} $, $ x_{1}x_{2} = 9 $. According to $ \\overrightarrow{AB} = 3\\overrightarrow{BC} $, find $ k $, solving gives answer. [She's solution] Let the equation of AB be $ y = k(x - 3) $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, $ C(0, -3k) $, $ \\overrightarrow{AB} = (x_{2} - x_{1}, y_{2} - y_{1}) $, $ \\overrightarrow{BC} = (-x_{2}, -3k - y_{2}) $. Since $ \\overrightarrow{AB} = 3\\overrightarrow{BC} $, $ x_{1} = 4x_{2} $. From $ \\begin{cases} y = k(x - 3) \\\\ y^{2} = 6x \\end{cases} $, we get $ k^{2}x^{2} - (6k^{2} + 6)x + 9k^{2} = 0 $, $ x_{1} + x_{2} = \\frac{6k^{2} + 6}{k^{2}} $, $ x_{1}x_{2} = 9 $, $ 9 $, $ x_{2} = \\frac{3}{2} $, $ \\frac{6k^{2} + 6}{k^{2}} = \\frac{15}{2} $, $ k^{2} = 4 $. $ \\begin{matrix} \\\\ AB = \\sqrt{(1 + k^{2})} \\end{matrix}^{2} \\frac{9\\sqrt{5}}{2} $"
  },
  {
    "text": "Given that the equation of the parabola $C$ is $y^{2}=4x$, $F$ is the focus of the parabola $C$, and a line $l$ with inclination angle $45^{\\circ}$ passes through point $F$ and intersects the parabola $C$ at points $A$ and $B$, then the length of segment $AB$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l: Line;Inclination(l) = ApplyUnit(45, degree);PointOnCurve(F, l);A: Point;B: Point;Intersection(l, C) = {A, B}",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 8], [29, 35], [67, 73]], [[2, 23]], [[25, 28], [62, 66]], [[25, 38]], [[56, 61]], [[39, 61]], [[56, 66]], [[74, 77]], [[78, 81]], [[56, 83]]]",
    "query_spans": "[[[85, 96]]]",
    "process": "The focus of the parabola $ C: y^{2} = 4x $ is $ F(1,0) $, and the equation of the directrix is $ x = -1 $. According to the given conditions, the equation of line $ l $ is $ y = x - 1 $. Solving the system \\begin{cases} y = x - 1 \\\\ y^{2} = 4x \\end{cases}, eliminating $ x $ and simplifying yields: $ x^{2} - 6x + 1 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = 6 $, thus we obtain $ |AB| = |AF| + |BF| = x_{1} + 1 + x_{2} + 1 = 8 $. Therefore, the length of segment $ AB $ is 8."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$, the focus is $F$, the intersection point of the directrix and the $x$-axis is $K$, point $A$ lies on the parabola, and $|AK|= \\sqrt{2} |AF|$, $O$ is the origin. Then $|OA|=$?",
    "fact_expressions": "C: Parabola;A: Point;K: Point;F: Point;O: Origin;Expression(C) = (y^2 = 8*x);Focus(C) = F;Intersection(Directrix(C), xAxis) = K;PointOnCurve(A, C);Abs(LineSegmentOf(A, K)) = sqrt(2)*Abs(LineSegmentOf(A, F))",
    "query_expressions": "Abs(LineSegmentOf(O, A))",
    "answer_expressions": "sqrt(5)*2",
    "fact_spans": "[[[2, 20], [48, 51]], [[43, 47]], [[39, 42]], [[24, 27]], [[76, 79]], [[2, 20]], [[2, 27]], [[2, 42]], [[43, 52]], [[54, 75]]]",
    "query_spans": "[[[86, 94]]]",
    "process": ""
  },
  {
    "text": "The point $p(x , y)$ is an arbitrary point on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$, $F_{1}$, $F_{2}$ are the two foci of the ellipse, and $\\angle F_{1} P F _{2} \\leq 90^{\\circ}$. Then the range of the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;p: Point;F1: Point;F2: Point;a>b;b>0;x1:Number;y1:Number;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(p) = (x1, y1);PointOnCurve(p, G);Focus(G) = {F1, F2};AngleOf(F1,p,F2)<=ApplyUnit(90,degree)",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(0, \\sqrt{2}/2]",
    "fact_spans": "[[[12, 66], [89, 91], [140, 142]], [[14, 66]], [[14, 66]], [[0, 11]], [[73, 80]], [[81, 88]], [[14, 66]], [[14, 66]], [[1, 11]], [[1, 11]], [[12, 66]], [[0, 11]], [[0, 72]], [[73, 96]], [[98, 137]]]",
    "query_spans": "[[[140, 153]]]",
    "process": ""
  },
  {
    "text": "Given that the line $x+2 y-3=0$ intersects the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ at points $A$ and $B$, and the midpoint of segment $AB$ lies on the line $3 x-4 y+1=0$, then the eccentricity of this ellipse is?",
    "fact_expressions": "H: Line;Expression(H) = (x + 2*y - 3 = 0);G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;A: Point;B: Point;Intersection(H, G) = {A, B};Z: Line;Expression(Z) = (3*x - 4*y + 1 = 0);PointOnCurve(MidPoint(LineSegmentOf(A, B)), Z)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 15]], [[2, 15]], [[16, 68], [112, 114]], [[16, 68]], [[18, 68]], [[18, 68]], [[18, 68]], [[18, 68]], [[71, 74]], [[75, 78]], [[2, 80]], [[93, 108]], [[93, 108]], [[82, 109]]]",
    "query_spans": "[[[112, 120]]]",
    "process": "Solving the system \\begin{cases}x+2y-3=0\\\\3x-4y+1=0\\end{cases}, we get x=1, y=1, hence the intersection point of the lines x+2y-3=0 and 3x-4y+1=0 is M(1,1), and the midpoint of segment AB is M(1,1). Let the intersection points of x+2y-3=0 and \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 be A(x_{1},y_{1}), B(x_{2},y_{2}) respectively, then x_{1}+x_{2}=2, y_{1}+y_{2}=2. The slope of the line x+2y-3=0 is k=-\\frac{1}{2}. Substituting A(x_{1},y_{1}), B(x_{2},y_{2}) into the ellipse equation \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0), we obtain \\begin{cases}\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{\\frac{1}{b}}=1\\\\\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{\\frac{2}{2}}=1\\end{cases}. Subtracting and simplifying the two equations yields \\frac{(y+y_{2})(y-y_{2})}{i.e.\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}x_{1}+x_{2}}=-\\frac{b^{2}}{a^{2}},\\frac{b}{2}=-\\frac{b}{a^{2}},a^{2}=2b^{2},a=\\sqrt{2}b=\\sqrt{2}c,e=\\frac{c}{a}=\\frac{\\sqrt{2}}{2},"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ has a focal distance of $4$. Then the value of $a$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2 + x^2/a^2 = 1);a: Number;a>0;FocalLength(G) = 4",
    "query_expressions": "a",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 39]], [[2, 39]], [[49, 52]], [[5, 39]], [[2, 46]]]",
    "query_spans": "[[[49, 56]]]",
    "process": "Since the focal distance of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ ($a>0$) is 4, we have $2c=2\\sqrt{a^{2}+b^{2}}=2\\sqrt{a^{2}+1}=4$, solving which gives $a=\\sqrt{3}$."
  },
  {
    "text": "The standard equation of a parabola with vertex at the origin, symmetry axis along the coordinate axes, and passing through the point $P(-4,-2)$ is?",
    "fact_expressions": "G: Parabola;P: Point;O: Origin;Coordinate(P) = (-4, -2);Vertex(G) = O;SymmetryAxis(G) = axis;PointOnCurve(P, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2 = -x, x^2 = -8*y}",
    "fact_spans": "[[[28, 31]], [[16, 27]], [[3, 5]], [[16, 27]], [[0, 31]], [[6, 31]], [[15, 31]]]",
    "query_spans": "[[[28, 38]]]",
    "process": "Based on the position of P(-4,-2), consider separately the cases where the parabola's focus lies on the negative x-axis and the negative y-axis, and thus determine the equations of the parabola. When the focus of the parabola is on the negative x-axis, let y^{2}=-2px (p>0); substituting point P(-4,-2) gives 4=8p, so p=\\frac{1}{2}, hence y^{2}=-x. When the focus of the parabola is on the negative y-axis, let x^{2}=-2py (p>0); substituting point P(-4,-2) gives 16=4p, so p=4, hence x^{2}=-8y."
  },
  {
    "text": "The hyperbola centered at the origin has asymptotes given by $y=\\pm \\sqrt{3} x$ and passes through the point $(\\sqrt{2}, \\sqrt{3})$. Then the standard equation of the hyperbola is?",
    "fact_expressions": "O: Origin;Center(G) = O;G: Hyperbola;Expression(Asymptote(G)) = (y = pm*(sqrt(3)*x));H: Point;Coordinate(H) = (sqrt(2), sqrt(3));PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[3, 5]], [[0, 9]], [[6, 9], [10, 11], [63, 66]], [[10, 35]], [[38, 61]], [[38, 61]], [[10, 61]]]",
    "query_spans": "[[[63, 73]]]",
    "process": "Since the asymptotes of the hyperbola are given by $ y = \\pm\\sqrt{3}x $, we assume the equation of the hyperbola is $ 3x^{2} - y^{2} = \\lambda $ ($ \\lambda \\neq 0 $). Substituting the point $ (\\sqrt{2},\\sqrt{3}) $ gives: $ 6 - 3 = \\lambda \\Rightarrow \\lambda = 3 $, so the equation of the hyperbola is $ x^{2} - \\frac{y^{2}}{3} = 1 $."
  },
  {
    "text": "Given that the standard equation of a parabola is $y=-8 x^{2}$, then the equation of the directrix of this parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = -8*x^2)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=1/32",
    "fact_spans": "[[[26, 29], [2, 5]], [[2, 23]]]",
    "query_spans": "[[[26, 36]]]",
    "process": "Convert the parabolic equation into standard form, then obtain the solution using the directrix equation of the parabola. According to the problem, the parabolic equation $ y = -8x^{2} $ can be written as $ x^{2} = -\\frac{1}{8}y $, so the directrix equation of this parabola is $ y = \\frac{1}{32} $."
  },
  {
    "text": "If the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ is $3$, and its asymptotes are tangent to the circle $x^{2}+y^{2}-6 y+m=0$, then $m$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(G) = 3;H: Circle;Expression(H) = (m - 6*y + x^2 + y^2 = 0);m: Number;IsTangent(Asymptote(G), H) = True",
    "query_expressions": "m",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 60], [69, 70]], [[1, 60]], [[4, 60]], [[4, 60]], [[4, 60]], [[4, 60]], [[1, 68]], [[74, 96]], [[74, 96]], [[100, 103]], [[69, 98]]]",
    "query_spans": "[[[100, 105]]]",
    "process": "Since the asymptotes of the hyperbola are $ bx \\pm ay = 0 $, and the standard equation of the circle is $ x^{2} + (y - 3)^{2} = 9 - m $, the center of the circle is $ C(0, 3) $, and $ r = \\sqrt{9 - m} $. Also, $ \\frac{c}{a} = 3 \\Rightarrow c = 3a $, $ b = 2\\sqrt{2}a $. From the given condition, we have $ \\frac{3a}{\\sqrt{a^{2} + b^{2}}} = \\frac{3a}{c} = 1 $, so $ \\sqrt{9 - m} = 1 $. Solving gives $ m = 8 $. Therefore, the answer to be filled in is $ 8 $."
  },
  {
    "text": "If the eccentricity of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is the reciprocal of the eccentricity of the ellipse $D$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, then what are the equations of the asymptotes of $C$?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;D: Ellipse;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(D) = (x^2/4 + y^2/3 = 1);InterReciprocal(Eccentricity(C),Eccentricity(D))",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[2, 63], [119, 122]], [[10, 63]], [[10, 63]], [[68, 110]], [[10, 63]], [[10, 63]], [[2, 63]], [[68, 110]], [[2, 116]]]",
    "query_spans": "[[[119, 130]]]",
    "process": ""
  },
  {
    "text": "Find a point $Q$ on the parabola $y^{2}=4 x$ such that the sum of its distance to the point $A(7,8)$ and its distance to the directrix is minimized. What are the coordinates of point $Q$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);Q: Point;PointOnCurve(Q, G);A: Point;Coordinate(A) = (7, 8);WhenMin(Distance(Q, A) + Distance(Q, Directrix(G)))",
    "query_expressions": "Coordinate(Q)",
    "answer_expressions": "(4, 4)",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[20, 23], [25, 26], [49, 53]], [[1, 23]], [[27, 36]], [[27, 36]], [[2, 47]]]",
    "query_spans": "[[[49, 58]]]",
    "process": "\\because the parabola equation is y^{2}=4x, \\therefore the focus of the parabola is F(1,0), and the directrix equation is x=-1. By the definition of the parabola, the distance from point Q to the directrix equals its distance to the focus. The minimum value sought is thus the minimum value of |QA|+|QF|. According to plane geometry knowledge, |QA|+|QF| is minimized when points A, Q, and F are collinear. Hence, the minimum value of |QA|+|QF| is the distance from A to the focus F(1,0). At this time, the slope of line AF is k=\\frac{4}{3}, and the equation of line AF is y=\\frac{4}{3}(x-1). Solving the system \\begin{cases}y=\\frac{4}{3}(x-1)\\\\y^{2}=4x\\end{cases}, we obtain \\begin{cases}x=4\\\\y=4\\end{cases} or \\begin{cases}x=\\frac{1}{4}\\\\y=-1\\end{cases} (discarded). Thus, the coordinates of point Q are (4,4)."
  },
  {
    "text": "Given that point $M$ lies on the parabola $y^{2}=2 p x$, the abscissa of point $M$ is $4$, and the distance from $M$ to the focus $F$ is $5$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;M: Point;PointOnCurve(M, G);XCoordinate(M) = 4;F: Point;Focus(G) = F;Distance(M, F) = 5",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[7, 23]], [[7, 23]], [[57, 60]], [[2, 6], [25, 29], [39, 42]], [[2, 24]], [[25, 37]], [[45, 48]], [[7, 48]], [[39, 55]]]",
    "query_spans": "[[[57, 64]]]",
    "process": "According to the problem, the directrix of the parabola is $ x = -\\frac{p}{2} $, then $ 4 + \\frac{p}{2} = 5 $, solving gives $ p = 2 $."
  },
  {
    "text": "Given that point $F$ is the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, a perpendicular is drawn from $F$ to an asymptote, with foot of the perpendicular at $A$. If the area of $\\triangle O A F$ (where point $O$ is the origin) is $2$, and the eccentricity $e$ of the hyperbola satisfies $e \\in[\\sqrt{17}, \\sqrt{65}]$, then what is the range of values for $a$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(G) = F;L: Line;PointOnCurve(F, L);IsPerpendicular(L, OneOf(Asymptote(G)));A: Point;FootPoint(L, OneOf(Asymptote(G))) = A;O: Origin;Area(TriangleOf(O, A, F)) = 2;e: Number;Eccentricity(G) = e;In(e, [sqrt(17), sqrt(65)])",
    "query_expressions": "Range(a)",
    "answer_expressions": "[sqrt(2)/2, 1]",
    "fact_spans": "[[[7, 63], [127, 130]], [[7, 63]], [[10, 63]], [[165, 168]], [[10, 63]], [[10, 63]], [[2, 6], [69, 72]], [[2, 67]], [], [[68, 81]], [[68, 81]], [[85, 88]], [[68, 88]], [[108, 112]], [[90, 126]], [[134, 163]], [[127, 163]], [[134, 163]]]",
    "query_spans": "[[[165, 175]]]",
    "process": "From the given conditions: the distance from point F to the asymptote is equal to _{d}=\\frac{|bc|}{c}=b', thus S_{\\triangleOAF}=\\frac{1}{2}ab=2, so ab=4. Also, e^{2}=1+\\frac{b^{2}}{a^{2}}\\in[17,65], therefore \\frac{b^{2}}{a^{2}}\\in[16,64], then \\frac{b}{a}\\in[4,8]. Since ab=4, it follows that \\frac{4}{a^{2}}\\in[4,8], solving gives a\\in[\\frac{\\sqrt{2}}{2}"
  },
  {
    "text": "The equation of the hyperbola with the focus of the parabola $y^{2}=4 x$ as its vertex, the vertex as its center, and eccentricity $2$ is?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 4*x);G: Hyperbola;Focus(H) = Vertex(G);Vertex(H) = Center(G);Eccentricity(G) = 2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[1, 15]], [[1, 15]], [[36, 39]], [[0, 39]], [[0, 39]], [[28, 39]]]",
    "query_spans": "[[[36, 43]]]",
    "process": "From the given conditions, the parabola $ y^{2} = 4x $ has focus at $ (1,0) $ and vertex at the origin. Therefore, for the hyperbola, $ a = 1 $, and the foci lie on the x-axis. Given the eccentricity is 2, we have $ c = 2 $. Thus, $ b^{2} = c^{2} - a^{2} = 3 $. Hence, the equation of the required hyperbola is $ x^{2} - \\frac{y^{2}}{3} = 1 $."
  },
  {
    "text": "Given that $AB$ is a chord of length $8$ on the parabola $x^{2}=4y$, when the midpoint of chord $AB$ is closest to the $x$-axis, what is the slope of line $AB$?",
    "fact_expressions": "G: Parabola;B: Point;A: Point;Expression(G) = (x^2 = 4*y);IsChordOf(LineSegmentOf(A, B),G);Length(LineSegmentOf(A, B)) = 8;WhenMin(Distance(MidPoint(LineSegmentOf(A,B)),xAxis))",
    "query_expressions": "Slope(LineOf(A,B))",
    "answer_expressions": "pm*1",
    "fact_spans": "[[[8, 22]], [[2, 7]], [[2, 7]], [[8, 22]], [[2, 33]], [[2, 31]], [[36, 52]]]",
    "query_spans": "[[[53, 65]]]",
    "process": "From the given conditions, the directrix of the parabola is $ l: y = -1 $. Draw $ AA_1 \\perp l $ at $ A_1 $, and draw $ BB_1 \\perp l $ at $ B_1 $. Let $ M $ be the midpoint of chord $ AB $, and draw $ MM_1 \\perp l $ at $ M_1 $. Then $ 2|MM_1| = |AA_1| + |BB_1| $. Let $ F $ be the focus of the parabola. Then $ |AF| + |BF| \\geqslant |AB| $, that is, $ |AA_1| + |BB_1| = |AF| + |BF| \\geqslant 8 $ (equality holds if and only if points $ A $, $ B $, $ F $ are collinear). Therefore, $ |AA_1| + |BB_1| = 2|MM_1| \\geqslant 8 $, solving gives $ |MM_1| \\geqslant 4 $, so the shortest distance from the midpoint of chord $ AB $ to the $ x $-axis is $ 4 - 1 = 3 $. Thus, the $ y $-coordinate of point $ M $ is $ (x_0, 3) $, $ A(x_1, y_1) $, $ B(x_2, y_2) $, $ F(0, 1) $, $ x_1^2 = 4y_1 $, $ x_2^2 = 4y_2 $. Therefore, the slope of line $ AB $ is $ k = \\frac{y_1 - y_2}{x_1 - x_2} = \\frac{x_1 + x_2}{4} = \\frac{x_0}{2} = \\frac{3 - 1}{x_0 - 0} $. Hence $ x_0 = \\pm 2 $, and at this time $ k = \\pm 1 $. When the midpoint of chord $ AB $ is closest to the $ x $-axis, the slope of line $ AB $ is $ \\pm 1 $."
  },
  {
    "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are given by $y=\\pm \\frac{4}{3} x$, then the eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Expression(Asymptote(G)) = (y = pm*x*4/3);e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "5/3",
    "fact_spans": "[[[1, 57], [87, 90]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 85]], [[94, 97]], [[87, 97]]]",
    "query_spans": "[[[87, 99]]]",
    "process": "From the given conditions, the asymptotes of the hyperbola are $ y = \\pm\\frac{b}{a}x = \\pm\\frac{4x}{3} $, so $ \\frac{b}{a} = \\frac{4}{3} $. Then, the eccentricity of the hyperbola is $ e = \\frac{c}{a} = \\sqrt{1+\\frac{b^{2}}{a^{2}}} = \\sqrt{1+\\frac{16}{9}} = \\frac{5}{3} $."
  },
  {
    "text": "Parabola $C$: $x^{2}=2 p y$, the distance from its focus to the directrix $l$ is $4$. Then, what is the length of the chord cut by the circle $x^{2}+y^{2}-6 x=0$ on the directrix $l$?",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 2*p*y);p: Number;l: Line;Directrix(C) = l;Distance(Focus(C), l) = 4;G: Circle;Expression(G) = (-6*x + x^2 + y^2 = 0)",
    "query_expressions": "Length(InterceptChord(l, G))",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[0, 21], [22, 23]], [[0, 21]], [[8, 21]], [[28, 31], [42, 45]], [[22, 31]], [[22, 38]], [[46, 66]], [[46, 66]]]",
    "query_spans": "[[[42, 73]]]",
    "process": "First, obtain the equation of the directrix $ y = \\pm 2 $. $ x^{2} + y^{2} \\cdot 6x = 0 \\Rightarrow (x-3)^{2} + y^{2} = 9 $. The distance from the circle center $ (3,0) $ to the directrix $ y = \\pm 2 $ is 2, so the chord length is $ 2\\sqrt{9 - 2^{2}} = 2\\sqrt{5} $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, a line passing through point $F$ intersects the parabola $C$ at two points $A$ and $B$, satisfying $\\overrightarrow{AF}=3\\overrightarrow{FB}$. Let $E$ be the midpoint of $AB$. Then, the distance from point $E$ to the directrix of the parabola is?",
    "fact_expressions": "C: Parabola;G: Line;A: Point;B: Point;F: Point;E: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F, G);Intersection(G, C) = {A, B};VectorOf(A, F) = 3*VectorOf(F, B);MidPoint(LineSegmentOf(A, B)) = E",
    "query_expressions": "Distance(E, Directrix(C))",
    "answer_expressions": "8/3",
    "fact_spans": "[[[2, 21], [38, 44], [130, 133]], [[35, 37]], [[52, 55]], [[56, 59]], [[25, 28], [31, 34]], [[111, 114], [125, 129]], [[2, 21]], [[2, 28]], [[29, 37]], [[35, 59]], [[63, 108]], [[111, 123]]]",
    "query_spans": "[[[125, 140]]]",
    "process": "By the given condition, we have $ F(1,0) $, and the directrix is $ x = -1 $. Let the equation of line $ AB $ be $ x = ty + 1 $. Substituting into $ y^2 = 4x $, we get $ y^2 - 4ty - 4 = 0 $. Let $ A(x_1, y_1) $, $ B(x_2, y_2) $, so $ E\\left(\\frac{x_1 + x_2}{2}, \\frac{y_1 + y_2}{2}\\right) $. Then $ y_1 + y_2 = 4t $, $ y_1 y_2 = -4 $. Since $ \\overrightarrow{AF} = 3\\overrightarrow{FB} $, we have $ (1 - x_1, -y_1) = 3(x_2 - 1, y_2) $, so $ -y_1 = 3y_2 $, i.e., $ y_1 = -3y_2 $. Thus $ -3y_2 + y_2 = 4t $, so $ y_2 = -2t $, $ y_1 = 6t $. Therefore $ (-2t) \\cdot 6t = -4 $, so $ t^2 = \\frac{1}{3} $. Hence $ x_1 + x_2 = t(y_1 + y_2) + 2 = 4t^2 + 2 = \\frac{4}{3} + 2 = \\frac{10}{3} $. Therefore, the distance from point $ E $ to the parabola's directrix is $ \\frac{x_1 + x_2}{2} - (-1) = \\frac{5}{3} + 1 = \\frac{8}{3} $."
  },
  {
    "text": "If $\\frac{x^{2}}{|m|-1}+\\frac{y^{2}}{m}=1$ represents a hyperbola, then the range of the real number $m$ is?",
    "fact_expressions": "G: Hyperbola;m: Real;Expression(G)=(x^2/(Abs(m) - 1) + y^2/m = 1)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-\\infty,-1)+(0,1)",
    "fact_spans": "[[[42, 45]], [[47, 52]], [[1, 45]]]",
    "query_spans": "[[[47, 57]]]",
    "process": "If $\\frac{x^2}{|m|-1} + \\frac{y^{2}}{m} = 1$ represents a hyperbola, then $(|m|-1) \\cdot m < 0$, clearly $m \\neq 0$. When $m < 0$, the inequality becomes $(-m-1) \\cdot m < 0$, solving this inequality yields $m < -1$ or $m > 0$, i.e., $m < -1$. When $m > 0$, the inequality becomes $(m-1) \\cdot m < 0$, solving this inequality yields $0 < m < 1$, i.e., $0 < m < 1$. In summary, the range of values of $m$ for which $\\frac{x^{2}}{|m|-1} + \\frac{y^{2}}{m} = 1$ represents a hyperbola is $(-\\infty, -1) \\cup (0, 1)$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, the foci of the ellipse are exactly the two vertices of the hyperbola, and the eccentricities of the ellipse and the hyperbola are reciprocals of each other. Then the equation of the ellipse is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(G) = (x^2/16 - y^2/9 = 1);Focus(H)=Vertex(G);InterReciprocal(Eccentricity(G),Eccentricity(H))",
    "query_expressions": "Expression(H)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 41], [50, 53], [62, 65]], [[42, 44], [59, 61], [75, 77]], [[2, 41]], [[42, 58]], [[59, 73]]]",
    "query_spans": "[[[75, 82]]]",
    "process": ""
  },
  {
    "text": "Given that the hyperbola $C$ is centered at the origin, its foci lie on the coordinate axes, the focal distance is $10$, and the point $P(2, 1)$ lies on an asymptote of $C$, then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;O: Origin;Center(C) = O;PointOnCurve(Focus(C),axis) = True;FocalLength(C) = 10;P: Point;Coordinate(P) = (2, 1);PointOnCurve(P, Asymptote(C))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "{(x^2/20-y^2/5=1),(y^2/5-x^2/20=1)}",
    "fact_spans": "[[[2, 8], [43, 46], [53, 56]], [[12, 14]], [[2, 14]], [[2, 22]], [[2, 30]], [[31, 42]], [[31, 42]], [[31, 51]]]",
    "query_spans": "[[[53, 61]]]",
    "process": ""
  },
  {
    "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a, b>0)$ with eccentricity $\\frac{5}{4}$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(G) = 5/4",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "3*x+pm*4*y=0",
    "fact_spans": "[[[18, 72]], [[18, 72]], [[21, 72]], [[21, 72]], [[21, 72]], [[21, 72]], [[0, 72]]]",
    "query_spans": "[[[18, 80]]]",
    "process": "\\because the eccentricity of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 is \\frac{5}{4}, that is, e=\\frac{c}{a}=\\frac{5}{4}, let c=5k (k>0), then a=4k, thus we obtain"
  },
  {
    "text": "What is the shortest distance from a point on the parabola $x^{2}=4 y$ to its focus?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);P:Point;PointOnCurve(P,G)",
    "query_expressions": "Min(Distance(P,Focus(G)))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 14], [18, 19]], [[0, 14]], [[16, 17]], [[0, 17]]]",
    "query_spans": "[[[16, 28]]]",
    "process": "The focus of the parabola \\( x^{2} = 4y \\) is \\( F(0,1) \\). Let point \\( P(t, \\frac{t^{2}}{4}) \\) be an arbitrary point on the parabola \\( x^{2} = 4y \\). Therefore, when \\( t = 0 \\), that is, when point \\( P \\) is the vertex of the parabola, \\( |PF| \\) attains the minimum value of 1."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with left focus $F$, a line $l$ passing through $F$ is tangent to the circle $x^{2}+y^{2}=a^{2}$ at point $T$, and the line $l$ intersects the right branch of the hyperbola $C$ at point $P$. If the eccentricity of the hyperbola $C$ is $\\frac{5}{3}$, then $|P T|$=?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;G: Circle;P: Point;T: Point;F: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = a^2);LeftFocus(C) = F;PointOnCurve(F, l);TangentPoint(l, G) = T;Intersection(l, RightPart(C)) = P;Eccentricity(C) = 5/3",
    "query_expressions": "Abs(LineSegmentOf(P, T))",
    "answer_expressions": "3",
    "fact_spans": "[[[77, 82], [112, 117]], [[2, 63], [118, 124], [135, 141]], [[10, 63]], [[10, 63]], [[83, 103]], [[129, 133]], [[106, 110]], [[68, 71], [73, 76]], [[10, 63]], [[10, 63]], [[2, 63]], [[83, 103]], [[2, 71]], [[72, 82]], [[77, 110]], [[112, 133]], [[135, 159]]]",
    "query_spans": "[[[161, 170]]]",
    "process": "Let the right focus of hyperbola C be G, and draw GH\\bot PF at H. By the midline theorem, |GH|=2|OT|=2a, |FH|=2|FT|=2b. Since e=\\frac{c}{a}=\\frac{5}{3}\\Rightarrow b=\\frac{4}{3}a. Let |PT|=\\lambda|FT|=3b (\\lambda>0). By the definition of hyperbola: |PG|=|PF|-2a=(\\frac{4}{3},-\\frac{2}{3})a. Also, |PH|=|PF|-|FH|=(\\lambda+1)b-2b=(\\lambda-1)b=\\frac{4}{3}(\\lambda-1)a. By the Pythagorean theorem: |PH|^2+|GH|^2=|PG|^2 \\Rightarrow \\frac{16}{9}(\\lambda-1)^2+4=(\\frac{4}{3}\\lambda-\\frac{2}{3})^2 \\Rightarrow \\lambda=3;"
  },
  {
    "text": "Let the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{16}=1  (a>0)$ pass through the point $(3,0)$. Then the equation of the asymptotes of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Point;a>0;Expression(G) = (-y^2/16 + x^2/a^2 = 1);Coordinate(H) = (3, 0);PointOnCurve(H, G)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 51], [64, 67]], [[4, 51]], [[53, 61]], [[4, 51]], [[1, 51]], [[53, 61]], [[1, 61]]]",
    "query_spans": "[[[64, 75]]]",
    "process": "Since the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{16}=1$ $(a>0)$ passes through the point $(3,0)$, we have $\\frac{9}{a^{2}}=1$, so $a=3$. Thus, the equation of the hyperbola is $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and its asymptotes are given by $y=\\pm\\frac{4}{3}x$."
  },
  {
    "text": "Let $A(-1,0)$, $B(1,0)$. If there exists a point $P$ on the line $y=kx$ $(k>0)$ such that $|PA|+|PB|=4$, and the distance from the incenter of $\\Delta PAB$ to the $x$-axis is $\\frac{2\\sqrt{21}}{21}$, then $k=$?",
    "fact_expressions": "A: Point;Coordinate(A) = (-1, 0);B: Point;Coordinate(B) = (1, 0);G: Line;Expression(G) = (y = k*x);k: Number;k>0;P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B)) = 4;Distance(Incenter(TriangleOf(P, A, B)), xAxis) = (2*sqrt(21))/21",
    "query_expressions": "k",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 10]], [[1, 10]], [[12, 20]], [[12, 20]], [[22, 38]], [[22, 38]], [[117, 120]], [[24, 38]], [[43, 46]], [[22, 46]], [[48, 63]], [[65, 115]]]",
    "query_spans": "[[[117, 122]]]",
    "process": "Since |PA| + |PB| = 4 > |AB| = 2, the point P lies on an ellipse with A and B as foci, and the major axis length of the ellipse is 4, with focal distance 2. Thus, 2a = 4, 2c = 2, that is, a = 2, c = 1, so b^{2} = a^{2} - c^{2} = 4 - 1 = 3. Therefore, the equation of the ellipse is \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1. Without loss of generality, let P(x_{0}, y_{0}), solve the system \\begin{cases} y = kx \\\\ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 \\end{cases}, we obtain \\begin{cases} x_{0} = \\frac{12}{3 + 4k^{2}} \\\\ |y_{0}| = \\sqrt{\\frac{12k^{2}}{3 + 4k^{2}}} \\end{cases}. Since the distance from the incenter of \\triangle PAB to the side is \\frac{3 + 4k^{2}}{21}, the inradius r of \\triangle PAB is \\frac{2\\sqrt{21}}{21}. Thus, the area of \\triangle PAB is \\frac{1}{2}(|PA| + |PB| + |AB|)r = \\frac{1}{2}(2a + 2c)r = (a + c)r = 3 \\times \\frac{2\\sqrt{21}}{21} = \\frac{2\\sqrt{21}}{7}. Also, the area of \\triangle PAB is \\frac{1}{2}|AB||y_{0}| = \\frac{1}{2} \\times 2 \\sqrt{\\frac{12k^{2}}{3 + 4k^{2}}} = \\sqrt{\\frac{12k^{2}}{3 + 4k^{2}}}. Setting this equal to \\frac{2\\sqrt{21}}{7}, we solve and get k^{2} = 1. Since k > 0, we have k = 1."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the right focus is $F$, the right vertex is $A$, and $O$ is the origin. If $|O F|=2|O A|$, then the asymptotes of $C$ have equations?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;O: Origin;F: Point;A: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;RightVertex(C)=A;Abs(LineSegmentOf(O, F)) = 2*Abs(LineSegmentOf(O, A))",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[2, 63], [104, 107]], [[10, 63]], [[10, 63]], [[80, 83]], [[68, 71]], [[76, 79]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 71]], [[2, 79]], [[88, 102]]]",
    "query_spans": "[[[104, 115]]]",
    "process": "\\because|OF|=2|OA|,\\thereforec=2a,\\becausea^{2}+b^{2}=c^{2}=4a^{2}, then we get \\frac{b}{a}=\\sqrt{3}, so the asymptotes of C are y=\\pm\\sqrt{3}x."
  },
  {
    "text": "Given the parabola $C$: $4 x^{2}+m y=0$ passes exactly through the center of the circle $M$: $(x-1)^{2}+(y-2)^{2}=1$, then the coordinates of the focus of the parabola $C$ are?",
    "fact_expressions": "C: Parabola;m: Number;M: Circle;Expression(C) = (m*y + 4*x^2 = 0);Expression(M) = ((x - 1)^2 + (y - 2)^2 = 1);PointOnCurve(Center(M),C)",
    "query_expressions": "Coordinate(Focus(C))",
    "answer_expressions": "(0,1/8)",
    "fact_spans": "[[[2, 25], [62, 68]], [[9, 25]], [[29, 57]], [[2, 25]], [[29, 57]], [[2, 60]]]",
    "query_spans": "[[[62, 75]]]",
    "process": "From the given condition, the center of circle M is (1,2). Substituting into $4x^{2}+my=0$ gives $m=-2$. Rewriting the equation of parabola C into standard form yields $x^{2}=\\frac{1}{2}y$, so the focus coordinates are $(0,\\frac{1}{8})$."
  },
  {
    "text": "It is known that one asymptote of a hyperbola is $y=\\frac{\\sqrt{3}}{2} x$, and the focal distance is $2 \\sqrt{7}$. Then, the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(OneOf(Asymptote(G))) = (y = x*(sqrt(3)/2));FocalLength(G) = 2*sqrt(7)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/4-y^2/3=1, y^2/3-x^2/4=1}",
    "fact_spans": "[[[2, 5], [58, 61]], [[2, 38]], [[2, 55]]]",
    "query_spans": "[[[58, 68]]]",
    "process": ""
  },
  {
    "text": "Given point $P(0 , 1)$, two points $A$, $B$ on the ellipse $\\frac{x^{2}}{4}+y^{2}=m$ $(m>1)$ satisfy $\\overrightarrow{A P}=2\\overrightarrow{PB}$. Then, when $m=$?, the absolute value of the horizontal coordinate of point $B$ is maximized.",
    "fact_expressions": "P: Point;Coordinate(P) = (0, 1);G: Ellipse;Expression(G) = (x^2/4 + y^2 = m);m: Number;m>1;A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);VectorOf(A, P) = 2*VectorOf(P, B);WhenMax(Abs(XCoordinate(B)))",
    "query_expressions": "m",
    "answer_expressions": "5",
    "fact_spans": "[[[2, 13]], [[2, 13]], [[14, 46]], [[14, 46]], [[104, 107]], [[16, 46]], [[49, 52]], [[53, 56], [111, 115]], [[14, 56]], [[14, 56]], [[58, 101]], [[111, 124]]]",
    "query_spans": "[[[104, 109]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. From $ \\overrightarrow{AP} = 2\\overrightarrow{PB} $, we get $ -x_{1} = 2x_{2} $, $ 1 - y_{1} = 2(y_{2} - 1) $, so $ -y_{1} = 2y_{2} - 3 $. Since $ A $ and $ B $ lie on the ellipse, $ \\frac{x_{1}^{2}}{4} + y_{1}^{2} = m $, $ \\frac{x_{2}^{2}}{4} + y_{2}^{2} = m $, and $ \\frac{4x_{2}^{2}}{4_{2}^{2}} + (2y_{2} - 3)^{2} = m $, so $ \\frac{x_{2}^{2}}{4} + (y_{2}\\frac{3}{2})^{2} - \\frac{m}{4} $. Subtracting corresponding equations gives $ y_{2} - \\frac{3 + m}{4}x_{2}^{2} - \\frac{1}{4}(m^{2} - 10m + 9) \\leqslant 4 $, with equality if and only if $ m = 5 $, at which point the maximum value is attained."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has its right focus at $F$. Draw tangents from point $F$ to the circle $x^{2}+y^{2}=b^{2}$. If the two tangents are perpendicular to each other, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Circle;F: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = b^2);RightFocus(C) = F;L1:Line;L2:Line;TangentOfPoint(F,G)={L1,L2};IsPerpendicular(L1,L2)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[2, 59], [109, 112]], [[9, 59]], [[9, 59]], [[74, 94]], [[64, 67], [69, 73]], [[9, 59]], [[9, 59]], [[2, 59]], [[74, 94]], [[2, 67]], [], [], [[68, 97]], [[68, 107]]]",
    "query_spans": "[[[109, 118]]]",
    "process": "By the given condition, the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, we can obtain the right focus as $F(c,0)$. Since a tangent line is drawn from point $F(c,0)$ to the circle $x^{2}+y^{2}=b^{2}$, we have $\\sqrt{2}b=c$, then $2b^{2}=c^{2}$, that is, $2(a^{2}-c^{2})=c^{2}$, which implies $2a^{2}=3c^{2}$. Thus, $\\frac{c^{2}}{a^{2}}=\\frac{2}{3}$, so $e=\\frac{c}{a}=\\frac{\\sqrt{6}}{3}$."
  },
  {
    "text": "Given that a line with slope $1$ passes through the right focus of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ and intersects the ellipse at points $A$ and $B$, find the length of chord $AB$.",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;Expression(G) = (x^2/4 + y^2 = 1);PointOnCurve(RightFocus(G), H);Slope(H) = 1;Intersection(H, G) = {A, B};IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "8/5",
    "fact_spans": "[[[12, 39], [45, 47]], [[9, 11]], [[48, 51]], [[52, 55]], [[12, 39]], [[9, 43]], [[2, 11]], [[9, 57]], [[9, 65]]]",
    "query_spans": "[[[59, 69]]]",
    "process": ""
  },
  {
    "text": "The ellipse with foci on the $x$-axis passes through the point $P(3 , 0)$, and the length of the major axis is 3 times the length of the minor axis. Then its standard equation is?",
    "fact_expressions": "G: Ellipse;PointOnCurve(Focus(G), xAxis) = True;P: Point;Coordinate(P) = (3, 0);PointOnCurve(P, G) = True;Length(MajorAxis(G)) = Length(MinorAxis(G))*3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9+y^2=1",
    "fact_spans": "[[[9, 11], [39, 40]], [[1, 11]], [[12, 23]], [[12, 23]], [[9, 23]], [[9, 37]]]",
    "query_spans": "[[[39, 46]]]",
    "process": ""
  },
  {
    "text": "$P$ is a moving point on the parabola $y=2 x^{2}$. When the distance from point $P$ to the line $2 x-y-4=0$ is shortest, what are the coordinates of point $P$?",
    "fact_expressions": "G: Parabola;H: Line;Expression(G) = (y = 2*x^2);Expression(H) = (2*x - y - 4 = 0);PointOnCurve(P, G);WhenMin(Distance(P,H));P:Point",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1/2,1/2)",
    "fact_spans": "[[[4, 18]], [[29, 42]], [[4, 18]], [[29, 42]], [[0, 22]], [[23, 48]], [[0, 3], [24, 28], [49, 53]]]",
    "query_spans": "[[[49, 58]]]",
    "process": "Let $ P(x, 2x^{2}) $, then the distance from point $ P $ to the line $ 2x - y - 4 = 0 $ is $ \\frac{|2x - 2x^{2} - 4|}{\\sqrt{5}} = \\frac{|2(x - \\frac{1}{2})^{2} + |}{\\sqrt{5}} $. Therefore, when $ x = \\frac{1}{2} $, the distance from point $ P $ to the line $ 2x - y - 4 = 0 $ is shortest, and at this time $ P(\\frac{1}{2}, \\frac{1}{2}) $."
  },
  {
    "text": "Given $a > b > 0$, $e_{1}$, $e_{2}$ are the eccentricities of the conic sections $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1$ and $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$, respectively. Let $m = \\lg e_{1} + \\lg e_{2}$. Then the range of values for $m$ is?",
    "fact_expressions": "a: Number;b: Number;e1: Number;e2: Number;m: Number;a>b;b>0;C1: ConicSection;C2: ConicSection;Expression(C1) = (x^2/a^2+y^2/b^2=1);Expression(C2) = (x^2/a^2-y^2/b^2=1);Eccentricity(C1) = e1;Eccentricity(C2) = e2;m = lg(e1) + lg(e2)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-oo, 0)",
    "fact_spans": "[[[2, 10]], [[2, 10]], [[12, 19]], [[21, 29]], [[154, 157]], [[2, 10]], [[2, 10]], [[32, 79]], [[80, 123]], [[32, 79]], [[80, 123]], [[12, 127]], [[12, 127]], [[129, 152]]]",
    "query_spans": "[[[154, 164]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $P$ is a point on the ellipse $C$, and $\\angle F_{1} P F_{2}=60^{\\circ}$, $S_{\\Delta P F_{1} F_{2}}=3 \\sqrt{3}$, then $b=$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C);AngleOf(F1, P, F2) = ApplyUnit(60, degree);Area(TriangleOf(P, F1, F2)) = 3*sqrt(3)",
    "query_expressions": "b",
    "answer_expressions": "3",
    "fact_spans": "[[[18, 75], [86, 91]], [[18, 75]], [[173, 176]], [[25, 75]], [[25, 75]], [[25, 75]], [[2, 9]], [[10, 17]], [[2, 80]], [[82, 85]], [[82, 95]], [[98, 131]], [[133, 170]]]",
    "query_spans": "[[[173, 178]]]",
    "process": "Let |PF_{1}|=t_{1}, |PF_{2}|=t_{2}, then by the definition of the ellipse we have: t_{1}+t_{2}=2a\\textcircled{1}. Applying the cosine law in \\triangle F_{1}PF_{2} gives: t_{1}^{2}+t_{2}^{2}-2t_{1}t_{2}\\cdot\\cos60^{\\circ}=4c^{2}\\textcircled{2}. Combining \\textcircled{1} and \\textcircled{2} yields 3t_{1}t_{2}=4a^{2}-4c^{2}=4b^{2}, so t_{1}t_{2}=\\frac{4b^{2}}{3}. Therefore, S_{\\triangle PF_{1}F_{2}}=3\\sqrt{3}=\\frac{1}{2}t_{1}t_{2}\\cdot\\sin60^{\\circ}=\\frac{1}{2}\\times\\frac{4}{3}b^{2}\\times\\frac{\\sqrt{3}}{2}, \\therefore b=3."
  },
  {
    "text": "The hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has left and right foci $F_{1}$ and $F_{2}$, respectively. A line $l$ passing through point $F_{2}$ intersects the right branch of the hyperbola at points $A$ and $B$, such that $|F_{1} F_{2}|=|F_{1} B|$ and $\\overrightarrow{A F_{2}}=2 \\overrightarrow{F_{2} B}$. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;F1: Point;F2: Point;B: Point;A: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F2, l);Intersection(l, RightPart(C)) = {A, B};Abs(LineSegmentOf(F1, F2)) = Abs(LineSegmentOf(F1, B));VectorOf(A, F2) = 2*VectorOf(F2, B)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[96, 101]], [[0, 61], [102, 105], [201, 204]], [[8, 61]], [[8, 61]], [[70, 77]], [[78, 85], [87, 95]], [[113, 116]], [[109, 112]], [[8, 61]], [[8, 61]], [[0, 61]], [[0, 85]], [[0, 85]], [[86, 101]], [[96, 118]], [[120, 145]], [[146, 199]]]",
    "query_spans": "[[[201, 210]]]",
    "process": "Let |F_{1}B|=|F_{1}F_{2}|=2c, then |F_{2}B|=2c-2a. According to the problem, |AF_{2}|=4(c-a), |AF_{1}|=4c-2a. In the isosceles \\triangle F_{1}F_{2}B, \\cos\\angle F_{1}F_{2}B=\\frac{\\frac{1}{2}|BF_{2}|}{|F_{1}F_{2}|}=\\frac{c-a}{2c}, while \\cos\\angle F_{1}F_{2}A=-\\cos\\angle F_{1}F_{2}B. In \\triangle F_{1}F_{2}A, by the law of cosines |AF_{1}|^{2}=|F_{1}F_{2}|^{2}+|AF_{2}|^{2}-2|F_{1}F_{2}||AF_{2}|\\cos\\angle F_{1}F_{2}A, we obtain: (4c-2a)^{2}=(2c)^{2}+16(c-a)^{2}-2\\cdot2c\\cdot4(c-a)(-\\frac{c-a}{2c}), simplifying yields: 3c^{2}-8ac+5a^{2}=0, i.e., 3e^{2}-8e+5=0, and since e>1, solving gives e=\\frac{5}{3}. Therefore, the eccentricity of the hyperbola is \\frac{5}{3}."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$, respectively, and $P$ is an arbitrary point on the left branch of the hyperbola, then the minimum value of $\\frac{|P F_{2}|^{2}}{|P F_{1}|}$ is?",
    "fact_expressions": "F2: Point;F1: Point;LeftFocus(G) = F1;RightFocus(G) = F2;G: Hyperbola;Expression(G) = (x^2/4 - y^2/3 = 1);P: Point;PointOnCurve(P, LeftPart(G))",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F2))^2/Abs(LineSegmentOf(P, F1)))",
    "answer_expressions": "16",
    "fact_spans": "[[[10, 17]], [[2, 9]], [[2, 64]], [[2, 64]], [[20, 58], [69, 72]], [[20, 58]], [[65, 68]], [[65, 79]]]",
    "query_spans": "[[[81, 120]]]",
    "process": ""
  },
  {
    "text": "Given that the center of an ellipse is at the origin, and one focus is $F(0,3 \\sqrt{3})$, the line $4 x+3 y-13=0$ intersects the ellipse at points $M$ and $N$, the horizontal coordinate of the midpoint of $M N$ is $1$, then the equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;H: Line;M: Point;N: Point;F: Point;O:Origin;Expression(H) = (4*x + 3*y - 13 = 0);Coordinate(F) = (0, 3*sqrt(3));Center(G)=O;OneOf(Focus(G))=F;Intersection(H,G)={M,N};XCoordinate(MidPoint(LineSegmentOf(M, N))) = 1",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/36 + x^2/9 = 1",
    "fact_spans": "[[[2, 4], [51, 52], [84, 86]], [[34, 50]], [[55, 58]], [[60, 63]], [[16, 33]], [[7, 9]], [[34, 50]], [[16, 33]], [[2, 9]], [[2, 33]], [[34, 65]], [[66, 81]]]",
    "query_spans": "[[[84, 91]]]",
    "process": "Let the equation of the ellipse be $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$ $(a>b>0)$, according to the given condition $c=3\\sqrt{3}$. Let $M(x_{1},y_{1})$, $N(x_{2},y_{2})$, we obtain $\\frac{y_{1}^{2}}{a^{2}}+\\frac{x_{1}^{2}}{b^{2}}=1$; $\\frac{y_{2}^{2}}{a^{2}}+\\frac{x_{2}^{2}}{b^{2}}=1$. Subtracting these two equations and simplifying yields: $\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=-\\frac{a^{2}}{b^{2}}\\cdot\\frac{x_{1}+x_{2}}{y_{1}+y_{2}}=-\\frac{4}{3}$. The line $4x+3y-13=0$ intersects it at points $M$ and $N$, and the horizontal coordinate of the midpoint of $MN$ is $1$, so $x_{1}+x_{2}=2$, then $y_{1}+y_{2}=6$, $\\therefore \\frac{a^{2}}{b^{2}}\\cdot\\frac{1}{3}=\\frac{4}{3}$, and $a^{2}-b^{2}=27$, solving gives $b^{2}=9$, $a^{2}=36$, $\\therefore$ the standard equation of the ellipse is: $\\frac{y^{2}}{36}+\\frac{x^{2}}{9}=1$"
  },
  {
    "text": "Given that the center of the ellipse is at the origin, the foci are on the $x$-axis, the major axis has a length of $12$, and the eccentricity is $\\frac{1}{3}$, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;Center(G) = O;O: Origin;PointOnCurve(Focus(G), xAxis) = True;Length(MajorAxis(G)) = 12;Eccentricity(G) = 1/3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/36+y^2/32=1",
    "fact_spans": "[[[2, 4], [48, 50]], [[2, 10]], [[8, 10]], [[2, 19]], [[2, 28]], [[2, 46]]]",
    "query_spans": "[[[48, 55]]]",
    "process": "Let the major axis be 2a, the minor axis be 2b, and the focal distance be 2c. Then 2a=12, a=6, e=\\frac{c}{a}=\\frac{c}{6}=\\frac{1}{3}, c=2, so b^{2}=a^{2}-c^{2}=32. Since the foci are on the x-axis, the equation of the ellipse is \\frac{x^{2}}{36}+\\frac{y^{2}}{32}=1"
  },
  {
    "text": "Given line $l_{1}$: $3x - 4y - 9 = 0$ and line $l_{2}$: $y = -\\frac{1}{4}$, the minimum value of the sum of distances from a moving point $P$ on the parabola $y = x^{2}$ to line $l_{1}$ and line $l_{2}$ is?",
    "fact_expressions": "G: Parabola;l1:Line;l2:Line;Expression(l1)=(3*x - 4*y - 9 = 0);Expression(l2)=(y = -1/4);Expression(G)=(y=x^2);PointOnCurve(P,G);P:Point",
    "query_expressions": "Min(Distance(P,l1)+Distance(P,l2))",
    "answer_expressions": "2",
    "fact_spans": "[[[55, 67]], [[2, 26], [75, 84]], [[27, 54], [85, 94]], [[2, 26]], [[27, 54]], [[55, 67]], [[55, 74]], [[71, 74]]]",
    "query_spans": "[[[71, 105]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $E$: $y^{2}=2 p x(p>0)$ with focus $F$, and two points $A$, $B$ in the first quadrant lying on $E$, if $|F A|=3$, $|F B|=5$, $|A B|=2 \\sqrt{2}$, then the inclination angle of line $AB$ is?",
    "fact_expressions": "E: Parabola;p: Number;B: Point;A: Point;F: Point;p>0;Expression(E) = (y^2 = 2*(p*x));Focus(E) = F;PointOnCurve(A,E);PointOnCurve(B,E);Quadrant(A)=1;Quadrant(B)=1;Abs(LineSegmentOf(F, A)) = 3;Abs(LineSegmentOf(F, B)) = 5;Abs(LineSegmentOf(A, B)) = 2*sqrt(2)",
    "query_expressions": "Inclination(LineOf(A,B))",
    "answer_expressions": "pi/4",
    "fact_spans": "[[[2, 28], [54, 57]], [[10, 28]], [[49, 52]], [[45, 48]], [[32, 35]], [[10, 28]], [[2, 28]], [[2, 35]], [[45, 58]], [[45, 58]], [[36, 52]], [[36, 52]], [[60, 69]], [[70, 79]], [[80, 98]]]",
    "query_spans": "[[[100, 113]]]",
    "process": "As shown in the figure, draw perpendiculars from A and B to the directrix, with feet C and D respectively, and draw AE\\bot BD from A meeting at E. Then |AF| = |AC| = 3, |BF| = |BD| = 5, so |BE| = 2. Since |AB| = 2\\sqrt{2}, it follows that \\sin\\angle EAB = \\frac{|BE|}{|AB|} = \\frac{2}{2\\sqrt{2}} = \\frac{\\sqrt{2}}{2}. Since \\angle EAB \\in (0, \\frac{\\pi}{2}), we have \\angle EAB = \\frac{\\pi}{4}. Therefore, the inclination angle of line AB is \\frac{\\pi}{2} - \\frac{\\pi}{4} = \\frac{\\pi}{4}."
  },
  {
    "text": "The focus of the parabola $y^{2}=4 x$ is $F$, and point $P(x, y)$ is a moving point on this parabola. Given point $A(-1,0)$, then the minimum value of $\\frac{|P F|}{|P A|}$ is?",
    "fact_expressions": "G: Parabola;P: Point;A: Point;F: Point;x1:Number;y1:Number;Expression(G) = (y^2 = 4*x);Coordinate(P) = (x1, y1);Coordinate(A) = (-1, 0);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F))/Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[0, 14], [34, 37]], [[22, 32]], [[43, 53]], [[18, 21]], [[23, 32]], [[23, 32]], [[0, 14]], [[22, 32]], [[43, 53]], [[0, 21]], [[22, 41]]]",
    "query_spans": "[[[55, 82]]]",
    "process": "According to the definition of a parabola, |PF| = x + 1, and |PA|, so $\\frac{|PF|}{|PA|} = \\frac{x+1}{\\sqrt{y^{2}+(x+1)^{2}}}$ ①. Since $y^{2} = 4x$, let $\\frac{2}{x+1} = t$, then equation ① can be simplified to $\\overline{\\sqrt{1}}\\underline{1}t^{2}+2t+1=$ where $t \\in (0,2]$, the minimum value is found to be $\\frac{\\sqrt{2}}{2}$, so the minimum value of $\\frac{|PF|}{|PA|}$ is $\\frac{\\sqrt{2}}{2}$."
  },
  {
    "text": "If $O$, $F$, and $B$ are the center, focus, and endpoint of the minor axis of an ellipse respectively, and $\\angle B F O = \\frac{\\pi}{3}$, then the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;F: Point;OneOf(Focus(G)) = F;B: Point;OneOf(Endpoint(MinorAxis(G)))= B;AngleOf(B, F, O) = pi/3;e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "1/2",
    "fact_spans": "[[[15, 17], [61, 63]], [[1, 4]], [[1, 29]], [[5, 8]], [[1, 29]], [[9, 12]], [[1, 29]], [[30, 58]], [[67, 70]], [[61, 70]]]",
    "query_spans": "[[[67, 72]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the hyperbola, the incenter of $\\Delta P F_{1} F_{2}$ is $I$, and it satisfies $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=\\overrightarrow{P I} \\cdot \\overrightarrow{P F_{1}}$, $P F_{2} \\perp F_{1} F_{2}$. Then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;I: Point;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) =F1;RightFocus(C)=F2;PointOnCurve(P, C);Center(InscribedCircle(TriangleOf(P,F1,F2)))=I;DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = DotProduct(VectorOf(P, I), VectorOf(P, F1));IsPerpendicular(LineSegmentOf(P,F2),LineSegmentOf(F1,F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)+1",
    "fact_spans": "[[[2, 53], [84, 87], [265, 271]], [[10, 53]], [[10, 53]], [[79, 83]], [[63, 70]], [[71, 78]], [[118, 121]], [[2, 53]], [[2, 78]], [[2, 78]], [[79, 88]], [[89, 121]], [[125, 234]], [[235, 262]]]",
    "query_spans": "[[[265, 277]]]",
    "process": "From \\overrightarrow{PF}\\cdot\\overrightarrow{PF_{2}}=\\overrightarrow{PI}\\cdot\\overrightarrow{PF}_{1}, that is, \\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}-\\overrightarrow{PI}\\cdot\\overrightarrow{PF_{1}}=0, namely \\overrightarrow{PF_{1}}\\cdot(\\overrightarrow{PF_{2}}-\\overrightarrow{PI})=\\overrightarrow{PF}\\cdot\\overrightarrow{IF_{2}}=0, so PF_{1}\\bot IF_{2}. Since IF_{2} is the angle bisector of \\angle PF_{2}F_{1}, it follows that F_{1}F_{2}=PF_{2}, i.e., 2c=\\frac{b^{2}}{a}, so 2ac=b^{2}. Hence c^{2}-a^{2}=2ac, leading to e^{2}-2e-1=0. Solving gives e=\\sqrt{2}+1 or e=-\\sqrt{2}+1 (discarded)."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$, the distance from the midpoint $P$ of chord $AB$ to the $y$-axis is $2$. Then, the maximum length of chord $AB$ is?",
    "fact_expressions": "C: Parabola;A: Point;B: Point;P: Point;Expression(C) = (y^2 = 4*x);IsChordOf(LineSegmentOf(A,B),C);MidPoint(LineSegmentOf(A,B))=P;Distance(P, yAxis) = 2",
    "query_expressions": "Max(Length(LineSegmentOf(A, B)))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 22]], [[24, 28]], [[24, 28]], [[31, 34]], [[2, 22]], [[2, 28]], [[24, 34]], [[31, 46]]]",
    "query_spans": "[[[49, 61]]]",
    "process": ""
  },
  {
    "text": "Given $ m \\in R $, if the moving line $ m x - y = 0 $ passing through fixed point $ A $ and the moving line $ x + m y + 1 = 0 $ passing through fixed point $ B $ intersect at point $ P(x_1, y_1) $, then the maximum value of $ PA + PB $ is?",
    "fact_expressions": "G: Line;H: Line;m: Real;P: Point;A: Point;B: Point;x1: Number;y1: Number;Expression(G) = (m*x - y = 0);Expression(H) = (m*y + x + 1 = 0);Coordinate(P) = (x1, y1);PointOnCurve(A, G);PointOnCurve(B, H);Intersection(G, H) = P",
    "query_expressions": "Max(LineSegmentOf(P, A) + LineSegmentOf(P, B))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[23, 32]], [[43, 54]], [[2, 11]], [[56, 68]], [[16, 19]], [[36, 39]], [[57, 68]], [[57, 68]], [[23, 32]], [[43, 54]], [[56, 68]], [[13, 32]], [[33, 54]], [[21, 68]]]",
    "query_spans": "[[[70, 85]]]",
    "process": "A(0,0), B(-1,0), the moving line mx - y = 0 and the moving line x + my + 1 = 0 are perpendicular to each other, so the trajectory of point P is a circle with AB as diameter. PA^{2}+PB^{2}=AB^{2}=1 \\therefore PA+PB\\leqslant2\\sqrt{\\frac{PA^{2}+PB^{2}}{2}}=\\sqrt{2}"
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, respectively, $P$ be any point on the ellipse, and the coordinates of point $M$ be $(6 , 4)$. Then the maximum value of $|P M|+| P F_{1}|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;M: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/16 = 1);Coordinate(M) = (6, 4);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "15",
    "fact_spans": "[[[19, 58], [69, 71]], [[65, 68]], [[76, 80]], [[1, 8]], [[9, 16]], [[19, 58]], [[76, 93]], [[1, 64]], [[1, 64]], [[65, 75]]]",
    "query_spans": "[[[95, 119]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the directrix of the parabola $y=-8 x^{2}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = -8*x^2)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y = 1/32",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 22]]]",
    "process": "Since the standard equation of the parabola $ y = -8x^{2} $ is $ x^{2} = -\\frac{1}{8}y $, it follows that $ 2p = \\frac{1}{8} $, i.e., $ p = \\frac{1}{16} $. Therefore, its directrix equation is: $ y = \\frac{1}{32} $."
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{\\frac{75}{4}}=1$, and $F_{1}$, $F_{2}$ are the foci of the ellipse. If $\\angle F_{1} P F_{2}=60^{\\circ}$, then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/25 + y^2/(75/4) = 1);PointOnCurve(P, G);Focus(G)={F1,F2};AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "25*sqrt(3)/4",
    "fact_spans": "[[[6, 55], [75, 77]], [[59, 66]], [[2, 5]], [[67, 74]], [[6, 55]], [[2, 58]], [[59, 80]], [[82, 115]]]",
    "query_spans": "[[[117, 144]]]",
    "process": "From the ellipse equation, $ a^{2}=25 $, $ b^{2}=\\frac{75}{4} $, so $ c^{2}=\\frac{25}{4} $, hence $ c=\\frac{5}{2} $, $ 2c=5 $. In $ \\triangle PF_{1}F_{2} $, $ |F_{1}F_{2}|^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}|\\cdot|PF_{2}|\\cos60^{\\circ} $, that is, $ 25=|PF_{1}|^{2}+|PF_{2}|^{2}-|PF_{1}|\\cdot|PF_{2}|\\textcircled{1} $. From the definition of the ellipse, $ 10=|PF_{1}|+|PF_{2}| $, so $ 100=|PF_{1}|^{2}+|PF_{2}|^{2}+2|PF_{1}|\\cdot|PF_{2}|\\textcircled{2} $. $ \\textcircled{2}-\\textcircled{1} $ gives $ 3|PF_{1}|\\cdot|PF_{2}|=75 $, so $ |PF_{1}|\\cdot|PF_{2}|=25 $, thus $ S_{\\triangle F_{1}PF_{2}}=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|\\sin60^{\\circ}=\\frac{25\\sqrt{3}}{4} $"
  },
  {
    "text": "Let the focus of the parabola $y^{2}=8x$ be $F$, the directrix be $l$, and let $P$ be a point on the parabola such that $PA \\perp l$, where $A$ is the foot of the perpendicular. If the slope of the line $AF$ is $-\\sqrt{2}$, then $|PF|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;l: Line;Directrix(G) = l;P: Point;PointOnCurve(P,G) = True;A: Point;IsPerpendicular(LineSegmentOf(P, A), l) = True;FootPoint(LineSegmentOf(P, A), l) = A;Slope(LineOf(A, F)) = -sqrt(2)",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 15], [36, 39]], [[1, 15]], [[19, 22]], [[1, 22]], [[26, 29]], [[1, 29]], [[32, 35]], [[32, 42]], [[59, 62]], [[43, 56]], [[43, 65]], [[68, 90]]]",
    "query_spans": "[[[95, 104]]]",
    "process": "The focus of the parabola is $ F(2,0) $, the equation of the directrix is $ x = -2 $, the equation of line $ AF $ is $ y = -\\sqrt{2}(x - 2) $. Solving together with the directrix equation $ x = -2 $, we obtain point $ A(-2, 4\\sqrt{2}) $, $ P(4, 4\\sqrt{2}) $. According to the definition of the parabola, we have $ |PF| = |PA| = 4 + 2 = 8 $."
  },
  {
    "text": "Given that the moving circle $M$ is externally tangent to the circle $C_{1}$: $(x+5)^{2}+y^{2}=16$ and internally tangent to the circle $C_{2}$: $(x-5)^{2}+y^{2}=16$, then the trajectory equation of the center of the moving circle is?",
    "fact_expressions": "M: Circle;C1: Circle;Expression(C1) = ((x + 5)^2 + y^2 = 16);IsOutTangent(M, C1) = True;C2: Circle;Expression(C2) = ((x - 5)^2 + y^2 = 16);IsInTangent(M, C2) = True",
    "query_expressions": "LocusEquation(Center(M))",
    "answer_expressions": "(x^2/16-y^2/9=1)&(x>0)",
    "fact_spans": "[[[4, 7], [74, 76]], [[8, 37]], [[8, 37]], [[4, 39]], [[41, 70]], [[41, 70]], [[4, 72]]]",
    "query_spans": "[[[74, 85]]]",
    "process": "Let the center of the moving circle be M(x, y), with radius r. Then we have  \n\\begin{cases}|MC_{1}|=4+r\\\\|MC_{2}|=r-4\\end{cases}.  \nEliminating r gives |MC_{1}|-|MC_{2}|=8. Thus, by the definition of a hyperbola, the locus of point M is the right branch of a hyperbola with foci C_{1}, C_{2} and real axis length 2a=8. Therefore, its trajectory equation is \\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1 (x>0).  \n【Method Insight】In this problem, setting the radius of the moving circle as r and using r as a parameter makes it easy to express |MC_{1}| and |MC_{2}|. Eliminating r yields |MC_{1}|-|MC_{2}|=8, revealing that the locus of point M is one branch of a hyperbola. Common student errors include failing to introduce the radius r of the moving circle as a parameter to bridge the solution, and mistakenly assuming the locus is the entire hyperbola rather than just one branch. This misunderstanding arises from an incomplete grasp of the definition of a hyperbola. This problem is beneficial for students’ understanding of the definition of a hyperbola and is a good question."
  },
  {
    "text": "The directrix of the parabola $y^{2}=16 x$ passes through a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{8}=1$. Then, the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;Expression(G) = (-y^2/8 + x^2/a^2 = 1);Expression(H) = (y^2 = 16*x);PointOnCurve(OneOf(Focus(G)), Directrix(H))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[20, 62], [69, 72]], [[23, 62]], [[0, 15]], [[20, 62]], [[0, 15]], [[0, 67]]]",
    "query_spans": "[[[69, 78]]]",
    "process": ""
  },
  {
    "text": "A chord $PQ$ lies on the left branch of the hyperbola $x^{2}-y^{2}=4$, passing through the left focus $F_{1}$. If $|PQ|=7$ and $F_{2}$ is the right focus of the hyperbola, then the perimeter of $\\triangle PF_{2} Q$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;Q: Point;F2: Point;F1:Point;Expression(G) = (x^2 - y^2 = 4);Abs(LineSegmentOf(P, Q)) = 7;RightFocus(G) = F2;LeftFocus(G)=F1;PointOnCurve(F1,LineSegmentOf(P,Q));IsChordOf(LineSegmentOf(P,Q),LeftPart(G))",
    "query_expressions": "Perimeter(TriangleOf(P, F2, Q))",
    "answer_expressions": "22",
    "fact_spans": "[[[1, 19], [63, 66]], [[34, 38]], [[34, 38]], [[55, 62]], [[23, 30]], [[1, 19]], [[44, 52]], [[55, 70]], [[1, 30]], [[0, 38]], [[1, 42]]]",
    "query_spans": "[[[72, 97]]]",
    "process": ""
  },
  {
    "text": "The sum of the distances from a point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$ to its two foci is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/9 = 1);F1:Point;F2:Point;P:Point;Focus(G)={F1,F2};PointOnCurve(P,G)",
    "query_expressions": "Distance(P,F1)+Distance(P,F2)",
    "answer_expressions": "6",
    "fact_spans": "[[[0, 37]], [[0, 37]], [], [], [], [[0, 44]], [[0, 40]]]",
    "query_spans": "[[[0, 51]]]",
    "process": "\\because4<9,\\thereforea^{2}=9,\\thereforea=3, using the definition of an ellipse, the sum of the distances from a point on the ellipse \\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1 to the two foci is 2a=6."
  },
  {
    "text": "The standard equation of an ellipse centered at the origin, with the length of the major axis being $\\sqrt{3}$ times the length of the minor axis, and one focus at $(0, \\sqrt{2})$, is?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O ;Length(MajorAxis(G)) = sqrt(3)*Length(MinorAxis(G));Coordinate(OneOf(Focus(G))) = (0, sqrt(2))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/3 + x^2 = 1",
    "fact_spans": "[[[48, 50]], [[1, 3]], [[0, 50]], [[7, 50]], [[27, 50]]]",
    "query_spans": "[[[48, 57]]]",
    "process": "According to the problem, let the equation of the ellipse be $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$ $(a>b>0)$. Since the center is at the origin, the length of the major axis is $\\sqrt{3}$ times the length of the minor axis, and one focus is at $(0,\\sqrt{2})$, we have $2a=2b\\times\\sqrt{3}$, $c=\\sqrt{2}$, and $a^{2}=b^{2}+c^{2}$. Solving these equations simultaneously gives $a^{2}=3$, $b^{2}=1$. Therefore, the standard equation of the ellipse is $\\frac{y^{2}}{3}+x^{2}=1$."
  },
  {
    "text": "The distance from the left focus of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ to the line $3 x-4 y-4=0$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1);H: Line;Expression(H) = (3*x - 4*y - 4 = 0)",
    "query_expressions": "Distance(LeftFocus(G),H)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 28]], [[0, 28]], [[33, 48]], [[33, 48]]]",
    "query_spans": "[[[0, 53]]]",
    "process": "Let the left focus be F(-2,0), then the distance from the left focus to the line 3x-4y-4=0 is d=\\frac{3\\times(-2)-4\\times0}{3^{2}+4^{2}}4\\times0-4"
  },
  {
    "text": "Draw a line through the focus $F$ of the parabola $x^{2}=8y$, intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $y_{1}+y_{2}=8$, then what is the length of segment $AB$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;x1:Number;y1:Number;x2:Number;y2:Number;Expression(G) = (x^2 = 8*y);Focus(G)=F;Coordinate(A) = (x1,y1);Coordinate(B)=(x2,y2);PointOnCurve(F,H);Intersection(H,G) = {B, A};y1 + y2 = 8;F:Point",
    "query_expressions": "Length(LineSegmentOf(A,B))",
    "answer_expressions": "12",
    "fact_spans": "[[[1, 15], [25, 28]], [[22, 24]], [[29, 47]], [[49, 66]], [[29, 46]], [[70, 85]], [[49, 66]], [[70, 85]], [[1, 15]], [[1, 21]], [[29, 46]], [[49, 66]], [[0, 24]], [[22, 68]], [[70, 85]], [[18, 21]]]",
    "query_spans": "[[[87, 98]]]",
    "process": "From the definition of the parabola, we have |AF| = y_{1} + \\frac{p}{2}, |BF| = y_{2} + \\frac{p}{2}, so |AB| = |AF| + |BF| = y + y + p = 12."
  },
  {
    "text": "Given that a line $l$ with slope $2$ intersects the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ at points $A$ and $B$, and that point $P(2,1)$ is the midpoint of segment $AB$, then the eccentricity of $C$ is equal to?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;B: Point;A: Point;P: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (2, 1);Slope(l)=2;Intersection(l,C) = {A, B};MidPoint(LineSegmentOf(A,B))=P",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[9, 14]], [[15, 76], [111, 114]], [[23, 76]], [[23, 76]], [[82, 85]], [[78, 81]], [[89, 98]], [[23, 76]], [[23, 76]], [[15, 76]], [[89, 98]], [[2, 14]], [[9, 87]], [[89, 109]]]",
    "query_spans": "[[[111, 121]]]",
    "process": "According to the point difference method, we obtain $a^{2}=b^{2}$; then by substituting $b^{2}=c^{2}-a^{2}$, the eccentricity can be derived. [f(x) Point $P(2,1)$ is the midpoint of segment $AB$, so $\\frac{2}{2}=0 \\Rightarrow \\frac{4}{a^{2}} - \\frac{2(y_{1}-y_{2})}{b^{2}(x_{1}-x_{2})} = 0$. Since the slope of line $l$ is 2, we have $a^{2}=b^{2}$, that is, $a^{2}=c^{2}-a^{2} \\Rightarrow \\frac{c}{a}=\\sqrt{2}$"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ and the line $3 x+4 y=0$, what is the maximum distance from a point on the ellipse to this line?",
    "fact_expressions": "G: Ellipse;H: Line;Expression(G) = (x^2/4 + y^2/3 = 1);Expression(H) = (3*x + 4*y = 0);P:Point;PointOnCurve(P,G)",
    "query_expressions": "Max(Distance(P,H))",
    "answer_expressions": "2*sqrt(21)/5",
    "fact_spans": "[[[2, 39], [55, 57]], [[40, 53], [62, 64]], [[2, 39]], [[40, 53]], [[58, 59]], [[55, 59]]]",
    "query_spans": "[[[58, 71]]]",
    "process": "Let $ P(2\\cos\\theta,\\sqrt{3}\\sin\\theta) $ be an arbitrary point on the ellipse, so the distance from point $ P $ to the line $ 3x+4y=0 $ is $ d $, then $ d = \\frac{|6\\cos\\theta + 4\\sqrt{3}\\sin\\theta|}{\\sqrt{3^{2}+4^{2}}} $. When $ |\\sin(\\theta+\\varphi)| = 1 $, $ d_{\\text{max}} = \\frac{2\\sqrt{21}|\\sin(\\theta+\\varphi)|}{} $ $ \\left( \\tan\\varphi = \\frac{\\sqrt{3}}{2} \\right) $."
  },
  {
    "text": "The line $l$ intersects the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ at points $A$ and $B$. If the midpoint of segment $AB$ has coordinates $(1, \\frac{1}{2})$, then the equation of line $l$ is?",
    "fact_expressions": "l: Line;G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);Intersection(l, G) = {A, B};A: Point;B: Point;Coordinate(MidPoint(LineSegmentOf(A,B))) = (1,1/2)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "2*x+2*y-3=0",
    "fact_spans": "[[[0, 5], [79, 84]], [[6, 33]], [[6, 33]], [[0, 43]], [[34, 37]], [[38, 41]], [[45, 76]]]",
    "query_spans": "[[[79, 89]]]",
    "process": "Let A(x₁,y₁), B(x₂,y₂), substituting into the ellipse equation gives \\frac{x_{1}^2}{2} + y_{1}^{2} = 1, \\frac{x_{2}^2}{2} + y_{2}^{2} = 1, subtracting these two equations yields \\frac{(x_{2}-x_{1})(x_{2}+x_{1})}{2} = -(y_{2}+y_{1})(y_{2}-y_{1}), i.e., (x_{2}-x_{1})\\frac{(x_{2}+x_{1})}{2} = -(y_{2}+y_{1})(y_{2}-y_{1}), thus k_{AB} = \\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = -1, so the equation of line l is y - \\frac{1}{2} = -(x - 1), i.e., 2x + 2y - 3 = 0. Method"
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{a^{2}}=1$ is $y=\\sqrt{2} x$, then the value of the real number $a$ is?",
    "fact_expressions": "G: Hyperbola;a: Real;Expression(G) = (x^2/2 - y^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (y = sqrt(2)*x)",
    "query_expressions": "a",
    "answer_expressions": "pm*2",
    "fact_spans": "[[[2, 44]], [[67, 72]], [[2, 44]], [[2, 65]]]",
    "query_spans": "[[[67, 76]]]",
    "process": "\\because\\frac{x^{2}}{2}-\\frac{y^{2}}{a^{2}}=1\\therefore the hyperbola \\frac{x^{2}}{2}-\\frac{y^{2}}{a^{2}}=1 has asymptotes y=\\pm\\frac{a}{\\sqrt{2}}x\\because one asymptote of the hyperbola \\frac{x^{2}}{2}-\\frac{y^{2}}{a^{2}}=1 is y=\\sqrt{2}x\\therefore\\pm\\frac{a}{\\sqrt{2}}=\\sqrt{2}, solving gives a=\\pm2"
  },
  {
    "text": "The left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a, b>0)$ are $F_{1}$ and $F_{2}$, respectively. The line passing through the focus $F_{2}$ and perpendicular to the $x$-axis intersects the hyperbola at points $A$ and $B$. If $\\overrightarrow{F_{1} A} \\cdot \\overrightarrow{F_{1} B}=0$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;a>0;b>0;F1: Point;F2: Point;H: Line;A: Point;B: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, H);IsPerpendicular(H, xAxis);Intersection(H, G) = {A, B};DotProduct(VectorOf(F1, A), VectorOf(F1, B)) = 0",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)+1",
    "fact_spans": "[[[0, 54], [101, 104], [179, 182]], [[3, 54]], [[3, 54]], [[3, 54]], [[3, 54]], [[63, 70]], [[71, 78], [82, 89]], [[98, 100]], [[107, 110]], [[111, 114]], [[0, 54]], [[0, 78]], [[0, 78]], [[79, 100]], [[90, 100]], [[98, 116]], [[118, 177]]]",
    "query_spans": "[[[179, 188]]]",
    "process": ""
  },
  {
    "text": "Two points $A$ and $B$ on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ are such that the lines connecting them to the center $O$ are mutually perpendicular. Then $\\frac{1}{|O A|^{2}}+\\frac{1}{|O B|^{2}}$=?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;O: Origin;A: Point;B: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(A,G);PointOnCurve(B,G);Center(G)=O;IsPerpendicular(LineSegmentOf(A,O),LineSegmentOf(B,O))",
    "query_expressions": "1/(Abs(LineSegmentOf(O, B))^2) + 1/(Abs(LineSegmentOf(O, A))^2)",
    "answer_expressions": "(a^2+b^2)/(a^2*b^2)",
    "fact_spans": "[[[0, 52]], [[2, 52]], [[2, 52]], [[65, 68]], [[55, 58]], [[59, 62]], [[2, 52]], [[2, 52]], [[0, 52]], [[0, 62]], [[0, 62]], [[0, 68]], [[55, 75]]]",
    "query_spans": "[[[77, 120]]]",
    "process": ""
  },
  {
    "text": "The line $l$ with slope $2$ is intersected by the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ such that the chord is exactly bisected at the point $M(2,1)$. Then the eccentricity of $C$ is?",
    "fact_expressions": "l: Line;Slope(l) = 2;C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;M: Point;Coordinate(M) = (2, 1);MidPoint(InterceptChord(l, C)) = M",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[7, 12]], [[0, 12]], [[13, 74], [93, 96]], [[13, 74]], [[21, 74]], [[21, 74]], [[21, 74]], [[21, 74]], [[80, 89]], [[80, 89]], [[7, 91]]]",
    "query_spans": "[[[93, 102]]]",
    "process": "Analysis: Let the coordinates of the two endpoints of the chord be given. After substituting into the hyperbola equation and subtracting, we obtain a relation between $a$ and $b$, thus finding the eccentricity. Specifically, let the two intersection points of line $l$ with the hyperbola be $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then\n\\[\n\\begin{cases}\n\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 \\\\\n\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{\\frac{2}{2}}=1\n\\end{cases}\n\\]\nSubtracting the two equations gives\n\\[\n\\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{x_{1}+x_{2}=4,y_{1}+y_{2}=2,\\frac{y_{1}+y_{2})(y_{1}-y_{2})}{x_{1}}y_{2}_{2}}=0,\n\\]\ni.e.,\n\\[\n\\frac{y_{1}-y_{2}}{x_{1}}=x_{2}=\\frac{b^{2}}{a^{2}}\\cdot\\frac{x_{1}+x_{2}^{2}}{y_{1}+y_{2}},\n\\]\nand from the given condition, we have $e=\\frac{c}{a}=\\sqrt{2}$."
  },
  {
    "text": "If the major axis of $m x^{2}+y^{2}=1$ is 2 times the minor axis, then $m$=?",
    "fact_expressions": "G: Curve;Expression(G) = (m*x^2 + y^2 = 1);MajorAxis(G) = 2*(MinorAxis(G));m: Number",
    "query_expressions": "m",
    "answer_expressions": "{1/4, 4}",
    "fact_spans": "[[[1, 18]], [[1, 18]], [[1, 29]], [[31, 34]]]",
    "query_spans": "[[[31, 36]]]",
    "process": ""
  },
  {
    "text": "The standard equation of the hyperbola that shares the same asymptotes as $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ and passes through the point $A(6,8)$ is?",
    "fact_expressions": "G: Hyperbola;H: Hyperbola;A: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Coordinate(A) = (6, 8);PointOnCurve(A, H);Asymptote(G) = Asymptote(H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2/64 - x^2/36 = 1",
    "fact_spans": "[[[1, 40]], [[61, 64]], [[51, 60]], [[1, 40]], [[51, 60]], [[50, 64]], [[0, 64]]]",
    "query_spans": "[[[61, 71]]]",
    "process": "Let the required hyperbola be \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=\\lambda(\\lambda\\neq0); substituting the point (6,8) gives \\frac{36}{9}-\\frac{128}{16}=\\lambda, solving yields \\lambda=4, \\therefore the standard equation of the required hyperbola is \\frac{y^{2}}{64}-\\frac{x^{2}}{36}=1."
  },
  {
    "text": "Given that the distance between the left vertex of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ and the focus of the parabola $y^{2}=2px$ is $4$, and the intersection point of one asymptote of the hyperbola and the directrix of the parabola has coordinates $(-2, -1)$, find the focal length of the hyperbola.",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;p: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 2*(p*x));Distance(LeftVertex(G),Focus(H))=4;Coordinate(Intersection(OneOf(Asymptote(G)),Directrix(H)))=(-2,-1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[2, 59], [92, 95], [125, 128]], [[5, 59]], [[5, 59]], [[64, 80], [102, 105]], [[67, 80]], [[5, 59]], [[5, 59]], [[2, 59]], [[64, 80]], [[2, 90]], [[92, 123]]]",
    "query_spans": "[[[125, 133]]]",
    "process": "By the given condition, the intersection point of one asymptote of the hyperbola and the directrix of the parabola is $(-2,-1)$, so the point $(-2,-1)$ lies on the directrix of the parabola, hence $-\\frac{p}{2} = -2$, yielding $p = 4$. Then the focus of the parabola has coordinates $(2,0)$. Thus, the left vertex of the hyperbola is $(-2,0)$, so $a = 2$. Since the point $(-2,-1)$ lies on the asymptote of the hyperbola, its asymptote equations are $y = \\pm\\frac{1}{2}x$. By the properties of the hyperbola, we obtain $b = 1$, and then $c = \\sqrt{5}$. Therefore, the focal length is $2\\sqrt{5}$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, let $l$ be its asymptote with positive slope. If there exist three distinct points on the curve $E$: $\\sqrt{4x - x^{2}}$ whose distance to $l$ is $1$, then the range of the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;l: Line;OneOf(Asymptote(C)) = l;Slope(l)>0;E: Curve;Expression(E) = (sqrt(-x^2 + 4*x));A: Point;B: Point;D: Point;Negation(A=B);Negation(A=D);Negation(B=D);PointOnCurve(A, E);PointOnCurve(B, E);PointOnCurve(D, E);Distance(A, l) = 1;Distance(B, l) = 1;Distance(D, l) = 1",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "[4*sqrt(15)/15, 2*sqrt(3)/3)",
    "fact_spans": "[[[2, 60], [122, 128]], [[2, 60]], [[10, 60]], [[10, 60]], [[10, 60]], [[10, 60]], [[70, 73], [110, 113]], [[2, 73]], [[61, 73]], [[75, 100]], [[75, 100]], [], [], [], [[75, 109]], [[75, 109]], [[75, 109]], [[75, 109]], [[75, 109]], [[75, 109]], [[75, 120]], [[75, 120]], [[75, 120]]]",
    "query_spans": "[[[122, 139]]]",
    "process": "Given $ l: y = \\frac{b}{a}x $, the curve $ E: y = \\sqrt{4x - x^{2}} $ represents the upper half (including endpoints) of a circle with center at point $ A(2,0) $ and radius 2. The distance from point $ A $ to line $ l $ is $ d_{1} = \\frac{2b}{\\sqrt{a^{2} + b^{2}}} $. The distance from an endpoint $ B(4,0) $ of curve $ E: y = \\sqrt{4x - x^{2}} $ to line $ l $ is $ d_{2} = \\frac{4b}{\\sqrt{a^{2} + b^{2}}} $. This is equivalent to $ 2 - d_{1} > 1 $ and $ d_{2} \\geqslant 1 $, i.e., $ \\frac{1}{4} \\leqslant \\frac{b}{\\sqrt{a^{2} + b^{2}}} < \\frac{1}{2} $. Solving the inequality gives the solution [Detailed Solution] From the given conditions: $ y = \\frac{b}{a}x $, curve $ E: y = \\sqrt{4x - x^{2}} $, i.e., $ (x - 2)^{2} + y^{2} = 4 $ ($ y \\geqslant 0 $), represents the upper half (including endpoints) of a circle with center at point $ A(2,0) $ and radius 2. The distance from point $ A $ to line $ l $ is $ d_{1} = \\frac{2b}{\\sqrt{a^{2} + b^{2}}} $. The distance from an endpoint $ B(4,0) $ of curve $ E $ to line $ l $ is $ d_{2} = \\frac{4b}{\\sqrt{a^{2} + b^{2}}} $. Since there exist three distinct points on curve $ E: y = \\sqrt{4x - x^{2}} $ whose distances to $ l $ are equal to 1, it follows that $ 2 - d_{1} > 1 $ and $ d_{2} \\geqslant 1 $. Rearranging yields $ \\frac{1}{4} \\leqslant \\frac{b}{\\sqrt{a^{2} + b^{2}}} < \\frac{1}{2} $, hence $ \\frac{1}{4} \\leqslant \\frac{1}{\\sqrt{1 + (\\frac{a}{b})^{2}}} < \\frac{1}{2} $, then $ 2 < \\sqrt{1 + \\frac{a^{2}}{b^{2}}} \\leqslant 4 $. Therefore, $ \\frac{1}{16} \\leqslant \\frac{b^{2}}{c^{2}} < \\frac{1}{4} $, i.e., $ \\frac{1}{16} \\leqslant 1 - \\frac{1}{e^{2}} < \\frac{1}{4} $, yielding $ \\frac{4\\sqrt{15}}{15} \\leqslant e < \\frac{2\\sqrt{3}}{3} $ ($ \\frac{\\sqrt{15}}{15}, \\frac{2\\sqrt{3}}{3} $)"
  },
  {
    "text": "If the two foci of an ellipse are $(-2,0)$, $(2,0)$, and the ellipse passes through the point $(2,3)$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Coordinate(F1) = (-2, 0);Coordinate(F2) = (2, 0);Focus(G) = {F1, F2};H: Point;Coordinate(H) = (2, 3);PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16+y^2/12=1",
    "fact_spans": "[[[1, 3], [28, 30], [42, 44]], [[8, 17]], [[18, 25]], [[1, 25]], [[1, 25]], [[1, 25]], [[31, 39]], [[31, 39]], [[28, 39]]]",
    "query_spans": "[[[42, 51]]]",
    "process": ""
  },
  {
    "text": "The length of the imaginary axis of the hyperbola $M$: $x^{2}-(y-1)^{2}=1$ is?",
    "fact_expressions": "M: Hyperbola;Expression(M) = (x^2 - (y - 1)^2 = 1)",
    "query_expressions": "Length(ImageinaryAxis(M))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 27]], [[0, 27]]]",
    "query_spans": "[[[0, 33]]]",
    "process": "Hyperbola M: x^{2}-(y-1)^{2}=1, imaginary semi-axis is 1, imaginary axis length is: 2."
  },
  {
    "text": "Given that the coordinates of point $M$ are $(1,1)$, $F_{1}$ is the left focus of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, and $P$ is a moving point on the ellipse, then the range of values of $|P F_{1}|+|P M|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;M: Point;Expression(G) = (x^2/9 + y^2/5 = 1);Coordinate(M) = (1, 1);LeftFocus(G) = F1;PointOnCurve(P, G)",
    "query_expressions": "Range(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "[6-sqrt(2),6+sqrt(2)]",
    "fact_spans": "[[[28, 65], [74, 76]], [[70, 73]], [[19, 27]], [[2, 6]], [[28, 65]], [[2, 18]], [[20, 69]], [[70, 80]]]",
    "query_spans": "[[[82, 106]]]",
    "process": "\\because|PF_{1}|+|PF_{2}|=2a=6，then |PF_{1}|=6-|PF_{2}|，then |PF_{1}|+|PM|=6-|PF_{2}|+|PM|=6+(|PM|-|PF_{2}|). According to the triangle inequality, when point P is at P_{1}, the difference |PM|-|PF_{2}| is minimized. At this time, the line connecting F_{2} and M intersects the ellipse at P_{1}, it is easy to obtain -|MF_{2}|=-\\sqrt{2}, and thus |PF_{1}|+|PM| reaches its minimum value of 6-\\sqrt{2}. When point P is at P_{2}, the difference |PM|-|PF_{2}| is maximized. At this time, the line connecting F_{2} and M intersects the ellipse at P_{2}, it is easy to obtain |MF_{2}|=\\sqrt{2}, and thus |PF_{1}|+|PM| reaches its maximum value of 6+\\sqrt{2}. Therefore, the required range is [6-\\sqrt{2},6+\\sqrt{2}]."
  },
  {
    "text": "The equation of the directrix of the parabola $x^{2}=\\frac{1}{4} y$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = y/4)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=-1/16",
    "fact_spans": "[[[0, 24]], [[0, 24]]]",
    "query_spans": "[[[0, 31]]]",
    "process": "This problem examines finding the directrix equation from the equation of a parabola, determining the direction of opening, obtaining the value of $ p $, and then deriving the directrix equation using the formula. For the parabola $ x^{2} = \\frac{1}{4}y $, where $ y \\geqslant 0 $, it is a parabola with the vertex at the origin and opening upwards. $ 2p = \\frac{1}{4} $. Therefore, the directrix equation of the parabola is $ y = -\\frac{1}{16} $."
  },
  {
    "text": "If one focus of the ellipse $x^{2}+m y^{2}=1$ coincides with the focus of the parabola $x^{2}=4 y$, then $m$=?",
    "fact_expressions": "G: Parabola;H: Ellipse;m: Number;Expression(G) = (x^2 = 4*y);Expression(H) = (m*y^2 + x^2 = 1);OneOf(Focus(H)) = Focus(G)",
    "query_expressions": "m",
    "answer_expressions": "1/2",
    "fact_spans": "[[[26, 40]], [[1, 20]], [[47, 50]], [[26, 40]], [[1, 20]], [[1, 45]]]",
    "query_spans": "[[[47, 52]]]",
    "process": "The focus of the parabola $x^{2}=4y$ is $(0,1)$. Given that one focus of the ellipse $x^{2}+my^{2}=1$ coincides with the focus of the parabola $x^{2}=4y$, we obtain $\\sqrt{\\frac{1}{m}-1}=1^{x}$, solving which gives $m=\\frac{1}{2}$."
  },
  {
    "text": "Draw a line through the focus of the parabola $y^{2}=4x$ intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $x_{1}+x_{2}=8$, then $|AB|$=?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;x1: Number;y1: Number;x2: Number;y2: Number;Expression(G) = (y^2 = 4*x);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);PointOnCurve(Focus(G), H);Intersection(H, G) = {A, B};x1 + x2 = 8",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "10",
    "fact_spans": "[[[1, 15], [22, 25]], [[19, 21]], [[26, 43]], [[46, 63]], [[26, 43]], [[26, 43]], [[46, 63]], [[46, 63]], [[1, 15]], [[26, 43]], [[46, 63]], [[0, 21]], [[19, 63]], [[66, 81]]]",
    "query_spans": "[[[83, 92]]]",
    "process": "The parabola $ y^{2} = 4x $, $ p = 2 $, let the focus of the parabola be $ F $, then $ |AB| = |AF| + |BF| = x_{1} + x_{2} + p = 10 $. This question examines the chord length of a parabola, tests knowledge of focal radii, and assesses students' computational and problem-solving abilities, belonging to basic problems."
  },
  {
    "text": "The equation of an ellipse that shares foci with the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{7}=1$ and has a major axis length of $8$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(G) = (x^2/2 - y^2/7 = 1);Focus(G)=Focus(H);Length(MajorAxis(H))=8",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2/16+y^2/7=1",
    "fact_spans": "[[[1, 39]], [[54, 56]], [[1, 39]], [[0, 56]], [[46, 56]]]",
    "query_spans": "[[[54, 60]]]",
    "process": "First, based on the given hyperbola equation, write down its foci coordinates, thereby determining the axis on which the ellipse's foci lie and obtaining c=3. Then, using the length of the major axis of the ellipse, obtain a=4. Using the relationship among a, b, and c in an ellipse, compute b^{2}=7, thus obtaining the ellipse's equation. Since the ellipse shares the same foci (\\pm3,0) with the hyperbola \\frac{x^{2}}{2}-\\frac{y^{2}}{7}=1, it follows that c=3. Given the major axis length is 8, we have 2a=8, so a=4. Therefore, b^{2}=a^{2}-c^{2}=16-9=7. Hence, the equation of the ellipse is \\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{8}+\\frac{y^{2}}{2}=1$ and the circle $M$: $(x+\\sqrt{2})^{2}+y^{2}=r^{2}$ $(0<r<\\sqrt{2})$, two tangents are drawn from the upper vertex $P$ of the ellipse $C$ to the circle $M$, intersecting the ellipse $C$ again at points $A$ and $B$ (distinct from $P$). Then, the product of the slopes of lines $PA$ and $PB$ is equal to?",
    "fact_expressions": "C: Ellipse;M: Circle;r: Number;P: Point;A: Point;B: Point;Expression(C) = (x^2/8 + y^2/2 = 1);0 < r;r < sqrt(2);Expression(M) = (y^2 + (x + sqrt(2))^2 = r^2);UpperVertex(C) = P;l1: Line;l2: Line;TangentOfPoint(P, M) = {l1, l2};Intersection(l1, C) = A;Intersection(l2, C) = B;Negation(A = P);Negation(B = P)",
    "query_expressions": "Slope(LineOf(P, A))*Slope(LineOf(P, B))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 44], [96, 101], [121, 126]], [[45, 94], [109, 113]], [[50, 94]], [[105, 108], [143, 147]], [[129, 132]], [[133, 136]], [[2, 44]], [[50, 94]], [[50, 94]], [[45, 94]], [[96, 108]], [], [], [[95, 118]], [[95, 138]], [[95, 138]], [[129, 148]], [[129, 148]]]",
    "query_spans": "[[[150, 173]]]",
    "process": "Circle M: $(x+\\sqrt{2})^{2}+y^{2}=r^{2}$ $(0<r<\\sqrt{2})$, from the ellipse equation $\\frac{x^{2}}{8}+\\frac{y^{2}}{2}=1$ we obtain its upper vertex $P(0,\\sqrt{2})$. Since $0<r<\\sqrt{2}$, the slopes of the two tangent lines from $P$ to circle $M$ exist. Let the equations of lines $PA$ and $PB$ be $y=kx+\\sqrt{2}$, then the distance from the center $M(-\\sqrt{2},0)$ to the line $kx-y+\\sqrt{2}=0$ equals the radius $r$, that is, $\\frac{|-\\sqrt{2}k+\\sqrt{2}|}{\\sqrt{k^{2}+1}}=r$. Simplifying yields $(2-r^{2})k^{2}-4k+2-r^{2}=0$, therefore $k_{PA}\\cdot k_{PB}=\\frac{2-r^{2}}{2-r^{2}}=1$,"
  },
  {
    "text": "Given that the directrix of the parabola $y = a x^{2}$ ($a \\in \\mathbb{R}$) is $y = -1$, then $a$ = ?",
    "fact_expressions": "G: Parabola;a: Real;Expression(G) = (y = a*x^2);Expression(Directrix(G)) = (y = -1)",
    "query_expressions": "a",
    "answer_expressions": "1/4",
    "fact_spans": "[[[2, 25]], [[5, 25], [40, 43]], [[2, 25]], [[2, 37]]]",
    "query_spans": "[[[40, 45]]]",
    "process": ""
  },
  {
    "text": "It is known that the asymptotes of a hyperbola with foci on the $x$-axis are given by $y = \\pm \\frac{1}{3} x$. What is the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), xAxis);Expression(Asymptote(G)) = (y = pm*(x/3))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)/3",
    "fact_spans": "[[[11, 14], [45, 48]], [[2, 14]], [[11, 42]]]",
    "query_spans": "[[[45, 54]]]",
    "process": "From the given condition, $\\frac{b}{a} = \\frac{1}{3} \\Rightarrow a : b : c = 3 : 1 : \\sqrt{10} \\Rightarrow e = \\frac{\\sqrt{10}}{3}$"
  },
  {
    "text": "The distance from the focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ to its asymptote is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1)",
    "query_expressions": "Distance(Focus(G), Asymptote(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 51]]]",
    "process": "Foci $(\\pm5,0)$, asymptotes $y=\\pm\\frac{4}{3}x$, i.e., $4x-3y=0$, then $d=\\frac{20}{5}=4$"
  },
  {
    "text": "The line $y = kx - 2$ intersects the parabola $y^2 = 8x$ at points $A$ and $B$. If the x-coordinate of the midpoint of $AB$ is $2$, then $|AB|$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);L: Line;k: Number;Expression(L) = (y = k*x - 2);A: Point;B: Point;Intersection(L,G) = {A,B};XCoordinate(MidPoint(LineSegmentOf(A, B))) = 2",
    "query_expressions": "Abs(LineSegmentOf(A,B))",
    "answer_expressions": "2*sqrt(15)",
    "fact_spans": "[[[11, 25]], [[11, 25]], [[0, 10]], [[2, 10]], [[0, 10]], [[26, 29]], [[30, 33]], [[0, 35]], [[37, 51]]]",
    "query_spans": "[[[56, 64]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>1, b>0)$ has a focal distance of $2c$. The line $l$ passes through the points $(a, 0)$ and $(0, b)$, and the sum $s$ of the distance from the point $(1,0)$ to the line $l$ and the distance from the point $(-1,0)$ to the line $l$ satisfies $s \\geq \\frac{4}{5} c$. Then, the range of the eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "l: Line;G: Hyperbola;b: Number;a: Number;c: Number;s: Number;H: Point;I: Point;J: Point;K: Point;e: Number;s>=(4/5)*c;a>1;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(H) = (a, 0);Coordinate(I) = (0, b);Coordinate(J) = (1, 0);Coordinate(K) = (-1, 0);FocalLength(G) = 2*c;PointOnCurve(H,l);PointOnCurve(I,l);Distance(J,l)+Distance(K,l) = s;Eccentricity(G) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "[sqrt(5)/2,sqrt(5)]",
    "fact_spans": "[[[66, 71], [101, 106], [120, 125]], [[0, 56], [154, 157]], [[3, 56]], [[3, 56]], [[60, 65]], [[130, 152]], [[72, 81]], [[82, 90]], [[92, 100]], [[110, 119]], [[160, 163]], [[130, 152]], [[3, 56]], [[3, 56]], [[0, 56]], [[72, 81]], [[82, 90]], [[92, 100]], [[110, 119]], [[0, 65]], [[66, 81]], [[66, 90]], [[92, 152]], [[154, 163]]]",
    "query_spans": "[[[160, 170]]]",
    "process": "Let the equation of line $ l $ be $ \\frac{x}{a} + \\frac{y}{b} = 1 $, that is, $ bx + ay - ab = 0 $. By the point-to-line distance formula, and given $ a > 1 $, the distance from point $ (1,0) $ to the line is $ d_{1} = \\frac{b(a-1)}{\\sqrt{a^{2}+b^{2}}} $, and the distance from point $ (-1,0) $ to the line is $ d_{2} = \\frac{b(a+1)}{\\sqrt{a^{2}+b^{2}}} $. Therefore, $ s = d_{1} + d_{2} = \\frac{2ab}{\\sqrt{a^{2}+b^{2}}} = \\frac{2ab}{c} $. From $ s \\geqslant \\frac{4}{5}c $, we get $ \\frac{2ab}{c} \\geqslant \\frac{4}{5}c $, that is, $ 5a\\sqrt{c^{2}-a^{2}} \\geqslant 2c^{2} $. Since $ e = \\frac{c}{a} $, it follows that $ 5\\sqrt{e^{2}-1} \\geqslant 2e^{2} $, so $ 25(e^{2}-1) \\geqslant 4e^{4} $, that is, $ 4e^{4} - 25e^{2} + 25 \\leqslant 0 $. Therefore, $ \\frac{5}{4} \\leqslant e^{2} \\leqslant 5 $ ($ e > 1 $), so $ \\frac{\\sqrt{5}}{2} \\leqslant e \\leqslant \\sqrt{5} $, that is, the range of $ e $ is $ \\left[ \\frac{\\sqrt{5}}{2}, \\sqrt{5} \\right] $."
  },
  {
    "text": "The line $x-2 y+2=0$ intersects the ellipse $x^{2}+4 y^{2}=4$ at points $A$ and $B$. Then $|A B|=$?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;Expression(G) = (x^2 + 4*y^2 = 4);Expression(H) = (x - 2*y + 2 = 0);Intersection(H, G) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[14, 33]], [[0, 13]], [[36, 39]], [[40, 43]], [[14, 33]], [[0, 13]], [[0, 45]]]",
    "query_spans": "[[[47, 56]]]",
    "process": ""
  },
  {
    "text": "The equation $\\frac{x^{2}}{m-3}+\\frac{y^{2}}{9-m}=1$ represents an ellipse; then the range of values for $m$ is?",
    "fact_expressions": "G: Ellipse;m:Number;Expression(G)=(x^2/(m-3)+y^2/(9-m)=1)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(3,9)&(Negation(m=6))",
    "fact_spans": "[[[43, 45]], [[47, 50]], [[0, 45]]]",
    "query_spans": "[[[47, 57]]]",
    "process": "Because $\\frac{x^{2}}{m-3}$ because $\\frac{m-3}{m-3}+\\frac{y^{2}m}{m-3}=1$ represents an ellipse, so $\\begin{cases}9-m>0\\\\m-3\\neq9-m\\end{cases}\\Rightarrow\\begin{cases}3<m<9\\\\m\\neq6\\end{cases}$ that is, the range of $m$ is $3<m<9$ and $m\\neq6$,"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. The line $l$ passes through point $F_{2}$ and intersects the right branch of the hyperbola at points $P$ and $Q$. If $|P F_{1}|=3|P F_{2}|$, $|P Q|=4|P F_{2} |$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;l: Line;PointOnCurve(F2, l);Intersection(l, RightPart(C)) = {P, Q};P: Point;Q: Point;Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, F2));Abs(LineSegmentOf(P, Q)) = 4*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(10)/2",
    "fact_spans": "[[[2, 66], [105, 108], [166, 172]], [[2, 66]], [[10, 66]], [[10, 66]], [[10, 66]], [[10, 66]], [[74, 81]], [[82, 89], [96, 104]], [[2, 89]], [[2, 89]], [[90, 95]], [[90, 104]], [[90, 120]], [[111, 114]], [[115, 118]], [[122, 144]], [[145, 164]]]",
    "query_spans": "[[[166, 178]]]",
    "process": "Let $|PF_{2}|=m$, then $|PF_{1}|=3m$, $|PQ|=4m$, implying $|QF_{2}|=3m$. By the definition of a hyperbola, and by determining that $\\triangle PQF_{1}$ is a right triangle, we obtain $(3a)^{2}+a^{2}=(2c)^{2}$, and solve for the eccentricity. Let $|PF_{2}|=m$, then $|PF_{1}|=3m$, $|PQ|=4m$, $\\therefore |QF_{2}|=3m$. By the definition of the hyperbola, we have\n\\[\n\\begin{cases}\n|PF_{1}|-|PF_{2}|=2m=2a \\\\\n|QF_{1}|-|QF_{2}|=|QF_{1}|-3m=2a\n\\end{cases}\n\\Rightarrow\n\\begin{cases}\n|QF_{1}|=5a \\\\\nm=a\n\\end{cases}\n\\]\nThen it satisfies $|PF_{1}|^{2}+|PQ|^{2}=|QF_{1}|^{2}$, $\\therefore \\triangle PQF_{1}$ is a right triangle with $\\angle QPF_{1}=90^{\\circ}$, $\\therefore |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2} \\Rightarrow (3a)^{2}+a^{2}=(2c)^{2}$, yielding $e=\\frac{\\sqrt{10}}{2}$."
  },
  {
    "text": "Given the ellipse $C_{1}$: $\\frac{x^{2}}{a_{1}^{2}}+\\frac{y^{2}}{b_{1}^{2}}=1(a_{1}>b_{1}>0)$ and the hyperbola $C_{2}$: $\\frac{x^{2}}{a_{2}^{2}}-\\frac{y^{2}}{b_{2}^{2}}=1(a_{2}>0, b_{2}>0)$ share common foci $F_{1}$, $F_{2}$, let $P$ be an intersection point of $C_{1}$ and $C_{2}$, and let the eccentricities of $C_{1}$ and $C_{2}$ be $e_{1}$, $e_{2}$ respectively. If $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$, then the minimum value of $e_{1} e_{2}$ is?",
    "fact_expressions": "C1: Ellipse;Expression(C1) = (x^2/a1^2+y^2/b1^2=1);a1: Number;b1: Number;a1>b1;b1>0;C2: Hyperbola;Expression(C2) = (x^2/a2^2-y^2/b2^2=1);a2: Number;b2: Number;a2>0;b2>0;F1: Point;F2: Point;Focus(C1) = {F1, F2};Focus(C2) = {F1, F2};P: Point;OneOf(Intersection(C1, C2)) = P;e1: Number;e2: Number;Eccentricity(C1) = e1;Eccentricity(C2) = e2;AngleOf(F1, P, F2) = pi/3",
    "query_expressions": "Min(e1*e2)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[2, 79], [188, 195], [209, 216]], [[2, 79]], [[12, 79]], [[12, 79]], [[12, 79]], [[12, 79]], [[80, 161], [196, 203], [217, 224]], [[80, 161]], [[91, 161]], [[91, 161]], [[91, 161]], [[91, 161]], [[167, 174]], [[175, 182]], [[2, 182]], [[2, 182]], [[184, 187]], [[184, 208]], [[231, 238]], [[240, 247]], [[209, 247]], [[209, 247]], [[249, 285]]]",
    "query_spans": "[[[287, 306]]]",
    "process": ""
  },
  {
    "text": "Given that the vertex of the parabola is at the origin, the focus is on the $y$-axis, and the distance from the point $P(a,-2)$ on the parabola to the focus is $3$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;P: Point;O: Origin;Coordinate(P) = (a,-2);Vertex(G) = O;PointOnCurve(Focus(G), yAxis);PointOnCurve(P, G);Distance(P, Focus(G)) = 3;a:Number",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2=-4*y",
    "fact_spans": "[[[2, 5], [21, 24], [48, 51]], [[26, 36]], [[9, 11]], [[26, 36]], [[2, 11]], [[2, 20]], [[21, 36]], [[21, 46]], [[27, 36]]]",
    "query_spans": "[[[48, 56]]]",
    "process": "Let the equation of the parabola be $x^{2} = -2py$ $(p > 0)$. By the definition of the parabola (or the focal radius formula), $p$ can be found. Let $x^{2} = -2py$ $(p > 0)$. The distance from point $P(a, -2)$ on the parabola to the focus is equal to the distance from point $P$ to the directrix $y = \\frac{p}{2}$, then $\\frac{p}{2} + 2 = 3$, solving gives $p = 2$. Thus, the equation of the parabola is $x^{2} = -4y$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ with focus $F$, and a point $M(2, m)$ on the parabola satisfying $|M F|=6$, then the equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;Expression(C)=(y^2=2*p*x);p:Number;p>0;M: Point;F: Point;Focus(C)=F;m: Number;Coordinate(M) = (2, m);Abs(LineSegmentOf(M,F))=6;PointOnCurve(M,C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=16*x",
    "fact_spans": "[[[2, 28], [36, 39], [64, 70]], [[2, 28]], [[10, 28]], [[10, 28]], [[42, 51]], [[32, 35]], [[2, 35]], [[42, 51]], [[42, 51]], [[53, 62]], [[36, 51]]]",
    "query_spans": "[[[64, 75]]]",
    "process": "Find the equation of the directrix of the parabola, and from the definition of the parabola derive $2+\\frac{p}{2}=6$, solve for $p$, and then obtain the equation of the parabola. [Detailed solution] $\\because$ parabola $C: y^{2}=2px$ $(p>0)$, $\\therefore$ the directrix of the parabola is $x=-\\frac{p}{2}$. Since the distance from a point $M(2,m)$ on the parabola to the focus is 6, by the definition of the parabola, the distance from point $M(2,m)$ to the directrix is also 6, i.e., $2+\\frac{p}{2}=6$, solving gives $p=8$, $\\therefore$ the equation of the parabola is $y^{2}=16x$."
  },
  {
    "text": "Given that one focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{16}=1$ $(a>0)$ is $F(3 , 0)$, then $a$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/16 + x^2/a^2 = 1);a: Number;a>0;F: Point;OneOf(Focus(G)) = F;Coordinate(F) = (3, 0)",
    "query_expressions": "a",
    "answer_expressions": "5",
    "fact_spans": "[[[2, 49]], [[2, 49]], [[66, 69]], [[4, 49]], [[54, 64]], [[2, 64]], [[54, 64]]]",
    "query_spans": "[[[66, 71]]]",
    "process": ""
  },
  {
    "text": "If point $P$ is any point on the parabola $y=x^{2}$, then the minimum distance from point $P$ to the line $y=x-2$ is?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Expression(G) = (y = x^2);Expression(H) = (y = x - 2);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, H))",
    "answer_expressions": "7*sqrt(2)/8",
    "fact_spans": "[[[6, 18]], [[30, 39]], [[1, 5], [25, 29]], [[6, 18]], [[30, 39]], [[1, 23]]]",
    "query_spans": "[[[25, 46]]]",
    "process": "When the distance from point P to the line y = x - 2 is minimized, P is the point of tangency on the parabola y = x² where the tangent line is parallel to the line y = x - 2. ∴ the required minimum distance d = \\frac{|\\frac{1}{2}-\\frac{\\therefore}{4}-2|}{\\sqrt{2}} = \\frac{1}{8}"
  },
  {
    "text": "The eccentricity $e$ of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ satisfies $e \\in[\\frac{\\sqrt{2}}{2}, 1)$; then the range of values for $m$ is?",
    "fact_expressions": "G: Ellipse;m: Number;e:Number;Expression(G) = (x^2/4 + y^2/m = 1);Eccentricity(G)=e;Range(e)=[sqrt(2)/2,1)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(0,2]+[8,+oo)",
    "fact_spans": "[[[0, 37]], [[73, 76]], [[41, 71]], [[0, 37]], [[0, 71]], [[41, 71]]]",
    "query_spans": "[[[73, 83]]]",
    "process": ""
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ are $F_{1}$ and $F_{2}$ respectively. There is a point $P$ on the ellipse such that $\\angle F_{1} P F_{2}=60^{\\circ}$. Then, the area of triangle $F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/16 + y^2/9 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(F1,P,F2))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 38], [62, 64]], [[46, 53]], [[68, 71]], [[54, 61]], [[0, 38]], [[0, 61]], [[0, 61]], [[62, 71]], [[74, 107]]]",
    "query_spans": "[[[109, 132]]]",
    "process": "\\because \\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1,\\therefore a=4,b=3,c=\\sqrt{7}. \\text{Also} \\because P \\text{is a point on the ellipse}, \\angle F_{1}PF_{2}=60^{\\circ}, F_{1},F_{2} \\text{are the left and right foci}, \\therefore |F_{1}P|+|PF_{2}|=2a=8, |F_{1}F_{2}|=2\\sqrt{7}, |F_{1}F_{2}|^{2}=(|PF_{1}|+|PF_{2}|)^{2}-2|F_{1}P|\\cdot|PF_{2}|-2|F_{1}P|\\cdot|PF_{2}|\\cos60^{\\circ}=64-3|F_{1}P|\\cdot|PF_{2}|=28, \\therefore |F_{1}P|\\cdot|PF_{2}|=12. \\therefore S_{\\Delta PF_{1}F_{2}}=\\frac{1}{2}|F_{1}P|\\cdot|PF_{2}|\\sin60^{\\circ}=3\\sqrt{3}. \\text{The answer is: } 3\\sqrt{3}"
  },
  {
    "text": "Given that the line $l$ passing through the focus $F$ of the parabola $C$: $x^{2}=8 y$ intersects $C$ at points $A$ and $B$, if the horizontal coordinate of point $A$ is $2$, then the distance from point $B$ to the directrix of $C$ is?",
    "fact_expressions": "l: Line;C: Parabola;F: Point;A: Point;B: Point;Expression(C) = (x^2 = 8*y);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};XCoordinate(A)=2",
    "query_expressions": "Distance(B, Directrix(C))",
    "answer_expressions": "10",
    "fact_spans": "[[[29, 34]], [[3, 22], [35, 38], [69, 72]], [[25, 28]], [[39, 42], [50, 54]], [[43, 46], [64, 68]], [[3, 22]], [[3, 28]], [[2, 34]], [[29, 48]], [[50, 62]]]",
    "query_spans": "[[[64, 80]]]",
    "process": "From the given, we have F(0,2), and the directrix equation is y = -2. Since the x-coordinate of point A is 2, the y-coordinate of point A is \\frac{1}{2}, so A(2,\\frac{1}{2}). Thus, the slope of line l is k = \\frac{2 - \\frac{1}{2}}{0 - 2} = -\\frac{3}{4}. Therefore, the equation of line l is y = -\\frac{3}{4}x + 2. Solving the system \\begin{cases} y = -\\frac{3}{4}x + 2 \\\\ x^{2} = 8y \\end{cases}, we obtain x^{2} + 6x - 16 = 0, which gives x = 2 or x = -8. Hence, the y-coordinate of point B is 8, so the distance from point B to the directrix of C is 8 + 2 = 10."
  },
  {
    "text": "The distance from a focus to the center of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{m-5}=1$ is $3$, then $m=$?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (-y^2/(m - 5) + x^2/m = 1);Distance(OneOf(Focus(G)),Center(G))=3",
    "query_expressions": "m",
    "answer_expressions": "{7,-2}",
    "fact_spans": "[[[0, 40]], [[58, 61]], [[0, 40]], [[0, 55]]]",
    "query_spans": "[[[58, 63]]]",
    "process": ""
  },
  {
    "text": "Given that the focus of the parabola $y=x^{2}$ is $F$, a line $l$ passing through point $F$ intersects the parabola at points $A$ and $B$. If $|A B|=3$, then the distance from the midpoint of segment $A B$ to the $x$-axis is?",
    "fact_expressions": "l: Line;G: Parabola;B: Point;A: Point;F: Point;Expression(G) = (y = x^2);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, B)) = 3",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A,B)), xAxis)",
    "answer_expressions": "5/4",
    "fact_spans": "[[[28, 33]], [[2, 14], [34, 37]], [[42, 45]], [[38, 41]], [[18, 21], [23, 27]], [[2, 14]], [[2, 21]], [[22, 33]], [[28, 47]], [[49, 58]]]",
    "query_spans": "[[[60, 80]]]",
    "process": "As shown in the figure, F is the focus, E is the midpoint of AB, and the equation of the directrix l of the parabola is $ y = -\\frac{1}{4} $. Perpendiculars are drawn from A, E, and B to l, intersecting l at points L, M, and N respectively. According to the properties of the parabola, $ |FA| = |AL| $, $ |FB| = |BN| $, $ \\therefore |AL| + |BN| = |FA| + |FB| = |AB| = 3 $, $ \\therefore |EM| = \\frac{3}{2} $, $ \\therefore d = |EM| - \\frac{1}{4} = \\frac{3}{2} - \\frac{1}{4} = \\frac{5}{4} $"
  },
  {
    "text": "Given that $F$ is the left focus of the ellipse $C$: $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$, and a line passing through the origin $O$ intersects the ellipse $C$ at points $P$ and $Q$. If $|PF| \\cdot |QF| = 9$, then $|PQ| =$?",
    "fact_expressions": "C: Ellipse;G: Line;P: Point;F: Point;Q: Point;O: Origin;Expression(C) = (x^2/16 + y^2/7 = 1);LeftFocus(C) = F;PointOnCurve(O, G);Intersection(G, C) = {P, Q};Abs(LineSegmentOf(P, F))*Abs(LineSegmentOf(Q, F)) = 9",
    "query_expressions": "Abs(LineSegmentOf(P, Q))",
    "answer_expressions": "2*sqrt(14)",
    "fact_spans": "[[[6, 48], [63, 68]], [[60, 62]], [[69, 72]], [[2, 5]], [[73, 76]], [[54, 59]], [[6, 48]], [[2, 52]], [[53, 62]], [[60, 78]], [[80, 102]]]",
    "query_spans": "[[[104, 112]]]",
    "process": ""
  },
  {
    "text": "$P(x_{0}, y_{0})(x_{0} \\neq \\pm a)$ is a point on the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, $M$ and $N$ are the left and right vertices of the ellipse $E$, respectively. The product of the slopes of lines $PM$ and $PN$ is $-\\frac{1}{5}$. Then the eccentricity of the ellipse is?",
    "fact_expressions": "P: Point;Coordinate(P) = (x0, y0);x0: Number;y0: Number;Negation(x0=pm*a);E: Ellipse;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>b;b>0;PointOnCurve(P,E) = True;M: Point;N: Point;LeftVertex(E) = M;RightVertex(E) = N;Slope(LineOf(P,M)) * Slope(LineOf(P,N)) = -1/5",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "2*sqrt(5)/5",
    "fact_spans": "[[[0, 35]], [[0, 35]], [[0, 35]], [[0, 35]], [[0, 35]], [[36, 93], [107, 112], [154, 156]], [[36, 93]], [[43, 93]], [[43, 93]], [[43, 93]], [[43, 93]], [[0, 96]], [[97, 100]], [[101, 104]], [[97, 118]], [[97, 118]], [[119, 152]]]",
    "query_spans": "[[[154, 162]]]",
    "process": "According to the problem, M(-a,0), N(a,0), \\frac{x^{2}}{a^{2}}+\\frac{y_{0}^{2}}{b^{2}}=1\\Rightarrow y_{0}^{2}=(a^{2}-x_{0}^{2})\\cdot\\frac{b^{2}}{a^{2}}, k_{PM}\\cdot k_{PN}=\\frac{y_{0}}{x_{0}+a}\\cdot\\frac{y_{0}}{x_{0}-a}=\\frac{y_{0}^{2}}{x_{0}^{2}-a^{2}}=-\\frac{b^{2}}{a^{2}}=-\\frac{1}{5}, \\frac{b^{2}}{a^{2}}=\\frac{1}{5}, e=\\frac{c}{a}=\\sqrt{1-\\frac{b^{2}}{a^{2}}}=\\frac{2\\sqrt{5}}{5}"
  },
  {
    "text": "The coordinates of the foci of the ellipse $a x^{2}+b y^{2}+a b=0(a<b<0)$ are?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;a < b;b < 0;Expression(G) = (a*b + a*x^2 + b*y^2 = 0)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, pm*(\\sqrt{b-a}))",
    "fact_spans": "[[[0, 32]], [[2, 32]], [[2, 32]], [[2, 32]], [[2, 32]], [[0, 32]]]",
    "query_spans": "[[[0, 39]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $\\frac{y^{2}}{2}-x^{2}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2 + y^2/2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "sqrt(2)*x+pm*y=0",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 36]]]",
    "process": "According to the definition of the asymptote equations of a hyperbola, solve as follows. [Detailed Solution] \\because the equation of the hyperbola is \\frac{y^{2}}{2}-x^{2}=1, the asymptote equations of the hyperbola are \\sqrt{2}x\\pm y=0"
  },
  {
    "text": "On the ellipse $\\frac{y^{2}}{4}+x^{2}=1$, there are two moving points $P$, $Q$, and a fixed point $E(0,1)$, such that $EP \\perp EQ$. Then the minimum value of $\\overrightarrow{EP} \\cdot \\overrightarrow{QP}$ is?",
    "fact_expressions": "G: Ellipse;E: Point;P: Point;Q: Point;Expression(G) = (x^2 + y^2/4 = 1);PointOnCurve(P,G);PointOnCurve(Q,G);Coordinate(E)=(0, 1);IsPerpendicular(LineSegmentOf(E, P), LineSegmentOf(E, Q))",
    "query_expressions": "Min(DotProduct(VectorOf(E, P), VectorOf(Q, P)))",
    "answer_expressions": "2/3",
    "fact_spans": "[[[1, 28]], [[45, 54]], [[34, 37]], [[40, 43]], [[1, 28]], [[1, 43]], [[1, 43]], [[45, 54]], [[58, 73]]]",
    "query_spans": "[[[75, 130]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+y^{2}=1(a>0)$ with foci $F_{1}$, $F_{2}$, and the parabola $y^{2}=2 p x$ with focus $F$, if $\\overrightarrow{F_{1} F}=3 \\overrightarrow{F F_{2}}$, and $z \\geq a^{2}-p^{2}$ always holds, then the range of values for $z$ is?",
    "fact_expressions": "H: Ellipse;Expression(H) = (y^2 + x^2/a^2 = 1);a: Number;a>0;F1: Point;F2: Point;G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;F: Point;Focus(G) = F;VectorOf(F1, F) = 3*VectorOf(F, F2);z: Number;z>=a^2-p^2;Focus(H)={F1,F2}",
    "query_expressions": "Range(z)",
    "answer_expressions": "[1, +\\infty)",
    "fact_spans": "[[[2, 38]], [[2, 38]], [[4, 38]], [[4, 38]], [[41, 48]], [[49, 56]], [[57, 73]], [[57, 73]], [[60, 73]], [[77, 80]], [[57, 80]], [[82, 135]], [[162, 165]], [[137, 157]], [2, 54]]",
    "query_spans": "[[[162, 172]]]",
    "process": "By the given condition $\\overrightarrow{F_{1}F}=3\\overrightarrow{FF_{2}}$, the points $F_{1}$, $F$, and $F_{2}$ are collinear, meaning the ellipse's foci lie on the x-axis. Since $a>1$, the foci of the ellipse are $F_{1}(-\\sqrt{a^{2}-1},0)$, $F_{2}(\\sqrt{a^{2}-1},0)$, and the focus of the parabola is $F(\\frac{p}{2},0)$. Expressing $\\overrightarrow{F_{1}F}=3\\overrightarrow{FF_{2}}$ in coordinates gives $(\\frac{p}{2}+\\sqrt{a^{2}-1},0)=3(\\sqrt{a^{2}-1}-\\frac{p}{2},0)$, leading to $p=\\sqrt{a^{2}-1}$, hence $a^{2}-p^{2}=1$. Thus $z\\geqslant a^{2}-p^{2}=1$, i.e., the range of $z$ is $[1,+\\infty)$."
  },
  {
    "text": "Let a line $ l $ with slope $ \\frac{\\sqrt{2}}{2} $ intersect the hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) at two distinct points $ P $ and $ Q $. If the projections of points $ P $ and $ Q $ onto the $ x $-axis are exactly the two foci of the hyperbola, then the eccentricity of the hyperbola is?",
    "fact_expressions": "l: Line;G: Hyperbola;b: Number;a: Number;P: Point;Q: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Slope(l)=sqrt(2)/2;Intersection(l, G) = {P, Q};F1:Point;F2:Point;Focus(G)={F1,F2};Projection(P,xAxis)=F1;Projection(Q,xAxis)=F2;Negation(P=Q)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[25, 30]], [[31, 87], [123, 126], [134, 137]], [[34, 87]], [[34, 87]], [[94, 97], [103, 107]], [[98, 101], [108, 111]], [[34, 87]], [[34, 87]], [[31, 87]], [[0, 30]], [[25, 101]], [], [], [[123, 131]], [[103, 131]], [[103, 131]], [86, 97]]",
    "query_spans": "[[[134, 143]]]",
    "process": "Substitute $ x = c $ into the hyperbola equation $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0 $, $ b > 0 $), we get: $ y = \\pm \\frac{b^{2}}{a} $. Therefore, the coordinates of points $ A $ and $ B $ are respectively: $ (c, \\frac{b^{2}}{a}) $, $ (-c, -\\frac{b^{2}}{a}) $. From the given condition:  \n$ \\frac{b^{2}}{ac} = \\frac{\\sqrt{2}}{2} $,  \n$ c^{2} - a^{2} = \\frac{\\sqrt{2}}{2}ac \\Rightarrow e^{2} - \\frac{\\sqrt{2}}{2}e - 1 = 0 $.  \nSolving this equation gives: $ e = \\sqrt{2} $ or $ e = -\\frac{\\sqrt{2}}{2} $ (discarded).  \nTherefore, the answer is: $ \\sqrt{2} $."
  },
  {
    "text": "If the distance from point $A(x_{0}, 2)$ on the parabola $y^{2}=2 p x(p>0)$ to the focus is twice the distance from point $A$ to the $y$-axis, then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;p>0;Expression(G) = (y^2 = 2*p*x);Coordinate(A) = (x0, 2);x0:Number;PointOnCurve(A,G);Distance(A,Focus(G))=2*Distance(A,yAxis)",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 22]], [[63, 66]], [[24, 38], [45, 49]], [[4, 22]], [[1, 22]], [[24, 38]], [[25, 38]], [[1, 38]], [[1, 61]]]",
    "query_spans": "[[[63, 68]]]",
    "process": "The equation of the directrix of the parabola is $x = -\\frac{p}{2}$. From the properties of the parabola, we have $x_{0} + \\frac{p}{2} = 2x_{0}$, so $x_{0} = \\frac{p}{2}$ $\\textcircled{1}$. Since point $A(x_{0}, 2)$ lies on the parabola $y^{2} = 2px$ $(p > 0)$, we have $4 = 2px_{0}$ $\\textcircled{2}$. From $\\textcircled{1}$ and $\\textcircled{2}$, we obtain $p = 2$."
  },
  {
    "text": "The coordinates of the left focus of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1)",
    "query_expressions": "Coordinate(LeftFocus(G))",
    "answer_expressions": "(-2, 0)",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 36]]]",
    "process": "The hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ gives $a=\\sqrt{3}$, $b=1$, then $c=2$, so the coordinates of the left focus of the hyperbola are $(-2,0)$."
  },
  {
    "text": "Given that the line $x - y - 1 = 0$ is tangent to the parabola $y = a x^{2}$, then $a = $?",
    "fact_expressions": "H: Line;Expression(H) = (x - y - 1 = 0);G: Parabola;Expression(G) = (y = a*x^2);a: Number;IsTangent(H, G)",
    "query_expressions": "a",
    "answer_expressions": "1/4",
    "fact_spans": "[[[2, 13]], [[2, 13]], [[14, 28]], [[14, 28]], [[32, 35]], [[2, 30]]]",
    "query_spans": "[[[32, 37]]]",
    "process": "The line x - y - 1 = 0 is tangent to the parabola y = ax^2, with the point of tangency (x_{0}, y_{0}). Then we have {y_{0} = ax_{0}^{2}}, yielding a = \\frac{1}{4}."
  },
  {
    "text": "A hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has an asymptote parallel to the line $x+2 y-1=0$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x + 2*y - 1 = 0);IsParallel(OneOf(Asymptote(G)), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[0, 56], [80, 83]], [[3, 56]], [[3, 56]], [[63, 76]], [[3, 56]], [[3, 56]], [[0, 56]], [[63, 76]], [[0, 78]]]",
    "query_spans": "[[[80, 89]]]",
    "process": "According to the problem, the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ are given by $y=\\pm\\frac{b}{a}x$. Since one asymptote of the hyperbola is parallel to the line $x+2y-1=0$, we have $-\\frac{b}{a}=-\\frac{1}{2}$, that is, $\\frac{b}{a}=\\frac{1}{2}$. Then $e=\\frac{c}{a}=\\sqrt{1+(\\frac{b}{a})^{2}}=\\frac{\\sqrt{5}}{2}$."
  },
  {
    "text": "If the equation of an ellipse is $\\frac{x^{2}}{10-a}+\\frac{y^{2}}{a-2}=1$, and the focal distance of this ellipse is $4$, then the real number $a$=?",
    "fact_expressions": "G: Ellipse;a: Real;Expression(G)=(x^2/(10-a)+y^2/(a-2)=1);FocalLength(G) = 4",
    "query_expressions": "a",
    "answer_expressions": "{4, 8}",
    "fact_spans": "[[[1, 3], [50, 52]], [[61, 66]], [[1, 47]], [[50, 59]]]",
    "query_spans": "[[[61, 68]]]",
    "process": "Since $\\frac{x^2}{10-a}+\\frac{y^{2}}{a-2}=1$ is the equation of an ellipse, it follows that $10-a>0$ and $a-2>0$, so $2<a<10$. From the equation of the ellipse, we obtain $c^{2}=|10-a-(a-2)|=|12-2a|$. Given $2c=4$, we have $|12-2a|=4$, solving which yields $a=4$ or $a=8$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with right focus $F$, if $M F \\perp x$-axis, the midpoint of $M F$ is $P$, and points $A$, $B$ are the vertices of the hyperbola, when $\\angle A P B$ is maximized, point $M$ lies exactly on the hyperbola. Then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(G) = F;M: Point;IsPerpendicular(LineSegmentOf(M, F), xAxis);P: Point;MidPoint(LineSegmentOf(M, F)) = P;A: Point;B: Point;Vertex(G) = {A, B};WhenMax(AngleOf(A, P, B));PointOnCurve(M, G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 58], [105, 108], [137, 140], [144, 147]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[63, 66]], [[2, 66]], [[130, 134]], [[68, 82]], [[92, 95]], [[83, 95]], [[96, 100]], [[101, 104]], [[96, 110]], [[111, 129]], [[130, 141]]]",
    "query_spans": "[[[144, 153]]]",
    "process": "Let P(c, y_{0}), y_{0} > 0, A be the left vertex, B be the right vertex. Therefore, when \\angle APB is maximum, \\tan\\angle APB is maximum. Since \\angle APB = \\angle APF - \\angle BPF, \\tan\\angle APF = \\frac{c+a}{y_{0}}, \\tan\\angle BPF = \\frac{c-a}{y_{0}}, \\therefore \\tan\\angle APB = \\frac{\\tan\\angle APF - \\tan\\angle BPF}{1 + \\tan\\angle APF \\cdot \\tan\\angle BPF} \\frac{b^{2}}{y_{0}} = \\frac{a}{b}, with equality if and only if y_{0} = \\frac{b^{2}}{y_{0}}, i.e., y_{0} = b. \\therefore At this time, the coordinates of point P are (c, b), and the coordinates of point M are (c, 2b). Substituting the coordinates of point M into the hyperbola equation, we get \\frac{c^{2}}{a^{2}} - \\frac{4b^{2}}{b^{2}} = 1, yielding \\frac{c}{a} = \\sqrt{5}. \\therefore When \\angle APB is maximum, the eccentricity of the hyperbola is \\sqrt{5}."
  },
  {
    "text": "Draw two tangents from a focus of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) to the circle $ x^{2} + y^{2} = a^{2} $, with points of tangency $ A $ and $ B $, respectively. If $ \\angle AOB = 120^\\circ $ (where $ O $ is the origin), then the eccentricity of hyperbola $ C $ is?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;G: Circle;A: Point;O: Origin;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = a^2);l1:Line;l2:Line;PointOnCurve(OneOf(Focus(C)),l1);PointOnCurve(OneOf(Focus(C)),l2);TangentOfPoint(OneOf(Focus(C)), G) = {l1, l2} ;TangentPoint(l1,G)=A;TangentPoint(l2,G)=B;AngleOf(A, O, B) = ApplyUnit(120,degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 62], [143, 149]], [[8, 62]], [[8, 62]], [[68, 90]], [[101, 104]], [[132, 135]], [[105, 108]], [[8, 62]], [[8, 62]], [[1, 62]], [[68, 90]], [], [], [[0, 95]], [[0, 95]], [[0, 95]], [[0, 95]], [[0, 108]], [[110, 131]]]",
    "query_spans": "[[[143, 155]]]",
    "process": ""
  },
  {
    "text": "The line passing through a focus $F_1$ of the ellipse $\\frac{x^{2}}{3}+y^{2}=1$ intersects the ellipse at points $A$ and $B$. What is the perimeter of $\\Delta A B F_{2}$ formed by $A$, $B$, and the other focus $F_2$ of the ellipse?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2/3 + y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(F1, H);Intersection(H, G) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A,B,F2))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[1, 28], [44, 46], [67, 69]], [[41, 43]], [[48, 51], [59, 62]], [[52, 55], [63, 66]], [[74, 81]], [[33, 40]], [[1, 28]], [[1, 81]], [[0, 43]], [[41, 57]]]",
    "query_spans": "[[[84, 107]]]",
    "process": "\\frac{x^{2}}{3}+y^{2}=1\\thereforea^{2}=3\\thereforea=\\sqrt{3}\\therefore4a=4\\sqrt{3}, by the definition of ellipse, the perimeter of \\triangleABF_{2} is 4a=4\\sqrt{3}"
  },
  {
    "text": "The focal distance of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{4}=1$ is equal to $2$, then the value of $m$ is?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (y^2/4 + x^2/m = 1);FocalLength(G) = 2",
    "query_expressions": "m",
    "answer_expressions": "{5,3}",
    "fact_spans": "[[[0, 37]], [[47, 50]], [[0, 37]], [[0, 45]]]",
    "query_spans": "[[[47, 54]]]",
    "process": ""
  },
  {
    "text": "Given circles $C_{1}$: $(x-2)^{2}+y^{2}=4$, $C_{2}$: $(x+r)^{2}+y^{2}=r^{2}$ $(r>0)$, if circle $C_{3}$ is externally tangent to both $C_{1}$ and $C_{2}$, and the eccentricity of the trajectory $\\Omega$ of the center of circle $C_{3}$ is $3$, then the set of values of $r$ is?",
    "fact_expressions": "C1: Circle;C2: Circle;C3: Circle;Omega: Curve;r: Number;r > 0;Expression(C1) = (y^2 + (x - 2)^2 = 4);Expression(C2) = ((x + r)^2 + y^2 = r^2);IsOutTangent(C3, C1);IsOutTangent(C3, C2);Locus(Center(C3))=Omega;Eccentricity(Omega) = 3",
    "query_expressions": "Range(r)",
    "answer_expressions": "{1, 4}",
    "fact_spans": "[[[2, 30], [80, 87]], [[33, 69], [88, 95]], [[71, 79], [100, 108]], [[114, 122]], [[132, 135]], [[33, 69]], [[2, 30]], [[33, 69]], [[71, 98]], [[71, 98]], [[100, 122]], [[114, 130]]]",
    "query_spans": "[[[132, 142]]]",
    "process": "From the positional relationship between circles, we have |C_{1}C_{3}|=2+R, |C_{2}C_{3}|=r+R. By classifying according to 0<r<2, r=2, r>2, and combining with the properties of hyperbolas, equations can be established to obtain the solution. According to the problem, circle C_{1} has center C_{1}(2,0) and radius 2, circle C_{2} has center C_{2}(-r,0) and radius r, and |C_{1}C_{2}|=2+r. Let the radius of circle C_{3} be R. Since circle C_{3} is externally tangent to both C_{1} and C_{2}, it follows that |C_{1}C_{3}|=2+R, |C_{2}C_{3}|=r+R. When 0<r<2, 0<|C_{1}C_{3}|-|C_{2}C_{3}|=2-r<2+r; at this time, the locus of the center of circle C_{3} is part of a hyperbola, so the eccentricity e of the locus of the center of circle C_{3} is e=\\frac{|C_{1}C_{2}|}{|C_{1}C_{3}|-|C_{2}C_{3}|}=\\frac{2+r}{2-r}=3, solving gives r=1. When r=2, |C_{1}C_{3}|-|C_{2}C_{3}|=2-r=0, which does not satisfy the condition; when r>2, 0<|C_{2}C_{3}|-|C_{1}C_{3}|=r-2<2+r; at this time, the locus of the center of circle C_{3} is part of a hyperbola, so the eccentricity e of the locus of the center of circle C_{3} is e=\\frac{|C_{1}C_{2}|}{|C_{2}C_{3}|-|C_{1}C_{3}|}=\\frac{2+r}{r-2}=3, solving gives r=4. Therefore, the set of values for r is {1,4}."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a, b>0)$, a line $l$ passing through $F_{1}$ is tangent to the circle $x^{2}+y^{2}=b^{2}$ at point $M$, and $|M F_{2}|=2|M F_{1}|$. What is the slope of line $l$?",
    "fact_expressions": "l: Line;G: Hyperbola;b: Number;a: Number;H: Circle;M: Point;F2: Point;F1: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + y^2 = b^2);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, l);TangentPoint(l, H) = M;Abs(LineSegmentOf(M, F2)) = 2*Abs(LineSegmentOf(M, F1))",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*sqrt(3)/2",
    "fact_spans": "[[[87, 92], [146, 151]], [[18, 72]], [[21, 72]], [[21, 72]], [[93, 113]], [[116, 120]], [[10, 17]], [[2, 9], [79, 86]], [[21, 72]], [[21, 72]], [[18, 72]], [[93, 113]], [[2, 77]], [[2, 77]], [[78, 92]], [[87, 120]], [[122, 144]]]",
    "query_spans": "[[[146, 156]]]",
    "process": "Let $F_{1}(-c,0)$, $F_{2}(c,0)$. Let the slope of line $l$ be $k$, then the equation of line $l$ is $y=k(x+c)$. Since the line $l$ passing through $F_{1}$ is tangent to the circle $x^{2}+y^{2}=b^{2}$, we have $\\frac{|kc|}{\\sqrt{1+k^{2}}}=b$. Squaring both sides gives $b^{2}(1+k^{2})=k^{2}c^{2}$, $\\textcircled{1}$. In the right triangle $OMF_{1}$, we obtain $|MF_{1}|=\\sqrt{c^{2}-b^{2}}=a$, so $|MF_{2}|=2|MF_{1}|=2a$. Since $OM$ is the median of triangle $MF_{1}F_{2}$, we have $(2|OM|)^{2}+(|F_{1}F_{2}|)^{2}=2(|MF_{1}|^{2}+|MF_{2}|^{2})$, which becomes $4b^{2}+4c^{2}=2(a^{2}+4a^{2})$, thus $10a^{2}=10(c^{2}-b^{2})=4b^{2}+4c^{2}$, leading to $3c^{2}=7b^{2}$. Substituting into $\\textcircled{1}$ yields $1+k^{2}=\\frac{7}{3}k^{2}$, solving gives $k=\\pm\\frac{\\sqrt{3}}{2}$. Using the condition for tangency between a line and a circle: $d=r$, as well as the Pythagorean theorem and the median length formula in plane geometry, this problem examines simplification and computational skills, classified as a medium-difficulty problem."
  },
  {
    "text": "Given that the distance from a moving point $P$ to the fixed point $(1,0)$ is equal to its distance to the fixed line $l$: $x=-1$, then the equation of the trajectory of point $P$ is?",
    "fact_expressions": "P: Point;G: Point;Coordinate(G) = (1, 0);l: Line;Expression(l) = (x = -1);Distance(P, G) = Distance(P, l)",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[4, 7], [22, 23], [43, 47]], [[10, 17]], [[10, 17]], [[27, 38]], [[27, 38]], [[4, 41]]]",
    "query_spans": "[[[43, 54]]]",
    "process": "Since the distance from the moving point P to the fixed point (1,0) is equal to its distance to the fixed line $ l: x = -1 $, by the definition of a parabola, the trajectory of point P is a parabola with focus (1,0) and directrix $ l: x = -1 $. Let the equation of the parabola be $ y^{2} = 2px $ ($ p > 0 $), then $ p = 2 $, so the required trajectory equation is $ y^{2} = 4x $."
  },
  {
    "text": "Given that $P$ is a point on the ellipse $4 x^{2}+y^{2}=4$, and $O$ is the origin, then the range of $|O P|$ is?",
    "fact_expressions": "G: Ellipse;O: Origin;P: Point;Expression(G) = (4*x^2 + y^2 = 4);PointOnCurve(P, G)",
    "query_expressions": "Range(Abs(LineSegmentOf(O, P)))",
    "answer_expressions": "[1,2]",
    "fact_spans": "[[[6, 25]], [[29, 32]], [[2, 5]], [[6, 25]], [[2, 28]]]",
    "query_spans": "[[[37, 51]]]",
    "process": "\\because the equation of the ellipse is 4x^{2}+y^{2}=4 \\therefore the standard form of the ellipse equation is x^{2}+\\frac{y^{2}}{4}=1 \\therefore a^{2}=4, b^{2}=1 \\because P is a point on the ellipse, O is the origin \\therefore the range of |OP| is [1,2]"
  },
  {
    "text": "Given that the foci of the ellipse $x^{2} \\sin \\alpha - y^{2} \\cos \\alpha = 1$ $(0 < \\alpha < 2\\pi)$ lie on the $y$-axis, what is the range of values for $\\alpha$?",
    "fact_expressions": "G: Ellipse;alpha: Number;0<alpha;alpha<2*pi;Expression(G) = (x^2*Sin(alpha) - y^2*Cos(alpha) = 1);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(alpha)",
    "answer_expressions": "(pi/2,3*pi/4)",
    "fact_spans": "[[[2, 59]], [[70, 78]], [[4, 59]], [[4, 59]], [[2, 59]], [[2, 68]]]",
    "query_spans": "[[[70, 85]]]",
    "process": "The ellipse $x^{2}\\sin\\alpha - y^{2}\\cos\\alpha = 1$ $(0 < \\alpha < 2\\pi)$ is converted into the standard form $\\frac{x^{2}}{\\sin\\alpha} + \\frac{y^{2}}{-\\cos\\alpha} = 1$. Since its foci lie on the $y$-axis, $0 < -\\cos\\alpha < \\sin\\alpha$, and $\\sin\\alpha + \\cos\\alpha = \\sqrt{2}\\sin(\\alpha + \\frac{\\pi}{4}) > 0$, so $0 < \\alpha < \\frac{3\\pi}{4}$. Combining these, we obtain $\\frac{\\pi}{2} < \\alpha < \\frac{3\\pi}{4}$, therefore $\\alpha \\in (\\frac{\\pi}{2}, \\frac{3\\pi}{4})$."
  },
  {
    "text": "Given that point $M$ is a moving point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$, draw $M D \\perp x$-axis, with foot of perpendicular at $D$. Point $P$ lies on segment $M D$ such that $\\frac{|D M|}{|D P|}=\\frac{3}{2}$. When point $M$ moves, what is the trajectory equation of point $P$?",
    "fact_expressions": "M: Point;PointOnCurve(M, G);G: Ellipse;Expression(G) = (x^2/4 + y^2/9 = 1);IsPerpendicular(LineSegmentOf(M, D), xAxis);D: Point;FootPoint(LineSegmentOf(M, D), xAxis) = D;P: Point;PointOnCurve(P, LineSegmentOf(M, D));Abs(LineSegmentOf(D, M))/Abs(LineSegmentOf(D, P)) = 3/2",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2+y^2=4",
    "fact_spans": "[[[2, 6], [124, 128]], [[2, 48]], [[7, 44]], [[7, 44]], [[50, 64]], [[69, 72]], [[50, 72]], [[74, 78], [132, 136]], [[74, 87]], [[89, 122]]]",
    "query_spans": "[[[132, 142]]]",
    "process": "Let $ P(x,y) $, $ M(x_{0},y_{0}) $, $ \\therefore \\frac{x_{0}^{2}}{4} + \\frac{y_{0}^{2}}{9} = 1 $; $ \\because MD \\perp x\\text{-axis} $, $ P $ lies on $ MD $ and $ \\frac{|DM|}{|DP|} = \\frac{3}{2} $, $ \\therefore \\begin{cases} x = x_{0} \\\\ \\frac{y_{0}}{y} = \\frac{3}{2} \\end{cases} $ $ \\Rightarrow \\begin{cases} x_{0} = x \\\\ y_{0} = \\frac{3}{2}y \\end{cases} $ $ \\therefore \\frac{x^{2}}{4} + \\frac{y^{2}}{4} = 1 $, i.e., the trajectory equation of point $ P $ is $ x^{2} + y^{2} = 4 $."
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=2x$, draw a line intersecting the parabola at points $A$ and $B$. If $|AB|=\\frac{25}{12}$ and $|AF|<|BF|$, then $|AF|=$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 2*x);Focus(G) =  F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 25/12;Abs(LineSegmentOf(A, F)) < Abs(LineSegmentOf(B, F))",
    "query_expressions": "Abs(LineSegmentOf(A, F))",
    "answer_expressions": "5/6",
    "fact_spans": "[[[1, 15], [25, 28]], [[22, 24]], [[29, 32]], [[34, 38]], [[18, 21]], [[1, 15]], [[1, 21]], [[0, 24]], [[22, 40]], [[42, 62]], [[64, 75]]]",
    "query_spans": "[[[77, 85]]]",
    "process": ""
  },
  {
    "text": "Given circle $M$: $x^{2}+(y-1)^{2}=1$, circle $N$: $x^{2}+(y+1)^{2}=1$, lines $l_{1}$, $l_{2}$ pass through the centers $M$, $N$ respectively, and $l_{1}$ intersects circle $M$ at points $A$, $B$, $l_{2}$ intersects circle $N$ at points $C$, $D$, $P$ is an arbitrary moving point on the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1$, then the minimum value of $\\overrightarrow{P A} \\cdot \\overrightarrow{P B}+\\overrightarrow{P C} \\cdot \\overrightarrow{P D}$ is?",
    "fact_expressions": "G: Ellipse;M: Circle;N: Circle;l1: Line;l2: Line;P: Point;A: Point;B: Point;C: Point;D: Point;Expression(G) = (x^2/3 + y^2/4 = 1);Expression(M) = (x^2 + (y - 1)^2 = 1);Expression(N) = (x^2 + (y + 1)^2 = 1);PointOnCurve(M1, l1);PointOnCurve(N1, l2);Intersection(l1, M) = {A, B};Intersection(l2, N) = {C, D};PointOnCurve(P, G);M1: Point;N1: Point;Center(M) = M1;Center(N) = N1",
    "query_expressions": "Min(DotProduct(VectorOf(P, A), VectorOf(P, B)) + DotProduct(VectorOf(P, C), VectorOf(P, D)))",
    "answer_expressions": "6",
    "fact_spans": "[[[140, 177]], [[2, 27], [94, 98]], [[28, 53], [121, 125]], [[54, 64], [86, 93]], [[66, 73], [113, 120]], [[136, 139]], [[101, 105]], [[107, 110]], [[128, 131]], [[132, 135]], [[140, 177]], [[2, 27]], [[28, 53]], [[54, 85]], [[54, 85]], [[86, 111]], [[113, 135]], [[136, 184]], [[78, 81]], [[82, 85]], [[2, 85]], [[2, 85]]]",
    "query_spans": "[[[186, 289]]]",
    "process": "It is easy to see that points $M$ and $N$ are the two foci of the ellipse, so $|M| + |PN| = 4$. Meanwhile, from the given figure, $MA + MB = 0$, $MA \\cdot MB = -1$, $NC + ND = 0$, $NC \\cdot ND = -1$, so $\\overset{\\rightarrow}{PA} = \\left(PM + \\frac{\\overset{\\rightarrow}{PC} \\cdot}{P_D} + MA\\right) \\cdot (PM + MB) + (\\overrightarrow{PN} + \\overrightarrow{NC}) \\cdot (\\overrightarrow{PN} + \\overrightarrow{ND}) = \\frac{\\sqrt{m}}{PM^{2}} + \\frac{\\sqrt[m]{A}}{PM} \\cdot (MA + MB) + MA \\cdot MB + PN^{2} + PN \\cdot (NC + ND) + NC \\cdot ND = |PM|^{2} + 0 - 1 + |PN|^{2} + 0 - 1 = |PM|^{2} + |PN|^{2} - 2$. Also, since $\\left(\\frac{x+y}{2}\\right)^{2} \\leqslant \\frac{x^{2} + y^{2}}{2}$, we have $|\\overrightarrow{PM}|^{2} + |\\overrightarrow{PN}|^{2} \\geqslant \\frac{(|PM| + |PN|)^{2}}{2} = 8$, so $|PM|^{2} + |PN|^{2} - 2 \\geqslant \\frac{(|M| + |N|)^{2}}{2} - 2 = 6$. Hence, $\\overrightarrow{PA} \\cdot \\overrightarrow{PB} + \\overrightarrow{PC} \\cdot \\overrightarrow{PD} \\geqslant 6$. Therefore, the minimum value of $\\frac{\\sqrt{1}}{PA} \\cdot \\overrightarrow{PB} + \\overrightarrow{PC} \\cdot \\overrightarrow{PD}$ is $6$."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ and a line $l$ passing through its focus $F$, intersecting the parabola at points $A$ and $B$, such that $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}=-\\frac{3}{4}$, where $O$ is the origin, then the minimum value of $|A F|+4|B F|$ is?",
    "fact_expressions": "G: Parabola;p: Number;l: Line;O: Origin;A: Point;B: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G)=F;PointOnCurve(F,l);Intersection(l,G) = {A, B};DotProduct(VectorOf(O, A), VectorOf(O, B)) = -3/4",
    "query_expressions": "Min(Abs(LineSegmentOf(A, F)) + 4*Abs(LineSegmentOf(B, F)))",
    "answer_expressions": "9/2",
    "fact_spans": "[[[2, 23], [25, 26]], [[5, 23]], [[32, 38]], [[116, 119]], [[40, 43]], [[44, 47]], [[28, 31]], [[5, 23]], [[2, 23]], [[25, 31]], [[24, 38]], [[2, 49]], [[51, 113]]]",
    "query_spans": "[[[126, 146]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, respectively, $P$ be any point on the ellipse, and the coordinates of point $M$ be $(6,4)$. Then the maximum value of $|PM|+|PF_{1}|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;M: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/16 = 1);Coordinate(M) = (6, 4);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, M))+Abs(LineSegmentOf(P,F1)))",
    "answer_expressions": "15",
    "fact_spans": "[[[19, 58], [69, 71]], [[65, 68]], [[76, 80]], [[1, 8]], [[9, 16]], [[19, 58]], [[76, 91]], [[1, 64]], [[1, 64]], [[65, 75]]]",
    "query_spans": "[[[92, 114]]]",
    "process": ""
  },
  {
    "text": "The left focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $F$, and points $A(-a, 0)$, $B(0, b)$, $C(0,-b)$ are its three vertices. The line $CF$ intersects $AB$ at point $D$. If the eccentricity of the ellipse is $e=\\frac{1}{3}$, then $\\tan \\angle BDC$=?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;C: Point;F: Point;A: Point;B: Point;D: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (-a, 0);Coordinate(B) = (0, b);Coordinate(C) = (0, -b);LeftFocus(G) = F;Intersection(LineOf(C,F),LineOf(A, B))=D;Eccentricity(G) = e;e=1/3;e:Number;Vertex(G)={A,B,C}",
    "query_expressions": "Tan(AngleOf(B, D, C))",
    "answer_expressions": "-8*sqrt(2)/5",
    "fact_spans": "[[[0, 52], [95, 96], [122, 124]], [[2, 52]], [[2, 52]], [[83, 92]], [[57, 60]], [[61, 71]], [[73, 82]], [[116, 120]], [[2, 52]], [[2, 52]], [[0, 52]], [[61, 71]], [[73, 82]], [[83, 92]], [[0, 60]], [[101, 120]], [[122, 143]], [[128, 143]], [[128, 143]], [59, 97]]",
    "query_spans": "[[[145, 166]]]",
    "process": "From the graph, it can be seen that: \\angleBDC=\\angleBAO+\\angleCFO. Using the tangent addition formula to express \\tan\\angleBDC, we have \\tan\\angleBAO=\\frac{b}{a}, \\tan\\angleCFO=\\frac{b}{c}. According to the eccentricity, we can find \\frac{b}{a}=\\frac{2\\sqrt{2}}{3}, \\frac{b}{c}=2\\sqrt{2}. Substituting into the tangent formula yields the result. From the graph, it is clear that: \\angleBDC=\\angleBAO+\\angleDFA=\\angleBAO+\\angleCFO, so \\tan\\angleBDC=\\tan(\\angleBAO+\\angleCFO)=\\frac{\\tan\\angleBAO+\\tan\\angleCFO}{1-\\tan\\angleBAO\\tan\\angleCFO}=\\frac{\\frac{b}{a}+\\frac{b}{c}}{1-\\frac{b}{a}\\cdot\\frac{b}{c}}. Since the eccentricity e=\\frac{c}{a}=\\frac{1}{3}, let a=3m, c=m, then b=2\\sqrt{2}m, thus \\frac{b}{a}=\\frac{2\\sqrt{2}}{3}, \\frac{b}{c}=2\\sqrt{2}. Substituting into the above expression gives \\frac{\\frac{2\\sqrt{2}}{3}+2\\sqrt{2}}{1-\\frac{2\\sqrt{2}}{3}\\times2\\sqrt{2}}=-\\frac{8\\sqrt{2}}{5}."
  },
  {
    "text": "Write the standard equation of a hyperbola that has the same asymptotes as the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1$ (different from the original hyperbola)?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/4 = 1);C:Hyperbola;Asymptote(G)=Asymptote(C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/18 - y^2/8 = 1",
    "fact_spans": "[[[5, 43]], [[5, 43]], [[49, 52]], [[4, 52]]]",
    "query_spans": "[[[49, 68]]]",
    "process": "The hyperbola with the same asymptotes as $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1$ can be written as $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=\\lambda$, $(\\lambda\\neq0)$. When $\\lambda=1$, it is identical to the original hyperbola, so we require $\\lambda\\neq1$. Thus, one obtains $\\frac{x^{2}}{9\\lambda}-\\frac{y^{2}}{4\\lambda}=1$ $(\\lambda\\neq0,\\lambda\\neq1)$. Setting $\\lambda=2$ gives the equation satisfying the condition: $\\frac{x^{2}}{18}-\\frac{y^{2}}{8}=1$."
  },
  {
    "text": "Point $P$ is a point on the ellipse $\\frac{y^{2}}{5}+\\frac{x^{2}}{4}=1$, $F_{1}$, $F_{2}$ are the foci, and $\\angle F_{1} P F_{2}=60^{\\circ}$, then the area of $\\triangle F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 + y^2/5 = 1);PointOnCurve(P, G);Focus(G)={F1,F2};AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "4*sqrt(3)/3",
    "fact_spans": "[[[5, 42]], [[47, 54]], [[0, 4]], [[55, 62]], [[5, 42]], [[0, 46]], [[5, 65]], [[67, 100]]]",
    "query_spans": "[[[102, 132]]]",
    "process": ""
  },
  {
    "text": "Given the equation of the hyperbola is $\\frac{x^{2}}{8}-\\frac{y^{2}}{2}=1$, then the equations of the asymptotes of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/8 - y^2/2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(1/2)*x",
    "fact_spans": "[[[2, 5], [46, 49]], [[2, 44]]]",
    "query_spans": "[[[46, 57]]]",
    "process": "From the hyperbola equation, we know: $ a=2\\sqrt{2}, b=\\sqrt{2} \\therefore $ the asymptotes are: $ y=\\pm\\frac{\\sqrt{2}}{2\\sqrt{2}}x=\\pm\\frac{1}{2}x $. The correct answer is: $ y=\\pm\\frac{1}{2}x $"
  },
  {
    "text": "The right focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $F$. One asymptote of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ intersects the ellipse $C$ at points $A$ and $B$, and $A F \\perp B F$. Then, the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "G: Hyperbola;C: Ellipse;a: Number;b: Number;A: Point;F: Point;B: Point;Expression(G) = (x^2 - y^2/3 = 1);a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;Intersection(OneOf(Asymptote(G)), C) = {A, B};IsPerpendicular(LineSegmentOf(A, F), LineSegmentOf(B, F))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3) - 1",
    "fact_spans": "[[[66, 94]], [[0, 57], [101, 106], [136, 141]], [[7, 57]], [[7, 57]], [[108, 111]], [[62, 65]], [[112, 115]], [[66, 94]], [[7, 57]], [[7, 57]], [[0, 57]], [[0, 65]], [[66, 117]], [[119, 134]]]",
    "query_spans": "[[[136, 147]]]",
    "process": "Let one asymptote of the hyperbola $ x^{2} - \\frac{y^{2}}{3} = 1 $ be $ y = \\sqrt{3}x $. Substituting into the ellipse equation, we obtain $ x^{2} = \\frac{a^{2}b^{2}}{b^{2} + 3a^{2}} $, so $ A\\left( \\frac{ab}{\\sqrt{b^{2} + 3a^{2}}}, \\frac{\\sqrt{3}ab}{\\sqrt{b^{2} + 3a^{2}}} \\right) $, $ B\\left( -\\frac{ab}{\\sqrt{b^{2} + 3a^{2}}}, \\frac{\\sqrt{3}ab}{\\sqrt{b^{2} + 3a^{2}}} \\right) $. From $ AF \\perp BF $, we get $ |OF| = \\frac{1}{2}|AB| $, that is, $ c = \\frac{2ab}{\\sqrt{b^{2} + 3a^{2}}} $, $ \\therefore c^{2}(4a^{2} - c^{2}) = 4a^{2}(4a^{2} - c^{2}) $, $ \\therefore c^{4} - 8a^{2}c^{2} + 4a^{4} = 0 $, $ \\therefore e^{4} - 8e^{2} + 4 = 0 $, $ \\therefore e^{2} = 4 - 2\\sqrt{3} $, $ \\therefore e = \\sqrt{3} - 1 $."
  },
  {
    "text": "The equation of the trajectory of the center $C$ of a moving circle tangent to the line $x = -2$ and passing through the point $(2, 0)$ is?",
    "fact_expressions": "G: Circle;H: Line;I: Point;Expression(H) = (x = -2);Coordinate(I) = (2, 0);Center(G)=C;PointOnCurve(I,G);IsTangent(H,G);C:Point",
    "query_expressions": "LocusEquation(C)",
    "answer_expressions": "y^2=8*x",
    "fact_spans": "[[[26, 28]], [[1, 9]], [[15, 25]], [[1, 9]], [[15, 25]], [[26, 33]], [[13, 28]], [[0, 28]], [[30, 33]]]",
    "query_spans": "[[[30, 40]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=4 x$, with focus $F$, $P$ is a point on the parabola, and fixed point $A(6 , 3)$. Then the minimum value of $|PA|+| PF |$ is?",
    "fact_expressions": "G: Parabola;P: Point;A: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (6, 3);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "7",
    "fact_spans": "[[[2, 16], [29, 32], [17, 18]], [[25, 28]], [[38, 48]], [[21, 24]], [[2, 16]], [[38, 48]], [[17, 24]], [[25, 35]]]",
    "query_spans": "[[[50, 69]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $2$, and one of its foci coincides with the focus of the parabola $y^{2}=16x$, then what is the equation of the asymptotes of this hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;Eccentricity(G) = 2;OneOf(Focus(G)) = Focus(H);H: Parabola;Expression(H) = (y^2 = 16*x)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "sqrt(3)*x+pm*y=0",
    "fact_spans": "[[[2, 48], [58, 59], [89, 92]], [[2, 48]], [[5, 48]], [[5, 48]], [[2, 56]], [[58, 85]], [[65, 80]], [[65, 80]]]",
    "query_spans": "[[[89, 100]]]",
    "process": "\\because one focus of the hyperbola coincides with the focus of the parabola y^{2}=16x, \\therefore the focus is (4,0), \\therefore c=4, \\because the eccentricity of the hyperbola is 2, \\therefore \\frac{c}{a}=2, \\therefore a=2, \\because c^{2}=a^{2}+b^{2}, \\therefore b=2\\sqrt{3}, \\therefore the asymptotes of the hyperbola are y=\\pm\\frac{b}{a}x=\\pm\\frac{2\\sqrt{3}}{2}x=\\pm\\sqrt{3}x, \\therefore the equations of the asymptotes of the hyperbola are \\sqrt{3}x\\pm y=0"
  },
  {
    "text": "Given that $P$ is a point on the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and $F_{1}$, $F_{2}$ are the left and right foci of $C$, respectively. If the diameter of the incircle of $\\triangle P F_{1} F_{2}$ is $a$, then the range of the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "P: Point;C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;PointOnCurve(P, C) = True;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;Diameter(InscribedCircle(TriangleOf(P, F1, F2))) = a",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(\\sqrt{5}/2,+\\infty)",
    "fact_spans": "[[[2, 5]], [[6, 67], [90, 93], [138, 144]], [[6, 67]], [[14, 67]], [[133, 136]], [[14, 67]], [[14, 67]], [[2, 71]], [[72, 79]], [[80, 87]], [[72, 98]], [[72, 98]], [[100, 136]]]",
    "query_spans": "[[[138, 155]]]",
    "process": "As shown in the figure, let |PF₁| = m, |PF₂| = n, and let P be a point on the right branch of the hyperbola, with P(s,t). By the second definition of the hyperbola: e = \\frac{|PF|}{d} (where d is the distance from P to the corresponding directrix of the focus F), we obtain m = es + a, n = es - a. From the area of triangle PF₁F₂, we have \\frac{1}{2} \\cdot 2c \\cdot |t| = \\frac{1}{2} \\cdot \\frac{a}{2} \\cdot (m + n + 2c), which simplifies to 2c|t| = a(es + c), that is, 2c|t| = cs + ca. Then the equation a = 2|t| - s has a solution. Without loss of generality, assume t > 0, so 2t - s > 0, giving t = \\frac{1}{2}s + \\frac{1}{2}a. Given the asymptotes are y = \\pm\\frac{b}{a}x, and since P(s,t) lies on the hyperbola, when \\frac{b}{a} > \\frac{1}{2}, the equation a = 2|t| - s has a solution. [This problem mainly examines the application of the definition of a hyperbola and the calculation of its eccentricity. To find the eccentricity (or its range), there are two common methods: \\textcircled{1} Using the definition method, compute the values of a and c, then substitute into the formula e = \\frac{c}{a}; \\textcircled{2} Obtain a homogeneous equation in terms of a, b, c, transform it into a homogeneous equation in a and c, and then convert it into an equation in e, thus obtaining the value (or range) of e. The problem emphasizes reasoning and computational skills and is of medium difficulty.]"
  },
  {
    "text": "Given that the ellipse $4 x^{2}+y^{2}=1$ and the line $y=x+m$ have common points, find the range of real values for $m$.",
    "fact_expressions": "G: Ellipse;H: Line;m: Real;Expression(G) = (4*x^2 + y^2 = 1);Expression(H) = (y = m + x);IsIntersect(G, H)",
    "query_expressions": "Range(m)",
    "answer_expressions": "[-sqrt(5)/2,sqrt(5)/2]",
    "fact_spans": "[[[2, 21]], [[22, 31]], [[37, 42]], [[2, 21]], [[22, 31]], [[2, 35]]]",
    "query_spans": "[[[37, 49]]]",
    "process": "From \\begin{cases}4x^{2}+y^{2}=1\\\\y=x+m\\end{cases}, we obtain $5x^{2}+2mx+m^{2}-1=0$. Since the line and the ellipse have common points, it follows that $\\Delta=4m^{2}-20(m^{2}-1)\\geqslant0$, i.e., $m^{2}\\leqslant\\frac{5}{4}$, solving yields $-\\frac{\\sqrt{5}}{2}\\leqslant m\\leqslant\\frac{\\sqrt{5}}{2}$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$, a circle with center $P(-2,0)$ and radius $5$ intersects the parabola $C$ at points $A$ and $B$. If $|O A|=\\sqrt{17}$ (where point $O$ is the origin), then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;G: Circle;P: Point;O: Origin;A: Point;B: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(P) = (-2, 0);Center(G)=P;Radius(G)=5;Intersection(G, C) = {A, B};Abs(LineSegmentOf(O, A)) = sqrt(17)",
    "query_expressions": "p",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 28], [52, 58]], [[101, 104]], [[50, 51]], [[30, 39]], [[89, 93]], [[60, 63]], [[64, 67]], [[10, 28]], [[2, 28]], [[30, 39]], [[29, 51]], [[43, 51]], [[50, 69]], [[71, 88]]]",
    "query_spans": "[[[101, 106]]]",
    "process": "In \\triangle AOP, use the cosine law to find \\cos\\angle AOP, then obtain \\cos\\angle AOx, and with |OA|=\\sqrt{17}, find the coordinates of point A, substitute into the parabola equation to solve. As shown: in \\triangle AOP, |AP|=5, |OP|=2, |OA|=\\sqrt{17}; by the cosine law, \\cos\\angle AOP=\\frac{OP^{2}+OA^{2}-AP^{2}}{2OP\\cdot OA}=-\\frac{\\sqrt{17}}{17}; thus \\cos\\angle AOx=\\frac{\\sqrt{17}}{17}; hence A(1,4), substitute into the equation y^{2}=2px (p>0). Solve to get p=8."
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x$ coincides with the right focus of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, then the value of $p$ is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;Expression(G) = (x^2 - y^2/3 = 1);Expression(H) = (y^2 = 2*(p*x));Focus(H) = RightFocus(G)",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[21, 49]], [[1, 17]], [[57, 60]], [[21, 49]], [[1, 17]], [[1, 55]]]",
    "query_spans": "[[[57, 64]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the left branch of the hyperbola such that $|P F_{1}|+|P F_{2}|=8$. Then $\\frac{\\sin \\angle P F_{1} F_{2}}{\\sin \\angle P F_{2} F_{1}}$=?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 - y^2/5 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, LeftPart(G));Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 8",
    "query_expressions": "Sin(AngleOf(P, F1, F2))/Sin(AngleOf(P, F2, F1))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 40], [70, 73]], [[65, 69]], [[49, 56]], [[57, 64]], [[2, 40]], [[2, 64]], [[2, 64]], [[65, 78]], [[79, 102]]]",
    "query_spans": "[[[104, 167]]]",
    "process": "According to the hyperbola equation, find $ a $, and by the definition of the hyperbola, we know $ |PF_{2}| - |PF_{1}| = 4 $, thus obtaining $ |PF_{1}| $, $ |PF_{2}| $, and then using the sine law to calculate; [Detailed Solution] Since the hyperbola is $ \\frac{x^{2}}{4} - \\frac{y^{2}}{5} = 1 $, so $ a = 2 $, $ c = 3 $. Since point $ P $ is a point on the left branch of the hyperbola and $ |PF_{1}| + |PF_{2}| = 8 $, therefore $ |PF_{2}| - |PF_{1}| = 4 $, hence $ |PF_{1}| = 2 $, $ |PF_{2}| = 6 $. In $ \\triangle PF_{1}F_{2} $, by the sine law we have $ \\frac{|PF_{1}|}{\\sin\\angle PF_{2}F_{1}} = \\frac{|PF_{2}|}{\\sin\\angle PF_{1}F_{2}} $, so $ \\frac{\\sin\\angle PF_{1}F_{2}}{\\sin\\angle PF_{2}F_{1}} = \\frac{|PF_{2}|}{|PF_{1}|} = 3 $;"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, a line $l$ passing through $M(4, 0)$ intersects $C$ at points $A$ and $B$. Let the areas of $\\Delta OAF$ and $\\Delta OBF$ be $S_{1}$ and $S_{2}$ respectively. Then the minimum value of $S_{1}+2S_{2}$ is?",
    "fact_expressions": "l: Line;C: Parabola;M: Point;O: Origin;A: Point;F: Point;B: Point;S1: Number;S2: Number;Expression(C) = (y^2 = 4*x);Coordinate(M) = (4, 0);Focus(C) = F;PointOnCurve(M, l);Intersection(l, C) = {A, B};Area(TriangleOf(O, A, F)) = S1;Area(TriangleOf(O, B, F)) = S2",
    "query_expressions": "Min(S1 + 2*S2)",
    "answer_expressions": "4*sqrt(2)",
    "fact_spans": "[[[41, 46]], [[2, 21], [47, 50]], [[30, 40]], [[62, 76]], [[51, 54]], [[25, 28]], [[55, 58]], [[99, 106]], [[107, 114]], [[2, 21]], [[30, 40]], [[2, 28]], [[29, 46]], [[41, 60]], [[62, 114]], [[62, 114]]]",
    "query_spans": "[[[116, 137]]]",
    "process": "From the parabola $ C: y^{2} = 4x $, the focus is $ F(1,0) $. Since line $ l $ intersects $ C $ at points $ A $ and $ B $, the slope of line $ l $ is not zero. Thus, let the equation of line $ l $ be $ x = my + 4 $, with $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Solving the system\n\\[\n\\begin{cases}\nx = my + 4 \\\\\ny^{2} = 4x\n\\end{cases}\n\\]\nyields $ y^{2} - 4my - 16 = 0 $. Then $ y_{1} + y_{2} = 4m $, $ y_{1}y_{2} = -16 $. Also,\n$ S_{\\Delta OAF} = S_{1} = \\frac{1}{2} \\cdot OF \\cdot |y_{1}| = \\frac{1}{2}|y_{1}| $,\n$ S_{\\Delta OBF} = S_{2} = \\frac{1}{2} \\cdot OF \\cdot |y_{2}| = \\frac{1}{2}|y_{2}| $. Moreover,\n$ \\frac{1}{2}|y_{1}| + |y_{2}| \\geqslant 2\\sqrt{\\frac{1}{2}|y_{1}| \\cdot |y_{2}|} = \\sqrt{\\frac{1}{2} \\times 16} = 4\\sqrt{2} $,\nwith equality if and only if $ \\frac{1}{2}|y_{1}| = |y_{2}| $, $ |y_{1}| = 4\\sqrt{2} $, $ |y_{2}| = 2\\sqrt{2} $. Therefore, the minimum value of $ S_{1} + 2S_{2} $ is $ 4\\sqrt{2} $."
  },
  {
    "text": "Point $A$ lies on the parabola $C$: $y^{2}=4x$, $F$ is the focus of $C$. The circle with diameter $AF$ intersects the $y$-axis at only one point $M$, and the coordinates of point $M$ are $(0,2)$. Then $|AF|=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);G: Circle;A: Point;F: Point;M: Point;Coordinate(M) = (0, 2);PointOnCurve(A, C);Focus(C) = F;IsDiameter(LineSegmentOf(A, F), G) ;NumIntersection(G, yAxis) = 1;Intersection(G, yAxis) = M",
    "query_expressions": "Abs(LineSegmentOf(A, F))",
    "answer_expressions": "5",
    "fact_spans": "[[[5, 24], [30, 33]], [[5, 24]], [[47, 48]], [[0, 4]], [[26, 29]], [[60, 63], [65, 69]], [[65, 80]], [[0, 25]], [[26, 36]], [[37, 48]], [[47, 60]], [[47, 63]]]",
    "query_spans": "[[[82, 91]]]",
    "process": "The coordinates of point A are $(\\frac{y^{2}}{4}, y)$. From $\\overrightarrow{MF} \\cdot \\overrightarrow{MA} = 0$, and according to the definition of the parabola, we can find $|AF|$. From the equation of the parabola $y^{2} = 4x$, we obtain the focus coordinates $F(1, 0)$ and the directrix equation $x = -1$. Let the coordinates of point A be $(\\frac{y^{2}}{4}, y)$, then $\\overrightarrow{MF} = (1, -2)$, $\\overrightarrow{MA} = (\\frac{y^{2}}{4}, y - 2)$. According to the problem, point M lies on the circle with AF as diameter, $\\therefore MF \\perp MA$, $\\therefore \\overrightarrow{MF} \\cdot \\overrightarrow{MA} = \\frac{y^{2}}{4} - 2(y - 2) = \\frac{y^{2}}{4} - 2y + 4 = 0$, simplifying gives $y^{2} - 8y + 16 = (y - 4)^{2} = 0$, solving yields $y = 4$. From the definition of the parabola, we get $|AF| = \\frac{y^{2}}{4} + 1 = 4 + 1 = 5$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, a line perpendicular to the $x$-axis is drawn through $F_{1}$ intersecting $C$ at points $A$ and $B$. If $|A B|=2|F_{1} F_{2}|$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;A: Point;B: Point;F1: Point;F2: Point;a>0;b>0;H: Line;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1,H);IsPerpendicular(H,xAxis);Intersection(C,H) = {A,B};Abs(LineSegmentOf(A,B)) = 2*Abs(LineSegmentOf(F1,F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)+1",
    "fact_spans": "[[[18, 79], [142, 145], [103, 106]], [[26, 79]], [[25, 79]], [[107, 110]], [[111, 114]], [[2, 9], [87, 94]], [[10, 17]], [[25, 79]], [[25, 79]], [], [[18, 79]], [[2, 85]], [[2, 85]], [[86, 102]], [[86, 102]], [[86, 116]], [[118, 140]]]",
    "query_spans": "[[[142, 149]]]",
    "process": "Let x = -c, substitute into the hyperbola equation to obtain y = \\pm\\frac{b^{2}}{a}, then |AB| = \\frac{2b^{2}}{a}. If |AB| = 2|F_{1}F_{2}|, we get \\frac{2b^{2}}{A} = 4c, i.e., c^{2} - a^{2} = 2ac. From e = \\frac{c}{a}, we obtain e^{2} - 2e - 1 = 0, e > 1, solving gives e = 1 + \\sqrt{2}, hence the answer is 1.5"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{6}+y^{2}=1$ with left and right foci $F_{1}$, $F_{2}$, the line $AB$ passes through point $F_{1}$ and intersects the ellipse at points $A$, $B$. Then the perimeter of $\\Delta A B F_{2}$ is?",
    "fact_expressions": "A: Point;B: Point;G: Ellipse;F2: Point;F1: Point;Expression(G) = (x^2/6 + y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1,LineOf(A, B));Intersection(LineOf(A, B), G) = {A,B}",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "4*sqrt(6)",
    "fact_spans": "[[[71, 74]], [[76, 80]], [[2, 29], [68, 70]], [[43, 50]], [[35, 42], [58, 66]], [[2, 29]], [[2, 50]], [[2, 50]], [[51, 66]], [[51, 82]]]",
    "query_spans": "[[[84, 107]]]",
    "process": ""
  },
  {
    "text": "Given a point $M(1, m)$ on the parabola $y^{2}=2 p x(p>0)$ such that the distance from $M$ to its focus is $5$, then $m=?$",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;M: Point;m: Number;Coordinate(M) = (1, m);PointOnCurve(M, G);Distance(M, Focus(G)) = 5",
    "query_expressions": "m",
    "answer_expressions": "pm*4",
    "fact_spans": "[[[2, 23], [36, 37]], [[2, 23]], [[5, 23]], [[5, 23]], [[26, 35]], [[48, 51]], [[26, 35]], [[2, 35]], [[26, 46]]]",
    "query_spans": "[[[48, 53]]]",
    "process": "Given a point M(1, m) on the parabola \\( y^{2} = 2px \\) (\\( p > 0 \\)) such that the distance from M to its focus is 5, by the definition of the parabola we obtain \\( 1 + \\frac{p}{2} = 5 \\), solving which gives \\( p = 8 \\), so \\( y^{2} = 16x \\). Substituting the point M(1, m) into the equation of the parabola \\( y^{2} = 16x \\), we get \\( m^{2} = 16 \\), solving which yields \\( m = \\pm 4 \\)."
  },
  {
    "text": "The left focus of the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{3}=1$ is $F_{1}$, and point $P$ lies on the ellipse. If the midpoint $M$ of segment $P F_{1}$ lies on the $y$-axis, then what is the ordinate of point $P$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/12 + y^2/3 = 1);F1: Point;LeftFocus(G) = F1;P: Point;PointOnCurve(P, G);M: Point;MidPoint(LineSegmentOf(P, F1)) = M;PointOnCurve(M, yAxis)",
    "query_expressions": "YCoordinate(P)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 38], [56, 58]], [[0, 38]], [[43, 50]], [[0, 50]], [[51, 55], [88, 92]], [[51, 59]], [[76, 79]], [[62, 79]], [[76, 85]]]",
    "query_spans": "[[[88, 98]]]",
    "process": ""
  },
  {
    "text": "Given that the line $l$ passing through the focus $F$ intersects the two asymptotes of the hyperbola $\\Gamma$ at points $A$ and $B$, and intersects the $x$-axis at point $C$, if $O$ is the origin, $O A = O C$, and $B C = 2 A B$, then the eccentricity of $\\Gamma$ is?",
    "fact_expressions": "l: Line;F: Point;PointOnCurve(F, l);Gamma: Hyperbola;A: Point;B: Point;Intersection(l, l1) = A;C: Point;Intersection(l, xAxis) = C;O: Origin;LineSegmentOf(O, A) = LineSegmentOf(O, C);LineSegmentOf(B, C) = 2*LineSegmentOf(A, B);l1: Line;l2: Line;Focus(Gamma) = F;Asymptote(Gamma) = {l1, l2};Intersection(l, l2) = B",
    "query_expressions": "Eccentricity(Gamma)",
    "answer_expressions": "4",
    "fact_spans": "[[[9, 14]], [[5, 8]], [[2, 14]], [[15, 26], [91, 99]], [[34, 37]], [[38, 41]], [[9, 43]], [[51, 55]], [[9, 55]], [[57, 60]], [[66, 75]], [[77, 88]], [], [], [[3, 26]], [[15, 32]], [[9, 43]]]",
    "query_spans": "[[[91, 105]]]",
    "process": "As shown in the figure, let the asymptotes of the hyperbola be given by $ y = \\pm\\frac{b}{a}x $, and the line $ l: x = my + c $, where $ C(c,0) $. Solving the system of equations  \n$$\n\\begin{cases}\nx = my + c \\\\\ny = \\frac{b}{a}x\n\\end{cases}\n$$\nyields $ x = \\frac{ac}{a - bm} $, $ y = \\frac{bc}{a - bm} $, so $ A\\left( \\frac{ac}{a - bm}, \\frac{bc}{a - bm} \\right) $. Similarly, we obtain $ B\\left( \\frac{ac}{a + bm}, -\\frac{bc}{a + bm} \\right) $. Since $ BC = 2AB $, it follows that $ 2x_A = x_B + x_C $, which simplifies to $ 3a = bm $. Also, since $ OA = OC $, we have  \n$$\n\\left( \\frac{ac}{a + bm} \\right)^2 + \\left( \\frac{bc}{a + bm} \\right)^2 = c^2,\n$$\nwhich simplifies to $ a^2 + b^2 = 16a^2 $, i.e., $ a^2 + c^2 - a^2 = 16a^2 $, so $ \\frac{c^2}{a^2} = 16 $. Therefore, the eccentricity of the hyperbola is $ e = \\frac{c}{a} = 4 $."
  },
  {
    "text": "If $A(-3,0)$, $B(0 ,\\sqrt{3})$, $O$ is the coordinate origin, point $C$ lies in the second quadrant, $\\angle {AOC}=60^{\\circ}$, and ${OC}=\\lambda {OA}+{OB}$, then the value of the real number $\\lambda$ is?",
    "fact_expressions": "A: Point;B: Point;O: Origin;C: Point;lambda: Real;Coordinate(A) = (-3, 0);Coordinate(B) = (0, sqrt(3));Quadrant(C)=2;AngleOf(A,O,C) = ApplyUnit(60, degree);LineSegmentOf(O, C) = lambda*LineSegmentOf(O, A) + LineSegmentOf(O, B)",
    "query_expressions": "lambda",
    "answer_expressions": "1/3",
    "fact_spans": "[[[1, 10]], [[11, 27]], [[28, 31]], [[37, 41]], [[101, 112]], [[1, 10]], [[11, 27]], [[37, 47]], [[49, 74]], [[75, 99]]]",
    "query_spans": "[[[101, 116]]]",
    "process": ""
  },
  {
    "text": "Draw two tangents from a focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ to the circle $x^{2}+y^{2}=a^{2}$, with points of tangency $A$ and $B$, respectively. If $\\angle A O B=120^{\\circ}$ ($O$ is the origin), then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;G: Circle;A: Point;O: Origin;B: Point;l1: Line;l2: Line;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = a^2);TangentOfPoint(OneOf(Focus(C)), G) = {l1, l2};TangentPoint(l1, G)= A ;TangentPoint(l2, G)= B ;AngleOf(A, O, B) = ApplyUnit(120, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 64], [149, 155]], [[8, 64]], [[8, 64]], [[70, 90]], [[102, 105]], [[138, 141]], [[106, 109]], [], [], [[8, 64]], [[8, 64]], [[1, 64]], [[70, 90]], [[0, 95]], [[0, 109]], [[0, 109]], [[111, 137]]]",
    "query_spans": "[[[149, 161]]]",
    "process": ""
  },
  {
    "text": "For the equation $\\frac{x^{2}}{m+9}+\\frac{y^{2}}{5-m}=1$ to represent an ellipse with foci on the $x$-axis, what is the range of real values for $m$?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (x^2/(m + 9) + y^2/(5 - m) = 1);PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-2, 5)",
    "fact_spans": "[[[53, 55]], [[57, 62]], [[1, 55]], [[44, 55]]]",
    "query_spans": "[[[57, 69]]]",
    "process": "From the given condition, we have $ m+9 > 5-m > 0 $, thus the range of real number $ m $ can be obtained. Since the equation $ \\frac{x^{2}}{m+9} + \\frac{y^{2}}{5-m} = 1 $ represents an ellipse with foci on the x-axis, it follows that $ m+9 > 5-m > 0 $. Solving this inequality yields $ -2 < m < 5 $. Therefore, the range of real number $ m $ is $ (-2, 5) $."
  },
  {
    "text": "If the line $y = kx - 1$ and the hyperbola $x^2 - y^2 = 4$ have only one common point, then the range of values for $k$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;k: Number;Expression(G) = (x^2 - y^2 = 4);Expression(H) = (y = k*x - 1);NumIntersection(H, G) = 1",
    "query_expressions": "Range(k)",
    "answer_expressions": "{-1, 1, -sqrt(5)/2, sqrt(5)/2}",
    "fact_spans": "[[[13, 31]], [[1, 12]], [[40, 43]], [[13, 31]], [[1, 12]], [[1, 38]]]",
    "query_spans": "[[[40, 50]]]",
    "process": ""
  },
  {
    "text": "The standard equation of the ellipse, whose minor axis endpoints are the major axis endpoints of the ellipse $9x^{2}+4y^{2}=36$, and passing through the point $(-4,1)$, is?",
    "fact_expressions": "G: Ellipse;C:Ellipse;H: Point;Expression(G) = (9*x^2 + 4*y^2 = 36);Coordinate(H) = (-4, 1);PointOnCurve(H, C);Endpoint(MajorAxis(G))=Endpoint(MinorAxis(C))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/18+y^2/9=1",
    "fact_spans": "[[[1, 23]], [[46, 48]], [[36, 45]], [[1, 23]], [[36, 45]], [[35, 48]], [[0, 48]]]",
    "query_spans": "[[[46, 54]]]",
    "process": ""
  },
  {
    "text": "From a focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a perpendicular to its asymptote, with foot at $M$. Extend $FM$ to intersect the $y$-axis at point $E$. If $FM = ME$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F: Point;M: Point;E: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);OneOf(Focus(G)) = F;L: Line;PointOnCurve(F, L);IsPerpendicular(L, Asymptote(G));FootPoint(L, Asymptote(G)) = M;Intersection(OverlappingLine(LineSegmentOf(F, M)), yAxis) = E;LineSegmentOf(F, M) = LineSegmentOf(M, E)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 57], [66, 67], [111, 114]], [[4, 57]], [[4, 57]], [[62, 65]], [[78, 81]], [[95, 99]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 65]], [], [[0, 74]], [[66, 74]], [[66, 81]], [[82, 99]], [[101, 110]]]",
    "query_spans": "[[[111, 120]]]",
    "process": ""
  },
  {
    "text": "It is known that hyperbola $C$ has the same asymptotes as the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ and passes through the point $A(\\sqrt{3},-3)$. Then, the standard equation of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Hyperbola;A: Point;Coordinate(A) = (sqrt(3), -3);Asymptote(C) = Asymptote(G);Expression(G) = (x^2/3 - y^2 = 1);PointOnCurve(A, C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/8 - x^2/24 = 1",
    "fact_spans": "[[[2, 8], [66, 72]], [[9, 37]], [[47, 64]], [[47, 64]], [[2, 44]], [[9, 37]], [[2, 64]]]",
    "query_spans": "[[[66, 79]]]",
    "process": ""
  },
  {
    "text": "$P$ is a point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, $A(1,0)$, $B(1,1)$, find the maximum value of $|P A|+|P B|$.",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;P: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Coordinate(A) = (1, 0);Coordinate(B) = (1, 1);PointOnCurve(P, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B)))",
    "answer_expressions": "4+sqrt(5)",
    "fact_spans": "[[[4, 41]], [[45, 53]], [[54, 62]], [[0, 3]], [[4, 41]], [[45, 53]], [[54, 62]], [[0, 44]]]",
    "query_spans": "[[[64, 83]]]",
    "process": "From $ a^{2}=4 $, $ b^{2}=3 $, the left focus of the ellipse is $ F(-1,0) $, $ |PA|+|PB|=2a-|PF|+|PB|=4+|PB|-|PF|\\leqslant4+|BF|=4+\\sqrt{5} $, so $ |PA|+|PB|<4+\\sqrt{5} $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with eccentricity $e \\in[\\frac{1}{2}, \\frac{\\sqrt{2}}{2}]$, the line $y=-x+1$ intersects the ellipse at points $M$ and $N$, $O$ is the origin, and $\\overrightarrow{O M} \\cdot \\overrightarrow{O N}=0$. Then, the minimum length of the minor axis of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;e: Number;Eccentricity(G) = e;In(e, [1/2, sqrt(2)/2]);H: Line;Expression(H) = (y = 1 - x);Intersection(H, G) = {M, N};M: Point;N: Point;O: Origin;DotProduct(VectorOf(O, M), VectorOf(O, N)) = 0",
    "query_expressions": "Min(Length(MinorAxis(G)))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 56], [113, 115], [189, 191]], [[2, 56]], [[4, 56]], [[4, 56]], [[4, 56]], [[4, 56]], [[60, 100]], [[2, 100]], [[60, 100]], [[102, 112]], [[102, 112]], [[102, 125]], [[116, 119]], [[120, 123]], [[126, 129]], [[136, 187]]]",
    "query_spans": "[[[189, 200]]]",
    "process": "Let M(x, y), N(x₂, y₂), from \\begin{cases} y = -x + 1 \\\\ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 \\end{cases} simplifying yields (a^{2} + b^{2})x^{2} - 2a^{2}x + a^{2} - a^{2}b^{2} = 0. \\frac{x_{1} + x_{2} = \\frac{2a^{2}}{a^{2} + b^{2}}, x_{1}x_{2}}{\\overrightarrow{OM} \\cdot \\overrightarrow{ON} = x_{1}x_{2} + y_{1}y_{2}} = \\frac{a^{2} - a^{2}b^{2}}{a_{1}x_{2} + b^{2}} - x_{2} + 1)) = 2x_{1}x_{2} - (x_{1} + x_{2}) + 1 = 0. Substituting gives a^{2} = \\frac{b^{2}}{2b^{2} - 1}, e^{2} = \\frac{c^{2}}{a^{2}} = 1 - \\frac{b^{2}}{a^{2}} = 2 - 2b^{2} \\in [\\frac{1}{4}, \\frac{1}{2}], \\frac{\\sqrt{3}}{2} \\leqslant b \\leqslant \\frac{\\sqrt{14}}{4}, \\frac{\\frac{y}{2}}{\\sqrt{3}} \\leqslant \\frac{\\sqrt{14}}{2}, so the minimum length of the minor axis of the ellipse is \\sqrt{3}."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{1}$ intersects the two asymptotes of the hyperbola at points $P$ and $Q$. If $P$ is the midpoint of segment $F_{1} Q$, and $Q F_{1} \\perp Q F_{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;Q: Point;F1: Point;F2: Point;P: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(F1,H);L1:Line;L2:Line;Asymptote(G)={L1,L2};Intersection(H,L1)=P;Intersection(H,L2)=Q;MidPoint(LineSegmentOf(F1,Q))=P;IsPerpendicular(LineSegmentOf(Q, F1), LineSegmentOf(Q, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 59], [101, 104], [166, 169]], [[5, 59]], [[5, 59]], [[96, 98]], [[116, 119]], [[68, 75], [88, 95]], [[78, 85]], [[111, 115], [121, 124]], [[5, 59]], [[5, 59]], [[2, 59]], [[2, 85]], [[2, 85]], [[87, 98]], [], [], [[101, 110]], [[96, 119]], [[96, 119]], [[121, 139]], [[141, 164]]]",
    "query_spans": "[[[166, 175]]]",
    "process": "From the figure and symmetry, OP is the perpendicular bisector of segment F_{1}P, and OQ is the median to the hypotenuse of right triangle F_{1}QF_{2}, so \\angle F_{1}OP = \\angle POQ = \\angle QOF_{2} = 60^{\\circ}, therefore e = \\sqrt{1 + \\tan^{2} 60^{\\circ}} = 2"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, point $P(x_{0}, y_{0})$ is an arbitrary point on the line $b x-a y+2 a=0$. If the circle $(x-x_{0})^{2}+(y-y_{0})^{2}=1$ has no common points with the right branch of hyperbola $C$, then the range of values for the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;G: Circle;x0: Number;y0: Number;H: Line;b: Number;a: Number;a>0;b>0;P: Point;Expression(G) = ((x - x0)^2 + (y - y0)^2 = 1);Expression(H) = (2*a - a*y + b*x = 0);Coordinate(P) = (x0, y0);Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, H);NumIntersection(G, RightPart(C))=0",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(1,2]",
    "fact_spans": "[[[2, 58], [135, 141], [151, 154]], [[102, 134]], [[60, 77]], [[60, 77]], [[78, 95]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[59, 77]], [[102, 134]], [[78, 95]], [[59, 77]], [[2, 58]], [[59, 100]], [[102, 149]]]",
    "query_spans": "[[[151, 164]]]",
    "process": "The hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) has an asymptote given by $ y = \\frac{b}{a}x $, or equivalently $ bx - ay = 0 $. Since $ P(x_{0}, y_{0}) $ is an arbitrary point on the line $ bx - ay + 2a = 0 $, the distance $ d $ between the lines $ bx - ay + 2a = 0 $ and $ bx - ay = 0 $ is $ d = \\frac{|2a|}{\\sqrt{a^{2} + b^{2}}} = \\frac{2a}{c} $. Since the circle $ (x - x_{0})^{2} + (y - y_{0})^{2} = 1 $ has no common points with the right branch of the hyperbola $ C $, it follows that $ d \\geqslant 1 $, i.e., $ \\frac{2a}{c} \\geqslant 1 $, which gives $ e = \\frac{c}{a} \\leqslant 2 $. Also, $ e > 1 $, therefore the range of $ e $ is $ (1, 2] $."
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{(a+3)^{2}}=1$ $(a>0)$ is $y=2x$, then $a=$?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2/(a + 3)^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (y = 2*x)",
    "query_expressions": "a",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 57]], [[75, 78]], [[5, 57]], [[2, 57]], [[2, 73]]]",
    "query_spans": "[[[75, 80]]]",
    "process": "] According to the standard equation of the hyperbola, the foci lie on the x-axis; thus, the asymptotes are given by $ y = \\frac{b}{a}x $, that is, $ \\frac{b}{a} = 2 $. Hence, we obtain $ a + 3 = b = 2a $, and solving this yields: since the asymptotes are $ y = 2x $ and the foci lie on the x-axis, $ \\frac{b}{a} = 2 $, i.e., $ b = 2a $. Since $ a > 0 $, it follows that $ a + 3 > 0 $, so $ a + 3 = b = 2a $, therefore $ a = 3 $."
  },
  {
    "text": "If the length of the imaginary axis of the hyperbola $\\frac{x^{2}}{6}-\\frac{y^{2}}{m}=1$ is $6 \\sqrt{2}$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/6 - y^2/m = 1);m: Number;Length(ImageinaryAxis(G)) = 6*sqrt(2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 39], [59, 62]], [[1, 39]], [[4, 39]], [[1, 56]]]",
    "query_spans": "[[[59, 68]]]",
    "process": "According to the geometric properties of the hyperbola equation, the result is obtained. From the given condition, $2b=6\\sqrt{2}$, so $b=3\\sqrt{2}$, and $a^{2}=6$, $m=b^{2}=18$, thus $c^{2}=6+18=24$, so $c=2\\sqrt{6}$. Therefore, the eccentricity of the hyperbola is $e=\\frac{c}{a}=\\frac{2\\sqrt{6}}{\\sqrt{6}}=2$."
  },
  {
    "text": "The line $l$ passes through the focus of the parabola $x = a y^{2}$ $(a > 0)$ and is perpendicular to the $x$-axis. If the segment of $l$ intercepted by the parabola has length $4$, then $a = $?",
    "fact_expressions": "l: Line;G: Parabola;a: Number;a>0;Expression(G) = (x = a*y^2);IsPerpendicular(l,xAxis);Length(InterceptChord(l,G))=4;PointOnCurve(Focus(G),l)",
    "query_expressions": "a",
    "answer_expressions": "1/4",
    "fact_spans": "[[[0, 5], [40, 43]], [[6, 25], [44, 47]], [[59, 62]], [[9, 25]], [[6, 25]], [[0, 38]], [[40, 57]], [[0, 28]]]",
    "query_spans": "[[[59, 64]]]",
    "process": "Rearranging the parabolic equation gives $ y^{2} = \\frac{1}{a}x $, the focus is $ \\left( \\frac{1}{4a}, 0 \\right) $. The length of the segment cut by line $ l $ on the parabola is the latus rectum length $ \\frac{1}{a} $, hence $ \\frac{1}{a} = 4 $, $ a = \\frac{1}{4} $;"
  },
  {
    "text": "Given the ellipse $C_{1}$: $\\frac{x^{2}}{36}+\\frac{y^{2}}{b^{2}}=1$ $(0<b<6)$ and the circle $C_{2}$: $x^{2}+y^{2}=9$, if from any point $P$ on the circle $C_{2}$, a tangent line to the circle intersects the ellipse at points $A$ and $B$ such that $\\angle A O B=\\frac{\\pi}{2}$ ($O$ being the origin), then the eccentricity of the ellipse $C_{1}$ is?",
    "fact_expressions": "C1:Ellipse;C2:Circle;A: Point;O: Origin;B: Point;Expression(C1) = (x^2/36 + y^2/b^2 = 1);Expression(C2)=(x^2+y^2=9);PointOnCurve(P,C2);Intersection(TangentOfPoint(P,C2),C1)={A,B};AngleOf(A, O, B) = pi/2;P:Point;b: Number;0 < b;b < 6",
    "query_expressions": "Eccentricity(C1)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[2, 60], [111, 113], [166, 175]], [[61, 86], [89, 97], [106, 107]], [[114, 117]], [[155, 158]], [[118, 121]], [[2, 60]], [[61, 86]], [[89, 105]], [[88, 123]], [[126, 154]], [[102, 105]], [[2, 60]], [[2, 60]], [[2, 60]]]",
    "query_spans": "[[[166, 183]]]",
    "process": "If the slope of line AB does not exist, then the equation of line AB is $x = \\pm 3$. Substituting into the ellipse equation gives: $y^{2} = b^{2}(1 - \\frac{9}{a^{2}})$. Since $\\angle AOB = \\frac{\\pi}{2}$, we have $b^{2}(1 - \\frac{9}{a^{2}}) = 9$, which simplifies to: $9(a^{2} + b^{2}) = a^{2}b^{2}$. If the slope of line AB exists, let the line equation be $y = kx + m$. Since line AB is tangent to circle $C_{2}: x^{2} + y^{2} = 9$, from $\\begin{cases} \\frac{|m|}{\\sqrt{1 + k^{2}}} = 3, \\\\ \\frac{y = k + m}{x^{2} + \\frac{y^{2}}{2}} = 1 \\end{cases}$, we obtain: $(a^{2}k^{2} + b^{2})x^{2} + 2a^{2}kmx + a^{2}m^{2} - a^{2}b^{2} = 0$, then $x_{1} + x_{2} = -\\frac{2a^{2}km}{a^{2}k^{2} + b^{2}}$, $x_{1}x_{2} = \\frac{a^{2}m^{2} - a^{2}b^{2}}{a^{2}k^{2} + b^{2}}$. Since $\\angle AOB = \\frac{\\pi}{2}$, we have $\\overrightarrow{OA} \\cdot \\overrightarrow{OB} = x_{1}x_{2} + y_{1}y_{2} = x_{1}x_{2} + (kx_{1} + m)(kx_{2} + m) = (1 + k^{2})x_{1}x_{2} + km(x_{1} + x_{2}) + m^{2}$, that is, $\\overrightarrow{OA} \\cdot \\overrightarrow{OB} = x_{1}x_{2} + y_{1}y_{2} = x_{1}x_{2} + (kx_{1} + m)(kx_{2} + m) = (1 + k^{2})x_{1}x_{2} + km(x_{1} + x_{2}) + m^{2} = 0$, i.e., $\\frac{(1 + k^{2})(a^{2}m^{2} - a^{2}b^{2})}{a^{2}k^{2} + b^{2}} - \\frac{2a^{2}k^{2}m^{2}}{a^{2}k^{2} + b^{2}} + m^{2} = 0$, simplifying to $m^{2}(a^{2} + b^{2}) = a^{2}b^{2}(1 + k^{2})$. Also, since $m^{2} = 9(1 + k^{2})$, we have $9(a^{2} + b^{2}) = a^{2}b^{2}$. In summary: $9(a^{2} + b^{2}) = a^{2}b^{2}$. Substituting $a = 6$ yields $b = 2\\sqrt{3}$, so $c = 2\\sqrt{6}$, then $e = \\frac{c}{a} = \\frac{\\sqrt{6}}{3}$."
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{25}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/25 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "5*x + pm*4*y = 0",
    "fact_spans": "[[[0, 40]], [[0, 40]]]",
    "query_spans": "[[[0, 48]]]",
    "process": "From \\frac{x^{2}}{16}-\\frac{y^{2}}{25}=1, the asymptotes are given by y=\\pm\\frac{5}{4}x, which simplifies to 5x\\pm4y=0"
  },
  {
    "text": "Given that the equation of the directrix of the parabola $x^{2}=2 a y$ is $y=4$, then the value of $a$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 2*(a*y));a: Number;Expression(Directrix(G)) = (y = 4)",
    "query_expressions": "a",
    "answer_expressions": "-8",
    "fact_spans": "[[[2, 18]], [[2, 18]], [[31, 34]], [[2, 29]]]",
    "query_spans": "[[[31, 38]]]",
    "process": "From the given $-\\frac{2a}{4}=4$, solve to get $a=-8$"
  },
  {
    "text": "A point $P$ on the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ has a distance of $7$ to the point $F_{1}(-5,0)$. Then, the distance from point $P$ to point $F_{2}(5,0)$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);P: Point;PointOnCurve(P, G);F1: Point;Coordinate(F1) = (-5, 0);Distance(P, F1) = 7;F2: Point;Coordinate(F2) = (5, 0)",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "13",
    "fact_spans": "[[[0, 39]], [[0, 39]], [[42, 45], [69, 73]], [[0, 45]], [[46, 60]], [[46, 60]], [[42, 67]], [[74, 87]], [[74, 87]]]",
    "query_spans": "[[[69, 92]]]",
    "process": "The hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ has $a=3$. From the given condition, $|PF_{1}|=7$. By the definition of a hyperbola, $||PF|-|PF_{2}||=2a=6$, so $|7-|PF_{2}||=6$. Solving gives $|PF_{2}|=13$ or $|PF_{2}|=1$. Since $|PF_{2}|\\geqslant c-a=2$, it follows that $|PF_{2}|=13$."
  },
  {
    "text": "Let $F$ be the focus of the parabola $y^{2}=4x$, and let $A$, $B$, $C$ be three points on this parabola such that $\\overrightarrow{F A}+\\overrightarrow{F B}+\\overrightarrow{F C}=\\overrightarrow{0}$. Then $|\\overrightarrow{F A}|+|\\overrightarrow{F B}|+|\\overrightarrow{F C}|$=?",
    "fact_expressions": "G: Parabola;F: Point;A: Point;B: Point;C: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(A, G);PointOnCurve(B, G);PointOnCurve(C, G);VectorOf(F, A) + VectorOf(F, B) + VectorOf(F, C) = 0",
    "query_expressions": "Abs(VectorOf(F, A)) + Abs(VectorOf(F, B)) + Abs(VectorOf(F, C))",
    "answer_expressions": "6",
    "fact_spans": "[[[5, 19], [36, 39]], [[1, 4]], [[23, 26]], [[27, 30]], [[31, 34]], [[5, 19]], [[1, 22]], [[23, 42]], [[23, 42]], [[23, 42]], [[44, 127]]]",
    "query_spans": "[[[129, 201]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, $ C(x_{3},y_{3}) $, and $ F(1,0) $. From $ \\overrightarrow{FA} + \\overrightarrow{FB} + \\overrightarrow{FC} = \\overrightarrow{0} $, we have $ (x_{1}-1, y_{1}) + (x_{2}-1, y_{2}) + (x_{3}-1, y_{3}) = (0,0) $. Thus, $ (x_{1}-1) + (x_{2}-1) + (x_{3}-1) = 0 $, which implies $ x_{1} + x_{2} + x_{3} = 3 $. Then, $ |\\overrightarrow{FA}| + |\\overrightarrow{FB}| + |\\overrightarrow{FC}| = x_{1} + x_{2} + x_{3} + \\frac{3}{2}p = 6 $. Hence, the answer is: 6."
  },
  {
    "text": "It is known that one asymptote of the hyperbola $x^{2}-my^{2}=1$ is perpendicular to the line $2x-y+1=0$. Then the real number $m$=?",
    "fact_expressions": "G: Hyperbola;m: Real;H: Line;Expression(G)=(-m*y^2+x^2=1);Expression(H) = (2*x - y + 1 = 0);IsPerpendicular(OneOf(Asymptote(G)),H)",
    "query_expressions": "m",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 21]], [[45, 50]], [[28, 41]], [[2, 21]], [[28, 41]], [[2, 43]]]",
    "query_spans": "[[[47, 52]]]",
    "process": ""
  },
  {
    "text": "The ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ has its left focus at point $F$. A line passing through the origin $O$ intersects the ellipse at points $P$ and $Q$. If $\\angle P F Q=120^{\\circ}$, $|O F|=\\sqrt{3}$, $|O P|=\\sqrt{7}$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;LeftFocus(C) = F;O: Origin;PointOnCurve(O, G) = True;G: Line;Intersection(G, C) = {P, Q};P: Point;Q: Point;AngleOf(P, F, Q) = ApplyUnit(120, degree);Abs(LineSegmentOf(O, F)) = sqrt(3);Abs(LineSegmentOf(O, P)) = sqrt(7)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[0, 57], [77, 79], [156, 161]], [[0, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[62, 66]], [[0, 66]], [[68, 73]], [[67, 76]], [[74, 76]], [[74, 90]], [[81, 84]], [[85, 88]], [[92, 118]], [[120, 136]], [[138, 154]]]",
    "query_spans": "[[[156, 167]]]",
    "process": "According to the problem, take the right focus of the ellipse as $F_{1}$, connect $PF_{1}$, $QF_{1}$, and draw the figure as follows: thus quadrilateral $FPF_{1}Q$ is a parallelogram, then $\\angle FPF_{1}=60^{\\circ}$. Let $|PF|=x$, by the definition of ellipse we have $|PF_{1}|=2a-x$. Applying the cosine law in $\\triangle FOP$ and $\\triangle F_{1}OP$, we obtain:  \n$$\nx^{2}=|OF|^{2}+|OP|^{2}-2|OF||OP|\\cos\\angle FOP\n$$\n$$\n(2a-x)^{2}=|OF_{1}|^{2}+|OP|^{2}-2|OF_{1}||OP|\\cos\\angle F_{1}OP\n$$\nSince $\\cos\\angle FOP = -\\cos\\angle F_{1}OP$, adding the above two equations yields:  \n$$\nx^{2}+(2a-x)^{2}=20, \\quad \\text{that is}, \\quad x^{2}-2ax+2a^{2}-10=0;\n$$\nIn $\\triangle FPF_{1}$, applying the cosine law gives:  \n$$\n|FF_{1}|^{2}=|PF|^{2}+|PF_{1}|^{2}-2|PF||PF_{1}|\\cos 60^{\\circ}\n$$\nThat is,  \n$$\n12=x^{2}+(2a-x)^{2}-x(2a-x), \\quad \\text{so} \\quad x(2a-x)=8, \\quad -x^{2}+2ax=8;\n$$\nHence we get $a^{2}=9$, so $a=3$, and $c=\\sqrt{3}$, therefore the eccentricity of the ellipse is $\\frac{c}{a}=\\frac{\\sqrt{3}}{3}$."
  },
  {
    "text": "Given the circle $C$: $x^{2} + y^{2}-6 x-4 y+8=0$. Taking the intersections of circle $C$ with the coordinate axes as a focus and a vertex of a hyperbola respectively, the standard equation of the hyperbola satisfying the above conditions is?",
    "fact_expressions": "C: Circle;Expression(C) = (-4*y - 6*x + x^2 + y^2 + 8 = 0);A: Point;B: Point;Intersection(C, axis) = {A, B};G: Hyperbola;OneOf(Focus(G)) = A;OneOf(Vertex(G)) = B",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/12 = 1",
    "fact_spans": "[[[2, 35], [38, 42]], [[2, 35]], [], [], [[38, 49]], [[53, 56], [73, 76]], [[36, 64]], [[36, 64]]]",
    "query_spans": "[[[73, 83]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. $P$ is a point on the directrix such that $P F _{1} \\perp PF_{2}$, and $\\overrightarrow {PF_{1}} \\cdot  \\overrightarrow {PF_{2}}=4 ab$. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F2: Point;F1: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P,Directrix(G));IsPerpendicular(LineSegmentOf(P,F1),LineSegmentOf(P,F2));DotProduct(VectorOf(P,F1),VectorOf(P,F2))=4*a*b",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 59], [185, 188]], [[5, 59]], [[5, 59]], [[84, 87]], [[76, 83]], [[68, 75]], [[5, 59]], [[5, 59]], [[2, 59]], [[2, 83]], [[2, 83]], [[2, 93]], [[95, 118]], [[120, 183]]]",
    "query_spans": "[[[185, 194]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the circle centered at the right focus of the ellipse $\\frac{x^{2}}{169}+\\frac{y^{2}}{144}=1$ and tangent to both asymptotes of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;C:Circle;Expression(G) = (x^2/9 - y^2/16 = 1);Expression(H) = (x^2/169 + y^2/144 = 1);Center(C)=RightFocus(H);IsTangent(Asymptote(G),C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "(x-5)^2+y^2=16",
    "fact_spans": "[[[52, 91]], [[1, 42]], [[101, 102]], [[52, 91]], [[1, 42]], [[0, 102]], [[51, 102]]]",
    "query_spans": "[[[101, 105]]]",
    "process": ""
  },
  {
    "text": "What is the focal length of the hyperbola $y^{2}-\\frac{x^{2}}{3}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/3 + y^2 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 34]]]",
    "process": "a=1,b=\\sqrt{3},\\because c^{2}=a^{2}+b^{2},\\therefore c=2, the distance between foci 2c=4"
  },
  {
    "text": "Given that $P(x, y)$ is a point on the ellipse $C$: $\\frac{x^{2}}{4}+y^{2}=1$, if the inequality $2 x-y+a \\geq 0$ always holds, then the range of values for $a$ is?",
    "fact_expressions": "P: Point;x1: Number;y1: Number;Coordinate(P) = (x1, y1);C: Ellipse;Expression(C) = (x^2/4 + y^2 = 1);PointOnCurve(P, C);a: Number;2*x1 - y1 + a >=0",
    "query_expressions": "Range(a)",
    "answer_expressions": "[\\sqrt{17}, +\\infty)",
    "fact_spans": "[[[2, 11]], [[2, 11]], [[2, 11]], [[2, 11]], [[12, 44]], [[12, 44]], [[2, 47]], [[73, 76]], [[52, 68]]]",
    "query_spans": "[[[73, 83]]]",
    "process": "Let $ P(2\\cos\\theta,\\sin\\theta) $, that is, $ x=2\\cos\\theta $, $ y=\\sin\\theta $. Substituting into the inequality gives:  \n$ 2x - y + a = 4\\cos\\theta - \\sin\\theta + a = \\sqrt{17}\\cos(\\theta+\\varphi) + a \\geqslant 0 $ ($ \\tan\\varphi = \\frac{1}{4} $) always holds,  \nwhich means $ -a \\leqslant \\sqrt{17}\\cos(\\theta+\\varphi) $ always holds.  \nSince $ -1 \\leqslant \\cos(\\theta+\\varphi) \\leqslant 1 $, we have $ -a \\leqslant -\\sqrt{17} $, thus $ a \\geqslant \\sqrt{17} $.  \nTherefore, the range of $ a $ is $ [\\sqrt{17}, +\\infty) $."
  },
  {
    "text": "For the equation $\\frac{x^{2}}{9-k}+\\frac{y^{2}}{4+k}=1$ to represent an ellipse with foci on the $x$-axis, what is the range of real values for $k$?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (x^2/(9 - k) + y^2/(k + 4) = 1);PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-4,5/2)",
    "fact_spans": "[[[53, 55]], [[57, 62]], [[1, 55]], [[44, 55]]]",
    "query_spans": "[[[57, 69]]]",
    "process": "The foci are on the x-axis, and the denominators are greater than 0, so $-k)>4+k\\begin{cases}9-k>0\\\\4+k>0\\end{cases}$ Solving gives the range of $k$ as $-4<k<\\frac{5}{2}$, that is, $k\\in(-4,\\frac{5}{2})$"
  },
  {
    "text": "Given a hyperbola with eccentricity $e$ and an ellipse with eccentricity $\\frac{\\sqrt{2}}{2}$ share the same foci $F_{1}$, $F_{2}$, and $P$ is a common point of the two curves. If $\\angle F_{1} P F_{2}=60^{\\circ}$, then $e=$?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;F1: Point;P: Point;F2: Point;e:Number;Eccentricity(G)=e;Eccentricity(H)=sqrt(2)/2;Focus(G)={F1,F2};Focus(H)={F1,F2};OneOf(Intersection(H,G))=P;AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "e",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[10, 13]], [[39, 41]], [[47, 54]], [[63, 66]], [[55, 62]], [[113, 116], [6, 9]], [[2, 13]], [[14, 41]], [[10, 62]], [[10, 62]], [[63, 76]], [[78, 111]]]",
    "query_spans": "[[[113, 118]]]",
    "process": "Let |PF₁| = x, |PF₂| = y, without loss of generality assume x > y, |F₁F₂| = 2c, the major axis of the ellipse is 2a, the real axis of the hyperbola is 2a', then  \n\\begin{cases}x+y=2a\\\\x-y=2a'\\end{cases}  \n\\begin{cases}x=a+a'\\\\y=a-a'\\end{cases},  \nfrom \\angle F₁PF₂ = 60^{\\circ}, we get (2c)^{2} = x^{2} + y^{2} - 2xy\\cos60^{\\circ},  \ni.e. 4c^{2} = x^{2} + y^{2} - xy,  \n4c^{2} = (a+a')^{2} + (a-a')^{2} - (a+a')(a-a'),  \nsimplifying yields 4c^{2} = a^{2} + 3a'^{2},  \n4 = \\left(\\frac{a}{c}\\right)^{2} + 3\\left(\\frac{a'}{c}\\right)^{2},  \ni.e. 4 = 2 + \\frac{3}{e'^{2}}, e' = \\frac{\\sqrt{6}}{2}"
  },
  {
    "text": "The vertex of the parabola is at the origin, and the focus lies on the line $x - 2y - 4 = 0$. Then the standard equation of the parabola is?",
    "fact_expressions": "G: Parabola;H: Line;O: Origin;Expression(H) = (x - 2*y - 4 = 0);Vertex(G) = O;PointOnCurve(Focus(G),H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=16*x,x^2=-8*y}",
    "fact_spans": "[[[0, 3], [29, 32]], [[13, 26]], [[7, 9]], [[13, 26]], [[0, 9]], [[0, 27]]]",
    "query_spans": "[[[29, 39]]]",
    "process": "The line x-2y-4=0 passes through points (4,0) and (0,-2). If (4,0) is the focus, then the parabola opens to the right, \\frac{p}{2}=4, 2p=16, and the equation of the parabola is y^{2}=16x. If (0,-2) is the focus, then the parabola opens downward, \\frac{p}{3}=2, 2p=8, and the equation of the parabola is x^{2}=-8y."
  },
  {
    "text": "The vertex of the parabola is at the origin, its axis of symmetry is a coordinate axis, and it passes through the point $P(-2,2 \\sqrt{2})$. Then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;P: Point;O: Origin;Coordinate(P) = (-2, 2*sqrt(2));Vertex(G) = O;SymmetryAxis(G) = axis;PointOnCurve(P, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2 = sqrt(2)*y, y^2 = -4*x}",
    "fact_spans": "[[[0, 3], [19, 20], [42, 45]], [[21, 40]], [[7, 9]], [[21, 40]], [[0, 9]], [[0, 17]], [[19, 40]]]",
    "query_spans": "[[[42, 50]]]",
    "process": ""
  },
  {
    "text": "A point $P$ on the hyperbola $x^{2}-\\frac{y^{2}}{15}=1$ is at a distance of $3$ from one of its foci. Then, the distance from point $P$ to the other focus is?",
    "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2 - y^2/15 = 1);PointOnCurve(P, G);F1:Point;F2:Point;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2);Distance(P, F1) = 3",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "5",
    "fact_spans": "[[[0, 29], [37, 38]], [[53, 57], [33, 36]], [[0, 29]], [[0, 36]], [], [], [[37, 43]], [[37, 63]], [[37, 63]], [[33, 51]]]",
    "query_spans": "[[[37, 68]]]",
    "process": "【Fen Mu Jin】According to the definition of hyperbola, the distance from point P to the other focus can be obtained. By comparing the value of c with the distance from P to one focus, the solution can be determined. For the hyperbola $ x^{2} - \\frac{y^{2}}{15} = 1 $, according to the definition of hyperbola, we have $ a = 1 $, $ b = \\sqrt{15} $, $ c = 4 $. $ ||PF_{1}| - |PF_{2}|| = 2a $, and $ ||PF_{1}| - |PF_{2}|| < |F_{1}F_{2}| $. Therefore, $ ||PF_{1}| - |PF_{2}|| = 2 $ and $ ||PF_{1}| - |PF_{2}|| < 8 $. Also, the distance from P to one of its foci is equal to 3. Let $ |PF_{1}| = 2 $, substituting into $ ||PF_{1}| - |PF_{2}|| = 2 $, we obtain $ |PF_{2}| = 5 $ or $ |PF_{2}| = 1 $. Since $ c > 3 $, the distance from P to one focus, it follows that $ |PF_{2}| > 3 $. Hence, $ |PF_{1}| = 5 $. This problem examines the definition and simple applications of hyperbola properties. Be sure to verify whether the obtained solutions satisfy the given conditions. It is a medium-level problem."
  },
  {
    "text": "A line with slope $1$ intersects the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ at two points $A$ and $B$. The midpoint $M(m, 1)$ of $AB$, then $m=$?",
    "fact_expressions": "G: Ellipse;l: Line;A: Point;B: Point;M: Point;m:Number;Expression(G) = (x^2/4 + y^2/3 = 1);Slope(l)=1;Intersection(l,G)={A,B};Coordinate(M) = (m, 1);MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "m",
    "answer_expressions": "-4/3",
    "fact_spans": "[[[10, 47]], [[7, 9]], [[50, 53]], [[55, 58]], [[68, 77]], [[79, 82]], [[10, 47]], [[0, 9]], [[7, 60]], [[68, 77]], [[61, 77]]]",
    "query_spans": "[[[79, 84]]]",
    "process": ""
  },
  {
    "text": "At the point on the parabola $y=x^2$ where the tangent line has an inclination angle of $\\frac{\\pi}{4}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2);P:Point;PointOnCurve(P,G);Inclination(TangentOnPoint(P,G))=pi/4",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "Slight",
    "fact_spans": "[[[1, 11]], [[1, 11]], [[13, 14]], [[0, 14]], [[0, 38]]]",
    "query_spans": "[[[13, 15]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the right focus of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ are?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1)",
    "query_expressions": "Coordinate(RightFocus(G))",
    "answer_expressions": "(4,0)",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 46]]]",
    "process": "From $\\frac{x2}{25}+\\frac{y2}{9}=1$, we get $a^{2}=25$, $b^{2}=9$, $\\cdot c=\\sqrt{a^{2}-b^{2}}=4$, so the coordinates of the right focus of the ellipse are $(4,0)$."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $C$: $y^{2}=2 p x$ ($p>0$) intersects the parabola at points $A$ and $B$. If $|A F|=8|O F|$ ($O$ is the origin), then $\\frac{|A F|}{|B F|}=$?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;A: Point;F: Point;O: Origin;B: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(F, G);Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, F)) = 8*Abs(LineSegmentOf(O, F))",
    "query_expressions": "Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(B, F))",
    "answer_expressions": "7",
    "fact_spans": "[[[1, 27], [38, 41]], [[8, 27]], [[34, 36]], [[42, 45]], [[30, 33]], [[69, 72]], [[46, 49]], [[8, 27]], [[1, 27]], [[1, 33]], [[0, 36]], [[34, 51]], [[54, 68]]]",
    "query_spans": "[[[80, 103]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, $ F(\\frac{p}{2},0) $, then by the definition of the parabola we have $ |AF| = x_{1} + \\frac{p}{2} = 8 \\times \\frac{p}{2} \\Rightarrow x_{1} = \\frac{7p}{2} $, then $ y_{1}^{2} = 2px_{1} \\Rightarrow y_{1} = \\sqrt{7}p $, so $ A(\\frac{7p}{2},\\sqrt{7}p) $, hence $ k_{AB} = \\frac{\\sqrt{7}p}{3p} = \\frac{\\sqrt{7}}{3} $, thus the equation of line AB is $ y = \\frac{\\sqrt{7}}{3}(x - \\frac{p}{2}) $. Substituting into the parabola equation and simplifying yields $ \\frac{7}{9}x^{2} - \\frac{25}{9}px + \\frac{7}{36}p^{2} = 0 $, then $ x_{1}x_{2} = \\frac{p^{2}}{4} \\Rightarrow x_{2} = \\frac{p}{14} $, then $ |BF| = x_{2} + \\frac{p}{2} = \\frac{4p}{7} $, so $ \\frac{|AF|}{|BF|} = 7 $, the answer to be filled is 7."
  },
  {
    "text": "The product of the distances from a point $P$ on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$ to its two foci is $m$. When $m$ takes its maximum value, what are the coordinates of point $P$?",
    "fact_expressions": "G: Ellipse;P: Point;F1:Point;F2:Point;m:Number;Expression(G) = (x^2/4 + y^2/2 = 1);PointOnCurve(P, G);Distance(P,F1)*Distance(P,F2)=m;WhenMax(m);Focus(G)={F1,F2}",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "{(0,-sqrt(2)),(0,sqrt(2))}",
    "fact_spans": "[[[0, 37]], [[41, 44], [69, 73]], [], [], [[55, 58], [60, 63]], [[0, 37]], [[0, 44]], [[0, 58]], [[59, 68]], [0, 45]]",
    "query_spans": "[[[69, 78]]]",
    "process": "By the definition of the ellipse, |PF_{1}| + |PF_{2}| = 2a = 4. Then, using the basic inequality, m = |PF_{1}||PF_{2}| \\leqslant \\frac{(|PF_{1}| + |PF_{2}|)^{2}}{4} = \\frac{(4)^{2}}{4} = 4. Let the two foci of the ellipse be F_{1} and F_{2}. By the definition of the ellipse: |PF_{1}| + |PF_{2}| = 2a = 4. Then m = |PF_{1}||PF_{2}| \\leqslant \\frac{(|PF_{1}| + |PF_{2}|)^{2}}{4} = 4, with equality if and only if |PF_{1}| = |PF_{2}| = 2, that is, when point P lies at the vertices on the minor axis of the ellipse, m attains its maximum value 4. Therefore, the coordinates of point P are (0, -\\sqrt{2}) or (0, \\sqrt{2})."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $x^{2}=4 y$. $O$ is the origin, $A$ is a point on $C$, and the circumcircle $\\odot B$ ($B$ being the center) of $\\triangle O F A$ is tangent to the directrix of $C$. What is the slope of the line passing through point $B$ and tangent to $C$?",
    "fact_expressions": "C: Parabola;B: Circle;B1:Point;O: Origin;F: Point;A: Point;Expression(C) = (x^2 = 4*y);Focus(C) = F;PointOnCurve(A, C);CircumCircle(TriangleOf(O,F,A))=B;Center(B)=B1;IsTangent(Directrix(C),B);H:Line;PointOnCurve(B,H);IsTangent(H,C)",
    "query_expressions": "Slope(H)",
    "answer_expressions": "pm*sqrt(2)/2",
    "fact_spans": "[[[6, 25], [42, 45], [87, 90], [103, 106]], [[69, 78]], [[79, 82], [98, 102]], [[29, 32]], [[2, 5]], [[38, 41]], [[6, 25]], [[2, 28]], [[38, 48]], [[49, 78]], [[69, 86]], [[69, 95]], [[109, 111]], [[97, 111]], [[102, 111]]]",
    "query_spans": "[[[109, 115]]]",
    "process": "The parabola C: x^{2} = 4y has focus F(0,1) and directrix equation y = -1. The circumcenter B of triangle OFA lies on the perpendicular bisector of OF, which is y = \\frac{1}{2}. Let B(a,\\frac{1}{2}). Since \\odot B is tangent to the directrix of C, the radius of \\odot B is \\frac{3}{2}. Therefore, |OB| = \\sqrt{a^{2} + \\frac{1}{4}} = \\frac{3}{2}, so a^{2} = 2, a = \\pm\\sqrt{2}. Since B(\\pm\\sqrt{2},\\frac{1}{2}) satisfies the equation of parabola C, and because a tangent line to a parabola intersects it at only one point, B is the point of tangency. For curve C: x^{2} = 4y, we have y = \\frac{1}{4}x^{2}, y' = \\frac{1}{2}x. Thus, the slope of the line tangent to C at point B is \\frac{\\sqrt{2}}{2}."
  },
  {
    "text": "Let $F$ be the focus of the parabola $y^{2}=4 x$, and let $A$, $B$, $C$ be three points on this parabola. If $\\overrightarrow {F A}+\\overrightarrow{F B}+\\overrightarrow{F C}=0$, then what is the value of $|\\overrightarrow{F A}|+|\\overrightarrow{F B}|+|\\overrightarrow{F C}|$?",
    "fact_expressions": "G: Parabola;F: Point;A: Point;B: Point;C: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(A,G);PointOnCurve(B,G);PointOnCurve(C,G);VectorOf(F, A) + VectorOf(F, B) + VectorOf(F, C) = 0",
    "query_expressions": "Abs(VectorOf(F, A)) + Abs(VectorOf(F, B)) + Abs(VectorOf(F, C))",
    "answer_expressions": "6",
    "fact_spans": "[[[5, 19], [40, 43]], [[1, 4]], [[23, 27]], [[29, 32]], [[35, 38]], [[5, 19]], [[1, 22]], [[23, 46]], [[23, 46]], [[23, 46]], [[48, 115]]]",
    "query_spans": "[[[117, 191]]]",
    "process": ""
  },
  {
    "text": "The standard equation of the hyperbola that has the same asymptotes as the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$ and passes through the point $(2 , 2)$ is?",
    "fact_expressions": "G:Hyperbola;H: Point;C:Hyperbola;Expression(G)=(x^2-y^2/2=1);Coordinate(H)=(2,2);Asymptote(G)=Asymptote(C);PointOnCurve(H,C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/2 - y^2/4 = 1",
    "fact_spans": "[[[1, 29]], [[39, 49]], [[50, 53]], [[1, 29]], [[39, 49]], [[0, 53]], [[38, 53]]]",
    "query_spans": "[[[50, 60]]]",
    "process": ""
  },
  {
    "text": "Draw a perpendicular line from a focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ to one of its asymptotes. If the foot of the perpendicular lies exactly on the perpendicular bisector of the segment $OF$ ($O$ being the origin), then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;O: Origin;F: Point;L1:Line;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);OneOf(Focus(G))=F;PointOnCurve(F,L1);IsPerpendicular(OneOf(Asymptote(G)),L1);PointOnCurve(FootPoint(OneOf(Asymptote(G)),L1),PerpendicularBisector(LineSegmentOf(O,F)))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 47], [95, 98]], [[4, 47]], [[4, 47]], [[78, 81]], [[52, 55]], [], [[1, 47]], [[1, 55]], [[0, 64]], [[0, 64]], [[0, 93]]]",
    "query_spans": "[[[95, 104]]]",
    "process": "Let the foot of the perpendicular be D. According to the hyperbola equation, one of the asymptotes is y = \\frac{b}{a}x, and the focus is F(\\sqrt{a^{2}+b^{2}},0). Therefore, the coordinates of point D are \\left( \\frac{\\sqrt{a^{2}+b^{2}}}{\\sqrt{1+\\frac{b^{2}}{a^{2}}}},\\ \\frac{b\\sqrt{a^{2}+b^{2}}}{a\\sqrt{1+\\frac{b^{2}}{a^{2}}}} \\right), so k_{DF} = -\\frac{b}{a}. Since OD \\perp DF, we have k_{DF} \\cdot k_{OD} = -1, therefore \\frac{b}{a} = \\frac{a}{b}, i.e., a = b. Hence, e = \\frac{c}{a} = \\frac{\\sqrt{a^{2}+b^{2}}}{a} = \\sqrt{2}."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the foci of the ellipse $\\frac{x^{2}}{3}+y^{2}=1$, and let points $A$ and $B$ lie on the ellipse. If $\\overrightarrow{F_{1} A}=5 \\overrightarrow{F_{2} B}$, then the coordinates of point $A$ are?",
    "fact_expressions": "G: Ellipse;F1: Point;A: Point;F2: Point;B: Point;Expression(G) = (x^2/3 + y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(A, G);PointOnCurve(B, G);VectorOf(F1, A) = 5*VectorOf(F2, B)",
    "query_expressions": "Coordinate(A)",
    "answer_expressions": "{(0, 1), (0, -1)}",
    "fact_spans": "[[[19, 46], [59, 61]], [[1, 8]], [[50, 54], [120, 124]], [[9, 16]], [[55, 58]], [[19, 46]], [[1, 49]], [[50, 62]], [[55, 62]], [[64, 117]]]",
    "query_spans": "[[[120, 129]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the right focus of the hyperbola $C$: $x^{2}-y^{2}=1$, $P$ is a point on the left branch of $C$, and $A(0, \\sqrt{2})$. When the perimeter of $\\triangle A P F$ is minimized, what is the area of this triangle?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2 = 1);F: Point;RightFocus(C) = F;P: Point;PointOnCurve(P, LeftPart(C));A: Point;Coordinate(A) = (0, sqrt(2));WhenMin(Perimeter(TriangleOf(A, P, F)))",
    "query_expressions": "Area(TriangleOf(A, P, F))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[6, 29], [38, 41]], [[6, 29]], [[2, 5]], [[2, 33]], [[34, 37]], [[34, 47]], [[48, 64]], [[48, 64]], [[65, 88]]]",
    "query_spans": "[[[90, 98]]]",
    "process": "If M is the left focus, then |PF| - |PM| = 2a = 2, and F(\\sqrt{2},0), A(0,\\sqrt{2}), so |AF| = 2. The perimeter of \\triangle APF is |AF| + |PF| + |AP| = |AF| + |PM| + |AP| + 2 = |PM| + |AP| + 4. The perimeter of \\triangle APF is minimized if and only if points A, P, M are collinear; at this time |AM| = |PM| + |AP| = 2, so \\triangle AFM is an isosceles right triangle with leg length 2. Let |AP| = x, then |PM| = 2 - x, hence |PF| = 4 - x, and |AF| = 2. In \\triangle APF, x^{2} + 4 = (4 - x)^{2}, solving gives x = \\frac{3}{2}, thus the area of the triangle is \\frac{1}{2} \\times 2 \\times \\frac{3}{2} = \\frac{3}{2}."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$, a line $l$ intersects the parabola $C$ at points $A$ and $B$. If the coordinates of the midpoint of segment $AB$ are $(2,2)$, then what is the equation of line $l$?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;Expression(C) = (y^2 = 4*x);Coordinate(MidPoint(LineSegmentOf(A,B))) = (2, 2);Intersection(l, C) = {A, B}",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x-y=0",
    "fact_spans": "[[[22, 27], [69, 74]], [[2, 21], [28, 34]], [[36, 39]], [[40, 43]], [[2, 21]], [[47, 67]], [[22, 45]]]",
    "query_spans": "[[[69, 79]]]",
    "process": ""
  },
  {
    "text": "If the equation of the hyperbola is $4 x^{2}-9 y^{2}=36$, then the length of its transverse axis is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (4*x^2 - 9*y^2 = 36)",
    "query_expressions": "Length(RealAxis(G))",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 4], [30, 31]], [[1, 28]]]",
    "query_spans": "[[[30, 36]]]",
    "process": ""
  },
  {
    "text": "Ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has left and right foci $F_{1}$, $F_{2}$, respectively. On the right directrix $l$ of $C$, there exists a point $P$ such that $\\angle F_{1} F_{2} P=120^{\\circ}$, $30^{\\circ} \\leq \\angle F_{1} PF_{2}<60^{\\circ}$. Then the range of the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C)= F2;l: Line;RightDirectrix(C)=l;P: Point;PointOnCurve(P,l);AngleOf(F1,F2,P)=ApplyUnit(120,degree);ApplyUnit(30,degree)<=AngleOf(F1,P,F2);AngleOf(F1,P,F2)<ApplyUnit(60,degree)",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "[\\sqrt{2}/2,1)",
    "fact_spans": "[[[0, 57], [84, 87], [191, 196]], [[0, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[66, 73]], [[75, 82]], [[0, 82]], [[0, 82]], [[91, 94]], [[84, 94]], [[99, 102]], [[83, 102]], [[104, 139]], [[141, 189]], [[141, 189]]]",
    "query_spans": "[[[191, 207]]]",
    "process": ""
  },
  {
    "text": "Given that $O$ is the coordinate origin, the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$, $P$ is a point on $C$, $PF$ is perpendicular to the $x$-axis, $Q$ is a point on the $x$-axis, and $PQ \\perp OP$. If $|FQ|=6$, then the equation of the directrix of $C$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;O: Origin;F: Point;Focus(C) = F;P: Point;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P, F), xAxis);Q: Point;PointOnCurve(Q, xAxis);IsPerpendicular(LineSegmentOf(P, Q), LineSegmentOf(O, P));Abs(LineSegmentOf(F, Q)) = 6",
    "query_expressions": "Expression(Directrix(C))",
    "answer_expressions": "x=-3/2",
    "fact_spans": "[[[11, 37], [49, 52], [110, 113]], [[11, 37]], [[19, 37]], [[19, 37]], [[2, 5]], [[41, 44]], [[11, 44]], [[45, 48]], [[45, 55]], [[56, 68]], [[69, 72]], [[69, 80]], [[82, 97]], [[99, 108]]]",
    "query_spans": "[[[110, 120]]]",
    "process": "The parabola $ C: y^{2} = 2px $ ($ p > 0 $) has focus $ F\\left(\\frac{p}{2}, 0\\right) $. Since $ P $ is a point on $ C $ and $ PF $ is perpendicular to the x-axis, the x-coordinate of $ P $ is $ \\frac{p}{2} $. Substituting into the parabola equation gives the y-coordinate of $ P $ as $ \\pm p $. Without loss of generality, assume $ P\\left(\\frac{p}{2}, p\\right) $. Since $ Q $ is a point on the x-axis and $ PQ \\perp OP $, $ Q $ lies to the right of $ F $. Given $ |FQ| = 6 $, we have $ Q\\left(6 + \\frac{p}{2}, 0\\right) $. Thus, $ \\overrightarrow{PQ} = (6, -p) $. Since $ PQ \\perp OP $, it follows that $ \\overrightarrow{PQ} \\cdot \\overrightarrow{OP} = \\frac{p}{2} \\times 6 - p^{2} = 0 $. As $ p > 0 $, we obtain $ p = 3 $. Therefore, the directrix of $ C $ is $ x = -\\frac{3}{2} $."
  },
  {
    "text": "If the focus of the parabola $y=a x^{2}$ has coordinates $(0,2)$, then what is the value of the real number $a$?",
    "fact_expressions": "G: Parabola;a: Real;Expression(G) = (y = a*x^2);Coordinate(Focus(G)) = (0, 2)",
    "query_expressions": "a",
    "answer_expressions": "1/8",
    "fact_spans": "[[[1, 15]], [[31, 36]], [[1, 15]], [[1, 28]]]",
    "query_spans": "[[[31, 40]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $x^{2}=4 y$, $A(0,3)$. If there exists a point $P(x_{0}, y_{0})$ $(x_{0} \\neq 0)$ on the parabola $C$ such that the tangent line $l$ at point $P$ is perpendicular to $PA$, and let $l$ intersect the $y$-axis at point $E$, then the area of $\\Delta A P E$ is?",
    "fact_expressions": "C: Parabola;A: Point;P: Point;E: Point;l:Line;x0:Number;y0:Number;Negation(x0=0);Expression(C) = (x^2 = 4*y);Coordinate(A) = (0, 3);Coordinate(P) = (x0, y0);PointOnCurve(P,C);PointOnCurve(P, l);TangentOfPoint(P,C)=l;IsPerpendicular(l,LineSegmentOf(P, A));Intersection(l, yAxis) = E",
    "query_expressions": "Area(TriangleOf(A, P, E))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 21], [33, 39]], [[23, 31]], [[78, 82], [42, 74]], [[111, 115]], [[101, 104]], [[43, 74]], [[43, 74]], [[43, 74]], [[2, 21]], [[23, 31]], [[42, 74]], [[33, 74]], [[77, 99]], [[33, 99]], [[85, 99]], [[101, 115]]]",
    "query_spans": "[[[117, 136]]]",
    "process": "First, differentiate the function to obtain $ y = \\frac{1}{2}x $, find the slope of line $ l $, use the slope formula to write the slope of line $ AP $, and since the two lines are perpendicular, we get $ \\frac{x_{0}}{2} \\cdot \\frac{y_{0}-?}{x_{0}}1' $ together with $ x_{0}^{2} = 4y_{0} $, solve simultaneously to find the coordinates of point $ P $, then compute the area of the triangle. Solution: From $ x^{2} = 4y $, we get $ y = \\frac{1}{4}x^{2} $, $ y' = \\frac{1}{2}x $, so the slope of line $ l $ is $ k = y'|_{x=x_{0}} = \\frac{1}{2}x_{0} $. The slope of line $ AP $ is $ \\frac{y_{0}-3}{x_{0}} $. Since tangent line $ l \\perp PA $, we have $ \\frac{x_{0}}{2} \\cdot \\frac{y_{0}-3}{x_{0}} = -1 $. Also, $ x_{0}^{2} = 4y_{0} $, solving gives $ x_{0} = \\pm2 $, $ y_{0} = 1 $. Without loss of generality, let $ P(2,1) $, then the equation of line $ l $ is $ y - 1 = x - 2 $, i.e., $ y = x - 1 $. Thus, $ E(0,-1) $, and the area of $ \\triangle APE $ is $ \\frac{1}{2} \\times 4 \\times 2 = 4 $."
  },
  {
    "text": "Given that the vertex of the parabola $C$ is at the origin and the focus is $F(1 , 0)$, the line $l$ intersects the parabola $C$ at points $A$ and $B$. If the midpoint of $AB$ is $(2 , 2)$, then what is the equation of the line $l$?",
    "fact_expressions": "l: Line;C: Parabola;F: Point;A: Point;B: Point;O: Origin;Coordinate(MidPoint(LineSegmentOf(A, B))) = (2, 2);Coordinate(F) = (1, 0);Vertex(C) = O;Intersection(l, C) = {A, B};Focus(C) = F",
    "query_expressions": "Expression(l)",
    "answer_expressions": "y = x",
    "fact_spans": "[[[32, 37], [81, 86]], [[3, 9], [38, 44]], [[21, 31]], [[47, 50]], [[53, 56]], [[13, 17]], [[61, 79]], [[21, 31]], [[3, 17]], [[32, 58]], [[3, 31]]]",
    "query_spans": "[[[81, 91]]]",
    "process": ""
  },
  {
    "text": "The line $x - y - 1 = 0$ intersects the parabola $y^2 = 4x$ at points $A$ and $B$. A perpendicular is drawn from the midpoint of segment $AB$ to the line $x = -1$, with foot $M$. Then $\\overrightarrow{M A} \\cdot \\overrightarrow{M B}$ = ?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;M: Point;l: Line;C:Line;Expression(G) = (y^2 = 4*x);Expression(H) = (x - y - 1 = 0);Expression(l) = (x = -1);Intersection(H,G)={A,B};PointOnCurve(MidPoint(LineSegmentOf(A,B)),C);IsPerpendicular(C,l);FootPoint(C,l)=M",
    "query_expressions": "DotProduct(VectorOf(M, A), VectorOf(M, B))",
    "answer_expressions": "0",
    "fact_spans": "[[[12, 26]], [[0, 11]], [[28, 31]], [[32, 35]], [[65, 68]], [[50, 58]], [], [[12, 26]], [[0, 11]], [[50, 58]], [[0, 37]], [[38, 61]], [[38, 61]], [[38, 68]]]",
    "query_spans": "[[[70, 121]]]",
    "process": "As shown in the figure, let A(x_{1},y_{1}), B(x_{2},y_{2}), then M(-1,\\frac{y_{1}+y_{2}}{2}), solving simultaneously \\begin{cases}x-y-\\\\v2=4x\\end{cases} so \\overrightarrow{MA}\\cdot\\overrightarrow{MB}=(x_{1}+1,\\frac{y_{1}-y_{2}}{2})(x_{2}+1,\\frac{y_{2}-y_{1}}{2})=x_{1}x_{2}+x_{1}+x_{2}+1-\\frac{(y_{1}-y_{2})^{2}}{4}, thus y_{1}+y_{2}=4, y_{1}y_{2}=-4, then x_{1}+x_{2}=y_{1}+y_{2}+2=6, x_{1}x_{2}=(y_{1}+1)(y_{2}+1)=y_{1}y_{2}+y_{1}+y_{2}+1=-4+4+1=1, (y_{1}-y_{2})^{2}=(y_{1}+y_{2})^{2}-4y_{1}y_{2}=16+16=32, hence \\overrightarrow{MA}\\cdot\\overrightarrow{MB}=x_{1}x_{2}+x_{1}+x_{2}+1-\\frac{(y_{1}-y_{2})^{2}}{4}=1+6+1-8=0"
  },
  {
    "text": "Given fixed points $A$, $B$ with $|AB|=8$, and a moving point $P$ satisfying $|PA|-|PB|=4$, then the minimum value of $|PA|$ is?",
    "fact_expressions": "A: Point;B: Point;P: Point;Abs(LineSegmentOf(A, B)) = 8;Abs(LineSegmentOf(P, A)) - Abs(LineSegmentOf(P, B)) = 4",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "6",
    "fact_spans": "[[[4, 7]], [[8, 11]], [[25, 28]], [[13, 22]], [[30, 45]]]",
    "query_spans": "[[[47, 60]]]",
    "process": "Since the moving point P satisfies |PA| - |PB| = 4 < |AB| = 8, the trajectory of point P is one branch of a hyperbola with foci at A and B. Thus, 2a = 4, 2c = 8, that is, a = 2, c = 4, b^{2} = 12. Without loss of generality, assume the foci lie on the x-axis; then the equation of the hyperbola is \\frac{x^{2}}{4} - \\frac{y^{2}}{12} = 1 (x \\geqslant 2). The left focus is A(-4,0), the right focus is B(4,0). Let P(x_{0}, y_{0}), (x_{0} \\geqslant 2), then \\frac{x_{0}^{2}}{4} - \\frac{y_{0}^{2}}{12} = 1. Therefore, the minimum value of |PA| is 6."
  },
  {
    "text": "$P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, and $M$, $N$ are points on the circles $(x+5)^{2}+y^{2}=9$ and $(x-5)^{2}+y^{2}=4$, respectively. Then the maximum value of $|P M|-|P N|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1);P: Point;PointOnCurve(P, RightPart(G));H: Circle;Expression(H) = (y^2 + (x + 5)^2 = 9);Z: Circle;Expression(Z) = (y^2 + (x - 5)^2 = 4);M: Point;PointOnCurve(M, H);N: Point;PointOnCurve(N, Z)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "13",
    "fact_spans": "[[[4, 43]], [[4, 43]], [[0, 3]], [[0, 49]], [[60, 80]], [[60, 80]], [[81, 100]], [[81, 100]], [[50, 53]], [[50, 103]], [[54, 57]], [[50, 103]]]",
    "query_spans": "[[[105, 124]]]",
    "process": "\\because\\frac{x^2}{16}-\\frac{y^{2}}{9}=1\\thereforea^{2}=16, b^{2}=9, then c^{2}=25, so the two foci of the hyperbola are F_{1}(-5,0),F_{2}(5,0). F_{1}(-5,0),F_{2}(5,0) are also the centers of the two circles respectively, with radii r_{1}=3,r_{2}=2. |PF_{1}|-|PF_{2}|=(|PF_{1}|-r_{1})+(r_{2}+|PF_{2}|)-r_{1}-r_{2}=2\times4-5=13."
  },
  {
    "text": "Given that point $P$ is a moving point on the ellipse $C$: $\\frac{x^{2}}{9}+y^{2}=1$, and point $Q$ is a moving point on the circle $E$: $x^{2}+(y-4)^{2}=3$, then the maximum value of $|PQ|$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/9 + y^2 = 1);P: Point;PointOnCurve(P, C);E: Circle;Expression(E) = (x^2 + (y - 4)^2 = 3);Q: Point;PointOnCurve(Q, E)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[7, 39]], [[7, 39]], [[2, 6]], [[2, 45]], [[51, 76]], [[51, 76]], [[46, 50]], [[46, 82]]]",
    "query_spans": "[[[84, 96]]]",
    "process": "From the circle $ E: x^{2} + (y - 4)^{2} = 3 $, we obtain the center $ E(0, 4) $. Since point $ Q $ lies on circle $ E $, $ |PQ| \\leq |EP| + |EQ| = |EP| + \\sqrt{3} $ (equality holds if and only if line $ PQ $ passes through point $ E $). Let $ P(x_{1}, y_{1}) $ be an arbitrary point on the ellipse $ C $, then $ \\frac{x^{2}}{9} + y_{1}^{2} = 1 $, i.e., $ x_{1}^{2} = 9 - 9y_{1}^{2} $. Since $ |EP|^{2} = x_{1}^{2} + (y_{1} - 4)^{2} = 9 - 9y_{1}^{2} + (y_{1} - 4)^{2} = -8(y_{1} + \\frac{1}{2})^{2} + 27 $, and $ y_{1} \\in [-1, 1] $, therefore when $ y_{1} = -\\frac{1}{2} $, $ |EP|^{2} $ attains its maximum value 27, i.e., $ |PQ| \\leq 3\\sqrt{3} + \\sqrt{3} = 4\\sqrt{3} $. Since the maximum value of $ |PQ| $ is $ 4\\sqrt{3} $."
  },
  {
    "text": "The line $l$ passes through the focus $F$ of the parabola $C$: $y^{2}=2 p x$ ($p>0$), intersects the parabola $C$ at points $A$ and $B$, and intersects its directrix at point $D$. If $|A F|=6$, $\\overrightarrow{D B}=2 \\overrightarrow{B F}$, then $p=$?",
    "fact_expressions": "l: Line;C: Parabola;p: Number;A: Point;F: Point;D: Point;B: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(F, l);Intersection(l,C)={A,B};Intersection(l,Directrix(C))=D;Abs(LineSegmentOf(A, F)) = 6;VectorOf(D, B) = 2*VectorOf(B, F)",
    "query_expressions": "p",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 5]], [[6, 32], [40, 46], [59, 60]], [[129, 132]], [[48, 51]], [[35, 38]], [[64, 68]], [[52, 55]], [[13, 32]], [[6, 32]], [[6, 38]], [[0, 38]], [[0, 57]], [[0, 68]], [[70, 79]], [[82, 127]]]",
    "query_spans": "[[[129, 134]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{n}-\\frac{y^{2}}{12-n}=1$ is $\\sqrt{3}$. Then $n$=?",
    "fact_expressions": "G: Hyperbola;n: Number;Expression(G) = (-y^2/(12 - n) + x^2/n = 1);Eccentricity(G) = sqrt(3)",
    "query_expressions": "n",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 43]], [[60, 63]], [[2, 43]], [[2, 58]]]",
    "query_spans": "[[[60, 65]]]",
    "process": ""
  },
  {
    "text": "Given that $O$ is the coordinate origin, the parabola $C$: $x^{2}=2py$ ($p>0$) has focus $F$, $|OF|=\\frac{3}{4}$, and a line $l$ passing through point $F$ with slope $\\sqrt{3}$ intersects the parabola $C$ at points $A$ and $B$. Then $|AB|=$?",
    "fact_expressions": "l: Line;C: Parabola;p: Number;O: Origin;F: Point;A: Point;B: Point;p>0;Expression(C) = (x^2 = 2*(p*y));Focus(C) = F;Abs(LineSegmentOf(O, F)) = 3/4;PointOnCurve(F, l);Slope(l)=sqrt(3);Intersection(l, C) = {A,B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "12",
    "fact_spans": "[[[86, 91]], [[11, 37], [92, 98]], [[18, 37]], [[2, 5]], [[41, 45], [67, 71]], [[100, 103]], [[104, 107]], [[18, 37]], [[11, 37]], [[11, 44]], [[46, 65]], [[66, 91]], [[72, 91]], [[86, 109]]]",
    "query_spans": "[[[111, 120]]]",
    "process": "From |OF| = \\frac{3}{4}, we know \\frac{p}{2} = \\frac{3}{4}, so p = \\frac{3}{2}, thus the equation of the parabola is x^{2} = 3y, and the focus is F(0, \\frac{3}{4}). Therefore, the line l passing through point F with slope \\sqrt{3} has the equation y = \\sqrt{3}x + \\frac{3}{4}. Solving the system \\begin{cases} y = \\sqrt{3}x + \\frac{3}{4}, \\\\ x^{2} = 3y \\end{cases} by eliminating x gives 16y^{2} - 168y + 9 = 0. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), then y_{1} + y_{2} = \\frac{21}{2}. Using the focal property of x^{2} = 3y, |AB| = y_{1} + y_{2} + p = \\frac{21}{2} + \\frac{3}{2} = 12, hence |AB| = 12."
  },
  {
    "text": "The distance from the focus of the parabola $y^{2}=4 x$ to the asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ is?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 4*x);G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1)",
    "query_expressions": "Distance(Focus(H), Asymptote(G))",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[18, 46]], [[18, 46]]]",
    "query_spans": "[[[0, 55]]]",
    "process": "The asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ are $\\sqrt{3}x-y=0$. The distance from the focus $(1,0)$ of $y^{2}=4x$ to the asymptote is $\\frac{\\sqrt{3}}{2}$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through $F_{1}$ is tangent to the circle $x^{2}+y^{2}=a^{2}$ at point $T$, and intersects the right branch of the hyperbola $C$ at point $P$. If $F_{1} P=3 F_{1} T$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;G: Circle;F1: Point;P: Point;T: Point;F2: Point;a > b;b > 0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = a^2);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, l);TangentPoint(l, G) = T;Intersection(l, RightPart(C)) = P;LineSegmentOf(F1, P) = 3*LineSegmentOf(F1, T)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(13)/2",
    "fact_spans": "[[[94, 99], [129, 134]], [[2, 60], [135, 141], [173, 179]], [[10, 60]], [[10, 60]], [[100, 120]], [[69, 76], [86, 93]], [[146, 150]], [[123, 127]], [[77, 84]], [[10, 60]], [[10, 60]], [[2, 60]], [[100, 120]], [[2, 84]], [[2, 84]], [[85, 99]], [[94, 127]], [[129, 150]], [[152, 171]]]",
    "query_spans": "[[[173, 185]]]",
    "process": "[Analysis] Draw the graph, set up equations based on the property of proportional line segments and the definition of a hyperbola to find the relationship among a, b, c, then simplify to obtain the eccentricity. As shown in the figure, from the given conditions we have |OF_{1}| + |OF_{2}| = c, |OT| = a, so |F_{1}T| = b. Since \\overrightarrow{F_{1}P} = 3\\overrightarrow{F_{1}T}, it follows that |TP| = 2b, |F_{1}P| = 3b. Also, because |PF_{1}| - |PF_{2}| = 2a, we have |PF_{2}| = 3b - 2a. Construct F_{2}M // OT, yielding |F_{2}M| = 2a, |TM| = b. Thus, |PM| = b. In \\triangle MPF_{2}, |PM|^{2} + |MF_{2}|^{2} = |PF_{2}|^{2}; that is, b^{2} + (2a)^{2} = (3b - 2a)^{2}, leading to 2b = 3a. Also, since c^{2} = a^{2} + b^{2}, simplifying gives 4c^{2} = 13a^{2}."
  },
  {
    "text": "The eccentricity of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 34]]]",
    "process": "\\because a=1, b=\\sqrt{3} \\therefore c=\\sqrt{a^{2}+b^{2}}=2, e=\\frac{c}{a}=2"
  },
  {
    "text": "$M$ is a point on the parabola $C$: $y^{2}=4x$, $F$ is the focus of the parabola $C$, $O$ is the origin, and $|MF|=2$. $K$ is the intersection point of the directrix of the parabola $C$ and the $x$-axis. Then $\\angle MKO = ?$",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);M: Point;PointOnCurve(M, C) = True;F: Point;Focus(C) = F;O: Origin;Abs(LineSegmentOf(M,F)) = 2;K: Point;Intersection(Directrix(C),xAxis) = K",
    "query_expressions": "AngleOf(M,K,O)",
    "answer_expressions": "ApplyUnit(45,degree)",
    "fact_spans": "[[[4, 23], [31, 37], [65, 71]], [[4, 23]], [[0, 3]], [[0, 26]], [[27, 30]], [[27, 40]], [[41, 44]], [[50, 59]], [[61, 64]], [[61, 82]]]",
    "query_spans": "[[[83, 99]]]",
    "process": "Analysis: Let $ M(x_{1},y_{1}) $ ($ y_{1}>0 $), then $ M(1,2) $, so $ \\overrightarrow{KM}=(2,2) $, $ \\overrightarrow{KO}=(1,0) $, and the angle $ \\angle MKO $ can be found using the vector angle formula. Due to the symmetry of the parabola, without loss of generality, let $ M(x_{1},y_{1}) $ ($ y_{1}>0 $), then $ x_{1}+1=2 $, giving $ M(1,2) $. Since $ K(-1,0) $, $ O(0,0) $, we have $ \\overrightarrow{KM}=(2,2) $, $ \\overrightarrow{KO}=(1,0) $, thus $ \\overrightarrow{KM}\\cdot\\overrightarrow{KO}=2 $, $ |\\overrightarrow{KM}|=2\\sqrt{2} $, $ |\\overrightarrow{KO}|=1 $, $ \\cos\\angle MKO=\\cos\\langle\\overrightarrow{KM},\\overrightarrow{KO}\\rangle=\\frac{\\overrightarrow{KM}\\cdot\\overrightarrow{KO}}{|\\overrightarrow{KM}||\\overrightarrow{KO}|}=\\frac{2}{2\\sqrt{2}\\cdot1}=\\frac{\\sqrt{2}}{2} $, so $ \\angle MKO=45^{\\circ} $."
  },
  {
    "text": "The coordinates of the foci of the ellipse $9 x^{2}+25 y^{2}=225$ are?",
    "fact_expressions": "G: Ellipse;Expression(G) = (9*x^2 + 25*y^2 = 225)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(pm*4, 0)",
    "fact_spans": "[[[0, 24]], [[0, 24]]]",
    "query_spans": "[[[0, 31]]]",
    "process": "Rewriting the ellipse equation into standard form gives $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$. Then $a^{2}=25$, $b^{2}=9$, hence $c^{2}=a^{2}-b^{2}=16$, so $c=4$. Since the foci of the ellipse lie on the x-axis, the coordinates of the foci are $(4,0)$, $(-4,0)$."
  },
  {
    "text": "The line passing through the focus $F$ of the parabola $y = \\frac{1}{4}x^{2}$ intersects the parabola at points $A$ and $B$. Then the equation of the locus of the midpoint of segment $AB$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2/4);F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);A: Point;B: Point;Intersection(H, G) = {A, B}",
    "query_expressions": "LocusEquation(MidPoint(LineSegmentOf(A, B)))",
    "answer_expressions": "y=(1/2)*x^2+1",
    "fact_spans": "[[[1, 25], [35, 38]], [[1, 25]], [[27, 30]], [[1, 30]], [[31, 33]], [[0, 33]], [[39, 42]], [[43, 47]], [[31, 49]]]",
    "query_spans": "[[[51, 66]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\Gamma$: $\\frac{x^{2}}{3}+y^{2}=1$, respectively, and let points $A$ and $B$ lie on the ellipse $\\Gamma$, but are not vertices of the ellipse. If $\\overrightarrow{F_{1} A}+\\lambda \\overrightarrow{F_{2} B}=\\overrightarrow{0}$ and $\\lambda>0$, then what is the value of the real number $\\lambda$?",
    "fact_expressions": "Gamma: Ellipse;F1: Point;A: Point;F2: Point;B: Point;lambda: Real;Expression(Gamma) = (x^2/3 + y^2 = 1);LeftFocus(Gamma) = F1;RightFocus(Gamma) = F2;PointOnCurve(A, Gamma);PointOnCurve(B, Gamma);Negation(A = Vertex(Gamma));Negation(B = Vertex(Gamma));lambda*VectorOf(F2, B) + VectorOf(F1, A) = 0;lambda > 0",
    "query_expressions": "lambda",
    "answer_expressions": "1",
    "fact_spans": "[[[19, 56], [72, 82], [87, 89]], [[1, 8]], [[63, 67]], [[9, 16]], [[68, 71]], [[188, 199]], [[19, 56]], [[1, 62]], [[1, 62]], [[63, 83]], [[63, 83]], [[63, 92]], [[63, 92]], [[95, 173]], [[175, 186]]]",
    "query_spans": "[[[188, 203]]]",
    "process": "Since $\\overrightarrow{F_{1}A}+\\lambda\\overrightarrow{F_{2}B}=\\overrightarrow{0}$, it follows that $\\overrightarrow{F_{1}A}=\\lambda\\overrightarrow{BF}$, so $F_{1}A//BF_{2}$. Also $\\lambda>0$, and $A,B$ are not vertices of the ellipse. By the symmetry of the ellipse, the quadrilateral $F_{1}AF_{2}B$ must be a parallelogram. As shown in the figure: thus $F_{1}A=F_{2}B$, so $\\overrightarrow{F_{1}A}=\\overrightarrow{BF}$, hence $\\lambda=1$."
  },
  {
    "text": "Given that point $P$ is a moving point on the parabola $y^{2}=4x$, $F$ is the focus of the parabola, and the coordinates of point $A$ are $(-1,0)$, then the minimum value of $\\frac{PF}{PA}$ is?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;A: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (-1, 0);PointOnCurve(P, G);Focus(G) = F",
    "query_expressions": "Min(LineSegmentOf(P, F)/LineSegmentOf(P, A))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[7, 21], [29, 32]], [[2, 6]], [[25, 28]], [[36, 40]], [[7, 21]], [[36, 52]], [[2, 24]], [[25, 35]]]",
    "query_spans": "[[[54, 77]]]",
    "process": "From the given conditions, the focus of the parabola $ y^{2} = 4x $ is $ F(1,0) $, and the equation of the directrix is $ x = -1 $. Draw $ PM $ perpendicular to the directrix, with $ M $ as the foot of the perpendicular, as shown in the figure. By the definition of a parabola, $ |PF| = |PM| $, so $ \\frac{PF}{PA} = \\frac{PM}{PA} = \\sin\\angle PAM $. Since $ \\angle PAM $ is acute, $ \\frac{PF}{PA} $ is minimized when $ \\angle PAM $ is minimized, which occurs when $ PA $ is tangent to the parabola, thus minimizing $ \\frac{PF}{PA} $. Let the slope of line $ PA $ be $ k $, so the equation of line $ PA $ is $ y = k(x+1) $. Solving the system of equations:\n$$\n\\begin{cases}\ny = k(x+1) \\\\\ny^{2} = 4x\n\\end{cases}\n$$\nwe obtain $ k^{2}x^{2} + (2k^{2}-4)x + k^{2} = 0 $. Since the line is tangent to the parabola, the equation has exactly one real root, so the discriminant $ \\Delta = (2k^{2}-4)^{2} - 4 \\times k^{2} \\times k^{2} = 0 $. Solving gives $ k^{2} = 1 $, so $ k = \\pm1 $. Without loss of generality, let $ k = 1 $, then $ \\angle PAx = 45^{\\circ} $, $ \\angle PAM = 45^{\\circ} $, so $ \\sin 45^{\\circ} = \\frac{\\sqrt{2}}{2} $."
  },
  {
    "text": "The line passing through the left vertex $A$ of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with slope $k$ intersects the ellipse $C$ at another point $B$, and the projection of point $B$ onto the $x$-axis is exactly the right focus $F$. If $\\frac{1}{3}<k<\\frac{1}{2}$, then the range of the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;A:Point;B: Point;F: Point;e: Number;k:Number;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(C)=A;PointOnCurve(A,G);Slope(G) = k;1/3<k;k<1/2;Intersection(G,C) = {A, B};Projection(B, xAxis) = F;RightFocus(C)=F;Eccentricity(C) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "(1/2,2/3)",
    "fact_spans": "[[[1, 58], [76, 81], [142, 144]], [[8, 58]], [[8, 58]], [[73, 75]], [[62, 65]], [[85, 88], [90, 94]], [[108, 111]], [[148, 151]], [[69, 72]], [[8, 58]], [[8, 58]], [[1, 58]], [[1, 65]], [[0, 75]], [[66, 75]], [[113, 140]], [[113, 140]], [[62, 88]], [[90, 111]], [[76, 111]], [[142, 151]]]",
    "query_spans": "[[[148, 158]]]",
    "process": "From the given conditions, the coordinates of point A are $(-a, 0)$, and the coordinates of point B are $(c, \\frac{b^{2}}{a})$. Therefore, the slope $k$ of line AB is $k = \\frac{\\frac{b^{2}}{a}}{c+a} = \\frac{b^{2}}{ac+a^{2}} = \\frac{a^{2}-c^{2}}{a^{2}+ac} = \\frac{1-e^{2}}{1+e}$. Since $\\frac{1}{3} < k < \\frac{1}{2}$, it follows that $\\frac{1}{3} < \\frac{1-e^{2}}{1+e} < \\frac{1}{2}$, thus the range of eccentricity $e$ is $(\\frac{1}{2}, \\frac{2}{3})$."
  },
  {
    "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ are intercepted by the circle $x^{2}+y^{2}-6x+5=0$ to form a chord of length $2$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (-6*x + x^2 + y^2 + 5 = 0);Length(InterceptChord(Asymptote(G),H))=2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[2, 58], [97, 100]], [[5, 58]], [[5, 58]], [[63, 85]], [[5, 58]], [[5, 58]], [[2, 58]], [[63, 85]], [[2, 94]]]",
    "query_spans": "[[[97, 106]]]",
    "process": "The standard equation of the circle is (x-3)^{2}+y^{2}=4, the center is (3,0), the radius is r=2, one asymptote equation is bx-ay=0, the distance from the center to the asymptote is d=\\frac{3b}{\\sqrt{a^{2}+b^{2}}}, since the chord length is 2, then (\\frac{3b}{\\sqrt{a^{2}+b^{2}}})^{2}=2^{2}-1^{2}, therefore e=\\frac{c}{a}=\\frac{\\sqrt{6}}{2}"
  },
  {
    "text": "A point $P$ on the hyperbola $\\frac{x^{2}}{36}-\\frac{y^{2}}{16}=1$, with foci $F_{1}$, $F_{2}$, satisfies $P F_{1} \\perp P F_{2}$. Then the area of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/36 - y^2/16 = 1);PointOnCurve(P,G);Focus(G)={F1,F2};IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "16",
    "fact_spans": "[[[0, 40], [47, 48]], [[43, 46]], [[51, 58]], [[60, 67]], [[0, 40]], [[0, 46]], [[47, 67]], [[70, 93]]]",
    "query_spans": "[[[95, 125]]]",
    "process": "Let $ PF_{1} = m $, $ PF_{2} = n $. By the definition of hyperbola and Pythagorean theorem, we have $ m - n = 12 $, $ m^{2} + n^{2} = 208 $, solve to get $ mn = 32 $, then the area can be obtained. Let $ P $ be a point on the right branch of the hyperbola, $ PF_{1} = m $, $ PF_{2} = n $. From the hyperbola equation, we get $ a = 6 $, $ b = 4 $, $ c = \\sqrt{a^{2} + b^{2}} = 2\\sqrt{13} $. Then by the definition of hyperbola, $ m - n = 2a = 12 $. Since $ PF_{1} \\bot PF_{2} $, $ \\therefore m^{2} + n^{2} = (2c)^{2} = 208 $. Then $ (m - n)^{2} = m^{2} + n^{2} - 2mn = 208 - 2mn = 144 $, solving gives $ mn = 32 $. $ \\therefore S_{\\Delta PF_{1}F_{2}} = \\frac{1}{2}mn = 16 $."
  },
  {
    "text": "Let $M$ be a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and let $F_{1}$, $F_{2}$ be the left and right foci of the ellipse, respectively. If $|M F_{1}|=4|M F_{2}|$, then the coordinates of point $M$ are?",
    "fact_expressions": "G: Ellipse;M: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(M, G);LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(M, F1)) = 4*Abs(LineSegmentOf(M, F2))",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(5, 0)",
    "fact_spans": "[[[5, 44], [68, 70]], [[1, 4], [102, 106]], [[49, 56]], [[57, 64]], [[5, 44]], [[1, 48]], [[49, 76]], [[49, 76]], [[78, 100]]]",
    "query_spans": "[[[102, 111]]]",
    "process": ""
  },
  {
    "text": "Given circle $C$: $(x+1)^{2}+y^{2}=25$ and point $A(1,0)$, $Q$ is a point on the circle, the perpendicular bisector of $AQ$ intersects $CQ$ at $M$, then the equation of the trajectory of point $M$ is?",
    "fact_expressions": "C: Circle;Expression(C) = (y^2 + (x + 1)^2 = 25);A: Point;Coordinate(A) = (1, 0);Q: Point;PointOnCurve(Q, C) = True;Intersection(PerpendicularBisector(LineSegmentOf(A, Q)), LineSegmentOf(C, Q)) = M;M: Point",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "4*x^2/25+4*y^2/21=1",
    "fact_spans": "[[[2, 27], [43, 44]], [[2, 27]], [[28, 37]], [[28, 37]], [[39, 42]], [[39, 47]], [[48, 67]], [[64, 67], [69, 73]]]",
    "query_spans": "[[[69, 80]]]",
    "process": "From the given condition, |MA| + |MC| = |MQ| + |MC| = R = 5 > |AC|, therefore the trajectory of point M is an ellipse with foci at A and C. Thus, a = \\frac{5}{2}, c = 1, therefore b^{2} = \\frac{21}{4}, and the equation is \\frac{4x^{2}}{25} + \\frac{4y^{2}}{21} = 1"
  },
  {
    "text": "The line $l$ passes through the focus $F$ of the parabola $y^{2}=4x$, intersecting the parabola at points $A$ and $B$. Then $\\frac{1}{|A F|}+\\frac{1}{|B F|}$=?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B}",
    "query_expressions": "1/Abs(LineSegmentOf(B, F)) + 1/Abs(LineSegmentOf(A, F))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 5]], [[7, 21], [29, 32]], [[33, 36]], [[24, 27]], [[37, 40]], [[7, 21]], [[7, 27]], [[0, 27]], [[0, 42]]]",
    "query_spans": "[[[44, 79]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a moving point on the parabola $y^{2}=2x$, $A\\left(\\frac{7}{2}, 4\\right)$, if the distance from point $P$ to the $y$-axis is $d_{1}$, and the distance from point $P$ to point $A$ is $d_{2}$, then the minimum value of $d_{1}+d_{2}$ is?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;Expression(G) = (y^2 = 2*x);Coordinate(A) = (7/2, 4);PointOnCurve(P, G);Distance(P, yAxis) = d1;Distance(P, A) = d2;d1:Number;d2:Number",
    "query_expressions": "Min(d1 + d2)",
    "answer_expressions": "9/2",
    "fact_spans": "[[[6, 20]], [[24, 43], [71, 75]], [[45, 49], [2, 5], [66, 70]], [[6, 20]], [[24, 43]], [[2, 23]], [[45, 65]], [[66, 86]], [[58, 65]], [[79, 86]]]",
    "query_spans": "[[[88, 107]]]",
    "process": "From the equation of the parabola and the coordinates of point A, it can be determined that point A lies outside the parabola. By the definition of a parabola, we obtain $ d_{1} = PF - \\frac{1}{2} $, and thus $ d_{1} + d_{2} = PF + PA - \\frac{1}{2} $. From the figure, it is clear that the minimum value occurs when points P, F, and A are collinear, which equals $ AF - \\frac{1}{2} $. The result can then be calculated using the distance formula between two points. Since $ y^{2} = 2 \\times \\frac{7}{2} = 7 < 4^{2} $, point A lies outside the parabola. Because point P lies on the parabola, $ d_{1} = PF - \\frac{1}{2} $ (where point F is the focus of the parabola), so $ d_{1} + d_{2} = PF + PA - \\frac{1}{2} \\geqslant AF - \\frac{1}{2} = \\sqrt{(\\frac{7}{2} - \\frac{1}{2})^{2} + 4^{2}} - \\frac{1}{2} = 5 - \\frac{1}{2} = \\frac{9}{2} $, with equality holding if and only if point P is the intersection point of segment AF and the parabola. This problem primarily examines the equation and definition of a parabola, testing analytical solving and transformation abilities, and is a basic-level question. When finding the distance from a point on the parabola to the directrix, note the mutual conversion with the distance from the point on the parabola to the focus."
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the ellipse $C$: $\\frac{x^{2}}{4}+y^{2}=1$, and let $M$ be a point on $C$. The ordinate of the incenter $I$ of $\\Delta M F_{1} F_{2}$ is $2-\\sqrt{3}$. Then the cosine value of $\\angle F_{1} M F_{2}$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2 = 1);F1: Point;F2: Point;Focus(C) = {F1,F2};M: Point;PointOnCurve(M,C);I: Point;Incenter(TriangleOf(M,F1,F2)) = I;YCoordinate(I) = 2-sqrt(3)",
    "query_expressions": "Cos(AngleOf(F1, M, F2))",
    "answer_expressions": "0",
    "fact_spans": "[[[17, 49], [58, 61]], [[17, 49]], [[1, 8]], [[9, 16]], [[1, 54]], [[54, 57]], [[54, 63]], [[89, 92]], [[64, 92]], [[89, 109]]]",
    "query_spans": "[[[111, 139]]]",
    "process": "As shown in the figure, from the given conditions, the inradius of triangle $ AMF_{1}F_{2} $ is $ 2-\\sqrt{3} $. Also, since the inradius of a triangle is given by $ r = \\frac{2S}{\\text{perimeter}} $, we have $ S = \\frac{1}{2}(2-\\sqrt{3})(4+2\\sqrt{3}) = (2-\\sqrt{3})(2+\\sqrt{3}) = 1 $. Furthermore, the area of the focal triangle is $ S = b^{2}\\tan\\left(\\frac{1}{2}\\angle F_{1}MF_{2}\\right) = \\tan\\left(\\frac{1}{2}\\angle F_{1}MF_{2}\\right) $, so $ \\tan\\left(\\frac{1}{2}\\angle F_{1}MF_{2}\\right) = 1 $, hence $ \\angle F_{1}MF_{2} = \\frac{\\pi}{2} $, therefore $ \\cos\\angle F_{1}MF_{2} = ( $"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$, a line $l$ passing through point $F$ intersects $C$ at points $A$ and $B$. If $|AF| \\cdot |BF| \\leq 32$, then what is the range of the inclination angle of line $l$?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;F: Point;B: Point;Expression(C) = (y^2 = 8*x);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, F))*Abs(LineSegmentOf(B, F)) <= 32",
    "query_expressions": "Range(Inclination(l))",
    "answer_expressions": "[\\pi/4, 3\\pi/4]",
    "fact_spans": "[[[35, 40], [85, 90]], [[2, 21], [41, 44]], [[46, 49]], [[25, 28], [30, 34]], [[50, 53]], [[2, 21]], [[2, 28]], [[29, 40]], [[35, 55]], [[57, 83]]]",
    "query_spans": "[[[85, 98]]]",
    "process": "From the equation of the parabola, we obtain F(2,0). When the slope of the line does not exist, A(2,4), B(2,-4), and |AF|·|BF|=4×4=16, which satisfies the condition. When the slope of the line exists, let the line l: y=k(x-2), A(x_{1},y_{1}), B(x_{2},y_{2}). From \\begin{cases}y=k(x-2)\\\\y^{2}=8x\\end{cases}, we get k^{2}x^{2}-4(k^{2}+2)x+4k^{2}=0. Then Δ=64k^{2}+64>0 and x_{1}+x_{2}=4, x_{1}x_{2}=4. Hence |AF|=x_{1}+2, |BF|=x_{2}+2, |AF|·|BF|=x_{1}x_{2}+2(x_{1}+x_{2})+4=\\frac{16}{k^{2}}+16\\leqslant32. Solving gives k\\geqslant1 or k\\leqslant-1. Therefore, the range of the inclination angle of the line is [\\frac{\\pi}{4},\\frac{3\\pi}{4}]."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$, a line passing through point $F$ intersects the parabola at points $A$ and $B$, and intersects the directrix $l$ of the parabola at point $D$. If $B$ is the midpoint of $AD$, then the length of chord $|AB|$ is?",
    "fact_expressions": "C: Parabola;G: Line;A: Point;D: Point;B: Point;F: Point;l: Line;Expression(C) = (y^2 = 8*x);Focus(C) = F;PointOnCurve(F, G);Intersection(G, C) = {A, B};Directrix(C) = l;Intersection(G, l) = D;MidPoint(LineSegmentOf(A, D)) = B;IsChordOf(LineSegmentOf(A, B), C)",
    "query_expressions": "Length(Abs(LineSegmentOf(A,B)))",
    "answer_expressions": "9",
    "fact_spans": "[[[2, 21], [38, 41], [52, 55]], [[35, 37]], [[42, 46]], [[61, 65]], [[47, 50], [67, 70]], [[25, 28], [30, 34]], [[57, 60]], [[2, 21]], [[2, 28]], [[29, 37]], [[35, 50]], [[52, 60]], [[35, 65]], [[67, 79]], [[52, 89]]]",
    "query_spans": "[[[82, 93]]]",
    "process": "According to the problem, the parabola has $ p = 4 $, focus $ F(2,0) $, and the equation of the directrix $ l $ is: $ x = -2 $. Since line $ AB $ passes through $ F(2,0) $ and intersects the $ y $-axis, the slope of line $ AB $ exists. Let the slope of line $ AB $ be $ k $, then the equation of line $ AB $ is $ y = k(x - 2) $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, with $ x_{1} > 0 $, $ x_{2} > 0 $. From\n$$\n\\begin{cases}\ny = k(x - 2) \\\\\ny^{2} = 8x\n\\end{cases}\n$$\neliminating $ y $ and simplifying yields $ k^{2}x^{2} - (4k^{2} + 8)x + 4k^{2} = 0 $, so\n$$\n\\begin{cases}\ny^{2} = 8x \\\\\nx_{1} + x_{2} = \\frac{4k^{2} + 8}{k^{2}} \\\\\nx_{1} \\cdot x_{2} = 4\n\\end{cases}\n$$\nDraw $ AM \\perp l $ from $ A $, intersecting $ l $ at $ M $; draw $ BN \\perp l $ from $ B $, intersecting $ l $ at $ N $. According to the definition of the parabola, $ |AF| = |AM| = x_{1} + \\frac{p}{2} = x_{1} + 2 $, $ |BF| = |BN| = x_{2} + \\frac{p}{2} = x_{2} + 2 $. Since $ B $ is the midpoint of segment $ AD $, $ BN $ is the midline of triangle $ DAM $, so $ |AM| = 2|BN| $, i.e., $ x_{1} + 2 = 2(x_{2} + 2) \\Rightarrow x_{1} = 2x_{2} + 2 $. Substituting into $ x_{1} \\cdot x_{2} = 4 $ gives $ (2x_{2} + 2) \\cdot x_{2} = 4 $, solving yields $ x_{2} = 1 $ (negative root discarded), so $ x_{1} = 2x_{2} + 2 = 4 $. Therefore, $ |AB| = x_{1} + x_{2} + p = 4 + 1 + 4 = 9 $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, and point $M$ lies on $C$, then the maximum value of $|M F_{1}| \\cdot|M F_{2}|$ is?",
    "fact_expressions": "C: Ellipse;M: Point;F1: Point;F2: Point;Expression(C) = (x^2/16 + y^2/9 = 1);Focus(C) = {F1, F2};PointOnCurve(M, C)",
    "query_expressions": "Max(Abs(LineSegmentOf(M, F1))*Abs(LineSegmentOf(M, F2)))",
    "answer_expressions": "16",
    "fact_spans": "[[[18, 61], [72, 75]], [[67, 71]], [[2, 9]], [[10, 17]], [[18, 61]], [[2, 66]], [[67, 76]]]",
    "query_spans": "[[[78, 110]]]",
    "process": "By the definition of an ellipse, we have: |MF_{1}| + |MF_{2}| = 2a = 8. By the AM-GM inequality, |MF_{1}| \\cdot |MF_{2}| \\leqslant \\left( \\frac{|MF_{1}| + |MF_{2}|}{2} \\right)^{2} = 16, and the equality holds if and only if |MF_{1}| = |MF_{2}|. Therefore, the maximum value of |MF_{1}| \\cdot |MF_{2}| is 16."
  },
  {
    "text": "Let the parabola $C$: $y^{2}=4 x$ have focus $F$. The line passing through the point $(-2,0)$ with slope $\\frac{2}{3}$ intersects $C$ at points $M$ and $N$. Then $\\overrightarrow{F M} \\cdot  \\overrightarrow{F N}$=?",
    "fact_expressions": "C: Parabola;G: Line;H: Point;F: Point;M: Point;N: Point;Expression(C) = (y^2 = 4*x);Coordinate(H) = (-2, 0);Focus(C)=F;PointOnCurve(H, G);Slope(G) = 2/3;Intersection(G, C) = {M, N}",
    "query_expressions": "DotProduct(VectorOf(F, M), VectorOf(F, N))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 20], [59, 62]], [[56, 58]], [[29, 38]], [[24, 27]], [[64, 67]], [[68, 71]], [[1, 20]], [[29, 38]], [[1, 27]], [[28, 58]], [[39, 58]], [[56, 73]]]",
    "query_spans": "[[[75, 127]]]",
    "process": "The focus of the parabola $ C: y^{2} = 4x $ is $ F(1,0) $. The line passing through the point $ (-2,0) $ with slope $ \\frac{2}{3} $ is: $ 3y = 2x + 4 $. Combining the line and the parabola $ C: y^{2} = 4x $, eliminating $ x $ yields: $ y^{2} \\cdot 6y + 8 = 0 $. Solving gives $ y_{1} = 2 $, $ y_{2} = 4 $. Let $ M(1,2) $, $ N(4,4) $, then $ \\overrightarrow{FM} = (0,2) $, $ \\overrightarrow{FN} = (3,4) $, so $ \\overrightarrow{FM} \\cdot \\overrightarrow{FN} = (0,2) \\cdot (3,4) = 8 $."
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the left and right foci of $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, $M \\in E$, and $M F_{2} \\perp F_{1} F_{2}$, $\\angle M F_{1} F_{2}=30^{\\circ}$, then the eccentricity $e$ of $E$ is?",
    "fact_expressions": "M: Point;F2: Point;F1: Point;E:Curve;e: Number;Expression(E)=(x^2/a^2-y^2/b^2=1);In(M, E) = True;LeftFocus(E)=F1;RightFocus(E)=F2;IsPerpendicular(LineSegmentOf(M, F2), LineSegmentOf(F1, F2));AngleOf(M, F1, F2) = ApplyUnit(30, degree);Eccentricity(E) = e;b:Number;a:Number",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[69, 78]], [[8, 15]], [[0, 7]], [[16, 64], [143, 146]], [[150, 153]], [[16, 64]], [[69, 78]], [[0, 68]], [[0, 68]], [[80, 107]], [[108, 141]], [[143, 153]], [[16, 64]], [[16, 64]]]",
    "query_spans": "[[[150, 155]]]",
    "process": ""
  },
  {
    "text": "The two endpoints $A$ and $B$ of a line segment $AB$ of length $a$ both slide along the parabola $y^{2}=2 P x$ $(P>0, a>2 P)$. Then, the shortest distance from the midpoint $M$ of segment $AB$ to the $y$-axis is?",
    "fact_expressions": "G: Parabola;P: Number;A: Point;B: Point;M: Point;a:Number;P>0;a>2*P;Expression(G) = (y^2 = 2*P*x);Endpoint(LineSegmentOf(A,B))={A,B};Length(LineSegmentOf(A,B))=a;PointOnCurve(A, G);PointOnCurve(B,G);MidPoint(LineSegmentOf(A,B))=M",
    "query_expressions": "Min(Distance(M, yAxis))",
    "answer_expressions": "(a-p)/2",
    "fact_spans": "[[[28, 56]], [[31, 56]], [[19, 22]], [[23, 26]], [[72, 75]], [[3, 6]], [[31, 56]], [[31, 56]], [[28, 56]], [[7, 26]], [[0, 14]], [[19, 57]], [[19, 57]], [[61, 75]]]",
    "query_spans": "[[[72, 87]]]",
    "process": ""
  },
  {
    "text": "Draw a line $l$ through the focus of the parabola $y^{2}=4x$, intersecting the parabola at points $A$ and $B$. If the x-coordinate of the midpoint of segment $AB$ is $4$, then $|AB|$ equals?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);l: Line;PointOnCurve(Focus(G), l);A: Point;B: Point;Intersection(l, G) = {A, B};XCoordinate(MidPoint(LineSegmentOf(A, B))) = 4",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "10",
    "fact_spans": "[[[1, 15], [26, 29]], [[1, 15]], [[19, 24]], [[0, 24]], [[30, 33]], [[34, 37]], [[19, 39]], [[41, 58]]]",
    "query_spans": "[[[60, 70]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Since the horizontal coordinate of the midpoint of segment AB is 4, we have: $ \\frac{x_{1}+x_{2}}{2}=4 $, so $ x_{1}+x_{2}=8 $. By the focal chord length formula of the parabola, we get: $ |AB|=x_{1}+x_{2}+p=8+2=10 $. Hence, fill in: 10. [Note] This question mainly examines the focal chord length formula of a parabola $ |AB|=x_{1}+x_{2}+p $ and the midpoint coordinate formula $ \\frac{x_{1}+x_{2}}{2}=x+ $"
  },
  {
    "text": "What is the equation of the line passing through point $A(4,-1)$ and the right focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$?",
    "fact_expressions": "G: Hyperbola;H: Line;A: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Coordinate(A) = (4, -1);PointOnCurve(A,H);PointOnCurve(RightFocus(G),H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x-y-5=0",
    "fact_spans": "[[[12, 51]], [[55, 57]], [[1, 11]], [[12, 51]], [[1, 11]], [[0, 57]], [[0, 57]]]",
    "query_spans": "[[[55, 61]]]",
    "process": ""
  },
  {
    "text": "It is known that $F$ is the focus of the parabola $y = \\frac{1}{8}x^{2}$, and $P$ is a moving point on this parabola. Then the equation of the trajectory of the midpoint of $PF$ is?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;Expression(G) = (y = x^2/8);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "LocusEquation(MidPoint(LineSegmentOf(P, F)))",
    "answer_expressions": "x^2-4*y+4=0",
    "fact_spans": "[[[6, 30], [39, 42]], [[34, 37]], [[2, 5]], [[6, 30]], [[2, 33]], [[34, 46]]]",
    "query_spans": "[[[48, 62]]]",
    "process": "Convert $ y = \\frac{1}{8}x^2 $ into $ x^{2} = 8y $, so $ F(0,2) $. Let point $ P(a,b) $ be on the parabola, and the midpoint of $ PF $ be $ M(x,y) $, then $ a = 2x $, $ b = 2y - 2 $, then $ (2x)^{2} = 8(2y - 2) $, thus the trajectory equation of the midpoint of $ PF $ is $ x^2 - 4y + 4 = 0 $."
  },
  {
    "text": "The right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $F_{1}(2 \\sqrt{2}, 0)$, the coordinates of point $A$ are $(0,1)$, and point $P$ is a moving point on the left branch of the hyperbola. The minimum perimeter of $\\Delta A P F_{1}$ is $8$. Then, the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F1: Point;A: Point;P: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (0, 1);Coordinate(F1) = (2*sqrt(2), 0);RightFocus(G) = F1;PointOnCurve(P, LeftPart(G));Min(Perimeter(TriangleOf(A, P, F1))) = 8",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[0, 56], [105, 108], [146, 149]], [[3, 56]], [[3, 56]], [[61, 83]], [[84, 88]], [[100, 104]], [[3, 56]], [[3, 56]], [[0, 56]], [[84, 99]], [[61, 83]], [[0, 83]], [[100, 114]], [[116, 144]]]",
    "query_spans": "[[[146, 155]]]",
    "process": "Let the left focus of the hyperbola be $ F_{2}(-2\\sqrt{2},0) $, and $ |AF_{1}| = 3 $. Therefore, the perimeter of $ \\triangle APF $ is $ l = |AF_{1}| + |PF_{1}| + |AP| = 3 + |PF_{1}| + |AP| $. By the definition of the hyperbola, $ |PF_{1}| - |PF_{2}| = 2a $, so $ |PF_{1}| = |PF_{2}| + 2a $, and thus $ l = 3 + 2a + |PF_{2}| + |AP| $. When points $ A $, $ P $, and $ F_{2} $ are collinear, the perimeter $ l $ reaches its minimum value, at which point $ |PF_{2}| + |AP| = |AF_{2}| = 3 $, so $ 3 + 2a + 3 = 8 $. Solving gives $ a = 1 $, hence $ e = \\frac{c}{a} = 2\\sqrt{2} $."
  },
  {
    "text": "If the ellipse $C$ has coordinate axes as its axes of symmetry, foci on the $y$-axis, an eccentricity of $\\frac{4}{5}$, and an area of $20\\pi$, then what is the standard equation of the ellipse $C$?",
    "fact_expressions": "C: Ellipse;SymmetryAxis(C) = axis;PointOnCurve(Focus(C), yAxis);Eccentricity(C) = 4/5;Area(C) = 20*pi",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/(100/3) + x^2/12 = 1",
    "fact_spans": "[[[1, 6], [25, 30], [62, 67]], [[1, 14]], [[1, 23]], [[25, 48]], [[25, 60]]]",
    "query_spans": "[[[62, 74]]]",
    "process": ""
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ has two foci $F_{1}$, $F_{2}$. A line passes through $F_{1}$ and intersects the ellipse at points $P$, $Q$. Then, what is the perimeter of $\\triangle P Q F_{2}$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);F2: Point;F1: Point;Focus(G) = {F1,F2};L: Line;PointOnCurve(F1,L) = True;Intersection(L,G) = {P,Q};P: Point;Q: Point",
    "query_expressions": "Perimeter(TriangleOf(P, Q, F2))",
    "answer_expressions": "20",
    "fact_spans": "[[[0, 38], [75, 77]], [[0, 38]], [[53, 60]], [[43, 50], [67, 74]], [[0, 60]], [[64, 66]], [[64, 74]], [[64, 85]], [[78, 81]], [[82, 85]]]",
    "query_spans": "[[[87, 113]]]",
    "process": ""
  },
  {
    "text": "Given that the major axis of ellipse $C$ is $4$ and the minor axis is $3$, what is the eccentricity of $C$?",
    "fact_expressions": "C: Ellipse;Length(MajorAxis(C))=4;Length(MinorAxis(C))=3",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(7)/4",
    "fact_spans": "[[[2, 7], [25, 28]], [[2, 15]], [[2, 23]]]",
    "query_spans": "[[[25, 34]]]",
    "process": "Since $\\frac{b}{a}=\\frac{2b}{2a}=\\frac{3}{4}$, it follows that $e=\\sqrt{1-(\\frac{b}{a})^{2}}=\\sqrt{1-\\frac{9}{16}}=\\frac{\\sqrt{7}}{4}$"
  },
  {
    "text": "Given that the parabola $C$ has focus $F(0 , 1)$ and the $x$-axis as directrix, then the equation of this parabola is?",
    "fact_expressions": "C: Parabola;F: Point;Coordinate(F) = (0, 1);Focus(C) = F;Directrix(C) = xAxis",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 = 2*y - 1",
    "fact_spans": "[[[2, 8], [33, 36]], [[9, 19]], [[9, 19]], [[2, 22]], [[2, 30]]]",
    "query_spans": "[[[33, 41]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has an eccentricity of $\\frac{\\sqrt{3}}{3}$, and a line $l$ intersects the ellipse $C$ at points $A$ and $B$, with the midpoint of segment $AB$ being $M(3,2)$. Then, what is the slope of line $l$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Eccentricity(C) = sqrt(3)/3;l: Line;Intersection(l, C) = {A, B};A: Point;B: Point;M: Point;Coordinate(M) = (3, 2);MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Slope(l)",
    "answer_expressions": "-1",
    "fact_spans": "[[[2, 59], [91, 96]], [[2, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[2, 84]], [[85, 90], [129, 134]], [[85, 107]], [[98, 101]], [[102, 105]], [[119, 127]], [[119, 127]], [[108, 127]]]",
    "query_spans": "[[[129, 139]]]",
    "process": "From the given conditions, we have $ e = \\frac{c}{a} = \\sqrt{1 - \\frac{b^{2}}{a^{2}}} = \\frac{\\sqrt{3}}{3} $, simplifying yields $ a = \\frac{\\sqrt{6}}{2}b $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ \\frac{x_{1}^{2}}{a^{2}} + \\frac{y_{1}^{2}}{b^{2}} = 1 $, subtracting the two equations gives $ \\frac{(x_{1} - x_{2})(x_{1} + x_{2})}{a^{2}} + \\frac{(y_{1} - y_{2})(y_{1} + y_{2})}{b^{2}} = 0 $. Since the midpoint of $ AB $ is $ M(3, 2) $, $ \\therefore x_{1} + x_{2} = 6 $, $ y_{1} + y_{2} = 4 $, then the slope of the line $ k = \\frac{y_{1} - y_{2}}{x_{1} - x_{2}} = -\\frac{b^{2}}{a^{2}} \\cdot \\frac{x_{1} + x_{2}}{y_{1} + y_{2}} = -\\frac{b^{2}}{a^{2}} \\cdot \\frac{6}{4} = -\\frac{2}{3} \\cdot \\frac{6}{4} = -1 $."
  },
  {
    "text": "The eccentricity $e$ of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ satisfies $e \\in[\\frac{\\sqrt{2}}{2} , 1)$, then the range of values for $m$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/m = 1);m: Number;Eccentricity(G) = e;In(e, [sqrt(2)/2, 1));e: Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(0, 2]+[8, +\\infty)",
    "fact_spans": "[[[0, 37]], [[0, 37]], [[74, 77]], [[0, 72]], [[41, 72]], [39, 68]]",
    "query_spans": "[[[74, 84]]]",
    "process": ""
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{a^{2}}=1$ and the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{2}=1$ have the same foci, then $a$=?",
    "fact_expressions": "H: Ellipse;Expression(H) = (x^2/4 + y^2/a^2 = 1);G: Hyperbola;Expression(G) = (-y^2/2 + x^2/a^2 = 1);a: Number;Focus(H) = Focus(G)",
    "query_expressions": "a",
    "answer_expressions": "{1,-1}",
    "fact_spans": "[[[0, 41]], [[0, 41]], [[42, 84]], [[42, 84]], [[91, 94]], [[0, 89]]]",
    "query_spans": "[[[91, 96]]]",
    "process": ""
  },
  {
    "text": "The equation of the directrix of the parabola $y=a x^{2}$ is $y=\\frac{1}{32}$. Then the value of $a$ is?",
    "fact_expressions": "G: Parabola;a: Number;Expression(G) = (y = a*x^2);Expression(Directrix(G)) = (y = 1/32)",
    "query_expressions": "a",
    "answer_expressions": "-8",
    "fact_spans": "[[[0, 14]], [[38, 41]], [[0, 14]], [[0, 36]]]",
    "query_spans": "[[[38, 45]]]",
    "process": "First, rewrite the parabola equation into standard form, then derive the directrix equation, and subsequently obtain the directrix equation based on the properties of the parabola. Specifically, rearranging the parabola equation yields $ x^{2} = \\frac{1}{a}y $, $\\therefore$ the directrix equation is $ p = \\frac{1}{4a} = \\frac{1}{32} $. Since the parabola opens downward, $\\therefore$ the parameter value is $-8$."
  },
  {
    "text": "Given that $M$ is a moving point on the circle $x^{2}+y^{2}=9$, $A(6,0)$ is a fixed point, and the moving point $P$ satisfies $\\overrightarrow{A P}=2 \\overrightarrow{P M}$, then the trajectory equation of the moving point $P$ is?",
    "fact_expressions": "G: Circle;A: Point;P: Point;M: Point;Expression(G) = (x^2 + y^2 = 9);Coordinate(A) = (6, 0);PointOnCurve(M, G);VectorOf(A, P) = 2*VectorOf(P, M)",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(x-2)^2+y^2=4",
    "fact_spans": "[[[6, 22]], [[27, 35]], [[42, 45], [96, 99]], [[2, 5]], [[6, 22]], [[27, 35]], [[2, 26]], [[47, 92]]]",
    "query_spans": "[[[96, 106]]]",
    "process": "Let point M(x,y), P(a,b), then x^{2}+y^{2}=9, \\overrightarrow{AP}=2\\overrightarrow{PM}, that is, (a-6,b)=2(x-a,y-b).  \n\\begin{cases}a-6=2x-2a\\\\b=2y-2b\\end{cases}, solving yields \\begin{cases}x=\\frac{3a-6}{2}\\\\y=\\frac{3b}{2}\\end{cases}. Substituting into x^{2}+y^{2}=9, (\\frac{3a-6}{2})^{2}+(\\frac{3b}{2})^{2}=9, simplifying gives: (a-2)^{2}+b^{2}=4, i.e., the moving point P(a,b) satisfies (a-2)^{2}+b^{2}=4. Therefore, the trajectory equation of the moving point P is (x-2)^{2}+y^{2}=4."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$, a line $l$ passing through $F_{2}$ with slope $k$ ($k<-1$) intersects the right branch of the hyperbola at points $P$ and $Q$. If the area of $\\Delta F_{1} P Q$ is $4$, then the coordinates of the intersection point of line $l$ with the $y$-axis are?",
    "fact_expressions": "l: Line;G: Hyperbola;F1: Point;P: Point;Q: Point;F2: Point;Expression(G) = (x^2/2 - y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, l);Slope(l) = k;k:Number;k<-1;Intersection(l, RightPart(G)) = {P, Q};Area(TriangleOf(F1, P, Q)) = 4",
    "query_expressions": "Coordinate(Intersection(l, yAxis))",
    "answer_expressions": "(0,\\sqrt{6})",
    "fact_spans": "[[[75, 80], [125, 130]], [[18, 46], [81, 84]], [[2, 9]], [[87, 90]], [[91, 94]], [[10, 17], [54, 61]], [[18, 46]], [[2, 52]], [[2, 52]], [[53, 80]], [[62, 80]], [[65, 74]], [[65, 74]], [[75, 96]], [[98, 123]]]",
    "query_spans": "[[[125, 142]]]",
    "process": "By the given conditions and the properties of the hyperbola, we have $F_{1}(-\\sqrt{3},0)$, $F_{2}(\\sqrt{3},0)$. Let the equation of line $l$ be $x=my+\\sqrt{3}$, where $m=\\frac{1}{k}\\in(-1,0)$, and let $P(x_{1},y_{1})$, $Q(x_{2},y_{2})$. By solving the system of equations and using Vieta's formulas, we can obtain $y_{1}+y_{2}$ and $y_{1}y_{2}$. Then, using $S_{\\triangle F_{1}PQ}=\\frac{1}{2}|F_{1}F_{2}|\\cdot|y_{1}-y_{2}|$, we can solve for $m$, thus obtaining the solution.  \nSolution: From the given conditions, $F_{1}(-\\sqrt{3},0)$, $F_{2}(\\sqrt{3},0)$. Let the equation of line $l$ be $x=my+\\sqrt{3}$, where $m=\\frac{1}{k}\\in(-1,0)$. Let $P(x_{1},y_{1})$, $Q(x_{2},y_{2})$. Then  \n$$\n\\begin{cases}\nx=my+\\sqrt{3} \\\\\n\\frac{x^{2}}{2}-y^{2}=1\n\\end{cases}\n$$\nEliminating $x$, we get $(m^{2}-2)y^{2}+2\\sqrt{3}my+1=0$, $4>0$, so $y_{y_{5}}+y_{2}e=\\frac{2}{2}|\\frac{\\sqrt{3}m}{2FFF_{2}},\\sqrt{x}|y_{y}=\\frac{1}{m_{2}-2}\\times\\sqrt{3}|y_{1}-y_{2}|=\\sqrt{3}.\\sqrt{(\\frac{2\\sqrt{3}}{1})^{1}y_{2}^{2}-4yy}.2$ Therefore, $\\frac{\\sqrt{m^{2}+1}}{\\sqrt{(m^{2}-2)^{2}}}=\\frac{\\sqrt{6}}{3}$, i.e., $2(m^{2}-2)^{2}=3(m^{2}+1)$. Solving gives $m^{2}=\\frac{1}{2}$ or $m^{2}=5$ (discarded). Thus, $m=-\\frac{\\sqrt{2}}{2}$ or $m=\\frac{\\sqrt{2}}{2}$ (discarded). So $x=-\\frac{\\sqrt{2}}{2}y+\\sqrt{3}$, i.e., $y=-\\sqrt{2}x+\\sqrt{6}$. Therefore, the intersection point of line $l$ with the $y$-axis is $(0,\\sqrt{6})$."
  },
  {
    "text": "Given that the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ passes through the point $(2 , 1)$, what is the range of values for $a$?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(H) = (2, 1);PointOnCurve(H, G)",
    "query_expressions": "Range(a)",
    "answer_expressions": "(\\sqrt{5}, +\\infty)",
    "fact_spans": "[[[2, 54]], [[4, 54]], [[67, 70]], [[55, 65]], [[4, 54]], [[4, 54]], [[2, 54]], [[55, 65]], [[2, 65]]]",
    "query_spans": "[[[67, 77]]]",
    "process": ""
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ are $F_{1}$ and $F_{2}$, respectively. If there exists a point $P$ on the ellipse such that the segment $P F_{1}$ is tangent to the circle having the minor axis of the ellipse as its diameter, and the point of tangency is exactly the midpoint of the segment $P F_{1}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);H: Circle;IsDiameter(MinorAxis(G), H);IsTangent(LineSegmentOf(P, F1), H);TangentPoint(LineSegmentOf(P, F1), H) = MidPoint(LineSegmentOf(P, F1))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 52], [78, 80], [134, 136], [101, 103]], [[0, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[61, 68]], [[69, 76]], [[0, 76]], [[0, 76]], [[83, 87]], [[78, 87]], [[109, 110]], [[100, 110]], [[88, 112]], [[88, 131]]]",
    "query_spans": "[[[134, 142]]]",
    "process": "\\frac{\\sqrt{5}}{3} Let the midpoint of segment PF_{1} be M, then OM = b, and since OM is the midline of \\triangle F_{1}PF_{2}, \\therefore OM = \\frac{1}{2}PF_{2} \\Rightarrow PF_{2} = 2b. By the definition of the ellipse, PF_{1} = 2a - 2b, MF_{1} = \\frac{1}{2}PF_{1} = a - b. In right triangle Rt\\triangle OMF_{1}, (a - b)^{2} + b^{2} = c^{2}. \\because a^{2} = b^{2} + c^{2} \\Rightarrow 2a = 3b \\Rightarrow 4a^{2} = 9b^{2} = 9(a^{2} - c^{2}), we obtain e = \\frac{\\sqrt{5}}{3}. Key point: eccentricity of the ellipse"
  },
  {
    "text": "$P$ is a point on the hyperbola $3 x^{2}-5 y^{2}=15$, $F_{1}$, $F_{2}$ are its two foci, and the area of $\\Delta F_{1} P F_{2}$ is $3 \\sqrt{2}$. Then $\\angle F_{1} P F_{2}$=?",
    "fact_expressions": "G: Hyperbola;F1: Point;P: Point;F2: Point;Expression(G) = (3*x^2 - 5*y^2 = 15);PointOnCurve(P, G);Focus(G) = {F1, F2};Area(TriangleOf(F1, P, F2)) = 3*sqrt(2)",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "pi/3",
    "fact_spans": "[[[4, 27], [47, 48]], [[31, 38]], [[0, 3]], [[39, 46]], [[4, 27]], [[0, 30]], [[31, 52]], [[54, 92]]]",
    "query_spans": "[[[94, 118]]]",
    "process": "The hyperbola $3x^{2}-5y^{2}=15$ can be rewritten as: $\\frac{x^2}{5}-\\frac{y^{2}}{3}=1$, $a=\\sqrt{5}$, $b=\\sqrt{3}$, $c=2\\sqrt{2}$. Let $\\angle F_{1}PF_{2}=\\alpha$, $|PF_{1}|=m$, $|PF_{2}|=n$, $m>n$, then $m-n=2\\sqrt{5}$ $\\textcircled{1}$. Since the area of $\\triangle F_{1}PF_{2}$ is $3\\sqrt{3}$, we have $\\frac{1}{2}mn\\sin\\alpha=3\\sqrt{3}$ $\\textcircled{2}$. Also, $32=m^{2}+n^{2}-2mn\\cos\\alpha$ $\\textcircled{3}$. From $\\textcircled{1}$, $\\textcircled{2}$, $\\textcircled{3}$, we obtain $\\alpha=\\frac{\\pi}{3}$."
  },
  {
    "text": "Point $P$ is a point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, and $F_{1}$, $F_{2}$ are its foci. If $\\angle F_{1} P F_{2}=90^{\\circ}$, what is the area of $\\Delta F_{1} P F_{2}$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/9 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "9",
    "fact_spans": "[[[5, 43], [63, 64]], [[5, 43]], [[0, 4]], [[0, 46]], [[47, 54]], [[55, 62]], [[47, 66]], [[68, 101]]]",
    "query_spans": "[[[103, 129]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$, the point $(\\frac{p}{2}, 1)$ lies on the parabola. Then, the distance from the point on the parabola with ordinate $3$ to the directrix is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;H: Point;Coordinate(H) = (p/2, 1);PointOnCurve(H, G);K: Point;PointOnCurve(K, G);YCoordinate(K) = 3",
    "query_expressions": "Distance(K, Directrix(G))",
    "answer_expressions": "5",
    "fact_spans": "[[[2, 23], [44, 47], [52, 55]], [[2, 23]], [[5, 23]], [[5, 23]], [[24, 43]], [[24, 43]], [[24, 50]], [[64, 65]], [[52, 65]], [[56, 65]]]",
    "query_spans": "[[[52, 73]]]",
    "process": "Since the point $(\\frac{p}{2},1)$ lies on the parabola, we find the equation of the parabola and then use the definition of a parabola to solve. Because the point $(\\frac{p}{2},1)$ lies on the parabola, $1 = 2p \\times \\frac{p}{2}$, solving gives $p = 1$, so the equation of the parabola is $y^{2} = 2x$. Letting $y = 3$, we get $x = \\frac{9}{2}$. Therefore, the distance from the point on the parabola with ordinate 3 to the directrix is $\\frac{9}{2} + \\frac{p}{2} = \\frac{9}{2} + \\frac{1}{2} = 5$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, with left and right foci $F_{1}(-c, 0)$ and $F_{2}(c, 0)$, respectively. Its asymptotes are given by $y=\\pm 2x$, and the focal length is $2 \\sqrt{10}$. Point $P$ lies on the hyperbola $C$ and is in the first quadrant. If the area of $\\Delta P F_{1} F_{2}$ is $4 \\sqrt{10}$, then the coordinates of point $P$ are?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;F1: Point;F2: Point;P: Point;c: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);RightFocus(C) = F2;LeftFocus(C) = F1;Expression(Asymptote(C)) = (y = pm*(2*x));FocalLength(C) = 2*sqrt(10);PointOnCurve(P, C);Quadrant(P) = 1;Area(TriangleOf(P, F1, F2)) = 4*sqrt(10)",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(\\sqrt{6}, 4)",
    "fact_spans": "[[[2, 63], [103, 104], [145, 151]], [[10, 63]], [[10, 63]], [[72, 86]], [[89, 102]], [[140, 144], [203, 207]], [[89, 102]], [[10, 63]], [[10, 63]], [[2, 63]], [[72, 86]], [[89, 102]], [[2, 102]], [[2, 102]], [[103, 121]], [[103, 138]], [[140, 152]], [[140, 160]], [[162, 201]]]",
    "query_spans": "[[[203, 212]]]",
    "process": "From the given, we have $\\frac{b}{a}=2$, $c=\\sqrt{10}$, the hyperbola equation can be found. Since $S_{\\triangle PF_{1}F_{2}}=\\frac{1}{2}\\times2c\\times y_{0}=4\\sqrt{10}$, the result can be calculated. Because the foci of the hyperbola lie on the $x$-axis, $\\frac{b}{a}=2$, thus $b=2a$, and $c=\\sqrt{10}$. Therefore, $10=a^{2}+4a^{2}$, so $a^{2}=2$, $b^{2}=8$. Hence, the hyperbola equation is $\\frac{x^{2}}{2}-\\frac{y^{2}}{8}=1$. Let point $P(x_{0},y_{0})$ where $x_{0}>0$, $y_{0}>0$. From the condition, $\\frac{1}{2}\\times2c\\times y_{0}=\\frac{1}{2}\\times2\\sqrt{10}\\times y_{0}=4\\sqrt{10}$, solving gives $y_{0}=4$. Substituting point $P(x_{0},4)$ into the hyperbola equation $\\frac{x^{2}}{2}-\\frac{y^{2}}{8}=1$, we get $x_{0}=\\sqrt{6}$. Therefore, the coordinates of point $P$ are $(\\sqrt{6},4)$."
  },
  {
    "text": "If point $O$ and point $F_{1}(-\\sqrt{2}, 0)$ are respectively the center and the left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1  (a>0)$, and point $P$ is an arbitrary point on the right branch of the hyperbola, then the range of values of $\\frac{|P F_{1}|^{2}}{|O P|^{2}+1}$ is?",
    "fact_expressions": "G: Hyperbola;a: Number;F1: Point;P: Point;O: Origin;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Coordinate(F1) = (-sqrt(2), 0);Center(G)=O;LeftFocus(G)=F1;PointOnCurve(P, RightPart(G))",
    "query_expressions": "Range(Abs(LineSegmentOf(P, F1))^2/(Abs(LineSegmentOf(O, P))^2 + 1))",
    "answer_expressions": "(1,3/2+sqrt(2)]",
    "fact_spans": "[[[31, 70], [83, 86]], [[34, 70]], [[6, 28]], [[78, 82]], [[1, 5]], [[34, 70]], [[31, 70]], [[6, 28]], [[1, 77]], [[1, 77]], [[78, 94]]]",
    "query_spans": "[[[96, 138]]]",
    "process": "Since point $O$ and point $F_{1}(-\\sqrt{2},0)$ are the center and left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$, respectively, it follows that $c=\\sqrt{2}$, then $c^{2}=a^{2}+1=2$, so $a^{2}=1$, i.e., the equation of the hyperbola is $x^{2}-y^{2}=1$. Let $P(x,y)$, then $x\\geqslant1$, then $\\frac{|PF|^{2}}{|OP|^{2}+1}=\\frac{(x+\\sqrt{2})^{2}+y^{2}}{x^{2}+y^{2}+1}=\\frac{x^{2}+2\\sqrt{2}x+2+x^{2}-1}{x^{2}+x^{2}-1+1}=\\frac{2x^{2}+2\\sqrt{2}x+1}{2x^{2}}=1+\\frac{\\sqrt{2}}{x}+\\frac{1}{2}\\left(\\frac{1}{x}\\right)^{2}$. Since $x\\geqslant1$, we have $1+\\frac{\\sqrt{2}}{x}+\\frac{1}{2}\\left(\\frac{1}{x}\\right)^{2}>1$. Also, $1+\\frac{\\sqrt{2}}{x}+\\frac{1}{2}\\left(\\frac{1}{x}\\right)^{2}=\\frac{1}{2}\\left(\\frac{1}{x}+\\sqrt{2}\\right)^{2}$. Since $x\\geqslant1$, we have $0<\\frac{1}{x}\\leqslant1$, thus when $\\frac{1}{x}=1$, $1+\\frac{\\sqrt{2}}{x}+\\frac{1}{2}\\left(\\frac{1}{x}\\right)^{2}=\\frac{1}{2}\\left(\\frac{1}{x}+\\sqrt{2}\\right)^{2}$ attains its maximum value $\\frac{1}{2}(1+\\sqrt{2})^{2}=\\frac{3}{2}+\\sqrt{2}$. Therefore, the range of $\\frac{|PF_{1}|^{2}}{|OP|^{2}+1}$ is $\\left(1,\\frac{3}{2}+\\sqrt{2}\\right]$."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, a line $l$ is drawn through point $F$ intersecting the parabola at points $A$ and $B$. If point $M(t, 2)$ in the first quadrant satisfies $\\overrightarrow{OM} = \\frac{1}{2}(\\overrightarrow{OA} + \\overrightarrow{OB})$ (where $O$ is the origin), then $|AB|=$?",
    "fact_expressions": "l: Line;G: Parabola;M: Point;O: Origin;A: Point;B: Point;F: Point;t:Number;Expression(G) = (y^2 = 4*x);Coordinate(M) = (t, 2);Quadrant(M) = 1;Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};VectorOf(O, M) = (VectorOf(O, A) + VectorOf(O, B))/2",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[30, 35]], [[2, 16], [36, 39]], [[60, 69]], [[152, 155]], [[43, 46]], [[47, 50]], [[26, 29]], [[60, 69]], [[2, 16]], [[60, 69]], [[54, 69]], [[2, 23]], [[24, 35]], [[30, 52]], [[72, 149]]]",
    "query_spans": "[[[163, 172]]]",
    "process": "From the condition, we have $ F(1,0) $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, and the equation of line $ AB $ be: $ x = my + 1 $, $ m \\in \\mathbb{R} $. Solving the system\n\\[\n\\begin{cases}\nx = my + 1 \\\\\ny^2 = 4x\n\\end{cases},\n\\]\nwe obtain $ y^{2} - 4my - 4 = 0 $, and $ y_{1} + y_{2} = 4m $, $ y_{1}y_{2} = -4 $. From the condition, we know $ y_{1} + y_{2} = 4m = 4 $, solving gives $ m = 1 $, $ t = \\frac{x_{1} + x_{2}}{2} = \\frac{m(y_{1} + y_{2}) + 2}{2} = 3 $. Therefore, $ |AB| = 2(3 + 1) = 8 $."
  },
  {
    "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ have no common points with the circle $x^{2}+(y+2)^{2}=1$, find the range of values for the eccentricity of this hyperbola.",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Circle;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + (y + 2)^2 = 1);NumIntersection(Asymptote(G), H) = 0",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1,2)",
    "fact_spans": "[[[2, 58], [91, 94]], [[5, 58]], [[5, 58]], [[63, 83]], [[5, 58]], [[5, 58]], [[2, 58]], [[63, 83]], [[2, 88]]]",
    "query_spans": "[[[91, 105]]]",
    "process": "Let the center of the circle be $ C(0,-2) $, and let the equation of one asymptote of the hyperbola be $ bx + ay = 0 $. Then the distance $ d $ from the center $ C(0,-2) $ to the asymptote $ bx + ay = 0 $ is $ d = \\frac{|0 - 2a|}{\\sqrt{a^{2} + b^{2}}} = \\frac{2a}{c} > 1 $. Solving this yields $ 1 < e < 2 $. Therefore, the range of the eccentricity of the hyperbola is $ (1,2) $. The answer should be filled in as: $ (1,2) $."
  },
  {
    "text": "Draw two tangents from $C(-1,-1)$ to $x^{2}=4 y$, with points of tangency $A$ and $B$ respectively. If the equation of line $AB$ is $\\frac{x}{a}+\\frac{y}{b}=1$, then $a+b=$?",
    "fact_expressions": "G: Curve;B: Point;A: Point;C: Point;a:Number;b:Number;Expression(G) = (x^2 = 4*y);Coordinate(C) = (-1, -1);L1:Line;L2:Line;TangentOfPoint(C,G)={L1,L2};TangentPoint(L1,G)=A;TangentPoint(L2,G)=B;Expression(LineOf(A,B)) = (y/b + x/a = 1)",
    "query_expressions": "a + b",
    "answer_expressions": "3",
    "fact_spans": "[[[12, 23]], [[38, 41]], [[34, 37]], [[1, 11]], [[83, 88]], [[83, 88]], [[12, 23]], [[1, 11]], [], [], [[0, 28]], [[0, 41]], [[0, 41]], [[43, 81]]]",
    "query_spans": "[[[83, 90]]]",
    "process": "From $x^{2}=4y$, that is, $y=\\frac{1}{4}x^{2}$, so $y'=\\frac{1}{2}x$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then the tangent line at $A$ is $y-y_{1}=\\frac{1}{2}x_{1}(x-x_{1})$, i.e., $x_{1}x-2y-2y_{1}=0$. Substituting $C(-1,-1)$ gives $-x_{1}+2-2y_{1}=0$. The tangent line at $B$ is $y-y_{2}=\\frac{1}{2}x_{2}(x-x_{2})$, i.e., $x_{2}x-2y-2y_{2}=0$. Substituting $C(-1,-1)$ gives $-x_{2}+2-2y_{2}=0$. Thus, both $A$ and $B$ satisfy $-x-2y+2=0$, so $AB: x+2y=2$, i.e., $\\frac{x}{2}+y=1$, so $a=2$, $b=1$, hence $a+b=3$;"
  },
  {
    "text": "Given points $A(-1,0)$, $B(1,0)$ and the parabola $y^{2}=2 x$, if point $P$ on the parabola satisfies $|P A|=m|P B|$, then the maximum value of $m$ is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;P: Point;Expression(G) = (y^2 = 2*x);Coordinate(A) = (-1, 0);Coordinate(B) = (1, 0);PointOnCurve(P,G);Abs(LineSegmentOf(P,A))=m*Abs(LineSegmentOf(P,B));m:Number",
    "query_expressions": "Max(m)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[23, 37], [39, 42]], [[2, 12]], [[13, 22]], [[43, 47]], [[23, 37]], [[2, 12]], [[13, 22]], [[39, 47]], [[49, 63]], [[65, 68]]]",
    "query_spans": "[[[65, 74]]]",
    "process": "Let $ P\\left(\\frac{y^{2}}{2}, y\\right) $. According to the problem, we have $ m^{2} = \\frac{PA^{2}}{PB^{2}} = \\frac{\\left(\\frac{y}{2}+1\\right)^{2} + y^{2}}{\\frac{y+44+8y}{y+4}} = 1 + \\frac{8y^{2}}{y^{4+4}} \\leqslant 1 + \\frac{8y^{2}}{2\\sqrt{4v4}} = 3 $. Therefore, $ m \\leqslant \\sqrt{3} $, and equality holds if and only if $ y^{2} = 2 $."
  },
  {
    "text": "The equation $\\frac{x^{2}}{k+1}+\\frac{y^{2}}{k-2}=1$ represents a hyperbola; then the range of real values for $k$ is?",
    "fact_expressions": "G: Hyperbola;k: Real;Expression(G)=(x^2/(k + 1) + y^2/(k - 2) = 1)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-1,2)",
    "fact_spans": "[[[43, 46]], [[48, 53]], [[0, 46]]]",
    "query_spans": "[[[48, 60]]]",
    "process": "\\because the equation \\frac{x^{2}}{k+1}+\\frac{y^{2}}{k-2}=1 represents a hyperbola, \\therefore (k+1)(k-2)<0, \\therefore -1<k<2."
  },
  {
    "text": "Given point $P(0,1)$, two points $A$, $B$ on the ellipse $\\frac{x^{2}}{4 m}+\\frac{y^{2}}{m}=1$ $(m>1)$, and a point $E$ distinct from $P$, $A$, $B$ such that $\\overrightarrow{E P}=\\frac{1}{4} \\overrightarrow{E A}+\\frac{3}{4} \\overrightarrow{E B}$, find the range of the horizontal coordinate of point $B$.",
    "fact_expressions": "G: Ellipse;m: Number;P: Point;E: Point;A: Point;B: Point;m>1;Expression(G) = (x^2/(4*m) + y^2/m = 1);Coordinate(P) = (0, 1);PointOnCurve(A,G);PointOnCurve(B,G);Negation(P=E);Negation(A=E);Negation(B=E);VectorOf(E, P) = VectorOf(E, A)/4 + (3/4)*VectorOf(E, B)",
    "query_expressions": "Range(XCoordinate(B))",
    "answer_expressions": "[-1,1]",
    "fact_spans": "[[[12, 56]], [[14, 56]], [[2, 11], [71, 74]], [[83, 87]], [[75, 78], [59, 62]], [[180, 184], [79, 82], [63, 66]], [[14, 56]], [[12, 56]], [[2, 11]], [[12, 66]], [[12, 66]], [[69, 87]], [[69, 87]], [[69, 87]], [[90, 178]]]",
    "query_spans": "[[[180, 195]]]",
    "process": "From $\\overrightarrow{EP}=\\frac{1}{4}\\overrightarrow{EA}+\\frac{3}{4}\\overrightarrow{EB}$ we obtain $\\frac{1}{4}\\overrightarrow{EP}-\\frac{1}{4}\\overrightarrow{EA}=\\frac{3}{4}\\overrightarrow{EB}-\\frac{3}{4}\\overrightarrow{EP}$, that is, $\\frac{1}{4}\\overrightarrow{AP}=\\frac{3}{4}\\overrightarrow{PB}$. $\\overrightarrow{AP}=3\\overrightarrow{PB}$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $\\overrightarrow{AP}=(-x_{1},1-y_{1})$, $\\overrightarrow{PB}=(x_{2},y_{2}-1)$, $1=3x_{2}$, $x_{1}=-3x_{2}$, $1-y_{1}=3(y_{2}-1)$. That is, $\\begin{cases}3\\\\y\\end{cases}$ $\\therefore=4-3y$. Also, points $A$, $B$ are both on the ellipse, then $\\begin{cases}\\frac{x^{2}}{4}+y_{1}= m\\\\\\frac{x^{2}}{4}+y_{2}= m\\end{cases}$, that is, $\\begin{cases}9x_{2}^{2}\\\\\\frac{4}{}\\end{cases}$ $-3y_{2}^{2}=m$, solving gives $y_{2}=\\frac{m+2}{3}$, while $x_{2}^{2}=4(m-y_{2}^{2})=\\left(\\frac{y_{2}^{2}}{3}\\right)^{2}]=-\\frac{4}{9}(m-\\frac{5}{2})^{2}+1$, and $m>1$, $\\therefore x_{2}^{2}\\leqslant1$, $x_{2}\\in[-1,1]$"
  },
  {
    "text": "Given that the asymptotes of hyperbola $C$ are $y = \\pm 3x$, write a standard equation for hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;Expression(Asymptote(C)) = (y = pm*(3*x))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2/9 = 1 (the answer is not unique)",
    "fact_spans": "[[[2, 8], [29, 35]], [[2, 26]]]",
    "query_spans": "[[[29, 43]]]",
    "process": "When the foci of hyperbola C are on the x-axis, let the equation be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0). Therefore, according to the given condition, \\frac{b}{a}=3. Without loss of generality, let a=1, then b=3. Thus, the equation is x^{2}-\\frac{y^{2}}{9}=1"
  },
  {
    "text": "Given the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ has left vertex $A_{1}$ and right focus $F_{2}$. If $P$ is a point on the right branch of the hyperbola, then the minimum value of $\\overrightarrow{P A_{1}} \\cdot \\overrightarrow{P F_{2}}$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;A1: Point;F2: Point;Expression(G) = (x^2 - y^2/3 = 1);LeftVertex(G) = A1;RightFocus(G)=F2;PointOnCurve(P,RightPart(G))",
    "query_expressions": "Min(DotProduct(VectorOf(P, A1), VectorOf(P, F2)))",
    "answer_expressions": "-2",
    "fact_spans": "[[[2, 30], [60, 63]], [[56, 59]], [[35, 42]], [[48, 55]], [[2, 30]], [[2, 42]], [[2, 55]], [[56, 68]]]",
    "query_spans": "[[[70, 132]]]",
    "process": ""
  },
  {
    "text": "Given the equation $\\frac{x^{2}}{3+k}+\\frac{y^{2}}{2-k}=1$ represents an ellipse with foci on the $y$-axis, what is the range of values for $k$?",
    "fact_expressions": "G: Ellipse;Expression(G)=(x^2/(k + 3) + y^2/(2 - k) = 1);PointOnCurve(Focus(G),yAxis);k:Number",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-3,-1/2)",
    "fact_spans": "[[[54, 56]], [[2, 56]], [[46, 56]], [[58, 61]]]",
    "query_spans": "[[[58, 68]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x$ coincides with the left focus of the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;H: Ellipse;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (x^2/6 + y^2/2 = 1);Focus(G) = LeftFocus(H)",
    "query_expressions": "p",
    "answer_expressions": "-4",
    "fact_spans": "[[[1, 17]], [[66, 69]], [[21, 58]], [[1, 17]], [[21, 58]], [[1, 64]]]",
    "query_spans": "[[[66, 73]]]",
    "process": ""
  },
  {
    "text": "If the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{n}=1$ $(m>n>0)$ and the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{b}=1$ $(a>0, b>0)$ have the same foci $F_{1}$, $F_{2}$, and $P$ is a common point of the two curves, then what is the value of $|PF_{1}| \\cdot |PF_{2}|$?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Ellipse;m: Number;n: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b + x^2/a = 1);m > n;n > 0;Expression(H) = (y^2/n + x^2/m = 1);Focus(G)={F1,F2};Focus(H)={F1,F2};Intersection(H,G)=P",
    "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "m-a",
    "fact_spans": "[[[46, 95]], [[49, 95]], [[49, 95]], [[1, 45]], [[3, 45]], [[3, 45]], [[120, 123]], [[100, 107]], [[110, 117]], [[49, 95]], [[49, 95]], [[46, 95]], [[3, 45]], [[3, 45]], [[1, 45]], [[1, 118]], [[1, 118]], [[120, 132]]]",
    "query_spans": "[[[134, 163]]]",
    "process": ""
  },
  {
    "text": "The vertex of the parabola is at the origin, the axis of symmetry is a coordinate axis, and the focus lies on the line $x - y + 2 = 0$. Then the equation of this parabola is?",
    "fact_expressions": "G: Parabola;H: Line;O: Origin;Expression(H) = (x - y + 2 = 0);Vertex(G) = O;SymmetryAxis(G)=axis;PointOnCurve(Focus(G), H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=-8*x,x^2=8*y}",
    "fact_spans": "[[[0, 3], [37, 40]], [[22, 33]], [[7, 9]], [[22, 33]], [[0, 9]], [[0, 17]], [[0, 34]]]",
    "query_spans": "[[[37, 44]]]",
    "process": "The vertex of the parabola is at the origin, the axis of symmetry is the coordinate axis, and the focus lies on the line $x - y + 2 = 0$. Therefore, the focus of the parabola is $(-2, 0)$ or $(0, 2)$. Thus, $\\frac{p}{2} = -2$ or $\\frac{p}{2} = 2$. The equation of the parabola is: $y^2 = -8x$ or $x^{2} = 8y$. This question examines finding the equation of a parabola from its focus, and is a simple problem."
  },
  {
    "text": "If the curve represented by the equation $(k-1) x^{2}+(2 k-5) y^{2}=1$ is a hyperbola, then the range of real values for $k$ is?",
    "fact_expressions": "G: Hyperbola;H: Curve;k: Real;Expression(H) = (x^2*(k - 1) + y^2*(2*k - 5) = 1);G=H",
    "query_expressions": "Range(k)",
    "answer_expressions": "(1,5/2)",
    "fact_spans": "[[[38, 41]], [[35, 37]], [[43, 48]], [[1, 37]], [[35, 41]]]",
    "query_spans": "[[[43, 55]]]",
    "process": "Since the equation $(k-1)x^{2}+(2k-5)y^{2}=1$ represents a hyperbola, $(k-1)(2k-5)<0$, solving gives $k\\in(1,\\frac{5}{3})$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{2}$ intersects the ellipse $C$ at points $A$ and $B$, and $\\angle A F_{1} B=90^{\\circ}$. Circle $M$ is tangent to the extension of $F_{1} A$, the extension of $F_{1} B$, and the line $AB$. Then, the radius of circle $M$ is?",
    "fact_expressions": "C: Ellipse;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;G: Line;PointOnCurve(F2, G);A: Point;B: Point;Intersection(G, C) = {A, B};AngleOf(A, F1, B) = ApplyUnit(90, degree);M: Circle;IsTangent(M, OverlappingLine(LineSegmentOf(F1, A)));IsTangent(M, OverlappingLine(LineSegmentOf(F1, B)));IsTangent(M, LineOf(A, B));Expression(C)=(x^2/2+y^2=1)",
    "query_expressions": "Radius(M)",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 34], [71, 76]], [[43, 50]], [[51, 58], [60, 67]], [[2, 58]], [[2, 58]], [[68, 70]], [[59, 70]], [[77, 80]], [[81, 84]], [[68, 86]], [[88, 117]], [[119, 123], [164, 168]], [[119, 162]], [[119, 162]], [[119, 162]], [2, 32]]",
    "query_spans": "[[[164, 173]]]",
    "process": ""
  },
  {
    "text": "If a focus of the hyperbola is $(\\sqrt{10}, 0)$, and the equations of the asymptotes are $y = \\pm 3x$, then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Coordinate(OneOf(Focus(G))) = (sqrt(10), 0);Expression(Asymptote(G)) = (y = pm*(3*x))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/9 = 1",
    "fact_spans": "[[[1, 4], [48, 51]], [[1, 26]], [[1, 45]]]",
    "query_spans": "[[[48, 56]]]",
    "process": "Since the asymptotes of the hyperbola are given by $ y = \\pm 3x $, and one focus is at $ (\\sqrt{10}, 0) $, we can assume the equation of the hyperbola is $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ $ (a > 0, b > 0) $. Since $ c = \\sqrt{10} $, we have $ a^{2} + b^{2} = 10 $. Also, since $ \\frac{b}{a} = 3 $, solving gives $ a = 1 $, $ b = 3 $. Therefore, the equation of the hyperbola is $ x^{2} - \\frac{y^{2}}{9} = 1 $."
  },
  {
    "text": "Let the equation of the ellipse be $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, with two foci $F_{1}$ and $F_{2}$. Let point $P$ be any point on the ellipse. Then the maximum value of $\\frac{|P F_{1}|^{2}}{|P F_{2}|}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Focus(G) = {F1, F2};Expression(G) = (x^2/4 + y^2/3 = 1);PointOnCurve(P, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, F1))^2/Abs(LineSegmentOf(P, F2)))",
    "answer_expressions": "9",
    "fact_spans": "[[[1, 40], [67, 69]], [[62, 66]], [[46, 53]], [[54, 61]], [[1, 61]], [[1, 40]], [[62, 74]]]",
    "query_spans": "[[[76, 117]]]",
    "process": "From the given conditions: $ a=2, b=\\sqrt{3}, c=1, |PF_{1}|+|PF_{2}|=4 $, so $ |PF_{1}|=4-|PF_{2}| $. Therefore, $ \\frac{|PF_{1}|^{2}}{|PF_{2}|} = \\frac{(4-|PF_{2}|)^{2}}{|PF_{2}|} = |PF_{2}| + \\frac{16}{|PF_{2}|} - 8 $. Since $ a-c \\leqslant |PF_{2}| \\leqslant a+c $, that is, $ 1 \\leqslant |PF_{2}| \\leqslant 3 $, let $ t = |PF_{2}| $, $ t \\in [1,3] $. Then $ y = \\frac{|PF_{1}|^{2}}{|PF_{2}|} = t + \\frac{16}{t} - 8 $, $ y' = 1 - \\frac{16}{t^{2}} = \\frac{t^{2}-16}{t^{2}} $. If $ t \\in [1,3] $, $ y' < 0 $, so the function $ y = t + \\frac{16}{t} - 8 $ is decreasing on $ [1,3] $. Thus, $ y_{\\max} = 1 + \\frac{16}{1} - 8 = 9 $."
  },
  {
    "text": "Point $P$ moves on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and points $Q$ and $R$ move on the circles $(x+1)^{2}+y^{2}=1$ and $(x-1)^{2}+y^{2}=1$, respectively. Then the maximum value of $| P Q|+| P R| $ is?",
    "fact_expressions": "G: Ellipse;H: Circle;C:Circle;P: Point;Q: Point;R: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Expression(H) = (y^2 + (x + 1)^2 = 1);Expression(C) = ((x-1)^2+y^2=1);PointOnCurve(P,G);PointOnCurve(Q,H);PointOnCurve(R,C)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, Q)) + Abs(LineSegmentOf(P, R)))",
    "answer_expressions": "6",
    "fact_spans": "[[[5, 42]], [[58, 77]], [[78, 97]], [[0, 4]], [[46, 49]], [[50, 53]], [[5, 42]], [[58, 77]], [[78, 97]], [[0, 45]], [[46, 100]], [[46, 100]]]",
    "query_spans": "[[[102, 124]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ be the left focus of the ellipse $5 x^{2} + 9 y^{2} = 45$, and let $P$ be a moving point on the ellipse, $A(1,0)$. Then the minimum value of $|P A| + |P F_{1}|$ is?",
    "fact_expressions": "G: Ellipse;A: Point;P: Point;F1: Point;Expression(G) = (5*x^2 + 9*y^2 = 45);Coordinate(A) = (1, 0);LeftFocus(G) = F1;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F1)))",
    "answer_expressions": "5",
    "fact_spans": "[[[9, 31], [40, 42]], [[47, 55]], [[36, 39]], [[1, 8]], [[9, 31]], [[47, 55]], [[1, 35]], [[36, 46]]]",
    "query_spans": "[[[57, 80]]]",
    "process": "First, convert the ellipse equation into standard form, from which the coordinates of the two foci are obtained. Using the definition of an ellipse, express |PF_{1}| as 2a - |PF_{2}|, then apply numerical and geometric reasoning to find the result. From the given information, the standard equation of the ellipse is \\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1, so c^{2}=a^{2}-b^{2}=9-5=4, \\therefore c=2. \\therefore the coordinates of the foci are F_{1}(-2,0) and (2,0). By the definition of an ellipse, |PF_{1}| + |PF_{2}| = 2a = 6, \\therefore |PF_{1}| = 6 - |PF_{2}|. \\therefore |PA| + |PF_{1}| = 6 + |PA| - |PF_{2}|. Using geometric properties, it follows that when point P is at the left endpoint of the ellipse, |PA| + |PF| attains its minimum value, which is 6 - |AF_{2}| = 6 - 1 = 5."
  },
  {
    "text": "Given that an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is perpendicular to the line $2 x+y+1=0$, find the eccentricity $e$ of the hyperbola.",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;e: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (2*x + y + 1 = 0);IsPerpendicular(OneOf(Asymptote(G)), H);Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 48], [72, 75]], [[5, 48]], [[5, 48]], [[55, 68]], [[79, 82]], [[2, 48]], [[55, 68]], [[2, 70]], [[72, 82]]]",
    "query_spans": "[[[79, 84]]]",
    "process": ""
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$ and the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{b^{2}}=1$ have common foci $F_{1}$, $F_{2}$. Then the asymptotes of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;b: Number;H: Ellipse;F1: Point;F2: Point;Expression(G) = (x^2/3 - y^2/b^2 = 1);Expression(H) = (x^2/6 + y^2/2 = 1);Focus(H) = {F1, F2};Focus(G) = {F1, F2}",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(3)/3)*x",
    "fact_spans": "[[[38, 80], [103, 106]], [[41, 80]], [[0, 37]], [[86, 93]], [[94, 101]], [[38, 80]], [[0, 37]], [[0, 101]], [[0, 101]]]",
    "query_spans": "[[[103, 114]]]",
    "process": "From the given conditions, the coordinates of the foci are (\\pm2,0), so b^{2}+3=4 \\Rightarrow b^{2}=1, therefore the asymptotes of the hyperbola are y=\\pm\\frac{\\sqrt{3}}{3}x"
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2 p x(p>0)$ be $F$, and the directrix be $l$. From point $A(3, y_{0})$ on the parabola, draw a perpendicular to $l$, with foot $B$. Let $C(\\frac{7}{2} p, 0)$. If $AF$ and $BC$ intersect at point $E$, and $|F E|=2|A E|$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(G) = F;l: Line;Directrix(G) = l;A: Point;Coordinate(A) = (3, y0);y0: Number;PointOnCurve(A, G);L: Line;PointOnCurve(A, L);IsPerpendicular(L, l);FootPoint(L, l) = B;B: Point;C: Point;Coordinate(C) = ((7/2)*p, 0);Intersection(LineSegmentOf(A, F), LineSegmentOf(B, C)) = E;E: Point;Abs(LineSegmentOf(F, E)) = 2*Abs(LineSegmentOf(A, E))",
    "query_expressions": "p",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 22], [38, 41]], [[1, 22]], [[130, 133]], [[4, 22]], [[26, 29]], [[1, 29]], [[33, 36], [57, 60]], [[1, 36]], [[42, 56]], [[42, 56]], [[43, 56]], [[37, 56]], [], [[37, 63]], [[37, 63]], [[37, 70]], [[67, 70]], [[73, 94]], [[73, 94]], [[95, 113]], [[109, 113]], [[114, 128]]]",
    "query_spans": "[[[130, 137]]]",
    "process": "From the given conditions: AB // CF, then $\\triangle ABE \\sim \\triangle FCE$, and combined with $|FE| = 2|AE|$, it follows that: $\\frac{|AB|}{|CF|} = \\frac{|AE|}{|FE|} = \\frac{1}{2}$. According to the problem: $|AB| = 3 + \\frac{p}{2}$, $|CF| = \\frac{7}{2}p - \\frac{p}{2} = 3p$. Thus we have: $\\frac{3 + \\frac{p}{2}}{3p} = \\frac{1}{2}$. Solving this equation for the real number $p$ yields: $p = 3$."
  },
  {
    "text": "It is known that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively. A circle with diameter $F_{1}F_{2}$ intersects an asymptote of the hyperbola at a common point $P$. If $|P F_{1}|=2|P F_{2}|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;F1: Point;F2: Point;P: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;IsDiameter(LineSegmentOf(F1, F2), G);OneOf(Intersection(G,Asymptote(C)))=P;Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[20, 81], [108, 111], [151, 154]], [[28, 81]], [[28, 81]], [[106, 107]], [[2, 9]], [[10, 17]], [[122, 125]], [[28, 81]], [[28, 81]], [[20, 81]], [[2, 87]], [[2, 87]], [[88, 107]], [[106, 125]], [[127, 149]]]",
    "query_spans": "[[[151, 160]]]",
    "process": "\\because the center of the circle with diameter F_{1}F_{2} is (0,0), and the radius is: c; hence the standard equation of the circle is: x^{2}+y^{2}=c^{2}. Also \\because one of the asymptotes of the hyperbola is: y=\\frac{b}{a}x. Without loss of generality, assume P lies in the first quadrant. Solving the system \\begin{cases}y=\\frac{b}{a}x\\\\x^{2}+y^{2}=c^{2}\\end{cases}, we obtain: \\begin{cases}x=a\\\\y=b\\end{cases}. \\because F_{1}(-c,0), F_{2}(c,0). Using the distance formula between two points, we get: \\sqrt{(a+c)^{2}+b^{2}}=2\\sqrt{(a-c)^{2}+b^{2}}+c^{2+b^{2}}c_{n}=2\\sqrt{2c^{2}-2ac}\\cdot3c=5a, i.e., \\frac{c}{a}=\\frac{5}{2}. \\therefore e=\\frac{5}{3}. The eccentricity of the hyperbola is: \\frac{5}{3}."
  },
  {
    "text": "What is the length of the real axis of the hyperbola $25 x^{2}-16 y^{2}=400$?",
    "fact_expressions": "E: Hyperbola;Expression(E) = (25*x^2 - 16*y^2 = 400)",
    "query_expressions": "Length(RealAxis(E))",
    "answer_expressions": "8",
    "fact_spans": "[[[0, 3]], [[0, 27]]]",
    "query_spans": "[[[0, 34]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with left and right foci $F_{1}$, $F_{2}$, the length of the real axis is $6$, the asymptotes are given by $y=\\pm \\frac{1}{3} x$, point $M$ moves on the left branch of the hyperbola, and point $N$ lies on the circle $E$: $x^{2}+(y+\\sqrt{6})^{2}=1$. Then the minimum value of $|M N|+|M F_{2}|$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;E: Circle;M: Point;N: Point;F2: Point;F1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(E) = (x^2 + (y + sqrt(6))^2 = 1);LeftFocus(C) =F1;RightFocus(C)=F2;Length(RealAxis(C)) = 6;Expression(Asymptote(C)) = (y = pm*(x/3));PointOnCurve(M, LeftPart(C));PointOnCurve(N, E)",
    "query_expressions": "Min(Abs(LineSegmentOf(M, F2)) + Abs(LineSegmentOf(M, N)))",
    "answer_expressions": "9",
    "fact_spans": "[[[2, 63], [132, 135]], [[10, 63]], [[10, 63]], [[144, 176]], [[128, 131]], [[139, 143]], [[81, 88]], [[71, 79]], [[10, 63]], [[10, 63]], [[2, 63]], [[144, 176]], [[2, 88]], [[2, 88]], [[2, 97]], [[2, 125]], [[128, 138]], [[139, 179]]]",
    "query_spans": "[[[181, 204]]]",
    "process": "From the given conditions, we have 2a=6, so a=3. The asymptotes are y=\\pm\\frac{1}{3}x, which gives \\frac{b}{a}=\\frac{1}{3}, hence b=1. Thus, the hyperbola equation is \\frac{x^{2}}{9}-y^{2}=1, with foci at F_{1}(-\\sqrt{10},0), F_{2}(\\sqrt{10},0). By the definition of a hyperbola, |MF_{2}|=2a+|MF_{1}|=6+|MF_{1}|. From the circle E: x^{2}+(y+\\sqrt{6})^{2}=1, we obtain E(0,-\\sqrt{6}) and radius r=1. Then |MN|+|MF_{2}|=6+|MN|+|MF_{1}|. Connecting EF_{1}, intersecting the hyperbola at M and the circle at N, |MN|+|MF_{1}| attains its minimum value, which is |EF_{1}|=\\sqrt{6+10}=4. Therefore, the minimum value of |MN|+|MF_{2}| is 6+4-1=9."
