[ { "text": "From the left focus $F_{1}$ of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{25}=1$, draw a tangent to the circle $x^{2}+y^{2}=16$, with the point of tangency $T$. Extend $F_{1} T$ to intersect the right branch of the hyperbola at point $P$. Let $M$ be the midpoint of segment $F_{1} P$, and let $O$ be the origin. Then $|M O|-|M T|$=?", "fact_expressions": "G: Hyperbola;H: Circle;F1: Point;Z: Line;T: Point;P: Point;M: Point;Expression(G) = (x^2/16 - y^2/25 = 1);LeftFocus(G) = F1;Expression(H) = (x^2 + y^2 = 16);TangentOfPoint(F1, H) = Z;TangentPoint(Z, H) = T;Intersection(OverlappingLine(LineSegmentOf(F1, T)), RightPart(G)) = P;MidPoint(LineSegmentOf(F1, P)) = M;O: Origin", "query_expressions": "Abs(LineSegmentOf(M, O)) - Abs(LineSegmentOf(M, T))", "answer_expressions": "1", "fact_spans": "[[[1, 41], [93, 96]], [[53, 70]], [[45, 52]], [], [[77, 80]], [[99, 103]], [[105, 108]], [[1, 41]], [[1, 52]], [[53, 70]], [[0, 73]], [[0, 80]], [[81, 103]], [[105, 123]], [[124, 127]]]", "query_spans": "[[[134, 149]]]", "process": "Let F' be the right focus of the hyperbola, connect PF'. Since M and O are the midpoints of F_{1}P and F_{1}F' respectively, then |MO| = \\frac{1}{2}|PF'|. By the definition of the hyperbola, |F_{1}P| - |PF'| = 8. Hence, |MO| - |MT| = \\frac{1}{2}|PF| - |MF_{1}| + |F_{1}T| = \\frac{1}{2}(|PF'| - |F_{1}P|) + |F_{1}T| = -4 + 5 = 1" }, { "text": "Given the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ with left and right foci $F_{1}$, $F_{2}$, point $P$ is a moving point on the ellipse. Then the range of values of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ is?", "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G)", "query_expressions": "Range(DotProduct(VectorOf(P, F1), VectorOf(P, F2)))", "answer_expressions": "[-2, 1]", "fact_spans": "[[[2, 29], [57, 59]], [[52, 56]], [[36, 43]], [[44, 51]], [[2, 29]], [[2, 51]], [[2, 51]], [[52, 62]]]", "query_spans": "[[[64, 128]]]", "process": "Let P(x,y) be an arbitrary point on the ellipse, then \\overrightarrow{PF_{1}}=(-\\sqrt{3}-x,-y), \\overrightarrow{PF_{2}}=(\\sqrt{3}-x,-y). Therefore, \\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=(-\\sqrt{3}-x,-y)\\cdot(\\sqrt{3}-x,-y)=x^{2}+y^{2}-3=x^{2}+1-\\frac{x^{2}}{4}=\\frac{3}{4}x^{2}-2. Since P lies on the ellipse, -2\\leqslant x \\leqslant 2, so -2\\leqslant \\frac{3}{4}x^{2}-2 \\leqslant 1, that is, the range of \\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}} is [-2,1]." }, { "text": "The coordinates of the focus of the parabola $y^{2}=4 x$ are?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(1, 0)", "fact_spans": "[[[0, 14]], [[0, 14]]]", "query_spans": "[[[0, 21]]]", "process": "The focus of the parabola y^{2}=4x lies on the x-axis, and p=2, \\therefore\\frac{p}{2}=1, so the focus coordinates of the parabola y^{2}=4x are (1,0)." }, { "text": "It is known that the focus $F$ of the parabola $y^{2}=4x$ is exactly the right vertex of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and the equations of the asymptotes are $y=\\pm \\sqrt{3}x$. Then the equation of the hyperbola is?", "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 4*x);F: Point;Focus(H) = F;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;RightVertex(G) = F;Expression(Asymptote(G)) = (y = pm*sqrt(3)*x)", "query_expressions": "Expression(G)", "answer_expressions": "x^2-y^2/3=1", "fact_spans": "[[[2, 16]], [[2, 16]], [[19, 22]], [[2, 22]], [[25, 81], [113, 116]], [[25, 81]], [[28, 81]], [[28, 81]], [[28, 81]], [[28, 81]], [[19, 85]], [[25, 111]]]", "query_spans": "[[[113, 120]]]", "process": "" }, { "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{n}+\\frac{y^{2}}{12-n}=-1 (n>0)$ is $\\sqrt{3}$, then $n=?$", "fact_expressions": "G: Hyperbola;Expression(G) = (y^2/(12 - n) + x^2/n = -1);n: Number;n>0;Eccentricity(G) = sqrt(3)", "query_expressions": "n", "answer_expressions": "Preserve all mathematical expressions, LaTeX formulas, symbols, and equations EXACTLY. \nDo not add explanations, do not rewrite formulas, do not omit any content. \n\nText:", "fact_spans": "[[[2, 51]], [[2, 51]], [[68, 71]], [[5, 51]], [[2, 66]]]", "query_spans": "[[[68, 73]]]", "process": "" }, { "text": "Given that the asymptotes of the hyperbola $\\Gamma$ are $y=\\pm \\sqrt{3} x$, and that point $P(\\sqrt{6}, 3)$ lies on $\\Gamma$, then the standard equation of the hyperbola $\\Gamma$ is?", "fact_expressions": "Gamma: Hyperbola;P: Point;Coordinate(P) = (sqrt(6), 3);Expression(Asymptote(Gamma))=(y=pm*sqrt(3)*x);PointOnCurve(P,Gamma)", "query_expressions": "Expression(Gamma)", "answer_expressions": "x^2/3-y^2/9=1", "fact_spans": "[[[2, 13], [57, 65], [71, 82]], [[40, 56]], [[40, 56]], [[2, 38]], [[40, 69]]]", "query_spans": "[[[71, 89]]]", "process": "When the foci of the hyperbola lie on the x-axis, from the asymptote equations $ y = \\pm\\sqrt{3}x $, we know that $ \\frac{b}{a} = \\sqrt{3} $, so $ b = \\sqrt{3}a $. Assume the hyperbola equation is $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{3a^{2}} = 1 $. Substituting point $ P(\\sqrt{6}, 3) $ gives $ \\frac{6}{a^2} - \\frac{9}{3a^{2}} = 1 $. Solving yields $ a^{2} = 3 $, $ b^{2} = 9 $. Hence, the hyperbola equation is $ \\frac{x^{2}}{3} - \\frac{y^{2}}{9} = 1 $. When the foci of the hyperbola lie on the y-axis, from the asymptote equations $ y = \\pm\\sqrt{3}x $, we know that $ \\frac{a}{b} = \\sqrt{3} $, so $ a = \\sqrt{3}b $. Assume the hyperbola equation is $ \\frac{y^{2}}{3b^{2}} - \\frac{x^{2}}{b^{2}} = 1 $. Substituting point $ P(\\sqrt{6}, 3) $ gives $ \\frac{9}{3b^{2}} - \\frac{6}{b^{2}} = 1 $, which has no solution. Therefore, fill in: $ \\frac{x_{2}}{2} - \\frac{y^{2}}{a} = 1 = 1 $. This question primarily examines finding the standard equation of a hyperbola given its asymptote equations and a point on the hyperbola, and belongs to basic problems." }, { "text": "Given the circle $(x+1)^{2}+y^{2}=36$ with center $M$, let $A$ be any point on the circle, and let $N(1 , 0)$. The perpendicular bisector of segment $AN$ intersects $MA$ at point $P$. Then the trajectory equation of the moving point $P$ is?", "fact_expressions": "G: Circle;Expression(G) = (y^2 + (x + 1)^2 = 36);Center(G) = M;M: Point;A: Point;PointOnCurve(A, G);N: Point;Coordinate(N) = (1, 0);Intersection(PerpendicularBisector(LineSegmentOf(A,N)),LineSegmentOf(M,A)) = P;P: Point", "query_expressions": "LocusEquation(P)", "answer_expressions": "x^2/9+y^2/8=1", "fact_spans": "[[[2, 23], [36, 37]], [[2, 23]], [[2, 30]], [[27, 30]], [[32, 35]], [[32, 41]], [[43, 54]], [[43, 54]], [[55, 79]], [[75, 79], [83, 86]]]", "query_spans": "[[[83, 93]]]", "process": "Since the circle $(x+2)^{2}+y^{2}=36$ has center $M(-1,0)$ and radius $r=6$, let point $P(x,y)$. Since the perpendicular bisector of segment $AN$ intersects $MA$ at point $P$, we have $|PN|=|PA|$. Therefore, $|PM|+|PN|=|PM|+|PA|=|MA|=6>|MN|=2$. Hence, the trajectory of point $P$ is an ellipse with foci at $M(-1,0)$ and $N(1,0)$, and major axis length $2a=6$, so $b=\\sqrt{a^{2}-c^{2}}=\\sqrt{9-1}=2\\sqrt{2}$. Thus, the equation of the trajectory of moving point $P$ is $\\frac{x^{2}}{9}+\\frac{y^{2}}{8}=1$." }, { "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$ has foci at $F$, and the endpoints of the minor axis are $P$. If the line $PF$ is tangent to the circle $O$: $x^{2}+y^{2}=R^{2}$ $(R>0)$, then what is the radius of circle $O$?", "fact_expressions": "G: Ellipse;O: Circle;R: Number;F: Point;P: Point;Expression(G) = (x^2/4 + y^2/2 = 1);R>0;Expression(O) = (x^2 + y^2 = R^2);OneOf(Focus(G)) = F;OneOf(Endpoint(MinorAxis(G)))=P;IsTangent(LineOf(P,F),O)", "query_expressions": "Radius(O)", "answer_expressions": "1", "fact_spans": "[[[2, 39]], [[65, 95], [99, 103]], [[70, 95]], [[43, 46]], [[52, 55]], [[2, 39]], [[70, 95]], [[65, 95]], [[2, 46]], [[2, 55]], [[57, 97]]]", "query_spans": "[[[99, 108]]]", "process": "The foci of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$ are denoted by $F$, and the endpoints of the minor axis are denoted by $P$. Then $c=\\sqrt{a^{2}-b^{2}}=\\sqrt{2}$. Without loss of generality, take $F(\\sqrt{2},0)$, $P(0,\\sqrt{2})$. Then the equation of line $PF$ is: $y - x + \\sqrt{2} = 0$. Since the line $PF$ is tangent to the circle $O: x^{2}+y^{2}=R^{2}$ ($R>0$), it follows that $R = \\frac{|0 - 0 + \\sqrt{2}|}{\\sqrt{1^{2}+(-1)^{2}}} = 1$. The answer is 1. This problem examines the simple geometric properties of an ellipse and the point-to-line distance formula, testing basic computational skills, and is a fundamental problem." }, { "text": "If the equation $\\frac{x^{2}}{5-k}+\\frac{y^{2}}{k-3}=1$ represents an ellipse, then what is the range of values for $k$?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(5 - k) + y^2/(k - 3) = 1);k: Number", "query_expressions": "Range(k)", "answer_expressions": "(3,4)+(4,5)", "fact_spans": "[[[44, 46]], [[1, 46]], [[48, 51]]]", "query_spans": "[[[48, 58]]]", "process": "Since the equation \\frac{x^{2}}{k>0}+\\frac{y^{2}}{k-3}=1 represents an ellipse," }, { "text": "Let the parabola $C$: $y^{2}=4x$ have focus $F$, and let line $l$ passing through $F$ intersect the parabola at points $A$ and $B$. Let $M$ be the intersection point of the directrix of parabola $C$ and the $x$-axis. If $\\tan \\angle AMB = 2\\sqrt{2}$, then $|AB| = $?", "fact_expressions": "l: Line;C: Parabola;A: Point;M: Point;B: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Intersection(Directrix(C), xAxis) = M;Tan(AngleOf(A, M, B)) = 2*sqrt(2)", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "8", "fact_spans": "[[[33, 38]], [[1, 20], [39, 42], [58, 64]], [[44, 47]], [[54, 57]], [[48, 51]], [[24, 27], [29, 32]], [[1, 20]], [[1, 27]], [[28, 38]], [[33, 53]], [[54, 75]], [[77, 107]]]", "query_spans": "[[[109, 118]]]", "process": "Test Analysis: According to symmetry, as shown in the figure below, let $ my+1 $, $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. From\n\\[\n\\begin{cases}\ny^{2}=4x \\\\\nx=my+1\n\\end{cases}\n\\Rightarrow y^{2}-4my-4=0,\n\\]\nthus $ y_{1}+y_{2}=4m $, $ y_{1}y_{2}=-4 $, $ x_{1}x_{2}=\\frac{y_{1}^{2}}{4}\\cdot\\frac{y_{2}^{2}}{4}=1 $, $ x_{1}+x_{2}=m(y_{1}+y_{2})+2=4m^{2}+2 $. Also, since $ \\tan\\angle AMB = \\tan(\\angle AMF + \\angle BMF) $, $ \\frac{\\frac{y_{2}}{+1}}{-y_{2}}=2\\sqrt{2} \\Rightarrow \\frac{y_{1}(my_{2}+2)-y_{2}(my_{1}+2)}{(x_{1}+1)(x_{2}+1)+y_{1}y_{2}}=2\\sqrt{2} \\Rightarrow y_{1}-y_{2}=4\\sqrt{2}m^{2} $, thus $ 4\\sqrt{m^{2}+1}=4\\sqrt{2}m^{2} \\Rightarrow m^{2}=1 $, therefore $ AB=|AF|+|BF|=x_{1}+1+x_{2}+1=4m^{2}+4=8 $. Hence, fill in: 8." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola, respectively, and point $P$ lies on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, satisfying $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, and $|P F_{1}|=\\sqrt{3}|P F_{2}|$, then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Abs(LineSegmentOf(P, F1)) = sqrt(3)*Abs(LineSegmentOf(P, F2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)+1", "fact_spans": "[[[20, 23], [35, 91], [193, 196]], [[38, 91]], [[38, 91]], [[30, 34]], [[2, 9]], [[10, 17]], [[38, 91]], [[38, 91]], [[35, 91]], [[2, 29]], [[2, 29]], [[30, 97]], [[100, 159]], [[161, 190]]]", "query_spans": "[[[193, 201]]]", "process": "" }, { "text": "Given the ellipse equation $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$, a line passing through the point $M(1, \\frac{1}{2})$ intersects the ellipse at points $P$ and $Q$. If point $M$ is exactly the midpoint of segment $PQ$, then the equation of line $PQ$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/2 = 1);M: Point;Coordinate(M) = (1, 1/2);PointOnCurve(M, H) = True;H: Line;Intersection(H, G) = {P, Q};Q: Point;P: Point;MidPoint(LineSegmentOf(P, Q)) = M", "query_expressions": "Expression(LineOf(P,Q))", "answer_expressions": "2*x+2*y-3=0", "fact_spans": "[[[2, 4], [68, 70]], [[2, 42]], [[44, 64], [84, 88]], [[44, 64]], [[43, 67]], [[65, 67]], [[65, 82]], [[77, 80]], [[73, 76]], [[84, 100]]]", "query_spans": "[[[102, 114]]]", "process": "Let P(x_{1},y_{1}), Q(x_{2},y_{2}), the slope of line PQ is k. From the given conditions we have: \\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1, \\frac{x_{2}^{2}}{4}+\\frac{y_{2}^{2}}{2}=1. Subtracting these two equations yields: \\underline{(x_{1}+x_{2})(x_{1}-x_{2})} + (y_{1}+y_{2})(y_{1}-y_{2}) = 0, that is, \\frac{x_{1}+x_{2}}{4}+\\frac{y_{1}+y_{2}}{2}\\cdot\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=0. Since x_{1}+x_{2}=2, y_{1}+y_{2}=1, k=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}, it follows that \\frac{2}{4}+\\frac{1}{2}k=0. Solving gives: k=-1. Therefore, the equation of line PQ is 2x+2y-3=0." }, { "text": "Given that $F(1,0)$ is the focus of the parabola $P$: $y^{2}=2 p x(p>0)$, a line $l$ passing through point $F$ with slope $k$ intersects curve $P$ at points $B$ and $C$. The line passing through $O$ and the midpoint $M$ of $BC$ intersects curve $P$ at point $N$. Then the range of $\\frac {S_{\\Delta O M C}} {S_{\\Delta O B N}}$ is?", "fact_expressions": "l: Line;P: Parabola;p: Number;B: Point;C: Point;F: Point;O: Origin;M: Point;N: Point;p>0;Expression(P) = (y^2 = 2*(p*x));Coordinate(F) = (1, 0);Focus(P) = F;PointOnCurve(F, l);k:Number;Slope(l) = k;Intersection(l, P) = {B, C};L:Line;PointOnCurve(O,L);MidPoint(LineSegmentOf(B,C))=M;PointOnCurve(M,L);Intersection(L,P)=N", "query_expressions": "Range(Area(TriangleOf(O,M,C))/Area(TriangleOf(O,B,N)))", "answer_expressions": "(0,1/2)", "fact_spans": "[[[54, 59]], [[11, 37], [60, 65], [96, 101]], [[18, 37]], [[67, 70]], [[71, 74]], [[2, 10], [42, 46]], [[78, 81]], [[89, 92]], [[103, 107]], [[18, 37]], [[11, 37]], [[2, 10]], [[2, 40]], [[41, 59]], [[50, 53]], [[47, 59]], [[54, 76]], [[93, 95]], [[77, 95]], [[82, 92]], [[77, 95]], [[93, 107]]]", "query_spans": "[[[109, 162]]]", "process": "From the coordinates of the focus, the equation of the parabola is obtained as $ y^{2} = 4x $. By solving simultaneously with the equation of line $ l $, the coordinates of point $ M $ are found, and then the equation of line $ OM $ is determined. Solving this equation simultaneously with the parabola's equation gives the vertical coordinate of point $ N $, leading to a functional expression of $ \\frac{|OM|}{|ON|} $ in terms of $ k $, from which the range of values is derived—this range corresponds to the desired range of the ratio of areas.\n\nSince $ F(1,0) $ is the focus of the parabola $ P: y^2 = 2px $ ($ p > 0 $), it follows that $ \\frac{p}{2} = 1 $, so $ p = 2 $. Thus, the equation of parabola $ P $ is $ y^{2} = 4x $, denoted as $ \\textcircled{1} $. The line $ l $ passing through point $ F $ with slope $ k $ has the equation $ y = k(x - 1) $, denoted as $ \\textcircled{2} $. Combining $ \\textcircled{1} $ and $ \\textcircled{2} $, eliminating $ y $, and simplifying yields:\n$$\nk^{2}x^{2} - 2(k^{2} + 2)x + k^{2} = 0.\n$$\nThe x-coordinate of $ M $ is given by:\n$$\nx_{M} = \\frac{x_{B} + x_{C}}{2} = \\frac{k^{2} + 2}{k^{2}},\n$$\nand the y-coordinate of $ M $ is:\n$$\ny_{M} = k(x_{M} - 1) = \\frac{2}{k}.\n$$\nThus,\n$$\n\\frac{y_{M}}{x_{M}} = \\frac{2k}{k^{2} + 2}, \\quad \\text{denoted as } \\textcircled{3}.\n$$\nCombining $ \\textcircled{3} $ with $ \\textcircled{1} $, eliminating $ x $, we get:\n$$\ny = \\frac{2k}{k^{2} + 2} \\cdot \\frac{y^{2}}{4}.\n$$\nSince triangles $ \\triangle OMC $ and $ \\triangle OBN $ have $ OM $ and $ ON $ as their respective bases, and the heights are the distances from points $ C $ and $ B $ to line $ OM $, and since $ M $ is the midpoint of $ BC $, the heights are equal. The problem concerning the ratio of the areas of triangles involves the standard equation of a parabola, intersection points between a line and a parabola, midpoint coordinates—it is a moderately difficult problem. The key is transforming the ratio of the areas of the two triangles into the ratio of the base lengths $ |OM| $ and $ |ON| $. Using the simultaneous equations of the line and the parabola, the coordinates of $ M $ are found, and then the expression for the vertical coordinate of $ N $ in terms of $ k $ is obtained, leading to the functional expression of $ \\frac{|OM|}{|ON|} $ in terms of $ k $." }, { "text": "Given the parabola $y^{2}=2 p x(p>0)$ has focus $F$, directrix $l$, a line passing through point $F$ with an inclination angle of $60^{\\circ}$ intersects the parabola at point $M$ ($M$ is in the first quadrant), $M N \\perp l$, with foot of perpendicular at $N$. If the area of $\\Delta M N F$ is $4 \\sqrt{3}$, then the equation of the parabola is?", "fact_expressions": "G: Parabola;p: Number;l: Line;H:Line;M: Point;N: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;Directrix(G)=l;PointOnCurve(F, H);Inclination(H)=ApplyUnit(60,degree);Intersection(H,G)=M;Quadrant(M) = 1;IsPerpendicular(LineSegmentOf(M,N),l);FootPoint(LineSegmentOf(M,N),l)=N;Area(TriangleOf(M, N, F)) = 4*sqrt(3)", "query_expressions": "Expression(G)", "answer_expressions": "y^2=4*x", "fact_spans": "[[[2, 23], [64, 67], [139, 142]], [[5, 23]], [[34, 37]], [[61, 63]], [[68, 72], [73, 76]], [[102, 105]], [[27, 30], [39, 43]], [[5, 23]], [[2, 23]], [[2, 30]], [[2, 37]], [[38, 63]], [[44, 63]], [[61, 72]], [[73, 81]], [[84, 98]], [[84, 105]], [[107, 137]]]", "query_spans": "[[[139, 147]]]", "process": "Analysis: As shown in the figure, \\angle NMF=60^{\\circ}, MN\\bot l, with foot of perpendicular at N. Connect NF, and let A be the intersection point of the directrix and the x-axis. Therefore, MN = FM, so triangle NMF is an equilateral triangle. Given that the area of \\triangle MNF is 4\\sqrt{3}, we have S_{\\triangle MNF}=\\frac{1}{2}|NF|^{2}\\sin60^{\\circ}=4\\sqrt{3}, thus |NF|=4. In the right triangle \\triangle NAF, \\angle ANF=30^{\\circ}, so |AF|=p=2. Hence, the equation of the parabola is y^{2}=4x." }, { "text": "If the equation $\\frac{x^{2}}{m}+\\frac{y^{2}}{2-m}=1$ represents an ellipse, then the range of real values for $m$ is?", "fact_expressions": "G: Ellipse;m: Real;Expression(G)=(y^2/(2 - m) + x^2/m = 1)", "query_expressions": "Range(m)", "answer_expressions": "(0,1)+(1,2)", "fact_spans": "[[[42, 44]], [[46, 51]], [[1, 44]]]", "query_spans": "[[[46, 58]]]", "process": "Since the equation \\frac{x^{2}}{m}+\\frac{y^{2}}{2-m}=1 represents an ellipse, we have \\begin{cases}m>0\\\\2-m>0\\\\m\\neq2-m\\end{cases}, which yields 00)$, and $A_{1}$, $A_{2}$ are respectively the left and right vertices of the hyperbola, with $\\angle A_{2} P A_{1}=2 \\angle P A_{1} A_{2}$. Then $\\angle P A_{1} A_{2}$=?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = a^2);a: Number;a>0;P: Point;PointOnCurve(P, RightPart(G));A1: Point;A2: Point;LeftVertex(G) = A1;RightVertex(G) = A2;AngleOf(A2, P, A1) = 2*AngleOf(P, A1, A2)", "query_expressions": "AngleOf(P, A1, A2)", "answer_expressions": "pi/8", "fact_spans": "[[[7, 34], [57, 60]], [[7, 34]], [[10, 34]], [[10, 34]], [[2, 6]], [[2, 38]], [[39, 46]], [[47, 54]], [[39, 65]], [[39, 65]], [[67, 112]]]", "query_spans": "[[[114, 138]]]", "process": "Let \\angle A_{2}PA_{1}=\\alpha, \\angle PA_{1}A_{2}=2\\alpha, P(x,y), A_{1}(-a,0), A_{2}(a,0). Then k_{PA_{1}}=\\tan\\alpha=\\frac{y}{x+a}, k_{PA_{2}}=\\tan3\\alpha=\\frac{y}{x-a}. Point P lies on the right branch of the hyperbola x^{2}-y^{2}=a^{2} (a>0). Then k_{PA_{1}}\\cdot k_{PA_{2}}=\\frac{y}{x+a}\\cdot\\frac{y}{x-a}=\\frac{y^{2}}{x^{2}-a^{2}}=1, that is, \\tan\\alpha\\cdot\\tan3\\alpha=1. Therefore, \\tan3\\alpha=\\frac{1}{\\tan\\alpha}=\\tan(\\frac{\\pi}{2}-\\alpha). Since \\angle PA_{2}x=3\\alpha is an acute angle, it must hold that 3\\alpha=\\frac{\\pi}{2}-\\alpha. Solving gives \\alpha=\\frac{\\pi}{8}, that is, \\angle A_{2}PA_{1}=\\frac{\\pi}{8}." }, { "text": "The asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is given by $y=\\frac{3}{4} x$. What is the eccentricity of the hyperbola?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;Expression(OneOf(Asymptote(G))) = (y = (3/4)*x)", "query_expressions": "Eccentricity(G)", "answer_expressions": "5/4", "fact_spans": "[[[0, 46], [74, 77]], [[0, 46]], [[3, 46]], [[3, 46]], [[0, 72]]]", "query_spans": "[[[74, 83]]]", "process": "" }, { "text": "The coordinates of the left focus of the hyperbola are $(-2,0)$, and it passes through the point $(\\sqrt{2}, \\sqrt{3})$. Then the standard equation of this hyperbola is?", "fact_expressions": "G: Hyperbola;P:Point;Coordinate(P)=(sqrt(2),sqrt(3));Coordinate(LeftFocus(G))=(-2,0);PointOnCurve(P,G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2-y^2/3=1", "fact_spans": "[[[2, 5], [49, 52]], [[23, 46]], [[23, 46]], [[2, 20]], [[2, 46]]]", "query_spans": "[[[49, 59]]]", "process": "Since the coordinates of the left focus of the hyperbola are (-2,0), the foci of the hyperbola lie on the x-axis and the coordinates of the right focus are (2,0). Let the standard equation of the hyperbola be: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0). Since the point (\\sqrt{2},\\sqrt{3}) lies on the hyperbola, it follows that a=1, hence b^{2}=4-1=3. Therefore, the equation of the hyperbola is: x^{2}-\\frac{y^{2}}{3}=1" }, { "text": "If the point $(3,1)$ is the midpoint of a chord of the parabola $y^{2}=2 p x(p>0)$ and the slope of the chord is $2$, then $p=$?", "fact_expressions": "G: Parabola;p: Number;H: Point;L:LineSegment;p>0;Expression(G) = (y^2 = 2*p*x);Coordinate(H) = (3, 1);IsChordOf(L,G);MidPoint(L)=H;Slope(L)=2", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[10, 31]], [[50, 53]], [[1, 9]], [], [[13, 31]], [[10, 31]], [[1, 9]], [[10, 35]], [[1, 38]], [[10, 48]]]", "query_spans": "[[[50, 55]]]", "process": "Let the two endpoints of the chord be A(x_{1},y_{1}), B(x_{2},y_{2}), substitute into the parabola equation, using the point difference method we can obtain the answer. Let the two endpoints of the chord be A(x_{1},y_{1}), B(x_{2},y_{2}), given that point (3,1) is the midpoint of AB. Therefore, k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=2 and y_{1}+y_{2}=2, clearly x_{1}\\neq x_{2}, then \\begin{cases}y_{1}^{2}=2px_{1}\\\\y_{2}^{2}=2px_{2}\\end{cases}. Subtracting the two equations, we get y_{1}^{2}-y_{2}^{2}=2p(x_{1}-x_{2}), so \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{2p}{y_{1}+y_{2}}. Hence, 2=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{2p}{y_{1}+y_{2}}=\\frac{2p}{2}, then p=2." }, { "text": "Given $A(-1,0)$, $B$ is a moving point on the circle $C$: $(x-1)^{2}+y^{2}=8$, the perpendicular bisector of segment $AB$ intersects $BC$ at $P$, then the trajectory equation of the moving point $P$ is?", "fact_expressions": "C: Circle;B: Point;A: Point;P: Point;Expression(C) = (y^2 + (x - 1)^2 = 8);Coordinate(A) = (-1, 0);PointOnCurve(B, C);Intersection(PerpendicularBisector(LineSegmentOf(A,B)), LineSegmentOf(B, C)) = P", "query_expressions": "LocusEquation(P)", "answer_expressions": "x^2/2+y^2=1", "fact_spans": "[[[18, 42]], [[14, 17]], [[2, 11]], [[67, 70], [74, 77]], [[18, 42]], [[2, 11]], [[14, 46]], [[47, 70]]]", "query_spans": "[[[74, 84]]]", "process": "As shown in the figure, the circle C: (x-1)^{2}+y^{2}=8 has center coordinates C(1,0) and radius r=|CB|=2\\sqrt{2}. Since P is a point on the perpendicular bisector of segment AB, |PA|=|PB|. Then |AP|+|PC|=|CB|=2\\sqrt{2}. According to the definition of an ellipse, the locus of point P is an ellipse with foci at A and C. Here, a=\\sqrt{2}, c=1, so b=\\sqrt{a^{2}-c^{2}}=1. Therefore, the equation of the locus of point P is \\frac{x^{2}}{2}+y^{2}=1." }, { "text": "The distance from the right focus of the hyperbola $C$: $\\frac{x^{2}}{3}-\\frac{y^{2}}{3}=1$ to its asymptotes is?", "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/3 - y^2/3 = 1)", "query_expressions": "Distance(RightFocus(C), Asymptote(C))", "answer_expressions": "sqrt(3)", "fact_spans": "[[[0, 43], [48, 49]], [[0, 43]]]", "query_spans": "[[[0, 57]]]", "process": "Because the hyperbola equation is \\frac{x^{2}}{3}-\\frac{y^{2}}{3}=1, so c^{2}=3+3=6, therefore c=\\sqrt{6}, thus the coordinates of the right focus F are (\\sqrt{6},0). Let \\frac{x^{2}}{3}-\\frac{y^{2}}{3}=0, so y=\\pm x. Take one of the asymptotes l: x-y=0. Therefore, the distance from F to the line l is: \\frac{|\\sqrt{6}-0|}{\\sqrt{1^{2}+(-1)^{2}}}=\\sqrt{3}." }, { "text": "Let the focus of the parabola $y^{2}=4x$ be $F$, points $A$ and $B$ lie on the parabola, the line $AB$ passes through the focus $F$. If $|BF|-|AF|=\\frac{3}{2}$, then the value of $\\frac{|AF|}{|BF|}$ is?", "fact_expressions": "G: Parabola;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(A, G);PointOnCurve(B, G);PointOnCurve(F,LineOf(A,B));-Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F)) = 3/2", "query_expressions": "Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(B, F))", "answer_expressions": "1/2", "fact_spans": "[[[1, 15], [32, 35]], [[23, 27]], [[28, 31]], [[19, 22], [47, 50]], [[1, 15]], [[1, 22]], [[23, 36]], [[23, 36]], [[37, 50]], [[52, 77]]]", "query_spans": "[[[79, 104]]]", "process": "Let the equation of line AB be x = ty + 1, A(x₁, y₁), B(x₂, y₂). Solving the system of equations \n\\begin{cases}x=ty+1\\\\y^{2}=4x\\end{cases} \nwe obtain y² - 4ty - 4 = 0. Therefore, \n\\begin{cases}\\triangle>0\\\\y_{1}+y_{2}=4t\\\\y_{1}y_{2}=-4\\end{cases} \nThus, \nx₁·x₂ = (ty₁ + 1)(ty₂ + 1) = t²y₁y₂ + t(y₁ + y₂) + 1 = 1. \nSince |BF| - |AF| = \\frac{3}{2}, by the definition of a parabola, we get: x₂ - x₁ = \\frac{3}{2}. \nTherefore, x₁ = \\frac{1}{2}, x₂ = 2. Hence, \n\\frac{|AF|}{|BF|} = \\frac{1}{2+1} = \\frac{1}{2}." }, { "text": "Given that the circle $x^{2}+y^{2}-6 x-7=0$ is tangent to the directrix of the parabola $y^{2}=2 p x(p>0)$, find the equation of the directrix of the parabola.", "fact_expressions": "H: Circle;Expression(H) = (-6*x + x^2 + y^2 - 7 = 0);G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;IsTangent(H, Directrix(G)) = True", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "x = -1", "fact_spans": "[[[2, 24]], [[2, 24]], [[25, 46], [53, 56]], [[25, 46]], [[28, 46]], [[28, 46]], [[2, 51]]]", "query_spans": "[[[53, 63]]]", "process": "The equation of the circle is (x-3)^{2}+y^{2}=16. The directrix of the parabola y^{2}=2px is x=-\\frac{p}{2}, \\therefore 3-(-\\frac{p}{2})=4, \\therefore p=2. Hence, the equation of the directrix of the parabola is x=-1. [Note: This problem examines the method of finding the equation of a parabola, and in solving it, attention should be paid to the flexible application of the properties of a circle.]" }, { "text": "Through the focus of the parabola $y^{2}=8x$, draw a chord $AB$, with point $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$, and $x_{1}+x_{2}=10$, then $|AB|$=?", "fact_expressions": "G: Parabola;A: Point;B: Point;x1:Number;y1:Number;x2:Number;y2:Number;Expression(G) = (y^2 = 8*x);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);PointOnCurve(Focus(G),LineSegmentOf(A,B));IsChordOf(LineSegmentOf(A,B),G);x1 + x2 = 10", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "14", "fact_spans": "[[[1, 15]], [[26, 44]], [[46, 63]], [[26, 44]], [[26, 44]], [[46, 63]], [[46, 63]], [[1, 15]], [[26, 44]], [[46, 63]], [[0, 25]], [[1, 25]], [[65, 81]]]", "query_spans": "[[[83, 92]]]", "process": "" }, { "text": "If the ellipse $C$: $\\frac{x^{2}}{m+3}+\\frac{y^{2}}{m}=1(m>0)$ has an eccentricity of $\\frac{\\sqrt{3}}{2}$, then what is the length of its major axis?", "fact_expressions": "C: Ellipse;m: Number;m>0;Expression(C) = (x^2/(m + 3) + y^2/m = 1);Eccentricity(C) = sqrt(3)/2", "query_expressions": "Length(MajorAxis(C))", "answer_expressions": "4", "fact_spans": "[[[1, 50], [77, 78]], [[8, 50]], [[8, 50]], [[1, 50]], [[1, 75]]]", "query_spans": "[[[77, 83]]]", "process": "According to the eccentricity formula, set up an equation to solve for the parameter. From the ellipse equation, we obtain a^{2}=m+3, b^{2}=m, c^{2}=3. Using the eccentricity formula \\frac{c}{a}=\\frac{\\sqrt{3}}{2}=\\frac{\\sqrt{3}}{\\sqrt{m+3}}, solving gives m=1, then a=2, 2a=4." }, { "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=4 x$, $E$ is the intersection point of its directrix and the $x$-axis, a line passing through $F$ intersects the parabola $C$ at points $A$ and $B$, $M$ is the midpoint of segment $AB$, and $|M E|=\\sqrt{11}$, then $|A B|$=?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;E: Point;Intersection(Directrix(C), xAxis) = E;G: Line;PointOnCurve(F, G);A: Point;B: Point;Intersection(G, C) = {A, B};M: Point;MidPoint(LineSegmentOf(A, B)) = M;Abs(LineSegmentOf(M, E)) = sqrt(11)", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "6", "fact_spans": "[[[6, 25], [33, 34], [53, 59]], [[6, 25]], [[2, 5], [46, 49]], [[2, 28]], [[29, 32]], [[29, 44]], [[50, 52]], [[45, 52]], [[60, 63]], [[64, 67]], [[50, 69]], [[70, 73]], [[70, 84]], [[86, 103]]]", "query_spans": "[[[105, 114]]]", "process": "Analysis: To solve this problem, we need to find the coordinates of points F and E, then set up the equation of the line, solve it together with the parabola's equation, eliminate variables to write the coordinates of M, apply the distance formula between two points to find the value of $ k^{2} $, and use the focal chord length formula to obtain the result. \nDetailed solution: According to the given conditions, the slope of the line exists. The focus of the parabola has coordinates $ F(1,0) $. Let the line $ l: y = k(x - 1) $. Solving the system of equations: \n$$\n\\begin{cases}\ny^{2} = 4x \\\\\ny = k(x - 1)\n\\end{cases}\n$$\nEliminating variables yields: \n$ k^{2}x^{2} - (2k^{2} + 4)x + k^{2} = 0 $, \nfrom which we get: \n$ x_{1} + x_{2} = \\frac{2k^{2} + 4}{k^{2}} $, \nthus we find $ M\\left(\\frac{k^{2} + 2}{k^{2}}, \\frac{2}{k}\\right) $, and $ E(-1, 0) $. Given $ |ME| = \\sqrt{11} $, we have: \n$ \\left(\\frac{k^{2} + 2}{12} + 1\\right) + \\frac{4}{12} = 11 $, \nsolving gives $ k^{2} = 2 $, and $ |AB| = x_{1} + x_{2} + p = 2 + \\frac{4}{2} + 2 = 6 $. Therefore, the answer is 6." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are denoted by $F_{1}$ and $F_{2}$ respectively. The line $x-c=0$ intersects the hyperbola $C$ at point $P$, and intersects one asymptote of the hyperbola $C$ at point $Q$. Let $O$ be the origin. If $\\overrightarrow{O P}=\\frac{1}{3} \\overrightarrow{O F_{2}}+\\frac{2}{3} \\overrightarrow{O Q}$, then the eccentricity of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;G: Line;Expression(G) = (-c + x = 0);c: Number;P: Point;OneOf(Intersection(G, C)) = P;Q: Point;Intersection(G, OneOf(Asymptote(C))) = Q;O: Origin;VectorOf(O, P) = (1/3)*VectorOf(O, F2) + (2/3)*VectorOf(O, Q)", "query_expressions": "Eccentricity(C)", "answer_expressions": "3*sqrt(5)/5", "fact_spans": "[[[2, 63], [99, 105], [117, 123], [240, 246]], [[2, 63]], [[10, 63]], [[9, 63]], [[10, 63]], [[10, 63]], [[73, 80]], [[81, 88]], [[2, 88]], [[2, 88]], [[89, 98]], [[89, 98]], [[91, 98]], [[111, 115]], [[89, 115]], [[131, 135]], [[89, 135]], [[136, 139]], [[146, 238]]]", "query_spans": "[[[240, 252]]]", "process": "As shown in the figure, since \\overrightarrow{OP}=\\frac{1}{3}\\overrightarrow{OF}_{2}+\\frac{2}{3}\\overrightarrow{OQ}, it follows that \\overrightarrow{OP}-\\overrightarrow{OF_{2}}=\\frac{1}{3}\\overrightarrow{OF_{2}}+\\frac{2}{3}\\overrightarrow{OQ}-\\overrightarrow{OF}_{2}, i.e., \\overrightarrow{F_{2}P}=\\frac{2}{3}\\overrightarrow{F_{2}Q}. Therefore, points F_{2}, P, and Q are collinear. Substituting x=c into the hyperbola gives y_{P}=-\\frac{b^{2}}{a}; the asymptotes of the hyperbola are y=-\\frac{b}{a}x'. Substituting x=c into the asymptote yields y_{Q}=-\\frac{bc}{a}. Hence, |F_{2}P|=\\frac{b^{2}}{a}, |F_{2}Q|=\\frac{bc}{a}. Thus, \\frac{b^{2}}{a}=\\frac{2bc}{3a}. Solving gives: 3b=2c. Since c^{2}=a^{2}+b^{2}, solving yields: \\frac{c}{a}=\\frac{3\\sqrt{5}}{5}." }, { "text": "Given that the line $y = kx - 1$ always has common points with the ellipse $C$: $\\frac{x^{2}}{4} + \\frac{y^{2}}{b^{2}} = 1$ $(b > 0)$ whose foci lie on the $x$-axis, what is the range of values for the eccentricity of ellipse $C$?", "fact_expressions": "C: Ellipse;b: Number;G: Line;k: Number;b>0;Expression(C) = (x^2/4 + y^2/b^2 = 1);Expression(G) = (y = k*x - 1);PointOnCurve(Focus(C),xAxis);IsIntersect(G,C)", "query_expressions": "Range(Eccentricity(C))", "answer_expressions": "(0,sqrt(3)/2]", "fact_spans": "[[[23, 74], [81, 86]], [[30, 74]], [[2, 13]], [[4, 13]], [[30, 74]], [[23, 74]], [[2, 13]], [[14, 74]], [[2, 79]]]", "query_spans": "[[[81, 96]]]", "process": "Since the foci are on the x-axis, we have $ b^{2} < 4 $. Furthermore, since the line $ y = kx - 1 $ always intersects the ellipse, we obtain $ \\frac{0}{4} + \\frac{1}{b^{2}} \\leqslant 1 $. Solving this inequality gives $ 1 \\leqslant b < 2 $. Finally, we compute the result using the eccentricity formula. Because the ellipse's foci lie on the x-axis, $ b^{2} < 4 $. Since $ b > 0 $, it follows that $ 0 < b < 2 $. Because the line $ y = kx - 1 $ always has common points with the ellipse, we have $ \\frac{0}{4} + \\frac{(-1)^{2}}{b^{2}} \\leqslant 1 $. Since $ b > 0 $, we get $ b \\geqslant 1 $. Combining these results, $ 1 \\leqslant b < 2 $. Then, $ e = \\frac{c}{a} = \\sqrt{1 - \\frac{b^{2}}{a^{2}}} = \\sqrt{1 - \\frac{b^{2}}{4}} \\in \\left(0, \\frac{\\sqrt{3}}{2}\\right) $." }, { "text": "Given that point $M$ is a point on the parabola $y^{2}=4x$, $F$ is the focus of the parabola, and $A$ lies on the circle $C$: $(x-4)^{2}+(y-1)^{2}=1$, then the minimum value of $|MA|+|MF|$ is?", "fact_expressions": "M: Point;G: Parabola;Expression(G) = (y^2 = 4*x);PointOnCurve(M, G) = True;F: Point;Focus(G) = F;A: Point;C: Circle;Expression(C) = ((x - 4)^2 + (y - 1)^2 = 1);PointOnCurve(A, C) = True", "query_expressions": "Min(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F)))", "answer_expressions": "4", "fact_spans": "[[[2, 6]], [[7, 21], [29, 32]], [[7, 21]], [[2, 24]], [[25, 28]], [[25, 35]], [[36, 39]], [[40, 68]], [[40, 68]], [[36, 69]]]", "query_spans": "[[[71, 88]]]", "process": "" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{7}=1(a>0)$ has eccentricity $\\frac{4}{3}$, then $a=?$", "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2/7 + x^2/a^2 = 1);Eccentricity(G) = 4/3", "query_expressions": "a", "answer_expressions": "3", "fact_spans": "[[[2, 49]], [[69, 72]], [[5, 49]], [[2, 49]], [[2, 67]]]", "query_spans": "[[[69, 74]]]", "process": "" }, { "text": "Point $P$ is a moving point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, $M(1,4)$, $N(2,0)$, then the maximum value of $|P M|+|P N|$ is?", "fact_expressions": "G: Ellipse;M: Point;N: Point;P: Point;Expression(G) = (x^2/9 + y^2/5 = 1);Coordinate(M) = (1, 4);Coordinate(N) = (2, 0);PointOnCurve(P, G)", "query_expressions": "Max(Abs(LineSegmentOf(P, M)) + Abs(LineSegmentOf(P, N)))", "answer_expressions": "11", "fact_spans": "[[[5, 42]], [[48, 56]], [[59, 67]], [[0, 4]], [[5, 42]], [[48, 56]], [[59, 67]], [[0, 47]]]", "query_spans": "[[[69, 88]]]", "process": "Analysis: By the definition of an ellipse, we have |PN| = 2a - |PF|, then |PM| + |PN| = 2a + |PM| - |PF| ≤ 2a + |MF|, and thus the result follows. From the given information, point N is the right focus of the ellipse; let the left focus of the ellipse be F(-2,0), so |PN| = 2a - |PF|. Therefore, |PM| + |PN| = 2a + |PM| - |PF| ≤ 2a + |MF|. Moreover, since |MF| = √(3² + 4²) = 5, the maximum value of |PM| + |PN| is 2 × 3 + 5 = 11." }, { "text": "Given that $P$ is a moving point on the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{2}=1$, $A(-2,1)$, $B(2,-1)$, let lines $AP$ and $BP$ intersect the line $x=4$ at points $M$ and $N$ respectively. If the areas of $\\Delta ABP$ and $\\Delta MNP$ are equal, then what is the length of segment $OP$?", "fact_expressions": "E: Ellipse;Expression(E) = (x^2/8 + y^2/2 = 1);P: Point;PointOnCurve(P, E);A: Point;B: Point;Coordinate(A) = (-2, 1);Coordinate(B) = (2, -1);l: Line;Expression(l) = (x=4);M: Point;N: Point;Intersection(LineOf(A, P), l) = M;Intersection(LineOf(B, P), l) = N;Area(TriangleOf(A, B, P)) = Area(TriangleOf(M, N, P));O: Origin", "query_expressions": "Length(LineSegmentOf(O, P))", "answer_expressions": "sqrt(107)/4", "fact_spans": "[[[6, 43]], [[6, 43]], [[2, 5]], [[2, 49]], [[50, 59]], [[62, 71]], [[50, 59]], [[62, 71]], [[89, 96]], [[89, 96]], [[98, 101]], [[102, 105]], [[73, 107]], [[73, 107]], [[109, 143]], [[147, 152]]]", "query_spans": "[[[145, 156]]]", "process": "" }, { "text": "The ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{b^{2}}=1(00 , b>0)$, point $F$ is its right focus, point $B(0, b)$, if the line $BF$ is perpendicular to one of the asymptotes of the hyperbola, then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;B: Point;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(B) = (0, b);RightFocus(G) = F;IsPerpendicular(OverlappingLine(LineSegmentOf(B,F)),OneOf(Asymptote(G)))", "query_expressions": "Eccentricity(G)", "answer_expressions": "(sqrt(5)+1)/2", "fact_spans": "[[[2, 59], [65, 66], [92, 95], [108, 111]], [[5, 59]], [[5, 59]], [[70, 80]], [[60, 64]], [[5, 59]], [[5, 59]], [[2, 59]], [[70, 80]], [[60, 69]], [[82, 105]]]", "query_spans": "[[[108, 117]]]", "process": "According to the problem: B(0, b), F(c, 0), then k_{BF} = -\\frac{b}{c}. The line BF is perpendicular to one of the asymptotes of the hyperbola, hence -\\frac{b}{c} \\cdot \\frac{b}{a} = -1, that is, b^{2} = ac. Therefore, c^{2} - a^{2} = ac, e^{2} - e - 1 = 0. Solving gives e = \\frac{\\sqrt{5}+1}{2} or e = \\frac{-\\sqrt{5}+1}{2} (discarded)." }, { "text": "The number of intersection points between the hyperbola $x^{2}-9 y^{2}=9$ and the line $2 x-6 y-21=0$ is?", "fact_expressions": "G: Hyperbola;H: Line;Expression(G) = (x^2 - 9*y^2 = 9);Expression(H) = (2*x - 6*y - 21 = 0)", "query_expressions": "NumIntersection(G, H)", "answer_expressions": "1", "fact_spans": "[[[0, 20]], [[21, 37]], [[0, 20]], [[21, 37]]]", "query_spans": "[[[0, 44]]]", "process": "From the line equation 2x-6y-21=0, we get x=3y+\\frac{21}{2}; substituting into the hyperbola yields y=-\\frac{3}{28}. Therefore, the number of intersection points is 1." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0, b>0)$ has a focus at $(m, 0) (m>0)$, and point $P(m, 2 m)$ lies on the hyperbola, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;m: Number;m>0;Coordinate(OneOf(Focus(G))) = (m, 0);P: Point;Coordinate(P) = (m, 2*m);PointOnCurve(P, G)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)+1", "fact_spans": "[[[2, 60], [97, 100], [103, 106]], [[2, 60]], [[5, 60]], [[5, 60]], [[5, 60]], [[5, 60]], [[67, 82]], [[67, 82]], [[2, 82]], [[84, 96]], [[84, 96]], [[84, 101]]]", "query_spans": "[[[103, 112]]]", "process": "" }, { "text": "Given that $F$ is the focus of the parabola $C$: $x^{2}=4 y$, a line $l$ passing through point $F$ intersects the parabola $C$ at two distinct points $A$ and $B$. The tangents to the parabola $C$ at points $A$ and $B$ are $l_{1}$ and $l_{2}$, respectively, and $l_{1}$, $l_{2}$ intersect at point $P$. Then the minimum value of $|P F|+\\frac{32}{|A B|}$ is?", "fact_expressions": "F: Point;C: Parabola;Expression(C) = (x^2 = 4*y);Focus(C) = F;PointOnCurve(F, l);l: Line;Intersection(l, C) = {A, B};A: Point;B: Point;l1: Line;l2: Line;TangentOnPoint(A, C) = l1;TangentOnPoint(B, C) = l2;Intersection(l1, l2)= P;P: Point;Negation(A=B)", "query_expressions": "Min(Abs(LineSegmentOf(P, F)) + 32/Abs(LineSegmentOf(A, B)))", "answer_expressions": "6", "fact_spans": "[[[2, 5], [30, 34]], [[6, 25], [41, 47], [63, 70]], [[6, 25]], [[2, 28]], [[29, 40]], [[35, 40]], [[35, 62]], [[55, 58], [71, 74]], [[59, 62], [75, 78]], [[87, 94], [105, 112]], [[96, 103], [114, 121]], [[64, 103]], [[64, 103]], [[105, 128]], [[124, 128]], [50, 61]]", "query_spans": "[[[130, 159]]]", "process": "Let the equation of line $ l $ be: $ y = kx + 1 $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. \nSolving the system \n\\[\n\\begin{cases}\ny = kx + 1 \\\\\nx^{2} = 4y\n\\end{cases}\n\\]\nleads to: $ x^{2} - 4kx - 4 = 0 $, from which we obtain: \n$ x_{1} + x_{2} = 4k $, $ x_{1}x_{2} = -4 $, \n$ |AB| = y_{1} + y_{2} + p = k(x_{1} + x_{2}) + 4 = 4k^{2} + 4 $. \n\nDifferentiating both sides of $ x^{2} = 4y $ gives: $ y' = \\frac{1}{2}x $. \nThus, the equation of tangent line $ PA $ is: \n$ y - y_{1} = \\frac{x_{1}}{2}(x - x_{1}) $; \nthe equation of tangent line $ PB $ is: \n$ y - y_{2} = \\frac{x_{2}}{2}(x - x_{2}) $. \n\nSolving these two equations simultaneously yields: \n$ x = \\frac{1}{2}(x_{1} + x_{2}) = 2k $, \n$ y = \\frac{1}{4}x_{1}x_{2} = -1 $. \nTherefore, $ P(2k, -1) $, and $ |PF| = \\sqrt{4k^{2} + 4} $. \n\nHence, \n$ |PF| + \\frac{\\sqrt{4}}{|AB|} = \\sqrt{4k^{2} + 4} + \\frac{32}{4k^{2} + 4} $. \nLet $ \\sqrt{4k^{2} + 4} = t \\geqslant 2 $. \nThen, \n$ |PF| + \\frac{32}{|AB|} = t + \\frac{32}{t^{2}} = f(t) $, \n$ f'(t) = 1 - \\frac{64}{t^{3}} = \\frac{(t - 4)(t^{2} + 4t + 16)}{t^{3}} $. \n\nWhen $ t > 4 $, $ f'(t) > 0 $; when $ 2 \\leqslant t < 4 $, $ f'(t) < 0 $. \nIt follows that when $ t = 4 $, the function $ f(t) $ attains a local minimum, which is also the minimum value, $ f(4) = 6 $. \nThe equality holds if and only if $ k = \\pm\\sqrt{3} $." }, { "text": "Given the equation $\\frac{x^{2}}{6+m}+\\frac{y^{2}}{4-m}=1$ represents an ellipse with foci on the $y$-axis, find the range of real values for $m$.", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(m + 6) + y^2/(4 - m) = 1);m: Real;PointOnCurve(Focus(G), yAxis)", "query_expressions": "Range(m)", "answer_expressions": "(-6, -1)", "fact_spans": "[[[54, 56]], [[2, 56]], [[58, 63]], [[45, 56]]]", "query_spans": "[[[58, 70]]]", "process": "Since the equation \\frac{x^2}{6+m} + \\frac{y^{2}}{4-m} = 1 represents an ellipse with foci on the y-axis, we have \\begin{cases} 4 - m > 6 \\\\ 6 + m > 0 \\end{cases} > 6 + \\frac{+}{m}1, solving which yields -6 < m < -1" }, { "text": "Given that $O$ is the coordinate origin, the distance from point $M$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ to the left focus $F_{1}$ is $4$, and $N$ is the midpoint of $M F_{1}$, then the value of $O N$ equals?", "fact_expressions": "G: Ellipse;M: Point;F1: Point;O: Origin;N: Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(M, G);LeftFocus(G) = F1;Distance(M, F1) = 4;MidPoint(LineSegmentOf(M, F1)) = N", "query_expressions": "LineSegmentOf(O, N)", "answer_expressions": "3", "fact_spans": "[[[11, 50]], [[52, 56]], [[60, 67]], [[2, 5]], [[77, 80]], [[11, 50]], [[11, 56]], [[11, 67]], [[52, 74]], [[77, 93]]]", "query_spans": "[[[95, 105]]]", "process": "As shown in the figure, connect ON. Since N is the midpoint of MF_{1} and O is the origin, it follows that ON = \\frac{1}{2}MF_{2}. By the definition of an ellipse, MF_{1} + MF_{2} = 2a = 10. Given that MF_{1} = 4, we have MF_{2} = 6. Therefore, ON = 3." }, { "text": "Given that $A B$ is a chord passing through the left focus $F_{1}$ of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and $|A F_{2}|+|B F_{2}|=8$, where $F_{2}$ is the right focus of the ellipse, then the length of chord $A B$ is?", "fact_expressions": "G: Ellipse;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2/25 + y^2/16 = 1);LeftFocus(G) = F1;PointOnCurve(F1, LineSegmentOf(A, B));IsChordOf(LineSegmentOf(A,B),G);Abs(LineSegmentOf(A, F2)) + Abs(LineSegmentOf(B, F2)) = 8;RightFocus(G) = F2", "query_expressions": "Length(LineSegmentOf(A,B))", "answer_expressions": "12", "fact_spans": "[[[9, 48], [96, 98]], [[2, 7]], [[2, 7]], [[88, 95]], [[51, 58]], [[9, 48]], [[9, 58]], [[2, 60]], [[2, 60]], [[62, 85]], [[88, 102]]]", "query_spans": "[[[105, 114]]]", "process": "According to the definition of an ellipse, (|AF₁| + |AF₂|) + (|BF₁| + |BF₂|) = 4a = 20, from which we obtain |AB| = 20 - (|AF₂| + |BF₂|) = 12, yielding the answer to this problem. \n∵ The equation of the ellipse is \\frac{x^2}{25} + \\frac{y^{2}}{16} = 1, ∴ a = 5, b = 4, and we get c = \\sqrt{a^{2} - b^{2}} = 3. \nAccording to the definition of an ellipse, we have |AF₁| + |AF₂| = |BF₁| + |BF₂| = 2a = 10, thus (|AF₁| + |AF₂|) + (|BF₁| + |BF₂|) = 20. \n∵ AB is a chord passing through the left focus F₁ of the ellipse \\frac{x^{2}}{25} + \\frac{y^{2}}{16} = 1, we have |AF₁| + |BF₁| = |AB|. \n∴ |AB| = 20 - (|AF₂| + |BF₂|) = 20 - 8 = 12" }, { "text": "Given that $A$, $B$, $P$ are three distinct points on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, and the line connecting $A$ and $B$ passes through the origin. If the product of the slopes of lines $PA$ and $PB$ is $k_{PA} \\cdot k_{PB}=3$, then what is the eccentricity of this hyperbola?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;A: Point;B: Point;P: Point;PointOnCurve(A, G);PointOnCurve(B, G);PointOnCurve(P, G);Negation(A=B);Negation(A=P);Negation(B=P);O: Origin;PointOnCurve(O, LineSegmentOf(A, B));k1: Number;k2: Number;Slope(LineOf(P, A)) = k1;Slope(LineOf(P, B)) = k2;k1*k2=3", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[14, 60], [132, 135]], [[14, 60]], [[17, 60]], [[17, 60]], [[2, 5], [68, 71]], [[6, 9], [72, 75]], [[10, 13]], [[2, 66]], [[2, 66]], [[2, 66]], [[2, 66]], [[2, 66]], [[2, 66]], [[79, 83]], [[68, 83]], [[104, 129]], [[104, 129]], [[85, 129]], [[85, 129]], [[104, 129]]]", "query_spans": "[[[132, 141]]]", "process": "Analysis: According to the symmetry of the hyperbola, A and B are symmetric about the origin. Let the coordinates of A, B, P be set as A(x_{1},y), B(-x_{1},-y), P(x,y). Then \\frac{x_{1}^2}{a^{2}}-\\frac{y_{1}^{2}}{b^{2}}=1, k_{PA}\\cdot k_{PB}=\\frac{b^{2}}{a^{2}}=3, therefore e=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=2. Hence, the answer is 2." }, { "text": "If the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ has an asymptote with an inclination angle of $\\frac{2 \\pi}{3}$ and eccentricity $e$, then the minimum value of $\\frac{a^{2}+e^{2}}{2 b}$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Inclination(OneOf(Asymptote(C))) = (2*pi)/3;Eccentricity(C)=e;e:Number", "query_expressions": "Min((a^2 + e^2)/(2*b))", "answer_expressions": "2*sqrt(3)/3", "fact_spans": "[[[1, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[1, 63]], [[1, 91]], [[1, 99]], [[96, 99]]]", "query_spans": "[[[101, 132]]]", "process": "From the inclination angle of the asymptote, obtain \\frac{b}{a}, then find the eccentricity e. Express \\frac{a^2+e^{2}}{2b} as an expression in terms of a, and use the basic inequality to find the minimum value. \\frac{b}{a}=\\sqrt{3}, e^{2}=\\frac{c^{2}}{a^{2}}=1+\\frac{b^{2}}{a^{2}}=4, so e=2, \\frac{a^{2}+e^{2}}{2b}=\\frac{a^{2}+4}{2\\sqrt{3}a}=\\frac{a}{2\\sqrt{3}}+\\frac{2}{\\sqrt{3}a}\\geqslant\\frac{2\\sqrt{3}}{3} (equality holds if and only if a=2)." }, { "text": "If the line $x-2 y+2=0$ passes through one focus and one vertex of an ellipse, then the standard equation of the ellipse with the focus on the $x$-axis is?", "fact_expressions": "G: Ellipse;l: Line;Expression(l) = (x - 2*y + 2 = 0);PointOnCurve(OneOf(Focus(G)),l);PointOnCurve(OneOf(Vertex(G)),l);PointOnCurve(Focus(G),xAxis)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/5+y^2=1", "fact_spans": "[[[16, 18], [38, 40]], [[1, 14]], [[1, 14]], [[1, 23]], [[1, 28]], [[30, 40]]]", "query_spans": "[[[38, 46]]]", "process": "Since the line x-2y+2=0 passes through the points (-2,0) and (0,1), and the foci of the ellipse lie on the x-axis, we have c=2, b=1, so a^{2}=1+4=5, b^{2}=1. Therefore, the equation of the ellipse is: \\frac{x^{2}}{5}+y^{2}=1." }, { "text": "The distance from a point $P$ on the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ to the point $F_{1}(-5,0)$ is $9$. What is the distance from point $P$ to the point $F_{2}(5,0)$?", "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;P: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Coordinate(F1) = (-5, 0);Coordinate(F2) = (5, 0);PointOnCurve(P, G);Distance(P, F1) = 9", "query_expressions": "Distance(P, F2)", "answer_expressions": "3, 15", "fact_spans": "[[[0, 39]], [[46, 60]], [[74, 87]], [[42, 45], [69, 73]], [[0, 39]], [[46, 60]], [[74, 87]], [[0, 45]], [[42, 67]]]", "query_spans": "[[[69, 91]]]", "process": "First, find the coordinates of the foci based on the hyperbola equation, then use the definition of the hyperbola to obtain ||PF_{1}|-|PF_{2}||=2a, thereby solving for the value of |PF_{2}| and obtaining the answer. \\because the hyperbola \\frac{x2}{9}-\\frac{y^{2}}{16}=1 \\therefore a=3, b=4, c=5, F_{1}(-5,0) and F_{2}(5,0) are the two foci of the hyperbola \\because point P lies on the hyperbola \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1, ||PF_{1}|-|PF_{2}||=|9-|PF_{2}||=6, solving gives |PF_{2}|=3 or 15, \\cdot |PF_{2}|\\geqslant c-a=2, \\therefore |PF_{2}|=3 or 15," }, { "text": "If the focal distance of the ellipse $x^{2}+\\frac{y^{2}}{m}=1$ is $4$, then $m=?$", "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2 + y^2/m = 1);FocalLength(G) = 4", "query_expressions": "m", "answer_expressions": "5", "fact_spans": "[[[1, 28]], [[37, 40]], [[1, 28]], [[1, 35]]]", "query_spans": "[[[37, 42]]]", "process": "From the definition of the ellipse, we know that a > c = 2, so the foci lie on the y-axis, thus we obtain the answer. From the definition of the ellipse, we have 2c = 4 \\Rightarrow c = 2, and since a > c, the foci lie on the y-axis, \\therefore m - 1 = 2^{2}, \\therefore m = 5." }, { "text": "Let $F_{1}$, $F_{2}$ be the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$. $O$ is the coordinate origin. Draw a perpendicular from $F_{2}$ to an asymptote of $C$, with foot $P$. If $|P F_{1}|=\\sqrt{5} a$, then the eccentricity of $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;L:Line;PointOnCurve(F2,L);IsPerpendicular(L,OneOf(Asymptote(C)));FootPoint(L,OneOf(Asymptote(C)))=P;Abs(LineSegmentOf(P, F1)) = sqrt(5)*a ;O:Origin", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[17, 78], [102, 105], [148, 151]], [[24, 78]], [[24, 78]], [[118, 121]], [[1, 8]], [[9, 16], [94, 101]], [[24, 78]], [[24, 78]], [[17, 78]], [[1, 84]], [[1, 84]], [], [[93, 114]], [[93, 114]], [[93, 121]], [[123, 145]], [[85, 88]]]", "query_spans": "[[[148, 157]]]", "process": "First, find the lengths of the sides of triangles $\\triangle OPF_{1}$ and $\\triangle OPF_{2}$. Then use $\\cos\\angle POF_{2} = -\\cos\\angle POF_{1}$ to find an equation relating $a$, $b$, and $c$, and finally determine the eccentricity. According to the problem: $|PF_{1}| = \\sqrt{5}a$. Since $|PF_{2}|$ equals the distance from point $F_{2}(c,0)$ to the asymptote $bx - ay = 0$, we have $|PF_{2}| = \\frac{bc}{\\sqrt{a^{2}+b^{2}}} = b$. Therefore, $|OP| = a$. In $\\triangle OPF_{1}$, using the law of cosines:\n\n$$\n\\cos\\angle POF_{1} = \\frac{|OP|^{2} + |OF_{1}|^{2} - |PF_{1}|^{2}}{2 \\times |OP| \\times |OF_{1}|} = \\frac{a^{2} + c^{2} - 5a^{2}}{2ac} = \\frac{c^{2} - 4a^{2}}{2ac}\n$$\n\nIn $\\triangle OPF_{2}$, $\\cos\\angle POF_{2} = \\frac{a}{c}$. Because $\\cos\\angle POF_{2} = -\\cos\\angle POF_{1}$, we have:\n\n$$\n\\frac{c^{2} - 4a^{2}}{2ac} = -\\frac{a}{c}\n$$\n\nRearranging and simplifying yields $c^{2} = 2a^{2}$. Thus, the eccentricity of $C$ is $e = \\frac{c}{a} = \\sqrt{2}$." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{3}-y^{2}=1$, the left and right foci are denoted as $F_{1}$ and $F_{2}$, respectively. Point $P(x, y)$ lies on the right branch of $C$. If $|P F_{2}|=2$, then $|P F_{1}|$=?", "fact_expressions": "C: Hyperbola;P: Point;F2: Point;F1: Point;x1:Number;y1:Number;Expression(C) = (x^2/3 - y^2 = 1);Coordinate(P) = (x1, y1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, RightPart(C));Abs(LineSegmentOf(P, F2)) = 2", "query_expressions": "Abs(LineSegmentOf(P, F1))", "answer_expressions": "2+2*sqrt(3)", "fact_spans": "[[[2, 35], [70, 73]], [[59, 69]], [[51, 58]], [[43, 50]], [[60, 69]], [[60, 69]], [[2, 35]], [[59, 69]], [[2, 58]], [[2, 58]], [[59, 76]], [[78, 91]]]", "query_spans": "[[[93, 106]]]", "process": "Solution: From the hyperbola equation C: \\frac{x^{2}}{3}-y^{2}=1, we obtain a=\\sqrt{3}. Since P(x,y) lies on the right branch of C, if |PF_{2}|=2, then |PF_{1}|-|PF_{2}|=2a=2\\sqrt{3}, which gives: |PF_{1}|=2+2\\sqrt{3}" }, { "text": "Given that two tangents are drawn from point $T(-1,1)$ to the parabola $C$: $y^{2}=2 p x$, with points of tangency $A$ and $B$, and the line $AB$ passes through the focus $F$ of the parabola $C$, then $|T A|^{2}+|T B|^{2}$=?", "fact_expressions": "T: Point;Coordinate(T) = (-1, 1);C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;L1: Line;L2: Line;TangentOfPoint(T, C) = {L1, L2};TangentPoint(L1, C) = A;TangentPoint(L2, C) = B;A: Point;B: Point;Focus(C) = F;F: Point;PointOnCurve(F, LineOf(A, B)) = True", "query_expressions": "Abs(LineSegmentOf(T, A))^2 + Abs(LineSegmentOf(T, B))^2", "answer_expressions": "25", "fact_spans": "[[[3, 13]], [[3, 13]], [[14, 35], [61, 67]], [[14, 35]], [[22, 35]], [], [], [[2, 40]], [[2, 51]], [[2, 51]], [[44, 47]], [[48, 51]], [[61, 73]], [[70, 73]], [[52, 73]]]", "query_spans": "[[[75, 98]]]", "process": "Let A(x₁, y₁) be a point on the parabola C: y² = 2px. The slope of the line tangent to the parabola at the point of tangency A is k. By solving the system of equations and using the discriminant, we obtain k = p/y₁. Thus, the equation of the tangent line at point A(x₁, y₁) is yy₁ = p(x + x₁). Similarly, the equation of the tangent line at point B(x₂, y₂) is yy₂ = p(x + x₂). Given that T(-1, 1) lies on both tangent lines, the equation of chord AB is y = p(x - 1). Since chord AB passes through the focus F of the parabola, we find F(1, 0), and thus the equation of the parabola is y² = 4x. The equation of line AB is y = 2(x - 1). Solving the system of equations yields the answer.\n\n[Detailed Solution] \nSolution: Let A(x₁, y₁) be a point on the parabola C: y² = 2px. The slope of the line tangent to the parabola at the point of tangency A is k. Then the equation of the tangent line at A(x₁, y₁) is y - y₁ = k(x - x₁). \nSolving this together with the parabola C: y² = 2px, we form the system: \n\\begin{cases} y - y₁ = k(x - x₁) \\\\ y^2 = 2px \\end{cases} \nThis leads to: \ny² - \\frac{2p}{k}y + \\frac{2py₁}{k} - 2px₁ = 0 \nSo Δ_A = \\left(-\\frac{2p}{k}\\right)^2 - 4\\left(\\frac{2py₁}{k} - 2px₁\\right) = 0 \nSimplifying gives: \n4p² - 8kpy₁ + 4k²y₁² = 0 \nThus, (2p - 2ky₁)² = 0, solving gives k = \\frac{p}{y₁}. \nTherefore, the equation of the tangent line at A(x₁, y₁) is y - y₁ = \\frac{p}{y₁}(x - x₁), which simplifies to yy₁ = p(x + x₁). \nSimilarly, let B(x₂, y₂) be a point on the parabola C: y² = 2px. The equation of the tangent line at B is yy₂ = p(x + x₂). \nSince T(-1, 1) lies on both tangent lines yy₁ = p(x + x₁) and yy₂ = p(x + x₂), we have: \ny₁ = p(-1 + x₁), y₂ = p(-1 + x₂) \nThus, the equation of line AB is y = p(x - 1). \nSince line AB passes through the focus F of parabola C, setting y = 0 gives x = 1, so F(1, 0). \nHence, the equation of the parabola is y² = 4x, and the equation of line AB is y = 2(x - 1). \nSolving the system: \n\\begin{cases} y^2 = 4x \\\\ y = 2(x - 1) \\end{cases} \nWe get: x² - 3x + 1 = 0 or y² - 2y - 4 = 0. \nThus, x₁ + x₂ = 3, x₁x₂ = 1, y₁ + y₂ = 2, y₁y₂ = -4. \nThen |TA|² + |TB|² = (x₁ + 1)² + (y₁ - 1)² + (x₂ + 1)² + (y₂ - 1)² \n= x₁² + x₂² + 2(x₁ + x₂) + y₁² + y₂² - 2(y₁ + y₂) + 4 \n= (x₁ + x₂)² - 2x₁x₂ + 2(x₁ + x₂) + (y₁ + y₂)² - 2y₁y₂ - 2(y₁ + y₂) + 4 \nFinal answer: 25" }, { "text": "The coordinates of the focus of the parabola $y=\\frac{1}{m} x^{2}$ are?", "fact_expressions": "G: Parabola;m: Number;Expression(G) = (y = x^2/m)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0, m/4)", "fact_spans": "[[[0, 24]], [[3, 24]], [[0, 24]]]", "query_spans": "[[[0, 31]]]", "process": "" }, { "text": "Let $F_{1}$, $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $A$ be the upper vertex of the ellipse, and point $M$ satisfies $\\overrightarrow{A M}=\\overrightarrow{M F_{2}}$ and $\\overrightarrow{M F_{1}} \\cdot \\overrightarrow{A F_{2}}=0$. Then the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;M: Point;F2: Point;F1: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;UpperVertex(G)=A;VectorOf(A, M) = VectorOf(M, F2);DotProduct(VectorOf(M, F1), VectorOf(A, F2)) = 0", "query_expressions": "Eccentricity(G)", "answer_expressions": "1/2", "fact_spans": "[[[17, 69], [81, 83], [205, 207]], [[19, 69]], [[19, 69]], [[76, 79]], [[88, 92]], [[9, 16]], [[1, 8]], [[19, 69]], [[19, 69]], [[17, 69]], [[1, 75]], [[1, 75]], [[76, 87]], [[94, 141]], [[143, 202]]]", "query_spans": "[[[205, 213]]]", "process": "" }, { "text": "The parabola $y^{2}=8 \\sqrt{3} x$ has focus $F$ as its right focus, and the hyperbola with asymptotes $x \\pm \\sqrt{3} y=0$ is?", "fact_expressions": "G: Hyperbola;H: Parabola;F:Point;Expression(H) = (y^2 = 8*(sqrt(3)*x));Focus(H) = F;RightFocus(G)=F;Expression(Asymptote(G))=(x + pm*sqrt(3)*y = 0)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/9 - y^2/3 = 1", "fact_spans": "[[[64, 67]], [[1, 24]], [[27, 30]], [[1, 24]], [[1, 30]], [[0, 67]], [[36, 67]]]", "query_spans": "[[[64, 71]]]", "process": "" }, { "text": "Given the hyperbola equation $2 x^{2}-y^{2}=k$, and the focal distance is $6$, then the value of $k$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (2*x^2 - y^2 = k);FocalLength(G) = 6;k:Number", "query_expressions": "k", "answer_expressions": "pm*6", "fact_spans": "[[[2, 5]], [[2, 25]], [[2, 32]], [[34, 37]]]", "query_spans": "[[[34, 41]]]", "process": "Knowing the focal length is 6, we have c=3. If the foci lie on the m-axis, then the equation can be written as \\frac{x^{2}}{2}-\\frac{y^{2}}{k}=1, that is, \\frac{k}{2}+k=9, solving gives k=6; if the foci lie on the y-axis, then the equation can be written as \\frac{y^{2}}{-k}-\\frac{x2}{-k}=1, that is, -\\frac{k}{2}-k=9, thus k=6. In summary, the value of k is 6 or 6. The final answer is: +6" }, { "text": "Let a point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ have a distance of $10$ to the left directrix, and let $F$ be the left focus of the ellipse. If point $M$ satisfies $\\overrightarrow{O M}=\\frac{1}{2}(\\overrightarrow{O P}+\\overrightarrow{O F})$, then $|\\overrightarrow{O M}|=?$", "fact_expressions": "G: Ellipse;O: Origin;M: Point;P: Point;F: Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(P, G);Distance(P, LeftDirectrix(G)) = 10;LeftFocus(G) = F;VectorOf(O, M) = (VectorOf(O, F) + VectorOf(O, P))/2", "query_expressions": "Abs(VectorOf(O, M))", "answer_expressions": "2", "fact_spans": "[[[1, 40], [65, 67]], [[158, 182]], [[73, 77]], [[43, 46]], [[60, 63]], [[1, 40]], [[1, 46]], [[1, 58]], [[60, 71]], [[79, 156]]]", "query_spans": "[[[158, 184]]]", "process": "Solution: From the equation of the ellipse, we obtain a=5, b=4, c=3, e=\\frac{3}{5}. According to the definition of the ellipse, \\frac{PF}{10}=e=\\frac{3}{5}, so PF=6. Let F_{1} be the right focus of this ellipse. By the definition of the ellipse, PF_{1}+PF=2a=10, then PF_{1}=10-6=4. Since \\overrightarrow{OM}=\\frac{1}{2}(\\overrightarrow{OP}+\\overrightarrow{OF}), point M is the midpoint of segment PF. Therefore, OM/\\!/PF_{1} and OM=\\frac{1}{2}PF_{1}, so OM=\\frac{1}{2}\\times4=2, which means |\\overrightarrow{OM}|=2. The final answer is 2. This problem examines the definition of an ellipse and the geometric meaning of vector addition, focusing on computational ability, and is a medium-level question." }, { "text": "The chord of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$ is bisected by the point $(1,1)$. What is the equation of the line on which this chord lies?", "fact_expressions": "G: Ellipse;H: LineSegment;Expression(G) = (x^2/4 + y^2/2 = 1);IsChordOf(H,G);Coordinate(MidPoint(H))=(1,1)", "query_expressions": "Expression(OverlappingLine(H))", "answer_expressions": "x+2*y-3=0", "fact_spans": "[[[0, 37]], [], [[0, 37]], [[0, 39]], [[0, 50]]]", "query_spans": "[[[0, 64]]]", "process": "Let the line passing through point A(1,1) intersect the ellipse at two points E(x_{1},y_{1}) and F(x_{2},y_{2}). From the midpoint coordinate formula, we have: \n\\begin{cases}\\frac{x_{1}+x_{2}}{2}=1\\\\\\frac{y_{1}+y_{2}}{2}=1\\end{cases} \nThen \n\\begin{cases}\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1\\\\\\frac{x^{2}}{4}+\\frac{y}{2}=1\\therefore\\frac{y}{2}=-\\frac{1}{2},\\end{cases} \nSubtracting the two equations gives: \n\\frac{(x_{1}+x)x_{1}-x_{2}}{4}+\\frac{y_{+}y_{2}y_{2}-y_{2}}{2}=0 \n\\therefore The slope of line EF is k=\\frac{y_{1}-y_{2}}{x_{1}-x_{0}}= \n\\therefore The equation of line EF is: y-1=\\frac{1}{3}(x-1), which simplifies to: 2y+x-3=0." }, { "text": "Let $F_{1}$, $F_{2}$ be the two foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, with eccentricity $\\frac {\\sqrt {5}}{2}$. $P$ is a point on the hyperbola such that $\\angle F_{1} P F_{2}=90^{\\circ} $, and $S\\triangle{F_{1} P F_{2}}=1$. Then what is the equation of the asymptotes of the hyperbola? What is the equation of this hyperbola?", "fact_expressions": "G: Hyperbola;a: Number;b: Number;F1: Point;P: Point;F2: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Focus(G) = {F1, F2};Eccentricity(G)=sqrt(5)/2;PointOnCurve(P, G);AngleOf(F1, P,F2) = ApplyUnit(90, degree);Area(TriangleOf(F1,P,F2))=1", "query_expressions": "Expression(Asymptote(G));Expression(G)", "answer_expressions": "y=pm*(1/2)*x\nx^2/4-y^2=1", "fact_spans": "[[[17, 63], [102, 105], [178, 181], [190, 193]], [[20, 63]], [[20, 63]], [[1, 8]], [[98, 101]], [[9, 16]], [[17, 63]], [[1, 68]], [[17, 95]], [[98, 108]], [[110, 145]], [[147, 176]]]", "query_spans": "[[[178, 189]], [[189, 197]]]", "process": "" }, { "text": "The equation of the circle with its center at the focus of the parabola $y^{2}=4 x$ and passing through the origin is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Circle;Focus(G) = Center(H);O: Origin;PointOnCurve(O, H)", "query_expressions": "Expression(H)", "answer_expressions": "x^2+y^2-2*x=0", "fact_spans": "[[[1, 15]], [[1, 15]], [[29, 30]], [[0, 30]], [[24, 28]], [[23, 30]]]", "query_spans": "[[[29, 35]]]", "process": "Solution: Since the parabola is $ y^{2} = 4x $, the focus is $ (1, 0) $. Therefore, the center of the required circle is $ (1, 0) $. Also, since the required circle passes through the origin, the radius $ R $ of the required circle is $ 1 $. The equation of the required circle is $ (x - 1)^{2} + y^{2} = 1 $, or $ x^{2} - 2x + y^{2} = 0 $." }, { "text": "Through a point $M$ on the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ ($a>0$), draw a line $l$ intersecting the two asymptotes of the hyperbola at points $P$ and $Q$, respectively, such that $M$ is the midpoint of segment $PQ$. If the area of triangle $POQ$ ($O$ being the origin) is $2$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2 + x^2/a^2 = 1);a: Number;a>0;M: Point;PointOnCurve(M,G) = True;l: Line;PointOnCurve(M,l) =True;l1: Line;l2: Line;Asymptote(G) = {l1,l2};P: Point;Q: Point;Intersection(l,l1) = P;Intersection(l,l2) = Q;MidPoint(LineSegmentOf(P,Q)) = M;Area(TriangleOf(P,O,Q)) = 2;O: Origin", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)/2", "fact_spans": "[[[1, 38], [127, 130], [52, 55]], [[1, 38]], [[4, 38]], [[4, 38]], [[41, 44], [74, 77]], [[1, 44]], [[45, 50]], [[0, 50]], [], [], [[52, 61]], [[65, 68]], [[69, 72]], [[45, 72]], [[45, 72]], [[74, 88]], [[90, 125]], [[108, 111]]]", "query_spans": "[[[127, 136]]]", "process": "According to the given conditions, the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ ($a>0$) has two asymptotes with equations $y=\\pm\\frac{1}{a}x$. Let the coordinates of points $P$ and $Q$ be $P(x_{1},\\frac{1}{a}x_{1})$, $Q(x_{2},-\\frac{1}{a}x_{2})$. Using the midpoint coordinate formula, express the coordinates of the midpoint $M$. Substitute the coordinates of point $M$ into the hyperbola equation. Then use the triangle area formula to express the area of $\\triangle POQ$, find the value of $a$, and then determine the eccentricity. According to the given conditions, the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ ($a>0$) has two asymptotes with equations $y=\\pm\\frac{1}{a}x$. Let $P(x_{1},\\frac{1}{a}x_{1})$, $Q(x_{2},-\\frac{1}{a}x_{2})$, then $M\\left(\\frac{x_{1}+x_{2}}{2},\\frac{1}{2a}(x_{1}-x_{2})\\right)$. Since point $M$ lies on the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$, we have $\\frac{(x_{1}+x_{2})^{2}}{4a^{2}}-\\frac{(x_{1}-x_{2})^{2}}{4a^{2}}=1$, which gives $x_{1}x_{2}=a^{2}$. From the equations of the two asymptotes of the hyperbola, we get $\\tan\\frac{\\angle POQ}{2}=\\frac{1}{a}$, $\\sin\\angle POQ=2\\sin\\frac{\\angle POQ}{2}\\cos\\frac{\\angle POQ}{2}=\\frac{2\\sin}{\\sin^{2}\\angle}$. Therefore, $S_{\\triangle POQ}=\\frac{1}{2}\\sqrt{x_{1}^{2}+\\frac{1}{a^{2}}x_{1}^{2}}\\sqrt{x_{2}^{2}+\\frac{1}{a^{2}}x_{2}^{2}}\\sin\\angle POQ=\\frac{1}{2}\\times\\frac{a^{2}+1}{a^{2}}\\times x_{1}x_{2}\\times\\frac{\\frac{2}{a}}{1+\\frac{1}{a^{2}}}=a$. But $S_{\\triangle POQ}=2$, so $a=2$, and since $b=1$, we have $c=\\sqrt{5}$, and the eccentricity is $e=\\frac{\\sqrt{5}}{2}$." }, { "text": "If the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ and the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ have common foci $F_{1}$, $F_{2}$, and $P$ is an intersection point of the two curves, then the value of $\\cos \\angle F_{1} P F_{2}$ is?", "fact_expressions": "H: Ellipse;Expression(H) = (x^2/25 + y^2/16 = 1);G: Hyperbola;Expression(G) = (x^2/4 - y^2/5 = 1);F1: Point;F2: Point;Focus(H) = {F1, F2};Focus(G) = {F1, F2};P: Point;OneOf(Intersection(H, G)) = P", "query_expressions": "Cos(AngleOf(F1, P, F2))", "answer_expressions": "11/21", "fact_spans": "[[[1, 40]], [[1, 40]], [[41, 79]], [[41, 79]], [[85, 92]], [[93, 100]], [[1, 100]], [[1, 100]], [[101, 104]], [[101, 113]]]", "query_spans": "[[[115, 146]]]", "process": "Let F_{1}(-c,0), F_{2}(c,0). Due to the symmetry of the ellipse and hyperbola, assume without loss of generality that |PF_{1}| > |PF_{2}|. According to the definitions of the ellipse and hyperbola, we have: |PF_{1}| + |PF_{2}| = 10, |PF_{1}| - |PF_{2}| = 4. Solving these two equations simultaneously gives: |PF_{1}| = 7, |PF_{2}| = 3. From the equation of the ellipse, we obtain c = \\sqrt{25 - 16} = 3, so |F_{1}F_{2}| = 2c = 6. In \\triangle F_{1}PF_{2}, by the law of cosines, \\cos\\angle F_{1}PF_{2} = \\frac{|PF_{1}|^{2} + |PF_{2}|^{2} - |F_{1}F_{2}|^{2}}{2|PF_{1}||PF_{2}|} = \\frac{49 + 9 - 36}{2 \\times 7 \\times 3} = \\frac{22}{42} = \\frac{11}{21}." }, { "text": "Let $F_{1}$, $F_{2}$ be the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$. A line $l$ passing through $F_{2}$ intersects the right branch of hyperbola $C$ at points $A$, $B$, and satisfies $\\overrightarrow{A F_{1}} \\cdot \\overrightarrow{A F_{2}}=0$, $3\\overrightarrow{A F_{2}}=\\overrightarrow{F_{2} B}$. Then the eccentricity of the hyperbola is?", "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;A: Point;F1: Point;F2: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F2, l);Intersection(l, RightPart(C)) = {A, B};DotProduct(VectorOf(A, F1), VectorOf(A, F2)) = 0;3*VectorOf(A, F2) = VectorOf(F2, B)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(10)/2", "fact_spans": "[[[95, 100]], [[17, 79], [101, 107], [236, 239]], [[25, 79]], [[25, 79]], [[111, 114]], [[1, 8]], [[9, 16], [87, 94]], [[115, 118]], [[25, 79]], [[25, 79]], [[17, 79]], [[1, 85]], [[1, 85]], [[86, 100]], [[95, 120]], [[122, 181]], [[182, 234]]]", "query_spans": "[[[236, 245]]]", "process": "Solution: According to the given conditions, let |AF₂| = m. Since 3\\overrightarrow{AF}_{2} = \\overrightarrow{F_{2}B}, then |BF₂| = 3m. Therefore, |AF₁| = 2a + m, |BF₁| = 2a + 3m. Since \\overrightarrow{AF_{1}} \\cdot \\overrightarrow{AF_{2}} = 0, it follows that AF₂ ⊥ AF₁. Hence, (2a + 3m)² = (2a + m)² + (4m)². Therefore, m = a. Because (2c)² = (2a + m)² + (m)², that is, (2c)² = (3a)² + a², which implies 4c² = 10a². Thus, e = \\frac{c}{a} = \\frac{\\sqrt{10}}{2}." }, { "text": "$P$ is a point on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ with foci $F_{1}$ and $F_{2}$. If $\\angle PF_{1} F_{2}=\\alpha$, $\\angle PF_{2} F_{1}=\\beta$, and $\\cos \\alpha=\\frac{\\sqrt{5}}{5}$, $\\sin (\\alpha+\\beta)=\\frac{3}{5}$, then what is the eccentricity of this ellipse?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1,F2};AngleOf(P, F1, F2) = alpha;AngleOf(P, F2, F1) = beta;alpha: Number;beta: Number;Cos(alpha) = sqrt(5)/5;Sin(alpha + beta) = 3/5", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)/7", "fact_spans": "[[[26, 78], [217, 219]], [[26, 78]], [[28, 78]], [[28, 78]], [[28, 78]], [[28, 78]], [[0, 3]], [[0, 84]], [[5, 12]], [[15, 22]], [[4, 78]], [[86, 115]], [[117, 144]], [[181, 214]], [[181, 214]], [[146, 178]], [[181, 214]]]", "query_spans": "[[[217, 225]]]", "process": "Problem Analysis: \\cos\\alpha=\\frac{\\sqrt{5}}{5}\\Rightarrow\\sin\\alpha=\\frac{2\\sqrt{5}}{5}, so \\sin\\beta=\\sin[(\\alpha+\\beta)-\\alpha]=\\sin(\\alpha+\\beta)\\cos\\alpha-\\cos(\\alpha+\\beta)\\sin\\alpha=\\frac{3}{5}\\cdot\\frac{\\sqrt{5}}{5}\\pm\\frac{4}{5}\\cdot\\frac{2\\sqrt{5}}{5}=\\frac{11\\sqrt{5}}{25} or -\\frac{\\sqrt{5}}{5} (discarded). Let PF_{1}=r_{1}, PF_{2}=r_{2}, by the law of sines: \\frac{r_{1}}{\\frac{11\\sqrt{5}}{25}}=\\frac{r_{2}}{\\frac{2\\sqrt{5}}{5}}=\\frac{2c}{5}\\Rightarrow\\frac{r_{1}+r_{2}}{\\frac{21\\sqrt{5}}{25}}=\\frac{2c}{5}\\Rightarrow e=\\frac{c}{a}=\\frac{\\sqrt{5}}{7}" }, { "text": "The foci of the ellipse $x^{2}+m y^{2}=1$ lie on the $y$-axis, and the length of the major axis is twice the length of the minor axis. What is the value of $m$?", "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (m*y^2 + x^2 = 1);PointOnCurve(Focus(G), yAxis);Length(MajorAxis(G)) = 2*Length(MinorAxis(G))", "query_expressions": "m", "answer_expressions": "1/4", "fact_spans": "[[[0, 19]], [[41, 44]], [[0, 19]], [[0, 28]], [[0, 39]]]", "query_spans": "[[[41, 48]]]", "process": "The ellipse equation can be written as: $x^{2}+\\frac{y^{2}}{m}=1$. Since the foci of the ellipse lie on the y-axis, $\\therefore a=\\sqrt{\\frac{1}{m}}, b=1$. $\\therefore 2a=2\\times2b$, that is, $a=2b$. $\\therefore \\sqrt{\\frac{1}{m}}=2$, solving gives: $m=\\frac{1}{4}$. The correct answer to this problem: $\\frac{1}{4}$. [Analysis] This problem examines the issue of finding parameter values based on the ellipse equation and geometric properties. A common mistake is ignoring the axis on which the foci lie, leading to incorrect solutions. This belongs to routine problems." }, { "text": "The moving point $P$ has equal tangent lengths to the two circles $x^{2}+y^{2}-2=0$ and $x^{2}+y^{2}-8 x+10=0$. Then, the trajectory equation of the moving point $P$ is?", "fact_expressions": "G: Circle;Expression(G) = (x^2 + y^2 - 2 = 0);C:Circle;Expression(C)=(x^2+y^2-8*x+10=0);Length(TangentOfPoint(P,G))=Length(TangentOfPoint(P,C));P:Point", "query_expressions": "LocusEquation(P)", "answer_expressions": "x=3/2", "fact_spans": "[[[8, 25]], [[8, 25]], [[26, 48]], [[26, 48]], [[2, 56]], [[2, 5], [60, 63]]]", "query_spans": "[[[60, 70]]]", "process": "As shown in the figure, $x^{2}+y^{2}-2=0 \\Leftrightarrow x^{2}+y^{2}=2$, center of circle $O_{1}(0,0)$; \n$x^{2}+y^{2}-8x+10=0 \\Rightarrow (x-4)^{2}+y^{2}=6$, center of circle $O_{2}(4,0)$. \nLet $P(x,y)$. In right triangles $\\triangle PAO_{1}$ and $\\triangle PAO_{2}$, \nsince $|PA|^{2}=|PO_{1}|^{2}-|AO_{1}|^{2}$, $|PB|^{2}=|PO_{2}|^{2}-|BO_{2}|^{2}$, \nand $|PA|^{2}=|PB|^{2}$, \nso $x^{2}+y^{2}-2=(x-4)^{2}+y^{2}-6$, \nsimplifying gives: $x=\\frac{3}{2}$, hence fill in: $x=\\frac{3}{2}$" }, { "text": "Given that $F_{1}$, $F_{2}$ are the two foci of the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, point $A$ lies on the ellipse $E$, and $\\angle F_{1} A F_{2}=120^{\\circ}$, $|A F_{1}|=2|A F_{2}|$, then what is the eccentricity of the ellipse?", "fact_expressions": "F1: Point;F2: Point;Focus(E) = {F1, F2};E: Ellipse;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;A: Point;PointOnCurve(A, E) = True;AngleOf(F1, A, F2) = ApplyUnit(120, degree);Abs(LineSegmentOf(A,F1)) = 2*Abs(LineSegmentOf(A,F2))", "query_expressions": "Eccentricity(E)", "answer_expressions": "sqrt(7)/3", "fact_spans": "[[[2, 9]], [[10, 17]], [[2, 80]], [[18, 75], [86, 91], [154, 156]], [[18, 75]], [[25, 75]], [[25, 75]], [[25, 75]], [[25, 75]], [[81, 85]], [[81, 92]], [[94, 128]], [[130, 152]]]", "query_spans": "[[[154, 161]]]", "process": "First, from |AF₁| = 2|AF₂|, according to the definition of an ellipse, find |AF₁| and |AF₂|. Then, by the law of cosines and given ∠F₁AF₂ = 120°, set up an equation to solve for the eccentricity. Since point A lies on the ellipse E: x²/a² + y²/b² = 1 (a > b > 0), we have |AF₁| + |AF₂| = 2a. Also, |AF₁| = 2|AF₂|, so \n\\begin{cases} \n|AF₁| = \\frac{4}{3}a \\\\ \n|AF₂| = \\frac{2}{3}a \n\\end{cases} \nSince |F₁F₂| = 2c, in triangle AF₁F₂, given ∠F₁AF₂ = 120°, by the law of cosines we get \ncos∠F₁AF₂ = \\frac{|AF₁|² + |AF₂|² - |F₁F₂|²}{2|AF₁||AF₂|} = \\frac{\\frac{16}{9}a² + \\frac{4}{9}a² - 4c²}{\\frac{16}{9}a²} = \\frac{5}{4} - \\frac{9}{4}e² = -\\frac{1}{2} \nSolving gives e = \\frac{\\sqrt{7}}{3} (the negative value is discarded)." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+y^{2}=1$, the left and right foci are $F_{1}$, $F_{2}$ respectively. A line passing through point $F_{1}$ intersects the ellipse at points $A$, $B$. Then the maximum area of the incircle of triangle $ABF_{2}$ is?", "fact_expressions": "C: Ellipse;H: Line;A: Point;B: Point;F2: Point;F1: Point;Expression(C) = (x^2/4 + y^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, H);Intersection(H, C) = {A, B}", "query_expressions": "Max(Area(InscribedCircle(TriangleOf(A, B, F2))))", "answer_expressions": "pi/4", "fact_spans": "[[[2, 34], [72, 74]], [[69, 71]], [[75, 78]], [[79, 82]], [[51, 58]], [[43, 50], [60, 68]], [[2, 34]], [[2, 58]], [[2, 58]], [[59, 71]], [[69, 84]]]", "query_spans": "[[[86, 119]]]", "process": "Solution: The slope of line AB cannot be 0, but it may be undefined. Let the equation of line AB be $ x = ty - \\sqrt{3} $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. From \n$$\n\\begin{cases}\nx = ty - \\sqrt{3} \\\\\n\\frac{x^{2}}{4} + y^{2} = 1\n\\end{cases}\n$$\nwe obtain $ (t^{2} + 4)y^{2} - 2\\sqrt{3}ty - 1 = 0 $, \n$ y_{1} + y_{2} = \\frac{2\\sqrt{3}t}{t^{2} + 4} $, \n$ y_{1}y_{2} = -\\frac{1}{t^{2} + 4} $, \nand $ 2 $ (equality holds if and only if $ t = \\pm\\sqrt{2} $). \nLet $ r $ be the inradius of triangle $ \\triangle ABF_{2} $, and $ |AF_{2}| + |BF_{2}| + |AB| = 4a = 8 $. Then \n$$\n\\frac{1}{2}(|AF_{2}| + |BF_{2}| + |AB|) \\cdot r \\leqslant 2, \\quad r \\leqslant \\frac{1}{2}\n$$\nThus, the maximum area of the incircle of $ \\triangle ABF_{2} $ is $ \\pi \\times \\left(\\frac{1}{2}\\right)^{2} = \\frac{\\pi}{4} $." }, { "text": "The standard equation of a hyperbola with foci on the $x$-axis, a conjugate axis length of $8$, and eccentricity $e=\\frac{5}{3}$ is?", "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G),xAxis);Length(ImageinaryAxis(G))=8;Eccentricity(G)=e;e:Number;e=5/3", "query_expressions": "Expression(G)", "answer_expressions": "x^2/9 - y^2/16 = 1", "fact_spans": "[[[37, 40]], [[0, 40]], [[9, 40]], [[18, 40]], [[21, 36]], [[21, 36]]]", "query_spans": "[[[37, 47]]]", "process": "According to the given conditions, the equation of the hyperbola can be written as \\frac{x2}{a^{2}}-\\frac{y^{2}}{b^{2}}=1. From 2b=8, \\frac{c}{a}=\\frac{5}{3}, and c^{2}=a^{2}+b^{2}, we obtain a^{2}=9, b^{2}=16. Therefore, the standard equation of the required hyperbola is \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1." }, { "text": "Given the parabola $ T $: $ y^{2} = 8x $, let the focus be $ F $. Two mutually perpendicular lines $ l_{1} $ and $ l_{2} $ passing through $ F $ intersect the parabola at points $ A $, $ B $ and $ C $, $ D $, respectively. The midpoints of segments $ AB $ and $ CD $ are $ P $ and $ Q $, respectively. Then the minimum value of $ |FP| \\cdot |FQ| $ is?", "fact_expressions": "T: Parabola;Expression(T) = (y^2 = 8*x);F: Point;Focus(T) = F;l1: Line;l2: Line;PointOnCurve(F, l1);PointOnCurve(F, l2);IsPerpendicular(l1, l2);A: Point;B: Point;C: Point;D: Point;Intersection(l1, T) = {A, B};Intersection(l2, T) = {C, D};P: Point;Q: Point;MidPoint(LineSegmentOf(A, B)) = P;MidPoint(LineSegmentOf(C, D)) = Q", "query_expressions": "Min(Abs(LineSegmentOf(F, P))*Abs(LineSegmentOf(F, Q)))", "answer_expressions": "32", "fact_spans": "[[[3, 22], [54, 57]], [[3, 22]], [[25, 28]], [[3, 28]], [[34, 43]], [[44, 51]], [[2, 43]], [[2, 51]], [[29, 51]], [[58, 62]], [[63, 66]], [[67, 71]], [[72, 75]], [[34, 75]], [[34, 75]], [[95, 98]], [[99, 102]], [[76, 102]], [[76, 102]]]", "query_spans": "[[[104, 128]]]", "process": "From the given conditions, the slopes of lines $l_{1}$ and $l_{2}$ both exist and are non-zero, and $F(2,0)$. Therefore, we can assume the equation of line $l_{1}$ is $x=ky+2$, then the equation of line $l_{2}$ is $x=-\\frac{1}{k}y+2$. From \n$$\n\\begin{cases}\ny^2=8x \\\\\nx=ky+2\n\\end{cases},\n$$\neliminating $x$, we obtain $y^{2}-8ky-16=0$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $y_{1}+y_{2}=8k$, so $y_{P}=\\frac{y_{1}+y_{2}}{2}=4k$. Substituting into the equation of line $l_{1}$, we get $x_{P}=4k^{2}+2$. Hence, point $P(4k^{2}+2,4k)$, and thus $|PF|=4\\sqrt{k^{2}(1+k^{2})}$. Similarly, we obtain $|QF|=\\frac{4\\sqrt{1+k^{2}}}{k^{2}}$. Therefore,\n$$\n|FP|\\cdot|FQ|=4\\sqrt{k^{2}(1+k^{2})}\\cdot\\frac{4\\sqrt{1+k^{2}}}{k^{2}}=16\\cdot\\frac{1+k^{2}}{|k|}=16(|k|+\\frac{1}{|k|})\\geqslant16\\times2\\sqrt{|k|\\cdot\\frac{1}{|k|}}=32,\n$$\nwith equality if and only if $|k|=\\frac{1}{|k|}$, that is, when $k=\\pm1$. Therefore, the minimum value of $|FP|\\cdot|FQ|$ is 32." }, { "text": "Given two fixed points $A(-2,0)$, $B(1,0)$, if a moving point $P$ satisfies $|PA|=2|PB|$, then the trajectory equation of point $P$ is?", "fact_expressions": "A: Point;Coordinate(A) = (-2, 0);B: Point;Coordinate(B) = (1, 0);P: Point;Abs(LineSegmentOf(P, A)) = 2*Abs(LineSegmentOf(P, B))", "query_expressions": "LocusEquation(P)", "answer_expressions": "x^2+y^2-4*x=0", "fact_spans": "[[[5, 14]], [[5, 14]], [[16, 25]], [[16, 25]], [[30, 33], [49, 53]], [[35, 47]]]", "query_spans": "[[[49, 60]]]", "process": "" }, { "text": "The line passing through a focus $F_{1}$ of the ellipse $4x^{2}+y^{2}=1$ intersects the ellipse at points $A$ and $B$. What is the perimeter of the triangle formed by $A$, $B$, and the other focus $F_{2}$ of the ellipse?", "fact_expressions": "G: Ellipse;l: Line;F1: Point;A: Point;B: Point;F2:Point;Expression(G) = (4*x^2 + y^2 = 1);Focus(G)={F1,F2};PointOnCurve(F1,l);Intersection(l,G)={A,B}", "query_expressions": "Perimeter(TriangleOf(A,B,F2))", "answer_expressions": "4", "fact_spans": "[[[1, 20], [59, 61]], [[33, 35]], [[25, 32]], [[40, 43], [51, 54]], [[44, 47], [55, 58]], [[67, 74]], [[1, 20]], [[0, 74]], [[0, 35]], [[33, 49]]]", "query_spans": "[[[51, 85]]]", "process": "Solution: The equation of the ellipse can be rewritten as y^{2}+\\frac{x^{2}}{1}=1. Clearly, the foci lie on the y-axis, and a=1. According to the definition of an ellipse, |AF_{1}|+|AF_{2}|=2a and |BF_{1}|+|BF_{2}|=2a. Therefore, the perimeter of \\triangle ABF_{2} is |AF_{1}|+|AF_{2}|+|BF_{1}|+|BF_{2}|=4a=4." }, { "text": "Given that hyperbola $S$ has the same foci as the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{34}=1$, and if $y=\\frac{3}{4} x$ is an asymptote of hyperbola $S$, then the equation of hyperbola $S$ is?", "fact_expressions": "S: Hyperbola;G: Ellipse;Expression(G) = (x^2/9 + y^2/34 = 1);Focus(S)=Focus(G);Expression(OneOf(Asymptote(S)))=(y=(3/4)*x)", "query_expressions": "Expression(S)", "answer_expressions": "y^2/9 - x^2/16 = 1", "fact_spans": "[[[2, 8], [73, 79], [87, 93]], [[9, 47]], [[9, 47]], [[2, 52]], [[55, 85]]]", "query_spans": "[[[87, 98]]]", "process": "\\because the ellipse equation is \\frac{x^{2}}{9}+\\frac{y^{2}}{34}=1, and hyperbola S has the same foci as the ellipse \\frac{x^{2}}{9}+\\frac{y^{2}}{34}=1, \\therefore the foci coordinates of hyperbola S are (0,\\pm5). Let the hyperbola equation be \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1 (a>0,b>0), then c=5. \\because y=\\frac{3}{4}x is an asymptote of hyperbola S, \\therefore \\frac{a}{b}=\\frac{3}{4}, \\because c^{2}=a^{2}-b^{2}, \\therefore a=3, b=4. \\therefore the equation of hyperbola S is \\frac{y^{2}}{a}-\\frac{x^{2}}{-\\frac{1}{2}}=1" }, { "text": "The asymptotic equations of the hyperbola $x^{2}-4 y^{2}=1$ are?