  },
  {
    "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1$ are (in general form)?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/4 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "{(2*x - 3*y = 0), (2*x + 3*y = 0)}",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 53]]]",
    "process": "Let $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=0$, solve to get $2x-3y=0$ and $2x+3y=0$"
  },
  {
    "text": "The number of intersection points between the curve $\\frac{y^{2}}{9}-\\frac{x |x|}{4}=1$ and the line $y=x+3$ is?",
    "fact_expressions": "G: Line;H: Curve;Expression(G) = (y = x + 3);Expression(H) = (-x*Abs(x)/4 + y^2/9 = 1)",
    "query_expressions": "NumIntersection(H, G)",
    "answer_expressions": "3",
    "fact_spans": "[[[38, 47]], [[0, 37]], [[38, 47]], [[0, 37]]]",
    "query_spans": "[[[0, 54]]]",
    "process": ""
  },
  {
    "text": "The minimum distance from the point $M(10,0)$ to a point on the parabola $y^{2}=10 x$ is?",
    "fact_expressions": "G: Parabola;M: Point;P: Point;Expression(G) = (y^2 = 10*x);Coordinate(M) = (10, 0);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(M, P))",
    "answer_expressions": "5*sqrt(3)",
    "fact_spans": "[[[11, 26]], [[0, 10]], [[28, 29]], [[11, 26]], [[0, 10]], [[11, 29]]]",
    "query_spans": "[[[0, 38]]]",
    "process": "Let the point $ N(x_{0},y_{0}) $ be on the parabola, then $ MN = \\sqrt{(x_{0}-10)^{2}+y_{0}^{2}} = \\sqrt{(x_{0}-10)^{2}+10x_{0}} = \\sqrt{x_{0}^{2}-10x_{0}+100} = \\sqrt{(x_{0}-5)^{2}+75} $. Hence, when $ x_{0} = 5 $, $ MN $ takes the minimum value $ \\sqrt{75} = 5\\sqrt{3} $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ has eccentricity $\\frac{\\sqrt{5}}{2}$, then the asymptotes of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = sqrt(5)/2",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(1/2)",
    "fact_spans": "[[[2, 61], [88, 91]], [[5, 61]], [[5, 61]], [[5, 61]], [[5, 61]], [[2, 61]], [[2, 86]]]",
    "query_spans": "[[[88, 99]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$, $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, a line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. If $|F_{2} A|+|F_{2} B|=12$, then $|A B|$=?",
    "fact_expressions": "G: Ellipse;H: Line;F2: Point;A: Point;B: Point;F1: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(G) ={F1,F2} ;PointOnCurve(F1, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(F2, A)) + Abs(LineSegmentOf(F2, B)) = 12",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[22, 60], [78, 80]], [[75, 77]], [[13, 21]], [[81, 84]], [[85, 88]], [[2, 11]], [[22, 60]], [[2, 65]], [[66, 77]], [[75, 90]], [[93, 117]]]",
    "query_spans": "[[[119, 128]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{b}=1$, and point $A$ lies on the hyperbola $C$. If $A F_{1} \\perp A F_{2}$ and $\\angle A F_{1} F_{2}=30^{\\circ}$, then $b$=?",
    "fact_expressions": "C: Hyperbola;b: Number;A: Point;F1: Point;F2: Point;Expression(C) = (x^2/4 - y^2/b = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(A, C);IsPerpendicular(LineSegmentOf(A, F1), LineSegmentOf(A, F2));AngleOf(A, F1, F2) = ApplyUnit(30, degree)",
    "query_expressions": "b",
    "answer_expressions": "12+8*sqrt(3)",
    "fact_spans": "[[[17, 60], [70, 76]], [[139, 142]], [[65, 69]], [[1, 8]], [[9, 16]], [[17, 60]], [[1, 64]], [[1, 64]], [[65, 77]], [[79, 102]], [[104, 137]]]",
    "query_spans": "[[[139, 144]]]",
    "process": "According to the definition of hyperbola, find |F_{1}F_{2}|, then obtain the value of parameter b. From the given condition, c=\\sqrt{4+b}, |F_{1}F_{2}|=2c=2\\sqrt{4+b}. Since AF_{1}\\bot AF_{2}, and \\angle AF_{1}F_{2}=30^{\\circ}, therefore |AF_{1}|=|F_{1}F_{2}|\\cos\\angle PF_{1}F_{2}=2c\\cos30^{\\circ}=\\sqrt{3}c. Similarly, |AF_{2}|= so |AF_{1}|-|BF_{1}|=\\sqrt{3}c-c=2\\times\\sqrt{4}=4, c=\\frac{4}{\\sqrt{3}}\\frac{1}{4}=2(\\sqrt{3}+1). Thus b=[2(\\sqrt{3}+1)]^{2}-4=12+8\\sqrt{3}"
  },
  {
    "text": "If a point $P$ on the hyperbola $\\frac{x^{2}}{64}-\\frac{y^{2}}{36}=1$ is at a distance of $8$ from its right focus, then what is the distance from $P$ to its right directrix?",
    "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2/64 - y^2/36 = 1);PointOnCurve(P, G);Distance(P, RightFocus(G)) = 8",
    "query_expressions": "Distance(P, RightDirectrix(G))",
    "answer_expressions": "32/5",
    "fact_spans": "[[[2, 42], [49, 50], [68, 69]], [[45, 48]], [[2, 42]], [[2, 48]], [[45, 61]]]",
    "query_spans": "[[[64, 78]]]",
    "process": "By the second definition of a hyperbola, the distance from P to the right directrix is $\\frac{8}{e}=8\\times\\frac{8}{10}=\\frac{32}{5}$"
  },
  {
    "text": "Given the parabola $C$: $x^{2}=8 y$ with focus $F$, a line $l$ passing through point $P(0,-1)$ with slope $k$ ($k>0$) intersects the parabola $C$ at points $A$ and $B$. The midpoint $Q$ of $A B$ has a distance of $3$ to the $x$-axis. If $M$ is a moving point on line $l$ and $E(3,0)$, then the maximum value of $||M F|-| M E||$ is?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;P: Point;E: Point;M: Point;F: Point;Q: Point;Expression(C) = (x^2 = 8*y);Coordinate(P) = (0, -1);Coordinate(E) = (3, 0);Focus(C) = F;PointOnCurve(P,l);Slope(l) = k;Intersection(l, C) = {A, B};MidPoint(LineSegmentOf(A,B))=Q;Distance(Q, xAxis) = 3;PointOnCurve(M, l);k:Number;k>0",
    "query_expressions": "Max(Abs(-Abs(LineSegmentOf(M, E)) + Abs(LineSegmentOf(M, F))))",
    "answer_expressions": "1",
    "fact_spans": "[[[52, 57], [105, 110]], [[2, 21], [58, 64]], [[66, 69]], [[70, 73]], [[30, 40]], [[117, 125]], [[101, 104]], [[25, 28]], [[84, 87]], [[2, 21]], [[30, 40]], [[117, 125]], [[2, 28]], [[29, 57]], [[40, 57]], [[52, 75]], [[76, 87]], [[84, 99]], [[101, 116]], [[43, 51]], [[43, 51]]]",
    "query_spans": "[[[127, 149]]]",
    "process": "Let the equation of line $ l $ be $ y = kx - 1 $ ($ k > 0 $), $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Solving the system  \n\\[\n\\begin{cases}\ny = kx - 1 \\\\\nx^{2} = 8y\n\\end{cases}\n\\]  \ngives $ x^{2} - 8kx + 8 = 0 $. Thus, $ x_{1} + x_{2} = 8k $, so $ x_{Q} = \\frac{x_{1} + x_{2}}{2} = 4k $, and $ y_{Q} = kx_{Q} - 1 = 4k^{2} - 1 $. Since the distance from the midpoint $ Q $ of $ AB $ to the $ x $-axis is 3, we have $ 4k^{2} - 1 = 3 $, $ k > 0 $, so $ k = 1 $. Then the equation of line $ l $ is $ y = x - 1 $. Let the symmetric point of point $ F $ with respect to line $ l $ be $ F'(a, b) $. Then $ \\frac{2 + b}{2} = \\frac{a}{2} - 1 $, and $ \\frac{b - 2}{a} = -1 $. Solving gives $ a = 3 $, $ b = -1 $. Thus, the symmetric point of $ F $ with respect to line $ l $ is $ F'(3, -1) $. Therefore, $ ||MF| - |ME|| = ||MF'| - |ME|| \\leqslant |EF'| = 1 $, and equality holds when $ M $ is at the intersection point of ray $ F'E $ and line $ l $."
  },
  {
    "text": "Let $F$ be a focus of a hyperbola, and let point $P$ lie on the hyperbola such that the midpoint of segment $PF$ is exactly an endpoint of the imaginary axis of the hyperbola. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "P: Point;F: Point;G: Hyperbola;OneOf(Focus(G)) = F;PointOnCurve(P, G);MidPoint(LineSegmentOf(P,F))=OneOf(Endpoint(ImageinaryAxis(G)))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[14, 18]], [[1, 4]], [[5, 8], [19, 22], [36, 39], [48, 51]], [[1, 13]], [[14, 23]], [[25, 46]]]",
    "query_spans": "[[[48, 57]]]",
    "process": "Assume $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, $F(c,0)$, then point $P(-c,\\pm2b)$, thus we have $\\frac{c^{2}}{a^{2}}-\\frac{4b^{2}}{b^{2}}=1 \\Rightarrow \\frac{c^{2}}{a^{2}}=5 \\Rightarrow e=\\sqrt{5}$."
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x$ coincides with the left focus of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, then the equation of the parabola is?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 2*(p*x));p: Number;G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1);Focus(H) = LeftFocus(G)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2=-8*x",
    "fact_spans": "[[[1, 17], [57, 60]], [[1, 17]], [[4, 17]], [[21, 49]], [[21, 49]], [[1, 55]]]",
    "query_spans": "[[[57, 64]]]",
    "process": "The coordinates of the left focus of $\\frac{x^{2}}{3}-y^{2}=1$ are $(-2,0)$, then $\\frac{p}{2}=-2$, solving gives: $p=-4$, so the equation of the parabola is $y^2=-8x$."
  },
  {
    "text": "A value of $m$ that shows \"the curve of the equation $(m-1) x^{2}+(3-m) y^{2}=(m-1)(3-m)$ is an ellipse\" is?",
    "fact_expressions": "G: Ellipse;H: Curve;Expression(H) = ((m - 1)*x^2 + (3 - m)*y^2 = (m - 1)*(3 - m));m: Number;H = G",
    "query_expressions": "OneOf(m)",
    "answer_expressions": "3/2",
    "fact_spans": "[[[47, 49]], [[44, 46]], [[5, 46]], [[53, 56]], [[44, 49]]]",
    "query_spans": "[[[53, 60]]]",
    "process": ""
  },
  {
    "text": "Let point $F$ be the focus of the parabola $C$: $y^{2}=2 p x  (p>0)$, and point $A(0,2)$. If the midpoint $B$ of segment $A F$ lies on the parabola, then $|B F|$=?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;A: Point;F: Point;B: Point;p>0;Coordinate(A) = (0, 2);Focus(C) = F;PointOnCurve(B, C);MidPoint(LineSegmentOf(A,F))=B",
    "query_expressions": "Abs(LineSegmentOf(B, F))",
    "answer_expressions": "3*sqrt(2)/4",
    "fact_spans": "[[[6, 34], [64, 67]], [[6, 34]], [[14, 34]], [[38, 47]], [[1, 5]], [[60, 63]], [[14, 34]], [[38, 47]], [[1, 37]], [[60, 68]], [[50, 63]]]",
    "query_spans": "[[[70, 79]]]",
    "process": "The coordinates of the focus F are $(\\frac{p}{2},0)$, then the midpoint B of segment AF has coordinates $(\\frac{p}{4},1)$. Since point B lies on the parabola, substituting the coordinates of point B into the parabola equation yields $1=\\frac{p^{2}}{2}$, solving gives $p=\\sqrt{2}$ or $-\\sqrt{2}$ (discarded), thus the coordinates of point F are $(\\frac{\\sqrt{2}}{2},0)$, and the coordinates of point B are $(\\frac{\\sqrt{2}}{4},1)$. Using the distance formula between two points, we obtain $|BF|=(\\frac{\\sqrt{2}}{2}-\\frac{\\sqrt{2}}{4})^{2}+(0-1)^{2}=\\frac{3\\sqrt{2}}{4}$"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, and two points $A$, $B$ on the parabola satisfying $|AF|=2$, $|BF|=6$, what is the distance from the midpoint of chord $AB$ to the $y$-axis?",
    "fact_expressions": "C: Parabola;A: Point;B: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(A, C);PointOnCurve(B, C);Abs(LineSegmentOf(A, F)) = 2;Abs(LineSegmentOf(B, F)) = 6;IsChordOf(LineSegmentOf(A, B), C)",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A, B)), yAxis)",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 21], [29, 32]], [[36, 39]], [[40, 43]], [[25, 28]], [[2, 21]], [[2, 28]], [[29, 43]], [[29, 43]], [[45, 54]], [[56, 65]], [[29, 73]]]",
    "query_spans": "[[[68, 85]]]",
    "process": "According to the focal radius formula of the parabola, find the coordinates of points A and B respectively, then use the midpoint formula to find the x-coordinate of the midpoint of chord AB, thus obtaining the answer. Let points A(x_{1},y_{1}), B(x_{2},y_{2}), and focus F(1,0). From the problem, we have x_{1}+1=2, x_{2}+1=6, solving gives x_{1}=1, x_{2}=5. Therefore, the x-coordinate of the midpoint of chord AB is \\frac{x_{1}+x_{2}}{2}=\\frac{1+5}{2}=3, hence the distance from the midpoint of chord AB to the y-axis is 3."
  },
  {
    "text": "The focal length of an ellipse is equal to its minor axis length; then the eccentricity is?",
    "fact_expressions": "G: Ellipse;FocalLength(G)=Length(MinorAxis(G))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[0, 2], [8, 9]], [[0, 13]]]",
    "query_spans": "[[[8, 20]]]",
    "process": "From the given condition, 2c = 2b, that is, b = c, so a^{2} = c^{2} + b^{2} = 2c^{2}, that is, a = \\sqrt{2}c, therefore e = \\frac{c}{a} = \\frac{\\sqrt{2}}{2}"
  },
  {
    "text": "The standard equation of the hyperbola that shares the same asymptotes as $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ and passes through the point $A(6 , 8 \\sqrt{2})$ is?",
    "fact_expressions": "G: Hyperbola;C:Hyperbola;A: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Coordinate(A) = (6, 8*sqrt(2));Asymptote(G) = Asymptote(C);PointOnCurve(A, C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/64 - x^2/36 = 1",
    "fact_spans": "[[[1, 40]], [[72, 75]], [[51, 71]], [[1, 40]], [[51, 71]], [[0, 75]], [[50, 75]]]",
    "query_spans": "[[[72, 82]]]",
    "process": ""
  },
  {
    "text": "Given that $(4,0)$ is the right focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$, then the equations of the asymptotes of this hyperbola are?",
    "fact_expressions": "G: Hyperbola;b: Number;H: Point;Expression(G) = (x^2/4 - y^2/b^2 = 1);Coordinate(H) = (4, 0);RightFocus(G) = H",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*sqrt(3)*x",
    "fact_spans": "[[[10, 52], [59, 62]], [[13, 52]], [[2, 9]], [[10, 52]], [[2, 9]], [[2, 56]]]",
    "query_spans": "[[[59, 70]]]",
    "process": "From the given conditions and the properties of a hyperbola, we obtain $4 + b^{2} = 4^{2}$, solving gives $b^{2} = 12$, then the asymptotes of the hyperbola are $y = \\pm\\sqrt{\\frac{b^{2}}{4}}x = \\pm\\sqrt{3}x$"
  },
  {
    "text": "The equation $x^{2}+(k-1) y^{2}=k+1$ represents a hyperbola with foci on the $x$-axis. Then the range of real values for $k$ is?",
    "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (x^2 + y^2*(k - 1) = k + 1);PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-1,1)",
    "fact_spans": "[[[36, 39]], [[41, 46]], [[0, 39]], [[27, 39]]]",
    "query_spans": "[[[41, 53]]]",
    "process": "Convert the hyperbola equation into standard form, then use the equation $ x^{2}+(k-1)y^{2}=k+1 $ representing a hyperbola with foci on the x-axis to construct a system of inequalities, thereby determining the range of real values for $ k $.  \nThe hyperbola equation can be rewritten as:  \n$ \\frac{x^{2}}{k+1}-\\frac{y^{2}}{\\frac{k+1}{1-k}}=1 $.  \nSince the equation $ x^{2}+(k-1)y^{2}=k+1 $ represents a hyperbola with foci on the x-axis,  \nwe have  \n$ \\begin{cases} k+1>0 \\\\ \\frac{k+1}{1-k}>0 \\end{cases} $,  \nwhich gives $ -1<k<1 $.  \nHence, the range of real values for $ k $ is"
  },
  {
    "text": "Given that the distance from the focus of the parabola $C_{1}$: $x^{2}=2 p y(p>0)$ to the asymptote of the hyperbola $C_{2}$: $\\frac{y^{2}}{9}-\\frac{x^{2}}{16}=1$ is $\\frac{\\sqrt{5}}{5}$, find the value of the real number $p$.",
    "fact_expressions": "C1: Parabola;Expression(C1) = (x^2 = 2*p*y);p: Real;p>0;C2: Hyperbola;Expression(C2) = (y^2/9 - x^2/16 = 1);Distance(Focus(C1), Asymptote(C2)) = sqrt(5)/5",
    "query_expressions": "p",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 32]], [[2, 32]], [[114, 119]], [[14, 32]], [[36, 84]], [[36, 84]], [[2, 112]]]",
    "query_spans": "[[[114, 123]]]",
    "process": "Calculate the focus at (0,\\frac{p}{2}), asymptote equation 3x-4y=0, using the point-to-line distance formula to obtain the answer. The parabola C_{1}:x^{2}=2py(p>0) has focus (0,\\frac{p}{2}). One asymptote of the hyperbola C_{2}:\\frac{y^{2}}{9}-\\frac{x^{2}}{16}=1 is y=\\frac{3}{4}x, i.e., 3x-4y=0. Hence, d=\\frac{|-2p|}{\\sqrt{3^{2}+4^{2}}}=\\frac{\\sqrt{5}}{5}, so p=\\frac{\\sqrt{5}}{2}."
  },
  {
    "text": "Let point $P$ be a moving point on the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$. Then the minimum distance from point $P$ to the line $\\frac{x}{3}+\\frac{y}{4}=1$ is?",
    "fact_expressions": "C: Ellipse;G: Line;P: Point;Expression(C) = (x^2/2 + y^2 = 1);Expression(G) = (x/3 + y/4 = 1);PointOnCurve(P, C)",
    "query_expressions": "Min(Distance(P, G))",
    "answer_expressions": "(12 - sqrt(41))/5",
    "fact_spans": "[[[5, 37]], [[49, 78]], [[0, 4], [44, 48]], [[5, 37]], [[49, 78]], [[0, 42]]]",
    "query_spans": "[[[44, 87]]]",
    "process": "Let the line $ l: \\frac{x}{3} + \\frac{y}{4} = m $ be parallel to $ \\frac{x}{3} + \\frac{y}{4} = 1 $. When $ t $ is tangent to the ellipse $ C $, we have:\n\\[\n\\begin{cases}\n\\frac{x}{3} + \\frac{y}{4} = m \\\\\nx^{2} + 2y^{2} = 2\n\\end{cases}\n\\]\nThus,\n\\[\n41x^{2} - 64 \\times 3m x + 32 \\times 9m^{2} - 18 = 0\n\\]\nSo,\n\\[\n\\triangle = (64 \\times 3)^{2} m^{2} - 4 \\times 41 \\times 9 (32m^{2} - 2) = 0\n\\]\nHence,\n\\[\nm = \\pm \\frac{\\sqrt{41}}{12}\n\\]\nAccording to the problem, take $ m = \\frac{\\sqrt{41}}{12} $, then the distance from the line $ t: 4x + 3y - \\sqrt{41} = 0 $ to the line $ \\frac{x}{3} + \\frac{y}{4} = 1 $ is smaller. At this time, the distance between $ t: 4x + 3y - \\sqrt{41} = 0 $ and $ \\frac{x}{3} + \\frac{y}{4} = 1 $ (i.e., $ l: 4x + 3y - 12 = 0 $) is\n\\[\nd = \\frac{|12 - \\sqrt{41}|}{\\sqrt{3^{2} + 4^{2}}} = \\frac{12 - \\sqrt{41}}{5}\n\\]\nTherefore, the minimum distance from point $ P $ to the line $ \\frac{x}{3} + \\frac{y}{4} = 1 $ is $ \\frac{12 - \\sqrt{41}}{5} $."
  },
  {
    "text": "Given that the vertex of the parabola is at the origin and the focus is $F(1,0)$, a line passing through the focus intersects the parabola at points $A$ and $B$. A perpendicular from the midpoint $M$ of $AB$ to the directrix intersects the parabola at point $P$. If $|AB| = 6$, then the coordinates of point $P$ are?",
    "fact_expressions": "G: Parabola;O: Origin;Vertex(G) = O;F: Point;Coordinate(F) = (1, 0);Focus(G) = F;H: Line;PointOnCurve(Focus(G), H);A: Point;B: Point;Intersection(H, G) = {A, B};M: Point;MidPoint(LineSegmentOf(A, B)) = M;P: Point;Z: Line;PointOnCurve(M, Z);IsPerpendicular(Z, Directrix(G));Intersection(Z, G) = P;Abs(LineSegmentOf(A, B)) = 6",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1/2, pm*sqrt(2))",
    "fact_spans": "[[[2, 5], [31, 34], [65, 68]], [[9, 11]], [[2, 11]], [[15, 23]], [[15, 23]], [[2, 23]], [[28, 30]], [[2, 30]], [[36, 39]], [[40, 43]], [[28, 45]], [[55, 58]], [[47, 58]], [[87, 91], [87, 91]], [], [[46, 64]], [[31, 64]], [[31, 74]], [[76, 85]]]",
    "query_spans": "[[[87, 96]]]",
    "process": "The vertex of the parabola is at the origin, and the focus is $ F(1,0) $. Thus, the equation of the parabola is: $ y^{2}=4x $, $ p=2 $. A line passing through the focus intersects the parabola at points $ A $ and $ B $. From the midpoint $ M $ of $ AB $, a perpendicular is drawn to the directrix, intersecting the parabola at point $ P $, and $ |AB|=6 $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, $ |AB|=6=x_{1}+x_{2}+p $, so $ x_{1}+x_{2}=4 $. The line passing through the focus is set as $ y=k(x-1) $, then:\n\\[\n\\begin{cases}\ny^{2}=4x \\\\\ny=k(x-1)\n\\end{cases}\n\\]\nThis leads to $ k^{2}x^{2}-(2k^{2}+4)x+k^{2}=0 $, $ x_{1}+x_{2}=\\frac{2k^{2}+4}{k^{2}}=4 $, solving gives $ k=\\pm\\sqrt{2} $. $ y_{1}+y_{2}=\\pm\\sqrt{2}(x_{1}+x_{2}-2)=\\pm2\\sqrt{2} $, the ordinate of the midpoint is: $ \\pm\\sqrt{2} $, substituting into the parabola equation gives: $ x=\\frac{1}{2} $. Then the coordinates of point $ P $ are: $ (\\frac{1}{2},\\pm\\sqrt{2}) $"
  },
  {
    "text": "Given the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$, the right focus is $F$. A line passing through point $F$ and parallel to one of the asymptotes of the hyperbola intersects the hyperbola at points $P$ and $M$ lies on the line $PF$ such that $\\overrightarrow{O M} \\cdot \\overrightarrow{P F}=0$. Then $\\frac{|P M|}{|P F|}$=?",
    "fact_expressions": "G: Hyperbola;P: Point;F: Point;O: Origin;M: Point;Expression(G) = (x^2 - y^2/4 = 1);RightFocus(G) = F;H:Line;PointOnCurve(F,H);IsParallel(H,OneOf(Asymptote(G)));Intersection(H,G)=P;PointOnCurve(M,LineOf(P,F));DotProduct(VectorOf(O, M), VectorOf(P, F)) = 0",
    "query_expressions": "Abs(LineSegmentOf(P, M))/Abs(LineSegmentOf(P, F))",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 30], [48, 51], [61, 64]], [[66, 70]], [[35, 38], [40, 44]], [[87, 138]], [[71, 74]], [[2, 30]], [[2, 38]], [[58, 60]], [[39, 60]], [[45, 60]], [[58, 70]], [[71, 83]], [[87, 138]]]",
    "query_spans": "[[[140, 163]]]",
    "process": "The hyperbola $ x^{2} - \\frac{y^{2}}{4} = 1 $ has focus $ F(\\sqrt{5}, 0) $, and its asymptotes are given by $ y = \\pm 2x $. Let the line passing through point $ F $ and parallel to one asymptote of the hyperbola be $ y = 2(x - \\sqrt{5}) $. Substituting into the equation of the hyperbola yields $ x = \\frac{3\\sqrt{5}}{5} $, so $ P\\left( \\frac{3\\sqrt{5}}{5}, -\\frac{4\\sqrt{5}}{5} \\right) $. From the line $ OM: y = -\\frac{1}{2}x $ and the line $ y = 2(x - \\sqrt{5}) $, we obtain $ M\\left( \\frac{4\\sqrt{5}}{5}, -\\frac{2\\sqrt{5}}{5} \\right) $. Thus, $ \\frac{|PM|}{|PF|} = \\frac{|\\sqrt{5} - \\frac{3\\sqrt{5}}{5}|}{|\\sqrt{5} - \\frac{3\\sqrt{5}}{5}|} = \\frac{1}{2} $."
  },
  {
    "text": "If point $P$ lies on an ellipse with foci $F_{1}$, $F_{2}$, $PF_{2} \\perp F_{1} F_{2}$, $\\tan \\angle P F_{1} F_{2}=\\frac{3}{4}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;P: Point;F2: Point;F1: Point;PointOnCurve(P, G);Focus(G)={F1,F2};IsPerpendicular(LineSegmentOf(P, F2), LineSegmentOf(F1, F2));Tan(AngleOf(P, F1, F2)) = 3/4",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[28, 30], [102, 104]], [[1, 5]], [[17, 24]], [[7, 14]], [[1, 31]], [[6, 30]], [[32, 58]], [[60, 100]]]",
    "query_spans": "[[[102, 110]]]",
    "process": ""
  },
  {
    "text": "Given that the midpoint $M$ of chord $AB$ of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$ has coordinates $(2,1)$, then the equation of $AB$ is?",
    "fact_expressions": "G: Ellipse;M:Point;A: Point;B: Point;Expression(G) = (x^2/16 + y^2/4 = 1);Coordinate(M) = (2, 1);MidPoint(LineSegmentOf(A,B))=M;IsChordOf(LineSegmentOf(A,B),G)",
    "query_expressions": "Expression(LineSegmentOf(A, B))",
    "answer_expressions": "x-2*y-4=0",
    "fact_spans": "[[[2, 40]], [[50, 53]], [[42, 47]], [[42, 47]], [[2, 40]], [[50, 64]], [[42, 53]], [[2, 47]]]",
    "query_spans": "[[[66, 76]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), and let the slope of line AB be k. Given \\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1 and \\frac{x_{2}^{2}}{16}+\\frac{y_{2}^{2}}{4}=1, subtracting these two equations yields \\frac{1}{16}(x_{1}^{2}-x_{2}^{2})+\\frac{1}{4}(y_{1}^{2}-y_{2}^{2})=0, so \\frac{1}{16}+\\frac{1}{4}\\cdot\\frac{y_{1}^{2}-y_{2}^{2}}{x_{1}^{2}-x_{2}^{2}}=0. Thus, \\frac{1}{16}+\\frac{1}{4}\\cdot k \\cdot \\frac{y_{M}}{x_{M}}=0. Given x_{M}=2, y_{M}=1, we have k=-\\frac{1}{2}. Since the line passes through (2,1), the equation of the line is x+2y-4=0."
  },
  {
    "text": "Given the line $y=k(x+2) ,(k>0)$ intersects the parabola $y^{2}=8 x$ at points $A$ and $B$, and $F$ is the focus of the parabola. If $F A=2 F B$, then $k=$?",
    "fact_expressions": "G: Parabola;H: Line;F: Point;A: Point;B: Point;k:Number;Expression(G) = (y^2 = 8*x);k>0;Expression(H) = (y = k*(x + 2));Intersection(H, G) = {A, B};Focus(G) = F;LineSegmentOf(F, A) = 2*LineSegmentOf(F, B)",
    "query_expressions": "k",
    "answer_expressions": "pm*2*sqrt(2)/3",
    "fact_spans": "[[[22, 36], [53, 56]], [[2, 21]], [[49, 52]], [[39, 42]], [[43, 46]], [[74, 77]], [[22, 36]], [[4, 21]], [[2, 21]], [[2, 48]], [[49, 59]], [[61, 72]]]",
    "query_spans": "[[[74, 79]]]",
    "process": "Solve the system of equations formed by the line and the parabola, then use the relationship between roots and coefficients, along with the definition of the parabola and |FA|=2|FB|, to set up a system of equations, from which the result can be obtained. [Detailed Solution] According to the problem, let A(x_{1},y_{1}), B(x_{2},y_{2}). By the definition of the parabola, we have |FA|=x_{1}+2, |FB|=x_{2}+2. Since |FA|=2|FB|, it follows that x_{1}+2=2(x_{2}+2), i.e., x_{1}=2x_{2}+2\\textcircled{1}. Solving the system \\begin{cases}y=k(x+2)\\\\y^{2}=8x\\end{cases} yields k^{2}x^{2}+(4k^{2}-8)x+4k^{2}=0. Thus, \\Delta=(4k^{2}-8)^{2}-16k^{4}>0, so -1<k<1. Also, \\begin{cases}x_{1}+x_{2}=\\frac{8-4k^{2}}{k^{2}}\\textcircled{2}\\\\x_{1}x_{2}=4\\textcircled{3}\\end{cases}. Solving equations \\textcircled{1}, \\textcircled{2}, \\textcircled{3} gives k=\\pm\\frac{2\\sqrt{2}}{3}, which satisfies the conditions."
  },
  {
    "text": "Given that point $P$ is any point on the ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{16}=1$, and the line $l$: $x-y+1=0$ intersects the two coordinate axes at points $M$ and $N$ respectively, then the maximum area of $\\Delta P M N$ is?",
    "fact_expressions": "l: Line;C: Ellipse;P: Point;M: Point;N: Point;Expression(C) = (x^2/9 + y^2/16 = 1);PointOnCurve(P, C);Expression(l)=(x-y+1=0);Intersection(l,xAxis)=M;Intersection(l,yAxis)=N",
    "query_expressions": "Max(Area(TriangleOf(P, M, N)))",
    "answer_expressions": "3",
    "fact_spans": "[[[56, 72]], [[7, 50]], [[2, 6]], [[81, 84]], [[85, 88]], [[7, 50]], [[2, 55]], [[56, 72]], [[56, 90]], [[56, 90]]]",
    "query_spans": "[[[92, 114]]]",
    "process": "By the given conditions, M(-1,0), N(0,1), then |MN| = \\sqrt{2}. P is an arbitrary point on the ellipse C: \\frac{x^{2}}{9} + \\frac{y^{2}}{16} = 1, so the coordinates of point P can be set as (3\\cos\\theta, 4\\sin\\theta). Then the distance from point P to line l is d = \\frac{|3\\cos\\theta - 4\\sin\\theta + 1|}{\\sqrt{2}} \\leqslant \\frac{5 + 1}{\\sqrt{2}} = 3\\sqrt{2}, yielding S_{\\triangle PMN} = \\frac{1}{2} \\cdot |MN| \\cdot d \\leqslant \\frac{1}{2} \\cdot \\sqrt{2} \\cdot 3\\sqrt{2} = 3. Hence the answer is: 3"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$ and the moving line $l$: $ y=\\frac{3}{2} x+m $ intersecting at points $A$ and $B$, what is the range of real values for $m$?",
    "fact_expressions": "l: Line;C: Ellipse;A: Point;B: Point;m: Real;Expression(C) = (x^2/4 + y^2/9 = 1);Expression(l)=(y = m + (3/2)*x);Intersection(C, l) = {A, B}",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-3*sqrt(2),3*sqrt(2))",
    "fact_spans": "[[[48, 72]], [[2, 44]], [[75, 78]], [[79, 82]], [[86, 91]], [[2, 44]], [[48, 72]], [[2, 84]]]",
    "query_spans": "[[[86, 98]]]",
    "process": ""
  },
  {
    "text": "If on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the distance from the point with abscissa $\\frac{a}{3}$ to the left focus is greater than its distance to the right directrix, then the range of the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;a > b;b > 0;P:Point;e:Number;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);XCoordinate(P)=a/3;Distance(P,LeftFocus(G))>Distance(P,RightDirectrix(G));Eccentricity(G)=e;PointOnCurve(P,G)",
    "query_expressions": "Range(e)",
    "answer_expressions": "(\\sqrt{7}-2,1)",
    "fact_spans": "[[[1, 53], [92, 94]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[72, 73], [82, 83]], [[97, 100]], [[1, 53]], [[54, 73]], [[1, 90]], [[92, 100]], [[1, 73]]]",
    "query_spans": "[[[97, 107]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{25}-\\frac{y^{2}}{9}=1$ has two foci $F_{1}$, $F_{2}$. A point $P$ on the hyperbola satisfies $|P F_{1}|=12$, then $|P F_{2}|$=?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 - y^2/9 = 1);Focus(G)={F1,F2};PointOnCurve(P,G);Abs(LineSegmentOf(P,F1))=12",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "{2, 22}",
    "fact_spans": "[[[0, 39], [61, 64]], [[67, 70]], [[45, 52]], [[53, 60]], [[0, 39]], [[0, 60]], [[61, 70]], [[72, 86]]]",
    "query_spans": "[[[88, 101]]]",
    "process": "According to the definition of a hyperbola, solve directly. From the definition of the hyperbola, we have $||PF_{1}|-|PF_{2}||=10$, that is, $|12-|PF_{2}||=10$, so $|PF_{2}|=2$ or $|PF_{2}|=22$, both satisfy $|PF|\\geqslant c-a=\\sqrt{34}-5$."
  },
  {
    "text": "The hyperbola $m x^{2}-n y^{2}=1$ has left and right foci $F_{1}$, $F_{2}$ respectively, and left and right vertices $A$, $B$. Let $P$ be a point on an asymptote of the hyperbola. If the circle with diameter $F_{1} F_{2}$ passes through point $P$, and $\\angle A P B = \\frac{\\pi}{3}$, then what is the equation of the asymptotes of this hyperbola?",
    "fact_expressions": "G: Hyperbola;m: Number;n: Number;H: Circle;F1: Point;F2: Point;A: Point;P: Point;B: Point;Expression(G) = (m*x^2 - n*y^2 = 1);LeftVertex(G) = A;RightVertex(G) = B;LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, Asymptote(G));IsDiameter(LineSegmentOf(F1, F2), H);PointOnCurve(P, H);AngleOf(A, P, B) = pi/3",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*(2*sqrt(3)/3)*x",
    "fact_spans": "[[[0, 22], [64, 67], [134, 137]], [[3, 22]], [[3, 22]], [[93, 94]], [[29, 36]], [[37, 44]], [[52, 55]], [[60, 63], [96, 100]], [[56, 59]], [[0, 22]], [[0, 59]], [[0, 59]], [[0, 44]], [[0, 44]], [[60, 73]], [[75, 94]], [[93, 100]], [[102, 130]]]",
    "query_spans": "[[[134, 145]]]",
    "process": ""
  },
  {
    "text": "The equation of the ellipse with the foci of $\\frac{y^{2}}{12}-\\frac{x^{2}}{4}=1$ as vertices and the vertices as foci is?",
    "fact_expressions": "G: Ellipse;Z: Curve;Expression(Z) = (-x^2/4 + y^2/12 = 1);Focus(Z) = Vertex(G);Vertex(Z) = Focus(G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/16 + x^2/4 = 1",
    "fact_spans": "[[[50, 52]], [[1, 37]], [[1, 37]], [[0, 52]], [[0, 52]]]",
    "query_spans": "[[[50, 56]]]",
    "process": ""
  },
  {
    "text": "If the equation $\\frac{x^{2}}{k-3}+\\frac{y^{2}}{k+3}=1  (k \\in R)$ represents a hyperbola, then the range of $k$ is?",
    "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (x^2/(k - 3) + y^2/(k + 3) = 1)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-3, 3)",
    "fact_spans": "[[[55, 58]], [[60, 63], [3, 53]], [[1, 58]]]",
    "query_spans": "[[[60, 68]]]",
    "process": ""
  },
  {
    "text": "The number of common points between the line $y=-2x-3$ and the curve $\\frac{y^{2}}{9}-\\frac{x|x|}{4}=1$ is?",
    "fact_expressions": "G: Line;H: Curve;Expression(G) = (y = -2*x - 3);Expression(H) = (-x*Abs(x)/4 + y^2/9 = 1)",
    "query_expressions": "NumIntersection(G, H)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 12]], [[13, 49]], [[0, 12]], [[13, 49]]]",
    "query_spans": "[[[0, 58]]]",
    "process": "Analysis: First, discuss by cases to simplify the equation of the curve $\\frac{y^{2}}{9}-\\frac{x|x|}{4}=1$, then analyze the number of intersection points by combining numerical and graphical methods. When $x\\geqslant0$, the curve $\\frac{y^{2}}{9}-\\frac{x|x|}{4}=1$ becomes $\\frac{y^{2}}{9}-\\frac{x^{2}}{4}=1$. When $x<0$, the curve $\\frac{y^{2}}{9}-\\frac{x|x|}{4}=1$ becomes $\\frac{y^{2}}{9}+\\frac{x^{2}}{4}=1$. Therefore, the curve $\\frac{y^{2}}{9}-\\frac{x|x|}{4}=1$ is a graph composed of half a hyperbola and half an ellipse. Since the asymptotes of $\\frac{y^{2}}{9}-\\frac{x^{2}}{4}=1$ are $y=\\pm\\frac{3}{2}x$, and the slope of the line $y=-2x-3$ is $-2<-\\frac{3}{2}$, by combining numerical and graphical analysis, the number of intersection points between the line $y=-2x-3$ and the curve $\\frac{y^{2}}{9}-\\frac{x|x|}{4}=1$ is 2."
  },
  {
    "text": "Given the line $y=x-1$ and the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{m-1}=1  (m>1)$ intersect at points $A$ and $B$. If the circle with diameter $AB$ passes through the left focus $F$ of the ellipse, then the value of the real number $m$ is?",
    "fact_expressions": "G: Ellipse;m: Real;H: Line;A: Point;B: Point;m>1;Expression(G) = (y^2/(m - 1) + x^2/m = 1);Expression(H) = (y = x - 1);Intersection(H, G) = {A, B};C:Circle;IsDiameter(LineSegmentOf(A,B),C);F:Point;LeftFocus(G)=F;PointOnCurve(F,C)",
    "query_expressions": "m",
    "answer_expressions": "2+sqrt(3)",
    "fact_spans": "[[[12, 58], [82, 84]], [[94, 99]], [[2, 11]], [[60, 63]], [[64, 67]], [[14, 58]], [[12, 58]], [[2, 11]], [[2, 69]], [[80, 81]], [[71, 81]], [[89, 92]], [[82, 92]], [[80, 92]]]",
    "query_spans": "[[[94, 103]]]",
    "process": ""
  },
  {
    "text": "If a focus of the hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{9}=1$ $(a>0)$ is $(0, \\sqrt{13})$, then what is the equation of the asymptotes of this hyperbola?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-x^2/9 + y^2/a^2 = 1);Coordinate(OneOf(Focus(G))) = (0, sqrt(13))",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(2/3)*x",
    "fact_spans": "[[[1, 48], [73, 76]], [[4, 48]], [[4, 48]], [[1, 48]], [[1, 70]]]",
    "query_spans": "[[[73, 84]]]",
    "process": "The hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{9}=1$ ($a>0$) has a focus at $(0,\\sqrt{13})$, so $a^{2}+9=13$, solving gives $a=2$. Thus, the asymptotes of the hyperbola $\\frac{y^{2}}{4}-\\frac{x^{2}}{9}=1$ are $y=\\pm\\frac{2}{3}x$."
  },
  {
    "text": "The distance from a moving point $P$ in the plane to the point $F(0,2)$ is equal to its distance to the line $l$: $y=-2$. Then, what is the trajectory equation of the moving point $P$?",
    "fact_expressions": "l: Line;F: Point;P: Point;Coordinate(F) = (0, 2);Expression(l)=(y=-2);Distance(P,F)=Distance(P,l)",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2=8*y",
    "fact_spans": "[[[23, 36]], [[9, 18]], [[5, 8], [45, 48]], [[9, 18]], [[23, 36]], [[5, 41]]]",
    "query_spans": "[[[45, 56]]]",
    "process": "According to the given conditions, the trajectory of the point is a parabola with focus F(0,2) and directrix y=-2, where p=4, so the equation is x^{2}=8y. 【Analysis】This problem mainly examines the definition and standard equation of a parabola, and is a medium-difficulty problem."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ and the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ share common foci $F_{1}$, $F_{2}$, and $P$ is one of their intersection points such that $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Ellipse;F1: Point;P: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2/25 + y^2/9 = 1);Focus(G)={F1,F2};Focus(H)={F1,F2};OneOf(Intersection(G,H))=P;AngleOf(F1, P, F2) = pi/3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "4*sqrt(13)/13",
    "fact_spans": "[[[41, 97], [169, 172]], [[44, 97]], [[44, 97]], [[2, 40]], [[102, 109]], [[118, 121]], [[110, 117]], [[44, 97]], [[44, 97]], [[41, 97]], [[2, 40]], [[2, 117]], [[2, 117]], [[118, 129]], [[131, 167]]]",
    "query_spans": "[[[169, 178]]]",
    "process": "The semi-major axis of the ellipse is 5, and the semi-real axis of the hyperbola is $ a $. According to the definitions of the ellipse and hyperbola: $ |PF_{1}| + |PF_{2}| = 10 $, $ |PF_{1}| - |PF_{2}| = 2a $. Thus, $ |PF_{1}| = 5 + a $, $ |PF_{2}| = 5 - a $, $ |F_{1}F_{2}| = 8 $, $ \\angle F_{1}PF_{2} = \\frac{\\pi}{3} $. By the law of cosines, we obtain $ 64 = (5 + a)^{2} + (5 - a)^{2} - 2(5 + a)(5 - a)\\cos\\frac{\\pi}{3} $. Simplifying yields $ a^{2} = 13 $, $ e = \\frac{c}{a} = \\frac{4}{\\sqrt{13}} = \\frac{4\\sqrt{13}}{13} $."
  },
  {
    "text": "Given that $AB$ is a chord of the parabola $y^{2}=4x$ passing through its focus $F$, and $P$ is a moving point on the directrix of this parabola, then the minimum value of $\\overrightarrow{P A} \\cdot \\overrightarrow{P B}$ is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;P: Point;F:Point;Expression(G) = (y^2 = 4*x);PointOnCurve(P, Directrix(G));IsChordOf(LineSegmentOf(A,B),G);PointOnCurve(F,LineSegmentOf(A,B));Focus(G) = F",
    "query_expressions": "Min(DotProduct(VectorOf(P, A), VectorOf(P, B)))",
    "answer_expressions": "0",
    "fact_spans": "[[[9, 23], [36, 39]], [[2, 7]], [[2, 7]], [[31, 34]], [[25, 28]], [[9, 23]], [[31, 45]], [[2, 30]], [[2, 30]], [[9, 28]]]",
    "query_spans": "[[[47, 102]]]",
    "process": "The focus of the parabola $ y^{2} = 4x $ is $ F(1,0) $, so the equation of line $ AB $ can be written as $ x = ty + 1 $. Substituting into the parabola equation gives $ y^{2} - 4ty - 4 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ y_{1} + y_{2} = 4t $, $ y_{1}y_{2} = -4 $. Also, since $ P $ is a moving point on the directrix of the parabola, we can set $ P(-1, m) $. Then $ \\overrightarrow{PA} = (x_{1} + 1, y_{1} - m) = (ty_{1} + 2, y_{1} - m) $, $ \\overrightarrow{PB} = (x_{2} + 1, y_{2} - m) = (ty_{2} + 2, y_{2} - m) $. Therefore, $ \\overrightarrow{PA} \\cdot \\overrightarrow{PB} = (ty_{1} + 2)(ty_{2} + 2) + (y_{1} - m)(y_{2} - m) = (t^{2} + 1)y_{1}y_{2} + (2t - m)(y_{1} + y_{2}) + 4 + m^{2} = (2t - m)^{2} \\geqslant 0 $"
  },
  {
    "text": "Given point $M(-1,2)$ and parabola $C$: $y^{2}=4x$, the line $l$ passing through the focus $F$ of $C$ intersects $C$ at points $A$ and $B$. If $\\angle AMB=90^{\\circ}$, then the equation of line $l$ is?",
    "fact_expressions": "l: Line;C: Parabola;M: Point;A: Point;B: Point;F: Point;Expression(C) = (y^2 = 4*x);Coordinate(M) = (-1, 2);Focus(C)=F;PointOnCurve(F, l);Intersection(l, C) = {A, B};AngleOf(A, M, B) = ApplyUnit(90, degree)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x-y-1=0",
    "fact_spans": "[[[44, 49], [93, 98]], [[13, 32], [34, 37], [50, 53]], [[2, 12]], [[55, 58]], [[59, 62]], [[40, 43]], [[13, 32]], [[2, 12]], [[34, 43]], [[33, 49]], [[44, 64]], [[66, 91]]]",
    "query_spans": "[[[93, 103]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ lies on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, a line passing through point $P$ intersects the circle $x^{2}+y^{2}=a^{2}+b^{2}$ at points $A$ and $B$. If there exists a point $P$ such that $|P A| \\cdot|P B|=a^{2}-b^{2}$, then the range of values for the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;H: Circle;Expression(H) = (x^2 + y^2 = a^2 + b^2);L:Line;P: Point;A: Point;B: Point;a > b;b > 0;PointOnCurve(P, G);Intersection(L,H)={A,B};Abs(LineSegmentOf(P, A))*Abs(LineSegmentOf(P, B)) = (a^2 - b^2);PointOnCurve(P,L)",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[\\sqrt{2}/2,1)",
    "fact_spans": "[[[7, 59], [155, 157]], [[7, 59]], [[9, 59]], [[9, 59]], [[74, 100]], [[74, 100]], [[71, 73]], [[2, 6], [64, 68], [116, 120]], [[103, 106]], [[107, 110]], [[9, 59]], [[9, 59]], [[2, 62]], [[71, 112]], [[123, 153]], [[63, 73]]]",
    "query_spans": "[[[155, 167]]]",
    "process": ""
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$ are $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{2}$ intersects the ellipse at points $A$ and $B$. Then, the perimeter of $\\triangle A F_{1} B$ is?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;F1: Point;B: Point;F2: Point;Expression(G) = (x^2/36 + y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, H);Intersection(H, G) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, F1, B))",
    "answer_expressions": "24",
    "fact_spans": "[[[0, 38], [74, 76]], [[71, 73]], [[77, 80]], [[46, 53]], [[81, 84]], [[54, 61], [63, 70]], [[0, 38]], [[0, 61]], [[0, 61]], [[62, 73]], [[71, 86]]]",
    "query_spans": "[[[88, 113]]]",
    "process": "From $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$, we get $a=6$, $b=3$. According to the problem, the perimeter of $\\triangle AF_{1}B$ is $|AB|+|AF_{1}|+|BF_{1}|=|AF_{2}|+|BF_{2}|+|AF_{1}|+|BF_{1}|=4a=24$."
  },
  {
    "text": "Given that the line $y=\\frac{x}{2}$ intersects the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ at two points, what is the range of values for the eccentricity of this hyperbola?",
    "fact_expressions": "H: Line;Expression(H) = (y = x/2);G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;NumIntersection(H, G) = 2",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(\\sqrt{5}/2,+\\infty)",
    "fact_spans": "[[[2, 19]], [[2, 19]], [[20, 76], [83, 86]], [[20, 76]], [[23, 76]], [[23, 76]], [[23, 76]], [[23, 76]], [[2, 80]]]",
    "query_spans": "[[[83, 97]]]",
    "process": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ are given by $y=\\pm\\frac{b}{a}x$. From the problem, we know that $\\frac{b}{a}>\\frac{1}{2}$, $\\therefore e=\\frac{c}{a}=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{\\frac{a^{2}+b^{2}}{a^{2}}}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}>\\frac{\\sqrt{5}}{2}$"
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $\\frac{y^{2}}{3}-\\frac{x^{2}}{4}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/4 + y^2/3 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*((sqrt(3)/2)*x)",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 46]]]",
    "process": ""
  },
  {
    "text": "Given that $O$ is the coordinate origin, points $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, $A$ is a point on the ellipse $C$ such that $A F_{2} \\perp F_{1} F_{2}$, and $A F_{1}$ intersects the $y$-axis at point $B$, then $|O B|$=?",
    "fact_expressions": "C: Ellipse;A: Point;F1: Point;F2: Point;O: Origin;B: Point;Expression(C) = (x^2/4 + y^2/3 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(A, C);IsPerpendicular(LineSegmentOf(A, F2), LineSegmentOf(F1, F2));Intersection(LineSegmentOf(A, F1), yAxis) = B",
    "query_expressions": "Abs(LineSegmentOf(O, B))",
    "answer_expressions": "3/4",
    "fact_spans": "[[[30, 72], [83, 88]], [[79, 82]], [[11, 19]], [[20, 27]], [[2, 5]], [[140, 144]], [[30, 72]], [[11, 78]], [[11, 78]], [[79, 92]], [[94, 121]], [[123, 144]]]",
    "query_spans": "[[[146, 155]]]",
    "process": "Since $AF_{2}\\bot F_{1}F_{2}$, the length of $AF_{2}$ is half the latus rectum of the ellipse $C: \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, that is, $|AF_{2}|=\\frac{3}{2}$. Since $|F_{1}O|=|F_{2}O|$, $OB$ is the midline of triangle $AF_{1}F_{2}$, so $|OB|=\\frac{1}{2}|AF_{2}|=\\frac{3}{4}$."
  },
  {
    "text": "A line is drawn through the focus of the parabola $y^{2}=8x$, intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $x_{1}+x_{2}=6$, then the distance from the midpoint $M$ of segment $AB$ to the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;H: Line;B: Point;A: Point;M: Point;x1:Number;x2:Number;y1:Number;y2:Number;Expression(G) = (y^2 = 8*x);PointOnCurve(Focus(G),H);Coordinate(A)=(x1,y1);Coordinate(B)=(x2,y2);Intersection(H, G) = {A,B};x1+x2=6;MidPoint(LineSegmentOf(A,B)) = M",
    "query_expressions": "Distance(M,Directrix(G))",
    "answer_expressions": "5",
    "fact_spans": "[[[1, 15], [22, 25], [95, 98]], [[19, 21]], [[45, 62]], [[26, 44]], [[91, 94]], [[27, 44]], [[45, 62]], [[27, 44]], [[45, 62]], [[1, 15]], [[0, 21]], [[26, 44]], [[45, 62]], [[19, 62]], [[64, 79]], [[81, 94]]]",
    "query_spans": "[[[91, 105]]]",
    "process": ""
  },
  {
    "text": "Let the left focus of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ be $F$, and let point $P$ lie on the right branch of the hyperbola such that $PF$ is tangent to the circle $x^{2}+y^{2}=16$ at point $N$. Let $M$ be the midpoint of segment $PF$, and $O$ be the origin. Then $|MN|-|MO|=$?",
    "fact_expressions": "G: Hyperbola;H: Circle;F: Point;P: Point;M: Point;N: Point;O: Origin;Expression(G) = (x^2/16 - y^2/9 = 1);Expression(H) = (x^2 + y^2 = 16);LeftFocus(G) = F;PointOnCurve(P, RightPart(G));TangentPoint(LineSegmentOf(P,F),H)=N;MidPoint(LineSegmentOf(P,F))=M",
    "query_expressions": "Abs(LineSegmentOf(M, N)) - Abs(LineSegmentOf(M, O))",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 40], [54, 57]], [[71, 88]], [[45, 48]], [[49, 53]], [[96, 99]], [[91, 95]], [[111, 114]], [[1, 40]], [[71, 88]], [[1, 48]], [[49, 63]], [[65, 95]], [[96, 110]]]",
    "query_spans": "[[[121, 136]]]",
    "process": "From the given conditions: the hyperbola $\\frac{x^2}{16}-\\frac{y^{2}}{9}=1$ has its foci on the $x$-axis, $a=4$, $b=3$, $c=5$. Let the right focus of the hyperbola be $F_{1}(5,0)$, and the left focus be $F(-5,0)$. Since $OM$ is the midline of $\\triangle PFF_{1}$, then $|OM|=\\frac{1}{2}|PF_{1}|$. Since $PF$ is tangent to the circle $x^{2}+y^{2}=16$ at point $N$, $\\triangle ONF$ is a right triangle. Therefore, $|NF|^{2}=|OF|^{2}-|ON|^{2}=25-16=9$, so $|NF|=3$. Hence, $|MN|=|MF|-|NF|=|MF|-3$. Since $|MF|=\\frac{1}{2}|PF|$, it follows that $|MN|\\cdot|MO|=\\frac{1}{2}|PF||PF_{1}|=\\frac{1}{2}(|PF|\\cdot|PF_{1}|)\\cdot3=\\frac{1}{2}\\times2a\\cdot3=1$. Therefore, $|MN|\\cdot|MO|=1$."
  },
  {
    "text": "Given an ellipse and a hyperbola with the same foci $F_{1}$, $F_{2}$ intersecting at point $P$, $|P O|=|F_{1} F_{2}|$, the eccentricities of the ellipse and hyperbola are $e_{1}$, $e_{2}$ respectively, then what is $\\frac{1}{e_{1}^{2}}+\\frac{1}{e_{2}^{2}}$? (Point $O$ is the origin)",
    "fact_expressions": "H: Ellipse;G: Hyperbola;F1: Point;F2: Point;Focus(H) = {F1, F2};Focus(G) = {F1, F2};P: Point;Intersection(H, G) = P;Abs(LineSegmentOf(P, O)) = Abs(LineSegmentOf(F1, F2));e1: Number;e2: Number;Eccentricity(H) = e1;Eccentricity(G) = e2;O: Origin",
    "query_expressions": "1/e2^2 + 1/e1^2",
    "answer_expressions": "5",
    "fact_spans": "[[[23, 25], [59, 61]], [[26, 29], [62, 65]], [[7, 14]], [[15, 22]], [[2, 29]], [[2, 29]], [[31, 35]], [[23, 35]], [[37, 58]], [[72, 79]], [[82, 89]], [[59, 89]], [[59, 89]], [[136, 140]]]",
    "query_spans": "[[[92, 135]]]",
    "process": "Let the semi-major axis of the ellipse be $a_{1}$, the semi-transverse axis of the hyperbola be $a_{2}$, and their semi-focal distances both be $c$. Let $|PF_{1}|=m$, $|PF_{2}|=n$. According to the definitions of the ellipse and hyperbola, we have $m+n=2a_{1}$, $m-n=2a_{2}$. In $\\triangle POF_{1}$, by the cosine law, $|PF_{1}|^{2}=|OF_{1}|^{2}+|OP|^{2}-2|OF_{1}||OP|\\cos\\angle POF_{1}$, that is, $m^{2}=c^{2}+4c^{2}-2c\\times2c\\cos\\angle POF_{1}$. In $\\triangle POF_{2}$, by the cosine law, $|PF_{2}|^{2}=|OF_{2}|^{2}+|OP|^{2}-2|OF_{2}||OP|\\cos\\angle POF_{2}$, that is, $n^{2}=c^{2}+4c^{2}-2c\\times2c\\cos\\angle POF_{1}$. Since $\\angle POF_{1}=\\pi-\\angle POF_{2}$, adding the two equations gives $m^{2}+n^{2}=10c^{2}$. Also, $m^{2}+n^{2}=(m+n)^{2}-2mn=4a_{1}^{2}-2mn$, but from earlier, $m+n=2a_{1}$, $m-n=2a_{2}$, so $m^{2}+n^{2}=2a_{1}^{2}+2a_{2}^{2}$. Therefore, $2a_{1}^{2}+2a_{2}^{2}=10c^{2} \\Rightarrow a_{1}^{2}+a_{2}^{2}=5c^{2}$. Hence, $\\frac{a_{1}^{2}}{c^{2}}+\\frac{a_{2}^{2}}{c^{2}}=5$, i.e., $\\frac{1}{e_{1}^{2}}+\\frac{1}{e_{2}^{2}}=5$. This problem mainly examines the definitions and geometric properties of the ellipse and hyperbola. The key to the solution lies in using the definitions of the ellipse and hyperbola, and applying the cosine law in $\\triangle POF_{1}$ and $\\triangle POF_{2}$, then adding the two equations to obtain $m^{2}+n^{2}=10c^{2}$. It emphasizes the ability to analyze and solve problems, and belongs to the medium-difficulty category of problems."
  },
  {
    "text": "The point $P(x, y)$ moves along the curve $4(x-2)^{2}+y^{2}=4$. Then the maximum value of $\\frac{y}{2 x}$ is?",
    "fact_expressions": "P: Point;x1: Number;y1: Number;Coordinate(P) = (x1, y1);G: Curve;Expression(G) = (y^2 + 4*(x - 2)^2 = 4);PointOnCurve(P, G)",
    "query_expressions": "Max(y1/(2*x1))",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[0, 10]], [[1, 10]], [[1, 10]], [[0, 10]], [[11, 33]], [[11, 33]], [[0, 36]]]",
    "query_spans": "[[[38, 59]]]",
    "process": "According to the problem, $\\frac{y}{2x}$ represents half of the slope of the line connecting a point on the ellipse and the origin. Let $y = kx$. By determining the range of $k$ for which $y = kx$ intersects $(x-2)^{2} + \\frac{y^{2}}{4} = 1$, we can find the maximum value of $\\frac{y}{2x}$. Since $4(x-2)^{2} + y^{2} = 4$, it follows that $(x-2)^{2} + \\frac{y^{2}}{4} = 1$, representing an ellipse centered at $(2,0)$ with its major axis on $x=2$, minor axis on the $x$-axis, and vertices at $(1,0)$, $(3,0)$, $(2,2)$, $(2,-2)$. The expression $\\frac{y}{2x}$ can be viewed as half of the slope of the line joining the origin and a point $P(x,y)$ on the ellipse. Let $y = kx$. Since $y = kx$ intersects $(x-2)^{2} + \\frac{y^{2}}{4} = 1$, solving the system\n$$\n\\begin{cases}\ny = kx \\\\\n4(x-2)^{2} + y^{2} = 4\n\\end{cases}\n$$\nby eliminating $y$ gives $4(x-2)^{2} + k^{2}x^{2} = 4$, which simplifies to $(4 + k^{2})x^{2} - 16x + 12 = 0$. Therefore, $\\Delta = 16^{2} - 4 \\times (4 + k^{2}) \\times 12 \\geqslant 0$, solving this yields $-\\frac{2\\sqrt{3}}{3} \\leqslant k \\leqslant \\frac{2\\sqrt{3}}{3}$. Thus, $k_{\\max} = \\frac{2\\sqrt{3}}{3}$, so $\\left(\\frac{y}{2x}\\right)_{\\max} = \\frac{1}{2}k_{\\max} = \\frac{\\sqrt{3}}{3}$."
  },
  {
    "text": "The parabola $y^{2}=x$, a point $P$ on it such that the sum of the distance from $P$ to $A(3, -1)$ and to the focus is minimized; then the coordinates of point $P$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = x);P: Point;PointOnCurve(P, G);A: Point;Coordinate(A) = (3, -1);WhenMin(Distance(P, A) + Distance(P, Focus(G)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1,-1)",
    "fact_spans": "[[[0, 12], [13, 14]], [[0, 12]], [[17, 20], [44, 48]], [[13, 20]], [[21, 31]], [[21, 31]], [[13, 42]]]",
    "query_spans": "[[[44, 52]]]",
    "process": "Since point A(-3,1) lies inside the parabola, as shown in the figure, let point P be such that PQ\\botl at Q, and draw AB\\botl at B from A. Then |PA|+|PF|=|PA|+|PQ|\\geqslant|AB|. Thus, |PA|+|PF| is minimized if and only if P, A, B are collinear. At this time, the coordinates of point P are P(x_{0},-1). Substituting into y^{2}=x gives x_{0}=1. Hence, the coordinates of point P are (1,-1)."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, and point $P$ lies on $C$, then the perimeter of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/9 + y^2/5 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C)",
    "query_expressions": "Perimeter(TriangleOf(P, F1, F2))",
    "answer_expressions": "10",
    "fact_spans": "[[[18, 60], [73, 76]], [[68, 72]], [[2, 9]], [[10, 17]], [[18, 60]], [[2, 67]], [[2, 67]], [[68, 77]]]",
    "query_spans": "[[[79, 109]]]",
    "process": "From the ellipse equation, we know that $ a = 3 $, $ c = \\sqrt{9 - 5} = 2 $, and $ P $ lies on the ellipse. Therefore, $ |PF_{1}| + |PF_{2}| + |F_{1}F_{2}| = 2a + 2c = 2 \\times 3 + 2 \\times 2 = 10 $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with directrix $l$, a line passing through $M(1,0)$ with slope $k$ intersects $l$ at point $A$ and intersects the parabola $C$ at point $B$. If $\\overrightarrow{AM}=2\\overrightarrow{MB}$, then $k=$?",
    "fact_expressions": "C: Parabola;G: Line;M: Point;A: Point;B: Point;l: Line;k: Number;Expression(C) = (y^2 = 4*x);Coordinate(M) = (1, 0);Directrix(C) = l;PointOnCurve(M,G);Slope(G)=k;Intersection(l,G)=A;OneOf(Intersection(G, C)) = B;VectorOf(A, M) = 2*VectorOf(M, B)",
    "query_expressions": "k",
    "answer_expressions": "pm*2*sqrt(2)",
    "fact_spans": "[[[2, 21], [61, 67]], [[47, 49]], [[31, 39]], [[56, 60]], [[73, 76]], [[25, 29], [50, 53]], [[126, 129], [43, 46]], [[2, 21]], [[31, 39]], [[2, 29]], [[30, 49]], [[40, 49]], [[47, 60]], [[47, 76]], [[79, 124]]]",
    "query_spans": "[[[126, 131]]]",
    "process": "As shown in the figure: according to the problem, the distance from $M$ to the directrix is 2. Since $\\overrightarrow{AM} = 2\\overrightarrow{MB}$, the horizontal coordinate of $B$ is 2. Substituting into the parabola $C: y^{2} = 4x$, we get $y = \\pm 2\\sqrt{2}$. Therefore, the coordinates of $B$ are $(2, \\pm 2\\sqrt{2})$. Hence, $k = \\frac{\\pm 2\\sqrt{2}}{2 - 1} = \\pm 2\\sqrt{2}$."
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ has two foci $F_{1}$, $F_{2}$. A line passing through $F_{1}$ intersects the ellipse at points $A$, $B$. Then the perimeter of $\\triangle A B F_{2}$ is?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2/9 + y^2/4 = 1);Focus(G) = {F1, F2};PointOnCurve(F1, H);Intersection(H, G) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "12",
    "fact_spans": "[[[0, 37], [70, 72]], [[67, 69]], [[73, 76]], [[77, 80]], [[50, 57]], [[42, 49], [59, 66]], [[0, 37]], [[0, 57]], [[58, 69]], [[67, 82]]]",
    "query_spans": "[[[84, 110]]]",
    "process": "Slight"
  },
  {
    "text": "What is the eccentricity of the hyperbola $\\frac{y^{2}}{6}-\\frac{x^{2}}{3}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/3 + y^2/6 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 44]]]",
    "process": "The hyperbola $\\frac{y^{2}}{6}-\\frac{x^{2}}{3}=1$ gives $a=\\sqrt{6}$, $b=\\sqrt{3}$, then $c=3$, so $e=\\frac{c}{a}=\\frac{3}{\\sqrt{6}}=\\frac{\\sqrt{6}}{2}$"
  },
  {
    "text": "Let the hyperbola centered at the origin share common foci with the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, and let their eccentricities be reciprocals of each other. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;O: Origin;Expression(H) = (x^2/2 + y^2 = 1);Center(G) = O;Focus(G) = Focus(H);InterReciprocal(Eccentricity(G), Eccentricity(H))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "2*x^2-2*y^2=1",
    "fact_spans": "[[[7, 10], [59, 62]], [[11, 38]], [[4, 6]], [[11, 38]], [[1, 10]], [[7, 44]], [[46, 56]]]",
    "query_spans": "[[[59, 67]]]",
    "process": "The ellipse $\\frac{x^{2}}{2}+y^{2}=1$ has $c=1$. Since a hyperbola centered at the origin shares common foci with the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, the hyperbola has $c=1$. The eccentricity of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ is $\\frac{c}{a}=\\frac{\\sqrt{2}}{2}$, and the eccentricities of the ellipse and hyperbola are reciprocals of each other. Therefore, the eccentricity of the hyperbola is $\\sqrt{2}$. Thus, for the hyperbola, $a=\\frac{\\sqrt{2}}{2}$, $b^{2}=c^{2}-a^{2}=\\frac{1}{2}$, $b=\\frac{\\sqrt{2}}{2}$. Hence, the equation of the hyperbola is $2x^{2}-2y^{2}=1$."
  },
  {
    "text": "The directrix equation of the parabola $y=-\\frac{3}{4} x^{2}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = -3/4*x^2)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=1/3",
    "fact_spans": "[[[0, 25]], [[0, 25]]]",
    "query_spans": "[[[0, 32]]]",
    "process": ""
  },
  {
    "text": "The standard equation of a parabola with vertex at the origin and passing through the point $(-4,4)$ is?",
    "fact_expressions": "G: Parabola;O: Origin;Vertex(G) = O;H: Point;Coordinate(H) = (-4, 4);PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=-4*x, x^2=4*y}",
    "fact_spans": "[[[18, 21]], [[3, 5]], [[0, 21]], [[8, 17]], [[8, 17]], [[7, 21]]]",
    "query_spans": "[[[18, 28]]]",
    "process": "When the parabola opens upward, let the equation of the parabola be $x^{2}=2py$, $(p>0)$; substituting the point $(-4,4)$ gives $2p=4$, so the equation of the parabola is $x^{2}=4y$. When the parabola opens to the left, let the equation of the parabola be $y^{2}=2px$, $(p>0)$; substituting the point $(-4,4)$ gives $2p=-4$, so the equation of the parabola is $y^{2}=-4x$. In conclusion, the required equations of the parabola are $y^{2}=-4x$ or $x^{2}=4y$. Solution. Equation of the parabola"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, the two foci are $F_{1}$ and $F_{2}$. A line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. If $|A F_{2}|-|B F_{1}|=4$, then $|A B|=$?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;F2: Point;B: Point;F1: Point;Expression(G) = (x^2/25 + y^2/16 = 1);Focus(G)={F1,F2};PointOnCurve(F1, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, F2)) - Abs(LineSegmentOf(B, F1)) = 4",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 41], [75, 77]], [[72, 74]], [[78, 81]], [[55, 62]], [[82, 85]], [[47, 54], [64, 71]], [[2, 41]], [[2, 62]], [[63, 74]], [[72, 87]], [[89, 112]]]",
    "query_spans": "[[[114, 123]]]",
    "process": "According to the definition of an ellipse, we have  \n\\begin{cases}|AF_{1}|+|AF_{2}|=10\\\\|BF_{1}|+|BF_{2}|=10\\end{cases}  \nSubtracting the two equations gives  \n|AF_{2}|-|BF_{1}|+|AF_{1}|-|BF_{2}|=0,  \nthat is,  \n|AF_{2}|-|BF_{1}|=|BF_{2}|-|AF_{1}|=4,  \nwhich means  \n|AF_{2}|=4+|BF_{1}|, \\quad |BF_{2}|=4+|AF_{1}|\\textcircled{1}.  \nAdding the two equations yields  \n|AF_{1}|+|AF_{2}|+|BF_{1}|+|BF_{2}|=20,  \nthat is,  \n|AB|+|AF_{2}|+|BF_{2}|=20.  \nSubstituting \\textcircled{1} into the above equation gives  \n|AB|+|AF_{1}|+|BF_{1}|+8=20,  \nthus  \n2|AB|=12, \\quad |AB|=6."
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the hyperbola $x^{2}-\\frac{y^{2}}{24}=1$, and let $P$ be a point on the hyperbola such that $|P F_{1}|+|P F_{2}|=14$. Then the area of $\\Delta P F_{1} F_{2}$ equals?",
    "fact_expressions": "F1: Point;F2: Point;Focus(G) = {F1, F2};G: Hyperbola;Expression(G) = (x^2 - y^2/24 = 1);P: Point;PointOnCurve(P, G) = True;Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 14",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "24",
    "fact_spans": "[[[1, 8]], [[9, 16]], [[1, 51]], [[17, 46], [56, 59]], [[17, 46]], [[52, 55]], [[52, 62]], [[64, 88]]]",
    "query_spans": "[[[90, 118]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=4 x$, with focus $F$ and directrix $l$, let $P$ be a point on the parabola, $P A \\perp l$, where $A$ is the foot of the perpendicular. If the slope of the line $A F$ is $-\\sqrt{3}$, then what is the area of $\\triangle P A F$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;l: Line;Directrix(G) = l;P: Point;PointOnCurve(P, G);A: Point;IsPerpendicular(LineSegmentOf(P, A), l);FootPoint(LineSegmentOf(P, A), l) = A;Slope(LineOf(A, F)) = -sqrt(3)",
    "query_expressions": "Area(TriangleOf(P, A, F))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[2, 16], [36, 39]], [[2, 16]], [[20, 23]], [[2, 23]], [[27, 30]], [[2, 30]], [[32, 35]], [[32, 42]], [[59, 62]], [[43, 56]], [[43, 65]], [[68, 90]]]",
    "query_spans": "[[[93, 115]]]",
    "process": "Analysis: First, according to the given equation of the parabola, find the equation of the directrix and the coordinates of the focus. Let the coordinates of point A be given, and using the two-point slope formula, obtain $ y_{A} = 2\\sqrt{3} $, thus getting $ y_{P} = 2\\sqrt{3} $. Substitute into the equation of the parabola to find the corresponding horizontal coordinate, then find the lengths of the corresponding segments, and finally use the area formula to find the area of the triangle. Since $ y^{2} = 4x $, the directrix is $ l: x = -1 $, and $ F(1, 0) $. Because $ PA \\perp l $ with foot at A, let $ A(-1, y_{A}) $. Since $ k_{AF} = -\\sqrt{3} $, we have $ \\frac{y_{A} - 0}{-1 - 1} = -\\sqrt{3} $, so $ y_{A} = 2\\sqrt{3} $, hence $ y_{P} = 2\\sqrt{3} $. Substituting $ y_{P} = 2\\sqrt{3} $ into $ y^{2} = 4x $, we get $ x_{P} = 3 $. Therefore, $ |AP| = 3 - (-1) = 4 $, so $ S_{\\triangle PAF} = \\frac{1}{2} \\times |AP| \\times |y_{P}| = \\frac{1}{2} \\times 4 \\times 2\\sqrt{3} = 4\\sqrt{3} $. Hence, the answer is $ 4\\sqrt{3} $."
  },
  {
    "text": "Draw a tangent from the left focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ to the circle $x^{2}+y^{2}=a^{2}$, with tangent point $E$. Extend $FE$ to intersect the hyperbola at point $P$. Let $O$ be the origin. If $\\overrightarrow{O E}=\\frac{1}{2}(\\overrightarrow{O F}+\\overrightarrow{O P})$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;LeftFocus(G) = F;H: Circle;Expression(H) = (x^2 + y^2 = a^2);Z: Line;TangentOfPoint(F, H) = Z;E: Point;TangentPoint(Z, H) = E;P: Point;Intersection(OverlappingLine(LineSegmentOf(F, E)), G) = P;O: Origin;VectorOf(O, E) = (1/2)*(VectorOf(O, F) + VectorOf(O, P))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[1, 57], [103, 106], [201, 204]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[61, 64]], [[1, 64]], [[65, 85]], [[65, 85]], [], [[0, 88]], [[92, 95]], [[0, 95]], [[107, 111]], [[96, 111]], [[112, 115]], [[122, 199]]]",
    "query_spans": "[[[201, 210]]]",
    "process": "According to the problem, for the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$, the foci lie on the $x$-axis. Let the focus be $F(c,0)$, then $|OF|=c$, $|OE|=a$, so $|EF|=b$. Since $\\overrightarrow{OE}=\\frac{1}{2}(\\overrightarrow{OF}+\\overrightarrow{OP})$, point $E$ is the midpoint of $PF$, and $OE$ is the midline of triangle $PFF'$, thus $|PF|=2|EF|=2b$, $|PF'|=2a$. By the definition of the hyperbola, $|PF|-|PF'|=2a$, then $b=2a$. Therefore, the eccentricity of the hyperbola is $e=\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{5}$. [Comment] This problem mainly examines the application of the standard equation of a hyperbola and its simple geometric properties. The key to solving it lies in reasonably using the given conditions, combining the definition of the hyperbola and the midline theorem of a triangle to find the relationship between $a$ and $b$. It emphasizes the application of number-shape combination and transformation ideas, and belongs to a medium-difficulty problem."
  },
  {
    "text": "The length of the curve $y=-\\sqrt{4-x^{2}}(x \\leq 1)$ is?",
    "fact_expressions": "G: Curve;Expression(G) = (y = -sqrt(4 - x^2))&(x <= 1)",
    "query_expressions": "Length(G)",
    "answer_expressions": "4*pi/3",
    "fact_spans": "[[[0, 31]], [[0, 31]]]",
    "query_spans": "[[[0, 36]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of a certain hyperbola are $3x \\pm 2y = 0$, and the hyperbola passes through the point $(2, -3\\sqrt{2})$, what is the length of the real axis of the hyperbola?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (2, -3*sqrt(2));Expression(Asymptote(G)) = (3*x + pm*2*y = 0);PointOnCurve(H, G)",
    "query_expressions": "Length(RealAxis(G))",
    "answer_expressions": "6",
    "fact_spans": "[[[3, 6], [31, 34], [57, 60]], [[36, 54]], [[36, 54]], [[3, 28]], [[31, 54]]]",
    "query_spans": "[[[57, 66]]]",
    "process": "Since the asymptotes of the hyperbola are $3x \\pm 2y = 0$ and it passes through the point $(2, -3\\sqrt{2})$, it can be determined that the foci of the hyperbola lie on the $y$-axis. Let the standard equation of this hyperbola be $\\frac{y^{2}}{a^{2}} - \\frac{x^{2}}{b^{2}} = 1$ $(a > 0, b > 0)$. Then $\\frac{a}{b} = \\frac{3}{2}$, $\\frac{18}{a^{2}} - \\frac{4}{b^{2}} = 1$, yielding $a = 3$. Therefore, the length of the real axis of this hyperbola is $6$."
  },
  {
    "text": "A point $P$ on the hyperbola $\\frac{y^{2}}{64}-\\frac{x^{2}}{16}=1$ is at a distance of $1$ from one of its foci. What is the distance from point $P$ to the other focus?",
    "fact_expressions": "G: Hyperbola;P: Point;F1:Point;F2:Point;Expression(G) = (-x^2/16 + y^2/64 = 1);PointOnCurve(P, G);OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2);Distance(P,F1) = 1",
    "query_expressions": "Distance(P,F2)",
    "answer_expressions": "17",
    "fact_spans": "[[[0, 40], [47, 48]], [[43, 46], [64, 68]], [], [], [[0, 40]], [[0, 46]], [[47, 53]], [[47, 74]], [[47, 74]], [[43, 61]]]",
    "query_spans": "[[[47, 80]]]",
    "process": "From the equation of the hyperbola, we obtain the real semi-axis length $ a = 8 $ and the imaginary semi-axis length $ b = 4 $, hence $ c = \\sqrt{80} = 4\\sqrt{5} $. Since the distance from point $ P $ to one focus is equal to 1, and $ a + c = 8 + 4\\sqrt{5} > 1 $, point $ P $ lies on the same side (above or below) of the x-axis as that focus. Therefore, the distance from point $ P $ to the other focus is $ 1 + 2a = 17 $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1(0<b<2)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Let $P$ be a point on the ellipse such that $|P F_{1}|=3$, $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$. Then $b$=?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/b^2 = 1);b: Number;0 < b;b < 2;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 3;AngleOf(F1, P, F2) = pi/3",
    "query_expressions": "b",
    "answer_expressions": "3/2",
    "fact_spans": "[[[2, 55], [85, 87]], [[2, 55]], [[144, 147]], [[9, 55]], [[9, 55]], [[64, 71]], [[72, 79]], [[2, 79]], [[2, 79]], [[81, 84]], [[81, 90]], [[91, 104]], [[106, 142]]]",
    "query_spans": "[[[144, 149]]]",
    "process": "Draw the graph, use the definition of the ellipse to obtain |PF_{2}|, use the cosine theorem to find the value of c, and then find the value of b. According to the definition of the ellipse: |PF_{2}| = 2a - 3 = 1, 4c^{2} = |F_{1}F_{2}|^{2} = |PF_{1}|^{2} + |PF_{2}|^{2} - 2|PF_{1}| \\cdot |PF_{2}| \\cos\\frac{\\pi}{3} = 7 \\therefore c^{2} = \\frac{7}{4}, then b^{2} = a^{2} - c^{2} = 4 - \\frac{7}{4} = \\frac{9}{4}, so, b = \\frac{3}{2}"
  },
  {
    "text": "Given the ellipse $ C $: $ \\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1 $ $ (a > b > 0) $, the upper and lower foci are $ F_{1} $, $ F_{2} $, respectively, and the left and right vertices are $ B_{1} $, $ B_{2} $, respectively. Point $ P $ lies on the ellipse, and $ \\overrightarrow{F_{2} P} = \\frac{1}{2} \\overrightarrow{F_{1} B_{1}} $. Then the eccentricity of ellipse $ C $ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/b^2 + y^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;UpperFocus(C) = F1;LowerFocus(C) = F2;B1: Point;B2: Point;LeftVertex(C) = B1;RightVertex(C) = B2;P: Point;PointOnCurve(P, C) ;VectorOf(F2, P) = VectorOf(F1, B1)/2",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[2, 59], [113, 115], [186, 191]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[68, 75]], [[76, 83]], [[2, 83]], [[2, 83]], [[92, 99]], [[100, 107]], [[2, 107]], [[2, 107]], [[108, 112]], [[108, 116]], [[117, 184]]]",
    "query_spans": "[[[186, 197]]]",
    "process": "It is easy to see that the points are $F_{1}(0,c)$, $F_{2}(0,-c)$, $B_{1}(-b,0)$, $\\overrightarrow{F_{2}P}=(x,y+c)$, $\\overrightarrow{F_{1}B_{1}}=(-b,-$ From $\\overrightarrow{F_{2}P}=\\frac{1}{2}\\overrightarrow{F_{1}B_{1}}$, we obtain $\\begin{cases}x=-\\frac{b}{2}\\\\y+c=-\\frac{c}{2}\\end{cases}$, which gives $\\begin{cases}x=-\\frac{b}{2}\\\\y=-\\frac{3c}{2}\\end{cases}$, so the point $P(-\\frac{b}{2},-\\frac{3c}{2})$,"
  },
  {
    "text": "Let the left and right foci of the ellipse be $F_{1}$ and $F_{2}$, respectively, and the upper vertex be $B$. If $B F_{2}=F_{1} F_{2}=2$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;B: Point;F2: Point;F1: Point;LeftFocus(G) = F1;RightFocus(G) = F2;UpperVertex(G) = B;LineSegmentOf(B,F2)=LineSegmentOf(F1,F2);LineSegmentOf(F1, F2) = 2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 + y^2/3 = 1",
    "fact_spans": "[[[1, 3], [63, 65]], [[32, 35]], [[20, 27]], [[12, 19]], [[1, 27]], [[1, 27]], [[1, 35]], [[37, 60]], [[37, 60]]]",
    "query_spans": "[[[63, 72]]]",
    "process": "Since the left and right foci of the ellipse are $F_{1}$ and $F_{2}$, respectively, and the upper vertex is $B$, if $BF_{2}=F_{1}F_{2}=2$, then $2c=2$, solving gives $c=1$, $a^{2}=4$, thus the equation of the ellipse is $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, a line passing through $F$ intersects the parabola at points $A$ and $B$. If point $M(-1,1)$ and $MA \\perp MB$, then the length of chord $AB$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;G: Line;PointOnCurve(F, G);A: Point;B: Point;Intersection(G, C) = {A, B};M: Point;Coordinate(M) = (-1, 1);IsPerpendicular(LineSegmentOf(M, A), LineSegmentOf(M, B));IsChordOf(LineSegmentOf(A, B), C)",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "5",
    "fact_spans": "[[[2, 21], [37, 40]], [[2, 21]], [[25, 28], [30, 33]], [[2, 28]], [[34, 36]], [[29, 36]], [[41, 44]], [[45, 48]], [[34, 50]], [[52, 62]], [[52, 62]], [[64, 79]], [[37, 87]]]",
    "query_spans": "[[[82, 92]]]",
    "process": "It is easy to see that point $ F(1,0) $. Let points $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. If line $ AB $ coincides with the $ x $-axis, then this line intersects the parabola $ C $ at only one point, which does not meet the requirement. Therefore, line $ AB $ cannot coincide with the $ x $-axis. Let the equation of line $ AB $ be $ x = my + 1 $. Solving simultaneously:\n$$\n\\begin{cases}\nx = my + 1 \\\\\ny^2 = 4x\n\\end{cases}\n$$\nEliminating $ x $ and simplifying yields $ y^{2} - 4my - 4 = 0 $. The discriminant $ \\Delta = 16m^{2} + 16 > 0 $. By Vieta's formulas, we have $ y_{1} + y_{2} = 4m $, $ y_{1}y_{2} = -4 $. $ \\overrightarrow{MA} = (x_{1}+1, y_{1}-1) = (my_{1}+2, y_{1}-1) $. Similarly, $ \\overrightarrow{MB} = (my_{2}+2, y_{2}-1) $. Since $ MA \\perp MB $, then $ \\overrightarrow{MA} \\cdot \\overrightarrow{MB} = (my_{1}+2)(my_{2}+2) + (y_{1}-1)(y_{2}-1) = 0 $. That is, $ (m^{2}+1)y_{1}y_{2} + (2m-1)(y_{1}+y_{2}) + 5 = 0 $. Simplifying gives $ 4m^{2} - 4m + 1 = 0 $, solving yields $ m = \\frac{1}{2} $. Therefore, $ |AB| = x_{1} + x_{2} + 2 = m(y_{1}+y_{2}) + 4 = 4m^{2} + 4 = 5 $."
  },
  {
    "text": "Given that the line $l$ intersects the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$ at points $A$ and $B$, and the point $(4 , 2)$ is the midpoint of segment $AB$, then the equation of line $l$ is?",
    "fact_expressions": "l: Line;G: Ellipse;Expression(G) = (x^2/36 + y^2/9 = 1);A: Point;B: Point;Intersection(l, G) = {A, B};I: Point;Coordinate(I) = (4, 2);MidPoint(LineSegmentOf(A, B)) = I",
    "query_expressions": "Expression(l)",
    "answer_expressions": "Answer missing",
    "fact_spans": "[[[2, 7], [81, 86]], [[8, 46]], [[8, 46]], [[48, 51]], [[52, 55]], [[2, 57]], [[58, 68]], [[58, 68]], [[58, 79]]]",
    "query_spans": "[[[81, 91]]]",
    "process": ""
  },
  {
    "text": "Given the line $l_{1}$: $4 x-3 y+11=0$ and the line $l_{2}$: $x=-1$, find the minimum value of the sum of distances from a moving point $P$ on the parabola $y^{2}=4 x$ to the lines $l_{1}$ and $l_{2}$.",
    "fact_expressions": "G: Parabola;l1:Line;l2:Line;P:Point;Expression(G) = (y^2 = 4*x);Expression(l1)=(4*x-3*y+11=0);Expression(l2)=(x=-1);PointOnCurve(P,G)",
    "query_expressions": "Min(Distance(P,l1)+Distance(P,l2))",
    "answer_expressions": "3",
    "fact_spans": "[[[46, 60]], [[2, 27], [68, 77]], [[28, 45], [78, 87]], [[64, 67]], [[46, 60]], [[2, 27]], [[28, 45]], [[46, 67]]]",
    "query_spans": "[[[64, 98]]]",
    "process": "\\because the focus of the parabola \\( y^{2} = 4x \\) is \\( F(1,0) \\), and the distance from \\( P \\) to the line \\( l_{2} \\) equals the distance from \\( P \\) to the focus of the parabola, \\( \\therefore \\) the minimum value of the sum of the distances from \\( P \\) to the lines \\( l_{1} \\) and \\( l_{2} \\) is the distance from \\( F(1,0) \\) to the line \\( l_{1} \\), which is \\( \\frac{|4+11|}{5} = 3 \\)."
  },
  {
    "text": "If the distance from point $P$ on the parabola $C$: $x^{2}=2 p y(p>0)$ to the focus is $8$, and the distance to the $x$-axis is $6$, then the equation of parabola $C$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 2*(p*y));p: Number;p>0;P: Point;PointOnCurve(P, C);Distance(P, Focus(C)) = 8;Distance(P, xAxis) = 6",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2=8*y",
    "fact_spans": "[[[1, 27], [58, 64]], [[1, 27]], [[9, 27]], [[9, 27]], [[29, 33]], [[1, 33]], [[1, 43]], [[29, 56]]]",
    "query_spans": "[[[58, 69]]]",
    "process": "According to the definition of a parabola, the result can be obtained. By the definition of a parabola, $\\frac{p}{2}=8-6=2$, solving gives $p=4$. Hence, the equation of parabola $C$ is $x^{2}=8y$."
  },
  {
    "text": "Given that point $P$ lies on the parabola $y^{2}=4x$, let $d_{1}$ be the distance from point $P$ to the directrix of the parabola, and $d_{2}$ be the distance from $P$ to a moving point $Q$ on the circle $(x+3)^{2}+(y-3)^{2}=1$. What is the minimum value of $d_{1}+d_{2}$?",
    "fact_expressions": "G: Parabola;H: Circle;P: Point;Q:Point;Expression(G) = (y^2 = 4*x);Expression(H) = ((x + 3)^2 + (y - 3)^2 = 1);PointOnCurve(P, G);Distance(P,Directrix(G)) = d1;PointOnCurve(Q, H);Distance(P,Q)=d2;d1:Number;d2:Number",
    "query_expressions": "Min(d1 + d2)",
    "answer_expressions": "4",
    "fact_spans": "[[[7, 21], [31, 34]], [[49, 73]], [[2, 6], [26, 30]], [[77, 80]], [[7, 21]], [[49, 73]], [[2, 24]], [[26, 47]], [[49, 80]], [[26, 91]], [[40, 47]], [[84, 91]]]",
    "query_spans": "[[[93, 112]]]",
    "process": "Connect the focus of the parabola to the center of the circle. By the definition of the parabola, the length of the line segment between these two points minus the radius of the circle gives the minimum distance we are seeking. \\because the focus of the parabola is (1,0), and the center of the circle is (-3,3), \\therefore the minimum value of $d_{1}+d_{2}$ is 4"
  },
  {
    "text": "Let the two foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{a+1}=1$ be $F_{1}$ and $F_{2}$, and let point $P$ lie on the hyperbola. If $P F_{1} \\perp P F_{2}$, then the minimum distance from point $P$ to the origin $O$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/(a + 1) + x^2/a^2 = 1);a: Number;F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));O: Origin",
    "query_expressions": "Min(Distance(P, O))",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[1, 45], [72, 75]], [[1, 45]], [[4, 45]], [[51, 58]], [[59, 66]], [[1, 66]], [[67, 71], [103, 107]], [[67, 76]], [[78, 101]], [[108, 115]]]",
    "query_spans": "[[[103, 124]]]",
    "process": "Using the given condition PF_{1}\\bot PF_{2}, the distance from point P to the coordinate origin O is c; transform and solve for the minimum value of c. [Detailed solution] The two foci of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{a+1}=1 are F_{1}, F_{2}, and point P lies on the hyperbola. If PF_{1}\\bot PF_{2}, then the distance from point P to the coordinate origin O is c. Therefore, c=\\sqrt{a^{2}+a+1}=\\sqrt{(a+\\frac{1}{2})^{2}+\\frac{3}{4}}\\geqslant\\frac{\\sqrt{3}}{2}, with equality holding if and only if a=-\\frac{1}{2}, achieving the minimum value: \\frac{\\sqrt{3}}{2}."
  },
  {
    "text": "The equation of the hyperbola $C^{\\prime}$ passing through the point $(2,4)$ and having the same asymptotes as the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{8}=1$ is?",
    "fact_expressions": "C: Hyperbola;G: Point;C1:Hyperbola;Expression(C) = (x^2/4 - y^2/8 = 1);Coordinate(G) = (2, 4);PointOnCurve(G, C1);Asymptote(C1)=Asymptote(C)",
    "query_expressions": "Expression(C1)",
    "answer_expressions": "y^2/8 - x^2/4 = 1",
    "fact_spans": "[[[11, 54]], [[1, 9]], [[61, 76]], [[11, 54]], [[1, 9]], [[0, 76]], [[10, 76]]]",
    "query_spans": "[[[61, 81]]]",
    "process": "Let the equation of the hyperbola C be \\frac{x^{2}}{4}-\\frac{y^{2}}{8}=\\lambda (\\lambda\\neq0). Substituting (2,4) gives 1-2=-1=\\lambda, so the equation of the hyperbola C is \\frac{y^{2}}{8}-\\frac{x^{2}}{4}=1."
  },
  {
    "text": "Given that $P$ is a point on the parabola $N$: $x^{2}=8 y$, and points $A(-1,3)$, $B(0,2)$, then the minimum perimeter of $\\triangle P A B$ is?",
    "fact_expressions": "N: Parabola;A: Point;B: Point;P: Point;Expression(N) = (x^2 = 8*y);Coordinate(A) = (-1, 3);Coordinate(B) = (0, 2);PointOnCurve(P, N)",
    "query_expressions": "Min(Perimeter(TriangleOf(P, A, B)))",
    "answer_expressions": "5+sqrt(2)",
    "fact_spans": "[[[6, 25]], [[29, 39]], [[40, 48]], [[2, 5]], [[6, 25]], [[29, 39]], [[40, 48]], [[2, 28]]]",
    "query_spans": "[[[50, 75]]]",
    "process": "It is easy to see that B(0,2) is the focus of the parabola N, AB = \\sqrt{2}, and the perimeter of \\triangle PAB is |PA| + |PB| + |AB| = |PA| + |PB| + \\sqrt{2}. Combining this with the definition of a parabola, the minimum value of |PA| + |PB| is the distance from point A(-1,3) to the directrix y = -2 of the parabola N, which is 3 - (-2) = 5. Therefore, the minimum perimeter of \\triangle PAB is 5 + \\sqrt{2}."
  },
  {
    "text": "Given the line $l_{1}$: $\\sqrt{3} x+y+5=0$, the parabola $C$: $x^{2}=4 y$ has focus $F$ and directrix $l_{2}$. Let $A$ be a point on the parabola $C$, and let $d_{1}$, $d_{2}$ be the distances from $A$ to $l_{1}$ and $l_{2}$, respectively. Then the minimum value of $d_{1}+d_{2}$ is?",
    "fact_expressions": "C: Parabola;l1: Line;F: Point;l2: Line;A: Point;Expression(C) = (x^2 = 4*y);Expression(l1)=(sqrt(3)*x + y + 5 = 0);Focus(C) = F;Directrix(C) = l2;PointOnCurve(A, C);d1:Number;d2:Number;Distance(A,l1)=d1;Distance(A,l2)=d2",
    "query_expressions": "Min(d1+ d2)",
    "answer_expressions": "3",
    "fact_spans": "[[[32, 51], [75, 81]], [[2, 31], [90, 97]], [[55, 58]], [[62, 69], [100, 107]], [[71, 74], [86, 89]], [[32, 51]], [[2, 31]], [[32, 58]], [[32, 69]], [[71, 85]], [[113, 120]], [[122, 130]], [[86, 131]], [[86, 131]]]",
    "query_spans": "[[[133, 152]]]",
    "process": "As shown in the figure, let M be the projection of A on $ l_{2} $, and N be the projection of A on $ l_{1} $. By the definition of a parabola, we have $ AM = AF $, so $ d_{1} + d_{2} = AN + AF $. Therefore, when points A, F, and N are collinear, $ d_{1} + d_{2} $ reaches its minimum value, which is the distance from F to the line $ l_{1} $, $ \\frac{|1+5|}{2} = 3 $."
  },
  {
    "text": "Given the line $y = kx + m$ ($k > 0$) intersects the parabola $C$: $y^2 = 4x$ and its directrix at points $M$ and $N$ respectively, $F$ is the focus of the parabola. If $3 \\overrightarrow{FM} = \\overrightarrow{MN}$, then what is the value of $k$?",
    "fact_expressions": "G: Line;Expression(G) = (y = k*x + m);k: Number;k>0;m: Number;C: Parabola;Expression(C) = (y^2 = 4*x);M: Point;Intersection(G, C) = M;N: Point;Intersection(G, Directrix(C)) = N;F: Point;Focus(C) = F;3*VectorOf(F, M) = VectorOf(M, N)",
    "query_expressions": "k",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 18]], [[2, 18]], [[116, 119]], [[4, 18]], [[4, 18]], [[19, 38], [39, 40], [60, 63]], [[19, 38]], [[46, 49]], [[2, 55]], [[50, 53]], [[2, 55]], [[56, 59]], [[56, 66]], [[68, 114]]]",
    "query_spans": "[[[116, 122]]]",
    "process": "The parabola $ C: y^{2} = 4x $ has focus $ F(1,0) $. The line $ l: y = kx + m $ passes through the focus of the parabola, so $ k + m = 0 $. Draw $ NN' \\perp $ the directrix $ x = -1 $, with foot $ N' $. By the definition of the parabola, $ |NN'| = |NF| $. Since $ \\angle NNM $ is equal to the inclination angle of line $ l $, and given $ 3\\overrightarrow{FM} = \\overrightarrow{MN} $, then $ \\cos\\angle NNM = \\frac{|NN'|}{|MN|} = \\frac{1}{3} $, thus $ \\tan\\angle NNM = \\pm 2\\sqrt{2} $. Because $ k > 0 $, the slope of line $ l $ is $ k = 2\\sqrt{2} $."
  },
  {
    "text": "If the hyperbola $\\frac{x^{2}}{m^{2}}-y^{2}=1$ $(m>0)$ has eccentricity $2$, then $m=$?",
    "fact_expressions": "G: Hyperbola;m: Number;m>0;Expression(G) = (-y^2 + x^2/m^2 = 1);Eccentricity(G) = 2",
    "query_expressions": "m",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[1, 38]], [[48, 51]], [[4, 38]], [[1, 38]], [[1, 46]]]",
    "query_spans": "[[[48, 53]]]",
    "process": "The hyperbola $\\frac{x^{2}}{m^{2}}-y^{2}=1$ $(m>0)$ has eccentricity $e=\\frac{c}{a}=\\frac{\\sqrt{m^{2}+1}}{m}=2$, $\\therefore m=\\frac{\\sqrt{3}}{3}$, so the answer is $\\frac{\\sqrt{3}}{3}$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a, b>0)$, respectively, a line $l$ passing through $F_{2}$ intersects the right branch of the hyperbola at points $A$ and $B$. The inradius of $\\Delta A F_{1} F_{2}$ is $r_{1}$, and the inradius of $\\Delta B F_{1} F_{2}$ is $r_{2}$. If $r_{1}=2 r_{2}$, then what is the slope of line $l$?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;A: Point;F1: Point;F2: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F2, l);Intersection(l, RightPart(C)) = {A, B};Radius(InscribedCircle(TriangleOf(A,F1,F2)))=r1;Radius(InscribedCircle(TriangleOf(B,F1,F2)))=r2;r1:Number;r2:Number;r1 = 2*r2",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*2*sqrt(2)",
    "fact_spans": "[[[95, 100], [215, 220]], [[20, 79], [101, 104]], [[28, 79]], [[28, 79]], [[111, 114]], [[2, 9]], [[10, 17], [87, 94]], [[115, 118]], [[28, 79]], [[28, 79]], [[20, 79]], [[2, 85]], [[2, 85]], [[86, 100]], [[95, 120]], [[121, 157]], [[160, 196]], [[150, 157]], [[189, 196]], [[198, 213]]]",
    "query_spans": "[[[215, 225]]]",
    "process": "Fully utilize $^{2}$|AM|=|AN|, |F_{1}M|=|F_{1}E|, |F_{2}N|=|F_{2}E|, and combine with the definition of hyperbola to obtain |F_{1}E|-|F_{2}E|=2a, thus the x-coordinate of the incenter of $\\triangle AF_{1}F_{2}$ is a, implying CD\\bot x-axis. In $\\triangle CEF_{2}$, $\\triangle DEF_{2}$, using knowledge of solving right triangles, the definition of tangent function, and double angle formula, the slope of the line can be simplified. [Solution] Denote the incenter of $\\triangle AF_{1}F_{2}$ as C. Clearly, C and E have equal x-coordinates, so |AM|=|AN|, |F_{1}M|=|F_{1}E|, |F_{2}N|=|F_{2}E|. From |AF_{1}|-|AF_{2}|=2a, i.e., |AM|+|MF_{1}|-(|AN|+|NF_{2}|)=2a, we get |MF_{1}|-|NF_{2}|=2a, that is, |F_{1}E|-|F_{2}E|=2a. Let the x-coordinate of C be $x_{0}$, then E$(x_{0},0)$, thus $x_{0}+c-(c-x_{0})=2a$, yielding $x_{0}=a$. Similarly, the x-coordinate of incenter D is also a, hence CD\\bot x-axis. Let the inclination angle of the line be $\\theta$, then $\\angle OF_{2}D=\\frac{\\theta}{2}$, $\\angle CF_{2}O=90^{\\circ}-\\frac{\\theta}{2}$. In $\\triangle CEF_{2}$, $\\tan\\angle CF_{2}O=\\tan(90^{\\circ}-\\frac{\\theta}{2})=\\frac{r_{1}}{|EF|}$. In $\\triangle DEF_{2}$, $\\tan\\angle DF_{2}O=\\tan\\frac{\\theta}{2}=\\frac{r_{2}}{|EF|}$. From $r_{1}=2r_{2}$, we get $2\\tan\\frac{\\theta}{2}=\\tan(90^{\\circ}-\\frac{\\theta}{2})=\\cot\\frac{\\theta}{2}$. Solving gives $\\tan\\frac{\\theta}{2}=\\frac{\\sqrt{2}}{2}$, then the slope of the line is $\\tan\\theta=\\frac{2\\tan\\frac{\\theta}{2}}{1-\\tan^{2}\\frac{\\theta}{2}}=\\frac{\\sqrt{2}}{1-\\frac{1}{2}}=2\\sqrt{2}$. By symmetry, the slope of line l is $\\pm2\\sqrt{2}$."
  },
  {
    "text": "Given that the center of the ellipse is at the origin and it passes through the point $P(3,0)$, with $a=3b$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;P: Point;Coordinate(P) = (3, 0);PointOnCurve(P, G);a: Number;b: Number;a=3*b",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2/81+x^2/9=1, x^2/9+y^2=1}",
    "fact_spans": "[[[2, 4], [34, 36]], [[8, 10]], [[2, 10]], [[14, 23]], [[14, 23]], [[2, 23]], [[25, 32]], [[25, 32]], [[25, 32]]]",
    "query_spans": "[[[34, 43]]]",
    "process": "When the foci are on the x-axis, let the ellipse equation be \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a_{1}>b_{1}>0). Since the ellipse passes through point P(3,0), we know a_{1}=3b_{1}. Solving gives b_{1}^{2}=1, a_{1}^{2}=9. Thus, the standard equation of the ellipse is \\frac{x^{2}}{9}+y^{2}=1. When the foci are on the y-axis, let the ellipse equation be \\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1 (a_{2}>b_{2}>0). Since the ellipse passes through point P(3,0), we know \\frac{0}{a_{2}^{2}}+\\frac{4}{l}\\frac{9}{2}=1 and a_{2}=3b_{2}. Solving gives a_{2}^{2}=81, b_{2}^{2}=9. Thus, the standard equation of the ellipse is \\frac{y^{2}}{81}+\\frac{x^{2}}{9}=1. In summary, the standard equations of the ellipse are \\frac{y^{2}}{81}+\\frac{x^{2}}{0}=1 or \\frac{x^{2}}{9}+y^{2}=1. [Note: This problem examines solving for the standard equation of an ellipse. The key is to find a, b, c, and also pay attention to the position of the ellipse's foci. It tests computational and problem-solving abilities and is a basic-level question.]"
  },
  {
    "text": "Let the parabola $C$: $y^{2}=2 p x(p>0)$ have focus $F$, and let points $A$ and $B$ in the first quadrant lie on $C$, with $O$ being the origin. If $\\angle A F O = \\angle A F B = \\frac{\\pi}{3}$ and $|A B| = 2 \\sqrt{7}$, then the coordinates of point $A$ are?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;A: Point;B: Point;Quadrant(A) = 1;Quadrant(B) = 1;PointOnCurve(A, C);PointOnCurve(B, C);O: Origin;AngleOf(A, F, O) = AngleOf(A, F, B);AngleOf(A, F, B) = pi/3;Abs(LineSegmentOf(A, B)) = 2*sqrt(7)",
    "query_expressions": "Coordinate(A)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 27], [52, 55]], [[1, 27]], [[9, 27]], [[9, 27]], [[31, 34]], [[1, 34]], [[41, 44], [130, 134]], [[45, 48]], [[35, 44]], [[35, 48]], [[41, 56]], [[41, 56]], [[57, 60]], [[67, 108]], [[67, 108]], [[110, 128]]]",
    "query_spans": "[[[130, 139]]]",
    "process": "According to the given conditions, draw perpendiculars from points A and B to the x-axis, with feet at D and M respectively. Let A(x_{1},y_{1}), and given F(\\frac{p}{2},0) and \\angle AFO = \\frac{\\pi}{3}, combining with the focal radius formula of the parabola, we obtain |DF| = \\frac{p}{2} - x_{1} = \\frac{|AF|}{2} = \\frac{x_{1} + \\frac{p}{2}}{2}, thus solving |AF| = \\frac{2p}{3}, |BF| = 2p. Then solving \\triangle ABF gives the solution. As shown in the figure, draw perpendiculars from points A and B to the x-axis, with feet at D and M. Let A(x_{1},y_{1}). Given F(\\frac{p}{2},0) and \\angle AFO = \\frac{\\pi}{3}, so |DF| = \\frac{|AF|}{2}, \\frac{p}{2} - x_{1} = \\frac{|AF|}{2} = \\frac{x_{1} + \\frac{p}{2}}{2}. Thus x_{1} = \\frac{p}{6}, so |AF| = \\frac{2p}{3}. Similarly, |BF| = 2p. Hence in \\triangle ABF, \\cos\\angle AFB = \\frac{|AF|^{2} + |BF|^{2} - |AB|^{2}}{2|AF|\\cdot|BF|} = \\frac{\\frac{4}{9}p^{2} + 4p^{2} - 28}{2 \\times \\frac{2}{3}p \\times 2p} = \\cos\\frac{\\pi}{3} = \\frac{1}{2}. Solving gives p = 3, so x_{1} = \\frac{1}{2}, y_{1} = \\sqrt{3}, therefore A(\\frac{1}{2},\\sqrt{3})."
  },
  {
    "text": "Given that $F$ is a focus of the hyperbola $C$: $x^{2}-m y^{2}=3 m$ ($m>0$), what is the distance from point $F$ to an asymptote of $C$?",
    "fact_expressions": "C: Hyperbola;m: Number;F: Point;m>0;Expression(C) = (-m*y^2 + x^2 = 3*m);OneOf(Focus(C)) = F",
    "query_expressions": "Distance(F, OneOf(Asymptote(C)))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[6, 38], [50, 53]], [[13, 38]], [[2, 5], [45, 49]], [[13, 38]], [[6, 38]], [[2, 43]]]",
    "query_spans": "[[[45, 64]]]",
    "process": "Analysis: Convert the hyperbola equation into standard form, find the coordinates of the foci and the equation of one asymptote, then use the point-to-line distance formula to obtain the conclusion. The hyperbola C: $x^{2}-my^{2}=3m$ ($m>0$) can be rewritten as $\\frac{x^{2}}{3m}-\\frac{y^{2}}{3}=1$, so one focus is $(\\sqrt{3m+3},0)$, and one asymptote equation is $x+\\sqrt{m}y=0$. Therefore, the distance from point F to one asymptote of C is $\\frac{\\sqrt{3m+3}}{\\sqrt{1+m}}=\\sqrt{3}$."
  },
  {
    "text": "Given that line $l$ intersects the parabola $C$: $y^{2}=x$ at points $A$ and $B$, and the midpoint of segment $AB$ lies on the line $y=1$. If $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}=0$ ($O$ being the origin), then the area of $\\triangle A O B$ is?",
    "fact_expressions": "l: Line;C: Parabola;Expression(C) = (y^2 = x);Intersection(l, C) = {A, B};A: Point;B: Point;l1: Line;Expression(l1) = (y = 1);PointOnCurve(MidPoint(LineSegmentOf(A,B)),l1) = True;DotProduct(VectorOf(O, A), VectorOf(O, B)) = 0;O: Origin",
    "query_expressions": "Area(TriangleOf(A, O, B))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 7]], [[8, 25]], [[8, 25]], [[2, 36]], [[27, 30]], [[31, 34]], [[49, 56]], [[49, 56]], [[38, 57]], [[59, 110]], [[111, 114]]]",
    "query_spans": "[[[122, 144]]]",
    "process": "From the given conditions, the slope of line $ l $ is not zero, so let the equation of line $ l $ be $ x = my + t $ ($ t \\neq 0 $). From \n\\[\n\\begin{cases}\nx = my + t \\\\\ny^2 = x\n\\end{cases},\n\\]\neliminating $ x $ gives $ y^2 - my - t = 0 $. Let $ A(x_1, y_1) $, $ B(x_2, y_2) $, then $ y_1 + y_2 = m $, $ y_1 y_2 = -t $, and $ \\Delta = m^2 + 4t > 0 $. Since the midpoint of segment $ AB $ lies on the line $ y = 1 $, we have $ y_1 + y_2 = 2 $, so $ m = 2 $. Because $ \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = 0 $, it follows that\n\\[\nx_1 x_2 + y_1 y_2 = y_1 y_2 + (my_1 + t)(my_2 + t) = y_1 y_2 + m^2 y_1 y_2 + mt(y_1 + y_2) + t^2 = y_1 y_2 + 4y_1 y_2 + 2t \\times 2 + t^2 = t^2 - t = 0,\n\\]\nsolving gives $ t = 1 $ or $ t = 0 $ (discarded). Thus, $ y_1 + y_2 = 2 $, $ y_1 y_2 = -1 $, and the equation of line $ l $ is $ x = 2y + 1 $. Therefore,\n\\[\n|AB| = \\sqrt{1 + 2^2} \\sqrt{(y_1 + y_2)^2 - 4y_1 y_2} = 2\\sqrt{10}.\n\\]\nThe distance from the origin $ O $ to line $ l $ is\n\\[\nd = \\frac{|0 - 0 - 1|}{\\sqrt{5}} = \\frac{\\sqrt{5}}{5}.\n\\]\nThus, the area of $ \\triangle AOB $ is\n\\[\n\\frac{1}{2} \\times |AB| \\times d = \\frac{1}{2} \\times 2\\sqrt{10} \\times \\frac{\\sqrt{5}}{5} = \\sqrt{2}.\n\\]"
  },
  {
    "text": "Points $A$ and $B$ are two points on the right branch of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$. The distance from the midpoint of $AB$ to the $y$-axis is $4$. Then the maximum value of $AB$ is?",
    "fact_expressions": "G: Hyperbola;A: Point;B: Point;Expression(G) = (x^2/4 - y^2/5 = 1);PointOnCurve(A, RightPart(G));PointOnCurve(B, RightPart(G));Distance(MidPoint(LineSegmentOf(A,B)),yAxis)=4",
    "query_expressions": "Max(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[9, 47]], [[0, 4]], [[5, 8]], [[9, 47]], [[0, 53]], [[0, 53]], [[54, 72]]]",
    "query_spans": "[[[74, 84]]]",
    "process": "Let the right focus be F, the midpoint of AB be M, then AB ≤ AF + BF = e(x_{A} - \\frac{a^{2}}{c} + x_{B} - \\frac{a^{2}}{c}) = e(2x_{M} - \\frac{2a^{2}}{c}) = \\frac{3}{2}(2 \\times 4 - \\frac{2 \\times 4}{3}) = 8"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. If there exists a point $P$ on the ellipse such that $|P F_{1}|=e|P F_{2}|$, then the range of values for the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;e: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;Abs(LineSegmentOf(P,F1))=e*Abs(LineSegmentOf(P,F2));Eccentricity(G) = e;PointOnCurve(P,G)",
    "query_expressions": "Range(e)",
    "answer_expressions": "(\\sqrt{2}-1,1)",
    "fact_spans": "[[[2, 54], [83, 85], [120, 122]], [[4, 54]], [[4, 54]], [[90, 93]], [[64, 72]], [[74, 81]], [[126, 129]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 81]], [[2, 81]], [[94, 116]], [[120, 129]], [81, 90]]",
    "query_spans": "[[[126, 136]]]",
    "process": ""
  },
  {
    "text": "Given that $A$ and $F$ are the lower vertex and left focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{3}=1$ $(a>\\sqrt{3})$, respectively, a line $l$ passing through $A$ with an inclination angle of $60^{\\circ}$ intersects the ellipse $C$ at point $M$ (distinct from point $A$), and the perimeter of $\\Delta F A M$ is $4 a$, then the area of $\\triangle F A M$ is?",
    "fact_expressions": "l: Line;C: Ellipse;a: Number;F: Point;A: Point;M: Point;a>sqrt(3);Expression(C) = (y^2/3 + x^2/a^2 = 1);LowerVertex(C) = A;LeftFocus(C) = F;PointOnCurve(A, l);Inclination(l) = ApplyUnit(60, degree);Intersection(l, C) = M;Negation(M=A);Perimeter(TriangleOf(F, A, M)) = 4*a",
    "query_expressions": "Area(TriangleOf(F, A, M))",
    "answer_expressions": "8*sqrt(3)/5",
    "fact_spans": "[[[100, 105]], [[12, 69], [106, 111]], [[18, 69]], [[6, 9]], [[2, 5], [79, 82], [119, 123]], [[112, 116]], [[18, 69]], [[12, 69]], [[2, 77]], [[2, 77]], [[78, 105]], [[83, 105]], [[100, 116]], [[112, 124]], [[126, 149]]]",
    "query_spans": "[[[151, 173]]]",
    "process": "As shown in the figure, let the right focus be F, and A, M lie on the ellipse, then |FA| + |FA| = 2a, |FM| + |FM| = 2a, hence |FA| + |FM| + |FA| + |FM| = 4a. Also, the perimeter of \\triangle FAM is 4a, \\therefore |AM| = |FA| + |FM|, i.e., points A, F, M are collinear. Since the inclination angle of line l is 60^{\\circ}, \\therefore the slope of line l is \\sqrt{3}. Given A(0, -\\sqrt{3}), F(c, 0), we have \\frac{-\\sqrt{3}-0}{0-c} = \\sqrt{3}, then c = 1. Thus a = 2, so the equation of the ellipse is \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1. Solving the system \\begin{cases} y = \\sqrt{3}x - \\sqrt{3} \\\\ y = \\sqrt{3}x - \\sqrt{3} \\end{cases}, we obtain A(0, -\\sqrt{3}), M(\\frac{8}{5}, \\frac{3\\sqrt{3}}{5}). \\therefore s_{\\Delta FAM} = \\frac{1}{2} \\times 2 \\times (\\frac{3\\sqrt{3}}{5} + \\sqrt{3}) = \\frac{8\\sqrt{3}}{5}"
  },
  {
    "text": "The line $l$ passing through the focus $F$ of the parabola $C$: $y^{2}=4x$ intersects $C$ at points $A$ and $B$. Let the projections of $A$ and $B$ onto the $y$-axis be $A^{\\prime}$ and $B^{\\prime}$, respectively. If $|AB|=\\frac{3}{2}(|AA^{\\prime}|+|BB^{\\prime}|)$, then what is the slope of line $l$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, C) = {A, B};A1: Point;B1: Point;Projection(A, yAxis) = A1;Projection(B, yAxis) = B1;Abs(LineSegmentOf(A, B)) = (3/2)*(Abs(LineSegmentOf(A, A1)) + Abs(LineSegmentOf(B, B1)))",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*sqrt(2)",
    "fact_spans": "[[[1, 20], [33, 36]], [[1, 20]], [[23, 26]], [[1, 26]], [[27, 32], [147, 152]], [[0, 32]], [[37, 40], [48, 51]], [[41, 44], [52, 55]], [[27, 46]], [[67, 79]], [[81, 93]], [[48, 93]], [[48, 93]], [[95, 145]]]",
    "query_spans": "[[[147, 157]]]",
    "process": "From the definition of the parabola, we have: |AB| = |AF| + |BF| = |AA| + 1 + |BB| + 1 = |AA| + |BB| + 2 = \\frac{3}{2}(|AA| + |BB|) \\therefore |AA| + |BB| = 4, \\therefore |AB| = 6. Let the inclination angle of line l be \\alpha, then |AB| = \\frac{4}{\\sin^{2}\\alpha} = 6, \\therefore \\sin^{2}\\alpha = \\frac{2}{3}. \\tan\\alpha = \\pm\\sqrt{2}, that is, the slope of line l is \\pm\\sqrt{2}."
  },
  {
    "text": "Given that the moving circle $P$ is internally tangent to the fixed circle $B$: $x^{2}+y^{2}+2 \\sqrt{5} x-31=0$, and that the moving circle $P$ passes through a fixed point $A(\\sqrt{5}, 0)$. What is the equation of the locus $E$ of the center of the moving circle $P$?",
    "fact_expressions": "P: Circle;B: Circle;Expression(B) = (x^2+y^2+2*sqrt(5)*x-31=0);IsInTangent(P,B);A: Point;Coordinate(A) = (sqrt(5), 0);PointOnCurve(A,P);P1:Point ;Center(P)=P1;Locus(P1)=E;E:Curve",
    "query_expressions": "Expression(E)",
    "answer_expressions": "x^2/9+y^2/4=1",
    "fact_spans": "[[[4, 7], [52, 55], [79, 81]], [[10, 46]], [[10, 46]], [[4, 48]], [[60, 76]], [[60, 76]], [[52, 76]], [[83, 86]], [[79, 86]], [[83, 92]], [[89, 92]]]",
    "query_spans": "[[[89, 97]]]",
    "process": "From $ B: x^{2} + y^{2} + 2\\sqrt{5}x - 31 = 0 $, we obtain $ (x + \\sqrt{5})^{2} + y^{2} = 36 $. Therefore, the center of the circle is $ B(-\\sqrt{5}, 0) $, and the radius $ r = 6 $. Since the moving circle $ P $ is internally tangent to the fixed circle $ B: x^{2} + y^{2} + 2\\sqrt{5}x - 31 = 0 $ and passes through $ A(\\sqrt{5}, 0) $, it follows that $ |PA| + |PB| = 6 $. Hence, the locus $ E $ of the center $ P $ of the moving circle is an ellipse with foci at $ B $ and $ A $, and major axis length $ 6 $. Let the equation of the ellipse be $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $). Then $ 2a = 6 $, so $ a = 3 $, $ c = \\sqrt{5} $, and thus $ b^{2} = a^{2} - c^{2} = 4 $. Therefore, the equation of the ellipse is $ \\frac{x^{2}}{9} + \\frac{y^{2}}{4} = 1 $."
  },
  {
    "text": "Given that the minor axis of ellipse $E$ is $6$ and the distance from focus $F$ to an endpoint of the major axis is $9$, then the eccentricity of ellipse $E$ is equal to?",
    "fact_expressions": "E: Ellipse;F: Point;Focus(E)=F;Length(MinorAxis(E))=6;Distance(F,Endpoint(MajorAxis(E)))=9",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "4/5",
    "fact_spans": "[[[2, 7], [35, 40]], [[18, 21]], [[2, 21]], [[2, 15]], [[2, 33]]]",
    "query_spans": "[[[35, 47]]]",
    "process": ""
  },
  {
    "text": "If the foci of a hyperbola are $(\\sqrt{5}, 0)$ and the asymptotes are given by $y = \\pm \\frac{x}{2}$, then what is the standard equation of this hyperbola?",
    "fact_expressions": "G: Hyperbola;Coordinate(OneOf(Focus(G))) = (sqrt(5), 0);Expression(Asymptote(G)) = (y = pm*(x/2))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4-y^2=1",
    "fact_spans": "[[[1, 4], [52, 55]], [[1, 23]], [[1, 49]]]",
    "query_spans": "[[[52, 62]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of the hyperbola $k x^{2}-2 y^{2}=k$ is $\\sqrt{3}$, what is the value of the real number $k$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (k*x^2 - 2*y^2 = k);k: Real;Eccentricity(G) = sqrt(3)",
    "query_expressions": "k",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 24]], [[2, 24]], [[41, 46]], [[2, 39]]]",
    "query_spans": "[[[41, 50]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{m^{2}}=1$, with its foci on the $x$-axis, then its focal distance is?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/8 + y^2/m^2 = 1);PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(8-m^2)",
    "fact_spans": "[[[2, 43], [44, 45], [55, 56]], [[4, 43]], [[2, 43]], [[44, 53]]]",
    "query_spans": "[[[55, 60]]]",
    "process": "\\because the ellipse equation is \\frac{x^{2}}{8}+\\frac{y^{2}}{m^{2}}=1, with foci on the x-axis, \\therefore c=\\sqrt{8-m^{2}}, \\therefore the focal length of this ellipse equation is 2\\sqrt{8-m^{2}}"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has an eccentricity of $2$, and the hyperbola $C$ shares the same foci with the ellipse $\\frac{x^{2}}{5}+y^{2}=1$. Point $P$ lies on the hyperbola $C$. From point $P$, perpendiculars are drawn to the two asymptotes of hyperbola $C$, with feet of the perpendiculars denoted as $A$ and $B$. Then the minimum value of $|A B|$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(C) = 2;G: Ellipse;Expression(G) = (x^2/5 + y^2 = 1);Focus(C) = Focus(G);P: Point;PointOnCurve(P, C);Z1: Line;Z2: Line;L1: Line;L2: Line;Asymptote(C) = {L1, L2};PointOnCurve(P, Z1);PointOnCurve(P, Z2);IsPerpendicular(Z1, L1);IsPerpendicular(Z2, L2);A: Point;B: Point;FootPoint(Z1, L1) = A;FootPoint(Z2, L2) = B",
    "query_expressions": "Min(Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[2, 63], [73, 79], [120, 126], [136, 142]], [[2, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[2, 71]], [[80, 107]], [[80, 107]], [[73, 113]], [[114, 119], [129, 133]], [[115, 127]], [], [], [], [], [[136, 147]], [[128, 150]], [[128, 150]], [[128, 150]], [[128, 150]], [[156, 159]], [[160, 163]], [[128, 163]], [[128, 163]]]",
    "query_spans": "[[[165, 178]]]",
    "process": "From the given conditions, we have\n\\begin{cases}\ne = \\frac{c}{a} = 2, \\\\\nc = \\sqrt{5 - 1} = 2,\n\\end{cases}\nthen $ a^{2} = 1 $, $ b^{2} = 4 - 1 = 3 $, so the equation of hyperbola $ C $ is $ x^{2} - \\frac{y^{2}}{3} = 1 $, and its asymptotes are $ \\sqrt{3}x \\pm y = 0 $. Let point $ P(x_{0}, y_{0}) $, $ |PA| = m $, $ |PB| = n $, then $ m = \\frac{|\\sqrt{3}x_{0} + y_{0}|}{2} $, $ n = \\frac{|\\sqrt{3}x_{0} - y_{0}|}{2} $. Since point $ P $ lies on hyperbola $ C $, we have $ x_{0}^{2} - \\frac{y_{0}^{2}}{3} = 1 $, thus $ mn = \\frac{3}{4} $. Because the angle between asymptotes $ \\sqrt{3}x - y = 0 $ is $ \\angle AOB = \\frac{2\\pi}{3} $, hence $ \\angle APB = \\frac{\\pi}{3} $. In triangle $ \\triangle APB $, by the law of cosines, we get $ |AB|^{2} = m^{2} + n^{2} - 2mn\\cos\\frac{\\pi}{3} = m^{2} + n^{2} - mn \\geqslant mn = \\frac{3}{4} $, with equality if and only if $ m = n $, then $ |AB| \\geqslant \\frac{\\sqrt{3}}{2} $, i.e., the minimum value of $ |AB| $ is $ \\frac{\\sqrt{3}}{2} $."
  },
  {
    "text": "Given the parabola equation $x^{2}=4 y$, a line passing through the point $M(0 , m)$ intersects the parabola at points $A(x_{1} , y_{1})$ and $B(x_{2} , y_{2})$, and $x_{1} x_{2}=-4$. Then the value of $m$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);H: Line;M: Point;m: Number;Coordinate(M) = (0, m);PointOnCurve(M, H);A: Point;B: Point;Intersection(H, G) = {A,B};x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);x1*x2 = -4",
    "query_expressions": "m",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 5], [36, 39]], [[2, 19]], [[33, 35]], [[21, 32]], [[101, 104]], [[21, 32]], [[20, 35]], [[40, 58]], [[61, 79]], [[33, 81]], [[40, 58]], [[61, 79]], [[40, 58]], [[61, 79]], [[40, 58]], [[61, 79]], [[83, 99]]]",
    "query_spans": "[[[101, 108]]]",
    "process": ""
  },
  {
    "text": "Through the focus $F$ of the parabola $E$: $y^{2}=4x$, two mutually perpendicular lines $l_{1}$ and $l_{2}$ are arbitrarily drawn, intersecting the parabola $E$ at points $A$, $B$ and $C$, $D$ respectively. Then the minimum value of $|AB|+4|CD|$ is?",
    "fact_expressions": "E: Parabola;Expression(E) = (y^2 = 4*x);F: Point;Focus(E) = F;l1: Line;l2: Line;PointOnCurve(F, l1) = True;PointOnCurve(F, l2) = True;IsPerpendicular(l1, l2);Intersection(l1, E) = {A, B};Intersection(l2, E) = {C, D};A: Point;B: Point;C: Point;D: Point",
    "query_expressions": "Min(Abs(LineSegmentOf(A, B)) + 4*Abs(LineSegmentOf(C, D)))",
    "answer_expressions": "36",
    "fact_spans": "[[[1, 20], [58, 64]], [[1, 20]], [[23, 26]], [[1, 26]], [[35, 45]], [[47, 54]], [[0, 54]], [[0, 54]], [[30, 54]], [[35, 85]], [[35, 85]], [[66, 69]], [[70, 73]], [[76, 79]], [[80, 83]]]",
    "query_spans": "[[[87, 107]]]",
    "process": ""
  },
  {
    "text": "The distance from a point $M$ on the parabola $x^{2}=2 p y(p>0)$ to the focus is $a(a>\\frac{p}{2})$. What is the vertical coordinate of point $M$?",
    "fact_expressions": "G: Parabola;p: Number;M: Point;p>0;a: Number;a > p/2;Expression(G) = (x^2 = 2*(p*y));PointOnCurve(M, G);Distance(M, Focus(G)) = a",
    "query_expressions": "YCoordinate(M)",
    "answer_expressions": "a - 2*p",
    "fact_spans": "[[[0, 21]], [[3, 21]], [[24, 27], [54, 58]], [[3, 21]], [[34, 52]], [[34, 52]], [[0, 21]], [[0, 27]], [[0, 52]]]",
    "query_spans": "[[[54, 64]]]",
    "process": "According to the problem, the equation of the directrix of the parabola is $ y = -\\frac{p}{2} $. By the definition of a parabola, the distance from point M to the focus is equal to the distance from M to the directrix. Therefore, the vertical coordinate of point M is $ a - \\frac{p}{2} $. [Note: This question primarily examines the definition of a parabola and the method for finding coordinates of points on a parabola, belonging to basic problems.]"
  },
  {
    "text": "If the equation $\\frac{x^{2}}{7-m}+\\frac{y^{2}}{4}=1$ represents an ellipse, then the range of real values for $m$ is?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (x^2/(7 - m) + y^2/4 = 1)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-oo, 3)+(3, 7)",
    "fact_spans": "[[[42, 44]], [[46, 51]], [[1, 44]]]",
    "query_spans": "[[[46, 58]]]",
    "process": "From the given conditions, we have \\begin{cases}7-m>0\\\\7-m\\neq4\\end{cases}, solving yields: m<7 and m\\neq3, so the range of real number m is (-\\infty,3)\\cup(3,7)"
  },
  {
    "text": "Given the ellipse $ C $: $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{4} = 1 $ ($ a > 2 $), the left and right foci are $ F_{1} $ and $ F_{2} $, respectively. If there exists a point $ P $ on $ C $ such that $ P F_{1} \\perp P F_{2} $, then the range of the eccentricity of the hyperbola $ \\Gamma $: $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{8} = 1 $ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/4 + x^2/a^2 = 1);a: Number;a>2;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C) = True;IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2)) = True;Gamma: Hyperbola;Expression(Gamma) = (x^2/a^2 - y^2/8 = 1)",
    "query_expressions": "Range(Eccentricity(Gamma))",
    "answer_expressions": "(1, \\sqrt{2}]",
    "fact_spans": "[[[2, 53], [79, 82]], [[2, 53]], [[8, 53]], [[8, 53]], [[62, 69]], [[70, 77]], [[2, 77]], [[2, 77]], [[85, 89]], [[79, 89]], [[91, 114]], [[116, 168]], [[116, 168]]]",
    "query_spans": "[[[116, 179]]]",
    "process": "Since there exists a point P on C such that PF_{1}\\bot PF_{2}, it follows that \\angle F_{1}PF_{2} \\geqslant 90^{\\circ}. For the ellipse, c_{\\text{ell}}^{2} \\geqslant b^{2} = 4 \\Rightarrow a^{2} = 4 + c_{\\text{ell}}^{2} \\geqslant 8. Therefore, e_{\\text{hyp}}^{2} = \\frac{a^{2} + 8}{a^{2}} = 1 + \\frac{8}{a^{2}} \\leqslant 2 \\Rightarrow e_{\\text{hyp}} \\in (1, \\sqrt{2}]"
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, $F_{1}$, $F_{2}$ are the foci, and the cosine of $\\angle F_{1} P F_{2}$ when it reaches its maximum value is $\\frac{1}{3}$, then the eccentricity of this ellipse is?",
    "fact_expressions": "P: Point;PointOnCurve(P, G);G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;F1: Point;F2: Point;Focus(G) = {F1, F2};WhenMax(AngleOf(F1, P, F2));Cos(AngleOf(F1, P, F2)) = 1/3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[2, 5]], [[2, 54]], [[6, 51], [123, 125]], [[6, 51]], [[8, 51]], [[8, 51]], [[55, 62]], [[63, 70]], [[6, 73]], [[74, 101]], [[74, 119]]]",
    "query_spans": "[[[123, 131]]]",
    "process": "First, use the law of cosines and the basic inequality to determine that when |PF₁| = |PF₂| = a, ∠F₁PF₂ reaches its maximum value, and the cosine value is minimized at $\\frac{2b^{2}}{a^{2}} - 1 = \\frac{1}{3}$, solving gives $\\frac{b^{2}}{a^{2}} = \\frac{2}{3}$. Then use $e = \\sqrt{1 - \\frac{b^{2}}{a^{2}}}$ to calculate the eccentricity. [Solution] According to the problem, |PF₁| + |PF₂| = 2a, |F₁F₂| = 2c, $a^{2} = b^{2} + c^{2}$. When ∠F₁PF₂ reaches its maximum value, i.e., $\\cos\\angle F_{1}PF_{2}$ is minimized, the minimum value of $\\cos\\angle F_{1}PF_{2}$ is $\\frac{1}{3}$. While $\\cos\\angle F_{1}PF_{2} = \\frac{|PF_{1}|^{2} + |PF_{2}|^{2} - |F_{1}F_{2}|^{2}}{2|PF_{1}||PF_{2}|} = \\frac{(|PF_{1}| + |PF_{2}|)^{2} - 2|PF_{1}|\\cdot|PF_{2}| - 4c^{2}}{2|PF_{1}|\\cdot|PF_{2}|} = \\frac{4a^{2} - 2|PF_{1}|\\cdot|PF_{2}| - 4c^{2}}{2|PF_{1}|\\cdot|PF_{2}|}$. And $|PF_{1}|\\cdot|PF_{2}| \\leqslant \\frac{(|PF_{1}| + |PF_{2}|)^{2}}{4} = a^{2}$, with equality if and only if |PF₁| = |PF₂| = a. Hence $\\cos\\angle F_{1}PF_{2} \\geqslant \\frac{2b^{2}}{a^{2}} - 1$, with equality if and only if |PF₁| = |PF₂| = a. Therefore, the minimum value of $\\cos\\angle F_{1}PF_{2}$ is $\\frac{2b^{2}}{a^{2}} - 1 = \\frac{1}{3}$, so $\\frac{b^{2}}{a^{2}} = \\frac{2}{3}$. Thus $e = \\frac{c}{a} = \\sqrt{\\frac{a^{2} - b^{2}}{a^{2}}} = \\sqrt{1 - \\frac{b^{2}}{a^{2}}} = \\sqrt{1 - \\frac{2}{3}} = \\frac{\\sqrt{3}}{3}$."
  },
  {
    "text": "Given that the ellipse with foci on the $x$-axis, $x^{2}+\\frac{y^{2}}{m}=1$, has a focal distance of $\\sqrt{3}$, then the value of $m$ is?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2 + y^2/m = 1);PointOnCurve(Focus(G), xAxis);FocalLength(G)=sqrt(3)",
    "query_expressions": "m",
    "answer_expressions": "1/4",
    "fact_spans": "[[[11, 38]], [[54, 57]], [[11, 38]], [[2, 38]], [[11, 52]]]",
    "query_spans": "[[[54, 61]]]",
    "process": "The ellipse with foci on the x-axis is given by $x^{2}+\\frac{y^{2}}{m}=1$, and its focal distance is $\\sqrt{3}$. Thus, $a=1$, $b=\\sqrt{m}$, $c=\\frac{\\sqrt{3}}{2}$. Since $a^{2}=b^{2}+c^{2}$, it follows that $1^{2}=\\left(\\sqrt{m}\\right)^{2}+\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}$. Solving gives $m=\\frac{1}{4}$."
  },
  {
    "text": "Given that $O$ is the coordinate origin, $F$ is the focus of the parabola $C$: $y^{2}=2 x$, the line $l$: $y=m(2 x-1)$ intersects the parabola $C$ at points $A$ and $B$, point $A$ lies in the first quadrant, and $|A F|=2|B F|$, then the value of $m$ is?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;F: Point;B: Point;O: Origin;m:Number;Expression(C) = (y^2 = 2*x);Focus(C) = F;Expression(l)=(y=m*(2*x-1));Intersection(l, C) = {A, B};Quadrant(A)=1;Abs(LineSegmentOf(A, F)) = 2*Abs(LineSegmentOf(B, F))",
    "query_expressions": "m",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[38, 57]], [[15, 34], [58, 64]], [[66, 69], [76, 80]], [[11, 14]], [[70, 73]], [[2, 5]], [[103, 106]], [[15, 34]], [[11, 37]], [[38, 57]], [[38, 75]], [[76, 85]], [[87, 101]]]",
    "query_spans": "[[[103, 110]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Using the focal radius formula and substituting $ |AF| = 2|BF| $, then solving simultaneously with the parabola equation, we find the coordinates of points $ A $ and $ B $, and then substitute into the slope formula to obtain the value of $ m $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, line $ l $ passes through the focus $ F(\\frac{1}{2},0) $ of parabola $ C $. Since $ |AF| = 2|BF| $, $ \\therefore \\overrightarrow{AF} = 2\\overrightarrow{FB} $, so $ x_{1} - \\frac{1}{2} = 2(\\frac{1}{2} - x_{2}) $, $ y_{1} = -2y_{2} $. $ \\therefore x_{1} = \\frac{3}{2} - 2x_{2} $. From $ y_{1}^{2} = 2x_{1} $, $ y_{2}^{2} = 2x_{2} $, we get $ \\begin{cases} 4y_{2}^{2} = 3 - 4x_{2} \\\\ y_{2}^{2} = 2x_{2} \\end{cases} $. $ \\therefore x_{2} = \\frac{1}{4} $, $ y_{2}^{2} = \\frac{1}{2} $, $ y_{2} = -\\frac{\\sqrt{2}}{2} $, $ \\therefore 2m = \\frac{0 + \\frac{\\sqrt{2}}{2}}{\\frac{1}{2} - \\frac{1}{4}} = 2\\sqrt{2} $, $ m = \\sqrt{2} $."
  },
  {
    "text": "Given point $M(-2,-3)$, point $F(2,0)$ is the focus of the parabola $C$: $y^{2}=2 p x$ $(p>0)$, and point $P$ in the first quadrant lies on the parabola $C$. Then the maximum value of $\\frac{|P M|}{|P F|}$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (-2, -3);C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Coordinate(F) = (2, 0);Focus(C) = F;P: Point;Quadrant(P) = 1;PointOnCurve(P, C)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, M))/Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 13]], [[2, 13]], [[24, 50], [65, 71]], [[24, 50]], [[32, 50]], [[32, 50]], [[14, 23]], [[14, 23]], [[14, 53]], [[60, 64]], [[54, 64]], [[60, 72]]]",
    "query_spans": "[[[74, 101]]]",
    "process": "From the given, we have p = 4, so the equation of parabola C is y^{2} = 8x, and the directrix l: x = -2. Let P(x, y) (x > 0, y > 0). In right triangle PMH, \\frac{|PM|}{|PH|} = \\frac{\\sqrt{(x+2)^{2}+(y+3)^{2}}}{x+2} = \\sqrt{1+(\\frac{y+3}{x+2})^{2}} \\frac{1+(y+3)^{2}}{\\frac{y^{2}+2}{8}+\\frac{1}{2}}{^{2}} Thus |PF| = |1+|\\frac{(-(-3))^{2}+2}{By the basic inequality we obtain +\\frac{25}{1}-6 \\geqslant 2\\sqrt{x}x^{2}5}-6 = 4 (equality holds if and only if t = 5). Therefore \\frac{|\\frac{PM}{1+2}}{|PFF|}"
  },
  {
    "text": "Given that the hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ passes through the focus of the parabola $y^{2}=8x$, then the asymptote equations of this hyperbola are?",
    "fact_expressions": "G: Hyperbola;m: Number;H: Parabola;Expression(G) = (-y^2 + x^2/m = 1);Expression(H) = (y^2 = 8*x);PointOnCurve(Focus(H),G)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=(pm*1/2)*x",
    "fact_spans": "[[[2, 30], [51, 54]], [[5, 30]], [[31, 45]], [[2, 30]], [[31, 45]], [[2, 48]]]",
    "query_spans": "[[[51, 62]]]",
    "process": "The focus of $ y^{2}=8x $ is $ (2,0) \\therefore \\sqrt{m}=2 \\therefore m=4 \\therefore $ the asymptotes are $ y=\\pm\\frac{1}{2}x $"
  },
  {
    "text": "What are the coordinates of the foci of the ellipse $16x^{2}+9 y^{2}=144$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (16*x^2 + 9*y^2 = 144)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,pm*sqrt(7))",
    "fact_spans": "[[[0, 23]], [[0, 23]]]",
    "query_spans": "[[[0, 29]]]",
    "process": ""
  },
  {
    "text": "If the vertex of a parabola is at the origin and its focus is a focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, then what is the distance from the focus to the directrix of this parabola?",
    "fact_expressions": "G: Parabola;H: Ellipse;O: Origin;Expression(H) = (x^2/4 + y^2/3 = 1);Vertex(G) = O;Focus(G)=OneOf(Focus(H))",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 4], [61, 64]], [[16, 53]], [[8, 12]], [[16, 53]], [[1, 12]], [[1, 58]]]",
    "query_spans": "[[[61, 75]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1$ has an asymptote $\\sqrt{3} x+y=0$, then the focal distance of $C$ is?",
    "fact_expressions": "C: Hyperbola;a: Number;Expression(C) = (-y^2/3 + x^2/a^2 = 1);Expression(OneOf(Asymptote(C))) = (sqrt(3)*x + y = 0)",
    "query_expressions": "FocalLength(C)",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 47], [72, 75]], [[8, 47]], [[0, 47]], [[0, 70]]]",
    "query_spans": "[[[72, 80]]]",
    "process": "The hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{3} = 1 $ has asymptotes $ y = \\pm \\frac{\\sqrt{3}}{a}x $. Since one asymptote of the hyperbola is $ \\sqrt{3}x + y = 0 \\Rightarrow y = -\\sqrt{3}x $, it follows that $ \\frac{\\sqrt{3}}{|a|} = \\sqrt{3} \\Rightarrow a = \\pm 1 $. Then $ c^{2} = a^{2} + b^{2} = 1 + 3 = 4 \\Rightarrow c = 2 $. The focal length of $ C $ is $ 2c = 4 $."
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{4+k}+\\frac{y^{2}}{2-k}=1$ represents an ellipse, what is the range of real values for $k$?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G)=(x^2/(k + 4) + y^2/(2 - k) = 1)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-4,-1)+(-1,2)",
    "fact_spans": "[[[45, 47]], [[49, 54]], [[2, 47]]]",
    "query_spans": "[[[49, 61]]]",
    "process": "Solve according to the requirements of the standard equation of an ellipse. $4+k>0$\n$$\n\\begin{cases}\n4+k>0 \\\\\n2-k>0 \\\\\nk+4+2\n\\end{cases}\n$$\n($k+4 \\neq 2-k$,\n$$\n\\begin{cases}\nk>-4 \\\\\nk<2 \\\\\nk \\neq -1\n\\end{cases}\n$$\n, that is, $k \\in (-4,-1) \\cup (-1,2)$"
  },
  {
    "text": "If the distance between the two directrices of a hyperbola is $\\frac{3}{5}$ of the focal distance, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;L1: Line;L2: Line;Directrix(G) = {L1, L2};Distance(L1, L2) = (3/5)*FocalLength(G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(15)/3",
    "fact_spans": "[[[1, 4], [31, 34]], [], [], [[1, 8]], [[1, 29]]]",
    "query_spans": "[[[31, 40]]]",
    "process": "The distance between the two directrices of the hyperbola is: $2\\frac{a^{2}}{c}$, the distance between the two foci is: $2c$. According to the problem, we have: $\\frac{2a^{2}}{c} = \\frac{3}{5} \\times 2c$, which simplifies to: $5a^{2} = 3c^{2}$, solving gives: $e = \\frac{\\sqrt{15}}{3}$, so the answer is: $\\frac{\\sqrt{15}}{3}$"
  },
  {
    "text": "If the equation $\\frac{x^{2}}{2}+\\frac{y^{2}}{a}=1$ represents an ellipse with foci on the $x$-axis, then the range of real values for $a$ is?",
    "fact_expressions": "G: Ellipse;a: Real;Expression(G) = (x^2/2 + y^2/a = 1);PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Range(a)",
    "answer_expressions": "(0,2)",
    "fact_spans": "[[[49, 51]], [[53, 58]], [[1, 51]], [[40, 51]]]",
    "query_spans": "[[[53, 65]]]",
    "process": ""
  },
  {
    "text": "It is known that the hyperbola has the same foci as the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{6}=1$, and the asymptotes of the hyperbola are given by $y=\\pm \\frac{1}{2} x$. Then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/16 + y^2/6 = 1);Focus(G) = Focus(H);Expression(Asymptote(G)) = (y = pm*(x/2))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/8 - y^2/2 = 1",
    "fact_spans": "[[[2, 5], [52, 55], [86, 89]], [[6, 44]], [[6, 44]], [[2, 50]], [[52, 83]]]",
    "query_spans": "[[[86, 93]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola $y^{2}=2 p x(p>0)$ is $F$, points $A$ and $B$ lie on the parabola such that $\\angle A F B=\\frac{\\pi}{2}$, and the midpoint $M$ of chord $AB$ has its projection $N$ on the directrix. Then the maximum value of $\\frac{|MN |}{|AB|}$ is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;F: Point;M: Point;N: Point;p>0;Expression(G) = (y^2 = 2*p*x);Focus(G) = F;PointOnCurve(B, G);PointOnCurve(A, G);AngleOf(A, F, B) = pi/2;MidPoint(LineSegmentOf(A, B)) = M;Projection(M, Directrix(G)) = N;IsChordOf(LineSegmentOf(A, B),G)",
    "query_expressions": "Max(Abs(LineSegmentOf(M, N))/Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 21], [37, 40], [85, 86]], [[3, 21]], [[29, 32]], [[33, 36]], [[25, 28]], [[81, 84]], [[93, 96]], [[3, 21]], [[0, 21]], [[0, 28]], [[33, 41]], [[29, 41]], [[43, 71]], [[73, 84]], [[81, 96]], [[37, 78]]]",
    "query_spans": "[[[98, 124]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the endpoints of the minor axis are $P(0, b)$, $Q(0,-b)$, and one endpoint of the major axis is $M$. Let $AB$ be a chord passing through the center of the ellipse and not lying on the coordinate axes. If the product of the slopes of $PA$ and $PB$ equals $-\\frac{1}{4}$, then what is the distance from $P$ to the line $QM$?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;M: Point;Q: Point;A: Point;B: Point;P: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (0, b);Coordinate(Q) = (0, -b);Endpoint(MinorAxis(G)) = {P, Q};OneOf(Endpoint(MajorAxis(G))) = M;IsChordOf(LineSegmentOf(A, B), G);PointOnCurve(Center(G), LineSegmentOf(A, B));Negation(OverlappingLine(LineSegmentOf(A, B), axis));Slope(LineSegmentOf(P, A))*Slope(LineSegmentOf(P, B)) = -1/4",
    "query_expressions": "Distance(P,LineOf(Q,M))",
    "answer_expressions": "2*sqrt(5)/5",
    "fact_spans": "[[[2, 54], [101, 103]], [[4, 54]], [[4, 54]], [[88, 92]], [[70, 79]], [[93, 98]], [[93, 98]], [[59, 68], [153, 156]], [[4, 54]], [[4, 54]], [[2, 54]], [[59, 68]], [[70, 79]], [[2, 79]], [[2, 91]], [[93, 116]], [[93, 116]], [[93, 116]], [[118, 151]]]",
    "query_spans": "[[[153, 169]]]",
    "process": "Let the ellipse be given with points P(0,b), A(x_{0},y_{0}), then the coordinates of point B are (x_{0}-,y_{0}), then \\frac{y_{0}-b}{x_{0}}\\times\\frac{y_{0}-b}{x_{0}}=\\frac{1}{4}, since \\frac{x_{0}^{2}}{a^{2}}+\\frac{y_{0}^{2}}{b^{2}}=1, then -\\frac{b^{2}}{a^{2}}=-\\frac{1}{4}, thus \\frac{b}{a}=\\frac{1}{2}. Let M(a,0), the equation of line QM is bx-ay-ab=0, then the distance from P to line QM is d=\\frac{|2ab|}{\\sqrt{a^{2}+b^{2}}}=\\frac{2\\frac{b}{a}}{\\sqrt{1+(\\frac{b}{a})^{2}}}=\\frac{2\\sqrt{5}}{5}"
  },
  {
    "text": "Draw a line through the focus of the parabola $y^{2}=2 p x$ ($p>0$) intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $x_{1}+x_{2}=3 p$, then $|A B|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;H: Line;PointOnCurve(Focus(G), H);A: Point;B: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Intersection(H, G) = {A, B};x1 + x2 = 3*p",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "4*p",
    "fact_spans": "[[[1, 22], [29, 32]], [[1, 22]], [[4, 22]], [[4, 22]], [[26, 28]], [[0, 28]], [[33, 50]], [[51, 68]], [[33, 50]], [[51, 68]], [[33, 50]], [[51, 68]], [[33, 50]], [[51, 68]], [[26, 70]], [[72, 89]]]",
    "query_spans": "[[[91, 100]]]",
    "process": "Solve directly using the definition of a parabola. Let the focus of the parabola be $ F $, then by the definition of the parabola we have $ |AB| = |AF| + |BF| = x_{1} + \\frac{p}{2} + x_{2} + \\frac{p}{2} = x_{1} + x_{2} + p = 3p + p = 4p $."
  },
  {
    "text": "Draw a line with slope $-\\frac{1}{2}$ passing through the point $M(2,1)$, intersecting the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ at points $A$ and $B$. If $M$ is the midpoint of segment $AB$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;A: Point;B: Point;M: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(M) = (2, 1);PointOnCurve(M, G);Slope(G) = -1/2;Intersection(G, C) = {A, B};MidPoint(LineSegmentOf(A,B)) = M",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[32, 89], [119, 124]], [[39, 89]], [[39, 89]], [[29, 31]], [[92, 95]], [[96, 99]], [[1, 10], [103, 106]], [[39, 89]], [[39, 89]], [[32, 89]], [[1, 10]], [[0, 31]], [[11, 31]], [[29, 101]], [[103, 117]]]",
    "query_spans": "[[[119, 130]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, $ \\frac{x_{1}^{2}}{a^{2}} + \\frac{y_{1}^{2}}{b^{2}} = 1 $ \\textcircled{1}, $ \\frac{x_{2}^{2}}{a^{2}} + \\frac{y_{2}^{2}}{b^{2}} = 1 $ \\textcircled{2}. Since $ M $ is the midpoint of segment $ AB $, $ \\therefore \\frac{x_{1}+x_{2}}{2} $, $ \\frac{y_{1}+y_{2}}{2} $. Subtracting \\textcircled{2} from \\textcircled{1} gives $ \\frac{x_{1}^{2}-x_{2}^{2}}{a^{2}} + \\frac{y_{1}^{2}-y_{2}^{2}}{b^{2}} = 0 $. $ \\therefore a = 2b $, $ \\therefore e = \\frac{c}{a} = \\frac{\\sqrt{3}}{2} $."
  },
  {
    "text": "The equation of the directrix of the parabola $x^{2}=4 y$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=-1",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of hyperbola $C$ is $2$, and one of its foci is the focus of the parabola $x^{2}=8 y$, then the standard equation of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Parabola;Expression(G) = (x^2 = 8*y);Eccentricity(C) = 2;OneOf(Focus(C)) = Focus(G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2 - x^2/3 = 1",
    "fact_spans": "[[[2, 8], [17, 18], [43, 49]], [[24, 38]], [[24, 38]], [[2, 16]], [[17, 41]]]",
    "query_spans": "[[[43, 56]]]",
    "process": "The focus of the parabola $x^{2}=8y$ is $F(0,2)$, so in the hyperbola, $c=2$, $e=\\frac{c}{a}=\\frac{2}{a}=2$, $\\therefore a=1$, $b^{2}=c^{2}-a^{2}=3$, so the required hyperbola equation is: $y^{2}-\\frac{x^{2}}{3}=1$"
  },
  {
    "text": "A line passing through the focus of the parabola $C$: $y^{2}=2 p x$ ($p>0$) with slope $2$ intersects $C$ at points $A$ and $B$. The circle with diameter $AB$ intersects the directrix of $C$ at a point $M$. If the ordinate of point $M$ is $2$, then the value of $p$ is?",
    "fact_expressions": "C: Parabola;p: Number;G: Circle;H: Line;A: Point;B: Point;M:Point;p>0;Expression(C) = (y^2 = 2*(p*x));PointOnCurve(Focus(C), H);Slope(H) = 2;Intersection(H, C) = {A, B};IsDiameter(LineSegmentOf(A,B),G);Intersection(Directrix(C),G)=M;YCoordinate(M) = 2",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 27], [41, 44], [68, 71]], [[97, 100]], [[66, 67]], [[38, 40]], [[46, 49]], [[50, 53]], [[78, 81], [83, 87]], [[9, 27]], [[1, 27]], [[0, 40]], [[31, 40]], [[38, 55]], [[56, 67]], [[66, 81]], [[83, 95]]]",
    "query_spans": "[[[97, 104]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, and the midpoint of $ AB $ be $ N $. Analysis shows that the circle with $ AB $ as diameter is tangent to the directrix of $ C $. Then, using the point difference method to find the ordinate of point $ M $, the value of $ p $ can be obtained. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, and the midpoint of $ AB $ be $ N\\left(\\frac{x_{1}+x_{2}}{2},\\frac{y_{1}+y_{2}}{2}\\right) $, then $ AB = x_{1} + x_{2} + p $, hence the radius is $ \\frac{x_{1}+x_{2}+p}{2} $. Also, the distance from the midpoint $ N\\left(\\frac{x_{1}+x_{2}}{2},\\frac{y_{1}+y_{2}}{2}\\right) $ to the directrix $ x = -\\frac{p}{2} $ is $ \\frac{x_{1}+x_{2}+p}{2} $. Therefore, the circle with $ AB $ as diameter is tangent to the directrix of $ C $, and $ M\\left(-\\frac{p}{2},\\frac{y_{1}+y_{2}^{2}}{2}\\right) $ is the point of tangency. Hence $ \\frac{y_{1}+y_{2}}{2} = 2 $, i.e., $ y_{1} + y_{2} = 4 $. Also,\n$$\n\\begin{cases}\ny_{1}^{2} = 2px_{1} \\\\\ny_{2}^{2} = 2px_{2}\n\\end{cases}\n\\Rightarrow y_{1}^{2} - y_{2}^{2} = 2p(x_{1} - x_{2}) \\Rightarrow \\frac{y_{1} - y_{2}}{x_{1} - x_{2}} = \\frac{2p}{y_{1} + y_{2}},\n$$\nand since the slope of the line is 2, $ y_{1} + y_{2} = 4 $, we have $ 2 = \\frac{2p}{4} \\Rightarrow p = 4 $. This problem mainly examines the point difference method for solving chord midpoints, and also tests properties of focal chords and the directrix, belonging to medium-level problems."
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{2}=1$ are $F_{1}$, $F_{2}$, and point $P$ lies on the ellipse. If $|P F_{1}|=4$, what is the measure of $\\angle F_{1} P F_{2}$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 + y^2/2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 4",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "ApplyUnit(120,degree)",
    "fact_spans": "[[[0, 37], [62, 64]], [[57, 61]], [[41, 48]], [[49, 56]], [[0, 37]], [[0, 56]], [[57, 65]], [[67, 80]]]",
    "query_spans": "[[[81, 108]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola $y^{2}=4 x$ is $F$, and $P(4 , y)$ lies on the parabola. Then $|P F|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;P: Point;y0: Number;Coordinate(P) = (4, y0);PointOnCurve(P, G)",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "5",
    "fact_spans": "[[[0, 14], [35, 38]], [[0, 14]], [[18, 21]], [[0, 21]], [[23, 34]], [[23, 34]], [[23, 34]], [[23, 39]]]",
    "query_spans": "[[[41, 50]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $x^{2}=2 p y(p>0)$ with focus $F$, if a point $M(2, y_{0})$ on the parabola $C$ satisfies $|M O|=|M F|$ ($O$ is the origin), then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;M: Point;O: Origin;F: Point;p>0;y0:Number;Expression(C) = (x^2 = 2*p*y);Coordinate(M) = (2,y0);Focus(C) = F;PointOnCurve(M,C);Abs(LineSegmentOf(M,O))=Abs(LineSegmentOf(M,F))",
    "query_expressions": "p",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 28], [37, 43]], [[87, 90]], [[45, 59]], [[76, 79]], [[32, 35]], [[9, 28]], [[46, 59]], [[2, 28]], [[45, 59]], [[2, 35]], [[37, 59]], [[61, 74]]]",
    "query_spans": "[[[87, 92]]]",
    "process": "Analysis: According to the distance formula between two points and |MO| = |MF|, we obtain p^{2} - 4py_{0} = 0. Then, since point M(2, y_{0}) lies on the parabola, we have 2^{2} = 2py_{0}. Solving these equations simultaneously gives p. From the problem: \\sqrt{(2-0)^{2} + (y_{0}-0)^{2}} = \\sqrt{(2-0)^{2} + (y_{0}-\\frac{P}{2})^{2}}, so p^{2} - 4py_{0} = 0. Also, since 2^{2} = 2py_{0}, p > 0, we have p^{2} - 4p \\times \\frac{2}{p} = 0, thus p = -2\\sqrt{2} (discarded) or p = 2\\sqrt{2}."
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{n}-y^{2}=1(n>1)$ has two foci $F_{1}$, $F_{2}$. Point $P$ lies on the hyperbola and satisfies $|PF_{1}|+|P F_{2}|=2 \\sqrt{n+2}$. Then the area of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;n: Number;P: Point;F1: Point;F2: Point;n>1;Expression(G) = (-y^2 + x^2/n = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 2*sqrt(n + 2)",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 33], [63, 66]], [[3, 33]], [[58, 62]], [[39, 46]], [[49, 56]], [[3, 33]], [[0, 33]], [[0, 57]], [[59, 67]], [[71, 104]]]",
    "query_spans": "[[[106, 133]]]",
    "process": ""
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=2x$, draw a line intersecting the parabola at points $A$ and $B$. If $|AB|=\\frac{25}{12}$, $|AF|<|BF|$, then $|AF|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);F: Point;Focus(G) = F;Z: Line;PointOnCurve(F, Z);A: Point;B: Point;Intersection(Z, G) = {A, B};Abs(LineSegmentOf(A, B)) = 25/12;Abs(LineSegmentOf(A, F))<Abs(LineSegmentOf(B, F))",
    "query_expressions": "Abs(LineSegmentOf(A, F))",
    "answer_expressions": "5/6",
    "fact_spans": "[[[1, 15], [25, 28]], [[1, 15]], [[18, 21]], [[1, 21]], [[22, 24]], [[0, 24]], [[29, 32]], [[33, 36]], [[22, 38]], [[40, 61]], [[63, 76]]]",
    "query_spans": "[[[79, 88]]]",
    "process": "Let |AF| = m, |BF| = n, then \\frac{1}{m} + \\frac{1}{n} = \\frac{2}{p}. Also, |AB| = \\frac{25}{12}, so m + n = \\frac{25}{12}, mn = \\frac{25}{24}, thus m = \\frac{5}{6}, n = \\frac{5}{4}. [Knowledge Point] This question mainly examines the simple properties of parabolas and the relationship between parabolas and lines. When dealing with problems involving focal chords of parabolas, it is common to set up the line equation based on the focus, solve it together with the parabola equation, and combine Vieta's formulas with the definition of the parabola to solve the problem, which is considered difficult."
  },
  {
    "text": "It is known that the right focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ coincides with the focus of the parabola $y^{2}=12 x$. Then, the length of the imaginary axis of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;H: Parabola;Expression(G) = (x^2/4 - y^2/b^2 = 1);Expression(H) = (y^2 = 12*x);RightFocus(G) = Focus(H)",
    "query_expressions": "Length(ImageinaryAxis(G))",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[2, 44], [72, 75]], [[5, 44]], [[49, 64]], [[2, 44]], [[49, 64]], [[2, 69]]]",
    "query_spans": "[[[72, 81]]]",
    "process": "The focus of the parabola $ y^{2} = 12x $ is $ F(3,0) $. Since the right focus of the hyperbola $ \\frac{x^{2}}{4} - \\frac{y^{2}}{b^{2}} = 1 $ coincides with the focus of the parabola $ y^{2} = 12x $, we have $ 4 + b^{2} = c^{2} = 9 $. Solving gives $ b = \\sqrt{5} $, so $ 2b = 2\\sqrt{5} $."
  },
  {
    "text": "Draw two tangents from point $M(-1,2)$ to the parabola $y^{2}=4x$, with points of tangency $A$ and $B$. Then the distance from the midpoint of $AB$ to the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;M: Point;l1: Line;l2:Line;Expression(G) = (y^2 = 4*x);Coordinate(M) = (-1, 2);TangentOfPoint(M,G)={l1,l2};TangentPoint(l1,G)=A;TangentPoint(l2,G)=B",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A,B)),Directrix(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[12, 26], [57, 60]], [[37, 40]], [[41, 44]], [[1, 11]], [], [], [[12, 26]], [[1, 11]], [[0, 31]], [[0, 46]], [[0, 46]]]",
    "query_spans": "[[[48, 67]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, the upper vertex is $D$, and $\\angle F_{1} D F_{2}=120^{\\circ}$. If points $A$ and $B$ in the first quadrant lie on $C$, with $|A F_{2}|=2$, $|B F_{2}|=4$, $|A B|=3$, then the slope of line $A B$ is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;B: Point;F1: Point;D: Point;F2: Point;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;UpperVertex(G) = D;AngleOf(F1, D, F2) = ApplyUnit(120, degree);PointOnCurve(B, C);PointOnCurve(A, C);Quadrant(A) = 1;Quadrant(B) = 1;Abs(LineSegmentOf(A, F2)) = 2;Abs(LineSegmentOf(B, F2)) = 4;Abs(LineSegmentOf(A, B)) = 3",
    "query_expressions": "Slope(LineOf(A,B))",
    "answer_expressions": "-sqrt(11)/4",
    "fact_spans": "[[[2, 47], [131, 134]], [[4, 47]], [[4, 47]], [[122, 126]], [[127, 130]], [[56, 63]], [[76, 79]], [[64, 71]], [[2, 47]], [[2, 71]], [[2, 71]], [[2, 79]], [[81, 115]], [[127, 135]], [[122, 135]], [[117, 126]], [[117, 130]], [[136, 149]], [[151, 164]], [[166, 175]]]",
    "query_spans": "[[[177, 189]]]",
    "process": "The ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ has left and right foci $F_{1}$, $F_{2}$ respectively, and the upper vertex is $D$. Given $\\angle F_{1}DF_{2}=120^{\\circ}$, so $\\angle F_{1}DO=60^{\\circ}$. From the geometric properties of the ellipse, we know $|DF_{1}|=a$, $|OF_{1}|=c$, and the eccentricity of the ellipse is $e=\\frac{c}{a}=\\sin60^{\\circ}=\\frac{\\sqrt{3}}{2}$. Let points $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $0<x_{1}<a$, $0<x_{2}<a$, $|AF_{2}|=|ex_{1}-a|=a-ex_{1}=2$. Similarly, $|BF_{2}|=a-ex_{2}=4$, so $|BF_{2}|-|AF_{2}|=e(x_{1}-x_{2})=\\frac{\\sqrt{3}}{2}(x_{1}-x_{2})=2$. Solving gives $x_{1}-x_{2}=\\frac{4\\sqrt{3}}{3}$. Let the slope of line $AB$ be $k$, then by the chord length formula, $|AB|=\\sqrt{1+k^{2}}|x_{1}-x_{2}|=\\frac{4\\sqrt{3}}{3}\\sqrt{1+k^{2}}=3$. Solving gives $k=\\pm\\frac{\\sqrt{11}}{4}$. Since points $A$, $B$ are both in the first quadrant, then $k<0$, hence $k=-\\frac{\\sqrt{11}}{4}$."
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ are $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{1}$ with an inclination angle of $45^{\\circ}$ intersects the ellipse at a point $M$. If $M F_{2}$ is perpendicular to the $x$-axis, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;M: Point;F2: Point;F1: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1,H);Inclination(H)=ApplyUnit(45,degree);OneOf(Intersection(H,G))=M;IsPerpendicular(LineSegmentOf(M,F2),xAxis)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)-1",
    "fact_spans": "[[[0, 52], [106, 108], [137, 139]], [[2, 52]], [[2, 52]], [[103, 105]], [[114, 117]], [[69, 76]], [[78, 85], [61, 68]], [[2, 52]], [[2, 52]], [[0, 52]], [[0, 76]], [[0, 76]], [[77, 105]], [[86, 105]], [[103, 117]], [[119, 135]]]",
    "query_spans": "[[[137, 145]]]",
    "process": ""
  },
  {
    "text": "Let $AB$ be the major axis of ellipse $\\Gamma$, point $C$ lies on $\\Gamma$, and $\\angle CBA = \\frac{\\pi}{4}$. If $AB = 4$, $BC = \\sqrt{2}$, then the distance between the two foci of $\\Gamma$ is?",
    "fact_expressions": "Gamma: Ellipse;A: Point;B: Point;C: Point;F1: Point;F2: Point;MajorAxis(Gamma) = LineSegmentOf(A, B);PointOnCurve(C, Gamma);AngleOf(C, B, A) = pi/4;LineSegmentOf(A, B) = 4;LineSegmentOf(B, C) = sqrt(2);Focus(Gamma) = {F1, F2}",
    "query_expressions": "Distance(F1, F2)",
    "answer_expressions": "4*sqrt(6)/3",
    "fact_spans": "[[[6, 16], [88, 96], [25, 33]], [[1, 5]], [[1, 5]], [[20, 24]], [], [], [[1, 19]], [[20, 34]], [[36, 62]], [[64, 70]], [[73, 86]], [[88, 101]]]",
    "query_spans": "[[[88, 108]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, and the point $(a, b)$ lies on the line $y=2x$. Then, the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;H: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y = 2*x);Coordinate(H) = (a, b);PointOnCurve(H, G)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 64], [87, 93]], [[66, 74]], [[66, 74]], [[75, 84]], [[65, 74]], [[10, 63]], [[10, 63]], [[2, 64]], [[75, 84]], [[65, 74]], [[65, 84]]]",
    "query_spans": "[[[87, 99]]]",
    "process": "Since the point (a, b) lies on the line y = 2x, we have b = 2a, that is, \\frac{b}{a} = 2, then the eccentricity is \\frac{c}{a} = \\sqrt{1 + (\\frac{b}{a})^{2}} = \\sqrt{5}"
  },
  {
    "text": "Given that the directrix of the parabola $C$: $y^{2}=a x$ is $x=-1$, then $a$=?",
    "fact_expressions": "C: Parabola;a: Number;Expression(C) = (y^2 = a*x);Expression(Directrix(C)) = (x = -1)",
    "query_expressions": "a",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 21]], [[35, 38]], [[2, 21]], [[2, 33]]]",
    "query_spans": "[[[35, 40]]]",
    "process": "The equation of the directrix of the parabola \\( y^{2} = ax \\) is \\( x = -\\frac{a}{4} \\), so \\( -\\frac{a}{4} = -1 \\), solving gives \\( a = 4 \\)."
  },
  {
    "text": "Through the right vertex $A$ of the hyperbola $G$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$), draw a line $m$ with slope $1$, intersecting the two asymptotes at points $B$ and $C$, respectively. If $|AB|=2|AC|$, then the eccentricity of the hyperbola $G$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;A: Point;RightVertex(G) = A;m: Line;Slope(m) = 1;PointOnCurve(A, m);Z1: Line;Z2: Line;Asymptote(G) = {Z1, Z2};B: Point;Intersection(m, Z1) = B;C: Point;Intersection(m, Z2) = C;Abs(LineSegmentOf(A, B)) = 2*Abs(LineSegmentOf(A, C))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{sqrt(10), sqrt(10)/3}",
    "fact_spans": "[[[1, 62], [117, 123]], [[1, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[66, 69]], [[1, 69]], [[77, 82]], [[70, 82]], [[0, 82]], [], [], [[1, 90]], [[92, 95]], [[1, 101]], [[96, 99]], [[1, 101]], [[103, 115]]]",
    "query_spans": "[[[117, 129]]]",
    "process": ""
  },
  {
    "text": "$F$ is the focus of the parabola $y^{2}=4x$. A line $l$ passing through $F$ intersects the parabola at points $A$ and $B$. $O$ is the origin. If $|AF|=10$, then the area of $\\triangle OAB$ is?",
    "fact_expressions": "F: Point;G: Parabola;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, l);l: Line;Intersection(l, G) = {A, B};A: Point;B: Point;O: Origin;Abs(LineSegmentOf(A, F)) = 10",
    "query_expressions": "Area(TriangleOf(O, A, B))",
    "answer_expressions": "10/3",
    "fact_spans": "[[[0, 3], [23, 26]], [[4, 18], [33, 36]], [[4, 18]], [[0, 21]], [[22, 32]], [[27, 32]], [[27, 46]], [[37, 40]], [[41, 44]], [[47, 50]], [[57, 67]]]",
    "query_spans": "[[[69, 91]]]",
    "process": "Let point A be a point in the first quadrant, let points A(x_{1},y_{1}), B(x_{2},y_{2}). Using the definition of the parabola, the coordinates of point A can be found, and the equation of line AB can be determined. By solving simultaneously the equation of line AB and the equation of the parabola, listing Vieta's formulas, and finding the value of |y_{1}-y_{2}|, the area of \\triangle OAB can thus be obtained. Let point A be a point in the first quadrant, let points A(x_{1},y_{1}), B(x_{2},y_{2}). The directrix of the parabola y^{2}=4x is x=-1. By the definition of the parabola, |AF|=x_{1}+1=10, solving gives x_{1}=9. Since point A lies in the first quadrant, y_{1}>0, we obtain y_{1}=\\sqrt{4x_{1}}=6, so point A is (9,6). The slope of line AF is k_{AF}=\\frac{6}{9-1}=\\frac{3}{4}. Therefore, the equation of line AB is y=\\frac{3}{4}(x-1), or equivalently x=\\frac{4}{3}y+1. Solving simultaneously:\n\\begin{cases}x=\\frac{4}{3}y+1\\\\y^{2}=4x\\end{cases},\neliminating x and simplifying yields y^{2}-\\frac{16}{3}y-4=0. By Vieta's formulas, y_{1}+y_{2}=\\frac{16}{3}, \\therefore y_{2}=\\frac{16}{3}-y_{1}=\\frac{16}{3}-6=-\\frac{2}{3}. Thus, S_{\\triangle OAB}=\\frac{1}{2}|OF|\\cdot|y_{1}-y_{2}|=\\frac{1}{2}\\times1\\times\\left(6+\\frac{2}{3}\\right)=\\frac{10}{3}."
  },
  {
    "text": "Given that the line $y = kx - 1$ intersects each of the left and right branches of the hyperbola $x^2 - y^2 = 1$ at one point, then the range of values for $k$ is?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*x - 1);k: Number;G: Hyperbola;Expression(G) = (x^2 - y^2 = 1);NumIntersection(H, LeftPart(G)) = 1;NumIntersection(H, RightPart(G)) = 1",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-1, 1)",
    "fact_spans": "[[[2, 13]], [[2, 13]], [[44, 47]], [[14, 32]], [[14, 32]], [[2, 42]], [[2, 42]]]",
    "query_spans": "[[[44, 54]]]",
    "process": "Let the coordinates of the two intersection points be A(x_{1},y_{1}) and B(x_{2},y_{2}). Solving simultaneously the line and the hyperbola  \n\\begin{cases}y=kx-1\\\\x^{2}-y^{2}=1\\end{cases},  \nsimplifying gives (1-k^{2})x^{2}+2kx-2=0 (1-k^{2}\\neq0). Since the line y=kx-1 intersects each of the left and right branches of the hyperbola x^{2}-y^{2}=1 at one point, the x-coordinates of the two intersection points have opposite signs, i.e., x_{1}\\cdot x_{2}=\\frac{-2}{1-k^{2}}<0. Solving the inequality yields -1<k<1. Therefore, the range of k is (-1,1). [Note: This problem examines the positional relationship between a line and a hyperbola, and coordinate relationships between intersection points, belonging to basic problems.]"
  },
  {
    "text": "The equation $\\frac{x^{2}}{16-m}+\\frac{y^{2}}{m+4}=1$ represents an ellipse with foci on the $x$-axis. Then, the range of real values for $m$ is?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (x^2/(16 - m) + y^2/(m + 4) = 1);PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-4, 6)",
    "fact_spans": "[[[53, 55]], [[57, 62]], [[0, 55]], [[44, 55]]]",
    "query_spans": "[[[57, 69]]]",
    "process": "According to the problem, the equation $\\frac{x^2}{16-m}+\\frac{y^{2}}{m+4}=1$ represents an ellipse with foci on the x-axis. Therefore, the range of real number $m$ is $(-4,6)$."
  },
  {
    "text": "Let the parabola $C$: $y=\\frac{1}{4} x^{2}$ have focus $F$, and let line $l$ pass through focus $F$, intersecting the parabola $C$ at points $A$ and $B$, with $|A F|=3$. Then $\\frac{S_{\\Delta A O F}}{S_{\\Delta B O F}}$=?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;F: Point;O: Origin;B: Point;Expression(C) = (y = x^2/4);Focus(C) = F;PointOnCurve(F,l);Intersection(l,C)={A,B};Abs(LineSegmentOf(A, F)) = 3",
    "query_expressions": "Area(TriangleOf(A,O,F))/Area(TriangleOf(B,O,F))",
    "answer_expressions": "2",
    "fact_spans": "[[[38, 43]], [[52, 58], [1, 30]], [[60, 63]], [[34, 37], [46, 49]], [[81, 124]], [[64, 67]], [[1, 30]], [[1, 37]], [[38, 49]], [[38, 69]], [[70, 79]]]",
    "query_spans": "[[[81, 126]]]",
    "process": "The focus of the parabola is (0,1). Since the line intersects the parabola at two points, the slope of the line exists. According to the definition of the parabola, |AF| = y_{A} + 1 = 3, so the y-coordinate of A is 2 and the x-coordinate is \\pm2\\sqrt{2}. Without loss of generality, assume A(-2\\sqrt{2}, 2), then the equation of line l is y = -\\frac{\\sqrt{2}}{4}x + 1. Solving the system of the line and the parabola equations, simplifying yields x^{2} + \\sqrt{2}x - 4 = 0. The solutions are x_{A} = -2\\sqrt{2}, x_{B} = \\sqrt{2}, hence y_{B} = \\frac{1}{4} \\times (\\sqrt{2})^{2} = \\frac{1}{2}. Therefore, \\frac{S_{AAOF}}{S_{ABOF}} = \\frac{|x_{A}|}{|x_{B}|} = \\frac{2\\sqrt{2}}{\\sqrt{2}} = 2."
  },
  {
    "text": "If the coordinates of one focus of a hyperbola are $(5,0)$ and the length of the real axis is $6$, then its standard equation is?",
    "fact_expressions": "G: Hyperbola;Coordinate(OneOf(Focus(G))) = (5, 0);Length(RealAxis(G)) = 6",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/16 = 1",
    "fact_spans": "[[[1, 4], [29, 30]], [[1, 19]], [[1, 27]]]",
    "query_spans": "[[[29, 37]]]",
    "process": "Since one focus of the hyperbola is at (5,0), the foci of the hyperbola lie on the x-axis, c=5, the length of the real axis is 2a=6, a=3, b=\\sqrt{c^{2}-a^{2}}=4, so the standard equation of the hyperbola is \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1"
  },
  {
    "text": "If the focus of the parabola $y^{2}=\\frac{4}{m} x$ coincides with the left focus of the ellipse $\\frac{x^{2}}{7}+\\frac{y^{2}}{3}=1$, then the value of $m$ is?",
    "fact_expressions": "G: Parabola;m: Number;H: Ellipse;Expression(G) = (y^2 = (4/m)*x);Expression(H) = (x^2/7 + y^2/3 = 1);Focus(G) = LeftFocus(H)",
    "query_expressions": "m",
    "answer_expressions": "-1/2",
    "fact_spans": "[[[1, 25]], [[74, 77]], [[29, 66]], [[1, 25]], [[29, 66]], [[1, 72]]]",
    "query_spans": "[[[74, 81]]]",
    "process": ""
  },
  {
    "text": "If the equation $\\frac{x^{2}}{k-2}+\\frac{y^{2}}{3-k}=1$ represents an ellipse, then what is the range of values for $k$?",
    "fact_expressions": "G: Ellipse;k: Number;Expression(G) = (x^2/(k - 2) + y^2/(3 - k) = 1)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(2, 5/2) + (5/2, 3)",
    "fact_spans": "[[[45, 47]], [[49, 52]], [[2, 47]]]",
    "query_spans": "[[[49, 59]]]",
    "process": ""
  },
  {
    "text": "Given that the distance between the foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $4$, and it passes through the point $A(\\sqrt{2}, \\sqrt{3})$, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (sqrt(2), sqrt(3));FocalLength(G) = 4;PointOnCurve(A,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/8+y^2/4=1",
    "fact_spans": "[[[2, 54], [90, 92]], [[4, 54]], [[4, 54]], [[64, 88]], [[4, 54]], [[4, 54]], [[2, 54]], [[64, 88]], [[2, 61]], [[2, 88]]]",
    "query_spans": "[[[90, 97]]]",
    "process": "By the given condition, $\\frac{x^{2}}{2}+\\frac{y^{2}}{b^{2}}=1$, and the ellipse passes through point $A(\\sqrt{2},\\sqrt{3})$, so $=1 \\Rightarrow 2b^{2}+3(b^{2}+4)=(b^{2}+4)b^{2}$. Simplifying yields $b^{4}-b^{2}-12=0$, hence $b^{2}=4$, $a^{2}=4+4=8$. Therefore, the equation of the ellipse is $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$."
  },
  {
    "text": "Given real numbers $x$, $y$ satisfying $\\frac{x|x|}{4}+y|y|=1$, then the range of $|x+2 y-4|$ is?",
    "fact_expressions": "x_: Real;y_: Real;y_*Abs(y_) + (x_*Abs(x_))/4 = 1",
    "query_expressions": "Range(Abs(x_ + 2*y_ - 4))",
    "answer_expressions": "[4-2*sqrt(2), 4)",
    "fact_spans": "[[[2, 7]], [[9, 12]], [[14, 37]]]",
    "query_spans": "[[[39, 57]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has eccentricity $\\frac{2 \\sqrt{3}}{3}$, focal length $2 c$, and $2 a^{2}=3 c$. On the hyperbola, a point $P$ satisfies $\\overrightarrow{PF_{1}} \\cdot \\overrightarrow{P F_{2}}=2$ ($F_{1}$, $F_{2}$ are the left and right foci), then $|\\overrightarrow{PF_{1}}| \\cdot |\\overrightarrow{PF_{2}} |$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(G) = (2*sqrt(3))/3;FocalLength(G) = 2*c;c: Number;2*a^2 = 3*c;P: Point;PointOnCurve(P,G) = True;DotProduct(VectorOf(P,F1), VectorOf(P,F2)) = 2;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2",
    "query_expressions": "Abs(VectorOf(P,F1))*Abs(VectorOf(P,F2))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 58], [110, 113]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 85]], [[2, 94]], [[89, 94]], [[96, 109]], [[117, 120]], [[110, 120]], [[122, 180]], [[181, 188]], [[190, 197]], [[110, 204]], [[110, 204]]]",
    "query_spans": "[[[206, 268]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with foci $F_{1}$, $F_{2}$, and $P$ a point on the hyperbola such that $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$. If the circumradius and inradius of $\\Delta F_{1} P F_{2}$ are $R$ and $r$ respectively, and $R=4 r$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;F1: Point;P: Point;F2: Point;a>0;b>0;R: Number;r: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2};AngleOf(F1, P, F2) = pi/3;R = 4*r;Radius(CircumCircle(TriangleOf(F1, P, F2))) = R;Radius(InscribedCircle(TriangleOf(F1, P, F2))) = r",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(2/7)*sqrt(21)",
    "fact_spans": "[[[2, 58], [82, 85], [184, 187]], [[5, 58]], [[5, 58]], [[62, 69]], [[78, 81]], [[70, 77]], [[5, 58]], [[5, 58]], [[165, 168]], [[170, 173]], [[2, 58]], [[78, 88]], [[2, 77]], [[90, 126]], [[175, 182]], [[129, 173]], [[129, 173]]]",
    "query_spans": "[[[184, 193]]]",
    "process": "The foci of the hyperbola are $F_{1}(-c,0)$, $F_{2}(c,0)$, $|F_{1}F_{2}|=2c$. In $\\triangle F_{1}PF_{2}$, by the law of sines:  \n$2R = \\frac{|F_{1}F_{2}|}{\\sin\\angle F_{1}PF_{2}} = \\frac{2c}{\\sin\\frac{\\pi}{3}} = \\frac{4\\sqrt{3}}{3}c$, solving gives $R = \\frac{2\\sqrt{3}}{3}c$, $r = \\frac{1}{4}R = \\frac{\\sqrt{3}}{6}c$.  \nLet $|PF_{1}| = m$, $|PF_{2}| = n$. In $\\triangle F_{1}PF_{2}$, by the law of cosines:  \n$4c^{2} = m^{2} + n^{2} - 2mn\\cos\\frac{\\pi}{3} = (m-n)^{2} + mn$, solving gives $mn = 4(c^{2}-a^{2})$.  \nSo $S_{\\triangle F_{1}PF_{2}} = \\frac{1}{2}mn\\sin\\frac{\\pi}{3} = \\sqrt{3}(c^{2}-a^{2})$.  \nSince $(m+n)^{2} = (m-n)^{2} + 4mn = 4a^{2} + 16(c^{2}-a^{2}) = 16c^{2}-12a^{2}$.  \nAlso $s_{\\Delta F_{1}PF_{2}} = \\frac{1}{2}(m+n+2c)r = \\frac{\\sqrt{3}c(m+n+2c)}{12}$.  \nSo $\\sqrt{3}(c^{2}-a^{2}) = \\frac{\\sqrt{3}c(m+n+2c)}{12}$, then $m+n = \\frac{10c^{2}-12a^{2}}{c}$.  \nTherefore $(m+n)^{2} = \\left(\\frac{10c^{2}-12a^{2}}{c}\\right)^{2} = 16c^{2}-12a^{2}$.  \nRearranging gives $21c^{4} + 36a^{4} - 57a^{2}c^{2} = 0$, then $(c^{2}-a^{2})(2lc^{2}-36a^{2}) = 0$.  \nSolving gives $e = \\frac{c}{a} = \\frac{2\\sqrt{21}}{7}$ or $e = 1$ (discarded)."
  },
  {
    "text": "If point $P$ lies on the curve $C_{1}$: $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, point $Q$ lies on the curve $C_{2}$: $(x-5)^{2}+y^{2}=1$, and point $R$ lies on the curve $C_{3}$: $(x+5)^{2}+y^{2}=1$, then the maximum value of $|P Q|-|P R|$ is?",
    "fact_expressions": "C1: Curve;C2: Curve;C3: Curve;P: Point;Q: Point;R: Point;Expression(C1) = (x^2/16-y^2/9=1);Expression(C2) = ((x-5)^2+y^2=1);Expression(C3) = ((x+5)^2+y^2=1);PointOnCurve(P,C1);PointOnCurve(Q,C2);PointOnCurve(R,C3)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, Q)) - Abs(LineSegmentOf(P, R)))",
    "answer_expressions": "10",
    "fact_spans": "[[[6, 53]], [[60, 89]], [[96, 125]], [[1, 5]], [[55, 59]], [[91, 95]], [[6, 53]], [[60, 89]], [[96, 125]], [[1, 54]], [[55, 90]], [[91, 126]]]",
    "query_spans": "[[[128, 147]]]",
    "process": "According to the problem, points $ F_{1}(-5,0) $, $ F_{2}(5,0) $ are the left and right foci of hyperbola $ C_{1} $, respectively. Therefore, $ |PQ| - |PR| \\leqslant (|PF_{2}| + 1)(|PF_{1}| - 1)|||PF_{2}| - |PF_{1}| + 2 = 2 \\times 4 + 2 = 10 $. Hence, the maximum value of $ |PQ| - |PR| $ is 10."
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[0, 27]], [[0, 27]]]",
    "query_spans": "[[[0, 33]]]",
    "process": "From the equation of the ellipse, we have: a=2, b=1, hence c=\\sqrt{a^{2}-b^{2}}=\\sqrt{3}, so the eccentricity of the ellipse is e=\\frac{c}{a}=\\frac{\\sqrt{3}}{2}"
  },
  {
    "text": "It is known that the hyperbola $C$ shares the same foci with the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{25}=1$, and the sum of their eccentricities is $\\frac{14}{5}$. Then, the equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Ellipse;Expression(G) = (x^2/9 + y^2/25 = 1);Focus(C) = Focus(G);Eccentricity(C)+Eccentricity(G)=14/5",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/4 - x^2/12 = 1",
    "fact_spans": "[[[2, 8], [80, 86]], [[9, 47]], [[9, 47]], [[2, 53]], [[55, 78]]]",
    "query_spans": "[[[80, 91]]]",
    "process": "Since hyperbola C shares the same foci with the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{25}=1$, we have $c=\\sqrt{25-9}=4$, and the foci lie on the y-axis. Let the equation of the hyperbola be $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0,b>0)$. Given that the sum of the eccentricities is $\\frac{14}{5}$, it follows that $\\frac{c}{a}+\\frac{4}{5}=\\frac{14}{5}$, solving which gives $a=2$, so $b^{2}=16-4=12$. Therefore, the equation of hyperbola C is $\\frac{y^{2}}{4}-\\frac{x^{2}}{12}=1$."
  },
  {
    "text": "The length of the real axis of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$ is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/2 = 1)",
    "query_expressions": "Length(RealAxis(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 45]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of a hyperbola with asymptotes $y=\\pm \\sqrt{3} x$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;Expression(H) = (y = pm*(sqrt(3)*x));Asymptote(G) = H",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{2,2*sqrt(3)/3}",
    "fact_spans": "[[[26, 29]], [[1, 21]], [[1, 21]], [[0, 29]]]",
    "query_spans": "[[[26, 35]]]",
    "process": "When the foci are on the x-axis: the lines $y=\\pm\\sqrt{3}x$ are asymptotes, then $\\frac{b}{a}=\\sqrt{3} \\therefore c=2a \\therefore e=2$. When the foci are on the y-axis: the lines $y=\\pm\\sqrt{3}x$ are asymptotes, then $\\frac{b}{a}=\\frac{\\sqrt{3}}{3} \\therefore c=\\frac{2\\sqrt{3}}{3}a \\therefore e=\\frac{2\\sqrt{3}}{3}$. In summary: the eccentricity is $2$ or $\\frac{2\\sqrt{3}}{3}$."
  },
  {
    "text": "What is the length of the major axis of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{25}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/25 = 1)",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "10",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 43]]]",
    "process": "From the given condition, for the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{25}=1$, we have $a=5$, $b=3$, so the length of the major axis of the ellipse is $2a=10$."
  },
  {
    "text": "Given that the distance from point $P$ to point $F(-3 , 0)$ is greater by $1$ than its distance to the line $x=2$, what equation does point $P$ satisfy?",
    "fact_expressions": "G: Line;F: Point;P: Point;Expression(G) = (x = 2);Coordinate(F) = (-3, 0);Distance(P, F) = Distance(P, G) + 1",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "y^2 = -12*x",
    "fact_spans": "[[[25, 32]], [[7, 19]], [[2, 6], [23, 24], [41, 45]], [[25, 32]], [[7, 19]], [[2, 39]]]",
    "query_spans": "[[[41, 52]]]",
    "process": ""
  },
  {
    "text": "The point $(2,3)$ lies on the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and the focal length of $C$ is $4$. What is its eccentricity?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;G: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(G) = (2, 3);PointOnCurve(G, C);FocalLength(C) = 4",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[11, 71], [73, 76], [85, 86]], [[18, 71]], [[18, 71]], [[2, 10]], [[18, 71]], [[18, 71]], [[11, 71]], [[2, 10]], [[2, 72]], [[73, 83]]]",
    "query_spans": "[[[85, 92]]]",
    "process": ""
  },
  {
    "text": "What is the focal length of the hyperbola $\\frac{x^{2}}{32}-\\frac{y^{2}}{18}=-2$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/32 - y^2/18 = -2)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "20",
    "fact_spans": "[[[0, 41]], [[0, 41]]]",
    "query_spans": "[[[0, 47]]]",
    "process": ""
  },
  {
    "text": "Given a hyperbola centered at the origin with foci on the $x$-axis, its eccentricity is $\\frac{3}{2}$ and the length of the real axis is $4$. What is the equation of the hyperbola?",
    "fact_expressions": "G: Hyperbola;O: Origin;Center(G) = O;PointOnCurve(Focus(G), xAxis);Length(RealAxis(G)) = 4;Eccentricity(G) = 3/2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/5 = 1",
    "fact_spans": "[[[18, 21], [51, 54]], [[5, 7]], [[2, 21]], [[9, 21]], [[18, 48]], [[18, 39]]]",
    "query_spans": "[[[51, 59]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{4}+\\frac{y^{2}}{k}=1$ is $e=3$, then the value of $k$ is?",
    "fact_expressions": "G: Hyperbola;k: Number;Expression(G) = (x^2/4 + y^2/k = 1);Eccentricity(G) = e;e:Number;e = 3",
    "query_expressions": "k",
    "answer_expressions": "-32",
    "fact_spans": "[[[0, 38]], [[49, 52]], [[0, 38]], [[0, 47]], [[42, 47]], [[42, 47]]]",
    "query_spans": "[[[49, 56]]]",
    "process": ""
  },
  {
    "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $\\frac{\\sqrt{3}}{2}$, then what is the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Ellipse;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a > b;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);Eccentricity(H) = sqrt(3)/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[82, 139]], [[4, 54]], [[4, 54]], [[2, 54]], [[85, 138]], [[4, 54]], [[82, 139]], [[4, 54]], [[2, 54]], [[2, 79]]]",
    "query_spans": "[[[82, 144]]]",
    "process": "Using the eccentricity of the ellipse, the value of \\frac{b^{2}}{a^{2}} can be obtained, and then the eccentricity of the hyperbola can be determined. Let the semi-focal length of the ellipse \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0) be c_{1}, and the semi-focal length of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0) be c_{2}. The eccentricity of the ellipse is e_{1}=\\frac{c_{1}}{a}=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{\\frac{a^{2}-b^{2}}{a^{2}}}=\\sqrt{1-\\frac{b^{2}}{a^{2}}}=\\frac{\\sqrt{3}}{2}, \\therefore \\frac{b^{2}}{a^{2}}=\\frac{1}{4},"
  },
  {
    "text": "It is known that an asymptote of the hyperbola $k x^{2}-y^{2}=1$ is perpendicular to the line $2 x+y+1=0$. Then, the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;k: Number;H: Line;Expression(G) = (k*x^2 - y^2 = 1);Expression(H) = (2*x + y + 1 = 0);IsPerpendicular(OneOf(Asymptote(G)), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 22], [47, 50]], [[5, 22]], [[29, 42]], [[2, 22]], [[29, 42]], [[2, 44]]]",
    "query_spans": "[[[47, 56]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively. Point $P$ lies on $C$. If $\\angle P F_{1} F_{2} = \\angle F_{1} P F_{2} = 30^{\\circ}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);AngleOf(P,F1,F2)=AngleOf(F1,P,F2);AngleOf(F1,P,F2)=ApplyUnit(30,degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(\\sqrt{3}+1)/2",
    "fact_spans": "[[[19, 80], [92, 95], [154, 157]], [[27, 80]], [[27, 80]], [[87, 91]], [[1, 8]], [[9, 16]], [[27, 80]], [[27, 80]], [[19, 80]], [[1, 86]], [[1, 86]], [[87, 96]], [[98, 152]], [[98, 152]]]",
    "query_spans": "[[[154, 163]]]",
    "process": "From the given conditions: $F_{1}F_{2}=PF_{2}=2c$, $PF_{1}=2\\sqrt{3}c$, $\\because PF_{1}-PF_{2}=2a$, $\\therefore PF_{1}-PF_{2}=2\\sqrt{3}c-2c=2a$, hence $e=\\frac{1+\\sqrt{3}}{2}$"
  },
  {
    "text": "Given $F_{1}(-3 , 0)$, $F_{2}(3 , 0)$, and point $M$ satisfies $|M F_{1}|+|M F_{2}|=10$, what is the equation of the trajectory of $M$?",
    "fact_expressions": "F1: Point;F2: Point;M: Point;Coordinate(F1) = (-3, 0);Coordinate(F2) = (3, 0);Abs(LineSegmentOf(M, F1)) + Abs(LineSegmentOf(M, F2)) = 10",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2/25+y^2/16=1",
    "fact_spans": "[[[2, 17]], [[20, 34]], [[35, 39], [68, 71]], [[2, 18]], [[20, 34]], [[41, 65]]]",
    "query_spans": "[[[68, 78]]]",
    "process": ""
  },
  {
    "text": "Given that the equations of the two asymptotes of a hyperbola are $3x \\pm 4y = 0$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (3*x + pm*4*y = 0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16 - y^2/9 = 1",
    "fact_spans": "[[[2, 5], [31, 34]], [[2, 29]]]",
    "query_spans": "[[[31, 38]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola $y^{2}=8 x$ is $F$, $P$ is a moving point on the parabola, and $A(3,2)$ is a fixed point. Then the minimum value of $|P A|+|P F|$ is?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 8*x);Coordinate(A) = (3, 2);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "5",
    "fact_spans": "[[[0, 14], [25, 28]], [[36, 44]], [[21, 24]], [[17, 20]], [[0, 14]], [[36, 44]], [[0, 20]], [[21, 33]]]",
    "query_spans": "[[[46, 65]]]",
    "process": "The directrix is $x = -2$. Draw a perpendicular from point $P$ to the directrix $l$, with foot $M$. Then $|PM| = |PF|$, so $|PF| + |PA| = |PM| + |PA|$. It is clear that $|PM| + |PA|$ reaches its minimum value when points $M$, $P$, and $A$ are collinear, and this minimum value is $3 - (-2) = 5$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F(2,0)$, then the equation of parabola $C$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Coordinate(F) = (2, 0);Focus(C) = F",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=8*x",
    "fact_spans": "[[[2, 28], [42, 48]], [[2, 28]], [[10, 28]], [[10, 28]], [[32, 40]], [[32, 40]], [[2, 40]]]",
    "query_spans": "[[[42, 53]]]",
    "process": "Since the focus of the parabola $ C: y^{2} = 2px $ ($ p > 0 $) is $ F(2,0) $, that is, $ \\frac{p}{2} = 2 $, so $ p = 4 $, therefore the equation of the parabola $ C $ is $ y^{2} = 8x $."
  },
  {
    "text": "Given a line with inclination angle $60^{\\circ}$ passing through the focus $F$ of the curve $C$: $y=2 x^{2}$, and intersecting $C$ at two distinct points $A$, $B$ ($A$ in the first quadrant), then $|A F|$=?",
    "fact_expressions": "G: Line;C: Curve;A: Point;B:Point;F: Point;Expression(C) = (y = 2*x^2);Inclination(G)=ApplyUnit(60,degree);PointOnCurve(F,G);Intersection(G,C)={A,B};Negation(A=B);Quadrant(A)=1;Focus(C)=F",
    "query_expressions": "Abs(LineSegmentOf(A, F))",
    "answer_expressions": "(2+sqrt(3))/2",
    "fact_spans": "[[[19, 21]], [[22, 40], [49, 52]], [[60, 63], [69, 72]], [[64, 67]], [[43, 46]], [[22, 40]], [[2, 21]], [[19, 46]], [[19, 67]], [[55, 67]], [[69, 77]], [[22, 46]]]",
    "query_spans": "[[[80, 89]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $y^{2}-m x^{2}=1$ has its foci on the $y$-axis and an eccentricity of $\\frac{\\sqrt{5}}{2}$. Then, the value of $m$ is?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (-m*x^2 + y^2 = 1);PointOnCurve(Focus(G), yAxis);Eccentricity(G) = sqrt(5)/2",
    "query_expressions": "m",
    "answer_expressions": "4",
    "fact_spans": "[[[9, 29]], [[56, 59]], [[9, 29]], [[0, 29]], [[9, 54]]]",
    "query_spans": "[[[56, 63]]]",
    "process": "The standard equation of the hyperbola is $ y^{2}-\\frac{x^{2}}{m}=1 $. From the given conditions, we have $ m>0 $, then $ a=1 $, $ b=\\frac{1}{\\sqrt{m}} $, $ c=\\sqrt{a^{2}+b^{2}}=\\sqrt{1+\\frac{1}{m}} $. Therefore, $ e=\\frac{c}{a}=\\sqrt{1+\\frac{1}{m}}=\\frac{\\sqrt{5}}{2} $, solving for $ m $ gives $ m=4 $."
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x$ coincides with the right focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, then the value of the real number $p$ is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Real;Expression(G) = (x^2/4 - y^2/5 = 1);Expression(H) = (y^2 = 2*(p*x));Focus(H) = RightFocus(G)",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[21, 59]], [[1, 17]], [[67, 72]], [[21, 59]], [[1, 17]], [[1, 65]]]",
    "query_spans": "[[[67, 76]]]",
    "process": "Since the right focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ is $(3,0)$, it follows that $\\frac{p}{2}=3$, $p=6$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$, a line $l$ passes through the point $P(2,1)$ and intersects the parabola $C$ at two points $M$ and $N$. If the midpoint of the segment $MN$ is exactly the point $P$, then what is the slope of the line $l$?",
    "fact_expressions": "l: Line;C: Parabola;M: Point;N: Point;P: Point;Expression(C) = (y^2 = 4*x);Coordinate(P) = (2, 1);PointOnCurve(P, l);Intersection(l, C) = {M, N};MidPoint(LineSegmentOf(M, N)) = P",
    "query_expressions": "Slope(l)",
    "answer_expressions": "2",
    "fact_spans": "[[[22, 28], [82, 87]], [[2, 21], [42, 48]], [[50, 53]], [[54, 57]], [[29, 38]], [[2, 21]], [[29, 38]], [[23, 38]], [[23, 59]], [[62, 79]]]",
    "query_spans": "[[[82, 92]]]",
    "process": "Let M(x_{1},y_{1}), N(x_{2},y_{2}). \\therefore y_{1}=4x_{1}-\\textcircled{1}, y_{2}=4x_{2}-\\textcircled{2}. From \\textcircled{1}-\\textcircled{2} we get: (y_{1}+y_{2})(y_{1}-y_{2})=4(x_{1}-x_{2}). \\because the midpoint of segment MN is exactly point P, \\therefore according to the midpoint coordinate formula we obtain: y_{1}+y_{2}=2. \\therefore k=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=2"
  },
  {
    "text": "If the line $y=kx-1$ and the hyperbola $x^{2}-y^{2}=4$ have only one common point, then $k=$?",
    "fact_expressions": "G: Hyperbola;H: Line;k: Number;Expression(G) = (x^2-y^2=4);Expression(H) = (y=k*x - 1);NumIntersection(H, G) = 1",
    "query_expressions": "k",
    "answer_expressions": "{pm*(sqrt(5)/2),pm*1}",
    "fact_spans": "[[[12, 30]], [[1, 11]], [[39, 42]], [[12, 30]], [[1, 11]], [[1, 37]]]",
    "query_spans": "[[[39, 44]]]",
    "process": ""
  },
  {
    "text": "What is the distance from the focus to the directrix of the parabola $y=x^{2}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "1/2",
    "fact_spans": "[[[0, 12]], [[0, 12]]]",
    "query_spans": "[[[0, 23]]]",
    "process": "From the analytical expression of the parabola, we can find p. By transforming $ y = x^{2} $ into $ x^{2} = y $, the focus of the parabola lies on the positive half-axis of $ y $, $ 2p = 1 $, $ p = \\frac{1}{2} $. Therefore, the distance from the focus to the directrix of the parabola $ y = x^{2} $ is $ p = \\frac{1}{2} $."
  },
  {
    "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ have its right focus at $F$. Draw a line through $F$ perpendicular to the $x$-axis, intersecting $C$ at points $A$ and $B$. If the circle with diameter $AB$ is tangent to the asymptotes of $C$, then the equation of the asymptotes of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(C) = F;H: Line;PointOnCurve(F,H) = True;IsPerpendicular(H,xAxis) = True;Intersection(H,C) = {A,B};A: Point;B: Point;IsDiameter(LineSegmentOf(A,B),G) = True;G: Circle;IsTangent(G,Asymptote(C)) = True",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "x + pm*y = 0",
    "fact_spans": "[[[1, 62], [87, 90], [115, 118], [126, 132]], [[1, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[67, 70], [72, 75]], [[1, 70]], [[84, 86]], [[71, 86]], [[76, 86]], [[84, 100]], [[91, 94]], [[95, 98]], [[101, 114]], [[113, 114]], [[113, 124]]]",
    "query_spans": "[[[126, 140]]]",
    "process": "According to the problem, the equation of line AB is x = c. Substituting into C: \\frac{x^2}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1, we get y = \\pm\\frac{b^{2}}{a}, so the center of the circle with segment AB as diameter is (c, 0), and the radius is \\frac{b^{2}}{a}. One asymptote of the hyperbola C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 (a > 0, b > 0) is bx - ay = 0. Since the asymptote is tangent to the circle, \\frac{|bc|}{\\sqrt{a^{2} + b^{2}}} = b = \\frac{b^{2}}{a}. Simplifying gives a = b, so the asymptote equation is x \\pm y = 0."
  },
  {
    "text": "Given fixed points $A(0,2)$, $B(0,-2)$, $C(3,2)$, an ellipse passing through points $A$ and $B$ is drawn with $C$ as one focus. What is the trajectory equation of the other focus $F$ of the ellipse?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;C: Point;Coordinate(A) = (0, 2);Coordinate(B) = (0, -2);Coordinate(C) = (3, 2);OneOf(Focus(G)) = C;PointOnCurve(A, G);PointOnCurve(B, G);Focus(G) = {C, F};F: Point",
    "query_expressions": "LocusEquation(F)",
    "answer_expressions": "(y^2 - x^2/3 = 1)&(y <= -1)",
    "fact_spans": "[[[57, 59], [61, 63]], [[4, 12], [47, 50]], [[15, 25], [51, 54]], [[26, 35], [37, 40]], [[4, 12]], [[15, 25]], [[26, 35]], [[36, 59]], [[46, 59]], [[46, 59]], [[36, 72]], [[69, 72]]]",
    "query_spans": "[[[69, 79]]]",
    "process": "\\because A,B lie on an ellipse with foci C,F, \\therefore |AC|+|AF|=|BC|+|BF|, \\therefore |AF|-|BF|=|BC|-|AC|=\\sqrt{3^{2}+4^{2}}-\\sqrt{3^{2}+0^{2}}=2. Thus, the trajectory of F is the lower branch of a hyperbola with foci A,B. Let the hyperbola equation be \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1 (y\\leqslant -a). Then 2a=2, so a=1, c=2, \\therefore b^{2}=c^{2}-a^{2}=3. The trajectory equation of focus F is y^{2}-\\frac{x^{2}}{3}=1 (y\\leqslant -1)."
  },
  {
    "text": "The focus of the parabola $C$: $y^{2}=2 p x(p>0)$ is $F$, and point $P(2, m)$ lies on $C$. If $|P F|=3$, then $m$=?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(C) = F;P: Point;Coordinate(P) = (2, m);m: Number;PointOnCurve(P, C);Abs(LineSegmentOf(P, F)) = 3",
    "query_expressions": "m",
    "answer_expressions": "pm*2*sqrt(2)",
    "fact_spans": "[[[0, 26], [45, 48]], [[0, 26]], [[8, 26]], [[8, 26]], [[30, 33]], [[0, 33]], [[34, 44]], [[34, 44]], [[64, 67]], [[35, 51]], [[53, 62]]]",
    "query_spans": "[[[64, 69]]]",
    "process": "According to the definition of a parabola, solve by substitution. [Detailed solution] The directrix equation of parabola $ C: y^{2} = 2px $ is: $ x = -\\frac{p}{2} $. Since $ |PF| = 3 $, we have $ 2 - (-\\frac{p}{2}) = 3 \\Rightarrow p = 2 $. Substituting $ P(2, m) $ into the parabola equation gives $ m^{2} = 2 \\times 2 \\times 2 \\Rightarrow m = \\pm 2\\sqrt{2} $."
  },
  {
    "text": "The coordinates of the focus of the parabola $y^{2}=2 p x(p>0)$ are $(3,0)$. Then the value of $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(Focus(G)) = (3, 0)",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[0, 21]], [[36, 39]], [[3, 21]], [[0, 21]], [[0, 34]]]",
    "query_spans": "[[[36, 43]]]",
    "process": "Since the focus of the parabola \\( y^{2} = 2px \\) (\\( p > 0 \\)) is at \\( (3, 0) \\), we have \\( \\frac{p}{2} = 3 \\), solving gives \\( p = 6 \\)."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has its right focus at $F(2,0)$. The line passing through point $F$ and perpendicular to an asymptote is tangent to the circle $x^{2}+y^{2}+4 x=0$. Then, the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;Coordinate(F) = (2, 0);RightFocus(C) = F;PointOnCurve(F, H);IsPerpendicular(H, Asymptote(C));H: Line;G: Circle;Expression(G) = (4*x + x^2 + y^2 = 0);IsTangent(H, G)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 63], [117, 123]], [[2, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[68, 76], [78, 82]], [[68, 76]], [[2, 76]], [[77, 91]], [[2, 91]], [[89, 91]], [[93, 113]], [[93, 113]], [[89, 115]]]",
    "query_spans": "[[[117, 129]]]",
    "process": "From $x^{2}+y^{2}+4x=0$, we obtain $(x+2)^{2}+y^{2}=4$, yielding the center $(-2,0)$ and radius $r=2$. Since $F(2,0)$, the angle of inclination of the line passing through point $F$ and tangent to the circle $x^{2}+y^{2}+4x=0$ is $150^{\\circ}$, so the slope is $-\\frac{\\sqrt{3}}{3}$. Thus, the slope of the asymptote is $\\sqrt{3}$, that is, $\\frac{b}{a}=\\sqrt{3}$, so the eccentricity $e=\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{4}=2$."
  },
  {
    "text": "The equation of the hyperbola that shares the same asymptotes as $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ and passes through $A(-3 , 4 \\sqrt{2})$ is?",
    "fact_expressions": "G1: Hyperbola;Expression(G1) = (x^2/9 - y^2/16 = 1);G2: Hyperbola;Asymptote(G1) = Asymptote(G2);A: Point;Coordinate(A) = (-3, 4*sqrt(2));PointOnCurve(A, G2) = True",
    "query_expressions": "Expression(G2)",
    "answer_expressions": "y^2/16 - x^2/9 = 1",
    "fact_spans": "[[[1, 40]], [[1, 40]], [[70, 73]], [[0, 73]], [[49, 69]], [[49, 69]], [[48, 73]]]",
    "query_spans": "[[[70, 77]]]",
    "process": ""
  },
  {
    "text": "Let $O$ be the origin, and the line $x = a$ intersects the two asymptotes of the hyperbola $C$: $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ ($a > 0$, $b > 0$) at points $D$ and $E$, respectively. If the area of $\\triangle O D E$ is $8$, then the minimum value of the focal length of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;O: Origin;D: Point;E: Point;a>0;b>0;L1:Line;L2:Line;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x = a);Asymptote(C)={L1,L2};Intersection(G,L1)=D;Intersection(G,L2)=E;Area(TriangleOf(O, D, E)) = 8",
    "query_expressions": "Min(FocalLength(C))",
    "answer_expressions": "8",
    "fact_spans": "[[[18, 79], [126, 129]], [[26, 79]], [[26, 79]], [[10, 17]], [[1, 4]], [[89, 92]], [[93, 96]], [[26, 79]], [[26, 79]], [], [], [[18, 79]], [[10, 17]], [[18, 85]], [[10, 98]], [[10, 98]], [[100, 124]]]",
    "query_spans": "[[[126, 138]]]",
    "process": "The asymptotes of hyperbola $ C $ are given by $ y = \\pm\\frac{b}{a}x $. Without loss of generality, let point $ D $ be in the first quadrant and $ E $ in the fourth quadrant. From \n\\[\n\\begin{cases}\nx = a \\\\\ny = \\frac{b}{a}x\n\\end{cases}\n\\]\nwe obtain $ D(a, b) $, and similarly $ E(a, -b) $. Then $ |ED| = 2b $, so the area of $ \\triangle ODE $ is $ S_{\\triangle ODE} = \\frac{1}{2}a \\cdot 2b = ab = 8 $. Thus, the focal distance of hyperbola $ C $ is $ 2c = 2\\sqrt{a^{2} + b^{2}} > 2\\sqrt{2ab} = 2\\sqrt{16} = 8 $, with equality if and only if $ a = b = 2\\sqrt{2} $. Therefore, the minimum value of the focal distance of $ C $ is $ 8 $. The answer is $ 8 $."
  },
  {
    "text": "Given that the directrix of the parabola $y^{2}=4 x$ is $l$, and $l$ intersects the two asymptotes of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ at points $A$ and $B$ respectively, then the length of segment $AB$ is?",
    "fact_expressions": "A: Point;B: Point;G: Hyperbola;H: Parabola;Expression(G) = (x^2/4 - y^2 = 1);Expression(H) = (y^2 = 4*x);l:Line;Directrix(H)=l;L1:Line;L2:Line;Asymptote(G)={L1,L2};Intersection(L1,l)=A;Intersection(L2,l)=B",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "1",
    "fact_spans": "[[[66, 69]], [[70, 73]], [[28, 56]], [[2, 16]], [[28, 56]], [[2, 16]], [[20, 23], [24, 27]], [[2, 23]], [], [], [[28, 62]], [[24, 75]], [[24, 75]]]",
    "query_spans": "[[[77, 88]]]",
    "process": "According to the problem, write the equation of the directrix of the parabola and the equations of the asymptotes of the hyperbola. Then solve the system of equations with the line to find the coordinates of points A and B, from which the result can be obtained. [Detailed solution] The directrix $ l $ of the parabola $ y^2 = 4x $ has the equation: $ x = -1 $. The two asymptotes of the hyperbola $ \\frac{x^{2}}{4} - y^{2} = 1 $ are: $ x \\pm 2y = 0 $. Solving $ x = -1 $ together with $ x \\pm 2y = 0 $, we obtain: $ A(-1, \\frac{1}{2}) $, $ B(-1, -\\frac{1}{2}) $. Therefore, the length of segment AB is: 1."
  },
  {
    "text": "The distance from the focus to the directrix of the parabola $y^{2}=8 x$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 25]]]",
    "process": "From $ y^{2} = 2px = 8x $, we know $ p = 4 $. Since the distance from the focus to the directrix is $ p $, the distance from the focus to the directrix is 4."
  },
  {
    "text": "A line $l$ passing through the point $M(2,1)$ intersects the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$ at points $A$ and $B$, and $M$ is the midpoint of $AB$. Then the equation of line $l$ is?",
    "fact_expressions": "l: Line;G: Hyperbola;A: Point;B: Point;M: Point;Expression(G) = (x^2 - y^2/2 = 1);Coordinate(M) = (2, 1);PointOnCurve(M, l);Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Expression(l)",
    "answer_expressions": "4*x-y-7=0",
    "fact_spans": "[[[12, 17], [72, 77]], [[18, 46]], [[47, 50]], [[51, 54]], [[2, 11], [58, 61]], [[18, 46]], [[2, 11]], [[0, 17]], [[12, 56]], [[58, 70]]]",
    "query_spans": "[[[72, 82]]]",
    "process": "Let point A(x_{1},y_{1}), point B(x_{2},y_{2}), substitute into the hyperbola equation, subtract the two equations, and use the midpoint coordinate formula to find the slope of the line. The solution can then be obtained using the point-slope form. Let point A(x_{1},y_{1}), point B(x_{2},y_{2}), M(x_{0},y_{0}), then 2x_{1}^{2}-y_{1}^{2}=2,\\cdots\\cdots\\textcircled{1} 2x_{2}^{2}-y_{2}^{2}=2,\\cdots\\cdots\\textcircled{2} \\textcircled{1}-\\textcircled{2} yields 2(x_{1}+x_{2})(x_{1}-x_{2})-(y_{1}+y_{2})(y_{1}-y_{2})=0, 2\\times2x_{0}-2y_{0}\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=0, so 8-2k=0, thus k=4, so y-1=4(x-2). Therefore, the equation of line l is 4x-y-7=0."
  },
  {
    "text": "The coordinates of the focus of the parabola $y=\\frac{1}{m} x^{2}(m<0)$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2/m);m: Number;m<0",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,m/4)",
    "fact_spans": "[[[0, 29]], [[0, 29]], [[3, 29]], [[3, 29]]]",
    "query_spans": "[[[0, 36]]]",
    "process": "The standard equation of the parabola \\( y = \\frac{1}{m}x^{2} \\) (\\( m < 0 \\)) is \\( x^{2} = my \\), and the coordinates of the focus are \\( \\left(0, \\frac{m}{4}\\right) \\)."
  },
  {
    "text": "What is the standard equation of the hyperbola whose vertices are the foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{15}=1$, and whose foci are the vertices of this ellipse?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/25 + y^2/15 = 1);Vertex(H)=Focus(G);Focus(H)=Vertex(G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/10 - y^2/15 = 1",
    "fact_spans": "[[[53, 56]], [[1, 40]], [[1, 40]], [[0, 56]], [[0, 56]]]",
    "query_spans": "[[[53, 63]]]",
    "process": "The ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{15}=1$ has foci at $(\\pm\\sqrt{10},0)$ and vertices at $(\\pm5,0)$. Let the hyperbola equation be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$, with focal distance $2c$. Then $a=\\sqrt{10}$, $c=5$, $\\therefore b=\\sqrt{25-10}=\\sqrt{15}$. The required hyperbola equation is $\\frac{x^{2}}{10}-\\frac{y^{2}}{15}=1$."
  },
  {
    "text": "Given the circle $C$: $(x+2)^{2}+(y-1)^{2}=\\frac{5}{2}$ and the ellipse $\\Gamma$: $x^{2}+4 y^{2}=4 b^{2}$ intersect at points $A$ and $B$, if $AB$ is a diameter of the circle $C$, then the equation of the ellipse $\\Gamma$ is?",
    "fact_expressions": "Gamma:Ellipse;C: Circle;A: Point;B: Point;b:Number;Expression(Gamma) = (x^2+4*y^2=4*b^2);Expression(C) = ((x + 2)^2 + (y - 1)^2 = 5/2);Intersection(Gamma, C)={A,B};IsDiameter(LineSegmentOf(A, B), C)",
    "query_expressions": "Expression(Gamma)",
    "answer_expressions": "x^2/12+y^2/3=1",
    "fact_spans": "[[[41, 76], [105, 115]], [[2, 40], [96, 100]], [[79, 82]], [[83, 86]], [[52, 76]], [[41, 76]], [[2, 40]], [[2, 88]], [[90, 103]]]",
    "query_spans": "[[[105, 120]]]",
    "process": "Let the intersection points be $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, use the point difference method to find the slope of line $AB$, then solve the system of equations of the line and the ellipse, use the chord length $|AB|=\\sqrt{10}$ to establish a relationship and solve for parameter $b$. The ellipse is $F:\\frac{x^{2}}{4b^{2}}+\\frac{y^{2}}{b^{2}}=1$, so $a^{2}=4b^{2}$, with foci on the $x$-axis. According to the problem, $C(-2,1)$ is the midpoint of $AB$, let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, $\\frac{12}{b^{2}}+\\frac{y_{1}^{2}}{b^{2}}=1$, thus $k_{AB}\\cdot k_{CO}=-\\frac{(y_{1}-y_{2})(y_{1}+y_{2})}{(x_{1}-x_{2})(x_{1}+x_{2})}=-\\frac{1}{4}$, and $k_{CO}=\\frac{1}{-2}=-\\frac{1}{2}$, then $k_{AB}=\\frac{1}{2}$. Hence, the equation of line $AB$ is: $y-1=\\frac{1}{2}(x+2)$. Solve the system:  \n$$\n\\begin{cases}\ny-1=\\frac{1}{2}(x+2)\\\\\nx^{2}+4y^{2}=4b^{2}\n\\end{cases}\n$$\nobtain: $x^{2}+4[\\frac{1}{2}(x+2)+1]^{2}=4b^{2}$, i.e., $x^{2}+(x+4)^{2}=4b^{2}$, i.e., $2x^{2}+8x+16-4b^{2}=0$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $x_{1}+x_{2}=4$, $x_{1}\\cdot x_{2}=\\frac{16-4b^{2}}{2}$, i.e., $\\frac{5}{4}\\cdot(8b^{2}-16)=10$, i.e., $8b^{2}-16=8$, $b^{2}=3$, then $a^{2}=4b^{2}=12$, the equation of ellipse $T$ is: $\\frac{x^{2}}{12}+\\frac{y^{2}}{3}=1$"
  },
  {
    "text": "If the distance from point $P$ to the line $x=-2$ is equal to the distance from $P$ to the point $F(2,0)$, then what is the equation of the trajectory of point $P$?",
    "fact_expressions": "G: Line;F: Point;P: Point;Expression(G) = (x=-2);Coordinate(F) = (2, 0);Distance(P,G)=Distance(P,F)",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "y^2=8*x",
    "fact_spans": "[[[6, 14]], [[20, 29]], [[1, 5], [1, 5]], [[6, 14]], [[20, 29]], [[1, 32]]]",
    "query_spans": "[[[34, 45]]]",
    "process": "By the given condition: the distance from point P to the line x = -2 is equal to the distance from point P to the point F(2,0), then the trajectory of point P is a parabola with vertex at the origin, opening to the right, where x = -2 is its directrix and F(2,0) is its focus. Let its equation be y^{2} = 2px, \\frac{p}{2} = 2, solving gives p = 4, so its equation is: y^{2} = 8x."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the common foci of the ellipse $C_{1}$: $\\frac{x^{2}}{a_{1}^{2}}+\\frac{y^{2}}{b_{1}^{2}}=1$ $(a_{1}>b_{1}>0)$ and the hyperbola $C_{2}$: $\\frac{x^{2}}{a_{2}^{2}}-\\frac{y^{2}}{b_{2}^{2}}=1$ $(a_{2}>0, b_{2}>0)$. They intersect at point $M$ in the first quadrant, $\\angle F_{1} M F_{2}=90^{\\circ}$, and if the eccentricity of the hyperbola $C_{2}$ is $e_{2}=\\frac{3 \\sqrt{2}}{2}$, then what is the eccentricity $e_{1}$ of the ellipse $C_{1}$?",
    "fact_expressions": "C1: Ellipse;Expression(C1) = (x^2/a1^2+y^2/b1^2=1);a1: Number;b1: Number;a1>b1;b1>0;F1: Point;F2: Point;C2: Hyperbola;Expression(C2) = (x^2/a2^2-y^2/b2^2=1);a2: Number;b2: Number;a2>0;b2>0;Focus(C1) = {F1, F2};Focus(C2) = {F1, F2};M: Point;Quadrant(M) = 1;Intersection(C1, C2) = M;AngleOf(F1, M, F2) = ApplyUnit(90, degree);e1: Number;e2: Number;Eccentricity(C1) = e1;Eccentricity(C2) = e2;e2 = (3*sqrt(2))/2",
    "query_expressions": "e1",
    "answer_expressions": "3/4",
    "fact_spans": "[[[19, 96], [280, 289]], [[19, 96]], [[30, 96]], [[30, 96]], [[30, 96]], [[30, 96]], [[1, 8]], [[9, 16]], [[97, 178], [236, 246]], [[97, 178]], [[109, 178]], [[109, 178]], [[109, 178]], [[109, 178]], [[1, 183]], [[1, 183]], [[194, 198]], [[187, 199]], [[184, 198]], [[201, 234]], [[293, 300]], [[250, 278]], [[280, 300]], [[236, 278]], [[250, 278]]]",
    "query_spans": "[[[293, 304]]]",
    "process": "By the definitions of the ellipse and hyperbola, we have |MF_{1}|+|MF_{2}|=2a_{1}, |MF_{1}|-|MF_{2}|=2a_{2}, so |MF_{1}|=a_{1}+a_{2}, |MF_{2}|=a_{1}-a_{2}. Since \\angle F_{1}MF_{2}=90^{\\circ}, it follows that |MF_{1}|^{2}+|MF_{2}|^{2}=4c^{2}, i.e., (a_{1}+a_{2})^{2}+(a_{1}-a_{2})^{2}=4c^{2}, which gives a_{1}^{2}+a_{2}^{2}=2c^{2}. Also, from e_{1}=\\frac{c}{a_{1}}, e_{2}=\\frac{c}{a_{2}}, we obtain \\frac{1}{e_{1}^{2}}+\\frac{1}{e_{2}^{2}}=2. Since e_{2}=\\frac{3\\sqrt{2}}{2}, it follows that \\frac{1}{e_{1}^{2}}=\\frac{16}{9}, so e_{1}=\\frac{3}{4}."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, with left and right foci $F_{1}$ and $F_{2}$ respectively. The line $l$ passing through $F_{2}$ intersects the right branch of $C$ at points $A$ and $B$, such that $\\overrightarrow{A B} \\cdot \\overrightarrow{A F_{1}}=0$ and $12|\\overrightarrow{A B}|=5|\\overrightarrow{A F_{1}}|$. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;l: Line;PointOnCurve(F2, l);A: Point;B: Point;Intersection(l, RightPart(C)) = {A, B};DotProduct(VectorOf(A, B), VectorOf(A, F1)) = 0;12*Abs(VectorOf(A, B)) = 5*Abs(VectorOf(A, F1))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(37)/5",
    "fact_spans": "[[[2, 63], [103, 106], [233, 236]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[72, 79]], [[80, 87], [89, 96]], [[2, 87]], [[2, 87]], [[97, 102]], [[88, 102]], [[110, 113]], [[114, 117]], [[97, 119]], [[121, 176]], [[177, 231]]]",
    "query_spans": "[[[233, 242]]]",
    "process": "From the given condition, $ AB \\perp AF_{1} $, let $ |AF_{1}| = 12m $, it is easy to obtain $ |AB| = 5m $. According to the definition of hyperbola, $ |AF_{2}| = |AF_{1}| - 2a $, $ |BF_{2}| = |BF_{1}| - 2a $. Since $ |AF_{2}| = |AB| - |BF_{2}| $, the quantitative relationship between hyperbola parameter $ a $ and $ m $ can be found. In right triangle $ \\triangle F_{1}AF_{2} $, apply the Pythagorean theorem to construct a homogeneous equation in $ a $ and $ c $, thus obtaining the eccentricity. Given $ |\\overrightarrow{AF}| = 5|\\overrightarrow{AF}| $, then $ |AB| = 5m $, so in right triangle $ \\Delta F_{1}AB $, $ |BF_{1}| = \\sqrt{|AF_{1}|^{2} + |AB|^{2}} = 13m $. Also, $ |AF_{2}| = |AB| - |BF_{2}| = |AF_{1}| - 2a $ and $ |BF_{2}| = |BF_{1}| - 2a = 13m - 2a $. Therefore, $ |AF_{2}| = 2a - 8m = 12m - 2a $, which gives $ m = \\frac{a}{5} $, so $ |AF_{1}| = \\frac{12a}{5} $, $ |AF_{2}| = \\frac{2a}{5} $. Hence, in right triangle $ \\triangle F_{1}AF_{2} $, $ |F_{1}F_{2}|^{2} = |AF_{1}|^{2} + |AF_{2}|^{2} = 4c^{2} $, i.e., $ \\frac{148a^{2}}{25} = 4c^{2} $, thus $ e = \\frac{c}{a} = \\frac{\\sqrt{37}}{5} $."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=2 p x(p>0)$ with an inclination angle of $\\frac{\\pi}{4}$ intersects the parabola at points $A$ and $B$. If the perpendicular bisector of chord $AB$ passes through the point $(0,2)$, then what is the value of $p$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);Inclination(H) = pi/4;A: Point;B: Point;Intersection(H, G) = {A, B};IsChordOf(LineSegmentOf(A, B), G);I: Point;Coordinate(I) = (0, 2);PointOnCurve(I, PerpendicularBisector(LineSegmentOf(A, B)))",
    "query_expressions": "p",
    "answer_expressions": "4/5",
    "fact_spans": "[[[1, 22], [53, 56]], [[1, 22]], [[93, 96]], [[4, 22]], [[25, 28]], [[1, 28]], [[50, 52]], [[0, 52]], [[30, 52]], [[58, 61]], [[62, 65]], [[50, 67]], [[53, 75]], [[83, 91]], [[83, 91]], [[70, 91]]]",
    "query_spans": "[[[93, 99]]]",
    "process": "By the given condition, the focus of the parabola $ y^{2} = 2px $ ($ p > 0 $) is $ F\\left(\\frac{p}{2}, 0\\right) $. Then the line passing through the focus $ F $ with an inclination angle of $ \\frac{\\pi}{4} $ has the equation $ y = x - \\frac{p}{2} $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. From \n\\[\n\\begin{cases}\nx = y + \\frac{p}{2} \\\\\ny^{2} = 2px\n\\end{cases}\n\\]\nwe obtain $ y^{2} - 2py - p^{2} = 0 $. Therefore, $ y_{1} + y_{2} = 2p $, $ x_{1} + x_{2} = 3p $. The midpoint of chord $ AB $ is $ \\left(\\frac{3p}{2}, p\\right) $. The perpendicular bisector of chord $ AB $ has the equation $ y - 2 = -x $. Since the midpoint of chord $ AB $ lies on this line, $ p - 2 = -\\frac{3p}{2} $, solving gives $ p = \\frac{4}{5} $. This problem mainly examines the standard equation of a parabola and the application of its geometric properties, as well as the relationship between a line and a parabola, emphasizing reasoning and computational ability, and is a basic problem."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$, the line $l$: $y=2 x+b$ passes through the focus of the parabola $C$ and intersects $C$ at points $A$ and $B$. If $|A B|=5$, then $p=$?",
    "fact_expressions": "l: Line;C:Parabola;p: Number;A: Point;B: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Expression(l) = (y = b + 2*x);PointOnCurve(Focus(C), l);Intersection(l,C)={A,B};Abs(LineSegmentOf(A, B)) = 5;b:Number",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[29, 45]], [[2, 28], [47, 53], [59, 62]], [[88, 91]], [[65, 68]], [[69, 72]], [[10, 28]], [[2, 28]], [[29, 45]], [[29, 56]], [[29, 74]], [[77, 86]], [[36, 45]]]",
    "query_spans": "[[[88, 93]]]",
    "process": "Method 1: First, using the fact that the line passes through the focus, we obtain b = -p; then, by solving the system of the line and the parabola equation simultaneously, and using the relationship between roots and coefficients to express |AB| = x_{1} + x_{2} + p, solve for p.  \nMethod 2: From the given \\tan\\theta = 2, find the value of \\sin\\theta, then use the chord length formula |AB| = \\frac{2p}{\\sin^{2}\\theta} to find the value of p.  \n\nMethod 1: According to the problem, the line l: y = 2x + b, i.e., y = 2(x + \\frac{b}{2}).  \nSince line l passes through the focus of the parabola C: y^{2} = 2px (p > 0),  \nwe have -\\frac{b}{2} = \\frac{p}{2}, so b = -p.  \nThus, the equation of line l is y = 2x - p.  \nLet A(x_{1}, y_{1}), B(x_{2}, y_{2}), solve the system  \n\\begin{cases} y = 2x - p \\\\ y^{2} = 2px \\end{cases},  \neliminate y and simplify to get 4x^{2} - 6px + p^{2} = 0.  \nBy Vieta's formulas, x_{1} + x_{2} = \\frac{3p}{2}. Since |AB| = 5,  \nwe have x_{1} + x_{2} + p = \\frac{5}{2}p = 5, thus p = 2.  \n\nMethod 2: Let the angle of inclination of the line be \\theta, then \\tan\\theta = k = 2, so \\sin\\theta = \\frac{2\\sqrt{5}}{5},  \ntherefore p = 2."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{2-m}+\\frac{y^{2}}{|m|-3}=1$ represents a hyperbola, then the range of real values for $m$ is?",
    "fact_expressions": "G: Hyperbola;m: Real;Expression(G)=(x^2/(2 - m) + y^2/(Abs(m) - 3) = 1 )",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-3,2)+(3,+\\infty)",
    "fact_spans": "[[[46, 49]], [[51, 56]], [[1, 49]]]",
    "query_spans": "[[[51, 63]]]",
    "process": "Since the equation $\\frac{x^{2}}{2-m}+\\frac{y^{2}}{|m|-3}=1$ represents a hyperbola, we have $(2-m)(|m|-3)<0$, which implies $\\begin{cases}m-2<0\\\\|m|-3<0\\end{cases}$ or $\\begin{cases}m-2>0\\\\|m|-3>0\\end{cases}$. Solving gives $-3<m<2$ or $m>3$. Therefore, the range of real values for $m$ is $(-3,2)\\cup(3,+\\infty)$."
  },
  {
    "text": "Given an ellipse $\\frac{x^{2}}{m}+y^{2}=1$ $(m>1)$ and a hyperbola $\\frac{x^{2}}{n}-y^{2}=1$ $(n>0)$ with the same foci $F_{1}$, $F_{2}$, and point $P$ is one of their intersection points, then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2 + x^2/m = 1);H: Hyperbola;Expression(H) = (-y^2 + x^2/n = 1);m: Number;n: Number;m>1;n>0;F1: Point;F2: Point;Focus(G) = {F1, F2};Focus(H) = {F1, F2};P: Point;OneOf(Intersection(G, H)) = P",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[23, 55]], [[23, 55]], [[56, 89]], [[56, 89]], [[25, 55]], [[59, 89]], [[25, 55]], [[59, 89]], [[7, 14]], [[15, 22]], [[2, 89]], [[2, 89]], [[90, 94]], [[90, 102]]]",
    "query_spans": "[[[104, 133]]]",
    "process": "Let |PF_{1}|=s, |PF_{2}|=t. From the definitions of the ellipse and hyperbola, we obtain  \n\\begin{cases}s^{2}+t^{2}=2m+2n\\\\st=m-n\\end{cases}  \nSince the foci coincide, we have m-n=2. Combining with the cosine law, it can be proved that \\angleF_{1}PF_{2}=90^{\\circ}, thus the area can be found. As shown in the figure, without loss of generality, assume the intersection point P of the two curves lies on the right branch of the hyperbola. Let |PF_{1}|=s, |PF_{2}|=t. From the definitions of the hyperbola and ellipse, we get  \n\\begin{cases}s+t=2\\sqrt{m}\\\\s-t=2\\sqrt{n},\\end{cases}  \nwhich simplifies to  \n\\begin{cases}s^{2}+t^{2}=2m+2n\\\\st=m-n\\end{cases}  \n\\because m-1=n+1, \\therefore m-n=2, \\therefore \\cos\\angleF_{1}PF_{2}=0, \\therefore \\angleF_{1}PF_{2}=90^{\\circ} \\therefore the area of \\triangleF_{1}PF_{2} is \\frac{1}{2}st=1. Original] This problem examines the definition of the ellipse, the definition of the hyperbola, and the cosine law, and belongs to a medium-difficulty problem."
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{2}=1$ are $F_{1}$, $F_{2}$, point $P$ lies on the ellipse. If $|P F_{1}|=4$, then what is $|P F_{2}|$? What is the area $S_{\\Delta P F_{1} F_{2}}$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 + y^2/2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 4",
    "query_expressions": "Abs(LineSegmentOf(P, F2));Area(TriangleOf(P,F1,F2))",
    "answer_expressions": "2\n4*sqrt(3)",
    "fact_spans": "[[[0, 37], [62, 64]], [[57, 61]], [[41, 48]], [[49, 56]], [[0, 37]], [[0, 56]], [[57, 65]], [[67, 80]]]",
    "query_spans": "[[[82, 95]], [[97, 127]]]",
    "process": "The first question uses the definition method: given $ PF_{1} + |PF_{2}| = 6 $ and $ |PF| = 4 $, it is easy to obtain $ PF_{2} $. For the second question, as shown in the figure: the three sides of the triangle containing the angle have been found; using the cosine law yields $ \\angle F_{1}PF_{2} = 120^{\\circ} $, then applying the area formula gives $ \\because |PF_{1}| + |PF_{2}| = 2a = 6, \\therefore |PF_{2}| = 6 - |PF_{1}| = 2. \\therefore \\cos\\angle F_{1}PF_{2} = -\\frac{1}{2}, \\therefore \\angle F_{1}PF_{2} = 120^{\\circ} \\therefore S_{APF_{1}F_{2}} = \\frac{1}{2} \\times 2 \\times 4 \\times \\frac{\\sqrt{3}}{2} = 4\\sqrt{3} $"
  },
  {
    "text": "Given that $P$ is a point on the parabola $y=2 x^{2}$, if the distance from point $P$ to the line $l$: $4 x - y - 6 = 0$ is minimized, then the coordinates of point $P$ are?",
    "fact_expressions": "P: Point;PointOnCurve(P, G);G: Parabola;Expression(G) = (y = 2*x^2);l: Line;Expression(l) = (4*x - y - 6 = 0);WhenMin(Distance(P, l))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1, 2)",
    "fact_spans": "[[[2, 5], [25, 29], [55, 59]], [[2, 23]], [[6, 20]], [[6, 20]], [[30, 48]], [[30, 48]], [[25, 53]]]",
    "query_spans": "[[[55, 64]]]",
    "process": "Let a point on the parabola $ y = 2x^{2} $ be $ A(x_{0}, 2x_{0}^{2}) $. The distance $ d $ from point $ A(x_{0}, 2x_{0}^{2}) $ to the line $ 4x - y - 6 = 0 $ is $ d = \\frac{|4x_{0} - 2x_{0}^{2} - 6|}{\\sqrt{17}} = \\frac{1}{\\sqrt{17}} | -2(x_{0} - 1)^{2} - 4 | $. When $ x_{0} = 1 $, i.e., when $ A(1, 2) $, the distance from a point on the parabola $ y = 2x^{2} $ to the line $ l: 4x - y - 6 = 0 $ is the shortest."
  },
  {
    "text": "If one focus of the ellipse $3 x^{2}-t y^{2}=6$ is $F(0,2)$, then the real number $t$=?",
    "fact_expressions": "G: Ellipse;t: Real;F: Point;Expression(G) = (-t*y^2 + 3*x^2 = 6);Coordinate(F) = (0, 2);OneOf(Focus(G)) = F",
    "query_expressions": "t",
    "answer_expressions": "-1",
    "fact_spans": "[[[1, 22]], [[38, 43]], [[28, 36]], [[1, 22]], [[28, 36]], [[1, 36]]]",
    "query_spans": "[[[38, 45]]]",
    "process": "First, convert the ellipse equation into standard form, then solve based on its focus at F(0,2). The standard form of the ellipse 3x^{2}-ty^{2}=6 is: \\frac{x^{2}}{2}+\\frac{y^{2}}{-\\frac{6}{t}}=1. Since one focus is at F(0,2), we have a^{2}=-\\frac{6}{4}, b^{2}=2, so -\\frac{6}{1}-2=4, solving gives t=-1. The answer is: 1"
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, respectively, $P$ a point on the ellipse, $M$ the midpoint of segment $P F_{1}$, and $|O M|=2$ ($O$ being the coordinate origin). Then $|P F_{1}|=$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;O: Origin;M: Point;F2: Point;Expression(G) = (x^2/25 + y^2/16 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);MidPoint(LineSegmentOf(P,F1)) = M;Abs(LineSegmentOf(O, M)) = 2",
    "query_expressions": "Abs(LineSegmentOf(P, F1))",
    "answer_expressions": "6",
    "fact_spans": "[[[19, 58], [69, 71]], [[65, 68]], [[1, 8]], [[104, 107]], [[75, 78]], [[9, 16]], [[19, 58]], [[1, 64]], [[1, 64]], [[65, 74]], [[75, 93]], [[94, 103]]]",
    "query_spans": "[[[115, 128]]]",
    "process": "The ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, then $|PF_{1}|+|PF_{2}|=10$, $|OM|=2$ $\\therefore |PF_{2}|=2|OM|=4$ $\\therefore |PF_{1}|=6$"
  },
  {
    "text": "Given that the left vertex of ellipse $C$ is $A$, the right focus is $F$, and a circle centered at $A$ with radius $AF$ intersects $C$ at point $M$, where $|AM|=2|MF|$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;G: Circle;A: Point;F: Point;M: Point;LeftVertex(C) = A;RightFocus(C) = F;Center(G) = A;Radius(G) = LineSegmentOf(A,F);Intersection(G, C) = M;Abs(LineSegmentOf(A, M)) = 2*Abs(LineSegmentOf(M, F))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "4/7",
    "fact_spans": "[[[2, 7], [43, 46], [71, 76]], [[41, 42]], [[12, 15], [25, 28]], [[20, 23]], [[49, 53]], [[2, 15]], [[2, 23]], [[24, 42]], [[32, 42]], [[41, 53]], [[55, 69]]]",
    "query_spans": "[[[71, 82]]]",
    "process": "Let the left focus of ellipse C be E. According to the given conditions, |AM| = |AF| = a + c, |MF| = \\frac{1}{2}|AM| = \\frac{a+c}{2}. Let circle A intersect the x-axis at another point N (not coinciding with point F), then MF \\bot MN, so \\cos\\angle MFN = \\frac{|MF|}{|NF|} = \\frac{1}{4}. By the definition of the ellipse, |ME| = 2a - |MF| = \\frac{3a - c}{2}. By the law of cosines, |ME|^{2} = |MF|^{2} + |EF|^{2} - 2|MF| \\cdot |EF| \\cos\\angle MFE. That is, (\\frac{3a - c}{2})^{2} = (\\frac{a + c}{2})^{2} + 4c^{2} - 2 \\times \\frac{a + c}{2} \\times 2c \\times \\frac{1}{4}, which simplifies to 7c^{2} + 3ac - 4a^{2} = 0, or 7e^{2} + 3e - 4 = 0. Since 0 < e < 1, solving gives e = \\frac{4}{7}. Hence, the answer is: \\frac{4}{7}"
  },
  {
    "text": "It is known that the hyperbola $C$ is centered at the origin, $F(-2,0)$ is a focus, and a line $l$ passing through $F$ intersects the hyperbola $C$ at points $A$ and $B$. The midpoint of $AB$ is $N(-3,-1)$. Then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;O: Origin;Center(C) = O;F: Point;Coordinate(F) = (-2, 0);OneOf(Focus(C)) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, C) = {A, B};N: Point;Coordinate(N) = (-3, -1);MidPoint(LineSegmentOf(A, B)) = N",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/3-y^2=1",
    "fact_spans": "[[[2, 8], [41, 47], [81, 84]], [[12, 14]], [[2, 14]], [[15, 24], [31, 34]], [[15, 24]], [[2, 29]], [[35, 40]], [[30, 40]], [[49, 52]], [[53, 56]], [[35, 58]], [[69, 79]], [[69, 79]], [[60, 79]]]",
    "query_spans": "[[[81, 89]]]",
    "process": "From the coordinates of F and N, we get $k_{l}=1$. Let the hyperbola equation be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$, then $a^{2}+b^{2}=4$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $x_{1}+x_{2}=-6$, $y_{1}+y_{2}=2$, $\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=k_{l}=1$. From $\\frac{x_{1}^{2}}{a^{2}}-\\frac{y_{1}^{2}}{b^{2}}=1$, $\\frac{x_{2}^{2}}{a^{2}}-\\frac{y_{2}^{2}}{b^{2}}=1$, $\\frac{y_{2}^{2}}{x_{1}+x_{2}}=k_{1}=1$, i.e., $\\frac{-6}{a^{2}}+\\frac{2k_{l}}{b^{2}}=0$, $a^{2}=3b$. Thus $a^{2}=3$, $b^{2}=1$. So the equation of $C$ is $\\frac{x^{2}}{3}-y^{2}=1$."
  },
  {
    "text": "Let the foci of the ellipse be $F_{1}$ and $F_{2}$. A circle with diameter $F_{1} F_{2}$ intersects the ellipse at a point $P$. If $|F_{1} F_{2}| = 2|P F_{2}|$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;H: Circle;F1: Point;F2: Point;P: Point;Focus(G)={F1,F2};IsDiameter(LineSegmentOf(F1,F2),H);OneOf(Intersection(H,G))=P;Abs(LineSegmentOf(F1, F2)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[1, 3], [48, 50], [89, 91]], [[46, 47]], [[7, 14]], [[17, 25]], [[56, 59]], [[1, 25]], [[27, 47]], [[46, 59]], [[61, 87]]]",
    "query_spans": "[[[89, 97]]]",
    "process": ""
  },
  {
    "text": "It is known that line $l$ passes through the focus of the parabola $C$: $y^{2}=2 p x$ ($p>0$) and is perpendicular to the axis of symmetry of $C$. Line $l$ intersects $C$ at points $A$ and $B$, and point $P$ lies on the directrix of $C$ such that $S_{\\triangle A B P}=36$. Then, what is the minimum length of a chord of the parabola $C$ passing through its focus?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;l: Line;PointOnCurve(Focus(C), l);IsPerpendicular(l, SymmetryAxis(C));A: Point;B: Point;Intersection(l, C) = {A, B};P: Point;PointOnCurve(P, Directrix(C));Area(TriangleOf(A, B, P)) = 36;H: LineSegment;PointOnCurve(Focus(C), H);IsChordOf(H, C)",
    "query_expressions": "Min(Length(H))",
    "answer_expressions": "12",
    "fact_spans": "[[[8, 34], [39, 42], [53, 56], [72, 75], [110, 116]], [[8, 34]], [[16, 34]], [[16, 34]], [[2, 7], [49, 52]], [[2, 37]], [[2, 48]], [[58, 61]], [[62, 65]], [[49, 67]], [[68, 71]], [[68, 81]], [[83, 107]], [], [[109, 121]], [[109, 121]]]",
    "query_spans": "[[[109, 128]]]",
    "process": "From the given conditions, AB is the latus rectum of the parabola, ∴ |AB| = 2p. Also, S_{\\triangle ABP} = \\frac{1}{2}|AB| \\cdot p = p^{2} = 36, ∴ p = 6. Since |AB| = 12, and the shortest chord passing through the focus of parabola C is the latus rectum, i.e., |AB|, ∴ the minimum length of a chord passing through the focus of parabola C is: 12"
  },
  {
    "text": "If the line $l$: $y=x+b$ is tangent to the parabola $C$: $x^{2}=4y$ at point $A$, then what is the standard equation of the circle centered at point $A$ and tangent to the directrix of the parabola $C$?",
    "fact_expressions": "l: Line;C: Parabola;G: Circle;A: Point;Expression(C) = (x^2 = 4*y);Expression(l)=(y = b + x) ;TangentPoint(l,C)=A;Center(G)=A;b:Number;IsTangent(Directrix(C),G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "(x-2)^2+(y-1)^2=4",
    "fact_spans": "[[[1, 15]], [[16, 35], [54, 60]], [[66, 67]], [[38, 42], [45, 49]], [[16, 35]], [[1, 15]], [[1, 42]], [[44, 67]], [[7, 15]], [53, 66]]",
    "query_spans": "[[[66, 74]]]",
    "process": "y=\\frac{1}{4}x^{2},y=\\frac{1}{2}x=1,x=2, hence the tangent point is (2,1), the distance to the directrix y=-1 is 2, thus the radius is 2, the equation of the circle is (x-2)^{2}+(y-1)^{2}=4."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$ and directrix $l$. A line passing through point $F$ with an inclination angle of $120^{\\circ}$ intersects the directrix $l$ at point $A$, and the segment $A F$ intersects the parabola $C$ at point $B$, with $|A B|=\\frac{4}{3}$. Then the standard equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;l: Line;Directrix(C) = l;G: Line;Inclination(G) = ApplyUnit(120, degree);PointOnCurve(F, G);A: Point;Intersection(G, l) = A;B: Point;Intersection(LineSegmentOf(A, F), C) = B;Abs(LineSegmentOf(A, B)) = 4/3",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=2*x",
    "fact_spans": "[[[2, 28], [92, 98], [128, 134]], [[2, 28]], [[10, 28]], [[10, 28]], [[32, 35], [45, 49]], [[2, 35]], [[39, 42], [73, 76]], [[2, 42]], [[68, 70]], [[50, 70]], [[44, 70]], [[79, 83]], [[68, 83]], [[101, 105]], [[84, 105]], [[107, 126]]]",
    "query_spans": "[[[128, 141]]]",
    "process": "The equation of line AF is given by $ y = -\\sqrt{3}\\left(x - \\frac{p}{2}\\right) $, so $ A\\left(-\\frac{p}{2}, \\sqrt{3}p\\right) $. From the system \n$$\n\\begin{cases}\ny^2 = 2px \\\\\ny = -\\sqrt{3}\\left(x - \\frac{p}{2}\\right)\n\\end{cases}\n$$\neliminating $ x $, we obtain $ \\sqrt{3}y^{2} + 2py - \\sqrt{3}p^{2} = 0 $, solving which gives $ y = \\frac{\\sqrt{3}}{3} $, and from $ \\frac{n}{m}|AB| = \\frac{4}{3} $, we get $ \\sqrt{\\left(\\frac{-\\sqrt{3}p}{p + \\frac{1}{p}}\\right)^{2} + \\left(\\frac{\\sqrt{3}}{3}p - \\sqrt{3}p\\right)} \\frac{p}{\\frac{4}{3}} $, solving yields $ p = 1 $, thus the standard equation of parabola $ C $ is $ y^{2} = 2x $."
  },
  {
    "text": "Given that $M$ is a point on the parabola $y^{2}=4x$, $F$ is its focus, and point $A$ lies on the circle $C$: $(x-6)^{2}+(y+1)^{2}=1$, then the minimum value of $|MA|+|MF|$ is?",
    "fact_expressions": "G: Parabola;C: Circle;M: Point;A: Point;F: Point;Expression(G) = (y^2 = 4*x);Expression(C) = ((x - 6)^2 + (y + 1)^2 = 1);PointOnCurve(M, G);PointOnCurve(A, C);Focus(G)=F",
    "query_expressions": "Min(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F)))",
    "answer_expressions": "6",
    "fact_spans": "[[[6, 20], [28, 29]], [[37, 65]], [[2, 5]], [[32, 36]], [[24, 27]], [[6, 20]], [[37, 65]], [[2, 23]], [[32, 66]], [[24, 31]]]",
    "query_spans": "[[[68, 87]]]",
    "process": "According to the parabolic equation, the directrix equation is obtained. Draw MN perpendicular from point M to the directrix at N. By the definition of a parabola, |MN| = |MF|. The problem is transformed into finding the minimum value of |MA| + |MN|. Since A lies on circle C, it follows that |MA| + |MN| reaches its minimum when points M, N, and C are collinear. Further calculation yields: M is a point on the parabola y^{2} = 4x, and the directrix of the parabola is x = -1. Draw MN perpendicular from point M to the directrix at N, then |MN| = |MF|, so |MA| + |MF| = |MA| + |MN|. Since point A lies on circle C: (x-6)^{2} + (y+1)^{2} = 1 with center C(6, -1) and radius 1, |MA| + |MN| reaches its minimum value of 6 when M, N, and C are collinear."
  },
  {
    "text": "Given that $M$ is a point on $y=\\frac{1}{4} x^{2}$, $F$ is the focus of the parabola, and $A$ lies on the circle $C$: $(x-1)^{2}+(y-4)^{2}=1$, find the minimum value of $|M A|+|M F|$.",
    "fact_expressions": "G: Parabola;M: Point;A: Point;F: Point;C: Curve;Expression(G)=(y = x^2/4);PointOnCurve(M, G);Focus(G) = F;Expression(C)=((x - 1)^2 + (y - 4)^2 = 1);PointOnCurve(A, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F)))",
    "answer_expressions": "4",
    "fact_spans": "[[[6, 27], [35, 38]], [[2, 5]], [[41, 44]], [[31, 34]], [[45, 72]], [[6, 27]], [[2, 30]], [[31, 40]], [[45, 72]], [[41, 73]]]",
    "query_spans": "[[[75, 93]]]",
    "process": "\\because the parabola y=\\frac{1}{4}x^2 can be rewritten in standard form as x^{2}=4y, \\because the directrix of the parabola is l: y=-1, draw MN\\bot l from point M intersecting at N, |MN|=|MF|, \\therefore |MA|+|MF|=|MA|+|MN|, \\because A moves on the circle C: (x-1)^{2}+(y-4)^{2}=1 with center C(1,4) and radius r=1, \\therefore |MA|+|MF| is minimized when points N, M, C are collinear, as shown in the figure. Draw CN_{0}\\bot l from C intersecting circle C, the x-axis, and line l at points A_{0}, M_{0}, N_{0} respectively, \\therefore |MA|+|MF|_{\\min}=(|MA|+|MN|)_{\\min}=|CN_{0}|-r=5-1=4, thus the minimum value of |MA|+|MF| is 4"
  },
  {
    "text": "Given that the curve $\\frac{x^{2}}{a}-\\frac{y^{2}}{b}=1$ intersects the line $x+y-1=0$ at points $P$ and $Q$, and $\\overrightarrow{O P} \\cdot \\overrightarrow{O Q}=0$ ($O$ is the origin), then $\\frac{1}{a}-\\frac{1}{b}$=?",
    "fact_expressions": "G: Line;H: Curve;b: Number;a: Number;O: Origin;P: Point;Q: Point;Expression(G) = (x + y - 1 = 0);Expression(H) = (-y^2/b + x^2/a = 1);Intersection(H, G) = {P, Q};DotProduct(VectorOf(O, P), VectorOf(O, Q)) = 0",
    "query_expressions": "-1/b + 1/a",
    "answer_expressions": "2",
    "fact_spans": "[[[40, 51]], [[2, 39]], [[4, 39]], [[4, 39]], [[117, 120]], [[54, 57]], [[58, 61]], [[40, 51]], [[2, 39]], [[2, 63]], [[65, 116]]]",
    "query_spans": "[[[126, 153]]]",
    "process": "Substitute $ y = 1 - x $ into $ \\frac{x^2}{a} - \\frac{y^2}{b} = 1 $, we get $ (b - a)x^2 + 2ax - (a + ab) = 0 $. Let $ P(x_1, y_1) $, $ Q(x_2, y_2) $, then $ x_1 + x_2 = \\frac{2a}{a - b} $, $ x_1 x_2 = \\frac{a + ab}{a - b} $. Since $ x_1 x_2 + (1 - x_1)(1 - x_2) = 2x_1 x_2 - (x_1 + x_2) + 1 $, so $ \\frac{2a + 2ab}{a - b} - \\frac{2a}{a - b} + 1 = 0 $, that is $ 2a + 2ab - 2a + a - b = 0 $, i.e., $ b - a = 2ab $, therefore $ \\frac{1}{a} - \\frac{1}{b} = 2 $."
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ has two foci $F_{1}$ and $F_{2}$. Point $P$ lies on the hyperbola. If $P F_{1} \\perp P F_{2}$, then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Focus(G) = {F1,F2};PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "16",
    "fact_spans": "[[[0, 39], [66, 69]], [[61, 65]], [[45, 52]], [[53, 60]], [[0, 39]], [[0, 60]], [[61, 70]], [[72, 95]]]",
    "query_spans": "[[[97, 124]]]",
    "process": "According to the problem, $F_{1}(-5,0)$, $F_{2}(5,0)$, since $PF_{1}\\perp PF_{2}$, we have $|PF_{1}|^{2}+|PF_{2}|^{2}=4c^{2}=100$. By the definition of hyperbola, $||PF_{1}|-|PF_{2}||=2a=6$, so $|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}|\\cdot|PF_{2}|=36$. Thus, $|PF_{1}|\\cdot|PF_{2}|=32$. Therefore, $S_{\\triangle PF_{1}F_{2}}=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|=16$."
  },
  {
    "text": "If the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1$ $(a>0)$ is $\\frac{\\sqrt{5}}{2}$, then $a$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/4 + x^2/a^2 = 1);a: Number;a>0;Eccentricity(G) = sqrt(5)/2",
    "query_expressions": "a",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 48]], [[1, 48]], [[75, 78]], [[4, 48]], [[1, 73]]]",
    "query_spans": "[[[75, 80]]]",
    "process": "Analysis: According to the eccentricity formula $ e = \\frac{c}{a} $, and the relationship among $ a $, $ b $, $ c $ in a hyperbola, we can set up a system of equations to solve for parameter $ a $. In the hyperbola, $ c = \\sqrt{a^{2} + b^{2}} = \\sqrt{a^{2} + 4} $, and $ e = \\frac{c}{a} = \\frac{\\sqrt{5}}{2} $, therefore $ \\frac{\\sqrt{a^{2} + 4}}{a} = \\frac{\\sqrt{5}}{2} $, $ \\frac{a^{2} + 4}{a^{2}} = \\frac{5}{4} $."
  },
  {
    "text": "Given the ellipse $C$: $x^{2}+\\frac{y^{2}}{4}=1$, and a moving point $P(0, m)$ on the $y$-axis. If there exists a circle $P$ centered at point $P$ that intersects the ellipse $C$ at four distinct common points, then the range of values for $m$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2 + y^2/4 = 1);P: Point;m: Number;Coordinate(P) = (0, m);PointOnCurve(P, yAxis);P1: Circle;Center(P1) = P;NumIntersection(P1, C) = 4",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-3/2, 3/2)",
    "fact_spans": "[[[2, 34], [71, 76]], [[2, 34]], [[36, 45], [58, 62]], [[87, 90]], [[36, 45]], [[36, 54]], [[66, 70]], [[57, 70]], [[66, 85]]]",
    "query_spans": "[[[87, 97]]]",
    "process": "Let the equation of circle P be $ x^{2} + (y - m)^{2} = r^{2} $. By combining it with the ellipse equation $ x^{2} + \\frac{y^{2}}{4} = 1 $, we obtain a quadratic equation in $ y $. Using the condition that the discriminant is 0, and that circle P passes through the point $ (0, -2) $, we find that when the circle and the ellipse have three intersection points, $ m = -\\frac{3}{2} $. Similarly, when the circle passes through $ (0, 2) $ and has three intersection points with the ellipse, $ m = \\frac{3}{2} $. Combining numerical and geometric analysis, the range of values for $ m $ can be determined. Solution: According to the problem, let the equation of circle P be $ x^{2} + (y - m)^{2} = r^{2} $. Combining it with the ellipse $ C: x^{2} + \\frac{y^{2}}{4} = 1 $, we get: $ 3y^{2} - 8my - 4 + 4m^{2} - 4r^{2} = 0 $. From $ \\Delta = 64m^{2} - 12(4 + 4m^{2} - 4r^{2}) = 16m^{2} + 48r^{2} - 48 = 0 $, we obtain: $ m^{2} + 3r^{2} - 3 = 0 $ ①. Since circle P passes through the points $ (0, -2) $ and $ (0, 2) $, we have: $ (-2 - m)^{2} = r^{2} $ ②. From ① and ②, we get $ m = -\\frac{3}{2} $. Similarly, when circle P passes through $ (0, 2) $, we get $ m = \\frac{3}{2} $. As shown in the figure, by analyzing the graph, we see that when $ m \\in \\left(-\\frac{3}{2}, \\frac{3}{2}\\right) $, circles centered at P intersect the ellipse C at four distinct points."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the circle $O$: $x^{2}+y^{2}=b^{2}$, if there exists a point $P$ on $C$ such that two tangents drawn from $P$ to circle $O$ touch the circle at points $A$ and $B$ respectively, satisfying $\\angle APB=60^{\\circ}$, then the range of eccentricity of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;O: Circle;A: Point;P: Point;B: Point;L1: Line;L2: Line;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(O) = (x^2 + y^2 = b^2);PointOnCurve(P, C);TangentOfPoint(P, O) = {L1,L2};TangentPoint(L1,O)=A;TangentPoint(L2,O)=B;AngleOf(A, P, B) = ApplyUnit(60, degree)",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 60], [88, 91], [159, 164]], [[9, 60]], [[9, 60]], [[61, 86], [107, 111]], [[122, 125]], [[94, 98], [102, 106]], [[126, 129]], [], [], [[9, 60]], [[9, 60]], [[2, 60]], [[61, 86]], [[88, 98]], [[101, 116]], [[101, 129]], [[101, 129]], [[132, 157]]]",
    "query_spans": "[[[159, 175]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$, let the left focus of $C$ be $F$. A line $l$, perpendicular to one asymptote of $C$, is drawn through $F$, with foot of perpendicular at $A$. Let $B$ be the intersection point of $l$ and the other asymptote of $C$. If $A$ is the midpoint of segment $F B$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;LeftFocus(C) = F;L1: Line;L2: Line;OneOf(Asymptote(C))=L1;OneOf(Asymptote(C))=L2;Negation(L1=L2);l: Line;PointOnCurve(F, l);IsPerpendicular(l, L1);A: Point;FootPoint(l, L1)=A;B: Point;Intersection(l, L2)=B;MidPoint(LineSegmentOf(F, B))=A",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 64], [78, 81], [106, 109], [141, 147]], [[2, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[69, 72], [74, 77]], [[2, 72]], [], [], [[78, 87]], [[106, 116]], [[78, 116]], [[90, 93], [102, 105]], [[73, 93]], [[78, 93]], [[97, 100], [125, 128]], [[78, 100]], [[120, 123]], [[102, 123]], [[125, 139]]]",
    "query_spans": "[[[141, 153]]]",
    "process": "The hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) has its left focus at $ F(-c, 0) $. Draw a line $ l $ perpendicular to an asymptote of $ C $ through $ F $, with foot of perpendicular at $ A $. Thus, the equation of $ AF $ is $ y = \\frac{a}{b}(x + c) $. Solving simultaneously with $ bx + ay = 0 $ gives $ A\\left(-\\frac{a^{2}}{c}, \\frac{ab}{c}\\right) $. The intersection point of $ l $ with the other asymptote of $ C $ is $ B $. If $ A $ is the midpoint of segment $ FB $, then $ B\\left(\\frac{c^{2} - 2a^{2}}{c}, \\frac{2ab}{c}\\right) $. Substituting into $ bx - ay = 0 $ yields $ c^{2} = 4a^{2} $. Therefore, the eccentricity of hyperbola $ C $ is $ e = 2 $."
  },
  {
    "text": "Given that the standard equation of the parabola is $y^{2}=8 x$, what is the equation of the directrix of this parabola?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=-2",
    "fact_spans": "[[[2, 5], [26, 29]], [[2, 22]]]",
    "query_spans": "[[[26, 36]]]",
    "process": "Since the standard equation of the parabola is y^{2}=8x, the focus of the parabola lies on the positive x-axis, and p=4, so the directrix equation of the parabola is: x=-\\frac{p}{2}=-2."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with left and right foci $F_{1}$, $F_{2}$. Draw a line $l$ perpendicular to the line $y=-\\frac{b}{a}x$ passing through $F_{2}$, with foot of perpendicular at $Q$. The line $l$ intersects the left branch of the hyperbola at point $P$. If $|F_{2} P|=2|F_{2} Q|$, then the eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;F2: Point;P: Point;Q: Point;F1: Point;e: Number;l:Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = -x*b/a);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, l);FootPoint(l, H)=Q;Intersection(l,LeftPart(G)) = P;Abs(LineSegmentOf(F2, P)) = 2*Abs(LineSegmentOf(F2, Q));Eccentricity(G) = e;IsPerpendicular(H, l)",
    "query_expressions": "e",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 58], [130, 133], [167, 170]], [[5, 58]], [[5, 58]], [[90, 110]], [[72, 79], [82, 89]], [[137, 141]], [[120, 123]], [[64, 71]], [[174, 177]], [[113, 116], [126, 129]], [[5, 58]], [[5, 58]], [[2, 58]], [[90, 110]], [[2, 79]], [[2, 79]], [[81, 116]], [[90, 123]], [[126, 141]], [[143, 165]], [[167, 177]], [[90, 116]]]",
    "query_spans": "[[[174, 179]]]",
    "process": "Let $ F_{2}(c,0) $, then the equation of line $ PQ $ is $ y = \\frac{a}{b}(x - c) $. Substituting into the asymptote equation of the hyperbola $ y = -\\frac{b}{a}x $, we get $ Q\\left( \\frac{a^{2}}{c}, -\\frac{ab}{c} \\right) $. From $ |F_{2}P| = 2|F_{2}Q| $, i.e., $ \\overrightarrow{F_{2}P} = \\overrightarrow{F_{2}Q} $, we obtain $ P\\left( -\\frac{c^{2} + 2a^{2}}{3c}, -\\frac{2ab}{3c} \\right) $. Substituting the coordinates of point $ P $ into the hyperbola equation $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $, we have $ \\frac{(c^{2} + 2a^{2})^{2}}{9c^{2}a^{2}} - \\frac{4a^{2}}{9c^{2}} = 1 $. Simplifying yields $ c = \\sqrt{5}a $, so the eccentricity is $ e = \\frac{c}{a} = \\sqrt{5} $."
  },
  {
    "text": "The focus of a parabola lies on the line $x+4y-1=0$, then the standard equation of the parabola is?",
    "fact_expressions": "G: Parabola;H: Line;Expression(H) = (x + 4*y - 1 = 0);PointOnCurve(Focus(G), H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=4*x,x^2=y}",
    "fact_spans": "[[[0, 3], [23, 26]], [[7, 20]], [[7, 20]], [[0, 21]]]",
    "query_spans": "[[[23, 33]]]",
    "process": "The intersections of x+4y-1=0 with the coordinate axes are (1,0) and (0,\\frac{1}{4}). When the focus of the parabola is (1,0), \\frac{P}{2}=1, P=2. Hence the parabola is y^{2}=4x. Hence the parabola is x^{2}=y"
  },
  {
    "text": "Let the foci of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{b^{2}}=1(b>0)$ be $F_{1}$ and $F_{2}$, and let $P$ be a point on this hyperbola. If $|P F_{1}|=5$, then $|P F_{2}|$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/b^2 = 1);b: Number;b>0;F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P,G) = True;Abs(LineSegmentOf(P, F1)) = 5",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "11",
    "fact_spans": "[[[1, 48], [73, 76]], [[1, 48]], [[4, 48]], [[4, 48]], [[52, 59]], [[60, 67]], [[1, 67]], [[68, 71]], [[68, 80]], [[82, 95]]]",
    "query_spans": "[[[97, 110]]]",
    "process": "From the hyperbola equation $\\frac{x^2}{9}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, we obtain $a=3$. According to the definition of a hyperbola, $|PF_{1}|-|PF_{2}|=\\pm2a=\\pm6$. Given that $|PF_{1}|=5$, it follows that $|PF_{2}|=11$."
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $y=\\sqrt{3}x$, and one of its foci lies on the directrix of the parabola $y^{2}=8x$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x);H: Parabola;Expression(H) = (y^2 = 8*x);PointOnCurve(OneOf(Focus(G)), Directrix(H))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[2, 58], [83, 84], [110, 113]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 81]], [[90, 104]], [[90, 104]], [[83, 108]]]",
    "query_spans": "[[[110, 118]]]",
    "process": ""
  },
  {
    "text": "Let the upper and lower vertices of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ be $A$ and $B$, respectively, the right focus be $F$, the other intersection point of line $AF$ with the ellipse be $P$, and connect $BP$. When the slope of line $BP$ reaches its maximum value, what is the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;F: Point;P: Point;B: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);UpperVertex(G) = A;LowerVertex(G) = B;RightFocus(G) = F;Intersection(LineOf(A,F), G) = {A,P};WhenMax(Slope(LineOf(B,P)))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[1, 53], [86, 88], [123, 125]], [[3, 53]], [[3, 53]], [[62, 65]], [[74, 77]], [[94, 97]], [[66, 69]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 69]], [[1, 69]], [[1, 77]], [[62, 97]], [[106, 122]]]",
    "query_spans": "[[[123, 131]]]",
    "process": "From the given conditions, we have: A(0,b), B(0,-b), F(c,0). Therefore, the equation of line AF is \\frac{x}{c}+\\frac{y}{b}=1. From \\begin{cases}\\frac{x}{c}+\\frac{y}{b}=1\\\\\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1\\end{cases}, eliminating y yields (c^{2}+a^{2})x^{2}-2a^{2}cx=0, so x_{P}=\\frac{2a2c}{c^{2}+a^{2}} and y_{P}=b(1-\\frac{x_{P}}{c})=\\frac{(c^{2}-a^{2})b}{c^{2}+a^{2}}, i.e., P(\\frac{2a2c}{c^{2}+a^{2}},\\frac{(c^{2}-a^{2})b}{c^{2}+a^{2}}). Thus, kBP=\\frac{\\frac{(c^{2}-a2)b}{\\frac{c^{2}+a^{2}}{2a^{2}c}}}=\\frac{bc}{a^{2}}=\\frac{bc}{b^{2}+c^{2}}\\leqslant\\frac{bc}{2bc}=\\frac{1}{2}, with equality if and only if b=c. At this time, the eccentricity of the ellipse is \\frac{c}{a}=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{\\frac{c^{2}}{b^{2}+c}}=\\frac{\\sqrt{2}}{2}."
  },
  {
    "text": "The equation of a parabola with vertex at the origin, focus on the coordinate axis, and passing through the point $M(-2,-4)$ is?",
    "fact_expressions": "G: Parabola;M: Point;O:Origin;Vertex(G)=O;Coordinate(M)=(-2,-4);PointOnCurve(Focus(G),axis);PointOnCurve(M, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=-8*x,x^2=-y}",
    "fact_spans": "[[[32, 35]], [[20, 31]], [[1, 5]], [[0, 35]], [[20, 31]], [[9, 35]], [[18, 35]]]",
    "query_spans": "[[[32, 39]]]",
    "process": "If the focus of the parabola lies on the x-axis, the standard equation of the parabola can be written as $ y^{2} = mx $. Substituting the coordinates of point M into this equation gives $ -2m = (-4)^{2} $, solving which yields $ m = -8 $. At this time, the standard equation of the parabola is $ y^{2} = -8x $. If the focus of the parabola lies on the y-axis, the standard equation of the parabola can be written as $ x^{2} = ny $. Substituting the coordinates of point M into this equation gives $ -4n = (-2)^{2} $, solving which yields $ n = -1 $. At this time, the standard equation of the parabola is $ x^{2} = -y $. In conclusion, the standard equation of the parabola is $ y^{2} = -8x $ or $ x^{2} = -y $."
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F_{1}$, $F_{2}$ are the two foci of the ellipse, and the sum of the distances from $F_{1}$, $F_{2}$ to the line $\\frac{x}{a}+\\frac{y}{b}=1$ is $\\sqrt{3} b$. Then the eccentricity $e=$?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y/b + x/a = 1);Focus(G)={F1,F2};Distance(F1,H)+Distance(F2,H)=sqrt(3)*b;Eccentricity(G)=e;e:Number;F1: Point;F2: Point",
    "query_expressions": "e",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[0, 52], [71, 73]], [[2, 52]], [[2, 52]], [[95, 124]], [[2, 52]], [[2, 52]], [[0, 52]], [[95, 124]], [[54, 78]], [[79, 142]], [[71, 150]], [[147, 150]], [[79, 86], [54, 61]], [[87, 94], [63, 70]]]",
    "query_spans": "[[[147, 152]]]",
    "process": "F_{1},F_{2} are the two foci of the ellipse with coordinates (c,0),(-c,0). Let the distances from F_{1},F_{2} to the line \\frac{x}{a}+\\frac{y}{b}=1 be d_{1},d_{2} respectively. Then d_{1}+d_{2}=\\frac{|\\frac{c}{a}-1|}{\\sqrt{\\frac{1}{a^{2}}+\\frac{1}{b^{2}}}}+\\frac{|-\\frac{c}{a}-1|}{\\sqrt{\\frac{1}{a^{2}}+\\frac{1}{b^{2}}}}=\\frac{2}{\\sqrt{\\frac{1}{a^{2}}+\\frac{1}{b^{2}}}}=\\sqrt{3}b. Squaring both sides and simplifying yields a^{2}=3b^{2}. Since b^{2}=a^{2}-c^{2}, it follows that 2a^{2}=3c^{2}, so e^{2}=\\frac{2}{3}, hence e=\\frac{\\sqrt{6}}{3}."
  },
  {
    "text": "What is the length of the chord passing through the focus of the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{4}=1$ and perpendicular to the $x$-axis?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2/4 = 1);L: LineSegment;IsChordOf(L,G) = True;PointOnCurve(Focus(G),L) = True;IsPerpendicular(L,xAxis) = True",
    "query_expressions": "Length(L)",
    "answer_expressions": "8*sqrt(3)/3",
    "fact_spans": "[[[1, 39]], [[1, 39]], [], [[1, 52]], [[0, 52]], [[0, 52]]]",
    "query_spans": "[[[0, 57]]]",
    "process": "Slight"
  },
  {
    "text": "Given that $A$ is a moving point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, and $MN$ is a diameter of the circle $(x-1)^{2}+y^{2}=1$, then the maximum value of $\\overrightarrow{A M} \\cdot \\overrightarrow{A N}$ is?",
    "fact_expressions": "A: Point;G: Ellipse;Expression(G) = (x^2/9 + y^2/5 = 1);PointOnCurve(A, G);M: Point;N: Point;H: Circle;Expression(H) = (y^2 + (x - 1)^2 = 1);IsDiameter(LineSegmentOf(M, N), H) = True",
    "query_expressions": "Max(DotProduct(VectorOf(A, M), VectorOf(A, N)))",
    "answer_expressions": "15",
    "fact_spans": "[[[2, 5]], [[6, 43]], [[6, 43]], [[2, 47]], [[48, 53]], [[48, 53]], [[54, 74]], [[54, 74]], [[48, 79]]]",
    "query_spans": "[[[81, 136]]]",
    "process": "Because the circle $(x-1)^{2}+y^{2}=1$ has center $C(1,0)$ and radius 1, let $A(x,y)$, $-3\\leqslant x \\leqslant 3$. Since $\\overrightarrow{AM}=\\overrightarrow{AC}+\\overrightarrow{CM}$, $\\overrightarrow{AN}=\\overrightarrow{AC}+\\overrightarrow{CN}=\\overrightarrow{AC}-\\overrightarrow{CM}$, so $\\overrightarrow{AM}\\cdot\\overrightarrow{AN}=(\\overrightarrow{AC}+\\overrightarrow{CM})\\cdot(\\overrightarrow{AC}-\\overrightarrow{CM})=\\overrightarrow{AC}^{2}-\\overrightarrow{CM}^{2}=\\overrightarrow{AC}^{2}-1=(x-1)^{2}+y^{2}-1$. Since $A(x,y)$ lies on $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, we have $y^{2}=5-\\frac{5}{9}x^{2}$. Therefore, $\\overrightarrow{AM}\\cdot\\overrightarrow{AN}=(x-1)^{2}+5-\\frac{5}{9}x^{2}-1=\\frac{4}{9}x^{2}-2x+5$, $-3\\leqslant x \\leqslant 3$. For the function $y=\\frac{4}{9}x^{2}-2x+5$, the axis of symmetry is $x=\\frac{9}{4}$, and when $x=-3$, $\\overrightarrow{AM}\\cdot\\overrightarrow{AN}$ attains its maximum value of 15."
  },
  {
    "text": "Given the parabola $C$: $x^{2}=2 p y(p>0)$, a line $l$ with inclination angle $\\frac{\\pi}{4}$ passes through the focus $F$ of the parabola and intersects the parabola at points $A$ and $B$, $O$ is the origin, $|A B|=8$, then the area of $\\triangle A O B$ is?",
    "fact_expressions": "l: Line;C: Parabola;p: Number;A: Point;O: Origin;B: Point;p>0;Expression(C) = (x^2 = 2*p*y);Inclination(l)=pi/4;PointOnCurve(F,l);Intersection(l,C)={A,B};Abs(LineSegmentOf(A, B)) = 8;F:Point;Focus(C)=F",
    "query_expressions": "Area(TriangleOf(A, O, B))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[49, 54]], [[2, 28], [55, 58], [67, 70]], [[10, 28]], [[73, 76]], [[83, 86]], [[77, 80]], [[10, 28]], [[2, 28]], [[29, 54]], [[49, 64]], [[49, 82]], [[92, 101]], [[61, 64]], [[55, 64]]]",
    "query_spans": "[[[103, 125]]]",
    "process": "According to the problem, $ F(0,\\frac{p}{2}) $. Let the equation of line $ l $ be $ y = x + \\frac{p}{2} $. Substituting into the parabola gives $ x^{2} - 2px - p^{2} = 0 $. Thus, $ |AB| = y_{1} + y_{2} + p = x_{1} + x_{2} + 2p = 4p = 8 $. Solving yields $ p = 2 $, so the equation of line $ l $ is $ y = x + 1 $. Also, the distance from the origin $ O $ to line $ l $ is $ d = \\frac{1}{\\sqrt{2}} = \\frac{\\sqrt{2}}{2} $. Therefore, $ S_{\\triangle AOB} = \\frac{1}{2}|AB| \\cdot d = \\frac{1}{2} \\times 8 \\times \\frac{\\sqrt{2}}{2} = 2\\sqrt{2} $."
  },
  {
    "text": "Let $P$ be a moving point on the curve $y^{2}=4x$. Then, the minimum value of the sum of the distance from point $P$ to point $A(-1, 2)$ and the distance from point $P$ to the line $x=-1$ is?",
    "fact_expressions": "G: Curve;A: Point;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (-1, 2);PointOnCurve(P, G);l:Line;Expression(l)=(x=-1)",
    "query_expressions": "Min(Distance(P,A)+Distance(P,l))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[5, 18]], [[31, 43]], [[26, 30], [1, 4], [47, 51]], [[5, 18]], [[31, 43]], [[1, 24]], [[52, 58]], [[52, 58]]]",
    "query_spans": "[[[26, 69]]]",
    "process": ""
  },
  {
    "text": "A moving circle passes through the point $(1,0)$ and is tangent to the line $x=-1$. Then, the equation of the locus of the center of the moving circle is?",
    "fact_expressions": "G: Circle;H: Line;I: Point;Expression(H) = (x = -1);Coordinate(I) = (1, 0);PointOnCurve(I, G);IsTangent(G,H)",
    "query_expressions": "LocusEquation(Center(G))",
    "answer_expressions": "y^2 = 4*x",
    "fact_spans": "[[[0, 2], [26, 28]], [[14, 22]], [[3, 11]], [[14, 22]], [[3, 11]], [[0, 11]], [[0, 24]]]",
    "query_spans": "[[[26, 38]]]",
    "process": "Let the coordinates of the center of the moving circle be (x, y). The moving circle passes through the fixed point P(1, 0) and is tangent to the fixed line x = -1, so the distance from the center to the fixed point P and to the line is equal to the radius. According to the distance formula between two points, (x-1)^{2}+y^{2}=(x+1)^{2}. Simplifying yields y^{2}=4x."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the right focus is $F$, the focal distance is $8$, the left vertex is $A$, and on the $y$-axis there is a point $B(0, b)$ such that $\\overrightarrow{B A} \\cdot \\overrightarrow{B F}=2 a$. Then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(G) = F;FocalLength(G) = 8;A: Point;LeftVertex(G) = A;B: Point;Coordinate(B) = (0, b);PointOnCurve(B, yAxis) = True;DotProduct(VectorOf(B, A), VectorOf(B, F)) = 2*a",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 58], [159, 162]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[63, 66]], [[2, 66]], [[2, 73]], [[78, 81]], [[2, 81]], [[91, 100]], [[91, 100]], [[82, 100]], [[103, 156]]]",
    "query_spans": "[[[159, 170]]]",
    "process": "Analysis: Using the dot product formula of vectors, we obtain \\cdot 4a + b^{2} = 2a, that is, 16\\cdota^{2} = 6a, from which the value of a can be obtained, and thus the eccentricity of the hyperbola can be found. Details: From the given conditions, A(-a,0), F(4,0), B(0,b), \\therefore \\overrightarrow{BA}=(-a,-b), \\overrightarrow{BF}=(4,-b). \\because \\overrightarrow{BA}\\cdot\\overrightarrow{BF}=2a, \\therefore (-a,-b)\\cdot(4,-b)=2a, \\therefore 4a + b^{2} = 2a, \\therefore b^{2} = 6a, \\therefore 16\\cdota^{2} = 6a, \\therefore a = 2, \\therefore e = \\frac{c}{a} = \\frac{4}{2} = 2,"
  },
  {
    "text": "Given the parabola $y^{2}=4 x$, a line passing through the point $P(4 , 0)$ intersects the parabola at two points $A(x_{1} , y_{1})$ and $B(x_{2} , y_{2})$. Then the minimum value of $y_{1}^{2}+y_{2}^{2}$ is?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Coordinate(P) = (4, 0);PointOnCurve(P, H);x1: Number;y1: Number;x2: Number;y2: Number ;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Intersection(H,G)={A,B}",
    "query_expressions": "Min(y1^2 + y2^2)",
    "answer_expressions": "32",
    "fact_spans": "[[[2, 16], [33, 36]], [[30, 32]], [[18, 29]], [[39, 57]], [[59, 78]], [[2, 16]], [[18, 29]], [[17, 32]], [[39, 57]], [[39, 57]], [[59, 78]], [[59, 78]], [[39, 57]], [[59, 78]], [[30, 80]]]",
    "query_spans": "[[[82, 110]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, let $F_{1}$ and $F_{2}$ be its left and right foci respectively, and let $P$ be a point on the hyperbola such that $|P F_{1}|=5$. Then the value of $|P F_{2}|$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 - y^2/16 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 5",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "11",
    "fact_spans": "[[[2, 41], [60, 61], [72, 75]], [[68, 71]], [[42, 49]], [[50, 57]], [[2, 41]], [[42, 67]], [[42, 67]], [[68, 78]], [[80, 93]]]",
    "query_spans": "[[[95, 110]]]",
    "process": ""
  },
  {
    "text": "A point $P$ on the ellipse $\\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1$ is such that the lines connecting $P$ to the two foci $F_{1}$, $F_{2}$ of the ellipse are perpendicular to each other. Then the area of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/49 + y^2/24 = 1);P: Point;PointOnCurve(P,G) = True;F1: Point;F2: Point;Focus(G) = {F1,F2}\t;IsPerpendicular(LineSegmentOf(P,F1),LineSegmentOf(P,F2)) = True",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "24",
    "fact_spans": "[[[0, 39], [46, 48]], [[0, 39]], [[41, 45]], [[0, 45]], [[53, 60]], [[61, 68]], [[46, 68]], [[42, 75]]]",
    "query_spans": "[[[77, 107]]]",
    "process": "Let PF_{1}=x, PF_{2}=14-x, 2c=10. According to the Pythagorean theorem, x^{2}+(14-x)^{2}=100. Solving gives x=8, so S=\\frac{1}{2}\\times8\\times6=24"
  },
  {
    "text": "Given the parabola $C$: $y=\\frac{1}{4} x^{2}$ with focus $F$, its directrix $l$ intersects the $y$-axis at point $A$. Point $M$ lies on the parabola $C$. When $\\frac{|M A|}{|M F|}=\\sqrt{2}$, the area of $\\Delta A M F$ is?",
    "fact_expressions": "C: Parabola;A: Point;M: Point;F: Point;l:Line;Expression(C) = (y = x^2/4);Focus(C) = F;Directrix(C)=l;Intersection(l, yAxis) = A;PointOnCurve(M, C);Abs(LineSegmentOf(M,A))/Abs(LineSegmentOf(M,F))=sqrt(2)",
    "query_expressions": "Area(TriangleOf(A, M, F))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 31], [39, 40], [62, 68]], [[52, 56]], [[57, 61]], [[35, 38]], [[42, 45]], [[2, 31]], [[2, 38]], [[39, 45]], [[42, 56]], [[57, 69]], [[71, 101]]]",
    "query_spans": "[[[103, 122]]]",
    "process": "Through point $M$, draw $MM' \\perp l$, with the foot of the perpendicular at $M'$, as shown in the figure. By the definition of a parabola, we have: $|MF| = |MM'|$. Since $\\frac{|MA|}{|MF|} = \\sqrt{2}$, it follows that $\\sin\\angle MAM' = \\frac{|MM'|}{|MA|} = \\frac{\\sqrt{2}}{2}$, so $\\angle MAM' = 45^\\circ$. Therefore, triangle $AM'M$ is an isosceles right triangle, i.e., $|MM'| = |MA| = |MF|$. In triangle $AMF$, $\\angle MAF = 45^{\\circ}$. By the law of sines: $\\frac{|MF|}{\\sin\\angle MAF} = \\frac{|MA|}{\\sin\\angle MFA}$. Thus, $\\sin\\angle MFA = \\frac{|MA|}{|MF|} \\sin\\angle MAF = \\sqrt{2} \\times \\frac{\\sqrt{2}}{2} = 1$, so $\\angle MFA = 90^{\\circ}$. Hence, $MF \\parallel M'A$, and since $|MF| = |MA| = |MM'|$, quadrilateral $AM'MF$ is a square, so $|AF| = |MF| = 2$. The area of triangle $AMF$ is: $S = \\frac{1}{2} \\times 2 \\times 2 = 2$. The correct answer to this problem: $2$."
  },
  {
    "text": "Find the minimum distance from a point on the parabola $y=-x^{2}$ to the line $4 x+3 y-8=0$?",
    "fact_expressions": "G: Parabola;H: Line;P0: Point;Expression(G) = (y = -x^2);Expression(H) = (4*x + 3*y - 8 = 0);PointOnCurve(P0, G)",
    "query_expressions": "Min(Distance(P0, H))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[1, 14]], [[18, 33]], [[16, 17]], [[1, 14]], [[18, 33]], [[1, 17]]]",
    "query_spans": "[[[16, 41]]]",
    "process": "Let a point on the parabola $ y = -x^{2} $ be $ (m, -m^{2}) $. The distance from this point to the line $ 4x + 3y - 8 = 0 $ is $ \\frac{|4m - 3m^{2} - 8|}{5} $. When $ m = \\frac{2}{3} $, the minimum value $ \\frac{4}{\\frac{4}{2}} $ is attained."
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=4x$, draw a line $l$ intersecting the parabola at points $A$ and $B$, and let $l$ intersect the directrix at point $C$. If $\\overrightarrow{A B}=2 \\overrightarrow{B C}$, then $\\frac{|A F|}{|B F|}=$?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;C: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};Intersection(l,Directrix(G))=C;VectorOf(A, B) = 2*VectorOf(B, C)",
    "query_expressions": "Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(B, F))",
    "answer_expressions": "3",
    "fact_spans": "[[[22, 27], [43, 46]], [[1, 15], [28, 31]], [[32, 35]], [[36, 39]], [[51, 55]], [[18, 21]], [[1, 15]], [[1, 21]], [[0, 27]], [[22, 41]], [[28, 55]], [[57, 102]]]",
    "query_spans": "[[[104, 127]]]",
    "process": "Using the definition of a parabola and similar triangles to solve the ratio relationship, draw AH perpendicular to the directrix with foot H, and draw BM perpendicular to the directrix with foot M, as shown in the figure below: Since $\\overrightarrow{AB}=2\\overrightarrow{BC}$, it follows that $AB=2BC$, so $|BC|=\\frac{1}{3}AC$. Because $\\triangle CMB \\sim \\triangle CHA$ and by the definition of a parabola, $|BF|=|BM|=\\frac{1}{3}|AB|=\\frac{1}{3}|AF|$. Hence, we obtain $\\frac{|AF|}{|BF|}=3$. Answer: 3. This problem examines the definition of a parabola, involving similar triangles, and is a basic question."
  },
  {
    "text": "$P$ is a point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$, $Q(2,0)$, then the minimum length of the segment $P Q$ is?",
    "fact_expressions": "G: Ellipse;P: Point;Q: Point;Expression(G) = (x^2/16 + y^2/4 = 1);Coordinate(Q) = (2, 0);PointOnCurve(P, G)",
    "query_expressions": "Min(Length(LineSegmentOf(P,Q)))",
    "answer_expressions": "2*sqrt(6)/3",
    "fact_spans": "[[[4, 42]], [[0, 3]], [[46, 54]], [[4, 42]], [[46, 54]], [[0, 45]]]",
    "query_spans": "[[[56, 71]]]",
    "process": "Let $ P(x,y) $, then $ y^{2} = 4 - \\frac{x^{2}}{4} $ $ (-4 \\leqslant x \\leqslant 4) $, $ |PQ| = \\sqrt{(x-2)^{2} + y^{2}} = \\sqrt{\\frac{3}{4}x^{2} - 4x + 8} = \\sqrt{\\frac{3}{4}\\left(x - \\frac{8}{3}\\right)^{2} + \\frac{24}{9}} \\geqslant \\frac{2\\sqrt{6}}{3} $, that is, the minimum length of segment $ PQ $ is $ \\frac{2\\sqrt{6}}{3} $."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=8x$ is $F$, $P$ is a point on the directrix of the parabola, $Q$ is an intersection point of the line $PF$ and the parabola, and $\\overrightarrow{PQ}=\\sqrt{2} \\overrightarrow{QF}$, then the equation of the line $PF$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;P: Point;PointOnCurve(P, Directrix(G)) = True;OneOf(Intersection(LineOf(P, F), G)) = Q;Q: Point;VectorOf(P, Q) = sqrt(2)*VectorOf(Q, F)",
    "query_expressions": "Expression(LineOf(P, F))",
    "answer_expressions": "{(x+y-2=0),(x-y-2=0)}",
    "fact_spans": "[[[2, 16], [28, 31], [49, 52]], [[2, 16]], [[20, 23]], [[2, 23]], [[24, 27]], [[24, 36]], [[37, 57]], [[37, 40]], [[59, 111]]]",
    "query_spans": "[[[113, 125]]]",
    "process": "From the given conditions, we have $ F(2,0) $. Let $ P(-2,t) $, $ Q(m,n) $, then $ \\overrightarrow{PQ} = (m+2, n-t) $, $ \\overrightarrow{QF} = (2-m, -n) $. From $ \\overrightarrow{PQ} = \\sqrt{2} \\overrightarrow{QF} $, we obtain\n$$\n\\begin{cases}\nm+2 = \\sqrt{2}(2-m) \\\\\nn-t = -\\sqrt{2}n\n\\end{cases}\n$$\nSolving gives $ m = 6 - 4\\sqrt{2} $. Substituting into $ y^2 = 8x $ yields $ n = 4(\\sqrt{2}-1) $ or $ n = -4(\\sqrt{2}-1) $. Hence,\n$$\nk_{QF} = \\frac{n}{m-2} = \\frac{4(\\sqrt{2}-1)}{4-4\\sqrt{2}} = -1 \\quad \\text{or} \\quad k_{QF} = \\frac{n}{m-2} = \\frac{-4(\\sqrt{2}-1)}{4-4\\sqrt{2}} = 1\n$$\nTherefore, the equation of line $ PF $ is $ x+y-2=0 $ or $ x-y-2=0 $. The answer to be filled in is $ x+y-2=0 $ or $ x-y-2=0 $."
  },
  {
    "text": "The line passing through the point $P(2,2)$ with slope $-1$ intersects the parabola $y^{2}=x$ at points $A$ and $B$. Then $|P A|+|P B|=$?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;A: Point;B: Point;Expression(G) = (y^2 = x);Coordinate(P) = (2, 2);PointOnCurve(P, H);Slope(H) = -1;Intersection(H, G) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B))",
    "answer_expressions": "5*sqrt(2)",
    "fact_spans": "[[[22, 34]], [[19, 21]], [[1, 10]], [[36, 39]], [[40, 43]], [[22, 34]], [[1, 10]], [[0, 21]], [[11, 21]], [[19, 45]]]",
    "query_spans": "[[[47, 62]]]",
    "process": "Let points A(x_{1},y_{1}) and B(x_{2},y_{2}), the equation of line AB is y-2=-(x-2), that is, y=-x+4. Solving the system \\begin{cases}y=-x\\\\y^{2}=x\\end{cases} (+4, eliminating y yields x^{2}-9x+16=0, solving gives x_{1}=\\frac{9+\\sqrt{17}}{2}, x_{2}=\\frac{9-\\sqrt{17}}{2}. Therefore, |PA|+|PB|=\\sqrt{1+(-1)^{2}}(|x_{1}-2|+|x_{2}-2|)=5\\sqrt{2}."
  },
  {
    "text": "Given that $P$ is a moving point on the plane, $A(-1,0)$, $B(1,0)$ are two fixed points on the plane, and $\\overrightarrow{P A} \\cdot \\overrightarrow{P B}=0$, then the trajectory equation of the moving point $P$ is?",
    "fact_expressions": "A: Point;B: Point;P: Point;Coordinate(A) = (-1, 0);Coordinate(B) = (1, 0);DotProduct(VectorOf(P, A), VectorOf(P, B)) = 0",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2+y^2=1",
    "fact_spans": "[[[13, 23]], [[25, 33]], [[2, 5], [98, 101]], [[13, 23]], [[25, 33]], [[43, 94]]]",
    "query_spans": "[[[98, 108]]]",
    "process": "Let P(x,y), then set up the equation directly according to the given conditions. [Detailed solution] Let P(x,y), then \\overrightarrow{PA}=(-1-x,-y), \\overrightarrow{PB}=(1-x,-y). Since \\overrightarrow{PA}\\cdot\\overrightarrow{PB}=0, we have (-1-x)(1-x)+y^{2}=0. Simplifying gives x^{2}+y^{2}=1. Therefore, the trajectory equation of the moving point P is x^{2}+y^{2}=1."
  },
  {
    "text": "If $P$ is a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and $F_{1}$, $F_{2}$ are the two foci, if $\\angle F_{1} P F_{2}=90^{\\circ}$, then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(P, G);Focus(G)={F1,F2};AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "16",
    "fact_spans": "[[[5, 44]], [[49, 56]], [[1, 4]], [[57, 64]], [[5, 44]], [[1, 48]], [[5, 69]], [[71, 104]]]",
    "query_spans": "[[[106, 133]]]",
    "process": ""
  },
  {
    "text": "Let $O$ be the origin of coordinates, and let $F_{1}$, $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. If there exists a point $P$ on the ellipse such that $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$ and $|OP|=\\frac{\\sqrt{3}}{2} a$, then the eccentricity of this ellipse is?",
    "fact_expressions": "O: Origin;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;G: Ellipse;Expression(G) = (x^2/a^2 + y^2/b^2 = 1);a: Number;b: Number;a>b;b>0;P: Point;PointOnCurve(P,G) = True;AngleOf(F1,P,F2) = pi/3;Abs(LineSegmentOf(O,P)) = a*sqrt(3)/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[1, 4]], [[10, 17]], [[20, 27]], [[10, 86]], [[10, 86]], [[28, 80], [89, 91], [168, 170]], [[28, 80]], [[30, 80]], [[30, 80]], [[30, 80]], [[30, 80]], [[94, 98]], [[89, 98]], [[100, 135]], [[137, 165]]]",
    "query_spans": "[[[168, 176]]]",
    "process": ""
  },
  {
    "text": "Let $A$, $B$ be the vertices of the major axis of ellipse $\\Gamma$, $E$, $F$ be the two foci of $\\Gamma$, $|A B|=4$, $|A F|=2+\\sqrt{3}$, and $P$ be a point on $\\Gamma$ such that $|P E| \\cdot |P F|=2$. Then the area of $\\triangle P E F$ is?",
    "fact_expressions": "Gamma: Ellipse;P: Point;E: Point;F: Point;A: Point;B: Point;Endpoint(MajorAxis(Gamma))={A,B};Focus(Gamma) = {E, F};Abs(LineSegmentOf(A, B)) = 4;Abs(LineSegmentOf(A, F)) = sqrt(3) + 2;PointOnCurve(P, Gamma);Abs(LineSegmentOf(P, E))*Abs(LineSegmentOf(P, F)) = 2",
    "query_expressions": "Area(TriangleOf(P, E, F))",
    "answer_expressions": "1",
    "fact_spans": "[[[9, 19], [33, 41], [83, 91]], [[78, 82]], [[25, 28]], [[29, 32]], [[1, 4]], [[5, 8]], [[1, 24]], [[25, 46]], [[47, 56]], [[58, 76]], [[79, 94]], [[97, 117]]]",
    "query_spans": "[[[119, 141]]]",
    "process": "Let the standard equation of the ellipse be $\\Gamma:\\frac{x^2}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. According to the given conditions, we have $2a=|AB|=4$, $|AF|=a+c=a+\\sqrt{a^{2}-b^{2}}=2+\\sqrt{3}$. Solving gives $a=2$, $b=1$, so $c^{2}=a^{2}-b^{2}=3$, hence $c=\\sqrt{3}$, and thus $|EF|=2\\sqrt{3}$. By the definition of the ellipse, $|PE|+|PF|=2a=4$. Since $|PE|\\cdot|PF|=2$, we obtain $|PE|+|PF|^{2}=(|PE|+|PF|)^{2}-2|PE|\\cdot|PF|=12$. Also, since $|EF|=2\\sqrt{3}$, we have $|EF|^{2}=12$, so $|PE|+|PF|^{2}=|EF|^{2}$, hence $\\angle EPF$ is a right angle. Therefore, the area of $\\triangle PEF$ is $S_{\\triangle PEF}=\\frac{1}{2}\\cdot|PE|\\cdot|PF|=1$."
  },
  {
    "text": "A point $P$ on the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$ is at a distance of $5$ from one of its foci. What is the distance from point $P$ to the other focus?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/9 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2);Distance(P, F1) = 5",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "9",
    "fact_spans": "[[[0, 38], [45, 46]], [[0, 38]], [[41, 44], [62, 66]], [[0, 44]], [], [], [[45, 51]], [[45, 72]], [[45, 72]], [[41, 59]]]",
    "query_spans": "[[[45, 78]]]",
    "process": "From the hyperbola equation, we know: a=2, b=3, and c=\\sqrt{13}, so the minimum distance from a point on the hyperbola to a focus is \\sqrt{13}-2. Let the distance from P to the other focus be m. According to the definition of a hyperbola, we have: |5-m|=2a=4, solving gives m=9, m=1<\\sqrt{13}-2, \\therefore m=9. This problem examines the definition of a hyperbola; it is easy to overlook the minimum distance from a point on the hyperbola to a focus."
  },
  {
    "text": "If the focal distance of a hyperbola is $2 \\sqrt{10}$, and the equations of the asymptotes are $y = \\pm 3x$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;FocalLength(G) = 2*sqrt(10);Expression(Asymptote(G)) = (y = pm*3*x)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/9 = \\pm 1",
    "fact_spans": "[[[1, 4], [42, 45]], [[1, 21]], [[1, 40]]]",
    "query_spans": "[[[42, 52]]]",
    "process": "The focal distance of the hyperbola is $2\\sqrt{10}$, so $c=\\sqrt{10}$. When the foci of the hyperbola lie on the x-axis, the asymptotes are given by $y=\\pm3x$, so $\\frac{b}{a}=3$, $a^{2}+b^{2}=10$, hence $a=1$, $b=3$. When the foci of the hyperbola lie on the y-axis, $\\frac{a}{b}=3$, $a^{2}+b^{2}=10$, so $b=1$, $a=3$. Therefore, the required hyperbola equation is: $x^{2}-\\frac{y^{2}}{9}=\\pm1$"
  },
  {
    "text": "Given the curve $x^{2}-4 y^{2}=4$, find the equation of the line on which the chord $M N$ lies, passing through the point $A(3,1)$ and bisected by the point $A$.",
    "fact_expressions": "H: Curve;Expression(H) = (x^2 - 4*y^2 = 4);A: Point;Coordinate(A) = (3, 1);PointOnCurve(A, OverlappingLine(LineSegmentOf(M, N)));M: Point;N: Point;IsChordOf(LineSegmentOf(M, N), H);MidPoint(LineSegmentOf(M, N)) = A",
    "query_expressions": "Expression(OverlappingLine(LineSegmentOf(M, N)))",
    "answer_expressions": "3*x-4*y-5=0",
    "fact_spans": "[[[2, 21]], [[2, 21]], [[23, 32], [34, 38]], [[23, 32]], [[22, 52]], [[42, 47]], [[42, 47]], [[2, 47]], [[34, 47]]]",
    "query_spans": "[[[42, 56]]]",
    "process": "Let the coordinates of the two intersection points be M(x_{1},y_{1}), N(x_{2},y_{2}). Thus, x_{1}^{2}-4y_{1}^{2}=4, x_{2}^{2}-4y_{2}^{2}=4. Subtracting these two equations gives (x_{1}+x_{2})(x_{1}-x_{2})=4(y_{1}+y_{2})(y_{1}-y_{2}). Since the midpoint of M(x_{1},y_{1}), N(x_{2},y_{2}) is A(3,1), we have \\frac{x_{1}+x_{2}}{2}=3, \\frac{y_{1}+y_{2}}{2}=1. Therefore, \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{3}{4}. Hence, the equation of the line is y-1=\\frac{3}{4}(x-3), that is, 3x-4y-5=0. Since point A(3,1) lies inside the hyperbola, the line equation satisfies the condition. Therefore, the equation of the line MN is 3x-4y-5=0."
  },
  {
    "text": "Let a focus of the hyperbola be $F$, and an endpoint of the imaginary axis be $B$. If the line $FB$ is perpendicular to an asymptote of the hyperbola, then the eccentricity of this hyperbola is?",
    "fact_expressions": "F: Point;B: Point;G: Hyperbola;OneOf(Focus(G)) = F;OneOf(Endpoint(ImageinaryAxis(G))) = B;IsPerpendicular(LineOf(F, B), OneOf(Asymptote(G)))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(1 + sqrt(5))/2",
    "fact_spans": "[[[10, 13]], [[23, 26]], [[1, 4], [37, 40], [52, 55]], [[1, 13]], [[1, 26]], [[29, 48]]]",
    "query_spans": "[[[52, 61]]]",
    "process": ""
  },
  {
    "text": "What is the standard equation of a parabola with the directrix $x=-2$?",
    "fact_expressions": "G: Parabola;H: Line;Expression(H) = (x = -2);Directrix(G) = H",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = 8*x",
    "fact_spans": "[[[13, 16]], [[1, 9]], [[1, 9]], [[0, 16]]]",
    "query_spans": "[[[13, 23]]]",
    "process": ""
  },
  {
    "text": "The equation of the ellipse with the foci of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$ as vertices and the vertices as foci is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/12 = 1);H: Ellipse;Focus(G) = Vertex(H);Vertex(G) = Focus(H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2/16+y^2/12=1",
    "fact_spans": "[[[1, 40]], [[1, 40]], [[53, 55]], [[0, 55]], [[0, 55]]]",
    "query_spans": "[[[53, 59]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the right vertex of the hyperbola is $A$. A circle is drawn with center $A$ and radius $b$, intersecting an asymptote of the hyperbola at points $M$ and $N$. If $\\angle M A N=60^{\\circ}$, then $\\frac{a^{2}}{b^{2}}$=?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;M: Point;A: Point;N: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightVertex(C) = A;Center(G)=A;Radius(G)=b;Intersection(OneOf(Asymptote(C)),G)={M,N};AngleOf(M, A, N) = ApplyUnit(60, degree)",
    "query_expressions": "a^2/b^2",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 63], [90, 93]], [[81, 84]], [[10, 63]], [[88, 89]], [[101, 104]], [[68, 71], [74, 77]], [[105, 108]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 71]], [[73, 89]], [[81, 89]], [[88, 110]], [[112, 137]]]",
    "query_spans": "[[[139, 162]]]",
    "process": "From the given conditions, the center of the circle A is at (a,0), the line MN is the asymptote y=\\frac{b}{a}x', and \\angle MAN=60^{\\circ}, so \\triangle AMN is an equilateral triangle with side length b. Therefore, the distance from the center to the line is \\frac{\\sqrt{3}}{2}b=\\frac{|b|}{\\sqrt{1+\\frac{b^{2}}{a^{2}}}}, hence \\frac{a^{2}}{b^{2}}=3."
  },
  {
    "text": "The left and right foci of hyperbola $C$ are $(-2,0)$ and $(2,0)$, respectively, and it passes through the point $(2, \\sqrt{2})$. Then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;F1: Point;LeftFocus(C) = F1;Coordinate(F1) = (-2,0);F2:Point;RightFocus(C) = F2;Coordinate(F2) = (2,0);P: Point;PointOnCurve(P,C) = True;Coordinate(P) = (2,sqrt(2))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/2 - y^2/2 = 1",
    "fact_spans": "[[[0, 6], [52, 55]], [[14, 22]], [[0, 31]], [[14, 22]], [[24, 31]], [[0, 31]], [[24, 31]], [[34, 50]], [[0, 50]], [[34, 50]]]",
    "query_spans": "[[[52, 60]]]",
    "process": "According to the problem, the foci of hyperbola C lie on the x-axis. Let its equation be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0). Then c=2, and we have \\begin{cases}c^{2}=4=a^{2}+b^{2}\\\\\\frac{4}{a^{2}}-\\frac{2}{b^{2}}=1\\end{cases}^{2}. Solving gives a^{2}=2, b^{2}=2. Thus, the equation of C is \\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1."
  },
  {
    "text": "The number of intersection points of the line $y=x+3$ and the curve $-\\frac{x| x |}{4}+\\frac{y^{2}}{9}=1$?",
    "fact_expressions": "G: Line;H: Curve;Expression(G) = (y = x + 3);Expression(H) = (y^2/9 - x*Abs(x)/4 = 1)",
    "query_expressions": "NumIntersection(G, H)",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 9]], [[10, 49]], [[0, 9]], [[10, 49]]]",
    "query_spans": "[[[0, 56]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of a hyperbola are $y = \\pm \\frac{2}{3}x$ and the focal distance is $2\\sqrt{26}$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*(2/3)*x);FocalLength(G)=2*sqrt(26)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/18 - y^2/8 = 1, y^2/8 - x^2/18 = 1}",
    "fact_spans": "[[[2, 5], [51, 54]], [[2, 32]], [[2, 49]]]",
    "query_spans": "[[[51, 61]]]",
    "process": "When the foci of the hyperbola are on the x-axis, from $\\frac{b}{a}=\\frac{2}{3}$ and $c^{2}=a^{2}+b^{2}=26$, solving these two equations simultaneously gives $a^{2}=18$, $b^{2}=8$, so the standard equation of the required hyperbola is $\\frac{x^{2}}{18}-\\frac{y^{2}}{8}=1$; when the foci of the hyperbola are on the y-axis, from $\\frac{a}{b}=\\frac{2}{3}$ and $c^{2}=a^{2}+b^{2}=26$ solving these two equations simultaneously gives $a^{2}=8$, $b^{2}=18$, so the standard equation of the required hyperbola is $\\frac{y^{2}}{8}-\\frac{x^{2}}{18}=1$. In summary, the standard equation of the required hyperbola is $\\frac{x^{2}}{18}-\\frac{y^{2}}{8}=1$ or $\\frac{y^{2}}{8}-\\frac{x^{2}}{18}=1$."
  },
  {
    "text": "If the line $m x+n y-3=0$ and the circle $x^{2}+y^{2}=3$ have no common points, then taking $(m, n)$ as the coordinates of point $P$, the number of common points between a line passing through point $P$ and the ellipse $\\frac{x^{2}}{7}+\\frac{y^{2}}{3}=1$ is?",
    "fact_expressions": "G: Ellipse;H: Circle;I: Line;m: Number;n: Number;P: Point;Expression(G) = (x^2/7 + y^2/3 = 1);Expression(H) = (x^2 + y^2 = 3);Expression(I) = (m*x + n*y - 3 = 0);Coordinate(P) = (m, n);NumIntersection(I,H)=0;L:Line;PointOnCurve(P,L)",
    "query_expressions": "NumIntersection(L,G)",
    "answer_expressions": "2",
    "fact_spans": "[[[69, 106]], [[17, 33]], [[1, 16]], [[41, 49]], [[41, 49]], [[50, 54], [59, 63]], [[69, 106]], [[17, 33]], [[1, 16]], [[40, 57]], [[1, 38]], [[66, 68]], [[58, 68]]]",
    "query_spans": "[[[66, 114]]]",
    "process": "Since the line $ mx + ny - 3 = 0 $ and the circle $ x^{2} + y^{2} = 3 $ have no common points, the distance from the center of the circle to the line is greater than the radius. Thus, we obtain $ \\frac{3}{\\sqrt{m^{2} + n^{2}}} > \\sqrt{3} $, which implies $ m^{2} + n^{2} < 3 $. Therefore, the trajectory of point $ P $ with coordinates $ (m, n) $ is the interior of a circle centered at the origin with radius $ \\sqrt{3} $ (excluding points on the circumference). Clearly, all points within this circular region lie inside the ellipse, so any line passing through point $ P $ intersects the ellipse at two points."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A tangent line is drawn from $F_{1}$ to the circle $x^{2}+y^{2}=a^{2}$, intersecting the right branch of the hyperbola at point $M$. If $\\angle F_{1} M F_{2}=45^\\circ$, then the asymptotes of the hyperbola have equations?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Circle;F1: Point;M: Point;F2: Point;l: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + y^2 = a^2);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, l);TangentOfPoint(F1,H)=l;Intersection(l,RightPart(G))=M;AngleOf(F1, M, F2) = ApplyUnit(45,degree)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*sqrt(2)*x",
    "fact_spans": "[[[2, 58], [118, 121], [158, 161]], [[5, 58]], [[5, 58]], [[93, 113]], [[67, 74], [85, 92]], [[124, 128]], [[75, 82]], [], [[5, 58]], [[5, 58]], [[2, 58]], [[93, 113]], [[2, 82]], [[2, 82]], [[84, 116]], [[83, 116]], [[84, 128]], [[130, 156]]]",
    "query_spans": "[[[158, 169]]]",
    "process": "As shown in the figure: the point of tangency is A, connect OA, draw BF₂ ⊥ F₁M at B. O is the midpoint of F₁F₂, OA ∥ BF₂ ⇒ BF₂ = 2OA = 2a. ∠F₁MF₂ = 45° ⇒ MF = 2√2 a, BM = 2a ⇒ BF₁ = 2√2 a. In right triangle △BF₁F₂, by the Pythagorean theorem: (2√2 a)² + (2a)² = (2c)² ⇒ b² = 2a². The equations of the asymptotes are: y = ±√2 x"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$, and $O$ is the coordinate origin. If a point $P$ on the right branch satisfies $P O \\perp P F_{2}$ and $\\angle F_{1} P O=30^{\\circ}$, then the asymptotes of $C$ have equations?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;O: Origin;F2: Point;F1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P,RightPart(C));IsPerpendicular(LineSegmentOf(P,O),LineSegmentOf(P,F2));AngleOf(F1, P, O) = ApplyUnit(30, degree)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*sqrt(6)*x",
    "fact_spans": "[[[20, 84], [163, 166]], [[28, 84]], [[28, 84]], [[106, 109]], [[91, 94]], [[10, 17]], [[2, 9]], [[28, 84]], [[28, 84]], [[20, 82]], [[2, 90]], [[2, 90]], [[20, 109]], [[111, 130]], [[132, 161]]]",
    "query_spans": "[[[163, 174]]]",
    "process": "Take point Q as the symmetric point of point P with respect to the origin O. By the symmetry of the hyperbola, Q lies on the hyperbola. Therefore, quadrilateral $PF_{1}QF_{2}$ is a parallelogram, so $F_{1}Q\\bot QP$. In right triangle $\\triangle PQF_{1}$, $\\angle F_{1}PQ=30^{\\circ}$, so $|PF_{1}|=2|F_{1}Q|=2|PF_{2}|$. By the definition of the hyperbola: $|PF_{1}|-|PF_{2}|=|PF_{2}|=2a$, thus $|PF_{1}|=2|PF_{2}|=4a$, $|OQ|=\\frac{1}{2}|PQ|=\\sqrt{3}a$. In right triangle $\\triangle OQF_{1}$, $|OF_{1}|^{2}=|OQ|^{2}+|QF_{1}|^{2}$, so $c^{2}=3a^{2}+4a^{2}$. Therefore, $b^{2}=6a^{2}$, and the asymptotes are given by: $y=\\pm\\sqrt{6}x$."
  },
  {
    "text": "Let the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ have its left vertex at point $A$, upper vertex at point $B$, $AB = \\sqrt{13}$, and the eccentricity of the ellipse is $\\frac{\\sqrt{5}}{3}$. Find the equation of the line $l$ passing through the right focus $F_2$ of ellipse $C$ and parallel to the line $AB$.",
    "fact_expressions": "l: Line;C: Ellipse;b: Number;a: Number;B: Point;A: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(C) = A;UpperVertex(C)=B;LineSegmentOf(A, B) = sqrt(13);Eccentricity(C) = sqrt(5)/3;F2:Point;RightFocus(C)=F2;PointOnCurve(F2,l);IsParallel(l,LineOf(A,B))",
    "query_expressions": "Expression(l)",
    "answer_expressions": "2*x-3*y-2*sqrt(5)=0",
    "fact_spans": "[[[154, 159]], [[1, 58], [95, 97], [126, 131]], [[8, 58]], [[8, 58]], [[72, 75]], [[63, 66]], [[8, 58]], [[8, 58]], [[1, 58]], [[1, 66]], [[1, 75]], [[77, 92]], [[95, 122]], [[135, 142]], [[126, 142]], [[125, 159]], [[143, 159]]]",
    "query_spans": "[[[154, 164]]]",
    "process": "From the given conditions, \\begin{cases}a^{2}+b^{2}=13\\\\\\frac{c}{a}=\\frac{\\sqrt{5}}{3}\\end{cases}, \\therefore a=3, b=2, c=\\sqrt{5}. Therefore, the coordinates of the right focus of the ellipse are (\\sqrt{5},0). From the given conditions, the slope of line AB is \\frac{2}{3}, \\therefore l:y-0=\\frac{2}{3}(x-\\sqrt{5}), \\therefore l:2x-3y-2\\sqrt{5}=0."
  },
  {
    "text": "The coordinates of the foci of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/3 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "{(sqrt(7),0) , (-sqrt(7),0)}",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 45]]]",
    "process": "Since $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=$, we have $a=2$, $b=\\sqrt{3}$, and the foci of the hyperbola lie on the x-axis, so $c=\\sqrt{4+3}=\\sqrt{7}$. Hence, the coordinates of the foci are $(\\sqrt{7},0)$, $(-\\sqrt{7},0)$."
  },
  {
    "text": "The ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ has left and right vertices $A_{1}$ and $A_{2}$, respectively. Point $P$ is a point on curve $C$ distinct from $A_{1}$ and $A_{2}$. The slopes of lines $P A_{1}$ and $P A_{2}$ are $k_{1}$ and $k_{2}$, respectively. Then $k_{1} \\cdot k_{2}$=?",
    "fact_expressions": "C: Ellipse;P: Point;A1: Point;A2: Point;k1: Number;k2: Number;Expression(C) = (x^2/4 + y^2/3 = 1);LeftVertex(C) = A1;RightVertex(C) = A2;PointOnCurve(P, C);Negation(P = A1);Negation(P = A2);Slope(LineOf(P, A1)) = k1;Slope(LineOf(P, A2)) = k2",
    "query_expressions": "k1*k2",
    "answer_expressions": "-3/4",
    "fact_spans": "[[[0, 42], [72, 77]], [[67, 71]], [[51, 58], [80, 87]], [[59, 66], [88, 95]], [[128, 135]], [[136, 143]], [[0, 42]], [[0, 66]], [[0, 66]], [[67, 98]], [[67, 98]], [[67, 98]], [[99, 143]], [[99, 143]]]",
    "query_spans": "[[[145, 166]]]",
    "process": "From the given information, the coordinates of $ A_{1} $ and $ A_{2} $ can be written. Let $ P(x_{0}, y_{0}) $, then $ k_{1} \\cdot k_{2} = \\frac{y_{0}^{2}}{x_{0}^{2} - 4} $. Since $ P $ lies on the ellipse $ C $, we have $ y_{0}^{2} = \\frac{3}{4}(4 - x_{0}^{2}) $, thus $ k_{1} \\cdot k_{2} $ can be found. From the problem: $ A_{1}(-2, 0) $, $ A_{2}(2, 0) $, let $ P(x_{0}, y_{0}) $, then  \n$ k_{1} \\cdot k_{2} = \\frac{y_{0}}{x_{0} + 2} \\cdot \\frac{y_{0}}{x_{0} - 2} = \\frac{y_{0}^{2}}{x_{0}^{2} - 4} $.  \nSince $ P $ lies on the ellipse $ C $, i.e., $ y_{0}^{2} = \\frac{3}{4}(4 - x_{0}^{2}) $,  \n$ \\therefore k_{1} \\cdot k_{2} = -\\frac{3}{4} $."
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the left and right foci of the ellipse $C$: $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$, point $P$ lies on $C$, $O$ is the origin, and $|O P|=3$. Then the area of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;O: Origin;Expression(C) = (x^2/16 + y^2/7 = 1);LeftFocus(C) =F1;RightFocus(C)=F2;PointOnCurve(P, C);Abs(LineSegmentOf(O, P)) = 3",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "7",
    "fact_spans": "[[[17, 60], [72, 75]], [[67, 71]], [[1, 8]], [[9, 16]], [[77, 80]], [[17, 60]], [[1, 66]], [[1, 66]], [[67, 76]], [[87, 96]]]",
    "query_spans": "[[[98, 128]]]",
    "process": "From the given conditions, $ a=4 $, $ c=3 $, $ |OP|=3=\\frac{1}{2}|F_{1}F_{2}| $, so $ P $ lies on the circle with segment $ F_{1}F_{2} $ as diameter, thus $ PF_{1}\\bot PF_{2} $, hence $ |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}=36 $ ①. By the definition of an ellipse, $ |PF_{1}|+|PF_{2}|=8 $ ②. From ① and ②, solving gives $ |PF_{1}|\\cdot|PF_{2}|=14 $, therefore $ S_{\\Delta PF_{1}F_{2}}=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|=7 $."
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, $F_{1}$ and $F_{2}$ are the two foci of the ellipse, $\\angle F_{1} P F_{2}=60^{\\circ}$, then the area of $\\triangle F_{1} PF_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/25 + y^2/9 = 1);PointOnCurve(P, G);Focus(G) = {F1,F2};AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "3*sqrt(3)",
    "fact_spans": "[[[6, 44], [66, 68]], [[48, 55]], [[2, 5]], [[58, 65]], [[6, 44]], [[2, 47]], [[48, 73]], [[74, 107]]]",
    "query_spans": "[[[109, 138]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$, the focus is $F$, and $K$ is the intersection point of the directrix $l$ of $C$ and the $x$-axis. A line passing through point $K$ with an inclination angle of $45^{\\circ}$ intersects $C$ at exactly one point $P(3, t)$. Then $t=?$",
    "fact_expressions": "C: Parabola;p: Number;P: Point;F: Point;l: Line;K:Point;H:Line;t:Number;p>0;Expression(C) = (y^2 = 2*p*x);Coordinate(P) = (3, t);Focus(C) = F;Directrix(C)=l;Intersection(l,xAxis)=K;PointOnCurve(K,H);Inclination(H)=ApplyUnit(45,degree);Intersection(H,C)=P",
    "query_expressions": "t",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 28], [40, 43], [84, 87]], [[10, 28]], [[94, 103]], [[32, 35]], [[46, 49]], [[36, 39], [59, 63]], [[81, 83]], [[105, 108]], [[10, 28]], [[2, 28]], [[94, 103]], [[2, 35]], [[40, 49]], [[36, 57]], [[58, 83]], [[64, 83]], [[81, 103]]]",
    "query_spans": "[[[105, 110]]]",
    "process": "Since the parabola $ C: y^{2} = 2px $ ($ p > 0 $) has focus $ F $, and $ K $ is the intersection point of the directrix $ l $ of $ C $ and the $ x $-axis, then $ K\\left(-\\frac{p}{2}, 0\\right) $. Since the line passing through point $ K $ has an inclination angle of $ 45^{\\circ} $, let the equation of the line be $ y = x + \\frac{p}{2} $. From \n\\[\n\\begin{cases}\ny = x + \\frac{p}{2} \\\\\ny^{2} = 2px\n\\end{cases},\n\\]\nwe obtain $ y^{2} - 2py + p^{2} = 0 $, that is, $ (y - p)^{2} = 0 $, so $ y = p $. Also, $ p = 3 + \\frac{p}{2} $, solving gives $ p = 6 $, thus $ t = 6 $."
  },
  {
    "text": "If $P$ is any point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{15}=1$, and $EF$ is any diameter of the circle $N$: $(x-1)^{2}+y^{2}=4$, then the range of values of $\\overrightarrow{PE} \\cdot \\overrightarrow{PF}$ is?",
    "fact_expressions": "P: Point;G: Ellipse;Expression(G) = (x^2/16 + y^2/15 = 1);PointOnCurve(P, G);E: Point;F: Point;IsDiameter(LineSegmentOf(E, F), N) ;N: Circle;Expression(N) = (y^2 + (x - 1)^2 = 4)",
    "query_expressions": "Range(DotProduct(VectorOf(P, E), VectorOf(P, F)))",
    "answer_expressions": "[5,21]",
    "fact_spans": "[[[1, 4]], [[5, 44]], [[5, 44]], [[1, 49]], [[50, 55]], [[50, 55]], [[50, 87]], [[56, 80]], [[56, 80]]]",
    "query_spans": "[[[89, 145]]]",
    "process": "The center of circle N is at N(1,0) with radius 2. Since $\\overrightarrow{PE}\\cdot\\overrightarrow{PF}=(\\overrightarrow{NE}-\\overrightarrow{NP})\\cdot(\\overrightarrow{NF}-\\overrightarrow{NP})=\\overrightarrow{NE}\\cdot\\overrightarrow{NF}-\\overrightarrow{NP}\\cdot(\\overrightarrow{NE}+\\overrightarrow{NF})+\\overrightarrow{NP}^{2}=-|\\overrightarrow{NE}|\\cdot|\\overrightarrow{NF}|\\cdot\\cos\\pi-0+|\\overrightarrow{NP}|^{2}=-4+|\\overrightarrow{NP}|^{2}$. Also, for the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{15}=1$, we have $a=4$, $b=\\sqrt{15}$, $c=1$, so N(1,0) is the right focus of the ellipse. Let $P(x_{0},y_{0})$, $\\frac{x_{0}^{2}}{16}=15-\\frac{15x_{0}^{2}}{1}$, thus $|NP|\\in[3,5]$, $|NP|^{2}\\in[9,25]$, and $\\overrightarrow{PE}\\cdot\\overrightarrow{PF}\\in[5,21]$."
  },
  {
    "text": "The hyperbola $C$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0, b>0)$ has an asymptote $y=\\sqrt{2} x$. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-x^2/b^2 + y^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Expression(OneOf(Asymptote(C))) = (y = sqrt(2)*x)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[0, 61], [84, 87]], [[0, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[0, 82]]]",
    "query_spans": "[[[84, 93]]]",
    "process": "From the hyperbola equation $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$, it follows that its foci lie on the $y$-axis. Since one of its asymptotes is $y=\\sqrt{2}x$, we have $\\frac{a}{b}=\\sqrt{2}$, $\\frac{b}{a}=\\frac{\\sqrt{2}}{2}$, and $e=\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\frac{\\sqrt{6}}{2}$."
  },
  {
    "text": "Let the hyperbola $C$ pass through the point $(1 , 3)$ and have the same asymptotes as $\\frac{y^{2}}{3}-x^{2}=1$. Then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;H:Curve;G: Point;Coordinate(G) = (1, 3);PointOnCurve(G, C);Expression(H)=(y^2/3-x^2=1);Asymptote(C)=Asymptote(H)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/6 - x^2/2 = 1",
    "fact_spans": "[[[1, 7], [56, 59]], [[22, 47]], [[9, 19]], [[9, 19]], [[1, 19]], [[22, 47]], [[1, 54]]]",
    "query_spans": "[[[56, 64]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of an ellipse is equal to $\\frac{\\sqrt{3}}{3}$, and it has the same focal distance as the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$. Then the standard equation of the ellipse is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(G) = (x^2/16 - y^2/9 = 1);Eccentricity(H) = sqrt(3)/3;FocalLength(G) = FocalLength(H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "{x^2/75+y^2/50=1,y^2/75+x^2/50=1}",
    "fact_spans": "[[[31, 70]], [[0, 2], [78, 80]], [[31, 70]], [[0, 28]], [[0, 76]]]",
    "query_spans": "[[[78, 87]]]",
    "process": ""
  },
  {
    "text": "Draw tangents from the point $(1, \\frac{1}{2})$ to the circle $x^{2}+y^{2}=1$, with points of tangency $A$ and $B$. The line $AB$ passes exactly through the right focus and the upper vertex of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$. Then the standard equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Circle;A: Point;B: Point;J: Point;L1:Line;L2:Line;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + y^2 = 1);Coordinate(J) = (1, 1/2);TangentOfPoint(J, H)={L1,L2};TangentPoint(L1,H)=A;TangentPoint(L2,H)=B;PointOnCurve(RightFocus(G),LineOf(A,B));PointOnCurve(UpperVertex(G),LineOf(A,B))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5+y^2/4=1",
    "fact_spans": "[[[65, 119], [130, 132]], [[67, 119]], [[67, 119]], [[21, 37]], [[46, 49]], [[50, 53]], [[1, 20]], [], [], [[67, 119]], [[67, 119]], [[65, 119]], [[21, 37]], [[1, 20]], [[0, 40]], [[0, 53]], [[0, 53]], [[54, 123]], [[54, 127]]]",
    "query_spans": "[[[130, 139]]]",
    "process": "Let the tangent line to the circle $x^{2}+y^{2}=1$ passing through the point $(1,\\frac{1}{2})$ be $l: y-\\frac{1}{2}=k(x-1)$, i.e., $kx-y-k+\\frac{1}{2}=0$ ①. When the line $l$ is perpendicular to the $x$-axis, $k$ does not exist, and the equation of the line is $x=1$, which is exactly tangent to the circle $x^{2}+y^{2}=1$ at point $A(1,0)$ ②. When the line $l$ is not perpendicular to the $x$-axis, the distance from the origin to line $l$ is: $d=\\frac{|-k+\\frac{1}{2}|}{\\sqrt{k^{2}+1}}=1$, solving gives $k=-\\frac{3}{4}$, then the equation of line $l$ is $y=-\\frac{3}{4}x+\\frac{5}{4}$, which is tangent to the circle $x^{2}+y^{2}=1$ at point $B(\\frac{3}{5},\\frac{4}{5})$; thus, the slope of line $AB$ is $k_{1}=\\frac{0-\\frac{4}{5}}{1-\\frac{3}{5}}=-2$, and the equation of line $AB$ is $y=-2(x-1)$. Therefore, line $AB$ intersects the $x$-axis at point $A(1,0)$ and the $y$-axis at point $C(0,2)$. The right focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ is $(1,0)$, and the upper vertex is $(0,2)$. So $c=1$, $b=2$, giving $a^{2}=b^{2}+c^{2}=5$, and the equation of the ellipse is $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$."
  },
  {
    "text": "Let $P$ be a moving point on the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{3}=1$. Then the sum of the distances from $P$ to the two foci of this ellipse is?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/5 + y^2/3 = 1);PointOnCurve(P, G);J1: Point;J2: Point;Focus(G) = {J1, J2}",
    "query_expressions": "Distance(P, J1) + Distance(P, J2)",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[5, 42], [53, 55]], [[1, 4], [48, 51]], [[5, 42]], [[1, 46]], [], [], [[53, 60]]]",
    "query_spans": "[[[48, 67]]]",
    "process": "From the ellipse equation, find $a$, then use the definition of the ellipse to obtain the result. From $\\frac{x^{2}}{5}+\\frac{y^{2}}{3}=1$, we get $a^{2}=5$, so $a=\\sqrt{5}$. By the definition of the ellipse, the sum of the distances from point $P$ to the two foci of the ellipse is $2a=2\\sqrt{5}$."
  },
  {
    "text": "Given the parabola equation $y^{2} = -4x$, and the line $l$ with equation $2x + y - 4 = 0$. There is a moving point $A$ on the parabola, the distance from point $A$ to the $y$-axis is $m$, the distance from point $A$ to line $l$ is $n$, then the minimum value of $m + n$ is?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;Expression(G) = (y^2 = -4*x);Expression(l) = (2*x + y - 4 = 0);PointOnCurve(A,G);Distance(A, yAxis) = m;Distance(A, l) = n;m:Number;n:Number",
    "query_expressions": "Min(m + n)",
    "answer_expressions": "6*sqrt(5)/5-1",
    "fact_spans": "[[[21, 26], [78, 83]], [[2, 5], [44, 47]], [[52, 55], [56, 60], [73, 77]], [[2, 20]], [[21, 41]], [[44, 55]], [[56, 72]], [[73, 90]], [[69, 72]], [[87, 90]]]",
    "query_spans": "[[[92, 103]]]",
    "process": "By the given condition, the distance from point A to the directrix is equal to the distance from point A to the focus F, so the distance from A to the y-axis is equal to the distance from point A to the focus F minus 1. Draw a perpendicular line from focus F to the line $2x + y \\cdot 4 = 0$, at which point $m + n = |AF| + n \\cdot 1$ is minimized. Since $F(-1, 0)$, then $(|AF| + n)_{\\min} = \\frac{|-2 + 0 - 4|}{\\sqrt{5}} = \\frac{6\\sqrt{5}}{5}$, hence the minimum value of $m + n$ is $\\frac{6\\sqrt{5}}{5} - 1$."
  },
  {
    "text": "Given that the center of the ellipse is at the origin, one vertex is the focus of the parabola $y^{2}=8 x$, and the eccentricity is $\\frac{1}{2}$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Parabola;H: Ellipse;O:Origin;Expression(G) = (y^2 = 8*x);Center(H)=O;OneOf(Vertex(H))=Focus(G);Eccentricity(H)=1/2",
    "query_expressions": "Expression(H)",
    "answer_expressions": "{x^2/4+y^2/3=1,x^2/4+3*y^2/16=1}",
    "fact_spans": "[[[15, 29]], [[2, 4], [53, 55]], [[7, 9]], [[15, 29]], [[2, 9]], [[2, 32]], [[2, 51]]]",
    "query_spans": "[[[53, 61]]]",
    "process": "The focus of the parabola $y^{2}=8x$ is $F(2,0)$, so a vertex of the ellipse is $A(2,0)$. If the foci of the ellipse lie on the $x$-axis, then $a=2$, and the eccentricity is $e=\\frac{c}{a}=\\frac{1}{2}$, thus $c=1$, $b^{2}=a^{2}-c^{2}=3$, and the standard equation of the ellipse is $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$. If the foci of the ellipse lie on the $y$-axis, then $b=2$, and the eccentricity is $e=\\frac{c}{a}=\\frac{1}{2}$, so $b=2$, $b^{2}=a^{2}-c^{2}=3c^{2}=4$, solving gives $c^{2}=\\frac{4}{3}$, $a^{2}=\\frac{16}{3}$, thus the standard equation of the ellipse is $\\frac{x^{2}}{4}+\\frac{3y^{2}}{16}=1$. In conclusion, the required standard equations of the ellipse are $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ or $\\frac{x^{2}}{4}+\\frac{3y^{2}}{16}=1$."
  },
  {
    "text": "The coordinates of the focus of the parabola $C$: $x=4 y^{2}$ are?",
    "fact_expressions": "C: Parabola;Expression(C) = (x = 4*y^2)",
    "query_expressions": "Coordinate(Focus(C))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 19]], [[0, 19]]]",
    "query_spans": "[[[0, 26]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ has eccentricity $e=2$, then the length of the imaginary axis is?",
    "fact_expressions": "C: Hyperbola;b: Number;b>0;e:Number;Expression(C) = (x^2/4 - y^2/b^2 = 1);Eccentricity(C) = e ;e= 2",
    "query_expressions": "Length(ImageinaryAxis(C))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[2, 54]], [[10, 54]], [[10, 54]], [[58, 63]], [[2, 54]], [[2, 63]], [[58, 63]]]",
    "query_spans": "[[[2, 70]]]",
    "process": "According to the problem, $ a=2 $, $ \\frac{c}{a}=2 $, $ c=4 $, $ b=\\sqrt{c^{2}-a^{2}}=2\\sqrt{3} $, so the length of the imaginary axis is $ 2b=4\\sqrt{3} $."
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/8 + y^2/4 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 43]]]",
    "process": "Given $ c=2 $, $ b=2 $, $ a=2\\sqrt{2} $, the eccentricity of the required ellipse equation is $ \\frac{2}{2\\sqrt{2}} = \\frac{\\sqrt{2}}{2} $."
  },
  {
    "text": "Let $P$ be a point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ in the first quadrant, and let $F_{1}$, $F_{2}$ be the left and right foci respectively. If $|P F_{1}|-|P F_{2}|=\\frac{8}{3}$, then the standard equation of the circle with center $P$ and radius $|P F_{2}|$ is?",
    "fact_expressions": "G: Ellipse;H: Circle;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 + y^2/5 = 1);PointOnCurve(P,G);Quadrant(P)=1;LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(P, F1))-Abs(LineSegmentOf(P, F2)) = 8/3;Center(H)=P;Radius(H)=Abs(LineSegmentOf(P,F2))",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x-2)^2+(y-5/3)^2=25/9",
    "fact_spans": "[[[5, 42]], [[136, 137]], [[1, 4], [114, 117]], [[53, 60]], [[61, 68]], [[5, 42]], [[1, 52]], [[1, 52]], [[5, 76]], [[5, 76]], [[78, 111]], [[113, 137]], [[121, 137]]]",
    "query_spans": "[[[136, 144]]]",
    "process": "\\because P is a point in the first quadrant on the ellipse \\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1, F_{1}, F_{2} are the left and right foci respectively \\therefore |F_{1}F_{2}|=4, |PF_{1}|+|PF_{2}|=6 \\because |PF_{1}|-|PF_{2}|=\\frac{8}{3} \\therefore |PF_{1}|=\\frac{13}{3}, |PF_{2}|=\\frac{5}{3} \\because |PF_{2}|^{2}+|F_{1}F_{2}|^{2}=\\frac{25}{9}+16=\\frac{169}{9}, |PF_{1}|^{2}=\\frac{169}{9} \\therefore |PF_{2}|^{2}+|F_{1}F_{2}|^{2}=|PF_{1}|^{2}, i.e. \\angle PF_{2}F_{1}=90^{\\circ} \\therefore P(2,\\frac{5}{3}) \\therefore the standard equation of the circle with center P and radius |PF_{2}| is (x-2)^{2}+(y-\\frac{5}{3})^{2}=\\frac{25}{9}"
  },
  {
    "text": "Given a line $l$ with slope $k$ ($k \\neq 0$) that intersects the ellipse $\\frac{x^{2}}{3}+y^{2}=1$ at two distinct points $M$ and $N$, and the distances from $M$ and $N$ to the point $A(0,-1)$ are equal, then the range of values for $k$ is?",
    "fact_expressions": "l: Line;G: Ellipse;A: Point;M: Point;k:Number;Negation(k=0);Expression(G) = (x^2/3 + y^2 = 1);Coordinate(A) = (0, -1);Slope(l)=k;Intersection(l,G)={M,N};Negation(M=N);Distance(M,A)=Distance(N,A);N:Point",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-1,0)+(0,1)",
    "fact_spans": "[[[21, 26]], [[28, 55]], [[80, 90]], [[62, 66], [72, 75]], [[97, 100]], [[7, 20]], [[28, 55]], [[80, 90]], [[4, 26]], [[21, 70]], [[57, 70]], [[72, 95]], [[76, 79], [76, 79]]]",
    "query_spans": "[[[97, 107]]]",
    "process": "Let the equation of line $ l $ be $ y = kx + m $ ($ k \\neq 0 $), and let $ P $ be the midpoint of $ MN $, then we have $ AP \\perp MN $. Combining the line $ l $ with the ellipse, we obtain $ (1 + 3k^{2})x^{2} + 6mkx + 3m^{2} - 3 = 0 $, so $ x_{1} + x_{2} = -\\frac{6mk}{1 + 3k^{2}} $, and thus $ P $ is $ \\left( -\\frac{3mk}{1 + 3k^{2}}, \\frac{m}{1 + 3k^{2}} \\right) $. From $ k_{AP} = -\\frac{1}{k} $, we get $ m = \\frac{3k^{2} + 1}{2} $. Using $ a > 0 $, we can find the range of $ k $. Solution: According to the problem, let the equation of line $ l $ be $ y = kx + m $ ($ k \\neq 0 $), and let $ P $ be the midpoint of $ MN $, then we have $ AP \\perp MN $. Solving the system\n$$\n\\begin{cases}\ny = kx + m \\\\\n\\frac{x^{2}}{3} + y^{2} = 1\n\\end{cases}\n$$\nby eliminating $ y $, we get $ (1 + 3k^{2})x^{2} + 6mkx + 3m^{2} - 3 = 0 $. Let $ M(x_{1}, y_{1}) $, $ N(x_{2}, y_{2}) $, so $ x_{1} + x_{2} = -\\frac{6mk}{1 + 3k^{2}} $. Since $ AP \\perp MN $, we have $ \\frac{3k^{2} + m + 1}{3mk} = -\\frac{1}{k} $ ($ k \\neq 0 $), solving gives $ m = \\frac{3k^{2} + 1}{2} $. Since $ \\Delta = 36m^{2}k^{2} - 4(1 + 3k^{2})(3m^{2} - 3) = 36 \\times \\frac{3k^{2} + 1}{2} k^{2} - 4(1 + 3k^{2}) \\left( 3 \\left( \\frac{3k^{2} + 1}{2} \\right)^{2} - 3 \\right) = 9(1 + 3k^{2})(1 - k^{2}) > 0 $, solving yields $ -1 < k < 1 $ and $ k \\neq 0 $, hence $ k \\in (-1, 0) \\cup (0, 1) $."
  },
  {
    "text": "Given the parabola $y^{2}=4 x$, with focus $F$, point $A(2,2)$, and $P$ a point on the parabola, find the minimum value of $|P A|+|P F|$.",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (2, 2);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 16], [40, 43]], [[26, 34]], [[36, 39]], [[21, 24]], [[2, 16]], [[26, 34]], [[2, 24]], [[36, 46]]]",
    "query_spans": "[[[49, 69]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $A$, $B$ are the two endpoints of the major axis of $C$, and point $M$ is a point on $C$ satisfying $\\angle M A B=30^{\\circ}$, $\\angle M B A=45^{\\circ}$. Let the eccentricity of ellipse $C$ be $e$, then $e^{2}$=?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;A: Point;B: Point;Endpoint(MajorAxis(C)) = {A, B};M: Point;PointOnCurve(M, C);AngleOf(M, A, B) = ApplyUnit(30, degree);AngleOf(M, B, A) = ApplyUnit(45, degree);e: Number;Eccentricity(C) = e",
    "query_expressions": "e^2",
    "answer_expressions": "1-(\\sqrt{3}/3)",
    "fact_spans": "[[[2, 60], [71, 74], [88, 91], [152, 157]], [[2, 60]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[62, 65]], [[66, 70]], [[61, 82]], [[83, 87]], [[83, 95]], [[98, 123]], [[124, 150]], [[162, 165]], [[152, 165]]]",
    "query_spans": "[[[167, 176]]]",
    "process": "Let M(x_{0},y_{0}), A(-a,0), B(a,0). Since \\angle MAB = 30^{\\circ}, \\angle MBA = 45^{\\circ}, it follows that k_{BM} = \\frac{y_{0}}{x_{0}-a} = -1, k_{AM} = \\frac{y_{0}}{x_{0}+a} = \\frac{\\sqrt{3}}{3}, \\frac{x_{0}^{2}}{a^{2}} + \\frac{y_{0}^{2}}{b^{2}} = 1. Solving these three equations simultaneously and eliminating x_{0}, y_{0} yields \\frac{b^{2}}{a^{2}} = \\frac{\\sqrt{3}}{3} = 1 - e^{2}, e^{2} = 1 - \\frac{\\sqrt{3}}{3}"
  },
  {
    "text": "The eccentricity of the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$ is equal to?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/4 - y^2/2 = 1)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[0, 43]], [[0, 43]]]",
    "query_spans": "[[[0, 50]]]",
    "process": "Given from the problem: $a^{2}=4$, $b^{2}=2$, so $c^{2}=a^{2}+b^{2}=6$, that is, $c=\\sqrt{6}$; thus, the eccentricity is $e=\\frac{c}{a}=\\frac{\\sqrt{6}}{2}$."
  },
  {
    "text": "Given a point $M(2,-2)$ on the parabola $y^{2}=2x$, points $A$ and $B$ are two moving points on the parabola $C$ distinct from $M$, and $\\overrightarrow{M A} \\cdot \\overrightarrow{M B}=0$. Then, the maximum distance from point $M$ to the line $AB$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*x);M: Point;Coordinate(M) = (2, -2);PointOnCurve(M, C);A: Point;B: Point;PointOnCurve(A, C);PointOnCurve(B, C);Negation(M=A);Negation(M=B);DotProduct(VectorOf(M, A), VectorOf(M, B)) = 0",
    "query_expressions": "Max(Distance(M, LineOf(A, B)))",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[2, 16], [38, 44]], [[2, 16]], [[19, 28], [109, 113], [47, 50]], [[19, 28]], [[2, 28]], [[29, 33]], [[34, 37]], [[29, 54]], [[29, 54]], [[29, 54]], [[29, 54]], [[56, 107]]]",
    "query_spans": "[[[109, 130]]]",
    "process": "The parabola $ y^{2} = 2x $, $ A, B $ are two moving points on the parabola $ C $ distinct from $ M $. Let $ A\\left(\\frac{y_{1}^{2}}{2}, y_{1}\\right) $, $ B\\left(\\frac{y_{2}^{2}}{2}, y_{2}\\right) $. Let the equation of line $ AB $ be $ x = my + n $. Then  \n$$\n\\begin{cases}\nx = my + n \\\\\ny^{2} = 2x\n\\end{cases}\n$$  \nSimplifying yields $ y^{2} - 2my - 2n = 0 $. So $ y_{1} + y_{2} = 2m $, $ y_{1} \\cdot y_{2} = -2n $, $ \\Delta = 4m^{2} + 8n > 0 $. Since $ M(2, -2) $, then  \n$ \\overrightarrow{MA} = \\left(\\frac{y_{1}^{2}}{2} - 2, y_{1} + 2\\right) $, $ \\overrightarrow{MB} = \\left(\\frac{y_{2}^{2}}{2} - 2, y_{2} + 2\\right) $.  \nSince $ \\overrightarrow{MA} \\cdot \\overrightarrow{MB} = 0 $ (because $ MA \\perp MB $),  \n$$\n\\left(\\frac{y_{1}^{2}}{2} - 2\\right)\\left(\\frac{y_{2}^{2}}{2} - 2\\right) + (y_{1} + 2)(y_{2} + 2) = 0\n$$  \nSimplifying gives  \n$$\n\\frac{1}{4}(y_{1} + 2)(y_{2} + 2)\\left[(y_{1} - 2)(y_{2} - 2) + 4\\right] = 0\n$$  \nSo $ (y_{1} + 2)(y_{2} + 2) = 0 $ or $ (y_{1} - 2)(y_{2} - 2) + 4 = 0 $. Expanding and simplifying yields  \n$ y_{1}y_{2} + 2(y_{1} + y_{2}) + 4 = 0 $ or $ y_{1}y_{2} - 2(y_{1} + y_{2}) + 8 = 0 $.  \nSubstituting $ y_{1} + y_{2} = 2m $, $ y_{1} \\cdot y_{2} = -2n $ gives  \n$ 2m - n + 2 = 0 $ or $ 2m + n - 4 = 0 $, i.e., $ n = 2m + 2 $ or $ n = -2m + 4 $.  \nSince $ \\Delta = 4m^{2} + 8n > 0 $ must hold. When $ n = 2m + 2 $, substituting gives $ \\Delta = 4(m + 2)^{2} $, when $ m = -2 $, $ \\Delta = 0 $, not strictly greater than 0, so discard. When $ n = -2m + 4 $, substituting gives $ \\Delta = 4(m - 2)^{2} + 16 > 0 $, always holds. So $ n = -2m + 4 $. Then the equation of line $ AB $ is $ x = my - 2m + 4 $, i.e., $ x - 4 = m(y - 2) $. Thus line $ AB $ passes through the fixed point $ N(4, 2) $. When $ MN $ is perpendicular to line $ AB $, the distance from point $ M $ to line $ AB $ is maximized, and the maximum distance is $ |MN| = \\sqrt{(4 - 2)^{2} + (2 + 2)^{2}} = 2\\sqrt{5} $."
  },
  {
    "text": "Write the standard equation of a hyperbola whose asymptotes are given by $y=\\pm x$?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*x)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 = 1",
    "fact_spans": "[[[20, 23]], [[4, 23]]]",
    "query_spans": "[[[20, 28]]]",
    "process": "Assume without loss of generality that the hyperbola equation has foci on the x-axis, and the asymptotes are given by $ y = \\pm x $, then $ \\frac{b}{a} = 1 $."
  },
  {
    "text": "If a point $P$ on the ellipse $\\frac{x^{2}}{64}+\\frac{y^{2}}{36}=1$ has a distance of $4$ to one of its foci, then what is the distance from point $P$ to the other focus?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/64 + y^2/36 = 1);PointOnCurve(P, G);F1:Point;F2:Point;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2);Distance(P, F1) = 4",
    "query_expressions": "Distance(P,F2)",
    "answer_expressions": "12",
    "fact_spans": "[[[1, 40], [47, 48]], [[43, 46], [64, 68]], [[1, 40]], [[1, 46]], [], [], [[47, 53]], [[47, 74]], [[47, 74]], [[43, 61]]]",
    "query_spans": "[[[47, 80]]]",
    "process": "Since by the definition of an ellipse, for a point P on the ellipse $\\frac{x^{2}}{64}+\\frac{y^{2}}{36}=1$, the distance from P to one of its foci is 4, then the distance from point P to the other focus equals $2a-4=2*8-4=12$, so fill in 12."
  },
  {
    "text": "In $\\triangle {A B C}$, ${B C}={A B}$, $\\angle{A B C}=120°$, then the eccentricity of the hyperbola with foci at $A$, $B$ and passing through point $C$ is?",
    "fact_expressions": "A: Point;B: Point;C: Point;LineSegmentOf(B, C) = LineSegmentOf(A, B);AngleOf(A, B, C) = ApplyUnit(120, degree);E: Hyperbola;Focus(E) = {A, B};PointOnCurve(C, E)",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "(\\sqrt{3} + 1) / 2",
    "fact_spans": "[[[64, 67]], [[68, 71]], [[77, 82]], [[25, 38]], [[39, 60]], [[84, 87]], [[62, 87]], [[76, 87]]]",
    "query_spans": "[[[84, 93]]]",
    "process": ""
  },
  {
    "text": "Given that the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{5}=1$ is $(3,0)$, then the eccentricity of this hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Point;Expression(G) = (-y^2/5 + x^2/a^2 = 1);Coordinate(H) = (3, 0);RightFocus(G) = H",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3/2",
    "fact_spans": "[[[2, 44], [59, 62]], [[5, 44]], [[49, 56]], [[2, 44]], [[49, 56]], [[2, 56]]]",
    "query_spans": "[[[59, 69]]]",
    "process": "Since the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{5}=1$ is $(3,0)$, then from $c^{2}=a^{2}+b^{2}$, we get $a^{2}=9-5=4$, $e=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\frac{3}{2}$, choose C."
  },
  {
    "text": "The coordinates of the focus of the parabola $x^{2}=\\frac{1}{2} y$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = y/2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,-1/8)",
    "fact_spans": "[[[0, 24]], [[0, 24]]]",
    "query_spans": "[[[0, 31]]]",
    "process": ""
  },
  {
    "text": "Given that $P(x , y)$ is a moving point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, then the maximum value of $x+y$ is?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/16 + y^2/9 = 1);Coordinate(P) = (x1, y1);x1:Number;y1:Number;PointOnCurve(P, G)",
    "query_expressions": "Max(x1 + y1)",
    "answer_expressions": "5",
    "fact_spans": "[[[13, 51]], [[2, 12]], [[13, 51]], [[2, 12]], [[2, 12]], [[2, 12]], [[2, 57]]]",
    "query_spans": "[[[59, 70]]]",
    "process": "Let $ x + y = t $, then the problem is transformed into finding the range of $ t $ for which the line $ x + y = t $ and the ellipse have common points. From  \n\\[\n\\begin{cases}\ny = -x + t \\\\\n\\frac{x^{2}}{16} + \\frac{y^{2}}{9} = 1\n\\end{cases}\n\\]\neliminating $ y $ gives $ 25x^{2} - 32tx + 16t^{2} - 144 = 0 $,  \n$ \\therefore A = (-32t)^{2} - 100(16t^{2} - 144) = -576t^{2} + 14400 \\geqslant 0 $, $ < t < $,  \n$ \\therefore $ the maximum value of $ x + y $ is $ 5 $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{3}+y^{2}=1$, a line $l$ with slope $1$ intersects the ellipse $C$ at points $A$ and $B$, and $|A B|=\\frac{3 \\sqrt{2}}{2}$. Then the equation of line $l$ is?",
    "fact_expressions": "l: Line;C: Ellipse;A: Point;B: Point;Expression(C) = (x^2/3 + y^2 = 1);Slope(l) = 1;Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, B)) = 3*sqrt(2)/2",
    "query_expressions": "Expression(l)",
    "answer_expressions": "y = x + pm*1",
    "fact_spans": "[[[41, 46], [95, 100]], [[2, 33], [47, 52]], [[54, 57]], [[58, 61]], [[2, 33]], [[34, 46]], [[41, 63]], [[65, 93]]]",
    "query_spans": "[[[95, 105]]]",
    "process": "Let the equation of the line be $ y = x + b $. Solving the system  \n\\[\n\\begin{cases}\ny = x + b \\\\\n\\frac{x^{2}}{3} + y^{2} = 1\n\\end{cases}\n\\]  \nyields $ 4x^{2} + 6bx + 3b^{2} - 3 = 0 $.  \n$ x_{1} + x_{2} = -\\frac{6b}{4} $, $ x_{1}x_{2} = \\frac{3b^{2} - 3}{4} $.  \nSince $ |AB| = \\sqrt{1 + k^{2}} |x_{1} - x_{2}| $, if the point: the positional relationship between the line and the ellipse is intersection."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right vertices are denoted as $A_{1}$ and $A_{2}$ respectively. Point $M$ is a point on the ellipse $C$ distinct from $A_{1}$ and $A_{2}$. If the product of the slopes of lines $M A_{1}$ and $M A_{2}$ equals $-\\frac{1}{2}$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;A1: Point;A2: Point;LeftVertex(C) = A1;RightVertex(C) = A2;M: Point;PointOnCurve(M, C);Negation(M = A1);Negation(M = A2);Slope(LineOf(M, A1)) * Slope(LineOf(M, A2)) = -1/2",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 59], [88, 93], [163, 168]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[67, 74], [97, 104]], [[75, 82], [105, 112]], [[2, 82]], [[2, 82]], [[83, 87]], [[83, 115]], [[83, 115]], [[83, 115]], [[117, 161]]]",
    "query_spans": "[[[163, 174]]]",
    "process": "From the equation of the ellipse, we have A(-a,0), B(a,0). Let M(x_{0},y_{0}), so k_{AM}=\\frac{y_{0}}{x_{0}+a}, k_{BM}=\\frac{y_{0}}{x_{0}-a}. Then \\frac{y_{0}}{x_{0}+a}\\cdot\\frac{y_{0}}{x_{0}-a}=-\\frac{1}{2}, simplifying to: \\frac{y_{0}^{2}}{x_{0}^{2}-a^{2}}=-\\frac{1}{2}, \\textcircled{1} also \\frac{x_{0}^{2}}{a^{2}}+\\frac{y_{0}^{2}}{b^{2}}=1, yielding y_{0}^{2}=\\frac{b^{2}}{a^{2}}(a^{2}-x_{0}^{2}), that is, \\frac{y_{0}^{2}}{x_{0}^{2}-a^{2}}=-\\frac{b^{2}}{a^{2}}, \\textcircled{2} combining \\textcircled{1} and \\textcircled{2}, we get -\\frac{b^{2}}{a^{2}}=-\\frac{1}{2}, i.e., \\frac{a^{2}-c^{2}}{a^{2}}=\\frac{1}{2}, solving gives e=\\frac{\\sqrt{2}}{2}."