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - 4*y^2 = 1)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "pm*x - 2*y = 0", "fact_spans": "[[[0, 20]], [[0, 20]]]", "query_spans": "[[[0, 28]]]", "process": "Convert the hyperbola equation into standard form, find a and b, then substitute them into the asymptote equation of the hyperbola. [Detailed solution] x^{2}-4y^{2}=1 \\Rightarrow x^{2}-\\frac{y^{2}}{\\frac{1}{4}}=1 \\Rightarrow a=1, b=\\frac{1}{2}, therefore the asymptote equations of the hyperbola x^{2}-4y^{2}=1 are y=\\pm\\frac{b}{a}x \\Rightarrow y=\\pm\\frac{1}{2}x \\Rightarrow \\pm x-2y=0" }, { "text": "If the hyperbola $C$: $x^{2}-\\frac{y^{2}}{b^{2}}=1(b>0)$ has a focal distance of $2 \\sqrt{5}$, then $b$=?", "fact_expressions": "C: Hyperbola;b: Number;b>0;Expression(C) = (x^2 - y^2/b^2 = 1);FocalLength(C) = 2*sqrt(5)", "query_expressions": "b", "answer_expressions": "2", "fact_spans": "[[[1, 43]], [[61, 64]], [[9, 43]], [[1, 43]], [[1, 59]]]", "query_spans": "[[[61, 66]]]", "process": "" }, { "text": "The length of the chord cut from the parabola $C$: $y^{2}=4 x$ by the line $l$: $2 x-y-1=0$ is?", "fact_expressions": "C: Parabola;l: Line;Expression(C) = (y^2 = 4*x);Expression(l) = (2*x - y - 1 = 0)", "query_expressions": "Length(InterceptChord(l,C))", "answer_expressions": "sqrt(15)", "fact_spans": "[[[0, 18]], [[19, 36]], [[0, 18]], [[19, 36]]]", "query_spans": "[[[0, 43]]]", "process": "" }, { "text": "Given the parabola equation $y^{2}=4 x$, the line $l$ has the equation $x-y+5=0$. On the parabola, there is a moving point $P$ whose distance to the $y$-axis is $d_{1}$ and whose distance to the line $l$ is $d_{2}$. Find the minimum value of $d_{1}+d_{2}$.", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);l: Line;Expression(l) = (x - y + 5 = 0);P: Point;PointOnCurve(P, G) = True;d1: Number;Distance(P, yAxis) = d1;d2: Number;Distance(P,l) = d2", "query_expressions": "Min(d1 + d2)", "answer_expressions": "3*sqrt(2) - 1", "fact_spans": "[[[2, 5], [40, 43]], [[2, 19]], [[20, 25], [69, 74]], [[20, 38]], [[48, 51]], [[39, 51]], [[60, 67]], [[48, 67]], [[78, 85]], [[48, 85]]]", "query_spans": "[[[87, 106]]]", "process": "According to the problem, draw the figure: Based on the definition of a parabola, transform the problem into the distance from the focus to the line $ l $ minus 1; use the point-to-line distance formula to solve. According to the problem, draw the figure: \n$\\because$ The equation of the parabola is $ y^{2} = 4x $, the equation of line $ l $ is $ x - y + 5 = 0 $, \n$\\therefore F(1,0) $, the directrix is $ x = -1 $, \n$\\because$ On the parabola, there is a moving point $ P $ whose distance to the $ y $-axis is $ d_{1} $, and whose distance to line $ l $ is $ d_{2} $, \n$\\therefore$ By the definition of the parabola, the minimum value of $ d_{1} + d_{2} $ is the distance from the focus to line $ l $ minus 1, \n$\\therefore$ the minimum value is $ \\frac{|1 - 0 + 5|}{\\sqrt{1^{2} + (-1)^{2}}} - 1 = 3\\sqrt{2} - 1 $." }, { "text": "Given the line $y = -k x + k$ and the curve $y = x^{2} - 2x$. When the length of the segment cut from the line by the curve is $\\sqrt{10}$, what is the equation of the line?", "fact_expressions": "G: Line;k: Number;H: Curve;Expression(G) = (y = -k*x + k);Expression(H) = (y = x^2 - 2*x);Length(InterceptChord(G,H))=sqrt(10)", "query_expressions": "Expression(G)", "answer_expressions": "{y=x-1,y=-x+1}", "fact_spans": "[[[2, 14], [33, 35], [58, 60]], [[4, 14]], [[15, 30], [36, 38]], [[2, 14]], [[15, 30]], [[33, 56]]]", "query_spans": "[[[58, 64]]]", "process": "Substituting the line $ y = -kx + k $ into the curve $ y = x^{2} - 2x $ gives $ x^{2} + (k - 2)x - k = 0 $. Let $ x_{1} + x_{2} = 2 - k $, $ x_{1}x_{2} = -k $, $ \\Delta \\geqslant 0 $. The length of the line segment intercepted by the curve is $ \\sqrt{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}} \\cdot \\sqrt{1 + k^{2}} = \\sqrt{1 + k^{2}} \\cdot \\sqrt{4 + k^{2}} $. Since the length of the line segment intercepted by the curve is $ \\sqrt{10} $, we have $ \\sqrt{1 + k^{2}} \\cdot \\sqrt{4 + k^{2}} = \\sqrt{10} $. Therefore, $ k^{4} + 5k^{2} - 6 = 0 $, so $ k^{2} = 1 $ (the negative value is discarded), thus $ k = \\pm 1 $, satisfying $ \\Delta \\geqslant 0 $. Hence, the equations of the lines are $ y = x - 1 $ or $ y = -x + 1 $." }, { "text": "Given real numbers $x$, $y$ satisfying $x|x|+\\frac{y|y|}{3}=1$, then the range of $|\\sqrt{3} x+y-4|$ is?", "fact_expressions": "x_: Real;y_: Real;x_*Abs(x_) + (y_*Abs(y_))/3 = 1", "query_expressions": "Range(Abs(sqrt(3)*x_ + y_ - 4))", "answer_expressions": "[4-sqrt(6), 4)", "fact_spans": "[[[2, 7]], [[9, 12]], [[14, 37]]]", "query_spans": "[[[39, 64]]]", "process": "When x>0, y>0, it represents the part of the ellipse x+\\frac{y^{2}}{3}=1 in the first quadrant; when x>0, y<0, it represents the part of the hyperbola x^{2}-\\frac{y^{2}}{3}=1 in the fourth quadrant; when x<0, y>0, it represents the part of the hyperbola -x^{2}+\\frac{y^{2}}{3}=1 in the second quadrant; when x<0, y<0, -x^{2}-\\frac{y^{2}}{3}=1 does not represent any graph; along with the two points (1,0), (0,\\sqrt{3}), the rough sketch is shown in the figure: The distance from a point on the curve to \\sqrt{3}x+y-4=0 is d=\\frac{|\\sqrt{3}x+y-4|}{2}. According to the hyperbola equation, the asymptotes of the hyperbolas in the second and fourth quadrants are both y=\\pm\\sqrt{3}x. The distance from \\sqrt{3}x+y-4=0 is 2, and the distance from points on the hyperbolas in the second and fourth quadrants to \\sqrt{3}x+y-4=0 is d=\\frac{|\\sqrt{3}x+y-4|}{2}, which is less than and infinitely close to 2. Consider an arbitrary point in the first quadrant of the curve denoted as P(\\cos\\theta,\\sqrt{3}\\sin\\theta), \\theta\\in(0,\\frac{\\pi}{2}] to \\sqrt{3}x+y-4=0, so d=\\frac{|\\sqrt{3}\\cos\\theta+\\sqrt{3}\\sin\\theta-4|}{2}=\\frac{|\\sqrt{6}\\sin(\\theta+\\frac{\\pi}{4})-4|}{2}\\geqslant\\frac{4-\\sqrt{6}}{2}, where equality holds when \\theta=\\frac{\\pi}{4}, then the range of |\\sqrt{3}x+y-4| is (4-\\sqrt{6},4)." }, { "text": "The standard equation of an ellipse with a vertex at $(0,2)$ and eccentricity $\\frac{1}{2}$ is?", "fact_expressions": "G: Ellipse;Eccentricity(G) = 1/2;Coordinate(OneOf(Vertex(G)))=(0,2)", "query_expressions": "Expression(G)", "answer_expressions": "{x^2/3+y^2/4=1, 3*x^2/16+y^2/4=1}", "fact_spans": "[[[32, 34]], [[14, 34]], [0, 32]]", "query_spans": "[[[32, 41]]]", "process": "" }, { "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=4x$, a line $l$ passing through point $F$ intersects the parabola $C$ at points $A$ and $B$, and intersects the directrix of the parabola $C$ at point $D$. If $F$ is the midpoint of $AD$, then $|FB|=$?", "fact_expressions": "l: Line;C: Parabola;A: Point;D: Point;F: Point;B: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Intersection(l, Directrix(C)) = D;MidPoint(LineSegmentOf(A, D)) = F", "query_expressions": "Abs(LineSegmentOf(F, B))", "answer_expressions": "4/3", "fact_spans": "[[[35, 40]], [[6, 25], [41, 47], [60, 66]], [[49, 52]], [[71, 75]], [[2, 5], [30, 34], [77, 80]], [[53, 56]], [[6, 25]], [[2, 28]], [[29, 40]], [[35, 58]], [[35, 75]], [[77, 89]]]", "query_spans": "[[[91, 101]]]", "process": "Draw the graph. Given that the focal-directrix distance is 4 and F is the midpoint of AD, the length of AM can be found. Using the definition of a parabola, the length of AF can be obtained, and thus it can be determined that \\angle MAD = \\angle EFD = 60^{\\circ}. In \\triangle EFD, using the definition, FB = BN, allowing the answer to be found. As shown in the figure: draw perpendiculars from points A, B, F to the directrix, intersecting the directrix at points M, N, E respectively. From the given conditions, FE = 2, and F is the midpoint of AD, so EF is the midline of \\triangle ADM. Therefore, AM = 4, then AF = DF = 4, so \\cos MAD = \\frac{AM}{AD} = \\frac{1}{2}, hence \\angle MAD = \\angle EFD = 60^{\\circ}. Furthermore, by the definition of a parabola: FB = BN, and BD = 2BN. Therefore, 3BF = DF = 4, so BF = \\frac{4}{\\frac{4}{2}}, the numerical answer is: 4" }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through point $F_{1}$ intersects the ellipse at points $P$ and $Q$. It is given that $P F_{2}=F_{1} F_{2} $, $3 P F_{1}=4 Q F_{1}$, then the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Line;PointOnCurve(F1, H);P: Point;Q: Point;Intersection(H, G) = {P, Q};LineSegmentOf(P, F2) = LineSegmentOf(F1, F2);3*LineSegmentOf(P, F1) = 4*LineSegmentOf(Q, F1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "5/7", "fact_spans": "[[[2, 54], [92, 94], [153, 155]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[63, 70], [80, 88]], [[71, 78]], [[2, 78]], [[2, 78]], [[89, 91]], [[79, 91]], [[95, 98]], [[99, 102]], [[89, 104]], [[106, 128]], [[130, 151]]]", "query_spans": "[[[153, 161]]]", "process": "PF_{2}=F_{1}F_{2}=2c, hence PF_{1}=2a-2c, 3PF_{1}=4QF_{1}, hence QF_{1}=\\frac{3}{2}(a-c), QF_{2}=2a-QF_{1}=\\frac{1}{2}a+\\frac{3}{2}c\\cos\\angle PF_{1}F_{2}=\\cos(\\pi-\\angle QF_{1}F_{2})=-\\cos\\angle QF_{1}F_{2}. Using the law of cosines: \\frac{4(a-c)^{2}+4c^{2}-4c^{2}}{2\\cdot2(a-c)\\cdot2c}=-\\frac{\\frac{9}{4}(a-c)^{2}+4c^{2}-(\\frac{1}{2}a)^{2}}{2\\cdot\\frac{3}{2}(a-c)\\cdot2c}. Simplifying yields: 5a^{2}-12ac+7c^{2}=0, i.e., 7e^{2}-12e+5=0. Solving gives e=\\frac{5}{7} or e=1 (discarded)." }, { "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/12 = 1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[0, 39]], [[0, 39]]]", "query_spans": "[[[0, 45]]]", "process": "Test analysis: e=\\sqrt{1+\\frac{12}{4}}=2" }, { "text": "Given the equation of the hyperbola is $\\frac{x^{2}}{3}-y^{2}=1$, then the distance from the focus of this hyperbola to its asymptote is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1)", "query_expressions": "Distance(Focus(G), Asymptote(G))", "answer_expressions": "1", "fact_spans": "[[[2, 5], [37, 40]], [[2, 34]]]", "query_spans": "[[[37, 52]]]", "process": "" }, { "text": "Given that point $P$ lies on the ellipse $\\frac{x^{2}}{45}+\\frac{y^{2}}{20}=1$ and is in the third quadrant, and the lines connecting $P$ to the two foci are perpendicular to each other. If the distance from point $P$ to the line $4x - 3y - 2m + 1 = 0$ is no greater than $3$, then the range of real values for $m$ is?", "fact_expressions": "P: Point;G: Ellipse;Expression(G) = (x^2/45 + y^2/20 = 1);PointOnCurve(P, G) = True;Quadrant(P) = 3;J1: Point;J2: Point;Focus(G) = {J1, J2};IsPerpendicular(LineSegmentOf(P, J1), LineSegmentOf(P, J2)) = True;H: Line;Expression(H) = (-2*m + 4*x - 3*y + 1 = 0);Negation(Distance(P, H) > 3);m: Real", "query_expressions": "Range(m)", "answer_expressions": "[-7,8]", "fact_spans": "[[[2, 6], [60, 61], [73, 77]], [[7, 46]], [[7, 46]], [[2, 58]], [[2, 58]], [], [], [[7, 65]], [[7, 71]], [[78, 97]], [[78, 97]], [[73, 106]], [[108, 113]]]", "query_spans": "[[[108, 120]]]", "process": "" }, { "text": "The line $x-2 y+2=0$ passes through one focus and one vertex of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$. What is the eccentricity of this ellipse?", "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x - 2*y + 2 = 0);PointOnCurve(OneOf(Focus(G)),H);PointOnCurve(OneOf(Vertex(G)),H)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2*sqrt(5)/5", "fact_spans": "[[[15, 67], [80, 82]], [[17, 67]], [[17, 67]], [[0, 13]], [[17, 67]], [[17, 67]], [[15, 67]], [[0, 13]], [[0, 72]], [[0, 77]]]", "query_spans": "[[[80, 88]]]", "process": "" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, both of its asymptotes are tangent to the circle $C$: $x^{2}+y^{2}-6 x+5=0$, and the right focus of the hyperbola is the center of circle $C$. Then the equation of this hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;C: Circle;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(C) = (-6*x + x^2 + y^2 + 5 = 0);IsTangent(Asymptote(G),C);Center(C)=RightFocus(G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/5-y^2/4=1", "fact_spans": "[[[2, 58], [97, 100], [115, 118]], [[5, 58]], [[5, 58]], [[66, 93], [105, 109]], [[5, 58]], [[5, 58]], [[2, 58]], [[66, 93]], [[2, 95]], [[97, 112]]]", "query_spans": "[[[115, 123]]]", "process": "Circle C: x^{2}+y^{2}-6x+5=0 is a circle with center (3,0) and radius 2. It follows that in the hyperbola, c=3. The asymptotes of the hyperbola have equations: y=\\pm\\frac{b}{a}x. According to the problem, the distance from point (3,0) to the asymptote y=\\frac{b}{a}x is 2. Applying the point-to-line distance formula, we obtain b=2, a=\\sqrt{5}. Therefore, the equation of the hyperbola is \\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, with left and right foci $F_{1}$, $F_{2}$ respectively. If a point $P$ on $C$ satisfies $|\\overrightarrow{P F_{1}}+\\overrightarrow{P F_{2}}|=|\\overrightarrow{F_{1}{F}_{2}}|$ and $|\\overrightarrow{P F_{1}}|=2|\\overrightarrow{P F_{2}}|$, then the asymptotes of hyperbola $C$ have equations?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P,C);Abs(VectorOf(P,F1)+VectorOf(P,F2))=Abs(VectorOf(F1,F2));Abs(VectorOf(P, F1)) = 2*Abs(VectorOf(P, F2))", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "y=pm*2*x", "fact_spans": "[[[2, 63], [88, 91], [244, 250]], [[10, 63]], [[10, 63]], [[94, 97]], [[71, 78]], [[79, 86]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 86]], [[2, 86]], [[88, 97]], [[99, 184]], [[186, 242]]]", "query_spans": "[[[244, 258]]]", "process": "From the given condition: $|\\overrightarrow{PF_{1}}+\\overrightarrow{PF_{2}}|=|2\\overrightarrow{OP}|=|\\overrightarrow{F_{1}F_{2}}|$, then $\\triangle F_{1}PF_{2}$ is a right triangle with point $P$ as the right-angle vertex. Let $|PF_{1}|=2m$, $|PF_{2}|=m$. By the definition of hyperbola: $|PF_{1}|-|PF_{2}|=2a$, so $a=\\frac{m}{2}$. By the Pythagorean theorem: $m^{2}+4m^{2}=4c^{2}$, so $m^{2}=\\frac{4}{5}c^{2}$. Altogether: $4a^{2}=\\frac{4}{5}c^{2}$, so $5a^{2}=a^{2}+b^{2}$, $\\frac{a^{2}}{b^{2}}=\\frac{1}{4}$. Then the asymptotes of the hyperbola are: $y=\\pm\\frac{b}{a}x=\\pm2x$." }, { "text": "Given that line $l$ passing through point $(1,0)$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$, point $C(-1,-1)$, and $CA \\perp CB$, then the area of $\\triangle ABC$ is?", "fact_expressions": "H: Point;Coordinate(H) = (1, 0);l: Line;PointOnCurve(H, l);G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;B: Point;Intersection(l, G) = {A, B};C: Point;Coordinate(C) = (-1, -1);IsPerpendicular(LineSegmentOf(C, A), LineSegmentOf(C, B))", "query_expressions": "Area(TriangleOf(A, B, C))", "answer_expressions": "(5*sqrt(5))/2", "fact_spans": "[[[4, 12]], [[4, 12]], [[13, 18]], [[2, 18]], [[19, 33]], [[19, 33]], [[36, 39]], [[40, 43]], [[13, 45]], [[46, 58]], [[46, 58]], [[60, 75]]]", "query_spans": "[[[77, 99]]]", "process": "According to the problem, we can assume the equation of line $ l $ is $ x = ky + 1 $, with points $ A\\left(\\frac{y_{1}^{2}}{4}, y_{1}\\right) $, $ B\\left(\\frac{y_{2}^{2}}{4}, y_{2}\\right) $. By solving the system with the parabola equation, we obtain $ y_{1} + y_{2} = 4k $, $ y_{1}y_{2} = -4 $. Combining $ \\overrightarrow{CA} \\perp \\overrightarrow{CB} $ and using the coordinate representation of perpendicular vectors, we can solve for $ k $. Applying the point-to-line distance formula and the focal chord length formula, we can find the area of $ \\triangle ABC $.\n\nAccording to the problem, let the equation of line $ l $ be $ x = ky + 1 $, intersecting the parabola $ y^{2} = 4x $ at points $ A\\left(\\frac{y_{1}^{2}}{4}, y_{1}\\right) $, $ B\\left(\\frac{y_{2}^{2}}{4}, y_{2}\\right) $. Therefore, $ y_1 $ and $ y_2 $ are the two roots of the equation $ y^{2} - 4ky - 4 = 0 $, so we have $ y_{1} + y_{2} = 4k $, $ y_{1}y_{2} = -4 $. Since $ \\overrightarrow{CA} \\perp \\overrightarrow{CB} $, by the coordinate representation of perpendicular vectors, we have:\n$$\n\\left(\\frac{y_{1}^{2}}{4} + 1\\right)\\left(\\frac{y_{2}^{2}}{4} + 1\\right) + (y_{1} + 1)(y_{2} + 1) = 0,\n$$\nwhich gives $ k = -\\frac{1}{2} $.\n\nLet $ h $ be the distance from point $ C $ to line $ AB $, then $ S_{\\triangle ABC} = \\frac{1}{2} h \\cdot |AB| $. Using the point-to-line distance formula:\n$$\nh = \\frac{|Ax_0 + By_0 + C|}{\\sqrt{A^2 + B^2}} = \\frac{\\left| \\frac{1}{2} + 1 + 1 \\right|}{\\sqrt{1 + 1}} = \\frac{\\left| \\frac{5}{2} \\right|}{\\sqrt{2}} = \\frac{5}{2\\sqrt{2}} = \\frac{5\\sqrt{2}}{4},\n$$\nbut there seems to be an error in the original text. Correcting based on standard form:\n\nActually, the line $ AB $: $ x = -\\frac{1}{2}y + 1 $, or $ 2x + y - 2 = 0 $. Then distance from $ C(-1, -1) $ to this line is:\n$$\nh = \\frac{|2(-1) + (-1) - 2|}{\\sqrt{2^2 + 1^2}} = \\frac{|-2 -1 -2|}{\\sqrt{5}} = \\frac{5}{\\sqrt{5}} = \\sqrt{5}.\n$$\n\nAnd $ |AB| = \\sqrt{1 + k^2} \\cdot |y_1 - y_2| $. Since $ (y_1 - y_2)^2 = (y_1 + y_2)^2 - 4y_1y_2 = (4k)^2 - 4(-4) = 16k^2 + 16 $. With $ k = -\\frac{1}{2} $, $ k^2 = \\frac{1}{4} $, so:\n$$\n(y_1 - y_2)^2 = 16 \\cdot \\frac{1}{4} + 16 = 4 + 16 = 20 \\Rightarrow |y_1 - y_2| = \\sqrt{20} = 2\\sqrt{5}.\n$$\nThen:\n$$\n|AB| = \\sqrt{1 + \\left(-\\frac{1}{2}\\right)^2} \\cdot 2\\sqrt{5} = \\sqrt{1 + \\frac{1}{4}} \\cdot 2\\sqrt{5} = \\sqrt{\\frac{5}{4}} \\cdot 2\\sqrt{5} = \\frac{\\sqrt{5}}{2} \\cdot 2\\sqrt{5} = 5.\n$$\n\nTherefore, $ S_{\\triangle ABC} = \\frac{1}{2} \\cdot \\sqrt{5} \\cdot 5 = \\frac{5\\sqrt{5}}{2} $." }, { "text": "Given the parabola $C$: $y^{2}=4x$, the focus is $F$, the directrix intersects the $x$-axis at point $M$, and point $N$ lies on the parabola $C$. It is given that $|\\overrightarrow{MN}| = \\lambda |\\overrightarrow{NF}|$. When $\\lambda$ takes its maximum value, what is the equation of the line $ON$?", "fact_expressions": "C: Parabola;N: Point;O: Origin;M: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;Intersection(Directrix(C), xAxis) = M;PointOnCurve(N, C);lambda:Number;Abs(VectorOf(M,N))=lambda*Abs(VectorOf(N,F));WhenMax(lambda)", "query_expressions": "Expression(LineOf(O,N))", "answer_expressions": "y=pm*2*x", "fact_spans": "[[[2, 21], [48, 54]], [[43, 47]], [[131, 136]], [[38, 42]], [[25, 28]], [[2, 21]], [[2, 28]], [[2, 42]], [[43, 55]], [[114, 123]], [[57, 111]], [[113, 128]]]", "query_spans": "[[[129, 141]]]", "process": "Solution: According to the given conditions, F(1,0), M(-1,0). Let N(4t^{2},4t), then |\\overrightarrow{NF}|^{2}=(1-4t^{2})^{2}+16t^{2}=16t^{4}+8t^{2}+1, |\\overrightarrow{MN}|^{2}=(4t^{2}+1)^{2}+16t^{2}=16t^{4}+24t^{2}+1. Hence, \\lambda^{2}=\\frac{|\\overrightarrow{MN}|^{2}}{|\\overrightarrow{NF}|^{2}}=\\frac{16t^{4}+24t^{2}+1}{16t^{4}+8t^{2}+1}=1+\\frac{-}{1}. When t=0, 2^{2}=1; when t\\neq0, \\lambda^{2}=1 if and only if \\frac{1}{12}=16t^{2}, that is, when t=\\pm\\frac{1}{2}, the equality holds. Therefore, when \\lambda takes its maximum value, N(1,2) or N(1,-2). Thus, the equation of line ON is y=\\pm2x." }, { "text": "The line $l$: $y=k(x+1)$ has only one common point with the parabola $y^{2}=x$, then the value of the real number $k$ is?", "fact_expressions": "l: Line;G: Parabola;k: Real;Expression(G) = (y^2 = x);Expression(l) = (y = k*(x + 1));NumIntersection(l,G)=1", "query_expressions": "k", "answer_expressions": "{0,pm*1/2}", "fact_spans": "[[[0, 17]], [[18, 30]], [[39, 44]], [[18, 30]], [[0, 17]], [[0, 37]]]", "query_spans": "[[[39, 48]]]", "process": "Problem analysis: \n\\begin{cases}y=k(x+1)\\\\y^{2}=x\\end{cases} \nEliminating $ x $, we obtain $ ky^{2}-y+k=0 $. According to the problem, we have $ k=0 $ or \n\\begin{cases}k\\neq0\\\\\\end{cases}" }, { "text": "Given the line $y=\\frac{\\sqrt{3}}{3} x-2$ intersects the right branch of the hyperbola $\\frac{x^{2}}{12}-\\frac{y^{2}}{3}=1$ at points $A$ and $B$, and there exists a point $C$ on the right branch of the hyperbola such that $\\overrightarrow{O A}+\\overrightarrow{O B}=t \\overrightarrow{O C}$, find the value of $t$?", "fact_expressions": "G: Hyperbola;H: Line;O: Origin;A: Point;B: Point;C: Point;Expression(G) = (x^2/12 - y^2/3 = 1);Expression(H) = (y = x*(sqrt(3)/3) - 2);Intersection(H, RightPart(G)) = {A, B};PointOnCurve(C, RightPart(G));VectorOf(O, A) + VectorOf(O, B) = t*VectorOf(O, C);t:Number", "query_expressions": "t", "answer_expressions": "4", "fact_spans": "[[[31, 70], [87, 90]], [[2, 30]], [[102, 168]], [[75, 78]], [[79, 82]], [[96, 100]], [[31, 70]], [[2, 30]], [[2, 84]], [[87, 100]], [[102, 168]], [[170, 173]]]", "query_spans": "[[[170, 176]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), C(x_{0},y_{0}) be such that \\overrightarrow{OA}+\\overrightarrow{OB}=t\\overrightarrow{OC}; then x_{1}+x_{2}=tx_{0}, y_{1}+y_{2}=ty_{0}. By solving the line equation y=\\frac{\\sqrt{3}}{3}x-2 together with the curve equation \\frac{x^{2}}{12}-\\frac{y^{2}}{3}=1, we obtain x^{2}-16\\sqrt{3}x+84=0. Therefore, x_{1}+x_{2}=16\\sqrt{3}, and \\frac{\\sqrt{3}}{3}\\times16\\sqrt{3}-4=12. Then \\frac{x_{0}}{y_{0}}=\\frac{4\\sqrt{3}}{3}. Solving \\begin{matrix}x&2\\\\\\frac{x_{0}}{12}&\\frac{y_{0}^{2}}{3}=1 gives x_{0}=4\\sqrt{3}, y_{0}=3. Hence, t=4." }, { "text": "Point $P$ lies on an ellipse with foci $F_{1}(0,-1)$, $F_{2}(0, 1)$, and a directrix $y=4$, such that $| P F_{1}|-|P F_{2}|=1$, $\\tan \\angle F_{1} P F_{2}$?", "fact_expressions": "G: Ellipse;F1: Point;F2: Point;P: Point;Coordinate(F1) = (0, -1);Coordinate(F2) = (0, 1);Focus(G)={F1,F2};Expression(OneOf(Directrix(G)))=(y=4);PointOnCurve(P,G);Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)) = 1", "query_expressions": "Tan(AngleOf(F1, P, F2))", "answer_expressions": "4/3", "fact_spans": "[[[50, 52]], [[8, 21]], [[23, 36]], [[0, 4]], [[8, 21]], [[23, 36]], [[5, 52]], [[39, 52]], [[0, 53]], [[55, 79]]]", "query_spans": "[[[80, 108]]]", "process": "" }, { "text": "If the focus of the parabola $y^{2}=2 p x$ coincides with the right focus of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, then what is the value of $p$?", "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;Expression(G) = (x^2/3 - y^2 = 1);Expression(H) = (y^2 = 2*(p*x));Focus(H) = RightFocus(G)", "query_expressions": "p", "answer_expressions": "4", "fact_spans": "[[[21, 49]], [[1, 17]], [[57, 60]], [[21, 49]], [[1, 17]], [[1, 55]]]", "query_spans": "[[[57, 64]]]", "process": "According to the given condition, the right focus of the hyperbola \\frac{x^2}{3} - y^2 = 1 is (2,0). Since the focus of the parabola y^2 = 2px coincides with the right focus of the hyperbola \\frac{x^2}{3} - y^2 = 1, the focus of the parabola y^2 = 2px is (2,0), which implies \\frac{p}{2} = 2. Solving gives p = 4. [Note] This problem mainly examines the application of standard equations and geometric properties of hyperbolas and parabolas. The key to solving lies in being familiar with the geometric properties of hyperbolas and parabolas. It is a basic problem." }, { "text": "The length of the imaginary axis of the hyperbola $x^{2}-4 y=1$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - 4*y = 1)", "query_expressions": "Length(ImageinaryAxis(G))", "answer_expressions": "1", "fact_spans": "[[[0, 16]], [[0, 16]]]", "query_spans": "[[[0, 22]]]", "process": "The hyperbola $x^{2}-4y^{2}=1$ can be rewritten in standard form as $x^{2}-\\frac{y^{2}}{4}=1$, therefore $a=1$, $b=\\frac{1}{2}$. Hence, the length of the minor axis is $2b=1$." }, { "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=4 x$, $A$, $B$ are two points on $C$, and the midpoint of segment $AB$ is $M(2 , 2)$, then the area of $\\Delta ABF$ equals?", "fact_expressions": "C: Parabola;A: Point;B: Point;M: Point;F: Point;Expression(C) = (y^2 = 4*x);Coordinate(M) = (2, 2);Focus(C) = F;PointOnCurve(A, C);PointOnCurve(B, C);MidPoint(LineSegmentOf(A, B)) = M", "query_expressions": "Area(TriangleOf(A, B, F))", "answer_expressions": "2", "fact_spans": "[[[6, 25], [37, 40]], [[29, 32]], [[33, 36]], [[57, 67]], [[2, 5]], [[6, 25]], [[57, 67]], [[2, 28]], [[29, 45]], [[29, 45]], [[46, 67]]]", "query_spans": "[[[69, 87]]]", "process": "" }, { "text": "The eccentricity of the hyperbola $C$: $x^{2}-y^{2}=\\lambda (\\lambda>0)$ is? The equations of the asymptotes are?", "fact_expressions": "C: Hyperbola;lambda: Number;lambda>0;Expression(C) = (x^2 - y^2 = lambda)", "query_expressions": "Eccentricity(C);Expression(Asymptote(C))", "answer_expressions": "sqrt(2);y=pm*x", "fact_spans": "[[[0, 42]], [[8, 42]], [[8, 42]], [[0, 42]]]", "query_spans": "[[[0, 48]], [[0, 55]]]", "process": "Test analysis: \\because x^{2}-y^{2}=\\lambda,\\therefore\\frac{x^{2}}{\\lambda}-\\frac{y^{2}}{\\lambda}=1,\\therefore a=\\sqrt{\\lambda},b=\\sqrt{\\lambda},c=\\sqrt{2},\\therefore e=\\frac{c}{a}=\\sqrt{2},\\therefore the asymptotes are y=\\pm\\frac{b}{a}x=\\pm x." }, { "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$ and directrix $l$, a line passing through point $F$ intersects $l$ at point $A$ and intersects the parabola at point $B$, such that $\\overrightarrow{FA} = -3 \\overrightarrow{FB}$. Then $|AB| = $?", "fact_expressions": "C: Parabola;G: Line;F: Point;A: Point;B: Point;l: Line;Expression(C) = (y^2 = 4*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(F, G);Intersection(G, l) = A;OneOf(Intersection(G, C)) = B;VectorOf(F, A) = -3*VectorOf(F, B)", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "32/3", "fact_spans": "[[[2, 21], [53, 56]], [[40, 42]], [[24, 27]], [[47, 51]], [[62, 65]], [[30, 33], [43, 46]], [[2, 21]], [[2, 27]], [[2, 33]], [[34, 42]], [[40, 51]], [[40, 65]], [[67, 113]]]", "query_spans": "[[[115, 124]]]", "process": "" }, { "text": "Let $A(-1,0)$, $B(1,0)$ be two fixed points in the plane, and point $P$ satisfies $|P A|+|P B|=6$. Then the trajectory equation of point $P$ is?", "fact_expressions": "A: Point;B: Point;P: Point;Coordinate(A) = (-1, 0);Coordinate(B) = (1, 0);Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B)) = 6", "query_expressions": "LocusEquation(P)", "answer_expressions": "x^2/9+y^2/8=1", "fact_spans": "[[[1, 10]], [[12, 20]], [[27, 31], [50, 54]], [[1, 10]], [[12, 20]], [[33, 48]]]", "query_spans": "[[[50, 61]]]", "process": "Problem Analysis: Since A and B are fixed points and |PA| + |PB| = 6 > |AB|, according to the definition of an ellipse, the moving point P traces an ellipse with A and B as foci and 6 as the major axis length. Therefore, a = 3, c = 1, and thus b^{2} = a^{2} - c^{2} = 8. Hence, the trajectory equation of the moving point P is \\frac{x^{2}}{9}+\\frac{y^{2}}{8}=1." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, with left and right foci $F_{1}$, $F_{2}$, and eccentricity $\\frac{\\sqrt{3}}{3}$. A line $l$ passing through $F_{2}$ intersects $C$ at points $A$, $B$. If the perimeter of $\\Delta A F_{1} B$ is $4$, then the equation of $C$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;Eccentricity(C) = sqrt(3)/3;l: Line;PointOnCurve(F2, l);A: Point;B: Point;Intersection(l, C) = {A, B};Perimeter(TriangleOf(A, F1, B)) = 4", "query_expressions": "Expression(C)", "answer_expressions": "x^2+y^2/(2/3)=1", "fact_spans": "[[[2, 59], [122, 125], [164, 167]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[66, 73]], [[74, 81], [108, 115]], [[2, 81]], [[2, 81]], [[2, 106]], [[116, 121]], [[107, 121]], [[126, 129]], [[130, 133]], [[116, 135]], [[137, 162]]]", "query_spans": "[[[164, 172]]]", "process": "\\because4a=4,\\frac{c}{a}=\\frac{\\sqrt{3}}{3}\\thereforea=1,c=\\frac{\\sqrt{3}}{3}\\thereforeb^{2}=\\frac{2}{3}\\therefore the equation of C is \\frac{x^{2}+\\frac{y^{2}}{3}}=1" }, { "text": "The segment connecting the focus $F$ of the parabola $y^{2}=4x$ and the point $M(0, 1)$ intersects the parabola at point $A$. Let point $O$ be the origin. Then the area of $\\angle OAM$ is?", "fact_expressions": "G: Parabola;M: Point;O: Origin;A: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(M) = (0, 1);Focus(G) = F;Intersection(LineSegmentOf(F, M), G) = A", "query_expressions": "Area(AngleOf(O, A, M))", "answer_expressions": "3/2 - sqrt(2)", "fact_spans": "[[[2, 16], [40, 43]], [[23, 34]], [[51, 55]], [[45, 49]], [[19, 22]], [[2, 16]], [[23, 34]], [[2, 22]], [[0, 49]]]", "query_spans": "[[[62, 79]]]", "process": "" }, { "text": "Given that one of the directrices of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ is $x=\\frac{3}{2}$, then what is the value of $a$, and what is the eccentricity of this hyperbola?", "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(OneOf(Directrix(G))) = (x = 3/2)", "query_expressions": "a;Eccentricity(G)", "answer_expressions": "sqrt(3)\n2*sqrt(3)/3", "fact_spans": "[[[2, 39], [72, 75]], [[64, 67]], [[5, 39]], [[2, 39]], [[2, 62]]]", "query_spans": "[[[64, 70]], [[72, 81]]]", "process": "" }, { "text": "Given that points $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, respectively, $P$ is a point on the right branch of the hyperbola, $I$ is the incenter of $\\triangle P F_{1} F_{2}$, and $S_{\\triangle I P F_2}=S_{\\triangle I P F_1}-\\lambda S_{\\triangle I F 1 F 2}$, then $\\lambda=?$", "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;I: Point;Expression(G) = (x^2/16 - y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));Incenter(TriangleOf(P, F1, F2)) = I;Area(TriangleOf(I,P,F2))=Area(TriangleOf(I,P,F1))-lambda*Area(TriangleOf(I,F1,F2));lambda:Number", "query_expressions": "lambda", "answer_expressions": "4/5", "fact_spans": "[[[21, 60], [71, 74]], [[67, 70]], [[2, 10]], [[11, 18]], [[80, 83]], [[21, 60]], [[2, 66]], [[2, 66]], [[67, 79]], [[80, 112]], [[114, 191]], [[193, 202]]]", "query_spans": "[[[193, 204]]]", "process": "" }, { "text": "Given the parabola equation $x^{2}=12 y$, a line passing through the focus of the parabola with an inclination angle of $60^{\\circ}$ intersects the parabola at points $A$ and $B$. Then $|A B|$=?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 12*y);H: Line;PointOnCurve(Focus(G), H);Inclination(H) = ApplyUnit(60, degree);A: Point;B: Point;Intersection(H, G) = {A, B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "48", "fact_spans": "[[[2, 5], [22, 25], [49, 52]], [[2, 20]], [[46, 48]], [[21, 48]], [[29, 48]], [[55, 58]], [[59, 62]], [[46, 64]]]", "query_spans": "[[[66, 75]]]", "process": "The focus of the parabola has coordinates (0,3), so the equation of line AB is y=\\sqrt{3}x+3. From \\begin{cases}y=\\sqrt{3}x+3\\\\x^{2}=12y\\end{cases}, eliminating y and simplifying gives x^{2}-12\\sqrt{3}x-36=0, so x_{1}+x_{2}=12\\sqrt{3}, x_{1}\\cdot x_{2}=-36, therefore |AB|" }, { "text": "Given fixed point $B(3,0)$, point $A$ moves on the circle $x^{2}+y^{2}=1$, and $M$ is the midpoint of segment $AB$. Then the trajectory equation of point $M$ is?", "fact_expressions": "B: Point;Coordinate(B) = (3, 0);A: Point;G: Circle;Expression(G) = (x^2 + y^2 = 1);PointOnCurve(A, G) = True;M: Point;MidPoint(LineSegmentOf(A,B)) = M", "query_expressions": "LocusEquation(M)", "answer_expressions": "(x - 3/2)^2 + y^2 = 1/4", "fact_spans": "[[[4, 12]], [[4, 12]], [[13, 17]], [[18, 34]], [[18, 34]], [[13, 35]], [[38, 41], [55, 59]], [[38, 53]]]", "query_spans": "[[[55, 66]]]", "process": "According to the given condition, let A(x_{0},y_{0}), M(x,y). Since M is the midpoint of line segment AB, we have \n\\therefore\\begin{cases}\\frac{x_{0}+3}{2}=x\\\\\\frac{y_{0}^{2}}{2}=y\\end{cases}, that is, \\therefore\\begin{cases}x_{0}=2x-3\\\\y_{0}=2y\\end{cases}-\\textcircled{1}. \nIt is also known that point A(x_{0},y_{0}) does not lie on the circle x^{2}+y^{2}=1. Substituting \\textcircled{1} into this equation gives: (2x-3)^{2}+(2y)^{2}=1. \nTherefore, simplifying yields the trajectory equation of point M as: (x-\\frac{3}{2})^{2}+y^{2}=\\frac{1}{4}." }, { "text": "Given the ellipse $ E $: $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $) has its right focus at $ F $, and eccentricity $ \\frac{\\sqrt{3}}{2} $. A line $ l $ passing through the origin $ O $ with an inclination angle of $ \\frac{\\pi}{3} $ intersects the ellipse $ E $ at points $ A $ and $ B $. If the perimeter of triangle $ \\triangle AFB $ is $ 4 + \\frac{8\\sqrt{13}}{13} $, then what is the equation of the ellipse?", "fact_expressions": "l: Line;E: Ellipse;b: Number;a: Number;A: Point;F: Point;B: Point;O:Origin;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(E) = F;Eccentricity(E)=sqrt(3)/2;PointOnCurve(O,l);Inclination(l)=pi/3;Intersection(l,E)={A,B};Perimeter(TriangleOf(A, F, B))=4+(8*sqrt(13)/13)", "query_expressions": "Expression(E)", "answer_expressions": "x^2/4+y^2=1", "fact_spans": "[[[119, 124]], [[2, 58], [125, 130], [193, 195]], [[8, 58]], [[8, 58]], [[133, 136]], [[63, 66]], [[138, 141]], [[93, 98]], [[8, 58]], [[8, 58]], [[2, 58]], [[2, 66]], [[2, 91]], [[92, 124]], [[99, 124]], [[119, 143]], [[145, 190]]]", "query_spans": "[[[193, 199]]]", "process": "Problem Analysis: From the eccentricity $\\frac{\\sqrt{3}}{2}$, we obtain $a=2b$. The ellipse equation can be rewritten as: $x^{2}+4y^{2}=a^{2}$. Substituting $l: y=\\sqrt{3}x$ into it gives $\\frac{|}{4}|=\\frac{\\sqrt{13}}{13}a$. By the symmetry of the ellipse, the perimeter of $\\triangle AFB$ is $2a+|AB|=2a+4x$, from which we get $a=2$. Therefore, the ellipse equation is" }, { "text": "Given that $P$ is a moving point on the parabola $y^{2}=4 x$, perpendiculars are drawn from $P$ to the $y$-axis and to the line $x-y+4=0$, with the feet of the perpendiculars being $A$ and $B$ respectively. \nThen, the minimum value of $PA$+$PB$ is?", "fact_expressions": "G: Parabola;H: Line;P: Point;A: Point;B: Point;L1:Line;L2:Line;Expression(G) = (y^2 = 4*x);Expression(H) = (x - y + 4 = 0);PointOnCurve(P, G);PointOnCurve(P,L1);PointOnCurve(P,L2);IsPerpendicular(L1,yAxis);IsPerpendicular(L2,H);FootPoint(L1,yAxis)=A;FootPoint(L2,H)=B", "query_expressions": "Min(LineSegmentOf(P, A)+LineSegmentOf(P, B))", "answer_expressions": "5*sqrt(2)/2-1", "fact_spans": "[[[6, 20]], [[37, 48]], [[2, 5], [26, 29]], [[57, 60]], [[61, 64]], [], [], [[6, 20]], [[37, 48]], [[2, 24]], [[25, 51]], [[25, 51]], [[25, 51]], [[25, 51]], [[25, 64]], [[25, 64]]]", "query_spans": "[[[67, 82]]]", "process": "" }, { "text": "The line passing through the left focus $F_{1}$ of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$ intersects the left branch of the hyperbola at points $A$ and $B$. The right focus is $F_{2}$. Then, the value of $|A F_{2}| + |B F_{2}|-|A B|$ is?", "fact_expressions": "G: Hyperbola;H: Line;A: Point;F2: Point;B: Point;F1: Point;Expression(G) = (x^2/4 - y^2/3 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(F1, H);Intersection(H, LeftPart(G)) = {A, B}", "query_expressions": "Abs(LineSegmentOf(A, F2)) + Abs(LineSegmentOf(B, F2)) - Abs(LineSegmentOf(A, B))", "answer_expressions": "8", "fact_spans": "[[[1, 39], [54, 57]], [[51, 53]], [[61, 64]], [[74, 81]], [[65, 68]], [[43, 50]], [[1, 39]], [[1, 50]], [[54, 81]], [[0, 53]], [[51, 70]]]", "query_spans": "[[[83, 116]]]", "process": "" }, { "text": "If a hyperbola $C$ centered at the origin has one focus at $F_{1}(0,-2)$ and one asymptote with equation $x-y=0$, then the equation of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;O: Origin;F1: Point;Center(C) = O;OneOf(Focus(C)) = F1;Coordinate(F1) = (0, -2);Expression(OneOf(Asymptote(C))) = (x - y = 0)", "query_expressions": "Expression(C)", "answer_expressions": "y^2/2 - x^2/2 = 1", "fact_spans": "[[[7, 13], [51, 57]], [[4, 6]], [[19, 32]], [[1, 13]], [[7, 32]], [[19, 32]], [[7, 49]]]", "query_spans": "[[[51, 62]]]", "process": "Test Analysis: Since the foci of the hyperbola lie on the y-axis, the asymptotes of the hyperbola are given by $ y = \\pm\\frac{a}{b}x $. Therefore, $ \\frac{a}{b} = 1 \\Rightarrow a = b $. Also, since $ c = 2 $ and $ a^{2} + b^{2} = 4 $, we have $ a^{2} = b^{2} = 2 $. Hence, the equation of the hyperbola is $ \\frac{y^{2}}{2} - \\frac{x^{2}}{2} = 1 $." }, { "text": "The asymptotes of the hyperbola $16 y^{2}-9 x^{2}=1$ are?", "fact_expressions": "G: Hyperbola;Expression(G) = (-9*x^2 + 16*y^2 = 1)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*3*x/4", "fact_spans": "[[[0, 23]], [[0, 23]]]", "query_spans": "[[[0, 31]]]", "process": "Rewrite the hyperbola equation into standard form: \\frac{y^{2}}{16}-\\frac{x^{2}}{1}=, we get a=\\frac{1}{4}, b=\\frac{1}{3}, hence its asymptote equations are: y=\\pm\\frac{a}{b}x=\\pm\\frac{3}{4}x," }, { "text": "Given that the asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{a}=1$ pass through the center of the circle $x^{2}+y^{2}-2 x-4 y-10=0$, then $a$=?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/a = 1);a: Number;H: Circle;Expression(H) = (-4*y - 2*x + x^2 + y^2 - 10 = 0);PointOnCurve(Center(H), Asymptote(G))", "query_expressions": "a", "answer_expressions": "4", "fact_spans": "[[[2, 30]], [[2, 30]], [[67, 70]], [[35, 62]], [[35, 62]], [[2, 65]]]", "query_spans": "[[[67, 72]]]", "process": "From the given information, (x-1)^{2}+(y-2)^{2}=15, the center is (1,2). Therefore, one asymptote equation of the hyperbola is y=2x, yielding \\underline{\\sqrt{a}}=2, so a=4" }, { "text": "Given that vertices $B$ and $C$ of $\\triangle ABC$ lie on the ellipse $\\frac{x^{2}}{3}+y^{2}=1$, vertex $A$ coincides with the focus $F_{1}$ of the ellipse, and the other focus $F_{2}$ of the ellipse lies on side $BC$, then what is the perimeter of $\\triangle ABC$?", "fact_expressions": "G: Ellipse;B: Point;C: Point;A: Point;Expression(G) = (x^2/3 + y^2 = 1);PointOnCurve(F2,LineSegmentOf(B,C));PointOnCurve(B, G);PointOnCurve(C, G);F1:Point;F2:Point;Focus(G)={F1,F2};A=F1", "query_expressions": "Perimeter(TriangleOf(A, B, C))", "answer_expressions": "4*sqrt(3)", "fact_spans": "[[[28, 55], [63, 65], [79, 81]], [[20, 23]], [[24, 27]], [[59, 62]], [[28, 55]], [[89, 103]], [[20, 56]], [[24, 56]], [[68, 75]], [[89, 96]], [[63, 96]], [[59, 77]]]", "query_spans": "[[[105, 125]]]", "process": "" }, { "text": "Given the parabola $ C $: $ y = \\frac{1}{4} x^{2} $ has focus $ F $. The line $ l $ passing through $ F $ with inclination angle $ \\frac{\\pi}{4} $ intersects the parabola $ C $ at points $ A $ and $ B $. Then, the distance from the midpoint of segment $ AB $ to the directrix of the parabola $ C $ is?", "fact_expressions": "l: Line;C: Parabola;B: Point;A: Point;F: Point;Expression(C) = (y = x^2/4);Focus(C) = F;PointOnCurve(F, l);Inclination(l)=pi/4;Intersection(l, C) = {A, B}", "query_expressions": "Distance(MidPoint(LineSegmentOf(A,B)),Directrix(C))", "answer_expressions": "4", "fact_spans": "[[[64, 69]], [[2, 31], [70, 76], [99, 105]], [[81, 84]], [[77, 80]], [[35, 38], [40, 43]], [[2, 31]], [[2, 38]], [[39, 69]], [[44, 69]], [[64, 86]]]", "query_spans": "[[[88, 113]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). By solving simultaneously the equations of the line and the parabola, we obtain Vieta's formulas, then use Vieta's formulas to calculate \\frac{y_{1}+1+y_{2}+1}{2} to get the solution. Solution: From the problem, we have x^{2}=4y, so the focus of the parabola is F(0,1). Thus, the equation of line l is y=x+1. Let A(x_{1},y_{1}), B(x_{2},y_{2}). Solving the system of equations: \n\\begin{cases}x^{2}=4y\\\\y=x+1\\end{cases}, \nwe get x^{2}-4x-4=0, so x_{1}+x_{2}=4, x_{1}x_{2}=-4. According to the problem, the distance from the midpoint of segment AB to the directrix of parabola C is |MN|, \n|MN|=\\frac{|AC|+|BD|}{2}=\\frac{y_{1}+1+y_{2}+1}{2}=1+\\frac{x_{1}+x_{2}}{2}=1+\\frac{-2x_{1}x_{2}}{2}=1+\\frac{16+8}{8}=4" }, { "text": "Given that hyperbola $C$ is centered at the origin, its right focus is at $F(3,0)$, and its eccentricity is $\\frac{3}{2}$, then what is the equation of hyperbola $C$? What are the equations of its asymptotes?", "fact_expressions": "C: Hyperbola;O: Origin;Center(C) = O;F: Point;Coordinate(F) = (3, 0);RightFocus(C) = F;Eccentricity(C) = 3/2", "query_expressions": "Expression(C);Expression(Asymptote(C))", "answer_expressions": "x^2/4 - y^2/5 = 1 \ny = pm*x*sqrt(5)/2", "fact_spans": "[[[2, 8], [16, 17], [49, 55]], [[10, 12]], [[2, 15]], [[21, 29]], [[21, 29]], [[16, 29]], [[16, 47]]]", "query_spans": "[[[49, 60]], [[49, 67]]]", "process": "Problem analysis: c=3, a=2, b^{2}=3^{2}-2^{2}=5, C: \\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1, the equations of the asymptotes are y=\\pm\\frac{\\sqrt{5}x}{2}" }, { "text": "Given the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{m-3}=1$, the distance from its right focus $F$ to one of its asymptotes is $3$. What is the value of the real number $m$?", "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (-y^2/(m - 3) + x^2/m = 1);RightFocus(G)=F;Distance(F,OneOf(Asymptote(G)))=3;F:Point", "query_expressions": "m", "answer_expressions": "12, -9", "fact_spans": "[[[2, 42], [50, 51]], [[64, 69]], [[2, 42]], [[2, 49]], [[46, 62]], [[46, 49]]]", "query_spans": "[[[64, 73]]]", "process": "Test analysis: From the given conditions, we have b=3, therefore \\begin{cases}m>3\\\\m-3=3^{2}\\end{cases} or \\begin{cases}m<0\\\\-m=3^{3}\\end{cases}, that is, the real number m is 12 or -9." }, { "text": "The focus of the parabola $y^{2}=2 p x(p>0)$ is $F$, points $A$ and $B$ lie on the parabola, and $\\angle A F B=120^{\\circ}$. Draw a perpendicular from the midpoint $M$ of chord $AB$ to the directrix $l$, with foot $M_{1}$. Then the maximum value of $\\frac{|M M_{1}|}{|A B|}$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(G) = F;A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);AngleOf(A, F, B) = ApplyUnit(120, degree);IsChordOf(LineSegmentOf(A, B), G);MidPoint(LineSegmentOf(A, B)) = M;M: Point;L: Line;PointOnCurve(M, L);IsPerpendicular(L, l);Directrix(G) = l;l: Line;FootPoint(L, l) = M1;M1: Point", "query_expressions": "Max(Abs(LineSegmentOf(M, M1))/Abs(LineSegmentOf(A, B)))", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[0, 21], [38, 41]], [[0, 21]], [[3, 21]], [[3, 21]], [[25, 28]], [[0, 28]], [[29, 33]], [[34, 37]], [[29, 42]], [[34, 42]], [[44, 70]], [[38, 78]], [[73, 83]], [[80, 83]], [], [[71, 92]], [[71, 92]], [[38, 89]], [[86, 89]], [[71, 103]], [[96, 103]]]", "query_spans": "[[[105, 136]]]", "process": "" }, { "text": "Given that point $P$ is a point on the parabola $C$: $y^{2}=2 p x$, the focus of $C$ is $F(1 , 0)$, and the coordinates of point $A$ are $(4 , 2)$. Then the minimum value of $|P F|+|P A|$ is?", "fact_expressions": "C: Parabola;p: Number;F: Point;P: Point;A: Point;Expression(C) = (y^2 = 2*(p*x));Coordinate(A) = (4, 2);Coordinate(F) = (1, 0);PointOnCurve(P, C);Focus(C) = F", "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))", "answer_expressions": "5", "fact_spans": "[[[7, 28], [32, 35]], [[15, 28]], [[39, 49]], [[2, 6]], [[50, 54]], [[7, 28]], [[50, 67]], [[39, 49]], [[2, 31]], [[32, 49]]]", "query_spans": "[[[69, 88]]]", "process": "From the given information, the equation of the parabola is $ y^{2} = 4x $, the directrix $ l: x = -1 $. Draw $ PM \\perp l $, with foot at $ M $, and draw $ AC \\perp l $, with foot at point $ C $ as shown in the figure. Then $ |PM| = |PF| $. Since $ |PF| + |PA| = |PM| + |PA| \\geqslant |AC| $, the required minimum value is $ |AC| = 4 - (-1) = 5 $. Thus, $ |PF| + |PA| $ attains the minimum value of 5." }, { "text": "The line $l$ passing through the point $(0,3b)$ is parallel to the asymptote of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) that has a positive slope. If the distance from any point on the right branch of the hyperbola $C$ to the line $l$ is always greater than $b$, then what is the maximum value of the eccentricity of the hyperbola $C$?", "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;P: Point;P0: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (0, 3*b);PointOnCurve(P, l);L1: Line;OneOf(Asymptote(C)) = L1;Slope(L1) > 0;IsParallel(l, L1);PointOnCurve(P0, RightPart(C));Distance(P0, l) > b", "query_expressions": "Max(Eccentricity(C))", "answer_expressions": "3", "fact_spans": "[[[12, 17], [107, 112]], [[18, 79], [94, 100], [123, 129]], [[118, 121]], [[26, 79]], [[1, 11]], [[105, 106]], [[26, 79]], [[26, 79]], [[18, 79]], [[1, 11]], [[0, 17]], [], [[18, 90]], [[18, 90]], [[12, 92]], [[94, 106]], [[105, 121]]]", "query_spans": "[[[123, 139]]]", "process": "From the given conditions, the equation of the asymptote with positive slope for the hyperbola is $ y = \\frac{b}{a}x $, or equivalently $ bx - ay = 0 $. Then the equation of line $ l $ is $ y = \\frac{b}{a}x + 3b $, that is, $ bx - ay + 3ab = 0 $. Since the distance from any point on the right branch of the hyperbola to line $ l $ is always greater than $ b $, the distance between the asymptote $ y = \\frac{b}{a}x $ and line $ l $ must be at least $ b $, i.e., \n$$\n\\frac{|3ab|}{\\sqrt{b^{2}+(-a)^{2}}} \\geqslant b\n$$\nCombining with $ c^{2} = a^{2} + b^{2} $, simplifying yields $ 9a^{2} \\geqslant c^{2} $. Therefore, \n$$\n1 < e = \\frac{c}{a} \\leqslant 3,\n$$\nthat is, the maximum value of the eccentricity of the hyperbola is." }, { "text": "The line $l$ with an inclination angle of $45^{\\circ}$ passes through the focus $F$ of the parabola $y^{2}=4x$ and intersects the parabola at points $A$ and $B$. What is the length of $AB$?", "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;F:Point;Expression(G) = (y^2 = 4*x);Inclination(l)=ApplyUnit(45,degree);Focus(G)=F;PointOnCurve(F,l);Intersection(l,G)={A,B}", "query_expressions": "Length(LineSegmentOf(A, B))", "answer_expressions": "8", "fact_spans": "[[[17, 22]], [[24, 38], [47, 50]], [[52, 55]], [[56, 59]], [[41, 44]], [[24, 38]], [[0, 22]], [[24, 44]], [[17, 44]], [[17, 61]]]", "query_spans": "[[[63, 72]]]", "process": "The equation of line l is y = x - 1. By solving simultaneously with the parabola equation, we obtain x^{2} - 6x + 1 = 0; thus, x_{A} + x_{B} = 6. Then, according to the definition of the parabola, the length of AB can be found. The focus F of the parabola y^{2} = 4x has coordinates (1, 0), so the equation of line l is y - 0 = \\tan 45^{\\circ}(x - 1), that is, y = x - 1. From \n\\begin{cases} y = x - 1 \\\\ y^{2} = 4x \\end{cases}, \nwe get x^{2} - 6x + 1 = 0, so x_{A} + x_{B} = 6. By the definition of the parabola, AB = x_{A} + x_{B} + p = 6 + 2 = 8. Therefore, the length of AB is 8." }, { "text": "The foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 ( a>b>0)$ are $F_{1}$, $F_{2}$, and the intersections of the two directrices with the $x$-axis are $M$, $N$. If $M N \\leq 2 F_{1} F_{2}$, then the range of values for the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;Focus(G) = {F1, F2};Z1: Line;Z2: Line;Directrix(G) = {Z1, Z2};M: Point;N: Point;Intersection(Z1, xAxis) = M;Intersection(Z2, xAxis) = N;LineSegmentOf(M, N) <= 2*LineSegmentOf(F1, F2)", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "[sqrt(2)/2, 1)", "fact_spans": "[[[0, 54], [125, 127]], [[0, 54]], [[2, 54]], [[2, 54]], [[2, 54]], [[2, 54]], [[58, 65]], [[66, 73]], [[0, 73]], [], [], [[0, 78]], [[89, 92]], [[93, 96]], [[0, 96]], [[0, 96]], [[98, 122]]]", "query_spans": "[[[125, 138]]]", "process": "" }, { "text": "Given that the equation of the parabola $ C $ is $ y^{2} = 4x $, a line passing through the point $ (4,0) $ intersects the parabola $ C $ at points $ A $ and $ B $. Tangents to the parabola are drawn at points $ A $ and $ B $ respectively. What is the trajectory equation of the intersection point of the two tangents?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);H: Point;Coordinate(H) = (4, 0);PointOnCurve(H, G) = True;G: Line;Intersection(G, C) = {A, B};A: Point;B: Point", "query_expressions": "LocusEquation(Intersection(TangentOfPoint(A, C),TangentOfPoint(B, C)))", "answer_expressions": "x = -4", "fact_spans": "[[[2, 8], [39, 45], [69, 72]], [[2, 24]], [[27, 35]], [[27, 35]], [[25, 38]], [[36, 38]], [[36, 56]], [[47, 50], [60, 64]], [[51, 54], [65, 68]]]", "query_spans": "[[[57, 91]]]", "process": "Let the equation of line $ l $ be $ x = my + 4 $. Let points $ A(x_1, y_1) $, $ B(x_2, y_2) $. Substitute the equation of line $ l $ into the equation of parabola $ C $ and simplify to obtain $ y^2 - 4my - 16 = 0 $. By Vieta's formulas, we get $ y_1 + y_2 = 4m $, $ y_1 y_2 = -16 $. The equation of the tangent line to parabola $ C $ at point $ A $ is $ y_1 y = 2(x + x_1) $, that is, $ y_1 y = 2x + \\frac{y_1^2}{2} $. The equation of the tangent line to parabola $ C $ at point $ B $ is $ y_2 y = 2(x + x_2) $, that is, $ y_2 y = 2x + \\frac{y_2^2}{2} $. Solving these two tangent equations simultaneously gives\n\\[\n\\begin{cases}\ny_1 y = 2x + \\frac{y_1^2}{2} \\\\\ny_2 y = 2x + \\frac{y_2^2}{2}\n\\end{cases}\n\\]\nSolving yields\n\\[\n\\begin{cases}\nx = \\frac{y_1 y_2}{4} = \\frac{-16}{4} = -4 \\\\\ny = \\frac{y_1 + y_2}{2} = m\n\\end{cases}\n\\]\nTherefore, the trajectory equation of point $ P $ is $ x = -4 $." }, { "text": "If the hyperbola $\\frac{x^{2}}{9 k^{2}}-\\frac{y^{2}}{4 k^{2}}=1$ and the circle $x^{2}+y^{2}=1$ have no common points, then the range of real values for $k$ is?", "fact_expressions": "G: Hyperbola;k: Real;H: Circle;Expression(G) = (-1/(4*k^2)*y^2 + x^2/((9*k^2)) = 1);Expression(H) = (x^2 + y^2 = 1);NumIntersection(G, H)=0", "query_expressions": "Range(k)", "answer_expressions": "{(1/3,+oo),(-oo,-1/3)}", "fact_spans": "[[[1, 51]], [[75, 80]], [[52, 68]], [[1, 51]], [[52, 68]], [[1, 73]]]", "query_spans": "[[[75, 87]]]", "process": "" }, { "text": "Let the ellipse $ C $: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 $ ($ a>b>0 $) have left and right foci $ F_{1} $, $ F_{2} $, respectively. A line passing through point $ F_{1} $ intersects the ellipse $ C $ at points $ M $, $ N $. If $ |M F_{2}| = |F_{1} F_{2}| $, and $ 7|M F_{1}| = 4|M N| $, then what is the eccentricity of ellipse $ C $?", "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;M: Point;F2: Point;F1: Point;N: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, G);Intersection(G, C) = {M, N};Abs(LineSegmentOf(M, F2)) = Abs(LineSegmentOf(F1, F2));7*Abs(LineSegmentOf(M, F1)) = 4*Abs(LineSegmentOf(M, N))", "query_expressions": "Eccentricity(C)", "answer_expressions": "5/7", "fact_spans": "[[[1, 58], [96, 101], [164, 169]], [[8, 58]], [[8, 58]], [[93, 95]], [[104, 107]], [[75, 82]], [[67, 74], [84, 92]], [[108, 111]], [[8, 58]], [[8, 58]], [[1, 58]], [[1, 82]], [[1, 82]], [[82, 95]], [[93, 113]], [[116, 141]], [[143, 162]]]", "query_spans": "[[[164, 175]]]", "process": "As shown in the figure, draw $F_{2}E\\bot MN$, with the foot of the perpendicular at $E$. Using the definition of the ellipse and $7|MF_{1}|=4|MN|$, compute $|F_{2}M|$, $|ME|$, $|NF_{2}|$, $|NE|$, and combine with the Pythagorean theorem to obtain a homogeneous equation in terms of $a$ and $c$, thereby obtaining the result. As shown in the figure, draw $F_{2}E\\bot MN$, with the foot of the perpendicular at $E$. Since $|F_{2}M|=|F_{1}F_{2}|=2c$, point $E$ is the midpoint of $F_{1}M$. Therefore, $|MF_{1}|=2a-2c$, $|ME|=a-c$. Also, since $7|MF_{1}|=4|MN|$, we have $|MN|=\\frac{7}{2}(a-c)$. Thus, $|NE|=\\frac{7}{2}(a-c)-(a-c)=\\frac{5}{2}(a-c)$, $|NF_{1}|=\\frac{3}{2}(a-c)$, $|NF_{2}|=2a-\\frac{3}{2}(a-c)=\\frac{1}{2}a+\\frac{3}{2}c$. By the Pythagorean theorem: $(\\frac{1}{2}a+\\frac{3}{2}c)^{2}-\\frac{25}{4}(a-c)^{2}=4c^{2}-(a-c)^{2}$. Simplifying yields: $7c^{2}-12ac+5a^{2}=0$, i.e., $7e^{2}-12e+5=0$. Solving gives: $e=1$ (discarded), $e=\\frac{5}{7}$." }, { "text": "$F_{1}$,$F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$. Point $P$ lies on the hyperbola and satisfies $|P F_{1}|-|P F_{2}|=32$. Then $\\angle F_{1} P F_{2}$=?", "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Focus(G)={F1,F2};PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)) = 32", "query_expressions": "AngleOf(F1, P, F2)", "answer_expressions": "ApplyUnit(90, degree)", "fact_spans": "[[[16, 55], [66, 69]], [[61, 65]], [[0, 7]], [[8, 15]], [[16, 55]], [[0, 60]], [[61, 70]], [[73, 97]]]", "query_spans": "[[[99, 123]]]", "process": "" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the right branch of the hyperbola, and $|P F_{1}|=4|P F_{2}|$. What is the maximum value of the eccentricity $e$ of the hyperbola?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F1)) = 4*Abs(LineSegmentOf(P, F2));e: Number;Eccentricity(G) = e", "query_expressions": "Max(e)", "answer_expressions": "5/3", "fact_spans": "[[[2, 58], [88, 91], [121, 124]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[67, 74]], [[75, 82]], [[2, 82]], [[2, 82]], [[83, 87]], [[83, 95]], [[97, 119]], [[128, 131]], [[121, 131]]]", "query_spans": "[[[128, 137]]]", "process": "Let $\\angle F_{1}PF_{2}=\\theta$, from $\\begin{cases}|PF_{1}|=4|PF_{2}|\\\\|PF_{1}|-|PF_{2}|=2a\\end{cases}$, we obtain $\\begin{cases}|PF_{1}|=\\frac{8a}{3}\\\\|PF_{2}|=\\frac{2a}{3}\\end{cases}$, by the cosine law we get $\\cos\\theta=\\frac{17a^{2}-9c^{2}}{8a^{2}}=\\frac{17}{8}-\\frac{9}{8}e^{2}$, since $\\theta\\in(0,\\pi]$, so $\\cos\\theta\\in[-1,1)$, that is $-1\\leqslant\\frac{17}{8}-\\frac{9}{8}e^{2}<1$. Also $e>1$, therefore $10)$ to the focus is $2 a$, then the ordinate of this point is?", "fact_expressions": "G: Parabola;a: Number;a>0;Expression(G) = (y^2 = 2*(a*x));H:Point;PointOnCurve(H,G);Distance(H,Focus(G))=2*a", "query_expressions": "YCoordinate(H)", "answer_expressions": "pm*sqrt(3)*a", "fact_spans": "[[[0, 20]], [[31, 36]], [[3, 20]], [[0, 20]], [[39, 40]], [[0, 24]], [[0, 36]]]", "query_spans": "[[[39, 46]]]", "process": "" }, { "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{3}+y^{2}=1$, respectively; points $A$ and $B$ lie on the ellipse. If $\\overrightarrow{F_{1} A}=5 \\overrightarrow{F_{2} B}$, then the coordinates of point $A$ are?", "fact_expressions": "G: Ellipse;F1: Point;A: Point;F2: Point;B: Point;Expression(G) = (x^2/3 + y^2 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(A,G);PointOnCurve(B,G);VectorOf(F1, A) = 5*VectorOf(F2, B)", "query_expressions": "Coordinate(A)", "answer_expressions": "(0,pm*1)", "fact_spans": "[[[19, 46], [62, 64]], [[1, 8]], [[53, 57], [122, 126]], [[9, 16]], [[58, 61]], [[19, 46]], [[1, 52]], [[1, 52]], [[53, 65]], [[53, 65]], [[67, 120]]]", "query_spans": "[[[122, 131]]]", "process": "" }, { "text": "Through the focus of the parabola $y^{2}=2 px(p>0)$, draw a line $l$ intersecting the parabola at points $A(x_{1} , y_{1})$ and $B(x_{2} , y_{2})$, then the value of $\\frac{y_{1} y_{2}}{x_{1} x_{2}}$ is?", "fact_expressions": "l: Line;p: Number;A: Point;B: Point;p>0;x1: Number;y1: Number;x2: Number;y2: Number;Expression(G) = (y^2 = 2*(p*x));Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);PointOnCurve(Focus(G), l);Intersection(l, G) = {A, B};G: Parabola", "query_expressions": "(y1*y2)/(x1*x2)", "answer_expressions": "-4", "fact_spans": "[[[28, 33]], [[5, 22]], [[38, 56]], [[59, 77]], [[5, 22]], [[38, 56]], [[38, 56]], [[59, 77]], [[59, 77]], [[2, 22]], [[38, 56]], [[59, 77]], [[0, 33]], [[28, 77]], [[2, 22], [34, 37]]]", "query_spans": "[[[79, 116]]]", "process": "" }, { "text": "From a point $P$ on the parabola $y^{2}=4x$, draw a perpendicular to the directrix of the parabola, with the foot of the perpendicular at $M$, and $|PM|=5$. Let $F$ be the focus of the parabola. Then the area of $\\triangle MPF$ is?", "fact_expressions": "G: Parabola;M: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(P, G);L: Line;PointOnCurve(P, L);IsPerpendicular(L, Directrix(G));FootPoint(L, Directrix(G)) = M;Abs(LineSegmentOf(P, M)) = 5;Focus(G) = F", "query_expressions": "Area(TriangleOf(M, P, F))", "answer_expressions": "10", "fact_spans": "[[[1, 15], [22, 25], [50, 53]], [[34, 37]], [[18, 21]], [[57, 60]], [[1, 15]], [[1, 21]], [], [[0, 30]], [[0, 30]], [[0, 37]], [[39, 48]], [[50, 60]]]", "query_spans": "[[[62, 84]]]", "process": "" }, { "text": "Given a moving point $P(x, y)$ (where $y \\geq 0$) such that its distance to the $x$-axis is less by 1 than its distance to the point $F(0,1)$, then the trajectory equation of the moving point $P$ is?", "fact_expressions": "P: Point;x1: Number;y1: Number;Coordinate(P) = (x1, y1);y1>=0;F: Point;Coordinate(F) = (0, 1);Distance(P, xAxis) = Distance(P, F)-1", "query_expressions": "LocusEquation(P)", "answer_expressions": "x^2=4*y", "fact_spans": "[[[4, 13], [58, 61], [36, 37]], [[4, 13]], [[4, 13]], [[4, 13]], [[16, 26]], [[38, 47]], [[38, 47]], [[4, 54]]]", "query_spans": "[[[58, 68]]]", "process": "Since the distance from the moving point P(x,y) to the x-axis is 1 less than its distance to the point F(0,1), the distance from the moving point P(x,y) to x = -1 is equal to its distance to the point F(0,1). According to the definition, the trajectory of the moving point P is a parabola, and F(0,1) is the focus, so \\frac{p}{2}=1. Therefore, the trajectory equation of the moving point P is x^{2}=4y. [Note: This problem examines a simple application of the definition of a parabola; the key is to transform the distances appropriately. It is a medium-difficulty problem." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, the left and right foci are $F_{1}$, $F_{2}$ respectively, and $O$ is the coordinate origin. Point $M$ is a point on the left branch of the hyperbola. If $|O M|=\\sqrt{a^{2}+b^{2}}$, $4|M F_{1}|=3|M F_{2}|$, then what is the eccentricity of the hyperbola?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;O: Origin;M: Point;PointOnCurve(M, LeftPart(G));Abs(LineSegmentOf(O, M)) = sqrt(a^2 + b^2);4*Abs(LineSegmentOf(M, F1)) = 3*Abs(LineSegmentOf(M, F2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "5", "fact_spans": "[[[2, 48], [87, 90], [152, 155]], [[2, 48]], [[5, 48]], [[5, 48]], [[57, 64]], [[65, 72]], [[2, 72]], [[2, 72]], [[73, 76]], [[82, 86]], [[82, 96]], [[98, 124]], [[127, 150]]]", "query_spans": "[[[152, 161]]]", "process": "Let the focal distance of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ be $2c$, then $a^{2}+b^{2}=c^{2}$. Since $|OM|=\\sqrt{a^{2}+b^{2}}=c=|OF_{1}|=|OF_{2}|$, it follows that $MF_{1}\\bot MF_{2}$. Because $4|MF_{1}|=3|MF_{2}|$, assume without loss of generality that $|MF_{1}|=3k$, $|MF_{2}|=4k$. By the definition of the hyperbola, we have $|MF_{2}|-|MF_{1}|=k=2a$. Therefore, $|MF_{1}|=6a$, $|MF_{2}|=8a$. By the Pythagorean theorem, $|MF_{1}|^{2}+|MF_{2}|^{2}=(6a)^{2}+(8a)^{2}=100a^{2}=|F_{1}F_{2}|^{2}=4c^{2}$. Hence, $\\frac{c^{2}}{a^{2}}=25$, so the eccentricity of the hyperbola is $e=5$." }, { "text": "Let the hyperbola $C$: $x^{2}-4 y^{2}+64=0$ have foci $F_{1}$, $F_{2}$, and let point $P$ be a point on $C$ such that $|P F_{1}|=6$, then what is $|P F_{2}|$?", "fact_expressions": "C: Hyperbola;P: Point;F1: Point;F2: Point;Expression(C) = (x^2 - 4*y^2 + 64 = 0);Focus(C) = {F1, F2};PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 6", "query_expressions": "Abs(LineSegmentOf(P, F2))", "answer_expressions": "14", "fact_spans": "[[[1, 29], [54, 57]], [[49, 53]], [[33, 40]], [[41, 48]], [[1, 29]], [[1, 48]], [[49, 60]], [[61, 74]]]", "query_spans": "[[[76, 89]]]", "process": "Convert $x^{2}-4y^{2}+64=0$ into $\\frac{y^{2}}{16}-\\frac{x^{2}}{64}=1$, so $a=4$, $2a=8$. By the definition of a hyperbola, we obtain: $||PF_{2}|-|PF_{1}||=8$, that is, $||PF_{2}|-6|=8$. Therefore, $|PF_{2}|=14$ or $|PF_{2}|=-2$ (discarded)." }, { "text": "Given foci $F_{1}$ and $F_{2}$, construct an ellipse such that the sum of distances from a point $P$ on the ellipse to $F_{1}$ and $F_{2}$ is $10$. Find the maximum value of $|P F_{1}| \\cdot|P F_{2}|$?", "fact_expressions": "F1: Point;F2: Point;Focus(G) = {F1, F2};G: Ellipse;P: Point;PointOnCurve(P, G);Distance(P, F1) + Distance(P, F2) = 10", "query_expressions": "Max(Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)))", "answer_expressions": "25", "fact_spans": "[[[1, 8], [32, 39]], [[9, 16], [40, 47]], [[0, 22]], [[20, 22], [23, 25]], [[28, 31]], [[23, 31]], [[28, 57]]]", "query_spans": "[[[59, 90]]]", "process": "Given the problem, |PF_{1}| + |PF_{2}| = 10. By the AM-GM inequality, we have |PF_{1}| \\cdot |PF_{2}| \\leqslant \\left(\\frac{|PF_{1}| + |PF_{2}|}{2}\\right)^{2} = 25. Moreover, when |PF_{1}| = |PF_{2}| = 5, |PF_{1}| \\cdot |PF_{2}| = 25. Therefore, the maximum value of |PF_{1}| \\cdot |PF_{2}| is 25." }, { "text": "The coordinates of the focus of the parabola $4 x^{2}=y$?", "fact_expressions": "G: Parabola;Expression(G) = (4*x^2 = y)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0,1/8)", "fact_spans": "[[[0, 14]], [[0, 14]]]", "query_spans": "[[[0, 20]]]", "process": "\\because the parabola 4x^{2}=y can be rewritten as the standard equation x^{2}=\\frac{1}{4}y, \\therefore the parabola x^{2}=\\frac{1}{4}y has its focus on the y-axis, and 2p=\\frac{1}{4}, \\therefore \\frac{p}{2}=\\frac{1}{8}, \\therefore the focus coordinates of the parabola \\frac{1}{4}x^{2}=y are (0,\\frac{1}{8})," }, { "text": "The asymptote equations of the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$ are?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/4 = 1)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*2*x", "fact_spans": "[[[0, 28]], [[0, 28]]]", "query_spans": "[[[0, 36]]]", "process": "Test Analysis: The asymptote equation is x^{2}-\\frac{y^{2}}{4}=0; after simplification, we obtain the asymptotes of the hyperbola. Solution: \\because the standard equation of the hyperbola is x^{2}-\\frac{y^{2}}{4}=1, its asymptote equation is x^{2}-\\frac{y^{2}}{4}=0, which simplifies to y=\\pm2x." }, { "text": "The standard equation of an ellipse whose right focus is at $(2,0)$ and that passes through the point $(-2,-\\sqrt{2})$ is?", "fact_expressions": "Coordinate(RightFocus(G)) = (2, 0);G: Ellipse;H: Point;Coordinate(H) = (-2, -sqrt(2));PointOnCurve(H, G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/8+y^2/4=1", "fact_spans": "[[[0, 37]], [[35, 37]], [[17, 34]], [[17, 34]], [[15, 37]]]", "query_spans": "[[[35, 44]]]", "process": "Problem Analysis: Let the equation of the ellipse be \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1. Therefore, a^{2}-b^{2}=4, \\frac{4}{a^{2}}+\\frac{2}{b^{2}}=1. Solving the equations gives a^{2}=8, b^{2}=4, so the equation is \\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1." }, { "text": "If the foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ lie on the $x$-axis, and tangents are drawn from the point $(1 , \\frac{1}{2})$ to the circle $x^{2}+y^{2}=1$, with points of tangency $A$ and $B$, respectively, and the line $AB$ passes exactly through the right focus and the upper vertex of the ellipse, then what is the equation of the ellipse?", "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Circle;B: Point;A: Point;J: Point;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + y^2 = 1);Coordinate(J) = (1, 1/2);PointOnCurve(Focus(G), xAxis);L1:Line;L2:Line;TangentOfPoint(J,H)={L1,L2};TangentPoint(L1,H)=A;TangentPoint(L2,H)=B;PointOnCurve(RightFocus(G),LineOf(A,B));PointOnCurve(UpperVertex(G),LineOf(A,B))", "query_expressions": "Expression(G)", "answer_expressions": "x^2/5+y^2/4=1", "fact_spans": "[[[1, 46], [122, 124], [134, 136]], [[3, 46]], [[3, 46]], [[78, 94]], [[107, 110]], [[103, 106]], [[57, 77]], [[1, 46]], [[78, 94]], [[57, 77]], [[1, 55]], [], [], [[56, 97]], [[56, 110]], [[56, 110]], [[111, 128]], [[111, 132]]]", "query_spans": "[[[134, 140]]]", "process": "\\because point (1,\\frac{1}{2}) is outside the circle, one tangent line passing through point (1,\\frac{1}{2}) and touching the circle is x=1, and line AB passes exactly through the right focus and the upper vertex of the ellipse, \\therefore the right focus of the ellipse is (1,0), i.e., c=1. Let point P(1,\\frac{1}{2}), connect OP, then OP\\bot AB, \\because k_{OP}=\\frac{1}{2}, \\therefore k_{AB}=-2. Also, line AB passes through point (1,0), \\therefore the equation of line AB is 2x+y-2=0, \\because point (0,b) lies on line AB, \\therefore b=2, and since c=1, \\therefore a^{2}=5, hence the equation of the ellipse is \\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1" }, { "text": "The line passing through the focus $F$ of the parabola $C$: $y^{2}=4x$ intersects the parabola $C$ at points $A$ and $B$. What is the minimum value of $|AF| + 4|BF|$?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;G: Line;PointOnCurve(F,G) = True;Intersection(G, C) = {A, B};A: Point;B: Point", "query_expressions": "Min(Abs(LineSegmentOf(A, F)) + 4*Abs(LineSegmentOf(B, F)))", "answer_expressions": "9", "fact_spans": "[[[1, 20], [30, 36]], [[1, 20]], [[23, 26]], [[1, 26]], [[27, 29]], [[0, 26]], [[27, 47]], [[38, 41]], [[42, 45]]]", "query_spans": "[[[49, 69]]]", "process": "" }, { "text": "The midpoint of chord $AB$ of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ is $P(1, \\frac{1}{2})$. Then the equation of the line on which chord $AB$ lies is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);P: Point;Coordinate(P) = (1, 1/2);MidPoint(LineSegmentOf(A, B)) = P", "query_expressions": "Expression(OverlappingLine(LineSegmentOf(A, B)))", "answer_expressions": "x + 2*y - 2 = 0", "fact_spans": "[[[0, 27]], [[0, 27]], [[29, 34]], [[29, 34]], [[0, 34]], [[38, 57]], [[38, 57]], [[29, 57]]]", "query_spans": "[[[60, 74]]]", "process": "Problem Analysis: Let A(x_{1},y_{1}) and B(x_{2},y_{2}), substitute into the ellipse and subtract to obtain \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{1}{2}\\therefore k=-\\frac{1}{2}, so the line is x+2y-2=0" }, { "text": "Through the right focus $F$ of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a perpendicular line to an asymptote, with the foot of the perpendicular at point $A$, intersecting the $y$-axis at point $B$. If $\\overrightarrow{F A}=2 \\overrightarrow{A B}$, then the eccentricity of $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;F: Point;A: Point;B: Point;L: Line;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;PointOnCurve(F, L);IsPerpendicular(L, Asymptote(C));FootPoint(L, Asymptote(C)) = A;Intersection(L, yAxis) = B;VectorOf(F, A) = 2*VectorOf(A, B)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3)", "fact_spans": "[[[1, 62], [144, 147]], [[9, 62]], [[9, 62]], [[66, 69]], [[80, 84]], [[91, 95]], [], [[9, 62]], [[9, 62]], [[1, 62]], [[1, 69]], [[0, 76]], [[0, 76]], [[0, 84]], [[0, 95]], [[97, 142]]]", "query_spans": "[[[144, 153]]]", "process": "" }, { "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$, and the directrix $l$ intersects the $x$-axis at point $A$. $P$ is a point on the parabola $C$ such that $P F \\perp x$-axis. If the chord intercepted on the line $A P$ by the circle with diameter $A F$ has length $2$, then the value of the real number $p$ is?", "fact_expressions": "C: Parabola;p: Real;G: Circle;l:Line;A: Point;P: Point;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;Directrix(C)=l;Intersection(l,xAxis)=A;PointOnCurve(P,C);IsPerpendicular(LineSegmentOf(P,F),xAxis);IsDiameter(LineSegmentOf(A,F),G);Length(InterceptChord(LineOf(A,P),G))=2", "query_expressions": "p", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[2, 28], [58, 64]], [[116, 121]], [[96, 97]], [[38, 41]], [[50, 53]], [[54, 57]], [[32, 35]], [[10, 28]], [[2, 28]], [[2, 35]], [[2, 41]], [[38, 53]], [[54, 67]], [[69, 83]], [[86, 97]], [[96, 114]]]", "query_spans": "[[[116, 125]]]", "process": "According to the given conditions, $ F\\left(\\frac{p}{2},0\\right) $, $ A\\left(-\\frac{p}{2},0\\right) $. Assume point $ P $ lies in the first quadrant, then $ P\\left(\\frac{p}{2},p\\right) $, $ k_{AP} = \\frac{p}{p} = 1 $, so the equation of line $ AP $ is $ x - y + \\frac{p}{2} = 0 $. The circle with diameter $ AF $ has center $ O(0,0) $ and radius $ R = \\frac{p}{2} $. Then the distance from $ O $ to line $ AP $ is $ d = \\frac{\\frac{p}{2}}{\\sqrt{2}} = \\frac{\\sqrt{2}p}{4} $. Thus, the chord length intercepted by circle $ O $ on line $ AP $ is $ l_1 = 2\\sqrt{R^{2} - d^{2}} = 2\\sqrt{\\frac{p^{2}}{4} - \\left(\\frac{\\sqrt{2}p}{4}\\right)^{2}} $, solving this yields $ p = \\sqrt{2} $. This problem examines properties of parabolas, properties of circles, and calculations of chord lengths in circles, and belongs to a medium-difficulty level problem." }, { "text": "The focal distance of the hyperbola $\\frac{x^{2}}{7}-\\frac{y^{2}}{9}=-1$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/7 - y^2/9 = -1)", "query_expressions": "FocalLength(G)", "answer_expressions": "8", "fact_spans": "[[[0, 39]], [[0, 39]]]", "query_spans": "[[[0, 44]]]", "process": "" }, { "text": "If the line $l$: $y = kx + 1$ and the curve $C$: $x^2 + y|y| = 1$ have two common points, then the range of real values for $k$ is?", "fact_expressions": "l: Line;C: Curve;k: Real;Expression(C) = (x^2 + y*Abs(y) = 1);Expression(l) = (y = k*x + 1);NumIntersection(l, C) = 2", "query_expressions": "Range(k)", "answer_expressions": "[-1, 0) + (0, 1] + {-sqrt(2), sqrt(2)}", "fact_spans": "[[[1, 17]], [[18, 39]], [[47, 52]], [[18, 39]], [[1, 17]], [[1, 45]]]", "query_spans": "[[[47, 59]]]", "process": "From the curve $ C: x^{2} + y|y| = 1 $, we obtain: when $ y \\geqslant 0 $, $ x^{2} + y^{2} = 1 $; when $ y < 0 $, $ x^{2} - y^{2} = 1 $. The line $ l: y = kx + 1 $ always passes through the point $ (0, 1) $. Therefore, the graph of the line and the curve is such that when the line $ y = kx + 1 $ is tangent to $ x^{2} + y^{2} = 1 $ ($ y \\geqslant 0 $), at this moment $ k = 0 $, and the line $ l $ has only one common point with the curve $ C $. When $ k = \\pm 1 $ in the line $ y = kx + 1 $, the line and the curve have two intersection points. When the line $ y = kx + 1 $ ($ k \\neq \\pm 1 $) is tangent to $ x^{2} - y^{2} = 1 $ ($ y < 0 $), solving the system of equations:\n\\[\n\\begin{cases}\ny = kx + 1 \\quad (k \\neq \\pm 1) \\\\\nx^{2} - y^{2} = 1 \\quad (y < 0)\n\\end{cases}\n\\]\nwe get $ x^{2} - (kx + 1)^{2} = 1 $, simplifying to $ (1 - k^{2})x^{2} + 2kx - 2 = 0 $. The discriminant is $ \\Delta = 4k^{2} + 8(1 - k^{2}) = 8 - 4k^{2} = 0 $, solving gives $ k = \\pm \\sqrt{2} $, at this moment the line $ l $ and the curve $ C $ have exactly two common points. For the line $ y = kx + 1 $ and the curve $ x^{2} + y|y| = 1 $ to have exactly two common points, we have $ k = \\pm \\sqrt{2} $, or $ -1 \\leqslant k < 0 $ or $ 0 < k \\leqslant 1 $." }, { "text": "Given the parabola $C$: $y^{2}=8x$ has focus $F$, directrix $l$, and point $P$ is a point on the parabola $C$ in the first quadrant. A perpendicular is drawn from point $P$ to $l$, with foot of perpendicular at $M$. The slope of line $PF$ is $\\sqrt{3}$. Then $|MF|=$?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 8*x);F: Point;Focus(C) = F;l: Line;Directrix(C) = l;P: Point;PointOnCurve(P, C) = True;Quadrant(P) = 1;PointOnCurve(P, G) = True;IsPerpendicular(G, l) = True;G: Line;FootPoint(G, l) = M;M: Point;Slope(LineOf(P, F)) = sqrt(3)", "query_expressions": "Abs(LineSegmentOf(M, F))", "answer_expressions": "8", "fact_spans": "[[[2, 21], [41, 47]], [[2, 21]], [[25, 28]], [[2, 28]], [[32, 35], [63, 66]], [[2, 35]], [[36, 40], [58, 62]], [[36, 56]], [[36, 56]], [[57, 69]], [[57, 69]], [], [[57, 76]], [[73, 76]], [[77, 98]]]", "query_spans": "[[[100, 109]]]", "process": "Since the slope of line PF is \\sqrt{3}, the inclination angle is 60^{\\circ}. By the definition of a parabola, it follows that \\triangle FPM is an equilateral triangle. Based on geometric relationships, the answer can be obtained. Because the slope of line PF is \\sqrt{3}, \\angle P F x = \\angle M P F = 60^{\\circ}. By the definition of the parabola: |PF| = |PM|, so \\triangle FPM is an equilateral triangle. Let l intersect the x-axis at point A, then |MF| = 2|AF| = 8." }, { "text": "If a point $P$ on the parabola $x^{2}=8 y$ is at a distance of $5$ from the focus, then what is the ordinate (y-coordinate) of point $P$?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 8*y);P: Point;PointOnCurve(P, G);Distance(P, Focus(G)) = 5", "query_expressions": "YCoordinate(P)", "answer_expressions": "3", "fact_spans": "[[[1, 15]], [[1, 15]], [[18, 21], [34, 38]], [[1, 21]], [[1, 31]]]", "query_spans": "[[[34, 44]]]", "process": "The focus of the parabola \\( x^{2} = 8y \\) is \\( F(0,2) \\), and the equation of the directrix is: \\( y = -2 \\). Let \\( P(x,y) \\), \\( y > 0 \\). Since a point \\( P \\) on the parabola is at a distance of 5 from the focus, by the definition of a parabola we have: \\( |PF| = |y - (-2)| = 5 \\), solving gives \\( y = 3 \\)." }, { "text": "The directrix of the parabola $y^{2}=2 p x(p>0)$ is given by the equation $x=-1$. Then, what is the value of $p$?", "fact_expressions": "G: Parabola;p: Number;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(Directrix(G)) = (x = -1)", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[2, 23]], [[37, 40]], [[5, 23]], [[2, 23]], [[2, 35]]]", "query_spans": "[[[37, 43]]]", "process": "Since the equation of the directrix of the parabola \\( y^{2} = 2px \\) (\\( p > 0 \\)) is \\( x = -\\frac{p}{2} \\), we have \\( -\\frac{p}{2} = -1 \\). Solving this gives \\( p = 2 \\)." }, { "text": "The coordinates of the focus of the parabola $y^{2}=12 x$ are?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 12*x)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(3, 0)", "fact_spans": "[[[0, 15]], [[0, 15]]]", "query_spans": "[[[0, 22]]]", "process": "Problem Analysis: The focus coordinates of $y^{2}=2px$ ($p>0$) are $(\\frac{p}{2},0)$, so $(3,0)$." }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+ \\frac{y^{2}}{b^{2}}=1(a>b>0)$, the left focus is $F_{1}$, $O$ is the coordinate origin, point $P$ lies on the ellipse, and point $Q$ lies on the right directrix of the ellipse. If $\\overrightarrow{PQ}=2 \\overrightarrow{F_{1}O}$, $\\overrightarrow{F_{1}Q}= \\lambda(\\frac{\\overrightarrow{F_{1}P}}{|\\overrightarrow{F_{1}P}|}+ \\frac{\\overrightarrow{F_{1} O}}{| \\overrightarrow{F_{1}O}|})(\\lambda>0)$, then what is the eccentricity of the ellipse?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;F1: Point;LeftFocus(G) = F1;O: Origin;P: Point;PointOnCurve(P, G);Q: Point;PointOnCurve(Q, RightDirectrix(G));VectorOf(P, Q) = 2*VectorOf(F1, O);lambda: Number;lambda>0;VectorOf(F1, Q) = lambda*(VectorOf(F1, O)/Abs(VectorOf(F1, O)) + VectorOf(F1, P)/Abs(VectorOf(F1, P)))", "query_expressions": "Eccentricity(G)", "answer_expressions": "(sqrt(5)-1)/2", "fact_spans": "[[[2, 55], [82, 84], [92, 94], [318, 320]], [[2, 55]], [[4, 55]], [[4, 55]], [[4, 55]], [[4, 55]], [[59, 66]], [[2, 66]], [[68, 71]], [[77, 81]], [[77, 85]], [[87, 91]], [[87, 99]], [[102, 149]], [[151, 316]], [[151, 316]], [[151, 316]]]", "query_spans": "[[[318, 326]]]", "process": "" }, { "text": "A line passing through the focus $F$ of the parabola $y^{2}=2 p x(p>0)$ intersects the parabola at points $P$ and $Q$. Perpendiculars $P H_{1}$ and $Q H_{2}$ are drawn from $P$ and $Q$ to the directrix, with feet $H_{1}$ and $H_{2}$ respectively. The midpoint of $H_{1} H_{2}$ is $M$. Let $|PF|=a$, $| QF |=b$, then $|MF|$=?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H) = True;Intersection(H, G) = {P, Q};P: Point;Q: Point;IsPerpendicular(LineSegmentOf(P, H1), Directrix(G)) ;IsPerpendicular(LineSegmentOf(P, H2), Directrix(G)) ;FootPoint(LineSegmentOf(P, H1), Directrix(G)) = H1;FootPoint(LineSegmentOf(P, H2), Directrix(G)) = H2;PointOnCurve(P, LineSegmentOf(P, H1));PointOnCurve(Q, LineSegmentOf(P, H2));H1: Point;H2: Point;MidPoint(LineSegmentOf(H1, H2)) = M;M: Point;Abs(LineSegmentOf(P, F)) = a;a: Number;Abs(LineSegmentOf(Q, F)) = b;b: Number", "query_expressions": "Abs(LineSegmentOf(M, F))", "answer_expressions": "Preserve all mathematical expressions, LaTeX formulas, symbols, and equations EXACTLY. \nDo not add explanations, do not rewrite formulas, do not omit any content. \n\nText:", "fact_spans": "[[[1, 22], [31, 34], [55, 56]], [[1, 22]], [[4, 22]], [[4, 22]], [[24, 27]], [[1, 27]], [[28, 30]], [[0, 30]], [[28, 43]], [[36, 39], [45, 48]], [[40, 43], [49, 52]], [[55, 80]], [[55, 80]], [[55, 102]], [[55, 102]], [[44, 80]], [[44, 80]], [[85, 92]], [[95, 102]], [[105, 125]], [[122, 125]], [[127, 135]], [[127, 135]], [[136, 146]], [[136, 146]]]", "query_spans": "[[[148, 156]]]", "process": "" }, { "text": "If the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{6}=1(a>0)$ has eccentricity $2$, then what is the focal length of the hyperbola $C$?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/6 + x^2/a^2 = 1);a: Number;a>0;Eccentricity(C) = 2", "query_expressions": "FocalLength(C)", "answer_expressions": "4*sqrt(2)", "fact_spans": "[[[1, 53], [63, 69]], [[1, 53]], [[9, 53]], [[9, 53]], [[1, 61]]]", "query_spans": "[[[63, 74]]]", "process": "According to the problem, e = \\frac{c}{a} = \\frac{\\sqrt{a^{2+6}}}{a} = 2, solving gives a^{2} = 2, so c = 2\\sqrt{2}, that is, the focal length of hyperbola C is 2c = 4\\sqrt{2}" }, { "text": "The chord of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ is bisected by the point $(\\frac{1}{2}, \\frac{1}{2})$, then the equation of the line containing this chord is?", "fact_expressions": "G: Ellipse;H: LineSegment;Expression(G) = (x^2/2 + y^2 = 1);Coordinate(MidPoint(H)) = (1/2, 1/2);IsChordOf(H,G) = True", "query_expressions": "Expression(OverlappingLine(H))", "answer_expressions": "2*x+4*y-3=0", "fact_spans": "[[[0, 27]], [], [[0, 27]], [[0, 61]], [[0, 29]]]", "query_spans": "[[[0, 75]]]", "process": "Let the endpoints of the chord be A(x_{1},y_{1}), B(x_{2},y_{2}) \\Rightarrow x_{1}+x_{2}=1, y_{1}+y_{2}=1. Also, since A and B lie on the ellipse \\Rightarrow \\frac{x_{1}^{2}}{2}+y_{1}^{2}=1, \\frac{x_{2}^{2}}{2}+y_{2}^{2}=1 \\Rightarrow \\frac{1}{2}(x_{1}+x_{2})(x_{1}-x_{2})+(y_{1}+y_{2})(y_{1}-y_{2})=0 \\Rightarrow \\frac{1}{2}+\\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{(x_{1}+x_{2})(x_{1}-x_{2})}=0 \\Rightarrow \\frac{(y_{1}-y_{2})}{(x_{1}-x_{2})}=-\\frac{1}{2} \\Rightarrow that is, the slope of line AB is -\\frac{1}{2} \\Rightarrow the equation of line AB is y-\\frac{1}{2}=-\\frac{1}{2}(x-\\frac{1}{2}) \\Rightarrow 2x+4y-3=0." }, { "text": "Let the line $2x + y - 3 = 0$ intersect the parabola $\\Gamma$: $y^2 = 8x$ at points $A$ and $B$. A circle passing through $A$ and $B$ intersects the parabola $\\Gamma$ at two other points $C$ and $D$. Then the slope $k$ of line $CD$ is?", "fact_expressions": "G: Line;Expression(G) = (2*x + y - 3 = 0);Gamma: Parabola;Expression(Gamma) = (y^2 = 8*x);A: Point;B: Point;Intersection(G, Gamma) = {A, B};H: Circle;PointOnCurve(A, H);PointOnCurve(B, H);C: Point;D: Point;Intersection(H, Gamma) = {C, D};k: Number;Slope(LineOf(C, D)) = k", "query_expressions": "k", "answer_expressions": "2", "fact_spans": "[[[1, 14]], [[1, 14]], [[15, 39], [62, 73]], [[15, 39]], [[41, 44], [52, 55]], [[45, 48], [56, 59]], [[1, 50]], [[60, 61]], [[51, 61]], [[51, 61]], [[79, 82]], [[83, 86]], [[60, 86]], [[98, 101]], [[88, 101]]]", "query_spans": "[[[98, 103]]]", "process": "Analysis: According to the symmetry of the circle and the parabola, line AB and line CD are symmetric about the x-axis, so their slopes are opposite in sign; thus, the result is obtained. \nDetailed explanation: Because of the symmetry of the circle and the parabola, line AB and line CD are symmetric about the x-axis, so the sum of the slopes of line AB and line CD is zero (i.e., their slopes are opposites). Since the slope of line AB is -2, the slope of line CD is 2." }, { "text": "Let $O$, $F$ be the vertex and focus of the parabola $y^{2}=2 x$, respectively, and let $M$ be a moving point on the parabline. Then the maximum value of $\\frac{|M O|}{|M F|}$ is?", "fact_expressions": "O: Origin;F: Point;Vertex(G) = O;Focus(G) = F;G: Parabola;Expression(G) = (y^2 = 2*x);M: Point;PointOnCurve(M, G)", "query_expressions": "Max(Abs(LineSegmentOf(M, O))/Abs(LineSegmentOf(M, F)))", "answer_expressions": "2*sqrt(3)/3", "fact_spans": "[[[1, 4]], [[5, 8]], [[1, 31]], [[1, 31]], [[11, 25], [36, 39]], [[11, 25]], [[32, 35]], [[32, 43]]]", "query_spans": "[[[45, 72]]]", "process": "Problem Analysis: Let the coordinates of point M be M(x, y). By the definition of a parabola, |MF| = x + \\frac{1}{2}. Then \\frac{|MO|}{|MF|} = \\frac{\\sqrt{x^{2}+y^{2}}}{x+\\frac{1}{2}} = \\frac{x^{2}+2x}{x+\\frac{1}{2}} = |\\frac{x^{2}+2x}{2}| = |1 + \\frac{x - \\frac{1}{4}}{x^{2} + x + \\frac{1}{4}}|. Let t = x - \\frac{1}{4}, then t > -\\frac{1}{16t} + \\frac{3}{2} \\leqslant \\sqrt{1 + \\frac{1}{3}} = \\frac{2\\sqrt{3}}{3}, with equality holding if and only if t = \\frac{3}{4}. If -\\frac{1}{4} < t < 0, then t + \\frac{\\frac{9}{16}}{t} + \\frac{3}{2} is monotonically decreasing, hence y < -1, \\frac{1}{y} \\in (-1, 0). Therefore, the maximum value of \\frac{|MO|}{|MF|} is \\frac{2\\sqrt{3}}{3}." }, { "text": "If the focus of the parabola $y^{2}=2 p x(p>0)$ lies on the circle $x^{2}+y^{2}+2 x-3=0$, then $p=?$", "fact_expressions": "G: Parabola;p: Number;H: Circle;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (2*x + x^2 + y^2 - 3 = 0);PointOnCurve(Focus(G), H)", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[1, 22]], [[51, 54]], [[26, 48]], [[4, 22]], [[1, 22]], [[26, 48]], [[1, 49]]]", "query_spans": "[[[51, 56]]]", "process": "" }, { "text": "The real numbers $x$, $y$ satisfy $3 x^{2}+2 y^{2}=6 x$, then the maximum value of $\\sqrt{x^{2}+y^{2}}$ is?", "fact_expressions": "x: Real;y:Real;3*x^2 + 2*y^2 = 6*x", "query_expressions": "Max(sqrt(x^2 + y^2))", "answer_expressions": "2", "fact_spans": "[[[0, 6]], [[7, 10]], [[12, 33]]]", "query_spans": "[[[35, 61]]]", "process": "\\sqrt{x^{2}+y^{2}} represents the distance from a point on the curve 3x^{2}+2y^{2}=6x to the origin. Therefore, we can use the method of eliminating variables to convert it into a function in terms of y, and then find its maximum and minimum values. From the given condition, we obtain y^{2}=3x-\\frac{3}{2}x^{2}\\geqslant0, thus 0\\leqslant x \\leqslant 2. Hence, x^{2}+y^{2}=3x-\\frac{1}{2}x^{2}=-\\frac{1}{2}(x-3)^{2}+\\frac{9}{2}. Therefore, when x=2, x^{2}+y^{2} reaches its maximum value of 4, so the maximum value of \\sqrt{x^{2}+y^{2}} is 2. [Comments] This problem examines the method of eliminating variables and finding the extreme values of functions. It is necessary to master the geometric meanings of certain expressions in such problems: for example, (x-a)^{2}+(y-b)^{2} represents the square of the distance between a point (x,y) on the curve and the point (a,b); \\frac{y-b}{x-a} represents the slope of the line connecting the point (x,y) on the curve to the point (a,b); z=Ax+By requires relating the intercepts of the line z=Ax+By on the coordinate axes to z when solving problems. In summary, the key to solving extremum problems related to circles lies in skillfully using the idea of combining numbers and shapes, applying geometric knowledge to find extrema, and being adept at using transformation and reduction methods to convert extremum problems into finding the extrema of functions. Generally, based on the given conditions, an expression—the functional relationship—concerning the desired objective is established, and then according to the characteristics of this functional relationship, methods such as parameterization, completing the square, or discriminant method are selected. Apply the properties of inequalities to find the extrema. Especially, make use of the geometric properties of circles and solve based on the geometric meaning of expressions; this precisely demonstrates the application of the idea of combining numbers and shapes." }, { "text": "The foci of the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$ are $F_{1}$, $F_{2}$, and $P$ is a point on the ellipse. When $P F_{1} \\perp P F_{2}$, what is the area of $\\triangle F_{1} P F_{2}$?", "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Expression(G) = (x^2/36 + y^2/9 = 1);Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "9", "fact_spans": "[[[0, 38], [61, 63]], [[41, 48]], [[49, 56]], [[0, 38]], [[0, 56]], [[57, 60]], [[57, 67]], [[69, 92]]]", "query_spans": "[[[94, 124]]]", "process": "According to the definition of an ellipse, PF_{1}+PF_{2}=2a=12. Then, by the Pythagorean theorem, PF_{1}^{2}+PF_{2}^{2}=F_{1}F_{2}^{2}=4c^{2}=108. Furthermore, we can obtain PF_{1}\\times PF_{2}=18, thus obtaining the area of \\triangle F_{1}PF_{2}. According to the definition of the ellipse, PF_{1}+PF_{2}=2a=12\\textcircled{1}. Since PF_{1}\\bot PF_{2}, by the Pythagorean theorem, PF_{1}^{2}+PF_{2}^{2}=F_{1}F_{2}^{2}=4c^{2}=4\\times(36-9)=108\\textcircled{2}. From \\textcircled{1}^{2}-\\textcircled{2}, we get: 2PF_{1}\\times PF_{2}=144-108=36. Therefore, S_{\\triangle F_{1}PF_{2}}=\\frac{1}{2}PF_{1}\\times PF_{2}=\\frac{1}{2}\\times 18=9." }, { "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ intersect the circle $(x-\\sqrt{3})^{2}+y^{2}=2$, then the range of the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;H: Circle;Expression(H) = (y^2 + (x - sqrt(3))^2 = 2);IsIntersect(Asymptote(G), H)", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1, sqrt(3))", "fact_spans": "[[[1, 57], [94, 97]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[62, 89]], [[62, 89]], [[1, 91]]]", "query_spans": "[[[94, 108]]]", "process": "Analysis: According to the given conditions, first find the asymptotes of the hyperbola. By using the condition that the distance from an asymptote to a given line is less than the radius, obtain an inequality in terms of $a$ and $b$. Then, combining the definition and properties of the eccentricity of a hyperbola, the final result can be determined.\n\nDetailed Solution: The asymptotes of the hyperbola are $bx \\pm ay = 0$. These asymptotes intersect the circle $(x-\\sqrt{3})^{2}+y^{2}=2$, so the distance from the center of the circle to the asymptote is less than the radius, i.e., \n$$\n\\frac{\\sqrt{3}b}{\\sqrt{a^{2}+b^{2}}}<\\sqrt{2},\n$$\nwhich simplifies to: \n$$\nb^{2}<2a^{2},\n$$\n$\\therefore c^{2}<3a^{2}$. Thus, the eccentricity of the hyperbola $e=\\frac{c}{a}<\\sqrt{3}$. Since the eccentricity of a hyperbola satisfies $e>1$, it follows that the range of the eccentricity of this hyperbola is $(1,\\sqrt{3})$." }, { "text": "Let the focus of the parabola $y^{2}=2 p x$ ($p>0$) be $F$, and let line $l$ passing through point $F$ with inclination angle $\\frac{\\pi}{4}$ intersect the parabola at points $A$ and $B$. If the circle with diameter $AB$ passes through the point $(-\\frac{p}{2}, 2)$, then the equation of the parabola is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;l: Line;PointOnCurve(F, l) = True;Inclination(l) = pi/4;Intersection(l,G) = {A,B};A: Point;B: Point;IsDiameter(LineSegmentOf(A,B),H) = True;H: Circle;PointOnCurve(I,H) = True;I: Point;Coordinate(I) = (-p/2, 2)", "query_expressions": "Expression(G)", "answer_expressions": "y^2=4*x", "fact_spans": "[[[1, 22], [62, 65], [114, 117]], [[1, 22]], [[4, 22]], [[4, 22]], [[26, 29], [31, 35]], [[1, 29]], [[56, 61]], [[30, 61]], [[36, 61]], [[56, 77]], [[68, 71]], [[72, 75]], [[79, 90]], [[89, 90]], [[89, 111]], [[91, 111]], [[91, 111]]]", "query_spans": "[[[114, 122]]]", "process": "Analysis: Find the equation of line $ l $, use the properties of the parabola to determine the y-coordinate of the midpoint of $ AB $, then solve the system of equations formed by the line and the parabola, and apply Vieta's formulas to find $ p $, thereby obtaining the equation of the parabola. \nDetailed Solution: The focus of the parabola $ y^{2} = 2px $ ($ p > 0 $) is $ F $. The line $ l $ passing through $ F $ with an inclination angle of $ \\frac{\\pi}{4} $ intersects the parabola at points $ A $ and $ B $. The circle with diameter $ AB $ is tangent to the directrix of the parabola, and this circle passes through the point $ \\left(-\\frac{p}{2}, 2\\right) $, indicating that the y-coordinate of the midpoint of $ AB $ is 2. The equation of line $ l $ is: $ y = x - \\frac{p}{2} $, then \n$$\n\\begin{cases}\ny = x - \\frac{p}{2} \\\\\ny^{2} = 2px\n\\end{cases}\n$$\nwhich leads to $ y^{2} - 2py - p^{2} = 0 $. Thus, the y-coordinate of the midpoint of $ AB $ is $ \\frac{2p}{2} = 2 $, solving gives $ p = 2 $. The equation of the parabola is: $ y^{2} = 4x $." }, { "text": "The equation of the hyperbola that has the same asymptotes as the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ and passes through the point $A(2 \\sqrt{3},-3)$ is?", "fact_expressions": "G: Hyperbola;C:Hyperbola;A: Point;Expression(G) = (x^2/16 - y^2/9 = 1);Coordinate(A) = (2*sqrt(3), -3);PointOnCurve(A, C);Asymptote(C) = Asymptote(G)", "query_expressions": "Expression(C)", "answer_expressions": "(4/9)*x^2 - x^2/4 = 1", "fact_spans": "[[[1, 40]], [[69, 72]], [[49, 68]], [[1, 40]], [[49, 68]], [[48, 72]], [[0, 72]]]", "query_spans": "[[[69, 76]]]", "process": "" }, { "text": "Given that the asymptote of a hyperbola whose axes of symmetry are the coordinate axes is $2x - y = 0$, what is the eccentricity of this hyperbola?", "fact_expressions": "G: Hyperbola;SymmetryAxis(G) = axis;Expression(OneOf(Asymptote(G))) = (2*x-y=0)", "query_expressions": "Eccentricity(G)", "answer_expressions": "{sqrt(5),sqrt(5)/2}", "fact_spans": "[[[10, 13], [33, 36]], [[2, 13]], [[10, 29]]]", "query_spans": "[[[33, 42]]]", "process": "Problem Analysis: When the foci of a hyperbola lie on the x-axis, its standard equation can be written as \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0). In this case, the equations of the asymptotes are y=\\pm\\frac{b}{a}x. Comparing with the given asymptote equation, we obtain b=2a. Then, using the relationship between a, b, and c, we get c=\\sqrt{5}a. Using the formula, the eccentricity is e=\\frac{c}{a}=\\sqrt{5}. When the foci of the hyperbola lie on the y-axis, using a similar method, the eccentricity of the hyperbola is found to be \\frac{\\sqrt{5}}{2}. Thus, the correct answer can be obtained.