  },
  {
    "text": "Given that $P$ is a point on the parabola $y^{2}=x$ distinct from the origin $O$, $P Q \\perp x$-axis with foot $Q$, and a line parallel to the $x$-axis is drawn through the midpoint of $P Q$, intersecting the parabola at point $M$. The line $Q M$ intersects the $y$-axis at point $N$. Then $\\frac{|P Q|}{|N O|}$=?",
    "fact_expressions": "G: Parabola;Q: Point;M: Point;P: Point;O:Origin;Expression(G) = (y^2 = x);PointOnCurve(P,G);Negation(P=O);IsPerpendicular(LineSegmentOf(P,Q),xAxis);FootPoint(LineSegmentOf(P,Q),xAxis)=Q;L:Line;PointOnCurve(MidPoint(LineSegmentOf(P,Q)),L);IsParallel(L,xAxis);Intersection(L,G)=M;Intersection(LineOf(Q,M),yAxis)=N;N:Point",
    "query_expressions": "Abs(LineSegmentOf(P,Q))/Abs(LineSegmentOf(N,O))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[6, 18], [70, 73]], [[47, 50]], [[74, 78]], [[2, 5]], [[21, 26]], [[6, 18]], [[2, 28]], [[2, 28]], [[29, 43]], [[29, 50]], [], [[51, 69]], [[51, 69]], [[51, 78]], [[79, 96]], [[92, 96]]]",
    "query_spans": "[[[98, 121]]]",
    "process": "As shown in the figure, let $ P(t^{2},t) $, then $ Q(t^{2},0) $, so the midpoint $ H $ of $ PQ $ is $ (t^{2},\\frac{t}{2}) $. Let $ y=\\frac{t}{2} $ be substituted into the parabola $ y^{2}=x $, we get $ x=\\frac{t^{2}}{4} $, that is, point $ M(\\frac{t^{2}}{4},\\frac{t}{2}) $. Therefore, the equation of line $ MQ $ is: $ y=\\frac{\\frac{t}{2}-0}{\\frac{t^{2}}{4}-t^{2}}(x-t^{2}) $. Let $ x=0 $, we obtain $ y_{N}=\\frac{2t}{3} $, so $ \\frac{|PQ|}{|NO|}=\\frac{t}{\\frac{2t}{3}}=\\frac{3}{2} $. This problem mainly examines the standard equation of a parabola and the application of the relationship between a line and a parabola. The key to solving the problem lies in determining the coordinates of point $ M $, deriving the equation of line $ MQ $, and finding the coordinates of point $ N $. It primarily tests reasoning and computational ability, and belongs to a basic-level problem."
  },
  {
    "text": "Let the line $ l: x - 3y + c = 0 $ intersect the two asymptotes of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ $ (a > b > 0) $ at points $ M $ and $ N $, respectively. If the perpendicular bisector of segment $ MN $ passes through the right focus $ (c, 0) $ of the hyperbola $ C $, then the eccentricity of the hyperbola is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;N: Point;M: Point;a > b;b > 0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(l)=(x-3*y+c=0);l1:Line;l2:Line;Asymptote(C)={l1,l2};Intersection(l,l1)=M;Intersection(l,l2)=N;PointOnCurve(RightFocus(C),PerpendicularBisector(LineSegmentOf(M,N)));Coordinate(RightFocus(C)) = (c, 0);c:Number",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[1, 18]], [[19, 77], [111, 117], [131, 134]], [[27, 77]], [[27, 77]], [[91, 94]], [[87, 90]], [[27, 77]], [[27, 77]], [[19, 77]], [[1, 18]], [], [], [[19, 83]], [[1, 96]], [[1, 96]], [[98, 129]], [[111, 129]], [[121, 129]]]",
    "query_spans": "[[[131, 140]]]",
    "process": "Let $ M(x_{1},y_{1}) $, $ N(x_{2},y_{2}) $. Solving the system \n\\[\n\\begin{cases}\n\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=0 \\\\\nx-3y+c=0\n\\end{cases}\n\\]\nyields \n\\[\n(9b^{2}-a^{2})x^{2}-2ca^{2}x-a^{2}c^{2}=0.\n\\]\n\\[\nx_{1}+x_{2}=\\frac{2ca^{2}}{9b^{2}-a^{2}}, \\quad y_{1}+y_{2}=\\frac{1}{3}[(x_{1}+x_{2})+2c]=\\frac{6cb^{2}}{9b^{2}-a^{2}}.\n\\]\nThus, the midpoint of $ MN $ is \n\\[\n\\left( \\frac{ca^{2}}{9b^{2}-a^{2}}, \\frac{3cb^{2}}{9b^{2}-a^{2}} \\right).\n\\]\nSince the perpendicular bisector of segment $ MN $ passes through the right focus $ (c,0) $ of hyperbola $ C $, we have \n\\[\n\\frac{\\frac{3cb^{2}}{9b^{2}-a^{2}}}{\\frac{ca^{2}}{9b^{2}-a^{2}}-c} = -3,\n\\]\nsolving which gives \n\\[\n4b^{2}=a^{2}.\n\\]\nTherefore, \n\\[\n4c^{2}=5a^{2},\n\\]\ni.e., the eccentricity \n\\[\ne=\\frac{\\sqrt{5}}{2}.\n\\]"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $A$ lies on an asymptote of the hyperbola such that $A F_{1} \\perp A O$ (where $O$ is the origin), and line $A F_{1}$ intersects the hyperbola at point $B$, with $|A B|=|B F_{1}|$. Find the eccentricity of the hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;A: Point;F1: Point;O: Origin;B: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(A, Asymptote(G));IsPerpendicular(LineSegmentOf(A, F1), LineSegmentOf(A, O));Intersection(LineSegmentOf(A, F1), G) = B;Abs(LineSegmentOf(A, B)) = Abs(LineSegmentOf(B, F1))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 59], [90, 93], [143, 146], [172, 175]], [[5, 59]], [[5, 59]], [[85, 89]], [[69, 76]], [[123, 126]], [[147, 151]], [[77, 84]], [[5, 59]], [[5, 59]], [[2, 59]], [[2, 84]], [[2, 84]], [[85, 99]], [[101, 120]], [[133, 151]], [[153, 170]]]",
    "query_spans": "[[[172, 181]]]",
    "process": "By the symmetry of the hyperbola, assume without loss of generality that point A lies in the second quadrant, and let F_{1}(-c,0). Since AF_{1}\\bot AO, the distance from point F_{1} to the line bx+ay=0 is d=\\frac{|-bc|}{\\sqrt{b^{2}+a^{2}}}=b, so |AF_{1}|=b. Also, since |F_{1}O|=c, we have \\cos\\angle AF_{1}O=\\frac{b}{c}. Because |AB|=|BF_{1}|, it follows that |BF_{1}|=\\frac{1}{2}|AF_{1}|=\\frac{b}{2}. By the definition of the hyperbola, |BF_{2}|=|BF_{1}|+2a=2a+\\frac{b}{2}. In \\triangle BF_{1}F_{2}, by the law of cosines, \\cos\\angle AF_{1}O=\\frac{b}{c}=\\frac{b^{2}}{4}+4c^{2}-(2a+\\frac{b}{2})^{2}. Using c^{2}=a^{2}+b^{2}, simplifying yields b=a, so c=\\sqrt{2}a, hence the eccentricity e=\\frac{c}{a}=\\sqrt{2}."
  },
  {
    "text": "Point $P$ lies on the curve $C_{3}$: $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, point $Q$ lies on the curve $C_{1}$: $(x+5)^{2}+y^{2}=1$, and point $R$ lies on the curve $C_{2}$: $(x-5)^{2}+y^{2}=1$. Then the maximum value of $|P Q|-|P R|$ is?",
    "fact_expressions": "P: Point;C3: Curve;Expression(C3) = (x^2/16 - y^2/9 = 1);PointOnCurve(P, C3);Q: Point;C1: Curve;Expression(C1) = (y^2 + (x + 5)^2 = 1);PointOnCurve(Q, C1);R: Point;C2: Curve;Expression(C2) = (y^2 + (x - 5)^2 = 1);PointOnCurve(R, C2)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, Q)) - Abs(LineSegmentOf(P, R)))",
    "answer_expressions": "10",
    "fact_spans": "[[[0, 4]], [[5, 52]], [[5, 52]], [[0, 53]], [[54, 58]], [[59, 88]], [[59, 88]], [[54, 89]], [[90, 94]], [[95, 124]], [[95, 124]], [[90, 125]]]",
    "query_spans": "[[[127, 145]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\Gamma$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Let $P$ be a point on the right branch of $\\Gamma$, and $Q$ be a point on the extension of $PF_{2}$ such that $QF_{1} \\perp QF_{2}$. If $\\sin \\angle PF_{1}Q = \\frac{3}{5}$, then the range of the eccentricity of $\\Gamma$ is?",
    "fact_expressions": "Gamma: Hyperbola;Expression(Gamma) = (x^2/a^2 - y^2/b^2 = 1);a: Number;b: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(Gamma) = F1;RightFocus(Gamma) = F2;P: Point;PointOnCurve(P, RightPart(Gamma));Q: Point;PointOnCurve(Q, OverlappingLine(LineSegmentOf(P, F2)));IsPerpendicular(LineSegmentOf(Q, F1), LineSegmentOf(Q, F2));Sin(AngleOf(P, F1, Q)) = 3/5",
    "query_expressions": "Range(Eccentricity(Gamma))",
    "answer_expressions": "(1,2)",
    "fact_spans": "[[[2, 68], [97, 105], [198, 206]], [[2, 68]], [[15, 68]], [[15, 68]], [[15, 68]], [[15, 68]], [[77, 84]], [[85, 92]], [[2, 92]], [[2, 92]], [[93, 96]], [[93, 111]], [[112, 115]], [[112, 132]], [[134, 158]], [[161, 196]]]",
    "query_spans": "[[[198, 217]]]",
    "process": "Analysis: Let |PF_{1}| = m. Using the hyperbola definition and the given conditions, we can determine the lengths of the sides of the right triangle. Combining the Pythagorean theorem and the triangle inequality that the sum of two sides is greater than the third side, we can find the range of the eccentricity of the hyperbola. Specifically, let |PF_{1}| = m, then |PF_{2}| = m - 2a, |PQ| = \\frac{3}{5}m, |QF_{2}| = 2a - \\frac{2}{5}m, |QF_{1}| = \\frac{4}{5}m. Therefore, (2c)^{2} = \\frac{16}{25}m^{2} + (2a - \\frac{2}{5}m)^{2}, which simplifies to m^{2} - 2ma + 5a^{2} - 5c^{2} = 0. Also, |QF_{1}| + |QF_{2}| > |F_{1}F_{2}| implies \\frac{4}{5}m + 2a - \\frac{2}{5}m > 2c, yielding m > 5(c - a). Thus, the equation m^{2} - 2ma + 5a^{2} - 5c^{2} = 0 has a root greater than 5(c - a). Hence, 25(c - a)^{2} - 2 \\times 5 \\times (c - a) \\times a + 5a^{2} - 5c^{2} < 0, leading to e < 2. Since e > 1, we have e \\in (1, 2)."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, its two foci are $F_{1}(-c, 0)$ and $F_{2}(c, 0)$, respectively. Let $M$ be a point on the ellipse such that $\\overrightarrow {F_{1} M} \\cdot \\overrightarrow {F_{2} M}=3 c^{2}$. Then the range of values for the eccentricity of this ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;c: Number;F1: Point;F2: Point;M: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);Focus(G) = {F1, F2};PointOnCurve(M, G);DotProduct(VectorOf(F1, M), VectorOf(F2, M)) = 3*c^2",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[sqrt(5)/5, 1/2]",
    "fact_spans": "[[[2, 54], [99, 101], [176, 178]], [[4, 54]], [[4, 54]], [[62, 76]], [[62, 76]], [[79, 92]], [[95, 98]], [[4, 54]], [[4, 54]], [[2, 54]], [[62, 76]], [[79, 92]], [[2, 92]], [[94, 104]], [[106, 173]]]",
    "query_spans": "[[[176, 188]]]",
    "process": "Let point M(x,y), \\overrightarrow{F_{1}M}\\cdot\\overrightarrow{F_{2}M}=3c^{2}, (x-c,y)*(x+c,y)=x^{2}-c^{2}+y^{2}=3c^{2} \\Rightarrow x^{2}+y^{2}=4c^{2}. Also, since b^{2}x^{2}+a^{2}y^{2}=a^{2}b^{2}, combining the two equations gives x^{2}=\\frac{5a^{2}c^{2}-a^{4}}{c^{2}}. Since 0\\leqslant x^{2}\\leqslant a^{2}, we obtain (\\frac{\\sqrt{5}}{5},\\frac{1}{2}]. Hence, |\\frac{\\sqrt{5}}{5},\\frac{1}{2}"
  },
  {
    "text": "If the left focus of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ lies on the directrix of the parabola $y^{2}=2 p x(p>0)$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;H: Ellipse;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (x^2/25 + y^2/16 = 1);PointOnCurve(LeftFocus(H), Directrix(G))",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[45, 66]], [[72, 75]], [[1, 40]], [[48, 66]], [[45, 66]], [[1, 40]], [[1, 70]]]",
    "query_spans": "[[[72, 79]]]",
    "process": "From the properties of an ellipse, it is known that the left focus of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ is $(-3,0)$. Since the left focus of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ lies on the directrix of the parabola $y^{2}=2px$ $(p>0)$, we have $-\\frac{p}{2}=-3$, solving which gives $p=6$."
  },
  {
    "text": "Given that $F_{1}$, $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and $P$ is a point on the ellipse such that $\\angle F_{1} PF_{2}=90^{\\circ}$, find the minimum value of the eccentricity of the ellipse.",
    "fact_expressions": "F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;P: Point;PointOnCurve(P, G) = True;AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Min(Eccentricity(G))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 9]], [[11, 18]], [[2, 76]], [[2, 76]], [[19, 71], [81, 83], [121, 123]], [[19, 71]], [[21, 71]], [[21, 71]], [[21, 71]], [[21, 71]], [[77, 80]], [[77, 86]], [[87, 119]]]",
    "query_spans": "[[[121, 132]]]",
    "process": "Since $\\angle F_{1}PF_{2}=90^{\\circ}$, it follows that $|PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}$. Since $|PF_{1}|+|PF_{2}|=2a$ and $|F_{1}F_{2}|=2c$, we can solve to get $|PF_{1}||PF_{2}|=2a^{2}-2c^{2}$. Because $2a=|PF_{1}|+|PF_{2}|\\geqslant2\\sqrt{|PF_{1}||PF_{2}|}=2\\sqrt{2a^{2}-2c^{2}}$, simplifying yields $2c^{2}\\geqslanta^{2}$, that is, $e^{2}=\\frac{c^{2}}{a^{2}}\\geqslant\\frac{1}{2}$, so $e\\geqslant\\frac{\\sqrt{2}}{2}$."
  },
  {
    "text": "Given points $A(-\\sqrt{5}, 0) $, $B(\\sqrt{5}, 0)$ , $C(-1,0) $, $D(1,0)$ , $P(x_1, y_1)$, if the product of the slopes of lines $P A$ and $P B$ is $-\\frac{4}{5}$, denote $\\angle P C D=\\alpha$ , $\\angle P D C=\\beta$, then $\\frac{\\sin \\alpha+\\sin \\beta}{\\sin (\\alpha+\\beta)}$=?",
    "fact_expressions": "A: Point;B:Point;P: Point;C: Point;D: Point;Coordinate(A) = (-sqrt(5), 0);Coordinate(B) = (sqrt(5), 0);Coordinate(C) = (-1, 0);Coordinate(D) = (1, 0);Coordinate(P) = (x1, y1);Slope(LineOf(P,A))*Slope(LineOf(P,B))=-4/5;AngleOf(P,C,D)=alpha ;AngleOf(P,D,C)=beta;beta:Number;alpha:Number;x1:Number;y1:Number",
    "query_expressions": "(Sin(alpha) + Sin(beta))/Sin(alpha + beta)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 21]], [[22, 39]], [[65, 78]], [[42, 52]], [[54, 62]], [[2, 21]], [[22, 39]], [[42, 52]], [[54, 62]], [[65, 78]], [[81, 116]], [[118, 140]], [[142, 162]], [[142, 162]], [[118, 139]], [[65, 78]], [[65, 78]]]",
    "query_spans": "[[[164, 218]]]",
    "process": "By the given condition, $k_{PA} \\cdot k_{PB} = \\frac{y-0}{x+\\sqrt{5}} \\cdot \\frac{y-0}{x-\\sqrt{5}} = -\\frac{4}{5}$ $(x \\neq \\pm\\sqrt{5})$, simplifying yields $\\frac{x^{2}}{5} + \\frac{y^{2}}{4} = 1$ $(x \\neq \\pm\\sqrt{5})$. In the ellipse $\\frac{x^{2}}{5} + \\frac{y^{2}}{4} = 1$, $a = \\sqrt{5}$, $b = 2$, $c = 1$, so $C$, $D$ are the two foci of the ellipse $\\frac{x^{2}}{5} + \\frac{y^{2}}{4} = 1$. Therefore, $\\frac{\\sin\\alpha + \\sin\\beta}{\\sin(\\alpha + \\beta)} = \\frac{|PD| + |PC|}{|CD|} = \\frac{2a}{2c} = \\sqrt{5}$."
  },
  {
    "text": "It is known that the vertex of the parabola is at the origin, the axis of symmetry is the $x$-axis, and the focus lies on the line $3x - 4y - 12 = 0$. Then the equation of this parabola is?",
    "fact_expressions": "G: Parabola;H: Line;O: Origin;Expression(H) = (3*x - 4*y - 12 = 0);Vertex(G) = O;SymmetryAxis(G)=xAxis;PointOnCurve(Focus(G),H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y**2=16*x",
    "fact_spans": "[[[2, 5], [44, 47]], [[24, 41]], [[9, 11]], [[24, 41]], [[2, 11]], [[2, 20]], [[2, 41]]]",
    "query_spans": "[[[44, 52]]]",
    "process": ""
  },
  {
    "text": "If the focal distance of an ellipse is $8$, the foci lie on the $x$-axis, and the lines connecting one focus to the two endpoints of the minor axis are perpendicular to each other, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;FocalLength(G) = 8;PointOnCurve(Focus(G), xAxis);F: Point;A: Point;B: Point;Endpoint(MinorAxis(G)) = {A, B};IsPerpendicular(LineSegmentOf(F, A), LineSegmentOf(F, B))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/32 + y^2/16 = 1",
    "fact_spans": "[[[1, 3], [39, 41]], [[1, 9]], [[1, 18]], [], [], [], [[1, 30]], [[1, 37]]]",
    "query_spans": "[[[39, 48]]]",
    "process": ""
  },
  {
    "text": "Draw a perpendicular line from the right focus $F$ of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ to an asymptote of $C$, with the foot of the perpendicular at $A$, intersecting the other asymptote at point $B$. If $\\overrightarrow{F B}= \\lambda \\overrightarrow{A F}$ and $3 \\leq \\lambda \\leq 4$, find the range of values for the eccentricity of $C$.",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;F: Point;B: Point;A: Point;l1: Line;l2: Line;l: Line;lambda: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;Asymptote(C) = {l1, l2};PointOnCurve(F, l);IsPerpendicular(l, l1);FootPoint(l, l1) = A;Intersection(l, l2) = B;VectorOf(F, B) = lambda*VectorOf(A, F);3 <= lambda;lambda <= 4",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "[sqrt(6)/2, 2*sqrt(10)/5]",
    "fact_spans": "[[[1, 62], [70, 73], [183, 186]], [[9, 62]], [[9, 62]], [[66, 69]], [[97, 101]], [[86, 89]], [], [], [], [[158, 181]], [[9, 62]], [[9, 62]], [[1, 62]], [[1, 69]], [[70, 96]], [[0, 82]], [[0, 82]], [[0, 89]], [[0, 101]], [[103, 155]], [[158, 181]], [[158, 181]]]",
    "query_spans": "[[[183, 197]]]",
    "process": "Let the right focus be $ F(c,0) $, and let the equation of one asymptote be $ y = \\frac{b}{a}x $, and the equation of the other asymptote be $ y = -\\frac{b}{a}x $. Since $ FA \\perp OA $, the equation of $ FA $ is: $ y = -\\frac{a}{b}(x - c) $.  \n$ y = \\frac{b}{a}x $, $ \\frac{a}{b}(x $, $ y = -\\frac{b}{a}x \\Rightarrow x = \\frac{a^{2}c}{a^{2}+b^{2}} = \\frac{a^{2}}{c} $, i.e., $ x_{A} = \\frac{a^{2}}{c} $  \n$ \\boxed{2} $  \n$ \\begin{cases} y = -\\frac{a}{b}(x - c) \\\\ \\end{cases} \\Rightarrow x = \\frac{a^{2}c}{a^{2}-b^{2}} = \\frac{ca^{2}}{2a^{2}-c^{2}} $, i.e., $ x_{B} = \\frac{ca^{2}}{2a^{2}-c^{2}} $  \nAlso $ \\overrightarrow{FB} = \\lambda \\overrightarrow{AF} $,  \n$ \\therefore \\frac{c a^{2}}{2a^{2}-c^{2}} - c = \\lambda \\left( c - \\frac{a^{2}}{c} \\right) \\Rightarrow \\frac{c(c^{2}-a^{2})}{2a^{2}-c^{2}} = \\lambda \\times \\frac{c^{2}-a^{2}}{c} $  \nAlso $ 3 \\leqslant \\frac{e^{2}}{2-e^{2}} \\leqslant 4 $, i.e.,  \n$ \\begin{cases} \\frac{e^{2}}{2-e^{2}} \\leqslant 4 \\\\ \\frac{e^{2}}{2-e^{2}} \\geqslant 3 \\end{cases} $  \nSolving gives: $ \\frac{6}{5} \\leqslant e^{2} \\leqslant \\frac{8}{5} $, i.e., $ \\frac{\\sqrt{6}}{2} \\leqslant e \\leqslant \\frac{2\\sqrt{10}}{5} $"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, point $P$ is a point on the right branch of the hyperbola, and $M$ is the incenter of $\\triangle P F_{1} F_{2}$, satisfying $S_{\\Delta M P F_{1}}=S_{\\Delta M P F_{2}}+\\lambda S_{\\triangle M F_{1} F_{2}}$. If the eccentricity of this hyperbola is $3$, then $\\lambda=$? (Note: $S_{\\triangle M P F_{1}}$, $S_{\\triangle M P F_{2}}$, $S_{\\triangle M F_{1} F_{2}}$ are the areas of $\\triangle M P F_{1}$, $\\triangle M P F_{2}$, and $\\triangle M F_{1} F_{2}$, respectively)",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;P: Point;F1: Point;F2: Point;M: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));Incenter(TriangleOf(P, F1, F2)) = M;lambda:Number;Area(TriangleOf(M,P,F1))=Area(TriangleOf(M,P,F2))+lambda*Area(TriangleOf(M,F1,F2));Eccentricity(G) = 3",
    "query_expressions": "lambda",
    "answer_expressions": "1/3",
    "fact_spans": "[[[20, 79], [91, 94], [217, 220]], [[23, 79]], [[23, 79]], [[86, 90]], [[2, 9]], [[10, 17]], [[100, 103]], [[23, 79]], [[23, 79]], [[20, 79]], [[2, 85]], [[2, 85]], [[86, 99]], [[100, 132]], [[230, 239]], [[135, 214]], [[217, 228]]]",
    "query_spans": "[[[230, 241]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, a line passing through point $F$ intersects the parabola $C$ at points $M(x_{1}, y_{1})$ and $N(x_{2}, y_{2})$. If point $P(x_{2},-y_{2})$ and $S_{\\Delta M P F}=10$, then the slope of line $MN$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;H: Line;PointOnCurve(F, H);x1: Number;x2: Number;y1: Number;y2: Number;M: Point;N: Point;Coordinate(M) = (x1, y1);Coordinate(N) = (x2, y2);Intersection(H, C) = {M, N};P: Point;Coordinate(P) = (x2, -y2);Area(TriangleOf(M, P, F)) = 10",
    "query_expressions": "Slope(LineOf(M, N))",
    "answer_expressions": "pm*(2/5)",
    "fact_spans": "[[[2, 21], [38, 44]], [[2, 21]], [[25, 28], [30, 34]], [[2, 28]], [[35, 37]], [[29, 37]], [[47, 64]], [[67, 84]], [[47, 64]], [[67, 84]], [[46, 64]], [[67, 84]], [[46, 64]], [[67, 84]], [[35, 84]], [[86, 104]], [[86, 104]], [[106, 127]]]",
    "query_spans": "[[[129, 141]]]",
    "process": "Let the slope of line MN be $ k $, then line MN: $ y = k(x - 1) $; combining  \n\\[\n\\begin{cases}\ny = k(x - 1) \\\\\ny^2 = 4x,\n\\end{cases}\n\\]  \neliminating $ y $ gives $ k^{2}x^{2} - 2(k^{2} + 2)x + k^{2} = 0 $, then $ x_{1} + x_{2} = 2 + \\frac{4}{k^{2}} $, $ x_{1}x_{2} = 1 $, hence $ |MF| = x_{1} + 1 $, $ |PF| = x_{2} + 1 $; let the inclination angle of line MN be $ \\alpha $, then $ \\tan\\alpha = k $, hence $ \\sin\\angle MFP = |\\sin(\\pi - 2\\alpha)| = |\\sin 2\\alpha| = \\left|\\frac{2\\tan\\alpha}{1 + \\tan^{2}\\alpha}\\right| = \\frac{2|k|}{k^{2} + 1} $; thus  \n\\[\nS_{\\Delta MPF} = \\frac{1}{2}(x_{1} + 1)(x_{2} + 1)|\\sin 2\\alpha| = \\frac{1}{2}[x_{1}x_{2} + (x_{1} + x_{2}) + 1] \\cdot \\frac{2|k|}{k^{2} + 1} = \\frac{4}{|k|};\n\\]  \nlet $ \\frac{4}{|k|} = 10 $, solving gives $ k = \\pm \\frac{2}{5} $."
  },
  {
    "text": "From the right focus $F_{2}$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a perpendicular line $l$ to one of its asymptotes, with foot at $P$. Line $l$ intersects the other asymptote at point $Q$. If $\\overrightarrow{F_{2} Q}=3 \\overrightarrow{F_{2} P}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F2: Point;Q: Point;P: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G)=F2;l:Line;PointOnCurve(F2,l);L1:Line;L2:Line;OneOf(Asymptote(G))=L1;OneOf(Asymptote(G))=L2;Negation(L1=L2);IsPerpendicular(l,L1);FootPoint(l,L1)=P;Intersection(l,L2)=Q;VectorOf(F2,Q)=3*VectorOf(F2,P)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[1, 57], [69, 70], [162, 165]], [[4, 57]], [[4, 57]], [[61, 68]], [[101, 105]], [[85, 88]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 68]], [[78, 81], [89, 92]], [[0, 81]], [], [], [[69, 75]], [[69, 99]], [[69, 99]], [[68, 81]], [[68, 88]], [[69, 105]], [[106, 159]]]",
    "query_spans": "[[[162, 171]]]",
    "process": "From the given conditions, the asymptotes of the hyperbola are $ y = \\pm\\frac{b}{a}x $. Let the right focus be $ F_{2}(c,0) $. Without loss of generality, assume that line $ l $ is perpendicular to the line $ y = \\frac{b}{a}x $, then the equation of line $ l $ is $ y = -\\frac{a}{b}(x - c) $. Solving the system\n$$\n\\begin{cases}\ny = \\frac{b}{a}x \\\\\ny = -\\frac{a}{b}(x - c)\n\\end{cases}\n$$\ngives point $ P\\left(\\frac{a^{2}c}{a^{2} + b^{2}}, \\frac{abc}{a^{2} + b^{2}}\\right) $. Since $ a^{2} + b^{2} = c^{2} $, point $ P $ becomes $ \\left(\\frac{a^{2}}{c}, \\frac{ab}{c}\\right) $. Solving the system\n$$\n\\begin{cases}\ny = -\\frac{b}{a}x \\\\\ny = -\\frac{a}{b}(x - c)\n\\end{cases}\n$$\ngives point $ Q\\left(\\frac{a^{2}c}{a^{2} - b^{2}}, -\\frac{abc}{a^{2} - b^{2}}\\right) $. Given $ \\overrightarrow{F_{2}Q} = 3\\overrightarrow{F_{2}P} $, we have $ -\\frac{abc}{a^{2} - b^{2}} = 3 \\cdot \\frac{ab}{c} $, which implies $ -c^{2} = 3a^{2} - 3b^{2} = 3a^{2} - 3(c^{2} - a^{2}) $. Thus, $ c^{2} = 3a^{2} $, so the eccentricity of the hyperbola is $ e = \\frac{c}{a} = \\sqrt{\\frac{c^{2}}{a^{2}}} = \\sqrt{3} $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0, b>0$) has eccentricity $2$, left focus at $F_{1}$, and point $Q(0, \\sqrt{3} c)$ ($c$ is the semi-focal length). $P$ is a moving point on the right branch of hyperbola $C$, and the minimum value of $|P F_{1}|+|P Q|$ is $6$. Then the equation of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;c:Number;Q: Point;P: Point;F1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(Q) = (0, sqrt(3)*c);Eccentricity(C) = 2;LeftFocus(C) = F1;HalfFocalLength(C)=c;PointOnCurve(P, RightPart(C));Min(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, Q))) = 6",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[2, 63], [118, 124], [161, 167]], [[10, 63]], [[10, 63]], [[104, 107]], [[84, 103]], [[114, 117]], [[76, 83]], [[10, 63]], [[10, 63]], [[2, 63]], [[84, 103]], [[2, 71]], [[2, 83]], [[2, 112]], [[114, 131]], [[133, 158]]]",
    "query_spans": "[[[161, 172]]]",
    "process": "Let the right focus of the hyperbola be $ F_{2} $, then $ |PF_{1}| - |PF_{2}| = 2a $, so $ |PF_{1}| + |PQ| = 2a + |PF_{2}| + |PQ| $. The minimum value of $ |PF_{2}| + |PQ| $ is $ |QF_{2}| = \\sqrt{c^{2} + (\\sqrt{3}c)^{2}} = 2c $, so the minimum value of $ |PF_{1}| + |PQ| $ is $ 2a + 2c = 6 $. Given $ \\frac{c}{a} = 2 $, solving gives $ a = 1 $, $ c = 2 $, thus $ b^{2} = 3 $, hence the equation of the hyperbola is $ x^2 - \\frac{y^{2}}{3} = 1 $. This question examines the equation of a hyperbola, the definition of a hyperbola, and the eccentricity of a hyperbola, as well as computational ability, and is a medium-difficulty problem."
  },
  {
    "text": "Given the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has its right focus at $F(3,0)$. A line passing through point $F$ intersects the ellipse $E$ at points $A$ and $B$. If the midpoint of $AB$ has coordinates $(1,-1)$, then the equation of $E$ is?",
    "fact_expressions": "E: Ellipse;H: Line;A: Point;B: Point;F: Point;a:Number;b:Number;a>b;b>0;Expression(E)=(x^2/a^2+y^2/b^2=1);Coordinate(F) = (3, 0);RightFocus(E) = F;PointOnCurve(F, H);Intersection(H, E) = {A,B};Coordinate(MidPoint(LineSegmentOf(A, B))) = (1,-1)",
    "query_expressions": "Expression(E)",
    "answer_expressions": "x^2/18+y^2/9=1",
    "fact_spans": "[[[2, 58], [81, 86], [121, 124]], [[78, 80]], [[87, 90]], [[91, 94]], [[63, 71], [73, 77]], [[8, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[2, 58]], [[63, 71]], [[2, 71]], [[72, 80]], [[78, 96]], [[99, 119]]]",
    "query_spans": "[[[121, 129]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the directrix of the parabola $x^{2}=8 y$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 8*y)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=-2",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": "From the given condition, we have p=4, so the equation of the directrix is y=-2. Fill in y=-2"
  },
  {
    "text": "For the equation $\\frac{x^{2}}{25-m}+\\frac{y^{2}}{16+m}=1$ to represent an ellipse with foci on the $y$-axis, what is the range of values for $m$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(25 - m) + y^2/(m + 16) = 1);PointOnCurve(Focus(G), yAxis);m: Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(9/2, 25)",
    "fact_spans": "[[[55, 57]], [[1, 57]], [[46, 57]], [[59, 62]]]",
    "query_spans": "[[[59, 69]]]",
    "process": "Since the equation $\\frac{x^{2}}{25-m}+\\frac{y^{2}}{16+m}=1$ represents an ellipse with foci on the $y$-axis, it follows that $16+m>25-m>0$, solving which yields $\\frac{9}{2}<m<25$."
  },
  {
    "text": "Given the two foci of a hyperbola $F_{1}(-\\sqrt{5}, 0)$, $F_{2}(\\sqrt{5}, 0)$, and $P$ is a point on this hyperbola such that $P F_{1} \\perp P F_{2}$ and $|P F_{1}| \\cdot |P F_{2}| = 2$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;F1: Point;Coordinate(F1) = (-sqrt(5), 0);F2: Point;Coordinate(F2) = (sqrt(5), 0);Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 5], [63, 66], [127, 130]], [[11, 32]], [[11, 32]], [[36, 56]], [[36, 56]], [[2, 56]], [[58, 61]], [[58, 70]], [[72, 95]], [[96, 124]]]",
    "query_spans": "[[[127, 136]]]",
    "process": "From $ PF_{1} \\perp PF_{2} $, by the Pythagorean theorem we obtain: $ |PF_{1}|^{2} + |PF_{2}|^{2} = |F_{1}F_{2}|^{2} = 20 $. Thus, $ (|PF_{1}| - |PF_{2}|)^{2} = |PF_{1}|^{2} + |PF_{2}|^{2} - 2|PF_{1}| \\cdot |PF_{2}| = 20 - 2|PF_{1}| \\cdot |PF_{2}| $. Since $ |PF_{1}| \\cdot |PF_{2}| = 2 $, it follows that $ (|PF_{1}| - |PF_{2}|)^{2} = 16 $, so $ |PF_{1}| - |PF_{2}| = 4 = 2a $. Hence, $ a = 2 $, $ c = \\sqrt{5} $. The eccentricity of this hyperbola is $ \\frac{c}{a} = \\frac{\\sqrt{5}}{2} $."
  },
  {
    "text": "Given $F_{1}$, $F_{2}$ are the left and right foci of the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. There exists a point $P$ on the ellipse $E$ such that $|P F_{2}|=|F_{1} F_{2}|$, and the distance from $F_{2}$ to the line $P F_{1}$ equals $b$. Then the eccentricity of the ellipse $E$ is?",
    "fact_expressions": "E: Ellipse;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(E) = F1;RightFocus(E) = F2;P: Point;PointOnCurve(P, E) = True;Abs(LineSegmentOf(P, F2)) = Abs(LineSegmentOf(F1, F2));Distance(F2, LineOf(P,F1)) = b",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[19, 76], [83, 88], [154, 159]], [[19, 76]], [[149, 152]], [[26, 76]], [[26, 76]], [[26, 76]], [[2, 9]], [[10, 17], [125, 132]], [[2, 81]], [[2, 81]], [[91, 95]], [[83, 95]], [[98, 123]], [[125, 152]]]",
    "query_spans": "[[[154, 165]]]",
    "process": "First, from the given conditions, $\\triangle F_{2}PF_{1}$ is an isosceles triangle. Let $M$ be the midpoint of $PF_{1}$, and connect $F_{2}M$. By the definition of the ellipse and the given conditions, $|F_{2}M|=b$, $|PM|=a-c$. Using the Pythagorean theorem to set up an equation, the result can be obtained. Denote the focal distance of the ellipse as $2c$, that is, $|PF_{2}|=|F_{1}F_{2}|=2c$, then $\\triangle F_{2}PF_{1}$ is an isosceles triangle, so $\\angle F_{2}PF_{1}$ is acute. Let $M$ be the midpoint of $PF_{1}$, connect $F_{2}M$, then $F_{2}M\\bot PF_{1}$. From the definition of the ellipse, $|PF_{1}|=2a-|PF_{2}|=2a-2c$. Since the distance from $F_{2}$ to line $PF_{1}$ equals $b$, then $|F_{2}M|=b$, $|PM|=a-c$. By the Pythagorean theorem, $|F_{2}M|^{2}+|PM|^{2}=|PF_{2}|^{2}$, that is, $b^{2}+(a-c)^{2}=4c^{2}$. Rearranging gives $2c^{2}+ac-a^{2}=0$, so $2e^{2}+e-1=0$. Solving yields $e=\\frac{1}{2}$ or $e=-1$ (discarded)."
  },
  {
    "text": "Given the parabola $C$: $x^{2}=2 y$, draw a line $l$ through its focus $F$ with slope $\\frac{1}{2}$, intersecting $C$ at points $A$ and $B$. Then $|A B|$=?",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 2*y);F: Point;Focus(C) = F;Slope(l) = 1/2;PointOnCurve(F, l) = True;l: Line;Intersection(l, C) = {A, B};A: Point;B: Point",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "5/2",
    "fact_spans": "[[[2, 21], [23, 24], [53, 56]], [[2, 21]], [[26, 29]], [[23, 29]], [[30, 52]], [[22, 52]], [[47, 52]], [[47, 66]], [[57, 60]], [[61, 64]]]",
    "query_spans": "[[[68, 77]]]",
    "process": "The focus of the parabola has coordinates $(0,\\frac{1}{2})$. The equation of line $l$ is $y-\\frac{1}{2}=\\frac{1}{2}x$, that is, $y=\\frac{1}{2}x+\\frac{1}{2}$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$. $\\therefore\\begin{cases}x^{2}=2y\\\\y=\\frac{1}{2}x+\\frac{1}{2}\\end{cases}$ Then $x^{2}=x+1$, that is, $x^{2}-x-1=(_{2}=1,x_{1}x_{2}=-1|AB|=\\sqrt{1+(\\frac{1}{2})^{2}}\\times\\sqrt{(x_{1}+})\\frac{\\sqrt{5}}{2}$"
  },
  {
    "text": "Given points $M(-3,0)$, $N(3, 0)$, $B(1, 0)$, circle $C$ is tangent to line $MN$ at point $B$. Two lines passing through $M$ and $N$ respectively and tangent to circle $C$ intersect at point $P$. Then the equation of the trajectory of point $P$ is?",
    "fact_expressions": "M: Point;N: Point;B: Point;Coordinate(M) = (-3, 0);Coordinate(N) = (3, 0);Coordinate(B) = (1, 0);C: Circle;TangentPoint(C, LineOf(M, N)) = B;l1: Line;l2: Line;TangentOfPoint(M, C) = l1;TangentOfPoint(N, C) = l2;Intersection(l1, l2) = P;P: Point",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(x^2-y^2/8=1)&(x>1)",
    "fact_spans": "[[[2, 12], [58, 61]], [[15, 25], [62, 65]], [[27, 37], [52, 56]], [[2, 12]], [[15, 25]], [[27, 37]], [[38, 42], [66, 70]], [[38, 56]], [], [], [[57, 76]], [[57, 76]], [[57, 83]], [[79, 83], [85, 89]]]",
    "query_spans": "[[[85, 96]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $2 x^{2}-y^{2}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (2*x^2 - y^2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(2)*x",
    "fact_spans": "[[[0, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 28]]]",
    "process": "The hyperbola equation is transformed into $\\frac{x^{2}}{\\frac{1}{2}}-y^{3}=1$, $\\therefore a^{2}=\\frac{1}{2}, b^{2}=1$, $\\therefore a=\\frac{\\sqrt{2}}{2}, b=1$, the asymptote equations are $y=\\pm\\sqrt{2}x$."
  },
  {
    "text": "The asymptotes of the hyperbola are given by $x \\pm 2 y=0$, and the focal distance is $10$. What is the equation of this hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (x+pm*2*y = 0);FocalLength(G) = 10",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/20 - y^2/5 = pm*1",
    "fact_spans": "[[[0, 3], [34, 37]], [[0, 23]], [[0, 31]]]",
    "query_spans": "[[[34, 42]]]",
    "process": ""
  },
  {
    "text": "Given that point $A(m, 1)$ lies on the parabola $x^{2}=2 p y$ ($p>0$), $F$ is the focus of the parabola, and $|A F|=3$, then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;F: Point;m: Number;p>0;Expression(G) = (x^2 = 2*(p*y));Coordinate(A) = (m, 1);PointOnCurve(A, G);Focus(G) = F;Abs(LineSegmentOf(A, F)) = 3",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[13, 34], [42, 45]], [[61, 64]], [[2, 12]], [[38, 41]], [[3, 12]], [[16, 34]], [[13, 34]], [[2, 12]], [[2, 37]], [[38, 48]], [[50, 59]]]",
    "query_spans": "[[[61, 66]]]",
    "process": "Since |AF| = 1 + \\frac{p}{2} = 3, it follows that p = 4. Hence the answer is: 4. This question examines the definition of a parabola and tests computational solving ability."
  },
  {
    "text": "The right focus of hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $F$, and $O$ is the coordinate origin. A circle centered at $F$ with radius $|O F|$ intersects $C$ and the asymptote of $C$ in the first quadrant at points $M$ and $N$, respectively. The midpoint of segment $M F$ is $P$. If $|O N|=|O P|$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C)=(x^2/a^2 - y^2/b^2 = 1);a: Number;b: Number;a>0;b>0;RightFocus(C) = F;O: Origin;D: Circle;F:Point;F = Center(D);M: Point;N: Point;Quadrant(M) = 1;Quadrant(N) = 1;Radius(D)=Abs(LineSegmentOf(O,F));Intersection(D,C)=M;Intersection(D,Asymptote(C))=N;P: Point;P=MidPoint(LineSegmentOf(M,F));Abs(LineSegmentOf(O, N)) = Abs(LineSegmentOf(O, P))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 61], [100, 103], [104, 107], [162, 165]], [[0, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[0, 69]], [[70, 73]], [[98, 99]], [[80, 83], [66, 69]], [[79, 99]], [[120, 123]], [[124, 127]], [[98, 129]], [[98, 129]], [[87, 99]], [[98, 129]], [[98, 129]], [[141, 144]], [[130, 144]], [[147, 160]]]",
    "query_spans": "[[[162, 171]]]",
    "process": "Find the distance from F to the asymptote, then use the Pythagorean theorem to find |ON|. Solve the system of equations of circle F and the hyperbola to obtain the coordinates of point M. Use the midpoint formula to find the coordinates of point N, then set up an equation using |ON| = |OP| to calculate the eccentricity e. The asymptote of the hyperbola in the first quadrant is y = \\frac{b}{a}x, i.e., bx - ay = 0. F(c,0). The distance from F to the asymptote is d = \\frac{|bc|}{\\sqrt{b^{2} + a^{2}}} = b, \\therefore |ON| = 2\\sqrt{c^{2} - d^{2}} = 2\\sqrt{c^{2} - b^{2}} = 2a. The equation of circle F is (x - c)^{2} + y^{2} = c^{2}. Solving \\begin{cases} \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 \\\\ (x - c)^{2} + y^{2} = c^{2} \\end{cases}, we get \\begin{cases} x = \\frac{a^{2} + ac}{c} \\\\ y = \\frac{b\\sqrt{a^{2} + 2ac}}{c} \\end{cases} (since point M lies in the first quadrant), i.e., M\\left(\\frac{a^{2} + ac}{c}, \\frac{b\\sqrt{a^{2} + 2ac}}{c}\\right). (x - c) + y^{2} = c - x. \\because |ON| = |OP|, i.e., |OP|^{2} = 4a^{2}, \\frac{(a^{2} + ac + c^{2})^{2}}{4c^{2}} + \\frac{b^{2}(a^{2} + 2ac)}{4c^{2}} = 4a^{2}, simplifying yields c^{2} + 4ac - 12a^{2} = 0, i.e., e^{2} + 4e - 12 = 0,"
  },
  {
    "text": "Given that an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ is $x+y=0$, then $a$=?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (x + y = 0)",
    "query_expressions": "a",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 39]], [[57, 60]], [[5, 39]], [[2, 39]], [[2, 55]]]",
    "query_spans": "[[[57, 62]]]",
    "process": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}} - y^{2} = 1$ ($a > 0$) are given by $\\frac{x}{a} \\pm y = 0$. Since one of the asymptotes of this hyperbola is $x + y = 0$, it follows that $\\frac{1}{a} = 1$, solving which yields $a = 1$."
  },
  {
    "text": "Given that $P$ is a point on the parabola $C$: $y^{2}=4x$, $F$ is the focus of the parabola $C$, its directrix intersects the $x$-axis at point $N$, the line $NP$ intersects the parabola at another point $Q$, and $|PF|=3|QF|$, then the coordinates of point $P$ are?",
    "fact_expressions": "C: Parabola;P: Point;N: Point;F: Point;Q: Point;Expression(C) = (y^2 = 4*x);PointOnCurve(P, C);Focus(C) = F;Intersection(Directrix(C), xAxis) = N;Intersection(LineOf(N,P),C)=Q;Abs(LineSegmentOf(P, F)) = 3*Abs(LineSegmentOf(Q, F))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(3,pm*2*sqrt(3))",
    "fact_spans": "[[[6, 25], [34, 40], [44, 45], [67, 70]], [[2, 5], [96, 100]], [[54, 58]], [[30, 33]], [[75, 78]], [[6, 25]], [[2, 29]], [[30, 43]], [[44, 58]], [[59, 78]], [[80, 94]]]",
    "query_spans": "[[[96, 104]]]",
    "process": ""
  },
  {
    "text": "It is known that hyperbola $C$ shares the same foci with the ellipse $x^{2}+4 y^{2}=64$, and the line $x+\\sqrt{3} y=0$ is an asymptote of hyperbola $C$. Then, the equation of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Ellipse;H: Line;Expression(G) = (x^2 + 4*y^2 = 64);Expression(H) = (x + sqrt(3)*y = 0);Focus(C) = Focus(G);OneOf(Asymptote(C))=H",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/36 - y^2/12 = 1",
    "fact_spans": "[[[2, 8], [56, 62], [70, 76]], [[9, 29]], [[37, 55]], [[9, 29]], [[37, 55]], [[2, 35]], [[37, 68]]]",
    "query_spans": "[[[70, 81]]]",
    "process": "Given: The ellipse $ x^{2} + 4y^{2} = 64 $, i.e., $ \\frac{x^{2}}{64} + \\frac{y^{2}}{16} = 1 $, has foci $ F_{1}(-4\\sqrt{3}, 0) $, $ F_{2}(4\\sqrt{3}, 0) $. Let the hyperbola $ C $ have equation $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $, then $ c = 4\\sqrt{3} $. The line $ x + \\sqrt{3}y = 0 $, i.e., $ y = -\\frac{\\sqrt{3}}{3}x $, is an asymptote of hyperbola $ C $, so $ \\frac{b}{a} = \\frac{\\sqrt{3}}{3} $, and $ a^{2} + b^{2} = 48 $. Solving gives: $ a = 6 $, $ b = 2\\sqrt{3} $. Therefore, the equation of hyperbola $ C $ is $ \\frac{x^{2}}{36} - \\frac{y^{2}}{12} = $."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ with directrix $l$, point $P$ lies on the parabola, $P Q \\perp l$ at point $Q$, $M(2,0)$ does not coincide with the focus of the parabola, and $|P Q|=|P M|$, $\\angle M P Q=120^{\\circ}$, then $p$=?",
    "fact_expressions": "G: Parabola;p: Number;P: Point;Q: Point;M: Point;l: Line;p>0;Expression(G) = (y^2 = 2*(p*x));Directrix(G) = l;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, Q),l);FootPoint(LineSegmentOf(P, Q),l)=Q;Coordinate(M) = (2, 0);Negation(M=Focus(G));Abs(LineSegmentOf(P, Q)) = Abs(LineSegmentOf(P, M));AngleOf(M, P, Q) = ApplyUnit(120, degree)",
    "query_expressions": "p",
    "answer_expressions": "4/5",
    "fact_spans": "[[[2, 23], [36, 39], [71, 74]], [[124, 127]], [[31, 35]], [[55, 60]], [[62, 70]], [[27, 30]], [[5, 23]], [[2, 23]], [[2, 30]], [[31, 40]], [[41, 54]], [[41, 60]], [[62, 70]], [[61, 80]], [[82, 95]], [[96, 122]]]",
    "query_spans": "[[[124, 129]]]",
    "process": "As shown in the figure, let the focus of the parabola be $ F $, connect $ PF $. By the definition of the parabola, we know $ |PQ| = |PF| $. Since $ |PQ| = |PM| $, it follows that $ |PF| = |PM| $. Given $ PQ \\bot l $ and $ \\angle MPQ = 120^{\\circ} $, we obtain $ \\angle PMF = 60^{\\circ} $, so $ \\triangle PFM $ is an equilateral triangle. Thus, $ |MF| = 2 - \\frac{p}{2} $, and the coordinates of point $ P $ are $ \\left(1 + \\frac{p}{4},\\ \\frac{\\sqrt{3}}{2}\\left(2 - \\frac{p}{2}\\right)\\right) $. Substituting into the parabola equation $ y^{2} = 2px $ ($ p > 0 $), we get $ \\frac{3}{4}\\left(2 - \\frac{p}{2}\\right)^{2} = 2p\\left(1 + \\frac{p}{4}\\right) $, which simplifies to $ 5p^{2} + 56p - 48 = 0 $. Solving yields $ p = \\frac{4}{5} $ or $ p = -12 $ (discarded)."
  },
  {
    "text": "The line $y = kx + k$ always has two common points with the ellipse $\\frac{x^{2}}{m} + \\frac{y^{2}}{4} = 1$ whose foci lie on the $y$-axis. Then the range of real values for $m$ is?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*x + k);k: Number;G: Ellipse;Expression(G) = (y^2/4 + x^2/m = 1);m: Real;PointOnCurve(Focus(G),yAxis) = True;NumIntersection(H,G) = 2",
    "query_expressions": "Range(m)",
    "answer_expressions": "(1,4)",
    "fact_spans": "[[[0, 11]], [[0, 11]], [[2, 11]], [[21, 58]], [[21, 58]], [[67, 72]], [[12, 58]], [[0, 65]]]",
    "query_spans": "[[[67, 79]]]",
    "process": "Analysis: From the equation of the line, it is known that the line always passes through the point $(-1,0)$. For the line $y = kx + k$ to always have two common points with the ellipse, the point $(-1,0)$ must lie inside the ellipse, thereby allowing us to find the range of $m$. The line $y = kx + k$ always passes through the point $(-1,0)$, and since the line $y = kx + k$ always has common points with the ellipse, $\\therefore (-1,0)$ lies inside the ellipse, $\\therefore \\frac{1}{m} < 1$, $\\therefore m > 1'ax-bm\\therefore$ the range of real number $m$ is $(1,4)$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, a point $A$ on the parabola $C$ lies in the first quadrant and satisfies $|AF|=3$. Then, the equation of the circle with center at point $A$ and radius $AF$ is?",
    "fact_expressions": "C: Parabola;G: Circle;A: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(A,C);Quadrant(A)=1;Abs(LineSegmentOf(A, F)) = 3;Center(G)=A;Radius(G)=LineSegmentOf(A,F)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "(x-2)^2 + (y - 2\\sqrt{2})^2 = 9",
    "fact_spans": "[[[2, 21], [29, 35]], [[80, 81]], [[38, 41], [63, 67]], [[25, 28]], [[2, 21]], [[2, 28]], [[29, 41]], [[38, 47]], [[51, 60]], [[62, 81]], [[71, 81]]]",
    "query_spans": "[[[80, 86]]]",
    "process": "By the given condition, the parabola $ y^{2} = 4x $ has focus $ F(1,0) $. Let $ A(x_{0},y_{0}) $ ($ x_{0} > 0 $, $ y_{0} > 0 $). According to the definition of a parabola, $ |AF| = x_{0} + 1 = 3 $, solving gives $ x_{0} = 2 $. Substituting into the equation $ y^{2} = 4x $, and using $ y_{0} > 0 $, we get $ y_{0} = 2\\sqrt{2} $, so $ A(2,2\\sqrt{2}) $. Since $ |AF| = 3 $, the equation of the circle is: $ (x-2)^{2} + (y-2\\sqrt{2})^{2} = 9 $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has its right focus at $F(c , 0)$. A line passing through $F$ and perpendicular to the $x$-axis intersects the ellipse at point $P$. The tangent line $l$ to the ellipse at point $P$ intersects the $x$-axis at point $A$. Then the coordinates of point $A$ are?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;F: Point;P: Point;A: Point;c:Number;a > b;b > 0;l:Line;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (c, 0);RightFocus(G) = F;PointOnCurve(F, H);Intersection(H,G) = P;TangentOnPoint(P,G)=l;Intersection(l,xAxis)=A;IsPerpendicular(H,xAxis)",
    "query_expressions": "Coordinate(A)",
    "answer_expressions": "(a^2/c,0)",
    "fact_spans": "[[[2, 54], [86, 88], [102, 104]], [[4, 54]], [[4, 54]], [[83, 85]], [[71, 74], [59, 69]], [[91, 95], [97, 101]], [[118, 122], [124, 128]], [[59, 69]], [[4, 54]], [[4, 54]], [[107, 110]], [[2, 54]], [[59, 69]], [[2, 69]], [[70, 85]], [[83, 95]], [[96, 110]], [[107, 122]], [[75, 85]]]",
    "query_spans": "[[[124, 133]]]",
    "process": ""
  },
  {
    "text": "Draw a line with slope $1$ passing through the left focus of the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$, intersecting the ellipse $C$ at points $A$ and $B$ respectively. $O$ is the coordinate origin. Then $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}$=?",
    "fact_expressions": "C: Ellipse;G: Line;O: Origin;A: Point;B: Point;Expression(C) = (x^2/2 + y^2 = 1);Slope(G)=1;PointOnCurve(LeftFocus(C),G);Intersection(G,C)={A,B}",
    "query_expressions": "DotProduct(VectorOf(O, A), VectorOf(O, B))",
    "answer_expressions": "-1/3",
    "fact_spans": "[[[1, 33], [48, 53]], [[45, 47]], [[66, 69]], [[57, 61]], [[62, 65]], [[1, 33]], [[38, 47]], [[0, 47]], [[45, 65]]]",
    "query_spans": "[[[76, 127]]]",
    "process": "Let the equation of the line be: y = x + 1. Substituting into the ellipse equation gives: 3x^{2} + 4x = 0, solving yields: x_{A} = 0, x_{B} = -\\frac{4}{3}, so A(0,1), B(-\\frac{4}{3}, -\\frac{1}{3}). \\therefore \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = -\\frac{1}{3}"
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{m-3}+\\frac{y^{2}}{m+5}=1$ is $\\frac{4}{3}$, what is the equation of the directrix of this hyperbola?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/(m - 3) + y^2/(m + 5) = 1);Eccentricity(G) = 4/3",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=pm*(9*sqrt(2))/8",
    "fact_spans": "[[[2, 44], [66, 69]], [[5, 44]], [[2, 44]], [[2, 62]]]",
    "query_spans": "[[[66, 76]]]",
    "process": ""
  },
  {
    "text": "Draw a line with slope $2$ passing through the right focus of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$, intersecting the ellipse at points $A$ and $B$, $O$ being the origin. Find the area of $\\triangle OAB$.",
    "fact_expressions": "G: Ellipse;H: Line;O: Origin;A: Point;B: Point;Expression(G) = (x^2/5 + y^2/4 = 1);PointOnCurve(RightFocus(G), H);Slope(H)=2;Intersection(H,G) = {B, A}",
    "query_expressions": "Area(TriangleOf(O, A, B))",
    "answer_expressions": "5/3",
    "fact_spans": "[[[1, 38], [55, 57]], [[52, 54]], [[69, 72]], [[59, 62]], [[63, 66]], [[1, 38]], [[0, 54]], [[45, 54]], [[52, 68]]]",
    "query_spans": "[[[79, 99]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{3}+y^{2}=1$, respectively, and let points $A$ and $B$ lie on the ellipse. If $\\overrightarrow{F_{1} A}=5 \\overrightarrow {F_{2} B}$, then the coordinates of point $A$ are?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/3 + y^2 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);VectorOf(F1, A) = 5*VectorOf(F2, B)",
    "query_expressions": "Coordinate(A)",
    "answer_expressions": "(0, pm*1)",
    "fact_spans": "[[[19, 46], [62, 64]], [[19, 46]], [[1, 8]], [[9, 16]], [[1, 52]], [[1, 52]], [[53, 57], [124, 128]], [[58, 61]], [[53, 65]], [[58, 65]], [[68, 122]]]",
    "query_spans": "[[[124, 133]]]",
    "process": "The ellipse $\\frac{x^2}{3}+y^{2}=1$ has its foci on the $x$-axis, $a=\\sqrt{3}$, $b=1$, $c=\\sqrt{2}$. $\\therefore$ the coordinates of the foci are $F_{1}(-\\sqrt{2},0)$, $F_{2}(\\sqrt{2},0)$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $\\overrightarrow{F_{1}A}=(x_{1}+\\sqrt{2},y_{1})$, $\\overrightarrow{F_{2}B}=(x_{2}-\\sqrt{2},y_{2})$. $\\because \\overrightarrow{F_{1}A}=5\\overrightarrow{F_{2}B}$, $\\begin{cases}x_{1}+\\sqrt{2}=5x_{2}-5\\sqrt{2}\\\\y_{1}=5y_{2}\\end{cases} \\Rightarrow \\begin{cases}x_{2}=\\frac{x_{1}+6\\sqrt{2}}{5}\\\\y_{2}=\\frac{y_{1}}{5}\\end{cases}$. Since points $A$, $B$ lie on the ellipse, $\\frac{(x_{1}+6\\sqrt{2})^2}{75}+(\\frac{y_{1}}{5})^2=1$, solving gives: $x_{1}=0$, $y_{1}=\\pm1$, $\\therefore$ the coordinates of point $A$ are $(0,\\pm1)$."
  },
  {
    "text": "Given the line $l$: $\\sqrt{3} x - y - \\sqrt{3} = 0$ intersects the parabola $y^{2} = 4x$ at points $A$ and $B$, and intersects the $x$-axis at point $P$. If $\\overrightarrow{O P} = m \\overrightarrow{O A} + n \\overrightarrow{O B}$ $(m \\leq n)$, then $\\frac{n}{m} = $?",
    "fact_expressions": "l: Line;G: Parabola;O: Origin;P: Point;A: Point;B: Point;m:Number;n:Number;m<=n;Expression(G) = (y^2 = 4*x);Expression(l) = (sqrt(3)*x - y - sqrt(3) = 0);Intersection(l, G) = {A, B};Intersection(l,xAxis)=P;VectorOf(O, P) = m*VectorOf(O, A) + n*VectorOf(O, B)",
    "query_expressions": "n/m",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 34]], [[35, 49]], [[77, 155]], [[70, 74]], [[52, 55]], [[56, 59]], [[158, 171]], [[158, 171]], [[77, 155]], [[35, 49]], [[2, 34]], [[2, 61]], [[2, 74]], [[77, 155]]]",
    "query_spans": "[[[158, 173]]]",
    "process": "The line $ l: \\sqrt{3}x - y - \\sqrt{3} = 0 $ passes through the focus $ F(1,0) $ of the parabola. Substituting the equation of the line into the equation of the parabola gives  \n\\[\n\\begin{cases}\n\\sqrt{3}x - y - \\sqrt{3} = 0 \\\\\ny^{2} = 4x\n\\end{cases}\n\\]  \nSolving yields  \n\\[\n\\begin{cases}\nx = 3 \\\\\ny = 2\\sqrt{3}\n\\end{cases}\n\\quad \\text{or} \\quad\n\\begin{cases}\nx = \\frac{1}{3} \\\\\ny = -\\frac{2\\sqrt{3}}{3}\n\\end{cases}\n\\]  \nLet $ A(3, 2\\sqrt{3}) $, $ B\\left(\\frac{1}{3}, -\\frac{2\\sqrt{3}}{3}\\right) $. Since $ \\overrightarrow{OP} = m\\overrightarrow{OA} + n\\overrightarrow{OB} $ ($ m \\leqslant n $), it follows that  \n$ (1, 0) = m(3, 2\\sqrt{3}) + n\\left(\\frac{1}{3}, -\\frac{2\\sqrt{3}}{3}\\right) $, then $ 0 = 2\\sqrt{3}m - \\frac{2\\sqrt{3}}{3}n $, so $ \\frac{n}{m} = 3 $."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, and let point $P$ lie on the hyperbola. When the area of $\\triangle F_{1} P F_{2}$ is $1$, what is the value of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$?",
    "fact_expressions": "G: Hyperbola;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 - y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);Area(TriangleOf(F1, P, F2)) = 1",
    "query_expressions": "DotProduct(VectorOf(P, F1), VectorOf(P, F2))",
    "answer_expressions": "0",
    "fact_spans": "[[[17, 45], [56, 59]], [[1, 8]], [[51, 55]], [[9, 16]], [[17, 45]], [[1, 50]], [[1, 50]], [[51, 60]], [[62, 93]]]",
    "query_spans": "[[[95, 156]]]",
    "process": "Using the constant area of $\\triangle F_{1}PF_{2}$ to find the coordinates of point $P$, and substituting into the formula for dot product yields the answer. Since the hyperbola equation is: $\\frac{x^{2}}{4}-y^{2}=1$, the coordinates of the two foci $F_{1}, F_{2}$ are: $(-\\sqrt{5},0), (\\sqrt{5},0)$, $\\therefore |F_{1}F_{2}|=2\\sqrt{5}$. When the area of $\\triangle F_{1}PF_{2}$ is $1$, let the coordinates of point $P$ be $(m,n)$. Hence $s=\\frac{1}{2}|F_{1}F_{2}|\\cdot|n|=1$, $\\therefore n=\\pm\\frac{\\sqrt{5}}{5}$. Without loss of generality, take $n=\\frac{\\sqrt{5}}{5}$, substitute $P(m,\\frac{\\sqrt{5}}{5})$ into the hyperbola equation to obtain: $m=\\pm\\frac{2\\sqrt{30}}{5}$. Take $m=\\frac{2\\sqrt{30}}{5}$, then $P(\\frac{2\\sqrt{30}}{5},\\frac{\\sqrt{5}}{5})$. $\\therefore \\overrightarrow{PF_{1}}=(-\\frac{2\\sqrt{30}}{5}-\\sqrt{5},-\\frac{\\sqrt{5}}{5})$, $\\overrightarrow{PF_{2}}=(-\\frac{2\\sqrt{30}}{5}+\\sqrt{5},-\\frac{\\sqrt{5}}{5})$. $\\therefore \\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=0$"
  },
  {
    "text": "The coordinates of the focus of the parabola $x=2 y^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (x = 2*y^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(1/8,0)",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": "According to the problem, the standard equation of the parabola is $ y^{2} = \\frac{1}{2}x $, so $ 2p = \\frac{1}{2} $, $ \\frac{p}{2} = \\frac{1}{8} $, and the parabola opens to the right with the focus on the positive x-axis; hence, the focus is $ \\left( \\frac{1}{8}, 0 \\right) $. [Note: This question mainly examines the method of finding the coordinates of the focus of a parabola and is a basic problem."
  },
  {
    "text": "The chord $AB$ of the parabola $y^{2}=4x$ is perpendicular to the $x$-axis. If $|AB|=4\\sqrt{3}$, then the distance from the focus $F$ to the line $AB$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);IsPerpendicular(LineSegmentOf(A, B), xAxis);Abs(LineSegmentOf(A, B)) = 4*sqrt(3);F: Point;Focus(G) = F",
    "query_expressions": "Distance(F, LineOf(A, B))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[60, 65]], [[60, 65]], [[0, 29]], [[16, 30]], [[32, 50]], [[54, 57]], [[0, 57]]]",
    "query_spans": "[[[54, 70]]]",
    "process": "Assume point A is above the x-axis. According to the problem, $ y_{A} = 2\\sqrt{3} $, then $ x_{A} = \\frac{12}{4} = 3 $. The focus of the parabola has coordinates $ F(1,0) $, so the distance from focus F to line AB is $ 3 - 1 = 2 $."
  },
  {
    "text": "If a point $M(x, y)$ always satisfies the relation $\\sqrt{x^{2}+(y-5)^{2}}-\\sqrt{x^{2}+(y+5)^{2}}=8$ during its motion, then what is the trajectory equation of point $M$?",
    "fact_expressions": "M: Point;Coordinate(M) = (x, y);x:Number;y:Number;sqrt(x^2+(y-5)^2)-sqrt(x^2+(y+5)^2)=8",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "{(y^2/16 - x^2/9 = 1) & (y < 0)}",
    "fact_spans": "[[[2, 12], [76, 80]], [[2, 12]], [[3, 12]], [[3, 12]], [[24, 73]]]",
    "query_spans": "[[[76, 88]]]",
    "process": "The equation $\\sqrt{x^{2}+(y-5)^{2}}-\\sqrt{x^{2}+(y+5)^{2}}$ represents the difference of distances from point $M(x,y)$ to the fixed points $(0,5)$, $(0,-5)$ being $8$. $\\because 8<10$, $\\therefore$ the locus of point $M(x,y)$ is the lower branch of a hyperbola with the two fixed points as foci. $\\because 2a=8$, $c=5$, $\\therefore b=3$, $\\therefore$ the equation of the locus of point $M$ is $\\frac{y^{2}}{16}-\\frac{x^{2}}{9}=1$ $(y<0)$."
  },
  {
    "text": "It is known that point $P$ lies on the right branch of the hyperbola $\\frac{x^{2}}{12}-\\frac{y^{2}}{4}=1$, and the area of the triangle with vertices at point $P$ and the foci $F_{1}$, $F_{2}$ is $4$. Then the coordinates of point $P$ are?",
    "fact_expressions": "G: Hyperbola;P: Point;F1:Point;F2:Point;Expression(G) = (x^2/12 - y^2/4 = 1);PointOnCurve(P,RightPart(G));Area(TriangleOf(P,F1,F2))=4;Focus(G)={F1,F2}",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(\\sqrt{15}, \\pm 1)",
    "fact_spans": "[[[7, 46]], [[2, 6], [55, 59], [93, 97]], [[62, 69]], [[70, 77]], [[7, 46]], [[2, 52]], [[55, 91]], [[7, 77]]]",
    "query_spans": "[[[93, 102]]]",
    "process": "Let P(x,y), x>0. From the given condition, c^{2}=a^{2}+b^{2}=12+4=16, so c=4, then F_{1}(-4,0), F_{2}(4,0). From the given condition, S_{\\triangle PF_{1}F_{2}}=\\frac{1}{2}|F_{1}F_{2}||y|=4|y|=4 \\Rightarrow y=\\pm1. Substituting y=\\pm1 into \\frac{x^{2}}{12}-\\frac{y^{2}}{4}=1, we get x=\\sqrt{15}. Therefore, the coordinates of point P are (\\sqrt{15},\\pm1)."
  },
  {
    "text": "Given the ellipse $E$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$, and a line $l$ intersecting the ellipse at points $A$ and $B$. If the midpoint of segment $AB$ has coordinates $(\\frac{1}{2},-1)$, then what is the general equation of line $l$?",
    "fact_expressions": "l: Line;E: Ellipse;B: Point;A: Point;Expression(E) = (x^2/4 + y^2/2 = 1);Intersection(l, E) = {A, B};Coordinate(MidPoint(LineSegmentOf(A,B)))=(1/2,-1)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "2*x-8*y-9=0",
    "fact_spans": "[[[45, 50], [45, 50]], [[2, 44], [51, 53]], [[58, 61]], [[54, 57]], [[2, 44]], [[45, 63]], [[65, 96]]]",
    "query_spans": "[[[98, 110]]]",
    "process": "Let A(x₁, y₁), B(x₂, y₂) be substituted into the ellipse equation to obtain \\frac{x_{1}}{4}+\\frac{y_{1}}{2}=1. Subtracting the two equations and simplifying yields \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{1}{2}\\cdot\\frac{x_{1}+x_{2}}{y_{1}+y_{2}}=\\frac{1}{4}. Thus, the line equation is y+1=\\frac{1}{4}(x-\\frac{1}{2}), which simplifies to 2x-8y-9=0."
  },
  {
    "text": "The distance from the focus of the parabola $y^{2}=8 x$ to the asymptotes of the hyperbola $\\frac{x^{2}}{12}-\\frac{y^{2}}{4}=1$ is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (x^2/12 - y^2/4 = 1);Expression(H) = (y^2 = 8*x)",
    "query_expressions": "Distance(Focus(H),Asymptote(G))",
    "answer_expressions": "1",
    "fact_spans": "[[[18, 57]], [[0, 14]], [[18, 57]], [[0, 14]]]",
    "query_spans": "[[[0, 66]]]",
    "process": ""
  },
  {
    "text": "The length of the imaginary axis of the hyperbola $mx^{2}+y^{2}=1$ is twice the length of the real axis, then $m$=?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (m*x^2 + y^2 = 1);Length(ImageinaryAxis(G)) = 2*Length(RealAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "-1/4",
    "fact_spans": "[[[0, 19]], [[34, 37]], [[0, 19]], [[0, 32]]]",
    "query_spans": "[[[34, 39]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$, $F_{2}$ are the common foci of the ellipse $C_{1}$: $\\frac{x^{2}}{3}+y^{2}=1$ and the hyperbola $C_{2}$, and $P$ is a common point of $C_{1}$, $C_{2}$. If $O P=O F$, then the asymptotes of $C_{2}$ are?",
    "fact_expressions": "C1: Ellipse;Expression(C1) = (x^2/3 + y^2 = 1);C2: Hyperbola;F1: Point;F2: Point;Focus(C1) = {F1, F2};Focus(C2) = {F1, F2};P: Point;Intersection(C1, C2) = P;O: Origin;LineSegmentOf(O, P) = LineSegmentOf(O, F)",
    "query_expressions": "Expression(Asymptote(C2))",
    "answer_expressions": "y=pm*x",
    "fact_spans": "[[[18, 54], [75, 82]], [[18, 54]], [[55, 65], [83, 90], [107, 114]], [[2, 9]], [[10, 17]], [[2, 70]], [[2, 70]], [[71, 74]], [[71, 94]], [[96, 105]], [[96, 105]]]",
    "query_spans": "[[[107, 122]]]",
    "process": "In the ellipse $ C_{1} $, there is $ |PF_{1}| = \\sqrt{3} + \\frac{\\sqrt{6}}{3}x_{p} $, $ |PF_{2}| = \\sqrt{3} - \\frac{\\sqrt{6}}{3}x_{p} $. Because $ OP = OF $, $ \\angle F_{1}PF_{2} = 90^{\\circ} $, so $ \\frac{1}{2}|PF_{1}|\\cdot|PF_{2}| = 1^{2}\\tan\\frac{90^{\\circ}}{2} \\Rightarrow x_{p}^{2} = \\frac{3}{2} $. According to the problem, in the hyperbola $ C_{2} $, $ c^{2} = 2 $, $ ||PF_{1}| - |PF_{2}|| = 2\\frac{\\sqrt{6}}{3}x_{p} = 2a \\Rightarrow a = 1 $, so $ b^{2} = c^{2} - a^{2} = 1 $, therefore the asymptotes of $ C_{2} $ are $ y = \\pm x $."
  },
  {
    "text": "The line $y = x + 2$ intersects the ellipse $\\frac{x^{2}}{m} + \\frac{y^{2}}{4} = 1$ at points $A$ and $B$. If $|AB| = 3\\sqrt{2}$, then the value of $m$ is?",
    "fact_expressions": "G: Ellipse;m: Number;H: Line;A: Point;B: Point;Expression(G) = (y^2/4 + x^2/m = 1);Expression(H) = (y = x + 2);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 3*sqrt(2)",
    "query_expressions": "m",
    "answer_expressions": "12",
    "fact_spans": "[[[10, 47]], [[79, 82]], [[0, 9]], [[48, 51]], [[52, 55]], [[10, 47]], [[0, 9]], [[0, 57]], [[59, 77]]]",
    "query_spans": "[[[79, 86]]]",
    "process": "From \\begin{cases}y=x+2\\\\\\frac{x^{2}}{m}+\\frac{y^{2}}{4}=1\\end{cases} we obtain (4+m)x^{2}+4mx=0, so x_{A}=0, x_{B}=\\frac{-4m}{4+m}. Also |AB|=\\sqrt{(x_{B}-x_{A})^{2}+(y_{B}-y_{A})^{2}}=\\sqrt{1+k^{2}}|x_{B}-x_{A}|=\\sqrt{2}|x_{B}|, hence \\sqrt{2}\\left|\\frac{4m}{4+m}\\right|=3\\sqrt{2}. Since m>0, it follows that \\frac{4m}{4+m}=3, thus m=12."
  },
  {
    "text": "Given that $F$ is the right focus of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and a line $l$ passing through the origin intersects $C$ at points $M$ and $N$, then the range of $\\frac{1}{|M F|}+\\frac{4}{|N F|}$ is?",
    "fact_expressions": "l: Line;C: Ellipse;M: Point;F: Point;N: Point;O:Origin;Expression(C) = (x^2/4 + y^2/3 = 1);RightFocus(C)=F;PointOnCurve(O,l);Intersection(l, C) = {M, N}",
    "query_expressions": "Range(4/Abs(LineSegmentOf(N, F)) + 1/Abs(LineSegmentOf(M, F)))",
    "answer_expressions": "[9/4,13/3]",
    "fact_spans": "[[[57, 62]], [[6, 48], [63, 66]], [[68, 71]], [[2, 5]], [[72, 75]], [[54, 56]], [[6, 48]], [[2, 52]], [[53, 62]], [[57, 77]]]",
    "query_spans": "[[[79, 119]]]",
    "process": "The ellipse $ C: \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $ has $ a = 2 $, $ b = \\sqrt{3} $, $ c = 1 $, and the left focus can be taken as $ F' $. Connecting $ MF' $, $ NF $ gives quadrilateral $ MFNF $ as a parallelogram, thus $ |MF| + |NF| = |MF| + |MF| = 2a = 4 $. Let $ |MF| = x $, $ x \\in [1, 3] $, then $ |NF| = 4 - x $, and $ \\frac{1}{|MF|} + \\frac{4}{|Nl} f'(x) = \\frac{-1}{x^{2}} + \\frac{4}{(x-4)^{2}} \\frac{4}{NF} = \\frac{(3x-4)^{4} - x}{x^{2}(x-4)^{2}} $. It follows that $ f(x) $ decreases on $ [1, \\frac{4}{3}] $ and increases on $ (\\frac{4}{3}, 3] $. The minimum value of $ f(x) $ is $ f(\\frac{4}{3}) = \\frac{9}{4} $, $ f(1) = \\frac{7}{3} $, $ f(3) = \\frac{13}{3} $, so the maximum value of $ f(x) $ is $ \\frac{13}{3} $. Thus, the range of $ \\frac{1}{|MF|} + \\frac{4}{|NF|} $ is $ [\\frac{9}{4}, \\frac{13}{3}] $."
  },
  {
    "text": "The line passing through the right vertex of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with slope $3$ intersects both the left and right branches of the hyperbola. Then the range of the eccentricity of this hyperbola is (expressed in interval notation)?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;PointOnCurve(RightVertex(G), L) = True;Slope(L) = 3;L: Line;IsIntersect(L, LeftPart(G)) = True;IsIntersect(L, RightPart(G)) = True",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(\\sqrt{10}, +\\infty)",
    "fact_spans": "[[[1, 57], [73, 76], [88, 91]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[0, 71]], [[62, 71]], [[69, 71]], [[69, 85]], [[69, 85]]]",
    "query_spans": "[[[88, 110]]]",
    "process": "According to the problem, construct the inequality relation between the slope of the asymptote and 3, then use $ e = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} $ to find the range of eccentricity. A line passing through the right focus and parallel to an asymptote intersects the hyperbola at one point, and the slope of one asymptote is $ \\frac{b}{a} $. If a line with slope 3 intersects both the left and right branches of the hyperbola, then $ \\frac{b}{a} > 3 $. Thus, the eccentricity $ e = \\frac{c}{a} = \\sqrt{\\frac{a^{2} + b^{2}}{a^{2}}} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} > \\sqrt{10} $."
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ has two foci $F_{1}$ and $F_{2}$. Point $P$ lies on the hyperbola. If $P F_{1} \\perp P F_{2}$, then the distance from point $P$ to the $x$-axis is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);F1: Point;F2: Point;Focus(G) = {F1,F2};P: Point;PointOnCurve(P, G) = True;IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2)) = True",
    "query_expressions": "Distance(P, xAxis)",
    "answer_expressions": "16/5",
    "fact_spans": "[[[0, 39], [66, 69]], [[0, 39]], [[45, 52]], [[53, 60]], [[0, 60]], [[61, 65], [97, 101]], [[61, 70]], [[72, 95]]]",
    "query_spans": "[[[97, 111]]]",
    "process": "Let point P(x, y), since F_{1}(-5, 0), F_{2}(5, 0), PF_{1} \\bot PF_{2}, \\frac{0}{5} = -1, the distance from P to the x-axis is \\frac{16}{5}."
  },
  {
    "text": "It is known that the hyperbola $C$ is centered at the origin and has coordinate axes as its axes of symmetry. One asymptote of $C$ intersects the parabola $y^{2}=8x$ with focus $F$ at point $P$, and $|P F|=4$. Then, the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;G: Parabola;O:Origin;P: Point;F: Point;Expression(G) = (y^2 = 8*x);Focus(G)=F;Center(C)=O;SymmetryAxis(C)=axis;Intersection(OneOf(Asymptote(C)),G)=P;Abs(LineSegmentOf(P, F)) = 4",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "{sqrt(5),sqrt(5)/2}",
    "fact_spans": "[[[2, 8], [23, 26], [73, 76]], [[40, 54]], [[12, 14]], [[56, 60]], [[36, 39]], [[40, 54]], [[33, 54]], [[2, 14]], [[2, 22]], [[23, 60]], [[62, 71]]]",
    "query_spans": "[[[73, 82]]]",
    "process": "Let the coordinates of point P be (x_{0}, y_{0}). Since |PF| = 4, it follows that x_{0} + 2 = 4, so x_{0} = 2, meaning the coordinates of point P are (2, \\pm4). Without loss of generality, we may denote P as P(2, 4). When the foci of hyperbola C lie on the x-axis, the asymptotes have equations y = \\pm\\frac{bx}{a}; thus, \\frac{b}{a} = 2, and the eccentricity e = \\frac{c}{a} = \\frac{\\sqrt{a^{2}+b^{2}}}{a} = \\sqrt{5}. When the foci of hyperbola C lie on the y-axis, the asymptotes have equations y = \\pm\\frac{ax}{b}, so \\frac{a}{b} = 2, and the eccentricity e = \\frac{c}{a} = \\frac{\\sqrt{a^{2}+b^{2}}}{a} = \\frac{\\sqrt{5}}{2}."
  },
  {
    "text": "The equation of the hyperbola that shares the same asymptotes as the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$ and passes through the point $(-1,4)$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/4 = 1);G1: Hyperbola;Asymptote(G) = Asymptote(G1);H: Point;Coordinate(H) = (-1, 4);PointOnCurve(H, G1)",
    "query_expressions": "Expression(G1)",
    "answer_expressions": "y^2/12 - x^2/3 = 1",
    "fact_spans": "[[[1, 29]], [[1, 29]], [[49, 52]], [[0, 52]], [[39, 48]], [[39, 48]], [[38, 52]]]",
    "query_spans": "[[[49, 56]]]",
    "process": "According to the problem, assume the required hyperbola equation is: $x^{2}-\\frac{y^{2}}{4}=\\lambda$ ($\\lambda\\neq0$). Substituting the point $(-1,4)$ gives: $1-4=\\lambda \\Rightarrow \\lambda=-3$. Therefore, the required hyperbola equation is: $x^{2}-\\frac{y^{2}}{4}=-3 \\Rightarrow \\frac{y^{2}}{12}-\\frac{x^{2}}{3}=1$."
  },
  {
    "text": "Through an arbitrary point $P$ on the hyperbola $\\frac{x^{2}}{25}-\\frac{y^{2}}{16}=1$, draw a line parallel to the $x$-axis intersecting the two asymptotes at points $Q$ and $R$. Then $P Q \\cdot P R = $?",
    "fact_expressions": "G: Hyperbola;P: Point;Q: Point;R: Point;Expression(G) = (x^2/25 - y^2/16 = 1);PointOnCurve(P, G);Z: Line;L1: Line;L2: Line;Asymptote(G) = {L1, L2};PointOnCurve(P, Z);IsParallel(Z, xAxis);Intersection(Z, L1) = Q;Intersection(Z, L2) = R",
    "query_expressions": "LineSegmentOf(P, Q)*LineSegmentOf(P, R)",
    "answer_expressions": "25",
    "fact_spans": "[[[1, 41]], [[46, 49]], [[65, 68]], [[69, 72]], [[1, 41]], [[1, 49]], [], [], [], [[1, 64]], [[0, 58]], [[0, 58]], [[0, 74]], [[0, 74]]]",
    "query_spans": "[[[76, 93]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Let $P$ be a point on the right branch of the hyperbola such that $P F_{1} \\perp P F_{2}$. The line $P F_{2}$ intersects the $y$-axis at point $Q$, and $\\overrightarrow{F_{2} P}=\\frac{2}{3} \\overrightarrow{P Q}$. Then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F2: Point;F1: Point;Q: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, RightPart(C));IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));Intersection(LineOf(P, F2), yAxis) = Q;VectorOf(F2, P) = (2/3)*VectorOf(P, Q)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 64], [93, 96], [210, 216]], [[9, 64]], [[9, 64]], [[89, 92]], [[81, 88]], [[73, 80]], [[143, 147]], [[9, 64]], [[9, 64]], [[2, 64]], [[2, 88]], [[2, 88]], [[89, 101]], [[102, 125]], [[126, 147]], [[149, 208]]]",
    "query_spans": "[[[210, 222]]]",
    "process": "Solution 1: From the given condition, |PF_{1}|-|PF_{2}|=2a, |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}=4c^{2}. Therefore, |PF_{1}|\\cdot|PF_{2}|=\\frac{4c^{2}-4a^{2}}{2}=2b^{2}. Let P(x_{0},y_{0}), then |PF_{1}|\\cdot|PF_{2}|=2c\\cdot|y_{0}|, so |y_{0}|=\\frac{b^{2}}{c}. Since \\overrightarrow{F_{2}P}=\\frac{2}{3}\\overrightarrow{PQ}, we have x_{0}=\\frac{3}{5}c. Substituting P(\\frac{3}{5}c,\\frac{b^{2}}{c}) into the hyperbola equation yields 9e^{4}-50e^{2}+25=0. Solving gives e^{2}=5 or e^{2}=\\frac{5}{9}. Since e>1, we have e=\\sqrt{5}. Solution 2: Let O be the coordinate origin. It is easy to see that \\triangle QOF_{2} \\sim \\triangle F_{1}PF_{2}, so \\frac{|QF_{2}|}{|F_{1}F_{2}|}=\\frac{|OF_{2}|}{|PF_{2}|}. Let |PF_{2}|=2x. Since \\overrightarrow{F_{2}P}=\\frac{2}{3}\\overrightarrow{PQ}, we have |QF_{2}|=5x. Then \\frac{5x}{2c}=\\frac{c}{2x}, giving c=\\sqrt{5}x. \\overset{-}{2c}=2x'|_{PF_{1}^{2}+|PF_{2}|}^{2}=|F_{1}F_{2}|^{2}=4c^{2}, so |PF_{1}|=\\sqrt{|F_{1}F_{2}|^{2}-|PF_{2}|^{2}}=4x. Thus |PF_{1}|-|PF_{2}|=4x-2x=2a, yielding a=x. Therefore, e=\\frac{c}{a}=\\sqrt{5}."
  },
  {
    "text": "The eccentricity $e \\in (1, 2)$ of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{k}=1$, then the range of values for $k$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/k = 1);k: Number;e:Number;Eccentricity(G) = e;In(e, (1, 2)) = True",
    "query_expressions": "Range(k)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 38]], [[0, 38]], [[58, 61]], [[42, 56]], [[0, 56]], [[42, 56]]]",
    "query_spans": "[[[58, 68]]]",
    "process": ""
  },
  {
    "text": "Let the focus of the parabola $y^{2}=8x$ be $F$, and let point $P$ lie on this parabola with an abscissa of $4$. Then $|PF|$ equals?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;Expression(G) = (y^2 = 8*x);Focus(G) = F;PointOnCurve(P, G);XCoordinate(P)=4",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 15], [28, 31]], [[22, 26]], [[18, 21]], [[1, 15]], [[1, 21]], [[22, 32]], [[22, 40]]]",
    "query_spans": "[[[42, 51]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a point on the ellipse $4 x^{2}+y^{2}=1$, and $F$ is one of its foci, then the minimum value of $P F$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F: Point;Expression(G) = (4*x^2 + y^2 = 1);PointOnCurve(P, G);OneOf(Focus(G))=F",
    "query_expressions": "Min(LineSegmentOf(P, F))",
    "answer_expressions": "1-sqrt(3)/2",
    "fact_spans": "[[[6, 25]], [[2, 5]], [[29, 32]], [[6, 25]], [[2, 28]], [[6, 39]]]",
    "query_spans": "[[[41, 52]]]",
    "process": "4x^{2}+y^{2}=1 is rewritten as \\frac{x^{2}}{4}+y^{2}=1, so the ellipse's foci lie on the y-axis, giving a=1, b=\\frac{1}{2}, thus c=\\sqrt{a^{2}-b^{2}}=\\frac{\\sqrt{3}}{2}. The minimum value of PF is the distance from the vertex on the major axis closer to F to the focus F, i.e., the minimum value of PF is a-c=1-\\frac{\\sqrt{3}}{2}."
  },
  {
    "text": "Draw two tangents from point $M(2, -2p)$ to the parabola $x^{2} = 2py$ ($p > 0$), with points of tangency $A$ and $B$. If the vertical coordinate of the midpoint of segment $AB$ is $6$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;M: Point;l1: Line;l2: Line;p>0;Expression(G) = (x^2 = 2*(p*y));Coordinate(M) = (2, -2*p);TangentOfPoint(M, G) = {l1, l2};TangentPoint(l1, G) = A;TangentPoint(l2, G) = B;YCoordinate(MidPoint(LineSegmentOf(A, B))) = 6",
    "query_expressions": "p",
    "answer_expressions": "{1, 2}",
    "fact_spans": "[[[15, 36]], [[76, 79]], [[47, 50]], [[51, 54]], [[1, 14]], [], [], [[18, 36]], [[15, 36]], [[1, 14]], [[0, 41]], [[0, 54]], [[0, 54]], [[56, 74]]]",
    "query_spans": "[[[76, 83]]]",
    "process": ""
  },
  {
    "text": "Given that the line $x - \\sqrt{3} y + \\sqrt{3} = 0$ passes through the left focus $F$ of the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1$ $(a > b > 0)$, intersects the ellipse at points $A$ and $B$, and intersects the $y$-axis at point $C$, with $\\overrightarrow{F A} = 2 \\overrightarrow{F C}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "H: Line;Expression(H) = (x - sqrt(3)*y + sqrt(3) = 0);G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;LeftFocus(G) = F;PointOnCurve(F, H);A: Point;B: Point;Intersection(H, G) = {A, B};C: Point;Intersection(H, yAxis) = C;VectorOf(F, A) = 2*VectorOf(F, C)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[2, 29]], [[2, 29]], [[30, 84], [93, 95], [165, 167]], [[30, 84]], [[32, 84]], [[32, 84]], [[32, 84]], [[32, 84]], [[88, 91]], [[30, 91]], [[2, 91]], [[96, 99]], [[100, 103]], [[2, 105]], [[112, 116]], [[2, 116]], [[117, 162]]]",
    "query_spans": "[[[165, 173]]]",
    "process": "Since the line $x - \\sqrt{3}y + \\sqrt{3} = 0$ passes through the left focus $F$ of the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1$ $(a > b > 0)$, we have $F(-\\sqrt{3}, 0)$. Let $A(x, y)$. Since $\\overrightarrow{FA} = 2\\overrightarrow{FC}$, from the given condition we get $x - (-\\sqrt{3}) = 2[0 - (-\\sqrt{3})]$, so $x = \\sqrt{3}$. Also, since $A(x, y)$ lies on the line $x - \\sqrt{3}y + \\sqrt{3} = 0$, we have $y = 2$, thus $A(\\sqrt{3}, 2)$. From the conditions, we obtain\n$$\n\\begin{cases}\na^2 - b^2} = c^2 = 3 \\\\\n\\frac{3}{a^2} + \\frac{4}{b^2} = 1\n\\end{cases}\n$$\nSolving gives\n$$\n\\begin{cases}\na = 3 \\\\\nb = \\sqrt{6}\n\\end{cases}\n$$\nTherefore, the eccentricity is $e = \\frac{c}{a} = \\frac{\\sqrt{3}}{3}$"
  },
  {
    "text": "Given that $P$ is a point on the parabola $C$: $y^{2}=2 p x(p>0)$, the distance from point $P$ to the focus of parabola $C$ is $7$, and the distance from $P$ to the $y$-axis is $5$. Then $p=?$",
    "fact_expressions": "C: Parabola;p: Number;P: Point;p>0;Expression(C) = (y^2 = 2*(p*x));PointOnCurve(P, C);Distance(P, Focus(C)) = 7;Distance(P,yAxis) = 5",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[6, 32], [41, 47]], [[72, 75]], [[2, 5], [36, 40]], [[14, 32]], [[6, 32]], [[2, 35]], [[36, 57]], [[36, 70]]]",
    "query_spans": "[[[72, 77]]]",
    "process": "By the given condition, |PF| = x_{P} + \\frac{p}{2} = 5 + \\frac{p}{2} = 7, solving yields p = 4."
  },
  {
    "text": "Given a point $P$ on the hyperbola $\\frac{y^{2}}{16}-\\frac{x^{2}}{20}=1$ such that the distance from $P$ to focus $F_{1}$ is $9$, then the distance from point $P$ to $F_{2}$ is equal to?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2:Point;Expression(G) = (-x^2/20 + y^2/16 = 1);PointOnCurve(P, G);OneOf(Focus(G)) = F1;Distance(P, F1) = 9",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "17",
    "fact_spans": "[[[2, 42]], [[45, 48], [68, 72]], [[51, 58]], [[73, 80]], [[2, 42]], [[2, 48]], [[2, 58]], [[2, 66]]]",
    "query_spans": "[[[68, 86]]]",
    "process": "If point $p$ is on the left branch, then by the definition of a hyperbola, $|PF_{2}| - |PF_{1}| = 2a$, solving gives $|PF_{2}| = 17$. If $p$ is on the right branch, then $|PF_{1}| \\geqslant a + c = 10$, which does not satisfy the condition. Therefore, $p$ must be on the left branch."
  },
  {
    "text": "It is known that the center of the hyperbola is at the origin, one focus is $F_{1}(-\\sqrt{2}, 0)$, point $P$ lies on the hyperbola, and the midpoint coordinates of segment $P F_{1}$ are $(0, \\frac{1}{2})$. Then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;O:Origin;F1: Point;P: Point;Center(G)=O;OneOf(Focus(G))=F1;Coordinate(MidPoint(LineSegmentOf(P,F1))) = (0, 1/2);Coordinate(F1) = (-sqrt(2), 0);PointOnCurve(P, G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 5], [43, 46], [87, 90]], [[8, 10]], [[16, 37]], [[38, 42]], [[2, 10]], [[2, 37]], [[49, 84]], [[16, 37]], [[38, 47]]]",
    "query_spans": "[[[87, 96]]]",
    "process": "Set the equation of the hyperbola, use the midpoint coordinate formula to find the coordinates of point P, substitute its coordinates into the hyperbola's equation, and form another equation using the relationship among a, b, c. Solve the two equations to obtain the values of a and c. Then the eccentricity of the hyperbola can be found. According to the given focus coordinates, determine that the foci lie on the x-axis. Let the equation of the hyperbola be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$. Since one focus is $F_{1}(-\\sqrt{2},0)$, we have $c=\\sqrt{2} \\Rightarrow a^{2}+b^{2}=2$ ①. Since the midpoint of segment $PF_{1}$ has coordinates $(0,\\frac{1}{2})$, the coordinates of point P are $(\\sqrt{2},1)$. Substituting $(\\sqrt{2},1)$ into the hyperbola's equation gives $\\frac{2}{a^{2}}-\\frac{1}{b^{2}}=1$ ②. From ① and ②, we get $a^{2}=1, b^{2}=1$, so $a=1$. The eccentricity of the hyperbola is: $e=\\frac{c}{a}=\\sqrt{2}$."
  },
  {
    "text": "$AB$ is a chord of the parabola $y = x^2$. If the distance from the midpoint of $AB$ to the $x$-axis is $1$, then the maximum length of the chord $AB$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);Distance(MidPoint(LineSegmentOf(A, B)), xAxis) = 1",
    "query_expressions": "Max(Length(LineSegmentOf(A, B)))",
    "answer_expressions": "5/2",
    "fact_spans": "[[[6, 18]], [[6, 18]], [[0, 5]], [[0, 5]], [[0, 22]], [[24, 44]]]",
    "query_spans": "[[[47, 61]]]",
    "process": ""
  },
  {
    "text": "If the coordinates of the foci of an ellipse are $F_{1}(-1,0)$, $F_{2}(1,0)$, the eccentricity is $\\frac{2}{3}$, and a line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$, then the perimeter of $\\triangle A B F_{2}$ is?",
    "fact_expressions": "G: Ellipse;H: Line;F1: Point;F2: Point;A: Point;B: Point;Coordinate(F1) = (-1, 0);Coordinate(F2) = (1, 0);Focus(G) = {F1, F2};Eccentricity(G) = 2/3;PointOnCurve(F1, H);Intersection(H, G) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 4], [68, 70]], [[65, 67]], [[10, 23], [57, 64]], [[25, 37]], [[71, 74]], [[75, 78]], [[10, 23]], [[25, 37]], [[2, 37]], [[2, 55]], [[56, 67]], [[65, 80]]]",
    "query_spans": "[[[82, 108]]]",
    "process": "Let the standard equation of the ellipse be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. Since the coordinates of the foci of the ellipse are $F_{1}(-1,0)$, $F_{2}(1,0)$, and the eccentricity is $\\frac{2}{3}$, we have $c=1$, and $\\frac{c}{a}=\\frac{2}{3}$, solving gives $a=\\frac{3}{2}$. By the definition of the ellipse, the perimeter of $ABF_{2}$ is $|AB|+|AF_{2}|+|BF_{2}|=|AF_{1}|+|AF_{2}|+|BF_{1}|+|BF_{2}|=4a=4\\times\\frac{3}{2}=6$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2x$, point $P$ lies on $C$ in the first quadrant. The tangent line to the parabola $C$ at point $P$ intersects its directrix at point $N$, and the focus of the parabola is $F$. If $\\angle PNF=60^{\\circ}$, then the coordinates of point $P$ are?",
    "fact_expressions": "C: Parabola;P: Point;N: Point;F: Point;L:Line;Focus(C)=F;Expression(C) = (y^2 = 2*x);PointOnCurve(P,C);PointOnCurve(P,L);Quadrant(P)=1;TangentOfPoint(P,C)=L;Intersection(L,Directrix(C))=N;AngleOf(P, N, F) = ApplyUnit(60, degree)",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(3/2, sqrt(3))",
    "fact_spans": "[[[2, 21], [27, 30], [63, 66], [44, 50], [54, 55]], [[22, 26], [39, 43], [102, 106]], [[58, 62]], [[70, 73]], [], [[63, 73]], [[2, 21]], [[22, 31]], [[38, 53]], [[22, 37]], [[38, 53]], [[38, 62]], [[75, 100]]]",
    "query_spans": "[[[102, 111]]]",
    "process": "By the given conditions, the directrix of the parabola is $ x = -\\frac{1}{2} $ and $ F\\left(\\frac{1}{2}, 0\\right) $. If $ P(m, \\sqrt{2m}) $ with $ m > 0 $, and in the first quadrant $ y = \\sqrt{2x} $, then $ y' = \\frac{1}{\\sqrt{2x}} $, so the tangent line at $ P $ is $ y - \\sqrt{2m} = \\frac{1}{\\sqrt{2m}}(x - m) $. Letting $ x = -\\frac{1}{2} $, we have $ y - \\sqrt{2m} = -\\frac{\\sqrt{2}}{\\sqrt{2m}}\\left(\\frac{1}{2} + m\\right) $, that is, $ y = \\frac{2m - 1}{2\\sqrt{2m}} $, so $ N\\left(-\\frac{1}{2}, \\frac{2m - 1}{2\\sqrt{2m}}\\right) $. In summary, $ \\overrightarrow{NP} = \\left(m + \\frac{1}{2}, \\frac{2m + 1}{2\\sqrt{2m}}\\right) $, $ \\overrightarrow{NF} = \\left(1, \\frac{1 - 2m}{2\\sqrt{2m}}\\right) $, and since $ \\angle PNF = 60^{\\circ} $, $ \\cos\\angle PNF = \\frac{\\overrightarrow{NP} \\cdot \\overrightarrow{NF}}{|\\overrightarrow{NP}||\\overrightarrow{NF}|} = \\frac{}{\\sqrt{(m+}}\\frac{m + \\frac{1}{2} + \\frac{1 - 4m^{2}}{8m}}{1^{2} + \\frac{(2m + 1)^{2}}{11 + \\frac{(1 - 2m)^{2}}{2}}} = \\frac{1}{2} $, solving gives $ m = \\frac{3}{2} $, so $ P\\left(\\frac{3}{2}, \\sqrt{3}\\right) $."
  },
  {
    "text": "If a line passing through the focus of the parabola $y^{2}=8 x$ with an inclination angle of $\\frac{\\pi}{3}$ intersects the parabola at points $A$ and $B$, then $|A B|$=?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;Expression(G) = (y^2 = 8*x);PointOnCurve(Focus(G), H);Inclination(H)=pi/3;Intersection(H, G) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "32/3",
    "fact_spans": "[[[2, 16], [44, 47]], [[41, 43]], [[48, 51]], [[52, 55]], [[2, 16]], [[1, 43]], [[21, 43]], [[41, 55]]]",
    "query_spans": "[[[57, 66]]]",
    "process": "From the given condition, the focus of the parabola is (2,0). Therefore, the equation of line AB is y−0=√3(x−2), that is, y=√3x−2√3. Substituting y=√3x−2√3 into y²=8x yields 3x²−20x+12=0, so AB=√(1+3)."
  },
  {
    "text": "Given the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, a line passing through the point $P(2,1)$ intersects the hyperbola at points $A$ and $B$, such that $P$ is the midpoint of $AB$. What is the slope of the line $AB$?",
    "fact_expressions": "G: Hyperbola;H: Line;B: Point;A: Point;P: Point;Expression(G) = (x^2 - y^2/3 = 1);Coordinate(P) = (2, 1);PointOnCurve(P, H);Intersection(H, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Slope(LineOf(A,B))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 30], [47, 50]], [[44, 46]], [[55, 58]], [[51, 54]], [[33, 42], [63, 66]], [[2, 30]], [[33, 42]], [[32, 46]], [[44, 60]], [[63, 75]]]",
    "query_spans": "[[[77, 89]]]",
    "process": "Let points $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then \n$$\n\\begin{cases}\n\\frac{x_{1}+x_{2}}{2}=2 \\\\\n\\frac{y_{1}+y_{2}}{2}=1\n\\end{cases}\n$$\n, i.e.,\n$$\n\\begin{cases}\nx_{1}+x_{2}=4 \\\\\ny_{1}+y_{2}=2\n\\end{cases}\n$$\nFrom the given conditions we get\n$$\n\\begin{cases}\nx_{2}-\\frac{y^{2}}{3}=1 \\\\\nx_{2}^{2}-\\frac{y^{2}}{3}=1\n\\end{cases}\n$$\n, subtracting the two equations yields \n$ (x_{1}^{2}-x_{2}^{2})-\\frac{y_{2}-y_{2}}{3}=0 $, i.e., $ (x_{1}+x_{2})(x_{1}-x_{2})= $ So, the slope of line $ AB $ is $ k_{AB}=\\frac{y_{1}-y}{x_{1}-x}\\frac{3}{3} $"
  },
  {
    "text": "Given that the focus of the parabola $C$: $y^{2}=2 p x$ coincides with the right focus of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, then the standard equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);Focus(C) = RightFocus(G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[2, 23], [62, 68]], [[2, 23]], [[10, 23]], [[27, 54]], [[27, 54]], [[2, 60]]]",
    "query_spans": "[[[62, 75]]]",
    "process": "From the given information: the coordinates of the right focus of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ are $(1,0)$, so the focus of the parabola is $(1,0)$, and the standard equation of the parabola $C$ is $y^{2}=4x$."
  },
  {
    "text": "A line with inclination angle $\\frac{\\pi}{4}$ passes through the focus of the hyperbola $C$: $\\frac{x^{2}}{3}-y^{2}=1$ and intersects the hyperbola $C$ at points $A$ and $B$. Then $|A B|$=?",
    "fact_expressions": "G: Line;Inclination(G) = pi/4;C: Hyperbola;Expression(C) = (x^2/3 - y^2 = 1);PointOnCurve(Focus(C), G);A: Point;B: Point;Intersection(G, C) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[20, 22]], [[0, 22]], [[23, 56], [62, 68]], [[23, 56]], [[20, 59]], [[70, 73]], [[74, 77]], [[20, 79]]]",
    "query_spans": "[[[81, 90]]]",
    "process": "Let the equation of the line be combined with the hyperbola equation. From the standard equation of hyperbola C: \\frac{x^{2}}{3}-y^{2}=1, we know: a=\\sqrt{3}, b=1, so c=\\sqrt{a^{2}+b^{2}}=\\sqrt{3+1}=2. Thus, the coordinates of the foci are (\\pm2,0). By the symmetry of the hyperbola, assume without loss of generality that line AB passes through the right focus (2,0). Therefore, the equation of line AB is y-0=(\\tan\\frac{\\pi}{4})\\cdot(x-2) \\Rightarrow y=x-2. Combining with the hyperbola gives: \\begin{cases}\\frac{x^{2}}{3}-y^{2}=1\\\\y=x-2\\end{cases} \\Rightarrow 2x^{2}-12x+15. Hence, x_{1}+x_{2}=6, x_{1}x_{2}=\\frac{15}{2}. Let A(x_{1},y_{1}), B(x_{2},y_{2}). Then |AB|=\\sqrt{1+(\\tan\\frac{\\pi}{4})^{2}}|x_{1}-x_{2}|=\\sqrt{2}\\cdot\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\sqrt{2}\\times\\sqrt{6^{2}-4\\times\\frac{15}{2}}=2\\sqrt{3}"
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=2x$ is $F$, point $P$ is a moving point on the parabola, and there is a point $A(3, 2)$, then when $|PA|+|PF|$ takes the minimum value, the coordinates of point $P$ are?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 2*x);Coordinate(A) = (3, 2);Focus(G) = F;PointOnCurve(P, G);WhenMin(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2, 2)",
    "fact_spans": "[[[2, 16], [29, 32]], [[39, 50]], [[24, 28], [68, 72]], [[20, 23]], [[2, 16]], [[39, 50]], [[2, 23]], [[24, 36]], [[52, 68]]]",
    "query_spans": "[[[68, 77]]]",
    "process": ""
  },
  {
    "text": "The number of intersection points between the line $y=2x+3$ and the curve $\\frac{y^{2}}{9}-\\frac{x|x|}{4}=1$ is?",
    "fact_expressions": "G: Line;H: Curve;Expression(G) = (y = 2*x + 3);Expression(H) = (-x*Abs(x)/4 + y^2/9 = 1)",
    "query_expressions": "NumIntersection(G, H)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 11]], [[12, 48]], [[0, 11]], [[12, 48]]]",
    "query_spans": "[[[0, 55]]]",
    "process": "Analyze the problem, discuss the range of values for $ x $, and derive the expressions of the curve $ \\frac{y^{2}}{9} - \\frac{x|x|}{4} = 1 $ when $ x \\geqslant 0 $ and $ x < 0 $, respectively. Combine the line $ y = 2x + 3 $ with the curve equation; the number of intersection points can be determined by the solutions of the system of equations. If $ x \\geqslant 0 $, from $ \\begin{cases} y = 2x + 3 \\\\ \\frac{y^{2}}{9} - \\frac{x^{2}}{4} = 1 \\end{cases} $, we obtain $ 7x^{2} + 48x = 0 $, solving gives $ x_{1} = 0 $, $ x_{2} = -\\frac{48}{7} $ (discarded), thus the line intersects the semi-hyperbola at only one point. If $ x < 0 $, from $ \\begin{cases} y = 2x + 3 \\\\ \\frac{y^{2}}{9} + \\frac{3}{2} \\end{cases} $ leads to $ 25x^{2} + 48x = 0 $, we get $ x_{1} = 0 $, $ x_{2} = -\\frac{48}{25} $, thus the line intersects the semi-ellipse at only one point, (1 where $ x_{1} = 0 $ when $ t $"
  },
  {
    "text": "Given a fixed point $M(3,4)$, $F$ is the focus of the parabola $y^{2}=8x$, and point $P$ moves on this parabola. When $|PM|+|PF|$ takes the minimum value, what are the coordinates of point $P$?",
    "fact_expressions": "M: Point;Coordinate(M) = (3, 4);F: Point;Focus(G) = F;G: Parabola;Expression(G) = (y^2 = 8*x);P: Point;PointOnCurve(P, G) = True;WhenMin(Abs(LineSegmentOf(P,M)) + Abs(LineSegmentOf(P,F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2,4)",
    "fact_spans": "[[[4, 12]], [[4, 12]], [[14, 17]], [[14, 35]], [[18, 32], [42, 45]], [[18, 32]], [[36, 40], [69, 73]], [[36, 48]], [[49, 68]]]",
    "query_spans": "[[[69, 78]]]",
    "process": "From the parabola equation, the focus is known to be F(2,0), and the directrix equation is x = -2. Let P' be the projection of point P on the directrix of the parabola. Since the distance from point P to the directrix is equal to the distance from P to the focus, |PM| + |PF| = |PM| + |PP'|. When x = 3, substituting into the parabola equation yields y = \\pm2\\sqrt{6}. Therefore, point M lies inside the parabola. When points P', P, and M are collinear, |PM| + |PP'| attains its minimum value, at which time |PM| + |PP'| = |MP'| = 5. At this moment, the y-coordinate of P is 4, x = 2, so the coordinates of P are (2,4)."
  },
  {
    "text": "The length of the chord formed by the intersection of the directrix of the parabola $y^{2}=4 x$ and the circle $x^{2}+y^{2}-4 y=0$ is?",
    "fact_expressions": "G: Parabola;H: Circle;Expression(G) = (y^2 = 4*x);Expression(H) = (-4*y + x^2 + y^2 = 0)",
    "query_expressions": "Length(InterceptChord(Directrix(G),H))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[0, 14]], [[18, 38]], [[0, 14]], [[18, 38]]]",
    "query_spans": "[[[0, 48]]]",
    "process": "The directrix of the parabola \\( y^{2} = 4x \\) is \\( x = -1 \\). The standard equation of the circle \\( x^{2} + y^{2} - 4y = 0 \\) is \\( x^{2} + (y - 2)^{2} = 4 \\), with center at \\( (0, 2) \\) and radius 2. The distance from the center to the directrix is 1, so the chord length is \\( 2\\sqrt{2^{2} - 1} = 2\\sqrt{3} \\)."
  },
  {
    "text": "It is known that the vertex of the parabola is at the origin, and the focus of the parabola coincides with the right focus of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$. Then the equation of the parabola is?",
    "fact_expressions": "H: Parabola;Vertex(H) = O;O: Origin;G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1);Focus(H) = RightFocus(G)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2 = 8*x",
    "fact_spans": "[[[2, 5], [12, 15], [55, 58]], [[2, 11]], [[9, 11]], [[19, 47]], [[19, 47]], [[12, 53]]]",
    "query_spans": "[[[55, 63]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has eccentricity $\\frac{\\sqrt{3}}{2}$, the line $y=\\frac{1}{2}x+1$ intersects the ellipse at points $A$ and $B$, and point $M$ lies on the ellipse such that $OM = \\frac{1}{2}OA + \\frac{\\sqrt{3}}{2}OB$. Then $b=$?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;O: Origin;M: Point;A: Point;B: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = x/2 + 1);Eccentricity(G) = sqrt(3)/2;Intersection(H, G) = {A, B};PointOnCurve(M, G);LineSegmentOf(O, M) = LineSegmentOf(O, A)/2 + (sqrt(3)/2)*LineSegmentOf(O, B)",
    "query_expressions": "b",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 54], [103, 105], [124, 126]], [[178, 181]], [[4, 54]], [[81, 102]], [[131, 175]], [[119, 123]], [[108, 111]], [[112, 115]], [[4, 54]], [[4, 54]], [[2, 54]], [[81, 102]], [[2, 80]], [[81, 117]], [[118, 127]], [[131, 175]]]",
    "query_spans": "[[[178, 183]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A circle with diameter $F_{1} F_{2}$ intersects one asymptote at point $A$. If the midpoint $B$ of $A F_{1}$ lies exactly on the other asymptote, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;F1: Point;F2: Point;A: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;IsDiameter(LineSegmentOf(F1, F2), G);L1: Line;L2: Line;Asymptote(C) = {L1, L2};Intersection(G, L1) = A;MidPoint(LineSegmentOf(A, F1)) = B;PointOnCurve(B, L2)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 63], [145, 151]], [[10, 63]], [[10, 63]], [[105, 106]], [[71, 78]], [[79, 86]], [[113, 117]], [[131, 134]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 86]], [[2, 86]], [[87, 106]], [], [], [[2, 142]], [[2, 117]], [[119, 134]], [[2, 143]]]",
    "query_spans": "[[[145, 157]]]",
    "process": "The positional relationship between the circle and the asymptotes of the hyperbola is shown in the figure. From the given conditions, $\\angle AOB = \\angle F_{1}OB$. By the property of the hyperbola, $\\angle AOF_{2} = \\angle F_{1}OB$, so $\\angle AOF_{2} = 60^{\\circ}$. Therefore, $\\frac{b}{a} = \\tan 60^{\\circ} = \\sqrt{3}$, thus the eccentricity of the hyperbola is $e = \\sqrt{1 + \\left(\\frac{b}{a}\\right)^{2}} = 2$."
  },
  {
    "text": "Given that a point $P$ on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{3}=1(a>0)$ also lies on the parabola $y^{2}=\\frac{9}{4} x$, let the focus of the parabola be $F$. If $|P F|=\\frac{25}{16}$, then $a$=?",
    "fact_expressions": "H: Ellipse;Expression(H) = (y^2/3 + x^2/a^2 = 1);a: Number;a>0;P: Point;PointOnCurve(P, H);G: Parabola;Expression(G) = (y^2 = (9/4)*x);PointOnCurve(P, G);F: Point;Focus(G) = F;Abs(LineSegmentOf(P, F)) = 25/16",
    "query_expressions": "a",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 48]], [[2, 48]], [[118, 121]], [[4, 48]], [[52, 55]], [[2, 55]], [[57, 81], [84, 87]], [[57, 81]], [[52, 82]], [[90, 93]], [[84, 93]], [[95, 116]]]",
    "query_spans": "[[[118, 123]]]",
    "process": "Since |PF| = \\frac{25}{16}, we have |PF| = x_{P} + \\frac{p}{2} = \\frac{25}{16} and p = \\frac{9}{8}. Solving gives: x_{P} = 1, so y_{P}^{2} = \\frac{9}{4}, thus y_{P} = \\pm\\frac{3}{2}. Substituting P(1, \\frac{3}{2}) into the ellipse equation yields: \\frac{1}{a^{2}} + \\frac{3}{4} = 1 (a > 0), solving gives: a = 2."
  },
  {
    "text": "Given that the coordinates of points $A$ and $B$ are $(-1,0)$ and $(1,0)$, respectively. Lines $AM$ and $BM$ intersect at point $M$, and the sum of their slopes is $2$. Then the equation of the trajectory of point $M$ is?",
    "fact_expressions": "A: Point;M: Point;B: Point;Intersection(LineOf(A,M),LineOf(B,M))=M;Coordinate(A) = (-1, 0);Coordinate(B) = (1, 0);Slope(LineOf(A,M))+Slope(LineOf(B,M))=2",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x**2-x*y-1=0",
    "fact_spans": "[[[2, 6]], [[52, 56], [71, 75]], [[7, 10]], [[35, 56]], [[2, 33]], [[2, 33]], [[58, 69]]]",
    "query_spans": "[[[71, 82]]]",
    "process": "Let M(x,y). Since the slopes of AM and BM exist, we have x ≠ 1. Also, since k_{AM} = \\frac{y}{x+1}, k_{BM} = \\frac{y}{x-1}, then from k_{AM} + k_{BM} = 2 we get: \\frac{y}{x+1} + \\frac{y}{x-1} = 2. Rearranging yields: x^{2} - xy - 1 = 0. Therefore, the trajectory equation of point M is: x^{2} - xy - 1 = 0."
  },
  {
    "text": "Let the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ have left and right foci $F_{1}$, $F_{2}$ respectively, right vertex $A$, and upper vertex $B$. Given that $|A B|=\\frac{\\sqrt{3}}{2}|F_{1} F_{2}|$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;A: Point;B: Point;RightVertex(C)=A;UpperVertex(C)=B;Abs(LineSegmentOf(A, B)) = (sqrt(3)/2)*Abs(LineSegmentOf(F1, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[1, 58], [142, 145]], [[1, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[67, 74]], [[75, 82]], [[1, 82]], [[1, 82]], [[87, 90]], [[95, 98]], [[1, 90]], [[1, 98]], [[101, 140]]]",
    "query_spans": "[[[142, 151]]]",
    "process": "From the given condition, |AB|^{2}=a^{2}+b^{2}, |F_{1}F_{2}|^{2}=4c^{2}. \\because AB=\\frac{\\sqrt{3}}{2}F_{1}F_{2} \\therefore a^{2}+b^{2}=\\frac{3}{4}\\cdot4c^{2}, i.e., a^{2+a^{2}-c^{2}}=3c^{2}, i.e., a^{2}=2c^{2}, hence e=\\frac{c}{a}=\\frac{\\sqrt{2}}{2}"
  },
  {
    "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1(b>0)$ are given by $y = \\frac{1}{2} x$, then what is the value of $b$?",
    "fact_expressions": "G: Hyperbola;b: Number;b>0;Expression(G) = (x^2/4 - y^2/b^2 = 1);Expression(OneOf(Asymptote(G))) = (y = x/2)",
    "query_expressions": "b",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 48]], [[77, 80]], [[4, 48]], [[1, 48]], [[1, 75]]]",
    "query_spans": "[[[77, 83]]]",
    "process": ""
  },
  {
    "text": "The distance from the focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$ to its asymptote is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/9 = 1)",
    "query_expressions": "Distance(Focus(G), Asymptote(G))",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 51]]]",
    "process": "In the hyperbola \\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1, the real semi-axis is a=2, the imaginary semi-axis is b=3, then the semi-focal distance is c=\\sqrt{a^{2}+b^{2}}=\\sqrt{2^{2}+3^{2}}=\\sqrt{13}, so the foci of the hyperbola are F(\\pm\\sqrt{13},0), and the asymptotes are y=\\pm\\frac{3}{2}x, or equivalently 3x\\pm2y=0. Using the point-to-line distance formula, the required distance is \\frac{|3\\cdot(\\pm\\sqrt{13})\\pm2\\cdot0|}{\\sqrt{3^{2}+(\\pm2)^{2}}}=\\frac{3\\sqrt{13}}{\\sqrt{13}}=3"
  },
  {
    "text": "Given that $M(x_{0}, y_{0})$ is a point on the hyperbola $C$: $\\frac{x^{2}}{2}-y^{2}=1$, and $F_{1}$, $F_{2}$ are the two foci of $C$. If $\\overrightarrow{M F}_{1} \\cdot \\overrightarrow{M F}_{2}<0$, then what is the range of values for $y_{0}$?",
    "fact_expressions": "M: Point;Coordinate(M) = (x0, y0);x0: Number;y0: Number;C: Hyperbola;Expression(C) = (x^2/2 - y^2 = 1);PointOnCurve(M, C) = True;F1: Point;F2: Point;Focus(C) = {F1, F2};DotProduct(VectorOf(M, F1), VectorOf(M, F2)) < 0",
    "query_expressions": "Range(y0)",
    "answer_expressions": "(-sqrt(3)/3, sqrt(3)/3)",
    "fact_spans": "[[[2, 19]], [[2, 19]], [[2, 19]], [[146, 153]], [[20, 53], [74, 77]], [[20, 53]], [[2, 57]], [[58, 65]], [[66, 73]], [[58, 83]], [[85, 144]]]",
    "query_spans": "[[[146, 160]]]",
    "process": "By the given condition, $\\overrightarrow{MF_{1}}\\cdot\\overrightarrow{MF_{2}}=(-\\sqrt{3}-x_{0},-y_{0})\\cdot(\\sqrt{3}-x_{0},-y_{0})=x_{0}^{2}-3+y_{0}^{2}=3y_{0}^{2}-1<0$, $\\therefore-\\frac{\\sqrt{3}}{3}<y_{0}<\\frac{\\sqrt{3}}{3}$"
  },
  {
    "text": "The focus of the parabola $y^{2}=2 a x(a>0)$ is $F$, and its directrix intersects the hyperbola $\\frac{y^{2}}{4}-\\frac{x^{2}}{9}=1$ at points $M$ and $N$. If $\\angle M F N=120^{\\circ}$, then $a=$?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 2*(a*x));a: Number;a>0;F: Point;Focus(H) = F;G: Hyperbola;Expression(G) = (-x^2/9 + y^2/4 = 1);M: Point;N: Point;Intersection(Directrix(H), G) = {M, N};AngleOf(M, F, N) = ApplyUnit(120, degree)",
    "query_expressions": "a",
    "answer_expressions": "3*sqrt(26)/13",
    "fact_spans": "[[[0, 21], [29, 30]], [[0, 21]], [[113, 116]], [[3, 21]], [[25, 28]], [[0, 28]], [[33, 71]], [[33, 71]], [[74, 77]], [[78, 81]], [[29, 83]], [[85, 111]]]",
    "query_spans": "[[[113, 118]]]",
    "process": "Based on the given conditions, a sketch can be drawn to show that angle MFO is 60 degrees. Using trigonometric values, we obtain $\\underline{2\\sqrt{1+\\frac{a^{2}}{36}}}=\\sqrt{3}$. Solving gives $a=\\frac{2\\sqrt{26}}{13}$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ passes through the point $(1, \\sqrt{2})$, and the range of its major axis length is $[4 , 6]$, then what is the range of the eccentricity of this ellipse?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(H) = (1, sqrt(2));PointOnCurve(H, G);Range(Length(MajorAxis(G)))=[4,6]",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[sqrt(3)/3,sqrt(3)/2]",
    "fact_spans": "[[[2, 54], [95, 97], [73, 74]], [[4, 54]], [[4, 54]], [[55, 71]], [[4, 54]], [[4, 54]], [[2, 54]], [[55, 71]], [[2, 71]], [[73, 92]]]",
    "query_spans": "[[[95, 107]]]",
    "process": "Substituting the point $(1,\\sqrt{2})$, we obtain $\\frac{b^{2}}{a^{2}}=\\frac{2}{a^{2}-1}$, and from $e=\\sqrt{1-\\frac{b^{2}}{a^{2}}}=\\sqrt{1-\\frac{2}{a^{2}-1}}$, combining with the range of $a$, we can solve. From the given condition, we have $\\frac{1}{a^{2}}+\\frac{2}{b^{2}}=1$, so $b^{2}=\\frac{2a^{2}}{a^{2}-1}$, thus $\\frac{b^{2}}{a^{2}}=\\frac{2}{a^{2}-1}$. Also, $e=\\frac{c}{a}=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{1-\\frac{b^{2}}{a^{2}}}=\\sqrt{1-\\frac{2}{a^{2}-1}}$, so $\\frac{1}{4}\\leqslant\\frac{2}{a^{2}-1}\\leqslant\\frac{2}{3}$, which implies $\\frac{\\sqrt{3}}{3}\\leqslante$. The range of eccentricity of this ellipse is $\\left[\\frac{\\sqrt{3}}{3},\\frac{\\sqrt{3}}{}\\right]$"
  },
  {
    "text": "A focus of the hyperbola $8 kx^{2}-ky^{2}=8$ is $(0 , 3)$, then what is the value of $k$?",
    "fact_expressions": "G: Hyperbola;k: Number;Expression(G) = (8*(k*x^2) - k*y^2 = 8);Coordinate(OneOf(Focus(G)))=(0,3)",
    "query_expressions": "k",
    "answer_expressions": "-1",
    "fact_spans": "[[[0, 22]], [[39, 42]], [[0, 22]], [[0, 37]]]",
    "query_spans": "[[[39, 46]]]",
    "process": ""
  },
  {
    "text": "The line $l$ passing through the point $M(0,1)$ with slope $1$ intersects the two asymptotes of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ at points $A$ and $B$, and $\\overrightarrow{B M}=2 \\overrightarrow{A M}$. The focal length of the hyperbola is $2 \\sqrt{10}$. Then the value of $b$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;M: Point;B: Point;A: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(M) = (0, 1);PointOnCurve(M, l);Slope(l) = 1;VectorOf(B, M) = 2*VectorOf(A, M);FocalLength(C) = 2*sqrt(10);L1:Line;L2:Line;Asymptote(C)={L1,L2};Intersection(l,L1)=A;Intersection(l,L2)=B",
    "query_expressions": "b",
    "answer_expressions": "1",
    "fact_spans": "[[[18, 23]], [[24, 85], [150, 153]], [[173, 176]], [[32, 85]], [[1, 10]], [[97, 100]], [[92, 96]], [[32, 85]], [[32, 85]], [[24, 85]], [[1, 10]], [[0, 23]], [[11, 23]], [[103, 148]], [[150, 170]], [-1, -1], [-1, -1], [24, 87], [24, 97], [24, 97]]",
    "query_spans": "[[[173, 180]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). From \\overline{BM}=2\\overline{A}\\overline{M}, we get x_{2}=2x_{1}.\\textcircled{1}. From the given condition, the equation of the line is y=x+1, and the asymptotes of \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 are y=\\pm\\frac{b}{a}x. Solving the line l and the asymptote equations simultaneously, we obtain x_{1}=-\\frac{a}{a+b}, x_{2}=\\frac{a}{b-a}, thus -\\frac{2a}{a+b}=\\frac{a}{b-a}, which simplifies to a=3b. Given that the focal distance of the hyperbola is 2\\sqrt{10}, we have a^{2}+b^{2}=c^{2}=10, so 10b^{2}=10, solving gives b=1."
  },
  {
    "text": "Given that $F$ is the right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, a perpendicular line $F D$ is drawn from $F$ to an asymptote of $C$, with $D$ being the foot of the perpendicular, and $|F D|=\\frac{\\sqrt{3}}{2}|O F|$ ($O$ is the origin). Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;F: Point;D: Point;O: Origin;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C)=F;PointOnCurve(F,LineSegmentOf(F,D));IsPerpendicular(LineSegmentOf(F,D),Asymptote(C));FootPoint(LineSegmentOf(F,D),Asymptote(C))=D;Abs(LineSegmentOf(F, D)) = (sqrt(3)/2)*Abs(LineSegmentOf(O, F))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[6, 67], [77, 80], [146, 149]], [[14, 67]], [[14, 67]], [[2, 5], [73, 76]], [[95, 98]], [[135, 138]], [[14, 67]], [[14, 67]], [[6, 67]], [[2, 71]], [[72, 92]], [[77, 92]], [[77, 101]], [[103, 134]]]",
    "query_spans": "[[[146, 155]]]",
    "process": "Find the distance from the focus to the asymptote to obtain an equation relating a, b, c, and thus determine the eccentricity. According to the problem, F(c,0), one asymptote equation is y=\\frac{b}{a}x, i.e., bx-ay=0, \\therefore|FD|=\\frac{|bc|}{\\sqrt{b^{2}+a^{2}}}=b, from FD=\\frac{\\sqrt{3}}{2}\\begin{matrix}\\sqrt{3}&\\\\\\end{matrix}OF we get b=\\frac{\\sqrt{3}}{2}c', \\therefore b^{2}=\\frac{3}{4}c^{2}=c^{2}-a^{2}, c^{2}=4a^{2}, \\therefore e=\\frac{c}{a}=2."
  },
  {
    "text": "The equation of the directrix of the parabola $y=a x^{2}$ is $y=-\\frac{1}{2}$, then $a$=?",
    "fact_expressions": "G: Parabola;a: Number;Expression(G) = (y = a*x^2);Expression(Directrix(G)) = (y = -1/2)",
    "query_expressions": "a",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 14]], [[38, 41]], [[0, 14]], [[0, 36]]]",
    "query_spans": "[[[38, 43]]]",
    "process": ""
  },
  {
    "text": "The vertices of $\\triangle ABC$ are $A(-5,0)$, $B(5,0)$, and the incenter of $\\triangle ABC$ lies on the line $x=3$. Then the equation of the trajectory of vertex $C$ is?",
    "fact_expressions": "H: Line;A: Point;B: Point;C: Point;Expression(H) = (x = 3);Coordinate(A) = (-5, 0);Coordinate(B) = (5, 0);PointOnCurve(Center(InscribedCircle(TriangleOf(A, B, C))), H)",
    "query_expressions": "LocusEquation(C)",
    "answer_expressions": "x^2/9 - y^2/16 = 1 & (x > 3)",
    "fact_spans": "[[[59, 66]], [[18, 27]], [[28, 36]], [[71, 74]], [[59, 66]], [[18, 27]], [[28, 36]], [[37, 67]]]",
    "query_spans": "[[[71, 81]]]",
    "process": ""
  },
  {
    "text": "The two endpoints of a line segment $AB$ of fixed length $4$ move along the parabola $y^{2}=x$. Let $M$ be the midpoint of $AB$. Then, the shortest distance from point $M$ to the $y$-axis is?",
    "fact_expressions": "A: Point;B: Point;LineSegmentOf(A,B) = 4;Endpoint(LineSegmentOf(A,B)) = {A,B};G: Parabola;Expression(G) = (y^2=x);MidPoint(LineSegmentOf(A,B)) = M;M: Point;PointOnCurve(Endpoint(LineSegmentOf(A,B)),G)",
    "query_expressions": "Min(Distance(M, yAxis))",
    "answer_expressions": "7/4",
    "fact_spans": "[[[9, 14], [37, 42]], [[9, 14], [37, 42]], [[0, 14]], [[7, 19]], [[20, 32]], [[20, 32]], [[37, 49]], [[46, 49], [51, 55]], [0, 34]]",
    "query_spans": "[[[51, 67]]]",
    "process": "Let the projections of points A, B, M on the y-axis be A_{1}, B_{1}, M_{1}, respectively. Let the focus of the parabola be F. By the definition of a parabola: AF = AA_{1} + \\frac{p}{2}, BF = BB_{1} + \\frac{p}{2}, so AF + BF = AA_{1} + BB_{1} + p = AA_{1} + BB_{1} + \\frac{1}{2}, thus MM_{1} = \\frac{AA_{1} + BB_{1}}{2} = \\frac{AF + BF - \\frac{1}{2}}{2} = \\frac{AF + BF}{2} - \\frac{1}{4} \\geqslant \\frac{AB}{2} - \\frac{1}{4} = \\frac{7}{4}. Equality holds when A, B, F are collinear, i.e., AB passes through point F. Therefore, the shortest distance from M to the v-axis is \\frac{7}{1}. Hence, the answer is: 7"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$, the distance from the focus $F$ to the directrix is $2$. Then the equation of curve $C$ is?",
    "fact_expressions": "C: Parabola;p: Number;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;Distance(F,Directrix(C)) = 2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[2, 28], [46, 51]], [[10, 28]], [[31, 34]], [[10, 28]], [[2, 28]], [[2, 34]], [[2, 44]]]",
    "query_spans": "[[[46, 56]]]",
    "process": "The distance from the focus F to the directrix is 2, so p=2, hence the equation of curve C is y^{2}=4x"
  },
  {
    "text": "Given that point $F$ is the focus of the parabola $y^{2}=4x$, and $M$, $N$ are two points on this parabola such that $|M F|+|N F|=6$, then the horizontal coordinate of the midpoint of segment $M N$ is?",
    "fact_expressions": "G: Parabola;M: Point;N: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(M, G);PointOnCurve(N, G);Abs(LineSegmentOf(M, F)) + Abs(LineSegmentOf(N, F)) = 6",
    "query_expressions": "XCoordinate(MidPoint(LineSegmentOf(M,N)))",
    "answer_expressions": "2",
    "fact_spans": "[[[7, 21], [34, 37]], [[25, 28]], [[29, 32]], [[2, 6]], [[7, 21]], [[2, 24]], [[25, 40]], [[25, 40]], [[41, 56]]]",
    "query_spans": "[[[58, 74]]]",
    "process": "\\because the parabola y^{2}=4x, \\therefore the equation of the directrix is x=-1. From |MF|+|NF|=6, we obtain x_{M}+1+x_{N}+1=6, that is, x_{M}+x_{N}=4, \\therefore the horizontal coordinate of the midpoint of MN is \\frac{x_{M}+x_{N}}{2}=2,"
  },
  {
    "text": "Given that the line $AB$ passing through the right focus $F_2$ of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ intersects the ellipse at points $A$ and $B$, and $F_1$ is the left focus of the ellipse, then the perimeter of $\\triangle A F_{1} B$ equals?",
    "fact_expressions": "G: Ellipse;B: Point;A: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/9 = 1);RightFocus(G)=F2;PointOnCurve(F2,LineOf(A,B));Intersection(LineOf(A,B),G) = {A, B};LeftFocus(G) = F1",
    "query_expressions": "Perimeter(TriangleOf(A, F1, B))",
    "answer_expressions": "20",
    "fact_spans": "[[[4, 42], [62, 64], [83, 85]], [[69, 72]], [[65, 68]], [[75, 82]], [[46, 53]], [[4, 42]], [[4, 53]], [[2, 61]], [[54, 74]], [[75, 89]]]",
    "query_spans": "[[[92, 119]]]",
    "process": "\\because the line AB passing through the right focus F_{2} of the ellipse \\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1 intersects the ellipse at points A and B, and F_{1} is the left focus of the ellipse, \\therefore the perimeter of AAF_{1}B = |AB|+|AF|+|BF| = |AF_{2}|+|AF|+|BF|+|BF_{2}| = 4a = 20. The answer is 20"
  },
  {
    "text": "Suppose a hyperbola shares the same foci with the ellipse $\\frac{x^{2}}{27}+\\frac{y^{2}}{36}=1$ and intersects the ellipse, with one intersection point at $(\\sqrt{15}, 4)$. Then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/27 + y^2/36 = 1);IsIntersect(G, H);Focus(G) = Focus(H);Coordinate(OneOf(Intersection(G, H))) = (sqrt(15), 4)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/4 - x^2/5 = 1",
    "fact_spans": "[[[2, 5], [88, 91]], [[6, 45], [54, 56]], [[6, 45]], [[2, 58]], [[2, 51]], [[2, 85]]]",
    "query_spans": "[[[88, 98]]]",
    "process": "It is given that the ellipse $\\frac{x^{2}}{27}+\\frac{y^{2}}{36}=1$ has its foci on the $y$-axis, and $c^{2}=36-27=9$, so the coordinates of the foci are $(0,\\pm3)$. By the definition of a hyperbola, we have $2a=\\sqrt{(\\sqrt{15}-0)^{2}+(4-3)^{2}}-\\sqrt{(\\sqrt{15}-0)^{2}+(4+3)^{2}}=4$, hence $a=2$, $b^{2}=9-4=5$. Therefore, the standard equation of the required hyperbola is $\\frac{y^{2}}{4}-\\frac{x^{2}}{5}=1$."
  },
  {
    "text": "Given that $F$ is the left focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, $A(1,\\ 4)$, and $P$ is a moving point on the right branch of the hyperbola, then the minimum value of $|PF|+|PA|$ is?",
    "fact_expressions": "G: Hyperbola;A: Point;P: Point;F: Point;Expression(G) = (x^2/4 - y^2/12 = 1);Coordinate(A) = (1, 4);LeftFocus(G) = F;PointOnCurve(P, RightPart(G))",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "9",
    "fact_spans": "[[[6, 45], [66, 69]], [[50, 60]], [[62, 65]], [[2, 5]], [[6, 45]], [[50, 60]], [[2, 49]], [[62, 75]]]",
    "query_spans": "[[[77, 95]]]",
    "process": ""
  },
  {
    "text": "If the curve $y=\\sqrt{x^{2}-4}$ and the line $y=k(x-2)+3$ have two distinct common points, then the range of real values for $k$ is?",
    "fact_expressions": "H: Curve;Expression(H) = (y = sqrt(x^2 - 4));G: Line;Expression(G) = (y = k*(x - 2) + 3);k: Real;NumIntersection(H, G) = 2",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-1,3/4]",
    "fact_spans": "[[[1, 21]], [[1, 21]], [[22, 36]], [[22, 36]], [[47, 52]], [[1, 45]]]",
    "query_spans": "[[[47, 59]]]",
    "process": "The line $ y = k(x - 2) + 3 $ passes through the fixed point $ P(2, 3) $. The curve $ y = \\sqrt{x^{2} - 4} $ represents the upper part (on and above the x-axis) of the hyperbola $ x^{2} - y^{2} = 4 $ where $ y \\geqslant 0 $. The asymptotes of the hyperbola $ x^{2} - y^{2} = 4 $ are $ y = \\pm x $, and the left and right vertices are $ A_{1}(-2, 0) $, $ A_{2}(2, 0) $, respectively, as shown in the figure. Draw lines $ l_{1} $ and $ l_{2} $ through point $ P $ parallel to the two asymptotes. Rotate line $ l_{1} $ clockwise around point $ P $. When it passes through point $ A_{1} $, the condition is satisfied, at which time $ k_{A_{1}P} = \\frac{3 - 0}{2 - (-2)} = \\frac{3}{4} $. According to the graphical characteristics of the hyperbola, as shown in the figure, when line $ l_{1} $ rotates clockwise from position $ PA_{1} $ to position $ l_{2} $, it satisfies having two intersection points with the curve $ y = \\sqrt{x^{2} - 4} $. Therefore, the slope satisfies $ -1 < k \\leqslant \\frac{3}{4} $."
  },
  {
    "text": "If one of the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is given by $x+3 y=0$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G)))=(x+3*y=0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)/3",
    "fact_spans": "[[[1, 47], [68, 71]], [[4, 47]], [[4, 47]], [[1, 47]], [[1, 65]]]",
    "query_spans": "[[[68, 77]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{m+3}+\\frac{y^{2}}{3}=1$ $(m>0)$ has eccentricity $e=\\frac{1}{3}$, then the value of $m$ is equal to?",
    "fact_expressions": "G: Ellipse;m: Number;m>0;Expression(G) = (x^2/(m + 3) + y^2/3 = 1);e: Number;Eccentricity(G) = e;e = 1/3",
    "query_expressions": "m",
    "answer_expressions": "3/8",
    "fact_spans": "[[[2, 48]], [[70, 73]], [[4, 48]], [[2, 48]], [[52, 67]], [[2, 67]], [[52, 67]]]",
    "query_spans": "[[[70, 78]]]",
    "process": "When $ m > 0 $, $ m + 3 > 3 $, the foci of the ellipse lie on the x-axis, then $ a = \\sqrt{m + 3} $, $ b = \\sqrt{3} $, $ c = \\sqrt{m} $, so $ e = \\frac{c}{a} = \\frac{\\sqrt{m}}{\\sqrt{m + 3}} = \\frac{1}{3} $, solving gives $ m = \\frac{3}{8} $."
  },
  {
    "text": "If the line passing through the two points $A(a , 0)$ and $B(0 , a)$ does not intersect the parabola $y=x^{2}-2 x-3$, then what is the range of real values for $a$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;a: Real;Expression(G) = (y = x^2 - 2*x - 3);Coordinate(A) = (a, 0);Coordinate(B) = (0, a);PointOnCurve(A,H);PointOnCurve(B,H);NumIntersection(H,G)=0",
    "query_expressions": "Range(a)",
    "answer_expressions": "(-\\infty,-13/4)",
    "fact_spans": "[[[30, 48]], [[27, 29]], [[5, 15]], [[16, 26]], [[55, 60]], [[30, 48]], [[5, 15]], [[16, 26]], [[2, 29]], [[2, 29]], [[27, 52]]]",
    "query_spans": "[[[55, 67]]]",
    "process": ""
  },
  {
    "text": "It is known that the eccentricity of an ellipse with foci on the $x$-axis is $0.5$, and its major axis length equals the radius of the circle $C$: $x^{2}+y^{2}-2 x-15=0$. Then, the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;PointOnCurve(Focus(G), xAxis);Eccentricity(G) = 0.5;C: Circle;Expression(C) = (-2*x + x^2 + y^2 - 15 = 0);Length(MajorAxis(G)) = Radius(C)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 + y^2/3 = 1",
    "fact_spans": "[[[11, 13], [25, 26], [65, 67]], [[2, 13]], [[11, 23]], [[32, 60]], [[32, 60]], [[25, 63]]]",
    "query_spans": "[[[65, 74]]]",
    "process": "Complete the square to simplify $ x^{2}+y^{2}-2x-15=0 $, obtaining a circle with radius 4. Thus, the major axis of the ellipse is 4. Find $ c $ using the eccentricity, and find $ b $ using the Pythagorean theorem, then obtain the equation of the ellipse. Detailed solution: $ \\because x^{2}+y^{2}-2x-15=0 $, $ (x-1)^{2}+y^{2}=16 $, $ \\therefore r=4=2a $, $ \\therefore a=2 $, $ \\because e=0.5 $, $ \\therefore c=1 $, $ \\therefore b^{2}=3 $. The equation is: $ \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1 $."
  },
  {
    "text": "The focus of the parabola $y=\\frac{1}{16} x^{2}$ coincides with the upper focus of the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{m}=-1$, then $m$=?",
    "fact_expressions": "H: Parabola;Expression(H) = (y = x^2/16);G: Hyperbola;Expression(G) = (x^2/3 - y^2/m = -1);m: Number;Focus(H) = UpperFocus(G)",
    "query_expressions": "m",
    "answer_expressions": "13",
    "fact_spans": "[[[0, 25]], [[0, 25]], [[29, 68]], [[29, 68]], [[76, 79]], [[0, 74]]]",
    "query_spans": "[[[76, 81]]]",
    "process": "The focus of the parabola \\( y = \\frac{1}{16}x^{2} \\) is \\( (0, 4) \\), so \\( m + 3 = 4^{2} \\Rightarrow m = 13 \\)."
  },
  {
    "text": "The left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a, b>0)$ are $F_{1}$ and $F_{2}$, respectively. The line passing through focus $F_{2}$ and perpendicular to the $x$-axis intersects the hyperbola at points $A$ and $B$. If $\\overrightarrow{F_{1} A} \\cdot \\overrightarrow{F_{1} B}=0$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Line;PointOnCurve(F2, H);IsPerpendicular(H, xAxis) ;Intersection(H, G) = {A, B};A: Point;B: Point;DotProduct(VectorOf(F1, A), VectorOf(F1, B)) = 0",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2) + 1",
    "fact_spans": "[[[0, 54], [105, 108], [185, 188]], [[0, 54]], [[3, 54]], [[3, 54]], [[3, 54]], [[3, 54]], [[63, 70]], [[73, 80], [85, 92]], [[0, 80]], [[0, 80]], [[101, 103]], [[82, 103]], [[93, 103]], [[101, 122]], [[111, 114]], [[116, 120]], [[124, 183]]]",
    "query_spans": "[[[185, 194]]]",
    "process": ""
  },
  {
    "text": "If the line $y=kx+1$ always has common points with the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$ whose foci lie on the $x$-axis, then the range of real number $m$ is?",
    "fact_expressions": "G: Ellipse;m: Real;H: Line;k: Number;Expression(G) = (x^2/5 + y^2/m = 1);Expression(H) = (y = k*x + 1);PointOnCurve(Focus(G),xAxis);IsIntersect(H,G)",
    "query_expressions": "Range(m)",
    "answer_expressions": "[1,5)",
    "fact_spans": "[[[22, 59]], [[66, 71]], [[1, 12]], [[3, 12]], [[22, 59]], [[1, 12]], [[13, 59]], [[1, 64]]]",
    "query_spans": "[[[66, 78]]]",
    "process": "Since the foci of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$ lie on the $x$-axis, we have $0<m<5$. Since the line $y=kx+l$ passes through the fixed point $P(0,1)$, and the line always has common points with the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$, the point $P$ lies on or inside the ellipse, that is, $\\frac{0^{2}}{5}+\\frac{1^{2}}{m}\\leqslant1$. Solving this gives $1\\leqslant m$. Combining these results, we get $1\\leqslant m<5$. The answer is: $1\\leqslant m<5$."
  },
  {
    "text": "What is the equation of the directrix of the parabola $y=2 p x^{2}(p>0)$?",
    "fact_expressions": "G: Parabola;p: Number;p>0;Expression(G) = (y = 2*(p*x^2))",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=-1/(8*p)",
    "fact_spans": "[[[0, 21]], [[3, 21]], [[3, 21]], [[0, 21]]]",
    "query_spans": "[[[0, 28]]]",
    "process": ""
  },
  {
    "text": "Given that the distance from point $A(m, 2)$ on the parabola $C$: $y = a x^{2}$ $(a > 0)$ to its directrix is $3$, then $a =$?",
    "fact_expressions": "C: Parabola;a: Number;A: Point;a>0;Expression(C) = (y = a*x^2);Coordinate(A) = (m, 2);PointOnCurve(A, C);Distance(A, Directrix(C)) = 3;m:Number",
    "query_expressions": "a",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 27], [40, 41]], [[52, 55]], [[29, 39]], [[9, 27]], [[2, 27]], [[29, 39]], [[2, 39]], [[29, 50]], [[30, 39]]]",
    "query_spans": "[[[52, 57]]]",
    "process": "The parabola $ y = ax^{2} $, that is, $ x^{2} = \\frac{1}{a}y' $, has the directrix equation $ y = -\\frac{1}{4a} $. The distance from point $ A(m, 2) $ to the focus is 3, so $ 2 - (-\\frac{1}{4a}) = 3 $, solving gives $ a = \\frac{1}{4} $."
  },
  {
    "text": "Let $A(0,-\\sqrt{5})$, $B(0, \\sqrt{5})$, $|P A|-|P B|=4$, then the trajectory equation of the moving point $P$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (0, -sqrt(5));B: Point;Coordinate(B) = (0, sqrt(5));P: Point;Abs(LineSegmentOf(P, A)) - Abs(LineSegmentOf(P, B)) = 4",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(y^2/4-x^2=1)&(y>=2)",
    "fact_spans": "[[[1, 17]], [[1, 17]], [[19, 35]], [[19, 35]], [[55, 58]], [[36, 51]]]",
    "query_spans": "[[[55, 65]]]",
    "process": "Since |PA| - |PB| = 4 < 2\\sqrt{5}, the trajectory of the moving point P is the upper branch of a hyperbola with foci A and B and a real axis length of 4. Because \\frac{y^{2}}{4} - x^{2} = 1 (y \\geqslant 2), 2c = 2\\sqrt{5}, it follows that a = 2, c = \\sqrt{5}, b^{2} = c^{2} - a^{2} = 1. Therefore, the equation of the trajectory of the moving point P is"
  },
  {
    "text": "The maximum distance from a point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$ to the line $x-2 y-12=0$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/12 = 1);P: Point;PointOnCurve(P, G);H: Line;Expression(H) = (x - 2*y - 12 = 0)",
    "query_expressions": "Max(Distance(P, H))",
    "answer_expressions": "4*sqrt(5)",
    "fact_spans": "[[[0, 39]], [[0, 39]], [[41, 42]], [[0, 42]], [[43, 57]], [[43, 57]]]",
    "query_spans": "[[[41, 65]]]",
    "process": "Let the parametric equations of the ellipse be \\begin{cases}x=4\\cos\\theta\\\\y=2\\sqrt{3}\\sin\\theta\\end{cases}d=\\frac{|4\\cos\\theta-4\\sqrt{3}\\sin\\theta-12|}{\\sqrt{5}}=\\frac{4\\sqrt{5}}{5}|\\cos\\theta-\\sqrt{3}\\sin\\theta-3|=\\frac{4\\sqrt{5}}{5}|2\\cos(\\theta+\\frac{\\pi}{3})-3|, when \\cos(\\theta+\\frac{\\pi}{3})=1, d_{\\min}=\\frac{4\\sqrt{5}}{5}, at this time the required point is (2,-3)"
  },
  {
    "text": "Given that the line $MN$ passes through the left focus $F$ of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, intersecting the ellipse at points $M$ and $N$. The line $PQ$ passes through the origin $O$ and is parallel to $MN$, and intersects the ellipse at points $P$ and $Q$. Then $\\frac{|P Q|^{2}}{|MN|}$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);F: Point;LeftFocus(G) = F;PointOnCurve(F, LineOf(M, N)) = True;Intersection(LineOf(M, N), G) = {M, N};N: Point;M: Point;O: Origin;PointOnCurve(O, LineOf(P, Q)) = True;IsParallel(LineOf(P, Q), LineSegmentOf(M, N)) = True;Intersection(LineSegmentOf(P, Q), G) = {P, Q};Q: Point;P: Point",
    "query_expressions": "Abs(LineSegmentOf(P, Q))^2/Abs(LineSegmentOf(M, N))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[10, 37], [46, 48], [90, 92]], [[10, 37]], [[41, 44]], [[10, 44]], [[2, 44]], [[2, 59]], [[54, 57]], [[50, 53]], [[69, 74]], [[61, 74]], [[61, 82]], [[84, 103]], [[98, 101]], [[94, 97]]]",
    "query_spans": "[[[105, 131]]]",
    "process": "F(-1,0) When the slope of line MN does not exist, |MN| = \\frac{2b^{2}}{a} = \\sqrt{2}, |PQ| = 2b = 2, then \\frac{|PQ|^{2}}{|MN|} = 2\\sqrt{2}; when the slope of line MN exists, let the slope of MN be k, then the equation of MN is y = k(x+1), M(x_{1},y_{1}), N(x_{2},y_{2}). Solving the system of equations: \\begin{cases} y = k(x+1) \\\\ \\frac{x^{2}}{2} + y^{2} = 1 \\end{cases} yields (2k^{2}+1)x^{2} + 4k^{2}x + 2k^{2}-2 = 0. By Vieta's formulas, x_{1} + x_{2} = -\\frac{4k^{2}}{2k^{2}+1}, x_{1}x_{2} = \\frac{2k^{2}-3}{2k^{2}+1}, |MN| = \\sqrt{1+k^{2}} \\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}} = \\frac{2\\sqrt{2}(1+k^{2})}{2k^{2}+1}. Then the equation of line PQ is y = kx, P(x_{3},y_{3}), Q(x_{4},y_{4}), then \\begin{cases} y = kx \\\\ \\frac{x^{2}}{2} + y^{2} = 1 \\end{cases} solving gives x^{2} = \\frac{2}{1+2k^{2}}, y^{2} = \\frac{2k^{2}}{1+2k^{2}}, then |OP|^{2} = x^{2} + y^{2} = \\frac{2(1+k^{2})}{1+2k^{2}}, then |PQ| = 2|OP|, so |PQ|^{2} = 4|OP|^{2} = \\frac{8(1+k^{2})}{1+2k^{2}}, \\therefore \\frac{|PQ|^{2}}{|MN|} = 2\\sqrt{2}, hence fill in 2\\sqrt{2}."
  },
  {
    "text": "A line $l$ passing through the focus of the parabola $y^{2}=4x$ intersects the parabola at points $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$. If $x_{1}+x_{2}=4$, then $|PQ|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);l: Line;PointOnCurve(Focus(G), l);P: Point;x1: Number;y1: Number;Coordinate(P) = (x1, y1);Q: Point;x2: Number;y2: Number;Coordinate(Q) = (x2, y2);Intersection(l, G) = {P, Q};x1+x2=4",
    "query_expressions": "Abs(LineSegmentOf(P, Q))",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 15], [25, 28]], [[1, 15]], [[19, 24]], [[0, 24]], [[29, 46]], [[29, 46]], [[29, 46]], [[29, 46]], [[48, 65]], [[48, 65]], [[48, 65]], [[48, 65]], [[19, 67]], [[70, 85]]]",
    "query_spans": "[[[87, 96]]]",
    "process": ""
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, $P$ is a moving point on the parabola, and $A(1,1)$ is a fixed point, then the minimum perimeter of $\\triangle PAF$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;P: Point;PointOnCurve(P, G);A: Point;Coordinate(A) = (1, 1)",
    "query_expressions": "Min(Perimeter(TriangleOf(P, A, F)))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 16], [28, 31]], [[2, 16]], [[20, 23]], [[2, 23]], [[24, 27]], [[24, 35]], [[38, 46]], [[38, 46]]]",
    "query_spans": "[[[48, 72]]]",
    "process": "Find the minimum perimeter of APAF, which is to find the minimum value of |PA| + |PF|. Let point D be the projection of point P on the directrix. According to the definition of a parabola, |PF| = |PD|. Therefore, the minimum value of |PA| + |PF| is the minimum value of |PA| + |PD|. Based on plane geometry, |PA| + |PD| is minimized when points D, P, and A are collinear. Thus, the minimum value is $ x_{A} - (-1) = 1 + 1 = 2 $. Since $|AF| = 1$, the minimum perimeter of APAF is $ 2 + 1 = 3 $."
  },
  {
    "text": "Given that an asymptote of the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{3}=1$ is $y= \\sqrt{3} x$, then the distance from a point $M(2 , y_{0})$ on the parabola $y^{2}=4 a x$ to the focus $F$ of the parabola is?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;M: Point;y0:Number;Expression(G) = (-y^2/3 + x^2/a = 1);Expression(H) = (y^2 = 4*(a*x));Coordinate(M) = (2, y0);Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x);PointOnCurve(M,H);Focus(H)=F;F:Point",
    "query_expressions": "Distance(M,F)",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 40]], [[69, 82]], [[66, 82], [101, 104]], [[85, 99]], [[85, 99]], [[2, 40]], [[66, 82]], [[85, 99]], [[2, 64]], [[66, 99]], [[101, 109]], [[106, 109]]]",
    "query_spans": "[[[85, 114]]]",
    "process": ""
  },
  {
    "text": "It is known that the focus of the parabola $y^{2}=-12 x$ coincides with one focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1$, then $a$=?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;Expression(G) = (-y^2/4 + x^2/a^2 = 1);Expression(H) = (y^2 = -12*x);Focus(H) = OneOf(Focus(G))",
    "query_expressions": "a",
    "answer_expressions": "pm*sqrt(5)",
    "fact_spans": "[[[22, 64]], [[73, 76]], [[2, 18]], [[22, 64]], [[2, 18]], [[2, 71]]]",
    "query_spans": "[[[73, 78]]]",
    "process": "The focus of the parabola \\( y^{2} = -12x \\) is \\( (-3, 0) \\), which coincides with one focus of the hyperbola \\( \\frac{x^{2}}{a} - \\frac{y^{2}}{4} = 1 \\). We obtain \\( 3 = \\sqrt{a^{2} + 4} \\), solving gives \\( a = \\pm\\sqrt{5} \\)"
  },
  {
    "text": "Let $B$ be the upper vertex of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. If any point $P$ on $C$ satisfies $|P B| \\leq 2 b$, then what is the range of the eccentricity of $C$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;B: Point;UpperVertex(C) = B;P: Point;PointOnCurve(P,C);Abs(LineSegmentOf(P,B)) <= 2*b",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(0, \\sqrt{2}/2]",
    "fact_spans": "[[[5, 62], [68, 71], [101, 104]], [[5, 62]], [[11, 62]], [[11, 62]], [[11, 62]], [[11, 62]], [[1, 4]], [[1, 66]], [[77, 80]], [[68, 80]], [[83, 99]]]",
    "query_spans": "[[[101, 114]]]",
    "process": "Let $ P(x,y) $, then $ |PB| = \\sqrt{x^{2} + (y - b)^{2}} = \\sqrt{ -\\frac{c^{2}}{b^{2}}y^{2} - 2by + a^{2} + b^{2} } = \\sqrt{ -\\frac{c^{2}}{b^{2}}\\left(y + \\frac{b^{3}}{c^{2}}\\right)^{2} + a^{2} + b^{2} + \\frac{b^{4}}{c^{2}} } \\leqslant 2b $. Since $ y \\in [-b, b] $, when $ -\\frac{b^{3}}{c^{2}} > -b $, i.e., $ a^{2} < 2c^{2} $, $ |PA|_{\\max} = \\sqrt{a^{2} + b^{2} + \\frac{b^{4}}{c^{2}}} \\leqslant 2b $, so $ a^{2} + b^{2} + \\frac{b^{4}}{c^{2}} \\leqslant 4b^{2} $. Simplifying yields: $ a^{4} - 4a^{2}c^{2} + 4c^{4} \\leqslant 0 \\Rightarrow (a^{2} - 2c^{2})^{2} \\leqslant 0 $, which clearly does not hold. When $ -\\frac{b^{3}}{c^{2}} \\leqslant -b $, i.e., $ a^{2} \\geqslant 2c^{2} $, $ |PA|_{\\max} = \\sqrt{4a^{2} - 4c^{2}} \\leqslant 2b $, which always holds. From $ a^{2} \\geqslant 2c^{2} $, we get $ \\frac{c^{2}}{a^{2}} \\leqslant \\frac{1}{2} $, so $ 0 < e \\leqslant \\frac{\\sqrt{2}}{2} $. In conclusion, the range of eccentricity is $ \\left(0, \\frac{\\sqrt{2}}{2}\\right] $."
  },
  {
    "text": "Let point $P(x_{1}, y_{1})$ lie on the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{2}=1$, and point $Q(x_{2}, y_{2})$ lie on the line $x+2 y-8=0$. Then the minimum value of $|x_{2}-x_{1}|+|y_{2}-y_{1}|$ is?",
    "fact_expressions": "P: Point;Coordinate(P) = (x1, y1);x1: Number;y1: Number;G: Ellipse;Expression(G) = (x^2/8 + y^2/2 = 1);PointOnCurve(P, G) = True;Q: Point;Coordinate(Q) = (x2, y2);x2: Number;y2: Number;H: Line;Expression(H) = (x + 2*y - 8 = 0);PointOnCurve(Q, H) = True",
    "query_expressions": "Min(Abs(-x1 + x2) + Abs(-y1 + y2))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 19]], [[1, 19]], [[2, 19]], [[2, 19]], [[20, 57]], [[20, 57]], [[1, 58]], [[59, 77]], [[59, 77]], [[60, 77]], [[60, 77]], [[78, 91]], [[78, 91]], [[59, 92]]]",
    "query_spans": "[[[94, 129]]]",
    "process": "Let $x_{1}=2\\sqrt{2}\\cos\\theta$, $y_{1}=\\sqrt{2}\\sin\\theta$ and $\\theta\\in[0,2\\pi)$. $|x_{2}-x_{1}|+|y_{2}-y_{1}|=|x_{2}-2\\sqrt{2}\\cos\\theta|+|y_{2}-\\sqrt{2}\\sin\\theta|=\\frac{1}{2}(2|x_{2}-2\\sqrt{2}\\cos\\theta|+2|y_{2}-\\sqrt{2}\\sin\\theta|\\geqslant\\frac{1}{2}(|x_{2}-2\\sqrt{2}\\cos\\theta|+2|y_{2}-\\sqrt{2}\\sin\\theta|)\\geqslant\\frac{1}{2}|x_{2}+2y_{2}-2\\sqrt{2}\\cos\\theta-2\\sqrt{2}\\sin\\theta=\\frac{1}{2}|8-2\\sqrt{2}(\\cos\\theta+\\sin\\theta)|=\\frac{1}{2}|8-4\\sin(\\theta+\\frac{\\pi}{4})|\\geqslant2$, equality holds if and only if $\\sin(\\theta+\\frac{\\pi}{4})=1$ and $x_{2}-2\\sqrt{2}\\cos\\theta=0$. Hence the answer is: $2$"
  },
  {
    "text": "The standard equation of the hyperbola with foci on the $y$-axis passing through points $P_{1}(3, -4 \\sqrt{2})$ and $P_{2}(\\frac{9}{4}, 5)$ is?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), yAxis) = True;P1: Point;Coordinate(P1) = (3, -4*sqrt(2));PointOnCurve(P1,G) = True;P2: Point;Coordinate(P2) = (9/4,5);PointOnCurve(P2,G) = True",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/16 - x^2/9 = 1",
    "fact_spans": "[[[63, 66]], [[0, 66]], [[11, 35]], [[11, 35]], [[10, 66]], [[38, 62]], [[38, 62]], [[10, 66]]]",
    "query_spans": "[[[63, 73]]]",
    "process": ""
  },
  {
    "text": "Let $F$ be the right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and $O$ be the coordinate origin. The circle with diameter $OF$ intersects the circle $x^{2}+y^{2}=a^{2}$ at points $P$ and $Q$. If $|PQ|=|OF|$, then the asymptotes of $C$ have equations?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;H: Circle;O: Origin;F: Point;P: Point;Q: Point;D:Circle;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(D) = (x^2 + y^2 = a^2);RightFocus(C)=F;IsDiameter(LineSegmentOf(O,F),H);Intersection(H,D)={P,Q};Abs(LineSegmentOf(P, Q)) = Abs(LineSegmentOf(O, F))",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*x",
    "fact_spans": "[[[5, 65], [140, 143]], [[12, 65]], [[12, 65]], [[90, 91]], [[70, 73]], [[1, 4]], [[114, 117]], [[118, 121]], [[92, 112]], [[12, 65]], [[12, 65]], [[5, 65]], [[92, 112]], [[1, 69]], [[79, 91]], [[90, 123]], [[125, 138]]]",
    "query_spans": "[[[140, 151]]]",
    "process": "According to the geometric properties of hyperbolas and circles, the relationship between the segments is derived as |OP|^{2}+|PF|^{2}=|OF|, leading to a^{2}=b^{2}, from which the equations of the asymptotes of the hyperbola can be obtained. From the given conditions, the figure below is obtained: from the properties of hyperbola C, F(c,0) is obtained. Also, the circle with OF as diameter intersects the circle x^{2}+y^{2}=a^{2} at points P and Q, and |PQ|=|OF|, so PQ is the diameter of the circle with OF as diameter, OP=a, PF=OP; thus |OP|^{2}+|PF|^{2}=|OF|, then a^{2}+a^{2}=c^{2}, 2a^{2}=c^{2}, so a^{2}=b^{2}, therefore the asymptotes of C are y=\\pm x. This problem mainly examines the geometric properties of hyperbolas and circles, with the key being to derive segment relationships from their geometric properties and establish connections with a, b, c of the hyperbola, making it a medium-difficulty problem."
  },
  {
    "text": "If the symmetry axes of an ellipse are the coordinate axes, the sum of the lengths of the major axis and minor axis is $18$, and the coordinates of one focus are $(3 , 0)$, then what is the standard equation of the ellipse?",
    "fact_expressions": "G: Ellipse;SymmetryAxis(G)=axis;Length(MajorAxis(G))+Length(MinorAxis(G))=18;Coordinate(OneOf(Focus(G))) = (3, 0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/25+y^2/16=1",
    "fact_spans": "[[[1, 3], [46, 48]], [[1, 11]], [[1, 26]], [[1, 44]]]",
    "query_spans": "[[[46, 55]]]",
    "process": "From the given conditions, we have  \n\\begin{cases}2a+2b=18\\\\c=3\\\\\\end{cases}  \nSolving gives  \n\\begin{cases}a=\\frac{1}{5}\\\\b=4\\end{cases}  \n=b^{2}+c^{2}  \nSince the foci of the ellipse lie on the x-axis, the standard equation of the ellipse is  \n\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1"
  },
  {
    "text": "The equation of the directrix of the parabola $x^{2}=4 y$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=-1",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": ""
  },
  {
    "text": "The line $2x + y - 10 = 0$ passes through a focus of the hyperbola $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ $(a > 0, b > 0)$ and is perpendicular to one of the asymptotes of the hyperbola. Then the equation of the hyperbola is?",
    "fact_expressions": "H: Line;Expression(H) = (2*x + y - 10 = 0);G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;PointOnCurve(OneOf(Focus(G)),H) = True;IsPerpendicular(H,OneOf(Asymptote(G))) = True",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/20 - y^2/5 = 1",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[15, 71], [80, 83], [94, 97]], [[15, 71]], [[18, 71]], [[18, 71]], [[18, 71]], [[18, 71]], [[0, 76]], [[0, 91]]]",
    "query_spans": "[[[94, 102]]]",
    "process": "Determine the position of the foci from the hyperbola equation, and find $ c $ according to the given conditions. Then, since the line is perpendicular to an asymptote, we have $ \\frac{b}{a} = \\frac{1}{2} $, and thus can find $ a^{2} $, $ b^{2} $, and write the hyperbola equation. From the problem, the foci of the hyperbola lie on the x-axis, so one focus is $ (5,0) $, i.e., $ c=5 $. Also, $ \\because $ the asymptotes of the hyperbola are $ y = \\pm \\frac{b}{a}x $, and the line $ 2x + y - 10 = 0 $ is perpendicular to one asymptote, $ \\therefore \\frac{b}{a} = \\frac{1}{2} $, $ a^{2} + b^{2} = c^{2} = 25 $, yielding $ a^{2} = 20 $, $ b^{2} = 5 $. $ \\therefore $ the equation of the hyperbola is $ \\frac{x^{2}}{20} - \\frac{y^{2}}{5} = 1 $."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=4 x$ be $F$, and the directrix be $l$. Then, the equation of the circle with center $F$ and tangent to $l$ is?",
    "fact_expressions": "G: Parabola;H: Circle;F: Point;l: Line;Expression(G) = (y^2 = 4*x);Focus(G) = F;Directrix(G) = l;Center(H)=F;IsTangent(l,H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x-1)^2+y^2=4",
    "fact_spans": "[[[1, 15]], [[48, 49]], [[19, 22], [33, 36]], [[26, 29], [42, 45]], [[1, 15]], [[1, 22]], [[1, 29]], [[32, 49]], [[41, 49]]]",
    "query_spans": "[[[48, 54]]]",
    "process": "From the parabolic equation, the coordinates of the focus (i.e., the center of the circle) can be obtained, and the distance from the focus to the directrix is the radius, thus the result can be found. [Detailed Solution] For the parabola $ y^{2} = 4x $, we have $ 2p = 4 $, $ p = 2 $, the focus $ F(1,0) $, and the equation of the directrix $ l $ is $ x = -1 $. The equation of the circle centered at $ F $ and tangent to $ l $ is $ (x-1)^{2} + y^{2} = 2^{2} $, that is, $ (x-1)^{2} + y^{2} = 4 $. [Note] This problem mainly examines the focus coordinates of a parabola, the equation of the directrix of a parabola, the necessary and sufficient conditions for a line to be tangent to a circle, and other knowledge, aiming to develop students' transformation ability and computational solving ability."
  },
  {
    "text": "If a line $l$ passing through the point $P(-2 , 0)$ intersects the parabola $y^{2}=8 x$ at exactly one common point, then the equation of line $l$ is?",
    "fact_expressions": "l: Line;G: Parabola;P: Point;Expression(G) = (y^2 = 8*x);Coordinate(P) = (-2, 0);PointOnCurve(P,l);NumIntersection(l,G)=1",
    "query_expressions": "Expression(l)",
    "answer_expressions": "{y=0,x-y+2=0,x+y+2=0}",
    "fact_spans": "[[[15, 20], [15, 20]], [[21, 35]], [[2, 14]], [[21, 35]], [[2, 14]], [[1, 20]], [[15, 42]]]",
    "query_spans": "[[[44, 54]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse, respectively, $A$ is the bottom vertex of the ellipse, and the line $A F_{2}$ intersects the ellipse at another point $P$. If $|P F_{1}|=|P A|$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F2: Point;A: Point;P: Point;F1: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) =F2;LowerVertex(G)=A;Intersection(LineOf(A,F2),G) ={A,P};Abs(LineSegmentOf(P, F1)) = Abs(LineSegmentOf(P, A))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[2, 54], [73, 75], [86, 88], [105, 107], [135, 137]], [[4, 54]], [[4, 54]], [[63, 70]], [[82, 85]], [[111, 114]], [[55, 62]], [[4, 54]], [[4, 54]], [[2, 54]], [[55, 81]], [[55, 81]], [[82, 92]], [[82, 114]], [[116, 133]]]",
    "query_spans": "[[[135, 143]]]",
    "process": "As shown in the figure, point P lies on the ellipse, so |PF₁| + |PF₂| = 2a. From |PF₁| = |PA| = |PF₂| + |AF₂| and |AF₂| = a, substituting into the above equation gives |PF₁| = \\frac{3a}{2}, |PF₂| = \\frac{a}{2}. In \\triangle APF₁, \\cos\\angle PAF₁ = \\frac{|AF₁|^{2} + |AP|^{2} - |PF₁|^{2}}{2|AF₁||AP|} = \\frac{a^{2} + (\\frac{3a}{2})^{2} - (\\frac{3a}{2})^{2}}{2a \\times \\frac{3a}{2}} = \\frac{1}{3}. Also, \\cos\\angle PAF₁ = 1 - 2\\sin^{2}\\angle OAF₁ = \\frac{1}{3}, so \\sin\\angle OAF₁ = \\frac{\\sqrt{3}}{3}, that is, \\sin\\angle OAF₁ = \\frac{c}{a} = e = \\frac{\\sqrt{3}}{3}"
  },
  {
    "text": "The equation of the hyperbola with foci at the endpoints of the major axis of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ and vertices at the foci of the ellipse is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/25 + y^2/9 = 1);Focus(G)=Endpoint(MajorAxis(H));Focus(H)=Vertex(G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16 - y^2/9 = 1",
    "fact_spans": "[[[58, 61]], [[1, 39], [49, 51]], [[1, 39]], [[0, 61]], [[48, 61]]]",
    "query_spans": "[[[58, 65]]]",
    "process": "The ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ has left and right vertices at $(\\pm5,0)$ and foci at $(\\pm4,0)$. Therefore, the desired hyperbola has its foci on the $x$-axis, and we can set the standard equation of the hyperbola as $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$. Then $a=4$, the semi-focal distance $c=5$, so $b^{2}=9$. Thus, the hyperbola equation is $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$."
  },
  {
    "text": "On the parabola $y^{2}=2 p x$, the distance from the point with abscissa $2$ to the focus of the parabola is $3$. Then $p$=?",
    "fact_expressions": "G: Parabola;p: Number;P: Point;Expression(G) = (y^2 = 2*(p*x));PointOnCurve(P, G);XCoordinate(P)=2;Distance(P, Focus(G)) = 3",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 17], [29, 32]], [[43, 46]], [[27, 28]], [[1, 17]], [[1, 28]], [[19, 28]], [[27, 41]]]",
    "query_spans": "[[[43, 48]]]",
    "process": ""
  },
  {
    "text": "Given that point $M(x, y)$ moves on the curve $x^{2}+2 y^{2}=4$, and point $A$ is $(8,2)$, then the trajectory equation of the midpoint $P$ of $MA$ is?",
    "fact_expressions": "G: Curve;M: Point;A: Point;P: Point;x1:Number;y1:Number;Expression(G) = (x^2 + 2*y^2 = 4);Coordinate(Z) = (8, 2);Coordinate(M) = (x1, y1);PointOnCurve(M, G);MidPoint(LineSegmentOf(M, A)) = P;Z: Point;A = Z",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2+2*y^2-8*x-4*y+17=0",
    "fact_spans": "[[[13, 32]], [[2, 12]], [[36, 40]], [[57, 60]], [[3, 12]], [[3, 12]], [[13, 32]], [[41, 48]], [[2, 12]], [[2, 33]], [[50, 60]], [[41, 48]], [[36, 48]]]",
    "query_spans": "[[[57, 67]]]",
    "process": "Let P(m,n). Since point P is the midpoint of MA, and point M(x,y), A(8,2), we have \\begin{cases}m=\\frac{x+8}{2}\\\\n=\\frac{y+2}{2}\\end{cases}, so \\begin{cases}x=2m-8\\\\y=2n-2\\end{cases}. Since point M(x,y) lies on the curve x^{2}+2y^{2}=4, it follows that (2m-8)^{2}+2(2n-2)^{2}=4, thus m^{2}+2n^{2}-8m-4n+18=0. Therefore, the trajectory equation of the midpoint P of MA is x^{2}+2y^{2}-8x-4y+17=0."
  },
  {
    "text": "Given points $A$ and $B$ are the left and right vertices of the ellipse $C$: $\\frac{x^{2}}{4}+y^{2}=1$, point $M$ is a point on the $x$-axis, a perpendicular line through $M$ to the $x$-axis intersects the ellipse $C$ at points $P$ and $Q$, a perpendicular line through $M$ to $AP$ intersects $BQ$ at point $N$, then $\\frac{S_\\triangle{B M N}}{S_\\triangle{B M Q}}$=?",
    "fact_expressions": "C: Ellipse;A: Point;P: Point;B: Point;Q: Point;M: Point;N: Point;Expression(C) = (x^2/4 + y^2 = 1);LeftVertex(C) = A;RightVertex(C) = B;PointOnCurve(M, xAxis);L2:Line;PointOnCurve(M,L2);IsPerpendicular(L2,xAxis);Intersection(C, L2) = {P, Q};L1:Line;PointOnCurve(M,L1);IsPerpendicular(L1,LineSegmentOf(A, P));Intersection(L1, LineSegmentOf(B, Q)) = N",
    "query_expressions": "Area(TriangleOf(B, M, N))/Area(TriangleOf(B, M, Q))",
    "answer_expressions": "4/5",
    "fact_spans": "[[[11, 43], [76, 81]], [[2, 6]], [[82, 85]], [[7, 10]], [[86, 89]], [[93, 96], [50, 54], [64, 67]], [[112, 116]], [[11, 43]], [[2, 49]], [[2, 49]], [[50, 62]], [], [[63, 75]], [[63, 75]], [[63, 91]], [], [[92, 105]], [[92, 105]], [[92, 116]]]",
    "query_spans": "[[[118, 167]]]",
    "process": "As shown in the figure, let $ P(m,n) $, then $ M(m,0) $, $ Q(m,-n) $. From the given conditions, $ m \\neq \\pm 2 $ and $ n \\neq 0 $. The slope of line $ AP $ is $ k_{AP} = \\frac{n}{m+2} $, and the slope of line $ MN $ is $ k_{MN} = -\\frac{m+2}{n} $. Therefore, the equation of line $ MN $ is $ y = -\\frac{m+2}{n}(x - m) $, and the equation of line $ BQ $ is $ y = \\frac{n}{2 - m}(x - 2) $. Solving the system  \n$$\n\\begin{cases}\ny = -\\frac{m+2}{n}(x - m) \\\\\ny = \\frac{n}{2 - m}(x - 2)\n\\end{cases}\n$$\nyields $ y_N = -\\frac{n(4 - m)}{4 - m^2 + n^2} $. Since point $ P $ lies on the ellipse $ C $, we have $ 4 - m^2 = 4n^2 $, thus $ y_N = -\\frac{4}{5}n $. Also, $ S_{ABMN} = \\frac{1}{2}|BM| \\cdot |y_N| = \\frac{2}{5}|BM| \\cdot |n| $, and $ S_{ABMQ} = \\frac{1}{2}|BM| \\cdot |n| $. Therefore, $ \\frac{S_{ABMN}}{S_{ABMQ}} = \\frac{4}{5} $."
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{\\lambda-2}-\\frac{y^{2}}{\\lambda-4}=1$ has eccentricity $e=\\frac{2}{\\sqrt{3}}$, then the equations of its asymptotes are?",
    "fact_expressions": "G: Hyperbola;lambda: Number;Expression(G) = (x^2/(lambda - 2) - y^2/(lambda - 4) = 1);e: Number;Eccentricity(G) = e;e = 2/sqrt(3)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "{(y=pm*sqrt(3)*x/3), (y=pm*sqrt(3)*x)}",
    "fact_spans": "[[[0, 54], [82, 83]], [[3, 54]], [[0, 54]], [[58, 80]], [[0, 80]], [[58, 80]]]",
    "query_spans": "[[[82, 90]]]",
    "process": "From the given condition: $(\\lambda-2)(\\lambda-4)>0$, solving yields: $\\lambda>4$ or $\\lambda<2$. When $\\lambda>4$, $c^{2}=\\lambda-2+\\lambda-4=2\\lambda-6$, then $e=\\frac{\\sqrt{22-6}}{\\sqrt{2-2}}=\\frac{2}{\\sqrt{3}}$, solving gives: $\\lambda=5$, at this time the hyperbola equation is: $\\frac{x2}{3}-y^{2}=1$, asymptote equations are: $y=\\pm\\frac{\\sqrt{3}}{3}x$; when $\\lambda<2$, $c^{2}=2-\\lambda+4-\\lambda=6-2\\lambda$, then $e=\\frac{\\sqrt{6-2\\lambda}}{\\sqrt{4-\\lambda}}=\\frac{2}{\\sqrt{3}}$, solving gives: $\\lambda=1$, at this time the hyperbola equation is: $\\frac{y^{2}}{3}-x^{2}=1$, asymptote equations are: $y=\\pm\\sqrt{3}x$."
  },
  {
    "text": "Given that $A$ and $B$ are the intersection points of a line passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) with the parabola, $O$ is the origin, satisfying $\\overrightarrow{A F}=2 \\overrightarrow{F B}$, $S_{\\triangle O A B}=\\sqrt{3}|A B|$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;A: Point;F: Point;B: Point;O: Origin;p>0;Expression(G) = (y^2 = 2*p*x);Focus(G)=F;PointOnCurve(F,H);Intersection(H,G)={A,B};VectorOf(A, F) = 2*VectorOf(F, B);Area(TriangleOf(O,A,B))=sqrt(3)*Abs(LineSegmentOf(A,B))",
    "query_expressions": "p",
    "answer_expressions": "3*sqrt(6)",
    "fact_spans": "[[[11, 32], [41, 44]], [[144, 147]], [[38, 40]], [[2, 5]], [[34, 37]], [[6, 9]], [[48, 51]], [[14, 32]], [[11, 32]], [[11, 37]], [[10, 40]], [[2, 47]], [[59, 104]], [[106, 141]]]",
    "query_spans": "[[[144, 151]]]",
    "process": "Let the angle of inclination of the focal chord be $\\alpha$. From the focal radius formula of a parabola's focal chord, we have: $|AF| = \\frac{p}{1 - \\cos\\alpha}$, $|BF| = \\frac{p}{1 + \\cos\\alpha}$. Hence, $\\frac{p}{1 - \\cos\\alpha} = 2 \\times \\frac{p}{1 + \\cos\\alpha}$, solving gives: $\\cos\\alpha = \\frac{1}{3}$, so $\\sin\\alpha = \\frac{2\\sqrt{2}}{3}$. Let the distance from the origin to line $AB$ be $h$, then $S_{\\triangle OAB} = \\sqrt{3}|AB| = \\frac{1}{2} \\times |AB| \\times h$, $\\therefore h = 2\\sqrt{2}$. By the definition of trigonometric functions: $\\sin\\alpha =$ that is, $\\frac{2\\sqrt{2}}{3} = \\frac{4\\sqrt{3}}{p}$, solving gives: $p = 3\\sqrt{6}$."
  },
  {
    "text": "The distance from the focus of the parabola $y^{2}=8 x$ to the asymptotes of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (x^2/3 - y^2 = 1);Expression(H) = (y^2 = 8*x)",
    "query_expressions": "Distance(Focus(H),Asymptote(G))",
    "answer_expressions": "1",
    "fact_spans": "[[[18, 46]], [[0, 14]], [[18, 46]], [[0, 14]]]",
    "query_spans": "[[[0, 55]]]",
    "process": "The coordinates of the focus of $ y^{2}=8x $ are $ (2,0) $. An asymptote of the hyperbola $ \\frac{x^{2}}{3}-y^{2}=1 $ is $ y=\\frac{\\sqrt{3}}{3}x $, that is, $ \\sqrt{3}x-3y=0 $. The distance from the point $ (2,0) $ to the line $ \\sqrt{3}x-3y=0 $ is $ d=\\frac{|2\\sqrt{3}-0|}{\\sqrt{3+9}}=\\frac{2\\sqrt{3}}{2\\sqrt{3}}=1 $."
  },
  {
    "text": "The shortest distance from a point on the parabola $x^{2}=2 p y(p>0)$ to the line $y=x-5$ is $\\sqrt{2}$, then the value of the positive number $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;P0: Point;p>0;Expression(G) = (x^2 = 2*(p*y));Expression(H) = (y = x - 5);PointOnCurve(P0, G);Min(Distance(P0, H)) = sqrt(2)",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[0, 21]], [[52, 57]], [[25, 34]], [[23, 24]], [[3, 21]], [[0, 21]], [[25, 34]], [[0, 24]], [[23, 50]]]",
    "query_spans": "[[[52, 61]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and $P$ is an arbitrary point on the right branch, if the maximum value of $\\frac{|P F_{1}|^{2}}{|P F_{2}|^{2}+4 a^{2}}$ is $2$, then the range of the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, RightPart(C));Max(Abs(LineSegmentOf(P, F1))^2/(4*a^2 + Abs(LineSegmentOf(P, F2))^2)) = 2",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(1, 3]",
    "fact_spans": "[[[20, 81], [156, 162]], [[20, 81]], [[28, 81]], [[28, 81]], [[28, 81]], [[28, 81]], [[2, 9]], [[10, 17]], [[2, 87]], [[2, 87]], [[88, 91]], [[20, 99]], [[101, 154]]]",
    "query_spans": "[[[156, 172]]]",
    "process": "According to the definition of hyperbola, |PF_{1}| - |PF_{2}| = 2a, that is, |PF_{1}| = |PF_{2}| + 2a. Let t = |PF_{2}|, then \\frac{|PF_{1}|^{2}}{|PF_{2}|^{2}+4a^{2}} = \\frac{(t+2a)^{2}}{t^{2}+4a^{2}} = \\frac{t^{2}+4at+4a^{2}}{t^{2}+4a^{2}} = 1 + \\frac{4a}{t+\\frac{4a^{2}}{t}} \\leqslant 2. The equality holds if and only when t = 2a, \\frac{|PF_{1}|^{2}}{|PF_{2}|^{2}+4a^{2}} reaches the maximum value 2, i.e., 2a \\geqslant c - a. Therefore, the range of eccentricity of hyperbola C is (1,3]."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with asymptotes given by $y=\\pm x$, and the right vertex at point $(1,0)$. If a line passing through point $P(0,-1)$ intersects the right branch of hyperbola $C$ at two distinct points $M$, $N$, then the range of the intercept $t$ on the $y$-axis of the perpendicular bisector $l$ of segment $MN$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;M: Point;N: Point;P: Point;F1:Point;l:Line;t:Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (1, 0);RightVertex(C)=F1;Coordinate(P) = (0, -1);PointOnCurve(P,G);Intersection(G,RightPart(C))={M,N};Negation(M=N);Expression(Asymptote(C)) = (y = pm*x);PerpendicularBisector(LineSegmentOf(M,N))=l;Intercept(l,yAxis)=t",
    "query_expressions": "Range(t)",
    "answer_expressions": "(2,+oo)",
    "fact_spans": "[[[2, 63], [111, 117]], [[10, 63]], [[10, 63]], [[108, 110]], [[127, 130]], [[131, 134]], [[97, 107]], [[84, 92]], [[147, 150]], [[158, 161]], [[10, 63]], [[10, 63]], [[2, 63]], [[84, 92]], [[2, 92]], [[97, 107]], [[95, 110]], [[108, 134]], [[122, 134]], [[2, 79]], [[136, 150]], [[147, 161]]]",
    "query_spans": "[[[158, 168]]]",
    "process": "From the given conditions, the hyperbola equation is $x^{2}-y^{2}=1$. Let the line equation be $y=kx-1$. Combining  \n\\[\n\\begin{cases}\nx^{2}-y^{2}=1 \\\\\ny=kx-1\n\\end{cases}\n\\]\nand eliminating $y$, we obtain $(1-k^{2})x^{2}+2kx-2=0$. According to the problem, this equation has two positive roots, so  \n\\[\n\\frac{4=4k^{2}+8(1-k}{1-k^{2}}>0 \\quad \\frac{-2}{1-k^{2}>0}-k^{2})>0\n\\]  \nSolving gives $1<k<\\sqrt{2}$. The midpoint of $MN$ is $\\left(\\frac{-k}{1-k^{2}}, \\frac{-1}{1-k^{2}}\\right)$, so the perpendicular bisector of segment $MN$ is  \n\\[\ny+\\frac{1}{1-k^{2}}=-\\frac{1}{k}\\left(x+\\frac{k}{1-k^{2}}\\right).\n\\]  \nSetting $x=0$, the intercept is $t=\\frac{-2}{1-12}=\\frac{2}{12}>2$. Answer: $(2,+\\infty)$. This problem examines the standard equation and properties of hyperbolas, and the relationship between a hyperbola and a line; it is a medium-difficulty problem."
  },
  {
    "text": "The left and right foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ are $F_{1}$ and $F_{2}$, respectively. If point $P$ lies on the hyperbola and $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, then $|\\overrightarrow{P F_{1}}+\\overrightarrow{P F_{2}}|$=?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/16 - y^2/9 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P, G);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0",
    "query_expressions": "Abs(VectorOf(P, F1) + VectorOf(P, F2))",
    "answer_expressions": "10",
    "fact_spans": "[[[0, 39], [69, 72]], [[64, 68]], [[47, 54]], [[55, 62]], [[0, 39]], [[0, 62]], [[0, 62]], [[64, 73]], [[74, 133]]]",
    "query_spans": "[[[135, 190]]]",
    "process": "Connect PO, then we have |\\overrightarrow{PF_{1}}+\\overrightarrow{PF_{2}}|=2|\\overrightarrow{PO}|=2c, thus obtain correctly connecting PO. Since O is the midpoint of F_{1}, F_{2}, it follows that \\overrightarrow{PF}_{1}+\\overrightarrow{PF_{2}}=2\\overrightarrow{PO}, so |\\overrightarrow{PF_{1}}+\\overrightarrow{PF_{2}}|=2|\\overrightarrow{PO}|. And \\overrightarrow{PF}\\cdot\\overrightarrow{PF_{2}}=0, hence \\triangle PF_{2}F_{1} is a right triangle with right angle at P, therefore |\\overrightarrow{PF_{1}}+\\overrightarrow{PF_{2}}|=2|\\overrightarrow{PO}|=2|\\overrightarrow{F_{1}O}|=10"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+y^{2}=1$, and a moving point $P(a, 0)$ on the $x$-axis. If there exists a circle $O$ centered at point $P$ such that the ellipse $C$ and the circle $O$ have four distinct common points, then what is the range of values for $a$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2 = 1);P: Point;a: Number;Coordinate(P) = (a, 0);PointOnCurve(P, xAxis);O: Circle;Center(O) = P;NumIntersection(C, O) = 4",
    "query_expressions": "Range(a)",
    "answer_expressions": "(-3/2, 3/2)",
    "fact_spans": "[[[2, 34], [73, 78]], [[2, 34]], [[36, 45], [58, 62]], [[94, 97]], [[36, 45]], [[36, 54]], [[66, 70], [79, 83]], [[57, 70]], [[73, 92]]]",
    "query_spans": "[[[94, 104]]]",
    "process": "Let an intersection point of the ellipse and the circle be $ Q(2\\cos\\theta,\\sin\\theta) $. Then  \n$ PQ^{2} = (2\\cos\\theta - a)^{2} + \\sin^{2}\\theta = 3\\cos^{2}\\theta - 4a\\cos\\theta + a^{2} + 1 = 3\\left(\\cos\\theta - \\frac{2a}{3}\\right)^{2} - \\frac{a^{2}}{3} + 1 $.  \nAccording to the problem, $ PQ $ is less than the distance from $ P $ to the left and right vertices, so the minimum value of the function $ f(\\theta) = 3\\left(\\cos\\theta - \\frac{2a}{3}\\right)^{2} - \\frac{a^{2}}{3} + 1 $ cannot occur at the left or right vertices. Hence, $ -1 < \\frac{2a}{3} < 1 $, that is, $ a \\in \\left(-\\frac{3}{2}, \\frac{3}{2}\\right) $."
  },
  {
    "text": "Let $P$ be a moving point on the curve $\\frac{x^{2}}{4}-y^{2}=1$, $O$ be the origin, and $M$ be the midpoint of the segment $OP$. Then the equation of the trajectory of point $M$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P, G) = True;G: Curve;Expression(G) = (x^2/4 - y^2 = 1);O: Origin;M: Point;MidPoint(LineSegmentOf(O, P)) = M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2-4*y^2=1",
    "fact_spans": "[[[1, 4]], [[1, 36]], [[5, 32]], [[5, 32]], [[37, 40]], [[46, 49], [61, 65]], [[46, 59]]]",
    "query_spans": "[[[61, 72]]]",
    "process": ""
  },
  {
    "text": "Given that point $F$ is the focus of the parabola $C$: $y^{2}=4x$, point $B$ lies on the parabola $C$, and point $A(5,4)$. When the perimeter of $\\triangle ABF$ is minimized, what is the area of this triangle?",
    "fact_expressions": "C: Parabola;A: Point;B: Point;F: Point;Expression(C) = (y^2 = 4*x);Coordinate(A) = (5, 4);Focus(C) = F;PointOnCurve(B, C);WhenMin(Perimeter(TriangleOf(A, B, F)))",
    "query_expressions": "Area(TriangleOf(A, B, F))",
    "answer_expressions": "2",
    "fact_spans": "[[[7, 26], [35, 41]], [[43, 51]], [[30, 34]], [[2, 6]], [[7, 26]], [[43, 51]], [[2, 29]], [[30, 42]], [[52, 75]]]",
    "query_spans": "[[[52, 85]]]",
    "process": "Draw a perpendicular line from point B to the directrix $ x = -1 $ of the parabola C, with the foot of the perpendicular being point $ B_{1} $. Since the perimeter $ L = |AF| + |AB| + |BF| = 4\\sqrt{2} + |AB| + |BB_{1}| $, the perimeter of $ \\triangle ABF $ is minimized when points A, B, and $ B_{1} $ are collinear. At this time, the coordinates of point B are $ (4, 4) $, and the area of $ \\triangle ABF $ is $ S = \\frac{1}{2} \\times 1 \\times 4 = 2 $."
  },
  {
    "text": "Given points $A(-1,0)$, $B(1 , 0)$, if point $C(x , y)$ satisfies $2 \\sqrt{(x-1)^{2}+y^{2}}=|x-4|$, then $|A C|+|B C|$=?",
    "fact_expressions": "A: Point;B: Point;C: Point;Coordinate(A) = (-1, 0);Coordinate(B) = (1, 0);Coordinate(C) = (x1, y1);x1:Number;y1:Number;2*sqrt(y1^2 + (x1 - 1)^2) = Abs(x1 - 4)",
    "query_expressions": "Abs(LineSegmentOf(A, C)) + Abs(LineSegmentOf(B, C))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 12]], [[13, 24]], [[26, 37]], [[2, 12]], [[13, 24]], [[26, 37]], [[27, 37]], [[27, 37]], [[39, 71]]]",
    "query_spans": "[[[73, 88]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse equation $\\frac{x^{2}}{4}+y^{2}=1$, $AB$ is a chord of the ellipse passing through the right focus $F$, then the minimum value of $|A F|+2|F B|$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);F: Point;RightFocus(G) = F;PointOnCurve(F, LineSegmentOf(A, B))",
    "query_expressions": "Min(Abs(LineSegmentOf(A, F)) + 2*Abs(LineSegmentOf(F, B)))",
    "answer_expressions": "(3+2*sqrt(2))/4",
    "fact_spans": "[[[2, 4], [41, 43]], [[2, 33]], [[35, 40]], [[35, 40]], [[35, 52]], [[47, 50]], [[41, 50]], [[35, 52]]]",
    "query_spans": "[[[54, 74]]]",
    "process": "Combining the second definition of the ellipse, $\\frac{|AF|}{|AM|}=e$, let the angle of inclination of $AB$ be $\\theta$, we obtain $|AM|=|CF|-|AF|\\cdot\\cos\\theta$, combining these gives $|AF|=\\frac{1}{2+\\sqrt{3}\\cos\\theta}$, similarly $|BF|=\\frac{1}{2-\\sqrt{3}\\cos\\theta}$, then simplifying yields $|AF|+2|FB|=\\frac{6+\\sqrt{3}\\cos\\theta}{4-3\\cos^{2}\\theta}$, let $t=\\cos\\theta$, combining completing the square and the AM-GM inequality solves it. As shown in Figure 4, from $\\frac{x^{2}}{4}+y^{2}=1=c^{2}=4-1=3\\Rightarrow c=\\sqrt{3}$, by the second definition of the ellipse we get $\\frac{|AF|}{|AM|}=e=\\frac{c}{a}=\\frac{\\sqrt{3}}{2}$, $\\textcircled{1}$ also $|AM|=|CF|-|AF|\\cdot\\cos\\theta$, that is $|AM|=\\frac{a^{2}}{c}-c-|AF|\\cos\\theta=\\frac{4\\sqrt{3}}{3}-\\sqrt{3}-|AF|\\cos\\theta$, combining $\\textcircled{1}$ and $\\textcircled{2}$ solves to $|AF|=\\frac{1}{2+\\sqrt{3}\\cos\\theta}$, similarly we can find $|AF|+2|FB|=\\frac{1}{2+\\sqrt{3}\\cos\\theta}+\\frac{2}{2-\\sqrt{3}\\cos\\theta}=\\frac{6+\\sqrt{3}\\cos\\theta}{4-3\\cos^{2}\\theta}$. Let $t=\\cos\\theta$, $t\\in[-1,1]$, then $|AF|+2|FB|=\\frac{6+\\sqrt{3}t}{4-3t^{2}}$, $t\\in[-1,1]$, $\\frac{6+\\sqrt{3}t}{4-3t^{2}}=\\frac{6+\\sqrt{3}t}{-(6+\\sqrt{3}t)^{2}+12(6+\\sqrt{3}t)-32}=\\frac{1}{-(6+\\sqrt{3}t)-\\frac{32}{(6+\\sqrt{3}t)}+12}$, by AM-GM inequality, $(6+\\sqrt{3}t)\\cdot\\frac{32}{(6+\\sqrt{3}t)}\\geq 2\\sqrt{(6+\\sqrt{3}t)\\cdot\\frac{32}{(6+\\sqrt{3}t)}}$, equality holds when $6+\\sqrt{3}t=\\frac{32}{6+\\sqrt{3}t}$, i.e., $t=\\frac{4\\sqrt{6}-6\\sqrt{3}}{3}$, so the minimum value of $|AF|+2|FB|$ is $\\frac{3+2\\sqrt{2}}{4}$."
  },
  {
    "text": "If the line $a x - y + 1 = 0$ passes through the focus of the parabola $y^{2} = 4 x$, then the real number $a =$?",
    "fact_expressions": "H: Line;Expression(H) = (a*x - y + 1 = 0);a: Real;G: Parabola;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), H)",
    "query_expressions": "a",
    "answer_expressions": "-1",
    "fact_spans": "[[[1, 14]], [[1, 14]], [[35, 40]], [[16, 30]], [[16, 30]], [[1, 33]]]",
    "query_spans": "[[[35, 42]]]",
    "process": ""
  },
  {
    "text": "Given points $A(-2,0)$, $B(2,0)$, $C(3, \\sqrt{11})$, the moving point $M$ is such that the distance from $M$ to $A$ is $2$ more than the distance from $M$ to $B$. Then, the minimum value of the sum of the distances from $M$ to $B$ and $C$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (-2, 0);B: Point;Coordinate(B) = (2, 0);C: Point;Coordinate(C) = (3, sqrt(11));M: Point;Distance(M, A) = Distance(M, B) + 2",
    "query_expressions": "Min(Distance(M, B)+Distance(M, C))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 12], [49, 52]], [[2, 12]], [[14, 22], [57, 60], [75, 78]], [[14, 22]], [[25, 42], [79, 82]], [[25, 42]], [[45, 48], [71, 74]], [[45, 67]]]",
    "query_spans": "[[[71, 95]]]",
    "process": "Points A(-2,0), B(2,0), and the moving point M satisfies that the distance from M to A is 2 greater than the distance from M to B, so |MA| - |MB| = 2 < |AB| = 4. Therefore, the trajectory of the moving point M is the right branch of a hyperbola. Then, the sum of distances from M to points B and C, |MB| + |MC| = |MA| + |MC| - 2 \\geqslant |AC| - 2 = \\sqrt{(3+2)^{2}+(\\sqrt{11}-0)^{2}} - 2 = 4, with equality if and only if points M, A, and C are collinear. Hence, the minimum value of the sum of distances from the moving point M to points B and C is 4."
  },
  {
    "text": "The coordinates of the point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ where the product of the distances to the two foci is minimized are?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);WhenMin(Distance(P, F1)*Distance(P, F2))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(pm*5, 0)",
    "fact_spans": "[[[0, 38]], [[0, 38]], [], [], [[0, 44]], [[51, 52]], [[0, 52]], [[0, 52]]]",
    "query_spans": "[[[51, 57]]]",
    "process": ""
  },
  {
    "text": "Through the focus $F$ of the parabola $M$: $y^{2}=8x$, draw two lines $l_{1}$, $l_{2}$ whose slopes have a product of $-2$. Line $l_{1}$ intersects $M$ at points $A$, $C$, and line $l_{2}$ intersects $M$ at points $B$, $D$. Then the minimum value of $|AC|+|BD|$ is?",
    "fact_expressions": "M: Parabola;l1:Line;l2:Line;A: Point;C: Point;B: Point;D: Point;F:Point;Expression(M) = (y^2 = 8*x);Focus(M)=F;PointOnCurve(F,l1);PointOnCurve(F,l2);Slope(l1)*Slope(l2)=-2;Intersection(l1,M)={A,C};Intersection(l2,M)={B,D}",
    "query_expressions": "Min(Abs(LineSegmentOf(A, C)) + Abs(LineSegmentOf(B, D)))",
    "answer_expressions": "24",
    "fact_spans": "[[[1, 20], [69, 72], [69, 72]], [[39, 49], [61, 68]], [[50, 58], [83, 90]], [[73, 76]], [[77, 80]], [[95, 98]], [[99, 102]], [[23, 26]], [[1, 20]], [[1, 26]], [[0, 58]], [[0, 58]], [[29, 58]], [[61, 82]], [[83, 104]]]",
    "query_spans": "[[[106, 125]]]",
    "process": "According to the problem, let $ y = k(x - 2) $. Substituting into $ y^2 = 8x $, we get $ k^2x^2 - (4k^2 + 8)x + 4k^2 = 0 $. $ x_A + x_C = \\frac{4k^2 + 8}{k^2} - \\frac{8}{k^2} $, so $ |AC| = x_A + x_C + p = 8 + \\frac{8}{k^2} $. Replacing $ k $ with $ -\\frac{2}{k} $, we obtain $ |BD| = 8 + \\frac{8}{(-\\frac{2}{2})^2} = 8 + 2k^2 $. Therefore, $ |AC| + |BD| = 16 + 2k^2 + \\frac{8}{k^2} \\geqslant 16 + 2\\sqrt{2k^2 \\cdot \\frac{8}{k^2}} = 24 $."
  },
  {
    "text": "If the asymptotes of a hyperbola are given by $y = \\pm 3x$, and one of its foci is $(\\sqrt{10}, 0)$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*(3*x));Coordinate(OneOf(Focus(G)))=(sqrt(10),0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/9 = 1",
    "fact_spans": "[[[1, 4], [23, 24], [49, 52]], [[1, 22]], [[23, 47]]]",
    "query_spans": "[[[49, 57]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y=x^{2}$, what is the equation of the directrix of parabola $C$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y = x^2)",
    "query_expressions": "Expression(Directrix(C))",
    "answer_expressions": "y=-1/4",
    "fact_spans": "[[[2, 19], [21, 27]], [[2, 19]]]",
    "query_spans": "[[[21, 33]]]",
    "process": "Parabola C: y = x^{2}, that is, x^{2} = y, \\therefore p = \\frac{1}{2}, opens upward, hence the directrix equation is y = -\\frac{1}{4}"
  },
  {
    "text": "Given that the hyperbola $D$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has asymptotes $y=\\pm \\sqrt{3} x$, and the left and right foci are $F_{1}$ and $F_{2}$ respectively. If $P$ is any point on the right branch of the hyperbola $D$, then the range of $\\frac{|P F_{1}|-|P F_{2}|}{|P F_{1}|+|P F_{2}|}$ is?",
    "fact_expressions": "D: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(D) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(D) = F1;RightFocus(D) = F2;PointOnCurve(P, RightPart(D));Expression(Asymptote(D)) = (y = pm*sqrt(3)*x)",
    "query_expressions": "Range((Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)))/(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2))))",
    "answer_expressions": "(0, 1/2]",
    "fact_spans": "[[[26, 87], [117, 123]], [[34, 87]], [[34, 87]], [[113, 116]], [[96, 103]], [[104, 111]], [[34, 87]], [[34, 87]], [[26, 87]], [[26, 111]], [[26, 111]], [[113, 130]], [[2, 87]]]",
    "query_spans": "[[[132, 188]]]",
    "process": "The hyperbola $ D: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) has asymptotes $ y = \\pm\\sqrt{3}x $, then $ \\frac{b}{a} = \\sqrt{3} $, $ \\frac{c}{a} = \\sqrt{1 + \\left( \\frac{b}{a} \\right)^{2}} = 2 $. Let $ P $ be any point on the right branch of the hyperbola $ D $, then $ |PF_{1}| - |PF_{2}| = 2a $, $ |PF_{1}| + |PF_{2}| \\geqslant |F_{1}F_{2}| = 2c $, therefore $ 0 < \\frac{|PF_{1}| - |PF_{2}|}{|PF_{1}| + |PF_{2}|} \\leqslant \\frac{2a}{2c} = \\frac{1}{2} $,"
  },
  {
    "text": "The number of intersection points between the curve $y=-\\sqrt{1-x^{2}}$ and the curve $y+|a x|=0(a \\in R)$ is ?",
    "fact_expressions": "G: Curve;Expression(G) = (y = -sqrt(1 - x^2));Z: Curve;Expression(Z) = (y + Abs(a*x) = 0);a: Real",
    "query_expressions": "NumIntersection(G, Z)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 21]], [[0, 21]], [[22, 44]], [[22, 44]], [[24, 44]]]",
    "query_spans": "[[[0, 51]]]",
    "process": "Using the idea of combining numbers and shapes, as shown in the figure: it can be seen from the figure that there are $2$ intersection points."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=8x$ intersects the parabola at points $A$ and $B$. If $|AF|=6$, then $|AB|=$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 8*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, F)) = 6",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "9",
    "fact_spans": "[[[1, 15], [26, 29]], [[22, 24]], [[30, 33]], [[18, 21]], [[34, 37]], [[1, 15]], [[1, 21]], [[0, 24]], [[22, 39]], [[41, 50]]]",
    "query_spans": "[[[52, 61]]]",
    "process": "By the given condition, the focus of the parabola $ y^{2} = 8x $ is $ F(2,0) $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. According to the definition of the parabola, we have $ |AF| = x_{1} + 2 = 6 $, so $ x_{1} = 4 $, and thus $ y_{1} = \\pm 4\\sqrt{2} $. Without loss of generality, assume point $ A $ lies in the first quadrant, so $ A(4, 4\\sqrt{2}) $. Therefore, $ k_{AF} = 2\\sqrt{2} $, and the equation of line $ AB $ is $ y = 2\\sqrt{2}(x - 2) $. Solving the system of equations\n\\[\n\\begin{cases}\ny = 2\\sqrt{2}(x - 2) \\\\\ny^{2} = 8x\n\\end{cases}\n\\]\nyields $ x^{2} - 5x + 4 = 0 $, so $ x_{1} + x_{2} = 5 $. By the definition of the parabola, $ |AB| = x_{1} + x_{2} + p = 5 + 4 = 9 $."
  },
  {
    "text": "The equation of the chord of the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$ with midpoint at point $P(4,2)$ is?",
    "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/36 + y^2/9 = 1);Coordinate(P) = (4, 2);Z: LineSegment;MidPoint(Z) = P;IsChordOf(Z, G)",
    "query_expressions": "Expression(Z)",
    "answer_expressions": "x + 2*y - 8 = 0",
    "fact_spans": "[[[0, 38]], [[39, 48]], [[0, 38]], [[39, 48]], [], [[0, 53]], [[0, 53]]]",
    "query_spans": "[[[0, 58]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the midpoint of the chord cut by the parabola $y^{2}=-2 x$ from the line $y=x+1$ are?",
    "fact_expressions": "H: Line;Expression(H) = (y = x + 1);G: Parabola;Expression(G) = (y^2 = -2*x)",
    "query_expressions": "Coordinate(MidPoint(InterceptChord(H, G)))",
    "answer_expressions": "(-2, -1)",
    "fact_spans": "[[[0, 9]], [[0, 9]], [[10, 25]], [[10, 25]]]",
    "query_spans": "[[[0, 38]]]",
    "process": ""
  },
  {
    "text": "What is the eccentricity of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 43]]]",
    "process": ""
  },
  {
    "text": "What is the focal distance of the hyperbola $\\frac{x^{2}}{10}-\\frac{y^{2}}{2}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/10 - y^2/2 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 44]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x$ coincides with the right focus of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, then the equation of the directrix of this parabola is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;Expression(G) = (x^2 - y^2/3 = 1);Expression(H) = (y^2 = 2*p*x);Focus(H) = RightFocus(G)",
    "query_expressions": "Expression(Directrix(H))",
    "answer_expressions": "x=-2",
    "fact_spans": "[[[21, 49]], [[1, 17], [58, 61]], [[4, 17]], [[21, 49]], [[1, 17]], [[1, 55]]]",
    "query_spans": "[[[58, 68]]]",
    "process": "\\because the standard form of the hyperbola is: x^{2}-\\frac{y^{2}}{3}=1, \\therefore c=2, the right focus of the hyperbola is F(2,0). \\because the focus of the parabola y^{2}=2px (p>0) coincides with the right focus of the hyperbola x^{2}-\\frac{y^{2}}{3}=1. \\therefore \\frac{p}{2}=2, we obtain p=4."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, there exists a line passing through the left focus $F$ intersecting the ellipse $C$ at points $A$ and $B$ such that $\\frac{A F}{B F}=2$. What is the minimum eccentricity of the ellipse $C$?",
    "fact_expressions": "C: Ellipse;a: Number;b: Number;G: Line;A: Point;F: Point;B: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;PointOnCurve(F, G);Intersection(G, C) = {A, B};LineSegmentOf(A, F)/LineSegmentOf(B, F) = 2",
    "query_expressions": "Min(Eccentricity(C))",
    "answer_expressions": "1/3",
    "fact_spans": "[[[2, 59], [73, 78], [113, 118]], [[9, 59]], [[9, 59]], [[70, 72]], [[80, 83]], [[66, 69]], [[84, 87]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 69]], [[2, 72]], [[70, 89]], [[92, 111]]]",
    "query_spans": "[[[113, 127]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$. If there exists a point $P$ on the left branch of the hyperbola such that $|P F_{1}|=|F_{1} F_{2}|$ and the distance from $F_{1}$ to the line $P F_{2}$ is $\\sqrt{7} a$, then the eccentricity $e$ of the hyperbola is $?$.",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F2: Point;P: Point;F1: Point;e: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P,LeftPart(G));Abs(LineSegmentOf(P, F1)) = Abs(LineSegmentOf(F1, F2));Distance(F1,LineOf(P,F2)) = sqrt(7)*a;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "2",
    "fact_spans": "[[[19, 75], [83, 86], [163, 166]], [[22, 75]], [[22, 75]], [[9, 16]], [[91, 95]], [[1, 8], [125, 132]], [[170, 173]], [[22, 75]], [[22, 75]], [[19, 75]], [[1, 81]], [[1, 81]], [[83, 95]], [[98, 123]], [[125, 160]], [[163, 173]]]",
    "query_spans": "[[[170, 175]]]",
    "process": "Let the foot of the perpendicular from $ F_{1} $ to the line $ PF_{2} $ be $ M $. Since $ |PF_{1}| = |F_{1}F_{2}| = 2c $, $ M $ is the midpoint of $ |PF_{2}| $. From the given condition, $ |PF_{2}| = 2|MF_{2}| = 2\\sqrt{F_{1}F_{2}^{2} - MF_{1}^{2}} $. According to the definition of a hyperbola, $ |PF_{2}| - |PF_{1}| = 2a $, so $ |PF_{2}| = 2a + 2c $. Simplifying yields $ 3c^{2} - 2ac - 8a^{2} = 0 $, which factors as $ (3c + 4a)(c - 2a) = 0 $. Since $ e > 1 $, solving gives $ e = \\frac{c}{a} = 2 $."
  },
  {
    "text": "It is known that line $l$ passes through the left focus $F$ of a hyperbola and is tangent to the circle with the real axis as diameter. If line $l$ is exactly parallel to one of the asymptotes of the hyperbola, then the eccentricity of the hyperbola is?",
    "fact_expressions": "l: Line;G: Hyperbola;H: Circle;F: Point;LeftFocus(G) = F;PointOnCurve(F, l);IsDiameter(RealAxis(G),H);IsTangent(l,H);IsParallel(l, OneOf(Asymptote(G)))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 7], [33, 38]], [[8, 11], [39, 42], [55, 58]], [[28, 29]], [[15, 18]], [[8, 18]], [[2, 18]], [[8, 29]], [[2, 31]], [[33, 52]]]",
    "query_spans": "[[[55, 64]]]",
    "process": ""
  },
  {
    "text": "Let line $l$ pass through a focus of hyperbola $C$ and be perpendicular to the line containing the real axis of $C$. The line $l$ intersects $C$ at points $A$ and $B$. If $|AB|$ is twice the length of the imaginary axis of $C$, then the eccentricity of $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;PointOnCurve(OneOf(Focus(C)), l) = True;IsPerpendicular(l, OverlappingLine(RealAxis(C))) = True;Intersection(l, C) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(A, B)) = Length(ImageinaryAxis(C))*2",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[1, 6], [34, 37]], [[7, 13], [21, 24], [38, 41], [62, 65], [76, 79]], [[1, 18]], [[1, 33]], [[34, 52]], [[43, 46]], [[47, 50]], [[54, 74]]]",
    "query_spans": "[[[76, 85]]]",
    "process": "|AB| = \\frac{2b^{2}}{a} = 2 \\times 2b, b = 2a \\therefore c = \\sqrt{5}a, i.e. e = \\sqrt{5}"
  },
  {
    "text": "It is known that the center of the hyperbola is at the origin, the foci are on the $x$-axis, and one asymptote has the equation $y=\\frac{3}{4} x$. What is the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;O: Origin;Center(G) = O;PointOnCurve(Focus(G), xAxis);Expression(OneOf(Asymptote(G))) = (y = (3/4)*x)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/4",
    "fact_spans": "[[[2, 5], [49, 52]], [[9, 11]], [[2, 11]], [[2, 20]], [[2, 46]]]",
    "query_spans": "[[[49, 58]]]",
    "process": "Let the equation of the hyperbola be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0). Since one asymptote is y=\\frac{3}{4}x, \\therefore \\frac{b}{a}=\\frac{3}{4}, \\therefore \\frac{b^{2}}{a^{2}}=\\frac{c^{2}-a^{2}}{a^{2}}=\\frac{9}{16}. \\therefore The eccentricity of the hyperbola e=\\frac{c}{a}=\\frac{5}{4}."
  },
  {
    "text": "Given point $A(0,2)$, the focus of the parabola $y^{2}=2 p x(p>0)$ is $F$, the directrix is $l$, the line segment $FA$ intersects the parabola at point $B$, and from point $B$ a perpendicular is drawn to the directrix $l$, with foot of the perpendicular at $M$. If $AM \\perp MF$, then $p=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;A: Point;Coordinate(A) = (0, 2);l: Line;Directrix(G) = l;B: Point;Intersection(LineSegmentOf(F, A), G) = B;L: Line;PointOnCurve(B, L);IsPerpendicular(L, l);M: Point;FootPoint(L, l) = M;IsPerpendicular(LineSegmentOf(A, M), LineSegmentOf(M, F))",
    "query_expressions": "p",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[12, 33], [56, 59]], [[12, 33]], [[106, 109]], [[15, 33]], [[37, 40]], [[12, 40]], [[2, 11]], [[2, 11]], [[44, 47], [73, 76]], [[12, 47]], [[60, 64], [66, 70]], [[48, 64]], [], [[65, 79]], [[65, 79]], [[83, 86]], [[65, 86]], [[89, 104]]]",
    "query_spans": "[[[106, 111]]]",
    "process": "As shown in the figure: since AM\\botMF, point B is the midpoint of segment FA. Given point A(0,2), it follows that B(\\frac{p}{4},1). Since point B lies on the parabola, we have 1=2p\\times\\frac{p}{4}. Solving gives p=\\sqrt{2}."
  },
  {
    "text": "Let point $P$ be an arbitrary point on the circle $x^{2}+y^{2}=1$. From point $P$, draw a perpendicular line $P P_{0}$ to the $x$-axis, with foot at $P_{0}$, and $\\overrightarrow{M P_{0}}=\\frac{\\sqrt{3}}{2} \\overrightarrow{P P_{0}}$. Find the equation of the trajectory $C$ of point $M$?",
    "fact_expressions": "G: Circle;P: Point;P0: Point;M: Point;Expression(G) = (x^2 + y^2 = 1);PointOnCurve(P, G);PointOnCurve(P,LineSegmentOf(P,P0));IsPerpendicular(xAxis,LineSegmentOf(P,P0));FootPoint(xAxis,LineSegmentOf(P,P0))=P0;VectorOf(M, P0) = (sqrt(3)/2)*VectorOf(P, P0);Locus(M)=C;C:Curve",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 + y^2/(3/4) = 1",
    "fact_spans": "[[[6, 22]], [[29, 33], [1, 5]], [[54, 61]], [[135, 139]], [[6, 22]], [[1, 27]], [[28, 50]], [[28, 50]], [[28, 61]], [[63, 133]], [[135, 145]], [[142, 145]]]",
    "query_spans": "[[[142, 149]]]",
    "process": "Let M(x,y), P(m,n), then draw a perpendicular from point P to the x-axis, denoted as PP_{0}, with foot at P_{0}, and \\overrightarrow{MP_{0}} = \\frac{\\sqrt{3}}{2}\\overrightarrow{PP_{0}}, hence P_{0}(m,0), (m-x,-y) = \\frac{\\sqrt{3}}{2}(0,-n) \\therefore m = x, n = \\frac{2}{\\sqrt{3}}y. Since point P lies on the circle x^{2} + y^{2} = 1, \\therefore x^{2} + \\frac{4}{3}y^{2} = 1 \\therefore x^{2} + \\frac{y^{2}}{3} = 1"
  },
  {
    "text": "If the curves $\\frac{x^{2}}{4}+\\frac{y|y|}{9}=1$ and $k x+y-3=0$ have three distinct intersection points, then the range of values for $k$ is?",
    "fact_expressions": "G: Curve;C:Curve;k: Number;Expression(G) = (x^2/4 + (y*Abs(y))/9 = 1);Expression(C) = (k*x + y - 3 = 0);NumIntersection(G, C) = 3",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-3*sqrt(2)/2,-3/2)+(3/2,3*sqrt(2)/2)",
    "fact_spans": "[[[1, 37]], [[38, 51]], [[61, 64]], [[1, 37]], [[38, 51]], [[1, 59]]]",
    "query_spans": "[[[61, 71]]]",
    "process": "From the given conditions, when $ y \\geqslant 0 $, $ \\frac{x^{2}}{4} + \\frac{y^{2}}{9} = 1 $; when $ y < 0 $, $ \\frac{x^{2}}{4} - \\frac{y^{2}}{9} = 1 $, and the asymptotes are $ y = \\pm \\frac{3}{2}x $. As shown in the figure, from  \n$$\n\\begin{cases}\nkx + y - 3 = 0 \\\\\n\\frac{x^{2}}{4} - \\frac{y^{2}}{9} = 1\n\\end{cases}\n$$\nwe obtain $ (9 - 4k^{2})x^{2} + 24kx - 72 = 0 $. Therefore, $ A = (24k)^{2} + 288(9 - 4k^{2}) = 0 $, solving gives $ k = \\pm \\frac{3\\sqrt{2}}{2} $. Combining with the graph, the range of real values for $ k $ is"
  },
  {
    "text": "The distance from the focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ to its asymptote is equal to?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1)",
    "query_expressions": "Distance(Focus(G), Asymptote(G))",
    "answer_expressions": "b",
    "fact_spans": "[[[0, 59]], [[3, 59]], [[3, 59]], [[3, 59]], [[3, 59]], [[0, 59]]]",
    "query_spans": "[[[0, 72]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus is $F$, the right vertex is $A$, and point $B$ lies on the ellipse such that $B F \\perp x$-axis. The line $A B$ intersects the $y$-axis at point $P$. If $\\overrightarrow{A P}=2 \\overrightarrow{P B}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;B: Point;F: Point;P:Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F;RightVertex(G)=A;PointOnCurve(B,G);IsPerpendicular(LineSegmentOf(B,F),xAxis);Intersection(LineOf(A,B),yAxis)=P;VectorOf(A,P)=2*VectorOf(P,B)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 56], [79, 81], [167, 169]], [[4, 56]], [[4, 56]], [[70, 73]], [[74, 78]], [[61, 64]], [[112, 116]], [[4, 56]], [[4, 56]], [[2, 56]], [[2, 64]], [[2, 73]], [[74, 82]], [[84, 98]], [[99, 116]], [[120, 165]]]",
    "query_spans": "[[[167, 175]]]",
    "process": ""
  },
  {
    "text": "Let a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ be $F$, and let an endpoint of the imaginary axis be $B$. The line segment $BF$ intersects an asymptote of the hyperbola at point $A$. If $\\overrightarrow{F A}=2 \\overrightarrow{A B}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;B: Point;F: Point;A: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);OneOf(Focus(G))=F;OneOf(Endpoint(ImageinaryAxis(G)))=B;Intersection(LineSegmentOf(B,F),OneOf(Asymptote(G)))= A;VectorOf(F, A) = 2*VectorOf(A, B)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 60], [90, 93], [90, 93]], [[4, 60]], [[4, 60]], [[78, 81]], [[66, 69]], [[101, 105]], [[4, 60]], [[4, 60]], [[1, 60]], [[1, 69]], [[1, 81]], [[82, 105]], [[107, 152]]]",
    "query_spans": "[[[154, 163]]]",
    "process": "From $\\overrightarrow{FA}=2\\overrightarrow{AB}$, we get $\\overrightarrow{OA}=\\frac{1}{3}(\\overrightarrow{OF}+2\\overrightarrow{OB})$, thus finding the coordinates of point $A$. Then, since point $A$ lies on the asymptote $y=\\frac{b}{a}x$, the eccentricity of the hyperbola can be determined. Let point $F(c,0)$, $B(0,b)$. From $\\overrightarrow{FA}=2\\overrightarrow{AB}$, we have $\\overrightarrow{OA}-\\overrightarrow{OF}=2(\\overrightarrow{OB}-\\overrightarrow{OA})$. Therefore, $\\overrightarrow{OA}=\\frac{1}{3}(\\overrightarrow{OF}+2\\overrightarrow{OB})$, so $A\\left(\\frac{c}{3},\\frac{2b}{3}\\right)$. Since point $A$ lies on the asymptote $y=\\frac{b}{a}x$, we have $\\frac{2b}{3}=\\frac{b}{a}\\times\\frac{c}{3}$. Solving gives $e=\\frac{c}{a}=2$."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. If $\\overrightarrow{F A}=2 \\overrightarrow{B F}$, then what is the slope of the line $AB$?",
    "fact_expressions": "G: Parabola;H: Line;B: Point;A: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G)=F;PointOnCurve(F, H);Intersection(H, G) = {A, B};VectorOf(F, A) = 2*VectorOf(B, F)",
    "query_expressions": "Slope(LineOf(A,B))",
    "answer_expressions": "pm*2*sqrt(2)",
    "fact_spans": "[[[1, 15], [25, 28]], [[22, 24]], [[34, 37]], [[30, 33]], [[18, 21]], [[1, 15]], [[1, 21]], [[0, 24]], [[22, 39]], [[41, 86]]]",
    "query_spans": "[[[88, 100]]]",
    "process": "\\because the parabola C: y^{2} = 4x has focus F(1,0) and directrix x = -1, then the equation of line AB is y = k(x - 1). Solving the system \\begin{cases} y = k(x - 1) \\\\ y^{2} = 4x \\end{cases} yields k^{2}x^{2} - 2(2 + k^{2})x + k^{2} = 0. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), then x_{1} + x_{2} = \\frac{2(2 + k^{2})}{k^{2}}, x_{1} \\cdot x_{2} = 1, y_{1} + y_{2} = k(x_{1} + x_{2} - 2) = \\frac{4}{k} \\textcircled{1}. \\therefore \\overrightarrow{FA} = (1 - x_{1}, -y_{1}), \\overrightarrow{BF} = (x_{2} - 1, y_{2}), so \\begin{cases} 1 - x_{1} = 2(x_{2} - 1) \\\\ y_{1} = -2y_{2} \\end{cases} \\Rightarrow \\begin{cases} x_{1} = 3 - 2x_{2} \\\\ y_{1} = -2y_{2} \\end{cases} \\textcircled{2}. Combining \\textcircled{1} and \\textcircled{2} gives x_{2} = \\frac{k^{2} - 4}{k^{2}}, y_{2} = \\frac{4}{k}. Substituting into the parabola equation y^{2} = 4x yields k^{2} = 8, \\because k = \\pm 2\\sqrt{2}"
  },
  {
    "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ are given by $y=\\pm \\frac{3}{2} x$, then the eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Expression(Asymptote(G)) = (y = pm*x*3/2);e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(13)/2",
    "fact_spans": "[[[1, 58], [88, 91]], [[1, 58]], [[4, 58]], [[4, 58]], [[4, 58]], [[4, 58]], [[1, 86]], [[95, 98]], [[88, 98]]]",
    "query_spans": "[[[95, 100]]]",
    "process": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) are given by $y=\\pm\\frac{b}{a}x=\\pm\\frac{3}{2}x$. Therefore, the foci of the hyperbola lie on the $x$-axis, and $\\frac{b}{a}=\\frac{3}{2}$. Thus, the eccentricity of the hyperbola is $e=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{1+\\frac{9}{4}}=\\frac{\\sqrt{13}}{2}$."
  },
  {
    "text": "Given that the equation of the directrix of the parabola $y=a x^{2}$ is $y=-\\frac{1}{4}$, then $a$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = a*x^2);a: Number;Expression(Directrix(G)) = (y = -1/4)",
    "query_expressions": "a",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[40, 43]], [[2, 38]]]",
    "query_spans": "[[[40, 45]]]",
    "process": "Convert the equation of the parabola into standard form, and the value of $ a $ can be obtained by combining with the equation of the directrix. [Detailed solution] Since the parabola $ y = ax^{2} $ is converted into standard form as $ x^{2} = \\frac{1}{a}y $, we get $ p = \\frac{1}{2a} $, so the equation of the directrix is $ y = -\\frac{1}{4a} $, that is, $ -\\frac{1}{4a} = -\\frac{1}{4} $, solving gives $ a = $"
  },
  {
    "text": "$AB$ is a chord passing through the focus of the parabola $y^{2}=4x$, given that $AB=20$, then the equation of the line $AB$ is?",
    "fact_expressions": "A: Point;B: Point;G: Parabola;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), LineSegmentOf(A, B)) = True;IsChordOf(LineSegmentOf(A, B), G) = True;LineSegmentOf(A, B) = 20",
    "query_expressions": "Expression(LineOf(A, B))",
    "answer_expressions": "{(x+2*y-1=0),(x-2*y+1=0)}",
    "fact_spans": "[[[0, 4]], [[0, 4]], [[6, 20]], [[6, 20]], [[0, 26]], [[0, 26]], [[29, 36]]]",
    "query_spans": "[[[38, 49]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{m}=1$ are $y=\\pm \\frac{\\sqrt{2}}{2} x$, then $m$=?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/4 - y^2/m = 1);Expression(Asymptote(G)) = (y = pm*x*(sqrt(2)/2))",
    "query_expressions": "m",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 40]], [[77, 80]], [[2, 40]], [[2, 75]]]",
    "query_spans": "[[[77, 82]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of ellipse $E$, and parabola $C$ has $F_{1}$ as its vertex and $F_{2}$ as its focus. Let $P$ be an intersection point of the ellipse and the parabola. If the eccentricity $e$ of the ellipse satisfies $|PF_{1}|=e|PF_{2}|$, then what is the value of $e$?",
    "fact_expressions": "C: Parabola;E: Ellipse;P: Point;F1: Point;F2: Point;LeftFocus(E) = F1;RightFocus(E) = F2;Vertex(C)=F1;Focus(C)=F2;OneOf(Intersection(C,E))=P;Eccentricity(E)=e;e:Number;Abs(LineSegmentOf(P,F1))=e*Abs(LineSegmentOf(P,F2))",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[31, 37], [68, 71]], [[18, 23], [65, 67], [79, 81]], [[61, 64]], [[2, 9], [38, 45]], [[10, 17], [49, 56]], [[2, 30]], [[2, 30]], [[31, 48]], [[31, 59]], [[61, 76]], [[79, 87]], [[84, 87], [111, 114]], [[89, 109]]]",
    "query_spans": "[[[111, 118]]]",
    "process": ""
  },
  {
    "text": "Given that the vertex of the parabola is the center of the hyperbola $16 x^{2}-9 y^{2}=144$, and the focus is the left vertex of the hyperbola, then the equation of the parabola is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (16*x^2 - 9*y^2 = 144);Vertex(H) = Center(G);Focus(H)=LeftVertex(G)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2 = -12x",
    "fact_spans": "[[[42, 45], [9, 34]], [[2, 5], [51, 54]], [[9, 34]], [[2, 37]], [[2, 49]]]",
    "query_spans": "[[[51, 59]]]",
    "process": "Since the hyperbola is $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, the left vertex is $(-3,0)$, so the focus of the parabola is $(-3,0)$, therefore the equation of the parabola is $y^2=-12x$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, the line $2x+y+1=0$ intersects the $x$-axis at point $Q$, and $P$ is a moving point on the parabola. Then the maximum value of $\\frac{|PQ|}{|PF|}$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;G: Line;Expression(G) = (2*x + y + 1 = 0);Q: Point;Intersection(G, xAxis) = Q;P: Point;PointOnCurve(P, C)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, Q))/Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "2*sqrt(10)/5",
    "fact_spans": "[[[2, 21], [58, 61]], [[2, 21]], [[25, 28]], [[2, 28]], [[29, 42]], [[29, 42]], [[49, 53]], [[29, 53]], [[54, 57]], [[54, 67]]]",
    "query_spans": "[[[69, 96]]]",
    "process": "First, according to the distance formula between two points, we have $\\frac{|PQ|}{|PF|}=\\frac{\\sqrt{(x+\\frac{1}{x+1})^{2}+y^{2}}{}=\\frac{\\sqrt{x^{2}+5x+\\frac{1}{4}}}(x\\geqslant0)^{,}{}{-\\frac{115}{4t^{2}}+\\frac{3}{t}+1}$, then by $\\frac{1}{1}=u$ we get $\\frac{|PQ|}{|PF|}=\\sqrt{-\\frac{15}{4}(u-\\frac{2}{5})^{2}+\\frac{8}{5},}$ thus the maximum value can be found. From the problem, point $F(1,0)$, $Q(-\\frac{1}{2},0)$, let point $P(x,y)$, then $|PF|=x+1$, so $\\frac{|PQ|}{|PF|}=\\underline{\\sqrt{(}}\\frac{+x+\\frac{1}{4}+4x}{x+1}=\\frac{\\sqrt{x^{2}+5x+\\frac{1}{4}}}{x+1}(x\\geqslant0$. Let $x+1=t(t\\geqslant1)$, then $x=t-1$, so $\\frac{|PQ|}{|PF|}=\\frac{\\sqrt{(t-}}{}\\frac{-1)+\\frac{1}{4}}{t}=\\frac{\\sqrt{t^{2}+3t-\\frac{15}{4}}}{t}=\\sqrt{-\\frac{15}{4t^{2}}+\\frac{3}{t}+1}$. Let $\\frac{1}{t}=u$, then $0<u\\leqslant1$, so $\\frac{|PQ|}{|PF|}=\\sqrt{-\\frac{15}{4}u^{2}+3u+1}=\\sqrt{-\\frac{15}{4}(u-\\frac{2}{5})^{2}+\\frac{8}{5}}$. Therefore, when $u=\\frac{2}{5}$, i.e., $x=\\frac{3}{2}$, $(\\frac{|PQ|}{|PF|})_{\\max}=\\sqrt{\\frac{8}{5}}=\\frac{2\\sqrt{10}}{5}$."
  },
  {
    "text": "If the distance from a moving point $P$ to the point $F(2 , 0)$ is equal to its distance to the line $x+2=0$, then what is the equation of the trajectory of point $P$?",
    "fact_expressions": "P: Point;F: Point;Coordinate(F) = (2, 0);G: Line;Expression(G) = (x + 2 = 0);Distance(P, F) = Distance(P, G)",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "y^2=8*x",
    "fact_spans": "[[[3, 6], [22, 23], [40, 44]], [[7, 18]], [[7, 18]], [[24, 33]], [[24, 33]], [[3, 38]]]",
    "query_spans": "[[[40, 51]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has its right focus at $F$. A circle centered at $F$ with radius $a$ intersects an asymptote of the hyperbola $C$ at points $A$ and $B$. If $\\overrightarrow{O A}=2 \\overrightarrow{O B}$ ($O$ is the origin), then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;G: Circle;O: Origin;A: Point;B: Point;F: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;Center(G)=F;Radius(G)=a;Intersection(G, OneOf(Asymptote(C))) = {A, B};VectorOf(O, A) = 2*VectorOf(O, B)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(17)/3",
    "fact_spans": "[[[2, 63], [90, 96], [172, 178]], [[81, 84]], [[9, 63]], [[88, 89]], [[161, 164]], [[104, 107]], [[108, 111]], [[68, 71], [73, 76]], [[9, 63]], [[9, 63]], [[2, 63]], [[2, 71]], [[72, 89]], [[80, 89]], [[88, 113]], [[115, 160]]]",
    "query_spans": "[[[172, 184]]]",
    "process": "Through F, draw FE\\bot AB. Using the distance from a point to a line, we obtain |FE|=b. Then by the Pythagorean theorem, |OE|=a|BE|=\\sqrt{a^{2}-b^{2}}. From \\overrightarrow{OA}=2\\overrightarrow{OB}, we get |OE|=3|BE|. Thus, a relationship can be established and solved. As shown in the figure, through F, draw FE\\bot AB, then E is the midpoint of AB. Let the asymptote be y=\\frac{b}{a}x, then |FE|=\\frac{|bc|}{\\sqrt{a^{2}+b^{2}}}=b. In right triangle OEF, |OE|=\\sqrt{c^{2}-b^{2}}=a. In right triangle BEF, |BE|=\\sqrt{a^{2}-b^{2}}. Since \\overrightarrow{OA}=2\\overrightarrow{OB}, then |OE|=3|BE|, so a=3\\sqrt{a^{2}-b^{2}}, thus 8a^{2}=9b^{2}. Then 8a^{2}=9(c^{2}-a^{2}), so 17a^{2}=9c^{2}. Therefore, e=\\frac{c}{a}=\\frac{\\sqrt{17}}{3}"
  },
  {
    "text": "The focus of the parabola $C$: $y^{2}=8 x$ is $F$, and point $A$ is a point on $C$. If $|F A|=8$, then what is the inclination angle of the line $F A$?",
    "fact_expressions": "C: Parabola;F: Point;A: Point;Expression(C) = (y^2 = 8*x);Focus(C) = F;PointOnCurve(A, C);Abs(LineSegmentOf(F, A)) = 8",
    "query_expressions": "Inclination(LineOf(F,A))",
    "answer_expressions": "pi/3, 2*pi/3",
    "fact_spans": "[[[0, 19], [32, 35]], [[23, 26]], [[27, 31]], [[0, 19]], [[0, 26]], [[27, 39]], [[41, 50]]]",
    "query_spans": "[[[52, 65]]]",
    "process": "From the parabola equation, we know: $F(2,0)$. Let $A(x_{0},y_{0})$, then $|FA|=x_{0}+2=8$, solving gives: $x_{0}=6$. $\\therefore y_{0}^{2}=48$, solving gives: $y_{0}=\\pm4\\sqrt{3}$. $\\therefore$ the slope of line $FA$ is $\\frac{4\\sqrt{3}-0}{6-2}=\\sqrt{3}$ or $\\frac{-4\\sqrt{3}-0}{6-2}=-\\sqrt{3}$. $\\because$ the range of the inclination angle of a line is $[0,\\pi)$, $\\therefore$ the inclination angle of line $FA$ is $\\frac{\\pi}{3}$ or $\\frac{2\\pi}{3}$."
  },
  {
    "text": "Let the parabola $ C $: $ y^{2} = 2 p x $ ($ p > 0 $) have focus $ F $, point $ M $ lies on $ C $, $ |MF| = 5 $. If the circle with $ MF $ as diameter passes through the point $ (0,2) $, then the equation of $ C $ is?",
    "fact_expressions": "C: Parabola;p: Number;G: Circle;H: Point;M: Point;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(H) = (0, 2);Focus(C) = F;PointOnCurve(M, C);Abs(LineSegmentOf(M, F)) = 5;IsDiameter(LineSegmentOf(M, F),G);PointOnCurve(H, G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "{y^2=4*x,y^2=16*x}",
    "fact_spans": "[[[1, 26], [39, 42], [75, 78]], [[8, 26]], [[63, 64]], [[65, 73]], [[34, 38]], [[30, 33]], [[8, 26]], [[1, 26]], [[65, 73]], [[1, 33]], [[34, 43]], [[44, 52]], [[54, 64]], [[63, 73]]]",
    "query_spans": "[[[75, 83]]]",
    "process": ""
  },
  {
    "text": "Let $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$ be two distinct points on the parabola $C$: $x^{2}=2 p y(p>0)$, and let the perpendicular bisector of segment $A B$ be $y=x+b$. If $x_{1}+x_{2}=-\\frac{1}{2}$, then $p=?$",
    "fact_expressions": "A: Point;Coordinate(A) = (x1, y1);x1: Number;y1: Number;B: Point;Coordinate(B) = (x2, y2);x2: Number;y2: Number;C: Parabola;Expression(C) = (x^2 = 2*(p*y));p: Number;p>0;PointOnCurve(A, C) = True;PointOnCurve(B, C) = True;Negation(A=B);Expression(PerpendicularBisector(LineSegmentOf(A,B))) = (y=x+b);b: Number;x1+x2=-1/2",
    "query_expressions": "p",
    "answer_expressions": "1/4",
    "fact_spans": "[[[1, 18]], [[1, 18]], [[1, 18]], [[1, 18]], [[21, 38]], [[21, 38]], [[21, 38]], [[21, 38]], [[39, 65]], [[39, 65]], [[123, 126]], [[47, 65]], [[1, 71]], [[1, 71]], [[1, 71]], [[72, 93]], [[86, 93]], [[95, 121]]]",
    "query_spans": "[[[123, 128]]]",
    "process": "From the equation of the perpendicular bisector of segment AB, the slope of line AB can be determined. Using the point difference method, an equation involving p can be obtained, and thus the real value of p can be found. [Detailed solution] From the given, $x_{1}^{2}=2py_{1}$, $x_{2}^{2}=2py_{2}$. Subtracting these two equations gives $(x_{1}-x_{2})(x_{1}+x_{2})=2p(y_{1}-y_{2})$. Therefore, $\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{x_{1}+x_{2}}{2p}=k_{AB}$. Given that $k_{AB}=-1$, it follows that $x_{1}+x_{2}=-2p=-\\frac{1}{2}$, so $p=\\frac{1}{4}$."
  },
  {
    "text": "Given that the point $(1,2)$ lies on an asymptote of the hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0, b>0)$, then its eccentricity is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Point;a>0;b>0;Expression(G) = (-x^2/b^2 + y^2/a^2 = 1);Coordinate(H) = (1, 2);PointOnCurve(H, Asymptote(G))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[11, 67], [75, 76]], [[14, 67]], [[14, 67]], [[2, 10]], [[14, 67]], [[14, 67]], [[11, 67]], [[2, 10]], [[2, 73]]]",
    "query_spans": "[[[75, 81]]]",
    "process": "Since the point (1,2) lies on the asymptote of the hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0,b>0)$, the equation of the asymptote is $y=2x$, so $\\frac{a}{b}=2$. Therefore, $e=\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\frac{\\sqrt{5}}{2}$."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, point $A$ lies on the $y$-axis, and the midpoint $B$ of segment $AF$ lies on the parabola, then $|AF|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;A: Point;PointOnCurve(A, yAxis);B: Point;MidPoint(LineSegmentOf(A, F)) = B;PointOnCurve(B, G)",
    "query_expressions": "Abs(LineSegmentOf(A, F))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 16], [49, 52]], [[2, 16]], [[20, 23]], [[2, 23]], [[24, 28]], [[24, 34]], [[45, 48]], [[35, 48]], [[45, 53]]]",
    "query_spans": "[[[55, 64]]]",
    "process": "By the given conditions, the focus of the parabola $ y^{2} = 4x $ is $ F(1,0) $, point $ A $ lies on the $ y $-axis, and the midpoint $ B $ of segment $ AF $ lies on the parabola. By the midpoint formula, the horizontal coordinate of $ B $ is $ x = \\frac{1}{2} $, then the vertical coordinate of $ B $ is $ y = \\pm\\sqrt{2} $, so $ A(0, \\pm2\\sqrt{2}) $. Thus, $ |AF| = \\sqrt{1^{2} + (\\pm2\\sqrt{2})^{2}} = 3 $."
  },
  {
    "text": "It is known that the eccentricity of ellipse $C$ is $e=\\frac{\\sqrt{3}}{2}$, and its foci coincide with the foci of the hyperbola $x^{2}-2 y^{2}=4$. Then the equation of ellipse $C$ is?",
    "fact_expressions": "G: Hyperbola;C: Ellipse;e: Number;Expression(G) = (x^2 - 2*y^2 = 4);Eccentricity(C) = e;e = sqrt(3)/2;Focus(C) = Focus(G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/8+y^2/2=1",
    "fact_spans": "[[[40, 60]], [[2, 7], [35, 36], [67, 72]], [[11, 33]], [[40, 60]], [[2, 33]], [[11, 33]], [[35, 65]]]",
    "query_spans": "[[[67, 77]]]",
    "process": ""
  },
  {
    "text": "The minimum value of the sum of distances from a moving point $M$ on the parabola $y=-\\frac{1}{4} x^{2}$ to the two fixed points $(0,-1)$ and $(1,-3)$ is?",
    "fact_expressions": "G: Parabola;H: Point;I: Point;M:Point;Expression(G) = (y = -x^2/4);Coordinate(H) = (0, -1);Coordinate(I) = (1, -3);PointOnCurve(M,G)",
    "query_expressions": "Min(Distance(M,H)+Distance(M,I))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 25]], [[36, 44]], [[45, 53]], [[29, 32]], [[0, 25]], [[36, 44]], [[45, 53]], [[0, 32]]]",
    "query_spans": "[[[29, 64]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola $y^{2}=4 x$ is the right vertex of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, and one of the asymptotes of the hyperbola is given by $y=\\sqrt{3} x$. What is the eccentricity of the hyperbola?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 4*x);G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Focus(H) = RightVertex(G);Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[20, 76], [81, 84], [111, 114]], [[20, 76]], [[23, 76]], [[23, 76]], [[23, 76]], [[23, 76]], [[2, 80]], [[81, 109]]]",
    "query_spans": "[[[111, 120]]]",
    "process": ""
  },
  {
    "text": "The length of the imaginary axis of the hyperbola $m x^{2}+y^{2}=81$ is twice the length of the real axis, then $m$=?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (m*x^2 + y^2 = 81);Length(ImageinaryAxis(G)) = 2*Length(RealAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "-1/4",
    "fact_spans": "[[[0, 21]], [[36, 39]], [[0, 21]], [[0, 34]]]",
    "query_spans": "[[[36, 41]]]",
    "process": "According to the condition that the length of the imaginary axis is twice the length of the real axis, set up an equation to find the value of m. [Detailed solution] The hyperbola mx^{2}+y^{2}=81 can be rewritten as \\frac{y^{2}}{81}-\\frac{x^{2}}{\\frac{81}{m}}=1, where a=9, b=\\sqrt{\\frac{81}{m}}. Since the length of the imaginary axis is twice the length of the real axis, we have 2b=2\\times2a, so b=2a, and thus \\sqrt{\\frac{81}{m}}=2\\times9 \\Rightarrow m=-\\frac{1}{4}"
  },
  {
    "text": "Given that point $P$ is a moving point on the parabola $y^{2}=4x$, $A(1,\\ 0)$, $B(4,\\ 2)$, then the minimum value of $|PA|+|PB|$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G);A: Point;B: Point;Coordinate(A) = (1, 0);Coordinate(B) = (4, 2)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B)))",
    "answer_expressions": "5",
    "fact_spans": "[[[7, 21]], [[7, 21]], [[2, 6]], [[2, 24]], [[25, 35]], [[37, 47]], [[25, 35]], [[37, 47]]]",
    "query_spans": "[[[49, 66]]]",
    "process": ""
  },
  {
    "text": "If the distance from the focus to the directrix of the parabola $y=a x^{2}$ is $2$, then the real number $a$=?",
    "fact_expressions": "G: Parabola;a: Real;Expression(G) = (y = a*x^2);Distance(Focus(G), Directrix(G)) = 2",
    "query_expressions": "a",
    "answer_expressions": "pm*1/4",
    "fact_spans": "[[[1, 15]], [[30, 35]], [[1, 15]], [[1, 28]]]",
    "query_spans": "[[[30, 37]]]",
    "process": "The parabola $ y = ax^{2} $, that is, $ x^{2} = \\frac{1}{a}y' $, so $ 2p = \\pm\\frac{1}{a} $ ($ p > 0 $), thus when $ a > 0 $, $ p = \\frac{1}{2a} $, when $ a < 0 $, $ p = -\\frac{1}{2a} $. Since the distance from the focus to the directrix is 2, $ \\frac{1}{2a} = 2 $ or $ -\\frac{1}{2a} = 2 $, solving gives $ a = \\pm\\frac{1}{4} $, therefore the answer is:"
  },
  {
    "text": "The line $l$: $y = kx + 1$ has exactly one common point with the hyperbola $C$: $x^2 - y^2 = 1$, then $k = $?",
    "fact_expressions": "l: Line;C: Hyperbola;k: Number;Expression(C) = (x^2 - y^2 = 1);Expression(l) = (y = k*x + 1);NumIntersection(l, C) = 1",
    "query_expressions": "k",
    "answer_expressions": "{pm*1, pm*sqrt(2)}",
    "fact_spans": "[[[0, 16]], [[17, 40]], [[51, 54]], [[17, 40]], [[0, 16]], [[0, 49]]]",
    "query_spans": "[[[51, 56]]]",
    "process": ""
  },
  {
    "text": "From the right focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a perpendicular line $l$ to the asymptote, with foot of perpendicular at $M$. The line $l$ intersects the $y$-axis at point $P$. If $\\overrightarrow{F M}=\\lambda \\overrightarrow{M P}$, and the eccentricity of the hyperbola is $\\sqrt{3}$, then the value of $\\lambda$ is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F: Point;M: Point;P: Point;l: Line;lambda: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;PointOnCurve(F, l);IsPerpendicular(Asymptote(G), l);FootPoint(Asymptote(G), l) = M;Intersection(l, yAxis) = P;VectorOf(F, M) = lambda*VectorOf(M, P);Eccentricity(G) = sqrt(3)",
    "query_expressions": "lambda",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 57], [153, 156]], [[4, 57]], [[4, 57]], [[61, 64]], [[78, 81]], [[94, 98]], [[83, 87], [71, 74]], [[173, 182]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 64]], [[0, 74]], [[1, 74]], [[1, 81]], [[83, 98]], [[100, 151]], [[153, 171]]]",
    "query_spans": "[[[173, 186]]]",
    "process": "Given the hyperbola's eccentricity is $\\sqrt{3}$, find the asymptote equations. Since the right focus $F$ and line $l$ are perpendicular to the asymptote, set up the line equation, find $|FM|$ and $|FP|$, then from $\\overrightarrow{FM} = \\lambda\\overrightarrow{MP}$, obtain $\\frac{|FM|}{|FP|} = \\frac{\\lambda}{\\lambda+1}$, thereby solving for $\\lambda$. According to the problem, the hyperbola's eccentricity is $\\sqrt{3}$, that is, $\\frac{c}{a} = \\frac{\\sqrt{a^{2}+b^{2}}}{a} = \\sqrt{1+(\\frac{b}{a})^{2}} = \\sqrt{3}$, solving gives $\\frac{b}{a} = \\sqrt{2}$. Let one asymptote of the hyperbola be: $y = \\sqrt{2}x$. The right focus of the hyperbola is $F(c,0)$, and line $l$ is perpendicular to the asymptote. So set when $x = |FM| = \\frac{|\\sqrt[c]{2} + (\\frac{\\sqrt{2}}{2}c})| = \\sqrt{(\\sqrt{2})^{2}+1} = \\frac{\\sqrt{6}}{3}c$, from $\\overrightarrow{FM} = \\lambda\\overrightarrow{MP}$, get $\\frac{|FM|}{|FP|} = \\frac{2}{2+1} = \\frac{\\sqrt{6}}{\\frac{\\sqrt{6}}{6}}c = \\frac{2}{3}$, solving gives $\\lambda = 2$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the right focus is $F$. Draw tangents from point $F$ to the circle $x^{2}+y^{2}=b^{2}$. If the two tangents are perpendicular to each other, then $\\frac{a}{b}=$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;RightFocus(C) = F;G: Circle;Expression(G) = (x^2 + y^2 = b^2);Z1: Line;Z2: Line;TangentOfPoint(F, G) = {Z1, Z2};IsPerpendicular(Z1, Z2)",
    "query_expressions": "a/b",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 59]], [[2, 59]], [[109, 122]], [[109, 122]], [[9, 59]], [[9, 59]], [[64, 67], [69, 73]], [[2, 67]], [[74, 94]], [[74, 94]], [], [], [[68, 97]], [[68, 107]]]",
    "query_spans": "[[[109, 124]]]",
    "process": "Let the tangent points be A and B. Since the two tangents are perpendicular, quadrilateral OBFA is a square, |OF| = c, |OA| = b. Thus, $\\frac{c}{b}=\\sqrt{2}$. From $a^{2}=b^{2}+c^{2}$, we obtain $\\frac{a}{b}=\\sqrt{3}$. Hence, fill in: $\\sqrt{3}$"
  },
  {
    "text": "The parabola $C$: $y^{2}=4 x$ has focus $F$, point $P$ moves on parabola $C$, and point $A(-1,0)$. When $\\frac{|P A|}{|P F|}$ reaches its maximum value, what is the equation of line $A P$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;P: Point;PointOnCurve(P, C);A: Point;Coordinate(A) = (-1, 0);WhenMax(Abs(LineSegmentOf(P, A))/Abs(LineSegmentOf(P, F)))",
    "query_expressions": "Expression(LineOf(A, P))",
    "answer_expressions": "{x+y+1=0, x-y+1=0}",
    "fact_spans": "[[[0, 19], [33, 39]], [[0, 19]], [[23, 26]], [[0, 26]], [[29, 32]], [[29, 40]], [[41, 51]], [[41, 51]], [[52, 80]]]",
    "query_spans": "[[[81, 93]]]",
    "process": "Let the coordinates of point P be $(4t^{2}, 4t)$. Since $F(1,0)$, $A(-1,0)$, we have $|PF|^{2}=(4t^{2}-1)^{2}+16t^{2}=16t^{4}+8t^{2}+1$, $|PA|^{2}=(4t^{2}+1)^{2}+16t^{2}=16t^{4}+24t^{2}+1$, so $\\left(\\frac{|PA|}{|PF|}\\right)^{2}=\\frac{16t^{4}+8t^{2}+1}{16t^{4}+24t^{2}+1}=1-\\frac{16}{16t^{2}+\\frac{1}{t^{2}}+24}\\geqslant1-\\frac{16}{2\\sqrt{16t^{2}\\cdot\\frac{1}{t^{2}}}+24}=1-\\frac{16}{40}=\\frac{3}{5}$. The equality holds if and only if $16t^{2}=\\frac{1}{t^{2}}$, i.e., $t=\\pm\\frac{1}{2}$, at which point the coordinates of P are $(1,2)$ and $(1,-2)$. At this time, the equations of line AP are $y=\\pm(x+1)$, i.e., $x+y+1=0$ or $x-y+1=0$."
  },
  {
    "text": "The standard equation of the hyperbola that shares the same asymptotes as the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$ and passes through the point $(2 , 2)$ is?",
    "fact_expressions": "C: Hyperbola;D: Hyperbola;Expression(C) = (x^2 - y^2/4 = 1);Asymptote(C) = Asymptote(D);P: Point;Coordinate(P) = (2, 2);PointOnCurve(P, D)",
    "query_expressions": "Expression(D)",
    "answer_expressions": "x^2/3 - y^2/12 = 1",
    "fact_spans": "[[[1, 4]], [[54, 57]], [[1, 30]], [[0, 57]], [[41, 53]], [[41, 53]], [[40, 57]]]",
    "query_spans": "[[[54, 64]]]",
    "process": ""
  },
  {
    "text": "If a hyperbola with foci on the $X$-axis intersects the $X$-axis at a point $(2 , 0)$, and one of its asymptotes has the equation $y=-\\frac{\\sqrt{3}}{2} x$, then what are the coordinates of the foci of the hyperbola?",
    "fact_expressions": "G: Hyperbola;Coordinate(OneOf(Intersection(G,xAxis))) = (2, 0);PointOnCurve(Focus(G), xAxis);Expression(OneOf(Asymptote(G)))=(y=(-sqrt(3)/2)*x)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "{(sqrt(7),0),(-sqrt(7),0)}",
    "fact_spans": "[[[10, 13], [14, 15], [72, 75]], [[14, 36]], [[1, 13]], [[14, 70]]]",
    "query_spans": "[[[72, 82]]]",
    "process": ""
  },
  {
    "text": "Given that the equation of the parabola $C$ is $y^{2}=2 p x$ ($p>0$), the line $l$ passes through the focus $F$ of the parabola and intersects the parabola at points $A$ and $B$, with $|A F|=3|B F|$. What is the inclination angle of the line $l$?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;F: Point;B: Point;p:Number;p>0;Expression(C) = (y^2 = 2*p*x);Focus(C) = F;PointOnCurve(F,l);Intersection(l,C)={A,B};Abs(LineSegmentOf(A, F)) = 3*Abs(LineSegmentOf(B, F))",
    "query_expressions": "Inclination(l)",
    "answer_expressions": "{pi/3,2*pi/3}",
    "fact_spans": "[[[31, 36], [80, 85]], [[2, 8], [37, 40], [48, 51]], [[53, 56]], [[43, 46]], [[57, 60]], [[12, 30]], [[12, 30]], [[2, 30]], [[37, 46]], [[31, 46]], [[31, 62]], [[64, 78]]]",
    "query_spans": "[[[80, 91]]]",
    "process": "Let the parabola $ y^{2} = 2px $ ($ p > 0 $) have the directrix $ l': x = -\\frac{p}{2} $, as shown in the figure.  \n① When the inclination angle of line $ AB $ is acute, draw $ AM \\perp l' $, $ BN \\perp l' $ from points $ A $, $ B $, with feet of perpendiculars $ M $, $ N $, respectively. Draw $ BC \\perp AM $ intersecting at point $ C $. Then $ |AM| = |AF| $, $ |BN| = |BF| $.  \n$$\n\\cdots\n|AM\n\\begin{matrix}\n|AF| & 3|BF| \\\\\n|AM| - |BN| = |AC| & |AF| - |BF|\n\\end{matrix}\n= \\frac{1}{2}|AB|\n$$\nIn right triangle $ \\triangle ABC $, from $ |AC| = \\frac{1}{2}|AB| $, we obtain $ \\angle BAC = \\frac{\\pi}{3} $.  \nSince $ AM \\parallel x $-axis, $ \\angle BAC = \\angle AFx = \\frac{\\pi}{2} $.  \nTherefore, $ k_{AB} = \\tan\\frac{\\pi}{3} = \\sqrt{3} $.  \n② When the inclination angle of line $ AB $ is obtuse, the inclination angle of the line is $ \\frac{2\\pi}{3} $."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$, a line passing through the point $M(2 p , 0)$ intersects the parabola at points $A$, $B$, $\\overrightarrow{O A} \\cdot \\overrightarrow {O B}$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;M: Point;Coordinate(M) = (2*p, 0);H: Line;PointOnCurve(M, H);A: Point;B: Point;Intersection(H, G) = {A, B};O: Origin",
    "query_expressions": "DotProduct(VectorOf(O, A), VectorOf(O, B))",
    "answer_expressions": "0",
    "fact_spans": "[[[2, 23], [42, 45]], [[2, 23]], [[5, 23]], [[5, 23]], [[25, 38]], [[25, 38]], [[39, 41]], [[24, 41]], [[48, 51]], [[53, 56]], [[39, 56]], [[58, 108]]]",
    "query_spans": "[[[58, 110]]]",
    "process": ""
  },
  {
    "text": "What is the eccentricity of the hyperbola $\\frac{x^{2}}{25}-\\frac{y^{2}}{11}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/25 - y^2/11 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "6/5",
    "fact_spans": "[[[0, 40]], [[0, 40]]]",
    "query_spans": "[[[0, 46]]]",
    "process": "The hyperbola $\\frac{x^{2}}{25}-\\frac{y^{2}}{11}=1$, $a=5$, $b=\\sqrt{11}$, $\\therefore c=6$, $\\therefore$ the eccentricity of the hyperbola $\\frac{x^{2}}{25}-\\frac{y^{2}}{11}=1$ is $e=\\frac{c}{a}=\\frac{6}{5}$,"
  },
  {
    "text": "It is known that an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ passes through the point $(1,2)$. Then, the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;PointOnCurve(H, OneOf(Asymptote(G)));H: Point;Coordinate(H) = (1, 2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 58], [77, 80]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 74]], [[66, 74]], [[66, 74]]]",
    "query_spans": "[[[77, 86]]]",
    "process": ""
  },
  {
    "text": "If the line $2 x+y+4=0$ passes through the focus of the parabola $y^{2}=a x$, then the real number $a$=?",
    "fact_expressions": "H: Line;Expression(H) = (2*x + y + 4 = 0);G: Parabola;Expression(G) = (y^2 = a*x);a: Real;PointOnCurve(Focus(G),H) = True",
    "query_expressions": "a",
    "answer_expressions": "-8",
    "fact_spans": "[[[1, 14]], [[1, 14]], [[16, 30]], [[16, 30]], [[35, 40]], [[1, 33]]]",
    "query_spans": "[[[35, 42]]]",
    "process": "Since the focus of the parabola $ y^{2} = ax $ lies on the x-axis, and the intersection point of the line $ 2x + y + 4 = 0 $ with the x-axis is $ (-2, 0) $, the focus of the parabola $ y^{2} = ax $ is $ (-2, 0) $. Also, the focus of the parabola $ y^{2} = ax $ is $ \\left( \\frac{a}{4}, 0 \\right) $. Therefore, $ \\frac{a}{4} = -2 $, so $ a = -8 $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left vertex is $M$, the upper vertex is $N$, and the right focus is $F$. If $\\overrightarrow{N M} \\cdot \\overrightarrow{N F}=0$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;M: Point;LeftVertex(G) = M;N: Point;UpperVertex(G) = N;F: Point;RightFocus(G) = F;DotProduct(VectorOf(N, M), VectorOf(N, F)) = 0",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(\\sqrt{5}-1)/2",
    "fact_spans": "[[[2, 54], [133, 135]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[59, 62]], [[2, 62]], [[67, 70]], [[2, 70]], [[75, 78]], [[2, 78]], [[80, 131]]]",
    "query_spans": "[[[133, 141]]]",
    "process": "The left vertex of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $M(-a,0)$, the upper vertex is $N(0,b)$, and the right focus is $F(c,0)$. If $\\overrightarrow{NM}\\cdot\\overrightarrow{NF}=0$, then $(-a,-b)\\cdot(c,-b)=b^{2}-ac=0$. Also, $a^{2}=b^{2}+c^{2}$, so $a^{2}-c^{2}=ac$, which implies $e^{2}+e-1=0$. Solving gives $e=\\frac{\\sqrt{5}-1}{2}$."
  },
  {
    "text": "If the eccentricity of a hyperbola is $2$, then the standard equation of this hyperbola might be?",
    "fact_expressions": "G: Hyperbola;Eccentricity(G) = 2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[1, 4], [15, 18]], [[1, 12]]]",
    "query_spans": "[[[15, 27]]]",
    "process": "From the eccentricity $ e = 2 = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} $, we obtain $ 3a^{2} = b^{2} $. Therefore, the standard equation of this hyperbola could be $ x^{2} - \\frac{y^{2}}{3} = 1 $."
  },
  {
    "text": "Given the ellipse $ C $: $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ $ (a > b > 0) $ has foci $ F_{1} $, $ F_{2} $. If there exists a point $ P $ on the ellipse $ C $ such that $ \\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}} = 0 $ and the area of $ \\Delta P F_{1} F_{2} $ equals $ 4 $, then the range of values for $ a b $ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Area(TriangleOf(P, F1, F2)) = 4",
    "query_expressions": "Range(a*b)",
    "answer_expressions": "[4*sqrt(2),+oo)",
    "fact_spans": "[[[2, 59], [81, 86]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[63, 70]], [[71, 78]], [[2, 78]], [[91, 94]], [[81, 94]], [[97, 156]], [[158, 188]]]",
    "query_spans": "[[[190, 202]]]",
    "process": "Let $ P(x,y) $, $ F_{1}(-c,0) $, $ F_{2}(c,0) $. Since there exists a point $ P $ on the ellipse $ C $ such that $ \\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}} = 0 $, then $ \\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}} = x^{2} + y^{2} - c^{2} = 0 $, that is, $ x^{2} + y^{2} = c^{2} $, we obtain $ c \\geqslant b $. Since the area of $ \\triangle PF_{1}F_{2} $ equals 4, we have $ \\frac{1}{2} \\cdot 2c \\cdot y_{P} = 4 $, that is, $ y_{P} = \\frac{4}{c} $. Solving the ellipse and the circle simultaneously:\n\\[\n\\begin{cases}\nx^{2} + y^{2} = c^{2} \\\\\n\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1\n\\end{cases}\n\\]\nwe get $ y_{P} = \\frac{b^{2}}{c} $. Therefore, $ b^{2} = 4 $, that is, $ b = 2 $. Since $ c \\geqslant b $, $ a^{2} = b^{2} + c^{2} $, we have $ a^{2} > 2b^{2} = 8 $, that is, $ a > 2\\sqrt{2} $. Hence, $ ab > 4\\sqrt{2} $."
  },
  {
    "text": "Given the ellipse $ C $: $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ $ (a > b > 0) $, the left and right foci are $ F_{1} $ and $ F_{2} $, respectively. A line $ l $ with slope $ k $ passes through $ F_{1} $, intersects the ellipse $ C $ at points $ A $ and $ B $, and intersects the $ y $-axis at point $ M $. Point $ B $ is the midpoint of segment $ M F_{1} $. If $ |k| \\leq \\frac{\\sqrt{14}}{2} $, then the range of the eccentricity $ e $ of ellipse $ C $ is?",
    "fact_expressions": "l: Line;C: Ellipse;b: Number;a: Number;F1: Point;M: Point;A: Point;B: Point;F2: Point;e: Number;k: Number;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;Slope(l) = k;PointOnCurve(F1, l);Intersection(l, C) = {A, B};Intersection(l, yAxis) = M;MidPoint(LineSegmentOf(M, F1)) = B;Abs(k) <= sqrt(14)/2;Eccentricity(C) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "[\\sqrt{2}/2, 1)",
    "fact_spans": "[[[91, 96]], [[2, 59], [107, 112], [195, 200]], [[8, 59]], [[8, 59]], [[68, 75], [97, 104]], [[136, 139]], [[116, 119]], [[121, 125], [140, 143]], [[76, 83]], [[204, 207]], [[87, 90]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 83]], [[2, 83]], [[84, 96]], [[91, 104]], [[91, 125]], [[91, 139]], [[140, 158]], [[161, 191]], [[195, 207]]]",
    "query_spans": "[[[204, 214]]]",
    "process": "Let the equation of line $ l $ be $ y = k(x + c) $, then $ M(0, kc) $, $ B\\left(-\\frac{c}{2}, \\frac{kc}{2}\\right) $. Since $ B $ lies on the ellipse $ C $, $ \\therefore \\frac{c^{2}}{4a^{2}} + \\frac{k^{2}c^{2}}{4b^{2}} = 1 $, i.e., $ \\frac{c^{2}}{4a^{2}} + \\frac{k^{2}c^{2}}{4(a^{2}-c^{2})} = 1 $, rearranging yields $ e^{2} + \\frac{k^{2}e^{2}}{1-e^{2}} = 4 $, thus $ k^{2} = \\frac{(4-e^{2})(1-e^{2})}{e^{2}} \\leqslant \\frac{7}{2} \\therefore 2e^{4} - 17e^{2} + 8 \\leqslant 0 $, solving gives $ \\frac{1}{2} \\leqslant e^{2} \\leqslant 8 $. Also $ e^{2} < 1 $, $ \\therefore \\frac{1}{2} \\leqslant e^{2} < 1 $, thereby obtaining $ \\frac{\\sqrt{2}}{2} \\leqslant e < 1 $. The range of eccentricity $ e $ for the ellipse $ C $ is $ \\left[\\frac{\\sqrt{2}}{2}, 1\\right) $."
  },
  {
    "text": "Given the foci of an ellipse are $F_{1}(-1,0)$, $F_{2}(1,0)$, and $P$ is a point on the ellipse in the second quadrant such that $2|F_{1} F_{2}|=|P F_{1}|+|P F_{2}|$, $\\angle F_{2} F_{1} P=120^{\\circ}$, then the area of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;P: Point;Coordinate(F1) = (-1, 0);Coordinate(F2) = (1, 0);Focus(G) = {F1,F2};PointOnCurve(P, G);Quadrant(P)=2;2*Abs(LineSegmentOf(F1, F2)) = Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2));AngleOf(F2, F1, P) = ApplyUnit(120, degree)",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "3*sqrt(3)/5",
    "fact_spans": "[[[2, 4], [42, 44]], [[9, 22]], [[23, 36]], [[38, 41]], [[9, 22]], [[23, 36]], [[2, 36]], [[38, 51]], [[38, 51]], [[54, 90]], [[91, 125]]]",
    "query_spans": "[[[128, 158]]]",
    "process": "According to the problem, c=1, from 2|F_{1}F_{2}|=|PF_{1}|+|PF_{2}|=2a, a=2c=2, hence b=\\sqrt{4-1}=\\sqrt{3}, let |PF_{1}|=x, |PF_{2}|=4-x, \\cos120^{\\circ}=\\frac{x^{2}+4-(4-x)^{2}}{2\\cdot x\\cdot 2}, solving gives x=1.2, thus |PH|=1.2\\sin60^{\\circ}=\\frac{3\\sqrt{3}}{5}, so s=c|PH|=|PH|=\\frac{3\\sqrt{3}}{5}"
  },
  {
    "text": "If the equation $\\frac{x^{2}}{1+k}+\\frac{y^{2}}{1-k}=1$ represents a hyperbola, then the range of real values for $k$ is?",
    "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (x^2/(k + 1) + y^2/(1 - k) = 1)",
    "query_expressions": "Range(k)",
    "answer_expressions": "{(-oo, -1), (1, +oo)}",
    "fact_spans": "[[[44, 47]], [[49, 54]], [[1, 47]]]",
    "query_spans": "[[[49, 61]]]",
    "process": ""
  },
  {
    "text": "The equation $\\frac{x^{2}}{k-4}+\\frac{y^{2}}{10-k}=1$ represents an ellipse with foci on the $y$-axis. Then, the range of real values for $k$ is?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (x^2/(k - 4) + y^2/(10 - k) = 1);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(4,7)",
    "fact_spans": "[[[53, 55]], [[57, 62]], [[0, 55]], [[44, 55]]]",
    "query_spans": "[[[57, 69]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$), one of its asymptotes is parallel to the line $l$: $2x+y-3=0$. What is the eccentricity of the hyperbola $C$?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(l)=(2*x+y-3=0);IsParallel(OneOf(Asymptote(C)),l)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[70, 88]], [[2, 63], [92, 98]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[2, 63]], [[70, 88]], [[2, 90]]]",
    "query_spans": "[[[92, 104]]]",
    "process": "From the given conditions, one asymptote of hyperbola C is given by the equation $ y = -\\frac{b}{a}x $, so $ -\\frac{b}{a} = -2 $, which implies $ \\frac{b}{a} = 2 $. Then $ e^{2} = \\frac{c^{2}}{a^{2}} = \\frac{a^{2} + b^{2}}{a^{2}} = 1 + \\frac{b^{2}}{a^{2}} $. Therefore, the eccentricity of hyperbola C is $ e = \\sqrt{\\frac{b^{2}}{a^{2}} + 1} = \\sqrt{5} $."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. Then $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}$=?",
    "fact_expressions": "G: Parabola;H: Line;O: Origin;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G)=F;PointOnCurve(F, H);Intersection(H, G) = {A, B}",
    "query_expressions": "DotProduct(VectorOf(O, A), VectorOf(O, B))",
    "answer_expressions": "-3",
    "fact_spans": "[[[1, 15], [25, 28]], [[22, 24]], [[41, 90]], [[30, 33]], [[34, 37]], [[18, 21]], [[1, 15]], [[1, 21]], [[0, 24]], [[22, 39]]]",
    "query_spans": "[[[41, 92]]]",
    "process": "From the given conditions, the focus of the parabola $ y^{2} = 4x $ is at $ (1, 0) $. Therefore, the equation of line $ AB $ is $ y = k(x - 1) $. From \n\\[\n\\begin{cases}\ny^2 = 4x \\\\\ny = k(x - 1)\n\\end{cases}\n\\]\nwe obtain $ k^{2}x^{2} - (2k^{2} + 4)x + k^{2} = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = \\frac{2k^{2} + 4}{k^{2}} $, $ x_{1} \\cdot x_{2} = 1 $, $ y_{1} \\cdot y_{2} = k(x_{1} - 1) \\cdot k(x_{2} - 1) = k^{2}[x_{1} \\cdot x_{2} - (x_{1} + x_{2}) + 1] $. Therefore, $ \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = x_{1} \\cdot x_{2} + y_{1} \\cdot y_{2} = 1 + k^{2}\\left(1 - \\frac{2k^{2} + 4}{k^{2}} + 1\\right) = -3 $."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=4x$ be $F$, and the directrix be $l$. A line passing through the focus intersects the parabola at points $A$ and $B$. Perpendiculars from $A$ and $B$ to $l$ have feet $C$ and $D$, respectively. If $|AF|=2|BF|$, then the area of triangle $CDF$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;C: Point;D: Point;l: Line;Expression(G) = (y^2 = 4*x);Focus(G) = F;Directrix(G) = l;PointOnCurve(F,H);Intersection(H, G) = {A, B};L1:Line;L2:Line;PointOnCurve(A,L1);PointOnCurve(B,L2);IsPerpendicular(L1,l);IsPerpendicular(L2,l);FootPoint(L1,l)=C;FootPoint(L2,l)=D;Abs(LineSegmentOf(A, F)) = 2*Abs(LineSegmentOf(B, F))",
    "query_expressions": "Area(TriangleOf(C, D, F))",
    "answer_expressions": "3*sqrt(2)",
    "fact_spans": "[[[1, 15], [38, 41]], [[35, 37]], [[42, 45], [55, 58]], [[19, 22]], [[46, 49], [59, 62]], [[72, 75]], [[76, 79]], [[26, 29], [63, 66]], [[1, 15]], [[1, 22]], [[1, 29]], [[1, 37]], [[35, 51]], [], [], [[52, 69]], [[52, 69]], [[52, 69]], [[52, 69]], [[52, 79]], [[52, 79]], [[82, 96]]]",
    "query_spans": "[[[98, 113]]]",
    "process": "As shown in the figure: The focus of the parabola $ y^{2}=4x $ is $ F(1,0) $, and the directrix is $ l $. Let the equation of line $ l $ be $ y=k(x-1) $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Solving the system \n\\[\n\\begin{cases}\ny=k(x-1) \\\\\ny^{2}=4x\n\\end{cases}\n\\]\nyields $ k^{2}x^{2}-(2k^{2}+4)x+k^{2}=0 $, so $ x_{1}x_{2}=1 $ ①. Since $ |AF|=2|BF| $, we have $ x_{1}+1=2(x_{2}+1) $ ②. Solving ① and ② gives $ x_{1}=2 $, $ x_{2}=\\frac{1}{2} $, so $ y_{1}=2\\sqrt{2} $, $ y_{2}=-\\sqrt{2} $, thus $ |CD|=y_{1}-y_{2}=3\\sqrt{2} $. Since $ |FG|=1+1=2 $, we have $ S_{ACDE}=\\frac{1}{2}\\times|CD|\\times|FG|=\\frac{1}{2}\\times3\\sqrt{2}\\times2=3\\sqrt{2} $. Hence, the answer is $ 3\\sqrt{2} $."
  },
  {
    "text": "Let $P$ be the intersection point of the line $y=\\frac{b}{3 a} x$ and the left branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, and let $F_{1}$ be the left focus. If $P F_{1}$ is perpendicular to the $x$-axis, then the eccentricity $e$ of the hyperbola is $?$.",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;P: Point;F1: Point;e: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = x*(b/(3*a)));Intersection(H, LeftPart(G)) = P;LeftFocus(G) = F1;IsPerpendicular(LineSegmentOf(P, F1), xAxis);Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "3*sqrt(2)/4",
    "fact_spans": "[[[27, 85], [121, 124]], [[30, 85]], [[30, 85]], [[5, 26]], [[1, 4]], [[91, 98]], [[128, 131]], [[30, 85]], [[30, 85]], [[27, 85]], [[5, 26]], [[1, 90]], [[27, 102]], [[103, 119]], [[121, 131]]]",
    "query_spans": "[[[128, 133]]]",
    "process": ""
  },
  {
    "text": "A line passing through the right focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>b>0)$ intersects the two asymptotes at points $A$ and $B$, $\\angle O A B=90^{\\circ}$, $O$ is the origin, and the inradius of $\\triangle O A B$ is $\\frac{a}{3}$. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;O: Origin;A: Point;B: Point;F:Point;RightFocus(G)=F;L1:Line;L2:Line;L3:Line;Asymptote(G)={L2,L3};a > b;b > 0;PointOnCurve(F,L1);Intersection(L1,L2)=A;Intersection(L1,L3)=B;AngleOf(O, A, B) = ApplyUnit(90, degree);Radius(InscribedCircle(TriangleOf(O,A,B)))=a/3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[1, 54], [153, 156]], [[1, 54]], [[4, 54]], [[4, 54]], [[105, 108]], [[69, 72]], [[73, 76]], [[57, 60]], [[1, 60]], [[61, 63]], [], [], [[1, 68]], [[4, 54]], [[4, 54]], [[0, 63]], [[61, 78]], [[61, 78]], [[79, 104]], [[115, 151]]]",
    "query_spans": "[[[153, 162]]]",
    "process": "Let the incenter be M, draw MN\\botOA at N and MT\\botAB at T. It is easy to see that MTAN is a square, thus |MN|=\\frac{1}{3}a, |NO|=\\frac{2}{3}a. Combining with \\tan\\angleAOF=\\frac{|MN|}{|NO|}=\\frac{b}{a}, we obtain the homogeneous expression of hyperbola parameters, hence the eccentricity of the hyperbola can be found. Let the incenter be M, then M lies on the angle bisector Ox of \\angleAOB. Draw MN\\botOA at N and MT\\botAB at T. Since FA\\botOA, quadrilateral MTAN is a square. Given that the distance from the focus to the asymptote is b, we have FA=b, and OF=c, \\therefore OA=a, then |NA|=|MN|=\\frac{1}{3}a, so |NO|=\\frac{2}{3}a,"
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C$: $5 x^{2}-4 y^{2}=m^{2}$, respectively. A line $l$ passing through $F_{2}$ intersects the right branch of hyperbola $C$ at points $A$ and $B$, and satisfies $S_{\\triangle B F_{1} F_{2}}=14 S_{\\triangle A O F_{1}}$ ($O$ is the origin). Then the slope of line $l$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;m: Number;B: Point;F1: Point;A: Point;O: Origin;F2: Point;Expression(C) = (5*x^2 - 4*y^2 = m^2);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F2, l);Intersection(l, RightPart(C)) = {A, B};Area(TriangleOf(B, F1, F2)) = 14*Area(TriangleOf(A, O ,F1))",
    "query_expressions": "Slope(l)",
    "answer_expressions": "{sqrt(3), -sqrt(3)}",
    "fact_spans": "[[[66, 71], [165, 170]], [[19, 50], [72, 78]], [[26, 50]], [[87, 90]], [[1, 8]], [[83, 86]], [[154, 157]], [[9, 16], [58, 65]], [[19, 50]], [[1, 56]], [[1, 56]], [[57, 71]], [[66, 92]], [[96, 153]]]",
    "query_spans": "[[[165, 175]]]",
    "process": "_As shown in the figure, let A be above the x-axis. Since O is the midpoint of F_{1} and F_{2}, then S_{\\triangle BF_{1}F_{2}} = 14S_{\\triangle AOF} is equivalent to BF_{2} = 7AF_{2}. 5x^{2} - 4y^{2} = m^{2}, which is \\frac{x^{2}}{\\frac{m^{2}}{5}} - \\frac{y^{2}}{\\frac{m^{2}}{4}} = 1, so a^{2} = \\frac{m^{2}}{5}, b^{2} = \\frac{m^{2}}{4}, e = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\frac{3}{2}. Let AF_{2} = x, then AF_{1} = 2a + x, and the inclination angle of line l is \\theta. In \\triangle AF_{1}F_{2}, by the law of cosines: (2a + x)^{2} = x^{2} + (2c)^{2} - 2 \\times 2c \\cdot x \\cdot \\cos(\\pi - \\theta), we get b^{2} = (a - c \\cdot \\cos\\theta)x, so x = \\frac{b^{2}}{a - c\\cos\\theta}. Similarly, let BF_{2} = y, in \\triangle BF_{1}F_{2}, we obtain y = \\frac{b^{2}}{a + c\\cos\\theta}. Therefore, \\frac{y}{x} = \\frac{a - c\\cos\\theta}{a + c\\cos\\theta} = \\frac{1 - e\\cos\\theta}{1 + e\\cos\\theta} = 7. Since e = \\frac{3}{2}, substituting gives \\cos\\theta = -\\frac{1}{2}, so \\theta = \\frac{2\\pi}{3}, hence \\tan\\theta = -\\sqrt{3}, i.e., k = -\\sqrt{3}. Due to symmetry, A can also be below the x-axis, so the slope is \\pm\\sqrt{3}_"
  },
  {
    "text": "Ellipses and hyperbolas share common foci $F_{1}$, $F_{2}$, their intersection point is $P$, and $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$. If the eccentricity of the ellipse is $\\frac{\\sqrt{3}}{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Focus(G) = {F1, F2};Focus(H) = {F1, F2};F1: Point;F2: Point;Intersection(G ,H) = P;P: Point;AngleOf(F1, P, F2) = pi/3;Eccentricity(H) = sqrt(3)/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3*sqrt(2)/4",
    "fact_spans": "[[[3, 6], [106, 109]], [[0, 2], [77, 79]], [[0, 26]], [[0, 26]], [[9, 16]], [[19, 26]], [[28, 37]], [[34, 37]], [[39, 75]], [[77, 104]]]",
    "query_spans": "[[[106, 115]]]",
    "process": "Let point $ P $ be a point in the first quadrant, and suppose the foci of the ellipse and hyperbola both lie on the x-axis. Let the major axis length of the ellipse be $ 2a_{1} $, the real axis length of the hyperbola be $ 2a_{2} $, and the focal distance of both curves be $ 2c $. The eccentricities of the ellipse and hyperbola are $ e_{1} $ and $ e_{2} $, respectively. Using the cosine law and the definitions of the ellipse and hyperbola, we obtain $ a_{1}^{2} + 3a_{2}^{2} = 4c^{2} $, which leads to $ \\frac{1}{e_{1}^{2}} + \\frac{3}{e_{2}^{2}} = 4 $. Combining with $ e_{1} = \\frac{\\sqrt{3}}{2} $, the value of $ e_{2} $ can be found. Suppose the foci of the ellipse and hyperbola both lie on the x-axis, let the major axis length of the ellipse be $ 2a_{1} $, the real axis length of the hyperbola be $ 2a_{2} $, and the focal distance of both curves be $ 2c $. The eccentricities of the ellipse and hyperbola are $ e_{1} $ and $ e_{2} $, respectively. Solving equations $ \\textcircled{1} $ and $ \\textcircled{2} $ together yields $ |PF_{1}| = a_{1} + a_{2} $, $ |PF_{2}| = a_{1} - a_{2} $. In $ \\triangle PF_{1}F_{2} $, by the cosine law:  \n$$\n(2c)^{2} = |PF_{1}|^{2} + |PF_{2}|^{2} - 2|PF_{1}| \\cdot |PF_{2}| \\cos\\frac{\\pi}{3}\n$$  \nthat is,  \n$$\n4c^{2} = (a_{1}+a_{2})^{2} + (a_{1}-a_{2})^{2} - 2(a_{1}+a_{2})(a_{1}-a_{2}) \\cdot \\frac{1}{2}\n$$  \nwhich simplifies to  \n$$\n4c^{2} = 2(a_{1}^{2} + a_{2}^{2}) - (a_{1}^{2} - a_{2}^{2}) = a_{1}^{2} + 3a_{2}^{2},\n$$  \ni.e.,  \n$$\n4 = \\frac{a_{1}^{2}}{c^{2}} + \\frac{3a_{2}^{2}}{c^{2}},\n$$  \nso  \n$$\n\\frac{1}{e_{1}^{2}} + \\frac{3}{e_{2}^{2}} = 4.\n$$  \nSince the eccentricity of the ellipse is $ e_{1} = \\frac{\\sqrt{3}}{2} $, the eccentricity of the hyperbola is $ e_{2} = \\frac{3\\sqrt{2}}{4}. \n$$"
  },
  {
    "text": "Given that line $l$ passes through point $A(1,3)$, and line $l$ intersects the curve $y=2x^{2}$ at points $M$ and $N$. If point $A$ is exactly the midpoint of $M$ and $N$, then the equation of line $l$ is?",
    "fact_expressions": "l: Line;A: Point;Coordinate(A) = (1, 3);G: Curve;Expression(G) = (y = 2*x^2);PointOnCurve(A, l);M: Point;N: Point;Intersection(l, G) = {M, N};MidPoint(LineSegmentOf(M, N)) = A",
    "query_expressions": "Expression(l)",
    "answer_expressions": "4*x-y-1=0",
    "fact_spans": "[[[20, 25], [72, 77], [20, 25]], [[8, 17], [53, 57]], [[8, 17]], [[26, 39]], [[26, 39]], [[2, 17]], [[60, 63], [41, 44]], [[45, 48], [64, 67]], [[20, 50]], [[53, 70]]]",
    "query_spans": "[[[72, 82]]]",
    "process": ""
  },
  {
    "text": "If a hyperbola passes through the point $k(3, -\\sqrt{2})$, and its asymptotes are given by $y = \\pm \\frac{1}{3}x$, then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;k: Point;Coordinate(k) = (3, -sqrt(2));PointOnCurve(k, G);Expression(Asymptote(G)) = (y = pm*(x/3))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 - x^2/9 = 1",
    "fact_spans": "[[[1, 4], [57, 60]], [[6, 24]], [[6, 24]], [[1, 24]], [[1, 53]]]",
    "query_spans": "[[[57, 65]]]",
    "process": ""
  },
  {
    "text": "Let the hyperbola share the same foci as the ellipse $\\frac{x^{2}}{27}+\\frac{y^{2}}{36}=1$, and intersect the ellipse at a point with coordinates $(\\sqrt{15}, 4)$. Then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/27 + y^2/36 = 1);Focus(G) = Focus(H);IsIntersect(G, H) = True;Coordinate(OneOf(Intersection(G, H))) = (sqrt(15), 4)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/4 - x^2/5 = 1",
    "fact_spans": "[[[1, 4], [86, 89]], [[5, 44], [53, 55]], [[5, 44]], [[1, 50]], [[1, 57]], [[1, 83]]]",
    "query_spans": "[[[86, 94]]]",
    "process": "Since the foci of the ellipse $\\frac{x^{2}}{27}+\\frac{y^{2}}{36}=1$ are $(0,3)$ or $(0,-3)$, let the equation of the hyperbola be $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0,b>0)$. From the given conditions, we have $\\begin{cases}\\frac{16}{a^{2}}-\\frac{15}{b^{2}}=1\\\\a^{2}+b^{2}=9\\end{cases}$, solving which yields $\\begin{cases}a=2\\\\b=\\sqrt{5}\\end{cases}$, so $\\frac{y^{2}}{4}-\\frac{x^{2}}{5}=1$."
  },
  {
    "text": "The asymptote equations of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$ are?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2/3 = 1)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y = pm*sqrt(3)*x",
    "fact_spans": "[[[0, 33]], [[0, 33]]]",
    "query_spans": "[[[0, 41]]]",
    "process": "Let $x^{2}-\\frac{y^{2}}{3}=0$, solving gives: $y=\\pm\\sqrt{3}x$, so the asymptotes of the hyperbola $C: x^{2}-\\frac{y^{2}}{3}=1$ are $y=\\pm\\sqrt{3}x$."
  },
  {
    "text": "If the left and right foci of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$ are $F_{1}$, $F_{2}$ respectively, and $P$ is a point on the hyperbola such that $P F_{1}=3$, then $P F_{2}$=?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 - y^2/3 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P,G);LineSegmentOf(P, F1) = 3",
    "query_expressions": "LineSegmentOf(P, F2)",
    "answer_expressions": "7",
    "fact_spans": "[[[1, 39], [67, 70]], [[63, 66]], [[47, 54]], [[55, 62]], [[1, 39]], [[1, 62]], [[1, 62]], [[63, 73]], [[74, 85]]]",
    "query_spans": "[[[87, 98]]]",
    "process": "From the equation, we have $a^{2}=4$, $\\therefore 2a=4$, $\\therefore |PF_{1}-PF_{2}|=2a=4$, $\\therefore PF_{2}=7$"
  },
  {
    "text": "The focus of the parabola $x^{2}=4 y$ is $F$, and a line passing through the intersection point $Q$ of its directrix and the $y$-axis is tangent to the parabola at point $P$. Then, the standard equation of the circumcircle of $\\Delta F P Q$ is?",
    "fact_expressions": "G: Parabola;H: Line;F: Point;P: Point;Q: Point;Expression(G) = (x^2 = 4*y);Focus(G) = F;Intersection(Directrix(G),yAxis)=Q;PointOnCurve(Q,H);TangentPoint(H,G)=P",
    "query_expressions": "Expression(CircumCircle(TriangleOf(F,P,Q)))",
    "answer_expressions": "{(x-1)^2+y^2=2,(x+1)^2+y^2=2}",
    "fact_spans": "[[[0, 14], [24, 25], [42, 45]], [[39, 41]], [[18, 21]], [[47, 51]], [[35, 38]], [[0, 14]], [[0, 21]], [[24, 38]], [[22, 41]], [[39, 51]]]",
    "query_spans": "[[[53, 77]]]",
    "process": "By the given condition, F(0,1), let P(x_{0},\\frac{1}{4}x_{0}^{2}). Since y=\\frac{1}{2}x, the tangent line equation is y-\\frac{1}{4}x_{0}^{2}=\\frac{1}{2}x_{0}(x-x_{0}). Substituting (0,-1) yields x_{0}=\\pm2, so P(2,1) or P(-2,1). Thus PF\\botFQ, so the circumcircle of AFPQ has PQ as diameter, hence (x-1)^{2}+y^{2}=2 or (x+1)^{2}+y^{2}=2."
  },
  {
    "text": "If the line $l$ passes through the fixed point $M(1 , 2)$ and has exactly one common point with the parabola $y=2 x^{2}$, then the equation of line $l$ is?",
    "fact_expressions": "l: Line;M: Point;Coordinate(M) = (1, 2);PointOnCurve(M, l);G: Parabola;Expression(G) = (y = 2*x^2);NumIntersection(l, G) = 1",
    "query_expressions": "Expression(l)",
    "answer_expressions": "{(x=1),(y=4*x-2)}",
    "fact_spans": "[[[2, 7], [48, 53]], [[10, 20]], [[10, 20]], [[2, 20]], [[22, 36]], [[22, 36]], [[2, 45]]]",
    "query_spans": "[[[48, 58]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the two foci of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$, and let point $P$ lie on the ellipse. If the midpoint of segment $PF_{1}$ lies on the $y$-axis, then $|PF_{1}|$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/2 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);PointOnCurve(MidPoint(LineSegmentOf(P, F1)), yAxis)",
    "query_expressions": "Abs(LineSegmentOf(P, F1))",
    "answer_expressions": "3",
    "fact_spans": "[[[18, 55], [66, 68]], [[18, 55]], [[1, 8]], [[10, 17]], [[1, 60]], [[61, 65]], [[61, 69]], [[71, 90]]]",
    "query_spans": "[[[92, 104]]]",
    "process": "From the given conditions, we have $ F_{1}(-\\sqrt{2},0) $, $ F_{2}(\\sqrt{2},0) $. Let $ M $ be the midpoint of $ PF_{1} $. According to the problem, $ MO \\parallel PF_{2} $, therefore $ PF_{2} \\perp x $-axis. In the equation $ \\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1 $, let $ x=\\sqrt{2} $, we obtain $ y^{2}=1 $, so $ |PF_{2}|=1 $. Hence, $ |PF_{1}|=4-|PF_{2}|=3 $. Answer: 3"
  },
  {
    "text": "A point $P$ on the hyperbola $4 x^{2}-9 y^{2}=36$ is at a distance of $3$ from the right focus. What is the distance from $P$ to the left directrix?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (4*x^2 - 9*y^2 = 36);P: Point;PointOnCurve(P, G);Distance(P, RightFocus(G)) = 3",
    "query_expressions": "Distance(P, LeftDirectrix(G))",
    "answer_expressions": "27*sqrt(13)/13",
    "fact_spans": "[[[0, 23]], [[0, 23]], [[26, 29], [43, 46]], [[0, 29]], [[0, 40]]]",
    "query_spans": "[[[0, 55]]]",
    "process": ""
  },
  {
    "text": "If the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ and the hyperbola $\\frac{x^{2}}{a_{1}^{2}}-\\frac{y^{2}}{b_{1}^{2}}=1(a_{1}>0, b_{1}>0)$ have the same foci $F_{1}$, $F_{2}$, point $P$ is an intersection point of the two curves, $\\angle F_{1} P F_{2}=\\frac{\\pi}{2}$, the eccentricity of the ellipse is $e_{1}$, the eccentricity of the hyperbola is $e_{2}$, and $e_{1} e_{2}=2$, then $e_{1}^{2}+e_{2}^{2}$=?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Ellipse;a1: Number;b1: Number;F1: Point;P: Point;F2: Point;e1:Number;e2:Number;a1>0;b1>0;Expression(G) = (-y^2/b1^2 + x^2/a1^2 = 1);a > b;b > 0;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);Focus(H)=Focus(G);Focus(H)={F1,F2};Focus(G)={F1,F2};OneOf(Intersection(H, G)) = P;AngleOf(F1, P, F2) = pi/2;Eccentricity(H) = e1;Eccentricity(G) = e2;e1*e2 = 2",
    "query_expressions": "e1^2 + e2^2",
    "answer_expressions": "8",
    "fact_spans": "[[[54, 126], [215, 218]], [[3, 53]], [[3, 53]], [[1, 53], [200, 202]], [[57, 126]], [[57, 126]], [[132, 139]], [[148, 152]], [[140, 147]], [[207, 214]], [[223, 230]], [[57, 126]], [[57, 126]], [[54, 126]], [[3, 53]], [[3, 53]], [[1, 53]], [[0, 147]], [[0, 147]], [[0, 147]], [[148, 162]], [[163, 199]], [[200, 214]], [[215, 230]], [[231, 246]]]",
    "query_spans": "[[[248, 271]]]",
    "process": "Let PF_{1}=s, PF_{2}=t. By the definition of the ellipse, we have s+t=2a; by the definition of the hyperbola, we have s-t=2a_{1}. Using the Pythagorean theorem and the eccentricity formula, we obtain 4=\\frac{2}{e_{1}}+\\frac{2}{e_{2}}, and simplifying yields the conclusion. Without loss of generality, assume P lies in the first quadrant, and let PF_{1}=s, PF_{2}=t. By the definition of the ellipse, we have s+t=2a; by the definition of the hyperbola, we have s-t=2a_{1}. Solving gives s=a+a_{1}, t=a-a_{1}. Since \\angle F_{1}PF_{2}=\\frac{\\pi}{2}, in triangle F_{1}PF_{2}, using the Pythagorean theorem we get 4c^{2}=s^{2}+t^{2}=(a+a_{1})^{2}+(a-a_{1})^{2}=2a^{2}+2a_{1}^{2}, \\therefore 4=\\frac{2}{e_{1}^{2}}+\\frac{2}{e_{2}^{2}}, simplifying \\frac{1}{e_{1}^{2}}+\\frac{1}{e_{2}^{2}}=\\frac{2}{e_{1}^{2}e_{2}^{2}}=2. Given e_{1}e_{2}=2, it follows that e_{1}^{2}+e_{2}^{2}=2e_{1}^{2}e_{2}^{2}=8."
  },
  {
    "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ are tangent to the circle $(x-\\sqrt{2})^{2}+y^{2}=1$. Then, the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 + (x - sqrt(2))^2 = 1);IsTangent(Asymptote(G),H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 59], [96, 99]], [[3, 59]], [[3, 59]], [[64, 91]], [[3, 59]], [[3, 59]], [[0, 59]], [[64, 91]], [[0, 93]]]",
    "query_spans": "[[[96, 105]]]",
    "process": "Since the asymptotes of the hyperbola are $ y = \\pm\\frac{b}{a}x $, the distance $ d $ from the center of the circle $ C(\\sqrt{2},0) $ to the asymptote is $ d = \\frac{|\\sqrt{2}b|}{\\sqrt{a^{2}+b^{2}}} = \\frac{\\sqrt{2}b}{c} = 1 $, which implies $ 2b^{2} = c^{2} \\Rightarrow 2c^{2} - 2a^{2} = c^{2} $. Solving this gives $ e = \\sqrt{2} $. The answer to be filled in is $ \\sqrt{2} $."
  },
  {
    "text": "What is the equation of the directrix of the parabola $x^{2}=4 ay(a \\neq 0)$?",
    "fact_expressions": "G: Parabola;a: Number;Negation(a=0);Expression(G) = (x^2 = 4*a*y)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y = -a",
    "fact_spans": "[[[0, 25]], [[3, 25]], [[3, 25]], [[0, 25]]]",
    "query_spans": "[[[0, 32]]]",
    "process": ""
  },
  {
    "text": "Given a fixed point $B(-1 , 0)$ and two moving points $P$, $Q$ on the parabola $y = x^{2} - 1$, when $P$ moves along the parabola, $BP \\perp PQ$. Then the range of the horizontal coordinate of point $Q$ is?",
    "fact_expressions": "G: Parabola;B: Point;P: Point;Q: Point;Expression(G) = (y = x^2 - 1);Coordinate(B) = (-1, 0);PointOnCurve(P, G);PointOnCurve(Q, G);IsPerpendicular(LineSegmentOf(B, P), LineSegmentOf(P, Q));PointOnCurve(B, G)",
    "query_expressions": "Range(XCoordinate(Q))",
    "answer_expressions": "(-oo, 3] + [1, +oo)",
    "fact_spans": "[[[2, 16], [49, 52]], [[20, 31]], [[36, 39], [45, 48]], [[40, 43], [73, 77]], [[2, 16]], [[20, 31]], [[45, 53]], [[2, 43]], [[57, 71]], [[2, 31]]]",
    "query_spans": "[[[73, 88]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/12 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(3)*x",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 47]]]",
    "process": "The asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$ are given by $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=0$, which simplifies to $y=\\pm\\sqrt{3}x$."
  },
  {
    "text": "The left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $F_{1}$. A line with slope $\\sqrt{2}$ passing through point $F_{1}$ intersects the $y$-axis and the right branch of the hyperbola at points $A$ and $B$, respectively. If $\\overrightarrow{F_{1} A}=\\overrightarrow{A B}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;LeftFocus(G) = F1;Slope(H) = sqrt(2);PointOnCurve(F1, H) = True;H: Line;Intersection(H, yAxis) = A;Intersection(H, RightPart(G)) = B;A: Point;B: Point;VectorOf(F1, A) = VectorOf(A, B)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)+sqrt(2)",
    "fact_spans": "[[[0, 56], [101, 104], [171, 174]], [[0, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[61, 68], [70, 78]], [[0, 68]], [[79, 95]], [[69, 95]], [[93, 95]], [[93, 120]], [[93, 120]], [[111, 114]], [[115, 118]], [[122, 169]]]",
    "query_spans": "[[[171, 180]]]",
    "process": "Let the right focus of the hyperbola be $ F_{2} $, connect $ BF_{2} $, and since $ |F_{1}A| = |AB| $, we obtain that $ A $ is the midpoint of $ F_{1}B $. Thus, $ BF_{2} \\bot x $-axis. From the given condition, $ \\tan\\angle BF_{1}F_{2} = \\frac{BF_{2}}{F_{2}F_{1}} = \\sqrt{2} $, so $ |BF_{2}| = 2\\sqrt{2}c $, and we get $ |BF_{1}| = \\sqrt{8c^{2}+}4c^{2} = 2\\sqrt{3}c $. By the definition of the hyperbola, $ |BF_{1}| - |BF_{2}| = 2\\sqrt{3}c - 2\\sqrt{2}c = 2a $. $ \\frac{1}{3} - \\sqrt{2} = \\sqrt{3} + \\sqrt{2} $"
  },
  {
    "text": "The hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1(b>0)$ has an asymptote given by $3 x+2 y=0$. What is the focal length of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;b: Number;b>0;Expression(C) = (x^2/4 - y^2/b^2 = 1);Expression(OneOf(Asymptote(C)))=(3*x+2*y=0)",
    "query_expressions": "FocalLength(C)",
    "answer_expressions": "2*sqrt(13)",
    "fact_spans": "[[[0, 52], [74, 80]], [[8, 52]], [[8, 52]], [[0, 52]], [[0, 72]]]",
    "query_spans": "[[[74, 85]]]",
    "process": "Since the asymptotes of the hyperbola are given by $ bx \\pm 2y = 0 $, combining with the given conditions we obtain $ b = 3 $, hence the focal distance is $ 2c = 2\\sqrt{a^{2}+b^{2}} = 2\\sqrt{3^{2}+4} = 2\\sqrt{13} $."
  },
  {
    "text": "The length of the line segment cut by the parabola $y^{2}=4x$ from the line $y=x-1$ is?",
    "fact_expressions": "G: Parabola;H: Line;Expression(G) = (y^2 = 4*x);Expression(H) = (y = x - 1)",
    "query_expressions": "Length(InterceptChord(H, G))",
    "answer_expressions": "8",
    "fact_spans": "[[[10, 24]], [[0, 9]], [[10, 24]], [[0, 9]]]",
    "query_spans": "[[[0, 31]]]",
    "process": "Let the line $ y = x - 1 $ intersect the parabola $ y^2 = 4x $ at points $ A(x_1, y_1) $, $ B(x_2, y_2) $. The focus of the parabola $ y^2 = 4x $ is $ F(1, 0) $, and the line $ y = x - 1 $ passes through point $ F $. Solving the system\n$$\n\\begin{cases}\ny = x - 1 \\\\\ny^2 = 4x\n\\end{cases}\n$$\neliminating $ y $ gives $ x^2 - 6x + 1 = 0 $, $ \\Delta = 36 - 4 > 0 $, so $ x_1 + x_2 = 6 $. By the focal chord length formula of the parabola, we obtain $ |AB| = x_1 + x_2 + 2 = 8 $."
  },
  {
    "text": "The standard equation of the hyperbola with foci $F_{1}(-3 , 0)$, $F_{2}(3 , 0)$ and asymptotes given by $y=\\pm \\sqrt{2} x$ is?",
    "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;Coordinate(F1) = (-3, 0);Coordinate(F2) = (3, 0);Focus(G) = {F1, F2};Expression(Asymptote(G)) = (y = pm*(sqrt(2)*x))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/3 - y^2/6 = 1",
    "fact_spans": "[[[62, 65]], [[1, 16]], [[19, 33]], [[1, 16]], [[19, 33]], [[0, 65]], [[37, 65]]]",
    "query_spans": "[[[62, 72]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{8}=1$, with left and right foci $F_{1}$ and $F_{2}$ respectively. A line passing through point $F_{2}$ intersects the right branch of hyperbola $C$ at points $P$ and $Q$, such that $\\angle F_{1} PQ=90^{\\circ}$. Then the inradius $r$ of triangle $\\triangle F_{1} PQ$ is $?$.",
    "fact_expressions": "C: Hyperbola;H: Line;F1: Point;P: Point;Q: Point;F2: Point;Expression(C) = (x^2/9 - y^2/8 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F2, H);Intersection(H, RightPart(C)) = {P, Q};AngleOf(F1, P, Q) = ApplyUnit(90, degree);r:Number;Radius(InscribedCircle(TriangleOf(F1,P,Q)))=r",
    "query_expressions": "r",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 45], [84, 90]], [[81, 83]], [[54, 61]], [[96, 99]], [[100, 103]], [[62, 69], [71, 79]], [[2, 45]], [[2, 69]], [[2, 69]], [[70, 83]], [[81, 105]], [[108, 136]], [[165, 168]], [[138, 168]]]",
    "query_spans": "[[[165, 170]]]",
    "process": "By the properties of the hyperbola, we know $ F_{1}F_{2}=2\\sqrt{17} $, $ PF_{1}-PF_{2}=QF_{1}-QF_{2}=6 $. Since $ \\angle F_{1}PQ=90^{\\circ} $, it follows that $ PF_{1}^{2}+PF_{2}^{2}=F_{1}F_{2}^{2} $. Therefore, $ PF_{1}+PF_{2}=\\sqrt{2(PF_{1}^{2}+PF_{2}^{2})-(PF_{1}-PF_{2})^{2}}=\\sqrt{2\\times4\\times17-6^{2}}=10 $. Thus, the inradius of the right triangle $ \\triangle F_{1}PQ $ is $ r=\\frac{1}{2}(F_{1}P+PQ-F_{1}Q)=\\frac{1}{2}(F_{1}P+PF_{2})-\\frac{1}{2}(QF_{1}-QF_{2})=5-3=2 $. Hence, fill in 2."
  },
  {
    "text": "What is the standard equation of a parabola with its focus at the right focus of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (x^2/3 - y^2 = 1);RightFocus(G)=Focus(H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "y^2=8*x",
    "fact_spans": "[[[1, 29]], [[37, 40]], [[1, 29]], [[0, 40]]]",
    "query_spans": "[[[37, 46]]]",
    "process": ""
  },
  {
    "text": "The distance from the focus to the directrix of the parabola $x^{2}=a y(a>0)$ is $\\frac{5}{2}$, then the value of $a$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = a*y);a: Number;a>0;Distance(Focus(G), Directrix(G)) = 5/2",
    "query_expressions": "a",
    "answer_expressions": "5",
    "fact_spans": "[[[0, 19]], [[0, 19]], [[44, 47]], [[3, 19]], [[0, 42]]]",
    "query_spans": "[[[44, 51]]]",
    "process": "First, find the focus coordinates and the directrix equation of the parabola, then solve for the parabola $ x^{2} = ay $ ($ a > 0 $) given that the distance from the focus to the directrix is $ p $. The focus of the parabola $ x^{2} = ay $ ($ a > 0 $) is $ \\left(0, \\frac{a}{4}\\right) $, and the directrix equation is: $ y = -\\frac{a}{4} $. Since the distance from the focus to the directrix of the parabola $ x^{2} = ay $ ($ a > 0 $) is $ \\frac{5}{2} $, it follows that $ \\frac{a}{4} \\times 2 = \\frac{5}{2} $, solving gives $ a = 5 $."
  },
  {
    "text": "If the left focus of the hyperbola $\\frac{x^{2}}{3}-\\frac{16 y^{2}}{p^{2}}=1$ lies on the directrix of the parabola $y^{2}=2 p x$, then the value of $p$ is?",
    "fact_expressions": "G: Hyperbola;p: Number;H: Parabola;Expression(G) = (x^2/3 - 16*y^2/p^2 = 1);Expression(H) = (y^2 = 2*p*x);PointOnCurve(LeftFocus(G),Directrix(H))",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 46]], [[73, 76]], [[51, 67]], [[1, 46]], [[51, 67]], [[1, 71]]]",
    "query_spans": "[[[73, 80]]]",
    "process": "First, express the coordinates of the left focus according to the hyperbola's equation, then express the directrix equation according to the parabola's equation. Finally, since the left focus of the hyperbola $\\frac{x^{2}}{3}-\\frac{16y^{2}}{p^{2}}=1$ lies on the directrix of the parabola $y^{2}=2px$, we obtain the equation $-\\sqrt{3+\\frac{p^{2}}{16}}=-\\frac{p}{2}$. Solving for $p$ gives: $p=4$. The coordinates of the left focus of the hyperbola are: $(-\\sqrt{3+\\frac{p^{2}}{16}},0)$, and the directrix equation of the parabola $y^{2}=2px$ is $x=-\\frac{p}{2}$, hence $-\\sqrt{3+\\frac{p^{2}}{16}}=-\\frac{p}{2}$, solving yields: $p=4$."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$. A line passes through $F_{1}$ and intersects the ellipse at points $A$ and $B$. Then, what is the perimeter of $\\Delta A F_{2} B$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};H: Line;PointOnCurve(F1, H);A: Point;B: Point;Intersection(H, G) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, F2, B))",
    "answer_expressions": "20",
    "fact_spans": "[[[17, 56], [73, 75]], [[17, 56]], [[1, 8], [65, 72]], [[9, 16]], [[1, 61]], [[62, 64]], [[62, 72]], [[76, 79]], [[80, 83]], [[62, 85]]]",
    "query_spans": "[[[87, 110]]]",
    "process": ""
  },
  {
    "text": "If the asymptotes of the hyperbola $x^{2} + m y^{2}=1$ pass through the point $(-1,2)$, then what is the length of the imaginary axis of this hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G)=(m*y^2+x^2=1);m: Number;H: Point;Coordinate(H) = (-1, 2);PointOnCurve(H, Asymptote(G))",
    "query_expressions": "Length(ImageinaryAxis(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 23], [40, 43]], [[1, 23]], [[4, 23]], [[28, 37]], [[28, 37]], [[1, 37]]]",
    "query_spans": "[[[40, 50]]]",
    "process": ""
  },
  {
    "text": "Point $A(1,0)$, point $B$ is a moving point on the $x$-axis, the midpoint $E$ of segment $PB$ lies on the $y$-axis, and $AE$ is perpendicular to $PB$. Then the trajectory equation of point $P$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (1, 0);B: Point;PointOnCurve(B, xAxis);P: Point;E: Point;MidPoint(LineSegmentOf(P, B)) = E;PointOnCurve(E, yAxis);IsPerpendicular(LineSegmentOf(A, E), LineSegmentOf(P, B))",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(y^2=4*x)&(Negation(x=0))",
    "fact_spans": "[[[0, 9]], [[0, 9]], [[10, 14]], [[10, 23]], [[59, 63]], [[34, 37]], [[24, 37]], [[34, 43]], [[45, 57]]]",
    "query_spans": "[[[59, 70]]]",
    "process": "Let P(x,y), B(m,0), then the coordinates of the midpoint E are E(\\frac{x+m}{2},\\frac{y}{2}). From \\frac{x+m}{2}=0, we get m=-x, so E(0,\\frac{y}{2}). Since AE\\bot PB, if x\\neq0, then k_{AE}\\cdot k_{PB}=\\frac{\\frac{y}{2}}{0-1}\\times\\frac{y}{2x}, which gives y^{2}=4x. If x=0, then m=0, and points P and B coincide, so line PB does not exist. Therefore, the trajectory equation is y^{2}=4x (x\\neq0)."
  },
  {
    "text": "Given that in $\\triangle ABC$, $AB=3$, $\\frac{CA}{CB}=\\frac{1}{2}$, then the maximum area of $\\triangle ABC$ is?",
    "fact_expressions": "A: Point;B: Point;C: Point;LineSegmentOf(A, B) = 3;LineSegmentOf(C, A)/LineSegmentOf(C, B) = 1/2",
    "query_expressions": "Max(Area(TriangleOf(A, B, C)))",
    "answer_expressions": "3",
    "fact_spans": "[[[30, 59]], [[30, 59]], [[30, 59]], [[21, 28]], [[29, 59]]]",
    "query_spans": "[[[61, 86]]]",
    "process": "Solution 1: Let $ AC = x $. From $ \\frac{CA}{CB} = \\frac{1}{2} $, we get $ BC = 2x $. By the law of cosines, $ \\cos A = \\frac{x^{2} + 3^{2} - (2x)^{2}}{2 \\times 3x} = \\frac{3 - x^{2}}{2x} $. So $ \\sin A = \\sqrt{1 - \\cos^{2} A} = \\sqrt{1 - \\left( \\frac{3 - x^{2}}{2x} \\right)^{2}} $. Therefore, $ S_{\\triangle ABC} = \\frac{1}{2} AC \\cdot AB \\sin A = \\frac{3x}{2} \\cdot \\sqrt{1 - \\left( \\frac{3 - x^{2}}{2x} \\right)^{2}} = \\frac{3}{2} \\cdot \\sqrt{x^{2} - \\left( \\frac{3 - x^{2}}{2} \\right)^{2}} = \\frac{3}{2} \\cdot \\sqrt{ -\\frac{1}{4}x^{4} + \\frac{5}{2}x^{2} - \\frac{9}{4} } = \\frac{3}{2} \\cdot \\sqrt{ -\\frac{1}{4}(x^{2} - 5)^{2} + 4 } $. From $ \\begin{cases} 2x + x > 3, \\\\ 2x - x < 3, \\end{cases} $ we get $ 1 < x < 3 $. Thus, when $ x^{2} = 5 $, the maximum area of $ \\triangle ABC $ is $ 3 $.\n\nSolution 2: Take point $ A $ as the origin and the line $ AB $ as the $ x $-axis to establish a Cartesian coordinate system as shown in the figure. Then $ A(0,0) $, $ B(3,0) $. Let $ C(x,y) $. From $ \\frac{|CA|}{|CB|} = \\frac{1}{2} $, we get $ \\frac{\\sqrt{x^{2} + y^{2}}}{\\sqrt{(x - 3)^{2} + y^{2}}} = \\frac{1}{2} $, which simplifies to $ (x + 1)^{2} + y^{2} = 4 $. Thus, the locus of point $ C $ is a circle centered at $ M(-1,0) $ with radius $ 2 $ (excluding the two points collinear with $ AB $). Therefore, $ S_{\\triangle ABC} = \\frac{1}{2} AB \\cdot |y_{C}| = \\frac{3}{2} |y_{C}| \\leqslant 3 $. Hence, the maximum area of $ \\triangle ABC $ is $ 3 $."
  },
  {
    "text": "Given that the hyperbolas $C_{1}$ and $C_{2}$ share the same vertices, the equation of $C_{1}$ is $\\frac{x^{2}}{4}-y^{2}=1$. If the slope of one asymptote of $C_{2}$ is twice the slope of one asymptote of $C_{1}$, then the equation of $C_{2}$ is?",
    "fact_expressions": "C1: Hyperbola;C2: Hyperbola;Vertex(C1) = Vertex(C2);Expression(C1) = (x^2/4 - y^2 = 1);Slope(OneOf(Asymptote(C2))) = 2*Slope(OneOf(Asymptote(C1)))",
    "query_expressions": "Expression(C2)",
    "answer_expressions": "x^2/4 - y^2/4 = 1",
    "fact_spans": "[[[2, 12], [26, 33], [81, 88]], [[13, 20], [64, 71], [104, 111]], [[2, 25]], [[26, 62]], [[64, 102]]]",
    "query_spans": "[[[104, 116]]]",
    "process": "The equation of $ C_{1} $ is $ \\frac{x^{2}}{4} - y^{2} = 1 $. An equation of one asymptote is $ y = \\frac{x}{2} $. Since the slope of one asymptote of $ C_{2} $ is twice the slope of one asymptote of $ C_{1} $, an equation of one asymptote of $ C_{2} $ is $ y = x $. Because the vertices of hyperbolas $ C_{1} $ and $ C_{2} $ coincide, the equation of $ C_{2} $ is $ \\frac{x^{2}}{4} - \\frac{y^{2}}{4} = 1 $."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $C$: $x^{2}=2 p y$ ($p>0$) intersects the parabola at points $A$ and $B$. If $4|A F|=|B F|$, then $\\frac{|A F|}{|O F |}$=?",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 2*p*y);p: Number;p>0;F: Point;Focus(C) = F;G: Line;PointOnCurve(F, G) = True;Intersection(G, C) = {A, B};A: Point;B: Point;4*Abs(LineSegmentOf(A, F)) = Abs(LineSegmentOf(B, F));O: Origin",
    "query_expressions": "Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(O, F))",
    "answer_expressions": "5/4",
    "fact_spans": "[[[1, 27], [38, 41]], [[1, 27]], [[9, 27]], [[9, 27]], [[30, 33]], [[1, 33]], [[34, 36]], [[0, 36]], [[34, 51]], [[42, 45]], [[46, 49]], [[53, 67]], [[69, 91]]]",
    "query_spans": "[[[69, 93]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a, b>0)$ is $\\sqrt{3} x+3 y=0$, find the eccentricity of the hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Expression(OneOf(Asymptote(G))) = (sqrt(3)*x + 3*y = 0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 56], [85, 88]], [[2, 56]], [[5, 56]], [[5, 56]], [[5, 56]], [[5, 56]], [[2, 83]]]",
    "query_spans": "[[[85, 94]]]",
    "process": "\\because the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a,b>0) has an asymptote equation \\sqrt{3}x+3y=0, \\therefore \\frac{b}{a}=\\frac{\\sqrt{3}}{3}, i.e., b=\\frac{\\sqrt{3}}{3}a \\therefore c=\\sqrt{a^{2}+b^{2}}=\\frac{2\\sqrt{3}}{3}a \\therefore the eccentricity of the hyperbola is e=\\frac{c}{a}=\\frac{2\\sqrt{3}}{3}"
  },
  {
    "text": "The equation $\\frac{x^{2}}{3-k}+\\frac{y^{2}}{2+k}=1$ represents an ellipse; then the range of values for $k$ is?",
    "fact_expressions": "G: Ellipse;Expression(G)=(x^2/(3 - k) + y^2/(k + 2) = 1 );k:Number",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-2, 1/2) + (1/2, 3)",
    "fact_spans": "[[[43, 45]], [[0, 45]], [[47, 50]]]",
    "query_spans": "[[[47, 57]]]",
    "process": "Since $\\frac{x^2}{3-k} + \\frac{y^{2}}{k^{2}+k} = 1$ represents an ellipse, we have $(3 - \\frac{1}{2})$, solving gives $\\begin{cases} -2 \\\\ x \\neq \\frac{1}{7} \\end{cases}$, the range of $k$ is $(-2, \\frac{1}{2}) \\cup (\\frac{1}{2}, 3)$, thus the answer is $\\underline{1} U (\\frac{1}{2}, 3)$. This question mainly examines the standard equation of an ellipse; during the solution process, exclude cases where the equation represents a circle. It is a basic problem."
  },
  {
    "text": "Given the parabola $ C $: $ y^{2} = 2 p x $ ($ p > 0 $), a line with inclination angle $ \\frac{\\pi}{4} $ intersects the parabola $ C $ at points $ A $ and $ B $, and the ordinate of the midpoint of segment $ AB $ is $ 1 $. Then the equation of the directrix of the parabola $ C $ is?",
    "fact_expressions": "C: Parabola;Expression(C) =(y^2 = 2*(p*x));p:Number;p>0;L:Line;Inclination(L)=pi/4;Intersection(L,C)={A,B};A: Point;B: Point;YCoordinate(MidPoint(LineSegmentOf(A,B))) = 1",
    "query_expressions": "Expression(Directrix(C))",
    "answer_expressions": "x=-1/2",
    "fact_spans": "[[[2, 28], [52, 58], [89, 95]], [[2, 28]], [[10, 28]], [[10, 28]], [[49, 51]], [[29, 51]], [[49, 68]], [[59, 62]], [[63, 66]], [[70, 87]]]",
    "query_spans": "[[[89, 102]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), then we have y_{1}=2px_{1}, y_{2}=2px_{2}. Subtracting these two equations gives: (y_{1}-y_{2})(y_{1}+y_{2})=2p(x_{1}-x_{2}). Since the slope of the line is 1, \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=1, hence y_{1}+y_{2}=2p. The y-coordinate of the midpoint of segment AB is 1, so y_{1}+y_{2}=2, thus p=1. Therefore, the equation of the directrix of the parabola is x=-\\frac{1}{2}."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and $P$ is a point on the directrix of the ellipse $C$. If $P F_{1}=2 P F_{2}$, then the range of the eccentricity of ellipse $C$ is?",
    "fact_expressions": "F1: Point;F2: Point;Focus(C) = {F1, F2};C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;P: Point;PointOnCurve(P, Directrix(C)) = True;LineSegmentOf(P, F1) = 2*LineSegmentOf(P, F2)",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "[\\sqrt{3}/3,1)",
    "fact_spans": "[[[2, 9]], [[10, 17]], [[2, 78]], [[18, 75], [83, 88], [117, 122]], [[18, 75]], [[25, 75]], [[25, 75]], [[25, 75]], [[25, 75]], [[79, 82]], [[79, 94]], [[96, 115]]]",
    "query_spans": "[[[117, 133]]]",
    "process": "Since |PF_{1}|=2|PF_{2}|, |PF_{1}|+2|PF_{2}|=2a, so |PF_{1}|=\\frac{2a}{3}, and a-c\\leqslant|PF_{1}|\\leqslant a+c, hence a-c\\leqslant\\frac{2a}{3}\\leqslanta+c, therefore \\frac{c}{a}\\geqslant\\frac{1}{3}, that is, the range of eccentricity is [\\frac{1}{3},1), fill in \\frac{1}{3},1"
  },
  {
    "text": "Given that $A$ and $B$ are the two intersection points of a line $l$ (not passing through the origin $O$) with the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $E$ is the midpoint of $AB$, and the slopes of lines $AB$ and $OE$ are denoted by $k_{AB}$ and $k_{OE}$ respectively. If $k_{AB} \\cdot k_{OE} = -\\frac{1}{2}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "l: Line;C: Ellipse;b: Number;a: Number;A: Point;B: Point;O: Origin;E: Point;k1: Number;k2: Number;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Negation(PointOnCurve(O, l));Intersection(l, C) = {A, B};MidPoint(LineSegmentOf(A, B)) = E;Slope(LineOf(A, B)) = k1;Slope(LineOf(O, E)) = k2;k1*k2 = -1/2",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[18, 23]], [[24, 81], [185, 187]], [[31, 81]], [[31, 81]], [[2, 5]], [[6, 9]], [[12, 17]], [[87, 90]], [[122, 131]], [[134, 143]], [[31, 81]], [[31, 81]], [[24, 81]], [[10, 23]], [[2, 86]], [[87, 98]], [[100, 143]], [[100, 143]], [[146, 182]]]",
    "query_spans": "[[[185, 193]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, $ E(x_{0},y_{0}) $. Since the slope of line $ AB $ exists, we have $ x_{1} \\neq x_{2} $. From the given conditions, we obtain\n\\[\n\\begin{cases}\n\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 \\\\\n\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{\\frac{2}{2}} = 1\n\\end{cases}\n\\]\nSubtracting these two equations yields\n\\[\n\\frac{(x_{1}-x_{2})(x_{1}+x_{2})}{a^{2}} + \\frac{(y_{1}-y_{2})(y_{1}+y_{2})}{b^{2}} = 0.\n\\]\nAlso, $ x_{1}+x_{2} = 2x_{0}' $, $ y_{1}+y_{2} = 2y_{0} $, so\n\\[\n\\frac{b^{2}}{a^{2}} + \\frac{y_{1}-y_{2}}{x_{1}} \\cdot \\frac{y_{0}}{x_{0}} = 0,\n\\]\nthus\n\\[\n\\frac{a^{2}}{a^{2}} + k_{AB} \\cdot k_{OE} = 0.\n\\]\nSince $ k_{AB} \\cdot k_{OE} = -\\frac{1}{2} $, we have $ \\frac{a}{a^{2}} = \\frac{1}{2} $, hence\n\\[\n\\frac{a^{2}-c^{2}}{a^{2}} = \\frac{1}{2},\n\\]\nthat is,\n\\[\n\\frac{c^{2}}{a^{2}} = \\frac{1}{2}.\n\\]\nTherefore, the eccentricity of the ellipse is $ e = \\frac{\\sqrt{2}}{2} $."
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{2}=-3$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/2 - y^2/2 = -3)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*x",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 47]]]",
    "process": "\\frac{x^{2}}{2}-\\frac{y^{2}}{2}=-3, simplifying yields \\frac{y^{2}}{6}-\\frac{x^{2}}{6}=1, its asymptotes are given by y=\\pm x"
  },
  {
    "text": "Given the ellipse $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$ $(a>b>0)$ has its right vertex at $A(1 , 0)$, and the length of the chord passing through its focus and perpendicular to the major axis is $1$, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;H:LineSegment;a > b;b > 0;Expression(G) = (x^2/b^2 + y^2/a^2 = 1);Coordinate(A) = (1, 0);RightVertex(G)=A;PointOnCurve(Focus(G),H);IsPerpendicular(H,MajorAxis(G));Length(H)=1;IsChordOf(l,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/4 + x^2 = 1",
    "fact_spans": "[[[2, 54], [71, 72], [89, 91]], [[4, 54]], [[4, 54]], [[59, 69]], [], [[4, 54]], [[4, 54]], [[2, 54]], [[59, 69]], [[2, 69]], [[70, 82]], [[71, 82]], [[70, 87]], [68, 79]]",
    "query_spans": "[[[89, 95]]]",
    "process": "From the given, we have $ b=1 $, $ \\frac{2b^{2}}{a}=1 $. Calculations yield $ a $ and $ b $, thus the result. Since the right vertex of the ellipse $ \\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1 $ is $ A(1,0) $, it follows that $ b=1 $, and the foci are at $ (0,c) $. Because the chord passing through the focus and perpendicular to the major axis has length 1, $ \\frac{2b^{2}}{a}=1 $, so $ a=2 $. Therefore, the equation of the ellipse is $ \\frac{y^{2}}{4}+x^{2}=1 $."
  },
  {
    "text": "What is the focal distance of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "10",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 44]]]",
    "process": "The foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ are $(-5,0)$ and $(5,0)$, so the focal distance is $10$."
  },
  {
    "text": "Given a line with slope $\\sqrt{3}$ passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$), intersecting the parabola at points $A$ and $B$, and $|A F|>|B F|$, then $\\frac{|A F|}{|B F|}$=?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;A: Point;F: Point;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(F, H);Slope(H) = sqrt(3);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, F)) > Abs(LineSegmentOf(B, F))",
    "query_expressions": "Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(B, F))",
    "answer_expressions": "3",
    "fact_spans": "[[[3, 24], [48, 51]], [[6, 24]], [[45, 47]], [[53, 56]], [[27, 30]], [[57, 60]], [[6, 24]], [[3, 24]], [[3, 30]], [[2, 47]], [[31, 47]], [[45, 62]], [[64, 77]]]",
    "query_spans": "[[[79, 102]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of an ellipse is $\\frac{\\sqrt{7}}{4}$, and the distance from one endpoint of the minor axis to the right focus is $4$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;Eccentricity(G) = sqrt(7)/4;Distance(OneOf(Endpoint(MinorAxis(G))),RightFocus(G)) = 4",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16+y^2/9=1",
    "fact_spans": "[[[2, 4], [49, 51]], [[2, 29]], [[2, 47]]]",
    "query_spans": "[[[49, 58]]]",
    "process": "Let the equation of the ellipse be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. From the given conditions, we have $e=\\frac{c}{a}=\\frac{\\sqrt{7}}{4}$, $\\sqrt{c^{2}+b^{2}}=a=4$, $a^{2}=b^{2}+c^{2}$. It follows that $a=4$, $c=\\sqrt{7}$, $b=3$. Thus, the standard equation of the ellipse is $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$."
  },
  {
    "text": "Given that $F$ is the right focus of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{8}=1$, $P$ is a point on the left branch of $C$, and $A(0, \\frac{36}{5})$, when the perimeter of $\\triangle A P F$ is minimized, then the coordinates of point $P$ are?",
    "fact_expressions": "C: Hyperbola;A: Point;P: Point;F: Point;Expression(C) = (x^2 - y^2/8 = 1);Coordinate(A) = (0, 36/5);RightFocus(C) = F;PointOnCurve(P, LeftPart(C));WhenMin(Perimeter(TriangleOf(A,P,F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(-11/7, 24/7)",
    "fact_spans": "[[[6, 39], [48, 51]], [[58, 78]], [[44, 47], [105, 109]], [[2, 5]], [[6, 39]], [[58, 78]], [[2, 43]], [[44, 56]], [[79, 103]]]",
    "query_spans": "[[[105, 114]]]",
    "process": "Let the left focus of the hyperbola be F. From the hyperbola $ C: x^{2}-\\frac{y^{2}}{8}=1 $, we obtain $ a=1 $, $ b=2\\sqrt{2} $, $ c=3 $, so $ F(3,0) $ and $ F(-3,0) $. Thus, $ |AF|=\\sqrt{(3-0)^{2}+(0-\\frac{36}{5})^{2}}=\\sqrt{\\frac{1521}{25}}=\\frac{39}{5} $ is constant. By the definition of the hyperbola, $ |PF|-|PF|=2a=2 $, so $ |PA|+|PF|=|PA|+|PF|+2 $. The perimeter of $ \\triangle PFA $ is $ |PA|+|PF|+|AF| $. Therefore, when point P moves along the left branch such that A, P, F are collinear, $ |PA|+|PF| $ attains its minimum value $ |AF| $, and thus the perimeter of $ \\triangle APF $ reaches its minimum. The equation of line AF is $ \\frac{x}{-3}+\\frac{y}{\\frac{36}{5}}=1 $, or $ 12x-5y+36=0 $. Solving simultaneously with the hyperbola equation:\n$$\n\\begin{cases}\n12x-5y+36=0 \\\\\nx^{2}-\\frac{y^{2}}{8}=1\n\\end{cases}\n$$\nwe obtain the coordinates of point P as $ \\left(-\\frac{11}{7},\\frac{24}{7}\\right) $."
  },
  {
    "text": "What is the distance from the focus to the directrix of the parabola $y=\\sqrt{2} x^{2}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = sqrt(2)*x^2)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "sqrt(2)/4",
    "fact_spans": "[[[0, 21]], [[0, 21]]]",
    "query_spans": "[[[0, 32]]]",
    "process": "Analysis: According to the given conditions, transform the equation of the parabola into standard form, then find its focus coordinates and directrix equation; based on this, calculate the distance from the focus to the directrix to obtain the answer. Details: According to the given conditions, the standard equation of the parabola $ y = \\sqrt{2}x^{2} $ is $ x^{2} = \\frac{\\sqrt{2}}{2}y $, its focus coordinates are $ (0, \\frac{\\sqrt{2}}{8}) $, and the directrix equation is $ y = -\\frac{\\sqrt{2}}{8} $. Then the distance from the focus to the directrix is $ \\underline{\\sqrt{2}} $."
  },
  {
    "text": "The focus of the parabola $y=a x^{2}$ is the intersection point of the line $x+y-1=0$ with the coordinate axes. Then, the equation of the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;a: Number;H: Line;Expression(G) = (y = a*x^2);Expression(H) = (x + y - 1 = 0);Focus(G) = Intersection(H,axis)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=-1",
    "fact_spans": "[[[0, 14], [37, 40]], [[3, 14]], [[18, 29]], [[0, 14]], [[18, 29]], [[0, 35]]]",
    "query_spans": "[[[37, 46]]]",
    "process": "Since the focus of the parabola $ y = ax^2 $ lies on the vertical axis, and the intersection point of the line $ x + y - 1 = 0 $ with the vertical axis has coordinates $ (0, 1) $, the directrix equation of the parabola is $ y = -1 $. (This question examines the directrix equation of a parabola; correctly finding the intersection point of the line with the vertical axis is key to solving the problem.)"
  },
  {
    "text": "It is known that the hyperbola $C$ passes through the point $(\\sqrt{2}, 3)$, and its asymptotes have equations $y = \\pm \\sqrt{3} x$. Then, the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Point;Coordinate(G) = (sqrt(2), 3);PointOnCurve(G, C) = True;Expression(Asymptote(C)) = (y = pm*sqrt(3)*x)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 8], [26, 27], [53, 59]], [[9, 25]], [[9, 25]], [[2, 25]], [[26, 51]]]",
    "query_spans": "[[[53, 65]]]",
    "process": "Since the asymptotes of hyperbola $ C $ are given by $ y = \\pm\\sqrt{3}x $, we can assume $ C: y^{2} - (\\sqrt{3}x)^{2} = \\lambda $ ($ \\lambda \\neq 0 $), that is, $ y^{2} - 3x^{2} = \\lambda $ ($ \\lambda \\neq 0 $). Since hyperbola $ C $ passes through the point $ (\\sqrt{2}, 3) $, we have $ 3^{2} - 3 \\times (\\sqrt{2})^{2} = \\lambda $. Solving gives $ \\lambda = 3 $. Therefore, the standard equation of hyperbola $ C $ is $ \\frac{y^{2}}{3} - x^{2} = 1 $, and its eccentricity is $ e = \\sqrt{\\frac{3+1}{3}} = \\frac{2\\sqrt{3}}{3} $."
  },
  {
    "text": "Given that $O$ is the coordinate origin, the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has eccentricity $\\frac{3 \\sqrt{5}}{5}$. From the right focus $F$ of hyperbola $C$, a perpendicular is drawn to an asymptote, with foot of the perpendicular at $A$. If the area of $\\triangle A F O$ is $\\sqrt{5}$, then the equation of hyperbola $C$ is?",
    "fact_expressions": "O: Origin;C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(C) = (3*sqrt(5))/5;F: Point;RightFocus(C) = F;L: Line;PointOnCurve(F, L);IsPerpendicular(L, Asymptote(C));A: Point;FootPoint(L, Asymptote(C)) = A;Area(TriangleOf(A, F, O)) = sqrt(5)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/5 - y^2/4 = 1",
    "fact_spans": "[[[2, 5]], [[11, 72], [101, 107], [163, 169]], [[11, 72]], [[18, 72]], [[18, 72]], [[18, 72]], [[18, 72]], [[11, 99]], [[111, 114]], [[101, 114]], [], [[100, 121]], [[100, 121]], [[125, 128]], [[100, 128]], [[130, 161]]]",
    "query_spans": "[[[163, 174]]]",
    "process": "As shown in the figure, |AF| = \\frac{|bc|}{\\sqrt{b^{2}+(-a)^{2}}} = b, so |OA| = a. Therefore, S_{\\triangle AFO} = \\frac{1}{2}ab = \\sqrt{5} \\Rightarrow ab = 2\\sqrt{5} \\textcircled{1}. Also, e = \\frac{c}{a} = \\frac{3\\sqrt{5}}{5} \\textcircled{2}, c^{2} = a^{2} + b^{2} \\textcircled{3}. Thus, from \\textcircled{1}\\textcircled{2}\\textcircled{3}, we get: a = \\sqrt{5}, b = 2. Hence, the equation of the hyperbola is: \\frac{x^{2}}{5} - \\frac{y^{2}}{4} = 1"
  },
  {
    "text": "10. Given that line $l_{1}$ is the directrix of the parabola $C$: $y^{2}=8x$, and $P$ is a moving point on $C$, then the minimum value of the sum of the distances from $P$ to line $l_{1}$ and line $l_{2}$: $3x-4y+24=0$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 8*x);l1: Line;Directrix(C) = l1;P: Point;PointOnCurve(P, C);l2: Line;Expression(l2) = (3*x - 4*y + 24 = 0)",
    "query_expressions": "Min(Distance(P, l1) + Distance(P, l2))",
    "answer_expressions": "6",
    "fact_spans": "[[[18, 25], [49, 52]], [[18, 37]], [[6, 16]], [[6, 41]], [[42, 46]], [[43, 58]], [[78, 88]], [[78, 104]]]",
    "query_spans": "[[[59, 115]]]",
    "process": ""
  },
  {
    "text": "Let the line $ l $ passing through the focus $ F $ of the parabola $ C: y^2 = 2px $ ($ p > 0 $) intersect the parabola at points $ M $ and $ N $ (where point $ M $ lies in the first quadrant). If $ \\overrightarrow{MN} = 3 \\overrightarrow{FN} $, then the slope of line $ l $ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;l: Line;PointOnCurve(F, l);M: Point;N: Point;Quadrant(M) = 1;Intersection(l, C) = {M, N};VectorOf(M, N) = 3*VectorOf(F, N)",
    "query_expressions": "Slope(l)",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[1, 27], [40, 43]], [[1, 27]], [[9, 27]], [[9, 27]], [[30, 33]], [[1, 33]], [[34, 39], [116, 121]], [[0, 39]], [[45, 48], [57, 61]], [[49, 52]], [[57, 66]], [[34, 54]], [[69, 114]]]",
    "query_spans": "[[[116, 126]]]",
    "process": "Let the angle of inclination be $\\theta$, then $\\cos\\theta=\\frac{MF-FN}{MN}=\\frac{1}{3}\\therefore\\tan\\theta=2\\sqrt{2}$"
  },
  {
    "text": "Given that one asymptote of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1(b>0)$ has the equation $y=2x$, then $b$=?",
    "fact_expressions": "G: Hyperbola;b: Number;b>0;Expression(G) = (x^2 - y^2/b^2 = 1);Expression(OneOf(Asymptote(G))) = (y = 2*x)",
    "query_expressions": "b",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 39]], [[58, 61]], [[5, 39]], [[2, 39]], [[2, 56]]]",
    "query_spans": "[[[58, 63]]]",
    "process": "From the hyperbola equation, it is known that the asymptote equation is $ y = \\pmb{x} $, so $ b = 2 $."
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ are $F_{1}$ and $F_{2}$ respectively, $P$ is a point on the ellipse, and $\\angle P F_{1} F_{2}=\\frac{\\pi}{6}$, $\\angle P F_{2} F_{1}=\\frac{\\pi}{3}$. Then the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;e: Number;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(P,G);AngleOf(P, F1, F2) = pi/6;AngleOf(P, F2, F1) = pi/3;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[0, 45], [73, 75], [156, 158]], [[2, 45]], [[2, 45]], [[69, 72]], [[53, 60]], [[61, 68]], [[162, 165]], [[0, 45]], [[0, 68]], [[0, 68]], [[69, 78]], [[80, 116]], [[117, 154]], [[156, 165]]]",
    "query_spans": "[[[162, 167]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through the right focus $F_{2}$ intersects the right branch of the hyperbola at points $A$ and $B$. If $|A B|=5$, then the perimeter of $\\triangle A B F_{1}$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;l: Line;PointOnCurve(F2,l) = True;Intersection(l, RightPart(G)) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(A, B)) = 5",
    "query_expressions": "Perimeter(TriangleOf(A, B, F1))",
    "answer_expressions": "26",
    "fact_spans": "[[[2, 41], [84, 87]], [[2, 41]], [[50, 57]], [[58, 65], [70, 77]], [[2, 65]], [[2, 65]], [[78, 83]], [[66, 83]], [[78, 100]], [[91, 94]], [[95, 98]], [[102, 111]]]",
    "query_spans": "[[[114, 140]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{m}=1$ is $\\frac{5}{4}$, then $m$ equals?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/m = 1);m: Number;Eccentricity(G) = 5/4",
    "query_expressions": "m",
    "answer_expressions": "9",
    "fact_spans": "[[[0, 39]], [[0, 39]], [[59, 62]], [[0, 57]]]",
    "query_spans": "[[[59, 65]]]",
    "process": "Using the eccentricity formula of a hyperbola $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} $, we can obtain: since the hyperbola $ \\frac{x^{2}}{16} - \\frac{y^{2}}{m} = 1 $ gives $ a^{2} = 16 $, $ b^{2} = m $, and the eccentricity is $ \\frac{5}{4} $, then $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{1 + \\frac{m}{16}} = \\frac{5}{4} $, solving yields $ m = 9 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{8}-\\frac{y^{2}}{8}=1$ has left focus $F$, point $M$ lies on the right branch of hyperbola $C$, $A(0,4)$, when the perimeter of $\\triangle M A F$ is minimized, the area of $\\triangle M A F$ is?",
    "fact_expressions": "C: Hyperbola;A: Point;M: Point;F: Point;Expression(C) = (x^2/8 - y^2/8 = 1);Coordinate(A) = (0, 4);LeftFocus(C) = F;PointOnCurve(M, RightPart(C));WhenMin(Perimeter(TriangleOf(M,A,F)))",
    "query_expressions": "Area(TriangleOf(M, A, F))",
    "answer_expressions": "12",
    "fact_spans": "[[[2, 45], [59, 65]], [[70, 78]], [[54, 58]], [[50, 53]], [[2, 45]], [[70, 78]], [[2, 53]], [[54, 69]], [[79, 103]]]",
    "query_spans": "[[[104, 126]]]",
    "process": "The perimeter of $\\triangle MAF$ is $|MA|+|MF|+|AF|$, where $|AF|=4\\sqrt{2}$ is a constant, so it suffices to find $|MA|+|MF|$. Using the definition, we obtain $|MF|=|MF|+4\\sqrt{2}$, so the perimeter is $|MA|+|MF|+8\\sqrt{2}$. The perimeter is minimized when points $M$, $A$, and $F$ are collinear. The area is obtained by area partitioning. As shown in the figure, let the right focus of hyperbola $C$ be $F$. From the given conditions, $a=2\\sqrt{2}$, $F(-4,0)$, $F(4,0)$. Since point $M$ lies on the right branch, $|MF|-|MF|=2a=4\\sqrt{2}$, so $|MF|=|MF|+4\\sqrt{2}$. Then the perimeter of $\\triangle MAF$ is $|MA|+|MF|+|AF|=|MA|+|MF|+8\\sqrt{2}\\geqslant|AF|+8\\sqrt{2}=12\\sqrt{2}$. That is, when $M$ is at $M$, the perimeter of $\\triangle MAF$ is minimized. At this time, the equation of line $AF$ is $y=-x+4$. Solving simultaneously\n$$\n\\begin{cases}\ny=-x+4 \\\\\n\\frac{x^{2}}{8}-\\frac{y^{2}}{8}=1\n\\end{cases}\n$$\nwe obtain $y-1=0$, so $y_{M}=1$. Hence, the area of $\\triangle MAF$ is $\\frac{1}{2}|FF'||OA|-\\frac{1}{2}|FF||y_{M}|=\\frac{1}{2}\\times8\\times(4-1)=12$"
  },
  {
    "text": "The equation of an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $y=2 x$, and the length of the imaginary axis is $2$. Then the coordinates of the foci of this hyperbola are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Expression(OneOf(Asymptote(G))) = (y = 2*x);Length(ImageinaryAxis(G)) = 2",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(pm*sqrt(5)/2, 0)",
    "fact_spans": "[[[0, 56], [84, 87]], [[0, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[0, 73]], [[0, 81]]]",
    "query_spans": "[[[84, 94]]]",
    "process": "Since the asymptote equation is y=2x, it follows that \\frac{b}{a}=2^{i}; since the length of the imaginary axis is 2, it follows that 2b=2; and since a^{2}+b^{2}=c^{2}, solving gives a=\\frac{1}{2}, b=1, c=\\frac{\\sqrt{5}}{2}; thus the coordinates of the foci are \\pm\\frac{\\sqrt{5}}{2},0"
  },
  {
    "text": "The equation of the locus of the centers of circles externally tangent to both circles $x^{2}+y^{2} + 8 x+7=0$ and $x^{2} + y^{2}-8 x+12=0$ is?",
    "fact_expressions": "G: Circle;Expression(G) = (x^2 + y^2/(8*x) + 7 = 0);C:Circle;Expression(C) = (x^2/y^2 - 8*x + 12 = 0);H:Circle;IsOutTangent(H,G);IsOutTangent(H,C)",
    "query_expressions": "LocusEquation(Center(H))",
    "answer_expressions": "(4*x^2-4*y^2/63=1)&(x>0)",
    "fact_spans": "[[[1, 25]], [[1, 25]], [[26, 51]], [[26, 51]], [[55, 56]], [[0, 56]], [[0, 56]]]",
    "query_spans": "[[[55, 66]]]",
    "process": ""
  },
  {
    "text": "The line $ l $: $ x - y - 2 = 0 $ is tangent to a circle at point $ M(3,1) $, and the center of the circle lies on the directrix of the parabola $ y^{2} = -4x $. Then the standard equation of the circle is?",
    "fact_expressions": "l: Line;Expression(l) = (x-y-2=0);M: Point;Coordinate(M) = (3, 1);H: Circle;TangentPoint(l, H) = M;G: Parabola;Expression(G) = (y^2 = -4*x);PointOnCurve(Center(H), Directrix(G))",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x-1)^2+(y-3)^2=8",
    "fact_spans": "[[[0, 16]], [[0, 16]], [[21, 30]], [[21, 30]], [[17, 18], [56, 57]], [[0, 30]], [[35, 50]], [[35, 50]], [[17, 54]]]",
    "query_spans": "[[[56, 64]]]",
    "process": "From the parabola equation, it is known that the directrix equation is $x=1$. Let the center of the circle be $(1,b)$. According to the problem, $(1-b)^{2}=\\frac{|1-b-2|}{\\sqrt{2}}$, solving yields $b=3$, $r=2\\sqrt{2}$. Therefore, the standard equation of the circle is $(x-1)^{2}+(y-3)^{2}=8$."
  },
  {
    "text": "It is known that the two asymptotes of a hyperbola passing through the point $(\\sqrt{3}, 4)$ are $2x \\pm y = 0$, then the equation of this hyperbola is?",
    "fact_expressions": "H: Point;Coordinate(H) = (sqrt(3), 4);PointOnCurve(H, G);G: Hyperbola;Expression(Asymptote(G)) = (2*x + pm*y = 0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/4-x^2=1",
    "fact_spans": "[[[3, 19]], [[3, 19]], [[2, 23]], [[20, 23], [46, 49]], [[20, 43]]]",
    "query_spans": "[[[46, 54]]]",
    "process": "Assume $\\frac{y^{2}}{4}-x^{2}=\\lambda$ ($\\lambda\\neq0$). Since it passes through $(\\sqrt{3},4)$, we have $\\frac{4^{2}}{4}-(\\sqrt{3})^{2}=\\lambda$, so $\\lambda=1$. Therefore, the equation of the hyperbola is $\\frac{y^{2}}{4}-x^{2}=1$. Fill in $\\frac{y^{2}}{4}-x^{2}=1$."