\n\nSolution: \n(1) When the foci of the hyperbola lie on the x-axis, let its standard equation be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0). \nSince one asymptote of the hyperbola is 2x-y=0, \nthe asymptote equations of the hyperbola are y=\\pm\\frac{b}{a}x, i.e., y=\\pm2x. \nThus, \\frac{b}{a}=2 \\Rightarrow b=2a. \n\n(2) When the foci of the hyperbola lie on the y-axis, let its standard equation be \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1 (a>0,b>0). \nUsing a method similar to (1), we obtain \\frac{b}{a}=\\frac{1}{2}. \nThen, c=\\sqrt{a^{2}+b^{2}}=\\sqrt{a^{2}+\\left(\\frac{1}{2}a\\right)^{2}}=\\sqrt{a^{2}+\\frac{1}{4}a^{2}}=\\sqrt{\\frac{5}{4}a^{2}}=\\frac{\\sqrt{5}}{2}a. \nTherefore, the eccentricity of the hyperbola is e=\\frac{c}{a}=\\frac{\\sqrt{5}}{2}. \n\nIn summary, the eccentricity of the hyperbola is \\sqrt{5} or \\frac{\\sqrt{5}}{2}." }, { "text": "Given that one of the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $y=\\frac{1}{2} x$, what is the eccentricity of this hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (y = x/2)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)/2", "fact_spans": "[[[2, 58], [88, 91]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 84]]]", "query_spans": "[[[88, 97]]]", "process": "From the asymptote equation, we get \\frac{b}{a}=\\frac{1}{2}, thus the eccentricity can be obtained. \\because one asymptote equation is y=\\frac{1}{2}x, \\therefore\\frac{b}{a}=\\frac{1}{2} \\therefore e=\\frac{c}{a}=\\sqrt{\\frac{a^2+b^{2}}{a^{2}}}=\\sqrt{1+(\\frac{b}{a})^{2}}=\\frac{\\sqrt{5}}{2}" }, { "text": "The center of the hyperbola is at the coordinate origin, the eccentricity equals $2$, and the coordinates of one focus are $(2 , 0)$. Then what is the equation of this hyperbola?", "fact_expressions": "G: Hyperbola;O: Origin;Center(G) = O;Eccentricity(G) = 2;Coordinate(OneOf(Focus(G))) = (2, 0)", "query_expressions": "Expression(G)", "answer_expressions": "x^2 - y^2/3 = 1", "fact_spans": "[[[0, 3], [43, 46]], [[7, 11]], [[0, 11]], [[0, 20]], [[0, 40]]]", "query_spans": "[[[43, 50]]]", "process": "" }, { "text": "If the focus of the parabola $y=x^{2} + 2 m x+m^{2}-m$ has coordinates $(2001 , 2002)$, then what is its vertex coordinates?", "fact_expressions": "G: Parabola;m: Number;Expression(G) = (y = -m + m^2 + 2*(m*x) + x^2);Coordinate(Focus(G)) = (2001, 2002)", "query_expressions": "Coordinate(Vertex(G))", "answer_expressions": "(2001,2001*(3/4))", "fact_spans": "[[[1, 29], [51, 52]], [[4, 29]], [[1, 29]], [[1, 49]]]", "query_spans": "[[[51, 59]]]", "process": "" }, { "text": "If a focus of the hyperbola $x^{2}-\\frac{y^{2}}{k}=1$ is $(3,0)$, then the real number $k$=?", "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (x^2 - y^2/k = 1);Coordinate(OneOf(Focus(G))) = (3, 0)", "query_expressions": "k", "answer_expressions": "8", "fact_spans": "[[[1, 29]], [[44, 49]], [[1, 29]], [[1, 42]]]", "query_spans": "[[[44, 51]]]", "process": "Since one focus of the hyperbola is (3,0), the foci of the hyperbola lie on the x-axis, a^{2}=1, b^{2}=k,\\therefore c^{2}=a^{2}+b^{2}=1+k=9,\\therefore k=8" }, { "text": "The left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ are $F_{1}$ and $F_{2}$, respectively. If there exists a point $P$ on the hyperbola such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=-2 a^{2}$, then the range of values for the eccentricity of the hyperbola is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;LeftFocus(C) = F1;RightFocus(C) = F2;F2: Point;F1: Point;PointOnCurve(P,C) = True;P: Point;DotProduct(VectorOf(P,F1),VectorOf(P,F2)) = -2*a^2", "query_expressions": "Range(Eccentricity(C))", "answer_expressions": "[sqrt(3),+oo)", "fact_spans": "[[[0, 61], [87, 90], [167, 170]], [[0, 61]], [[7, 61]], [[99, 165]], [[7, 61]], [[7, 61]], [[0, 85]], [[0, 85]], [[78, 85]], [[70, 77]], [[87, 97]], [[93, 97]], [[100, 165]]]", "query_spans": "[[[167, 180]]]", "process": "" }, { "text": "The standard equation of the ellipse, using the vertices and foci of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ as foci and two vertices respectively, is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/5 = 1);Vertex(G) = Focus(H);Focus(G) = Vertex(H);H: Ellipse", "query_expressions": "Expression(H)", "answer_expressions": "x^2/9+y^2/5=1", "fact_spans": "[[[1, 39]], [[1, 39]], [[0, 58]], [[0, 58]], [[56, 58]]]", "query_spans": "[[[56, 64]]]", "process": "" }, { "text": "Given the ellipse $ C $: $ \\frac{x^{2}}{25} + \\frac{y^{2}}{9} = 1 $, the left focus of the ellipse is $ F $. Point $ M $ is a point on the ellipse $ C $, point $ N $ is the midpoint of $ M F $, $ O $ is the center of the ellipse, and $ O N = 4 $. Find the distance from point $ M $ to the left directrix of the ellipse $ C $?", "fact_expressions": "C: Ellipse;M: Point;F: Point;O: Point;N: Point;Expression(C) = (x^2/25 + y^2/9 = 1);LeftFocus(C) = F;PointOnCurve(M, C);MidPoint(LineSegmentOf(M, F)) = N;Center(C) = O;LineSegmentOf(O, N) = 4", "query_expressions": "Distance(M, LeftDirectrix(C))", "answer_expressions": "5/2", "fact_spans": "[[[2, 45], [59, 64], [86, 88], [106, 111]], [[54, 58], [101, 105]], [[50, 53]], [[82, 85]], [[68, 72]], [[2, 45]], [[2, 53]], [[54, 67]], [[68, 81]], [[82, 91]], [[92, 99]]]", "query_spans": "[[[101, 120]]]", "process": "Test Analysis: Let the right focus be F'. Then by the definition of the ellipse, MF + MF' = 10. According to the given condition, we obtain ON = \\frac{1}{2}MF' = 4, that is, MF' = 8, so MF = 2. By the second definition of the ellipse, we get \\frac{2}{d} = e = \\frac{4}{5}, hence d = \\frac{10}{4} = \\frac{5}{2}. The answer to be filled in is \\frac{5}{2}." }, { "text": "The standard equation of an ellipse is given as $\\frac{x^{2}}{25}+\\frac{y^{2}}{m^{2}}=1(m>0)$, and the focal distance is $6$. What is the value of the real number $m$?", "fact_expressions": "G: Ellipse;m: Real;m>0;Expression(G) = (x^2/25 + y^2/m^2 = 1);FocalLength(G)=6", "query_expressions": "m", "answer_expressions": "{4, sqrt(34)}", "fact_spans": "[[[2, 4]], [[66, 71]], [[10, 55]], [[2, 55]], [[2, 64]]]", "query_spans": "[[[66, 75]]]", "process": "Since 2c=6, we have c=3. (1) When the foci are on the x-axis, from the standard equation of the ellipse, 25−m^{2}=9, solving gives m=4. (2) When the foci are on the y-axis, from the standard equation of the ellipse, m^{2}−25=9, solving gives m=\\sqrt{34}. In conclusion, the solutions are m=4 or \\sqrt{34}." }, { "text": "If the focus of the parabola $y^{2}=2 px$ coincides with the right focus of the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{3}=1$, then what is the value of $p$?", "fact_expressions": "G: Parabola;p: Number;H: Ellipse;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (x^2/12 + y^2/3 = 1);Focus(G) = RightFocus(H)", "query_expressions": "p", "answer_expressions": "6", "fact_spans": "[[[1, 16]], [[66, 69]], [[20, 58]], [[1, 16]], [[20, 58]], [[1, 64]]]", "query_spans": "[[[66, 73]]]", "process": "" }, { "text": "Given that the eccentricity of the ellipse $M$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $\\frac{\\sqrt{2}}{2}$, a line passing through its left focus $F(-c, 0)$ intersects the ellipse $M$ at points $A$ and $B$. If the midpoint of chord $AB$ is $D(-4,2)$, then what is the equation of the ellipse $M$?", "fact_expressions": "M: Ellipse;b: Number;a: Number;G: Line;A: Point;B: Point;F: Point;D: Point;a > b;b > 0;Expression(M) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (-c, 0);Coordinate(D) = (-4, 2);Eccentricity(M) = sqrt(2)/2;LeftFocus(M)=F;PointOnCurve(F, G);Intersection(G, M) = {A, B};IsChordOf(LineSegmentOf(A, B),M);MidPoint(LineSegmentOf(A, B)) = D;c:Number", "query_expressions": "Expression(M)", "answer_expressions": "x^2/72+y^2/36=1", "fact_spans": "[[[2, 59], [104, 109], [143, 148], [86, 87]], [[8, 59]], [[8, 59]], [[101, 103]], [[110, 113]], [[114, 117]], [[90, 100]], [[132, 141]], [[8, 59]], [[8, 59]], [[2, 59]], [[90, 100]], [[132, 141]], [[2, 84]], [[86, 100]], [[85, 103]], [[101, 119]], [[104, 128]], [[123, 141]], [[90, 100]]]", "query_spans": "[[[143, 153]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). By the midpoint coordinate formula, we obtain x_{1}+x_{2}=-8, y_{1}+y_{2}=4. Substituting the coordinates of A and B into the equation of M respectively, we get \\begin{cases}\\frac{x_{2}}{a^{2}}+\\frac{y_{2}^{2}}{b^{2}}=1\\\\\\frac{x_{2}}{a^{2}}+\\frac{y_{2}^{2}}{b^{2}}=1\\end{cases}. Subtracting these two equations and simplifying yields \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{2b^{2}}{a^{2}}. Since points A, B, D, F are collinear, it follows that \\frac{2-0}{c-4}=\\frac{y_{1}}{x_{1}-x_{2}}=\\frac{2b^{2}}{a^{2}}, so a^{2}=b^{2}(c-4). Solving the system \\begin{cases}a^{2}=b^{2}(c-4)\\\\\\frac{c^{2}}{a^{2}}=\\frac{1}{2}\\\\b^{2}+c^{2}=a^{2}\\end{cases}, we obtain \\begin{cases}a^{2}=72\\\\b^{2}=36\\\\c=6\\end{cases}. Therefore, the equation of ellipse M is \\frac{x^{2}}{72}+\\frac{y^{2}}{36}=1." }, { "text": "Given point $A(4,0)$, the parabola $C$: $y^{2}=2 p x (00$), the left and right foci are denoted by $F_{1}$ and $F_{2}$, respectively, and the asymptotes have equations $\\sqrt{7} x \\pm 3 y=0$. Point $A$ lies on the circle $C_{2}$: $x^{2}+y^{2}=16$. If $\\overrightarrow{F_{1} A}=\\overrightarrow{A B}$ and point $B$ lies on the right branch of the hyperbola $C_{1}$, then the tangent value of $\\angle B F_{2} F_{1}$ is?", "fact_expressions": "C1: Hyperbola;C2: Circle;A: Point;B: Point;F2: Point;F1: Point;m: Number;m > 0;Expression(C1) = (x^2/9 - y^2/m = 1);LeftFocus(C1) = F1;RightFocus(C1) = F2;Expression(Asymptote(C1)) = (sqrt(7)*x + pm*3*y = 0);PointOnCurve(A, C2);Expression(C2) = (x^2 + y^2 = 16);VectorOf(F1, A) = VectorOf(A, B);PointOnCurve(B, RightPart(C1))", "query_expressions": "Tan(AngleOf(B, F2, F1))", "answer_expressions": "-7*sqrt(15)/17", "fact_spans": "[[[2, 54], [194, 204]], [[113, 138]], [[108, 112]], [[189, 193]], [[71, 78]], [[63, 70]], [[14, 54]], [[14, 54]], [[2, 54]], [[2, 78]], [[2, 78]], [[2, 107]], [[108, 139]], [[113, 138]], [[141, 188]], [[189, 209]]]", "query_spans": "[[[211, 239]]]", "process": "From the given conditions, we know that the asymptotes of the hyperbola are $\\sqrt{7}x \\pm 3y = 0$, i.e., $y = \\pm\\frac{\\sqrt{7}}{3}x$. Also, since the equation of the hyperbola is $\\frac{x^{2}}{9} - \\frac{y^{2}}{m} = 1$ ($m > 0$), it follows that $m = 7$, then $c^{2} = 9 + 7 = 16$, so the foci are $F_{1}(-4,0)$, $F_{2}(4,0)$. The circle's equation implies $\\angle F_{1}AF_{2} = 90^{\\circ}$, and since $\\overrightarrow{F_{1}A} = \\overrightarrow{AB}$, point $A$ is the midpoint of segment $F_{1}B$, $\\triangle F_{1}F_{2}B$ is an isosceles triangle, so $F_{2}B = F_{1}F_{2} = 8$, and point $B$ lies on the right branch of the hyperbola. By the definition of a hyperbola, $BF_{1} - BF_{2} = 6$, thus $BF_{1} = 14$, $F_{1}A = \\frac{1}{2}BF_{1} = 7$. In the right triangle $\\triangle F_{1}AF_{2}$, by the Pythagorean theorem, $AF_{2} = \\sqrt{8^{2} - 7^{2}} = \\sqrt{15}$, $\\tan\\angle F_{1}F_{2}A = \\frac{7\\sqrt{15}}{15}$, then $\\tan\\angle BF_{2}F_{1} = \\tan 2\\angle F_{1}F_{2}A = \\frac{2 \\times \\frac{7\\sqrt{15}}{15}}{1 - \\frac{49}{15}} = -\\frac{7\\sqrt{15}}{17}$." }, { "text": "Given $F_{1}$, $F_{2}$ are the left and right foci of the hyperbola $C$: $x^{2}-y^{2}=1$, point $P$ lies on $C$, $\\angle F_{1} P F_{2}=90^{\\circ}$, then the distance from point $P$ to the $x$-axis is?", "fact_expressions": "F1: Point;F2: Point;C: Hyperbola;Expression(C) = (x^2 - y^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C) = True;AngleOf(F1, P, F2) = ApplyUnit(90, degree)", "query_expressions": "Distance(P, xAxis)", "answer_expressions": "sqrt(2) / 2", "fact_spans": "[[[2, 9]], [[10, 17]], [[18, 41], [53, 56]], [[18, 41]], [[2, 47]], [[2, 47]], [[48, 52], [94, 98]], [[48, 57]], [[58, 91]]]", "query_spans": "[[[94, 108]]]", "process": "[Score] Let the coordinates of point P be denoted. According to the coordinate representation formula of the dot product of planar vectors, an equation can be obtained. Then, since point P lies on C, another equation can be obtained. By solving the system of equations, the ordinate of point P can be found, and thus the distance from point P to the x-axis can be determined. From the given information: $ F_{1}(-\\sqrt{2},0), F_{2}(\\sqrt{2},0) $. Let the coordinates of point P be $ (x_{0}, y_{0}) $. Therefore, we have $ \\overrightarrow{F_{1}P} = (x_{0} + \\sqrt{2}, y_{0}) $, $ \\overrightarrow{F_{2}P} = (x_{0} - \\sqrt{2}, y_{0}) $. Since $ \\angle F_{1}PF_{2} = 90^{\\circ} $, it follows that $ \\overrightarrow{F_{1}P} \\cdot \\overrightarrow{F_{2}P} = 0 \\Rightarrow (x_{0} + \\sqrt{2})(x_{0} - \\sqrt{2}) + y_{0}^{2} = 0 \\Rightarrow x_{0}^{2} + y_{0}^{2} = 2 $. While point P lies on C, we also have $ x_{0}^{2} - y_{0}^{2} = 1 $. Solving these equations yields $ y_{0} = \\pm \\frac{\\sqrt{2}}{2} $, so the distance from point P to the x-axis is $ \\frac{\\sqrt{2}}{2} $." }, { "text": "Given the parabola $y^{2}=2 p x(p>0)$ with a chord $AB$ passing through the focus $F$, if $|AB|=8$ and the abscissa of the midpoint of $AB$ is $3$, then the equation of the parabola is?", "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;F: Point;p>0;Expression(G) = (y^2 = 2*p*x);IsChordOf(LineSegmentOf(A,B),G);Focus(G) = F;PointOnCurve(F,LineSegmentOf(A,B));Abs(LineSegmentOf(A, B)) = 8;XCoordinate(MidPoint(LineSegmentOf(A, B))) = 3", "query_expressions": "Expression(G)", "answer_expressions": "y^2=4*x", "fact_spans": "[[[2, 23], [66, 69]], [[5, 23]], [[25, 30]], [[25, 30]], [[33, 36]], [[5, 23]], [[2, 23]], [[2, 30]], [[2, 36]], [[25, 36]], [[38, 47]], [[49, 64]]]", "query_spans": "[[[66, 74]]]", "process": "From the definition of the parabola, we obtain |AB| = x_{A} + x_{B} + p = 2x_{mid} + p = 6 + p = 8, therefore p = 2. The equation of the parabola is y^{2} = 4x." }, { "text": "Given $A(-1 , 0)$, $B(1 , 0)$, and point $C(x , y)$ satisfies: $\\frac{\\sqrt{(x-1)^{2}+y^{2}}}{|x-4|}=\\frac{1}{2}$, then $|AC|+|BC|$=?", "fact_expressions": "A: Point;Coordinate(A) = (-1, 0);B: Point;Coordinate(B) = (1, 0);C: Point;Coordinate(C) = (x1, y1);x1: Number;y1: Number;sqrt(y1^2 + (x1 - 1)^2)/Abs(x1 - 4) = 1/2", "query_expressions": "Abs(LineSegmentOf(A, C)) + Abs(LineSegmentOf(B, C))", "answer_expressions": "4", "fact_spans": "[[[2, 13]], [[2, 13]], [[16, 26]], [[16, 26]], [[27, 38]], [[27, 38]], [[28, 38]], [[28, 38]], [[41, 91]]]", "query_spans": "[[[93, 106]]]", "process": "" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}(-1,0)$, $F_{2}(1,0)$ respectively. Point $A$ is a moving point on the line $x+y-2=0$. If point $A$ lies on the ellipse $C$, then the maximum value of the eccentricity of ellipse $C$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;Coordinate(F1) = (-1, 0);Coordinate(F2) = (1, 0);LeftFocus(C) = F1;RightFocus(C) = F2;G: Line;Expression(G) = (x + y - 2 = 0);A: Point;PointOnCurve(A, G);PointOnCurve(A, C)", "query_expressions": "Max(Eccentricity(C))", "answer_expressions": "sqrt(10)/5", "fact_spans": "[[[2, 59], [121, 126], [129, 134]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[66, 79]], [[81, 93]], [[66, 79]], [[81, 93]], [[2, 93]], [[2, 93]], [[99, 110]], [[99, 110]], [[94, 98], [116, 120]], [[94, 114]], [[116, 127]]]", "query_spans": "[[[129, 144]]]", "process": "From the given conditions, it is easy to know: the line x+y-2=0 and the ellipse C have common points. Solving the system of equations gives: (a^{2}+b^{2})x^{2}-4a^{2}x+4a^{2}-a^{2}b^{2}=0 \n\\therefore \\Delta=(-4a^{2})^{2}-4(a^{2}+b^{2})(4a^{2}-a^{2}b^{2})\\geqslant0 \n\\therefore a^{2}\\geqslant\\frac{5}{2}, i.e., a\\geqslant\\sqrt{\\frac{5}{2}} \n\\therefore the eccentricity of ellipse C satisfies e\\geqslant\\frac{\\sqrt{10}}{5} \n\\therefore the maximum value of the eccentricity of ellipse C is \\frac{\\sqrt{10}}{6} \n\nThere are two commonly used methods for finding eccentricity: \n(1) Obtain the values of a and c, then directly substitute into the formula e=\\frac{c}{a} to solve. \n(2) Set up a homogeneous equation (or inequality) in terms of a, b, c, and then use b^{2}=a^{2}-c^{2} to eliminate b, converting it into an equation (or inequality) in terms of e to solve." }, { "text": "Given a point $M(x_{0}, 2 \\sqrt{3})$ on the parabola $y^{2}=4 x$, then the distance from point $M$ to the focus of the parabola is equal to?", "fact_expressions": "G: Parabola;M: Point;x0:Number;Expression(G) = (y^2 = 4*x);Coordinate(M) = (x0, 2*sqrt(3));PointOnCurve(M,G)", "query_expressions": "Distance(M, Focus(G))", "answer_expressions": "4", "fact_spans": "[[[2, 16], [48, 51]], [[19, 41], [43, 47]], [[19, 41]], [[2, 16]], [[19, 41]], [[2, 41]]]", "query_spans": "[[[43, 59]]]", "process": "Analysis: Substitute the point M(x_{0},2\\sqrt{3}) into the parabola equation to solve for x_{0}. Using the definition of a parabola, the distance from point M to the focus of the parabola is x_{0}+1. \nDetailed solution: Substituting the point M(x_{0},2\\sqrt{3}) into the parabola equation gives: (2\\sqrt{3})^{2}=4x_{0}, solving yields x_{0}=3. Therefore, the distance from point M to the focus of the parabola is x_{0}+1=4." }, { "text": "If point $P$ is a point on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ with foci $F_{1}$ and $F_{2}$, satisfying $P F_{1} \\perp P F_{2}$, and $|P F_{1}|=2|P F_{2}|$, then the eccentricity of this hyperbola is?", "fact_expressions": "P: Point;F1: Point;F2: Point;Focus(G) = {F1, F2};G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;PointOnCurve(P, G) = True;IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2)) = True;Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[1, 5]], [[7, 14]], [[15, 22]], [[6, 72]], [[26, 72], [128, 131]], [[26, 72]], [[29, 72]], [[29, 72]], [[1, 75]], [[78, 101]], [[103, 125]]]", "query_spans": "[[[128, 137]]]", "process": "Test analysis: By the definition of a hyperbola, we have |PF_{1}|-|PF_{2}|=2a. Since |PF_{1}|=2|PF_{2}|, it follows that |PF_{1}|=4a, |PF_{2}|=2a. Also, since PF_{1}\\bot PF_{2}, we have |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}, that is, (4a)^{2}+(2a)^{2}=(2c)^{2}. Simplifying yields c^{2}=5a^{2}, so e=\\frac{c}{a}=\\sqrt{5}." }, { "text": "The moving point $P$ satisfies $|PF_{1}|-|PF_{2}|=6$ with respect to the points $F_{1}(0 , 5)$ and $F_{2}(0 ,-5)$. What is the trajectory equation of point $P$?", "fact_expressions": "P: Point;F1: Point;F2: Point;Coordinate(F1) = (0, 5);Coordinate(F2) = (0, -5);Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)) = 6", "query_expressions": "LocusEquation(P)", "answer_expressions": "(y^2/9-x^2/16=1)&(y<=-3)", "fact_spans": "[[[2, 5], [62, 66]], [[6, 21]], [[22, 37]], [[6, 21]], [[22, 37]], [[39, 60]]]", "query_spans": "[[[62, 73]]]", "process": "" }, { "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ have its right focus at $F$, and left and right vertices at $A_{1}$, $A_{2}$ respectively. A line $l$ passing through $F$ and parallel to one asymptote of the hyperbola $C$ intersects the other asymptote at point $P$. If $P$ lies exactly on the circle with $A_{1}A_{2}$ as diameter, then the eccentricity of the hyperbola is?", "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;G: Circle;A1: Point;A2: Point;F: Point;P: Point;a > b;b > 0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;LeftVertex(C) = A1;RightVertex(C) = A2;PointOnCurve(F, l);L1: Line;L2: Line;OneOf(Asymptote(C)) = L1;OneOf(Asymptote(C)) = L2;Negation(L1=L2);IsParallel(l, L1);Intersection(l, L2) = P;IsDiameter(LineSegmentOf(A1, A2), G);PointOnCurve(P, G)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[112, 117]], [[1, 59], [97, 103], [160, 163]], [[9, 59]], [[9, 59]], [[156, 157]], [[75, 82]], [[83, 90]], [[64, 67], [92, 95]], [[127, 130], [132, 135]], [[9, 59]], [[9, 59]], [[1, 59]], [[1, 67]], [[1, 90]], [[1, 90]], [[91, 117]], [], [], [[97, 109]], [[97, 124]], [[97, 124]], [[96, 117]], [[97, 130]], [[138, 157]], [[132, 158]]]", "query_spans": "[[[160, 169]]]", "process": "" }, { "text": "Given that the hyperbola $C$ is centered at the origin of the coordinate system, one focus coincides with the focus of the parabola $y=\\frac{x^{2}}{12}$, and one of its asymptotes has the equation $y=\\frac{2 \\sqrt{5}}{5} x$, find the equation of $C$.", "fact_expressions": "C: Hyperbola;O: Origin;Center(C) = O;G: Parabola;Expression(G) = (y = x^2/12);OneOf(Focus(C)) = Focus(G);Expression(OneOf(Asymptote(C))) = (y = x*(2*sqrt(5)/5))", "query_expressions": "Expression(C)", "answer_expressions": "y^2/4 - x^2/5 = 1", "fact_spans": "[[[2, 8], [52, 53], [90, 93]], [[12, 16]], [[2, 16]], [[22, 45]], [[22, 45]], [[2, 50]], [[52, 88]]]", "query_spans": "[[[90, 98]]]", "process": "According to the given conditions, the equation of hyperbola C can be set as \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1 (a>0,b>0). Since one asymptote of the hyperbola is y=\\frac{2\\sqrt{5}}{5}x, then \\frac{a}{b}=\\frac{2\\sqrt{5}}{5}\\textcircled{1}; also, since the focus of the parabola y=\\frac{x^{2}}{12} is (0,3), the semi-focal length c of the hyperbola is 3, so a^{2}+b^{2}=c^{2}=9\\textcircled{2}. Solving \\textcircled{1} and \\textcircled{2}, we obtain a^{2}=4, b^{2}=5, thus the equation of C is \\frac{y^{2}}{4}-\\frac{x^{2}}{5}=1." }, { "text": "The focus $F$ of the parabola $y^{2}=4x$ is exactly the right vertex of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and the asymptotes are given by $y=\\pm \\sqrt{3}x$. Then the equation of the hyperbola is?", "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 4*x);F: Point;Focus(H) = F;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;RightVertex(G) = F;Expression(Asymptote(G)) = (y = pm*sqrt(3)*x)", "query_expressions": "Expression(G)", "answer_expressions": "x^2 - y^2/3 = 1", "fact_spans": "[[[2, 16]], [[2, 16]], [[19, 22]], [[2, 22]], [[25, 82], [114, 117]], [[25, 82]], [[28, 82]], [[28, 82]], [[28, 82]], [[28, 82]], [[19, 86]], [[25, 112]]]", "query_spans": "[[[114, 121]]]", "process": "" }, { "text": "Given the parabola $y^{2}=2 p x$ has focus $F$, a line passing through point $F$ intersects the parabola at points $A$ and $B$. If $\\overrightarrow{A F}=3 \\overrightarrow{F B}$, then the slope of line $A B$ is?", "fact_expressions": "G: Parabola;p: Number;H: Line;B: Point;A: Point;F: Point;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};VectorOf(A, F) = 3*VectorOf(F, B)", "query_expressions": "Slope(LineOf(A,B))", "answer_expressions": "pm*sqrt(3)", "fact_spans": "[[[2, 18], [35, 38]], [[5, 18]], [[32, 34]], [[43, 46]], [[39, 42]], [[22, 25], [27, 31]], [[2, 18]], [[2, 25]], [[26, 34]], [[32, 48]], [[50, 95]]]", "query_spans": "[[[97, 109]]]", "process": "The focus of the parabola $ y^{2}=2px $ is $ F(\\frac{p}{2},0) $. According to the given conditions, the slope of line $ AB $ exists. Let the equation of the line be $ y=k(x-\\frac{p}{2}) $. Solving the system \n\\[\n\\begin{cases}\ny^{2}=2px \\\\\ny=k(x-\\frac{p}{2})\n\\end{cases}\n\\]\nyields $ ky^{2}-2py-kp^{2}=0 $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, then $ y_{1}+y_{2}=\\frac{2p}{k} $, $ y_{1}y_{2}=-p^{2} $. From $ \\overrightarrow{AF}=3\\overrightarrow{FB} $, we obtain $ \\frac{y_{1}}{y_{2}}=-3 $. Therefore,\n\\[\n\\frac{(y_{1}+y_{2})^{2}}{y_{1}y_{2}} = \\frac{y_{1}}{y_{2}} + \\frac{y_{2}}{y_{1}} + 2 = -3 - \\frac{1}{3} + 2 = \\frac{4p^{2}}{-p^{2}} = -\\frac{4}{k^{2}}\n\\]\nSolving gives: $ k=\\pm\\sqrt{3} $" }, { "text": "The coordinates of a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ are $(\\sqrt{3}, 0)$, then the real number $a$=?", "fact_expressions": "G: Hyperbola;a: Real;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Coordinate(OneOf(Focus(G))) = (sqrt(3), 0)", "query_expressions": "a", "answer_expressions": "sqrt(2)", "fact_spans": "[[[0, 37]], [[62, 67]], [[3, 37]], [[0, 37]], [[0, 60]]]", "query_spans": "[[[62, 69]]]", "process": "The hyperbola \\frac{x^{2}}{a^{2}}-y^{2}=1 (a>0) has a focus coordinate (\\sqrt{3},0), 1+a^{2}=3, \\therefore a=\\sqrt{2}," }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, the left vertex and upper vertex are denoted as $A$ and $B$, respectively, and the left and right foci are $F_{1}$ and $F_{2}$. On the line segment $AB$, there exists exactly one point $P$ such that $PF_{1} \\perp PF_{2}$. Then, the square of the eccentricity of the ellipse is?", "fact_expressions": "C: Ellipse;b: Number;a: Number;B: Point;A: Point;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(C) = A;UpperVertex(C) = B;LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P,LineSegmentOf(A,B));IsPerpendicular(LineSegmentOf(P,F1),LineSegmentOf(P,F2))", "query_expressions": "Eccentricity(C)^2", "answer_expressions": "(3-sqrt(5))/2", "fact_spans": "[[[2, 59], [148, 150]], [[9, 59]], [[9, 59]], [[74, 77]], [[70, 73]], [[117, 121]], [[86, 93]], [[94, 101]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 77]], [[2, 77]], [[2, 101]], [[2, 101]], [[103, 121]], [[123, 146]]]", "query_spans": "[[[148, 159]]]", "process": "From the given conditions, we have A(-a,0), B(0,b); then the equation of line AB is \\frac{x}{-a}+\\frac{y}{b}=1, that is, bx-ay+ab=0. Since PF_{1}\\bot PF_{2}, the trajectory of point P is a circle centered at the origin with radius c, and its equation is x^{2}+y^{2}=c^{2}. Since there is exactly one point P on segment AB satisfying PF_{1}\\bot PF_{2}, the line AB is tangent to the circle x^{2}+y^{2}=c^{2}. Therefore, \\frac{ab}{\\sqrt{b^{2}+a^{2}}}=c. Simplifying yields b^{2}=ac. Then a^{2}-c^{2}=ac, which leads to e^{2}+e-1=0. Given 00 , b>0)$, and the focal length of $C$ is $4$. What is its eccentricity?", "fact_expressions": "G: Point;Coordinate(G) = (2, 3);PointOnCurve(G, C);C: Hyperbola;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;FocalLength(C) = 4", "query_expressions": "Eccentricity(C)", "answer_expressions": "2", "fact_spans": "[[[2, 12]], [[2, 12]], [[2, 78]], [[13, 77], [79, 82], [91, 92]], [[13, 77]], [[21, 77]], [[21, 77]], [[21, 77]], [[21, 77]], [[79, 89]]]", "query_spans": "[[[91, 98]]]", "process": "" }, { "text": "It is known that $F$ is the focus of the parabola $C$: $y^{2}=4 x$. A line passing through $F$ with slope $\\sqrt{3}$ intersects $C$ at points $A$ and $B$. Given $|F A|>|F B|$, then the value of $\\frac{|F A|}{|F B |}$ equals?", "fact_expressions": "C: Parabola;G: Line;F: Point;A: Point;B: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F, G);Slope(G) = sqrt(3);Intersection(G, C) = {A, B};Abs(LineSegmentOf(F, A)) > Abs(LineSegmentOf(F, B))", "query_expressions": "Abs(LineSegmentOf(F,A))/Abs(LineSegmentOf(F,B))", "answer_expressions": "3", "fact_spans": "[[[6, 25], [51, 54]], [[48, 50]], [[2, 5], [30, 33]], [[55, 58]], [[59, 62]], [[6, 25]], [[2, 28]], [[29, 50]], [[34, 50]], [[48, 64]], [[67, 80]]]", "query_spans": "[[[82, 109]]]", "process": "" }, { "text": "The parabola $C$: $x^{2}=2 y$ has focus $F$. The tangent line $l$ at a point $P(1 , y_{0})$ on $C$ intersects the $y$-axis at $A$. Then $|AF|=$?", "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 2*y);F: Point;Focus(C) = F;P: Point;y0: Number;Coordinate(P) = (1, y0);PointOnCurve(P, C);A: Point;l: Line;TangentOnPoint(P,C)=l;Intersection(l, yAxis) = A", "query_expressions": "Abs(LineSegmentOf(A, F))", "answer_expressions": "1", "fact_spans": "[[[0, 19], [28, 31]], [[0, 19]], [[23, 26]], [[0, 26]], [[34, 48]], [[34, 48]], [[34, 48]], [[27, 48]], [[61, 64]], [[51, 54]], [[27, 54]], [[51, 64]]]", "query_spans": "[[[66, 74]]]", "process": "" }, { "text": "If the equation $\\frac{x^{2}}{3+k}+\\frac{y^{2}}{2-k}=1$ represents an ellipse, then what is the range of values for $k$?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(k + 3) + y^2/(2 - k) = 1);k:Number", "query_expressions": "Range(k)", "answer_expressions": "(-3,-1/2) + (-1/2,2)", "fact_spans": "[[[44, 46]], [[1, 46]], [[48, 51]]]", "query_spans": "[[[48, 58]]]", "process": "From the form of the equation, a system of inequalities about $k$ can be obtained, thereby determining the range of values for $k$. According to the given conditions, we have \n\\begin{cases}3+k>0\\\\2-k>0\\\\3+k\\neq2-k\\end{cases}, \nsolving which yields $k\\in(-3,-\\frac{1}{2})\\cup(-\\frac{1}{2},2)$." }, { "text": "Given $P(4,-1)$, $F$ is the focus of the parabola $y^{2}=8 x$, $M$ is a point on this parabola such that $|M P|+|M F|$ is minimized. Then the coordinates of point $M$ are?", "fact_expressions": "P: Point;Coordinate(P) = (4, -1);G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;M: Point;PointOnCurve(M, G);WhenMin(Abs(LineSegmentOf(M, P)) + Abs(LineSegmentOf(M, F)))", "query_expressions": "Coordinate(M)", "answer_expressions": "(1/8,-1)", "fact_spans": "[[[2, 11]], [[2, 11]], [[17, 31], [40, 43]], [[17, 31]], [[13, 16]], [[12, 34]], [[35, 38], [68, 72]], [[35, 46]], [[49, 66]]]", "query_spans": "[[[68, 77]]]", "process": "\\therefore the parabola is y^{2}=8x,\\therefore 2p=8, yielding \\frac{p}{2}=2, so the focus is F(2,0), and the equation of the directrix l is: x=-2. Draw MN\\bot l from point M, with foot N; then according to the definition of the parabola, |MN|=|MF|. By plane geometry knowledge, |MP|+|MF| achieves the minimum value if and only if points M, N, P are collinear; at this time, M(\\frac{1}{2},-1)." }, { "text": "Let $F$ be a focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$. If there exists a point $P$ on $C$ such that the midpoint of the segment $PF$ is exactly one endpoint of its imaginary axis, then the eccentricity of $C$ is?", "fact_expressions": "F: Point;C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;OneOf(Focus(C)) = F;P: Point;PointOnCurve(P, C);MidPoint(LineSegmentOf(P, F)) = OneOf(Endpoint(ImageinaryAxis(C)))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[1, 4]], [[5, 67], [74, 77], [98, 99], [108, 111]], [[5, 67]], [[12, 67]], [[12, 67]], [[12, 67]], [[12, 67]], [[1, 72]], [[80, 84]], [[74, 84]], [[86, 106]]]", "query_spans": "[[[108, 117]]]", "process": "Problem Analysis: By symmetry, without loss of generality, assume F(c,0), and the endpoint of the minor axis is (0,b); thus, the point (-c,2b) lies on the hyperbola \\cdot\\frac{c^{2}}{a^{2}}-\\frac{4b^{2}}{b^{2}}=1\\Rightarrow e=\\frac{c}{a}=\\sqrt{5}" }, { "text": "Given that one focus of the ellipse $5 x^{2}+ky^{2}=5$ is $(0 , 2)$, what is the value of the real number $k$?", "fact_expressions": "G: Ellipse;Expression(G) = (k*y^2 + 5*x^2 = 5);k: Real;H: Point;Coordinate(H) = (0, 2);OneOf(Focus(G)) = H", "query_expressions": "k", "answer_expressions": "1", "fact_spans": "[[[2, 22]], [[2, 22]], [[39, 44]], [[28, 37]], [[28, 37]], [[2, 37]]]", "query_spans": "[[[39, 48]]]", "process": "" }, { "text": "Given point $M(m, 0)$ $(m>0)$ and parabola $C$: $y^{2}=4x$, a line passing through the focus $F$ of $C$ intersects $C$ at points $A$ and $B$. If $\\overrightarrow{AF}=2\\overrightarrow{FB}$, and $|\\overrightarrow{MF}|=|\\overrightarrow{MA}|$, then $m=?$", "fact_expressions": "C: Parabola;G: Line;M: Point;A: Point;F: Point;B: Point;m>0;Expression(C) = (y^2 = 4*x);Coordinate(M) = (m, 0);Focus(C)=F;PointOnCurve(F, G);Intersection(G, C) = {A, B};VectorOf(A, F) = 2*VectorOf(F, B);Abs(VectorOf(M, F)) = Abs(VectorOf(M, A));m:Number", "query_expressions": "m", "answer_expressions": "11/2", "fact_spans": "[[[20, 39], [41, 44], [54, 57]], [[51, 53]], [[2, 19]], [[59, 62]], [[47, 50]], [[63, 66]], [[3, 19]], [[20, 39]], [[2, 19]], [[41, 50]], [[40, 53]], [[51, 68]], [[70, 115]], [[117, 164]], [[166, 169]]]", "query_spans": "[[[166, 171]]]", "process": "Draw the graph, use the given conditions to find the coordinates of A and B, and determine the value of m through vector relationships. From the problem, we know: F(1,0). According to the definition of the parabola, A(x_{1},y_{1}), B(x_{2},y_{2}). Since \\overrightarrow{AF}=2\\overrightarrow{FB}, we obtain: 2((x_{2}-1,y_{2}))=(1-x_{1},-y_{1}), which gives y_{2}=-\\frac{y_{1}}{2}, x_{2}=\\frac{3-x_{1}}{2}. \n\\begin{cases}y_{1}=4x_{1}\\\\(-\\frac{y_{1}}{2})^{2}=4\\times\\frac{3-x_{1}}{2}\\end{cases}, solving yields x_{1}=2, y_{1}=|\\overrightarrow{MF}|=|\\overrightarrow{MA}|, so |m-1|=\\sqrt{(m-2)^{2}+=\\pm2\\sqrt{2}.1|=\\sqrt{(m-2)^{2}+(0\\pm2\\sqrt{2})^{2}}, solving gives m=\\frac{11}{2}." }, { "text": "Given that $F_{1}$ is the left focus of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, $P$ is a moving point on this ellipse, and $A(-1,3)$ is a fixed point, then the maximum value of $|P A|+|P F_{1}|$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);A: Point;P: Point;F1: Point;Coordinate(A) = (-1, 3);LeftFocus(G) = F1;PointOnCurve(P, G)", "query_expressions": "Max(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F1)))", "answer_expressions": "15", "fact_spans": "[[[10, 49]], [[10, 49]], [[66, 75]], [[54, 57]], [[2, 9]], [[66, 75]], [[2, 53]], [[54, 65]]]", "query_spans": "[[[81, 104]]]", "process": "From the problem, we know: F_{1}(-3,0), F_{2}(3,0). Using the definition of an ellipse, we obtain: |PF_{1}| + |PF_{2}| = 2a = 10. Using the triangle inequality, we get |PA| + |PF| = |PA| + 10 - |PF_{2}| \\leqslant 10 + |AF_{2}| = 15" }, { "text": "The equation of a hyperbola with the coordinate axes as axes of symmetry, two asymptotes perpendicular to each other, and a distance of $2$ between the two directrices is?", "fact_expressions": "G: Hyperbola;SymmetryAxis(G) = axis;Z1: Line;Z2: Line;Asymptote(G) = {Z1, Z2};IsPerpendicular(Z1, Z2);K1: Line;K2: Line;Directrix(G) = {K1, K2};Distance(K1, K2) = 2", "query_expressions": "Expression(G)", "answer_expressions": "{x^2/2-y^2/2=1, y^2/2-x^2/2=1}", "fact_spans": "[[[29, 32]], [[0, 32]], [], [], [[9, 32]], [[9, 32]], [], [], [[18, 32]], [[18, 32]]]", "query_spans": "[[[29, 36]]]", "process": "" }, { "text": "Given that point $P$ is a moving point on the parabola $y^{2}=8x$, $F$ is the focus of the parabola, and the coordinates of point $A$ are $(4,1)$, then the minimum value of $|PA|+|PF|$ is?", "fact_expressions": "G: Parabola;P: Point;A: Point;F: Point;Expression(G) = (y^2 = 8*x);Coordinate(A) = (4, 1);PointOnCurve(P, G);Focus(G) = F", "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))", "answer_expressions": "6", "fact_spans": "[[[7, 21], [29, 32]], [[2, 6]], [[36, 40]], [[25, 28]], [[7, 21]], [[36, 51]], [[2, 24]], [[25, 35]]]", "query_spans": "[[[53, 72]]]", "process": "Solution: From the given conditions, we have F(2,0), and the equation of the directrix is x = -2. Let PM \\bot directrix l, with M being the foot of the perpendicular. By the definition of the parabola, we obtain |PA| + |PF| = |PA| + |PM|. Therefore, when points P, A, and M are collinear, |PA| + |PM| reaches its minimum value, which is |AM| = 4 - (-2) = 6. Hence, the minimum value of |PA| + |PF| is 6." }, { "text": "The standard equation of a parabola whose foci are the intersection points of the line $2x - 3y - 6 = 0$ with the coordinate axes is?", "fact_expressions": "G: Parabola;H: Line;Expression(H) = (2*x - 3*y - 6 = 0);Focus(G)=Intersection(H,axis)", "query_expressions": "Expression(G)", "answer_expressions": "{y^2=12*x,x^2=-8*y}", "fact_spans": "[[[26, 29]], [[3, 18]], [[3, 18]], [[0, 29]]]", "query_spans": "[[[26, 36]]]", "process": "(1) The line 2x-3y-6=0 intersects the x-axis at point (3,0) and intersects the y-axis at point (0,-2). When the focus is at (3,0), \\frac{p}{2}=3, that is, p=6; at this time, the parabola equation is y^{2}=12x. When the focus is at (0,-2), \\frac{p}{2}=2, that is, p=4; at this time, the parabola equation is x^{2}=-8y. Therefore, the required standard equations of the parabola are y^{2}=12x or x^{2}=-8y. This problem examines determining the standard equation of a parabola. In solving it, one must skillfully apply the properties of parabolas and use case analysis to separately compute and solve for cases where the parabola opens to the right and downward." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, let the left vertex, upper vertex, and right focus be $A$, $B$, $F$ respectively. Then $\\overrightarrow{A B} \\cdot \\overrightarrow{A F}$=?", "fact_expressions": "C: Ellipse;A: Point;B: Point;F: Point;Expression(C) = (x^2/4 + y^2/3 = 1);LeftVertex(C) = A;UpperVertex(C) = B;RightFocus(C) = F", "query_expressions": "DotProduct(VectorOf(A, B), VectorOf(A, F))", "answer_expressions": "6", "fact_spans": "[[[2, 44]], [[59, 62]], [[63, 66]], [[67, 70]], [[2, 44]], [[2, 70]], [[2, 70]], [[2, 70]]]", "query_spans": "[[[72, 123]]]", "process": "From the ellipse equation, we know A(-2,0), B(0,\\sqrt{3}), C(1,0), \\overrightarrow{AB}=(2,\\sqrt{3}), \\overrightarrow{AF}=(3,0), so \\overrightarrow{AB}\\cdot\\overrightarrow{AF}=6, therefore fill in 6." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{5}-\\frac{y^{2}}{b^{2}}=1$ ($b>0$), the right vertex of the hyperbola is point $A$. Taking point $A$ as the center and $b$ as the radius, a circle is drawn. This circle intersects one asymptote of the hyperbola $C$ at points $M$ and $N$. If $\\overrightarrow{O M}=\\frac{3}{2} \\overrightarrow{O N}$ (where $O$ is the origin), then what is the standard equation of the hyperbola $C$?", "fact_expressions": "C: Hyperbola;b: Number;A: Point;O: Origin;M: Point;N: Point;b>0;A0: Circle;Expression(C) = (x^2/5 - y^2/b^2 = 1);RightVertex(C) = A;Center(A0) = A;Radius(A0) = b;Intersection(A0, OneOf(Asymptote(C))) = {M, N};VectorOf(O, M) = (3/2)*VectorOf(O, N)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/5 - y^2 = 1", "fact_spans": "[[[2, 54], [88, 94], [180, 186]], [[72, 75]], [[59, 62], [64, 68]], [[169, 172]], [[102, 105]], [[106, 109]], [[10, 54]], [[82, 87], [79, 80]], [[2, 54]], [[2, 62]], [[63, 80]], [[72, 80]], [[82, 111]], [[113, 168]]]", "query_spans": "[[[180, 193]]]", "process": "As shown in the figure, without loss of generality, assume circle A intersects one asymptote of the hyperbola $ y = \\frac{b}{\\sqrt{5}}x $ at points M and N. Draw AB perpendicular from point A to this asymptote, meeting it at point B, and connect AN. First, compute $ |OA| = \\sqrt{5} $, $ AB = \\frac{\\sqrt{5}b}{c} $, $ |OB| = \\frac{5b^{2}}{c} $. Then, from the given condition, we obtain $ \\frac{25b^{4}}{c^{2}} + \\frac{5b^{2}}{c^{2}} = 5 $, solving which gives $ b = 1 $, thus obtaining the standard equation of the hyperbola C. From the equation of hyperbola C: $ \\frac{x^{2}}{5} - \\frac{y^{2}}{b^{2}} = 1 $ ($ b > 0 $), we know $ a = \\sqrt{5} $. Without loss of generality, assume circle A intersects one asymptote of the hyperbola $ y = \\frac{b}{\\sqrt{5}}x $ at points M and N. Draw AB perpendicular from point A to this asymptote, meeting it at point B, and connect AN, as shown in the figure. The distance from point $ A(\\sqrt{5}, 0) $ to the asymptote $ bx - \\sqrt{5}y = 0 $ is $ |AB| = \\frac{|ab|}{\\sqrt{b^{2} + a^{2}}} = \\frac{\\sqrt{5}b}{c} $. Since $ |AN| = r = b $, $ |a| \\cdot \\overrightarrow{OM} = \\frac{3}{2}\\overrightarrow{ON} = \\overrightarrow{ON} + 2\\overrightarrow{NB} $, therefore $ \\overrightarrow{ON} = 4\\overrightarrow{NB} $, hence $ \\overrightarrow{OB} = 5\\overrightarrow{NB} $. In right triangle $ \\triangle ABO $, $ |OA| = \\sqrt{5} $, $ AB = \\frac{\\sqrt{5}b}{c} $, $ |OB| = \\frac{5b^{2}}{c} $, and $ |OB|^{2} + |AB|^{2} = |OA|^{2} $. That is, $ \\frac{25b^{4}}{c^{2}} + \\frac{5b^{2}}{c^{2}} = 5 $, so $ 25b^{4} + 5b^{2} = 5c^{2} $, thus $ 25b^{4} = 5c^{2} - 5b^{2} = 5(c^{2} - b^{2}) = 5a^{2} = 25 $, therefore $ b^{2} = 1 $. Hence, the standard equation of hyperbola C is $ \\frac{x^{2}}{5} - y^{2} = 1 $." }, { "text": "The left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $F(-\\sqrt{5}, 0)$, the coordinates of point $A$ are $(0,2)$, and point $P$ is a moving point on the right branch of the hyperbola such that the minimum perimeter of $\\triangle P A F$ is $8$. Then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;F: Point;P: Point;A: Point;a > b;b > 0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (0, 2);Coordinate(F) = (-sqrt(5), 0);LeftFocus(G) = F;PointOnCurve(P, RightPart(G));Min(Perimeter(TriangleOf(P, A, F))) = 8", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[0, 53], [97, 100], [137, 140]], [[3, 53]], [[3, 53]], [[58, 75]], [[92, 96]], [[76, 80]], [[3, 53]], [[3, 53]], [[0, 53]], [[76, 91]], [[58, 75]], [[0, 75]], [[92, 106]], [[108, 135]]]", "query_spans": "[[[137, 146]]]", "process": "Given |AF| = \\sqrt{5+4} = 3, and the minimum perimeter of triangle APF is 8, we obtain that the minimum value of |PA| + |PF| is 5. Since F is the right focus of the hyperbola, we have |PF| = |PF| + 2a. When points A, P, and F are collinear, |PA| + |PF| reaches its minimum value, which is |AF| = 3. Thus, 3 + 2a = 5, so a = 1, c = \\sqrt{5}, and therefore e = \\frac{c}{a} = \\sqrt{5}. This problem examines the definition, equation, and properties of hyperbolas, primarily the method for finding eccentricity. It tests the property that the minimum value occurs when three points are collinear, as well as equation-solving skills and computational ability, and is of medium difficulty." }, { "text": "The equation $k x^{2}+4 y^{2}=4 k$ represents an ellipse with foci on the $x$-axis. Then, the range of real values for $k$ is?", "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (k*x^2 + 4*y^2 = 4*k);PointOnCurve(Focus(G), xAxis)", "query_expressions": "Range(k)", "answer_expressions": "(0, 4)", "fact_spans": "[[[34, 36]], [[38, 43]], [[0, 36]], [[25, 36]]]", "query_spans": "[[[38, 50]]]", "process": "When k=0, it does not satisfy the condition; when k≠0, from kx^{2}+4y^{2}=4k we obtain \\frac{x^{2}}{4}+\\frac{y^{2}}{k}=1. Since the equation kx^{2}+4y^{2}=4k represents an ellipse with foci on the x-axis, it follows that 0 0. Solving yields k < \\frac{1}{3} and k \\neq 0. Also, x_{1} + x_{2} = \\frac{4 - 6k}{k^{2}}, so the midpoint of AB is \\left(\\frac{2 - 3k}{k^{2}}, \\frac{2}{k}\\right). The perpendicular bisector of segment AB passes through the point (4,0) and has equation y = -\\frac{1}{k}(x - 4). Since it passes through the midpoint \\left(\\frac{2 - 3k}{k^{2}}, \\frac{2}{k}\\right), we substitute: \\frac{2}{k} = -\\frac{1}{k}\\left(\\frac{2 - 3k}{k^{2}} - 4\\right), which simplifies to 2k^{2} + 3k - 2 = 0. Solving gives k = -2 or k = \\frac{1}{2} (discarded). Thus, H is at (2, -1), |H'H'| = 2 + 1 = 3, and therefore |AF| + |BF| = |AA'| + |BB'| = 2|H'H'| = 6." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, and $P$ is a point on the ellipse such that $\\angle F_{1} P F_{2}=60^{\\circ}$, then the minimum value of the eccentricity is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;P: Point;F2: Point;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree)", "query_expressions": "Min(Eccentricity(G))", "answer_expressions": "1/2", "fact_spans": "[[[18, 63], [72, 74]], [[20, 63]], [[20, 63]], [[2, 9]], [[68, 71]], [[10, 17]], [[18, 63]], [[2, 67]], [[68, 77]], [[79, 112]]]", "query_spans": "[[[72, 123]]]", "process": "" }, { "text": "The hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ has a focus at $F(3,0)$, then $m=?$", "fact_expressions": "G: Hyperbola;m: Number;F: Point;Expression(G) = (-y^2 + x^2/m = 1);Coordinate(F) = (3, 0);OneOf(Focus(G)) = F", "query_expressions": "m", "answer_expressions": "8", "fact_spans": "[[[0, 28]], [[43, 46]], [[33, 41]], [[0, 28]], [[33, 41]], [[0, 41]]]", "query_spans": "[[[43, 48]]]", "process": "Test analysis: From the focus F(3,0), we know that c=3, therefore m+1=c^{2}=9, thus m=8" }, { "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{b^{2}}=1$ is $e=\\sqrt{2}$, then the equations of the asymptotes of the hyperbola are?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/b^2 = 1);b: Number;e: Number;Eccentricity(G) = e;e = sqrt(2)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*x", "fact_spans": "[[[2, 45], [63, 66]], [[2, 45]], [[5, 45]], [[49, 61]], [[2, 61]], [[49, 61]]]", "query_spans": "[[[63, 74]]]", "process": "Since $ e = \\frac{c}{a} = \\sqrt{2} $, we have $ c = \\sqrt{2}a $, that is, $ c^{2} = 2a^{2} $. Also, for a hyperbola, $ c^{2} = a^{2} + b^{2} $. Substituting gives $ a^{2} = b^{2} $, that is, $ a = b $. Therefore, the asymptotes are given by $ y = \\pm\\frac{b}{a}x $, so $ y = \\pm x $." }, { "text": "Given the line $y = a x + 1$ intersects the hyperbola $3 x^{2} - y^{2} = 1$ at points $A$ and $B$, then the range of values for $a$ is?", "fact_expressions": "G: Hyperbola;H: Line;a: Number;A: Point;B: Point;Expression(G) = (3*x^2 - y^2 = 1);Expression(H) = (y = a*x + 1);Intersection(H, G) = {A, B}", "query_expressions": "Range(a)", "answer_expressions": "(-sqrt(6),sqrt(6))&Negation(a=pm*sqrt(3))", "fact_spans": "[[[14, 34]], [[2, 13]], [[47, 50]], [[36, 39]], [[40, 43]], [[14, 34]], [[2, 13]], [[2, 45]]]", "query_spans": "[[[47, 57]]]", "process": "From \\begin{cases}y=ax+1\\\\3x^2-y^2=1\\end{cases}, we obtain (3-a^{2})x^{2}-2ax-2=0. Since the line intersects the hyperbola at two points, we have \\begin{cases}3-a^2\\neq0\\\\\\Delta=4a^2-4(3-a^2)\\times(-2)>0\\end{cases}. Solving gives: -\\sqrt{6}2." }, { "text": "It is known that the sine of the inclination angle of line $l$ (with a slope greater than $0$) is $\\frac{\\sqrt{2}}{2}$, and its intercept on the $x$-axis is $-2$. Line $l$ intersects the parabola $C$: $x^{2}=2 p y(p>0)$ at points $A$ and $B$. If $|A B|=16$, then $p=$?", "fact_expressions": "l: Line;Slope(l) > 0;Sin(Inclination(l)) = sqrt(2)/2;Intercept(l, xAxis) = -2;Intersection(l, C) = {A, B};A: Point;B: Point;C: Parabola;Expression(C) = (x^2 = 2*p*y);p: Number;p>0;Abs(LineSegmentOf(A, B)) = 16", "query_expressions": "p", "answer_expressions": "4", "fact_spans": "[[[2, 7], [61, 66]], [[2, 16]], [[2, 45]], [[2, 60]], [[61, 104]], [[95, 98]], [[99, 102]], [[67, 93]], [[67, 93]], [[117, 120]], [[75, 93]], [[105, 115]]]", "query_spans": "[[[117, 122]]]", "process": "According to the problem, the inclination angle of line l is 45^{\\circ}, the slope k=1, and the equation of line l is: y=x+2. From \\begin{cases}y=x+2\\\\x^{2}=2py\\end{cases}, we obtain x^{2}-2px-4p=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), then x_{1}+x_{2}=2p, x_{1}x_{2}=-4p. Thus, |AB|=\\sqrt{1+1^{2}}|x_{1}-x_{2}|=\\sqrt{2}\\cdot\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\sqrt{2}\\cdot\\sqrt{4p^{2}+16p}=16, which implies p^{2}+4p-32=0. Since p>0, solving gives p=4." }, { "text": "Given the ellipse $\\frac{x^{2}}{4+m^{2}}+\\frac{y^{2}}{m^{2}}=1$ and the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ share the same foci, and the eccentricity of the hyperbola is $2$, then the equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Ellipse;m: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2/(m^2 + 4) + y^2/m^2 = 1);Focus(H) = Focus(G);Eccentricity(G) = 2", "query_expressions": "Expression(G)", "answer_expressions": "x^2-y^2/3=1", "fact_spans": "[[[50, 96], [104, 107], [118, 121]], [[53, 96]], [[53, 96]], [[2, 49]], [[4, 49]], [[50, 96]], [[2, 49]], [[2, 102]], [[104, 115]]]", "query_spans": "[[[118, 126]]]", "process": "Since the ellipse \\frac{x^{2}}{4+m^{2}}+\\frac{y^{2}}{m^{2}}=1 and the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 have common foci, from (14+m^{2})-m^{2}=4, we obtain a^2+b^{2}=4, that is, c=2. Because the eccentricity of the hyperbola is \\frac{c}{a}=2=\\frac{2}{a}, \\therefore a=1, then b=\\sqrt{3}. Therefore, the equation of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 is x^{2}-\\frac{y^{2}}{3}=1." }, { "text": "Given that hyperbola $C$ shares the same asymptotes as the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1$, then the eccentricity of hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;G:Hyperbola;Expression(G)=(x^2/9-y^2/4=1);Asymptote(C)=Asymptote(G)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(13)/3, sqrt(13)/2", "fact_spans": "[[[2, 8], [56, 62]], [[9, 47]], [[9, 47]], [[2, 54]]]", "query_spans": "[[[56, 68]]]", "process": "It is easy to know that the asymptotes of the hyperbola \\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1 are given by y=\\pm\\frac{2}{3}x. Therefore, if the foci of hyperbola C lie on the x-axis, then \\frac{b}{a}=\\frac{2}{3}, so the eccentricity is e=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{1+\\frac{4}{9}}=\\frac{\\sqrt{13}}{3}" }, { "text": "For any real number $k$, the line $y=kx+1$ always intersects the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{n}=1$ at two points; then what is the range of values for $n$?", "fact_expressions": "G: Ellipse;n: Number;H: Line;k: Real;Expression(G) = (x^2/4 + y^2/n = 1);Expression(H) = (y = k*x + 1);NumIntersection(H, G) = 2", "query_expressions": "Range(n)", "answer_expressions": "{(1,4)+(4,oo)}", "fact_spans": "[[[22, 59]], [[67, 70]], [[10, 21]], [[4, 9]], [[22, 59]], [[10, 21]], [[10, 65]]]", "query_spans": "[[[67, 76]]]", "process": "" }, { "text": "Given the ellipse $\\Gamma$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the line $x+y=1$ intersects the ellipse $\\Gamma$ at points $M$ and $N$. The circle with segment $MN$ as diameter passes through the origin. If the eccentricity of the ellipse $\\Gamma$ is at most $\\frac{\\sqrt{3}}{2}$, then what is the range of values for $a$?", "fact_expressions": "Gamma: Ellipse;H: Circle;I: Line;N: Point;M: Point;O:Origin;Expression(I) = (x + y = 1);Expression(Gamma) = (y^2/b^2 + x^2/a^2 = 1);Intersection(I, Gamma) = {M, N};IsDiameter(LineSegmentOf(M,N),H);PointOnCurve(O,H);Negation(Eccentricity(Gamma)>sqrt(3)/2);a:Number;b:Number;a>b;b>0", "query_expressions": "Range(a)", "answer_expressions": "(1,sqrt(10)/2]", "fact_spans": "[[[3, 64], [75, 85], [117, 127]], [[109, 110]], [[65, 74]], [[91, 94]], [[87, 90]], [[112, 114]], [[65, 74]], [[2, 64]], [[65, 96]], [[97, 110]], [[109, 114]], [[117, 154]], [[156, 159]], [[14, 64]], [[14, 64]], [[14, 64]]]", "query_spans": "[[[156, 166]]]", "process": "Let M(x_{1},y_{1}), N(x_{2},y_{2}). By solving simultaneously the line MN and the ellipse equation \n\\begin{cases}\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1\\\\x+y=1\\end{cases}, \neliminating y gives \n(a^{2}+b^{2})x^{2}-2a^{2}x+a^{2}-a^{2}b^{2}=0, \nx_{1}\\cdot x_{2}=\\frac{a^{2}-a^{2}b^{2}}{a^{2}+b^{2}}, and since a>0, simplifying yields a^{2}+b^{2}>1. \nSolving simultaneously the line MN and the ellipse equation \n\\begin{cases}\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1\\\\x+y=1\\end{cases}, \neliminating x gives \n(a^{2}+b^{2})y^{2}-2b^{2}y+b^{2}-a^{2}b^{2}=0, \nthus y_{1}\\cdot y_{2}=\\frac{b^{2}-a^{2}b^{2}}{a^{2}+b^{2}}. \nSince the circle with diameter MN passes through the origin, \n\\overrightarrow{OM}\\cdot\\overrightarrow{ON}=0, hence x_{1}\\cdot x_{2}+y_{1}\\cdot y_{2}=0, \ni.e., \\frac{a^{2}-a^{2}b^{2}}{a^{2}+b^{2}}+\\frac{b^{2}-a^{2}b^{2}}{a^{2}+b^{2}}=0, \nsimplifying gives b^{2}=\\frac{a^{2}}{2a^{2}-1}. \nAlso, the eccentricity of the ellipse e=\\sqrt{1-\\frac{b^{2}}{a^{2}}} is not greater than \\frac{\\sqrt{3}}{2}, \nhence \\sqrt{1-\\frac{b^{2}}{a^{2}}}\\leqslant\\frac{\\sqrt{3}}{2}, \ni.e., a^{2}>b^{2}\\geqslant\\frac{1}{4}a^{2}, \nso a^{2}>\\frac{a^{2}}{2a^{2}-1}\\geqslant\\frac{1}{4}a^{2} \\Rightarrow 1<2a^{2}-1\\leqslant4 \\Rightarrow 10, b>0)$, its asymptotes are tangent to the circle $M$: $(x-2)^{2}+y^{2}=3$. What is the eccentricity $e$ of this hyperbola?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;M: Circle;e: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(M) = (y^2 + (x - 2)^2 = 3);IsTangent(Asymptote(C),M);Eccentricity(C) = e", "query_expressions": "e", "answer_expressions": "2", "fact_spans": "[[[2, 63], [96, 99]], [[10, 63]], [[10, 63]], [[68, 92]], [[103, 106]], [[10, 63]], [[10, 63]], [[2, 63]], [[68, 92]], [[2, 94]], [[96, 106]]]", "query_spans": "[[[103, 108]]]", "process": "From the problem, we know that one of the asymptotes of the hyperbola has equation $ y = \\frac{b}{a}x $. Since it is tangent to the circle $ M: (x-2)^{2} + y^{2} = 3 $, we have: \n$ \\sqrt{3} = \\frac{|2b|}{\\sqrt{a^{2} + b^{2}}} $. \nSolving this gives $ \\frac{b}{a} = \\sqrt{3} $, then the eccentricity $ e = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = 2 $." }, { "text": "The coordinates of the foci of the hyperbola $C$: $3 x^{2}-4 y^{2}=12$ are?", "fact_expressions": "C: Hyperbola;Expression(C) = (3*x^2 - 4*y^2 = 12)", "query_expressions": "Coordinate(Focus(C))", "answer_expressions": "(pm*sqrt(7), 0)", "fact_spans": "[[[0, 28]], [[0, 28]]]", "query_spans": "[[[0, 35]]]", "process": "Simplifying gives \\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1, so a^{2}=4, b^{2}=3, c^{2}=a^{2}+b^{2}=7, hence the coordinates of the foci are: (\\pm\\sqrt{7},0)" }, { "text": "Given that the eccentricity of the hyperbola $x^{2}+m y^{2}=1$ is $2$, then the value of the real number $m$ is?", "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (m*y^2 + x^2 = 1);Eccentricity(G) = 2", "query_expressions": "m", "answer_expressions": "-1/3", "fact_spans": "[[[2, 22]], [[32, 37]], [[2, 22]], [[2, 30]]]", "query_spans": "[[[32, 41]]]", "process": "a=1, e=\\frac{c}{a}=2, so c=2, from b^{2}=c^{2}-a^{2}, we get -\\frac{1}{m}=3, m=-\\frac{1}{3}, hence fill in -\\frac{1}{3}." }, { "text": "Given the hyperbola $C$: $\\frac{y^{2}}{16}-\\frac{x^{2}}{9}=1$ has its right focus at point $F$, then the distance from point $F$ to the asymptote is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-x^2/9 + y^2/16 = 1);F: Point;RightFocus(C) = F", "query_expressions": "Distance(F, Asymptote(C))", "answer_expressions": "3", "fact_spans": "[[[2, 46]], [[2, 46]], [[51, 54], [56, 60]], [[2, 54]]]", "query_spans": "[[[2, 69]]]", "process": "The hyperbola $ C: \\frac{y^{2}}{16}-\\frac{x^{2}}{9}=1 $ has its foci on the y-axis, so the right focus is $ F(0,5) $. The asymptotes are $ 3y \\pm 4x = 0 $. The distance from point $ F $ to the asymptote is $ \\frac{15}{5} = 3 $. The answer is: $ 3 $." }, { "text": "Given that the major axis $AB$ of an ellipse has length $4$, and $N$ is a point on the ellipse such that $|NA|=1$, $\\angle NAB=60^{\\circ}$, then what is the eccentricity of the ellipse?", "fact_expressions": "G: Ellipse;MajorAxis(G) = LineSegmentOf(A, B);Length(LineSegmentOf(A, B)) = 4;A: Point;B: Point;N: Point;PointOnCurve(N, G) ;Abs(LineSegmentOf(N, A)) = 1;TriangleOf(N, A, B) = ApplyUnit(60, degree)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2*sqrt(7)/7", "fact_spans": "[[[2, 4], [24, 26], [70, 72]], [[2, 11]], [[6, 17]], [[6, 11]], [[6, 11]], [[19, 23]], [[20, 29]], [[32, 41]], [[43, 68]]]", "query_spans": "[[[70, 78]]]", "process": "" }, { "text": "Given that the distance from point $M$ on the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ to the right focus is $2$, then what is the distance from point $M$ to the left directrix?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);M: Point;PointOnCurve(M, C);Distance(M, RightFocus(C)) = 2", "query_expressions": "Distance(M, LeftDirectrix(C))", "answer_expressions": "4", "fact_spans": "[[[2, 44]], [[2, 44]], [[46, 50], [63, 67]], [[2, 50]], [[2, 61]]]", "query_spans": "[[[2, 76]]]", "process": "Since the distance from point M on the ellipse $ C: \\frac{x^2}{4} + \\frac{y^{2}}{3} = 1 $ to the right focus is 2, the distance from M to the left focus is $ 4 - 2 = 2 $, so the x-coordinate of M is 0, that is, the distance from point M to the left directrix $ l: x = -4 $ is 4." }, { "text": "Through the focus of the parabola $y^{2}=4 x$, draw two mutually perpendicular chords $A B$ and $C D$. Then $\\frac{1}{|A B|}+\\frac{1}{|C D|}$=?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;B: Point;C: Point;D: Point;PointOnCurve(Focus(G), LineSegmentOf(A, B));PointOnCurve(Focus(G), LineSegmentOf(C, D));IsPerpendicular(LineSegmentOf(A, B), LineSegmentOf(C, D));IsChordOf(LineSegmentOf(A, B), G);IsChordOf(LineSegmentOf(C, D), G)", "query_expressions": "1/Abs(LineSegmentOf(C, D)) + 1/Abs(LineSegmentOf(A, B))", "answer_expressions": "1/4", "fact_spans": "[[[1, 15]], [[1, 15]], [[27, 32]], [[27, 32]], [[34, 39]], [[34, 39]], [[0, 39]], [[0, 39]], [[1, 39]], [[1, 32]], [[1, 39]]]", "query_spans": "[[[41, 76]]]", "process": "From the given conditions, we know: the focus of the parabola is F(1,0), the equation of the directrix is x=-1, and the slopes of the two lines must exist. Let l_{AB}: y=k(x-1) with k≠0, then l_{CD}: y=-\\frac{1}{k}(x-1). Solving simultaneously:\n\\begin{cases}y^{2}=4x\\\\y=k(x-1)\\end{cases}\nwe obtain: k^{2}x^{2}-(2k^{2}+4)x+k^{2}=0. Then \\triangle=(2k^{2}+4)^{2}-4k^{4}>0'=x_{A}+x_{B}+2=\\frac{2k^{2}+4}{k^{2}}+2=4+\\frac{4}{k^{2}}. Similarly, |CD|=4+4k^{2}-|=\\frac{k^{2}}{4k^{2}+4}\\frac{1}{4k^{2}}=\\frac{1}{4}" }, { "text": "Given points $A$ and $B$ are the left and right vertices of the ellipse $C$: $\\frac{x^{2}}{4}+y^{2}=1$, point $M$ is a point on the $x$-axis, a line perpendicular to the $x$-axis through $M$ intersects the ellipse $C$ at points $P$ and $Q$, a line through $M$ perpendicular to $AP$ intersects $BQ$ at point $N$, then $\\frac{S_\\Delta{ B M N}}{S_\\Delta { B M Q}}$=?", "fact_expressions": "C: Ellipse;A: Point;P: Point;B: Point;Q: Point;M: Point;N: Point;Expression(C) = (x^2/4 + y^2 = 1);LeftVertex(C) = A;RightVertex(C) = B;PointOnCurve(M, xAxis);l:Line;PointOnCurve(M,l);Intersection(l, C) = {P, Q};l1:Line;PointOnCurve(M, l1);IsPerpendicular(l1,LineSegmentOf(A, P));Intersection(l1, LineSegmentOf(B, Q)) = N;IsPerpendicular(l,xAxis)", "query_expressions": "Area(TriangleOf(B,M,N))/Area(TriangleOf(B,M,Q))", "answer_expressions": "4/5", "fact_spans": "[[[11, 43], [75, 80]], [[2, 6]], [[81, 84]], [[7, 10]], [[85, 88]], [[49, 53], [92, 95], [63, 66]], [[111, 115]], [[11, 43]], [[2, 48]], [[2, 48]], [[49, 61]], [], [[62, 74]], [[62, 90]], [], [[91, 104]], [[91, 104]], [[91, 115]], [[62, 74]]]", "query_spans": "[[[117, 164]]]", "process": "Let P(m,n). Then M(m,0), Q(m,-n). From the given conditions, we know m\\neq\\pm2 and n\\neq0. The slope of line AP is k_{AP}=\\frac{n}{m+2}, the slope of line MN is k_{MN}=-\\frac{m+2}{n}. Therefore, the equation of line MN is y=-\\frac{m+2}{n}(x-m), and the equation of line BQ is y=\\frac{n}{2-m}(x-2). Solving the system \n\\begin{cases}\ny=-\\frac{m+2}{n}(x-m)\\\\\ny=\\frac{n}{2-m}(x-2)\n\\end{cases}, \nwe obtain y_{N}=-\\frac{n(4-m^{2})}{4-m^{2}+n^{2}}. Since point P lies on the ellipse C, we have 4-m^{2}=4n^{2}, thus y_{N}=-\\frac{4}{5}n. Also, S_{\\DeltaBMN}=\\frac{1}{2}|BM|\\cdot|y_{N}|=\\frac{2}{5}|BM|\\cdot|n|, S_{\\DeltaBMQ}=\\frac{1}{2}|BM|\\cdot|n|. Therefore, \\frac{S_{\\DeltaBMN}}{S_{ABMO}}=\\frac{4}{5}S_{\\DeltaBMQ}." }, { "text": "The standard equation of a parabola with focus at $(0,-3)$ is?", "fact_expressions": "G: Parabola;P:Point;Coordinate(P)=(0,-3);Focus(G)=P", "query_expressions": "Expression(G)", "answer_expressions": "x^2 = -12 * y", "fact_spans": "[[[12, 15]], [[0, 11]], [[0, 11]], [[0, 15]]]", "query_spans": "[[[12, 22]]]", "process": "Since the focus of the parabola is (0, -3), the parabola opens downward, and \\frac{p}{2}=3, hence 2p=12. The equation of the parabola is x^{2}=-12y. [Solution] This question mainly examines the geometric properties of a parabola and the method for finding the standard equation of a parabola. It is a basic-level problem." }, { "text": "If one of the asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{m}=1$ is given by $y=\\sqrt{3} x$, then what is the value of $m$?", "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2 - y^2/m = 1);Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x)", "query_expressions": "m", "answer_expressions": "3", "fact_spans": "[[[1, 29]], [[54, 57]], [[1, 29]], [[1, 52]]]", "query_spans": "[[[54, 60]]]", "process": "The asymptotes of the hyperbola \\( x^{2} - \\frac{y^{2}}{m} = 1 \\) are given by \\( y = \\pm\\sqrt{m}x \\). It is also known that one asymptote is \\( y = \\sqrt{3}x \\), so \\( \\sqrt{m} = \\sqrt{3} \\), hence \\( m = 3 \\)." }, { "text": "Let the line with slope $2$ pass through the focus $F$ of the parabola $y^{2}=a x(a>0)$, and intersect the $y$-axis at point $A$. If the area of $\\triangle OAF$ ($O$ being the origin) is $4$, then the equation of the parabola is?", "fact_expressions": "H: Line;Slope(H) = 2;G: Parabola;Expression(G) = (y^2 = a*x);a: Number;a>0;F: Point;Focus(G) = F;PointOnCurve(F, H) = True;Intersection(H, yAxis) = A;O: Origin;A: Point;Area(TriangleOf(O, A, F)) = 4", "query_expressions": "Expression(G)", "answer_expressions": "y^2=8*x", "fact_spans": "[[[8, 10]], [[1, 10]], [[11, 30], [86, 89]], [[11, 30]], [[14, 30]], [[14, 30]], [[33, 36]], [[11, 36]], [[8, 36]], [[8, 49]], [[67, 70]], [[45, 49]], [[51, 84]]]", "query_spans": "[[[86, 93]]]", "process": "" }, { "text": "If the center is at the origin, and the axis of symmetry is the coordinate axis, one focus of the ellipse is $(2,0)$, and the length of the major axis is $\\sqrt{3}$ times the length of the minor axis, then the standard equation is?", "fact_expressions": "Center(G) = O;O: Origin;SymmetryAxis(G) = axis;G: Ellipse;Length(MajorAxis(G)) = Length(MinorAxis(G))*sqrt(3);Coordinate(OneOf(Focus(G)))=(2,0)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/6+y^2/2=1", "fact_spans": "[[[1, 17]], [[4, 6]], [[7, 17]], [[15, 17]], [[15, 51]], [15, 29]]", "query_spans": "[[[15, 59]]]", "process": "From the given \n\\begin{cases}c=2\\\\2a=2\\sqrt{3}b\\\\a^{2}=b^{2}+c^{2}\\end{cases}, \nsolving yields $ b^{2}=2 $, $ a^{2}=6 $. Hence, the standard equation is $ \\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1 $." }, { "text": "The line $ l $: $ y = 2x + m $ is tangent to the parabola $ y = x^2 $ at point $ A $. The intersection point of $ l $ and the $ y $-axis is $ B $, and $ O $ is the origin. Then $ \\overrightarrow{OA} \\cdot \\overrightarrow{AB} $ = ?", "fact_expressions": "l: Line;G: Parabola;O: Origin;A: Point;B: Point;Expression(G) = (y = x^2);Expression(l) = (y=2*x+m);TangentPoint(l,G)=A;Intersection(l, yAxis) = B;m: Number", "query_expressions": "DotProduct(VectorOf(O, A), VectorOf(A, B))", "answer_expressions": "-3", "fact_spans": "[[[0, 16], [38, 41]], [[17, 29]], [[55, 58]], [[31, 36]], [[50, 53]], [[17, 29]], [[0, 16]], [[0, 36]], [[37, 53]], [[0, 16]]]", "query_spans": "[[[63, 114]]]", "process": "Differentiate the line, let the tangent point be (x,y), y=2x=2\\Rightarrow x=1, y=1, substitute the tangent point into the line to get m=-1, L: y=2x-1, B(0,-1), A(1,1), \\overrightarrow{OA}\\cdot\\overrightarrow{AB}=(1,1)\\cdot(-1,-2)=-3." }, { "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/5 = 1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "3/2", "fact_spans": "[[[0, 38]], [[0, 38]]]", "query_spans": "[[[0, 44]]]", "process": "Problem Analysis: According to the given equation, obtain the values of a, b, c, and then find the eccentricity of the hyperbola. \nSolution: Since the equation of the hyperbola is \\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1, we have a^{2}=4, a=2, b^{2}=5, so c^{2}=9, c=3, thus the eccentricity e=\\frac{3}{2}." }, { "text": "If the ellipse is $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$, then what is the range of real values for $x$?", "fact_expressions": "G: Ellipse;x: Real;Expression(G) = (x^2/5 + y^2/4 = 1)", "query_expressions": "Range(x)", "answer_expressions": "[-sqrt(5),sqrt(5)]", "fact_spans": "[[[1, 38]], [[40, 45]], [[1, 38]]]", "query_spans": "[[[40, 52]]]", "process": "Test analysis: From the equation, it is known that $0\\leqslant\\frac{x^{2}}{5}\\leqslant1$, therefore $0\\leqslant x^{2}\\leqslant5$, thus $-\\sqrt{5}\\leqslant x\\leqslant\\sqrt{5}$." }, { "text": "Given that the midpoint coordinates of two points $A$ and $B$ on the asymptote of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ are $(2 , 2)$, what is the slope of the line $AB$?", "fact_expressions": "G: Hyperbola;B: Point;A: Point;Expression(G) = (x^2/4 - y^2 = 1);PointOnCurve(A,Asymptote(G));PointOnCurve(B,Asymptote(G));Coordinate(MidPoint(LineSegmentOf(A,B)))=(2,2)", "query_expressions": "Slope(LineOf(A,B))", "answer_expressions": "1/4", "fact_spans": "[[[2, 30]], [[41, 44]], [[37, 40]], [[2, 30]], [[2, 44]], [[2, 44]], [[37, 59]]]", "query_spans": "[[[61, 73]]]", "process": "Hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, $a=2$, $b=1$, the asymptotes are $y=\\pm\\frac{1}{2}x$. Let the equation of line $AB$ be $y-2=k(x-2)$, $k\\neq\\pm\\frac{1}{2}$. From\n$$\n\\begin{cases}\ny-2=k(x-2)\\\\\ny=\\frac{1}{2}x\\\\\ny-2=k(x-2)\\\\\ny=-\\frac{1}{1+2k}\n\\end{cases}\n$$\nwe get $\\frac{4-4k}{1-2k}+\\frac{4k-4}{1+2k}=2\\times2=4 \\Rightarrow k=\\frac{1}{4}$. Therefore, the slope of line $AB$ is $\\frac{1}{4}$." }, { "text": "The standard equation of the parabola with focus at the left focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ is?", "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (x^2/4 - y^2/5 = 1);LeftFocus(G)=Focus(H)", "query_expressions": "Expression(H)", "answer_expressions": "y^2=-12*x", "fact_spans": "[[[1, 39]], [[47, 50]], [[1, 39]], [[0, 50]]]", "query_spans": "[[[47, 56]]]", "process": "Let the equation of the parabola be $ y^{2} = -2px $, with focus at $ \\left(-\\frac{p}{2}, 0 \\right) $. Since the left focus of the hyperbola $ \\frac{x^{2}}{4} - \\frac{y^{2}}{5} = 1 $ is $ (-3, 0) $, we have $ -\\frac{p}{2} = -3 $, so $ p = 6 $. Therefore, the equation of the parabola is $ y^2 = -12x $." }, { "text": "$P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, $F_{1}$ and $F_{2}$ are the left and right foci respectively, then the horizontal coordinate of the incenter of $\\triangle P F_{1} F_{2}$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);P: Point;PointOnCurve(P, RightPart(G));F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2", "query_expressions": "XCoordinate(Center(InscribedCircle(TriangleOf(P, F1, F2))))", "answer_expressions": "3", "fact_spans": "[[[4, 43]], [[4, 43]], [[0, 3]], [[0, 49]], [[50, 57]], [[58, 65]], [[4, 73]], [[4, 73]]]", "query_spans": "[[[75, 111]]]", "process": "" }, { "text": "If the vertex of the parabola is $(0 , 0)$, the axis of symmetry is the $x$-axis, and the focus lies on $3 x-4 y-12=0$, then the equation of the parabola is?", "fact_expressions": "G: Parabola;L:Curve;H: Point;Coordinate(H) = (0, 0);Vertex(G)=H;SymmetryAxis(G)=xAxis;Expression(L)=(3*x - 4*y - 12 = 0);PointOnCurve(Focus(G), L)", "query_expressions": "Expression(G)", "answer_expressions": "y^2=16*x", "fact_spans": "[[[1, 4], [47, 50]], [[29, 43]], [[7, 16]], [[7, 16]], [[1, 16]], [[1, 25]], [[29, 43]], [[1, 44]]]", "query_spans": "[[[47, 55]]]", "process": "" }, { "text": "Given that the line $x-3 y+1=0$ intersects the ellipse $\\Gamma$: $\\frac{x^{2}}{2}+y^{2}=1$ at points $A$ and $B$, if there exists a point $C$ on the ellipse such that $\\angle A C B=90^{\\circ}$, then what are the coordinates of point $C$?", "fact_expressions": "Gamma: Ellipse;H: Line;A: Point;C: Point;B: Point;Expression(H) = (x - 3*y + 1 = 0);Expression(Gamma)=(x^2/2+y^2=1);Intersection(H,Gamma)={A,B};PointOnCurve(C,Gamma);AngleOf(A, C, B) = ApplyUnit(90, degree)", "query_expressions": "Coordinate(C)", "answer_expressions": "{(0,-1),(-12/11,-7/11)}", "fact_spans": "[[[16, 52], [67, 69]], [[2, 15]], [[56, 59]], [[72, 76], [106, 110]], [[60, 63]], [[2, 15]], [[16, 52]], [[2, 65]], [[67, 76]], [[79, 104]]]", "query_spans": "[[[106, 115]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). From \\begin{cases}x-3y+1=0\\\\\\frac{x^{2}}{2}+y^{2}=1\\end{cases}, eliminating and simplifying yields 11y^{2}-6y-1=0, 4>0. Then y_{1}+y_{2}=\\frac{6}{11}, y_{1}y_{2}=-\\frac{1}{11}. Therefore, x_{1}+x_{2}=3(y_{1}+y_{2})-2=-\\frac{4}{11}, |AB|=\\sqrt{1+9}\\cdot\\sqrt{(y_{1}+y_{2})^{2}}-4y_{1}y_{2}=20\\sqrt{2}. Also, \\angle ACB=90^{\\circ}, so point C lies on the circle with AB as diameter (not coinciding with A or B), i.e., point C lies on the circle (x+\\frac{2}{11})^{2}+(y-\\frac{3}{11})^{2}=\\frac{200}{121}. From \\begin{cases}(x+\\frac{2}{11})^{2}+(y-\\frac{3}{11})^{2}=\\frac{200}{121}\\\\\\frac{x^{2}}{2}+y^{2}=1\\\\x-3y+1\\neq0\\end{cases}, we obtain \\begin{cases}x=0\\\\y=-1\\end{cases} or \\begin{cases}x=-\\frac{12}{11}\\\\y=-\\frac{7}{11}\\end{cases}." }, { "text": "Given that the vertex of the parabola is at the origin, the axis of symmetry is a coordinate axis, and the focus lies on the line $x-2 y-4=0$, then the equation of the parabola is?", "fact_expressions": "G: Parabola;H: Line;O: Origin;Expression(H) = (x - 2*y - 4 = 0);Vertex(G) = O;PointOnCurve(Focus(G), H);SymmetryAxis(G)=axis", "query_expressions": "Expression(G)", "answer_expressions": "{x^2=-8*y,y^2=16*x}", "fact_spans": "[[[2, 5], [39, 42]], [[23, 36]], [[9, 11]], [[23, 36]], [[2, 11]], [[2, 37]], [[2, 19]]]", "query_spans": "[[[39, 47]]]", "process": "The focus is (4,0) or (0,-2) according to the problem, and the corresponding equations of the parabola are y^{2}=16x or x^{2}=-83" }, { "text": "The standard equation of the hyperbola that shares foci with the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{5}=1$ and passes through the point $(-3 \\sqrt{2}, 0)$ is?", "fact_expressions": "H: Ellipse;Expression(H) = (x^2/25 + y^2/5 = 1);G: Hyperbola;Focus(H) = Focus(G);I: Point;Coordinate(I) = (-3*sqrt(2), 0);PointOnCurve(I, G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/18-y^2/2=1", "fact_spans": "[[[1, 39]], [[1, 39]], [[64, 67]], [[0, 67]], [[44, 63]], [[44, 63]], [[43, 67]]]", "query_spans": "[[[64, 74]]]", "process": "The ellipse $\\frac{x^2}{25}+\\frac{y^{2}}{5}=1$, its foci are $F_{1}(-2\\sqrt{5},0)$, $F_{2}(2\\sqrt{5},0)$. Let the required hyperbola equation be: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, then $a^{2}+b^{2}=20$. Since the hyperbola passes through the point $(-3\\sqrt{2},0)$, then $a=3\\sqrt{2}$, so $a^2=18$, $b^{2}=2$. Therefore, the required hyperbola equation is: $\\frac{x^{2}}{18}-\\frac{y^{2}}{2}=1$." }, { "text": "The equation of the trajectory of a point whose distance to the point $(4,0)$ is equal to its distance to the $y$-axis is?", "fact_expressions": "G: Point;P:Point;Coordinate(G) = (4, 0);Distance(P,G)=Distance(P,yAxis)", "query_expressions": "LocusEquation(P)", "answer_expressions": "y^2-8*x+16=0", "fact_spans": "[[[4, 12]], [[2, 3], [17, 18], [30, 31]], [[4, 12]], [[2, 26]]]", "query_spans": "[[[30, 38]]]", "process": "" }, { "text": "The line with slope $\\frac{\\sqrt{2}}{2}$ intersects the ellipse $x^{2}+\\frac{y^{2}}{b^{2}}=1(b>0)$, which has foci on the $x$-axis, at two distinct points $P$ and $Q$. If the projections of points $P$ and $Q$ onto the $x$-axis are exactly the two foci of the ellipse, then what is the focal distance of the ellipse?", "fact_expressions": "L: Line;Slope(L) = sqrt(2)/2;G: Ellipse;Expression(G) = (x^2 + y^2/b^2 = 1);b: Number;b>0;PointOnCurve(Focus(G), xAxis);P: Point;Q: Point;Negation(P=Q);Intersection(L, G) = {P, Q};F1: Point;F2: Point;Focus(G) = {F1, F2};Projection(P, xAxis) = F1;Projection(Q, xAxis) = F2", "query_expressions": "FocalLength(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[24, 26]], [[0, 26]], [[36, 72], [109, 111], [118, 120]], [[36, 72]], [[38, 72]], [[38, 72]], [[27, 72]], [[79, 82], [89, 93]], [[83, 86], [94, 97]], [[74, 86]], [[24, 86]], [], [], [[109, 115]], [[89, 115]], [[89, 115]]]", "query_spans": "[[[118, 125]]]", "process": "Problem Analysis: According to the problem, points P and Q are symmetric about the origin. Let P(-c,-n), then Q(c,n). Thus, we have \\frac{n-(-n)}{c-(-c)}=\\frac{\\sqrt{2}}{2}, which implies n=\\frac{\\sqrt{2}}{c}c'. Substituting the coordinates of point Q into the ellipse equation gives \\frac{c^2}{a^2}+\\frac{1}{b^{2}}=1. Since b^{2}=1-c^{2}, substituting this into the previous equation yields c^{2}=\\frac{1}{2} (discarding c^{2}=2), so c=\\frac{\\sqrt{2}}{2}, and the focal distance is 2c=\\sqrt{2}." }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{9}=1 (a>3)$ has two foci $F_{1}$, $F_{2}$, and $| F_{1} F_{2} |=8$. The chord $AB$ passes through point $F_{1}$. Then the perimeter of $\\triangle ABF_{2}$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/9 + x^2/a^2 = 1);a: Number;a>3;F2: Point;F1: Point;Focus(G) = {F1,F2};A: Point;B: Point;IsChordOf(LineSegmentOf(A,B),G) = True;Abs(LineSegmentOf(F1,F2)) = 8;PointOnCurve(F1, LineSegmentOf(A, B))", "query_expressions": "Perimeter(TriangleOf(A,B,F2))", "answer_expressions": "20", "fact_spans": "[[[2, 50]], [[2, 50]], [[4, 50]], [[4, 50]], [[64, 71]], [[56, 63], [99, 107]], [[2, 71]], [[94, 98]], [[94, 98]], [[2, 98]], [[73, 92]], [[93, 107]]]", "query_spans": "[[[109, 133]]]", "process": "" }, { "text": "The x-coordinate of point $A$ on the parabola $y^{2}=2 \\sqrt{5} x$ is $\\frac{3}{2}\\sqrt{5}$, then the distance from $A$ to its focus $F$ is?", "fact_expressions": "G: Parabola;A: Point;F:Point;Expression(G) = (y^2 = 2*sqrt(5)*x);XCoordinate(A) = 3*sqrt(5)/2;PointOnCurve(A,G);Focus(G)=F", "query_expressions": "Distance(A,F)", "answer_expressions": "2*sqrt(5)", "fact_spans": "[[[0, 23], [60, 61]], [[25, 29], [56, 59]], [[63, 66]], [[0, 23]], [[25, 55]], [[0, 29]], [[60, 66]]]", "query_spans": "[[[56, 71]]]", "process": "Solution: The directrix equation of the parabola $ y^{2}=2\\sqrt{5}x $ is: $ x=-\\frac{\\sqrt{5}}{2} $. The abscissa of point A on the parabola $ y^{2}=2\\sqrt{5}x $ is $ \\frac{3}{2}\\sqrt{5} $. Then the distance from A to its focus F is: $ \\frac{3}{3}\\sqrt{5}+\\frac{\\sqrt{5}}{2}=2\\sqrt{5} $." }, { "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $\\frac{5}{3}$, what is the equation of the asymptotes of the hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = 5/3", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y = pm*(4/3)*x", "fact_spans": "[[[2, 48], [68, 71]], [[5, 48]], [[5, 48]], [[2, 48]], [[2, 66]]]", "query_spans": "[[[68, 79]]]", "process": "From the given information, the hyperbola's foci lie on the x-axis, $ e = \\frac{c}{a} = \\frac{5}{3} $. Also, for a hyperbola, $ c^{2} = a^{2} + b^{2} $, so $ e^{2} = \\frac{c^{2}}{a^{2}} = 1 + \\frac{b^{2}}{a^{2}} = \\frac{25}{9} $. Hence, $ \\frac{b}{a} = \\frac{4}{3} $. Therefore, the asymptotes of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ are given by $ y = \\pm\\frac{b}{a}x = \\pm\\frac{4}{3}x $." }, { "text": "If one of the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{9-k}=1$ is given by $x+2 y=0$, then $k=?$", "fact_expressions": "G: Hyperbola;k: Number;Expression(G) = (x^2/4 - y^2/(9 - k) = 1);Expression(OneOf(Asymptote(G))) = (x + 2*y = 0)", "query_expressions": "k", "answer_expressions": "8", "fact_spans": "[[[1, 41]], [[61, 64]], [[1, 41]], [[1, 59]]]", "query_spans": "[[[61, 66]]]", "process": "Given: The hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{9-k}=1$, $a=2$, $b=\\sqrt{9-k}$, the foci lie on the x-axis. Since one asymptote equation is $x+2y=0$, it follows that $\\frac{b}{a}=\\frac{\\sqrt{9-k}}{2}=\\frac{1}{2}$, solving gives $k=8$." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{8}+\\frac{y^{2}}{6}=1$, a line $l$ is drawn through point $P(-2,1)$ intersecting the ellipse $C$ at points $A$ and $B$. If point $P$ is exactly the midpoint of segment $AB$, then what is the slope of line $l$?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/8 + y^2/6 = 1);P: Point;Coordinate(P) = (-2, 1);l: Line;PointOnCurve(P, l);A: Point;B: Point;Intersection(l, C) = {A, B};MidPoint(LineSegmentOf(A, B)) = P", "query_expressions": "Slope(l)", "answer_expressions": "3/2", "fact_spans": "[[[2, 44], [63, 68]], [[2, 44]], [[46, 56], [80, 84]], [[46, 56]], [[57, 62], [99, 104]], [[45, 62]], [[70, 73]], [[74, 78]], [[57, 78]], [[80, 97]]]", "query_spans": "[[[99, 109]]]", "process": "Using the point-difference method to solve for the slope of the line [Detailed Solution] Let A(x_{1},y_{1}), B(x_{2},y_{2}), then \\frac{x_{1}^{2}}{8}+\\frac{y_{1}^{2}}{6}=1,\\frac{x_{2}^{2}}{8}+\\frac{y_{2}^{2}}{6}=1, subtracting the two equations gives \\frac{x_{1}^{2}-x_{2}^{2}}{8}+\\frac{y_{1}^{2}-y_{2}^{2}}{6}=0, that is, the slope of line l is \\frac{3}{7}" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$, $F_{1}$, $F_{2}$ are two foci of the ellipse $C$. If point $P$ is a point on the ellipse such that $|PF_{2}|=|F_{1} F_{2}|$, and the distance from $F_{2}$ to the line $PF_{1}$ equals the length of the minor axis of the ellipse, then the eccentricity of ellipse $C$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C);Abs(LineSegmentOf(P, F2)) = Abs(LineSegmentOf(F1, F2));Distance(F2, LineOf(P, F1)) = Length(MinorAxis(C))", "query_expressions": "Eccentricity(C)", "answer_expressions": "5/7", "fact_spans": "[[[2, 65], [84, 89], [101, 103], [160, 162], [168, 173]], [[2, 65]], [[9, 65]], [[9, 65]], [[9, 65]], [[9, 65]], [[67, 74]], [[76, 83], [137, 144]], [[67, 94]], [[96, 100]], [[96, 106]], [[111, 135]], [[137, 166]]]", "query_spans": "[[[168, 179]]]", "process": "" }, { "text": "The left focus $F_{1}$ of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ is used to draw a line intersecting the ellipse at points $A$ and $B$, and $F_{2}$ is the right focus. Then the perimeter of $\\triangle A F_{2} B$ is?", "fact_expressions": "G: Ellipse;H: Line;A: Point;F2: Point;B: Point;F1: Point;Expression(G) = (x^2/25 + y^2/16 = 1);LeftFocus(G)=F1;PointOnCurve(F1, H);Intersection(H, G)={A,B};RightFocus(G) = F2", "query_expressions": "Perimeter(TriangleOf(A, F2, B))", "answer_expressions": "20", "fact_spans": "[[[2, 41], [56, 58]], [[53, 55]], [[59, 62]], [[69, 76]], [[63, 66]], [[45, 52]], [[2, 41]], [[2, 52]], [[0, 55]], [[53, 68]], [[56, 80]]]", "query_spans": "[[[82, 108]]]", "process": "Since the ellipse equation is \\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1, we have a=5. According to the definition of an ellipse: AF_{1}+AF_{2}=2a, BF_{1}+BF_{2}=2a. The perimeter of quadrilateral AF_{2}B is 4a=20." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1(a>0)$, the right focus is $F$. A perpendicular is drawn from point $F$ to one of the asymptotes, with foot of perpendicular at $P$. The area of $\\triangle O P F$ is $\\sqrt{2}$. Then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;a: Number;O: Origin;P: Point;F: Point;L:Line;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);RightFocus(G) = F;PointOnCurve(F,L);IsPerpendicular(OneOf(Asymptote(G)),L);FootPoint(OneOf(Asymptote(G)),L)=P;Area(TriangleOf(O, P, F)) = sqrt(2)", "query_expressions": "Eccentricity(G)", "answer_expressions": "3*sqrt(2)/4", "fact_spans": "[[[2, 39], [106, 109]], [[5, 39]], [[71, 89]], [[66, 69]], [[44, 47], [49, 53]], [], [[5, 39]], [[2, 39]], [[2, 47]], [[2, 62]], [[2, 62]], [[2, 69]], [[72, 103]]]", "query_spans": "[[[106, 115]]]", "process": "Let the line $ l $ pass through the right focus $ F(c,0) $ and be perpendicular to an asymptote $ x - ay = 0 $. Then the equation of $ l $ can be written as: $ y = -a(x - c) $. \nFrom \n\\[\n\\begin{cases}\ny = \\frac{1}{a}x \\\\\ny = -a(x - c)\n\\end{cases}\n\\] \nwe obtain: \n\\[\n\\begin{cases}\nx = \\frac{a^2c}{1 + a^{2}} \\\\\ny = \\frac{ac}{1 + a^2}\n\\end{cases}\n\\], \ni.e., the coordinates of point $ P $ are $ \\left( \\frac{a^{2}c}{1 + a^{2}}, \\frac{ac}{1 + a^{2}} \\right) $. \nFrom the equation of the hyperbola, we know: $ c^{2} = 1 + a^{2} $, \n$ \\therefore P\\left( \\frac{a^{2}}{c}, \\frac{a}{c} \\right) $, \n$ \\therefore S_{\\Delta OPF} = \\frac{1}{2} OF \\cdot |y_{P}| = \\frac{1}{2} c \\cdot \\frac{a}{c} = \\sqrt{2} $, \n$ \\frac{1}{2} |OF| \\times y_{P} = \\frac{1}{2} c \\times \\frac{a}{c} = \\sqrt{2} $, \n$ \\therefore a = 2\\sqrt{2} $. \n$ \\therefore c^{2} = a^{2} + 1 = 8 + 1 = 9 $, \n$ \\therefore e = \\frac{3}{2}, = \\frac{3\\sqrt{2}}{4} $" }, { "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. Point $P$ is a point on the ellipse such that $P F_{1} \\perp P F_{2}$. What is the area of $\\triangle F_{1} P F_{2}$?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/5 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "5", "fact_spans": "[[[2, 39], [69, 71]], [[2, 39]], [[48, 55]], [[56, 63]], [[2, 63]], [[2, 63]], [[64, 68]], [[64, 75]], [[77, 100]]]", "query_spans": "[[[102, 132]]]", "process": "From the ellipse equation, we know that a=3, c=2. Let |PF_{1}|=m, |PF_{2}|=n. According to the given conditions, we have \n\\begin{cases}m+n=2a=6\\\\m^{2}+n^{2}=(2c)^{2}=16\\end{cases}. \nTherefore, (m+n)^{2}-2mn=16, 6^{2}-2mn=16, mn=10. Hence, the area of triangle F_{1}PF_{2} is \\frac{1}{2}mn=5." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{16}=1(a>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the ellipse $C$, and the perimeter of $\\triangle P F_{1} F_{2}$ is $16$. Find $a=?$", "fact_expressions": "C: Ellipse;a: Number;P: Point;F1: Point;F2: Point;a>0;Expression(C) = (y^2/16 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);Perimeter(TriangleOf(P, F1, F2)) = 16", "query_expressions": "a", "answer_expressions": "5", "fact_spans": "[[[2, 54], [85, 90]], [[127, 130]], [[80, 84]], [[64, 71]], [[72, 79]], [[9, 54]], [[2, 54]], [[2, 79]], [[2, 79]], [[80, 91]], [[92, 125]]]", "query_spans": "[[[127, 132]]]", "process": "Let the focal length be 2c. Since the perimeter of triangle PF₁F₂ is 16, we have 2a + 2c = 16, which simplifies to a + c = 8 ①. Also, a² - c² = 16, so (a + c)(a - c) = 16. Hence, a - c = 2 ②. From ① and ②, solving gives a = 5." }, { "text": "Given that the focus of the parabola $y^{2}=4 x$ is $F$, the intersection point of the directrix and the $x$-axis is $M$, if $P$ is any point on the parabola, and the midpoint of $P F$ is $Q$, then the maximum value of the slope of the line $Q M$ equals?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;M: Point;Intersection(Directrix(G), xAxis) = M;P: Point;PointOnCurve(P, G);Q: Point;MidPoint(LineSegmentOf(P, F)) = Q", "query_expressions": "Max(Slope(LineOf(Q, M)))", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[2, 16], [44, 47]], [[2, 16]], [[20, 23]], [[2, 23]], [[35, 38]], [[2, 38]], [[40, 43]], [[40, 52]], [[62, 65]], [[53, 65]]]", "query_spans": "[[[67, 84]]]", "process": "According to the problem, F(1,0), M(-1,0). Let P(x_{0},y_{0}), then Q(\\frac{x_{0}+1}{2},\\frac{y_{0}}{2}), y_{0}^{2}=4x_{0}, so the slope k of line QM is \\frac{y_{0}}{\\frac{y_{0}^{2}}{4}+3} \\leqslant \\frac{y_{0}}{2\\cdot\\frac{y_{0}}{2}} = \\frac{\\sqrt{3}}{3}. \\sqrt{3} The equality holds if and only if \\frac{y_{0}}{2}=\\sqrt{3}, that is, when y_{0}=2\\sqrt{3}. Hence, the maximum value of the slope of line QM equals \\frac{\\sqrt{3}}{3}." }, { "text": "Given that the directrix of the parabola $y^{2}=4 \\sqrt{3} x$ passes through one focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, and the major axis length of the ellipse is twice the minor axis length, then the equation of the ellipse is?", "fact_expressions": "G: Parabola;H: Ellipse;b: Number;a: Number;Expression(G) = (y^2 = 4*(sqrt(3)*x));a>0;b>0;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(OneOf(Focus(H)), Directrix(G));Length(MajorAxis(H)) = 2*Length(MinorAxis(H))", "query_expressions": "Expression(H)", "answer_expressions": "x^2/4+y^2=1", "fact_spans": "[[[2, 25]], [[29, 84], [90, 92], [109, 111]], [[31, 84]], [[31, 84]], [[2, 25]], [[31, 84]], [[31, 84]], [[29, 84]], [[2, 89]], [[90, 105]]]", "query_spans": "[[[109, 116]]]", "process": "Problem Analysis: The directrix of the parabola is $ x = -\\sqrt{3} $, so the semi-focal distance of the ellipse is $ c = \\sqrt{3} $, and $ a = 2b $. Combining with $ a^{2} = b^{2} + c^{2} $, solving yields $ a^{2} = 4 $, $ b^{2} = 1 $. Therefore, the equation is $ \\frac{x^{2}}{4} + y^{2} = 1 $." }, { "text": "Given that the foci of ellipse $C$ are $F_{1}$ and $F_{2}$, a line passing through point $F_{1}$ intersects ellipse $C$ at points $A$ and $B$. If $|A F_{1}|=2|F_{1} B|$ and $|A B|=|B F_{2}|$, then the eccentricity of ellipse $C$ is?", "fact_expressions": "C: Ellipse;G: Line;A: Point;F1: Point;B: Point;F2: Point;Focus(C) ={F1,F2};PointOnCurve(F1,G);Abs(LineSegmentOf(A, F1)) = 2*Abs(LineSegmentOf(F1, B));Intersection(G,C) = {B, A};Abs(LineSegmentOf(A, B)) = Abs(LineSegmentOf(B, F2))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[2, 7], [40, 45], [100, 105]], [[37, 39]], [[47, 50]], [[11, 18], [28, 36]], [[51, 54]], [[19, 26]], [[2, 26]], [[27, 39]], [[57, 79]], [[37, 56]], [[81, 98]]]", "query_spans": "[[[100, 111]]]", "process": "According to the problem, draw the figure as follows: Let |BF₁| = x, then |AF₁| = 2x, |BF₂| = |AB| = 3x. By the definition of the ellipse, |AF₁| + |AF₂| = 2a, |BF₁| + |BF₂| = x + 3x = 4x = 2a. Since |AF₁| = 2x, it follows that |AF₂| = 2x. In △ABF₂, by the law of cosines, we have \ncos∠ABF₂ = (|AB|² + |BF₂|² − |AF₂|²) / (2|AB|⋅|BF₂|) = ((3x)² + (3x)² − (2x)²) / (2⋅3x⋅3x) = 7/9. \nIn △BF₁F₂, by the law of cosines, we have \n|F₁F₂|² = |BF₁|² + |BF₂|² − 2|BF₁|⋅|BF₂|⋅cos∠ABF₂, \nthat is, |F₁F₂|² = x² + (3x)² − 2⋅x⋅3x⋅(7/9), solving gives |F₁F₂| = (4√3)/3 x. \nTherefore, 2a = 4x, 2c = (4√3)/3 x. \nThus, the eccentricity of the ellipse e = c/a = ((2√3)/3 x) / (2x) = √3/3." }, { "text": "The eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $\\frac{\\sqrt{3}}{2}$. A line with slope $1$ passes through $M(b , 0)$ and intersects the ellipse at points $A$ and $B$. $O$ is the origin. If $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}=\\frac{32}{5 \\tan \\angle A O B}$, then the standard equation of the ellipse is?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Eccentricity(G) = sqrt(3)/2;H: Line;Slope(H) = 1;M: Point;Coordinate(M) = (b, 0);PointOnCurve(M, H);A: Point;B: Point;Intersection(H, G) = {A, B};O: Origin;DotProduct(VectorOf(O, A), VectorOf(O, B)) = 32/(5*Tan(AngleOf(A, O, B)))", "query_expressions": "Expression(G)", "answer_expressions": "x^2/16+y^2/4=1", "fact_spans": "[[[0, 52], [100, 102], [206, 208]], [[0, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[0, 77]], [[85, 87]], [[78, 87]], [[88, 98]], [[88, 98]], [[85, 98]], [[104, 107]], [[108, 111]], [[85, 113]], [[114, 117]], [[124, 204]]]", "query_spans": "[[[206, 215]]]", "process": "From the given condition $ e = \\frac{\\sqrt{3}}{2} = \\frac{c}{a}, b^{2} = a^{2} - c^{2} $, we obtain $ a^{2} = 4b^{2} $. Therefore, the equation of the ellipse is: \n$ \\frac{x^{2}}{4b^{2}} + \\frac{y^{2}}{b^{2}} = 1 $. \nFrom the given condition, the equation of line AB is: $ y = x - b $. \nSolving the system \n$$\n\\begin{cases}\ny = x - b \\\\\n\\frac{x^{2}}{4b^{2}} + \\frac{y^{2}}{b^{2}} = 1\n\\end{cases}\n$$ \nwe get \n$$\n\\begin{cases}\nx = 0 \\\\\ny = -b\n\\end{cases}\n\\quad \\text{or} \\quad\nx = \\frac{8}{5}b, \\quad y = \\frac{3}{5}b.\n$$ \nSo we set $ \\overrightarrow{OA} = (0, -b) $, $ \\overrightarrow{OB} = \\left( \\frac{8}{5}b, \\frac{3}{5}b \\right) $, \n$$\n\\frac{ \\overrightarrow{OA} \\cdot \\overrightarrow{OB} }{ |\\overrightarrow{OA}| \\cdot |\\overrightarrow{OB}| } = \\frac{ -\\frac{3}{5}b^{2} }{ b \\sqrt{ \\frac{64}{25}b^{2} + \\frac{9}{25}b^{2} } } = \\frac{ -\\frac{3}{5} }{ \\sqrt{ \\frac{73}{25} } } = -\\frac{3}{\\sqrt{73}}.\n$$ \nThen $ \\cos \\angle AOB = -\\frac{3}{\\sqrt{73}} $, and $ \\sin \\angle AOB = \\sqrt{ 1 - \\left( -\\frac{3}{\\sqrt{73}} \\right)^2 } = \\sqrt{ \\frac{73 - 9}{73} } = \\frac{8}{\\sqrt{73}} $. \nThus, $ b^{2} = 4 $. Therefore, the equation of the ellipse is: \n$ \\frac{x^{2}}{16} + \\frac{y^{2}}{4} = 1 $." }, { "text": "Given circle $C_{1}$: $(x+2)^{2}+y^{2}=81$ and circle $C_{2}$: $(x-2)^{2}+y^{2}=9$, a moving circle $M$ is internally tangent to circle $C_{1}$ and externally tangent to circle $C_{2}$. Then the trajectory equation of the center $M$ of the moving circle is?", "fact_expressions": "C1: Circle;Expression(C1) = ((x + 2)^2 + y^2 = 81);C2: Circle;Expression(C2) = ((x - 2)^2 + y^2 = 9);M: Circle;IsInTangent(M, C1);IsOutTangent(M, C2);M1: Point;Center(M) = M1", "query_expressions": "LocusEquation(M1)", "answer_expressions": "x^2/36+y^2/32=1", "fact_spans": "[[[2, 31], [67, 75]], [[2, 31]], [[32, 60], [79, 87]], [[32, 60]], [[63, 66], [91, 93]], [[63, 77]], [[63, 89]], [[95, 99]], [[91, 99]]]", "query_spans": "[[[95, 106]]]", "process": "Analysis: According to the given conditions, |MC₁| + |MC₂| = 12 > |C₁C₂|, which implies that the locus of the center M of the moving circle is an ellipse with foci at C₁ and C₂. From the problem, the center of circle C₁ is C₁(-2,0) and its radius is r₁ = 9; the center of circle C₂ is C₂(2,0) and its radius is r₂ = 3. Let the radius of the moving circle M be r. Since the moving circle M is internally tangent to circle C₁, it follows from the condition that circle M lies inside circle C₁, so |MC₁| = 9 – r. Because the moving circle M is externally tangent to circle C₂, we have |MC₂| = 3 + r. Then |MC₁| + |MC₂| = 12 > |C₁C₂|. Therefore, the locus of the center M is an ellipse with foci at C₁ and C₂, so a = 6, c = 2, and thus b² = a² – c² = 32. Hence, the equation of the locus of the center M is \\frac{x^{2}}{36}+\\frac{y^{2}}{32}=1." }, { "text": "Given that the circle $x^{2}+y^{2}-6 x-7=0$ is tangent to the directrix of the parabola $y^{2}=2 px$ ($p>0$), then $p=?