  },
  {
    "text": "If the point $P(t, 1)$ lies on the parabola $x^{2}=2 p y(p>0)$ and its distance to the directrix is $2$, then $t=?$",
    "fact_expressions": "G: Parabola;p: Number;P: Point;p>0;Expression(G)=(x^2 = 2*p*y);Coordinate(P)=(t,1);PointOnCurve(P,G);Distance(P,Directrix(G))=2;t:Number",
    "query_expressions": "t",
    "answer_expressions": "pm*2",
    "fact_spans": "[[[12, 33], [37, 38]], [[15, 33]], [[1, 11]], [[15, 33]], [[12, 33]], [[1, 11]], [[1, 34]], [[1, 47]], [[49, 52]]]",
    "query_spans": "[[[49, 54]]]",
    "process": "(Analysis) From the equation of the parabola, find its directrix equation; then, since point $ P(t,1) $ lies on the parabola and the distance from point $ P(t,1) $ to the directrix is 2, set up a system of equations to solve. From $ x^{2} = 2py $ ($ p > 0 $), the directrix equation is $ y = -\\frac{p}{2} $. Since point $ P(t,1) $ lies on the parabola and its distance to the directrix is 2, we have:\n$$\n\\begin{cases}\n1 - \\left(-\\frac{p}{2}\\right) = 2 \\\\\nt^{2} = 2p \\quad (p > 0)\n\\end{cases}\n$$\nSolving gives:\n$$\n\\begin{cases}\np = 2 \\\\\nt = \\pm 2\n\\end{cases}\n$$"
  },
  {
    "text": "If a point $P$ on the parabola $y^{2}=8x$ has a horizontal coordinate of $1$, then the distance from point $P$ to the focus $F$ of the parabola is?",
    "fact_expressions": "G: Parabola;P:Point;F:Point;Expression(G) = (y^2 = 8*x);Focus(G)=F;PointOnCurve(P,G);XCoordinate(P)=1",
    "query_expressions": "Distance(P,F)",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 15], [37, 40]], [[18, 21], [31, 35]], [[43, 46]], [[1, 15]], [[37, 46]], [[1, 21]], [[18, 29]]]",
    "query_spans": "[[[31, 51]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{k}=1$ is $e=3$, then the value of $k$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/k = 1);k: Number;e: Number;Eccentricity(G) = e;e = 3",
    "query_expressions": "k",
    "answer_expressions": "-32",
    "fact_spans": "[[[0, 38]], [[0, 38]], [[49, 52]], [[42, 47]], [[0, 47]], [[42, 47]]]",
    "query_spans": "[[[49, 56]]]",
    "process": ""
  },
  {
    "text": "Given $\\sin \\theta+\\cos \\theta=\\frac{1}{5}$, and the foci of the hyperbola $x^{2} \\sin \\theta+y^{2} \\cos \\theta=1$ lie on the $y$-axis, then the eccentricity $e$ of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;theta:Number;Sin(theta)+Cos(theta)=1/5;Expression(C)=(x^2*Sin(theta)+y^2*Cos(theta)=1);PointOnCurve(Focus(C),yAxis);Eccentricity(C)=e;e:Number",
    "query_expressions": "e",
    "answer_expressions": "sqrt(21)/3",
    "fact_spans": "[[[40, 82], [93, 99]], [[2, 39]], [[2, 39]], [[40, 82]], [[40, 91]], [[93, 106]], [[103, 106]]]",
    "query_spans": "[[[103, 108]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, the upper vertex is $A$, and the line $A F_{1}$ intersects the ellipse $C$ at another point $B$. Then the area of $\\triangle A B F_{2}$ is?",
    "fact_expressions": "C: Ellipse;F1: Point;A: Point;B: Point;F2: Point;Expression(C) = (x^2/2 + y^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;UpperVertex(C) = A;Intersection(LineOf(A,F1),C)={A,B}",
    "query_expressions": "Area(TriangleOf(A, B, F2))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[2, 34], [79, 84]], [[43, 50]], [[63, 66]], [[91, 94]], [[51, 58]], [[2, 34]], [[2, 58]], [[2, 58]], [[2, 66]], [[63, 94]]]",
    "query_spans": "[[[96, 122]]]",
    "process": "From the given conditions, we have A(0,1), F₁(-1,0), F₂(1,0). The equation of line AF₁ is y = x + 1. Solving the system of equations  \n\\begin{cases}\\frac{x^{2}}{2}+y^{2}=1\\\\y=x+1\\end{cases}  \nyields  \n\\begin{cases}x=0\\\\y=1\\end{cases}  \nor  \n\\begin{cases}x=-\\frac{4}{3}\\\\y=-\\frac{1}{3}\\end{cases},  \nso B(-\\frac{4}{3},-\\frac{1}{3}). Therefore, S_{\\triangle ABF} = \\frac{1}{2} \\times 2 \\times (1 + \\frac{1}{3}) = \\frac{4}{3}. The answer is."
  },
  {
    "text": "Given that on the line $l$: $x - m y + \\sqrt{3} m = 0$ there exists a point $M$ such that the product of the slopes $k_{MA}$ and $k_{MB}$ of the lines connecting $M$ to the points $A(-1,0)$ and $B(1,0)$ is $3$, then the range of real values for $m$ is?",
    "fact_expressions": "l: Line;Expression(l) = (x - m*y + sqrt(3)*m = 0);m: Real;M: Point;PointOnCurve(M, l);A: Point;B: Point;Coordinate(A) = (-1, 0);Coordinate(B) = (1, 0);k1: Number;k2: Number;Slope(LineSegmentOf(M, A)) = k1;Slope(LineSegmentOf(M, B)) = k2;k1*k2 = 3",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-oo, -sqrt(3)/3)+(-sqrt(3)/3, -sqrt(6)/6]+[sqrt(6)/6, sqrt(3)/3)+(sqrt(3)/3, +oo)",
    "fact_spans": "[[[2, 29]], [[2, 29]], [[93, 98]], [[32, 36]], [[2, 36]], [[41, 50]], [[53, 61]], [[41, 50]], [[53, 61]], [[66, 75]], [[76, 85]], [[32, 85]], [[32, 85]], [[66, 91]]]",
    "query_spans": "[[[93, 105]]]",
    "process": "Let point M(x,y). According to k_{MA}\\cdotk_{MB}=3, the trajectory equation of point M can be found. Then, since line l and the trajectory of point M have a common point, solve the system of equations of line l and the trajectory of point M. From A\\geqslant0, an inequality in terms of m can be obtained, from which the range of real values of m can be determined. Let point M(x,y). From k_{MA}\\cdotk_{MB}=3, we get \\frac{y}{x+1}\\cdot\\frac{y}{x-1}=3. Simplifying yields y^{2}=3x^{2}-3 (x\\neq\\pm1). According to the problem, line l intersects the curve y^{2}=3x^{2}-3 (x\\neq\\pm1). Solving the system: \\begin{cases}x-my+\\sqrt{3}m=0\\\\y^{2}=3x^{2}-3\\end{cases}, eliminating x gives (3m^{2}-1)y^{2}-6\\sqrt{3}m^{2}y+9m^{2}-3=0,\\textcircled{1} When 3m^{2}-1=0, we obtain m^{2}=\\frac{1}{3}, at which point equation \\textcircled{1} becomes -2\\sqrt{3}y=0, yielding y=0, which does not satisfy the conditions. When 3m^{2}-1\\neq0, A=108m^{4}-4(3m^{2}-1)(9m^{2}-3)\\geqslant0, simplifying gives 6m^{2}-1\\geqslant0, so m^{2}\\geqslant\\frac{1}{6} and m^{2}\\neq\\frac{1}{3}, leading to m\\in(-\\infty,-\\frac{\\sqrt{6}}{6}]\\cup[\\frac{\\sqrt{6}}{6},\\frac{\\sqrt{3}}{3})\\cup(\\frac{\\sqrt{3}}{3},+\\infty)"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left focus is $F_{1}$, and $P$ is a point on the hyperbola. $P F_{1}$ is parallel to an asymptote of the hyperbola $C$, and $|P O|=|F_{1} O|$, where $O$ is the origin. Then the eccentricity $e$ of the hyperbola $C$ is $?$.",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F1: Point;O: Origin;e: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, C);LeftFocus(C)=F1;IsParallel(LineSegmentOf(P, F1), Asymptote(C));Abs(LineSegmentOf(P, O)) = Abs(LineSegmentOf(F1, O));Eccentricity(C) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 63], [80, 83], [97, 103], [141, 147]], [[10, 63]], [[10, 63]], [[76, 79]], [[68, 75]], [[131, 134]], [[151, 154]], [[10, 63]], [[10, 63]], [[2, 63]], [[76, 86]], [[2, 75]], [[87, 109]], [[111, 128]], [[141, 154]]]",
    "query_spans": "[[[151, 156]]]",
    "process": "From the given conditions, $F_{1}(-c,0)$, the asymptotes of the hyperbola are $y=\\pm\\frac{b}{a}x$. Without loss of generality, take the equation of line $PF_{1}$ as $y=-\\frac{b}{a}(x+c)$. Solving the system  \n$$\n\\begin{cases}\ny=-\\frac{b}{a}(x+c)\\\\\n\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1\n\\end{cases},\n$$\nwe obtain $x=-\\frac{a^{2}+c^{2}}{2c}$, $y=-\\frac{b(c^{2}-a^{2})}{2ac}$, so $P\\left(-\\frac{a^{2}+c^{2}}{2c}, -\\frac{b(c^{2}-a^{2})}{2ac}\\right)$. Let $M$ be the midpoint of $PF_{1}$, and connect $OM$, then $M\\left(-\\frac{a^{2}+3c^{2}}{4c}, -\\frac{b(c^{2}-a^{2})}{4ac}\\right)$. Since $|PO|=|F_{1}O|$, it follows that $OM\\perp PF_{1}$, thus  \n$$\n\\frac{-\\frac{b(c^{2}-a^{2})}{4ac}}{-\\frac{a^{2}+3c^{2}}{4c}} \\cdot \\left(-\\frac{b}{a}\\right) = -1.\n$$\nSimplifying yields $c=\\sqrt{5}a$, so the eccentricity is $e=\\frac{c}{a}=\\sqrt{5}$."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$, its directrix intersects the curve $4 x^{2}-\\frac{3 y^{2}}{4}=1(y>0)$ at point $P$, $F$ is the focus of the parabola, and the inclination angle of line $P F$ is $135^{\\circ}$. Then $p=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;I: Curve;Expression(I) = And((4*x^2 - 3*y^2/4 = 1), (y > 0));Intersection(Directrix(G), I) = P;P: Point;F: Point;Focus(G) = F;Inclination(LineOf(P,F)) = ApplyUnit(135, degree)",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 23], [74, 77]], [[2, 23]], [[107, 110]], [[5, 23]], [[27, 63]], [[27, 63]], [[2, 69]], [[65, 69]], [[70, 73]], [[70, 79]], [[80, 105]]]",
    "query_spans": "[[[107, 112]]]",
    "process": "Let $ P\\left(\\frac{p}{2}-t,t\\right) $ be substituted into the parabola, yielding $ t^{2}+2pt-p^{2}=0 \\Rightarrow t=-p-\\sqrt{2}p \\Rightarrow P\\left(\\frac{3}{2}p+\\sqrt{2}p,-p-\\sqrt{2}p\\right) $. Substituting into the hyperbola gives $ p=2 $."
  },
  {
    "text": "If the distance from point $P(m, n)$ on the parabola $x^{2}=4 y$ to its focus $F$ is $3$, then the value of $n$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);P: Point;Coordinate(P) = (m, n);m: Number;n: Number;PointOnCurve(P, G) = True;F: Point;Focus(G) = F;Distance(P, F) = 3",
    "query_expressions": "n",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 15], [28, 29]], [[1, 15]], [[17, 27]], [[17, 27]], [[18, 27]], [[43, 46]], [[1, 27]], [[31, 34]], [[28, 34]], [[17, 41]]]",
    "query_spans": "[[[43, 50]]]",
    "process": "The focus of the parabola $x^{2}=4y$ is $(0,1)$, and the equation of the directrix is $y=-1$. According to the problem, there is a point $P(m,n)$ on the parabola $x^{2}=4y$ whose distance to the focus is 3. Using the definition of the parabola, we obtain $n+1=3$, solving gives $n=2$."
  },
  {
    "text": "Given the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the right focus of the ellipse is $F$, point $M$ is any point on the ellipse $E$, point $A(0, b)$. If the maximum perimeter of triangle $\\triangle A M F$ is $6b$, then the eccentricity of the ellipse $E$ is?",
    "fact_expressions": "E: Ellipse;b: Number;a: Number;A: Point;M: Point;F: Point;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (0, b);RightFocus(E) = F;PointOnCurve(M, E);Max(Perimeter(TriangleOf(A, M, F))) = 6*b",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[2, 59], [73, 78], [128, 133]], [[9, 59]], [[9, 59]], [[84, 94]], [[68, 72]], [[64, 67]], [[9, 59]], [[9, 59]], [[2, 59]], [[84, 94]], [[2, 67]], [[68, 83]], [[96, 126]]]",
    "query_spans": "[[[128, 139]]]",
    "process": "According to the definition of the ellipse, |AM| + |MF| + |AF| = 3a + |AM| - |MF|. Using plane geometric properties to find the maximum value and establish a relationship between a and b, the eccentricity can be solved. Let F be the left focus of the ellipse, ∵ A(0, b). ∴ |AF| = |AF| = a, the perimeter of △AMF is |AM| + |MF| + |AF| = 3a + |AM| - |MF|. ∴ when points A, F, M are collinear, (|AM| - |MF|)_{\\max} = |AF| = a. ∴ the maximum perimeter of △AMF is 4a = 6b. ∴ 4a^{2} = 9b^{2} = 9(a^{2} - c^{2}), ∴ \\frac{c^{2}}{a^{2}} = \\frac{5}{9}, ∴ e = \\frac{\\sqrt{5}}{3}"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, one of its asymptotes is $y=\\frac{\\sqrt{5}}{2} x$, and it passes through the point $(2,2 \\sqrt{5})$. Then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Expression(OneOf(Asymptote(C))) = (y = x*(sqrt(5)/2));G: Point;Coordinate(G) = (2, 2*sqrt(5));PointOnCurve(G,C) = True",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/15 - x^2/12 = 1",
    "fact_spans": "[[[2, 63], [119, 122]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 96]], [[100, 117]], [[100, 117]], [[2, 117]]]",
    "query_spans": "[[[119, 127]]]",
    "process": "According to the problem, we can assume the standard equation of the hyperbola as \\frac{x^{2}}{4}-\\frac{y^{2}}{5}=\\lambda(\\lambda\\neq0). Then substitute the coordinates of the point (2,2\\sqrt{5}) into the equation of hyperbola C to find the value of \\lambda, thereby obtaining the equation of hyperbola C. Solution: From the given information, one asymptote equation of hyperbola C is \\frac{x}{2}-\\frac{y}{\\sqrt{5}}=0. Assume the equation of hyperbola C is \\frac{x^{2}}{4}-\\frac{y^{2}}{5}=\\lambda(\\lambda\\neq0). Substituting the coordinates of the point (2,2\\sqrt{5}) into the equation of hyperbola C gives \\lambda=\\frac{2^{2}}{4}-\\frac{(2\\sqrt{5})^{2}}{5}=-3. Therefore, the equation of hyperbola C is \\frac{x^{2}}{4}-\\frac{y^{2}}{5}=-3, which is equivalent to \\frac{y^{2}}{15}-\\frac{x^{2}}{12}=1."
  },
  {
    "text": "Given that point $P$ lies on the parabola $y^{2}=4x$, find the coordinates of point $P$ when the sum of the distance from point $P$ to point $Q(2,-2)$ and the distance from point $P$ to the focus of the parabola is minimized.",
    "fact_expressions": "G: Parabola;Q: Point;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(Q) = (2, -2);PointOnCurve(P, G);WhenMin(Distance(P,Q)+Distance(P,Focus(G)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1,-2)",
    "fact_spans": "[[[7, 21], [49, 52]], [[30, 40]], [[25, 29], [2, 6], [44, 48], [65, 69]], [[7, 21]], [[30, 40]], [[2, 22]], [[25, 64]]]",
    "query_spans": "[[[65, 74]]]",
    "process": "From the given conditions, point Q lies inside the parabola. Draw a perpendicular line from Q to the directrix, with foot N, intersecting the parabola at point P. Let the focus of the parabola be F(1,0), and the equation of the directrix be x = -1. Then |PQ| + |PF| = |PQ| + |PN| \\geqslant |QN|, with equality if and only if points P, Q, N are collinear. This means the sum of the distance from point P to point Q(2,-2) and the distance from point P to the focus of the parabola is minimized. Therefore, the y-coordinate of P is -2. Substituting y = -2 into the equation of the parabola gives 4 = 4x_{P}, so the coordinates of P are (1,-2)."
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2 x$ be $F$, and let two lines $l_{1}$, $l_{2}$ passing through $F$ intersect the parabola at points $A$, $B$, $C$, $D$, respectively. The slopes $k_{1}$, $k_{2}$ of $l_{1}$, $l_{2}$ satisfy $k_{1}^{2}+k_{2}^{2}=2$. Then the minimum value of $|A B|+|C D|$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);F: Point;Focus(G) = F;l1: Line;l2: Line;A: Point;B: Point;C: Point;D: Point;PointOnCurve(F, l1);PointOnCurve(F, l2);Intersection(l1, G) = {A, B};Intersection(l2, G) = {C, D};k1: Number;k2: Number;Slope(l1) = k1;Slope(l2) = k2;k1^2+k2^2 = 2",
    "query_expressions": "Min(Abs(LineSegmentOf(A, B)) + Abs(LineSegmentOf(C, D)))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 15], [52, 55]], [[1, 15]], [[19, 22], [24, 27]], [[1, 22]], [[30, 39], [74, 81]], [[42, 49], [84, 91]], [[56, 60]], [[61, 64]], [[65, 68]], [[69, 72]], [[23, 49]], [[23, 49]], [[30, 72]], [[30, 72]], [[94, 101]], [[104, 111]], [[74, 111]], [[74, 111]], [[113, 136]]]",
    "query_spans": "[[[138, 157]]]",
    "process": "The focus of the parabola has coordinates $F\\left(\\frac{1}{2},0\\right)$. Let line $l_{1}: y = k_{1}\\left(x - \\frac{1}{2}\\right)$ ($k_{1} \\neq 0$), line $l_{2}: y = k_{2}\\left(x - \\frac{1}{2}\\right)$ ($k_{2} \\neq 0$). Solving the system $\\begin{cases} y^{2} = 2x \\\\ y = k_{1}\\left(x - \\frac{1}{2}\\right) \\end{cases}$, we obtain $4k_{1}^{2}x^{2} - (4k_{1}^{2} + 8)x + k_{1}^{2} = 0$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $x_{1} + x_{2} = 1 + \\frac{2}{k_{1}^{2}}$. Let $C(x_{3},y_{3})$, $D(x_{4},y_{4})$, similarly we get $x_{3} + x_{4} = 1 + \\frac{2}{k^{2}}$. By the properties of the parabola: $|AB| = 2 + \\frac{2}{k_{1}}$, $|CD| = 2 + \\frac{2}{k_{2}}$, and since $k_{1}^{2} + k_{2}^{2} = 2$, $|AB| + |CD| = 4 + \\frac{2(k_{1}^{2} + k_{2}^{2})}{k_{1}^{2}k_{2}^{2}} = 4 + \\frac{4}{k_{1}^{2}k_{2}^{2}} = 4 + \\frac{2}{\\left(\\frac{k_{1}^{2} + k_{2}^{2}}{2}\\right)^{2}} = 4 + 4 = 8$, with equality holding if and only if $|k_{1}| = |k_{2}| = 1$. Therefore, the minimum value of $|AB| + |CD|$ is $8$."
  },
  {
    "text": "Given the equation of the parabola is $y=2 x^{2}$, then the coordinates of the focus of this parabola are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 2*x^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,1/8)",
    "fact_spans": "[[[2, 5], [23, 26]], [[2, 20]]]",
    "query_spans": "[[[23, 33]]]",
    "process": "According to the given condition, $x^{2}=\\frac{y}{2}$, hence its focus lies on the positive half of the y-axis, $p=\\frac{1}{4}$, so the coordinates of the focus are $(0,\\frac{1}{8})$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $A$ lies on the right branch of $C$, and $A F_{1}$ intersects $C$ at point $B$. If $\\overrightarrow{F_{2} A} \\cdot \\overrightarrow{F_{2} B}=0$, $|\\overrightarrow{F_{2} A}|=|\\overrightarrow{F_{2} B}|$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;A: Point;PointOnCurve(A, RightPart(C)) = True;Intersection(LineSegmentOf(A, F1), C) = B;B: Point;DotProduct(VectorOf(F2, A), VectorOf(F2, B)) = 0;Abs(VectorOf(F2, A)) = Abs(VectorOf(F2, B))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 63], [93, 96], [111, 114], [239, 242]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[72, 79]], [[80, 87]], [[2, 87]], [[2, 87]], [[88, 92]], [[88, 100]], [[101, 120]], [[116, 120]], [[122, 181]], [[182, 237]]]",
    "query_spans": "[[[239, 248]]]",
    "process": "Because $\\overrightarrow{F_{2}A}\\cdot\\overrightarrow{F_{2}B}=0$, $|\\overrightarrow{F_{2}A}|=|\\overrightarrow{F_{2}B}|$, so $\\triangle ABF_{2}$ is an isosceles right triangle. Let $|AF_{2}|=|BF_{2}|=m$, then $|AB|=\\sqrt{2}m$. By the definition of the hyperbola, we have $|AF_{1}|-|AF_{2}|=2a$, $|BF_{2}|-|BF_{1}|=2a$. So $|AF_{1}|=2a+m$, $|BF_{1}|=m-2a$. Because $|AB|=|AF_{1}|-|BF_{1}|=(m+2a)-(m-2a)=\\sqrt{2}m$, so $m=2\\sqrt{2}a$. Therefore $|AF_{1}|=2a+2\\sqrt{2}a=(2+2\\sqrt{2})a$, $|AF_{2}|=2\\sqrt{2}a$. In $\\triangle AF_{1}F_{2}$, by the law of cosines, $|F_{1}F_{2}|^{2}=|AF_{1}|^{2}+|AF_{2}|^{2}-2|AF_{1}||AF_{2}|\\cos\\angle F_{1}AF_{2}$, so $4c^{2}=[(2\\sqrt{2}+2)a]^{2}+(2\\sqrt{2}a)^{2}-2(2\\sqrt{2}+2)a\\cdot2\\sqrt{2}a\\cdot\\frac{\\sqrt{2}}{2}=12a^{2}$, so $c^{2}=3a^{2}$, giving $c=\\sqrt{3}a$, so the eccentricity is $e=\\frac{c}{a}=\\sqrt{3}$."
  },
  {
    "text": "The equation of the ellipse passing through the point $(3,-2)$ and having the same foci as the ellipse $4 x^{2}+9 y^{2}=36$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (4*x^2 + 9*y^2 = 36);H: Point;Coordinate(H) = (3, -2);Z: Ellipse;PointOnCurve(H, Z);Focus(Z) = Focus(G)",
    "query_expressions": "Expression(Z)",
    "answer_expressions": "x^2/15+y^2/10=1",
    "fact_spans": "[[[12, 34]], [[12, 34]], [[1, 10]], [[1, 10]], [[40, 42]], [[0, 42]], [[11, 42]]]",
    "query_spans": "[[[40, 46]]]",
    "process": ""
  },
  {
    "text": "Given fixed points $A(-2,0)$, $B(2,0)$, if a moving point $M$ satisfies $|M A|+|M B|=8$, then the range of $|M A|$ is?",
    "fact_expressions": "A: Point;B: Point;Coordinate(A) = (-2, 0);Coordinate(B) = (2, 0);M: Point;Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, B)) = 8",
    "query_expressions": "Range(Abs(LineSegmentOf(M, A)))",
    "answer_expressions": "[2, 6]",
    "fact_spans": "[[[4, 13]], [[16, 24]], [[4, 13]], [[16, 24]], [[28, 31]], [[33, 48]]]",
    "query_spans": "[[[50, 64]]]",
    "process": "By the given condition, the moving point M satisfies |MA| + |MB| = 8 ≥ |AB| = 4. According to the definition of an ellipse, the trajectory of point M is an ellipse with A and B as foci, where 2a = 8, 2c = 4. Solving gives a = 4, c = 2. By the properties of an ellipse, |MA| ∈ [a - c, a + c], that is, |MA| ∈ [2, 6]."
  },
  {
    "text": "Given that the vertex of the parabola is at the origin and the focus is at $(0,-3)$, the standard equation of the parabola is?",
    "fact_expressions": "G: Parabola;O:Origin;Vertex(G)=O;Coordinate(Focus(G))=(0,-3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2=-12*y",
    "fact_spans": "[[[2, 5], [30, 33]], [[9, 13]], [[2, 13]], [[2, 27]]]",
    "query_spans": "[[[30, 40]]]",
    "process": "Since the vertex of the parabola is at the origin and the focus is at (0, -3), we have \\frac{p}{2} = 3, solving gives p = 6, and the standard equation of the parabola is x^{2} = -12y"
  },
  {
    "text": "If the line $y=k x-\\frac{3}{5}$ intersects the ellipse $E$: $\\frac{x^{2}}{4}+y^{2}=1$ at two points $P$ and $Q$, then the range of the intercept on the $x$-axis of the perpendicular bisector $l$ of segment $P Q$ is?",
    "fact_expressions": "E: Ellipse;G: Line;k: Number;Q: Point;P: Point;l:Line;Expression(E) = (x^2/4 + y^2 = 1);Expression(G) = (y = k*x - 3/5);Intersection(G, E) = {P, Q};PerpendicularBisector(LineSegmentOf(P,Q))=l",
    "query_expressions": "Range(Intercept(l,xAxis))",
    "answer_expressions": "[-9/20,9/20]",
    "fact_spans": "[[[23, 55]], [[1, 22]], [[3, 22]], [[60, 63]], [[56, 59]], [[78, 81]], [[23, 55]], [[1, 22]], [[1, 65]], [[67, 81]]]",
    "query_spans": "[[[78, 97]]]",
    "process": "Let $ P(x_{1},y_{1}) $, $ Q(x_{2},y_{2}) $, and the midpoint of segment $ PQ $ be $ T(x_{0},y_{0}) $.  \n① When $ k=0 $, it is easy to see that the intercept on the x-axis of the perpendicular bisector $ l $ of segment $ PQ $ is 0;  \n② When $ k\\neq0 $, from  \n$$\n\\begin{cases}\ny=kx-\\frac{3}{5} \\\\\n\\frac{x^{2}}{4}+y^{2}=1\n\\end{cases}\n$$\nwe obtain  \n$$\n(4k^{2}+1)x^{2}-\\frac{24}{5}kx-\\frac{64}{25}=0,\n$$\nso  \n$$\nx_{1}+x_{2}=\\frac{24}{5}\\cdot\\frac{k}{4k^{2}+1},\n$$\n$$\ny_{1}+y_{2}=k(x_{1}+x_{2})-\\frac{6}{5}=k\\left(\\frac{24}{5}\\cdot\\frac{k}{4k^{2}+1}\\right)-\\frac{6}{5}=\\frac{-6}{5(4k^{2}+1)}.\n$$\nThus,  \n$$\nx_{0}=\\frac{12}{5}\\cdot\\frac{k}{4k^{2}+1}=\\frac{12}{5}\\cdot\\frac{4k^{2}+1}{4k+\\frac{1}{k}},\n\\quad\ny_{0}=\\frac{-3}{5(4k^{2}+1)}.\n$$\nSince $ 4k+\\frac{1}{k}\\in(-\\infty,-4]\\cup[4,+\\infty) $, we have $ x_{0}\\in[-\\frac{3}{5},0)\\cup(0,\\frac{3}{5}] $.  \nBecause $ x_{0}=\\frac{12}{5}\\cdot\\frac{k}{4k^{2}+1} $, $ y_{0}=\\frac{-3}{5(4k^{2}+1)} $, so $ ky_{0}=-\\frac{1}{4}x_{0} $.  \nSince the equation of line $ l $ is $ y-y_{0}=-\\frac{1}{k}(x-x_{0}) $, the intercept of line $ l $ on the x-axis is  \n$$\nky_{0}+x_{0}=\\frac{3}{4}x_{0}\\in[-\\frac{9}{20},0)\\cup(0,\\frac{9}{20}].\n$$\nIn conclusion, the range of the intercept on the x-axis of the perpendicular bisector $ l $ of segment $ PQ $ is $ [-\\frac{9}{20},\\frac{9}{20}] $."
  },
  {
    "text": "Given the two foci of a hyperbola are $F_{1}(-\\sqrt{10}, 0)$, $F_{2}(\\sqrt{10}, 0)$, and the asymptotes are $y=\\pm \\frac{1}{2} x$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;Coordinate(F1) = (-sqrt(10), 0);Coordinate(F2) = (sqrt(10), 0);Expression(Asymptote(G)) = (y = pm*(x/2));Focus(G) = {F1, F2}",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/8 - y^2/2 = 1",
    "fact_spans": "[[[2, 5], [84, 87]], [[11, 33]], [[35, 56]], [[11, 33]], [[35, 56]], [[2, 82]], [[2, 56]]]",
    "query_spans": "[[[84, 94]]]",
    "process": ""
  },
  {
    "text": "What is the length of the real axis of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{16}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/16 = 1)",
    "query_expressions": "Length(RealAxis(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 45]]]",
    "process": "From the given, we can obtain that the length of the real axis is 2a = 4."
  },
  {
    "text": "Given that a line passing through the focus $F$ of the parabola $C$: $y^2 = 3x$ intersects the parabola $C$ at points $A$ and $B$, and the line $OA$ ($O$ being the origin) intersects the directrix of the parabola $C$ at point $D$, then the minimum area of $\\triangle ABD$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 3*x);F: Point;Focus(C) = F;G: Line;PointOnCurve(F, G);A: Point;B: Point;Intersection(G, C) = {A, B};O: Origin;D: Point;Intersection(LineOf(O, A), Directrix(C)) = D",
    "query_expressions": "Min(Area(TriangleOf(A, B, D)))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[3, 22], [32, 38], [70, 76]], [[3, 22]], [[25, 28]], [[3, 28]], [[29, 31]], [[2, 31]], [[41, 44]], [[45, 48]], [[29, 50]], [[59, 62]], [[82, 86]], [[51, 86]]]",
    "query_spans": "[[[88, 113]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $E$: $x^{2}=4 y$, draw two tangents from point $P(2,-1)$ to $E$, with points of tangency $A$ and $B$. Then $|A B|=$?",
    "fact_expressions": "E: Parabola;P: Point;A: Point;B: Point;l1: Line;l2: Line;Expression(E) = (x^2 = 4*y);Coordinate(P) = (2, -1);TangentOfPoint(P, E) = {l1, l2};TangentPoint(l1, E) = A;TangentPoint(l2, E) = B",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 21], [34, 37]], [[23, 33]], [[48, 51]], [[52, 55]], [], [], [[2, 21]], [[23, 33]], [[22, 42]], [[22, 55]], [[22, 55]]]",
    "query_spans": "[[[57, 66]]]",
    "process": "The slope of the tangent line clearly exists. Let the equation of the tangent line be $ y+1 = k(x-2) $, that is, $ y = kx - 2k - 1 $. Combining  \n\\[\n\\begin{cases}\ny = kx - 2k - 1 \\\\\nx^{2} = 4y\n\\end{cases}\n\\]  \nand eliminating $ y $, we obtain $ x^{2} - 4kx + 8k + 4 = 0 $. Therefore, $ \\Delta = (-4k)^{2} - 4(8k + 4) = 0 $, which simplifies to $ k^{2} - 2k - 1 = 0 $. Thus, $ k = 1 - \\sqrt{2} $ or $ k = 1 + \\sqrt{2} $. Let the slopes of tangents $ PA $ and $ PB $ be $ k_{1} $ and $ k_{2} $, respectively, and let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Then $ k_{1} = 1 - \\sqrt{2} $, $ k_{2} = 1 + \\sqrt{2} $. Substituting $ k_{1} = 1 - \\sqrt{2} $ into $ x^{2} - 4kx + 8k + 4 = 0 $, we get  \n$ x^{2} - 4(1 - \\sqrt{2})x + 8(1 - \\sqrt{2}) + 4 = 0 $,  \nthat is, $ (x - 2 + 2\\sqrt{2})^{2} = 0 $, yielding $ x = 2 - 2\\sqrt{2} $, so $ x_{1} = 2 - 2\\sqrt{2} $. Hence,  \n$ y_{1} = \\frac{x_{1}^{2}}{4} = \\frac{(2 - 2\\sqrt{2})^{2}}{4} = 3 - 2\\sqrt{2} $,  \nso $ A(2 - 2\\sqrt{2}, 3 - 2\\sqrt{2}) $. Similarly, we obtain $ B(2 + 2\\sqrt{2}, 3 + 2\\sqrt{2}) $."
  },
  {
    "text": "Draw a line $l$ through the focus $F$ of the parabola $W$: $x^{2}=8 y$, intersecting the parabola at points $A$ and $B$. Then, when the sum of the distances from points $A$ and $B$ to the line $x-2 y-4=0$ is minimized, the length of segment $A B$ is?",
    "fact_expressions": "W: Parabola;Expression(W) = (x^2 = 8*y);F: Point;Focus(W) = F;l: Line;PointOnCurve(F,l) = True;Intersection(l, W) = {A, B};A: Point;B: Point;l1: Line;Expression(l1) = (x-2*y-4=0);WhenMin(Distance(A,l1) + Distance(B,l1)) = True",
    "query_expressions": "Length(LineSegmentOf(A,B))",
    "answer_expressions": "17/2",
    "fact_spans": "[[[1, 20], [33, 36]], [[1, 20]], [[23, 26]], [[1, 26]], [[27, 32]], [[0, 32]], [[27, 47]], [[38, 41], [50, 54]], [[42, 45], [55, 58]], [[59, 72]], [[59, 72]], [[49, 80]]]",
    "query_spans": "[[[81, 93]]]",
    "process": "From the parabola $ W: x^{2} = 8y $, we obtain $ F(0,2) $. Let the equation of line $ l $ be $ y = kx + 2 $. From \n\\[\n\\begin{cases}\nx^{2} = 8y \\\\\ny = kx + 2\n\\end{cases},\n\\]\nwe get $ x^{2} - 8kx - 16 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = 8k $, so $ y_{1} + y_{2} = k(x_{1} + x_{2}) + 4 = 8k^{2} + 4 $. Thus, the midpoint $ M $ of segment $ AB $ has coordinates $ M(4k, 4k^{2} + 2) $. The distance from $ M $ to the line $ x - 2y - 4 = 0 $ is $ d = \\frac{|4k|}{\\sqrt{5}} $, then the sum of distances from points $ A $, $ B $ to the line $ x - 2y - 4 = 0 $ is $ 2d = \\frac{2|8k^{2} - 4k + 8|}{\\sqrt{5}} = \\frac{8(2k^{2} - k + 2)}{\\sqrt{5}} $. Therefore, when $ k = \\frac{1}{4} $, $ 2d $ takes the minimum value. At this time, $ |AB| = y_{1} + y_{2} + p = 8k^{2} + 4 + p = 8 \\times \\left(\\frac{1}{4}\\right)^{2} + 4 + 4 = \\frac{17}{2} $. Hence, the answer is:"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ has eccentricity $e=\\frac{\\sqrt{3}}{2}$, then the value of $m$ is?",
    "fact_expressions": "E: Ellipse;m: Number;e: Number;Expression(E) = (x^2/4 + y^2/m = 1);Eccentricity(E) = e ;e = sqrt(3)/2",
    "query_expressions": "m",
    "answer_expressions": "{1, 16}",
    "fact_spans": "[[[2, 39]], [[67, 70]], [[43, 65]], [[2, 39]], [[2, 65]], [[43, 65]]]",
    "query_spans": "[[[67, 74]]]",
    "process": "\\because the ellipse equation is \\frac{x^2}{4} + \\frac{y^2}{m} = 1, \\therefore \\textcircled{1} when the foci of the ellipse are on the x-axis, a^{2}=4, b^{2}=m, we get c=\\sqrt{4-m}; eccentricity e=\\frac{\\sqrt{4-m}}{2}=\\frac{\\sqrt{3}}{2}, solving gives m=1; \\textcircled{2} when the foci of the ellipse are on the y-axis, a^{2}=m, b^{2}=4, we get c=\\sqrt{m-4}; eccentricity e=\\frac{\\sqrt{m-4}}{\\sqrt{m}}=\\frac{\\sqrt{3}}{2}, solving gives m=16. In summary, m=16 or m=1. Hence, choose D."
  },
  {
    "text": "Given $F_{1}(-3,0)$, $F_{2}(3,0)$, a moving point $M$ satisfies $|M F_{1}|+|M F_{2}|=10$, then what is the trajectory equation of the moving point $M$?",
    "fact_expressions": "F1: Point;Coordinate(F1) = (-3, 0);F2: Point;Coordinate(F2) = (3, 0);M: Point;Abs(LineSegmentOf(M,F1)) + Abs(LineSegmentOf(M,F2)) = 10",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2/25 + y^2/16 = 1",
    "fact_spans": "[[[2, 15]], [[2, 15]], [[17, 30]], [[17, 30]], [[32, 35], [65, 68]], [[37, 61]]]",
    "query_spans": "[[[65, 74]]]",
    "process": "Since |MF_{1}| + |MF_{2}| = 10 > |F_{1}F_{2}| = 6, the trajectory of point M is an ellipse with foci F_{1} and F_{2}, where 2a = 10, so a = 5, c = 3, and b = \\sqrt{a^{2} - c^{2}} = 4. Therefore, the equation of the trajectory of moving point M is \\frac{x^{2}}{25} + \\frac{y^{2}}{16} = 1."
  },
  {
    "text": "It is known that the right focus of the ellipse $E$ centered at the origin coincides with the focus of the parabola $C$: $y^{2}=4x$, and the ellipse $E$ intersects the directrix of the parabola $C$ at points $A$ and $B$. If $AB=3$, then what is the length of the minor axis of the ellipse $E$?",
    "fact_expressions": "C: Parabola;E: Ellipse;A: Point;B: Point;O:Origin;Expression(C) = (y^2 = 4*x);Center(E)=O;RightFocus(E) = Focus(C);Intersection(E, Directrix(C)) = {A, B};LineSegmentOf(A, B) = 3",
    "query_expressions": "Length(MinorAxis(E))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[20, 39], [51, 57]], [[10, 15], [45, 50], [83, 88]], [[62, 65]], [[66, 69]], [[5, 9]], [[20, 39]], [[2, 15]], [[10, 44]], [[45, 71]], [[74, 81]]]",
    "query_spans": "[[[83, 94]]]",
    "process": "First, find the focus of the parabola; then the right focus of the ellipse is known. Next, set up the equation of the ellipse and rewrite $ a^{2} $ in the equation into an expression form. According to $ AB = 3 $, solve for the value of $ b^{2} $, so the length of the minor axis $ 2b $ of the ellipse can be found. Since the parabola $ C: y^{2} = 4x $ has focus $ (1,0) $, the right focus of the ellipse is $ (1,0) $. Let the ellipse equation be $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $), so $ a^{2} = b^{2} + 1 $, thus the ellipse equation becomes $ \\frac{x^{2}}{b^{2}+1} $. The directrix of the parabola is $ x = -1 $, so $ y^{2} = b^{2}\\left(1 - \\frac{1}{b^{2}+1}\\right) $, so $ y = $. Also, since $ AB = 3 $, we have $ \\frac{2b^{2}}{\\sqrt{b^{2}+1}} = 3 $, so $ b^{2} = 3 $, hence $ 2b = 2\\sqrt{3} $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left vertex is $A$, the left focus is $F$. If the distance from the upper vertex of the ellipse to the focus is $2$, and the eccentricity is $e=\\frac{1}{2}$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;F: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(G) = A;LeftFocus(G) = F;Distance(UpperVertex(G), Focus(G)) = 2;e: Number;Eccentricity(G) = e;e = 1/2",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2/4+y^2/3=1",
    "fact_spans": "[[[2, 54], [73, 75], [110, 112]], [[4, 54]], [[4, 54]], [[59, 62]], [[67, 70]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 62]], [[2, 70]], [[73, 89]], [[93, 108]], [[73, 108]], [[93, 108]]]",
    "query_spans": "[[[110, 119]]]",
    "process": "Since the distance from the upper vertex of the ellipse to the focus is 2, we have a=2. Because the eccentricity is e=\\frac{1}{2}, it follows that c=1, b=\\sqrt{a^{2}-c^{2}}=\\sqrt{3}. Thus, the equation of the ellipse is \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and point $P$ lies on the ellipse. (1) If the distance from point $P$ to focus $F_{1}$ is equal to $1$, then what is the distance from point $P$ to focus $F_{2}$? (2) A line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$; what is the perimeter of $\\triangle ABF_{2}$? (3) If $\\angle PF_{1}F_{2}=120^{\\circ}$, then what is the distance from point $P$ to focus $F_{1}$?",
    "fact_expressions": "F1: Point;F2: Point;Focus(G) = {F1, F2};G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);P: Point;PointOnCurve(P, G);Distance(P, F1) = 1;H: Line;PointOnCurve(F1, H);A: Point;B: Point;Intersection(H, G) = {A, B};AngleOf(P, F1, F2) = ApplyUnit(120, degree)",
    "query_expressions": "Distance(P, F2);Perimeter(TriangleOf(A, B, F2));Distance(P, F1)",
    "answer_expressions": "3\n8\n6/5",
    "fact_spans": "[[[2, 9], [83, 90], [125, 132], [222, 229]], [[10, 17], [107, 114]], [[2, 60]], [[18, 55], [66, 68], [136, 138]], [[18, 55]], [[61, 65], [76, 80], [100, 104], [215, 219]], [[61, 69]], [[76, 98]], [[133, 135]], [[124, 135]], [[140, 143]], [[144, 147]], [[133, 149]], [[181, 213]]]",
    "query_spans": "[[[100, 119]], [[151, 175]], [[215, 234]]]",
    "process": "(1) The distance from point P to focus F_{2} can be calculated using the fact that the sum of the distances from point P to focus F_{1} and focus F_{2} is 2a. (2) The value of |AB|+|AF_{2}|+|BF_{2}| can be calculated using the fact that the sum of the distances from point P to focus F_{1} and focus F_{2} is 2a. (3) It can be solved using the cosine law and the definition of an ellipse. [Detailed solution] From the standard equation of the ellipse, we have: a^{2}=4, b^{2}=3, hence a=2, b=\\sqrt{3}, c=\\sqrt{a^{2}-b^{2}}=\\sqrt{4-3}=1. (1) By the definition of the ellipse, |PF_{1}|+|PF_{2}|=2a, and since |PF_{1}|=1, it follows that |PF_{2}|=4-1=3. (2) The perimeter of \\triangle ABF_{2}, L_{\\Delta ABF_{2}} = |AB|+|AF_{2}|+|BF_{2}| = |AF_{1}|+|BF_{1}|+|AF_{2}|+|BF_{2}| = (|AF_{1}|+|AF_{2}|)+(|BF_{1}|+|BF_{2}|) = 2a+2a=4a=8. (3) In \\triangle PF_{1}F_{2}, by the cosine law, we have |PF_{2}|^{2}=|PF_{1}|^{2}+|F_{1}F_{2}|^{2}-2|PF_{1}||F_{1}F_{2}|\\cos\\angle PF_{1}F_{2}, which gives |PF_{2}|^{2}=|PF_{1}|^{2}+4+2|PF_{1}|. By the definition of the ellipse, |PF_{1}|+|PF_{2}|=2a. Solving these two equations simultaneously yields |PF_{1}|=\\frac{6}{5}."
  },
  {
    "text": "A line $l$ passing through the right focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ intersects the ellipse at points $A$ and $B$. If the midpoint of chord $AB$ is $M\\left(\\frac{4}{7}, -\\frac{3}{7}\\right)$, then $|AB|=$?",
    "fact_expressions": "l: Line;G: Ellipse;A: Point;B: Point;M: Point;Expression(G) = (x^2/4 + y^2/3 = 1);PointOnCurve(RightFocus(G),l);Intersection(l, G) = {A, B};Coordinate(M) = (4/7, -3/7);MidPoint(LineSegmentOf(A, B)) = M;IsChordOf(LineSegmentOf(A,B),G)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "24/7",
    "fact_spans": "[[[43, 48]], [[1, 38], [49, 51]], [[53, 56]], [[57, 60]], [[72, 102]], [[1, 38]], [[0, 48]], [[43, 62]], [[72, 102]], [[65, 102]], [[49, 69]]]",
    "query_spans": "[[[104, 114]]]",
    "process": ""
  },
  {
    "text": "If hyperbola $C$ shares the same asymptotes with the hyperbola $\\frac{x^{2}}{12}-\\frac{y^{2}}{8}=1$ and passes through the point $A(3 , \\sqrt{2})$, then the equation of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Hyperbola;A: Point;Coordinate(A) = (3, sqrt(2));Expression(G) = (x^2/12-y^2/8=1);Asymptote(C)=Asymptote(G);PointOnCurve(A,C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/6 - y^2/4 = 1",
    "fact_spans": "[[[1, 7], [75, 81]], [[8, 48]], [[55, 73]], [[55, 73]], [[8, 48]], [[1, 52]], [[1, 73]]]",
    "query_spans": "[[[75, 86]]]",
    "process": ""
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ has two foci $F_{1}$, $F_{2}$. A line passes through $F_{1}$ and intersects the ellipse at points $P$, $Q$. Then, what is the perimeter of $\\triangle P Q F_{2}$?",
    "fact_expressions": "G: Ellipse;H: Line;P: Point;Q: Point;F2: Point;F1: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(G)={F1,F2};PointOnCurve(F1, H);Intersection(H, G) = {P, Q}",
    "query_expressions": "Perimeter(TriangleOf(P, Q, F2))",
    "answer_expressions": "20",
    "fact_spans": "[[[0, 38], [71, 73]], [[60, 62]], [[74, 77]], [[78, 81]], [[51, 59]], [[43, 50], [63, 70]], [[0, 38]], [[0, 59]], [[60, 70]], [[60, 81]]]",
    "query_spans": "[[[83, 109]]]",
    "process": ""
  },
  {
    "text": "The ellipse and hyperbola have the same foci $F_{1}(-c, 0)$, $F_{2}(c, 0)$. A minor axis endpoint of the ellipse is $B$, and the line $F_{1} B$ is parallel to an asymptote of the hyperbola. If the eccentricities of the ellipse and hyperbola are $e_{1}$, $e_{2}$ respectively, then the minimum value of $3 e_{1}^{2}+e_{2}^{2}$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;F1: Point;B: Point;F2: Point;OneOf(Endpoint(MinorAxis(H)))=B;Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);Focus(G) = {F1, F2};Focus(H) = {F1, F2};IsParallel(LineOf(F1, B), OneOf(Asymptote(G)));e1: Number;e2: Number;Eccentricity(G) = e2;Eccentricity(H) = e1;c:Number",
    "query_expressions": "Min(3*e1^2 + e2^2)",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[3, 6], [68, 71], [84, 87]], [[0, 2], [42, 44], [81, 83]], [[12, 26]], [[52, 55]], [[28, 41]], [[42, 55]], [[12, 26]], [[28, 41]], [[0, 41]], [[0, 41]], [[56, 79]], [[94, 101]], [[103, 110]], [[81, 110]], [[81, 110]], [[12, 26]]]",
    "query_spans": "[[[112, 141]]]",
    "process": "From the given conditions, the foci of the hyperbola lie on the x-axis. Let the major axis of the ellipse be $2a$, the minor axis be $2b$, the real axis of the hyperbola be $2a'$, and the imaginary axis be $2b'$. Since one endpoint of the minor axis of the ellipse is $B$, and the line $F_{1}B$ is parallel to an asymptote of the hyperbola, we have $k_{F_{1}B} = \\frac{b'}{a}$, that is, $\\frac{b}{c} = \\frac{a^{3}}{b^{2}}$. Squaring both sides yields $\\frac{b^{2}}{a^{2}} = \\frac{b^{2}}{c^{2}}$, leading to $\\frac{c^{2}-a^{2}}{a^{2}} = \\frac{a^{2}-c^{2}}{c^{2}}$, i.e., $\\frac{c^{2}}{a^{2}} = \\frac{a^{2}}{c^{2}}$, hence $(\\frac{c}{a})^{2} = (\\frac{a}{c})^{2}$. Given $e_{1} = \\frac{a}{c}$, $e_{2} = \\frac{a^{2}}{a^{2}}$, we obtain $e_{1} \\cdot e_{2} = 1$. Since $e_{1}, e_{2}$ are both positive, $3e_{1}^{2} + e_{2}^{2} > 2\\sqrt{3e^{2} \\cdot e_{1}^{2}} = 2\\sqrt{3}$. The equality holds if and only if $3e_{1}^{2} = e_{2}^{2}$, i.e., $e_{2} = \\sqrt{3}e_{1}$, $e_{1} = \\frac{\\sqrt{3}}{3}$, $e_{2} = \\sqrt{3}$. Therefore, the minimum value of $3e_{1}^{2} + e_{2}^{2}$ is $2\\sqrt{3}$."
  },
  {
    "text": "If the ellipse $x^{2}+4(y-a)^{2}=4$ and the parabola $x^{2}=2 y$ have common points, then the range of real number $a$ is?",
    "fact_expressions": "G: Parabola;H: Ellipse;a: Real;Expression(G) = (x^2 = 2*y);Expression(H) = (x^2 + 4*(-a + y)^2 = 4);IsIntersect(H, G)",
    "query_expressions": "Range(a)",
    "answer_expressions": "[-1,17/8]",
    "fact_spans": "[[[24, 38]], [[1, 23]], [[44, 49]], [[24, 38]], [[1, 23]], [[1, 42]]]",
    "query_spans": "[[[44, 56]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$, point $P$ is a moving point on the parabola $C$. From point $P$, draw tangents to the circle $M$: $(x-3)^{2}+y^{2}=4$, with points of tangency $A$ and $B$. Then, the range of values for the length of segment $AB$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);P: Point;PointOnCurve(P, C);M: Circle;Expression(M) = (y^2 + (x - 3)^2 = 4);L1: Line;L2: Line;TangentOfPoint(P, M) = {L1, L2};TangentPoint(L1, M) = A;TangentPoint(L2, M) = B;A: Point;B: Point",
    "query_expressions": "Range(Length(LineSegmentOf(A, B)))",
    "answer_expressions": "[2*sqrt(2),4)",
    "fact_spans": "[[[2, 21], [27, 33]], [[2, 21]], [[22, 26], [39, 43]], [[22, 37]], [[44, 68]], [[44, 68]], [], [], [[38, 71]], [[38, 84]], [[38, 84]], [[77, 80]], [[81, 84]]]",
    "query_spans": "[[[86, 102]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, with foci $F_{1}$ and $F_{2}$. A line perpendicular to the $x$-axis is drawn through $F_{2}$, intersecting the hyperbola at points $A$ and $B$, and the inradius of triangle $\\triangle ABF_{1}$ is $a$. Then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;Focus(G) = {F1, F2};Z: Line;PointOnCurve(F2, Z);IsPerpendicular(Z, xAxis);A: Point;B: Point;Intersection(Z, G) = {A, B};Radius(InscribedCircle(TriangleOf(A, B, F1))) = a",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(1+sqrt(5))/2",
    "fact_spans": "[[[2, 58], [96, 99], [143, 146]], [[2, 58]], [[5, 58]], [[137, 140]], [[5, 58]], [[5, 58]], [[63, 70]], [[71, 78], [80, 87]], [[2, 78]], [], [[79, 95]], [[79, 95]], [[100, 103]], [[104, 107]], [[79, 109]], [[111, 140]]]",
    "query_spans": "[[[143, 152]]]",
    "process": ""
  },
  {
    "text": "The equation $\\frac{x^{2}}{5-m} + \\frac{y^{2}}{m-3}=1$ represents an ellipse; what is the range of real values for $m$?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G)=(x^2/(5-m)+y^2/(m-3)=1)",
    "query_expressions": "Range(m)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[45, 47]], [[49, 54]], [[0, 47]]]",
    "query_spans": "[[[49, 60]]]",
    "process": ""
  },
  {
    "text": "$F_{1}$ is the left focus of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, $P$ is a moving point on the ellipse, and $A(1,1)$ is a fixed point. Then the minimum value of $|P A|+|P F_{1}|$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/5 = 1);F1: Point;LeftFocus(G) = F1;P: Point;PointOnCurve(P, G);A: Point;Coordinate(A) = (1, 1)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F1)))",
    "answer_expressions": "6-sqrt(2)",
    "fact_spans": "[[[8, 45], [54, 56]], [[8, 45]], [[0, 7]], [[0, 49]], [[50, 53]], [[50, 60]], [[61, 69]], [[61, 69]]]",
    "query_spans": "[[[74, 97]]]",
    "process": "The ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ has $a=3$, $b=\\sqrt{5}$, $c=2$. As shown in the figure, let the right focus of the ellipse be $F'(2,0)$; then $|PF|+|PF'|=2a=6$; therefore $|PA|+|PF|=|PA|+6-|PF'|=6+|PA|-|PF'|$. From the graph, when $P$ lies on the line $AF'$, $|PA|-|PF'|=|AF'|=\\sqrt{2}$. When $P$ does not lie on the line $AF'$, by the triangle inequality that the difference of two sides is less than the third side, we have $|PA|-|PF'|<|AF'|=\\sqrt{2}$. When $P$ lies on the extension of $F'A$, $|PA|-|PF'|$ attains its minimum value $-\\sqrt{2}$. Hence, the minimum value of $|PA|+|PF|$ is $6-\\sqrt{2}$."
  },
  {
    "text": "A moving circle intersects the lines $3x - y = 0$ and $3x + y = 0$, producing chord lengths of $8$ and $4$, respectively. Then the trajectory equation of the center of the moving circle is?",
    "fact_expressions": "G: Circle;H: Line;C:Line;Expression(H)=(3*x - y = 0);Expression(C)=(3*x+y=0);Length(InterceptChord(H,G))=8;Length(InterceptChord(C,G))=4",
    "query_expressions": "LocusEquation(Center(G))",
    "answer_expressions": "x*y=10",
    "fact_spans": "[[[1, 3], [44, 46]], [[4, 15]], [[16, 25]], [[4, 15]], [[16, 25]], [[1, 41]], [[1, 41]]]",
    "query_spans": "[[[44, 55]]]",
    "process": "The moving circle intersects the lines $3x - y = 0$ and $3x + y = 0$, producing chord lengths of 8 and 4 respectively. Using the point-to-line distance formula, $|MA|^{2}$ and $|MC|^{2}$ can be found. By the perpendicular chord theorem, $|MA|^{2} + |AB|^{2} = |MC|^{2} + |EC|^{2}$, which can then be simplified. As shown in the figure, let point $M(x, y)$, and from the conditions we have $AB = 4$, $EC = 2$. Using the point-to-line distance formula, we obtain $|MA|^{2} = \\frac{(3x - y)^{2}}{10}$, $|MC|^{2} = \\frac{(3x + y)^{2}}{10}$. By the perpendicular chord theorem: $|MA|^{2} + |AB|^{2} = |MC|^{2} + |EC|^{2}$. $\\therefore \\frac{(3x - y)^{2}}{10} + 6 = \\frac{(3x + y)^{2}}{10} + 4$, simplifying yields $xy = 10$, $\\therefore$ the trajectory equation of point $M$ is $xy = 10$."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the two foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$. A circle with diameter $F_{1} F_{2}$ intersects the hyperbola at a point $P$. If $P F_{1}=2 P F_{2}$, then the equations of the two asymptotes of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;F2: Point;F1: Point;P: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Focus(G) = {F1, F2};IsDiameter(LineSegmentOf(F1,F2),H);OneOf(Intersection(H,G))=P;LineSegmentOf(P, F1) = 2*LineSegmentOf(P, F2)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*2*x",
    "fact_spans": "[[[17, 74], [102, 105], [137, 140]], [[20, 74]], [[20, 74]], [[100, 101]], [[9, 16]], [[1, 8]], [[111, 114]], [[20, 74]], [[20, 74]], [[17, 74]], [[1, 79]], [[80, 101]], [[100, 114]], [[116, 135]]]",
    "query_spans": "[[[137, 150]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y=x^{2}$ and the line $x-y-2=0$, what is the shortest distance from a point on the parabola to this line?",
    "fact_expressions": "G: Parabola;H: Line;P0: Point;Expression(G) = (y = x^2);Expression(H) = (x - y - 2 = 0);PointOnCurve(P0, G)",
    "query_expressions": "Min(Distance(P0, H))",
    "answer_expressions": "7*sqrt(2)/8",
    "fact_spans": "[[[2, 14], [28, 31]], [[15, 26], [36, 38]], [[33, 34]], [[2, 14]], [[15, 26]], [[28, 34]]]",
    "query_spans": "[[[33, 44]]]",
    "process": "Let A(x_{0},x_{0}^{2}) be an arbitrary point on the parabola, then the distance from A to the line x-y-2=0 is d=\\frac{|x_{0}-x_{0}^{2}-2|}{\\sqrt{2}}=\\frac{|-x_{0}^{2}+x_{0}-2|}{\\sqrt{2}}=\\frac{-(x_{0}-\\frac{1}{2})^{2}+\\frac{7}{4}}{\\sqrt{2}}. Therefore, when x_{0}=\\frac{1}{2}, the distance from A to the line attains the minimum value \\frac{7}{4\\sqrt{2}}=\\frac{7\\sqrt{2}}{8}."
  },
  {
    "text": "Given that the line $l$: $y=2x$ is perpendicular to one of the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, and the distance from the right focus to the line $l$ is $2$, then the standard equation of the hyperbola is?",
    "fact_expressions": "l: Line;G: Hyperbola;b: Number;a: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(l) = (y=2*x);IsPerpendicular(l,OneOf(Asymptote(G)));Distance(RightFocus(G),l)=2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4-y^2=1",
    "fact_spans": "[[[2, 16], [77, 82]], [[17, 63], [91, 94]], [[20, 63]], [[20, 63]], [[17, 63]], [[2, 16]], [[2, 71]], [[17, 89]]]",
    "query_spans": "[[[91, 101]]]",
    "process": "Since line $ l $ is perpendicular to an asymptote, we have $ \\frac{b}{a} $. The distance from the right focus to line $ l $ gives $ c $, thus allowing us to find $ a $ and $ b $. Given the line $ l: y = 2x $ is perpendicular to one asymptote, the slope of this asymptote is $ -\\frac{1}{2} $, so $ \\frac{b}{a} = \\frac{1}{2} $ ①. The right focus is $ F_2(c, 0) $, then $ \\frac{2c}{\\sqrt{5}} = 2 $, hence $ c = \\sqrt{5} $ ②. Also, $ a^2 + b^2 = c^2 $ ③. Solving ①, ②, and ③ together yields $ a = 2 $, $ b = 1 $. The equation of the hyperbola is $ \\frac{x^2}{4} - y^2 = 1 $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left vertex is $A$, the left focus is $F$, and point $P$ is any point on the ellipse. If the distance from the upper vertex of the ellipse to the focus is $2$, and the eccentricity is $e=\\frac{1}{2}$, then the range of values of $\\overrightarrow{A P} \\cdot \\overrightarrow{F P}$ is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;P: Point;F: Point;e: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(G) = A;LeftFocus(G) = F;PointOnCurve(P, G);Distance(UpperVertex(G), Focus(G))=2;Eccentricity(G) = e;e = 1/2",
    "query_expressions": "Range(DotProduct(VectorOf(A, P), VectorOf(F, P)))",
    "answer_expressions": "[0, 12]",
    "fact_spans": "[[[2, 54], [77, 79], [87, 89]], [[4, 54]], [[4, 54]], [[59, 62]], [[71, 75]], [[67, 70]], [[107, 122]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 62]], [[2, 70]], [[71, 84]], [[87, 103]], [[87, 122]], [[107, 122]]]",
    "query_spans": "[[[124, 180]]]",
    "process": "Since the distance from the upper vertex of the ellipse to the focus is 2, we have $ a = 2 $. Since the eccentricity is $ e = \\frac{1}{2} $, it follows that $ c = 1 $, $ b = \\sqrt{a^{2} - c^{2}} = \\sqrt{3} $. Then the equation of the ellipse is $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $. Therefore, the coordinates of point $ A $ are $ (-2, 0) $, and the coordinates of point $ F $ are $ (-1, 0) $. Let $ P(x, y) $, $ -2 \\leqslant x \\leqslant 2 $, then $ \\overrightarrow{AP} \\cdot \\overrightarrow{FP} = (x+2, y) \\cdot (x+1, y) = x^{2} + 3x + 2 + y^{2} $. From the equation of the ellipse, we get $ y^{2} = 3 - \\frac{3}{4}x^{2} $, so $ \\overrightarrow{AP} \\cdot \\overrightarrow{FP} = x^{2} + 3x - \\frac{3}{4}x^{2} + 5 = \\frac{1}{4}(x+6)^{2} - 4 $. Since $ x \\in [-2, 2] $, we have $ \\overrightarrow{AP} \\cdot \\overrightarrow{FP} \\in [0, 12] $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A point $P$ on the ellipse satisfies $|P F_{1}|-|P F_{2}|=2$. Then the area of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/2 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)) = 2",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 39], [64, 66]], [[2, 39]], [[48, 55]], [[56, 63]], [[2, 63]], [[2, 63]], [[68, 72]], [[64, 72]], [[74, 97]]]",
    "query_spans": "[[[99, 126]]]",
    "process": "By the definition of an ellipse, |PF_{1}| + |PF_{2}| = 2 \\times 2 = 4. From |PF_{1}| - |PF_{2}| = 2, we get |PF_{1}| = 3, |PF_{2}| = 1. Since |F_{1}F_{2}| = 2\\sqrt{4-2} = 2\\sqrt{2}, it follows that 3^{2} = (2\\sqrt{2})^{2} + 1, so triangle PF_{1}F_{2} is a right triangle, and its area is \\frac{1}{2} \\times 2\\sqrt{2} \\times 1 = \\sqrt{2}."
  },
  {
    "text": "If one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1(a>0)$ is $y=\\sqrt{2} x$, then $a$=?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (y = sqrt(2)*x)",
    "query_expressions": "a",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[1, 38]], [[61, 64]], [[4, 38]], [[1, 38]], [[1, 59]]]",
    "query_spans": "[[[61, 66]]]",
    "process": "From the standard equation of the hyperbola, we know its asymptotes are given by $ y = \\pm\\frac{1}{a}x' $. Since one asymptote of the hyperbola $ \\frac{x^{2}}{a^{2}} - y^{2} = 1 $ ($ a > 0 $) is $ y = \\sqrt{2}x $, therefore $ \\frac{1}{a} = \\sqrt{2} $, hence $ a = \\frac{\\sqrt{2}}{2} $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has the directrix equation $x=-2$, focus $F$, and point $A$ is the intersection of the directrix and the $x$-axis. Let $B$ be a point on the parabola $C$ satisfying $\\sqrt{5}|B F|=2|A B|$, then $|B F|=$?",
    "fact_expressions": "C: Parabola;p: Number;B: Point;F: Point;A: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Expression(Directrix(C)) = (x = -2);Focus(C) = F;Intersection(Directrix(C), xAxis) = A;PointOnCurve(B, C);sqrt(5)*Abs(LineSegmentOf(B, F)) = 2*Abs(LineSegmentOf(A, B))",
    "query_expressions": "Abs(LineSegmentOf(B, F))",
    "answer_expressions": "16 + pm*8*sqrt(3)",
    "fact_spans": "[[[2, 28], [67, 73]], [[10, 28]], [[63, 66]], [[44, 47]], [[59, 62]], [[10, 28]], [[2, 28]], [[2, 40]], [[2, 47]], [[2, 62]], [[63, 76]], [[80, 102]]]",
    "query_spans": "[[[104, 113]]]",
    "process": "From $-\\frac{p}{2}=-2$, solve for $p=4$, yielding the parabola equation $y^{2}=8x$. Using the definition of a parabola, $\\cos\\angle BAF=\\frac{2\\sqrt{5}}{5}$ can be found, and then the answer can be obtained using the cosine law. From the given: parabola $C: y^{2}=2px$ $(p>0)$, directrix equation $x=-2$, then $F(2,0)$, and $-\\frac{p}{2}=-2$, $\\therefore p=4$, $\\therefore$ the parabola equation is $C: y^{2}=8x$. $\\because \\sqrt{5}|BF|=2|AB|$, draw $BQ\\bot$ directrix, intersecting at point $Q$. By the definition of a parabola: $|BF|=|BQ|$, $\\therefore \\sqrt{5}|BQ|=2|AB|$. Let $\\angle ABQ=\\theta$, then $\\angle BAF=\\theta$ $(0<\\theta<\\frac{\\pi}{2})$, $\\therefore \\cos\\theta=\\frac{|BQ|}{|AB|}=\\frac{2}{\\sqrt{5}}=\\frac{2\\sqrt{5}}{5}$. In triangle $ABF$, $AF=4$, $\\sqrt{5}|BF|=2|AB|$, $\\cos\\angle BAF=\\frac{2\\sqrt{5}}{5}$. By the cosine law: $|BF|^{2}=\\frac{\\sqrt{5}}{2}|BF|+4^{2}-2\\times\\frac{\\sqrt{5}}{2}|BF|\\times4\\times\\frac{\\sqrt{5}}{5}$, solving gives $|BF|=16\\pm8\\sqrt{3}$."
  },
  {
    "text": "Given a fixed point $A(3 , 4)$, point $P$ is a moving point on the parabola $y^{2}=4 x$, and the distance from point $P$ to the line $x=1$ is $d$. Then the minimum value of $|PA|+d$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;P: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (x = 1);Coordinate(A) = (3, 4);PointOnCurve(P, G);Distance(P, H) = d;d:Number",
    "query_expressions": "Min(d + Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[20, 34]], [[44, 51]], [[4, 14]], [[15, 19], [39, 43]], [[20, 34]], [[44, 51]], [[4, 14]], [[15, 38]], [[39, 58]], [[55, 58]]]",
    "query_spans": "[[[60, 74]]]",
    "process": ""
  },
  {
    "text": "Given points $A(x_{1}, y_{1})$ and $C(x_{2}, y_{2})$ are two moving points on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, with $x_{1}+x_{2}=8$, and the equation of the perpendicular bisector of $AC$ is $y=k x+m$. Then, what is the range of values for $m$?",
    "fact_expressions": "G: Ellipse;A: Point;C: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Coordinate(A) = (x1, y1);Coordinate(C) = (x2, y2);x1:Number;y1:Number;x2:Number;y2:Number;PointOnCurve(A, G);PointOnCurve(B, G);x1 + x2 = 8;Expression(PerpendicularBisector(LineSegmentOf(A, C))) = (y = k*x + m);k:Number;m:Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-16/5,16/5)",
    "fact_spans": "[[[40, 78]], [[2, 19]], [[22, 39]], [[40, 78]], [[2, 19]], [[22, 39]], [[2, 20]], [[2, 20]], [[22, 39]], [[22, 39]], [[2, 82]], [[2, 82]], [[83, 98]], [[100, 124]], [[115, 124]], [[126, 129]]]",
    "query_spans": "[[[126, 136]]]",
    "process": "Let the midpoint of $A(x_{1},y_{1})$, $C(x_{2},y_{2})$ be $M(x_{0},y_{0})$. By the point difference method and using the midpoint coordinate relation, we obtain $\\frac{4}{25}+\\frac{y_{0}(y_{1}-y_{2})}{9(x_{1}-x_{2})}=0$. Analyzing this, when $k\\neq0$, $k=\\frac{25}{36}y_{0}'$. Substituting the midpoint gives $m=-\\frac{16}{9}y_{0}'$. Since point $M(x_{0},y_{0})$ lies inside the ellipse, $|y_{0}|<\\frac{9}{5}$, leading to the solution. [Sheep's solution] Let the midpoint of $A(x_{1},y_{1})$, $C(x_{2},y_{2})$ be $M(x_{0},y_{0})$, then $x_{0}=\\frac{x_{1}+x_{2}}{2}=4$. Letting $x=4$, we get $y^{2}=9(1-\\frac{16}{25})=\\frac{81}{25}$, so $|y_{0}|<\\frac{9}{5}$. From $\\begin{cases}\\frac{x_{1}^{2}}{25}+\\frac{y_{1}^{2}}{9}=1\\\\\\frac{x_{2}^{2}}{25}+\\frac{y_{2}^{2}}{9}=1\\end{cases}$, subtracting the two equations yields $\\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{25}+\\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{9}=0$, giving $\\frac{8}{25}+\\frac{2y_{0}(y_{1}-y_{2})}{9(x_{1}-x_{2})}=0$. When $k=0$, the slope of $AC$ does not exist, then it must pass through the origin, which does not satisfy $x_{1}+x_{2}=8$. When $k\\neq0$, $\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{1}{k}$, yielding $k=\\frac{25}{36}y_{0}'$. Also, since $M(4,y_{0})$ lies on the line $y=kx+m$, we have $y_{0}=4k+m$, thus $m=-\\frac{16}{9}y_{0}\\in(-\\frac{16}{5},\\frac{16}{5})$."
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the left and right foci of an ellipse. There exists a point $P$ on the ellipse such that $P F_{1} = 2 P F_{2}$. What is the range of the eccentricity?",
    "fact_expressions": "F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;G: Ellipse;PointOnCurve(P, G);P: Point;LineSegmentOf(P, F1) = LineSegmentOf(P, F2)*2",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[1/3,1)",
    "fact_spans": "[[[0, 7]], [[8, 15]], [[0, 24]], [[0, 24]], [[16, 18], [26, 28]], [[25, 35]], [[31, 35]], [[37, 56]]]",
    "query_spans": "[[[26, 64]]]",
    "process": "Analysis: From the definition of an ellipse, we obtain $ e\\left(x+\\frac{a^{2}}{c}\\right) = 2e\\left(\\frac{a^{2}}{c}-x\\right) $, solving which gives $ x = \\frac{a}{3e} $. According to the given condition, $ -a \\leqslant \\frac{a}{3e} \\leqslant a $, solving this inequality yields the range of eccentricity $ e $. Let the horizontal coordinate of point $ P $ be $ x $. Since $ |PF_{1}| = 2|PF_{2}| $, from the definition of an ellipse we have $ e\\left(x+\\frac{a^{2}}{c}\\right) = 2e\\left(\\frac{a^{2}}{c}-x\\right) $, thus $ x = \\frac{a}{3e} $. From the given condition, $ -a \\leqslant \\frac{a}{3e} \\leqslant a $, therefore $ \\frac{1}{3} \\leqslant e < 1 $."
  },
  {
    "text": "What is the standard equation of the parabola passing through point $A(-3,2)$?",
    "fact_expressions": "G: Parabola;A: Point;Coordinate(A) = (-3, 2);PointOnCurve(A,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2=(9/2)*y,y^2=(-4/3)*x}",
    "fact_spans": "[[[12, 15]], [[1, 11]], [[1, 11]], [[0, 15]]]",
    "query_spans": "[[[12, 22]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. If there exists a point $P$ on the ellipse such that $|P F_{1}|=2|P F_{2}|$, then the range of the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P,G);Abs(LineSegmentOf(P,F1))=2*Abs(LineSegmentOf(P,F2))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[1/3, 1)",
    "fact_spans": "[[[2, 54], [80, 82], [117, 119]], [[4, 54]], [[4, 54]], [[87, 90]], [[63, 70]], [[71, 78]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 78]], [[2, 78]], [[80, 90]], [[92, 114]]]",
    "query_spans": "[[[117, 129]]]",
    "process": "According to the definition of an ellipse, the lengths of |PF_{1}| and |PF_{2}| can be determined. Based on the geometric properties of a triangle, the answer can be obtained. From the definition of an ellipse, |PF_{1}| + |PF_{2}| = 2a, and given |PF_{1}| = 2|PF_{2}|, it follows that |PF_{1}| = \\frac{4a}{3}, |PF_{2}| = \\frac{2a}{3}. In an ellipse, |PF_{1}| - |PF_{2}| \\leqslant 2c. Therefore, \\frac{2a}{3} \\leqslant 2c, which implies e = \\frac{c}{a} \\geqslant \\frac{1}{3}. Also, since 2a > 2c, we have e = \\frac{c}{a} < 1. Hence, the range of eccentricity for this ellipse is [\\frac{1}{3}, 1)."
  },
  {
    "text": "The distance from point $M(5,3)$ to the directrix of the parabola $x^{2}=a(a>0)$ is $6$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;a: Number;M: Point;a>0;Expression(G) = (x^2 = a);Coordinate(M) = (5, 3);Distance(M, Directrix(G)) = 6",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 = 12*y",
    "fact_spans": "[[[10, 27], [40, 43]], [[13, 27]], [[0, 9]], [[13, 27]], [[10, 27]], [[0, 9]], [[0, 37]]]",
    "query_spans": "[[[40, 48]]]",
    "process": ""
  },
  {
    "text": "What are the coordinates of the points on the parabola $y^{2}=12 x$ whose distance to the focus is equal to $9$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 12*x);P:Point;PointOnCurve(P, G);Distance(P, Focus(G)) = 9",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(6, pm*6*sqrt(2))",
    "fact_spans": "[[[0, 15]], [[0, 15]], [[28, 29]], [[0, 29]], [[0, 29]]]",
    "query_spans": "[[[28, 34]]]",
    "process": ""
  },
  {
    "text": "$F$ is the focus of the parabola $y^{2}=4x$, point $P$ lies on the parabola, and $Q$ is a point on the circle $(x-2)^{2}+(y-1)^{2}=1$. Then the minimum value of $|PQ|+|PF|$ is?",
    "fact_expressions": "G: Parabola;H: Circle;P: Point;Q: Point;F: Point;Expression(G) = (y^2 = 4*x);Expression(H) = ((x - 2)^2 + (y - 1)^2 = 1);Focus(G) = F;PointOnCurve(P, G);PointOnCurve(Q, H)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F)) + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "2",
    "fact_spans": "[[[4, 18], [27, 30]], [[36, 60]], [[22, 26]], [[32, 35]], [[0, 3]], [[4, 18]], [[36, 60]], [[0, 21]], [[22, 31]], [[32, 63]]]",
    "query_spans": "[[[65, 83]]]",
    "process": "Let the distance from point P to the directrix of the parabola be d. According to the definition of a parabola, |PQ| + |PF| = |PQ| + d. Based on geometric properties of circles and plane geometry knowledge, the minimum value of |PQ| + d is the difference between the distance from the center of the circle to the directrix and the radius, that is, |PQ| + |PF| = |PQ| + d \\geqslant 3 - 1 = 2."
  },
  {
    "text": "Let point $P$ be a moving point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$, and let $F$ be the left focus of the ellipse. Then the range of $| P F |$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/7 = 1);P: Point;PointOnCurve(P, G);F: Point;LeftFocus(G) = F",
    "query_expressions": "Range(Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "[1, 7]",
    "fact_spans": "[[[6, 44], [54, 56]], [[6, 44]], [[1, 5]], [[1, 49]], [[50, 53]], [[50, 60]]]",
    "query_spans": "[[[62, 78]]]",
    "process": ""
  },
  {
    "text": "The foci of the hyperbola lie on the $x$-axis, the length of the imaginary axis is $12$, and the eccentricity is $\\frac{5}{4}$. Then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), xAxis);Length(ImageinaryAxis(G)) = 12;Eccentricity(G)=5/4",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/64 - y^2/36 = 1",
    "fact_spans": "[[[0, 3], [41, 44]], [[0, 12]], [[0, 21]], [[0, 39]]]",
    "query_spans": "[[[41, 51]]]",
    "process": ""
  },
  {
    "text": "Given that the focus of the parabola $x^{2}=4 y$ is $F$, the intersection point of the directrix and the $y$-axis is $M$, and $N$ is an arbitrary point on the parabola satisfying $|N F|=\\frac{\\sqrt{3}}{2}|M N|$, then $\\angle N M F$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);F: Point;Focus(G) = F;M: Point;Intersection(Directrix(G), yAxis) = M;N: Point;PointOnCurve(N, G);Abs(LineSegmentOf(N, F)) = (sqrt(3)/2)*Abs(LineSegmentOf(M, N))",
    "query_expressions": "AngleOf(N, M, F)",
    "answer_expressions": "pi/6",
    "fact_spans": "[[[2, 16], [43, 46]], [[2, 16]], [[20, 23]], [[2, 23]], [[35, 38]], [[2, 38]], [[39, 42]], [[39, 51]], [[55, 86]]]",
    "query_spans": "[[[88, 103]]]",
    "process": "Analysis: By the definition of the parabola, we have |NF| = y + 1. From $\\sin\\left(\\frac{\\pi}{2}-\\theta\\right) = \\frac{NH}{MN} = \\frac{NF}{MN} = \\frac{\\sqrt{3}}{2}$, we can find the value of $\\frac{\\pi}{2}-\\theta$, and thus determine the size of the acute angle $\\theta$. Given the parabola's equation $x^{2} = 4y$, we obtain the directrix equation $y = -1$. Let $\\angle NMF = \\theta$. Draw NH perpendicular to the directrix of the parabola from point N, with H being the foot of the perpendicular. Then, by the definition of the parabola, we have $NH = NF$. In right triangle MNH, $\\angle NMH = \\frac{\\pi}{2} - \\theta$. From the side-angle relationship in the right triangle, we get $\\sin\\left(\\frac{\\pi}{2}-\\theta\\right) = \\frac{NH}{MN} = \\frac{NF}{MN} = \\frac{\\sqrt{3}}{2}$, hence $\\frac{\\pi}{2} - \\theta = \\frac{\\pi}{3}$."
  },
  {
    "text": "Let $F$ be the right focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, $P$ a moving point on the ellipse, and $A$ a moving point on the line $3x-4y-12=0$. Then the minimum value of $|PA|-|PF|$ is?",
    "fact_expressions": "F: Point;RightFocus(G) = F;G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);P: Point;PointOnCurve(P, G) = True;A: Point;PointOnCurve(A, H) = True;H: Line;Expression(H) = (3*x - 4*y - 12 = 0)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) - Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "-1",
    "fact_spans": "[[[1, 4]], [[1, 47]], [[5, 42], [52, 54]], [[5, 42]], [[48, 51]], [[48, 58]], [[59, 62]], [[59, 82]], [[63, 79]], [[63, 79]]]",
    "query_spans": "[[[84, 103]]]",
    "process": "\\because F is the right focus of the ellipse \\frac{x^2}{4} + \\frac{y^{2}}{3} = 1, a^{2} = 4, b^{2} = 3 \\therefore c^{2} = a^{2} - b^{2} = 1, i.e., F(1,0). Let F be the left focus of the ellipse \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1, then F(-1,0). According to the definition of the ellipse, |PF| + |PF| = 2a = 4. \\therefore |PF| = 4 - |PF|, |PA| - |PF| = |PA| - (4 - |PF|) = |PA| + |PF| - 4. Then the minimum value of |PA| - |PF| is equivalent to finding the minimum value of |PA| + |PF|. According to the graph, the minimum value of |PA| + |PF| is the value of |AF| when AF is perpendicular to the line 3x - 4y - 12 = 0. Then |AF| = \\frac{|-3 - 12|}{\\sqrt{3^{2} + 4^{2}}} = 3. Then at this time |PA| + |PF| - 4 = |AF| - 4 = 3 - 4 = -1. Therefore, the minimum value of |PA| - |PF| is -1."
  },
  {
    "text": "Given that $F(-4,0)$ is a focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, a line $l$ passing through $F$ intersects the ellipse at points $A$ and $B$, and the midpoint of segment $AB$ has coordinates $(-3,1)$. Then the eccentricity of the ellipse is?",
    "fact_expressions": "l: Line;G: Ellipse;b: Number;a: Number;B: Point;A: Point;F: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(MidPoint(LineSegmentOf(A,B))) = (-3, 1);Coordinate(F) = (-4, 0);OneOf(Focus(G)) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B}",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[75, 80]], [[12, 64], [82, 84], [119, 121]], [[14, 64]], [[14, 64]], [[89, 92]], [[85, 88]], [[2, 11], [71, 74]], [[14, 64]], [[14, 64]], [[12, 64]], [[95, 116]], [[2, 11]], [[2, 69]], [[70, 80]], [[75, 94]]]",
    "query_spans": "[[[119, 127]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Since $ A $ and $ B $ lie on the ellipse, we have\n\\[\n\\begin{cases}\n\\frac{x_{1}^{2}}{a^{2}} + \\frac{y_{1}^{2}}{b^{2}} = 1 \\\\\n\\frac{x_{2}^{2}}{a^{2}} + \\frac{y_{2}^{2}}{b^{2}} = 1\n\\end{cases}\n\\]\nThus,\n\\[\n\\frac{x_{1}^{2} - x_{2}^{2}}{a^{2}} + \\frac{y_{1}^{2} - y_{2}^{2}}{b^{2}} = 0,\n\\]\nso\n\\[\n\\frac{y_{1} - y_{2}}{x_{1} - x_{2}} \\cdot \\frac{y_{1} + y_{2}}{x_{1} + x_{2}} = -\\frac{b^{2}}{a^{2}}.\n\\]\nSince the midpoint of segment $ AB $ is $ (-3,1) $ and $ F(-4,0) $, we have\n\\[\nx_{1} + x_{2} = -3 \\times 2 = -6, \\quad y_{1} + y_{2} = 1 \\times 2 = 2,\n\\]\nand\n\\[\nk_{AB} = \\frac{0 - 1}{-4 - (-3)} = 1.\n\\]\nThus,\n\\[\n1 \\cdot \\frac{2}{-6} = -\\frac{b^{2}}{a^{2}},\n\\]\nso\n\\[\n\\frac{b^{2}}{a^{2}} = \\frac{a^{2} - c^{2}}{a^{2}} = 1 - e^{2} = \\frac{1}{3},\n\\]\nand since $ e \\in (0,1) $, it follows that\n\\[\ne = \\frac{\\sqrt{6}}{3}.\n\\]"
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1$ $(a>0)$ is $\\frac{5}{4}$, then the coordinates of the right focus of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2/9 + x^2/a^2 = 1);Eccentricity(G) = 5/4",
    "query_expressions": "Coordinate(RightFocus(G))",
    "answer_expressions": "(5,0)",
    "fact_spans": "[[[2, 49], [69, 72]], [[5, 49]], [[5, 49]], [[2, 49]], [[2, 67]]]",
    "query_spans": "[[[69, 80]]]",
    "process": "The value of the real number $ a $ can be found from the eccentricity of the hyperbola, from which the coordinates of the right focus of the hyperbola can be determined. According to the problem, the eccentricity of the hyperbola is $ e = \\sqrt{\\frac{a^{2}+9}{a^{2}}} = \\frac{5}{4} $, solving gives $ a = 4 $. Therefore, the standard equation of the hyperbola is $ \\frac{x^{2}}{16} - \\frac{y^{2}}{9} = 1 $, then $ c = \\sqrt{16+9} = 5 $. Thus, the coordinates of the right focus of the hyperbola are $ (5, 0) $."
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. Points $M$ and $N$ are moving points on the asymptote and the left branch of the hyperbola, respectively. When $|M N|+|N F_{2}|$ attains its minimum value, the area of $\\Delta M F_{1} F_{2}$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2/3 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;M: Point;N: Point;PointOnCurve(M, Asymptote(C));PointOnCurve(N, LeftPart(C));WhenMin(Abs(LineSegmentOf(M,N))+Abs(LineSegmentOf(N,F2)))",
    "query_expressions": "Area(TriangleOf(M, F1, F2))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 35], [76, 79]], [[2, 35]], [[45, 52]], [[53, 60]], [[2, 60]], [[2, 60]], [[61, 65]], [[66, 69]], [[61, 85]], [[61, 85]], [[86, 110]]]",
    "query_spans": "[[[111, 137]]]",
    "process": "From the given conditions, we know $F_{1}(-2,0)$, $F_{2}(2,0)$, and without loss of generality, take one asymptote $y=\\sqrt{3}x$. By the definition of a hyperbola, $|NF_{2}|-|NF_{1}|=2a=2$, so $|NF_{2}|=2+|NF_{1}|$, thus $|MN|+|NF_{2}|=|MN|+|NF_{1}|+2$. Therefore, when points $M$, $N$, $F_{1}$ are collinear and $F_{1}M$ is perpendicular to the asymptote $y=\\sqrt{3}x$, $|MN|+|NF_{2}|$ reaches its minimum value. At this time, the equation of line $F_{1}M$ is $y=-\\frac{1}{\\sqrt{3}}(x+2)$. Solving\n$$\n\\begin{cases}\ny=\\sqrt{3}x \\\\\ny=-\\frac{1}{\\sqrt{3}}(x+2)\n\\end{cases}\n$$\ngives\n$$\n\\begin{cases}\nx=-\\frac{2}{2} \\\\\ny=-\\frac{\\sqrt{3}}{2}\n\\end{cases}\n$$\nHence, point $M\\left(-\\frac{1}{2},-\\frac{\\sqrt{3}}{2}\\right)$. $\\therefore S_{\\Delta MF_{1}F_{2}}=\\frac{1}{2}|F_{1}F_{2}|\\cdot\\left|-\\frac{\\sqrt{3}}{2}\\right|=\\frac{1}{2}\\times4\\times\\frac{\\sqrt{3}}{2}=\\sqrt{3}$"
  },
  {
    "text": "Given the parabola $y^{2}=4 x$, with focus $F$, and the three vertices of $\\Delta ABC$ lying on the parabola. If $\\overrightarrow{F A}+\\overrightarrow{F B}+\\overrightarrow{F C}=\\overrightarrow{0}$, then $|FA|+|FB|+|FC|$=?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;C: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(Vertex(TriangleOf(A, B, C)), G);VectorOf(F, A) + VectorOf(F, B) + VectorOf(F, C) = 0",
    "query_expressions": "Abs(LineSegmentOf(F, A)) + Abs(LineSegmentOf(F, B)) + Abs(LineSegmentOf(F, C))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 16], [43, 46]], [[25, 37]], [[25, 37]], [[25, 37]], [[21, 24]], [[2, 16]], [[2, 24]], [[25, 47]], [[49, 132]]]",
    "query_spans": "[[[135, 153]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $x^{2}-4 y^{2}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - 4*y^2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "x \\pm 2y = 0",
    "fact_spans": "[[[0, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 28]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{2}+\\frac{y^{2}}{3}=1$, $F_{1}$, $F_{2}$ are its foci, and $A B$ is a chord passing through $F_{1}$, then the perimeter of $\\triangle A B F_{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2/3 = 1);F1: Point;F2: Point;Focus(G) = {F1,F2};A: Point;B: Point;IsChordOf(LineSegmentOf(A,B),G) = True ;PointOnCurve(F1,LineSegmentOf(A,B)) = True",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 39], [60, 61]], [[2, 39]], [[42, 50], [72, 79]], [[52, 59]], [[42, 64]], [[65, 70]], [[65, 70]], [[60, 81]], [[65, 79]]]",
    "query_spans": "[[[83, 109]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/2 - y^2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(2)/2)*x",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 36]]]",
    "process": "The asymptotes of a hyperbola with foci on the x-axis are given by $ y = \\pm\\frac{b}{a}x $, and since $ a = \\sqrt{2} $, $ b = 1 $, the equations of the asymptotes for the hyperbola $ \\frac{x^{2}}{2} - y^{2} = 1 $ are $ y = \\pm\\frac{\\sqrt{2}}{2}x $."
  },
  {
    "text": "The line passing through the right focus of the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a, b>0$) with slope $2$ intersects the right branch of $E$ at two distinct points. Find the range of values for the eccentricity of the hyperbola.",
    "fact_expressions": "E: Hyperbola;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;G: Line;PointOnCurve(RightFocus(E), G);Slope(G) = 2;NumIntersection(G, RightPart(E)) = 2",
    "query_expressions": "Range(Eccentricity(E))",
    "answer_expressions": "(1, \\sqrt{5})",
    "fact_spans": "[[[1, 60], [76, 79], [93, 96]], [[1, 60]], [[8, 60]], [[8, 60]], [[8, 60]], [[8, 60]], [[73, 75]], [[0, 75]], [[66, 75]], [[73, 91]]]",
    "query_spans": "[[[93, 106]]]",
    "process": "According to the properties of hyperbola asymptotes, the range of the asymptote slopes gives the range of eccentricity. [Detailed solution] From the given condition, $\\frac{b}{a}<2$, $\\therefore b^{2}<4a^{2}$, $c^{2}-a^{2}<4a^{2}$, $\\therefore e^{2}<5$. $\\because e>1$, $\\therefore 1<e<\\sqrt{5}$"
  },
  {
    "text": "Given that the hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ has an eccentricity of $2$. If the distance from the focus of the parabola $C_{2}$: $x^{2}=2 p y$ $(p>0)$ to the asymptotes of the hyperbola $C_{1}$ is $2$, then what is the equation of the parabola $C_{2}$?",
    "fact_expressions": "C1: Hyperbola;Expression(C1) = (x^2/a^2 - y^2/b^2 = 1);a: Number;b: Number;a>0;b>0;Eccentricity(C1) = 2;C2: Parabola;Expression(C2) = (x^2 = 2*(p*y));p: Number;p>0;Distance(Focus(C2), Asymptote(C1)) = 2",
    "query_expressions": "Expression(C2)",
    "answer_expressions": "x^2=16*y",
    "fact_spans": "[[[2, 68], [113, 123]], [[2, 68]], [[14, 68]], [[14, 68]], [[14, 68]], [[14, 68]], [[2, 76]], [[79, 109], [136, 146]], [[79, 109]], [[91, 109]], [[91, 109]], [[79, 134]]]",
    "query_spans": "[[[136, 151]]]",
    "process": "\\because hyperbola C_{1}:\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0,b>0) has eccentricity 2, \\therefore \\frac{c}{a}=\\frac{\\sqrt{a^{2}+b^{2}}}{a}=2, \\therefore b=\\sqrt{3}a, \\therefore the asymptotes of the hyperbola are \\sqrt{3}x\\pm y=0, \\therefore the distance from the focus (0,\\frac{p}{2}) of parabola C_{2}:x^{2}=2py(p>0) to the asymptote of the hyperbola is \\frac{|\\sqrt{3}\\times0\\pm\\frac{p}{2}|}{2}=2, \\therefore p=8, \\therefore the required parabola equation is x^{2}=16y."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$, the line $l$: $x=2$ intersects the parabola $C$ at points $P$ and $Q$, and $O P \\perp O Q$. Then the equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;l: Line;Expression(l) = (x = 2);P: Point;Q: Point;Intersection(l, C) = {P, Q};O: Origin;IsPerpendicular(LineSegmentOf(O, P), LineSegmentOf(O, Q))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=2*x",
    "fact_spans": "[[[2, 28], [42, 48], [77, 83]], [[2, 28]], [[10, 28]], [[10, 28]], [[29, 41]], [[29, 41]], [[49, 52]], [[53, 56]], [[29, 58]], [[60, 75]], [[60, 75]]]",
    "query_spans": "[[[77, 88]]]",
    "process": "Substitute the line $ l: x = 2 $ into the parabola $ C: y^{2} = 2px $ ($ p > 0 $), we get $ P(2, 2\\sqrt{p}) $, $ Q(2, -2\\sqrt{p}) $, so $ \\overrightarrow{OP} = (2, 2\\sqrt{p}) $, $ \\overrightarrow{OQ} = (2, -2\\sqrt{p}) $. Since $ OP \\perp OQ $, it follows that $ \\overrightarrow{OP} \\cdot \\overrightarrow{OQ} = (2, 2\\sqrt{p}) \\cdot (2, -2\\sqrt{p}) = 4 - 4p = 0 $, $ p = 1 $. Therefore, the equation of the parabola $ C $ is $ y^{2} = 2x $."
  },
  {
    "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{2}+\\frac{y^{2}}{m}=1$ is $\\frac{1}{2}$, then $m$=?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/2 + y^2/m = 1);Eccentricity(G) = 1/2",
    "query_expressions": "m",
    "answer_expressions": "{3/2, 8/3}",
    "fact_spans": "[[[1, 38]], [[58, 61]], [[1, 38]], [[1, 56]]]",
    "query_spans": "[[[58, 63]]]",
    "process": "Discuss the cases where the foci lie on the x-axis and y-axis separately, and derive the expression using the eccentricity to solve. ① When the foci of the ellipse lie on the x-axis, it follows from the given condition that $\\frac{\\sqrt{2-m}}{\\sqrt{2}}=\\frac{1}{2}$, solving which yields $m=\\frac{3}{2}$; in summary, $m=\\frac{3}{2}$ or $\\frac{8}{3}$."
  },
  {
    "text": "Given point $A(5 , 0)$, from a point $P$ on the parabola $y^{2}=8 x$, draw a perpendicular to the line $x=-2$, with foot of perpendicular at $B$. If $|P B|=|P A|$, then the horizontal coordinate of $P$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;P: Point;B: Point;Z: Line;Expression(G) = (y^2 = 8*x);Expression(H) = (x = -2);Coordinate(A) = (5, 0);PointOnCurve(P, G);PointOnCurve(P, Z);IsPerpendicular(H, Z);FootPoint(H, Z) = B;Abs(LineSegmentOf(P, B)) = Abs(LineSegmentOf(P, A))",
    "query_expressions": "XCoordinate(P)",
    "answer_expressions": "7/2",
    "fact_spans": "[[[15, 29]], [[36, 44]], [[2, 13]], [[32, 35], [72, 75]], [[51, 54]], [], [[15, 29]], [[36, 44]], [[2, 13]], [[15, 35]], [[14, 47]], [[14, 47]], [[14, 54]], [[57, 70]]]",
    "query_spans": "[[[72, 81]]]",
    "process": "The line $x = -2$ is the directrix of the parabola $y^2 = 8x$, and the focus of the parabola is $F(2,0)$. According to the problem, $|PB| = |PA| = |PF|$. Draw a perpendicular from $P$ to the $x$-axis, with the foot of the perpendicular being the midpoint of $AF$. Therefore, the $x$-coordinate of $P$ is $\\frac{7}{3}$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, and $P Q$ is a chord passing through the focus $F_{1}$ with an inclination angle of $60^{\\circ}$, then the value of $|P F_{2}|+|Q F_{2}|-|P Q|$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;Q: Point;F2: Point;F1: Point;Expression(G) = (x^2/16 - y^2/9 = 1);Focus(G) = {F1, F2};PointOnCurve(F1,LineSegmentOf(P, Q));IsChordOf(LineSegmentOf(P, Q),G);Inclination(LineSegmentOf(P,Q))=ApplyUnit(60, degree)",
    "query_expressions": "Abs(LineSegmentOf(P, F2)) + Abs(LineSegmentOf(Q, F2)) - Abs(LineSegmentOf(P, Q))",
    "answer_expressions": "16",
    "fact_spans": "[[[18, 57]], [[61, 66]], [[61, 66]], [[10, 17]], [[2, 9], [70, 77]], [[18, 57]], [[2, 60]], [[61, 79]], [[18, 79]], [[81, 103]]]",
    "query_spans": "[[[106, 137]]]",
    "process": "From the hyperbola equation $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, we know $2a=8$. By the definition of a hyperbola, $|PF_{2}|-|PF_{1}|=2a=8$ \\textcircled{1}, $|QF_{2}|-|QF_{1}|=2a=8$ \\textcircled{2}. Adding \\textcircled{1} and \\textcircled{2} gives $|PF_{2}|+|QF_{2}|-(|QF_{1}|+|PF_{1}|)=16$, $\\therefore |PF_{2}|+|QF_{2}|-|PQ|=16$. Therefore, the answer should be: 16."
  },
  {
    "text": "Given that the foci of the ellipse $4 x^{2}+n y^{2}=1$ lie on the $x$-axis, and the length of the major axis is $2$ times the length of the minor axis, then the value of $n$ is?",
    "fact_expressions": "G: Ellipse;n: Number;Expression(G) = (n*y^2 + 4*x^2 = 1);PointOnCurve(Focus(G), xAxis);Length(MajorAxis(G)) = 2*Length(MinorAxis(G))",
    "query_expressions": "n",
    "answer_expressions": "16",
    "fact_spans": "[[[2, 23]], [[47, 50]], [[2, 23]], [[2, 32]], [[2, 45]]]",
    "query_spans": "[[[47, 54]]]",
    "process": "The foci of the ellipse $4x^{2}+ny^{2}=1$ lie on the x-axis, $\\therefore \\frac{x^{2}}{4}+\\frac{y^{2}}{n}=1$, $a=\\frac{1}{2}$, $b=\\frac{\\sqrt{n}}{1}$, $\\because$ the length of the major axis is twice the length of the minor axis. $1=\\frac{4\\sqrt{n}}{n}$, solving gives $n=16$. Hence answer. This problem examines the standard equation and properties of an ellipse, belonging to basic questions."
  },
  {
    "text": "The equation of the hyperbola with asymptotes $y=\\pm \\sqrt{3} x$ and passing through the point $(1,1)$ is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (1, 1);Expression(Asymptote(G)) = (y = pm*sqrt(3)*x);PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/(2/3)-y^2/2=1",
    "fact_spans": "[[[34, 37]], [[25, 33]], [[25, 33]], [[0, 37]], [[24, 37]]]",
    "query_spans": "[[[34, 41]]]",
    "process": "\\because one asymptote of the hyperbola is y=\\pm\\sqrt{3}x, \\therefore let 3x^{2}-y^{2}=k be the equation of the hyperbola. \\because the point (1,1) lies on the hyperbola, substituting gives 3-1=k-2. \\therefore the standard equation is \\frac{x^{2}}{2}-\\frac{y^{2}}{2}="
  },
  {
    "text": "Given $A(3 , 0)$, $B(9 , 5)$, and $P$ an arbitrary point on the right branch of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, then the minimum value of $|PA|+|PB|$ is?",
    "fact_expressions": "G: Hyperbola;A: Point;B: Point;P: Point;Expression(G) = (x^2/4 - y^2/5 = 1);Coordinate(A) = (3, 0);Coordinate(B) = (9, 5);PointOnCurve(P, RightPart(G))",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B)))",
    "answer_expressions": "9",
    "fact_spans": "[[[30, 68]], [[2, 13]], [[15, 25]], [[26, 29]], [[30, 68]], [[2, 13]], [[15, 25]], [[26, 76]]]",
    "query_spans": "[[[78, 95]]]",
    "process": ""
  },
  {
    "text": "If the line $l$: $y=-\\frac{x}{2}+m$ and the curve $C$: $y=\\frac{1}{2} \\sqrt{|4-x^{2}|}$ have exactly three intersection points, then the range of real values for $m$ is?",
    "fact_expressions": "l: Line;C: Curve;m: Real;Expression(l)=(y=-x/2+m);Expression(C) = (y = sqrt(Abs(4 - x^2))/2);NumIntersection(l,C)=3",
    "query_expressions": "Range(m)",
    "answer_expressions": "(1, \\sqrt{2})",
    "fact_spans": "[[[1, 25]], [[26, 65]], [[75, 80]], [[1, 25]], [[26, 65]], [[1, 73]]]",
    "query_spans": "[[[75, 87]]]",
    "process": "\\because y=\\frac{1}{2}\\sqrt{|4-x^{2}|},\\therefore 2y=\\sqrt{|4-x^{2}|} \\text{ i.e. } 4y^{2}=|4-x^{2}| (y\\geqslant0). \\text{ When } 4-x^{2}\\geqslant0, \\text{ we get } 4y^{2}=4-x^{2}, \\text{ i.e. } \\frac{x^{2}}{4}+y^{2}=1 (y\\geqslant0). \\text{ When } 4-x^{2}<0, \\text{ we get } 4y^{2}=x^{2}-4, \\text{ i.e. } \\frac{x^{2}}{4}-y^{2}=1 (y>0). \\text{ From the above, the graph of curve } C \\text{ is the upper half of an ellipse and the upper half of a hyperbola, as shown in the figure. } \\because \\text{ line } l: y=-\\frac{x}{2}+m \\text{ is parallel to the asymptotes of the hyperbola } \\frac{x^{2}}{4}-y^{2}=1, \\therefore \\text{ when line } l \\text{ passes through point } A(2,0), \\text{ it intersects curve } C \\text{ at two points, at which } m=1. \\text{ Shifting upward gives three intersection points. Also, from } \\begin{cases} y=-\\frac{x}{2}+m \\\\ \\frac{x^{2}}{4}+y^{2}=1 \\end{cases}, \\text{ eliminating variables yields } x^{2}-2mx+2m^{2}-2=0. \\text{ Setting } \\Delta=4m^{2}-4(2m^{2}-2)=8-4m^{2}=0, \\text{ we obtain } m=\\sqrt{2} \\text{ or } m=-\\sqrt{2} \\text{ (discarded). Thus, the range of real number } m \\text{ is } (1,\\sqrt{2})."
  },
  {
    "text": "Given that $A$ and $B$ are two distinct points on the parabola $C$: $y^{2}=2 p x$ ($p>0$) with ordinates $a$ and $b$ ($a b \\neq 0$), respectively. If the tangents to $C$ at points $A$ and $B$ intersect at point $M$, then what is the ordinate of point $M$? (Express the result in terms of $a$, $b$)",
    "fact_expressions": "C: Parabola;p: Number;p>0;Expression(C) = (y^2 = 2*p*x);A:Point;B:Point;M:Point;a:Number;b:Number;Negation(a*b=0);PointOnCurve(A,C);PointOnCurve(B,C);YCoordinate(A)=a;YCoordinate(B)=b;Negation(A=B);Intersection(TangentOnPoint(A,C),TangentOnPoint(B,C))=M",
    "query_expressions": "YCoordinate(M)",
    "answer_expressions": "(a+b)/2",
    "fact_spans": "[[[10, 35], [69, 72]], [[17, 35]], [[17, 35]], [[10, 35]], [[2, 5], [73, 76]], [[6, 9], [77, 80]], [[89, 93], [95, 99]], [[42, 45]], [[47, 62]], [[47, 62]], [[2, 67]], [[2, 67]], [[2, 62]], [[2, 62]], [[2, 67]], [[69, 93]]]",
    "query_spans": "[[[95, 105]]]",
    "process": "Let the tangent line equation be set, and by solving the system of equations of the tangent and the parabola, the slopes of the two tangents can be found respectively. Solving the system of tangent equations gives the coordinates of the intersection point. Since the ordinate of point A is a, the coordinates of point A are (\\frac{a^{2}}{2p},a). According to the problem, the slope of the tangent passing through point A must exist and is not zero, denoted as k (k\\neq0). Then the tangent equation through point A is y-a=k(x-\\frac{a^{2}}{2p}). Combining this with y^{2}=2px and eliminating x yields ky^{2}-2py+2pa-ka^{2}=0. Since the line is tangent to the parabola, the discriminant \\Delta=4p^{2}-4k(2pa-ka^{2})=0. Solving gives k=\\frac{p}{a}. Thus, the tangent equation through point A is y-a=\\frac{p}{a}(x-\\frac{a^{2}}{2p}), which simplifies to y=\\frac{p}{a}x+\\frac{a}{2}. Similarly, the tangent equation through point B is y=\\frac{p}{b}x+\\frac{b}{2}. Solving these two tangent equations simultaneously, the ordinate of point M is \\frac{a+b}{2}."
  },
  {
    "text": "The equation of a hyperbola with foci on the $x$-axis has an asymptote given by $y=2x$, and the focal length is $10$. What is the equation of this hyperbola?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), xAxis);Expression(OneOf(Asymptote(G))) = (y = 2*x);FocalLength(G) = 10",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5 - y^2/20 = 1",
    "fact_spans": "[[[8, 11], [37, 40]], [[0, 11]], [[8, 27]], [[8, 35]]]",
    "query_spans": "[[[37, 45]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola $y^{2}=2 p x(p>0)$ is $F$, and $M$ is a point on the parabola. Let $A\\left(\\frac{7}{2} p, 0\\right)$. If $|A F|=2|M F|$ and the area of $\\Delta A M F$ is $\\frac{27 \\sqrt{2}}{2}$, then the value of $p$ is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;F: Point;M: Point;p>0;Expression(G) = (y^2 = 2*p*x);Focus(G)=F;PointOnCurve(M,G);Coordinate(A) = ((7/2)*p, 0);Abs(LineSegmentOf(A, F)) = 2*Abs(LineSegmentOf(M, F));Area(TriangleOf(A,M,F))=27*sqrt(2)/2",
    "query_expressions": "p",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 21], [33, 36]], [[124, 127]], [[41, 62]], [[25, 28]], [[29, 32]], [[3, 21]], [[0, 21]], [[0, 28]], [[29, 39]], [[41, 62]], [[64, 78]], [[81, 122]]]",
    "query_spans": "[[[124, 131]]]",
    "process": "\\because A(\\frac{7p}{2},0), \\therefore |AF|=3p \\text{ then } |MF|=\\frac{3}{2}p \\therefore \\text{ the abscissa of point } M \\text{ is } p, \\text{ substitute to obtain } y^{2}=2p^{2}, y=\\pm\\sqrt{2}p \\times 3P \\times \\sqrt{2}P = \\frac{27\\sqrt{2}}{2}"
  },
  {
    "text": "If a focus of the hyperbola $x^{2}-\\frac{y^{2}}{m}=1$ is $(\\sqrt{10}, 0)$, then what is the equation of the asymptotes of this hyperbola?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2 - y^2/m = 1);Coordinate(OneOf(Focus(G))) = (sqrt(10), 0)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*3*x",
    "fact_spans": "[[[1, 29], [54, 57]], [[4, 29]], [[1, 29]], [[1, 51]]]",
    "query_spans": "[[[54, 65]]]",
    "process": "From the coordinates of the focus, we know $ c = \\sqrt{10} $, thus $ 1 + m = (\\sqrt{10})^2 $, solving for $ m $ and substituting into the hyperbola equation yields its asymptotes. Solution: $ \\because \\sqrt{1 + m} = \\sqrt{10}, \\therefore m = 9, \\therefore $ the asymptotes of this hyperbola are $ y = \\pm 3x $."
  },
  {
    "text": "A line passes through the point $(\\frac{1}{4}, 0)$ and intersects the parabola $y^{2}=x$ at points $A$ and $B$. If $|A B|=4$, then the distance from the midpoint of chord $A B$ to the line $x+\\frac{1}{2}=0$ is equal to?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = x);H: Line;I: Point;Coordinate(I) = (1/4, 0);PointOnCurve(I, H);A: Point;B: Point;Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 4;IsChordOf(LineSegmentOf(A, B), G);L: Line;Expression(L) = (x + 1/2 = 0)",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A, B)), L)",
    "answer_expressions": "9/4",
    "fact_spans": "[[[27, 39]], [[27, 39]], [[2, 4]], [[5, 24]], [[5, 24]], [[2, 24]], [[41, 44]], [[45, 48]], [[2, 50]], [[53, 62]], [[27, 70]], [[74, 93]], [[74, 93]]]",
    "query_spans": "[[[65, 99]]]",
    "process": "As shown in the figure, the focus $ F $ of the parabola $ y^{2} = x $ is $ \\left( \\frac{1}{4}, 0 \\right) $, so the distance from the midpoint of chord $ AB $ to the directrix $ x = -\\frac{1}{4} $ is $ \\frac{|AB|}{2} = 2 $. Then, the distance from the midpoint of chord $ AB $ to the line $ x + \\frac{1}{2} = 0 $ is equal to $ 2 + \\frac{1}{4} = \\frac{9}{4} $."
  },
  {
    "text": "Given that point $M$ is a point on the parabola $C$: $y^{2}=2 p x(p>0)$. If the sum of the distances from point $M$ to two fixed points $A(p, p)$ and $F(\\frac{p}{2}, 0)$ is minimized, then the coordinates of point $M$ are?",
    "fact_expressions": "M: Point;PointOnCurve(M, C);C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;A: Point;Coordinate(A) = (p, p);F: Point;Coordinate(F) = (p/2, 0);WhenMin(Distance(M, A) + Distance(M, F))",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(p/2, p)",
    "fact_spans": "[[[2, 6], [38, 42], [85, 89]], [[2, 36]], [[7, 33]], [[7, 33]], [[15, 33]], [[15, 33]], [[46, 55]], [[46, 55]], [[57, 76]], [[57, 76]], [[38, 83]]]",
    "query_spans": "[[[85, 94]]]",
    "process": "Draw a perpendicular line from point M to the directrix of the parabola, with foot at B. By the definition of the parabola, the distance from point M to the focus $ F\\left(\\frac{p}{2},0\\right) $ is equal to the distance from point M to the directrix, that is, $ |MF| = |MB| $. Therefore, $ |MF| + |MA| = |MB| + |MA| $. It is clear that $ |MB| + |MA| $ reaches its minimum value when points A, B, and M are collinear. Thus, $ (|MF| + |MA|)_{\\min} = |AB| = \\frac{3p}{2} $, and at this time, the coordinates of point M are $ \\left(\\frac{p}{2}, p\\right) $."
  },
  {
    "text": "The line $l$ passing through the focus $F$ of the parabola $C$: $y^{2}=4x$ intersects $C$ at points $A$ and $B$. Given point $M(-1,2)$, if $\\overrightarrow{M A} \\cdot \\overrightarrow{M B}=0$, then the slope $k$ of line $l$ is?",
    "fact_expressions": "l: Line;C: Parabola;M: Point;A: Point;B: Point;F: Point;k: Number;Expression(C) = (y^2 = 4*x);Coordinate(M) = (-1, 2);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};DotProduct(VectorOf(M, A), VectorOf(M, B)) = 0;Slope(l) = k",
    "query_expressions": "k",
    "answer_expressions": "1",
    "fact_spans": "[[[27, 32], [112, 117]], [[1, 20], [33, 36]], [[47, 57]], [[37, 40]], [[41, 44]], [[23, 26]], [[120, 123]], [[1, 20]], [[47, 57]], [[1, 26]], [[0, 32]], [[27, 46]], [[59, 110]], [[112, 123]]]",
    "query_spans": "[[[120, 125]]]",
    "process": "From the given conditions, the coordinates of the focus are $ F(1,0) $. Let the intersection points be $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Substituting the line $ l: y = k(x - 1) $ into $ y^{2} = 4x $, we obtain $ k^{2}x^{2} - (2k^{2} + 4)x + k^{2} = 0 $. Then, $ x_{1} + x_{2} = \\frac{2k^{2} + 4}{k^{2}} $, $ x_{1}x_{2} = 1 $, $ y_{1} = k(x_{1} - 1) $, $ y_{2} = k(x_{2} - 1) $. Hence, $ y_{1}y_{2} = k^{2}(x_{1} - 1)(x_{2} - 1) = k^{2}[x_{1}x_{2} - (x_{1} + x_{2}) + 1] = -4 $; $ y_{1} + y_{2} = k(x_{1} + x_{2} - 2) = \\frac{4}{k} $. Since $ \\overrightarrow{MA} = (x_{1} + 1, y_{1} - 2) $, $ \\overrightarrow{MB} = (x_{2} + 1, y_{2} - 2) $, we have $ (x_{1} + 1)(x_{2} + 1) + (y_{1} - 2)(y_{2} - 2) = 0 $, that is, $ x_{1}x_{2} + x_{1} + x_{2} + 1 - 4 - 2 \\times \\frac{4}{k} + 4 = 0 $, which simplifies to $ 1 + \\frac{1}{12} - \\frac{2}{k} = 0 $. Solving gives $ k = 1 $. The answer to be filled in is $ 1 $."
  },
  {
    "text": "The length of the imaginary axis of the hyperbola $\\frac{x^{2}}{5}-\\frac{y^{2}}{9}=1$ is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/5 - y^2/9 = 1)",
    "query_expressions": "Length(ImageinaryAxis(G))",
    "answer_expressions": "6",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 45]]]",
    "process": ""
  },
  {
    "text": "Given the line $l$: $x - y + 8 = 0$ and two points $A(2,0)$, $B(-2,-4)$, find a point $P$ on line $l$ such that $|PA| + |PB|$ is minimized. Then the coordinates of point $P$ are?",
    "fact_expressions": "l: Line;Expression(l) = (x - y + 8 = 0);A: Point;B: Point;Coordinate(A) = (2, 0);Coordinate(B) = (-2, -4);PointOnCurve(P,l);WhenMin(Abs(LineSegmentOf(P,A))+Abs(LineSegmentOf(P,B)));P:Point",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(-5,3)",
    "fact_spans": "[[[43, 48], [2, 18]], [[2, 18]], [[21, 29]], [[31, 41]], [[21, 29]], [[31, 41]], [[42, 55]], [[58, 73]], [[52, 55], [75, 79]]]",
    "query_spans": "[[[75, 83]]]",
    "process": "First determine whether the two points are on the same side or opposite sides of the line. Since $(2-0+8)(-2+4+8)>0$, points $A$ and $B$ are on the same side of the line. Find the symmetric point $A'(-8,10)$ of point $A$ with respect to the line. Connect $A'$ and $B$ to intersect the line at point $P$, which is the desired point. Write the equation of line $A'B$: $7x+3y+26=0$. Solve simultaneously to obtain the coordinates of point $P$: $(-5,3)$."
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and point $P$ moves on the ellipse. Then the maximum value of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(P, G)",
    "query_expressions": "Max(DotProduct(VectorOf(P, F1), VectorOf(P, F2)))",
    "answer_expressions": "1",
    "fact_spans": "[[[18, 45], [56, 58]], [[51, 55]], [[0, 7]], [[10, 17]], [[18, 45]], [[0, 50]], [[0, 50]], [[51, 61]]]",
    "query_spans": "[[[64, 127]]]",
    "process": ""
  },
  {
    "text": "The equation $\\frac{x^{2}}{m-3} +\\frac{y^{2}}{9-m}=1$ represents an ellipse; then the range of values for $m$ is?",
    "fact_expressions": "G: Ellipse;m:Number;Expression(G)=(x^2/(m-3)+y^2/(9-m)=1)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(3,9)&(Negation(m=6))",
    "fact_spans": "[[[44, 46]], [[48, 51]], [[0, 46]]]",
    "query_spans": "[[[48, 58]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, let $A$ be a point on the $x$-axis different from the origin $O$. A circle centered at $A$ intersects the asymptotes of the hyperbola at points $P$ and $Q$. If $\\overrightarrow{O P}=3 \\overrightarrow{O Q}$ and $\\overrightarrow{A P} \\cdot \\overrightarrow{A Q}=0$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;O: Origin;P: Point;Q: Point;A: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(A, xAxis);Negation(A = O);Center(G) = A;Intersection(G, Asymptote(C)) = {P, Q};VectorOf(O, P) = 3*VectorOf(O, Q);DotProduct(VectorOf(A, P), VectorOf(A, Q)) = 0",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 63], [98, 101], [219, 225]], [[10, 63]], [[10, 63]], [[96, 97]], [[77, 84]], [[107, 110]], [[111, 114]], [[65, 68], [89, 92]], [[10, 63]], [[10, 63]], [[2, 63]], [[65, 87]], [[65, 87]], [[88, 97]], [[96, 116]], [[118, 163]], [[166, 217]]]",
    "query_spans": "[[[219, 231]]]",
    "process": ""
  },
  {
    "text": "If the equation $\\frac{x^{2}}{m+2}-\\frac{y^{2}}{m-1}=1$ represents a hyperbola with foci on the $x$-axis, then the range of real values for $m$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(m + 2) - y^2/(m - 1) = 1);m: Real;PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(1, +\\infty)",
    "fact_spans": "[[[53, 56]], [[1, 56]], [[58, 63]], [[44, 56]]]",
    "query_spans": "[[[58, 70]]]",
    "process": "This problem first uses the fact that the equation $\\frac{x^{2}}{m+2}-\\frac{y^{2}}{m-1}=1$ represents a hyperbola with foci on the $x$-axis, from which the signs of the two denominators can be determined, and then the range of $m$ can be obtained by calculation. Since the equation $\\frac{x^{2}}{m+2}-\\frac{y^{2}}{m-1}=1$ represents a hyperbola with foci on the $x$-axis, we have $\\begin{cases}m+2>0\\\\m-1>0\\end{cases}$, solving gives $m>1$."
  },
  {
    "text": "Given that the distance from point $P$ to point $F(3,0)$ is greater by $1$ than its distance to the line $x=-2$, then the equation satisfied by point $P$ is?",
    "fact_expressions": "G: Line;F: Point;P: Point;Expression(G) = (x = -2);Coordinate(F) = (3, 0);Distance(P, F) = Distance(P, G) + 1",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "y^2=12*x",
    "fact_spans": "[[[22, 30]], [[7, 16]], [[2, 6], [20, 21], [39, 43]], [[22, 30]], [[7, 16]], [[2, 37]]]",
    "query_spans": "[[[39, 50]]]",
    "process": ""
  },
  {
    "text": "Given two fixed points $A(-2,0)$ and $B(2,0)$, a moving point $P(x_1, y_1)$ moves along the line $l$: $y = x + 3$, and an ellipse $C$ has foci at $A$ and $B$ and passes through point $P$. Then, the maximum eccentricity of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;A: Point;Coordinate(A) = (-2, 0);B: Point;Coordinate(B) = (2, 0);P: Point;Coordinate(P) = (x1, y1);l: Line;Expression(l) = (y = x + 3);PointOnCurve(P,l) = True;Focus(C) = {A,B};PointOnCurve(P,C) = True;x1: Number;y1: Number",
    "query_expressions": "Max(Eccentricity(C))",
    "answer_expressions": "2*sqrt(26)/13",
    "fact_spans": "[[[56, 61], [81, 86]], [[5, 14], [62, 65]], [[5, 14]], [[15, 23], [66, 69]], [[15, 23]], [[26, 37], [75, 79]], [[26, 37]], [[38, 52]], [[38, 52]], [[26, 53]], [[56, 72]], [[56, 79]], [[26, 37]], [[26, 37]]]",
    "query_spans": "[[[81, 96]]]",
    "process": "According to the diagram and using the definition of an ellipse along with symmetry properties, the minimum value of $ 2a $ can be found, thus obtaining the maximum value of eccentricity. As shown in the figure, let the symmetric point of $ B $ about line $ l $ be $ B_{1}(a,b) $, so  \n$$\n\\begin{cases}\n\\frac{b-0}{a-2}=-1 \\\\\n\\frac{b+0}{2}=\\frac{a+2}{2}+3\n\\end{cases},\n$$\nso  \n$$\n\\begin{cases}\na=-3 \\\\\nb=5\n\\end{cases},\n$$\nthus $ B_{1}(-3,5) $, so  \n$$\n(|PA|+|PB|)_{\\min}=(|PA|+|PB_{1}|)_{\\min}=|AB_{1}|=\\sqrt{(-3-(-2))^{2}+5^{2}}.\n$$\nBy the definition of an ellipse: $ |PA|+|PB|=2a \\geqslant \\sqrt{26} $, so $ a \\geqslant \\frac{\\sqrt{26}}{2} $. Also, $ l_{AB_{1}}: 5x+y+10=0 $, so equality holds when  \n$$\n\\begin{cases}\ny=-5x- \\\\\ny=x+3\n\\end{cases}\n10,\n$$\nat this time $ P(-\\frac{13}{6},\\frac{5}{6}) $. So $ e=\\frac{c}{a} \\leqslant \\frac{2}{\\sqrt{26}} = \\frac{\\sqrt[2]{26}}{13} $, therefore the maximum eccentricity is $ \\frac{2\\sqrt{26}}{13} $."
  },
  {
    "text": "A line passing through the focus $F$ of a parabola intersects the parabola at points $A$ and $B$. If the projections of $A$ and $B$ onto the directrix of the parabola are $A_{1}$ and $B_{1}$ respectively, then $\\angle A_{1} FB_{1}$=?",
    "fact_expressions": "G: Parabola;H: Line;A1: Point;F: Point;B1: Point;A: Point;PointOnCurve(F, H);Intersection(H, G) = {A,B};Projection(A,Directrix(G))=A1;Projection(B,Directrix(G))=B1;B:Point;Focus(G)=F",
    "query_expressions": "AngleOf(A1,F,B1)",
    "answer_expressions": "ApplyUnit(90,degree)",
    "fact_spans": "[[[1, 4], [13, 16], [41, 44]], [[10, 12]], [[53, 60]], [[6, 9]], [[61, 68]], [[31, 34], [18, 21]], [[0, 12]], [[10, 29]], [[31, 68]], [[31, 68]], [[24, 27], [37, 40]], [[1, 9]]]",
    "query_spans": "[[[69, 92]]]",
    "process": ""
  },
  {
    "text": "The standard equation of the hyperbola, with foci at the endpoints of the major axis of the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{5}=1$ and passing through the point $(3 , \\sqrt{10})$, is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;I: Point;Expression(H) = (x^2/8 + y^2/5 = 1);Coordinate(I) = (3, sqrt(10));PointOnCurve(I, G);Endpoint(MajorAxis(H))=Focus(G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/3 - y^2/5 = 1",
    "fact_spans": "[[[69, 72]], [[1, 38]], [[50, 68]], [[1, 38]], [[50, 68]], [[48, 72]], [[0, 72]]]",
    "query_spans": "[[[69, 79]]]",
    "process": ""
  },
  {
    "text": "A focus of the hyperbola $8 k x^{2}-k y^{2}=8$ is $(0,3)$, then what is the value of $k$?",
    "fact_expressions": "G: Hyperbola;k: Number;Expression(G) = (8*(k*x^2) - k*y^2 = 8);Coordinate(OneOf(Focus(G))) = (0, 3)",
    "query_expressions": "k",
    "answer_expressions": "-1",
    "fact_spans": "[[[0, 24]], [[39, 42]], [[0, 24]], [[0, 37]]]",
    "query_spans": "[[[39, 46]]]",
    "process": "8kx^{2}-ky^{2}=8 is transformed into \\frac{y^{2}}{-\\frac{8}{k}}-\\frac{x^{2}}{-\\frac{1}{k}}=1 \\therefore (-\\frac{8}{k})+(-\\frac{1}{k})=9 \\therefore k=-1"
  },
  {
    "text": "The line $l$: $x - y = 0$ intersects the ellipse $\\frac{x^{2}}{2} + y^{2} = 1$ at points $A$ and $B$, and point $C$ is a moving point on the ellipse. Then the maximum area of $\\triangle A B C$ is?",
    "fact_expressions": "l: Line;Expression(l) = (x - y = 0);G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);A: Point;B: Point;Intersection(l, G) = {A, B};C: Point;PointOnCurve(C, G)",
    "query_expressions": "Max(Area(TriangleOf(A, B, C)))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[15, 42], [60, 62]], [[15, 42]], [[44, 47]], [[49, 52]], [[0, 54]], [[55, 59]], [[55, 66]]]",
    "query_spans": "[[[68, 93]]]",
    "process": ""
  },
  {
    "text": "It is known that one focus of the hyperbola $x^{2}-\\frac{y^{2}}{m}=1$ coincides with the focus of the parabola $8 x+y^{2}=0$. Then, the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;m: Number;H: Parabola;Expression(G) = (x^2 - y^2/m = 1);Expression(H) = (8*x + y^2 = 0);OneOf(Focus(G)) = Focus(H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 30], [60, 63]], [[5, 30]], [[36, 52]], [[2, 30]], [[36, 52]], [[2, 57]]]",
    "query_spans": "[[[60, 69]]]",
    "process": "Let the focus of the parabola be $ F $, $\\because 8x + y^2 = 0 \\Rightarrow y^2 = -8x \\Rightarrow F(-2, 0) \\therefore 1 + m = 2^{2} \\Rightarrow m = 3, \\therefore e = \\frac{c}{a} = \\frac{2}{1} = 2$, all answers are correct."
  },
  {
    "text": "A line passing through point $E(-\\frac{p}{2}, 0)$ intersects the parabola $y^{2}=2 p x(p>0)$ at points $A$ and $B$, $F$ is the focus of the parabola. If $A$ is the midpoint of segment $E B$ and $|A F|=3$, then $p$=?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;E: Point;B: Point;A: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(E) = (-p/2, 0);PointOnCurve(E, H);Intersection(H, G) = {A, B};Focus(G) = F;MidPoint(LineSegmentOf(E,B)) = A;Abs(LineSegmentOf(A, F)) = 3",
    "query_expressions": "p",
    "answer_expressions": "4",
    "fact_spans": "[[[26, 47], [63, 66]], [[98, 101]], [[23, 25]], [[1, 22]], [[53, 56]], [[49, 52], [71, 74]], [[59, 62]], [[29, 47]], [[26, 47]], [[1, 22]], [[0, 25]], [[23, 58]], [[59, 69]], [[71, 85]], [[87, 96]]]",
    "query_spans": "[[[98, 103]]]",
    "process": "Let the coordinates of points A and B be $(x_{1},y_{1})$, $(x_{2},y_{2})$, $\\therefore|AF|=x_{1}+\\frac{p}{2}$, and $|AF|=3$, hence $x_{1}=3-\\frac{p}{2}$. By the midpoint coordinate formula, we have $\\begin{cases}x_{1}=\\frac{x_{2}-\\frac{p}{2}}{2}\\\\y=\\frac{y_{2}}{2}\\end{cases}$, that is, $x_{2}=6-\\frac{p}{2}$, $y_{2}=2y_{1}$, $\\therefore y_{2}^{2}=4y_{1}^{2}$, i.e., $2p(6-\\frac{y}{2})=4\\times2p(3-\\frac{p}{2})$, and since $p>0$, we get: $p=4$."
  },
  {
    "text": "Let point $A(-2 , \\sqrt{3})$, the right focus of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$ be $F$, and point $P$ moves on the ellipse. When $|P A|+2|P F|$ takes the minimum value, what are the coordinates of point $P$?",
    "fact_expressions": "G: Ellipse;A: Point;P: Point;F: Point;Expression(G) = (x^2/16 + y^2/12 = 1);Coordinate(A) = (-2, sqrt(3));RightFocus(G) = F;PointOnCurve(P, G);WhenMin(Abs(LineSegmentOf(P, A)) + 2*Abs(LineSegmentOf(P, F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2*sqrt(3), sqrt(3))",
    "fact_spans": "[[[21, 60], [74, 76]], [[1, 20]], [[69, 73], [101, 105]], [[65, 68]], [[21, 60]], [[1, 20]], [[21, 68]], [[69, 79]], [[80, 100]]]",
    "query_spans": "[[[101, 110]]]",
    "process": ""
  },
  {
    "text": "The standard equation of the hyperbola with asymptote $x+2 y=0$ and intercepting the line $x-y-3=0$ to form a chord of length $\\frac{8 \\sqrt{3}}{3}$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;L:Line;Expression(L) = (x + 2*y = 0);Expression(H) = (x - y - 3 = 0);OneOf(Asymptote(G)) = L;Length(InterceptChord(H,G))=8*sqrt(3)/3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4-y^2=1",
    "fact_spans": "[[[58, 61]], [[19, 30]], [[1, 12]], [[1, 12]], [[19, 30]], [[0, 61]], [[18, 61]]]",
    "query_spans": "[[[58, 68]]]",
    "process": "Given that one asymptote of the hyperbola is $x + 2y = 0$, we can assume the hyperbola has the form $x^{2} - 4y^{2} = k$ ($k \\neq 0$). Substituting $y = x - 3$ into the hyperbola gives: $3x^{2} - 24x + k + 36 = 0$. If the line intersects the hyperbola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$, then $x_{1} + x_{2} = 8$, $x_{1}x_{2} = \\frac{k+3}{3}64.\\frac{4(k+36)}{3} = \\frac{8\\sqrt{3}}{3}$, solving yields: $k = 4$, thus the equation of the hyperbola is $\\frac{x^{2}}{4} - y^{2} = 1$."
  },
  {
    "text": "It is known that the focus of the parabola $y^{2}=8x$ is the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1  (a>0)$. Then, the equation of the right directrix of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;a>0;Expression(G) = (-y^2/3 + x^2/a^2 = 1);Expression(H) = (y^2 = 8*x);Focus(H) = RightFocus(G)",
    "query_expressions": "Expression(RightDirectrix(G))",
    "answer_expressions": "x=1/2",
    "fact_spans": "[[[20, 69], [75, 78]], [[23, 69]], [[2, 16]], [[23, 69]], [[20, 69]], [[2, 16]], [[2, 73]]]",
    "query_spans": "[[[75, 86]]]",
    "process": "The focus of the parabola \\( y^{2} = 8x \\) is \\( (2, 0) \\), so \\( a^{2} + 3 = 4 \\), \\( a^{2} = 1 \\), and the equation of the right directrix of the hyperbola is \\( x = \\frac{a^{2}}{c} \\), \\( x = \\frac{1}{2} \\)."
  },
  {
    "text": "If the two foci of an ellipse are $F_{1}(-4 , 0)$, $F_{2}(4 , 0)$, and the chord $A B$ of the ellipse passes through point $F_{1}$, with the perimeter of $\\triangle A B F_{2}$ being $20$, then what is the equation of this ellipse?",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;F1: Point;F2: Point;Coordinate(F1) = (-4, 0);Coordinate(F2) = (4, 0);Focus(G) = {F1, F2};IsChordOf(LineSegmentOf(A, B), G);PointOnCurve(F1, LineSegmentOf(A, B));Perimeter(TriangleOf(A, B, F2)) = 20",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/25 + y^2/9 = 1",
    "fact_spans": "[[[1, 3], [42, 44], [95, 97]], [[46, 51]], [[46, 51]], [[9, 24]], [[26, 41]], [[9, 24]], [[26, 41]], [[1, 41]], [[42, 51]], [[46, 60]], [[62, 91]]]",
    "query_spans": "[[[95, 102]]]",
    "process": ""
  },
  {
    "text": "The length of the minor axis of the ellipse $\\sqrt{x^{2}+(y+1)^{2}}=\\frac{|x-2 y+3|}{5}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (sqrt(x^2 + (y + 1)^2) = Abs(x - 2*y + 3)/5)",
    "query_expressions": "Length(MinorAxis(G))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 46]], [[0, 46]]]",
    "query_spans": "[[[0, 52]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the right focus is $F$, the left vertex is $A$. A perpendicular is drawn from $F$ to an asymptote of $C$, with foot of the perpendicular at $M$. If $\\tan \\angle M A F = \\frac{1}{2}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;M: Point;A: Point;LeftVertex(C)=A;F: Point;RightFocus(C)=F;PointOnCurve(F,l);IsPerpendicular(l, OneOf(Asymptote(C)));FootPoint(l,OneOf(Asymptote(C))) = M;Tan(AngleOf(M, A, F)) = 1/2;l:Line",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[2, 61], [85, 88], [139, 142]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[101, 104]], [[76, 79]], [[2, 79]], [[68, 71], [81, 84]], [[2, 71]], [[80, 97]], [[80, 97]], [[80, 104]], [[106, 137]], []]",
    "query_spans": "[[[139, 148]]]",
    "process": "According to the problem, one of the asymptotes of the hyperbola is given by $ y = -\\frac{b}{a}x $. Then, the equation of the line passing through point $ F(c,0) $ and perpendicular to $ y = -\\frac{b}{a}x $ is $ y = \\frac{a}{b}(x - c) $. Solving the system of equations:\n$$\n\\begin{cases}\ny = \\frac{a}{b}(x - c) \\\\\ny = -\\frac{b}{a}x\n\\end{cases}\n$$\nyields $ x = \\frac{a^2}{c} $, $ y = -\\frac{ab}{c} $, so point $ M\\left( \\frac{a^2}{c}, -\\frac{ab}{c} \\right) $. Given $ \\tan\\angle MAF = \\frac{1}{2} $, we have:\n$$\n\\frac{|-\\frac{ab}{c}|}{|\\frac{a^{2}}{c}-(-a)|} = \\frac{1}{2}\n$$\nwhich simplifies to $ \\frac{b}{a + c} = \\frac{1}{2} $, hence $ a + c = 2b $. Therefore,\n$$\n(a + c)^2 = 4b^2 = 4(c^2 - a^2)\n$$\nSimplifying gives $ 3c^2 - 2ac - 5a^2 = 0 $, or $ 3e^2 - 2e - 5 = 0 $, where $ e $ is the eccentricity. Solving yields $ e = \\frac{5}{3} $ or $ e = -1 $ (discarded). Thus, the eccentricity of the hyperbola is $ \\frac{5}{3} $."
  },
  {
    "text": "Given that the foci of the ellipse are the vertices of the hyperbola, and the foci of the hyperbola are the vertices of the major axis of the ellipse, if the eccentricities of the two curves are $e_{1}$, $e_{2}$ respectively, then $e_{1} \\cdot e_{2}$=?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Focus(H) = Vertex(G);Focus(G) = Endpoint(MajorAxis(H));e1: Number;e2: Number;Eccentricity(G) = e2;Eccentricity(H) = e1",
    "query_expressions": "e1*e2",
    "answer_expressions": "1",
    "fact_spans": "[[[8, 11], [15, 18]], [[2, 4], [22, 24]], [[2, 14]], [[15, 29]], [[41, 48]], [[50, 57]], [[31, 57]], [[31, 57]]]",
    "query_spans": "[[[60, 81]]]",
    "process": ""
  },
  {
    "text": "The equation of the hyperbola with asymptotes $y=\\pm x$ and passing through the point $(2,0)$ is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (2, 0);Expression(Asymptote(G))=(y=pm*x);PointOnCurve(H,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/4 = 1",
    "fact_spans": "[[[26, 29]], [[17, 25]], [[17, 25]], [[0, 29]], [[15, 29]]]",
    "query_spans": "[[[26, 33]]]",
    "process": "Hyperbolas with asymptotes $ y = \\pm x $ are rectangular hyperbolas, and their equation can be written as $ x^{2} - y^{2} = \\lambda $ ($ \\lambda \\neq 0 $). Substituting the point $ (2, 0) $ gives $ \\lambda = 4 $, $ \\therefore x^{2} - y^{2} = 4 $, $ \\therefore \\frac{x^{2}}{4} - \\frac{y^{2}}{4} = 1 $."
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1(a>0)$ is $\\sqrt{5}$, then the equation of the directrix of the parabola $y=a x^{2}$ is?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(H) = (y = a*x^2);Eccentricity(G) = sqrt(5)",
    "query_expressions": "Expression(Directrix(H))",
    "answer_expressions": "y=-1/2",
    "fact_spans": "[[[2, 39]], [[5, 39]], [[56, 70]], [[5, 39]], [[2, 39]], [[56, 70]], [[2, 54]]]",
    "query_spans": "[[[56, 77]]]",
    "process": "\\because the eccentricity of the hyperbola e=\\frac{c}{a}=\\sqrt{5}, c=\\sqrt{a^{2}+1}\\frac{a^{2+1}}{a}=\\sqrt{5}, solving gives a=\\frac{1}{2}, \\cdot v=\\frac{1}{3}x^{2}x^{2}=2v, so the directrix is y=-"
  },
  {
    "text": "Given points $A(-2,0)$, $B(3,0)$, a moving point $P(x, y)$ satisfies $\\overrightarrow{P A} \\cdot \\overrightarrow{P B}=x^{2}-6$, then what is the trajectory of point $P$?",
    "fact_expressions": "A: Point;B: Point;P: Point;Coordinate(A) = (-2, 0);Coordinate(B) = (3, 0);Coordinate(P) = (x, y);x: Number;y: Number;DotProduct(VectorOf(P, A), VectorOf(P, B)) = x^2 - 6",
    "query_expressions": "Locus(P)",
    "answer_expressions": "Parabola",
    "fact_spans": "[[[2, 12]], [[14, 23]], [[26, 35], [98, 101]], [[2, 12]], [[14, 23]], [[26, 35]], [[26, 35]], [[26, 35]], [[37, 94]]]",
    "query_spans": "[[[98, 106]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has its right focus at $F$ $(c, 0)$. There exists a line $l$ passing through point $F$ that intersects the ellipse at points $A$ and $B$ such that $OA \\perp OB$. Then the range of the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/a^2 + y^2/b^2 = 1);a: Number;b: Number;a > b;b > 0;F: Point;c:Number;Coordinate(F) = (c, 0);RightFocus(G) = F;l: Line;Intersection(l,G) = {A,B};A: Point;B: Point;IsPerpendicular(LineSegmentOf(O, A), LineSegmentOf(O, B)) = True;PointOnCurve(F, l) = True;O:Origin",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[(sqrt(5)-1)/2,1)",
    "fact_spans": "[[[2, 54], [89, 91], [122, 124]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[59, 70], [76, 80]], [[59, 71]], [[59, 71]], [[2, 71]], [[83, 88]], [[83, 101]], [[92, 95]], [[96, 99]], [[104, 119]], [[74, 88]], [[104, 119]]]",
    "query_spans": "[[[122, 135]]]",
    "process": "First, from the given conditions, $ l $ is not a horizontal line. Let the equation of line $ l $ be $ x = ty + c $, and let the coordinates of points $ A $ and $ B $ be $ (x_{1}, y_{1}) $ and $ (x_{2}, y_{2}) $, respectively. By combining the equations of the line and the ellipse, and using Vieta's formulas together with the coordinate representation of the dot product of vectors, we obtain  \n$$\n\\overrightarrow{OA} \\cdot \\overrightarrow{OB} = \\frac{-t^{2}b^{2}a^{2} - b^{4} + a^{2}c^{2}}{b^{2}t^{2} + a^{2}}.\n$$\nThen, since $ OA \\perp OB $, it follows that  \n$$\nt^{2} = \\frac{a^{2}c^{2} - b^{4}}{b^{2}a^{2}}.\n$$\nFrom this, an inequality can be established to obtain the result.  \n\nSince there exists a line $ l $ passing through point $ F $ intersecting the ellipse at points $ A $ and $ B $ such that $ OA \\perp OB $, clearly $ l $ is not a horizontal line. Let the equation of line $ l $ be $ x = ty + c $, and let the coordinates of points $ A $ and $ B $ be $ (x_{1}, y_{1}) $ and $ (x_{2}, y_{2}) $. From  \n$$\n\\begin{cases}\n\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 \\\\\nx = ty + c\n\\end{cases}\n$$\neliminating $ x $, we obtain  \n$$\n(b^{2}t^{2} + a^{2})y^{2} + 2tb^{2}cy + b^{2}(c^{2} - a^{2}) = 0.\n$$\nBy Vieta's formulas,  \n$$\n\\begin{cases}\ny_{1} + y_{2} = -\\frac{2tb^{2}c}{b^{2}t^{2} + a^{2}}, \\\\\ny_{1}y_{2} = \\frac{b^{2}(c^{2} - a^{2})}{b^{2}t^{2} + a^{2}}\n\\end{cases}\n$$\n$$\n\\overrightarrow{OA} \\cdot \\overrightarrow{OB} = x_{1}x_{2} + y_{1}y_{2} = (ty_{1} + c)(ty_{2} + c) + y_{1}y_{2} = (t^{2} + 1)y_{1}y_{2} + tc(y_{1} + y_{2}) + c^{2} = \\frac{-t^{2}b^{2}a^{2} - b^{4} + a^{2}c^{2}}{b^{2}t^{2} + a^{2}}.\n$$\nSince $ OA \\perp OB $, we have $ \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = 0 $, so  \n$$\nt^{2} = \\frac{a^{2}c^{2} - b^{4}}{b^{2}a^{2}}.\n$$\nSince $ t^{2} \\geqslant 0 $, it follows that $ a^{2}c^{2} - b^{4} \\geqslant 0 $, i.e.,  \n$$\na^{2}c^{2} - (a^{2} - c^{2})^{2} \\geqslant 0,\n$$\nwhich simplifies to  \n$$\nc^{4} - 3a^{2}c^{2} + a^{4} \\leqslant 0.\n$$\nThus,  \n$$\n\\left(\\frac{c}{a}\\right)^{4} - 3\\left(\\frac{c}{a}\\right)^{2} + 1 \\leqslant 0,\n$$\ni.e.,  \n$$\ne^{4} - 3e^{2} + 1 \\leqslant 0.\n$$\nSolving this inequality yields  \n$$\n\\frac{3 - \\sqrt{5}}{2} \\leqslant e^{2} \\leqslant \\frac{3 + \\sqrt{5}}{2},\n$$\ni.e.,  \n$$\n\\frac{6 - 2\\sqrt{5}}{4} \\leqslant e^{2} \\leqslant \\frac{6 + 2\\sqrt{5}}{4}.\n$$\nSince the eccentricity of an ellipse satisfies $ 0 < e < 1 $, we conclude  \n$$\n\\frac{\\sqrt{5} - 1}{2} \\leqslant e < 1.\n$$"
  },
  {
    "text": "Given the coordinates of two points $A(-4,0)$, $B(4,0)$, if $|P A|+|P B|=10$, then the equation of the trajectory of point $P$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (-4, 0);B: Point;Coordinate(B) = (4, 0);P: Point;Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B)) = 10",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "x^2/25 + y^2/9 = 1",
    "fact_spans": "[[[6, 15]], [[6, 15]], [[18, 26]], [[18, 26]], [[46, 50]], [[28, 44]]]",
    "query_spans": "[[[46, 56]]]",
    "process": "Since A(-4,0), B(4,0), |PA| + |PB| = 10 and 10 > 8, the sum of the distances from point P to two fixed points is constant, therefore the locus of point P is an ellipse with foci at A and B. Since 2a = 10, 2c = 8, then b = 3, so the standard equation of the ellipse is \\frac{x^2}{25} + \\frac{y^{2}}{9} = 1."
  },
  {
    "text": "Given the hyperbola $\\frac{y^{2}}{4}-x^{2}=1$, its two asymptotes intersect the directrix of the parabola $y^{2}=2 p x$ ($p>0$) at points $A$ and $B$ respectively, and $O$ is the origin. If the area of $\\triangle A O B$ is $1$, then the value of $p$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2 + y^2/4 = 1);L1:Line;L2: Line;Asymptote(G) = {L1,L2};H: Parabola;Expression(H) = (y^2 = 2*p*x);p: Number;p>0;Intersection(L1, Directrix(H)) = A;Intersection(L2, Directrix(H)) = B;A: Point;O: Origin;B: Point;Area(TriangleOf(A, O, B)) = 1",
    "query_expressions": "p",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 30]], [[2, 30]], [], [], [[2, 36]], [[39, 60]], [[39, 60]], [[111, 114]], [[42, 60]], [[2, 74]], [[2, 74]], [[65, 68]], [[75, 78]], [[69, 72]], [[85, 109]]]",
    "query_spans": "[[[111, 118]]]",
    "process": "From the given conditions, the asymptotes are $ y = \\pm 2x $. Solving simultaneously with the parabola: $ 4x^{2} = 2px $, we get $ x = 0 $ or $ x = \\frac{p}{2} $. When $ x = 0 $, $ y = 0 $; when $ x = -\\frac{p}{2} $, $ y = \\pm p $. Then the area of $ \\triangle AOB $ is $ \\frac{1}{2} \\times \\frac{p}{2} \\times 2p = 1 $. Since $ p > 0 $, it follows that $ p = \\sqrt{2} $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ with focus $F(2,0)$, a line passing through $F$ with slope $1$ intersects the hyperbola at exactly one point. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;F: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (2, 0);OneOf(Focus(G)) = F;PointOnCurve(F, H);Slope(H) = 1;NumIntersection(H, G) = 1",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/2 - y^2/2 = 1",
    "fact_spans": "[[[2, 48], [75, 78], [88, 91]], [[5, 48]], [[5, 48]], [[72, 74]], [[51, 59], [61, 64]], [[2, 48]], [[51, 59]], [[2, 59]], [[60, 74]], [[65, 74]], [[72, 86]]]",
    "query_spans": "[[[88, 96]]]",
    "process": "\\because the line intersects the hyperbola at exactly one point, and the focus is F(2,0), c=2, \\therefore the line is parallel to the asymptote of the hyperbola, \\therefore \\frac{b}{a}=1, i.e., a=b, \\therefore a^{2}=b^{2}=c^{2}-a^{2}, i.e., c^{2}=2a^{2}=4, \\therefore a=b=\\sqrt{2}; then the equation of the hyperbola is \\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1"
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, the intersection point of the directrix and the $x$-axis is $M$, and $N$ is a point on the parabola satisfying $|NF|=\\frac{\\sqrt{3}}{2}|MN|$, then $\\angle NMF$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;Intersection(Directrix(G),xAxis) = M;M: Point;PointOnCurve(N,G);N: Point;Abs(LineSegmentOf(N, F)) = (sqrt(3)/2)*Abs(LineSegmentOf(M, N))",
    "query_expressions": "AngleOf(N, M, F)",
    "answer_expressions": "ApplyUnit(30, degree)",
    "fact_spans": "[[[2, 16], [43, 46]], [[2, 16]], [[20, 23]], [[2, 23]], [[2, 38]], [[35, 38]], [[39, 50]], [[39, 42]], [[54, 85]]]",
    "query_spans": "[[[87, 103]]]",
    "process": "Draw a perpendicular line from point N to the directrix, with foot E. According to the geometric definition of the parabola: NF = NE, so |NE| = \\frac{\\sqrt{3}}{2}|MN|, hence \\angle EMN = 60^{\\circ}, then \\angle NMF = 30^{\\circ}"
  },
  {
    "text": "Let point $P$ be a moving point on the ellipse $x^{2}+4 y^{2}=36$, and let $F$ be the left focus of the ellipse. Then the maximum value of $|P F|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F: Point;Expression(G) = (x^2 + 4*y^2 = 36);PointOnCurve(P, G);LeftFocus(G) = F",
    "query_expressions": "Max(Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "6+3*sqrt(3)",
    "fact_spans": "[[[6, 26], [35, 37]], [[1, 5]], [[31, 34]], [[6, 26]], [[1, 30]], [[31, 41]]]",
    "query_spans": "[[[43, 56]]]",
    "process": "The standard equation of the ellipse is $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$, so $a=6$, $c=3\\sqrt{3}$. By the property of the ellipse, when point $P$ is the right vertex of the ellipse, $|PF|$ reaches its maximum value, and $|PF|_{\\max}=a+c=6+3\\sqrt{3}$. Answer: $6+3\\sqrt{3}$"
  },
  {
    "text": "Given that the distances from one focus of an ellipse to the two vertices on the major axis are $3 \\sqrt{3}$ and $\\sqrt{3}$, respectively, what is the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;A:Point;B:Point;Endpoint(MajorAxis(G))={A,B};Distance(OneOf(Focus(G)),A)=3*sqrt(3);Distance(OneOf(Focus(G)),B)=sqrt(3)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 4], [49, 51]], [], [], [[2, 15]], [[2, 47]], [[2, 47]]]",
    "query_spans": "[[[49, 58]]]",
    "process": ""
  },
  {
    "text": "Given that the line $y=k(x+2)$ ($k>0$) intersects the parabola $C$: $y^{2}=8x$ at points $A$ and $B$, and $F$ is the focus of $C$. If $|FA|=2|FB|$, then $k=$?",
    "fact_expressions": "C: Parabola;G: Line;F: Point;A: Point;B: Point;k:Number;Expression(C) = (y^2 = 8*x);k>0;Expression(G) = (y = k*(x + 2));Intersection(G, C) = {A, B};Focus(C) = F;Abs(LineSegmentOf(F, A)) = 2*Abs(LineSegmentOf(F, B))",
    "query_expressions": "k",
    "answer_expressions": "2*sqrt(2)/3",
    "fact_spans": "[[[20, 39], [56, 59]], [[2, 19]], [[52, 55]], [[42, 45]], [[46, 49]], [[80, 83]], [[20, 39]], [[4, 19]], [[2, 19]], [[2, 51]], [[52, 62]], [[64, 78]]]",
    "query_spans": "[[[80, 85]]]",
    "process": ""
  },
  {
    "text": "A line perpendicular to the $x$-axis intersects the parabola $y^{2}=4 x$ at points $A$ and $B$. If the length of $A B$ is $4 \\sqrt{3}$, then the distance from the focus to the line $A B$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;Expression(G) = (y^2 = 4*x);IsPerpendicular(H,xAxis);Intersection(H, G) = {A, B};Length(LineSegmentOf(A, B)) = 4*sqrt(3)",
    "query_expressions": "Distance(Focus(G), LineOf(A,B))",
    "answer_expressions": "2",
    "fact_spans": "[[[11, 25]], [[8, 10]], [[28, 31]], [[32, 35]], [[11, 25]], [[0, 10]], [[8, 37]], [[40, 60]]]",
    "query_spans": "[[[11, 77]]]",
    "process": "The focus of the parabola $ y^{2}=4x $ has coordinates $ (1,0) $. Since the length of the line segment $ AB $, perpendicular to the x-axis, is $ 4\\sqrt{3} $, and point $ A $ lies in the first quadrant, the y-coordinate of point $ A $ is $ 2\\sqrt{3} $. Substituting $ y=2\\sqrt{3} $ into $ y^{2}=4x $ gives $ 12=4x $, solving for $ x $ yields $ x=3 $, so $ A(3,2\\sqrt{3}) $. Therefore, the distance from the focus to the line $ AB $ is $ 3-1=2 $."
  },
  {
    "text": "Given the ellipse $ C $: $ \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 $ $ (a > b > 0) $ with eccentricity $ e $, the line $ l $: $ y = e x + a $ intersects the $ x $-axis and $ y $-axis at points $ A $ and $ B $, respectively. $ M $ is a common point of the line $ l $ and the ellipse $ C $. Suppose $ |A M| = e |A B| $, then the eccentricity $ e $ of the ellipse is?",
    "fact_expressions": "l: Line;C: Ellipse;b: Number;a: Number;A: Point;M: Point;B: Point;e: Number;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Eccentricity(C) = e;Expression(l)=(y=e*x+a);Intersection(l,xAxis)=A;Intersection(l,yAxis)=B;OneOf(Intersection(l,C))=M;Abs(LineSegmentOf(A, M)) = e*Abs(LineSegmentOf(A, B))",
    "query_expressions": "e",
    "answer_expressions": "(\\sqrt{5}-1)/2",
    "fact_spans": "[[[68, 84], [111, 116]], [[2, 59], [117, 122], [147, 149]], [[9, 59]], [[9, 59]], [[98, 102]], [[107, 110]], [[103, 106]], [[64, 67], [153, 156]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 67]], [[68, 84]], [[68, 106]], [[68, 106]], [[107, 128]], [[130, 144]]]",
    "query_spans": "[[[147, 158]]]",
    "process": "Since the line $ l: y = ex + a $ intersects the x-axis and y-axis at points $ A $ and $ B $ respectively, we have $ A\\left(-\\frac{a}{e}, 0\\right) $, $ B(0, a) $. From \n$$\n\\begin{cases}\ny = ex + a \\\\\n\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1\n\\end{cases},\n$$\neliminating $ y $ yields $ x^{2} + 2cx + c^{2} = 0 $, so $ M(-c, a - ec) $. From $ |AM| = e|AB| $, we know $ \\overrightarrow{AM} = e\\overrightarrow{AB} $, that is, \n$$\n\\left(-c + \\frac{a}{e}, a - ec\\right) = e\\left(\\frac{a}{e}, a\\right).\n$$\nThus $ a - ec = ae $, i.e., $ 1 - e^{2} = e $, solving gives $ e = \\frac{\\sqrt{5} - 1}{2} $ or $ \\frac{-\\sqrt{5} - 1}{2} $ (discarded)."
  },
  {
    "text": "The line $l$ passing through the focus $F$ of the parabola $C$: $y^{2}=4x$ intersects the parabola $C$ at points $P$ and $Q$, and intersects the directrix at point $M$, with $\\overrightarrow{F M}=3 \\overrightarrow{F P}$. Then $|\\overrightarrow{F P}|$=?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;PointOnCurve(F, l) = True;l: Line;Intersection(l, C) = {P, Q};P: Point;Q: Point;Intersection(Directrix(C), l) = M;M: Point;VectorOf(F, M) = 3*VectorOf(F, P)",
    "query_expressions": "Abs(VectorOf(F, P))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[1, 20], [33, 39]], [[1, 20]], [[23, 26]], [[1, 26]], [[0, 32]], [[27, 32]], [[27, 50]], [[41, 44]], [[45, 48]], [[27, 60]], [[56, 60]], [[62, 107]]]",
    "query_spans": "[[[109, 135]]]",
    "process": "Transform $\\overrightarrow{FM}=3\\overrightarrow{FP}$ into $|PM|=2|PF|$, then use the definition of the parabola and triangle similarity to obtain the answer. Draw $PH$ perpendicular to the directrix at $H$. From $\\overrightarrow{FM}=3\\overrightarrow{FP}$, we get $|PM|=2|PF|$, as shown in the figure. Also, by the definition of the parabola, $|PF|=|PH|$, so $|PM|=2|PH|$. By triangle similarity, $\\frac{|PH|}{P}=\\frac{|PH|}{2}=\\frac{|MP|}{|MF|}=\\frac{2}{3}$, so $|PH|=\\frac{4}{3}$, hence $|\\overrightarrow{FP}|=\\frac{4}{3}$."
  },
  {
    "text": "In the plane, the line segment $|GH| = 4$. If vertex $P$ of triangle $GPH$ always satisfies $|\\overrightarrow{PG}| = 2|\\overrightarrow{PH}|$, then as $P$ moves, the maximum area of triangle $GPH$ equals?",
    "fact_expressions": "G: Point;H: Point;Abs(LineSegmentOf(G, H)) = 4;P: Point;Abs(VectorOf(P, G)) = 2*Abs(VectorOf(P, H))",
    "query_expressions": "Max(TriangleOf(G, P, H))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[20, 25]], [[20, 25]], [[5, 14]], [[29, 32], [89, 92]], [[36, 84]]]",
    "query_spans": "[[[96, 113]]]",
    "process": "Let $ P(x,y) $, and establish a rectangular coordinate system with the midpoint of $ GH $ as the origin, $ GH $ as the $ x $-axis, and the perpendicular bisector of $ GH $ as the $ y $-axis. Based on the given conditions, the locus of point $ P $ is a circle centered at $ \\left(\\frac{10}{3}, 0\\right) $ with radius $ \\frac{8}{3} $ (excluding the two intersection points with the $ x $-axis). Then, computation yields the solution.  \n**Solution**: Let $ P(x,y) $, take the midpoint of $ GH $ as the origin, $ GH $ as the $ x $-axis, and the perpendicular bisector of $ GH $ as the $ y $-axis to establish a rectangular coordinate system. Since the segment $ |GH| = 4 $ in the plane, and vertex $ P $ of triangle $ GPH $ always satisfies $ |\\overrightarrow{PG}| = 2|\\overrightarrow{PH}| $, we have $ G(-2,0) $, $ H(2,0) $, and  \n$$\n\\sqrt{(x+2)^{2}+y^{2}} = 2\\sqrt{(x-2)^{2}+y^{2}} \\quad (y \\neq 0)\n$$  \nRearranging gives:  \n$$\nx^{2}+y^{2}-\\frac{20}{3}x+4=0\n$$  \nConverting to standard form yields:  \n$$\n\\left(x-\\frac{10}{3}\\right)^{2}+y^{2}=\\frac{64}{9}, \\quad (y \\neq 0)\n$$  \nThus, the locus of point $ P $ is a circle centered at $ \\left(\\frac{10}{3}, 0\\right) $ with radius $ \\frac{8}{3} $ (excluding the two intersection points with the $ x $-axis). Therefore, the maximum area of triangle $ GPH $ is  \n$$\n\\frac{1}{2} \\times 4 \\times \\frac{8}{3} = \\frac{16}{3}\n$$"
  },
  {
    "text": "The asymptotes of the hyperbola are given by $3 x \\pm 2 y=0$, then the eccentricity of this hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (pm*2*y + 3*x = 0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{sqrt(13)/2, sqrt(13)/3}",
    "fact_spans": "[[[0, 3], [28, 31]], [[0, 25]]]",
    "query_spans": "[[[28, 38]]]",
    "process": "For the two cases where the foci of the hyperbola lie on the x-axis and y-axis respectively, the result can be obtained using the relationship between the asymptote slope and the hyperbola parameters a, b, c. Let the hyperbola equation be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0,b>0), then \\frac{b}{a}=\\frac{3}{2}, \\therefore e=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{1+\\frac{9}{4}}=\\frac{\\sqrt{13}}{2}; in summary, the eccentricity of the hyperbola is \\frac{\\sqrt{13}}{2} or \\frac{\\sqrt{13}}{3}."
  },
  {
    "text": "Let $m$ be a constant. If the point $F(0,5)$ is a focus of the hyperbola $\\frac{y^{2}}{m}-\\frac{x^{2}}{9}=1$, then $m=$?",
    "fact_expressions": "G: Hyperbola;m: Number;F: Point;Expression(G) = (-x^2/9 + y^2/m = 1);Coordinate(F) = (0, 5);OneOf(Focus(G)) = F",
    "query_expressions": "m",
    "answer_expressions": "16",
    "fact_spans": "[[[19, 57]], [[1, 4], [64, 67]], [[9, 18]], [[19, 57]], [[9, 18]], [[9, 62]]]",
    "query_spans": "[[[64, 69]]]",
    "process": "The coordinates of the hyperbola's focus are F(0,5), so the focus lies on the y-axis. From c^{2}=a^{2}+b^{2}, we get 25=m+9, m=16"
  },
  {
    "text": "Given that the focal distance of the ellipse $\\frac{x^{2}}{20}+\\frac{y^{2}}{k}=1$ is $6$, what is the value of $k$?",
    "fact_expressions": "G: Ellipse;k: Number;Expression(G) = (x^2/20 + y^2/k = 1);FocalLength(G) = 6",
    "query_expressions": "k",
    "answer_expressions": "{11,29}",
    "fact_spans": "[[[2, 40]], [[49, 52]], [[2, 40]], [[2, 47]]]",
    "query_spans": "[[[49, 56]]]",
    "process": "When the foci of the ellipse are on the x-axis, $a^{2}=20$, $b^{2}=k$, so $c^{2}=a^{2}-b^{2}=20-k$, thus $c=\\sqrt{20-k}$. Since the focal length of the ellipse $\\frac{x^{2}}{20}+\\frac{y^{2}}{k}=1$ is 6, we have $c=\\sqrt{20-k}=3$, so $k=11$. When the foci of the ellipse are on the y-axis, $a^{2}=k$, $b^{2}=20$, so $c^{2}=a^{2}-b^{2}=k-20$, thus $c=\\sqrt{k-20}$. Since the focal length of the ellipse $\\frac{x^{2}}{20}+\\frac{y^{2}}{k}=1$ is 6, we have $c=\\sqrt{k-20}=3$, so $k=29$. Therefore, the answer should be 11 or 29."
  },
  {
    "text": "It is known that the line $x - y - 2 = 0$ passes through a focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ $(a > 0, b > 0)$, and is perpendicular to one of the asymptotes of the hyperbola. Then the length of the real axis of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x - y - 2 = 0);PointOnCurve(OneOf(Focus(C)), G);IsPerpendicular(G, OneOf(Asymptote(C)))",
    "query_expressions": "Length(RealAxis(C))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[14, 75], [83, 86], [96, 99]], [[22, 75]], [[22, 75]], [[2, 13]], [[22, 75]], [[22, 75]], [[14, 75]], [[2, 13]], [[2, 80]], [[2, 94]]]",
    "query_spans": "[[[96, 105]]]",
    "process": "From the equation of the line, the coordinates of the foci of the hyperbola can be determined. Then, using the perpendicular relationship, the relationship between $a$ and $b$ is obtained. Combining this with the value of $c$, the length of the real axis can be solved. (Detailed explanation) Since $x - y - 2 = 0$ passes through the points $(2, 0)$ and $(0, -2)$, and the foci of the hyperbola lie on the $x$-axis, we have $c = 2$. Also, since $x - y - 2 = 0$ is perpendicular to one asymptote, $\\frac{b}{a} = 1$. Therefore, $a = \\sqrt{c^{2} - b^{2}} = \\sqrt{4 - a^{2}}$, so $a = \\sqrt{2}$, hence the length of the real axis is $2\\sqrt{2}$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $P$ is a point on its asymptote in the first quadrant, point $Q$ lies on the hyperbola, and satisfies $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, $\\overrightarrow{P F_{2}}=4 \\overrightarrow{P Q}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;Q: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P,Asymptote(G));Quadrant(P)=1;PointOnCurve(Q,G);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;VectorOf(P,F2)=4*VectorOf(P,Q)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[18, 74], [85, 86], [103, 106], [224, 227]], [[21, 74]], [[21, 74]], [[81, 84]], [[2, 9]], [[10, 17]], [[98, 102]], [[21, 74]], [[21, 74]], [[18, 74]], [[2, 80]], [[2, 80]], [[81, 97]], [[81, 97]], [[98, 107]], [[111, 170]], [[173, 222]]]",
    "query_spans": "[[[224, 233]]]",
    "process": "From the given conditions, $\\triangle PF_{1}F_{2}$ is a right triangle, so $OP = OF_{1} = OF_{2}$. Let the coordinates of point $P$ be $P(x, y)$ $(x > 0, y > 0)$. Since point $P$ lies on the asymptote $y = \\frac{b}{a}x$, we have:\n\\[\n\\begin{cases}\ny = \\frac{b}{a}x \\\\\nx^{2} + y^{2} = c^{2}\n\\end{cases}\n\\]\nSolving gives:\n\\[\n\\begin{cases}\nx = a \\\\\ny = b\n\\end{cases}\n\\]\nThus, $P(a, b)$ and $F_{2}(c, 0)$. Let $Q(m, n)$. From the given conditions:\n$\\overrightarrow{PF_{2}} = (c - a, -b)$, $\\overrightarrow{PQ} = (m - a, n - b)$, then:\n$(c - a, -b) = 4(m - a, n - b)$,\n\\[\n\\begin{matrix}\nm = \\frac{c}{4} + \\frac{3a}{4} & \\text{then } \\left(\\frac{c}{4} + \\frac{3a}{4}, -\\frac{3b}{4}\\right) \\text{ lies on the hyperbola:} \\\\\nn = -\\frac{3b}{4} & \\text{that is: } \\frac{1}{16}(e + 3)^{2} - \\frac{9}{16} = 1, \\text{ so: } (e + 3)^{2} = 25\n\\end{matrix}\n\\]\nThus, the eccentricity of the hyperbola is $2$."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$ has focus $F$, directrix $l$, and circle $\\odot C$: $(x-a)^{2}+(y-2 \\sqrt{3})^{2}=16$ passes through point $F$ and is tangent to $l$, then $p=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;l: Line;Directrix(G) = l;C: Circle;Expression(C) = ((x-a)^2 + (y-2*sqrt(3))^2 = 16);PointOnCurve(F, C);IsTangent(C, l)",
    "query_expressions": "p",
    "answer_expressions": "{2, 6}",
    "fact_spans": "[[[2, 23]], [[2, 23]], [[96, 99]], [[5, 23]], [[27, 30], [83, 87]], [[2, 30]], [[34, 37], [89, 92]], [[2, 37]], [[39, 82]], [[39, 82]], [[39, 87]], [[39, 94]]]",
    "query_spans": "[[[96, 101]]]",
    "process": "Substituting $F(\\frac{p}{2},0)$ into the circle equation gives one equation, and the distance from the center to line $l$ equaling the radius gives another equation; solving the system yields that $F(\\frac{p}{2},0)$ lies on $(x-a)^{2}+(y-2\\sqrt{3})^{2}=16$, so $(\\frac{p}{2}-a)^{2}+(0-2\\sqrt{3})^{2}=16$, namely $|\\frac{p}{2}-a|=2$ (1). $(x-a)^{2}+(y-2\\sqrt{3})^{2}=16$ is tangent to $l$, so $|a+\\frac{p}{2}|=4$ (2). From (1) and (2), it follows that $p=2$ or $p=6$."
  },
  {
    "text": "Draw a line $l$ through the left focus $F_{1}$ of the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$, intersecting the left branch of the hyperbola at points $P$ and $Q$. If $|P Q|=4$ and $F_{2}$ is the right focus of the hyperbola, then the perimeter of $\\triangle P F_{2} Q$ is?",
    "fact_expressions": "l: Line;G: Hyperbola;P: Point;F1:Point;F2: Point;Q: Point;Expression(G) = (x^2 - y^2/4 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(F1, l);Intersection(l, LeftPart(G)) = {P, Q};Abs(LineSegmentOf(P, Q)) = 4",
    "query_expressions": "Perimeter(TriangleOf(P, F2, Q))",
    "answer_expressions": "12",
    "fact_spans": "[[[43, 48]], [[1, 29], [49, 52], [85, 88]], [[55, 58]], [[33, 40]], [[77, 84]], [[59, 62]], [[1, 29]], [[1, 40]], [[77, 92]], [[0, 48]], [[43, 64]], [[66, 75]]]",
    "query_spans": "[[[94, 120]]]",
    "process": "By the given condition and according to the definition of a hyperbola, we have |PF_{2}|-|PF_{1}|=2 and |QF_{2}|-|QF_{1}|=2. Since |PF_{1}|+|QF_{1}|=|PQ|=4, it follows that |PF_{2}|+|QF_{2}|-4=4, solving gives |PF_{2}|+|QF_{2}|=8, therefore the perimeter of \\triangle PF_{2}Q is |PF_{2}|+|QF_{2}|+|PQ|=8+4=12."
  },
  {
    "text": "The standard equation of a parabola with the center of the ellipse $x^{2}+2 y^{2}=1$ as its vertex and the right vertex as its focus is?",
    "fact_expressions": "H: Ellipse;Expression(H) = (x^2 + 2*y^2 = 1);Vertex(G) = Center(H);Focus(G) = RightVertex(H);G: Parabola",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[1, 20]], [[1, 20]], [[0, 36]], [[0, 36]], [[33, 36]]]",
    "query_spans": "[[[33, 43]]]",
    "process": "Since the right vertex of \\( x^{2} + 2y^{2} = 1 \\) is \\( (1, 0) \\), the standard equation of the parabola is \\( y^{2} = 4x \\)."
  },
  {
    "text": "Given the hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ and the hyperbola $C_{2}$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{16}=1$ have the same asymptotes; and the right focus of $C_{1}$ is $F(\\sqrt{5}, 0)$, then $a=$? $b=$?",
    "fact_expressions": "C1: Hyperbola;C2: Hyperbola;F: Point;a:Number;b:Number;a>0;b>0;Expression(C1) = (x^2/a^2-y^2/b^2=1);Expression(C2) = (x^2/4-y^2/16=1);Asymptote(C1)=Asymptote(C2);RightFocus(C1)=F;Coordinate(F) = (sqrt(5), 0)",
    "query_expressions": "a;b",
    "answer_expressions": "1\n2",
    "fact_spans": "[[[2, 67], [125, 132]], [[68, 116]], [[137, 153]], [[156, 159]], [[161, 164]], [[14, 67]], [[14, 67]], [[2, 67]], [[68, 116]], [[2, 123]], [[125, 153]], [[137, 153]]]",
    "query_spans": "[[[156, 161]], [[161, 166]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the foci of the curve $C_{1}$: $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$, and let $P$ be an intersection point of the curves $C_{2}$: $\\frac{x^{2}}{3}-y^{2}=1$ and $C_{1}$. Then the area of $\\triangle PF_{1}F_{2}$ is?",
    "fact_expressions": "C1: Curve;C2: Curve;P: Point;F1: Point;F2: Point;Expression(C1) = (x^2/6 + y^2/2 = 1);Expression(C2) = (x^2/3 - y^2 = 1);Focus(C1) = {F1, F2};OneOf(Intersection(C1, C2)) = P",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[19, 65], [110, 117]], [[73, 117]], [[69, 72]], [[1, 8]], [[11, 18]], [[19, 65]], [[73, 117]], [[1, 68]], [[69, 122]]]",
    "query_spans": "[[[124, 152]]]",
    "process": "From $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$\\textcircled{1}, $\\frac{x^{2}}{3}-y^{2}=1$\\textcircled{2}, solve to get $x^{2}=\\frac{9}{2}$, $y^{2}=\\frac{1}{2}$, without loss of generality, let $P(\\frac{3\\sqrt{2}}{2},\\frac{\\sqrt{2}}{2})$, so the area of $\\triangle PF_{1}F_{2}$ is $\\frac{1}{2}|F_{1}F_{2}|\\times|y_{P}|=\\frac{1}{2}\\times4\\times\\frac{\\sqrt{2}}{2}=\\sqrt{2}$"
  },
  {
    "text": "Let point $P$ be a moving point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$, and let $F$ be the left focus of the ellipse. Then the range of values for $|P F|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F: Point;Expression(G) = (x^2/16 + y^2/7 = 1);PointOnCurve(P, G);LeftFocus(G) = F",
    "query_expressions": "Range(Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "[1,7]",
    "fact_spans": "[[[6, 44], [54, 56]], [[1, 5]], [[50, 53]], [[6, 44]], [[1, 49]], [[50, 60]]]",
    "query_spans": "[[[62, 76]]]",
    "process": "Ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$. Let $P(x_{0},y_{0})$, we get $|PF|=\\sqrt{(x_{0}-3)^{2}+y_{0}^{2}}=\\frac{3}{4}x_{0}+4\\cdot -4\\leqslant x_{0}\\leqslant 4 \\therefore -3\\leqslant \\frac{3}{4}x_{0}\\leqslant 3, 1\\leqslant \\frac{3}{4}x_{0}+4\\leqslant 7$, hence $1\\leqslant |PF|\\leqslant 7$"
  },
  {
    "text": "Given point $P(1,-1)$, $F$ is the right focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and $M$ is a point on the ellipse such that $|MP|+2|MF|$ is minimized. Then the point $M$ is?",
    "fact_expressions": "P: Point;Coordinate(P) = (1, -1);G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);F: Point;RightFocus(G) = F;M: Point;PointOnCurve(M, G);WhenMin(Abs(LineSegmentOf(M, P)) + 2*Abs(LineSegmentOf(M, F)))",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(26/3, -1)",
    "fact_spans": "[[[2, 12]], [[2, 12]], [[18, 55], [64, 66]], [[18, 55]], [[14, 17]], [[14, 59]], [[60, 63], [90, 94]], [[60, 69]], [[72, 88]]]",
    "query_spans": "[[[90, 96]]]",
    "process": ""
  },
  {
    "text": "The line $l$ passing through the point $M(1, 1)$ intersects the curve $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$ at points $A$ and $B$. If point $M$ is the midpoint of chord $AB$, then the equation of line $l$ is?",
    "fact_expressions": "l: Line;C: Curve;A: Point;B: Point;M: Point;Expression(C) = (x^2/4 + y^2/9 = 1);Coordinate(M) = (1, 1);PointOnCurve(M, l);Intersection(l, C) = {A, B};IsChordOf(LineSegmentOf(A, B), C);MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Expression(l)",
    "answer_expressions": "9*x + 4*y - 13 = 0",
    "fact_spans": "[[[13, 18], [91, 96]], [[19, 61]], [[64, 67]], [[70, 73]], [[1, 12], [77, 81]], [[19, 61]], [[1, 12]], [[0, 18]], [[13, 75]], [[19, 87]], [[77, 90]]]",
    "query_spans": "[[[91, 101]]]",
    "process": ""
  },
  {
    "text": "It is known that the focus of the parabola $y^{2}=8 x$ is exactly the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{2}=1(a>0)$. Then, the eccentricity of the hyperbola is?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 8*x);G: Hyperbola;Expression(G) = (-y^2/2 + x^2/a^2 = 1);a: Number;a>0;Focus(H) = RightFocus(G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[22, 69], [76, 79]], [[22, 69]], [[25, 69]], [[25, 69]], [[2, 73]]]",
    "query_spans": "[[[76, 85]]]",
    "process": "Find the focus of the parabola to obtain the value of c. From the hyperbola equation, obtain the value of a, and then find the eccentricity of the hyperbola. It is easy to see that the focus of the parabola y^{2}=8x is: (2,0). Thus, the right focus of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{2}=1 (a>0) is (2,0), so c=2. Then: a^{2}+2=2^{2}, a=\\sqrt{2}. Therefore, the eccentricity of the hyperbola is: e=\\frac{c}{a}=\\frac{2}{\\sqrt{2}}=\\sqrt{2}."
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has two foci $F_{1}$, $F_{2}$, and $P$ is a point on the ellipse such that $\\angle F_{1} P F_{2}=60^{\\circ}$. Then the range of the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;F2:Point;P: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Focus(G)={F1,F2};PointOnCurve(P,G);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[1/2,1)",
    "fact_spans": "[[[0, 52], [78, 80], [119, 121]], [[2, 52]], [[2, 52]], [[58, 65]], [[66, 73]], [[74, 77]], [[2, 52]], [[2, 52]], [[0, 52]], [[0, 73]], [[74, 83]], [[84, 117]]]",
    "query_spans": "[[[119, 131]]]",
    "process": ""
  },
  {
    "text": "What is the length of the major axis of the ellipse $16 x^{2}+9 y^{2}=144$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (16*x^2 + 9*y^2 = 144)",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "8",
    "fact_spans": "[[[0, 24]], [[0, 24]]]",
    "query_spans": "[[[0, 30]]]",
    "process": ""
  },
  {
    "text": "Given that the curve represented by the equation $\\frac{x^{2}}{m}+y^{2}=1$ is an ellipse with foci on the $x$-axis and eccentricity $\\frac{1}{2}$, then $m$=?",
    "fact_expressions": "G: Ellipse;H: Curve;G = H;Expression(H) = (x^2/m + y^2 = 1);PointOnCurve(Focus(G), xAxis);Eccentricity(G) = 1/2;m: Number",
    "query_expressions": "m",
    "answer_expressions": "4/3",
    "fact_spans": "[[[62, 64]], [[32, 34]], [[32, 64]], [[4, 34]], [[35, 64]], [[44, 64]], [[66, 69]]]",
    "query_spans": "[[[66, 71]]]",
    "process": "The eccentricity of the ellipse equation $\\frac{x^{2}}{m}+y^{2}=1$ with foci on the x-axis is $\\frac{1}{2}$, then $a=\\sqrt{m}>1$, $b=1$, $c=\\sqrt{1-m}$, $\\frac{\\sqrt{1-m}}{\\sqrt{m}}=\\frac{1}{2}$, solving gives $m=\\frac{4}{3}$"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P(x_{0},\\frac{5}{2})$ lies on the hyperbola. If the inradius of $\\Delta P F_{1} F_{2}$ is $1$, and the distance from the incenter $G$ to the origin $O$ is $\\sqrt{5}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;x0:Number;P: Point;F1: Point;F2: Point;G: Point;O: Origin;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;Coordinate(P)=(x0,5/2);PointOnCurve(P, C);Radius(InscribedCircle(TriangleOf(P,F1,F2)))=1;Center(InscribedCircle(TriangleOf(P,F1,F2)))=G;Distance(G, O) = sqrt(5)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3/2",
    "fact_spans": "[[[2, 58], [107, 110], [176, 179]], [[5, 58]], [[5, 58]], [[84, 106]], [[83, 106]], [[67, 74]], [[75, 82]], [[151, 154]], [[155, 160]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 82]], [[2, 82]], [[83, 106]], [[83, 113]], [[115, 147]], [[115, 154]], [[151, 174]]]",
    "query_spans": "[[[176, 185]]]",
    "process": "Let P be a point in the first quadrant, and let the points of tangency of the circle with $ F_{1}F_{2} $, $ PF_{1} $, $ PF_{2} $ be $ A' $, $ B $, $ D $, respectively. $ \\because |PF_{1}| - |PF_{2}| = 2a $, $ |PD| = |PB| $, $ |DF_{1}| = |A'F_{1}| $, $ |BF_{2}| = |A'F_{2}| $, it follows that $ |PD| + |DF_{1}| - |PB| - |BF_{2}| = |DF_{1}| - |BF_{2}| = |A'F_{1}| - |A'F_{2}| = 2a $, and $ |A'F_{1}| + |A'F_{2}| = 2c $, we obtain $ |A'F_{2}| = c - a $, so $ A $ coincides with $ A' $, hence $ |OA'| = |OA| = a $, thus $ \\sqrt{a^{2} + 1} = \\sqrt{5} $, i.e., $ a = 2 $. Also, the area $ S $ of $ \\triangle PF_{1}F_{2} $ is $ \\frac{1}{2} \\times \\frac{5}{2} \\times |2c| = \\frac{1}{2}(|F_{1}F_{2}| + |PF_{1}| + |PF_{2}|) \\times 1 $, $ \\therefore |PF_{1}| + |PF_{2}| = 3c $, $ \\because |PF_{1}| - |PF_{2}| = 2a $, $ \\therefore |PF_{1}| = \\frac{3c + 2a}{2} $, $ |PF_{2}| = \\frac{3c - 2a}{2} $, $ \\because |PF_{1}| = \\sqrt{(x_{0} + c)^{2} + \\frac{25}{4}} $, $ |PF_{2}| = \\sqrt{(x_{0} - c)^{2} + \\frac{25}{4}} $, solving simultaneously and simplifying yields $ x_{0} = 3 $. Substituting P into the hyperbola equation and solving simultaneously gives $ b = \\sqrt{5} $, $ c = \\sqrt{4 + 5} = 3 $, thus the eccentricity of the hyperbola is $ e = \\frac{c}{a} = \\frac{3}{2} $."
  },
  {
    "text": "What is the equation of the locus of the midpoints of the lines joining each point on the parabola $y^{2}=2 x$ to its focus?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);P: Point;PointOnCurve(P,G) = True",
    "query_expressions": "LocusEquation(MidPoint(LineSegmentOf(P,Focus(G))))",
    "answer_expressions": "y^2=(x-1)/4",
    "fact_spans": "[[[0, 14]], [[0, 14]], [], [[0, 17]]]",
    "query_spans": "[[[0, 32]]]",
    "process": "Let $ P(x_{0},y_{0}) $ be an arbitrary point on the parabola $ y^{2} = 2x $. The midpoint $ M(x,y) $ of the line segment joining point $ P $ and the focus $ F\\left(\\frac{1}{2},0\\right) $ satisfies $ \\frac{y_{0}}{0} $, i.e., $ 3_{0} = 2x - \\frac{1}{2} $, $ y_{0} = 2y $, and since $ y_{0}^{2} = 2x_{0} $, the trajectory equation of the midpoint $ M(x,y) $ is $ -\\frac{1}{2} $, i.e., $ y^{2} = x - \\frac{1}{4} $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, with left and right foci $F_{1}$ and $F_{2}$, respectively. A point $P$ on it satisfies $P F_{1}=5 P F_{2}$. Then, the distance from point $P$ to the right directrix is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);LineSegmentOf(P, F1) = 5*LineSegmentOf(P, F2)",
    "query_expressions": "Distance(P, RightDirectrix(G))",
    "answer_expressions": "8/5",
    "fact_spans": "[[[2, 41], [64, 65]], [[2, 41]], [[48, 55]], [[56, 63]], [[2, 63]], [[2, 63]], [[68, 71], [94, 98]], [[64, 71]], [[73, 92]]]",
    "query_spans": "[[[64, 107]]]",
    "process": "Let the distance from point P to the right directrix be d. According to the definition of the hyperbola, $ PF_{1} - PF_{2} = 8 $, solving gives $ PF_{2} = 2 $. By the second definition of the hyperbola, $ \\frac{PF_{2}}{d} = e = \\frac{5}{4} $, solving gives $ d = \\frac{8}{5} $."
  },
  {
    "text": "If one focus of the ellipse $\\frac{x^{2}}{k+4}+\\frac{y^{2}}{9}=1$ is $(0,-2)$, then $k=$?",
    "fact_expressions": "G: Ellipse;k: Number;H: Point;Expression(G) = (x^2/(k + 4) + y^2/9 = 1);Coordinate(H) = (0, -2);OneOf(Focus(G)) = H",
    "query_expressions": "k",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 40]], [[56, 59]], [[46, 54]], [[1, 40]], [[46, 54]], [[1, 54]]]",
    "query_spans": "[[[56, 61]]]",
    "process": "Since the foci of the ellipse are (0, -2), we have \\begin{cases}9>k+4>0\\\\2=\\sqrt{9-(k+4)}\\end{cases}\\Rightarrow k=1,"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, let the right focus be $F$. Draw a perpendicular from $F$ to an asymptote, with foot of the perpendicular at $P$. If the midpoint of segment $PF$ lies on this hyperbola, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;P: Point;F: Point;b: Number;a: Number;L: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;PointOnCurve(F,L);IsPerpendicular(L,Asymptote(G));FootPoint(L,Asymptote(G))=P;PointOnCurve(MidPoint(LineSegmentOf(P, F)), G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 59], [73, 74], [100, 103], [107, 110]], [[84, 87]], [[64, 67], [69, 72]], [[5, 59]], [[5, 59]], [], [[5, 59]], [[5, 59]], [[2, 59]], [[2, 67]], [[68, 80]], [[68, 80]], [[68, 87]], [[89, 104]]]",
    "query_spans": "[[[107, 116]]]",
    "process": ""
  },
  {
    "text": "Given that hyperbola $C_{1}$ and hyperbola $C_{2}$ share the same foci, the equation of $C_{1}$ is $\\frac{x^{2}}{3}-y^{2}=1$. If the inclination angle of one asymptote of $C_{2}$ is twice the inclination angle of one asymptote of $C_{1}$, then the equation of $C_{2}$ is?",
    "fact_expressions": "Focus(C1) = Focus(C2);C1: Hyperbola;C2: Hyperbola;Expression(C1) = (x^2/3 - y^2 = 1);Inclination(OneOf(Asymptote(C2))) = Inclination(OneOf(Asymptote(C1)))*2",
    "query_expressions": "Expression(C2)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[2, 28]], [[2, 12], [29, 36], [85, 92]], [[13, 23], [67, 74], [109, 116]], [[29, 65]], [[67, 107]]]",
    "query_spans": "[[[109, 121]]]",
    "process": "By the given condition, the foci of $ C_{1} $ are $ (\\pm2,0) $, so the foci of hyperbola $ C_{2} $ are $ (\\pm2,0) $, i.e., $ c=2 $. One asymptote of $ C_{1} $ is $ y=\\frac{\\sqrt{3}}{3}x $, with slope $ k=\\tan\\alpha=\\frac{\\sqrt{3}}{3} $, so the inclination angle $ \\alpha $ of one asymptote of $ C_{1} $ is $ \\frac{\\pi}{6} $. The inclination angle of one asymptote of $ C_{2} $ is twice that of one asymptote of $ C_{1} $, so the inclination angle of one asymptote of $ C_{2} $ is $ 2\\alpha=\\frac{\\pi}{3} $, with slope $ k=\\sqrt{3} $, i.e., one asymptote of $ C_{2} $ is $ y=\\sqrt{3}x=\\frac{b}{a}x $, so $ \\frac{b}{a}=\\sqrt{3} $. Since $ a^{2}+b^{2}=c^{2} $, solving gives $ a=1 $, $ b=\\sqrt{3} $. Therefore, the equation of $ C_{2} $ is $ x^{2}-\\frac{y^{2}}{3}=1 $.  \n[Note] This problem mainly examines the application of the standard equation of a hyperbola and its simple geometric properties. The key to solving lies in memorizing the standard equation of a hyperbola and reasonably applying its geometric properties, emphasizing reasoning and computational ability. It is a basic problem."
  },
  {
    "text": "The equation $\\frac{x^{2}}{2-k}+\\frac{y^{2}}{k-1}=1$ represents a hyperbola; then the range of real values for $k$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(2 - k) + y^2/(k - 1) = 1);k: Real",
    "query_expressions": "Range(k)",
    "answer_expressions": "{(-oo,1),(2,+oo)}",
    "fact_spans": "[[[43, 46]], [[0, 46]], [[48, 53]]]",
    "query_spans": "[[[48, 60]]]",
    "process": "\\because the equation \\frac{x^{2}}{2-k}+\\frac{y^{2}}{k-1}=1 represents a hyperbola, \\therefore (2-k)(k-1)<0 \\therefore k<1 or k>2 \\therefore the range of real number k is k<1 or k>2. Answer: k<1 or k>2. [Note] This problem examines the standard equation of a hyperbola; the key to solving it is determining that the coefficients of the squared terms in the standard equation of a hyperbola have opposite signs."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=8x$ is $F$, a line passing through $F$ intersects the parabola at points $A$ and $B$, and $\\overrightarrow{A F}=2 \\overrightarrow{F B}$, then $|\\overrightarrow{A F}|$=?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 8*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};VectorOf(A, F) = 2*VectorOf(F, B)",
    "query_expressions": "Abs(VectorOf(A, F))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 16], [32, 35]], [[29, 31]], [[36, 39]], [[20, 23], [25, 28]], [[40, 43]], [[2, 16]], [[2, 23]], [[24, 31]], [[29, 45]], [[47, 92]]]",
    "query_spans": "[[[94, 120]]]",
    "process": "It is easy to see that the focus of the parabola $ y^{2} = 8x $ is $ F(2,0) $, and the directrix is $ x = -2 $. As shown in the figure, let $ C $ be the midpoint of $ AF $, and draw perpendiculars from $ A, C, F, B $ to the directrix, with feet at $ M, Q, P, N $, respectively. By the definition of a parabola, $ |AM| = |AF| $, $ |BN| = |BF| $, so $ |AM| = 2|BN| $. Let $ |BN| = a $, then $ |AM| = 2a $. Also, $ |PF| = 4 $, so $ |CQ| = 8 - a $. Moreover, $ |PF| + |AM| = 2|CQ| $, that is, $ 4 + 2a = 2(8 - a) $, solving gives $ a = 3 $. Therefore, $ |\\overrightarrow{AF}| = 2 \\times 3 = 6 $."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the two foci of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{3}=1$, and let $P$ be a point on the ellipse in the first quadrant such that $\\frac{\\sin \\angle P F_{1} F_{2}+\\sin \\angle P F_{2} F_{1}}{\\sin \\angle F_{1} P F_{2}}=2$. Then the value of the positive number $m$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/3 + x^2/m = 1);m: Number;m>0;F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;Quadrant(P) = 1;PointOnCurve(P, G);(Sin(AngleOf(P, F1, F2)) + Sin(AngleOf(P, F2, F1)))/Sin(AngleOf(F1, P, F2)) = 2",
    "query_expressions": "m",
    "answer_expressions": "{4, 9/4}",
    "fact_spans": "[[[19, 56], [72, 74]], [[19, 56]], [[170, 175]], [[170, 175]], [[1, 8]], [[9, 16]], [[1, 61]], [[62, 65]], [[62, 77]], [[62, 77]], [[79, 168]]]",
    "query_spans": "[[[170, 179]]]",
    "process": "When the foci are on the x-axis, $2=\\frac{2a}{2c}=\\frac{2\\sqrt{m}}{2\\sqrt{m-3}}$, solving gives: $m=4$; when the foci are on the y-axis, $2=\\frac{2a}{2c}=\\frac{\\sqrt{2\\sqrt{3}}}{2\\sqrt{3-m}}$, solving gives: $m=\\frac{9}{4}$."
  },
  {
    "text": "The right focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ coincides with the focus of the parabola $y^{2}=8 \\sqrt{2} x$. Then the equations of the asymptotes of this hyperbola are?",
    "fact_expressions": "G: Hyperbola;b: Number;H: Parabola;Expression(G) = (x^2/4 - y^2/b^2 = 1);Expression(H) = (y^2 = 8*sqrt(2)*x);RightFocus(G) = Focus(H)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*x",
    "fact_spans": "[[[0, 42], [78, 81]], [[3, 42]], [[47, 70]], [[0, 42]], [[47, 70]], [[0, 75]]]",
    "query_spans": "[[[78, 90]]]",
    "process": "The focus of the parabola $ y^{2}=8\\sqrt{2}x $ is $ (2\\sqrt{2},0) $. From the given condition, we obtain for the hyperbola $ c=2\\sqrt{2} $, which gives $ 4+b^{2}=c^{2}=8 $, solving yields $ b=2 $, thus the equation of the hyperbola is $ x^{2}-y^{2}=4 $, and the asymptotes are $ y=\\pm x $."
  },
  {
    "text": "Given that the line $2x - 3y = 0$ is an asymptote of the hyperbola $C$: $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ $(a > 0, b > 0)$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;G: Line;Expression(G) = (2*x - 3*y = 0);OneOf(Asymptote(C)) = G",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(13)/3",
    "fact_spans": "[[[16, 77], [85, 88]], [[16, 77]], [[24, 77]], [[24, 77]], [[24, 77]], [[24, 77]], [[2, 15]], [[2, 15]], [[2, 83]]]",
    "query_spans": "[[[85, 94]]]",
    "process": "Since the line $2x-3y=0$ is an asymptote of the hyperbola $C: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$, we have $\\frac{b}{a}=\\frac{2}{3}$. Therefore, the eccentricity of $C$ is $e=\\frac{c}{a}=\\sqrt{1+(\\frac{b}{a})^{2}}=\\frac{\\sqrt{13}}{3}$."
  },
  {
    "text": "Draw a chord $P Q$ perpendicular to the real axis through one focus $F_{2}$ of a hyperbola, and let $F_{1}$ be the other focus. If $\\angle P F_{1} Q = \\frac{\\pi}{2}$, then the eccentricity $e$ of the hyperbola equals?",
    "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;OneOf(Focus(G)) = F1;OneOf(Focus(G)) = F2;Negation(F1=F2);P: Point;Q: Point;PointOnCurve(F2, LineSegmentOf(P, Q));IsPerpendicular(LineSegmentOf(P, Q), RealAxis(G));IsChordOf(LineSegmentOf(P, Q), G);AngleOf(P, F1, Q) = pi/2;e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(2)+1",
    "fact_spans": "[[[1, 4], [79, 82]], [[31, 38]], [[9, 16]], [[1, 43]], [[1, 16]], [[1, 43]], [[24, 29]], [[24, 29]], [[0, 29]], [[1, 29]], [[1, 29]], [[45, 77]], [[86, 89]], [[79, 89]]]",
    "query_spans": "[[[86, 92]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the focus of the parabola $y=\\frac{1}{4} x^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2/4)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,1)",
    "fact_spans": "[[[0, 24]], [[0, 24]]]",
    "query_spans": "[[[0, 31]]]",
    "process": ""
  },
  {
    "text": "The ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ has two foci $F_{1}$, $F_{2}$, and $P$ is a point on $C$. Then $|P F_{1}|+|P F_{2}|$=?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/4 + y^2/3 = 1);Focus(C)={F1,F2};PointOnCurve(P,C)",
    "query_expressions": "Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 42], [67, 70]], [[63, 66]], [[47, 54]], [[55, 62]], [[0, 42]], [[0, 62]], [[63, 74]]]",
    "query_spans": "[[[76, 99]]]",
    "process": "In ellipse C, a=2, by the definition of an ellipse, |PF_{1}|+|PF_{2}|=2a=4"
  },
  {
    "text": "The equation of the hyperbola passing through the point $(\\sqrt{2},-2)$ and having common asymptotes with the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$ is?",
    "fact_expressions": "G: Hyperbola;C:Hyperbola;H: Point;Expression(G) = (x^2/2 - y^2 = 1);Coordinate(H) = (sqrt(2), -2);Asymptote(G)=Asymptote(C);PointOnCurve(H,C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2/3 - x^2/6 = 1",
    "fact_spans": "[[[19, 47]], [[54, 57]], [[1, 17]], [[19, 47]], [[1, 17]], [[18, 57]], [[0, 57]]]",
    "query_spans": "[[[54, 61]]]",
    "process": "Analysis: Let the equation of the desired hyperbola be $\\frac{x^{2}}{2}-y^{2}=m$ $(m\\neq0)$; substitute $(\\sqrt{2},2)$ to find $m$. Specifically, let the hyperbola equation be $\\frac{x^{2}}{2}-y^{2}=m$ $(m\\neq0)$; substituting $(\\sqrt{2},2)$ yields $m=-3$, hence the desired hyperbola equation is $\\frac{y^{2}}{3}-\\frac{x^{2}}{6}=1$."
  },
  {
    "text": "Let $P$ be a moving point on the parabola $y^{2}=4x$. Then the minimum value of the sum of the distance from point $P$ to point $A(-1,1)$ and the distance from point $P$ to the line $x=-1$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;P: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (x = -1);Coordinate(A) = (-1, 1);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P,A)+Distance(P,H))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[5, 19]], [[51, 59]], [[32, 42]], [[1, 4], [27, 31], [46, 50]], [[5, 19]], [[51, 59]], [[32, 42]], [[1, 25]]]",
    "query_spans": "[[[27, 70]]]",
    "process": "Let the focus of the parabola $ y^{2} = 4x $ be $ F(1,0) $. Since the line $ x = -1 $ is the directrix of this parabola, the distance from point $ P $ to the line $ x = -1 $ equals $ PF $. Therefore, when points $ A $, $ P $, and $ F $ are collinear, the sum of the distance from point $ P $ to point $ A(-1,1) $ and the distance from point $ P $ to the line $ x = -1 $ is minimized. The minimum value is $ AF = \\sqrt{(1 - (-1))^{2} + (0 - 1)^{2}} = \\sqrt{5} $."
  },
  {
    "text": "It is known that the foci and the endpoints of the real axis of hyperbola $C$ are exactly the endpoints of the major axis and the foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, respectively. Then the equation of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);Focus(C) = Endpoint(MajorAxis(G));Endpoint(RealAxis(C)) = Focus(G)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/9 - y^2/16 = 1",
    "fact_spans": "[[[2, 8], [69, 75]], [[19, 58]], [[19, 58]], [[2, 67]], [[2, 67]]]",
    "query_spans": "[[[69, 80]]]",
    "process": "The ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ has vertices at $(\\pm5,0)$ and foci at $(\\pm3,0)$, so the hyperbola has foci at $(\\pm5,0)$ and real axis endpoints at $(\\pm3,0)$. Let the equation of the hyperbola be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, thus $c=5$, $a=3$, $b=4$, so the asymptotes are: $y=\\pm\\frac{4}{3}x$, or equivalently: $4x\\pm3y=0$."
  },
  {
    "text": "Given that the vertex of the parabola is at the origin, the axis of symmetry is the $x$-axis, and the distance from point $P(-5,2 \\sqrt{5})$ to the focus is $6$, then its standard equation is?",
    "fact_expressions": "G: Parabola;P: Point;O: Origin;Coordinate(P) = (-5, 2*sqrt(5));Vertex(G) = O;SymmetryAxis(G) = xAxis;Distance(P, Focus(G)) = 6",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=-4*x,y^2=-36*x}",
    "fact_spans": "[[[2, 5], [51, 52]], [[20, 39]], [[8, 10]], [[20, 39]], [[2, 10]], [[2, 19]], [[2, 49]]]",
    "query_spans": "[[[51, 58]]]",
    "process": "Let the focus be $ F(a,0) $, $ |PF| = \\sqrt{(a+5)^{2}+20} = 6 $, which gives $ a^{2}+10a+9=0 $. Solving yields $ a = -1 $ or $ a = -9 $. When the focus is $ F(-1,0) $, the parabola opens to the left, and its equation is $ y^{2} = -4x $. When the focus is $ F(-9,0) $, the parabola opens to the left, and its equation is $ y^{2} = -36x $."
  },
  {
    "text": "If a hyperbola passes through the point $(1, \\sqrt{3})$ and its asymptotes are given by $y = \\pm 2x$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (1, sqrt(3));PointOnCurve(H, G);Expression(Asymptote(G)) = (y = pm*(2*x))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "4*x^2-y^2=1",
    "fact_spans": "[[[1, 4], [43, 46], [23, 24]], [[6, 22]], [[6, 22]], [[1, 22]], [[23, 41]]]",
    "query_spans": "[[[43, 51]]]",
    "process": "From the given conditions, ① if the foci of the hyperbola lie on the x-axis, then we can set $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$, so $\\frac{1}{a^{2}}-\\frac{3}{b^{2}}=1$ and $\\frac{b}{a}=2$. Solving these equations simultaneously gives $a=\\frac{1}{2}, b=1$, so the standard equation of the hyperbola is $4x^{2}-y^{2}=1$. ② If the foci of the hyperbola lie on the y-axis, then we can set $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0,b>0)$, so $\\frac{3}{a^{2}}-\\frac{1}{b^{2}}=1$ and $\\frac{a}{b}=2$, but this system has no solution. Therefore, the equation of the hyperbola is $4x^{2}-y^{2}=1$."
  },
  {
    "text": "Given that point $P$ lies on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and $F_{1}$, $F_{2}$ are the left and right foci of the ellipse respectively. It is known that $\\angle F_{1} P F_{2}=120^{\\circ}$ and $|P F_{1}|=3|P F_{2}|$. Find the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;P: Point;PointOnCurve(P, G);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;AngleOf(F1, P, F2) = ApplyUnit(120, degree);Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(13)/4",
    "fact_spans": "[[[7, 59], [82, 84], [153, 155]], [[7, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[2, 6]], [[2, 63]], [[64, 71]], [[72, 79]], [[64, 90]], [[64, 90]], [[93, 127]], [[129, 151]]]",
    "query_spans": "[[[153, 161]]]",
    "process": "Let |PF₂| = x, |PF₁| = 3x, 2a = 4x. By the law of cosines, (2c)² = 13x², so c/a = √13/4. Hence, fill in √13/4."
  },
  {
    "text": "If the two foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ are $F_{1}$, $F_{2}$, and $P$ is a point on the hyperbola such that $|P F_{1}|=3|P F_{2}|$, then the range of values for the eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Focus(G)={F1,F2};PointOnCurve(P,G);Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, F2));Eccentricity(G)=e;e:Number",
    "query_expressions": "Range(e)",
    "answer_expressions": "(1,2]",
    "fact_spans": "[[[1, 57], [85, 88], [118, 121]], [[4, 57]], [[4, 57]], [[81, 84]], [[65, 72]], [[73, 80]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 80]], [[81, 91]], [[93, 115]], [[118, 127]], [[124, 127]]]",
    "query_spans": "[[[124, 134]]]",
    "process": "According to the given conditions,  \n\\begin{cases}|PF_{1}|=3|PF_{2}|\\\\|PF_{1}|-|PF_{2}|=2a\\end{cases}  \nSolving this yields |PF_{2}|=a, |PF_{1}|=3a.  \nSince |PF_{1}|+|PF_{2}|\\geqslant|F_{1}F_{2}|, we have 4a\\geqslant2c, i.e., c\\leqslant2a, so e=\\frac{c}{a}\\leqslant2.  \nAlso, e>1, therefore the range of eccentricity e is (1,2].  \nAnswer: (1,2)."
  },
  {
    "text": "Draw a line $l$ through the focus of the parabola $y^{2}=4x$, intersecting the parabola at points $A$ and $B$. If the x-coordinate of the midpoint of segment $AB$ is $3$, then $|AB|$ equals?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);l: Line;PointOnCurve(Focus(G), l);A: Point;B: Point;Intersection(l, G) = {A, B};XCoordinate(MidPoint(LineSegmentOf(A, B))) = 3",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 15], [26, 29]], [[1, 15]], [[19, 24]], [[0, 24]], [[30, 33]], [[34, 37]], [[19, 39]], [[41, 58]]]",
    "query_spans": "[[[60, 70]]]",
    "process": ""
  },
  {
    "text": "Let the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ be $F$, and the left and right vertices be $A_{1}$, $A_{2}$, respectively. Draw a line through point $F$ perpendicular to the $x$-axis, intersecting the hyperbola at points $B$ and $C$. If $A_{1} B \\perp A_{2} C$, then the slope of the asymptotes of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;A1: Point;B: Point;A2: Point;C: Point;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;LeftVertex(G) = A1;RightVertex(G) = A2;L:Line;PointOnCurve(F,L);IsPerpendicular(L,xAxis);Intersection(L,G)={B,C};IsPerpendicular(LineSegmentOf(A1, B), LineSegmentOf(A2, C))",
    "query_expressions": "Slope(Asymptote(G))",
    "answer_expressions": "pm*1",
    "fact_spans": "[[[1, 57], [104, 107], [146, 149]], [[4, 57]], [[4, 57]], [[74, 81]], [[109, 112]], [[82, 89]], [[113, 116]], [[62, 65], [91, 95]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 65]], [[1, 89]], [[1, 89]], [], [[90, 103]], [[90, 103]], [[90, 118]], [[120, 143]]]",
    "query_spans": "[[[146, 158]]]",
    "process": "Assume point B is in the first quadrant, then A_{1}(-a,0), B(c,\\frac{b^{2}}{a}), A_{2}(a,0), C(c,-\\frac{b^{2}}{a}), so \\overrightarrow{A_{1}B}=(a+c,\\frac{b^{2}}{a}), \\overrightarrow{A_{2}C}=(c-a,-\\frac{b^{2}}{a}) B_{1}B=(a+c,\\frac{a}{a}), A_{2}C=(c-a,-\\frac{a}{a}) so the slope of the asymptote of this hyperbola is k=+\\frac{b}{1}=+1. The answer is: +1"
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $2$, and one of its foci coincides with the focus of the parabola $y^{2}=8x$, then what are the coordinates of the foci of the hyperbola? What is the equation of its asymptotes?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 8*x);Eccentricity(G) = 2;OneOf(Focus(G))=Focus(H)",
    "query_expressions": "Coordinate(Focus(G));Expression(Asymptote(G))",
    "answer_expressions": "(pm*2,0)\nsqrt(3)*x+pm*y=0",
    "fact_spans": "[[[2, 48], [57, 58], [86, 89]], [[5, 48]], [[5, 48]], [[64, 78]], [[2, 48]], [[64, 78]], [[2, 56]], [[57, 83]]]",
    "query_spans": "[[[86, 96]], [[86, 103]]]",
    "process": ""
  },
  {
    "text": "Given that the hyperbola $\\Gamma$ passes through the point $(2, \\sqrt{3})$ and has the same asymptotes as the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, then the standard equation of the hyperbola $\\Gamma$ is?",
    "fact_expressions": "Gamma: Hyperbola;A:Point;Coordinate(A)=(2,sqrt(3));PointOnCurve(A,Gamma);C:Hyperbola;Expression(C) = (x^2/4 - y^2 = 1);Asymptote(Gamma) = Asymptote(C)",
    "query_expressions": "Expression(Gamma)",
    "answer_expressions": "y^2/2 - x^2/8 = 1",
    "fact_spans": "[[[2, 13], [70, 81]], [[14, 30]], [[14, 30]], [[2, 30]], [[33, 61]], [[33, 61]], [[2, 68]]]",
    "query_spans": "[[[70, 88]]]",
    "process": "Let the hyperbola equation that has the same asymptotes as $\\frac{x^2}{4}-y^{2}=1$ be $\\frac{x^{2}}{4}-y^{2}=\\lambda\\ (\\lambda\\neq0)$. Substituting the point $(2,\\sqrt{3})$ into the equation gives $\\frac{4}{4}-3=\\lambda$, so $\\lambda=-2$. Then the required hyperbola equation is $\\frac{y^{2}}{2}-\\frac{x^{2}}{8}=1$."
  },
  {
    "text": "For any two points $P$, $Q$ on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, if $O P \\perp O Q$, then the minimum value of the product $|O P| \\cdot|O Q|$ is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;O: Origin;P: Point;Q: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P,G);PointOnCurve(Q,G);IsPerpendicular(LineSegmentOf(O, P), LineSegmentOf(O, Q))",
    "query_expressions": "Min(Abs(LineSegmentOf(O, P))*Abs(LineSegmentOf(O, Q)))",
    "answer_expressions": "2*a^2*b^2/a^2+b^2",
    "fact_spans": "[[[0, 52]], [[2, 52]], [[2, 52]], [[66, 81]], [[57, 60]], [[61, 64]], [[2, 52]], [[2, 52]], [[0, 52]], [[0, 64]], [[0, 64]], [[66, 81]]]",
    "query_spans": "[[[85, 109]]]",
    "process": "Let $ P(|OP|\\cos\\theta,|OP|\\sin\\theta) $, $ Q(|OQ|\\cos(\\theta\\pm\\frac{\\pi}{2}),|OQ|\\sin(\\theta\\pm\\frac{\\pi}{2})) $. Since $ P, Q $ lie on the ellipse, we have $ \\frac{\\cos^{2}\\theta}{a^{2}}+\\frac{\\sin^{2}\\theta}{b^{2}} \\textcircled{1} \\frac{1}{1^{2}} = \\frac{\\sin^{2}\\theta}{a^{2}}+\\frac{\\cos^{2}\\theta}{b^{2}} \\textcircled{2} \\frac{1}{20}|^{2} = \\frac{1}{a^{2}}+\\frac{1}{b^{2}} $. Thus, when $ |OP| = |OQ| = \\sqrt{\\frac{2a^{2}b^{2}}{a^{2}+b^{2}}} $, $ |OP||OQ| $ reaches the minimum value $ \\frac{2a^{2}b^{2}}{a^{2}+b^{2}} $."
  },
  {
    "text": "Let $F$ be the focus of the parabola $y^{2}=4 x$, and let $A$, $B$, $C$ be three points on this parabola. If $\\overrightarrow{F A}+\\overrightarrow{F B}+\\overrightarrow{F C}=0$, then $|\\overrightarrow{F A}|+| \\overrightarrow{F B}|+| \\overrightarrow{F C} |$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;A: Point;B: Point;C: Point;PointOnCurve(A, G);PointOnCurve(B, G);PointOnCurve(C, G);VectorOf(F, A) + VectorOf(F, B) + VectorOf(F, C) = 0",
    "query_expressions": "Abs(VectorOf(F, A)) + Abs(VectorOf(F, B)) + Abs(VectorOf(F, C))",
    "answer_expressions": "6",
    "fact_spans": "[[[5, 19], [40, 43]], [[5, 19]], [[1, 4]], [[1, 22]], [[23, 26]], [[29, 32]], [[35, 38]], [[23, 46]], [[23, 46]], [[23, 46]], [[48, 114]]]",
    "query_spans": "[[[116, 191]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4 a^{2}}=1  (a>0)$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/a^2 - y^2/(4*a^2) = 1);a: Number;a>0",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[0, 55]], [[0, 55]], [[3, 55]], [[3, 55]]]",
    "query_spans": "[[[0, 61]]]",
    "process": ""
  },
  {
    "text": "It is known that the center of the hyperbola is at the origin $O$, the foci lie on the $x$-axis, its conjugate axis has length $2$, and the focal distance is twice the distance between the two directrices. Then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;O: Origin;Center(G)=O;PointOnCurve(Focus(G), xAxis);Length(ImageinaryAxis(G)) = 2;L1:Line;L2:Line;Directrix(G)={L1,L2};FocalLength(G)=2*Distance(L1,L2)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 = 1",
    "fact_spans": "[[[2, 5], [24, 25], [52, 55]], [[9, 14]], [[2, 14]], [[2, 23]], [[24, 33]], [], [], [[24, 41]], [[24, 49]]]",
    "query_spans": "[[[52, 60]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the foci of the ellipse $\\frac{x^{2}}{3}+y^{2}=1$ are?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/3 + y^2 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(pm*sqrt(2),0)",
    "fact_spans": "[[[0, 27]], [[0, 27]]]",
    "query_spans": "[[[0, 34]]]",
    "process": "From the given conditions, we have $ a = \\sqrt{3}, b = 1, \\therefore c = \\sqrt{a^{2} - b^{2}} = \\sqrt{3 - 1} = \\sqrt{2} $. Therefore, the coordinates of the foci of the ellipse $ \\frac{x^{2}}{3} + y^{2} = 1 $ are $ (\\pm \\sqrt{2}, 0) $."
  },
  {
    "text": "Given that $F$ is the left focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, $A(1,4)$, and $P$ is a moving point on the right branch of the hyperbola, then the minimum value of $|P F|+|P A|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/12 = 1);F: Point;LeftFocus(G) = F;A: Point;Coordinate(A) = (1, 4);P: Point;PointOnCurve(P, RightPart(G))",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "9",
    "fact_spans": "[[[6, 45], [64, 67]], [[6, 45]], [[2, 5]], [[2, 49]], [[50, 58]], [[50, 58]], [[60, 63]], [[60, 73]]]",
    "query_spans": "[[[75, 94]]]",
    "process": "Draw the graph, let the right focus of the hyperbola be M. According to the definition of the hyperbola, we have |PF| = 4 + |PM|. Thus, |PF| + |PA| = |PM| + |PA| + 4. The minimum value of |PF| + |PA| occurs when points A, P, and M are collinear, which gives the solution. For the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, we have $a=2$, $b=2\\sqrt{3}$, $c=4$, as shown in the figure below. By the definition of the hyperbola, |PF| - |PM| = 4, so |PF| = 4 + |PM|. Therefore,  \n|PF| + |PA| = |PM| + |PA| + 4 \\geqslant |AM| + 4 = \\sqrt{(1-4)^{2}+(4-0)^{2}} + 4 = 9.  \nThe equality holds if and only if points A, P, and M are collinear. Hence, the minimum value of |PF| + |PA| is 9."
  },
  {
    "text": "Given that $F$ is the right focus of the ellipse $C$: $\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1$, $P$ is a point on the ellipse $C$, and $A(1,2 \\sqrt{2})$, then the maximum value of $|P A|+|P F|$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/3 + y^2/2 = 1);F: Point;RightFocus(C) = F;P: Point;PointOnCurve(P, C);A: Point;Coordinate(A) = (1, 2*sqrt(2))",
    "query_expressions": "Max(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[6, 48], [57, 62]], [[6, 48]], [[2, 5]], [[2, 52]], [[53, 56]], [[53, 65]], [[66, 83]], [[66, 83]]]",
    "query_spans": "[[[85, 104]]]",
    "process": "The standard equation can represent points, and by specializing |PF|, then constructing an inequality relationship based on the triangle formation principle to find the maximum. According to the problem, let the left focus of the ellipse be F, and the equation of the ellipse be \\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1, where a=\\sqrt{3}, and P is a point on the ellipse C, then |PF|+|PF|=2a=2\\sqrt{3}, so c=\\sqrt{3-2}=1, thus F(1,0), F(-1,0), then |PF|=2a-|PF|=2\\sqrt{3}-|PF|, and clearly point A lies inside the ellipse, so |PA|+|PF|=|PA|+2\\sqrt{3}-|PF|=2\\sqrt{3}+|PA|-|PF|. It can be analyzed that: |PA|-|PF|\\leqslant|AF|=2\\sqrt{3}, (triangle formation principle) equality holds when points P, A, F are collinear, then the maximum value of |PA|+|PF| is 4\\sqrt{3}."
  },
  {
    "text": "Given: The eccentricity of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{k}=1$ is $e=\\frac{\\sqrt{10}}{5}$, then the value of the real number $k$ is?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (x^2/5 + y^2/k = 1);Eccentricity(G) = e ;e: Number;e= sqrt(10)/5",
    "query_expressions": "k",
    "answer_expressions": "25/3, 3",
    "fact_spans": "[[[4, 41]], [[70, 75]], [[4, 41]], [[4, 68]], [[45, 68]], [[45, 68]]]",
    "query_spans": "[[[70, 79]]]",
    "process": "When $ K>5 $, solve for $ K $ using $ e=\\frac{c}{a}=\\sqrt{\\frac{k-5}{k}}=\\frac{\\sqrt{10}}{5} $. When $ 0<K<5 $, solve for $ K $ using $ e=\\frac{c}{a}\\sqrt{\\frac{5-k}{5}}=\\frac{\\sqrt{10}}{5} $. When $ K>5 $, $ e=\\frac{c}{a}\\sqrt{\\frac{k-5}{k}}=\\frac{\\sqrt{10}}{5} $, $ K=\\frac{25}{3} $. In conclusion, $ K=\\frac{25}{3} $ or $ 3 $."
  },
  {
    "text": "The coordinates of the foci of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(pm*2, 0)",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 35]]]",
    "process": "According to the hyperbola equation, find $ c^{2} $ and then directly determine the coordinates of the foci. From the given conditions, $ a^{2}=3 $, $ b^{2}=1 $, then $ c^{2}=a^{2}+b^{2}=4 $, so $ c=2 $, and the foci lie on the x-axis; therefore, the coordinates of the foci of the hyperbola are $ (2,0) $, $ (-2,0) $."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=16 x$, $P$ is a point on $C$, and the extension of $P F$ intersects the $y$-axis at point $Q$. If $\\overrightarrow{P F}=\\frac{2}{3} \\overrightarrow{P Q}$, then $|F Q|$=?",
    "fact_expressions": "C: Parabola;P: Point;F: Point;Q: Point;Expression(C) = (y^2 = 16*x);Focus(C) = F;PointOnCurve(P, C);Intersection(OverlappingLine(LineSegmentOf(P,F)),yAxis)=Q;VectorOf(P,F)=(2/3)*VectorOf(P,Q)",
    "query_expressions": "Abs(LineSegmentOf(F, Q))",
    "answer_expressions": "8",
    "fact_spans": "[[[6, 26], [34, 37]], [[30, 33]], [[2, 5]], [[56, 60]], [[6, 26]], [[2, 29]], [[30, 40]], [[41, 60]], [[61, 116]]]",
    "query_spans": "[[[118, 127]]]",
    "process": "From the parabola $ C: y^{2} = 16x $, we know $ F(4,0) $, so $ OF = 4 $ ($ O $ is the origin). Draw a perpendicular from point $ P $ to the $ y $-axis, with foot $ N $. By similar triangles, we have $ \\frac{|OF|}{|PN|} = \\frac{|FQ|}{|QP|} = \\frac{1}{3} $, so $ |PN| = 3|FO| = 12 $. Hence, $ |PF| = 12 + 4 = 16 $, and therefore $ |FQ| = \\frac{1}{2}|PF| = 8 $."
  },
  {
    "text": "The distance from a point on the hyperbola $\\frac{x^{2}}{25}-\\frac{y^{2}}{9}=1$ to one focus is $12$, then the distance to the other focus is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/25 - y^2/9 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;OneOf(Focus(G)) = F1;OneOf(Focus(G)) = F2;Negation(F1=F2);Distance(P, F1) = 12",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "{2, 22}",
    "fact_spans": "[[[0, 39]], [[0, 39]], [[41, 42]], [[0, 42]], [], [], [[0, 47]], [[0, 63]], [[0, 63]], [[0, 55]]]",
    "query_spans": "[[[0, 68]]]",
    "process": "Let the left and right foci of the hyperbola $\\frac{x^{2}}{25}-\\frac{y^{2}}{9}=1$ be $F_{1}, F_{2}$ respectively. Using the definition of a hyperbola, $||PF_{1}|-|PF_{2}||=2a=10$, the answer can be obtained. [Solution] Let the left and right foci of the hyperbola be $F_{1}, F_{2}$ respectively, then $a=5$, $b=3$, $c=\\sqrt{34}$. Without loss of generality, let $|PF_{1}|=12$ ($12>a+c=5+\\sqrt{34}$). Therefore, point $P$ may lie on the left branch or the right branch. From $||PF_{1}|-|PF_{2}||=2a=10$, we get: $|12-|PF_{2}||=10$, so $|PF_{2}|=22$ or $2$. Therefore, the distance from point $P$ to the other focus is $22$ or $2$. The answer is: $2$ or $22$. [Comment] This problem examines the simple properties of hyperbolas; careful reading and accurate, rigorous reasoning are required in solving it."
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{2}=1$ with foci on the $y$-axis is $\\frac{1}{2}$. Then the value of $m$ is?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (y^2/2 + x^2/m = 1);PointOnCurve(Focus(G), yAxis);Eccentricity(G) = 1/2",
    "query_expressions": "m",
    "answer_expressions": "32",
    "fact_spans": "[[[9, 46]], [[66, 69]], [[9, 46]], [[0, 46]], [[9, 64]]]",
    "query_spans": "[[[66, 73]]]",
    "process": "Given the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{2}=1$ has eccentricity $\\frac{1}{2}$. Since the foci of the ellipse lie on the $y$-axis, $0<m<2$. Therefore, $a=\\sqrt{2}$, $b=\\sqrt{m}$, $c=\\sqrt{2-m}$. Thus, $e=\\frac{c}{a}=\\frac{\\sqrt{2-m}}{\\sqrt{2}}=\\frac{1}{2}$. Solving gives: $m=\\frac{3}{3}$"
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$. Point $P$ moves on the ellipse. What is the maximum value of $|P F_{1}| \\cdot |P F_{2}|$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)))",
    "answer_expressions": "4",
    "fact_spans": "[[[16, 43], [55, 57]], [[50, 54]], [[0, 7]], [[8, 15]], [[16, 43]], [[0, 49]], [[0, 49]], [[50, 60]]]",
    "query_spans": "[[[62, 95]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{m^{2}}=1$ $(0<m<3)$ is equal to $\\frac{\\sqrt{3}}{3}$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;m: Number;0 < m;m < 3;Expression(G) = (x^2/9 + y^2/m^2 = 1);Eccentricity(G) = sqrt(3)/3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9+y^2/6=1",
    "fact_spans": "[[[0, 48], [76, 78]], [[2, 48]], [[2, 48]], [[2, 48]], [[0, 48]], [[0, 74]]]",
    "query_spans": "[[[76, 85]]]",
    "process": "According to the basic concept of an ellipse, combining the given conditions we obtain $ a=3 $, $ c=\\sqrt{3} $, thus getting $ b^{2}=6 $. Then, based on the position of the foci of the ellipse, the standard equation of this ellipse can be determined. Solution: $ \\because $ the ellipse $ \\frac{x^{2}}{9}+\\frac{y^{2}}{m^{2}}=1 $ ($ 0<m<3 $) has a major axis of length 6 and eccentricity $ \\frac{\\sqrt{3}}{3} $, with foci on the x-axis, $ \\therefore 2a=6 $, $ e=\\frac{c}{a}=\\frac{\\sqrt{3}}{3} $, solving gives $ a=3 $, $ c=\\sqrt{3} $, $ b^{2}=a^{2}-c^{2}=6 $. Since the foci of the ellipse lie on the x-axis, its equation is $ \\frac{x^{2}}{9}+\\frac{y^{2}}{6}=1 $."
  },
  {
    "text": "Given that point $A(-2 , 3)$ lies on the directrix of the parabola $C$: $y^{2}=2 p x$, and denoting the focus of $C$ as $F$, what is the slope of line $A F$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;A: Point;Coordinate(A) = (-2, 3);PointOnCurve(A, Directrix(C));F: Point;Focus(C) = F",
    "query_expressions": "Slope(LineOf(A, F))",
    "answer_expressions": "-3/4",
    "fact_spans": "[[[15, 36], [42, 45]], [[15, 36]], [[22, 36]], [[2, 14]], [[2, 14]], [[2, 40]], [[49, 52]], [[42, 52]]]",
    "query_spans": "[[[54, 66]]]",
    "process": "Since point A(-2,3) lies on the directrix of the parabola $ C: y^2 = 2px $, we have $ \\frac{p}{2} = 2 $, so $ p = 4 $, and $ F(2,0) $. Therefore, the slope of line AF is $ \\frac{3}{-2 - 2} = -\\frac{3}{4} $."
  },
  {
    "text": "The equation of the hyperbola that shares the same asymptotes as $\\frac{x^{2}}{5}-\\frac{y^{2}}{20}=1$ and passes through the point $(-5,2 \\sqrt{15})$ is?",
    "fact_expressions": "G: Hyperbola;C:Hyperbola;H: Point;Expression(G) = (x^2/5 - y^2/20 = 1);Coordinate(H) = (-5, 2*sqrt(15));PointOnCurve(H, C);Asymptote(G)=Asymptote(C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/10 - y^2/40 = 1",
    "fact_spans": "[[[1, 40]], [[71, 74]], [[51, 70]], [[1, 40]], [[51, 70]], [[49, 74]], [[0, 74]]]",
    "query_spans": "[[[71, 78]]]",
    "process": "From the given conditions, the asymptotes are $ y = \\pm 2x $. Let the desired hyperbola equation be $ 4x^{2} - y^{2} = \\lambda $, $ \\lambda \\in \\mathbb{R} $. Since $ (-5, 2\\sqrt{15}) $ lies on the hyperbola, then $ \\lambda = 4 \\times (-5)^{2} - (2\\sqrt{15})^{2} = 100 - 60 = 40 $. The desired hyperbola equation is $ \\frac{x^{2}}{10} - \\frac{y^{2}}{40} = 1 $."
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{2}=1$ are $F_{1}$, $F_{2}$, and point $P$ lies on the ellipse. If $|P F_{1}|=4$, what is the measure of $\\angle F_{1} P F_{2}$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 + y^2/2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 4",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "ApplyUnit(120,degree)",
    "fact_spans": "[[[0, 37], [62, 64]], [[57, 61]], [[41, 48]], [[49, 56]], [[0, 37]], [[0, 56]], [[57, 65]], [[67, 80]]]",
    "query_spans": "[[[83, 110]]]",
    "process": ""
  },
  {
    "text": "From a point $A(x_{0}, y_{0})$ $(x_{0}>\\frac{p}{2})$ on the parabola $y^{2}=2 p x$ $(p>0)$ with focus $F$, draw a perpendicular to its directrix, and let the foot of the perpendicular be $B$. If $|A F|=4$ and the distance from $B$ to the line $A F$ is $\\sqrt{7}$, then the equation of this parabola is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;F: Point;p>0;x0:Number;y0:Number;x0>p/2;Expression(G) = (y^2 = 2*(p*x));Coordinate(A) = (x0,y0);Focus(G)=F;PointOnCurve(A,G);L:Line;PointOnCurve(A,L);IsPerpendicular(L,Directrix(G));FootPoint(L,Directrix(G))=B;Abs(LineSegmentOf(A, F)) = 4 ;Distance(B,LineOf(A, F)) = sqrt(7)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=2*x",
    "fact_spans": "[[[8, 31], [72, 73], [126, 129]], [[11, 31]], [[35, 71]], [[82, 85], [98, 101]], [[4, 7]], [[11, 31]], [[35, 71]], [[35, 71]], [[35, 71]], [[8, 31]], [[35, 71]], [[1, 31]], [[8, 71]], [], [[0, 78]], [[0, 78]], [[0, 85]], [[88, 97]], [[98, 123]]]",
    "query_spans": "[[[126, 134]]]",
    "process": "Draw BC ⊥ AF from B at point C, then BC = \\sqrt{7}, since |AF| = |AB| = 4, so \\sin\\angle BAF = \\frac{\\sqrt{7}}{4}, then \\cos\\angle BAF = \\frac{3}{4}. Draw FD ⊥ AB from F at point D, then |AD| = |AF|\\cos\\angle BAF = 3, so p + |AD| = 4, yielding p = 1, thus v2 = 2x."
  },
  {
    "text": "$P$ is a moving point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$. Draw $P D \\perp y$-axis, where $D$ is the foot of the perpendicular. Then the equation of the locus of the midpoint of $P D$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/9 = 1);P: Point;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P,D), yAxis) = True;FootPoint(LineSegmentOf(P, D), yAxis) = D;D: Point",
    "query_expressions": "LocusEquation(MidPoint(LineSegmentOf(P, D)))",
    "answer_expressions": "x^2/4+y^2/9=1",
    "fact_spans": "[[[4, 42]], [[4, 42]], [[0, 3]], [[0, 46]], [[48, 62]], [[48, 69]], [[63, 66]]]",
    "query_spans": "[[[71, 85]]]",
    "process": "Let the coordinates of point P be $(x_{0},y_{0})$, then $\\frac{x_{0}^{2}}{16}+\\frac{y_{0}^{2}}{9}=1$. Since $PD\\bot y$-axis and D is the foot of the perpendicular, then $D(0,y_{0})$. Let M$(x,y)$ be the midpoint of PD, then $\\begin{cases}x=\\frac{x_{0}}{2}\\\\y=y_{0}\\end{cases}$, which gives $\\begin{cases}x_{0}=2x\\\\y_{0}=y\\end{cases}$. Substituting $\\begin{cases}x_{0}=2x\\\\y_{0}=y\\end{cases}$ into the equation $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ yields $\\frac{(2x)^{2}}{16}+\\frac{y^{2}}{9}=1$, that is, $\\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$."
  },
  {
    "text": "Let $P(x_1, y_1)$ be a point on the curve $C$: $\\sqrt{\\frac{x^{2}}{25}}+\\sqrt{\\frac{y^{2}}{9}}=1$, and let $F_{1}(-4,0)$, $F_{2}(4,0)$. Then the maximum value of $|P F_{1}|+|P F_{2}|$ is?",
    "fact_expressions": "C: Curve;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/25 + y^2/9 = 1);Coordinate(P) = (x1, y1);Coordinate(F1) = (-4, 0);Coordinate(F2)= (4, 0);PointOnCurve(P, C);x1:Number;y1:Number",
    "query_expressions": "Max(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)))",
    "answer_expressions": "10",
    "fact_spans": "[[[13, 70]], [[1, 12]], [[74, 87]], [[89, 102]], [[13, 70]], [[1, 12]], [[74, 87]], [[89, 102]], [[1, 73]], [[1, 12]], [[1, 12]]]",
    "query_spans": "[[[104, 131]]]",
    "process": "The equation of curve $ C $ is $ \\frac{|x|}{5} + \\frac{|y|}{3} = 1 $. The graphs of the ellipse $ \\frac{x^{2}}{25} + \\frac{y^{2}}{9} = 1 $ and curve $ C $ are shown in the figure below. Let points $ F_{1} $ and $ F_{2} $ be the left and right foci of the ellipse $ \\frac{x^{2}}{25} + \\frac{y^{2}}{9} = 1 $, respectively. By the definition of an ellipse, $ |MF_{1}| + |MF_{2}| = 10 $. Extending $ F_{1}P $ to intersect the ellipse at point $ M $, when point $ P $ is not on the coordinate axes, by the triangle inequality, $ |MP| + |MF_{2}| > |PF_{2}| $. Therefore, $ |PF_{1}| + |PF_{2}| < |PF_{1}| + |MP| + |MF_{2}| = |MF_{1}| + |MF_{2}| = 10 $. When point $ P $ is a vertex of curve $ C $, $ |PF_{1}| + |PF_{2}| = 10 $. In summary, $ |PF_{1}| + |PF_{2}| \\leqslant 10 $. Thus, the maximum value of $ |PF_{1}| + |PF_{2}| $ is $ 10 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left focus is $F$, and point $P$ lies on the right branch of the hyperbola. If the midpoint of segment $PF$ lies on the circle centered at the origin $O$ with radius $|OF|$, and the slope of line $PF$ is $\\frac{\\sqrt{5}}{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;LeftFocus(G) = F;P: Point;PointOnCurve(P, RightPart(G)) = True;H: Circle;Center(H) = O;O: Origin;Radius(H) = Abs(LineSegmentOf(O,F));PointOnCurve(MidPoint(LineSegmentOf(P,F)),H) = True;Slope(LineOf(P,F)) = sqrt(5)/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 58], [72, 75], [150, 153]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[63, 66]], [[2, 66]], [[67, 71]], [[67, 78]], [[112, 113]], [[91, 113]], [[92, 97]], [[101, 113]], [[80, 114]], [[116, 147]]]",
    "query_spans": "[[[150, 159]]]",
    "process": "Let the right focus of the hyperbola be $ F $, the midpoint of segment $ PF $ be $ M $, and connect $ OM $, $ MF' $, $ PF' $. Then $ |OM| = c $, $ |PF| = 2c $, and the slope of line $ PF $ is $ \\frac{\\sqrt{5}}{2} $. Therefore, in right triangle $ FMF' $, $ \\tan\\angle MFF' = \\frac{\\sqrt{5}}{2} $, $ \\cos\\angle MFF' = \\frac{2}{3} $, $ |FF'| = 2c $. Thus, $ |FM| = \\frac{4c}{3} $, $ |FP| = \\frac{8c}{3} $, so $ |FP| - |PF| = \\frac{8c}{3} - 2c = 2a $, which gives $ c = 3a $, hence $ e = \\frac{c}{a} = 3 $."
  },
  {
    "text": "The standard equation of a parabola with the distance from focus to directrix being $\\frac{3}{2}$ is?",
    "fact_expressions": "G: Parabola;Distance(Focus(G), Directrix(G)) = 3/2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{(y^2 = 3*x), (y^2 = -3*x), (x^2 = 3*y), (x^2 = -3*y)}",
    "fact_spans": "[[[23, 26]], [[0, 26]]]",
    "query_spans": "[[[23, 33]]]",
    "process": "According to the problem, \\Square, so the equation of the parabola is: region or interval or region"
  },
  {
    "text": "The major axis of an ellipse is $10$ and the minor axis is $8$. What is the range of distances from points on the ellipse to the center of the ellipse?",
    "fact_expressions": "G: Ellipse;Length(MajorAxis(G)) = 10;Length(MinorAxis(G)) = 8;D: Point;PointOnCurve(D, G)",
    "query_expressions": "Range(Distance(D, Center(G)))",
    "answer_expressions": "[4,5]",
    "fact_spans": "[[[0, 2], [21, 23], [27, 29]], [[0, 11]], [[0, 19]], [[25, 26]], [[21, 26]]]",
    "query_spans": "[[[25, 41]]]",
    "process": ""
  },
  {
    "text": "It is known that the directrix of the parabola $y^{2}=8x$ passes through the left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and the line segment intercepted by the hyperbola has length $6$. Then the asymptotes of the hyperbola have equations?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 8*x);G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;PointOnCurve(LeftFocus(G), Directrix(H));Length(InterceptChord(Directrix(H), G)) = 6",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(3)*x)",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[20, 76], [83, 86], [98, 101]], [[20, 76]], [[23, 76]], [[23, 76]], [[23, 76]], [[23, 76]], [[2, 80]], [[2, 96]]]",
    "query_spans": "[[[98, 109]]]",
    "process": "From the given conditions, the directrix of the parabola is $x = -2$, so for the hyperbola, $c = 2$, meaning the left focus of the hyperbola is $F(-2, 0)$. Let the line $x = -2$ intersect the hyperbola at points $A$ and $B$, then $|AB| = \\frac{2b^{2}}{a}$, so $\\frac{2b^{2}}{a} = 6$, hence $b^{2} = 3a$. Also, $c^{2} = a^{2} + b^{2}$, that is, $4 = a^{2} + b^{2} = a^{2} + 3a \\Rightarrow a^{2} + 3a - 4 = 0$. Solving gives $a = 1$, $b = \\sqrt{3}$. Therefore, the asymptotes of the hyperbola are $y = \\pm\\sqrt{3}x$."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{2 m}+\\frac{y^{2}}{m-2}=1$ represents a hyperbola with foci on the $x$-axis, then the range of values for $m$ is?",
    "fact_expressions": "G: Hyperbola;m:Number;Expression(G) = (y^2/(m - 2) + x^2/(2*m) = 1);PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(0,2)",
    "fact_spans": "[[[53, 56]], [[58, 61]], [[1, 56]], [[44, 56]]]",
    "query_spans": "[[[58, 68]]]",
    "process": "The equation $\\frac{x^{2}}{2m}+\\frac{y^{2}}{m-2}=1$ represents a hyperbola with foci on the $x$-axis, we obtain: $\\begin{cases}\\\\\\end{cases}$, solving gives $m\\in(0,2)$. $m-2<0$"
  },
  {
    "text": "The length of the major axis of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{a}=1$ is $2$ times the length of the minor axis, then the value of $a$ is?",
    "fact_expressions": "G: Ellipse;a: Number;Expression(G) = (x^2/a^2 + y^2/a = 1);Length(MajorAxis(G)) = 2 * Length(MinorAxis(G))",
    "query_expressions": "a",
    "answer_expressions": "{4,1/4}",
    "fact_spans": "[[[0, 41]], [[56, 59]], [[0, 41]], [[0, 54]]]",
    "query_spans": "[[[56, 63]]]",
    "process": "When $a^{2}>a$ and $a>0$, i.e., $a>1$, the major axis is twice the minor axis, so $2a=2\\times2\\sqrt{a}$, solving gives $a=4$; when $0<a<1$, the major axis is twice the minor axis, so $2\\sqrt{a}=2\\times2a$, solving gives $a=\\frac{1}{4}$. Therefore, the real number $a$ is $4$ or $\\frac{1}{4}$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$, and point $M$ lies on $C$, then the maximum value of $\\overrightarrow{M F_{1}} \\cdot \\overrightarrow{M F_{2}}$ is?",
    "fact_expressions": "C: Ellipse;M: Point;F1: Point;F2: Point;Expression(C) = (x^2/4 + y^2/2 = 1);Focus(C) = {F1, F2};PointOnCurve(M, C)",
    "query_expressions": "Max(DotProduct(VectorOf(M, F1), VectorOf(M, F2)))",
    "answer_expressions": "2",
    "fact_spans": "[[[18, 60], [71, 74]], [[66, 70]], [[2, 9]], [[10, 17]], [[18, 60]], [[2, 65]], [[66, 75]]]",
    "query_spans": "[[[77, 140]]]",
    "process": "Let the coordinates of point M be given. According to the dot product operation rules of vectors, express the required quantity as a function of the x-coordinate of point M, then find the maximum value of this function. Solution: According to the problem, the coordinates of $ F_{1} $ and $ F_{2} $ are $ (-\\sqrt{2},0) $ and $ (\\sqrt{2},0) $, respectively. Let the coordinates of M be $ (x,y) $. Then we have:  \n$ \\overrightarrow{F}_{1} = (-\\sqrt{2}-x, -y) $, $ \\overrightarrow{MF_{2}} = (\\sqrt{2}-x, -y) $,  \n$ \\overrightarrow{MF} \\cdot \\overrightarrow{MF_{2}} = x^{2} - 2 + y^{2} $.  \nSince point M lies on the ellipse, we have:  \n$ \\frac{x^{2}}{4} + \\frac{y^{2}}{2} = 1 $ $ (-2 \\leqslant x \\leqslant 2) $,  \n$ \\therefore \\overrightarrow{MF_{1}} \\cdot \\overrightarrow{MF_{2}} = x^{2} - 2 + y^{2} = x^{2} - 2 + 2 - \\frac{x^{2}}{2} $.  \nThe maximum value of $ \\overrightarrow{MF_{1}} \\cdot \\overrightarrow{MF_{2}} $ is $ \\frac{2^{2}}{2} = 2 $."
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{k-4}+\\frac{y^{2}}{10-k}=1$ represents an ellipse, what is the range of real values for $k$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(k - 4) + y^2/(10 - k) = 1);k: Real",
    "query_expressions": "Range(k)",
    "answer_expressions": "(4,7)+(7,10)",
    "fact_spans": "[[[46, 48]], [[2, 48]], [[50, 55]]]",
    "query_spans": "[[[50, 62]]]",
    "process": "According to the problem, the equation $\\frac{x^{2}}{k-4}+\\frac{y^{2}}{10-k}=1$ represents an ellipse, which gives the system:  \n$$\n\\begin{cases}\nk-4>0 \\\\\n10-k>0 \\\\\nk-4 \\neq 10-k\n\\end{cases}\n$$\nSolving yields: $4<k<10$ and $k \\neq 7$ (since $k-4 \\neq 10-k$). Therefore, the range of real values for $k$ is $(4,7) \\cup (7,10)$."
  },
  {
    "text": "$A B$ is a diameter of the circle $(x+2)^{2}+(y-1)^{2}=\\frac{5}{2}$. If the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{3}=1$ passes through points $A$ and $B$, then the equation of line $A B$ is?",
    "fact_expressions": "G: Ellipse;H: Circle;A: Point;B: Point;Expression(G) = (x^2/12 + y^2/3 = 1);Expression(H) = ((x + 2)^2 + (y - 1)^2 = 5/2);IsDiameter(LineSegmentOf(A, B), H);PointOnCurve(A,G);PointOnCurve(B,G)",
    "query_expressions": "Expression(LineOf(A,B))",
    "answer_expressions": "x-2*y+4=0",
    "fact_spans": "[[[47, 85]], [[6, 40]], [[87, 90]], [[91, 94]], [[47, 85]], [[6, 40]], [[0, 45]], [[47, 96]], [[47, 96]]]",
    "query_spans": "[[[98, 109]]]",
    "process": "According to the properties of a circle, the midpoint of AB is (-2,1). Using the point-difference method in the ellipse, $k_{AB}$ can be found, and the point-slope form of the line can be rearranged to obtain the required line equation. -2,1)"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, with focal length $2c$. $P$ is a point on the hyperbola $C$. If the perimeter of $\\Delta F_{1} P F_{2}$ is $6a+2c$ and the area is $2ac$, then the asymptotes of the hyperbola $C$ have equations?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;FocalLength(C) = 2*c;c: Number;P: Point;PointOnCurve(P, C);Perimeter(TriangleOf(F1, P, F2)) = 6*a + 2*c;Area(TriangleOf(F1, P, F2)) = 2*a*c",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y = pm*sqrt(2)*x",
    "fact_spans": "[[[2, 64], [89, 90], [105, 111], [164, 170]], [[2, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[73, 80]], [[81, 88]], [[2, 88]], [[2, 88]], [[89, 98]], [[93, 98]], [[101, 104]], [[101, 114]], [[116, 151]], [[116, 162]]]",
    "query_spans": "[[[164, 178]]]",
    "process": "Without loss of generality, let point $ P $ be a point on the right branch of hyperbola $ C $. According to the definition of the hyperbola and the given conditions, we have $ |PF_{1}| = 4a $, $ |PF_{2}| = 2a $, $ \\sin\\angle PF_{1}F_{2} = \\frac{1}{2} $. Since $ |PF_{1}| > |PF_{2}| $, in $ \\triangle F_{1}PF_{2} $, by the law of cosines, we obtain $ 3a^{2} - 2\\sqrt{3}ac + c^{2} = 0 $, solving which gives $ c = \\sqrt{3}a $, then $ b^{2} = 2a^{2} $, leading to the answer. [Solution] Without loss of generality, let point $ P $ be a point on the right branch of hyperbola $ C $. By the definition of the hyperbola, we have: $ |PF_{1}| - |PF_{2}| = 2a $, and $ |PF_{1}| + |PF_{2}| = 6a $, so $ |PF_{1}| = 4a $, $ |PF_{2}| = 2a $. The area of $ \\Delta F_{1}PF_{2} $ is $ S = \\frac{1}{2} \\times 4a \\times 2c \\times \\sin\\angle PF_{1}F_{2} = 2ac $, thus $ \\sin\\angle PF_{1}F_{2} = \\frac{1}{2} $. Since $ |PF_{1}| > |PF_{2}| $, $ \\angle PF_{1}F_{2} $ is an acute angle, so $ \\cos\\angle PF_{1}F_{2} = \\frac{\\sqrt{3}}{2} $. In $ \\triangle F_{1}PF_{2} $, by the law of cosines, $ \\cos\\angle PF_{1}F_{2} = \\frac{(4a)^{2} + (2c)^{2} - (2a)^{2}}{2 \\times 4a \\times 2c} = \\frac{\\sqrt{3}}{2} $, hence $ 3a^{2} - 2\\sqrt{3}ac + c^{2} = 0 $, i.e., $ (\\sqrt{3}a - c)^{2} = 0 $, yielding $ c = \\sqrt{3}a $, so $ c^{2} = 3a^{2} $. Since $ a^{2} + b^{2} = c^{2} $, we get $ b^{2} = 2a^{2} $. Therefore, the asymptotes of the hyperbola are $ y = \\pm\\sqrt{2}x $."
  },
  {
    "text": "It is known that the center of the ellipse is at the origin, the foci $F_{1}$ and $F_{2}$ are on the $x$-axis, the point $P(2,\\ 3)$ lies on the ellipse, and $P F_{2} \\perp x$-axis. Then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;P: Point;O: Origin;F1: Point;F2: Point;Coordinate(P) = (2, 3);Center(G) = O;Focus(G)={F1,F2};PointOnCurve(F1, xAxis);PointOnCurve(F2, xAxis);PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P,F2),xAxis)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16 + y^2/12 = 1",
    "fact_spans": "[[[2, 4], [47, 49], [72, 74]], [[35, 46]], [[8, 10]], [[13, 20]], [[21, 28]], [[35, 46]], [[2, 10]], [[2, 28]], [[13, 34]], [[21, 34]], [[35, 50]], [[52, 70]]]",
    "query_spans": "[[[72, 81]]]",
    "process": ""
  },
  {
    "text": "Given $A(0,7)$, $B(0,-7)$, $C(12, 2)$, an ellipse passing through $A$, $B$ is drawn with $C$ as one focus. Find the equation of the trajectory of the other focus of the ellipse.",
    "fact_expressions": "G: Ellipse;A: Point;B: Point;C: Point;Coordinate(A) = (0, 7);Coordinate(B) = (0, -7);Coordinate(C) = (12, 2);OneOf(Focus(G))=C;PointOnCurve(A,G);PointOnCurve(B,G);F:Point;OneOf(Focus(G)) = F;Negation(C=F)",
    "query_expressions": "LocusEquation(F)",
    "answer_expressions": "(y^2-x^2/48=1)&(y<0)",
    "fact_spans": "[[[57, 59], [61, 63]], [[2, 10], [47, 50]], [[12, 21], [53, 56]], [[23, 35], [37, 40]], [[2, 10]], [[12, 21]], [[23, 35]], [[37, 59]], [[46, 59]], [[46, 59]], [], [[37, 68]], [61, 67]]",
    "query_spans": "[[[61, 75]]]",
    "process": ""
  },
  {
    "text": "Given that $A$ and $B$ are two moving points on the parabola $x^{2}=y$, and the two tangents at $A$ and $B$ intersect at point $P$, if $\\angle A P B=90^{\\circ}$, then what is the $y$-coordinate of point $P$?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;B: Point;Expression(G) = (x^2 = y);PointOnCurve(A, G);PointOnCurve(B, G);l1: Line;l2: Line;TangentOnPoint(A, G) = l1;TangentOnPoint(B, G) = l2;Intersection(l1, l2) = P;AngleOf(A, P, B) = ApplyUnit(90, degree)",
    "query_expressions": "YCoordinate(P)",
    "answer_expressions": "-1/4",
    "fact_spans": "[[[10, 22]], [[2, 5], [30, 33]], [[44, 48], [77, 81]], [[6, 9], [34, 37]], [[10, 22]], [[2, 28]], [[2, 28]], [], [], [[2, 42]], [[2, 42]], [[2, 48]], [[50, 75]]]",
    "query_spans": "[[[77, 87]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $y=a x^{2}$ has coordinates $(0,1)$, then $a$=?",
    "fact_expressions": "G: Parabola;a: Number;Expression(G) = (y = a*x^2);Coordinate(Focus(G)) = (0,1)",
    "query_expressions": "a",
    "answer_expressions": "1/4",
    "fact_spans": "[[[1, 15]], [[30, 33]], [[1, 15]], [[1, 28]]]",
    "query_spans": "[[[30, 35]]]",
    "process": "The standard equation of the parabola y=ax^{2} is x^{2}=\\frac{1}{a}y'. Since the focus of the parabola y=ax^{2} is (0,1), \\therefore\\frac{1}{4a}=1, \\therefore a=\\frac{1}{4},"
  },
  {
    "text": "Let $a, b \\in R$, $a^{2}+2 b^{2}=6$, then the maximum value of $\\frac{b}{a-3}$ is?",
    "fact_expressions": "a: Real;b: Real;a^2 + 2*b^2 = 6",
    "query_expressions": "Max(b/(a - 3))",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 13]], [[1, 13]], [[15, 32]]]",
    "query_spans": "[[[34, 55]]]",
    "process": "According to the problem, since $ a, b \\in \\mathbb{R} $ and $ a^{2} + 2b^{2} = 6 $, the point $ (a, b) $ lies inside the ellipse. The goal is to find the range of the slope between the point $ (a, b) $ and $ (3, 0) $. The maximum slope occurs when the line is tangent to the ellipse. Let the line equation be $ y = k(x - 3) $. By combining this with the ellipse equation and setting the discriminant to zero, we obtain $ k = 1 $. Thus, the maximum value of $ \\frac{b}{a - 3} $ is 1."
  },
  {
    "text": "Given a line passing through the point $P(2,1)$ intersects the parabola $y^{2}=2x$ at points $A$ and $B$, and $P$ is the midpoint of chord $AB$, then the slope of line $AB$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);H: Line;P: Point;Coordinate(P) = (2, 1);PointOnCurve(P, H);A: Point;B: Point;Intersection(H, G) = {A, B};IsChordOf(LineSegmentOf(A, B), G);MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Slope(LineOf(A, B))",
    "answer_expressions": "1",
    "fact_spans": "[[[18, 32]], [[18, 32]], [[15, 17]], [[5, 14], [44, 47]], [[5, 14]], [[4, 17]], [[34, 37]], [[38, 41]], [[15, 43]], [[18, 54]], [[44, 57]]]",
    "query_spans": "[[[59, 71]]]",
    "process": "Point $ P(2,1) $ lies inside the parabola. A line passing through $ P $ with nonzero slope intersects the parabola at two points. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ y_{1}+y_{2}=2 $. From the given conditions:  \n$$\n\\begin{cases}\ny_{1}=2x_{1} \\\\\ny_{2}=2x_{2}\n\\end{cases}\n$$  \n$ x_{1} $, then $ (y_{1}+y_{2})(y_{1}-y_{2})=2(x_{1}-x_{2}) $, i.e., $ 2(y_{1}-y_{2})=2(x_{1}-x_{2}) $, $ k_{AB}=\\frac{y_{1}}{x_{1}}\\frac{-y_{2}}{x_{2}} $"
  },
  {
    "text": "Given the hyperbola $x^{2}-\\frac{y^{2}}{a^{2}}=1$ $(a>0)$, its asymptotes are given by $y=\\pm 2 x$. What is the value of $a$?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (x^2 - y^2/a^2 = 1);Expression(Asymptote(G)) = (y = pm*(2*x))",
    "query_expressions": "a",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 39], [40, 41]], [[62, 65]], [[5, 39]], [[2, 39]], [[40, 59]]]",
    "query_spans": "[[[62, 69]]]",
    "process": ""
  },
  {
    "text": "What is the slope of the tangent line at a point $P(x_{0}, y_{0})  (y_{0} \\neq 0)$ on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;x0:Number;y0:Number;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (x0, y0);H:Line;PointOnCurve(P,G);Negation(y0=0);TangetOnPoint(P,G) = H",
    "query_expressions": "Slope(H)",
    "answer_expressions": "-b^2*x0/a^2*y0",
    "fact_spans": "[[[1, 46]], [[3, 46]], [[3, 46]], [[49, 82]], [[49, 82]], [[49, 82]], [[1, 46]], [[49, 82]], [], [[1, 82]], [[49, 82]], [0, 81]]",
    "query_spans": "[[[0, 90]]]",
    "process": ""
  },
  {
    "text": "The right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $F(5,0)$, and one asymptote has the equation $y=\\frac{4}{3} x$. What is the equation of this hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (5, 0);RightFocus(G) = F;Expression(OneOf(Asymptote(G)))=(y=(4/3)*x)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/16 = 1",
    "fact_spans": "[[[0, 56], [99, 102]], [[3, 56]], [[3, 56]], [[61, 69]], [[3, 56]], [[3, 56]], [[0, 56]], [[61, 69]], [[0, 69]], [[0, 96]]]",
    "query_spans": "[[[99, 107]]]",
    "process": "From the given conditions, we have  \n\\begin{cases}\\frac{b}{a}=\\frac{4}{3}\\\\\\sqrt{a^{2}+b^{2}}=5\\end{cases}  \nSolving gives \\begin{cases}a=3\\\\b=4\\end{cases}, therefore, the equation of the hyperbola is \\frac{x2}{9}-\\frac{y^{2}}{16}=1"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, point $P$ lies on the hyperbola, and $\\angle F_{1} P F_{2}=60^{\\circ}$, then $|P F_{1}|$=?",
    "fact_expressions": "G: Hyperbola;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 - y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Abs(LineSegmentOf(P, F1))",
    "answer_expressions": "{2*sqrt(2)-2,2*sqrt(2)+2}",
    "fact_spans": "[[[18, 46], [57, 60]], [[2, 9]], [[52, 56]], [[10, 17]], [[18, 46]], [[2, 51]], [[53, 61]], [[63, 96]]]",
    "query_spans": "[[[98, 112]]]",
    "process": "The hyperbola $\\frac{x^2}{4}-y^{2}=1$ has $a=2$, $b=1$, $c=\\sqrt{5}$. Let $|PF_{1}|=m$, $|PF_{2}|=n$. By the definition of the hyperbola, we have $|m-n|=2a=4$, $\\textcircled{1}$. In $\\triangle PF_{1}F_{2}$, $\\angle F_{1}PF_{2}=60^{\\circ}$, so we get $4c^{2}=m^{2}+n^{2}-2mn\\cos60^{\\circ}=m^{2}+n^{2}-mn=(m-n)^{2}+mn$, which leads to $mn+16=20$, hence $mn=4$, $\\textcircled{2}$. From $\\textcircled{1}$ and $\\textcircled{2}$, solving gives $m=2\\sqrt{2}-2$ or $2\\sqrt{2}+2$."
  },
  {
    "text": "Given the ellipse $E$: $\\frac{x^{2}}{4}+y^{2}=1$, let $P$ be any point on the major axis of $E$. A line $l$ with slope $\\frac{1}{2}$ passing through point $P$ intersects $E$ at points $M$ and $N$. Then the value of $|P M|^{2}+|P N|^{2}$ is?",
    "fact_expressions": "l: Line;E: Ellipse;P: Point;M: Point;N: Point;Expression(E) = (x^2/4 + y^2 = 1);PointOnCurve(P, MajorAxis(E));PointOnCurve(P, l);Slope(l) = 1/2;Intersection(l, E) = {M, N}",
    "query_expressions": "Abs(LineSegmentOf(P, M))^2 + Abs(LineSegmentOf(P, N))^2",
    "answer_expressions": "5",
    "fact_spans": "[[[76, 81]], [[2, 34], [41, 44], [82, 85]], [[54, 58], [37, 40]], [[87, 90]], [[91, 94]], [[2, 34]], [[37, 52]], [[53, 81]], [[59, 81]], [[76, 96]]]",
    "query_spans": "[[[98, 123]]]",
    "process": "Let $ P(m,0) $ where $ -2 \\leqslant m \\leqslant 2 $, the equation of line $ l $ is $ y = \\frac{1}{2}(x - m) $, and let $ M(x_{1},y_{1}) $, $ N(x_{2},y_{2}) $. Substituting the line equation into the ellipse equation and simplifying yields $ 2x^{2} - 2mx + m^{2} - 4 = 0 $, from which we obtain $ x_{1} + x_{2} = m $, $ x_{1}x_{2} = \\frac{m^{2} - 4}{2} $. Therefore, $ |MP|^{2} + |PN|^{2} = (x_{1} - m)^{2} + y_{1}^{2} + (x_{2} - m)^{2} + y_{2}^{2} $,  \n$ |MP|^{2} + |PN|^{2} = \\frac{5}{4}[(x_{1} + x_{2})^{2} - 2m(x_{1} + x_{2}) - 2x_{1}x_{2} + 2m^{2}] $,  \n$ |MP|^{2} + |PN|^{2} = \\frac{5}{4}[m^{2} - 2m^{2} - (m^{2} - 4) + 2m^{2}] = 5 $."
  },
  {
    "text": "The equation of the directrix of the parabola $x^{2}=a y$ is $y=2$, then $a$=?",
    "fact_expressions": "G: Parabola;a: Number;Expression(G) = (x^2 = a*y);Expression(Directrix(G)) = (y = 2)",
    "query_expressions": "a",
    "answer_expressions": "-8",
    "fact_spans": "[[[0, 14]], [[27, 30]], [[0, 14]], [[0, 25]]]",
    "query_spans": "[[[27, 32]]]",
    "process": "From the parabola equation, we know that $2p = a \\therefore \\frac{p}{2} = \\frac{a}{4} \\therefore -\\frac{a}{4} = 2 \\therefore a = -8$"
  },
  {
    "text": "The line $l$ intersects the circle $x^{2}+y^{2}+2 x-4 y+a=0$ $(a<3)$ at two points $A$, $B$, and the midpoint of chord $AB$ is $(0,1)$. Then the equation of line $l$ is?",
    "fact_expressions": "l: Line;G: Circle;Expression(G) = (a - 4*y + 2*x + x^2 + y^2 = 0);a: Number;a<3;Intersection(l,G) = {A,B};IsChordOf(LineSegmentOf(A,B),G) = True;Coordinate(MidPoint(LineSegmentOf(A,B))) = (0,1);A: Point;B: Point",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x-y+1=0",
    "fact_spans": "[[[0, 5], [72, 77]], [[6, 37]], [[6, 37]], [[7, 37]], [[7, 37]], [[0, 51]], [[6, 59]], [[54, 70]], [[42, 45]], [[48, 51]]]",
    "query_spans": "[[[72, 82]]]",
    "process": "Let the center of the circle be O, the slope of line l be k, the midpoint of chord AB be P, and the slope of PO be k_{op}. Since k_{op} = \\frac{2-1}{-1-0}, then l \\perp PO, so k \\cdot k_{op} = k \\cdot (-1) = -1. Therefore, k = 1. By point-slope form, y = x + 1."
  },
  {
    "text": "Given that $P$ is a moving point on the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{4}=1$, and $F_{1}$, $F_{2}$ are the two foci of the ellipse, then the range of values of $\\overrightarrow {PF_{1}} \\cdot \\overrightarrow {PF_{2}}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/12 + y^2/4 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2}",
    "query_expressions": "Range(DotProduct(VectorOf(P, F1),VectorOf(P, F2)))",
    "answer_expressions": "[-4,4]",
    "fact_spans": "[[[6, 44], [65, 67]], [[2, 5]], [[49, 56]], [[57, 64]], [[6, 44]], [[2, 48]], [[49, 72]]]",
    "query_spans": "[[[74, 138]]]",
    "process": ""
  },
  {
    "text": "Given that a line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$, $|AF|=5$, $|BF|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);A: Point;B: Point;Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, F)) = 5",
    "query_expressions": "Abs(LineSegmentOf(B, F))",
    "answer_expressions": "5/4",
    "fact_spans": "[[[3, 17], [28, 31]], [[3, 17]], [[20, 23]], [[3, 23]], [[24, 26]], [[2, 26]], [[32, 35]], [[36, 39]], [[24, 41]], [[42, 51]]]",
    "query_spans": "[[[53, 62]]]",
    "process": ""
  },
  {
    "text": "Point $A$ is an intersection point of the ellipse $C_{1}$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ and the hyperbola $C_{2}$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, and points $F_{1}$, $F_{2}$ are the two foci of the ellipse $C_{1}$. Then the value of $|A F_{1}| \\cdot|A F_{2}|$ is?",
    "fact_expressions": "C1: Ellipse;Expression(C1) = (x^2/25 + y^2/16 = 1);C2: Hyperbola;Expression(C2) = (x^2/4 - y^2/5 = 1);A: Point;OneOf(Intersection(C1, C2)) = A;Focus(C1) = {F1, F2};F1: Point;F2: Point",
    "query_expressions": "Abs(LineSegmentOf(A, F1))*Abs(LineSegmentOf(A, F2))",
    "answer_expressions": "21",
    "fact_spans": "[[[5, 53], [124, 133]], [[5, 53]], [[54, 101]], [[54, 101]], [[0, 4]], [[0, 106]], [[107, 138]], [[107, 115]], [[116, 123]]]",
    "query_spans": "[[[140, 170]]]",
    "process": "First determine that the ellipse and hyperbola have the same focal coordinates. Let |AF_{1}| = m, |AF_{2}| = n, without loss of generality assume 0 < n < m. Using the definitions of the ellipse and hyperbola, solve for m and n. For ellipse C_{1}: the foci lie on the x-axis, c^{2} = a^{2} - b^{2} = 25 - 16 = 9. For hyperbola C_{2}: the foci lie on the x-axis, c^{2} = a^{2} + b^{2} = 4 + 5 = 9. Thus, the ellipse and hyperbola have the same focal coordinates. Let |AF_{1}| = m, |AF_{2}| = n, without loss of generality assume 0 < n < m. Using the definitions of the ellipse and hyperbola, we obtain \\begin{cases} m + n = 10 \\\\ m - n = 4 \\end{cases}. Therefore, mn = 21. Hence, the value of |AF_{1}| \\cdot |AF_{2}| is 21."
  },
  {
    "text": "$P$ is a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and $M$, $N$ are points on the circles $(x+3)^{2}+y^{2}=4$ and $(x-3)^{2}+y^{2}=1$, respectively. Then the range of values of $|P M|+|P N|$ is?",
    "fact_expressions": "G: Ellipse;H1: Circle;H2: Circle;P: Point;M: Point;N: Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(P,G) = True;Expression(H1) = (y^2 + (x + 3)^2 = 4);PointOnCurve(M,H1) = True;Expression(H2) = (y^2 + (x - 3)^2 = 1);PointOnCurve(N,H2) = True",
    "query_expressions": "Range(Abs(LineSegmentOf(P, M)) + Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "[7,13]",
    "fact_spans": "[[[4, 43]], [[58, 77]], [[78, 97]], [[0, 3]], [[47, 50]], [[51, 54]], [[4, 43]], [[0, 46]], [[58, 77]], [[47, 100]], [[78, 97]], [[47, 100]]]",
    "query_spans": "[[[102, 122]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C_{1}: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ has eccentricity $\\sqrt{5}$, one asymptote is $l$, the parabola $C_{2}: y^{2} = 8x$ has focus $F$, and point $P$ is the intersection point of line $l$ and parabola $C_{2}$ other than the origin, then $|PF|$ = ?",
    "fact_expressions": "l: Line;C1: Hyperbola;C2: Parabola;P: Point;F: Point;O:Origin;Expression(C1)=(x^2/a^2-y^2/b^2=1);Expression(C2)=(y^2=8*x);Eccentricity(C1)=sqrt(5);OneOf(Asymptote(C1))=l;Focus(C2)=F;Intersection(l,C2)=P;Negation(P=O);a: Number;b: Number",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "4",
    "fact_spans": "[[[78, 81], [117, 122]], [[2, 56]], [[82, 104], [123, 133]], [[112, 116]], [[108, 111]], [[135, 137]], [[2, 56]], [[82, 104]], [[2, 71]], [[2, 81]], [[82, 111]], [[112, 140]], [[112, 137]], [[2, 56]], [[2, 56]]]",
    "query_spans": "[[[142, 151]]]",
    "process": "In the hyperbola, $\\frac{c}{a}=\\sqrt{5}$, that is, $\\frac{c^{2}}{a^{2}}=\\frac{a^{2}+b^{2}}{a^{2}}=5$, $\\frac{b}{a}=2$, without loss of generality, assume the equation is $y=2x$. From $\\begin{cases}y=2x\\\\y=8x\\end{cases}$, we get $\\begin{cases}x=0,\\\\y=0\\end{cases}$ or $\\begin{cases}x=\\\\y=\\end{cases}2$, i.e., $P(2,4)$. In the parabola $y^{2}=8x$, $p=4$, $\\therefore |PF|=2+\\frac{4}{2}=4$"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1$ $(0<b<2)$, with left and right foci $F_{1}$, $F_{2}$. A tangent line to the ellipse is drawn through the point $M(2,2)$, with point of tangency $P$, where $P$ does not lie on the $x$-axis. Then $S_{\\Delta P M F_{1}}+S_{\\Delta P M F_{2}}$=?",
    "fact_expressions": "G: Ellipse;b: Number;M: Point;P: Point;l: Line;0 < b;b < 2;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2/b^2 = 1);Coordinate(M) = (2, 2);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(M, l);TangentOfPoint(M, G) = l;TangentPoint(G, l) = P;Negation(PointOnCurve(P, xAxis))",
    "query_expressions": "Area(TriangleOf(P, M, F1)) + Area(TriangleOf(P, M, F2))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 50], [84, 86]], [[4, 50]], [[74, 83]], [[93, 96], [98, 102]], [], [[4, 50]], [[4, 50]], [[57, 64]], [[65, 72]], [[2, 50]], [[74, 83]], [[2, 72]], [[2, 72]], [[73, 89]], [[73, 89]], [[73, 96]], [[98, 109]]]",
    "query_spans": "[[[111, 156]]]",
    "process": "According to the problem, the slope of the tangent line exists. Let it be $ k $, then the equation of the tangent line is $ y - 2 = k(x - 2) $. Combining\n\\[\n\\begin{cases}\ny = kx + 2 - 2k \\\\\n\\frac{x^{2}}{4} + \\frac{y^{2}}{b^{2}} = 1\n\\end{cases}\n\\]\nand eliminating $ y $, we obtain\n\\[\n(b^{2} + 4k^{2})x^{2} + 8k(2 - 2k)x + 4(2 - 2k)^{2} - 4b^{2} = 0.\n\\]\nFrom the problem, we know\n\\[\n\\Delta = 64k^{2}(2 - 2k)^{2} - 4(b^{2} + 4k^{2})[4(2 - 2k)^{2} - 4b^{2}] = 0,\n\\]\nwhich simplifies to $ b^{2} = 4 - 8k $. Substituting this into\n\\[\n(b^{2} + 4k^{2})x^{2} + 8k(2 - 2k)x + 4(2 - 2k)^{2} - 4b^{2} = 0,\n\\]\nwe get $ x = \\frac{2k}{k - 1} $. Substituting into $ y = kx + 2 - 2k $, we get $ y = \\frac{4k - 2}{k - 1} $, so $ P\\left(\\frac{2k}{k - 1}, \\frac{4k - 2}{k - 1}\\right) $. Also, $ c^{2} = 4 - b^{2} = 8k $, so $ c = \\sqrt{8k} = 2\\sqrt{2k} $, thus $ F_{1}(-2\\sqrt{2k}, 0) $, $ F_{2}(2\\sqrt{2k}, 0) $.  \n\\[\n|PM| = \\sqrt{\\left(\\frac{2k}{k - 1} - 2\\right)^{2} + \\left(\\frac{4k - 2}{k - 1} - 2\\right)^{2}} = \\frac{2\\sqrt{1 + k^{2}}}{|1 - k|}.\n\\]\nThe distance from $ F_{1} $ to the line $ y = kx + 2 - 2k $ is\n\\[\nd_{1} = \\frac{|-2k\\sqrt{2k} + 2 - 2k|}{\\sqrt{k^{2} + 1}},\n\\]\nand the distance from $ F_{2} $ to the line $ y = kx + 2 - 2k $ is\n\\[\nd_{2} = \\frac{|2k\\sqrt{2k} + 2 - 2k|}{\\sqrt{k^{2} + 1}}.\n\\]\nSince $ 0 < b^{2} = 4 - 8k < 4 $, we have $ 0 < k < \\frac{1}{2} $, so $ 2k\\sqrt{2k} + 2 - 2k > 0 $. Since $ F_{1}, F_{2} $ lie on the same side of the line $ y = kx + 2 - 2k $, we have $ -2k\\sqrt{2k} + 2 - 2k > 0 $. Thus,\n\\[\nd_{2} = \\frac{2k\\sqrt{2k} + 2 - 2k}{\\sqrt{k^{2} + 1}}, \\quad d_{1} + d_{2} = \\frac{4(1 - k)}{\\sqrt{k^{2} + 1}}.\n\\]\nThen,\n\\[\n\\frac{1}{2}(d_{1} + d_{2}) \\cdot |PM| = \\frac{1}{2} \\cdot \\frac{4(1 - k)}{\\sqrt{k^{2} + 1}} \\cdot \\frac{2\\sqrt{k^{2} + 1}}{1 - k} = 4.\n\\]"
  },
  {
    "text": "If the line $k x - y + 3 = 0$ and the ellipse $\\frac{x^{2}}{16} + \\frac{y^{2}}{4} = 1$ have two common points, then the range of real values for $k$ is?",
    "fact_expressions": "G: Ellipse;H: Line;k: Real;Expression(G) = (x^2/16 + y^2/4 = 1);Expression(H) = (k*x - y + 3 = 0);NumIntersection(H, G)= 2",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-oo, -sqrt(5)/4) + (sqrt(5)/4, +oo)",
    "fact_spans": "[[[15, 53]], [[1, 14]], [[61, 66]], [[15, 53]], [[1, 14]], [[1, 59]]]",
    "query_spans": "[[[61, 73]]]",
    "process": "The equation of the line is y = kx + 3. Substituting into the ellipse equation gives x^{2} + 4(kx + 3)^{2} - 16 = 0, which simplifies to (1 + 4k^{2})x^{2} + 24kx + 20 = 0. Since there are two roots, we have _{4} = (24k)^{2} - 4 \\cdot 20 \\cdot (1 + 4k^{2}) > 0. Solving yields the range of k as (-\\infty, -\\frac{\\sqrt{5}}{4}) \\cup (\\frac{\\sqrt{5}}{4}, \\infty)."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$), respectively. If there exists a point $P$ on the right branch of the hyperbola such that $|O P|=|O F_{1}|$ ($O$ is the origin) and $|P F_{1}|=\\sqrt{3}|P F_{2}|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;O: Origin;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, RightPart(C));Abs(LineSegmentOf(O, P)) = Abs(LineSegmentOf(O, F1));Abs(LineSegmentOf(P, F1)) = sqrt(3)*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)+1",
    "fact_spans": "[[[19, 80], [88, 91], [163, 166]], [[27, 80]], [[27, 80]], [[121, 124]], [[98, 101]], [[1, 8]], [[9, 16]], [[27, 80]], [[27, 80]], [[19, 80]], [[1, 86]], [[1, 86]], [[88, 101]], [[103, 120]], [[132, 161]]]",
    "query_spans": "[[[163, 172]]]",
    "process": "\\because|OF_{1}|=|OF_{2}|=|OP|\\therefore\\angle F_{1}PF_{2}=90^{\\circ}, let |PF_{2}|=t, then |F_{1}P|=\\sqrt{3}t, a=\\frac{\\sqrt{3}t-t}{2}, t^{2}+3t^{2}=4c^{2}, t=c,\\therefore e=\\frac{c}{a}=\\sqrt{3}+1"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, with left and right foci $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through $F_{1}$ is tangent to the circle $x^{2}+y^{2}=a^{2}$ at point $T$, and intersects the right branch of hyperbola $C$ at point $P$. If $F_{1} P=4 F_{1} T$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;G: Circle;F1: Point;F2:Point;P: Point;T: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = a^2);LeftFocus(C) =F1;RightFocus(C)=F2;PointOnCurve(F1, l);TangentPoint(l,G)=T;Intersection(l, RightPart(C)) = P;LineSegmentOf(F1, P) = 4*LineSegmentOf(F1, T)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[99, 104], [134, 139]], [[2, 63], [140, 146], [178, 184]], [[9, 63]], [[9, 63]], [[105, 125]], [[71, 78], [91, 98]], [[81, 89]], [[151, 155]], [[128, 132]], [[9, 63]], [[9, 63]], [[2, 63]], [[105, 125]], [[2, 89]], [[2, 89]], [[90, 104]], [[99, 132]], [[134, 155]], [[157, 176]]]",
    "query_spans": "[[[178, 190]]]",
    "process": "According to the problem, draw the figure, combine with the first definition of hyperbola, then convert all side length relationships into the right triangle $MPF_{2}$, and simplify to find the value. As shown in the figure, from the given conditions we have $|OF_{1}|=|OF_{2}|=c$, $|OT|=a$, then $|F_{1}T|=b$. Also, $\\because\\overrightarrow{F_{1}P}=4\\overrightarrow{F_{1}T}$, $\\therefore|TP|=3b$, $\\therefore|F_{1}P|=4b$. Also, $\\because|PF_{1}|-|PF_{2}|=2a$, $\\therefore|PF_{2}|=4b-2a$. Draw $F_{2}M//OT$, we get $|F_{2}M|=2a$, $|TM|=b$, then $|PM|=2b$. In $\\triangle MPF_{2}$, $|PM|^{2}+|MF_{2}|^{2}=|PF_{2}|^{2}$, that is $c^{2}=(2b-a)^{2}$, $2b=a+c$. Also, $\\because c^{2}=a^{2}+b^{2}$, simplifying gives $3c^{2}-2ac-5a^{2}=0$, dividing through by $a^{2}$, we get $3e^{2}-2e-5=0$. Solving yields $e=\\frac{5}{3}$. The eccentricity of the hyperbola is $\\frac{5}{3}$."
  },
  {
    "text": "$O$ is the coordinate origin, $F$ is the focus of the parabola $C$: $y^{2}=4x$, $P$ is a point on $C$, if $|PF|=4$, then the area of $\\triangle POF$ is?",
    "fact_expressions": "C: Parabola;P: Point;O: Origin;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(P, C);Abs(LineSegmentOf(P, F)) = 4",
    "query_expressions": "Area(TriangleOf(P, O, F))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[13, 31], [39, 42]], [[35, 38]], [[0, 3]], [[9, 12]], [[13, 31]], [[9, 34]], [[35, 45]], [[47, 56]]]",
    "query_spans": "[[[58, 80]]]",
    "process": "According to the problem, the focus of the parabola C is $ F(1,0) $, and the equation of the directrix is $ x = -1 $. Given $ |PF| = 4 $, let $ P(x,y) $, then $ x + 1 = 4 $, so $ x = 3 $, hence $ y = \\pm 2\\sqrt{3} $, i.e., the coordinates of point P are $ (3, \\pm 2\\sqrt{3}) $. Then the area of $ \\triangle POF $ is $ S = \\frac{1}{2} \\times 1 \\times 2\\sqrt{3} = \\sqrt{3} $."
  },
  {
    "text": "Given the standard equation of a hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0,b>0)$, $F_{1}$ and $F_{2}$ are its left and right foci. If $P$ is a point on the right branch of the hyperbola, and $\\tan \\angle P F_{1} F_{2}=\\frac{1}{2}$, $\\tan \\angle P F_{2} F_{1}=2$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a:Number;b:Number;a>0;b>0;P: Point;F1: Point;F2: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(P, RightPart(G));Tan(AngleOf(P, F1, F2)) = 1/2;Tan(AngleOf(P, F2, F1)) = 2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 5], [94, 97], [83, 84], [179, 182]], [[10, 62]], [[10, 62]], [[10, 62]], [[10, 62]], [[90, 93]], [[65, 72]], [[75, 82]], [[2, 62]], [[65, 88]], [[65, 88]], [[90, 103]], [[105, 145]], [[147, 176]]]",
    "query_spans": "[[[179, 188]]]",
    "process": "Let P(m,n), we have \\frac{m^2}{a^{2}}-\\frac{n^{2}}{b^{2}}=1, F_{1}(-c,0), F_{2}(c,0) being its left and right foci. Using the slope formula of a line, solving the equations gives m,n. Then using b^{2}=c^{2}-a^{2}, e=\\frac{c}{a}, we obtain an equation in e; solving it yields the desired eccentricity. Solution: Let P(m,n), we have \\frac{m^2}{a^{2}}-\\frac{n^{2}}{b^{2}}=1, F_{1}(-c,0), F_{2}(c,0) as its left and right foci. The slope of line PF_{1} is k_{1}=\\frac{n}{m+c}, the slope of line PF_{2} is k_{2}=\\frac{n}{m-c}. Given k_{2}=-2, k_{1}=\\frac{1}{2}, we have \\frac{n}{m+c}=\\frac{1}{2}, \\frac{n}{m-c}=-2. Solving gives m=\\frac{3}{5}c, n=\\frac{4}{5}c. Then \\frac{9c^{2}}{25a^{2}}-\\frac{16c^{2}}{25b^{2}}=1. Using b^{2}=c^{2}-a^{2}, e=\\frac{c}{a}, we get 9e^{2}-\\frac{16e^{2}}{e^{2}-1}=25, which simplifies to 9e^{4}-50e^{2}+25=0, giving e^{2}=5 (discarding \\frac{5}{9}<1). Thus e=\\sqrt{5}."
  },
  {
    "text": "For the equation $\\frac{x^{2}}{7-k}+\\frac{y^{2}}{k-1}=1$ to represent an ellipse with foci on the $y$-axis, what is the range of values for $k$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(7 - k) + y^2/(k - 1) = 1);PointOnCurve(Focus(G), yAxis) = True;k: Number",
    "query_expressions": "Range(k)",
    "answer_expressions": "(4,7)",
    "fact_spans": "[[[53, 55]], [[1, 55]], [[44, 55]], [[57, 60]]]",
    "query_spans": "[[[57, 67]]]",
    "process": ""
  },
  {
    "text": "Given $A(0,-4)$, $B(3,2)$, the shortest distance from a point on the parabola $y^{2}=8 x$ to the line $A B$ is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;P: Point;Expression(G) = (y^2 = 8*x);Coordinate(A) = (0, -4);Coordinate(B) = (3, 2);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, LineOf(A,B)))",
    "answer_expressions": "0",
    "fact_spans": "[[[22, 36]], [[2, 11]], [[13, 21]], [[38, 39]], [[22, 36]], [[2, 11]], [[13, 21]], [[22, 39]]]",
    "query_spans": "[[[38, 54]]]",
    "process": "\\because A(0,-4), B(3,2), \\therefore the slope of line AB is 2. \\therefore the equation of line AB is: y = 2x - 4, i.e., 2x - y - 4 = 0. Solving the line equation and parabola equation simultaneously: (2x - 4)^{2} = 8x. Simplifying yields: x^{2} - 6x + 4 = 0, whose discriminant is \\Delta = 36 - 16 = 20 > 0. Therefore, the line intersects the parabola, hence the shortest distance from a point on the parabola to line AB is 0."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through $F_{2}$ intersects the right branch of the hyperbola at points $A$ and $B$. Point $M$ satisfies $\\overrightarrow{A B}+\\overrightarrow{A F_{1}}=2 \\overrightarrow{A M}$, and $\\overrightarrow{A M} \\cdot \\overrightarrow{B F_{1}}=0$. If $\\cos \\angle A F_{1} B=\\frac{1}{3}$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;l: Line;PointOnCurve(F2, l) = True;Intersection(l, RightPart(C)) = {A, B};A: Point;B: Point;M: Point;VectorOf(A, B) + VectorOf(A, F1) = 2*VectorOf(A, M);DotProduct(VectorOf(A, M), VectorOf(B, F1)) = 0;Cos(AngleOf(A, F1, B)) = 1/3",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(33)/3",
    "fact_spans": "[[[2, 63], [104, 107], [294, 300]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[73, 80]], [[81, 88], [90, 97]], [[2, 88]], [[2, 88]], [[98, 103]], [[89, 103]], [[98, 120]], [[111, 114]], [[115, 118]], [[121, 125]], [[127, 197]], [[199, 254]], [[257, 292]]]",
    "query_spans": "[[[294, 306]]]",
    "process": "From $\\overrightarrow{AB}+\\overrightarrow{AF}=2\\overrightarrow{AM}$, we get that $M$ is the midpoint of side $BF_{1}$, and since $\\overrightarrow{AM}\\cdot\\overrightarrow{BF}_{1}=0$, it follows that $AM\\bot BF_{1}$. Therefore, triangle $\\triangle ABF$ is isosceles, $|AB|=|AF_{1}|$, $\\angle AF_{1}B=\\angle ABF_{1}$ implies $|AB|=|AF_{1}|=2a+m$, so $|BF_{2}|=2a$, $|BF_{1}|=4a$. In triangle $\\triangle BFF$, by the law of cosines, $\\cos\\angle F_{1}BF_{2}=\\frac{|BF_{1}|^{2}+|BF_{2}|^{2}-|F_{1}F_{2}|^{2}}{2|BF_{1}|\\cdot|BF_{2}|}$, that is, $\\frac{1}{2}=\\frac{16a^{2}+4a^{2}-4c^{2}}{2\\times4a\\times2a}$, simplifying yields $\\frac{c^{2}}{a^{2}}=\\frac{11}{3}$, thus the eccentricity $e=\\frac{c}{a}=\\frac{\\sqrt{33}}{3}$."
  },
  {
    "text": "If the curve $C$: $\\frac{x^{2}}{2+m}+\\frac{y^{2}}{2-m}=1$ is a hyperbola with foci on the $x$-axis, then what is the range of values for $m$?",
    "fact_expressions": "C: Curve;Expression(C) = (x^2/(m + 2) + y^2/(2 - m) = 1);m: Number;PointOnCurve(Focus(G), xAxis);G: Hyperbola;C = G",
    "query_expressions": "Range(m)",
    "answer_expressions": "(2, +oo)",
    "fact_spans": "[[[1, 47]], [[1, 47]], [[62, 65]], [[48, 60]], [[57, 60]], [[1, 60]]]",
    "query_spans": "[[[62, 71]]]",
    "process": "\\because the equation C: \\frac{x^{2}}{2+m} + \\frac{y^{2}}{2-m} = 1 represents a hyperbola with foci on the x-axis, \\therefore \\begin{cases} m+2 > 0 \\\\ 2-m < 0 \\end{cases} \\therefore m > 2."
  },
  {
    "text": "If the line $2 x - y + c = 0$ is a tangent to the parabola $x^{2} = 4 y$, then $c =$?",
    "fact_expressions": "G: Parabola;H: Line;c: Number;Expression(G) = (x^2 = 4*y);Expression(H) = (c + 2*x - y = 0);IsTangent(H, G)",
    "query_expressions": "c",
    "answer_expressions": "-4",
    "fact_spans": "[[[15, 29]], [[1, 14]], [[36, 39]], [[15, 29]], [[1, 14]], [[1, 34]]]",
    "query_spans": "[[[36, 41]]]",
    "process": "Solving the line and parabola simultaneously gives  \n\\begin{cases}2x-y+c=0\\\\x^2=4y\\end{cases} \\Rightarrow x^{2}-8x-4c=0 \\Rightarrow \\Delta=0 \\Rightarrow c=-4."
  },
  {
    "text": "Given that $O$ is the coordinate origin, $F$ is the focus of the parabola $C$: $y^{2}=4 x$, $P$ is a point on the parabola $C$, and $|P F|=4$, then the area of $\\triangle P O F$ is?",
    "fact_expressions": "C: Parabola;P: Point;O: Origin;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(P, C);Abs(LineSegmentOf(P, F)) = 4",
    "query_expressions": "Area(TriangleOf(P, O, F))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[15, 34], [42, 48]], [[38, 41]], [[2, 5]], [[11, 14]], [[15, 34]], [[11, 37]], [[38, 51]], [[53, 62]]]",
    "query_spans": "[[[64, 86]]]",
    "process": "Let P(x,y), F(1,0), according to the focal radius formula: x+1=4, x=3, substitute into the parabola equation, we get: y=\\pm2\\sqrt{3}, \\therefores=\\frac{1}{2}|OF|\\cdot|y|=\\frac{1}{2}\\times1\\times2\\sqrt{3}=\\sqrt{3}."
  },
  {
    "text": "Given that the line $y = kx + m$ intersects the parabola $y^2 = 4x$ at points $P$ and $Q$, and the midpoint of segment $PQ$ has coordinates $(x_0, 2)$, then $k$ equals?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Line;Expression(H) = (y = k*x + m);k: Number;m: Number;P: Point;Q: Point;Intersection(H, G) = {P, Q};Coordinate(MidPoint(LineSegmentOf(P,Q))) = (x0, 2);x0: Number",
    "query_expressions": "k",
    "answer_expressions": "1",
    "fact_spans": "[[[14, 28]], [[14, 28]], [[2, 13]], [[2, 13]], [[68, 71]], [[4, 13]], [[31, 34]], [[35, 38]], [[2, 40]], [[41, 66]], [[54, 66]]]",
    "query_spans": "[[[68, 74]]]",
    "process": "By the given condition, let $ P(x_{1}, y_{1}) $, $ Q(x_{2}, y_{2}) $. Substituting into the equation of the parabola, we obtain  \n$$\n\\begin{cases}\ny_{1}^{2} = 4x_{1} \\\\\ny_{2}^{2} = 4x_{2}\n\\end{cases}\n$$\nSubtracting the two equations gives  \n$ (y_{1} + y_{2})(y_{1} - y_{2}) = 4(x_{1} - x_{2}) $,  \nso  \n$ \\frac{y_{1} - y_{2}}{x_{1} - x_{2}} = \\frac{4}{y_{1} + y_{2}} = \\frac{4}{4} = 1 $,  \nhence $ k = 1 $."
  },
  {
    "text": "A line $l$ with inclination angle $\\frac{\\pi}{4}$ is drawn through the focus of the parabola $y^{2}=8x$. The line $l$ intersects the parabola at points $A$ and $B$. Then, the length of chord $AB$ is?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;Expression(G) = (y^2 = 8*x);PointOnCurve(Focus(G),l);Inclination(l) = pi/4;Intersection(l, G) = {A, B};IsChordOf(LineSegmentOf(A,B),G)",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "16",
    "fact_spans": "[[[38, 43], [44, 49]], [[1, 15], [50, 53]], [[56, 59]], [[60, 63]], [[1, 15]], [[0, 43]], [[19, 43]], [[44, 65]], [[50, 73]]]",
    "query_spans": "[[[68, 77]]]",
    "process": "The focus of the parabola $ y^{2}=8x $ is $ F(2,0) $. The angle of inclination $ \\frac{\\pi}{4} $ indicates a slope of 1, so the line equation is $ y=x-2 $. Solving the system of equations with $ y^{2}=8x $, eliminating $ y $ yields: $ x^{2}-12x+4=0 $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ x_{1}+x_{2}=12 $, and $ |AB|=x_{1}+x_{2}+P=12+4=16 $."
  },
  {
    "text": "Given that point $A$ is the intersection of the axis of symmetry and the directrix of the parabola $x^{2}=8y$, point $B$ is the focus of the parabola, and point $P$ lies on the parabola. In $\\triangle PAB$, $\\sin \\angle PAB = m \\sin \\angle PBA$ ($m \\in \\mathbf{R}$). When $m$ takes its minimum value, point $P$ lies exactly on an ellipse with foci at $A$ and $B$. Then, the eccentricity of the ellipse is?",
    "fact_expressions": "A: Point;G: Parabola;Expression(G) = (x^2 = 8*y);Intersection(SymmetryAxis(G), Directrix(G)) = A;B: Point;Focus(G) = B;P: Point;PointOnCurve(P, G);Sin(TriangleOf(P, A, B)) = m * Sin(TriangleOf(P, B, A));m: Real;WhenMin(m);PointOnCurve(P, H) = True;Focus(H) = {A, B};H: Ellipse",
    "query_expressions": "Eccentricity(H)",
    "answer_expressions": "sqrt(2)-1",
    "fact_spans": "[[[2, 6], [150, 153]], [[7, 21], [37, 40], [48, 51]], [[7, 21]], [[2, 31]], [[32, 36], [154, 157]], [[32, 43]], [[44, 47], [142, 146]], [[44, 52]], [[74, 131]], [[133, 136]], [[132, 141]], [[142, 164]], [[149, 163]], [[161, 163], [166, 168]]]",
    "query_spans": "[[[166, 174]]]",
    "process": "As shown in the figure, draw PH perpendicular to the directrix at H. Since $\\sin\\angle PAB = m \\sin\\angle PBA$, then $|PB| = m|PA| \\Rightarrow m = \\frac{|PB|}{|PA|} = \\sin\\angle PAH$. When $m$ is minimized, $\\sin\\angle PAH$ is minimized; thus, $\\angle PAH$ is minimized when line AP is tangent to the parabola. It is easy to know point $A(0,-2)$. Let the equation of line AP be $y = kx - 2$. Solving simultaneously:\n$$\n\\begin{cases}\nx^{2} = 8y \\\\\ny = kx - 2\n\\end{cases}\n\\Rightarrow x^{2} - 8kx + 16 = 0 \\Rightarrow \\Delta = 64k^{2} - 64 = 0 \\Rightarrow k = \\pm1,\n$$\nso $P(\\pm4,2)$, $|PA| = 4\\sqrt{2}$, $|PH| = |PB| = 4$. At this time, in the ellipse, $2a = |PA| + |PB| = 4\\sqrt{2} + 4$, $2c = 4 \\Rightarrow$ the ellipse's eccentricity $e = \\frac{c}{a} = \\sqrt{2} - 1$."
  },
  {
    "text": "Given that the distance from the right focus to the right directrix of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is equal to $\\frac{1}{3}$ of the focal length, then the eccentricity is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Distance(RightFocus(G),RightDirectrix(G))=(1/3)*FocalLength(G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 48]], [[5, 48]], [[5, 48]], [[2, 48]], [[2, 77]]]",
    "query_spans": "[[[2, 84]]]",
    "process": ""
  },
  {
    "text": "A circle centered at the vertex of the parabola $C$: $y^{2}=2 p x(p>0)$ intersects $C$ at points $A$ and $B$, and intersects the directrix of $C$ at points $D$ and $E$. Given that $|A B|=2 \\sqrt{6}$, $|D E|=2 \\sqrt{10}$, then what is the value of $p$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;G: Circle;Center(G) = Vertex(C);A: Point;B: Point;Intersection(G, C) = {A, B};D: Point;E: Point;Intersection(G, Directrix(C)) = {D, E};Abs(LineSegmentOf(A, B)) = 2*sqrt(6);Abs(LineSegmentOf(D, E)) = 2*sqrt(10)",
    "query_expressions": "p",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 27], [36, 39], [51, 54]], [[1, 27]], [[111, 114]], [[9, 27]], [[34, 35]], [[0, 35]], [[40, 43]], [[44, 47]], [[34, 49]], [[58, 61]], [[62, 65]], [[34, 67]], [[70, 88]], [[90, 109]]]",
    "query_spans": "[[[111, 117]]]",
    "process": "As shown: |AB|=2\\sqrt{6}, |AM|=\\sqrt{6}, |DE|=2\\sqrt{10}, |DN|=\\sqrt{10}, |ON|=\\frac{p}{2}, \\therefore x_{A}=\\frac{(\\sqrt{6})^{2}}{2p}=\\frac{3}{p}, \\because |OD|=|OA|, \\frac{1}{20}x^{2}\\sqrt{|ON|} \\therefore \\frac{p^{2}}{4}+10=\\frac{9}{p^{2}}+6, solve to get:"
  },
  {
    "text": "The equation of the ellipse with foci at $(0, 4)$ and $(0, -4)$, passing through the point $(\\sqrt{5}, -3\\sqrt{3})$, is?",
    "fact_expressions": "G: Ellipse;H: Point;I: Point;J: Point;Coordinate(H) = (0, 4);Coordinate(I) = (0, -4);Focus(G) = {H, I};Coordinate(J) = (sqrt(5), -3*sqrt(3));PointOnCurve(J, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/20 + y^2/36 = 1",
    "fact_spans": "[[[50, 52]], [[3, 12]], [[13, 21]], [[24, 49]], [[3, 12]], [[13, 21]], [[0, 52]], [[24, 49]], [[23, 52]]]",
    "query_spans": "[[[50, 56]]]",
    "process": ""
  },
  {
    "text": "It is known that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $y=\\sqrt{3}x$, and one of its foci coincides with the focus of the parabola $y^{2}=16x$. Find the equation of the hyperbola.",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 16*x);Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x);OneOf(Focus(G))=Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/12 = 1",
    "fact_spans": "[[[2, 58], [83, 84], [112, 115]], [[5, 58]], [[5, 58]], [[90, 105]], [[5, 58]], [[5, 58]], [[2, 58]], [[90, 105]], [[2, 81]], [[83, 110]]]",
    "query_spans": "[[[112, 120]]]",
    "process": ""
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the common foci of the ellipse $C_{1}$ and the hyperbola $C_{2}$, $e_{1}$, $e_{2}$ are the eccentricities of the curves $C_{1}$, $C_{2}$ respectively, $P$ is a common point of the curves $C_{1}$, $C_{2}$, if $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$, and $e_{2} \\in[\\sqrt{3}, 2]$, then $e_{1} \\in$?",
    "fact_expressions": "F1: Point;F2: Point;C1: Ellipse;C2: Hyperbola;Focus(C1) = {F1, F2};Focus(C2) = {F1, F2};e1: Number;e2: Number;Eccentricity(C1) = e1;Eccentricity(C2) = e2;OneOf(Intersection(C1, C2)) = P;P: Point;AngleOf(F1, P, F2) = pi/3;In(e2, [sqrt(3), 2])",
    "query_expressions": "Range(e1)",
    "answer_expressions": "[2*sqrt(13)/13, sqrt(3)/3]",
    "fact_spans": "[[[0, 7]], [[8, 15]], [[16, 25], [62, 71], [88, 97]], [[26, 36], [72, 79], [98, 105]], [[0, 41]], [[0, 41]], [[42, 49]], [[51, 59]], [[42, 83]], [[42, 83]], [[84, 111]], [[84, 87]], [[113, 150]], [[152, 176]]]",
    "query_spans": "[[[178, 190]]]",
    "process": ""
  },
  {
    "text": "Given point $A(-2,1)$, the focus of $y^{2}=-4x$ is $F$, and $P$ is a point on $y^{2}=-4x$. To minimize $|PA|+|PF|$, what are the coordinates of point $P$?",
    "fact_expressions": "A: Point;Coordinate(A) = (-2, 1);G: Curve;Expression(G) = (y^2 = -4*x);F: Point;Focus(G) = F;P: Point;PointOnCurve(P, G);WhenMin(Abs(LineSegmentOf(P, A))+Abs(LineSegmentOf(P, F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(-1/4, 1)",
    "fact_spans": "[[[2, 12]], [[2, 12]], [[14, 26], [38, 50]], [[14, 26]], [[30, 33]], [[14, 33]], [[34, 37], [73, 77]], [[34, 53]], [[56, 72]]]",
    "query_spans": "[[[73, 82]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a point on the hyperbola $C$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and point $M$ satisfies $|\\overrightarrow{O M}|=1$ and $\\overrightarrow{O M} \\cdot \\overrightarrow{P M}=0$, what is the distance from the point $P$ to the asymptotes of the hyperbola $C$ when $|\\overrightarrow{P M}|$ attains its minimum value?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/9 - y^2/16 = 1);P: Point;PointOnCurve(P, C);M: Point;O: Origin;Abs(VectorOf(O, M)) = 1;DotProduct(VectorOf(O, M), VectorOf(P, M)) = 0;WhenMin(Abs(VectorOf(P, M)))",
    "query_expressions": "Distance(P, Asymptote(C))",
    "answer_expressions": "12/5",
    "fact_spans": "[[[6, 50], [178, 184]], [[6, 50]], [[2, 5], [173, 177]], [[2, 53]], [[54, 58]], [[60, 86]], [[60, 86]], [[88, 139]], [[141, 172]]]",
    "query_spans": "[[[173, 193]]]",
    "process": "As shown in the figure: $\\overrightarrow{OM}\\cdot\\overrightarrow{PM}=0$, so $PM$ is tangent to the circle $x^{2}+y^{2}=1$. Since $|OP|^{2}=|OM|^{2}+|MP|^{2}$, $|MP|$ is minimized when $|OP|$ is minimized. When $|OP|$ takes its minimum value, $P$ is located at one of the left or right vertices; without loss of generality, take $P(3,0)$ and the asymptote as $4x-3y=0$. Therefore, the distance from $P$ to the asymptote is: $\\frac{|4\\times3|}{\\sqrt{2}+2^{2}}=\\frac{12}{5}$."
  },
  {
    "text": "The ellipse $C$ has its center at the origin, and its right focus $F$ is the intersection point of the line $x - 2y - 2 = 0$ and the $x$-axis. Among all chords passing through the point $F$, the shortest chord has length $\\frac{10}{3}$. Then the equation of $C$ is?",
    "fact_expressions": "C: Ellipse;G: Line;F: Point;O: Origin;Expression(G) = (x - 2*y - 2 = 0);Center(C)=O;RightFocus(G)=F;Intersection(G, xAxis) = F;H:LineSegment;PointOnCurve(F,H);IsChordOf(H,C);Min(Length(H))=10/3",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/9+y^2/5=1",
    "fact_spans": "[[[2, 7], [82, 85]], [[23, 36]], [[19, 22], [49, 53]], [[11, 15]], [[23, 36]], [[2, 15]], [[2, 22]], [[19, 44]], [], [[2, 57]], [[2, 57]], [[2, 80]]]",
    "query_spans": "[[[82, 90]]]",
    "process": "From the given, F(2,0), let C: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0), then \\begin{cases}c=2,\\\\\\frac{2b^{2}}{a}=\\frac{10}{3},\\\\a^{2}=b^{2}+c^{2}\\end{cases} solving gives a=3, b=\\sqrt{5}, c=2, so the equation of C is \\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1"
  },
  {
    "text": "Given that for any point on the parabola $y^{2}=16x$, the distance to the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is greater by $1$ than the distance to the left directrix, find $a^{2}$.",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 16*x);D: Point;PointOnCurve(D, H) ;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;Distance(D, RightFocus(G)) = Distance(D, LeftDirectrix(G)) + 1",
    "query_expressions": "a^2",
    "answer_expressions": "12",
    "fact_spans": "[[[2, 17]], [[2, 17]], [], [[2, 22]], [[23, 69]], [[23, 69]], [[26, 69]], [[26, 69]], [[2, 87]]]",
    "query_spans": "[[[89, 98]]]",
    "process": "(Analysis) Using the parabola equation, find the coordinates of the focus and the equation of the directrix. According to the given conditions, obtain the coordinates of the right focus and the equation of the left directrix of the hyperbola, thereby finding $c$ and $a^{2}$. For the parabola $y^{2}=16x$, $p=8$, the focus is $P(4,0)$, and the directrix equation is $x=-4$; based on the definition of a parabola: the distance from any point on the parabola to the focus equals its distance to the directrix, it follows that for the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, the right focus is $F(4,0)$, and the left directrix equation is $x=-3$, $\\therefore c=4$, and $-\\frac{a^{2}}{c}=-3$, solving gives $a^{2}=12$."
  },
  {
    "text": "Given that an asymptote of the hyperbola $kx^{2}-y^{2}=1$ is perpendicular to the line $2x+y+1=0$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (k*x^2 - y^2 = 1);k: Number;H: Line;Expression(H) = (2*x + y + 1 = 0);IsPerpendicular(OneOf(Asymptote(G)), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 21], [46, 49]], [[2, 21]], [[5, 21]], [[28, 41]], [[28, 41]], [[2, 43]]]",
    "query_spans": "[[[46, 55]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the common foci of an ellipse and a hyperbola, $P$ is one of their common points, and $|P F_{2}| > |P F_{1}|$, the eccentricity of the ellipse is $e_{1}$, the eccentricity of the hyperbola is $e_{2}$. If $|P F_{1}| = |F_{1} F_{2}|$, then the minimum value of $\\frac{3}{e_{1}} + \\frac{e_{2}}{3}$ is?",
    "fact_expressions": "H: Ellipse;G: Hyperbola;F1: Point;F2: Point;Focus(H) = {F1, F2};Focus(G) = {F1, F2};P: Point;OneOf(Intersection(H, G)) = P;Abs(LineSegmentOf(P, F2)) > Abs(LineSegmentOf(P, F1));e1: Number;Eccentricity(H) = e1;e2: Number;Eccentricity(G) = e2;Abs(LineSegmentOf(P, F1)) = Abs(LineSegmentOf(F1, F2))",
    "query_expressions": "Min(e2/3 + 3/e1)",
    "answer_expressions": "8",
    "fact_spans": "[[[18, 20], [66, 68]], [[21, 24], [81, 84]], [[2, 9]], [[10, 17]], [[2, 29]], [[2, 29]], [[30, 33]], [[30, 42]], [[44, 65]], [[73, 80]], [[66, 80]], [[89, 96]], [[81, 96]], [[98, 123]]]",
    "query_spans": "[[[125, 164]]]",
    "process": ""
  },
  {
    "text": "Given that the distance between the left vertex of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ and the focus of the parabola $y^{2}=2px$ $(p>0)$ is $4$, and the intersection point of one asymptote of the hyperbola and the directrix of the parabola is $(-2,-1)$, find the focal length of the hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;p: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);p>0;Expression(H) = (y^2 = 2*(p*x));Coordinate(Intersection(OneOf(Asymptote(G)),Directrix(H))) = (-2, -1);Distance(LeftVertex(G),Focus(H)) = 4",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[2, 59], [97, 100], [130, 133]], [[5, 59]], [[5, 59]], [[64, 85], [107, 110]], [[67, 85]], [[5, 59]], [[5, 59]], [[2, 59]], [[67, 85]], [[64, 85]], [[97, 128]], [[2, 95]]]",
    "query_spans": "[[[130, 138]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $y=x$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (y = x)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 58], [74, 77]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 72]]]",
    "query_spans": "[[[74, 83]]]",
    "process": "From the asymptotic line equation of the hyperbola $ y = x $, we get $ \\frac{b}{a} = 1 $, so the eccentricity $ e = \\frac{c}{a} = \\sqrt{1 + \\left( \\frac{b}{a} \\right)^2} = \\sqrt{2} $."
  },
  {
    "text": "The x-coordinate of the point on the parabola $y^{2}=4 x$ whose distance to its focus is equal to $6$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P : Point;PointOnCurve(P, G);Distance(P, Focus(G)) = 6",
    "query_expressions": "XCoordinate(P)",
    "answer_expressions": "5",
    "fact_spans": "[[[0, 14], [16, 17]], [[0, 14]], [[28, 29]], [[0, 29]], [[0, 29]]]",
    "query_spans": "[[[28, 35]]]",
    "process": "The abscissa can be obtained directly from the definition of a parabola. The directrix equation of the parabola $ y^{2} = 4x $ is $ x = -1 $. Since the distance from a point on the parabola $ y^{2} = 4x $ to the focus is equal to 6, according to the property that the distance from any point on a parabola to the focus equals the distance from that point to the directrix, the abscissa of the desired point is 5. The answer is: 5"
  },
  {
    "text": "The equation $\\frac{x^{2}}{m^{2}+1}+\\frac{y^{2}}{m-2}=1$ represents a hyperbola; then the range of real values for $m$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(m^2 + 1) + y^2/(m - 2) = 1);m: Real",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-oo, 2)",
    "fact_spans": "[[[47, 50]], [[0, 50]], [[52, 57]]]",
    "query_spans": "[[[52, 64]]]",
    "process": "Convert to the standard form equation $\\frac{x^2}{m^{2}+1}-\\frac{y^{2}}{2-m}=1$. From $2-m>0$, we get $m<2$, hence $m\\in(-\\infty,2)$"
  },
  {
    "text": "The distance from the focus of the parabola $y^{2}=16 x$ to the asymptotes of the hyperbola $\\frac{x^{2}}{12}-\\frac{y^{2}}{4}=1$ is?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 16*x);G: Hyperbola;Expression(G) = (x^2/12 - y^2/4 = 1)",
    "query_expressions": "Distance(Focus(H), Asymptote(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 15]], [[0, 15]], [[19, 58]], [[19, 58]]]",
    "query_spans": "[[[0, 67]]]",
    "process": "The focus of the parabola is $ F(4,0) $, and the asymptotes of the hyperbola are given by $ y = \\pm\\frac{\\sqrt{3}}{3}x $, or equivalently $ x \\pm \\sqrt{3}y = 0 $. Then the distance from the point to the line is $ d = \\frac{4}{\\sqrt{1^{2} + (\\pm\\sqrt{3})^{2}}} = 2 $."
  },
  {
    "text": "Given that the distance between the directrix of the parabola $y^{2}=m x$ ($m>0$) and the line $x=1$ is $3$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;m: Number;H: Line;m>0;Expression(G) = (y^2 = m*x);Expression(H) = (x = 1);Distance(Directrix(G) , H) = 3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = 8*x",
    "fact_spans": "[[[1, 20], [41, 44]], [[4, 20]], [[24, 31]], [[4, 20]], [[1, 20]], [[24, 31]], [[1, 38]]]",
    "query_spans": "[[[41, 48]]]",
    "process": "According to the parabolic equation, obtain the equation of the directrix. Combine with the given distance between the directrix and $x=1$ to form an equation, solve the equation to find the value of $m$, and thus obtain the parabolic equation. [Detailed Solution] Since $m>0$, the equation of the directrix of the parabola is $x=-\\frac{m}{4}$. According to the problem, the distance between the directrix $x=-\\frac{m}{4}$ and the line $x=1$ is 3, so $\\frac{m}{4}+1=3$, solving gives $m=8$. Therefore, the equation of the parabola is $y^{2}=8x$. Hence, fill in: $y^{2}=8x$. [Insight] This question mainly examines the directrix equation of a parabola and is a basic problem."
  },
  {
    "text": "If the curve $\\frac{x^{2}}{4+k}+\\frac{y^{2}}{1-k}=1$ represents an ellipse, then the range of values for $k$ is?",
    "fact_expressions": "G: Ellipse;H: Curve;k: Number;Expression(H) = (x^2/(k + 4) + y^2/(1 - k) = 1);H = G",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-4,-3/2)+(-3/2,1)",
    "fact_spans": "[[[44, 46]], [[1, 42]], [[48, 51]], [[1, 42]], [[1, 46]]]",
    "query_spans": "[[[48, 58]]]",
    "process": "From the given conditions, we have (4+k)(1-k)>0 and 4+k≠1-k. Solving these inequalities yields -4<k<1 and k≠-3/2. Therefore, the answer should be k∈(-4,-3/2)∪(-3/2,1)."
  },
  {
    "text": "If one of the asymptotes of the hyperbola $x^{2}+ky^{2}=1$ is $y=\\frac{1}{2} x$, then the value of the real number $k$ is?",
    "fact_expressions": "G: Hyperbola;k: Real;Expression(G)=(k*y^2+x^2=1);Expression(OneOf(Asymptote(G))) = (y = x/2)",
    "query_expressions": "k",
    "answer_expressions": "-4",
    "fact_spans": "[[[1, 20]], [[48, 53]], [[1, 20]], [[1, 46]]]",
    "query_spans": "[[[48, 57]]]",
    "process": ""
  },
  {
    "text": "The equation of the parabola with the origin as its focus and the line $x+y-1=0$ as its directrix is?",
    "fact_expressions": "G: Parabola;H: Line;O:Origin;Expression(H) = (x + y - 1 = 0);Focus(G)=O;Directrix(G) = H",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2+y^2-2*x*y+2*x+2*y-1=0",
    "fact_spans": "[[[25, 28]], [[10, 21]], [[1, 5]], [[10, 21]], [[0, 28]], [[9, 28]]]",
    "query_spans": "[[[25, 32]]]",
    "process": ""
  },
  {
    "text": "If on the hyperbola $\\frac{x^{2}}{m+2}-\\frac{y^{2}}{2-m}=1$ $(-2<m<2)$, the absolute value of the difference of distances from a point to $A(-2,0)$ and $B(2,0)$ is $2 \\sqrt{3}$, then what is the length of the imaginary axis of the hyperbola?",
    "fact_expressions": "G: Hyperbola;m: Number;A: Point;B:Point;P:Point;m>-2;m < 2;Expression(G) = (x^2/(m + 2) - y^2/(2 - m) = 1);Coordinate(A) = (-2, 0);Coordinate(B) = (2, 0);PointOnCurve(P, G);Abs(Distance(P,A)-Distance(P,B))=2*sqrt(3)",
    "query_expressions": "Length(ImageinaryAxis(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 51], [102, 105]], [[4, 51]], [[55, 65]], [[67, 75]], [[53, 54]], [[4, 51]], [[4, 51]], [[1, 51]], [[55, 64]], [[67, 75]], [[1, 54]], [[1, 99]]]",
    "query_spans": "[[[102, 111]]]",
    "process": "From the given condition, c^{2}=m+2+2-m=4, then A(-2,0), B(2,0) are the left and right foci of the hyperbola, respectively. Thus, 2a=2\\sqrt{m+2}=2\\sqrt{3}, solving gives m=1, hence b^{2}=1, and the length of the imaginary axis is 2b=2."
  },
  {
    "text": "Given that one focus of the hyperbola $x^{2}-\\frac{y^{2}}{8}=k$ is $(0,-3)$, what is the value of $k$? What are the equations of the asymptotes?",
    "fact_expressions": "G: Hyperbola;k: Number;Expression(G) = (x^2 - y^2/8 = k);Coordinate(OneOf(Focus(G))) = (0, -3)",
    "query_expressions": "k;Expression(Asymptote(G))",
    "answer_expressions": "-1\ny=pm*2*sqrt(2)*x",
    "fact_spans": "[[[2, 30]], [[46, 49]], [[2, 30]], [[2, 44]]]",
    "query_spans": "[[[46, 53]], [[2, 60]]]",
    "process": "Convert the curve equation into standard form; since one focus of the hyperbola is (0, -3), solve for k. Using the standard equation of the hyperbola, the asymptotes can then be found. [Detailed solution] The hyperbola $ x^{2} - \\frac{y^{2}}{8} = k $ is converted into standard form $ \\frac{x^{2}}{k} - \\frac{y^{2}}{8k} = 1 $. Since one focus of the hyperbola is $ (0, -3) $, $ \\therefore k < 0 $, and the standard form becomes $ \\frac{y^{2}}{-8k} - \\frac{x^{2}}{-k} = 1 $. $ \\therefore -8k - k = (-3)^{2} $, solving gives $ k = -1 $; $ \\therefore $ the hyperbola $ \\frac{y^{2}}{8} - x^{2} = 1 $, its asymptotes are $ y = \\pm 2\\sqrt{2}x $."
  },
  {
    "text": "If the equation of an ellipse is $16 x^{2}+25 y^{2}=400$, then the focal distance of this ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (16*x^2 + 25*y^2 = 400)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 3], [33, 35]], [[1, 30]]]",
    "query_spans": "[[[33, 40]]]",
    "process": "The equation $16x^{2}+25y^{2}=400$, that is, $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, so $a^{2}=25$, $b^{2}=16$, then $c^{2}=a^{2}-b^{2}=9$, solving gives $c=3$, hence the focal length $2c$ is $6$."
  },
  {
    "text": "Given the line $l$: $4x - 3y + 6 = 0$, and a moving point on the parabola $C$: $y^2 = 4x$, find the minimum value of the sum of the distance from the point to the line $l$ and the distance to the $y$-axis.",
    "fact_expressions": "l: Line;Expression(l) = (4*x - 3*y + 6 = 0);C: Parabola;Expression(C) = (y^2 = 4*x);P: Point;PointOnCurve(P, C)",
    "query_expressions": "Min(Distance(P, l)+Distance(P, yAxis))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 22], [50, 55]], [[2, 22]], [[23, 42]], [[23, 42]], [], [[23, 49]]]",
    "query_spans": "[[[23, 71]]]",
    "process": "Let the distance from a point on the parabola to the line 4x - 3y + 6 = 0 be d_{1}, the distance to the directrix be d_{2}, and the distance to the y-axis be d_{3}. The distance to the focus is equal, d_{2} = |PF|, and d_{1} + d_{3} = d_{1} + d_{2} - 1 = d_{1} + |PF| - 1. As shown in the figure, the minimum value of d_{1} + |PF| is the distance from the focus F to the line 4x - 3y + 6 = 0. The distance from the focus F to the line 4x - 3y + 6 = 0 is d = \\frac{|4 \\times 1 - 0 + 6|}{\\sqrt{4^{2} + 3^{2}}} = 2. Therefore, the minimum value of d_{1} + |PF| - 1 is 1."
  },
  {
    "text": "Let $P$ be a point on the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and let $M$, $N$ be points on the two circles $(x-5)^{2}+y^{2}=4$ and $(x+5)^{2}+y^{2}=1$, respectively. Then the maximum value of $|P M|-|P N|$ is?",
    "fact_expressions": "P: Point;G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);PointOnCurve(P, G);H1: Circle;Expression(H1) = (y^2 + (x - 5)^2 = 4);H2: Circle;Expression(H2) = (y^2 + (x + 5)^2 = 1);M: Point;N: Point;PointOnCurve(M, H1)  ;PointOnCurve(N,H2)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "9",
    "fact_spans": "[[[1, 4]], [[5, 44]], [[5, 44]], [[1, 47]], [[61, 80]], [[61, 80]], [[81, 100]], [[81, 100]], [[48, 51]], [[52, 55]], [[48, 103]], [[48, 103]]]",
    "query_spans": "[[[105, 124]]]",
    "process": "[F Analysis] According to the given conditions and the equation of the circle, using geometric knowledge, when points P, M, and B are collinear, |PM| - |PN| reaches its maximum value. Let the centers of the two circles (x-5)^{2}+y^{2}=4 and (x+5)^{2}+y^{2}=1 be A and B, respectively. Then A and B are exactly the two foci of a hyperbola. Thus, |PM| - |PN| \\leqslant |PA| + 2 - (|PB| - 1) = |PA| - |PB| + 3 = 2a + 3 = 6 + 3 = 9. Hence, the maximum value is 9."
  },
  {
    "text": "If a point on the parabola $x=\\frac{1}{12} y^{2}$ is at a distance of $9$ from the focus, then the coordinates of this point are?",
    "fact_expressions": "G: Parabola;Expression(G) = (x = y^2/12);P: Point;PointOnCurve(P,G);Distance(Focus(G),P) = 9",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(6, pm*6*sqrt(2))",
    "fact_spans": "[[[1, 26]], [[1, 26]], [], [[1, 29]], [[1, 39]]]",
    "query_spans": "[[[1, 48]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, let the left vertex be $A$, and $O$ be the coordinate origin. If there exists a point $M$ on the ellipse such that $OM \\perp MA$, then the range of the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;O: Origin;M: Point;A: Point;e: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(G) = A;PointOnCurve(M, G);IsPerpendicular(LineSegmentOf(O, M), LineSegmentOf(M, A));Eccentricity(G) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "(\\sqrt{2}/2, 1)",
    "fact_spans": "[[[2, 54], [72, 74], [99, 101]], [[4, 54]], [[4, 54]], [[62, 65]], [[77, 81]], [[58, 61]], [[105, 108]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 61]], [[72, 81]], [[82, 97]], [[99, 108]]]",
    "query_spans": "[[[105, 115]]]",
    "process": "The trajectory equation of M is: $(x+\\frac{a}{2})^{2}+y^{2}=\\frac{a^{2}}{4},(y\\neq0)$, combining equations and simplifying yields $\\frac{a2-b^{2}}{a^{2}}x^{2}+ax+b^{2}=0$, based on the symmetry axis of the corresponding function, it is concluded that there exists a point M on the ellipse such that $OM\\bot MA$, i.e., the trajectory equation of M is: $(x+\\frac{a}{2})^{2}+y^{2}=\\frac{a^{2}}{4},(y\\neq0)$, $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ simplifies to $\\frac{a2-b^{2}}{a^{2}}x^{2}+ax+b^{2}=0$, $(x+\\frac{a}{2})^{2}+y^{2}=\\frac{a^{2}}{4}$, it is clear that $x=-a$ is a solution of the equation, and when $c=0$, the equation has a solution in $(-a,0)$, only need to satisfy: $0>\\frac{a-b^{2}}{a^{2}x+a}+b^{2}>0.2\\frac{a^{2}-b^{2}}{a}$, solving gives $\\frac{c}{a}=e>\\frac{\\sqrt{2}}{2}$."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $E$: $y^{2}=2 p x(p>0)$, a line $l$ with inclination angle $30^{\\circ}$ is drawn through $F$, intersecting the parabola $E$ at points $A$ and $B$. Perpendiculars are drawn from $A$ and $B$ to the directrix of $E$, with feet at $C$ and $D$, respectively. Let $M$ be the midpoint of $C D$. Then $|M F|$=?",
    "fact_expressions": "E: Parabola;Expression(E) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(E) = F;l: Line;Inclination(l) = ApplyUnit(30, degree);PointOnCurve(F, l);A: Point;B: Point;Intersection(l, E) = {A, B};L1: Line;L2: Line;C: Point;D: Point;PointOnCurve(A, L1);IsPerpendicular(L1, Directrix(E));FootPoint(L1, Directrix(E)) = C;PointOnCurve(B, L2);IsPerpendicular(L2, Directrix(E));FootPoint(L2, Directrix(E)) = D;M: Point;MidPoint(LineSegmentOf(C, D)) = M",
    "query_expressions": "Abs(LineSegmentOf(M, F))",
    "answer_expressions": "2*p",
    "fact_spans": "[[[6, 32], [64, 70], [91, 94]], [[6, 32]], [[14, 32]], [[14, 32]], [[2, 5], [37, 40]], [[2, 35]], [[58, 63]], [[41, 63]], [[36, 63]], [[72, 75], [83, 86]], [[76, 79], [87, 90]], [[58, 81]], [], [], [[106, 109]], [[110, 113]], [[82, 100]], [[82, 100]], [[82, 113]], [[82, 100]], [[82, 100]], [[82, 113]], [[125, 128]], [[116, 128]]]",
    "query_spans": "[[[130, 139]]]",
    "process": "Analysis: First, write the equation of the line, then solve it simultaneously with the equation of the parabola to obtain the coordinates of the midpoint N of AB, and then use the Pythagorean theorem to find |MF|. Let the equation of the line be $ y-0=\\frac{\\sqrt{3}}{3}(x-\\frac{p}{2})=\\frac{\\sqrt{3}}{3}x-\\frac{\\sqrt{3}}{6}p $. Solving the line and parabola equations simultaneously yields $ 4x^{2}-28px+p^{2}=0 $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ x_{1}+x_{2}=7p $, so the horizontal coordinate of the midpoint N of AB is $ \\frac{x_{1}+x_{2}}{2}=\\frac{7}{2}p $, and thus the vertical coordinate of N is $ y=\\frac{\\sqrt{3}}{3}\\times\\frac{7}{2}p-\\frac{\\sqrt{3}}{6}p=\\sqrt{3}p $. Therefore, $ |MF|=\\sqrt{3p^{2}+p^{2}}=2p $. Hence, fill in $ 2p $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ with focus $F$, and directrix $l$ intersecting the $x$-axis at point $K$. Draw a line through $F$ with inclination angle $\\alpha$ intersecting $C$ at points $A$ and $B$. If $\\angle A K B=60^{\\circ}$, then $\\sin \\alpha=?$",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(C) = F;l: Line;Directrix(C) = l;K: Point;Intersection(l, xAxis) = K;G: Line;PointOnCurve(F, G);A: Point;B: Point;Intersection(G, C) = {A, B};alpha: Number;Slope(G) = alpha;AngleOf(A, K, B) = ApplyUnit(60, degree)",
    "query_expressions": "Sin(alpha)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[2, 28], [73, 76]], [[2, 28]], [[10, 28]], [[10, 28]], [[32, 35], [53, 56]], [[2, 35]], [[38, 41]], [[2, 41]], [[47, 51]], [[38, 51]], [[70, 72]], [[52, 72]], [[78, 81]], [[82, 85]], [[70, 87]], [[61, 69]], [[57, 72]], [[89, 114]]]",
    "query_spans": "[[[116, 131]]]",
    "process": "According to the problem, $ F(\\frac{p}{2},0) $, $ K(-\\frac{p}{2},0) $, let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, as shown in the figure, $ A $ is in the first quadrant. Let the equation of line $ AB $ be $ x = my + \\frac{p}{2} $, from $ \\begin{cases} x = my + \\frac{p}{2} \\\\ y^{2} = 2px \\end{cases} $, eliminating $ x $ gives $ y^{2} - 2pmy - p^{2} = 0 $, so $ y_{1} + y_{2} = 2pm $, $ y_{1}y_{2} = -p^{2} $, $ \\Delta = 4p^{2}m^{2} + 4p^{2} > 0 $, $ \\frac{y_{1} - 0}{x_{1} + \\frac{p}{2}} = \\frac{y_{1}}{x_{1} + \\frac{p}{2}} $, $ k_{KB} = \\frac{y_{2}}{x_{2} + \\frac{p}{2}} $, $ \\angle AKB = 60^{\\circ} $, so $ \\tan \\angle AKB = \\sqrt{3} $, $ \\frac{k_{KA} - k_{KB}}{1 + k_{KA} \\cdot k_{KB}} = \\sqrt{3} $, $ \\frac{1}{2} + \\sqrt{3}pm(y_{1} + y_{2}) + \\sqrt{3}p^{2} $, $ y_{1}y_{2} + \\sqrt{3}pm(y_{1} + y_{2}) + \\sqrt{3}p^{2} - \\sqrt{3}(m^{2} + 1)p^{2} + 2\\sqrt{3}p^{2}m^{2} + \\sqrt{3}p^{2} $, eliminating $ p^{2} $ and simplifying gives $ 2\\sqrt{m^{2}+1} = \\sqrt{3}m^{2} $, i.e., $ 3m^{4} - 4m^{2} - 4 = 0 $, solving yields $ m^{2} = 2 $, discard the negative root. So $ \\tan^{2}\\alpha = \\frac{1}{m^{2}} = \\frac{1}{2} = \\frac{\\sin^{2}\\alpha}{\\cos^{2}\\alpha} = \\frac{\\sin^{2}\\alpha}{1 - \\sin^{2}\\alpha} $, $ \\sin^{2}\\alpha = \\frac{1}{3} $, since $ \\alpha \\in [0,\\pi) $, so $ \\sin\\alpha = \\frac{\\sqrt{3}}{3} $."
  },
  {
    "text": "Given point $A(3,1)$, the focus of the parabola $y^{2}=4x$ is $F$, and point $P$ lies on the parabola. Then the minimum value of $|PF| + |PA|$ is?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (3, 1);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "4",
    "fact_spans": "[[[12, 26], [39, 42]], [[2, 11]], [[34, 38]], [[30, 33]], [[12, 26]], [[2, 11]], [[12, 33]], [[34, 43]]]",
    "query_spans": "[[[45, 64]]]",
    "process": "From the given conditions, the equation of the directrix of the parabola is $ x = -1 $. Draw a perpendicular line from point $ P $ to the directrix of the parabola, with foot of perpendicular at point $ D $, then $ |PD| = |PF| $. It is clear that when points $ A $, $ P $, and $ D $ are collinear, $ |PD| + |PA| $ attains the minimum value $ 3 - (-1) = 4 $. Therefore, the minimum value of $ |PF| + |PA| $ is $ 4 $."
  },
  {
    "text": "Given that the eccentricity of hyperbola $C$ is $2$, and one of its foci is $(0,2)$, what is the standard equation of hyperbola $C$? What are the equations of its asymptotes?",
    "fact_expressions": "C: Hyperbola;Eccentricity(C) = 2;Coordinate(OneOf(Focus(C)))=(0,2)",
    "query_expressions": "Expression(C);Expression(Asymptote(C))",
    "answer_expressions": "y^2 - x^2/3 = 1; y = pm*(sqrt(3)/3)*x",
    "fact_spans": "[[[2, 8], [18, 19], [34, 40]], [[2, 16]], [18, 31]]",
    "query_spans": "[[[34, 47]], [[34, 55]]]",
    "process": "From the given conditions, c=2, e=\\frac{c}{a}=2\\Rightarrow a=\\frac{c}{2}=1\\Rightarrow b=\\sqrt{c^{2}-a^{2}}=\\sqrt{3}, thus the standard equation of the hyperbola is y^2-\\frac{x^{2}}{3}=1, and the asymptotes are y=\\pm\\frac{a}{\\frac{a}{3}x}=\\pm\\frac{\\sqrt{3}}{3}x"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the asymptotes of which are tangent to the circle $E$: $(x-5)^{2}+y^{2}=9$, then the eccentricity of the hyperbola $C$ is equal to?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;E: Circle;Expression(E) = ((x-5)^2 + y^2 = 9);IsTangent(Asymptote(C), E) = True",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "5/4",
    "fact_spans": "[[[2, 63], [96, 102]], [[2, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[68, 92]], [[68, 92]], [[2, 94]]]",
    "query_spans": "[[[96, 109]]]",
    "process": "According to the asymptote of the hyperbola being tangent to the circle $ E: (x-5)^{2}+y^{2}=9 $, the distance from the center $ (5,0) $ to the asymptote equals the radius $ r=3 $. Using the point-to-line distance formula and the eccentricity formula, we can derive the result. Let one asymptote be $ bx-ay=0 $. The given circle has center $ (5,0) $ and radius $ r=3 $. Since the asymptote is tangent to the circle $ E: (x-5)^{2}+y^{2}=9 $, we have $ \\frac{|5b|}{\\sqrt{b^{2}+a^{2}}}=3 $, thus $ \\frac{9}{16}a^{2}=b^{2} $, therefore $ e=\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\frac{5}{4} $."
  },
  {
    "text": "Given the parabola $x^{2}=2 p y(p>0)$ with focus $F$, points $P$ and $Q$ lie on the parabola such that $\\angle P F Q=\\frac{5 \\pi}{6}$. A perpendicular is drawn from the midpoint $M$ of chord $P Q$ to the directrix $l$, with foot of the perpendicular at $M_{1}$. Then the minimum value of $\\frac{|P Q|}{|M M_{1}|}$ is?",
    "fact_expressions": "G: Parabola;p: Number;P: Point;Q: Point;F: Point;M: Point;M1: Point;p>0;Expression(G) = (x^2 = 2*(p*y));Focus(G) = F;PointOnCurve(P, G);PointOnCurve(Q, G);AngleOf(P, F, Q) = 5*pi/6;MidPoint(LineSegmentOf(P, Q)) = M;L:Line;PointOnCurve(M,L);Directrix(G)=l;IsPerpendicular(L,l);FootPoint(L,l)=M1;IsChordOf(LineSegmentOf(P,Q),G);l:Line",
    "query_expressions": "Min(Abs(LineSegmentOf(P, Q))/Abs(LineSegmentOf(M, M1)))",
    "answer_expressions": "(\\sqrt{6}+\\sqrt{2})/2",
    "fact_spans": "[[[2, 23], [40, 43]], [[5, 23]], [[31, 35]], [[36, 39]], [[27, 30]], [[87, 90]], [[103, 110]], [[5, 23]], [[2, 23]], [[2, 30]], [[31, 44]], [[36, 44]], [[46, 76]], [[79, 90]], [], [[77, 99]], [[40, 96]], [[77, 99]], [[77, 110]], [[40, 84]], [[93, 96]]]",
    "query_spans": "[[[112, 143]]]",
    "process": "Analysis: Draw perpendiculars PA and QB from P and Q to the directrix, with feet A and B respectively. Let |PF| = 2a, |QF| = 2b. Then |MM₁| = a + b. By the law of cosines, PQ² = 4a² + 4b² + 4√3ab. Then, using the basic inequality, find the range of |AB|, thereby obtaining the solution to this problem. Details as follows: Draw perpendiculars PA and QB from P and Q to the directrix, with feet A and B respectively. Let |PF| = 2a, |QF| = 2b. By the definition of parabola, |PF| = |PA|, |QF| = |QB|. In trapezoid ABQP, 2|MM| = |PA| + |QF| = 2a + 2b. ∴ |MM₁| = a + b. By the law of cosines, PQ² = 4a² + 4b² - 8ab cos(5/6)π = 4a² + 4b² + 4√3ab = 4(a + b)² + (4√3 - 8)ab ≥ 4(a + b)² + (4√3 - 8) × ((a + b)/2)² = (2 + √3)(a + b)². Then the minimum value of |PQ| / |MM| is √(2 + √3) = (√2 + √6)/2."
  },
  {
    "text": "On the parabola $y=2 x^{2}$, there is a moving chord $A B$ with midpoint $M$, and the length of chord $A B$ is $3$. Then the minimum value of the $y$-coordinate of point $M$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 2*x^2);A: Point;B: Point;IsChordOf(LineSegmentOf(A,B),G) = True;MidPoint(LineSegmentOf(A,B)) = M;M: Point;Length(LineSegmentOf(A,B)) = 3",
    "query_expressions": "Min(YCoordinate(M))",
    "answer_expressions": "11/8",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[34, 39]], [[34, 39]], [[0, 24]], [[19, 31]], [[47, 51], [28, 31]], [[34, 45]]]",
    "query_spans": "[[[47, 61]]]",
    "process": ""
  },
  {
    "text": "The line $m$ intersects the ellipse $\\frac{x^{2}}{4^{2}}+y^{2}=1$ at points $P_{1}$ and $P_{2}$, with the midpoint of segment $P_{1}P_{2}$ being $P$. Let the slope of line $m$ be $k_{1}$ $(k_{1} \\neq 0)$, and the slope of line $OP$ (where $O$ is the origin) be $k_{2}$. Then $k_{1}k_{2}=$?",
    "fact_expressions": "m: Line;G: Ellipse;O: Origin;P: Point;P1: Point;P2: Point;Expression(G) = (x^2/4^2 + y^2 = 1);Intersection(m, G) = {P1, P2};MidPoint(LineSegmentOf(P1,P2)) = P;Slope(m) = k1;Slope(LineOf(O,P))=k2;k1:Number;k2:Number;Negation(k1=0)",
    "query_expressions": "k1*k2",
    "answer_expressions": "-1/16",
    "fact_spans": "[[[0, 5], [81, 86]], [[6, 37]], [[122, 125]], [[76, 79]], [[39, 46]], [[47, 54]], [[6, 37]], [[0, 56]], [[57, 79]], [[81, 111]], [[112, 142]], [[90, 111]], [[135, 142]], [[90, 111]]]",
    "query_spans": "[[[144, 159]]]",
    "process": "Let $P_{1}(x_{1},y_{1}), P_{2}(x_{2},y_{2}), P(x_{0},y_{0})$, then  \n\\[\n\\begin{cases}\n\\frac{x_{1}^{2}}{4^{2}} + y_{1}^{2} = 1 \\textcircled{1} \\\\\n\\frac{x_{2}^{2}}{2} + y_{2}^{2} = 1 \\textcircled{2}\n\\end{cases}\n\\]\n$\\textcircled{1} - \\textcircled{2}$ yields:  \n$\\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = -\\frac{1}{4^{2}} \\cdot \\frac{x_{0}}{y_{0}} = -\\frac{1}{16} \\cdot \\frac{x_{0}}{y_{0}}$,  \ni.e., $k_{1} \\cdot \\frac{y_{0}}{x_{0}} = -\\frac{1}{16}$,  \ni.e., $k_{1} \\cdot k_{2} = -\\frac{1}{16}$"
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b}=1$ $(a>0, b>0)$, respectively. Point $P$ lies on $E$. If the midpoint of segment $P F_{2}$ lies on the $y$-axis and $\\angle P F_{2} F_{1}=60^{\\circ}$, then the eccentricity of $E$ is?",
    "fact_expressions": "E: Hyperbola;b: Number;a: Number;F2: Point;P: Point;F1: Point;a>0;b>0;Expression(E) = (-y^2/b + x^2/a^2 = 1);LeftFocus(E) = F1;RightFocus(E) = F2;PointOnCurve(P, E);PointOnCurve(MidPoint(LineSegmentOf(P,F2)), yAxis);AngleOf(P, F2, F1) = ApplyUnit(60, degree)",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(3)+2",
    "fact_spans": "[[[19, 79], [91, 94], [153, 156]], [[27, 79]], [[27, 79]], [[9, 16]], [[86, 90]], [[1, 8]], [[27, 79]], [[27, 79]], [[19, 79]], [[1, 85]], [[1, 85]], [[86, 95]], [[97, 117]], [[118, 151]]]",
    "query_spans": "[[[153, 162]]]",
    "process": "Since the midpoint of segment $PF_{2}$ lies on the $y$-axis, it follows that $PF_{1}//y$-axis, then $PF_{1}\\bot x$-axis. So $_{PF_{1}}=\\frac{b^{2}}{a}$. Because $\\angle PF_{2}F_{1}=60^{\\circ}$, so $\\tan60^{\\circ}=\\frac{PF_{1}}{F_{1}F_{2}}=\\frac{b^{2}}{2ac}=\\sqrt{3}$, i.e., $c^{2}-2\\sqrt{3}ac-a^{2}=0$, so $e^{2}-2\\sqrt{3}e-1=0$, because $e>1$, solving gives: $e=\\sqrt{3}+2$."
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and $M$, $N$ are points on the circles $(x+3)^{2}+y^{2}=1$ and $(x-3)^{2}+y^{2}=4$ respectively, then the minimum value of $|PM|+|PN|$ is?",
    "fact_expressions": "G: Ellipse;H: Circle;C:Circle;P: Point;M: Point;N: Point;Expression(G) = (x^2/25 + y^2/16 = 1);Expression(H) = (y^2 + (x + 3)^2 = 1);Expression(C) = (y^2 + (x - 3)^2 = 4);PointOnCurve(P, G);PointOnCurve(M, H);PointOnCurve(N, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, M)) + Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "7",
    "fact_spans": "[[[6, 45]], [[60, 80]], [[81, 101]], [[2, 5]], [[50, 53]], [[54, 57]], [[6, 45]], [[60, 80]], [[81, 101]], [[2, 49]], [[50, 104]], [[50, 104]]]",
    "query_spans": "[[[106, 123]]]",
    "process": ""
  },
  {
    "text": "The length of the chord cut by the line $y=x+1$ on the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$?",
    "fact_expressions": "G: Hyperbola;H: Line;Expression(G) = (x^2 - y^2/4 = 1);Expression(H) = (y = x + 1)",
    "query_expressions": "Length(InterceptChord(H,G))",
    "answer_expressions": "8*sqrt(2)/3",
    "fact_spans": "[[[10, 38]], [[0, 9]], [[10, 38]], [[0, 9]]]",
    "query_spans": "[[[0, 44]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with right vertex $A$, and the parabola $C$: $y^{2}=16 a x$ with focus $F$. If there exists a point $P$ on an asymptote of the hyperbola such that $\\overrightarrow{P A} \\cdot \\overrightarrow{P F}=0$, then the range of the eccentricity of hyperbola $E$ is?",
    "fact_expressions": "E: Hyperbola;b: Number;a: Number;C: Parabola;P: Point;A: Point;F: Point;a>0;b>0;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);Expression(C) = (y^2 = 16*(a*x));RightVertex(E) = A;Focus(C) = F;PointOnCurve(P, Asymptote(E));DotProduct(VectorOf(P, A), VectorOf(P, F)) = 0",
    "query_expressions": "Range(Eccentricity(E))",
    "answer_expressions": "(1, 5/4]",
    "fact_spans": "[[[2, 63], [104, 107], [175, 181]], [[9, 63]], [[9, 63]], [[71, 93]], [[116, 119]], [[67, 70]], [[97, 100]], [[9, 63]], [[9, 63]], [[2, 63]], [[71, 93]], [[2, 70]], [[71, 100]], [[103, 119]], [[122, 173]]]",
    "query_spans": "[[[175, 192]]]",
    "process": "Hyperbola $ E:\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 $ ($ a>0,b>0 $) has right vertex $ A(a,0) $, asymptotes given by $ y=\\pm\\frac{b}{a}x $. Parabola $ C:y^{2}=16ax $ has focus $ F(4a,0) $. Let $ P(m,\\frac{b}{a}m) $, then $ \\overrightarrow{AP}=(m-a,\\frac{b}{a}m) $, $ \\overrightarrow{PF}=(m-4a,\\frac{b}{a}m) $. Since $ \\overrightarrow{PA}\\cdot\\overrightarrow{PF}=0 $, we have $ (m-a)(m-4a)+\\frac{b^{2}}{a^{2}}m^{2}=0 $, i.e., $ (1+\\frac{b^{2}}{a^{2}})m^{2}-5am+4a^{2}=0 $. Since there exists a point $ P $ on the asymptote of the hyperbola such that $ \\overrightarrow{PA}\\cdot\\overrightarrow{PF}=0 $, the equation $ (1+\\frac{b^{2}}{a^{2}})m^{2}-5am+4a^{2}=0 $ in $ m $ has a solution. Therefore, $ \\Delta=25a^{2}-16(1+\\frac{b^{2}}{a^{2}})a^{2}\\geqslant0 $, i.e., $ 1+\\frac{b^{2}}{a^{2}}\\leqslant\\frac{25}{16} $. Then $ e=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}\\leqslant\\sqrt{\\frac{25}{16}}=\\frac{5}{4} $. Also $ e>1 $, so $ e\\in(1,\\frac{5}{4}] $."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, respectively. If there exists a point $A$ on the hyperbola such that $\\angle F_{1} A F_{2}=90^{\\circ}$ and $|AF_{1}|=3|AF_{2}|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F1: Point;A: Point;F2: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(A, G);AngleOf(F1, A, F2) = ApplyUnit(90, degree);Abs(LineSegmentOf(A, F1)) = 3*Abs(LineSegmentOf(A, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)/2",
    "fact_spans": "[[[19, 65], [73, 76], [142, 145]], [[22, 65]], [[22, 65]], [[1, 8]], [[79, 83]], [[9, 16]], [[19, 65]], [[1, 71]], [[1, 71]], [[73, 83]], [[85, 118]], [[120, 140]]]",
    "query_spans": "[[[142, 151]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, its two asymptotes intersect the directrix of the parabola $y^{2}=2px$ $(p>0)$ at points $A$ and $B$, respectively, and $O$ is the origin. If the eccentricity of the hyperbola is $2$, and the area of $\\triangle AOB$ is $\\sqrt{3}$, then $p=$?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Parabola;p: Number;A: Point;O: Origin;B: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);p>0;Expression(H) = (y^2 = 2*(p*x));L1:Line;L2:Line;Asymptote(G)={L1,L2};Intersection(L1,Directrix(H))=A;Intersection(L2,Directrix(H))=B;Eccentricity(G) = 2;Area(TriangleOf(A, O, B)) = sqrt(3)",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 61], [117, 120]], [[5, 61]], [[5, 61]], [[68, 89]], [[162, 165]], [[96, 99]], [[106, 109]], [[100, 103]], [[5, 61]], [[5, 61]], [[2, 61]], [[71, 89]], [[68, 89]], [], [], [[2, 67]], [[2, 105]], [[2, 105]], [[117, 128]], [[129, 160]]]",
    "query_spans": "[[[162, 167]]]",
    "process": "Find the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ and the directrix of the parabola $y^{2}=2px$ $(p>0)$, then find the coordinates of points $A$ and $B$. Given that the eccentricity of the hyperbola is $2$ and the area of $\\triangle AOB$ is $\\sqrt{3}$, set up an equation and solve for the value of $p$.  \n$\\because$ the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$,  \n$\\therefore$ the asymptotes of the hyperbola are $y=\\pm\\frac{b}{a}x$.  \nAlso, the directrix of the parabola $y^{2}=2px$ $(p>0)$ is $x=-\\frac{p}{2}$, so the $y$-coordinates of points $A$ and $B$ are $y=\\pm\\frac{pb}{2a}$.  \nSince the eccentricity of the hyperbola is $2$, we have $\\frac{c}{a}=2$. The $y$-coordinates of points $A$ and $B$ are $y=\\pm\\frac{pb}{2a}$.  \nGiven that the area of $\\triangle AOB$ is $\\sqrt{3}$ and the $x$-axis is the angle bisector of $\\angle AOB$,  \n$\\therefore \\frac{1}{2}\\times\\sqrt{3}p\\times\\frac{p}{2}\\sqrt{3}$, solving gives $p=2$."
  },
  {
    "text": "The standard equation of the ellipse passing through the points $P(\\sqrt{3},-2)$ and $Q(-2 \\sqrt{3}, 1)$ is?",
    "fact_expressions": "G: Ellipse;P: Point;Q: Point;Coordinate(P) = (sqrt(3), -2);Coordinate(Q) = (-2*sqrt(3), 1);PointOnCurve(Q, G);PointOnCurve(P, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/25 + y^2/5 = 1",
    "fact_spans": "[[[44, 46]], [[1, 18]], [[20, 41]], [[1, 18]], [[20, 41]], [[0, 46]], [[0, 46]]]",
    "query_spans": "[[[44, 52]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ lies on the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, point $A$ satisfies $\\overrightarrow{P A}=(t-1) \\overrightarrow{O P}(t \\in R)$, and $\\overrightarrow{O A} \\cdot \\overrightarrow{O P}=60$, $\\overrightarrow{O B}=(0,1)$, then the maximum value of $|\\overrightarrow{O B} \\cdot \\overrightarrow{O A}|$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;A: Point;O: Origin;B: Point;t: Real;Expression(G) = (x^2/9 - y^2/16 = 1);PointOnCurve(P, G);VectorOf(P, A) = VectorOf(O, P)*(t - 1);DotProduct(VectorOf(O, A), VectorOf(O, P)) = 60;Coordinate(VectorOf(O, B)) = (0, 1)",
    "query_expressions": "Max(Abs(DotProduct(VectorOf(O, B), VectorOf(O, A))))",
    "answer_expressions": "8",
    "fact_spans": "[[[7, 46]], [[2, 6]], [[48, 52]], [[54, 112]], [[169, 197]], [[54, 112]], [[7, 46]], [[2, 47]], [[54, 112]], [[114, 166]], [[169, 197]]]",
    "query_spans": "[[[199, 256]]]",
    "process": "\\because\\overrightarrow{PA}=(t-1)\\overrightarrow{OP}=t\\overrightarrow{OP}-\\overrightarrow{OP},\\therefore\\overrightarrow{OA}-\\overrightarrow{OP}=t\\overrightarrow{OP}-\\overrightarrow{OP}, then \\overrightarrow{OA}=t\\overrightarrow{OP},\\therefore|\\overrightarrow{OA}|=|t|\\cdot|\\overrightarrow{OP}|, let A(x_{A},y_{A}),P(x_{P},y_{P}),\\therefore(x_{A},y_{A})=t(x_{P},y_{P}) then \\begin{cases}x_{A}=tx_{p}\\\\y_{A}=ty_{P}\\end{cases}, i.e. \\begin{cases}x_{p}=\\frac{x_{4}}{y}\\\\y_{p}=\\frac{y_{1}}{t},\\end{cases} substitute point (\\frac{x_{1}}{t},\\frac{y_{1}}{t}) into the hyperbola to obtain: \\frac{x_{A^{2}}}{9t^{2}}-\\frac{y_{n}^{2}}{16t^{2}}=1,\\therefore x_{A^{2}}=\\frac{9y_{1}^{2}}{16}+9t^{2}\\cdots\\textcircled{1}, =1(\\frac{x_{1}2}{f_{2}}+\\frac{y_{A}^{2}}{\\frac{1}{12}})=60\\cdots\\textcircled{2}, from \\textcircled{1}\\textcircled{2} we get 60=|t|\\cdot(\\frac{9y_{A}}{16t^{2}}+\\frac{y_{2}}{t^{2}}+9)=|t|\\cdot(\\frac{25y_{2}}{16t^{2}}+9)=\\frac{25y_{2}}{16|t|}+9|t|\\geqslant\\frac{15}{2}|y_{A}| \\therefore|y_{A}|\\leqslant8,\\therefore|\\overrightarrow{OB}\\cdot\\overrightarrow{OA}|=|y|\\leqslant8. Then the maximum value of |\\overrightarrow{OB}\\cdot\\overrightarrow{OA}| is 8."
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{2}=1$ are $F_{1}$, $F_{2}$, and point $P$ lies on the ellipse. If $|P F_{1}|=4$, then $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/2 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G) = True;Abs(LineSegmentOf(P, F1)) = 4",
    "query_expressions": "DotProduct(VectorOf(P, F1), VectorOf(P, F2))",
    "answer_expressions": "-4",
    "fact_spans": "[[[0, 37], [62, 64]], [[0, 37]], [[41, 48]], [[49, 56]], [[0, 56]], [[57, 61]], [[57, 65]], [[67, 80]]]",
    "query_spans": "[[[82, 141]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ is a moving point on the line $y = x + 1$, and point $Q$ is a moving point on the parabola $y = x^{2}$. Let point $M$ be the midpoint of segment $PQ$, and $O$ be the origin. Then the minimum value of $|OM|$ is?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Q: Point;O: Origin;M: Point;Expression(G) = (y = x^2);Expression(H) = (y = x + 1);PointOnCurve(P, H);PointOnCurve(Q, G);MidPoint(LineSegmentOf(P, Q)) = M",
    "query_expressions": "Min(Abs(LineSegmentOf(O, M)))",
    "answer_expressions": "3*sqrt(2)/16",
    "fact_spans": "[[[26, 38]], [[7, 16]], [[2, 6]], [[21, 25]], [[60, 63]], [[44, 48]], [[26, 38]], [[7, 16]], [[2, 20]], [[21, 42]], [[44, 59]]]",
    "query_spans": "[[[68, 81]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the hyperbola $C$: $\\frac{x^{2}}{3}-y^{2}=1$ is? If the parabola $y^{2}=2 m x$ has the same foci as the hyperbola $C$, then $m$=?",
    "fact_expressions": "C: Hyperbola;G: Parabola;m: Number;Expression(C) = (x^2/3 - y^2 = 1);Expression(G) = (y^2 = 2*(m*x));Focus(G) = Focus(C)",
    "query_expressions": "Eccentricity(C);m",
    "answer_expressions": "sqrt(3)*2/3\npm*4",
    "fact_spans": "[[[0, 33], [57, 63]], [[40, 56]], [[71, 74]], [[0, 33]], [[40, 56]], [[40, 69]]]",
    "query_spans": "[[[0, 39]], [[71, 76]]]",
    "process": "From the equation of the hyperbola $ C: \\frac{x^{2}}{3} - y^{2} = 1 $, we know: $ a^{2} = 3 $, $ b^{2} = 1 $, so $ c^{2} = a^{2} + b^{2} = 4 $. Therefore, the eccentricity of the hyperbola is: $ e = \\frac{c}{a} = \\frac{2}{\\sqrt{3}} = \\frac{2\\sqrt{3}}{3} $; since the coordinates of the foci of the hyperbola are: $ (-2, 0) $ and $ (2, 0) $. If the focus of the parabola $ y^{2} = 2mx $ is $ (-2, 0) $, then $ \\frac{m}{2} = -2 $, thus $ m = -4 $; if the focus of the parabola $ y^{2} = 2mx $ is $ (2, 0) $, then $ \\frac{m}{2} = 2 $, thus $ m = 4 $. In conclusion, the answer to the first blank should be $ \\frac{2}{3}\\sqrt{3} $, and the answer to the second blank should be $ \\pm4 $."
  },
  {
    "text": "The length of the chord cut from the ellipse $x^{2}+4 y^{2}=16$ by the line $y=x+1$ is?",
    "fact_expressions": "G: Ellipse;H: Line;Expression(G) = (x^2 + 4*y^2 = 16);Expression(H) = (y = x + 1)",
    "query_expressions": "Length(InterceptChord(H,G))",
    "answer_expressions": "4*sqrt(38)/5",
    "fact_spans": "[[[0, 20]], [[21, 30]], [[0, 20]], [[21, 30]]]",
    "query_spans": "[[[0, 37]]]",
    "process": ""
  },
  {
    "text": "The equation of the line passing through the fixed point $P(0 , 1)$ and having only one common point with the parabola $y^{2}=2 x$ is?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Expression(G) = (y^2 = 2*x);Coordinate(P) = (0, 1);PointOnCurve(P,H);NumIntersection(G, H) = 1",
    "query_expressions": "Expression(H)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[16, 30]], [[38, 40]], [[3, 13]], [[16, 30]], [[3, 13]], [[0, 40]], [[15, 40]]]",
    "query_spans": "[[[38, 44]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has its right focus at $F(2,0)$. A line passing through point $F$ is perpendicular to one asymptote of the hyperbola $C$, with foot of perpendicular at point $E$, and $O$ is the origin. When the area of $\\triangle O E F$ reaches its maximum value, what is the eccentricity of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;F: Point;O: Origin;E: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (2, 0);RightFocus(C) = F;l:Line;PointOnCurve(F,l);IsPerpendicular(l,OneOf(Asymptote(C)));FootPoint(l,OneOf(Asymptote(C)))=E;WhenMax(Area(TriangleOf(O,E,F)))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 63], [86, 92], [146, 152]], [[10, 63]], [[10, 63]], [[68, 76], [78, 82]], [[109, 112]], [[104, 108]], [[10, 63]], [[10, 63]], [[2, 63]], [[68, 76]], [[2, 76]], [[83, 85]], [[77, 85]], [[83, 100]], [[83, 108]], [[118, 145]]]",
    "query_spans": "[[[146, 158]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$, and $P$ is a point on $C$. If $P$ lies in the first quadrant and $|PF|=8$, then what are the coordinates of point $P$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 8*x);F: Point;Focus(C) = F;P: Point;PointOnCurve(P, C);Quadrant(P) = 1;Abs(LineSegmentOf(P, F)) = 8",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(6, 4*sqrt(3))",
    "fact_spans": "[[[2, 21], [33, 36]], [[2, 21]], [[25, 28]], [[2, 28]], [[29, 32], [41, 44], [61, 65]], [[29, 39]], [[41, 49]], [[50, 59]]]",
    "query_spans": "[[[61, 70]]]",
    "process": "The focus of the parabola is $F(2,0)$. Let the coordinates of point $P$ be $(x_{0},y_{0})$, then $|PF|=x_{0}+2=8$, $x_{0}=6$, $y_{0}=\\sqrt{8\\times6}=4\\sqrt{3}$, so $P(6,4\\sqrt{3})$."
  },
  {
    "text": "Given that point $M$ lies on the parabola $y^{2}=4x$, $F$ is the focus of the parabola, and $A$ lies on the circle $C$: $(x-4)^{2}+(y-1)^{2}=1$, then the minimum value of $|MA|+|MF|$ is?",
    "fact_expressions": "M: Point;G: Parabola;Expression(G) = (y^2 = 4*x);PointOnCurve(M, G) = True;F: Point;Focus(G) = F;A: Point;C: Circle;Expression(C) = ((x - 4)^2 + (y - 1)^2 = 1);PointOnCurve(A, C) = True",
    "query_expressions": "Min(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F)))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 6]], [[7, 21], [30, 33]], [[7, 21]], [[2, 25]], [[26, 29]], [[26, 36]], [[37, 40]], [[41, 69]], [[41, 69]], [[37, 70]]]",
    "query_spans": "[[[72, 91]]]",
    "process": "The directrix of the parabola $ y^{2}=4x $ is $ x=-1 $. Draw $ MN \\perp $ directrix, with foot of perpendicular at $ N $. Since point $ M $ lies on the parabola $ y^{2}=4x $, and $ F $ is the focus of the parabola, $ |MN|=|MF| $. Therefore, $ |MA|+|MF|=|MA|+|MN| $. Since $ A $ lies on the circle $ C: (x-4)^{2}+(y-1)^{2}=1 $ with center $ C(4,1) $, $ |MA|+|MF| $ is minimized when points $ N $, $ M $, $ C $ are collinear. Thus, $ (|MA|+|MF|)_{\\min} = (|MA|+|MN|)_{\\min} = |CN|-r = 5-1=4 $. Therefore, $ (|MA|+|MF|)_{\\min}=4 $."
  },
  {
    "text": "The minimum distance from a point on the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$ to the line $x-y+6=0$ is?",
    "fact_expressions": "G: Ellipse;H: Line;P0: Point;Expression(G) = (x^2/8 + y^2/4 = 1);Expression(H) = (x - y + 6 = 0);PointOnCurve(P0, G)",
    "query_expressions": "Min(Distance(P0, H))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[0, 37]], [[41, 52]], [[39, 40]], [[0, 37]], [[41, 52]], [[0, 40]]]",
    "query_spans": "[[[39, 61]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, point $P$ is a point on the hyperbola $C$ such that $\\angle F_{1} P F_{2}=\\frac{\\pi}{2}$, and the area of $\\Delta F_{1} P F_{2}$ is $a^{2}$, then the asymptotic equations of the hyperbola $C$ are?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;F1: Point;P: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);AngleOf(F1, P, F2) = pi/2;Area(TriangleOf(F1, P, F2)) = a^2",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "pm*y+x=0",
    "fact_spans": "[[[20, 81], [93, 99], [178, 184]], [[27, 81]], [[27, 81]], [[2, 9]], [[88, 92]], [[10, 17]], [[27, 81]], [[27, 81]], [[20, 81]], [[2, 87]], [[2, 87]], [[88, 102]], [[104, 140]], [[142, 176]]]",
    "query_spans": "[[[178, 192]]]",
    "process": "\\because||PF_{1}|-|PF_{2}||=2a,|PF_{1}|^{2}+|PF_{2}|^{2}=4c^{2}, then 2|PF_{1}|\\cdot|PF_{2}|=|PF_{1}|^{2}+|PF_{2}|^{2}-||PF_{1}|-|PF_{2}||^{2}=4c^{2}-4a^{2}=4b^{2}, so |PF_{1}|\\cdot|PF_{2}|=2b^{2}. Because \\angle F_{1}PF_{2}=\\frac{\\pi}{2}, so S_{\\triangle F_{1}PF_{2}}=\\frac{1}{2}|PF_{1}||PF_{2}|=b^{2}=a^{2}, we get a=b. Therefore, the asymptotes of hyperbola C are y=\\pm\\frac{b}{a}x=\\pm x, i.e., x\\pm y=0."
  },
  {
    "text": "Given points $A(0 , 2)$, $O(0 , 0)$, if there exists a point $M$ on the circle $C$: $(x-a)^{2}+(y-a+2)^{2}=1$ such that $\\overrightarrow{M A} \\cdot \\overrightarrow{M O}=3$, then the range of the horizontal coordinate $a$ of the center $C$ is?",
    "fact_expressions": "C: Circle;a: Number;A: Point;O: Origin;M: Point;Expression(C) = ((-a + x)^2 + (-a + y + 2)^2 = 1);Coordinate(A) = (0, 2);Coordinate(O) = (0, 0);PointOnCurve(M, C);DotProduct(VectorOf(M, A), VectorOf(M, O)) = 3;C1: Point;Center(C) = C1;XCoordinate(C1) = a",
    "query_expressions": "Range(a)",
    "answer_expressions": "[0, 3]",
    "fact_spans": "[[[28, 58]], [[129, 132]], [[2, 14]], [[16, 26]], [[61, 65]], [[28, 58]], [[2, 14]], [[16, 26]], [[28, 65]], [[67, 118]], [[122, 125]], [[28, 125]], [[122, 132]]]",
    "query_spans": "[[[129, 139]]]",
    "process": ""
  },
  {
    "text": "The chord length intercepted by the circle $x^{2}+y^{2}-4 x+4 \\sqrt{3} y=0$ from a line passing through the focus of the parabola $y^{2}=4 x$ and inclined at an angle of $60^{\\circ}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);L: Line;PointOnCurve(Focus(G), L);Inclination(L) = ApplyUnit(60, degree);H: Circle;Expression(H) = (4*(sqrt(3)*y) - 4*x + x^2 + y^2 = 0)",
    "query_expressions": "Length(InterceptChord(L, H))",
    "answer_expressions": "sqrt(37)",
    "fact_spans": "[[[1, 15]], [[1, 15]], [[36, 38]], [[0, 38]], [[19, 38]], [[39, 72]], [[39, 72]]]",
    "query_spans": "[[[36, 79]]]",
    "process": "From the given, the equation of line $ l $ is $ \\sqrt{3}x - y - \\sqrt{3} = 0 $, the center of the circle is $ (2, -2\\sqrt{3}) $, radius $ r = 4 $, the distance from the center to the line is $ d = \\frac{\\sqrt{3} \\times 2 + 2\\sqrt{3} - \\sqrt{3}}{\\sqrt{(3)^{2} + (-1)^{2}}} = \\frac{3\\sqrt{3}}{2} $, so by the chord length formula we get"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, the left and right vertices are denoted as $A_{1}$ and $A_{2}$ respectively. A line $l$ passing through $A_{1}$ intersects $C$ at another point $P$. If the slope of line $l$ is $2$, then what is the slope of line $P A_{2}$?",
    "fact_expressions": "l: Line;C: Ellipse;P: Point;A2: Point;A1: Point;Expression(C) = (x^2/4 + y^2/3 = 1);LeftVertex(C)=A1;RightVertex(C)=A2;PointOnCurve(A1,l);Intersection(l,C)={A1, P};Slope(l)=2",
    "query_expressions": "Slope(LineOf(P, A2))",
    "answer_expressions": "-3/8",
    "fact_spans": "[[[80, 85], [99, 104]], [[2, 44], [86, 89]], [[94, 97]], [[62, 70]], [[53, 60], [72, 79]], [[2, 44]], [[2, 69]], [[2, 69]], [[71, 85]], [[72, 97]], [[99, 111]]]",
    "query_spans": "[[[113, 129]]]",
    "process": "Let $ P(x_{0},y_{0}) $, then $ \\frac{x_{0}^{2}}{4} + \\frac{y_{0}^{2}}{3} = 1 $, $ k_{PA_{1}} \\cdot k_{PA_{2}} = \\frac{y_{0}}{x_{0}+2} \\cdot \\frac{y_{0}}{x_{0}-2} = \\frac{y_{0}^{2}}{x_{0}^{2}-4} = \\frac{3(1-\\frac{x_{0}^{2}}{4})}{x_{0}^{2}-4} = -\\frac{3}{4} = 2 \\times k_{PA_{2}} $, so $ k_{PA_{2}} = -\\frac{3}{8} $."
  },
  {
    "text": "The line $ l $: $ y = k(x - \\sqrt{2}) $ intersects the curve $ x^{2} - y^{2} = 1 $ ($ x > 0 $) at two points $ A $ and $ B $. Then, what is the range of the inclination angle of the line $ l $?",
    "fact_expressions": "l: Line;Expression(l) = (y = k*(x - sqrt(2)));G: Curve;Expression(G) = ((x^2 - y^2 = 1)&(x>0));A: Point;B: Point;Intersection(l, G) = {A, B};k: Number",
    "query_expressions": "Range(Inclination(l))",
    "answer_expressions": "(pi/4, pi/2)+(pi/2, 3*pi/4)",
    "fact_spans": "[[[0, 24], [61, 66]], [[0, 24]], [[25, 47]], [[25, 47]], [[50, 53]], [[54, 57]], [[0, 59]], [[0, 24]]]",
    "query_spans": "[[[61, 77]]]",
    "process": "The asymptotes of the curve $x^{2}-y^{2}=1$ ($x>0$) are: $y=\\pm x$. Since the line $l: y=k(x-\\sqrt{2})$ intersects the curve at two points $A$ and $B$, the slope $k$ of line $l$ satisfies $k>1$ or $k<-1$, i.e., $\\alpha\\in(\\frac{\\pi}{4},\\frac{3\\pi}{4})$. Moreover, the slope of line $l$ exists, meaning the inclination angle $\\alpha\\neq\\frac{\\pi}{2}$. Therefore, the range of the inclination angle of line $l$ is $(\\frac{\\pi}{4},\\frac{\\pi}{2})\\cup(\\frac{\\pi}{2},\\frac{3\\pi}{4})$. Hence, the answer is $(\\frac{\\pi}{4},\\frac{\\pi}{2})\\cup(\\frac{\\pi}{2},\\frac{3\\pi}{4})$."
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1$, what is the length of the real axis? What is the length of the imaginary axis? What are the coordinates of the foci? What is the eccentricity? What are the equations of the asymptotes?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/5 - y^2/4 = 1)",
    "query_expressions": "Length(RealAxis(G));Length(ImageinaryAxis(G));Coordinate(Focus(G));Eccentricity(G);Expression(Asymptote(G))",
    "answer_expressions": "2*sqrt(5)\n4\n(pm*3,0)\n3*sqrt(5)/5\ny=pm*(2*sqrt(5)/5)*x",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 45]], [[0, 51]], [[0, 57]], [[0, 62]], [[0, 69]]]",
    "process": "From the hyperbola equation $\\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1$, we obtain: $a=\\sqrt{5}$, $b=2$, $c=3$, and the foci of the hyperbola lie on the $x$-axis; then the length of the real axis is $2\\sqrt{5}$, the length of the imaginary axis is $4$, the coordinates of the foci are $F_{1}(-3,0)$, $F_{2}(3,0)$, the eccentricity is $e=\\frac{c}{a}=\\frac{3\\sqrt{5}}{5}$, and the equations of the asymptotes are $y=\\pm\\frac{b}{a}x=\\pm\\frac{2}{5}\\sqrt{5}x$."
  },
  {
    "text": "If one of the asymptotes of the hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1(a>0, b>0)$ is given by $x-2 y=0$, then its eccentricity $e$=?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-x^2/b^2 + y^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (x - 2*y = 0);Eccentricity(G)=e;e:Number",
    "query_expressions": "e",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[1, 57], [77, 78]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 75]], [[77, 85]], [[82, 85]]]",
    "query_spans": "[[[82, 87]]]",
    "process": "Since one asymptote of the hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ ($a>0$, $b>0$) is given by $x-2y=0$, $\\therefore a=2b$, $\\therefore c=\\sqrt{5}b$, so the eccentricity of the hyperbola is $e=\\frac{c}{a}=\\frac{\\sqrt{5}}{2}$,"
  },
  {
    "text": "Draw a line through the focus of the parabola $y^{2}=4 x$ with an inclination angle of $\\frac{\\pi}{3}$, intersecting the parabola at points $A$ and $B$. Then $|A B|=$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), H);Inclination(H) = pi/3;Intersection(H, G) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 15], [42, 45]], [[39, 41]], [[47, 51]], [[52, 55]], [[1, 15]], [[0, 41]], [[19, 41]], [[39, 55]]]",
    "query_spans": "[[[57, 66]]]",
    "process": ""
  },
  {
    "text": "The line $l$ passes through the focus of the parabola $y^{2}=x$, and intersects the parabola at points $A$ and $B$. If $|AB|=4$, then what is the distance from the midpoint of chord $AB$ to the $y$-axis?",
    "fact_expressions": "l: Line;G: Parabola;Expression(G) = (y^2 = x);PointOnCurve(Focus(G), l);A: Point;B: Point;Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, B)) = 4;IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A, B)), yAxis)",
    "answer_expressions": "7/4",
    "fact_spans": "[[[0, 5], [23, 26]], [[6, 18], [27, 30]], [[6, 18]], [[0, 21]], [[32, 35]], [[36, 39]], [[23, 41]], [[43, 52]], [[27, 61]]]",
    "query_spans": "[[[56, 74]]]",
    "process": ""
  },
  {
    "text": "Point $M$ is a point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and $F_{1}$, $F_{2}$ are the left and right foci of the ellipse, respectively. Find the coordinates of point $M$ satisfying $|MF_{1}|=3|MF_{2}|$.",
    "fact_expressions": "G: Ellipse;M: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2/3 = 1);PointOnCurve(M, G);LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(M, F1)) = 3*Abs(LineSegmentOf(M, F2))",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(2, 0)",
    "fact_spans": "[[[5, 42], [65, 67]], [[0, 4], [97, 101]], [[47, 54]], [[55, 62]], [[5, 42]], [[0, 46]], [[47, 71]], [[47, 71]], [[75, 95]]]",
    "query_spans": "[[[97, 105]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a, b>0)$ intersects the line $y=2 x$; then the range of its eccentricity is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;H: Line;Expression(H) = (y = 2*x);IsIntersect(G, H)",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(\\sqrt{5}, +\\infty)",
    "fact_spans": "[[[0, 54], [69, 70]], [[0, 54]], [[3, 54]], [[3, 54]], [[3, 54]], [[3, 54]], [[55, 64]], [[55, 64]], [[0, 67]]]",
    "query_spans": "[[[69, 81]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ is $2x+y=0$, and one focus is $(\\sqrt{5}, 0)$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Expression(OneOf(Asymptote(G))) = (2*x + y = 0);H: Point;Coordinate(H) = (sqrt(5), 0);OneOf(Focus(G)) = H",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/4 = 1",
    "fact_spans": "[[[2, 59], [99, 102]], [[2, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[2, 75]], [[82, 97]], [[82, 97]], [[2, 97]]]",
    "query_spans": "[[[99, 107]]]",
    "process": "From the given conditions: $ c=\\sqrt{5},\\frac{b}{a}=2,\\ c^{2}=a^{2}+b^{2} $, solving yields $ a=1,b=2 $. Then the equation of the hyperbola is: $ x^{2}-\\frac{y^{2}}{4}=1 $"
  },
  {
    "text": "The distance from the vertex of the parabola $y^{2}=4 x$ to its directrix is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x)",
    "query_expressions": "Distance(Vertex(G),Directrix(G))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 14], [18, 19]], [[0, 14]]]",
    "query_spans": "[[[0, 26]]]",
    "process": "Find p=2, then find the distance from the vertex coordinates to the directrix as \\frac{p}{2}=1. [Detailed solution] In y^{2}=4x, 2p=4, so p=2, hence the equation of the directrix is x=-\\frac{p}{2}=-1, and the distance from the vertex coordinates to the directrix is \\frac{p}{2}=1"
  },
  {
    "text": "Let the focus of the parabola $y^{2}=8x$ be $F$, and let $A$ be a point on the parabola such that $|AF|=6$. Then the coordinates of point $A$ are?",
    "fact_expressions": "G: Parabola;A: Point;F: Point;Expression(G) = (y^2 = 8*x);Focus(G) = F;PointOnCurve(A, G);Abs(LineSegmentOf(A, F))=6",
    "query_expressions": "Coordinate(A)",
    "answer_expressions": "(4, pm*4*sqrt(2))",
    "fact_spans": "[[[1, 15], [27, 30]], [[23, 26], [47, 51]], [[19, 22]], [[1, 15]], [[1, 22]], [[23, 34]], [[36, 45]]]",
    "query_spans": "[[[47, 56]]]",
    "process": "From the given conditions, let A(x_{0},y_{0}). According to the definition of the parabola, |AF| = x_{0} + \\frac{p}{2} = x_{0} + 2 = 6 \\Rightarrow x_{0} = 4. Substituting into the equation of the parabola yields y_{0} = \\pm4\\sqrt{2}. Therefore, the coordinates of point A are (4, \\pm4\\sqrt{2})."
  },
  {
    "text": "The focus of the parabola $y^{2}=4 x$ is $F$. A line passing through the point $(0,3)$ intersects the parabola at points $A$ and $B$. The perpendicular bisector of segment $A B$ intersects the $x$-axis at point $D$. If $|A F|+|B F|=6$, then the horizontal coordinate of point $D$ is?",
    "fact_expressions": "G: Parabola;H: Line;Q: Point;A: Point;B: Point;F: Point;D: Point;Expression(G) = (y^2 = 4*x);Coordinate(Q) = (0, 3);Focus(G) = F;PointOnCurve(Q, H);Intersection(H, G) = {A, B};Intersection(PerpendicularBisector(LineSegmentOf(A, B)), xAxis) = D;Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F)) = 6",
    "query_expressions": "XCoordinate(D)",
    "answer_expressions": "4",
    "fact_spans": "[[[0, 14], [35, 38]], [[32, 34]], [[23, 31]], [[40, 43]], [[44, 47]], [[18, 21]], [[69, 73], [92, 96]], [[0, 14]], [[23, 31]], [[0, 21]], [[22, 34]], [[32, 49]], [[50, 73]], [[75, 90]]]",
    "query_spans": "[[[92, 102]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), the line AB has equation y = kx + 3. Solving the system \\begin{cases} y^{2} = 4x \\\\ y = kx + 3 \\end{cases}, we obtain k^{2}x^{2} + (6k - 4)x + 9 = 0, \\therefore x_{1} + x_{2} = \\frac{4 - 6k}{k^{2}}. From the property of the parabola, |AF| + |BF| = x_{1} + x_{2} + p = 6, \\therefore x_{1} + x_{2} = 4. Therefore, \\frac{4 - 6k}{k^{2}} = 4, solving gives k = \\frac{1}{2} or k = -2. From the figure, we see k = -2. Thus, the equation of AB is y = -2x + 3, the midpoint of AB is (2, -1), the perpendicular bisector of segment AB is y + 1 = \\frac{1}{2}(x - 2). Letting y = 0, we get x = 4."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, line $l$ is a tangent to the circle $O$: $x^{2}+y^{2}=b^{2}$. Denote the eccentricity of ellipse $C$ as $e$. If the inclination angle of line $l$ is $\\frac{\\pi}{3}$ and it passes exactly through the right vertex of the ellipse, then what is the value of $e$?",
    "fact_expressions": "l: Line;C: Ellipse;b: Number;a: Number;O: Circle;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(O) = (x^2 + y^2 = b^2);IsTangent(l, O);e:Number;Eccentricity(C) = e;Inclination(l) = pi/3;PointOnCurve(RightVertex(C),l)",
    "query_expressions": "e",
    "answer_expressions": "1/2",
    "fact_spans": "[[[59, 64], [113, 118]], [[2, 58], [97, 102], [144, 146]], [[8, 58]], [[8, 58]], [[65, 90]], [[8, 58]], [[8, 58]], [[2, 58]], [[65, 90]], [[59, 95]], [[107, 110], [152, 155]], [[97, 110]], [[113, 138]], [[113, 150]]]",
    "query_spans": "[[[152, 159]]]",
    "process": "Let line $ l $ be tangent to circle $ O $ at point $ C $, and let the right vertex of the ellipse be $ D $. Then from the given conditions, $ AOCD $ is a right triangle with $ OC = b $, $ OD = a $, $ \\angle ODC = \\frac{\\pi}{3} $. Therefore, $ CD = \\sqrt{OD^{2} - OC^{2}} = \\sqrt{a^{2} - b^{2}} = c $ (where $ c $ is the semi-focal length of the ellipse). Thus, the eccentricity of the ellipse is $ e = \\frac{c}{a} = \\cos\\frac{\\pi}{3} = \\frac{1}{2} $."
  },
  {
    "text": "The line $ l $: $ x - t y + 1 = 0 $ ($ t > 0 $) and the parabola $ C $: $ y^2 = 4x $ intersect at two distinct points $ A $, $ B $. Let $ M $ be the midpoint of $ AB $, and let $ F $ be the focus of the parabola $ C $. The circle with diameter $ MF $ intersects the line $ l $ at another point $ N $, and satisfies $ |MN| = 3\\sqrt{3}|NF| $. Then the equation of the line $ l $ is?",
    "fact_expressions": "l: Line;C: Parabola;G: Circle;t: Number;A: Point;B: Point;M: Point;F: Point;N: Point;Expression(C) = (y^2 = 4*x);t>0;Expression(l) = (-t*y + x + 1 = 0);Intersection(l, C) = {A, B};Negation(A=B);MidPoint(LineSegmentOf(A, B)) = M;Focus(C) = F;IsDiameter(LineSegmentOf(M,F),G);Intersection(l,G)={M,N};Abs(LineSegmentOf(M,N))=3*sqrt(3)*Abs(LineSegmentOf(N,F))",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x-sqrt(3)*y+1=0",
    "fact_spans": "[[[0, 23], [98, 103], [144, 149]], [[24, 43], [72, 78]], [[96, 97]], [[6, 23]], [[50, 53]], [[54, 57]], [[68, 71]], [[82, 85]], [[109, 112]], [[24, 43]], [[6, 23]], [[0, 23]], [[0, 57]], [[46, 57]], [[59, 71]], [[72, 85]], [[86, 97]], [[68, 112]], [[116, 141]]]",
    "query_spans": "[[[144, 154]]]",
    "process": "The focus of $ y^{2}=4x $ is $ F(1,0) $. Solving $ x\\cdotty+1=0 $ and $ y^{2}=4x $ together, we obtain $ y^{2}\\cdot4ty+4=0 $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ y_{1}+y_{2}=4t $, so the midpoint $ M(2t^{2}-1,2t) $. Let $ N(ty_{0}\\cdot1,y_{0}) $. Since $ NF\\botl $, we have $ \\frac{y_{0}}{ty_{0}-2}=-t $, thus $ y_{0}=\\frac{2t}{1+t^{2}} $. From $ |MN|=3\\sqrt{3}|NF| $, we get $ |MN|^{2}=27|NF|^{2} $, that is, $ (2t^{2}-ty_{0})^{2}+(2t-y_{0})^{2}=27[(ty_{0}-2)^{2}+y_{0}^{2}] $. Combining $ y_{0}=\\frac{2t}{1+t^{2}} $, simplifying yields $ t^{6}=27 $, solving gives $ t=\\sqrt{3} $, hence the equation of the line is $ x-\\sqrt{3}y+1=0 $."
  },
  {
    "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{n}-\\frac{y^{2}}{5-n}=1$ are given by $y=\\pm 2 x$, then $n$=?",
    "fact_expressions": "G: Hyperbola;n: Number;Expression(G) = (-y^2/(5 - n) + x^2/n = 1);Expression(Asymptote(G)) = (y = pm*(2*x))",
    "query_expressions": "n",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 40]], [[60, 63]], [[0, 40]], [[0, 58]]]",
    "query_spans": "[[[60, 65]]]",
    "process": "From the given condition, we have \\frac{5-n}{n}=4, solving for n gives: n=1"
  },
  {
    "text": "What is the eccentricity of the hyperbola $\\frac{x^{2}}{8}-\\frac{y^{2}}{4}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/8 - y^2/4 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 44]]]",
    "process": "\\because a=2\\sqrt{2}, c=2\\sqrt{3}, \\therefore e=\\frac{2\\sqrt{3}}{2}=\\frac{\\sqrt{6}}{2}"
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{100}+\\frac{y^{2}}{36}=1$, point $P$ lies on the ellipse and satisfies $\\angle F_{1} P F_{2}=90^{\\circ}$, then the area of triangle $F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/100 + y^2/36 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Area(TriangleOf(F1,P,F2))",
    "answer_expressions": "36",
    "fact_spans": "[[[16, 56], [67, 69]], [[0, 7]], [[62, 66]], [[8, 15]], [[16, 56]], [[0, 61]], [[62, 70]], [[74, 107]]]",
    "query_spans": "[[[109, 132]]]",
    "process": "From the ellipse $\\frac{x^{2}}{100}+\\frac{y^{2}}{36}=1$, we get $a=10$, $b=6$, $c=8$. Let $F_{1}$, $F_{2}$ be the two foci of the ellipse, and point $P$ lies on the ellipse, so $|PF_{1}|+|PF_{2}|=20$ ①. Since $\\angle F_{1}PF_{2}=90^{\\circ}$, it follows that $|PF_{1}|^{2}+|PF_{2}|^{2}=(2c)^{2}=256$ ②. Squaring equation ① gives $|PF_{1}|^{2}+|PF_{2}|^{2}+2|PF_{1}|\\cdot|PF_{2}|=256+2|PF_{1}|\\cdot|PF_{2}|=400$. Rearranging yields $|PF_{1}|\\cdot|PF_{2}|=72$. Therefore, $S_{\\Delta F_{1}PF_{2}}=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|=36$."
  },
  {
    "text": "Given that the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has an eccentricity of $2$, and its foci coincide with the foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, what are the coordinates of the foci of the hyperbola? What is the equation of its asymptotes?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Ellipse;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2/25 + y^2/9 = 1);Eccentricity(G) = 2;Focus(G) = Focus(H)",
    "query_expressions": "Coordinate(Focus(G));Expression(Asymptote(G))",
    "answer_expressions": "(pm*4, 0)\nsqrt(3)*x + pm*y = 0",
    "fact_spans": "[[[2, 48], [106, 109]], [[5, 48]], [[5, 48]], [[60, 98]], [[2, 48]], [[60, 98]], [[2, 56]], [[2, 103]]]",
    "query_spans": "[[[106, 116]], [[106, 123]]]",
    "process": ""
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, its directrix intersects the $x$-axis at point $H$, and point $P$ lies on the parabola such that $|PH|=\\sqrt{2}|PF|$, then the horizontal coordinate of point $P$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;H: Point;Intersection(Directrix(G), xAxis) = H;P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, H)) = sqrt(2)*Abs(LineSegmentOf(P, F))",
    "query_expressions": "XCoordinate(P)",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 16], [44, 47], [24, 25]], [[2, 16]], [[20, 23]], [[2, 23]], [[34, 38]], [[24, 38]], [[39, 43], [73, 77]], [[39, 48]], [[50, 71]]]",
    "query_spans": "[[[73, 83]]]",
    "process": "Let $ P(x_{0},y_{0}) $, then $ |PF| = x_{0} + 1 $, $ |PH| = \\sqrt{(x_{0}+1)^{2} + y_{0}^{2}} $. According to the problem, $ \\sqrt{(x_{0}+1)^{2} + y_{0}^{2}} = \\sqrt{2}(x_{0}+1) $, so $ y_{0}^{2} = (x_{0}+1)^{2} $, $ 4x_{0} = (x_{0}+1)^{2} $, solving gives $ x_{0} = 1 $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola, respectively, point $P$ lies on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, satisfying $(\\overrightarrow{O P}+\\overrightarrow{O F_{2}}) \\cdot \\overrightarrow{F_{2} P}=0$ ($O$ is the coordinate origin), and $|P F_{1}|=\\sqrt{3}|P F_{2}|$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, RightPart(G));O: Origin;DotProduct((VectorOf(O, F2) + VectorOf(O, P)), VectorOf(F2, P)) = 0;Abs(LineSegmentOf(P, F1)) = sqrt(3)*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)+1",
    "fact_spans": "[[[20, 23], [35, 91], [227, 230]], [[35, 91]], [[38, 91]], [[38, 91]], [[38, 91]], [[38, 91]], [[2, 9]], [[10, 17]], [[2, 29]], [[2, 29]], [[30, 34]], [[30, 97]], [[184, 187]], [[100, 182]], [[195, 224]]]",
    "query_spans": "[[[227, 235]]]",
    "process": ""
  },
  {
    "text": "There are two moving points $M$ and $N$ on the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$. If $K(2,0)$ is a fixed point and $\\overrightarrow{K M} \\cdot \\overrightarrow{K N}=0$, then the minimum value of $\\overrightarrow{K M} \\cdot \\overrightarrow{N M}$ is?",
    "fact_expressions": "G: Ellipse;K: Point;M: Point;N: Point;Expression(G) = (x^2/36 + y^2/9 = 1);Coordinate(K) = (2, 0);DotProduct(VectorOf(K, M), VectorOf(K, N)) = 0;PointOnCurve(M, G);PointOnCurve(N, G)",
    "query_expressions": "Min(DotProduct(VectorOf(K, M), VectorOf(N, M)))",
    "answer_expressions": "23/3",
    "fact_spans": "[[[1, 39]], [[54, 62]], [[45, 48]], [[49, 52]], [[1, 39]], [[54, 62]], [[67, 118]], [[0, 52]], [[0, 52]]]",
    "query_spans": "[[[120, 175]]]",
    "process": "Since point M lies on the ellipse $\\frac{x^2}{36} + \\frac{y^{2}}{9} = 1$, we can set $M(6\\cos\\alpha, 3\\sin\\alpha)$ $(0 \\leqslant \\alpha < 2\\pi)$. Then $\\overrightarrow{KM} \\cdot \\overrightarrow{NM} = \\overrightarrow{KM} \\cdot (\\overrightarrow{KM} - \\overrightarrow{KN}) = K\\overrightarrow{M}^{2} - \\overrightarrow{KM} \\cdot \\overrightarrow{KN} = K\\overrightarrow{M}^{2}$. Given $K(2,0)$, we have $KM^{2} = |\\overrightarrow{KM}|^{2} = (6\\cos\\alpha - 2)^{2} + (3\\sin\\alpha)^{2} = 27\\cos^{2}\\alpha - 24\\cos\\alpha + 13 = 27(\\cos\\alpha - \\frac{4}{9})^{2} + \\frac{23}{3}$. When $\\cos\\alpha = \\frac{4}{9}$, $K\\overrightarrow{M}^{2}$ attains the minimum value $\\frac{23}{3}$."
  },
  {
    "text": "The focus of the parabola is the right focus of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$, and the vertex is at the center of the ellipse. Then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;H: Ellipse;Expression(H) = (x^2/5 + y^2/4 = 1);Focus(G) = RightFocus(H);Vertex(G)=Center(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[0, 3], [59, 62]], [[7, 44], [52, 54]], [[7, 44]], [[0, 48]], [[0, 57]]]",
    "query_spans": "[[[59, 66]]]",
    "process": "From the ellipse equation, the right focus of the ellipse is (1,0). Let the parabola equation be: $ y^{2} = 2px $, then $ \\frac{p}{3} = 1 \\therefore p = 2 \\therefore $ the parabola equation is: $ y^{2} = 4x $"
  },
  {
    "text": "Given the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, one of its foci has coordinates $(2,0)$, and the distance from this focus to an asymptote of the hyperbola is $1$. Then, the standard equation of the hyperbola $E$ is?",
    "fact_expressions": "E: Hyperbola;a: Number;b: Number;a>0;b>0;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(OneOf(Focus(E))) = (2, 0);Distance(OneOf(Focus(E)),Asymptote(E))=1",
    "query_expressions": "Expression(E)",
    "answer_expressions": "x^2/3-y^2=1",
    "fact_spans": "[[[2, 63], [84, 87], [99, 105]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 78]], [[80, 97]]]",
    "query_spans": "[[[99, 112]]]",
    "process": "Since the right focus of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) is $ F(2,0) $, we have $ c = 2 $. Also, since the distance from point $ F $ to one asymptote $ bx - ay = 0 $ of the hyperbola $ C $ is 1, it follows that $ \\frac{|2b|}{\\sqrt{b^{2} + a^{2}}} = \\frac{2b}{c} = \\frac{2b}{2} = b = 1 $. Thus, $ a^{2} = c^{2} - b^{2} = 3 $. Therefore, the standard equation of the hyperbola $ C $ is $ \\frac{x^{2}}{3} - y^{2} = 1 $."
  },
  {
    "text": "Let the parabola $C$: $y^{2}=2 px$ ($p>0$) have focus $F$, and let $Q$ be the intersection point of its directrix with the $x$-axis. Draw a line through point $F$ intersecting the parabola $C$ at points $A$ and $B$. If $\\angle QBF=90^{\\circ}$, then $|AF|-|BF|$=?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;Q: Point;B: Point;F: Point;A: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(F, G);Intersection(Directrix(C), xAxis) = Q;Intersection(G, C) = {A, B};AngleOf(Q, B, F) = ApplyUnit(90, degree)",
    "query_expressions": "Abs(LineSegmentOf(A, F)) - Abs(LineSegmentOf(B, F))",
    "answer_expressions": "2*p",
    "fact_spans": "[[[1, 25], [33, 34], [58, 64]], [[8, 25]], [[55, 57]], [[45, 48]], [[71, 74]], [[50, 54], [29, 32]], [[65, 68]], [[8, 25]], [[1, 25]], [[1, 32]], [[49, 57]], [[33, 48]], [[55, 76]], [[78, 101]]]",
    "query_spans": "[[[103, 116]]]",
    "process": ""
  },
  {
    "text": "Let the line $y=kx$ intersect the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ at points $A$ and $B$. Perpendiculars are drawn from $A$ and $B$ to the $x$-axis. If the feet of the perpendiculars coincide exactly with the two foci of the hyperbola, then the real number $k=$?",
    "fact_expressions": "G: Hyperbola;H: Line;k: Real;A: Point;B: Point;l1:Line;l2:Line;Expression(G) = (x^2 - y^2/3 = 1);Expression(H) = (y = k*x);Intersection(H, G) = {A, B};IsPerpendicular(l1,xAxis);IsPerpendicular(l2,xAxis);PointOnCurve(A,l1);PointOnCurve(B,l2);FootPoint(l1,xAxis)=F1;FootPoint(l2,xAxis)=F2;Focus(G)={F1,F2};F1:Point;F2:Point",
    "query_expressions": "k",
    "answer_expressions": "pm*(3/2)",
    "fact_spans": "[[[11, 39], [76, 79]], [[1, 10]], [[86, 91]], [[42, 45], [55, 58]], [[46, 49], [59, 62]], [], [], [[11, 39]], [[1, 10]], [[1, 51]], [[54, 70]], [[54, 70]], [[54, 70]], [[54, 70]], [[54, 84]], [[54, 84]], [[76, 84]], [], []]",
    "query_spans": "[[[86, 93]]]",
    "process": "Solve the system of equations of the line and the hyperbola: $ y = kx $, $ x^{2} - \\frac{y^{2}}{3} = 1 $. Simplifying and rearranging gives $ (3 - k^{2})x^{2} = 3 $ (*). Since perpendicular lines are drawn from points A and B perpendicular to the x-axis, and the feet of these perpendicular lines exactly coincide with the two foci of the hyperbola, the two roots of the equation are $ \\pm 1 $. Substituting into equation (*), we get $ k = \\pm \\frac{3}{2} $, so the answer is $ \\pm \\frac{3}{2} $."
  },
  {
    "text": "Let point $P$ be a moving point on the parabola $y^{2}=x$. A tangent line is drawn from point $P$ to the circle $M$: $(x-3)^{2}+y^{2}=1$, with the point of tangency being $A$. Then the minimum value of $\\overrightarrow{P A} \\cdot \\overrightarrow{P M}$ is?",
    "fact_expressions": "P: Point;G: Parabola;Expression(G) = (y^2 = x);PointOnCurve(P, G);M: Circle;Expression(M) = (y^2 + (x - 3)^2 = 1);l1: Line;A: Point;TangentOfPoint(P, M) = l1;TangentPoint(l1, M) = A",
    "query_expressions": "Min(DotProduct(VectorOf(P, A), VectorOf(P, M)))",
    "answer_expressions": "7/4",
    "fact_spans": "[[[0, 4], [23, 27]], [[5, 17]], [[5, 17]], [[0, 21]], [[28, 52]], [[28, 52]], [], [[61, 64]], [[22, 57]], [[22, 64]]]",
    "query_spans": "[[[66, 121]]]",
    "process": "From the given, it is easy to obtain $\\overrightarrow{PA}\\cdot\\overrightarrow{PM}=|\\overrightarrow{PA}|^{2}=|\\overrightarrow{PM}|^{2}-1$. Let point $P(y^{2},y)$, then $|\\overrightarrow{PM}|^{2}-1=(y^{2}-3)^{2}+y^{2}-1=y^{4}-5y^{2}+8=(y^{2}-\\frac{5}{2})^{2}+\\frac{7}{4}\\geqslant\\frac{7}{4}$. When $y^{2}=\\frac{5}{2}$, $\\overrightarrow{PA}\\cdot\\overrightarrow{PM}=|\\overrightarrow{PM}|^{2}-1$ reaches the minimum value $\\frac{7}{4}$."
  },
  {
    "text": "$A$, $B$ are two points on the parabola $y=2x^{2}$, and line $l$ is the perpendicular bisector of segment $AB$. When the slope of line $l$ is $\\frac{1}{2}$, what is the range of the $y$-intercept of line $l$?",
    "fact_expressions": "l: Line;G: Parabola;B: Point;A: Point;Expression(G) = (y = 2*x^2);PointOnCurve(A, G);PointOnCurve(B, G);PerpendicularBisector(LineSegmentOf(A,B))=l;Slope(l) = 1/2",
    "query_expressions": "Range(Intercept(l,yAxis))",
    "answer_expressions": "(5/4,+oo)",
    "fact_spans": "[[[27, 32], [48, 53], [73, 78]], [[8, 22]], [[4, 7]], [[0, 3]], [[8, 22]], [[0, 26]], [[0, 26]], [[27, 46]], [[48, 70]]]",
    "query_spans": "[[[73, 93]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, a line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. If $|F_{2} A|+|F_{2} B|=12$, then $|A B|=$?",
    "fact_expressions": "G: Ellipse;H: Line;F2: Point;A: Point;B: Point;F1: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(G) = {F1, F2};PointOnCurve(F1, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(F2, A)) + Abs(LineSegmentOf(F2, B)) = 12",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[18, 56], [74, 76]], [[71, 73]], [[10, 17]], [[77, 80]], [[81, 84]], [[2, 9], [63, 70]], [[18, 56]], [[2, 61]], [[62, 73]], [[71, 86]], [[88, 112]]]",
    "query_spans": "[[[115, 124]]]",
    "process": ""
  },
  {
    "text": "Given that the equation $\\frac{x^{2}}{4-m}+\\frac{y^{2}}{6+m}=1$ represents an ellipse, find the range of real values for $m$?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G)=(x^2/(4 - m) + y^2/(m + 6) = 1)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-6,-1)+(-1,4)",
    "fact_spans": "[[[45, 47]], [[49, 54]], [[2, 47]]]",
    "query_spans": "[[[49, 60]]]",
    "process": "The equation $\\frac{x2}{4-m}+\\frac{y2}{6+m}=1$ represents an ellipse, so $_{6+m>0}^{+}m_{m>0}\\Rightarrow-6<m<-1$ or $-1<m<4$, ($4-m\\neq6+m$), therefore the range of real number $m$ is $(-6,-1)\\cup(-1,4)$"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, respectively, and $A B$ is a chord passing through the point $F_{1}$ (with both $A$ and $B$ on the left branch of the hyperbola). If the perimeter of $\\triangle A B F_{2}$ is $30$, then $|A B|=$?",
    "fact_expressions": "G: Hyperbola;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2/9 - y^2/16 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;IsChordOf(LineSegmentOf(A,B),G);PointOnCurve(F1,LineSegmentOf(A,B));PointOnCurve(A,LeftPart(G));PointOnCurve(B,LeftPart(G));Perimeter(TriangleOf(A, B, F2)) = 30",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "9",
    "fact_spans": "[[[20, 59], [96, 99]], [[87, 90]], [[91, 94]], [[10, 17]], [[2, 9], [73, 81]], [[20, 59]], [[2, 65]], [[2, 65]], [[20, 85]], [[66, 81]], [[87, 103]], [[87, 103]], [[106, 135]]]",
    "query_spans": "[[[137, 146]]]",
    "process": "The hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ gives $a=3$. Since points $A$ and $B$ are both on the left branch of the hyperbola, $|AF_{2}|-|AF_{1}|=2a$, $|BF_{2}|-|BF_{1}|=2a$. Then the perimeter of $\\triangle ABF_{2}$ is $|AF_{2}|+|BF_{2}|+|AB|=(|AF|+2a)+(|BF|+2a)+|AB|=2|AB|+4a$. Therefore, $2|AB|+4\\times3=30$, so $|AB|=9$. Hence the answer is $9$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and points $A$ and $B$ are the right vertex and upper vertex of the ellipse $E$, respectively. If there exists a point $P$ on the line $AB$ such that $PF_{1} \\perp PF_{2}$, then the range of the eccentricity $e$ of the ellipse $E$ is?",
    "fact_expressions": "E: Ellipse;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(E) = F1;RightFocus(E) = F2;A: Point;B: Point;RightVertex(E) = A;UpperVertex(E) = B;P: Point;PointOnCurve(P, LineOf(A, B));IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));e: Number;Eccentricity(E) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "[(sqrt(5)-1)/2, 1)",
    "fact_spans": "[[[20, 77], [94, 99], [151, 156]], [[20, 77]], [[27, 77]], [[27, 77]], [[27, 77]], [[27, 77]], [[2, 9]], [[10, 17]], [[2, 82]], [[2, 82]], [[83, 87]], [[88, 91]], [[83, 107]], [[83, 107]], [[119, 123]], [[109, 123]], [[126, 149]], [[160, 163]], [[151, 163]]]",
    "query_spans": "[[[160, 170]]]",
    "process": "As shown in the figure, from the given conditions we have A(a,0), B(0,b), F_{1}(-c,0), F_{2}(c,0). Point P lies on line AB, let the coordinates of point P be (x,-\\frac{b}{a}x+b), then \\overrightarrow{PF_{1}}=(-c-x,\\frac{b}{a}x-b), \\overrightarrow{PF_{2}}=(c-x,\\frac{b}{a}x-b). Since PF_{1}\\bot PF_{2}, we have \\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=0, i.e., (-c-x,\\frac{b}{a}x-b)(c-x,\\frac{b}{a}x-b)=0, which gives x^{2}-c^{2}+\\frac{b^{2}}{a^{2}}x^{2}-\\frac{2b^{2}}{a}x+b^{2}=0. Simplifying yields (1+\\frac{b^{2}}{a^{2}})x^{2}-\\frac{2b^{2}}{a}x+2b^{2}-a^{2}=0. The existence of point P on line AB such that PF_{1}\\bot PF_{2} means equation (1) has a solution. Therefore, \\Delta=\\frac{4b^{4}}{a^{2}}-4(1+\\frac{b^{2}}{a^{2}})(2b^{2}-a^{2})\\geqslant0. Simplifying yields a^{4}-b^{2}a^{2}-b^{4}\\geqslant0, i.e., a^{4}-(a^{2}-c^{2})a^{2}-(a^{2}-c^{2})^{2}\\geqslant0. Simplifying further gives a^{4}+c^{4}-3a^{2}c^{2}\\leqslant0, i.e., (\\frac{c}{a})^{4}-\\frac{3c^{2}}{a^{2}}+1\\leqslant0, i.e., e^{4}-3e^{2}+1\\leqslant0. Solving yields: \\frac{3-\\sqrt{5}}{2}\\leqslante^{2}\\leqslant\\frac{3+\\sqrt{5}}{2}, i.e., \\frac{6-2\\sqrt{5}}{4}\\leqslante^{2}\\leqslant\\frac{6+2\\sqrt{5}}{4}, i.e., (\\frac{\\sqrt{5}-1}{2})^{2}\\leqslante^{2}\\leqslant(\\frac{\\sqrt{5}+1}{2})^{2}, i.e., \\frac{\\sqrt{5}-1}{2}\\leqslante\\leqslant\\frac{\\sqrt{5}+1}{2}. Also, for an ellipse 0<e<1, thus \\frac{\\sqrt{5}-1}{2}\\leqslante<1"
  },
  {
    "text": "Given that the line $l$ passing through point $P(1, 0)$ with an inclination angle of $60^{\\circ}$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$, then the chord length $|AB|=$?",
    "fact_expressions": "P: Point;Coordinate(P) = (1, 0);l: Line;PointOnCurve(P, l);Inclination(l) = ApplyUnit(60, degree);G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;B: Point;Intersection(l, G) = {A, B};IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[3, 14]], [[3, 14]], [[32, 37]], [[2, 37]], [[15, 37]], [[38, 52]], [[38, 52]], [[54, 57]], [[60, 63]], [[32, 65]], [[38, 75]]]",
    "query_spans": "[[[69, 77]]]",
    "process": ""
  },
  {
    "text": "Let $P(x, y)$ be a point on $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$. Then the minimum value of $\\frac{2}{5} x+\\frac{3}{4} y$ is?",
    "fact_expressions": "P: Point;G:Curve;x1:Number;y1:Number;Coordinate(P) = (x1, y1);Expression(G)=(x^2/25 + y^2/16 = 1);PointOnCurve(P,G)",
    "query_expressions": "Min((2/5)*x1 + (3/4)*y1)",
    "answer_expressions": "-sqrt(13)",
    "fact_spans": "[[[1, 10]], [[11, 48]], [[1, 10]], [[1, 10]], [[1, 10]], [[11, 48]], [[1, 51]]]",
    "query_spans": "[[[53, 88]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(b>a>0)$, with focal distance $2 c$, and a line $l$ passing through the points $(a, 0)$ and $(0, b)$. If the distance from $(-a, 0)$ to the line $l$ is $\\frac{2 \\sqrt{2}}{3} c$, then the eccentricity is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;b > a;a > 0;c: Number;FocalLength(G) = 2*c;l: Line;Q: Point;W: Point;Coordinate(Q) = (a, 0);Coordinate(W) = (0, b);PointOnCurve(Q, l);PointOnCurve(W, l);E: Point;Coordinate(E) = (-a, 0);Distance(E, l) = c*(2*sqrt(2)/3)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 55]], [[2, 55]], [[5, 55]], [[5, 55]], [[5, 55]], [[5, 55]], [[59, 64]], [[2, 64]], [[65, 70], [102, 107]], [[72, 81]], [[82, 90]], [[72, 81]], [[82, 90]], [[65, 81]], [[65, 90]], [[92, 101]], [[92, 101]], [[92, 135]]]",
    "query_spans": "[[[2, 142]]]",
    "process": "The equation of line $ l $ is $ \\frac{x}{a} + \\frac{y}{b} = 1 $, that is, $ bx + ay - ab = 0 $, $ c^{2} = a^{2} + b^{2} $, the distance from $ (-a, 0) $ to line $ l $ is $ \\frac{2\\sqrt{2}}{3}c $, we obtain: $ \\frac{2ab}{\\sqrt{a^{2} + b^{2}}} = \\frac{2\\sqrt{2}}{3}c $. Thus, $ 3ab = \\sqrt{2}c^{2} $, that is, $ 9a^{2}b^{2} = 2c^{4} $, that is, $ 9a^{2}(c^{2} - a^{2}) = 2c^{4} $, $ 9a^{2c2 - 9a^{4} - 2c^{4}} = 0 $, since $ e = \\frac{c}{a} $, then $ 2e^{4} - 9e^{2} + 9 = 0 $, solving gives $ e^{2} = 3 $ or $ e^{2} = \\frac{3}{2} $, since $ 0 < a < b $, i.e., $ a^{2} < b^{2} $, thus $ c^{2} > 2a^{2} $, hence $ e^{2} > 2 $, therefore $ e = \\sqrt{3} $"
  },
  {
    "text": "Given that the point $P$ on the parabola $y^{2}=4x$ is at a distance $d_{1}$ from the directrix of the parabola and at a distance $d_{2}$ from the line $3x-4y+9=0$, what is the minimum value of $d_{1}+d_{2}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G);d1: Number;Distance(P, Directrix(G)) = d1;H: Line;Expression(H) = (3*x - 4*y + 9 = 0);d2: Number;Distance(P, H) = d2",
    "query_expressions": "Min(d1 + d2)",
    "answer_expressions": "12/5",
    "fact_spans": "[[[2, 16], [23, 26]], [[2, 16]], [[18, 22]], [[2, 22]], [[33, 40]], [[18, 40]], [[42, 57]], [[42, 57]], [[61, 68]], [[18, 68]]]",
    "query_spans": "[[[70, 89]]]",
    "process": ""
  },
  {
    "text": "If the distance from the focus of the hyperbola $x^{2}-\\frac{y^{2}}{k}=1$ to its asymptote is $2 \\sqrt {2}$, then the value of the real number $k$ is?",
    "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (x^2 - y^2/k = 1);Distance(Focus(G),Asymptote(G))=2*sqrt(2)",
    "query_expressions": "k",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 29]], [[55, 60]], [[1, 29]], [[1, 53]]]",
    "query_spans": "[[[55, 64]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is an arbitrary point on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and $F_{1}$, $F_{2}$ are the two foci of the ellipse, then the maximum value of $|P F_{1}| \\cdot|P F_{2}|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2}",
    "query_expressions": "Max(Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)))",
    "answer_expressions": "4",
    "fact_spans": "[[[6, 33], [55, 57]], [[2, 5]], [[39, 46]], [[47, 54]], [[6, 33]], [[2, 38]], [[39, 62]]]",
    "query_spans": "[[[64, 96]]]",
    "process": ""
  },
  {
    "text": "Given that point $A$ is the intersection of the axis of symmetry and the directrix of the parabola $y = \\frac{1}{4} x^{2}$, point $F$ is the focus of this parabola, and point $P$ lies on the parabola such that $|P F| = m|P A|$, then the minimum value of $m$ is?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;A: Point;Expression(G) = (y = x^2/4);Focus(G) = F;Intersection(SymmetryAxis(G),Directrix(G))=A;PointOnCurve(P,G);Abs(LineSegmentOf(P,F))=m*Abs(LineSegmentOf(P,A));m:Number",
    "query_expressions": "Min(m)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[7, 31], [48, 51], [60, 63]], [[55, 59]], [[42, 46]], [[2, 6]], [[7, 31]], [[42, 54]], [[2, 41]], [[55, 64]], [[67, 81]], [[83, 86]]]",
    "query_spans": "[[[83, 92]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1  (a>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, with eccentricity $2$. Let $P$ be a point on the right branch of the hyperbola such that $|P F_{1}|^{2}-|P F_{2}|^{2}=4$. Then the perimeter of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;a: Number;P: Point;F1: Point;F2: Point;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;Eccentricity(G)=2;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F1))^2 - Abs(LineSegmentOf(P, F2))^2 = 4",
    "query_expressions": "Perimeter(TriangleOf(P, F1, F2))",
    "answer_expressions": "10*sqrt(3)/3",
    "fact_spans": "[[[2, 41], [78, 81]], [[5, 41]], [[74, 77]], [[50, 57]], [[58, 65]], [[5, 41]], [[2, 41]], [[2, 65]], [[2, 65]], [[2, 73]], [[74, 86]], [[90, 121]]]",
    "query_spans": "[[[123, 150]]]",
    "process": "First, obtain $ a $ from the eccentricity, and from the definition of the hyperbola obtain $ |PF_{1}| - |PF_{2}| $, finally the perimeter can be found from the given conditions. According to the problem, $ \\frac{\\sqrt{a^{2}+1}}{a} = 2 $, $ a = \\frac{\\sqrt{3}}{3} $, $ c = \\sqrt{a^{2}+1} = \\frac{2\\sqrt{3}}{3} $. Since $ P $ is a point on the right branch of the hyperbola, $ |PF_{1}| - |PF_{2}| = 2a = \\frac{2\\sqrt{3}}{3} $. Because $ |PF_{1}|^{2} - |PF_{2}|^{2} = (|PF_{1}| - |PF_{2}|)(|PF_{1}| + |PF_{2}|) = 4 $, so $ |PF_{1}| + |PF_{2}| = 2\\sqrt{3} $. Therefore, the perimeter of $ \\triangle PF_{1}F_{2} $ is $ |PF_{1}| + |PF_{2}| + |F_{1}F_{2}| = 2\\sqrt{3} + \\frac{4\\sqrt{3}}{3} = \\frac{10\\sqrt{3}}{3} $."
  },
  {
    "text": "Let the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ have its left focus at $F$, and let the line $x=m$ intersect the ellipse $C$ at points $A$ and $B$. When the perimeter of $\\triangle A B F$ is maximized, the area of $\\triangle A B F$ is $b^{2}$. Then the eccentricity $e$ of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;LeftFocus(C) = F;G: Line;Expression(G) = (x = m);m: Number;A: Point;B: Point;Intersection(G, C) = {A, B};WhenMax(Perimeter(TriangleOf(A, B, F)));Area(TriangleOf(A, B, F)) = b^2;e: Number;Eccentricity(C) = e",
    "query_expressions": "e",
    "answer_expressions": "1/2",
    "fact_spans": "[[[1, 58], [75, 80], [151, 156]], [[1, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[63, 66]], [[1, 66]], [[67, 74]], [[67, 74]], [[69, 74]], [[83, 86]], [[87, 90]], [[67, 92]], [[94, 118]], [[120, 148]], [[160, 163]], [[151, 163]]]",
    "query_spans": "[[[160, 165]]]",
    "process": "First, based on the definition of an ellipse, analyze the position of line AB when the perimeter of $\\triangle ABF$ is maximized, then find the area of $\\triangle ABF$ to obtain the eccentricity of the ellipse. Let the right focus of the ellipse be $F$, $|AF| + |BF| \\geqslant |AB|$, and equality holds when line AB passes through the right focus $F$. Therefore, the perimeter $l$ of $\\triangle ABF = |AF| + |BF| + |AB| \\leqslant |AF| + |BF| + |AF| + |BF| = 4a$. At this time, line AB passes through the right focus, $|AB| = \\frac{2b^{2}}{a}$, $S_{\\Delta ABF} = \\frac{1}{2} \\times \\frac{2b^{2}}{a} \\times 2c = b^{2}$, yielding $e = \\frac{c}{a} = \\frac{1}{2}$."
  },
  {
    "text": "Given that $A$, $B$, $C$ are any three points on the parabola $y=4x$, $F$ is the focus of the parabola, and $FA + FB + FC = 0$, then $|FA| + |FB| + |FC| =$?",
    "fact_expressions": "A: Point;B: Point;C: Point;PointOnCurve(A, G);PointOnCurve(B, G);PointOnCurve(C, G);G: Parabola;Expression(G) = (y = 4*x);F: Point;Focus(G) = F;LineSegmentOf(F, A) + LineSegmentOf(F, B) + LineSegmentOf(F, C) = 0",
    "query_expressions": "Abs(LineSegmentOf(F, A)) + Abs(LineSegmentOf(F, B)) + Abs(LineSegmentOf(F, C))",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 5]], [[6, 9]], [[10, 13]], [[2, 29]], [[2, 29]], [[2, 29]], [[14, 24], [34, 37]], [[14, 24]], [[30, 33]], [[30, 40]], [[41, 56]]]",
    "query_spans": "[[[58, 85]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $x^{2}=2 p y(p>0)$, $P$ and $Q$ are any two points on $C$, and the point $M(0,-1)$ satisfies $\\overrightarrow{M P} \\cdot \\overrightarrow{M Q} \\geq 0$. Then the range of values for $p$ is?",
    "fact_expressions": "C: Parabola;p: Number;M: Point;P: Point;Q: Point;p>0;Expression(C) = (x^2 = 2*p*y);Coordinate(M) = (0, -1);PointOnCurve(P,C);PointOnCurve(Q, C);DotProduct(VectorOf(M, P), VectorOf(M, Q)) >= 0",
    "query_expressions": "Range(p)",
    "answer_expressions": "(0,2]",
    "fact_spans": "[[[2, 28], [40, 43]], [[119, 122]], [[49, 59]], [[31, 34]], [[36, 39]], [[10, 28]], [[2, 28]], [[49, 59]], [[31, 48]], [[31, 48]], [[61, 117]]]",
    "query_spans": "[[[119, 129]]]",
    "process": "When the lines MQ and MP are tangent to the parabola, the angle between the two vectors is maximized. Let the slope of line MQ be k. Then, when $ k \\geqslant 1 $, it always holds that $ \\overrightarrow{MP} \\cdot \\overrightarrow{MQ} \\geqslant 0 $. The equation of line MQ is $ y = kx - 1 $. Combining this with $ x^{2} = 2py $ gives $ x^{2} - 2pkx + 2p = 0 $. From $ \\triangle = 0 $, we obtain $ k^{2} = \\frac{p}{2} \\geqslant 1 $, which implies $ p \\leqslant 2 $. Therefore, the range of values for p is $ (0, 2] $."
  },
  {
    "text": "Let $P$ be a point on the line $y=2x$ located in the first quadrant. If the product of the distances from point $P$ to the two asymptotes of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ is $27$, then the coordinates of point $P$ are?",
    "fact_expressions": "H: Line;Expression(H) = (y = 2*x);P: Point;Quadrant(P) = 1;PointOnCurve(P, H);G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1);L1: Line;L2: Line;Asymptote(G) = {L1, L2};Distance(P, L1)*Distance(P, L2) = 27",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(3, 6)",
    "fact_spans": "[[[5, 14]], [[5, 14]], [[1, 4], [28, 32], [79, 83]], [[1, 26]], [[1, 18]], [[33, 61]], [[33, 61]], [], [], [[33, 67]], [[28, 77]]]",
    "query_spans": "[[[79, 88]]]",
    "process": "Let $ P(x_{0},2x_{0}) $ ($ x_{0}>0 $), the two asymptotes of the hyperbola $ \\frac{x^{2}}{4}-y^{2}=1 $ are $ y=\\pm\\frac{1}{2}x $, i.e., $ x\\pm2y=0 $. Then the product of the distances from point $ P $ to the two asymptotes of the hyperbola is $ \\frac{|x_{0}+4x_{0}|}{\\sqrt{5}} \\cdot \\frac{|x_{0}-4x_{0}|}{\\sqrt{5}} = 27 $, solving gives: $ x_{0}=3 $. Therefore, the coordinates of point $ P $ are $ (3,6) $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, let $M$ be a moving point on its right branch, $F$ its right focus, and point $A(3 , 1)$. Then the minimum value of $|M A|+|M F|$ is?",
    "fact_expressions": "G: Hyperbola;A: Point;M: Point;F: Point;Expression(G) = (x^2/3 - y^2 = 1);Coordinate(A) = (3, 1);RightFocus(G) = F;PointOnCurve(M,RightPart(G))",
    "query_expressions": "Min(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F)))",
    "answer_expressions": "sqrt(26)-2*sqrt(3)",
    "fact_spans": "[[[2, 30], [37, 38], [49, 50]], [[54, 65]], [[33, 36]], [[45, 48]], [[2, 30]], [[54, 65]], [[45, 53]], [[33, 44]]]",
    "query_spans": "[[[67, 86]]]",
    "process": "Let the left focus of the hyperbola be F, then |PA| + |PF| = |PF| - 2a + |PA|. \\therefore when points P, F, and A are collinear, it has the minimum value. \\because the equation of the hyperbola is \\frac{x^{2}}{3} - y^{2} = 1, \\therefore F(-2,0), A(3,1). \\therefore |PF| + |PA| = |AF| = \\sqrt{26}. \\because 2a = 2\\sqrt{3}, \\therefore the minimum value of |MA| + |MF| is \\sqrt{26} - 2\\sqrt{3}."
  },
  {
    "text": "Given that the asymptotes of a hyperbola are $x \\pm y=0$, and it passes through the point $P(-1,-2)$, then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;P: Point;Coordinate(P) = (-1, -2);Expression(Asymptote(G)) = (x + pm*y = 0);PointOnCurve(P, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/3 - x^2/3 = 1",
    "fact_spans": "[[[2, 5], [40, 43]], [[26, 37]], [[26, 37]], [[2, 23]], [[2, 37]]]",
    "query_spans": "[[[40, 50]]]",
    "process": "According to the problem, one asymptote of the hyperbola is given by $ x \\pm y = 0 $, so we may assume the equation of the hyperbola is $ x^{2} - y^{2} = \\lambda $ ($ \\lambda \\neq 0 $). Since the hyperbola passes through the point $ P(-1, -2) $, we have $ 1 - 4 = \\lambda $, so $ \\lambda = -3 $. Therefore, the required equation of the hyperbola is $ \\frac{y^{2}}{3} - \\frac{x^{2}}{3} = 1 $."
  },
  {
    "text": "It is known that $F$ is the left focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $A$ is the right vertex of the ellipse $C$, and $P$ is a point on the ellipse $C$ such that $PF$ is perpendicular to the $x$-axis. If the slope of the line $PA$ is $\\frac{\\sqrt{5}}{5}$, then what is the eccentricity of the ellipse $C$?",
    "fact_expressions": "C: Ellipse;Expression(C)=(x^2/a^2 + y^2/b^2 = 1);b:Number;a>b;F: Point;LeftFocus(C)=F;A:Point;RightVertex(C)=A;PointOnCurve(P,C);IsPerpendicular(LineSegmentOf(P,F),xAxis);Slope(LineOf(P,A))=sqrt(5)/5;b>0;a:Number;P:Point",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "1-sqrt(5)/5",
    "fact_spans": "[[[6, 63], [72, 77], [86, 91], [146, 152]], [[6, 63]], [[13, 63]], [[13, 63]], [[2, 5]], [[2, 67]], [[68, 71]], [[68, 81]], [[82, 94]], [[96, 108]], [[110, 143]], [[13, 63]], [[12, 63]], [[82, 85]]]",
    "query_spans": "[[[146, 158]]]",
    "process": "Let the semi-focal length of ellipse C be $ c $, then $ F(-c,0) $, $ A(a,0) $, and since $ PF \\perp x $-axis, from $ x = -c $ we obtain the ordinate of point P as $ y_{P} = \\pm\\frac{b^{2}}{a} $. Given that the slope of line PA is $ \\frac{\\sqrt{5}}{5} $, the coordinates of point P are $ (-c, -\\frac{b^{2}}{a}) $. Thus we have $ \\frac{b^{2}}{a+c} = \\frac{\\sqrt{5}}{5} $, that is, $ \\frac{a^{2}-c^{2}}{a(a+c)} = \\frac{\\sqrt{5}}{5} $. Simplifying yields: $ 1 - \\frac{c}{a} = \\frac{\\sqrt{5}}{5} $, hence $ e = \\frac{c}{a} = 1 - \\frac{\\sqrt{5}}{5} $. Therefore, the eccentricity of ellipse C is $ 1 - \\frac{\\sqrt{5}}{5} $."
  },
  {
    "text": "Given that a line $l$ passing through the left focus $F$ of the ellipse $E$: $\\frac{x^{2}}{2}+y^{2}=1$ intersects $E$ at points $A$ and $B$, what is the minimum value of $|A F|+2|B F|$?",
    "fact_expressions": "l: Line;E: Ellipse;A: Point;F: Point;B: Point;Expression(E) = (x^2/2 + y^2 = 1);LeftFocus(E) = F;PointOnCurve(F, l);Intersection(l, E) = {A, B}",
    "query_expressions": "Min(Abs(LineSegmentOf(A, F)) + 2*Abs(LineSegmentOf(B, F)))",
    "answer_expressions": "1+3*sqrt(2)/4",
    "fact_spans": "[[[43, 48]], [[3, 35], [49, 52]], [[53, 56]], [[39, 42]], [[57, 60]], [[3, 35]], [[3, 42]], [[2, 48]], [[43, 62]]]",
    "query_spans": "[[[64, 84]]]",
    "process": "From the given conditions, F(-1,0). Let the equation of line $ l $ be $ x = my - 1 $. From  \n$$\n\\begin{cases}\nx = my - 1 \\\\\n\\frac{x^{2}}{2} + y^{2} = 1\n\\end{cases},\n$$\nwe obtain $ (m^{2} + 2)y^{2} - 2my - 1 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ y_{1} + y_{2} $. Without loss of generality, assume $ y_{1} > 0 $, $ y_{2} < 0 $, then $ \\frac{|AF|}{|BF|} = \\frac{|}{|}.\\frac{1}{3} $. Since $ |AF| + |BF| = |AB| $, we have $ \\frac{1}{t} = \\frac{y_{1} - y_{2}}{|AB|} $. Therefore,  \n$$\n\\frac{M}{|AF|} + \\frac{1}{|BF|} = \\frac{1}{t \\cdot y_{1}} + \\frac{1}{-t \\cdot y_{2}} = \\frac{1}{t} \\cdot \\frac{y_{2} - y_{1}}{y_{1}y_{2}} = \\frac{y_{1} - y_{2}}{|AB|} \\cdot \\frac{y_{2} - y_{1}}{y_{1}y_{2}} = \\frac{y_{1} - y_{2}}{\\sqrt{1 + m^{2}} \\cdot (y_{1} - y_{2})} \\cdot \\frac{y_{2} - y_{1}}{y_{1}y_{2}} = \\frac{1}{\\sqrt{1 + m^{2}}} \\cdot \\frac{y_{2} - y_{1}}{y_{1}y_{2}} = 2\\sqrt{2} = \\frac{-\\frac{\\sqrt{4}}{m^{2} + 2}}{\\sqrt{1 + m^{2}} \\cdot \\frac{1}{m^{2} + 2}}\n$$\nThus,  \n$$\n2\\sqrt{2} \\cdot (|AF| + 2|BF|) = \\left( \\frac{1}{|AF|} + \\frac{1}{|BF|} \\right) \\cdot (|AF| + 2|BF|) = \\frac{2|BF|}{|AF|} + \\frac{|AF|}{|BF|} + 3 \\geqslant 2\\sqrt{2} + 3,\n$$\nso $ |AF| + 2|BF| \\geqslant \\frac{2\\sqrt{2} + 3}{2\\sqrt{2}} = 1 + \\frac{3\\sqrt{2}}{4} $, with equality if and only if $ \\frac{2|BF|}{|AF|} = \\frac{|AF|}{|BF|} $, i.e., $ |AF| = \\sqrt{2}|BF| $. Also, when line $ l $ is $ y = 0 $, it is easy to get $ |AF| + 2|BF| = |AB| + |BF| = 3\\sqrt{2} \\pm 1 $. Therefore, the minimum value of $ |AF| + 2|BF| $ is $ 1 + \\frac{3\\sqrt{2}}{4} $."
  },
  {
    "text": "The coordinates of the point on the parabola $y=x^{2}$ that is closest to the line $2 x-y-4=0$ are?",
    "fact_expressions": "G: Parabola;H: Line;Expression(G) = (y = x^2);Expression(H) = (2*x - y - 4 = 0);P: Point;PointOnCurve(P, G);WhenMin(Distance(P, H))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1, 1)",
    "fact_spans": "[[[0, 12]], [[14, 27]], [[0, 12]], [[14, 27]], [[33, 34]], [[0, 34]], [[13, 34]]]",
    "query_spans": "[[[33, 39]]]",
    "process": "The derivative of $ y = x^{2} $ is $ y' = 2x $. Let the desired point be $ P(x_{0}, y_{0}) $, then $ 2x_{0} = 2 $, $ \\therefore x_{0} = 1 $, $ \\therefore P(1, 1) $."
  },
  {
    "text": "It is known that an ellipse $C$ with foci on the $x$-axis passes through the point $A(2,1)$ and has eccentricity $\\frac{\\sqrt{2}}{2}$. Then the equation of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;A: Point;Coordinate(A) = (2, 1);PointOnCurve(Focus(C), xAxis);PointOnCurve(A,C);Eccentricity(C)=sqrt(2)/2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/6+y^2/3=1",
    "fact_spans": "[[[11, 16], [55, 60]], [[18, 27]], [[18, 27]], [[2, 16]], [[11, 27]], [[11, 53]]]",
    "query_spans": "[[[55, 65]]]",
    "process": "According to the problem, let the equation of ellipse C be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, we have $\\left|\\frac{a^{2}}{a}=\\frac{b^{2}+c^{2}}{\\frac{\\sqrt{2}}{2}}\\right|$, solving gives $b^{2}=3$, $a^{2}=6$, $\\frac{4}{2}+\\frac{1}{2}=1$. Since the equation of ellipse C is $\\frac{x^{2}}{6}+\\frac{y^{2}}{3}=1$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ with focus $F$, and a point $M(3, y_{0})(y_{0}>0)$ on the parabola such that the distance from $M$ to the focus is $|M F|=4$, then the coordinates of point $M$ are?",
    "fact_expressions": "C: Parabola;p: Number;M: Point;F: Point;p>0;y0: Number;y0 > 0;Expression(C) = (y^2 = 2*(p*x));Coordinate(M) = (3, y0);Focus(C) = F;PointOnCurve(M, C);Distance(M, Focus(C)) = Abs(LineSegmentOf(M, F));Abs(LineSegmentOf(M, F)) = 4",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(3, 2*sqrt(3))",
    "fact_spans": "[[[2, 28], [36, 39]], [[10, 28]], [[41, 64], [81, 85]], [[32, 35]], [[10, 28]], [[42, 64]], [[42, 64]], [[2, 28]], [[41, 64]], [[2, 35]], [[36, 64]], [[36, 79]], [[70, 79]]]",
    "query_spans": "[[[81, 90]]]",
    "process": "The coordinates of point F are (\\frac{p}{2},0), the equation of the directrix of the parabola is x=-\\frac{p}{2}. By the definition of a parabola, |MF|=3+\\frac{p}{2}=4, so p=2. Therefore, the equation of the parabola is y^{2}=4x. Substituting x=3 into the equation gives y_{0}=2\\sqrt{3} (the negative value is discarded), so the coordinates of point M are (3,2\\sqrt{3}). The answer is: (3.,2)"
  },
  {
    "text": "Point $P$ lies on the graph of the parabola $x^{2}=4 y$, $F$ is the focus of the parabola, and point $A(-1 , 3)$. If $|PF|+|PA|$ is minimized, then the coordinates of the corresponding point $P$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);P: Point;PointOnCurve(P, G);F: Point;Focus(G) = F;A: Point;Coordinate(A) = (-1, 3);WhenMin(Abs(LineSegmentOf(P, F))+Abs(LineSegmentOf(P, A)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(-1, 1/4)",
    "fact_spans": "[[[5, 19], [28, 31]], [[5, 19]], [[0, 4], [67, 71]], [[0, 23]], [[24, 27]], [[24, 34]], [[35, 47]], [[35, 47]], [[50, 63]]]",
    "query_spans": "[[[67, 76]]]",
    "process": ""
  },
  {
    "text": "Point $P$ lies on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the eccentricity of the hyperbola is $\\frac{5}{4}$, $F_{1}$, $F_{2}$ are its foci, $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, if the area of $\\Delta F_{1} P F_{2}$ is $18$, what is the value of $a+b$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;P: Point;PointOnCurve(P, G);Eccentricity(G) = 5/4;F1: Point;F2: Point;Focus(G) = {F1, F2};DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Area(TriangleOf(F1, P, F2)) = 18",
    "query_expressions": "a + b",
    "answer_expressions": "7*sqrt(2)",
    "fact_spans": "[[[5, 61], [65, 68], [105, 106]], [[5, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[0, 4]], [[0, 64]], [[65, 86]], [[88, 95]], [[97, 104]], [[88, 108]], [[109, 168]], [[171, 201]]]",
    "query_spans": "[[[203, 213]]]",
    "process": "Without loss of generality, assume point P lies on the right branch of the hyperbola. Let |PF_{1}| = m, |PF_{2}| = n. According to the definition of a hyperbola, we have m - n = 2a\\textcircled{1}. From \\overrightarrow{PF}_{1} \\cdot \\overrightarrow{PF_{2}} = 0, we know \\overrightarrow{PF_{1}} \\bot \\overrightarrow{PF_{2}}, so \\frac{1}{2}mn = 18\\textcircled{2}, and m^{2} + n^{2} = 4c^{2}\\textcircled{3}. Solving \\textcircled{1}\\textcircled{2}\\textcircled{3} simultaneously, we obtain c^{2} - a^{2} = 18\\textcircled{4}. Given the eccentricity of the hyperbola is \\frac{5}{4}, we have e = \\frac{c}{a} = \\frac{5}{4}\\textcircled{5}. Using the relationship among a, b, c in a hyperbola, we have a^{2} + b^{2} = c^{2}\\textcircled{6}. Solving \\textcircled{4}\\textcircled{5}\\textcircled{6} simultaneously, we get a = 4\\sqrt{2}, b = 3\\sqrt{2}, c = 5\\sqrt{2}. Hence, a + b = 4\\sqrt{2} + 3\\sqrt{2} = 7\\sqrt{2}. Therefore, fill in: 7\\sqrt{2}."
  },
  {
    "text": "It is known that an ellipse and a hyperbola share common foci $F_{1}$, $F_{2}$, and $P$, $Q$ are their intersection points in the first quadrant and third quadrant respectively, with $\\angle Q F_{2} P = 60^{\\circ}$. Let the eccentricities of the ellipse and the hyperbola be $e_{1}$, $e_{2}$ respectively. Then $\\frac{3}{e_{1}^{2}} + \\frac{1}{e_{2}^{2}}$ equals?",
    "fact_expressions": "H: Ellipse;G: Hyperbola;F1: Point;F2: Point;Focus(H) = {F1,F2};Focus(G) = {F1,F2};P: Point;Q: Point;Quadrant(P) = 1;Quadrant(Q) = 3;AngleOf(Q, F2, P) = ApplyUnit(60, degree);e1: Number;e2: Number;e1 = Eccentricity(H);e2 = Eccentricity(G);Intersection(H,G) = {P,Q}",
    "query_expressions": "1/(e2^2) + 3/(e1^2)",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 4], [88, 90]], [[5, 8], [91, 94]], [[14, 21]], [[22, 29]], [[2, 29]], [[2, 29]], [[30, 33]], [[34, 37]], [[30, 55]], [[30, 55]], [[57, 86]], [[101, 108]], [[110, 117]], [[88, 117]], [[88, 117]], [30, 54]]",
    "query_spans": "[[[119, 163]]]",
    "process": "Let the semi-major axis of the ellipse be $ a_{1} $, the semi-transverse axis of the hyperbola be $ a_{2} $, $ F_{1}(-c,0) $, $ F_{2}(c,0) $, $ P $ be the intersection point of the two curves in the first quadrant, and $ Q $ be the intersection point of the two curves in the third quadrant. From the definitions of the ellipse and hyperbola, we have: $ |PF_{1}| + |PF_{2}| = 2a_{1} $, $ |PF_{1}| - |PF_{2}| = 2a_{2} $. Therefore, $ |PF_{1}| = a_{1} + a_{2} $, $ |PF_{2}| = a_{1} - a_{2} $. By the symmetry of the ellipse and hyperbola, quadrilateral $ PF_{1}OF_{2} $ is a parallelogram. Since $ \\angle QF_{2}P = 60^{\\circ} $, $ \\angle F_{1}PF_{2} = 120^{\\circ} $. Thus, $ |F_{1}F_{2}|^{2} = |PF_{1}|^{2} + |PF_{2}|^{2} - 2|PF_{1}||PF_{2}|\\cos\\angle F_{1}PF_{2} $, that is, $ 4c^{2} = (a_{1}+a_{2})^{2} + (a_{1}-a_{2})^{2} + (a_{1}+a_{2})(a_{1}-a_{2}) = 3a_{1}^{2} + a_{2}^{2} $, $ \\frac{\\sqrt{2}}{2} + \\frac{1}{e^{2}} = \\frac{3a_{1}^{2}}{c^{2}} + \\frac{a_{2}^{2}}{c^{2}} = 4 $."
  },
  {
    "text": "If the chord length intercepted by the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ from a line passing through its focus and perpendicular to the $x$-axis is $a$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;a: Number;b: Number;H: Line;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(Focus(G), H);IsPerpendicular(H, xAxis);Length(InterceptChord(H, G)) = a",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 54], [69, 71], [83, 85]], [[77, 80]], [[4, 54]], [[66, 68]], [[4, 54]], [[4, 54]], [[2, 54]], [[1, 68]], [[58, 68]], [[66, 80]]]",
    "query_spans": "[[[83, 92]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=4x$, a line $l$ passing through point $F$ intersects the parabola $C$ at points $A$ and $B$. If $|AB|=10$, then the distance from the midpoint $M$ of segment $AB$ to the line $x+1=0$ is?",
    "fact_expressions": "l: Line;C: Parabola;G: Line;B: Point;A: Point;F: Point;M: Point;Expression(C) = (y^2 = 4*x);Expression(G) = (x + 1 = 0);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, B)) = 10;MidPoint(LineSegmentOf(A,B)) = M",
    "query_expressions": "Min(Distance(M, G))",
    "answer_expressions": "5",
    "fact_spans": "[[[35, 40]], [[6, 25], [41, 47]], [[85, 94]], [[52, 55]], [[48, 51]], [[2, 5], [30, 34]], [[81, 84]], [[6, 25]], [[85, 94]], [[2, 28]], [[29, 40]], [[35, 57]], [[59, 69]], [[71, 84]]]",
    "query_spans": "[[[81, 99]]]",
    "process": "As shown in the figure, F is the focus of the parabola C: y^{2}=4x. The line l passing through point F intersects the parabola C at points A and B. The equation of the directrix of the parabola is x+1=0. Perpendiculars are drawn from points A, B, and M to the directrix, with feet at C, D, and N respectively. Then |AB| = |AF| + |BF| = |AC| + |BD| = 10. Since M is the midpoint of AB, it follows that |MN| = \\frac{1}{2}(|AC| + |BD|) = \\frac{1}{2}|AB| = 5. Thus, the distance from the midpoint M of segment AB to the line x+1=0 is 5."
  },
  {
    "text": "Given the parabola $C$: $x^{2}=-4 y$ with focus $F$, a point $A$ on the parabola $C$ satisfies $|A F|=3$. What is the length of the chord intercepted on the $x$-axis by the circle centered at point $A$ with radius $A F$?",
    "fact_expressions": "C: Parabola;G: Circle;A: Point;F: Point;Expression(C) = (x^2 = -4*y);Focus(C) = F;PointOnCurve(A,C);Abs(LineSegmentOf(A,F))=3;Center(G)=A;Radius(G)=LineSegmentOf(A,F)",
    "query_expressions": "Length(InterceptChord(xAxis,G))",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[2, 22], [30, 36]], [[73, 74]], [[56, 60], [39, 42]], [[26, 29]], [[2, 22]], [[2, 29]], [[30, 42]], [[44, 53]], [[55, 74]], [[64, 74]]]",
    "query_spans": "[[[73, 85]]]",
    "process": "By the given condition, the parabola $ x^{2} = -4y $ has focus $ F(0, -1) $. Let $ A(x_{0}, y_{0}) $. According to the definition of a parabola, $ |AF| = -y_{0} + 1 = 3 $, solving gives $ y_{0} = -2 $, so the distance from $ A $ to the $ x $-axis is $ d = 2 $. Therefore, the chord length intercepted by the circle on the $ x $-axis is $ 2\\sqrt{R^{2} - d^{2}} = 2\\sqrt{3^{2} - 2^{2}} = 2\\sqrt{5} $."
  },
  {
    "text": "Given that point $P$ lies on the parabola $y^{2}=4x$, when the distance from $P$ to the line $y=x+4$ is shortest, what are the coordinates of point $P$?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (y = x + 4);PointOnCurve(P, G);WhenMin(Distance(P,H))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1,2)",
    "fact_spans": "[[[7, 21]], [[28, 37]], [[2, 6], [24, 27], [44, 48]], [[7, 21]], [[28, 37]], [[2, 22]], [[23, 43]]]",
    "query_spans": "[[[44, 53]]]",
    "process": "Let $ P\\left(\\frac{y^{2}}{4}, y\\right) $, then the distance $ d $ from point $ P $ to the line $ y = x + 4 $ is $ d = \\frac{\\left| \\frac{y^{2}}{4} - y + 4 \\right|}{\\sqrt{2}} = \\frac{1}{4\\sqrt{2}}(y - 2)^{2} + \\frac{3}{\\sqrt{2}} $. When $ y = 2 $, $ d $ attains its minimum value. Substituting $ y = 2 $ into $ y^{2} = 4x $, we get $ x = 1 $, so the coordinates of point $ P $ are $ (1, 2) $."
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ are $F_{1}$ and $F_{2}$, respectively. A point $M$ on the ellipse satisfies: $\\angle F_{1} M F_{2}=\\frac{2 \\pi}{3}$ and $\\overrightarrow{M F_{1}} \\cdot \\overrightarrow{M F_{2}}=-2$, then $b$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;M: Point;PointOnCurve(M, G) = True;AngleOf(F1, M, F2) = 2*pi/3;DotProduct(VectorOf(M, F1), VectorOf(M, F2)) = -2",
    "query_expressions": "b",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 52], [77, 79]], [[0, 52]], [[189, 192]], [[2, 52]], [[2, 52]], [[2, 52]], [[61, 68]], [[69, 76]], [[0, 76]], [[0, 76]], [[81, 85]], [[77, 85]], [[88, 126]], [[127, 187]]]",
    "query_spans": "[[[189, 194]]]",
    "process": "First, from the dot product operation we obtain |MF_{1}||MF_{2}|=4; then, combining the definition of the ellipse and the cosine law, we get b=1. Since \\angle F_{1}MF_{2}=\\frac{2\\pi}{3} and \\overrightarrow{MF_{1}}\\cdot\\overrightarrow{MF_{2}}=-2, it follows that |MF_{1}||MF_{2}|=4. From the definition of the ellipse, |MF_{1}|+|MF_{2}|=2a, hence |MF_{1}|^{2}+|MF_{2}|^{2}+2|MF_{1}||MF_{2}|=4a^{2}. In \\triangle F_{1}MF_{2}, by the cosine law, \\cos\\angle F_{1}MF_{2}=\\frac{|MF_{1}|^{2}+|MF_{2}|^{2}-4c^{2}}{2|MF_{1}||MF_{2}|}. Substituting the data yields -\\frac{1}{2}=\\frac{4a^{2}-4c^{2}-8}{8}=\\frac{4b^{2}-8}{8}, solving gives: b=1"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the hyperbola $x^{2}-4 y^{2}=4$, and $P$ is a point on the hyperbola such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, then the area of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "F1: Point;F2: Point;G: Hyperbola;Expression(G) = (x^2 - 4*y^2 = 4);Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G) = True;DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 9]], [[10, 17]], [[19, 39], [49, 52]], [[19, 39]], [[2, 44]], [[45, 48]], [[45, 56]], [[60, 119]]]",
    "query_spans": "[[[121, 148]]]",
    "process": "\\because hyperbola x^{2}-4y^{2}=4, \\therefore standard equation of the hyperbola: \\frac{x^{2}}{4}-y^{2}=1, \\therefore a=2, b=1, c=\\sqrt{5}. Let PF_{1}=m, PF_{2}=n, by the definition of hyperbola we have: |m-n|=4 \\textcircled{1} \\because \\overrightarrow{PF}_{1} \\cdot \\overrightarrow{PF}_{2}=0, \\therefore PF_{1} \\bot PF_{2}, by Pythagorean theorem we have: m^{2}+n^{2}=(2\\sqrt{5})^{2} \\textcircled{2} Square \\textcircled{1}, then substitute into \\textcircled{2}, solve to get mn=2, \\therefore area of \\triangle PF_{1}F_{2} is S=\\frac{1}{2}mn=1,"
  },
  {
    "text": "If the circle $C$: $(x+\\frac{1}{2})^{2}+y^{2}=2$ intersects the parabola $E$: $y^{2}=2 p y$ ($p>0$) at points $A$ and $B$, and the chord $AB$ passes through the focus $F$ of the parabola, then $p=$?",
    "fact_expressions": "C: Circle;Expression(C) = (y^2 + (x + 1/2)^2 = 2);E: Parabola;Expression(E) = (y^2 = 2*(p*y));p: Number;p>0;A: Point;B: Point;Intersection(C, E) = {A, B};IsChordOf(LineSegmentOf(A, B), C);IsChordOf(LineSegmentOf(A, B), E);F: Point;Focus(E) = F;PointOnCurve(F, LineSegmentOf(A, B))",
    "query_expressions": "p",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 35]], [[1, 35]], [[36, 62], [83, 86]], [[36, 62]], [[94, 97]], [[44, 62]], [[65, 68]], [[69, 72]], [[1, 74]], [[1, 82]], [[36, 82]], [[89, 92]], [[83, 92]], [[77, 92]]]",
    "query_spans": "[[[94, 99]]]",
    "process": "According to the problem, the parabola $ E: y^{2} = 2px $ has focus $ F\\left(\\frac{p}{2}, 0\\right) $, $ AB \\perp x $-axis, and the $x$-coordinates of points $ A $ and $ B $ are both $ \\frac{p}{2} $. Without loss of generality, assume $ A $ is in the first quadrant. Substituting $ \\frac{p}{2} $ into the parabola $ E: y^{2} = 2px $, we solve to get $ y = \\pm p $, so $ A\\left(\\frac{p}{2}, p\\right) $, $ B\\left(\\frac{p}{2}, -p\\right) $, hence $ |AB| = 2p $. In right triangle $ \\triangle ACF $, $ AF^{2} + CF^{2} = AC^{2} $, that is, $ p^{2} + \\left(\\frac{1}{2} + \\frac{p}{2}\\right)^{2} = 2 $, and $ p > 0 $, solving gives $ p = 1 $."
  },
  {
    "text": "Given that point $P$ is a moving point on the ellipse $C$: $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$, $F_{1}$ and $F_{2}$ are the left and right foci respectively, and $O$ is the origin, then the range of $\\frac{||PF_{1}|-|PF_{2}||}{|OP|}$ is?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;O: Origin;Expression(C) = (x^2/8 + y^2/4 = 1);PointOnCurve(P, C);LeftFocus(C)=F1;RightFocus(C)=F2",
    "query_expressions": "Range(Abs(Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)))/Abs(LineSegmentOf(O, P)))",
    "answer_expressions": "[0,sqrt(2)]",
    "fact_spans": "[[[7, 49]], [[2, 6]], [[54, 61]], [[62, 69]], [[78, 81]], [[7, 49]], [[2, 53]], [[7, 77]], [[7, 77]]]",
    "query_spans": "[[[88, 129]]]",
    "process": "From the definition and properties of an ellipse, it is known that $|PF_{1}|+|PF_{2}|=4\\sqrt{2}$. Therefore, $\\frac{|PF|-|PF_{2}|}{|op|}$ expressed in coordinates, combined with functions, yields its range as $0,\\sqrt{2}$."
  },
  {
    "text": "The minimum distance from a moving point $Q$ on the parabola $y^{2}=2 p x(p>0)$ to the focus is $1$, then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;Q: Point;p>0;Expression(G) = (y^2 = 2*(p*x));PointOnCurve(Q, G);Min(Distance(Q, Focus(G))) = 1",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 21]], [[44, 47]], [[25, 28]], [[3, 21]], [[0, 21]], [[0, 28]], [[0, 42]]]",
    "query_spans": "[[[44, 49]]]",
    "process": ""
  },
  {
    "text": "Given that the directrix of the parabola $y^{2}=4 x$ passes through the focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{2}=1$, then the equation of the directrix of the ellipse is?",
    "fact_expressions": "G: Parabola;H: Ellipse;a: Number;Expression(G) = (y^2 = 4*x);Expression(H) = (y^2/2 + x^2/a^2 = 1);PointOnCurve(Focus(H), Directrix(G))",
    "query_expressions": "Expression(Directrix(H))",
    "answer_expressions": "x=pm*3",
    "fact_spans": "[[[2, 16]], [[20, 61], [66, 68]], [[22, 61]], [[2, 16]], [[20, 61]], [[2, 64]]]",
    "query_spans": "[[[66, 75]]]",
    "process": ""
  },
  {
    "text": "Given the standard equation of an ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{k^{2}}=1(k>0)$, and the focal distance is $8$, then the value of the real number $k$ is?",
    "fact_expressions": "G: Ellipse;k: Real;k > 0;Expression(G) = (x^2/36 + y^2/k^2 = 1);FocalLength(G) = 8",
    "query_expressions": "k",
    "answer_expressions": "{2*sqrt(5), 2*sqrt(13)}",
    "fact_spans": "[[[2, 4]], [[66, 71]], [[10, 55]], [[2, 55]], [[2, 64]]]",
    "query_spans": "[[[66, 75]]]",
    "process": "Since $2c=8$, we have $c=4$.  \n① When the foci are on the $x$-axis, from the standard equation of the ellipse, we know $a^{2}=36$, $b^{2}=k^{2}$, and $a^{2}=b^{2}+c^{2}$.  \nThus, $36=k^{2}+4^{2}$, so $k^{2}=20$. Since $k>0$, we have $k=2\\sqrt{5}$.  \n② When the foci are on the $y$-axis, from the standard equation of the ellipse, we know $a^{2}=k^{2}$, $b^{2}=36$, and $a^{2}=b^{2}+c^{2}$.  \nThus, $k^{2}=36+4^{2}$, so $k^{2}=52$. Since $k>0$, we have $k=2\\sqrt{13}$.  \nIn conclusion, $k=2\\sqrt{5}$ or $2\\sqrt{13}$."
  },
  {
    "text": "The vertex of the parabola is $O$, the focus is $F$, and $M$ is a moving point on the parabola. Then the maximum value of $\\frac{MO}{MF}$ is?",
    "fact_expressions": "G: Parabola;M: Point;O: Origin;F: Point;Vertex(G) = O;Focus(G)=F;PointOnCurve(M, G)",
    "query_expressions": "Max(LineSegmentOf(M, O)/LineSegmentOf(M, F))",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[0, 3], [21, 24]], [[17, 20]], [[6, 9]], [[13, 16]], [[0, 9]], [[0, 16]], [[17, 28]]]",
    "query_spans": "[[[30, 51]]]",
    "process": ""
  },
  {
    "text": "A moving circle is externally tangent to the circle $x^{2}+y^{2}+6 x+5=0$ and internally tangent to the circle $x^{2}+y^{2}-6 x-91=0$. Then, the equation of the trajectory of the center of the moving circle is?",
    "fact_expressions": "G: Circle;C:Circle;H:Circle;Expression(G) = (6*x + x^2 + y^2 + 5 = 0);Expression(C) = (-6*x + x^2 + y^2 - 91 = 0);IsOutTangent(H,G);IsInTangent(H,C)",
    "query_expressions": "LocusEquation(Center(H))",
    "answer_expressions": "x^2/36+y^2/27=1",
    "fact_spans": "[[[4, 26]], [[32, 55]], [[1, 3], [59, 61]], [[4, 26]], [[32, 55]], [[1, 28]], [[1, 57]]]",
    "query_spans": "[[[59, 70]]]",
    "process": "The circle $x^{2}+y^{2}+6x+5=0$ has center $A(-3,0)$ and radius $2$; the circle $x^{2}+y^{2}-6x-91=0$ has center $B(3,0)$ and radius $10$; let the moving circle have center $M(x,y)$ and radius $r$; then $MA=2+r$, $MB=10-r$; thus $MA+MB=12>AB=6$. Therefore, the locus of the moving circle's center $M$ is an ellipse with foci at $A(-3,0)$, $B(3,0)$, and major axis length $12$, where $a=6$, $c=3$, $b^{2}=a^{2}-c^{2}=27$. Hence, the equation of the locus of $M$ is $\\frac{x^{2}}{36}+\\frac{y^{2}}{27}=1$."
  },
  {
    "text": "Given the ellipse equation $\\frac{x^{2}}{16}+\\frac{y^{2}}{m^{2}}=1(m>0)$, the line $y=\\frac{\\sqrt{2}}{2} x$ intersects the ellipse at a point whose projection on the $x$-axis is exactly the right focus of the ellipse. Find $m=?$",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/m^2 = 1);m: Number;m>0;Z: Line;Expression(Z) = (y = (sqrt(2)/2)*x)\t;Projection(OneOf(Intersection(Z, G)), xAxis) = RightFocus(G)",
    "query_expressions": "m",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 4], [81, 83], [100, 102]], [[2, 52]], [[108, 111]], [[7, 52]], [[53, 79]], [[53, 79]], [[53, 106]]]",
    "query_spans": "[[[108, 113]]]",
    "process": "\\because the line y=\\frac{\\sqrt{2}}{2}x^{ intersects} one point M of the ellipse, and the projection of M on the x-axis is exactly the right focus of the ellipse, \\therefore M(c,\\frac{b^{2}}{a}). \\therefore \\frac{b^{2}}{a}=\\frac{\\sqrt{2}}{2}c'; simultaneously a^{2}=b^{2}+c^{2}, a^{2}=16, b^{2}=m^{2} \\therefore m^{4}+8m^{2}-128=0, solving gives m^{2}=8, m>0, \\therefore m=2\\sqrt{2}"
  },
  {
    "text": "$P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and $M$, $N$ are points on the circles $(x+5)^{2}+y^{2}=4$ and $(x-5)^{2}+y^{2}=1$, respectively. Then the maximum value of $|P M| - |P N|$ is?",
    "fact_expressions": "G: Hyperbola;H: Circle;Z: Circle;P: Point;M: Point;N: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Expression(H) = (y^2 + (x + 5)^2 = 4);PointOnCurve(P, RightPart(G));Expression(Z) = (y^2 + (x - 5)^2 = 1);PointOnCurve(M, H);PointOnCurve(N, Z)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "9",
    "fact_spans": "[[[4, 43]], [[62, 82]], [[83, 102]], [[0, 3]], [[50, 53]], [[55, 58]], [[4, 43]], [[62, 82]], [[0, 49]], [[83, 102]], [[50, 105]], [[50, 105]]]",
    "query_spans": "[[[107, 128]]]",
    "process": ""
  },
  {
    "text": "Draw a line with slope $1$ passing through the right focus of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, intersecting the ellipse at points $A$ and $B$, and let $O$ be the origin. Then the area of $\\triangle A O B$ is?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;O: Origin;B: Point;Expression(G) = (x^2/2 + y^2 = 1);PointOnCurve(RightFocus(G), H);Slope(H) = 1;Intersection(H, G) = {A, B}",
    "query_expressions": "Area(TriangleOf(A, O, B))",
    "answer_expressions": "2/3",
    "fact_spans": "[[[1, 28], [45, 47]], [[42, 44]], [[49, 52]], [[59, 62]], [[53, 56]], [[1, 28]], [[0, 44]], [[35, 44]], [[42, 58]]]",
    "query_spans": "[[[69, 91]]]",
    "process": ""
  },
  {
    "text": "Ellipse $M$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ has left and right foci $F_{1}$, $F_{2}$ respectively, $P(\\frac{a}{3},\\frac{b}{2})$. If $P F_{1} \\perp P F_{2}$, then the eccentricity of $M$ is?",
    "fact_expressions": "M: Ellipse;Expression(M) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;P: Point;Coordinate(P) = (a/3, b/2);F1: Point;F2: Point;LeftFocus(M) = F1;RightFocus(M) = F2;IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(M)",
    "answer_expressions": "sqrt(65)/15",
    "fact_spans": "[[[0, 57], [138, 141]], [[0, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[83, 111]], [[83, 111]], [[66, 73]], [[74, 81]], [[0, 81]], [[0, 81]], [[113, 136]]]",
    "query_spans": "[[[138, 147]]]",
    "process": "According to the problem, the coordinates of $ F_{1} $ and $ F_{2} $ can be obtained. Since $ PF_{1} \\perp PF_{2} $, then $ k_{PF_{1}} \\cdot k_{PF_{2}} = -1 $, which gives the relationship between $ a $ and $ c $, allowing us to find the ellipse's eccentricity.  \n$ \\because \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $, $ \\therefore F_{1}(-c,0), F_{2}(c,0) $  \n$ \\because P\\left(\\frac{a}{3}, \\frac{b}{2}\\right), PF_{1} \\perp PF_{2} $, so $ \\frac{\\frac{b}{2}}{\\frac{a}{3} + c} \\times \\frac{\\frac{b}{2}}{\\frac{a}{3} - c} = -1 $, then $ \\frac{a^{2}-c^{2}}{4} + \\frac{a^{2}}{9} - c^{2} = 0 $, i.e., $ 13a^{2} = 45c^{2} $,  \n$ \\therefore e^{2} = \\frac{c^{2}}{a^{2}} = \\frac{13}{45} $, so $ e = \\frac{\\sqrt{65}}{6} $"
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{20}=1$, and point $P$ lies on the hyperbola. If the distance from point $P$ to focus $F_{1}$ is equal to $9$, then the distance from point $P$ to focus $F_{2}$ is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/20 = 1);F1: Point;F2: Point;Focus(G) = {F1,F2};P: Point;PointOnCurve(P, G) = True;Distance(P, F1) = 9",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "17",
    "fact_spans": "[[[16, 56], [65, 68]], [[16, 56]], [[0, 7], [78, 85]], [[8, 15], [101, 108]], [[0, 59]], [[60, 64], [71, 75], [95, 99]], [[60, 69]], [[71, 93]]]",
    "query_spans": "[[[95, 114]]]",
    "process": ""
  },
  {
    "text": "Let $P$ be any point on the ellipse $M$: $\\frac{x^{2}}{2}+y^{2}=1$, and let $EF$ be any diameter of the circle $N$: $x^{2}+(y-2)^{2}=1$. Then the maximum value of $\\overrightarrow{P E} \\cdot \\overrightarrow{P F}$ is?",
    "fact_expressions": "M: Ellipse;N: Circle;E: Point;F: Point;P: Point;Expression(M) = (x^2/2 + y^2 = 1);Expression(N) = (x^2 + (y - 2)^2 = 1);PointOnCurve(P, M);IsDiameter(LineSegmentOf(E, F), N)",
    "query_expressions": "Max(DotProduct(VectorOf(P, E), VectorOf(P, F)))",
    "answer_expressions": "8",
    "fact_spans": "[[[5, 36]], [[48, 73]], [[42, 47]], [[42, 47]], [[1, 4]], [[5, 36]], [[48, 73]], [[1, 41]], [[42, 79]]]",
    "query_spans": "[[[81, 136]]]",
    "process": ""
  },
  {
    "text": "The moving circle $M$ passes through the point $(0,-1)$ and is tangent to the line $y=1$. Then, the equation of the trajectory of the center $M$ is?",
    "fact_expressions": "M: Circle;H: Point;Coordinate(H) = (0, -1);G: Line;Expression(G) = (y = 1);M1: Point;Center(M) = M1;PointOnCurve(H,M);IsTangent(M,G)",
    "query_expressions": "LocusEquation(M1)",
    "answer_expressions": "x^2 = -4*y",
    "fact_spans": "[[[2, 5]], [[6, 15]], [[6, 15]], [[17, 24]], [[17, 24]], [[30, 33]], [[0, 33]], [0, 14], [0, 25]]",
    "query_spans": "[[[30, 40]]]",
    "process": "Let the center of the moving circle be M(x, y). Using the given conditions to set up an equation, we can solve it. Let the center of the moving circle be M(x, y). Since the moving circle M passes through the point (0, -1) and is tangent to the line y = 1, we obtain \\sqrt{x^{2}+(y+1)^{2}}=|1-y|. Rearranging gives x^{2}+4y=0, so the trajectory of the center M of the moving circle is x^{2}=-4y."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, a line $l$ passing through point $F_{1}$ intersects the left branch of hyperbola $E$ at points $P$ and $Q$. If $|P F_{1}|=2|F_{1} Q|$ and $F_{2} Q \\perp P Q$, then the eccentricity of $E$ is?",
    "fact_expressions": "l: Line;E: Hyperbola;b: Number;a: Number;F2: Point;Q: Point;P: Point;F1: Point;a>0;b>0;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(E) = F1;RightFocus(E) = F2;PointOnCurve(F1,l);Intersection(l,LeftPart(E))={P, Q};Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(F1, Q));IsPerpendicular(LineSegmentOf(F2, Q), LineSegmentOf(P, Q))",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(17)/3",
    "fact_spans": "[[[96, 101]], [[18, 79], [102, 108], [169, 172]], [[26, 79]], [[26, 79]], [[10, 17]], [[117, 120]], [[113, 116]], [[2, 9], [87, 95]], [[26, 79]], [[26, 79]], [[18, 79]], [[2, 85]], [[2, 85]], [[86, 101]], [[96, 122]], [[124, 146]], [[148, 167]]]",
    "query_spans": "[[[169, 178]]]",
    "process": "As shown in the figure, according to the given conditions, let $ QF_{1} = m $, $ PF_{1} = 2m $. By the definition of the hyperbola, we obtain: $ PF_{2} = 2a + 2m $, $ QF_{2} = 2a + m $. In $ \\triangle PQF_{2} $, by the Pythagorean theorem: $ (3m)^{2} + (2a + m)^{2} = (2a + 2m)^{2} $, from which we get: $ m = \\frac{2}{3}a $. In $ \\triangle QF_{1}F_{2} $, $ QF_{1} = m = \\frac{2}{3}a $, $ QF_{2} = 2a + m = \\frac{8}{3}a $. By the Pythagorean theorem: $ \\left(\\frac{2}{3}a\\right)^{2} + \\left(\\frac{8}{3}a\\right)^{2} = (2c)^{2} $, from which we get: $ e^{2} = \\frac{c^{2}}{a^{2}} = \\frac{17}{9} $, $ \\therefore e = \\frac{\\sqrt{17}}{3} $."
  },
  {
    "text": "The length of the imaginary axis of the hyperbola $x^{2}+\\frac{y^{2}}{m}=1$ is twice the length of the real axis, then $m$=?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2 + y^2/m = 1);Length(ImageinaryAxis(G)) = 2*Length(RealAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "-4",
    "fact_spans": "[[[0, 28]], [[43, 46]], [[0, 28]], [[0, 41]]]",
    "query_spans": "[[[43, 48]]]",
    "process": "Since the length of the imaginary axis of the hyperbola $x^{2}+\\frac{y^{2}}{m}=1$ is twice the length of the real axis, we have $a^{2}=1$, $b^{2}=-m$, so $2\\sqrt{-m}=2\\times2$, solving gives $m=-c$"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+y^{2}=1  (a>0)$ with foci $F_{1}$, $F_{2}$, and the parabola $y^{2}=2 x$ with focus $F$, if $\\overrightarrow{F_{1} F}=3 \\overrightarrow{F F_{2}}$, then $a$=?",
    "fact_expressions": "G: Parabola;H: Ellipse;a: Number;F1: Point;F: Point;F2: Point;Expression(G) = (y^2 = 2*x);a>0;Expression(H) = (y^2 + x^2/a^2 = 1);Focus(H) = {F1, F2};Focus(G) = F;VectorOf(F1, F) = 3*VectorOf(F, F2)",
    "query_expressions": "a",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[59, 73]], [[2, 40]], [[137, 140]], [[43, 50]], [[77, 80]], [[51, 58]], [[59, 73]], [[4, 40]], [[2, 40]], [[2, 58]], [[59, 80]], [[82, 135]]]",
    "query_spans": "[[[137, 142]]]",
    "process": "From the standard equation of the parabola, we obtain its focus coordinate as $ F\\left(\\frac{1}{2},0\\right) $. Let the foci of the ellipse be: $ F_{1}(-c,0), F_{2}(c,0) $. Then: $ \\overrightarrow{F_{1}F} = \\left(\\frac{1}{2}+c, 0\\right), \\overrightarrow{FF_{2}} = \\left(c-\\frac{1}{2}, 0\\right) $. From the given condition: $ \\left(\\frac{1}{2}+c, 0\\right) = 3\\left(c-\\frac{1}{2}, 0\\right) $, it follows that: $ \\frac{1}{2}+c = 3\\left(c-\\frac{1}{2}\\right) $. Solving this equation for $ c $ yields: $ c=1 $, then: $ a = \\sqrt{b^{2}+c^{2}} = \\sqrt{2} $"
  },
  {
    "text": "Given that $F(2 , 0)$ is the right focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and the length of the chord passing through $F$ and perpendicular to the $x$-axis is $6$. If $A(-2, \\sqrt{2})$, and point $M$ is any point on the ellipse, then the maximum value of $|M F|+| M A |$ is?",
    "fact_expressions": "G: Ellipse;a: Number;b: Number;F: Point;A: Point;M: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (2, 0);Coordinate(A) = (-2, sqrt(2));RightFocus(G) = F;l: LineSegment;IsChordOf(l, G);IsPerpendicular(l, xAxis);PointOnCurve(F, l);Length(l) = 6;PointOnCurve(M, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(M, F)) + Abs(LineSegmentOf(M, A)))",
    "answer_expressions": "8 + sqrt(2)",
    "fact_spans": "[[[13, 65], [116, 118]], [[15, 65]], [[15, 65]], [[2, 12], [71, 74]], [[93, 110]], [[111, 115]], [[15, 65]], [[15, 65]], [[13, 65]], [[2, 12]], [[93, 110]], [[0, 69]], [], [[13, 84]], [[13, 84]], [[13, 91]], [[13, 91]], [[111, 122]]]",
    "query_spans": "[[[124, 145]]]",
    "process": "Obtain the ellipse equation. According to the definition of an ellipse, |MF| + |MA| = 8 + |MA| \\cdot |MF_{1}|. When M lies on the extension line of AF_{1}, |MA| \\cdot |MF_{1}| = |F_{1}A|, and thus the answer can be obtained. Given c = 2, \\frac{2b^{2}}{a} = \\frac{2(a^{2} - c^{2})}{a} = 6, simplifying yields: a^{2} \\cdot 3a \\cdot 4 = 0, solving gives: a = 4, b^{2} = 12, \\therefore the equation of the ellipse: \\frac{x^{2}}{16} + \\frac{y^{2}}{12} = 1. The two foci of ellipse C are respectively F(2,0), F_{1}(-2,0), |MF| + |MF_{1}| = 2a = 8, |MF| + |MA| = 8 + |MA| \\cdot |MF_{1}|. When points M, A, F_{1} are not collinear, |MA| \\cdot |MF| < |F_{1}A|. When M lies on the extension line of AF_{1}, |MA| \\cdot |MF_{1}| = |F_{1}A|, |F_{1}A| = \\sqrt{2}. Therefore, the maximum value of |MF| + |MA| is 8 + \\sqrt{2}."
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 43]]]",
    "process": "The ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ has a semi-major axis $a=3$ and a semi-minor axis $b=2$, then the semi-focal distance is $c=\\sqrt{9-4}=\\sqrt{5}$. Thus, the eccentricity of the ellipse is: $e=\\frac{c}{a}=\\frac{\\sqrt{5}}{3}$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. If a point $P$ on the ellipse satisfies $|P F_{1}|=2|P F_{2}|$, then the measure of $\\angle F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1);F1: Point;F2: Point;P: Point;PointOnCurve(P,G) = True;LeftFocus(G) = F1;RightFocus(G) = F2;Abs((LineSegmentOf(P,F1))) = Abs((LineSegmentOf(P,F2)))*2",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "pi/2",
    "fact_spans": "[[[2, 39], [65, 67]], [[2, 39]], [[48, 55]], [[56, 63]], [[69, 73]], [[65, 73]], [[2, 63]], [[2, 63]], [[75, 97]]]",
    "query_spans": "[[[99, 126]]]",
    "process": "According to the problem, for the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, we have $a=3$, $b=2$, $c=\\sqrt{a^{2}-b^{2}}=\\sqrt{5}$, $|F_{1}F_{2}|=2c=2\\sqrt{5}$. Thus,  \n$$\n\\begin{cases}\n|PF_{1}|=2|PF_{2}| \\\\\n|PF_{1}|+|PF_{2}|=2a=6\n\\end{cases}\n$$\nSolving gives $|PF_{1}|=4$, $|PF_{2}|=2$, so $|PF_{1}|^{2}+|PF_{2}|^{2}=4^{2}+2^{2}=20=|F_{1}F_{2}|^{2}$. Therefore, triangle $F_{1}PF_{2}$ is a right triangle, and $\\angle F_{1}PF_{2}=\\frac{\\pi}{2}$."
  },
  {
    "text": "A moving circle passes through a fixed point $A(-4,0)$ and is externally tangent to a fixed circle $B$: $(x-4)^{2}+y^{2}=16$. What is the equation of the locus of the center of the moving circle?",
    "fact_expressions": "B: Circle;A: Point;C: Circle;Expression(B) = (y^2 + (x - 4)^2 = 16);Coordinate(A) = (-4, 0);PointOnCurve(A, C);IsOutTangent(C, B)",
    "query_expressions": "LocusEquation(Center(C))",
    "answer_expressions": "(x^2/4 - y^2/12 = 1)&(x <= -2)",
    "fact_spans": "[[[20, 44]], [[6, 15]], [[1, 3], [49, 51]], [[20, 44]], [[6, 15]], [[0, 15]], [[0, 47]]]",
    "query_spans": "[[[49, 60]]]",
    "process": "Let the center of the moving circle be point P, then |PB| = |PA| + 4, that is, |PB| - |PA| = 4 < |AB| = 8. Therefore, the trajectory of point P is the left branch of a hyperbola with foci at A(-4,0) and B(4,0), where 2a = 4, a = 2. Also, since 2c = 8, c = 4. Therefore, b^{2} = c^{2} - a^{2} = 12. Thus, the equation of the trajectory of the center of the moving circle is \\frac{x^{2}}{4} - \\frac{y^{2}}{12} = 1 (x \\leqslant -2)."
  },
  {
    "text": "A straight line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $P$ and $Q$. If the length of segment $PF$ is $3$, then what is the length of segment $FQ$?",
    "fact_expressions": "G: Parabola;H: Line;F: Point;P: Point;Q: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(F,H);Focus(G) = F;Intersection(H, G) = {P, Q};Length(LineSegmentOf(P,F)) = 3",
    "query_expressions": "Length(LineSegmentOf(F,Q))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[1, 15], [26, 29]], [[23, 25]], [[18, 21]], [[30, 33]], [[34, 37]], [[1, 15]], [[0, 25]], [[1, 21]], [[23, 39]], [[41, 54]]]",
    "query_spans": "[[[56, 67]]]",
    "process": "Let P(x_{1},y_{1}). Since the length of segment PF is 3, \\therefore x_{1}+\\frac{p}{2}=3, thus x_{1}=2, \\therefore P(2,2\\sqrt{2}). Also F(1,0), \\therefore the equation of line PQ is: y=2\\sqrt{2}(x-1). Substituting into the parabola equation gives (2\\sqrt{2}(x-1))^{2}=4x, i.e., \\frac{2x^{2}-5x+2=0}{(\\frac{1}{2}-1)^{2}+(-\\sqrt{2}-0)^{2}}=\\frac{3}{2}, solving yields x=2 or x=\\frac{1}{2}."
  },
  {
    "text": "Let a point $P$ on the parabola $y^{2}=6x$ be at a distance of $\\frac{9}{2}$ from its focus $F$, and let $O$ be the coordinate origin. Then the area of $\\triangle POF$ is?",
    "fact_expressions": "G: Parabola;P: Point;O: Origin;F: Point;Expression(G) = (y^2 = 6*x);PointOnCurve(P, G);Focus(G) = F;Distance(P, F) = 9/2",
    "query_expressions": "Area(TriangleOf(P, O, F))",
    "answer_expressions": "9*sqrt(2)/4",
    "fact_spans": "[[[1, 15], [22, 23]], [[18, 21]], [[48, 51]], [[25, 28]], [[1, 15]], [[1, 21]], [[22, 28]], [[18, 45]]]",
    "query_spans": "[[[58, 80]]]",
    "process": "Let P(x_{0},y_{0}). From the standard equation of the parabola, we know that the directrix is x=-\\frac{3}{2}, and the focus is F(\\frac{3}{2},0). By the definition of the parabola: x_{0}-(-\\frac{3}{2})=\\frac{9}{2} \\Rightarrow x_{0}=3. Substituting into the standard equation of the parabola gives y_{0}^{2}=6\\times3 \\Rightarrow |y_{0}|=3\\sqrt{2}. Therefore, the area of \\triangle POF is: \\frac{1}{2}\\times\\frac{3}{2}\\times3\\sqrt{2}=\\frac{9\\sqrt{2}}{4}"
  },
  {
    "text": "Given that $A$ and $B$ are the two endpoints of the major axis of the ellipse $C$: $\\frac{x^{2}}{m+4}+\\frac{y^{2}}{m}=1$, and $P$ is a moving point on the ellipse $C$, and the maximum value of $\\angle A P B$ is $\\frac{2 \\pi}{3}$, then the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;m: Number;A: Point;P: Point;B: Point;Expression(C) = (x^2/(m + 4) + y^2/m = 1);Endpoint(MajorAxis(C))={A,B};PointOnCurve(P, C);Max(AngleOf(A, P, B)) = (2*pi)/3",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[10, 54], [67, 72], [116, 121]], [[17, 54]], [[2, 5]], [[63, 66]], [[6, 9]], [[10, 54]], [[2, 62]], [[63, 76]], [[78, 114]]]",
    "query_spans": "[[[116, 127]]]",
    "process": "P is a moving point on the ellipse C, and the maximum value of ∠APB is \\frac{2\\pi}{3}; at this time, P is the endpoint of the minor axis. Using the tangent function property of a right triangle, we obtain the solution. Since P is a moving point on the ellipse C, when the maximum value of ∠APB is \\frac{2\\pi}{3}, by the property of the ellipse, P is the endpoint of the minor axis and ∠APO = 60^{\\circ}. Therefore, \\tan60^{\\circ} = \\frac{\\sqrt{m+4}}{\\sqrt{m}}, solving gives m = 2. \\therefore a^{2} = 6, b^{2} = 2, c^{2} = 4. \\therefore e = \\frac{c}{a} = \\frac{2}{\\sqrt{6}} = \\frac{\\sqrt{6}}{3}"
  },
  {
    "text": "If the equation $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1, a \\in[1,5], b \\in[2,4]$ represents an ellipse with foci on the $x$-axis and eccentricity less than $\\frac{\\sqrt{3}}{2}$, then the maximum value of $z=a+b$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/a^2 + y^2/b^2 = 1);a: Number;In(a, [1, 5]);In(b, [2, 4]);b: Number;PointOnCurve(Focus(G), xAxis);Eccentricity(G) < sqrt(3)/2;z: Number;z = a + b",
    "query_expressions": "Max(z)",
    "answer_expressions": "9",
    "fact_spans": "[[[107, 109]], [[1, 109]], [[3, 70]], [[3, 70]], [[3, 70]], [[3, 70]], [[72, 109]], [[81, 109]], [[111, 118]], [[111, 118]]]",
    "query_spans": "[[[111, 124]]]",
    "process": "Find the inequality relations satisfied by a and b in the problem, then represent the plane region corresponding to the point (a,b) in the rectangular coordinate system, draw the line x+y=0, and use the idea of linear programming to obtain its minimum value. Solution: From the problem, a > b, \\frac{\\sqrt{a2-b^{2}}}{a} < \\frac{\\sqrt{3}}{2}, simplifying yields a < 2b, and 1 \\leqslant a \\leqslant 5, 2 \\leqslant b \\leqslant 4. Therefore, the region represented by the point (a,b) in the coordinate plane is the shaded part as shown in the figure. Draw the line l: x+y=0, where z = x+y represents the y-intercept of the line. By translating this line, when it passes through point C(5,4), z attains its maximum value of z = 5+4 = 9."
  },
  {
    "text": "It is known that one asymptote of a hyperbola has the equation $y=\\sqrt{3} x$, and one of its foci lies on the directrix of the parabola $y^{2}=24 x$. Then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(H) = (y^2 = 24*x);Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x);PointOnCurve(OneOf(Focus(G)), Directrix(H))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/27 = 1",
    "fact_spans": "[[[2, 5], [29, 30], [57, 60]], [[36, 51]], [[36, 51]], [[2, 28]], [[29, 55]]]",
    "query_spans": "[[[57, 67]]]",
    "process": "From the given conditions, the directrix of the parabola $ y^{2} = 24x $ is $ x = -6 $, and one focus of the hyperbola is $ (-6, 0) $, so $ c = 6 $. Also, $ \\frac{b}{a} = \\sqrt{3} $, $ 36 = a^{2} + b^{2} = 4a^{2} $, $ a^{2} = 9 $, $ b^{2} = 27 $. The equation of the required hyperbola is $ \\frac{x^{2}}{9} - \\frac{y^{2}}{27} = $"
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{10}+y^{2}=1$, $F_{1}$ and $F_{2}$ are the two foci of the ellipse, and when $\\angle F_{1} P F_{2}=\\frac{2 \\pi}{3}$, then the area of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/10 + y^2 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2};AngleOf(F1,P,F2)=2*pi/3",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[6, 34], [55, 57]], [[2, 5]], [[39, 46]], [[47, 54]], [[6, 34]], [[2, 38]], [[39, 62]], [[64, 102]]]",
    "query_spans": "[[[105, 132]]]",
    "process": "Let |PF_{1}| = m, |PF_{2}| = n, then m + n = 2a = 2\\sqrt{10}. In triangle PF_{1}F_{2}, by the cosine law: F_{1}F_{2}^{2} = m^{2} + n^{2} - 2mn\\cos\\angle F_{1}PF_{2}, that is: 36 = (m+n)^{2} - 2mn - 2mn\\cos\\frac{2\\pi}{3} = 40 - mn. Solving gives: mn = 4. Therefore, S_{\\Delta PF_{1}F_{2}} = \\frac{1}{2}mn\\sin\\frac{2\\pi}{3} = \\sqrt{3}. The correct answer: \\sqrt{3}"
  },
  {
    "text": "A focus of the hyperbola $2 x^{2}-y^{2}=m$ is $(0 , \\sqrt{3})$, then the value of $m$ is?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (2*x^2 - y^2 = m);Coordinate(OneOf(Focus(G)))=(0,sqrt(3))",
    "query_expressions": "m",
    "answer_expressions": "-2",
    "fact_spans": "[[[0, 20]], [[44, 47]], [[0, 20]], [0, 41]]",
    "query_spans": "[[[44, 51]]]",
    "process": ""
  },
  {
    "text": "It is known that the center of the hyperbola is at the origin, and the coordinates of the two foci $F_{1}$ and $F_{2}$ are $(\\sqrt{5}, 0)$ and $(-\\sqrt{5}, 0)$ respectively. Point $P$ lies on the hyperbola such that $P F_{1} \\perp P F_{2}$, and the area of $\\triangle P F_{1} F_{2}$ is $1$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;O: Origin;Coordinate(F1) = (sqrt(5), 0);Coordinate(F2) = (-sqrt(5), 0);Focus(G) = {F1, F2};Center(G) = O;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));Area(TriangleOf(P, F1, F2)) = 1",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2 = 1",
    "fact_spans": "[[[2, 5], [77, 80], [142, 145]], [[72, 76]], [[16, 23]], [[24, 31]], [[9, 11]], [[16, 71]], [[16, 71]], [[2, 31]], [[2, 11]], [[72, 81]], [[83, 106]], [[108, 140]]]",
    "query_spans": "[[[142, 150]]]",
    "process": "From the given conditions, we have\n\\[\n\\begin{cases}\n|PF_{1}|\\cdot|PF_{2}|=2, \\\\\n|PF_{1}|+|PF_{2}|=2\\sqrt{5}\n\\end{cases}\n\\Rightarrow (|PF_{1}|-|PF_{2}|)^{2}=16,\n\\]\nthus $ 2a=4 $, solving gives $ a=2 $. Also $ c=\\sqrt{5} $, so $ b=1 $. Therefore, the equation of the hyperbola is $ \\frac{x^{2}}{4}-y^{2}=1 $."
  },
  {
    "text": "Given the ellipse $x^{2}+\\frac{y^{2}}{b^{2}}=1$ $(0<b<1)$, the left focus is $F$, the left and right vertices are $A$ and $C$ respectively, and the upper vertex is $B$. A circle $P$ passes through points $F$, $B$, and $C$, with center $P$ at coordinates $(m, n)$. When $m+n>0$, what is the range of values for the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + y^2/b^2 = 1);b: Number;0 < b;b < 1;F: Point;LeftFocus(G) = F;A: Point;C: Point;LeftVertex(G) = A;RightVertex(G) = C;B: Point;UpperVertex(G) = B;PointOnCurve(F, P1) = True;PointOnCurve(B, P1) = True;PointOnCurve(C, P1) = True;P1: Circle;Center(P1) = P;P: Point;Coordinate(P) = (m, n);m: Number;n: Number;m + n > 0",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(0, \\sqrt{2}/2)",
    "fact_spans": "[[[2, 40], [126, 128]], [[2, 40]], [[4, 40]], [[4, 40]], [[4, 40]], [[45, 48], [77, 80]], [[2, 48]], [[58, 61]], [[62, 65], [85, 88]], [[2, 65]], [[2, 65]], [[71, 74], [81, 84]], [[2, 74]], [[76, 93]], [[76, 93]], [[76, 93]], [[89, 93]], [[89, 102]], [[99, 102]], [[99, 114]], [[106, 114]], [[106, 114]], [[117, 124]]]",
    "query_spans": "[[[126, 138]]]",
    "process": "Find the perpendicular bisector of BC and the perpendicular bisector of FC; their intersection is the center of the circle P. Then solve using $ m+n>0 $. According to the problem, $ F(-c,0) $, $ B(0,b) $, $ C(1,0) $. Let M be the midpoint of BC, $ M\\left(\\frac{1}{2},\\frac{b}{2}\\right) $, and $ k_{BC} = -b $, then the perpendicular bisector of BC is $ y - \\frac{b}{2} = \\frac{1}{b}\\left(x - \\frac{1}{2}\\right) $, and the perpendicular bisector of FC is $ x = \\frac{1-c}{2} $. Solving the system  \n$$\n\\begin{cases}\ny - \\frac{b}{2} = \\frac{1}{b}\\left(x - \\frac{1}{2}\\right) \\\\\nx = \\frac{1-c}{2}\n\\end{cases}\n$$\ngives  \n$$\n\\begin{cases}\nx = \\frac{1-c}{2} \\\\\ny = \\frac{b^{2}-c}{2b}\n\\end{cases}\n$$\nSince $ m+n > 0 $, that is, $ \\frac{1-c}{2} + \\frac{b^{2}-c}{2b} > 0 $,  \nso $ b - bc + b^{2} - c > 0 $, i.e., $ (1+b)(b-c) > 0 $, thus $ b > c $, i.e., $ b^{2} > c^{2} $, so $ a^{2} = b^{2} + c^{2} > 2c^{2} $, hence $ e^{2} = \\frac{c^{2}}{a^{2}} < \\frac{1}{2} $, then $ 0 < e < \\frac{\\sqrt{2}}{2} $."
  },
  {
    "text": "It is known that the asymptotes of a hyperbola are given by $y = \\pm x$, and the right focus coincides with the focus of the parabola $y^{2} = 4x$. Then, the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;Expression(H) = (y^2 = 4*x);Expression(Asymptote(G)) = (y = pm*x);RightFocus(G) = Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "2*x^2-2*y^2=1",
    "fact_spans": "[[[2, 5], [50, 53]], [[27, 41]], [[27, 41]], [[2, 21]], [[2, 46]]]",
    "query_spans": "[[[50, 58]]]",
    "process": "\\because the focus of the parabola y^{2}=4x is: (1,0) \\therefore the right focus of the hyperbola is: (1,0), i.e., c=1 \\because the asymptotes of the hyperbola are given by y=\\pm x. \\therefore the equation of the hyperbola can be written as: x^{2}-y^{2}=\\lambda, \\lambda>0, i.e., \\frac{x^{2}}{\\lambda}-\\frac{y^{2}}{\\lambda}=1, \\therefore a^{2}=b^{2}=\\lambda. From a^{2}+b^{2}=c^{2} we obtain: \\lambda+\\lambda=1, \\therefore \\lambda=\\frac{1}{2}. The equation of the hyperbola is 2x^{2}-2y^{2}=1."
  },
  {
    "text": "Points $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. Points $A$, $B$, $C$ lie on the ellipse $E$, satisfying $\\overrightarrow{A F_{1}}=\\overrightarrow{F_{1} B}$, $\\overrightarrow{A F_{2}}=2 \\overrightarrow{F_{2} C}$. Then the eccentricity of the ellipse $E$ is?",
    "fact_expressions": "E: Ellipse;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(E) = F1;RightFocus(E) = F2;A: Point;B: Point;C: Point;PointOnCurve(A, E);PointOnCurve(B, E);PointOnCurve(C, E);VectorOf(A, F1) = VectorOf(F1, B);VectorOf(A, F2) = 2*VectorOf(F2, C)",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(5)/5",
    "fact_spans": "[[[19, 78], [98, 103], [214, 219]], [[19, 78]], [[26, 78]], [[26, 78]], [[26, 78]], [[26, 78]], [[0, 8]], [[9, 16]], [[0, 84]], [[0, 84]], [[85, 89]], [[90, 93]], [[94, 97]], [[85, 104]], [[90, 104]], [[94, 104]], [[107, 158]], [[159, 212]]]",
    "query_spans": "[[[214, 225]]]",
    "process": "As shown in the figure, since $\\overrightarrow{AF}=\\overrightarrow{F_{1}B}$, by the symmetry of the ellipse, we have $AB\\bot F_{1}F_{2}$. Without loss of generality, assume point $A$ is above the $x$-axis, then $x_{A}=-c$, hence $y_{A}=\\frac{b^{2}}{a}$. Since $\\overrightarrow{AF_{2}}=2\\overrightarrow{F_{2}C}$, it follows that $x_{C}-c=\\frac{1}{2}\\times 2c$, so $x_{C}=2c$, and $y_{C}=-\\frac{b^{2}}{2a}$. Substituting the coordinates of point $C$ into the ellipse equation gives: $\\frac{4c^{2}}{a^{2}}+\\frac{b^{2}}{4a^{2}}=1$. Simplifying yields: $4e^{2}+\\frac{1-e^{2}}{4}=1$, solving gives $e=\\frac{\\sqrt{5}}{5}$."
  },
  {
    "text": "Given the parabola $x^{2}=4 y$, point $M(t,-2), t \\in(-1,1)$, two tangents $MA$, $MB$ are drawn from $M$ to the parabola, where $A$, $B$ are the points of tangency, and line $AB$ intersects the $y$-axis at point $P$. Then the range of $\\frac{|PA|}{|PB|}$ is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;M: Point;P: Point;Expression(G) = (x^2 = 4*y);Coordinate(M) = (t, -2);t:Number;In(t, (-1, 1));TangentOfPoint(M,G)={LineOf(M,A),LineOf(M,B)};TangentPoint(LineOf(M,A),G)=A;TangentPoint(LineOf(M,B),G)=B;Intersection(LineOf(A,B), yAxis) = P",
    "query_expressions": "Range(Abs(LineSegmentOf(P, A))/Abs(LineSegmentOf(P, B)))",
    "answer_expressions": "[1/2, 2]",
    "fact_spans": "[[[2, 16], [46, 49]], [[69, 72]], [[73, 76]], [[17, 40], [42, 45]], [[94, 98]], [[2, 16]], [[17, 40]], [[18, 40]], [[18, 40]], [[41, 66]], [[41, 79]], [[41, 79]], [[80, 98]]]",
    "query_spans": "[[[101, 129]]]",
    "process": "Let the tangent points be $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. From the parabola $ y = \\frac{1}{4}x^{2} $, $ y' = \\frac{1}{2}x $. Therefore, the tangent line $ MA: x_{1}x = 2y_{1} + 2y $, similarly, the tangent line $ MB: x_{2}x = 2y_{2} + 2y $. Since point $ M $ is the intersection of the two tangents, $ x_{1}t = 2y_{1} - 4 $, $ x_{2}t = 2y_{2} - 4 $. Thus, the equation of line $ AB $ is $ tx = -4 + 2y $, i.e., $ y - 2 = -\\frac{tx}{2} $, this line always passes through $ P(0,2) $, then $ \\frac{|PA|}{|PB|} = \\frac{\\sqrt{x_{1}^{2} + (y_{1}-2)^{2}}}{\\sqrt{x_{2}^{2} + (y_{2}-2)^{2}}} = \\frac{\\sqrt{x_{1}^{2} + (-\\frac{tx_{1}}{2})^{2}}}{\\sqrt{x_{2}^{2} + (-\\frac{tx_{2}}{2})^{2}}} = -\\frac{x_{1}}{x_{2}} $. Solving $ \\begin{cases} y = \\frac{tx}{2} + 2 \\\\ x^{2} = 4y \\end{cases} $, eliminating $ y $, we get $ x^{2} - 2tx - 8 = 0 $, so $ x_{1} + x_{2} = 2t $, $ x_{1}x_{2} = -8 $. Since $ \\frac{(x_{1}+x_{2})^{2}}{x_{1}x_{2}} = \\frac{x_{1}}{x_{2}} + \\frac{x_{2}}{x_{1}} + 2 = -\\frac{t^{2}}{2} $. Because $ t \\in (-1,1) $, $ -\\frac{t^{2}}{2} \\in [-\\frac{1}{2}, 0] $, i.e., $ -\\frac{1}{2} \\leqslant \\frac{x_{1}}{x_{2}} + \\frac{x_{2}}{x_{1}} + 2 \\leqslant 0 $, i.e., $ \\begin{cases} -\\frac{1}{2} \\leqslant m + \\frac{1}{m} + 2 \\\\ m + \\frac{1}{m} + 2 < 0 \\end{cases} $, solving gives $ -2 \\leqslant m \\leqslant -\\frac{1}{2} $, $ |m + \\frac{1}{m} + 2| \\leqslant 0 $, i.e., $ \\frac{x_{1}}{x_{2}} \\in [-2, -\\frac{1}{2}] $, $ -\\frac{x_{1}}{x_{2}} \\in [\\frac{1}{2}, 2] $."
  },
  {
    "text": "The minor axis has length $\\sqrt{5}$, and the eccentricity is $e = \\frac{2}{3}$. The foci of the ellipse are $F_{1}$ and $F_{2}$. A line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. Find the perimeter of $\\triangle ABF_{2}$.",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;F1: Point;F2:Point;e:Number;Length(MinorAxis(G))=sqrt(5);Eccentricity(G)=e;e=2/3;Focus(G)={F1,F2};PointOnCurve(F1, H);Intersection(H, G) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, B,F2))",
    "answer_expressions": "12",
    "fact_spans": "[[[34, 36], [71, 73]], [[68, 70]], [[74, 77]], [[78, 81]], [[41, 48], [60, 67]], [[51, 58]], [[18, 33]], [[0, 36]], [[15, 36]], [[18, 33]], [[34, 58]], [[59, 70]], [[68, 83]]]",
    "query_spans": "[[[85, 110]]]",
    "process": ""
  },
  {
    "text": "The line $ l $ with slope $ 1 $ intersects the ellipse $ \\frac{x^{2}}{4} + y^{2} = 1 $ at points $ A $ and $ B $. Then the maximum value of $ AB $ is?",
    "fact_expressions": "Slope(l) = 1;l: Line;G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);Intersection(l, G) = {A, B};A: Point;B: Point",
    "query_expressions": "Max(LineSegmentOf(A, B))",
    "answer_expressions": "4*sqrt(10)/5",
    "fact_spans": "[[[0, 12]], [[7, 12]], [[13, 40]], [[13, 40]], [[7, 52]], [[43, 46]], [[47, 50]]]",
    "query_spans": "[[[54, 65]]]",
    "process": ""
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$, intersects the directrix at point $P$, and intersects the $y$-axis at point $Q$. If $\\overrightarrow{P Q}=\\overrightarrow{F B}$, then the chord length $|A B|$=?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Q: Point;F: Point;B: Point;A: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Intersection(H,Directrix(G))=P;Intersection(H,yAxis)=Q;VectorOf(P, Q) = VectorOf(F, B);IsChordOf(LineSegmentOf(A,B),G)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "9/2",
    "fact_spans": "[[[1, 15], [24, 27]], [[21, 23]], [[41, 45]], [[52, 56]], [[17, 20]], [[33, 36]], [[28, 32]], [[1, 15]], [[1, 20]], [[0, 23]], [[21, 36]], [[21, 45]], [[21, 56]], [[58, 101]], [[24, 112]]]",
    "query_spans": "[[[105, 114]]]",
    "process": "Let the projections of points B and F on the directrix be points G and K, respectively. It is calculated that \\frac{|KF|}{|GB|}=\\frac{|PF|}{|PB|}=\\frac{2}{3}, yielding the coordinates of B as (2,2\\sqrt{2}), and the x-coordinate of A as \\frac{1}{2}, leading to the answer. [Solution] Let the projections of points B and F on the directrix be points G and K, respectively. According to the definition of a parabola, the origin O is the midpoint of segment KF, so Q is the midpoint of segment PF, and |PQ|=|QF|. Given \\overrightarrow{PQ}=\\overrightarrow{FB}, it follows that \\frac{|PF|}{|PB|}=\\frac{2}{3}, thus \\frac{|KF|}{|GB|}=\\frac{|PF|}{|PB|}=\\frac{2}{3}. Since |KF|=2, then |GB|=3, hence the coordinates of point B are (2,2\\sqrt{2}) (point B must lie in the first quadrant). Therefore, the equation of line AB is y=2\\sqrt{2}(x-1). Substituting into y^{2}=4x, the x-coordinate of point A is found to be \\frac{1}{2}. Thus, |AF|=\\frac{1}{2}+1=\\frac{3}{2}, and |AB|=|AF|+|BF|=\\frac{3}{2}+3=\\frac{9}{2}."
  },
  {
    "text": "Given the parabola $C$: $x^{2}=8 y$, $F$ is the focus, and point $A(-2,4)$. If point $P$ lies on the parabola and the value of $|P F|+|P A|$ is minimized, then the coordinates of point $P$ are?",
    "fact_expressions": "C: Parabola;A: Point;P: Point;F: Point;Expression(C) = (x^2 = 8*y);Coordinate(A) = (-2, 4);Focus(C) = F;PointOnCurve(P, C);WhenMin(Abs(LineSegmentOf(P, F)) + Abs(LineSegmentOf(P, A)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(-2, 1/2)",
    "fact_spans": "[[[2, 21], [48, 51]], [[31, 41]], [[43, 47], [73, 77]], [[23, 27]], [[2, 21]], [[31, 41]], [[2, 30]], [[43, 52]], [[54, 71]]]",
    "query_spans": "[[[73, 82]]]",
    "process": "When point P is at the intersection of the line passing through A and perpendicular to the directrix with the parabola C, the value of |PF| + |PA| is minimized, and at this time x_{p} = -2. Substituting into the parabola equation gives y_{p} = \\frac{1}{2}. Draw a perpendicular PN from point P to the directrix y = -2 of the parabola, then |PF| = |PN|. Therefore, |PF| + |PA| = |PA| + |PN|. Hence, when points P, A, N are collinear, the value of |PN| + |PA| is minimized. Clearly, the x-coordinate of point P is x_{P} = -2. Substituting into the parabola equation yields y_{n} = \\frac{1}{2}."
  },
  {
    "text": "Given that the line $AB$ passing through the right focus $F_2$ of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ intersects the ellipse at points $A$ and $B$, and $F_1$ is the left focus of the ellipse, then the perimeter of $\\triangle A F_{1} B$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);A: Point;B: Point;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2,LineOf(A,B));Intersection(LineOf(A,B),G) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, F1, B))",
    "answer_expressions": "20",
    "fact_spans": "[[[4, 43], [63, 65], [84, 86]], [[4, 43]], [[66, 69]], [[70, 73]], [[76, 83]], [[47, 54]], [[76, 90]], [[4, 54]], [[2, 62]], [[55, 75]]]",
    "query_spans": "[[[92, 118]]]",
    "process": "Using the definition of an ellipse gives the answer. As shown in the figure: $|AF_{1}|+|AF_{2}|=2a=10$, $|BF_{1}|+|BF_{2}|=2a=10$, so the perimeter of $\\triangle AF_{1}B$ is $4a=20$."
  },
  {
    "text": "The length of the minor axis of the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{m}=1$ is $8$, then the real number $m$=?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (x^2/36 + y^2/m = 1);Length(MinorAxis(G))=8",
    "query_expressions": "m",
    "answer_expressions": "16",
    "fact_spans": "[[[0, 38]], [[48, 53]], [[0, 38]], [[0, 46]]]",
    "query_spans": "[[[48, 55]]]",
    "process": "From the given information, since the length of the minor axis of the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{m}=1$ is 8, the foci cannot lie on the y-axis; thus, for the ellipse, $2b=8$, $b=4$, $m=b^{2}=16$. Therefore, fill in 16. This is a basic problem examining fundamental quantities of an ellipse; note that when the focus coordinates are not provided, one should consciously consider and discuss the possibilities."
  },
  {
    "text": "In triangle $A B C$, $B C=4$, and $A B=\\sqrt{3} A C$. What is the maximum area of triangle $A B C$?",
    "fact_expressions": "A: Point;B: Point;C: Point;LineSegmentOf(B, C) = 4;LineSegmentOf(A, B) = sqrt(3)*LineSegmentOf(A, C)",
    "query_expressions": "Max(Area(TriangleOf(A, B, C)))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[3, 10]], [[3, 10]], [[3, 10]], [[12, 19]], [[21, 39]]]",
    "query_spans": "[[[41, 58]]]",
    "process": "Set up a rectangular coordinate system as shown in the figure, then: B(-2,0), C(2,0). Let the coordinates of point A be A(x,y). According to the given conditions, we can obtain: (x-4)^{2}+y^{2}=12. Combining the properties of triangles, the trajectory equation of point A is a circle centered at (4,0) with radius 2\\sqrt{3}, excluding its intersection points with the x-axis. Based on this, the maximum area of triangle ABC is S=\\frac{1}{2}\\times4\\times2\\sqrt{3}=4\\sqrt{3}."
  },
  {
    "text": "Given that the center of a hyperbola is at the origin, one focus is at $(\\sqrt{10}, 0)$, and the eccentricity is $\\sqrt{5}$, then the equations of the asymptotes of this hyperbola are?",
    "fact_expressions": "G: Hyperbola;O:Origin;Center(G)=O;Coordinate(OneOf(Focus(G))) = (sqrt(10), 0);Eccentricity(G)=sqrt(5)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*2*x",
    "fact_spans": "[[[2, 5], [57, 60]], [[8, 12]], [[2, 12]], [[2, 38]], [[2, 54]]]",
    "query_spans": "[[[57, 68]]]",
    "process": ""
  },
  {
    "text": "Given that a line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$, and $|AF|=2$, then $|BF|=$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 4*x);Focus(G)=F;PointOnCurve(F,H);Intersection(H,G) = {B, A};Abs(LineSegmentOf(A, F)) = 2",
    "query_expressions": "Abs(LineSegmentOf(B, F))",
    "answer_expressions": "2",
    "fact_spans": "[[[3, 17], [28, 31]], [[24, 26]], [[32, 35]], [[20, 23]], [[38, 41]], [[3, 17]], [[3, 23]], [[2, 26]], [[24, 43]], [[44, 52]]]",
    "query_spans": "[[[55, 63]]]",
    "process": ""
  },
  {
    "text": "If the sum of the distances from two points $A$ and $B$ on the parabola $y^{2}=4x$ to the focus is $6$, then what is the distance from the midpoint of segment $AB$ to the $y$-axis?",
    "fact_expressions": "G: Parabola;B: Point;A: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(A, G);PointOnCurve(B,G);Distance(A,Focus(G))+Distance(B,Focus(G))=6",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A,B)),yAxis)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 15]], [[23, 26]], [[19, 22]], [[1, 15]], [[1, 26]], [[1, 26]], [[1, 38]]]",
    "query_spans": "[[[40, 60]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), the midpoint M(x_{0},y_{0}), and the focus be F; then |AF| + |BF| = x_{1} + 1 + x_{2} + 1 = 2x_{0} + 2 = 6, solving gives x_{0} = 2, that is, the distance from the midpoint of segment AB to the y-axis is 2."
  },
  {
    "text": "Given that the equation of the directrix of a certain parabola is $y=1$, then the standard equation of this parabola is?",
    "fact_expressions": "G: Parabola;Expression(Directrix(G)) = (y = 1)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 = -4*y",
    "fact_spans": "[[[3, 6], [20, 23]], [[3, 17]]]",
    "query_spans": "[[[20, 30]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{20}=1$ ($a>0$) is $y=2x$, then the focal distance of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/20 + x^2/a^2 = 1);a: Number;a>0;Expression(OneOf(Asymptote(G))) = (y = 2*x)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "10",
    "fact_spans": "[[[2, 50], [69, 72]], [[2, 50]], [[5, 50]], [[5, 50]], [[2, 66]]]",
    "query_spans": "[[[69, 77]]]",
    "process": "From the hyperbola equation given in the problem, we know its asymptotes are $ y = \\pm\\frac{\\sqrt{20}}{a}x $, then $ \\frac{\\sqrt{20}}{a} = 2 \\Rightarrow a^{2} = 5 $, hence $ c = \\sqrt{5 + 20} = 5 $, so the focal distance is $ 2c = 2 \\times 5 = 10 $. The answer to be filled is 10."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x$ ($p>0$) with focus $F$ and directrix equation $x=-1$. Let point $M$ lie on this parabola such that $|M F|=3$. If $O$ is the origin, then the area of $\\triangle O F M$ is?",
    "fact_expressions": "G: Parabola;p: Number;O: Origin;F: Point;M: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;Expression(Directrix(G)) = (x = -1);PointOnCurve(M, G);Abs(LineSegmentOf(M, F)) = 3",
    "query_expressions": "Area(TriangleOf(O, F, M))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 23], [51, 54]], [[5, 23]], [[68, 71]], [[27, 30]], [[45, 49]], [[5, 23]], [[2, 23]], [[2, 30]], [[2, 42]], [[45, 55]], [[57, 66]]]",
    "query_spans": "[[[78, 100]]]",
    "process": "The equation of the directrix of the parabola is x = -1, then the focus is F(1,0), p = 2, the equation of the parabola is y^{2} = 4x. |MF| = x_{M} + 1 = 3, x_{M} = 2, so y_{M}^{2} = 4x_{M} = 8, |y_{M}| = 2\\sqrt{2}. S_{\\DeltaOMF} = \\frac{1}{2}|OF||y_{M}| = \\frac{1}{2} \\times 1 \\times 2\\sqrt{2} = \\sqrt{2}."
  },
  {
    "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 34]]]",
    "process": "The hyperbola $\\frac{x^2}{3} \\cdot y^{2}=1$ has $a=\\sqrt{3}$, $b=1$,"
  },
  {
    "text": "Given points $A(-2,0)$, $B(3,0)$, a moving point $P(x, y)$ satisfies $\\overrightarrow{A P} \\cdot \\overrightarrow{B P}=x^{2}$, then the trajectory equation of point $P$ is?",
    "fact_expressions": "A: Point;P: Point;B: Point;Coordinate(A) = (-2,0);Coordinate(B)=(3,0);Coordinate(P) = (x1, y1);x1:Number;y1:Number;DotProduct(VectorOf(A, P), VectorOf(B, P)) = x1^2",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "y^2=x+6",
    "fact_spans": "[[[2, 12]], [[25, 34], [95, 98]], [[13, 22]], [[2, 12]], [[14, 22]], [[25, 34]], [[25, 34]], [[25, 34]], [[36, 91]]]",
    "query_spans": "[[[95, 105]]]",
    "process": "According to $\\overrightarrow{AP}\\cdot\\overrightarrow{BP}=x^2$, set up and simplify the equation. Since $\\overrightarrow{AP}\\cdot\\overrightarrow{BP}=x^2$, then $(x+2,y)\\cdot(x-3,y)=x^{2}\\Rightarrow x^{2}-x-6+y^{2}=x^{2}$, which simplifies to $y^2=x+6$."
  },
  {
    "text": "What is the length of the chord passing through the focus of the parabola $y^{2}=4 x$ and perpendicular to the axis of symmetry?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);L: LineSegment;IsChordOf(L, G) = True;PointOnCurve(Focus(G), L) = True;IsPerpendicular(L, SymmetryAxis(G)) = True",
    "query_expressions": "Length(L)",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 15]], [[1, 15]], [], [[1, 27]], [[0, 27]], [[0, 27]]]",
    "query_spans": "[[[1, 30]]]",
    "process": "The focus of the parabola $ y^{2} = 4x $ is $ (1, 0) $. We obtain: $ y^{2} = 4 $, solving gives $ y = \\pm 2 $. We obtain: the length of the chord perpendicular to the axis of symmetry is:"
  },
  {
    "text": "The hyperbola $C$ passes through the point $A(8,3)$ and shares the same asymptotes with the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$. Then the equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;G:Hyperbola;A: Point;Coordinate(A) = (8, 3);Expression(G)=(x^2/16-y^2/9=1);PointOnCurve(A, C);Asymptote(G)=Asymptote(C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/48 - y^2/27 = 1",
    "fact_spans": "[[[0, 6], [67, 73]], [[19, 58]], [[7, 16]], [[7, 16]], [[19, 58]], [[0, 16]], [[0, 65]]]",
    "query_spans": "[[[67, 78]]]",
    "process": "Since hyperbola $ C $ has the same asymptotes as the hyperbola $ \\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1 $, let the equation of hyperbola $ C $ be $ \\frac{x^{2}}{16}-\\frac{y^{2}}{9}=\\lambda $ ($ \\lambda \\neq 0 $). Since hyperbola $ C $ passes through point $ A(8,3) $, we have $ \\frac{8^{2}}{16}-\\frac{3^{2}}{9}=\\lambda $. Solving gives $ \\lambda=3 $, so $ \\frac{x^{2}}{16}-\\frac{y^{2}}{9}=3 $. Therefore, the equation of hyperbola $ C $ is $ \\frac{x^{2}}{48}-\\frac{y^{2}}{27}=1 $."
  },
  {
    "text": "The equations of the two asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*sqrt(3)*x",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 39]]]",
    "process": "From the given conditions, we have $ a^{2}=1 $, $ b^{2}=3 $, then the asymptotes of the hyperbola are $ y=\\pm\\frac{b}{a}x=\\pm\\sqrt{3}x $."
  },
  {
    "text": "The parabola $C$: $y^{2}=4 x$ has focus $F$ and directrix $l$. A line passing through $F$ intersects $l$ at point $A$ and intersects $C$ at a point $B$. If $F$ is the midpoint of segment $A B$, then $|A B|$=?",
    "fact_expressions": "C: Parabola;G: Line;B: Point;A: Point;F: Point;l: Line;Expression(C) = (y^2 = 4*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(F,G);Intersection(l, G) = A;OneOf(Intersection(G,C))=B;MidPoint(LineSegmentOf(A,B)) = F",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[0, 19], [57, 60]], [[42, 44]], [[66, 69]], [[51, 55]], [[23, 26], [38, 41], [71, 74]], [[30, 33], [45, 48]], [[0, 19]], [[0, 26]], [[0, 33]], [[36, 44]], [[42, 55]], [[42, 69]], [[71, 85]]]",
    "query_spans": "[[[87, 96]]]",
    "process": "According to the definition of a parabola, the graph is drawn as shown below, where |BD| is the distance from point B to the directrix. Since F is the midpoint of AB and EF // BD, it follows that |BD| = 2|FE| = 4. By the definition of a parabola, |BF| = |BD| = 4, hence |AB| = 2|BF| = 8."
  },
  {
    "text": "If a point $P$ on the parabola $x^{2}=2 y$ is at a distance of $2$ from its focus $F$, and $O$ is the origin, then the area of $\\triangle O F P$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 2*y);P: Point;PointOnCurve(P, G);F: Point;Focus(G) = F;Distance(P, F) = 2;O: Origin",
    "query_expressions": "Area(TriangleOf(O, F, P))",
    "answer_expressions": "9/32",
    "fact_spans": "[[[1, 15], [22, 23]], [[1, 15]], [[18, 21]], [[1, 21]], [[25, 28]], [[22, 28]], [[18, 35]], [[36, 39]]]",
    "query_spans": "[[[46, 68]]]",
    "process": "By the definition of a parabola, |PF| = x_{P} + \\frac{1}{2} = 2, so x_{P} = \\frac{3}{2}, |y_{P}| = \\frac{9}{8}. Therefore, the area S of APFO is S = \\frac{1}{2}|OF| \\cdot |y_{P}| = \\frac{1}{2} \\times \\frac{1}{2} \\times \\frac{9}{8} = \\frac{9}{32}."
  },
  {
    "text": "The intersection point of the lines $2x - y + 4 = 0$ and $x - y + 5 = 0$ is $P$. The line $l$ passes through point $P$ and is parallel to the line $l_{1}$: $x + y - 6 = 0$. Find the equation of line $l$?",
    "fact_expressions": "l: Line;G: Line;l1: Line;H: Line;P: Point;Expression(G) = (2*x - y + 4 = 0);Expression(H) = (x - y + 5 = 0);Intersection(G, H) = P;PointOnCurve(P, l);IsParallel(l, l1);Expression(l1) = (x + y - 6 = 0)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x + y - 7 = 0",
    "fact_spans": "[[[31, 36], [67, 72]], [[0, 13]], [[43, 63]], [[14, 23]], [[27, 30], [37, 41]], [[0, 13]], [[14, 23]], [[0, 30]], [[31, 41]], [[31, 65]], [[43, 63]]]",
    "query_spans": "[[[67, 76]]]",
    "process": "(1) Let the equation of line $ l $ be: $ x + y + m = 0 $. Solving the system:  \n\\[\n\\begin{cases}\n2x - y + 4 = 0 \\\\\nx - y + 5 = 0\n\\end{cases}\n\\]  \nwe get $ P(1,6) $, $ \\therefore m = -1 - 6 = -7 $, $ \\therefore $ the equation of line $ l $ is: $ x + y - 7 = 0 $.  \n(2) From the problem, it is clear that: $ c = 2 $. Let the hyperbola equation be $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $,  \n\\[\n\\begin{cases}\na^{2} + b^{2} = 4 \\\\\n\\frac{9}{a^{2}} - \\frac{2}{b^{2}} = 1\n\\end{cases}\n\\]  \nSolving this system yields: $ a^{2} = 3 $, $ b^{2} = 1 $. The standard equation of the hyperbola is: $ \\frac{x^{2}}{3} - y^{2} = 1 $, $ e = \\frac{c}{a} = \\frac{2\\sqrt{3}}{3} $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has eccentricity $\\sqrt{2}$, then the distance from the point $(0,2)$ to the asymptotes of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;a: Number;b: Number;G: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(G) = (0, 2);Eccentricity(C) = sqrt(2)",
    "query_expressions": "Distance(G, Asymptote(C))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 63], [89, 95]], [[10, 63]], [[10, 63]], [[80, 88]], [[10, 63]], [[10, 63]], [[2, 63]], [[80, 88]], [[2, 78]]]",
    "query_spans": "[[[80, 104]]]",
    "process": "Since the eccentricity of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) is $ \\sqrt{2} $, we have $ \\frac{c}{a} = \\sqrt{2} $, that is, $ \\frac{a^{2} + b^{2}}{a^{2}} = 2 $, so $ a = b $. Therefore, the asymptotes of the hyperbola are $ y = \\pm x $, and the distance from the point $ (0, 2) $ to the asymptotes of the hyperbola $ C $ is $ \\frac{2}{\\sqrt{1^{2} + (-1)^{2}}} = \\sqrt{2} $."
  },
  {
    "text": "If the directrix of the parabola $y^{2}=ax$ is the line $x=-1$, then its focus coordinates are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = a*x);a: Number;Expression(Directrix(G)) = (x = -1);H: Line",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(1, 0)",
    "fact_spans": "[[[2, 15], [30, 31]], [[2, 15]], [[5, 15]], [[2, 27]], [[19, 27]]]",
    "query_spans": "[[[30, 38]]]",
    "process": "From the properties of a parabola, it is known that the x-coordinate of the directrix equation and the x-coordinate of the focus are opposite numbers, so the focus is (1,0)."
  },
  {
    "text": "If $F_{1}$ and $F_{2}$ are the two foci of an ellipse, a line passing through $F_{2}$ intersects the ellipse at points $P$ and $Q$, $P F_{1} \\perp P Q$, and $4|P F_{1}|=3|P Q|$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Focus(G) = {F1, F2};H: Line;PointOnCurve(F2, H) = True;Intersection(H, G) = {P, Q};P: Point;Q: Point;IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, Q)) = True;4*Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, Q))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[17, 19], [92, 94], [37, 39]], [[1, 8]], [[9, 16], [26, 33]], [[1, 24]], [[34, 36]], [[25, 36]], [[34, 49]], [[40, 43]], [[44, 47]], [[50, 69]], [[71, 90]]]",
    "query_spans": "[[[92, 100]]]",
    "process": "Let |PQ| = x, then |PF₁| = \\frac{3}{4}x, |QF₁| = 4a - \\frac{7}{4}x. Since PF₁ ⊥ PQ, it follows that |PF₁|² + |PQ|² = |QF₁|², so \\frac{9}{16}x² + x² = (4a - \\frac{7}{4}x)². Solving gives x = \\frac{4}{3}a or x = 8a (discarded). Thus, |PF₁| = a, |PF₂| = 2a - a = a, so |F₁F₂| = \\sqrt{2}a = 2c. Therefore, the eccentricity of the ellipse is \\frac{\\sqrt{2}}{2}."
  },
  {
    "text": "A line with slope $\\sqrt{3}$ passes through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) and intersects the parabola at points $A$ and $B$ (point $A$ is in the first quadrant). If $|A F|=2$, then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;A: Point;B: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Slope(H) = sqrt(3);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Quadrant(A) = 1;Abs(LineSegmentOf(A, F)) = 2",
    "query_expressions": "p",
    "answer_expressions": "1",
    "fact_spans": "[[[17, 38], [47, 50]], [[85, 88]], [[14, 16]], [[52, 55], [62, 66]], [[56, 59]], [[41, 44]], [[20, 38]], [[17, 38]], [[0, 16]], [[17, 44]], [[14, 44]], [[14, 61]], [[62, 71]], [[74, 83]]]",
    "query_spans": "[[[85, 90]]]",
    "process": "From the given condition, the directrix of the parabola is $ l: x = -\\frac{p}{2} $. As shown in the figure, draw $ AM \\perp l $ from point $ A $, with foot at $ M $, and draw $ FC \\perp AM $ from focus $ F $, with foot at $ C $. Then $ |AM| = |AF| $. Since the slope of line $ AB $ is $ \\sqrt{3} $, we have $ \\angle FAC = \\frac{\\pi}{3} $. Given $ |AF| = 2 $, it follows that $ |CA| = 1 $. Therefore, $ p = |OF| = |AM| - |CA| = 2 - 1 = 1 $."
  },
  {
    "text": "A point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ has a distance of $3$ to one of its foci. What is the distance from point $P$ to the other focus?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);P: Point;PointOnCurve(P, G);OneOf(Focus(G)) = F1;OneOf(Focus(G)) = F2;Negation(F1 = F2);F1: Point;F2: Point;Distance(P, F1) = 3",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "7",
    "fact_spans": "[[[0, 38], [45, 46]], [[0, 38]], [[41, 44], [62, 66]], [[0, 44]], [[45, 51]], [[45, 72]], [[45, 72]], [], [], [[41, 59]]]",
    "query_spans": "[[[45, 78]]]",
    "process": ""
  },
  {
    "text": "$P$ is a point in the fourth quadrant on $3 x^{2}-5 y^{2}=15$, $F_{1}$, $F_{2}$ are its two foci, and $S\\Delta F_{1} P F_{2}=2 \\sqrt{6}$. Find the coordinates of point $P$.",
    "fact_expressions": "F1: Point;F2: Point;H: Curve;Expression(H) = (3*x^2 - 5*y^2 = 15);P: Point;PointOnCurve(P, H);Quadrant(P) = 4;Focus(H) = {F1, F2};Area(TriangleOf(F1, P, F2)) = 2*sqrt(6)",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(\\sqrt{10}, -\\sqrt{3})",
    "fact_spans": "[[[33, 40]], [[42, 49]], [[4, 24], [50, 51]], [[4, 24]], [[0, 3], [92, 96]], [[0, 32]], [[0, 32]], [[33, 54]], [[56, 90]]]",
    "query_spans": "[[[92, 100]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$, a line passing through the focus $F$ with an inclination angle of $60^{\\circ}$ intersects $C$ at points $P$ and $Q$. The projections of $P$ and $Q$ onto the directrix of $C$ are $M$ and $N$, respectively. Then $|MN|=$?",
    "fact_expressions": "C: Parabola;G: Line;M: Point;N: Point;P: Point;Q: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F, G);Inclination(G) = ApplyUnit(60, degree);Intersection(G, C) = {P, Q};Projection(P, Directrix(C)) = M;Projection(Q, Directrix(C)) = N",
    "query_expressions": "Abs(LineSegmentOf(M, N))",
    "answer_expressions": "8*sqrt(3)/3",
    "fact_spans": "[[[2, 21], [49, 52], [72, 75]], [[46, 48]], [[85, 88]], [[89, 92]], [[54, 57], [64, 67]], [[58, 61], [68, 71]], [[25, 28]], [[2, 21]], [[2, 28]], [[22, 48]], [[29, 48]], [[46, 63]], [[64, 94]], [[64, 94]]]",
    "query_spans": "[[[96, 105]]]",
    "process": "From the parabola $ C: y^{2} = 4x $, it follows that the focus has coordinates $ F(1,0) $. Therefore, the equation of the line passing through the focus $ F $ with slope $ \\sqrt{3} $ is $ y = \\sqrt{3}(x - 1) $, which simplifies to $ x = \\frac{\\sqrt{3}}{3}y + 1 $. Let $ P(x_{1}, y_{1}) $, $ Q(x_{2}, y_{2}) $, then $ |MN| = |y_{1} - y_{2}| $. From \n\\[\n\\begin{cases}\nx = \\frac{\\sqrt{3}}{3}y + 1 \\\\\ny^{2} = 4x\n\\end{cases}\n\\]\nwe obtain $ y^{2} - \\frac{4\\sqrt{3}}{3}y - 4 = 0 $. Thus, $ y_{1} + y_{2} = \\frac{4\\sqrt{3}}{3} $, $ y_{1} + y_{2} = \\frac{4\\sqrt{3}}{3} $, $ y_{1} \\cdot y_{2} = -4 $. $ v = -4bqg $ for. $ 8\\sqrt{3} $"
  },
  {
    "text": "Real numbers $x$, $y$ satisfy $x|x| + y|y| = 1$. Then the range of the distance from the point $(x, y)$ to the line $x + y + 1 = 0$ is?",
    "fact_expressions": "G: Line;H: Point;x_: Real;y_:Real;Expression(G) = (x + y + 1 = 0);Coordinate(H) = (x_, y_);x_*Abs(x_)+y_*Abs(y_) = 1",
    "query_expressions": "Range(Distance(H, G))",
    "answer_expressions": "(\\sqrt{2}/2,1+\\sqrt{2}/2]",
    "fact_spans": "[[[36, 47]], [[26, 35]], [[0, 5]], [[6, 9]], [[36, 47]], [[26, 35]], [[11, 24]]]",
    "query_spans": "[[[26, 57]]]",
    "process": "Discussing piecewise to remove absolute values and determine the graph represented by $ x|x| + y|y| = 1 $, it can be concluded that the graph of $ x|x| + y|y| = 1 $ lies between $ y = -x $ and $ x + y - \\sqrt{2} = 0 $. Using the distance formula between parallel lines, we can find: since real numbers $ x, y $ satisfy $ x|x| + y|y| = 1 $, when $ x \\geqslant 0, y \\geqslant 0 $, the equation becomes $ x^2 + y^2 = 1 $, representing an arc; when $ x \\geqslant 0, y < 0 $, the equation becomes $ x^2 - y^2 = 1 $, representing part of a hyperbola; when $ x < 0, y \\geqslant 0 $, the equation becomes $ y^2 - x^2 = 1 $, representing part of a hyperbola; when $ x < 0, y < 0 $, the equation becomes $ x^2 + y^2 = -1 $, which does not represent any graph. Drawing the graph of $ x|x| + y|y| = 1 $, it is seen that one asymptote of the hyperbola is $ y = -x $, and suppose the line parallel to $ x + y + 1 = 0 $ and tangent to the circle $ x^2 + y^2 = 1 $ in the first quadrant is $ x + y + a = 0 $, then $ \\frac{|a|}{\\sqrt{2}} = 1 $, solving gives $ a = -\\sqrt{2} $, thus obtaining that the graph of $ x|x| + y|y| = 1 $ lies between $ y = -x $ and $ x + y - \\sqrt{2} = 0 $. The distance between $ y = -x $ and $ x + y + 1 = 0 $ is $ \\frac{1}{\\sqrt{2}} = \\frac{\\sqrt{2}}{2} $, and the distance between $ x + y - \\sqrt{2} = 0 $ and $ x + y + 1 = 0 $ is $ \\frac{|-\\sqrt{2} - 1|}{\\sqrt{2}} = 1 + \\frac{\\sqrt{2}}{2} $. Then, combining with the graph, the range of distances from point $ (x, y) $ to the line $ x + y + 1 = 0 $ is $ \\left( \\frac{\\sqrt{2}}{2}, 1 + \\frac{\\sqrt{2}}{2} \\right] $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. The line $l$: $7 x-24 y+28=0$ passing through point $F_{1}$ intersects the hyperbola $C$ at points $A$ and $B$. If $A F_{2} \\perp x$-axis, then the standard equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;l: Line;Expression(l) = (7*x - 24*y + 28 = 0);PointOnCurve(F1, l);A: Point;B: Point;Intersection(l, C) = {A, B};IsPerpendicular(LineSegmentOf(A, F2), xAxis)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/9 - y^2/7 = 1",
    "fact_spans": "[[[2, 63], [121, 127], [160, 166]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[72, 79], [89, 97]], [[80, 87]], [[2, 87]], [[2, 87]], [[98, 120]], [[98, 120]], [[88, 120]], [[129, 132]], [[133, 136]], [[98, 138]], [[140, 158]]]",
    "query_spans": "[[[160, 173]]]",
    "process": "Since the line $ l: 7x - 24y + 28 = 0 $ passes through $ F_{1} $, we have $ F_{1}(-4, 0) $, so $ c = 4 $. Because $ AF_{2} \\perp x $-axis, $ A(4, \\frac{b^{2}}{a}) $. Substituting into the line equation gives $ \\frac{b^{2}}{a} = \\frac{7}{3} $. Also, since $ a^{2} + b^{2} = c^{2} = 16 $, solving yields $ a^{2} = 9 $, $ b^{2} = 7 $. Therefore, the standard equation of hyperbola $ C $ is $ \\frac{x^{2}}{9} - \\frac{y^{2}}{7} = 1 $."
  },
  {
    "text": "Draw a line through the focus of the parabola $y^{2}=4 x$ intersecting the parabola at points $A$ and $B$. If the x-coordinate of the midpoint $M$ of segment $A B$ is $2$, then $|A B|$ equals?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;M:Point;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), H);Intersection(H, G) = {A, B};MidPoint(LineSegmentOf(A,B))=M;XCoordinate(M)=2",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 15], [24, 27]], [[21, 23]], [[28, 31]], [[32, 35]], [[49, 52]], [[1, 15]], [[0, 23]], [[21, 37]], [[39, 52]], [[49, 60]]]",
    "query_spans": "[[[62, 72]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since the directrix of the parabola is x = -1 and the focus is F(1,0), according to the definition of the parabola, |AF| = x_{1}+1, |BF| = x_{2}+1. Therefore, |AB| = x_{1}+1 + x_{2}+1 = 2x_{M}+2 = 2\\times2+2=6."
  },
  {
    "text": "The standard equation of an ellipse with foci on the $x$-axis, focal length equal to $4$, and passing through point $P(6,0)$ is?",
    "fact_expressions": "G: Ellipse;P: Point;Coordinate(P) = (6, 0);PointOnCurve(Focus(G), xAxis);FocalLength(G) = 4;PointOnCurve(P, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/36+y^2/32=1",
    "fact_spans": "[[[30, 32]], [[20, 29]], [[20, 29]], [[0, 32]], [[9, 32]], [[18, 32]]]",
    "query_spans": "[[[30, 38]]]",
    "process": "It is easy to know from the problem that $ c = 2 $, $ a = 6 $, then according to $ b^{2} = a^{2} - c^{2} $ we can find the value of $ b^{2} $, and finally write the equation of the ellipse. Since the focal length of the ellipse is 4, $ 2c = 4 $, so $ c = 2 $. Also, since the ellipse passes through the point $ P(6,0) $, $ a = 6 $, so $ b^{2} = a^{2} - c^{2} = 36 - 4 = 32 $. Therefore, the equation of the ellipse is $ \\frac{x^{2}}{36} + \\frac{y^{2}}{32} = 1 $."
  },
  {
    "text": "If the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{a^{2}}=1$ and the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{2}=1$ have the same foci, then the value of the real number $a$ is?",
    "fact_expressions": "G: Ellipse;Z: Ellipse;a: Real;Expression(G) = (x^2/4 + y^2/a^2 = 1);Expression(Z) = (y^2/2 + x^2/a^2 = 1);Focus(G) = Focus(Z)",
    "query_expressions": "a",
    "answer_expressions": "pm*sqrt(3)",
    "fact_spans": "[[[1, 42]], [[43, 84]], [[92, 97]], [[1, 42]], [[43, 84]], [[1, 90]]]",
    "query_spans": "[[[92, 101]]]",
    "process": "\\because the ellipse \\frac{x^{2}}{4}+\\frac{y^{2}}{a^{2}}=1 and the ellipse \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{2}=1 have the same foci; when the foci are on the x-axis, 4-a^{2}=a^{2}-2>0, \\therefore a=\\pm\\sqrt{3}; when the foci are on the y-axis, a^{2}-4=2-a^{2}>0, no solution. In conclusion, the real value of a is \\pm\\sqrt{3}."
  },
  {
    "text": "What is the equation of the parabola with the center of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ as its vertex and the right focus as its focus?",
    "fact_expressions": "G: Parabola;H: Ellipse;Expression(H) = (x^2/9 + y^2/5 = 1);Center(H)=Vertex(G);RightFocus(H)=Focus(G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=8*x",
    "fact_spans": "[[[52, 55]], [[1, 38]], [[1, 38]], [[0, 55]], [[0, 55]]]",
    "query_spans": "[[[52, 59]]]",
    "process": ""
  },
  {
    "text": "The distance from the point with ordinate $2$ on the parabola $x^{2}=2 p y(p>0)$ to the focus is $5$, then the equation of this parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 2*p*y);p: Number;p>0;D: Point;PointOnCurve(D, G) = True;YCoordinate(D) = 2;Distance(D, Focus(G)) = 5",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2=12*y",
    "fact_spans": "[[[0, 21], [43, 46]], [[0, 21]], [[3, 21]], [[3, 21]], [[30, 31]], [[0, 31]], [[22, 31]], [[0, 40]]]",
    "query_spans": "[[[43, 51]]]",
    "process": "By analyzing the definition of a parabola, it can be deduced that the distance from the point with ordinate 2 to the directrix is also 5, i.e., $2 - (-\\frac{p}{2}) = 5$. Solving for $p$ yields the value of $p$, and thus the answer. [Detailed solution] According to the problem, the directrix of the parabola $x^{2} = 2py$ is $y = -\\frac{p}{2}$. Since the distance from the point with ordinate 2 to the focus is 5, the distance from this point to the directrix is also 5. Therefore, $2 - (-\\frac{p}{2}) = 5$, solving gives $p = 6$. Hence, the equation of the parabola is $x^{2} = 12y$."
  },
  {
    "text": "A line $ l $ with slope $ k $ ($ k<0 $) passes through the point $ F(0,1) $, and intersects the curve $ y=\\frac{1}{4} x^{2} $ ($ x \\geqslant 0 $) and the line $ y=-1 $ at points $ A $ and $ B $, respectively. If $ |F B|=6|F A| $, then $ k=? $",
    "fact_expressions": "l: Line;G: Line;H: Curve;F: Point;B: Point;A: Point;k<0;Expression(G) = (y = -1);Expression(H) = And((y = x^2/4),x>=0);PointOnCurve(F,l);Coordinate(F) = (0, 1);Intersection(l, H) = A;Abs(LineSegmentOf(F, B)) = 6*Abs(LineSegmentOf(F, A));Slope(l)=k;k:Number;Intersection(l,G)=B",
    "query_expressions": "k",
    "answer_expressions": "-sqrt(6)/12",
    "fact_spans": "[[[12, 17]], [[69, 77]], [[30, 68]], [[18, 27]], [[85, 88]], [[81, 84]], [[3, 11]], [[69, 77]], [[30, 68]], [[12, 27]], [[18, 27]], [[12, 90]], [[92, 106]], [[0, 17]], [[3, 11], [108, 111]], [[12, 90]]]",
    "query_spans": "[[[108, 113]]]",
    "process": "Let $ A(x_{1},\\frac{x_{1}^{2}}{4}) $, $ B(x_{2},-1) $. From $ \\overrightarrow{FB}=6\\overrightarrow{FA} $, we obtain a system of equations in $ x_{1},x_{2} $. Solving it allows us to find the slope of the line. Let $ A(x_{1},\\frac{x_{1}^{2}}{4}) $, $ B(x_{2},-1) $. Since $ k<0 $ and $ |FB|=6|FA| $, we have $ \\overrightarrow{FB}=6\\overrightarrow{FA} $. And $ \\overrightarrow{FB}=(x_{2},-2) $, $ \\overrightarrow{FA}=(x_{1},\\frac{x_{1}^{2}}{4}-1) $, so  \n$$\n\\begin{cases}\nx_{2}=6x_{1} \\\\\n-2=6(\\frac{x^{2}}{4}-1) \\\\\nx_{1}>0\n\\end{cases}\n$$\nSolving gives  \n$$\n\\begin{cases}\nx_{1}=\\frac{2\\sqrt{6}}{3} \\\\\nx_{2}=4\\sqrt{6}\n\\end{cases}\n$$\nThus $ B(4\\sqrt{6},-1) $. Hence the slope $ k=\\frac{-1-1}{4\\sqrt{6}-0}=-\\frac{\\sqrt{6}}{12} $."
  },
  {
    "text": "Given a point $P$ on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ such that the distance from $P$ to the left focus $F_{1}$ of the ellipse is $3$, and point $M$ is the midpoint of $P F_{1}$, then what is the distance from point $M$ to the origin $O$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;M: Point;Expression(G) = (x^2/16 + y^2/9 = 1);PointOnCurve(P,G);LeftFocus(G)=F1;Distance(P,F1)=3;MidPoint(LineSegmentOf(P, F1)) = M;O:Origin",
    "query_expressions": "Distance(M, O)",
    "answer_expressions": "5/2",
    "fact_spans": "[[[2, 40], [47, 49]], [[43, 46]], [[53, 60]], [[68, 72], [87, 91]], [[2, 40]], [[2, 46]], [[47, 60]], [[43, 67]], [[68, 85]], [[92, 99]]]",
    "query_spans": "[[[87, 104]]]",
    "process": "\\because ellipse \\frac{x^2}{16}+\\frac{y^{2}}{9}=1, a=4, \\therefore |PF_{1}|+|PF_{2}|=2a=8, combining with |PF_{1}|=3, we get |PF_{2}|=2a-|PF_{1}|=8-3=5, \\because OM is the median line of \\triangle PF_{1}F_{2}, |OM|=\\frac{1}{2}|PF_{2}|=\\frac{1}{2}\\times5=\\frac{5}{2}"
  },
  {
    "text": "Let $O$ be the coordinate origin, $F$ be the focus of the parabola $y^{2}=4 x$, and $A$ be an arbitrary point on the parabola. If $\\overrightarrow{O A} \\cdot \\overrightarrow{A F}=-4$, then what are the coordinates of point $A$?",
    "fact_expressions": "G: Parabola;O: Origin;A: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(A, G);DotProduct(VectorOf(O, A), VectorOf(A, F)) = -4",
    "query_expressions": "Coordinate(A)",
    "answer_expressions": "(1, \\pm 2)",
    "fact_spans": "[[[14, 28], [36, 39]], [[1, 4]], [[32, 35], [100, 104]], [[10, 13]], [[14, 28]], [[10, 31]], [[32, 44]], [[46, 98]]]",
    "query_spans": "[[[100, 109]]]",
    "process": "From the given conditions, we know F(1,0). Let A(\\frac{y_{0}^{2}}{4},y_{0}), then \\overrightarrow{OA}=(\\frac{y_{0}^{2}}{4},y_{0}), \\overrightarrow{AF}=(1-\\frac{y_{0}^{2}}{4},-y_{0}), from \\overrightarrow{OA}\\cdot\\overrightarrow{AF}=-4, we obtain y_{0}=\\pm2, so A(1,\\pm2)."
  },
  {
    "text": "Given that point $P$ moves on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{3}=1$, and point $Q$ moves on the circle $(x-1)^{2}+y^{2}=\\frac{5}{8}$, then the minimum value of $|P Q|$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P, G) = True;G: Ellipse;Expression(G) = (x^2/9 + y^2/3 = 1);Q: Point;PointOnCurve(Q, H) = True;H: Circle;Expression(H) = (y^2 + (x - 1)^2 = 5/8)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "sqrt(10)/4",
    "fact_spans": "[[[2, 6]], [[2, 47]], [[7, 44]], [[7, 44]], [[48, 52]], [[48, 86]], [[53, 83]], [[53, 83]]]",
    "query_spans": "[[[88, 101]]]",
    "process": "Let point P be $(x_{0},y_{0})$, $x_{0}\\in[-3,3]$, then $\\frac{x^{2}}{9}+\\frac{y_{0}^{2}}{3}=1$, then $y_{0}=3-\\frac{x^{2}}{3}$. Let the center of the circle $(x-1)^{2}+y^{2}=\\frac{5}{8}$ be M, then the coordinates of M are $(1,0)$. The minimum value of $|PQ|$ is the difference between the minimum value of $|MP|$ and the radius $\\frac{\\sqrt{10}}{4}$ of the circle $(x-1)^{2}+y^{2}=\\frac{5}{8}$. Then $|MP|=\\sqrt{(x_{0}-1)^{2}+y_{0}^{2}}=\\sqrt{\\frac{2}{3}(x_{0}^{2}-3x_{0})+4}=\\sqrt{\\frac{2}{3}(x_{0}-\\frac{3}{2})^{2}+\\frac{5}{2}}$. When $x_{0}\\in[-3,3]$, $|MP|\\geqslant\\frac{\\sqrt{10}}{2}$, with equality if and only if $x_{0}=\\frac{3}{2}$. Hence $|PQ|\\geqslant\\frac{\\sqrt{10}}{2}-\\frac{\\sqrt{10}}{4}=\\frac{\\sqrt{10}}{4}$."
  },
  {
    "text": "Given that one focus of the hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{5}=1$ $(a>0)$ is $(0,3)$, what is the value of $a$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/5 + y^2/a^2 = 1);a: Number;a>0;H: Point;Coordinate(H) = (0, 3);OneOf(Focus(G))=H",
    "query_expressions": "a",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 49]], [[2, 49]], [[64, 67]], [[5, 49]], [[55, 62]], [[55, 62]], [[2, 62]]]",
    "query_spans": "[[[64, 71]]]",
    "process": "According to the intersection points of the hyperbola, $ c = 3 $. The value of $ a $ can be obtained from the parameter relationship of the hyperbola. [Detailed solution] From the given condition: $ a^{2} + b^{2} = c^{2} = 9 $, that is, $ a^{2} + 5 = 9 $, and since $ a > 0 $, we get $ a = 2 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left vertex is $A$, the right focus is $F$, the point $B(0, b)$, and $\\overrightarrow{B A} \\cdot \\overrightarrow{B F}=0$. Then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/a^2-y^2/b^2=1);a: Number;b: Number;a>0;b>0;A: Point;LeftVertex(C) = A;F: Point;RightFocus(C) = F;B: Point;Coordinate(B) = (0, b);DotProduct(VectorOf(B, A), VectorOf(B, F)) = 0",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(\\sqrt{5}+1)/2",
    "fact_spans": "[[[2, 63], [145, 151]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 71]], [[2, 71]], [[76, 79]], [[2, 79]], [[80, 90]], [[80, 90]], [[92, 143]]]",
    "query_spans": "[[[145, 157]]]",
    "process": "From the given conditions: A(-a,0), F(c,0), so \\overrightarrow{BA}=(-a,-b), \\overrightarrow{BF}=(c,-b). Since \\overrightarrow{BA}\\cdot\\overrightarrow{BF}=0, we have b^{2}-ac=0. Because b^{2}=c^{2}-a^{2}, it follows that c^{2}-ac-a^{2}=0. Dividing both sides by a^{2} gives e^{2}-e-1=0. Solving this equation yields: e=\\frac{1-\\sqrt{5}}{2} (discarded) or e=\\frac{1+\\sqrt{5}}{2}."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{m+2}-\\frac{y^{2}}{m+1}=1$ represents an ellipse, then the range of real values for $m$ is?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G)=(x^2/(m + 2) - y^2/(m + 1) = 1)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-2,-3/2)+(-3/2,-1)",
    "fact_spans": "[[[44, 46]], [[48, 53]], [[1, 46]]]",
    "query_spans": "[[[48, 60]]]",
    "process": "From the equation of the ellipse, we have \\begin{cases}m+2>0\\\\m+1<0\\\\m+2+m+1\\neq0\\end{cases}. Solving the inequalities, the range of real values for $ m $ is $(-2,-\\frac{3}{2})\\cup(-\\frac{3}{2},-1)$."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=4x$, $E$ is the intersection point of its directrix and the $x$-axis, a line passing through $F$ intersects the parabola $C$ at points $A$ and $B$, $M$ is the midpoint of segment $AB$, and $|M E|=\\sqrt{20}$, then $|A B|=$?",
    "fact_expressions": "C: Parabola;G: Line;A: Point;B: Point;M: Point;E: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;Intersection(Directrix(C), xAxis) = E;PointOnCurve(F, G);Intersection(G, C) = {A, B};MidPoint(LineSegmentOf(A, B)) = M;Abs(LineSegmentOf(M, E)) = sqrt(20)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[6, 25], [33, 34], [53, 59]], [[50, 52]], [[60, 63]], [[64, 67]], [[70, 73]], [[29, 32]], [[2, 5], [46, 49]], [[6, 25]], [[2, 28]], [[29, 44]], [[45, 52]], [[50, 69]], [[70, 84]], [[86, 103]]]",
    "query_spans": "[[[105, 114]]]",
    "process": "Analysis: Obtain the focus and directrix equations of the parabola, yielding the coordinates of E. Let the line passing through F be $ y = k(x - 1) $, substitute into the parabola equation $ y^2 = 4x $, apply Vieta's formulas and the midpoint coordinate formula to find the coordinates of M. Use the distance formula between two points to solve for $ k $, then apply the focal chord formula of the parabola to compute the desired value. Detailed solution: $ F(1,0) $ is the focus of the parabola $ C: y^2 = 4x $, and $ E(-1,0) $ is the intersection point of its directrix with the x-axis. Let the line through F be $ y = k(x - 1) $, substituting into the parabola equation $ y^2 = 4x $ yields $ k^2x^2 - (2k^2 + 4)x + k^2 = 0 $. Let $ A(x_1, y_1), B(x_2, y_2) $, then $ x_1 + x_2 = 2 + \\frac{4}{k^2} $, midpoint $ M\\left(1 + \\frac{2}{k^2}, \\frac{2}{k}\\right) \\Rightarrow \\sqrt{\\left(2 + \\frac{2}{k^2}\\right)^2 + \\frac{4}{k^2}} = \\sqrt{20} $, solving gives $ k^2 = 1 $, then $ x_1 + x_2 = 6 $, by the definition of the parabola, $ |AB| = x_1 + x_2 + 2 = 8 $."
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{4}=1$ has the equation $y=x$, then the distance between the two directrices of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/4 + x^2/m = 1);m: Number;Expression(OneOf(Asymptote(G))) = (y = x);L1: Line;L2: Line;Directrix(G) = {L1,L2}",
    "query_expressions": "Distance(L1,L2)",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 40], [58, 61]], [[2, 40]], [[5, 40]], [[2, 55]], [], [], [[58, 65]]]",
    "query_spans": "[[[58, 70]]]",
    "process": ""
  },
  {
    "text": "If the distance from point $P(2,0)$ to one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ is $1$, then $a=$?",
    "fact_expressions": "G: Hyperbola;a: Number;P: Point;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Coordinate(P) = (2, 0);Distance(P,OneOf(Asymptote(G)))=1",
    "query_expressions": "a",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[11, 48]], [[63, 66]], [[1, 10]], [[14, 48]], [[11, 48]], [[1, 10]], [[1, 61]]]",
    "query_spans": "[[[63, 68]]]",
    "process": "The asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ is given by: $x+ay=0$. The distance from point $P(2,0)$ to an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ is 1, which gives: $\\frac{|2+0|}{\\sqrt{1+a^{2}}}=1$, solving yields $a=\\sqrt{3}$."
  },
  {
    "text": "Given that the eccentricity of the ellipse is $\\frac{3}{5}$, and one endpoint of the minor axis is $(0, -8)$, then the standard equation of this ellipse is?",
    "fact_expressions": "G: Ellipse;Coordinate(OneOf(Endpoint((MinorAxis((G)))))) = (0, -8);Eccentricity(G) = 3/5",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/100 + y^2/64=1",
    "fact_spans": "[[[2, 4], [43, 45]], [[2, 40]], [[2, 21]]]",
    "query_spans": "[[[43, 52]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C_{1}: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the circle $C_{2}: x^{2}+y^{2}=\\frac{4 b^{2}}{5}$, if there does not exist a point $P$ on the ellipse $C_{1}$ such that the two tangents drawn from point $P$ to the circle $C_{2}$ are perpendicular to each other, then the range of eccentricity of the ellipse $C_{1}$ is?",
    "fact_expressions": "C1: Ellipse;Expression(C1) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>b;b>0;C2: Circle;Expression(C2) = (x^2 + y^2 = (4*b^2)/5);P: Point;Negation(PointOnCurve(P, C1));Z1: Line;Z2: Line;TangentOfPoint(P, C2) = {Z1, Z2};IsPerpendicular(Z1, Z2)",
    "query_expressions": "Range(Eccentricity(C1))",
    "answer_expressions": "(0, \\sqrt{6}/4)",
    "fact_spans": "[[[2, 62], [107, 116], [154, 163]], [[2, 62]], [[12, 62]], [[12, 62]], [[12, 62]], [[12, 62]], [[63, 104], [135, 143]], [[63, 104]], [[120, 124], [128, 132]], [[106, 124]], [], [], [[127, 148]], [[127, 152]]]",
    "query_spans": "[[[154, 174]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola, respectively, and point $P$ lies on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, satisfying $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$ and $|P F_{1}|=\\sqrt{3}|P F_{2}|$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) =F2;PointOnCurve(P, RightPart(G));DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Abs(LineSegmentOf(P, F1)) = sqrt(3)*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)+1",
    "fact_spans": "[[[37, 93], [22, 25], [195, 198]], [[40, 93]], [[40, 93]], [[32, 36]], [[2, 9]], [[12, 19]], [[40, 93]], [[40, 93]], [[37, 93]], [[2, 31]], [[2, 31]], [[32, 99]], [[102, 161]], [[163, 192]]]",
    "query_spans": "[[[195, 203]]]",
    "process": "\\because PF_{1} \\perp PF_{2}, in Rt\\triangle PF_{1}F_{2}, let |PF_{1}| = \\sqrt{3} |PF_{2}| = \\sqrt{3}, then 2c = |F_{1}F_{2}| = 2, e = \\frac{c}{a} = \\frac{2c}{2a} = \\frac{2}{\\sqrt{3}-1} = \\sqrt{3}+1."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$, $A$ is the left vertex, $B$ is one endpoint of the minor axis, $F$ is the right focus, and $AB \\perp BF$. Then the eccentricity of this ellipse is equal to?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;A: Point;LeftVertex(G) = A;B: Point;OneOf(Vertex(MinorAxis(G))) = B;F: Point;RightFocus(G) = F;IsPerpendicular(LineSegmentOf(A, B), LineSegmentOf(B, F))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 56], [102, 104]], [[2, 56]], [[4, 56]], [[4, 56]], [[4, 56]], [[4, 56]], [[57, 60]], [[2, 64]], [[65, 68]], [[2, 74]], [[75, 78]], [[2, 82]], [[83, 98]]]",
    "query_spans": "[[[102, 111]]]",
    "process": ""
  },
  {
    "text": "If a point $M(4, m)$ on the parabola $y^{2}=2 p x(p>0)$ has a distance of $8$ to the focus $F$ of the parabola, then what is the equation of the parabola?",
    "fact_expressions": "G: Parabola;p: Number;M: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(M) = (4, m);PointOnCurve(M, G);Distance(M, F) = 8;Focus(G) = F;F: Point;m: Number",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = 16*x",
    "fact_spans": "[[[1, 22], [36, 39], [55, 58]], [[4, 22]], [[25, 34]], [[4, 22]], [[1, 22]], [[25, 34]], [[1, 34]], [[25, 52]], [[36, 45]], [[42, 45]], [[25, 34]]]",
    "query_spans": "[[[55, 63]]]",
    "process": "From the given condition, |MF| = 8, then 4 + \\frac{p}{2} = 8, p = 8, so the equation of the parabola is v^2 = 16x"
  },
  {
    "text": "The asymptotes of a hyperbola, whose foci are the endpoints of the major axis of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{25}=1$ and whose directrix passes through the foci of the ellipse, have slopes equal to?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/9 + y^2/25 = 1);Endpoint(MajorAxis(H))=Focus(G);PointOnCurve(Focus(H),Directrix(G))",
    "query_expressions": "Slope(Asymptote(G))",
    "answer_expressions": "pm*2",
    "fact_spans": "[[[57, 60]], [[1, 39], [52, 54]], [[1, 39]], [[0, 60]], [[49, 60]]]",
    "query_spans": "[[[57, 69]]]",
    "process": ""
  },
  {
    "text": "For any point $Q$ on the parabola $y^{2}=4 x$, the point $P(a, 0)$ satisfies $|P Q| \\geq|a|$. Then the range of values for $a$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);Q: Point;PointOnCurve(Q, G);P: Point;a: Number;Coordinate(P) = (a, 0);Abs(LineSegmentOf(P, Q)) >= Abs(a)",
    "query_expressions": "Range(a)",
    "answer_expressions": "(-oo,2]",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[21, 24]], [[2, 24]], [[25, 35]], [[55, 58]], [[25, 35]], [[38, 53]]]",
    "query_spans": "[[[55, 65]]]",
    "process": "Let $ Q\\left(\\frac{t^{2}}{4}, t\\right) $. From $ |PQ| \\geqslant |a| $, we obtain $ \\left(\\frac{t^{2}}{4} - a\\right)^{2} + t^{2} \\geqslant a^{2} $, which simplifies to $ t^{2}(t^{2} + 16 - 8a) \\geqslant 0 $. When $ t = 0 $, the inequality always holds; when $ t \\neq 0 $, $ t^{2} + 16 - 8a \\geqslant 0 $, so $ t^{2} \\geqslant 8a - 16 $ must always hold, which implies $ 8a - 16 \\leqslant 0 $, hence $ a \\leqslant 2 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A perpendicular is drawn from point $F_{2}$ to an asymptote of the hyperbola $C$, with foot of perpendicular at $H$, intersecting the hyperbola at point $M$, and $\\overrightarrow{F_{2} M}=2 \\overrightarrow{M H}$. Then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;Z: Line;PointOnCurve(F2, Z);IsPerpendicular(Z, OneOf(Asymptote(C)));H: Point;FootPoint(Z, OneOf(Asymptote(C))) = H;M: Point;Intersection(Z, C) = M;VectorOf(F2, M) = 2*VectorOf(M, H)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 63], [98, 104], [122, 125], [182, 188]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[72, 79]], [[80, 87], [89, 97]], [[2, 87]], [[2, 87]], [], [[88, 113]], [[88, 113]], [[117, 120]], [[88, 120]], [[126, 130]], [[88, 130]], [[131, 180]]]",
    "query_spans": "[[[182, 194]]]",
    "process": "From the given conditions, the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a, b > 0 $) has an asymptote with equation $ y = \\frac{b}{a}x $. Then the equation of line $ F_{2}H $ is $ y - 0 = -\\frac{a}{b}(x - c) $. Substituting into the asymptote equation $ y = \\frac{b}{a}x $ yields $ \\left( \\frac{a^{2}}{c}, \\frac{ab}{c} \\right) $. Thus, the midpoint $ M\\left( \\frac{c + \\frac{a^{2}}{c}}{2}, \\frac{ab}{2c} \\right) $ lies on the hyperbola, so $ \\frac{\\left( c + \\frac{a^{2}}{c} \\right)^{2}}{4a^{2}} - \\frac{a^{2}b^{2}}{4b^{2}c^{2}} = 1 $. Therefore, $ \\frac{c^{2}}{a^{2}} = 2 $, so $ \\frac{c}{a} = \\sqrt{2} $, and thus the eccentricity of hyperbola $ C $ is $ \\sqrt{2} $."
  },
  {
    "text": "On the line $l$: $x+y-4=0$, take an arbitrary point $M$. Construct an ellipse passing through $M$ with foci at the foci of $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$. Then, what is the minimum length of the major axis of such an ellipse?",
    "fact_expressions": "l: Line;G: Ellipse;C:Curve;Expression(l)=(x+y-4=0);Expression(C)=(x^2/16+y^2/12=1);Focus(C)=Focus(G);PointOnCurve(M,l);PointOnCurve(M,G);M:Point",
    "query_expressions": "Min(Length(MajorAxis(G)))",
    "answer_expressions": "2*sqrt(10)",
    "fact_spans": "[[[1, 17]], [[75, 77], [81, 83]], [[31, 68]], [[1, 17]], [[31, 68]], [[30, 77]], [[1, 24]], [[25, 77]], [[21, 24], [26, 29]]]",
    "query_spans": "[[[81, 93]]]",
    "process": "The foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$ are $F_{1}(-2,0)$ and $F_{2}(2,0)$. Let the point symmetric to $F_{2}(2,0)$ with respect to the line $l: x+y-4=0$ be $P(x,y)$, then $\\frac{\\frac{x+2}{2}+\\frac{y}{2}-4}{\\frac{y}{x-2}}=1$, solving gives $P(4,2)$. Connect $PF_{1}$ intersecting line $l$ at point $M$. The equation of line $PF_{1}$ is $x-3y+2=0$. Solving the system of equations $\\begin{cases}x+y-4=0\\\\x-3y+2=0\\end{cases}$ yields $M(\\frac{5}{2},\\frac{3}{2})$, which minimizes the major axis length of the desired ellipse. Otherwise, taking any point $Q$ on line $l$ different from $M(\\frac{5}{2},\\frac{3}{2})$, we have $|QF_{1}|+|QF_{2}|>|PF_{1}|=|MF_{1}|+|MF_{2}|$. Let the major axis length of the desired ellipse be $2a$, then $2a=|PF_{1}|=2\\sqrt{10}$."
  },
  {
    "text": "Given the hyperbola equation $x^{2}-\\frac{y^{2}}{2}=1$, a line passing through the fixed point $P(2 , 1)$ intersects the hyperbola at points $P_{1}$ and $P_{2}$, such that $P(2 , 1)$ is the midpoint of $P_{1} P_{2}$. Then the equation of this line is?",
    "fact_expressions": "G: Hyperbola;H: Line;P1: Point;P2: Point;P: Point;Coordinate(P) = (2, 1);Expression(G) = (x^2 - y^2/2 = 1);PointOnCurve(P, H);Intersection(H, G) = {P1, P2};MidPoint(LineSegmentOf(P1, P2)) = P",
    "query_expressions": "Expression(H)",
    "answer_expressions": "4*x-y-7=0",
    "fact_spans": "[[[2, 5], [51, 54]], [[48, 50], [105, 107]], [[55, 62]], [[63, 70]], [[37, 47], [75, 85]], [[37, 47]], [[2, 33]], [[34, 50]], [[48, 72]], [[75, 102]]]",
    "query_spans": "[[[105, 111]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is an arbitrary point on the right branch of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, $F$ is the right focus of the hyperbola, and the fixed point $A$ has coordinates $(3, \\sqrt{3})$, then the minimum value of $|P F|+|P A|$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P, RightPart(G)) = True;RightFocus(G) = F;F: Point;G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1);A: Point;Coordinate(A) = (3, sqrt(3))",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "2*sqrt(7)-2",
    "fact_spans": "[[[2, 5]], [[2, 43]], [[44, 55]], [[44, 47]], [[6, 35], [48, 51]], [[6, 35]], [[58, 61]], [[58, 80]]]",
    "query_spans": "[[[82, 101]]]",
    "process": "By the given conditions, in the hyperbola $ a=1 $, $ b=\\sqrt{3} $, $ c=2 $. Let $ F_{1} $ be the left focus of the hyperbola, so $ F_{1}(-2,0) $. Since $ A $ is on the right side, $ |PF|=|PF_{1}|-2 $, therefore $ |PF|+|PA|=|PF_{1}|+|PA|-2 \\geqslant |AF_{1}|-2 = \\sqrt{(3+2)^{2}+(\\sqrt{3}-0)^{2}}-2 = 2\\sqrt{7}-2 $. Equality holds if and only if $ F_{1} $, $ P $, $ A $ are collinear. Therefore, the minimum value of $ |PF|+|PA| $ is $ 2\\sqrt{7}-2 $."
  },
  {
    "text": "The line $y = kx + 2$ has exactly one common point with the parabola $y^2 = 8x$, then the value of $k$ is?",
    "fact_expressions": "G: Parabola;H: Line;k: Number;Expression(G) = (y^2 = 8*x);Expression(H) = (y = k*x + 2);NumIntersection(H, G) = 1",
    "query_expressions": "k",
    "answer_expressions": "{0,1}",
    "fact_spans": "[[[12, 26]], [[0, 11]], [[37, 40]], [[12, 26]], [[0, 11]], [[0, 35]]]",
    "query_spans": "[[[37, 45]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, $F_{1}$, $F_{2}$ are the foci, and $\\angle F_{1} P F_{2}=60^{\\circ}$, then $S_{\\Delta F_{1} P F_{2}}$=?",
    "fact_expressions": "P: Point;G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);PointOnCurve(P, G) = True;F1: Point;F2: Point;Focus(G) = {F1, F2};AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 5]], [[6, 43]], [[6, 43]], [[2, 46]], [[47, 54]], [[55, 62]], [[6, 65]], [[66, 99]]]",
    "query_spans": "[[[101, 129]]]",
    "process": "Using the law of cosines and the definition of an ellipse, we obtain |PF_{1}|\\cdot|PF_{2}|=4. Then, using the triangle area formula, the result follows. From the given information, a=2, b=\\sqrt{3}, so c=\\sqrt{a^{2}-b^{2}}=\\sqrt{4-3}=1, thus |F_{1}F_{2}|=2c=2. In \\triangle F_{1}PF_{2}, |F_{1}F_{2}|^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}|\\cdot|PF_{2}|\\cos60^{\\circ}, that is, 4=|PF_{1}|^{2}+|PF_{2}|^{2}-|PF_{1}|\\cdot|PF_{2}|,\\textcircled{1} From the definition of the ellipse, |PF_{1}|+|PF_{2}|=4, so 16=|PF_{1}|^{2}+|PF_{2}|^{2}+2|PF_{1}|\\cdot|PF_{2}|,\\textcircled{2} From \\textcircled{1}\\textcircled{2}, we get |PF_{1}|\\cdot|PF_{2}|=4. Therefore, S_{\\triangle F_{1}PF_{2}}=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|\\sin60^{\\circ}=\\sqrt{3}."
  },
  {
    "text": "The standard equation of an ellipse with foci on the $x$-axis, minor axis length equal to $16$, and eccentricity equal to $\\frac{3}{5}$ is?",
    "fact_expressions": "G: Ellipse;PointOnCurve(Focus(G), xAxis);Eccentricity(G) = 3/5;Length(MinorAxis(G))=16",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/100+y^2/64=1",
    "fact_spans": "[[[38, 40]], [[0, 40]], [[19, 40]], [[9, 40]]]",
    "query_spans": "[[[38, 47]]]",
    "process": "From the given conditions: 2b=16, solving gives: b=8. Also, \\begin{cases}e=\\frac{3}{5}=\\frac{c}{a}\\\\a^{2}=b^{2}+c^{2}\\end{cases}, solving gives: a^{2}=100. Therefore, the standard equation of the required ellipse is \\frac{x^{2}}{100}+\\frac{y^{2}}{64}=1"
  },
  {
    "text": "It is known that the foci of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{m}=1$ lie on the line $x-3 y-3=0$, then the value of the real number $m$ is?",
    "fact_expressions": "G: Hyperbola;m: Real;H: Line;Expression(G) = (x^2/4 - y^2/m = 1);Expression(H) = (x - 3*y - 3 = 0);PointOnCurve(Focus(G), H)",
    "query_expressions": "m",
    "answer_expressions": "5",
    "fact_spans": "[[[2, 40]], [[60, 65]], [[44, 57]], [[2, 40]], [[44, 57]], [[2, 58]]]",
    "query_spans": "[[[60, 69]]]",
    "process": ""
  },
  {
    "text": "The standard equation of the hyperbola passing through points $P(-3,2 \\sqrt{7})$ and $Q(-6 \\sqrt{2},-7)$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;Q: Point;Coordinate(P) = (-3, 2*sqrt(7));Coordinate(Q) = (-6*sqrt(2), -7);PointOnCurve(P,G);PointOnCurve(Q,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/25 - x^2/75 = 1",
    "fact_spans": "[[[42, 45]], [[2, 21]], [[22, 41]], [[2, 21]], [[22, 41]], [[0, 45]], [[0, 45]]]",
    "query_spans": "[[[42, 52]]]",
    "process": "Let the equation of the hyperbola be $mx^{2}+ny^{2}=1$ ($mn<0$), then $\\begin{cases}9m+28n=1\\\\72m+49n=1\\end{cases}$, solving gives $\\begin{cases}m=-\\frac{1}{75}\\\\n=\\frac{1}{25}\\end{cases}$, hence the standard equation of the hyperbola is $\\frac{y^{2}}{25}-\\frac{x^{2}}{75}=1$."
  },
  {
    "text": "The left focus of the hyperbola $x^{2}-\\frac{8 y^{2}}{p^{2}}=1  (p>0)$ lies on the directrix of the parabola ${y}^{2}=2 px$. Then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;p: Number;H: Parabola;p>0;Expression(G) = (x^2 - 8*y^2/p^2 = 1);Expression(H) = (y^2 = 2*p*x);PointOnCurve(LeftFocus(G), Directrix(H))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 41], [70, 73]], [[3, 41]], [[46, 63]], [[3, 41]], [[0, 41]], [[46, 63]], [[0, 67]]]",
    "query_spans": "[[[70, 79]]]",
    "process": ""
  },
  {
    "text": "Given that the parabola $y^{2}=4 x$ intersects an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ at point $M$, and $F$ is the focus of the parabola, if $|M F|=3$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;M: Point;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 4*x);Intersection(H,OneOf(Asymptote(G)))=M;Focus(H)=F;Abs(LineSegmentOf(M, F)) = 3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[17, 74], [112, 115]], [[20, 74]], [[20, 74]], [[2, 16], [92, 95]], [[84, 87]], [[88, 91]], [[20, 74]], [[20, 74]], [[17, 74]], [[2, 16]], [[2, 87]], [[88, 98]], [[100, 109]]]",
    "query_spans": "[[[112, 121]]]",
    "process": "Let M(m,n), then by the definition of the parabola, |MF| = m + 1 = 3, ∴ m = 2, ∴ n² = 4 × 2, ∴ n = ±2√2. Substituting point M(2, ±2√2) into the asymptote equation of the hyperbola y = ±(b/a)x, ∴ b/a = √2, ∴ (c² - a²)/a² = 2, ∴ e = √3,"
  },
  {
    "text": "$P$ is a point on the right branch of the hyperbola $x^{2}-\\frac{y^{2}}{15}=1$, and $M$, $N$ are points on the circles $(x+4)^{2}+y^{2}=4$ and $(x-4)^{2}+y^{2}=1$, respectively. Then the maximum value of $|P M|-|P N|$ is?",
    "fact_expressions": "G: Hyperbola;H: Circle;D: Circle;P: Point;M: Point;N: Point;Expression(G) = (x^2 - y^2/15 = 1);Expression(H) = (y^2 + (x + 4)^2 = 4);Expression(D) = (y^2 + (x - 4)^2 = 1);PointOnCurve(P, RightPart(G));PointOnCurve(M, H);PointOnCurve(N, D)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "5",
    "fact_spans": "[[[4, 33]], [[51, 71]], [[72, 91]], [[0, 3]], [[39, 42]], [[45, 48]], [[4, 33]], [[51, 71]], [[72, 91]], [[0, 38]], [[39, 94]], [[39, 94]]]",
    "query_spans": "[[[96, 115]]]",
    "process": ""
  },
  {
    "text": "If the center of circle $C$: $x^{2}+(y+1)^{2}=n$ is a focus of the hyperbola $M$: $m y^{2}-2 x^{2}=1$, and circle $C$ passes through the other focus of $M$, then $\\frac{n}{m}$=?",
    "fact_expressions": "C: Circle;Expression(C) = (x^2 + (y + 1)^2 = n);n: Number;M: Hyperbola;Expression(M) = (m*y^2 - 2*x^2 = 1);m: Number;F1: Point;F2: Point;OneOf(Focus(M)) = F1;OneOf(Focus(M)) = F2;Negation(F1=F2);Center(C) = F1;PointOnCurve(F2, C)",
    "query_expressions": "n/m",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 26], [64, 68]], [[1, 26]], [[7, 26]], [[30, 57], [70, 73]], [[30, 57]], [[38, 57]], [], [], [[30, 62]], [[70, 79]], [[30, 79]], [[1, 62]], [[64, 79]]]",
    "query_spans": "[[[81, 96]]]",
    "process": "From the given information, the center of circle C: $x^{2}+(y+1)^{2}=n$ is $(0,-1)$. The hyperbola M: $my^{2}-2x^{2}=1$ can be rewritten in standard form as $\\frac{y^{2}}{\\frac{1}{m}}-\\frac{x^{2}}{\\frac{1}{2}}=1$. Since the center of circle C: $x^{2}+(y+1)^{2}=n$ is a focus of the hyperbola M: $my^{2}-2x^{2}=1$, the foci of hyperbola M: $my^{2}-2x^{2}=1$ lie on the y-axis, and $\\frac{1}{m}+\\frac{1}{2}=1$. Solving gives $m=2$. Therefore, circle C passes through the other focus of M, so the radius of the circle is 2, hence $n=4$, and thus $\\frac{n}{m}=2$."
  },
  {
    "text": "Given that $O$ is the coordinate origin, a line $l$ passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) intersects the parabola at points $M$ and $N$, $T$ is the midpoint of chord $MN$, $|M F|=4|O F|$, the area of $\\Delta M F O$ is $4 \\sqrt{3}$, connect $O T$ and extend it to intersect the parabola at point $S$, then $\\frac{|O T|}{|O S|}$=?",
    "fact_expressions": "O: Origin;G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;l: Line;PointOnCurve(F, l);M: Point;N: Point;Intersection(l, G) = {M, N};T: Point;IsChordOf(LineSegmentOf(M, N), G);MidPoint(LineSegmentOf(M, N)) = T;Abs(LineSegmentOf(M, F)) = 4*Abs(LineSegmentOf(O, F));Area(TriangleOf(M, F, O)) = 4*sqrt(3);S: Point;Intersection(OverlappingLine(LineSegmentOf(O, T)), G) = S",
    "query_expressions": "Abs(LineSegmentOf(O, T))/Abs(LineSegmentOf(O, S))",
    "answer_expressions": "1/5",
    "fact_spans": "[[[2, 5]], [[12, 33], [46, 49], [133, 136]], [[12, 33]], [[15, 33]], [[15, 33]], [[36, 39]], [[12, 39]], [[40, 45]], [[11, 45]], [[50, 53]], [[54, 57]], [[40, 59]], [[60, 63]], [[46, 70]], [[60, 73]], [[74, 88]], [[90, 120]], [[137, 141]], [[121, 141]]]",
    "query_spans": "[[[143, 166]]]",
    "process": "Let M(x_{1},y_{1}), since |OF| = \\frac{p}{2}, |MF| = 4|OF|, therefore |MF| = 2p. By the definition of the parabola, we have x_{1} + \\frac{p}{2} = 2p, so x_{1} = \\frac{3}{2}p, thus y_{1} = \\pm\\sqrt{3}p. Therefore, S_{\\triangle MFO} = \\frac{1}{2} \\times \\frac{p}{2} \\times \\sqrt{3}p = 4\\sqrt{3}, solving gives p = 4. Hence, the equation of the parabola is y^{2} = 8x, F(2,0), M(6,\\pm4\\sqrt{3}). By symmetry, without loss of generality, take M(6,4\\sqrt{3}), then the equation is y = \\frac{4\\sqrt{3}-0}{6-2}(x-2), i.e., y = \\sqrt{3}(x-2). Substituting into y^{2} = 8x, we get 3x^{2} - 20x + 12 = 0, solving gives x_{N} = \\frac{2}{3}, so N(\\frac{2}{3}, -\\frac{4\\sqrt{3}}{3}). Using the midpoint formula, we obtain T(\\frac{10}{3}, \\frac{4\\sqrt{3}}{3}), then the equation of line OT is y = \\frac{2\\sqrt{3}}{5}x."
  },
  {
    "text": "The directrix of the parabola $y^{2}=16 x$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 16*x)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x = -4",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 20]]]",
    "process": ""
  },
  {
    "text": "Suppose the curve $y=x^{2}$ is tangent to the line $2 x-y-a=0$, then $a$=?",
    "fact_expressions": "H: Curve;Expression(H) = (y = x^2);G: Line;Expression(G) = (-a + 2*x - y = 0);a: Number;IsTangent(H,G)",
    "query_expressions": "a",
    "answer_expressions": "1",
    "fact_spans": "[[[1, 12]], [[1, 12]], [[13, 26]], [[13, 26]], [[30, 33]], [[1, 28]]]",
    "query_spans": "[[[30, 35]]]",
    "process": ""
  },
  {
    "text": "The equation $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ represents an ellipse with foci on the $y$-axis. What are the coordinates of its foci?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/m = 1);PointOnCurve(Focus(G), yAxis);m: Number",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, pm*sqrt(m - 4))",
    "fact_spans": "[[[48, 50], [51, 52]], [[0, 50]], [[39, 50]], [[2, 37]]]",
    "query_spans": "[[[51, 58]]]",
    "process": "According to the equation $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ representing an ellipse with foci on the $y$-axis, determine $a^{2}=m$, $b^{2}=4$, then find $c$ from the relationship among $a$, $b$, and $c$, and write down the coordinates. Since the equation $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ represents an ellipse with foci on the $y$-axis, we have $a^{2}=m$, $b^{2}=4$, so $c=\\sqrt{a^{2}-b^{2}}=\\sqrt{m-4}$, thus the foci coordinates are: $(0,\\pm\\sqrt{m-4})$."
  },
  {
    "text": "Given that the slope of a line $l$ with inclination angle $\\alpha$ is equal to the eccentricity of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, then $\\sin 2 \\alpha$=?",
    "fact_expressions": "alpha: Number;l: Line;Inclination(l) = alpha;G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1);Slope(l) = Eccentricity(G)",
    "query_expressions": "Sin(2*alpha)",
    "answer_expressions": "4/5",
    "fact_spans": "[[[6, 14]], [[15, 20]], [[2, 20]], [[25, 53]], [[25, 53]], [[15, 57]]]",
    "query_spans": "[[[59, 76]]]",
    "process": "The eccentricity of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ is $e=\\frac{\\sqrt{1+3}}{1}=2$, so $\\tan\\alpha=2$, $\\sin2\\alpha=\\frac{2\\sin\\alpha\\cos\\alpha}{\\sin^{2}\\alpha+\\cos^{2}\\alpha}=\\frac{2\\tan\\alpha}{\\tan^{2}\\alpha+1}=\\frac{4}{5}$"
  },
  {
    "text": "Given a parabola $C$: $y^{2}=x$ with a moving point $P$, then the range of the difference of distances from the moving point $P$ to two fixed points $A(-1,0)$ and $B(1,0)$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = x);PointOnCurve(P,C) = True;P: Point;A: Point;Coordinate(A) = (-1, 0);B: Point;Coordinate(B) = (1, 0)",
    "query_expressions": "Range(Distance(P,A) - Distance(P,B))",
    "answer_expressions": "[0,2)",
    "fact_spans": "[[[2, 19]], [[2, 19]], [[2, 27]], [[24, 27], [31, 34]], [[35, 45]], [[35, 45]], [[47, 55]], [[47, 55]]]",
    "query_spans": "[[[31, 69]]]",
    "process": "When point $ P $ is at the origin, $ PA = PB $, so the difference in distances from point $ P $ to the two fixed points $ A(-1,0) $, $ B(1,0) $ is $ 0 $. When point $ P $ is not at the origin, points $ P $, $ A $, $ B $ form a triangle, then $ PA - PB < AB = 2 $, or $ PB - PA < AB = 2 $. Therefore, the range of the difference in distances from moving point $ P $ to the two fixed points $ A(-1,0) $, $ B(1,0) $ is $ [0,2) $."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, a fixed point $A\\left(\\frac{1}{2}, 1\\right)$, and $P$ is a moving point on the parabola, then the minimum value of $|PA|+|PF|$ is?",
    "fact_expressions": "G: Parabola;P: Point;A: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(P,G);Coordinate(A)=(1/2,1)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[2, 16], [53, 56]], [[49, 52]], [[26, 46]], [[20, 23]], [[2, 16]], [[2, 23]], [[49, 60]], [[26, 46]]]",
    "query_spans": "[[[62, 79]]]",
    "process": ""
  },
  {
    "text": "Given a point $A(m , 2 \\sqrt{2})$ on the parabola $y^{2}=2 p x(p>0)$ with focus $F$. If the circle centered at $A$ with radius $|A F|$ is intersected by the $y$-axis to form a chord of length $2 \\sqrt{5}$, then $m=?$",
    "fact_expressions": "G: Parabola;p: Number;A: Point;F: Point;m:Number;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(A) = (m, 2*sqrt(2));Focus(G)=F;PointOnCurve(A,G);C:Circle;Center(C)=A;Radius(C)=Abs(LineSegmentOf(A,F));Length(InterceptChord(yAxis,C))=2*sqrt(5)",
    "query_expressions": "m",
    "answer_expressions": "2",
    "fact_spans": "[[[9, 30]], [[12, 30]], [[33, 52], [55, 58]], [[5, 8]], [[102, 105]], [[12, 30]], [[9, 30]], [[33, 52]], [[2, 30]], [[9, 52]], [[73, 77]], [[54, 77]], [[62, 77]], [[73, 100]]]",
    "query_spans": "[[[102, 107]]]",
    "process": "Since the chord length intercepted by the y-axis on circle A is $2\\sqrt{5}$, we have $\\sqrt{m^{2}+5}=|AF|=m+\\frac{p}{2}$.$\\textcircled{1}$ Also, since point $A(m,2\\sqrt{2})$ lies on the parabola, we have $8=2pm$.$\\textcircled{2}$ From $\\textcircled{1}$ and $\\textcircled{2}$, we obtain $p=2$, $m=2$."
  },
  {
    "text": "The equation of the parabola with vertex at the origin, focus on the $x$-axis, and chord length $3 \\sqrt{5}$ intercepted by the line $2 x - y - 4 = 0$ is?",
    "fact_expressions": "G: Parabola;H: Line;O: Origin;Expression(H) = (2*x - y - 4 = 0);Vertex(G) = O;PointOnCurve(Focus(G), xAxis);Length(InterceptChord(H, G)) = 3*sqrt(5)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2 = 4*x, y^2 = -36*x}",
    "fact_spans": "[[[47, 50]], [[16, 29]], [[3, 5]], [[16, 29]], [[0, 50]], [[6, 50]], [[15, 50]]]",
    "query_spans": "[[[47, 54]]]",
    "process": "Let the equation of the parabola be $ y^{2} = 2px $ ($ p \\neq 0 $). Substitute the line equation $ y = 2x \\cdot 4 $ into it, and rearrange to obtain $ 2x^{2} - (8 + p)x + 8 = 0 $. Let the two roots of the equation be $ x_{1}, x_{2} $; then by Vieta's formulas, $ x_{1} + x_{2} = \\frac{p + 8}{2} $, $ x_{1}x_{2} = 4 $. Using the chord length formula, we get $ (3\\sqrt{5})^{2} = (1 + 2^{2})[(x_{1} + x_{2})^{2} - 4x_{1}x_{2}] $. That is, $ 9 = \\left( \\frac{8 + p}{2} \\right)^{2} - 16 $. Rearranging gives $ p^{2} + 16p - 36 = 0 $, solving which yields $ p = 2 $, or $ p = -18 $, and at this time $ \\Delta > 0 $. Hence, the required equations of the parabola are $ y^{2} = 4x $, or $ y^{2} = -36x $."
  },
  {
    "text": "The distance between the two directrices of the hyperbola $x^{2}-4 y^{2}=4$ is?",
    "fact_expressions": "G: Hyperbola;L1: Line;L2: Line;Expression(G) = (x^2 - 4*y^2 = 4);Directrix(G) = {L1, L2}",
    "query_expressions": "Distance(L1, L2)",
    "answer_expressions": "8*sqrt(5)/5",
    "fact_spans": "[[[0, 20]], [], [], [[0, 20]], [[0, 24]]]",
    "query_spans": "[[[0, 31]]]",
    "process": ""
  },
  {
    "text": "The distance from a focus of the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{6}=1$ to one of its asymptotes is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2/6 = 1)",
    "query_expressions": "Distance(OneOf(Focus(G)),OneOf(Asymptote(G)))",
    "answer_expressions": "sqrt(6)",
    "fact_spans": "[[[0, 38], [44, 45]], [[0, 38]]]",
    "query_spans": "[[[0, 56]]]",
    "process": "The distance $ d $ from a focus $ (3,0) $ to an asymptote $ y = \\sqrt{2}x $ is $ \\frac{|3\\sqrt{2}|}{\\sqrt{1+2}} = \\sqrt{6} $."
  },
  {
    "text": "Let one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ be $y=2x$, and one focus coincide with the focus of the parabola $y^{2}=4x$. Then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 4*x);Expression(OneOf(Asymptote(G)))=(y=2*x);OneOf(Focus(G))=Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "5*x^2-5*y^2/4=1",
    "fact_spans": "[[[1, 57], [100, 103]], [[4, 57]], [[4, 57]], [[78, 92]], [[4, 57]], [[4, 57]], [[1, 57]], [[78, 92]], [[1, 71]], [[1, 97]]]",
    "query_spans": "[[[100, 108]]]",
    "process": "Since the focus of the parabola is (1,0), we have \\begin{cases}c=1\\\\\\frac{b}{a}=2\\\\c^{2}=a^{2}+b^{2}\\end{cases}, solving gives \\begin{cases}a^{2}=\\frac{1}{5}\\\\b^{2}=\\frac{4}{5}\\end{cases}, the equation of the hyperbola is 5x^{2}-\\frac{5y^{2}}{4}=1"
  },
  {
    "text": "Given fixed points $A$ and $B$, with $|AB|=4$, and a moving point $P$ satisfying $|PA|-|PB|=3$, then the minimum value of $|PA|$ is?",
    "fact_expressions": "A: Point;B: Point;P: Point;Abs(LineSegmentOf(A, B)) = 4;Abs(LineSegmentOf(P, A)) - Abs(LineSegmentOf(P, B)) = 3",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "7/2",
    "fact_spans": "[[[4, 7]], [[8, 11]], [[26, 29]], [[13, 22]], [[31, 46]]]",
    "query_spans": "[[[49, 62]]]",
    "process": "According to the definition of a hyperbola, the trajectory of point P is the right branch of a hyperbola with $2c=4$ and $2a=3$, as shown in the figure below. When P moves to point C, $|PA|$ is minimized, and the minimum value is $a+c=\\frac{7}{2}$."
  },
  {
    "text": "If $P$ is a point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{6}=1$, $F_{1}$ and $F_{2}$ are the two foci of the ellipse, and $\\angle F_{1}PF_{2}=60^{\\circ}$, then the value of $|P F_{1}| \\cdot|P F_{2}|$ is?",
    "fact_expressions": "P: Point;G: Ellipse;Expression(G) = (x^2/9 + y^2/6 = 1);PointOnCurve(P, G) = True;F1: Point;F2: Point;Focus(G) = {F1, F2};AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 4]], [[5, 42], [62, 64]], [[5, 42]], [[1, 45]], [[46, 53]], [[54, 61]], [[46, 69]], [[70, 101]]]",
    "query_spans": "[[[103, 133]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the distance from the right focus to an asymptote is $3$, and for a point $P$ on the right branch of the hyperbola, the difference of the distances from $P$ to the two foci is $\\frac{4}{3}$ times the length of the imaginary axis. Find the standard equation of the hyperbola $C$.",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Distance(RightFocus(C),Asymptote(C))=3;P: Point;F1: Point;F2: Point;Focus(C) = {F1, F2};PointOnCurve(P, RightPart(C));Distance(P, F1) - Distance(P, F2) = (4/3)*Length(ImageinaryAxis(C))",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/16 - y^2/9 = 1",
    "fact_spans": "[[[2, 63], [80, 83], [122, 128]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 78]], [[89, 92]], [], [], [[80, 96]], [[80, 92]], [[80, 120]]]",
    "query_spans": "[[[122, 135]]]",
    "process": ""
  },
  {
    "text": "Given that the distance from a moving point $M(x, y)$ to the fixed point $(2, 0)$ is less by $1$ than its distance to the line $x = -3$, then the equation of the trajectory of point $M$ is?",
    "fact_expressions": "M: Point;x1: Number;y1: Number;Coordinate(M) = (x1, y1);H: Point;Coordinate(H) = (2, 0);G: Line;Expression(G) = (x = -3);Distance(M, H) = Distance(M, G) - 1",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "y^2=8*x",
    "fact_spans": "[[[4, 13], [49, 52]], [[4, 13]], [[4, 13]], [[4, 13]], [[16, 25]], [[16, 25]], [[30, 38]], [[30, 38]], [[4, 45]]]",
    "query_spans": "[[[49, 59]]]",
    "process": ""
  },
  {
    "text": "The equation of the directrix of the parabola $4 x=y^{2}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (4*x = y^2)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "y=-1",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": ""
  },
  {
    "text": "From a point $M$ on the parabola $y^{2}=12x$ with focus $F$, draw a perpendicular to its directrix, with foot $N$. If the slope of line $NF$ is $-\\frac{\\sqrt{3}}{3}$, then $|MF|=$?",
    "fact_expressions": "G: Parabola;H: Line;N: Point;F: Point;M: Point;Expression(G) = (y^2 = 12*x);Focus(G)=F;PointOnCurve(M,G);PointOnCurve(M,H);IsPerpendicular(H,Directrix(G));FootPoint(H,Directrix(G))=N;Slope(LineOf(N,F)) = -sqrt(3)/3",
    "query_expressions": "Abs(LineSegmentOf(M, F))",
    "answer_expressions": "4",
    "fact_spans": "[[[8, 23], [30, 31]], [], [[40, 43]], [[4, 7]], [[26, 29]], [[8, 23]], [[1, 23]], [[8, 29]], [[0, 36]], [[0, 36]], [[0, 43]], [[45, 77]]]",
    "query_spans": "[[[79, 88]]]",
    "process": "Let the intersection point of the directrix and the x-axis be A. Given the length |FA| and the angle \\angle NFA, we can find |NF| and \\angle NMF. Then, using |MN| = |MF| and side-angle relationships, we can obtain |MF|. [Solution] From the problem, let the intersection point of the directrix and the x-axis be A. Since the slope of line NF is -\\frac{\\sqrt{3}}{3}, it follows that \\angle NFA = \\frac{\\pi}{6}. Because |FA| = p = 6, we have |NF| = \\frac{|FA|}{\\cos\\angle NFA} = 4\\sqrt{3}. Since \\angle NMF = \\frac{2\\pi}{3} and |MN| = |MF|, it follows that |MF| = \\frac{\\frac{|FN|}{2}}{\\sin\\frac{1}{2}\\angle NMF} = 4."
  },
  {
    "text": "Let an ellipse centered at the origin share common foci with the hyperbola $2 x^{2}-2 y^{2}=1$, and let their eccentricities be reciprocals of each other. Then the equation of the ellipse is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;O: Origin;Expression(G) = (2*x^2 - 2*y^2 = 1);Center(H) = O;Focus(H) = Focus(G);InterReciprocal(Eccentricity(G), Eccentricity(H))",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2/2+y^2=1",
    "fact_spans": "[[[10, 32]], [[7, 9], [53, 55]], [[4, 6]], [[10, 32]], [[1, 9]], [[7, 38]], [[40, 50]]]",
    "query_spans": "[[[53, 60]]]",
    "process": "In the hyperbola, $a = b = \\frac{\\sqrt{2}}{2}$, $\\therefore F(\\pm1, 0)$, $e = \\frac{c}{a} = \\sqrt{2}$, $\\therefore$ the foci of the ellipse are $(\\pm1, 0)$, eccentricity is $\\frac{\\sqrt{2}}{2}$, $\\therefore$ the semi-major axis length is $\\sqrt{2}$, semi-minor axis length is $1$, $\\therefore$ the equation is $\\frac{x^{2}}{2} + y^{2} = 1$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{24}=1$, respectively, and $P$ is a point on the hyperbola such that $|P F_{1}|=2|P F_{2}|$, then $|P F_{1}|$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/24 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G) = True;Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Abs(LineSegmentOf(P, F1))",
    "answer_expressions": "16",
    "fact_spans": "[[[20, 60], [71, 74]], [[20, 60]], [[2, 9]], [[10, 17]], [[2, 66]], [[2, 66]], [[67, 70]], [[67, 78]], [[82, 104]]]",
    "query_spans": "[[[106, 119]]]",
    "process": "It is easy to see that $ a=4 $, and point $ P $ lies on the right branch of the hyperbola, so $ |PF_{1}| - |PF_{2}| = 8 $. Since $ |PF_{1}| = 2|PF_{2}| $, it follows that $ |PF_{1}| = 16 $."
  },
  {
    "text": "The equation of a curve is $\\sqrt{(x-1)^{2}+y^{2}}+\\sqrt{(x+1)^{2}+y^{2}}=2$. If the line $l$: $y=k x+1-\\frac{1}{2} k$ intersects the curve, then the range of real values for $k$ is?",
    "fact_expressions": "l: Line;G: Curve;k: Real;Expression(G) = (sqrt(y^2 + (x - 1)^2) + sqrt(y^2 + (x + 1)^2) = 2);Expression(l) = (y = k*x + 1 - k/2);IsIntersect(l, G)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-oo, -2] + [2/3, +oo)",
    "fact_spans": "[[[58, 88]], [[1, 3], [90, 92]], [[98, 103]], [[1, 56]], [[58, 88]], [[58, 96]]]",
    "query_spans": "[[[98, 110]]]",
    "process": "From the given conditions, the curve is the line segment $F_{1}F_{2}$ (where $F_{1}(-1,0)$, $F_{2}(1,0)$). Note that line $l$ passes through the fixed point $(\\frac{1}{2},1)$ and has slope $k$. Using the idea of combining algebra and geometry, observe the change in the inclination angle of line $l$ when it intersects the line segment $F_{1}F_{2}$, then the range of real values for $k$ can be determined. Solution: $\\because \\sqrt{(x-1)^{2}+y^{2}}+\\sqrt{(x+1)^{2}+y^{2}}=2$, it follows that for any point $P$ on this curve, the sum of distances to points $F_{1}(-1,0)$ and $F_{2}(1,0)$ is $2 = |F_{1}F_{2}|$. Thus, the curve is the line segment $F_{1}F_{2}$. Moreover, line $l: y = kx + 1 - \\frac{1}{2}k$ is a line passing through the fixed point $P(\\frac{1}{2},1)$ with slope $k$, as shown in the figure below: The slope of line $PF_{1}$ is $k_{1} = \\frac{1-0}{\\frac{1}{2}+1} = \\frac{2}{3}$, and the slope of line $PF_{2}$ is $k_{2} = \\frac{1-0}{\\frac{1}{2}-1} = -2$. Therefore, when line $l$ intersects the line segment $F_{1}F_{2}$, the range of real values for $k$ is $(-\\infty,-2] \\cup [\\frac{2}{3},+\\infty)$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{1}$ intersects the two asymptotes of $C$ at points $A$ and $B$. If $\\overrightarrow{F_{1} {A}}=2 \\overrightarrow{A B}$ and $\\overrightarrow{F_{1} {A}}\\cdot \\overrightarrow{A O}=0$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;H: Line;A: Point;B: Point;O: Origin;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);F1:Point;F2:Point;LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1,H);L1:Line;L2:Line;Asymptote(C)={L1,L2};Intersection(H,L1)=A;Intersection(H,L2)=B;VectorOf(F1,A) = 2*VectorOf(A,B);DotProduct(VectorOf(F1,A),VectorOf(A,O)) = 0",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(6)",
    "fact_spans": "[[[2, 63], [104, 107], [240, 243]], [[9, 63]], [[9, 63]], [[101, 103]], [[117, 120]], [[121, 124]], [[182, 238]], [[9, 63]], [[9, 63]], [[2, 63]], [[72, 79], [92, 100]], [[81, 88]], [[2, 88]], [[2, 88]], [[91, 103]], [], [], [[104, 113]], [[101, 126]], [[101, 126]], [[128, 180]], [[182, 238]]]",
    "query_spans": "[[[240, 249]]]",
    "process": "Draw the figure according to the given conditions. From the known information, we have $F_{1}A\\bot OA$. Write the equation of $F_{1}A$, solve it together with $y=\\frac{b}{a}x$ to obtain the coordinates of point $B$, and solve it together with $y=-\\frac{b}{a}x$ to obtain the coordinates of point $A$. Then, using $\\overrightarrow{F_{1}A}=2\\overrightarrow{AB}$, we get $\\frac{y_{B}}{y_{A}}=\\frac{3}{2}$, from which the eccentricity can be found. Draw the figure according to the given conditions. Since the hyperbola $C:\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$, the asymptotes are $y=\\pm\\frac{b}{a}x$. $F_{1}(-c,0)$. A line passing through $F_{1}$ intersects the two asymptotes of $C$ at points $A$ and $B$ respectively. Given $\\overrightarrow{F_{1}A}\\cdot\\overrightarrow{AO}=0$, then $\\overrightarrow{F_{1}A}\\bot\\overrightarrow{AO}$, i.e., $F_{1}A\\bot AO$, so $k_{F_{1}A}\\cdot k_{AO}=-1$. Since $k_{AO}=-\\frac{b}{a}$, then $k_{F_{1}A}=\\frac{a}{b}$. Solving the system:  \n$$\n\\begin{cases}\ny=\\frac{a}{b}(x+c) \\\\\ny=\\frac{b}{a}x\n\\end{cases}\n$$\ngives $B\\left(\\frac{a^{2}c}{b^{2}-a^{2}},\\frac{abc}{b^{2}-a^{2}}\\right)$. Solving the system:\n$$\n\\begin{cases}\ny=\\frac{b}{a}x \\\\\ny=\\frac{a}{b}(x+c)\n\\end{cases}\n$$\ngives $A\\left(\\frac{-a^{2}c}{b^{2}+a^{2}},\\frac{abc}{b^{2}+a^{2}}\\right)$. Since $\\overrightarrow{F_{1}A}=2\\overrightarrow{AB}$, then $2\\times\\frac{abc}{b^{2}-a^{2}}=3\\times\\frac{abc}{b^{2}+a^{2}}$, thus $b^{2}=5a^{2}$, i.e., $6a^{2}=c^{2}$, so $\\frac{c^{2}}{a^{2}}=6$, hence $e=\\frac{c}{a}=\\sqrt{6}$."
  },
  {
    "text": "If the ellipse $C$: $\\frac{x^{2}}{m+1}+\\frac{y^{2}}{m}=1$ $(m>0)$ has eccentricity $\\frac{\\sqrt{3}}{3}$, then $m=$?",
    "fact_expressions": "C: Ellipse;m: Number;m>0;Expression(C) = (x^2/(m + 1) + y^2/m = 1);Eccentricity(C) = sqrt(3)/3",
    "query_expressions": "m",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 50]], [[77, 80]], [[8, 50]], [[1, 50]], [[1, 75]]]",
    "query_spans": "[[[77, 82]]]",
    "process": "Since the ellipse $ C: \\frac{x^2}{m+1} + \\frac{y^{2}}{m} = 1 $ ($ m > 0 $) has eccentricity $ \\frac{\\sqrt{3}}{3} $, it follows that $ \\frac{1}{\\sqrt{m+1}} = \\frac{\\sqrt{3}}{3} \\Rightarrow m = 2 $."
  },
  {
    "text": "If the angle of inclination of an asymptote of the hyperbola $x^{2}+\\frac{y^{2}}{m}=1$ is $60^{\\circ}$, then $m=$?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2 + y^2/m = 1);Inclination(OneOf(Asymptote(G)))=ApplyUnit(60,degree)",
    "query_expressions": "m",
    "answer_expressions": "-3",
    "fact_spans": "[[[1, 29]], [[54, 57]], [[1, 29]], [[1, 52]]]",
    "query_spans": "[[[54, 59]]]",
    "process": "From the given condition, the asymptotes of the hyperbola are given by $ y = \\pm\\sqrt{-m}x $. Since the inclination of one asymptote is $ 60^{\\circ} $, $ \\sqrt{-m} = \\sqrt{3} $, hence $ m = -3 $."
  },
  {
    "text": "Given that $c$ is the semi-focal distance of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, then the range of $\\frac{b+c}{a}$ is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;c: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);HalfFocalLength(G)=c",
    "query_expressions": "Range((b + c)/a)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[6, 58]], [[8, 58]], [[8, 58]], [[2, 5]], [[8, 58]], [[8, 58]], [[6, 58]], [[2, 62]]]",
    "query_spans": "[[[64, 86]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has eccentricity $e=\\frac{\\sqrt{3}}{2}$, $A$, $B$ are the left and right vertices of the ellipse, $P$ is a point on the ellipse different from $A$ and $B$, the inclinations of lines $PA$, $PB$ are $\\alpha$, $\\beta$ respectively, then $\\frac{\\cos (\\alpha-\\beta)}{\\cos (\\alpha+\\beta)}$=?",
    "fact_expressions": "G:Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;e:Number;Eccentricity(G)=e;e=sqrt(3)/2;A:Point;B:Point;LeftVertex(G)=A;RightVertex(G)=B;PointOnCurve(P,G);Negation(P=A);Negation(P=B);Inclination(LineOf(P, A)) = alpha;Inclination(LineOf(P,B)) = beta;alpha:Number;beta:Number;P:Point",
    "query_expressions": "Cos(alpha - beta)/Cos(alpha + beta)",
    "answer_expressions": "3/5",
    "fact_spans": "[[[2, 54], [94, 96], [107, 109]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[58, 82]], [[2, 82]], [[58, 82]], [[83, 87], [113, 116]], [[90, 93], [117, 120]], [[84, 102]], [[84, 102]], [[103, 123]], [[103, 123]], [[103, 123]], [[124, 160]], [[124, 160]], [[142, 151]], [[153, 160]], [[103, 106]]]",
    "query_spans": "[[[162, 213]]]",
    "process": ""
  },
  {
    "text": "Let point $P(x_{1}, y_{1})$ lie on the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{2}=1$, and point $Q(x_{2}, y_{2})$ lie on the line $x+2 y-8=0$. Then the minimum value of $3|x_{2}-x_{1}|+5|y_{2}-y_{1}|$ is?",
    "fact_expressions": "G: Ellipse;H: Line;P: Point;Q: Point;x1:Number;y1:Number;x2:Number;y2:Number;Expression(G) = (x^2/8 + y^2/2 = 1);Expression(H) = (x + 2*y - 8 = 0);Coordinate(P) = (x1, y1);Coordinate(Q) = (x2, y2);PointOnCurve(P, G);PointOnCurve(Q, H)",
    "query_expressions": "Min(3*Abs(x2-x1)+5*Abs(y2-y1))",
    "answer_expressions": "10",
    "fact_spans": "[[[20, 57]], [[78, 91]], [[1, 19]], [[59, 77]], [[94, 125]], [[94, 125]], [[94, 125]], [[94, 125]], [[20, 57]], [[78, 91]], [[1, 19]], [[59, 77]], [[1, 58]], [[59, 92]]]",
    "query_spans": "[[[94, 131]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x(p>0)$ coincides with the right focus of the hyperbola $\\frac{x^{2}}{6}-\\frac{y^{2}}{10}=1$, then the value of the real number $p$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/6 - y^2/10 = 1);H: Parabola;Expression(H) = (y^2 = 2*(p*x));p: Real;p>0;Focus(H) = RightFocus(G)",
    "query_expressions": "p",
    "answer_expressions": "8",
    "fact_spans": "[[[26, 65]], [[26, 65]], [[1, 22]], [[1, 22]], [[73, 78]], [[4, 22]], [[1, 71]]]",
    "query_spans": "[[[73, 82]]]",
    "process": ""
  },
  {
    "text": "Let $F$ be the right focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$. A line passing through the center of the ellipse intersects the ellipse at points $P$ and $Q$. When the area of triangle $P F Q$ is maximized, what is the value of $P F \\cdot Q F$?",
    "fact_expressions": "G: Ellipse;H: Line;P: Point;F: Point;Q: Point;Expression(G) = (x^2/4 + y^2/3 = 1);RightFocus(G) = F;PointOnCurve(Center(G), H);Intersection(H, G) = {P, Q};WhenMax(Area(TriangleOf(P,F,Q)))",
    "query_expressions": "LineSegmentOf(P, F)*LineSegmentOf(Q, F)",
    "answer_expressions": "-2",
    "fact_spans": "[[[5, 42], [48, 50], [57, 59]], [[54, 56]], [[61, 64]], [[1, 4]], [[65, 68]], [[5, 42]], [[1, 46]], [[47, 56]], [[54, 70]], [[71, 88]]]",
    "query_spans": "[[[89, 108]]]",
    "process": ""
  },
  {
    "text": "The length of the chord cut by the line $x-2 y+4=0$ on the ellipse $x^{2}+4 y^{2}=16$ is?",
    "fact_expressions": "G: Ellipse;H: Line;Expression(G) = (x^2 + 4*y^2 = 16);Expression(H) = (x - 2*y + 4 = 0)",
    "query_expressions": "Length(InterceptChord(H,G))",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[0, 20]], [[21, 34]], [[0, 20]], [[21, 34]]]",
    "query_spans": "[[[0, 41]]]",
    "process": "From \\begin{cases}x^{2}+4y^{2}=16\\\\x-2y+4=0\\end{cases}, eliminating $y$ gives $x^{2}+4x=0$, $\\therefore x=0$ or $x=-4$. When $x=0$, $y=2$; when $x=-4$, $y=0$. Therefore, the two intersection points of the line and the ellipse are $(0,2)$ and $(-4,0)$. Thus, the length of the chord cut by the line $x-2y+4=0$ on the ellipse $x^{2}+4y^{2}=16$ is $\\sqrt{(0+4)^{2}+(2-0)^{2}}=2\\sqrt{5}$."
  },
  {
    "text": "Given that the focus of the parabola $x^{2}=p y$ coincides with the upper focus of the hyperbola $\\frac{y^{2}}{3}-x^{2}=1$, then the value of $p$ is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;Expression(G) = (-x^2 + y^2/3 = 1);Expression(H) = (x^2 = p*y);UpperFocus(G)=Focus(H)",
    "query_expressions": "p",
    "answer_expressions": "8",
    "fact_spans": "[[[19, 47]], [[1, 15]], [[56, 59]], [[19, 47]], [[1, 15]], [[1, 53]]]",
    "query_spans": "[[[56, 63]]]",
    "process": "Given that the focus of the parabola is $(0,\\frac{p}{4})$, and the focus of the hyperbola is $(0,2)$, we have $\\frac{p}{4}=2$, so $p=8$."
  },
  {
    "text": "The equation of an ellipse that has the same foci as the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ and passes through the point $(-3 , \\frac{7}{4})$ is?",
    "fact_expressions": "G: Ellipse;C:Ellipse;H: Point;Expression(G) = (x^2/25 + y^2/16 = 1);Coordinate(H) = (-3, 7/4);Focus(G)=Focus(C);PointOnCurve(H,C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/16+y^2/7=1",
    "fact_spans": "[[[1, 40]], [[70, 72]], [[48, 69]], [[1, 40]], [[48, 69]], [[0, 72]], [[47, 72]]]",
    "query_spans": "[[[70, 76]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, point $M$ lies on the hyperbola $C$, point $I$ is the incenter of $\\triangle M F_{1} F_{2}$, and $S_{\\Delta I M F_{1}}+S_{\\Delta I M F_{2}}=\\frac{3}{2} S_{\\Delta I F_{1} F_{2}}$, $|M F_{1}|=2|M F_{2}|$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;M: Point;F1: Point;F2: Point;I: Point;a: Number;b: Number;a > 0;b > 0;Expression(C) = (x^2/a^2 - y^2/b^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(M, C);Incenter(TriangleOf(M, F1, F2)) = I;Area(TriangleOf(I, M, F1)) + Area(TriangleOf(I, M, F2)) = (3/2)*Area(TriangleOf(I, F1, F2));Abs(LineSegmentOf(M, F1)) = 2*Abs(LineSegmentOf(M, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[18, 79], [90, 96], [238, 244]], [[85, 89]], [[2, 9]], [[10, 17]], [[98, 102]], [[26, 79]], [[26, 79]], [[26, 79]], [[26, 79]], [[18, 79]], [[2, 84]], [[2, 84]], [[85, 97]], [[98, 131]], [[133, 213]], [[214, 236]]]",
    "query_spans": "[[[238, 250]]]",
    "process": ""
  },
  {
    "text": "Given that the directrix of the parabola $y = a x^{2}$ is $y = -\\frac{1}{2}$, then the real number $a$ = ?",
    "fact_expressions": "G: Parabola;a: Real;Expression(G) = (y = a*x^2);Expression(Directrix(G)) = (y = -1/2)",
    "query_expressions": "a",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 16]], [[40, 45]], [[2, 16]], [[2, 38]]]",
    "query_spans": "[[[40, 47]]]",
    "process": "From the given conditions, the standard equation of the parabola $ y = ax^{2} $ is $ x^{2} = \\frac{1}{a}y' $, so its directrix equation is $ y = -\\frac{1}{4a} = -\\frac{1}{2} $, therefore $ a = \\frac{1}{2} $."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $M$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively. The line passing through point $F_{1}$ and perpendicular to the $x$-axis intersects the hyperbola $M$ at points $A$ and $B$. If point $F_{2}$ satisfies $\\overrightarrow{F_{2} A} \\cdot \\overrightarrow{F_{2} B}=0$, then the eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "M: Hyperbola;b: Number;a: Number;G: Line;F2: Point;A: Point;B: Point;F1: Point;e: Number;a>0;b>0;Expression(M) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(M) = F1;RightFocus(M) = F2;PointOnCurve(F1,G);IsPerpendicular(G,xAxis);Intersection(G, M) = {A, B};DotProduct(VectorOf(F2, A), VectorOf(F2, B)) = 0;Eccentricity(M) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(2)+1",
    "fact_spans": "[[[19, 79], [107, 113], [197, 200]], [[26, 79]], [[26, 79]], [[104, 106]], [[9, 16], [126, 134]], [[115, 118]], [[119, 122]], [[1, 8], [87, 95]], [[204, 207]], [[26, 79]], [[26, 79]], [[19, 79]], [[1, 85]], [[1, 85]], [[86, 106]], [[96, 106]], [[104, 124]], [[136, 195]], [[197, 207]]]",
    "query_spans": "[[[204, 209]]]",
    "process": ""
  },
  {
    "text": "Let $A$, $B$ be two points on the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$, and $O$ be the coordinate origin. If $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}=0$, then the minimum area of $\\triangle AOB$ is?",
    "fact_expressions": "A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);G: Hyperbola;Expression(G) = (x^2 - y^2/2 = 1);O: Origin;DotProduct(VectorOf(O, A), VectorOf(O, B)) = 0",
    "query_expressions": "Min(Area(TriangleOf(A, O, B)))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 4]], [[7, 10]], [[1, 43]], [[1, 43]], [[12, 40]], [[12, 40]], [[44, 47]], [[55, 106]]]",
    "query_spans": "[[[108, 131]]]",
    "process": ""
  },
  {
    "text": "Given that $M$ is a point on the graph of the parabola $y^{2}=4x$, $F$ is the focus of the parabola, and $\\angle MFO = 120^{\\circ}$, then $|FM|=$?",
    "fact_expressions": "G: Parabola;M: Point;F: Point;O: Origin;Expression(G) = (y^2 = 4*x);PointOnCurve(M, G);Focus(G) = F;AngleOf(M, F, O) = ApplyUnit(120, degree)",
    "query_expressions": "Abs(LineSegmentOf(F, M))",
    "answer_expressions": "4",
    "fact_spans": "[[[6, 20], [31, 34]], [[2, 5]], [[27, 30]], [[39, 65]], [[6, 20]], [[2, 26]], [[27, 37]], [[39, 65]]]",
    "query_spans": "[[[67, 76]]]",
    "process": "From the parabola $ y^{2} = 4x $, we have $ F(1,0) $, $ \\angle MFO = 120^{\\circ} $, so the inclination angle of line $ MF $ is $ 60^{\\circ} $. Draw $ MH $ perpendicular to the directrix $ x = -1 $ of the parabola, with $ H $ as the foot of the perpendicular. Let $ G $ be the point where the directrix intersects the $ x $-axis. Draw $ ME \\perp x $-axis with $ E $ as the foot. By the definition of the parabola, we obtain $ |MF| = |MH| = |EG| = |FG| + |EF| = 2 + |MF|\\cos60^{\\circ} $, that is, $ |MF| = 2 + \\frac{1}{2}|MF| $, so $ |MF| = 4 $."
  },
  {
    "text": "The eccentricity of the ellipse $C$: $\\frac{y^{2}}{3}+\\frac{x^{2}}{2}=1$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/2 + y^2/3 = 1)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[0, 42]], [[0, 42]]]",
    "query_spans": "[[[0, 48]]]",
    "process": "\\because the ellipse is \\frac{y^{2}}{3}+\\frac{x^{2}}{2}=1, \\therefore a=\\sqrt{3}, c=\\sqrt{3-2}=1, \\therefore e=\\frac{c}{a}=\\frac{\\sqrt{3}}{3}"
  },
  {
    "text": "If a point $P(x_{0}, y_{0})$ on the parabola $y^{2}=4x$ is at a distance of $4$ from its focus $F$, then $x_{0}=$?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;x0:Number;y0:Number;Expression(G) = (y^2 = 4*x);Coordinate(P) = (x0,y0);PointOnCurve(P, G);Focus(G) = F;Distance(P, F) = 4",
    "query_expressions": "x0",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 15], [36, 37]], [[18, 35]], [[39, 42]], [[52, 59]], [[18, 35]], [[1, 15]], [[18, 35]], [[1, 35]], [[36, 42]], [[18, 50]]]",
    "query_spans": "[[[52, 61]]]",
    "process": ""
  },
  {
    "text": "Given that the coordinates of the right vertex of a hyperbola are $(2,0)$, and one asymptote equation is $2x - y = 0$, then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Coordinate(RightVertex(G))=(2,0);Expression(OneOf(Asymptote(G))) = (2*x - y = 0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/16 = 1",
    "fact_spans": "[[[2, 5], [39, 42]], [[2, 18]], [[2, 36]]]",
    "query_spans": "[[[39, 49]]]",
    "process": "From the vertex coordinates of the hyperbola, we get a=2; the asymptote equation 2x=y implies \\frac{b}{a}=2, so b=4. From c^{2}=a^{2}+b^{2} \\Rightarrow c=2\\sqrt{5}. Thus, the equation is: \\frac{x^{2}}{4}-\\frac{y^{2}}{16}=1."
  },
  {
    "text": "Given that $m$ is a real number, the number of intersection points between the line $m x+y-1=0$ and the ellipse $\\frac{x^{2}}{m^{2}}+y^{2}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2 + x^2/m^2 = 1);m: Real;H: Line;Expression(H) = (m*x + y - 1 = 0)",
    "query_expressions": "NumIntersection(H, G)",
    "answer_expressions": "2",
    "fact_spans": "[[[23, 54]], [[23, 54]], [[2, 5]], [[9, 22]], [[9, 22]]]",
    "query_spans": "[[[9, 61]]]",
    "process": "(Analysis) According to the equation of the line, it is easy to see that the line passes through the fixed point (0,1). Since this fixed point lies on the ellipse and $ m \\neq 0 $, the line is not parallel to the x-axis; therefore, the line intersects the ellipse. Because the equation of the line is $ mx + y - 1 = 0 $, the line passes through the fixed point (0,1), which lies on the ellipse. Since $ m \\neq 0 $, the line is not parallel to the x-axis, so the line intersects the ellipse, resulting in 2 intersection points."
  },
  {
    "text": "The moving chord $AB$ of the parabola $y^{2}=2 p x$ $(p>0)$ has length $a$ $(a \\geq 2 p)$. Then, the shortest distance from the midpoint $M$ of chord $AB$ to the $y$-axis is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);a: Number;a>=2*p;Length(LineSegmentOf(A, B)) = a;M: Point;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Min(Distance(M, yAxis))",
    "answer_expressions": "a/2 - p/2",
    "fact_spans": "[[[0, 21]], [[0, 21]], [[3, 21]], [[3, 21]], [[24, 29]], [[24, 29]], [[0, 29]], [[32, 47]], [[32, 47]], [[24, 47]], [[58, 61]], [[50, 61]]]",
    "query_spans": "[[[58, 73]]]",
    "process": "According to the problem, the directrix of the parabola is $ l: x = -\\frac{p}{2} $. Draw $ AC \\perp l $, $ BD \\perp l $, and $ MH \\perp l $ from points $ A $, $ B $, and $ M $ respectively, with perpendicular feet $ C $, $ D $, and $ H $. In the right trapezoid, we have $ MH = \\frac{1}{2}(AC + BD) $. By the definition of the parabola, $ AC = AF $, $ BD = BF $, so $ MH = \\frac{1}{2}(AC + BD) \\geqslant \\frac{AB}{2} = \\frac{a}{2} $. Thus, the minimum distance from the midpoint $ M $ of $ AB $ to the directrix of the parabola is $ \\frac{a}{2} $. Therefore, the shortest distance from the midpoint $ M $ of the segment to the $ y $-axis is $ \\frac{a}{2} - \\frac{p}{2} $."
  },
  {
    "text": "The equation $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{2+m}=1$ represents an ellipse with foci on the $y$-axis. Then the range of $m$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/(m + 2) + x^2/m^2 = 1);m: Number;PointOnCurve(Focus(G), yAxis) = True",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-1,0)+(0,2)",
    "fact_spans": "[[[54, 56]], [[0, 56]], [[58, 61]], [[45, 56]]]",
    "query_spans": "[[[58, 66]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, a line passing through the origin with an inclination angle of $\\frac{\\pi}{6}$ intersects the left and right branches of the hyperbola at points $P$ and $Q$, respectively. The circle with diameter $PQ$ passes through the right focus $F$. Find the eccentricity of the hyperbola.",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;L: Line;Q: Point;P: Point;F: Point;O:Origin;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(O,L);Inclination(L)=pi/6;Intersection(L,LeftPart(G))=P;Intersection(L,RightPart(G))=Q;IsDiameter(LineSegmentOf(P,Q),H);PointOnCurve(F,G);RightFocus(G)=F",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 58], [90, 93], [130, 133]], [[5, 58]], [[5, 58]], [[120, 121]], [[85, 87]], [[102, 105]], [[98, 101]], [[125, 128]], [[60, 62]], [[5, 58]], [[5, 58]], [[2, 58]], [[59, 87]], [[65, 87]], [[85, 107]], [[85, 107]], [[108, 121]], [[120, 128]], [[90, 128]]]",
    "query_spans": "[[[130, 138]]]",
    "process": "Let $ P(x_{1},y_{1}) $, $ Q(x_{2},y_{2}) $, the equation of line $ PQ $ is $ y = \\frac{\\sqrt{3}}{3}x $. Substituting into the hyperbola equation, it simplifies to $ (3b^{2}-a^{2})x^{2}-3a^{2}b^{2}=0 $, $ x_{1}+x_{2}=0 $, $ x_{1}x_{2}=-\\frac{3a^{2}b^{2}}{3b^{2}-a^{2}} $. Then $ y_{1}+y_{2}=0 $, $ y_{1}y_{2}=\\frac{1}{3}x_{1}x_{2}=-\\frac{a^{2}b^{2}}{3b^{2}-a^{2}} $. Let the right focus be $ F(c,0) $. From the condition we have $ \\overrightarrow{FP} \\bot \\overrightarrow{FQ} $, i.e., $ \\overrightarrow{FP} \\cdot \\overrightarrow{FQ} = 0 $. $ \\overrightarrow{FP}=(x_{1}-c,y_{1}) $, $ \\overrightarrow{FQ}=(x_{2}-c,y_{2}) $. Thus $ (x_{1}-c)(x_{2}-c)+y_{1}y_{2}=0 $, that is $ x_{1}x_{2}-c(x_{1}+x_{2})+c^{2}+y_{1}y_{2}=0 $. Substituting $ b^{2}=c^{2}-a^{2} $ and simplifying yields $ 3c^{4}-8a^{2}c^{2}+4a^{4}=0 $. Dividing both sides by $ a^{4} $ gives $ 3e^{4}-8e^{2}+4=0 $, i.e., $ (e^{2}-2)(3e^{2}-2)=0 $. Since $ e>1 $, we have $ e^{2}=2 $, thus $ e=\\sqrt{2} $."
  },
  {
    "text": "What is the distance from the point with x-coordinate $2$ on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$ to its right focus?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/7 = 1);P: Point;PointOnCurve(P,G) = True;XCoordinate(P) = 2",
    "query_expressions": "Distance(P,RightFocus(G))",
    "answer_expressions": "5/2",
    "fact_spans": "[[[0, 38]], [[0, 38]], [[47, 48]], [[0, 48]], [[39, 48]]]",
    "query_spans": "[[[0, 57]]]",
    "process": "Find the coordinates of the right focus, find the coordinates of the point on the ellipse with x-coordinate 2, and calculate using the distance formula between two points. From the given conditions, a=4, b=\\sqrt{7}, c=\\sqrt{16-7}=3, the right focus is F(3,0). From x=2 we get \\frac{7}{7}=1-\\frac{16}{2}, |\\frac{3}{2}, y=\\frac{\\sqrt{21}}{2}, so P(2,\\pm\\frac{\\sqrt{2}}{2})^{n}, there are two points."
  },
  {
    "text": "Given the hyperbola equation $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, what is the equation of the asymptotes of this hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "4*x+pm*3*y=0",
    "fact_spans": "[[[2, 5], [47, 50]], [[2, 44]]]",
    "query_spans": "[[[47, 58]]]",
    "process": "It is easy to see that the foci of the hyperbola lie on the x-axis, and $ a=3 $, $ b=4 $, so the asymptotes of the hyperbola are $ y=\\pm\\frac{4}{3}x $, i.e., $ 4x\\pm3y=0 $."
  },
  {
    "text": "From the right focus $F_{2}$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a perpendicular line $l$ to one of its asymptotes, with foot at $P$. Let $l$ intersect the other asymptote at point $Q$. If $\\overrightarrow{F_{2} Q}=2 \\overrightarrow{F_{2} P}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F2: Point;l:Line;Q: Point;P: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G)=F2;PointOnCurve(F2,l);L1:Line;L2:Line;IsPerpendicular(L1,l);FootPoint(L1,l)=P;Intersection(L2,l)=Q;VectorOf(F2, Q) = 2*VectorOf(F2, P);OneOf(Asymptote(G))=L1;OneOf(Asymptote(G))=L2;Negation(L1=L2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 57], [69, 70], [166, 169]], [[4, 57]], [[4, 57]], [[61, 68]], [[78, 81], [91, 94]], [[103, 107]], [[85, 88]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 68]], [[0, 81]], [], [], [[68, 81]], [[68, 88]], [[69, 107]], [[110, 163]], [67, 72], [67, 98], [67, 98]]",
    "query_spans": "[[[166, 175]]]",
    "process": "From the given conditions, the asymptotes of the hyperbola are $ y = \\pm\\frac{b}{a}x $. Let the right focus be $ F_{2}(c,0) $. Without loss of generality, assume the line is perpendicular to the line $ y = \\frac{b}{a}x $, then the equation of the line is $ y = -\\frac{a}{b}(x - c) $. Solving the system\n$$\n\\begin{cases}\ny = \\frac{b}{a}x, \\\\\ny = -\\frac{a}{b}(x - c)\n\\end{cases}\n$$\nyields\n$$\n\\begin{cases}\nx = \\frac{a^{2}c}{a^{2} + b^{2}} \\\\\ny = \\frac{abc}{a^{2} + b^{2}}\n\\end{cases},\n$$\ni.e., the point is $ \\left( \\frac{a^{2}c}{a^{2} + b^{2}}, \\frac{abc}{a^{2} + b^{2}} \\right) $. Since $ a^{2} + b^{2} = c^{2} $, the point $ P $ is $ \\left( \\frac{a^{2}}{c}, \\frac{ab}{c} \\right) $. Therefore,\n$ \\overrightarrow{F_{2}Q} = \\left( \\frac{a^{2}c}{a^{2} + b^{2}} - c, -\\frac{abc}{a^{2} + b^{2}} \\right) $,\n$ \\overrightarrow{F_{2}P} = \\left( \\frac{a^{2}}{c} - c, \\frac{ab}{c} \\right) $. Hence, $ c^{2} = 4a^{2} $, so the eccentricity of the hyperbola is $ e = \\frac{c}{a} = \\sqrt{\\frac{c^{2}}{a^{2}}} = 2 $."
  },
  {
    "text": "What is the focal length of the hyperbola $\\frac{x^{2}}{64}-\\frac{y^{2}}{36}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/64 - y^2/36 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "20",
    "fact_spans": "[[[0, 40]], [[0, 40]]]",
    "query_spans": "[[[0, 45]]]",
    "process": "First, from the hyperbola equation $\\frac{x^2}{64}-\\frac{y^{2}}{36}=1$, we obtain $a^{2}=64$, $b^{2}=36$, then use $c=\\sqrt{a^{2}+b^{2}}$ to solve. Since the hyperbola equation is $\\frac{x^{2}}{64}-\\frac{y^{2}}{36}=1$, so $a^{2}=64$, $b^{2}=36$, thus $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{64+36}=10$, therefore the focal distance of this hyperbola is $2c=20$. Hence the answer is: 20. This question mainly examines the geometric properties of hyperbolas, as well as computational and problem-solving abilities, and is a basic problem."
  },
  {
    "text": "The tangent to the curve $x^{2}=4 y$ at the point $P(m, n)$ is perpendicular to the line $2 x+y-1=0$, then $m=$?",
    "fact_expressions": "G: Line;H: Curve;P: Point;Expression(G) = (2*x + y - 1 = 0);Expression(H) = (x^2 = 4*y);Coordinate(P) = (m, n);m:Number;n:Number;IsPerpendicular(TangentOnPoint(P,H),G)",
    "query_expressions": "m",
    "answer_expressions": "1",
    "fact_spans": "[[[29, 42]], [[0, 13]], [[14, 24]], [[29, 42]], [[0, 13]], [[14, 24]], [[46, 49]], [[15, 24]], [[0, 44]]]",
    "query_spans": "[[[46, 51]]]",
    "process": "According to the problem, the slope of the tangent line is $\\frac{1}{2}$, $y=\\frac{x^{2}}{4}$, $y=\\frac{x}{2}=\\frac{1}{2}$, $x=1$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively, $P$ is a point on the ellipse such that $P F_{2}$ is perpendicular to the $x$-axis, and $|F_{1} F_{2}|=2|P F_{2}|$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F2), xAxis);Abs(LineSegmentOf(F1, F2)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(\\sqrt{5}-1)/2",
    "fact_spans": "[[[20, 72], [137, 139], [83, 85]], [[20, 72]], [[22, 72]], [[22, 72]], [[22, 72]], [[22, 72]], [[2, 9]], [[10, 17]], [[2, 78]], [[2, 78]], [[79, 82]], [[79, 88]], [[90, 106]], [[108, 134]]]",
    "query_spans": "[[[137, 145]]]",
    "process": "By the given condition, let |F_{1}F_{2}| = 2c. Since |F_{1}F_{2}| = 2|PF_{2}|, it follows that |PF_{2}| = c. Then in the right triangle \\triangle PF_{1}F_{2}, we have |PF_{1}|^{2} = |PF_{2}|^{2} + |F_{1}F_{2}|^{2} = 5c^{2}, so |PF_{1}| = \\sqrt{5}c. Therefore, the eccentricity of the ellipse is e = \\frac{2c}{2a} = \\frac{2c}{\\sqrt{5}c + c} = \\frac{\\sqrt{5}-1}{2}. Thus, the answer is \\underline{\\sqrt{5}-1}."
  },
  {
    "text": "Given a hyperbola with foci on the $x$-axis, its asymptotes are given by $y = \\pm \\frac{1}{2} x$, and the focal distance is $2 \\sqrt{5}$. Then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G),xAxis);Expression(Asymptote(G))=(y=pm*(1/2)*x);FocalLength(G) = 2*sqrt(5)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4-y^2=1",
    "fact_spans": "[[[11, 14], [15, 16], [62, 65]], [[2, 14]], [[15, 43]], [[15, 59]]]",
    "query_spans": "[[[62, 72]]]",
    "process": "From the given conditions, we have: $\\frac{b}{a}=\\frac{1}{2}, c=\\sqrt{5}$. Therefore, from $a^{2}+b^{2}=c^{2}=5$, we obtain $a^{2}=4, b^{2}=1$. Since the foci lie on the x-axis, the standard equation of the hyperbola is $\\frac{x^{2}}{4}-y^{2}=1$."
  },
  {
    "text": "The coordinates of the foci of the conic section $\\frac{(x-1)^{2}}{16}-\\frac{y^{2}}{9}=1$ are?",
    "fact_expressions": "G: ConicSection;Expression(G) = (-y^2/9 + (x - 1)^2/16 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "{(-4,0),(6,0)}",
    "fact_spans": "[[[0, 44]], [[0, 44]]]",
    "query_spans": "[[[0, 51]]]",
    "process": ""
  },
  {
    "text": "Let $O$ be the coordinate origin, and $F$ be the focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. A line $l$ passing through $F$ intersects the two asymptotes of $C$ at points $A$ and $B$, respectively. If $\\overrightarrow{O A} \\cdot \\overrightarrow{F A}=0$, and the inradius of $\\triangle O A B$ is $\\frac{a}{3}$, then the eccentricity of $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;O: Origin;A: Point;B: Point;F: Point;a > b;b > 0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1) ;OneOf(Focus(C)) = F;PointOnCurve(F, l);L1:Line;L2:Line;Asymptote(C)={L1,L2};Intersection(l,L1) = A;Intersection(l,L2 ) = B;DotProduct(VectorOf(O, A), VectorOf(F, A)) = 0;Radius(InscribedCircle(TriangleOf(O,A,B)))=a/3",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[80, 85]], [[14, 71], [86, 89], [205, 208]], [[21, 71]], [[190, 203]], [[1, 4]], [[99, 102]], [[103, 106]], [[76, 79], [10, 13]], [[21, 71]], [[21, 71]], [[14, 71]], [[10, 74]], [[75, 85]], [], [], [[86, 95]], [[80, 108]], [[80, 108]], [[111, 162]], [[165, 203]]]",
    "query_spans": "[[[205, 214]]]",
    "process": "\\because a>b>0 \\therefore the asymptotes OA and OB of the hyperbola are as shown, symmetric about the x-axis. Let M be the incenter of \\triangle OAB, then M lies on the bisector Ox of \\angle AOB. Draw MN \\bot ON at N and MT \\bot AB at T. Since FA \\bot OA, quadrilateral MTAN is a square. The distance from the focus to the asymptote is b, so |FA| = b. Also |OF| = c, \\therefore |OA| = a, and |NA| = |MN| = \\frac{a}{3}, \\therefore |NO| = \\frac{2a}{3}, \\therefore \\frac{b}{a} = \\tan \\angle AOF = \\frac{|MN|}{|NO|} = \\frac{1}{2}, then e = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\frac{\\sqrt{5}}{2}"
  },
  {
    "text": "What is the distance from the focus to the directrix of the parabola $y^{2}=x$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = x)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "1/2",
    "fact_spans": "[[[0, 12]], [[0, 12]]]",
    "query_spans": "[[[0, 23]]]",
    "process": "2p=1, so p=\\frac{1}{2}, so the distance from the focus to the directrix of the parabola is \\frac{1}{2}."
  },
  {
    "text": "Given that line $l$ intersects the two asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ at points $A$ and $B$, intersects the $x$-axis at point $C$, and $O$ is the origin. If $A$ is the midpoint of segment $BC$, and $OA=OC$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "l: Line;G: Hyperbola;b: Number;a: Number;B: Point;C: Point;O: Origin;A: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);L1:Line;L2:Line;Asymptote(G)={L1,L2};Intersection(L1,l)=A;Intersection(L2,l) = B;Intersection(xAxis,l) = C;MidPoint(LineSegmentOf(B,C))=A;LineSegmentOf(O, A) = LineSegmentOf(O, C)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 7]], [[8, 64], [132, 135]], [[11, 64]], [[11, 64]], [[79, 82]], [[90, 94]], [[95, 98]], [[74, 78], [105, 108]], [[11, 64]], [[11, 64]], [[8, 64]], [], [], [[8, 70]], [[2, 82]], [[2, 82]], [[2, 94]], [[105, 119]], [[121, 130]]]",
    "query_spans": "[[[132, 141]]]",
    "process": "Let OC = m. Draw a line through B parallel to the x-axis, intersecting the extension of OA at D. It follows that BD = OC = m. By symmetry, OB = 2m. Let θ be the inclination angle of line OB, then cosθ = (½BD)/OB. Then using tanθ = b/a, the answer can be obtained. Let OC = m. Draw a line through B parallel to the x-axis, intersecting the extension of OA at D. Then BD ∥ OC. Since A is the midpoint of BC, BD = OC = m, AD = AO = m, ∴ OD = 2m. By symmetry, OB = 2m. Let θ be the inclination angle of line OB, then cosθ = (½BD)/OB = 1/4. The equation of line OB is y = (b/a)x. Also tanθ = b/a, hence cosθ = a/c = 1/4, ∴ e = 4. This problem examines the method of finding the eccentricity of a hyperbola. The key step is drawing a line through B parallel to the x-axis intersecting the extension of OA at D, testing students' ability to analyze and solve problems."
  },
  {
    "text": "Through the focus of the parabola $y^{2}=4x$, draw a line intersecting the parabola at points $P$ and $Q$. What is the trajectory equation of the midpoint of segment $PQ$?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Q: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), H);Intersection(H, G) = {P,Q}",
    "query_expressions": "LocusEquation(MidPoint(LineSegmentOf(P,Q)))",
    "answer_expressions": "y^2=2*x-2",
    "fact_spans": "[[[1, 15], [23, 26]], [[19, 21]], [[28, 31]], [[32, 35]], [[1, 15]], [[0, 21]], [[19, 37]]]",
    "query_spans": "[[[40, 56]]]",
    "process": ""
  },
  {
    "text": "If point $P$ is a moving point on the parabola $y^{2}=2 x$, then the minimum value of the sum of the distance from point $P$ to the line $3 x-4 y+\\frac{7}{2}=0$ and the distance from $P$ to the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Expression(G) = (y^2 = 2*x);Expression(H) = (3*x - 4*y + 7/2 = 0);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P,H)+Distance(P,Directrix(G)))",
    "answer_expressions": "1",
    "fact_spans": "[[[6, 20], [67, 70]], [[33, 58]], [[1, 5], [28, 32], [62, 65]], [[6, 20]], [[33, 58]], [[1, 26]]]",
    "query_spans": "[[[28, 84]]]",
    "process": "As shown in the figure, from point P, draw perpendiculars PQ and PA to the parabola's directrix and the line $3x - 4y + \\frac{7}{2} = 0$, with feet of perpendiculars Q and A, respectively. By the definition of a parabola, we have $|PQ| = |PF|$. Therefore, $|PA| + |PQ| = |PA| + |PF| \\geqslant |AF|_{\\min}$. When points A, P, and F are collinear, $|AF|$ achieves its minimum value, which is the distance from point F to the line $3x - 4y + \\frac{7}{2} = 0$, that is, $|AF|_{\\min} = \\frac{|3 \\times \\frac{1}{2} + \\frac{7}{2}|}{\\sqrt{3^{2} + (-4)^{2}}} = 1$."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) intersects the parabola at points $A$ and $B$, and intersects its directrix $l$ at point $C$. If point $F$ is the midpoint of $AC$ and $AF=4$, then what is the length of segment $AB$?",
    "fact_expressions": "A: Point;B: Point;G: Parabola;p: Number;H: Line;C: Point;F: Point;l:Line;p>0;Expression(G) = (y^2 = 2*p*x);Focus(G)=F;PointOnCurve(F,H);Intersection(H,G) = {A,B};Directrix(G)=l;Intersection(H,l)=C;MidPoint(LineSegmentOf(A, C)) = F;LineSegmentOf(A, F) = 4",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[36, 40]], [[43, 46]], [[1, 22], [32, 35], [48, 49]], [[4, 22]], [[29, 31]], [[55, 59]], [[61, 65]], [[51, 54]], [[4, 22]], [[1, 22]], [[1, 28]], [[0, 31]], [[29, 46]], [[48, 54]], [[29, 59]], [[61, 73]], [[75, 82]]]",
    "query_spans": "[[[84, 94]]]",
    "process": "Let a line passing through the focus $ F\\left(\\frac{p}{2},0\\right) $ of the parabola $ y^{2}=2px $ ($ p>0 $) intersect the parabola at points $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, and intersect its directrix $ l: x=-\\frac{p}{2} $ at $ C\\left(-\\frac{p}{2},y_{3}\\right) $. Since $ F $ is the midpoint of $ AC $, and $ AF=4 $, then $ -\\frac{p}{2}+x_{1}=\\frac{p}{2}\\times2 $, solving gives $ \\begin{cases} p=2 \\\\ x_{1}=3 \\end{cases} $, so $ F(1,0) $, $ A(3,2\\sqrt{3}) $, then the equation of $ AF $ is $ y=\\sqrt{3}(x-1) $. Solving the system $ \\begin{cases} \\frac{\\frac{1}{2}}=4 \\\\ y^{2}=4x \\\\ y=\\sqrt{3}(x-1) \\end{cases} $ yields $ 3x^{2}-10x+3=0 $, solving gives $ x_{2}=\\frac{1}{3} $, so $ |AB|=|AF|+|BF|=4+\\frac{1}{3}+1=\\frac{16}{3} $."
  },
  {
    "text": "Given that the point $(m, n)$ lies on the ellipse $4 x^{2}+9 y^{2}=36$, then the range of $\\frac{m}{3}+\\frac{n}{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (4*x^2 + 9*y^2 = 36);H: Point;Coordinate(H) = (m, n);PointOnCurve(H, G);m: Number;n: Number",
    "query_expressions": "Range(m/3 + n/2)",
    "answer_expressions": "[-sqrt(2), sqrt(2)]",
    "fact_spans": "[[[12, 34]], [[12, 34]], [[2, 11]], [[2, 11]], [[2, 35]], [[3, 11]], [[3, 11]]]",
    "query_spans": "[[[37, 69]]]",
    "process": ""
  },
  {
    "text": "Given points $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$. In $\\Delta P F_{1} F_{2}$, point $P$ is a point on the ellipse. Then $\\frac{\\sin \\angle P F_{1} F_{2}+\\sin \\angle P F_{2} F_{1}}{\\sin \\angle F_{1} P F_{2}}$=?",
    "fact_expressions": "F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);P: Point;PointOnCurve(P, C)",
    "query_expressions": "(Sin(AngleOf(P, F1, F2)) + Sin(AngleOf(P, F2, F1)))/Sin(AngleOf(F1, P, F2))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 10]], [[11, 18]], [[2, 66]], [[2, 66]], [[19, 61], [97, 99]], [[19, 61]], [[92, 96]], [[92, 102]]]",
    "query_spans": "[[[104, 193]]]",
    "process": "First, determine a, b, c from the ellipse equation, then use the law of sines to convert angles into sides, and solve in combination with the definition of the ellipse. Since the ellipse equation is \\frac{x^2}{4}+\\frac{y^{2}}{3}=1, it follows that a=2,b=\\sqrt{3},c=1, so \\frac{\\sin\\anglePF_{1}F_{2}+\\sin\\anglePF_{2}F_{1}}{\\sin\\angleF_{1}PF_{2}}=\\frac{|PF_{2}|+|PF_{1}|}{|F_{1}F_{2}|}=\\frac{2a}{2c}=2"
  },
  {
    "text": "Let the left focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ be $F$. A tangent line is drawn from a point $A$ on the ellipse, intersecting the $y$-axis at point $Q$. If $\\angle Q F O=\\frac{\\pi}{4}$, $\\angle Q F A=\\frac{\\pi}{6}$, then the eccentricity of this ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;Q: Point;F: Point;O: Origin;A: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F;PointOnCurve(A, G);Intersection(TangentOfPoint(A,G), yAxis) = Q;AngleOf(Q, F, O) = pi/4;AngleOf(Q, F, A) = pi/6",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[1, 53], [72, 74], [150, 152], [63, 65]], [[3, 53]], [[3, 53]], [[83, 87]], [[58, 61]], [[89, 117]], [[68, 71]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 61]], [[63, 71]], [[62, 87]], [[89, 117]], [[119, 147]]]",
    "query_spans": "[[[150, 158]]]",
    "process": ""
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{m^{2}+1}+\\frac{y^{2}}{m^{2}}=1(m>0)$ are $F_{1}$, $F_{2}$, and the upper vertex is $A$. If $\\angle F_{1} A F_{2}=\\frac{\\pi}{3}$, then $m=$?",
    "fact_expressions": "G: Ellipse;m: Number;F1: Point;A: Point;F2: Point;m>0;Expression(G) = (x^2/(m^2 + 1) + y^2/m^2 = 1);Focus(G) = {F1, F2};UpperVertex(G) = A;AngleOf(F1, A, F2) = pi/3",
    "query_expressions": "m",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[0, 52]], [[119, 122]], [[56, 63]], [[76, 79]], [[64, 71]], [[2, 52]], [[0, 52]], [[0, 71]], [[0, 79]], [[81, 117]]]",
    "query_spans": "[[[119, 124]]]",
    "process": "According to the given condition, the ellipse $\\frac{x^2}{m^{2}+1}+\\frac{y^{2}}{m^{2}}=1$ $(m>0)$, we have $a^{2}=m^{2}+1$, $b^{2}=m^{2}$. Then $c^{2}=a^{2}-b^{2}=1$, so $F_{1}(-1,0)$, $F_{2}(1,0)$, and the upper vertex $A(0,m)$. As shown in the figure, since $\\angle F_{1}AF_{2}=\\frac{\\pi}{3}$, we obtain $\\angle F_{1}AO=\\frac{\\pi}{6}$, then $\\tan\\angle F_{1}AO=\\frac{1}{m}=\\frac{\\sqrt{3}}{3}$, solving gives $m=\\sqrt{3}$."
  },
  {
    "text": "Given that a hyperbola passes through the point $A(1,2)$ and its asymptotes are given by $y = \\pm x$, then the focal distance of the hyperbola is?",
    "fact_expressions": "A: Point;Coordinate(A) = (1, 2);PointOnCurve(A, G);G: Hyperbola;Expression(Asymptote(G)) = (y = pm*x)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(6)",
    "fact_spans": "[[[5, 14]], [[5, 14]], [[1, 14]], [[15, 16], [1, 4], [33, 36]], [[15, 31]]]",
    "query_spans": "[[[33, 41]]]",
    "process": "Let the hyperbola equation be $x^{2}-y^{2}=k$. Since the hyperbola passes through point $A(1,2)$, $\\therefore k=1^{2}-2^{2}=-3$. $\\therefore$ The hyperbola equation is $x^{2}-y^{2}=-3$, i.e., $\\frac{y^{2}}{3}-\\frac{x^{2}}{3}=1$. $a^{2}=b^{2}=3$, $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{6}$. $\\therefore$ The focal distance is $2\\sqrt{6}$."
  },
  {
    "text": "If one focus of the ellipse $2 k x^{2}+k y^{2}=1$ is $(0 ,-4)$, then the value of $k$ is?",
    "fact_expressions": "G: Ellipse;k: Number;Expression(G) = (2*(k*x^2) + k*y^2 = 1);Coordinate(OneOf(Focus(G))) = (0, -4)",
    "query_expressions": "k",
    "answer_expressions": "1/32",
    "fact_spans": "[[[1, 24]], [[41, 44]], [[1, 24]], [[1, 39]]]",
    "query_spans": "[[[41, 48]]]",
    "process": "Analysis: It is clear that $ k \\neq 0 $. The equation $ 2kx^{2} + ky^{2} = 1 $ can be rewritten as $ \\frac{y^{2}}{k} + \\frac{x^{2}}{2k} = 1 $. Since the foci lie on the $ y $-axis, we have $ \\frac{1}{k} - \\frac{1}{2k} = 16 $. Solving this gives $ k = \\frac{1}{32} $."
  },
  {
    "text": "Given the parabola $x^{2}=4 y$, a fixed point $A(12,39)$, point $P$ is a moving point on this parabola, $F$ is the focus of the parabola, find the minimum value of $|P A|+|P F|$?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (x^2 = 4*y);Coordinate(A) = (12, 39);PointOnCurve(P, G);Focus(G) = F",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "40",
    "fact_spans": "[[[2, 16], [37, 40], [51, 54]], [[20, 30]], [[31, 35]], [[46, 49]], [[2, 16]], [[20, 30]], [[31, 45]], [[46, 57]]]",
    "query_spans": "[[[59, 77]]]",
    "process": "Substitute $ x = 12 $ into $ x^{2} = 4y $, we get $ y = 36 < 39 $. Therefore, point $ A(12, 39) $ lies inside the parabola. The focus of the parabola is $ (0, 1) $, and the directrix $ l $ is $ y = -1 $. Draw $ PB \\perp l $ from $ P $ to point $ B $, then $ |PA| + |PF| = |PA| + |PB| $. From the figure, it is clear that $ |PA| + |PB| $ is minimized when points $ P $, $ A $, and $ B $ are collinear. Thus, the minimum value of $ |PA| + |PB| $ is $ 39 + 1 = 40 $. That is, the minimum value of $ |PA| + |PF| $ is $ 40 $."
  },
  {
    "text": "Given that the coordinates of point $A$ are $(-1,0)$, and point $B$ is a moving point on the circle $(x-1)^{2}+y^{2}=16$ with center $C$. The perpendicular bisector of segment $AB$ intersects $BC$ at point $M$. What is the trajectory equation of the moving point $M$?",
    "fact_expressions": "G: Circle;A: Point;B: Point;C: Point;M: Point;Expression(G) = (y^2 + (x - 1)^2 = 16);Coordinate(A) = (-1, 0);Center(G)=C;PointOnCurve(B,G);Intersection(PerpendicularBisector(LineSegmentOf(A,B)), LineSegmentOf(B, C)) = M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2/4 + y^2/3 = 1",
    "fact_spans": "[[[31, 52]], [[2, 6]], [[19, 23]], [[27, 30]], [[77, 81], [85, 88]], [[31, 52]], [[2, 18]], [[24, 52]], [[19, 56]], [[57, 81]]]",
    "query_spans": "[[[85, 95]]]",
    "process": "Using the definition of an ellipse, determine that the trajectory of point M is an ellipse with foci at A and C, find the values of a and b, thus obtaining the equation of the ellipse. According to the problem, the center of the circle is C(1,0) and the radius is 4. Connect MA, then |MA| = |MB|, ∴ |MC| + |MA| = |MC| + |MB| = |BC| = 4 > |AC| = 2, hence the trajectory of point M is an ellipse with foci at A and C, 2a = 4, so a = 2"
  },
  {
    "text": "Given the two foci of a hyperbola are $(-3,0)$, $(3,0)$, then the focal distance of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;Coordinate(F1) = (-3, 0);Coordinate(F2) = (3, 0);Focus(G) = {F1, F2}",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "6",
    "fact_spans": "[[[2, 5], [29, 32]], [[11, 19]], [[20, 27]], [[2, 27]], [[2, 27]], [[2, 27]]]",
    "query_spans": "[[[29, 37]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+y^{2}=1$ $(a>1)$ with two foci $F_{1}$, $F_{2}$, and a point $P$ on the ellipse such that $\\angle F_{1} P F_{2}=60^{\\circ}$, what is the value of $|P F_{1}| \\cdot |P F_{2}|$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2 + x^2/a^2 = 1);a: Number;a>1;F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[2, 38], [65, 67]], [[2, 38]], [[4, 38]], [[4, 38]], [[44, 51]], [[52, 59]], [[2, 59]], [[61, 64]], [[61, 70]], [[73, 106]]]",
    "query_spans": "[[[109, 139]]]",
    "process": ""
  },
  {
    "text": "If $C(-\\sqrt{3} , 0)$, $D(\\sqrt{3} , 0)$, and $M$ is a moving point on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, then the minimum value of $\\frac{1}{|M C|}+\\frac{1}{|M D|}$ is?",
    "fact_expressions": "G: Ellipse;C: Point;D: Point;M: Point;Expression(G)=(x^2/4 + y^2 = 1);Coordinate(C)=(-sqrt(3), 0);Coordinate(D)=(sqrt(3), 0);PointOnCurve(M, G)",
    "query_expressions": "Min(1/Abs(LineSegmentOf(M, D)) + 1/Abs(LineSegmentOf(M, C)))",
    "answer_expressions": "1",
    "fact_spans": "[[[44, 71]], [[1, 19]], [[20, 38]], [[40, 43]], [[44, 71]], [[1, 19]], [[20, 38]], [[40, 75]]]",
    "query_spans": "[[[77, 116]]]",
    "process": "From the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, we know $c^{2}=4-1=3$, $\\therefore c=\\sqrt{3}$, $\\therefore C, D$ are the two foci of the ellipse. Let $|MC|=r_{1}$, $|MD|=r_{2}$, then $r_{1}+r_{2}=2a=4$. $\\because \\frac{1}{|MC|}+\\frac{1}{|MD|}=\\frac{1}{r_{1}}+\\frac{1}{r_{2}}=$, and since $r_{1}r_{2}\\leqslant\\frac{(r_{1}+r_{2})^{2}}{4}=\\frac{16}{4}=4$, $\\frac{1}{|MC|}+\\frac{1}{|MD|}=\\frac{4}{r_{1}r_{2}}\\geqslant 1$. The equality holds if and only if $r_{1}=r_{2}$. Hence, the minimum value of $\\frac{1}{|MC|}+\\frac{1}{|MD|}$ is $1$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x$ ($p>0$) passes through the point $(3,6)$, and the line $l$ passes through the point $M(2,2)$ and intersects the parabola $C$ at points $A$ and $B$. If the midpoint of segment $AB$ is $M$, and $F$ is the focus of the parabola $C$, then the perimeter of $\\triangle A B F$ is?",
    "fact_expressions": "l: Line;C: Parabola;p: Number;B: Point;A: Point;H: Point;M: Point;F: Point;p>0;Expression(C) = (y^2 = 2*p*x);Coordinate(H) = (3, 6);Coordinate(M) = (2, 2);PointOnCurve(H, C);PointOnCurve(M, l);Intersection(l, C)={A, B};MidPoint(LineSegmentOf(A,B))=M;Focus(C)=F",
    "query_expressions": "Perimeter(TriangleOf(A, B, F))",
    "answer_expressions": "10+(20*sqrt(2)/3)",
    "fact_spans": "[[[39, 44]], [[2, 28], [57, 63], [96, 102]], [[10, 28]], [[69, 72]], [[65, 68]], [[30, 38]], [[46, 55], [88, 91]], [[92, 95]], [[10, 28]], [[2, 28]], [[30, 38]], [[46, 55]], [[2, 38]], [[39, 55]], [[39, 74]], [[77, 91]], [[92, 105]]]",
    "query_spans": "[[[107, 129]]]",
    "process": "Substituting the point (3,6) into $ y^{2}=2px $ gives $ p=6 $, so the equation of the parabola $ C $ is $ y^{2}=12x $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. From the problem, the slope of line $ l $ exists and is not zero, hence $ x_{1}\\neq x_{2} $. Then $ y_{1}^{2}=12x_{1} $, $ y_{2}^{2}=12x_{2} $. Subtracting these two equations yields $ (y_{1}+y_{2})(y_{1}-y_{2})=12(x_{1}-x_{2}) $. Since the midpoint of $ AB $ is $ M(2,2) $, we have $ y_{1}+y_{2}=4 $. Substituting $ y_{1}+y_{2}=4 $ into the above equation gives the slope of line $ l $ as $ k=3 $. Thus, the equation of line $ AB $ is $ y-2=3(x-2) $, or $ y=3x-4 $. Solving the system\n\\[\n\\begin{cases}\ny^{2}=12x, \\\\\ny=3x-4,\n\\end{cases}\n\\]\neliminating $ y $ leads to $ 9x^{2}-36x+16=0 $, $ \\Delta>0 $. By the relationship between roots and coefficients, $ x_{1}+x_{2}=4 $, $ x_{1}x_{2}=\\frac{16}{9} $. By the definition of the parabola, $ |AF|+|BF|=x_{1}+x_{2}+p=4+6=10 $. While $ |AB|=\\sqrt{1+k^{2}}\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\sqrt{3^{2}+1}\\cdot\\sqrt{4^{2}-4\\times\\frac{16}{9}}=\\frac{20\\sqrt{2}}{3} $. Therefore, the perimeter of $ \\triangle ABF $ is $ 10+\\frac{20\\sqrt{2}}{3} $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ has eccentricity $\\sqrt{3}$. If one of its directrices coincides with the directrix of the parabola $y^{2}=4 x$. Let $P$ be an intersection point of the hyperbola and the parabola, and let $F$ be the focus of the parabola. Then $|P F|$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;Eccentricity(G) = sqrt(3);H: Parabola;Expression(H) = (y^2 = 4*x);OneOf(Directrix(G)) = Directrix(H);OneOf(Intersection(G, H)) = P;P: Point;Focus(H) = F;F: Point",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 59], [76, 77], [105, 108]], [[2, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[2, 74]], [[83, 97], [109, 112], [122, 125]], [[83, 97]], [[76, 102]], [[105, 121]], [[118, 121]], [[122, 132]], [[129, 132]]]",
    "query_spans": "[[[134, 143]]]",
    "process": ""
  },
  {
    "text": "The eccentricity $e<2$ of the hyperbola $\\frac{x^{2}}{k}+\\frac{y^{2}}{4}=1$, then the range of values for $k$ is?",
    "fact_expressions": "G: Hyperbola;k: Number;e:Number;Expression(G) = (y^2/4 + x^2/k = 1);Eccentricity(G) = e;e<2",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-12,0)",
    "fact_spans": "[[[0, 38]], [[49, 52]], [[42, 47]], [[0, 38]], [[0, 47]], [[42, 47]]]",
    "query_spans": "[[[49, 59]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, point $F$ is the right focus of hyperbola $C$, and point $A$ is the right vertex of hyperbola $C$. A line perpendicular to the $x$-axis is drawn through point $F$, intersecting the hyperbola at points $M$ and $N$. If $\\tan \\angle M A N = -\\frac{3}{4}$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(C) = F;A: Point;RightVertex(C) = A;L: Line;PointOnCurve(F, L) = True;IsPerpendicular(L, xAxis) = True;Intersection(L, C) = {M, N};M: Point;N: Point;Tan(AngleOf(M, A, N)) = -3/4",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 63], [69, 75], [84, 90], [110, 113], [159, 165]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[64, 68], [96, 100]], [[64, 79]], [[80, 83]], [[80, 94]], [], [[95, 108]], [[95, 108]], [[95, 123]], [[114, 117]], [[118, 121]], [[125, 157]]]",
    "query_spans": "[[[159, 171]]]",
    "process": "First, draw the graph according to the given conditions. Let $\\angle MAN = 2\\theta$, $\\theta \\in (0, \\frac{\\pi}{2})$. Based on $\\tan \\angle MAN = -\\frac{3}{4}$, solve for $\\tan \\theta = 3$, then use $\\frac{b^{2}}{c-a} = 3$ to find the result. According to the problem, let $\\angle MAN = 2\\theta$, $\\theta \\in (0, \\frac{\\pi}{2})$, then $\\tan 2\\theta = -\\frac{3}{4} = \\frac{2 \\tan \\theta}{1 - \\tan^{2} \\theta}$. Solving gives $\\tan \\theta = 3$, that is, $\\frac{b^{2}}{c-a} = 3$. Rearranging yields $c^{2} + 2a^{2} - 3ac = 0$, or $e^{2} + 2 - 3e = 0$, $e > 1$. Solving gives $e = 2$."
  },
  {
    "text": "The line $l$ passes through the focus $F$ of the parabola $C$: $y^{2}=2 p x$ ($p>0$) and intersects $C$ at points $A$ and $B$. The midpoint $M$ of $AB$ has coordinates $(3,2)$. Then the equation of the parabola $C$ is?",
    "fact_expressions": "l: Line;C: Parabola;p:Number;p>0;A: Point;B: Point;M: Point;F:Point;Expression(C)=(y^2=2*p*x);Coordinate(M) = (3, 2);Focus(C) = F;PointOnCurve(F,l);Intersection(l, C) = {A, B};MidPoint(LineSegmentOf(A,B))=M",
    "query_expressions": "Expression(C)",
    "answer_expressions": "{y^2=4*x,y^2=8*x}",
    "fact_spans": "[[[0, 5]], [[6, 32], [40, 43], [81, 87]], [[13, 32]], [[13, 32]], [[46, 49]], [[50, 53]], [[65, 68]], [[35, 38]], [[6, 32]], [[65, 79]], [[6, 38]], [[0, 38]], [[0, 55]], [[57, 68]]]",
    "query_spans": "[[[81, 92]]]",
    "process": "By the point difference method, we get $y_{1}^{2}=2px_{1}, y_{2}^{2}=2px_{2} \\Rightarrow y_{1}^{2}-y_{2}^{2}=2p(x_{1}-x_{2}) \\Rightarrow (y_{1}+y_{2})k_{AB}=2p \\Rightarrow 4k_{AB}=2p \\Rightarrow k_{AB}=\\frac{p}{2}$, old point: midpoint of parabola chord"
  },
  {
    "text": "Given that the length of segment $AB$ is $3$, with its two endpoints $A$ and $B$ sliding on the $x$-axis and $y$-axis respectively, and point $M$ satisfies $2 \\overrightarrow{A M} = \\overrightarrow{M B}$. Then the trajectory equation of point $M$ is?",
    "fact_expressions": "B: Point;A: Point;M: Point;Length(LineSegmentOf(A,B)) = 3;Endpoint(LineSegmentOf(A,B))={A,B};PointOnCurve(A,xAxis);PointOnCurve(B,yAxis);2*VectorOf(A, M) = VectorOf(M, B)",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2/4+y^2=1",
    "fact_spans": "[[[26, 29]], [[22, 25]], [[45, 49], [99, 103]], [[2, 16]], [[2, 29]], [[22, 44]], [[22, 44]], [[51, 96]]]",
    "query_spans": "[[[99, 110]]]",
    "process": "Let M(x,y), A(a,0), B(0,b). From 2\\overrightarrow{AM}=\\overrightarrow{MB}, we have 2(x-a,y)=(-x,b-y), yielding \\begin{cases}a=\\frac{3x}{2}\\\\b=3y\\end{cases}, so A(\\frac{3x}{2},0), B(0,3y). From |AB|=3, we get: \\frac{9x^{2}}{4}+9y^{2}=9. Therefore, the equation of the trajectory C of point M is \\frac{x^{2}}{4}+y^{2}=1."
  },
  {
    "text": "One focus of the ellipse $5 x^{2}-k y^{2}=5$ is $(0 , 2)$, then $k=$?",
    "fact_expressions": "G: Ellipse;k: Number;H: Point;Expression(G) = (-k*y^2 + 5*x^2 = 5);Coordinate(H) = (0, 2);OneOf(Focus(G)) = H",
    "query_expressions": "k",
    "answer_expressions": "-1",
    "fact_spans": "[[[0, 21]], [[39, 42]], [[27, 36]], [[0, 21]], [[27, 36]], [[0, 36]]]",
    "query_spans": "[[[39, 44]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $x^{2}=2 p y(p>0)$ with focus $F$, directrix $l$, and point $P$ on $C$. A perpendicular line from $P$ to $l$ intersects $l$ at point $E$, and $\\angle P F E=60^{\\circ}$, $|P F|=4$. Then the equation of the parabola $C$ is?",
    "fact_expressions": "C: Parabola;p: Number;P: Point;F: Point;E: Point;l: Line;p>0;Expression(C) = (x^2 = 2*(p*y));Focus(C) = F;Directrix(C) = l;PointOnCurve(P, C);L:Line;PointOnCurve(P, L);IsPerpendicular(l,L);Intersection(l, L) = E;AngleOf(P, F, E) = ApplyUnit(60, degree);Abs(LineSegmentOf(P, F)) = 4",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2=4*y",
    "fact_spans": "[[[2, 28], [48, 51], [113, 119]], [[9, 28]], [[43, 47], [54, 58]], [[32, 35]], [[70, 74]], [[39, 42], [59, 62], [66, 69]], [[9, 28]], [[2, 28]], [[2, 35]], [[2, 42]], [[43, 52]], [], [[53, 65]], [[53, 65]], [[53, 74]], [[76, 101]], [[102, 111]]]",
    "query_spans": "[[[113, 124]]]",
    "process": "As shown in the figure, construct PE\\bot l, \\angle PFE=60^{\\circ}. By the definition of the parabola, \\triangle PFE is an equilateral triangle. Then draw FM\\bot PE from the focus F, intersecting PE at M, so M is the midpoint of PE. Therefore, |PM|=|ME|=2, meaning the distance from the focus to the directrix is p=2, thus the equation of the parabola can be found. Parabola C: x^{2}=2py (p>0), focus F(0,\\frac{p}{2}), directrix l: y=-\\frac{p}{2}. As shown in the figure, PE\\bot l, \\angle PFE=60^{\\circ}, |PF|=4. By the definition of the parabola, |PF|=|PE|=4, hence \\triangle PFE is an equilateral triangle. Draw FM\\bot PE from the focus F, intersecting PE at M, then M is the midpoint of PE, so |PM|=|ME|=2, meaning the distance from the focus to the directrix is p=2."
  },
  {
    "text": "The standard equation of an ellipse passing through the point $(\\sqrt{3}, -\\sqrt{5})$ and having the same foci as the ellipse $\\frac{y^{2}}{25}+\\frac{x^{2}}{9}=1$ is?",
    "fact_expressions": "G: Ellipse;PointOnCurve(H,G) = True;H: Point;Coordinate(H) = (sqrt(3), -sqrt(5));G1: Ellipse;Expression(G1) = (x^2/9 + y^2/25 = 1);Focus(G) = Focus(G1)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/20 + x^2/4 = 1",
    "fact_spans": "[[[72, 74]], [[0, 74]], [[1, 25]], [[1, 25]], [[28, 66]], [[28, 66]], [[27, 74]]]",
    "query_spans": "[[[72, 81]]]",
    "process": "The desired ellipse has the same foci as the ellipse $\\frac{y^{2}}{25}+\\frac{x^{2}}{9}=1$, so its foci lie on the $y$-axis, and the semi-focal distance $c$ satisfies $c^{2}=25-9=16$. Let its standard equation be $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$ $(a>b>0)$, thus we obtain $a^{2}-b^{2}=16$. Additionally, the point $(\\sqrt{3},-\\sqrt{5})$ lies on the desired ellipse, that is, $\\frac{5}{a^{2}}+\\frac{3}{b^{2}}=1$. Combining the two equations gives $\\frac{5}{b^{2}+16}+\\frac{3}{b^{2}}=1$, which simplifies to $(b^{2})^{2}+8b^{2}-48=0$. Solving yields $b^{2}=4$, then $a^{2}=20$. Therefore, the standard equation of the desired ellipse is $\\frac{y^{2}}{20}+\\frac{x^{2}}{4}=1$."
  },
  {
    "text": "The line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$, and point $O$ is the origin. Then the minimum value of $|AF| \\cdot |BF|$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;O: Origin;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B}",
    "query_expressions": "Min(Abs(LineSegmentOf(A, F))*Abs(LineSegmentOf(B, F)))",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 15], [25, 28]], [[22, 24]], [[29, 32]], [[18, 21]], [[33, 36]], [[39, 43]], [[1, 15]], [[1, 21]], [[0, 24]], [[22, 38]]]",
    "query_spans": "[[[50, 74]]]",
    "process": "Let the inclination angle of line AB be $\\theta$, we obtain $|AF|=\\frac{2}{1-\\cos\\theta}$, $|BF|=\\frac{2}{1+\\cos\\theta}$, compute $|AF||BF|=\\frac{4}{\\sin^{2}\\theta}$ to get the answer. Let the inclination angle of line AB be $\\theta$, we obtain $|AF|=\\frac{}{1-}\\frac{2}{\\cos\\theta}$, $|BF|=\\frac{2}{1+\\cos\\theta}$, then $|AF||BF|=\\frac{2}{1-\\cos\\theta}\\times\\frac{}{1+}\\frac{2}{\\cos\\theta}=\\frac{4}{\\sin^{2}\\theta}\\geqslant4$, equality holds when $\\theta=\\frac{\\pi}{2}$."
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x(p>0)$ is also a focus of the hyperbola $x^{2}-y^{2}=8$, then $p$=?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;Expression(G) = (x^2 - y^2 = 8);p>0;Expression(H) = (y^2 = 2*(p*x));Focus(H) = OneOf(Focus(G))",
    "query_expressions": "p",
    "answer_expressions": "8",
    "fact_spans": "[[[27, 45]], [[1, 22]], [[52, 55]], [[27, 45]], [[4, 22]], [[1, 22]], [[1, 50]]]",
    "query_spans": "[[[52, 57]]]",
    "process": "\\because the hyperbola x^{2}-y^{2}=8, \\therefore the foci of the hyperbola x^{2}-y^{2}=8 are (4,0) and (-4,0). Since the focus of the parabola y^{2}=2px (p>0), which is (\\frac{p}{2},0), is also a focus of the hyperbola x^{2}-y^{2}=8, \\therefore \\frac{p}{2}=4, p=8,"
  },
  {
    "text": "Given the hyperbola $C$: $x^{2}-y^{2}=1$, what is the distance from the point $(2,0)$ to the asymptotes of $C$?",
    "fact_expressions": "C: Hyperbola;G: Point;Expression(C) = (x^2 - y^2 = 1);Coordinate(G) = (2, 0)",
    "query_expressions": "Distance(G,Asymptote(C))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 25], [36, 39]], [[27, 35]], [[2, 25]], [[27, 35]]]",
    "query_spans": "[[[27, 48]]]",
    "process": "The asymptotes of the hyperbola $ C: x^{2} - y^{2} = 1 $ are given by $ y = \\pm x $. Without loss of generality, by symmetry, take the asymptote equation as $ x - y = 0 $. Then the distance from the point $ (2, 0) $ to the asymptote $ x - y = 0 $ is $ \\frac{2}{\\sqrt{2}} = \\sqrt{2} $."
  },
  {
    "text": "Draw a line through the focus of the parabola $y^{2}=4x$ intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $|AB|=12$, then $x_{1}+x_{2}$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Line;PointOnCurve(Focus(G), H);A: Point;B: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 12",
    "query_expressions": "x1 + x2",
    "answer_expressions": "10",
    "fact_spans": "[[[1, 15], [22, 25]], [[1, 15]], [[19, 21]], [[0, 21]], [[26, 44]], [[46, 64]], [[26, 44]], [[46, 64]], [[26, 44]], [[46, 64]], [[26, 44]], [[46, 64]], [[19, 66]], [[68, 78]]]",
    "query_spans": "[[[81, 96]]]",
    "process": ""
  },
  {
    "text": "Let a point $P$ on the parabola $y^{2}=16 x$ be at a distance of $12$ from the $x$-axis. Then the distance $|P F|$ from point $P$ to the focus $F$ is $?$.",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 16*x);P: Point;PointOnCurve(P, G);Distance(P, xAxis) = 12;F: Point;Focus(G) = F;Distance(P, F) = Abs(LineSegmentOf(P, F))",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "16",
    "fact_spans": "[[[1, 16]], [[1, 16]], [[19, 22], [37, 41]], [[1, 22]], [[19, 35]], [[44, 47]], [[1, 47]], [[37, 57]]]",
    "query_spans": "[[[50, 59]]]",
    "process": "From the parabola equation, we know that $2p=16$, $\\therefore \\frac{p}{2}=4$, so the distance from $P$ to the directrix is 16. By definition, the distance from point $P$ to the focus $F$ is $|PF|=16$."
  },
  {
    "text": "Given that $F$ is the focus of $y^{2}=2 x$, and $A$, $B$ are two points on the parabola such that $|A F|+|B F|=3$, then the distance from the midpoint of segment $A B$ to the directrix of this parabola is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;F: Point;Expression(G)=(y^2=2*x);Focus(G)=F;PointOnCurve(A, G);PointOnCurve(B, G);Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F)) = 3",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A,B)),Directrix(G))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[6, 17], [29, 32], [66, 69]], [[21, 24]], [[25, 28]], [[2, 5]], [[6, 17]], [[2, 20]], [[21, 36]], [[21, 36]], [[37, 52]]]",
    "query_spans": "[[[54, 76]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), obtain x_{1}+x_{2} using the focal radius formula, thus the horizontal coordinate of the midpoint of AB can be found. Then combining with F(\\frac{1}{2},0), directrix x=-\\frac{1}{2}, let A(x_{1},y_{1}), B(x_{2},y_{2}), |AF|+|BF|=x_{1}+\\frac{1}{2}+x_{2}+\\frac{1}{2}=x_{1}+x_{2}+1=3 \\therefore x_{1}+x_{2}=2, then the horizontal coordinate of the midpoint of segment AB is 1, and the distance from the midpoint of segment AB to the directrix of this parabola is \\frac{3}{2}."
  },
  {
    "text": "Let $P$ be an intersection point of ellipse $C_{1}$ and hyperbola $C_{2}$ sharing common foci $F_{1}$, $F_{2}$, and $P F_{1} \\perp P F_{2}$. If the eccentricity of ellipse $C_{1}$ is $e_{1}$, and the eccentricity of hyperbola $C_{2}$ is $e_{2}$, then the minimum value of $9 e_{1}^{2}+e_{2}^2$ is?",
    "fact_expressions": "C2: Hyperbola;C1: Ellipse;P: Point;F1: Point;F2: Point;e1: Number;e2: Number;Focus(C1) = {F1, F2};Focus(C2) = {F1, F2};OneOf(Intersection(C1, C2)) = P;IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));Eccentricity(C1) = e1;Eccentricity(C2) = e2;Focus(C1) = Focus(C2)",
    "query_expressions": "Min(9*e1^2 + e2^2)",
    "answer_expressions": "8",
    "fact_spans": "[[[36, 46], [100, 110]], [[26, 35], [78, 87]], [[1, 4]], [[10, 18]], [[18, 25]], [[92, 99]], [[115, 122]], [[8, 35]], [[5, 46]], [[1, 51]], [[53, 76]], [[78, 99]], [[100, 122]], [[5, 46]]]",
    "query_spans": "[[[124, 151]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{2}=1$ $(a>0)$ is $y=\\sqrt{2} x$, then the real number $a$=?",
    "fact_expressions": "G: Hyperbola;a: Real;a>0;Expression(G) = (-y^2/2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (y = sqrt(2)*x)",
    "query_expressions": "a",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 49]], [[72, 77]], [[5, 49]], [[2, 49]], [[2, 70]]]",
    "query_spans": "[[[72, 79]]]",
    "process": "From the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{2}=1$ ($a>0$), it follows that the foci lie on the $x$-axis, so one asymptote equation is $y=\\frac{\\sqrt{2}}{a}x=\\sqrt{2}x$, hence $a=1$. 【Analysis】This problem examines the asymptote equations of a hyperbola; the result can be directly obtained according to its definition, and is relatively basic."
  },
  {
    "text": "The line $x+2 y-2=0$ passes through one focus and one vertex of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$. Then the eccentricity of the ellipse is equal to?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x + 2*y - 2 = 0);PointOnCurve(OneOf(Focus(G)),H);PointOnCurve(OneOf(Vertex(G)),H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(5)/5",
    "fact_spans": "[[[15, 69], [82, 84]], [[17, 69]], [[17, 69]], [[0, 13]], [[17, 69]], [[17, 69]], [[15, 69]], [[0, 13]], [[0, 74]], [[0, 79]]]",
    "query_spans": "[[[82, 91]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ is a moving point on a circle, draw $P Q \\perp x$-axis at point $Q$. Let $\\overrightarrow{O M}=\\overrightarrow{O P}+\\overrightarrow{O Q}$. Then the trajectory equation of point $M$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P, G);G: Circle;Q: Point;PointOnCurve(P, LineSegmentOf(P, Q));IsPerpendicular(LineSegmentOf(P, Q), xAxis);FootPoint(LineSegmentOf(P, Q), xAxis) = Q;O: Origin;M: Point;VectorOf(O, M) = VectorOf(O, P) + VectorOf(O, Q)",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2/4+y^2=1",
    "fact_spans": "[[[2, 6], [16, 20]], [[2, 14]], [[7, 8]], [[36, 40]], [[15, 35]], [[21, 35]], [[21, 40]], [[42, 106]], [[108, 112]], [[42, 106]]]",
    "query_spans": "[[[108, 119]]]",
    "process": ""
  },
  {
    "text": "The foci of a hyperbola with center at the origin and eccentricity $\\frac{5}{3}$ lie on the $y$-axis. What is its asymptote equation?",
    "fact_expressions": "G: Hyperbola;O: Origin;Eccentricity(G) = 5/3;PointOnCurve(Focus(G), yAxis);Center(G) = O",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*(3/4)*x",
    "fact_spans": "[[[26, 29], [40, 41]], [[3, 7]], [[8, 29]], [[26, 38]], [[0, 29]]]",
    "query_spans": "[[[40, 49]]]",
    "process": "According to the problem, the hyperbola is centered at the origin, has eccentricity $\\frac{5}{3}$, and its foci lie on the $y$-axis. Thus, $\\frac{c}{a} = \\frac{5}{3}$, so $\\frac{c^{2}}{a^{2}} = \\frac{a^{2}+b^{2}}{a^{2}} = \\frac{25}{9}$. Rearranging gives $\\frac{b^{2}}{a^{2}} = \\frac{16}{9}$, solving yields $\\frac{b}{a} = \\frac{4}{3}$, hence $\\frac{a}{b} = \\frac{3}{4}$. Therefore, the asymptotes of the hyperbola are $y = \\pm\\frac{a}{b}x = \\pm\\frac{3}{4}x$."
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $\\frac{5}{4}$, and it passes through the point $A(-4 \\sqrt{2}, 3)$, then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(G) = 5/4;A: Point;Coordinate(A) = (-4*sqrt(2), 3);PointOnCurve(A, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16 - y^2/9 = 1",
    "fact_spans": "[[[2, 58], [102, 105]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 76]], [[79, 99]], [[79, 99]], [[2, 99]]]",
    "query_spans": "[[[102, 110]]]",
    "process": "Let $ c = 5t $ ($ t > 0 $), we get $ a = 4t $, then $ b = \\sqrt{c^{2} - a^{2}} = 3t $, so the equation of the hyperbola can be written as $ \\frac{x^{2}}{16t^{2}} - \\frac{y^{2}}{9t^{2}} = 1 $. Substituting the coordinates of point A into the hyperbola equation gives $ \\frac{32}{16t^{2}} - \\frac{9}{9t^{2}} = 1 $, solving yields $ t = 1 $. Therefore, the equation of the hyperbola is $ \\frac{x^{2}}{16} - \\frac{y^{2}}{9} = 1 $."
  },
  {
    "text": "Given the ellipse $ r $: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has its right focus at $ F(0,1) $, and eccentricity $ \\frac{1}{2} $. The three vertices of triangle $ A B C $ lie on the ellipse $ r $. Let the midpoints of its three sides $ A B $, $ B C $, $ A C $ be $ D $, $ E $, $ M $, respectively, and the slopes of the lines containing the three sides be $ k_{1} $, $ k_{2} $, $ k_{3} $, respectively, where $ k_{1} $, $ k_{2} $, $ k_{3} $ are all non-zero. Let $ O $ be the origin. If the sum of the slopes of lines $ O D $, $ O E $, $ O M $ is $ 1 $, then $ \\frac{1}{k_{1}}+\\frac{1}{k_{2}}+\\frac{1}{k_{3}} = $?",
    "fact_expressions": "r: Ellipse;a: Number;b: Number;O: Origin;D: Point;A: Point;B: Point;C: Point;E: Point;M: Point;F: Point;a > b;b > 0;Expression(r) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (0, 1);RightFocus(r) = F;Eccentricity(r) = 1/2;PointOnCurve(Vertex(TriangleOf(A, B, C)), r);MidPoint(LineSegmentOf(A, B)) = D;MidPoint(LineSegmentOf(B, C)) = E;MidPoint(LineSegmentOf(A, C)) = M;k1: Number;k2: Number;k3: Number;Slope(OverlappingLine(LineSegmentOf(A, B))) = k1;Slope(OverlappingLine(LineSegmentOf(B, C))) = k2;Slope(OverlappingLine(LineSegmentOf(A, C))) = k3;Negation(k1 = 0);Negation(k2 = 0);Negation(k3 = 0);Slope(LineOf(O, D)) + Slope(LineOf(O, E))+ Slope(LineOf(O, M)) = 1",
    "query_expressions": "1/k2 + 1/k1 + 1/k3",
    "answer_expressions": "-4/3",
    "fact_spans": "[[[2, 59], [109, 114]], [[9, 59]], [[9, 59]], [[239, 242]], [[149, 152]], [[95, 102]], [[95, 102]], [[95, 102]], [[153, 156]], [[157, 160]], [[64, 72]], [[9, 59]], [[9, 59]], [[2, 59]], [[64, 72]], [[2, 72]], [[2, 91]], [[92, 115]], [[122, 160]], [[122, 160]], [[122, 160]], [[174, 181], [203, 210]], [[183, 192], [213, 220]], [[194, 201], [223, 230]], [[162, 201]], [[162, 201]], [[162, 201]], [[203, 236]], [[203, 236]], [[203, 236]], [[249, 281]]]",
    "query_spans": "[[[284, 335]]]",
    "process": ""
  },
  {
    "text": "The length of the line segment $A B$ cut by the curve $y=\\frac{1}{2} x^{2}-1$ from the line $y=x+1$ is?",
    "fact_expressions": "G: Line;H: Curve;B: Point;A: Point;Expression(G) = (y = x + 1);Expression(H) = (y = x^2/2 - 1);InterceptChord(H,G)=LineSegmentOf(A,B)",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "2*sqrt(10)",
    "fact_spans": "[[[0, 9]], [[10, 35]], [[40, 45]], [[40, 45]], [[0, 9]], [[10, 35]], [[0, 45]]]",
    "query_spans": "[[[38, 49]]]",
    "process": "Solve the system of equations \\begin{cases}y=x+1\\\\y=\\frac{1}{2}x^{2}-1\\end{cases}, rearranging yields x^{2}-2x-4=0, solving gives x=1+\\sqrt{5} or x=1-\\sqrt{5}. \\therefore the coordinates of the intersection points where the line y=x+1 is intercepted by the curve y=\\frac{1}{2}x^{2}-1 are A(1+\\sqrt{5},2+\\sqrt{5}), B(1-\\sqrt{5},2-\\sqrt{5}). \\therefore |AB|=\\sqrt{(1+\\sqrt{5}-1+\\sqrt{5})^{2}+(2+\\sqrt{5}-2+\\sqrt{5})^{2}}=2\\sqrt{10}."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$, point $A(0,1)$, and $P$ a moving point on the ellipse, then the maximum value of $|P A|$ is?",
    "fact_expressions": "G: Ellipse;A: Point;P: Point;Expression(G) = (x^2/4 + y^2/2 = 1);Coordinate(A) = (0, 1);PointOnCurve(P, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "sqrt(6)",
    "fact_spans": "[[[2, 39], [56, 58]], [[40, 49]], [[52, 55]], [[2, 39]], [[40, 49]], [[52, 62]]]",
    "query_spans": "[[[64, 77]]]",
    "process": "Let point P(x,y), then \\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1, we obtain x^{2}=4-2y^{2}, where -\\sqrt{2}\\leqslant y \\leqslant \\sqrt{2}, |PA|=\\sqrt{x^{2}+(y-1)^{2}}=\\sqrt{4-2y^{2}+y^{2}-2y+1}=\\sqrt{-y^{2}-2y+5}=\\sqrt{-(y+1)^{2}+6}\\leqslant\\sqrt{6}. The maximum value of |PA| is \\sqrt{6}, achieved if and only if y=-1."
  },
  {
    "text": "Given $A(-2,0)$, $B(2,0)$, on a line $l$ with slope $k$, there exist two distinct points $M$, $N$ satisfying $|M A|-|M B|=2 \\sqrt{3}$, $|N A|-|N B|=2 \\sqrt{3}$, and the midpoint of segment $M N$ is $(6,1)$. Then the value of $k$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (-2, 0);B: Point;Coordinate(B) = (2, 0);l: Line;k:Number;Slope(l) = k;M: Point;N: Point;PointOnCurve(M,l) = True;PointOnCurve(N,l) = True;Negation(M = N);Abs(LineSegmentOf(M,A)) - Abs(LineSegmentOf(M,B)) = 2*sqrt(3);Abs(LineSegmentOf(N, A)) - Abs(LineSegmentOf(N, B)) = 2*sqrt(3);Coordinate(MidPoint(LineSegmentOf(M,N))) = (6,1)",
    "query_expressions": "k",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 11]], [[2, 11]], [[13, 21]], [[13, 21]], [[29, 34]], [[25, 28], [123, 126]], [[22, 34]], [[42, 45]], [[46, 49]], [[29, 49]], [[29, 49]], [[37, 49]], [[51, 75]], [[77, 101]], [[103, 121]]]",
    "query_spans": "[[[123, 130]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola is the left focus of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, and the vertex is at the center of the ellipse. Then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;H: Ellipse;Expression(H) = (x^2/9 + y^2/4 = 1);Focus(G) = LeftFocus(H);Vertex(G) = Center(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 = -4 \\sqrt{5} x",
    "fact_spans": "[[[0, 3], [58, 61]], [[7, 44], [52, 54]], [[7, 44]], [[0, 48]], [[0, 56]]]",
    "query_spans": "[[[58, 65]]]",
    "process": "\\frac{x2}{9}+\\frac{y^{2}}{4}=1 has its left focus at (-\\sqrt{5},0), \\therefore y^{2}=-4\\sqrt{5}x"
  },
  {
    "text": "The point $(0,-2)$ is a focus of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{m^{2}-2}=1$. Then the value of the real number $m$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/(m^2 - 2) + x^2/m = 1);m: Real;H: Point;Coordinate(H) = (0, -2);OneOf(Focus(G)) = H",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[10, 53]], [[10, 53]], [[60, 65]], [[0, 9]], [[0, 9]], [[0, 58]]]",
    "query_spans": "[[[60, 69]]]",
    "process": "According to the problem, the foci of the ellipse lie on the y-axis, ∴a^{2}=m^{2}-2, b^{2}=m and m^{2}-2>m>0 ∴c^{2}=a^{2}-b^{2}=m^{2}-2-m=4, solving gives m=-2 (discarded) or m=3, ∴m=3."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$, and $|AB|=4$. If the origin $O$ is the orthocenter of $\\triangle ABC$, then what are the coordinates of point $C$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;C: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 4;O: Origin;Orthocenter(TriangleOf(A, B, C)) = O",
    "query_expressions": "Coordinate(C)",
    "answer_expressions": "(-3, 0)",
    "fact_spans": "[[[1, 15], [25, 28]], [[21, 23]], [[29, 32]], [[33, 36]], [[81, 85]], [[17, 20]], [[1, 15]], [[1, 20]], [[0, 23]], [[21, 38]], [[40, 50]], [[53, 58]], [[53, 79]]]",
    "query_spans": "[[[81, 90]]]",
    "process": "Clearly, the slope of line AB is not zero. According to the problem, let the equation of line AB be: $x = my + 1$, and let $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$. Combining the equations of line AB and the parabola:\n\\[\n\\begin{cases}\nx = my + 1 \\\\\ny^{2} = 4x\n\\end{cases}\n\\]\nSimplifying yields $y^{2} - 4my - 4 = 0$, so $y_{1} + y_{2} = 4m$, and thus $x_{1} + x_{2} = 4m^{2} + 2$. By the property of the parabola, $|AB| = x_{1} + x_{2} + 2 = 4m^{2} + 4$. According to the problem, $4m^{2} + 4 = 4$, so $m = 0$, meaning line AB is perpendicular to the x-axis. Therefore, $A(1, 2)$, $B(1, -2)$. Since the origin O is the orthocenter of $\\triangle ABC$, point C lies on the x-axis. Let $C(a, 0)$, then $AO \\perp BC$, i.e., $\\overrightarrow{AO} \\cdot \\overrightarrow{BC} = 0$, which gives $(1, 2) \\cdot (1 - a, -2) = 0$. Simplifying yields: $1 - a + 4 = 0$, solving gives $a = -3$."
  },
  {
    "text": "The coordinates of the point on the parabola $y^{2}=x$ that is closest to the line $x-2 y+4=0$ are?",
    "fact_expressions": "G: Parabola;H: Line;P:Point;Expression(G) = (y^2 = x);Expression(H) = (x - 2*y + 4 = 0);PointOnCurve(P, G);WhenMin(Distance(P,H))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1, 1)",
    "fact_spans": "[[[0, 12]], [[14, 27]], [[33, 34]], [[0, 12]], [[14, 27]], [[0, 34]], [[13, 34]]]",
    "query_spans": "[[[33, 39]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has its right focus at $F(c, 0)$. The line passing through $F$ and perpendicular to the asymptotes of $C$ is tangent to the circle $x^{2}+y^{2}+2 c x=0$. Then, the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;c: Number;H: Line;F: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (2*(c*x) + x^2 + y^2 = 0);Coordinate(F) = (c, 0);RightFocus(C) = F;PointOnCurve(F,H);IsPerpendicular(H,Asymptote(C));IsTangent(H,G)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 63], [86, 89], [124, 130]], [[10, 63]], [[10, 63]], [[98, 120]], [[68, 77]], [[94, 96]], [[68, 77], [79, 83]], [[10, 63]], [[10, 63]], [[2, 63]], [[98, 120]], [[68, 77]], [[2, 77]], [[78, 96]], [[83, 96]], [[94, 122]]]",
    "query_spans": "[[[124, 136]]]",
    "process": "The asymptotes of hyperbola C are given by $ y = \\pm\\frac{b}{a}x $. The equation of the line passing through point F and perpendicular to the line $ y = \\frac{b}{a}x $ is $ y = -\\frac{a}{b}(x - c) $, or $ ax + by - bc = 0 $. The standard equation of the circle $ x^{2} + y^{2} + 2cx = 0 $ is $ (x + c)^{2} + y^{2} = c^{2} $, with center $ (-c, 0) $ and radius $ c $. According to the problem, the line $ ax + by - ac = 0 $ is tangent to the circle $ (x + c)^{2} + y^{2} = c^{2} $. Therefore, $ \\frac{2ac}{\\sqrt{a^{2} + b^{2}}} = 2a = c $, so the eccentricity of the hyperbola is $ e = \\frac{c}{a} = 2 $."
  },
  {
    "text": "Given the ellipse $ C $: $ \\frac{x^{2}}{a^{2}} + y^{2} = 1 $ ($ a > 1 $) with foci $ F_{1} $, $ F_{2} $, and a circle $ \\odot O $ centered at the origin with diameter equal to the focal distance of the ellipse. The circle $ \\odot O $ intersects the ellipse $ C $ at point $ P $. Then $ S_{\\Delta P F_{1} F_{2}} $ = ?",
    "fact_expressions": "C: Ellipse;a: Number;Q: Circle;P: Point;F1: Point;F2: Point;a>1;O: Origin;Expression(C) = (y^2 + x^2/a^2 = 1);Focus(C) = {F1, F2};Center(Q) = O ;Diameter(Q) = FocalLength(C);Intersection(Q, C) = P",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 43], [70, 72], [89, 94]], [[9, 43]], [[79, 88]], [[96, 100]], [[47, 54]], [[55, 62]], [[9, 43]], [[64, 66]], [[2, 43]], [[2, 62]], [[63, 88]], [[70, 88]], [[79, 100]]]",
    "query_spans": "[[[102, 130]]]",
    "process": "The angle subtended by a diameter on the circumference is a right angle, hence $S_{APF_{1}F_{2}}=b^{2}\\tan\\frac{\\theta}{2}=1\\times\\tan\\frac{\\pi}{4}=1$"
  },
  {
    "text": "An ellipse $5 x^{2}+k y^{2}=5$ has a focus at $(0 , 2)$, then what is $k=?$",
    "fact_expressions": "G: Ellipse;k: Number;H: Point;Expression(G) = (k*y^2 + 5*x^2 = 5);Coordinate(H) = (0, 2);OneOf(Focus(G)) = H",
    "query_expressions": "k",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 21]], [[39, 42]], [[27, 36]], [[0, 21]], [[27, 36]], [[0, 36]]]",
    "query_spans": "[[[39, 44]]]",
    "process": ""
  },
  {
    "text": "Given the curve $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a \\cdot b \\neq 0$, and $a \\neq b$) intersects the line $x+y-1=0$ at two points $P$, $Q$, and $\\overrightarrow{O P} \\cdot \\overrightarrow{O Q}=0$ ($O$ is the origin), then the value of $\\frac{1}{a}-\\frac{1}{b}$ is?",
    "fact_expressions": "H: Curve;Expression(H) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;Negation((a*b)=0);Negation(a=b);G: Line;Expression(G) = (x + y - 1 = 0);P: Point;Q: Point;Intersection(H, G) = {P, Q};O: Origin;DotProduct(VectorOf(O, P), VectorOf(O, Q)) = 0",
    "query_expressions": "-1/b + 1/a",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 75]], [[2, 75]], [[162, 187]], [[162, 187]], [[4, 75]], [[4, 75]], [[76, 87]], [[76, 87]], [[90, 93]], [[94, 97]], [[2, 99]], [[153, 156]], [[101, 152]]]",
    "query_spans": "[[[162, 191]]]",
    "process": ""
  },
  {
    "text": "Given that there are two points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$ on the parabola $y^{2}=3x$, if the slope of line $AB$ is $2$ and $y_{1}=4$, then $y_{2}=$?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;x1: Number;x2: Number;y1: Number;y2: Number;Expression(G) = (y^2 = 3*x);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);PointOnCurve(A, G);PointOnCurve(B, G);Slope(LineOf(A, B)) = 2;y1 = 4",
    "query_expressions": "y2",
    "answer_expressions": "-5/2",
    "fact_spans": "[[[2, 16]], [[21, 38]], [[41, 58]], [[21, 38]], [[41, 58]], [[76, 85]], [[87, 94]], [[2, 16]], [[21, 38]], [[41, 58]], [[2, 58]], [[2, 58]], [[60, 74]], [[76, 85]]]",
    "query_spans": "[[[87, 96]]]",
    "process": "Let the equation of line AB be y=2x+b, A(x_{1},y_{1}), B(x_{2},y_{2}). Solving the system of equations \\begin{cases}y=2x+b\\\\y^{2}=3x\\end{cases} gives \\frac{2}{3}y^{2}-y+b=0. Therefore, y_{1}+y_{2}=\\frac{3}{2}. Since y_{1}=4, it follows that y_{2}=\\frac{3}{2}-4=-\\frac{5}{2}."
  },
  {
    "text": "If the directrix of the parabola $y^{2}=2 p x(p>0)$ passes through the left vertex of the hyperbola $x^{2}-y^{2}=1$, then $p$=?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 2*(p*x));p: Number;p>0;G: Hyperbola;Expression(G) = (x^2 - y^2 = 1);PointOnCurve(LeftVertex(G), Directrix(H))",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 22]], [[1, 22]], [[52, 55]], [[4, 22]], [[27, 45]], [[27, 45]], [[1, 49]]]",
    "query_spans": "[[[52, 57]]]",
    "process": ""
  },
  {
    "text": "Given that the center of the ellipse is at the origin, one focus is $F(-2 \\sqrt{3} , 0)$, and the length of the major axis is twice the length of the minor axis, then the standard equation of the ellipse is?",
    "fact_expressions": "E: Ellipse;O: Origin;Center(E) = O;F: Point;OneOf(Focus(E)) = F;Coordinate(F) = (-2*sqrt(3), 0);Length(MajorAxis(E)) = 2*Length(MinorAxis(E))",
    "query_expressions": "Expression(E)",
    "answer_expressions": "x^2/16+y^2/4=1",
    "fact_spans": "[[[2, 4], [52, 54]], [[7, 9]], [[2, 9]], [[15, 35]], [[2, 35]], [[15, 35]], [[2, 49]]]",
    "query_spans": "[[[52, 61]]]",
    "process": ""
  },
  {
    "text": "Given that the line $y = kx$ ($k \\neq 0$) intersects the hyperbola $C$: $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ ($a > 0$, $b > 0$) at points $A$ and $B$, and the circle with diameter $AB$ passes exactly through the right focus $F$ of the hyperbola $C$. If the area of $\\triangle OAF$ is $4a^{2}$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*x);k: Number;Negation(k=0);C: Hyperbola;Expression(C) = (x^2/a^2 - y^2/b^2 = 1);a: Number;b: Number;a>0;b>0;A: Point;B: Point;Intersection(H, C) = {A, B};E: Circle;IsDiameter(LineSegmentOf(A, B), E);F: Point;RightFocus(C) = F;PointOnCurve(F, E);Area(TriangleOf(O, A, F)) = 4*a^2;O: Origin",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 21]], [[2, 21]], [[4, 21]], [[4, 21]], [[22, 83], [110, 116], [157, 163]], [[22, 83]], [[30, 83]], [[30, 83]], [[30, 83]], [[30, 83]], [[85, 88]], [[89, 92]], [[2, 94]], [[105, 106]], [[95, 106]], [[120, 123]], [[110, 123]], [[105, 123]], [[125, 155]], [[125, 142]]]",
    "query_spans": "[[[157, 169]]]",
    "process": "Since the circle with diameter AB passes exactly through the right focus F of the hyperbola, the equation of the circle with diameter AB is $x^{2}+y^{2}=c^{2}$, and the circle also passes through the left focus $F_{1}$. Since AB is equal to $F_{1}F$ and bisects it, the quadrilateral $AF_{1}BF$ is a rectangle, so $|AF|=|BF_{1}|$. Let $|AF|=m$, $|BF|=n$, then $|AF|-|BF|=|BF_{1}|-|BF|=m-n=2a$. $\\because$ the area of $\\triangle OAF$ is $4a^{2}$, $\\therefore$ the area of $\\Delta ABF$ is $S_{\\Delta ABF}=2S_{\\Delta OAF}=\\frac{1}{2}m\\cdot n=8a^{2}$, and $m^{2}+n^{2}=|AB|^{2}=4c^{2}$. Solving the three equations:  \n$$\n\\begin{cases}\nm-n=2a\\\\\nmn=16a^{2}\\\\\nm^{2}+n^{2}=4c^{2}\n\\end{cases}\n$$\nwe obtain $4c^{2}=4a^{2}+32a^{2}$, $\\therefore c^{2}=9a^{2}$, i.e., $e=3$."
  },
  {
    "text": "Given that the line $l$ passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$, if $|A F|=4$, then $|B F|=$?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, F)) = 4",
    "query_expressions": "Abs(LineSegmentOf(B, F))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[23, 28]], [[3, 17], [29, 32]], [[35, 38]], [[19, 22]], [[39, 42]], [[3, 17]], [[3, 22]], [[2, 28]], [[23, 44]], [[46, 55]]]",
    "query_spans": "[[[57, 66]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ has an eccentricity of $\\sqrt{3}$. Then the equations of the asymptotes of this hyperbola are?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(C) = sqrt(3)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*sqrt(2)*x",
    "fact_spans": "[[[0, 63], [81, 84]], [[0, 63]], [[7, 63]], [[7, 63]], [[7, 63]], [[7, 63]], [[0, 78]]]",
    "query_spans": "[[[81, 92]]]",
    "process": ""
  },
  {
    "text": "The equation of an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $4 x-3 y=0$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G)))=(4*x-3*y=0)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[0, 56], [78, 81]], [[3, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[0, 56]], [[0, 76]]]",
    "query_spans": "[[[78, 87]]]",
    "process": "Since one of the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) is given by $4x-3y=0$, $\\therefore y=\\frac{4}{3}x$, $\\therefore \\frac{4}{3}=\\frac{b}{a}$, $\\therefore e=\\frac{5}{3}$"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{y^{2}}{4}-\\frac{x^{2}}{4}=1$, $P$ is a point on the asymptote of the hyperbola in the first quadrant, $O$ is the origin, and $|O P|=2 \\sqrt{2}$. Then the coordinates of point $P$ are?",
    "fact_expressions": "C: Hyperbola;O: Origin;P: Point;Expression(C) = (-x^2/4 + y^2/4 = 1);Quadrant(P)=1;PointOnCurve(P,Asymptote(C));Abs(LineSegmentOf(O, P)) = 2*sqrt(2)",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2,2)",
    "fact_spans": "[[[2, 44], [51, 54]], [[66, 69]], [[47, 50], [96, 100]], [[2, 44]], [[46, 65]], [[46, 65]], [[76, 94]]]",
    "query_spans": "[[[96, 105]]]",
    "process": "First, find the equation of the asymptote of the hyperbola in the first quadrant, $ y = x $, and use the distance formula between two points to solve. The equation of the asymptote of the hyperbola $ y^{2} - x^{2} = 4 $ in the first quadrant is $ y = x $. Let $ P(x, x) $, $ x > 0 $. Since $ |OP| = 2\\sqrt{2} $, the coordinates of point $ P $ are $ (2,"
  },
  {
    "text": "Let the parabola $C$: $y^{2}=2 p x(p>0)$ have focus $F$, and let points $A$ and $B$ in the first quadrant lie on $C$, with $O$ being the origin. If $\\angle A F O = \\angle A F B = \\frac{\\pi}{3}$ and the area of $\\triangle A F B$ is $3 \\sqrt{3}$, then the coordinates of point $A$ are?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2=2*p*x);p: Number;p>0;F: Point;Focus(C) = F;A: Point;B: Point;Quadrant(A) = 1;Quadrant(B) = 1;PointOnCurve(A,C) = True;PointOnCurve(B,C) = True;O: Origin;AngleOf(A,F,O)=AngleOf(A,F,B);AngleOf(A,F,B) = pi/3;Area(TriangleOf(A,F,B)) = 3*sqrt(3)",
    "query_expressions": "Coordinate(A)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 27], [52, 55]], [[1, 27]], [[9, 27]], [[9, 27]], [[31, 34]], [[1, 34]], [[41, 44], [145, 149]], [[45, 48]], [[35, 44]], [[35, 50]], [[41, 56]], [[41, 56]], [[57, 60]], [[67, 108]], [[67, 108]], [[110, 143]]]",
    "query_spans": "[[[145, 154]]]",
    "process": ""
  },
  {
    "text": "A focus of the hyperbola is $(0,5)$, and its asymptotes are given by $y=\\pm \\frac{4}{3} x$. Then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (0, 5);OneOf(Focus(G)) = H;Expression(Asymptote(G)) = (y = pm*x*4/3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/16 - x^2/9 = 1",
    "fact_spans": "[[[0, 3], [17, 18], [47, 50]], [[9, 16]], [[9, 16]], [[0, 16]], [[17, 45]]]",
    "query_spans": "[[[47, 57]]]",
    "process": "Given: one focus of the hyperbola is (0,5), and its asymptotes are y=\\pm\\frac{4}{3}x. Therefore, the focus lies on the y-axis. Let the standard equation be \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1, (a>0,b>0), with \\frac{a}{b}=\\frac{4}{3}, a^{2}+b^{2}=25. Solving gives: a=4, b=3. Thus, the standard equation of the hyperbola is \\frac{y^{2}}{16}-\\frac{x^{2}}{9}=1"
  },
  {
    "text": "The hyperbola passes through the points $(4, \\sqrt{3})$ and $(3, \\frac{\\sqrt{5}}{2})$. What is the standard equation of the hyperbola?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (4, sqrt(3));PointOnCurve(H, G);P:Point;Coordinate(P)=(3,sqrt(5)/2);PointOnCurve(P,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4-y^2=1",
    "fact_spans": "[[[0, 3], [50, 53]], [[4, 20]], [[4, 20]], [[0, 20]], [[22, 47]], [[22, 47]], [[0, 47]]]",
    "query_spans": "[[[50, 60]]]",
    "process": "Since the position of the foci of the hyperbola is unknown, the equation of the hyperbola can be set as \\frac{x^{2}}{m}-\\frac{y^{2}}{n}=1, mn>0. Substituting the two given points yields the solution. Since the position of the foci of the hyperbola is unknown, set the equation of the hyperbola as \\frac{x2}{m}-\\frac{y^{2}}{n}=1, mn>0. Substituting the coordinates of the two points gives \\begin{cases}\\frac{16}{m}-\\frac{3}{n}=1\\\\\\frac{9}{m}-\\frac{5}{4n}=1\\end{cases}, solving which yields \\begin{cases}m=4\\\\n=1\\end{cases}. Therefore, the standard equation of the hyperbola is \\frac{x^{2}}{4}-y2=1."
  },
  {
    "text": "The focus of the parabola $y^{2}=8 x$ is $F$, and point $P(x_0, y_0)$ is a moving point on this parabola. Given point $A(-2,0)$, what is the range of $\\frac{|P A|}{|P F|}$?",
    "fact_expressions": "G: Parabola;P: Point;A: Point;F: Point;x0: Number;y0: Number;Expression(G) = (y^2 = 8*x);Coordinate(P) = (x0, y0);Coordinate(A) = (-2, 0);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Range(Abs(LineSegmentOf(P, A))/Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "[1, sqrt(2)]",
    "fact_spans": "[[[0, 14], [36, 39]], [[22, 34]], [[48, 58]], [[18, 21]], [[23, 34]], [[23, 34]], [[0, 14]], [[22, 34]], [[48, 58]], [[0, 21]], [[22, 43]]]",
    "query_spans": "[[[60, 88]]]",
    "process": "Draw a perpendicular line from point P to the directrix of the parabola, with foot of perpendicular at M, then PF₁ = |PM|. According to the symmetry of the parabola, only consider the case when point P(x, y) lies above the x-axis. ∵ the focus of the parabola y² = 8x is F(2, 0), and point A(-2, 0), ∴ |PA| / |PF| = 1 / sin∠MAP. Let the tangent line from A to the parabola be y = k(x + 2). Substituting into the parabola equation gives k²x² + (4k² - 8)x + 4k² = 0. ∴ Δ = (4k² - 8)² - 16k⁴ = 0, ∴ k = ±1. It is easy to see: ∠MAP ∈ [π/4, π/2]. ∴ 1 / sin(∠MAP) ∈ [1, √2]"
  },
  {
    "text": "Given the circle $O$: $x^{2}+y^{2}=1$ and point $A(-2,0)$, if vertex $B(b, 0)$ $(b \\neq-2)$ and constant $\\lambda$ satisfy: for any point $M$ on circle $O$, $|M B|=\\lambda|M A|$, then $\\lambda-b$=?",
    "fact_expressions": "O: Circle;A: Point;B: Point;M: Point;b:Number;lambda:Number;Negation(b=-2);Expression(O) = (x^2 + y^2 = 1);Coordinate(A) = (-2, 0);Coordinate(B) = (b, 0);PointOnCurve(M,O);Abs(LineSegmentOf(M,B))=lambda*Abs(LineSegmentOf(M,A));Vertex(O)=B",
    "query_expressions": "-b + lambda",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 22], [74, 78]], [[23, 33]], [[37, 58]], [[83, 86]], [[111, 122]], [[61, 70]], [[37, 58]], [[2, 22]], [[23, 33]], [[37, 58]], [[74, 86]], [[89, 109]], [[2, 58]]]",
    "query_spans": "[[[111, 124]]]",
    "process": "Let M(x,y), |MB| = \\lambda|MA|, (x-b)^{2} + y^{2} = \\lambda^{2}(x+2)^{2} + \\lambda^{2}y^{2}. Substituting (1,0) and (-1,0) into the equation, we obtain (1-b)^{2} = \\lambda^{2}(1+2)^{2}, (-1-b)^{2} = \\lambda^{2}(-1+2)^{2}. Solving gives b = -\\frac{1}{2}, \\lambda = \\frac{1}{2}, \\lambda - b = 1."
  },
  {
    "text": "A chord of the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$ is bisected by the point $A(4,2)$. What is the equation of the line containing this chord?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/36 + y^2/9 = 1);Z: LineSegment;IsChordOf(Z, G);A: Point;Coordinate(A) = (4, 2);MidPoint(Z) = A",
    "query_expressions": "Expression(OverlappingLine(Z))",
    "answer_expressions": "x+2*y-8=0",
    "fact_spans": "[[[0, 38]], [[0, 38]], [], [[0, 42]], [[43, 52]], [[43, 52]], [[0, 54]]]",
    "query_spans": "[[[0, 69]]]",
    "process": "By the point difference method, we obtain $\\frac{x_{midpoint}}{36}+\\frac{y_{midpoint}k_{chord}}{9}=0\\Rightarrow\\frac{4}{36}+\\frac{2k_{chord}}{9}=0\\Rightarrow k_{chord}=-\\frac{1}{2}$, therefore the equation of the line containing this chord is $y-2=-\\frac{1}{2}(x-4)\\Rightarrow x+2y-8=0$"
  },
  {
    "text": "The segment connecting the focus $F$ of the parabola $y^{2}=4x$ and the point $M(0,1)$ intersects the parabola at point $A$. Let point $O$ be the origin. Then the area of $\\triangle OAM$ is?",
    "fact_expressions": "G: Parabola;M: Point;O: Origin;A: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(M) = (0, 1);Focus(G) =F;Intersection(LineSegmentOf(F,M),G)=A",
    "query_expressions": "Area(TriangleOf(O, A, M))",
    "answer_expressions": "3/2 - sqrt(2)",
    "fact_spans": "[[[2, 16], [38, 41]], [[23, 32]], [[49, 53]], [[43, 47]], [[19, 22]], [[2, 16]], [[23, 32]], [[2, 22]], [[0, 47]]]",
    "query_spans": "[[[60, 82]]]",
    "process": "The focus $ F $ of the parabola $ y^{2} = 4x $ is $ (1,0) $; then the equation of line $ MF $ is: $ x + y = 1 $. Solving the system  \n\\[\n\\begin{cases}\nx + y = 1 \\\\\ny^{2} = 4x\n\\end{cases}\n\\]  \nyields $ x^{2} - 6x + 1 = 0 $, solving which gives $ x = 3 + 2\\sqrt{2} $ (discarded) or $ x = 3 - 2\\sqrt{2} $. Therefore, the area $ S $ of $ \\triangle OAM $ is $ \\frac{1}{2} \\times |OM| \\times (3 - 2\\sqrt{2}) = \\frac{3}{2} - \\sqrt{2} $."
  },
  {
    "text": "Let the asymptotes of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ be tangent to the circle $(x-a)^{2}+y^{2}=4$, then $a$=?",
    "fact_expressions": "G: Hyperbola;H: Circle;a: Number;Expression(G) = (x^2/4 - y^2 = 1);Expression(H) = (y^2 + (-a + x)^2 = 4);IsTangent(G, Asymptote(H))",
    "query_expressions": "a",
    "answer_expressions": "pm*2*sqrt(5)",
    "fact_spans": "[[[1, 29]], [[34, 54]], [[58, 61]], [[1, 29]], [[34, 54]], [[1, 56]]]",
    "query_spans": "[[[58, 63]]]",
    "process": ""
  },
  {
    "text": "If the parabola $y^{2}=2 p x(p>0)$ passes through the point $(2,1)$, then $p$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;H: Point;Coordinate(H) = (2, 1);PointOnCurve(H, G)",
    "query_expressions": "p",
    "answer_expressions": "1/4",
    "fact_spans": "[[[1, 22]], [[1, 22]], [[34, 37]], [[4, 22]], [[24, 32]], [[24, 32]], [[1, 32]]]",
    "query_spans": "[[[34, 39]]]",
    "process": "Substitute the point into the parabola to solve: the parabola $ y^{2} = 2px $ ($ p > 0 $) passes through the point $ (2,1) $, so $ 1 = 4p $, that is, $ p = \\frac{1}{4} $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus is $F_{1}$, the right vertex is $A$, and the circle with diameter $F_{1}A$ intersects the ellipse at three common points. Find the range of values for the eccentricity of the ellipse.",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;LeftFocus(G) = F1;A: Point;RightVertex(G) = A;H: Circle;IsDiameter(LineSegmentOf(F1, A), H);NumIntersection(H, G) = 3",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1/2, 1)",
    "fact_spans": "[[[2, 54], [90, 92], [100, 102]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[58, 65]], [[2, 65]], [[70, 73]], [[2, 73]], [[88, 89]], [[74, 89]], [[88, 98]]]",
    "query_spans": "[[[100, 112]]]",
    "process": "Let the common points be $ M(x_{0},y_{0}) $, $ N(x_{0},-y_{0}) $, then $ \\cdot\\frac{y_{0}}{x_{0}-a}=-1 $, $ (x_{0}+c)(x_{0}-a)+y_{0}^{2}=0 $, $ \\frac{x_{0}+c\\cdotx_{0}-a}{\\frac{x_{0}^{2}}{a^{2}}+\\frac{y_{0}^{2}}{b^{2}}}=1 \\Rightarrow x_{0}^{2}+(c-a)x_{0}-ac+b^{2}-\\frac{b^{2}x_{0}^{2}}{a^{2}}=0 $, i.e., $ \\frac{c^{2}}{a^{2}}x_{0}^{2}+(c-a)x_{0}-ac+b^{2}=0 $. Then the two roots of the equation $ \\frac{c^{2}}{a^{2}}x^{2}+(c-a)x-ac+b^{2}=0 $ are $ x_{0} $, $ a $. Hence $ ax_{0}=\\frac{b^{2}-ac}{c^{2}} \\Rightarrow x_{0}=\\frac{a(b^{2}-ac)}{c^{2}}<a \\Rightarrow b^{2}-ac<c^{2} $, which transforms into $ 2e^{2}+e-1>0 \\Rightarrow \\frac{1}{2}<e<1 $"
  },
  {
    "text": "Given that the line $y = kx - 1$ intersects each of the left and right branches of the hyperbola $x^2 - y^2 = 4$ at one common point, then the range of values for $k$ is?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*x - 1);k: Number;G: Hyperbola;Expression(G) = (x^2 - y^2 = 4);NumIntersection(H, LeftPart(G)) = 1;NumIntersection(H, RightPart(G)) = 1",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-1, 1)",
    "fact_spans": "[[[2, 13]], [[2, 13]], [[46, 49]], [[14, 32]], [[14, 32]], [[2, 44]], [[2, 44]]]",
    "query_spans": "[[[46, 56]]]",
    "process": "From \\begin{cases}y=kx-1\\\\x^{2}-y^{2}=4\\end{cases} \\Rightarrow (1-k^{2})x^{2}+2kx-5=0, according to the problem we have \\begin{cases}1-k^{2}\\neq0\\\\\\end{cases} \\frac{-5}{1-k^{2}}<0 \\Rightarrow -1<k<1."
  },
  {
    "text": "If the coordinates of point $A$ are $(3,2)$, $F$ is the focus of the parabola $y^{2}=2 x$, and point $P$ moves along this parabola, then what are the coordinates of point $P$ such that $|PA|+|P F|$ is minimized?",
    "fact_expressions": "A: Point;Coordinate(A) = (3, 2);G: Parabola;Expression(G) = (y^2 = 2*x);F: Point;Focus(G) = F;P: Point;PointOnCurve(P, G);WhenMin(Abs(LineSegmentOf(P, A))+Abs(LineSegmentOf(P, F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2, 2)",
    "fact_spans": "[[[1, 5]], [[1, 16]], [[22, 36], [46, 49]], [[22, 36]], [[18, 21]], [[18, 39]], [[40, 44], [75, 79]], [[40, 52]], [[56, 73]]]",
    "query_spans": "[[[75, 84]]]",
    "process": "The directrix of the parabola is given by $ x = -\\frac{1}{2} $. Draw a perpendicular from point $ P $ to the directrix, with foot at $ B $. By the definition of a parabola, we have $ |PF| = |PB| $. Then $ |PA| + |PF| = |PA| + |PB| \\geqslant |AB| = 4 $, with equality if and only if points $ P $, $ A $, and $ B $ are collinear. At this moment, point $ P(2,2) $."
  },
  {
    "text": "Given that the right focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{a}=1$ is $(\\sqrt{13}, 0)$, what is the equation of the asymptotes of this hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/a = 1);a: Number;Coordinate(RightFocus(G)) = (sqrt(13), 0)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*2*x/3",
    "fact_spans": "[[[2, 40], [64, 67]], [[2, 40]], [[5, 40]], [[2, 61]]]",
    "query_spans": "[[[64, 75]]]",
    "process": "Since the right focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{a}=1$ is $(\\sqrt{13},0)$, we have $c=\\sqrt{13}$, $9+a=13$, so $a=4$, therefore the asymptotes of the hyperbola are given by $y=\\pm\\frac{2}{3}x$."
  },
  {
    "text": "Given circle $M$: $(x+1)^{2}+y^{2}=1$, circle $N$: $(x-1)^{2}+y^{2}=9$, a moving circle $P$ is externally tangent to circle $M$ and internally tangent to circle $N$, the locus of the center of circle $P$ is curve $C$, then the equation of curve $C$ is?",
    "fact_expressions": "M: Circle;Expression(M) = (y^2 + (x + 1)^2 = 1);N: Circle;Expression(N) = ((x-1)^2 + y^2 = 9);P: Circle;IsOutTangent(P, M) = True;IsInTangent(P, N) = True;C: Curve;P1: Point;Center(P) = P1;Locus(P1) = C",
    "query_expressions": "Expression(C)",
    "answer_expressions": "(x^2/4+y^2/3=1)&Negation(x=-2)",
    "fact_spans": "[[[2, 26], [58, 62]], [[2, 26]], [[27, 51], [67, 71]], [[27, 51]], [[54, 57]], [[54, 64]], [[54, 73]], [[83, 88], [90, 95]], [[76, 79]], [[54, 79]], [[76, 88]]]",
    "query_spans": "[[[90, 100]]]",
    "process": "From circle M: $(x+1)^{2}+y^{2}=1$, circle N: $(x-1)^{2}+y^{2}=9$, we obtain M$(-1,0)$, radius $r_{1}=1$, N$(1,0)$, radius $r_{2}=3$. Let the radius of moving circle P be $R$. Since circle M is inside circle N, the moving circle can only be internally tangent to circle N within N, and N cannot be inside the moving circle, so $R<3$. Since moving circle P is externally tangent to circle M, $PM=1+R$. Since moving circle P is internally tangent to circle N, $PN=3-R$. Therefore, $PM+PN=4$, i.e., the sum of distances from P to M and from P to N is constant. Hence, P is an ellipse with M and N as foci, where $a=2$, $c=1$, so $b=\\sqrt{4-1}=\\sqrt{3}$. Therefore, the trajectory equation of the center P of the moving circle is $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$. Also, circle N passes through point $(-2,0)$, and the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ also passes through point $(-2,0)$. However, point P is clearly not on circle N, so the required trajectory equation is: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ ($x\\neq-2$)."
  },
  {
    "text": "Parabola $C$: $y^{2}=4x$, a point $P$ on it has the minimal sum of distance to point $A(3,4\\sqrt{2})$ and to the directrix. Then the coordinates of point $P$ are?",
    "fact_expressions": "C: Parabola;A: Point;P: Point;Expression(C) = (y^2 = 4*x);Coordinate(A) = (3, 4*sqrt(2));PointOnCurve(P, C);WhenMin(Distance(P, A)+Distance(P, Directrix(C)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2,2*sqrt(2))",
    "fact_spans": "[[[0, 19]], [[26, 44]], [[22, 25], [56, 60]], [[0, 19]], [[26, 44]], [[0, 25]], [[0, 54]]]",
    "query_spans": "[[[56, 65]]]",
    "process": "Given that F(1,0), as shown in the figure, connect PF. By the definition of the parabola, the distance from point P to the directrix equals |PF|, i.e., |PP'| = |PF|. Therefore, |PP'| + |PA| = |PF| + |PA| \\geqslant |AF|, with equality holding if and only if points A, P, F are collinear, which occurs when point P is at position P_{1}. At this moment, the equation of line AF is y = \\frac{4\\sqrt{2}-0}{3-1}(x-1), i.e., y = 2\\sqrt{2}x - 2\\sqrt{2}. Thus, solving the system of equations \\begin{cases} y = 2\\sqrt{2}x - 2\\sqrt{2} \\\\ y^{2} = 4x \\end{cases} yields 2x^{2} - 5x + 2 = 0, whose solutions are x = 2 or x = \\frac{1}{2}. According to the conditions, x_{P_{1}} > 1, so x = 2, y = 2\\sqrt{2}. Hence, the coordinates of point P are (2, 2\\sqrt{2})."
  },
  {
    "text": "On the parabola $C$: $y^{2}=4x$, a point $Q$ has the minimal sum of distances to the point $B(4,1)$ and to the focus $F$. Then the coordinates of point $Q$ are?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);Q: Point;PointOnCurve(Q, C);B: Point;Coordinate(B) = (4, 1);F:Point;Focus(C)=F;WhenMin(Distance(Q,B)+Distance(Q,F))",
    "query_expressions": "Coordinate(Q)",
    "answer_expressions": "(1/4,1)",
    "fact_spans": "[[[0, 19]], [[0, 19]], [[22, 25], [50, 54]], [[0, 25]], [[26, 35]], [[26, 35]], [[39, 42]], [[0, 42]], [[0, 48]]]",
    "query_spans": "[[[50, 59]]]",
    "process": "According to the definition of a parabola, the minimum sum of the distance from point Q to point B(4,1) and to the focus F is equal to the distance from point B to the directrix, from which the coordinates of point Q can be determined. By the definition of a parabola, QF = QM, therefore QB + QF = QB + QM. As shown in the figure, when M, Q, B are collinear, i.e., BM is perpendicular to the directrix, QB + QM reaches its minimum value. At this time, y_{Q} = 1, substituting into the parabola equation gives x_{Q} = \\frac{1}{4}, so Q(\\frac{1}{4},1)."
  },
  {
    "text": "If a line passing through the point $M(2,0)$ with slope $\\sqrt{3}$ intersects the directrix $l$ of the parabola $C$: $y^{2}=a x$ ($a>0$) at point $B$, and intersects $C$ at a point $A$, and if $\\overrightarrow{B M}=\\overrightarrow{M A}$, then $a=$?",
    "fact_expressions": "M: Point;Coordinate(M) = (2, 0);G: Line;PointOnCurve(M, G);Slope(G) = sqrt(3);C: Parabola;Expression(C) = (y^2 = a*x);a: Number;a>0;l: Line;Directrix(C) = l;A: Point;B: Point;Intersection(G, l) = B;OneOf(Intersection(G, C)) = A;VectorOf(B, M) = VectorOf(M, A)",
    "query_expressions": "a",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 11]], [[2, 11]], [[26, 28]], [[1, 28]], [[12, 28]], [[29, 53], [68, 71]], [[29, 53]], [[127, 130]], [[37, 53]], [[56, 59]], [[29, 59]], [[77, 80]], [[62, 66]], [[26, 66]], [[26, 80]], [[82, 125]]]",
    "query_spans": "[[[127, 132]]]",
    "process": "The directrix of the parabola $ C: y^{2} = ax $ ($ a > 0 $) is $ l: x = -\\frac{a}{4} $. The equation of the line passing through point $ M(2,0) $ with slope $ \\sqrt{3} $ is $ y = \\sqrt{3}(x - 2) $. Solving the system of equations  \n\\[\n\\begin{cases}\ny = \\sqrt{3}(x - 2) \\\\\nx = -\\frac{a}{4}\n\\end{cases}\n\\]\ngives the intersection point $ B $ with coordinates $ \\left(-\\frac{a}{4}, \\frac{-\\sqrt{3}(a+8)}{4}\\right) $. Let the coordinates of point $ A $ be $ (x_{0}, y_{0}) $. Since $ \\overrightarrow{BM} = \\overrightarrow{MA} $, point $ M $ is the midpoint of segment $ AB $. Solving gives $ A\\left(4 + \\frac{a}{4}, \\frac{\\sqrt{3}(a+8)}{4}\\right) $. Substituting $ A\\left(4 + \\frac{a}{4}, \\frac{\\sqrt{3}(a+8)}{4}\\right) $ into the parabola equation yields $ \\left(\\frac{\\sqrt{3}(a+8)}{4}\\right)^{2} = a\\left(4 + \\frac{a}{4}\\right) $. Since $ a > 0 $, solving gives $ a = 8 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through the right focus $F_{2}$ with an inclination angle of $\\frac{\\pi}{4}$ intersects the hyperbola at points $M$ and $N$ (point $M$ lies in the first quadrant). The inradius of $\\Delta M F_{1} F_{2}$ is $R_{1}$, and the inradius of $\\Delta N F_{1} F_{2}$ is $R_{2}$. Then $\\frac{R_{1}}{R_{2}}$ is?",
    "fact_expressions": "l: Line;G: Hyperbola;M: Point;F1: Point;F2: Point;N: Point;Expression(G) = (x^2/4 - y^2/2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(F2,l);Inclination(l)=pi/4;Intersection(l,G)={M,N};Quadrant(M)=1;Radius(InscribedCircle(TriangleOf(M,F1,F2)))=R1;Radius(InscribedCircle(TriangleOf(N,F1,F2)))=R2;R1:Number;R2:Number",
    "query_expressions": "R1/R2",
    "answer_expressions": "3+2*sqrt(2)",
    "fact_spans": "[[[97, 102]], [[2, 40], [104, 107]], [[109, 112], [120, 124]], [[49, 56]], [[57, 64], [69, 76]], [[113, 116]], [[2, 40]], [[2, 64]], [[2, 64]], [[65, 102]], [[77, 102]], [[97, 118]], [[120, 130]], [[132, 168]], [[170, 207]], [[161, 168]], [[200, 207]]]",
    "query_spans": "[[[209, 232]]]",
    "process": ""
  },
  {
    "text": "If the distance from the vertex of the hyperbola $C: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$ to its asymptote is $\\frac{b}{2}$, then what is the minimum value of $\\frac{b^{2}+1}{\\sqrt{3} a}$?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;a > 0;b > 0;Expression(C) = (x**2/a**2 + (-y**2)/b**2 = 1);Distance(Vertex(C), Asymptote(C)) = b/2",
    "query_expressions": "Min((b**2 + 1)/((sqrt(3)*a)))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 62]], [[4, 62]], [[4, 62]], [[4, 62]], [[4, 62]], [[1, 62]], [[1, 86]]]",
    "query_spans": "[[[88, 121]]]",
    "process": ""
  },
  {
    "text": "If the foci of the ellipse $\\frac{x^{2}}{3 m}+\\frac{y^{2}}{2 m+1}=1$ lie on the $y$-axis, then the range of the real number $m$ is?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (y^2/(2*m + 1) + x^2/((3*m)) = 1);PointOnCurve(Focus(G), yAxis) = True",
    "query_expressions": "Range(m)",
    "answer_expressions": "(0,1)",
    "fact_spans": "[[[1, 44]], [[55, 60]], [[1, 44]], [[1, 53]]]",
    "query_spans": "[[[55, 67]]]",
    "process": "From the given conditions, $a^{2}=2m+1$, $b^{2}=3m$ $\\therefore\\begin{cases}2m+1>0\\\\3m>0\\\\2m+1>3m\\end{cases}$ Solving yields: $0<m<1$. Thus, the range of real number $m$ is $0<m<1$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0, a \\neq b)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A perpendicular is drawn from $F_{2}$ to an asymptote, with foot of perpendicular at $P$, and $\\tan \\angle P F_{1} F_{2}=\\frac{1}{3}$. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Negation(a = b);Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F2;LeftFocus(G) = F1;l: Line;PointOnCurve(F2, l);IsPerpendicular(l, Asymptote(G));FootPoint(l, Asymptote(G)) = P;Tan(AngleOf(P, F1, F2)) = 1/3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 68], [160, 163]], [[5, 68]], [[5, 68]], [[114, 117]], [[76, 83]], [[85, 92], [96, 103]], [[5, 68]], [[5, 68]], [[5, 68]], [[2, 68]], [[2, 92]], [[2, 92]], [], [[2, 110]], [[2, 110]], [[2, 117]], [[119, 158]]]",
    "query_spans": "[[[160, 169]]]",
    "process": "Let $\\angle PF_{1}F_{2} = \\alpha$, draw a perpendicular line from point $P$ to the $x$-axis, with foot $E$. It follows that $\\triangle POE$ is similar to $\\triangle F_{2}OP$, $|PF_{2}| = b$, $|OP| = a$, $|OF_{2}| = c$, $|PE| = \\frac{|OP||PF_{2}|}{|OF_{2}|}$, i.e., $|PE| = \\frac{ab}{c}$, $\\frac{|OE|}{|PF_{2}|} = \\frac{|OP|}{|OF_{2}|}$, so $|OE| = \\frac{a^{2}}{c}$, then $\\tan\\alpha = \\frac{1}{3} = \\frac{|PE|}{|F_{1}E|} = \\frac{|PE|}{|F_{1}O| + |OE|} = \\frac{\\frac{ab}{c}}{c + \\frac{a^{2}}{c}}$, and $a \\neq b$, so $2a^{2} - 3ab + b^{2} = 0$, $\\therefore b = 2a$ or $b = a$ (discarded). Thus $c^{2} - a^{2} = 4a^{2}$, $\\therefore e = \\sqrt{5}$."
  },
  {
    "text": "The standard equation of an ellipse with focal length $2$, minor axis length $4$, and foci on the $x$-axis is?",
    "fact_expressions": "G: Ellipse;FocalLength(G) = 2;Length(MinorAxis(G)) = 4;PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5 + y^2/4 = 1",
    "fact_spans": "[[[25, 27]], [[0, 27]], [[7, 27]], [[16, 27]]]",
    "query_spans": "[[[25, 34]]]",
    "process": "Since the focal distance of the ellipse is 2 and the length of the minor axis is 4, we have c=1, b=2, thus a^{2}=b^{2}+c^{2}=5. Moreover, since the foci of the ellipse lie on the x-axis, the standard equation of the ellipse is \\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has foci $F_{1}$ and $F_{2}$. If there exists a point $P$ on the ellipse such that $\\angle F_{1} P F_{2}=60^{\\circ}$, then the range of the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;P: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Focus(G)={F1,F2};PointOnCurve(P,G);AngleOf(F1,P,F2)=ApplyUnit(60,degree)",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 54], [78, 80], [124, 126]], [[4, 54]], [[4, 54]], [[60, 67]], [[85, 88]], [[68, 75]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 75]], [[78, 88]], [[90, 122]]]",
    "query_spans": "[[[124, 136]]]",
    "process": ""
  },
  {
    "text": "A focus of the hyperbola $8 k x^{2}-k y^{2}=8$ is $(0 , 3)$, then what is the value of $k$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (8*k*x^2 - k*y^2 = 8);k: Number;Coordinate(OneOf(Focus(G)))=(0,3)",
    "query_expressions": "k",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 24]], [[0, 24]], [[42, 45]], [0, 38]]",
    "query_spans": "[[[42, 49]]]",
    "process": ""
  },
  {
    "text": "The point $P(\\frac{3}{2}, y_{0})$ on the parabola $y^{2}=a x(a>0)$ has a distance of $2$ to the focus $F$. Then $a$=?",
    "fact_expressions": "G: Parabola;a: Number;P: Point;F: Point;a>0;y0:Number;Expression(G) = (y^2 = a*x);Coordinate(P) = (3/2, y0);PointOnCurve(P, G);Focus(G) = F;Distance(P, F) = 2",
    "query_expressions": "a",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 19]], [[60, 63]], [[21, 45]], [[48, 51]], [[3, 19]], [[22, 45]], [[0, 19]], [[21, 45]], [[0, 45]], [[0, 51]], [[21, 58]]]",
    "query_spans": "[[[60, 65]]]",
    "process": "\\because the distance from a point P(\\frac{3}{2},y) on the parabola y^{2}=ax(a>0) to the focus F is 2, \\therefore the distance from this point to the directrix is 2. The equation of the directrix of the parabola is x=-\\frac{a}{4}, \\therefore \\frac{3}{2}+\\frac{a}{4}=2, solving gives a=2,"
  },
  {
    "text": "Given that $M$ is a point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, $F_{1}$ and $F_{2}$ are the two foci of the ellipse, $I$ is the incenter of $\\Delta M F_{1} F_{2}$, and the extension of $MI$ intersects $F_{1} F_{2}$ at $N$, then $\\frac{|MI|}{|NI|}$ equals?",
    "fact_expressions": "G: Ellipse;M: Point;I: Point;F1: Point;F2: Point;N: Point;Expression(G) = (x^2/9 + y^2/5 = 1);PointOnCurve(M, G);Focus(G) = {F1, F2};Incenter(TriangleOf(M, F1, F2)) = I;Intersection(OverlappingLine(LineSegmentOf(M, I)), LineSegmentOf(F1, F2)) = N",
    "query_expressions": "Abs(LineSegmentOf(M, I))/Abs(LineSegmentOf(N, I))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[6, 43], [63, 65]], [[2, 5]], [[71, 74]], [[47, 54]], [[55, 62]], [[122, 125]], [[6, 43]], [[2, 46]], [[47, 70]], [[71, 100]], [[101, 125]]]",
    "query_spans": "[[[127, 149]]]",
    "process": ""
  },
  {
    "text": "The line $y = k(x - 2) + 4$ and the curve $y = 1 + \\sqrt{4 - x^{2}}$ have only one common point; then the range of real values for $k$ is?",
    "fact_expressions": "G: Line;Expression(G) = (y = k*(x - 2) + 4);k: Real;H: Curve;Expression(H) = (y = sqrt(4 - x^2) + 1);NumIntersection(G, H) = 1",
    "query_expressions": "Range(k)",
    "answer_expressions": "(3/4, +oo)+{5/12}",
    "fact_spans": "[[[0, 14]], [[0, 14]], [[46, 51]], [[15, 37]], [[15, 37]], [[0, 44]]]",
    "query_spans": "[[[46, 58]]]",
    "process": "As shown in the figure, from the given condition, the curve $ y = 1 + \\sqrt{4 - x^{2}} $, i.e., $ x^{2} + (y - 1)^{2} = 4 $, represents a semicircle with center at $ (0,1) $ and radius $ 2 $, located above the line $ y = 1 $. The line $ y = k(x - 2) + 4 $ always passes through the point $ (2,4) $. Since the line intersects the curve at only one point, the distance from the center to the line equals the radius, giving $ \\frac{|2k - 3|}{\\sqrt{1 + k^{2}}} = 2 $. Solving yields $ k = \\frac{5}{12} $. From the figure, when the line passes through the point $ (-2,1) $, the slope of the line is $ \\frac{4 - 1}{2 - (-2)} = \\frac{3}{4} $. When the line passes through the point $ (2,1) $, the slope of the line does not exist. In conclusion, the range of real number $ k $ is $ k = \\frac{5}{12} $, or $ k > \\frac{3}{4} $."
  },
  {
    "text": "Let the vertex of the parabola $C$: $y^{2}=4x$ be $O$, and let a line passing through the focus of the parabola $C$ and perpendicular to the $x$-axis intersect the parabola $C$ at points $A$ and $B$. Then $|\\overrightarrow{O A}+\\overrightarrow{O B}|=$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);O:Origin;Vertex(C) = O;G: Line;PointOnCurve(Focus(C), G) = True;IsPerpendicular(G, xAxis) = True;Intersection(G, C) = {A, B};A: Point;B: Point",
    "query_expressions": "Abs(VectorOf(O, A) + VectorOf(O, B))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 20], [30, 36], [51, 57]], [[1, 20]], [[24, 27]], [[1, 27]], [[48, 50]], [[28, 50]], [[40, 50]], [[48, 68]], [[59, 62]], [[63, 66]]]",
    "query_spans": "[[[70, 117]]]",
    "process": "The focus of the parabola $ C: y^{2} = 4x $ is $ (1,0) $. The line passing through the focus of the parabola $ C $ and perpendicular to the x-axis intersects the parabola $ C $ at points $ A $ and $ B $. Then $ A(1,2) $, $ B(1,-2) \\Rightarrow \\overrightarrow{OA} + \\overrightarrow{OB} = (2,0) $, so $ |\\overrightarrow{OA} + \\overrightarrow{OB}| = 2 $."
  },
  {
    "text": "On the line $l$: $x - y + 9 = 0$, take a point $M$. Through $M$, draw an ellipse with foci $F_{1}(-3,0)$, $F_{2}(3,0)$. When $|M F_{1}| + |M F_{2}|$ is minimized, the standard equation of the ellipse is?",
    "fact_expressions": "l: Line;Expression(l) = (x - y + 9 = 0);M: Point;PointOnCurve(M, l);G: Ellipse;F1: Point;F2: Point;Focus(G) = {F1, F2};Coordinate(F1) = (-3, 0);Coordinate(F2) = (3, 0);PointOnCurve(M, G);WhenMin(Abs(LineSegmentOf(M, F1)) + Abs(LineSegmentOf(M, F2)))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/45+y^2/36=1",
    "fact_spans": "[[[1, 17]], [[1, 17]], [[20, 23], [26, 29]], [[1, 23]], [[62, 64], [92, 94]], [[31, 44]], [[46, 58]], [[30, 64]], [[31, 44]], [[46, 58]], [[25, 64]], [[66, 91]]]",
    "query_spans": "[[[92, 101]]]",
    "process": ""
  },
  {
    "text": "Given an ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with focal length $2$, the maximum distance from a moving point $P$ on the ellipse $C$ to one of its foci is equal to $3$. Now, a line $l$ passes through the point $Q(1,\\ 1)$ and intersects the ellipse $C$ at points $A$ and $B$, such that point $Q$ is exactly the midpoint of $AB$. Then, the area of $\\Delta AOB$ is?",
    "fact_expressions": "l: Line;C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);FocalLength(C)=2;b: Number;a: Number;A: Point;B: Point;Q: Point;O: Origin;P: Point;a > b;b > 0;Coordinate(Q) = (1, 1);PointOnCurve(P, C);Max(Distance(P,OneOf(Focus(C))) = 3);PointOnCurve(Q, l);Intersection(l, C) = {A, B};MidPoint(LineSegmentOf(A, B)) = Q",
    "query_expressions": "Area(TriangleOf(A,O,B))",
    "answer_expressions": "sqrt(105)/6",
    "fact_spans": "[[[99, 104]], [[9, 66], [67, 72], [117, 122]], [[9, 66]], [[2, 66]], [[15, 66]], [[15, 66]], [[125, 128]], [[129, 132]], [[105, 116], [136, 140]], [[152, 166]], [[76, 79]], [[15, 66]], [[15, 66]], [[105, 116]], [[67, 79]], [[67, 94]], [[99, 116]], [[99, 134]], [[136, 150]]]",
    "query_spans": "[[[152, 171]]]",
    "process": "The maximum distance from a point P on the ellipse C to one of its foci is 3; then  \n\\begin{cases}2c=2\\\\a+c=3\\end{cases}, solving gives \\begin{cases}a=2\\\\c=1\\end{cases}, so $b^{2}=a^{2}-c^{2}=4-1=3$, thus the equation of the ellipse C is: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $\\frac{x_{1}^2}{4}+\\frac{y_{1}^{2}}{3}=1$ \\textcircled{1}, $\\frac{x_{2}^2}{4}+\\frac{y_{2}^2}{3}=1$ \\textcircled{2}, \\textcircled{1}$-$\\textcircled{2} yields: $\\frac{x_{1}^2-x_{2}^2}{4}+\\frac{y_{1}^2-y_{2}^2}{3}=0$, i.e., $\\frac{(x_{1}-x_{2})(x_{1}+x_{2})}{4}+\\frac{(y_{1}-y_{2})(y_{1}+y_{2})}{3}=0$. Since $x_{1}+x_{2}=2$, $y_{1}+y_{2}=2$, we have $\\frac{2(x_{1}-x_{2})}{4}+\\frac{2(y_{1}-y_{2})}{3}=0$, so the slope of line $l$ is $\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{3}{4}$, thus the equation of line $l$ is $y-1=-\\frac{3}{4}(x-1)$, i.e., $3x+4y-7=0$. From  \n\\begin{cases}\\frac{x^2}{4}+\\frac{y^{2}}{3}=1\\\\3x+4y-7=0\\end{cases}  \nwe obtain: $21x^{2}-42x+1=0$, so $x_{1}+x_{2}=2$, $x_{1}x_{2}=\\frac{1}{21}$. Thus the chord length $|AB|=\\sqrt{1+\\left(-\\frac{3}{4}\\right)^2}|x_{1}-x_{2}|=\\sqrt{1+\\left(-\\frac{3}{4}\\right)^2}\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\sqrt{\\frac{25}{16}}\\sqrt{4-4\\times\\frac{1}{21}}=\\frac{5}{4}\\times\\sqrt{\\frac{80}{21}}=\\frac{5\\sqrt{105}}{21}$. The distance from the origin $O(0,0)$ to the line $3x+4y-7=0$ is $d=\\frac{|-7|}{\\sqrt{3^{2}+4^{2}}}=\\frac{7}{5}$, so the area of $\\triangle AOB$ is $\\frac{1}{2}\\times|AB|\\times d=\\frac{1}{2}\\times\\frac{5\\sqrt{105}}{21}\\times\\frac{7}{5}=\\frac{\\sqrt{105}}{6}$."
  },
  {
    "text": "Given that the eccentricity of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{9}=1$ is equal to $\\frac{1}{3}$, then the real number $m$=?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (y^2/9 + x^2/m = 1);Eccentricity(G) = 1/3",
    "query_expressions": "m",
    "answer_expressions": "{8,81/8}",
    "fact_spans": "[[[2, 39]], [[60, 65]], [[2, 39]], [[2, 58]]]",
    "query_spans": "[[[60, 67]]]",
    "process": "Discuss the position of the foci of the ellipse. Find the eccentricity of the ellipse in two cases, then find the value of the real number $ m $. When the foci of the ellipse are on the x-axis, $ a^{2}=m $, $ b^{2}=9 $, $ c^{2}=m-9 $, so $ \\frac{m-9}{m}=\\frac{1}{9} $, solving gives: $ m=\\frac{81}{8} $. When the foci of the ellipse are on the y-axis, $ a^{2}=9 $, $ b^{2}=m $, $ c^{2}=9-m $, so $ \\frac{9-m}{9}=\\frac{1}{9} $, solving gives: $ m=8 $."
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. A point $M$ on the ellipse satisfies $\\angle M F_{1} F_{2}=30^{\\circ}$, $\\angle M F_{2} F_{1}=105^{\\circ}$. Then the eccentricity of the ellipse is?",
    "fact_expressions": "F1: Point;F2: Point;G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;LeftFocus(G) = F1;RightFocus(G) = F2;M: Point;PointOnCurve(M, G) = True;AngleOf(M, F1, F2) = ApplyUnit(30, degree);AngleOf(M, F2, F1) = ApplyUnit(105, degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(sqrt(2)+2-sqrt(6))/2",
    "fact_spans": "[[[0, 7]], [[8, 15]], [[16, 68], [75, 77], [157, 159]], [[16, 68]], [[18, 68]], [[18, 68]], [[18, 68]], [[18, 68]], [[0, 74]], [[0, 74]], [[80, 83]], [[75, 83]], [[85, 118]], [[121, 155]]]",
    "query_spans": "[[[157, 165]]]",
    "process": ""
  },
  {
    "text": "A line passing through the point $P(1,0)$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$, and $|AP|=2$. Then the equation of the line $AB$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(P) = (1, 0);PointOnCurve(P, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, P)) = 2",
    "query_expressions": "Expression(LineOf(A,B))",
    "answer_expressions": "NO ANSWER",
    "fact_spans": "[[[14, 28]], [[11, 13]], [[31, 34]], [[35, 38]], [[1, 10]], [[14, 28]], [[1, 10]], [[0, 13]], [[11, 40]], [[41, 50]]]",
    "query_spans": "[[[52, 64]]]",
    "process": ""
  },
  {
    "text": "Given that the vertex of the parabola is at the origin and the focus is $F(1 , 0)$, a line passing through point $F$ intersects the parabola at points $A$ and $B$, with $|AB|=4$. Then, the distance from the midpoint $M$ of segment $AB$ to the line $x=-2$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;F: Point;O: Origin;M: Point;L: Line;Expression(L) = (x = -2);Coordinate(F) = (1, 0);Vertex(G) = O;Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 4;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Distance(M, L)",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 5], [34, 37]], [[31, 33]], [[39, 42]], [[43, 46]], [[26, 30], [14, 24]], [[9, 11]], [[72, 75]], [[76, 84]], [[76, 84]], [[14, 24]], [[2, 11]], [[2, 24]], [[25, 33]], [[31, 48]], [[50, 60]], [[62, 75]]]",
    "query_spans": "[[[72, 89]]]",
    "process": "According to the problem, the directrix of the parabola is the line $x = -1$. Let the distances from points $A$ and $B$ to the line $x = -1$ be $d_{1}$ and $d_{2}$, respectively. By the definition of a parabola, we have $|AB| = |AF| + |BF| = 4$. Therefore, the distance from the midpoint $M$ of segment $AB$ to the line $x = -2$ is $\\frac{d_{1}+d_{2}}{2}+1=3$."
  },
  {
    "text": "If a point $M$ on the parabola $y^{2}=4x$ is at a distance of $4$ from the focus $F$, then what is the horizontal coordinate of point $M$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);M: Point;PointOnCurve(M, G);F: Point;Focus(G) = F;Distance(M, F) = 4",
    "query_expressions": "XCoordinate(M)",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 15]], [[1, 15]], [[18, 21], [36, 40]], [[1, 21]], [[24, 27]], [[1, 27]], [[18, 34]]]",
    "query_spans": "[[[36, 46]]]",
    "process": "According to the property that the distance from a point on a parabola to its focus equals the distance from the point to the directrix, we can establish a relationship and obtain the result. It is evident that the directrix of the parabola $ y^{2} = 4x $ is $ x = -1 $. Let $ M(x_{0}, y_{0}) $. Then the distance from $ M $ to the directrix is $ x_{0} + 1 $, which equals the distance from $ M $ to the focus $ F $, given as 4. Thus, $ x_{0} + 1 = 4 $, so $ x_{0} = 3 $, meaning the horizontal coordinate of point $ M $ is 3."
  },
  {
    "text": "Given that an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is parallel to the line $2 x+y+3=0$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;H: Line;Expression(H) = (2*x + y + 3 = 0);IsParallel(OneOf(Asymptote(G)), H) = True",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 58], [83, 86]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[65, 78]], [[65, 78]], [[2, 80]]]",
    "query_spans": "[[[83, 92]]]",
    "process": "Since one asymptote of the hyperbola is parallel to the line $2x + y + 3 = 0$, we obtain $\\frac{b}{a} = 2$, and then find the eccentricity of the hyperbola. According to the problem, the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ $(a > 0, b > 0)$ are given by $y = \\pm\\frac{b}{a}x$. Since one asymptote of the hyperbola is parallel to the line $2x + y + 3 = 0$, we have $-\\frac{b}{a} = -2$, that is, $\\frac{b}{a} = 2$, then $e = \\sqrt{1 + \\left(\\frac{b}{a}\\right)^{2}} = \\sqrt{5}$."
  },
  {
    "text": "Let the hyperbola $\\frac{y^{2}}{9}-\\frac{x^{2}}{a^{2}}=1(a>0)$ have foci $F_{1}$, $F_{2}$, and let $P$ be a point on this hyperbola. If $|P F_{2}|=2$ and $\\sin \\angle P F_{2} F_{1}=\\lambda \\sin \\angle P F_{1} F_{2}$, then $\\lambda=$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (y^2/9 - x^2/a^2 = 1);a: Number;a>0;F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, F2)) = 2;lambda: Number;Sin(AngleOf(P, F2, F1)) = lambda*Sin(AngleOf(P, F1, F2))",
    "query_expressions": "lambda",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 48], [73, 76]], [[1, 48]], [[4, 48]], [[4, 48]], [[52, 59]], [[60, 67]], [[1, 67]], [[68, 71]], [[68, 80]], [[82, 95]], [[160, 169]], [[97, 158]]]",
    "query_spans": "[[[160, 171]]]",
    "process": "First, use the definition of the hyperbola to find |PF_{1}|=8, then apply the sine law to obtain the answer. Since the hyperbola is \\frac{y^{2}}{9}-\\frac{x^{2}}{a^{2}}=1 (a>0), the foci of the hyperbola lie on the y-axis, and the length of the real axis is 2a=2\\times3=6. Since P is a point on this hyperbola, ||PF_{1}|-|PF_{2}||=6. Given that |PF_{2}|=2, it follows that |PF_{1}|=8 or |PF_{1}|=-4 (discarded). Let R be the circumradius of triangle PF_{2}F_{1}. Since \\sin\\angle PF_{2}F_{1}=\\lambda\\sin\\angle PF_{1}F_{2}, by the sine law we have \\frac{|PF_{1}|}{2R}=\\lambda\\frac{|PF_{2}|}{2R} \\Rightarrow \\lambda=\\frac{|PF_{1}|}{|PF_{2}|}=\\frac{8}{2}=4."
  },
  {
    "text": "Draw a line with slope $\\frac{1}{2}$ passing through the point $M(-2,1)$, intersecting the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ at two distinct points $A$ and $B$. If $M$ is the midpoint of $AB$, then the eccentricity $e$ of the ellipse is $?$.",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;A: Point;B: Point;M: Point;e: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(M) = (-2, 1);PointOnCurve(M, H);Slope(H) = 1/2;Intersection(H, G) = {A, B};Negation(A = B);MidPoint(LineSegmentOf(A, B)) = M;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[32, 84], [116, 118]], [[34, 84]], [[34, 84]], [[29, 31]], [[87, 90]], [[91, 94]], [[1, 11], [101, 104]], [[122, 125]], [[34, 84]], [[34, 84]], [[32, 84]], [[1, 11]], [[0, 31]], [[12, 31]], [[29, 99]], [[87, 99]], [[101, 113]], [[116, 125]]]",
    "query_spans": "[[[122, 127]]]",
    "process": "Let points A(x_{1},y_{1}) and B(x_{2},y_{2}), then \\begin{cases}x_{1}+x_{2}\\\\y_{1}+y_{2}\\end{cases}\\frac{2}{2}=2, from the given we have k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{1}{2}. According to the problem, we have \\begin{cases}\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1\\\\\\frac{x^{2}}{\\frac{y}{2}+\\frac{y^{2}}{2}}=1\\end{cases}, subtracting the two equations gives \\frac{x^{2}-x^{2}}{a^{2}}+\\frac{y^{2}-y^{2}}{b^{2}}=0. Therefore, \\frac{b^{2}}{a^{2}}=-\\frac{y_{2}}{x_{2}^{2}-x_{2}^{2}}=-\\frac{(y-y_{2})(y_{1}+y_{2})}{(x_{1}-x_{2})(x_{1}+x_{2})}=\\frac{1}{4},"
  },
  {
    "text": "If the asymptotes of hyperbola $C$ are given by $y = \\pm 2x$, and it passes through the point $(2, 2\\sqrt{2})$, then the standard equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Point;Coordinate(G) = (2, 2*sqrt(2));Expression(Asymptote(C)) = (y = pm*(2*x));PointOnCurve(G,C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 7], [48, 51]], [[29, 46]], [[29, 46]], [[1, 25]], [[1, 46]]]",
    "query_spans": "[[[48, 58]]]",
    "process": "From the asymptotes, the hyperbola is given by $4x^{2}-y^{2}=m$. Substituting the point $(2,2\\sqrt{2})$ gives $m=8$, so $4x^{2}-y^{2}=8$, thus $\\frac{x^{2}}{2}-\\frac{y^{2}}{8}=1$. Therefore, $a=\\sqrt{2}$, $b=2\\sqrt{2}$, $c=\\sqrt{10}$, and the directrices are $x=\\pm\\frac{a^{2}}{c}=\\pm\\frac{2}{\\sqrt{10}}=\\pm\\frac{\\sqrt{10}}{5}$."
  },
  {
    "text": "If a point $M$ on the parabola ${y}^{2}=2 x$ is at a distance of $\\sqrt{3}$ from the origin $O$, then what is the distance from point $M$ to the focus of the parabola?",
    "fact_expressions": "G: Parabola;M: Point;O: Origin;Expression(G) = (y^2 = 2*x);PointOnCurve(M, G);Distance(M, O) = sqrt(3)",
    "query_expressions": "Distance(M, Focus(G))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[1, 17], [55, 58]], [[21, 24], [49, 53]], [[25, 32]], [[1, 17]], [[1, 24]], [[49, 65]]]",
    "query_spans": "[[[49, 65]]]",
    "process": "Let point $ M\\left(\\frac{y^{2}}{2}, y\\right) $, since $ |MO| = \\sqrt{3} \\Rightarrow \\left(\\frac{y^{2}}{2} - 0\\right)^{2} + (y - 0)^{2} = 3 \\Rightarrow y^{2} = 2 $ or $ y^{2} = -6 $ (discarded), $ \\therefore x = \\frac{y^{2}}{2} = 1 $. The distance $ d $ from $ M $ to the directrix $ x = \\frac{1}{2} $ of the parabola $ y^{2} = 2x $ is $ 1 - \\frac{1}{2} = \\frac{3}{2} $. Since the distance from point $ M $ to the focus of the parabola equals the distance from point $ M $ to the directrix of the parabola $ y^{2} = 2x $, the distance from point $ M $ to the focus of this parabola is $ \\frac{3}{2} $."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $x^{2}-\\frac{y^{2}}{9}=1$, respectively. If point $P$ lies on the hyperbola such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, then $|\\overrightarrow{P F_{1}}+\\overrightarrow{P F_{2}}|$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/9 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G) = True;DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0",
    "query_expressions": "Abs(VectorOf(P, F1) + VectorOf(P, F2))",
    "answer_expressions": "2*sqrt(10)",
    "fact_spans": "[[[19, 47], [60, 63]], [[19, 47]], [[1, 8]], [[9, 16]], [[1, 53]], [[1, 53]], [[55, 59]], [[55, 64]], [[66, 125]]]",
    "query_spans": "[[[127, 182]]]",
    "process": ""
  },
  {
    "text": "The left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are $F_{1}$ and $F_{2}$, respectively. A perpendicular is drawn from $F_{1}$ to one of the asymptotes, intersecting the right branch of the hyperbola at point $P$. If $\\angle F_{1} P F_{2}=\\frac{\\pi}{4}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;F1: Point;P: Point;F2: Point;H:Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;IsPerpendicular(H,OneOf(Asymptote(G)));PointOnCurve(F1,H);Intersection(H,RightPart(G))=P;AngleOf(F1, P, F2) = pi/4",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[0, 56], [103, 106], [103, 106]], [[3, 56]], [[3, 56]], [[65, 72], [83, 90]], [[110, 114]], [[73, 80]], [], [[3, 56]], [[3, 56]], [[0, 56]], [[0, 80]], [[0, 80]], [[0, 101]], [[0, 101]], [[0, 114]], [[116, 152]]]",
    "query_spans": "[[[154, 163]]]",
    "process": "As shown in the figure, $ PF_{1} $ is perpendicular to an asymptote, then $ k_{PF_{1}} = \\frac{a}{b} $. Draw $ F_{2}A \\perp PF_{1} $ from $ F_{2} $, so $ \\frac{|AF_{2}|}{|AF|} = \\frac{a}{b} $. Also, $ |AF_{2}| + |AF_{1}|^{2} = |F_{1}F_{2}|^{2} = 4c^{2} $, $ \\therefore |AF_{2}| = 2a $, $ |AF| = 2b $. Moreover, in right triangle $ \\triangle PAF_{2} $, $ \\angle F_{1}PF_{2} = \\frac{\\pi}{4} $, hence $ |PA| = |AF_{2}| = 2a $, $ |PF_{2}| = 2\\sqrt{2}a $. By the definition of hyperbola: $ |PF_{1}| - |PF_{2}| = 2b + 2a - 2\\sqrt{2}a = 2a $, then $ b = \\sqrt{2}a $. $ e = \\frac{c}{a} = \\frac{\\sqrt{a^{2} + b^{2}}}{a} = \\sqrt{3} $"
  },
  {
    "text": "It is known that $F$ is the right focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $A$ is the left vertex of the ellipse $C$, and $P$ is a point on the ellipse $C$ such that $PF$ is perpendicular to the $x$-axis. If the angle of inclination of the line $PA$ is $30^{\\circ}$, then the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;A: Point;P: Point;F: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(C)=F;LeftVertex(C)=A;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P,F),xAxis);Inclination(LineOf(P,A))=ApplyUnit(30, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "1-(sqrt(3)/3)",
    "fact_spans": "[[[6, 63], [72, 77], [86, 91], [136, 141]], [[12, 63]], [[12, 63]], [[68, 71]], [[82, 85]], [[2, 5]], [[12, 63]], [[12, 63]], [[6, 63]], [[2, 67]], [[68, 81]], [[82, 94]], [[96, 108]], [[110, 134]]]",
    "query_spans": "[[[136, 147]]]",
    "process": "By the given conditions, since PF ⊥ x-axis, F is the focus, and P lies on the ellipse, let x = c, then we have $\\frac{c^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, solving gives $y=\\frac{b^{2}}{a}$ (the negative value is discarded since P is in the first quadrant), so $|PF|=\\frac{b^{2}}{a}$. Also, $|AF|=a+c$, $\\angle PAF=30^{\\circ}$, thus $\\frac{b^{2}}{a}=(a+c)\\tan 30^{\\circ}$, $\\sqrt{3}b^{2}=a(a+c)$, $3b^{4}=a^{2}(a+c)^{2}$, i.e., $3(a^{2}-c^{2})^{2}=a^{2}(a+c)^{2}$, $3(a-c)^{2}=a^{2}$, $3(1-e)^{2}=1$, $e=1-\\frac{\\sqrt{3}}{3}$."
  },
  {
    "text": "The hyperbola shares the same asymptotes as $\\frac{y^{2}}{16}-\\frac{x^{2}}{9}=1$ and passes through the point $A(-3 , 2\\sqrt {3})$. What is the distance from one focus of this hyperbola to one of its asymptotes?",
    "fact_expressions": "G: Hyperbola;A: Point;C:Hyperbola;Expression(G) = (-x^2/9 + y^2/16 = 1);Coordinate(A) = (-3,2*sqrt(3));Asymptote(G) = Asymptote(C);PointOnCurve(A,C)",
    "query_expressions": "Distance(OneOf(Focus(C)),OneOf(Asymptote(C)))",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 40]], [[51, 72]], [[73, 76]], [[1, 40]], [[51, 72]], [[0, 76]], [[49, 76]]]",
    "query_spans": "[[[73, 93]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has eccentricity $\\frac{\\sqrt{3}}{2}$ and focal length $2 \\sqrt{3}$, then the equation of the ellipse is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Eccentricity(C) = sqrt(3)/2;FocalLength(C) = 2*sqrt(3)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4+y^2=1",
    "fact_spans": "[[[2, 59], [102, 104]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[2, 84]], [[2, 100]]]",
    "query_spans": "[[[102, 109]]]",
    "process": "According to the problem, determine the values of $a$ and $b$ to obtain the equation of ellipse $C$. Let the semi-focal length of ellipse $C$ be $c$ ($c>0$), then $2c=2\\sqrt{3} \\Rightarrow c=\\sqrt{3}$. The eccentricity of ellipse $C$ is $e=\\frac{c}{a}=\\frac{\\sqrt{3}}{a}=\\frac{\\sqrt{3}}{2}$, which gives $a=2$, $\\therefore b=\\sqrt{a^{2}-c^{2}}=1$. Therefore, the equation of ellipse $C$ is $\\frac{x^{2}}{4}+y^{2}=1$."
  },
  {
    "text": "If the chord $AB$ of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ is bisected by the point $P(1,1)$, then the equation of the line containing $AB$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);A: Point;B: Point;IsChordOf(LineSegmentOf(A,B),G) = True;MidPoint(LineSegmentOf(A,B)) = P;Coordinate(P) = (1,1);P: Point",
    "query_expressions": "Expression(OverlappingLine(LineSegmentOf(A,B)))",
    "answer_expressions": "3*x+4*y-7=0",
    "fact_spans": "[[[1, 38]], [[1, 38]], [[40, 45]], [[40, 45]], [[1, 45]], [[40, 57]], [[46, 55]], [[46, 55]]]",
    "query_spans": "[[[59, 73]]]",
    "process": "Let the intersection points of the line and the ellipse be A(x_{1},y_{1}), B(x_{2},y_{2}). Since (1,1) is the midpoint of AB, we have x_{1}+x_{2}=2, y_{1}+y_{2}=2. Since points A and B lie on the ellipse, subtracting the equations yields \\underline{(x_{1}+x_{2})(x_{1}-x_{2})}, then \\frac{y_{1}-y_{2}}{x_{2}} = -\\frac{3(x_{1}^{4}+x_{2})}{4(y_{1}+y_{2})} = -\\frac{3}{4}; hence the required equation of the line is y-1 = -\\frac{3}{4}(x-1), or 3x+4y-7=0."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{10-m}+\\frac{y^{2}}{m-2}=1$, with the major axis on the $y$-axis. If the focal distance is $2 \\sqrt{2}$, then what is the value of $m$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(10 - m) + y^2/(m - 2) = 1);m: Number;OverlappingLine(MajorAxis(G),yAxis);FocalLength(G) = 2*sqrt(2)",
    "query_expressions": "m",
    "answer_expressions": "7",
    "fact_spans": "[[[2, 44]], [[2, 44]], [[72, 75]], [[2, 53]], [[2, 70]]]",
    "query_spans": "[[[72, 78]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, and a line $l$ with slope $-1$ intersecting $C$ at points $M$ and $N$. If $|MF| + |NF| = 8$, then the equation of line $l$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;Slope(l) = -1;l: Line;Intersection(l, C) = {M, N};M: Point;N: Point;Abs(LineSegmentOf(M, F)) + Abs(LineSegmentOf(N, F)) = 8",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x+y-1=0",
    "fact_spans": "[[[2, 21], [43, 46]], [[2, 21]], [[25, 28]], [[2, 28]], [[29, 42]], [[37, 42], [76, 79]], [[37, 57]], [[50, 53]], [[54, 57]], [[59, 74]]]",
    "query_spans": "[[[76, 84]]]",
    "process": "]Let the equation of line l be set and solved together with the parabola equation. Solve using the definition of the parabola. The parabola C: y^{2}=4x has focus F(1,0) and directrix equation x=-1. Since line l with slope -1 intersects C at points M and N, let the equation of line l be y=-x+m. Thus, we have \\begin{cases}y^2=4x\\\\y=-x+m\\end{cases}\\Rightarrow x^{2}-(2m+4)x+m^{2}=0. Let M(x_{1},y_{1}), N(x_{2},y_{2}). Therefore, x_{1}+x_{2}=2m+4. From |MF|+|NF|=8 \\Rightarrow x_{1}-(-1)+x_{2}-(-1)=8 \\Rightarrow 2m+4=6 \\Rightarrow m=1. Hence, the equation of line l is y=-x+1 \\Rightarrow x+y-1=0"
  },
  {
    "text": "Given the parabola $C$: $x^{2}=8 y$, with focus $F$, a line $l$ passing through $F$ intersects $C$ at points $A$ and $B$. The tangents to the parabola $C$ at $A$ and $B$ are drawn, and these two tangents intersect at point $P$. Then the equation of the locus of point $P$ is?",
    "fact_expressions": "l: Line;C: Parabola;F: Point;A: Point;B: Point;P:Point;Expression(C) = (x^2 = 8*y);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};TangentOnPoint(A,C)=l1;TangentOnPoint(B,C)=l2;Intersection(l1,l2)=P",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "y=-2",
    "fact_spans": "[[[34, 39]], [[2, 21], [40, 43], [57, 63]], [[25, 28], [30, 33]], [[44, 47], [64, 67]], [[48, 51], [68, 71]], [[82, 86], [82, 86]], [[2, 21]], [[2, 28]], [[29, 39]], [[34, 53]], [54, 74], [54, 74], [54, 85]]",
    "query_spans": "[[[88, 99]]]",
    "process": "\\because x^{2}=8y \\therefore F(0,2), by the problem condition: the slope of line l exists, let the equation of line l be: y=kx+2, A(x_{1},y_{1}), B(x_{2},y_{2}). Solving simultaneously: \\begin{cases}x^{2}=8y\\\\y=kx+2\\end{cases}, we get: x^{2}-8kx-16=0, x_{1}+x_{2}=8k, x_{1}x_{2}=-16. Also \\because y=\\frac{1}{4}x, \\therefore the slopes of the tangents at A and B are respectively: \\frac{1}{4}x_{1}, \\frac{1}{4}x_{2}. Hence the tangent lines at points A and B are: y-y_{1}=\\frac{1}{4}x_{1}(x-x_{1}), y-y_{2}=\\frac{1}{4}x_{2}(x-x_{2}). Solving: \\begin{cases}y-y_{1}=\\frac{1}{4}x_{1}(x-x_{1})\\\\y-y_{2}=\\frac{1}{4}x_{2}(x-x_{2})\\end{cases}. y_{1}=\\frac{1}{8}x_{1}^{2}, solving gives: x=\\frac{x_{2}-x_{1}}{2}, y=\\frac{x_{1}x_{2}}{8}=\\frac{-16}{8}=-2"
  },
  {
    "text": "If the curve $2 x=\\sqrt{4+y^{2}}$ and the line $y=m(x+1)$ have a common point, then the range of real values for $m$ is?",
    "fact_expressions": "G: Line;H: Curve;m: Real;Expression(G) = (y = m*(x + 1));Expression(H) = (2*x = sqrt(y^2 + 4));IsIntersect(H, G)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-2, 2)",
    "fact_spans": "[[[24, 36]], [[1, 23]], [[42, 47]], [[24, 36]], [[1, 23]], [[1, 40]]]",
    "query_spans": "[[[42, 54]]]",
    "process": "The curve $2x=\\sqrt{4+y^{2}}$, after rearranging, becomes $x^{2}-\\frac{y^{2}}{4}=1$ $(x>0)$, representing the right branch of a hyperbola with focus at $(\\sqrt{5},0)$. Solve by using the relationship between the line $y=m(x+1)$ and the asymptotes of the hyperbola. As shown in the figure, the curve $2x=\\sqrt{4+y^{2}}$, which is $x^{2}-\\frac{y^{2}}{4}=1$ $(x>0)$, represents the right branch of a hyperbola with focus at $(\\sqrt{5},0)$. At this time, the asymptotes of the hyperbola are $y=-2x$ and $y=2x$. Since $y=m(x+1)$ passes through the fixed point $(-1,0)$, for the line $y=m(x+1)$ to intersect the right branch of the hyperbola, its slope must lie between the two asymptotes, thus $-2<m<2$."
  },
  {
    "text": "Suppose the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ intersect the parabola $y=x^{2}+1$ at exactly two points. Then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = x^2 + 1);NumIntersection(Asymptote(G),H)=2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[1, 57], [85, 88]], [[4, 57]], [[4, 57]], [[62, 76]], [[4, 57]], [[4, 57]], [[1, 57]], [[62, 76]], [[1, 82]]]",
    "query_spans": "[[[85, 94]]]",
    "process": "By the symmetry of the asymptotes of the hyperbola, obtain the equation of one asymptote of the hyperbola, combine it with the parabola equation, eliminate y, and then based on the discriminant equal to 0, obtain b=2a. Then, using c=\\sqrt{a^{2}+b^{2}} and e=\\frac{c}{a}, the required eccentricity can be found. [Detailed solution] One asymptote of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0, b>0) is y=\\frac{b}{a}x. From the system of equations \\begin{cases}y=\\frac{b}{a}x\\\\v=1+x2\\end{cases}, eliminating y yields x^{2}-\\frac{b}{a}x+1=0 having a unique solution. Therefore, \\triangle=(\\frac{b}{a})^{2}-4=0, so \\frac{b}{a}=2, and e=\\frac{c}{a}=\\sqrt{1+(\\frac{b}{a})^{2}}=\\sqrt{1+4}=\\sqrt{5}. Similarly, the other asymptote of the hyperbola is also tangent to the parabola, yielding e=\\sqrt{5}."
  },
  {
    "text": "Given $F_{1}(-c , 0)$, $F_{2}(c , 0)$ are the two foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, and $P$ is a point on the ellipse such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=c^{2}$, then the range of the eccentricity of this ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;F1: Point;F2: Point;c: Number;Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = c^2",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[sqrt(3)/3, sqrt(2)/2]",
    "fact_spans": "[[[34, 79], [89, 91], [161, 163]], [[34, 79]], [[36, 79]], [[36, 79]], [[2, 17]], [[19, 33]], [[2, 17]], [[2, 17]], [[19, 33]], [[2, 84]], [[85, 88]], [[85, 94]], [[95, 158]]]",
    "query_spans": "[[[161, 173]]]",
    "process": ""
  },
  {
    "text": "The line $l$ passes through the point $P(1,1)$ and intersects the parabola $y^{2}=4x$ at points $A$ and $B$. If $P$ is exactly the midpoint of segment $AB$, then the slope of line $l$ is?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(P) = (1, 1);PointOnCurve(P, l);Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A,B)) = P",
    "query_expressions": "Slope(l)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 5], [60, 65]], [[16, 30]], [[32, 35]], [[36, 39]], [[43, 46], [6, 15]], [[16, 30]], [[6, 15]], [[0, 15]], [[0, 41]], [[43, 58]]]",
    "query_spans": "[[[60, 70]]]",
    "process": "According to the problem, using the \"point-difference method\" for the chord of a parabola passing through its midpoint, we can solve it. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then we have  \n$$\n\\begin{cases}\ny_{1} = 4x \\\\\ny_{2}^{2} = 4\n\\end{cases}\n$$\nSubtracting the two equations gives $ y_{1}^{2} - y_{2}^{2} = 4(x_{1} - x_{2}) $, that is, $ (y_{1} + y_{2})(y_{1} - y_{2}) = 4(x_{1} - x_{2}) $, we obtain $ \\frac{y_{1} - y_{2}}{x_{1} - x_{2}} = \\frac{4}{y_{1} + y_{2}} $. Since point $ P(1,1) $ is the midpoint of $ AB $, we have $ y_{1} + y_{2} = 2 $, thus $ \\frac{y_{1} - y_{2}}{x_{1} - x_{2}} = \\frac{4}{2} = 2 $. Therefore, the slope of line $ l $ is $ k = 2 $."
  },
  {
    "text": "Given two points $A(-2,0)$, $B(0,2)$, and point $P$ is any point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, then the maximum distance from point $P$ to line $AB$ is?",
    "fact_expressions": "A: Point;B: Point;G: Ellipse;P: Point;Expression(G) = (x^2/16 + y^2/9 = 1);Coordinate(A) = (-2, 0);Coordinate(B) = (0, 2);PointOnCurve(P, G)",
    "query_expressions": "Max(Distance(P, LineOf(A, B)))",
    "answer_expressions": "7*sqrt(2)/2",
    "fact_spans": "[[[4, 13]], [[15, 23]], [[30, 68]], [[25, 29], [75, 79]], [[30, 68]], [[4, 13]], [[15, 23]], [[25, 73]]]",
    "query_spans": "[[[75, 94]]]",
    "process": ""
  },
  {
    "text": "Given an ellipse centered at the origin with coordinate axes as its axes of symmetry, one focus at $F(2,0)$, and the midpoint of the chord intercepted by the line $l$: $y = x + 3$ having an $x$-coordinate of $-2$, then the standard equation of this ellipse is?",
    "fact_expressions": "l: Line;G: Ellipse;F: Point;O: Origin;OneOf(Focus(G))=F;Coordinate(F) = (2, 0);Center(G) = O;SymmetryAxis(G)=axis;XCoordinate(MidPoint(InterceptChord(l,G)))=-2;Expression(l)=(y=x+3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/8+y^2/4=1",
    "fact_spans": "[[[40, 54]], [[16, 18], [37, 39], [73, 75]], [[28, 36]], [[5, 7]], [[16, 36]], [[28, 36]], [[2, 18]], [[8, 18]], [[37, 70]], [[40, 54]]]",
    "query_spans": "[[[73, 82]]]",
    "process": "Let the equation of the ellipse be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. From $\\begin{cases}\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1\\\\y=x+3\\end{cases}$, we obtain $(a^{2}+b^{2})x^{2}+6a^{2}x+9a^{2}-a^{2}b^{2}=0$. Thus, $x_{1}+x_{2}=-\\frac{6a^{2}}{a^{2}+b^{2}}$. According to the condition, $-\\frac{6a^{2}}{a^{2}+b^{2}}=-2\\times2$, so $a^{2}=2b^{2}$. Also, $c=2$, hence $a^{2}-b^{2}=b^{2}=c^{2}=4$, $a^{2}=8$. The equation of the ellipse is $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$."
  },
  {
    "text": "Let real numbers $a$, $b$ satisfy $0 \\leq a$, $b \\leq 8$, and $b^{2}=16+a^{2}$. Then the maximum value of $b-a$ is?",
    "fact_expressions": "a: Real;b:Real;0<=a;b<=8;b^2=16+a^2",
    "query_expressions": "Max(-a + b)",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 6]], [[8, 11]], [[13, 23]], [[25, 35]], [[37, 53]]]",
    "query_spans": "[[[55, 66]]]",
    "process": "b^{2}=16+a^{2}, which is \\frac{b^{2}}{16}-\\frac{a^{2}}{16}=1, or equivalently \\frac{b^{2}}{16}-\\frac{a^{2}}{16}=1. Taking the a-axis and b-axis as the horizontal and vertical axes respectively, the curve represented by the equation is the portion of a hyperbola with foci on the vertical axis located in the first quadrant (including points on the vertical axis). The intersection point of this arc with the vertical axis is (0,4). Let the objective function be b-a=t. By translating the line b-a=0, when it passes through the point (0,4), the value of t is maximized. That is, when the objective function passes through the point (0,4), t achieves its maximum value. Thus, t=b-a=4-0=4. Therefore, the maximum value of b-a is 4."
  },
  {
    "text": "If the right focus of the ellipse $C$: $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$ is $F$, and it intersects the line $l$: $x-\\sqrt{3} y+2=0$ at points $P$ and $Q$, then what is the perimeter of $\\triangle P Q F$?",
    "fact_expressions": "C: Ellipse;l: Line;P: Point;Q: Point;F: Point;Expression(C) = (x^2/8 + y^2/4 = 1);Expression(l) = (x - sqrt(3)*y + 2 = 0);RightFocus(C) = F;Intersection(C,l) = {P,Q}",
    "query_expressions": "Perimeter(TriangleOf(P, Q, F))",
    "answer_expressions": "8*sqrt(2)",
    "fact_spans": "[[[1, 43]], [[54, 79]], [[81, 84]], [[85, 88]], [[48, 51]], [[1, 43]], [[54, 79]], [[1, 51]], [[1, 90]]]",
    "query_spans": "[[[92, 114]]]",
    "process": "Find the coordinates of the left focus, and since the line passes through the left focus of the ellipse, use the definition of the ellipse to find the perimeter of the triangle. From the problem, the left focus $ F $ of ellipse $ C $ is $ (-2,0) $, so the line $ l: x - \\sqrt{3}y + 2 = 0 $ passes through the left focus $ F $. Therefore, the perimeter of $ \\Delta PQF $ is $ |PQ| + |PF| + |QF| = |PF| + |PF| + |QF| + |QF| = 4a = 8\\sqrt{2} $."
  },
  {
    "text": "The line passing through the left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ and perpendicular to the $x$-axis intersects the hyperbola at points $M$ and $N$. The circle with $MN$ as diameter passes exactly through the right vertex of the hyperbola. Then, the eccentricity of the hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;PointOnCurve(LeftFocus(G), L) = True;IsPerpendicular(L, xAxis) = True;L: Line;Intersection(L, G) = {M, N};M: Point;N: Point;IsDiameter(LineSegmentOf(M, N), V) = True;PointOnCurve(RightVertex(G), V) = True;V: Circle",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 60], [76, 79], [105, 108], [114, 117]], [[1, 60]], [[4, 60]], [[4, 60]], [[4, 60]], [[4, 60]], [[0, 75]], [[65, 75]], [[73, 75]], [[73, 91]], [[82, 85]], [[85, 88]], [[92, 102]], [[101, 112]], [[101, 102]]]",
    "query_spans": "[[[114, 124]]]",
    "process": ""
  },
  {
    "text": "Given that $M(x_{0}, y_{0})$ is a point on the parabola $y^{2}=4 x$, $F$ is the focus of the parabola, and the point $P(-1,0)$ satisfies $\\overrightarrow{M F} \\cdot \\overrightarrow{M P}<0$, then what is the range of values for $x_{0}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);x0: Number;y0: Number;M: Point;Coordinate(M) = (x0, y0);PointOnCurve(M, G);F: Point;Focus(G) = F;P: Point;Coordinate(P) = (-1, 0);DotProduct(VectorOf(M, F), VectorOf(M, P)) < 0",
    "query_expressions": "Range(x0)",
    "answer_expressions": "[0, \\sqrt{5}-2)",
    "fact_spans": "[[[20, 34], [42, 45]], [[20, 34]], [[115, 122]], [[2, 19]], [[2, 19]], [[2, 19]], [[2, 37]], [[38, 41]], [[38, 48]], [[50, 60]], [[50, 60]], [[62, 113]]]",
    "query_spans": "[[[115, 129]]]",
    "process": ""
  },
  {
    "text": "Draw a line $l$ passing through the point $M(1 , 1)$ intersecting the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$ at points $A$ and $B$, such that $M$ is the midpoint of $AB$. Then the equation of line $l$ is?",
    "fact_expressions": "l: Line;M: Point;Coordinate(M) = (1, 1);PointOnCurve(M, l);G: Ellipse;Expression(G) = (x^2/4 + y^2/9 = 1);A: Point;B: Point;Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Expression(l)",
    "answer_expressions": "4*x+9*y-13=0",
    "fact_spans": "[[[14, 19], [82, 87]], [[2, 13], [69, 72]], [[2, 13]], [[1, 19]], [[20, 57]], [[20, 57]], [[58, 61]], [[62, 65]], [[14, 67]], [[69, 80]]]",
    "query_spans": "[[[82, 91]]]",
    "process": ""
  },
  {
    "text": "Let $P$ be a point on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1$ $(a>0)$, and one asymptote of the hyperbola is $3x-2y=0$. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola, respectively. If $|P F_{1}|=3$, then $|P F_{2}|=$?",
    "fact_expressions": "G: Hyperbola;a: Number;P: Point;F1: Point;F2: Point;a>0;Expression(G) = (-y^2/9 + x^2/a^2 = 1);PointOnCurve(P, G);Expression(OneOf(Asymptote(G))) = (3*x - 2*y = 0);LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(P, F1)) = 3",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "7",
    "fact_spans": "[[[5, 52], [56, 59], [101, 104]], [[8, 52]], [[1, 4]], [[82, 89]], [[91, 98]], [[8, 52]], [[5, 52]], [[1, 55]], [[56, 79]], [[82, 110]], [[82, 110]], [[112, 125]]]",
    "query_spans": "[[[127, 140]]]",
    "process": "By the given condition, point $P$ lies on the hyperbola $\\frac{x^{2}}{a^2}-\\frac{y^{2}}{9}=1$, and one asymptote of the hyperbola is $3x-2y=0$. From this, we obtain $\\frac{3}{a}=\\frac{3}{2}$, solving gives $a=2$, then $c=\\sqrt{13}$. Since $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola respectively, and $PF_{1}=3$, it follows that point $P$ lies on the left branch of the hyperbola. According to the definition of a hyperbola, we have $PF_{2}=2a+|PF_{1}|=2\\times2+3=7$."
  },
  {
    "text": "Given that one asymptote of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ has the equation $y=2 x$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Expression(OneOf(Asymptote(C))) = (y = 2*x)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 65], [84, 90]], [[2, 65]], [[9, 65]], [[9, 65]], [[9, 65]], [[9, 65]], [[2, 82]]]",
    "query_spans": "[[[84, 96]]]",
    "process": ""
  },
  {
    "text": "It is known that the focus of the parabola $y^{2}=m x$ coincides with the right vertex of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$. Then the value of the real number $m$ is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;m: Real;Expression(G) = (x^2/4 - y^2/3 = 1);Expression(H) = (y^2 = m*x);Focus(H) = RightVertex(G)",
    "query_expressions": "m",
    "answer_expressions": "8",
    "fact_spans": "[[[20, 58]], [[2, 16]], [[66, 71]], [[20, 58]], [[2, 16]], [[2, 64]]]",
    "query_spans": "[[[66, 75]]]",
    "process": "According to the given conditions, the focus of the parabola is $(\\frac{m}{4},0)$, and the right vertex of the hyperbola is $(2,0)$, so $\\frac{m}{4}=2$, that is, $m=8$."
  },
  {
    "text": "The asymptotes of the hyperbola are given by $3 x \\pm 4 y=0$, and the coordinates of the foci are $(\\pm 5,0)$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Coordinate(Focus(G)) = (pm*5, 0);Expression(Asymptote(G)) = (3*x + pm*4*y = 0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16 - y^2/9 = 1",
    "fact_spans": "[[[0, 3], [44, 47]], [[0, 42]], [[0, 25]]]",
    "query_spans": "[[[44, 52]]]",
    "process": "\\because the asymptotes of the hyperbola are given by 3x\\pm4y=0, we obtain 3a=4b, b=\\frac{3}{4}a \\begin{cases}b=\\frac{3}{4}a\\\\a^{2}+b^{2}=c^{2}\\end{cases} solving yields \\begin{cases}a^{2}=16\\\\b^{2}=9\\end{cases} the equation of the hyperbola is \\frac{x2}{16}-\\frac{y^{2}}{9}=1"
  },
  {
    "text": "Given the parabola $E$: $y^{2}=4x$, and the line $l$: $y=k(x-1)$ $(k>0)$. The line $l$ intersects the parabola $E$ at points $A$ and $B$. The extension of $AB$ intersects the directrix of the parabola $E$ at point $C$. If $S_{\\triangle OAB}=S_{\\triangle OBC}$ (where $O$ is the origin), then $k=$?",
    "fact_expressions": "l: Line;E: Parabola;A: Point;B: Point;O: Origin;C:Point;Expression(l) = (y = k*(x - 1));Expression(E) = (y^2 = 4*x);Intersection(l, E) = {A, B};Intersection(OverlappingLine(LineSegmentOf(A,B)),Directrix(E))=C;Area(TriangleOf(O,A,B))=Area(TriangleOf(O,B,C));k:Number;k>0",
    "query_expressions": "k",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[22, 44], [45, 50]], [[2, 21], [51, 57], [81, 87]], [[60, 63]], [[64, 67]], [[141, 144]], [[91, 95]], [[22, 44]], [[2, 21]], [[45, 69]], [[71, 95]], [[96, 138]], [[152, 155]], [[29, 44]]]",
    "query_spans": "[[[152, 157]]]",
    "process": "From $ S_{AOAB} = S_{AOBC} $, we get that $ B $ is the midpoint of $ AC $, so $ B(x_{1}, y_{1}) $, $ x_{2} = 2x_{1} + 1 $, $ y_{1} < 0 $. From $ y^{2} = 4x $, $ y = k(x - 1) $, we get $ k^{2}x^{2} - (4 + 2k)x + k^{2} = 0 $. $ \\therefore x_{1}x_{2} = 1 $. $ \\therefore x_{1}(2x_{1} + 1) = 1 $. $ \\therefore x_{1} = \\frac{1}{2} $. $ \\therefore y_{1} = -\\sqrt{2} $, $ k = \\frac{-\\sqrt{2}}{2} = 2\\sqrt{2} $."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=2x$ intersects the parabola at points $A$ and $B$. A perpendicular is drawn from point $F$ to the segment $AB$, intersecting the directrix of the parabola at point $G$. Given that $|FG|=\\frac{3}{2}$ and $O$ is the origin, then $S_{\\Delta AOB}=$?",
    "fact_expressions": "C: Parabola;H: Line;B: Point;A: Point;F: Point;G: Point;O: Origin;Expression(C) = (y^2 = 2*x);Focus(C)=F;PointOnCurve(F, H);Intersection(H, C) = {A, B};L:Line;PointOnCurve(F,L);IsPerpendicular(LineSegmentOf(A,B),L);Intersection(L,Directrix(C))=G;Abs(LineSegmentOf(F, G)) = 3/2",
    "query_expressions": "Area(TriangleOf(A,O,B))",
    "answer_expressions": "3/4",
    "fact_spans": "[[[1, 15], [25, 28], [59, 62]], [[21, 23]], [[34, 37]], [[30, 33]], [[17, 20], [42, 46]], [[66, 70]], [[92, 95]], [[1, 15]], [[1, 20]], [[0, 23]], [[21, 39]], [], [[41, 57]], [[41, 57]], [[41, 70]], [[72, 91]]]",
    "query_spans": "[[[102, 122]]]",
    "process": "From the properties of the parabola, we have |FG| = \\frac{1}{2}|y_{A}-y_{B}|, from which the value of |y_{A}-y_{B}| can be obtained. Substituting this gives the value of S_{AAOB}. According to the problem, a line passing through the focus F of the parabola y^{2} = 2x intersects the parabola at points A and B. Then, a perpendicular is drawn from point F to segment AB, intersecting the directrix of the parabola at point G. Thus, p = 1, F(\\frac{1}{2}, 0). Given |FG| = \\frac{1}{2}|y_{A}-y_{B}| and |FG| = \\frac{3}{2}, we obtain |y_{A}-y_{B}| = 3. Therefore, S_{AAOB} = \\frac{1}{2} \\times |OF| \\times |y_{A}-y_{B}| = \\frac{1}{2} \\times \\frac{1}{2} \\times 3 = \\frac{3}{4}"
  },
  {
    "text": "Given the line $l$: $y = m x - 4$ and the parabola $C$: $y^{2} = 8 x$, if $l$ and $C$ have exactly one common point, then the value of the real number $m$ is?",
    "fact_expressions": "l: Line;C: Parabola;m: Real;Expression(C) = (y^2 = 8*x);Expression(l)=(y=m*x-4);NumIntersection(l, C) = 1",
    "query_expressions": "m",
    "answer_expressions": "{0,-1/2}",
    "fact_spans": "[[[2, 18], [40, 43]], [[19, 38], [44, 47]], [[58, 63]], [[19, 38]], [[2, 18]], [[40, 56]]]",
    "query_spans": "[[[58, 67]]]",
    "process": "When the slope $ m = 0 $, the line $ l: y = mx - 4 $ is parallel to the $ x $-axis and intersects the parabola $ y^{2} = 8x $ at exactly one point. When the slope is not equal to 0, substituting $ y = mx - 4 $ into the parabola $ y^{2} = 8x $ yields $ m^{2}x^{2} + (-8m - 8)x + 16 = 0 $. According to the problem, this equation has a unique solution, so the discriminant $ A = (-8m - 8)^{2} - 4 \\times 16m^{2} = 0 $, hence $ m = -\\frac{1}{2} $."
  },
  {
    "text": "If a point $P$ on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$ is at a distance of $\\frac{3}{2}$ from one focus, then the distance from point $P$ to the other focus is?",
    "fact_expressions": "G: Ellipse;P: Point;F1:Point;F2:Point;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2);Expression(G) = (x^2/4 + y^2/9 = 1);PointOnCurve(P, G);Distance(P, F1) = 3/2",
    "query_expressions": "Distance(P, F2)",
    "answer_expressions": "9/2",
    "fact_spans": "[[[1, 38]], [[41, 44], [68, 72]], [], [], [[1, 49]], [[1, 77]], [[1, 77]], [[1, 38]], [[1, 44]], [[1, 66]]]",
    "query_spans": "[[[1, 82]]]",
    "process": "From the ellipse equation, we get $ a = 3 $, $ |PF_{1}| + |PF_{2}| = 6 $. Given $ |PF_{1}| = \\frac{3}{2} $, then $ |PF_{2}| = 6 - \\frac{3}{2} = \\frac{9}{2} $."
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ lie on the $y$-axis. Then the range of values for $m$ is?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/4 + y^2/m = 1);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(4, +oo)",
    "fact_spans": "[[[0, 37]], [[48, 51]], [[0, 37]], [[0, 46]]]",
    "query_spans": "[[[48, 58]]]",
    "process": "\\because the foci of the ellipse \\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1 are on the y-axis, and according to the standard equation of an ellipse, \\therefore m>4"
  },
  {
    "text": "Given that a moving circle $C$ passes through the point $F(0,1)$ and is tangent to the line $y=-1$. If the line $3x-4y+20=0$ intersects the circle $C$, then the minimum area of circle $C$ is?",
    "fact_expressions": "C: Circle;F: Point;Coordinate(F) = (0, 1);PointOnCurve(F, C);G: Line;Expression(G) = (y = -1);IsTangent(C, G);L: Line;Expression(L) = (3*x - 4*y + 20 = 0);IsIntersect(L, C)",
    "query_expressions": "Min(Area(C))",
    "answer_expressions": "4*pi",
    "fact_spans": "[[[4, 7], [61, 65], [51, 55]], [[9, 18]], [[9, 18]], [[4, 18]], [[21, 29]], [[21, 29]], [[4, 31]], [[34, 50]], [[34, 50]], [[34, 59]]]",
    "query_spans": "[[[61, 74]]]",
    "process": "The trajectory of the known center C is a parabola with focus F and directrix y = -1, its equation being x^{2} = 4y. Since the line 3x - 4y + 20 = 0 and circle C have common points, it follows that \\frac{|3a - 4b + 20|}{\\sqrt{3^{2} + (-4)^{2}}} \\leqslant r = \\frac{a^{2}}{4} + 1. Because circle C is tangent to the line y = -1, in order for the radius of the circle to be minimized, C(a, b) and O(0, 0) must lie on the same side within the plane region represented by the inequality 3x - 4y + 20 > 0. Thus we obtain \\frac{-a^{2} + 3a + 20}{5} \\leqslant \\frac{a^{2}}{4} + 1, solving which gives a \\leqslant -2 or a \\geqslant \\frac{10}{3}. When a = -2, b = 1, r = 2; when a = \\frac{10}{3}, r > 2, which does not meet the condition. Therefore, the minimum area of the circle is S = \\pi \\times 2^{2} = 4\\pi."
  },
  {
    "text": "A line $l$ with slope $k>0$ passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$, and intersects the directrix $q$ at point $M$. Given $|A F|=\\frac{3}{2}|B M|$, find $k=?$",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;l: Line;PointOnCurve(F,l) = True;Slope(l) = k;k: Number;k>0;Intersection(l,G) = {A,B};A: Point;B: Point;Directrix(G) = q;q: Line;Intersection(q,l) = M;M: Point;Abs(LineSegmentOf(A, F)) = (3/2)*Abs(LineSegmentOf(B, M))",
    "query_expressions": "k",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[1, 15], [37, 40]], [[1, 15]], [[18, 21]], [[1, 21]], [[22, 27]], [[0, 27]], [[22, 36]], [[90, 93]], [[30, 35]], [[22, 51]], [[42, 45]], [[46, 49]], [[37, 58]], [[55, 58]], [[22, 63]], [[59, 63]], [[64, 88]]]",
    "query_spans": "[[[90, 95]]]",
    "process": ""
  },
  {
    "text": "If a moving point $M(x, y)$ on the plane satisfies: $\\sqrt{(x+3)^{2}+y^{2}}=10-\\sqrt{(x-3)^{2}+y^{2}}$, then what is the standard equation of the trajectory of point $M$?",
    "fact_expressions": "M: Point;x1:Number;y1:Number;Coordinate(M) = (x1, y1);sqrt(y1^2 + (x1 + 3)^2) = 10 - sqrt(y1^2 + (x1 - 3)^2)",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2/25 + y^2/16 = 1",
    "fact_spans": "[[[7, 16], [73, 76]], [[7, 16]], [[7, 16]], [[7, 16]], [[19, 69]]]",
    "query_spans": "[[[73, 86]]]",
    "process": "The given equation is: $\\sqrt{(x+3)^{2}+y^{2}}+\\sqrt{(x-3)^{2}+y^{2}}=10$. Combining with the point-to-point distance formula, the geometric meaning of this expression is that the sum of the distances from point $M$ to the fixed points $(-3,0)$ and $(3,0)$ is a constant $10$. Since $\\frac{10}{2}>3$, the trajectory equation represents an ellipse with foci on the $x$-axis, where $a=5$, $c=3$, hence $b=4$. Therefore, the standard equation of the trajectory of the moving point $M$ is $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$."
  },
  {
    "text": "$P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, and $M$, $N$ are points on the circles $(x+3)^{2}+y^{2}=2$ and $(x-3)^{2}+y^{2}=1$, respectively. Then the maximum value of $|P M|-|P N|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/5 = 1);H: Circle;Expression(H) = (y^2 + (x + 3)^2 = 2);C: Circle;Expression(C) = (y^2 + (x - 3)^2 = 1);P: Point;M: Point;N: Point;PointOnCurve(P, RightPart(G));PointOnCurve(M, H);PointOnCurve(N, C)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)))",
    "answer_expressions": "5 + sqrt(2)",
    "fact_spans": "[[[4, 42]], [[4, 42]], [[59, 79]], [[59, 79]], [[80, 99]], [[80, 99]], [[0, 3]], [[49, 52]], [[53, 56]], [[0, 48]], [[49, 102]], [[49, 102]]]",
    "query_spans": "[[[104, 123]]]",
    "process": "Let the left and right foci of the hyperbola be $ F_{1} $ and $ F_{2} $, then $ |PF_{1}| - |PF_{2}| = 2a = 4 $. The circle $ (x+3)^{2} + y^{2} = 2 $ has center $ F_{1}(-3,0) $ and radius $ r_{1} = \\sqrt{2} $. The circle $ (x-3)^{2} + y^{2} = 1 $ has center $ F_{2}(3,0) $ and radius $ r_{2} = 1 $. By the symmetry of the circles, we have $ |PF_{1}| - r_{1} \\leqslant |PM| \\leqslant |PF_{1}| + r_{1} $, $ |PF_{2}| - r_{2} \\leqslant |PN| \\leqslant |PF_{2}| + r_{2} $. Therefore, $ |PM| - |PN| \\leqslant |PF_{1}| + r_{1} - |PF_{2}| + r_{2} = 5 + \\sqrt{2} $, that is, the maximum value of $ |PM| - |PN| $ is $ 5 + \\sqrt{2} $."
  },
  {
    "text": "If a point $P$ on the parabola $y^{2}=4x$ is at a distance of $10$ from the focus $F$, then what is the abscissa of point $P$?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(P, G);Focus(G) = F;Distance(P, F) = 10",
    "query_expressions": "XCoordinate(P)",
    "answer_expressions": "9",
    "fact_spans": "[[[1, 15]], [[18, 21], [37, 41]], [[24, 27]], [[1, 15]], [[1, 21]], [[1, 27]], [[18, 35]]]",
    "query_spans": "[[[37, 47]]]",
    "process": "From the analytical expression of the parabola, it is known that the equation of the directrix is $x = -1$. According to the definition of the parabola, the distance from point $P$ to the directrix is $10$, so the horizontal coordinate of point $P$ is $-1 + 10 = 9$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$, what is the length of the chord passing through its focus and perpendicular to the $x$-axis?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;H: LineSegment;IsChordOf(H, G);PointOnCurve(Focus(G), H);IsPerpendicular(H, xAxis)",
    "query_expressions": "Length(H)",
    "answer_expressions": "2*b^2/a",
    "fact_spans": "[[[2, 59], [62, 63]], [[2, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [], [[62, 76]], [[61, 76]], [[62, 76]]]",
    "query_spans": "[[[62, 79]]]",
    "process": "Use the substitution method to solve: let a focus be $ F(c,0) $, where $ c^{2} = a^{2} + b^{2} $. Let chord $ AB $ pass through $ F $ and be perpendicular to the $ x $-axis, then $ A(c,y_{0}) $. Since $ A(c,y_{0}) $ lies on the hyperbola, $ \\frac{c^{2}}{a^{2}} - \\frac{y_{0}^{2}}{b^{2}} = 1 \\cdots y_{0} = \\pmb{\\sqrt{\\frac{c^{2}}{a^{2}} - 1}} = \\pm\\frac{b^{2}}{a} \\cdots |AB| = 2|y_{0}| = \\frac{2b^{2}}{a} $"
  },
  {
    "text": "Let $P$ be a moving point on the circle $x^{2}+y^{2}=9$, $PD$ is perpendicular to the $x$-axis, with foot at $D$, and $M$ is a point on the segment $PD$ such that $\\overrightarrow{P M}=\\frac{1}{3} \\overrightarrow{P D}$. When point $P$ moves on the circle, the trajectory equation of the moving point $M$ is?",
    "fact_expressions": "G: Circle;P: Point;D: Point;M: Point;Expression(G) = (x^2 + y^2 = 9);PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, D), xAxis);FootPoint(LineSegmentOf(P, D), xAxis) = D;PointOnCurve(M, LineSegmentOf(P, D));VectorOf(P, M) = (1/3)*VectorOf(P, D)",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2/9 + y^2/4 = 1",
    "fact_spans": "[[[5, 21], [124, 125]], [[1, 4], [119, 123]], [[42, 45]], [[46, 49], [132, 135]], [[5, 21]], [[1, 25]], [[26, 38]], [[26, 45]], [[46, 60]], [[63, 118]]]",
    "query_spans": "[[[132, 142]]]",
    "process": "Let the coordinates of point M be $M(x, y)$ and the coordinates of point P be $P(x_{1}, y_{1})$. Since point D is the projection of P onto the x-axis and M is a point on the segment PD such that $\\overrightarrow{PM} = \\frac{1}{3}\\overrightarrow{PD}$, we have $(x - x_{1}, y - y_{1}) = \\frac{1}{3}(0, -y_{1})$, which implies\n\\[\n\\begin{cases}\nx_{1} = x \\\\\ny_{1} = \\frac{3}{2}y\n\\end{cases}\n\\]\nSince $P(x_{1}, y_{1})$ lies on the circle $x^{2} + y^{2} = 9$, it follows that $x^{2} + \\left(\\frac{3}{2}y\\right)^{2} = 9$. Simplifying gives $\\frac{x^{2}}{9} + \\frac{y^{2}}{4} = 1$."
  },
  {
    "text": "The equation of the parabola is $x=2 y^{2}$, then the coordinates of the focus of the parabola are?",
    "fact_expressions": "G: Parabola;Expression(G) = (x = 2*y^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(1/8, 0)",
    "fact_spans": "[[[0, 3], [20, 23]], [[0, 18]]]",
    "query_spans": "[[[20, 30]]]",
    "process": "x=2y^2 is transformed into y^{2}=\\frac{1}{2}x \\therefore 2p=\\frac{1}{2} \\therefore \\frac{p}{2}=\\frac{1}{8}, the focus is (\\frac{1}{8}, 0)"
  },
  {
    "text": "What is the focal distance of the hyperbola $y^{2}-\\frac{x^{2}}{2}=1$? What are the equations of its asymptotes?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/2 + y^2 = 1)",
    "query_expressions": "FocalLength(G);Expression(Asymptote(G))",
    "answer_expressions": "2*sqrt(3)\ny=pm*(sqrt(2)/2)*x",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 33]], [[0, 40]]]",
    "process": "\\because c=\\sqrt{a^{2}+b^{2}}=\\sqrt{3},\\therefore the focal distance is 2\\sqrt{3}, the asymptotes are y=\\pm\\sqrt{\\frac{b}{a}}x=\\pm\\frac{\\sqrt{2}}{2}x"
  },
  {
    "text": "Given that point $M$ is any point on the ellipse $\\frac{x^{2}}{27}+\\frac{y^{2}}{26}=1$, and $AB$ is a diameter of the circle $(x-1)^{2}+y^{2}=8$, then the sum of the maximum and minimum values of $\\overrightarrow{M A} \\cdot \\overrightarrow{M B}$ is?",
    "fact_expressions": "G: Ellipse;H: Circle;A: Point;B: Point;M: Point;Expression(G) = (x^2/27 + y^2/26 = 1);Expression(H) = (y^2 + (x - 1)^2 = 8);PointOnCurve(M, G);IsDiameter(LineSegmentOf(A, B), H)",
    "query_expressions": "Min(DotProduct(VectorOf(M, A), VectorOf(M, B)))+Max(DotProduct(VectorOf(M, A), VectorOf(M, B)))",
    "answer_expressions": "40",
    "fact_spans": "[[[7, 46]], [[58, 78]], [[52, 57]], [[52, 57]], [[2, 6]], [[7, 46]], [[58, 78]], [[2, 51]], [[52, 83]]]",
    "query_spans": "[[[85, 146]]]",
    "process": "Because the circle $(x-1)^{2}+y^{2}=8$ has center $C(1,0)$ and radius $2\\sqrt{2}$, let $M(x,y)$, $-3\\sqrt{3}\\leqslant x \\leqslant 3\\sqrt{3}$. Since $\\overrightarrow{MA}=\\overrightarrow{MC}+\\overrightarrow{CA}$, $\\overrightarrow{MB}=\\overrightarrow{MC}+\\overrightarrow{CB}=\\overrightarrow{MC}-\\overrightarrow{BC}=\\overrightarrow{MC}-\\overrightarrow{CA}$, so $\\overrightarrow{MA}\\cdot\\overrightarrow{MB}=(\\overrightarrow{MC}+\\overrightarrow{CA})\\cdot(\\overrightarrow{MC}-\\overrightarrow{CA})=(\\overrightarrow{MC})^{2}-(\\overrightarrow{CA})^{2}=(\\overrightarrow{MC})^{2}-8=(x-1)^{2}+y^{2}-8$. Because $A(x,y)$ lies on $\\frac{x^{2}}{27}+\\frac{y^{2}}{26}=1$, we have $y^{2}=26-\\frac{26}{27}x^{2}$. Thus, $\\overrightarrow{MA}\\cdot\\overrightarrow{MB}=(x-1)^{2}+26-\\frac{26}{27}x^{2}-8=\\frac{1}{27}x^{2}-2x+19$, $-3\\sqrt{3}\\leqslant x \\leqslant 3\\sqrt{3}$. The function $y=\\frac{1}{27}x^{2}-2x+19$ is a parabola opening upwards with axis of symmetry $x=27$. When $x=-3\\sqrt{3}$, $\\overrightarrow{MA}\\cdot\\overrightarrow{MB}$ attains its maximum value $6\\sqrt{3}+20$; when $x=3\\sqrt{3}$, $\\overrightarrow{MA}\\cdot\\overrightarrow{MB}$ attains its minimum value $-6\\sqrt{3}+20$. Therefore, the sum of the maximum and minimum values is $40$."
  },
  {
    "text": "From a point $P$ on the directrix $l$ of the parabola $x^{2}=4 y$, two tangent lines $P A$ and $P B$ are drawn to the parabola, where $A$ and $B$ are the points of tangency. If the inclination angle of the line $A B$ is $\\frac{\\pi}{6}$, then the horizontal coordinate of point $P$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);l: Line;Directrix(G) = l;P: Point;PointOnCurve(P, l) = True;TangentOfPoint(P, G) = {LineOf(P, A), LineOf(P, B)};TangentPoint(LineOf(P, A), G) = A;TangentPoint(LineOf(P, B), G) = B;B: Point;A: Point;Inclination(LineOf(A, B)) = pi/6",
    "query_expressions": "XCoordinate(P)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[1, 15], [28, 31]], [[1, 15]], [[18, 21]], [[1, 21]], [[24, 27], [92, 96]], [[1, 27]], [[0, 49]], [[28, 61]], [[28, 61]], [[55, 58]], [[51, 54]], [[63, 90]]]",
    "query_spans": "[[[92, 102]]]",
    "process": "Let point $ P(t,-1) $, and points $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Differentiating the function $ y=\\frac{x^{2}}{4} $ gives $ y'=\\frac{x}{2} $. Thus, the equation of line $ PA $ is $ y-y_{1}=\\frac{x_{1}}{2}(x-x_{1}) $, which simplifies to $ y-y_{1}=\\frac{x_{1}x}{2}-\\frac{x_{1}^{2}}{2} $, or $ y=\\frac{x_{1}x}{2}-y_{1} $. Similarly, the equation of line $ PB $ is $ y=\\frac{x_{2}x}{2}-y_{2} $. Since point $ P $ is the common point of lines $ PA $ and $ PB $, we have\n$$\n\\begin{cases}\ntx_{1}-2y_{1}+2=0 \\\\\ntx_{2}-2y_{2}+2=0\n\\end{cases}\n$$\nThus, the coordinates of points $ A $ and $ B $ satisfy the equation $ tx-2y+2=0 $, so the equation of line $ AB $ is $ tx-2y+2=0 $. From the given condition, $ \\tan\\frac{\\pi}{6}=\\frac{t}{2}=\\frac{\\sqrt{3}}{3} $, solving yields $ t=\\frac{2\\sqrt{3}}{3} $."
  },
  {
    "text": "Given that one asymptote of the hyperbola $x^{2}-\\frac{y^{2}}{m}=1$ is $x=2 y$, then $m=$?",
    "fact_expressions": "G: Hyperbola;Expression(G)=(x^2-y^2/m=1);Expression(OneOf(Asymptote(G)))=(x=2*y);m:Number",
    "query_expressions": "m",
    "answer_expressions": "1/4",
    "fact_spans": "[[[2, 30]], [[2, 30]], [[2, 46]], [[48, 51]]]",
    "query_spans": "[[[48, 53]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus of $C$ is $F$. The line passing through the origin intersects $C$ at points $A$ and $B$. Connect $AF$ and $BF$. If $|AB|=10$, $|BF|=8$, $\\cos \\angle ABF=\\frac{4}{5}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;a: Number;b: Number;G: Line;A: Point;F: Point;B: Point;O: Origin;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;PointOnCurve(O, G);Intersection(C, G) = {A, B};Abs(LineSegmentOf(A, B)) = 10;Abs(LineSegmentOf(B, F)) = 8;Cos(AngleOf(A, B, F)) = 4/5",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "5/7",
    "fact_spans": "[[[2, 59], [70, 73], [162, 165]], [[9, 59]], [[9, 59]], [[78, 80]], [[83, 86]], [[64, 67]], [[88, 92]], [[75, 77]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 67]], [[74, 80]], [[70, 94]], [[109, 118]], [[120, 129]], [[131, 160]]]",
    "query_spans": "[[[162, 171]]]",
    "process": ""
  },
  {
    "text": "The line $ l $ passes through $ M(-1, 0) $ and intersects the parabola $ y^{2} = 4x $ at points $ A $ and $ B $. The focus of the parabola is $ F $, and $ |BF| = \\frac{\\sqrt{3}}{2}|BM| $. Then, what is the distance from the midpoint of $ AB $ to the directrix of the parabola?",
    "fact_expressions": "G: Parabola;l: Line;A: Point;B: Point;M: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(M) = (-1, 0);PointOnCurve(M,l);Intersection(l, G) = {A,B};Focus(G) = F;Abs(LineSegmentOf(B, F)) = (sqrt(3)/2)*Abs(LineSegmentOf(B, M))",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A, B)), Directrix(G))",
    "answer_expressions": "6",
    "fact_spans": "[[[18, 32], [42, 45], [91, 94]], [[0, 5]], [[33, 36]], [[38, 41]], [[6, 17]], [[48, 52]], [[18, 32]], [[6, 17]], [[0, 17]], [[0, 41]], [[42, 52]], [[53, 82]]]",
    "query_spans": "[[[84, 101]]]",
    "process": "As shown in the figure, from the parabola $ y^{2} = 4x $, we obtain the focus $ F(1,0) $, and the directrix equation is $ x = -1 $. Draw a perpendicular line $ BG $ from $ B $ to the directrix. Since $ |BF| = \\frac{\\sqrt{3}}{2}|BM| $, it follows that $ |BG| = \\frac{\\sqrt{3}}{2}|BM| $, then $ \\angle BMF = 30^{\\circ} $. Therefore, the slope of line $ l $ is $ \\underline{\\sqrt{3}} $, and the equation of line $ l $ is $ y = \\frac{\\sqrt{3}}{3}(x+1) $. Solving the system \n\\[\n\\begin{cases}\ny = \\frac{\\sqrt{3}}{3}(x+1),\n\\end{cases}\n\\]\nwe obtain $ x^{2} - 10x + 1 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = 10 $, so the horizontal coordinate of the midpoint of $ AB $ is 5. Therefore, the distance from the midpoint of $ AB $ to the directrix of the parabola is $ 5 - (-1) = 6 $."
  },
  {
    "text": "Given the ellipse $ C $: $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ $ (a > b > 0) $, the left vertex is $ A $, the right focus is $ F $, point $ P $ lies on the line $ x = a $, and the line $ PA $ intersects the ellipse at point $ Q $. If $ \\overrightarrow{A Q} = 2 \\overrightarrow{Q P} $, $ \\overrightarrow{A Q} \\cdot \\overrightarrow{Q F} = 0 $, then the eccentricity of ellipse $ C $ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;A: Point;P: Point;Q: Point;F: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x = a);LeftVertex(C)=A;RightFocus(C)=F;PointOnCurve(P,G);Intersection(LineOf(P,A),C) = Q;VectorOf(A, Q) = 2*VectorOf(Q, P);DotProduct(VectorOf(A, Q), VectorOf(Q, F)) = 0",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(\\sqrt{33}-3)/4",
    "fact_spans": "[[[2, 59], [98, 100], [208, 213]], [[9, 59]], [[9, 59]], [[81, 88]], [[64, 67]], [[76, 80]], [[101, 105]], [[72, 75]], [[9, 59]], [[9, 59]], [[2, 59]], [[81, 88]], [[2, 67]], [[2, 75]], [[76, 89]], [[90, 105]], [[107, 152]], [[155, 206]]]",
    "query_spans": "[[[208, 219]]]",
    "process": "From the given conditions: A(-a,0), F(c,0), let P(a,m), Q(x_{0},y_{0}). From \\overrightarrow{AQ}=2\\overrightarrow{QP}, we get x_{0}=\\frac{-a+2a}{1+2}=\\frac{a}{3}. Substituting gives: \\frac{(\\frac{a}{3})^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1, solving yields y_{0}^{2}=\\frac{8}{9}b^{2}. \\overrightarrow{AQ}\\cdot\\overrightarrow{QF}=\\frac{4}{3}a(c-\\frac{a}{3})-y_{0}^{2}=\\frac{4}{3}a(c-\\frac{a}{3})-\\frac{8}{9}b^{2}=0. Rearranging gives: 2c^{2}+3ac-3a^{2}=0, so 2e^{2}+3e-3=0, thus e=\\frac{-3+\\sqrt{33}}{4} or e=\\frac{-3-\\sqrt{33}}{4} (discarded)."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$, $M$ a point on $C$, and point $N(0,4)$. If $\\angle MNF=90^{\\circ}$, then $|MF|$=?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 8*x);F: Point;Focus(C) = F;M: Point;PointOnCurve(M, C) = True;N: Point;Coordinate(N) = (0, 4);AngleOf(M, N, F) = ApplyUnit(90, degree)",
    "query_expressions": "Abs(LineSegmentOf(M, F))",
    "answer_expressions": "10",
    "fact_spans": "[[[2, 21], [33, 36]], [[2, 21]], [[25, 28]], [[2, 28]], [[29, 32]], [[29, 39]], [[40, 49]], [[40, 49]], [[51, 76]]]",
    "query_spans": "[[[78, 87]]]",
    "process": "Let $ M\\left(\\frac{t^{2}}{8}, t\\right) $, by $ \\angle MNF = 90^{\\circ} $ we can solve for $ t $, and then obtain the result using the focal radius formula. From $ \\angle MNF = 90^{\\circ} $, we get $ 0 = \\overrightarrow{NM} \\cdot \\overrightarrow{NF} = \\frac{t^{2}}{4} - 4t + 16 $, which implies $ (t - 8)^{2} = 0 $, solving gives $ t = 8 $. Therefore, $ M(8, 8) $, hence $ |MF| = x_{M} + 2 = 8 + 2 = 10 $."
  },
  {
    "text": "Given that $F$ is the focus of the parabola $y^{2}=8x$, $A$, $B$ are two points on the parabola, $O$ is the origin, and $\\overrightarrow{O F}=\\frac{1}{2}(\\overrightarrow{O A}+\\overrightarrow{O B})$, then $S_\\triangle{O A B}$=?",
    "fact_expressions": "G: Parabola;O: Origin;F: Point;A: Point;B: Point;Expression(G) = (y^2 = 8*x);Focus(G) = F;PointOnCurve(A, G);PointOnCurve(B, G);VectorOf(O, F) = (VectorOf(O, A) + VectorOf(O, B))/2",
    "query_expressions": "Area(TriangleOf(O, A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[6, 20], [32, 35]], [[39, 42]], [[2, 5]], [[24, 27]], [[28, 31]], [[6, 20]], [[2, 23]], [[24, 38]], [[24, 38]], [[49, 126]]]",
    "query_spans": "[[[129, 151]]]",
    "process": "\\because\\overrightarrow{OF}=\\frac{1}{2}(\\overrightarrow{OA}+\\overrightarrow{OB}). then 2\\overrightarrow{OF}=\\overrightarrow{OA}+\\overrightarrow{OB}, \\overrightarrow{OF}-\\overrightarrow{OA}=\\overrightarrow{OB}-\\overrightarrow{OF}, \\therefore\\overrightarrow{AF}=\\overrightarrow{FB}, \\because F is a common point, then points A, B, F are collinear. From the problem, F(2,0), then A(2,4), B(2,-4). S_{AOAB}=\\frac{1}{2}\\times2\\times8=8,"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the distance from a point $P$ on the curve to the origin is $b$, and $\\sin \\angle P F_{2} F_{1}=2 \\sin \\angle P F_{1} F_{2}$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;a: Number;b: Number;P: Point;F2: Point;F1: Point;O:Origin;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);Distance(P, O) = b;Sin(AngleOf(P, F2, F1)) = 2*Sin(AngleOf(P, F1, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(22)/2",
    "fact_spans": "[[[20, 76], [82, 84], [160, 163]], [[23, 76]], [[97, 100]], [[86, 90]], [[10, 17]], [[2, 9]], [[91, 93]], [[23, 76]], [[23, 76]], [[20, 76]], [[2, 81]], [[82, 90]], [[86, 100]], [[102, 157]]]",
    "query_spans": "[[[160, 169]]]",
    "process": "Let the focal distance be $2c$. Since $\\sin\\angle PF_{2}F_{1}=2\\sin\\angle PF_{1}F_{2}$, and $\\frac{1}{2}c\\cdot|PF_{1}|\\sin\\angle PF_{1}F_{2}=\\frac{1}{2}c\\cdot|PF_{2}|\\sin\\angle PF_{2}F_{1}$, it follows that $|PF_{1}|=2|PF_{2}|$. Also, $|PF_{1}|-|PF_{2}|=2a$, so $|PF_{2}|=2a$, $|PF_{1}|=4a$. Because $\\cos\\angle POF_{1}=\\frac{b^{2}+c^{2}-16a^{2}}{2bc}$, $\\cos\\angle POF_{2}=\\frac{b^{2}+c^{2}-4a^{2}}{2bc}$, and $\\angle POF_{1}+\\angle POF_{2}=180^{\\circ}$, we have $\\frac{b^{2}+c^{2}-16a^{2}}{2bc}=-\\frac{b^{2}+c^{2}-4a^{2}}{2bc}$. Combining with $b^{2}=c^{2}-a^{2}$ and simplifying yields $\\frac{c^{2}}{a^{2}}=\\frac{11}{2}$, i.e., $e=\\frac{c}{a}=\\frac{\\sqrt{22}}{2}$."
  },
  {
    "text": "Given that the left focus of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{m^{2}}=1$ $(m>0)$ is $F_{1}(-4,0)$, then $m=$?",
    "fact_expressions": "G: Ellipse;m: Number;F1: Point;m>0;Expression(G) = (x^2/25 + y^2/m^2 = 1);Coordinate(F1) = (-4, 0);LeftFocus(G) = F1",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 49]], [[69, 72]], [[54, 67]], [[4, 49]], [[2, 49]], [[54, 67]], [[2, 67]]]",
    "query_spans": "[[[69, 74]]]",
    "process": "The ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{m^{2}}=1$ $(m>0)$ has its left focus at $F_{1}(-4,0)$, giving $a=5$, $b=m$, $c=4$, leading to $25=m^{2}+16$, solving for $m=3$."
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, respectively. $P$ is a point on the right branch of the hyperbola, $I$ is the incenter of $\\Delta PF_{1} F_{2}$, and $S_{\\triangle I P F_{2}}=S_{\\Delta I P F_{1}}-\\lambda S_{\\Delta I F_{1} F_{2}}$. Then $\\lambda=?$",
    "fact_expressions": "G: Hyperbola;P: Point;F2: Point;I: Point;lambda:Number;F1: Point;Expression(G) = (x^2/16 - y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));Incenter(TriangleOf(P,F1,F2))=I;Area(TriangleOf(I,P,F2))=Area(TriangleOf(I,P,F1))-lambda*Area(TriangleOf(I,F1,F2))",
    "query_expressions": "lambda",
    "answer_expressions": "4/5",
    "fact_spans": "[[[18, 57], [68, 71]], [[64, 67]], [[8, 15]], [[77, 80]], [[188, 197]], [[0, 7]], [[18, 57]], [[0, 63]], [[0, 63]], [[64, 76]], [[77, 105]], [[107, 186]]]",
    "query_spans": "[[[188, 199]]]",
    "process": ""
  },
  {
    "text": "Given that $O$ is the coordinate origin, the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$, $P$ is a point on $C$, $PF$ is perpendicular to the $x$-axis, $Q$ is a point on the $y$-axis, and $PQ \\perp OP$. If $|PQ|=\\sqrt{5}$, then the equation of the directrix of the parabola $C$ is?",
    "fact_expressions": "O: Origin;C: Parabola;F: Point;P: Point;p: Number;Q: Point;p>0;Expression(C) = (y^2 = 2*p*x);Focus(C) = F;PointOnCurve(P, C) = True;IsPerpendicular(LineSegmentOf(P, F), xAxis) = True;PointOnCurve(Q, yAxis);IsPerpendicular(LineSegmentOf(P, Q), LineSegmentOf(O, P)) = True;Abs(LineSegmentOf(P, Q)) = sqrt(5)",
    "query_expressions": "Expression(Directrix(C))",
    "answer_expressions": "x = -2",
    "fact_spans": "[[[2, 5]], [[11, 37], [49, 52], [117, 123]], [[41, 44]], [[45, 48]], [[18, 37]], [[69, 72]], [[18, 37]], [[11, 37]], [[11, 44]], [[45, 55]], [[56, 68]], [[69, 80]], [[82, 97]], [[99, 115]]]",
    "query_spans": "[[[117, 130]]]",
    "process": "The parabola $ C: y^{2} = 2px $ ($ p > 0 $) has focus $ F\\left(\\frac{p}{2}, 0\\right) $. By the symmetry of the parabola $ C $, we may take $ P\\left(\\frac{p}{2}, p\\right) $, then $ k_{OP} = \\frac{p - 0}{\\frac{p}{2} - 0} = 2 $. Since $ PQ \\perp OP $, we have $ k_{PQ} = -\\frac{1}{2} $. The equation of line $ PQ $ can be written as $ y = -\\frac{1}{2}\\left(x - \\frac{p}{2}\\right) + p = -\\frac{1}{2}x + \\frac{5}{4}p $, so $ Q\\left(0, \\frac{5}{4}p\\right) $. Thus, $ PQ = \\sqrt{\\left(\\frac{p}{2} - 0\\right)^{2} + \\left(\\frac{5p}{4} - p\\right)^{2}} = \\frac{\\sqrt{5}}{4}p = \\sqrt{5} $, hence $ p = 4 $. Therefore, the directrix equation of the parabola $ C $ is $ x = -2 $."
  },
  {
    "text": "The minimum distance from a point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ to the right focus is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);P: Point;PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, RightFocus(G)))",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 38]], [[0, 38]], [[41, 44]], [[0, 44]]]",
    "query_spans": "[[[0, 57]]]",
    "process": "The minimum distance from point P to the right focus occurs when P is at the right vertex of the ellipse. From the equation of the ellipse, we have a=5, b=3, c=4. When point P is at the right vertex of the ellipse, the minimum distance to the right focus is a-c=1."
  },
  {
    "text": "A line passing through point $Q(2,0)$ intersects the parabola $C$: $y^{2}=2x$ at point $A$, and intersects the $y$-axis at point $D$, with $\\overrightarrow{DA}=2\\overrightarrow{DQ}$. $P$ is a moving point on the parabola $C$. Then the minimum value of $\\overrightarrow{PA} \\cdot \\overrightarrow{PD}$ is?",
    "fact_expressions": "Q: Point;Coordinate(Q) = (2, 0);G: Line;PointOnCurve(Q, G) = True;C: Parabola;Expression(C) = (y^2 = 2*x);OneOf(Intersection(G, C)) = A;A: Point;Intersection(G, yAxis) = D;D: Point;VectorOf(D, A) = 2*VectorOf(D, Q);P: Point;PointOnCurve(P, C) = True",
    "query_expressions": "Min(DotProduct(VectorOf(P, A), VectorOf(P, D)))",
    "answer_expressions": "-9",
    "fact_spans": "[[[1, 10]], [[1, 10]], [[11, 13]], [[0, 13]], [[14, 33], [108, 114]], [[14, 33]], [[11, 42]], [[39, 42]], [[11, 54]], [[50, 54]], [[56, 101]], [[104, 107]], [[104, 118]]]",
    "query_spans": "[[[120, 175]]]",
    "process": "By the given condition, let $ P(x,y) $, $ A(x_{1},y_{1}) $, $ D(0,m) $, without loss of generality, assume $ y_{1}>0 $. Since $ \\overrightarrow{DA}=2\\overrightarrow{DQ} $, it follows that point $ Q $ is the midpoint of $ AD $. By the midpoint formula, we have $ \\frac{x_{1}+0}{2}=2 $, solving gives $ x_{1}=4 $. From the parabola $ C: y^{2}=2x $, we obtain $ y_{1}=\\sqrt{2x_{1}}=\\sqrt{2\\times4}=2\\sqrt{2} $, thus $ A(4,2\\sqrt{2}) $. At this point, $ \\underline{2\\sqrt{2}+m}=0 $, solving gives $ m=-2\\sqrt{2} $, so $ D(0,-2\\sqrt{2}) $. Therefore, $ \\overrightarrow{PA}=(4-x,2\\sqrt{2}-y) $, $ \\overrightarrow{PD}=(-x,-2\\sqrt{2}-y) $. Hence,  \n$$\n\\overrightarrow{PA}\\cdot\\overrightarrow{PD}=(4-x,2\\sqrt{2}-y)\\cdot(-x,-2\\sqrt{2}-y)=x^{2}-4x+y^{2}-8.\n$$\nSince $ y^{2}=2x $, we have  \n$$\n\\overrightarrow{PA}\\cdot\\overrightarrow{PD}=x^{2}-2x-8=(x-1)^{2}-9.\n$$\nTherefore, when $ x=1 $, $ \\overrightarrow{PA}\\cdot\\overrightarrow{PD} $ attains the minimum value $ -9 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has a focal distance of $4 \\sqrt{3}$ and a real axis length of $4 \\sqrt{2}$, then the asymptote equations of hyperbola $C$ are?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);FocalLength(C) = 4*sqrt(3);Length(RealAxis(C)) = 4*sqrt(2)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*(sqrt(2)/2)*x",
    "fact_spans": "[[[2, 63], [98, 104]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 79]], [[2, 96]]]",
    "query_spans": "[[[98, 112]]]",
    "process": "Since $2c=4\\sqrt{3}$, $2a=4\\sqrt{2}$, it follows that $b=\\sqrt{c^{2}-a^{2}}=2$, $\\frac{b}{a}=\\frac{\\sqrt{2}}{2}$, so the asymptotes of hyperbola $C$ are $y=\\pm\\frac{\\sqrt{2}}{2}x$."
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, respectively, and $P$ is a point on the ellipse such that $\\angle F_{1} P F_{2}=60^{\\circ}$. Then the inradius of $\\Delta F_{1} P F_{2}$ equals?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Radius(InscribedCircle(TriangleOf(F1, P, F2)))",
    "answer_expressions": "2*sqrt(3)/3 - 1",
    "fact_spans": "[[[18, 45], [57, 59]], [[18, 45]], [[0, 7]], [[8, 15]], [[0, 51]], [[0, 51]], [[52, 55]], [[52, 62]], [[64, 97]]]",
    "query_spans": "[[[99, 130]]]",
    "process": "By the given condition, |F_{1}P|+|PF_{2}|=4, |F_{1}F_{2}|=2\\sqrt{3}, and by the cosine law we obtain |F_{1}P||PF_{2}|=\\frac{4}{3}. Using the area formula S=\\frac{1}{2}|F_{1}P||PF_{2}|\\cdot\\sin60^{\\circ}=\\frac{1}{2}(|F_{1}P|+|PF_{2}|+|F_{1}F_{2}|)\\cdot r, the solution can be found. Since F_{1} and F_{2} are the left and right foci of the ellipse \\frac{x^{2}}{4}+y^{2}=1, and P is a point on this ellipse, we have |F_{1}P|+|PF_{2}|=4, |F_{1}F_{2}|=2\\sqrt{3}. Then by the cosine law, |F_{1}F_{2}|^{2}=|F_{1}P|^{2}+|PF_{2}|^{2}-2|F_{1}P||PF_{2}|\\cos60^{\\circ}, so 12=(|F_{1}P|+|PF_{2}|)^{2}-2|F_{1}P||PF_{2}|\\cos60^{\\circ}-2|F_{1}P||PF_{2}|, which simplifies to 12=16-3|F_{1}P||PF_{2}|, thus |F_{1}P||PF_{2}|=\\frac{4}{3}. Therefore, the area of triangle PF_{1}F_{2} is S=\\frac{1}{2}|F_{1}P||PF_{2}|\\cdot\\sin60^{\\circ}=\\frac{\\sqrt{3}}{3}. Let r be the inradius of triangle F_{1}PF_{2}, then S=\\frac{1}{2}(|F_{1}P|+|PF_{2}|+|F_{1}F_{2}|)\\cdot r=\\frac{1}{2}(4+2\\sqrt{3})\\cdot r=\\frac{\\sqrt{3}}{3}. Solving gives r=\\frac{2\\sqrt{3}}{3}-1."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=16 x$, the intersection point of its axis of symmetry and directrix is $M$. The line $l$: $y=k x-4 k$ intersects $C$ at points $A$ and $B$. If $|A M|=4|B M|$, then the real number $k$=?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;M: Point;B: Point;k: Real;Expression(C) = (y^2 = 16*x);Expression(l) = (y = k*x - 4*k);Intersection(SymmetryAxis(C),Directrix(C))=M;Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, M)) = 4*Abs(LineSegmentOf(B, M))",
    "query_expressions": "k",
    "answer_expressions": "pm*4/3",
    "fact_spans": "[[[37, 55]], [[2, 22], [56, 59]], [[61, 64]], [[33, 36]], [[65, 68]], [[89, 94]], [[2, 22]], [[37, 55]], [[2, 36]], [[37, 70]], [[72, 86]]]",
    "query_spans": "[[[89, 96]]]",
    "process": ""
  },
  {
    "text": "If the distance from point $A$ on the parabola $y=4 x^{2}$ to the focus is $\\frac{33}{16}$, then what is the distance from $A$ to the $x$-axis?",
    "fact_expressions": "G: Parabola;A: Point;Expression(G) = (y = 4*x^2);PointOnCurve(A, G);Distance(A, Focus(G)) = 33/16",
    "query_expressions": "Distance(A, xAxis)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 15]], [[17, 21], [45, 48]], [[1, 15]], [[1, 21]], [[1, 43]]]",
    "query_spans": "[[[45, 58]]]",
    "process": "The focus of the parabola $ y = 4x^2 $ has coordinates $ F(0, \\frac{1}{16}) $, and the equation of the directrix is $ y = -\\frac{1}{16} $. Since the distance from point $ A $ on the parabola $ y = 4x^2 $ to the focus is $ \\frac{33}{16} $, by the definition of a parabola, the distance from point $ A $ to the directrix $ y = -\\frac{1}{16} $ is $ \\frac{33}{16} $. Then, the distance from point $ A $ to the x-axis is 2."
  },
  {
    "text": "Given fixed point $A(4,0)$ and a moving point $B$ on the curve $x^{2}+y^{2}=1$, what is the trajectory equation of the midpoint $P$ of segment $AB$?",
    "fact_expressions": "G: Curve;B: Point;A: Point;P: Point;Expression(G) = (x^2 + y^2 = 1);Coordinate(A) = (4, 0);PointOnCurve(B, G);MidPoint(LineSegmentOf(A,B)) = P",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "(x-2)^2+y^2=1/4",
    "fact_spans": "[[[13, 30]], [[34, 37]], [[4, 12]], [[48, 51]], [[13, 30]], [[4, 12]], [[13, 37]], [[39, 51]]]",
    "query_spans": "[[[48, 58]]]",
    "process": "Let the midpoint of segment AB be P(x, y), and B(m, n), then m = 2x \\cdot 4, n = 2y. Since the moving point B moves on the circle x^{2} + y^{2} = 1, \\therefore m^{2} + n^{2} = 1, \\therefore (2x \\cdot 4)^{2} + (2y)^{2} = 1, \\therefore (x - 2)^{2} + y^{2} = \\frac{1}{4}."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a, b>0)$, respectively, $P$ is a point on the hyperbola such that $\\overrightarrow{P F}_{1} \\cdot(\\overrightarrow{O F_{1}}+\\overrightarrow{O P})=0$ ($O$ being the origin), and $|\\overrightarrow{P F_{1}}|=2|\\overrightarrow{P F_{2}}|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;O: Origin;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);DotProduct(VectorOf(P, F1),(VectorOf(O,F1)+VectorOf(O,P)))=0;Abs(VectorOf(P, F1)) = 2*Abs(VectorOf(P, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[20, 74], [85, 88], [245, 248]], [[23, 74]], [[23, 74]], [[81, 84]], [[175, 178]], [[2, 9]], [[10, 17]], [[23, 74]], [[23, 74]], [[20, 74]], [[2, 80]], [[2, 80]], [[81, 91]], [[93, 174]], [[186, 242]]]",
    "query_spans": "[[[245, 254]]]",
    "process": ""
  },
  {
    "text": "If the point $P(m, 2 \\sqrt{3})$ lies on the parabola $y^{2}=4 x$ with focus $F$, then $|P F|$ equals?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;m:Number;Expression(G) = (y^2 = 4*x);Coordinate(P) = (m, 2*sqrt(3));PointOnCurve(P, G);Focus(G) = F",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "4",
    "fact_spans": "[[[29, 43]], [[1, 20]], [[22, 25]], [[2, 20]], [[29, 43]], [[1, 20]], [[1, 44]], [[21, 43]]]",
    "query_spans": "[[[46, 56]]]",
    "process": "Analysis: According to the given conditions, first find the coordinates of point P, then use the definition of the parabola to solve for |PF|. Since point P(m, 2\\sqrt{3}) lies on the parabola y^{2}=4x, \\therefore (2\\sqrt{3})^{2}=4m, solving gives m=3, \\therefore the coordinates of point P are (3, 2\\sqrt{3}). The directrix equation of the parabola is x=-1, \\therefore |PF|=3+1=4. [Master Teacher"
  },
  {
    "text": "Given point $F$, $A$ are the left focus and right vertex of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively, and point $B(0, b)$ satisfies $\\overrightarrow{F B} \\cdot \\overrightarrow{A B}=0$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;B: Point;F: Point;A: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(B) = (0, b);LeftFocus(C) = F;RightVertex(C) = A;DotProduct(VectorOf(F, B), VectorOf(A, B)) = 0",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(1+sqrt(5))/2",
    "fact_spans": "[[[15, 79], [154, 157]], [[23, 79]], [[23, 79]], [[88, 99]], [[2, 6]], [[8, 12]], [[23, 79]], [[23, 79]], [[15, 79]], [[88, 99]], [[2, 87]], [[2, 87]], [[101, 152]]]",
    "query_spans": "[[[154, 163]]]",
    "process": ""
  },
  {
    "text": "A chord $AB$ passing through the left focus $F_1$ of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ has length $6$. What is the perimeter of $\\triangle A B F_{2}$ ($F_{2}$ being the right focus)?",
    "fact_expressions": "G: Hyperbola;A: Point;B: Point;F1:Point;F2: Point;Expression(G) = (x^2/16 - y^2/9 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;IsChordOf(LineSegmentOf(A,B),G);PointOnCurve(F1,LineSegmentOf(A,B));Length(LineSegmentOf(A,B))=6",
    "query_expressions": "Perimeter(TriangleOf(A,B,F2))",
    "answer_expressions": "28",
    "fact_spans": "[[[1, 40]], [[52, 56]], [[52, 56]], [[43, 50]], [[85, 92]], [[1, 40]], [[1, 50]], [[1, 97]], [[1, 56]], [[0, 56]], [[52, 61]]]",
    "query_spans": "[[[63, 103]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$, the focus is $F$, point $M$ lies on the parabola $C$, and point $N$ lies on the directrix $l$ of the parabola $C$. If $3 \\overrightarrow{M F}+\\overrightarrow{F N}=\\overrightarrow{0}$ and $2|M F|-p=4$, then the distance from $F$ to $l$ is?",
    "fact_expressions": "C: Parabola;p: Number;M: Point;F: Point;N: Point;p>0;l:Line;Expression(C) = (y^2 = 2*p*x);Focus(C) = F;PointOnCurve(M, C);Directrix(C)=l;PointOnCurve(N,l);VectorOf(F, N) + 3*VectorOf(M, F) = 0;-p + 2*Abs(LineSegmentOf(M, F)) = 4",
    "query_expressions": "Distance(F, l)",
    "answer_expressions": "12",
    "fact_spans": "[[[2, 28], [41, 47], [54, 60]], [[10, 28]], [[36, 40]], [[32, 35], [149, 152]], [[49, 53]], [[10, 28]], [[63, 66], [153, 156]], [[2, 28]], [[2, 35]], [[36, 48]], [[54, 66]], [[49, 67]], [[69, 133]], [[135, 147]]]",
    "query_spans": "[[[149, 161]]]",
    "process": "From $3\\overrightarrow{MF}+\\overrightarrow{FN}=\\overrightarrow{0}$, it follows that M is the trisection point on segment FN closer to F. Draw $MM' \\bot l$ from point M, intersecting at point M'. Therefore, $|MM'|=|MF|=\\frac{2}{3}p$. Since $2|MF|-p=4$, we have $\\frac{4}{3}p-p=4$, solving gives $p=12$."
  },
  {
    "text": "The curve of the equation $4 x^{2} + k y^{2}=1$ is an ellipse with foci on the $y$-axis. Then the range of values for $k$ is?",
    "fact_expressions": "G: Ellipse;H: Curve;k:Number;Expression(H)=(4*x^2+k*y^2=1);G=H;PointOnCurve(Focus(G),yAxis)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(0,4)",
    "fact_spans": "[[[36, 38]], [[24, 26]], [[40, 43]], [[0, 26]], [[24, 38]], [[27, 38]]]",
    "query_spans": "[[[40, 50]]]",
    "process": ""
  },
  {
    "text": "The distance from the focus to the directrix of the parabola $C$: $x^{2}=2 p y  (p>0)$ is $4$. Then, the equation of the directrix of the parabola is?",
    "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 2*p*y);p: Number;p>0;Distance(Focus(C), Directrix(C)) = 4",
    "query_expressions": "Expression(Directrix(C))",
    "answer_expressions": "y = -2",
    "fact_spans": "[[[0, 27], [42, 45]], [[0, 27]], [[7, 27]], [[7, 27]], [[0, 40]]]",
    "query_spans": "[[[42, 52]]]",
    "process": ""
  },
  {
    "text": "Given that $A$ and $B$ are the left and right vertices of the hyperbola $C$: $\\frac{x^{2}}{2 b^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, respectively, and point $P$ lies on the branch of the hyperbola in the first quadrant. Let the slopes of $PA$, $PB$, and $PO$ be $k_{1}$, $k_{2}$, $k_{3}$, respectively. Then the range of $\\frac{k_{3}}{k_{1} k_{2}}$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;P: Point;A: Point;B: Point;O: Origin;b>0;k1: Number;k2: Number;k3: Number;Expression(C) = (x^2/(2*b^2) - y^2/b^2 = 1);LeftVertex(C) = A;RightVertex(C) = B;Quadrant(P) = 1;PointOnCurve(P, C);Slope(LineSegmentOf(P, A)) = k1;Slope(LineSegmentOf(P, B)) = k2;Slope(LineSegmentOf(P, O)) = k3",
    "query_expressions": "Range(k3/(k1*k2))",
    "answer_expressions": "(0, \\sqrt{2})",
    "fact_spans": "[[[12, 70], [88, 91]], [[20, 70]], [[77, 81]], [[2, 5]], [[6, 9]], [[109, 114]], [[20, 70]], [[120, 127]], [[128, 135]], [[136, 143]], [[12, 70]], [[2, 76]], [[2, 76]], [[77, 86]], [[77, 92]], [[94, 143]], [[94, 143]], [[94, 143]]]",
    "query_spans": "[[[145, 179]]]",
    "process": "Let point $ P(x_{0},y_{0}) $. From the problem, we know: $ A(-\\sqrt{2}b,0) $, $ B(\\sqrt{2}b,0) $, so $ k_{1}k_{2} = \\frac{y_{0}}{x_{0}+\\sqrt{2}b} \\cdot \\frac{y_{0}}{x_{0}-\\sqrt{2}b} = \\frac{y_{0}^{2}}{x_{0}^{2}-2b^{2}} $. Also, $ \\frac{x_{0}^{2}}{2b^{2}} - \\frac{y_{0}^{2}}{b^{2}} = 1 \\Rightarrow y_{0}^{2} = b^{2}\\left( \\frac{x_{0}^{2}}{2b^{2}} - 1 \\right) = \\frac{x_{0}^{2}-2b^{2}}{2} $, so $ k_{1}k_{2} = \\frac{1}{2} $. Therefore, $ \\frac{k_{3}}{k_{1}k_{2}} = 2k_{3} $. Since the asymptotes of the hyperbola are given by $ y = \\pm \\frac{\\sqrt{2}}{2}x $ and $ P $ lies in the first quadrant, we have $ 0 < k_{3} < \\frac{\\sqrt{2}}{2} $, so $ \\frac{k_{3}}{k_{1}k_{2}} = 2k_{3} \\in (0,\\sqrt{2}) $."
  },
  {
    "text": "The focus of the parabola is the lower focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$, and the vertex is at the center of the ellipse. Then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;H: Ellipse;Expression(H) = (x^2/4 + y^2/9 = 1);Focus(G) = LowerFocus(H);Vertex(G) = Center(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 = -4\\sqrt{5}y",
    "fact_spans": "[[[0, 3], [58, 61]], [[7, 44], [52, 54]], [[7, 44]], [[0, 48]], [[0, 56]]]",
    "query_spans": "[[[58, 65]]]",
    "process": "From the ellipse equation, we know that $a^{2}=9$, $b^{2}=4$, the foci lie on the y-axis, and the lower focus has coordinates $(0,-c)$, where $c=\\sqrt{a^{2}-b^{2}}=\\sqrt{5}$. Therefore, the focus of the parabola is at $(0,-\\sqrt{5})$, and the equation of the parabola is $x^{2}=-4\\sqrt{5}y$."
  },
  {
    "text": "If the coordinates of the focus $F$ of the parabola $y = a x^{2}$ are $(0, -1)$, then the value of the real number $a$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = a*x^2);a: Real;F: Point;Coordinate(F) = (0, -1);Focus(G) = F",
    "query_expressions": "a",
    "answer_expressions": "-1/4",
    "fact_spans": "[[[1, 15]], [[1, 15]], [[35, 40]], [[18, 21]], [[18, 33]], [[1, 21]]]",
    "query_spans": "[[[35, 44]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the two foci of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{27}=1$, and let $A$ be a point on the hyperbola. If $|A F_{1}|=8$, then $|A F_{2}|=$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/27 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};A: Point;PointOnCurve(A, G);Abs(LineSegmentOf(A, F1)) = 8",
    "query_expressions": "Abs(LineSegmentOf(A, F2))",
    "answer_expressions": "14",
    "fact_spans": "[[[17, 56], [66, 69]], [[17, 56]], [[1, 8]], [[9, 16]], [[1, 61]], [[62, 65]], [[62, 73]], [[75, 88]]]",
    "query_spans": "[[[90, 103]]]",
    "process": "From $\\frac{x^{2}}{9}-\\frac{y^{2}}{27}=1$, we get $a^{2}=9$, $\\therefore a=3$, the length of the transverse axis of the hyperbola is $2a=6$. According to the definition of a hyperbola, $|AF_{1}|-|AF_{2}|=\\pm6$, $\\therefore |AF_{2}|=|AF_{1}|\\pm6=14$ or $2$. When $|AF_{2}|=2$, it does not satisfy $|AF_{2}|+|AF_{1}|\\geqslant2c=12$, $\\therefore$ only $|AF_{2}|=14$ is valid."
  },
  {
    "text": "What are the coordinates of the focus of the parabola $y=6 x^{2}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 6*x^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,1/24)",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": "Rewrite the equation of the parabola into standard form. According to the geometric properties of the parabola, the answer can be obtained. Rewrite the equation of the parabola $ y = 6x^2 $ into standard form to get $ x^{2} = \\frac{1}{6}y $. Then the focus of this parabola lies on the positive half of the y-axis, with coordinates $ (0, \\frac{1}{24}) $."
  },
  {
    "text": "Parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$. A line $l$ passing through point $F$ intersects parabola $C$ at points $A$ and $B$, and intersects the directrix $m$ of parabola $C$ at point $D$. If $|B D|=2|F B|$, $|F A|=2$, then $p=$?",
    "fact_expressions": "l: Line;C: Parabola;B: Point;D: Point;F: Point;A: Point;m: Line;p: Number;p > 0;Expression(C) = (y^2 = 2*p*x);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Directrix(C) = m;Intersection(l, m) = D;Abs(LineSegmentOf(B, D)) = 2*Abs(LineSegmentOf(F, B));Abs(LineSegmentOf(F, A)) = 2",
    "query_expressions": "p",
    "answer_expressions": "{1, 3}",
    "fact_spans": "[[[40, 45]], [[0, 26], [46, 52], [64, 70]], [[57, 60]], [[77, 81]], [[30, 33], [35, 39]], [[53, 56]], [[73, 76]], [[109, 112]], [[8, 26]], [[0, 26]], [[0, 33]], [[34, 45]], [[40, 62]], [[64, 76]], [[40, 81]], [[83, 97]], [[99, 108]]]",
    "query_spans": "[[[109, 114]]]",
    "process": "Analysis: According to the problem, discuss two cases: point A above and point B below, or point B above and point A below, to find the real value of $ p $. As shown in the figure, when point B is below point A, from geometric relations we have: $ x_{B} = -\\frac{p}{2} + \\frac{2p}{3} = \\frac{p}{6} $. Similarly, we obtain: $ 12x^2 - 20px + 3p^{2} = 0 $. Then: $ x_{B} = \\frac{3}{1}p $, $ x_{A} = \\frac{1}{6}p_{1} $. Combining with $ F\\left(\\frac{p}{2}, 0\\right) $, we get: $ |AF| = \\sqrt{\\frac{1}{9}p^{2} + \\frac{1}{3}p^{2}} = \\frac{2}{3}p $, so $ \\frac{2}{3}p = 2 $, $ p = 3 $. In conclusion, the value of $ p $ is 1 or 3."
  },
  {
    "text": "The directrix of the parabola $y=a x^{2}$ is given by $y=1$. Then the coordinates of the focus are?",
    "fact_expressions": "G: Parabola;a: Number;Expression(G) = (y = a*x^2);Expression(Directrix(G)) = (y = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,-1)",
    "fact_spans": "[[[0, 14]], [[3, 14]], [[0, 14]], [[0, 25]]]",
    "query_spans": "[[[0, 33]]]",
    "process": "y=ax^{2}\\Rightarrow x^{2}=\\frac{1}{a}y\\Rightarrow y=-\\frac{1}{4a}, so -\\frac{1}{4a}=1, F(0,\\frac{1}{4a}) i.e. F(0,-1)"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ with focus $F$ and directrix $l$. A line passing through the focus $F$ intersects $C$ at points $A$ and $B$. $A A^{\\prime} \\perp l$, $B B^{\\prime} \\perp l$, with foots of perpendiculars $A^{\\prime}$ and $B^{\\prime}$, respectively. If $|A^{\\prime} F|=2 \\sqrt{3}$, $|B^{\\prime} F|=2$, then $p=?$",
    "fact_expressions": "C: Parabola;p: Number;G: Line;A: Point;B: Point;F: Point;l: Line;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;Directrix(C) = l;PointOnCurve(F, G);Intersection(G, C) = {A, B};A1: Point;B1: Point;IsPerpendicular(LineSegmentOf(A, A1), l);IsPerpendicular(LineSegmentOf(B, B1), l);FootPoint(LineSegmentOf(A, A1), l) = A1;FootPoint(LineSegmentOf(B, B1), l) = B1;Abs(LineSegmentOf(A1, F)) = 2*sqrt(3);Abs(LineSegmentOf(B1, F)) = 2",
    "query_expressions": "p",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 28], [53, 56]], [[198, 201]], [[50, 52]], [[58, 61]], [[62, 65]], [[32, 35], [46, 49]], [[39, 42]], [[10, 28]], [[2, 28]], [[2, 35]], [[2, 42]], [[43, 52]], [[50, 67]], [[121, 134]], [[136, 148]], [[68, 91]], [[93, 115]], [[68, 148]], [[68, 148]], [[150, 177]], [[178, 196]]]",
    "query_spans": "[[[198, 203]]]",
    "process": "As shown in the figure, let the directrix $ l $ of parabola $ C $ intersect the $ x $-axis at point $ D $. By the property of the parabola, $ |AA'| = |AF| $, $ |BF| = |BB'| $. Since $ AA' \\parallel BB' \\parallel x $-axis, $ \\angle A'AF = \\angle AFD $, $ \\angle FBB' = \\angle BFD $. Therefore, $ \\angle AFB = \\angle A'AF + \\angle BB'F = 90^{\\circ} $. In right triangle $ \\triangle AFB $, by the Pythagorean theorem, $ |AB| = \\sqrt{|AF|^{2} + |BF|^{2}} = 4 $, so $ |AF| \\cdot |BF| = |FD| \\cdot |AB| $, hence $ p = |FD| = \\sqrt{3} $."
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $\\frac{y^{2}}{9}-\\frac{x^{2}}{16}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/16 + y^2/9 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*(3/4)*x",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 47]]]",
    "process": "\\because the hyperbola \\frac{y^{2}}{9}-\\frac{x^{2}}{16}=1, \\therefore the asymptotes are given by \\frac{y^{2}}{9}-\\frac{x^{2}}{16}=0, that is, y=\\pm\\frac{3}{4}x,"
  },
  {
    "text": "The line $x+2 y-2=0$ passes through one focus and one vertex of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$. Then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x + 2*y - 2 = 0);PointOnCurve(OneOf(Focus(G)),H);PointOnCurve(OneOf(Vertex(G)),H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/5+y^2=1",
    "fact_spans": "[[[15, 67], [80, 82]], [[17, 67]], [[17, 67]], [[0, 13]], [[17, 67]], [[17, 67]], [[15, 67]], [[0, 13]], [[0, 72]], [[0, 77]]]",
    "query_spans": "[[[80, 88]]]",
    "process": "For the line $x+2y-2=0$, let $x=0$, solve to get $y=1$; let $y=0$, solve to get $x=2$. Thus, the right focus of the ellipse has coordinates $(2,0)$, and the upper vertex has coordinates $(0,1)$. Then $c=2$, $b=1$, so $a=\\sqrt{b^{2}+c^{2}}=\\sqrt{5}$. Hence, the equation of the ellipse is $\\frac{x^{2}}{5}+y^{2}=1$."
  },
  {
    "text": "Given a moving point $P$ on the hyperbola $\\Gamma$, the distances from $P$ to points $F_{1}(-1,0)$ and $F_{2}(1,0)$ are $d_{1}$ and $d_{2}$ respectively, $\\angle F_{1} P F_{2}=2 \\theta$, and $d_{1} \\cdot d_{2} \\cdot \\sin ^{2} \\theta=\\frac{1}{3}$. Then the equation of the hyperbola $\\Gamma$ is?",
    "fact_expressions": "Gamma: Hyperbola;P: Point;PointOnCurve(P, Gamma) = True;F1: Point;Coordinate(F1) = (-1, 0);F2: Point;Coordinate(F2) = (1, 0);d1: Number;d2: Number;Distance(P, F1) = d1;Distance(P, F2) = d2;theta: Number;AngleOf(F1, P, F2) = 2*theta;d1*d2*Sin(theta)^2 = 1/3",
    "query_expressions": "Expression(Gamma)",
    "answer_expressions": "x^2/(2/3)-y^2/(1/3)=1",
    "fact_spans": "[[[2, 13], [161, 172]], [[17, 20]], [[2, 20]], [[21, 35]], [[21, 35]], [[36, 48]], [[36, 48]], [[54, 61]], [[62, 69]], [[17, 70]], [[17, 70]], [[72, 103]], [[72, 103]], [[105, 159]]]",
    "query_spans": "[[[161, 177]]]",
    "process": "In $\\triangle PF_{1}F_{2}$, by the law of cosines: $|F_{1}F_{2}|=4=d_{1}^{2}+d_{2}^{2}-2d_{1}d_{2}\\cos2\\theta=(d_{1}-d_{2})^{2}+4d_{1}d_{2}\\sin^{2}\\theta$. Since $d_{1} \\cdot d_{2} \\cdot \\sin^{2}\\theta=\\frac{1}{3}$, $\\therefore (d_{1}-d_{2})^{2}=\\frac{8}{3}=(2a)^{2}$, $\\therefore a^{2}=\\frac{2}{3}$, $b^{2}=c^{2}-a^{2}=\\frac{1}{3}$, then the hyperbola equation is $\\frac{x^{2}}{3}-\\frac{y^{2}}{3}=$"
  },
  {
    "text": "What is the equation of the directrix of the parabola $y^{2}=-12 x$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -12*x)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=3",
    "fact_spans": "[[[0, 16]], [[0, 16]]]",
    "query_spans": "[[[0, 23]]]",
    "process": "From the parabola equation $ y^{2} = -12x $, we know $ p = 6 $, its graph opens to the left, and the focus lies on the negative half of the x-axis. Therefore, its directrix equation is $ x = \\frac{p}{2} = 3 $, that is, $ x = 3 $."
  },
  {
    "text": "The equation of one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a, b>0)$ is $y=2 x$, and the distance from a focus to the asymptote is $2$. Then, the length of the real axis of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (y = 2*x);Distance(Focus(G), Asymptote(G)) = 2;a>0",
    "query_expressions": "Length(RealAxis(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 54], [86, 89]], [[3, 54]], [[3, 54]], [[3, 54]], [[0, 54]], [[0, 70]], [[0, 84]], [[3, 54]]]",
    "query_spans": "[[[86, 95]]]",
    "process": "The focus is at (c,0). From the given conditions, \\begin{cases}\\frac{b}{a}=2\\\\\\frac{2c}{\\sqrt{5}}=2\\end{cases}, and since c^{2}=a^{2}-b^{2}, solving yields \\begin{cases}c=\\sqrt{5}\\\\a=1\\\\b=2\\end{cases}. The length of the real axis is 2a=2."
  },
  {
    "text": "Given that a focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has coordinates $(2,0)$, and one of its asymptotes is perpendicular to the line $l$: $x+\\sqrt{3} y=0$, then the standard equation of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Coordinate(OneOf(Focus(C))) = (2, 0);l: Line;Expression(l) = (x + sqrt(3)*y = 0);IsPerpendicular(OneOf(Asymptote(C)),l)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2/3 = 1",
    "fact_spans": "[[[2, 63], [80, 81], [115, 121]], [[2, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[2, 78]], [[88, 111]], [[88, 111]], [[80, 113]]]",
    "query_spans": "[[[115, 128]]]",
    "process": "From the given conditions, the coordinates of one focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ are $(2,0)$, so $c=2$. Since one of its asymptotes is perpendicular to the line $l: x+\\sqrt{3}y=0$, we have $\\frac{b}{a}\\cdot(-\\frac{1}{\\sqrt{3}})=-1$, which implies $b=\\sqrt{3}a$. Also, since $c^{2}=a^{2}+b^{2}$, solving gives $a=1$, $b=\\sqrt{3}$. Therefore, the standard equation of the hyperbola is $x^{2}-\\frac{y^{2}}{3}=1$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{3}+\\frac{y^{2}}{m}=1$ has a major axis length of $4$, then the focal distance of $C$ is?",
    "fact_expressions": "C: Ellipse;m: Number;Expression(C) = (x^2/3 + y^2/m = 1);Length(MajorAxis(C)) = 4",
    "query_expressions": "FocalLength(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 44], [54, 57]], [[8, 44]], [[2, 44]], [[2, 52]]]",
    "query_spans": "[[[54, 62]]]",
    "process": "Since the major axis of the ellipse $ C: \\frac{x^{2}}{3} + \\frac{y^{2}}{m} = 1 $ is 4, we have $ 2\\sqrt{m} = 4 $, solving gives $ m = 4 $. Thus, $ c^{2} = 4 - 3 = 1 $, that is, $ c = 1 $, so the focal length of $ C $ is $ 2c = 2 $."
  },
  {
    "text": "If the foci of the ellipse $a x^{2}+y^{2}=6$ are $(0, \\pm \\sqrt{3})$, then the value of the real number $a$ is?",
    "fact_expressions": "G: Ellipse;a: Real;Expression(G) = (a*x^2 + y^2 = 6);Coordinate(Focus(G)) = (0, sqrt(3)*pm)",
    "query_expressions": "a",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 20]], [[47, 52]], [[1, 20]], [[1, 45]]]",
    "query_spans": "[[[47, 56]]]",
    "process": ""
  },
  {
    "text": "Let the origin be $O$, and suppose the parabola $y^{2}=4x$ intersects a line passing through the point $(m,\\ 0)$ at points $A$ and $B$. If $OA \\cdot OB = -3$, then what is the value of $m$?",
    "fact_expressions": "G: Parabola;H: Line;I: Point;O: Origin;A: Point;B: Point;m: Number;Expression(G) = (y^2 = 4*x);Coordinate(I) = (m, 0);PointOnCurve(I, H);Intersection(G, H) = {A, B};LineSegmentOf(O, A)*LineSegmentOf(O, B) = -3",
    "query_expressions": "m",
    "answer_expressions": "{1, 3}",
    "fact_spans": "[[[10, 24]], [[37, 39]], [[26, 36]], [[6, 9]], [[41, 44]], [[47, 50]], [[73, 76]], [[10, 24]], [[26, 36]], [[25, 39]], [[10, 52]], [[54, 70]]]",
    "query_spans": "[[[73, 80]]]",
    "process": ""
  },
  {
    "text": "Take a moving point $A(a , a^{2})$ on $y=x^{2}$, where $a \\in(0 , 5]$, and take a point $M(0 , \\frac{1}{a^{2}+a+4})$ on the $y$-axis. What is the maximum area of $\\Delta OAM$?",
    "fact_expressions": "G: Curve;Expression(G) = (y = x^2);Coordinate(A) = (a, a^2);A: Point;PointOnCurve(A, G) = True;a: Number;In(a, (0, 5]);M: Point;Coordinate(M) = (0, 1/(a^2 + a + 4));PointOnCurve(M, yAxis) = True;O: Origin",
    "query_expressions": "Max(Area(TriangleOf(O, A, M)))",
    "answer_expressions": "1/10",
    "fact_spans": "[[[1, 10]], [[1, 10]], [[14, 29]], [[14, 29]], [[0, 29]], [[31, 45]], [[31, 45]], [[54, 82]], [[54, 82]], [[46, 82]], [[85, 98]]]",
    "query_spans": "[[[85, 107]]]",
    "process": ""
  },
  {
    "text": "It is known that the focus of the parabola lies on the $y$-axis, the point $M(m,-3)$ is a point on the parabola, and the distance from $M$ to the focus is $5$. Then what is the value of $m$? What is the standard equation of the parabola? What is the equation of the directrix?",
    "fact_expressions": "G: Parabola;M: Point;m:Number;PointOnCurve(Focus(G),yAxis);Coordinate(M)=(m,-3);PointOnCurve(M, G);Distance(M,Focus(G))=5",
    "query_expressions": "m;Expression(G);Expression(Directrix(G))",
    "answer_expressions": "pm*2*sqrt(6);x^2=-8*y;y=2",
    "fact_spans": "[[[2, 5], [26, 29], [56, 59]], [[15, 25], [34, 37]], [[49, 52]], [[2, 14]], [[15, 25]], [[15, 33]], [[26, 47]]]",
    "query_spans": "[[[49, 56]], [[56, 66]], [[56, 72]]]",
    "process": "Let the equation of the parabola be $x^{2}=-2py$ $(p>0)$. Since the distance from $M$ to the focus is 5, we have $3+\\frac{p}{2}=5$, so $p=4$. Therefore, the standard equation of the parabola is $x^{2}=-8y$, and the equation of the directrix is $y=2$; since $m^{2}=-8\\times(-3)$, $\\therefore m=\\pm2\\sqrt{6}$"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{1}$ intersects the left branch at points $A$ and $B$. If $|A F_{2}|+|B F_{2}|=2|A B|$, then $|A B|=$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Line;PointOnCurve(F1, H);A: Point;B: Point;Intersection(H, LeftPart(G)) = {A, B};Abs(LineSegmentOf(A, F2)) + Abs(LineSegmentOf(B, F2)) = 2*Abs(LineSegmentOf(A, B))",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "4*a",
    "fact_spans": "[[[2, 58]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[67, 74], [85, 92]], [[75, 82]], [[2, 82]], [[2, 82]], [[93, 95]], [[83, 95]], [[101, 104]], [[106, 109]], [[2, 111]], [[114, 142]]]",
    "query_spans": "[[[145, 154]]]",
    "process": ""
  },
  {
    "text": "Point $A(x_{0},y_{0})$ lies on the right branch of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{32}=1$. If the distance from point $A$ to the right focus is equal to $2 x_{0}$, then $x_{0}=$?",
    "fact_expressions": "A: Point;Coordinate(A) = (x0, y0);x0: Number;y0: Number;G: Hyperbola;Expression(G) = (x^2/4 - y^2/32 = 1);PointOnCurve(A, RightPart(G)) = True;Distance(A, RightFocus(G)) = 2*x0",
    "query_expressions": "x0",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 17], [63, 67]], [[0, 17]], [[87, 94]], [[1, 17]], [[18, 57]], [[18, 57]], [[0, 61]], [[18, 85]]]",
    "query_spans": "[[[87, 96]]]",
    "process": ""
  },
  {
    "text": "Let point $P$ lie on the ellipse $x^{2}+\\frac{y^{2}}{m}=1$ $(m>0)$, and point $Q$ lie on the line $y=x+4$. If the minimum value of $PQ$ is $\\sqrt{2}$, then $m=$?",
    "fact_expressions": "P: Point;G: Ellipse;Expression(G) = (x^2 + y^2/m = 1);m: Number;m>0;PointOnCurve(P, G);H: Line;Expression(H) = (y = x + 4);Q: Point;PointOnCurve(Q, H);Min(LineSegmentOf(P, Q)) = sqrt(2)",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 5]], [[6, 38]], [[6, 38]], [[78, 81]], [[8, 38]], [[1, 39]], [[45, 54]], [[45, 54]], [[40, 44]], [[40, 55]], [[57, 76]]]",
    "query_spans": "[[[78, 83]]]",
    "process": "Analysis: Find the equation of the line parallel to the line $ y = x + 4 $ and at a distance of $ \\sqrt{2} $, use the condition that this line is tangent to the ellipse, set $ A = 0 $, thereby solving for the value of $ m $. According to the problem, the line equations parallel to $ y = x + 4 $ and at a distance of $ \\sqrt{2} $ are $ y = x + 2 $ or $ y = x + 6 $ (discarded). Solving the system \n\\[\n\\begin{cases}\ny = x + 2, \\\\\nx^{2} + \\frac{y^{2}}{m} = 1,\n\\end{cases}\n\\]\nwe obtain $ (m+1)x^{2} + 4x + 4 - m = 0 $. Let $ d = 16 - 4(m+1)(4-m) = 0 $, solving gives $ m = 0 $ or $ m = 3 $. Since $ m > 0 $, therefore $ m = 3 $."
  },
  {
    "text": "The hyperbola $16 x^{2}-9 y^{2}=144$, with left and right foci denoted as $F_{1}$ and $F_{2}$ respectively, has a point $P$ on the hyperbola such that $|PF_{1}| \\cdot |PF_{2}| = 64$. Find the area of $\\Delta PF_{1} F_{2}$.",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (16*x^2 - 9*y^2 = 144);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 64",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "16*sqrt(3)",
    "fact_spans": "[[[0, 25], [60, 63]], [[55, 59]], [[37, 44]], [[47, 54]], [[0, 25]], [[0, 54]], [[0, 54]], [[55, 64]], [[65, 92]]]",
    "query_spans": "[[[95, 121]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$ and directrix $l$, let $P$ be a point on $l$, and the extension of $PF$ intersects the parabola at point $Q$. If $2 \\overrightarrow{F P}+3 \\overrightarrow{F Q}=\\overrightarrow{0}$, then $|Q F|$=?",
    "fact_expressions": "C: Parabola;P: Point;F: Point;Q: Point;l: Line;Expression(C) = (y^2 = 4*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(P, l);Intersection(OverlappingLine(LineSegmentOf(P, F)), C) = Q;2*VectorOf(F, P) + 3*VectorOf(F, Q) =0",
    "query_expressions": "Abs(LineSegmentOf(Q, F))",
    "answer_expressions": "10/3",
    "fact_spans": "[[[2, 21], [58, 61]], [[37, 40]], [[25, 28]], [[62, 66]], [[41, 44], [41, 44]], [[2, 21]], [[2, 28]], [[2, 35]], [[37, 47]], [[48, 66]], [[68, 134]]]",
    "query_spans": "[[[136, 145]]]",
    "process": "From the given condition, $\\frac{|PF|}{|QF|}=\\frac{3}{2}$. Then, by drawing the figure and using similar triangles, combined with the definition of a parabola, $|QF|$ is obtained. According to the problem, $p=2$, $2\\overrightarrow{FP}+3\\overrightarrow{FQ}=\\overrightarrow{0}$, $\\frac{|PF|}{|QF|}=\\frac{3}{2}$. As shown in the figure, draw a perpendicular line from point $Q$ to the directrix $l$, with foot $Q'$, and let the intersection point of $l$ and the $x$-axis be $F$. Based on the given conditions and combining with the definition of a parabola, $\\frac{|FF|}{|QQ|}=\\frac{|PF|}{|PQ|}=\\frac{3}{5}$, so $|QQ|=\\frac{10}{3}$. Using the ratio of similar triangles and combining with the definition of a parabola, transform the distance from a point on the parabola to the focus into the distance to the directrix for solving."
  },
  {
    "text": "The moving line $l$ passing through point $P(3,4)$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. Perpendiculars are drawn from $A$ and $B$ to the $x$-axis and $y$-axis, respectively. Find the trajectory equation of the intersection point $M$ of these two perpendiculars.",
    "fact_expressions": "l: Line;P: Point;A: Point;B: Point;Coordinate(P) = (3, 4);PointOnCurve(P, l);Intersection(l, xAxis) = A;Intersection(l, yAxis) = B;L1: Line;L2: Line;M: Point;PointOnCurve(A, L1);PointOnCurve(B, L2);IsPerpendicular(L1, xAxis);IsPerpendicular(L2, yAxis);Intersection(L1, L2) = M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "3/x + 4/y = 1",
    "fact_spans": "[[[15, 18]], [[1, 11]], [[34, 37], [44, 47]], [[38, 41], [48, 51]], [[1, 11]], [[0, 18]], [[15, 41]], [[15, 41]], [], [], [[73, 76]], [[42, 66]], [[42, 66]], [[42, 66]], [[42, 66]], [[43, 76]]]",
    "query_spans": "[[[73, 83]]]",
    "process": ""
  },
  {
    "text": "Draw a line through the point $(1 , 0)$ with inclination angle $\\frac{2 \\pi}{3}$, intersecting $y^{2}=4 x$ at points $A$ and $B$. Then the chord length $A B$ is?",
    "fact_expressions": "H: Point;Coordinate(H) = (1, 0);G: Line;Inclination(G) = (2*pi)/3;PointOnCurve(H, G);A: Point;B: Point;Z: Curve;Expression(Z) = (y^2 = 4*x);Intersection(G, Z) = {A, B};IsChordOf(LineSegmentOf(A, B), Z)",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 11]], [[1, 11]], [[34, 36]], [[12, 36]], [[0, 36]], [[50, 53]], [[54, 57]], [[37, 48]], [[37, 48]], [[34, 57]], [[37, 66]]]",
    "query_spans": "[[[37, 69]]]",
    "process": ""
  },
  {
    "text": "The distance between the directrix of the parabola $x^{2}=m y$ and the line $y=2$ is $3$. Then the equation of this parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = m*y);m: Number;Z: Line;Expression(Z) = (y = 2);Distance(Directrix(G), Z) = 3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2 = -20*y, x^2 = 4*y}",
    "fact_spans": "[[[0, 14], [35, 38]], [[0, 14]], [[3, 14]], [[18, 25]], [[18, 25]], [[0, 32]]]",
    "query_spans": "[[[35, 43]]]",
    "process": "Let the equation of the directrix be $ y = -\\frac{m}{4} $, $\\therefore \\left| -\\frac{m}{4} - 2 \\right| = 3$, $\\therefore m = -20 $ or $ m = 4 $, $\\therefore x^{2} = -20y^{2} $, $ x^{2} = 4y $."
  },
  {
    "text": "Given the lines $l_{1}$: $x=-1$, $l_{2}$: $y=x+1$, and a point $P$ on the parabola $C$: $y^{2}=4x$, find the minimum value of the sum of the distances from $P$ to these two lines.",
    "fact_expressions": "C: Parabola;l1:Line;l2:Line;P: Point;Expression(C) = (y^2 = 4*x);Expression(l2) = (y=x+1);Expression(l1) = (x = -1);PointOnCurve(P, C)",
    "query_expressions": "Min(Distance(P,l1)+Distance(P,l2))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[45, 64]], [[2, 20]], [[22, 39]], [[40, 44], [69, 72]], [[45, 64]], [[22, 39]], [[2, 20]], [[41, 67]]]",
    "query_spans": "[[[69, 88]]]",
    "process": "Through point P, draw PM\\bot l_{1}, PN\\bot l_{2}, with M and N the feet of the perpendiculars respectively. The parabola C: y^2=4x has focus F(1,0), and the line l_{1}: x=-1 is the directrix of the parabola. By the definition of a parabola, |PM|=|PF|. Hence, |PM|+|PN|=|PN|+|PF|. When points N, P, F are collinear, |PN|+|PF| reaches its minimum value. Thus, the minimum value is the distance from point F to line l_{2}: \\frac{|1-0+1|}{\\sqrt{2}+\\frac{1}{2}}=\\sqrt{2}."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{m}-y^{2}=1$ $(m>0)$ has an asymptote $\\sqrt{3} x+m y=0$, then the length of the real axis of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2 + x^2/m = 1);m: Number;m>0;Expression(OneOf(Asymptote(C))) = (m*y + sqrt(3)*x = 0)",
    "query_expressions": "Length(RealAxis(C))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[2, 40], [67, 73]], [[2, 40]], [[10, 40]], [[10, 40]], [[2, 65]]]",
    "query_spans": "[[[67, 79]]]",
    "process": "From the asymptote equation $\\sqrt{3}x + my = 0$, simplifying yields $y = -\\frac{\\sqrt{3}}{m}x$, so $\\frac{b}{a} = \\frac{\\sqrt{3}}{m}$. Squaring both sides gives $\\frac{b^{2}}{a^{2}} = \\frac{3}{m^{2}}$. Also, for the hyperbola, $a^{2} = m$, $b^{2} = 1$. Hence, $\\frac{3}{m^{2}} = \\frac{1}{m}$, solving gives $m = 3$, $m = 0$ (discarded). Therefore, $a^{2} = 3$, $a = \\sqrt{3}$, and the length of the real axis is $2a = 2\\sqrt{3}$. Hence, the answer is: $2\\sqrt{3}$."
  },
  {
    "text": "Draw two tangents from a focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ to the circle $x^{2}+y^{2}=a^{2}$, with points of tangency $A$ and $B$. If $\\angle A O B=120^{\\circ}$ ($O$ is the origin), then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;G: Circle;Expression(G) = (x^2 + y^2 = a^2);L1: Line;L2: Line;TangentOfPoint(OneOf(Focus(C)), G) = {L1, L2};TangentPoint(L1, G) = A;TangentPoint(L2, G) = B;A: Point;B: Point;AngleOf(A, O, B) = ApplyUnit(120, degree);O: Origin",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 63], [148, 154]], [[1, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[69, 89]], [[69, 89]], [], [], [[0, 94]], [[0, 107]], [[0, 107]], [[100, 103]], [[104, 107]], [[110, 136]], [[137, 140]]]",
    "query_spans": "[[[148, 160]]]",
    "process": ""
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=4x$, draw a line $l$ intersecting it at points $A$ and $B$. If $|AF|=4$, then $|BF|=$?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 4*x);Focus(G)= F;PointOnCurve(F,l);Intersection(l,G) = {A, B};Abs(LineSegmentOf(A, F)) = 4",
    "query_expressions": "Abs(LineSegmentOf(B, F))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[22, 27]], [[1, 15], [28, 29]], [[31, 34]], [[18, 21]], [[35, 38]], [[1, 15]], [[1, 21]], [[0, 27]], [[22, 40]], [[42, 51]]]",
    "query_spans": "[[[53, 62]]]",
    "process": "Given: $ y^{2} = 4x $, the focus coordinates are $ (1,0) $, $ |AF| = 4 $. Using the definition of a parabola, we obtain $ A(3,2\\sqrt{3}) $. The equation of line $ l $ can be found as $ y = \\sqrt{3}(x-1) $. Solving the system of equations yields: $ 3x^{2} - 10x + 3 = 0 $, $ x_{1} + x_{2} = \\frac{10}{3} $, $ x_{1} = 3 $, $ x_{2} = \\frac{1}{3} $, $ |BF| = \\frac{1}{3} + \\frac{p}{2} = \\frac{4}{3} $"
  },
  {
    "text": "Given that $O$ is the coordinate origin, the focus of the parabola $C$: $y^{2}=8x$ is $F$, $P$ is a point on $C$, $PF$ is perpendicular to the $x$-axis, and $Q$ is a point on the $x$-axis. If $P$ lies on the circle with diameter $OQ$, then the coordinates of point $Q$ are?",
    "fact_expressions": "O: Origin;C: Parabola;Expression(C) = (y^2 = 8*x);F: Point;Focus(C) = F;P: Point;PointOnCurve(P, C) ;IsPerpendicular(LineSegmentOf(P, F), xAxis) ;Q: Point;PointOnCurve(Q, xAxis) ;PointOnCurve(P, G) ;IsDiameter(LineSegmentOf(O, Q), G);G: Circle",
    "query_expressions": "Coordinate(Q)",
    "answer_expressions": "(10, 0)",
    "fact_spans": "[[[2, 5]], [[11, 30], [42, 45]], [[11, 30]], [[34, 37]], [[11, 37]], [[38, 41], [75, 78]], [[38, 48]], [[49, 61]], [[62, 65], [95, 99]], [[62, 73]], [[75, 93]], [[79, 92]], [[91, 92]]]",
    "query_spans": "[[[95, 104]]]",
    "process": "Point F(2,0), let P(2,4), set Q(t,0). From OP\\botPQ we get: \\frac{4}{2}\\cdot\\frac{0-4}{t-2}=-1, solving gives t=10"
  },
  {
    "text": "Given the parabola $y^{2}=2 p x$ ($p>0$) with focus $F$, and two moving points $A$, $B$ on the parabola always satisfying $\\angle AFB=60^{\\circ}$, a perpendicular $HN$ is drawn from the midpoint $H$ of chord $AB$ to the directrix of the parabola, with foot $N$. Then the range of $\\frac{|H N|}{|A B|}$ is?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;H: Point;N: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(A,G);PointOnCurve(B,G);AngleOf(A, F, B) = ApplyUnit(60, degree);IsChordOf(LineSegmentOf(A,B),G);MidPoint(LineSegmentOf(A,B))=H;PointOnCurve(H,LineOf(H,N));IsPerpendicular(Directrix(G),LineOf(H,N));FootPoint(Directrix(G),LineOf(H,N))=N",
    "query_expressions": "Range(Abs(LineSegmentOf(H, N))/Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "(0, 1]",
    "fact_spans": "[[[2, 23], [31, 34], [90, 93]], [[5, 23]], [[40, 43]], [[46, 49]], [[86, 89]], [[107, 110]], [[27, 30]], [[5, 23]], [[2, 23]], [[2, 30]], [[31, 49]], [[31, 49]], [[53, 76]], [[31, 83]], [[79, 89]], [[77, 103]], [[90, 103]], [[90, 110]]]",
    "query_spans": "[[[112, 140]]]",
    "process": "Using the properties of a parabola, we obtain HN = \\frac{a+b}{2}. Combining the law of cosines and expressing |AB| in terms of a and b, we establish an equation. By using the inequality a + b \\geqslant 2\\sqrt{ab}, we compute the range and thus obtain the answer. Draw perpendiculars AQ and BP from A and B to the directrix of the parabola, with Q and P being the feet of the perpendiculars. Let |AF| = a, |BF| = b. Then by the definition of a parabola, |AQ| = a, |BP| = b, so |HN| = \\frac{a+b}{2}. In triangle ABF, by the law of cosines, |AB|^{2} = a^{2} + b^{2} - 2ab\\cos60^{0} = a^{2} + b^{2} - ab. Therefore, \\frac{|HN|}{|AB|} = \\frac{\\frac{a+b}{2}}{\\sqrt{a^{2}+b^{2}-ab}}. Since a + b \\geqslant 2\\sqrt{ab}, it follows that \\frac{1}{2\\sqrt{1-\\frac{3ab}{(a+b)^{2}}}} \\leqslant 1, with equality if and only if a = b. Hence, the range of \\frac{|HN|}{|AB|} is (0,1]."
  },
  {
    "text": "It is known that hyperbola $C_{1}$ has the same asymptotes as hyperbola $C_{2}$: $\\frac{x^{2}}{2}-\\frac{y^{2}}{6}=1$, and the focal distance of hyperbola $C_{1}$ is $8$. Then the equation of hyperbola $C_{1}$ is?",
    "fact_expressions": "C1: Hyperbola;C2: Hyperbola;Expression(C2)=(x^2/2-y^2/6=1);Asymptote(C1)=Asymptote(C2);FocalLength(C1) = 8",
    "query_expressions": "Expression(C1)",
    "answer_expressions": "{x^2/4-y^2/12=1,y^2/12-x^2/4=1}",
    "fact_spans": "[[[2, 12], [68, 78], [87, 97]], [[13, 60]], [[13, 60]], [[2, 66]], [[68, 85]]]",
    "query_spans": "[[[87, 102]]]",
    "process": "Let the equation of hyperbola $ C_{1} $ be $ \\frac{x^{2}}{2}-\\frac{y^{2}}{6}=\\lambda $. According to the focal distance calculation, we obtain the answer. Then $ 2\\lambda+6\\lambda=16 $ or $ -2\\lambda-6\\lambda=16 $, solving gives $ \\lambda=2 $ or $ \\lambda=-2 $. Hence, the equation of hyperbola $ C_{1} $ is $ \\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1 $ or $ \\frac{y^{2}}{12}-\\frac{x^{2}}{4}=1 $."
  },
  {
    "text": "The asymptote equations of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(1/2)*x",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 35]]]",
    "process": "First determine the coordinate axis on which the foci of the hyperbola lie, then determine the lengths of the real and imaginary axes of the hyperbola, and finally determine the equations of the asymptotes of the hyperbola. [Detailed explanation] \\because the hyperbola \\frac{x^{2}}{4}-y^{2}=1 has a=2, b=1, with foci on the x-axis, and the asymptotes of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{k^{2}}=1 are given by y=\\pm\\frac{b}{a}x, \\therefore the asymptotes of the hyperbola \\frac{x^{2}}{4}-y^{2}=1 are y=\\pm\\frac{1}{2}x"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, let $l$ be its asymptote with positive slope. If there are exactly three distinct points on the curve $E$: $(x-2)^{2}+y^{2}=4$ whose distance to $l$ is $1$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;E: Curve;l: Line;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(E) = (y^2 + (x - 2)^2 = 4);OneOf(Asymptote(C)) = l;Slope(l) > 0;F1: Point;F2: Point;F3: Point;PointOnCurve(F1, E);PointOnCurve(F2, E);PointOnCurve(F3, E);Distance(F1, l) = 1;Distance(F2, l) = 1;Distance(F3, l) = 1;Negation(F1 = F2);Negation(F2 = F3);Negation(F1 = F3)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 63], [125, 131]], [[10, 63]], [[10, 63]], [[78, 103]], [[73, 76], [113, 116]], [[10, 63]], [[10, 63]], [[2, 63]], [[78, 103]], [[2, 76]], [[64, 76]], [], [], [], [[78, 112]], [[78, 112]], [[78, 112]], [[78, 123]], [[78, 123]], [[78, 123]], [[78, 112]], [[78, 112]], [[78, 112]]]",
    "query_spans": "[[[125, 137]]]",
    "process": "The curve $ E: (x-2)^{2}+y^{2}=4 $ is a circle with center $ E(2,0) $ and radius $ r=2 $. The line $ l $ is given by $ y=\\frac{b}{a}x $, or equivalently $ bx-ay=0 $. From the problem, the distance from $ E $ to $ l $ is $ r-1=1 $, hence $ \\frac{2b}{\\sqrt{b^{2}+a^{2}}}=1 \\Rightarrow \\frac{b^{2}}{a^{2}}=\\frac{1}{3} $. Therefore, $ e=\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{1+\\frac{1}{3}}=\\frac{2\\sqrt{3}}{3} $."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{7-m}+\\frac{y^{2}}{m-1}=1$ represents an ellipse, then the range of real values for $m$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(7 - m) + y^2/(m - 1) = 1);m: Real",
    "query_expressions": "Range(m)",
    "answer_expressions": "(1, 4)+(4, 7)",
    "fact_spans": "[[[44, 46]], [[1, 46]], [[48, 53]]]",
    "query_spans": "[[[48, 60]]]",
    "process": "According to the standard form of the ellipse equation, the condition for the equation $\\frac{x^2}{7-m}+\\frac{y^{2}}{m-1}=1$ to represent an ellipse is:  \n\\begin{cases}7-m>0\\\\m-1>0\\\\7-m\\neq m\\end{cases}  \nSolving gives $1<m<7$ and $m\\neq4$, so the range of real number $m$ is $(1,4)\\cup(4,7)$. Hence, the answer is: $(1,4)\\cup(4,7)$."
  },
  {
    "text": "Given the line $l$: $a x + b y = 0$ intersects the ellipse $x^{2} + \\frac{y^{2}}{9} = 1$ at points $A$ and $B$, if $C(5,5)$, then the range of $\\overrightarrow{C A} \\cdot \\overrightarrow{C B}$ is?",
    "fact_expressions": "l: Line;G: Ellipse;a:Number;b:Number;C: Point;A: Point;B: Point;Expression(G) = (x^2 + y^2/9 = 1);Expression(l) = (a*x + b*y = 0);Coordinate(C) = (5, 5);Intersection(l, G) = {A, B}",
    "query_expressions": "Range(DotProduct(VectorOf(C, A), VectorOf(C, B)))",
    "answer_expressions": "[41,49]",
    "fact_spans": "[[[2, 20]], [[21, 48]], [[9, 20]], [[9, 20]], [[61, 69]], [[50, 53]], [[54, 57]], [[21, 48]], [[2, 20]], [[61, 69]], [[2, 59]]]",
    "query_spans": "[[[71, 127]]]",
    "process": "The line $ l: ax + by = 0 $ passes through the origin. Combining this with the symmetry of the ellipse, points $ A $ and $ B $ are symmetric about the origin.  \nMethod 1: Let $ A(x_0, y_0) $, $ B(-x_0, -y_0) $, then  \n$$\n\\overrightarrow{CA} \\cdot \\overrightarrow{CB} = (x_0 - 5, y_0 - 5) \\cdot (-x_0 - 5, -y_0 - 5) = 50 - (x_0^2 + y_0^2)\n$$\n$$\n|OA| = \\sqrt{x_0^2 + y_0^2} \\in [b, a], \\text{ i.e., } \\sqrt{x_0^2 + y_0^2} \\in [1, 3],\n$$\n$$\n\\therefore \\overrightarrow{CA} \\cdot \\overrightarrow{CB} \\in [41, 49]\n$$\nMethod 2: Using parametric equations, let $ A(\\cos\\theta, 3\\sin\\theta) $, $ B(-\\cos\\theta, -3\\sin\\theta) $, then  \n$$\n\\overrightarrow{CA} \\cdot \\overrightarrow{CB} = 50 - \\cos 2\\theta - 9\\sin^2\\theta = 49 - 8\\sin^2\\theta \\in [41, 49].\n$$\n[Note] This problem examines the range of the dot product of vectors formed by a point and two points on an ellipse that are symmetric about the origin. In the solution process, the condition that the two points are symmetric about the origin is crucial, and parametric equations can be used to express the coordinates of the points."
  },
  {
    "text": "An endpoint of the imaginary axis of a hyperbola is $M$, and the two foci are $F_{1}$, $F_{2}$, with $\\angle F_{1} M F_{2}=120^{\\circ}$. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;F1: Point;M: Point;F2: Point;OneOf(Endpoint(ImageinaryAxis(G)))=M;Focus(G) = {F1,F2};AngleOf(F1, M, F2) = ApplyUnit(120, degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[0, 3], [76, 79]], [[20, 27]], [[11, 14]], [[30, 37]], [[0, 14]], [[0, 37]], [[40, 74]]]",
    "query_spans": "[[[76, 85]]]",
    "process": ""
  },
  {
    "text": "Given that the directrix of the parabola is $x = -1$, then its standard equation is?",
    "fact_expressions": "G: Parabola;Expression(Directrix(G))=(x=-1)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[2, 5], [17, 18]], [[2, 15]]]",
    "query_spans": "[[[17, 24]]]",
    "process": "Since the directrix of the parabola is x = -1, the axis of symmetry of the parabola is the x-axis, and it opens to the right. Therefore, assume the equation of the parabola is y^{2} = 2px (p > 0). According to the given condition, we have \\frac{p}{2} = 1, \\therefore p = 2. Thus, the standard equation of the parabola is y^{2} = 4x. Hence, fill in y^{2} = 4x. The key to this problem is determining the form of the parabola's equation."
  },
  {
    "text": "The curve $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1(b>0)$ always has common points with the line $l$: $k x-y+k+2=0$, then the range of values for $b$ is?",
    "fact_expressions": "C: Curve;Expression(C) = (x^2/4 + y^2/b^2 = 1);b: Number;b>0;l: Line;Expression(l) = (k + k*x - y + 2 = 0);k: Number;IsIntersect(C, l) = True",
    "query_expressions": "Range(b)",
    "answer_expressions": "[4*sqrt(3)/3,+oo)",
    "fact_spans": "[[[0, 51]], [[0, 51]], [[79, 82]], [[7, 51]], [[52, 72]], [[52, 72]], [[59, 72]], [[0, 77]]]",
    "query_spans": "[[[79, 89]]]",
    "process": "The line $ l: kx - y + k + 2 = 0 \\therefore k(x+1) + (-y+2) = 0. \\begin{cases} x+1=0 \\\\ -y+2=0 \\end{cases} $ solving gives $ \\begin{cases} x=-1 \\\\ y=2 \\end{cases} $, so the line $ l $ always passes through $ (-1,2) $. The curve $ C: \\frac{x^{2}}{4} + \\frac{y^{2}}{b^{2}} = 1 $ ($ b>0 $) always has common points with the line $ l: kx - y + k + 2 = 0 $, which yields $ \\frac{1}{4} + \\frac{4}{b^{2}} \\leqslant 1 $. Since $ b>0 $, $ \\therefore b \\geqslant \\frac{4\\sqrt{3}}{3} $."
  },
  {
    "text": "The equation of the circle with center at the right focus $F$ of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ and passing through the endpoints of the minor axis of the ellipse is?",
    "fact_expressions": "G: Ellipse;H: Circle;Expression(G)=(x^2/4+y^2/3=1);RightFocus(G)=F;Center(H)=F;F:Point;PointOnCurve(Endpoint(MinorAxis(G)),H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x-1)^2+y^2=4",
    "fact_spans": "[[[1, 38], [53, 55]], [[61, 62]], [[1, 38]], [[1, 45]], [[0, 62]], [[42, 45]], [[52, 62]]]",
    "query_spans": "[[[61, 67]]]",
    "process": "The standard equation of the ellipse is: \\frac{x^2}{4}+\\frac{y^{2}}{3}=1, so a=2, b=\\sqrt{3}, and c=\\sqrt{a^{2}-b^{2}}=1, thus the coordinates of the right focus are (1,0). The center of the required circle is at (1,0). By the symmetry of the ellipse, it is known that the distances from the right focus to the endpoints of the minor axis of the ellipse are equal. Therefore, the distance from the right focus to the upper vertex of the ellipse can be calculated as \\sqrt{(1-0)^{2}+(0-\\sqrt{3})^{2}}=2, so the radius of the required circle is 2. Hence, the standard equation of the required circle is (x-1)^{2}+y^{2}=4."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{12}=1$ $(a>0)$, one of whose asymptotes is $\\sqrt{3} x - y = 0$, and the left focus is $F$. When point $M$ lies on the right branch of the hyperbola and point $N$ moves on the circle $x^{2} + (y - 3)^{2} = 4$, then the minimum value of $|MN| + |MF|$ is?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Circle;M: Point;N: Point;F: Point;a>0;Expression(G) = (-y^2/12 + x^2/a^2 = 1);Expression(H) = (x^2 + (y - 3)^2 = 4);Expression(OneOf(Asymptote(G))) = (sqrt(3)*x - y = 0);LeftFocus(G) = F;PointOnCurve(M, RightPart(G));PointOnCurve(N, H)",
    "query_expressions": "Min(Abs(LineSegmentOf(M, F)) + Abs(LineSegmentOf(M, N)))",
    "answer_expressions": "7",
    "fact_spans": "[[[2, 50], [90, 93]], [[5, 50]], [[102, 122]], [[85, 89]], [[97, 101]], [[80, 83]], [[5, 50]], [[2, 50]], [[102, 122]], [[2, 75]], [[2, 83]], [[84, 96]], [[97, 126]]]",
    "query_spans": "[[[128, 147]]]",
    "process": "From the hyperbola equation $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{12}=1$, we get $b=2\\sqrt{3}$, so the asymptotes are $y=\\pm\\frac{\\sqrt[2]{3}}{a}x$. Comparing with the equation $\\sqrt{3}x-y=0$, we obtain $a=2$. Thus, the hyperbola equation is $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$. The point $F(-4,0)$, denote the right focus of the hyperbola as $F'(4,0)$, and since point $M$ lies on the right branch of the hyperbola, $|MF|=4+|MF'|$. Therefore, $|MN|+|MF|=|MN|+|MF'|+4$. By the shortest distance between two points being a straight line, the minimum value of $|MN|+|MF'|+4$ is $|F'N|+4$. Since point $N$ moves on the circle $x^{2}+(y-3)^{2}=4$, the minimum value of $|F'N|$ is the distance from point $F'$ to the center $(0,3)$ minus the radius $2$. Hence, $|F'N|_{\\min}=5-2=3$. Therefore, the minimum value of $|MN|+|MF|$ is $7$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$, the eccentricity of the hyperbola is $2$, point $P$ lies on the right branch of hyperbola $C$, and the midpoint $N$ of $P F_{1}$ lies on the circle $O$: $x^{2}+y^{2}=c^{2}$, where $c$ is the semi-focal length of the hyperbola, then $\\sin \\angle F_{1} P F_{2}$=?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;O: Circle;c: Number;P: Point;F1: Point;F2: Point;N: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(O) = (x^2 + y^2 = c^2);LeftFocus(C) = F1;RightFocus(C) = F2;Eccentricity(C) = 2;PointOnCurve(P, RightPart(C));MidPoint(LineSegmentOf(P, F1)) = N;PointOnCurve(N, O);HalfFocalLength(C) = c",
    "query_expressions": "Sin(AngleOf(F1, P, F2))",
    "answer_expressions": "sqrt(7)/4",
    "fact_spans": "[[[18, 82], [89, 92], [106, 112], [167, 170]], [[26, 82]], [[26, 82]], [[134, 159]], [[163, 166]], [[101, 105]], [[2, 9]], [[10, 17]], [[130, 133]], [[26, 82]], [[26, 82]], [[18, 82]], [[134, 159]], [[2, 88]], [[2, 88]], [[89, 100]], [[101, 116]], [[118, 133]], [[130, 160]], [[163, 174]]]",
    "query_spans": "[[[176, 205]]]",
    "process": "As shown in the figure, from the given conditions we have $ OF_{1} = ON = c $. Since $ O $ is the midpoint of $ F_{1}F_{2} $, it follows that $ ON = \\frac{1}{2}PF_{2} $, so $ PF_{2} = 2c $, $ PF_{1} = 2a + 2c $. $ \\because $ the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) has eccentricity 2, $ \\therefore c = 2a $. Hence, in $ \\triangle F_{1}PF_{2} $, $ PF_{1} = 6a $, $ PF_{2} = F_{1}F_{2} = 4a $, $ \\sin\\angle F_{1}PF_{2} = \\frac{F_{2}N}{PF_{2}} = \\frac{\\sqrt{16a^{2} - 9a^{2}}}{4a} = \\frac{\\sqrt{7}}{4} $"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, if a line passing through the right focus $F$ with an inclination angle of $30^{\\circ}$ intersects the right branch of the hyperbola at two points, then the range of the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(G) = F;PointOnCurve(F, H) = True;Inclination(H) = ApplyUnit(30, degree);H: Line;NumIntersection(H, RightPart(G)) = 2",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1, 2*sqrt(3)/3)",
    "fact_spans": "[[[2, 60], [91, 94], [105, 108]], [[2, 60]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[67, 70]], [[2, 70]], [[63, 90]], [[71, 90]], [[88, 90]], [[88, 102]]]",
    "query_spans": "[[[105, 118]]]",
    "process": "To have two intersection points between the line and the hyperbola, the slope of one asymptote of the hyperbola must be less than the slope of the line, that is, $\\frac{b}{a}<\\frac{\\sqrt{3}}{3}$; obtain the inequality relation between $a$ and $b$, then use $b=\\sqrt{c^{2}-a^{2}}$ to convert it into an inequality relation between $a$ and $c$, find a range for the eccentricity, and finally combine with the fact that the eccentricity of a hyperbola is greater than 1 to obtain the range of $e$. To have two intersection points between the line and the hyperbola, the slope of one asymptote of the hyperbola must be less than the slope of the line, that is, $\\frac{b}{a}<\\tan30^{\\circ}=\\frac{\\sqrt{3}}{3}$, $\\therefore b<\\frac{\\sqrt{3}}{3}$, $\\because b=\\sqrt{c^{2}-a^{2}}$, $\\therefore \\sqrt{c^{2}-a^{2}}<\\frac{\\sqrt{3}}{3}$, simplifying yields $c<\\frac{2\\sqrt{3}}{3}$, $\\therefore e=\\frac{c}{a}<\\frac{2\\sqrt{3}}{3}$, $\\because$ in a hyperbola $e>1$, the range of $e$ is $(1,\\underline{2}$"
  },
  {
    "text": "Given points $O(0,0)$, $A(1,2)$, and a moving point $P$ satisfying $|\\overrightarrow{O P}+\\overrightarrow{A P}|=2$, what is the trajectory equation of point $P$?",
    "fact_expressions": "O: Point;A: Point;P: Point;Coordinate(O) = (0, 0);Coordinate(A) = (1, 2);Abs(VectorOf(A, P) + VectorOf(O, P)) = 2",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "4*x^2+4*y^2-4*x-8*y+1=0",
    "fact_spans": "[[[2, 11]], [[13, 21]], [[24, 27], [78, 82]], [[2, 11]], [[13, 21]], [[29, 76]]]",
    "query_spans": "[[[78, 89]]]",
    "process": "Let the coordinates of point P be (x, y), then \\overrightarrow{OP}=(x,y), \\overrightarrow{AP}=(x-1,y-2), \\overrightarrow{OP}+\\overrightarrow{AP}=(2x-1,2y-2). Since |\\overrightarrow{OP}+\\overrightarrow{AP}|=2, it follows that (2x-1)^{2}+(2y-2)^{2}=4, which simplifies to 4x^{2}+4y^{2}-4x-8y+1=0."
  },
  {
    "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 46]]]",
    "process": ""
  },
  {
    "text": "The right focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$ is $F$, and $M$, $N$ are two moving points on the $y$-axis. If $\\overrightarrow{M F} \\cdot \\overrightarrow{N F}=0$, then the minimum area of $\\Delta M N F$ is?",
    "fact_expressions": "G: Ellipse;M: Point;N: Point;F: Point;Expression(G) = (x^2/4 + y^2/2 = 1);RightFocus(G) = F;PointOnCurve(M, yAxis);PointOnCurve(N, yAxis);DotProduct(VectorOf(M, F), VectorOf(N, F)) = 0",
    "query_expressions": "Min(Area(TriangleOf(M, N, F)))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 37]], [[46, 49]], [[50, 53]], [[42, 45]], [[0, 37]], [[0, 45]], [[46, 64]], [[46, 64]], [[66, 117]]]",
    "query_spans": "[[[119, 141]]]",
    "process": "Let point M(0,m), N(0,n) (m > n). From the problem, F(\\sqrt{2},0), and \\overrightarrow{MF} \\cdot \\overrightarrow{NF} = 0, (\\sqrt{2},-m)(\\sqrt{2},-n) = 0, i.e., mn = -2, so m > 0. Let the area of \\triangle MNF be S, mn = -2, then n = -\\frac{2}{m} (m > 0). The minimum area of \\triangle MNF is 2."
  },
  {
    "text": "Given $F_{1}$, $F_{2}$ are the left and right foci of the hyperbola $C$: $x^{2}-y^{2}=1$, point $P$ lies on $C$, $\\angle F_{1} P F_{2}=60^{\\circ}$, then $|P F_{1}| \\cdot|P F_{2}|$=?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2 = 1);F1: Point;P: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "4",
    "fact_spans": "[[[18, 41], [53, 56]], [[18, 41]], [[2, 9]], [[48, 52]], [[10, 17]], [[2, 47]], [[2, 47]], [[48, 57]], [[58, 91]]]",
    "query_spans": "[[[93, 121]]]",
    "process": "From the equation of the hyperbola, we have a=1, b=1, c=\\sqrt{2}. By the cosine law, \\therefore\\frac{1}{2}=\\frac{F_{2}^{2}+2|PF_{1}^{2}+|PF_{1}^{2}|FF_{P}|_{PFPF_{1}PF_{2}}^{2}}{2|PF_{1}PF_{1}-2\\sqrt{2}}|,\\therefore|PF_{1}|PF_{1}=4"
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. If there exists a point $P$ on the ellipse such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, then what is the range of values for the eccentricity of the ellipse?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F2: Point;F1: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P,G);DotProduct(VectorOf(P,F1),VectorOf(P,F2))=0",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[\\sqrt{2}/2,1)",
    "fact_spans": "[[[17, 69], [77, 79], [149, 151]], [[19, 69]], [[19, 69]], [[82, 86]], [[9, 16]], [[1, 8]], [[19, 69]], [[19, 69]], [[17, 69]], [[1, 75]], [[1, 75]], [[77, 86]], [[88, 147]]]",
    "query_spans": "[[[149, 161]]]",
    "process": ""
  },
  {
    "text": "Given the line $ l $: $ 2x + 4y + 3 = 0 $, $ P $ is a moving point on $ l $, $ O $ is the origin, and point $ Q $ lies on the segment $ OP $ such that $ \\frac{|\\overrightarrow{OQ}|}{|\\overrightarrow{OP}|} = \\frac{1}{3} $. What is the trajectory equation of point $ Q $?",
    "fact_expressions": "l: Line;O: Origin;P: Point;Q: Point;Expression(l) = (2*x + 4*y + 3 = 0);PointOnCurve(P, l);PointOnCurve(Q, LineSegmentOf(O,P));Abs(VectorOf(O, Q))/Abs(VectorOf(O, P)) = 1/3",
    "query_expressions": "LocusEquation(Q)",
    "answer_expressions": "2*x+4*y+1=0",
    "fact_spans": "[[[2, 22], [28, 31]], [[36, 39]], [[24, 27]], [[45, 49], [129, 133]], [[2, 22]], [[24, 35]], [[45, 58]], [[60, 127]]]",
    "query_spans": "[[[129, 140]]]",
    "process": "Let Q(x, y). Since the line l: 2x + 4y + 3 = 0, P is a moving point on l, O is the origin, and point Q lies on the segment OP such that \\frac{|\\overrightarrow{OQ}|}{|\\overrightarrow{OP}|} = \\frac{1}{3}, then P(3x, 3y) satisfies 6x + 12y + 3 = 0, which simplifies to 2x + 4y + 1 = 0. Thus, the trajectory equation of point Q is: 2x + 4y + 1 = 0."
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{27}=1$, respectively. $A$ is a point on the ellipse, and $\\overrightarrow{O B}=\\frac{1}{2}(\\overrightarrow{O A}+\\overrightarrow{O F_{1}})$, $\\overrightarrow{O C}=\\frac{1}{2}(\\overrightarrow{O A}+\\overrightarrow{O F_{2}})$. Then $|\\overrightarrow{O B}|+|\\overrightarrow{O C}|$=?",
    "fact_expressions": "G: Ellipse;O: Origin;B: Point;A: Point;F1: Point;C: Point;F2: Point;Expression(G) = (x^2/36 + y^2/27 = 1);LeftFocus(G)= F1;RightFocus(G) = F2;PointOnCurve(A, G);VectorOf(O, B) = (VectorOf(O, A) + VectorOf(O, F1))/2;VectorOf(O, C) = 1/2*(VectorOf(O,A)+VectorOf(O,F2))",
    "query_expressions": "Abs(VectorOf(O, B)) + Abs(VectorOf(O, C))",
    "answer_expressions": "6",
    "fact_spans": "[[[18, 57], [68, 70]], [[75, 156]], [[75, 156]], [[64, 67]], [[0, 7]], [[159, 241]], [[8, 15]], [[18, 57]], [[0, 63]], [[0, 63]], [[64, 73]], [[75, 156]], [[159, 241]]]",
    "query_spans": "[[[243, 292]]]",
    "process": "From the ellipse equation $\\frac{x^{2}}{36}+\\frac{y^{2}}{27}=1$, we obtain $a=6$. By the definition of an ellipse, $|AF_{1}|+|AF_{2}|=2a=12$. Since $OB=\\frac{1}{2}(OA+OF_{1})$, point $B$ is the midpoint of $AF_{1}$; $OC=\\frac{1}{2}(OA+OF_{2})$, so point $C$ is the midpoint of $AF_{2}$. Since $O$ is the midpoint of $F_{1}F_{2}$, $|OB|=\\frac{1}{2}|AF_{2}|$, $|OC|=\\frac{1}{2}|AF_{1}|$, thus $|OB|+|OC|=\\frac{1}{2}(|AF_{1}|+|AF_{2}|)=6$. Therefore, the answer is $6$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, if there exists a point $P$ on the hyperbola $C$ such that $P F_{1} \\perp P F_{2}$ and $|P F_{1}|=3|P F_{2}|$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Focus(C)={F1,F2};PointOnCurve(P,C);IsPerpendicular(LineSegmentOf(P,F1),LineSegmentOf(P,F2));Abs(LineSegmentOf(P,F1))=3*Abs(LineSegmentOf(P,F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(10)/2",
    "fact_spans": "[[[19, 80], [87, 93], [150, 156]], [[27, 80]], [[27, 80]], [[96, 100]], [[2, 9]], [[11, 18]], [[27, 80]], [[27, 80]], [[19, 80]], [[2, 85]], [[87, 100]], [[102, 125]], [[126, 148]]]",
    "query_spans": "[[[150, 162]]]",
    "process": "There exists a point P on hyperbola C such that PF_{1}\\bot PF_{2} and |PF_{1}|=3|PF_{2}|. It follows that P is a point on the right branch. Let |PF_{1}|=m, |PF_{2}|=n, then m-n=2a, m=3n, and m^{2}+n^{2}=4c^{2}. We obtain (3a)^{2}+a^{2}=4c^{2}, i.e., c=\\frac{\\sqrt{10}}{2}a, thus e=\\frac{c}{a}=\\frac{\\sqrt{10}}{2}"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, point $P$ lies on the ellipse, and $P F_{2} \\perp x$-axis, then the area of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F2), xAxis)",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[18, 45], [56, 58]], [[18, 45]], [[2, 9]], [[10, 17]], [[2, 50]], [[51, 55]], [[51, 59]], [[60, 78]]]",
    "query_spans": "[[[80, 110]]]",
    "process": "Without loss of generality, assume $F_{1}(\\sqrt{3},0), F_{2}(\\sqrt{3},0)$. Since $PF_{2} \\perp x$-axis, $P(\\sqrt{3},\\pm\\frac{1}{2})$. Since the area of $\\triangle PF_{1}F_{2} = \\frac{1}{2}|PF_{2}||F_{1}F_{2}| = \\frac{1}{2} \\times \\frac{1}{2} \\times 2\\sqrt{3} = \\frac{\\sqrt{3}}{2}$"
  },
  {
    "text": "Given a point $P$ on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ and the point $A(0,2)$, what is the maximum value of $|P A|$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);P: Point;PointOnCurve(P, G);A: Point;Coordinate(A) = (0, 2)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "(2*sqrt(21))/3",
    "fact_spans": "[[[2, 29]], [[2, 29]], [[34, 37]], [[2, 37]], [[38, 47]], [[38, 47]]]",
    "query_spans": "[[[49, 62]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the right focus of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ are? The distance from the focus to the asymptote is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1)",
    "query_expressions": "Coordinate(RightFocus(G));Distance(Focus(G), Asymptote(G))",
    "answer_expressions": "(2, 0)\nsqrt(3)",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 36]], [[0, 47]]]",
    "process": "According to the equation of the hyperbola, solve for the coordinates of the foci, then use the point-to-line distance formula to find the distance from the focus to the asymptote. Hyperbola: $x^{2}-\\frac{y^{2}}{3}=1$, $\\therefore a^{2}=1$, $b^{2}=3$, $\\therefore c^{2}=a^{2}+b^{2}=4$, $\\because a=2$, $\\therefore$ the coordinates of the right focus of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ are $(2,0)$. $\\therefore$ the asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ are $y=\\pm\\sqrt{3}x$, i.e., $\\pm\\sqrt{3}x-y=0$, $\\therefore$ the distance from the focus to the asymptote is $d=\\frac{|\\pm2\\sqrt{3}+0|}{2}=\\sqrt{3}$."
  },
  {
    "text": "Given that a asymptote of the hyperbola $x^{2}-y^{2}=1$ is intercepted by the circle $C$: $(x-2)^{2}+y^{2}=r^{2}(r>0)$ to form a line segment of length $2 \\sqrt{2}$, then $r$=?",
    "fact_expressions": "G: Hyperbola;C: Circle;r: Number;Expression(G) = (x^2 - y^2 = 1);r>0;Expression(C) = (y^2 + (x - 2)^2 = r^2);Length(InterceptChord(OneOf(Asymptote(G)),C))=2*sqrt(2)",
    "query_expressions": "r",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 20]], [[27, 60]], [[81, 84]], [[2, 20]], [[32, 60]], [[27, 60]], [[2, 79]]]",
    "query_spans": "[[[81, 86]]]",
    "process": "Since the hyperbola is rectangular, its asymptotes are $ y = \\pm x $. Without loss of generality, assume the asymptote is $ x - y = 0 $. The center of the circle is $ (2, 0) $, and the radius is $ r $. The distance from the center to the line is $ \\frac{2}{\\sqrt{2}} = \\sqrt{2} $. Thus, the chord length is $ 2\\sqrt{r^2 - (\\sqrt{2})^2} = 2\\sqrt{2} $, solving gives $ r = 2 $."
  },
  {
    "text": "Given the parabola with focus $F$ has equation $y^{2}=4 x$, point $Q$ has coordinates $(3,4)$, and point $P$ lies on the parabola. Then the minimum value of the sum of the distance from point $P$ to the $y$-axis and the distance from point $P$ to point $Q$ is?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;Q: Point;Coordinate(Q) = (3, 4);Focus(G) = F;Expression(G) = (y^2 = 4*x);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, yAxis) + Distance(P, Q))",
    "answer_expressions": "2*sqrt(5) - 1",
    "fact_spans": "[[[9, 12], [49, 52]], [[44, 48], [55, 59]], [[5, 8]], [[28, 32], [69, 73]], [[28, 43]], [[2, 12]], [[9, 27]], [[44, 53]]]",
    "query_spans": "[[[55, 84]]]",
    "process": "The coordinates of the focus F are (1,0), and the directrix of the parabola is x = -1. As shown in the figure, draw PA perpendicular to the directrix at A, intersecting the y-axis at B. |PB| + |PQ| = (|PA| - 1) + |PQ| = |PF| + |PQ| - 1 \\geqslant |FQ| - 1 = \\sqrt{(3-1)^{2}+4^{2}}-1=2"
  },
  {
    "text": "Let a circle centered at the origin intersect the $x$-axis at points $A$ and $B$. If an ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with foci at $A$ and $B$ always has common points with the circle, then what is the range of values for the eccentricity of the ellipse?",
    "fact_expressions": "H: Circle;O: Origin;Center(H) = O;Intersection(H, xAxis) = {A, B};A: Point;B: Point;Focus(G) = {A, B};G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;IsIntersect(H, G) = True",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[\\sqrt{2}/2,1)",
    "fact_spans": "[[[8, 9], [92, 93]], [[2, 4]], [[1, 9]], [[8, 24]], [[15, 18], [28, 31]], [[19, 22], [32, 35]], [[27, 91]], [[39, 91], [101, 103]], [[39, 91]], [[41, 91]], [[41, 91]], [[41, 91]], [[41, 91]], [[39, 98]]]",
    "query_spans": "[[[101, 113]]]",
    "process": "Obtain the distance $ d \\in [b, a] $ from any point on the ellipse to the center of the circle, we can get $ b \\leqslant c < a $, thus obtaining the solution. Let any point on the ellipse be $ P(m, n) $, then $ \\frac{m^{2}}{a^{2}} + \\frac{n^{2}}{b^{2}} = 1 $ ($ a > b > 0 $), then the distance from point $ P(m, n) $ to the center of the circle is:  \n$ d = \\sqrt{m^{2} + n^{2}} = \\sqrt{m^{2} + b^{2}\\left(1 - \\frac{m^{2}}{a^{2}}\\right)} = \\sqrt{\\left(1 - \\frac{b^{2}}{a^{2}}\\right)m^{2} + b^{2}} = \\sqrt{e^{2}m^{2} + b^{2}} $.  \nAccording to the problem, $ b \\leqslant c < a $, so $ \\sqrt{a^{2} - c^{2}} \\leqslant c < a $, solving gives $ e = \\frac{c}{a} \\in \\left[\\frac{\\sqrt{2}}{2}, 1\\right) $."
  },
  {
    "text": "The range of the y-coordinates of points on the ellipse $x^{2}+4 y^{2}=16$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 4*y^2 = 16);P:Point;PointOnCurve(P,G)",
    "query_expressions": "Range(YCoordinate(P))",
    "answer_expressions": "[-2,2]",
    "fact_spans": "[[[0, 20]], [[0, 20]], [[21, 22]], [[0, 22]]]",
    "query_spans": "[[[21, 33]]]",
    "process": "The standard equation of the ellipse \\( x^{2} + 4y^{2} = 16 \\) is \\( \\frac{x^{2}}{16} + \\frac{y^{2}}{4} = 1 \\), thus the range of the y-coordinates of the points is \\([-2, 2]\\)."
  },
  {
    "text": "A circle centered at the right focus $F_{2}$ of an ellipse passes exactly through the center of the ellipse and intersects the ellipse at points $M$ and $N$. The left focus of the ellipse is $F_{1}$, and the line $M F_{1}$ is tangent to this circle. Then, what is the eccentricity $e$ of the ellipse?",
    "fact_expressions": "M: Point;N:Point;F1: Point;G: Ellipse;H: Circle;e: Number;F2:Point;Center(H)=F2;PointOnCurve(Center(G),H);Intersection(H,G)={M,N};LeftFocus(G)=F1;RightFocus(G)=F2;IsTangent(LineOf(M,F1),H);Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[32, 36]], [[38, 42]], [[50, 57]], [[1, 3], [22, 24], [29, 31], [43, 45], [77, 79]], [[18, 19], [72, 73]], [[83, 86]], [[7, 14]], [[0, 19]], [[18, 27]], [[18, 42]], [[43, 57]], [[1, 14]], [[59, 75]], [[77, 86]]]",
    "query_spans": "[[[83, 88]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ with focus $F$, and the line $l$: $y=x+m$ intersecting $C$ at points $M$ and $N$. If $|M F|=2$, $|N F|=3$, then $|M N|$=?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(C) = F;l: Line;Expression(l) = (y = m + x);m: Number;Intersection(l, C) = {M, N};M: Point;N: Point;Abs(LineSegmentOf(M, F)) = 2;Abs(LineSegmentOf(N, F)) = 3",
    "query_expressions": "Abs(LineSegmentOf(M, N))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 28], [51, 54]], [[2, 28]], [[10, 28]], [[10, 28]], [[32, 35]], [[2, 35]], [[36, 50]], [[36, 50]], [[43, 50]], [[36, 65]], [[56, 59]], [[60, 63]], [[67, 76]], [[78, 87]]]",
    "query_spans": "[[[89, 98]]]",
    "process": "Let the x-coordinates of M and N be $x_{M}$, $x_{N}$ respectively. From the definition of the parabola, we have $|NF|-|MF|=x_{N}+\\frac{p}{2}-(x_{M}+\\frac{p}{2})=x_{N}-x_{M}=1$. From the line $l: y=x+m$, we know the inclination angle is $\\frac{\\pi}{4}$, so $x_{N}-x_{M}=|MN|\\cos\\frac{\\pi}{4}=1$. Solving gives $|MN|=\\sqrt{2}$."
  },
  {
    "text": "Given point $P$ lies on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola respectively, and $I$ is the incenter of $\\Delta P F_{1} F_{2}$. If $S_{\\Delta I P F_{1}}=S_{\\Delta I P F_{2}}+\\lambda S_{\\Delta I F_{1} F_{2}}$ holds, then what is the value of $\\lambda$?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;I: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, RightPart(G));LeftFocus(G) = F1;RightFocus(G) = F2;Incenter(TriangleOf(P, F1, F2))=I;Area(TriangleOf(I,P, F1))=Area(TriangleOf(I,P,F2)) +lambda*Area(TriangleOf(I,F1,F2));lambda:Number",
    "query_expressions": "lambda",
    "answer_expressions": "a/(sqrt(a^2+b^2))",
    "fact_spans": "[[[7, 63], [88, 91]], [[10, 63]], [[10, 63]], [[2, 6]], [[70, 77]], [[78, 85]], [[98, 101]], [[10, 63]], [[10, 63]], [[7, 63]], [[2, 69]], [[70, 97]], [[70, 97]], [[98, 127]], [[129, 205]], [[209, 218]]]",
    "query_spans": "[[[209, 222]]]",
    "process": ""
  },
  {
    "text": "Given that the line $y=x+\\frac{p}{2}$ intersects the parabola $C$: $x^{2}=2 p y(p>0)$ at points $A$ and $B$, and $|A B|=8$, then the equation of the directrix of parabola $C$ is?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;A: Point;B: Point;p>0;Expression(C) = (x^2 = 2*(p*y));Expression(G) = (y = p/2 + x);Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, B)) = 8",
    "query_expressions": "Expression(Directrix(C))",
    "answer_expressions": "y = -1",
    "fact_spans": "[[[22, 48], [73, 79]], [[30, 48]], [[2, 21]], [[51, 54]], [[55, 58]], [[30, 48]], [[22, 48]], [[2, 21]], [[2, 60]], [[62, 71]]]",
    "query_spans": "[[[73, 86]]]",
    "process": "The line $ y = x + \\frac{p}{2} $ intersects the parabola $ C: x^{2} = 2py $ ($ p > 0 $) at points $ A $ and $ B $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Solving the line $ y = x + \\frac{p}{2} $ and the parabola $ C: x^{2} = 2py $ ($ p > 0 $) simultaneously yields $ x_{1} + x_{2} = 2p $. Thus, $ y_{1} + y_{2} = 2p + p = 3p $. $ |AB| = y_{1} + y_{2} + p = 8 $, so $ 4p = 8 $, solving gives $ p = 2 $. Therefore, the directrix equation of parabola $ C $ is: $ y = -\\frac{2}{2} = -1 $."
  },
  {
    "text": "The vertex of the parabola is at the origin, with the $y$-axis as the axis of symmetry. The length of the chord passing through the focus and perpendicular to the $y$-axis is $16$. Then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;O: Origin;Vertex(G) = O;SymmetryAxis(G) = yAxis;L: LineSegment;PointOnCurve(Focus(G), L) = True;IsPerpendicular(L, yAxis) = True;IsChordOf(L, G) = True;Length(L) = 16",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2=pm*16*y",
    "fact_spans": "[[[0, 3], [42, 45]], [[6, 10]], [[0, 10]], [[0, 20]], [], [[0, 34]], [[0, 34]], [[0, 34]], [[0, 40]]]",
    "query_spans": "[[[42, 49]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$), respectively, and there exists a point $P$ on the right branch of the hyperbola such that $(\\overrightarrow{O P}+\\overrightarrow{O F_{2}}) \\cdot \\overrightarrow{F_{2} P}=0$ ($O$ being the coordinate origin), and $|P F_{1}| \\geq \\sqrt{3}|P F_{2}|$, then the range of the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, RightPart(G));O: Origin;DotProduct((VectorOf(O, F2) + VectorOf(O, P)), VectorOf(F2, P)) = 0;Abs(LineSegmentOf(P, F1)) >= sqrt(3)*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1, 1+sqrt(3)]",
    "fact_spans": "[[[20, 77], [85, 88], [232, 235]], [[20, 77]], [[23, 77]], [[23, 77]], [[23, 77]], [[23, 77]], [[2, 9]], [[10, 17]], [[2, 83]], [[2, 83]], [[96, 99]], [[85, 99]], [[185, 188]], [[102, 184]], [[196, 230]]]",
    "query_spans": "[[[232, 246]]]",
    "process": "(\\overrightarrow{OP}+\\overrightarrow{OF}_{2})\\cdot\\overrightarrow{F_{2}P}=0, which is (\\overrightarrow{OP}+\\overrightarrow{OF}_{2})\\cdot(\\overrightarrow{OP}-\\overrightarrow{OF_{2}})=0, which simplifies to \\overrightarrow{OP}^{2}=\\overrightarrow{OF_{2}}, yielding |OP|=c, implying \\angle F_{1}PF_{2}=90^{\\circ}. Let |PF_{1}|=m, |PF_{2}|=n, then we have m \\cdot n=2a, and m^{2}+n^{2}=4c^{2}. Let m=kn, so n=\\frac{2a}{k-1}, m=\\frac{2ka}{k-1}. In \\triangle PF_{1}F_{2}, by the Pythagorean theorem, |PF_{1}|^{2}+|PF_{2}|^{2}=4c^{2}, 1+\\frac{2}{k-2+\\frac{1}{1}}\\leqslant1+\\frac{\\sqrt{3}}{}=4+2\\sqrt{3}, thus 1<e\\leqslant1+\\sqrt{3}."
  },
  {
    "text": "Given the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has its right focus at $F(c, 0)$. If the distance from point $F$ to the line $bx - ay = 0$ is $\\frac{\\sqrt{3}}{3}c$, then the eccentricity of $E$ is?",
    "fact_expressions": "E: Ellipse;b: Number;a: Number;G: Line;F: Point;c:Number;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (-a*y + b*x = 0);Coordinate(F) = (c, 0);RightFocus(E) = F;Distance(F, G) = (sqrt(3)/3)*c",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 59], [121, 124]], [[9, 59]], [[9, 59]], [[80, 93]], [[64, 73], [75, 79]], [[64, 73]], [[9, 59]], [[9, 59]], [[2, 59]], [[80, 93]], [[64, 73]], [[2, 73]], [[75, 119]]]",
    "query_spans": "[[[121, 130]]]",
    "process": "From the point-to-line distance formula, we obtain the equation $ a^{2}=2b^{2} $. Using $ a^{2}=b^{2}+c^{2} $, we can solve as follows: according to the given condition, $ \\frac{bc}{\\sqrt{a2+b^{2}}}=\\frac{\\sqrt{3}c}{3} $, we get $ a^{2}=2b^{2} $. Since $ b^{2}=a^{2}-c^{2} $, it follows that $ a^{2}=2c^{2} $, hence $ e=\\frac{c}{a}=\\frac{\\sqrt{2}}{2} $."
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $\\frac{y^{2}}{4}-\\frac{x^{2}}{8}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/8 + y^2/4 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(2)/2)*x",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 46]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{24}=1$, respectively, and let $P$ be a point on the hyperbola $C$ in the first quadrant. If $\\frac{|P F_{1}|}{|P F_{2}|}=\\frac{5}{4}$, then the area of the incircle of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;C: Hyperbola;Expression(C) = (x^2 - y^2/24 = 1);P: Point;PointOnCurve(P, C) = True;Quadrant(P) = 1;Abs(LineSegmentOf(P, F1))/Abs(LineSegmentOf(P, F2)) = 5/4",
    "query_expressions": "Area(InscribedCircle(TriangleOf(P, F1, F2)))",
    "answer_expressions": "48*pi/7",
    "fact_spans": "[[[1, 8]], [[9, 16]], [[1, 59]], [[1, 59]], [[19, 53], [64, 70]], [[19, 53]], [[60, 63]], [[60, 79]], [[60, 79]], [[81, 122]]]",
    "query_spans": "[[[124, 157]]]",
    "process": "Hyperbola $ C: x^{2} - \\frac{y^{2}}{24} = 1 $ has $ a = 1 $, $ b = 2\\sqrt{6} $, $ c = \\sqrt{a^{2} + b^{2}} = 5 $. By the definition of a hyperbola, $ |PF_{1}| - |PF_{2}| = 2a = 2 $. Since $ \\frac{|PF_{1}|}{|PF_{2}|} = \\frac{5}{4} $, solving gives $ |PF_{1}| = 10 $, $ |PF_{2}| = 8 $, $ |F_{1}F_{2}| = 2c = 10 $. The height on side $ PF_{2} $ is $ \\sqrt{100 - 16} = 2\\sqrt{21} $. Using the equal area method, $ \\frac{1}{2} \\times 2\\sqrt{21} \\times 8 = \\frac{1}{2} \\times (10 + 10 + 8)r $, so $ r = \\frac{4\\sqrt{21}}{7} $. Therefore, the area of the incircle of $ \\triangle PF_{1}F_{2} $ is $ \\frac{48}{7}\\pi $."
  },
  {
    "text": "Through the center of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, draw an arbitrary line intersecting the ellipse at points $P$ and $Q$, and let $F$ be the left focus of the ellipse. Then the minimum value of the perimeter of $\\triangle P F Q$ is?",
    "fact_expressions": "G: Ellipse;H: Line;P: Point;F: Point;Q: Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(Center(G), H);Intersection(H, G) = {P, Q};LeftFocus(G) = F",
    "query_expressions": "Min(Perimeter(TriangleOf(P, F, Q)))",
    "answer_expressions": "18",
    "fact_spans": "[[[1, 40], [49, 51], [66, 68]], [[46, 48]], [[52, 55]], [[62, 65]], [[56, 59]], [[1, 40]], [[0, 48]], [[46, 61]], [[62, 72]]]",
    "query_spans": "[[[74, 100]]]",
    "process": "Let the right focus be $F'$. From the given conditions, $a=5$, $b=4$. Since $P$ and $Q$ are symmetric about the origin, $|PF|$ equals the distance from $Q$ to the right focus $F'$, so $|PF| + |QF| = |QF| + |QF'| = 2a = 10$. The minimum length of segment $PQ$ is the minor axis length $2b = 8$. Therefore, the minimum perimeter of $\\triangle PFQ$ is $10 + 8 = 18$."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=8x$ is $F$, and $P$ is a point on the parabola such that $|PF|=3$, then the horizontal coordinate of point $P$ is?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;Expression(G) = (y^2 = 8*x);Focus(G) = F;PointOnCurve(P, G);Abs(LineSegmentOf(P, F)) = 3",
    "query_expressions": "XCoordinate(P)",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 16], [28, 31]], [[24, 27], [47, 51]], [[20, 23]], [[2, 16]], [[2, 23]], [[24, 34]], [[36, 45]]]",
    "query_spans": "[[[47, 57]]]",
    "process": "It can be solved using the focal radius formula. From the parabola equation, we get 2p = 8, so p = 4, then |PF| = x_{P} + \\frac{p}{2} = x_{P} + 2 = 3, solving gives x_{P} = 1"
  },
  {
    "text": "Given that $F_{1}(-c, 0)$, $F_{2}(c, 0)$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, if there exists a point $P$ on the ellipse $C$ such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=3 c^{2}$, then the range of the eccentricity $e$ of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;c: Number;F1: Point;F2: Point;Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 3*c^2;e: Number;Eccentricity(C) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "[sqrt(5)/5, 1/2]",
    "fact_spans": "[[[32, 89], [97, 102], [179, 184]], [[32, 89]], [[39, 89]], [[39, 89]], [[39, 89]], [[39, 89]], [[2, 16]], [[2, 16]], [[18, 31]], [[2, 16]], [[18, 31]], [[2, 95]], [[2, 95]], [[107, 110]], [[97, 110]], [[112, 177]], [[188, 191]], [[179, 191]]]",
    "query_spans": "[[[188, 198]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $\\frac{\\sqrt{2}}{2}$, and the line $l$ intersects the ellipse at points $A$ and $B$. When the midpoint of $AB$ is $M(1,1)$, what is the equation of the line $l$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;Eccentricity(G) = sqrt(2)/2;l: Line;A: Point;B: Point;Intersection(l, G) = {A, B};M: Point;Coordinate(M) = (1, 1);MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Expression(l)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 54], [86, 88]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[2, 79]], [[80, 85], [120, 125]], [[90, 93]], [[94, 97]], [[80, 99]], [[110, 118]], [[110, 118]], [[101, 118]]]",
    "query_spans": "[[[120, 130]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $x^{2}-3 y^{2}=3$, what is the equation of the asymptotes of the hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - 3*y^2 = 3)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=(pm*sqrt(3)/3)*x",
    "fact_spans": "[[[2, 22], [24, 27]], [[2, 22]]]",
    "query_spans": "[[[24, 35]]]",
    "process": "x^{2}-3y^{2}=3 is rewritten as \\frac{x^{2}}{3}-y^{2}=1, so the asymptotes are y=\\pm\\frac{\\sqrt{3}}{3}x"
  },
  {
    "text": "Draw a straight line through the center of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ intersecting the ellipse at points $P$ and $Q$, and let $F$ be a focus of the ellipse. Then the minimum perimeter of $\\triangle P F Q$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;PointOnCurve(Center(G), H) = True;H: Line;Intersection(H, G) = {P, Q};P: Point;Q: Point;F: Point;OneOf(Focus(G)) = F",
    "query_expressions": "Min(Perimeter(TriangleOf(P, F, Q)))",
    "answer_expressions": "2*a+2*b",
    "fact_spans": "[[[1, 53], [61, 63], [78, 80]], [[1, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[0, 60]], [[58, 60]], [[58, 73]], [[64, 67]], [[68, 71]], [[74, 77]], [[74, 85]]]",
    "query_spans": "[[[87, 113]]]",
    "process": "As shown in the figure, by the definition of an ellipse, |PF| + |PF_{1}| = 2a. By the symmetry of the ellipse, |QF| = |PF_{1}|. Thus, |PF| + |QF| = 2a, and the minimum value of |PQ| is 2b. Therefore, the minimum perimeter of \\triangle PFQ is 2a + 2b."
  },
  {
    "text": "The equation of the hyperbola sharing foci with $\\frac{x^{2}}{16}-\\frac{y^{2}}{4}=1$ and passing through the point $(3 \\sqrt{2}, 2)$ is?",
    "fact_expressions": "G: Hyperbola;Z: Hyperbola;H: Point;Expression(G) = (x^2/16 - y^2/4 = 1);Coordinate(H) = (3*sqrt(2), 2);PointOnCurve(H, Z);Focus(G) = Focus(Z)",
    "query_expressions": "Expression(Z)",
    "answer_expressions": "x^2/12 - y^2/8 = 1",
    "fact_spans": "[[[1, 40]], [[66, 69]], [[46, 65]], [[1, 40]], [[46, 65]], [[45, 69]], [[0, 69]]]",
    "query_spans": "[[[66, 73]]]",
    "process": ""
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$), draw a line $l$ intersecting the parabola at points $A$ and $B$, and intersecting its directrix at point $C$. If $\\overrightarrow{C B}=3 \\overrightarrow{B F}$, then the slope of line $l$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(G) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, G) = {A, B};C: Point;Intersection(l, Directrix(G)) = C;VectorOf(C, B) = 3*VectorOf(B, F)",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*2*sqrt(2)",
    "fact_spans": "[[[1, 22], [36, 39], [51, 52]], [[1, 22]], [[4, 22]], [[4, 22]], [[25, 28]], [[1, 28]], [[29, 34], [108, 113]], [[0, 34]], [[40, 43]], [[44, 47]], [[29, 49]], [[55, 59]], [[29, 59]], [[61, 106]]]",
    "query_spans": "[[[108, 118]]]",
    "process": "Draw BB₁ perpendicular to the directrix at point B₁, then by the definition of the parabola, BB₁ = BF, so cos∠CBB₁ = 1/3, ∴ tan∠CBB₁ = 2√2, therefore the slope k of line l is ±2√2."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $P$ is a point on the hyperbola $C$ such that $\\overrightarrow{P F_{1}} \\perp \\overrightarrow{P F_{2}}$, and the area of $\\Delta P F_{1} F_{2}$ is $16$, then $b=$?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Focus(C) = {F1, F2};PointOnCurve(P, C);IsPerpendicular(VectorOf(P, F2),VectorOf(P, F1));Area(TriangleOf(P, F1, F2)) = 16",
    "query_expressions": "b",
    "answer_expressions": "4",
    "fact_spans": "[[[18, 79], [89, 95]], [[191, 194]], [[26, 79]], [[85, 88]], [[2, 9]], [[10, 17]], [[26, 79]], [[26, 79]], [[18, 79]], [[2, 84]], [[85, 98]], [[100, 157]], [[159, 189]]]",
    "query_spans": "[[[191, 196]]]",
    "process": "Let |PF_{1}| = m, |PF_{2}| = n, \\overrightarrow{PF}_{1} \\bot \\overrightarrow{PF}_{2}, then \\angle F_{1}PF_{2} = 90^{\\circ}, \\therefore m^{2} + n^{2} = 4c^{2}. The area of \\triangle PF_{1}F_{2} is 16, \\therefore mn = 32. \\therefore 4a^{2} = (m - n)^{2} = 4c^{2} - 64, \\therefore b^{2} = c^{2} - a^{2} = 16, \\therefore b = 4."
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$), draw two mutually perpendicular chords $AB$ and $CD$. If the minimum value of $|AB|+|CD|$ is $16$, then the equation of the parabola is?",
    "fact_expressions": "F: Point;Focus(G) = F;G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;A: Point;B: Point;C: Point;D: Point;IsChordOf(LineSegmentOf(A, B), G);IsChordOf(LineSegmentOf(C, D), G);PointOnCurve(F, LineSegmentOf(A, B));PointOnCurve(F, LineSegmentOf(C, D));IsPerpendicular(LineSegmentOf(A, B), LineSegmentOf(C, D));Min(Abs(LineSegmentOf(A, B)) + Abs(LineSegmentOf(C, D))) = 16",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[25, 28]], [[1, 28]], [[1, 22], [76, 79]], [[1, 22]], [[4, 22]], [[4, 22]], [[37, 42]], [[37, 42]], [[45, 50]], [[45, 50]], [[0, 50]], [[0, 50]], [[0, 50]], [[0, 50]], [[0, 50]], [[52, 74]]]",
    "query_spans": "[[[76, 84]]]",
    "process": "Let the equation of line AB be: $ y = k(x - \\frac{p}{2}) $, then the equation of line CD is $ y = -\\frac{1}{k}(x - \\frac{p}{2}) $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, $ C(x_{3}, y_{3}) $, $ D(x_{4}, y_{4}) $.  \nSolving the system  \n\\[\n\\begin{cases}\ny = k(x - \\frac{p}{2}) \\\\\ny^2 = 2px\n\\end{cases}\n\\]  \nyields $ k^{2}x^{2} - (pk^{2} + 2p)x + \\frac{k^{2}p^{2}}{4} = 0 $.  \n$ \\therefore x_{1} + x_{2} = \\frac{pk^{2} + 2p}{k^{2}} $,  \n$ x_{3} + x_{4} = p + \\frac{2p}{k^{2}} + p + 2pk^{2} = 2p(k^{2} + \\frac{1}{k^{2}} + 1) $,  \n$ \\therefore |AB| + |CD| $ has a minimum value of $ 8p $,  \n$ \\therefore 8p = 16 $, then $ p = 2 $, so the equation of the parabola is $ y^{2} = 4x $."
  },
  {
    "text": "Given that the vertex of the parabola $C$ is at the origin and the focus is $F(1 , 0)$, the line $l$ intersects the parabola $C$ at points $A$ and $B$. If the midpoint of $AB$ is $(2 , 2)$, then what is the equation of the line $l$?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;F: Point;O: Origin;Coordinate(F) = (1, 0);Focus(C)=F;Vertex(C) = O;Intersection(l,C)={A,B};Coordinate(MidPoint(LineSegmentOf(A,B)))=(2,2)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "y=x",
    "fact_spans": "[[[32, 37], [76, 81]], [[3, 9], [38, 44]], [[47, 50]], [[51, 54]], [[21, 31]], [[13, 17]], [[21, 31]], [[3, 31]], [[3, 17]], [[32, 56]], [[57, 74]]]",
    "query_spans": "[[[76, 86]]]",
    "process": ""
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $\\sqrt{2}$, then the asymptotes of this hyperbola are?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = sqrt(2)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*x",
    "fact_spans": "[[[2, 58], [76, 79]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 73]]]",
    "query_spans": "[[[76, 87]]]",
    "process": "From the given conditions:  \n\\begin{cases}e=\\frac{c}{a}=\\sqrt{2}\\\\c^{2}=a^{2}+b^{2}\\end{cases} \\Rightarrow a=b,  \nthe asymptotes of the hyperbola are given by y=\\pm x"
  },
  {
    "text": "Write the equation of a hyperbola centered at the origin with foci on the coordinate axes and asymptotes given by $y=\\pm 2 x$.",
    "fact_expressions": "G: Hyperbola;O:Origin;Expression(Asymptote(G)) = (y = pm*(2*x));Center(G)=O;PointOnCurve(Focus(G),axis)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/4 = 1",
    "fact_spans": "[[[39, 42]], [[8, 12]], [[21, 42]], [[5, 42]], [[13, 42]]]",
    "query_spans": "[[[39, 45]]]",
    "process": "The hyperbola with center at the origin, foci on the coordinate axes, and asymptotes given by $ y = \\pm 2x $ has the equation $ x^{2} - \\frac{y^{2}}{4} = \\lambda $ ($ \\lambda \\neq 0 $)."
  },
  {
    "text": "Given that the line $x=1$ intersects the parabola $C$: $y^{2}=2 p x(p>0)$ at point $M$, and intersects the $x$-axis at point $N$. Points $A$ and $B$ lie on the parabola $C$, and the slope of line $AB$ is $2$. The distances from point $N$ to lines $MA$ and $MB$ are equal. Then the value of $p$ is?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;B: Point;A: Point;M: Point;N: Point;p>0;Expression(C) = (y^2 = 2*p*x);Expression(G) = (x = 1);Intersection(G, C) = M;Intersection(G,xAxis)=N;PointOnCurve(A, C);PointOnCurve(B, C);Slope(LineOf(A,B)) = 2;Distance(N,LineOf(M,A))=Distance(N,LineOf(M,B))",
    "query_expressions": "p",
    "answer_expressions": "8",
    "fact_spans": "[[[10, 36], [65, 71]], [[114, 117]], [[2, 9]], [[60, 63]], [[55, 59]], [[38, 42]], [[50, 54], [89, 93]], [[18, 36]], [[10, 36]], [[2, 9]], [[2, 42]], [[2, 54]], [[55, 72]], [[55, 72]], [[74, 88]], [[89, 112]]]",
    "query_spans": "[[[114, 121]]]",
    "process": "Solution 1: Let $ M\\left(\\frac{y_{0}^{2}}{2p}, y_{0}\\right) $, $ A\\left(\\frac{y_{1}^{2}}{2p}, y_{1}\\right) $, $ B\\left(\\frac{y_{2}^{2}}{2p}, y_{2}\\right) $. Express $ k_{MA} $, $ k_{MB} $, $ k_{AB} $. Since the distances from point $ N $ to lines $ MA $ and $ MB $ are equal, $ k_{MA} + k_{MB} = 0 $. Rearranging and combining with $ k_{AB} = 2 $, $ p $ can be solved.  \nSolution 2: Let $ M\\left(\\frac{y_{0}^{2}}{2p}, y_{0}\\right) $, $ A\\left(\\frac{y_{1}^{2}}{2p}, y_{1}\\right) $, $ B\\left(\\frac{y_{2}^{2}}{2p}, y_{2}\\right) $. The equation of line $ AB $ is $ y = 2x + b $. Combining with $ y^{2} = 2px $, a quadratic equation in $ y $ is obtained, so $ y_{1} + y_{2} = p $. Since the distances from point $ N $ to lines $ MA $ and $ MB $ are equal, $ k_{MA} + k_{MB} = 0 $. Substituting according to the problem gives the answer.  \nDetailed Solution:  \nSolution 1: Let $ M\\left(\\frac{y_{0}^{2}}{2p}, y_{0}\\right) $, $ A\\left(\\frac{y_{1}^{2}}{2p}, y_{1}\\right) $, $ B\\left(\\frac{y_{2}^{2}}{2p}, y_{2}\\right) $, then $ k_{MA} = \\frac{2p}{y_{1} + y_{0}} $, $ k_{MB} = \\frac{2p}{y_{2} + y_{0}} $. Since $ k_{MA} + k_{MB} = 0 $, we have $ \\frac{2p}{y_{1} + y_{0}} + \\frac{2p}{y_{2} + y_{0}} = 0 $. Rearranging yields $ y_{1} + y_{2} = -2y_{0} $. Thus, $ k_{AB} = \\frac{2p}{y_{1} + y_{2}} = \\frac{2p}{-2y_{0}} = 2 $, so $ y_{0} = -\\frac{p}{2} $. Hence, $ \\frac{y_{0}^{2}}{2p} = \\frac{\\left(-\\frac{p}{2}\\right)^{2}}{2p} = 1 $, therefore $ p = 8 $.  \nSolution 2: Let $ M\\left(\\frac{y_{0}^{2}}{2p}, y_{0}\\right) $, $ A\\left(\\frac{y_{1}^{2}}{2p}, y_{1}\\right) $, $ B\\left(\\frac{y_{2}^{2}}{2p}, y_{2}\\right) $. The equation of line $ AB $ is $ y = 2x + b $. Combining with $ y^{2} = 2px $, rearranging gives $ y^{2} - py + pb = 0 $, so $ y_{1} + y_{2} = p $. Since the distances from point $ N $ to lines $ MA $ and $ MB $ are equal, $ k_{MA} + k_{MB} = 0 $. Also, $ k_{MA} = \\frac{y_{1} - y_{0}}{\\frac{y_{1}^{2}}{2p} - \\frac{y_{0}^{2}}{2p}} = \\frac{2p}{y_{1} + y_{0}} $. Similarly, $ k_{MB} = \\frac{2p}{y_{2} + y_{0}} $. Therefore, $ \\frac{2p}{y_{1} + y_{0}} + \\frac{2p}{y_{2} + y_{0}} = 0 $. Rearranging yields $ y_{1} + y_{2} = -2y_{0} $, so $ -2y_{0} = p $, $ y_{0} = -\\frac{p}{2} $. Hence, $ \\frac{y_{0}^{2}}{2p} = \\frac{\\left(-\\frac{p}{2}\\right)^{2}}{2p} = 1 $, therefore $ p = 8 $."
  },
  {
    "text": "Given that an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ is $x-2 y=0$, then the eccentricity $e$ of this hyperbola is $?$.",
    "fact_expressions": "G: Hyperbola;a: Number;e: Number;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (x - 2*y = 0);Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 34], [55, 58]], [[5, 34]], [[62, 65]], [[2, 34]], [[2, 52]], [[55, 65]]]",
    "query_spans": "[[[62, 67]]]",
    "process": "According to the asymptote equation, find a, then use the eccentricity $ e = \\frac{c}{a} = \\frac{\\sqrt{a^{2}+b^{2}}}{a} $ to solve. [Detailed solution] The hyperbola $ \\frac{x^{2}}{a^{2}} - y^{2} = 1 $ has asymptote equations: $ y = \\pm\\frac{x}{a} $. Given that one asymptote equation is $ x - 2y = 0 $, we obtain $ a = 2 $. Since $ b = 1 $, it follows that $ e = \\frac{c}{a} = \\frac{\\sqrt{a^{2}+b^{2}}}{a} = \\frac{\\sqrt{5}}{2} $."
  },
  {
    "text": "The equation $\\frac{x^{2}}{1+k}+\\frac{y^{2}}{1-k}=1$ represents a hyperbola with foci on the $y$-axis. Then, the range of real values for $k$ is?",
    "fact_expressions": "E: Hyperbola;k: Real;Expression(E) = (x**2/(k + 1) + y**2/(1 - k) = 1);PointOnCurve(Focus(E), yAxis)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-oo, -1)",
    "fact_spans": "[[[52, 55]], [[57, 62]], [[0, 55]], [[43, 55]]]",
    "query_spans": "[[[57, 69]]]",
    "process": ""
  },
  {
    "text": "The distance from the origin to the line passing through a focus and an endpoint of the minor axis of the ellipse $\\frac{x^{2}}{a^{2}} +\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $\\frac{b}{2}$. What is the eccentricity of this ellipse?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;O:Origin;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(OneOf(Focus(G)),H);PointOnCurve(Endpoint(MinorAxis(G)),H);Distance(OneOf(Focus(G)),H)=b/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 55], [91, 93]], [[4, 55]], [[4, 55]], [[66, 68]], [[69, 71]], [[4, 55]], [[4, 55]], [[2, 55]], [[0, 68]], [[0, 68]], [[66, 88]]]",
    "query_spans": "[[[91, 99]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $x^{2}-y^{2}=1$, point $P$ lies on $C$, $\\angle F_{1} P F_{2}=60^{\\circ}$, then $|P F_{1}| \\cdot |P F_{2}|$=?",
    "fact_expressions": "C: Hyperbola;F1: Point;P: Point;F2: Point;Expression(C) = (x^2 - y^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "4",
    "fact_spans": "[[[18, 41], [53, 56]], [[2, 9]], [[48, 52]], [[10, 17]], [[18, 41]], [[2, 47]], [[2, 47]], [[48, 57]], [[58, 89]]]",
    "query_spans": "[[[91, 119]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$ has a semi-focal width of $c$, if $b^{2}-4 a c<0$, then what is the range of values for its eccentricity?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;c: Number;HalfFocalLength(G) = c;-4*a*c + b^2 < 0",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1, 2+sqrt(5))",
    "fact_spans": "[[[2, 61], [88, 89]], [[2, 61]], [[5, 61]], [[5, 61]], [[5, 61]], [[5, 61]], [[66, 69]], [[2, 69]], [[71, 86]]]",
    "query_spans": "[[[88, 101]]]",
    "process": ""
  },
  {
    "text": "The line $l$ passes through the point $A(t, 0)$ and is tangent to the curve $y = x^{2}$. If the angle of inclination of the line $l$ is $45^{\\circ}$, then $t = $?",
    "fact_expressions": "l: Line;G: Curve;A: Point;Expression(G) = (y = x^2);Coordinate(A) = (t, 0);Inclination(l) = ApplyUnit(45, degree);PointOnCurve(A,l);IsTangent(l, G);t:Number",
    "query_expressions": "t",
    "answer_expressions": "1/4",
    "fact_spans": "[[[0, 5], [35, 40]], [[20, 31]], [[7, 17]], [[20, 31]], [[7, 17]], [[35, 57]], [[0, 17]], [[0, 33]], [[59, 62]]]",
    "query_spans": "[[[59, 64]]]",
    "process": "If the inclination angle of line $ l $ is $ 45^{\\circ} $, then the slope of the line is $ 1 $, so $ y = x - t $. Solving the system $ \\begin{cases} y = x - t \\\\ y = x^{2} \\end{cases} $, eliminating $ y $ gives: $ x^{2} - x + t = 0 $. Since the line is tangent to the curve, $ \\triangle = (-1)^{2} - 4t = 0 $, $ \\therefore t = \\frac{1}{4} $"
  },
  {
    "text": "Given that $P$ is any point on the right branch of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, $F$ is the right focus of the hyperbola, and the fixed point $A$ has coordinates $(3,\\sqrt{3})$, then the minimum value of $|P F|+|P A|$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;F: Point;A: Point;Expression(G) = (x^2 - y^2/3 = 1);Coordinate(A) = (3, sqrt(3));PointOnCurve(P, RightPart(G));RightFocus(G) = F",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "2*sqrt(7)-2",
    "fact_spans": "[[[6, 34], [47, 50]], [[2, 5]], [[43, 46]], [[57, 60]], [[6, 34]], [[57, 78]], [[2, 42]], [[43, 54]]]",
    "query_spans": "[[[80, 99]]]",
    "process": ""
  },
  {
    "text": "What are the coordinates of the focus of the parabola $y=2 x^{2}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 2*x^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,1/8)",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 20]]]",
    "process": "From the given condition, we know that $x^{2}=\\frac{1}{2}y$, so the focus of the parabola lies on the positive y-axis, and its coordinates are $(0,\\frac{1}{8})$."
  },
  {
    "text": "A line with slope $2$ passes through the focus of the parabola $C$: $y^{2}=4 x$, and intersects $C$ at points $A$ and $B$. Then $|A B|=$?",
    "fact_expressions": "Slope(G) = 2;G: Line;C: Parabola;Expression(C) = (y^2 = 4*x);PointOnCurve(Focus(C), G) =  True;Intersection(G, C) = {A, B};A: Point;B: Point",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "5",
    "fact_spans": "[[[0, 9]], [[7, 9]], [[10, 29], [35, 38]], [[10, 29]], [[7, 32]], [[7, 49]], [[40, 43]], [[44, 47]]]",
    "query_spans": "[[[51, 60]]]",
    "process": "Write the equation of the line, solve it together with the parabola, and use Vieta's formulas combined with the chord length formula |AB| = x_{1} + x_{2} + 2 to obtain the result. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), then the equation of line AB is y = 2(x - 1). Solving the system of equations  \n\\begin{cases} y = 2(x - 1) \\\\ y^{2} = 4x \\end{cases}  \nyields x^{2} - 3x + 1 = 0, hence x_{1} + x_{2} = 3, and |AB| = x_{1} + x_{2} + 2 = 5, so |AB| = 5."
  },
  {
    "text": "If one of the asymptotes of the hyperbola $C$: $\\frac{x^{2}}{m}-\\frac{y^{2}}{9}=1$ $(m>0)$ is given by $y=\\frac{1}{3} x$, then $m=?$",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/9 + x^2/m = 1);m: Number;m>0;Expression(OneOf(Asymptote(C))) = (y = x/3)",
    "query_expressions": "m",
    "answer_expressions": "81",
    "fact_spans": "[[[1, 49]], [[1, 49]], [[77, 80]], [[9, 49]], [[1, 75]]]",
    "query_spans": "[[[77, 82]]]",
    "process": "According to the problem, $\\frac{3}{\\sqrt{m}}=\\frac{1}{3}$, solving gives $m=81$."
  },
  {
    "text": "Given $ k \\in [-2, -1] $, the range of the eccentricity of the hyperbola $ x^{2} + k y^{2} = 1 $ is?",
    "fact_expressions": "G: Hyperbola;k: Number;Expression(G) = (k*y^2 + x^2 = 1);In(k,[-2,-1])",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[sqrt(6)/2,sqrt(2)]",
    "fact_spans": "[[[18, 38]], [[2, 16]], [[18, 38]], [[2, 16]]]",
    "query_spans": "[[[18, 49]]]",
    "process": "From the given condition, the equation of the hyperbola can be rewritten as $x^{2}-\\frac{y^{2}}{-\\frac{1}{k}}=1$, $\\because e^{2}=\\frac{1-\\frac{1}{k}}{1}\\in[\\frac{3}{2},2]$, $\\therefore$ the eccentricity $e\\in[\\frac{\\sqrt{6}}{2},\\sqrt{2}]$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{6}+\\frac{y^{2}}{3}=1$ with a point $P(2,1)$ on it. If the line $l$: $y=-x+b$ intersects the ellipse $C$ at two distinct points $A$, $B$, and $PA \\perp PB$, then $b=$?",
    "fact_expressions": "l: Line;C: Ellipse;P: Point;A: Point;B: Point;b:Number;Expression(C) = (x^2/6 + y^2/3 = 1);PointOnCurve(P,C);Coordinate(P) = (2, 1);Expression(l)=(y=-x+b);Intersection(l, C) = {A,B};IsPerpendicular(LineSegmentOf(P, A), LineSegmentOf(P, B));Negation(A=B)",
    "query_expressions": "b",
    "answer_expressions": "1/3",
    "fact_spans": "[[[58, 73]], [[2, 44], [74, 79]], [[48, 56]], [[85, 88]], [[89, 92]], [[111, 114]], [[2, 44]], [[2, 56]], [[48, 56]], [[58, 73]], [[58, 92]], [[94, 109]], [[80, 92]]]",
    "query_spans": "[[[111, 116]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Solving the system \n\\[\n\\begin{cases}\ny = -x + b \\\\\nx^{2} + 2y^{2} = 6\n\\end{cases}\n\\]\nyields $ 3x^{2} - 4bx + 2b^{2} - 6 = 0 $. The discriminant $ \\Delta = 16b^{2} - 12(2b^{2} - 6) > 0 $ gives $ -3 < b < 3 $. By Vieta's formulas, $ x_{1} + x_{2} = \\frac{4b}{3} $, $ x_{1} \\cdot x_{2} = \\frac{2b^{2} - 6}{3} $. Then $ y_{1} + y_{2} = 2b - (x_{1} + x_{2}) = \\frac{2b}{3} $, $ y_{1} \\cdot y_{2} = (b - x_{1})(b - x_{2}) = b^{2} - b(x_{1} + x_{2}) + x_{1} \\cdot x_{2} = \\frac{b^{2} - 6}{3} $. Since $ PA \\perp PB $, we have $ \\overrightarrow{PA} \\cdot \\overrightarrow{PB} = (x_{1} - 2)(x_{2} - 2) + (y_{1} - 1)(y_{2} - 1) = x_{1} \\cdot x_{2} - 2(x_{1} + x_{2}) + 4 + y_{1} \\cdot y_{2} - (y_{1} + y_{2}) + 1 = \\frac{2b^{2} - 6}{3} - 2 \\cdot \\frac{4b}{3} + \\frac{b^{2} - 6}{3} - \\frac{2b}{3} + 5 = 0 $. Solving yields $ b = 3 $ or $ \\frac{1}{3} $. Since $ -3 < b < 3 $, it follows that $ b = \\frac{1}{3} $."
  },
  {
    "text": "If a point $P$ on the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$ is at a distance of $2$ from the right focus of the hyperbola, then what is the distance from point $P$ to the $y$-axis?",
    "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2/4 - y^2/2 = 1);PointOnCurve(P, G);Distance(P, RightFocus(G)) = 2",
    "query_expressions": "Distance(P, yAxis)",
    "answer_expressions": "4*sqrt(6)/3",
    "fact_spans": "[[[2, 40], [47, 50]], [[43, 46], [63, 67]], [[2, 40]], [[2, 46]], [[43, 60]]]",
    "query_spans": "[[[63, 77]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a moving point on the ellipse $\\frac{x^{2}}{10}+y^{2}=1$, and line $l$ passes through the center of the circle $x^{2}+(y-6)^{2}=2$ and intersects the circle at points $A$ and $B$, then the maximum value of $\\overrightarrow{P A} \\cdot \\overrightarrow{P B}$ is?",
    "fact_expressions": "l: Line;G: Ellipse;H:Circle;P: Point;A: Point;B: Point;Expression(G) = (x^2/10 + y^2 = 1);Expression(H) = (x^2 + (y - 6)^2 = 2);PointOnCurve(P, G);PointOnCurve(Center(H),l);Intersection(l, H) = {A, B}",
    "query_expressions": "Max(DotProduct(VectorOf(P, A), VectorOf(P, B)))",
    "answer_expressions": "48",
    "fact_spans": "[[[40, 45]], [[6, 34]], [[46, 66], [70, 71]], [[2, 5]], [[74, 77]], [[78, 81]], [[6, 34]], [[46, 66]], [[2, 39]], [[40, 69]], [[40, 83]]]",
    "query_spans": "[[[85, 140]]]",
    "process": "Denote the center of the circle $x^{2}+(y-6)^{2}=2$ as $C(0,6)$. Since line $l$ passes through the center of the circle $x^{2}+(y-6)^{2}=2$ and intersects the circle at points $A,B$, then $\\overrightarrow{PA}\\cdot\\overrightarrow{PB}=\\frac{(\\overrightarrow{PA}+\\overline{P}}{}$. Let $P(x_{0},y_{0})$ be a moving point on the ellipse $\\frac{x^{2}}{10}+y^{2}=1$, then $x_{0}^{2}=10-10y_{0}^{2}$, $|y_{0}|\\leqslant1$, so $PC^{2}=x_{0}^{2}+(y_{0}-6)^{2}=-9y_{0}^{2}-12y_{0}+46$. When $y_{0}=-\\frac{2}{3}$, $PC_{\\max}^{2}=50$, therefore the maximum value of $\\overrightarrow{PA}\\cdot\\overrightarrow{PB}$ is $PC_{\\max}^{2}-2=48$."
  },
  {
    "text": "The left vertex of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $A$, the right focus is $F$, the upper vertex is $B$, and the lower vertex is $C$. If the intersection point of line $AB$ and line $CF$ is $(3a,16)$, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;B: Point;A: Point;F: Point;C: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(G) = A;RightFocus(G) = F;UpperVertex(G) = B;LowerVertex(G) = C;T:Point;Coordinate(T)=(3*a,16);Intersection(LineOf(A,B),LineOf(C,F))=T",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/25+y^2/16=1",
    "fact_spans": "[[[0, 52], [117, 119]], [[2, 52]], [[2, 52]], [[73, 76]], [[57, 60]], [[65, 68]], [[81, 84]], [[2, 52]], [[2, 52]], [[0, 52]], [[0, 60]], [[0, 68]], [[0, 76]], [[0, 84]], [[106, 115]], [[106, 115]], [[86, 115]]]",
    "query_spans": "[[[117, 126]]]",
    "process": "Given the coordinates of the left vertex of the ellipse as A(-a,0), the upper and lower vertices as B(0,b), C(0,-b), and the right focus as F(c,0), then the equation of line AB is y=\\frac{b}{a}x+b, and the equation of line CF is y=\\frac{b}{c}x-b. Since the intersection point of line AB and line CF is (3a,16), substituting the point (3a,16) into the equations of the lines gives  \n\\begin{matrix}16=\\frac{b}{a}\\times3a+b&\\\\16=\\frac{b}{c}\\times3a-b&\\end{matrix}  \nSolving yields b=4 and 3a=5c."
  },
  {
    "text": "The center of ellipse $E$ is at the origin, with foci on the $x$-axis. The minimum distance from a point on the ellipse to a focus is $2 \\sqrt{2}-2$, and the eccentricity is $\\frac{\\sqrt{2}}{2}$. Then the equation of ellipse $E$ is?",
    "fact_expressions": "E: Ellipse;O: Origin;Center(E) = O;PointOnCurve(Focus(E), xAxis);K: Point;PointOnCurve(K, E);Min(Distance(K, Focus(E))) = 2*sqrt(2) - 2;Eccentricity(E) = sqrt(2)/2",
    "query_expressions": "Expression(E)",
    "answer_expressions": "x^2/8 + y^2/4 = 1",
    "fact_spans": "[[[2, 7], [23, 25], [78, 83]], [[11, 13]], [[2, 13]], [[2, 22]], [], [[23, 28]], [[23, 51]], [[23, 76]]]",
    "query_spans": "[[[78, 88]]]",
    "process": "\\because the minimum distance from a point on the ellipse to the focus is $a - c$, $\\therefore a - c = 2\\sqrt{2} - 2$. $\\because$ the eccentricity $e = \\frac{\\sqrt{2}}{2}$, $\\therefore \\frac{c}{a} = \\frac{\\sqrt{2}}{2}$.  \n\\begin{matrix} \\frac{c}{a} = \\frac{\\sqrt{2}}{2} \\\\ a - c = 2\\sqrt{2} - 2 & \\text{solve to get} \\begin{cases} a = 2\\sqrt{2} \\\\ b = 2 \\\\ c = 2 \\end{cases} \\end{matrix}  \n$\\therefore$ the required ellipse equation is $E: \\frac{x^{2}}{8} + \\frac{y^{2}}{4} = 1$.  \n【Note】This problem examines standard concepts related to ellipses, such as the distance from a point on the ellipse to the focus and the eccentricity. It is not difficult and belongs to the category of simple problems."
  },
  {
    "text": "There is an ellipse $\\Gamma$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $A$, $B$ are the endpoints of the major axis of the ellipse $\\Gamma$, $C$, $D$ are the endpoints of the minor axis of the ellipse $\\Gamma$, a moving point $M$ satisfies $\\frac{|M A|}{|M B|}=2$, the maximum area of $\\Delta M A B$ is $8$, the minimum area of $\\triangle M C D$ is $1$, then the eccentricity of the ellipse $\\Gamma$ is?",
    "fact_expressions": "Gamma: Ellipse;M: Point;A: Point;B: Point;C: Point;D: Point;a:Number;b:Number;a>b;b>0;Expression(Gamma) = (x^2/a^2+y^2/b^2=1);Endpoint(MajorAxis(Gamma))={A,B};Endpoint(MinorAxis(Gamma))={C,D};Abs(LineSegmentOf(M, A))/Abs(LineSegmentOf(M, B)) = 2;Max(Area(TriangleOf(M, A, B))) = 8;Min(Area(TriangleOf(M, C, D))) = 1",
    "query_expressions": "Eccentricity(Gamma)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[1, 63], [75, 85], [99, 109], [203, 213]], [[117, 120]], [[65, 69]], [[71, 74]], [[91, 94]], [[95, 98]], [[12, 63]], [[12, 63]], [[12, 63]], [[12, 63]], [[1, 63]], [[65, 90]], [[91, 114]], [[122, 145]], [[147, 172]], [[173, 201]]]",
    "query_spans": "[[[203, 219]]]",
    "process": ""
  },
  {
    "text": "The sum of the distances from a point $P$ on the ellipse $\\frac{y^{2}}{a^{2}} + \\frac{x^{2}}{b^{2}}=1(a>b>0)$ to the two foci $F_{1}$, $F_{2}$ is $6$, then $a=?$",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/b^2 + y^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};Distance(P, F1) + Distance(P, F2) = 6",
    "query_expressions": "a",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 54]], [[0, 54]], [[2, 54]], [[90, 93]], [[2, 54]], [[2, 54]], [[57, 60]], [[0, 60]], [[64, 71]], [[72, 79]], [[0, 79]], [[0, 88]]]",
    "query_spans": "[[[90, 95]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$ has eccentricity $e$, and the parabola $y^{2}=2 p x$ with focus $F$ intersects the line $y=k(x-\\frac{p}{2})$ at points $A$ and $B$, and $\\frac{|A F|}{|F B|}=e$, then the value of $k$ is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;I: Line;A: Point;F: Point;B: Point;k:Number;e:Number;Expression(G) = (x^2/4 - y^2/12 = 1);Expression(H) = (y^2 = 2*(p*x));Expression(I) = (y = k*(-p/2 + x));Eccentricity(G) = e;Focus(H)=F;Intersection(H, I) = {A, B};Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(F, B)) = e",
    "query_expressions": "k",
    "answer_expressions": "pm*2*sqrt(2)",
    "fact_spans": "[[[2, 41]], [[57, 73]], [[60, 73]], [[74, 96]], [[98, 101]], [[53, 56]], [[102, 105]], [[134, 137]], [[46, 49]], [[2, 41]], [[57, 73]], [[74, 96]], [[2, 49]], [[50, 73]], [[57, 107]], [[109, 132]]]",
    "query_spans": "[[[134, 141]]]",
    "process": ""
  },
  {
    "text": "The asymptote of a hyperbola centered at the origin with foci on the $x$-axis is given by $y = \\frac{3}{4}x$, and the distance from a focus to the asymptote is $6$.  \nThen the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;O: Origin;Center(G) = O;PointOnCurve(Focus(G), xAxis);Expression(OneOf(Asymptote(G))) = (y = (3/4)*x);Distance(Focus(G), Asymptote(G)) = 6",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/64 - y^2/36 = 1",
    "fact_spans": "[[[14, 17], [60, 63]], [[3, 5]], [[0, 17]], [[6, 17]], [[14, 43]], [[14, 57]]]",
    "query_spans": "[[[60, 67]]]",
    "process": ""
  },
  {
    "text": "The equation of the hyperbola that shares the same asymptotes as $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ and passes through the point $(-3 , 2 \\sqrt{3})$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);G1: Hyperbola;Asymptote(G) = Asymptote(G1);H: Point;Coordinate(H) = (-3, 2*sqrt(3));PointOnCurve(H, G1) = True",
    "query_expressions": "Expression(G1)",
    "answer_expressions": "4*x^2/9 - y^2/4 = 1",
    "fact_spans": "[[[1, 40]], [[1, 40]], [[72, 75]], [[0, 75]], [[51, 71]], [[51, 71]], [[50, 75]]]",
    "query_spans": "[[[72, 79]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a point on the parabola $y^{2}=4x$, $A(2, 2)$ is a fixed point in the plane, and $F$ is the focus of the parabola, find the coordinates of point $P$ such that $|PA|+|PF|$ is minimized.",
    "fact_expressions": "P: Point;PointOnCurve(P, G);G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;Coordinate(A) = (2, 2);F: Point;Focus(G) = F;WhenMin(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1, 2)",
    "fact_spans": "[[[2, 5], [56, 60]], [[2, 24]], [[6, 20], [48, 51]], [[6, 20]], [[25, 35]], [[25, 35]], [[44, 47]], [[44, 54]], [[67, 80]]]",
    "query_spans": "[[[56, 64]]]",
    "process": ""
  },
  {
    "text": "Given that $e_{1}$, $e_{2}$ are the eccentricities of the ellipse $C_{1}$ and hyperbola $C_{2}$ respectively, $F_{1}$, $F_{2}$ are their common foci, $M$ is one of their common points, and $\\angle F_{1} M F_{2}=60^{\\circ}$, then the maximum value of $\\frac{1}{e_{1}}+\\frac{1}{e_{2}}$ is?",
    "fact_expressions": "C2: Hyperbola;C1: Ellipse;F1: Point;M: Point;F2: Point;e1: Number;e2: Number;Eccentricity(C1) = e1;Eccentricity(C2) = e2;Focus(C1) = {F1, F2};Focus(C2) = {F1, F2};OneOf(Intersection(C1, C2)) = M;AngleOf(F1, M, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Max(1/e2 + 1/e1)",
    "answer_expressions": "4*sqrt(3)/3",
    "fact_spans": "[[[32, 42]], [[22, 31]], [[47, 54]], [[71, 74]], [[55, 62]], [[2, 10]], [[12, 19]], [[2, 46]], [[2, 46]], [[47, 70]], [[47, 70]], [[71, 83]], [[85, 118]]]",
    "query_spans": "[[[120, 159]]]",
    "process": "Analysis: Let the semi-major axis of the ellipse be $ a $, the semi-transverse axis of the hyperbola be $ a_{1} $ ($ a>a_{1} $), the semi-focal distance be $ c $, and let $ |MF_{1}|=m $, $ |MF_{2}|=n $, $ |F_{1}F_{2}|=2c $. Since $ \\angle F_{1}MF_{2}=\\frac{\\pi}{3} $, by the law of cosines we have $ 4c^{2}=m^{2}+n^{2}-2mn\\cos\\frac{\\pi}{3} $, $\\textcircled{1}$ In the ellipse, $ m+n=2a $, simplifying $\\textcircled{1}$ gives $ 4c^{2}=4a^{2}-3mn $, i.e., $ \\frac{3mn}{4c^{2}}=\\frac{1}{e^{2}}-1 $, $\\textcircled{2}$ In the hyperbola, $ |m-n|=2a_{1} $, simplifying $\\textcircled{1}$ gives $ 4c^{2}=4a_{1}^{2}+mn $, i.e., $ \\frac{mn}{4c^{2}}=1-\\frac{1}{e_{2}^{2}} $, $\\textcircled{3}$ Combining $\\textcircled{2}$ and $\\textcircled{3}$ yields $ \\frac{1}{e_{1}^{2}}+\\frac{3}{e_{2}^{2}}=4 $, i.e., $ \\frac{1}{4e_{1}^{2}}+\\frac{3}{4e_{2}^{2}}=1 $. Let $ \\frac{1}{2e_{1}}=\\cos\\alpha $, $ \\frac{\\sqrt{3}}{2e_{2}}=\\sin\\alpha $, $ \\alpha\\in(0,\\frac{\\pi}{2}) $, then $ \\frac{1}{e_{1}}+\\frac{1}{e_{2}}=2\\cos\\alpha+\\frac{2\\sqrt{3}}{3}\\sin\\alpha=\\frac{4\\sqrt{3}}{3}\\sin(\\alpha+\\frac{\\pi}{3})\\leqslant\\frac{4\\sqrt{3}}{3} $, with equality if and only if $ \\alpha=\\frac{\\pi}{6} $, i.e., $ e_{1}=\\frac{\\sqrt{3}}{3} $, $ e_{2}=\\sqrt{3} $."
  },
  {
    "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ are given by $x \\pm 2 y=0$, then the focal distance is?",
    "fact_expressions": "G: Hyperbola;a: Number;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(Asymptote(G)) = (pm*2*y + x = 0)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(5)",
    "fact_spans": "[[[0, 32]], [[3, 32]], [[0, 32]], [[0, 52]]]",
    "query_spans": "[[[0, 58]]]",
    "process": "From the asymptote equations $x \\pm 2y = 0$, we get $\\frac{b}{a} = \\frac{1}{2}$; since $b = 1$, then $a = 2$, so $a^{2} + b^{2} = 5$, hence $c = \\sqrt{5}$, therefore the focal distance is $2\\sqrt{5}$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1(a>0)$ has its right focus at $F$, then what is the distance from point $F$ to the asymptotes of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;a: Number;F: Point;a>0;Expression(C) = (-y^2/3 + x^2/a^2 = 1);RightFocus(C) = F",
    "query_expressions": "Distance(F, Asymptote(C))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 54], [69, 75]], [[10, 54]], [[59, 62], [64, 68]], [[10, 54]], [[2, 54]], [[2, 62]]]",
    "query_spans": "[[[64, 83]]]",
    "process": "Let F(c,0), so that a^{2}+b^{2}=a^{2}+3=c^{2}. Let the equation of one asymptote of the hyperbola be \\sqrt{3}x-ay=0. Then the distance from F to the asymptote is d=\\frac{\\sqrt{3}c}{\\sqrt{3+a^{2}}}=\\frac{\\sqrt{3}c}{c}=\\sqrt{3}"
  },
  {
    "text": "Given that the two asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ intersect the parabola $y^{2}=4x$ at points $A$ and $B$, respectively, both distinct from the origin $O$, and $F$ is the focus of the parabola $y^{2}=4x$. It is known that $\\angle AFB = \\frac{2\\pi}{3}$. Then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;L1: Line;L2: Line;Asymptote(G) = {L1, L2};H: Parabola;Expression(H) = (y^2 = 4*x);O: Origin;A: Point;Intersection(L1, H) = A;B: Point;Intersection(L2, H) = B;Negation(O=A);Negation(O=B);F: Point;Focus(H) = F;AngleOf(A, F, B) = (2*pi)/3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{sqrt(21)/3, sqrt(13)}",
    "fact_spans": "[[[2, 48], [151, 154]], [[2, 48]], [[5, 48]], [[5, 48]], [], [], [[2, 54]], [[55, 69], [98, 112]], [[55, 69]], [[76, 81]], [[84, 87]], [[2, 92]], [[89, 92]], [[2, 92]], [[74, 92]], [[74, 92]], [[94, 97]], [[94, 115]], [[118, 148]]]",
    "query_spans": "[[[151, 160]]]",
    "process": "Let $ A(m,n) $, then $ n = \\frac{b}{a}m $, $ n^{2} = 4m \\Rightarrow m = \\frac{4a^{2}}{b^{2}} $, $ n = \\frac{4a}{b} $. Because $ \\angle AFB = \\frac{2\\pi}{3} $, so $ \\frac{4a}{b} = \\sqrt{3} \\cdot \\left|1 - \\frac{4a^{2}}{b^{2}}\\right| \\Rightarrow 48\\left(\\frac{a^{2}}{b^{2}}\\right)^{2} - 40 \\cdot \\frac{a^{2}}{b^{2}} + 3 = 0 \\Rightarrow \\frac{a^{2}}{b^{2}} = \\frac{1}{12} $ or $ \\frac{a^{2}}{b^{2}} = \\frac{3}{4} \\Rightarrow \\frac{a^{2}}{c^{2}} = \\frac{1}{13} $ or $ \\frac{a^{2}}{c^{2}} = \\frac{3}{7} \\Rightarrow e = \\sqrt{13} $ or $ e = \\frac{\\sqrt{21}}{3} $"
  },
  {
    "text": "Given the hyperbola $C_{1}$: $\\frac{x^{2}}{a_{1}^{2}}-\\frac{y^{2}}{b_{1}^{2}}=1(a_{1}>0, b_{1}>0)$ and $C_{2}$: $\\frac{y^{2}}{a_{2}^{2}}-\\frac{x^{2}}{b_{2}^{2}}=1(a_{2}>0, b_{2}>0)$ have the same asymptotes. If the eccentricity of $C_{1}$ is $2$, then the eccentricity of $C_{2}$ is?",
    "fact_expressions": "C1: Hyperbola;Expression(C1) = (x^2/a1^2 - y^2/b1^2 = 1);a1: Number;b1: Number;a1>0;b1>0;C2: Hyperbola;Expression(C2) = (y^2/a2^2 - x^2/b2^2 = 1);a2: Number;b2: Number;a2>0;b2>0;Asymptote(C1) = Asymptote(C2);Eccentricity(C1) = 2",
    "query_expressions": "Eccentricity(C2)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 83], [171, 178]], [[2, 83]], [[14, 83]], [[14, 83]], [[14, 83]], [[14, 83]], [[84, 162], [188, 195]], [[84, 162]], [[93, 162]], [[93, 162]], [[93, 162]], [[93, 162]], [[2, 169]], [[171, 186]]]",
    "query_spans": "[[[188, 201]]]",
    "process": ""
  },
  {
    "text": "If the eccentricity of the ellipse $x^{2}+\\frac{y^{2}}{m}=1$ with foci on the $x$-axis is $\\frac{\\sqrt{3}}{2}$, then what is the value of $m$?",
    "fact_expressions": "C: Ellipse;PointOnCurve(Focus(C), xAxis);m: Number;Expression(C) = (x**2 + y**2/m = 1);Eccentricity(C) = sqrt(3)/2",
    "query_expressions": "m",
    "answer_expressions": "1/4",
    "fact_spans": "[[[10, 37]], [[1, 37]], [[64, 67]], [[10, 37]], [[10, 62]]]",
    "query_spans": "[[[64, 71]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x$ coincides with the right focus of the hyperbola $\\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1$, then what is the value of $p$?",
    "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;Expression(G) = (x^2/6 - y^2/3 = 1);Expression(H) = (y^2 = 2*(p*x));Focus(H) = RightFocus(G)",
    "query_expressions": "p",
    "answer_expressions": "6",
    "fact_spans": "[[[21, 59]], [[1, 17]], [[67, 70]], [[21, 59]], [[1, 17]], [[1, 65]]]",
    "query_spans": "[[[67, 73]]]",
    "process": "According to the problem, for the hyperbola $\\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1$, we have $a^{2}=6$, $b^{2}=3$, $c^{2}=a^{2}+b^{2}=9$, $\\therefore c=3$, and the coordinates of the right focus are $(3,0)$. Therefore, the focus of the parabola $y^{2}=2px$ is $(\\frac{p}{2},0)=(3,0)$, $\\therefore \\frac{p}{2}=3$, $\\therefore p=6$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on this ellipse. If $|P F_{1}|-|P F_{2}|=1$, then the area of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2/3 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)) = 1",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[2, 39], [70, 72]], [[64, 68]], [[48, 55]], [[56, 63]], [[2, 39]], [[2, 63]], [[2, 63]], [[64, 73]], [[75, 98]]]",
    "query_spans": "[[[100, 127]]]",
    "process": "From the definition of the ellipse, we know |PF_{1}| + |PF_{2}| = 2a = 4. Solving simultaneously with |PF_{1}| - |PF_{2}| = 1, we can find |PF_{1}| and |PF_{2}|. Given |F_{1}F_{2}| = 2c = 2, we obtain that triangle PF_{1}F_{2} is a right triangle, so the area can be calculated. Since \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1, we have a = 2, c = 1. From the definition of the ellipse, |PF_{1}| + |PF_{2}| = 2a = 4 and |PF_{1}| - |PF_{2}| = 1. Solving gives |PF_{1}| = \\frac{5}{2}, |PF_{2}| = \\frac{3}{2}. Since |F_{1}F_{2}| = 2c = 2, we have |PF_{1}|^{2} = |F_{1}F_{2}|^{2} + |PF_{2}|^{2}, so the area of triangle PF_{1}F_{2} is \\frac{1}{2} \\times |F_{1}F_{2}| \\times |PF_{2}| = \\frac{1}{2} \\times 2 \\times \\frac{3}{2} = \\frac{3}{2}."
  },
  {
    "text": "The line $l$ passes through the focus $F$ of the parabola $y^{2}=4x$, and intersects the parabola at points $A$ and $B$. If $\\overrightarrow{A F}=4 \\overrightarrow{F B}$, then the slope of line $l$ is?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};VectorOf(A, F) = 4*VectorOf(F, B)",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*4/3",
    "fact_spans": "[[[0, 5], [93, 98]], [[7, 21], [30, 33]], [[35, 38]], [[24, 27]], [[39, 42]], [[7, 21]], [[7, 27]], [[0, 27]], [[0, 44]], [[46, 91]]]",
    "query_spans": "[[[93, 103]]]",
    "process": "The focus of the parabola $ y^{2} = 4x $ is $ (1,0) $. Let line $ l $ be $ x = my + 1 $, and let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Then  \n$$\n\\begin{cases}\nx = my + 1 \\\\\ny^{2} = 4x\n\\end{cases}\n$$\nRearranging gives $ y^{2} - 4my - 4 = 0 $, so $ y_{1} + y_{2} = 4m $, $ y_{1}y_{2} = -4 $. If $ \\overrightarrow{AF} = 4\\overrightarrow{FB} $, then $ y_{1} = -4y_{2} $, solving yields  \n$$\n\\begin{cases}\ny_{1} = 4 \\\\\ny_{2} = -1\n\\end{cases}\n\\quad \\text{or} \\quad\n\\begin{cases}\ny_{1} = -4 \\\\\ny_{2} = 1\n\\end{cases}\n$$\nThus $ m = \\frac{1}{4}(-4 + 1) = -\\frac{3}{4} $, or $ \\frac{1}{4}(4 - 1) = \\frac{3}{4} $, so the slope is $ \\pm\\frac{4}{3} $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1$, a line passing through $F_{1}$ intersects the ellipse $C$ at points $M$ and $N$. Then, what is the perimeter of $\\Delta F_{2} M N$?",
    "fact_expressions": "C: Ellipse;G: Line;F2: Point;M: Point;N: Point;F1: Point;Expression(C) = (x^2/3 + y^2/4 = 1);Focus(C) = {F1, F2};PointOnCurve(F1, G);Intersection(G, C) = {M,N}",
    "query_expressions": "Perimeter(TriangleOf(F2, M, N))",
    "answer_expressions": "8",
    "fact_spans": "[[[18, 60], [78, 83]], [[75, 77]], [[10, 17]], [[85, 88]], [[89, 92]], [[2, 9], [67, 74]], [[18, 60]], [[2, 65]], [[66, 77]], [[75, 94]]]",
    "query_spans": "[[[96, 119]]]",
    "process": "Using the definition of an ellipse, we know that $|F_{1}M| + |F_{2}M|$ and $|F_{1}N| + |F_{2}N|$ can be determined, and by adding these four distances, the result can be obtained. Using the definition of an ellipse, we have $|F_{1}M| + |F_{2}M| = 2a = 4$, $|F_{1}N| + |F_{2}N| = 2a = 4$. Therefore, the perimeter of $\\triangle MNF_{2}$ is $|F_{1}M| + |F_{2}M| + |F_{1}N| + |F_{2}N| = 4 + 4 = 8$."
  },
  {
    "text": "It is known that the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1  (a>0)$ coincides with the focus of the parabola $y^{2}=8 x$. Then, what is the equation of the asymptotes of this hyperbola?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 8*x);RightFocus(G) = Focus(H)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(3)/3)*x",
    "fact_spans": "[[[2, 41], [67, 70]], [[5, 41]], [[46, 60]], [[5, 41]], [[2, 41]], [[46, 60]], [[2, 64]]]",
    "query_spans": "[[[67, 78]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively, a line $l$ passing through point $F_{1}$ intersects $E$ at exactly one point, and $l$ is tangent to the circle $\\odot O$: $x^{2}+y^{2}=2 a^{2}$. Then the eccentricity of this hyperbola is?",
    "fact_expressions": "F1: Point;F2: Point;LeftFocus(E) = F1;RightFocus(E) = F2;E: Hyperbola;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;l: Line;PointOnCurve(F1, l);NumIntersection(l, E) = 1;O1: Circle;Expression(O1) = (x^2 + y^2 = 2*a^2);IsTangent(l,O1) = True",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 9], [89, 97]], [[10, 17]], [[2, 87]], [[2, 87]], [[20, 81], [104, 107], [157, 160]], [[20, 81]], [[28, 81]], [[28, 81]], [[28, 81]], [[28, 81]], [[98, 103], [116, 119]], [[88, 103]], [[98, 114]], [[120, 152]], [[120, 152]], [[116, 154]]]",
    "query_spans": "[[[157, 166]]]",
    "process": "The hyperbola $ E: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) has asymptotes given by $ y = \\pm \\frac{b}{a}x $. The line $ l $ passing through point $ F_{1} $ intersects $ E $ at exactly one point, so line $ l $ is parallel to the asymptotes, and thus the equation of line $ l $ is $ y = \\pm \\frac{b}{a}(x + c) $. Since $ l $ is tangent to the circle $ \\odot O: x^{2} + y^{2} = 2a^{2} $, it follows that $ \\frac{bc}{\\sqrt{a^{2}+b^{2}}} = b = \\sqrt{2}a $. Then the eccentricity of the hyperbola is $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{3} $."
  },
  {
    "text": "$P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $F_{1}$ and $F_{2}$ are the left and right foci respectively, and the focal distance is $2c$. Then the horizontal coordinate of the incenter of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P, RightPart(G));G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;FocalLength(G) = 2*c;c: Number",
    "query_expressions": "XCoordinate(Center(InscribedCircle(TriangleOf(P, F1, F2))))",
    "answer_expressions": "a",
    "fact_spans": "[[[0, 3]], [[0, 66]], [[4, 60]], [[4, 60]], [[7, 60]], [[7, 60]], [[7, 60]], [[7, 60]], [[67, 74]], [[75, 82]], [[4, 90]], [[4, 90]], [[4, 100]], [[95, 100]]]",
    "query_spans": "[[[102, 137]]]",
    "process": "As shown in the figure: $F_{1}(-c,0)$, $F_{2}(c,0)$, let the point where the incircle touches the x-axis be point $H$, and let the points where $PF_{1}$, $PF_{2}$ touch the incircle be $M$, $N$ respectively. By the definition of a hyperbola, $|PF_{1}| - |PF_{2}| = 2a$. By the tangent segment theorem of a circle, $|PM| = |PN|$, so $|MF_{1}| - |NF_{2}| = 2a$, that is, $|HF_{1}| - |HF_{2}| = 2a$. Let the x-coordinate of the center of the incircle be $x$, then the x-coordinate of point $H$ is also $x$. Therefore, $(x + c) - (c - x) = 2a$, yielding $x = a$."
  },
  {
    "text": "The coordinates of the focus of the parabola $(y+1)^{2}=4(x-2)$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = ((y + 1)^2 = 4*(x - 2))",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(3, -1)",
    "fact_spans": "[[[0, 21]], [[0, 21]]]",
    "query_spans": "[[[0, 28]]]",
    "process": ""
  },
  {
    "text": "A point $(a, b)$ on the left branch of the hyperbola $x^{2}-y^{2}=1$ is at a distance of $\\sqrt{2}$ from its asymptote $y=x$. What is the value of $a+b$?",
    "fact_expressions": "G: Hyperbola;H: Point;a: Number;b: Number;Expression(G) = (x^2 - y^2 = 1);Coordinate(H) = (a, b);PointOnCurve(H, LeftPart(G));Expression(L) = (y = x);Distance(H, L) = sqrt(2);OneOf(Asymptote(G)) = L;L:Line",
    "query_expressions": "a + b",
    "answer_expressions": "-1/2",
    "fact_spans": "[[[0, 18], [33, 34]], [[23, 32]], [[23, 32]], [[23, 32]], [[0, 18]], [[23, 32]], [[0, 32]], [[37, 42]], [[23, 57]], [[33, 42]], [[37, 42]]]",
    "query_spans": "[[[59, 68]]]",
    "process": ""
  },
  {
    "text": "Given that one focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ coincides with the focus of the parabola $y^{2}=4x$, and the distance from the focus to an asymptote is $\\frac{\\sqrt{3}}{2}$, what is the eccentricity of the hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;H: Parabola;Expression(H) = (y^2 = 4*x);OneOf(Focus(G)) = Focus(H);Distance(Focus(G), Asymptote(G)) = sqrt(3)/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 58], [118, 121]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[64, 78]], [[64, 78]], [[2, 83]], [[2, 115]]]",
    "query_spans": "[[[118, 127]]]",
    "process": "Draw the figure according to the given conditions, then find the coordinates of the focus from the parabola equation, obtain the foci coordinates of the hyperbola, set up an equation using the distance from the focus to one asymptote of the hyperbola, and solve for the eccentricity. As shown in the figure, from the parabola equation $ y^{2} = 4x $, we obtain the focus coordinates of the parabola $ F(1,0) $, which is also the coordinates of the right focus of the hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $). The asymptotes of the hyperbola are $ y = \\pm\\frac{b}{a}x $. Take $ y = \\frac{b}{a}x $, rewritten in general form: $ bx - ay = 0 $. Then $ \\frac{|b|}{\\sqrt{a^{2} + b^{2}}} = \\frac{\\sqrt{3}}{2} $, i.e., $ 4b^{2} = 3a^{2} + 3b^{2} $. Also, $ a^{2} = 1 - b^{2} $, solving these equations simultaneously gives: $ a^{2} = \\frac{1}{4} $, $ \\therefore a = \\frac{1}{2} $. Thus, the eccentricity of the hyperbola is: $ e = \\frac{c}{a} = \\frac{1}{\\frac{1}{2}} = 2 $."
  },
  {
    "text": "Let $P$ be a point on the curve $C$: $x=\\sqrt{4-4 y^{2}}$, $A(-\\sqrt{3}, 0)$, $B(\\sqrt{3}, 0)$, $|P B|=1$, then the inradius of $\\triangle P A B$ is?",
    "fact_expressions": "C: Curve;A: Point;B: Point;P: Point;Expression(C) = (x = sqrt(4 - 4*y^2));Coordinate(A) = (-sqrt(3), 0);Coordinate(B) = (sqrt(3), 0);PointOnCurve(P, C);Abs(LineSegmentOf(P, B)) = 1",
    "query_expressions": "Radius(IsInscribedCircle(TriangleOf(P,A,B)))",
    "answer_expressions": "2*sqrt(2)-sqrt(6)",
    "fact_spans": "[[[5, 32]], [[36, 53]], [[54, 70]], [[1, 4]], [[5, 32]], [[36, 53]], [[54, 70]], [[1, 35]], [[71, 80]]]",
    "query_spans": "[[[82, 108]]]",
    "process": "It is easy to see that curve $ C $ represents the right half of the ellipse $ \\frac{x^{2}}{4} + y^{2} = 1 $, and $ A $, $ B $ are the left and right foci of the ellipse respectively. By the definition of an ellipse, $ |PA| + |PB| = 2a = 4 $, then $ |PA| = 3 $, thus $ \\cos\\angle APB = \\frac{1 + 9 - 12}{2 \\times 1 \\times 3} = -\\frac{1}{3} $, $ S_{APAB} = \\frac{1}{2} \\times 1 \\times 3 \\times \\frac{\\sqrt{2}}{3} = \\sqrt{2} $, then the inradius of $ APAB $ is $ \\frac{2\\sqrt{2}}{1 + 3 + 2\\sqrt{3}} = 2\\sqrt{2} - \\sqrt{6} $."
  },
  {
    "text": "Given $\\tan \\alpha = -2$, the focus of the parabola $y^2 = 2px$ ($p > 0$) is $F(-\\sin \\alpha \\cos \\alpha, 0)$, the line $l$ passes through point $F$ and intersects the parabola at points $A$ and $B$, and $|AB| = 4$, then the distance from the midpoint of segment $AB$ to the line $x = -\\frac{1}{2}$ is?",
    "fact_expressions": "l: Line;G: Parabola;p: Number;H:Line;A: Point;B: Point;F: Point;p>0;alpha:Number;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (x = -1/2);Coordinate(F) = (-Sin(alpha)*Cos(alpha), 0);Tan(alpha) = -2;Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, B)) = 4",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A,B)), H)",
    "answer_expressions": "21/10",
    "fact_spans": "[[[77, 82]], [[19, 40], [90, 93]], [[22, 40]], [[128, 146]], [[95, 98]], [[99, 102]], [[84, 88], [44, 76]], [[22, 40]], [[2, 18]], [[19, 40]], [[128, 146]], [[44, 76]], [[2, 18]], [[19, 76]], [[77, 88]], [[77, 104]], [[106, 115]]]",
    "query_spans": "[[[117, 151]]]",
    "process": "Since $\\tan\\alpha=-2$, and the focus of the parabola $y^{2}=2px$ $(p>0)$ is $F(-\\sin\\alpha\\cos\\alpha,0)$, we have $F\\left(\\frac{2}{5},0\\right)$, so $p=\\frac{4}{5}$. Since the line $l$ passes through point $F$ and intersects the parabola at points $A$ and $B$ with $AB=4$, we have $x_{1}+x_{2}+\\frac{4}{5}=4$, thus $x_{1}+x_{2}=\\frac{16}{5}$. Therefore, the distance from the midpoint of segment $AB$ to the line $x=-\\frac{1}{2}$ is $\\frac{8}{5}+\\frac{1}{2}=\\frac{21}{10}$."
  },
  {
    "text": "If the eccentricity of the hyperbola $C$: $\\frac{x^{2}}{2}-\\frac{y^{2}}{\\lambda}=1$ is not greater than $\\sqrt{2 \\lambda-5}$, then the range of the length of the imaginary axis of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/2 - y^2/lambda = 1);lambda: Number;Negation(Eccentricity(C) > sqrt(2*lambda - 5))",
    "query_expressions": "Range(Length(ImageinaryAxis(C)))",
    "answer_expressions": "[4, +oo)",
    "fact_spans": "[[[1, 50], [79, 82]], [[1, 50]], [[9, 50]], [[1, 77]]]",
    "query_spans": "[[[79, 93]]]",
    "process": "Since $a^{2}=2$, $b^{2}=\\lambda$, it follows that $c^{2}=2+\\lambda$, so $e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{2+\\lambda}{2}$, hence $e=\\sqrt{\\frac{2+\\lambda}{2}}\\leqslant\\sqrt{2\\lambda-5}$. Solving gives $\\lambda\\geqslant4$, then $b^{2}\\geqslant4$, thus the length of the imaginary axis $2b\\geqslant4$."
  },
  {
    "text": "Let the line $l$ passing through point $K(-1, 0)$ intersect the parabola $C$: $y^2 = 4x$ at points $A$ and $B$, and let $F$ be the focus of the parabola. If $|BF| = 2|AF|$, then $\\cos \\angle AFB = ?$",
    "fact_expressions": "K: Point;Coordinate(K) = (-1, 0);PointOnCurve(K, l) = True;l: Line;C: Parabola;Expression(C) = (y^2 = 4*x);Intersection(l, C) = {A, B};A: Point;B: Point;F: Point;Focus(C) = F;Abs(LineSegmentOf(B, F)) = 2*Abs(LineSegmentOf(A, F))",
    "query_expressions": "Cos(AngleOf(A, F, B))",
    "answer_expressions": "7/9",
    "fact_spans": "[[[2, 14]], [[2, 14]], [[1, 20]], [[15, 20]], [[21, 40], [56, 59]], [[21, 40]], [[15, 51]], [[42, 45]], [[46, 49]], [[52, 55]], [[52, 62]], [[64, 78]]]",
    "query_spans": "[[[80, 101]]]",
    "process": "Let the equation of line $ l $ be $ x = my - 1 $. Substituting this into the parabola $ C: y^{2} = 4x $, we obtain $ y^{2} - 4my + 4 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, and $ 0 < y_{1} < y_{2} $. Thus, $ y_{1} + y_{2} = 4m $, $ y_{1} \\cdot y_{2} = 4 $. By the definition of the parabola, $ |AF| = x_{1} + 1 $, $ |BF| = x_{2} + 1 $. From $ |BF| = 2|AF| $, we have $ 2(x_{1} + 1) = x_{2} + 1 $, then $ 2y_{1} = y_{2} $. Solving gives: $ y_{1} = \\sqrt{2} $, $ x_{1} = \\frac{1}{2} $, $ y_{2} = 2\\sqrt{2} $, $ x_{2} = 2 $, $ m = \\frac{3\\sqrt{2}}{4} $. The coordinates of points $ A $ and $ B $ are $ (\\frac{1}{2}, \\sqrt{2}) $, $ (2, 2\\sqrt{2}) $, respectively. Then $ \\overrightarrow{FA} = (-\\frac{1}{2}, \\sqrt{2}) $, $ \\overrightarrow{FB} = (1, 2\\sqrt{2}) $, so $ \\cos\\angle AFB = \\frac{\\overrightarrow{FA} \\cdot \\overrightarrow{FB}}{|FA| \\cdot |FB|} = \\frac{-\\frac{1}{2} + 4}{\\frac{3}{2} \\times 3} = \\frac{7}{9} $."
  },
  {
    "text": "What is the eccentricity of the hyperbola $3 x^{2}-y^{2}=12$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (3*x^2 - y^2 = 12)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 21]], [[0, 21]]]",
    "query_spans": "[[[0, 27]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ is a point on the ellipse such that $\\angle P F_{1} F_{2}=30^{\\circ}$, $\\angle P F_{2} F_{1}=60^{\\circ}$. Then the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;F2: Point;P: Point;e: Number;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);AngleOf(P, F1, F2) = ApplyUnit(30, degree);AngleOf(P, F2, F1) = ApplyUnit(60, degree);Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[2, 47], [76, 78], [152, 154]], [[4, 47]], [[4, 47]], [[55, 62]], [[63, 70]], [[71, 75]], [[158, 161]], [[2, 47]], [[2, 70]], [[2, 70]], [[71, 81]], [[83, 116]], [[117, 150]], [[152, 161]]]",
    "query_spans": "[[[158, 163]]]",
    "process": ""
  },
  {
    "text": "Let $P$ be a moving point on the parabola $y^{2}=4 x$, $F$ be the focus of the parabola, and point $A(6,3)$. Then the minimum value of $|P A|+|P F|$ is?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (6, 3);PointOnCurve(P, G);Focus(G) = F",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "7",
    "fact_spans": "[[[5, 19], [30, 33]], [[37, 46]], [[1, 4]], [[26, 29]], [[5, 19]], [[37, 46]], [[1, 25]], [[26, 36]]]",
    "query_spans": "[[[48, 67]]]",
    "process": "Let the projection of point P on the directrix be D. Then, according to the definition of a parabola, |PF| = |PD|, and thus the problem is transformed into finding the minimum value of |PA| + |PD|. It can then be deduced that |PA| + |PD| is minimized when points D, P, and A are collinear, from which the answer follows. Let the projection of point P on the directrix be D. Then, according to the definition of a parabola, |PF| = |PD|. Therefore, to minimize |PA| + |PF|, we need to minimize |PA| + |PD|. When D, P, and A are collinear, |PA| + |PD| is minimized as 6 - (-1) = 7."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $P$ is a point on the ellipse $C$, and $\\overrightarrow{P F_{1}} \\perp \\overrightarrow{P F_{2}}$. If the area of $\\Delta P F_{1} F_{2}$ is $9$, then $b=$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C);IsPerpendicular(VectorOf(P, F2), VectorOf(P, F1));Area(TriangleOf(P, F1, F2)) = 9",
    "query_expressions": "b",
    "answer_expressions": "3",
    "fact_spans": "[[[18, 77], [87, 92]], [[18, 77]], [[188, 191]], [[25, 77]], [[25, 77]], [[25, 77]], [[2, 9]], [[10, 17]], [[2, 82]], [[83, 86]], [[83, 95]], [[97, 154]], [[157, 186]]]",
    "query_spans": "[[[188, 193]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through point $F_{1}$ intersects the left and right branches of the hyperbola at points $A$ and $B$ respectively. If $|A B|=|A F_{2}|$ , $\\cos \\angle B A F_{2}=\\frac{7}{8}$, then the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F2: Point;F1: Point;LeftFocus(C) = F1;RightFocus(C) = F2;G: Line;PointOnCurve(F1, G);A: Point;B: Point;Intersection(G,LeftPart(C))=A;Intersection(G,RightPart(C))=B;Abs(LineSegmentOf(A, B)) = Abs(LineSegmentOf(A, F2));Cos(AngleOf(B, A, F2)) = 7/8",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2*sqrt(6)/3",
    "fact_spans": "[[[2, 66], [104, 107], [185, 191]], [[2, 66]], [[10, 66]], [[10, 66]], [[10, 66]], [[10, 66]], [[83, 90]], [[75, 82], [92, 100]], [[2, 90]], [[2, 90]], [[101, 103]], [[91, 103]], [[117, 120]], [[121, 124]], [[101, 126]], [[101, 126]], [[128, 145]], [[148, 183]]]",
    "query_spans": "[[[185, 197]]]",
    "process": ""
  },
  {
    "text": "It is known that the center of the ellipse is at the origin, and the foci are on the $y$-axis. If its eccentricity is $\\frac{1}{2}$ and the focal distance is $8$, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;O: Origin;Center(G)=O;PointOnCurve(Focus(G), yAxis);Eccentricity(G)=1/2;FocalLength(G) = 8",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/64 + x^2/48 = 1",
    "fact_spans": "[[[2, 4], [21, 22], [49, 51]], [[8, 10]], [[2, 10]], [[2, 19]], [[21, 39]], [[21, 46]]]",
    "query_spans": "[[[49, 56]]]",
    "process": "Since the foci of the ellipse lie on the y-axis, let the equation of the ellipse be: \\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1. Given that its eccentricity is \\frac{1}{2} and the focal distance is 8, we have \\frac{c}{a}=\\frac{1}{2}, 2c=8.\\cdots\\textcircled{1} Also, a^{2}=b^{2}+c^{2}\\textcircled{1}\\textcircled{2}. Solving these equations simultaneously gives a=8, b=4\\sqrt{3}. Therefore, the equation of the ellipse is \\frac{x^{2}}{48}+\\frac{y^{2}}{64}=1"
  },
  {
    "text": "Given the parabola $C$: $x^{2}=4 y$ with focus $F$, let $A$ be a point on $C$, and the extension of segment $F A$ intersects the $x$-axis at point $B$. If the distance $d$ from point $A$ to the line $l$: $y=-1$ equals the distance from $A$ to $B$, then $|F B|$=?",
    "fact_expressions": "C: Parabola;A: Point;F: Point;B: Point;l: Line;Expression(C) = (x^2 = 4*y);Focus(C) = F;PointOnCurve(A, C);Intersection(OverlappingLine(LineSegmentOf(F, A)), xAxis) = B;Expression(l) = (y = -1);d: Number;Distance(A, l) = d;Distance(A, B) = d",
    "query_expressions": "Abs(LineSegmentOf(F, B))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 21], [33, 36]], [[29, 32], [63, 67], [87, 90]], [[25, 28]], [[57, 61], [91, 94]], [[68, 79]], [[2, 21]], [[2, 28]], [[29, 39]], [[40, 61]], [[68, 79]], [[82, 85]], [[63, 85]], [[82, 97]]]",
    "query_spans": "[[[99, 108]]]",
    "process": "According to the definition of a parabola, |AB| = |AF|, so A is the midpoint of BF. Let B(x_{0},0), then A(\\frac{x_{0}}{2},\\frac{1}{2}). Since A lies on the parabola, x_{0} can be solved accordingly, leading to the solution. By the given conditions, F(0,1). Since A lies on the parabola and the directrix of the parabola is y = -1, the distance from point A to point F equals the distance from A to y = -1. Also, the distance d from A to the line l: y = -1 equals the distance from A to B, so |AB| = |AF|. Hence, A is the midpoint of BF. Let B(x_{0},0), so A(\\frac{x_{0}}{2},\\frac{1}{2}). Therefore, (\\frac{x_{0}}{2})^{2} = 4 \\times \\frac{1}{2}, solving gives x_{0} = \\pm 2\\sqrt{2}, so |FB| = \\sqrt{(\\pm 2\\sqrt{2})^{2} + 1^{2}} = 3"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1$ with an asymptote equation $2 x+3 y=0$, and $F_{1}$, $F_{2}$ being the left and right foci of the hyperbola $C$, respectively. Point $P$ lies on the hyperbola $C$ such that $|P F_{1}|=7$, then $|P F_{2}|$=?",
    "fact_expressions": "C: Hyperbola;a: Number;P: Point;F1: Point;F2: Point;Expression(C) = (-y^2/4 + x^2/a^2 = 1);Expression(OneOf(Asymptote(C))) = (2*x + 3*y = 0);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 7",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "{1, 13}",
    "fact_spans": "[[[2, 49], [91, 97], [109, 115]], [[10, 49]], [[104, 108]], [[71, 78]], [[81, 88]], [[2, 49]], [[2, 69]], [[70, 103]], [[70, 103]], [[104, 116]], [[118, 131]]]",
    "query_spans": "[[[133, 146]]]",
    "process": "Given that one asymptote of the hyperbola $ C: \\frac{x^2}{a^{2}} - \\frac{y^{2}}{4} = 1 $ is $ 2x + 3y = 0 $, i.e., $ y = -\\frac{2}{3}x $, then $ \\frac{b}{a} = \\frac{2}{3} $. Since $ b = 2 $, $ \\therefore a = 3 $. By the definition of a hyperbola, $ ||PF_{1}| - |PF_{2}|| = 2a = 6 $. $ \\because |PF_{1}| = 7 $, $ \\therefore |PF_{2}| = 1 $ or $ |PF_{2}| = 13 $."
  },
  {
    "text": "Point $P$ moves on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{25}=1$, and points $A$ and $B$ move on $x^{2}+(y-4)^{2}=16$ and $x^{2}+(y+4)^{2}=4$, respectively. Then the maximum value of $|P A|+|P B|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;A: Point;B: Point;C:Curve;D:Curve;Expression(G) = (x^2/9 + y^2/25 = 1);PointOnCurve(P, G);PointOnCurve(A, C);PointOnCurve(B, D);Expression(C)=(x^2+(y-4)^2=16);Expression(D)=(x^2+(y+4)^2=4)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B)))",
    "answer_expressions": "16",
    "fact_spans": "[[[5, 43]], [[0, 4]], [[47, 51]], [[52, 55]], [[58, 78]], [[79, 98]], [[5, 43]], [[0, 44]], [[47, 99]], [[47, 99]], [[58, 78]], [[79, 98]]]",
    "query_spans": "[[[103, 122]]]",
    "process": "According to the problem: the two foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{25}=1$, which are $(0,\\pm4)$, are respectively the centers of the circles $x^{2}+(y-4)^{2}=16$ and $x^{2}+(y+4)^{2}=4$. The sum of the distances from $P$ to the two foci is a constant $2\\times5=10$. The radii of the two circles are $4$ and $2$ respectively. Therefore, when $P$ is the lower vertex of the ellipse and $A$, $B$ are the points on the respective circles with the maximum $y$-coordinates, the maximum value of $|PA|+|PB|$ is $2\\times5+2+4=16$."
  },
  {
    "text": "Let $F$ be a focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$. If there exists a point $P$ on $C$ such that the midpoint of segment $PF$ is $(0,-\\sqrt{3} b)$, then the eccentricity of $C$ is?",
    "fact_expressions": "F: Point;OneOf(Focus(C)) = F;C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;P: Point;PointOnCurve(P,C) = True;Coordinate(MidPoint(LineSegmentOf(P,F))) = (0,-sqrt(3)*b)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(13)",
    "fact_spans": "[[[1, 4]], [[1, 71]], [[5, 66], [73, 76], [115, 118]], [[5, 66]], [[13, 66]], [[13, 66]], [[13, 66]], [[13, 66]], [[79, 83]], [[73, 83]], [[85, 113]]]",
    "query_spans": "[[[115, 124]]]",
    "process": "Without loss of generality, assume point F is the focus of the hyperbola; then point P lies on the left branch. From the equation of the hyperbola, we know F(c,0). Using the midpoint coordinate formula, we obtain: P(-c,-2\\sqrt{3}b). From the general term formula, we get: \\frac{b^{2}}{a}=\\frac{2\\sqrt{3}b,\\therefore b=2\\sqrt{3}a,\\frac{b}{a}}=2\\sqrt{3}, then the eccentricity of the hyperbola e=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{1+(2\\sqrt{3})^{2}}=\\sqrt{13}."
  },
  {
    "text": "The distance from the focus to the directrix of the parabola $x^{2}=4 y$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 25]]]",
    "process": "From the parabola equation $x^{2}=4y$, we know $2p=4$, $p=2$, so the distance from the focus to the directrix is 2. Qing] This question mainly examines the equation and geometric properties of a parabola, and belongs to an easy problem."
  },
  {
    "text": "Given fixed points $A(-3,0)$, $B(3,0)$, and a moving point $P$ on the parabola $y^{2}=2x$, then the minimum value of $\\overrightarrow{P A} \\cdot \\overrightarrow{P B}$ is equal to?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;B: Point;Expression(G) = (y^2 = 2*x);Coordinate(A) = (-3, 0);Coordinate(B) = (3,0);PointOnCurve(P, G)",
    "query_expressions": "Min(DotProduct(VectorOf(P, A), VectorOf(P, B)))",
    "answer_expressions": "-9",
    "fact_spans": "[[[31, 45]], [[4, 13]], [[27, 30]], [[16, 24]], [[31, 45]], [[4, 13]], [[16, 24]], [[27, 48]]]",
    "query_spans": "[[[50, 106]]]",
    "process": "Let $ P(x,y) $, then $ y^{2} = 2x $. Since $ A(-3,0) $, $ B(3,0) $, we have $ \\overrightarrow{PA} \\cdot \\overrightarrow{PB} = \\overrightarrow{AP} \\cdot \\overrightarrow{BP} = (x+3,y) \\cdot (x-3,y) = x^{2} + y^{2} - 9 = x^{2} + 2x - 9 = (x+1)^{2} - 10 $ ($ x \\geqslant 0 $). Hence, when $ x = 0 $, the minimum value is $ -9 $.  \n【Analysis】This question mainly examines points on a curve, satisfying the curve equation, vector dot product operations, and methods for finding extrema of quadratic functions. It is a medium-difficulty problem. In the solving process, one needs to boldly substitute into the vector dot product operation. There are two forms for the vector dot product: one expressed by magnitude and angle, the other by coordinates."
  },
  {
    "text": "Given points $F_{1}(-4 , 0)$ and $F_{2}(4 , 0)$, a moving point $P$ on a curve such that the absolute value of the difference of distances from $P$ to $F_{1}$ and $F_{2}$ is $6$, what is the equation of this curve?",
    "fact_expressions": "F1: Point;Coordinate(F1) = (-4, 0);F2: Point;Coordinate(F2) = (4, 0);G: Curve;P: Point;PointOnCurve(P, G) = True;Abs(Distance(P, F1) - Distance(P, F2)) = 6",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/7 = 1",
    "fact_spans": "[[[2, 18], [45, 52]], [[2, 18]], [[19, 33], [53, 60]], [[19, 33]], [[35, 37], [75, 77]], [[41, 44]], [[35, 44]], [[41, 73]]]",
    "query_spans": "[[[75, 81]]]",
    "process": "Find the equation of the curve according to the definition of a hyperbola. \\because|F_{1}F_{2}|=8,||PF_{1}|-|PF_{2}||=6,\\therefore the locus of point P is a hyperbola with foci F_{1},F_{2} and a transverse axis length of 6. 2a=6,a=3, and c=4,\\therefore b^{2}=c^{2}-a^{2}=7\\therefore the equation of the curve is \\frac{x^{2}}{9}-\\frac{y^{2}}{7}=1"
  },
  {
    "text": "Given that the line $y = -x + 1$ intersects the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1\\ (a > b > 0)$ at points $A$ and $B$, and the midpoint $M$ of segment $AB$ lies on the line $x - 2y = 0$, then the eccentricity of the ellipse is?",
    "fact_expressions": "H: Line;Expression(H) = (y = 1 - x);G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;A: Point;B: Point;Intersection(H, G) = {A, B};M: Point;MidPoint(LineSegmentOf(A, B)) = M;Z: Line;Expression(Z) = (x - 2*y = 0);PointOnCurve(M, Z)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 12]], [[2, 12]], [[13, 65], [107, 109]], [[13, 65]], [[15, 65]], [[15, 65]], [[15, 65]], [[15, 65]], [[68, 71]], [[72, 75]], [[2, 77]], [[89, 92]], [[79, 92]], [[93, 104]], [[93, 104]], [[89, 105]]]",
    "query_spans": "[[[107, 115]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). From \\begin{cases} y = -x + 1 \\\\ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 \\end{cases}, we obtain (a^{2} + b^{2})x^{2} - 2a^{2}x + a^{2} - a^{2}b^{2} = 0. By Vieta's formulas, x_{1} + x_{2} = \\frac{2a^{2}}{a^{2} + b^{2}}, y_{1} + y_{2} = \\frac{2b^{2}}{a^{2} + b^{2}}, and A = a^{2} + b^{2} - 1 > 0. Therefore, the midpoint M of segment AB is \\left( \\frac{a^{2}}{a^{2} + b^{2}}, \\frac{b^{2}}{a^{2} + b^{2}} \\right). Since M lies on the line x - 2y = 0, we have \\frac{a^{2}}{a^{2} + b^{2}} - \\frac{2b^{2}}{a^{2} + b^{2}} = 0, which implies a^{2} = 2b^{2} = 2a^{2} - 2c^{2}. Thus, a^{2} = 2c^{2}, solving for e gives e = \\frac{\\sqrt{2}}{2}."
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and $P$ a point on the ellipse. Then the perimeter of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G)",
    "query_expressions": "Perimeter(TriangleOf(F1, P, F2))",
    "answer_expressions": "18",
    "fact_spans": "[[[17, 55], [63, 65]], [[1, 8]], [[59, 62]], [[9, 16]], [[17, 55]], [[1, 58]], [[59, 68]]]",
    "query_spans": "[[[70, 97]]]",
    "process": "According to the definition of an ellipse, the perimeter of $AF_{1}PF_{2}$ is $2a+2c$, so the perimeter of $AF_{1}PF_{2}$ is $2\\times5+2\\times4=18$."
  },
  {
    "text": "The length of the imaginary axis is $2$, and the hyperbola has eccentricity $e=3$. The two foci of the hyperbola are $F_{1}$ and $F_{2}$. A line passing through $F_{1}$ intersects the hyperbola at points $A$ and $B$, with $|AB|=8$. Then, the perimeter of $\\triangle ABF_{2}$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;A: Point;B: Point;F2: Point;F1: Point;Length(ImageinaryAxis(G)) = 2;Eccentricity(G)=e;e:Number;e=3;Focus(G) = {F1, F2};PointOnCurve(F1, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 8",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "16+sqrt(2)",
    "fact_spans": "[[[17, 20], [52, 55]], [[49, 51]], [[56, 59]], [[60, 63]], [[32, 39]], [[24, 31], [41, 48]], [[0, 20]], [[8, 20]], [[11, 16]], [[11, 16]], [[17, 39]], [[40, 51]], [[49, 65]], [[67, 76]]]",
    "query_spans": "[[[78, 104]]]",
    "process": "According to the problem,  \n\\begin{cases}2b=2\\\\\\frac{c}{a}=3\\\\c^2=a^2+b^{2}\\end{cases},  \nsolving gives $ a=\\frac{\\sqrt{2}}{4}, b=1, c=\\frac{3\\sqrt{2}}{4} $.  \nSince  \n\\begin{cases}|AF_{2}|-|AF_{1}|=2a\\\\|BF_{2}|-|BF_{1}|=2a\\end{cases},  \nadding the two equations yields  \n$ |AF_{2}|+|BF_{2}|-(|AF_{1}|+|BF_{1}|)=|AF_{2}|+|BF_{2}|-|AB|=4a $,  \nso $ |AF_{2}|+|BF_{2}|=4a+|AB| $,  \nthus the perimeter of $ \\triangle ABF_{2} $ is $ 4a+2|AB|=4\\times\\frac{\\sqrt{2}}{4}+2\\times8=16+\\sqrt{2} $."
  },
  {
    "text": "Given that point $P$ is any point on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ except the vertices, $F_{1}$ and $F_{2}$ are the left and right foci respectively, $c$ is the semi-focal length, and the incircle of $\\triangle PF_{1} F_{2}$ touches $F_{1}F_{2}$ at point $M$, then $|F_{1} M| \\cdot |F_{2} M |$=?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;M: Point;c:Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);Negation(P=Vertex(G));LeftFocus(G) = F1;RightFocus(G) = F2;HalfFocalLength(G)=c;TangentPoint(InscribedCircle(TriangleOf(P,F1,F2)),LineSegmentOf(F1,F2))=M",
    "query_expressions": "Abs(LineSegmentOf(F1,M))*Abs(LineSegmentOf(F2,M))",
    "answer_expressions": "b^2",
    "fact_spans": "[[[7, 53]], [[10, 53]], [[10, 53]], [[2, 6]], [[64, 71]], [[72, 79]], [[139, 143]], [[88, 91]], [[7, 53]], [[2, 63]], [[2, 63]], [[7, 87]], [[7, 87]], [[7, 95]], [[96, 143]]]",
    "query_spans": "[[[145, 175]]]",
    "process": ""
  },
  {
    "text": "Let point $P$ be a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and let points $M$ and $N$ be moving points on the circles $(x+3)^{2}+y^{2}=4$ and $(x-3)^{2}+y^{2}=1$, respectively. Then the range of values of $PM + PN$ is?",
    "fact_expressions": "G: Ellipse;H: Circle;D: Circle;P: Point;M: Point;N: Point;Expression(G) = (x^2/25 + y^2/16 = 1);Expression(H) = (y^2 + (x + 3)^2 = 4);Expression(D) = (y^2 + (x - 3)^2 = 1);PointOnCurve(P, G);PointOnCurve(M, H);PointOnCurve(N, D)",
    "query_expressions": "Range(LineSegmentOf(P, M) + LineSegmentOf(P, N))",
    "answer_expressions": "[7, 13]",
    "fact_spans": "[[[5, 44]], [[58, 78]], [[79, 98]], [[0, 4]], [[48, 51]], [[52, 55]], [[5, 44]], [[58, 78]], [[79, 98]], [[0, 47]], [[48, 102]], [[48, 102]]]",
    "query_spans": "[[[104, 120]]]",
    "process": "The ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ has foci $F_{1}(-3,0)$, $F_{2}(3,0)$, and the circles $(x+3)^{2}+y^{2}=4$ and $(x-3)^{2}+y^{2}=1$ have centers at $(-3,0)$, $(3,0)$ respectively. Thus, the sum of distances from point $P$ to the centers of the two circles is 10. Since $M$, $N$ are moving points on the circles $(x+3)^{2}+y^{2}=4$ and $(x-3)^{2}+y^{2}=1$ respectively, $(PM+PN)_{\\min}=10-r_{1}-r_{2}=7$, $(PM+PN)_{\\max}=10+r_{1}+r_{2}=13$. Therefore, the range of $PM+PN$ is $[7,13]$."
  },
  {
    "text": "The coordinates of the focus of the parabola $y=-\\frac{1}{2} x^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = -x^2/2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,-1/2)",
    "fact_spans": "[[[0, 25]], [[0, 25]]]",
    "query_spans": "[[[0, 32]]]",
    "process": ""
  },
  {
    "text": "If the asymptotes of a hyperbola are given by $y = \\pm 2x$, and one of its foci coincides with the focus of the parabola $y^{2} = 4\\sqrt{5}x$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*2*x);H: Parabola;Expression(H) = (y^2 = 4*(sqrt(5)*x));OneOf(Focus(G)) = Focus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/1 - y^2/4 = 1",
    "fact_spans": "[[[1, 4], [23, 24], [60, 63]], [[1, 22]], [[30, 53]], [[30, 53]], [[23, 58]]]",
    "query_spans": "[[[60, 70]]]",
    "process": "Since the asymptotes of the hyperbola are given by $ y = \\pm 2x $, it follows that $ b = 2a $. Since the focus of the parabola $ y^2 = 4\\sqrt{5}x $ is $ (\\sqrt{5}, 0) $, we have $ c = \\sqrt{5} $. Therefore, $ a^{2} + b^{2} = 5a^{2} = 5 $, which gives $ a^{2} = 1 $, $ b^{2} = 4 $. Hence, the standard equation of the hyperbola is $ \\frac{x^{2}}{1} - \\frac{y^{2}}{4} = 1 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has an asymptote equation $y=k x(k>0)$ and eccentricity $2$, find the value of $k$?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;k:Number;a>0;b>0;k>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G)))=(y=k*x);Eccentricity(G)=2",
    "query_expressions": "k",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 58]], [[5, 58]], [[5, 58]], [[89, 92]], [[5, 58]], [[5, 58]], [[67, 79]], [[2, 58]], [[2, 79]], [[2, 87]]]",
    "query_spans": "[[[89, 96]]]",
    "process": "Since y = kx (k > 0) is an asymptote of the hyperbola, then k = \\frac{b}{a}. Also, e = \\frac{c}{a} = \\sqrt{1 + \\left(\\frac{b}{a}\\right)^2} = 2, so \\frac{b}{a} = \\sqrt{3}, then k = \\sqrt{3}. The answer is: ,,--"
  },
  {
    "text": "If the eccentricity of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{m}=1$ is not greater than $\\sqrt{2 m-8}$, then the range of the length of the imaginary axis of $C$ is?",
    "fact_expressions": "C: Hyperbola;m: Number;Expression(C) = (x^2 - y^2/m = 1);Negation(Eccentricity(C) > sqrt(2*m - 8))",
    "query_expressions": "Range(Length(ImageinaryAxis(C)))",
    "answer_expressions": "[6, +oo)",
    "fact_spans": "[[[1, 34], [57, 60]], [[41, 55]], [[1, 34]], [[1, 55]]]",
    "query_spans": "[[[57, 71]]]",
    "process": ""
  },
  {
    "text": "It is known that the focus $F$ of the parabola $y^{2}=4x$ is exactly the right vertex of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and the asymptotes of the hyperbola are given by $y=\\pm \\sqrt{3}x$. Then the equation of the hyperbola is?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 4*x);F: Point;Focus(H) = F;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;RightVertex(G) = F;Expression(Asymptote(G)) = (y = pm*sqrt(3)*x)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[19, 22]], [[2, 22]], [[25, 82], [88, 91], [118, 121]], [[25, 82]], [[28, 82]], [[28, 82]], [[28, 82]], [[28, 82]], [[19, 86]], [[88, 116]]]",
    "query_spans": "[[[118, 125]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ and the hyperbola $\\frac{x^{2}}{m^{2}}-\\frac{y^{2}}{n^{2}}=1  (m>0 , n>0)$ have the same foci $F_{1}$, $F_{2}$, let a point of intersection of the two curves be $Q$, and $\\angle Q F_{1} F_{2}=90^{\\circ}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;n: Number;m: Number;H: Ellipse;Q: Point;F1: Point;F2: Point;m>0;n>0;Expression(G) = (-y^2/n^2 + x^2/m^2 = 1);Expression(H) = (x^2/25 + y^2/16 = 1);Focus(G)={F1,F2};Focus(H)={F1,F2};OneOf(Intersection(G,H))=Q;AngleOf(Q, F1, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[42, 101], [176, 179]], [[45, 101]], [[45, 101]], [[2, 41]], [[135, 139]], [[108, 115]], [[116, 123]], [[45, 101]], [[45, 101]], [[42, 101]], [[2, 41]], [[2, 123]], [[2, 123]], [[125, 139]], [[141, 174]]]",
    "query_spans": "[[[176, 185]]]",
    "process": ""
  },
  {
    "text": "If the line $y=k(x-2)$ intersects the parabola $y^{2}=8x$ at points $A$ and $B$, and the x-coordinate of the midpoint of segment $AB$ is $3$, then $|AB|=$?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*(x - 2));k: Number;G: Parabola;Expression(G) = (y^2 = 8*x);Intersection(H, G) = {A, B};B: Point;A: Point;XCoordinate(MidPoint(LineSegmentOf(A, B))) = 3",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "10",
    "fact_spans": "[[[1, 13]], [[1, 13]], [[3, 13]], [[14, 28]], [[14, 28]], [[1, 39]], [[34, 37]], [[30, 33]], [[41, 59]]]",
    "query_spans": "[[[61, 70]]]",
    "process": "According to the problem, solve the system of equations of the line and the parabola:  \n\\begin{cases}y=k(x-2)\\\\y^2=8x\\end{cases},  \neliminate $ y $ to obtain $ k^{2}(x-2)^{2}=8x $,  \nrearrange to get $ k^{2}x^{2}-(4k^{2}+8)x+4k^{2}=0 $,  \nso $ \\Delta = [-(4k^{2}+8)]^{2}-4\\times4k^{2}\\times k^{2}=64k^{2}+64>0 $.  \nLet $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, so $ x_{1}+x_{2}=\\frac{1}{k^{2}}(4k^{2}+8) $.  \nSince the horizontal coordinate of the midpoint of segment $ AB $ is 3,  \nthen $ x_{1}+x_{2}=\\frac{1}{k^{2}}(4k^{2}+8)=2\\times3 $,  \nsolve to get $ k=\\pm2 $,  \nso $ |AB|=x_{1}+x_{2}+p=6+4=10 $."
  },
  {
    "text": "Given that point $P$ lies on the parabola $x^{2}=8 y$, point $A(-2,4)$, and $F$ is the focus, then the minimum value of $|P F|+|P A|$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 8*y);P: Point;PointOnCurve(P, G);A: Point;Coordinate(A) = (-2, 4);F: Point;Focus(G) = F",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "6",
    "fact_spans": "[[[7, 21]], [[7, 21]], [[2, 6]], [[2, 22]], [[23, 33]], [[23, 33]], [[35, 38]], [[7, 41]]]",
    "query_spans": "[[[43, 62]]]",
    "process": "Since $(-2)^2 < 8 \\times 4$, point $A$ lies inside the parabola. As shown in the figure, draw perpendiculars from points $P$ and $A$ to the directrix $l$, with feet at $Q$ and $B$ respectively. Then $|PF| = |PQ|$. It is clear that $|PF| + |PA|$ is minimized when points $A$, $P$, and $Q$ are collinear, which gives the minimum value as $|AB|$. The distance from point $A$ to the directrix is easily found to be $4 - \\left(\\frac{p}{2}\\right) = 4 - (-2) = 6$. Hence, the minimum value of $|PF| + |PA|$ is $6$."
  },
  {
    "text": "Given that one focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{2}=1(a>0)$ is $(2,0)$, then $a$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/2 + x^2/a^2 = 1);a: Number;a>0;H: Point;Coordinate(H) = (2, 0);OneOf(Focus(G)) = H",
    "query_expressions": "a",
    "answer_expressions": "sqrt(6)",
    "fact_spans": "[[[2, 48]], [[2, 48]], [[63, 66]], [[4, 48]], [[54, 61]], [[54, 61]], [[2, 61]]]",
    "query_spans": "[[[63, 68]]]",
    "process": "According to the standard equation of the ellipse, $a^2 = b^{2} + c^{2}$, the solution is obtained. Since one focus of the ellipse is $(2,0)$, the foci lie on the x-axis. Also, the standard equation of the ellipse satisfies: $a^{2} = b^{2} + c^{2}$, hence $a^{2} = 6$, $a > 0$, $\\therefore a = \\sqrt{6}$."
  },
  {
    "text": "The line passing through the point $P(2,1)$ intersects the hyperbola $\\frac{y^{2}}{2}-x^{2}=1$ at points $A$ and $B$. What is the slope of the line containing chord $AB$ with point $P$ as its midpoint?",
    "fact_expressions": "G: Hyperbola;H: Line;A: Point;B: Point;P: Point;Expression(G) = (-x^2 + y^2/2 = 1);Coordinate(P) = (2, 1);PointOnCurve(P, H);Intersection(H, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = P;IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "Slope(OverlappingLine(LineSegmentOf(A, B)))",
    "answer_expressions": "4",
    "fact_spans": "[[[14, 42]], [[11, 13]], [[44, 47]], [[48, 51]], [[1, 10], [56, 60]], [[14, 42]], [[1, 10]], [[0, 13]], [[11, 53]], [[55, 70]], [[14, 70]]]",
    "query_spans": "[[[65, 78]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then \n\\[\n\\begin{cases}\n\\frac{y_{2}}{2}-x_{1}=1 \\\\\n\\frac{y_{2}^{2}-x_{2}}{2}=1 \\\\\n\\text{Subtracting the two equations yields: } x_{2}-y_{1}=2 \\\\\n\\frac{x_{1}+x_{2}}{y_{1}+y}=\\frac{4}{1}=4,\n\\end{cases}\n\\]\ni.e., $ k=4 $. Upon verification, when $ k=4 $, the line intersects the hyperbola. Therefore, fill in $ 4 $. This question mainly examines the positional relationship between a line and a hyperbola, the point-difference method, and is a medium-level problem."
  },
  {
    "text": "A focus of the hyperbola $8 m x^{2}-m y^{2}=8$ is $(3,0)$, then what is the value of $m$?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (8*(m*x^2) - m*y^2 = 8);Coordinate(OneOf(Focus(G))) = (3, 0)",
    "query_expressions": "m",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 24]], [[40, 43]], [[0, 24]], [[0, 37]]]",
    "query_spans": "[[[40, 47]]]",
    "process": "The hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{\\frac{8}{m}}=1$ has a focus at $(3,0)$, so $\\frac{1}{m}+\\frac{8}{m}=9$, solving gives: $m=1$"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. If a line $l$ passing through $F_{1}$ intersects the ellipse at points $A$ and $B$, then the perimeter of $\\triangle A B F_{2}$ is equal to?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;F2: Point;F1: Point;l: Line;PointOnCurve(F1,l) = True;Intersection(G,l) = {A,B};A: Point;B: Point",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 39], [83, 85]], [[2, 39]], [[2, 65]], [[2, 65]], [[58, 65]], [[50, 57], [69, 76]], [[77, 82]], [[67, 82]], [[77, 97]], [[88, 91]], [[92, 95]]]",
    "query_spans": "[[[99, 126]]]",
    "process": "From \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1 we get a=2, then the perimeter of \\triangle ABF_{2} equals AB+AF_{2}+BF_{2}=AF_{1}+BF_{1}+AF_{2}+BF_{2}=4a=8"
  },
  {
    "text": "Let $A$ be a point on the parabola $C$: $y^{2}=12 x$. If the distance from $A$ to the focus of $C$ is $10$, then the distance from $A$ to the $y$-axis is?",
    "fact_expressions": "A: Point;C: Parabola;Expression(C) = (y^2 = 12*x);PointOnCurve(A, C);Distance(A, Focus(C)) = 10",
    "query_expressions": "Distance(A, yAxis)",
    "answer_expressions": "7",
    "fact_spans": "[[[1, 4], [30, 33], [50, 53]], [[5, 25], [34, 37]], [[5, 25]], [[1, 28]], [[30, 48]]]",
    "query_spans": "[[[50, 63]]]",
    "process": "Let the focus of the parabola be F. Since the distance from point A to the focus of C is 10, by the definition of the parabola, |AF| = x_{A} + \\frac{p}{2} = x_{A} + 3 = 10. Solving gives x_{A} = 7, so the distance from point A to the v-axis is 7. The answer is"
  },
  {
    "text": "If the ellipse $C$ passes through the points $(2, \\sqrt{2})$, $(\\sqrt{2}, \\sqrt{3})$, then what is the eccentricity of the ellipse $C$?",
    "fact_expressions": "C: Ellipse;G: Point;Coordinate(G) = (2, sqrt(2));PointOnCurve(G, C);H: Point;Coordinate(H) = (sqrt(2), sqrt(3));PointOnCurve(H, C)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[1, 6], [49, 54]], [[7, 23]], [[7, 23]], [[1, 23]], [[25, 47]], [[25, 47]], [[1, 47]]]",
    "query_spans": "[[[49, 60]]]",
    "process": "Let the ellipse $ C: mx^{2} + ny^{2} = 1 $ ($ m > 0, n > 0, m \\neq n $), then  \n$$\n\\begin{cases}\n4m + 2n = 1 \\\\\n2m + 3n = 1\n\\end{cases}\n\\quad \\text{then} \\quad\n\\begin{cases}\nm = \\frac{1}{8} \\\\\nn = \\frac{1}{4}\n\\end{cases},\n$$\nhence the ellipse $ C: \\frac{x^{2}}{8} + \\frac{y^{2}}{4} = 1 $, thus the eccentricity $ e = \\frac{c}{a} = \\sqrt{1 - \\frac{b^{2}}{a^{2}}} = \\frac{\\sqrt{2}}{2} $."
  },
  {
    "text": "The standard equation of the parabola passing through the point $P(4, -2)$ is?",
    "fact_expressions": "P: Point;Coordinate(P) = (4,-2);PointOnCurve(P, G);G: Parabola",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=x,x^2=-8*y}",
    "fact_spans": "[[[2, 13]], [[2, 13]], [[0, 17]], [[14, 17]]]",
    "query_spans": "[[[14, 24]]]",
    "process": ""
  },
  {
    "text": "Given that the center of a hyperbola is at the origin and one focus is $F(\\sqrt{7} , 0)$, the line $y = x - 1$ intersects it at points $M$ and $N$, and the x-coordinate of the midpoint of $MN$ is $-\\frac{2}{3}$, then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;M: Point;N: Point;F: Point;O:Origin;Center(G)=O;Coordinate(F) = (sqrt(7), 0);OneOf(Focus(G))=F;L: Line;Expression(L) = (y = x-1);Intersection(L, G) = {M, N};XCoordinate(MidPoint(LineSegmentOf(M, N))) = -2/3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/2 - y^2/5 = 1",
    "fact_spans": "[[[2, 5], [90, 93], [47, 48]], [[51, 54]], [[55, 58]], [[16, 33]], [[8, 10]], [[2, 10]], [[16, 33]], [[2, 33]], [[35, 46]], [[35, 46]], [[35, 60]], [[61, 87]]]",
    "query_spans": "[[[90, 98]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the foci of the hyperbola $\\frac{y^{2}}{3}-\\frac{x^{2}}{2}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/2 + y^2/3 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,pm*sqrt(5))",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 45]]]",
    "process": "According to the standard equation of the hyperbola, solve directly. From $\\frac{y^{2}}{3}-\\frac{x^{2}}{2}=1$, it can be seen that the foci lie on the $y$-axis with $a^{2}=3$, $b^{2}=2$, so $c^{2}=a^{2}+b^{2}=5$, that is, $c=\\sqrt{5}$. Therefore, the coordinates of the foci of the hyperbola are $(0,\\pm\\sqrt{5})$."
  },
  {
    "text": "Given that the parabola $\\Gamma$ has the origin $O$ as its vertex and the $x$-axis as its axis of symmetry, and that $\\Gamma$ intersects the circle $C$: $x^{2}+y^{2}-4 y=0$ at two points, if the distance between these two points is $3$, then what is the distance from the focus of $\\Gamma$ to its directrix?",
    "fact_expressions": "Gamma:Parabola;C: Circle;O:Origin;Vertex(Gamma)=O;SymmetryAxis(Gamma)=xAxis;P:Point;Q:Point;Expression(C) = (-4*y + x^2 + y^2 = 0);Intersection(Gamma,C)={P,Q};Distance(P,Q)=3",
    "query_expressions": "Distance(Focus(Gamma), Directrix(Gamma))",
    "answer_expressions": "27*sqrt(7)/56",
    "fact_spans": "[[[2, 13], [34, 42], [87, 95], [99, 100]], [[43, 68]], [[14, 19]], [[2, 22]], [[2, 32]], [], [], [[43, 68]], [[34, 73]], [[34, 85]]]",
    "query_spans": "[[[87, 107]]]",
    "process": "Analysis: Use the Pythagorean theorem to find the distance from the center of the circle $ C(0,2) $ to the line $ OA $, apply the point-to-line distance formula to determine the slope of line $ OA $, find the equation of the line passing through $ C $ and perpendicular to $ OA $, the intersection point of the two lines is the midpoint of $ OA $, thus obtaining the coordinates of point $ A $. Substituting into the parabola equation yields $ P=\\frac{27\\sqrt{7}}{56} $, thereby obtaining the result. Both the circle and the parabola pass through the origin; let the other intersection point be $ A $, then the distance from the center $ C(0,2) $ to the line $ OA $ is $ \\sqrt{4-\\frac{9}{4}}=\\frac{\\sqrt{7}}{2} $. Let line $ OA: y=kx $, then $ \\frac{2}{\\sqrt{1+k^{2}}}=\\frac{\\sqrt{7}}{2} $, solving gives $ k=\\frac{3}{\\sqrt{7}} $ (discard $ k=-\\frac{3}{\\sqrt{7}} $). The equation of the line passing through $ C $ and perpendicular to $ OA $ is $ y=-\\frac{\\sqrt{7}}{3}x+2 $, ①. Line $ OA: y=\\frac{3\\sqrt{7}}{7}x $, ②. Solving ① and ② gives $ \\begin{cases}x=\\frac{3}{8}\\sqrt{7}\\\\y=\\frac{9}{8}\\end{cases} $, so the midpoint of $ OA $ has coordinates $ (\\frac{3}{8}\\sqrt{7},\\frac{9}{8}) $. Coordinates of point $ A $ are $ (\\frac{3}{4}\\sqrt{7},\\frac{9}{4}) $. Substituting into $ y^{2}=2px $, we get $ P=\\frac{27\\sqrt{7}}{56} $, thus the distance from the focus of $ T $ to its directrix is $ \\frac{27\\sqrt{7}}{56} $."
  },
  {
    "text": "It is known that the directrix of the parabola $y^{2}=2 p x(p>0)$ is tangent to the circle $x^{2}+y^{2}-6 x-7=0$. What is the value of $p$?",
    "fact_expressions": "G: Parabola;p: Number;H: Circle;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (-6*x + x^2 + y^2 - 7 = 0);IsTangent(Directrix(G),H)",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 23]], [[53, 56]], [[27, 49]], [[5, 23]], [[2, 23]], [[27, 49]], [[2, 51]]]",
    "query_spans": "[[[53, 60]]]",
    "process": "First, express the equation of the directrix. Then, since the directrix of the parabola $ y^{2} = 2px $ ($ p > 0 $) is tangent to the circle $ (x - 3)^{2} + y^{2} = 16 $, the distance from the center of the circle to the directrix equals the radius, allowing us to find the value of $ p $. The directrix equation of the parabola $ y^{2} = 2px $ ($ p > 0 $) is $ x = -\\frac{p}{2} $. Because the directrix of the parabola $ y^{2} = 2px $ ($ p > 0 $) is tangent to the circle $ (x - 3)^{2} + y^{2} = 16 $, we have $ 3 + \\frac{p}{2} = 4 $, solving which gives $ p = 2 $."
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. Perpendiculars are drawn from $A$ and $B$ to the directrix, with feet $C$ and $D$, respectively. If $|A F|=4|B F|$, then $|C D|=$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;C: Point;D: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};L1:Line;L2:Line;PointOnCurve(A,L1);PointOnCurve(B,L2);IsPerpendicular(L1,Directrix(G));IsPerpendicular(L2,Directrix(G));Abs(LineSegmentOf(A, F)) = 4*Abs(LineSegmentOf(B, F));FootPoint(L1,Directrix(G))=C;FootPoint(L2,Directrix(G))=D",
    "query_expressions": "Abs(LineSegmentOf(C, D))",
    "answer_expressions": "5",
    "fact_spans": "[[[1, 15], [24, 27]], [[21, 23]], [[28, 31], [41, 44]], [[17, 20]], [[32, 35], [45, 48]], [[60, 63]], [[64, 67]], [[1, 15]], [[1, 20]], [[0, 23]], [[21, 37]], [], [], [[24, 54]], [[24, 54]], [[24, 54]], [[24, 54]], [[70, 84]], [[24, 67]], [[24, 67]]]",
    "query_spans": "[[[86, 95]]]",
    "process": "Let the angle of inclination of line AB be $\\theta$, and assume $\\theta$ is acute. Since $|AF|=4|BF|$, we have $\\frac{2}{1-\\cos\\theta}=\\frac{4\\times2}{1+\\cos\\theta}$, solving gives $\\cos\\theta=\\frac{3}{5}$, then $\\sin\\theta=\\frac{4}{5}$. By the focal chord length formula of the parabola, we get $AB=\\frac{4}{\\sin2\\theta}=\\frac{4}{(4)}$. Therefore, $|CD|=|AB|\\sin\\theta=\\frac{25}{4}\\times\\frac{4}{5}=5$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right vertices are $A_{1}$ and $A_{2}$ respectively. If the circle with diameter $A_{1}A_{2}$ is tangent to the line $bx - ay + 2ab = 0$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Circle;H: Line;A1: Point;A2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (2*a*b - a*y + b*x = 0);LeftVertex(C) = A1;RightVertex(C) = A2;IsDiameter(LineSegmentOf(A1,A2),G);IsTangent(H,G)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[2, 59], [130, 135]], [[9, 59]], [[9, 59]], [[105, 106]], [[107, 126]], [[68, 75]], [[76, 83]], [[9, 59]], [[9, 59]], [[2, 59]], [[107, 126]], [[2, 83]], [[2, 83]], [[85, 106]], [[105, 128]]]",
    "query_spans": "[[[130, 141]]]",
    "process": "Ellipse $ C: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $) has left and right vertices $ A_{1}(-a,0) $, $ A_{2}(a,0) $. The circle with diameter $ A_{1}A_{2} $ has center at $ (0,0) $ and radius $ a $. Since the line is tangent to the circle, the distance from the center to the line equals the radius, so $ \\frac{|2ab|}{\\sqrt{a^{2}+b^{2}}} = a $, that is, $ 4a^{2}b^{2} = a^{2}(a^{2}+b^{2}) $, yielding $ a^{2} = 3b^{2} $. The eccentricity of the ellipse is $ e = \\frac{c}{a} = \\sqrt{1 - \\frac{b^{2}}{a^{2}}} = \\sqrt{\\frac{2}{3}} = \\frac{\\sqrt{6}}{3} $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$. If there exists a circle centered on one asymptote of the hyperbola and tangent to the other asymptote and the $x$-axis, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;a>0;b>0;L1:Line;L2:Line;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);OneOf(Asymptote(G))=L1;OneOf(Asymptote(G))=L2;Negation(L1=L2);PointOnCurve(Center(H),L1);IsTangent(H,xAxis);IsTangent(H,L2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 58], [65, 68], [65, 68]], [[5, 58]], [[5, 58]], [[76, 77]], [[5, 58]], [[5, 58]], [], [], [[2, 58]], [[65, 74]], [[65, 85]], [[65, 85]], [[62, 77]], [[65, 93]], [[65, 93]]]",
    "query_spans": "[[[95, 104]]]",
    "process": ""
  },
  {
    "text": "Given that the line $l$: $x+3 y-2 b=0$ passes through the right focus $F$ of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, then the asymptotes of the hyperbola are?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;F:Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(l)=(x+3*y-2*b=0);RightFocus(C)=F;PointOnCurve(F,l)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "x + pm*sqrt(3)*y = 0",
    "fact_spans": "[[[2, 22]], [[23, 84], [93, 96]], [[30, 84]], [[30, 84]], [[88, 91]], [[30, 84]], [[30, 84]], [[23, 84]], [[2, 22]], [[23, 91]], [[2, 91]]]",
    "query_spans": "[[[93, 104]]]",
    "process": "Since the coordinates of the right focus $ F $ of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) are $ (c, 0) $, and from the given condition, $ c - 2b = 0 $. Also, $ \\because c^{2} = a^{2} + b^{2} $, $ \\therefore (2b)^{2} = a^{2} + b^{2} \\Rightarrow \\frac{b}{a} = \\frac{\\sqrt{3}}{3} $. Thus, the equations of the asymptotes of the hyperbola are $ y = \\pm \\frac{b}{a}x = \\pm \\frac{\\sqrt{3}}{3}x $, i.e., $ x \\pm \\sqrt{3}y = 0 $. Therefore, the answer should be filled in as: $ x \\pm \\sqrt{3}y = 0 $."
  },
  {
    "text": "The line passing through the focus of the parabola $y^{2}=4x$ with an inclination angle of $\\frac{\\pi}{6}$ intersects the parabola at points $A$ and $B$. Then $|AB|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G),H) = True;Inclination(H) = pi/6;H: Line;A: Point;B: Point;Intersection(H, G) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16",
    "fact_spans": "[[[1, 15], [41, 44]], [[1, 15]], [[0, 40]], [[18, 40]], [[38, 40]], [[45, 48]], [[49, 52]], [[38, 54]]]",
    "query_spans": "[[[56, 65]]]",
    "process": ""
  },
  {
    "text": "It is known that the center of the hyperbola is at the origin, the foci $F_{1}$ and $F_{2}$ lie on the coordinate axes, one asymptote has the equation $y = x$, and the hyperbola passes through the point $(4, -\\sqrt{10})$. The point $A$ has coordinates $(0, 2)$. Then, the coordinates of the point on the hyperbola closest to point $A$ are?",
    "fact_expressions": "G: Hyperbola;H: Point;O: Origin;F1: Point;F2:Point;A:Point;B:Point;Coordinate(H)=(4,-sqrt(10));Coordinate(A)=(0,2);Center(G)=O;Focus(G)={F1,F2};PointOnCurve(F1,axis);PointOnCurve(F2,axis);Expression(OneOf(Asymptote(G)))=(y=x);PointOnCurve(H,G);PointOnCurve(B,G);WhenMin(Distance(A,B))",
    "query_expressions": "Coordinate(B)",
    "answer_expressions": "(pm*sqrt(7),1)",
    "fact_spans": "[[[2, 5], [94, 97]], [[53, 74]], [[9, 11]], [[14, 21]], [[24, 31]], [[76, 80], [99, 103]], [[108, 109]], [[53, 73]], [[76, 92]], [[2, 11]], [[2, 31]], [[14, 36]], [[14, 36]], [[2, 50]], [[2, 73]], [[94, 109]], [[94, 109]]]",
    "query_spans": "[[[108, 114]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ is any point on the circle $(x-6)^{2}+(y-8)^{2}=1$, and $F_{1}$, $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, find the range of values of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$?",
    "fact_expressions": "G: Ellipse;H: Circle;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Expression(H) = ((x - 6)^2 + (y - 8)^2 = 1);PointOnCurve(P, H);Focus(G) = {F1, F2}",
    "query_expressions": "Range(DotProduct(VectorOf(P, F1), VectorOf(P, F2)))",
    "answer_expressions": "[80, 120]",
    "fact_spans": "[[[54, 91]], [[7, 31]], [[2, 6]], [[36, 43]], [[44, 51]], [[54, 91]], [[7, 31]], [[2, 35]], [[36, 96]]]",
    "query_spans": "[[[98, 161]]]",
    "process": "Given the problem, the foci of the ellipse are $F_{1}(-1,0)$, $F_{2}(1,0)$. Let point $P(6+\\cos\\theta,8+\\sin\\theta)$, then $\\overrightarrow{PF_{1}}=(-7-\\cos\\theta,-8-\\sin\\theta)$, $\\overrightarrow{PF_{2}}=(-5-\\cos\\theta,-8-\\sin\\theta)$. Therefore, $\\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=(-7-\\cos\\theta)(-5-\\cos\\theta)+(-8-\\sin\\theta)(-8-\\sin\\theta)=100+12\\cos\\theta+16\\sin\\theta=100+20\\sin(\\theta+\\varphi)$, $\\tan\\varphi=\\frac{3}{4}$. Since $\\sin(\\theta+\\varphi)\\in[-1,1]$, it follows that $\\overrightarrow{PF}\\cdot\\overrightarrow{PF}_{2}\\in[80,120]$."
  },
  {
    "text": "Given three points $A(2,2)$, $B$, $C$ on the parabola $y^{2}=2 p x$, and the lines $AB$, $AC$ are two tangent lines to the circle $(x-2)^{2}+y^{2}=1$, then the equation of line $BC$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;H: Circle;Expression(H) = (y^2 + (x - 2)^2 = 1);A: Point;Coordinate(A) = (2, 2);B: Point;C: Point;PointOnCurve(A,G);PointOnCurve(B,G);PointOnCurve(C,G)\t;IsTangent(LineOf(A,B),H);IsTangent(LineOf(A,C),H)",
    "query_expressions": "Expression(LineOf(B,C))",
    "answer_expressions": "3*x + 6*y + 4 = 0",
    "fact_spans": "[[[2, 18]], [[2, 18]], [[5, 18]], [[55, 75]], [[55, 75]], [[21, 29]], [[21, 29]], [[31, 34]], [[36, 39]], [[2, 39]], [[2, 39]], [[2, 39]], [[40, 80]], [[40, 80]]]",
    "query_spans": "[[[82, 94]]]",
    "process": "A(2,2) lies on the parabola $ y^{2}=2px $, so $ 2^{2}=2p\\times2 $, i.e., $ p=1 $, and the equation of the parabola is $ y^{2}=2x $. Let the equation of the line passing through point A(2,2) and tangent to the circle $ (x-2)^{2}+y^{2}=1 $ be: $ y-2=k(x-2) $, i.e., $ kx-y+2-2k=0 $. Then the distance from the center of the circle (2,0) to the tangent line is $ d=\\frac{|2k-0+2-2k|}{\\sqrt{k^{2}+1}}=1 $, solving gives $ k=\\pm\\sqrt{3} $. As shown in the figure, line AB: $ y-2=\\sqrt{3}(x-2) $, line AC: $ y-2=-\\sqrt{3}(x-2) $. Solving simultaneously \n$$\n\\begin{cases}\ny-2=\\sqrt{3}(x-2)\\\\\ny^{2}=2x\n\\end{cases}\n$$\ngives $ 3x^{2}+(4\\sqrt{3}-14)x+16-8\\sqrt{3}=0 $. Solving simultaneously \n$$\n\\begin{cases}\ny-2=\\sqrt{2}\\\\\ny^{2}=2x\n\\end{cases}\n$$\nthus $ x_{n}x_{B}=\\frac{16-8\\sqrt{3}}{3} $, from $ x_{A}=2 $ we get $ x_{B}=\\frac{8-4\\sqrt{3}}{3} $, hence $ y_{B}=\\frac{2\\sqrt{3}-6}{3} $, thus $ x_{A}x_{C}=\\frac{16+8\\sqrt{3}}{3} $, from $ x_{A}=2 $ we get $ x_{C}=\\frac{8+4\\sqrt{3}}{3} $, hence $ y_{C}=\\frac{-2\\sqrt{3}-6}{3} $. Therefore, $ y_{B}+y_{c}=\\frac{2\\sqrt{3}-6}{3}+\\frac{-2\\sqrt{3}-6}{y_{B}-y_{c}}=\\frac{\\sqrt{3}-y_{c}}{\\frac{1}{2}y_{2}-\\frac{1}{2}y_{2}}=\\frac{C \\text{ lies on the parabola }}{y_{B}+y_{c}}=\\frac{2}{-4}=-\\frac{1}{2} $. Hence, the equation of line BC is $ y-\\frac{2\\sqrt{3}-6}{3}=-\\frac{1}{2}\\left(x-\\frac{8-4\\sqrt{3}}{3}\\right) $, i.e., $ 3x+6y+4=0 $."
  },
  {
    "text": "Given that $M$ is a point on the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{8}=1$, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola, and $|M F_{1}|=4$, then $|M F_{2}|$=?",
    "fact_expressions": "G: Hyperbola;M: Point;F1: Point;F2: Point;Expression(G) = (x^2/16 - y^2/8 = 1);PointOnCurve(M, G);LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(M, F1)) = 4",
    "query_expressions": "Abs(LineSegmentOf(M, F2))",
    "answer_expressions": "12",
    "fact_spans": "[[[6, 45], [65, 68]], [[2, 5]], [[49, 56]], [[57, 64]], [[6, 45]], [[2, 48]], [[49, 74]], [[49, 74]], [[76, 89]]]",
    "query_spans": "[[[91, 104]]]",
    "process": "The hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{8}=1$ has its foci on the $x$-axis, where $a=4$, $b=2\\sqrt{2}$, $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{16+8}=2\\sqrt{6}$. $|MF|=4<4+2\\sqrt{6}$, that is, $|MF_{1}|<a+c$, so $M$ is a point on the left branch of the hyperbola, then $|MF_{2}|=|MF|+2a=4+8=12$."
  },
  {
    "text": "If the eccentricity of the hyperbola $\\frac{x^{2}}{m}+\\frac{y^{2}}{2}=1$ equals $2$, then the real number $m$ equals?",
    "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (x^2/m + y^2/2 = 1);Eccentricity(G) = 2",
    "query_expressions": "m",
    "answer_expressions": "-6",
    "fact_spans": "[[[2, 40]], [[51, 56]], [[2, 40]], [[2, 49]]]",
    "query_spans": "[[[51, 59]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, let $M$ be a moving point on the ellipse and $F_{1}$ be the left focus of the ellipse. Then, what is the shape of the trajectory of the midpoint $P$ of the segment $M F_{1}$?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;M: Point;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(M, G);LeftFocus(G) = F1;MidPoint(LineSegmentOf(M,F1))=P;a>b;b>0;P:Point",
    "query_expressions": "Locus(P)",
    "answer_expressions": "Ellipse",
    "fact_spans": "[[[2, 54], [60, 62], [75, 77]], [[4, 54]], [[4, 54]], [[67, 74]], [[55, 59]], [[2, 54]], [[56, 66]], [[67, 81]], [[83, 100]], [[4, 54]], [[4, 54]], [96, 98]]",
    "query_spans": "[[[97, 107]]]",
    "process": "Let $F_{2}$ be the right focus of the ellipse. In $\\triangle MF_{1}F_{2}$, it is easy to see that $PO \\parallel MF_{2}$ and $|PO| = \\frac{1}{2}|MF_{2}|$, $|PF_{1}| = \\frac{1}{2}|MF_{1}|$. Since $|MF_{1}| + |MF_{2}| = 2a$, it follows that $|PO| + |PF_{1}| = a > |F_{1}O|$. Hence, by the definition of an ellipse, the locus of point $P$ is an ellipse."
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $\\frac{\\sqrt{2}}{2}$. If the $x$-coordinate of an intersection point between the line $y=k x$ and the ellipse is $b$, then what is the value of $k$?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;k: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 11);Expression(H) = (y = k*x);Eccentricity(G) = sqrt(2)/2;XCoordinate(OneOf(Intersection(H, G))) = b",
    "query_expressions": "k",
    "answer_expressions": "1/2",
    "fact_spans": "[[[0, 56], [93, 94]], [[103, 106]], [[2, 56]], [[83, 92]], [[108, 111]], [[2, 56]], [[2, 56]], [[0, 56]], [[83, 92]], [[0, 81]], [[83, 106]]]",
    "query_spans": "[[[108, 115]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola $y^{2}=2 p x(p>0)$ is $F$, points $A$ and $B$ lie on the parabola such that $\\angle A F B=\\frac{\\pi}{2}$, and the projection of the midpoint $M$ of chord $AB$ onto the directrix is $N$. Then the maximum value of $\\frac{|M N|}{|A B|}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);AngleOf(A, F, B) = pi/2;IsChordOf(LineSegmentOf(A, B), G);M: Point;MidPoint(LineSegmentOf(A, B)) = M;N: Point;Projection(M, Directrix(G)) = N",
    "query_expressions": "Max(Abs(LineSegmentOf(M, N))/Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[0, 21], [37, 40], [86, 87]], [[0, 21]], [[3, 21]], [[3, 21]], [[25, 28]], [[0, 28]], [[29, 32]], [[33, 36]], [[29, 41]], [[33, 41]], [[43, 71]], [[37, 78]], [[82, 85]], [[73, 85]], [[94, 97]], [[82, 97]]]",
    "query_spans": "[[[99, 126]]]",
    "process": ""
  },
  {
    "text": "The asymptotes of hyperbola $C$ passing through the point $(2 \\sqrt{2}, \\sqrt{3})$ are given by $y=\\pm \\frac{\\sqrt{3}}{2} x$. Let $P$ be a point on the right branch of hyperbola $C$, $F$ be the left focus of hyperbola $C$, and point $A(0,3)$. Then the minimum value of $|P A|+|P F|$ is?",
    "fact_expressions": "C: Hyperbola;G: Point;A: Point;P: Point;F: Point;PointOnCurve(G,C);Coordinate(G) = (2*sqrt(2), sqrt(3));Coordinate(A) = (0, 3);PointOnCurve(P,RightPart(C));LeftFocus(C) = F;Expression(Asymptote(C))=(y=pm*(sqrt(3)/2)*x)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "8",
    "fact_spans": "[[[27, 33], [74, 80], [91, 97]], [[1, 26]], [[102, 111]], [[70, 73]], [[86, 89]], [[0, 33]], [[1, 26]], [[102, 111]], [[70, 85]], [[86, 101]], [[27, 68]]]",
    "query_spans": "[[[114, 133]]]",
    "process": ""
  },
  {
    "text": "The line $y = kx - 2$ intersects the parabola $y^2 = 8x$ at points $A$ and $B$. If the horizontal coordinate of the midpoint of $AB$ is $2$, then $|AB| = $?",
    "fact_expressions": "G: Parabola;H: Line;k: Number;A: Point;B: Point;Expression(G) = (y^2 = 8*x);Expression(H) = (y = k*x - 2);Intersection(H, G) = {A, B};XCoordinate(MidPoint(LineSegmentOf(A, B))) = 2",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "2*sqrt(15)",
    "fact_spans": "[[[12, 26]], [[0, 11]], [[2, 11]], [[27, 30]], [[31, 34]], [[12, 26]], [[0, 11]], [[0, 36]], [[38, 52]]]",
    "query_spans": "[[[53, 62]]]",
    "process": ""
  },
  {
    "text": "If the maximum distance between two points on the ellipse $4 x^{2} + y^{2}=2$ is $8$, then the value of the real number $k$ is?",
    "fact_expressions": "G: Ellipse;k: Real;A:Point;B:Point;Expression(G) = (4*x^2 + y^2 = 2);PointOnCurve(A,G);PointOnCurve(B,G);Max(Distance(A,B))=8",
    "query_expressions": "k",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 22]], [[37, 42]], [], [], [[1, 22]], [[1, 25]], [[1, 25]], [[1, 35]]]",
    "query_spans": "[[[37, 46]]]",
    "process": ""
  },
  {
    "text": "What is the equation of a parabola with focus $(-2 , 0)$?",
    "fact_expressions": "G: Parabola;H: Point;Coordinate(H) = (-2, 0);Focus(G) = H",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=-8*x",
    "fact_spans": "[[[14, 17]], [[3, 13]], [[3, 13]], [[0, 17]]]",
    "query_spans": "[[[14, 21]]]",
    "process": ""
  },
  {
    "text": "The foci of the hyperbola lie on the $x$-axis, the length of the real axis is $6$, and the length of the imaginary axis is $8$. Then the standard equation of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;PointOnCurve(Focus(C), xAxis);Length(RealAxis(C)) = 6;Length(ImageinaryAxis(C)) = 8",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x**2/9-y**2/16=1",
    "fact_spans": "[[[0, 3], [36, 39]], [[0, 14]], [[0, 23]], [[0, 33]]]",
    "query_spans": "[[[36, 46]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the two foci of an ellipse are known to be $(5 , 0)$ and $(-5 , 0)$, and the sum of the distances from a point $P$ on the ellipse to the two foci is $26$. What is the equation of the ellipse?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Focus(G) = {F1,F2};Coordinate(F1) = (5, 0);Coordinate(F2) = (-5, 0);P: Point;PointOnCurve(P, G) = True;Distance(P,F1) + Distance(P,F2) = 26",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/169+y^2/144=1",
    "fact_spans": "[[[2, 4], [35, 37], [60, 62]], [[13, 22]], [[24, 34]], [[2, 34]], [[13, 22]], [[24, 34]], [[40, 43]], [[35, 43]], [[35, 58]]]",
    "query_spans": "[[[60, 67]]]",
    "process": "From the given conditions, we have: c=5, and the foci of the ellipse lie on the x-axis. Then, according to the definition of an ellipse, we obtain a=13, and further solve for b using the relationship among a, b, and c, thus obtaining the equation of the ellipse. \\because the coordinates of the two foci are (5,0), (-5,0) \\therefore the foci of the ellipse lie on the x-axis, and c=5, \\therefore from the definition of the ellipse, 2a=26, that is, a=13, \\therefore solving for b using the relationship among a, b, and c gives b=12, and the equation of the ellipse is \\frac{x^{2}}{169}+\\frac{y^{2}}{144}=1"
  },
  {
    "text": "Given that $P$ is an arbitrary point on the right branch (in the first quadrant) of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, $A_{1}$ and $A_{2}$ are its left and right vertices respectively, $O$ is the origin, and the slopes of the lines $PA_{1}$, $PO$, $PA_{2}$ are $k_{1}$, $k_{2}$, $k_{3}$ respectively, then the range of the product of the slopes $k_{1} k_{2} k_{3}$ is?",
    "fact_expressions": "P: Point;G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1);PointOnCurve(P, RightPart(G));Quadrant(P) = 1;A1: Point;A2: Point;LeftVertex(G) = A1;RightVertex(G) = A2;O: Origin;k1: Number;k2: Number;k3: Number;Slope(LineOf(P, A1)) = k1;Slope(LineOf(P, O)) = k2;Slope(LineOf(P, A2)) = k3",
    "query_expressions": "Range(k1*k2*k3)",
    "answer_expressions": "(0,1/8)",
    "fact_spans": "[[[2, 5]], [[6, 34], [72, 73]], [[6, 34]], [[2, 53]], [[2, 47]], [[54, 61]], [[62, 69]], [[54, 77]], [[54, 77]], [[78, 81]], [[121, 129]], [[131, 140]], [[141, 148]], [[87, 148]], [[87, 148]], [[87, 148]]]",
    "query_spans": "[[[152, 178]]]",
    "process": ""
  },
  {
    "text": "Given that point $F$ is the left focus of the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$, point $P$ is any point on the ellipse $C$, and point $Q$ has coordinates $(4,3)$, find the coordinates of point $P$ when $|P Q|+|P F|$ attains its maximum value.",
    "fact_expressions": "C: Ellipse;P: Point;Q: Point;F: Point;Expression(C) = (x^2/2 + y^2 = 1);Coordinate(Q) = (4, 3);LeftFocus(C) = F;PointOnCurve(P, C);WhenMax(Abs(LineSegmentOf(P, F)) + Abs(LineSegmentOf(P, Q)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(0,-1)",
    "fact_spans": "[[[7, 39], [49, 54]], [[44, 48], [96, 100]], [[60, 64]], [[2, 6]], [[7, 39]], [[60, 75]], [[2, 43]], [[44, 59]], [[77, 95]]]",
    "query_spans": "[[[96, 105]]]",
    "process": "The left focus of the ellipse is F(-1,0), and the right focus is E(1,0). According to the definition of an ellipse, |PF| = 2a - |PE|. Therefore, |PF| + |PQ| = |PQ| + 2a - |PE| = 2a + (|PQ| - |PE|). By the triangle inequality, |PQ| - |PE| \\leqslant |QE|, and equality holds when P is the intersection point of the extension of QE with the ellipse, which is (0,-1). Thus, the maximum value is 2a + |QE| = 2\\sqrt{2} + 3\\sqrt{2} = 5\\sqrt{2}."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has eccentricity $2$, and the distance from a focus to an asymptote is $\\sqrt{3}$, then the focal length of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(C) = 2;Distance(Focus(C),Asymptote(C))=sqrt(3)",
    "query_expressions": "FocalLength(C)",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 63], [94, 100]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 71]], [[2, 92]]]",
    "query_spans": "[[[94, 105]]]",
    "process": "Since the eccentricity of the hyperbola is 2, and the distance from the focus $(c,0)$ to the asymptote $bx-ay=0$ is $\\sqrt{3}$, we have $\\frac{|\\frac{c}{a}|}{\\sqrt{a^{2}+b^{2}}}=\\sqrt{3}$. Solving gives $b=\\sqrt{3}$, $a=1$, $c=2$, so the focal distance of the hyperbola is 4."
  },
  {
    "text": "It is known that an ellipse and a hyperbola share the same foci $F_{1}$, $F_{2}$, with eccentricities $e_{1}$, $e_{2}$ respectively. $P$ is one of their common points, and $\\angle F_{1} P F_{2}=\\frac{2 \\pi}{3}$. If $e_{1} e_{2}=\\sqrt{3}$, then $e_{2}$=?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;F1: Point;P: Point;F2: Point;e1: Number;e2: Number;Focus(H) = {F1, F2};Focus(G) = {F1, F2};Eccentricity(H) = e1;Eccentricity(G) = e2;OneOf(Intersection(H, G)) = P;AngleOf(F1, P, F2) = (2*pi)/3;e1*e2 = sqrt(3)",
    "query_expressions": "e2",
    "answer_expressions": "(sqrt(6) + sqrt(2))/2",
    "fact_spans": "[[[5, 8]], [[2, 4]], [[14, 21]], [[58, 61]], [[22, 29]], [[39, 46]], [[47, 55], [137, 144]], [[2, 29]], [[2, 29]], [[30, 55]], [[30, 55]], [[58, 70]], [[72, 110]], [[113, 135]]]",
    "query_spans": "[[[137, 146]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $y^{2}=4 x$ is $F$, and a line passing through $F$ with slope $1$ intersects the parabola at points $A$ and $B$, and a moving point $P$ lies on the curve $y^{2}=-4 x(y \\geq 0)$, then the minimum area of $\\triangle P A B$ is?",
    "fact_expressions": "G: Parabola;H: Line;I: Curve;P: Point;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 4*x);Expression(I) = And((y^2=-4*x), (y>=0));Focus(G) = F;PointOnCurve(F, H);Slope(H) = 1;Intersection(H, G) = {A, B};PointOnCurve(P, I)",
    "query_expressions": "Min(Area(TriangleOf(P, A, B)))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[1, 15], [38, 41]], [[35, 37]], [[58, 82]], [[54, 57]], [[42, 45]], [[46, 49]], [[19, 22], [24, 27]], [[1, 15]], [[58, 82]], [[1, 22]], [[23, 37]], [[28, 37]], [[35, 51]], [[54, 83]]]",
    "query_spans": "[[[85, 111]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ with left and right foci $F_{1}$, $F_{2}$, a line passing through $F_{2}$ intersects the right branch of the hyperbola at points $A$, $B$. Then the range of the sum of the areas of the incircles of $\\triangle A F_{1} F_{2}$ and $\\triangle B F_{1} F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;I: Line;A: Point;F1: Point;B: Point;F2: Point;Expression(G) = (x^2 - y^2/3 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, I);Intersection(I, RightPart(G)) = {A, B}",
    "query_expressions": "Range(Area(InscribedCircle(TriangleOf(A, F1, F2))) + Area(InscribedCircle(TriangleOf(B, F1, F2))))",
    "answer_expressions": "[2*pi, pi*10/3)",
    "fact_spans": "[[[2, 30], [64, 67]], [[61, 63]], [[71, 74]], [[36, 43]], [[75, 78]], [[44, 51], [53, 60]], [[2, 30]], [[2, 51]], [[2, 51]], [[52, 63]], [[61, 80]]]",
    "query_spans": "[[[82, 148]]]",
    "process": "Analysis: Let the incenter of $\\triangle AF_{1}F_{2}$ and $\\triangle BF_{1}F_{2}$ be $I_{1}, I_{2}$, tangent to the $x$-axis at $M, N$, then $|F_{1}M| = \\frac{|F_{1}F_{2}| + |F_{1}A| - |F_{2}A|}{2} = 3 = |F_{1}N|$, so $M, N$ coincide at the right vertex of the hyperbola. A line through $F_{2}$ intersects the right branch of the hyperbola at points $A, B$. Let $\\angle AF_{2}F_{1} = \\alpha \\in (\\frac{\\pi}{3}, \\frac{2\\pi}{3})$. The sum of the areas of the incircles is $S = \\pi r_{1}^{2} + \\pi r_{2}^{2} = \\pi(\\tan^{2}\\frac{\\alpha}{2} + \\cot^{2}\\frac{\\alpha}{2}) \\in [2\\pi, \\frac{10}{3}\\pi)$."
  },
  {
    "text": "The distance from a point $P$ on the hyperbola $\\frac{x^{2}}{64}-\\frac{y^{2}}{36}=1$ to the right focus of the hyperbola is $4$. What is the distance from $P$ to the left directrix?",
    "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2/64 - y^2/36 = 1);PointOnCurve(P, G);Distance(P, RightFocus(G)) = 4",
    "query_expressions": "Distance(P,LeftDirectrix(G))",
    "answer_expressions": "16",
    "fact_spans": "[[[0, 40], [47, 50]], [[43, 46], [63, 66]], [[0, 40]], [[0, 46]], [[43, 60]]]",
    "query_spans": "[[[47, 75]]]",
    "process": ""
  },
  {
    "text": "If the distance from the right vertex of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0, b>0)$ to one of its asymptotes is $\\frac{b}{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Distance(RightVertex(G), OneOf(Asymptote(G))) = b/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 59], [90, 93]], [[1, 59]], [[4, 59]], [[4, 59]], [[4, 59]], [[4, 59]], [[1, 88]]]",
    "query_spans": "[[[90, 99]]]",
    "process": "The right vertex is $(a,0)$, and one asymptote equation is $y=\\frac{b}{a}x$, or $bx-ay=0$. From the given condition, $\\frac{|ab|}{\\sqrt{b^{2}+a^{2}}}=\\frac{b}{2}$, which implies $\\frac{a}{c}=\\frac{1}{2}$, so $e=\\frac{c}{a}=2$."
  },
  {
    "text": "If the ellipse $m x^{2}+y^{2}=8$ has the same foci as the ellipse $C$: $9 x^{2}+25 y^{2}=100$, then the real number $m$=?",
    "fact_expressions": "C: Ellipse;G:Ellipse;m: Real;Expression(C) = (9*x^2 + 25*y^2 = 100);Expression(G)=(m*x^2+y^2=8);Focus(C)=Focus(G)",
    "query_expressions": "m",
    "answer_expressions": "9/17",
    "fact_spans": "[[[21, 50]], [[1, 20]], [[56, 61]], [[21, 50]], [[1, 20]], [[1, 54]]]",
    "query_spans": "[[[56, 63]]]",
    "process": "From $ C: 9x^{2} + 25y^{2} = 100 $, we get: $ \\frac{x^{2}}{\\frac{100}{9}} + \\frac{y^{2}}{4} = 1 $, then $ c^{2} = \\frac{100}{9} - 4 = \\frac{64}{9} $ and the foci lie on the x-axis. From $ mx^{2} + y^{2} = 8 $, we get: $ \\frac{x^{2}}{\\frac{8}{m}} + \\frac{y^{2}}{8} = 1 $. Since $ mx^{2} + y^{2} = 8 $ shares the same foci with $ C: 9x^{2} + 25y^{2} = 100 $, $ \\therefore c^{2} = \\frac{8}{m} - 8 $. Hence, $ \\frac{8}{m} - 8 = \\frac{64}{9} $, solving gives: $ m = \\frac{9}{17} $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $M$ is an endpoint of the minor axis of the ellipse, point $N$ lies on the ellipse, and $\\overrightarrow{M F_{2}}=3 \\overrightarrow{F_{2} N}$. If the area of $\\Delta M N F_{1}$ is $2$, then $a=?$",
    "fact_expressions": "F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Endpoint(MinorAxis(C)) = M;M: Point;PointOnCurve(N,C) = True;N: Point;VectorOf(M, F2) = 3*VectorOf(F2, N);Area(TriangleOf(M, N, F1)) = 2",
    "query_expressions": "a",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 9]], [[10, 17]], [[2, 83]], [[2, 83]], [[20, 77], [88, 90], [101, 103]], [[20, 77]], [[27, 77]], [[189, 192]], [[27, 77]], [[27, 77]], [[84, 95]], [[84, 87]], [[96, 104]], [[96, 100]], [[106, 159]], [[162, 187]]]",
    "query_spans": "[[[189, 194]]]",
    "process": "Since $\\overrightarrow{MF}_{2}=3\\overrightarrow{F_{2}N}$, the points $M$, $F_{2}$, and $N$ are collinear. Also, since $M$ is an endpoint of the minor axis of the ellipse, $MF_{2}=MF_{1}=a$, so $NF_{2}=\\frac{1}{3}a$, $MN=\\frac{4}{3}a$, then $NF_{1}=2a-NF_{2}=\\frac{5}{3}a$, thus $MN^{2}+MF_{1}^{2}=NF_{1}^{2}$, so $S_{n}NF_{1}=\\frac{1}{2}\\cdot a \\cdot \\frac{4}{3}a=\\frac{2}{3}a^{2}=2$, solving gives $a=\\sqrt{3}$ (the negative value is discarded)."
  },
  {
    "text": "Given that $M(4 , 2)$ is the midpoint of the line segment $AB$ intercepted by the line $l$ on the ellipse $x^{2}+4 y^{2}=36$, then the equation of the line $l$ is?",
    "fact_expressions": "l: Line;G: Ellipse;A: Point;B: Point;M: Point;Expression(G) = (x^2 + 4*y^2 = 36);InterceptChord(l,G)=LineSegmentOf(A,B);MidPoint(LineSegmentOf(A,B))=M;Coordinate(M) = (4, 2)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "x+2*y-8=0",
    "fact_spans": "[[[13, 18], [55, 60]], [[19, 39]], [[45, 50]], [[45, 50]], [[2, 12]], [[19, 39]], [[13, 50]], [[2, 53]], [[2, 12]]]",
    "query_spans": "[[[55, 65]]]",
    "process": "Let the equation of line $ l $ be $ y - 2 = k(x - 4) $. Substituting into the ellipse equation and simplifying, from $ x_{1} + x_{2} = \\frac{32k^{2} - 16k}{1 + 4k^{2}} = 8 $, solve for $ k $, thus obtaining the equation of line $ l $. According to the problem, the slope exists, denoted as $ k $, then the equation of line $ l $ is $ y - 2 = k(x - 4) $, i.e., $ kx - y + 2 - 4k = 0 $. Substituting into the ellipse equation and simplifying yields $ (1 + 4k^{2})x^{2} + (16k - 32k^{2})x + 64k^{2} - 64k - 20 = 0 $. Therefore, $ x_{1} + x_{2} = \\frac{32k^{2} - 16k}{1 + 4k^{2}} = 8 $, solving gives $ k = -\\frac{1}{2} $. Hence, the equation of line $ l $ is $ x + 2y - 8 = 0 $."
  },
  {
    "text": "If the line segment $AB = 2$, and points $A$ and $B$ move on the two coordinate axes, then the trajectory equation of the midpoint $M$ of $AB$ is?",
    "fact_expressions": "A: Point;B: Point;LineSegmentOf(A, B) = 2;PointOnCurve(A, axis);PointOnCurve(B, axis);M: Point;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2 + y^2 = 1",
    "fact_spans": "[[[12, 15]], [[16, 19]], [[1, 10]], [[12, 27]], [[12, 27]], [[36, 39]], [[29, 39]]]",
    "query_spans": "[[[36, 46]]]",
    "process": "Let M(x,y), then A(2x,0), B(0,2y). Since |AB| = 2, we have \\sqrt{4x^{2}+4y^{2}}=2, that is, x^{2}+y^{2}=1. Therefore, the trajectory equation of the midpoint M of AB is x^{2}+y^{2}=1."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and $P$ is a point on the ellipse such that $\\angle F_{1} P F_{2}=60^{\\circ}$, then the area of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);AngleOf(F1, P, F2)=ApplyUnit(60,degree)",
    "query_expressions": "Area(TriangleOf(P,F1,F2))",
    "answer_expressions": "3*sqrt(3)",
    "fact_spans": "[[[18, 56], [66, 68]], [[62, 65]], [[2, 9]], [[10, 17]], [[18, 56]], [[2, 61]], [[62, 71]], [[73, 106]]]",
    "query_spans": "[[[107, 137]]]",
    "process": ""
  },
  {
    "text": "The standard equation of a hyperbola passing through the point $A(-2,0)$ with a focal distance of $6$ is?",
    "fact_expressions": "G: Hyperbola;A: Point;Coordinate(A) = (-2, 0);PointOnCurve(A,G);FocalLength(G)=6",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4 - y^2/5 = 1",
    "fact_spans": "[[[20, 23]], [[2, 12]], [[2, 12]], [[0, 23]], [[13, 23]]]",
    "query_spans": "[[[20, 30]]]",
    "process": ""
  },
  {
    "text": "The equation of the hyperbola with vertices at the foci of the ellipse $3 x^{2}+13 y^{2}=39$ and asymptotes $y = \\pm \\frac{1}{2} x$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (3*x^2 + 13*y^2 = 39);Focus(H)=Vertex(G);Expression(Asymptote(G)) = (y=pm*(1/2)*x)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/10 - y^2/(5/2) = 1",
    "fact_spans": "[[[60, 63]], [[1, 24]], [[1, 24]], [[0, 63]], [[31, 63]]]",
    "query_spans": "[[[60, 67]]]",
    "process": "The foci of the ellipse $3x^{2}+13y^{2}=39$ are $(\\pm\\sqrt{10},0)$, so the vertices of the hyperbola are $(\\pm\\sqrt{10},0)$, giving $a=\\sqrt{10}$. For the hyperbola with asymptotes $y=\\pm\\frac{1}{2}x$, we obtain $b=\\frac{\\sqrt{10}}{2}$. The required equation of the hyperbola is: $\\frac{x^{2}}{10}-\\frac{y^{2}}{\\frac{5}{5}}=1$."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$, point $M(x_{0}, 2 \\sqrt{2})(x_{0}>\\frac{p}{2})$ is a point on parabola $C$, a circle centered at point $M$ with radius $\\sqrt{3}$ intersects the line $x=\\frac{p}{2}$ at points $E$ and $G$, if $|E G|=2 \\sqrt{2}$, then the equation of parabola $C$ is?",
    "fact_expressions": "C: Parabola;p: Number;Z: Circle;H: Line;M: Point;E: Point;G: Point;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Expression(H) = (x = p/2);Coordinate(M) = (x0, 2*sqrt(2));x0:Number;x0>p/2;Focus(C) = F;PointOnCurve(M, C);Center(Z)=M;Radius(Z)=sqrt(3);Intersection(Z,H)={E,G};Abs(LineSegmentOf(E, G)) = 2*sqrt(2)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2 = 4*x",
    "fact_spans": "[[[2, 28], [79, 85], [163, 169]], [[10, 28]], [[111, 112]], [[113, 130]], [[36, 78], [90, 94]], [[132, 135]], [[136, 139]], [[32, 35]], [[10, 28]], [[2, 28]], [[113, 130]], [[36, 78]], [[37, 78]], [[37, 78]], [[2, 35]], [[36, 88]], [[89, 112]], [[97, 112]], [[111, 141]], [[143, 161]]]",
    "query_spans": "[[[163, 174]]]",
    "process": "According to the problem, use the Pythagorean theorem to find the coordinates of point M, substitute into the parabolic equation to find p, and thus obtain the parabolic equation. From the problem, draw MN\\bot x=\\frac{p}{2}, intersecting x=\\frac{p}{2} at point N. It is known that |MN|=|ME|^{2}-|NE|, so |MN|=1. Then M(1+\\frac{p}{2},2\\sqrt{2}), substitute into the parabola y^{2}=2px, yielding 8=2p(1+\\frac{p}{2}). Rearranging gives p^{2}+2p-8=0, solving yields p=2 or p=-4 (discarded). Therefore, the equation of the parabola is y^{2}=4x."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, respectively. What are the coordinates of points $F_{1}$ and $F_{2}$?",
    "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/16 = 1);LeftFocus(G) = F1;RightFocus(G) = F2",
    "query_expressions": "Coordinate(F1);Coordinate(F2)",
    "answer_expressions": "(-3,0);(3,0)",
    "fact_spans": "[[[19, 58]], [[1, 8], [66, 74]], [[9, 16], [75, 82]], [[19, 58]], [[1, 64]], [[1, 64]]]",
    "query_spans": "[[[66, 89]], [[66, 89]]]",
    "process": "From the standard equation of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, we obtain $a^{2}=25$, $b^{2}=16$, so $c^{2}=a^{2}-b^{2}=25-16=9$, that is, $c=3$. Then the coordinates of points $F_{1}$, $F_{2}$ are $(-3,0)$, $(3,0)$ respectively."
  },
  {
    "text": "Given the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$ with foci $F_{1}$ and $F_{2}$, and a point $M$ on the hyperbola such that $\\overrightarrow{M F_{1}} \\cdot \\overrightarrow{M F_{2}}=0$, then the distance from point $M$ to the $x$-axis is equal to?",
    "fact_expressions": "G: Hyperbola;M: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2/2 = 1);Focus(G) = {F1, F2};PointOnCurve(M, G);DotProduct(VectorOf(M, F1), VectorOf(M, F2)) = 0",
    "query_expressions": "Distance(M, xAxis)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 30], [55, 58]], [[50, 54], [121, 125]], [[34, 41]], [[42, 49]], [[2, 30]], [[2, 49]], [[50, 59]], [[60, 119]]]",
    "query_spans": "[[[121, 136]]]",
    "process": "According to the problem, the area $ S $ of $ AF_{1}MF_{2} $ is $ b^{2}\\tan\\frac{\\pi}{2} = 2 \\cdot 1 = 2 = \\frac{1}{2} \\cdot |MF_{1}| \\cdot |MF_{2}| $, $ |MF_{1}| \\cdot |MF_{2}| = 4 = \\frac{1}{2}|F_{1}F_{2}|d $, so the required distance is $ d = \\frac{4}{2\\sqrt{3}} = \\frac{2\\sqrt{3}}{3} $."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $E$: $x^{2}+\\frac{y^{2}}{b^{2}}=1$ $(0<b<1)$, respectively. A line $l$ passing through $F_{1}$ intersects the ellipse $E$ at points $A$ and $B$, and $2|A B|=|A F_{2}|+|B F_{2}|$. Then the length of $A B$ is?",
    "fact_expressions": "l: Line;E: Ellipse;b: Number;A: Point;B: Point;F2: Point;F1: Point;0 < b;b < 1;Expression(E) = (x^2 + y^2/b^2 = 1);LeftFocus(E) = F1;RightFocus(E) = F2;PointOnCurve(F1, l);Intersection(l, E) = {A, B};2*Abs(LineSegmentOf(A, B)) = Abs(LineSegmentOf(A, F2)) + Abs(LineSegmentOf(B, F2))",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[80, 85]], [[19, 64], [86, 91]], [[25, 64]], [[94, 97]], [[98, 101]], [[9, 16]], [[1, 8], [72, 79]], [[25, 64]], [[25, 64]], [[19, 64]], [[1, 70]], [[1, 70]], [[71, 85]], [[80, 103]], [[105, 133]]]",
    "query_spans": "[[[135, 144]]]",
    "process": ""
  },
  {
    "text": "Given that $F$ is the left focus of the hyperbola $C$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and $P$, $Q$ are points on $C$. If the length of $PQ$ is equal to twice the length of the imaginary axis, and point $A(5,0)$ lies on segment $PQ$, then the perimeter of $\\triangle PQF$ is?",
    "fact_expressions": "P: Point;Q: Point;C: Hyperbola;A: Point;F: Point;Expression(C) = (x^2/9 - y^2/16 = 1);Coordinate(A) = (5, 0);LeftFocus(C) = F;PointOnCurve(P, C);PointOnCurve(Q, C);2*Length(ImageinaryAxis(C)) = Length(LineSegmentOf(P, Q));PointOnCurve(A, LineSegmentOf(P, Q))",
    "query_expressions": "Perimeter(TriangleOf(P, Q, F))",
    "answer_expressions": "44",
    "fact_spans": "[[[55, 58]], [[59, 62]], [[63, 66], [6, 50]], [[88, 97]], [[2, 5]], [[6, 50]], [[88, 97]], [[2, 54]], [[55, 69]], [[59, 69]], [[6, 87]], [[88, 105]]]",
    "query_spans": "[[[107, 127]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{5}=1$ $(a>0)$ has foci at $(3,0)$, what is the equation of the asymptotes of hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;a: Number;a>0;Expression(C) = (-y^2/5 + x^2/a^2 = 1);Coordinate(OneOf(Focus(C))) = (3, 0)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*(sqrt(5)/2)*x",
    "fact_spans": "[[[2, 54], [69, 75]], [[10, 54]], [[10, 54]], [[2, 54]], [[2, 67]]]",
    "query_spans": "[[[69, 83]]]",
    "process": "From the given condition, $ c = \\sqrt{a^{2} + 5} = 3 $, solving gives $ a = 2 $. Also, $ b = \\sqrt{5} $, so the asymptotes of hyperbola $ C $ are $ y = \\pm\\frac{b}{a}x = \\pm\\frac{\\sqrt{5}}{2}x $."
  },
  {
    "text": "Given the ellipse equation $\\frac{x^{2}}{2}+y^{2}=1$, find the equation of the line on which the chord passing through point $P(\\frac{1}{2}, \\frac{1}{2})$ and bisected by $P$ lies.",
    "fact_expressions": "G: Ellipse;H: LineSegment;P: Point;Coordinate(P) = (1/2, 1/2);Expression(G) = (x^2/2 + y^2 = 1);IsChordOf(H,G);MidPoint(H)=P;PointOnCurve(P,H)",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "2*x+4*y-3=0",
    "fact_spans": "[[[2, 4]], [], [[36, 66], [68, 71]], [[36, 66]], [[2, 32]], [[2, 75]], [[2, 75]], [[2, 75]]]",
    "query_spans": "[[[2, 84]]]",
    "process": "Let the chord intersect the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ at points $A(x_{1},y_{1})$ and $B(x_{2},y_{2})$. By the midpoint formula, we have $x_{1}+x_{2}=1$, $y_{1}+y_{2}=1$. Substituting $A(x_{1},y_{1})$ and $B(x_{2},y_{2})$ into $\\frac{x^{2}}{2}+y^{2}=1$, and subtracting the equations, we obtain $(x_{1}-x_{2})+2(y_{1}-y_{2})=0$, $\\therefore k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{1}{2}$. Therefore, the equation of the line containing this chord is $y-\\frac{1}{2}=-\\frac{1}{2}(x-\\frac{1}{2})$, that is, $2x+4y-3=0$."
  },
  {
    "text": "The focal distance of the ellipse $C$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/25 + y^2/9 = 1)",
    "query_expressions": "FocalLength(C)",
    "answer_expressions": "8",
    "fact_spans": "[[[0, 42]], [[0, 42]]]",
    "query_spans": "[[[0, 47]]]",
    "process": "From the given conditions, we have: $a^{2}=25$, $b^{2}=9$, thus $c^{2}=a^{2}-b^{2}=25-9=16$, that is, $c=4$, so the focal distance is $2c=8$."
  },
  {
    "text": "Let a point $P$ on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{3}=1(a>0)$ with foci $F_{1}$, $F_{2}$ also lie on the parabola $y^{2}=\\frac{9}{4} x$, where the focus of the parabola is $F_{3}$. If $|P F_{3}|=\\frac{25}{16}$, then the perimeter of $\\triangle P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Parabola;H: Ellipse;a: Number;P: Point;F1: Point;F2: Point;F3: Point;a>0;Expression(H) = (y^2/3 + x^2/a^2 = 1);Focus(H)={F1,F2};PointOnCurve(P,H);PointOnCurve(P,G);Focus(G) = F3;Abs(LineSegmentOf(P, F3)) = 25/16;Expression(G)=(y^2=(9/4)*x)",
    "query_expressions": "Perimeter(TriangleOf(P, F1, F2))",
    "answer_expressions": "6",
    "fact_spans": "[[[75, 99], [101, 104]], [[20, 66]], [[22, 66]], [[70, 73]], [[4, 11]], [[12, 19]], [[107, 114]], [[22, 66]], [[20, 66]], [[1, 66]], [[20, 73]], [[70, 100]], [[101, 114]], [[116, 141]], [[75, 99]]]",
    "query_spans": "[[[143, 173]]]",
    "process": "Let $ P(x_{0},y_{0}) $, then $ x_{0} + \\frac{9}{16} = \\frac{25}{16} $, so $ x_{0} = 1 $. Substituting into the parabola equation gives $ y_{0} = \\pm\\frac{3}{2} $. Without loss of generality, let the coordinates of point $ P $ be $ (1, \\frac{3}{2}) $. Substituting into the ellipse equation yields $ a^{2} = 4 $. Therefore, the perimeter of $ \\triangle PF_{1}F_{2} $ is $ 2a + 2c = 4 + 2 = 6 $."
  },
  {
    "text": "What is the length of the minor axis of the ellipse $9 x^{2}+4 y^{2}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (9*x^2 + 4*y^2 = 1)",
    "query_expressions": "Length(MinorAxis(G))",
    "answer_expressions": "2/3",
    "fact_spans": "[[[0, 21]], [[0, 21]]]",
    "query_spans": "[[[0, 27]]]",
    "process": "The standard equation of the ellipse is \\frac{x^{2}}{9}+\\frac{y^{2}}{1}=1, then a=\\frac{1}{2}, b=\\frac{1}{3}, therefore, the length of the minor axis of the ellipse 9x^{2}+4y^{2}=1 is 2b=\\frac{2}{3}."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{4}-y^{2}=1$, one of its asymptotes intersects the circle $E$: $x^{2}+y^{2}-2 x=0$ at points $A$ and $B$. Then $|A B|=$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/4 - y^2 = 1);E: Circle;Expression(E) = (-2*x + x^2 + y^2 = 0);A: Point;B: Point;Intersection(OneOf(Asymptote(C)), E) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "4*sqrt(5)/5",
    "fact_spans": "[[[2, 35]], [[2, 35]], [[42, 67]], [[42, 67]], [[70, 73]], [[74, 77]], [[2, 79]]]",
    "query_spans": "[[[81, 90]]]",
    "process": "Assume without loss of generality that one asymptote of the hyperbola $ C: \\frac{x^{2}}{4} - y^{2} = 1 $ is $ y = \\frac{1}{2}x $. The standard equation of the circle $ E: x^{2} + y^{2} - 2x = 0 $ is $ (x-1)^{2} + y^{2} = 1 $, hence $ |AB| = 2\\sqrt{r - d^{2}} = 2 $."
  },
  {
    "text": "Given that $F$ is the right focus of the hyperbola $C$: $x^{2}-y^{2}=1$, and $P$ is a point on the left branch of $C$, point $A(0, \\sqrt{2})$. Then the minimum perimeter of $\\triangle APF$ is?",
    "fact_expressions": "C: Hyperbola;A: Point;P: Point;F: Point;Expression(C) = (x^2 - y^2 = 1);Coordinate(A) = (0, sqrt(2));RightFocus(C) = F;PointOnCurve(P, LeftPart(C))",
    "query_expressions": "Min(Perimeter(TriangleOf(A, P, F)))",
    "answer_expressions": "6",
    "fact_spans": "[[[6, 29], [38, 41]], [[49, 66]], [[34, 37]], [[2, 5]], [[6, 29]], [[49, 66]], [[2, 33]], [[34, 47]]]",
    "query_spans": "[[[69, 92]]]",
    "process": "Let the left focus of the hyperbola be F'. From the hyperbola $ C: x^{2} - y^{2} = 1 $, we obtain $ a = 1 $, $ b = 1 $, $ c = \\sqrt{2} $, so $ F(\\sqrt{2}, 0) $, $ F'(-\\sqrt{2}, 0) $. The perimeter of $ \\triangle APF $ is $ |PA| + |PF| + |AF| = |PA| + |PF| + 2 $. By the definition of the hyperbola, $ |PF| - |PF'| = 2a = 2 $, thus $ |PA| + |PF| = |PA| + |PF'| + 2 $. When point P moves along the left branch to make A, P, F' collinear, $ |PA| + |PF'| $ attains its minimum value $ |AF'| = 2 $. Therefore, the minimum perimeter of $ \\triangle APF $ is $ 2 + 2 + 2 = 6 $."
  },
  {
    "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ is $\\sqrt{5}$, then $a$=?",
    "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Eccentricity(G) = sqrt(5)",
    "query_expressions": "a",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 39]], [[57, 60]], [[5, 39]], [[2, 39]], [[2, 54]]]",
    "query_spans": "[[[57, 62]]]",
    "process": "Using the eccentricity formula of a hyperbola and $c^{2}=a^{2}+b^{2}$, we can solve for $a$. The eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ is $\\sqrt{5}$, and since $b=1$, we have $\\frac{c}{a}=\\frac{\\sqrt{a^{2}+b^{2}}}{a}=\\frac{\\sqrt{a^{2}+1}}{a}=\\sqrt{5}$. Solving gives $a=\\frac{1}{2}$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the hyperbola $\\frac{x^{2}}{m^{2}}-\\frac{y^{2}}{n^{2}}=1$ $(m>0, n>0)$ share common foci $F_{1}$, $F_{2}$, point $P$ is an intersection point of the two curves. If $|P F_{1}| \\cdot |P F_{2}|=2$, then the value of $b^{2}+n^{2}$ is?",
    "fact_expressions": "G: Hyperbola;m: Number;n: Number;H: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;m>0;n>0;Expression(G) = (-y^2/n^2 + x^2/m^2 = 1);a > b;b > 0;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);Focus(H) = {F1, F2};Focus(G) = {F1, F2};OneOf(Intersection(H,G))=P;Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 2",
    "query_expressions": "b^2 + n^2",
    "answer_expressions": "2",
    "fact_spans": "[[[55, 111]], [[58, 111]], [[58, 111]], [[2, 54]], [[4, 54]], [[4, 54]], [[132, 136]], [[116, 123]], [[124, 131]], [[58, 111]], [[58, 111]], [[55, 111]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 131]], [[2, 131]], [[132, 145]], [[147, 175]]]",
    "query_spans": "[[[177, 194]]]",
    "process": ""
  },
  {
    "text": "A line $l$ with slope $\\frac{1}{2}$ passes through the focus $F$ of the ellipse $C$: $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$ $(a>b>0)$, intersecting the ellipse at points $A$ and $B$. If $\\overrightarrow{A F}=\\frac{2}{3} \\overrightarrow{A B}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "l: Line;C: Ellipse;b: Number;a: Number;A: Point;F: Point;B: Point;a > b;b > 0;Expression(C) = (x^2/b^2 + y^2/a^2 = 1);Focus(C) = F;PointOnCurve(F, l);Slope(l) = 1/2;Intersection(l, C) = {A, B};VectorOf(A, F) = (2/3)*VectorOf(A, B)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[17, 22]], [[23, 80], [88, 90], [160, 162]], [[30, 80]], [[30, 80]], [[91, 94]], [[83, 86]], [[95, 98]], [[30, 80]], [[30, 80]], [[23, 80]], [[23, 86]], [[17, 86]], [[0, 22]], [[17, 100]], [[102, 157]]]",
    "query_spans": "[[[160, 168]]]",
    "process": ""
  },
  {
    "text": "Given that the tangent to the parabola $x^{2}=12 y$ is perpendicular to the line $x-y=0$, then the equation of the tangent line is?",
    "fact_expressions": "G: Parabola;H: Line;C:Line;Expression(G) = (x^2 = 12*y);Expression(H) = (x - y = 0);IsTangent(C,G);IsPerpendicular(C,H)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x+y+3=0",
    "fact_spans": "[[[2, 17]], [[23, 32]], [], [[2, 17]], [[23, 32]], [[2, 20]], [[2, 32]]]",
    "query_spans": "[[[2, 41]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the ellipse $x^{2}+4 y^{2}=16$ is? The equation of the hyperbola that shares the same foci with this ellipse and has an asymptote $x+\\sqrt{3} y=0$ is?",
    "fact_expressions": "H: Ellipse;Expression(H) = (x^2 + 4*y^2 = 16);G: Hyperbola;Focus(G) = Focus(H);Expression(OneOf(Asymptote(G))) = (x + sqrt(3)*y = 0)",
    "query_expressions": "Eccentricity(H);Expression(G)",
    "answer_expressions": "sqrt(3)/2\nx^2/9-y^2/3=1",
    "fact_spans": "[[[0, 20], [29, 31]], [[0, 20]], [[61, 64]], [[27, 64]], [[38, 64]]]",
    "query_spans": "[[[0, 27]], [[61, 68]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has left and right foci $F_{1}(-2,0)$ and $F_{2}(2,0)$, respectively. $M$ is a point on the right branch of $C$, and $MF_1$ intersects the $y$-axis at point $P$. The incircle of $\\Delta M P F_{2}$ touches side $P F_{2}$ at point $Q$. If $|P Q|=\\sqrt{2}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F2: Point;F1: Point;M: Point;Q: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-2, 0);Coordinate(F2) = (2, 0);LeftFocus(C) = F1;RightFocus(C)=F2;PointOnCurve(M, RightPart(C));Intersection(LineSegmentOf(M, F1), yAxis) = P;TangentPoint(InscribedCircle(TriangleOf(M,P,F2)),LineSegmentOf(P,F2))=Q;Abs(LineSegmentOf(P, Q)) = sqrt(2)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 61], [105, 108], [195, 198]], [[8, 61]], [[8, 61]], [[127, 132]], [[86, 98]], [[70, 83]], [[101, 104]], [[172, 175]], [[8, 61]], [[8, 61]], [[0, 61]], [[70, 83]], [[86, 98]], [[0, 98]], [[0, 98]], [[101, 114]], [[115, 132]], [[134, 175]], [[177, 193]]]",
    "query_spans": "[[[195, 204]]]",
    "process": "According to the tangent segment theorem, calculate $MF_{1}\\cdot MF_{2}$ to find $a$, thereby obtaining the hyperbola's eccentricity. Let the incircle of $\\triangle MPF_{2}$ touch $MF_{1}$ and $MF_{2}$ at points $A$ and $B$, respectively. By the tangent segment theorem, we have $MA=MB$, $PA=PQ$, $BF_{2}=QF_{2}$. Also, $PF_{1}=PF_{2}$. Therefore, $MF_{1}\\cdot MF_{2}=(MA+AP+PF_{1})\\cdot(MB+BF_{2})=PQ+PF_{2}\\cdot QF_{2}=2PQ$. By the definition of the hyperbola, $MF_{1}\\cdot MF_{2}=2a$, thus $a=PQ=\\sqrt{2}$. Since $c=2$, the eccentricity of the hyperbola is $e=\\frac{c}{a}=\\sqrt{2}$. This problem examines properties of a triangle's incircle, the tangent segment theorem, students' computational ability, and using the hyperbola's definition for transformation is key to solving it."
  },
  {
    "text": "Given that the parabola $y = a x^{2}$ intersects the line $y = k x + 1$ at two points, one of which has coordinates $(1,\\ 4)$, then the coordinates of the other point are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = a*x^2);a: Number;H: Line;Expression(H) = (y = k*x + 1);k: Number;A: Point;Coordinate(A) = (1, 4);B: Point;Negation(A=B);NumIntersection(G, H) = 2;OneOf(Intersection(G, H)) = A;OneOf(Intersection(G, H)) = B",
    "query_expressions": "Coordinate(B)",
    "answer_expressions": "(-1/4, 1/4)",
    "fact_spans": "[[[2, 16]], [[2, 16]], [[5, 16]], [[17, 28]], [[17, 28]], [[19, 28]], [], [[2, 49]], [], [[2, 55]], [[2, 32]], [[2, 37]], [[2, 55]]]",
    "query_spans": "[[[2, 60]]]",
    "process": ""
  },
  {
    "text": "On the hyperbola $x^{2}-y^{2}=a^{2}$, a point $P$ is such that the lines connecting $P$ to the two foci $F_{1}$ and $F_{2}$ are perpendicular to each other. Then the area of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;a: Number;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2 = a^2);Focus(G)={F1,F2};IsPerpendicular(LineSegmentOf(P,F1),LineSegmentOf(P,F2));PointOnCurve(P,G)",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "a^2",
    "fact_spans": "[[[0, 22]], [[3, 22]], [[25, 28]], [[32, 39]], [[40, 47]], [[0, 22]], [[0, 47]], [[0, 54]], [[0, 28]]]",
    "query_spans": "[[[56, 83]]]",
    "process": "First, from the given condition we have $ c=\\sqrt{2}a $. Using the Pythagorean theorem, we obtain $ |PF_{1}|^{2}+|PF_{2}|^{2}=(2\\sqrt{2}a)^{2} $. Combining this with $ ||PF_{1}|-|PF_{2}||=2a $, we get $ |PF_{1}||PF_{2}|=2a^{2} $. Then, calculating the area of $ \\triangle PF_{1}F_{2} $ gives the hyperbola $ \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{a^{2}}=1 $, $ c=\\sqrt{a^{2}+a^{2}}=\\sqrt{2}a $. Since $ PF_{1}\\perp PF_{2} $, it follows that $ |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}=(2c)^{2}=(2\\sqrt{2}a)^{2} $ ①. Also, since $ ||PF_{1}|-|PF_{2}||=2a $, we have $ (|PF_{1}|-|PF_{2}|)^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}||PF_{2}|=4a^{2} $ ②. Subtracting ② from ① yields: $ 2|PF_{1}||PF_{2}|=4a^{2} $, that is, $ |PF_{1}||PF_{2}|=2a^{2} $. Therefore, $ S_{\\Delta PF_{1}F_{2}}=\\frac{1}{2}|PF_{1}||PF_{2}|=a^{2} $."
  },
  {
    "text": "Let the parabola $C$: $y^{2}=4x$ have its directrix intersecting the axis of symmetry at point $P$. Draw a tangent line to the parabola $C$ from point $P$. What is the equation of the tangent line?",
    "fact_expressions": "C: Parabola;P: Point;L: Line;Expression(C) = (y^2 = 4*x);Intersection(Directrix(C),SymmetryAxis(C))=P;TangentOfPoint(P, C) = L",
    "query_expressions": "Expression(L)",
    "answer_expressions": "x \\pm y + 1 = 0",
    "fact_spans": "[[[1, 20], [41, 47]], [[30, 34], [36, 40]], [], [[1, 20]], [[1, 34]], [[35, 50]]]",
    "query_spans": "[[[35, 57]]]",
    "process": "The directrix of the parabola $ y^{2}=4x $ is $ x=-1 $, and the axis of symmetry is the $ x $-axis, so the coordinates of point $ P $ are $ (-1,0) $. Let the equation of the tangent line be $ y=k(x+1) $. Substituting into the parabola $ y^{2}=4x $, we obtain $ k^{2}(x+1)^{2}-4x=0 $. When the slope of the tangent line is $ -1 $, the equation of the tangent line is $ y-0=-1\\cdot(x+1) $, that is, $ x+y+1=0 $. When the slope of the tangent line is $ 1 $, the equation of the tangent line is $ y-0=1\\cdot(x+1) $, that is, $ x-y+1=0 $."
  },
  {
    "text": "The equation of the directrix of the parabola $y^{2}=8-4 x$ is? The equation of the circle centered at the vertex of this parabola and tangent to its directrix is?",
    "fact_expressions": "G: Parabola;H: Circle;Expression(G) = (y^2 = 8 - 4*x);Center(H)=Vertex(G);IsTangent(Directrix(G),H)",
    "query_expressions": "Expression(Directrix(G));Expression(H)",
    "answer_expressions": "x=3\n(x-2)^2+y^2=1",
    "fact_spans": "[[[0, 16], [27, 30], [35, 36]], [[41, 42]], [[0, 16]], [[23, 42]], [[34, 42]]]",
    "query_spans": "[[[0, 23]], [[41, 47]]]",
    "process": ""
  },
  {
    "text": "If the line $a x - y + 1 = 0$ passes through the focus of the parabola $y^{2} = 4 x$, then the real number $a$ = ?",
    "fact_expressions": "G: Parabola;a: Real;l:Line;Expression(G) = (y^2 = 4*x);Expression(l) = (a*x - y + 1 = 0);PointOnCurve(Focus(G), l)",
    "query_expressions": "a",
    "answer_expressions": "-1",
    "fact_spans": "[[[16, 30]], [[35, 40]], [[1, 14]], [[16, 30]], [[1, 14]], [[1, 33]]]",
    "query_spans": "[[[35, 42]]]",
    "process": "The focus of the parabola $ y^{2}=4x $ is $ (1,0) $; substituting into the line gives $ a=-1 $."
  },
  {
    "text": "It is known that one focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{2}=1$ coincides with the focus of the parabola $y^{2}=8 x$. Then, what is the eccentricity of this ellipse?",
    "fact_expressions": "H: Ellipse;Expression(H) = (y^2/2 + x^2/a^2 = 1);a: Number;G: Parabola;Expression(G) = (y^2 = 8*x);OneOf(Focus(H)) = Focus(G)",
    "query_expressions": "Eccentricity(H)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[2, 43], [71, 73]], [[2, 43]], [[4, 43]], [[49, 63]], [[49, 63]], [[2, 68]]]",
    "query_spans": "[[[71, 79]]]",
    "process": "The focus of the parabola $ y^{2}=8x $ is $ (2,0) $. Then one focus of the ellipse $ \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{2}=1 $ is $ (2,0) $, so $ c=2 $. Also $ b=\\sqrt{2} $, thus $ a=\\sqrt{b^{2}+c^{2}}=\\sqrt{6} $. The eccentricity is $ e=\\frac{c}{a}=\\frac{2}{\\sqrt{6}}=\\frac{\\sqrt{6}}{3} $."
  },
  {
    "text": "Given that the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=3(a>b>0)$ and the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ have the same foci, what is the equation of the asymptotes of the hyperbola?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;a1: Number;b1: Number;b1 > 0;a1 > b1;a2: Number;b2: Number;a2 > 0;b2 > 0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2/b^2 + x^2/a^2 = 3);Focus(G) = Focus(H)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(2)/2)*x",
    "fact_spans": "[[[55, 111], [118, 121]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[58, 111]], [[58, 111]], [[58, 111]], [[58, 111]], [[55, 111]], [[2, 54]], [[2, 116]]]",
    "query_spans": "[[[118, 129]]]",
    "process": ""
  },
  {
    "text": "Given the equation of the parabola is $x^{2}=4 y$, $F$ is the focus, point $A(5,4)$, and $P$ is any point on the parabola. Find the minimum value of the sum of the distance from point $P$ to point $A$ and the distance from point $P$ to the directrix of the parabola.",
    "fact_expressions": "G: Parabola;A: Point;F: Point;P: Point;Coordinate(A) = (5, 4);Expression(G) = (x^2 = 4*y);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, A) + Distance(P, Directrix(G)))",
    "answer_expressions": "sqrt(34)",
    "fact_spans": "[[[2, 5], [47, 50], [75, 78]], [[31, 40], [62, 66]], [[23, 26]], [[43, 46], [57, 61], [70, 74]], [[31, 40]], [[2, 20]], [[2, 29]], [[43, 55]]]",
    "query_spans": "[[[57, 91]]]",
    "process": "Substituting $ x=5 $ into $ x^{2}=4y $, we get $ y=\\frac{25}{4}>4 $, so point A is outside the parabola. The focus of the parabola is $ F(0,1) $, and the directrix is $ l: y=-1 $. As shown in the figure, draw $ PB \\bot l $ from point P, intersecting at point B. Then $ PA+PB=PA+PF $. From the figure, it can be seen that when points A, P, and F are collinear, the value of $ PA+PF $ is minimized. Therefore, the minimum value of $ PA+PF $ is $ FA=\\sqrt{5^{2}+(4-1)^{2}}=\\sqrt{34} $. Hence, the minimum sum of the distance from point P to point A and the distance from point P to the directrix of the parabola is $ \\sqrt{34} $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and point $M(4,4)$. If point $P$ is a moving point on the ellipse $C$, then the minimum value of $|P M|-|P F_{1}|$ is?",
    "fact_expressions": "C: Ellipse;M: Point;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/4 + y^2/3 = 1);Coordinate(M) = (4, 4);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, F1)))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 44], [85, 90]], [[69, 78]], [[80, 84]], [[53, 60]], [[61, 68]], [[2, 44]], [[69, 78]], [[2, 68]], [[2, 68]], [[80, 96]]]",
    "query_spans": "[[[98, 121]]]",
    "process": "According to the given condition, it can be transformed into |PM| - |PF_{1}| = |PM| + |PF_{2}| - 4, then the answer is obtained by the fact that the shortest distance between two points is a straight line when three points are collinear. From the given, a^{2} = 4, b^{2} = 3, c^{2} = a^{2} - b^{2} = 1, F_{2}(1,0). Since |PF_{2}| + |PF_{1}| = 2a = 4, we have |PF_{1}| = 4 - |PF_{2}|. Therefore, |PM| - |PF_{1}| = |PM| - (4 - |PF_{2}|) = |PM| + |PF_{2}| - 4. Thus, when points M, P, F_{2} are collinear, |PM| + |PF_{2}| - 4 is minimized, that is, |PM| + |PF_{2}| - 4 = |MF_{2}| - 4 = \\sqrt{3^{2} + 4^{2}} - 4 = 1."
  },
  {
    "text": "Let point $F$ be the focus of the parabola $y^{2}=2 p x(p>0)$. Draw a perpendicular from a point $P$ on the parabola to its directrix, with foot $Q$. Given that line $F Q$ intersects the $y$-axis at point $M(0,2)$ and the area of $\\triangle P Q F$ is $10$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;Q: Point;F: Point;M: Point;P: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(P, G);PointOnCurve(P, H);IsPerpendicular(Directrix(G), H);FootPoint(Directrix(G), H) = Q;Coordinate(M) = (0, 2);Intersection(LineOf(F, Q), yAxis) = M;Area(TriangleOf(P, Q, F)) = 10",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2 = 4*x, y^2 = 16*x}",
    "fact_spans": "[[[6, 27], [32, 35], [42, 43], [109, 112]], [[9, 27]], [], [[52, 55]], [[1, 5]], [[71, 80]], [[38, 41]], [[9, 27]], [[6, 27]], [[1, 30]], [[32, 41]], [[31, 48]], [[31, 48]], [[31, 55]], [[71, 80]], [[58, 80]], [[81, 106]]]",
    "query_spans": "[[[109, 117]]]",
    "process": "According to the problem, construct the figure as shown: where $ F\\left(\\frac{p}{2},0\\right) $, $ QE $ is the directrix of the hyperbola, and the equation of the directrix is $ x = -\\frac{p}{2} $, $ PQ \\perp QE $, $ A(0,2) $. Let $ P(x_{0},y_{0}) $, then $ Q\\left(-\\frac{p}{2},y_{0}\\right) $, $ |PQ| = x_{0} + \\frac{p}{2} $. In quadrilateral $ AQEF $, $ O $ is the midpoint of $ EF $, so $ A $ is the midpoint of $ QF $, thus $ |QE| = 4 $, $ y_{0} = 4 $. Since the area of $ APQF $ is 10, $ \\frac{1}{2}\\left(x_{0} + \\frac{p}{2}\\right) \\times 4 = 10 $, so $ x_{0} = 5 - \\frac{p}{2} $. $ x_{1} = x_{2} $, therefore $ 4^{2} = 2p\\left(5 - \\frac{p}{2}\\right) $, i.e., $ p^{2} - 10p + 16 = 0 $. Hence, $ p = 2 $ or $ p = 8 $. Therefore, the equation of the parabola is $ y^{2} = 4x $ or $ y^{2} = 16x $."
  },
  {
    "text": "Given $F_{1}(-1 , 0)$, $F_{2}(1 , 0)$ are the two foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$. If a point $P$ on the ellipse satisfies $| P F_{1}|+| P F_{2} |=4$, then the eccentricity $e$ of the ellipse =?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;F2: Point;P: Point;e: Number;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-1, 0);Coordinate(F2) = (1, 0);Focus(G) = {F1, F2};PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 4;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "1/2",
    "fact_spans": "[[[35, 80], [87, 89], [125, 127]], [[37, 80]], [[37, 80]], [[2, 17]], [[19, 34]], [[92, 95]], [[131, 134]], [[35, 80]], [[2, 17]], [[19, 34]], [[2, 85]], [[87, 95]], [[97, 123]], [[125, 134]]]",
    "query_spans": "[[[131, 136]]]",
    "process": ""
  },
  {
    "text": "The equation of an ellipse that has the same foci as the ellipse $9 x^{2} + 4 y^{2} = 36$ and a minor axis length of $4 \\sqrt{5}$ is?",
    "fact_expressions": "G: Ellipse;C:Ellipse;Expression(G) = (9*x^2 + 4*y^2 = 36);Length(MinorAxis(C))= 4*sqrt(5);Focus(G)=Focus(C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/20+y^2/25=1",
    "fact_spans": "[[[1, 23]], [[48, 50]], [[1, 23]], [[30, 50]], [[0, 50]]]",
    "query_spans": "[[[48, 54]]]",
    "process": ""
  },
  {
    "text": "The line $a x + y - 4 = 0$ and the parabola $y^{2} = 2 p x$ ($p > 0$) intersect at the point $(1, 2)$. Then, the distance from the focus of the parabola to this line is equal to?",
    "fact_expressions": "H: Line;Expression(H) = (a*x + y - 4 = 0);a: Number;G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;Coordinate(OneOf(Intersection(H, G))) = (1, 2)",
    "query_expressions": "Distance(Focus(G), H)",
    "answer_expressions": "(2*sqrt(5))/5",
    "fact_spans": "[[[0, 13], [60, 62]], [[0, 13]], [[2, 13]], [[14, 35], [52, 55]], [[14, 35]], [[17, 35]], [[17, 35]], [[0, 50]]]",
    "query_spans": "[[[52, 68]]]",
    "process": "From the given conditions, we have $ a+2-4=0 $, solving gives $ a=2 $, $ 4=2p $, solving gives $ p=2 $. Therefore, the distance from the focus of the parabola $ (1,0) $ to the line $ 2x+y-4=0 $ is equal to $ \\frac{|2+0-4|}{\\sqrt{5}} = \\frac{2\\sqrt{5}}{5} $."
  },
  {
    "text": "The endpoints of a line segment $MN$ with fixed length $4$ move along the parabola $y^{2}=x$. Let point $P$ be the midpoint of segment $MN$. Then the minimum distance from point $P$ to the $y$-axis is?",
    "fact_expressions": "G: Parabola;M: Point;N: Point;P1: Point;P2: Point;P: Point;Expression(G) = (y^2 = x);Length(LineSegmentOf(M, N)) = 4;Endpoint(LineSegmentOf(M, N)) = {P1, P2};PointOnCurve(P1, G);PointOnCurve(P2, G);MidPoint(LineSegmentOf(M, N)) = P",
    "query_expressions": "Min(Distance(P, yAxis))",
    "answer_expressions": "7/4",
    "fact_spans": "[[[19, 31]], [[9, 14]], [[9, 14]], [], [], [[36, 40], [53, 57]], [[19, 31]], [[0, 14]], [[7, 18]], [[7, 34]], [[7, 34]], [[36, 51]]]",
    "query_spans": "[[[53, 70]]]",
    "process": "Let $ M(x_{1},y_{1}) $, $ N(x_{2},y_{2}) $, and let $ F $ be the focus of the parabola $ y^{2}=x $. The directrix of the parabola is $ x=-\\frac{1}{4} $. The distance sought is:  \n$ S = \\frac{x_{1}+x_{2}}{2} = \\frac{x_{1}+\\frac{1}{4}+x_{2}+\\frac{1}{4}}{2}-\\frac{1}{4} = \\frac{|MF|+|NF|}{2}-\\frac{1}{4} $ (equality holds when the sum of two sides is greater than the third side and points $ M $, $ N $, $ F $ are collinear)  \n$ \\therefore \\frac{|MF|+|NF|}{2}-\\frac{1}{4} \\geqslant \\frac{|MN|}{2}-\\frac{1}{4} = \\frac{7}{4} $"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{8}=1$ with right focus $F$, if a line $l$ passing through $F$ intersects the ellipse $C$ at points $A$ and $B$, then the range of $\\frac{|A F|}{|B F|}$ is?",
    "fact_expressions": "l: Line;C: Ellipse;A: Point;F: Point;B: Point;Expression(C) = (x^2/9 + y^2/8 = 1);RightFocus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B}",
    "query_expressions": "Range(Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(B, F)))",
    "answer_expressions": "[1/2, 2]",
    "fact_spans": "[[[59, 64]], [[2, 44], [65, 70]], [[72, 75]], [[49, 52], [55, 58]], [[76, 79]], [[2, 44]], [[2, 52]], [[54, 64]], [[59, 81]]]",
    "query_spans": "[[[83, 111]]]",
    "process": ""
  },
  {
    "text": "Given $P(-4,-4)$, point $Q$ is a moving point on the ellipse $x^{2}+my^{2}=16$ with eccentricity $\\frac{\\sqrt{2}}{2}$ and foci on the $x$-axis, and $M$ is a point on segment $PQ$ satisfying $PM = \\frac{1}{3} MQ$. Then the equation of the trajectory of moving point $M$ is?",
    "fact_expressions": "G: Ellipse;m: Number;P: Point;Q: Point;M: Point;Expression(G) = (m*y^2 + x^2 = 16);Coordinate(P) = (-4, -4);PointOnCurve(Q,G);Eccentricity(G)=sqrt(2)/2;PointOnCurve(Focus(G),xAxis);PointOnCurve(M, LineSegmentOf(P, Q));LineSegmentOf(P, M) = LineSegmentOf(M, Q)*(1/3)",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "(x+3)^2+2*(y+3)^2=1",
    "fact_spans": "[[[52, 71]], [[54, 71]], [[2, 12]], [[13, 17]], [[76, 79], [119, 122]], [[52, 71]], [[2, 12]], [[13, 75]], [[18, 71]], [[43, 71]], [[76, 90]], [[94, 115]]]",
    "query_spans": "[[[119, 129]]]",
    "process": ""
  },
  {
    "text": "The equation of a parabola with vertex at the origin, focus on the coordinate axis, and directrix given by the line $y=-1$ is?",
    "fact_expressions": "G: Parabola;H: Line;Expression(H) = (y = -1);O: Origin;Vertex(G) = O;PointOnCurve(Focus(G), axis);Directrix(G) = H",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 = 4*y",
    "fact_spans": "[[[27, 30]], [[15, 23]], [[15, 23]], [[3, 5]], [[0, 30]], [[6, 30]], [[14, 30]]]",
    "query_spans": "[[[27, 34]]]",
    "process": "According to the problem, the vertex of the parabola is at the origin, the focus lies on the coordinate axis, and the line $ y = -1 $ is the directrix. Therefore, the parabola opens upward. Let its equation be $ x^{2} = 2py $ ($ p > 0 $). Then $ -\\frac{p}{2} = -1 $, solving gives $ p = 2 $. Hence, the equation of the required parabola is $ x^{2} = 4y $."
  },
  {
    "text": "Given that the line $x - y - 1 = 0$ is tangent to the parabola $y = ax^{2}$, then $a = $?",
    "fact_expressions": "G: Parabola;a: Number;H: Line;Expression(G) = (y = a*x^2);Expression(H) = (x - y - 1 = 0);IsTangent(H, G)",
    "query_expressions": "a",
    "answer_expressions": "1/4",
    "fact_spans": "[[[14, 27]], [[31, 34]], [[2, 13]], [[14, 27]], [[2, 13]], [[2, 29]]]",
    "query_spans": "[[[31, 36]]]",
    "process": ""
  },
  {
    "text": "If the equation $\\frac{x^{2}}{m+1}+\\frac{y^{2}}{m+2}=1$ represents a hyperbola, then what is the range of values for $m$?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G)=(x^2/(m+1)+y^2/(m+2)=1)",
    "query_expressions": "Range(m)",
    "answer_expressions": "(-2, -1)",
    "fact_spans": "[[[45, 48]], [[51, 54]], [[2, 48]]]",
    "query_spans": "[[[51, 61]]]",
    "process": ""
  },
  {
    "text": "The vertex of the parabola is at the origin, and the focus lies on the $x$-axis. The distance from a point $P(-2, a)$ on the parabola to the focus is $3$. Then $a = $?",
    "fact_expressions": "G: Parabola;O: Origin;Vertex(G) = O;PointOnCurve(Focus(G), xAxis);P: Point;a: Number;Coordinate(P) = (-2, a);PointOnCurve(P, G);Distance(P, Focus(G)) = 3",
    "query_expressions": "a",
    "answer_expressions": "pm*2*sqrt(2)",
    "fact_spans": "[[[0, 3], [19, 22]], [[7, 9]], [[0, 9]], [[0, 18]], [[24, 35]], [[47, 50]], [[24, 35]], [[19, 35]], [[19, 45]]]",
    "query_spans": "[[[47, 52]]]",
    "process": "Let the equation of the parabola be $ y^{2} = -2px $ ($ p > 0 $). Since the distance from point $ P(-2, a) $ on the parabola to the focus is 3, we have $ 2 + \\frac{p}{2} = 3 $, $ \\therefore p = 2 $. Therefore, $ a^{2} = -4 \\times (-2) $, $ \\therefore a = \\pm 2\\sqrt{2} $."
  },
  {
    "text": "The standard equation of a hyperbola with foci on the $y$-axis, focal length $6$, and passing through the point $(0, \\sqrt{5})$ is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (0, sqrt(5));PointOnCurve(Focus(G), yAxis);FocalLength(G) = 6;PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/5 - x^2/4 = 1",
    "fact_spans": "[[[36, 39]], [[19, 35]], [[19, 35]], [[0, 39]], [[9, 39]], [[17, 39]]]",
    "query_spans": "[[[36, 46]]]",
    "process": "The focus is on the y-axis, with a focal length of 6, so c=3; passing through the point (0,\\sqrt{5}) gives a=\\sqrt{5}, and since c^{2}=a^{2}+b^{2}, it follows that b^{2}=4. The standard equation of the hyperbola is: \\frac{y^{2}}{5}-\\frac{x^{2}}{4}=1."
  },
  {
    "text": "What is the distance from a focus $F$ of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ to its asymptote?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/5 = 1);F: Point;OneOf(Focus(G)) = F",
    "query_expressions": "Distance(F, Asymptote(G))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[0, 38], [47, 48]], [[0, 38]], [[43, 46]], [[0, 46]]]",
    "query_spans": "[[[43, 56]]]",
    "process": "\\because the hyperbola equation is \\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1, \\therefore the coordinates of the foci of the hyperbola are (\\pm3,0), the asymptotes are y=\\pm\\frac{\\sqrt{5}}{2}x, i.e., \\frac{\\sqrt{5}}{2}x+y=0. The distance from focus F to its asymptote is d=\\frac{\\sqrt{5}}{\\sqrt{\\frac{5}{4}+1}}=\\sqrt{5}"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$, and point $P$ lies on $C$ such that $|P F_{1}|=2|P F_{2}|$, then $\\cos \\angle F_{1} P F_{2}$=?",
    "fact_expressions": "C: Hyperbola;P: Point;F1: Point;F2: Point;Expression(C) = (x^2 - y^2/3 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Cos(AngleOf(F1, P, F2))",
    "answer_expressions": "1/4",
    "fact_spans": "[[[18, 51], [63, 66]], [[58, 62]], [[2, 9]], [[10, 17]], [[18, 51]], [[2, 57]], [[2, 57]], [[58, 67]], [[68, 90]]]",
    "query_spans": "[[[92, 121]]]",
    "process": "Given $ a=1, b=\\sqrt{3}, c=2 $, and $ \\begin{cases} |PF_1|+|PF_2|=2 \\\\ |PF_1|=2|PF_2| \\end{cases} $, solving yields $ \\begin{cases} |PF_1|=4 \\\\ |PF_2|=2 \\end{cases} $. Then in $ \\triangle PF_1F_2 $, $ |F_1F_2|=2c=4 $, so $ \\cos\\angle F_1PF_2 = \\frac{16+4-16}{2\\times4\\times2} = \\frac{1}{4} $."
  },
  {
    "text": "The equation of the hyperbola that has the same foci as the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ and passes through the point $P(2,1)$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/4 + y^2 = 1);Focus(G) = Focus(H);P: Point;Coordinate(P) = (2, 1);PointOnCurve(P, G) = True",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/2-y^2=1",
    "fact_spans": "[[[46, 49]], [[1, 28]], [[1, 28]], [[0, 49]], [[36, 45]], [[36, 45]], [[35, 49]]]",
    "query_spans": "[[[46, 53]]]",
    "process": ""
  },
  {
    "text": "Given that the asymptotes of a hyperbola are $y=\\pm \\frac{3}{2} x$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*((3/2)*x))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(13)/2, sqrt(13)/3",
    "fact_spans": "[[[33, 36], [1, 4]], [[1, 30]]]",
    "query_spans": "[[[33, 42]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola $y^{2}=4 x$ is $F$, point $A(4,2)$, and $P$ is a point on the parabola such that $P$ does not lie on the line $A F$. Then the minimum perimeter of $\\triangle P A F$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;A: Point;Coordinate(A) = (4,2);P: Point;PointOnCurve(P,G) =True;Negation(PointOnCurve(P,LineOf(A,F)))",
    "query_expressions": "Min(Perimeter(TriangleOf(P, A, F)))",
    "answer_expressions": "5+sqrt(13)",
    "fact_spans": "[[[0, 14], [38, 41]], [[0, 14]], [[18, 21]], [[0, 21]], [[22, 31]], [[22, 31]], [[34, 37], [45, 48]], [[34, 44]], [[45, 58]]]",
    "query_spans": "[[[60, 85]]]",
    "process": "By the given condition, the directrix of the parabola is $ x = -1 $. By the definition of a parabola, $ PF $ equals the distance from $ P $ to the directrix, and $ AF = \\sqrt{13} $. Therefore, to minimize the perimeter of $ \\triangle PAF $, it suffices that the distance from $ A $ to the directrix equals $ AP + PF $, which occurs when $ P $ lies on the line passing through $ A $ and perpendicular to the directrix. At this point, $ l_{\\triangle PAF} = 5 + AF = 5 + \\sqrt{13} $."
  },
  {
    "text": "Pass a line $l$ with slope $2 \\sqrt{2}$ through the focus $F$ of the parabola $y^{2}=8 x$, intersecting the parabola at points $A$ and $B$. Then the standard equation of the circle with $AB$ as diameter is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;l: Line;Slope(l) = 2*sqrt(2);PointOnCurve(F, l);A: Point;B: Point;Intersection(l, G) = {A, B};H: Circle;IsDiameter(LineSegmentOf(A, B), H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "(x-(5/2))^2+(y-sqrt(2))^2=81/4",
    "fact_spans": "[[[1, 15], [44, 47]], [[1, 15]], [[18, 21]], [[1, 21]], [[38, 43]], [[22, 43]], [[0, 43]], [[48, 51]], [[52, 55]], [[38, 57]], [[69, 70]], [[59, 70]]]",
    "query_spans": "[[[69, 77]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), and the midpoint M(x_{0},y_{0}) of AB. From the problem, the equation of line l is y=2\\sqrt{2}(x-2). Substituting into the parabola equation y^{2}=8x and simplifying yields x^{2}-5x+4=0. Thus, x_{0}=\\frac{x_{1}+x_{2}}{2}=\\frac{5}{2}, so y_{0}=2\\sqrt{2}(\\frac{5}{2}-2)=\\sqrt{2}, |AB|=p+x_{1}+x_{2}=9. Therefore, the equation of the circle with AB as diameter is (x-\\frac{5}{2})^{2}+(y-\\sqrt{2})^{2}=\\frac{81}{4}."
  },
  {
    "text": "The equation of the hyperbola that has the same foci as the ellipse $\\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1$ and has $y=\\pm \\frac{4}{3} x$ as asymptotes is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/49 + y^2/24 = 1);Focus(G)=Focus(H);Expression(Asymptote(G))=(y=(pm*4/3)*x)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/16 = 1",
    "fact_spans": "[[[74, 77]], [[1, 40]], [[1, 40]], [[0, 77]], [[47, 77]]]",
    "query_spans": "[[[74, 81]]]",
    "process": ""
  },
  {
    "text": "If a conic section $C$ with center at the origin and coordinate axes as its axes of symmetry has eccentricity $\\sqrt{2}$ and passes through the point $(2,3)$, then what is the equation of the curve $C$?",
    "fact_expressions": "C: ConicSection;H: Point;O: Origin;Coordinate(H) = (2, 3);Center(C) = O;SymmetryAxis(C)=axis;Eccentricity(C)=sqrt(2);PointOnCurve(H,C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2 - x^2 = 5",
    "fact_spans": "[[[17, 24], [55, 60]], [[44, 52]], [[4, 6]], [[44, 52]], [[1, 24]], [[7, 24]], [[17, 40]], [[17, 52]]]",
    "query_spans": "[[[55, 65]]]",
    "process": ""
  },
  {
    "text": "A point $P$ on the hyperbola $\\frac{x^{2}}{64}-\\frac{y^{2}}{36}=1$ is at a distance of $4$ from the right focus of the hyperbola. What is the distance from point $P$ to the left directrix?",
    "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2/64 - y^2/36 = 1);PointOnCurve(P, G);Distance(P, RightFocus(G)) = 4",
    "query_expressions": "Distance(P,LeftDirectrix(G))",
    "answer_expressions": "16",
    "fact_spans": "[[[0, 40], [47, 50]], [[43, 46], [63, 67]], [[0, 40]], [[0, 46]], [[43, 60]]]",
    "query_spans": "[[[47, 76]]]",
    "process": ""
  },
  {
    "text": "The minimum distance from a point on the ellipse $x^{2}+\\frac{y^{2}}{3}=1$ to the line $x+y-4=0$ is?",
    "fact_expressions": "G: Ellipse;H: Line;P: Point;Expression(G) = (x^2 + y^2/3 = 1);Expression(H) = (x + y - 4 = 0);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, H))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[0, 27]], [[31, 42]], [[29, 30]], [[0, 27]], [[31, 42]], [[0, 30]]]",
    "query_spans": "[[[29, 51]]]",
    "process": "Let the equation of the line parallel to the line $x + y - 4 = 0$ be $x + y + m = 0$, so $y = -x - m$. Substituting into the ellipse equation gives $4x^{2} + 2mx + m^{2} - 3 = 0$. Let $\\Delta = 4m^{2} - 16(m^{2} - 3) = 0$, therefore $m = 2$ or $m = -2$. When $m = 2$, the distance between the parallel lines is $\\frac{2 + 4}{\\sqrt{1^{2} + 1^{2}}} = 3\\sqrt{2}$; when $m = -2$, the distance between the parallel lines is $\\frac{|-2 + 4|}{\\sqrt{1^{2} + 1^{2}}} = \\sqrt{2}$. So the minimum distance is $2\\sqrt{2}$."
  },
  {
    "text": "If the equation $\\frac{y^{2}}{4}-\\frac{x^{2}}{m-1}=1$ represents a hyperbola, then the range of real values for $m$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/(m - 1) + y^2/4 = 1);m: Real",
    "query_expressions": "Range(m)",
    "answer_expressions": "(1,+\\infty)",
    "fact_spans": "[[[42, 45]], [[1, 45]], [[47, 52]]]",
    "query_spans": "[[[47, 59]]]",
    "process": "From the given condition, we have: $m-1>0$, solving yields $m>1$."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. Point $P$ lies on the ellipse $C$. If the midpoint of segment $P F_{1}$ lies on the $y$-axis and $\\angle P F_{1} F_{2}=30^{\\circ}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "F1: Point;F2: Point;C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C);PointOnCurve(MidPoint(LineSegmentOf(P, F1)), yAxis);AngleOf(P, F1, F2) = ApplyUnit(30, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[1, 8]], [[9, 16]], [[19, 76], [88, 93], [152, 154]], [[19, 76]], [[26, 76]], [[26, 76]], [[26, 76]], [[26, 76]], [[1, 82]], [[1, 82]], [[83, 87]], [[83, 94]], [[96, 116]], [[117, 150]]]",
    "query_spans": "[[[152, 160]]]",
    "process": "As shown in the figure: since the midpoint of segment PF lies on the y-axis, PF₂ // y-axis, then PF₂ ⊥ x-axis, so PF₂ = b²/a. Because ∠PF₁F₂ = 30°, tan30° = PF₂ / F₁F₂ = b²/(2ac) = √3/3, thus 3c² + 2√3ac - 3a² = 0, so 3e² + 2√3e - 3 = 0, solving gives: e = √3/3."
  },
  {
    "text": "Given the ellipse equation $9 x^{2}+16 y^{2}=144$, what is its eccentricity?",
    "fact_expressions": "G: Ellipse;Expression(G) = (9*x^2 + 16*y^2 = 144)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(7)/4",
    "fact_spans": "[[[2, 4], [31, 32]], [[2, 29]]]",
    "query_spans": "[[[31, 38]]]",
    "process": "According to the problem, the equation of the ellipse is $9x^{2}+16y^{2}=144$, which is $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$. Therefore, $a^{2}=16$, $b^{2}=9$, that is, $a=4$, $b=3$. Also, $c^{2}=a^{2}-b^{2}$, so $c=\\sqrt{7}$. Thus, the eccentricity of the ellipse is $e=\\frac{c}{a}=\\frac{\\sqrt{7}}{4}$."
  },
  {
    "text": "Given that $P$ is any point on the right branch of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, $F$ is the right focus of the hyperbola, and the fixed point $A$ has coordinates $(3, \\sqrt{3})$, then the minimum value of $|P F|+|P A|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1);P: Point;PointOnCurve(P, RightPart(G));F: Point;RightFocus(G) = F;A: Point;Coordinate(A) = (3, sqrt(3))",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "2*sqrt(7)-2",
    "fact_spans": "[[[6, 34], [47, 50]], [[6, 34]], [[2, 5]], [[2, 42]], [[43, 46]], [[43, 54]], [[57, 60]], [[57, 79]]]",
    "query_spans": "[[[81, 100]]]",
    "process": "By the given conditions, in the hyperbola $ a=1 $, $ b=\\sqrt{3} $, $ c=2 $. Let $ F_{1} $ be the left focus of the hyperbola, so $ F_{1}(-2,0) $. Since $ A $ is on the right side, $ |PF|=|PF_{1}|-2 $, therefore $ |PF|+|PA|=|PF_{1}|+|PA|-2 \\geqslant |AF_{1}|-2 = \\sqrt{(3+2)^{2}+(\\sqrt{3}-0)^{2}}-2 = 2\\sqrt{7}-2 $. Equality holds if and only if $ F_{1} $, $ P $, $ A $ are collinear. Therefore, the minimum value of $ |PF|+|PA| $ is $ 2\\sqrt{7}-2 $."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of a hyperbola, respectively, and let $P$ be any point on the right branch of the hyperbola. If the minimum value of $\\frac{P F_{1}^{2}}{P F_{2}}$ is exactly $4$ times the length of the real axis, then the range of values for the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));Min(LineSegmentOf(P, F1)^2/LineSegmentOf(P, F2))=4*Length(RealAxis(G))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1, 3]",
    "fact_spans": "[[[17, 20], [31, 34], [86, 89]], [[27, 30]], [[1, 8]], [[9, 16]], [[1, 26]], [[1, 26]], [[27, 40]], [[31, 83]]]",
    "query_spans": "[[[86, 99]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}(-c , 0)$, $F_{2}(c, 0)$ be the two foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and let $P$ be a point of intersection of the circle with diameter $|F_{1}F_{2}|$ and the ellipse, such that $\\angle PF_{1} F_{2}=5 \\angle PF_{2} F_{1}$. Then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Circle;F1: Point;F2: Point;P: Point;c:Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);Focus(G)={F1,F2};IsDiameter(Abs(LineSegmentOf(F1,F2)),H);OneOf(Intersection(H,G))=P;AngleOf(P, F1, F2) = 5*AngleOf(P, F2, F1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[33, 85], [116, 118], [171, 173]], [[35, 85]], [[35, 85]], [[114, 115]], [[1, 17]], [[19, 32]], [[91, 94]], [[19, 32]], [[35, 85]], [[35, 85]], [[33, 85]], [[1, 17]], [[19, 32]], [[1, 90]], [[95, 115]], [[91, 123]], [[125, 168]]]",
    "query_spans": "[[[171, 179]]]",
    "process": ""
  },
  {
    "text": "Given point $P$ is any point on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$, and the distances from point $P$ to the two asymptotes of the hyperbola are $d_{1}$ and $d_{2}$ respectively. If $3 a^{2}+b^{2} \\leq 8 d_{1} d_{2}$, then what is the range of values for the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;a>0;b>0;L1:Line;L2:Line;d1:Number;d2:Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);Asymptote(G)={L1,L2};Distance(P,L1)=d1;Distance(P,L2)=d2;3*a^2+b^2<=8*d1*d2",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[\\sqrt{2},2]",
    "fact_spans": "[[[7, 64], [74, 77], [142, 145]], [[10, 64]], [[10, 64]], [[69, 73], [2, 6]], [[10, 64]], [[10, 64]], [], [], [[88, 95]], [[96, 103]], [[7, 64]], [[2, 68]], [[74, 82]], [[69, 103]], [[69, 103]], [[105, 139]]]",
    "query_spans": "[[[142, 155]]]",
    "process": "By the given condition, the asymptotes of the hyperbola are $ ay \\pm bx = 0 $. If $ P(x, y) $, without loss of generality, let $ d_{1} = \\frac{|ay + bx|}{\\sqrt{a^{2} + b^{2}}} $, $ d_{2} = \\frac{|ay - bx|}{\\sqrt{a^{2} + b^{2}}} $. Therefore, $ d_{1}d_{2} = \\frac{|ay + bx|}{\\sqrt{a^{2} + b^{2}}} \\cdot \\frac{|ay - bx|}{\\sqrt{a^{2} + b^{2}}} = \\frac{|a^{2}y^{2} - b^{2}x^{2}|}{a^{2} + b^{2}} $. Since $ P $ lies on the hyperbola and $ a^{2} + b^{2} = c^{2} $, we have $ 3a^{2} + b^{2} \\leqslant 8d $, $ d_{2} = \\frac{8a^{2}b^{2}}{c^{2}} $, simplifying yields $ 8a^{4} - 6a^{2}c^{2} + c^{4} \\leqslant 0 $, then $ e^{4} - 6e^{2} + 8 \\leqslant 0 $, solving gives $ 2 \\leqslant e^{2} \\leqslant 4 $, and since $ e > 1 $, $ \\therefore \\sqrt{2} < e < 2 $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, the right vertex of the ellipse is $A$, and point $M(2,4)$. Draw a perpendicular line from any point $P$ on the ellipse $C$ to the line $MA$, with the foot of the perpendicular being $H$. Then the minimum value of $2|PM|+|PH|$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);A: Point;RightVertex(C) = A;M: Point;Coordinate(M) = (2, 4);P: Point;PointOnCurve(P, C) ;L: Line;PointOnCurve(P, L) ;IsPerpendicular(L, LineOf(M, A)) ;FootPoint(L, LineOf(M, A)) = H;H: Point",
    "query_expressions": "Min(Abs(LineSegmentOf(P, H)) + 2*Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "2*sqrt(17) - 2",
    "fact_spans": "[[[2, 44], [66, 71]], [[2, 44]], [[49, 52]], [[2, 52]], [[54, 63]], [[54, 63]], [[76, 79]], [[66, 79]], [], [[65, 90]], [[65, 90]], [[65, 98]], [[95, 98]]]",
    "query_spans": "[[[101, 121]]]",
    "process": "In the ellipse, $a=2$, $c=1$, so the coordinates of the right focus are $F(2,0)$, and the equation of the right directrix is $x=4$. Draw a perpendicular line from point $P$ to the right directrix, and let the foot of the perpendicular be $G$. Then $|PH|=|PG|-2$. By the second definition of the ellipse, $e=\\frac{|PF|}{|PG|}=\\frac{1}{2}$, so $|PG|=2|PF|$. Therefore, $2|PM|+|PH|=2|PM|+2|PF|-2=2(|PM|+|PF|)-2\\geqslant2|MF|-2=2\\sqrt{17}-2$, with equality holding if and only if points $M$, $P$, and $F$ are collinear. Thus, the minimum value of $2|PM|+|PH|$ is $2\\sqrt{17}-2$. Answer: $2\\sqrt{17}-2$"
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $\\frac{y^{2}}{25}-\\frac{x^{2}}{16}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/16 + y^2/25 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(5/4)*x",
    "fact_spans": "[[[0, 40]], [[0, 40]]]",
    "query_spans": "[[[0, 48]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $ C: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $), let $ F_{1} $ and $ F_{2} $ be the left and right foci of the ellipse, respectively. If there exists a point $ P(x_{0}, y_{0}) $ on the ellipse $ C $ ($ x_{0} \\geq 0 $) such that $ \\angle P F_{1} F_{2} = 60^{\\circ} $, then what is the range of values for the eccentricity of the ellipse?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C) = True;Coordinate(P) = (x0, y0);x0: Number;y0: Number;x0 >= 0;AngleOf(P, F1, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(0, 1/2]",
    "fact_spans": "[[[2, 59], [81, 83], [91, 96], [168, 170]], [[2, 59]], [[8, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[61, 69]], [[71, 78]], [[61, 89]], [[61, 89]], [[99, 131]], [[91, 131]], [[99, 131]], [[100, 131]], [[100, 131]], [[100, 131]], [[133, 166]]]",
    "query_spans": "[[[168, 181]]]",
    "process": "According to the problem, when point P is at the upper (or lower) vertex of the ellipse, $\\angle PF_{1}F_{2}$ is maximized. To ensure that there exists a point $P(x_{0},y_{0})$ $(x_{0}\\geqslant0)$ on the ellipse $C$ such that $\\angle PF_{1}F_{2}=60^{\\circ}$, we must have $90^{\\circ}>(\\angle PF_{1}F_{2})_{\\max}\\geqslant60^{\\circ}$. Therefore, $\\cos90^{\\circ}<\\cos(\\angle PF_{1}F_{2})_{\\max}\\leqslant\\cos60^{\\circ}=\\frac{1}{2}$, that is: $0<\\frac{c}{a}\\leqslant\\frac{1}{2}$. Hence, the range of eccentricity of the ellipse is $0<e\\leqslant\\frac{1}{2}$."
  },
  {
    "text": "Given that point $P$ is a moving point on the parabola $x^{2}=4 y$, $F$ is the focus of the parabola, and the coordinates of point $A$ are $(0,-1)$, then the minimum value of $\\frac{P F}{P A}$ is?",
    "fact_expressions": "P: Point;F: Point;C: Parabola;Expression(C) = (x**2=4*y);PointOnCurve(P, C);Focus(C) = F;A: Point;Coordinate(A) = (0, -1)",
    "query_expressions": "Min(LineSegmentOf(P, F) / LineSegmentOf(P, A))",
    "answer_expressions": "sqrt(2) / 2",
    "fact_spans": "[[[2, 6]], [[25, 28]], [[7, 21], [29, 32]], [[7, 21]], [[2, 24]], [[25, 35]], [[36, 40]], [[36, 52]]]",
    "query_spans": "[[[54, 77]]]",
    "process": ""
  },
  {
    "text": "Let $m>0$, the hyperbola $M$: $\\frac{x^{2}}{4}-y^{2}=1$ is tangent to the circle $N$: $x^{2}+(y-m)^{2}=5$, $A(-\\sqrt{5}, 0)$, $B(\\sqrt{5}, 0)$. If there exists a point $P$ on the circle $N$ such that $|P A|-|P B|=4$, then the distance from point $P$ to the $x$-axis is?",
    "fact_expressions": "m: Number;m > 0;M: Hyperbola;Expression(M) = (x^2/4 - y^2 = 1);N: Circle;Expression(N) = (x^2 + (-m + y)^2 = 5);IsTangent(M, N);A: Point;Coordinate(A) = (-sqrt(5), 0);B: Point;Coordinate(B) = (sqrt(5), 0);P: Point;PointOnCurve(P, N);Abs(LineSegmentOf(P, A)) - Abs(LineSegmentOf(P, B)) = 4",
    "query_expressions": "Distance(P, xAxis)",
    "answer_expressions": "sqrt(5)/10",
    "fact_spans": "[[[1, 6]], [[1, 6]], [[7, 40]], [[7, 40]], [[41, 66], [106, 110]], [[41, 66]], [[7, 68]], [[69, 86]], [[69, 86]], [[88, 104]], [[88, 104]], [[115, 118], [137, 141]], [[106, 118]], [[120, 135]]]",
    "query_spans": "[[[137, 151]]]",
    "process": "According to the given conditions, in the hyperbola $ a = 2 $, $ c = \\sqrt{5} $, it is clear that points $ A $ and $ B $ are the left and right foci of the hyperbola. Since point $ P $ satisfies $ |PA| - |PB| = 4 = 2a $, point $ P $ is the tangent point of the hyperbola and the circle, and lies on the right branch of the hyperbola. From the circle equation, its center is $ C(0, m) $, with radius $ \\sqrt{5} $. From\n$$\n\\begin{cases}\n\\frac{x^{2}}{4} - y^2 = 1 \\\\\nx^{2} + (y - m)^{2} = 5\n\\end{cases}\n$$\neliminating $ x $ yields $ 5y^{2} - 2my + m^{2} - 1 = 0 $. By setting discriminant $ \\Delta = (-2m)^{2} - 4 \\times 5 \\times (m^{2} - 1) = 0 $, and since $ m > 0 $, solving gives $ m = \\frac{\\sqrt{5}}{2} $. Then $ 5y^2 - 2 \\cdot \\frac{\\sqrt{5}}{2} y + \\left(\\frac{\\sqrt{5}}{2}\\right)^{2} - 1 = 0 $, solving yields $ y = \\frac{\\sqrt{5}}{10} $, so the required distance is $ \\frac{\\sqrt{5}}{10} $."
  },
  {
    "text": "Given that $F$ is the left focus of the hyperbola $C$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and $P$, $Q$ are points on $C$. If the length of $PQ$ is equal to twice the length of the imaginary axis, and point $A(5,0)$ lies on segment $PQ$, then the perimeter of $\\triangle PQF$ is?",
    "fact_expressions": "C: Hyperbola;P: Point;Q: Point;A: Point;F: Point;Expression(C) = (x^2/9 - y^2/16 = 1);Coordinate(A) = (5, 0);LeftFocus(C) = F;PointOnCurve(P, C);PointOnCurve(Q, C);Length(LineSegmentOf(P,Q))=2*Length(ImageinaryAxis(C));PointOnCurve(A,LineSegmentOf(P,Q))",
    "query_expressions": "Perimeter(TriangleOf(P, Q, F))",
    "answer_expressions": "44",
    "fact_spans": "[[[6, 50], [63, 66]], [[55, 58]], [[59, 62]], [[89, 98]], [[2, 5]], [[6, 50]], [[89, 98]], [[2, 54]], [[55, 69]], [[55, 69]], [[63, 88]], [[89, 107]]]",
    "query_spans": "[[[109, 131]]]",
    "process": "By the given condition, since PQ passes through the right focus (5,0) of the hyperbola, both P and Q lie on the right branch of the hyperbola. Thus, |FP| - |PA| = 6 and |PQ| - |QA| = 6. Adding these two equations and using the definition of the hyperbola yields |FP| + |FQ| = 28. Therefore, the perimeter of $\\triangle PQF$ is |FP| + |FQ| + |PQ| = 28 + 4b = 28 + 16 = 44."
  },
  {
    "text": "The two foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are $F_{1}$ and $F_{2}$. If $P$ is a point on its right branch such that $|P F_{1}|=2|P F_{2}|$, then the range of the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;Focus(G) = {F1,F2};P: Point;PointOnCurve(P,RightPart(G)) = True;Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1,3]",
    "fact_spans": "[[[0, 56], [83, 84], [115, 118]], [[0, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[62, 69]], [[70, 77]], [[0, 77]], [[79, 82]], [[79, 89]], [[91, 113]]]",
    "query_spans": "[[[115, 128]]]",
    "process": "Let the horizontal coordinate of point P be x. According to $|PF_{1}| = 2|PF_{2}|$ and P lying on the right branch of the hyperbola ($x \\geqslant a$), using the second definition of the hyperbola, an expression for x in terms of e can be obtained, and thus the range of e can be determined based on the range of x. [Detailed solution] $\\because |PF_{1}| = 2|PF_{2}|$, P lies on the right branch of the hyperbola ($x \\geqslant a$). According to the second definition of the hyperbola, we have $2e(x - \\frac{a^{2}}{c}) = e(x + \\frac{a^{2}}{c})$. $\\therefore ex = 3a$. $\\because x \\geqslant a$, $\\therefore ex \\geqslant ea$. $3a \\geqslant ea$, $\\therefore e \\leqslant 3$. $\\because e > 1$, $\\therefore 1 < e \\leqslant 3$."
  },
  {
    "text": "The equation of the hyperbola that has the same asymptotes as $x^{2}-4 y^{2}=1$ and passes through $M(4 , \\sqrt{3})$ is?",
    "fact_expressions": "G: Hyperbola;C: Curve;Expression(C) = (x^2 - 4*y^2 = 1);Asymptote(C) = Asymptote(G);M: Point;Coordinate(M) = (4, sqrt(3));PointOnCurve(M, G) = True",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4-y^2=1",
    "fact_spans": "[[[46, 49]], [[1, 18]], [[1, 18]], [[0, 49]], [[28, 45]], [[28, 45]], [[27, 49]]]",
    "query_spans": "[[[46, 53]]]",
    "process": ""
  },
  {
    "text": "The line passing through the left vertex $A$ of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1\\ (a>b>0)$ with slope $1$ intersects the ellipse again at point $M$ and intersects the $y$-axis at point $B$. If $|AM|=|MB|$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;A: Point;M: Point;B: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(G)=A;Slope(H) = 1;Intersection(H,G) = {A, M};Intersection(H,yAxis)=B;Abs(LineSegmentOf(A, M)) = Abs(LineSegmentOf(M, B))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[1, 55], [73, 75], [116, 118]], [[3, 55]], [[3, 55]], [[70, 72]], [[59, 62]], [[82, 85]], [[95, 98]], [[3, 55]], [[3, 55]], [[1, 55]], [[1, 62]], [[63, 72]], [[59, 85]], [[70, 98]], [[100, 113]]]",
    "query_spans": "[[[116, 124]]]",
    "process": "From the given conditions, the coordinates of point A are (-a, 0), and the equation of line l is y = x + a. Therefore, the coordinates of point B are (0, a), so the coordinates of point M are $(-\\frac{a}{2}, \\frac{a}{2})$. Substituting into the ellipse equation gives $\\frac{a^2}{4a^{2}} + \\frac{a^{2}}{4b^{2}} = 1$, which simplifies to $a^{2} = 3b^{2}$, hence $c^{2} = 2b^{2}$. Therefore, $e = \\frac{c}{a} = \\frac{\\sqrt{2}b}{\\sqrt{3}b} = \\frac{\\sqrt{6}}{3}$."
  },
  {
    "text": "What is the eccentricity of the ellipse $9 x^{2}+16 y^{2}=144$?",
    "fact_expressions": "E: Ellipse;Expression(E) = (9*x^2 + 16*y^2 = 144)",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "sqrt(7)/4",
    "fact_spans": "[[[0, 24]], [[0, 24]]]",
    "query_spans": "[[[0, 30]]]",
    "process": ""
  },
  {
    "text": "Given that a line $l$ with slope $2$ passes through the focus $F$ of the parabola $y^{2}=ax$ ($a>0$) and intersects the $y$-axis at point $A$, if the area of $\\triangle OAF$ ($O$ being the origin) is $4$, then the equation of the parabola is?",
    "fact_expressions": "l: Line;Slope(l) = 2;G: Parabola;Expression(G) = (y^2 = a*x);a: Number;a>0;F: Point;Focus(G) = F;PointOnCurve(F, l);A: Point;Intersection(l, yAxis) = A;O: Origin;Area(TriangleOf(O, A, F)) = 4",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=8*x",
    "fact_spans": "[[[9, 14]], [[2, 14]], [[15, 34], [90, 93]], [[15, 34]], [[18, 34]], [[18, 34]], [[37, 40]], [[15, 40]], [[9, 40]], [[50, 54]], [[9, 54]], [[72, 75]], [[56, 88]]]",
    "query_spans": "[[[90, 97]]]",
    "process": ""
  },
  {
    "text": "The equation of a hyperbola is $\\frac{x^{2}}{4-k}+\\frac{y^{2}}{k-2}=1$, then what is the range of values for $k$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(4 - k) + y^2/(k - 2) = 1);k: Number",
    "query_expressions": "Range(k)",
    "answer_expressions": "{(4, +oo), (-oo, 2)}",
    "fact_spans": "[[[0, 3]], [[0, 45]], [[47, 50]]]",
    "query_spans": "[[[47, 57]]]",
    "process": "The equation of the hyperbola is \\frac{x^2}{4-k}+\\frac{y^{2}}{k-2}=1. If the foci are on the x-axis, we have 4-k>0, k-2<0, solving gives k<2; if the foci are on the y-axis, we have 4-k<0, k-2>0, solving gives k>4. In summary, the range of k is k>4 or k<2."
  },
  {
    "text": "Given the left and right foci of a hyperbola are $F_{1}$, $F_{2}$ respectively, and a point $P$ on the right branch of the hyperbola such that $|P F_{1}| = 3|P F_{2}|$, then the range of the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P,RightPart(G));Abs(LineSegmentOf(P,F1))=3*Abs(LineSegmentOf(P,F2))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1,2]",
    "fact_spans": "[[[2, 5], [29, 32], [66, 69]], [[37, 40]], [[12, 19]], [[20, 28]], [[2, 28]], [[2, 28]], [[29, 40]], [[42, 64]]]",
    "query_spans": "[[[66, 77]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is an arbitrary point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{8}=1$, and $AB$ is any diameter of the circle $T$: $(x+1)^{2}+y^{2}=1$, then the range of values of $\\overrightarrow{P A} \\cdot \\overrightarrow{P B}$ is?",
    "fact_expressions": "G: Ellipse;T: Circle;A: Point;B: Point;P: Point;Expression(G) = (x^2/9 + y^2/8 = 1);Expression(T) = (y^2 + (x + 1)^2 = 1);PointOnCurve(P, G);IsDiameter(LineSegmentOf(A, B), T)",
    "query_expressions": "Range(DotProduct(VectorOf(P, A), VectorOf(P, B)))",
    "answer_expressions": "[3,15]",
    "fact_spans": "[[[6, 43]], [[55, 79]], [[49, 54]], [[49, 54]], [[2, 5]], [[6, 43]], [[55, 79]], [[2, 48]], [[49, 86]]]",
    "query_spans": "[[[88, 144]]]",
    "process": "\\overrightarrow{PA}\\cdot\\overrightarrow{PB}=(\\overrightarrow{PT}+\\overrightarrow{TA})\\cdot(\\overrightarrow{PT}+\\overrightarrow{TB})=(\\overrightarrow{PT}+\\overrightarrow{TA})\\cdot(\\overrightarrow{PT}-\\overrightarrow{TA})=\\overrightarrow{PT}^{2}-\\overrightarrow{TA}^{2}=\\overrightarrow{PT}^{2}-1. Let P(x_{0},y_{0}), then \\overrightarrow{PT}^{2}=(x_{0}+1)^{2}+y_{0}^{2}=(x_{0}+1)^{2}+8(1-\\frac{x_{0}^{2}}{9})=\\frac{x_{0}^{2}}{9}+2x_{0}+9=\\frac{1}{9}(x_{0}+9)^{2}. Since x_{0}\\in[-3,3], it follows that \\overrightarrow{PT}^{2}\\in[4,16]. \\therefore \\overrightarrow{PA}\\cdot\\overrightarrow{PB}\\in[3,15]"
  },
  {
    "text": "The line $l$ passes through the point $M(1,1)$, intersects the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ at points $A$ and $B$, and if the midpoint of $AB$ is $M$, what is the equation of the line $l$?",
    "fact_expressions": "l: Line;G: Ellipse;A: Point;B: Point;M: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Coordinate(M) = (1, 1);PointOnCurve(M, l);MidPoint(LineSegmentOf(A, B)) = M;Intersection(l,G)={A,B}",
    "query_expressions": "Expression(l)",
    "answer_expressions": "4*x+3*y-7=0",
    "fact_spans": "[[[0, 5], [81, 86]], [[17, 54]], [[57, 60]], [[61, 64]], [[6, 15], [77, 80]], [[17, 54]], [[6, 15]], [[0, 15]], [[68, 80]], [0, 63]]",
    "query_spans": "[[[81, 90]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Since line $ l $ passes through point $ M(1,1) $ and intersects the ellipse $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $ at points $ A $ and $ B $, we have  \n\\[\n\\begin{cases}\n\\frac{x_{1}^{2}}{4} + \\frac{y_{1}^{2}}{3} = 1 \\\\\n\\frac{x_{2}^{2}}{4} + \\frac{y_{2}^{2}}{3} = 1\n\\end{cases}\n\\]  \nSubtracting the two equations gives:  \n$ 3(x_{1}^{2} - x_{2}^{2}) = -4(y_{1}^{2} - y_{2}^{2}) $,  \ni.e.,  \n$ 3(x_{1} - x_{2})(x_{1} + x_{2}) = -4(y_{1} - y_{2})(y_{1} + y_{2}) $.  \nClearly, the slope of the line exists, so  \n$ k = \\frac{y_{1} - y_{2}}{x_{1} - x_{2}} = -\\frac{4}{3} \\left( \\frac{x_{1} + x_{2}}{y_{1} + y_{2}} \\right) = -\\frac{4 \\times 2}{3 \\times 2} = -\\frac{4}{3} $.  \nTherefore, the equation of line $ l $ is  \n$ y - 1 = -\\frac{4}{3}(x - 1) $,  \ni.e.,  \n$ 4x + 3y - 7 = 0 $."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, $F$ is the right focus of the ellipse $C$. A line $l$ passing through the origin $O$ with nonzero slope intersects the ellipse $C$ at two distinct points $P$ and $Q$. What is the maximum area of $\\Delta P Q F$?",
    "fact_expressions": "l: Line;C: Ellipse;P: Point;Q: Point;F: Point;O: Origin;Expression(C) = (x^2/4 + y^2/3 = 1);RightFocus(C)=F;PointOnCurve(O, l);Negation(Slope(l) = 0);Intersection(l, C) = {P, Q};Negation(P=Q)",
    "query_expressions": "Max(Area(TriangleOf(P, Q, F)))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[75, 80]], [[2, 44], [50, 55], [81, 86]], [[93, 96]], [[97, 100]], [[46, 49]], [[61, 66]], [[2, 44]], [[46, 59]], [[60, 80]], [[67, 80]], [[75, 100]], [73, 97]]",
    "query_spans": "[[[102, 123]]]",
    "process": ""
  },
  {
    "text": "The line $y=b$ intersects the left and right branches of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ at points $B$ and $C$, respectively. If $O B \\perp O C$, where $O$ is the origin, then the equation of the asymptotes of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;O: Origin;B: Point;C: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = b);Intersection(H, LeftPart(G))=B;Intersection(H,RightPart(G))=C;IsPerpendicular(LineSegmentOf(O,B),LineSegmentOf(O,C))",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(2)*x",
    "fact_spans": "[[[8, 64], [112, 115]], [[11, 64]], [[11, 64]], [[0, 7]], [[101, 105]], [[73, 76]], [[77, 80]], [[11, 64]], [[11, 64]], [[8, 64]], [[0, 7]], [[0, 82]], [[0, 82]], [[84, 99]]]",
    "query_spans": "[[[112, 123]]]",
    "process": "The line $y = b$ intersects the left and right branches of the hyperbola $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ $(a > 0, b > 0)$ at points $B$ and $C$, respectively. Solving the system:  \n$$\n\\begin{matrix}\ny = b \\\\\n\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1\n\\end{matrix}\n$$  \nwe obtain $B = (\\sqrt{2}a, b)$, $C = (-\\sqrt{2}a, b)$. Since $OB \\perp OC$, it follows that $\\overrightarrow{OB} \\cdot \\overrightarrow{OC} = -2a^{2} + b^{2} = 0$, so $2a^{2} = b^{2}$, $b = \\pm\\sqrt{2}a$. Therefore, the asymptotes of the hyperbola are $y = \\pm\\frac{b}{a}x = \\pm\\sqrt{2}x$."
  },
  {
    "text": "The equation of the directrix of the parabola $y=ax^{2}$ is $y=\\frac{1}{2}$, then $a$=?",
    "fact_expressions": "G: Parabola;a: Number;Expression(G) = (y = a*x^2);Expression(Directrix(G)) = (y = 1/2)",
    "query_expressions": "a",
    "answer_expressions": "-2",
    "fact_spans": "[[[0, 13]], [[36, 39]], [[0, 13]], [[0, 34]]]",
    "query_spans": "[[[36, 41]]]",
    "process": ""
  },
  {
    "text": "Given that the hyperbola passes through the point $(2,0)$ and has asymptotes with equations $y=\\pm \\frac{1}{2} x$, then the standard equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (2, 0);PointOnCurve(H, G);Expression(Asymptote(G)) =(y=pm*(x/2))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/4-y^2=1",
    "fact_spans": "[[[2, 5], [46, 49]], [[6, 14]], [[6, 14]], [[2, 14]], [[2, 43]]]",
    "query_spans": "[[[46, 56]]]",
    "process": "Since the hyperbola passes through the point (2,0), the foci lie on the x-axis, and a=2. The asymptotes are given by y=\\pm\\frac{b}{a}x=\\pm\\frac{1}{2}x, \\therefore b=1. Therefore, the standard equation of the hyperbola is \\frac{x^{2}}{4}-"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$, the left and right foci are $F_{1}$, $F_{2}$ respectively, and $O$ is the coordinate origin. A point $P$ on the ellipse satisfies $|O P|=3$. Then the area of $\\Delta F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;O: Origin;Expression(G) = (x^2/16 + y^2/7 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);Abs(LineSegmentOf(O, P)) = 3",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "7",
    "fact_spans": "[[[2, 40], [74, 76]], [[49, 56]], [[79, 82]], [[57, 64]], [[65, 68]], [[2, 40]], [[2, 64]], [[2, 64]], [[74, 82]], [[84, 93]]]",
    "query_spans": "[[[95, 122]]]",
    "process": "From the given condition: $c^{2}=16-7=9$, solving gives: $c=3$, so $|F_{1}F_{2}|=6$. Let $P(m,n)$, then $\\begin{cases}\\frac{m^{2}}{16}+\\frac{n^{2}}{7}=1\\\\m^{2}+n^{2}=9\\end{cases}$, solving yields: $n=\\pm\\frac{7}{3}$, thus $S_{\\triangle F_{1}PF_{2}}=\\frac{1}{2}|F_{1}F_{2}|\\cdot|n|=\\frac{1}{2}\\times6\\times\\frac{7}{3}=7$"
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{49}+\\frac{y^{2}}{33}=1$, respectively, $P$ be any point on the ellipse, and the coordinates of point $M$ be $(8,5)$. Then the maximum value of $|P M|+|P F_{1}|$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/49 + y^2/33 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);M: Point;Coordinate(M) = (8, 5)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "14+sqrt(41)",
    "fact_spans": "[[[19, 58], [69, 71]], [[19, 58]], [[1, 8]], [[9, 16]], [[1, 64]], [[1, 64]], [[65, 68]], [[65, 75]], [[76, 80]], [[76, 91]]]",
    "query_spans": "[[[93, 116]]]",
    "process": "In the ellipse, $ c = \\sqrt{49 - 33} = \\sqrt{16} = 4 $, so the coordinates of the foci are $ F_{1}(-4,0) $, $ F_{2}(4,0) $. Point $ M $ lies outside the ellipse, then $ |PM| + |PF_{1}| = |PM| + 2a - |PF_{2}| = 14 + |PM| - |PF_{2}| \\leqslant 14 + |MF_{2}| = 14 + \\sqrt{(8-4)^{2} + 5^{2}} = 14 + \\sqrt{41} $, with equality if and only if points $ M $, $ F_{2} $, $ P $ are collinear."
  },
  {
    "text": "Given the parabola $y^{2}=4 x$, a line $l$ is drawn through its focus $F$, intersecting the parabola at points $A$ and $B$. Let $M$ be the intersection point of the directrix of the parabola and the $x$-axis. If $\\tan \\angle A M B = \\frac{4}{3}$, then $|A B| =$?",
    "fact_expressions": "l: Line;G: Parabola;A: Point;M: Point;B: Point;Expression(G) = (y^2 = 4*x);Focus(G)=F;PointOnCurve(F, l);Intersection(l, G) = {A, B};Intersection(Directrix(G), xAxis) = M;Tan(AngleOf(A, M, B)) = 4/3;F:Point",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16",
    "fact_spans": "[[[25, 30]], [[2, 16], [18, 19], [31, 34], [49, 52]], [[35, 38]], [[45, 48]], [[39, 42]], [[2, 16]], [[18, 24]], [[17, 30]], [[25, 44]], [[45, 63]], [[64, 95]], [21, 23]]",
    "query_spans": "[[[97, 106]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, line $ AB: y = k(x - 1) $. By the symmetry of the parabola, without loss of generality, assume $ A $ is in the first quadrant and $ B $ in the second quadrant. Given $ \\tan\\angle AMB = \\frac{4}{3} $, we obtain $ \\frac{2}{3} = k \\times \\frac{x_{1} - x_{2}}{(1 + k^{2})x_{1}x_{2} + (1 - k^{2})(x_{1} + x_{2}) + k^{2} + 1} $. Combining the line equation with the parabola equation and simplifying using Vieta's formulas yields $ k = \\pm\\frac{\\sqrt{3}}{3} $, thus $ |AB| $ can be calculated. Solution: From the given conditions, $ F(1,0) $, $ M(-1,0) $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. By the symmetry of the parabola, without loss of generality, assume $ A $ is in the first quadrant and $ B $ in the fourth quadrant. If the slope of line $ AB $ does not exist, then $ A(1,2) $, $ B(1,-2) $, so $ k_{AM} = 1 $, hence $ \\angle AMB = 90^{\\circ} $, contradicting the given condition. Let line $ AB: y = k(x - 1) $, then $ \\tan\\angle AMB = \\frac{4}{3} = \\frac{\\frac{y_{1}}{x_{1}+}}{1+\\underline{y}} $. Rearranging gives: $ \\frac{2}{3} = k \\times $. From $ \\begin{cases} y = k(x - 1) \\\\ y^{2} = 4x \\end{cases} $, we get $ k^{2}x^{2} - (2k^{2} + 4)x + k^{2} = 0 $, so $ x_{1}x_{2} = 1 $, $ x_{1} + x_{2} = 2 + \\frac{4}{k^{2}} $, $ |x_{1} - x_{2}| = \\frac{\\sqrt{16k^{2} + 16}}{k^{2}} = \\frac{4\\sqrt{k^{2} + 1}}{k^{2}} $, $ \\Delta = 16k^{2} + 16 > 0 $, hence $ \\frac{2}{3} = k \\times \\frac{4\\sqrt{k^{2} + 1}}{k^{2}(1 + k^{2}) \\cdot 1 + (1 - k^{2})(2 + \\frac{4}{k^{2}}) + k^{2} + 1} $. After simplification, $ \\frac{2}{3} = \\frac{k}{1} \\times \\frac{4\\sqrt{k^{2} + 1}}{k^{2}} \\times \\frac{1}{(1 + k^{2}) + (1 - k^{2})(2 + \\frac{4}{k^{2}}) + k^{2} + 1} $, leading to $ 9k^{4} + 9k^{2} - 4 = 0 $, solving gives $ k^{2} = \\frac{1}{3} $, so $ k = \\pm\\frac{\\sqrt{3}}{3} $. Hence $ |AB| = \\sqrt{1 + k^{2}}|x_{1} - x_{2}| = \\sqrt{1 + \\frac{1}{3}} \\cdot \\frac{4\\sqrt{\\frac{1}{3} + 1}}{\\frac{1}{3}} = \\frac{2}{\\sqrt{3}} \\cdot \\frac{4 \\cdot \\frac{2}{\\sqrt{3}}}{\\frac{1}{3}} = \\frac{2}{\\sqrt{3}} \\cdot \\frac{8}{\\sqrt{3}} \\cdot 3 = \\frac{16}{3} \\cdot 3 = 16 $"
  },
  {
    "text": "Given that point $P$ lies on the parabola $y^{2}=4x$, then the minimum value of the sum of the distance from point $P$ to point $Q(2, -1)$ and the distance from point $P$ to the focus of the parabola is?",
    "fact_expressions": "P: Point;PointOnCurve(P, G);G: Parabola;Expression(G) = (y^2 = 4*x);Q: Point;Coordinate(Q) = (2, -1)",
    "query_expressions": "Min(Distance(P, Q) + Distance(P, Focus(G)))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 6], [25, 29], [45, 49]], [[2, 22]], [[7, 21], [50, 53]], [[7, 21]], [[30, 41]], [[30, 41]]]",
    "query_spans": "[[[25, 65]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the ellipse $x^{2}+4 y^{2}=4$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 4*y^2 = 4)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[0, 19]], [[0, 19]]]",
    "query_spans": "[[[0, 25]]]",
    "process": "Convert $x^{2}+4y^{2}=4$ into the standard equation $\\frac{x^2}{4}+y^{2}=1$, $\\therefore a=2$, $b=1$, $c=\\sqrt{3}$. $\\therefore$ eccentricity $e=\\frac{c}{a}=\\frac{\\sqrt{3}}{2}$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$, the left focus is $F$. A line passing through $F$ with an inclination angle of $30^{\\circ}$ intersects $C$ at point $A$ in the first quadrant, and $|O F|=| O A |$, where $O$ is the origin. Then the eccentricity of this hyperbola is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F: Point;LeftFocus(C) = F;G: Line;PointOnCurve(F, G) = True;Inclination(G) = ApplyUnit(30, degree);A: Point;Intersection(G,C) = A;Quadrant(A) = 1;O: Origin;Abs(LineSegmentOf(O,F)) = Abs(LineSegmentOf(O,A))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)+1",
    "fact_spans": "[[[2, 66], [102, 105], [145, 148]], [[2, 66]], [[10, 66]], [[10, 66]], [[10, 66]], [[10, 66]], [[71, 74], [76, 79]], [[2, 74]], [[99, 101]], [[75, 101]], [[82, 101]], [[112, 116]], [[99, 116]], [[106, 116]], [[134, 137]], [[118, 133]]]",
    "query_spans": "[[[145, 154]]]",
    "process": "A line passing through F with an inclination angle of 30^{\\circ} intersects C at point A in the first quadrant, and |OF| = |OA| = c. Let the right focus be E. It is known that the focal triangle AFE is a right triangle, \\therefore \\angle AOx = 60^{\\circ}, and |AF| = \\sqrt{3}c, |AE| = c. From the definition of the hyperbola, we obtain |AF| - |AE| = 2a = \\sqrt{3}c - c, \\therefore \\frac{c}{a} = \\frac{2}{\\sqrt{3}-1} = \\sqrt{3}+1, i.e., e = 1+\\sqrt{3}."
  },
  {
    "text": "If a point $P$ on the parabola $y^{2}=4 x$ is at a distance of $4$ from its focus, then what are the coordinates of point $P$?",
    "fact_expressions": "G: Parabola;P: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(P, G);Distance(P, Focus(G)) = 4",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(3,pm*2*sqrt(3))",
    "fact_spans": "[[[1, 15], [22, 23]], [[18, 21], [35, 39]], [[1, 15]], [[1, 21]], [[1, 32]]]",
    "query_spans": "[[[35, 44]]]",
    "process": ""
  },
  {
    "text": "Given that point $P(1,2)$ lies on the parabola $E$: $y^{2}=2 p x$ ($p>0$), a line $l$ passing through point $M(1,0)$ intersects the parabola $E$ at points $A$ and $B$. If $\\overrightarrow{A M}=3 \\overrightarrow{M B}$, then the sine value of the inclination angle of line $l$ is?",
    "fact_expressions": "l: Line;E: Parabola;p: Number;P: Point;M: Point;A: Point;B: Point;p>0;Expression(E) = (y^2 = 2*(p*x));Coordinate(P) = (1, 2);Coordinate(M) = (1, 0);PointOnCurve(P, E);PointOnCurve(M, l);Intersection(l, E) = {A, B};VectorOf(A, M) = 3*VectorOf(M, B)",
    "query_expressions": "Sin(Inclination(l))",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[51, 56], [122, 127]], [[12, 38], [57, 63]], [[19, 38]], [[2, 11]], [[41, 50]], [[64, 67]], [[68, 71]], [[19, 38]], [[12, 38]], [[2, 11]], [[41, 50]], [[2, 39]], [[40, 56]], [[51, 73]], [[75, 120]]]",
    "query_spans": "[[[122, 137]]]",
    "process": "Since the point lies on the parabola $ E: y^{2} = 2px $ ($ p > 0 $), we have $ 4 = 2p \\times 1 $, so $ p = 2 $, hence $ y^{2} = 4x $. Let the equation of the line passing through point $ M(1, 0) $ be $ x = my + 1 $. Then  \n$$\n\\begin{cases}\nx = my + 1 \\\\\ny^{2} = 4x\n\\end{cases}\n$$  \nThus $ y^{2} - 4my - 4 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, so $ y_{1} + y_{2} = 4m $, $ y_{1}y_{2} = -4 $. Also, since $ \\overrightarrow{AM} = 3\\overrightarrow{MB} $, we have $ -y_{1} = 3y_{2} $. Therefore $ m = \\pm\\frac{\\sqrt{3}}{3} $. Since the slope of the line $ k = \\tan\\theta = \\pm\\sqrt{3} $ and $ \\theta \\in (0, \\pi) $, we get $ \\theta = \\frac{\\pi}{3} $ or $ \\frac{2\\pi}{3} $, so $ \\sin\\theta = \\frac{\\sqrt{3}}{2} $."
  },
  {
    "text": "Given that one asymptote of the hyperbola $x^{2}-\\frac{y^{2}}{a}=1$ is parallel to the line $x-2 y+3=0$, then the eccentricity $e$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/a = 1);a: Number;IsParallel(OneOf(Asymptote(G)),H) = True;H: Line;Expression(H) = (x - 2*y + 3 = 0);e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[2, 30]], [[2, 30]], [[5, 30]], [[2, 52]], [[37, 50]], [[37, 50]], [[57, 60]], [[2, 60]]]",
    "query_spans": "[[[57, 62]]]",
    "process": "The asymptotes of the given hyperbola are $x-\\frac{1}{\\sqrt{a}}y=0$ and $x+\\frac{1}{\\sqrt{a}}y=0$. Clearly, the line $x-\\frac{1}{\\sqrt{a}}y=0$ is parallel to the line $x-2y+3=0$, so $\\frac{1}{\\sqrt{a}}=2$, $a=\\frac{1}{4}$. Thus, the equation of the hyperbola is $x^{2}-\\frac{y^{2}}{\\frac{1}{4}}=1$, the length of the real semi-axis is $a=1$, the length of the imaginary semi-axis is $b'=\\frac{1}{2}$, and the semi-focal distance is $c=\\sqrt{1+\\frac{1}{4}}=\\frac{\\sqrt{5}}{2}$. Therefore, the eccentricity is $e=\\frac{c}{a}=\\frac{\\sqrt{5}}{2}$."
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, the intersection point of the directrix and the $x$-axis is $M$, and $N$ is a point on the parabola satisfying $|NF|=\\frac{\\sqrt{3}}{2}|MN|$, then what is $\\angle NMF$?",
    "fact_expressions": "G: Parabola;N: Point;F: Point;M: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;Intersection(Directrix(G),xAxis)=M;PointOnCurve(N,G);Abs(LineSegmentOf(N, F)) = (sqrt(3)/2)*Abs(LineSegmentOf(M, N))",
    "query_expressions": "AngleOf(N, M, F)",
    "answer_expressions": "pi/6",
    "fact_spans": "[[[2, 16], [43, 46]], [[39, 42]], [[20, 23]], [[35, 38]], [[2, 16]], [[2, 23]], [[2, 38]], [[39, 50]], [[54, 85]]]",
    "query_spans": "[[[87, 103]]]",
    "process": "Draw NH perpendicular from N to the directrix, with foot H, then |NF| = |NH|. Since |NF| = \\frac{\\sqrt{3}}{2}|MN|, it follows that |NH| = \\frac{\\sqrt{3}}{2}|MN|. Therefore, \\cos\\angle NMF = \\cos\\angle MNH = \\frac{|NH|}{|MN|} = \\frac{\\sqrt{3}}{2}, \\angle NMF = \\frac{\\pi}{7}."
  },
  {
    "text": "Let the coordinates of points $A$ and $B$ be $(-4,0)$ and $(4,0)$, respectively. Lines $AM$ and $BM$ intersect at point $M$, and the product of their slopes is $-\\frac{3}{4}$. Then the equation of the trajectory of point $M$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (-4, 0);B: Point;Coordinate(B) = (4, 0);M: Point;Intersection(LineOf(A, M), LineOf(B, M)) = M;Slope(LineOf(A, M))*Slope(LineOf(B, M))=-3/4",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "(x^2/16+y^2/12=1)&(Negation(x=pm*4))",
    "fact_spans": "[[[1, 5]], [[1, 30]], [[6, 9]], [[1, 30]], [[48, 52], [78, 82]], [[31, 52]], [[54, 76]]]",
    "query_spans": "[[[78, 89]]]",
    "process": ""
  },
  {
    "text": "Given that the focal distance of a hyperbola is $6$, and the difference between the distances from a point $P$ on the hyperbola to the two foci is $-4$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;P:Point;FocalLength(G)=6;PointOnCurve(P,G);F1:Point;F2:Point;Focus(G)={F1,F2};Distance(P,F1)-Distance(P,F2)=-4",
    "query_expressions": "Expression(G)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 5], [36, 39], [13, 14]], [[17, 20]], [[2, 12]], [[13, 20]], [], [], [[13, 24]], [[13, 34]]]",
    "query_spans": "[[[36, 46]]]",
    "process": "From the definition of a hyperbola, the length of the real axis is obtained, and given the focal distance, the length of the imaginary semi-axis $ b $ can be easily found. When writing the standard equation of the hyperbola, two cases must be discussed: foci on the $ x $-axis and foci on the $ y $-axis. Let the focal distance of the hyperbola be $ 2c $, the real axis length be $ 2a $, and the imaginary axis length be $ 2b $. According to the problem:  \n$$\n\\begin{cases}\n2c=6 \\\\\n2a=4\n\\end{cases},\n\\begin{cases}\nc=3 \\\\\na=2\n\\end{cases}\n$$\nThus, $ b^{2} = c^{2} - a^{2} = 3^{2} - 2^{2} = 5 $. Therefore, if the foci of the hyperbola are on the $ x $-axis, the standard equation is $ \\frac{x^{2}}{4} - \\frac{y^{2}}{5} = 1 $; if the foci are on the $ y $-axis, the standard equation is $ \\frac{y^{2}}{4} - \\frac{y^{2}}{5} = 1 $."
  },
  {
    "text": "Given the curve $C$: $\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1$, what is the focal distance of $C$?",
    "fact_expressions": "C: Curve;Expression(C) = (x^2/3 + y^2/2 = 1)",
    "query_expressions": "FocalLength(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 44], [46, 49]], [[2, 44]]]",
    "query_spans": "[[[46, 54]]]",
    "process": "From the curve equation, it can be seen that the curve is an ellipse with foci on the x-axis, and $ a^{2}=3 $, $ b^{2}=2 $, $ \\therefore c^{2}=3-2=1 $, $ \\therefore c=1 $, $ \\therefore $ the focal distance is $ 2c=2 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$, one of its asymptotes is perpendicular to the line $l$: $x+\\sqrt{3} y=0$, and the distance from a focus of $C$ to $l$ is $1$. Then the equation of $C$ is?",
    "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(l) = (x+sqrt(3)*y=0);IsPerpendicular(OneOf(Asymptote(C)),l);Distance(OneOf(Focus(C)), l) = 1",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[71, 94], [106, 109]], [[2, 64], [97, 100], [119, 122]], [[10, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[2, 64]], [[71, 94]], [[2, 96]], [[97, 116]]]",
    "query_spans": "[[[119, 127]]]",
    "process": "From the given, one asymptote equation is $\\sqrt{3}x - y = 0$, so $b = \\sqrt{3}a$. Also, $\\frac{c}{\\sqrt{3+1}} = 1$, thus $c = 2$, that is, $a^{2} + b^{2} = 4$. Solving gives $a = 1$, $b = \\sqrt{3}$. The hyperbola equation is $x^{2} - \\frac{y^{2}}{3} = 1$."
  },
  {
    "text": "Let a line passing through the focus $F$ of the parabola $C$: $y^{2}=2 p x$ ($p>0$) intersect the parabola $C$ at points $A$ and $B$. Given that $|A F|=2|B F|=2$, then $p=$?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;A: Point;F: Point;B: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(F, G);Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, F)) = 2*Abs(LineSegmentOf(B, F));2*Abs(LineSegmentOf(B, F)) = 2",
    "query_expressions": "p",
    "answer_expressions": "4/3",
    "fact_spans": "[[[1, 27], [37, 43]], [[73, 76]], [[34, 36]], [[45, 48]], [[30, 33]], [[49, 52]], [[9, 27]], [[1, 27]], [[1, 33]], [[0, 36]], [[34, 54]], [[55, 71]], [[55, 71]]]",
    "query_spans": "[[[73, 78]]]",
    "process": "As shown in the figure, let A(x_{1},y_{1}), B(x_{2},y_{2}), then |AF| = x_{1} + \\frac{p}{2} = 2, |BF| = x_{2} + \\frac{p}{2}, and \\frac{y_{1}}{y_{2}} = \\frac{|AF|}{|BF|} = 2, so y_{1} = -2y_{2}. Also, y_{1}^{2} = 2px_{1}, y_{2}^{2} = 2px_{2}; thus y_{1}^{2} - y_{2}^{2} = 2p(x_{1} - x_{2}), i.e., 3y_{2}^{2} = 2p; and since y_{2}^{2} = 2px_{2}, we have 6px_{2} = 2p, solving gives x_{2} = \\frac{1}{3}; therefore p = \\frac{4}{3}."
  },
  {
    "text": "If point $P$ lies on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ with foci $F_{1}$ and $F_{2}$, satisfying $PF_{1} \\perp PF_{2}$ and $|PF_{1}|=2|PF_{2}|$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);Focus(G) ={F1,F2};IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[28, 74], [126, 129]], [[31, 74]], [[31, 74]], [[1, 5]], [[7, 15]], [[17, 24]], [[28, 74]], [[1, 77]], [[6, 74]], [[80, 101]], [[103, 123]]]",
    "query_spans": "[[[126, 135]]]",
    "process": ""
  },
  {
    "text": "If point $P$ is a point on the parabola $y^{2}=10x$, then the minimum distance from point $P$ to the line $x+y+5=0$ is?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Expression(G) = (y^2 = 10*x);Expression(H) = (x + y + 5 = 0);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, H))",
    "answer_expressions": "5*sqrt(2)/4",
    "fact_spans": "[[[6, 21]], [[31, 42]], [[1, 5], [26, 30]], [[6, 21]], [[31, 42]], [[1, 24]]]",
    "query_spans": "[[[26, 50]]]",
    "process": "Let point $ P(x,y) $, then the distance from the point to the line $ x+y+5=0 $ is $ d = \\frac{|x+y+5|}{\\sqrt{2}} = \\frac{|\\frac{y^{2}}{10}+y+5|}{\\sqrt{2}} = \\frac{|y^2+10y+50|}{10\\sqrt{2}} = \\frac{|(y+5)^{2}+25|}{10\\sqrt{2}} $. Therefore, when $ y = -5 $, the minimum distance from point $ P $ to the line $ x+y+5=0 $ is $ \\frac{5}{2\\sqrt{2}} = \\frac{5\\sqrt{2}}{4} $."
  },
  {
    "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ have no common points with the circle $x^{2}+(y+2)^{2}=1$, find the range of values for the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + (y + 2)^2 = 1);NumIntersection(Asymptote(G),H)=0",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1, 2)",
    "fact_spans": "[[[2, 59], [93, 96]], [[5, 59]], [[5, 59]], [[64, 84]], [[5, 59]], [[5, 59]], [[2, 59]], [[64, 84]], [[2, 89]]]",
    "query_spans": "[[[93, 107]]]",
    "process": "Find the equations of the asymptotes of the hyperbola. From the condition that the distance from the center of the circle to the asymptote is greater than the radius of the circle, obtain an inequality involving a, b, c and the range of eccentricity. [Solution] The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) has asymptotes $y=\\pm\\frac{b}{a}x$, or $bx+ay=0$. These have no common points with the circle $x^{2}+(y+2)^{2}=1$. Then $\\frac{2a}{a^{2}+b^{2}}=\\frac{2a}{c}>1$, so $2a>c$. Therefore, the eccentricity $e=\\frac{c}{a}$ of the hyperbola satisfies $1<e=\\frac{c}{a}<2$, i.e., the range of the eccentricity of the hyperbola is $(1,2)$."
  },
  {
    "text": "Draw a line through the focus of the parabola $y^{2}=4x$ intersecting the parabola at points $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$. If $|AB|=7$, then the distance from the midpoint $M$ of $AB$ to the directrix of the parabola is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;M: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (x1,y1);Coordinate(B) = (x2,y2);PointOnCurve(Focus(G), H);Intersection(H, G) = {A,B};Abs(LineSegmentOf(A, B)) = 7;MidPoint(LineSegmentOf(A, B)) = M;x1:Number;x2:Number;y1:Number;y2:Number",
    "query_expressions": "Distance(M,Directrix(G))",
    "answer_expressions": "7/2",
    "fact_spans": "[[[1, 15], [22, 25], [88, 91]], [[19, 21]], [[26, 44]], [[46, 63]], [[84, 87]], [[1, 15]], [[26, 44]], [[46, 63]], [[0, 21]], [[19, 63]], [[65, 74]], [[76, 87]], [[27, 44]], [[46, 63]], [[27, 44]], [[46, 63]]]",
    "query_spans": "[[[84, 98]]]",
    "process": "Find the coordinates of the focus and the equation of the directrix of the parabola. Using |AB| = |AF| + |BF| = 7, find that x_{1} + x_{2} = 5, then the horizontal coordinate of M can be determined. According to the point-to-line distance formula, it is easy to find the distance from M to the directrix of the parabola. The focus of the parabola is F(\\frac{7}{2},0), and the equation of the directrix is x = -. By the definition of the parabola, |AB| = |AF| + |BF| = x_{1} + \\frac{p}{2} + x_{2} + \\frac{p}{2} = x_{1} + x_{2} + 2 = 7, so x_{1} + x_{2} = 5. Therefore, the horizontal coordinate of the midpoint M of AB is \\frac{5}{2}. Thus, the distance from point M to the directrix of the parabola is \\frac{5}{2} + 1 = \\frac{7}{2}."
  },
  {
    "text": "The directrix of the parabola $x^{2}=8 y$ intersects the $y$-axis at point $A$, and point $B$ lies on the axis of symmetry of the parabola. A line passing through $A$ intersects the parabola at points $M$ and $N$, such that $\\overrightarrow{B M}\\cdot \\overrightarrow{M N}=-\\frac{1}{2} \\overrightarrow{M N}^{2}$. Then the range of $|O B|$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 8*y);A: Point;Intersection(Directrix(G), yAxis) = A;B: Point;PointOnCurve(B, SymmetryAxis(G));H: Line;PointOnCurve(A, H);M: Point;N: Point;Intersection(H, G) = {M, N};DotProduct(VectorOf(B, M), VectorOf(M, N)) = (-1/2)*VectorOf(M, N)^2;O: Origin",
    "query_expressions": "Range(Abs(LineSegmentOf(O, B)))",
    "answer_expressions": "(6,+oo)",
    "fact_spans": "[[[0, 14], [35, 38], [52, 55]], [[0, 14]], [[24, 28], [44, 47]], [[0, 28]], [[30, 34]], [[30, 42]], [[49, 51]], [[43, 51]], [[56, 60]], [[61, 64]], [[49, 64]], [[67, 153]], [[155, 162]]]",
    "query_spans": "[[[155, 169]]]",
    "process": ""
  },
  {
    "text": "Let $O$ be the origin, and let $F_{1}$, $F_{2}$ be the foci of $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$. If there exists a point $P$ on the hyperbola such that $\\angle F_{1} P F_{2}=60^{\\circ}$ and $|O P|=\\sqrt{7} a$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "O: Origin;F1: Point;F2: Point;C: Curve;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Focus(C) = {F1,F2};a: Number;b: Number;a>0;b>0;G: Hyperbola;P: Point;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree);Abs(LineSegmentOf(O, P)) = sqrt(7)*a",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[1, 4]], [[10, 17]], [[18, 25]], [[26, 79]], [[26, 79]], [[10, 82]], [[26, 79]], [[26, 79]], [[26, 79]], [[26, 79]], [[85, 88], [154, 157]], [[91, 95]], [[84, 95]], [[98, 131]], [[133, 151]]]",
    "query_spans": "[[[154, 163]]]",
    "process": "Assume |PF_{1}|=x, establish equations based on the triangle median theorem and the cosine law to obtain c^{2}+5a^{2}=14a^{2}-2c^{2}, then find the relationship between a and c, thus determine the eccentricity of the hyperbola. Without loss of generality, assume P lies on the left branch, |PF_{1}|=x, then |PF_{2}|=2a+x. Since OP is the median of triangle F_{1}F_{2}P, by the triangle median theorem we have x^{2}+(2a+x)^{2}=2(c^{2}+7a^{2}). Simplify to get x(x+2a)=c^{2}+5a^{2}. By the cosine law, x^{2}+(2a+x)^{2}-2x(x+2a)\\cos60^{\\circ}=4c^{2}, simplify to get x(x+2a)=14a^{2}-2c^{2}. Therefore, c^{2}+5a^{2}=14a^{2}-2c^{2}, simplifying yields c^{2}=3a^{2}. Thus e=\\frac{c}{a}=\\sqrt{3}."
  },
  {
    "text": "The coordinates of the vertices of the hyperbola $\\frac{y^{2}}{4}-x^{2}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2 + y^2/4 = 1)",
    "query_expressions": "Coordinate(Vertex(G))",
    "answer_expressions": "(0,pm*2)",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 35]]]",
    "process": "From the hyperbola $\\frac{y^{2}}{4}-x^{2}=1$, it can be seen that the vertices are on the $y$-axis, i.e., the vertices are $(0,\\pm2)$,"
  },
  {
    "text": "Given a point $M(1, m)$ on the parabola $y^{2}=2 p x(p>0)$ is at a distance of $5$ from its focus, then the equation of the directrix of this parabola is?",
    "fact_expressions": "G: Parabola;p: Number;M: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(M) = (1, m);m:Number;PointOnCurve(M, G);Distance(M, Focus(G)) = 5",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x = -4",
    "fact_spans": "[[[2, 23], [36, 37], [49, 52]], [[5, 23]], [[26, 35]], [[5, 23]], [[2, 23]], [[26, 35]], [[26, 35]], [[2, 35]], [[26, 46]]]",
    "query_spans": "[[[49, 59]]]",
    "process": "According to the problem: the focus of the parabola is $ F\\left(\\frac{p}{2}, 0\\right) $, and the equation of the directrix is $ x = -\\frac{p}{2} $. Since the distance from point $ M(1, m) $ to its focus is 5, using the definition of a parabola, we obtain the equation, leading to the detailed directrix equation.  \n$ \\because $ The equation of the parabola is $ y^2 = 2px $,  \n$ \\therefore $ the focus is $ F\\left(\\frac{p}{2}, 0\\right) $, and the directrix equation is $ x = -\\frac{p}{2} $.  \nAlso, $ \\because $ the distance from point $ M(1, m) $ to its focus is 5,  \n$ \\therefore p > 0 $, by the definition of the parabola, we get $ 1 + \\frac{p}{2} = 5 $,  \n$ \\therefore p = 8 $,  \n$ \\therefore $ the directrix equation is $ x = -4 $."
  },
  {
    "text": "The coordinates of the focus of the parabola $C$: $y=-\\frac{x^{2}}{8}$ are?",
    "fact_expressions": "C: Parabola;Expression(C) = (y = -x^2/8)",
    "query_expressions": "Coordinate(Focus(C))",
    "answer_expressions": "(0,-2)",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 35]]]",
    "process": "The standard form of the parabola C: y = -\\frac{x^{2}}{8} is x^{2} = -8y, so the coordinates of the focus are: (0, -2)"
  },
  {
    "text": "If a point $P(a , b)$ on the right branch of the hyperbola $x^{2}-y^{2}=1$ is at a distance of $\\sqrt{3}$ from the line $y=x$, then $a+b=$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 1);P: Point;a: Number;b: Number;Coordinate(P) = (a, b);PointOnCurve(P, RightPart(G));H: Line;Expression(H) = (y = x);Distance(P, H) = sqrt(3)",
    "query_expressions": "a + b",
    "answer_expressions": "sqrt(6)/6",
    "fact_spans": "[[[1, 19]], [[1, 19]], [[24, 34]], [[24, 34]], [[24, 34]], [[24, 34]], [[1, 34]], [[35, 42]], [[35, 42]], [[24, 56]]]",
    "query_spans": "[[[58, 65]]]",
    "process": "P(a,b) lies on the hyperbola, then we have a^{2}-b^{2}=1, i.e., (a+b)(a-b)=1, d=\\frac{|a-b|}{\\sqrt{2}}=\\sqrt{3}, \\therefore |a-b|=\\sqrt{6}. Since P lies on the right branch, we have a>b, \\therefore a-b=\\sqrt{6}, \\therefore (a+b)\\times\\sqrt{6}=1, a+b=\\frac{\\sqrt{6}}{6}."
  },
  {
    "text": "Let the parabola $C$: $y^{2}=8x$ have focus $F$, and let line $l$ pass through $F$ and intersect the parabola at points $P$ and $Q$. If $|PQ| = \\frac{32}{3}$ and $|PF| > |QF|$, then $\\frac{|PF|}{|QF|} = $?",
    "fact_expressions": "l: Line;C: Parabola;P: Point;Q: Point;F: Point;Expression(C) = (y^2 = 8*x);Focus(C) = F;PointOnCurve(F,l);Intersection(l, C) = {P,Q};Abs(LineSegmentOf(P, Q)) = 32/3;Abs(LineSegmentOf(P, F)) > Abs(LineSegmentOf(Q, F))",
    "query_expressions": "Abs(LineSegmentOf(P, F))/Abs(LineSegmentOf(Q, F))",
    "answer_expressions": "3",
    "fact_spans": "[[[28, 33]], [[1, 20], [39, 42]], [[44, 47]], [[48, 51]], [[34, 37], [24, 27]], [[1, 20]], [[1, 27]], [[28, 37]], [[28, 53]], [[55, 75]], [[77, 90]]]",
    "query_spans": "[[[92, 115]]]",
    "process": "Since $|PQ| = \\frac{32}{3}$, it follows that $|PF| = \\frac{32}{3} - |QF|$. From the standard equation of the parabola, we know $p = 4$. Since line $l$ passes through the focus of the parabola $y^2 = 8x$, by the property of chords passing through the focus of a parabola, we have $\\frac{1}{|PF|} + \\frac{1}{|QF|} = \\frac{2}{p} = \\frac{1}{2}$. Since $|PF| > |OF|$, it follows that $|QF| = \\frac{8}{3}$, $|PF| = 8$. Therefore, $\\frac{|PF|}{|OF|} = 3$."
  },
  {
    "text": "Let the two foci of hyperbola $C$ be $(-\\sqrt{2}, 0)$ and $(\\sqrt{2}, 0)$, and one vertex be $(1, 0)$. Then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Point;H: Point;I: Point;Coordinate(G) = (-sqrt(2), 0);Coordinate(H) = (sqrt(2), 0);Coordinate(I) = (1, 0);Focus(C)={G,H};OneOf(Vertex(C))= I",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2 - y^2 = 1",
    "fact_spans": "[[[1, 7], [63, 66]], [[13, 29]], [[31, 47]], [[54, 61]], [[13, 29]], [[31, 47]], [[54, 61]], [[1, 46]], [[1, 61]]]",
    "query_spans": "[[[63, 71]]]",
    "process": "From the given conditions, we have $ c = \\sqrt{2} $, $ a = 1 $, so $ b^{2} = c^{2} - a^{2} = 1 $. Since the foci of the hyperbola lie on the x-axis, the equation of $ C $ is $ x^{2} - y^{2} = 1 $."
  },
  {
    "text": "Given that the vertex of the parabola is the center of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and the focus is the right focus of the ellipse, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;H: Ellipse;Expression(H) = (x^2/25 + y^2/16 = 1);Vertex(G) = Center(H);Focus(G) = RightFocus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=12*x",
    "fact_spans": "[[[2, 5], [63, 66]], [[9, 48], [55, 57]], [[9, 48]], [[2, 51]], [[2, 61]]]",
    "query_spans": "[[[63, 70]]]",
    "process": "Since the right focus of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ is $(3,0)$, the value of $p$ for the parabola is $6$. The vertex of the parabola is the center of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and its focus is the right focus of the ellipse; therefore, the equation of the parabola is: $y^{2}=12x$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. If there exists a point $P$ on the right branch of the hyperbola such that $|PF_{1}|=3|PF_{2}|$, then the range of the eccentricity $e$ of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;e: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, F2));Eccentricity(G) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "(1,2]",
    "fact_spans": "[[[2, 58], [86, 89], [125, 128]], [[5, 58]], [[5, 58]], [[97, 100]], [[67, 74]], [[76, 83]], [[132, 135]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 83]], [[2, 83]], [[86, 100]], [[103, 123]], [[125, 135]]]",
    "query_spans": "[[[132, 142]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $ C: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 $ ($ a>b>0 $) has a major axis length of $ 4 $. If point $ P $ is any point on the ellipse $ C $, and a line $ l $ passing through the origin intersects the ellipse at points $ M $ and $ N $, denote the slopes of lines $ PM $ and $ PN $ as $ K_{PM} $ and $ K_{PN} $, respectively. If $ K_{PM} \\cdot K_{PN} = -\\frac{1}{4} $, then what is the equation of the ellipse?",
    "fact_expressions": "l: Line;C: Ellipse;b: Number;a: Number;P: Point;M: Point;N: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Length(MajorAxis(C)) = 4;PointOnCurve(P, C);O:Origin;PointOnCurve(O,l);Intersection(l, C) = {M, N};Slope(LineSegmentOf(P, M)) = K1;Slope(LineSegmentOf(P, N)) = K2;K1*K2 = -1/4;K1:Number;K2:Number",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4+y^2=1",
    "fact_spans": "[[[89, 94]], [[2, 59], [74, 79], [95, 97], [190, 192]], [[8, 59]], [[8, 59]], [[69, 73]], [[99, 102]], [[103, 106]], [[8, 59]], [[8, 59]], [[2, 59]], [[2, 67]], [[69, 84]], [[86, 88]], [[85, 94]], [[89, 108]], [[110, 150]], [[110, 150]], [[152, 188]], [[129, 138]], [[141, 150]]]",
    "query_spans": "[[[190, 197]]]",
    "process": "Let $ P(x_{1},y_{1}) $. Since $ M $ and $ N $ are symmetric about the origin, let $ M(x_{2},y_{2}) $, $ N(-x_{2},-y_{2}) $. Then  \n$$\n\\begin{cases}\n\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 \\cdots\\cdots \\textcircled{1} \\\\\n\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 \\cdots\\cdots \\textcircled{2}\n\\end{cases}\n$$  \n$ \\textcircled{1} - \\textcircled{2} $ yields:  \n$ \\frac{y_{1}-y_{2}}{x_{1}-x_{2}} \\cdot \\frac{y_{1}+y_{2}}{x_{1}+x_{2}} = -\\frac{b^{2}}{a^{2}} = -\\frac{b^{2}}{4} $,  \ni.e., $ k_{PM} \\cdot k_{PN} = -\\frac{b^{2}}{4} $. Given $ k_{PM} \\cdot k_{PN} = -\\frac{1}{4} $, so $ b = 1 $. Therefore, the ellipse equation is: $ \\frac{x^{2}}{4} + y^{2} = 1 $."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the two foci are $F_{1}$ and $F_{2}$, respectively. $M$ is a point on an asymptote of the hyperbola $C$, $|M F_{1}|=2|M F_{2}|$. Point $N$ satisfies $\\overrightarrow{M N}=2 \\overrightarrow{O N}$, and $\\angle M F_{2} N=120^{\\circ}$. Then the eccentricity of this hyperbola equals?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;M: Point;F1: Point;F2: Point;N: Point;O: Origin;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Focus(C) = {F1, F2};PointOnCurve(M, Asymptote(C));Abs(LineSegmentOf(M, F1)) = 2*Abs(LineSegmentOf(M, F2));VectorOf(M, N) = 2*VectorOf(O, N);AngleOf(M, F2, N) = ApplyUnit(120, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(21)/3",
    "fact_spans": "[[[2, 63], [91, 97], [213, 216]], [[10, 63]], [[10, 63]], [[87, 90]], [[71, 78]], [[79, 86]], [[127, 131]], [[133, 178]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 86]], [[87, 103]], [[104, 126]], [[133, 178]], [[180, 210]]]",
    "query_spans": "[[[213, 223]]]",
    "process": "As shown in the figure, from $\\overrightarrow{MN}=2\\overrightarrow{ON}$, we obtain that $M$ and $N$ are symmetric about the origin. Also, $F_1$ and $F_2$ are symmetric about the origin, $\\therefore$ quadrilateral $MF_1NF_2$ is a parallelogram. $\\because \\angle MF_2N = 120^{\\circ}$, $\\therefore \\angle F_1MF_2 = 60^{\\circ}$. Also, $|MF_1| = 2|MF_2|$, let $|MF_2| = m$, then $|MF_1| = 2m$, $\\therefore |F_1F_2|^2 = m^2 + 4m^2 - 2 \\cdot m \\cdot 2m \\cdot \\frac{1}{2} = 3m^2$, we get $|F_1F_2|^2 + |MF_2|^2 = |MF_1|^2$, then $MF_2 \\perp F_1F_2$, so $\\tan \\angle MOF_2 = \\frac{|MF_2|}{|OF_2|} = \\frac{m}{\\frac{\\sqrt{3}}{3}m} = \\frac{2\\sqrt{3}}{3}$, $\\therefore \\frac{b}{a} = \\frac{2\\sqrt{3}}{3}$, i.e., $3b^2 = 4a^2$, $\\frac{c^2 - a^2}{a^2} = \\frac{4}{3}$, solving gives: $e = \\frac{c}{a} = \\frac{\\sqrt{21}}{3}$ ($e > 1$)"
  },
  {
    "text": "It is known that an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is perpendicular to the line $x+3 y+1=0$. Then, the eccentricity of the hyperbola equals?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;H: Line;Expression(H) = (x + 3*y + 1 = 0);IsPerpendicular(OneOf(Asymptote(G)), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[2, 58], [82, 85]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[65, 78]], [[65, 78]], [[2, 80]]]",
    "query_spans": "[[[82, 92]]]",
    "process": "\\because the hyperbola \\frac{x^2}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0) has one asymptote perpendicular to the line x+3y+1=0. \\therefore the asymptotes of the hyperbola are y=\\pm3x. \\therefore \\frac{b}{a}=3, thus b^{2}=9a^{2}, c^{2}-a^{2}=9a^{2}, at this time, the eccentricity e=\\frac{c}{a}=\\sqrt{10}"
  },
  {
    "text": "The distance from the focus of hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ to its asymptote is $2$, and the focal length of $C$ is equal to the focal length of the ellipse $\\frac{y^{2}}{25}+\\frac{x^{2}}{20}=1$. Then the equation of the asymptotes of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Ellipse;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2/20 + y^2/25 = 1);Distance(Focus(C),Asymptote(C))=2;FocalLength(C)=FocalLength(G)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y=pm*2*x",
    "fact_spans": "[[[0, 61], [65, 66], [78, 81], [131, 137]], [[8, 61]], [[8, 61]], [[85, 124]], [[8, 61]], [[8, 61]], [[0, 61]], [[85, 124]], [[0, 76]], [[78, 129]]]",
    "query_spans": "[[[131, 145]]]",
    "process": ""
  },
  {
    "text": "If the line $y=a x-1(a \\in R)$ always has common points with the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$ whose focus lies on the $x$-axis, then what is the range of values for $m$?",
    "fact_expressions": "G: Ellipse;m: Number;H: Line;a: Real;Expression(G) = (x^2/5 + y^2/m = 1);Expression(H) = (y = -1 + a*x);IsIntersect(G, H) ;PointOnCurve(Focus(G),xAxis)",
    "query_expressions": "Range(m)",
    "answer_expressions": "[1, 5)",
    "fact_spans": "[[[31, 68]], [[75, 78]], [[1, 21]], [[3, 21]], [[31, 68]], [[1, 21]], [[1, 73]], [[22, 68]]]",
    "query_spans": "[[[75, 85]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\Gamma$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$), points $A$ and $B$ lie on the left and right branches of the hyperbola respectively, satisfying $\\overrightarrow{A B}=\\lambda \\overrightarrow{F_{1} A}$ ($\\lambda$ is a constant). Point $C$ lies on the $x$-axis, $\\overrightarrow{C B}=3 \\overrightarrow{F_{2} A}$, and $\\frac{\\overrightarrow{B F_{2}} \\cdot \\overrightarrow{B F_{1}}}{|\\overrightarrow{B F_{1}}|}=\\frac{\\overrightarrow{B F_{2}} \\cdot \\overrightarrow{B C}}{|\\overrightarrow{B C}|}$. Then the eccentricity of the hyperbola $\\Gamma$ is?",
    "fact_expressions": "Gamma: Hyperbola;a: Number;b: Number;lambda:Number;A: Point;B: Point;F1: Point;C: Point;F2: Point;a>0;b>0;Expression(Gamma) = (x^2/a^2 - y^2/b^2 = 1);LeftFocus(Gamma) = F1;RightFocus(Gamma) = F2;PointOnCurve(A,LeftPart(Gamma));PointOnCurve(B,RightPart(Gamma));VectorOf(A,B)=lambda*VectorOf(F1,A);PointOnCurve(C, xAxis);VectorOf(C, B) = 3*VectorOf(F2, A);DotProduct(VectorOf(B, F2), VectorOf(B, F1))/Abs(VectorOf(B, F1)) = DotProduct(VectorOf(B, F2), VectorOf(B, C))/Abs(VectorOf(B, C))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(7)",
    "fact_spans": "[[[18, 84], [101, 104], [423, 434]], [[31, 84]], [[31, 84]], [[170, 180]], [[91, 94]], [[95, 98]], [[2, 9]], [[185, 189]], [[10, 17]], [[31, 84]], [[31, 84]], [[18, 84]], [[2, 90]], [[2, 90]], [[91, 110]], [[91, 110]], [[114, 169]], [[185, 195]], [[196, 245]], [[246, 421]]]",
    "query_spans": "[[[423, 440]]]",
    "process": "Analysis: $F_{1}F_{2}=2c$, $\\because\\overrightarrow{CB}=3\\overrightarrow{F_{2}A}$, so $F_{2}A \\parallel CB$ $\\therefore \\triangle F_{1}AF_{2} \\sim \\triangle F_{1}BC$, $\\therefore F_{2}C=4c$, let $AF_{2}=t$, then $BC=3t$. From $\\frac{\\overrightarrow{BF_{1}}\\cdot\\overrightarrow{BF_{2}}}{|\\overrightarrow{BF_{1}}|}=\\frac{\\overrightarrow{BF_{2}}\\cdot\\overrightarrow{BC}}{|\\overrightarrow{BC}|}$ it follows that $BF_{2}$ bisects $\\angle F_{1}BC$, by the angle bisector theorem we have $\\frac{BF_{1}}{BC}=\\frac{F_{1}F_{2}}{F_{2}C}=\\frac{2c}{4c}=\\frac{1}{2}$, $\\therefore BF_{1}=\\frac{3t}{2}$, $AF_{1}=\\frac{1}{3}BF_{1}=\\frac{t}{2}$, $AB=\\frac{2}{3}BF_{1}=t$, by the definition of hyperbola, $AF_{2}-AF_{1}=2a$, $\\therefore t-\\frac{t}{2}=2a$, i.e., $t=4a$ $\\textcircled{1}$, $BF_{1}-BF_{2}=2a$, $\\therefore BF_{2}=\\frac{3t}{2}-2a=t$, $BF_{2}=AB=AF_{2}=t$, thus $\\triangle ABF_{2}$ is an equilateral triangle, $\\therefore \\angle F_{2}BC=\\angle ABF_{2}=60^{\\circ}$, in $\\triangle F_{2}BC$, by the cosine law, $\\cos\\angle F_{2}BC=\\frac{BF_{2}^{2}+BC^{2}-F_{2}C^{2}}{2\\cdot BF_{2}\\cdot BC}$, i.e., $\\frac{1}{2}=\\frac{t^{2}+9t^{2}-16c^{2}}{2\\cdot t\\cdot 3t}$, simplifying yields $7t^{2}=16c^{2}$ $\\textcircled{2}$, from $\\textcircled{1}\\textcircled{2}$ we get $\\frac{c^{2}}{a^{2}}=7$, $\\therefore$ eccentricity $e=\\frac{c}{a}=\\sqrt{7}$"
  },
  {
    "text": "Given that the ellipse $\\frac{x^{2}}{3 m^{2}}+\\frac{y^{2}}{5 n^{2}}=1$ and the hyperbola $\\frac{x^{2}}{2 m^{2}}-\\frac{y^{2}}{3 n^{2}}=1$ have common foci, what is the equation of the asymptotes of the hyperbola?",
    "fact_expressions": "G: Hyperbola;n: Number;m: Number;H: Ellipse;Expression(G) = (-y^2/(3*n^2)+ x^2/(2*m^2) = 1);Expression(H) = (y^2/(5*n^2) + x^2/(3*m^2) = 1);Focus(H) = Focus(G)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=(pm*sqrt(3)/4)*x",
    "fact_spans": "[[[52, 102], [111, 114]], [[4, 51]], [[4, 51]], [[2, 51]], [[52, 102]], [[2, 51]], [[2, 108]]]",
    "query_spans": "[[[111, 123]]]",
    "process": ""
  },
  {
    "text": "Through the right vertex $A$ of the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a line with slope $-1$. This line intersects the asymptotes of $E$ at points $B$ and $C$. If $\\overrightarrow{B C}+2 \\overrightarrow{B A}=\\overrightarrow{0}$, then the equation of the asymptotes of hyperbola $E$ is?",
    "fact_expressions": "E: Hyperbola;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;A: Point;RightVertex(E) = A;G: Line;PointOnCurve(A, G) = True;Slope(G) = -1;Intersection(G, Asymptote(E)) = {B, C};B: Point;C: Point;2*VectorOf(B, A) + VectorOf(B, C) = 0",
    "query_expressions": "Expression(Asymptote(E))",
    "answer_expressions": "y=pm*2*x",
    "fact_spans": "[[[1, 62], [85, 88], [171, 177]], [[1, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[66, 69]], [[1, 69]], [[78, 80], [82, 84]], [[0, 80]], [[70, 80]], [[82, 103]], [[94, 97]], [[98, 101]], [[105, 169]]]",
    "query_spans": "[[[171, 185]]]",
    "process": "According to the problem, draw the figure below. Solving simultaneously the asymptote $ l_{1}: bx - ay = 0 $, we obtain $ B\\left(\\frac{a^{2}}{a+b}, \\frac{ab}{a+b}\\right) $. Since $ \\overrightarrow{BC} + 2\\overrightarrow{BA} = \\overrightarrow{0} $, it follows that $ \\overrightarrow{AC} = 3\\overrightarrow{AB} $, so $ \\frac{a^{2}}{a-b} - a = 3\\left(\\frac{a^{2}}{a+b} - a\\right) $. Hence, we get $ b = 2a $. The asymptotes of hyperbola $ E $ are: $ y = \\pm 2x $."
  },
  {
    "text": "Let the hyperbola $x^{2}-m y^{2}=1$ have an eccentricity not less than $\\sqrt{3}$. The minimum distance from a focus of this hyperbola to an asymptote is?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (-m*y^2 + x^2 = 1);Negation(Eccentricity(G)<sqrt(3))",
    "query_expressions": "Min(Distance(Focus(G), Asymptote(G)))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[1, 21], [39, 42]], [[4, 21]], [[1, 21]], [[1, 37]]]",
    "query_spans": "[[[39, 55]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ is a point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and point $Q(\\frac{1}{4}, 0)$, then the minimum value of $|P Q|$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);P: Point;PointOnCurve(P, G);Q: Point;Coordinate(Q) = (1/4, 0)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "3*sqrt(5)/4",
    "fact_spans": "[[[7, 44]], [[7, 44]], [[2, 6]], [[2, 48]], [[49, 69]], [[49, 69]]]",
    "query_spans": "[[[71, 84]]]",
    "process": "Let P(x,y), so when x=1, |PQ|=\\frac{3\\sqrt{5}}{4} is minimized"
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively. If there exists a point $P$ on the right branch of the hyperbola such that $OP=|OF_{1}|$ ($O$ is the origin) and $|PF_{1}|=\\sqrt{3} |PF_{2}|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;O: Origin;F1: Point;P: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P,RightPart(C));Abs(LineSegmentOf(O,P))= Abs(LineSegmentOf(O, F1));Abs(LineSegmentOf(P, F1)) = sqrt(3)*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[19, 83], [159, 162], [91, 94]], [[27, 83]], [[27, 83]], [[120, 123]], [[1, 8]], [[101, 104]], [[9, 16]], [[27, 83]], [[27, 83]], [[19, 83]], [[1, 89]], [[1, 89]], [[91, 104]], [[106, 119]], [[129, 157]]]",
    "query_spans": "[[[159, 168]]]",
    "process": ""
  },
  {
    "text": "The distance from the right focus $F(c, 0)$ of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ to the asymptotes of $C$ is $\\frac{\\sqrt{3}}{2} c$. Then the equations of the asymptotes of $C$ are?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;RightFocus(C) = F;F: Point;Coordinate(F) = (c, 0);c: Number;Distance(F, Asymptote(C)) = c*sqrt(3)/2",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "y = pm*sqrt(3)*x",
    "fact_spans": "[[[0, 60], [74, 77], [109, 112]], [[0, 60]], [[7, 60]], [[7, 60]], [[7, 60]], [[7, 60]], [[0, 73]], [[64, 73]], [[64, 73]], [[64, 73]], [[64, 107]]]",
    "query_spans": "[[[109, 119]]]",
    "process": "The asymptotes of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ are: $ y = \\pm \\frac{b}{a}x $, i.e., $ bx \\pm ay = 0 $. According to the given condition, $ \\frac{bc}{\\sqrt{a^{2}+b^{2}}} = \\frac{\\sqrt{3}}{2}c $, that is, $ \\frac{b}{\\sqrt{a^{2}+b^{2}}} = \\frac{\\sqrt{3}}{2} $, solving yields $ b = \\sqrt{3}a $. Therefore, the equations of the asymptotes of $ C $ are $ y = \\pm \\sqrt{3}x $."
  },
  {
    "text": "Given the parabola $x^{2}=8 y$ with focus $F$, the directrix $l$ intersects the $y$-axis at point $M$. From a point $P$ on the parabola, draw $P Q \\perp l$ intersecting $l$ at point $Q$. If $|P F|=\\frac{8}{3}$, then $\\angle Q P F$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 8*y);F: Point;Focus(G) = F;l: Line;Directrix(G) = l;M: Point;Intersection(l, yAxis) = M;P: Point;PointOnCurve(P, G) = True;IsPerpendicular(LineSegmentOf(P, Q), l) = True;Intersection(LineSegmentOf(P, Q), l) = Q;Q: Point;Abs(LineSegmentOf(P, F)) = 8/3;PointOnCurve(P,LineSegmentOf(P,Q))",
    "query_expressions": "AngleOf(Q, P, F)",
    "answer_expressions": "2*pi/3",
    "fact_spans": "[[[2, 16], [41, 44]], [[2, 16]], [[20, 23]], [[2, 23]], [[26, 29], [65, 68]], [[2, 29]], [[35, 39]], [[26, 39]], [[47, 50]], [[41, 50]], [[51, 64]], [[51, 73]], [[69, 73]], [[75, 94]], [40, 63]]",
    "query_spans": "[[[96, 112]]]",
    "process": "Using the definition of a parabola, find the coordinates of point P, then determine the coordinates of point Q, calculate the side lengths of $\\triangle PQF$, and use the cosine law to find the value of $\\angle QPF$. [Detailed solution] Let point $P(x_{0},y_{0})$. By the definition of the parabola, we have $|PQ|=|PF|=y_{0}+2=\\frac{8}{3}$, so $y_{0}=\\frac{2}{3}$, then $x_{0}=\\pm\\frac{4\\sqrt{3}}{3}$. Thus, point $Q\\left(\\pm\\frac{4\\sqrt{3}}{3},-2\\right)$. The focus of the parabola $x^{2}=8y$ is $F(0,2)$. In $\\triangle PQF$, by the cosine law, $\\cos\\angle QPF = \\frac{|PF|^{2}+|PQ|^{2}-|FQ|^{2}}{2|PF|\\cdot|PQ|} = -\\frac{1}{2}$. Since $|FQ| = \\sqrt{\\frac{16}{3}+(2+2)^{2}} = \\frac{8\\sqrt{3}}{3}$, and $0<\\angle QPF<\\pi$, therefore, $\\angle QPF = \\frac{2\\pi}{3}$."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively. If there exists a point $P$ on the right branch of the hyperbola such that $|O P|=|O F_{1}|$ ($O$ is the origin) and $|P F_{1}|=\\sqrt{3}|P F_{2}|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "F1: Point;F2: Point;C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, RightPart(C)) = True;Abs(LineSegmentOf(O, P)) = Abs(LineSegmentOf(O, F1));O: Origin;Abs(LineSegmentOf(P, F1)) = sqrt(3)*Abs(LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3) + 1",
    "fact_spans": "[[[1, 8]], [[9, 16]], [[19, 80], [88, 91], [162, 165]], [[19, 80]], [[26, 80]], [[26, 80]], [[26, 80]], [[26, 80]], [[1, 86]], [[1, 86]], [[98, 101]], [[88, 101]], [[103, 121]], [[122, 125]], [[131, 160]]]",
    "query_spans": "[[[162, 171]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the line $l$ passing through the point $A(0,2)$ and tangent to the parabola $C$: $y^{2}=6x$?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;Expression(C) = (y^2 = 6*x);Coordinate(A) = (0, 2);IsTangent(l,C);PointOnCurve(A,l)=True",
    "query_expressions": "Expression(l)",
    "answer_expressions": "{x=0,3*x+4*y+8=0}",
    "fact_spans": "[[[33, 38]], [[12, 30]], [[1, 10]], [[12, 30]], [[1, 10]], [[11, 38]], [0, 37]]",
    "query_spans": "[[[33, 42]]]",
    "process": "When the slope of line $ l $ does not exist, the line passing through point $ A(0,2) $ is $ x=0 $, which clearly satisfies the requirement. When the slope of line $ l $ exists, let the equation of $ l $ be $ y = kx + 2 $ ($ k \\neq 0 $). Solving the system of equations \n\\[\n\\begin{cases}\ny^{2} = 6x \\\\\ny = kx + 2\n\\end{cases}\n\\]\nand eliminating $ y $, we obtain $ k^{2}x^{2} + (4k - 6)x + 4 = 0 $. Since the desired line is tangent to the parabola $ C: y^{2} = 6x $, we have $ \\Delta = (4k - 6)^{2} - 16k^{2} = 0 $. Solving this gives $ k = \\frac{3}{4} $. At this time, $ l: y = \\frac{3}{4}x + 2 $, or equivalently $ 3x - 4y + 8 = 0 $. Therefore, the required equations of the lines are $ x = 0 $ and $ 3x - 4y + 8 = 0 $."
  },
  {
    "text": "Find the standard equation of a parabola with focus on the $y$-axis and distance from the focus to the directrix equal to $5$.",
    "fact_expressions": "G: Parabola;PointOnCurve(Focus(G),yAxis);Distance(Focus(G),Directrix(G)) = 5",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2=10*y,x^2=-10*y}",
    "fact_spans": "[[[23, 26]], [[0, 26]], [[10, 26]]]",
    "query_spans": "[[[23, 33]]]",
    "process": "Let the standard equation of the parabola be $x^{2}=2my$ ($m\\neq0$), then the distance from the focus to the directrix of the parabola is $|m|=5$, giving $m=\\pm5$. Therefore, the required standard equations of the parabola are $x^{2}=10y$ and $x^{2}=-10y$."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of ellipse $C$, and $P$ is a point on $C$. If $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$ and $|P F_{1}|=2|P F_{2}|$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Focus(C) = {F1, F2};PointOnCurve(P, C);DotProduct(VectorOf(P,F1),VectorOf(P,F2))=0;Abs(LineSegmentOf(P,F1))=2*Abs(LineSegmentOf(P,F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[18, 23], [33, 36], [33, 36]], [[29, 32]], [[2, 9]], [[10, 17]], [[2, 28]], [[29, 40]], [[43, 102]], [[103, 125]]]",
    "query_spans": "[[[127, 136]]]",
    "process": "By the definition of an ellipse, |PF_{1}| + |PF_{2}| = 2a, |F_{1}F_{2}| = 2c, and given |PF_{1}| = 2|PF_{2}|, then |PF_{1}| = \\frac{4a}{3}, |PF_{2}| = \\frac{2a}{3}. Also, \\overrightarrow{PF}_{1} \\cdot \\overrightarrow{PF_{2}} = 0, so |PF_{1}|^{2} + |PF_{2}|^{2} = |F_{1}F_{2}|^{2}, \\frac{a^{2}}{a} = 4c^{2}, \\cdot eccentricity e = \\frac{c}{a} = \\frac{\\sqrt{5}}{3}"
  },
  {
    "text": "What are the foci of the hyperbola $\\frac{(x-3)^{2}}{4}-\\frac{(y+2)^{2}}{9}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = ((x - 3)^2/4 - ((y + 2)^2)/9 = 1)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(pm*sqrt(13)+3,-2)",
    "fact_spans": "[[[0, 46]], [[0, 46]]]",
    "query_spans": "[[[0, 51]]]",
    "process": ""
  },
  {
    "text": "Draw a line $l$ with slope $\\frac{1}{2}$ passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$), intersecting the parabola at points $A$ and $B$. If the area of $\\triangle O A B$ is $2 \\sqrt{5}$ ($O$ being the origin), then $p=$?",
    "fact_expressions": "l: Line;G: Parabola;p: Number;O: Origin;A: Point;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G)=F;PointOnCurve(F,l);Slope(l)=1/2;Intersection(l,G)={A,B};Area(TriangleOf(O,A,B))=2*sqrt(5);F:Point",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[46, 51]], [[1, 22], [54, 57]], [[118, 121]], [[105, 108]], [[59, 62]], [[63, 66]], [[4, 22]], [[1, 22]], [[1, 28]], [[0, 51]], [[29, 51]], [[46, 68]], [[70, 104]], [[25, 28]]]",
    "query_spans": "[[[118, 123]]]",
    "process": "From the given conditions, the focus of the parabola is $ F(\\frac{p}{2},0) $, thus the equation of line $ l $ is $ x=2y+\\frac{p}{2} $. Substituting into the parabola equation yields $ y^{2}-4py-p^{2}=0 $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ y_{1}+y_{2}=4p $, $ y_{1}y_{2}=-p^{2} $. The area of $ \\triangle OAB $ is $ \\frac{1}{2}\\times\\frac{p}{2}|y_{1}-y_{2}|=\\frac{p}{4}\\sqrt{(y_{1}+y_{2})^{2}-4y_{1}y_{2}}=\\frac{\\sqrt{5}p^{2}}{2}=2\\sqrt{5} $, solving gives $ p=2 $. Hence the answer is: $ 2 $"
  },
  {
    "text": "Given the parabola $x^{2}=2 p y  (p>0)$ with focus $F$, and $O$ as the origin. A line $l$ passes through point $F$ and intersects the parabola at points $A$, $B$, and intersects the $x$-axis at $C(2p,0)$. If $|A B|=17$, then the area of $\\triangle O C F$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 2*(p*y));p: Number;p>0;F: Point;Focus(G) = F;O: Origin;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, G) = {A, B};C: Point;Coordinate(C) = (2*p, 0);Intersection(l, xAxis) = C;Abs(LineSegmentOf(A, B)) = 17",
    "query_expressions": "Area(TriangleOf(O, C, F))",
    "answer_expressions": "32",
    "fact_spans": "[[[2, 25], [52, 55]], [[2, 25]], [[5, 25]], [[5, 25]], [[28, 31], [47, 51]], [[2, 31]], [[32, 35]], [[41, 46]], [[41, 51]], [[57, 60]], [[61, 64]], [[41, 66]], [[74, 83]], [[74, 83]], [[41, 83]], [[85, 95]]]",
    "query_spans": "[[[97, 119]]]",
    "process": "\\because the line $ l $ passes through points $ F(0,\\frac{p}{2}) $ and $ C(2p,0) $, $\\therefore K_{FC} = \\frac{0 - \\frac{p}{2}}{2p - 0} = -\\frac{1}{4} $. Solving the system of equations of line $ l $ and the parabola:\n\\[\n\\begin{cases}\nx + \\frac{p}{2}, \\\\\ny = -\\frac{1}{4}x + \\frac{p}{2} \\\\\nx^2 = 2py\n\\end{cases}\n\\]\nwe obtain $ x^2 + \\frac{p}{2}x - p^2 = 0 $. Let $ A(x_1, y_1) $, $ B(x_2, y_2) $. By Vieta's formulas, from the chord length formula we get $ |AB| $: $ x_1 + x_2 = -\\frac{p}{2} $, $ x_1 x_2 = -p^2 $, $ \\sqrt{1 + (-\\frac{1}{4})^2} |x_1 - x_2| = \\frac{\\sqrt{17}}{4} \\sqrt{(-\\frac{p}{2})^2 - 4(-p^2)} = \\frac{17p}{8} = 17 $, $\\therefore p = 8$, $\\therefore S_{\\triangle OCF} = \\frac{1}{2} \\cdot OC \\cdot OF = \\frac{1}{2} \\cdot 2p \\cdot \\frac{p}{2} = 32 $. The final answer is $ 32 $."
  },
  {
    "text": "If an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is perpendicular to the line $x-3 y+1=0$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x - 3*y + 1 = 0);IsPerpendicular(OneOf(Asymptote(G)), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[1, 57], [82, 85]], [[4, 57]], [[4, 57]], [[64, 77]], [[4, 57]], [[4, 57]], [[1, 57]], [[64, 77]], [[1, 79]]]",
    "query_spans": "[[[82, 91]]]",
    "process": "According to the given condition, $-\\frac{b}{a}=-3$, then $\\frac{b}{a}=3$, so $e=\\sqrt{1+(\\frac{b}{a})^{2}}=\\sqrt{10}$"
  },
  {
    "text": "If the curve represented by the equation $\\frac{x^{2}}{m}+\\frac{y^{2}}{m-1}=1$ is an ellipse, then the range of values for $m$ is?",
    "fact_expressions": "G: Ellipse;H: Curve;m: Number;Expression(H) = (y^2/(m - 1) + x^2/m = 1);H = G",
    "query_expressions": "Range(m)",
    "answer_expressions": "(1, +\\infty)",
    "fact_spans": "[[[46, 48]], [[43, 45]], [[50, 53]], [[1, 45]], [[43, 48]]]",
    "query_spans": "[[[50, 60]]]",
    "process": ""
  },
  {
    "text": "If the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ passes through the point $P(0, \\sqrt{3})$, and the length of the major axis is twice the focal distance, then $a$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;P: Point;Coordinate(P) = (0, sqrt(3));a > b;b > 0;PointOnCurve(P,G);Length(MajorAxis(G))=2*FocalLength(G)",
    "query_expressions": "a",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 53], [74, 76]], [[1, 53]], [[3, 53]], [[88, 91]], [[55, 72]], [[55, 72]], [[3, 53]], [[3, 53]], [[1, 72]], [[74, 86]]]",
    "query_spans": "[[[88, 91]]]",
    "process": "From the given \\begin{cases}a=2c\\\\b=\\sqrt{3}\\\\a^{2}=b^{2}+c^{2}\\end{cases}, solve to get a=2"
  },
  {
    "text": "A line passing through the focus $F$ of the parabola $y^{2}=-x$ intersects the parabola at points $A$ and $B$, and the projections of $A$ and $B$ on the line $x=\\frac{1}{4}$ are $M$ and $N$, respectively. Then the measure of $\\angle M F N$ is?",
    "fact_expressions": "G: Parabola;Q: Line;M: Point;F: Point;N: Point;A: Point;B: Point;Z: Line;Expression(G) = (y^2 = -x);Expression(Z) = (x = 1/4);Focus(G) = F;PointOnCurve(F, Q);Intersection(Q, G) = {A, B};Projection(A, Z) = M;Projection(B, Z) = N",
    "query_expressions": "AngleOf(M, F, N)",
    "answer_expressions": "ApplyUnit(90, degree)",
    "fact_spans": "[[[1, 14], [24, 27]], [[21, 23]], [[71, 74]], [[17, 20]], [[75, 78]], [[28, 31], [39, 42]], [[32, 35], [43, 46]], [[47, 64]], [[1, 14]], [[47, 64]], [[1, 20]], [[0, 23]], [[21, 37]], [[39, 78]], [[39, 78]]]",
    "query_spans": "[[[80, 99]]]",
    "process": "As shown in the figure, by the definition of a parabola, we have AF = AM, ∴ ∠AFM = ∠AMF; since alternate interior angles are equal, ∠AMF = ∠MFK, ∴ ∠AFM = ∠MFK; similarly, it can be proved that ∠KFN = ∠NFB. Since ∠MFK + ∠KFN + ∠NFB = 180^{0}, ∴ ∠MFN = 90^{0}."
  },
  {
    "text": "The standard equation of a hyperbola that has the same foci as the ellipse $x^{2}+4 y^{2}=16$ and one asymptote given by $x+\\sqrt{3} y=0$ is?",
    "fact_expressions": "H: Ellipse;Expression(H) = (x^2 + 4*y^2 = 16);G: Hyperbola;Focus(G) = Focus(H);Expression(OneOf(Asymptote(G))) = (x + sqrt(3)*y = 0)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/3 = 1",
    "fact_spans": "[[[1, 21]], [[1, 21]], [[52, 55]], [[0, 55]], [[29, 55]]]",
    "query_spans": "[[[52, 62]]]",
    "process": "Given that the ellipse $x^{2}+4y^{2}=16$ has the same foci, the position of the foci of the hyperbola is on the x-axis and the value of $c$, and given the asymptotes of the hyperbola, it can be directly determined as $y=\\pm\\frac{b}{a}x$, or we can assume the hyperbola equation with asymptote $x+\\sqrt{3}y=0$ to be $\\frac{x^{2}}{3}-\\frac{y^{2}}{1}=\\lambda(\\lambda>0)$, find $\\lambda$ from $c$, and then find the hyperbola equation.  \n**Solution:**  \nMethod 1:  \n$x^{2}+4y^{2}=16$  \n$\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$  \n$\\therefore a^{2}=16, b^{2}=4$  \n$\\cdot c^{2}=16-4=12$  \n$\\because$ the hyperbola and the ellipse $x^{2}+4y^{2}=16$ have the same foci  \n$\\therefore$ the foci of the hyperbola lie on the x-axis, asymptotes are $y=\\pm\\frac{b}{a}x$  \nAlso $\\because$ one asymptote of the hyperbola is $x+\\sqrt{3}y=0$, i.e., $y=-\\frac{\\sqrt{3}}{3}$,  \n$\\therefore \\frac{b}{a}=\\frac{\\sqrt{3}}{3}$, i.e., $\\frac{b^{2}}{a^{2}}=\\frac{3}{9}$  \nAlso $\\because c^{2}=a^{2}+b^{2}=12$  \n$\\therefore a^{2}=9, b^{2}=3$  \n$\\therefore$ the equation of the hyperbola is $\\frac{x^{2}}{9}-\\frac{y^{2}}{3}=1$  \n\nMethod 2:  \n$x^{2}+4y^{2}=16$  \n$\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$  \n$\\therefore a^{2}=16, b^{2}=4$  \n$\\therefore c^{2}=16-4=12$  \nThe hyperbola and the ellipse $x^{2}+4y^{2}=16$ have the same foci  \n$\\therefore$ the foci of the hyperbola lie on the x-axis  \nAlso $\\because$ one asymptote of the hyperbola is $x+\\sqrt{3}y=0$  \n$\\therefore$ assume the hyperbola equation is $\\frac{x^{2}}{3}-\\frac{y^{2}}{1}=\\lambda(\\lambda>0)$  \n$\\frac{x^{2}}{3\\lambda}-\\frac{y^{2}}{\\lambda}=1(\\lambda>0)$  \n$a^{2}=3\\lambda, b^{2}=\\lambda$  \n$c^{2}=3\\lambda+\\lambda=4\\lambda=12$  \n$\\lambda=3$  \n$\\frac{x^{2}}{3}-\\frac{y^{2}}{1}=3$  \n$\\frac{x^{2}}{9}-\\frac{y^{2}}{3}=1$  \n$\\therefore$ the equation of the hyperbola is $\\frac{x^{2}}{9}-\\frac{y^{2}}{3}=1$"
  },
  {
    "text": "The trajectory equation of the center of a moving circle passing through the point $F(1 , 0)$ and tangent to the line $l$: $x=-1$ is?",
    "fact_expressions": "l: Line;G: Circle;F: Point;Expression(l)=(x=-1);Coordinate(F) = (1, 0);PointOnCurve(F,G);IsTangent(l,G)",
    "query_expressions": "LocusEquation(Center(G))",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[14, 27]], [[30, 32]], [[1, 12]], [[14, 27]], [[1, 12]], [[0, 32]], [[13, 32]]]",
    "query_spans": "[[[30, 41]]]",
    "process": ""
  },
  {
    "text": "Let $P$ be any point on the parabola $y^{2}=4 x$. Then the minimum value of the sum of the distance from $P$ to the focus of the parabola and the distance from $P$ to $Q(2 , 3)$ is?",
    "fact_expressions": "G: Parabola;Q: Point;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(Q) = (2, 3);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P,Focus(G))+Distance(P,Q))",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[5, 19], [27, 30]], [[34, 44]], [[1, 4], [25, 26]], [[5, 19]], [[34, 44]], [[1, 23]]]",
    "query_spans": "[[[25, 54]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $x^{2}-y^{2}=1$, and point $P$ lies on $C$ with $\\angle F_{1} P F_{2}=60^{\\circ}$, find the distance from $P$ to the $x$-axis.",
    "fact_expressions": "F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;C: Hyperbola;Expression(C) = (x^2 - y^2 = 1);P: Point;PointOnCurve(P, C);AngleOf(F1, P, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Distance(P, xAxis)",
    "answer_expressions": "sqrt(6)/2",
    "fact_spans": "[[[2, 9]], [[10, 17]], [[2, 47]], [[2, 47]], [[18, 41], [53, 56]], [[18, 41]], [[48, 52], [93, 96]], [[48, 57]], [[58, 91]]]",
    "query_spans": "[[[93, 106]]]",
    "process": "Let point $ P(x_{0},y_{0}) $. By the definition of the hyperbola, $ ||PF_{1}|-|PF_{2}||=2 $. By the law of cosines, $ \\cos\\angle F_{1}PF_{2}=\\frac{|PF_{1}|^{2}+|PF_{2}|^{2}-|F_{1}F_{2}|^{2}}{2|PF_{1}||PF_{2}|} $. Solving simultaneously gives $ |PF_{1}|\\cdot|PF_{2}|=4 $, and $ S_{\\triangle F_{1}PF_{2}}=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|\\sin\\angle F_{1}PF_{2}=\\frac{1}{2}|F_{1}F_{2}|y_{0} $. Solving yields $ y_{0}=\\frac{\\sqrt{6}}{2} $."
  },
  {
    "text": "Given that $(2,0)$ is a focus of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1  (b>0)$, then $b$=?",
    "fact_expressions": "G: Hyperbola;b: Number;b>0;Expression(G) = (x^2 - y^2/b^2 = 1);Coordinate(OneOf(Focus(G))) = (2, 0)",
    "query_expressions": "b",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[10, 49]], [[56, 59]], [[13, 49]], [[10, 49]], [[2, 54]]]",
    "query_spans": "[[[56, 61]]]",
    "process": "From the given conditions, we have c=2, a=1, b^{2}=c^{2}-a^{2}=3, so b=\\sqrt{3}"
  },
  {
    "text": "A line with slope $-1$ passes through the focus of the parabola $y^{2}=-4 x$ and intersects the parabola at points $A$ and $B$. Then $|A B|=$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;Expression(G) = (y^2 = -4*x);Slope(H)=-1;PointOnCurve(Focus(G),H);Intersection(H,G)={A,B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[12, 27], [32, 35]], [[8, 10]], [[38, 41]], [[42, 45]], [[12, 27]], [[0, 10]], [[8, 30]], [[8, 47]]]",
    "query_spans": "[[[49, 58]]]",
    "process": "[Solution and Explanation] Using the point-slope form, derive the equation of line AB. By solving it simultaneously with the parabola equation, obtain $x_{1}+x_{2}$, then apply the definition of the parabola to find the result. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, focus $F(-1,0)$, $p=2$, then the equation of line AB is $y=-(x+1)$. Solving simultaneously with the parabola equation:\n\\[\n\\begin{cases}\ny^{2}=-4x \\\\\ny=-x-1\n\\end{cases}\n\\]\nRearranging yields $x^{2}+6x+1=0$, so $x_{1}+x_{2}=-6$. By the definition of the parabola, we get $|AB|=|AF|+|BF|=\\frac{p}{2}-x_{1}+\\frac{p}{2}-x_{2}=8$"
  },
  {
    "text": "A point $M$ on the parabola $y^{2}=4 x$ is at a distance of $2$ from the focus. What is the horizontal coordinate of point $M$?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(M, G);Distance(M, Focus(G)) = 2",
    "query_expressions": "XCoordinate(M)",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 14]], [[18, 21], [33, 37]], [[0, 14]], [[0, 21]], [[0, 31]]]",
    "query_spans": "[[[33, 43]]]",
    "process": "The directrix equation of the parabola \\( y^{2} = 4x \\) is \\( x = -1 \\). Given that the distance from a point \\( M \\) on the parabola \\( y^{2} = 4x \\) to the focus is 2, and since the distance from any point on the parabola to the focus equals the distance from that point to the directrix, the horizontal coordinate of point \\( M \\) is: \\( 2 - 1 = 1 \\)."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=12x$ and the point $M(-3,4)$, a line passing through the focus of $C$ with slope $k$ intersects $C$ at points $A$ and $B$. If $\\overrightarrow{M A} \\cdot \\overrightarrow{M B}=0$, then the value of $k$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 12*x);M: Point;Coordinate(M) = (-3, 4);PointOnCurve(Focus(C), G) = True;Slope(G) = k;k: Number;G: Line;Intersection(G, C) = {A, B};A: Point;B: Point;DotProduct(VectorOf(M, A), VectorOf(M, B)) = 0",
    "query_expressions": "k",
    "answer_expressions": "3/2",
    "fact_spans": "[[[2, 22], [35, 38], [52, 55]], [[2, 22]], [[23, 33]], [[23, 33]], [[34, 51]], [[42, 51]], [[45, 48], [121, 124]], [[49, 51]], [[49, 66]], [[57, 60]], [[61, 64]], [[68, 119]]]",
    "query_spans": "[[[121, 128]]]",
    "process": ":\\because the parabola C: y^{2}=12x has focus F(3,0), \\therefore the line passing through points A and B is given by y=k(x-3). Solving the system \\begin{cases}y^{2}=12x\\\\y=k(x-3)\\end{cases}, we obtain k^{2}x^{2}-2(6+3k^{2})x+9k^{2}=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), then x_{1}+x_{2}=\\frac{12+6k^{2}}{k^{2}}, x_{1}x_{2}=9, \\therefore y_{1}+y_{2}=\\frac{12}{k}, y_{1}y_{2}=-36, \\because M(-3,4), \\therefore \\overrightarrow{MA}=(x_{1}+3,y_{1}-4), \\overrightarrow{MB}=(x_{2}+3,y_{2}-4), \\because \\overrightarrow{MA}\\cdot\\overrightarrow{MB}=0, \\therefore (x_{1}+3)(x_{2}+3)+(y_{1}-4)(y_{2}-4)=0. Substituting and simplifying yields k=\\frac{3}{2}"
  },
  {
    "text": "What is the eccentricity of the ellipse $3 x^{2}+4 y^{2}=12$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (3*x^2 + 4*y^2 = 12)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[0, 22]], [[0, 22]]]",
    "query_spans": "[[[0, 28]]]",
    "process": "Since $3x^{2}+4y^{2}=12$, it follows that $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, so $a=2$, $b=\\sqrt{3}$, $c=1$, therefore the eccentricity of the ellipse is $e=\\frac{1}{2}$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, let $P$ be an arbitrary point on the ellipse in the second quadrant. Draw a tangent line to the ellipse at $P$, intersecting the $y$-axis at point $Q$. Let $O$ be the origin. Draw $Q R \\perp O P$ with foot of perpendicular at $R$. Then the range of $|O R|+|O P|$ is?",
    "fact_expressions": "C: Ellipse;Q: Point;R: Point;O: Origin;P: Point;Expression(C) = (x^2/4 + y^2/3 = 1);PointOnCurve(P,C);Quadrant(P)=2;L:Line;TangentOnPoint(P,C)=L;Intersection(L,yAxis)=Q;PointOnCurve(Q,LineSegmentOf(Q,R));IsPerpendicular(LineSegmentOf(Q,R),LineSegmentOf(O,P));FootPoint(LineSegmentOf(Q,R),LineSegmentOf(O,P))=R",
    "query_expressions": "Range(Abs(LineSegmentOf(O, P)) + Abs(LineSegmentOf(O, R)))",
    "answer_expressions": "(2*sqrt(3),7/2)",
    "fact_spans": "[[[2, 44], [46, 48], [63, 65]], [[76, 80], [91, 95]], [[115, 118]], [[81, 84]], [[59, 62]], [[2, 44]], [[46, 62]], [[49, 62]], [], [[45, 68]], [[45, 80]], [[90, 111]], [[96, 111]], [[96, 118]]]",
    "query_spans": "[[[120, 140]]]",
    "process": "Let $ P(t,s) $ ($ t<0, s>0 $), then $ \\frac{t^{2}}{4} + \\frac{s^{2}}{3} = 1 $, i.e., $ 3t^{2} + 4s^{2} = 12 $. Obviously, the slope of tangent line $ PQ $ exists. Let the equation of $ PQ $ be: $ y - s = k(x - t) $. From \n\\[\n\\begin{cases}\ny = kx - (kt - s) \\\\\n3x^{2} + 4y^{2} = 12\n\\end{cases}\n\\]\neliminating $ y $ and simplifying yields: \n\\[\n(4k^{2} + 3)x^{2} - 8k(kt - s)x + 4(kt - s)^{2} - 12 = 0\n\\]\nThus, \n\\[\n\\Delta = 64k^{2}(kt - s)^{2} - 16(4k^{2} + 3)[(kt - s)^{2} - 3] = 0\n\\]\nSimplifying gives: \n\\[\n4k^{2} - (kt - s)^{2} + 3 = 0\n\\]\ni.e., \n\\[\n(4 - t^{2})k^{2} + 2tsk + 3 - s^{2} = 0\n\\]\nor equivalently, \n\\[\n\\frac{4}{3}s^{2}k^{2} + 2tsk + \\frac{3}{4}t^{2} = 0\n\\]\nSolving yields $ k = -\\frac{3t}{4s} $. Then the equation of line $ PQ $ is: \n\\[\ny - s = -\\frac{3t}{4s}(x - t)\n\\]\nWhen $ x = 0 $, $ y = \\frac{3}{5} $, i.e., point $ Q\\left(0, \\frac{3}{5}\\right) $. $ \\overrightarrow{OP} = (t, s) $, $ \\overrightarrow{OQ} = \\left(0, \\frac{3}{5}\\right) $. Since $ QR \\perp OP $, then \n\\[\n|OP| \\cdot |OR| = |OP| \\cdot |OQ| \\cos \\angle POQ = \\overrightarrow{OP} \\cdot \\overrightarrow{OQ} = 3\n\\]\nLet $ m = |OP| \\in (\\sqrt{3}, 2) $, then \n\\[\n|OR| + |OP| = \\frac{3}{|OP|} + |OP| = \\frac{3}{m} + m\n\\]\nThe function $ f(m) = \\frac{3}{m} + m $ is monotonically increasing on $ m \\in (\\sqrt{3}, 2) $, so $ f(\\sqrt{3}) < f(m) < f(2) $, i.e., \n\\[\n2\\sqrt{3} < f(m) < \\frac{7}{2}\n\\]\nHence, \n\\[\n2\\sqrt{3} < |OR| + |OP| < \\frac{7}{2}\n\\]\nTherefore, the range of $ |OR| + |OP| $ is $ \\left(2\\sqrt{3}, \\frac{7}{2}\\right) $."
  },
  {
    "text": "Given that the line $x=t$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$. If there exists a point $C$ on the parabola such that $AC \\perp BC$,  \nthen the range of values for $t$ is?",
    "fact_expressions": "H: Line;Expression(H) = (x = t);t: Number;G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;B: Point;Intersection(H, G) = {A, B};C: Point;PointOnCurve(C, G);IsPerpendicular(LineSegmentOf(A, C), LineSegmentOf(B, C))",
    "query_expressions": "Range(t)",
    "answer_expressions": "[4, +oo)",
    "fact_spans": "[[[2, 9]], [[2, 9]], [[68, 71]], [[10, 24], [37, 40]], [[10, 24]], [[25, 28]], [[29, 32]], [[2, 34]], [[43, 47]], [[37, 47]], [[50, 65]]]",
    "query_spans": "[[[68, 78]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x$ ($p>0$), the distance from the focus $F$ to the directrix is $4$. A line $l$ passing through point $F$ intersects the parabola $C$ at points $P$ and $Q$. If $|P F|=10$, then $|P Q|+|Q F|$=?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;Distance(F, Directrix(C)) = 4;l: Line;PointOnCurve(F, l);P: Point;Q: Point;Intersection(l, C) = {P, Q};Abs(LineSegmentOf(P, F)) = 10",
    "query_expressions": "Abs(LineSegmentOf(P, Q)) + Abs(LineSegmentOf(Q, F))",
    "answer_expressions": "15",
    "fact_spans": "[[[2, 29], [58, 64]], [[2, 29]], [[10, 29]], [[10, 29]], [[32, 35], [47, 51]], [[2, 35]], [[2, 45]], [[52, 57]], [[46, 57]], [[66, 69]], [[70, 73]], [[52, 75]], [[77, 87]]]",
    "query_spans": "[[[89, 104]]]",
    "process": "Since the distance from the focus F of the parabola to the directrix is 4, we have p=4, then the parabola C: y^{2}=8x. Let the coordinates of point P be (x_{1},y_{1}) and Q be (x_{2},y_{2}). Since |PF|=x_{1}+\\frac{p}{2}=10, we get x_{1}=8, then y_{1}=\\pm8, so k_{1}=\\frac{\\pm8}{8-2}=\\pm\\frac{4}{3}. Thus, the equation of line l is y=\\pm\\frac{4}{3}(x-2). Substituting into the parabola equation yields 2x^{2}-17x+8=0, hence x_{1}+x_{2}=\\frac{17}{2}, so x_{2}=\\frac{1}{2}. Therefore, |PQ|+|QF|=x_{1}+x_{2}+p+x_{2}+\\frac{p}{2}=15"
  },
  {
    "text": "What is the distance from a focus of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ to its asymptote?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1)",
    "query_expressions": "Distance(OneOf(Focus(G)),Asymptote(G))",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 39], [45, 46]], [[0, 39]]]",
    "query_spans": "[[[0, 54]]]",
    "process": ""
  },
  {
    "text": "Given that the line $y = a x - 3a$ and the curve $y = \\sqrt{4 - x^{2}}$ have two distinct intersection points, what is the range of real values for $a$?",
    "fact_expressions": "G: Line;a: Real;H: Curve;Expression(G) = (y = a*x - 3*a);Expression(H) = (y = sqrt(4 - x^2));NumIntersection(G, H) = 2",
    "query_expressions": "Range(a)",
    "answer_expressions": "(-2*sqrt(5)/5,0]",
    "fact_spans": "[[[2, 15]], [[46, 51]], [[16, 36]], [[2, 15]], [[16, 36]], [[2, 44]]]",
    "query_spans": "[[[46, 58]]]",
    "process": "According to the problem, the line $ y = ax - 3a = a(x - 3) $ passes through the fixed point $ P(3,0) $. The curve $ y = \\sqrt{4 - x^{2}} $ represents the upper half (including the intersection points with the x-axis) of a circle centered at the origin with radius 2. The graph is shown below. When the line $ y = ax - 3a $, i.e., the line $ ax - y - 3a = 0 $, is tangent to the circle, we have $ \\frac{|-3a|}{\\sqrt{a^{2} + 1}} = 2 $. Solving this gives $ a^{2} = \\frac{4}{5} $, so $ a = \\pm \\frac{2\\sqrt{5}}{5} $. From the graph, when the line intersects the circle at two distinct points, we have $ -\\frac{2\\sqrt{5}}{5} < a \\leqslant 0 $. Therefore, the range of real values of $ a $ is $ \\left( -\\frac{2\\sqrt{5}}{5},\\ 0 \\right] $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and point $P$ is a point on the right branch of the hyperbola such that $|P F_{2}|=|F_{1} F_{2}|$. Then the area of triangle $P F_{1} F_{2}$ equals?",
    "fact_expressions": "G: Hyperbola;P: Point;F2: Point;F1: Point;Expression(G) = (x^2/9 - y^2/16 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P,RightPart(G));Abs(LineSegmentOf(P, F2)) = Abs(LineSegmentOf(F1, F2))",
    "query_expressions": "Area(TriangleOf(P, F1,F2))",
    "answer_expressions": "48",
    "fact_spans": "[[[2, 41], [70, 73]], [[65, 69]], [[57, 64]], [[49, 56]], [[2, 41]], [[2, 64]], [[2, 64]], [[65, 78]], [[80, 105]]]",
    "query_spans": "[[[107, 131]]]",
    "process": "Slight"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, with left and right foci $F_{1}$ and $F_{2}$ respectively, the focus of the parabola $y^{2}=2 p x$ coincides with $F_{2}$. If point $P$ is a common point of the ellipse and the parabola, and $\\cos \\angle P F_{1} F_{2}=\\frac{7}{9}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;Focus(G) = F2;P: Point;OneOf(Intersection(C, G)) = P;Cos(AngleOf(P, F1, F2)) = 7/9",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(7+pm*sqrt(17))/16",
    "fact_spans": "[[[2, 59], [120, 122], [174, 176]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[68, 75]], [[76, 83], [104, 111]], [[2, 83]], [[2, 83]], [[84, 100], [123, 126]], [[84, 100]], [[87, 100]], [[84, 113]], [[115, 119]], [[115, 132]], [[133, 172]]]",
    "query_spans": "[[[174, 182]]]",
    "process": "Since point P lies on the parabola, we have \\cos\\angle PF_{1}F_{2} = \\frac{PF_{2}}{PF_{1}} = \\frac{7}{9}. \\because PF_{1} + PF_{2} = 2a, \\therefore PF_{1} = \\frac{9}{8}a, PF_{2} = \\frac{7}{8}a. By the law of cosines, \\cos\\angle PF_{1}F_{2} = \\frac{PF_{1}^{2} + F_{1}F_{2}^{2} - PF_{2}^{2}}{2PF_{1}F_{1}F_{2}} = \\frac{\\frac{81}{64}a^{2} - \\frac{49}{64}a^{2} + 4c^{2}}{2 \\cdot \\frac{9}{8}a \\cdot 2c} = \\frac{7}{9}, simplifying yields 8c^{2} - 7ac + a^{2} = 0. \\therefore 8e^{2} - 7e + 1 = 0. \\therefore e = \\frac{7 \\pm \\sqrt{17}}{16}."
  },
  {
    "text": "The coordinates of the focus of the parabola $y=\\frac{1}{2} x^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2/2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,1/2)",
    "fact_spans": "[[[0, 24]], [[0, 24]]]",
    "query_spans": "[[[0, 31]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, point $P$ lies on the hyperbola, and $\\angle F_{1} P F_{2}=90^{\\circ}$. Then the distance from point $P$ to the $x$-axis is?",
    "fact_expressions": "G: Hyperbola;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 - y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Distance(P, xAxis)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[17, 45], [56, 59]], [[1, 8]], [[97, 101], [51, 55]], [[9, 16]], [[17, 45]], [[1, 50]], [[1, 50]], [[51, 60]], [[62, 95]]]",
    "query_spans": "[[[97, 111]]]",
    "process": ""
  },
  {
    "text": "Given that the major axis length of an ellipse with foci on the $x$-axis is $12$ and the eccentricity is $\\frac{1}{3}$, what is the standard equation of this ellipse?",
    "fact_expressions": "G: Ellipse;PointOnCurve(Focus(G), xAxis);Length(MajorAxis(G))=12;Eccentricity(G)=1/3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/36+y^2/32=1",
    "fact_spans": "[[[11, 13], [44, 46]], [[2, 13]], [[11, 22]], [[11, 40]]]",
    "query_spans": "[[[44, 53]]]",
    "process": "According to the properties of the ellipse, find $ a $ and $ b $, then the standard equation of the ellipse can be obtained. From the given conditions:  \n\\begin{matrix}2a=12\\\\\\frac{c}{a}=\\frac{1}{3}\\end{matrix},  \nsolving gives $ a=6 $, $ c=2 $, $ b^{2}=a^{2}-c^{2}=36-4=32 $. Then the standard equation of the ellipse is $ \\frac{x^{2}}{36}+\\frac{y^{2}}{32}=1 $."
  },
  {
    "text": "Given that the ellipse $E$ has its center at the origin $O$, with foci on the $x$-axis, the minimum distance from a point on the ellipse to a focus is $\\sqrt{2}-1$, and the eccentricity is $\\frac{\\sqrt{2}}{2}$. If $A$, $B$, $C$ are three distinct points on the ellipse such that $\\overrightarrow{O A}+\\overrightarrow{O B}+\\overrightarrow{O C}=\\overrightarrow{0}$, then the area of $\\triangle A B C$ is?",
    "fact_expressions": "E: Ellipse;A: Point;B: Point;C: Point;O: Origin;Center(E) = O;PointOnCurve(Focus(E), xAxis);PointOnCurve(A, E);PointOnCurve(B, E);PointOnCurve(C, E);Negation(A=B);Negation(A=C);Negation(B=C);P:Point;PointOnCurve(P,G);Min(Distance(P,Focus(E)))=sqrt(2)-1;VectorOf(O, A) + VectorOf(O, B) + VectorOf(O, C) = 0;Eccentricity(E)=sqrt(2)/2",
    "query_expressions": "Area(TriangleOf(A, B, C))",
    "answer_expressions": "3*sqrt(6)/4",
    "fact_spans": "[[[2, 7], [26, 28], [91, 93]], [[79, 82]], [[83, 86]], [[87, 90]], [[11, 16]], [[2, 16]], [[2, 25]], [[79, 100]], [[79, 100]], [[79, 100]], [[79, 100]], [[79, 100]], [[79, 100]], [], [[26, 31]], [[26, 52]], [[102, 185]], [[26, 77]]]",
    "query_spans": "[[[187, 209]]]",
    "process": "According to the problem, the minimum distance from a point on ellipse $ E $ to its focus is $ \\sqrt{2}-1 $, and the eccentricity is $ \\frac{\\sqrt{2}}{2} $. We obtain the system:\n$$\n\\begin{cases}\na - c = \\sqrt{2} - 1 \\\\\n\\frac{c}{a} = \\frac{\\sqrt{2}}{2}\n\\end{cases}\n$$\nSolving gives\n$$\n\\begin{cases}\na = \\sqrt{2} \\\\\nc = 1\n\\end{cases}\n$$\nThen $ b^{2} = a^{2} - c^{2} = 1 $, so the ellipse equation is $ \\frac{x^{2}}{2} + y^{2} = 1 $.\n\nWhen the slope of line $ AB $ does not exist, let line $ AB: x = t $. Let $ A(t, \\sqrt{1 - \\frac{t^{2}}{2}}) $, $ B(t, -\\sqrt{1 - \\frac{t^{2}}{2}}) $. From $ \\overrightarrow{OA} + \\overrightarrow{OB} + \\overrightarrow{OC} = \\overrightarrow{0} $, we get $ x_{c} = -2t $, $ y_{c} = 0 $, so $ C(-2t, 0) $. Substituting $ (-2t, 0) $ into the ellipse equation yields $ t^{2} = \\frac{1}{2} $, so $ |t| = \\frac{\\sqrt{2}}{2} $. Then\n$$\nS_{\\Delta ABC} = \\frac{1}{2} \\times 2 \\times \\sqrt{1 - \\frac{t^{2}}{2}} \\times 3|t| = \\frac{3\\sqrt{6}}{4}\n$$\n\nWhen the slope of line $ AB $ exists, let line $ AB: y = kx + m $. Solve the system\n$$\n\\begin{cases}\ny = kx + m \\\\\nx^{2} + 2y^{2} = 2\n\\end{cases}\n$$\nwhich simplifies to $ (1 + 2k^{2})x^{2} + 4kmx + 2(m^{2} - 1) = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then\n$$\nx_{1} + x_{2} = -\\frac{4km}{1 + 2k^{2}}, \\quad x_{1}x_{2} = \\frac{2(m^{2} - 1)}{1 + 2k^{2}}\n$$\nLet $ C(x_{3}, y_{3}) $. From $ \\overrightarrow{OA} + \\overrightarrow{OB} + \\overrightarrow{OC} = \\overrightarrow{0} $, we get\n$$\nx_{3} = -(x_{1} + x_{2}) = \\frac{4km}{1 + 2k^{2}}, \\quad y_{3} = -(y_{1} + y_{2}) = -[k(x_{1} + x_{2}) + 2m] = -\\frac{2m}{1 + 2k^{2}}\n$$\nSubstituting into $ x^{2} + 2y^{2} = 2 $, we obtain $ 2 + 2k^{2} = 4m^{2} $. So\n$$\n|AB| = \\sqrt{1 + k^{2}} |x_{1} - x_{2}|, \\quad d = \\frac{|m|}{\\sqrt{1 + k^{2}}}\n$$\nwhere $ d $ is the distance from $ O $ to line $ AB $. Then\n$$\nS_{\\triangle OAB} = \\frac{1}{2} \\times d \\times |AB| = \\frac{1}{2} |m| \\times \\sqrt{ \\left( -\\frac{4km}{1 + 2k^{2}} \\right)^{2} - 4 \\times \\frac{2(m^{2} - 1)}{1 + 2k^{2}} } = \\frac{\\sqrt{2} |m|}{4m^{2}} \\times \\sqrt{3} |m| = \\frac{\\sqrt{6}}{4}\n$$\nThus\n$$\nS_{\\triangle ABC} = 3S_{\\triangle OAB} = \\frac{3\\sqrt{6}}{4}\n$$\nIn conclusion, the area of $ \\triangle ABC $ is $ \\frac{3\\sqrt{6}}{4} $."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and let the line $y=2x$ be an asymptote of the hyperbola $C$. Point $P$ lies on $C$, and if $|P F_{1}|-|P F_{2}|=4$, then what is the equation of the hyperbola $C$? What is its eccentricity?",
    "fact_expressions": "C: Hyperbola;a:Number;b:Number;a>0;b>0;G: Line;P: Point;F1: Point;F2: Point;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y = 2*x);LeftFocus(C) = F1;RightFocus(C) = F2;OneOf(Asymptote(C))=G;PointOnCurve(P, C);Abs(LineSegmentOf(P,F1))-Abs(LineSegmentOf(P,F2))=4",
    "query_expressions": "Expression(C);Eccentricity(C)",
    "answer_expressions": "x^2/4 - y^2/16 = 1\nsqrt(5)",
    "fact_spans": "[[[17, 78], [96, 102], [148, 154], [114, 117]], [[25, 78]], [[25, 78]], [[25, 78]], [[25, 78]], [[86, 95]], [[109, 113]], [[1, 8]], [[9, 16]], [[17, 78]], [[86, 95]], [[1, 84]], [[1, 84]], [[86, 108]], [[109, 119]], [[122, 145]]]",
    "query_spans": "[[[148, 159]], [[148, 164]]]",
    "process": "Since |PF| - |PF₂| = 4, we have 2a = 4, so a = 2. Given that the line y = 2x is an asymptote of hyperbola C, it follows that \\frac{b}{a} = 2, thus b = 4. Therefore, the equation of hyperbola C is \\frac{x^2}{4} - \\frac{y^{2}}{16} = 1; hence, the eccentricity is e = \\frac{c}{a} = \\frac{\\sqrt{4 + 16}}{2} = \\sqrt{5}."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through point $F_{1}$ and perpendicular to the $x$-axis intersects the ellipse $C$ at points $A$ and $B$. Then, the radius of the incircle of triangle $A B F_{2}$ is?",
    "fact_expressions": "C: Ellipse;H: Line;A: Point;B: Point;F1: Point;F2: Point;Expression(C) = (x^2/4 + y^2/3 = 1);LeftFocus(C) =F1;RightFocus(C)=F2;IsPerpendicular(xAxis,H);PointOnCurve(F1, H);Intersection(H, C) = {A, B}",
    "query_expressions": "Radius(InscribedCircle(TriangleOf(A,B,F2)))",
    "answer_expressions": "3/4",
    "fact_spans": "[[[2, 44], [89, 94]], [[86, 88]], [[96, 99]], [[100, 103]], [[53, 60], [70, 78]], [[61, 68]], [[2, 44]], [[2, 68]], [[2, 68]], [[78, 88]], [[69, 88]], [[86, 105]]]",
    "query_spans": "[[[107, 130]]]",
    "process": "From the equation of the ellipse, we have a=2, b=\\sqrt{3}, and since c^{2}=a^{2}-b^{2}, it follows that c=1. Thus, the left focus is F_{1}(-1,0) and the right focus is F_{2}(1,0). Since the line passing through point F_{1} and perpendicular to the x-axis intersects the ellipse C at points A and B, we have x_{A}=x_{B}=-1, y_{A}=-y_{B}=\\frac{b^{2}}{a}=\\frac{3}{2}, so A(-1,\\frac{3}{2}), B(-1,-\\frac{3}{2}). Therefore, S_{\\triangleABF}=\\frac{1}{2}|AB|\\cdot|F_{1}F_{2}|=\\frac{1}{2}\\times3\\times2=3. |AF_{2}|=|BF_{2}|=\\sqrt{2^{2}+(\\frac{3}{2})^{2}}=\\frac{5}{2}. Let the inradius be r, then \\frac{1}{2}(|AB|+|AF_{2}|+|BF_{2}|)\\cdot r=S_{\\triangleABF}, which gives \\frac{1}{2}(\\frac{5}{2}+\\frac{5}{2}+3)\\cdot r=3, so r=\\frac{3}{4}."
  },
  {
    "text": "Given that point $F$ is the right focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, a line passing through the origin intersects the ellipse at points $A$ and $P$, $PF$ is perpendicular to the $x$-axis, the line $AF$ intersects the ellipse at point $B$, and $PB \\perp PA$. Then the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;RightFocus(G) = F;H: Line;O: Origin;PointOnCurve(O, H);A: Point;P: Point;Intersection(H, G) = {A, P};IsPerpendicular(LineSegmentOf(P, F), xAxis);B: Point;Intersection(LineOf(A, F), G) = B;IsPerpendicular(LineSegmentOf(P, B), LineSegmentOf(P, A));e: Number;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[7, 59], [71, 73], [103, 105], [130, 132]], [[7, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[2, 6]], [[2, 63]], [[68, 70]], [[65, 67]], [[64, 70]], [[74, 78]], [[79, 83]], [[68, 83]], [[84, 95]], [[106, 110]], [[96, 110]], [[111, 127]], [[136, 139]], [[130, 139]]]",
    "query_spans": "[[[136, 141]]]",
    "process": ""
  },
  {
    "text": "The vertex of parabola $C$ is at the origin, and its focus lies on the coordinate axis. If $C$ passes through the point $(-2,3)$, then the equation of $C$ is?",
    "fact_expressions": "C: Parabola;Vertex(C) = O;O: Origin;PointOnCurve(Focus(C), axis) = True;G: Point;Coordinate(G) = (-2, 3);PointOnCurve(G, C) = True",
    "query_expressions": "Expression(C)",
    "answer_expressions": "{(y^2=-9*x/2),(x^2=4*y/3)}",
    "fact_spans": "[[[0, 6], [24, 27], [39, 42]], [[0, 14]], [[10, 14]], [[0, 22]], [[28, 37]], [[28, 37]], [[24, 37]]]",
    "query_spans": "[[[39, 47]]]",
    "process": "(1) When the vertex of the parabola is at the origin and the axis of symmetry is the x-axis, let its standard equation be \\( y^{2} = -2px \\) (\\( p > 0 \\)). Substituting the point \\((-2, 3)\\), we get \\( 9 = 4p \\), so \\( 2p = \\frac{9}{2} \\). Therefore, \\( y^{2} = -\\frac{9}{2}x \\). (2) When the axis of symmetry is the y-axis, let the equation of the parabola be \\( x^{2} = 2py \\) (\\( p > 0 \\)). Substituting the point \\((-2, 3)\\), we get \\( 4 = 6p \\), so \\( 2p = \\frac{4}{3} \\). Therefore, the equation of the parabola is \\( x^{2} = \\frac{4}{3}y \\). In conclusion, the standard equations of the parabola are \\( y^{2} = -\\frac{9}{2}x \\) or \\( x^{2} = \\frac{4}{3}y \\)."
  },
  {
    "text": "If the hyperbola $C$: $2x^{2}-y^{2}=m$ ($m>0$) intersects the directrix of the parabola $y^{2}=16x$ at points $A$ and $B$, and $|AB|=4\\sqrt{3}$, then the value of $m$ is?",
    "fact_expressions": "C: Hyperbola;m: Number;G: Parabola;A: Point;B: Point;m>0;Expression(C) = (2*x^2 - y^2 = m);Expression(G) = (y^2 = 16*x);Intersection(C, Directrix(G)) = {A,B};Abs(LineSegmentOf(A,B))=4*sqrt(3)",
    "query_expressions": "m",
    "answer_expressions": "20",
    "fact_spans": "[[[1, 31]], [[84, 87]], [[32, 47]], [[52, 55]], [[58, 61]], [[8, 31]], [[1, 31]], [[32, 47]], [[1, 63]], [[65, 82]]]",
    "query_spans": "[[[84, 91]]]",
    "process": "The directrix of $ y^{2}=16x $ is $ x=-4 $. Since $ C $ intersects the directrix $ x=-4 $ of the parabola $ y^{2}=16x $ at points $ A $ and $ B $, and $ |AB|=4\\sqrt{3} $, therefore $ A(-4,2\\sqrt{3}) $, $ B(-4,-2\\sqrt{3}) $. Substituting the coordinates of point $ A $ into the hyperbola equation gives $ 2(-4)^{2}-(2\\sqrt{3})^{2}=m $, thus $ m=20 $."
  },
  {
    "text": "Given point $A(0,2)$ and any point $P$ on the ellipse $\\frac{x^{2}}{4} + y^{2}=1$, the maximum value of $|P A|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;A: Point;Expression(G) = (x^2/4 + y^2 = 1);Coordinate(A) = (0, 2);PointOnCurve(P, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, A)))",
    "answer_expressions": "(2/3)*sqrt(21)",
    "fact_spans": "[[[12, 41]], [[46, 49]], [[2, 10]], [[12, 41]], [[2, 10]], [[12, 49]]]",
    "query_spans": "[[[51, 63]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $y^{2}=a x$ coincides with the left focus of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, then $a$=?",
    "fact_expressions": "G: Hyperbola;H: Parabola;a: Number;Expression(G) = (x^2/4 - y^2 = 1);Expression(H) = (y^2 = a*x);Focus(H) = LeftFocus(G)",
    "query_expressions": "a",
    "answer_expressions": "-4*sqrt(5)",
    "fact_spans": "[[[19, 47]], [[1, 15]], [[55, 58]], [[19, 47]], [[1, 15]], [[1, 53]]]",
    "query_spans": "[[[55, 60]]]",
    "process": "The left focus of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ is $(-\\sqrt{5},0)$. Since the focus of the parabola $y^{2}=ax$ coincides with the left focus of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, it follows that $a<0$, $\\frac{a}{4}=-\\sqrt{5}$, thus $a=-4\\sqrt{5}$."
  },
  {
    "text": "What is the standard equation of a parabola with its focus at the right focus of the ellipse $\\frac{x^{2}}{3}+y^{2}=1$?",
    "fact_expressions": "G: Parabola;H: Ellipse;Expression(H) = (x^2/3 + y^2 = 1);Focus(G)=RightFocus(H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=4*sqrt(2)*x",
    "fact_spans": "[[[36, 39]], [[1, 28]], [[1, 28]], [[0, 39]]]",
    "query_spans": "[[[36, 46]]]",
    "process": "From \\frac{x^{2}}{3}+y^{2}=1, the right focus is (\\sqrt{2},0), so the standard equation of the parabola is v2=4\\sqrt{2}x"
  },
  {
    "text": "If the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ and the line $y=\\sqrt{3} x$ have no intersection points, then the range of eccentricity $e$ is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = sqrt(3)*x);NumIntersection(G, H)=0;e: Number;Eccentricity(G) = e",
    "query_expressions": "Range(e)",
    "answer_expressions": "(1, 2]",
    "fact_spans": "[[[1, 58]], [[4, 58]], [[4, 58]], [[59, 75]], [[4, 58]], [[4, 58]], [[1, 58]], [[59, 75]], [[1, 78]], [[83, 86]], [[1, 86]]]",
    "query_spans": "[[[83, 93]]]",
    "process": "Since the asymptotes of the hyperbola are $ y = \\pm\\frac{b}{a}x $, in order for the line $ y = \\sqrt{3}x $ to have no intersection with the hyperbola, the line $ y = \\sqrt{3}x $ must lie between the two asymptotes. Therefore, we have $ \\frac{b}{a} \\leqslant \\sqrt{3} $, that is, $ b \\leqslant \\sqrt{3}a $, so $ b^{2} \\leqslant 3a^{2} $, $ c^{2} - a^{2} \\leqslant 3a^{2} $, that is, $ c^{2} \\leqslant 4a^{2} $, $ e^{2} \\leqslant 4 $, hence $ 1 < e \\leqslant 2 $."
  },
  {
    "text": "Ellipse $T$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ has two vertices $A(a, 0)$, $B(0, b)$. Perpendicular lines to $AB$ are drawn through $A$ and $B$, intersecting ellipse $T$ at points $D$ and $C$ (distinct from the vertices), respectively. If $BC=3AD$, then the eccentricity of ellipse $T$ is?",
    "fact_expressions": "T: Ellipse;Expression(T) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;A: Point;B: Point;In(A,Vertex(T));Coordinate(A) = (a, 0);Coordinate(B) = (0, b);In(B,Vertex(T));C: Point;D: Point;L1: Line;L2: Line;PointOnCurve(A,L1);IsPerpendicular(L1,LineSegmentOf(A,B));PointOnCurve(B,L2);IsPerpendicular(L2,LineSegmentOf(A,B));Intersection(L1,T) = D;Intersection(L2,T) = C;LineSegmentOf(B, C) = 3*LineSegmentOf(A, D);Negation(C = Vertex(T))",
    "query_expressions": "Eccentricity(T)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[0, 57], [104, 109], [139, 144]], [[0, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[62, 72], [85, 88]], [[74, 83], [89, 92]], [[0, 83]], [[62, 72]], [[74, 83]], [[0, 83]], [[114, 117]], [[110, 113]], [], [], [[84, 103]], [[84, 103]], [[84, 103]], [[84, 103]], [[84, 117]], [[84, 117]], [[126, 137]], [[104, 124]]]",
    "query_spans": "[[[139, 150]]]",
    "process": "This problem first gives, according to the conditions, the line BC: $ y = \\frac{a}{b}x + b $ and the line AD: $ y = \\frac{a}{b}(x - a) $. Solving simultaneously with the ellipse equation yields $ x_{C} = \\frac{-2a3b^{2}}{b^{4} + a^{4}} $, $ x_{D} = \\frac{a5 - ab}{b^{4} + a^{4}}\\frac{4}{4} $. Then, using $ BC = 3AD $, we obtain $ 3x_{D} - x_{C} = 3a $, i.e., $ a^{2} = 3b^{2} $, finally yielding the ellipse's eccentricity $ e = \\sqrt{1 - \\frac{b^{2}}{a^{2}}} = \\sqrt{1 - \\frac{1}{3}} = \\frac{\\sqrt{6}}{3} $. According to the conditions, we have $ k_{BC} = k_{AD} = -\\frac{1}{k_{AB}} = \\frac{a}{b} $. Since perpendiculars from A and B to AB intersect the ellipse T at D and C (distinct from vertices), the line BC: $ y = \\frac{a}{b}x + b $, and the line AD: $ y = \\frac{a}{b}(x - a) $. From \n\\[\n\\begin{cases}\ny = \\frac{a}{b}x + b \\\\\nb^{2}x^{2} + a^{2}y^{2} = a^{2}b^{2}\n\\end{cases}\n\\Rightarrow (b^{4} + a^{4})x^{2} + 2a^{3}b^{2}x = 0\n\\]\nso $ x_{C} + x_{B} = \\frac{-2a3k}{b^{4} + a}2 = \\frac{-2a3b^{2}}{4} $. From \n\\[\n\\begin{cases}\ny = \\frac{a}{b}(x - a) \\\\\nb^{2}x^{2} + a^{2}y^{2} = a^{2}b^{2}\n\\end{cases}\n\\Rightarrow (b^{4} + c(4))x^{2} - 2a5x + a^{6} - a^{2}b^{4} = 0\n\\]\nso $ x_{A} \\cdot x_{D} = \\frac{a^{6} - a_{2}b^{4}}{a4 + b^{4}} $, $ x_{D} = \\frac{a^{5} - ab^{4}}{b^{4} + a^{4}} $. Since $ CB = \\sqrt{1 + \\left(\\frac{a + b^{4}}{b}\\right)^{2}} \\cdot (0 - x_{C})^{x} $, $ AD = \\sqrt{1 + \\left(\\frac{a}{b}\\right)^{2}} \\cdot (a - x_{D}) $. From $ BC = 3AD $, we get $ 3x_{D} - x_{C} = 3a $, so $ a_{2} = 3b^{2} $, the eccentricity of ellipse T is $ e = \\sqrt{1 - \\frac{b^{2}}{a^{2}}} = \\sqrt{1 - \\frac{1}{3}} = \\frac{\\sqrt{6}}{3} $."
  },
  {
    "text": "Given $A(1, \\frac{1}{2})$, $B(-1, \\frac{1}{2})$, the difference between the slope of line $AM$ and the slope of line $BM$ is $1$. Point $F(0, \\frac{1}{2})$, $P$ is a point on the line $l$: $y=-\\frac{1}{2}$, $Q$ is the intersection point of line $PF$ and the trajectory $C$ of point $M$, and $\\overrightarrow{F P}=4 \\overrightarrow{F Q}$. Then $|Q F|$=?",
    "fact_expressions": "l: Line;M: Point;A: Point;B: Point;F: Point;P: Point;Q: Point;Coordinate(A) = (1, 1/2);Coordinate(B) = (-1, 1/2);Coordinate(F) = (0, 1/2);Slope(LineOf(A,M))-Slope(LineOf(B,M))=1;Expression(l)=(y = -1/2);PointOnCurve(P,l);C:Curve;Locus(M)=C;Intersection(LineOf(P,F),C)=Q;VectorOf(F, P) = 4*VectorOf(F, Q)",
    "query_expressions": "Abs(LineSegmentOf(Q, F))",
    "answer_expressions": "3/4",
    "fact_spans": "[[[98, 121]], [[138, 142]], [[2, 21]], [[23, 43]], [[72, 92]], [[94, 97]], [[126, 129]], [[2, 21]], [[23, 43]], [[72, 92]], [[44, 71]], [[98, 121]], [[94, 125]], [[145, 148]], [[138, 148]], [[126, 151]], [[153, 198]]]",
    "query_spans": "[[[200, 209]]]",
    "process": "Let M(x,y), then \\( k_{AM} - k_{BM} = \\frac{y-\\frac{1}{2}}{x-1} - \\frac{y-\\frac{1}{2}}{x+1} = 1 \\), leading to the equation of the trajectory C of point M being \\( x^{2} = 2y \\) (\\( x \\neq \\pm1 \\)), as shown in the figure below: Let point \\( P(t, -\\frac{1}{2}) \\), \\( Q(x_{0}, y_{0}) \\), \\( \\overrightarrow{FP} = (t, -1) \\), \\( \\overrightarrow{FQ} = (x_{0}, y_{0}-\\frac{1}{2}) \\). Since \\( \\overrightarrow{FP} = 4\\overrightarrow{FQ} \\), it follows that \\( 4(y_{0} - \\frac{1}{2}) = -1 \\), solving gives \\( y_{0} = \\frac{1}{4} \\). By the definition of the parabola, we obtain \\( |QF| = \\frac{1}{4} + \\frac{1}{2} = \\frac{3}{4} \\)."
  },
  {
    "text": "It is known that the right focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ coincides with the focus of the parabola $y^{2}=12 x$. Then, the distance from the focus of this hyperbola to its asymptote is equal to?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/b^2 = 1);b: Number;H: Parabola;Expression(H) = (y^2 = 12*x);RightFocus(G) = Focus(H)",
    "query_expressions": "Distance(Focus(G), Asymptote(G))",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 44], [72, 75], [79, 80]], [[2, 44]], [[5, 44]], [[49, 64]], [[49, 64]], [[2, 69]]]",
    "query_spans": "[[[72, 89]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $ C: \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $, the left focus is $ F $, the right vertex is $ A $, and point $ P $ is any point on the ellipse. Then the minimum value of $ \\overrightarrow{P F} \\cdot \\overrightarrow{P A} $ is?",
    "fact_expressions": "C: Ellipse;P: Point;F: Point;A: Point;Expression(C) = (x^2/4 + y^2/3 = 1);LeftFocus(C) = F;RightVertex(C)=A;PointOnCurve(P, C)",
    "query_expressions": "Min(DotProduct(VectorOf(P, F), VectorOf(P, A)))",
    "answer_expressions": "0",
    "fact_spans": "[[[2, 44], [66, 68]], [[61, 65]], [[49, 52]], [[57, 60]], [[2, 44]], [[2, 52]], [[2, 60]], [[61, 73]]]",
    "query_spans": "[[[75, 130]]]",
    "process": "Let the coordinates of point P be $(x_{0}, y_{0})$. From the given conditions, $F(-1, 0)$, $A(2, 0)$, then $-2 \\leqslant x_{0} \\leqslant 2$, $-\\sqrt{3} \\leqslant y_{0} \\leqslant \\sqrt{3}$, $\\overrightarrow{PF} = (-1 - x_{0}, -y_{0})$, $\\overrightarrow{PA} = (2 - x_{0}, -y_{0})$. From $\\frac{x_{0}^{2}}{4} + \\frac{y_{0}^{2}}{3} = 1$, we obtain $y_{0}^{2} = 3 - \\frac{3}{4}x_{0}^{2}$, so $\\overrightarrow{PF} \\cdot \\overrightarrow{PA} = x_{0}^{2} - x_{0} - 2 + y_{0}^{2} = \\frac{1}{4}x_{0}^{2} - x_{0} + 1 = \\frac{1}{4}(x_{0} - 2)^{2}$. Hence, when $x_{0} = 2$, $\\overrightarrow{PB}$, $\\overrightarrow{PA}$ attains the minimum value $0$."
  },
  {
    "text": "Let $F_{1}$, $F_{2}$ be the foci of the ellipse $16 x^{2}+25 y^{2}=400$, and $P$ be a point on the ellipse. Then what is the perimeter of $\\Delta P F_{1} F_{2}$? What is the maximum area of $\\triangle PF_{1} F_{2}$?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (16*x^2 + 25*y^2 = 400);Focus(G)={F1, F2};PointOnCurve(P, G)",
    "query_expressions": "Perimeter(TriangleOf(P,F1,F2));Max(Area(TriangleOf(P,F1,F2)))",
    "answer_expressions": "16\n12",
    "fact_spans": "[[[17, 42], [50, 52]], [[46, 49]], [[1, 8]], [[9, 16]], [[17, 42]], [[1, 45]], [[46, 56]]]",
    "query_spans": "[[[58, 86]], [[85, 119]]]",
    "process": ""
  },
  {
    "text": "Let $F$ be the focus of the parabola $C$: $y^{2}=4x$. A line $l$ passing through point $P(-1,0)$ intersects the parabola $C$ at points $A$ and $B$, and point $Q$ is the midpoint of segment $AB$. If $|FQ|=2$, then the slope of line $l$ equals?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;P: Point;F: Point;Q: Point;Expression(C) = (y^2 = 4*x);Coordinate(P) = (-1, 0);Focus(C) = F;PointOnCurve(P, l);Intersection(l, C) = {A, B};MidPoint(LineSegmentOf(A,B)) = Q;Abs(LineSegmentOf(F, Q)) = 2",
    "query_expressions": "Slope(l)",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[40, 45], [91, 96]], [[5, 24], [46, 52]], [[53, 56]], [[57, 60]], [[29, 39]], [[1, 4]], [[63, 67]], [[5, 24]], [[29, 39]], [[1, 27]], [[28, 45]], [[40, 62]], [[63, 78]], [[80, 89]]]",
    "query_spans": "[[[91, 102]]]",
    "process": "According to the problem, let $ l: y = k(x+1) $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, $ Q(x_{0}, y_{0}) $. Substituting the line equation $ l $ into the parabola yields $ k^{2}x^{2} + (2k^{2}-4)x + k^{2} = 0 $, then we obtain $ x_{0} = \\frac{2-k^{2}}{k^{2}} $, $ y_{0} = \\frac{2}{k} $. Then using $ |FQ| = 2 $, we can solve $ k = \\pm1 $. Finally, verifying $ k \\in (-1,1) $ gives the answer. Substitute $ l: y = k(x+1) $ into the parabola $ y^{2} = 4x $, simplifying gives $ k^{2}x^{2} + (2k^{2}-4)x + k^{2} = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, $ Q(x_{0}, y_{0}) $, so $ \\Delta = (2k^{2}-4)^{2} - 4k^{4} = -16k^{2} + 16 > 0 $, solving gives $ k \\in (-1,1) $. Then $ x_{1} + x_{2} = \\frac{4-2k^{2}}{k^{2}} $, $ y = k(x_{1} + x_{2} + 2) = \\frac{4}{k} $, $ \\therefore x_{0} = \\frac{2-k^{2}}{k^{2}} $, $ y_{0} = \\frac{2}{k} $. Since $ |FQ| = 2 $, we have $ \\sqrt{(x_{0}-1)^{2} + y_{0}^{2}} = 2 $, i.e., $ \\left( \\frac{2-k^{2}}{k^{2}} - 1 \\right) + \\left( \\frac{2}{k} \\right)^{2} = 4 $, solving gives $ k = \\pm1 $. But $ k = \\pm1 $ does not satisfy $ k \\in (-1,1) $, therefore the slope of line $ l $ does not exist."
  },
  {
    "text": "If the ellipse $m x^{2}+n y^{2}=1$ $(m>0, n>0)$ intersects the line $y=1-x$ at points $A$ and $B$, and the slope of the line passing through the origin and the midpoint of segment $AB$ is $\\frac{\\sqrt{2}}{2}$, then the value of $\\frac{n}{m}$ is?",
    "fact_expressions": "G: Ellipse;n: Number;m: Number;H: Line;B: Point;A: Point;O:Origin;m>0;n>0;Expression(G) = (m*x^2 + n*y^2 = 1);Expression(H) = (y = 1 - x);Intersection(G, H) = {A, B};PointOnCurve(O,LineSegmentOf(O,MidPoint(LineSegmentOf(A,B))));Slope(LineSegmentOf(O,MidPoint(LineSegmentOf(A,B))))=sqrt(2)/2",
    "query_expressions": "n/m",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 35]], [[3, 35]], [[3, 35]], [[36, 45]], [[51, 54]], [[47, 50]], [[58, 60]], [[3, 35]], [[3, 35]], [[1, 35]], [[36, 45]], [[1, 56]], [[57, 74]], [[57, 97]]]",
    "query_spans": "[[[99, 116]]]",
    "process": "Let the intersection points be A(x_{1},y_{1}), B(x_{2},y_{2}). Since the line intersects the ellipse, solve the system of equations:  \n\\begin{cases}mx^{2}+ny^{2}=1\\\\y=1-x\\end{cases},  \neliminating variables yields: (m+n)x^{2}-2nx+n-1=0,  \nso x_{1}+x_{2}=\\frac{2n}{m+n}, x_{1}x_{2}=\\frac{n-1}{m+n},  \ny_{1}+y_{2}=1-x_{1}+1-x_{2}=2-(x_{1}+x_{2})=2-\\frac{2n}{m+n}.  \nThe midpoint coordinates of AB are (\\frac{n}{m+n},1-\\frac{n}{m+n}), so the slope of the line connecting it to the origin is \\frac{1-\\frac{n}{m+n}}{\\frac{n}{m+n}}=\\frac{\\sqrt{2}}{2}, solving gives: \\frac{n}{m}=\\sqrt{2}, so the answer should be \\sqrt{2}."
  },
  {
    "text": "Given that the focal distance of the ellipse $\\frac{x^{2}}{20}+\\frac{y^{2}}{k}=1$ is $6$, what is the value of $k$?",
    "fact_expressions": "G: Ellipse;k: Number;Expression(G) = (x^2/20 + y^2/k = 1);FocalLength(G) = 6",
    "query_expressions": "k",
    "answer_expressions": "{11, 29}",
    "fact_spans": "[[[2, 40]], [[49, 52]], [[2, 40]], [[2, 47]]]",
    "query_spans": "[[[49, 56]]]",
    "process": "From the given $2c=6$, we get $c=3$. When the focus is on the x-axis, $20-k=9$, solving gives $k=11$. When the focus is on the y-axis, $k-20=9$, solving gives $k=29$. In conclusion, $k=11$ or $29$."
  },
  {
    "text": "The equation of the hyperbola that shares the same asymptotes as $\\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1$ and passes through the point $(2,2)$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/5 - y^2/4 = 1);C: Hyperbola;Asymptote(C) = Asymptote(G);H: Point;Coordinate(H) = (2, 2);PointOnCurve(H, C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "5*y^2/4-x^2=1",
    "fact_spans": "[[[1, 39]], [[1, 39]], [[57, 60]], [[0, 60]], [[48, 56]], [[48, 56]], [[47, 60]]]",
    "query_spans": "[[[57, 64]]]",
    "process": ""
  },
  {
    "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, with foci $F_{1}$, $F_{2}$, and $\\angle F_{1} P F_{2}=\\frac{\\pi}{2}$, then the ordinate of point $P$ is?",
    "fact_expressions": "P: Point;G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};AngleOf(F1, P, F2) = pi/2",
    "query_expressions": "YCoordinate(P)",
    "answer_expressions": "pm*9/4",
    "fact_spans": "[[[2, 5], [107, 111]], [[6, 44]], [[6, 44]], [[2, 47]], [[51, 58]], [[60, 67]], [[6, 67]], [[69, 105]]]",
    "query_spans": "[[[107, 117]]]",
    "process": ""
  },
  {
    "text": "The hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{n}=1$ has an eccentricity of $2$, and one of its foci coincides with the focus of the parabola $y^{2}=4 x$. Find the value of $mn$.",
    "fact_expressions": "G: Hyperbola;n: Number;m: Number;H: Parabola;Expression(G) = (-y^2/n + x^2/m = 1);Expression(H) = (y^2 = 4*x);Eccentricity(G) = 2;OneOf(Focus(G)) = Focus(H)",
    "query_expressions": "m*n",
    "answer_expressions": "3/16",
    "fact_spans": "[[[0, 38], [47, 48]], [[75, 79]], [[75, 79]], [[54, 68]], [[0, 38]], [[54, 68]], [[0, 46]], [[47, 73]]]",
    "query_spans": "[[[75, 83]]]",
    "process": ""
  },
  {
    "text": "Given that the center $C$ of a moving circle lies on the parabola $x^{2}=2 p y$ ($p>0$), the circle passes through the point $A(0, p)$ and intersects the $x$-axis at two points $M$ and $N$, then the maximum value of $\\sin \\angle M C N$ is?",
    "fact_expressions": "G: Parabola;p: Number;H: Circle;M: Point;C: Point;N: Point;p>0;Expression(G) = (x^2 = 2*p*y);Center(H)=C;PointOnCurve(C,G);Coordinate(A) = (0, p);PointOnCurve(A,H);Intersection(xAxis,H) ={M,N};A:Point",
    "query_expressions": "Max(Sin(AngleOf(M, C, N)))",
    "answer_expressions": "1",
    "fact_spans": "[[[11, 32]], [[14, 32]], [[2, 4], [35, 36]], [[59, 62]], [[7, 10]], [[63, 66]], [[14, 32]], [[11, 32]], [[2, 10]], [[7, 33]], [[38, 48]], [[35, 48]], [[35, 66]], [[38, 48]]]",
    "query_spans": "[[[68, 93]]]",
    "process": ""
  },
  {
    "text": "The line $l$ passing through the focus of the parabola $E$: $y^{2}=4x$ intersects $E$ at points $A$ and $B$, and intersects the directrix of $E$ at point $C$. If point $A$ lies in the first quadrant and $B$ is the midpoint of $AC$, then the slope of line $l$ is equal to?",
    "fact_expressions": "l: Line;E: Parabola;A: Point;B: Point;C:Point;Expression(E) = (y^2 = 4*x);PointOnCurve(Focus(E), l);Intersection(l, E) = {A, B};Intersection(l, Directrix(E)) = C;MidPoint(LineSegmentOf(A, C)) = B;Quadrant(A)=1",
    "query_expressions": "Slope(l)",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[25, 30], [89, 94]], [[2, 21], [31, 34], [48, 51]], [[37, 40], [63, 67]], [[41, 44], [75, 78]], [[56, 60]], [[2, 21]], [[0, 30]], [[25, 46]], [[25, 60]], [[75, 87]], [[63, 73]]]",
    "query_spans": "[[[89, 100]]]",
    "process": "Let the slope of line $ l $ be $ k $, so the equation of the line is $ y = k(x - 1) $. By solving the system of equations and using the relationship between roots and coefficients, we obtain $ x_{1}x_{2} = 1 $. Then, since $ B $ is the midpoint of $ A $ and $ C $, we get $ x_{1} = 2x_{2} + 1 $. Solving the system of equations, we find $ B\\left(\\frac{1}{2}, -\\sqrt{2}\\right) $. Then, using the slope formula, the solution can be obtained.  \n\n**Detailed Solution:** As shown in the figure, for the parabola $ E: y^{2} = 4x $, the focus coordinates are $ F(1, 0) $, and the directrix equation is $ x = -1 $. Let the slope of line $ l $ be $ k $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, $ C(-1, y_{3}) $. Thus, the equation of the line is $ y = k(x - 1) $, \\cdots\\cdots\\cdots\\cdots\\textcircled{1}  \n\n\\textcircled{2} Solving the system of equations  \n\\[\n\\begin{cases}\ny = k(x - 1) \\\\\ny^{2} = 4x\n\\end{cases}\n\\]\nwe obtain $ k^{2}x - (2k^{2} + 4)x + k^{2} = 0 $. Therefore, $ x_{1}x_{2} = 1 $. Since $ B $ is the midpoint of $ A $ and $ C $, we have $ x_{2} = \\frac{x_{1} + (-1)}{2} $, i.e., $ x_{1} = 2x_{2} + 1 $, \\cdots\\cdots\\cdots\\cdots\\textcircled{2}  \n\nSolving \\textcircled{1} and \\textcircled{2} together, we get $ x_{2} = \\frac{1}{2} $. Substituting into the equation of the parabola, we find $ y_{2} = -\\sqrt{2} $, so $ B\\left(\\frac{1}{2}, -\\sqrt{2}\\right) $. Hence, the slope of line $ BF $ is $ k = \\frac{0 - (-\\sqrt{2})}{1 - \\frac{1}{2}} = 2\\sqrt{2} $, i.e., the slope of line $ l $ is $ 2\\sqrt{2} $."
  },
  {
    "text": "Given the hyperbola $x^{2}-\\frac{y^{2}}{n}=1$ $(n>0)$, if its eccentricity is $5$, then the equation of the asymptotes of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;n: Number;n>0;Expression(G) = (x^2 - y^2/n = 1);Eccentricity(G)=5",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*2*sqrt(6)*x",
    "fact_spans": "[[[2, 35], [37, 38], [48, 51]], [[5, 35]], [[5, 35]], [[2, 35]], [[37, 45]]]",
    "query_spans": "[[[48, 59]]]",
    "process": "According to the problem, $\\frac{c}{a} = \\frac{\\sqrt{1+n}}{1} = 5$, solving gives $n = 24$, that is, $b = 2\\sqrt{6}$, so the equations of the asymptotes are $y = \\pm\\frac{b}{a}x = \\pm2\\sqrt{6}x$."
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ are $F_{1}$ and $F_{2}$, respectively. A line passing through $F_{2}$ with an inclination angle of $120^{\\circ}$ intersects the ellipse at a point $M$. If $M F_{1}$ is perpendicular to the $x$-axis, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Line;PointOnCurve(F2, H);Inclination(H) = ApplyUnit(120, degree);M: Point;OneOf(Intersection(H, G)) = M;IsPerpendicular(LineSegmentOf(M, F1), xAxis)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2-sqrt(3)",
    "fact_spans": "[[[0, 52], [107, 109], [138, 140]], [[0, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[61, 68]], [[69, 76], [78, 85]], [[0, 76]], [[0, 76]], [[104, 106]], [[77, 106]], [[86, 106]], [[115, 118]], [[104, 118]], [[120, 136]]]",
    "query_spans": "[[[138, 146]]]",
    "process": ""
  },
  {
    "text": "Given that the line $y=1-x$ intersects the asymptotes of the hyperbola $a x^{2}+b y^{2}=1$ $(a>0, b<0)$ at points $A$ and $B$, and the slope of the line passing through the origin and the midpoint of segment $AB$ is $-\\frac{\\sqrt{3}}{2}$, then $\\frac{a}{b}=$?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;A: Point;B: Point;L:Line;O:Origin;a>0;b<0;Expression(G) = (a*x^2 + b*y^2 = 1);Expression(H) = (y = 1 - x);Intersection(H, Asymptote(G)) = {A, B};PointOnCurve(O,L);PointOnCurve(MidPoint(LineSegmentOf(A,B)),L);Slope(L)=-sqrt(3)/2",
    "query_expressions": "a/b",
    "answer_expressions": "-sqrt(3)/2",
    "fact_spans": "[[[12, 44]], [[15, 44]], [[15, 44]], [[2, 11]], [[50, 53]], [[54, 57]], [[75, 77]], [[62, 64]], [[15, 44]], [[15, 44]], [[12, 44]], [[2, 11]], [[2, 59]], [[61, 77]], [[61, 77]], [[75, 102]]]",
    "query_spans": "[[[104, 119]]]",
    "process": "According to the hyperbola equation, express the asymptotes of the hyperbola. Solve simultaneously with the line equation to obtain the coordinates of points A and B. Use the midpoint formula to find the coordinates of midpoint M. Then use the slope formula to find \\frac{a}{b}. [Detailed Solution] The hyperbola is given by ax^{2}+by^{2}=1 (a>0, b<0), so its asymptotes are y=\\pm\\sqrt{\\frac{a}{-b}}\\cdot x. Since the line y=1-x intersects the asymptotes at points A and B, we have \\begin{cases}y=1-x\\\\y=\\pm\\sqrt{\\frac{a}{-b}}\\end{cases}. Thus, the coordinates of the two intersection points are A.\\frac{-}{1}. Let M be the midpoint of A and B; then by the midpoint formula, M(\\frac{1}{1+\\frac{a}{k}},\\frac{1}{1}). Given that k_{OM}=-\\frac{\\sqrt{3}}{2}, it follows that _{k_{OM}}=\\frac{y_{M}}{x_{M}}=\\frac{a}{b}=-\\frac{\\sqrt{3}}{2}"
  },
  {
    "text": "If the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ passes through the focus of the parabola $y^{2}=8 x$, and has the same foci as the hyperbola $x^{2}-y^{2}=1$, then the standard equation of this ellipse is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;I: Ellipse;b: Number;a: Number;Expression(G) = (x^2 - y^2 = 1);Expression(H) = (y^2 = 8*x);Expression(I) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(Focus(H), I);Focus(I) = Focus(G)",
    "query_expressions": "Expression(I)",
    "answer_expressions": "x^2/4 + y^2/2 = 1",
    "fact_spans": "[[[67, 85]], [[47, 61]], [[1, 46], [94, 96]], [[3, 46]], [[3, 46]], [[67, 85]], [[47, 61]], [[1, 46]], [[1, 64]], [[1, 91]]]",
    "query_spans": "[[[94, 103]]]",
    "process": "The focus of the parabola is (2,0), the foci of the hyperbola are (\\pm\\sqrt{2},0), so a=2, c=\\sqrt{2} \\therefore b^{2}=2, and the equation of the ellipse is \\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1"
  },
  {
    "text": "The standard equation of the hyperbola with the focus of the parabola $y^{2}=4 x$ as its vertex, the vertex of the parabola as its center, and eccentricity $2$ is?",
    "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 4*x);G: Hyperbola;Focus(H) = Vertex(G);Vertex(H) = Center(G);Eccentricity(G) = 2",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2 / 3 = 1",
    "fact_spans": "[[[1, 15]], [[1, 15]], [[36, 39]], [[0, 39]], [[0, 39]], [[28, 39]]]",
    "query_spans": "[[[36, 45]]]",
    "process": "From the given conditions: a=1, e=\\frac{c}{a}=2, c=2, b^{2}=3, the standard equation of the hyperbola is x^{2}-\\frac{y^{2}}{3}=1"
  },
  {
    "text": "The eccentricity of the ellipse $m x^{2}+y^{2}=1$ is $\\frac{\\sqrt{3}}{2}$, then what is its major axis length?",
    "fact_expressions": "G: Ellipse;Expression(G) = (m*x^2 + y^2 = 1);m: Number;Eccentricity(G) = sqrt(3)/2",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "{2,4}",
    "fact_spans": "[[[0, 19], [46, 47]], [[0, 19]], [[2, 19]], [[0, 44]]]",
    "query_spans": "[[[46, 53]]]",
    "process": "Transform the equation of the ellipse $ mx^{2} + y^{2} = 1 $ into: $ \\frac{\\frac{x^{2}}{1} + y^{2} = 1}{m} $. Consider two cases.  \n$ \\omega_{m}^{m} > 1 $: The eccentricity of the ellipse is $ \\frac{\\sqrt{3}}{2} $, then $ \\frac{1}{\\frac{m-1}{1}} = \\frac{3}{4} $, solving gives $ m = \\frac{1}{4} $, further yielding the major axis length as 4.  \n$ \\frac{m}{m} < 1 $: The eccentricity of the ellipse is $ \\frac{\\sqrt{3}}{3} $, then the major axis length is 2. The answers are 2 or 4."
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$), draw a line inclined at $45^{\\circ}$ to intersect the parabola at points $A$ and $B$. If the length of segment $AB$ is $8$, then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;H: Line;A: Point;B: Point;F: Point;p>0;Expression(G) = (y^2 = 2*p*x);Focus(G) = F;PointOnCurve(F, H);Inclination(H) = ApplyUnit(45, degree);Intersection(H, G) = {A, B};Length(LineSegmentOf(A,B))=8",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 22], [49, 52]], [[79, 82]], [[46, 48]], [[53, 56]], [[57, 60]], [[25, 28]], [[4, 22]], [[1, 22]], [[1, 28]], [[0, 48]], [[29, 48]], [[46, 62]], [[64, 77]]]",
    "query_spans": "[[[79, 84]]]",
    "process": ""
  },
  {
    "text": "Hyperbola $C_{2}$ shares the same asymptotes with hyperbola $C_{1}$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{5}=1$, and $C_{2}$ passes through point $M(3,5)$. Then the equation of hyperbola $C_{2}$ is?",
    "fact_expressions": "C1:Hyperbola;C2:Hyperbola;M: Point;Expression(C1)=(x^2/9-y^2/5=1);Asymptote(C1)=Asymptote(C2);Coordinate(M) = (3, 5);PointOnCurve(M, C2)",
    "query_expressions": "Expression(C2)",
    "answer_expressions": "y^2/20 - x^2/36 = 1",
    "fact_spans": "[[[11, 58]], [[0, 10], [67, 74], [86, 96]], [[75, 84]], [[11, 58]], [[0, 65]], [[75, 84]], [[67, 84]]]",
    "query_spans": "[[[86, 101]]]",
    "process": "Using the given conditions, set up the hyperbola equation and substitute M(3,5) to solve. Since hyperbola C_{2} shares the same asymptotes with hyperbola C_{1}: \\frac{x^{2}}{9}-\\frac{y^{2}}{5}=1, we can let C_{2}: \\frac{x^{2}}{9}-\\frac{y^{2}}{5}=\\lambda (\\lambda\\neq0). Since C_{2} passes through point M(3,5), we obtain: \\frac{9}{9}-\\frac{25}{5}=\\lambda=-4. Thus, C_{2}: \\frac{y^{2}}{20}-\\frac{x^{2}}{36}=1"
  },
  {
    "text": "Given that $F$ is the focus of the parabola $C$: $y=2 x^{2}$, and point $P(x , y)$ lies on the parabola $C$ with $x=1$, then $| P F |$=?",
    "fact_expressions": "C: Parabola;Expression(C) = (y = 2*x^2);F: Point;Focus(C) = F;P: Point;x0: Number;y0: Number;Coordinate(P) = (x0, y0);PointOnCurve(P, C);x0 = 1",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "17/8",
    "fact_spans": "[[[6, 25], [41, 47]], [[6, 25]], [[2, 5]], [[2, 28]], [[29, 40]], [[30, 40]], [[30, 40]], [[29, 40]], [[29, 48]], [[50, 55]]]",
    "query_spans": "[[[57, 68]]]",
    "process": "From $ y = 2x^2 $, we get $ x^2 = \\frac{1}{2}y $, then $ p = \\frac{1}{4} $; from $ x = 1 $, we get $ y = 2 $, and by the property of the parabola, we have $ |PF| = 2 + \\frac{p}{2} = 2 + \\frac{1}{8} = \\frac{17}{8} $."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the common foci of an ellipse and a hyperbola, $P$ is one of their common points, and $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$, the eccentricity of the ellipse is $e_{1}$, the eccentricity of the hyperbola is $e_{2}$, then $\\frac{1}{e_{1}^{2}}+\\frac{3}{e_{2}^{2}}$=?",
    "fact_expressions": "E: Hyperbola;C:Ellipse;P: Point;F1: Point;F2: Point;Focus(C)={F1,F2};Focus(E)={F1,F2};OneOf(Intersection(C,E))=P;AngleOf(F1,P,F2) = pi/3;Eccentricity(C) = e1;Eccentricity(E) = e2;e1:Number;e2:Number",
    "query_expressions": "3/(e2^2) + 1/(e1^2)",
    "answer_expressions": "4",
    "fact_spans": "[[[21, 24], [96, 99]], [[18, 20], [81, 83]], [[30, 33]], [[2, 9]], [[10, 17]], [[2, 29]], [[2, 29]], [[30, 42]], [[44, 80]], [[81, 95]], [[96, 110]], [[88, 95]], [[103, 110]]]",
    "query_spans": "[[[112, 155]]]",
    "process": ""
  },
  {
    "text": "It is known that the angle of inclination of one of the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $\\frac{\\pi}{3}$. Then the eccentricity $e$ of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;e: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Inclination(OneOf(Asymptote(G))) = pi/3;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 58], [89, 92]], [[5, 58]], [[5, 58]], [[96, 99]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 86]], [[89, 99]]]",
    "query_spans": "[[[96, 101]]]",
    "process": "From $\\frac{a}{b}=\\tan\\frac{\\pi}{3}$, we get $\\frac{b}{a}=\\frac{\\sqrt{3}}{3}$, so $e=\\sqrt{1+(\\frac{b}{a})^{2}}=\\frac{2\\sqrt{3}}{3}$."
  },
  {
    "text": "A point $M$ on the parabola $\\Gamma$: $y=x^{2}$ is at a distance of $1$ from the focus. What is the slope of the tangent line to the parabola $\\Gamma$ at point $M$?",
    "fact_expressions": "Gamma: Parabola;Expression(Gamma) = (y = x^2);M: Point;PointOnCurve(M, Gamma) = True;Distance(M, Focus(Gamma)) = 1",
    "query_expressions": "Slope(TangentOnPoint(M, Gamma))",
    "answer_expressions": "pm*sqrt(3)",
    "fact_spans": "[[[0, 22], [39, 50]], [[0, 22]], [[25, 28], [51, 55]], [[0, 28]], [[0, 38]]]",
    "query_spans": "[[[39, 64]]]",
    "process": "Let $ M(x_{0},y_{0}) $, the focus of the parabola $ r: y = x^{2} $ is $ (0,\\frac{1}{4}) $, the directrix is $ y = -\\frac{1}{4} $. The distance from $ M $ to the focus equals the distance to the directrix: $ y_{0} + \\frac{1}{4} = 1 $, so $ y_{0} = 1 - \\frac{1}{4} = \\frac{3}{4} $, $ \\therefore x_{0} = \\pm \\frac{\\sqrt{3}}{2} $. Since $ y = x^{2} $, $ \\therefore y' = 2x $, $ \\therefore $ the slope of the tangent line to the parabola $ r $ at point $ M $ is $ 2x_{0} = \\pm \\sqrt{3} $."
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $y^{2}-\\frac{x^{2}}{3}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/3 + y^2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(sqrt(3)/3)*x",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 36]]]",
    "process": "From the given condition, we have $ y^{2} - \\frac{x^{2}}{3} = 0 $, which gives $ y = \\pm \\frac{\\sqrt{3}}{3}x $. Therefore, the asymptotes of the hyperbola $ y^{2} - \\frac{x^{2}}{3} = 1 $ are $ y = \\pm \\frac{\\sqrt{3}}{3}x $."
  },
  {
    "text": "The coordinates of the focus of the parabola $x^{2}=4 \\sqrt{3} y$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*(sqrt(3)*y))",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, \\sqrt{3})",
    "fact_spans": "[[[0, 23]], [[0, 23]]]",
    "query_spans": "[[[0, 30]]]",
    "process": "The focus coordinates of the parabola $x^{2}=4\\sqrt{3}y$ are $(0,\\sqrt{3})$."
  },
  {
    "text": "The curve $x^{2}-\\frac{y^{2}}{3}=1$ and the line $y=k x+1$ have two intersection points; what is the range of values for $k$?",
    "fact_expressions": "G: Line;k: Number;H: Curve;Expression(G) = (y = k*x + 1);Expression(H) = (x^2 - y^2/3 = 1);NumIntersection(H, G) = 2",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-2,2)&Negation(k=pm*sqrt(3))",
    "fact_spans": "[[[28, 39]], [[46, 49]], [[0, 27]], [[28, 39]], [[0, 27]], [[0, 44]]]",
    "query_spans": "[[[46, 55]]]",
    "process": "Substituting $ y = kx + 1 $ into the hyperbola equation yields $ (3 - k^2)x^2 - 2kx - 4 = 0 $. When $ k = \\pm\\sqrt{3} $, the line and the curve intersect at one point, which does not satisfy the condition, so these values are discarded. When $ k \\neq \\pm\\sqrt{3} $, $ A = 4k^2 + 16(3 - k^2) = -12k^2 + 48 > 0 $, so $ -2 < k < 2 $. Therefore, the range of $ k $ is $ -2 < k < 2 $ and $ k \\neq \\pm\\sqrt{3} $."
  },
  {
    "text": "A moving circle is externally tangent to the circle $(x+3)^{2}+y^{2}=4$ and internally tangent to the circle $(x-3)^{2}+y^{2}=100$. Then, the trajectory equation of the center of the moving circle is?",
    "fact_expressions": "C1: Circle;C2: Circle;Expression(C2) = (y^2 + (x + 3)^2 = 4);IsOutTangent(C1,C2) = True;C3: Circle;Expression(C3) = (y^2 + (x - 3)^2 = 100);IsInTangent(C1,C3) = True",
    "query_expressions": "LocusEquation(Center(C1))",
    "answer_expressions": "x^2/36+y^2/27=1",
    "fact_spans": "[[[1, 3], [56, 58]], [[4, 24]], [[4, 24]], [[1, 26]], [[30, 52]], [[30, 52]], [[1, 54]]]",
    "query_spans": "[[[56, 67]]]",
    "process": "According to the condition of circles being tangent, use the definition of an ellipse to find the trajectory equation. The circle $(x+3)^{2}+y^{2}=4$ has center $F_{1}(-3,0)$ and radius $r=2$, the circle $(x-3)^{2}+y^{2}=100$ has center $F_{2}(3,0)$ and radius $R=10$, $|F_{1}F_{2}|=6$. Let the center of the moving circle be $P$, with radius $r_{0}$. Since circle $F_{1}$ is inside circle $F_{2}$, from the given conditions:  \n\\begin{cases}\n|PF_{1}|=r_{0}+2\\\\\n|PF_{2}|=10-r_{0} \\quad (2\\alpha=12\n\\end{cases},  \ni.e., $|PF_{1}|+|PF_{2}|=12>6$, $\\therefore$ the trajectory of point $P$ is an ellipse. $(2a=12$, $b^{2}=a^{2}-c^{2}=27$, so the trajectory equation of point $P$ is $\\frac{x^{2}}{36}+\\frac{y^{2}}{27}=1$.  \n\\begin{cases}\n2c=6\\\\\n\\end{cases}"
  },
  {
    "text": "Given that the line $y = kx + m$ intersects the two asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ $(a > 0, b > 0)$ at points $A$ and $B$, and intersects $y = \\frac{1}{k}x$ at point $N$. If $N$ is the midpoint of $AB$, then the eccentricity of the hyperbola equals?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;m: Number;k: Number;A: Point;B: Point;N: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = k*x + m);C:Curve;Expression(C)=(y=x/k);Intersection(H,Asymptote(G))={A,B};Intersection(H,C)=N;MidPoint(LineSegmentOf(A, B)) = N",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[14, 70], [128, 131]], [[17, 70]], [[17, 70]], [[2, 13]], [[4, 13]], [[4, 13]], [[78, 81]], [[82, 85]], [[108, 112], [114, 117]], [[17, 70]], [[17, 70]], [[14, 70]], [[2, 13]], [[89, 106]], [[89, 106]], [[2, 87]], [[2, 112]], [[114, 126]]]",
    "query_spans": "[[[128, 138]]]",
    "process": "The hyperbola $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ $(a>0, b>0)$ has two asymptotes given by $y = \\pm\\frac{b}{a}x$. Then  \n$$\n\\begin{cases}\ny = kx + m \\\\\ny = -\\frac{b}{a}x\n\\end{cases}\n\\Rightarrow x_{A} = \\frac{-am}{ka + b},\n$$\nsimilarly, $x_{B} = \\frac{am}{b - ka}$.  \nSolving  \n$$\n\\begin{cases}\ny = kx + m \\\\\ny = \\frac{1}{k}x\n\\end{cases}\n\\Rightarrow x_{N} = \\frac{km}{1 - k^{2}}.\n$$\nSince $N$ is the midpoint of $AB$, we have $x_{A} + x_{B} = 2x_{N}$, that is  \n$$\n\\frac{-am}{b + ka} + \\frac{am}{b - ka} = \\frac{2km}{1 - k^{2}}.\n$$\nSimplifying yields $\\frac{b^{2}}{a^{2}} = 1$, therefore $e = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{2}$."
  },
  {
    "text": "Given that the focus of the parabola $C$ is $F$, and a line passing through $F$ intersects the parabola $C$ at points $A$ and $B$, if $\\frac{1}{|A F|}+\\frac{1}{|B F|}=2$, then an equation of a parabola $C$ satisfying the condition is?",
    "fact_expressions": "C: Parabola;G: Line;A: Point;F: Point;B: Point;Focus(C) = F;PointOnCurve(F,G);Intersection(G, C) = {A, B};1/Abs(LineSegmentOf(B, F)) + 1/Abs(LineSegmentOf(A, F)) = 2",
    "query_expressions": "Expression(C)",
    "answer_expressions": "y^2=2*x",
    "fact_spans": "[[[2, 8], [2, 8], [2, 8]], [[21, 23]], [[32, 35]], [[12, 15], [12, 15]], [[36, 39]], [[2, 15]], [[16, 23]], [[21, 41]], [[43, 78]]]",
    "query_spans": "[[[85, 98]]]",
    "process": "According to the problem, we can assume one form of the parabola's equation as $ y^{2} = 2px $, and the equation of the line passing through $ F $ is $ y = k\\left(x - \\frac{p}{2}\\right) $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. From the system  \n$$\n\\begin{cases}\ny^{2} = 2px \\\\\ny = k\\left(x - \\frac{p}{2}\\right)\n\\end{cases}\n$$\neliminating $ y $, we obtain  \n$$\nk^{2}x^{2} - (pk^{2} + 2p)x + \\frac{p^{2}}{4}k^{2} = 0\n$$\nThus,  \n$$\nx_{1} + x_{2} = p + \\frac{2p}{k^{2}}, \\quad x_{1}x_{2} = \\frac{p^{2}}{4},\n$$\n$$\n\\frac{1}{|AF|} + \\frac{1}{|BF|} = \\frac{1}{x_{1} + \\frac{p}{2}} + \\frac{1}{x_{2} + \\frac{p}{2}} = \\frac{x_{1} + x_{2} + p}{x_{1}x_{2} + \\frac{p}{2}(x_{1} + x_{2}) + \\frac{p^{2}}{4}} = \\frac{p + \\frac{2p}{k^{2}} + p}{\\frac{p^{2}}{4} + \\frac{p}{2}\\left(p + \\frac{2p}{k^{2}}\\right) + \\frac{p^{2}}{4}} = \\frac{2p + \\frac{2p}{k^{2}}}{p^{2} + \\frac{p^{2}}{k^{2}}} = \\frac{2}{p}\n$$\nGiven that $ \\frac{1}{|AF|} + \\frac{1}{|BF|} = 2 $, we have $ \\frac{2}{p} = 2 $, so $ p = 1 $. Then, an equation of the parabola $ C $ is $ y^{2} = 2x $."
  },
  {
    "text": "If real numbers $x$, $y$ satisfy $\\frac{y}{2}=\\sqrt{1-x^{2}}$, then the range of $y-2x$ is?",
    "fact_expressions": "x: Real;y:Real;y/2 = sqrt(1 - x^2)",
    "query_expressions": "Range(-2*x + y)",
    "answer_expressions": "[-2,2*sqrt(2)]",
    "fact_spans": "[[[1, 6]], [[7, 10]], [[12, 40]]]",
    "query_spans": "[[[42, 56]]]",
    "process": ""
  },
  {
    "text": "Given that a focus of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{3 m}=1$ is $(2,0)$. If point $M(4,0)$ is known and point $N(x, y)$ is an arbitrary point on the hyperbola, then the minimum value of $M N$ is?",
    "fact_expressions": "G: Hyperbola;m: Number;M: Point;N: Point;Expression(G) = (-y^2/(3*m) + x^2/m = 1);Coordinate(M) = (4, 0);Coordinate(N) = (x1, y1);x1:Number;y1:Number;PointOnCurve(N, G);Coordinate(OneOf(Focus(G)))=(2,0)",
    "query_expressions": "Min(LineSegmentOf(M, N))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 42], [81, 84]], [[5, 42]], [[60, 69]], [[70, 80]], [[2, 42]], [[60, 69]], [[70, 80]], [[71, 80]], [[71, 80]], [[70, 90]], [2, 52]]",
    "query_spans": "[[[92, 103]]]",
    "process": "From the given condition, we have $ m + 3m = 4 $, $\\therefore m = 1$, $\\therefore$ the equation of the hyperbola is $ x^{2} - \\frac{y^{2}}{3} = 1 $. From $ x^{2} - \\frac{y^{2}}{3} = 1 $, we get $ y^{2} = 3x^{2} - 3 $, $\\therefore MN = \\sqrt{(x-4)^{2} + y^{2}} = \\sqrt{4x^{2} - 8x + 13} = \\sqrt{4(x-1)^{2} + 9} $. Also, $ x < -1 $ or $ x > 1 $, $\\therefore$ when $ x = 1 $, $ MN $ attains its minimum value, which is $ 3 $."
  },
  {
    "text": "A asymptote of the hyperbola $k x^{2}-y^{2}=1$ is perpendicular to the line $2 x-y+3=0$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;k: Number;H: Line;Expression(G) = (k*x^2 - y^2 = 1);Expression(H) = (2*x - y + 3 = 0);IsPerpendicular(OneOf(Asymptote(G)), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[0, 20], [44, 47]], [[3, 20]], [[27, 40]], [[0, 20]], [[27, 40]], [[0, 42]]]",
    "query_spans": "[[[44, 53]]]",
    "process": "The asymptotes of the hyperbola $ka^{2}-y^{2}-1=0$ are $y=\\pm\\sqrt{k}x$. One asymptote is perpendicular to the line $2x-y+3=0$, so the slope of the asymptote is $-\\frac{1}{2}$, thus $\\frac{b}{a}=\\frac{1}{2}$, so $\\frac{c^{2}-a^{2}}{a^{2}}=\\frac{1}{4}$, hence $e=\\frac{\\sqrt{5}}{2}$."
  },
  {
    "text": "Given that a point $P$ on the parabola $x^{2}=8 y$ is at a distance of $6$ from its focus, then the coordinates of point $P$ are?",
    "fact_expressions": "G: Parabola;P: Point;Expression(G) = (x^2 = 8*y);PointOnCurve(P, G);Distance(P, Focus(G)) = 6",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(pm*4*sqrt(2),4)",
    "fact_spans": "[[[2, 16], [23, 24]], [[19, 22], [35, 39]], [[2, 16]], [[2, 22]], [[19, 33]]]",
    "query_spans": "[[[35, 44]]]",
    "process": "According to the definition of the parabola, we have $ y_{P} + 2 = 6 $, solving gives $ y_{P} = 4 $. Substituting into the equation of the parabola, the coordinates of point $ P $ can be found. From the given condition, a point $ P $ on the parabola $ x^{2} = 8y $ is at a distance of 6 from its focus. By the definition of the parabola, $ y_{P} + 2 = 6 $, solving gives $ y_{P} = 4 $. Substituting into the parabola $ x^{2} = 8y $, we get $ x^{2} = 8 \\times 4 = 32 $, solving gives $ x = \\pm 4\\sqrt{2} $. Therefore, the coordinates of point $ P $ are $ P(\\pm 4\\sqrt{2}, 4) $."
  },
  {
    "text": "The coordinates of the midpoint of the line segment cut by the line $y=x+2$ on the ellipse $x^{2}+2 y^{2}=4$ are?",
    "fact_expressions": "G: Ellipse;H: Line;Expression(G) = (x^2 + 2*y^2 = 4);Expression(H) = (y = x + 2)",
    "query_expressions": "Coordinate(MidPoint(InterceptChord(H,G)))",
    "answer_expressions": "(-4/3,2/3)",
    "fact_spans": "[[[10, 29]], [[0, 9]], [[10, 29]], [[0, 9]]]",
    "query_spans": "[[[0, 41]]]",
    "process": ""
  },
  {
    "text": "A hyperbola shares common foci with the ellipse $\\frac{x^{2}}{27}+\\frac{y^{2}}{36}=1$, and the ordinate of one of their intersection points is $4$. Then, the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/27 + y^2/36 = 1);Focus(G) = Focus(H);YCoordinate(OneOf(Intersection(G, H))) = 4",
    "query_expressions": "Expression(G)",
    "answer_expressions": "-x^2/5 + y^2/4 = 4",
    "fact_spans": "[[[1, 4], [70, 73]], [[5, 44]], [[5, 44]], [[1, 49]], [[1, 66]]]",
    "query_spans": "[[[70, 78]]]",
    "process": ""
  },
  {
    "text": "The equation of the parabola is $x^{2}=6 y$, then the length of the shortest chord passing through the focus of this parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 6*y);H: LineSegment;IsChordOf(H, G);PointOnCurve(Focus(G), H)",
    "query_expressions": "Min(Length(H))",
    "answer_expressions": "6",
    "fact_spans": "[[[0, 3], [21, 24]], [[0, 17]], [], [[21, 31]], [[19, 31]]]",
    "query_spans": "[[[21, 34]]]",
    "process": "From the given conditions, it is known that a chord passing through the focus must have a slope, denoted as $ k $. The coordinates of the focus of the parabola $ x^{2} = 6y $ are $ \\left(0, \\frac{3}{2}\\right) $. Then the equation of the line containing the chord is: $ y = kx + \\frac{3}{2} $. Combining this with the parabola's equation gives:\n\\[\n\\begin{cases}\ny = kx + \\frac{3}{2} \\\\\nx^{2} = 6y\n\\end{cases}\n\\Rightarrow x^{2} - 6kx - 9 = 0\n\\]\nLet $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, thus we have $ x_{1} + x_{2} = 6k $, $ x_{1}x_{2} = -9 $,\n\\[\n|AB| = \\sqrt{1 + k^{2}} |x_{1} - x_{2}| = \\sqrt{1 + k^{2}} \\sqrt{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}} = \\sqrt{1 + k^{2}} \\sqrt{36k^{2} + 36} = 6(1 + k^{2})\n\\]\nClearly, when $ k = 0 $, $ AB $ is minimized, and the minimum value is 6."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with left focus at $F(-c, 0)$, point $P$ lies on the circle $F_{1}$: $x^{2}+y^{2}-2 c x=0$. If the segment $F P$ is perpendicularly bisected by an asymptote of $C$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F: Point;LeftFocus(C) = F;Coordinate(F) = (-c, 0);P: Point;F1: Circle;Expression(F1) = (-2*c*x + x^2 + y^2 = 0);PointOnCurve(P, F1) = True;PerpendicularBisector(LineSegmentOf(F,P)) =  OneOf(Asymptote(C));c:Number",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 63], [143, 146]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 78]], [[2, 78]], [[68, 78]], [[79, 83]], [[84, 115]], [[84, 115]], [[79, 116]], [[118, 141]], [[68, 78]]]",
    "query_spans": "[[[143, 152]]]",
    "process": "x^{2}+y^{2}-2cx=0,\\ (x-c)^{2}+y^{2}=c^{2}, the center is at (c,0), radius is c. The center F(c,0) is the right focus of C, |OF|=|OP|=|OF|=|PF|=c. Without loss of generality, assume point P lies in the first quadrant, triangle OPF is equilateral, then \\angle PFF=\\frac{\\pi}{6}, so the slope of line PF is k=\\tan\\frac{\\pi}{6}=\\frac{\\sqrt{3}}{3}. Thus -\\frac{b}{a}=-\\sqrt{3},\\ \\frac{b}{a}=\\sqrt{3}, therefore the eccentricity of C is e=\\frac{c}{a}=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{1+(\\frac{b}{a})^{2}}=\\sqrt{1+3}=2."
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has two foci $F_{1}$ and $F_{2}$. If $P$ is a point on it such that $|P F_{1}|=2|P F_{2}|$, then the range of values for the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Focus(G) = {F1,F2};PointOnCurve(P,G);Abs(LineSegmentOf(P,F1))=2*Abs(LineSegmentOf(P,F2))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[1/3,1)",
    "fact_spans": "[[[0, 52], [81, 82], [111, 113]], [[2, 52]], [[2, 52]], [[77, 80]], [[58, 65]], [[68, 75]], [[2, 52]], [[2, 52]], [[0, 52]], [[0, 75]], [[77, 85]], [[87, 109]]]",
    "query_spans": "[[[111, 123]]]",
    "process": "Let the horizontal coordinate of point P be x. Since |PF₁| = 2|PF₂|, according to the second definition of the ellipse, we have a + ex = 2(a − ex). Therefore, 3ex = a. Since x ≤ a, we have ex ≤ ea. Hence, (1/3)a ≤ ea, so e ≥ 1/3. Since 0 < e < 1, it follows that e ∈ [1/3, 1)."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus is $F$. A line passing through $F$ with an inclination angle of $60^{\\circ}$ intersects the ellipse $C$ at points $A$ and $B$. If $M(-5, \\sqrt{3})$ is the midpoint of segment $AB$, then the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;LeftFocus(C) = F;G: Line;PointOnCurve(F, G);Inclination(G) = ApplyUnit(60, degree);A: Point;B: Point;Intersection(G, C) = {A, B};M: Point;Coordinate(M) = (-5, sqrt(3));MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(10)/5",
    "fact_spans": "[[[2, 59], [95, 100], [143, 148]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[64, 67], [69, 72]], [[2, 67]], [[92, 94]], [[68, 94]], [[75, 94]], [[102, 105]], [[106, 109]], [[92, 111]], [[113, 130]], [[113, 130]], [[113, 141]]]",
    "query_spans": "[[[143, 154]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Since $ A $ and $ B $ lie on the ellipse, we have $ \\frac{x_{1}^{2}}{a^{2}} + \\frac{y_{1}^{2}}{b^{2}} = 1 $, $ \\frac{x_{2}^{2}}{a^{2}} + \\frac{y_{2}^{2}}{b^{2}} = 1 $. Subtracting these two equations gives $ \\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{a^{2}} + \\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{b^{2}} = 0 $. Given that $ M(-5,\\sqrt{3}) $ is the midpoint of segment $ AB $, we have $ x_{1}+x_{2} = -10 $, $ y_{1}+y_{2} = 2\\sqrt{3} $. Therefore, $ \\frac{-10}{a^{2}} + \\frac{2\\sqrt{3}}{b^{2}} = 0 $, which implies $ 3a^{2} = 5b^{2} $, or equivalently $ 2a^{2} = 5c^{2} $. Hence, $ e = \\frac{c}{a} = \\frac{\\sqrt{10}}{5} $."
  },
  {
    "text": "Let the focus of the parabola $x^{2}=4 y$ be $F$. A line $l$ passing through the point $P(1,5)$ intersects the parabola at points $A$ and $B$, and point $P$ is exactly the midpoint of segment $AB$. Then $|A F|+|B F|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);F: Point;Focus(G) = F;P: Point;Coordinate(P) = (1, 5);l: Line;PointOnCurve(P, l);A: Point;B: Point;Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F))",
    "answer_expressions": "12",
    "fact_spans": "[[[1, 15], [42, 45]], [[1, 15]], [[19, 22]], [[1, 22]], [[26, 35], [59, 63]], [[26, 35]], [[36, 41]], [[24, 41]], [[48, 51]], [[52, 55]], [[36, 57]], [[59, 75]]]",
    "query_spans": "[[[77, 92]]]",
    "process": ""
  },
  {
    "text": "If the slopes of the two asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are $k_{1}$, $k_{2}$, and $k_{1} k_{2}=-3$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;L1: Line;L2: Line;Asymptote(G) = {L1, L2};k1: Number;k2: Number;Slope(L1) = k1;Slope(L2) = k2;k1*k2 = -3",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 57], [105, 108]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [], [], [[1, 63]], [[68, 75]], [[77, 84]], [[1, 84]], [[1, 84]], [[86, 102]]]",
    "query_spans": "[[[105, 114]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{8}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{2}$ intersects the right branch of the hyperbola at points $A$ and $B$, and $\\angle F_{1} A B=\\frac{\\pi}{3}$. Then, the area of $\\triangle F_{1} A F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;F1: Point;A: Point;F2: Point;B: Point;L:Line;Expression(G) = (x^2/4 - y^2/8 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2,L);Intersection(F2,RightPart(G))={A,B};AngleOf(F1, A, B) = pi/3",
    "query_expressions": "Area(TriangleOf(F1, A, F2))",
    "answer_expressions": "8*sqrt(3)",
    "fact_spans": "[[[2, 40], [77, 80]], [[48, 55]], [[84, 87]], [[56, 63], [66, 73]], [[88, 91]], [[74, 76]], [[2, 40]], [[2, 63]], [[2, 63]], [[65, 76]], [[74, 93]], [[95, 127]]]",
    "query_spans": "[[[130, 160]]]",
    "process": "Let $|AF_{1}|=m$, $|AF_{2}|=n$. From the law of cosines, derive a relation between $m$ and $n$, then another from the definition of the hyperbola; combining both allows solving for $mn$, thus obtaining the triangle area. Given $a^{2}=4$, $b^{2}=8$, so $c=\\sqrt{4+8}=2\\sqrt{3}$, hence $F_{1}(-2\\sqrt{3},0)$, $F_{2}(2\\sqrt{3},0)$. Let $|AF_{1}|=m$, $|AF_{2}|=n$. Since $\\angle F_{1}AB=\\frac{\\pi}{3}$, it follows that $|F_{1}F_{2}|^{2}=m^{2}+n^{2}-2mn\\cos\\frac{\\pi}{3}=m^{2}+n^{2}-mn$. While $m-n=2a=4$, so $(m-n)^{2}+mn=48$, $mn=48-4^{2}=32$. $S_{\\Delta AF_{1}F_{2}}=\\frac{1}{2}mn\\sin\\frac{\\pi}{3}=\\frac{1}{2}\\times32\\times\\frac{\\sqrt{3}}{2}=8\\sqrt{3}$."
  },
  {
    "text": "From a point $P$ on the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$, draw a perpendicular to the $x$-axis, with foot at $Q$. Then the equation of the locus of the midpoint $M$ of segment $P Q$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/2 = 1);P: Point;PointOnCurve(P, G);Z: Line;PointOnCurve(P, Z);IsPerpendicular(Z, xAxis);Q: Point;FootPoint(Z, xAxis) = Q;M: Point;MidPoint(LineSegmentOf(P, Q)) = M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2-2*y^2=1",
    "fact_spans": "[[[1, 29]], [[1, 29]], [[32, 35]], [[1, 35]], [], [[0, 43]], [[0, 43]], [[47, 50]], [[0, 50]], [[61, 64]], [[52, 64]]]",
    "query_spans": "[[[61, 71]]]",
    "process": "By the given condition, let P(x_{0},y_{0}), M(x,y), then Q(x_{0},0), so \\begin{cases}x=x_{0}\\\\y=\\frac{1}{2}y_{0}\\end{cases}, that is, \\begin{cases}x_{0}=x\\\\y_{0}=2y\\end{cases}. Since x_{0}^{2}-\\frac{y_{0}^{2}}{2}=1, then x^{2}-2y^{2}=1, so the trajectory equation of M is x^{2}-2y^{2}=1."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1$, let $A$ be the left vertex, and $B$, $C$ be points on the left and right branches of the hyperbola respectively, with $BC \\parallel x$-axis. Perpendiculars to the lines $AB$ and $AC$ are drawn through $B$ and $C$ respectively, and these two perpendiculars intersect at point $D$. If $S_{\\triangle BCD} = \\frac{27\\sqrt{3}}{4}$, then $|BC| = $?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/4 = 1);A: Point;LeftVertex(G) = A;B: Point;C: Point;PointOnCurve(B, LeftPart(G));PointOnCurve(C, RightPart(G));IsParallel(LineSegmentOf(B, C), xAxis);L1: Line;L2: Line;PointOnCurve(B, L1);PointOnCurve(C, L2);IsPerpendicular(L1, LineOf(A, B));IsPerpendicular(L2, LineOf(A, C));D: Point;Intersection(L1, L2) = D;Area(TriangleOf(B, C, D)) = (27*sqrt(3))/4",
    "query_expressions": "Abs(LineSegmentOf(B, C))",
    "answer_expressions": "3*sqrt(6)",
    "fact_spans": "[[[2, 40], [58, 61]], [[2, 40]], [[44, 47]], [[2, 47]], [[48, 51], [86, 89]], [[52, 55], [90, 93]], [[48, 70]], [[48, 70]], [[72, 84]], [], [], [[85, 113]], [[85, 113]], [[85, 113]], [[85, 113]], [[120, 124]], [[85, 124]], [[126, 169]]]",
    "query_spans": "[[[171, 180]]]",
    "process": "It is easy to see that point $ A(-3,0) $. According to the problem, let point $ B(x_{0},y_{0}) $ with $ x_{0}<0 $, then point $ C(-x_{0},y_{0}) $, $ \\frac{x_{0}^{2}}{9}-\\frac{y_{0}^{2}}{4}=1 $ and $ y_{0}\\neq0 $. The slope of line $ AB $ is $ k_{AB}=\\frac{y_{0}}{x_{0}+3} $, then the equation of line $ BD $ is $ y-y_{0}=-\\frac{x_{0}+3}{y_{0}}(x-x_{0}) $. Similarly, the equation of line $ CD $ is $ y-y_{0}=\\frac{x_{0}-3}{y_{0}}(x+x_{0}) $. Solving simultaneously, we get \n\\[\n\\begin{cases}\ny-y_{0}=-\\frac{x_{0}+3}{y_{0}}(x-x_{0}), \\\\\ny-y_{0}=\\frac{x_{0}-3}{y_{0}}(x+x_{0}),\n\\end{cases}\n\\]\nyielding $ \\left(3,\\frac{13y_{0}}{4}\\right) $. Combining with $ |BC|=2|x_{0}| $, we obtain $ S_{\\triangle BCD}=\\frac{1}{2}\\times2|x_{0}|\\times\\left|\\frac{13y_{0}}{4}-y_{0}\\right|=\\frac{9}{4}|x_{0}y_{0}| $. Since $ S_{\\triangle BCD}=\\frac{27\\sqrt{3}}{4} $, it follows that $ \\left(\\frac{9}{4}|x_{0}y_{0}|\\right)^{2}=\\left(\\frac{27\\sqrt{3}}{4}\\right)^2 $. Also, since $ \\frac{x_{0}^{2}}{9}-\\frac{y_{0}^{2}}{4}=1 $, we have $ y_{0}^{4}+4y_{0}^{2}-12=0 $, yielding $ y_{0}^{2}=2 $, then $ |x_{0}|=\\frac{3\\sqrt{6}}{2} $, so $ |BC|=2|x_{0}|=3\\sqrt{6} $."
  },
  {
    "text": "Given that point $A(-2,3)$ lies on the directrix of the parabola $C$: $y^{2}=2 p x$, a line passing through point $A$ is tangent to $C$ at point $B$ in the first quadrant. Let $F$ be the focus of $C$. Then the slope of line $B F$ is?",
    "fact_expressions": "C: Parabola;p: Number;G: Line;F: Point;B: Point;A: Point;Expression(C) = (y^2 = 2*(p*x));Coordinate(A) = (-2, 3);PointOnCurve(A, Directrix(C));PointOnCurve(A, G);TangentPoint(G, C) = B;Quadrant(B) = 1;Focus(C) = F",
    "query_expressions": "Slope(LineOf(B, F))",
    "answer_expressions": "4/3",
    "fact_spans": "[[[13, 34], [48, 51], [65, 68]], [[20, 34]], [[45, 47]], [[72, 75]], [[59, 63]], [[40, 44], [2, 12]], [[13, 34]], [[2, 12]], [[2, 38]], [[39, 47]], [[45, 63]], [[51, 63]], [[65, 75]]]",
    "query_spans": "[[[77, 89]]]",
    "process": ":\\because point A(-2,3) lies on the directrix of the parabola C: y^{2}=2px, i.e., the directrix equation is x=2, \\therefore p>0, -\\frac{p}{2}=-2, so p=4, \\therefore parabola C: y^{2}=8x, the equation in the first quadrant is y=2\\sqrt{2}\\sqrt{x}, let the tangent point be B(m,n), then n=2\\sqrt{2}\\sqrt{m}, and the derivative y=2\\sqrt{2}\\cdot\\frac{1}{x}\\cdot\\frac{1}{\\sqrt{x}}, thus the slope at the tangent point is \\frac{\\sqrt{2}}{\\sqrt{m}}, \\therefore \\frac{n-3}{m+2}=\\frac{\\sqrt{2}}{\\sqrt{m}} i.e. \\sqrt{2}m+2\\sqrt{2}=2\\sqrt{2}m-3\\sqrt{m}, solving gives \\sqrt{m}=2\\sqrt{2}, \\therefore tangent point B(8,8), and F(2,0), \\therefore the slope of line BF is \\frac{8-0}{8-2}=\\frac{4}{3}"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$, let the lower vertex be $A$. If the line $x=t y+4$ intersects the ellipse at two distinct points $M$ and $N$, then for what value of $t$ is the horizontal coordinate of the circumcenter of $\\Delta A M N$ maximized?",
    "fact_expressions": "G: Ellipse;H: Line;t: Number;A: Point;M: Point;N: Point;Expression(G) = (x^2/16 + y^2/4 = 1);Expression(H) = (x = t*y + 4);LowerVertex(G) = A;Intersection(H, G) = {M, N};Negation(M = N);WhenMax(XCoordinate(Circumcenter(TriangleOf(A, M, N))))",
    "query_expressions": "t",
    "answer_expressions": "2 - 2*sqrt(2)",
    "fact_spans": "[[[2, 40], [62, 64]], [[50, 61]], [[81, 84]], [[45, 48]], [[71, 74]], [[75, 78]], [[2, 40]], [[50, 61]], [[2, 48]], [[50, 78]], [[66, 78]], [[88, 110]]]",
    "query_spans": "[[[81, 86]]]",
    "process": ""
  },
  {
    "text": "A point $M$ on the parabola $y^{2}=2 p x(p>0)$ has a distance of $2$ to its directrix, and the distance from $M$ to the vertex of the parabola is equal to the distance from $M$ to its focus. Then, what are the coordinates of the focus of this parabola?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;M: Point;PointOnCurve(M, G);Distance(M, Directrix(G)) = 2;Distance(M, Vertex(G)) = Distance(M, Focus(G))",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(4/3, 0)",
    "fact_spans": "[[[0, 21], [28, 29], [45, 48], [59, 60], [69, 72]], [[0, 21]], [[3, 21]], [[3, 21]], [[24, 27], [40, 43], [55, 58]], [[0, 27]], [[24, 38]], [[40, 66]]]",
    "query_spans": "[[[69, 79]]]",
    "process": "Let the distance from point M(x,y) to its directrix be 2, then $x+\\frac{p}{2}=2$. Since the distance from M to the vertex of this parabola equals the distance from M to its focus, then $x^{2}+y^{2}=(x-\\frac{p}{2})^{2}+y^{2}$. Solving gives $x=\\frac{p}{4}$, $\\frac{p}{4}+\\frac{p}{2}=2$, $p=\\frac{8}{3}$. Therefore, the coordinates of the focus are $(\\frac{4}{3},0)$."
  },
  {
    "text": "Let the two foci of hyperbola $C$ be $(-2 , 0)$ and $(2 , 0)$, and one vertex be $(\\sqrt{2} , 0)$. Then the equation of $C$ is?",
    "fact_expressions": "C: Hyperbola;G: Point;H: Point;Coordinate(G) = (-2, 0);Coordinate(H) = (2, 0);Coordinate(OneOf(Vertex(C))) = (sqrt(2), 0);Focus(C) = {G, H}",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/2 - y^2/2 = 1",
    "fact_spans": "[[[1, 7], [60, 63]], [], [], [[1, 36]], [[1, 36]], [[1, 58]], [[1, 12]]]",
    "query_spans": "[[[60, 68]]]",
    "process": "The two foci of hyperbola C are (-2,0) and (2,0), and one vertex is (\\sqrt{2},0). Thus, we obtain a=\\sqrt{2}, c=2, and then find the value of b, allowing us to derive the equation of the hyperbola. According to the problem, the foci of hyperbola C lie on the x-axis, so let its equation be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0). From the given information, a=\\sqrt{2}, c=2, so b^{2}=c^{2}-a^{2}=2, b=\\sqrt{2}. Therefore, the equation of C is \\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1."
  },
  {
    "text": "Point $P$ is a moving point on the parabola $C$: $y^{2}=4x$. The minimum value of the sum of the distance from point $P$ to the point $(6, 12)$ and the distance from $P$ to the $y$-axis is?",
    "fact_expressions": "C: Parabola;G: Point;P: Point;Expression(C) = (y^2 = 4*x);Coordinate(G) = (6, 12);PointOnCurve(P, C)",
    "query_expressions": "Min(Distance(P,G)+Distance(P,yAxis))",
    "answer_expressions": "12",
    "fact_spans": "[[[5, 24]], [[35, 46]], [[30, 34], [0, 4]], [[5, 24]], [[35, 46]], [[0, 28]]]",
    "query_spans": "[[[30, 66]]]",
    "process": ""
  },
  {
    "text": "If the eccentricity $e \\in (1,2)$ of the hyperbola $\\frac{y^{2}}{5}-\\frac{x^{2}}{m}=1$, then what is the range of values for $m$?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (y^2/5 - x^2/m = 1);Eccentricity(G) = e;e:Number;In(e,(1,2))",
    "query_expressions": "Range(m)",
    "answer_expressions": "(0,15)",
    "fact_spans": "[[[1, 39]], [[58, 61]], [[1, 39]], [[1, 55]], [[43, 55]], [[43, 55]]]",
    "query_spans": "[[[58, 68]]]",
    "process": ""
  },
  {
    "text": "Given that the point $(1 , 1)$ is the midpoint of a chord of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$, then the equation of the line containing this chord is?",
    "fact_expressions": "G: Ellipse;H: LineSegment;Expression(G) = (x^2/4 + y^2/2 = 1);Coordinate(MidPoint(H)) = (1,1);IsChordOf(H,G)",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "x+2*y-3=0",
    "fact_spans": "[[[15, 52]], [], [[15, 52]], [[2, 58]], [[15, 55]]]",
    "query_spans": "[[[15, 71]]]",
    "process": "Let the chord of the ellipse with midpoint A(1,1) intersect the ellipse at E(x_{1},y_{1}), F(x_{2},y_{2}). Since A(1,1) is the midpoint of EF, we have x_{1}+x_{2}=2, y_{1}+y_{2}=2. Substituting E(x_{1},y_{1}) and F(x_{2},y_{2}) into the ellipse equation \\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1, we obtain \\frac{x_{1}^{2}}{4}+\\frac{y_{1}^{2}}{2}=1 and \\frac{x_{2}^{2}}{4}+\\frac{y_{2}^{2}}{2}=1. Subtracting these two equations yields (x_{1}+x_{2})(x_{1}-x_{2}) + 2(y_{1}+y_{2})(y_{1}-y_{2})=0. Therefore, 2(x_{1}-x_{2}) + 4(y_{1}-y_{2})=0, so k=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{1}{2}. Hence, the equation of the line containing the chord of the ellipse with midpoint A(1,1) is: y-1=-\\frac{1}{2}(x-1). Simplifying, we get x+2y-3=0."
  },
  {
    "text": "Given that the directrix of the parabola $y^{2}=-4 x$ passes through the focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1(b>0)$, then $b$=?",
    "fact_expressions": "G: Parabola;H: Ellipse;b: Number;Expression(G) = (y^2 = -4*x);b>0;Expression(H) = (x^2/4 + y^2/b^2 = 1);PointOnCurve(Focus(H), Directrix(G))",
    "query_expressions": "b",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 17]], [[22, 68]], [[73, 76]], [[2, 17]], [[24, 68]], [[22, 68]], [[2, 71]]]",
    "query_spans": "[[[73, 78]]]",
    "process": "According to the problem, the directrix of the parabola $ y^{2} = -4x $ is $ x = 1 $. Since the foci of the ellipse are $ (\\pm\\sqrt{4 - b^{2}}, 0) $, it follows that $ \\sqrt{4 - b^{2}} = 1 $. Thus, $ b^{2} = 3 $, so $ b = \\sqrt{3} $."
  },
  {
    "text": "Given points $A(2,0)$, $B(4,0)$, and a moving point $P$ traveling along the parabola $y^{2}=-4x$, then the coordinates of point $P$ that minimize $\\overrightarrow{A P} \\cdot \\overrightarrow{B P}$ are?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;P: Point;Expression(G) = (y^2 = -4*x);Coordinate(A) = (2, 0);Coordinate(B) = (4, 0);PointOnCurve(P, G);WhenMin(DotProduct(VectorOf(A, P), VectorOf(B, P)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(0, 0)",
    "fact_spans": "[[[28, 43]], [[2, 11]], [[13, 21]], [[24, 27], [105, 109]], [[28, 43]], [[3, 11]], [[13, 21]], [[24, 44]], [[50, 104]]]",
    "query_spans": "[[[105, 114]]]",
    "process": "Let the coordinates of the moving point P be (x_{0}, y_{0}), so we have y_{0}^{2} = -4x_{0} (x_{0} \\leqslant 0), thus \\overrightarrow{AP} = (x_{0} - 2, y_{0}), \\overrightarrow{BP} = (x_{0} - 4, y_{0}). \\overrightarrow{AP} \\cdot \\overrightarrow{BP} = (x_{0} - 2, y_{0}) \\cdot (x_{0} - 4, y_{0}) = x_{0}^{2} - 6x_{0} + 8 + y_{0}^{2}. Since y_{0}^{2} = -4x_{0}, it follows that \\overrightarrow{AP} \\cdot \\overrightarrow{BP} = x_{0}^{2} - 10x_{0} + 8 = (x_{0} - 5)^{2} - 17. Because x_{0} \\leqslant 0, when x_{0} = 0, \\overrightarrow{AP} \\cdot \\overrightarrow{BP} attains its minimum value of 8, so the coordinates of point P are (0, 0)."
  },
  {
    "text": "Given that for $k \\in R$, the line $y - kx - 1 = 0$ always has common points with the ellipse $\\frac{x^{2}}{5} + \\frac{y^{2}}{m} = 1$, then the range of real number $m$ is?",
    "fact_expressions": "k: Real;H: Line;Expression(H) = (-k*x + y - 1 = 0);G: Ellipse;Expression(G) = (x^2/5 + y^2/m = 1);m: Real;IsIntersect(H, G)",
    "query_expressions": "Range(m)",
    "answer_expressions": "[1, 5)+(5, +\\infty)",
    "fact_spans": "[[[3, 12]], [[13, 26]], [[13, 26]], [[27, 64]], [[27, 64]], [[71, 76]], [[13, 69]]]",
    "query_spans": "[[[71, 83]]]",
    "process": "The line $y = kx + 1$ passes through the fixed point $(0,1)$ outside the ellipse $\\frac{x^{2}}{5} + \\frac{y^{2}}{m} = 1$. Then, according to the definition of the ellipse, we obtain $m \\neq 5$. Combining these, the answer is that the line $y = kx + 1$ passes through the fixed point $(0,1)$, as long as $(0,1)$ is not outside the ellipse $\\frac{x^{2}}{5} + \\frac{y^{2}}{m} = 1$, thus $m \\geqslant 1$. Moreover, since in the ellipse $\\frac{x^{2}}{5} + \\frac{y^{2}}{m} = 1$ we have $m \\neq 5$, the range of values for $m$ is $[1,5) \\cup (5,+\\infty)$."
  },
  {
    "text": "If real numbers $x$, $y$ satisfy $\\frac{x|x|}{4}+\\frac{y|y|}{9}=1$, and the maximum value of $|3 x+2 y-t|$ is $3 \\sqrt{2}$, then the value of the real number $t$ is?",
    "fact_expressions": "x_: Real;y_:Real;t: Real;(x_*Abs(x_))/4 + (y_*Abs(y_))/9 = 1;Max(Abs(-t + 3*x_ + 2*y_)) = 3*sqrt(2)",
    "query_expressions": "t",
    "answer_expressions": "3*sqrt(2)",
    "fact_spans": "[[[1, 6]], [[7, 10]], [[79, 84]], [[12, 45]], [[47, 77]]]",
    "query_spans": "[[[79, 88]]]",
    "process": "When $x>0, y>0$, the curve is the graph of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$ in the first quadrant; when $x>0, y<0$, the curve is the graph of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$ in the fourth quadrant; when $x<0, y>0$, the curve is the graph of the hyperbola $\\frac{y^{2}}{9}-\\frac{x^{2}}{4}=1$ in the second quadrant; when $x<0, y<0$, the original equation has no real solutions. Since the line $3x+2y=0$ is the asymptote of the hyperbolas $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$ and $\\frac{y^{2}}{9}-\\frac{x^{2}}{4}=1$, let $3x+2y-t=0$, then $\\frac{|3x+2y-t|}{\\sqrt{13}}$ represents the distance from a point on the curve to the line $3x+2y-t=0$. Since the maximum value of $|3x+2y-t|$ is $3\\sqrt{2}$, the maximum value of $\\frac{|3x+2y-t|}{\\sqrt{13}}$ is $\\frac{3\\sqrt{26}}{3}$. From the graph, the point on the curve farthest from the line $3x+2y=0$ lies on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$. Let the moving point on the ellipse be $(2\\cos\\theta, 3\\sin\\theta)$, $\\theta\\in(0,\\frac{\\pi}{2})$. Using the point-to-line distance formula, we get $\\frac{|6\\cos\\theta+6\\sin\\theta|}{\\sqrt{3^{2}+2^{2}}}=\\frac{|6\\sqrt{2}\\sin(\\theta+\\frac{\\pi}{4})|}{\\sqrt{13}}\\leqslant\\frac{6\\sqrt{26}}{13}$. Since $\\frac{6\\sqrt{26}}{BB}>\\frac{3\\sqrt{26}}{B}$, in order for $\\frac{13x+2y-t}{\\sqrt{13}}$ to achieve the maximum value $\\frac{3\\sqrt{26}}{16}$, the line $3x+2y=0$ must be shifted upward. Thus, $\\frac{|t|}{\\sqrt{13}}=\\frac{3\\sqrt{26}}{13}$, solving gives $t=3\\sqrt{2}$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, $A$, $C$ are the upper and lower vertices of the ellipse, $B$ is the left vertex, $F$ is the left focus, the lines $AB$ and $FC$ intersect at point $D$, then the cosine value of $\\angle BDF$ is?",
    "fact_expressions": "G: Ellipse;B: Point;A: Point;F: Point;C: Point;D: Point;Expression(G) = (x^2/4 + y^2/3 = 1);UpperVertex(G)=A;LowerVertex(G)=C;LeftVertex(G)=B;LeftFocus(G) = F;Intersection(LineOf(A,B), LineOf(F, C)) = D",
    "query_expressions": "Cos(AngleOf(B, D, F))",
    "answer_expressions": "sqrt(7)/14",
    "fact_spans": "[[[2, 39], [50, 52]], [[59, 62]], [[40, 43]], [[67, 70]], [[44, 47]], [[91, 95]], [[2, 39]], [[40, 58]], [[40, 58]], [[50, 66]], [[50, 74]], [[75, 95]]]",
    "query_spans": "[[[97, 117]]]",
    "process": "From the given conditions, A(0,\\sqrt{3}), B(-2,0), C(0,-\\sqrt{3}), F(-1,0), line AB: \\sqrt{3}x-2y+2\\sqrt{3}=0, CF: \\sqrt{3}x+y\\frac{\\overrightarrow{DB}\\cdot\\overrightarrow{DF}}{|\\overrightarrow{DB}|\\cdot|\\overrightarrow{DF}|}=\\frac{\\sqrt{7}}{14}."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, respectively, and point $P$ is any point on the ellipse $C$, let $m=|P F_{1}| \\cdot|P F_{2}|$. Then the maximum value of $m$ is?",
    "fact_expressions": "C: Ellipse;Expression(C)=(x^2/25+y^2/16=1);P: Point;F1: Point;F2: Point;LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(P,C);Abs(LineSegmentOf(P,F1))*Abs(LineSegmentOf(P,F2))=m;m: Number",
    "query_expressions": "Max(m)",
    "answer_expressions": "25",
    "fact_spans": "[[[20, 64], [76, 81]], [[20, 64]], [[71, 75]], [[2, 9]], [[10, 17]], [[2, 70]], [[2, 70]], [[71, 86]], [[88, 116]], [[118, 121]]]",
    "query_spans": "[[[118, 127]]]",
    "process": "According to the problem, by the definition of an ellipse, we have: |PF_{1}| + |PF_{2}| = 10, then m = |PF_{1}| \\cdot |PF_{2}| \\leqslant \\left( \\frac{|PF_{1}| + |PF_{2}|}{2} \\right)^{2} = 25, with equality if and only if |PF_{1}| = |PF_{2}| = 5. Therefore, the maximum value of m is 25."
  },
  {
    "text": "If the line $y=k(x+2)+1$ and the parabola $y^{2}=4x$ have only one common point, then the value of $k$ is?",
    "fact_expressions": "H: Line;Expression(H) = (y = k*(x + 2) + 1);G: Parabola;Expression(G) = (y^2 = 4*x);NumIntersection(H, G) = 1;k: Number",
    "query_expressions": "k",
    "answer_expressions": "{0,1/2,-1}",
    "fact_spans": "[[[1, 15]], [[1, 15]], [[16, 30]], [[16, 30]], [[1, 37]], [[39, 42]]]",
    "query_spans": "[[[39, 46]]]",
    "process": ""
  },
  {
    "text": "If a point $P$ on the left branch of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ has a distance of $8$ to the right focus, then what is the distance from $P$ to the left directrix?",
    "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2/4 - y^2/5 = 1);PointOnCurve(P, LeftPart(G));Distance(P, RightFocus(G)) = 8",
    "query_expressions": "Distance(P, LeftDirectrix(G))",
    "answer_expressions": "8/3",
    "fact_spans": "[[[1, 39]], [[44, 47], [60, 63]], [[1, 39]], [[1, 47]], [[1, 58]]]",
    "query_spans": "[[[1, 72]]]",
    "process": ""
  },
  {
    "text": "Let the parabola $ C $: $ y^{2} = 2 p x $ ($ p > 0 $) have focus $ F $, and let point $ M $ lie on $ C $ such that $ |M F| = 5 $. If the circle with diameter $ M F $ passes through the point $ (0, 2) $, then $ p = $?",
    "fact_expressions": "C: Parabola;p: Number;G: Circle;H: Point;M: Point;F: Point;p>0;Expression(C) = (y^2 = 2*p*x);Coordinate(H) = (0, 2);Focus(C) = F;PointOnCurve(M, C);Abs(LineSegmentOf(M, F)) = 5;IsDiameter(LineSegmentOf(M,F),G);PointOnCurve(H,G)",
    "query_expressions": "p",
    "answer_expressions": "{2,8}",
    "fact_spans": "[[[1, 27], [40, 43]], [[78, 81]], [[66, 67]], [[68, 76]], [[35, 39]], [[31, 34]], [[9, 27]], [[1, 27]], [[68, 76]], [[1, 34]], [[35, 44]], [[45, 54]], [[56, 67]], [[66, 76]]]",
    "query_spans": "[[[78, 83]]]",
    "process": "Let M(x,y), |MF|=5 ⇒ x + \\frac{p}{2}=5 ⇒ x=5-\\frac{p}{2}, y^{2}=2px=10p-p^{2}, let A(0,2) ∴ \\overrightarrow{AM}=(x,y-2), \\overrightarrow{AF}=(\\frac{P}{2},-2), \\overrightarrow{AM}\\cdot\\overrightarrow{AF}=0 ⇒ x\\cdot\\frac{p}{2}+4-2y=0 ⇒ \\frac{y^{2}}{4}+4-2y=0 ⇒ y=4 ⇒ 16=10p-p^{2} ⇒ p=2 or 8"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus is point $F$. A line $l$ passing through the origin with an inclination angle of $\\frac{\\pi}{3}$ intersects the ellipse $C$ at points $M$ and $N$. If $\\angle MFN = \\frac{2\\pi}{3}$, then the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;LeftFocus(C) = F;l: Line;O: Origin;PointOnCurve(O, l);Inclination(l) = pi/3;M: Point;N: Point;Intersection(l, C) = {M, N};AngleOf(M, F, N) = (2*pi)/3",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(3*sqrt(2)-sqrt(10))/2",
    "fact_spans": "[[[2, 59], [98, 103], [147, 152]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[64, 68]], [[2, 68]], [[92, 97]], [[70, 72]], [[69, 97]], [[72, 97]], [[106, 109]], [[110, 113]], [[92, 115]], [[117, 145]]]",
    "query_spans": "[[[147, 158]]]",
    "process": "Let the right focus be $ F $, let the equation of line $ l $ be $ y = \\sqrt{3}x $, let $ M(x_{0}, y_{0}) $, $ N(-x_{0}, -y_{0}) $. Using geometric properties, we obtain $ \\angle FMF' = \\frac{\\pi}{3} $. Combining the properties of focal triangles and the law of cosines, we get $ s_{\\triangle MFF'} = \\frac{\\sqrt{3}}{3}b^{2} = c|y_{0}| $. Substituting the coordinates of $ M $ into the ellipse equation allows us to find the eccentricity. Let the right focus be $ F $. From the given conditions, the equation of line $ l $ is $ y = \\sqrt{3}x $. Let $ M(x_{0}, y_{0}) $, $ N(-x_{0}, -y_{0}) $. Connect $ MF' $, $ NF $. Since $ \\angle MFN = \\frac{2\\pi}{3} $, quadrilateral $ FMF'N $ is a parallelogram, thus $ \\angle FMF' = \\frac{\\pi}{3} $. Therefore, $ 4c^{2} = |MF|^{2} + |MF'|^{2} - 2|MF||MF'|\\cos\\frac{\\pi}{3} $. Rearranging gives $ 4c^{2} = (|MF| + |MF'|)^{2} - 3|MF||MF'| $, so $ |MF||MF'| = \\frac{4b^{2}}{3} $. Hence, $ s_{\\Delta MFF'} = \\frac{1}{2} \\times \\frac{\\sqrt{3}}{2} \\times \\frac{4b^{2}}{3} = \\frac{\\sqrt{3}}{3}b^{2} = \\frac{1}{2} \\times 2c \\times |y_{0}| $, so $ y_{0} = \\frac{\\sqrt{3}b^{2}}{3c} $. Substituting into the equation of line $ l $ gives $ x_{0} = \\frac{b^{2}}{3c} $. Substituting the coordinates of $ M $ into the ellipse equation yields: $ \\frac{b^{4}}{9c^{2}} \\cdot \\frac{1}{a^{2}} + \\frac{b^{2}}{3c^{2}} = 1 $, rearranging gives: $ 4a^{4} - 14a^{2}c^{2} + c^{4} = 0 $, i.e., $ e^{4} - 14e^{2} + 4 = 0 $. Solving gives $ e^{2} = 7 \\pm 3\\sqrt{5} $. Since the eccentricity of the ellipse $ e \\in (0,1) $, we have $ e = \\sqrt{7 - 3\\sqrt{5}} = \\frac{3\\sqrt{2} - \\sqrt{10}}{2} $."
  },
  {
    "text": "If the line passing through the focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ and perpendicular to the $x$-axis is intercepted by the ellipse to form a segment of length $\\frac{a}{2}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;H: Line;PointOnCurve(Focus(G), H);IsPerpendicular(H, xAxis);Length(InterceptChord(H, G)) = a/2",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[3, 55], [70, 72], [95, 97]], [[3, 55]], [[5, 55]], [[5, 55]], [[5, 55]], [[5, 55]], [[67, 69]], [[2, 69]], [[59, 69]], [[67, 92]]]",
    "query_spans": "[[[95, 103]]]",
    "process": "Because the length of the chord passing through the focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and perpendicular to the $x$-axis is $\\frac{a}{2}$, from $\\begin{cases}\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1\\\\\\therefore\\frac{2b^{2}}{a}=\\frac{a}{2},4b^{2}=a^{2},\\end{cases}$, and $b^{2}=a^{2}-c^{2}$, $\\therefore3a^{2}=4c^{2}$, $\\frac{c^{2}}{a^{2}}=\\frac{3}{4}$, so the eccentricity of the ellipse is $e=\\frac{c}{a}=\\frac{\\sqrt{3}}{2}$."
  },
  {
    "text": "The distance from a point $A(m, \\frac{3}{4})$ on the parabola $y=2 a x^{2}(a>0)$ to its focus $F$ is $1$. Then the value of $a$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = 2*(a*x^2));a: Number;a>0;A: Point;m: Number;Coordinate(A) = (m, 3/4);PointOnCurve(A, G);F: Point;Focus(G) = F;Distance(A, F) = 1",
    "query_expressions": "a",
    "answer_expressions": "1/2",
    "fact_spans": "[[[0, 21], [44, 45]], [[0, 21]], [[59, 62]], [[3, 21]], [[24, 43]], [[24, 43]], [[24, 43]], [[0, 43]], [[47, 50]], [[44, 50]], [[24, 57]]]",
    "query_spans": "[[[59, 66]]]",
    "process": "】Convert the parabolic equation into standard form, use the definition of a parabola to transform the distance from a point on the parabola to the focus into the distance to the directrix, then solve using the point-to-line distance formula. Convert the parabola $ y = 2ax^2 $ ($ a > 0 $) into $ x^2 = \\frac{1}{2a}y $ ($ a > 0 $). By the definition of a parabola, the distance from point $ A(m, \\frac{3}{4}) $ to the directrix $ l: y = -\\frac{1}{8a} $ is 1, so $ \\frac{3}{4} + \\frac{1}{8a} = 1 $. Solving gives $ a = \\frac{1}{2} $."
  },
  {
    "text": "What is the maximum distance from a point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ to the line $2 x-y+10=0$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1);P:Point;PointOnCurve(P,G);H: Line;Expression(H) = (2*x - y + 10 = 0)",
    "query_expressions": "Max(Distance(P,G))",
    "answer_expressions": "2*sqrt(2)+2*sqrt(5)",
    "fact_spans": "[[[0, 37]], [[0, 37]], [[39, 40]], [[0, 40]], [[41, 55]], [[41, 55]]]",
    "query_spans": "[[[39, 62]]]",
    "process": ""
  },
  {
    "text": "Given that the equation of line $l$ is $x+2y=0$, and $P$ is an arbitrary point on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, then the maximum distance from point $P$ to line $l$ is?",
    "fact_expressions": "l: Line;G: Ellipse;P: Point;Expression(G) = (x^2/4 + y^2 = 1);Expression(l) = (x + 2*y = 0);PointOnCurve(P, G)",
    "query_expressions": "Max(Distance(P, l))",
    "answer_expressions": "2*sqrt(10)/5",
    "fact_spans": "[[[2, 7], [64, 69]], [[25, 52]], [[59, 63], [21, 24]], [[25, 52]], [[2, 20]], [[21, 57]]]",
    "query_spans": "[[[59, 78]]]",
    "process": "(1) Since P is an arbitrary point on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, we can set $P(2\\cos\\theta,\\sin\\theta)$, where $\\theta\\in[0,2\\pi)$. Since the distance from point P to the line $l: x+2y=0$ is $d=\\frac{|2\\cos\\theta+2\\sin\\theta|}{\\sqrt{5}}=\\frac{2\\sqrt{2}|\\sin(\\theta+\\frac{\\pi}{4})|}{\\sqrt{5}}$, the maximum value of $d$ is $\\frac{2\\sqrt{10}}{5}$ when $\\sin(\\theta+\\frac{\\pi}{4})=\\pm1$."
  },
  {
    "text": "It is known that the focus of the parabola $x^{2}=a y$ coincides exactly with the upper focus of the hyperbola $y^{2}-x^{2}=2$. Then the value of $a$ is?",
    "fact_expressions": "G: Hyperbola;H: Parabola;a: Number;Expression(G) = (-x^2 + y^2 = 2);Expression(H) = (x^2 = a*y);Focus(H) = UpperFocus(G)",
    "query_expressions": "a",
    "answer_expressions": "8",
    "fact_spans": "[[[22, 40]], [[2, 16]], [[46, 49]], [[22, 40]], [[2, 16]], [[2, 44]]]",
    "query_spans": "[[[46, 53]]]",
    "process": ""
  },
  {
    "text": "A line with slope $\\sqrt{3}$ passes through the focus of the parabola $C$: $y^{2}=4 x$ and intersects $C$ at points $A$ and $B$. Then $|A B|=$?",
    "fact_expressions": "Slope(G) = sqrt(3);G: Line;C: Parabola;Expression(C) = (y^2 = 4*x);PointOnCurve(Focus(C), G) = True;Intersection(G, C) = {A, B};A: Point;B: Point",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16/3",
    "fact_spans": "[[[0, 16]], [[14, 16]], [[17, 36], [42, 45]], [[17, 36]], [[14, 39]], [[14, 56]], [[47, 50]], [[51, 54]]]",
    "query_spans": "[[[58, 67]]]",
    "process": "\\because the equation of the parabola is y^{2}=4x, \\therefore the coordinates of the focus F of the parabola are F(1,0). \\because the line AB passes through the focus F and has a slope of \\sqrt{3}, \\therefore the equation of line AB is: y=\\sqrt{3}(x-1). Substituting into the parabola equation, eliminating y and simplifying yields 3x^{2}-10x+3=0. Solution 1: Solving gives x_{1}=\\frac{1}{3}, x_{2}=3. Therefore, |AB|=\\sqrt{1+k^{2}}|x_{1}-x_{2}|=\\sqrt{1+3}\\cdot|3-\\frac{1}{3}|=\\frac{16}{3}. Solution 2: \\Delta=100-36=64>0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), then x_{1}+x_{2}=\\frac{10}{3}. Draw perpendiculars from A and B to the directrix x=-1, with feet C and D respectively, as shown in the figure. |AB|=|AF|+|BF|=|AC|+|BD|=x_{1}+1+x_{2}+1=x_{1}+x_{2}+2=\\frac{16}{3}"
  },
  {
    "text": "Let the parabola $C$: $y^{2}=4x$ have focus $F$. The line $l$ passing through $F$ with slope $k$ ($k>0$) intersects $C$ at points $A$ and $B$, and $|AB|=8$. Then the equation of $l$ is?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F, l);Slope(l) = k;k:Number;k>0;Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, B)) = 8",
    "query_expressions": "Expression(l)",
    "answer_expressions": "y=x-1",
    "fact_spans": "[[[45, 50], [77, 80]], [[1, 20], [51, 54]], [[56, 59]], [[60, 63]], [[24, 27], [29, 32]], [[1, 20]], [[1, 27]], [[28, 50]], [[33, 50]], [[36, 44]], [[36, 44]], [[45, 65]], [[66, 75]]]",
    "query_spans": "[[[77, 85]]]",
    "process": "From the given conditions, F(1,0), the equation of line $ l $ is $ y = k(x - 1) $ ($ k > 0 $). Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. From  \n$$\n\\begin{cases}\ny = k(x - 1) \\\\\ny^{2} = 4x\n\\end{cases}\n$$  \nwe obtain $ k^{2}x^{2} - (2k^{2} + 4)x + k^{2} = 0 $, $ \\Delta = 16k^{2} + 16 > 0 $, hence $ x_{1} + x_{2} = \\frac{2k^{2} + 4}{k^{2}} $. Therefore,  \n$ |AB| = |AF| + |BF| = (x_{1} + 1) + (x_{2} + 1) = \\frac{4k^{2} + 4}{k^{2}} $.  \nAccording to the given condition, $ \\frac{4k^{2} + 4}{k^{2}} = 8 $, solving yields $ k = -1 $ (discarded), $ k = 1 $. Thus, the equation of $ l $ is $ y = x - 1 $."
  },
  {
    "text": "Point $P$ lies on the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1  (b>0)$, and $F_{1}$, $F_{2}$ are the left and right foci of the hyperbola, respectively. Given $|P F_{1}|+|P F_{2}|=6$ and $P F_{1} \\perp P F_{2}$, find the eccentricity of the hyperbola?",
    "fact_expressions": "G: Hyperbola;b: Number;P: Point;F1: Point;F2: Point;b>0;Expression(G) = (x^2 - y^2/b^2 = 1);PointOnCurve(P, G);LeftFocus(G)=F1;RightFocus(G)=F2;Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 6;IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[5, 44], [64, 67], [125, 128]], [[8, 44]], [[0, 4]], [[48, 55]], [[56, 63]], [[8, 44]], [[5, 44]], [[0, 47]], [[48, 73]], [[48, 73]], [[74, 97]], [[100, 123]]]",
    "query_spans": "[[[125, 134]]]",
    "process": "According to the problem, point $ P $ lies on the hyperbola $ x^{2} - \\frac{y^{2}}{b^{2}} = 1 $ ($ b > 0 $), so $ |PF_{1}| - |PF_{2}| = 2a = 2 $. Assume $ |PF_{1}| > |PF_{2}| $, then $ |PF_{1}| - |PF_{2}| = 2 $. Also given $ |PF_{1}| + |PF_{2}| = 6 $, solving yields: $ |PF_{1}| = 4 $, $ |PF_{2}| = 2 $. Since $ PF_{1} \\perp PF_{2} $, it follows that $ |PF_{1}|^{2} + |PF_{2}|^{2} = 4c^{2} = 20 $, so $ c = \\sqrt{5} $. Given $ a = 1 $, the eccentricity of the hyperbola is $ e = \\frac{c}{a} = \\sqrt{5} $."
  },
  {
    "text": "Given the line $l_{1}$: $x - y + 3 = 0$, the line $l_{2}$: $x = -3$, then the minimum value of the sum of the distance from a moving point $P$ on the parabola $x = \\frac{1}{8} y^{2}$ to the line $l_{1}$ and the distance to the line $l_{2}$ is?",
    "fact_expressions": "l1: Line;Expression(l1) = (x - y + 3 = 0);l2: Line;Expression(l2) = (x = -3);G: Parabola;Expression(G) = (x = y^2/8);P: Point;PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, l1)+Distance(P, l2))",
    "answer_expressions": "(5*sqrt(2)/2)+1",
    "fact_spans": "[[[2, 22], [75, 84]], [[2, 22]], [[23, 40], [90, 99]], [[23, 40]], [[42, 66]], [[42, 66]], [[71, 74], [88, 89]], [[42, 74]]]",
    "query_spans": "[[[71, 110]]]",
    "process": "The standard equation of the parabola is $ y^{2} = 8x $, with focus $ F(2,0) $. Therefore, the distance from point $ P $ to the line $ l_{2} $ equals $ |PF| + 1 $. Hence, the sum of the distances from point $ P $ to the lines $ l_{1} $ and $ l_{2} $ equals the sum of the distance from $ P $ to $ l_{1} $ and $ |PF| + 1 $, whose minimum value is $ \\frac{|1 \\times 2 - 1 \\times 0 + 3|}{\\sqrt{1^{2} + 1^{2}}} + 1 = \\frac{5\\sqrt{2}}{2} + $"
  },
  {
    "text": "The coordinates of the focus of the parabola $y=4 a x^{2}  (a \\in R$ and $a \\neq 0)$ are?",
    "fact_expressions": "G: Parabola;a: Real;Expression(G) = (y = 4*(a*x^2));Negation(a=0)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,1/(16*a))",
    "fact_spans": "[[[0, 38]], [[3, 38]], [[0, 38]], [[3, 38]]]",
    "query_spans": "[[[0, 45]]]",
    "process": "From the given conditions, the standard equation of the parabola is $x^{2}=\\frac{1}{4a}y$. Therefore, the focus of this parabola lies on the $y$-axis. When $a>0$, the focus lies on the positive half of the $y$-axis; when $a<0$, the focus lies on the negative half of the $y$-axis, with coordinates $F(0,\\frac{1}{16a})$. The answer to be filled in is $F(0,\\frac{1}{16a})$."
  },
  {
    "text": "Given that the ellipse is symmetric with respect to the coordinate axes, the major axis is twice the minor axis, and it passes through the point $P(3,0)$, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;P: Point;Coordinate(P) = (3, 0);SymmetryAxis(G)=axis;MajorAxis(G)=2*MinorAxis(G);PointOnCurve(P,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{(x^2/9+y^2=1),(x^2/9+y^2/81=1)}",
    "fact_spans": "[[[2, 4], [38, 40]], [[27, 36]], [[27, 36]], [[2, 12]], [[2, 24]], [[2, 36]]]",
    "query_spans": "[[[38, 45]]]",
    "process": "Let the minor axis of the ellipse be 2b (b>0), then the major axis is 6b. Therefore, the standard equation of the ellipse is \\frac{x^{2}}{(3b)^{2}}+\\frac{y^{2}}{b^{2}}=1 or \\frac{x^{2}}{b^{2}}+\\frac{y^{2}}{(3b)^{2}}=1. Substituting the coordinates of point P(3,0) into the equation \\frac{x^{2}}{(3b)^{2}}+\\frac{y^{2}}{b^{2}}=1, we solve to get b=1, hence the standard equation of the ellipse is \\frac{x^{2}}{9}+y^{2}=1. Substituting the coordinates of point P(3,0) into the equation \\frac{x^{2}}{b^{2}}+\\frac{y^{2}}{(3b)^{2}}=1, we solve to get b=3, hence the standard equation of the ellipse is \\frac{x^{2}}{9}+\\frac{y^{2}}{81}=1. In conclusion, the standard equations of the ellipse are \\frac{x^{2}}{9}+y^{2}=1 or \\frac{x^{2}}{9}+\\frac{y^{2}}{81}=1. Answer: \\frac{x^{2}}{9}+y^{2}=1 or \\frac{x^{2}}{9}+\\frac{y^{2}}{81}=1."
  },
  {
    "text": "A point $N$ on the parabola $y=-\\frac{1}{2} x^{2}$ is at a distance of $3$ from its focus $F$. Then, what is the distance from point $N$ to the line $y=1$?",
    "fact_expressions": "G: Parabola;H: Line;N: Point;F: Point;Expression(G) = (y = -x^2/2);Expression(H) = (y = 1);PointOnCurve(N, G);Focus(G) = F;Distance(N, F) = 3",
    "query_expressions": "Distance(N, H)",
    "answer_expressions": "7/2",
    "fact_spans": "[[[0, 25], [32, 33]], [[52, 59]], [[28, 31], [47, 51]], [[35, 38]], [[0, 25]], [[52, 59]], [[0, 31]], [[32, 38]], [[28, 45]]]",
    "query_spans": "[[[47, 65]]]",
    "process": ""
  },
  {
    "text": "The foci of the ellipse $x^{2}+my^{2}=1$ lie on the $y$-axis, and the length of the major axis is twice the length of the minor axis. Then the value of $m$ is?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (m*y^2 + x^2 = 1);PointOnCurve(Focus(G), yAxis);Length(MajorAxis(G))=2*Length(MinorAxis(G))",
    "query_expressions": "m",
    "answer_expressions": "1/4",
    "fact_spans": "[[[0, 18]], [[40, 43]], [[0, 18]], [[0, 27]], [[0, 38]]]",
    "query_spans": "[[[40, 47]]]",
    "process": ""
  },
  {
    "text": "The line $y = kx - 2$ intersects the ellipse $x^{2} + 4y^{2} = 80$ at two distinct points $P$ and $Q$. If the x-coordinate of the midpoint of $PQ$ is $2$, then the slope of the line equals?",
    "fact_expressions": "G: Ellipse;H: Line;k: Number;P: Point;Q: Point;Expression(G) = (x^2 + 4*y^2 = 80);Expression(H) = (y = k*x - 2);Intersection(H, G) = {P, Q};Negation(P=Q);XCoordinate(MidPoint(LineSegmentOf(P, Q))) = 2",
    "query_expressions": "Slope(H)",
    "answer_expressions": "1/2",
    "fact_spans": "[[[12, 32]], [[0, 11], [66, 68]], [[2, 11]], [[40, 43]], [[44, 47]], [[12, 32]], [[0, 11]], [[0, 47]], [[35, 47]], [[49, 64]]]",
    "query_spans": "[[[66, 74]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has a real axis length of $2$ and an eccentricity of $2$, then the coordinates of the foci of hyperbola $C$ are?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Length(RealAxis(C))=2;Eccentricity(C)=2",
    "query_expressions": "Coordinate(Focus(C))",
    "answer_expressions": "(pm*2,0)",
    "fact_spans": "[[[2, 64], [82, 88]], [[10, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[2, 64]], [[2, 72]], [[2, 80]]]",
    "query_spans": "[[[82, 95]]]",
    "process": ""
  },
  {
    "text": "The abscissa of a point $A$ on the parabola $y^{2}=4 x$ is $4$, then the distance between point $A$ and the focus of the parabola is?",
    "fact_expressions": "G: Parabola;A: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(A, G);XCoordinate(A) = 4",
    "query_expressions": "Distance(A, Focus(G))",
    "answer_expressions": "5",
    "fact_spans": "[[[0, 14], [35, 38]], [[17, 20], [30, 34]], [[0, 14]], [[0, 20]], [[17, 28]]]",
    "query_spans": "[[[30, 45]]]",
    "process": ""
  },
  {
    "text": "Let $A(x_{1}, y_{1})$ be a point on the ellipse $3 x^{2}+4 y^{2}=12$. A line $l$ with slope $-\\frac{3 x_{1}}{4 y_{1}}$ is drawn through point $A$. Let $d$ be the distance from the origin to the line $l$, and let $r_{1}$, $r_{2}$ be the distances from point $A$ to the two foci of the ellipse. Then $\\sqrt{r_{1} r_{2}} d$=?",
    "fact_expressions": "G: Ellipse;A: Point;Expression(G) = (3*x^2 + 4*y^2 = 12);Coordinate(A) = (x1, y1);PointOnCurve(A,G);x1:Number;y1:Number;d:Number;r1:Number;r2:Number;l:Line;PointOnCurve(A,l);O:Origin;Slope(l)=-(3*x1/4*y1);Distance(O,l)=d;F1:Point;F2:Point;Focus(G)={F1,F2};Distance(A,F1)=r1;Distance(A,F2)=r2",
    "query_expressions": "sqrt(r1*r2)*d",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[19, 41], [131, 133]], [[1, 18], [46, 50]], [[19, 41]], [[1, 18]], [[1, 44]], [[1, 18]], [[1, 18]], [[91, 94]], [[107, 114]], [[116, 123]], [[83, 88], [98, 103]], [[45, 88]], [[95, 97]], [[53, 88]], [[91, 106]], [], [], [[131, 136]], [[107, 139]], [[107, 139]]]",
    "query_spans": "[[[142, 166]]]",
    "process": "Transform the ellipse into the standard equation $\\frac{x^{2}}{4}+\\frac{12}{3}=1$, yielding the two foci of the ellipse $F_{1}(-1,0)$, $F_{2}(1,0)$. From the point-slope form, the line $l$: $y-y_{1}=-\\frac{3x_{1}}{4y_{1}}(x-x_{1})$, simplifying yields $l: 3x_{1}x+4y_{1}y-3x_{1}-4y_{1}^{2}=0$. Thus, $d=\\frac{|3x_{1}2+4y_{1}|}{\\sqrt{9x_{1}2+16y_{2}}}=\\underline{12}2=\\sqrt{(x_{1}}1)^{2}+y_{2}(2x_{1}2+2)+y_{4}(*)$, substituting into $(*)$, simplifying yields $^{12}=2\\sqrt{3}$;"
  },
  {
    "text": "Given that the hyperbola $C$ with center at the origin has a focal distance of $6$ and an eccentricity of $3$, what is the standard equation of the hyperbola $C$?",
    "fact_expressions": "C: Hyperbola;O: Origin;Center(C) = O;FocalLength(C) = 6;Eccentricity(C) = 3",
    "query_expressions": "Expression(C)",
    "answer_expressions": "{(x^2-y^2/8=1),(y^2-x^2/8=1)}",
    "fact_spans": "[[[10, 16], [35, 41]], [[5, 9]], [[2, 16]], [[10, 23]], [[10, 32]]]",
    "query_spans": "[[[35, 48]]]",
    "process": ""
  },
  {
    "text": "The coordinates of the focus of the parabola $y^{2}=2 p x(p>0)$ are $(1,0)$, then $p$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;Coordinate(Focus(G)) = (1, 0)",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 21]], [[0, 21]], [[35, 38]], [[3, 21]], [[0, 33]]]",
    "query_spans": "[[[35, 40]]]",
    "process": "\\because y^{2}=2px(p>0) has focus coordinates (\\frac{p}{2},0), \\therefore \\frac{p}{2}=1, i.e., p=2."
  },
  {
    "text": "If a point $P$ on the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ is at a distance of $4$ from the left focus, then what is the distance from point $P$ to the right focus?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);P: Point;PointOnCurve(P, G);Distance(P, LeftFocus(G)) = 4",
    "query_expressions": "Distance(P, RightFocus(G))",
    "answer_expressions": "10",
    "fact_spans": "[[[1, 40]], [[1, 40]], [[43, 46], [59, 63]], [[1, 46]], [[1, 57]]]",
    "query_spans": "[[[1, 72]]]",
    "process": "From the hyperbola equation, we know $ a^{2} = 9 \\therefore a = 3, 2a = 6 $. By the definition $ ||PF_{1}| - |PF_{2}|| = 2a $, we get $ |PF_{2}| = 10 $."
  },
  {
    "text": "The vertex of the parabola is at the origin, the focus is on the $y$-axis, and its directrix is given by the equation $y=3$. Then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;O: Origin;Vertex(G) = O;PointOnCurve(Focus(G), yAxis);Expression(Directrix(G)) = (y=3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 = -12 \\cdot y",
    "fact_spans": "[[[0, 3], [18, 19], [33, 36]], [[6, 8]], [[0, 8]], [[0, 17]], [[18, 30]]]",
    "query_spans": "[[[33, 41]]]",
    "process": "Since the vertex of the parabola is at the origin and the focus is on the y-axis, the equation of the parabola is in standard form. Because the directrix equation is y=3, we have \\frac{p}{2}=3, so p=6. Therefore, the equation of the parabola is: x^{2}=-12y."
  },
  {
    "text": "If the focal distance of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{4}=1  (m \\in R)$ is $2$, then $m$=?",
    "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (y^2/4 + x^2/m = 1);FocalLength(G) = 2",
    "query_expressions": "m",
    "answer_expressions": "{5,3}",
    "fact_spans": "[[[1, 49]], [[58, 61]], [[1, 49]], [[1, 56]]]",
    "query_spans": "[[[58, 63]]]",
    "process": ""
  },
  {
    "text": "If the parabola $y^{2}=m x$ and the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ have a common focus, then $m=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = m*x);m: Number;H: Ellipse;Expression(H) = (x^2/9 + y^2/5 = 1);Focus(G) = OneOf(Focus(H))",
    "query_expressions": "m",
    "answer_expressions": "pm*8",
    "fact_spans": "[[[1, 15]], [[1, 15]], [[63, 66]], [[16, 53]], [[16, 53]], [[1, 61]]]",
    "query_spans": "[[[63, 68]]]",
    "process": "Since the focus coordinates of the parabola $y^{2}=mx$ are $(\\frac{m}{4},0)$; the focus coordinates of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ are $(\\pm2,0)$, we obtain $\\frac{m}{4}=\\pm2$, solving gives $m=\\pm8$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1 ,(a>0)$ has eccentricity $\\sqrt{3}$, then the value of the real number $a$ is?",
    "fact_expressions": "G: Hyperbola;a: Real;a>0;Expression(G) = (-y^2/4 + x^2/a^2 = 1);Eccentricity(G) = sqrt(3)",
    "query_expressions": "a",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 51]], [[68, 73]], [[5, 51]], [[2, 51]], [[2, 66]]]",
    "query_spans": "[[[68, 77]]]",
    "process": ""
  },
  {
    "text": "Given point $M(2,2)$ and parabola $C$: $y^{2}=-8x$, a line passing through the focus of $C$ with slope $k$ intersects $C$ at points $A$ and $B$. If $\\overrightarrow{MA} \\perp \\overrightarrow{MB}$, then $k=?$",
    "fact_expressions": "M: Point;Coordinate(M) = (2, 2);C: Parabola;Expression(C) = (y^2 = -8*x);G: Line;k: Number;Slope(G)=k;PointOnCurve(Focus(C),G);A: Point;B: Point;Intersection(G,C)={A,B};IsPerpendicular(VectorOf(M,A),VectorOf(M,B))",
    "query_expressions": "k",
    "answer_expressions": "-2",
    "fact_spans": "[[[2, 11]], [[2, 11]], [[12, 32], [35, 38], [52, 55]], [[12, 32]], [[49, 51]], [[45, 48], [120, 123]], [[42, 51]], [[34, 51]], [[58, 61]], [[62, 65]], [[49, 67]], [[69, 118]]]",
    "query_spans": "[[[120, 125]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then \n\\[\n\\begin{cases}\ny_{1} = -8x_{1} \\\\\ny_{2} = -8x_{2}\n\\end{cases}\n\\],\nso $ y_{1} - y_{2} = 8x_{2} - 8x_{1} $, thus $ k = \\frac{y_{1} - y_{2}}{x_{1} - x_{2}} = -\\frac{8}{y_{1} + y_{2}} $. Take the midpoint $ M_{0}(x_{0}, y_{0}) $ of $ AB $, and draw perpendiculars from points $ A $, $ B $ to the directrix $ x = 2 $, with feet $ A_{1} $, $ B_{1} $, respectively. Since $ \\overrightarrow{MA} \\bot \\overrightarrow{MB} $, we have $ \\angle AMB = 90^{\\circ} $, so $ |MM_{0}| = \\frac{1}{2}|AB| = \\frac{1}{2}(|AF| + |BF|) = \\frac{1}{2}(|AA_{1}| + |BB_{1}|) $. Since $ M_{0} $ is the midpoint of $ AB $, $ MM_{0} $ is parallel to the $ x $-axis. Given $ M(2,2) $, we have $ y_{0} = 2 $, so $ y_{1} + y_{2} = 4 $, hence $ k = 2 $. The final answer is: --"
  },
  {
    "text": "If one of the foci of the ellipse $C$: $\\frac{x^{2}}{m}+\\frac{y^{2}}{m^{2}-1}=1$ has coordinates $(0,1)$, then what is the length of the major axis of $C$?",
    "fact_expressions": "C: Ellipse;m: Number;Expression(C) = (y^2/(m^2 - 1) + x^2/m = 1);Coordinate(OneOf(Focus(C))) = (0, 1)",
    "query_expressions": "Length(MajorAxis(C))",
    "answer_expressions": "2*sqrt(3)",
    "fact_spans": "[[[1, 49], [66, 69]], [[8, 49]], [[1, 49]], [[1, 64]]]",
    "query_spans": "[[[66, 75]]]",
    "process": "From the given conditions, the foci of the ellipse lie on the y-axis, and c=1, so a^{2}-b^{2}=m^{2}-1-m=1, and m>0, m^{2}-1>0. Solving gives m=2, so the standard equation of the ellipse is: \\frac{x^{2}}{2}+\\frac{y^{2}}{3}=1, thus a^{2}=3, that is, a=\\sqrt{3}, so the major axis length 2a=2\\sqrt{3}."
  },
  {
    "text": "If the standard equation of the ellipse is $\\frac{x^{2}}{6-s}+\\frac{y^{2}}{s+2}=1$, the foci lie on the $y$-axis, and the focal distance is $4$, then the real number $s$=?",
    "fact_expressions": "G: Ellipse;s: Real;Expression(G) = (x^2/(6 - s) + y^2/(s + 2) = 1);PointOnCurve(Focus(G), yAxis);FocalLength(G) = 4",
    "query_expressions": "s",
    "answer_expressions": "4",
    "fact_spans": "[[[1, 3]], [[67, 72]], [[1, 48]], [[1, 57]], [[1, 65]]]",
    "query_spans": "[[[67, 74]]]",
    "process": ""
  },
  {
    "text": "Given the parabola $y^{2}=16 x$, with focus $F$, and a fixed point $A(8,2)$ on the plane, let $P$ be a moving point on the parabola. Then the minimum value of $|P A|+|P F|$ is?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 16*x);Coordinate(A) = (8, 2);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "12",
    "fact_spans": "[[[2, 17], [48, 51]], [[27, 35]], [[44, 47]], [[21, 24]], [[2, 17]], [[27, 35]], [[2, 24]], [[44, 56]]]",
    "query_spans": "[[[58, 77]]]",
    "process": "The directrix of the parabola is given by: x = -4, and the focus is F(4,0). Draw a perpendicular from A to the directrix, with foot at B. Therefore, |PA| + |PF| > |AB| = 12. Hence, the answer is 12."
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the two foci of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, and point $P$ lies on the hyperbola such that $\\angle F_{1} P F_{2}=90^{\\circ}$. Then the area of $\\triangle F_{1} P F_{2}$ is?",
    "fact_expressions": "G: Hyperbola;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 - y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(90, degree)",
    "query_expressions": "Area(TriangleOf(F1, P, F2))",
    "answer_expressions": "1",
    "fact_spans": "[[[17, 45], [56, 59]], [[1, 8]], [[51, 55]], [[9, 16]], [[17, 45]], [[1, 50]], [[51, 60]], [[62, 95]]]",
    "query_spans": "[[[98, 128]]]",
    "process": ""
  },
  {
    "text": "There is a point $P$ on the parabola $y^{2}=8 x$, and its distance to the focus is $20$. Then the coordinates of point $P$ are?",
    "fact_expressions": "G: Parabola;P: Point;Expression(G) = (y^2 = 8*x);PointOnCurve(P, G);Distance(P, Focus(G)) = 20",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(18, \\pm 12)",
    "fact_spans": "[[[1, 15]], [[19, 22], [37, 41], [23, 24]], [[1, 15]], [[0, 22]], [[1, 35]]]",
    "query_spans": "[[[37, 45]]]",
    "process": "The directrix of the parabola is x = -2. The distance from P(x_{0}, y_{0}) to the focus equals the distance to the directrix, i.e., x_{0} + 2 = 20, so x_{0} = 18. Then y_{0}^{2} = 8 \\times 18 \\Rightarrow y_{0} = \\pm 12. Therefore, the coordinates of point P are (18, \\pm 12)."
  },
  {
    "text": "Given that the directrix of the parabola $y^{2}=2 p x(p>0)$ is $x=-1$, and the line $l$: $y=x-1$ intersects the parabola at points $A$ and $B$, then the chord length $AB=$?",
    "fact_expressions": "l: Line;G: Parabola;p: Number;A: Point;B: Point;p>0;Expression(G) = (y^2 = 2*p*x);Expression(Directrix(G)) = (x = -1);Expression(l)=(y=x-1);Intersection(l,G)={A,B};IsChordOf(LineSegmentOf(A,B),G)",
    "query_expressions": "Length(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[36, 50]], [[2, 23], [51, 54]], [[5, 23]], [[55, 58]], [[59, 62]], [[5, 23]], [[2, 23]], [[2, 35]], [[36, 50]], [[36, 64]], [[51, 73]]]",
    "query_spans": "[[[68, 75]]]",
    "process": "Since the directrix of the parabola $ y^{2} = 2px $ ($ p > 0 $) is $ x = -1 $, we have $ \\frac{p}{2} = 1 $. Solving gives $ p = 2 $, so the equation of the parabola is $ y^{2} = 4x $. Thus, the focus of the parabola has coordinates $ F(1, 0) $. Therefore, the line $ l: y = x - 1 $ passes through the focus. Solving the system \n\\[\n\\begin{cases}\ny = x - 1 \\\\\ny^{2} = 4x\n\\end{cases}\n\\]\nand eliminating $ y $ yields $ x^{2} - 6x + 1 = 0 $. Then $ x_{1} + x_{2} = 6 $, so $ |AB| = x_{1} + x_{2} + p = 6 + 2 = 8 $. The answer is 8."
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively. The line $l$ with slope $1$ passing through $F_{1}$ intersects the right branch of $C$ at point $P$. If $\\angle F_{1} F_{2} P=90^{\\circ}$, then the eccentricity of the hyperbola $C$ is equal to?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;l: Line;Slope(l) = 1;PointOnCurve(F1, l);P: Point;Intersection(l, RightPart(C)) = P;AngleOf(F1, F2, P) = ApplyUnit(90, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "1+sqrt(2)",
    "fact_spans": "[[[18, 79], [108, 111], [157, 163]], [[18, 79]], [[26, 79]], [[26, 79]], [[26, 79]], [[26, 79]], [[0, 7], [94, 101]], [[8, 15]], [[0, 85]], [[0, 85]], [[102, 107]], [[86, 107]], [[93, 107]], [[116, 120]], [[102, 120]], [[122, 155]]]",
    "query_spans": "[[[157, 170]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ is a point on the parabola $y^{2}=16 x$, the distance from $P$ to the axis of symmetry is $12$, and $F$ is the focus of the parabola, then $|PF|$=?",
    "fact_expressions": "G: Parabola;P: Point;F: Point;Expression(G) = (y^2 = 16*x);PointOnCurve(P, G);Distance(P,SymmetryAxis(G))=12;Focus(G) = F",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "13",
    "fact_spans": "[[[7, 22], [47, 50]], [[2, 6], [27, 28]], [[42, 46]], [[7, 22]], [[2, 26]], [[7, 41]], [[43, 53]]]",
    "query_spans": "[[[55, 63]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with left and right foci $F_{1}$, $F_{2}$, $|F_{1} F_{2}|=4 \\sqrt{5}$, and point $P$ a point on the ellipse. If the perimeter of $\\Delta P F_{1} F_{2}$ is $4 \\sqrt{5}+12$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;Abs(LineSegmentOf(F1, F2)) = 4*sqrt(5);P: Point;PointOnCurve(P, C);Perimeter(TriangleOf(P, F1, F2)) = 4*sqrt(5) + 12",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)/3",
    "fact_spans": "[[[2, 59], [115, 117], [164, 169]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[65, 72]], [[74, 81]], [[2, 81]], [[2, 81]], [[83, 109]], [[110, 114]], [[110, 120]], [[122, 162]]]",
    "query_spans": "[[[164, 175]]]",
    "process": "Let the semi-focal distance of the ellipse be $c$. According to the problem,  \n$$\n\\begin{cases}\n2a+2c=4\\sqrt{5}+12 \\\\\n2c=4\\sqrt{5}\n\\end{cases}\n\\Rightarrow\n\\begin{cases}\nc=2\\sqrt{5} \\\\\na=6\n\\end{cases},\n$$\nso $e=\\frac{c}{a}=\\frac{\\sqrt{5}}{3}$,"
  },
  {
    "text": "$F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, respectively. $P$ is a point on the right branch of the hyperbola. $\\odot A$ is the incircle of $\\Delta P F_{1} F_{2}$, and $\\odot A$ is tangent to the $x$-axis at point $M(m, 0)$. Then the value of $m$ is?",
    "fact_expressions": "G: Hyperbola;A: Circle;M: Point;P: Point;F1: Point;F2: Point;m:Number;Expression(G) = (x^2/16 - y^2/9 = 1);Coordinate(M) = (m, 0);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));IsInscribedCircle(A, TriangleOf(P,F1,F2));TangentPoint(xAxis,A)=M",
    "query_expressions": "m",
    "answer_expressions": "4",
    "fact_spans": "[[[18, 57], [67, 70]], [[114, 123], [77, 86]], [[131, 141]], [[63, 66]], [[0, 7]], [[8, 15]], [[143, 146]], [[18, 57]], [[131, 141]], [[0, 62]], [[0, 62]], [[63, 76]], [[77, 113]], [[114, 141]]]",
    "query_spans": "[[[143, 150]]]",
    "process": ""
  },
  {
    "text": "From a focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a perpendicular to its asymptote, with foot at $M$. Extend $F M$ to intersect the $y$-axis at point $E$. If $|F M|=|M E|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;OneOf(Focus(G)) = F;F: Point;L: Line;PointOnCurve(F, L) ;IsPerpendicular(L, Asymptote(G)) ;FootPoint(L, Asymptote(G)) = M;M: Point;Intersection(OverlappingLine(LineSegmentOf(F, M)), yAxis) = E;E: Point;Abs(LineSegmentOf(F, M)) = Abs(LineSegmentOf(M, E))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 57], [66, 67], [116, 119]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 65]], [[62, 65]], [], [[0, 74]], [[0, 74]], [[0, 81]], [[78, 81]], [[82, 99]], [[95, 99]], [[101, 114]]]",
    "query_spans": "[[[116, 125]]]",
    "process": "According to |FM| = |ME|, the asymptote is perpendicular to FE, implying |OE| = |OF|, OM is the angle bisector, the angle between the asymptote and the x-axis is obtained, then the relationship between a and b is found, thus the relationship between a and c is derived, and finally the eccentricity of the hyperbola is obtained. [Detailed solution] \\because |FM| = |ME|, the asymptote is perpendicular to FE, \\therefore |OE| = |OF|, \\therefore OM is the angle bisector, the angle between the asymptote and the x-axis is 45^{\\circ}, \\therefore \\frac{b}{a} = \\tan 45^{\\circ} = 1, \\therefore a = b, \\therefore c = \\sqrt{a^{2} + b^{2}} = \\sqrt{2}a, \\therefore e = \\frac{c}{a} = \\sqrt{2}"
  },
  {
    "text": "There is a point $M$ on the parabola $y^{2}=4 x$, and its distance to the line $y=x$ is $4 \\sqrt{2}$. If the coordinates of point $M$ are $(m, n)$ with $m>0$, $n>0$, then what is the value of $\\frac{m}{n}$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);M: Point;PointOnCurve(M, G);H: Line;Expression(H) = (y = x);Distance(M, H) = 4*sqrt(2);m: Number;n: Number;Coordinate(M) = (m, n);m>0;n>0",
    "query_expressions": "m/n",
    "answer_expressions": "2",
    "fact_spans": "[[[1, 15]], [[1, 15]], [[17, 21], [22, 23], [50, 54]], [[1, 21]], [[24, 31]], [[24, 31]], [[22, 48]], [[58, 66]], [[58, 66]], [[50, 66]], [[67, 72]], [[74, 80]]]",
    "query_spans": "[[[82, 99]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{20}=1$, and $P$ is a point on the hyperbola such that $|P F_{1}|=9$, then $|P F_{2}|=$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/20 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 9",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "17",
    "fact_spans": "[[[18, 58], [70, 73]], [[18, 58]], [[2, 9]], [[10, 17]], [[2, 64]], [[2, 64]], [[66, 69]], [[66, 76]], [[78, 91]]]",
    "query_spans": "[[[93, 106]]]",
    "process": "From the hyperbola equation, we know $ a=4 $, $ b=2\\sqrt{5} $, $ c=\\sqrt{16+20}=6 $, $ |PF_{2}|\\geqslant c-a=2 $, and $ ||PF_{1}|-|PF_{2}||=2a $. $ |9-|PF_{2}||=8 $, so $ |PF_{2}|=17 $ (1 is discarded)."
  },
  {
    "text": "Given the equation of the ellipse is $\\frac{y^{2}}{4}+\\frac{x^{2}}{2}=1$, then the eccentricity of this ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G)=(y^2/4+x^2/2=1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 4], [45, 47]], [[2, 43]]]",
    "query_spans": "[[[45, 53]]]",
    "process": "The equation of the ellipse is $\\frac{y^{2}}{4}+\\frac{x^{2}}{2}=1$, then $a=2$, $b=\\sqrt{2}$, $c=\\sqrt{a^{2}-b^{2}}=\\sqrt{2}$, $\\therefore e=\\frac{c}{a}=\\frac{\\sqrt{2}}{2}$. The answer is $\\frac{\\sqrt{2}}{2}$."
  },
  {
    "text": "The asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(2)*x",
    "fact_spans": "[[[0, 28]], [[0, 28]]]",
    "query_spans": "[[[0, 36]]]",
    "process": "In the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$, $a=1$, $b=\\sqrt{2}$, $\\therefore\\frac{b}{a}=\\sqrt{2}$, so the asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$ are $y=\\pm\\sqrt{2}x$."
  },
  {
    "text": "If the equation $x^{2}+k y^{2}=2$ represents an ellipse with foci on the $y$-axis, then what is the range of real values for $k$?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (k*y^2 + x^2 = 2);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(0,1)",
    "fact_spans": "[[[31, 33]], [[36, 41]], [[2, 33]], [[23, 33]]]",
    "query_spans": "[[[36, 48]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{m}+\\frac{y^{2}}{4}=1$, if there exists a point $M$ on $C$ such that $\\overrightarrow{M F_{1}} \\cdot \\overrightarrow{M F_{2}}=0$, then the range of real values for $m$ is?",
    "fact_expressions": "F1: Point;F2: Point;C: Ellipse;Expression(C) = (y^2/4 + x^2/m = 1);m: Real;Focus(C) = {F1, F2};M: Point;PointOnCurve(M, C) = True;DotProduct(VectorOf(M, F1), VectorOf(M, F2)) = 0",
    "query_expressions": "Range(m)",
    "answer_expressions": "(0,2]+[8,+oo)",
    "fact_spans": "[[[2, 9]], [[10, 17]], [[18, 60], [67, 70]], [[18, 60]], [[140, 145]], [[2, 65]], [[73, 77]], [[67, 77]], [[79, 138]]]",
    "query_spans": "[[[140, 152]]]",
    "process": ""
  },
  {
    "text": "Given points $P(x_{1}, y_{1})$, $Q(x_{2}, y_{2})$ on the parabola $y^{2}=4 x$, and $x_{1}+x_{2}+2=2|P Q|$, where $F$ is the focus, then the maximum value of $\\angle P F Q$ is?",
    "fact_expressions": "G: Parabola;P: Point;Q: Point;F: Point;x1:Number;y1:Number;x2:Number;y2:Number;Expression(G) = (y^2 = 4*x);Coordinate(P) = (x1, y1);Coordinate(Q) = (x2, y2);Focus(G)=F;PointOnCurve(P,G);PointOnCurve(Q, G);x1 + x2 + 2 = 2*Abs(LineSegmentOf(P, Q))",
    "query_expressions": "Max(AngleOf(P, F, Q))",
    "answer_expressions": "ApplyUnit(60,degree)",
    "fact_spans": "[[[39, 53]], [[2, 19]], [[21, 38]], [[83, 86]], [[2, 19]], [[2, 19]], [[21, 38]], [[21, 38]], [[39, 53]], [[2, 19]], [[21, 38]], [[39, 89]], [[2, 56]], [[2, 56]], [[58, 81]]]",
    "query_spans": "[[[91, 110]]]",
    "process": "Because $x_{1}+x_{2}+2=2|PQ|$, that is, $x_{1}+1+x_{2}+1=2|PQ|$, so $|PF|+|QF|=2|PQ|$, then $\\frac{|PF|+|QF|}{2}=|PQ|$. In $\\triangle PQF$, $\\cos\\angle PFQ = \\frac{|PF|^{2}+|QF|^{2}-|PQ|^{2}}{2|PF|\\cdot|OF|} = \\frac{|PF|^{2}+|Q|}{2}|PF|^{2}+\\frac{3}{4}|QF|^{2}-|PF|\\cdot|QF| = \\frac{\\frac{3}{4}|PF|^{2}+\\frac{3}{4}|QF|^{2}}{2|PF|\\cdot|OF|}-\\frac{1}{4} \\geqslant \\frac{\\frac{3}{2}|PF|\\cdot|QF|}{2|PF|\\cdot|OF|}-\\frac{1}{4} = \\frac{1}{2}$. The equality holds if and only if $|PF|=|QF|$, at this time $\\cos\\angle PFQ = \\frac{1}{2}$, so $\\cos\\angle PFQ \\geqslant \\frac{1}{2}$. Since $y=\\cos x$ is monotonically decreasing in $(0,\\pi)$, it follows that $\\angle PFQ \\in (0,60^{\\circ}]$. Hence, the maximum value of $\\angle PFQ$ is $60^{\\circ}$."
  },
  {
    "text": "The asymptote equations of the hyperbola $\\frac{y^{2}}{4}-\\frac{x^{2}}{3}=1$ are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/3 + y^2/4 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*(2/3)*sqrt(3)*x",
    "fact_spans": "[[[0, 38]], [[0, 38]]]",
    "query_spans": "[[[0, 46]]]",
    "process": "Given that, we have $ a=2 $, $ b=\\sqrt{3} $, therefore the asymptotes of the hyperbola $ \\frac{y^{2}}{4}-\\frac{x^{2}}{3}=1 $ are $ y=\\pm\\frac{2}{3}\\sqrt{3}x $."
  },
  {
    "text": "What is the focal length of the hyperbola $3 x^{2}-y^{2}=9$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (3*x^2 - y^2 = 9)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[0, 20]], [[0, 20]]]",
    "query_spans": "[[[0, 25]]]",
    "process": "From $3x^{2}-y^{2}=9$, we get: $\\frac{x^{2}}{3}-\\frac{y^{2}}{9}=1$, so $a^{2}=3$, $b^{2}=9$, thus $c^{2}=a^{2}+b^{2}=3+9=12$, so $c=2\\sqrt{3}$, therefore the focal distance $2c=4\\sqrt{3}$. This question examines the geometric properties of hyperbolas, and is a basic problem."
  },
  {
    "text": "Given that $P$ is a point on the left branch of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F_{1}$ and $F_{2}$ are the left and right foci of $C$, respectively, and $M$ is an endpoint of the imaginary axis. If the minimum value of $|M P|+|P F_{2}|$ is $|F_{1} F_{2}|$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;M: Point;P: Point;F2: Point;F1: Point;a > b;b > 0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, LeftPart(C));LeftFocus(C) = F1;RightFocus(C) = F2;OneOf(Endpoint(ImageinaryAxis(C)))=M;Min(Abs(LineSegmentOf(M, P)) + Abs(LineSegmentOf(P, F2))) = Abs(LineSegmentOf(F1, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(4+sqrt(6))/2",
    "fact_spans": "[[[6, 64], [88, 91], [150, 153]], [[14, 64]], [[14, 64]], [[98, 101]], [[2, 5]], [[78, 85]], [[70, 77]], [[14, 64]], [[14, 64]], [[6, 64]], [[2, 69]], [[70, 97]], [[70, 97]], [[88, 109]], [[111, 148]]]",
    "query_spans": "[[[150, 159]]]",
    "process": "The minimum value of |MP| + |PF₂| is |F₁F₂|, and |PF₂| - |PF₁| = 2a, thus  \n\\begin{matrix}  \n\\text{the minimum value of } |MP| + |PF₂| \\text{ is } |F₁F₂|, \\text{ and } |PF₂| - |PF₁| = 2a, \\\\  \n|MP| + |PF₂| = |MP| + |PF₁| + 2a \\geqslant |MF₁| + 2a = \\sqrt{b^{2} + c^{2}} + 2a = 2c, \\text{ i.e., } \\sqrt{2c^{2} - a^{2}} + 2a = 2c,  \n\\end{matrix}  \nsimplifying yields the solution. Solution: |MP| + |PF₂| = |MP| + |PF₁| + 2a \\geqslant |MF₁| + 2a = \\sqrt{b^{2} + c^{2}} + 2a = 2c, i.e., \\sqrt{2c^{2} - a^{2}} + 2a = 2c, simplifying gives 2c^{2} - 8ac + 5a^{2} = 0, i.e., 2e^{2} - 8e + 5 = 0, solving yields e = \\frac{4 + \\sqrt{6}}{2} or e = \\frac{4 - \\sqrt{6}}{2}; since e > 1, then e = \\frac{4 + \\sqrt{6}}{2}."
  },
  {
    "text": "The perpendicular bisector of chord $AB$ passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) intersects the $x$-axis at point $P$. Given that $|A B|=10$, then $|F P|$=?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;F: Point;P: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;IsChordOf(LineSegmentOf(A, B), G);Intersection(PerpendicularBisector(LineSegmentOf(A, B)), xAxis) = P;Abs(LineSegmentOf(A, B)) = 10;PointOnCurve(F, LineSegmentOf(A, B))",
    "query_expressions": "Abs(LineSegmentOf(F, P))",
    "answer_expressions": "5",
    "fact_spans": "[[[1, 22]], [[4, 22]], [[30, 35]], [[30, 35]], [[25, 28]], [[47, 51]], [[4, 22]], [[1, 22]], [[1, 28]], [[1, 35]], [[30, 51]], [[54, 64]], [[0, 35]]]",
    "query_spans": "[[[66, 75]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25} + \\frac{y^{2}}{9} = 1$, a line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. If $|F_{2} A| + |F_{2} B| = 12$, then $|A B| =$?",
    "fact_expressions": "G: Ellipse;H: Line;F2: Point;A: Point;B: Point;F1: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(G) = {F1, F2};PointOnCurve(F1, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(F2, A)) + Abs(LineSegmentOf(F2, B)) = 12",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[18, 56], [74, 76]], [[71, 73]], [[10, 17]], [[77, 80]], [[81, 84]], [[2, 9], [63, 70]], [[18, 56]], [[2, 61]], [[62, 73]], [[71, 86]], [[89, 114]]]",
    "query_spans": "[[[116, 125]]]",
    "process": ""
  },
  {
    "text": "What are the asymptotes of the hyperbola $4 x^{2}-y^{2}+16=0$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (4*x^2 - y^2 + 16 = 0)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*2*x",
    "fact_spans": "[[[0, 23]], [[0, 23]]]",
    "query_spans": "[[[0, 31]]]",
    "process": "Since $4x^{2}-y^{2}+16=0$, we have $\\frac{y^{2}}{16}-\\frac{x^{2}}{4}=1$, so the asymptotes of the hyperbola are given by $y=\\pm2x$."
  },
  {
    "text": "Given that point $F$ is the focus of the parabola $C$: $y^{2}=4x$, and line $l$ passes through point $F$ and intersects parabola $C$ at points $A$ and $B$, with point $A$ in the first quadrant, $M(-2,0)$. If $\\frac{3}{2} \\leq \\frac{S_{\\triangle M A F}}{S_{\\triangle M B F}} \\leq 2$ (where $S_{\\triangle M A F}$, $S_{\\triangle M B F}$ denote the areas of $\\triangle M A F$ and $\\triangle M B F$, respectively), then what is the range of values for the slope of line $l$?",
    "fact_expressions": "l: Line;C: Parabola;M: Point;A: Point;F: Point;B: Point;Expression(C) = (y^2 = 4*x);Coordinate(M) = (-2, 0);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Quadrant(A) = 1;3/2 <= Area(TriangleOf(M, A, F))/Area(TriangleOf(M, B, F));Area(TriangleOf(M, A, F))/Area(TriangleOf(M, B, F)) <= 2",
    "query_expressions": "Range(Slope(l))",
    "answer_expressions": "[2*sqrt(2), 2*sqrt(6)]",
    "fact_spans": "[[[29, 34], [246, 251]], [[7, 25], [41, 47]], [[69, 78]], [[59, 63], [49, 52]], [[2, 6], [35, 39]], [[53, 56]], [[7, 25]], [[69, 78]], [[2, 28]], [[29, 39]], [[29, 58]], [[59, 68]], [[80, 154]], [[80, 154]]]",
    "query_spans": "[[[246, 261]]]",
    "process": ""
  },
  {
    "text": "Given that the line $l$ passes through the point $(0,2)$ and intersects the parabola $y^{2}=4x$ at two points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$, then $\\frac{1}{y_{1}} + \\frac{1}{y_{2}}$=?",
    "fact_expressions": "l: Line;G: Parabola;H: Point;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Coordinate(H) = (0, 2);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);PointOnCurve(H, l);x1:Number;x2:Number;y1:Number;y2:Number;Intersection(l,G)={A,B}",
    "query_expressions": "1/y1 + 1/y2",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 7]], [[19, 33]], [[8, 16]], [[35, 52]], [[54, 71]], [[19, 33]], [[8, 16]], [[35, 52]], [[54, 71]], [[2, 16]], [[35, 52]], [[54, 71]], [[35, 52]], [[54, 71]], [[2, 73]]]",
    "query_spans": "[[[75, 112]]]",
    "process": ""
  },
  {
    "text": "Given point $P(-1, -1)$, and point $F$ is the focus of the parabola $C$: $y^{2}=2 p x$ $(p>0)$. The line $l$ passing through point $F$ with slope $-2$ intersects the parabola at points $A$ and $B$. If $\\overrightarrow{A P} \\cdot \\overrightarrow{B P}=0$, then $p=$?",
    "fact_expressions": "P: Point;Coordinate(P) = (-1, -1);F: Point;Focus(C) = F;C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;l: Line;PointOnCurve(F,l) = True;Slope(l) = -2;A: Point;B: Point;Intersection(C,l) = {A,B};DotProduct(VectorOf(A, P), VectorOf(B, P)) = 0",
    "query_expressions": "p",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 14]], [[2, 14]], [[16, 20], [54, 58]], [[16, 52]], [[21, 49], [74, 77]], [[21, 49]], [[143, 146]], [[29, 49]], [[67, 72]], [[53, 72]], [[59, 72]], [[79, 82]], [[83, 86]], [[67, 88]], [[90, 141]]]",
    "query_spans": "[[[143, 148]]]",
    "process": "\\because F(\\frac{p}{2},0), the line: y=2(x-\\frac{p}{2})=2x-p, solve the system \\begin{cases}y=2x-p\\\\y^2=2px\\end{cases} eliminating y gives 4x^{2}-6px+p^{2}=0, let A(x_{1},y_{1}), B(x_{2},y_{2}), then x_{1}+x_{2}=\\frac{3}{2}p, x_{1}x_{2}=\\frac{p^{2}}{4} \\therefore \\overrightarrow{AP} \\cdot \\overrightarrow{BP} = (-1-x_{1})(-1-x_{2}) + (-1-y_{1})(-1-y_{2}) = 1 + x_{1}x_{2} + x_{1} + x_{2} + (p+1)^{2} + 4x_{1}x_{2} - 2(p+1)(x_{1}+x_{2}) = 5x_{1}x_{2} + (-1-2p)(x_{1}+x_{2}) + 1 + (p+1)^{2} = \\frac{5p^{2}}{4} + (-1-2p)\\times\\frac{3}{2}p + 1 + (p+1)^{2} = 0, solving yields p=2. The answer is 2. This question examines properties of parabolas, classified as a medium-level problem."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{10-k}+\\frac{y^{2}}{k-2}=1$, with foci on the $y$-axis, if the focal distance equals $4$, then the real number $k$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(10 - k) + y^2/(k - 2) = 1);k: Real;PointOnCurve(Focus(G), yAxis);FocalLength(G) = 4",
    "query_expressions": "k",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 44]], [[2, 44]], [[64, 69]], [[2, 53]], [[2, 62]]]",
    "query_spans": "[[[64, 71]]]",
    "process": ""
  },
  {
    "text": "Given $P(0,2)$, the focus $F$ of the parabola $C$: $y^{2}=2 p x(p>0)$, the line segment $PF$ intersects the parabola $C$ at point $M$. From $M$, draw a perpendicular to the directrix of the parabola, with foot $Q$. If $\\angle P Q F=90^{\\circ}$, then $p=?$",
    "fact_expressions": "C: Parabola;p: Number;F: Point;P: Point;Q: Point;M: Point;l: Line;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(P) = (0, 2);Focus(C) = F;Intersection(LineSegmentOf(P, F), C) = M;PointOnCurve(M, l);IsPerpendicular(Directrix(C), l);FootPoint(Directrix(C), l) = Q;AngleOf(P, Q, F) = ApplyUnit(90, degree)",
    "query_expressions": "p",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[11, 37], [52, 58], [72, 75]], [[117, 120]], [[40, 43]], [[2, 10]], [[84, 87]], [[62, 65], [68, 71]], [], [[19, 37]], [[11, 37]], [[2, 10]], [[11, 43]], [[44, 65]], [[67, 80]], [[67, 80]], [[67, 87]], [[89, 114]]]",
    "query_spans": "[[[117, 122]]]",
    "process": ""
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $M$ is a point on the right branch of $C$. Let $\\angle F_{1} M F_{2}=\\theta$. If $\\overrightarrow{M F_{1}} \\cdot \\overrightarrow{M F_{2}}=4 b^{2}$, then $\\cos \\theta=?$",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;F1: Point;M: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(M, RightPart(C));Theta: Number;AngleOf(F1, M, F2) = Theta;DotProduct(VectorOf(M, F1), VectorOf(M, F2)) = 4*b^2",
    "query_expressions": "Cos(Theta)",
    "answer_expressions": "2/3",
    "fact_spans": "[[[2, 66], [95, 98]], [[10, 66]], [[10, 66]], [[75, 82]], [[91, 94]], [[83, 90]], [[10, 66]], [[10, 66]], [[2, 66]], [[2, 90]], [[2, 90]], [[91, 103]], [[204, 217]], [[105, 134]], [[137, 202]]]",
    "query_spans": "[[[204, 219]]]",
    "process": "|MF_{1}|-|MF_{2}|=2a,4c^{2}=|MF_{1}|^{2}+|MF_{2}|^{2}-2|MF_{1}|\\cdot|MF_{2}|\\cos\\theta=(|MF_{1}|-|MF_{2}|)^{2}+2|MF_{1}|\\cdot|MF_{2}|-8b^{2} implies 4c^{2}=4a^{2}+2|MF_{1}|\\cdot|MF_{2}|-8b^{2}, yielding |MF_{1}|\\cdot|MF_{2}|=6b^{2}\\cos\\theta=\\frac{\\overrightarrow{MF_{1}}\\cdot\\overrightarrow{MF_{2}}}{|MF_{1}||MF_{2}|}=\\frac{4b^{2}}{6b^{2}}=\\frac{2}{3}"
  },
  {
    "text": "Given that a line $l$ with slope $2$ passes through the focus $F$ of the parabola $y^{2}=a x$ and intersects the $y$-axis at point $A$, if the area of $\\Delta O A F$ (where $O$ is the origin) is $4$, then the equation of the parabola is?",
    "fact_expressions": "l: Line;G: Parabola;a: Number;O: Origin;A: Point;F: Point;Expression(G) = (y^2 = a*x);Slope(l)=2;Focus(G)=F;PointOnCurve(F,l);Intersection(l,yAxis)=A;Area(TriangleOf(O,A,F))=4",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=pm*8*x",
    "fact_spans": "[[[9, 14]], [[15, 29], [85, 88]], [[18, 29]], [[66, 69]], [[45, 49]], [[32, 35]], [[15, 29]], [[2, 14]], [[15, 35]], [[9, 35]], [[9, 49]], [[51, 83]]]",
    "query_spans": "[[[85, 92]]]",
    "process": ""
  },
  {
    "text": "The line $y=2x-3$ intersects the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$ at points $A$ and $B$. Then $|AB|=$?",
    "fact_expressions": "H: Line;Expression(H) = (y = 2*x - 3);G: Hyperbola;Expression(G) = (x^2/2 - y^2 = 1);A: Point;B: Point;Intersection(H, G) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "4*sqrt(5)/7",
    "fact_spans": "[[[0, 11]], [[0, 11]], [[12, 40]], [[12, 40]], [[43, 46]], [[47, 50]], [[0, 52]]]",
    "query_spans": "[[[54, 63]]]",
    "process": ""
  },
  {
    "text": "Given that the minor axis of ellipse $C$ is $6$ and the eccentricity is $\\frac{4}{5}$, find the distance from a focus $F$ of ellipse $C$ to one endpoint of the major axis.",
    "fact_expressions": "C: Ellipse;Length(MinorAxis(C)) = 6;Eccentricity(C) = 4/5;F: Point;OneOf(Focus(C)) = F",
    "query_expressions": "Distance(F, OneOf(Endpoint(MajorAxis(C))))",
    "answer_expressions": "{1, 9}",
    "fact_spans": "[[[2, 7], [35, 40]], [[2, 15]], [[2, 33]], [[43, 46]], [[35, 46]]]",
    "query_spans": "[[[35, 59]]]",
    "process": ""
  },
  {
    "text": "Given that the distance from the vertex of a hyperbola to its asymptote is $2$, and the distance from the focus to the asymptote is $6$, then the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Distance(Vertex(G), Asymptote(G)) = 2;Distance(Focus(G), Asymptote(G)) = 6",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 5], [36, 39]], [[2, 19]], [[2, 33]]]",
    "query_spans": "[[[36, 45]]]",
    "process": "As shown in the figure, perpendiculars are drawn from the vertex A and the focus F of a hyperbola to its asymptote, with the feet of the perpendiculars being B and C, respectively. Then: $\\frac{|OF|}{|OA|}=\\frac{|FC|}{|AB|}\\Rightarrow\\frac{c}{a}=\\frac{6}{2}=3$"
  },
  {
    "text": "Ellipse $M$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $A$ is a vertex on the major axis, $B$ is a vertex on the minor axis, $F$ is the right focus, and $AB \\perp BF$. Then the eccentricity of ellipse $M$ is?",
    "fact_expressions": "M: Ellipse;a: Number;b: Number;A: Point;B: Point;F: Point;a > b;b > 0;Expression(M) = (y^2/b^2 + x^2/a^2 = 1);OneOf(Vertex(MajorAxis(M)))=A;OneOf(Vertex(MinorAxis(M)))=B;RightFocus(M) = F;IsPerpendicular(LineSegmentOf(A, B), LineSegmentOf(B, F))",
    "query_expressions": "Eccentricity(M)",
    "answer_expressions": "(\\sqrt{5}-1)/2",
    "fact_spans": "[[[0, 57], [108, 113]], [[6, 57]], [[6, 57]], [[58, 61]], [[70, 73]], [[82, 85]], [[6, 57]], [[6, 57]], [[0, 57]], [[0, 69]], [[0, 81]], [[0, 89]], [[91, 106]]]",
    "query_spans": "[[[108, 119]]]",
    "process": ""
  },
  {
    "text": "$P$ is a fixed point on the ellipse, $F_{1}$ and $F_{2}$ are the two foci of the ellipse. If $\\angle PF_{1} F_{2}=60^{\\circ}$, $\\angle PF_{2} F_{1}=30^{\\circ}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;PointOnCurve(P, G);Focus(G)={F1,F2};AngleOf(P, F1, F2) = ApplyUnit(60, degree);AngleOf(P, F2, F1) = ApplyUnit(30, degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[29, 31], [29, 31], [29, 31]], [[0, 3]], [[11, 18]], [[21, 28]], [[0, 10]], [[11, 36]], [[38, 70]], [[71, 104]]]",
    "query_spans": "[[[106, 114]]]",
    "process": "In triangle $APF_{1}F_{2}$, by the law of sines, $\\frac{|PF_{2}|}{\\sin60^{0}}=\\frac{|PF_{1}|}{\\sin30^{0}}=\\frac{2c}{\\sin90^{\\circ}}$, hence $\\frac{|PF_{2}|+|PF_{1}|}{\\sin60^{0}+\\sin30^{0}}=\\frac{2c}{\\sin90^{\\circ}}=\\frac{2a}{\\sqrt{3}+1}$, hence $e=\\frac{2}{\\sqrt{3}+1}=\\sqrt{3}-1$"
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{5}+y^{2}=1$, the center of the ellipse is at the origin $O$, point $F$ is the right focus of the ellipse, point $A$ is one endpoint of the minor axis, a line $l$ passing through point $F$ intersects the ellipse at points $M$ and $N$, and intersects the line $OA$ at point $E$. If $\\overrightarrow{E M}=\\lambda_{1} \\overrightarrow{M F}$, $\\overrightarrow{E N}=\\lambda_{2} \\overrightarrow{N F}$, then $\\lambda_{1}+\\lambda_{2}$=?",
    "fact_expressions": "l: Line;G: Ellipse;O: Origin;A: Point;E: Point;M: Point;F: Point;N: Point;lambda1:Number;lambda2:Number;Expression(G) = (x^2/5 + y^2 = 1);Center(G)=O;RightFocus(G) = F;OneOf(Endpoint(MinorAxis(G)))=A;PointOnCurve(F, l);Intersection(l, G) = {M, N};Intersection(OverlappingLine(LineSegmentOf(O, A)),l) = E;VectorOf(E, M) = lambda1*VectorOf(M, F);VectorOf(E, N) = lambda2*VectorOf(N, F)",
    "query_expressions": "lambda1 + lambda2",
    "answer_expressions": "-10",
    "fact_spans": "[[[77, 82]], [[2, 29], [30, 32], [49, 51], [61, 63], [83, 85]], [[36, 43]], [[56, 60]], [[109, 113]], [[87, 90]], [[44, 48], [72, 76]], [[91, 94]], [[229, 254]], [[229, 254]], [[2, 29]], [[30, 43]], [[44, 55]], [[56, 70]], [[71, 82]], [[77, 96]], [[77, 113]], [[115, 170]], [[172, 227]]]",
    "query_spans": "[[[229, 256]]]",
    "process": "Let $ M(x_{1},y_{1}) $, $ N(x_{2},y_{2}) $ with $ (x_{1}<x_{2}) $. $ \\because\\overrightarrow{EM}=\\lambda_{1}\\overrightarrow{MF} $, $ \\therefore\\lambda_{1}=\\frac{|EM|}{|MF|} $, $ \\therefore\\lambda_{1}+1=\\frac{|EF|}{|MF|}\\cdots\\textcircled{1} $ $ \\because\\overrightarrow{EN}=\\lambda_{2}\\overrightarrow{NF} $, $ \\therefore\\lambda_{2}=\\frac{|EN|}{|NF|} $, $ \\therefore\\lambda_{2}+1=\\frac{|EF|}{|NF|}\\cdots\\textcircled{2} $ Subtracting $ \\textcircled{2} $ from $ \\textcircled{1} $ gives $ \\lambda_{1}+\\lambda_{2}+2=|EF|\\left(\\frac{1}{|MF|}-\\frac{1}{|NF|}\\right)=|EF|\\cdot\\frac{|NF|-|MF|}{|MF||NF|} $. From the ellipse equation $ \\frac{x^{2}}{5}+y^{2}=1 $, we know $ c=\\sqrt{5-1}=2 $, hence $ F(2,0) $. According to the problem, the slope of line $ l $ exists; without loss of generality, let the inclination angle of line $ l $ be $ \\alpha\\in(0,\\frac{\\pi}{2}) $. Then the parametric equations of line $ l $ are:\n$$\n\\begin{cases}\nx=2+t\\cos\\alpha\\\\\ny=t\\sin\\alpha\n\\end{cases}\n$$\n($ t $ is a parameter). Substituting into $ \\frac{x^{2}}{5}+y^{2}=1 $ and simplifying yields:\n$$\n(1+4\\sin^{2}\\alpha)t^{2}+4t\\cos\\alpha-1=0\n$$\n$ \\therefore t_{1}+t_{2}=\\frac{4\\cos\\alpha}{1+4\\sin^{2}\\alpha} $, $ t_{1}\\cdot t_{2}=\\frac{-1}{1+4\\sin^{2}\\alpha} $. By the geometric meaning of $ t $, we have $ |MF|\\cdot|NF|=|t_{1}\\cdot t_{2}|=\\frac{1}{1+4\\sin^{2}\\alpha} $, $ |NF|-|MF|=-|t_{1}+t_{2}|=-\\frac{4\\cos\\alpha}{1+4\\sin^{2}\\alpha} $. Converting the parametric equations of line $ l $ into the standard form gives $ y=\\tan\\alpha(x-2) $. Thus point $ E(0,-2\\tan\\alpha) $, $ \\therefore|EF|=\\sqrt{4+4\\tan^{2}\\alpha}=\\frac{2}{\\cos\\alpha} $. Therefore,\n$$\n\\lambda_{1}+\\lambda_{2}+2=\\frac{2}{\\cos\\alpha}\\cdot\\frac{-\\frac{4\\cos\\alpha}{1+4\\sin^{2}\\alpha}}{\\frac{1}{1+4\\sin^{2}\\alpha}}=-8,\n$$\nso $ \\lambda_{1}+\\lambda_{2}=-10 $."
  },
  {
    "text": "Given that an asymptote of the hyperbola $m x^{2}-n y^{2}=1$ is $y=\\frac{4}{3} x$, then the eccentricity of the hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;n: Number;m: Number;Expression(G) = (m*x^2 - n*y^2 = 1);Expression(OneOf(Asymptote(G))) = (y = (4/3)*x)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "{5/3, 5/4}",
    "fact_spans": "[[[2, 24], [52, 55]], [[5, 24]], [[5, 24]], [[2, 24]], [[2, 50]]]",
    "query_spans": "[[[52, 62]]]",
    "process": "If $ m > 0, n > 0 $, then the foci of the hyperbola lie on the x-axis, so we have $ \\frac{b}{a} = \\frac{4}{3} $, and thus the eccentricity $ e = \\frac{c}{a} = \\sqrt{1 + \\left( \\frac{b}{a} \\right)^2} = \\frac{5}{3} $. In conclusion, the eccentricity of the hyperbola equals $ \\frac{5}{3} $ or $ \\frac{5}{4} $."
  },
  {
    "text": "What is the focal length of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$?",
    "fact_expressions": "C:Ellipse;Expression(C)=(x^2/2+y^2=1)",
    "query_expressions": "FocalLength(C)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 27]], [[0, 27]]]",
    "query_spans": "[[[0, 33]]]",
    "process": "Since in the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, $a^{2}=2$, $b^{2}=1$, so $c^{2}=a^{2}-b^{2}=1$, therefore the focal distance is $2c=2$."
  },
  {
    "text": "Let the left and right foci of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$ be $F_{1}$ and $F_{2}$, respectively, and let $M$ be a point on the ellipse distinct from the endpoints of the major axis, $\\angle F_{1} M F_{2}=2 \\theta$, and let $I$ be the incenter of $\\Delta MF_{1} F_{2}$. Then $|M I| \\cos \\theta$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/5 + y^2/4 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;M: Point;PointOnCurve(M,G);Negation(M=Endpoint(MajorAxis(G)));theta: Number;AngleOf(F1, M, F2) = 2*theta;I: Point;Incenter(TriangleOf(M, F1, F2)) = I",
    "query_expressions": "Abs(LineSegmentOf(M, I))*Cos(theta)",
    "answer_expressions": "sqrt(5)-1",
    "fact_spans": "[[[1, 38], [67, 69]], [[1, 38]], [[47, 54]], [[55, 62]], [[1, 62]], [[1, 62]], [[63, 66]], [[63, 79]], [[63, 79]], [[80, 111]], [[80, 111]], [[138, 141]], [[113, 141]]]",
    "query_spans": "[[[143, 164]]]",
    "process": "By the given condition, $ MF_{1} + |MF_{2}| = 4 $, while $ |F_{1}F_{2}| = 2 $. Let the circle be tangent to $ MF_{1} $ and $ MF_{2} $ at points $ A $ and $ B $, respectively. According to the tangent segment theorem, $ |F_{1}F_{2}| = |F_{1}A| + |F_{2}B| = 2 $. Therefore, $ |M||\\cos\\theta = |MA| = |MB| = a - c = \\sqrt{5} - 1 $."
  },
  {
    "text": "The line $y=x+2$ passes through one focus and one vertex of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$. Then, the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = x + 2);PointOnCurve(OneOf(Focus(G)),H);PointOnCurve(OneOf(Vertex(G)),H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[11, 63], [75, 77]], [[13, 63]], [[13, 63]], [[0, 9]], [[13, 63]], [[13, 63]], [[11, 63]], [[0, 9]], [[0, 68]], [[0, 73]]]",
    "query_spans": "[[[75, 83]]]",
    "process": ""
  },
  {
    "text": "A moving line passing through the focus $F$ of the parabola $C$: $y^{2}=4x$ intersects $C$ at points $A$ and $B$. The midpoint of segment $AB$ is $N$, and point $P(12, 4)$. When the value of $|NA|+|NP|$ is minimized, what is the horizontal coordinate of point $N$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;G: Line;PointOnCurve(F, G);A: Point;B: Point;Intersection(G, C) = {A, B};N: Point;MidPoint(LineSegmentOf(A, B)) = N;P: Point;Coordinate(P) = (12, 4);WhenMin(Abs(LineSegmentOf(N, A))+Abs(LineSegmentOf(N, P)))",
    "query_expressions": "XCoordinate(N)",
    "answer_expressions": "9",
    "fact_spans": "[[[31, 34], [1, 20]], [[1, 20]], [[23, 26]], [[1, 26]], [[27, 30]], [[0, 30]], [[35, 38]], [[39, 42]], [[27, 44]], [[56, 59], [94, 98]], [[45, 59]], [[60, 72]], [[60, 72]], [[74, 93]]]",
    "query_spans": "[[[94, 104]]]",
    "process": ""
  },
  {
    "text": "Let $P$ be an arbitrary point on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, $O$ be the coordinate origin, $F$ be the left focus of the ellipse, and point $M$ satisfy $OM=\\frac{1}{2}(OP+OF)$. Then $|OM|+|MF|$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);P: Point;PointOnCurve(P, G);O: Origin;F: Point;LeftFocus(G) = F;M: Point;LineSegmentOf(O, M) = (1/2)*(LineSegmentOf(O, F) + LineSegmentOf(O, P))",
    "query_expressions": "Abs(LineSegmentOf(M, F)) + Abs(LineSegmentOf(O, M))",
    "answer_expressions": "2",
    "fact_spans": "[[[5, 32], [51, 53]], [[5, 32]], [[1, 4]], [[1, 37]], [[38, 41]], [[47, 50]], [[47, 57]], [[58, 62]], [[64, 88]]]",
    "query_spans": "[[[90, 103]]]",
    "process": ""
  },
  {
    "text": "A moving circle passes through the point $A(0 , 1)$, has its center on the parabola $x^{2}=4 y$, and is always tangent to a fixed line $l$. Then the equation of the line $l$ is?",
    "fact_expressions": "l: Line;G: Parabola;H: Circle;A: Point;Expression(G) = (x^2 = 4*y);Coordinate(A) = (0, 1);PointOnCurve(A, H);PointOnCurve(Center(H),G);IsTangent(l,H)",
    "query_expressions": "Expression(l)",
    "answer_expressions": "y=-1",
    "fact_spans": "[[[41, 44], [48, 53]], [[19, 33]], [[1, 3]], [[4, 15]], [[19, 33]], [[4, 15]], [[1, 15]], [[1, 34]], [[1, 46]]]",
    "query_spans": "[[[48, 58]]]",
    "process": ""
  },
  {
    "text": "Let $P$ be a moving point on the curve $x^{2}-4 y^{2}-4=0$, $O$ be the origin, and $M$ be the midpoint of segment $PO$. Then the equation of the trajectory of point $M$ is?",
    "fact_expressions": "P: Point;PointOnCurve(P,G) = True;G: Curve;Expression(G) = (x^2 - 4*y^2 - 4 = 0);O: Origin;M: Point;MidPoint(LineSegmentOf(P,O)) = M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2-4*y^2=1",
    "fact_spans": "[[[1, 4]], [[1, 30]], [[5, 26]], [[5, 26]], [[31, 34]], [[40, 43], [56, 60]], [[40, 54]]]",
    "query_spans": "[[[56, 67]]]",
    "process": "Let M(x, y). Since O is the coordinate origin and M is the midpoint of segment PO, we obtain P(2x, 2y). Substituting into the equation of the hyperbola yields (2x)^{2} - 4 \\times (2y)^{2} - 4 = 0. Simplifying gives x^{2} - 4y^{2} = 1, so the trajectory equation of point M is x^{2} - 4y^{2} = 1. This problem primarily examines solving for trajectory equations. The key to solving it lies in carefully reading the problem and reasonably applying the substitution method, emphasizing reasoning and computational skills. It is a basic-level problem."
  },
  {
    "text": "Given that the line $y = kx + 1$ intersects the right branch of the hyperbola $\\frac{x^{2}}{4} - \\frac{y^{2}}{3} = 1$ at two points, find the range of real values for $k$.",
    "fact_expressions": "G: Hyperbola;H: Line;k: Real;Expression(G) = (x^2/4 - y^2/3 = 1);Expression(H) = (y = k*x + 1);NumIntersection(H, RightPart(G)) = 2",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-1, -\\sqrt{3}/2)",
    "fact_spans": "[[[14, 52]], [[2, 13]], [[61, 66]], [[14, 52]], [[2, 13]], [[2, 59]]]",
    "query_spans": "[[[61, 73]]]",
    "process": "According to the problem, solve simultaneously the line and the hyperbola $ v_{2} = 12 = \\frac{1}{2}8kx - 16 = 0 $. D: (sometimes) This problem examines the intersection of a line and a hyperbola. The problem is solved using the idea of equations, or it can be solved by combining numerical and geometric thinking."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{25} + \\frac{y^{2}}{18}=1$ with left and right foci $F_{1}$, $F_{2}$, and a point $P$ on the ellipse such that $|P F_{1}|=6$, then $\\angle F_{1} P F_{2}$=?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/18 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 6",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "pi/3",
    "fact_spans": "[[[2, 43], [70, 72]], [[65, 69]], [[49, 56]], [[57, 64]], [[2, 43]], [[2, 64]], [[2, 64]], [[65, 73]], [[75, 88]]]",
    "query_spans": "[[[90, 114]]]",
    "process": ""
  },
  {
    "text": "Given a fixed point $A(-2, \\sqrt{3})$, and $F$ is the right focus of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$. Find a point $M$ on the ellipse such that $|A M|+|M F|$ is minimized. What are the coordinates of point $M$?",
    "fact_expressions": "G: Ellipse;A: Point;M: Point;F: Point;Expression(G) = (x^2/16 + y^2/12 = 1);Coordinate(A) = (-2, sqrt(3));RightFocus(G) = F;PointOnCurve(M, G);WhenMin(Abs(LineSegmentOf(A,M))+Abs(LineSegmentOf(M,F)))",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(-2,3)",
    "fact_spans": "[[[28, 67], [73, 75]], [[4, 21]], [[79, 82], [104, 108]], [[24, 27]], [[28, 67]], [[4, 21]], [[24, 71]], [[72, 82]], [[84, 103]]]",
    "query_spans": "[[[104, 112]]]",
    "process": "The ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$ has $a=4$, $c=2$, $e=\\frac{1}{2}$, let the left focus be $F(-2,0)$. By the definition of the ellipse, $|MF|+|MF|=2a=8$, hence $|MF|=8-|MF|$. $|AM|+|MF|=|AM|-|MF|+8\\geqslant8-|AF|$. When $A$, $M$, $F$ are collinear and $M$ is in the second quadrant, $|AM|+|MF|$ attains the minimum value. Substitute $x=-2$ into the ellipse equation to get $y=\\pm3$, since $M$ is in the second quadrant, $M(-2,3)$."
  },
  {
    "text": "Point $P$ is a moving point on the parabola $y^{2}=4x$. Then, the minimum value of the sum of the distance from point $P$ to point $A(0,-1)$ and the distance from $P$ to the line $x=-1$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;P: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (x = -1);Coordinate(A) = (0, -1);PointOnCurve(P, G)",
    "query_expressions": "Min(Distance(P, A) + Distance(P, H))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[5, 19]], [[48, 56]], [[30, 40]], [[0, 4], [25, 29], [44, 47]], [[5, 19]], [[48, 56]], [[30, 40]], [[0, 23]]]",
    "query_spans": "[[[25, 66]]]",
    "process": "\\because the distance from point P to the line x = -1 is equal to the distance from point P to the focus F of the parabola y^{2} = 4x, when point P lies on AF, the sum of the distances from point P to point A(0,-1) and to the line x = 1 is minimized; at this time |PA| + |PF| = |AF| = \\sqrt{2}."
  },
  {
    "text": "Given point $A(6,0)$, point $P$ moves along the parabola $y^{2}=16 x$, and point $B$ moves along the curve $(x-4)^{2}+y^{2}=1$, then the minimum value of $\\frac{|P A|^{2}}{|P B|}$ is?",
    "fact_expressions": "A: Point;Coordinate(A) = (6, 0);P: Point;PointOnCurve(P, G);G: Parabola;Expression(G) = (y^2 = 16*x);B: Point;PointOnCurve(B, H);H: Curve;Expression(H) = (y^2 + (x - 4)^2 = 1)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A))^2/Abs(LineSegmentOf(P, B)))",
    "answer_expressions": "2*sqrt(41)-6",
    "fact_spans": "[[[2, 11]], [[2, 11]], [[12, 16]], [[12, 35]], [[17, 32]], [[17, 32]], [[36, 40]], [[36, 65]], [[41, 62]], [[41, 62]]]",
    "query_spans": "[[[67, 98]]]",
    "process": "The focus of the parabola $ y^{2} = 16x $ is $ F(4,0) $. Let point $ P $ have coordinates $ (x,y) $, then $ |PF| = x + 4 $, $ |PA|^{2} = (x - 6)^{2} + y^{2} = (x - 6)^{2} + 16x = x^{2} + 4x + 36 $. According to the problem, when $ |PB| = |PF| + 1 = x + 5 $, $ \\frac{|PA|^{2}}{|PB|} = \\frac{x^{2}}{} $. Let $ x + 5 = t $, then $ x = t - 5 $, $ \\frac{|PA|^{2}}{|PB|} = \\frac{(}{ $. From the basic inequality, $ t + \\frac{41}{t} \\geqslant 2\\sqrt{41} $, with equality if and only if $ t = \\sqrt{41} $, $ \\frac{|PA|^{2}}{|PB|^{2} + 4(t - 5) + 36} = \\frac{t^{2} - 6t + 41}{l} = t + \\frac{41}{1} - 6 $. Hence, the minimum value of $ \\frac{|PA|^{2}}{|PB|} $ is $ 2\\sqrt{41} - 6 $."
  },
  {
    "text": "If the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has eccentricity $\\sqrt{10}$, then the value of $\\frac{b}{a}$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(C) = sqrt(10)",
    "query_expressions": "b/a",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 62]], [[80, 93]], [[80, 93]], [[9, 62]], [[9, 62]], [[1, 62]], [[1, 78]]]",
    "query_spans": "[[[80, 97]]]",
    "process": "The eccentricity $ e = \\frac{c}{a} = \\sqrt{10} $, so $ c = \\sqrt{10}a $. Also, $ c^{2} = a^{2} + b^{2} $, therefore $ b = \\sqrt{c^{2} - a^{2}} = 3a $. Hence, $ \\frac{b}{a} = 3 $. The correct answer is: $ 3 $."
  },
  {
    "text": "Given that $x$, $y$ satisfy $y=\\frac{1}{2} \\sqrt{4-x^{2}}$, then the range of $\\frac{y}{x+3}$ is?",
    "fact_expressions": "x_: Number;y_: Number;y_ = sqrt(4 - x_^2)/2",
    "query_expressions": "Range(y_/(x_ + 3))",
    "answer_expressions": "[0,sqrt(5)/5]",
    "fact_spans": "[[[2, 5]], [[7, 10]], [[12, 42]]]",
    "query_spans": "[[[44, 66]]]",
    "process": "Rewriting the given equation as \\frac{x^{2}}{4}+y^{2}=1 (y\\geqslant0), its graph is a semi-ellipse; transforming \\frac{y}{x+3} into the range of slopes of lines connecting points on the semi-ellipse to (-3,0); from the graph, the lower bound is 0, and the upper bound occurs when the line is tangent to the semi-ellipse; assuming the tangent line equation and solving simultaneously with the ellipse equation using A=0 to find the slope of the tangent, thus obtaining the desired range. From y=\\frac{1}{2}\\sqrt{4-x^{2}} we get: \\frac{x^{2}}{4}+y^{2}=1 (y\\geqslant0), so its graph is the semi-ellipse shown below. \\frac{y}{x+3} can be viewed as the slope of the line connecting a point (x,y) on the semi-ellipse to (-3,0). When the line l passing through (-3,0) is tangent to the ellipse as shown in the figure, assume the line l: y=k(x+3), k>0. Combining with the ellipse equation gives: (4k^{2}+1)x^{2}+24k^{2}x+36k^{2}-4=0. \\therefore A=576k^{4}-4(4k^{2}+1)(36k^{2}-4)=0, solving yields: k=\\frac{\\sqrt{5}}{5}. \\therefore The range of slopes of lines connecting points (x,y) on the semi-ellipse to (-3,0) is \\left[0,\\frac{\\sqrt{5}}{5}\\right]. \\therefore \\frac{y}{x+3}\\in\\left[0,\\frac{\\sqrt{5}}{5}\\right]"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with right focus $F$, the line $y=k x(k \\neq 0)$ intersects the hyperbola $C$ at points $A$ and $B$. If $\\angle A F B=90^{\\circ}$ and the area of $\\triangle O A F$ is $4 a^{2}$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(C) = F;G: Line;Expression(G) = (y = k*x);k: Number;Negation(k=0);Intersection(G, C) = {A, B};A: Point;B: Point;AngleOf(A, F, B) = ApplyUnit(90, degree);O: Origin;Area(TriangleOf(O, A, F)) = 4*a^2",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 63], [92, 98], [170, 176]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 71]], [[2, 71]], [[72, 91]], [[72, 91]], [[74, 91]], [[74, 91]], [[72, 109]], [[100, 103]], [[104, 107]], [[111, 136]], [[138, 155]], [[138, 168]]]",
    "query_spans": "[[[170, 182]]]",
    "process": "Let the left focus of the hyperbola be $ F_{1} $. Since $ \\angle AFB = 90^{\\circ} $, by the symmetry of the graph of a direct proportional function and the hyperbola, we know $ \\angle AF_{1}B = 90^{\\circ} $, quadrilateral $ AF_{1}BF $ is a rectangle, and $ |AF| = |BF_{1}| $. Let $ |AF| = m $, $ |BF| = n $, then $ |AF| - |BF| = |BF_{1}| - |BF| = m - n = 2a $. Since the area of $ \\triangle OAF $ is $ 4a^{2} $, the area of $ \\triangle ABF $ is $ S_{\\Delta ABF} = 2S_{\\Delta OAF} = \\frac{1}{2}m \\cdot n = 8a^{2} $, and $ m^{2} + n^{2} = |AB|^{2} = 4c^{2} $. Solving the system of three equations:  \n$$\n\\begin{cases}\nm - n = 2a \\\\\nmn = 16a^{2} \\\\\nm^{2} + n^{2} = 4c^{2}\n\\end{cases}\n$$\nwe obtain $ 4c^{2} = 4a^{2} + 32a^{2} $, thus $ c^{2} = 9a^{2} $, i.e., $ e = 3 $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{18}=1$ with left and right foci $F_{1}$, $F_{2}$, point $P$ lies on the ellipse, and $|PF_{1}|=6$, then $\\angle F_{1} P F_{2}$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/18 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 6",
    "query_expressions": "AngleOf(F1, P, F2)",
    "answer_expressions": "pi/3",
    "fact_spans": "[[[2, 41], [68, 70]], [[2, 41]], [[47, 54]], [[55, 62]], [[2, 62]], [[2, 62]], [[63, 67]], [[63, 71]], [[73, 85]]]",
    "query_spans": "[[[87, 111]]]",
    "process": "From the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{18}=1$, we obtain $|PF_{1}|+|PF_{2}|=2a=10$, $|F_{1}F_{2}|=2c=2\\sqrt{7}$. Since $|PF_{1}|=6$, it follows that $|PF_{2}|=4$. By the cosine law, $\\cos\\angle F_{1}PF_{2}=\\frac{|PF_{1}|^{2}+|PF_{2}|^{2}-|F_{1}F_{2}|^{2}}{2|PF_{1}||PF_{2}|}=\\frac{36+16-28}{2\\times6\\times4}=\\frac{1}{2}$. Therefore, $\\angle F_{1}PF_{2}=\\frac{\\pi}{3}$. [Comment] This problem mainly examines the application of the equation and definition of an ellipse, and using the cosine law to find an angle. It is a basic problem."
  },
  {
    "text": "Given point $P(2 , 2)$, circle $C$: $x^{2}+y^{2}-8 y=0$, a moving line $l$ passing through point $P$ intersects circle $C$ at points $A$ and $B$, the midpoint of segment $AB$ is $M$, and $O$ is the origin. Then the trajectory equation of point $M$ is?",
    "fact_expressions": "l: Line;C: Circle;B: Point;A: Point;P: Point;M: Point;O: Origin;Expression(C) = (-8*y + x^2 + y^2 = 0);Coordinate(P) = (2, 2);PointOnCurve(P,l);Intersection(l,C)={A,B};MidPoint(LineSegmentOf(A,B))=M",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "(x-1)^2+(y-3)^2=2",
    "fact_spans": "[[[49, 52]], [[14, 39], [53, 57]], [[63, 66]], [[59, 62]], [[2, 13], [41, 45]], [[80, 83], [94, 98]], [[84, 87]], [[14, 39]], [[2, 13]], [[40, 52]], [[49, 68]], [[69, 83]]]",
    "query_spans": "[[[94, 105]]]",
    "process": "Let M(x,y). By the perpendicular chord theorem, we have CM\\botMP, so the locus of point M is a circle with CP as diameter. The equation of circle C can be rewritten as x^{2}+(y-4)^{2}=16, so the center is C(0,4) and the radius is 4. Let M(x,y), and let M be the midpoint of segment AB. By the perpendicular chord theorem, we have CM\\botMP, so the locus of point M is a circle with CP as diameter. Since 2^{2}+2^{2}-2\\times8<0, point P lies inside circle C. The midpoint of CP is (1,3), and |CP|=\\sqrt{(0-2)^{2}+(4-2)^{2}}=2\\sqrt{2}. The equation of the circle with CP as diameter is (x-1)^{2}+(y-3)^{2}=2. Since point P lies inside circle C, the trajectory equation of point M is (x-1)^{2}+(y-3)^{2}=2."
  },
  {
    "text": "Let point $P$ be an intersection point of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ and the circle $x^{2}+y^{2}=2 a^{2}$, and let $F_{1}$, $F_{2}$ be the left and right foci of the hyperbola, respectively. If $P F_{1}=3 P F_{2}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + y^2 = 2*a^2);OneOf(Intersection(H,G))=P;LeftFocus(G) = F1;RightFocus(G) = F2;LineSegmentOf(P, F1) = 3*LineSegmentOf(P, F2)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[6, 62], [109, 112], [141, 144]], [[9, 62]], [[9, 62]], [[63, 85]], [[1, 5]], [[91, 98]], [[99, 106]], [[9, 62]], [[9, 62]], [[6, 62]], [[63, 85]], [[1, 90]], [[91, 118]], [[91, 118]], [[120, 139]]]",
    "query_spans": "[[[141, 150]]]",
    "process": "From \\begin{cases} PF_{1}-PF_{2}=2a \\\\ PF_{1}=3PF_{2} \\end{cases} we get PF_{1}=3a, PF_{2}=a_{1}. Let \\angle F_{1}OP = \\alpha, then \\angle POF_{2} = 180^{\\circ}-\\alpha. In \\triangle PF_{1}O, PF_{1}^{2} = OF_{1}^{2} + OP^{2} - 2OF_{1} \\cdot OP \\cdot \\cos\\alpha \\textcircled{1}. In \\triangle OPF_{2}, PF_{2}^{2} = OF_{2}^{2} + OP^{2} - 2OF_{2} \\cdot OP \\cdot \\cos(180^{\\circ}-\\alpha) \\textcircled{2}. Using \\cos(180^{\\circ}-\\alpha) = -\\cos\\alpha and OP = \\sqrt{2}a, \\textcircled{1}+\\textcircled{2} gives c^{2} = 3a^{2}, \\therefore e = \\frac{c}{a} = \\frac{\\sqrt{3}a}{a} = \\sqrt{3}. Answer: \\sqrt{3}"
  },
  {
    "text": "Let the parabola $C$: $y^{2}=4x$ have focus $F$, and let $M$ be a point on the parabola $C$, $N(2,2)$. Then the range of values of $|MF|+|MN|$ is?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;M: Point;PointOnCurve(M, C);N: Point;Coordinate(N) = (2, 2)",
    "query_expressions": "Range(Abs(LineSegmentOf(M, F)) + Abs(LineSegmentOf(M, N)))",
    "answer_expressions": "[3, +\\infty)",
    "fact_spans": "[[[1, 20], [32, 38]], [[1, 20]], [[24, 27]], [[1, 27]], [[28, 31]], [[28, 41]], [[42, 50]], [[42, 50]]]",
    "query_spans": "[[[52, 72]]]",
    "process": "Since point N(2,2) is inside the parabola C: y^{2}=4x and the directrix of parabola C: y^{2}=4x is x=-1, let d be the distance from point N to the directrix, then |MF|+|MN|=d+|MN|\\geqslant2-(-1)=3, achieving the minimum value; |MF|+|MN| does not attain a maximum value. Therefore, |MF|+|MN|\\in[3,+\\infty), the range of MF+MN is [3,+\\infty)."
  },
  {
    "text": "Point $P(x, y)$ lies on the parabola $y^{2}=4 x$. Then, the minimum value of the sum of the distance from point $P$ to $(0,3)$ and the distance from point $P$ to the directrix is?",
    "fact_expressions": "P: Point;x0: Number;y0: Number;Coordinate(P) = (x0, y0);G: Parabola;Expression(G) = (y^2 = 4*x);PointOnCurve(P, G);H: Point;Coordinate(H) = (0, 3)",
    "query_expressions": "Min(Distance(P, H) + Distance(P, Directrix(G)))",
    "answer_expressions": "sqrt(10)",
    "fact_spans": "[[[0, 10], [28, 32], [44, 48]], [[1, 10]], [[1, 10]], [[0, 10]], [[11, 25]], [[11, 25]], [[0, 26]], [[33, 40]], [[33, 40]]]",
    "query_spans": "[[[11, 61]]]",
    "process": "As shown in the figure, focus F(1,0), directrix $ l: x = -1 $, draw $ PM \\perp l $ from point P, with foot of perpendicular at M, then $ PM = PF $. Let $ Q(0,3) $. At this time, when points F, P, Q are collinear, $ |PF| + |PQ| $ reaches its minimum value. Hence: $ (|PF| + |PQ|) = |QF| = \\sqrt{3^{2} + 1^{2}} = \\sqrt{10} $,"
  },
  {
    "text": "Given the line $l$: $y=2 \\sqrt{3} x+m$ passes through the right vertex of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$, and $l$ intersects the two asymptotes of $C$ at points $A$ and $B$ respectively, then $|A B|=$?",
    "fact_expressions": "l: Line;Expression(l) = (y = m + 2*sqrt(3)*x);m: Number;C: Hyperbola;Expression(C) = (x^2 - y^2/3 = 1);PointOnCurve(RightVertex(C),l) = True;L1: Line;L2: Line;Asymptote(C) = {L1, L2};Intersection(l, L1) = A;Intersection(l, L2) = B;A: Point;B: Point",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "4*sqrt(13)/3",
    "fact_spans": "[[[2, 27], [68, 71]], [[2, 27]], [[9, 27]], [[29, 62], [72, 75]], [[29, 62]], [[2, 66]], [], [], [[72, 81]], [[68, 94]], [[68, 94]], [[85, 88]], [[89, 92]]]",
    "query_spans": "[[[96, 105]]]",
    "process": "Since the hyperbola equation is $x^2 - \\frac{y^{2}}{3} = 1$, the right vertex of curve $C$ is $(1,0)$, so $m = -2\\sqrt{3}$. The asymptotes of the hyperbola are $y = \\pm\\sqrt{3}x$. From $\\begin{cases} y = 2\\sqrt{3}x - 2 \\\\ y = \\pm\\sqrt{3}x \\end{cases} - 2\\sqrt{3}$, we get $x_{1} = 2$, $x_{2} = \\frac{2}{3}$. Then $|AB| = \\sqrt{1 + 12} \\cdot \\left|\\frac{2}{3}\\right| = \\frac{4\\sqrt{13}}{3}$."
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ are $F_{1}$ and $F_{2}$ respectively, the chord $AB$ passes through $F_{1}$, if the circumference of the incircle of $\\triangle ABF_{2}$ is $\\pi$, and the coordinates of points $A$ and $B$ are $(x_{1}, y_{1})$, $(x_{2}, y_{2})$, then $|y_{2}-y_{1}|=$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);PointOnCurve(F1, LineSegmentOf(A, B));Perimeter(InscribedCircle(TriangleOf(A, B, F2))) = pi;x1: Number;y1: Number;x2: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2)",
    "query_expressions": "Abs(-y1 + y2)",
    "answer_expressions": "5/3",
    "fact_spans": "[[[0, 39]], [[0, 39]], [[48, 55], [71, 78]], [[56, 63]], [[0, 63]], [[0, 63]], [[115, 118]], [[119, 122]], [[0, 70]], [[65, 78]], [[80, 114]], [[130, 146]], [[130, 146]], [[147, 163]], [[147, 163]], [[115, 163]], [[115, 163]]]",
    "query_spans": "[[[165, 182]]]",
    "process": ""
  },
  {
    "text": "The equation of the ellipse passing through the point $M(-\\sqrt{6},0)$ and having the same foci as the hyperbola $x^{2}-2 y^{2}=2$ is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;M: Point;Expression(G) = (x^2 - 2*y^2 = 2);Coordinate(M) = (-sqrt(6), 0);PointOnCurve(M, H);Focus(G)=Focus(H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2/6+y^2/3=1",
    "fact_spans": "[[[20, 40]], [[47, 49]], [[1, 18]], [[20, 40]], [[1, 18]], [[0, 49]], [[19, 49]]]",
    "query_spans": "[[[47, 53]]]",
    "process": ""
  },
  {
    "text": "The circle $x^{2}+(y-a)^{2}=9$ and the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ have common points; then the range of real values for $a$ is?",
    "fact_expressions": "H: Circle;Expression(H) = (x^2 + (-a + y)^2 = 9);G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);IsIntersect(H, G) = True;a: Real",
    "query_expressions": "Range(a)",
    "answer_expressions": "[-6,6]",
    "fact_spans": "[[[0, 20]], [[0, 20]], [[21, 59]], [[21, 59]], [[0, 63]], [[65, 70]]]",
    "query_spans": "[[[65, 77]]]",
    "process": "\\because the ellipse \\frac{x^2}{25}+\\frac{y^{2}}{9}=1 has foci on the x-axis, a=5, b=3 and |x|\\leqslant5, |y|\\leqslant4; the circle x^{2}+(y-a)^{2}=9 has center coordinates (0,a) and radius r=3. \\therefore if the ellipse \\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1 and the circle x^{2}+(y-a)^{2}=9 have common points, then the range of real number a is |a|\\leqslant6. \\text{【Note】This question examines the standard equations of ellipses and circles and their positional relationships, studying positional relationships and parameter values based on graphs, belonging to basic problems.}"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. $P$ is a point on the hyperbola such that $P F_{1} \\perp P F_{2}$, and the area of $\\Delta P F_{1} F_{2}$ is $2 a b$. Find the eccentricity $e$ of the hyperbola.",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;e: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P,F1),LineSegmentOf(P,F2));Area(TriangleOf(P, F1, F2)) = 2*a*b;Eccentricity(G) = e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 61], [90, 93], [160, 163]], [[5, 61]], [[5, 61]], [[86, 89]], [[69, 76]], [[77, 85]], [[167, 170]], [[5, 61]], [[5, 61]], [[2, 61]], [[2, 85]], [[2, 85]], [[86, 97]], [[99, 122]], [[123, 157]], [[160, 170]]]",
    "query_spans": "[[[168, 173]]]",
    "process": ""
  },
  {
    "text": "Let the focus of the parabola $y^{2}=8x$ be $F$, the directrix be $l$, and let $P$ be a point on the parabola. Let $PA \\perp l$, where $A$ is the foot of the perpendicular. If the slope of the line $AF$ is $-\\sqrt{3}$, then $|PF|=$?",
    "fact_expressions": "G: Parabola;A: Point;F: Point;P: Point;l: Line;Expression(G) = (y^2 = 8*x);Focus(G) = F;Directrix(G)=l;PointOnCurve(P,G);IsPerpendicular(LineSegmentOf(P,A),l);FootPoint(LineSegmentOf(P,A),l)=A;Slope(LineOf(A,F)) = -sqrt(3)",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 15], [35, 38]], [[56, 60]], [[19, 22]], [[31, 34]], [[26, 29]], [[1, 15]], [[1, 22]], [[1, 29]], [[31, 41]], [[42, 55]], [[42, 63]], [[66, 87]]]",
    "query_spans": "[[[90, 99]]]",
    "process": "Given the parabola equation $ y^{2} = 8x $, its focus is $ F(2,0) $, and the directrix equation is $ x = -2 $. Let $ B $ be the intersection point of the directrix and the $ x $-axis, then the coordinates of $ B $ are $ (-2,0) $, and $ BF = 4 $. Since the slope of line $ AF $ is $ -\\sqrt{3} $, we have $ \\angle AFB = 60^{\\circ} $, so $ AF = \\frac{BF}{\\cos 60^{\\circ}} = 8 $. Because $ AP \\parallel x $-axis, $ \\angle PAF = \\angle AFB = 60^{\\circ} $. Also, from the geometric properties of the parabola, $ PA = PF $, so triangle $ APF $ is equilateral. Hence, $ PA = PF = 8 $."
  },
  {
    "text": "Given that $P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{8}=1$, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola, respectively, $O$ is the origin, and points $M$, $N$ satisfy $\\overrightarrow{F_{1} P}=\\lambda \\overrightarrow{P M}(\\lambda>0)$, $\\overrightarrow{P N}=\\mu(\\frac{\\overrightarrow{P M}}{|\\overrightarrow{P M}|}+\\frac{\\overrightarrow{P F_{2}}}{|\\overrightarrow{P F_{2}}|})$, $\\overrightarrow{P N} \\cdot \\overrightarrow{F_{2} N}=0$. If $|\\overrightarrow{P F_{2}}|=3$, then what is the area of the circle with center $O$ and radius $ON$?",
    "fact_expressions": "G: Hyperbola;H: Circle;O: Origin;N: Point;F1: Point;P: Point;M: Point;F2: Point;Expression(G) = (x^2/16 - y^2/8 = 1);PointOnCurve(P, RightPart(G));LeftFocus(G) = F1;RightFocus(G) = F2;lambda: Number;mu: Number;lambda > 0;VectorOf(F1, P) = lambda*VectorOf(P, M);VectorOf(P, N) = mu*((VectorOf(P, F2)/Abs(VectorOf(P, F2))) + (VectorOf(P, M)/Abs(VectorOf(P, M))));DotProduct(VectorOf(P, N), VectorOf(F2, N)) = 0;VectorOf(P, F2) = 3;Center(H) = O;Radius(H) = LineSegmentOf(O, N)",
    "query_expressions": "Area(H)",
    "answer_expressions": "49*pi",
    "fact_spans": "[[[6, 45], [69, 72]], [[411, 412]], [[79, 82], [395, 398]], [[93, 96]], [[51, 58]], [[2, 5]], [[88, 92]], [[59, 66]], [[6, 45]], [[2, 50]], [[51, 78]], [[51, 78]], [[98, 164]], [[166, 305]], [[98, 164]], [[98, 164]], [[166, 305]], [[307, 362]], [[365, 393]], [[394, 412]], [[402, 412]]]",
    "query_spans": "[[[411, 417]]]",
    "process": "Analysis: From the given conditions, it follows that $PN$ is the angle bisector of $\\angle MPF_{2}$. Extend $F_{2}N$ to intersect $PM$ at $K$, then $PN$ is the angle bisector of $\\triangle PF_{2}K$, and further we obtain that $N$ is also the midpoint of $F_{2}K$. We find $|\\overrightarrow{ON}|=7$, thus the desired conclusion. Specifically, from $\\overrightarrow{PN}=\\mu\\left(\\frac{\\overrightarrow{PM}}{|\\overrightarrow{PM}|}+\\frac{\\overrightarrow{PF_{2}}}{|\\overrightarrow{PF_{2}}|}\\right)$, we know $PN$ is the angle bisector of $\\angle MPF_{2}$. Since $\\overrightarrow{PN}\\cdot\\overrightarrow{F_{2}N}=0$, extend $F_{2}N$ to intersect $PM$ at $K$, then $PN$ is the angle bisector of $\\triangle PF_{2}K$, so $\\triangle PF_{2}K$ is isosceles with $|PK|=|PF_{2}|=3$. Given $|\\overrightarrow{PF_{2}}|=3$, we have $|\\overrightarrow{PF_{1}}|=11$, hence $|\\overrightarrow{F_{1}K}|=14$. Noting that $N$ is still the midpoint of $F_{2}K$, $ON$ is the midline of $\\triangle F_{1}F_{2}K$, so $|\\overrightarrow{ON}|=\\frac{1}{2}|\\overrightarrow{F_{1}K}|=7$. Therefore, the area of the circle centered at $O$ with radius $ON$ is $49\\pi$."
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{4 a^{2}}-\\frac{y^{2}}{a^{2}}=1$ $(a>0)$, and $P$ is a point on the hyperbola such that $\\overrightarrow {P F_{1}} \\cdot \\overrightarrow {P F_{2}}=0$, and the area of $\\Delta F_{1} P F_{2}$ is $1$. Then the value of $a$ is?",
    "fact_expressions": "G: Hyperbola;a: Number;F1: Point;P: Point;F2: Point;a>0;Expression(G) = (x^2/(4*a^2) - y^2/a^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);DotProduct(VectorOf(P,F1),VectorOf(P,F2))=0;Area(TriangleOf(F1, P, F2)) = 1",
    "query_expressions": "a",
    "answer_expressions": "1",
    "fact_spans": "[[[16, 69], [79, 82]], [[181, 184]], [[0, 7]], [[75, 78]], [[8, 15]], [[19, 69]], [[16, 69]], [[0, 74]], [[75, 85]], [[86, 148]], [[150, 179]]]",
    "query_spans": "[[[181, 188]]]",
    "process": ""
  },
  {
    "text": "It is known that one asymptote of hyperbola $E$ has the equation $y=x$, and the focal distance is greater than $4$. Then, a possible standard equation of hyperbola $E$ is? (Write one)",
    "fact_expressions": "E: Hyperbola;Expression(OneOf(Asymptote(E))) = (y = x);FocalLength(E)>4",
    "query_expressions": "Expression(E)",
    "answer_expressions": "x^2/3 - y^2/3 = 1",
    "fact_spans": "[[[2, 8], [33, 39]], [[2, 22]], [[2, 31]]]",
    "query_spans": "[[[33, 48]]]",
    "process": "\\because one asymptote of hyperbola E is y=x, \\therefore assume the standard equation of hyperbola E is \\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1 (\\lambda\\neq0). When \\lambda>0, the focal distance of this hyperbola is 2\\sqrt{2}, i.e., 2\\sqrt{2\\lambda}>4, solving gives \\lambda>2; when \\lambda<0, the focal distance of this hyperbola is 2\\sqrt{-2\\lambda}, i.e., 2\\sqrt{-2\\lambda}>4, solving gives \\lambda<-2. \\therefore the standard equation of the hyperbola is \\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1 (\\lambda>2 or \\lambda<-2). Letting \\lambda=3 gives the standard equation of the hyperbola as \\frac{x^{2}}{3}-\\frac{y^{2}}{2}=1."
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, respectively. $P$ is a point on the right branch of the hyperbola, and $I$ is the incenter of $\\Delta P F_{1} F_{2}$. If $S_{\\triangle I P F_{2}}=S_{\\triangle I P F_{1}}-\\lambda S_{\\triangle I F_{1} F_{2}}$, then $\\lambda=?$",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;I: Point;Expression(G) = (x^2/16 - y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));Incenter(TriangleOf(P, F1, F2)) = I;Area(TriangleOf(I,P,F2))=Area(TriangleOf(I,P,F1))-lambda*Area(TriangleOf(I,F1,F2));lambda:Number",
    "query_expressions": "lambda",
    "answer_expressions": "4/5",
    "fact_spans": "[[[18, 57], [68, 71]], [[64, 67]], [[0, 7]], [[8, 15]], [[77, 80]], [[18, 57]], [[0, 63]], [[0, 63]], [[64, 76]], [[77, 107]], [[109, 194]], [[196, 205]]]",
    "query_spans": "[[[196, 207]]]",
    "process": ""
  },
  {
    "text": "The equation of the directrix of the parabola $y^{2}=4 x$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=-1",
    "fact_spans": "[[[0, 14]], [[0, 14]]]",
    "query_spans": "[[[0, 21]]]",
    "process": "The focus of the parabola $ y^{2} = 4x $ lies on the positive x-axis, with focus coordinates: $ F(1,0) $, so the directrix of the parabola is $ x = -1 $."
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $\\frac{y^{2}}{12}-\\frac{x^{2}}{4}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/4 + y^2/12 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "sqrt(3)*x + pm*y = 0",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 47]]]",
    "process": "From the hyperbola equation, it is known that the foci lie on the y-axis, and $ a^{2}=12 $, $ b^{2}=4 $, so $ a=2\\sqrt{3} $, $ b=2 $. Therefore, the asymptotes of the hyperbola are $ y=\\pm\\frac{a}{b}x $, that is, $ y=\\pm\\frac{2\\sqrt{3}}{3}x $, or $ \\sqrt{3}x\\pm y=0 $."
  },
  {
    "text": "The equation $x^{2}+k y^{2}=2$ represents an ellipse with foci on the $y$-axis. What is the range of real values for $k$?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (k*y^2 + x^2 = 2);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(0,1)",
    "fact_spans": "[[[29, 31]], [[34, 39]], [[0, 31]], [[21, 31]]]",
    "query_spans": "[[[34, 46]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a}+\\frac{y^{2}}{9}=1$ ($a>0$) with right focus $F$, point $M$ lies on $C$, point $N$ is the midpoint of segment $MF$, point $O$ is the origin, and $|MF|=2|ON|=4$. Then the eccentricity of $C$ is?",
    "fact_expressions": "C: Ellipse;a: Number;M: Point;F: Point;O: Origin;N: Point;a>0;Expression(C) = (y^2/9 + x^2/a = 1);RightFocus(C) = F;PointOnCurve(M, C);MidPoint(LineSegmentOf(M,F)) = N;Abs(LineSegmentOf(M,F)) = 2*Abs(LineSegmentOf(O,N));2*Abs(LineSegmentOf(O,N)) = 4",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(7)/4",
    "fact_spans": "[[[2, 49], [63, 66], [113, 116]], [[9, 49]], [[58, 62]], [[54, 57]], [[84, 88]], [[68, 72]], [[9, 49]], [[2, 49]], [[2, 57]], [[58, 67]], [[68, 83]], [[95, 111]], [[95, 111]]]",
    "query_spans": "[[[113, 122]]]",
    "process": "According to the definition of the ellipse and the triangle midline theorem, find the value of $ a $ to solve the problem. Let the left focus of ellipse $ C $ be $ F $. By the definition of the ellipse, $ |MF| + |MF'| = 2\\sqrt{a} $, that is, $ 4 + |MF'| = 2\\sqrt{a} $ (*). Since $ O $ is the midpoint of segment $ FF' $, and $ N $ is the midpoint of segment $ MF' $, by the property of the midline, $ |MF'| = 2|ON| = 4 $. Substituting into (*) yields $ a = 16 $, hence the eccentricity $ e = \\frac{\\sqrt{a - 9}}{\\sqrt{a}} = \\frac{\\sqrt{7}}{4} $."
  },
  {
    "text": "Given that $F$ and $B$ are respectively the focus and the endpoint of the imaginary axis of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$. If the midpoint of segment $F B$ lies on the hyperbola $C$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;B: Point;F: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);OneOf(Focus(C))=F;OneOf(Endpoint(ImageinaryAxis(C)))=B;PointOnCurve(MidPoint(LineSegmentOf(F,B)),C)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[12, 73], [95, 101], [104, 110]], [[19, 73]], [[19, 73]], [[6, 9]], [[2, 5]], [[19, 73]], [[19, 73]], [[12, 73]], [[2, 81]], [[2, 81]], [[84, 102]]]",
    "query_spans": "[[[104, 116]]]",
    "process": "Let F(c,0), B(0,b), then the midpoint of segment FB is (\\frac{c}{2},\\frac{b}{2}). Substituting this point into the hyperbola equation yields: \\frac{(c)}{a^{2}}-\\frac{(\\frac{b}{2})^{2}}{b^{2}}=1, solving gives \\frac{c^{2}}{a^{2}}=5, so e=\\frac{c}{a}=\\sqrt{5}"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has its right focus at $F$. A line passing through $F$ with slope $\\sqrt{3}$ intersects $C$ at points $A$ and $B$. If $\\overrightarrow{A F}=4 \\overrightarrow{F B}$, then the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;A: Point;F: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;PointOnCurve(F, G);Slope(G)=sqrt(3);Intersection(G, C) = {A,B};VectorOf(A, F) = 4*VectorOf(F, B)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "6/5",
    "fact_spans": "[[[2, 63], [94, 97], [158, 161]], [[10, 63]], [[10, 63]], [[91, 93]], [[98, 101]], [[68, 71], [73, 76]], [[104, 107]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 71]], [[72, 93]], [[77, 93]], [[91, 109]], [[111, 156]]]",
    "query_spans": "[[[158, 167]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{8}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/8 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/3",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 43]]]",
    "process": "According to the standard equation of the ellipse, write down the values of $a$, $b$, $c$, then substitute into $e = \\frac{c}{a}$, the answer of this problem can be obtained. $\\because$ the equation of the ellipse is $\\frac{x^{2}}{9} + \\frac{y^{2}}{8} = 1$, $\\therefore a = 3$, $b = 2\\sqrt{2}$, $c = 1$, $\\therefore e = \\frac{c}{a} = \\frac{1}{3}$"
  },
  {
    "text": "Through the focus $F(1,0)$ of the parabola $y^{2}=2 p x$ ($p>0$), draw a line $l$ intersecting the parabola at points $A$ and $B$. If $|A F|=3$, then $|B F|$ is?",
    "fact_expressions": "l: Line;G: Parabola;p: Number;F: Point;A: Point;B: Point;p>0;Expression(G) = (y^2 = 2*p*x);Coordinate(F) = (1, 0);Focus(G) = F;PointOnCurve(F,l);Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, F)) = 3",
    "query_expressions": "Abs(LineSegmentOf(B, F))",
    "answer_expressions": "3/2",
    "fact_spans": "[[[34, 39]], [[40, 43], [1, 22]], [[4, 22]], [[25, 33]], [[44, 47]], [[48, 51]], [[4, 22]], [[1, 22]], [[25, 33]], [[1, 33]], [[0, 39]], [[34, 53]], [[55, 64]]]",
    "query_spans": "[[[66, 75]]]",
    "process": "\\because y^{2}=2px has focus F(1,0), \\therefore \\frac{p}{2}=1, p=2, the parabola equation is y^{2}=4x. By the definition of the parabola, the distance from A to the focus equals the distance from A to the directrix. \\therefore |AF|=x_{A}+1=3, x_{A}=2, thus y_{A}=2\\sqrt{2}, k_{AF}=2\\sqrt{2}. The line AF equation is y=2\\sqrt{2}(x-1). Solving simultaneously with y^{2}=4x, we get 2x^{2}-5x+2=0, x_{A}+x_{B}=\\frac{5}{2}, x_{B}=\\frac{5}{2}-2=\\frac{1}{2}. |BF|=x_{B}+\\frac{p}{2}=\\frac{1}{2}+1=\\frac{3}{2},"
  },
  {
    "text": "The line $y = kx + 2$ intersects the right branch of the hyperbola $x^2 - y^2 = 6$ at two distinct points. What is the range of real values for $k$?",
    "fact_expressions": "G: Hyperbola;H: Line;k: Real;Expression(G) = (x^2 - y^2 = 6);Expression(H) = (y = k*x + 2);NumIntersection(H,RightPart(G))=2",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-sqrt(15)/3,-1)",
    "fact_spans": "[[[12, 30]], [[0, 11]], [[42, 47]], [[12, 30]], [[0, 11]], [[0, 40]]]",
    "query_spans": "[[[42, 54]]]",
    "process": ""
  },
  {
    "text": "Let $B$ be the upper vertex of the ellipse $C$: $\\frac{x^{2}}{4}+y^{2}=1$, and let point $P$ lie on $C$. Then the maximum value of $|P B|$ is?",
    "fact_expressions": "C: Ellipse;P: Point;B: Point;Expression(C) = (x^2/4 + y^2 = 1);UpperVertex(C)=B;PointOnCurve(P, C)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, B)))",
    "answer_expressions": "4*sqrt(3)/3",
    "fact_spans": "[[[5, 37], [47, 50]], [[42, 46]], [[1, 4]], [[5, 37]], [[1, 41]], [[42, 51]]]",
    "query_spans": "[[[53, 66]]]",
    "process": "According to the problem, it is easy to know that B(0,1), let P(x,y), then \\frac{x^{2}}{4}+y^{2}=1, that is, x=4-4y^{2},"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has its right focus at $F$. Draw tangents from point $F$ to the circle $x^{2}+y^{2}=b^{2}$. If the two tangents are perpendicular to each other, then $\\frac{a}{b}=$?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Circle;F: Point;L1:Line;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = b^2);RightFocus(C) = F;TangentOfPoint(F,G)={L1,L2};IsPerpendicular(L1,L2);L2:Line",
    "query_expressions": "a/b",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 59]], [[9, 59]], [[9, 59]], [[75, 95]], [[64, 67], [70, 74]], [], [[9, 59]], [[9, 59]], [[2, 59]], [[75, 95]], [[2, 67]], [[68, 98]], [[69, 109]], []]",
    "query_spans": "[[[112, 127]]]",
    "process": "As shown in the figure, given that the right focus of the ellipse $ C: \\frac{x^2}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $) is $ F $, and two tangents drawn from point $ F $ to the circle $ x^{2} + y^{2} = b^{2} $ are perpendicular to each other, we obtain $ \\sqrt{2}b = c $, then $ 2b^{2} = c^{2} $, $ a^{2} = b^{2} + c^{2} = 3b^{2} $, thus $ \\frac{a}{b} = \\sqrt{3} $."
  },
  {
    "text": "Given that $F_{1}$ is the left focus of the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$, and the line $l$: $y=x-1$ intersects the ellipse $C$ at points $A$ and $B$, then the value of $|F_{1} A|+|F_{1} B|$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/2 + y^2 = 1);F1: Point;LeftFocus(C) = F1;l: Line;Expression(l) = (y = x - 1);A: Point;B: Point;Intersection(l, C) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(F1, A)) + Abs(LineSegmentOf(F1, B))",
    "answer_expressions": "8*sqrt(2)/3",
    "fact_spans": "[[[10, 42], [62, 67]], [[10, 42]], [[2, 9]], [[2, 46]], [[47, 61]], [[47, 61]], [[69, 72]], [[73, 76]], [[47, 78]]]",
    "query_spans": "[[[80, 105]]]",
    "process": "Let points A(x_{1},y_{1}), B(x_{2},y_{2}). Solving the system of equations:  \n\\begin{cases}\\frac{x^{2}}{2}+y^{2}=1\\\\y=x-1\\end{cases}  \nEliminating y, we get 3x^{2}-4x=0, solving gives x_{1}=0, x_{2}=\\frac{4}{3}. Thus, point A(0,-1), B(\\frac{4}{3},\\frac{1}{3}). Also, point F_{1}(-1,0), therefore |F_{1}A|+|F_{1}B|=\\sqrt{1^{2}+(-1)^{2}}+\\sqrt{(7)^{2}+(\\frac{1}{3})^{2}}=\\frac{8\\sqrt{2}}{3}"
  },
  {
    "text": "Given that $A$ is the right vertex of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $B_{1}$ and $B_{2}$ are the two endpoints of the imaginary axis, and $F$ is the right focus. If $B_{2} F \\perp A B_{1}$, then the eccentricity of the hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;A: Point;RightVertex(C) = A;B1: Point;B2: Point;Endpoint(ImageinaryAxis(C)) = {B1, B2};F: Point;RightFocus(C) = F;IsPerpendicular(LineSegmentOf(B2, F), LineSegmentOf(A, B1))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(\\sqrt{5}+1)/2",
    "fact_spans": "[[[6, 67], [132, 138]], [[6, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[2, 5]], [[2, 71]], [[72, 79]], [[80, 87]], [[6, 97]], [[98, 101]], [[6, 105]], [[107, 130]]]",
    "query_spans": "[[[132, 144]]]",
    "process": "The hyperbola equation is $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$, we obtain $A(a,0)$, $F(c,0)$, $B_{1}(0,b)$, $B_{2}(0,-b)$. Since $\\overrightarrow{B_{2}F}=(c,b)$, $\\overrightarrow{AB_{1}}=(-a,b)$, from $B_{2}F\\perp AB_{1}$ we get $\\overrightarrow{B_{2}F}\\cdot\\overrightarrow{AB_{1}}=0$, i.e., $ac-b^{2}=0$, thus $b^{2}=ac$, that is, $c^{2}-ac-a^{2}=0$. Dividing both sides by $a^{2}$ yields $e^{2}-e-1=0$. Solving gives $e=\\frac{\\sqrt{5}+1}{2}$ (negative root discarded)."
  },
  {
    "text": "Draw a line through the right focus $F$ of the hyperbola $x^{2}-y^{2}=2$ with an inclination angle of $30^{\\circ}$, intersecting the hyperbola at points $P$ and $Q$. Then the value of $|PQ|$ is?",
    "fact_expressions": "G: Hyperbola;H: Line;P: Point;Q: Point;F:Point;Expression(G) = (x^2 - y^2 = 2);RightFocus(G)=F;PointOnCurve(F,H);Inclination(H)=ApplyUnit(30,degree);Intersection(H,G)={P,Q}",
    "query_expressions": "Abs(LineSegmentOf(P, Q))",
    "answer_expressions": "4*sqrt(2)",
    "fact_spans": "[[[1, 19], [48, 51]], [[44, 46]], [[52, 55]], [[56, 59]], [[23, 26]], [[1, 19]], [[1, 26]], [[0, 46]], [[27, 46]], [[44, 61]]]",
    "query_spans": "[[[63, 73]]]",
    "process": ""
  },
  {
    "text": "If the point $(x, y)$ lies on the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, then the minimum value of $3 x^{2}-2 y$ is?",
    "fact_expressions": "G: Hyperbola;H: Point;Expression(G) = (x^2/4 - y^2 = 1);Coordinate(H) = (x1, y1);PointOnCurve(H, G);x1:Number;y1:Number",
    "query_expressions": "Min(3*x1^2 - 2*y1)",
    "answer_expressions": "143/12",
    "fact_spans": "[[[11, 39]], [[1, 10]], [[11, 39]], [[1, 10]], [[1, 40]], [[2, 10]], [[2, 10]]]",
    "query_spans": "[[[42, 61]]]",
    "process": "The point (x, y) lies on the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, hence $\\frac{x^{2}}{4}=1+y^{2}$, and further we obtain: $3x^{2}-2y=3(1+y^{2})\\times4-2y=12y^{2}-2y+12$. The axis of symmetry of the quadratic function is $y=\\frac{1}{12}$. Combining the graph and properties of the quadratic function, the minimum value is the value corresponding to $y=\\frac{1}{12}$, which is $\\frac{143}{12}$."
  },
  {
    "text": "If the coordinate origin $O$ and the point $F(-2,0)$ are respectively the center and the left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1(a>0)$, and point $P$ is any point on the right branch of the hyperbola, then the minimum value of $\\overrightarrow{O P} \\cdot \\overrightarrow{F P}$ is?",
    "fact_expressions": "G: Hyperbola;a: Number;F: Point;O: Origin;P: Point;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Center(G)=O;LeftFocus(G)=F;Coordinate(F) = (-2, 0);PointOnCurve(P, RightPart(G))",
    "query_expressions": "Min(DotProduct(VectorOf(O, P), VectorOf(F, P)))",
    "answer_expressions": "3+2*sqrt(3)",
    "fact_spans": "[[[22, 59]], [[25, 59]], [[9, 19]], [[1, 8]], [[67, 71]], [[25, 59]], [[22, 59]], [[1, 66]], [[9, 66]], [[9, 19]], [[67, 83]]]",
    "query_spans": "[[[85, 140]]]",
    "process": "From the given conditions: since $F(-2,0)$ is the left focus of the known hyperbola, $\\therefore a^{2}+1=4$, that is, $a^{2}=3$. $\\therefore$ the equation of the hyperbola is $\\frac{x^{2}}{3}-y^{2}=1$. Let point $P(x_{0},y_{0})$, then $\\frac{x_{0}^{2}}{3}-y_{0}^{2}=1$ $(x_{0}\\geqslant\\sqrt{3})$, solving gives $y_{0}^{2}=\\frac{x_{0}^{2}}{3}-1$ $(x_{0}\\geqslant\\sqrt{3})$. $\\because \\overrightarrow{FP}=(x_{0}+2,y_{0})$, $\\overrightarrow{OP}=(x_{0},y_{0})$, $\\therefore \\overrightarrow{OP}\\cdot\\overrightarrow{FP}=x_{0}(x_{0}+2)+y_{0}^{2}=x_{0}(x_{0}+2)+\\frac{x_{0}^{2}}{3}-1=\\frac{4x_{0}^{2}}{3}+2x_{0}-1$. $\\because x_{0}\\geqslant\\sqrt{3}$, according to monotonicity analysis of quadratic functions, the function is monotonically increasing on $[\\sqrt{3},+\\infty)$. $\\therefore$ when $x_{0}=\\sqrt{3}$, $\\overrightarrow{OP}\\cdot\\overrightarrow{FP}$ attains the minimum value $\\frac{4}{3}\\times3+2\\sqrt{3}-1=3+2\\sqrt{3}$."
  },
  {
    "text": "Given that $F_{1}$, $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $P$ is a point on the ellipse such that $\\overrightarrow{P F_{1}} \\perp \\overrightarrow{P F_{2}}$, and the area of $\\triangle PF_{1} F_{2}$ is $9$, then $b=$?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Focus(C)={F1,F2};PointOnCurve(P, C);IsPerpendicular(VectorOf(P,F1),VectorOf(P,F2));Area(TriangleOf(P,F1,F2))=9",
    "query_expressions": "b",
    "answer_expressions": "3",
    "fact_spans": "[[[20, 77], [87, 89]], [[186, 189]], [[26, 77]], [[83, 86]], [[2, 9]], [[11, 19]], [[27, 77]], [[26, 77]], [[20, 77]], [[2, 82]], [[83, 93]], [[95, 152]], [[153, 184]]]",
    "query_spans": "[[[186, 191]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{8}+\\frac{y^{2}}{2}=1$ and the circle $M$: $x^{2}+y^{2}+2 \\sqrt{2} x+2-r^{2}=0$ $(0<r<\\sqrt{2})$, two tangents are drawn from the upper vertex $P$ of the ellipse $C$ to the circle $M$, intersecting the ellipse $C$ again at points $A$ and $B$ (distinct from $P$). Then, the product of the slopes of lines $PA$ and $PB$ is equal to?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/8 + y^2/2 = 1);M: Circle;Expression(M) = (-r^2 + 2*(sqrt(2)*x) + x^2 + y^2 + 2 = 0);r: Number;0 < r;r < sqrt(2);P: Point;UpperVertex(C) = P;Z1: Line;Z2: Line;TangentOfPoint(P, M) = {Z1, Z2};A: Point;Intersection(Z1, C) = A;B: Point;Intersection(Z2, C) = B;Negation(P=A);Negation(P=B)",
    "query_expressions": "Slope(LineOf(P, A))*Slope(LineOf(P, B))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 44], [103, 108], [128, 133]], [[2, 44]], [[45, 101], [116, 120]], [[45, 101]], [[51, 101]], [[51, 101]], [[51, 101]], [[112, 115], [150, 154]], [[103, 115]], [], [], [[102, 125]], [[136, 139]], [[102, 146]], [[140, 143]], [[102, 146]], [[136, 155]], [[136, 155]]]",
    "query_spans": "[[[157, 180]]]",
    "process": "The center of the circle is $ M(-\\sqrt{2},0) $, $ P(0,\\sqrt{2}) $. Let the tangent line be $ y = kx + \\sqrt{2} $, and the distance from the point to the line is $ d = \\frac{|-\\sqrt{2}k + \\sqrt{2}|}{\\sqrt{1 + k^{2}}} = r $, $ (2 - r)^{2}k^{2} - 4k + (2 - r) = 0 $, $ k_{1}k_{2} = 1 $,"
  },
  {
    "text": "If the distance from point $M$ on the parabola $y^{2}=4x$ to the $y$-axis is $9$, then what is the distance from $M$ to the focus?",
    "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(M, G);Distance(M, yAxis) = 9",
    "query_expressions": "Distance(M, Focus(G))",
    "answer_expressions": "10",
    "fact_spans": "[[[1, 15]], [[17, 21], [35, 38]], [[1, 15]], [[1, 21]], [[17, 33]]]",
    "query_spans": "[[[1, 46]]]",
    "process": "The focus of the parabola $ y^{2} = 4x $ is $ F(1,0) $, and the equation of the directrix is $ x = -1 $. Since the distance from point $ M $ to the $ y $-axis is 9, the distance from point $ M $ to the directrix $ x = -1 $ is 10. According to the definition of a parabola, the distance from $ M $ to the focus is 10. Answer: 10"
  },
  {
    "text": "The left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ are $F_{1}$ and $F_{2}$ respectively, $P$ is a point on the ellipse, and $\\angle P F_{1} F_{2}=\\frac{\\pi}{6}$, $\\angle P F_{2} F_{1}=\\frac{\\pi}{3}$. Then the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P,G);AngleOf(P, F1, F2) = pi/6;AngleOf(P,F2,F1)=pi/3;Eccentricity(G)=e;e:Number",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3)-1",
    "fact_spans": "[[[0, 45], [73, 75], [155, 157]], [[2, 45]], [[2, 45]], [[69, 72]], [[53, 60]], [[61, 68]], [[0, 45]], [[0, 68]], [[0, 68]], [[69, 78]], [[80, 116]], [[118, 154]], [[155, 164]], [[161, 164]]]",
    "query_spans": "[[[161, 166]]]",
    "process": ""
  },
  {
    "text": "The equation of the ellipse passing through the point $M(-3,2)$ and having the same foci as $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (-3, 2);Z: Curve;Expression(Z) = (x^2/9 + y^2/4 = 1);G: Ellipse;PointOnCurve(M, G);Focus(Z) = Focus(G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/15+y^2/10=1",
    "fact_spans": "[[[1, 11]], [[1, 11]], [[13, 48]], [[13, 48]], [[54, 56]], [[0, 56]], [[12, 56]]]",
    "query_spans": "[[[54, 61]]]",
    "process": "According to the given conditions, the coordinates of the foci of the ellipse are $(-\\sqrt{5},0)$ and $(\\sqrt{5},0)$, and the ellipse passes through the point $M(-3,2)$, so $\\frac{-3+\\sqrt{5}}{2}+4=\\_+\\sqrt{(\\sqrt{15}+\\sqrt{3})^{2}}=2\\sqrt{15}$, thus $a^{2}=15$, $\\sqrt{18-2\\sqrt{45}+\\sqrt{18}+2\\sqrt{45}}=\\sqrt{(3)}$, and the equation of the ellipse is $\\frac{x^{2}}{15}+\\frac{y^{2}}{10}=1$."
  },
  {
    "text": "Given point $M(\\sqrt{3}, 0)$, the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ and the line $y=k(x+\\sqrt{3})$ intersect at points $A$ and $B$, then the perimeter of $\\triangle A B M$ is?",
    "fact_expressions": "M: Point;Coordinate(M) = (sqrt(3), 0);G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);H: Line;Expression(H) = (y = k*(x + sqrt(3)));Intersection(G, H) = {A, B};A: Point;B: Point;k:Number",
    "query_expressions": "Perimeter(TriangleOf(A, B, M))",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 21]], [[2, 21]], [[22, 49]], [[22, 49]], [[50, 70]], [[50, 70]], [[22, 80]], [[72, 76]], [[77, 80]], [[50, 70]]]",
    "query_spans": "[[[82, 104]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left vertex is $A$, the upper vertex is $B$, and the right focus is $F$. If $\\angle A B F=90^{\\circ}$, then the eccentricity of ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;A: Point;LeftVertex(C) = A;B: Point;UpperVertex(C) = B;F: Point;RightFocus(C) = F;AngleOf(A, B, F) = ApplyUnit(90, degree)",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "(\\sqrt{5}-1)/2",
    "fact_spans": "[[[2, 59], [112, 117]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[64, 67]], [[2, 67]], [[72, 75]], [[2, 75]], [[80, 83]], [[2, 83]], [[85, 110]]]",
    "query_spans": "[[[112, 123]]]",
    "process": "First write the coordinates of A, B, F, then use the condition \\angle ABF=90^{\\circ} and the dot product of vectors being zero to obtain b^{2}=ac, then convert it into a homogeneous equation e^{2}+e-1=0 and solve the equation. According to the problem, A(-a,0), B(0,b), F(c,0). \\angle ABF=90^{\\circ}, \\therefore \\overrightarrow{BA}\\cdot\\overrightarrow{BF}=0, i.e., (-a,-b)\\cdot(c,-b)=0 \\therefore b^{2}=ac, also \\because c^{2}=a^{2}-b^{2}, \\therefore c^{2}-a^{2}+ac=0, dividing both sides by a^{2} gives (\\frac{c}{a})^{2}+\\frac{c}{a}-1=0, i.e., e^{2}+e-1=0. Solving the equation yields e=\\frac{\\sqrt{5}+1}{2} (discarded) or e=\\frac{\\sqrt{5}-1}{2}."
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $\\frac{x^{2}}{8}-\\frac{y^{2}}{16}=1$?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/8 - y^2/16 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(2)*x",
    "fact_spans": "[[[0, 39]], [[0, 39]]]",
    "query_spans": "[[[0, 47]]]",
    "process": "Let $\\frac{x^{2}}{8}-\\frac{y^{2}}{16}=0$, the asymptotic line equations are: $y=\\pm\\sqrt{2}x$. Hence fill in $y=\\pm\\sqrt{2}x$."
  },
  {
    "text": "If the focus of the parabola $N$: $y^{2}=2 p x$ ($p>0$) is $(3,0)$, and the line $x=m$ ($m>0$) intersects the parabola at points $A$ and $B$, such that $O A \\perp O B$ ($O$ being the origin), then the value of $m$ is?",
    "fact_expressions": "N: Parabola;Expression(N) = (y^2 = 2*p*x);p: Number;p>0;G: Line;Expression(G) = (x = m);m: Number;m>0;Intersection(G, N) = {A, B};A: Point;B: Point;O: Origin;IsPerpendicular(LineSegmentOf(O, A), LineSegmentOf(O, B)) = True;Coordinate(Focus(N))=(3,0)",
    "query_expressions": "m",
    "answer_expressions": "12",
    "fact_spans": "[[[1, 27], [52, 55]], [[1, 27]], [[9, 27]], [[9, 27]], [[39, 51]], [[39, 51]], [[95, 98]], [[41, 51]], [[39, 66]], [[57, 60]], [[61, 64]], [[84, 87]], [[68, 83]], [1, 36]]",
    "query_spans": "[[[95, 102]]]",
    "process": "Since the focus of the parabola \\( N: y^{2} = 2px \\) (\\( p > 0 \\)) is at \\( (3, 0) \\), it follows that \\( p = 6 \\). Thus, the equation of the parabola is \\( y^{2} = 12x \\). Given that the line \\( x = m \\) (\\( m > 0 \\)) intersects the parabola \\( N \\) at points \\( A \\) and \\( B \\), and \\( OA \\perp OB \\) (where \\( O \\) is the origin), we have \\( A(m, m) \\), \\( B(m, -m) \\). Therefore, \\( m^{2} = 12m \\), solving gives \\( m = 12 \\)."
  },
  {
    "text": "Given the ellipse equation $3 x^{2}+2 y^{2}=1$, then the length of the major axis of this ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (3*x^2 + 2*y^2 = 1)",
    "query_expressions": "Length(MajorAxis(G))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 4], [29, 31]], [[2, 26]]]",
    "query_spans": "[[[29, 37]]]",
    "process": "From the given problem, the equation of the ellipse is $\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1$, $\\therefore a^{2}=\\frac{1}{2}$, $\\therefore a=\\frac{\\sqrt{2}}{2}$, $\\therefore 2a=\\sqrt{2}$. Therefore, the major axis length of the ellipse is $\\sqrt{2}$."
  },
  {
    "text": "The line connecting the focus of the parabola $C_{1}$: $y=\\frac{1}{4 a} x^{2}$ ($a>0$) and the right focus of the hyperbola $C_{2}$: $\\frac{x^{2}}{3}-y^{2}=1$ intersects $C_{1}$ at point $M$ in the first quadrant. If the tangent to $C_{1}$ at point $M$ is parallel to an asymptote of $C_{2}$, then $a=$?",
    "fact_expressions": "C1:Parabola;C2:Hyperbola;a: Number;a>0;Expression(C2) = (x^2/3 - y^2 = 1);Expression(C1) = (y = x^2/(4*a));Intersection(LineSegmentOf(Focus(C1),RightFocus(C2)),C1)=M;Quadrant(M)=1;IsParallel(TangentOnPoint(M,C1),OneOf(Asymptote(C2)));M:Point",
    "query_expressions": "a",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[0, 40], [109, 116], [89, 96]], [[44, 81], [128, 135]], [[143, 146]], [[11, 40]], [[44, 81]], [[0, 40]], [[0, 106]], [[97, 106]], [[109, 141]], [[102, 106]]]",
    "query_spans": "[[[143, 148]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. The line $l$ passing through $F_{2}$ with an inclination angle of $\\frac{\\pi}{4}$ intersects the ellipse $C$ at points $A$ and $B$. Then, what is the inradius of $\\Delta F_{1} A B$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;l: Line;A: Point;B: Point;PointOnCurve(F2,l);Inclination(l) = pi/4;Intersection(l, C) = {A, B}",
    "query_expressions": "Radius(InscribedCircle(TriangleOf(F1, A, B)))",
    "answer_expressions": "(3*sqrt(2))/7",
    "fact_spans": "[[[2, 44], [104, 109]], [[2, 44]], [[53, 60]], [[61, 68], [70, 77]], [[2, 68]], [[2, 68]], [[98, 103]], [[110, 113]], [[114, 117]], [[69, 103]], [[78, 103]], [[98, 119]]]",
    "query_spans": "[[[121, 147]]]",
    "process": "Find the perimeter and area of \\triangle ABF_{1} to obtain the inradius. Since the ellipse C: \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 has left and right foci F_{1}(-1,0) and F_{2}(1,0), respectively, the line l is: y = \\tan\\frac{\\pi}{4}(x-1) = x-1. Substituting into the ellipse equation yields: 7y^{2} + 6y - 9 = 0. Let the coordinates of points A and B be (x_{1},y_{1}), (x_{2},y_{2}). Then y_{1} - y_{2} = \\frac{\\sqrt{6^{2} + 4 \\times 7 \\times 9}}{7} = \\frac{12\\sqrt{2}}{7}. Hence, the area S of \\triangle F_{1}AB is \\frac{1}{2} \\cdot 2c \\cdot |y_{1} - y_{2}| = \\frac{12\\sqrt{2}}{7}. The perimeter c of \\triangle F_{1}AB is 4a = . Thus, the inradius r of \\triangle F_{1}AB is r = \\frac{2S}{} = \\frac{3\\sqrt{2}}{}."
  },
  {
    "text": "The focus of the parabola $y^{2}=12 x$ is $F$, and point $M$ lies on the parabola (point $M$ is in the fourth quadrant), with $|M F|=7$. Then the coordinates of point $M$ are?",
    "fact_expressions": "G: Parabola;M: Point;F: Point;Expression(G) = (y^2 = 12*x);Focus(G) = F;PointOnCurve(M,G);Quadrant(M)=4;Abs(LineSegmentOf(M, F)) = 7",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(4,-4*sqrt(3))",
    "fact_spans": "[[[0, 15], [28, 31]], [[23, 27], [33, 37], [56, 60]], [[19, 22]], [[0, 15]], [[0, 22]], [[23, 32]], [[33, 42]], [[45, 54]]]",
    "query_spans": "[[[56, 65]]]",
    "process": "Let M(x_{0},y_{0}), (x_{0}>0, y_{0}<0) be on y^{2}=12x, then |MF|=x_{0}+3=7, so x_{0}=4. Substituting into y^{2}=12x gives y_{0}^{2}=48, hence y_{0}=-4\\sqrt{3} or 4\\sqrt{3} (discarded). Therefore, the coordinates of point M are (4,-4\\sqrt{3})."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$, the left and right foci are denoted as $F_{1}$ and $F_{2}$, respectively. A line $l$ passing through $F_{1}$ intersects the left branch of the hyperbola at points $A$ and $B$. Then, the minimum value of $|B F_{2}|+|A F_{2}|$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/4 - y^2/3 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;l: Line;PointOnCurve(F1, l);A: Point;B: Point;Intersection(l, LeftPart(C)) = {A, B}",
    "query_expressions": "Min(Abs(LineSegmentOf(A, F2)) + Abs(LineSegmentOf(B, F2)))",
    "answer_expressions": "11",
    "fact_spans": "[[[2, 45], [85, 88]], [[2, 45]], [[54, 61], [71, 78]], [[62, 69]], [[2, 69]], [[2, 69]], [[79, 84]], [[70, 84]], [[92, 95]], [[96, 99]], [[79, 101]]]",
    "query_spans": "[[[103, 130]]]",
    "process": "Let the equation of line $ l $ be $ x = my - \\sqrt{7} $, and let points $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. By solving simultaneously the equation of line $ l $ and the equation of hyperbola $ C $, find the minimum value of $ |AB| $, then combining with the definition of the hyperbola, the minimum value of $ |BF_{2}| + |AF_{2}| $ can be obtained. If line $ l $ coincides with the $ x $-axis, then line $ l $ cannot intersect the left branch of hyperbola $ C $ at two points. It is clear that $ F_{1}(-\\sqrt{7}, 0) $. Let the equation of line $ l $ be $ x = my - \\sqrt{7} $, and let points $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Solving the system\n$$\n\\begin{cases}\nx = my - \\sqrt{7} \\\\\n\\frac{x^{2}}{4} - \\frac{y^{2}}{3} = 1\n\\end{cases}\n$$\nby eliminating $ x $, we obtain $ (3m^{2} - 4)y^{2} - 6\\sqrt{7}my + 9 = 0 $. Since the line intersects the left branch of the hyperbola at two points $ A $ and $ B $, we have\n$$\n\\begin{cases}\n3m - 4 \\neq 0 \\\\\nm = 14(m + 1) > 0 \\\\\ny_{1}y_{2} = \\frac{m}{3m - 4} < 0\n\\end{cases}\n$$\nwhich gives $ m < \\frac{4}{3} $. $ \\frac{1}{2}9 $. From the definition of the hyperbola, we have $ |AF_{2}| - |AF_{1}| = 2a = 4 $, $ |BF_{2}| - |BF_{1}| = 2a = 4 $. Adding these two equations yields $ |AF_{2}| + |BF_{2}| - |AB| = 8 $, thus $ |AF_{2}| + |BF_{2}| = 8 + |AB| \\geqslant 8 + 3 = 11 $."
  },
  {
    "text": "The number of intersection points between the line $y=x+3$ and the curve $\\frac{y^{2}}{9}-\\frac{x \\cdot|x|}{4}=1$ is?",
    "fact_expressions": "G: Line;Expression(G) = (y = x + 3);H: Curve;Expression(H) = (y^2/9 - (x*Abs(x))/4= 1)",
    "query_expressions": "NumIntersection(G, H)",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 9]], [[0, 9]], [[10, 52]], [[10, 52]]]",
    "query_spans": "[[[0, 59]]]",
    "process": ""
  },
  {
    "text": "If the ellipse $C$: $m x^{2}+n y^{2}=1(m>0 , n>0 , m \\neq n)$ intersects the line $l$: $x+y-1=0$ at points $A$ and $B$, and the slope of the line passing through the origin and the midpoint of segment $A B$ is $\\frac{\\sqrt{2}}{2}$, then $\\frac{m}{n}=$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (m*x^2 + n*y^2 = 1);m: Number;n: Number;m>0;n>0;Negation(m=n);l: Line;Expression(l) = (x + y - 1 = 0);Intersection(l, C) = {A, B};A: Point;B: Point;O: Origin;L1: Line;PointOnCurve(O,L1) = True;PointOnCurve(MidPoint(LineSegmentOf(A,B)),L1) = True;Slope(L1) = sqrt(2)/2",
    "query_expressions": "m/n",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[1, 49]], [[1, 49]], [[8, 49]], [[8, 49]], [[8, 49]], [[8, 49]], [[8, 49]], [[50, 66]], [[50, 66]], [[1, 77]], [[68, 71]], [[72, 75]], [[79, 81]], [[92, 94]], [[78, 94]], [[78, 94]], [[92, 117]]]",
    "query_spans": "[[[119, 134]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, and the midpoint $ M $ of $ AB $ be $ (x_{0},y_{0}) $. Substituting points $ A $ and $ B $ into the ellipse equation gives  \n$$\n\\begin{cases}\nmx_{1}^{2}+ny_{1}^{2}=1 & \\textcircled{1} \\\\\nmx_{2}^{2}+ny_{2}^{2}=1 & \\textcircled{2}\n\\end{cases}\n$$\n$ \\textcircled{1}-\\textcircled{2} $ yields  \n$$\nm(x_{1}^{2}-x_{2}^{2})+n(y_{1}^{2}-y_{2}^{2})=0 \\quad \\textcircled{3}\n$$\nRearranging $ \\textcircled{3} $ gives: $ y_{2}^{2} = -\\frac{m}{n} $. Since $ k_{A}, \\frac{2}{2}, ky_{0} = \\frac{2}{}, $ so $ k_{AB} \\cdot k_{OM} = -\\frac{m}{n} $. Because points $ A $ and $ B $ lie on the line $ l: x+y-1=0 $, we have $ k_{AB} = -1 $. Since $ k_{OM} = \\frac{\\sqrt{2}}{2} $, it follows that $ \\frac{m}{n} = -k_{AB} \\cdot k_{OM} = \\frac{\\sqrt{2}}{2} $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2x$ with focus $F$, a line $l$ passes through $F$ and intersects $C$ at points $A$ and $B$. If $|AF|=2|BF|$, then what is the length of the chord cut on the $y$-axis by the circle with diameter $AB$?",
    "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*x);F: Point;Focus(C) = F;l: Line;PointOnCurve(F, l);Intersection(l, C) = {A, B};B: Point;A: Point;Abs(LineSegmentOf(A, F)) = 2*Abs(LineSegmentOf(B, F));IsDiameter(LineSegmentOf(A, B), G) = True;G: Circle",
    "query_expressions": "Length(InterceptChord(yAxis, G))",
    "answer_expressions": "sqrt(14)/2",
    "fact_spans": "[[[2, 21], [39, 42]], [[2, 21]], [[25, 28], [35, 38]], [[2, 28]], [[29, 34]], [[29, 38]], [[29, 53]], [[48, 51]], [[44, 47]], [[55, 69]], [[76, 89]], [[88, 89]]]",
    "query_spans": "[[[71, 96]]]",
    "process": "As shown in the figure, without loss of generality, assume point A is in the first quadrant. Draw perpendiculars from points A and B to the directrix of the parabola, with feet at A' and B' respectively. Let line AB intersect the directrix at point P. By the definition of a parabola, we have AF = AA', BF = BB'. Given that \\frac{PB}{PA} = \\frac{BB'}{AA'} = \\frac{BF}{AF} = \\frac{1}{2}, it follows that PB = AB = 3BF, hence \\frac{PF}{PA} = \\frac{2}{3} = \\frac{FK}{AA'}, and since FK = 1, we get AA' = AF = \\frac{3}{2}, BF = BB' = \\frac{3}{4}. Also, AB = 3BF = \\frac{9}{4}, so the radius is R = \\frac{1}{2}AB = \\frac{9}{8}. Let O be the midpoint of segment AB, then the distance from the center of the circle to the y-axis is d = \\frac{9}{8} - \\frac{1}{2} = \\frac{5}{8}. Therefore, the chord length MN = 2\\sqrt{R^{2} - d^{2}} = \\frac{\\sqrt{14}}{2}."
  },
  {
    "text": "Given that $C$ and $F$ are the left vertex and left focus of the ellipse $\\Gamma$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, respectively, and $A$, $B$ are the lower and upper vertices of the ellipse. Let $D$ be the intersection point of $AF$ and $BC$. If $\\overrightarrow{C D}=2 \\overrightarrow{D B}$, then the eccentricity of the ellipse $\\Gamma$ is?",
    "fact_expressions": "Gamma:Ellipse;A: Point;F: Point;B: Point;C: Point;D: Point;Expression(Gamma) = (x^2/a^2+y^2/b^2=1);LeftVertex(Gamma)=C;LeftFocus(Gamma)=F;LowerVertex(Gamma)=A;UpperVertex(Gamma)=B;Intersection(LineSegmentOf(A,F),LineSegmentOf(B,C))=D;VectorOf(C, D) = 2*VectorOf(D, B)",
    "query_expressions": "Eccentricity(Gamma)",
    "answer_expressions": "1/5",
    "fact_spans": "[[[12, 67], [84, 86], [160, 170]], [[76, 79]], [[6, 9]], [[80, 83]], [[2, 5]], [[107, 111]], [[12, 67]], [[2, 75]], [[2, 75]], [[76, 92]], [[76, 92]], [[94, 111]], [[113, 158]]]",
    "query_spans": "[[[160, 176]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ be the left focus of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, $O$ the origin, and $P$ a point on the ellipse. Then the maximum value of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P O}$ is?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;O: Origin;Expression(G) = (x^2/4 + y^2 = 1);LeftFocus(G) = F1;PointOnCurve(P, G)",
    "query_expressions": "Max(DotProduct(VectorOf(P, F1), VectorOf(P, O)))",
    "answer_expressions": "4+2*sqrt(3)",
    "fact_spans": "[[[9, 36], [55, 57]], [[50, 54]], [[1, 8]], [[41, 44]], [[9, 36]], [[1, 40]], [[50, 58]]]",
    "query_spans": "[[[60, 119]]]",
    "process": "According to the problem, by drawing the figure, it can be analyzed that when point P is the right endpoint of the ellipse, the value of $\\overrightarrow{PF_{2}}\\cdot\\overrightarrow{PO}$ is maximized, thus its maximum value can be obtained. $\\because F_{1}$ is the left focus of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, obviously, when point P is the right endpoint of the ellipse, $|\\overline{PO}|$ and $|\\overline{PF_{1}}|$ both reach their maximum values and $|\\overline{PO}|$ and $|\\frac{|PF_{1}|}{|}|$ are in the same direction, $\\cos<\\overrightarrow{PF},\\overrightarrow{PO}>=1$, which is also the maximum value, $\\therefore \\overrightarrow{PF}_{1}\\cdot\\overrightarrow{PO}$ reaches its maximum value $\\therefore |\\overrightarrow{PF_{1}}|=a+c=2+\\sqrt{3}$, $|\\overrightarrow{PO}|=2$, $\\therefore (\\overrightarrow{PF}_{1}\\cdot\\overrightarrow{PO})_{\\max}=(2+\\sqrt{3})\\times2\\cos0=4+2\\sqrt{3}$."
  },
  {
    "text": "Given the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, with $F_{1}(-c, 0)$ and $F_{2}(c, 0)$ being its left and right foci respectively. A circle with diameter $F_{1} F_{2}$ intersects the ellipse $E$ at point $P$ in the first quadrant and at point $Q$ in the third quadrant. If the area of $\\triangle P Q F_{2}$ is $\\frac{\\sqrt{3}}{2} c^{2}$, then $\\frac{c^{2}}{a^{2}+b^{2}}$=?",
    "fact_expressions": "E: Ellipse;b: Number;a: Number;G: Circle;F1: Point;F2: Point;P: Point;Q: Point;c: Number;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);LeftFocus(E) = F1;RightFocus(E) = F2;IsDiameter(LineSegmentOf(F1, F2), G);Intersection(G, E) = {P, Q};Quadrant(P) = 1;Quadrant(Q) = 3;Area(TriangleOf(P, Q, F2)) = c^2*(sqrt(3)/2)",
    "query_expressions": "c^2/(a^2 + b^2)",
    "answer_expressions": "(\\sqrt{3} - 1)/2",
    "fact_spans": "[[[2, 59], [95, 96], [122, 127]], [[9, 59]], [[9, 59]], [[120, 121]], [[62, 76]], [[78, 92]], [[134, 138]], [[146, 150]], [[62, 76]], [[9, 59]], [[9, 59]], [[2, 59]], [[62, 76]], [[78, 92]], [[62, 101]], [[62, 101]], [[102, 121]], [[120, 150]], [[128, 138]], [[140, 150]], [[153, 204]]]",
    "query_spans": "[[[206, 235]]]",
    "process": "As shown in the figure: by symmetry, PQ is the diameter of circle O, so PF\\botPF_{2}. Since |OF_{1}|=|OF_{2}|, |OP|=|OQ|, the quadrilateral PF_{1}QF_{2} is a rectangle, so S_{\\trianglePQF_{2}}=S_{\\trianglePF_{1}F_{2}}. Because |PF_{1}|+|PF_{2}|=2a, |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}=4c^{2}, so |PF_{1}|\\cdot|PF_{2}|=2b^{2}, S_{\\DeltaPF_{1}F_{2}}=\\frac{1}{2}|PF_{1}||PF_{2}|=b^{2}=S_{\\DeltaPQF_{2}}=\\frac{\\sqrt{3}}{2}c^{2}, a^{2}=b^{2}+c^{2}=(1+\\frac{\\sqrt{3}}{2})c^{2}"
  },
  {
    "text": "Let the line $y = x + 1$ intersect the ellipse $\\frac{x^{2}}{2} + y^{2} = 1$ at points $A$ and $B$. Then $|AB| =$?",
    "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;Expression(G) = (x^2/2 + y^2 = 1);Expression(H) = (y = x + 1);Intersection(H, G) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[11, 38]], [[1, 10]], [[41, 44]], [[45, 48]], [[11, 38]], [[1, 10]], [[1, 50]]]",
    "query_spans": "[[[52, 61]]]",
    "process": ""
  },
  {
    "text": "If from any point $P$ on the circle $x^{2}+y^{2}=1$, a perpendicular line segment is drawn to the $y$-axis, then the trajectory equation of the midpoint $M$ of this line segment is?",
    "fact_expressions": "G: Circle;H: LineSegment;Expression(G) = (x^2 + y^2 = 1);PointOnCurve(P,G);IsPerpendicular(H,yAxis);M:Point;MidPoint(H)=M;P:Point",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "4*x^2+y^2=1",
    "fact_spans": "[[[2, 18]], [], [[2, 18]], [[2, 26]], [[1, 35]], [[41, 44]], [[1, 44]], [[23, 26]]]",
    "query_spans": "[[[41, 51]]]",
    "process": "Let the coordinates of point M be (x, y) and the coordinates of point P be (x_{0}, y_{0}). Then from the given conditions, we have x = \\frac{x_{0}}{2}, y = y_{0}. Since P(x_{0}, y_{0}) lies on the circle x^{2} + y^{2} = 1, it follows that x_{0}^{2} + y_{0}^{2} = 1. Substituting x_{0} = 2x, y_{0} = y into the equation x_{0}^{2} + y_{0}^{2} = 1, we obtain 4x^{2} + y^{2} = 1. Therefore, the trajectory equation of point M is 4x^{2} + y^{2} = 1."
  },
  {
    "text": "What is the eccentricity of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/2 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[2, 39]], [[2, 39]]]",
    "query_spans": "[[[2, 44]]]",
    "process": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$, $a^{2}=4$, $b^{2}=2$ $\\Rightarrow c^{2}=2$ $\\Rightarrow e=\\frac{c}{a}=\\frac{\\sqrt{2}}{2}$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{2}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with eccentricity $e$, if $e \\in(\\sqrt{5}, \\sqrt{10})$, then the range of the distance from a focus of $C$ to an asymptote is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/2 - y^2/b^2 = 1);a: Number;b: Number;a>0;b>0;e: Number;Eccentricity(C) = e;In(e, (sqrt(5), sqrt(10)))",
    "query_expressions": "Range(Distance(Focus(C), OneOf(Asymptote(C))))",
    "answer_expressions": "(2*sqrt(2),3*sqrt(2))",
    "fact_spans": "[[[2, 59], [99, 102]], [[2, 59]], [[10, 59]], [[10, 59]], [[10, 59]], [[10, 59]], [[64, 67]], [[2, 67]], [[69, 97]]]",
    "query_spans": "[[[99, 121]]]",
    "process": "Since $ e = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{1 + \\frac{b^{2}}{2}} \\in (\\sqrt{5}, \\sqrt{10}) $, it follows that $ b \\in (2\\sqrt{2}, 3\\sqrt{2}) $, and since the distance from the focus of $ C $ to its asymptote is $ b $, the range of the distance is $ (2\\sqrt{2}, 3\\sqrt{2}) $."
  },
  {
    "text": "What is the equation of the directrix of the parabola $x=a y^{2}  (a \\neq 0)$?",
    "fact_expressions": "G: Parabola;Expression(G) = (x = a*y^2);a: Number;Negation(a=0)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=-1/(4*a)",
    "fact_spans": "[[[0, 26]], [[0, 26]], [[3, 26]], [[3, 26]]]",
    "query_spans": "[[[0, 33]]]",
    "process": "The standard form of the parabola $x = ay^2$ is $y^2 = \\frac{1}{a}x$, so the required directrix equation is $x = -\\frac{1}{4a}$. Therefore, the answer should be: $x = -\\frac{1}{4a}$"
  },
  {
    "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ with focus $F$ and directrix $l$, let point $P$ be on $l$. Draw two tangent lines from $P$ to the parabola $C$, with points of tangency $A$ and $B$. If $|P A|=3$, $|P B|=4$, then $|P F|=$?",
    "fact_expressions": "C: Parabola;p: Number;F: Point;l: Line;P: Point;p>0;Expression(C) = (y^2 = 2*p*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(P, l);L1: Line;L2: Line;TangentOfPoint(P, C) = {L1, L2};A: Point;B: Point;TangentPoint(L1, C) = A;TangentPoint(L2, C) = B;Abs(LineSegmentOf(P, A)) = 3;Abs(LineSegmentOf(P, B)) = 4",
    "query_expressions": "Abs(LineSegmentOf(P, F))",
    "answer_expressions": "12/5",
    "fact_spans": "[[[2, 28], [54, 60]], [[9, 28]], [[32, 35]], [[39, 42], [44, 47]], [[50, 53]], [[10, 28]], [[2, 28]], [[2, 35]], [[2, 42]], [[44, 53]], [], [], [[43, 65]], [[71, 74]], [[75, 78]], [[43, 78]], [[43, 78]], [[80, 89]], [[90, 99]]]",
    "query_spans": "[[[101, 110]]]",
    "process": "Let $ A(x_{1},y_{1}), B(x_{2},y_{2}), P(-\\frac{p}{2},t) (t\\neq0) $, then $ AP: y - y_{1} = k(x - x_{1}) $. Substituting $ x = \\frac{y^{2}}{2p} $ gives $ 2py - 2py_{1} = ky^{2} - 2pkx_{1} $, i.e., $ ky^{2} - 2py + 2py_{1} - ky_{1}^{2} = 0 $. Since line $ AP $ is tangent to the parabola, its discriminant is $ 4p^{2} - 4k(2py_{1} - ky_{1}^{2}) = 0 $, i.e., $ p^{2} - 2kpy_{1} + k^{2}y_{1}^{2} = 0 \\Rightarrow k_{AP} = \\frac{p}{y_{1}} $. Thus, the tangent equation is $ AP: y - y_{1} = \\frac{p}{y_{1}}(x - x_{1}) $, i.e., $ AP: y_{1}y - y_{1}^{2} = p(x - x_{1}) $. Similarly, $ k_{PB} = \\frac{p}{y_{2}} $. So $ BP: y - y_{2} = \\frac{p}{y_{2}}(x - x_{2}) $, i.e., $ BP: y_{2}y - y_{2}^{2} = p(x - x_{2}) $. Since both tangents pass through point $ P(-\\frac{p}{2}, t) $, we have\n$$\n\\begin{cases}\ny_{1}t - \\frac{1}{2}y_{1}^{2} = -\\frac{p^{2}}{2} \\\\\ny_{2}t - \\frac{1}{2}y_{2}^{2} = -\\frac{p^{2}}{2}\n\\end{cases}\n$$\nThen $ y_{1}, y_{2} $ are roots of the equation $ y^{2} - 2ty - p^{2} = 0 $, hence $ y_{1}y_{2} = -p^{2}, y_{1} + y_{2} = 2t $, so $ k_{AP} \\cdot k_{PB} = \\frac{p^{2}}{y_{1}y_{2}} = -1 \\Rightarrow AP \\perp BP $. Given $ |AP| = 3, |BP| = 4 \\Rightarrow |AB| = 5 $. Also, $ k_{AF} = \\frac{y_{1}}{x_{1} - \\frac{p}{2}} = \\frac{2py_{1}}{y_{1}^{2} - p^{2}} = \\frac{2py_{1}}{y_{1}^{2} + y_{1}y_{2}} = \\frac{2p}{y_{1} + y_{2}} $. Similarly, $ k_{BF} = k_{AF} = k_{BF} \\Rightarrow A, F, B $ are collinear. While $ k_{AB} = \\frac{y_{1} - y_{2}}{x_{1} - x_{2}} = \\frac{2p}{y_{1} + y_{2}} = \\frac{p}{t} $, $ k_{PF} $ satisfies $ k_{AB} \\cdot k_{PF} = -1 $, i.e., $ PF \\perp AB $. Hence in right triangle $ \\triangle APB $, the altitude $ \\begin{matrix} 2 & 2 \\\\ PF \\end{matrix} = \\frac{|AP| \\cdot |BP|}{|AB|} = \\frac{12}{5} $. The answer is $ \\frac{12}{5} $."
  },
  {
    "text": "Draw a line through the focus of the parabola $y^{2}=4x$ intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $x_{1}+x_{2}=6$, then $|AB|=$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Line;PointOnCurve(Focus(G), H) = True;Intersection(H, G) = {A, B};A: Point;B: Point;Coordinate(A) = (x1, y1);x1: Number;y1: Number;Coordinate(B) = (x2, y2);x2: Number;y2: Number;x1 + x2 = 6",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[1, 15], [22, 25]], [[1, 15]], [[19, 21]], [[0, 21]], [[19, 65]], [[26, 43]], [[46, 63]], [[26, 43]], [[26, 43]], [[26, 43]], [[46, 63]], [[46, 63]], [[46, 63]], [[68, 83]]]",
    "query_spans": "[[[85, 96]]]",
    "process": ""
  },
  {
    "text": "Point $P$ lies on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, and $F_{1}$, $F_{2}$ are the foci. If $\\angle P F_{1} F_{2}=75^{\\circ}$, $\\angle P F_{2} F_{1}=15^{\\circ}$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);Focus(G) = {F1,F2};AngleOf(P,F1,F2)=ApplyUnit(75,degree);AngleOf(P, F2, F1) = ApplyUnit(15, degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[5, 61], [157, 160]], [[8, 61]], [[8, 61]], [[0, 4]], [[65, 72]], [[73, 80]], [[8, 61]], [[8, 61]], [[5, 61]], [[0, 64]], [[5, 83]], [[86, 119]], [[121, 154]]]",
    "query_spans": "[[[157, 166]]]",
    "process": "From \\angle PF_{1}F_{2}=75^{\\circ}, \\angle PF_{2}F_{1}=15^{\\circ}, it follows that \\angle F_{1}PF_{2}=90^{\\circ}. \\therefore |PF_{1}|=2c\\cos75^{\\circ}, |PF_{2}|=2c\\sin75^{\\circ}. According to the definition of hyperbola, we have: 2c\\sin75^{\\circ}-2c\\cos75^{\\circ}=2a. \\therefore e=\\frac{c}{a}=\\frac{1}{\\sin75^{\\circ}-\\cos75^{\\circ}}=\\frac{}{\\sin(45^{\\circ}+30^{\\circ}}\\frac{=\\angle a'}{\\sin(45^{\\circ}-30^{\\circ})}=\\frac{1}{2\\cos45^{\\circ}\\sin30^{\\circ}}=\\sqrt{2}"
  },
  {
    "text": "Through the point $P(0, -a)$, draw a line $l$ intersecting the parabola $C$: $x^{2} = 4ay$ $(a > 0)$ at points $A$ and $B$. Let $F$ be the focus of $C$. If $|FA| = 2|FB|$, then what is the slope of line $l$?",
    "fact_expressions": "l: Line;C: Parabola;a: Number;P: Point;F: Point;A: Point;B: Point;a>0;Expression(C) = (x^2 = 4*(a*y));Coordinate(P) = (0, -a);PointOnCurve(P, l);Intersection(l, C) ={A,B};Focus(C) = F;Abs(LineSegmentOf(F, A)) = 2*Abs(LineSegmentOf(F, B))",
    "query_expressions": "Slope(l)",
    "answer_expressions": "pm*(3*sqrt(2))/4",
    "fact_spans": "[[[13, 18], [88, 93]], [[19, 44], [63, 66]], [[27, 44]], [[1, 12]], [[59, 62]], [[47, 50]], [[53, 56]], [[27, 44]], [[19, 44]], [[1, 12]], [[0, 18]], [[13, 58]], [[59, 69]], [[71, 86]]]",
    "query_spans": "[[[88, 98]]]",
    "process": ""
  },
  {
    "text": "The standard equation of the hyperbola that has the same asymptotes as $x^{2}-y^{2}=1$ and passes through the point $(1,2)$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 1);Z: Hyperbola;Asymptote(G) = Asymptote(Z);H: Point;Coordinate(H) = (1, 2);PointOnCurve(H, Z)",
    "query_expressions": "Expression(Z)",
    "answer_expressions": "y^2/3 - x^2/3 = 1",
    "fact_spans": "[[[1, 19]], [[1, 19]], [[38, 41]], [[0, 41]], [[29, 37]], [[29, 37]], [[28, 41]]]",
    "query_spans": "[[[38, 48]]]",
    "process": "According to the problem, let the hyperbola equation be: $x^{2}-y^{2}=\\lambda$ ($\\lambda\\neq0$), then we obtain $\\lambda=1^{2}-2^{2}=-3$, thus $x^{2}-y^{2}=-3$. Therefore, the standard equation of the hyperbola is $\\frac{y^{2}}{3}-\\frac{x^{2}}{3}=1$."
  },
  {
    "text": "Draw a line through the focus of the parabola $y^{2}=4x$ intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $x_{1}+x_{2}=10$, then $|AB|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Line;PointOnCurve(Focus(G), H);A: Point;B: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Intersection(H, G) = {A, B};x1 + x2 = 10",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "12",
    "fact_spans": "[[[1, 15], [22, 25]], [[1, 15]], [[19, 21]], [[0, 21]], [[26, 43]], [[44, 61]], [[26, 43]], [[26, 43]], [[44, 61]], [[44, 61]], [[26, 43]], [[44, 61]], [[19, 63]], [[66, 82]]]",
    "query_spans": "[[[85, 94]]]",
    "process": "The directrix of the parabola \\( y^{2} = 4x \\) is: \\( x = -1 \\). Let the focus of the parabola \\( y^{2} = 4x \\) be \\( F \\). By the definition of the parabola, we have: \\( |AB| = |AF| + |BF| = (x_{1} + 1) + (x_{2} + 1) = 12 \\). Therefore, \\( |AB| = 12 \\). The answers are: 1.2"
  },
  {
    "text": "Let the conic section $C$ have its center at the origin and coordinate axes as symmetry axes, with eccentricity $\\sqrt{2}$, and pass through the point $(5,4)$. Then its focal distance is?",
    "fact_expressions": "Center(C) = O;O: Origin;SymmetryAxis(C) = axis;C: ConicSection;Eccentricity(C) = sqrt(2);H: Point;Coordinate(H) = (5, 4);PointOnCurve(H, C) = True",
    "query_expressions": "FocalLength(C)",
    "answer_expressions": "6*sqrt(2)",
    "fact_spans": "[[[1, 25]], [[4, 8]], [[9, 25]], [[18, 25], [53, 54]], [[18, 40]], [[43, 51]], [[43, 51]], [[18, 51]]]",
    "query_spans": "[[[53, 58]]]",
    "process": ""
  },
  {
    "text": "Given that an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ is $2 x-y=0$, then the real number $a$=?",
    "fact_expressions": "G: Hyperbola;a: Real;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G)))=(2*x-y=0)",
    "query_expressions": "a",
    "answer_expressions": "1/2",
    "fact_spans": "[[[2, 39]], [[59, 64]], [[5, 39]], [[2, 39]], [[2, 57]]]",
    "query_spans": "[[[59, 66]]]",
    "process": "\\because the hyperbola \\frac{x^{2}}{a^{2}}-y^{2}=1(a>0) has an asymptote equation 2x-y=0, \\therefore \\frac{1}{a}=2, that is, a=\\frac{1}{2}"
  },
  {
    "text": "Point $P$ is a point on the ellipse $\\frac{x^{2}}{5} + \\frac{y^{2}}{4}=1$. The area of the triangle with vertices at point $P$ and the foci $F_{1}$, $F_{2}$ is equal to $1$. Then the coordinates of point $P$ are?",
    "fact_expressions": "G: Ellipse;P: Point;F1:Point;F2:Point;Expression(G) = (x^2/5 + y^2/4 = 1);PointOnCurve(P,G);Focus(G)={F1,F2};Area(TriangleOf(P,F1,F2))=1",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(\\sqrt{15}/2, 1),(\\sqrt{15}/2, -1),(-\\sqrt{15}/2, 1),(-\\sqrt{15}/2, -1)",
    "fact_spans": "[[[5, 44]], [[0, 4], [48, 52], [91, 95]], [[56, 63]], [[66, 73]], [[5, 44]], [[0, 47]], [[5, 73]], [[47, 88]]]",
    "query_spans": "[[[91, 100]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the circle $O$: $x^{2}+y^{2}=b^{2}$, from a point $P$ on the ellipse, two tangents are drawn to the circle $O$, with points of tangency $A$ and $B$. If $\\angle A P B=90^{\\circ}$, then the range of values for the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;O: Circle;A: Point;P: Point;B: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(O) = (x^2 + y^2 = b^2);L1:Line;L2:Line;TangentOfPoint(P,O)={L1,L2};TangentPoint(L1,O)=A;TangentPoint(L2,O)=B;AngleOf(A, P, B) = ApplyUnit(90, degree);PointOnCurve(P,G)",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[\\sqrt{2}/2,1)",
    "fact_spans": "[[[2, 56], [83, 85], [145, 147]], [[4, 56]], [[4, 56]], [[57, 81], [92, 96]], [[108, 111]], [[88, 91]], [[112, 115]], [[4, 56]], [[4, 56]], [[2, 56]], [[57, 81]], [], [], [[82, 101]], [[82, 115]], [[82, 115]], [[118, 143]], [80, 88]]",
    "query_spans": "[[[145, 157]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the asymptotes of the hyperbola $C$: $\\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1$?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/5 - y^2/4 = 1)",
    "query_expressions": "Expression(Asymptote(C))",
    "answer_expressions": "2*x + pm*sqrt(5)*y = 0",
    "fact_spans": "[[[0, 42]], [[0, 42]]]",
    "query_spans": "[[[0, 50]]]",
    "process": ""
  },
  {
    "text": "Through the focus $F$ of the parabola $y^{2}=-4x$, draw a line $l$ intersecting the parabola at points $A$ and $B$, and intersecting the directrix at point $C$. If $\\overrightarrow{F C}=4 \\overrightarrow{F B}$, then $|\\overrightarrow{A B}|$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -4*x);F: Point;Focus(G) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, G) = {A, B};C: Point;Intersection(l, Directrix(G)) = C;VectorOf(F, C) = 4*VectorOf(F, B)",
    "query_expressions": "Abs(VectorOf(A, B))",
    "answer_expressions": "9/2",
    "fact_spans": "[[[1, 16], [30, 33]], [[1, 16]], [[19, 22]], [[1, 22]], [[23, 28]], [[0, 28]], [[35, 38]], [[39, 42]], [[23, 44]], [[50, 54]], [[23, 54]], [[56, 101]]]",
    "query_spans": "[[[103, 129]]]",
    "process": "The coordinates of the focus are F(-1,0). Draw the graph as shown below. Draw BD perpendicular to the directrix x=1, intersecting the directrix at D. According to the definition of a parabola, BF=BD. Since \\overrightarrow{FC}=4\\overrightarrow{FB}, it follows that \\frac{BD}{BC}=\\frac{BF}{BC}=\\frac{1}{3}. Let the angle of inclination of line l be \\alpha, then \\cos\\alpha=\\frac{1}{3}, so \\tan\\alpha=\\frac{\\sqrt{1-\\cos2\\alpha}}{\\cos\\alpha}=2\\sqrt{2}. Thus, the slope of line l is 2\\sqrt{2}, and the equation of line l is y=2\\sqrt{2}(x+1). Substituting into the parabola equation and simplifying yields 2x^{2}+5x+2=0, so x_{A}+x_{B}=-\\frac{5}{2}. Therefore, |\\overrightarrow{AB}|=|x_{1}+x_{2}|+p=\\frac{5}{2}+2=\\frac{9}{2}"
  },
  {
    "text": "One focus of the ellipse $k x^{2}+2 y^{2}=2$ is $(1,0)$, then $k=$?",
    "fact_expressions": "G: Ellipse;k: Number;H: Point;Expression(G) = (k*x^2 + 2*y^2 = 2);Coordinate(H) = (1, 0);OneOf(Focus(G)) = H",
    "query_expressions": "k",
    "answer_expressions": "1",
    "fact_spans": "[[[0, 21]], [[37, 40]], [[27, 34]], [[0, 21]], [[27, 34]], [[0, 34]]]",
    "query_spans": "[[[37, 42]]]",
    "process": "The standard form of the ellipse $ kx^{2} + 2y^{2} = 2 $ is: $ \\frac{x^{2}}{k} + y^{2} = 1 $. Since one focus is $ (1,0) $, the foci lie on the x-axis, and $ 0 < k < 2 $, then $ a^{2} = \\frac{2}{k} $, $ b^{2} = 1 $. Therefore, $ \\frac{2}{k} - 1 = 1 $, solving gives $ k = 1 $."
  },
  {
    "text": "The coordinates of the focus of the parabola $y=-4 x^{2}$ are?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = -4*x^2)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,-1/16)",
    "fact_spans": "[[[0, 15]], [[0, 15]]]",
    "query_spans": "[[[0, 22]]]",
    "process": "From the given condition, $x^{2}=-\\frac{1}{4}y$, then the coordinates of the focus are $F(0,-\\frac{1}{16})$"
  },
  {
    "text": "The equation of the ellipse with the foci of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$ as vertices and the vertices as foci is?",
    "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(G) = (x^2/4 - y^2/12 = 1);Focus(G) = Vertex(H);Vertex(G) = Focus(H)",
    "query_expressions": "Expression(H)",
    "answer_expressions": "x^2/16+y^2/12=1",
    "fact_spans": "[[[1, 40]], [[53, 55]], [[1, 40]], [[0, 55]], [[0, 55]]]",
    "query_spans": "[[[53, 59]]]",
    "process": "Hyperbola foci $(\\pm4,0)$, vertices $(\\pm2,0)$, thus the ellipse's foci are $(\\pm2,0)$, vertices $(\\pm4,0)$. Answer: $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$"
  },
  {
    "text": "If the equation $k x^{2}+y^{2}=2$ represents an ellipse with foci on the $y$-axis, then what is the range of real values for $k$?",
    "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (k*x^2 + y^2 = 2);PointOnCurve(Focus(G), yAxis)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(1, +\\infty)",
    "fact_spans": "[[[32, 34]], [[37, 42]], [[2, 34]], [[23, 34]]]",
    "query_spans": "[[[37, 49]]]",
    "process": "The given equation can be rewritten as \\frac{x^2}{k} + \\frac{y^{2}}{2} = 1, which represents an ellipse with foci on the y-axis. Then 0 < \\frac{2}{k} < 2, solving yields k > 1."
  },
  {
    "text": "The asymptotes of a hyperbola are given by $2x \\pm 3y = 0$, and the focal distance is $2\\sqrt{13}$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (2*x + pm*3*y = 0);FocalLength(G) = 2*sqrt(13)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{x^2/9-y^2/4=1, y^2/4-x^2/9=1}",
    "fact_spans": "[[[2, 5], [47, 50]], [[2, 27]], [[2, 45]]]",
    "query_spans": "[[[47, 55]]]",
    "process": "According to the problem, assume the hyperbola equation is $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=\\lambda$ ($\\lambda\\neq0$). If $\\lambda>0$, then $a^{2}=9\\lambda$, $b^{2}=4\\lambda$, $c^{2}=a^{2}+b^{2}=13\\lambda$. From the given condition $2c=2\\sqrt{13}$, we get $\\lambda=1$, so the required hyperbola equation is $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1$. If $\\lambda<0$, then $a^{2}=-4\\lambda$, $b^{2}=-9\\lambda$, $c^{2}=a^{2}+b^{2}=-13\\lambda$. From $2c=2\\sqrt{13}$, we obtain $\\lambda=-1$, hence the required hyperbola equation is $\\frac{y^{2}}{4}-\\frac{x^{2}}{9}=1$. This problem mainly examines the solution of hyperbola equations and is a basic question."
  },
  {
    "text": "The focus of the parabola $y^{2}=2 p x(p>0)$ is $F$, and $A$, $B$ are two points on the parabola. The circle with diameter $AB$ passes through point $F$. From the midpoint $M$ of $AB$, draw a perpendicular $MN$ to the directrix of the parabola, with foot $N$. Then the maximum value of $\\frac{|M N|}{|AB|}$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;A: Point;B: Point;PointOnCurve(A, G) = True;PointOnCurve(B, G) = True;IsDiameter(LineSegmentOf(A, B), H) = True;H: Circle;PointOnCurve(F, H) = True;MidPoint(LineSegmentOf(A, B)) = M;M: Point;PointOnCurve(M, LineOf(M, N)) = True;IsPerpendicular(LineOf(M, N), Directrix(G)) = True;FootPoint(LineOf(M, N), Directrix(G)) = N;N: Point",
    "query_expressions": "Max(Abs(LineSegmentOf(M, N))/Abs(LineSegmentOf(A, B)))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[0, 21], [37, 40], [75, 78]], [[0, 21]], [[3, 21]], [[3, 21]], [[25, 28], [57, 61]], [[0, 28]], [[29, 32]], [[33, 36]], [[29, 44]], [[29, 44]], [[45, 56]], [[55, 56]], [[55, 61]], [[63, 74]], [[71, 74]], [[62, 89]], [[75, 89]], [[75, 96]], [[93, 96]]]",
    "query_spans": "[[[98, 124]]]",
    "process": "From the definition of the parabola, we obtain $\\frac{|MN|}{|AB|}=\\frac{|AF|+|BF|}{\\sqrt{|AF^{2}+|BF^{2}}}\\leqslant\\frac{|AF|+|BF|}{\\sqrt{|AF^{2}+|BF^{2}}}=\\frac{\\sqrt{2}}{2}$, that is, the maximum value of $\\frac{|MN|}{|AB|}$ is $\\frac{\\sqrt{2}}{2}$."
  },
  {
    "text": "Given that point $M$ is a moving point on the parabola $y^{2}=8x$, $F$ is the focus of the parabola, and $A$ is a moving point on the circle $C$: $x^{2}+y^{2}-8x-2y+16=0$, then the minimum value of $|MA|+|MF|$ is?",
    "fact_expressions": "G: Parabola;C: Circle;M: Point;A: Point;F: Point;Expression(G) = (y^2 = 8*x);Expression(C) = (-2*y - 8*x + x^2 + y^2 + 16 = 0);PointOnCurve(M, G);Focus(G) = F;PointOnCurve(A, C)",
    "query_expressions": "Min(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F)))",
    "answer_expressions": "5",
    "fact_spans": "[[[7, 21], [31, 34]], [[42, 74]], [[2, 6]], [[38, 41]], [[27, 30]], [[7, 21]], [[42, 74]], [[2, 26]], [[27, 37]], [[38, 78]]]",
    "query_spans": "[[[80, 99]]]",
    "process": "From $x^{2}+y^{2}-8x-2y+16=0$ we obtain $(x-4)^{2}+(y-1)^{2}=1$, so the center of circle $C$ is $C(4,1)$ and the radius is $r=1$. By the properties of a circle, $|MA|+|MF|\\geqslant|MC|-r+|MF|=|MC|+|MF|-1$. Also, the directrix of the parabola $y^{2}=8x$ is $x=-2$, and the focus is $F(2,0)$. Denote the intersection point of $x=-2$ and the $x$-axis as $D(-2,0)$, and draw $MB$ perpendicular to the line $x=-2$ at point $B$. By the definition of a parabola, $|MF|=|MB|$. Then $|MC|+|MF|=|MC|+|MB|\\geqslant|BC|$. Since point $B$ moves with point $M$, and $MB\\perp BD$ always holds, when point $M$ moves to point $O$, $B$ coincides exactly with point $D$, and at this moment $|MC|+|MB|=|BC|=|CD|=6$. Thus, $|MA|+|MF|\\geqslant|MC|-r+|MF|=|MC|+|MF|-1\\geqslant|CD|-1=5$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{5}+\\frac{y^{2}}{9}=1$, the two foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through $F_{2}$ intersects $C$ at points $A$ and $B$. Then, what is the perimeter of $\\triangle A F_{1} B$?",
    "fact_expressions": "l: Line;C: Ellipse;A: Point;F1: Point;B: Point;F2: Point;Expression(C) = (x^2/5 + y^2/9 = 1);Focus(C) = {F1,F2};PointOnCurve(F2, l);Intersection(l, C) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(A, F1, B))",
    "answer_expressions": "12",
    "fact_spans": "[[[79, 84]], [[2, 44], [85, 88]], [[89, 92]], [[52, 59]], [[93, 96]], [[61, 68], [71, 78]], [[2, 44]], [[2, 68]], [[70, 84]], [[79, 98]]]",
    "query_spans": "[[[100, 126]]]",
    "process": "Given that $ a = 3 $, it follows that $ |AF_{1}| + |AF_{2}| = 2a = 6 $, $ |BF_{1}| + |BF_{2}| = 2a = 6 $, so $ |AF_{1}| + |AF_{2}| + |BF_{1}| + |BF_{2}| = 12 $, hence $ |AF_{1}| + |BF_{1}| + |AB| = 12 $. Therefore, the perimeter of $ \\triangle AF_{1}B $ is 12."
  },
  {
    "text": "A line with slope $l$ passes through the focus of the parabola $y^{2}=4x$ and intersects the parabola at points $A$ and $B$. Then $|AB|=$?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;l:Number;Expression(G) = (y^2 = 4*x);Slope(H)=l;Intersection(H,G)={A,B};PointOnCurve(Focus(G),H)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[11, 25], [30, 33]], [[7, 9]], [[36, 39]], [[40, 43]], [[3, 6]], [[11, 25]], [[0, 9]], [[7, 45]], [[7, 28]]]",
    "query_spans": "[[[47, 56]]]",
    "process": "The focus of the parabola is (1,0), so the equation of the line is y = x - 1. Substituting into the parabola equation gives x^{2} \\cdot 6x + 1 = 0, \\therefore x_{1} + x_{2} = 6. According to the definition of the parabola, |AB| = x_{1} + \\frac{p}{2} + x_{2} + \\frac{p}{2} = x_{1} + x_{2} + p = 6 + 2 = 8"
  },
  {
    "text": "Given the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, a line passing through its right focus $F$ intersects $E$ at points $A$ and $B$, and intersects the $y$-axis at point $P$. Given that $\\overrightarrow{P A}=3 \\overrightarrow{A F}$, $\\overrightarrow{P B}=-\\frac{7}{3} \\overrightarrow{B F}$, then the eccentricity of $E$ is?",
    "fact_expressions": "E: Hyperbola;b: Number;a: Number;G: Line;P: Point;A: Point;F: Point;B: Point;a>0;b>0;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(E) = F;PointOnCurve(F, G);Intersection(G, E) = {A, B};Intersection(G, yAxis) = P;VectorOf(P, A) = 3*VectorOf(A, F);VectorOf(P, B) = (-7/3)*VectorOf(B, F)",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 63], [65, 66], [76, 79], [207, 210]], [[10, 63]], [[10, 63]], [[73, 75]], [[96, 100]], [[80, 83]], [[69, 72]], [[84, 87]], [[10, 63]], [[10, 63]], [[2, 63]], [[65, 72]], [[64, 75]], [[73, 89]], [[73, 100]], [[102, 147]], [[149, 205]]]",
    "query_spans": "[[[207, 216]]]",
    "process": "From $\\overrightarrow{PA}=3\\overrightarrow{AF},\\overrightarrow{PB}=-\\frac{7}{3}\\overrightarrow{BF}$, it follows that $\\overrightarrow{PF}=\\overrightarrow{PA}+\\overrightarrow{AF}=\\frac{4}{3}\\overrightarrow{PA}$, $\\overrightarrow{PF}=\\overrightarrow{PB}+\\overrightarrow{BF}=\\frac{4}{7}\\overrightarrow{PB}$. Let $P(0,m)$, $F(c,0)$, so $\\overrightarrow{PF}=(c,-m)$, $\\overrightarrow{PA}=(x_{A},y_{A}-m)$, $\\overrightarrow{PB}=(x_{B},y_{B}-m)$. Since $\\overrightarrow{PF}=\\frac{4}{3}\\overrightarrow{PA}$, $\\overrightarrow{PF}=\\frac{4}{7}\\overrightarrow{PB}$, we have $A(\\frac{2}{2}c,\\frac{1}{4}m)$, $B(\\frac{7}{4}c,-\\frac{3m}{4})$. Substituting into the equation of hyperbola $E$, we obtain $\\begin{cases}\\frac{9c^{2}}{16a^{2}}\\\\\\frac{49c^{2}}{2}\\end{cases}\\frac{9c^{2}}{6a^{2}}-\\frac{91}{16}$. Eliminating $\\frac{m^{2}}{b^{2}}$ yields $9\\times\\frac{9c^{2}}{16a^{2}}-\\frac{49c^{2}}{16a^{2}}=8$, i.e., $2e^{2}=8$, solving gives $'$"
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, a line passing through the left focus $F_{1}$ intersects the ellipse $C$ at points $A$ and $B$, such that $|A F_{1}|=2|B F_{1}|$ and $|A B|=|B F_{2}|$. Then the eccentricity of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;A: Point;F1: Point;B: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, G);Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, F1)) = 2*Abs(LineSegmentOf(B, F1));Abs(LineSegmentOf(A, B)) = Abs(LineSegmentOf(B, F2))",
    "query_expressions": "Eccentricity(C)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[18, 75], [97, 102], [157, 162]], [[25, 75]], [[25, 75]], [[94, 96]], [[104, 107]], [[2, 9], [86, 93]], [[108, 111]], [[10, 17]], [[25, 75]], [[25, 75]], [[18, 75]], [[2, 81]], [[2, 81]], [[82, 96]], [[94, 113]], [[115, 137]], [[138, 155]]]",
    "query_spans": "[[[157, 168]]]",
    "process": "Let |BF₁| = k, then |AF₁| = 2k, |BF₂| = 3k. From |BF₁| + |BF₂| = |AF₁| + |AF₂| = 2a we get 2a = 4k, |AF₂| = 2k. In △ABF₂, cos∠BAF₂ = 1/3. Also in AF₁AF₂, cos∠F₁AF₂ = ( (2k)² + (2k)² - (2c)² ) / (2 × 2k × 2k) = 1/3, we obtain 2c = (4 / √3)k. Hence the eccentricity e = c/a = √3 / 3"
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=m(m>0)$ is?",
    "fact_expressions": "G: Ellipse;m: Number;m>0;Expression(G) = (x^2/16 + y^2/7 = m)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3/4",
    "fact_spans": "[[[0, 43]], [[2, 43]], [[2, 43]], [[0, 43]]]",
    "query_spans": "[[[0, 49]]]",
    "process": "From the ellipse equation, obtain a, b, c, and directly calculate the eccentricity. Since the ellipse is $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=m$ $(m>0)$, then $a^{2}=16m$, $b^{2}=7m$, $c^{2}=9m$, so $c=3\\sqrt{m}$, $e=\\frac{c}{a}=\\frac{3\\sqrt{m}}{4\\sqrt{m}}=\\frac{3}{4}$"
  },
  {
    "text": "Given that the focal distance of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{5}=1$ is $8$, find the value of the real number $m$.",
    "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (-y^2/5 + x^2/m = 1);FocalLength(G) = 8",
    "query_expressions": "m",
    "answer_expressions": "11",
    "fact_spans": "[[[2, 40]], [[49, 54]], [[2, 40]], [[2, 47]]]",
    "query_spans": "[[[49, 58]]]",
    "process": "From the given condition, we have m+5=16, which can be solved directly. From the given condition, we have a^{2}=m, b^{2}=5, c=4, then from a^{2}+b^{2}=c^{2} we get m+5=16, solving gives m=11."
  },
  {
    "text": "Given that the parabola passes through the point $(-3,2)$, then the standard equation of the parabola is?",
    "fact_expressions": "G: Parabola;H: Point;Coordinate(H) = (-3, 2);PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2 = (-4/3)*x, x^2 = (9/2)*y}",
    "fact_spans": "[[[2, 5], [18, 21]], [[6, 15]], [[6, 15]], [[2, 15]]]",
    "query_spans": "[[[18, 28]]]",
    "process": "Let the equation of the required parabola be y^{2}=-2px or x^{2}=2py (p>0). Since it passes through the point (-3,2), we have 4=-2p\\times(-3) or 9=2p\\times2. Solving gives p=\\frac{2}{3} or p=\\frac{9}{4}. Hence, the equation of the required parabola is y^{2}=-\\frac{4}{3}x or x^{2}=\\frac{9}{2}y. This question examines the equation of a parabola. One needs to understand parabola equations and their related properties. Note that when the problem does not specify the coordinate axis on which the focus lies, the focus may lie either on the x-axis or the y-axis."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{16}=1$, and a line passing through $F_{2}$ intersects the ellipse at points $A$ and $B$. If $|A B|=5$, then the value of $|AF_{1}|+|BF_{1}|$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/16 = 1);H: Line;A: Point;B: Point;F1: Point;F2: Point;Focus(G) = {F1, F2};PointOnCurve(F2, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 5",
    "query_expressions": "Abs(LineSegmentOf(A, F1)) + Abs(LineSegmentOf(B, F1))",
    "answer_expressions": "11",
    "fact_spans": "[[[18, 56], [74, 76]], [[18, 56]], [[71, 73]], [[77, 81]], [[82, 85]], [[2, 9]], [[10, 17], [63, 70]], [[2, 61]], [[62, 73]], [[71, 85]], [[88, 97]]]",
    "query_spans": "[[[99, 122]]]",
    "process": ""
  },
  {
    "text": "Given that point $M(-1,1)$ is the midpoint of chord $AB$ of the parabola $y^{2}=-8x$, what is the equation of line $AB$?",
    "fact_expressions": "M: Point;Coordinate(M) = (-1, 1);G: Parabola;Expression(G) = (y^2 = -8*x);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Expression(LineOf(A, B))",
    "answer_expressions": "4*x+y+3=0",
    "fact_spans": "[[[2, 12]], [[2, 12]], [[13, 28]], [[13, 28]], [[30, 35]], [[30, 35]], [[13, 35]], [[2, 38]]]",
    "query_spans": "[[[39, 51]]]",
    "process": "Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Since point $ M(-1,1) $ is the midpoint of chord $ AB $, we have $ y_{1}+y_{2}=2 $. Substituting into the parabola $ y^{2}=-8x $, we get $ y_{1}^{2}=-8x_{1} $, $ y_{2}^{2}=-8x_{2} $. Subtracting these two equations gives $ y_{2}^{2}-y_{1}^{2}=-8x_{2}+8x_{1} $, that is, $ (y_{2}-y_{1})(y_{2}+y_{1})=-8(x_{2}-x_{1}) $. Thus, $ k=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{-8}{y_{1}+y_{2}}=-4 $. Therefore, the equation of line $ AB $ is $ y-1=-4(x+1) $, i.e., $ 4x+y+3=0 $."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $P$ is a point on the ellipse. If $|P F_{1}|=2|P F_{2}|$, then the range of values for the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2));LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(P,G)",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[1/3,1)",
    "fact_spans": "[[[2, 54], [83, 85], [114, 116]], [[4, 54]], [[4, 54]], [[79, 82]], [[63, 70]], [[71, 78]], [[4, 54]], [[4, 54]], [[2, 54]], [[90, 112]], [[2, 78]], [[2, 78]], [[79, 88]]]",
    "query_spans": "[[[114, 127]]]",
    "process": "By the definition of an ellipse, |PF_{1}| + |PF_{2}| = 2a. Since |PF_{1}| = 2|PF_{2}|, it follows that |PF_{2}| = \\frac{2}{3}a. Since |PF_{2}| \\geqslant a - c, we have a - c \\leqslant \\frac{2a}{3}, so \\frac{a}{3} \\leqslant c, hence \\frac{c}{a} \\geqslant \\frac{1}{3}. Therefore, the range of the eccentricity of the ellipse is [\\frac{1}{3},"
  },
  {
    "text": "Given that vertices $B$ and $C$ of $\\triangle ABC$ lie on the ellipse $\\frac{x^{2}}{3}+y^{2}=1$, vertex $A$ is one focus of the ellipse, and the other focus of the ellipse lies on side $BC$, then what is the perimeter of $\\triangle ABC$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/3 + y^2 = 1);A: Point;B: Point;C: Point;PointOnCurve(B, G);PointOnCurve(C, G);K: Point;OneOf(Focus(G)) = A;OneOf(Focus(G)) = K;Negation(A=K);PointOnCurve(K, LineSegmentOf(B, C))",
    "query_expressions": "Perimeter(TriangleOf(A, B, C))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[28, 55], [63, 65], [72, 74]], [[28, 55]], [[59, 62]], [[20, 23]], [[24, 27]], [[20, 56]], [[24, 56]], [], [[59, 70]], [[72, 81]], [[59, 81]], [[72, 89]]]",
    "query_spans": "[[[91, 113]]]",
    "process": ""
  },
  {
    "text": "Given an ellipse with foci on the $x$-axis, center at the origin, and the sum of distances from a point on the ellipse to the two foci is $6$. If the eccentricity of the ellipse is $\\frac{1}{3}$, then the equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;PointOnCurve(Focus(G), xAxis);O: Origin;Center(G) = O;F1: Point;F2: Point;Focus(G) = {F1,F2};P: Point;PointOnCurve(P,G);Distance(P,F1)+Distance(P,F2) = 6;Eccentricity(G) = 1/3",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 + y^2/8 = 1",
    "fact_spans": "[[[17, 19], [38, 40], [60, 62]], [[2, 19]], [[14, 16]], [[11, 19]], [], [], [[17, 26]], [], [[17, 22]], [[17, 35]], [[38, 58]]]",
    "query_spans": "[[[60, 67]]]",
    "process": "From the given conditions, we have $ e = \\frac{1}{3} $, $ 2a = 6 $, $ \\therefore a = 3 $, $ b = 2\\sqrt{2} $, so the equation of the ellipse is $ \\frac{x^{2}}{9} + \\frac{y^{2}}{8} = 1 $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ has eccentricity $\\frac{3}{2}$, then $b=$?",
    "fact_expressions": "G: Hyperbola;b: Number;b>0;Expression(G) = (x^2/4 - y^2/b^2 = 1);Eccentricity(G) = 3/2",
    "query_expressions": "b",
    "answer_expressions": "sqrt(5)",
    "fact_spans": "[[[2, 49]], [[69, 72]], [[5, 49]], [[2, 49]], [[2, 67]]]",
    "query_spans": "[[[69, 74]]]",
    "process": "Since the eccentricity of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ is $\\frac{3}{2}$, we have $e^{2}=\\frac{4+b^{2}}{4}=1+\\frac{b^{2}}{4}=\\frac{9}{4}$, which implies $b^{2}=5$, solving gives $b=\\sqrt{5}$, so $b=\\sqrt{5}$."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{k-2}+\\frac{y^{2}}{5-k}=1$ represents an ellipse with foci on the $x$-axis, then the range of values for $k$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(k - 2) + y^2/(5 - k) = 1);PointOnCurve(Focus(G), xAxis);k: Number",
    "query_expressions": "Range(k)",
    "answer_expressions": "(7/2, 5)",
    "fact_spans": "[[[55, 57]], [[1, 57]], [[46, 57]], [[59, 62]]]",
    "query_spans": "[[[59, 69]]]",
    "process": "From the given condition: k-2 > 5-k > 0, solving yields: \\frac{7}{2} < k < 5. Therefore, the range of k is (\\frac{7}{2}, 5)."
  },
  {
    "text": "The length of the moving chord $AB$ of the parabola $y^{2}=8x$ is $16$. What is the shortest distance from the midpoint $M$ of chord $AB$ to the $y$-axis?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;M: Point;Expression(G) = (y^2 = 8*x);IsChordOf(LineSegmentOf(A,B),G);Length(LineSegmentOf(A, B)) = 16;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Min(Distance(M, yAxis))",
    "answer_expressions": "6",
    "fact_spans": "[[[0, 14]], [[17, 22]], [[17, 22]], [[39, 42]], [[0, 14]], [[0, 22]], [[17, 29]], [[31, 42]]]",
    "query_spans": "[[[39, 54]]]",
    "process": "The directrix of the parabola $ y^{2} = 8x $ is $ x = -2 $, and the focus is $ F(2,0) $. Let $ AC \\perp l $, $ BD \\perp l $, and $ MH \\perp l $ be drawn from points $ A $, $ B $, and $ M $ respectively, with feet of perpendiculars at $ C $, $ D $, and $ H $. In the right trapezoid $ ABDC $, $ MH = \\frac{AC + BD}{2} $. By the definition of a parabola, $ AC = AF $, $ BD = BF $, thus $ MH = \\frac{AF + BF}{2} \\geqslant \\frac{AB}{2} = 8 $ (with equality if and only if points $ A $, $ B $, $ F $ are collinear). Therefore, the shortest distance from $ M $ to the $ y $-axis is $ 8 - 2 = 6 $."
  },
  {
    "text": "$F_{1}$, $F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$. A line perpendicular to the $x$-axis is drawn through point $F_{2}$, intersecting the hyperbola at points $A$, $B$. Then the perimeter of $\\Delta{F_{1} AB}$ is?",
    "fact_expressions": "G: Hyperbola;F1: Point;A: Point;B: Point;F2: Point;Expression(G) = (x^2/4 - y^2/3 = 1);Focus(G) = {F1, F2};L:Line;PointOnCurve(F2, L);IsPerpendicular(L,xAxis);Intersection(L, G) = {A, B}",
    "query_expressions": "Perimeter(TriangleOf(F1, A, B))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[16, 54], [78, 81]], [[0, 7]], [[82, 85]], [[86, 89]], [[8, 15], [61, 69]], [[16, 54]], [[0, 59]], [], [[60, 77]], [[60, 77]], [[60, 91]]]",
    "query_spans": "[[[93, 116]]]",
    "process": ""
  },
  {
    "text": "It is known that $F$ is the focus of the parabola $C$: $y^{2}=4x$. A line passing through $F$ with slope $\\sqrt{3}$ intersects $C$ at points $A$ and $B$. Given $|\\overrightarrow{FA}|<|\\overrightarrow{F B}|$, if $\\overrightarrow{F A}=\\lambda \\overrightarrow{F B}$, then the value of $\\lambda$ is?",
    "fact_expressions": "F: Point;C: Parabola;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F, G) = True;Slope(G) = sqrt(3);G: Line;Intersection(G, C) = {A, B};A: Point;B: Point;Abs(VectorOf(F, A)) < Abs(VectorOf(F, B));VectorOf(F, A) = lambda*VectorOf(F, B);lambda: Number",
    "query_expressions": "lambda",
    "answer_expressions": "-1/3",
    "fact_spans": "[[[2, 5], [30, 33]], [[6, 25], [51, 54]], [[6, 25]], [[2, 28]], [[29, 50]], [[34, 50]], [[48, 50]], [[48, 64]], [[55, 58]], [[59, 62]], [[67, 113]], [[116, 167]], [[169, 178]]]",
    "query_spans": "[[[169, 182]]]",
    "process": ""
  },
  {
    "text": "Given that the hyperbola $\\frac{x^{2}}{m+2}-\\frac{y^{2}}{m+1}=1$ has eccentricity $\\frac{\\sqrt{7}}{2}$, then $m=?$",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(m + 2) - y^2/(m + 1) = 1);m: Number;Eccentricity(G) = sqrt(7)/2",
    "query_expressions": "m",
    "answer_expressions": "{2, -5}",
    "fact_spans": "[[[2, 44]], [[2, 44]], [[72, 75]], [[2, 69]]]",
    "query_spans": "[[[72, 77]]]",
    "process": ""
  },
  {
    "text": "If the line $2x - cy + 1 = 0$ is a tangent to the parabola $x^2 = y$, then $c = $?",
    "fact_expressions": "G: Parabola;H: Line;c: Number;Expression(G) = (x^2 = y);Expression(H) = (-c*y + 2*x + 1 = 0);IsTangent(H, G)",
    "query_expressions": "c",
    "answer_expressions": "-1",
    "fact_spans": "[[[17, 29]], [[1, 16]], [[36, 39]], [[17, 29]], [[1, 16]], [[1, 34]]]",
    "query_spans": "[[[36, 41]]]",
    "process": "According to the problem, solving the system of equations by combining the line and the parabola gives  \n\\begin{cases}2x-c\\\\x^{2}=y\\end{cases}y+1=0"
  },
  {
    "text": "The hyperbola $C$: $y^{2}-\\frac{x^{2}}{b^{2}}=1(b>0)$ and the circle $M$: $(x+\\frac{5}{3})^{2}+y^{2}=\\frac{16}{9}$ have four intersection points. Then the range of the eccentricity of $C$ is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (y^2 - x^2/b^2 = 1);b: Number;b>0;M: Circle;Expression(M) = (y^2 + (x + 5/3)^2 = 16/9);NumIntersection(C, M) = 4",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(5*sqrt(7)/7, +oo)",
    "fact_spans": "[[[0, 42], [95, 98]], [[0, 42]], [[8, 42]], [[8, 42]], [[43, 88]], [[43, 88]], [[0, 93]]]",
    "query_spans": "[[[95, 109]]]",
    "process": "From \\begin{cases}y^{2}-\\frac{x^{2}}{b^{2}}=1\\\\(x+\\frac{5}{3})^{2}+y^{2}=\\frac{16}{9}\\end{cases}, we obtain (1+\\frac{1}{b^{2}})x^{2}+\\frac{10}{3}x+2=0. According to the problem, A=\\frac{100}{9}-8(1+\\frac{1}{b^{2}})>0, solving gives b^{2}>\\frac{18}{7}, then \\frac{9}{e=\\sqrt{1+}b^{2}}=\\sqrt{1+b^{2}}>\\sqrt{1+\\frac{18}{7}}=\\frac{5\\sqrt{7}}{7}."
  },
  {
    "text": "A asymptote of the hyperbola $t x^{2}-y^{2}-1=0$ is perpendicular to the line $2 x+y+1=0$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;t: Number;H: Line;Expression(G) = (t*x^2 - y^2 - 1 = 0);Expression(H) = (2*x + y + 1 = 0);IsPerpendicular(OneOf(Asymptote(G)), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(5)/2",
    "fact_spans": "[[[0, 22], [46, 49]], [[3, 22]], [[29, 42]], [[0, 22]], [[29, 42]], [[0, 44]]]",
    "query_spans": "[[[46, 55]]]",
    "process": "According to the given conditions, find the slope of the asymptote, and use the relationship among a, b, c to find the eccentricity of the hyperbola. \\because the asymptotes of the hyperbola $tc^{2}-y^{2}-1=0$ are $y=\\pm\\sqrt{t}x$, and one asymptote is perpendicular to the line $2x+y+1=0$, $\\therefore$ the slope of the asymptote is $\\frac{1}{2}$, $\\therefore\\frac{b}{a}=\\frac{1}{2}$, $\\therefore\\frac{c^{2}-a^{2}}{a^{2}}=\\frac{1}{4}$, $\\therefore e=\\frac{\\sqrt{5}}{2}$"
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. If there exists a point $P$ on the line $x=\\frac{a^{2}}{c}$ such that the perpendicular bisector of the segment $PF_{1}$ passes through the point $F_{2}$, then the range of the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Line;c: Number;Expression(H) = (x = a^2/c);P: Point;PointOnCurve(P, H);PointOnCurve(F2, PerpendicularBisector(LineSegmentOf(P, F1)))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "[\\sqrt{3}/3, 1)",
    "fact_spans": "[[[20, 72], [136, 138]], [[20, 72]], [[22, 72]], [[22, 72]], [[22, 72]], [[22, 72]], [[1, 8]], [[10, 17], [126, 134]], [[1, 78]], [[1, 78]], [[81, 102]], [[83, 102]], [[81, 102]], [[105, 109]], [[80, 109]], [[111, 134]]]",
    "query_spans": "[[[136, 149]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, let points $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse, point $A$ a point on the ellipse, and the incenter of $\\triangle A F_{1} F_{2}$ denoted by $M$. If $\\overrightarrow{M F_{1}}+2 \\overrightarrow{M F_{2}}+2 \\overrightarrow{M A}=0$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;F1: Point;F2: Point;M: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(A, G);Center(InscribedCircle(TriangleOf(A,F1,F2)))=M;VectorOf(M, F1) + 2*VectorOf(M, F2) + 2*VectorOf(M, A) = 0",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2/3",
    "fact_spans": "[[[2, 54], [72, 74], [85, 87], [209, 211]], [[4, 54]], [[4, 54]], [[80, 84]], [[55, 63]], [[64, 71]], [[124, 127]], [[4, 54]], [[4, 54]], [[2, 54]], [[55, 79]], [[55, 79]], [[80, 90]], [[91, 127]], [[129, 207]]]",
    "query_spans": "[[[209, 217]]]",
    "process": ""
  },
  {
    "text": "The horizontal coordinate of the midpoint of a chord passing through the focus of the parabola $y^{2}=4 x$ is $4$. Then the length of this chord is?",
    "fact_expressions": "G: Parabola;f: LineSegment;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), f);IsChordOf(f, G);XCoordinate(MidPoint(f)) = 4",
    "query_expressions": "Length(f)",
    "answer_expressions": "10",
    "fact_spans": "[[[1, 15]], [], [[1, 15]], [[0, 19]], [[0, 19]], [[0, 30]]]",
    "query_spans": "[[[0, 37]]]",
    "process": ""
  },
  {
    "text": "Given the parabola equation $x^{2}=y$, what are the coordinates of its focus?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = y)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0,1/4)",
    "fact_spans": "[[[2, 5], [19, 20]], [[2, 17]]]",
    "query_spans": "[[[19, 26]]]",
    "process": "The parabola $x^{2}=y$ is a parabola opening upwards with vertex at the origin, and since $2p=1$, it follows that $\\frac{p}{2}=\\frac{1}{4}$, so the focus coordinates are $(0,\\frac{1}{4})$."
  },
  {
    "text": "The distance from a point $P$ on the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ to the point $(5,0)$ is $10$. Then what is the distance from $P$ to the line $x=\\frac{16}{5}$?",
    "fact_expressions": "G: Hyperbola;H: Line;F: Point;P: Point;Expression(G) = (x^2/16 - y^2/9 = 1);Expression(H) = (x = 16/5);Coordinate(F)=(5,0);PointOnCurve(P, G);Distance(P,F) = 10",
    "query_expressions": "Distance(P, H)",
    "answer_expressions": "8",
    "fact_spans": "[[[0, 39]], [[69, 87]], [[46, 54]], [[41, 45], [65, 68]], [[0, 39]], [[69, 87]], [[46, 54]], [[0, 45]], [[41, 62]]]",
    "query_spans": "[[[65, 91]]]",
    "process": "In the hyperbola \\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1, a=4, b=3, c=\\sqrt{16+9}=5, eccentricity e=\\frac{c}{a}=\\frac{5}{4}, the right focus is (5,0), the right directrix is x=\\frac{a^{2}}{c}=\\frac{16}{5}. The distance from P to the point (5,0) is 10. For points on the hyperbola, the ratio of the distance to the focus and the distance to the corresponding directrix equals the eccentricity. Therefore, the distance from P to the line x=\\frac{16}{5} is \\frac{10}{e}=10\\times\\frac{4}{5}=8."
  },
  {
    "text": "The standard equation of a hyperbola with focal length $10$ and conjugate axis length $8$ is?",
    "fact_expressions": "G: Hyperbola;FocalLength(G) = 10;Length(ImageinaryAxis(G)) = 8",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{(x^2/9-y^2/16=1),(y^2/9-x^2/16=1)}",
    "fact_spans": "[[[16, 19]], [[0, 19]], [[8, 19]]]",
    "query_spans": "[[[16, 26]]]",
    "process": "From the given conditions, we have 2c=10, 2b=8, ∴ c=5, b=4, a^{2}=9, so the hyperbola equation is \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1 or \\frac{y^{2}}{9}-\\frac{x^{2}}{16}=1"
  },
  {
    "text": "Given that the asymptotes of a hyperbola are $y=\\pm \\sqrt{2} x$, and it passes through the point $(1,2)$, then the standard equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (1, 2);Expression(Asymptote(G)) = (y = pm*(sqrt(2)*x));PointOnCurve(H, G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/2 - x^2 = 1",
    "fact_spans": "[[[2, 5], [43, 46]], [[33, 41]], [[33, 41]], [[2, 29]], [[2, 41]]]",
    "query_spans": "[[[43, 52]]]",
    "process": "The asymptotes of the hyperbola are given by $ y = \\pm\\sqrt{2}x $. Thus, we can assume the equation of the hyperbola as $ y^2 - 2x^{2} = t $ ($ t \\neq 0 $). Substituting $ (1, 2) $, we get $ t = 4 - 2 = 2 $. Therefore, the equation of the hyperbola is $ \\frac{y^{2}}{3} - x^{2} = 1 $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$ and directrix $l$, let $P$ be a fixed point on $l$. The line $PF$ intersects $C$ at points $A$ and $B$. If $\\overrightarrow{PA}=3\\overrightarrow{AF}$, then $|AB|=$?",
    "fact_expressions": "C: Parabola;P: Point;F: Point;A: Point;B: Point;l: Line;Expression(C) = (y^2 = 8*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(P, l);Intersection(LineOf(P,F), C) = {A, B};VectorOf(P, A) = 3*VectorOf(A, F)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "9",
    "fact_spans": "[[[2, 21], [59, 62]], [[37, 40]], [[25, 28]], [[63, 66]], [[67, 70]], [[32, 35], [41, 44]], [[2, 21]], [[2, 28]], [[2, 35]], [[37, 50]], [[51, 72]], [[74, 119]]]",
    "query_spans": "[[[121, 130]]]",
    "process": "Draw the graph. From the parabola equation, obtain the coordinates of the focus and the length of EF (the distance from the focus to the directrix). By the definition of scalar multiplication of vectors, obtain the x-coordinate of point A, and thus its y-coordinate. Then derive the equation of line AB, substitute into the parabola equation to find the x-coordinate of point B, and finally use the focal chord length formula to reach the conclusion. [Detailed Solution] F(2,0). As shown in the figure, let the directrix l intersect the x-axis at point E, then EF=4. \\because PA=3AF, \\therefore x_{A}=1, y=2\\sqrt{2}, k_{AB}=k_{AF}=-2\\sqrt{2}, AB: y=-2\\sqrt{2}(x-2). Substituting into the equation of C and simplifying gives x^{2}-5x+4=0, x_{B}=4, |AB|=x_{A}+x_{B}+4=9."
  },
  {
    "text": "The foci of the ellipse $\\frac{x^{2}}{a^{2}}+y^{2}=1(a>0)$ lie on the $x$-axis, and the length of the major axis is twice the length of the minor axis. Then $a$=?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2 + x^2/a^2 = 1);a: Number;a>0;PointOnCurve(Focus(G), xAxis) = True;Length(MajorAxis(G)) = 2*Length(MinorAxis(G))",
    "query_expressions": "a",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 36]], [[0, 36]], [[60, 63]], [[2, 36]], [[0, 45]], [[0, 58]]]",
    "query_spans": "[[[60, 65]]]",
    "process": ""
  },
  {
    "text": "Given the equation of the parabola is $y^{2}=4 x$. A line $l$ passing through the fixed point $P(-2,-1)$ intersects the parabola $y^{2}=4 x$ at exactly one common point. Then, what is the set of possible values for the slope of line $l$?",
    "fact_expressions": "l: Line;G: Parabola;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(P) = (-2, -1);PointOnCurve(P,l);NumIntersection(l,G)=1",
    "query_expressions": "Range(Slope(l))",
    "answer_expressions": "{-1/2, 0, 1}",
    "fact_spans": "[[[35, 40], [67, 72]], [[2, 5], [41, 55]], [[24, 34]], [[2, 20]], [[24, 34]], [[21, 40]], [[35, 64]]]",
    "query_spans": "[[[67, 82]]]",
    "process": "According to the problem, assume the equation of the line is: $ y = k(x+2) - 1 $. Substituting into the parabola equation and simplifying yields $ k^{2}x^{2} + (4k^{2} - 2k - 4)x + 4k^{2} - 4k + 1 = 0 $ (*). The line and the parabola have only one common point if and only if (*) has exactly one root.  \n① When $ k = 0 $, $ y = 1 $ satisfies the condition;  \n② When $ k \\neq 0 $, $ \\Delta = (4k^{2} - 2k - 4)^{2} - 4k^{2}(4k^{2} - 4k + 1) = 0 $. Simplifying gives $ k = -\\frac{1}{2} $ or $ k = 1 $.  \nIn conclusion, when $ k = -\\frac{1}{2} $, $ k = 1 $, or $ k = 0 $, the line $ l $ and the parabola have only one common point."
  },
  {
    "text": "Given that the eccentricity of the hyperbola $a x^{2}-4 y^{2}=1$ is $\\sqrt{3}$, then the value of the real number $a$ is?",
    "fact_expressions": "G: Hyperbola;a: Real;Expression(G) = (a*x^2 - 4*y^2 = 1);Eccentricity(G) = sqrt(3)",
    "query_expressions": "a",
    "answer_expressions": "8",
    "fact_spans": "[[[2, 24]], [[41, 46]], [[2, 24]], [[2, 39]]]",
    "query_spans": "[[[41, 50]]]",
    "process": "ax^{2}-4y^{2}=1\\Rightarrow\\frac{x^{2}}{a}-\\frac{y^{2}}{4}=1, so \\frac{\\frac{1}{a}+\\frac{1}{4}}{\\frac{1}{a}}=\\sqrt{3}, solving gives a=8"
  },
  {
    "text": "Given that $AB$ is a chord passing through the focus of the parabola $2x^2 = y$, if $|AB| = 4$, then the y-coordinate of the midpoint of $AB$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (2*x^2 = y);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);PointOnCurve(Focus(G), LineSegmentOf(A, B));Abs(LineSegmentOf(A, B)) = 4",
    "query_expressions": "YCoordinate(MidPoint(LineSegmentOf(A, B)))",
    "answer_expressions": "15/8",
    "fact_spans": "[[[9, 23]], [[9, 23]], [[2, 7]], [[2, 7]], [[2, 28]], [[2, 28]], [[30, 39]]]",
    "query_spans": "[[[41, 55]]]",
    "process": "[Analysis] Let A(x_{1},y_{1}), B(x_{2},y_{2}). According to the definition of the parabola, |AB| = y_{1} + y_{2} + p, derive y_{1} + y_{2}, and finally obtain the vertical coordinate of the midpoint of AB from \\frac{y_{1}+y_{2}}{2}. Let A(x_{1},y_{1}), B(x_{2},y_{2}). From the parabola 2x^{2} = y, we get p = \\frac{1}{4}. Since |AB| = y_{1} + y_{2} + p = 4, then y_{1} + y_{2} = 4 - \\frac{1}{4} = \\frac{15}{4}. Therefore, the vertical coordinate of the midpoint of AB is \\frac{y_{1}+y_{2}}{2} = \\frac{15}{8}."
  },
  {
    "text": "What are the coordinates of the focus of the parabola $y=a x^{2}(a \\neq 0)$?",
    "fact_expressions": "G: Parabola;a: Number;Expression(G) = (y = a*x^2);Negation(a = 0)",
    "query_expressions": "Coordinate(Focus(G))",
    "answer_expressions": "(0, 1/(4*a))",
    "fact_spans": "[[[0, 24]], [[3, 24]], [[0, 24]], [[3, 24]]]",
    "query_spans": "[[[0, 31]]]",
    "process": ""
  },
  {
    "text": "Given that $F_{1}(-c, 0)$ is the left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and the line $y=k x$ intersects the hyperbola at points $A$ and $B$. If $|\\overrightarrow{A F_{1}}|=\\frac{c}{a}|\\overrightarrow{B F_{1}}|$, then the range of the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F1: Point;c: Number;LeftFocus(G) = F1;Coordinate(F1) = (-c, 0);H: Line;Expression(H) = (y = k*x);k: Number;A: Point;B: Point;Intersection(H, G) = {A, B};Abs(VectorOf(A, F1)) = (c/a)*Abs(VectorOf(B, F1))",
    "query_expressions": "Range(Eccentricity(G))",
    "answer_expressions": "(1, \\sqrt{2}+1]",
    "fact_spans": "[[[17, 73], [88, 91], [172, 175]], [[17, 73]], [[20, 73]], [[20, 73]], [[20, 73]], [[20, 73]], [[2, 16]], [[2, 16]], [[2, 77]], [[2, 16]], [[78, 87]], [[78, 87]], [[80, 87]], [[93, 96]], [[97, 100]], [[78, 102]], [[104, 170]]]",
    "query_spans": "[[[172, 186]]]",
    "process": ""
  },
  {
    "text": "Draw a line through the focus of the parabola $y=\\frac{1}{4} x^{2}$ intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $y_{1}+y_{2}=2 \\sqrt{2}$, then the length of the chord $|A B|$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = (1/4)*x^2);H: Line;PointOnCurve(Focus(G), H);x1: Number;x2: Number;y1: Number;y2: Number;A: Point;B: Point;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Intersection(H, G) = {A, B};y1 + y2 = 2*sqrt(2);IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "2+2*sqrt(2)",
    "fact_spans": "[[[1, 25], [32, 35]], [[1, 25]], [[29, 31]], [[0, 31]], [[36, 53]], [[56, 73]], [[36, 53]], [[56, 73]], [[36, 53]], [[56, 73]], [[36, 53]], [[56, 73]], [[29, 75]], [[77, 101]], [[32, 112]]]",
    "query_spans": "[[[105, 116]]]",
    "process": "The parabola $ y = \\frac{1}{4}x^{2} $, that is $ x^{2} = 4y $, opens upward, $ p = 2 $, the chord length $ |AB| = y_{1} + y_{2} + p = 2\\sqrt{2} + 2 $,"
  },
  {
    "text": "Given that point $F_{1}$ is the left focus of the hyperbola $C$: $\\frac{x^{2}}{4}-y^{2}=1$, and a line $l$ passing through the origin intersects the hyperbola $C$ at points $P$ and $Q$. If $|P F_{1}|=3$, then $|Q F_{1}|=$?",
    "fact_expressions": "F1: Point;LeftFocus(C) = F1;C: Hyperbola;Expression(C) = (x^2/4 - y^2 = 1);l: Line;O: Origin;PointOnCurve(O, l);P: Point;Q: Point;Intersection(l, C) = {P, Q};Abs(LineSegmentOf(P, F1)) = 3",
    "query_expressions": "Abs(LineSegmentOf(Q, F1))",
    "answer_expressions": "7",
    "fact_spans": "[[[2, 10]], [[2, 48]], [[11, 44], [59, 65]], [[11, 44]], [[53, 58]], [[50, 52]], [[49, 58]], [[68, 71]], [[72, 75]], [[53, 77]], [[80, 93]]]",
    "query_spans": "[[[95, 108]]]",
    "process": "By the symmetry of the hyperbola, we have |OP| = |OQ|, and |OF_{1}| = |OF_{2}|, so quadrilateral F_{1}PF_{2}Q is a parallelogram, hence |PF_{2}| = |QF_{1}|. For the left branch of the hyperbola, as shown in the figure below, by the definition of the hyperbola, |PF_{2}| - |PF_{1}| = 4. Since |PF_{1}| = 3, it follows that |QF_{1}| = |PF_{2}| = 7."
  },
  {
    "text": "Given the parabola $y^{2}=4x$, draw a line $l$ through point $A(1,2)$ intersecting the parabola at another point $B$. Let $Q$ be the midpoint of segment $AB$. Draw a line $l_{1}$ through $Q$ perpendicular to the $y$-axis, intersecting the parabola at point $C$. If point $P$ satisfies $\\overrightarrow{QC}=\\overrightarrow{CP}$, then the minimum value of $|OP|$ is?",
    "fact_expressions": "l: Line;G: Parabola;l1:Line;B: Point;A: Point;Q: Point;C: Point;P: Point;O: Origin;Expression(G) = (y^2 = 4*x);Coordinate(A) = (1, 2);PointOnCurve(A,l);Intersection(l,G)={A,B};MidPoint(LineSegmentOf(A,B))=Q;PointOnCurve(Q,l1);IsPerpendicular(l1,yAxis);Intersection(l1,G)=C;VectorOf(Q, C) = VectorOf(C, P)",
    "query_expressions": "Min(Abs(LineSegmentOf(O, P)))",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[28, 33]], [[2, 16], [34, 37], [84, 87]], [[73, 82]], [[41, 44]], [[18, 27]], [[45, 48], [61, 64]], [[88, 92]], [[94, 98]], [[145, 152]], [[2, 16]], [[18, 27]], [[17, 33]], [[18, 44]], [[45, 59]], [[60, 82]], [[65, 82]], [[73, 92]], [[100, 143]]]",
    "query_spans": "[[[145, 158]]]",
    "process": ""
  },
  {
    "text": "Given that one focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ coincides with the focus of the parabola $y^{2}=4x$, and the line passing through the point $M(-\\frac{1}{2}, \\frac{1}{2})$ with slope $\\frac{1}{2}$ intersects the ellipse $C$ at points $A$ and $B$. If $M$ is the midpoint of segment $AB$, then the equation of the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;OneOf(Focus(C)) = Focus(G);G: Parabola;Expression(G) = (y^2 = 4*x);H: Line;M: Point;Coordinate(M) = (-1/2, 1/2);PointOnCurve(M, H) ;Slope(H) = 1/2;Intersection(H, C) = {A, B};B: Point;A: Point;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/2 + y^2 = 1",
    "fact_spans": "[[[2, 59], [138, 143], [171, 176]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[2, 84]], [[65, 79]], [[65, 79]], [[135, 137]], [[86, 117], [155, 158]], [[86, 117]], [[85, 137]], [[118, 137]], [[135, 153]], [[148, 151]], [[144, 147]], [[155, 169]]]",
    "query_spans": "[[[171, 181]]]",
    "process": "According to the problem, the focus of the parabola $ y^{2} = 4x $ is $ (1,0) $, so the foci of the ellipse $ C $ lie on the x-axis, and $ c = 1 $. The standard equation of this ellipse can be written as: $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $, then $ a^{2} - b^{2} = 1 $, $\\textcircled{1}$ Let the coordinates of point A be $ (x_{1}, y_{1}) $, and the coordinates of point B be $ (x_{2}, y_{2}) $, we have $ \\frac{x_{1}^{2}}{a^{2}} + \\frac{y_{1}^{2}}{b^{2}} = 1 $ $\\textcircled{2}$, $ \\frac{x_{2}^{2}}{a^{2}} + \\frac{y_{2}^{2}}{b^{2}} = 1 $ $\\textcircled{3}$ Subtracting $\\textcircled{3}$ from $\\textcircled{2}$ gives: $ \\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{a^{2}} + \\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{b^{2}} = 0 $ $\\textcircled{4}$ Since the slope of line AB is $ \\frac{1}{2} $, then $ \\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = \\frac{1}{2} $. The midpoint M of AB has coordinates $ M(-\\frac{1}{2}, \\frac{1}{2}) $, so $ x_{1}+x_{2} = -1 $, $ y_{1}+y_{2} = 1 $, substituting into $\\textcircled{4}$, we get $ \\frac{1}{a^{2}} = \\frac{1}{2b^{2}} $. Also, since $ a^{2} - b^{2} = 1 $, then $ a^{2} = 2 $, $ b^{2} = 1 $. Therefore, the required standard equation of the ellipse is: $ \\frac{x^{2}}{2} + y^{2} = 1 $;"
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$ with foci $F_{1}$, $F_{2}$, and a point $P(x_{0},y_{0})$ satisfying $0<\\frac{x_{0}^{2}}{2}+y_{0}^{2}<1$, then the range of $|P F_{1}|+|P F_{2}|$ is? The number of common points between the line $\\frac{x_{0} x}{2}+y_{0} y=1$ and the ellipse $C$ is?",
    "fact_expressions": "C: Ellipse;G: Line;y0: Number;x0: Number;P: Point;F1: Point;F2: Point;Expression(C)=(x^2/2 + y^2 = 1);Expression(G)=((x0*x)/2+y0*y=1);Focus(C)={F1,F2};Coordinate(P)=(x0,y0);0<x0^2/2+y0^2;x0^2/2+y0^2<1",
    "query_expressions": "Range(Abs(LineSegmentOf(P, F1)) + (LineSegmentOf(P, F2)));NumIntersection(G, C)",
    "answer_expressions": "[2,2*sqrt(2))\n0",
    "fact_spans": "[[[2, 34], [171, 176]], [[139, 170]], [[56, 72]], [[56, 72]], [[55, 72]], [[39, 46]], [[47, 54]], [[2, 34]], [[139, 170]], [[2, 54]], [[55, 72]], [[74, 109]], [[74, 109]]]",
    "query_spans": "[[[111, 139]], [[139, 184]]]",
    "process": ""
  },
  {
    "text": "Given that point $P$ lies on the ellipse $\\frac{x^{2}}{45}+\\frac{y^{2}}{20}=1$ with foci $F_{1}$ and $F_{2}$, if $\\angle F_{1} P F_{2}=90^{\\circ}$, then the value of $|P F_{1}| \\cdot|P F_{2}|$ is?",
    "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/45 + y^2/20 = 1);AngleOf(F1, P, F2) = ApplyUnit(90, degree);Focus(G)={F1,F2};PointOnCurve(P,G)",
    "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "40",
    "fact_spans": "[[[26, 65]], [[10, 17]], [[2, 6]], [[18, 25]], [[26, 65]], [[68, 101]], [[7, 65]], [[2, 66]]]",
    "query_spans": "[[[103, 133]]]",
    "process": "In the ellipse $\\frac{x^2}{45}+\\frac{y^{2}}{20}=1$, $a=3\\sqrt{5}$, $b=2\\sqrt{5}$, then $c=\\sqrt{a^{2}-b^{2}}=5$, $|F_{1}F_{2}|=2c=10$. By the definition of the ellipse, we have $|PF_{1}|+|PF_{2}|=6\\sqrt{5}$. Since $\\angle F_{1}PF_{2}=90^{\\circ}$, then $|PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}=100$, so, $|PF_{1}|\\cdot|PF_{2}|=\\underline{(|PF_{1}|+|PF_{2}|)}\\frac{-(|PF_{1}|^{2}+|PF_{2}|^{2})}{2}=40$"
  },
  {
    "text": "It is known that the ellipse $C$ passes through the points $M(1, \\frac{3}{2})$ and $N(\\sqrt{3}, \\frac{\\sqrt{3}}{2})$. What is its standard equation?",
    "fact_expressions": "C: Ellipse;M: Point;N: Point;Coordinate(M) = (1, 3/2);Coordinate(N) = (sqrt(3), sqrt(3)/2);PointOnCurve(M,C);PointOnCurve(N,C)",
    "query_expressions": "Expression(C)",
    "answer_expressions": "x^2/4 + y^2/3 = 1",
    "fact_spans": "[[[2, 7], [66, 67]], [[9, 29]], [[30, 64]], [[9, 29]], [[30, 64]], [[2, 29]], [[2, 64]]]",
    "query_spans": "[[[66, 73]]]",
    "process": "Let the equation of the ellipse $ C $ be $ mx^{2} + ny^{2} = 1 $ ($ m, n > 0 $ and $ m \\neq n $), then  \n$$\n\\begin{cases}\nm + \\frac{9}{4}n = 1 \\\\\n3m + \\frac{3}{4}n = 1\n\\end{cases}\n$$\nSolving gives  \n$$\n\\begin{cases}\nm = \\frac{1}{4} \\\\\nn = \\frac{1}{3}\n\\end{cases}\n$$\nThus, its standard equation is $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $. Hence, fill in $ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $."
  },
  {
    "text": "Given the parabola $y^{2}=2 p x(p>0)$, let the endpoints of chord $AB$ passing through the focus be $A(x_{1} , y_{1})$ and $B(x_{2}, y_{2})$. Is the expression $\\frac{y_{1} y_{2}}{x_{1} x_{2}}$ a constant value?",
    "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;x1:Number;x2:Number;y1:Number;y2:Number;p>0;Expression(G) = (y^2 = 2*(p*x));IsChordOf(LineSegmentOf(A, B), G);PointOnCurve(Focus(G), LineSegmentOf(A, B));Endpoint(LineSegmentOf(A, B))={A,B};Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2)",
    "query_expressions": "(y1*y2)/(x1*x2)",
    "answer_expressions": "-4",
    "fact_spans": "[[[2, 23]], [[5, 23]], [[37, 55]], [[58, 75]], [[80, 113]], [[80, 113]], [[80, 113]], [[80, 113]], [[5, 23]], [[2, 23]], [[2, 33]], [[2, 33]], [[27, 75]], [[37, 55]], [[58, 75]]]",
    "query_spans": "[[[80, 117]]]",
    "process": ""
  },
  {
    "text": "Given that one focus of the hyperbola $x^{2}-k y^{2}=1$ is $(\\sqrt{5}, 0)$, then its asymptotes are?",
    "fact_expressions": "G: Hyperbola;k: Number;F:Point;OneOf(Focus(G))=F;Coordinate(F)=(sqrt(5),0);Expression(G) = (-k*y^2 + x^2 = 1)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[2, 22], [45, 46]], [[5, 22]], [[28, 43]], [[2, 43]], [[28, 43]], [[2, 22]]]",
    "query_spans": "[[[45, 53]]]",
    "process": "From the given, we know $ c=\\sqrt{5} $, $ a^{2}=1 $, $ b^{2}=\\frac{1}{k} $ $ (k>0) $, $ \\therefore 1+\\frac{1}{k}=5 $, solving gives $ k=\\frac{1}{4} $. Let $ x^{2}-\\frac{1}{1}v2=0 $, the asymptotes are $ y=\\pm2x $. Final answer: \\_."
  },
  {
    "text": "The standard equation of the ellipse passing through points $P(-3,0)$, $Q(0,-2)$ is?",
    "fact_expressions": "G: Ellipse;P: Point;Q: Point;Coordinate(P) = (-3,0);Coordinate(Q) = (0,-2);PointOnCurve(P,G);PointOnCurve(Q,G)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9+y^2/4=1",
    "fact_spans": "[[[25, 27]], [[2, 12]], [[15, 24]], [[2, 12]], [[15, 24]], [[0, 27]], [[0, 27]]]",
    "query_spans": "[[[25, 34]]]",
    "process": "First, assume the ellipse equation is \\frac{x^{2}}{m}+\\frac{y^{2}}{n}=1 (m>0,n>0,m\\neqn). Substitute the known points and solve to obtain the result. Let the ellipse equation be \\frac{x^{2}}{m}+\\frac{y^{2}}{n}=1 (m>0,n>0,m\\neqn). Since the ellipse passes through point P(-3,0) and Q(0,-2), we have \\begin{cases}\\frac{9}{m}=1\\\\\\frac{4}{n}=1\\end{cases}, solving gives \\begin{cases}m=9\\\\n=4\\end{cases}, thus the required ellipse equation is \\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1"
  },
  {
    "text": "Given that $F_{1}(-c, 0)$, $F_{2}(c, 0)$ are the two foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and point $P$ lies on the ellipse such that the area of $\\triangle P F_{1} F_{2}$ is $\\frac{\\sqrt{2}}{2} b^{2}$, then $\\cos \\angle F_{1} P F_{2} =$?",
    "fact_expressions": "G: Ellipse;a: Number;b: Number;F1: Point;P: Point;F2: Point;c:Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);Focus(G) = {F1,F2} ;PointOnCurve(P, G);Area(TriangleOf(P, F1, F2)) = b^2*(sqrt(2)/2)",
    "query_expressions": "Cos(AngleOf(F1, P, F2))",
    "answer_expressions": "1/3",
    "fact_spans": "[[[33, 85], [95, 97]], [[35, 85]], [[35, 85]], [[2, 16]], [[91, 94]], [[18, 32]], [[2, 17]], [[35, 85]], [[35, 85]], [[33, 85]], [[2, 16]], [[18, 32]], [[2, 90]], [[91, 98]], [[100, 155]]]",
    "query_spans": "[[[157, 187]]]",
    "process": "From the given conditions, using the definition of an ellipse and the law of cosines, set up a system of equations; then, using the area of the triangle and the law of sines, obtain $1\\cdot\\cos\\theta=\\sqrt{2}\\sin\\theta$. Using $\\sin^{2}\\theta+\\cos^{2}\\theta=1$, $\\cos\\theta$ can be found. Since $F_{1}(-c,0)$, $F_{2}(c,0)$ are the two foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and point $P$ lies on the ellipse with the area of $\\triangle PF_{1}F_{2}$ being $\\frac{\\sqrt{2}}{2}b^{2}$, we have\n$$\n\\begin{cases}\n|PF_{1}|^{2}+|PF_{2}|^{2}+2|PF_{1}|\\cdot|PF_{2}|=4a^{2} \\\\\n|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}|\\cdot|PF_{2}|\\cos\\angle F_{1}PF_{2}=4c^{2}\n\\end{cases}\n$$\nRearranging, we get $|PF_{1}|+|PF_{2}|$. $1-\\cos\\theta=\\sqrt{2}\\sin\\theta$. Since $\\sin^{2}\\theta+\\cos^{2}\\theta=1$, it follows that $\\cos\\theta=\\frac{1}{3}$."
  },
  {
    "text": "The point $P$ on the parabola $y^{2}=2x$ satisfies the condition that its distance to the line $x - y + 3 = 0$ is minimized. Then the coordinates of point $P$ are?",
    "fact_expressions": "G: Parabola;H: Line;P: Point;Expression(G) = (y^2 = 2*x);Expression(H) = (x - y + 3 = 0);PointOnCurve(P, G);WhenMin(Distance(P, H))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1/2, 1)",
    "fact_spans": "[[[1, 15]], [[24, 35]], [[17, 21], [42, 46]], [[1, 15]], [[24, 35]], [[1, 21]], [[17, 40]]]",
    "query_spans": "[[[42, 51]]]",
    "process": "Let $ P(x_{0},y_{0}) $ be any point on $ y^{2}=2x $, then the distance $ d $ from point $ P $ to the line $ / $ is $ d=\\frac{|x_{0}-y_{0}+3|}{\\sqrt{2}}=\\frac{|\\frac{y_{0}^{2}}{2}-y_{0}+3|}{\\sqrt{2}}=\\frac{|(y_{0}-1)^{2}+5|}{2\\sqrt{2}} $. When $ y_{0}=1 $, $ d_{\\min}=\\frac{5\\sqrt{2}}{4} $, $ \\therefore P(\\frac{1}{2},1) $."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through point $F_{2}$ and perpendicular to the $x$-axis intersects the hyperbola at a point $P$, and $\\angle P F_{1} F_{2}=\\frac{\\pi}{6}$. Then the asymptotes of the hyperbola are?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Line;PointOnCurve(F2, H);IsPerpendicular(H, xAxis) ;OneOf(Intersection(H, G)) = P;P: Point;AngleOf(P, F1, F2) = pi/6",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y=pm*sqrt(2)*x",
    "fact_spans": "[[[2, 59], [105, 108], [156, 159]], [[2, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[68, 75]], [[76, 83], [85, 93]], [[2, 83]], [[2, 83]], [[102, 104]], [[84, 104]], [[94, 104]], [[102, 116]], [[113, 116]], [[118, 154]]]",
    "query_spans": "[[[156, 167]]]",
    "process": ""
  },
  {
    "text": "Given that a vertex and a focus of an ellipse are the intersection points of the line $x+3y-6=0$ with the two coordinate axes, then the standard equation of the ellipse is?",
    "fact_expressions": "G: Ellipse;F: Point;P: Point;H: Line;Expression(H) = (x + 3*y - 6 = 0);Intersection(H, axis) = {F, P};OneOf(Vertex(G)) = F;OneOf(Focus(G)) = P",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{(x^2/36+y^2/40=1),(x^2/40+y^2/4=1)}",
    "fact_spans": "[[[2, 4], [40, 42]], [], [], [[17, 30]], [[17, 30]], [[2, 38]], [[2, 9]], [[2, 14]]]",
    "query_spans": "[[[40, 49]]]",
    "process": ""
  },
  {
    "text": "Let the focus of the parabola $y^{2}=2 p x(p>0)$ be $F$, the directrix be $l$, and point $A(0,2)$. The line segment $F A$ intersects the parabola at point $B$. Draw a perpendicular from $B$ to $l$, with foot $M$. If $A M \\perp M F$, then $p=$?",
    "fact_expressions": "G: Parabola;p: Number;F: Point;A: Point;M: Point;l: Line;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(A) = (0, 2);Focus(G) = F;Directrix(G) = l;Intersection(LineSegmentOf(F,A), G) = B;L:Line;PointOnCurve(B, L);IsPerpendicular(l,L);FootPoint(l,L) = M;IsPerpendicular(LineSegmentOf(A, M), LineSegmentOf(M, F))",
    "query_expressions": "p",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[1, 22], [55, 58]], [[102, 105]], [[26, 29]], [[37, 46]], [[80, 83]], [[33, 36]], [[66, 69], [60, 64]], [[4, 22]], [[1, 22]], [[37, 46]], [[1, 29]], [[1, 36]], [[47, 64]], [], [[65, 76]], [[65, 76]], [[65, 83]], [[85, 100]]]",
    "query_spans": "[[[102, 107]]]",
    "process": ""
  },
  {
    "text": "Given that the equation of hyperbola $C$ is $x^{2}-y^{2}=1$, then the distance from the right focus of $C$ to its asymptote is?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2 = 1)",
    "query_expressions": "Distance(RightFocus(C), Asymptote(C))",
    "answer_expressions": "1",
    "fact_spans": "[[[2, 8], [29, 32], [37, 38]], [[2, 27]]]",
    "query_spans": "[[[29, 47]]]",
    "process": "Find the foci and asymptotes of the hyperbola, then use the point-to-line distance formula to solve. From the problem: the coordinates of the foci are $(-\\sqrt{2},0)$, $(\\sqrt{2},0)$; the equations of the asymptotes are $y = \\pm x$, so the distance from a focus to an asymptote is $d = \\frac{|\\pm\\sqrt{2}|}{\\sqrt{1^{2}+(\\pm1)^{2}}} = 1$."
  },
  {
    "text": "The left vertex of hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $A$, one endpoint of the imaginary axis is $B$, and the distance from the right focus $F$ to the line $AB$ is $\\frac{3 b}{2}$. Then the eccentricity of hyperbola $E$ is?",
    "fact_expressions": "E: Hyperbola;b: Number;a: Number;B: Point;A: Point;F: Point;a>0;b>0;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);LeftVertex(E) = A;OneOf(Endpoint(ImageinaryAxis(E)))=B;RightFocus(E)=F;Distance(F, LineOf(A,B)) = (3*b)/2",
    "query_expressions": "Eccentricity(E)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 61], [117, 123]], [[8, 61]], [[8, 61]], [[78, 81]], [[66, 69]], [[85, 88]], [[8, 61]], [[8, 61]], [[0, 61]], [[0, 69]], [[0, 81]], [[0, 88]], [[85, 115]]]",
    "query_spans": "[[[117, 129]]]",
    "process": "Assume point B lies on the positive y-axis. From the standard equation of the hyperbola, we have: A(-a,0), B(0,b), and the coordinates of the right focus F are (c,0). The equation of line AB is: \\frac{x}{-a}+\\frac{y}{b}=1 \\Rightarrow bx-ay+ab=0. Since the distance from the right focus F to line AB is \\frac{3b}{2}, it follows that \\frac{|bc+ab|}{\\sqrt{b^{2}+(-a)^{2}}}=\\frac{3b}{2} \\Rightarrow \\frac{a+c}{c}=\\frac{3}{2} \\Rightarrow \\frac{a}{c}=\\frac{1}{2} \\Rightarrow \\frac{c}{a}=2, which means the eccentricity of the hyperbola is 2."
  },
  {
    "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $\\frac{\\sqrt{10}}{2}$, then the slope of its asymptotes is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = sqrt(10)/2",
    "query_expressions": "Slope(Asymptote(G))",
    "answer_expressions": "pm*sqrt(6)/2",
    "fact_spans": "[[[0, 56], [84, 85]], [[3, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[0, 56]], [[0, 82]]]",
    "query_spans": "[[[84, 93]]]",
    "process": "\\because_{e}=\\frac{\\sqrt{10}}{2}, and e=\\frac{c}{a}, \\therefore 4c^{2}=10a^{2}, 4a^{2}+4b^{2}=10a^{2}, i.e., 4b^{2}=6a^{2} \\therefore k=\\pm\\frac{b}{a}=\\pm\\frac{\\sqrt{6}}{2}"
  },
  {
    "text": "Given two points $A$ and $B$ on the parabola $y^{2}=4x$ with focus $F$ such that $\\overrightarrow{A F}=3 \\overrightarrow{F B}$, then the distance from the midpoint of chord $AB$ to the directrix is?",
    "fact_expressions": "G: Parabola;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(A, G);PointOnCurve(B, G);VectorOf(A, F) = 3*VectorOf(F, B);IsChordOf(LineSegmentOf(A, B), G)",
    "query_expressions": "Distance(MidPoint(LineSegmentOf(A,B)),Directrix(G))",
    "answer_expressions": "8/3",
    "fact_spans": "[[[10, 24]], [[28, 32]], [[33, 37]], [[3, 6]], [[10, 24]], [[2, 24]], [[10, 37]], [[10, 37]], [[39, 84]], [[10, 91]]]",
    "query_spans": "[[[10, 102]]]",
    "process": ""
  },
  {
    "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, respectively, $P$ be any point on the ellipse, and the coordinates of point $M$ be $(6 , 4)$. Then the maximum value of $|P M|+|P F_{1}|$ is?",
    "fact_expressions": "G: Ellipse;P: Point;M: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/16 = 1);Coordinate(M) = (6, 4);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G)",
    "query_expressions": "Max(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "15",
    "fact_spans": "[[[19, 58], [69, 71]], [[65, 68]], [[76, 80]], [[1, 8]], [[9, 16]], [[19, 58]], [[76, 93]], [[1, 64]], [[1, 64]], [[65, 75]]]",
    "query_spans": "[[[95, 118]]]",
    "process": ""
  },
  {
    "text": "Given that the foci of the ellipse $\\frac{x^{2}}{m-2}+\\frac{y^{2}}{20-m}=1$ lie on the $x$-axis and the focal distance is $4$, then $m$=?",
    "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/(m - 2) + y^2/(20 - m) = 1);PointOnCurve(Focus(G),xAxis);FocalLength(G) = 4",
    "query_expressions": "m",
    "answer_expressions": "13",
    "fact_spans": "[[[2, 44]], [[62, 65]], [[2, 44]], [[2, 52]], [[2, 60]]]",
    "query_spans": "[[[62, 67]]]",
    "process": "\\because the ellipse \\frac{x^2}{m-2} + \\frac{y^{2}}{20-m} = 1 has its major axis on the x-axis, \\therefore \\begin{cases} 20 - m > 0 \\\\ m - 2 > 0 \\\\ m - 2 > 20 - m \\end{cases} solving gives 11 < m < 20, \\because the focal distance is 4, \\therefore c^{2} = m - 2 - 20 + m = 4, solving gives m = 13."
  },
  {
    "text": "The line passing through the left vertex $A$ of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with slope $k$ intersects the ellipse $C$ at another point $B$, and the projection of point $B$ onto the $x$-axis is exactly the right focus $F$. If $k=\\frac{1}{2}$, then the eccentricity $e$ of the ellipse is?",
    "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;B: Point;F: Point;e: Number;k: Number;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(C) = A;A: Point;PointOnCurve(A, G);Slope(G) = k;Intersection(C, G) = {A, B};Projection(B, xAxis) = F;RightFocus(C) = F;k = 1/2;Eccentricity(C) = e",
    "query_expressions": "e",
    "answer_expressions": "(1/2, 2/3)",
    "fact_spans": "[[[1, 58], [76, 81], [130, 132]], [[8, 58]], [[8, 58]], [[73, 75]], [[85, 88], [90, 94]], [[108, 111]], [[136, 139]], [[69, 72]], [[8, 58]], [[8, 58]], [[1, 58]], [[1, 65]], [[62, 65]], [[0, 75]], [[66, 75]], [[62, 88]], [[90, 111]], [[76, 111]], [[113, 128]], [[130, 139]]]",
    "query_spans": "[[[136, 143]]]",
    "process": ""
  },
  {
    "text": "It is known that the center of ellipse $G$ is at the origin of the coordinate system, the major axis lies on the $x$-axis, the eccentricity is $\\frac{\\sqrt{3}}{2}$, and the sum of the distances from a point on $G$ to the two foci of $G$ is $12$. Then the equation of ellipse $G$ is?",
    "fact_expressions": "G: Ellipse;O:Origin;Center(G)=O;OverlappingLine(MajorAxis(G), xAxis);Eccentricity(G)=sqrt(3)/2;P:Point;F1:Point;F2:Point;PointOnCurve(P,G);Focus(G)={F1,F2};Distance(P,F1)+Distance(P,F2) = 12",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/36+y^2/9=1",
    "fact_spans": "[[[2, 7], [59, 62], [52, 55], [79, 84]], [[11, 15]], [[2, 15]], [[2, 24]], [[2, 50]], [], [], [], [[52, 58]], [[59, 67]], [[52, 77]]]",
    "query_spans": "[[[79, 89]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola $y^{2}=2 p x(p<0)$ is $F$, the directrix is $l$. A line passing through point $H(-\\frac{p}{2}, 0)$ is tangent to a circle centered at $F$ and passing through the origin, with point of tangency $N$. The line $F N$ intersects the directrix $l$ at point $M$, and intersects the parabola at points $A$ and $B$ ($A$ lies between $M$ and $N$). Then $\\frac{|M N|}{|A N|}$=?",
    "fact_expressions": "G:Parabola;p:Number;p<0;Expression(G) = (y^2 = 2*(p*x));F:Point;Focus(G)=F;l:Line;Directrix(G)=l;H:Point;Coordinate(H)=(-p/2,0);Z:Line;PointOnCurve(H,Z);C:Circle;Center(C)=F;O:Origin;PointOnCurve(O,C);N:Point;TangentPoint(Z,C)=N;M:Point;A:Point;B:Point;Intersection(LineOf(F,N),l)=M;Intersection(LineOf(F,N),G)={A,B};Between(A,M,N)",
    "query_expressions": "Abs(LineSegmentOf(M, N))/Abs(LineSegmentOf(A, N))",
    "answer_expressions": "9",
    "fact_spans": "[[[0, 21], [103, 106]], [[3, 21]], [[3, 21]], [[0, 21]], [[25, 28], [63, 66]], [[0, 28]], [[32, 35], [91, 96]], [[0, 35]], [[37, 58]], [[37, 58]], [[59, 61]], [[36, 61]], [[74, 75]], [[62, 75]], [[71, 73]], [[70, 75]], [[126, 129], [78, 82]], [[59, 82]], [[97, 101], [122, 125]], [[107, 110], [118, 121]], [[111, 114]], [[83, 101]], [[83, 117]], [[118, 131]]]",
    "query_spans": "[[[134, 157]]]",
    "process": "First, a line passing through point $ H(-\\frac{p}{2},0) $ is tangent to a circle centered at $ F $ and passing through the origin at point $ N $. The line $ FN $ intersects line $ l $ at point $ M $. Find the length of $ MN $, then solve the system of equations of line $ MN $ and the parabola to find the coordinates of point $ A $, determine the length of $ AM $, and then find the length of $ AN $, thus obtaining the result.  \n**Solution:** From the given conditions, we have $ F(\\frac{p}{2},0) $. Since a line passing through point $ H(-\\frac{p}{2},0) $ is tangent to the circle centered at $ F $ and passing through the origin at point $ N $, it follows that $ |FN| = -\\frac{p}{2} $, $ |HF| = -p $. Therefore, in the right triangle $ FNH $, we get $ |HN| = \\sqrt{|HF|^2 - |FN|^2} = \\frac{\\sqrt{3}}{2}p $, $ \\angle HFN = 60^{\\circ} $; hence, the equation of line $ MN $ is $ y = \\sqrt{3}(x - \\frac{p}{2}) $. Since line $ FN $ intersects line $ l $ at point $ M $, we have $ |MF| = 2|HF| = -2p $, thus $ |MN| = |MF| - |NF| = -\\frac{3}{2}p $. Solving the system:  \n$$\n\\begin{cases}\ny = \\sqrt{3}(x - \\frac{p}{2}) \\\\\ny^2 = 2px\n\\end{cases}\n$$\nwe obtain $ 3(x - \\frac{p}{2})^2 = 2px $, which simplifies to $ 12x^2 - 20px + 3p^2 = 0 $. Solving gives $ x = \\frac{p}{6} $ or $ x = \\frac{3p}{2} $. Since $ A $ lies between $ M $ and $ N $ and $ p < 0 $, we have $ x_A = \\frac{p}{6} $, thus $ y_A = -\\frac{\\sqrt{3}}{3}p $, i.e., $ A(\\frac{p}{6}, -\\frac{\\sqrt{3}}{3}p) $. Also, $ M(-\\frac{p}{2}, -\\sqrt{3}p) $, so  \n$$\n|AM| = \\sqrt{ \\left( \\frac{p}{6} + \\frac{p}{2} \\right)^2 + \\left( -\\frac{\\sqrt{3}}{3}p + \\sqrt{3}p \\right)^2 } = -\\frac{4}{3}p\n$$\nTherefore, $ |AN| = |MN| - |AM| = -\\frac{p}{6} $, so  \n$$\n\\frac{|MN|}{|AN|} = \\frac{-\\frac{3}{2}p}{-\\frac{p}{6}} = 9\n$$"
  },
  {
    "text": "The hyperbola passing through the point $P(2 , 1)$ shares foci with the ellipse $\\frac{x^{2}}{4}+y^{2}=1$. What is its asymptote equation?",
    "fact_expressions": "G: Hyperbola;P: Point;Coordinate(P) = (2, 1);PointOnCurve(P, G);H: Ellipse;Expression(H) = (x^2/4 + y^2 = 1);Focus(G) = Focus(H)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "Preserve all mathematical expressions exactly (including LaTeX format, symbols, and numbers).  \nDo not add explanations.",
    "fact_spans": "[[[13, 16], [49, 50]], [[1, 12]], [[1, 12]], [[0, 16]], [[17, 44]], [[17, 44]], [[13, 47]]]",
    "query_spans": "[[[49, 57]]]",
    "process": ""
  },
  {
    "text": "If the equation $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{a+6}=1$ represents an ellipse with foci on the $x$-axis, then the range of real values for $a$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/(a + 6) + x^2/a^2 = 1);a: Real;PointOnCurve(Focus(G), xAxis)",
    "query_expressions": "Range(a)",
    "answer_expressions": "{(3, +oo), (-6, -2)}",
    "fact_spans": "[[[55, 57]], [[1, 57]], [[59, 64]], [[46, 57]]]",
    "query_spans": "[[[59, 71]]]",
    "process": "\\because the equation \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{a+6}=1 represents an ellipse with foci on the x-axis, \\therefore a^{2}>a+6>0, \\therefore a>3 or -6<a<-2."
  },
  {
    "text": "Given that the moving point $M(x, y)$ satisfies $5 \\sqrt{(x-1)^{2}+(y-2)^{2}}=|3 x+4 y+12 |$, what is the trajectory curve of point $M$?",
    "fact_expressions": "M: Point;x_:Number;y_:Number;Coordinate(M) = (x_, y_);5*sqrt((x_ - 1)^2 + (y_ - 2)^2) = Abs(3*x_ + 4*y_ + 12)",
    "query_expressions": "Locus(M)",
    "answer_expressions": "Parabola",
    "fact_spans": "[[[4, 13], [61, 65]], [[4, 13]], [[4, 13]], [[4, 13]], [[15, 59]]]",
    "query_spans": "[[[61, 72]]]",
    "process": ""
  },
  {
    "text": "Let the parabola $ C $: $ y^{2} = 2 p x $ ($ p > 0 $) have focus $ F $, and let point $ M $ lie on the parabola $ C $ such that $ |M F| = 5 $. If there exists a point $ A(0,2) $ on the $ y $-axis such that $ A M \\perp A F $, then the value of $ p $ is?",
    "fact_expressions": "C: Parabola;p: Number;A: Point;M: Point;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(A) = (0, 2);Focus(C) = F;PointOnCurve(M, C);Abs(LineSegmentOf(M, F)) = 5;PointOnCurve(A, yAxis);IsPerpendicular(LineSegmentOf(A, M), LineSegmentOf(A, F))",
    "query_expressions": "p",
    "answer_expressions": "{2,8}",
    "fact_spans": "[[[1, 27], [40, 46]], [[95, 98]], [[66, 75]], [[35, 39]], [[31, 34]], [[9, 27]], [[1, 27]], [[66, 75]], [[1, 34]], [[35, 47]], [[48, 57]], [[59, 75]], [[78, 93]]]",
    "query_spans": "[[[95, 102]]]",
    "process": "From the given conditions: the circle with diameter MF passes through the point (0,2). Let M(x,y). By the property of the parabola, |MF| = x + \\frac{p}{2} = 5, so x = 5 - \\frac{p}{2}. Since the center of the circle is the midpoint of MF, using the midpoint coordinate formula, the x-coordinate of the center is \\underline{5 - \\frac{p}{2} + \\frac{p}{2}}{2} = \\frac{5}{2}. It is known that the radius of the circle is also \\frac{5}{2}, so the circle is tangent to the y-axis at the point (0,2). Therefore, the y-coordinate of the center is 2, and thus the y-coordinate of M is 4, i.e., M(5 - \\frac{p}{2}, 4). Substituting into the parabolic equation yields p^{2} - 10p + 16 = 0, so p = 2 or p = 8."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the right branch of the hyperbola. When the eccentricity of the hyperbola is $3$, the maximum value of $\\frac{|P F_{2}|}{|P F_{1}|^{2}}$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/b^2 = 1);b: Number;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, RightPart(G));Eccentricity(G) = 3",
    "query_expressions": "Max(Abs(LineSegmentOf(P, F2))/Abs(LineSegmentOf(P, F1))^2)",
    "answer_expressions": "1/16",
    "fact_spans": "[[[2, 49], [80, 83], [89, 92]], [[2, 49]], [[5, 49]], [[5, 49]], [[59, 66]], [[67, 74]], [[2, 74]], [[2, 74]], [[75, 79]], [[75, 87]], [[89, 100]]]",
    "query_spans": "[[[102, 141]]]",
    "process": "According to the problem, we have $ a = 2 $, thus obtain $ c = 6 $. Then, by the definition of the hyperbola, $ |PF_{1}| = |PF_{2}| + 4 $, so $ \\frac{|PF_{2}|}{|PF_{1}|^{2}} = \\frac{1}{|PF_{2}| + \\frac{16}{|PF_{2}|} + 8} $. The result can then be calculated using the basic inequality. [Detailed solution] Given that the eccentricity of the hyperbola is 3, and $ a = 2 $, we have $ c = 6 $. From the definition of the hyperbola, $ |PF_{1}| - |PF_{2}| = 4 $, so $ |PF_{1}| = |PF_{2}| + 4 $, and $ |PF_{2}| \\geqslant c - a = 4 $. $ \\frac{1}{16} $ is achieved if and only if $ |PF_{2}| = 4 $. Therefore, the maximum value of $ \\frac{|PF_{2}|}{|PF_{1}|^{2}} $ is $ \\frac{1}{16} $."
  },
  {
    "text": "If the equation $\\frac{x^{2}}{m+3}+\\frac{y^{2}}{4-m}=1$ represents a hyperbola with foci on the $x$-axis, then the range of values for $m$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(m + 3) + y^2/(4 - m) = 1);PointOnCurve(Focus(G), xAxis);m: Number",
    "query_expressions": "Range(m)",
    "answer_expressions": "(4, +oo)",
    "fact_spans": "[[[53, 56]], [[1, 56]], [[44, 56]], [[58, 61]]]",
    "query_spans": "[[[58, 68]]]",
    "process": "Preserve the sum and $\\Square$. According to the problem, we obtain $\\begin{cases}m+3>0\\\\4-m<0\\end{cases}$, solving gives the answer. From the problem, we get $\\begin{cases}4-m<0\\\\m+3>0\\\\4-m<0\\end{cases}$, solving gives $m>4$."
  },
  {
    "text": "Given the hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1  (a>0 , b>0)$, with left and right foci $F_{1}$ and $F_{2}$ respectively, and the parabola $C_{2}$ whose vertex is at the origin and whose directrix coincides with the left directrix of the hyperbola $C_{1}$. If the intersection point $P$ of the hyperbola $C_{1}$ and the parabola $C_{2}$ satisfies $PF_{2} \\perp F_{1} F_{2}$, then the eccentricity of the hyperbola $C_{1}$ is?",
    "fact_expressions": "C1: Hyperbola;Expression(C1) = (x^2/a^2 - y^2/b^2 = 1);a: Number;b: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C1) = F1;RightFocus(C1) = F2;O: Origin;C2: Parabola;Vertex(C2) = O;Directrix(C2) = LeftDirectrix(C1);P: Point;Intersection(C1, C2) = P;IsPerpendicular(LineSegmentOf(P,F2), LineSegmentOf(F1, F2)) = True",
    "query_expressions": "Eccentricity(C1)",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 70], [119, 129], [137, 147], [194, 204]], [[2, 70]], [[14, 70]], [[14, 70]], [[14, 70]], [[14, 70]], [[79, 86]], [[89, 96]], [[2, 96]], [[2, 96]], [[111, 113]], [[97, 107], [114, 115], [148, 158]], [[97, 113]], [[114, 135]], [[161, 164]], [[137, 164]], [[166, 192]]]",
    "query_spans": "[[[194, 210]]]",
    "process": ""
  },
  {
    "text": "A triangle is formed by connecting one endpoint of the minor axis of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ and its two foci. If the radius of the incircle of this triangle is $\\frac{b}{5}$, then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;b: Number;a: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);P: Point;F1: Point;F2: Point;OneOf(Endpoint(MinorAxis(G))) = P;Focus(G) = {F1, F2};Radius(InscribedCircle(TriangleOf(P, F1, F2))) = b/5",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "1/4",
    "fact_spans": "[[[0, 52], [101, 103]], [[2, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[0, 52]], [], [], [], [[0, 59]], [[0, 64]], [[0, 98]]]",
    "query_spans": "[[[101, 109]]]",
    "process": "The triangle formed by one endpoint of the minor axis and the two foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has area $3.2a+2c$. From the given condition, we have $S=bc=\\frac{1}{2}(2a+2c)\\cdot\\frac{b}{5}$, yielding $a+c=5c$, so $e=\\frac{c}{a}=\\frac{1}{4}$. Therefore, the eccentricity of the ellipse is $\\frac{1}{4}$."
  },
  {
    "text": "The eccentricity of the hyperbola $9 y^{2}-16 x^{2}=144$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-16*x^2 + 9*y^2 = 144)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "5/4",
    "fact_spans": "[[[0, 25]], [[0, 25]]]",
    "query_spans": "[[[0, 31]]]",
    "process": ""
  },
  {
    "text": "The line passing through the center $ M_0 $ of the circle $ M $: $ (x+1)^{2} + y^{2} = \\frac{7}{9} $ intersects the parabola $ C $: $ y^{2} = 4x $ at points $ A $ and $ B $, and $ \\overrightarrow{MB} = 3 \\overrightarrow{MA} $. Then the minimum distance from point $ A $ to any point on the circle $ M $ is?",
    "fact_expressions": "C: Parabola;M: Circle;G: Line;B: Point;A: Point;P0: Point;M0: Point;Expression(C) = (y^2 = 4*x);Expression(M) = (y^2 + (x + 1)^2 = 7/9);Center(M) = M0;PointOnCurve(M0, G);Intersection(G, C) = {A, B};VectorOf(M, B) = 3*VectorOf(M, A);PointOnCurve(P0, M)",
    "query_expressions": "Min(Distance(A, P0))",
    "answer_expressions": "sqrt(7)/3",
    "fact_spans": "[[[46, 65]], [[1, 35], [131, 135]], [[43, 45]], [[72, 75]], [[68, 71], [126, 130]], [], [[38, 42]], [[46, 65]], [[1, 35]], [[1, 42]], [[0, 45]], [[43, 77]], [[79, 124]], [[131, 140]]]",
    "query_spans": "[[[126, 149]]]",
    "process": "Let $ A\\left(\\frac{y_{2}}{4}, y_{1}\\right), B\\left(\\frac{y_{2}}{4}, y_{2}\\right) $, from the problem we have \n\\[\n\\begin{cases}\ny_{2} = 3y_{1} \\\\\nk_{MA} = k_{MB}\n\\end{cases}\n\\]\nWithout loss of generality, assume $ y_{1} > 0 $, so $ y_{1} = \\frac{2}{3}\\sqrt{3} $, thus $ A\\left(\\frac{1}{3}, \\frac{2}{3}\\sqrt{3}\\right) $, so $ MA = \\frac{+1}{1-(-1)} + \\frac{y_{2}}{y_{2}} + \\frac{2}{3} - \\frac{2}{3}\\sqrt{3} $. Therefore, the minimum distance from point $ A $ to any point on circle $ M $ is $ \\frac{2}{3}\\sqrt{7} - r = \\frac{2}{3}\\sqrt{7} - \\frac{1}{3}\\sqrt{7} = \\frac{1}{3}\\sqrt{7} $. Fill in the number $ \\frac{\\sqrt{7}}{3} $."
  },
  {
    "text": "Given that the hyperbola $\\frac{y^{2}}{2}-\\frac{x^{2}}{m}=1$ passes through the point $M(2,2)$, its eccentricity $e$=?",
    "fact_expressions": "G: Hyperbola;m: Number;M: Point;e:Number;Expression(G) = (y^2/2 - x^2/m = 1);Coordinate(M) = (2, 2);PointOnCurve(M, G);Eccentricity(G)=e",
    "query_expressions": "e",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[2, 40], [53, 54]], [[5, 40]], [[42, 51]], [[57, 60]], [[2, 40]], [[42, 51]], [[2, 51]], [[53, 60]]]",
    "query_spans": "[[[57, 62]]]",
    "process": "From point M lying on the hyperbola, we obtain \\frac{2^{2}}{2}-\\frac{2}{m}=1, solving gives m=4, so the equation of the hyperbola is \\frac{y^{2}}{2}-\\frac{x^{2}}{4}=1. Hence a=\\sqrt{2}, c=\\sqrt{2+4}=\\sqrt{6}, so e=\\frac{c}{a}=\\frac{\\sqrt{6}}{\\sqrt{2}}=\\sqrt{3}."
  },
  {
    "text": "Given the parabola $C$: $x^{2}=12 y$, a line $l$ passes through the point $(0,3)$ and intersects the parabola $C$ at points $A$ and $B$, with $|A B|=14$. Then the sine of the inclination angle $\\alpha$ of the line $l$ is?",
    "fact_expressions": "l: Line;C: Parabola;G: Point;A: Point;B: Point;Expression(C) = (x^2 = 12*y);Coordinate(G) = (0, 3);PointOnCurve(G,l);Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, B)) = 14;Inclination(l)=alpha;alpha:Number",
    "query_expressions": "Sin(alpha)",
    "answer_expressions": "sqrt(7)/7",
    "fact_spans": "[[[22, 27], [68, 73]], [[2, 21], [37, 43]], [[28, 36]], [[45, 48]], [[49, 52]], [[2, 21]], [[28, 36]], [[22, 36]], [[22, 54]], [[56, 66]], [[68, 84]], [[76, 84]]]",
    "query_spans": "[[[76, 90]]]",
    "process": "From the given conditions, the slope of line $ l $ exists. When the slope of the line is zero, since $ (0,3) $ is the focus of the parabola, it should follow that $ |AB| = 12 $. Therefore, the slope of the line exists and is not zero. Let the equation of line $ l $ be $ y = kx + 3 $ ($ k \\neq 0 $). From \n\\[\n\\begin{cases}\ny = kx + 3 \\\\\nx^{2} = 12y\n\\end{cases}\n\\]\neliminating $ x $, we obtain $ y^{2} - (12k^{2} + 6)y + 9 = 0 $. Thus, $ y_{1} + y_{2} = 12k^{2} + 6 $, so $ |AB| = y_{1} + y_{2} + 6 = 12k^{2} + 12 = 14 $, hence $ k = \\pm\\frac{\\sqrt{6}}{6} $, so $ \\tan\\alpha = \\pm\\frac{\\sqrt{6}}{6} $. Since $ \\alpha \\in [0,\\pi) $, it follows that $ \\sin\\alpha = \\frac{\\sqrt{7}}{7} $."
  },
  {
    "text": "Given that the directrix of the parabola $y = a x^{2}$ is $y = -2$, what is the value of the real number $a$?",
    "fact_expressions": "G: Parabola;a: Real;Expression(G) = (y = a*x^2);Expression(Directrix(G)) = (y = -2)",
    "query_expressions": "a",
    "answer_expressions": "1/8",
    "fact_spans": "[[[2, 16]], [[30, 35]], [[2, 16]], [[2, 28]]]",
    "query_spans": "[[[30, 39]]]",
    "process": "Transform $ y = ax^2 $ into $ x^{2} = \\frac{1}{a}y' $. From the given condition, we have $ \\frac{1}{4a} = 2 $, which implies $ a = \\frac{1}{8} $."
  },
  {
    "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{k+4}+\\frac{y^{2}}{12}=1$ is $\\frac{1}{2}$, then what is the value of the real number $k$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(k + 4) + y^2/12 = 1);k: Real;Eccentricity(G) = 1/2",
    "query_expressions": "k",
    "answer_expressions": "{5, 12}",
    "fact_spans": "[[[1, 41]], [[1, 41]], [[62, 67]], [[1, 59]]]",
    "query_spans": "[[[62, 71]]]",
    "process": "If the foci of the ellipse $\\frac{x^{2}}{k+4} + \\frac{y^{2}}{12} = 1$ lie on the $x$-axis, then $\\frac{k+4-12}{k+4} =$ so $k=12$; if the foci of the ellipse $\\frac{x^{2}}{k+4} + \\frac{y^{2}}{12} = 1$ lie on the $y$-axis, then $\\frac{12-k-4}{12} = \\frac{1}{4}$. So $k=5$. Therefore, the real value of $k$ is $5$ or $12$."
  },
  {
    "text": "The left focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ is $F$, and the line $x=m$ intersects the ellipse at points $A$ and $B$. When the perimeter of $\\triangle FAB$ is maximized, what is the area of $\\triangle F A B$?",
    "fact_expressions": "G: Ellipse;H: Line;m: Number;F: Point;A: Point;B: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Expression(H) = (x = m);LeftFocus(G) = F;Intersection(H, G) = {A, B};WhenMax(Perimeter(TriangleOf(F,A,B)))",
    "query_expressions": "Area(TriangleOf(F, A, B))",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 37], [54, 56]], [[46, 53]], [[48, 53]], [[42, 45]], [[59, 63]], [[64, 67]], [[0, 37]], [[46, 53]], [[0, 45]], [[46, 67]], [[68, 90]]]",
    "query_spans": "[[[91, 113]]]",
    "process": ""
  },
  {
    "text": "Given point $A(0 , b)$, and point $B$ is the intersection of the left directrix of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1  (a>b>0)$ with the $x$-axis. If the midpoint $C$ of segment $AB$ lies on the ellipse, then the eccentricity of the ellipse is?",
    "fact_expressions": "A: Point;B: Point;G: Ellipse;b: Number;a: Number;C: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (0, b);Intersection(LeftDirectrix(G), xAxis) = B;MidPoint(LineSegmentOf(A, B))=C;PointOnCurve(C, G)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[2, 13]], [[14, 17]], [[18, 72], [99, 101], [105, 107]], [[20, 72]], [[20, 72]], [[95, 98]], [[20, 72]], [[20, 72]], [[18, 72]], [[2, 13]], [[14, 84]], [[86, 98]], [[95, 102]]]",
    "query_spans": "[[[105, 113]]]",
    "process": ""
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=12 x$ is $F$, and a line $l$ passing through point $P(2 , 1)$ intersects the parabola at points $A$ and $B$, with point $P$ being exactly the midpoint of segment $A B$, then $|A F|+| B F |$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 12*x);F: Point;Focus(G) = F;P: Point;Coordinate(P) = (2, 1);l: Line;PointOnCurve(P, l);A: Point;B: Point;Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = P",
    "query_expressions": "Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F))",
    "answer_expressions": "10",
    "fact_spans": "[[[2, 17], [45, 48]], [[2, 17]], [[21, 24]], [[2, 24]], [[26, 37], [61, 65]], [[26, 37]], [[38, 43]], [[25, 43]], [[50, 53]], [[54, 57]], [[38, 59]], [[61, 78]]]",
    "query_spans": "[[[80, 97]]]",
    "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since P(2,1) is the midpoint of AB, \\therefore \\frac{x_{1}+x_{2}}{2}=2, i.e., x_{1}+x_{2}=4. Since F(3,0) is the focus of the parabola y^{2}=12x, \\therefore |AF|=x_{1}+3, |BF|=x_{2}+3, then |AF|+|BF|=x_{1}+x_{2}+3+3=10,"
  },
  {
    "text": "If the line $y - kx - 1 = 0$ ($k \\in \\mathbb{R}$) always has common points with the ellipse $\\frac{x^2}{5} + \\frac{y^2}{m} = 1$, then the range of values for $m$ is?",
    "fact_expressions": "H: Line;Expression(H) = (-k*x + y - 1 = 0);k: Real;G: Ellipse;Expression(G) = (x^2/5 + y^2/m = 1);m: Number;IsIntersect(H, G)",
    "query_expressions": "Range(m)",
    "answer_expressions": "[1, 5) + (5, +oo)",
    "fact_spans": "[[[1, 23]], [[1, 23]], [[3, 23]], [[24, 61]], [[24, 61]], [[68, 71]], [[1, 66]]]",
    "query_spans": "[[[68, 78]]]",
    "process": "Rearranging the line equation shows that the line always passes through the point (0,1). Therefore, it is sufficient to ensure that the point (0,1) lies inside or on the ellipse. Letting $ x = 0 $, we obtain $ y^{2} = m $. To ensure that the point (0,1) lies inside or on the ellipse, we require $ y \\geqslant 1 $, which implies $ y^{2} \\geqslant 1 $, thus obtaining $ m \\geqslant 1 $. Noting that in the ellipse equation $ m \\neq 5 $, combining these results gives the range of $ m $ as $ [1,5) \\cup (5,+\\infty) $. Rearranging the line equation yields $ y - 1 = kx $, hence the line always passes through (0,1). Therefore, it suffices to ensure that the point (0,1) lies inside or on the ellipse. Since this point lies on the $ y $-axis and the ellipse is symmetric about the origin, we only need to set $ x = 0 $, giving $ 5y^{2} = 5m $, so $ y^{2} = m $. To ensure that the point (0,1) lies inside or on the ellipse, we require $ y \\geqslant 1 $, i.e., $ y^{2} \\geqslant 1 $, yielding $ m \\geqslant 1 $. Since $ m \\neq 5 $ in the ellipse equation, the range of $ m $ is $ [1,5) \\cup (5,+\\infty) $."
  },
  {
    "text": "Given the parabola equation: $x=\\frac{1}{4} y^{2}$, what is its directrix equation?",
    "fact_expressions": "G: Parabola;Expression(G) = (x = y^2/4)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=-1",
    "fact_spans": "[[[2, 5], [31, 32]], [[2, 30]]]",
    "query_spans": "[[[31, 38]]]",
    "process": "The parabola equation is: $x=\\frac{1}{4}y^{2}$, simplifying yields $y^{2}=4x$, then $2p=4 \\Rightarrow p=2$, so the corresponding directrix of the parabola is $x=-\\frac{p}{2}=-1$."
  },
  {
    "text": "Given the parabola $y=\\frac{1}{4} x^{2}$ and the point $M(2,2)$, a line passing through $M$ intersects the parabola at points $A$ and $B$. The tangents to the parabola at points $A$ and $B$, denoted $l_{1}$ and $l_{2}$, intersect at point $P$. If $M$ is the midpoint of segment $AB$, then the area of $\\triangle A B P$ is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = (1/4)*x^2);M: Point;Coordinate(M) = (2, 2);H: Line;PointOnCurve(M, H);A: Point;B: Point;Intersection(H, G) = {A, B};l1: Line;l2: Line;TangentOnPoint(A, G) = l1;TangentOnPoint(B, G) = l2;P: Point;Intersection(l1, l2) = P;MidPoint(LineSegmentOf(A, B)) = M",
    "query_expressions": "Area(TriangleOf(A, B, P))",
    "answer_expressions": "4",
    "fact_spans": "[[[2, 26], [45, 48], [59, 62]], [[2, 26]], [[27, 36], [38, 41], [99, 102]], [[27, 36]], [[42, 44]], [[37, 44]], [[49, 52], [63, 67]], [[53, 56], [68, 71]], [[42, 58]], [[75, 82]], [[84, 91]], [[59, 91]], [[59, 91]], [[93, 97]], [[75, 97]], [[99, 113]]]",
    "query_spans": "[[[115, 137]]]",
    "process": "Let the equation of line AB be $ y = kx + 2 - 2k $, and by solving it together with $ y = \\frac{1}{4}x^2 $, we obtain $ x^{2} - 4kx + 8k - 8 = 0 $. Let $ A(x_{1},\\frac{x_{1}^{2}}{4}) $, $ B(x_{2},\\frac{x_{2}^{2}}{4}) $, then $ x_{1} + x_{2} = 4k $. Since M is the midpoint of segment AB, we have $ 4k = 4 $, so $ k = 1 $. Solving the equation $ x^{2} - 4kx + 8k - 8 = 0 $ gives $ x_{1} = 0 $, $ x_{2} = 4 $, thus $ A(0,0) $, $ B(4,4) $. From $ y = \\frac{1}{2}x $, the slopes of the tangent lines $ l_{1} $, $ l_{2} $ are 0 and 2 respectively. Therefore, the equations of tangents $ l_{1} $, $ l_{2} $ are $ y = 0 $, $ y = 2(x - 4) + 4 $, respectively. Solving them simultaneously yields $ P(2,0) $. The distance from point P to line AB: $ y = x $ is $ d = \\frac{|2 - 0|}{\\sqrt{1^{2} + 1^{2}}} = \\sqrt{2} $, $ |AB| = \\sqrt{4^{2} + 4^{2}} = 4\\sqrt{2} $, then $ S_{\\Delta ABP} = \\frac{1}{2}|AB|\\cdot d = \\frac{1}{2} \\times 4\\sqrt{2} \\times \\sqrt{2} = 4 $."
  },
  {
    "text": "If one asymptote of the hyperbola $C$: $\\frac{y^{2}}{m}-x^{2}=1$ $(m>0)$ is $x+\\sqrt{3} y=0$, then $m$=?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (-x^2 + y^2/m = 1);m: Number;m>0;Expression(OneOf(Asymptote(C))) = (x + sqrt(3)*y = 0)",
    "query_expressions": "m",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 39]], [[1, 39]], [[64, 67]], [[8, 39]], [[1, 62]]]",
    "query_spans": "[[[64, 69]]]",
    "process": "Since $\\frac{y^{2}}{m}-x^{2}=1$ $(m>0)$, its asymptotes are given by $y=\\pm\\sqrt{m}x$. Given that one of its asymptotes is $x+\\sqrt{3}y=0$, it follows that $m=3$."
  },
  {
    "text": "Let $P$ be a point on the right branch of the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$, $F$ be the left focus of the hyperbola $C$, and point $A(0,3)$. Then the minimum value of $|P A|+|P F|$ is?",
    "fact_expressions": "C: Hyperbola;A: Point;P: Point;F: Point;Coordinate(A) = (0, 3);LeftFocus(C) = F;Expression(C)=(x^2/4-y^2/3=1);PointOnCurve(P,RightPart(C))",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "8",
    "fact_spans": "[[[4, 47], [57, 63]], [[68, 77]], [[0, 3]], [[53, 56]], [[68, 77]], [[53, 67]], [[4, 47]], [[0, 52]]]",
    "query_spans": "[[[78, 98]]]",
    "process": "\\because F is the left focus of the hyperbola C: \\frac{x^{2}}{4} - \\frac{y^{2}}{3} = 1, \\therefore a = 2, b = 2\\sqrt{3}, c = 4, F(-\\sqrt{7}, 0), the right focus is H(\\sqrt{7}, 0). By the definition of the hyperbola, we have |PA| + |PF| = 2a + |PH| + |PA| \\geqslant 2a + |AH| = 4 + 5 = 9. Therefore, the answer should be filled as: 8."
  },
  {
    "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, a line $l$ passing through point $F_{1}$ with slope $2$ intersects the left branch of the hyperbola at point $P$. If the line $P F_{2} \\perp l$, then the equation of the asymptotes of the hyperbola is?",
    "fact_expressions": "l: Line;G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, l);Slope(l) = 2;Intersection(l, LeftPart(G)) = P;IsPerpendicular(LineOf(P, F2), l)",
    "query_expressions": "Expression(Asymptote(G))",
    "answer_expressions": "y = pm*2*x",
    "fact_spans": "[[[97, 102]], [[18, 74], [103, 106], [139, 142]], [[21, 74]], [[21, 74]], [[110, 114]], [[2, 9], [81, 89]], [[10, 17]], [[21, 74]], [[21, 74]], [[18, 74]], [[2, 80]], [[2, 80]], [[80, 102]], [[90, 102]], [[97, 114]], [[117, 136]]]",
    "query_spans": "[[[139, 150]]]",
    "process": "By the given condition, the equation of line $ l $ passing through point $ F_{1} $ with slope 2 is $ y = 2(x + c) $. Since $ PF_{2} \\perp l $, the slope of line $ PF_{1} $ is $ -\\frac{1}{2} $, so the equation of line $ PF_{1} $ is $ y = -\\frac{1}{2}(x - c) $. Solving the system of equations of the two lines, the coordinates of intersection point $ P $ are found to be $ \\left(-\\frac{3c}{5}, \\frac{4c}{5}\\right) $, as shown in the figure. Substituting point $ P $ into the hyperbola's equation yields $ \\underline{\\left(-\\frac{3c}{5}\\right)} - \\frac{\\left(\\frac{4c}{5}\\right)^{2}}{b^{2}} = 1 $. Simplifying gives $ 9b^{2}c^{2} - 16a^{2}c^{2} = 25a^{2}b^{2} $. Using $ b^{2} = c^{2} - a^{2} $ and substituting, we get $ 9(c^{2} - a^{2})c^{2} - 16a^{2}c^{2} = 25a^{2}(c^{2} - a^{2}) $. Simplifying yields $ 9c^{4} - 50a^{2}c^{2} + 25a^{4} = 0 $, solving which gives $ c^{2} = 5a^{2} $, thus $ b^{2} = 4a^{2} $, i.e., $ b = 2a $. Therefore, the equations of the asymptotes of the hyperbola are $ y = \\pm 2x $."
  },
  {
    "text": "If the right focus of the hyperbola $x^{2}-y^{2}=a^{2}(a>0)$ coincides with the focus of the parabola $y^{2}=4 x$, then $a$=?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = a^2);a>0;a: Number;H: Parabola;Expression(H) = (y^2 = 4*x);RightFocus(G) = Focus(H)",
    "query_expressions": "a",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[1, 28]], [[1, 28]], [[4, 28]], [[54, 57]], [[33, 47]], [[33, 47]], [[1, 52]]]",
    "query_spans": "[[[54, 59]]]",
    "process": "The right focus of the hyperbola \\( x^{2} - y^{2} = a^{2} \\) (\\( a > 0 \\)) is \\( (\\sqrt{2}a, 0) \\). The focus of the parabola \\( y^{2} = 4x \\) is \\( (1, 0) \\). Therefore, \\( \\sqrt{2}a = 1 \\), \\( a = \\frac{\\sqrt{2}}{2} \\)."
  },
  {
    "text": "If two lines $PA$ and $PB$ are drawn from a point $P(4,4)$ on the parabola $y^{2}=4x$, intersecting the parabola at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$ respectively, and if the sum of their slopes is $0$, then what is the slope of line $AB$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;Coordinate(P) = (4, 4);PointOnCurve(P, G) = True;PointOnCurve(P, LineOf(P, A)) = True;PointOnCurve(P, LineOf(P, B)) = True;Intersection(LineOf(P, A), G) = A;Intersection(LineOf(P, B), G) = B;A: Point;Coordinate(A) = (x1, y1);x1: Number;y1: Number;B: Point;Coordinate(B) = (x2, y2);x2: Number;y2: Number;Slope(LineOf(P, A))+Slope(LineOf(P, B)) = 0",
    "query_expressions": "Slope(LineOf(A, B))",
    "answer_expressions": "-1/2",
    "fact_spans": "[[[2, 16], [49, 52]], [[2, 16]], [[19, 27]], [[19, 27]], [[1, 27]], [[1, 46]], [[1, 46]], [[31, 92]], [[31, 92]], [[54, 71]], [[54, 71]], [[54, 71]], [[54, 71]], [[73, 90]], [[73, 90]], [[73, 90]], [[73, 90]], [[33, 105]]]",
    "query_spans": "[[[107, 118]]]",
    "process": "According to the problem, we have \\frac{y_{1}-4}{x_{1}-4}+\\frac{y_{2}-4}{x_{2}-4}=0, so k_{AB}-\\frac{1}{2},"
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, two tangents are drawn from its right focus $F$ to the circle $x^{2}+y^{2}=9$, with points of tangency denoted as $C$ and $D$. The right vertex of the hyperbola is $E$, and $\\angle C E D=150^{\\circ}$. Then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;H: Circle;C: Point;E: Point;D: Point;F: Point;b>0;Expression(G) = (x^2/9 - y^2/b^2 = 1);Expression(H) = (x^2 + y^2 = 9);L1:Line;L2:Line;TangentOfPoint(F, H) = {L1,L2};RightFocus(G)=F;TangentPoint(L1,H)=C;TangentPoint(L2,H)=D;RightVertex(G)=E;AngleOf(C, E, D) = ApplyUnit(150, degree)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2*sqrt(3)/3",
    "fact_spans": "[[[2, 49], [51, 52], [93, 96], [133, 136]], [[5, 49]], [[59, 75]], [[85, 88]], [[101, 104]], [[89, 92]], [[55, 58]], [[5, 49]], [[2, 49]], [[59, 75]], [], [], [[50, 80]], [[51, 58]], [[50, 92]], [[50, 92]], [[93, 104]], [[105, 131]]]",
    "query_spans": "[[[133, 142]]]",
    "process": ""
  },
  {
    "text": "Point $P$ lies on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, with its right focus at $F_{2}$. If the slope of line $PF_{2}$ is $\\sqrt{3}$, $M$ is the midpoint of segment $PF_{2}$, and $|OF_{2}|=|F_{2}M|$, then the eccentricity of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;P: Point;PointOnCurve(P, RightPart(G));F2: Point;RightFocus(G) = F2;Slope(LineOf(P, F2)) = sqrt(3);M: Point;MidPoint(LineSegmentOf(P, F2)) = M;O: Origin;Abs(LineSegmentOf(O, F2)) = Abs(LineSegmentOf(F2, M))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "(\\sqrt{3}+1)/2",
    "fact_spans": "[[[5, 61], [68, 69], [152, 155]], [[5, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[0, 4]], [[0, 67]], [[73, 80]], [[68, 80]], [[82, 107]], [[108, 111]], [[108, 126]], [[128, 149]], [[128, 149]]]",
    "query_spans": "[[[152, 161]]]",
    "process": "From the given conditions: |OF_{2}| = |F_{2}M| = c, \\angle OF_{2}M = 120^{\\circ}, \\therefore |OM| = \\sqrt{3}c. Let the left focus be F_{1}, connect PF_{1}, then OM is the midline of \\triangle PF_{1}F_{2}, \\therefore |PF_{1}| = 2\\sqrt{3}c, and |PF_{2}| = 2c. By the definition of hyperbola, |PF_{1}| - |PF_{2}| = 2a; 3, eccentricity of the hyperbola. [Approach"
  },
  {
    "text": "Let points $A_{1}$ and $A_{2}$ be the left and right vertices of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. If there exists a point $P$ on the ellipse $C$, distinct from $A_{1}$ and $A_{2}$, such that $P O \\perp P A_{2}$, where $O$ is the origin, then the range of the eccentricity of ellipse $C$ is?",
    "fact_expressions": "A2: Point;A1: Point;C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;LeftVertex(C) = A1;RightVertex(C) = A2;P: Point;PointOnCurve(P, C);Negation(P = A1);Negation(P = A2);IsPerpendicular(LineSegmentOf(P, O), LineSegmentOf(P, A2));O: Origin",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(\\sqrt{2}/2,1)",
    "fact_spans": "[[[10, 17], [104, 111]], [[1, 9], [95, 103]], [[20, 77], [85, 90], [151, 156]], [[20, 77]], [[27, 77]], [[27, 77]], [[27, 77]], [[27, 77]], [[1, 82]], [[1, 82]], [[112, 116]], [[85, 116]], [[93, 116]], [[93, 116]], [[119, 138]], [[141, 144]]]",
    "query_spans": "[[[151, 167]]]",
    "process": "By the given condition, the circle with diameter $ OA_{2} $ intersects the ellipse at a point other than $ A_{2} $. The equation of the circle is $ (x-\\frac{a}{2})^{2}+y^{2}=\\frac{a^{2}}{4} $. From \n$$\n\\begin{cases}\n(x-\\frac{a}{2})^{2}+y^{2}=\\frac{a^{2}}{4} \\\\\n\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1\n\\end{cases}\n$$\nwe obtain $ \\frac{c^{2}}{a^{2}}x^{2}-ax+b^{2}=0 $. One root of this equation is $ a $, and the other root is $ \\frac{ab^{2}}{c^{2}} $. Then $ \\frac{ab^{2}}{c^{2}}<a $, $ a^{2}-c^{2}<c^{2} $, $ \\frac{c^{2}}{a^{2}}>\\frac{1}{2} $, so $ \\frac{\\sqrt{2}}{2}<e<1 $."
  },
  {
    "text": "If a focus of the hyperbola $x^{2}-k y^{2}=1$ is $(3,0)$, then the real number $k$=?",
    "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (-k*y^2 + x^2 = 1);Coordinate(OneOf(Focus(G)))=(3,0)",
    "query_expressions": "k",
    "answer_expressions": "1/8",
    "fact_spans": "[[[1, 21]], [[36, 41]], [[1, 21]], [1, 33]]",
    "query_spans": "[[[36, 43]]]",
    "process": "First, determine the corresponding $ a^{2} $, $ b^{2} $, $ c^{2} $ of the equation according to the given conditions, then solve using their interrelationship. [Detailed Solution] From the problem: $ c=3 $, $ a^{2}=1 $, $ b^{2}=\\frac{1}{k} \\Rightarrow c^{2}=1+\\frac{1}{k}=9 \\Rightarrow k=\\frac{1}{8} $"
  },
  {
    "text": "If the asymptotes of a hyperbola are given by $y = \\pm 2x$, and one of its foci is $(\\sqrt{5}, 0)$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (sqrt(5), 0);Expression(Asymptote(G)) = (y = pm*2*x);OneOf(Focus(G))=H",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2 - y^2/4 = 1",
    "fact_spans": "[[[1, 4], [23, 24], [47, 50]], [[30, 45]], [[30, 45]], [[1, 22]], [[23, 45]]]",
    "query_spans": "[[[47, 55]]]",
    "process": "Since the hyperbola's focus is at $(\\sqrt{5},0)$, the focus of this hyperbola lies on the x-axis. Let the equation of the hyperbola be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=$ Since the asymptotes of the hyperbola are $y=\\pm2x$, then $\\begin{cases}\\frac{b}{a}=2\\\\\\\\a+b^{2}=5\\end{cases}$ Then the equation of the hyperbola is $x^{2}-\\frac{y^{2}}{=1}$."
  },
  {
    "text": "If the sum of the distances from any point on the ellipse $\\frac{x^{2}}{m+6}-\\frac{y^{2}}{m}=1(-6<m<-3)$ to its two foci is $4$, then what is the length of the minor axis of this ellipse?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(m + 6) - y^2/m = 1);m: Number;-6 < m;m < -3;H: Point;PointOnCurve(H, G);F1: Point;F2: Point;Focus(G) = {F1, F2};Distance(H, F1) + Distance(H, F2) = 4",
    "query_expressions": "Length(MinorAxis(G))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[1, 49], [67, 69]], [[1, 49]], [[3, 49]], [[3, 49]], [[3, 49]], [[51, 52]], [[1, 52]], [], [], [[1, 56]], [[1, 64]]]",
    "query_spans": "[[[67, 75]]]",
    "process": "According to the definition of an ellipse and combining with the given conditions, the values of $a$, $b$, and $c$ for the ellipse can be determined. From the problem: $2a = 4$, so $a = 2$. Since $-6 < m < -3$, we have $-m = a^2 = 4$, hence $m = -4$. Then $b^2 = m + 6 = 2$, so $b = \\sqrt{2}$. Therefore, the length of the minor axis is $2b = 2\\sqrt{2}$."
  },
  {
    "text": "A point $P$ on the ellipse $\\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1$ is connected to the two foci $F_{1}$ and $F_{2}$ of the ellipse, and the lines are mutually perpendicular. Then the area of $\\Delta P F_{1} F_{2}$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/49 + y^2/24 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))",
    "query_expressions": "Area(TriangleOf(P, F1, F2))",
    "answer_expressions": "24",
    "fact_spans": "[[[0, 39], [46, 48]], [[0, 39]], [[42, 45]], [[0, 45]], [[53, 60]], [[61, 68]], [[46, 68]], [[42, 75]]]",
    "query_spans": "[[[77, 104]]]",
    "process": "Let |PF_{1}|=m, |PF_{2}|=n. From PF_{1}\\bot PF_{2} and the definition of the ellipse, mn can be found, thus the area of the triangle is easily obtained. For the ellipse \\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1, a=7, b=2\\sqrt{6}, c=\\sqrt{49-24}=5. Let |PF_{1}|=m, |PF_{2}|=n. Since PF_{1}\\bot PF_{2}, then m^{2}+n^{2}=(2c)^{2}=100. Also, m+n=2a=14. {m^{2}+n^{2}=4c^{2}=100 | m+n=2a=14. mn=\\frac{(m+n)^{2}-(m^{2}+n^{2})}{2}=\\frac{14^{2}-100}{2}=48. \\therefore S_{\\Delta PF_{1}F_{2}}=\\frac{1}{2}mn=24."
  },
  {
    "text": "One asymptote of the hyperbola is $y=2x$, and it passes through the point $(4,0)$. What is the equation of the hyperbola?",
    "fact_expressions": "G: Hyperbola;Expression(OneOf(Asymptote(G))) = (y = 2*x);H: Point;Coordinate(H) = (4, 0);PointOnCurve(H, OneOf(Asymptote(G)))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/16 - y^2/64 = 1",
    "fact_spans": "[[[0, 3], [32, 35]], [[0, 19]], [[22, 30]], [[22, 30]], [[0, 30]]]",
    "query_spans": "[[[32, 40]]]",
    "process": "The solution process is omitted"
  },
  {
    "text": "The distance from the focus to the directrix of the parabola $x^{2}=y$ is equal to?",
    "fact_expressions": "G: Parabola;Expression(G) = (x^2 = y)",
    "query_expressions": "Distance(Focus(G), Directrix(G))",
    "answer_expressions": "1/2",
    "fact_spans": "[[[0, 12]], [[0, 12]]]",
    "query_spans": "[[[0, 24]]]",
    "process": "The distance from the focus to the directrix of the parabola \\( x^{2} = y \\) is \\( p \\), and from the standard equation we obtain \\( p = \\frac{1}{2} \\)."
  },
  {
    "text": "Given $F_{1}(-4,0)$, $F_{2}(4,0)$ are the two foci of the hyperbola $C$: $\\frac{x^{2}}{m}-\\frac{y^{2}}{4}=1$ $(m>0)$, and point $M$ is a point on the hyperbola $C$ such that $\\angle F_{1} M F_{2}=60^{\\circ}$, then the area of $\\Delta F_{1} M F_{2}$ is?",
    "fact_expressions": "C: Hyperbola;m: Number;F1: Point;F2: Point;M: Point;m>0;Expression(C) = (-y^2/4 + x^2/m = 1);Coordinate(F1) = (-4, 0);Coordinate(F2) = (4, 0);Focus(C) = {F1, F2};PointOnCurve(M, C);AngleOf(F1, M, F2) = ApplyUnit(60, degree)",
    "query_expressions": "Area(TriangleOf(F1, M, F2))",
    "answer_expressions": "4*sqrt(3)",
    "fact_spans": "[[[31, 79], [90, 96]], [[39, 79]], [[2, 15]], [[18, 30]], [[85, 89]], [[39, 79]], [[31, 79]], [[2, 15]], [[18, 30]], [[2, 84]], [[85, 99]], [[101, 134]]]",
    "query_spans": "[[[136, 163]]]",
    "process": "\\because F_{1}(-4,0), F_{2}(4,0) are the two foci of the hyperbola C: \\frac{x^{2}}{m} - \\frac{y^{2}}{4} = 1 (m > 0), \\therefore m + 4 = 16, m = 12. Let |MF_{1}| = m, |MF_{2}| = n. \\because point M is a point on the hyperbola and \\angle F_{1}MF_{2} = 60^{\\circ}, \\therefore m - n = 4\\sqrt{3} \\textcircled{1}, m^{2} + n^{2} - 2mn\\cos60^{\\circ} = 64 \\textcircled{2}. From \\textcircled{2} - \\textcircled{1}^{2} we get mn = 16, \\therefore the area S of \\triangle F_{1}MF_{2} = \\frac{1}{2}mn\\cos60^{\\circ} = 4\\sqrt{3}"
  },
  {
    "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, and a line passing through $F$ intersects the parabola at points $A$ and $B$. If $\\overrightarrow{AF}=2\\overrightarrow{FB}$, then the coordinates of point $A$ are?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};VectorOf(A, F) = 2*VectorOf(F, B)",
    "query_expressions": "Coordinate(A)",
    "answer_expressions": "(2,pm*2*sqrt(2))",
    "fact_spans": "[[[2, 16], [32, 35]], [[29, 31]], [[36, 39], [94, 98]], [[20, 23], [25, 28]], [[40, 43]], [[2, 16]], [[2, 23]], [[24, 31]], [[29, 45]], [[47, 92]]]",
    "query_spans": "[[[94, 103]]]",
    "process": "Let the equation of line AB be $x = my + 1$. By solving the system of equations of line AB and the parabola simultaneously, write Vieta's formulas. From the condition $\\overrightarrow{AF} = 2\\overrightarrow{FB}$, we obtain $y_{1} = -2y_{2}$. Combining this with Vieta's formulas, the coordinates of point A can be found. According to the problem, the focus of the parabola is $F(1,0)$. Let the equation of line AB be $x = my + 1$. Solving the system of equations:\n$$\n\\begin{cases}\nx = my + 1 \\\\\ny^2 = 4x\n\\end{cases}\n$$\nEliminating $x$ gives $y^{2} - 4my - 4 = 0$. Let $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$, so $y_{1} + y_{2} = 4m$, $y_{1}y_{2} = -4$. Since $\\overrightarrow{AF} = 2\\overrightarrow{FB}$, we have $y_{1} = -2y_{2}$, that is, $y_{2} = -\\frac{y_{1}}{2}$. Therefore, $y_{1}y_{2} = -\\frac{1}{2}y_{1}^{2} = -4$, solving gives $y_{1} = \\pm 2\\sqrt{2}$. Thus, $x_{1} = \\frac{y_{1}^{2}}{4} = 2$. Hence, the coordinates of point A are $(2, 2\\sqrt{2})$ or $(2, -2\\sqrt{2})$."
  },
  {
    "text": "Write an equation of an ellipse that shares common foci with the ellipse $C$: $\\frac{x^{2}}{5}+\\frac{y^{2}}{3}=1$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/5 + y^2/3 = 1);C1: Ellipse;Focus(C) = Focus(C1)",
    "query_expressions": "Expression(C1)",
    "answer_expressions": "x^2/10 + y^2/8 = 1 (the answer is not unique)",
    "fact_spans": "[[[5, 47]], [[5, 47]], [[53, 55]], [[4, 55]]]",
    "query_spans": "[[[53, 58]]]",
    "process": "From the given condition, the form of the ellipse should be $\\frac{x^{2}}{5+m}+\\frac{y^{2}}{3+m}=1$ $(m>-3$, and $m\\neq0$), and we can take $m=5$."
  },
  {
    "text": "The distance from the focus of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ to the asymptote of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1)",
    "query_expressions": "Distance(Focus(G), Asymptote(G))",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 39], [44, 47]], [[0, 39]]]",
    "query_spans": "[[[0, 54]]]",
    "process": "From the problem: the coordinates of the foci are (-5,0), (5,0), and the asymptotes are given by $ y = \\pm\\frac{3}{4}x $. Therefore, the distance from a focus to an asymptote is $ d = \\frac{\\frac{3}{4}\\times5}{\\sqrt{1+(\\frac{3}{4})^{2}}} = 3 $. Thus, the answer is 3."
  },
  {
    "text": "Given that a line passing through the focus $F$ of the parabola $x^{2}=4 y$ intersects the parabola at two distinct points $A$ and $B$, tangents to the parabola are drawn at $A$ and $B$ respectively, intersecting at point $C$. Then the minimum area of $\\triangle A B C$ is?",
    "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;C: Point;F: Point;Expression(G) = (x^2 = 4*y);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Intersection(TangentOnPoint(A,G),TangentOnPoint(B,G))=C",
    "query_expressions": "Min(Area(TriangleOf(A, B, C)))",
    "answer_expressions": "4",
    "fact_spans": "[[[3, 17], [27, 30], [56, 59]], [[24, 26]], [[31, 34], [46, 49]], [[35, 38], [50, 53]], [[66, 70]], [[20, 23]], [[3, 17]], [[3, 23]], [[2, 26]], [[24, 44]], [[45, 70]]]",
    "query_spans": "[[[72, 98]]]",
    "process": "According to the problem, let the line be $ y = kx + 1 $. Solving simultaneously \n\\[\n\\begin{cases}\ny = kx + 1 \\\\\nx^{2} = 4y\n\\end{cases}\n\\]\ngives $ C(2k, -1) $. Therefore, $ S = \\frac{1}{2}|AB|d = 4\\sqrt{1+k^{2}}(1+k^{2}) = 4(\\sqrt{1+k^{2}}) $. When $ k = 0 $, $ S_{\\min} = 4 $."
  },
  {
    "text": "Given that the circle $x^{2}+y^{2}-6 x-7=0$ is tangent to the directrix of the parabola $y^{2}=2 p x$ $(p>0)$, then the equation of the parabola is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;H: Circle;p>0;Expression(H) = (-6*x + x^2 + y^2 - 7 = 0);IsTangent(H,Directrix(G))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2=4*x",
    "fact_spans": "[[[25, 46], [53, 56]], [[25, 46]], [[28, 46]], [[2, 24]], [[28, 46]], [[2, 24]], [[2, 51]]]",
    "query_spans": "[[[53, 61]]]",
    "process": ""
  },
  {
    "text": "What is the equation of the line passing through the point $M(2,1)$ and bisected by this point within the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$?",
    "fact_expressions": "G: Ellipse;H: LineSegment;M: Point;Expression(G) = (x^2/16 + y^2/9 = 1);Coordinate(M) = (2, 1);IsChordOf(H, G);MidPoint(H) = M;PointOnCurve(M, H)",
    "query_expressions": "Expression(OverlappingLine(H))",
    "answer_expressions": "9*x + 8*y - 26 = 0",
    "fact_spans": "[[[0, 38]], [], [[41, 50]], [[0, 38]], [[41, 50]], [[0, 58]], [[0, 58]], [[0, 58]]]",
    "query_spans": "[[[0, 67]]]",
    "process": "Let the two intersection points of the line and the ellipse be A(x_{1},y_{1}), B(x_{2},y_{2}). Since A and B lie on the ellipse, we have \\begin{cases}\\frac{x_{1}}{16}+\\frac{y_{1}}{9}=1\\\\\\frac{x_{2}}{16}+\\frac{y_{2}}{9}=1\\end{cases}. Therefore, \\frac{x_{1}^{2}}{16}-\\frac{x_{2}^{2}}{16}=-\\left(\\frac{y_{1}^{2}}{9}-\\frac{y_{2}^{2}}{9}\\right), so \\frac{y_{1}^{2}-y_{2}^{2}}{x_{1}^{2}-x_{2}^{2}}=-\\frac{9}{16}, thus \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\\cdot\\frac{y_{1}+y_{2}}{x_{1}+x_{2}}=-\\frac{9}{16}. Hence, k_{AB}\\cdot\\frac{1\\times2}{2\\times2}=-\\frac{9}{16}, so k_{AB}=-\\frac{9}{8}. Therefore, the equation of AB is: y-1=-\\frac{9}{8}(x-2), that is, 9x+8y-26=0."
  },
  {
    "text": "Given the parabola $x^{2}=2 p y(p>0)$ with focus $F$, its directrix $l$ intersects the $y$-axis at point $Q$. If a point $P(4, t)$ on the parabola satisfies $P Q=\\sqrt{2} P F$, then the value of the real number $t$ is?",
    "fact_expressions": "G: Parabola;p: Number;P: Point;Q: Point;F: Point;t: Real;l:Line;p>0;Expression(G) = (x^2 = 2*p*y);Coordinate(P) = (4, t);Focus(G) = F;Directrix(G)=l;Intersection(l, yAxis) = Q;PointOnCurve(P,G);LineSegmentOf(P,Q)=sqrt(2)*LineSegmentOf(P,F)",
    "query_expressions": "t",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 23], [31, 32], [50, 53]], [[5, 23]], [[56, 65]], [[44, 48]], [[27, 30]], [[87, 92]], [[34, 37]], [[5, 23]], [[2, 23]], [[56, 65]], [[2, 30]], [[31, 37]], [[34, 48]], [[50, 65]], [[67, 85]]]",
    "query_spans": "[[[87, 96]]]",
    "process": ""
  },
  {
    "text": "Given that a chord of the ellipse $\\frac{y^{2}}{75}+\\frac{x^{2}}{25}=1$ has slope $3$, and its intersection with the line $x=\\frac{1}{2}$ is exactly the midpoint $M$ of this chord, then what are the coordinates of point $M$?",
    "fact_expressions": "G: Ellipse;H: Line;l: LineSegment;M: Point;Expression(G) = (x^2/25 + y^2/75 = 1);Expression(H) = (x = 1/2);Slope(l) = 3;IsChordOf(l, G);MidPoint(l) = M;Intersection(G, H) = M",
    "query_expressions": "Coordinate(M)",
    "answer_expressions": "(1/2, -1/2)",
    "fact_spans": "[[[2, 41], [53, 54]], [[55, 72]], [], [[83, 86], [88, 92]], [[2, 41]], [[55, 72]], [[2, 52]], [[2, 45]], [[77, 86]], [[53, 86]]]",
    "query_spans": "[[[88, 97]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, a line passing through the point $F(1,0)$ intersects the ellipse at points $A$ and $B$, and the midpoint of segment $AB$ is $M$. What is the maximum value of the $y$-coordinate of point $M$?",
    "fact_expressions": "C: Ellipse;G: Line;B: Point;A: Point;F: Point;M: Point;Expression(C) = (x^2/4 + y^2/3 = 1);Coordinate(F) = (1, 0);PointOnCurve(F, G);Intersection(G, C) = {A, B};MidPoint(LineSegmentOf(A,B))=M",
    "query_expressions": "Max(YCoordinate(M))",
    "answer_expressions": "sqrt(3)/4",
    "fact_spans": "[[[2, 44], [59, 61]], [[56, 58]], [[68, 71]], [[64, 67]], [[46, 55]], [[85, 88], [90, 94]], [[2, 44]], [[46, 55]], [[45, 58]], [[56, 73]], [[74, 88]]]",
    "query_spans": "[[[90, 104]]]",
    "process": "When the slope of the line is 0, the ordinate of the midpoint M of segment AB is 0. When the slope of the line is not 0, let the line passing through F(1,0) be x=ty+1. Then, by solving the system of equations of the line and the ellipse simultaneously and eliminating x, using the relationship between roots and coefficients, we obtain y_{M}=-\\frac{3t}{3t^{2}+4}. Clearly, when t<0, y_{M}>0. Then, using the basic inequality, we can solve it. When the slope of the line is 0, the line is y=0, and the ordinate of the midpoint M of segment AB is 0. When the slope of the line is not 0, let the line passing through F(1,0) be x=ty+1. Let A(x_{1},y_{1}), B(x_{2},y_{2}). From \\begin{cases}x=ty+1\\\\\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1\\end{cases}, we get (3t^{2}+4)y^{2}+6ty-9=0, then y_{1}+y_{2}=-\\frac{6t}{3t^{2}+4}. Therefore, the ordinate of the midpoint M of segment AB is y_{M}=-\\frac{3t}{3t^{2}+4}. When t=0, the ordinate of M is 0. When t<0, y_{M}=-\\frac{3t}{3t^{2}+4}=-\\frac{-}{(}\\frac{\\sqrt{1}}{2}\\frac{3}{4\\sqrt{3}}=\\frac{\\sqrt{3}}{4}, with equality if and only if -3t=-\\frac{4}{t}, that is, when t=-\\frac{2\\sqrt{3}}{3}, at which point the maximum value of y_{M} is \\frac{\\sqrt{3}}{4}. When t>0, y_{M}<0. In conclusion, the maximum value of y_{M} is \\frac{\\sqrt{3}}{4}."
  },
  {
    "text": "It is known that a hyperbola passes through the point $(3,2)$, and one of its asymptotes has the equation $y=\\frac{1}{3} x$. Then, what is the standard equation of the hyperbola?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (3, 2);PointOnCurve(H, G);Expression(OneOf(Asymptote(G))) = (y = x/3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/3 - x^2/27 = 1",
    "fact_spans": "[[[2, 5], [15, 16], [44, 47]], [[6, 14]], [[6, 14]], [[2, 14]], [[15, 42]]]",
    "query_spans": "[[[44, 54]]]",
    "process": "Let the equation of the desired hyperbola be $\\frac{x^{2}}{9}-y^{2}=\\lambda$, then we have $\\lambda=\\frac{3^{2}}{9}-2^{2}=-3$, so the equation of the hyperbola is $\\frac{x^{2}}{9}-y^{2}=-3$. Therefore, the standard equation of this hyperbola is $\\frac{y^{2}}{3}-\\frac{x^{2}}{27}=1$."
  },
  {
    "text": "Given the ellipse $C$: $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$, with left and right foci $F_{1}$ and $F_{2}$ respectively. Point $M$ is a point on the ellipse $C$, point $N$ is the midpoint of segment $M F_{1}$, $O$ is the origin. If $|M N|=4$, then $|O N|=$?",
    "fact_expressions": "C: Ellipse;F1: Point;F2: Point;M: Point;N: Point;O: Origin;Expression(C) = (x^2/36 + y^2/9 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(M, C);MidPoint(LineSegmentOf(M, F1)) = N;Abs(LineSegmentOf(M, N)) = 4",
    "query_expressions": "Abs(LineSegmentOf(O, N))",
    "answer_expressions": "2",
    "fact_spans": "[[[2, 44], [73, 78]], [[52, 59]], [[60, 67]], [[68, 72]], [[82, 86]], [[102, 105]], [[2, 44]], [[2, 67]], [[2, 67]], [[68, 81]], [[82, 101]], [[112, 121]]]",
    "query_spans": "[[[123, 132]]]",
    "process": ""
  },
  {
    "text": "If a hyperbola passes through the point $(3, \\sqrt{2})$ and has asymptotes given by $y = \\pm \\frac{1}{3}x$, then the equation of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (3, sqrt(2));PointOnCurve(H, G) = True;Expression(Asymptote(G)) = (y = pm*x/3)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2 - x^2/9 = 1",
    "fact_spans": "[[[1, 4], [55, 58]], [[6, 22]], [[6, 22]], [[1, 22]], [[1, 51]]]",
    "query_spans": "[[[55, 63]]]",
    "process": "According to the problem, assume the hyperbola's equation is $ y^{2} - \\frac{x^{2}}{9} = \\lambda $. Since it passes through the point $ (3, \\sqrt{2}) $, we have $ \\lambda = 2 - \\frac{9}{9} = 1 $, so $ y^{2} - \\frac{x^{2}}{9} = 1 $."
  },
  {
    "text": "Given that $F_{2}$ is the right focus of the hyperbola $C$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{3}=1$, point $A$ moves on the left branch of the hyperbola, and point $B$ lies on the circle $E$: $x^{2}+(y+2)^{2}=1$, then the minimum value of $|A B|+|A F_{2}|$ is?",
    "fact_expressions": "F2: Point;RightFocus(C) = F2;C: Hyperbola;Expression(C) = (x^2/9 - y^2/3 = 1);A: Point;PointOnCurve(A, LeftPart(C));B: Point;PointOnCurve(B, E);E: Circle;Expression(E) = (x^2 + (y + 2)^2 = 1)",
    "query_expressions": "Min(Abs(LineSegmentOf(A, B)) + Abs(LineSegmentOf(A, F2)))",
    "answer_expressions": "9",
    "fact_spans": "[[[2, 9]], [[2, 57]], [[10, 53], [64, 67]], [[10, 53]], [[60, 63]], [[58, 70]], [[71, 74]], [[71, 103]], [[75, 100]], [[75, 100]]]",
    "query_spans": "[[[105, 128]]]",
    "process": "Denote the left focus of hyperbola $ C: \\frac{x^2}{9} - \\frac{y^{2}}{3} = 1 $ as $ F_{1} $, then $ F_{1}(-2\\sqrt{3},0) $. According to the definition of the hyperbola, we have $ |AF_{2}| - |AF_{1}| = 2a = 6 $, so $ |AF_{2}| = |AF_{1}| + 6 $. Therefore, $ |AE| + |AF_{2}| = |AE| + |AF_{1}| + 6 \\geqslant |EF_{1}| + 6 $, where equality holds when points $ F_{1} $, $ A $, and $ E $ are collinear. Since $ E $ is the center of the circle $ E: x^{2} + (y+2)^{2} = 1 $, we have $ E(0,-2) $, and the radius of the circle is $ r = 1 $. Then $ |EF_{1}| = \\sqrt{(2\\sqrt{3})^{2} + 2^{2}} = 4 $, so $ |AE| + |AF_{2}| \\geqslant |EF_{1}| + 6 = 10 $. Since $ B $ is a point on the circle $ E: x^{2} + (y+2)^{2} = 1 $, by the property of the circle, we obtain $ |AB| + |AF_{2}| \\geqslant |AE| - r + |AF_{2}| \\geqslant 10 - 1 = 9 $, where equality holds when points $ F_{1} $, $ A $, $ B $, and $ E $ are collinear."
  },
  {
    "text": "Given the parabola $y=\\frac{1}{2} x^{2}$ has focus $F$, directrix $l$, and point $M$ lies on $l$. The line segment $MF$ intersects the parabola at point $N$. If $|MN| = \\sqrt{2}|NF|$, then $|MF| =$?",
    "fact_expressions": "G: Parabola;Expression(G) = (y = x^2/2);F: Point;Focus(G) = F;l: Line;Directrix(G) = l;M: Point;PointOnCurve(M, l);N: Point;Intersection(LineSegmentOf(M, F), G) = N;Abs(LineSegmentOf(M, N)) = sqrt(2)*Abs(LineSegmentOf(N, F))",
    "query_expressions": "Abs(LineSegmentOf(M, F))",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[2, 26], [58, 61]], [[2, 26]], [[30, 33]], [[2, 33]], [[37, 40], [45, 48]], [[2, 40]], [[41, 44]], [[41, 49]], [[63, 67]], [[50, 67]], [[69, 90]]]",
    "query_spans": "[[[92, 102]]]",
    "process": "As shown in the figure, draw a perpendicular line NH from point N to the directrix, with H as the foot of the perpendicular. According to the definition of the parabola, |NH| = |NF|. In right triangle NHM, |NM| = \\sqrt{2}|NH|, then \\angle NMH = 45^{\\circ}. In triangle MFK, \\angle FMK = 45^{\\circ}, so |MF| = \\sqrt{2}|FK|. Since |FK| = 1, it follows that |MF| = \\sqrt{2}."
  },
  {
    "text": "When the maximum area of a triangle with vertices at a point on the ellipse and the two foci of the ellipse is $1$, what is the minimum value of the major axis of the ellipse?",
    "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;PointOnCurve(P, G);Focus(G) = {F1, F2};Max(Area(TriangleOf(P, F1, F2))) = 1",
    "query_expressions": "Min(MajorAxis(G))",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[2, 4], [8, 10], [33, 35]], [], [], [], [[2, 7]], [[8, 13]], [[1, 31]]]",
    "query_spans": "[[[33, 43]]]",
    "process": ""
  },
  {
    "text": "If the focus of the parabola $y^{2}=2 p x$ has coordinates $(1,0)$, then the equation of the directrix is?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;Coordinate(Focus(G)) = (1, 0)",
    "query_expressions": "Expression(Directrix(G))",
    "answer_expressions": "x=-1",
    "fact_spans": "[[[1, 17]], [[1, 17]], [[4, 17]], [[1, 30]]]",
    "query_spans": "[[[1, 38]]]",
    "process": "From the parabola equation $ y^{2} = 2px $, we know the focus of the parabola is $ \\left( \\frac{p}{2}, 0 \\right) $ and the directrix is $ x = -\\frac{p}{2} $. Since the coordinates of the focus are $ (1, 0) $, it follows that $ \\frac{p}{2} = 1 $. Therefore, the equation of the directrix of the parabola is $ x = -\\frac{p}{2} = -1 $."
  },
  {
    "text": "Given that a moving point $P(x, y)$ lies on the ellipse $C$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, $F$ is the right focus of the ellipse $C$, and point $M$ satisfies $|MF|=1$ and $MP \\perp MF$, then the minimum value of $|PM|$ is?",
    "fact_expressions": "C: Ellipse;M: Point;P: Point;F: Point;x1: Number;y1: Number;Expression(C) = (x^2/25 + y^2/16 = 1);Coordinate(P) = (x1, y1);PointOnCurve(P, C);RightFocus(C) = F;Abs(LineSegmentOf(M, F)) = 1;IsPerpendicular(LineSegmentOf(M, P), LineSegmentOf(M, F))",
    "query_expressions": "Min(Abs(LineSegmentOf(P, M)))",
    "answer_expressions": "sqrt(3)",
    "fact_spans": "[[[15, 59], [65, 70]], [[76, 80]], [[4, 14]], [[61, 64]], [[4, 14]], [[4, 14]], [[15, 59]], [[4, 14]], [[4, 60]], [[61, 74]], [[82, 91]], [[94, 108]]]",
    "query_spans": "[[[110, 122]]]",
    "process": "From the given conditions, it is known that the moving point $ M $ moves on a circle with center $ F(3,0) $ and radius 1, and $ |PM| $ is a tangent to the circle. According to the tangent length theorem, when $ |PF| $ is minimized, the tangent length $ |PM| $ reaches its minimum value. It is clear that $ |PF| $ reaches its minimum when $ P $ is at the right vertex, at which time $ |PF| = 5 - 3 = 2 $. By the tangent length theorem, $ |PM| = \\sqrt{2^{2} - 1^{2}} = \\sqrt{3} $."
  },
  {
    "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, and the line $l$: $y=x-1$ intersecting the parabola at points $A$ and $B$, then $|AB|$ equals?",
    "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;Expression(l) = (y = x - 1);Intersection(l, C) = {A, B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "8",
    "fact_spans": "[[[29, 43]], [[2, 21], [44, 47]], [[48, 51]], [[52, 55]], [[25, 28]], [[2, 21]], [[2, 28]], [[29, 43]], [[29, 57]]]",
    "query_spans": "[[[59, 69]]]",
    "process": "From the given condition, F(1,0), so the line $ l $ passes through the focus; therefore, by the focal chord formula, $ |AB| = \\frac{2p}{\\sin^{2}\\theta} = \\frac{4}{\\sin^{2}45^{0}} = 8 $."
  },
  {
    "text": "Write the equation of a line that is parallel to the asymptotes of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{2}=1$ and has a positive slope.",
    "fact_expressions": "C: Hyperbola;G: Line;Expression(C) = (x^2 - y^2/2 = 1);IsParallel(Asymptote(C),G);Slope(G)>0",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y=sqrt(2)*x+1",
    "fact_spans": "[[[5, 38]], [[52, 54]], [[5, 38]], [[4, 54]], [[46, 54]]]",
    "query_spans": "[[[52, 58]]]",
    "process": "The asymptotes of hyperbola C are given by the equation $x^{2}-\\frac{y^{2}}{2}=0$, that is, $y=\\pm\\sqrt{2}x$. Therefore, the required line equation can be $y=\\sqrt{2}x+$"
  },
  {
    "text": "Given that $F$ is the left focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $P$ is a point on the ellipse such that $PF \\perp x$-axis, and $OP \\parallel AB$ ($O$ is the origin, $A$ is the right vertex, $B$ is the upper vertex), then the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;LeftFocus(G) = F;P: Point;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F), xAxis);O: Origin;A: Point;B: Point;RightVertex(G) = A;UpperVertex(G) = B;IsParallel(LineSegmentOf(O, P), LineSegmentOf(A, B))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(2)/2",
    "fact_spans": "[[[6, 58], [67, 69], [128, 130]], [[6, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[2, 5]], [[2, 62]], [[63, 66]], [[63, 72]], [[73, 87]], [[102, 105]], [[109, 112]], [[117, 120]], [[67, 116]], [[67, 124]], [[88, 101]]]",
    "query_spans": "[[[128, 136]]]",
    "process": "Let the semi-focal length of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ be $c$. Since $PF \\perp x$-axis, solving $\\begin{cases}x=-c \\\\ \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1\\end{cases}$ yields $|y|=\\frac{b^{2}}{a}$, that is, $|PF|=\\frac{b^{2}}{a}$. Since $OP \\parallel AB$ ($O$ is the origin, $A$ is the right vertex, $B$ is the upper vertex), then $\\triangle OPF \\sim \\triangle ABO$, so $\\frac{|PF|}{|OB|}=\\frac{|OF|}{|OA|}$. Therefore, $\\frac{\\frac{b^{2}}{a}}{b}=\\frac{c}{a}$, simplifying gives $b=c$, then $a=\\sqrt{2}c$, so the eccentricity of the ellipse is $e=\\frac{\\sqrt{2}}{2}$."
  },
  {
    "text": "Given point $A(3,1)$, find a point $P$ on the parabola $y^{2}=2x$ such that $|PF|+|PA|$ is minimized ($F$ being the focus of the parabola); then the coordinates of point $P$ are?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 2*x);Coordinate(A) = (3, 1);PointOnCurve(P, G);Focus(G)=F;WhenMin(Abs(LineSegmentOf(P,F))+Abs(LineSegmentOf(P,A)))",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(1/2,1)",
    "fact_spans": "[[[13, 27], [59, 62]], [[2, 11]], [[31, 34], [69, 73]], [[55, 58]], [[13, 27]], [[2, 11]], [[12, 34]], [[55, 65]], [[37, 54]]]",
    "query_spans": "[[[69, 78]]]",
    "process": "Draw PQ perpendicular to the directrix $ x = -\\frac{1}{2} $ from point P, with foot at Q, then $ |PF| = |PQ| $. From point A, draw AB perpendicular to the directrix, with foot at B, then $ |PA| + |PF| = |PA| + |PQ| \\geqslant |AB| $. The sum achieves its minimum when point P is the intersection of AB and the parabola, i.e., when points P, A, B are collinear. At this time, the y-coordinate of P is 1; substituting into the parabola gives the x-coordinate of P as $ \\frac{1}{2} $. Therefore, the coordinates of point P are $ \\left( \\frac{1}{2}, 1 \\right) $."
  },
  {
    "text": "Given that the distance from a moving point $M$ to the fixed point $(8,0)$ is twice the distance from $M$ to $(2,0)$, what is the trajectory equation of point $M$?",
    "fact_expressions": "G: Point;H: Point;M:Point;Coordinate(G) = (8, 0);Coordinate(H) = (2, 0);Distance(M,G)=2*Distance(M,H)",
    "query_expressions": "LocusEquation(M)",
    "answer_expressions": "x^2+y^2=16",
    "fact_spans": "[[[10, 17]], [[26, 33]], [[4, 7], [44, 48], [22, 25]], [[10, 17]], [[26, 33]], [[2, 41]]]",
    "query_spans": "[[[44, 54]]]",
    "process": "Let M(x, y). Since the distance from M to the fixed point (8, 0) is equal to twice the distance from M to (2, 0), we have \\sqrt{(x-8)^{2}+y^{2}}=2\\times\\sqrt{(x-2)^{2}+y^{2}}. Simplifying this yields x^{2}+y^{2}=16."
  },
  {
    "text": "It is known that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $y=\\sqrt{3} x$, and one of its foci lies on the directrix of the parabola $y^{2}=24 x$. Then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 24*x);Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x);PointOnCurve(OneOf(Focus(G)), Directrix(H))",
    "query_expressions": "Expression(G)",
    "answer_expressions": "x^2/9 - y^2/27 = 1",
    "fact_spans": "[[[2, 58], [82, 83], [110, 113]], [[5, 58]], [[5, 58]], [[89, 104]], [[5, 58]], [[5, 58]], [[2, 58]], [[89, 104]], [[2, 81]], [[82, 108]]]",
    "query_spans": "[[[110, 118]]]",
    "process": ""
  },
  {
    "text": "Let the endpoints of the major axis of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ be $M$ and $N$, and let point $P$ lie on the ellipse. Then the product of the slopes of $PM$ and $PN$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);Endpoint(MajorAxis(G)) = {M,N};M: Point;N: Point;P: Point;PointOnCurve(P, G) = True",
    "query_expressions": "Slope(LineSegmentOf(P,M))*Slope(LineSegmentOf(P,N))",
    "answer_expressions": "-3/4",
    "fact_spans": "[[[1, 38], [58, 60]], [[1, 38]], [[1, 52]], [[45, 48]], [[49, 52]], [[53, 57]], [[53, 61]]]",
    "query_spans": "[[[63, 81]]]",
    "process": ""
  },
  {
    "text": "The asymptotes of a hyperbola with foci on the $y$-axis are given by $y=\\pm \\frac{\\sqrt{5}}{2} x$. Then, the eccentricity of this hyperbola is?",
    "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), yAxis);Expression(Asymptote(G)) = (y = pm*(x*(sqrt(5)/2)))",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "3*sqrt(5)/5",
    "fact_spans": "[[[9, 12], [50, 53]], [[0, 12]], [[9, 47]]]",
    "query_spans": "[[[50, 59]]]",
    "process": "From the given conditions, it is known that the hyperbola's foci lie on the y-axis, and the equations of the asymptotes are $ y = \\pm\\frac{\\sqrt{5}}{2}x $. Then $ a = \\sqrt{5} $, $ b = 2 $, so we can obtain $ c = \\sqrt{a^{2}+b^{2}} = \\sqrt{5+4} = 3 $. Therefore, the eccentricity of the hyperbola is $ e = \\frac{c}{a} = \\frac{3}{\\sqrt{5}} = \\frac{3\\sqrt{5}}{5} $."
  },
  {
    "text": "Given that $P$ is a point on the hyperbola $\\frac{x^{2}}{64}-\\frac{y^{2}}{36}=1$, and $F_{1}$, $F_{2}$ are the two foci of the hyperbola. If $|P F_{1}|=17$, then the value of $|P F_{2}|$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/64 - y^2/36 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2};Abs(LineSegmentOf(P, F1)) = 17",
    "query_expressions": "Abs(LineSegmentOf(P, F2))",
    "answer_expressions": "33",
    "fact_spans": "[[[6, 46], [66, 69]], [[2, 5]], [[50, 57]], [[58, 65]], [[6, 46]], [[2, 49]], [[50, 74]], [[76, 90]]]",
    "query_spans": "[[[92, 107]]]",
    "process": "From the hyperbola equation $\\frac{x^{2}}{64}-\\frac{y^{2}}{36}=1$, we know $a=8$, $b=6$, then $c=\\sqrt{a^{2}+b^{2}}=10$. Since $P$ is a point on the hyperbola, $|PF_{1}|-|PF_{2}|=2a=16$. Given $|PF_{1}|=17$, so $|PF_{2}|=1$ or $|PF_{2}|=33$. Also, $|PF_{2}|\\geqslant c-a=2$, therefore $|PF_{2}|=33$."
  },
  {
    "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ are both tangent to $x^{2}+y^{2}-4 x+1=0$, then the eccentricity of this hyperbola is equal to?",
    "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Curve;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (-4*x + x^2 + y^2 + 1 = 0);IsTangent(Asymptote(G), H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "2",
    "fact_spans": "[[[0, 57], [89, 92]], [[3, 57]], [[3, 57]], [[63, 84]], [[3, 57]], [[3, 57]], [[0, 57]], [[63, 84]], [[0, 86]]]",
    "query_spans": "[[[89, 98]]]",
    "process": "The circle $x^{2}+y^{2}-4x+1=0$ can be rewritten as $(x-2)^{2}+y^{2}=3$, $\\therefore$ the center coordinates are $C(2,0)$, and the radius is $\\sqrt{3}$. $\\because$ the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{k^{2}}=1$ are given by $bx+ay=0$, and the asymptotes are tangent to the circle $x^{2}+y^{2}-4x+1=0$, $\\therefore$ $\\frac{|2b|}{\\sqrt{a^{2}+b^{2}}}=\\sqrt{3}$, $\\therefore$ $b^{2}=3a^{2}$, $\\therefore$ $c^{2}=4a^{2}$, $\\therefore$ the eccentricity of the hyperbola is $e=2$."
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{10}-\\frac{y^{2}}{6}=1$, the vertex of the parabola is at the origin, the axis of symmetry is the $x$-axis, and the focus is the left focus of the hyperbola. Then the standard equation of the parabola is?",
    "fact_expressions": "C: Hyperbola;G: Parabola;O: Origin;Expression(C) = (x^2/10 - y^2/6 = 1);Vertex(G)=O;SymmetryAxis(G)=xAxis;Focus(G)=LeftFocus(C)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^0 = -16 * x",
    "fact_spans": "[[[2, 45], [68, 71]], [[46, 49], [77, 80]], [[53, 55]], [[2, 45]], [[46, 55]], [[46, 64]], [[46, 75]]]",
    "query_spans": "[[[77, 87]]]",
    "process": ""
  },
  {
    "text": "If the standard equation of hyperbola $E$ is $\\frac{x^{2}}{4}-y^{2}=1$, then what is the equation of the asymptotes of hyperbola $E$?",
    "fact_expressions": "E: Hyperbola;Expression(E) = (x^2/4 - y^2 = 1)",
    "query_expressions": "Expression(Asymptote(E))",
    "answer_expressions": "y = pm*(1/2)*x",
    "fact_spans": "[[[1, 7], [40, 46]], [[1, 38]]]",
    "query_spans": "[[[40, 54]]]",
    "process": ""
  },
  {
    "text": "The ellipse and hyperbola have the same foci $F_{1}$, $F_{2}$, and $P$ is one of their intersection points, with $\\angle F_{1}PF_{2}=\\frac{\\pi}{3}$. Denote the eccentricities of the ellipse and hyperbola as $e_{1}$, $e_{2}$, respectively. Then the minimum value of $e_{1} e_{2}$ is?",
    "fact_expressions": "H: Ellipse;G: Hyperbola;F1: Point;F2: Point;Focus(H) = {F1, F2};Focus(G) = {F1, F2};P: Point;OneOf(Intersection(H, G)) = P;AngleOf(F1, P, F2) = pi/3;e1: Number;e2: Number;Eccentricity(H) = e1;Eccentricity(G) = e2",
    "query_expressions": "Min(e1*e2)",
    "answer_expressions": "sqrt(3)/2",
    "fact_spans": "[[[0, 2], [77, 79]], [[3, 6], [80, 83]], [[12, 19]], [[20, 27]], [[0, 27]], [[0, 27]], [[28, 31]], [[28, 39]], [[41, 75]], [[90, 97]], [[100, 107]], [[77, 107]], [[77, 107]]]",
    "query_spans": "[[[110, 129]]]",
    "process": ""
  },
  {
    "text": "The focal distance of the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{4}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2/6 + y^2/4 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "2*sqrt(2)",
    "fact_spans": "[[[0, 37]], [[0, 37]]]",
    "query_spans": "[[[0, 42]]]",
    "process": "Since the ellipse is $\\frac{x^{2}}{6}+\\frac{y^{2}}{4}=1$, then $c=\\sqrt{6-4}=\\sqrt{2}$, so $2c=2\\sqrt{2}$, thus the focal distance of the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{4}=1$ is $2\\sqrt{2}$."
  },
  {
    "text": "Given that the asymptotes of a hyperbola are $2x \\pm 3y = 0$, and the hyperbola passes through the point $P(\\sqrt{6}, 2)$, then the equation of the hyperbola is?",
    "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (2*x + pm*3*y = 0);P: Point;Coordinate(P) = (sqrt(6), 2);PointOnCurve(P, G) = True",
    "query_expressions": "Expression(G)",
    "answer_expressions": "y^2/(4/3)-x^2/3=1",
    "fact_spans": "[[[2, 5], [29, 32], [53, 56]], [[2, 27]], [[34, 51]], [[34, 51]], [[29, 51]]]",
    "query_spans": "[[[53, 61]]]",
    "process": "From the given information, the asymptotes of the hyperbola are $ y = \\pm\\frac{2}{3}x $. Thus, we can assume the equation of the hyperbola is $ \\frac{x^{2}}{9} - \\frac{y^{2}}{4} = \\lambda $ ($ \\lambda \\neq 0 $). Since the hyperbola passes through the point $ P(\\sqrt{6}, 2) $, we have $ \\frac{6}{9} - \\frac{4}{4} = \\lambda $, solving which gives $ \\lambda = -\\frac{1}{3} $. Therefore, the required hyperbola equation is $ \\frac{y^{2}}{4} - \\frac{x^{2}}{3} = 1 $."
  },
  {
    "text": "If the intersection point $A$, distinct from the origin $O$, of the parabola $C_{1}$: $y^{2}=4x$ and the parabola $C_{2}$: $x^{2}=2py$ $(p>0)$ is at a distance of $3$ from the focus of parabola $C_{1}$, then what is the equation of parabola $C_{2}$?",
    "fact_expressions": "C1: Parabola;Expression(C1) = (y^2=4*x);C2: Parabola;Expression(C2) = (x^2=2*p*y);p: Number;p>0;O: Origin;Intersection(C1,C2) = A;A: Point;Negation(A=O);Distance(A,Focus(C1)) = 3",
    "query_expressions": "Expression(C2)",
    "answer_expressions": "x^2=sqrt(2)*y",
    "fact_spans": "[[[1, 24], [69, 79]], [[1, 24]], [[25, 55], [91, 101]], [[25, 55]], [[37, 55]], [[37, 55]], [[57, 62]], [[1, 68]], [[65, 68]], [[55, 68]], [[65, 89]]]",
    "query_spans": "[[[91, 106]]]",
    "process": "According to the problem, draw the graph. By the definition of a parabola, the distance from any point on the curve to the focus is equal to the distance from that point to the directrix. Let A(x, y), x = 2, y = 2\\sqrt{2}. Substituting into curve C_{2}, we get p = \\frac{1}{\\sqrt{2}}. Thus, the equation is x^{2} = \\sqrt{2}y"
  },
  {
    "text": "Given that $P$ is a point on $\\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$, and $F_{1}$, $F_{2}$ are the left and right foci of the ellipse, then $P F_{2}+P F_{1}$=?",
    "fact_expressions": "G: Ellipse;P: Point;F2: Point;F1: Point;Expression(G)=(x^2/4+y^2/9=1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P,G)",
    "query_expressions": "LineSegmentOf(P, F1) + LineSegmentOf(P, F2)",
    "answer_expressions": "4",
    "fact_spans": "[[[6, 41], [62, 64]], [[2, 5]], [[54, 61]], [[46, 53]], [[6, 41]], [[46, 69]], [[46, 69]], [[2, 45]]]",
    "query_spans": "[[[71, 90]]]",
    "process": ""
  },
  {
    "text": "A focus of the hyperbola $2 x^{2}-y^{2}=m$ is $(0, \\sqrt{3})$, then the value of $m$ is?",
    "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (2*x^2 - y^2 = m);Coordinate(OneOf(Focus(G)))= (0, sqrt(3))",
    "query_expressions": "m",
    "answer_expressions": "-2",
    "fact_spans": "[[[0, 20]], [[43, 46]], [[0, 20]], [[0, 41]]]",
    "query_spans": "[[[43, 50]]]",
    "process": ""
  },
  {
    "text": "Given a moving point $P$ on the parabola $y^{2}=4x$ with focus $F$, and a point $A(2,2)$, the minimum value of $|PA|+|PF|$ is?",
    "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (2, 2);Focus(G) = F;PointOnCurve(P, G)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))",
    "answer_expressions": "3",
    "fact_spans": "[[[9, 23]], [[32, 41]], [[28, 31]], [[5, 8]], [[9, 23]], [[32, 41]], [[2, 23]], [[9, 31]]]",
    "query_spans": "[[[43, 62]]]",
    "process": "Draw the directrix $ l $, and from point $ P $, draw $ PM \\bot l $ with foot of perpendicular at $ M $. Use $ |PF| = |PM| $ to transform the problem into one where the minimum value can be found. As shown in the figure, let $ l $ be the directrix of the parabola, and draw $ PM \\bot l $ at $ M $, then $ |PF| = |PM| $, so $ |PA| + |PF| = |PA| + |PM| $. Given that the equation of the directrix is $ x = -1 $, clearly when points $ A $, $ P $, and $ M $ are collinear, $ |PA| + |PF| $ attains the minimum value $ 2 - (-1) = 3 $."
  },
  {
    "text": "Given that the focus of the parabola lies on the line $x-2 y-4=0$, then the standard equation of this parabola is?",
    "fact_expressions": "G: Parabola;H: Line;Expression(H) = (x - 2*y - 4 = 0);PointOnCurve(Focus(G), H)",
    "query_expressions": "Expression(G)",
    "answer_expressions": "{y^2=16*x, x^2=-8*x}",
    "fact_spans": "[[[2, 5], [26, 29]], [[9, 22]], [[9, 22]], [[2, 23]]]",
    "query_spans": "[[[26, 36]]]",
    "process": "Let x=0, then y=-2; let y=0, then x=4, so the focus of the parabola is (4,0) or (0,-2). When the focus is (4,0), \\frac{p}{2}=4, so p=8, and the equation of the parabola is y^{2}=16x; when the focus is (0,-2), \\frac{p}{2}=2, so p=4, and the equation of the parabola is x^{2}=-8y. Therefore, the required equation of the parabola is y^{2}=16x or x^{2}=-8y."
  },
  {
    "text": "Let the line $l$: $y = kx + 1$ intersect the hyperbola $C$: $\\frac{x^2}{2} - y^2 = 1$ at two distinct points $A$ and $B$. Then the range of values for $k$ is?",
    "fact_expressions": "C: Hyperbola;l: Line;k: Number;A: Point;B: Point;Expression(l) = (y = k*x + 1);Expression(C) = (x^2/2 - y^2 = 1);Intersection(l, C) = {A, B};Negation(A = B)",
    "query_expressions": "Range(k)",
    "answer_expressions": "(-1, -\\sqrt{2}/2) + (-\\sqrt{2}/2, \\sqrt{2}/2) + (\\sqrt{2}/2, 1)",
    "fact_spans": "[[[18, 51]], [[1, 17]], [[68, 71]], [[59, 62]], [[63, 66]], [[1, 17]], [[18, 51]], [[1, 66]], [[54, 66]]]",
    "query_spans": "[[[68, 78]]]",
    "process": "Solve the system \\begin{cases}y=kx+1\\\\\\frac{x^{2}}{2}-y^{2}=1\\end{cases} by eliminating $ y $: $(1-2k^{2})x^{2}-4kx-4=0$, $\\triangle=16k^{2}+16(1-2k^{2})>0$, yielding $-1<k<1$. Since the line $y=kx+1$ is not parallel to the asymptotes $y=\\pm\\frac{\\sqrt{2}}{2}x$, it follows that $k\\in(-1,-\\frac{\\sqrt{2}}{2})\\cup(-\\frac{\\sqrt{2}}{2},\\frac{\\sqrt{2}}{2})\\cup(\\frac{\\sqrt{2}}{2},1)$."
  },
  {
    "text": "Given that $P$ is a moving point on the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, $F_{1}$ and $F_{2}$ are its left and right foci respectively, and $O$ is the origin, then the range of $\\frac{|P F_{1}|+|P F_{2}|}{|P O|}$ is?",
    "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;O: Origin;Expression(G) = (x**2/4 - y**2/12=1);PointOnCurve(P, G) = True;LeftFocus(G) = F1;RightFocus(G) = F2",
    "query_expressions": "Range((Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)))/Abs(LineSegmentOf(P, O)))",
    "answer_expressions": "(2,4]",
    "fact_spans": "[[[6, 45], [68, 69]], [[2, 5]], [[50, 57]], [[58, 65]], [[75, 78]], [[6, 45]], [[2, 49]], [[50, 74]], [[50, 74]]]",
    "query_spans": "[[[85, 127]]]",
    "process": "Let P(x,y), without loss of generality assume x>0, then F_{1}(-4,0), F_{2}(4,0)\\frac{,0),F_{2}(4,0)}{\\frac{2+12(\\frac{x^{2}-1}{4})-1)}=2x+2\\sqrt{x^{2}+y^{2}} because y^{2}=3x^{2}-12,, so \\frac{|PF_{1}|+|PF_{2}|}{|PO|}= and since in the hyperbola x^{2}, similarly when x<0, it still follows that \\frac{|PF_{1}|+|PF_{2}|}{|OP|}=\\frac{4}{\\sqrt{4-\\frac{12}{2}}}\\in(2,4], therefore the range of \\frac{|PF|+|PF_{2}|}{|OP|} is (2,4]."
  },
  {
    "text": "Let the right focus of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ be $F$, and let a moving line $l$ passing through the origin $O$ intersect the ellipse $C$ at points $A$ and $B$. If $AF=1$, then $BF=$?",
    "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);RightFocus(C) = F;F: Point;O: Origin;l: Line;PointOnCurve(O,l) = True;Intersection(l,C) = {A,B};B: Point;A: Point;LineSegmentOf(A, F) = 1",
    "query_expressions": "LineSegmentOf(B, F)",
    "answer_expressions": "3",
    "fact_spans": "[[[1, 43], [66, 71]], [[1, 43]], [[1, 51]], [[48, 51]], [[53, 58]], [[62, 65]], [[52, 65]], [[62, 82]], [[77, 80]], [[73, 76]], [[84, 91]]]",
    "query_spans": "[[[94, 101]]]",
    "process": "According to the problem, line AB passes through the origin. By the symmetry of the ellipse, OA = OB. As shown in the figure, given that OF = OF, quadrilateral AFBF' is a parallelogram, so BF = AF'. By the definition of the ellipse, AF + AF' = 2a = 4, and AF = 1, therefore BF = AF' = 3."
  },
  {
    "text": "Given that the line $l$ passes through points $M(2,0)$, $N(3,1)$, and intersects the parabola $y^{2}=8x$ at points $A$ and $B$, then $|AB|=$?",
    "fact_expressions": "l: Line;G: Parabola;M: Point;N: Point;A: Point;B: Point;Expression(G) = (y^2 = 8*x);Coordinate(M) = (2, 0);Coordinate(N) = (3, 1);PointOnCurve(M, l);PointOnCurve(N,l);Intersection(l,G)={A,B}",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16",
    "fact_spans": "[[[2, 7]], [[31, 45]], [[8, 17]], [[20, 28]], [[47, 50]], [[51, 54]], [[31, 45]], [[8, 17]], [[20, 28]], [[2, 17]], [[2, 28]], [[2, 56]]]",
    "query_spans": "[[[58, 67]]]",
    "process": "From the given conditions, the equation of line $ l $ is $ y = x - 2 $. Substituting into the equation of the parabola yields $ x^{2} - 12x + 4 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. According to the relationship between the roots and coefficients of a quadratic equation, we have $ x_{1} + x_{2} = 12 $. Then, by the definition of the parabola, the answer can be obtained. From the given conditions, point $ M $ is exactly the focus of the parabola. The equation of line $ l $ is $ y = x - 2 $. Substituting into the equation of the parabola yields $ x^{2} - 12x + 4 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = 12 $, $ |AB| = x_{1} + x_{2} + 4 = 16 $. From the given conditions, point $ M $ is exactly the focus of the parabola. The slope of line $ l $ is $ k = \\frac{1 - 0}{3 - 2} = 1 $, hence its inclination angle $ \\theta = \\frac{\\pi}{4} $, then $ |AB| = \\frac{8}{\\sin^{2}\\frac{\\pi}{4}} = \\frac{8}{\\frac{1}{2}} = 16 $."
  },
  {
    "text": "Given the hyperbola equation $x^{2}-2 y^{2}=1$, what are the coordinates of its right focus?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - 2*y^2 = 1)",
    "query_expressions": "Coordinate(RightFocus(G))",
    "answer_expressions": "(\\sqrt{6}/2,0)",
    "fact_spans": "[[[2, 5], [27, 28]], [[2, 25]]]",
    "query_spans": "[[[27, 36]]]",
    "process": ""
  },
  {
    "text": "Given that two tangents are drawn from point $T(-2,2)$ to the parabola $C$: $y^{2}=2 p x$, with points of tangency $A$ and $B$, respectively, and that line $AB$ passes through the focus $F$ of parabola $C$, then $|T A|^{2}+|T B|^{2}$=?",
    "fact_expressions": "T: Point;Coordinate(T) = (-2, 2);TangentOfPoint(T,C) = {l1,l2};l1: Line;l2: Line;C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;TangentPoint(l1,C) = A;TangentPoint(l2,C) = B;A: Point;B: Point;Focus(C) = F;F: Point;PointOnCurve(F,LineOf(A,B)) = True",
    "query_expressions": "Abs(LineSegmentOf(T, A))^2 + Abs(LineSegmentOf(T, B))^2",
    "answer_expressions": "100",
    "fact_spans": "[[[3, 13]], [[3, 13]], [[2, 40]], [], [], [[14, 35], [63, 69]], [[14, 35]], [[22, 35]], [[2, 53]], [[2, 53]], [[46, 49]], [[50, 53]], [[63, 75]], [[72, 75]], [[54, 75]]]",
    "query_spans": "[[[77, 100]]]",
    "process": "Let $ A(x_{1},y_{1}) $ be on the parabola $ C: y^{2} = 2px $. The slope of the line tangent to the parabola at point $ A(x_{1},y_{1}) $ is $ k $. Then the equation of the tangent line at $ A(x_{1},y_{1}) $ is $ y - y_{1} = k(x - x_{1}) $. Solving the system of equations and simplifying yields $ y^{2} - \\frac{2p}{k}y + \\frac{2py_{1}}{k} - 2px_{1} = 0 $. Then $ \\Delta = \\left(-\\frac{2p}{k}\\right)^{2} - 4\\left(\\frac{2py_{1}}{k} - 2px_{1}\\right) = 0 $, simplifying gives $ 4p^{2} - 8kpy_{1} + 4k^{2}y_{1}^{2} = 0 $, so $ (2p - 2ky_{1})^{2} = 0 $, solving yields $ k = \\frac{p}{y_{1}} $. Therefore, the equation of the tangent line at $ A(x_{1},y_{1}) $ is $ y - y_{1} = \\frac{p}{y_{1}}(x - x_{1}) $, i.e., $ yy_{1} = p(x + x_{1}) $. Similarly, let $ B(x_{2},y_{2}) $ be on the parabola $ C: y^{2} = 2px $, the tangent line at $ B $ is $ yy_{2} = p(x + x_{2}) $. Since $ T(-2,2) $ lies on the tangent lines $ yy_{1} = p(x + x_{1}) $ and $ yy_{2} = p(x + x_{2}) $, we have $ 2y_{2} = p(-2 + x_{2}) $, $ 2y_{1} = p(-2 + x_{1}) $. Thus, the equation of line $ AB $ is $ 2y = p(x - 2) $. Since line $ AB $ passes through the focus $ F $ of the parabola, setting $ y = 0 $ gives $ x = 2 $, so $ F(2,0) $. Hence, the equation of the parabola is $ y^{2} = 8x $, and the equation of line $ AB $ is $ y = 2(x - 2) $. Solving the system and simplifying yields $ x^{2} - 6x + 4 = 0 $ or $ y^{2} - 4y - 16 = 0 $, so $ x_{1} + x_{2} = 6 $, $ x_{1}x_{2} = 4 $, $ y_{1} + y_{2} = 4 $, $ y_{1}y_{2} = -16 $, thus $ |TA|^{2} + |TB|^{2} = (x_{1} + 2)^{2} + (y_{1} - 2)^{2} + (x_{2} + 2)^{2} + (y_{2} - 2)^{2} = x_{1}^{2} + x_{2}^{2} + 4(x_{1} + x_{2}) + y_{1}^{2} + y_{2}^{2} - 4(y_{1} + y_{2}) + 16 = (x_{1} + x_{2})^{2} - 2x_{1}x_{2} + 4(x_{1} + x_{2}) + (y_{1} + y_{2})^{2} - 2y_{1}y_{2} - 4(y_{1} + y_{2}) + 16 = 36 - 8 + 4 \\times 6 + 16 - 2 \\times (-16) - 4 \\times 4 + 16 = 100 $"
  },
  {
    "text": "If the right focus of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{3}=1$ coincides with the focus of the parabola $y^{2}=12 x$, then $m=$?",
    "fact_expressions": "G: Hyperbola;m: Number;H: Parabola;Expression(G) = (-y^2/3 + x^2/m = 1);Expression(H) = (y^2 = 12*x);RightFocus(G) = Focus(H)",
    "query_expressions": "m",
    "answer_expressions": "6",
    "fact_spans": "[[[1, 39]], [[66, 69]], [[44, 59]], [[1, 39]], [[44, 59]], [[1, 64]]]",
    "query_spans": "[[[66, 71]]]",
    "process": "[Analysis] Since the foci of the hyperbola and the parabola coincide, we obtain the hyperbola parameter $ c = 3 $. Using $ a^{2} + b^{2} = c^{2} $, we can find $ m $. From the given condition: the focus of $ y^{2} = 12x $ is $ (3,0) $, and the right focus of the hyperbola $ \\frac{x^{2}}{m} - \\frac{y^{2}}{3} = 1 $ coincides with the focus of the parabola, therefore $ m + 3 = 9 $, so $ m = 6 $."
  },
  {
    "text": "The focus $F$ of the parabola $C$: $y^{2}=2 p x$ is the center of the circle $x^{2}+y^{2}-2 x=0$, and $P$ is a point on the parabola $C$ in the first quadrant such that $P F=3$. Then the coordinates of point $P$ are?",
    "fact_expressions": "C: Parabola;p: Number;G: Circle;P: Point;F: Point;Expression(C) = (y^2 = 2*(p*x));Expression(G) = (-2*x + x^2 + y^2 = 0);Quadrant(P) = 1;PointOnCurve(P, C);LineSegmentOf(P, F) = 3;Focus(C)=F;Center(G)=F",
    "query_expressions": "Coordinate(P)",
    "answer_expressions": "(2,2*sqrt(2))",
    "fact_spans": "[[[0, 21], [56, 62]], [[8, 21]], [[28, 48]], [[52, 55], [82, 86]], [[24, 27]], [[0, 21]], [[28, 48]], [[52, 71]], [[52, 71]], [[73, 80]], [[0, 27]], [[24, 51]]]",
    "query_spans": "[[[82, 91]]]",
    "process": "According to the problem, find p, then use the focal radius formula of the parabola to solve. From the circle equation $x^{2}+y^{2}-2x=0$, we obtain $(x-1)^{2}+y^{2}=1$, so the center is $(1,0)$, thus $F(1,0)$, that is, $\\frac{p}{2}=1$, solving gives $p=2$. The standard equation of parabola $C$ is $y^{2}=4x$. Let point $P(x_{p},y_{p})$, where $x_{p}>0$, $y_{p}>0$. Since $PF=3$, we have $x_{p}+\\frac{p}{2}=3$, solving gives $x_{p}=2$. Substituting into the parabola equation yields $y_{p}^{2}=8$, solving gives $y_{p}=2\\sqrt{2}$. Therefore, the coordinates of point $P$ are $(2,2\\sqrt{2})$."
  },
  {
    "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, a line passing through the origin with an inclination angle of $\\frac{\\pi}{6}$ intersects the ellipse at points $P$ and $Q$. The circle with diameter $PQ$ passes through the right focus $F$. Then, the eccentricity of the ellipse is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;O: Origin;PointOnCurve(O, L) = True;Inclination(L) = pi/6;Intersection(L, G) = {P, Q};Q: Point;P: Point;L: Line;IsDiameter(LineSegmentOf(P, Q), H) = True;H: Circle;F: Point;RightFocus(G) = F;PointOnCurve(F, H) = True",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[2, 54], [83, 85], [118, 120]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[56, 58]], [[55, 82]], [[61, 82]], [[80, 95]], [[90, 93]], [[86, 89]], [[80, 82]], [[96, 109]], [[108, 109]], [[113, 116]], [[83, 116]], [[108, 116]]]",
    "query_spans": "[[[118, 125]]]",
    "process": "From the given conditions, the equation of line PQ is $ y = \\tan\\frac{\\pi}{6}x = \\frac{\\sqrt{3}}{3}x $. Let $ P(x_{1}, y_{1}) $, $ Q(x_{2}, y_{2}) $. The circle with diameter PQ passes through the right focus $ F $, then $ FP \\perp FQ $, so $ \\overrightarrow{FP} \\cdot \\overrightarrow{FQ} = 0 $. Substituting $ y = \\frac{\\sqrt{3}}{3}x $ into $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ gives: $ (a^{2} + 3b^{2})x^{2} - 3a^{2}b^{2} = 0 $. Then $ x_{1} + x_{2} = 0 $, $ x_{1}x_{2} = \\frac{-3a^{2}b^{2}}{a^{2} + 3b^{2}} $. Therefore, $ \\overrightarrow{FP} \\cdot \\overrightarrow{FQ} = (c - x_{1})(c - x_{2}) + y_{1}y_{2} = c^{2} - c(x_{1} + x_{2}) + x_{1}x_{2} + y_{1}y_{2} = c^{2} - c(x_{1} + x_{2}) + \\frac{4}{3}x_{1}x_{2} = c^{2} + \\frac{4}{3} \\times \\frac{-3a^{2}b^{2}}{a^{2} + 3b^{2}} = 0 $. Rearranging yields: $ (a^{2} + 3b^{2})c^{2} - 4a^{2}b^{2} = 0 $. Since $ b^{2} = a^{2} - c^{2} $, we have $ [a^{2} + 3(a^{2} - c^{2})]c^{2} - 4a^{2}(a^{2} - c^{2}) = 0 $, that is, $ 3c^{4} - 8a^{2}c^{2} + 4a^{4} = 0 $, which factors as $ (3c^{2} - 2a^{2})(c^{2} - 2a^{2}) = 0 $. Solving gives $ \\frac{c^{2}}{a^{2}} = \\frac{2}{3} $ or $ \\frac{c^{2}}{a^{2}} = 2 $. Thus, $ e = \\frac{\\sqrt{6}}{3} $ or $ e = \\sqrt{2} $ (discarded)."
  },
  {
    "text": "Given that the parabola $y^{2}=4 x$ intersects the line $y=2 x-4$ at points $A$ and $B$, if there exists a point $C$ on the parabola such that $\\overrightarrow{O A}+\\overrightarrow{O B}=\\lambda \\overrightarrow{O C}$ (where $O$ is the origin), then the real number $\\lambda$=?",
    "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Line;Expression(H) = (y = 2*x - 4);A: Point;B: Point;Intersection(G, H) = {A, B};C: Point;PointOnCurve(C, G);O: Origin;lambda: Real;VectorOf(O, A) + VectorOf(O, B) = lambda*VectorOf(O, C)",
    "query_expressions": "lambda",
    "answer_expressions": "1/5",
    "fact_spans": "[[[2, 16], [46, 49]], [[2, 16]], [[17, 28]], [[17, 28]], [[30, 33]], [[36, 39]], [[2, 41]], [[52, 56]], [[46, 56]], [[132, 135]], [[143, 154]], [[59, 131]]]",
    "query_spans": "[[[143, 156]]]",
    "process": ""
  },
  {
    "text": "Given the ellipse $\\Gamma$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, through a point $P$ in the plane, two mutually perpendicular lines $l_{1}$, $l_{2}$ are drawn intersecting $\\Gamma$ at points $A$, $B$ and $C$, $D$ respectively. If $\\frac{1}{|A P| \\cdot|B P|}+\\frac{1}{|C P| \\cdot|D P|}=\\frac{41}{256}$, then the minimum value of $|O P|$ is?",
    "fact_expressions": "Gamma: Ellipse;Expression(Gamma) = (x^2/25 + y^2/16 = 1);P: Point;l1: Line;l2: Line;PointOnCurve(P,l1) = True;PointOnCurve(P,l2) = True;IsPerpendicular(l1,l2) = True;Intersection(l1,Gamma) = {A,B};A: Point;B: Point;Intersection(l2,Gamma) = {C,D};C: Point;D: Point;O: Origin;1/(Abs(LineSegmentOf(C, P))*Abs(LineSegmentOf(D, P))) + 1/(Abs(LineSegmentOf(A, P))*Abs(LineSegmentOf(B, P))) = 41/256",
    "query_expressions": "Min(Abs(LineSegmentOf(O, P)))",
    "answer_expressions": "12/5",
    "fact_spans": "[[[2, 51], [89, 97]], [[2, 51]], [[58, 61]], [[69, 78]], [[80, 88]], [[52, 88]], [[52, 88]], [[62, 88]], [[69, 117]], [[102, 105]], [[106, 109]], [[69, 117]], [[110, 113]], [[114, 117]], [[191, 198]], [[119, 189]]]",
    "query_spans": "[[[191, 204]]]",
    "process": ""
  },
  {
    "text": "The focus of the parabola $y=a x^{2}$ coincides with the upper focus of the hyperbola $y^{2}-x^{2}=2$. Then $a$=?",
    "fact_expressions": "G: Hyperbola;H: Parabola;a: Number;Expression(G) = (-x^2 + y^2 = 2);Expression(H) = (y = a*x^2);Focus(H) =UpperFocus(G)",
    "query_expressions": "a",
    "answer_expressions": "1/8",
    "fact_spans": "[[[20, 38]], [[0, 14]], [[44, 47]], [[20, 38]], [[0, 14]], [[0, 42]]]",
    "query_spans": "[[[44, 49]]]",
    "process": "Transforming the hyperbola into standard form, we obtain that its foci lie on the y-axis and $ a^{2} = b^{2} = 2 $, giving the upper focus coordinate $ (0,2) $. Transforming the parabola $ y = ax^{2} $ into standard form, we find its focus is $ F\\left(0, \\frac{1}{4a}\\right) $. According to the given condition, $ \\frac{1}{4a} = 2 $. Solving this equation yields the value of the real number $ a $.  \nSolution: The hyperbola $ y^{2} - x^{2} = 2 $ is transformed into standard form as $ \\frac{y^{2}}{2} - \\frac{x^{2}}{2} = 1 $. Therefore, the foci of the hyperbola lie on the y-axis with $ a^{2} = b^{2} = 2 $. Hence, the semi-focal length of the hyperbola is $ c = \\sqrt{a^{2} + b^{2}} = 2 $, giving the upper focus coordinate $ (0,2) $. Since the parabola $ y = ax^{2} $ can be written as $ x^{2} = \\frac{1}{a}y $, its focus is $ F\\left(0, \\frac{1}{4a}\\right) $, and $ F $ is one of the foci of the hyperbola, so $ \\frac{1}{4a} = 2 $, solving which gives $ a = \\frac{1}{8} $."
  },
  {
    "text": "Given that the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ and the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{7}=1$ have coinciding foci, what is the eccentricity of this hyperbola?",
    "fact_expressions": "G: Hyperbola;a: Number;H: Ellipse;Expression(G) = (-y^2/7 + x^2/a = 1);Expression(H) = (x^2/25 + y^2/9 = 1);Focus(G) = Focus(H)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "4/3",
    "fact_spans": "[[[41, 79], [86, 89]], [[44, 79]], [[2, 40]], [[41, 79]], [[2, 40]], [[2, 83]]]",
    "query_spans": "[[[86, 95]]]",
    "process": "Let the semi-focal distance of the ellipse be $ c $, then $ c = \\sqrt{25 - 9} = 4 $, and $ a + 7 = 16 $, so $ a = 9 $. Hence, the eccentricity of the hyperbola is $ \\frac{4}{\\sqrt{9}} = \\frac{4}{3} $."
  },
  {
    "text": "The minor axis has length $2 \\sqrt{5}$, and the eccentricity is $e=\\frac{2}{3}$. The foci of the ellipse are $F_{1}$ and $F_{2}$. A line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. Then, the perimeter of $\\triangle A B F_{2}$ is?",
    "fact_expressions": "Length(MinorAxis(G)) = 2*sqrt(5);e: Number;Eccentricity(G) = e;e = 2/3;G: Ellipse;F1: Point;F2: Point;Focus(G) = {F1, F2};PointOnCurve(F1, H) = True;H: Line;Intersection(H, G) = {A, B};A: Point;B: Point",
    "query_expressions": "Perimeter(TriangleOf(A, B, F2))",
    "answer_expressions": "12",
    "fact_spans": "[[[0, 38]], [[20, 35]], [[17, 38]], [[20, 35]], [[36, 38], [71, 73]], [[43, 50], [60, 67]], [[51, 58]], [[36, 58]], [[59, 70]], [[68, 70]], [[68, 83]], [[74, 77]], [[78, 81]]]",
    "query_spans": "[[[85, 110]]]",
    "process": "Since the minor axis length is $2\\sqrt{5}$ and the eccentricity is $e=\\frac{2}{3}$, we have $b=\\sqrt{5}$, $e=\\frac{c}{a}=\\frac{2}{3}$, and $a^{2}=b^{2}+c^{2}$. Solving gives $a=3$, so the perimeter of $\\triangle ABF_{2}$ is $l=AB+AF_{1}+BF_{1}=4a=12$."
  },
  {
    "text": "Given the curve $C$: $y^{2}=4x$, with focus $F$, and $P$ being any point on the parabola, what is the minimum value of the sum of the distances from point $P$ to the focus $F$ and to the point $A(2,1)$?",
    "fact_expressions": "C: Curve;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;G: Parabola;P: Point;PointOnCurve(P, G) = True;A: Point;Coordinate(A) = (2, 1)",
    "query_expressions": "Min(Distance(P, F) + Distance(P, A))",
    "answer_expressions": "3",
    "fact_spans": "[[[2, 20]], [[2, 20]], [[24, 27], [49, 52]], [[2, 27]], [[32, 35]], [[28, 31], [42, 46]], [[28, 40]], [[54, 63]], [[54, 63]]]",
    "query_spans": "[[[42, 74]]]",
    "process": ""
  },
  {
    "text": "Given that one asymptote of the hyperbola $x^{2}-\\frac{y^{2}}{m^{2}}=1$ $(m>0)$ is $x+\\sqrt{3} y=0$, then $m=$?",
    "fact_expressions": "G: Hyperbola;m: Number;m>0;Expression(G)=(x^2-y^2/m^2=1);Expression(OneOf(Asymptote(G)))=(x+sqrt(3)*y=0)",
    "query_expressions": "m",
    "answer_expressions": "sqrt(3)/3",
    "fact_spans": "[[[2, 39]], [[67, 70]], [[5, 39]], [[2, 39]], [[2, 64]]]",
    "query_spans": "[[[67, 72]]]",
    "process": "The asymptote equation of $x^{2}-\\frac{y^{2}}{m^{2}}=1$ is $x^{2}-\\frac{y^{2}}{m^{2}}=0$, so $\\frac{1}{m^{2}}=(\\sqrt{3})^{2}$. Since $m>0$, $\\therefore m=\\frac{\\sqrt{3}}{3}$."
  },
  {
    "text": "What is the focal distance of the ellipse $2 x^{2}+y^{2}=1$?",
    "fact_expressions": "G: Ellipse;Expression(G) = (2*x^2 + y^2 = 1)",
    "query_expressions": "FocalLength(G)",
    "answer_expressions": "sqrt(2)",
    "fact_spans": "[[[0, 19]], [[0, 19]]]",
    "query_spans": "[[[0, 24]]]",
    "process": "2x^{2}+y^{2}=1 can be rewritten as \\frac{x^{2}}{2}+y^{2}=1, let the focal distance be 2c, then c=\\sqrt{1-\\frac{1}{2}}=\\frac{\\sqrt{2}}{2}, thus the focal distance 2c=\\sqrt{2}"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{8}-y^{2}=1$ has its right focus at point $F$, and asymptotes $l_{1}$, $l_{2}$. A line $l$ passing through point $F$ intersects $l_{1}$, $l_{2}$ at points $A$, $B$, respectively. If $A B \\perp l_{2}$, then $|A B|$=?",
    "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/8 - y^2 = 1);RightFocus(C) = F;F: Point;Asymptote(C) = {l1,l2};l1: Line;l2: Line;PointOnCurve(F,l) = True;l: Line;Intersection(l,l1) = A;Intersection(l,l2) = B;A: Point;B: Point;IsPerpendicular(LineSegmentOf(A,B),l2) = True",
    "query_expressions": "Abs(LineSegmentOf(A, B))",
    "answer_expressions": "16/7",
    "fact_spans": "[[[2, 35]], [[2, 35]], [[2, 43]], [[40, 43], [66, 70]], [[2, 64]], [[48, 55], [77, 84]], [[57, 64], [86, 93]], [[65, 76]], [[71, 76]], [[71, 106]], [[71, 106]], [[99, 102]], [[103, 106]], [[109, 126]]]",
    "query_spans": "[[[128, 137]]]",
    "process": "Since the hyperbola equation is $\\frac{x^{2}}{8}-y^{2}=1$, and its asymptotes $l_{1}, l_{2}$ are $y=-\\frac{\\sqrt{2}}{4}x$, $y=\\frac{\\sqrt{2}}{4}x$, and line $AB \\perp l_{2}$, the slope of line $AB$ is $-2\\sqrt{2}$. The coordinates of its right focus $F$ are $(3,0)$. Thus, the equation of line $AB$ is $y=-2\\sqrt{2}(x-3)$. Solving simultaneously $y=-\\frac{\\sqrt{2}}{4}x$, $y=-2\\sqrt{2}(x-3)^{2}$ gives $A(\\frac{24}{7},-\\frac{6\\sqrt{2}}{7})$. From $y=\\frac{\\sqrt{2}}{4}x$, $y=-\\therefore |AB|=\\sqrt{(\\frac{24}{7}-\\frac{8}{3}}\\frac{x-3)}{6\\sqrt{2}}-\\frac{c\\sqrt{2}}{3})^{\\frac{8}{3},\\frac{2\\sqrt{2}}{3}}$"
  },
  {
    "text": "The ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{a^{2}}=1$ and the hyperbola $\\frac{y^{2}}{2}-\\frac{x^{2}}{a}=1$ have the same foci, then $a$=?",
    "fact_expressions": "H: Ellipse;Expression(H) = (x^2/4 + y^2/a^2 = 1);a: Number;G: Hyperbola;Expression(G) = (y^2/2 - x^2/a = 1);Focus(H) = Focus(G)",
    "query_expressions": "a",
    "answer_expressions": "3",
    "fact_spans": "[[[0, 41]], [[0, 41]], [[88, 91]], [[42, 80]], [[42, 80]], [[0, 86]]]",
    "query_spans": "[[[88, 93]]]",
    "process": "The ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{a^{2}}=1$ and the hyperbola $\\frac{y^{2}}{2}-\\frac{x^{2}}{a}=1$ have the same foci, so:  \n\\begin{cases}a^{2}>4\\\\a^{2}-4=2+a\\end{cases},  \nsolving gives $a=3$."
  },
  {
    "text": "If the distance from point $P$ to point $F(4,0)$ is greater than its distance to the line $l$: $x+3=0$ by $1$, then the equation of the trajectory of point $P$ is?",
    "fact_expressions": "l: Line;F: Point;P: Point;Coordinate(F) = (4, 0);Expression(l) = (x+3=0);Distance(P,F)=Distance(P,l)+1",
    "query_expressions": "LocusEquation(P)",
    "answer_expressions": "y^2=16*x",
    "fact_spans": "[[[21, 35]], [[6, 15]], [[1, 5], [19, 20], [44, 48]], [[6, 15]], [[21, 35]], [[1, 42]]]",
    "query_spans": "[[[44, 56]]]",
    "process": ""
  },
  {
    "text": "The eccentricity of the ellipse $\\frac{y^{2}}{3}+x^{2}=1$ is?",
    "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + y^2/3 = 1)",
    "query_expressions": "Eccentricity(G)",
    "answer_expressions": "sqrt(6)/3",
    "fact_spans": "[[[0, 27]], [[0, 27]]]",
    "query_spans": "[[[0, 33]]]",
    "process": "Find the values of a, b, and c, then determine the eccentricity of the ellipse \\frac{y^{2}}{3}+x^{2}=1. In the ellipse \\frac{y^{2}}{3}+x^{2}=1, a=\\sqrt{3}, b=1, c=\\sqrt{a^{2}-b^{2}}=\\sqrt{2}. Therefore, the eccentricity of the ellipse \\frac{y^{2}}{3}+x^{2}=1 is e=\\frac{c}{a}=\\frac{\\sqrt{2}}{\\sqrt{3}}=\\frac{\\sqrt{6}}{3}"
  },
  {
    "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, with left and right foci $F_{1}$, $F_{2}$ respectively, and $O$ as the coordinate origin. Let point $M$ be a point on the right branch of the hyperbola. If $|F_{1} F_{2}|=2|O M|$, and $\\tan \\angle M F_{2} F_{1} \\geq 2$, then the range of the eccentricity of hyperbola $C$ is?",
    "fact_expressions": "C: Hyperbola;b: Number;a: Number;F1: Point;F2: Point;O: Origin;M: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) =F1;RightFocus(C)=F2;PointOnCurve(M, RightPart(C));Abs(LineSegmentOf(F1, F2)) = 2*Abs(LineSegmentOf(O, M));Tan(AngleOf(M, F2, F1)) >= 2",
    "query_expressions": "Range(Eccentricity(C))",
    "answer_expressions": "(1,\\sqrt{5}]",
    "fact_spans": "[[[2, 63], [101, 104], [172, 178]], [[10, 63]], [[10, 63]], [[71, 78]], [[79, 86]], [[87, 90]], [[96, 100]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 86]], [[2, 86]], [[96, 109]], [[111, 133]], [[136, 170]]]",
    "query_spans": "[[[172, 189]]]",
    "process": "Method one, according to the properties of right triangles and the Pythagorean theorem, $\\angle F_{1}MF_{2}=\\frac{\\pi}{2}$, $4c^{2}=|MF_{1}|^{2}+|MF_{2}|^{2}$, $\\tan\\angle MF_{2}F_{1}=\\frac{|MF_{1}|}{|MF_{2}|}$, and from the definition of the hyperbola, $|MF_{1}|-|MF_{2}|=2a$, express the eccentricity as an expression in terms of $|MF_{1}|$, $|MF_{2}|$, then let $\\frac{|MF_{1}|}{|MF_{2}|}=t\\geqslant2$, then $e^{2}=1+\\frac{2}{t+\\frac{1}{t}-2}$, let $f(t)=t+\\frac{1}{t}$, study the monotonicity of the function on $[2,+\\infty)$ by taking derivative, thus the range of eccentricity can be found. Method two: let $|MF_{1}|=r_{1}$, $|MF_{2}|=r_{2}$, $\\angle MF_{2}F_{1}=\\theta$, $\\tan\\theta\\geqslant2$, $r_{1}=2c\\sin\\theta$, according to the properties of right triangles and the Pythagorean theorem, $\\angle F_{1}MF_{2}=\\frac{\\pi}{2}$, express the eccentricity as a trigonometric function of angle $\\theta$, transform it into a function of $\\tan\\theta$ using trigonometric identities, thus the range of eccentricity can be obtained. Solution: Method one: $\\because |F_{1}F_{2}|=2|OM|$, $\\therefore \\angle F_{1}MF_{2}=\\frac{\\pi}{2}$, $\\therefore 4c^{2}=|MF_{1}|^{2}+|MF_{2}|^{2}$, $\\tan\\angle MF_{2}F_{1}=\\frac{|MF_{1}|}{|MF_{2}|}$, $\\because |MF_{1}|-|MF_{2}|=2a$, $\\therefore e^{2}=\\frac{4c^{2}}{4a^{2}}=\\frac{|MF_{1}|^{2}+|MF_{2}|^{2}}{(|MF_{1}|-|MF_{2}|)^{2}}$. Let $\\frac{|MF_{1}|}{|MF_{2}|}=t\\geqslant2$, then $e^{2}=\\frac{t^{2}+1}{t^{2}-2t+1}$. Let $f(t)=t+\\frac{1}{t}$, $f'(t)=1-\\frac{1}{t^{2}}=\\frac{t^{2}-1}{t^{2}}$, so when $t>2$, $f'(t)>0$, $f(t)$ is increasing on $[2,+\\infty)$, $\\therefore t+\\frac{1}{t}\\geqslant2+\\frac{1}{2}=\\frac{5}{2}$, $\\therefore 1<e^{2}\\leqslant5$, $\\therefore 1<e\\leqslant\\sqrt{5}$. Method two: $\\because |F_{1}F_{2}|=2|OM|$, $\\therefore \\angle F_{1}MF_{2}=\\frac{\\pi}{2}$, let $|MF_{1}|=r_{1}$, $|MF_{2}|=r_{2}$, $\\angle MF_{2}F_{1}=\\theta$, $\\tan\\theta\\geqslant2$, $r_{1}=2c\\sin\\theta$, $r_{2}=2c\\cos\\theta$, $\\therefore 2a=r_{1}-r_{2}=2c(\\sin\\theta-\\cos\\theta)$, $e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{1}{(\\sin\\theta-\\cos\\theta)^{2}}=\\frac{1}{\\sin^{2}\\theta+\\cos^{2}\\theta-2\\sin\\theta\\cos\\theta}=\\frac{1}{1-\\sin 2\\theta}=\\frac{\\tan^{2}\\theta+1}{\\tan^{2}\\theta+1-2\\tan\\theta}=1+\\frac{2}{\\tan\\theta+\\frac{1}{\\tan\\theta}-2}\\leqslant5$, $\\therefore 1<e\\leqslant\\sqrt{5}$."
  },
  {
    "text": "If the distance from point $P$ on the parabola $x^{2}=8y$ to the focus is $12$, then what is the distance from $P$ to the $x$-axis?",
    "fact_expressions": "G: Parabola;P: Point;Expression(G) = (x^2 = 8*y);PointOnCurve(P, G);Distance(P, Focus(G)) = 12",
    "query_expressions": "Distance(P, xAxis)",
    "answer_expressions": "10",
    "fact_spans": "[[[1, 15]], [[17, 21], [34, 37]], [[1, 15]], [[1, 21]], [[1, 32]]]",
    "query_spans": "[[[34, 47]]]",
    "process": "Since the parabola is $x^{2}=8y$, the focus coordinates are $(0,2)$, and the directrix equation is $y=-2$. Because the distance from point $P$ to the focus is $12$, according to the definition of a parabola, the distance from $P$ to the directrix is also $12$. Therefore, the distance from point $P$ to the $x$-axis is $10$."
  },
  {
    "text": "Given the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, $P$ is a point on the right branch of the hyperbola, and the coordinates of point $Q$ are $(-2,3)$. Then the minimum value of $|P Q|+|P F_{1}|$ is?",
    "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, RightPart(G));Q: Point;Coordinate(Q) = (-2, 3)",
    "query_expressions": "Min(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, Q)))",
    "answer_expressions": "5+2*sqrt(3)",
    "fact_spans": "[[[2, 30], [58, 61]], [[2, 30]], [[38, 45]], [[46, 53]], [[2, 53]], [[2, 53]], [[54, 57]], [[54, 66]], [[67, 71]], [[67, 83]]]",
    "query_spans": "[[[85, 108]]]",
    "process": ""
  }
]