$", "fact_expressions": "G: Parabola;p: Number;H: Circle;p > 0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (-6*x + x^2 + y^2 - 7 = 0);IsTangent(H, Directrix(G))", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[25, 46]], [[53, 56]], [[2, 24]], [[28, 46]], [[25, 46]], [[2, 24]], [[2, 51]]]", "query_spans": "[[[53, 58]]]", "process": "" }, { "text": "The line $ l $ with positive slope passes through the focus $ F $ of the parabola $ y^{2} = 2px $ ($ p > 0 $) and intersects the parabola at points $ A $ and $ B $ (with point $ A $ in the first quadrant), and intersects its directrix at point $ C $. If $ |BC| = 3|BF| $, then the slope of line $ AB $ is?", "fact_expressions": "l: Line;G: Parabola;p: Number;A: Point;B: Point;C: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(F, l);Slope(l) > 0;Intersection(l, G) = {A, B};Quadrant(A) = 1;Intersection(l,Directrix(G)) = C;Abs(LineSegmentOf(B, C)) = 3*Abs(LineSegmentOf(B, F))", "query_expressions": "Slope(LineOf(A, B))", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[37, 42]], [[1, 22], [43, 46], [68, 69]], [[4, 22]], [[47, 51], [56, 60]], [[52, 55]], [[72, 76]], [[25, 28]], [[4, 22]], [[1, 22]], [[1, 28]], [[0, 42]], [[29, 42]], [[37, 55]], [[56, 65]], [[37, 76]], [[78, 92]]]", "query_spans": "[[[94, 106]]]", "process": "Draw BD perpendicular to l at D (where l is the directrix), we obtain |BC| = 3|BD|, and ∠CBD has the same inclination angle as line AB; thus, computing tan∠CBD yields the conclusion. As shown in the figure, draw BD ⊥ l at D (where l is the directrix), then |BD| = |BF|. According to the given condition, |BC| = 3|BD|, ∴ |CD| = 2√2|BD|, tan∠CBD = |CD| / |BD| = 2√2. Since BD ⊥ l, BD is parallel to the x-axis, and ∠CBD has the same inclination angle as line AB; therefore, the slope of AB is 2√2." }, { "text": "Let the parabola $C$: $y^{2}=4x$ have focus $F$, and let line $l$ passing through $F$ with slope $k$ ($k>0$) intersect $C$ at points $A$ and $B$, such that $AB=8$. Then the equation of line $l$ is?", "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;F: Point;k: Number;k > 0;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F, l);Slope(l) = k;Intersection(l, C) = {A, B};LineSegmentOf(A, B) = 8", "query_expressions": "Expression(l)", "answer_expressions": "y = x - 1", "fact_spans": "[[[45, 50], [76, 81]], [[1, 20], [51, 54]], [[56, 59]], [[60, 63]], [[24, 27], [29, 32]], [[36, 44]], [[36, 44]], [[1, 20]], [[1, 27]], [[28, 50]], [[33, 50]], [[45, 65]], [[67, 74]]]", "query_spans": "[[[76, 86]]]", "process": "From the given conditions, we have F(1,0), and the equation of line l is y = k(x - 1) (k > 0). Let A(x₁, y₁), B(x₂, y₂). From \n\\begin{cases} y = k(x - 1) \\\\ y^2 = 4x \\end{cases}, \nwe obtain \nk²x² - (2k² + 4)x + k² = 0. \nThe discriminant Δ = 16k² + 16 > 0, so \nx₁ + x₂ = (2k² + 4)/k², \nand thus \nAB = AF + BF = (x₁ + 1) + (x₂ + 1) = (4k² + 4)/k². \nAccording to the given condition, \n(4k² + 4)/k² = 8. \nSolving this equation yields k = -1 (discarded) or k = 1. \nTherefore, the equation of line l is y = x - 1." }, { "text": "If a point $M(a, b)$ on the parabola $C$: $y^{2}=4 x$ is at a distance of $5$ from the focus $F$, and a circle centered at $M$ passing through $F$ intersects the $y$-axis at points $A$ and $B$, then $|A B|=$?", "fact_expressions": "C: Parabola;G: Circle;M: Point;A: Point;B: Point;F: Point;a: Number;b: Number;Expression(C) = (y^2 = 4*x);Coordinate(M) = (a, b);PointOnCurve(M, C);Focus(C) = F;Distance(M, F) = 5;Center(G) = M;PointOnCurve(F, G);Intersection(G, yAxis) = {A, B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "6", "fact_spans": "[[[1, 20]], [[61, 62]], [[48, 51], [23, 32]], [[69, 72]], [[73, 76]], [[35, 38]], [[23, 32]], [[23, 32]], [[1, 20]], [[23, 32]], [[1, 32]], [[1, 38]], [[1, 45]], [[47, 62]], [[55, 62]], [[61, 78]]]", "query_spans": "[[[80, 89]]]", "process": "From the given condition, a+1=5, we get a=4, b=\\pm4, then |AB|=2\\times3=6." }, { "text": "If the left focus of the ellipse $\\frac{x^{2}}{5}+\\frac{16 y^{2}}{p^{2}}=1$ lies on the directrix of the parabola $y^{2}=2 p x$, then what is the value of $p$?", "fact_expressions": "H: Ellipse;Expression(H) = (x^2/5 + (16*y^2)/p^2 = 1);p: Number;G: Parabola;Expression(G) = (y^2 = 2*p*x);PointOnCurve(LeftFocus(H), Directrix(G)) = True", "query_expressions": "p", "answer_expressions": "4", "fact_spans": "[[[1, 45]], [[1, 45]], [[72, 75]], [[50, 66]], [[50, 66]], [[1, 70]]]", "query_spans": "[[[72, 79]]]", "process": "" }, { "text": "If the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{b^{2}}=1$ satisfies $b \\geq 9$, then the range of eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/b^2 = 1);b: Number;b >= 9", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "[sqrt(10), +oo)", "fact_spans": "[[[1, 43], [58, 61]], [[1, 43]], [[45, 55]], [[45, 55]]]", "query_spans": "[[[58, 71]]]", "process": "" }, { "text": "Given the parabola $C$: $y^{2}=6x$ with focus $F$ and directrix $l$, point $A$ lies on the parabola $C$ such that the distance from point $A$ to the directrix $l$ is $6$. The perpendicular bisector of $AF$ intersects the directrix $l$ at point $N$, and point $O$ is the origin. Then the area of $\\triangle OFN$ is?", "fact_expressions": "C: Parabola;A: Point;F: Point;O: Origin;N: Point;l: Line;Expression(C) = (y^2 = 6*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(A, C);Distance(A, l) = 6;Intersection(PerpendicularBisector(LineSegmentOf(A, F)), l) = N", "query_expressions": "Area(TriangleOf(O, F, N))", "answer_expressions": "9*sqrt(3)/4", "fact_spans": "[[[2, 21], [41, 47]], [[36, 40], [50, 54]], [[25, 28]], [[92, 96]], [[87, 91]], [[32, 35], [57, 60], [82, 85]], [[2, 21]], [[2, 28]], [[2, 35]], [[36, 48]], [[50, 67]], [[68, 91]]]", "query_spans": "[[[103, 125]]]", "process": "From the given conditions, we have $ F\\left(\\frac{3}{2},0\\right) $, $ l: x = -\\frac{3}{2} $. Let the coordinates of point A be $ (m,n) $. Since the distance from A to the directrix $ l $ is 6, we obtain $ m + \\frac{3}{2} = 6 $, so $ m = \\frac{9}{2} $. Substituting into the equation of the parabola, we get $ n = \\pm 3\\sqrt{3} $. By the symmetry of the parabola, we may assume $ A\\left(\\frac{9}{2}, 3\\sqrt{3}\\right) $. Then the slope of line AF is $ k_{AF} = \\frac{3\\sqrt{3}}{2} - \\frac{3}{2} = \\sqrt{3} $. Also, the midpoint of segment AF has coordinates $ \\left(3, \\frac{3\\sqrt{3}}{2}\\right) $. Thus, the equation of the perpendicular bisector of AF is $ y - \\frac{3\\sqrt{3}}{2} = -\\frac{\\sqrt{3}}{3}(x - 3) $. Setting $ x = -\\frac{3}{2} $, we find $ y = 3\\sqrt{3} $, so $ N\\left(-\\frac{3}{2}, 3\\sqrt{3}\\right) $. Therefore, the area of $ \\triangle OFN $ is $ \\frac{1}{2} \\times \\frac{3}{2} \\times 3\\sqrt{3} = \\frac{9\\sqrt{3}}{4} $." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{5}+y^{2}=1$, draw a line $l$ through the right focus $F$ of the ellipse $C$, intersecting the ellipse $C$ at points $A$ and $B$, and intersecting the $y$-axis at point $M$, such that point $B$ lies on segment $F M$. Then $\\frac{|M B|}{|B F|}-\\frac{|M A|}{|A F|}$=?", "fact_expressions": "l: Line;C: Ellipse;F: Point;M: Point;B: Point;A: Point;Expression(C) = (x^2/5 + y^2 = 1);RightFocus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Intersection(l, yAxis) =M;PointOnCurve(B, LineSegmentOf(F,M))", "query_expressions": "Abs(LineSegmentOf(M, B))/Abs(LineSegmentOf(B, F)) - Abs(LineSegmentOf(M, A))/Abs(LineSegmentOf(A, F))", "answer_expressions": "-10", "fact_spans": "[[[49, 54]], [[2, 34], [36, 41], [55, 60]], [[45, 48]], [[77, 81]], [[65, 68], [83, 87]], [[61, 64]], [[2, 34]], [[36, 48]], [[35, 54]], [[49, 70]], [[49, 81]], [[83, 96]]]", "query_spans": "[[[98, 141]]]", "process": "Let AB: y = kx - 2k, A(x₁, y₁), B(x₂, y₂). Then |MB|/|BF| - |MA|/|AF| = x₂/(2 - x₂) - . From \n\\begin{cases} y = kx - 2k \\\\ x^{2} + 5y^{2} = 5 \\end{cases} \nit follows that (1 + 5k²) x₁/(x₁ - 2) = (2x₁ + 2x₂ - 2x₁x₂)/(4 - 2x₁ - 2x₂ + x₁x₂). Solving with 2k₅ gives (1 + 5k²)x² - 20k²x + 20k² - 5 = 0. Therefore, \n|MB|/|BF| - |MA|/|AF| = (2 × (20k²)/(1 + 5k²) - 2 × (20k² - 5)/(1 + 5k²)) / (4 - 2 × (20k² - 20k² - 5)/(20k² + 4 - 40k² + 20k² - 5)) = -10. Hence, fill in -10." }, { "text": "Given the line $y = k(x - 4)$ intersects the circle $O$: $x^{2} + y^{2} = 36$ at points $M$ and $N$, what is the trajectory equation of the midpoint $G$ of segment $MN$?", "fact_expressions": "L: Line;O: Circle;M: Point;N: Point;G: Point;Expression(O) = (x^2 + y^2 = 36);Expression(L) = (y = k*(x - 4));Intersection(L, O) = {M, N};MidPoint(LineSegmentOf(M, N)) = G", "query_expressions": "LocusEquation(G)", "answer_expressions": "((x-2)^2 + y^2 = 4)&Negation(x = 4)", "fact_spans": "[[[2, 14]], [[15, 37]], [[39, 42]], [[43, 46]], [[60, 63]], [[15, 37]], [[2, 14]], [[2, 48]], [[50, 63]]]", "query_spans": "[[[60, 70]]]", "process": "Solution: Let the coordinates of the midpoint of segment MN be (x, y). Since the line y = k(x - 4) intersects the circle O: x^{2} + y^{2} = 36 at points M and N, then k_{MN} \\cdot k_{OM} = -1, that is, \\frac{y}{x} \\cdot \\frac{y}{x - 4} = -1. Rearranging gives y^{2} + x^{2} - 4x = 0, which converts to (x - 2)^{2} + y^{2} = 4. When x = 4, the slope of the line does not exist; since the slope of the line exists, this case is discarded. Therefore, the trajectory equation of the midpoint G of segment MN is (x - 2)^{2} + y^{2} = 4 (x \\neq 4)." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$ with foci $F_{1}$, $F_{2}$, and a point $P(x_{0} , y_{0})$ satisfying $0<\\frac{x_{0}^{2}}{2}+y_{0}^{2}<1$, then the range of values for $|P F_{1}|+| P F_{2}|$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/2 + y^2 = 1);F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;x0: Number;y0: Number;Coordinate(P) = (x0, y0);00)$. A line passing through $F$ with an inclination angle of $60^{\\circ}$ intersects $C$ at points $A$ and $B$. If $|A F|-|B F|=4$, then $|A B|$=?", "fact_expressions": "F: Point;Focus(C) = F;C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;p>0;G: Line;PointOnCurve(F, G);Inclination(G) = ApplyUnit(60, degree);Intersection(G, C) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(A, F)) - Abs(LineSegmentOf(B, F)) = 4", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "8", "fact_spans": "[[[1, 4], [36, 39]], [[1, 34]], [[5, 31], [60, 63]], [[5, 31]], [[13, 31]], [[13, 31]], [[57, 59]], [[35, 59]], [[40, 59]], [[57, 73]], [[64, 67]], [[68, 71]], [[75, 90]]]", "query_spans": "[[[92, 101]]]", "process": "Solution: Let A(x_{1},y_{1}), B(x_{2},y_{2}) (x_{1}>0, x_{2}>0), then |AF| - |BF| = (x_{1} + \\frac{p}{2}) - (x_{2} + \\frac{p}{2}) = x_{1} - x_{2} = 4. The equation of line AB is y = \\sqrt{3}(x - \\frac{p}{2}). From \\begin{cases} y = \\sqrt{3}(x - \\frac{p}{2}), \\\\ y^{2} = 2px \\end{cases}, we obtain 3x^{2} - 5px + \\frac{3}{4}p^{2} = 0. Therefore, x_{1} + x_{2} = \\frac{5}{3}p, x_{1}x_{2} = \\frac{1}{4}p^{2}. Thus, (x_{1} - x_{2})^{2} = (x_{1} + x_{2})^{2} - 4x_{1}x_{2} = \\frac{16}{9}p^{2} = 4^{2}. Since p > 0, we have p = 3. Therefore, |AB| = x_{1} + x_{2} + p = \\frac{8}{3}p = 8, 4kgxgarrow.g" }, { "text": "If the focus of the parabola $C$: $y^{2}=2 p x(p>0)$ lies on the line $x+2 y-2=0$, then what is the length of the chord intercepted by the line on the parabola?", "fact_expressions": "C: Parabola;p: Number;G: Line;p>0;Expression(C) = (y^2 = 2*(p*x));Expression(G) = (x + 2*y - 2 = 0);PointOnCurve(Focus(C), G)", "query_expressions": "Length(InterceptChord(G,C))", "answer_expressions": "40", "fact_spans": "[[[1, 27], [50, 53]], [[9, 27]], [[31, 44], [47, 49]], [[9, 27]], [[1, 27]], [[31, 44]], [[1, 45]]]", "query_spans": "[[[47, 58]]]", "process": "Analysis: Find the coordinates of the intersection point between the given line and the x-axis, which gives the focus of the parabola. Then determine the parameter p in the parabola equation. Solve simultaneously the line equation and the parabola equation to find the coordinates of the two intersection points. Finally, use the distance formula between two points to obtain the chord length. \nDetailed Solution: In $x + 2y - 2 = 0$, set $y = 0$ to get $x = 2$, $\\therefore \\frac{p}{2} = 2$, $p = 4$, that is, the parabola equation is $y^{2} = 8x$. From $0$, we solve and obtain $\\begin{cases}x_{1}=10-8\\sqrt{5}\\\\y_{1}=-4+4\\sqrt{5}\\end{cases}$ or $\\begin{cases}x_{2}=10+8\\sqrt{5}\\\\y_{2}=-4-4\\sqrt{5}\\end{cases}$, $x_{2}^{2}+(y_{-}-y)^{2}=\\sqrt{(-16\\sqrt{5})^{2}+(8\\sqrt{5})^{2}}=40$, $\\therefore$ the chord length is $\\sqrt{(x)}$" }, { "text": "Let $F_{1}$, $F_{2}$ be the two foci of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$, point $P$ lies on this hyperbola, $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$. If the eccentricity of this hyperbola equals $\\frac{\\sqrt{5}}{2}$, then the distance from point $P$ to the $x$-axis is equal to?", "fact_expressions": "G: Hyperbola;a: Number;P: Point;F2: Point;F1: Point;Expression(G) = (-y^2 + x^2/a^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);DotProduct(VectorOf(P, F2),VectorOf(P, F1)) = 0;Eccentricity(G) = sqrt(5)/2", "query_expressions": "Distance(P, xAxis)", "answer_expressions": "sqrt(5)/5", "fact_spans": "[[[17, 49], [61, 64], [130, 133]], [[20, 49]], [[55, 59], [162, 166]], [[9, 16]], [[1, 8]], [[17, 49]], [[1, 54]], [[55, 65]], [[66, 125]], [[129, 159]]]", "query_spans": "[[[162, 177]]]", "process": "" }, { "text": "The coordinates of the foci of the hyperbola $2 x^{2}-3 y^{2}=k$ $(k<0)$ are? (Express in terms of $k$)", "fact_expressions": "G: Hyperbola;k: Number;k<0;Expression(G) = (2*x^2 - 3*y^2 = k)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0,pm*sqrt(-5*k/6))", "fact_spans": "[[[0, 27]], [[3, 27]], [[3, 27]], [[0, 27]]]", "query_spans": "[[[0, 43]]]", "process": "Hyperbola 2x^{2}-3y^{2}=k (k<0), rewritten as \\frac{y^{2}}{-\\frac{k}{2}}-\\frac{x^{2}}{-\\frac{k}{2}}=1. According to the geometric meaning of the hyperbola equation, we have c=\\sqrt{-\\frac{k}{3}-\\frac{k}{2}}=\\sqrt{-\\frac{5k}{6}}. Therefore, the coordinates of the foci of the hyperbola are (0,\\pm\\sqrt{-\\frac{5k}{6}})." }, { "text": "In the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$, a chord passing through the point $P(1,1)$ is exactly bisected by the point $P$. Then, what is the equation of the line containing this chord?", "fact_expressions": "G: Ellipse;H: LineSegment;P: Point;Expression(G) = (x^2/4 + y^2/2 = 1);Coordinate(P) = (1, 1);IsChordOf(H, G);PointOnCurve(P, H);MidPoint(H) = P", "query_expressions": "Expression(OverlappingLine(H))", "answer_expressions": "x+2*y-3=0", "fact_spans": "[[[0, 37]], [], [[39, 48], [53, 57]], [[0, 37]], [[39, 48]], [[0, 50]], [[0, 50]], [[0, 59]]]", "query_spans": "[[[0, 72]]]", "process": "Let the coordinates of the two intersection points between the line and the ellipse be $(x_{1},y_{1})$; $(x_{2},y_{2})$, then \n\\begin{cases}\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1\\\\\\frac{x^{2}}{2}+\\frac{y^{2}}{2}=1\\end{cases}\nleads to $\\underline{(x_{1}+x_{2})}+\\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{=} \\because P(1$ is the midpoint, $2(y_{1}-y_{2})k=\\frac{y_{2}^{2}-y_{1}}{x_{2}-x_{1}}=-\\frac{1}{2} \\therefore$ the equation of the line containing this chord is $y-1=\\frac{1}{2}(x-1)$, that is, $x+2v-3=0$. Answer: $x+2v-3=0$" }, { "text": "If $(k^{2}+k-2) x^{2}+(k+3) y^{2}=1$ represents a hyperbola with foci on the $y$-axis, then the range of values for $k$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2*(k^2 + k - 2) + y^2*(k + 3) = 1);PointOnCurve(Focus(G), yAxis);k: Number", "query_expressions": "Range(k)", "answer_expressions": "(-2, 1)", "fact_spans": "[[[45, 48]], [[1, 48]], [[36, 48]], [[50, 53]]]", "query_spans": "[[[50, 60]]]", "process": "" }, { "text": "The equation of a circle with the vertex of the parabola $y^{2}=4 x$ as its center and the distance from the focus to the directrix as its radius is?", "fact_expressions": "G: Parabola;H: Circle;Expression(G) = (y^2 = 4*x);Center(H) = Vertex(G);Radius(H) = Distance(Focus(G), Directrix(G))", "query_expressions": "Expression(H)", "answer_expressions": "x^2+y^2=4", "fact_spans": "[[[1, 15]], [[34, 35]], [[1, 15]], [[0, 35]], [[0, 35]]]", "query_spans": "[[[34, 40]]]", "process": "" }, { "text": "Given that an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is parallel to the line $l$: $x-2 y-\\sqrt{5}=0$, and that a focus of the hyperbola lies on the line $l$, then the equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;l: Line;Expression(l) = (x - 2*y - sqrt(5) = 0);IsParallel(OneOf(Asymptote(G)), l) = True;PointOnCurve(OneOf(Focus(G)), l)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4-y^2=1", "fact_spans": "[[[2, 58], [94, 97], [111, 114]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[65, 90], [103, 108]], [[65, 90]], [[2, 92]], [[94, 109]]]", "query_spans": "[[[111, 119]]]", "process": "" }, { "text": "Draw a line passing through point $M(1,1)$ with slope $\\frac{1}{2}$, intersecting the hyperbola $\\Gamma$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ at points $A$ and $B$. If $M$ is the midpoint of segment $AB$, then the eccentricity of hyperbola $\\Gamma$ is?", "fact_expressions": "Gamma: Hyperbola;Expression(Gamma) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;M: Point;Coordinate(M) = (1, 1);H: Line;Slope(H) = 1/2;PointOnCurve(M, H);A: Point;B: Point;Intersection(H, Gamma) = {A, B};MidPoint(LineSegmentOf(A, B)) = M", "query_expressions": "Eccentricity(Gamma)", "answer_expressions": "sqrt(6)/2", "fact_spans": "[[[31, 87], [117, 128]], [[31, 87]], [[44, 87]], [[44, 87]], [[1, 10], [101, 104]], [[1, 10]], [[28, 30]], [[11, 31]], [[0, 30]], [[90, 93]], [[94, 97]], [[28, 99]], [[101, 115]]]", "query_spans": "[[[117, 134]]]", "process": "Solution: Let A(x_{1},y_{1}), B(x_{2},y_{2}), then \\frac{x_{1}^{2}}{a^{2}}-\\frac{y_{1}^{2}}{b^{2}}=1\\textcircled{1}, \\frac{x_{2}^{2}}{a^{2}}-\\frac{y_{2}^{2}}{b^{2}}=1\\textcircled{2}. Since M is the midpoint of segment AB, \\frac{x_{1}+x_{2}}{2}=1, \\frac{y_{1}+y_{2}}{2}=1. Since the equation of line AB is y=\\frac{1}{2}(x-1)+1, y_{1}-y_{2}=\\frac{1}{2}(x_{1}-x_{2}). Since a line with slope \\frac{1}{2} passing through point M(1,1) intersects the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0, b>0) at points A and B, and M is the midpoint of segment AB, subtracting equations \\textcircled{1} and \\textcircled{2} gives \\frac{x_{1}^{2}-x_{2}^{2}}{a^{2}}-\\frac{y_{1}^{2}-y_{2}^{2}}{b^{2}}=0, i.e., \\frac{b^{2}}{a^{2}}=\\frac{(y_{1}-y_{2})(y_{1}+y_{2})}{(x_{1}-x_{2})(x_{1}+x_{2})}=\\frac{1}{2}. Therefore, e=\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\frac{\\sqrt{6}}{2}." }, { "text": "Let $F_{1}$, $F_{2}$ be the left and right foci of the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and let $O$ be the coordinate origin. If there exists a point $A$ on $E$ such that $\\angle F_{1} A F_{2}=120^{\\circ}$ and $|O A|=b$, then the eccentricity of this hyperbola is?", "fact_expressions": "E: Hyperbola;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(E) = F1;RightFocus(E) = F2;O: Origin;A: Point;PointOnCurve(A, E);AngleOf(F1, A, F2) = ApplyUnit(120, degree);Abs(LineSegmentOf(O, A)) = b", "query_expressions": "Eccentricity(E)", "answer_expressions": "sqrt(10)/2", "fact_spans": "[[[17, 78], [156, 159], [95, 98]], [[17, 78]], [[25, 78]], [[25, 78]], [[25, 78]], [[25, 78]], [[1, 8]], [[9, 16]], [[1, 84]], [[1, 84]], [[85, 88]], [[101, 105]], [[95, 105]], [[108, 142]], [[144, 153]]]", "query_spans": "[[[156, 165]]]", "process": "As shown in the figure below: without loss of generality, assume point A lies in the first quadrant. In triangle $ \\triangle F_{1}AF_{2} $, by the law of cosines, we have $ |AF_{1}|^{2} + |AF_{2}|^{2} + |AF_{1}|\\cdot|AF_{2}| = |F_{1}F_{2}|^{2} = 4c^{2} $. Also, since $ \\overrightarrow{AF}_{1} + \\overrightarrow{AF_{2}} = 2\\overrightarrow{AO} $, it follows that $ (\\overrightarrow{AF}_{1} + \\overrightarrow{AF_{2}})^{2} = (2\\overrightarrow{AO})^{2} $. Therefore, $ |AF_{1}|^{2} + |AF_{2}|^{2} - |AF_{1}|\\cdot|AF_{2}| = |F_{1}F_{2}|^{2} = 4b^{2} $. Hence, $ 2|AF_{1}|\\cdot|AF_{2}| = 4c^{2} - 4b^{2} = 4a^{2} $, so $ |AF_{1}|\\cdot|AF_{2}| = 2a^{2} $. Also, since $ |AF_{1}| - |AF_{2}| = 2a $, we have $ |AF_{1}|^{2} + |AF_{2}|^{2} + |AF_{1}|\\cdot|AF_{2}| = (|AF_{1}| - |AF_{2}|)^{2} + 3|AF_{1}|\\cdot|AF_{2}| = 10a^{2} $. Thus, $ 10a^{2} = 4c^{2} $, so $ \\frac{c^{2}}{a^{2}} = \\frac{5}{2} $, and therefore $ e = \\frac{a}{c} = \\frac{\\sqrt{10}}{2} $." }, { "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{2-m^{2}}=1$ with foci on the $x$-axis are perpendicular to each other, then $m=?$", "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (-y^2/(2 - m^2) + x^2/m = 1);PointOnCurve(Focus(G),xAxis);L1:Line;L2:Line;Asymptote(G)={L1,L2};IsPerpendicular(L1,L2)", "query_expressions": "m", "answer_expressions": "1", "fact_spans": "[[[11, 55]], [[67, 70]], [[11, 55]], [[2, 55]], [], [], [[11, 61]], [[11, 65]]]", "query_spans": "[[[67, 72]]]", "process": "Solution: Since the foci of the hyperbola $\\frac{x^2}{m}-\\frac{y^{2}}{2-m^{2}}=1$ lie on the $x$-axis, \n$$\n\\begin{cases}\nm>0 \\\\\n2-m^2>0\n\\end{cases},\n$$\nwhich implies $0b>0)$ has left and right foci $F_{1}$, $F_{2}$ respectively. The line $y=k x$ intersects curve $C$ at points $A$, $B$, with $|A F_{1}|=3|B F_{1}|$, and $\\angle F_{1} A F_{2}=60^{\\circ}$. Then the eccentricity of hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;k: Number;A: Point;F1: Point;B: Point;F2: Point;a > b;b > 0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y = k*x);LeftFocus(C) = F1;RightFocus(C) = F2;Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, F1)) = 3*Abs(LineSegmentOf(B, F1));AngleOf(F1, A, F2) = ApplyUnit(60, degree)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(7)/2", "fact_spans": "[[[0, 58], [93, 98], [169, 175]], [[8, 58]], [[8, 58]], [[83, 92]], [[85, 92]], [[100, 103]], [[67, 74]], [[104, 107]], [[75, 82]], [[8, 58]], [[8, 58]], [[0, 58]], [[83, 92]], [[0, 82]], [[0, 82]], [[83, 109]], [[110, 132]], [[134, 167]]]", "query_spans": "[[[169, 181]]]", "process": "As shown in the figure: According to the symmetry of the hyperbola and the line y = kx, |BF_{1}| = |AF_{2}|. Let |AF_{2}| = m (m > 0), then |AF_{1}| = 3|BF_{1}| = 3m. In \\triangle F_{1}AF_{2}, (2c)^{2} = m^{2} + 9m^{2} - 2 \\times m \\times 3m \\times \\frac{1}{2}, solving gives m = \\frac{2}{\\sqrt{7}}c. Therefore, by the definition of the hyperbola, 3m - m = 2a, so a = \\frac{2}{\\sqrt{7}}c, hence e = \\frac{c}{a} = \\frac{\\sqrt{7}}{2}." }, { "text": "The focus lies on the line $3 x-4 y-12=0$, and the standard parabolic equation is?", "fact_expressions": "G: Parabola;H: Line;Expression(H) = (3*x - 4*y - 12 = 0);PointOnCurve(Focus(G), H)", "query_expressions": "Expression(G)", "answer_expressions": "{y^2=16*x,x^2=-12*y}", "fact_spans": "[[[27, 30]], [[3, 19]], [[3, 19]], [[0, 30]]]", "query_spans": "[[[27, 34]]]", "process": "" }, { "text": "Given that $A$, $B$, $P$ are three distinct points on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, and the line connecting $A$ and $B$ passes through the origin $O$. If the product of the slopes of lines $PA$ and $PB$ is $k_{PA} \\cdot k_{PB}=3$, then what is the eccentricity of the hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;A: Point;B: Point;O: Origin;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(A, G);PointOnCurve(B, G);PointOnCurve(P, G);Negation(A = B);Negation(A = P);Negation(B = P);PointOnCurve(O, LineSegmentOf(A, B));kPA: Number;kPB: Number;Slope(LineOf(P, A)) = kPA;Slope(LineOf(P, B)) = kPB;kPA*kPB = 3", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[14, 70], [144, 147]], [[17, 70]], [[17, 70]], [[10, 13]], [[2, 5], [78, 81]], [[6, 9], [82, 85]], [[89, 96]], [[17, 70]], [[17, 70]], [[14, 70]], [[2, 76]], [[2, 76]], [[2, 76]], [[2, 76]], [[2, 76]], [[2, 76]], [[78, 96]], [[117, 142]], [[117, 142]], [[98, 142]], [[98, 142]], [[117, 142]]]", "query_spans": "[[[144, 153]]]", "process": "A and B must be symmetric about the origin. Let A(x_{1},y_{1}), B(-x_{1},-y_{1}), P(x,y), then \\frac{x_{1}^{2}}{a^{2}}-\\frac{y_{1}^{2}}{b^{2}}=1, \\frac{x_{2}^{2}}{a^{2}}-\\frac{y_{2}^{2}}{b^{2}}=1, \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1, k_{PA}\\cdot k_{PB}=\\frac{y^{2}-y_{1}^{2}}{x^{2}-x_{1}^{2}}=\\frac{b^{2}}{a^{2}}=3, e=\\sqrt{1+\\frac{b^{2}}{a^{2}}} is 2." }, { "text": "Given point $M(\\frac{3}{2},-1)$, line $l$ passes through the focus of the parabola $C$: $x^{2}=4 y$ and intersects the parabola $C$ at points $A$ and $B$, and $AM$ is tangent to the parabola $C$. Then, what are the coordinates of the midpoint of segment $AB$?", "fact_expressions": "M: Point;Coordinate(M) = (3/2, -1);l:Line;C:Parabola;Expression(C)=(x^2=4*y);A: Point;B: Point;PointOnCurve(Focus(C),l);Intersection(l,C)={A,B};IsTangent(LineSegmentOf(A,M),C)", "query_expressions": "Coordinate(MidPoint(LineSegmentOf(A,B)))", "answer_expressions": "(3/2,17/8)", "fact_spans": "[[[2, 22]], [[2, 22]], [[23, 28]], [[29, 48], [52, 58], [77, 83]], [[29, 48]], [[59, 62]], [[63, 66]], [[23, 51]], [[23, 68]], [[70, 85]]]", "query_spans": "[[[88, 102]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), then x_{1}^{2}=4y_{1}, \\textcircled{1} the parabola C: y=\\frac{1}{4}x^{2}, its derivative is: y'=\\frac{x}{2}, so we have \\frac{x_{1}}{2}=\\frac{y_{1}+1}{x_{1}-\\frac{3}{2}}, \\textcircled{2} combining \\textcircled{1} and \\textcircled{2} gives: x_{1}=4 or x_{1}=-1, \\therefore y_{1}=4 or y_{1}=\\frac{1}{4}, \\therefore A(4,4) or A(-1,\\frac{1}{4}), the focus of the parabola is F(0,1), \\therefore k=\\frac{4-1}{4-0}=\\frac{3}{4}, or k=\\frac{\\frac{1}{4}-1}{-1-0}=\\frac{3}{4}. Therefore, the equation of line l is: y=\\frac{3}{4}x+1. Solving simultaneously the equations of line l and parabola C: \n\\begin{cases}\ny=\\frac{3}{4}x+1\\\\\nx^{2}=4y\n\\end{cases}\neliminating y yields: x^{2}-3x-4=0, so x_{1}+x_{2}=3, y_{1}+y_{2}=\\frac{3}{4}(x_{1}+x_{2})+2=\\frac{17}{4}. Therefore, \\frac{x_{1}+x_{2}}{2}=\\frac{3}{2}, \\frac{y_{1}+y_{2}}{2}=\\frac{17}{8}. Hence, the midpoint coordinates of AB are (\\frac{3}{2},\\frac{17}{8})." }, { "text": "Let the parabola be $y^{2}=4 x$, and let line $l$ passing through the point $(1 , 0)$ intersect the parabola at points $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$. Then, what is $x_{1} x_{2}+y_{1} y_{2}$=?", "fact_expressions": "G: Parabola;Expression(G) = (y^2=4*x);H: Point;Coordinate(H) = (1, 0);L: Line;PointOnCurve(H,L) = True;Intersection(L,G) = {P,Q};P: Point;Coordinate(P) = (x1, y1);x1: Number;y1: Number;Q: Point;Coordinate(Q) = (x2,y2);x2: Number;y2: Number", "query_expressions": "x1*x2 + y1*y2", "answer_expressions": "-3", "fact_spans": "[[[1, 4], [35, 38]], [[1, 16]], [[18, 28]], [[18, 28]], [[29, 34]], [[17, 34]], [[29, 79]], [[40, 57]], [[40, 57]], [[40, 57]], [[40, 57]], [[60, 77]], [[60, 77]], [[60, 77]], [[60, 77]]]", "query_spans": "[[[81, 108]]]", "process": "" }, { "text": "Given that $P$ is a moving point on the parabola $y^{2}=4 x$, point $Q$ is a moving point on the circle $C$: $(x+3)^{2}+(y-3)^{2}=1$, and point $R$ is the projection of point $P$ onto the $y$-axis, then the minimum value of $|P Q|+|P R|$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);C: Circle;Expression(C) = ((x + 3)^2 + (y - 3)^2 = 1);P: Point;Q: Point;R: Point;PointOnCurve(P, G);PointOnCurve(Q, C);Projection(P, yAxis) = R", "query_expressions": "Min(Abs(LineSegmentOf(P, Q)) + Abs(LineSegmentOf(P, R)))", "answer_expressions": "3", "fact_spans": "[[[6, 20]], [[6, 20]], [[30, 58]], [[30, 58]], [[2, 5], [68, 72]], [[25, 29]], [[63, 67]], [[2, 24]], [[25, 62]], [[63, 81]]]", "query_spans": "[[[83, 102]]]", "process": "According to the definition of a parabola, we know |PR| = |PF| - 1, and the minimum value of |PQ| is |PC| - 1. Therefore, the minimum value of |PQ| + |PR| is the minimum value of |PF| + |PC| - 2. When points C, P, F are collinear, |PF| + |FC| is minimized, and the minimum value is" }, { "text": "The two foci of ellipse $E$ are $F_{1}(-1,0)$ and $F_{2}(1,0)$, respectively. The point $C(1, \\frac{3}{2})$ lies on the ellipse $E$. If point $P$ lies on the ellipse $E$, and $t=\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$, then what is the range of real values for $t$?", "fact_expressions": "E: Ellipse;F1: Point;F2: Point;Focus(E) = {F1, F2};Coordinate(F1) = (-1, 0);Coordinate(F2) = (1, 0);C: Point;Coordinate(C) = (1, 3/2);PointOnCurve(C, E);P: Point;PointOnCurve(P, E);t: Real;t = DotProduct(VectorOf(P, F1), VectorOf(P, F2))", "query_expressions": "Range(t)", "answer_expressions": "[2, 3]", "fact_spans": "[[[2, 7], [64, 69], [77, 82]], [[15, 28]], [[30, 42]], [[2, 42]], [[15, 28]], [[30, 42]], [[43, 63]], [[43, 63]], [[43, 70]], [[72, 76]], [[72, 83]], [[146, 151]], [[85, 144]]]", "query_spans": "[[[146, 158]]]", "process": "First, use the method of undetermined coefficients to find the equation of the ellipse. Let P(x_{0},y_{0}), from \\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=t, derive x_{0}^{2}+y_{0}^{2}=t+1. Since point P lies on curve C, \\frac{x_{0}^{2}}{4}+\\frac{y_{0}^{2}}{3}=1, obtain y_{0}^{2}=t+1-x_{0}^{2}, then solve to get 0\\leqslant x_{0}^{2}\\leqslant4, and thus find 2\\leqslant t\\leqslant3. This gives the range of real values for t. Solution: According to the given conditions, assume the equation of ellipse E is \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0). From the given information, the semi-focal length c=1, so a^{2}-b^{2}=1. Since point C(1,\\frac{3}{2}) lies on ellipse E, we have \\frac{1}{a^{2}}+\\frac{9}{4b^{2}}=1. Solving yields a^{2}=4, b^{2}=3. Therefore, the equation of ellipse E is \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1. Let P(x_{0},y_{0}), from \\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=t, we get (-1-x_{0},-y_{0})\\cdot(1-x_{0},-y_{0})=t, which implies t=x_{0}^{2}+y_{0}^{2}-1. Since point P lies on E, \\frac{x_{0}^{2}}{4}+\\frac{y_{0}^{2}}{3}=1, therefore t=\\frac{1}{4}x_{0}^{2}+2. Given -2\\leqslant x_{0}\\leqslant2, it follows that 0\\leqslant x_{0}^{2}\\leqslant4, hence 2\\leqslant t\\leqslant3, i.e., the range of real values is [2,3]." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{8}+\\frac{y^{2}}{2}=1$ and the circle $M$: $x^{2}+y^{2}+2 \\sqrt{2} x-4 \\sqrt{2} y+10-r^{2}=0(00, then 4-m^{2}>0 \\therefore 0\\leqslant m^{2}<4. Thus a^{2}=m^{2}+12, b^{2}=4-m^{2}, c^{2}=a^{2}+b^{2}=16. Then the eccentricity e=\\frac{c}{a}=\\sqrt{\\frac{16}{m^{2}+12}}\\leqslant\\sqrt{\\frac{16}{12}}=\\frac{2\\sqrt{3}}{3}. The equality holds when m=0." }, { "text": "The asymptotes of the hyperbola $x^{2}-y^{2}=4$ are given by?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 4)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*x", "fact_spans": "[[[0, 18]], [[0, 18]]]", "query_spans": "[[[0, 26]]]", "process": "From $x^{2}-y^{2}=4$, we get $\\frac{x^{2}}{4}-\\frac{y^{2}}{4}=1$, so $a^{2}=4$, $b^{2}=4 \\Rightarrow a=2$, $b=2$. Therefore, the asymptotes are $y=\\pm x$." }, { "text": "The point $P(4,1)$ bisects a chord of the hyperbola $x^{2}-4 y^{2}=4$, then the equation of the line on which this chord lies is?", "fact_expressions": "G: Hyperbola;H: LineSegment;P: Point;Expression(G) = (x^2 - 4*y^2 = 4);Coordinate(P) = (4, 1);MidPoint(H) = P;IsChordOf(H, G)", "query_expressions": "Expression(OverlappingLine(H))", "answer_expressions": "x - y - 3 = 0", "fact_spans": "[[[11, 31]], [], [[0, 9]], [[11, 31]], [[0, 9]], [[0, 35]], [[11, 35]]]", "query_spans": "[[[11, 49]]]", "process": "" }, { "text": "The ellipse centered at the origin has an eccentricity of $\\frac{1}{2}$, with two foci $F_{1}$ and $F_{2}$ on the $x$-axis. $P$ is an arbitrary point on the ellipse. If $|P F_{1}|+|P F_{2}|=4$, then the standard equation of the ellipse is?", "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;O:Origin;Center(G)=O;PointOnCurve(P, G);Eccentricity(G)=1/2;Focus(G)={F1,F2};PointOnCurve(F1,xAxis);PointOnCurve(F2,xAxis);Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 4", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4+y^2/3=1", "fact_spans": "[[[8, 10], [11, 12], [61, 63], [61, 63]], [[56, 59]], [[34, 41]], [[42, 49]], [[3, 7]], [[0, 10]], [[56, 69]], [[11, 29]], [[11, 49]], [[34, 55]], [[34, 55]], [[71, 94]]]", "query_spans": "[[[96, 105]]]", "process": "From the given conditions, we have \n\\begin{cases}2a=4\\\\\\frac{c}{a}=\\frac{1}{2}\\end{cases} \n\\therefore a=2, c=1, \\therefore b=\\sqrt{2^{2}-1^{2}}=\\sqrt{3}. \nSince the two foci F_{1} and F_{2} lie on the x-axis, the standard equation of the ellipse is \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1." }, { "text": "If the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has exactly two common points with both the circle $C_{1}$: $x^{2}+y^{2}=9$ and the circle $C_{2}$: $x^{2}+y^{2}=8$, then what is the standard equation of the ellipse $C$?", "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;C1: Circle;Expression(C1) = (x^2 + y^2 = 9);NumIntersection(C,C1) = 2;C2: Circle;Expression(C2) = (x^2 + y^2 = 8);NumIntersection(C,C2) = 2", "query_expressions": "Expression(C)", "answer_expressions": "x^2/9+y^2/8=1", "fact_spans": "[[[1, 58], [122, 127]], [[1, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[59, 84]], [[59, 84]], [[1, 120]], [[85, 110]], [[85, 110]], [[1, 120]]]", "query_spans": "[[[122, 134]]]", "process": "According to the symmetry of the circle and the ellipse, circle $ C_{1} $ intersects ellipse $ C $ at the two endpoints of the major axis, and circle $ C_{2} $ intersects ellipse $ C $ at the two endpoints of the minor axis, thus the values of $ a $ and $ b $ can be found, and then the standard equation of ellipse $ C $ can be determined. According to the symmetry of the circle and the ellipse, circle $ C_{1} $ intersects ellipse $ C $ at the two endpoints of the major axis, and circle $ C_{2} $ intersects ellipse $ C $ at the two endpoints of the minor axis, so $ a=3 $, $ b=2\\sqrt{2} $, hence the standard equation of ellipse $ C $ is $ \\frac{x^{2}}{9}+\\frac{y^{2}}{8}=1 $." }, { "text": "If on the parabola $y^{2}=-2 p x(p>0)$ there is a point $M$ whose abscissa is $-9$, and its distance to the focus is $10$, then what are the coordinates of point $M$?", "fact_expressions": "G: Parabola;p: Number;M: Point;p>0;Expression(G) = (y^2 = -2*p*x);PointOnCurve(M, G);XCoordinate(M) = -9;Distance(M, Focus(G)) = 10", "query_expressions": "Coordinate(M)", "answer_expressions": "{(-9, 6), (-9, -6)}", "fact_spans": "[[[1, 23]], [[4, 23]], [[27, 30], [31, 32], [41, 42], [55, 59]], [[4, 23]], [[1, 23]], [[1, 30]], [[31, 40]], [[1, 53]]]", "query_spans": "[[[55, 64]]]", "process": "From the parabola equation $ y^{2} = -2px $ ($ p > 0 $), we obtain its focus coordinates $ F\\left(-\\frac{p}{2}, 0\\right) $, and the directrix equation $ x = \\frac{p}{2} $. Let $ d $ be the distance from point $ M $ to the directrix. Then $ d = |MF| = 10 $, that is, $ \\frac{p}{2} - (-9) = 10 $, solving gives $ p = 2 $. Hence, the parabola equation is $ y^{2} = -4x $. Since point $ M(-9, y) $ lies on the parabola, we get $ y = \\pm6 $. Therefore, the coordinates of point $ M $ are $ (-9, 6) $ or $ (-9, -6) $." }, { "text": "Hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has one asymptote intersecting circle $M$: $(x-3)^{2}+y^{2}=8$ at points $A$, $B$, with $|A B|=2 \\sqrt{2}$. Then the eccentricity of the hyperbola equals?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;M: Circle;A: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(M) = (y^2 + (x - 3)^2 = 8);Intersection(OneOf(Asymptote(C)),M) = {A, B};Abs(LineSegmentOf(A, B)) = 2*sqrt(2)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3)", "fact_spans": "[[[0, 61], [125, 128]], [[8, 61]], [[8, 61]], [[68, 92]], [[95, 98]], [[99, 102]], [[8, 61]], [[8, 61]], [[0, 61]], [[68, 92]], [[0, 104]], [[105, 123]]]", "query_spans": "[[[125, 135]]]", "process": "Let the equations of the asymptotes of the hyperbola be y = kx, where k = \\pm\\frac{b}{a}. The center of the circle (x-3)^{2}+y^{2}=8 is C(3,0), and the radius is r = 2\\sqrt{2}. The distance from the center C to the line y = kx is d = \\frac{|3k|}{\\sqrt{k^{2}+1}}. Since |AB| = 2\\sqrt{2}, by the Pythagorean theorem we have r^{2} = (\\frac{|AB|}{2})^{2} + d^{2}, that is, 2 + (\\frac{|3k|}{\\sqrt{k^{2}+1}})^{2} = 8. Solving gives k = \\pm\\sqrt{2}, therefore \\frac{b}{a} = \\sqrt{2}. Thus, the eccentricity of the hyperbola is e = \\frac{c}{a} = \\sqrt{\\frac{c^{2}}{a^{2}}} = \\sqrt{\\frac{a^{2}+b^{2}}{a^{2}}} = \\sqrt{1+(\\frac{b}{a})^{2}} = \\sqrt{3}." }, { "text": "The standard equation of a parabola with focus at the right vertex of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ is?", "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (x^2/16 - y^2/9 = 1);RightVertex(G)=Focus(H)", "query_expressions": "Expression(H)", "answer_expressions": "y^2=16*x", "fact_spans": "[[[1, 40]], [[48, 51]], [[1, 40]], [[0, 51]]]", "query_spans": "[[[48, 58]]]", "process": "Hyperbola \\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1, so the right vertex is (4,0). The focus of the parabola is also (4,0), so \\frac{p}{2}=4, p=8. The standard equation of the parabola is: y^2=16x" }, { "text": "Given the hyperbola: $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, what is its focal length? What are the equations of the asymptotes? What is the distance from a focus to an asymptote?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1)", "query_expressions": "FocalLength(G);Expression(Asymptote(G));Distance(Focus(G), Asymptote(G))", "answer_expressions": "10;y = pm*(4/3)*x;4", "fact_spans": "[[[2, 42], [44, 45]], [[2, 42]]]", "query_spans": "[[[44, 50]], [[44, 57]], [[44, 68]]]", "process": "a=3, b=4, c=5, so the focal distance is 2c=10, the asymptotes are y=\\pm\\frac{4}{3}x, and the distance from the focus to the directrix is b=4." }, { "text": "Given that point $P$ is a moving point on the right branch of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, and $F_{1}$, $F_{2}$ are the left and right foci of the hyperbola, respectively. The moving point $Q$ satisfies the following conditions: (1) $\\overrightarrow{Q F_{2}} \\cdot(\\frac{\\overrightarrow{P F_{1}}}{|\\overrightarrow{P F_{1}}|}+\\frac{\\overrightarrow{P F_{2}}}{|\\overrightarrow{P F_{2}}|})=0$, (2) $\\overrightarrow{Q P}+\\lambda(\\frac{\\overrightarrow{P F_{1}}}{|\\overrightarrow{P F_{1}}|}+\\frac{\\overrightarrow{P F_{2}}}{|\\overrightarrow{P F_{2}}|})=0$. Then the trajectory equation of point $Q$ is?", "fact_expressions": "P: Point;PointOnCurve(P, RightPart(G));G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;Q: Point;lambda: Number;DotProduct(VectorOf(Q, F2), (VectorOf(P, F1)/Abs(VectorOf(P, F1)) + VectorOf(P, F2)/Abs(VectorOf(P, F2)))) = 0;VectorOf(Q, P) + lambda*(VectorOf(P, F1)/Abs(VectorOf(P, F1)) + VectorOf(P, F2)/Abs(VectorOf(P, F2))) = 0", "query_expressions": "LocusEquation(Q)", "answer_expressions": "(x^2+y^2=1)&(Negation(y=0))", "fact_spans": "[[[2, 6]], [[2, 41]], [[7, 35], [58, 61]], [[7, 35]], [[42, 49]], [[50, 57]], [[42, 67]], [[42, 67]], [[70, 73], [401, 405]], [[246, 399]], [[85, 240]], [[246, 399]]]", "query_spans": "[[[401, 412]]]", "process": "Let the coordinates of the moving point Q be (x, y). Extend F_{2}Q to intersect PF at point A. According to the vector addition law and the condition that the dot product is 0, we obtain QF_{2}\\bot PQ. Using the definition of the hyperbola, we get |OQ| = \\frac{1}{2}|AF_{1}| = 1, which leads to the answer. Let the coordinates of the moving point Q be (x, y). Extend F_{2}Q to intersect PF_{1} at point A. From condition \\textcircled{2}, we know that point Q lies on the angle bisector of \\angle F_{1}PF_{2}. Combining condition \\textcircled{1}, we have QF_{2}\\bot PQ. Therefore, in \\triangle PF_{2}A, PQ\\bot F_{2}A. Also, since PQ bisects \\angle APF_{2}, \\triangle PF_{2}A is an isosceles triangle, i.e., |PA| = |PF_{2}|, |AQ| = |QF_{2}|. Since point P is a point on the hyperbola, we have |PF_{1}| - |PF_{2}| = 2, that is, |PA| + |AF_{1}| - |PF_{2}| = 2, so |AF_{1}| = 2. Also, in \\triangle F_{1}AF_{2}, Q is the midpoint of AF_{2}, O is the midpoint of F_{1}F_{2}, so |OQ| = \\frac{1}{2}|AF_{1}| = 1. Therefore, the trajectory of point Q is a circle with center O and radius 1. Thus, the equation of the trajectory of point Q is x^{2} + y^{2} = 1 (y \\neq 0)." }, { "text": "Given points $A(-1,0)$, $B(1,0)$, a line passing through $A$ intersects the parabola $y^{2}=4 x$ at points $P$, $Q$. If $P$ is the midpoint of $A Q$, then $\\frac{|P B|}{|Q B|}$=?", "fact_expressions": "G: Parabola;H: Line;A: Point;Q: Point;B: Point;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (-1, 0);Coordinate(B) = (1, 0);PointOnCurve(A, H);Intersection(H, G) = {P, Q};MidPoint(LineSegmentOf(A, Q)) = P", "query_expressions": "Abs(LineSegmentOf(P, B))/Abs(LineSegmentOf(Q, B))", "answer_expressions": "1/2", "fact_spans": "[[[31, 45]], [[28, 30]], [[2, 12], [24, 27]], [[52, 55]], [[14, 22]], [[48, 51], [60, 63]], [[31, 45]], [[2, 12]], [[14, 22]], [[23, 30]], [[28, 57]], [[60, 71]]]", "query_spans": "[[[73, 96]]]", "process": "It is easy to see that for the parabola $ y^{2}=4x $, the focus is $ B $, and the directrix is $ x=-1 $. Draw perpendicular segments from points $ P $ and $ Q $ to the directrix, with feet at points $ D $ and $ C $, respectively. According to the definition of a parabola, we have $ |PB|=|PD| $, $ |QB|=|QC| $. Since $ PD \\parallel QC $, and $ P $ is the midpoint of $ AQ $, it follows that $ PD $ is the midline of $ \\triangle AQC $, so $ |PD|=\\frac{1}{2}|QC| $, that is, $ |PB|=\\frac{1}{2}|QB| $. Therefore, $ \\frac{|PB|}{|OB|}=\\frac{1}{2} $." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$ with foci $F_{1}$, $F_{2}$, and a point $P(x_{0}, y_{0})$ satisfying $0<\\frac{x_{0}^{2}}{2}+y_{0}^{2}<1$, then the range of values for $|P F_{1}|+|P F_{2}|$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/2 + y^2 = 1);F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;x0: Number;y0: Number;Coordinate(P) = (x0, y0);0 < x0^2/2+y0^2;x0^2/2+y0^2 < 1", "query_expressions": "Range(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)))", "answer_expressions": "[2, 2*sqrt(2)]", "fact_spans": "[[[2, 34]], [[2, 34]], [[39, 46]], [[47, 54]], [[2, 54]], [[56, 74]], [[76, 112]], [[76, 112]], [[56, 74]], [[76, 112]], [[76, 112]]]", "query_spans": "[[[114, 142]]]", "process": "" }, { "text": "Given that the ellipse $m x^{2}+4 y^{2}=1$ with foci on the $y$-axis has eccentricity $\\frac{\\sqrt{2}}{2}$, then the real number $m$ equals?", "fact_expressions": "PointOnCurve(Focus(G), yAxis);G: Ellipse;Expression(G) = (m*x^2 + 4*y^2 = 1);m: Real;Eccentricity(G) = sqrt(2)/2", "query_expressions": "m", "answer_expressions": "8", "fact_spans": "[[[2, 32]], [[11, 32]], [[11, 32]], [[59, 64]], [[11, 57]]]", "query_spans": "[[[59, 67]]]", "process": "When the focus is on the y-axis, $a^{2}=\\frac{1}{4},b^{2}=\\frac{1}{m},e=\\frac{\\sqrt{2}}{2}\\Rightarrow a=\\sqrt{2}c=\\sqrt{2}b \\therefore \\frac{1}{4}=2\\times\\frac{1}{m}, m=8$" }, { "text": "The standard equation of the hyperbola with foci at $(0,6)$, $(0,-6)$ and passing through the point $(2,-5)$ is?", "fact_expressions": "G: Hyperbola;H: Point;I: Point;J: Point;Coordinate(H) = (0, 6);Coordinate(I) = (0, -6);Coordinate(J) = (2, -5);Focus(G) = {H,I};PointOnCurve(J,G)", "query_expressions": "Expression(G)", "answer_expressions": "y^2/20 - x^2/16 = 1", "fact_spans": "[[[31, 34]], [[3, 10]], [[11, 19]], [[21, 30]], [[3, 10]], [[11, 19]], [[21, 30]], [[0, 34]], [[19, 34]]]", "query_spans": "[[[31, 41]]]", "process": "" }, { "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$, respectively. If point $P$ lies on the ellipse and $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, then $|\\overrightarrow{P F_{1}}+\\overrightarrow{P F_{2}}|$=?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/7 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0", "query_expressions": "Abs(VectorOf(P, F1) + VectorOf(P, F2))", "answer_expressions": "6", "fact_spans": "[[[19, 57], [70, 72]], [[19, 57]], [[1, 8]], [[9, 16]], [[1, 63]], [[1, 63]], [[65, 69]], [[65, 73]], [[75, 134]]]", "query_spans": "[[[136, 191]]]", "process": "From the equation of the ellipse $\\frac{x^2}{16} + \\frac{y^2}{7} = 1$, we obtain $a = 4$, then $c^2 = a^2 - b^2 = 9$. Given that $\\overrightarrow{PF}_1 \\cdot \\overrightarrow{PF_2} = 0$, it follows that $|\\overrightarrow{PF_1}|^2 + |\\overrightarrow{PF_2}|^2 = 4c^2$, that is, $(\\overrightarrow{PF_1} + \\overrightarrow{PF_2})^2 - 2\\overrightarrow{PF_1} \\cdot \\overrightarrow{PF_2} = 36$. Hence, $4 \\times 16 - 2|\\overrightarrow{PF_1}||\\overrightarrow{PF_2}| = 36$, so $|\\overrightarrow{PF_1}||\\overrightarrow{PF_2}| = 14$. Therefore, $|\\overrightarrow{PF_1} + \\overrightarrow{PF_2}|^2 = |\\overrightarrow{PF_1}|^2 + |\\overrightarrow{PF_2}|^2 + 2\\overrightarrow{PF_1} \\cdot \\overrightarrow{PF_2} = (|\\overrightarrow{PF_1}| + |\\overrightarrow{PF_2}|)^2 - 2|\\overrightarrow{PF_1}||\\overrightarrow{PF_2}| = 4a^2 - 2 \\times 14 = 36$. Thus, $|\\overrightarrow{PF_1}| + |\\overrightarrow{PF_2}| = 6$." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are denoted by $F_{1}$ and $F_{2}$ respectively. If there exists a point $P$ on the right branch of the hyperbola such that $|PF_{1}|=3|PF_{2}|$, then the range of the eccentricity $e$ of the hyperbola is?", "fact_expressions": "G: Hyperbola;a: Number;b: Number;P: Point;F1: Point;F2: Point;e: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G)=F2;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, F2));Eccentricity(G) = e", "query_expressions": "Range(e)", "answer_expressions": "(1,2]", "fact_spans": "[[[2, 58], [85, 88], [124, 127]], [[5, 58]], [[5, 58]], [[96, 99]], [[67, 74]], [[75, 82]], [[131, 134]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 82]], [[2, 82]], [[85, 99]], [[102, 122]], [[124, 134]]]", "query_spans": "[[[131, 141]]]", "process": "" }, { "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a}+y^{2}=1$ is $x+\\sqrt{3} y=0$, then the equation of this hyperbola is?", "fact_expressions": "G: Hyperbola;a: Number;Expression(G) = (y^2 + x^2/a = 1);Expression(OneOf(Asymptote(G))) = (x + sqrt(3)*y = 0)", "query_expressions": "Expression(G)", "answer_expressions": "y^2 - x^2 / 3 = 1", "fact_spans": "[[[2, 30], [57, 60]], [[5, 30]], [[2, 30]], [[2, 55]]]", "query_spans": "[[[57, 64]]]", "process": "From the given conditions, we have a<0, so the standard equation is y^{2}+\\frac{x^{2}}{a}=1. Since the asymptote is y=\\frac{\\sqrt{3}}{3}x, it follows that a=-\\frac{1}{3}, and thus y^{2}-\\frac{x^{2}}{3}=1. Fill in y^{2}-\\frac{x^{2}}{3}=1." }, { "text": "The asymptotes of the hyperbola $x^{2}-y^{2}=10$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 10)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*x", "fact_spans": "[[[0, 19]], [[0, 19]]]", "query_spans": "[[[0, 26]]]", "process": "Test analysis: The slopes of the asymptotes of an equilateral hyperbola are \\pm1, so the answer is y=\\pmx." }, { "text": "The hyperbola $8 kx^{2}-ky^{2}=8$ has a focus at $(0,3)$; then the value of the real number $k$ is?", "fact_expressions": "G: Hyperbola;k: Real;Expression(G)=(8*k*x^2-k*y^2=8);Coordinate(OneOf(Focus(G)))=(0,3)", "query_expressions": "k", "answer_expressions": "-1", "fact_spans": "[[[0, 22]], [[37, 42]], [[0, 22]], [0, 34]]", "query_spans": "[[[37, 46]]]", "process": "" }, { "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$, directrix $l$, and point $P$ is a point on the parabola $C$. A perpendicular is drawn from point $P$ to $l$, with foot of perpendicular at $A$. Let $B$ be the intersection point of the directrix $l$ and the $x$-axis. If $\\angle B A F=30^{\\circ}$, and the area of $\\triangle A P F$ is $12 \\sqrt{3}$, then what is the standard equation of the circle with $P F$ as diameter?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;l: Line;Directrix(C) = l;P: Point;PointOnCurve(P, C);Z: Line;PointOnCurve(P, Z);IsPerpendicular(Z, l);A: Point;FootPoint(Z, l) = A;B: Point;Intersection(l, xAxis) = B;AngleOf(B, A, F) = ApplyUnit(30, degree);Area(TriangleOf(A, P, F)) = 12*sqrt(3);G: Circle;IsDiameter(LineSegmentOf(P, F), G)", "query_expressions": "Expression(G)", "answer_expressions": "(x-2*sqrt(3))^2+(y+pm*3)^2=12", "fact_spans": "[[[2, 28], [48, 54]], [[2, 28]], [[10, 28]], [[10, 28]], [[32, 35]], [[2, 35]], [[39, 42], [64, 67], [80, 83]], [[2, 42]], [[43, 47], [59, 63]], [[43, 57]], [], [[58, 70]], [[58, 70]], [[74, 77]], [[58, 77]], [[93, 96]], [[80, 96]], [[98, 123]], [[125, 159]], [[171, 172]], [[161, 172]]]", "query_spans": "[[[171, 179]]]", "process": "According to the problem, the figure is drawn as shown: since $\\angle BAF = 30^{\\circ}$, it follows that $\\angle AFB = 60^{\\circ} = \\angle PAF$. By the definition of a parabola, $|PA| = |PF|$, hence $\\triangle APF$ is an equilateral triangle. Since the area of $\\triangle APF$ is $12\\sqrt{3}$, it follows that $|PF| = |PA| = |AF| = 4\\sqrt{3}$. Therefore, $|BF| = \\frac{1}{2}|AF| = 2\\sqrt{3} = p$, so the equation of the parabola is $y^{2} = 4\\sqrt{3}x$, and point $F$ is $(\\sqrt{3}, 0)$. Thus, the x-coordinate of point $P$ is $|PA| - \\frac{|BF|}{2} = 3\\sqrt{3}$. Substituting into $y^{2} = 4\\sqrt{3}x$, we solve to get $y = \\pm6$, so point $P$ is $(3\\sqrt{3}, \\pm6)$. Therefore, the center of the required circle is $(2\\sqrt{3}, \\pm3)$, and the radius is $\\frac{|PF|}{2} = 2\\sqrt{3}$. Hence, the standard equation of the required circle is $(x - 2\\sqrt{3})^{2} + (y \\pm 3)^{2} = 12$." }, { "text": "Given $F_{1}(-c, 0)$, $F_{2}(c, 0)$ are the two foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, $P$ lies on the ellipse, and the area of $\\Delta P F_{1} F_{2}$ is $\\frac{\\sqrt{2}}{2} b^{2}$, then $\\cos \\angle F_{1} P F_{2}$=?", "fact_expressions": "G: Ellipse;a: Number;b: Number;c: Number;F1: Point;F2: Point;P: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);Focus(G) = {F1, F2};PointOnCurve(P, G);Area(TriangleOf(P, F1, F2)) = b^2*(sqrt(2)/2)", "query_expressions": "Cos(AngleOf(F1, P, F2))", "answer_expressions": "1/3", "fact_spans": "[[[32, 84], [94, 96]], [[34, 84]], [[34, 84]], [[18, 31]], [[2, 16]], [[18, 31]], [[90, 93]], [[34, 84]], [[34, 84]], [[32, 84]], [[2, 16]], [[18, 31]], [[2, 89]], [[90, 97]], [[99, 151]]]", "query_spans": "[[[153, 182]]]", "process": "Problem Analysis: Let |PF_{1}|=m, |PF_{2}|=n, \\angle F_{1}PF_{2}=\\alpha. From the given condition, \\frac{1}{2}mn\\sin\\alpha=\\frac{\\sqrt{2}}{2}b^{2}, so mn=\\frac{\\sqrt{2}b^{2}}{\\sin\\alpha}. Also, m+n=2a, and (2c)^{2}=m^{2}+n^{2}-2mn\\cos\\alpha=(m+n)^{2}-2mn-2mn\\cos\\alpha, which implies 2mn(1+\\cos\\alpha)=4a^{2}-4c^{2}=4b^{2}. Therefore, \\frac{2\\sqrt{2}b^{2}}{\\sin\\alpha}\\cdot(1+\\cos\\alpha)=4b^{2}, leading to 1+\\cos\\alpha=\\sqrt{2}\\sin\\alpha. Then, (1+\\cos\\alpha)^{2}=2\\sin^{2}\\alpha=2(1-\\cos^{2}\\alpha), so 3\\cos^{2}\\alpha+2\\cos\\alpha-1=0. Thus, \\cos\\alpha=\\frac{1}{3} (\\cos\\alpha=-1 is discarded)." }, { "text": "The line passing through a focus of the hyperbola $x^{2}-y^{2}=1$ and perpendicular to the $x$-axis intersects the two asymptotes of the hyperbola at points $P$ and $Q$. Then $|P Q |$=?", "fact_expressions": "G: Hyperbola;H: Line;P: Point;Q: Point;Expression(G) = (x^2 - y^2 = 1);PointOnCurve(OneOf(Focus(G)), H);IsPerpendicular(H, xAxis);L1:Line;L2:Line;Asymptote(G)={L1,L2};Intersection(H,L1)=P;Intersection(H,L2)=Q", "query_expressions": "Abs(LineSegmentOf(P, Q))", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[1, 19], [38, 41]], [[33, 35]], [[48, 51]], [[52, 55]], [[1, 19]], [[0, 35]], [[25, 35]], [-1, -1], [-1, -1], [37, 46], [33, 56], [33, 56]]", "query_spans": "[[[59, 69]]]", "process": "From the given conditions, a=1, b=1, c=\\sqrt{2}, the hyperbola is an equilateral hyperbola, so the angle between the two asymptotes is 90^{\\circ}. Because in a right triangle, the median to the hypotenuse is half the length of the hypotenuse, thus |PQ|=2\\sqrt{2}" }, { "text": "The line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$, and point $O$ is the coordinate origin. If $|AF|=5$, then the area of $\\triangle ABO$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);A: Point;B: Point;Intersection(H, G) = {A, B};O: Origin;Abs(LineSegmentOf(A, F)) = 5", "query_expressions": "Area(TriangleOf(A, B, O))", "answer_expressions": "5/2", "fact_spans": "[[[1, 15], [25, 28]], [[1, 15]], [[18, 21]], [[1, 21]], [[22, 24]], [[0, 24]], [[29, 32]], [[33, 36]], [[22, 38]], [[39, 43]], [[50, 59]]]", "query_spans": "[[[61, 83]]]", "process": "From the given, we have p=2. As shown in the figure, draw AA_{1}\\bot l, with foot at A_{1}. Therefore, x_{A}+\\frac{p}{2}=5, so x_{A}=4. Substituting into y^{2}=4x gives y_{A}=\\pm4. Hence, A(4,4) or A(4,-4). Without loss of generality, assume A(4,4), and F(1,0). The equation of line AB is \\frac{y-0}{4-0}=\\frac{x-1}{4-1}, which simplifies to x=\\frac{3}{4}y+1. Substituting into y^{2}=4x yields y^{2}=3y+4, so y_{B}=-1. Therefore, S_{\\DeltaAOB}=\\frac{1}{2}|OF|(|y_{A}|+|y_{B}|)=\\frac{1}{2}\\times1\\times(4+1)=\\frac{5}{2}." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) has a real axis length of $2$ and eccentricity of $2$, then the coordinates of the foci of hyperbola $C$ are?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Length(RealAxis(C))=2;Eccentricity(C) = 2", "query_expressions": "Coordinate(Focus(C))", "answer_expressions": "(pm*2,0)", "fact_spans": "[[[2, 64], [82, 88]], [[9, 64]], [[9, 64]], [[9, 64]], [[9, 64]], [[2, 64]], [[2, 72]], [[2, 80]]]", "query_spans": "[[[82, 95]]]", "process": "" }, { "text": "The focus of the parabola $y^{2}=2 p x(p>0)$ is $F$, points $A$ and $B$ lie on the parabola such that $\\angle A F B=\\frac{\\pi}{2}$, and the midpoint $M$ of chord $AB$ has its projection $N$ on the directrix. Then the maximum value of $\\frac{|M N|}{|A B|}$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);AngleOf(A, F, B) = pi/2;IsChordOf(LineSegmentOf(A, B), G);M: Point;MidPoint(LineSegmentOf(A, B)) = M;N: Point;Projection(M, Directrix(G)) = N", "query_expressions": "Max(Abs(LineSegmentOf(M, N))/Abs(LineSegmentOf(A, B)))", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[0, 21], [37, 40], [85, 86]], [[0, 21]], [[3, 21]], [[3, 21]], [[25, 28]], [[0, 28]], [[29, 32]], [[33, 36]], [[29, 41]], [[33, 41]], [[43, 71]], [[37, 78]], [[81, 84]], [[73, 84]], [[93, 96]], [[81, 96]]]", "query_spans": "[[[98, 125]]]", "process": "Problem Analysis: According to the given conditions, since the focus of the parabola $ y^{2} = 2px $ ($ p > 0 $) is $ F $, points $ A $ and $ B $ lie on the parabola, and the line $ AB: y = k\\left(x - \\frac{p}{2}\\right) $, solving this equation together with the parabola's equation, combined with $ \\angle AFB = \\frac{\\pi}{2} $, the midpoint $ M $ of chord $ AB $ has its projection $ N $ on the directrix; then the maximum value of $ \\frac{|MN|}{|AB|} $ is $ \\frac{\\sqrt{2}}{2} $." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1$, respectively, and the line $x-2 y+1=0$ intersects the ellipse at points $P$ and $Q$, then the perimeter of $\\triangle P Q F_{2}$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/3 + y^2/2 = 1);l: Line;P: Point;Q: Point;F2: Point;F1: Point;Expression(l) = (x - 2*y + 1 = 0);LeftFocus(C) = F1;RightFocus(C) = F2;Intersection(l, C) = {P, Q}", "query_expressions": "Perimeter(TriangleOf(P, Q, F2))", "answer_expressions": "4*sqrt(3)", "fact_spans": "[[[20, 62], [82, 84]], [[20, 62]], [[68, 81]], [[86, 89]], [[90, 93]], [[10, 17]], [[2, 9]], [[68, 81]], [[2, 67]], [[2, 67]], [[68, 95]]]", "query_spans": "[[[97, 123]]]", "process": "From the equation of the ellipse $ C: \\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1 $, we know: $ F_{1}(-1,0) $, $ F_{2}(1,0) $. The line $ x-2y+1=0 $, setting $ y=0 $, gives $ x=-1 $, so the line passes through the left focus of the ellipse. By the definition of the ellipse: $ PF_{1}+PF_{2}=2a=2\\sqrt{3} $, $ QF_{1}+QF_{2}=2a=2\\sqrt{3} $, $ PQ=PF_{1}+QF_{1} $, then the perimeter of $ \\triangle PQF_{2} $ is $ PQ+PF_{2}+QF_{2}=4a=4\\sqrt{3} $." }, { "text": "Given that point $P$ is a moving point on the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$, $F_{1}$ and $F_{2}$ are its left and right foci respectively, and $O$ is the origin, then the range of values of $\\frac{|P F_{1}|-|P F_{2}|}{|O P|}$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/8 + y^2/4 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;O: Origin", "query_expressions": "Range((Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)))/Abs(LineSegmentOf(O, P)))", "answer_expressions": "[0, sqrt(2)]", "fact_spans": "[[[7, 44], [67, 68]], [[7, 44]], [[2, 6]], [[2, 48]], [[49, 56]], [[57, 64]], [[49, 73]], [[49, 73]], [[74, 77]]]", "query_spans": "[[[84, 126]]]", "process": "" }, { "text": "The line $l$ passing through point $P(1,1)$ intersects the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ at points $A$ and $B$, and $\\overrightarrow{A P}=\\lambda \\overrightarrow{P B}$. Point $Q$ satisfies $\\overrightarrow{A Q}=-\\lambda \\overrightarrow{Q B}$. If $O$ is the coordinate origin and $Q(m, n)$, then the value of $\\frac{m}{4}+\\frac{n}{3}$ is?", "fact_expressions": "l: Line;G: Ellipse;P: Point;Q: Point;A: Point;B: Point;O: Origin;m: Number;n: Number;Expression(G) = (x^2/4 + y^2/3 = 1);Coordinate(P) = (1, 1);Coordinate(Q) = (m, n);PointOnCurve(P, l);Intersection(l, G) = {A, B};lambda: Number;VectorOf(A, P) = lambda*VectorOf(P, B);VectorOf(A, Q) = -lambda*VectorOf(Q, B)", "query_expressions": "m/4 + n/3", "answer_expressions": "1", "fact_spans": "[[[11, 16]], [[17, 54]], [[1, 10]], [[117, 121], [187, 196]], [[56, 60]], [[61, 64]], [[177, 180]], [[187, 196]], [[187, 196]], [[17, 54]], [[1, 10]], [[187, 196]], [[0, 16]], [[11, 64]], [[66, 117]], [[66, 117]], [[123, 175]]]", "query_spans": "[[[198, 227]]]", "process": "According to the given conditions, let A(x_{1},y_{1}), B(x_{2},y_{2}), point P(1,1), point Q(m,n). Given \\overrightarrow{AP}=\\lambda\\overrightarrow{PB}, \\overrightarrow{AQ}=-\\lambda\\overrightarrow{QB}, then 1-x_{1}=\\lambda(x_{2}-1), m-x_{1}=-\\lambda(x_{2}-m), that is, x_{1}+\\lambda x_{2}=1+\\lambda, x_{1}-\\lambda x_{2}=m(1-\\lambda). Multiplying these two equations yields: x_{1}^{2}-\\lambda^{2}x_{2}^{2}=m(1-\\lambda^{2}), i.e., \\frac{x_{1}^{2}}{4}-\\frac{\\lambda^{2}x_{2}^{2}}{4}=\\frac{m}{4}(1-\\lambda^{2})\\cdots\\cdots\\textcircled{1}. Similarly, we obtain: \\frac{y_{1}^{2}}{3}-\\frac{\\lambda^{2}y_{2}^{2}}{3}=\\frac{n}{3}(1-\\lambda^{2})\\cdots\\cdots\\textcircled{2}. Since points A(x_{1},y_{1}), B(x_{2},y_{2}) lie on the ellipse, we have \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1. Adding equations \\textcircled{1} and \\textcircled{2} gives: \\frac{x_{1}^{2}}{4}-\\frac{\\lambda^{2}x_{2}^{2}}{4}+\\frac{y_{1}^{2}}{3}-\\frac{\\lambda^{2}y_{2}^{2}}{3}=\\frac{m}{4}(1-\\lambda^{2})+\\frac{n}{3}(1-\\lambda^{2}), i.e., \\frac{x_{1}^{2}}{4}+\\frac{y_{1}^{2}}{3}-\\lambda^{2}\\left(\\frac{x_{2}^{2}}{4}+\\frac{y_{2}^{2}}{3}\\right)=(1-\\lambda^{2})\\left(\\frac{m}{4}+\\frac{n}{3}\\right). Hence, 1-\\lambda^{2}=(1-\\lambda^{2})\\left(\\frac{m}{4}+\\frac{n}{3}\\right). Therefore, \\frac{m}{4}+\\frac{n}{3}=|" }, { "text": "Given a point $P$ on the parabola $x^{2}=4 y$ such that the distance from $P$ to the $x$-axis is $8$, then the distance from point $P$ to the focus of the parabola is?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);P: Point;PointOnCurve(P, G);Distance(P, xAxis) = 8", "query_expressions": "Distance(P, Focus(G))", "answer_expressions": "9", "fact_spans": "[[[2, 16], [42, 45]], [[2, 16]], [[19, 22], [36, 40]], [[2, 22]], [[19, 34]]]", "query_spans": "[[[36, 52]]]", "process": "First draw the graph, then calculate using the definition of the parabola. The focus of the parabola $x^{2}=4y$ is $F(0,1)$, and the equation of the directrix is $y=-1$. As shown in the figure, $|PA|=8$, $|AB|=1$. By the definition of the parabola, we obtain: $|PF|=|PB|=|PA|+|AB|=9$" }, { "text": "Given that an ellipse and a hyperbola share the same foci $F_{1}$ and $F_{2}$, let the eccentricities of the ellipse and the hyperbola be $e_{1}$, $e_{2}$ respectively, and let $P$ be a common point on both curves such that $|\\overrightarrow{P F_{1}}-\\overrightarrow{P F_{2}}|=2|\\overrightarrow{P O}|$ ($O$ is the coordinate origin). If $e_{1} \\in\\left(\\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{3}}{2}\\right)$, then the range of values for $e_{2}$ is?", "fact_expressions": "H: Ellipse;G: Hyperbola;F1: Point;F2: Point;Focus(H) = {F1, F2};Focus(G) = {F1, F2};e1: Number;e2: Number;Eccentricity(H) = e1;Eccentricity(G) = e2;P: Point;OneOf(Intersection(H, G)) = P;Abs(VectorOf(P, F1) - VectorOf(P, F2)) = 2*Abs(VectorOf(P, O));O: Origin;In(e1, (sqrt(2)/2, sqrt(3)/2))", "query_expressions": "Range(e2)", "answer_expressions": "[sqrt(6)/2, +oo)", "fact_spans": "[[[2, 4], [31, 33]], [[5, 8], [34, 37]], [[14, 21]], [[22, 29]], [[2, 29]], [[2, 29]], [[44, 51]], [[53, 60], [223, 230]], [[31, 60]], [[31, 60]], [[62, 65]], [[62, 75]], [[77, 154]], [[157, 160]], [[170, 221]]]", "query_spans": "[[[223, 237]]]", "process": "Let the semi-major axis of the ellipse be $ a $, the semi-transverse axis of the hyperbola be $ a $, and their semi-focal distance be $ c $. Then we obtain $ a = \\frac{c}{e_{1}} $, $ a = \\frac{c}{e_{2}} $. By the symmetry of the ellipse and hyperbola, without loss of generality, assume the foci $ F_{1} $ and $ F_{2} $ lie on the x-axis, and point $ P $ lies to the right of the y-axis. From the definitions of the ellipse and hyperbola, we have:\n\\[\n\\begin{cases}\n|PF_{1}| + |PF_{2}| = 2a \\\\\n|PF_{1} - |PF_{1}| = 2a\n\\end{cases}\n\\]\nSolving gives $ |PF_{1}| = a + a $, $ |PF_{2}| = a - a $. Since $ |\\overrightarrow{PF_{1}} - \\overrightarrow{PF_{2}}| = 2|\\overrightarrow{PO}| $, i.e., $ |\\overrightarrow{F_{1}F_{2}}| = 2|\\overrightarrow{PO}| $, and $ O $ is the midpoint of segment $ F_{1}F_{2} $, it follows that $ \\angle F_{1}PF_{2} = 90^{\\circ} $. Then we have $ |PF_{1}|^{2} + |PF_{2}|^{2} = |F_{1}F_{2}| $, i.e.,\n\\[\n(a + a)^{2} + (a - a)^{2} = 4c^{2}\n\\]\nSimplifying yields $ a^{2} + a^{2} = 2c^{2} $, thus\n\\[\n\\left(\\frac{c}{e_{1}}\\right)^{2} + \\left(\\frac{c}{e_{2}}\\right)^{2} = 2c^{2}\n\\]\nwhich implies\n\\[\n\\frac{1}{e_{2}^{2}} = 2 - \\frac{1}{e_{1}^{2}}\n\\]\nGiven $ \\frac{\\sqrt{2}}{2} < e_{1} \\leqslant \\frac{\\sqrt{3}}{2} $, it follows that $ 0 < 2 - \\frac{1}{e_{1}^{2}} \\leqslant \\frac{2}{3} $, i.e., $ e_{2}^{2} \\geqslant \\frac{3}{2} $, solving gives $ e_{2} \\geqslant \\frac{\\sqrt{6}}{2} $, so the range of values for $ e_{2} $ is $ \\left[\\frac{\\sqrt{6}}{2}, +\\infty\\right) $" }, { "text": "The coordinates of the foci of the ellipse $2 x^{2}+4 y^{2}=1$ are?", "fact_expressions": "G: Ellipse;Expression(G) = (2*x^2 + 4*y^2 = 1)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(pm*(1/2),0)", "fact_spans": "[[[0, 21]], [[0, 21]]]", "query_spans": "[[[0, 28]]]", "process": "Solution: From $2x^{2}+4y^{2}=1$, converting to standard form gives: $\\frac{x^{2}}{\\frac{1}{2}}+\\frac{y^{2}}{\\frac{1}{4}}=1$. Therefore, the ellipse has its foci on the x-axis, with $a^{2}=\\frac{1}{2}$, $b^{2}=\\frac{1}{4}$, so $c^{2}=a^{2}-b^{2}=\\frac{1}{2}-\\frac{1}{4}=\\frac{1}{4}$, then $c=\\frac{1}{2}$. Therefore, the coordinates of the foci are $(\\frac{1}{2},0)$ and $(-\\frac{1}{2},0)$." }, { "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and $F_{1}$, $F_{2}$ are the left and right foci of the ellipse, then the minimum value of $\\frac{1}{|P F_{1}|}+\\frac{1}{|PF_{2}|}$ is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2", "query_expressions": "Min(1/Abs(LineSegmentOf(P,F1))+1/Abs(LineSegmentOf(P,F2)))", "answer_expressions": "2/a", "fact_spans": "[[[6, 58], [79, 81]], [[8, 58]], [[8, 58]], [[2, 5]], [[63, 70]], [[71, 78]], [[8, 58]], [[8, 58]], [[6, 58]], [[2, 62]], [[63, 87]], [[63, 87]]]", "query_spans": "[[[89, 135]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, find the equation of the line containing the chord for which $P(-1,1)$ is the midpoint.", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);P: Point;Coordinate(P) = (-1, 1);H: LineSegment;IsChordOf(H, G);MidPoint(H) = P", "query_expressions": "Expression(OverlappingLine(H))", "answer_expressions": "3*x-4*y+7=0", "fact_spans": "[[[2, 39]], [[2, 39]], [[42, 51]], [[42, 51]], [], [[2, 56]], [[2, 56]]]", "query_spans": "[[[2, 65]]]", "process": "Let the two endpoints of the chord be A(x_{1},y_{1}), B(x_{2},y_{2}) and subtract to obtain y-y. Also, since M(-1,1) is the midpoint of AB, therefore x_{1}+x_{2}=-2, y_{1}+y_{2}=2. Substituting into equation \\textcircled{1} gives \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{3}{4}, that is, k_{AB}=\\frac{3}{4}. Therefore, the equation of line AB is y-1=\\frac{3}{4}(x+1), or 3x-4y+7=0." }, { "text": "Given the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. A line passing through point $F_{1}$ intersects the ellipse at points $M$ and $N$. Then, what is the perimeter of $\\triangle MNF_{2}$?", "fact_expressions": "G: Ellipse;H: Line;M: Point;N: Point;F2: Point;F1: Point;Expression(G) = (x^2/16 + y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, H);Intersection(H, G) = {M, N}", "query_expressions": "Perimeter(TriangleOf(M, N, F2))", "answer_expressions": "16", "fact_spans": "[[[2, 40], [78, 80]], [[75, 77]], [[81, 84]], [[85, 88]], [[57, 64]], [[49, 56], [66, 74]], [[2, 40]], [[2, 64]], [[2, 64]], [[65, 77]], [[75, 90]]]", "query_spans": "[[[92, 116]]]", "process": "" }, { "text": "The equation of the directrix of the parabola $y=a x^{2}$ is $y=1$, then the standard equation of this parabola is?", "fact_expressions": "G: Parabola;a: Number;Expression(G) = (y = a*x^2);Expression(Directrix(G)) = (y = 1)", "query_expressions": "Expression(G)", "answer_expressions": "x^2=-4*y", "fact_spans": "[[[0, 14], [28, 31]], [[3, 14]], [[0, 14]], [[0, 25]]]", "query_spans": "[[[28, 38]]]", "process": "" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. Point $A$ is a point on the right branch of the hyperbola. If line $A F_{2}$ is parallel to the line $y=-\\frac{b}{a} x$, and the perimeter of $\\Delta A F_{1} F_{2}$ is $9 a$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;F2: Point;A: Point;F1: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = -(b/a)*x);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(A, RightPart(G));IsParallel(LineOf(A,F2),H);Perimeter(TriangleOf(A, F1, F2)) = 9*a", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[2, 58], [88, 91], [168, 171]], [[5, 58]], [[5, 58]], [[111, 131]], [[75, 82]], [[83, 87]], [[67, 74]], [[5, 58]], [[5, 58]], [[2, 58]], [[111, 131]], [[2, 82]], [[2, 82]], [[83, 97]], [[99, 133]], [[135, 166]]]", "query_spans": "[[[168, 177]]]", "process": "As shown in the figure, let AF₁ = r₁, AF₂ = r₂, then \n\\begin{cases}r_{1}-r_{2}=2a\\\\r_{1}+r_{2}=9a-2c\\end{cases}, \nsolving gives \n\\begin{cases}r_{1}=\\frac{11}{2}a-c\\\\r_{2}=\\frac{7}{2}a-c\\end{cases}. \nSince line AF₂ is parallel to the line y = -\\frac{b}{a}x, \nit follows that \\tan\\angle F_{1}F_{2}A = \\frac{b}{a}, \nso \\cos\\angle F_{1}F_{2}A = \\frac{a}{c} = \\frac{4c^{2}+r_{2}^{2}-r_{1}^{2}}{2\\times2c\\times r_{2}}, \nthus 4c^{2} + r_{2}^{2} - r_{1}^{2} = 4ar_{2}. \nSubstituting r₁ and r₂ into the above equation yields \n8a^{2} - 2ac - c^{2} = 0, \nso (2a - c)(4a + c) = 0, giving e = 2. Hence, fill in: 2." }, { "text": "The standard equation of a parabola passing through the point $A(-2,4)$, with vertex at the origin and axis of symmetry along the coordinate axes, is?", "fact_expressions": "A: Point;Coordinate(A) = (-2, 4);G: Parabola;PointOnCurve(A, G);O: Origin;Vertex(G) = O;SymmetryAxis(G) = axis", "query_expressions": "Expression(G)", "answer_expressions": "{x^2=y, y^2=-8*x}", "fact_spans": "[[[1, 11]], [[1, 11]], [[27, 30]], [[0, 30]], [[16, 18]], [[13, 30]], [[19, 30]]]", "query_spans": "[[[27, 37]]]", "process": "Since point A(-2,4) is in the second quadrant, assume the parabola equation is x^{2}=2py or y^{2}=-2p^{x}. Substituting point A gives p=\\frac{1}{2}, p'=4. Therefore, the required parabola equations are x^{2}=y or y^{2}=-8x." }, { "text": "The distance between the right focus and the left directrix of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{8}=1$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/8 = 1)", "query_expressions": "Distance(RightFocus(G),LeftDirectrix(G))", "answer_expressions": "8*sqrt(3)/3", "fact_spans": "[[[0, 38]], [[0, 38]]]", "query_spans": "[[[0, 53]]]", "process": "" }, { "text": "Through the right focus $F$ of the hyperbola $x^{2}-y^{2}=4$, draw a line with an inclination angle of $105^{\\circ}$, intersecting the hyperbola at points $P$ and $Q$. Then the value of $|F P|\\cdot|F Q|$ is?", "fact_expressions": "G: Hyperbola;H: Line;F: Point;P: Point;Q: Point;Expression(G) = (x^2 - y^2 = 4);RightFocus(G)=F;PointOnCurve(F,H);Inclination(H)=ApplyUnit(105,degree);Intersection(H,G)={P,Q}", "query_expressions": "Abs(LineSegmentOf(F, P))*Abs(LineSegmentOf(F, Q))", "answer_expressions": "8*sqrt(3)/3", "fact_spans": "[[[1, 19], [49, 52]], [[45, 47]], [[23, 26]], [[53, 56]], [[57, 60]], [[1, 19]], [[1, 26]], [[0, 47]], [[27, 47]], [[45, 62]]]", "query_spans": "[[[64, 85]]]", "process": "" }, { "text": "The length of the major axis of an ellipse is $2$ times the length of its minor axis. What is the eccentricity of this ellipse?", "fact_expressions": "G: Ellipse;Length(MajorAxis(G)) = 2*Length(MinorAxis(G))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)/2", "fact_spans": "[[[2, 4], [20, 22]], [[2, 17]]]", "query_spans": "[[[20, 28]]]", "process": "From the given information, we know: a=2b, then substitute into the eccentricity formula_{e}=\\sqrt{1-\\frac{b^{2}}{a^{2}}}. According to the problem: 2a=2\\times2b, that is, a=2b. e=\\frac{c}{a}=\\sqrt{\\frac{a2-b^{2}}{a^{2}}}=\\sqrt{1-\\frac{b^{2}}{a^{2}}}=\\sqrt{1-\\frac{1}{4}}=\\frac{\\sqrt{3}}{2}" }, { "text": "Given $F_{1}$, $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{16}-\\frac{y^{2}}{20}=1$, and point $P$ lies on the hyperbola such that $P F_{2}=5$, find $\\cos \\angle P F_{1} F_{2}$?", "fact_expressions": "C: Hyperbola;P: Point;F2: Point;F1: Point;Expression(C) = (x^2/16 - y^2/20 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);LineSegmentOf(P, F2) = 5", "query_expressions": "Cos(AngleOf(P, F1, F2))", "answer_expressions": "12/13", "fact_spans": "[[[18, 63], [74, 77]], [[70, 73]], [[10, 17]], [[2, 9]], [[18, 63]], [[2, 69]], [[2, 69]], [[70, 78]], [[80, 91]]]", "query_spans": "[[[93, 121]]]", "process": "" }, { "text": "A line with an inclination angle of $30^{\\circ}$ is drawn through the focus $F$ of the parabola $y^{2}=16 x$, intersecting the parabola at points $A$ and $B$, $O$ being the origin. Then the area of $\\triangle A O B$ is?", "fact_expressions": "G: Parabola;H: Line;A: Point;O: Origin;B: Point;F: Point;Expression(G) = (y^2 = 16*x);Focus(G) = F;PointOnCurve(F, H);Inclination(H) = ApplyUnit(30, degree);Intersection(H, G) = {A, B}", "query_expressions": "Area(TriangleOf(A, O, B))", "answer_expressions": "64", "fact_spans": "[[[1, 16], [43, 46]], [[40, 42]], [[47, 50]], [[57, 60]], [[51, 54]], [[19, 22]], [[1, 16]], [[1, 22]], [[0, 42]], [[23, 42]], [[40, 56]]]", "query_spans": "[[[67, 89]]]", "process": "The focus of the parabola $ y^2 = 16x $ is $ F(4,0) $. Therefore, the equation of line $ AB $ is $ y = \\frac{\\sqrt{3}}{3}(x - 4) $. Combining the equation of line $ AB $ with the parabola's equation, we obtain\n\\[\n\\begin{cases}\ny = \\frac{\\sqrt{3}}{3}(x - 4) \\\\\ny^2 = 16x\n\\end{cases}\n\\]\nEliminating variables yields $ y^2 - 16\\sqrt{3}y - 64 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ y_{1} + y_{2} = 16\\sqrt{3} $, $ y_{1}y_{2} = -64 $. $ = 2\\sqrt{(16\\sqrt{3})^{2}} + 64 \\times 4 = 64 $" }, { "text": "Find the standard equation of a hyperbola that shares the same asymptotes as the hyperbola $\\frac{y^{2}}{4}-\\frac{x^{2}}{3}=1$ and passes through the point $M(3,-2)$.", "fact_expressions": "G1: Hyperbola;G2: Hyperbola;Expression(G2) = (-x^2/3 + y^2/4 = 1);Asymptote(G1) = Asymptote(G2);M: Point;Coordinate(M) = (3, -2);PointOnCurve(M, G1) = True", "query_expressions": "Expression(G1)", "answer_expressions": "x^2/6-y^2/8=1", "fact_spans": "[[[61, 64]], [[2, 40]], [[2, 40]], [[1, 64]], [[50, 60]], [[50, 60]], [[49, 64]]]", "query_spans": "[[[61, 70]]]", "process": "According to the given condition: sharing the same asymptotes with the hyperbola \\frac{y^2}{4}-\\frac{x^{2}}{3}=1, assume the equation of the hyperbola is \\frac{y^{2}}{4m}-\\frac{x^{2}}{3m}=1. The curve passes through the point M(3,-2); substituting this point gives: solving yields m=-2. Therefore, the equation of this hyperbola is \\frac{x^2}{6}-\\frac{y^{2}}{8}=1." }, { "text": "Given that the circle $(x-2)^{2}+y^{2}=1$ passes through a vertex and a focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, find the eccentricity $e$ of this ellipse.", "fact_expressions": "H: Circle;Expression(H) = (y^2 + (x - 2)^2 = 1);G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;PointOnCurve(OneOf(Vertex(G)), H);PointOnCurve(OneOf(Focus(G)), H);e: Number;Eccentricity(G) = e", "query_expressions": "e", "answer_expressions": "1/3", "fact_spans": "[[[2, 22]], [[2, 22]], [[24, 76], [89, 91]], [[24, 76]], [[26, 76]], [[26, 76]], [[26, 76]], [[26, 76]], [[2, 81]], [[2, 86]], [[95, 98]], [[89, 98]]]", "query_spans": "[[[95, 100]]]", "process": "Problem Analysis: From (x-2)^{2}+y^{2}=1, it can be seen that the curve passes through points (3,0) and (1,0). Therefore, a=3, c=1. Hence, e=\\frac{c}{a}=\\frac{1}{3}" }, { "text": "If the focus of the parabola $C$: $y^{2}=2 p x$ lies on the line $x+2 y-4=0$, then the equation of the directrix of $C$ is?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;G: Line;Expression(G) = (x + 2*y - 4 = 0);PointOnCurve(Focus(C), G)", "query_expressions": "Expression(Directrix(C))", "answer_expressions": "x=-4", "fact_spans": "[[[1, 22], [42, 45]], [[1, 22]], [[9, 22]], [[26, 39]], [[26, 39]], [[1, 40]]]", "query_spans": "[[[42, 52]]]", "process": "Since the focus of the parabola C: y^{2}=2px lies on the x-axis and also on the line x+2y-4=0, let y=0, we obtain x=4, \\therefore\\frac{p}{2}=4,\\therefore p=8, the directrix equation of C is x=-4," }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has foci $F_{1}$, $F_{2}$ and eccentricity $\\sqrt{3}$. If a point $P$ on $C$ satisfies $|P F_{1}|-|PF_{2} |=2 \\sqrt{3}$, then the equation of $C$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;Focus(C) = {F1, F2};Eccentricity(C) = sqrt(3);P: Point;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)) = 2*sqrt(3)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/3-y^2/6=1", "fact_spans": "[[[2, 63], [100, 103], [145, 148]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[67, 74]], [[75, 82]], [[2, 82]], [[2, 97]], [[106, 109]], [[100, 109]], [[111, 143]]]", "query_spans": "[[[145, 153]]]", "process": "From the definition of the hyperbola, we know that $ a = \\sqrt{3} $. From $ e = \\frac{c}{a} = \\sqrt{3} $, we obtain $ c = 3 $, then $ b^{2} = c^{2} - a^{2} = 6 $, so the equation of hyperbola $ C $ is $ \\frac{x^{2}}{3} - \\frac{y^{2}}{6} = 1 $." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1$ centered at the origin, its right focus coincides with the focus of the parabola $y^{2}=16 x$. What is the eccentricity of this hyperbola?", "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;O: Origin;Expression(G) = (-y^2/9 + x^2/a^2 = 1);Expression(H) = (y^2 = 16*x);RightFocus(G) = Focus(H);Center(G) = O", "query_expressions": "Eccentricity(G)", "answer_expressions": "4*sqrt(7)/7", "fact_spans": "[[[2, 44], [77, 80]], [[5, 44]], [[54, 69]], [[47, 49]], [[2, 44]], [[54, 69]], [[2, 74]], [[2, 49]]]", "query_spans": "[[[77, 86]]]", "process": "" }, { "text": "It is known that point $P$ is a moving point on the parabola $y^{2}=4 x$, $F$ is the focus of the parabola, and the coordinates of point $A$ are $(4 , a)$. When $|a|<4$, what is the minimum value of $|P A|+|P F|$?", "fact_expressions": "P: Point;G: Parabola;Expression(G) = (y^2=4*x);PointOnCurve(P,G) = True;F: Point;Focus(G) = F;A: Point;Coordinate(A) = (4,a);a: Number;Abs(a) < 4", "query_expressions": "Min(Abs(LineSegmentOf(P,A)) + Abs(LineSegmentOf(P,F)))", "answer_expressions": "5", "fact_spans": "[[[2, 6]], [[7, 21], [31, 34]], [[7, 21]], [[2, 25]], [[26, 29]], [[26, 37]], [[38, 42]], [[38, 55]], [[58, 65]], [[58, 65]]]", "query_spans": "[[[67, 86]]]", "process": "Draw a perpendicular line from point P to the directrix x = -1, with foot of perpendicular at E. Then |PF| = |PE|. Hence |PA| + |PF| = |PA| + |PE| \\geqslant AE \\geqslant 5, and equality holds if and only if points A, P, E are collinear. Therefore, the desired minimum value is 5; fill in 5." }, { "text": "If the line $y=k x+1$ intersects the curve $x=\\sqrt {y^{2}+1}$ at two distinct points, then what is the range of values for $k$?", "fact_expressions": "G: Line;Expression(G) = (y = k*x + 1);k: Number;H: Curve;Expression(H) = (x = sqrt(y^2 + 1));NumIntersection(G, H) = 2", "query_expressions": "Range(k)", "answer_expressions": "(-sqrt(2),-1)", "fact_spans": "[[[1, 12]], [[1, 12]], [[44, 47]], [[13, 34]], [[13, 34]], [[1, 42]]]", "query_spans": "[[[44, 54]]]", "process": "" }, { "text": "Given the parabola $C$: $y^{2}=4x$, a line $l$ intersects it at points $A$ and $B$, $O$ is the origin, the slopes of lines $OA$ and $OB$ are $k_{1}$ and $k_{2}$ respectively. If $k_{1} \\cdot k_{2} = -2$, then the minimum area of $\\triangle AOB$ is?", "fact_expressions": "l: Line;C: Parabola;A: Point;O: Origin;B: Point;k1: Number;k2: Number;Expression(C) = (y^2 = 4*x);Intersection(C, l) = {A, B};Slope(LineOf(O, A)) = k1;Slope(LineOf(O, B)) = k2;k1*k2 = -2", "query_expressions": "Min(Area(TriangleOf(A, O, B)))", "answer_expressions": "4*sqrt(2)", "fact_spans": "[[[22, 27]], [[2, 21]], [[29, 32]], [[39, 42]], [[33, 36]], [[68, 75]], [[77, 85]], [[2, 21]], [[2, 38]], [[48, 85]], [[48, 85]], [[87, 109]]]", "query_spans": "[[[111, 136]]]", "process": "According to the given conditions, the equation of line AB can be set as: $x = my + b$. Solving simultaneously \n\\[\n\\begin{cases}\nx = my + b \\\\\ny^2 = 4x\n\\end{cases}\n\\]\nyields $y^{2} - 4my - 4b = 0$, therefore $y_{1} + y_{2} = 4m$, $y_{1}y_{2} = -4b$. Since the slopes of lines OA and OB are $k_{1}, k_{2}$ respectively, and $k_{1} \\cdot k_{2} = -2$, we have $\\frac{y_{1}y_{2}}{x_{1}x_{2}} = -2$. Thus $y_{1}y_{2} = -8$, so $-4b = -8$, hence $b = 2$. Therefore, line AB passes through the fixed point $M(2, 0)$. The area of $\\triangle AOB$ is $S = \\frac{1}{2} \\times 2 \\times |y_{1} - y_{2}| = \\sqrt{(y_{1} + y_{2})^{2} - 4y_{1}y_{2}} = \\sqrt{16m^{2} + 32}$." }, { "text": "The coordinates of the focus of the parabola $x^{2}=-8 y$ are?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = -8*y)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0,-2)", "fact_spans": "[[[0, 15]], [[0, 15]]]", "query_spans": "[[[0, 22]]]", "process": "According to the problem, by analyzing the equation of the parabola, determine the direction of its opening and the value of p, then find the coordinates of its focus. Solution: According to the problem, the parabola $ x^{2} = -8y $ opens downward, where $ p = 4 $, then its focus coordinates are $ (0, -2) $." }, { "text": "It is known that a moving circle $M$ passes through point $A(3 , 0)$ and is tangent to the line $l$: $x=-3$. Find the trajectory equation of the center $M$ of the moving circle?", "fact_expressions": "l: Line;M: Circle;M1:Point;A: Point;Coordinate(A) = (3, 0);PointOnCurve(A, M);Expression(l)=(x=-3);Center(M)=M1;IsTangent(M, l) = True", "query_expressions": "LocusEquation(M1)", "answer_expressions": "y^2=12*x", "fact_spans": "[[[23, 36]], [[40, 42], [4, 7]], [[44, 47]], [[9, 20]], [[9, 20]], [[4, 20]], [[23, 36]], [[40, 47]], [[4, 38]]]", "query_spans": "[[[44, 54]]]", "process": "" }, { "text": "Given that a point $M$ on the parabola $y^{2}=4x$ is at a distance of $5$ from the focus, and that point $M$ lies in the first quadrant, then the coordinates of $M$ are?", "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(M, G);Distance(M, Focus(G)) = 5;Quadrant(M)=1", "query_expressions": "Coordinate(M)", "answer_expressions": "(4, 4)", "fact_spans": "[[[2, 16]], [[20, 23], [35, 39], [46, 49]], [[2, 16]], [[2, 23]], [[2, 33]], [[35, 44]]]", "query_spans": "[[[46, 54]]]", "process": "" }, { "text": "A circle is centered at a point $A$ on the parabola $C$: $y^{2}=8x$. If this circle passes through both the vertex and the focus of the parabola $C$, then what is the equation of the circle?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 8*x);A: Point;PointOnCurve(A, C) ;G: Circle;Center(G) = A;PointOnCurve(Vertex(C), G) ;PointOnCurve(Focus(C), G)", "query_expressions": "Expression(G)", "answer_expressions": "(x - 1)^2 + (y + pm*2*sqrt(2))^2 = 9", "fact_spans": "[[[1, 20], [38, 44]], [[1, 20]], [[24, 27]], [[1, 27]], [[31, 32], [35, 36], [54, 55]], [[24, 32]], [[35, 47]], [[35, 50]]]", "query_spans": "[[[54, 60]]]", "process": "Let A(x,y), and y^{2}=8x. \\therefore focus (2,0), vertex (0,0). \\because A is the center of a circle passing through the focus and the vertex, \\therefore A(1,\\pm2\\sqrt{2}). \\therefore R=3 \\therefore (x-1)^{2}+(y-2\\sqrt{2})^{2}=9 or (x-1)^{2}+(y+2\\sqrt{2})^{2}=9" }, { "text": "The eccentricity $e$ of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ is? The distance from the focus to the asymptote is?", "fact_expressions": "G: Hyperbola;e: Number;Expression(G) = (x^2/9 - y^2/16 = 1);Eccentricity(G) = e", "query_expressions": "e;Distance(Focus(G),Asymptote(G))", "answer_expressions": "5/3\n4", "fact_spans": "[[[0, 39]], [[43, 46]], [[0, 39]], [[0, 46]]]", "query_spans": "[[[43, 48]], [[0, 59]]]", "process": "" }, { "text": "In the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a, b>0)$, if a line passing through the left vertex $A$ of the hyperbola with slope $1$ intersects the right branch at point $B$, and the projection of point $B$ onto the $x$-axis is exactly the right focus $F$ of the hyperbola, then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;A: Point;B: Point;F: Point;a > 0;b > 0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftVertex(G) = A;Slope(H) = 1;Intersection(H, RightPart(G)) = B;Projection(B, xAxis) = F;RightFocus(G) = F;PointOnCurve(A, H)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[1, 55], [59, 62], [101, 104], [114, 117]], [[4, 55]], [[4, 55]], [[75, 77]], [[65, 68]], [[81, 85], [86, 90]], [[108, 111]], [[4, 55]], [[4, 55]], [[1, 55]], [[59, 68]], [[68, 77]], [[59, 85]], [[86, 111]], [[101, 111]], [[58, 77]]]", "query_spans": "[[[114, 123]]]", "process": "Problem Analysis: According to the given conditions, the equation of line AB is y = x + a. Since the projection of point B onto the x-axis is exactly the right focus F of the hyperbola, we have x_{B} = c. Substituting into the line equation gives y_{B} = c + a. Substituting the point (x_{B}, y_{B}) into the hyperbola equation \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 yields \\frac{c^{2}}{a^{2}} - \\frac{(c+a)^{2}}{b^{2}} = 1. Using b^{2} = c^{2} - a^{2} and substituting, we obtain c^{3} - 3a^{2}c - 2a^{3} = 0, which simplifies to e^{3} - 3e - 2 = 0. Factoring gives (e+1)^{2}(e-2) = 0, so e = 2. Therefore, the answer is 2." }, { "text": "Given a point $A(m, 4)$ on the parabola $x^{2}=4 y$, then the distance from point $A$ to the focus of the parabola is?", "fact_expressions": "G: Parabola;A: Point;Expression(G) = (x^2 = 4*y);Coordinate(A) = (m, 4);PointOnCurve(A,G);m:Number", "query_expressions": "Distance(A, Focus(G))", "answer_expressions": "5", "fact_spans": "[[[2, 16], [35, 38]], [[19, 28], [30, 34]], [[2, 16]], [[19, 28]], [[2, 28]], [[19, 28]]]", "query_spans": "[[[30, 45]]]", "process": "Solution: From the given condition, the directrix of the parabola \\( x^{2} = 4y \\) is \\( y = -1 \\). Therefore, the distance from point A to the directrix is \\( 4 - (-1) = 5 \\). According to the definition of a parabola, the distance from point A to the parabola is equal to the distance from point A to the directrix of the parabola. Hence, the distance from point A to the focus of the parabola is 5." }, { "text": "Given that the hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has an eccentricity of $2$, then what is the eccentricity of the hyperbola $C_{2}$: $\\frac{y^{2}}{b^{2}}-\\frac{x^{2}}{a^{2}}=1$?", "fact_expressions": "C1: Hyperbola;b: Number;a: Number;C2: Hyperbola;a>0;b>0;Expression(C1) = (x^2/a^2 - y^2/b^2 = 1);Expression(C2) = (y^2/b^2 - x^2/a^2 = 1);Eccentricity(C1) = 2", "query_expressions": "Eccentricity(C2)", "answer_expressions": "2*sqrt(3)/3", "fact_spans": "[[[2, 67]], [[14, 67]], [[14, 67]], [[77, 132]], [[14, 67]], [[14, 67]], [[2, 67]], [[77, 132]], [[2, 75]]]", "query_spans": "[[[77, 138]]]", "process": "" }, { "text": "Given the parabola $y=\\frac{1}{16} x^{2}$, $A$, $B$ are two points on this parabola, and $|A B|=24$. When the midpoint $P$ of segment $A B$ is closest to the $x$-axis, what is the $y$-coordinate of the point?", "fact_expressions": "G: Parabola;B: Point;A: Point;P:Point;Expression(G) = (y = x^2/16);PointOnCurve(A,G);PointOnCurve(B,G);Abs(LineSegmentOf(A, B)) = 24;MidPoint(LineSegmentOf(A,B))=P;WhenMin(Distance(P,xAxis))", "query_expressions": "YCoordinate(P)", "answer_expressions": "8", "fact_spans": "[[[2, 27], [37, 40]], [[32, 35]], [[28, 31]], [[67, 70], [79, 80]], [[2, 27]], [[28, 43]], [[28, 43]], [[45, 55]], [[57, 70]], [[67, 78]]]", "query_spans": "[[[79, 86]]]", "process": "" }, { "text": "Through a focus of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, draw a line $l$ with an inclination angle of $45^{\\circ}$, intersecting the ellipse at points $A$ and $B$. Let $O$ be the coordinate origin. Then $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}$=?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2 = 1);l: Line;PointOnCurve(OneOf(Focus(G)), l);Inclination(l) = ApplyUnit(45, degree);A: Point;B: Point;Intersection(l, G) = {A, B};O: Origin", "query_expressions": "DotProduct(VectorOf(O, A), VectorOf(O, B))", "answer_expressions": "-1/3", "fact_spans": "[[[2, 29], [59, 61]], [[2, 29]], [[52, 57]], [[1, 57]], [[35, 57]], [[62, 65]], [[66, 69]], [[52, 71]], [[73, 76]]]", "query_spans": "[[[83, 134]]]", "process": "\\because in the ellipse \\frac{x^2}{2}+y^2=1, a=\\sqrt{2}, b=1, the foci of the ellipse are F(\\pm1,0). Without loss of generality, assume the line l with inclination angle 45^{\\circ} passes through the focus (1,0), so the line l: y=x-1. Solving the system \n\\begin{cases}\ny=x-1 \\\\\n\\frac{x^{2}}{2}+y^{2}=1\n\\end{cases}\neliminating y gives 3x^{2}-4x=0. Solving the equation yields x_{1}=0, x_{2}=\\frac{4}{3}. Substituting into the line y=x-1 gives y_{1}=-1, y_{2}=\\frac{1}{3}. Then \\overrightarrow{OA}\\cdot\\overrightarrow{OB}=x_{1}x_{2}+y_{1}y_{2}=-\\frac{1}{3}. Similarly, when the line l passes through the focus (-1,0), \\overrightarrow{OA}\\cdot\\overrightarrow{OB}=-\\frac{1}{3}." }, { "text": "Given point $P$ on the circle $C$: $(x-4)^{2}+y^{2}=4$, and point $Q$ moving on the ellipse $x^{2}+\\frac{y^{2}}{4}=1$, then the maximum value of $|P Q|$ is?", "fact_expressions": "C: Circle;Expression(C) = (y^2 + (x - 4)^2 = 4);G: Ellipse;Expression(G) = (x^2 + y^2/4 = 1);P: Point;Q: Point;PointOnCurve(P,C) = True;PointOnCurve(Q,G) = True", "query_expressions": "Max(Abs(LineSegmentOf(P,Q)))", "answer_expressions": "7", "fact_spans": "[[[6, 30]], [[6, 30]], [[36, 63]], [[36, 63]], [[2, 6]], [[31, 35]], [[2, 31]], [[31, 64]]]", "query_spans": "[[[67, 80]]]", "process": "Let the coordinates of an arbitrary point Q on the ellipse $x^{2}+\\frac{y^{2}}{4}=1$ be $(x,y)$, then $4x^{2}+y^{2}=4$, and the distance from point Q to the center of the circle $(4,0)$ is $2+5=7$." }, { "text": "If the distance between the directrix of the parabola $y^{2}=m x$ and the line $x=1$ is $3$, then the equation of the parabola is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = m*x);m: Number;H: Line;Expression(H) = (x = 1);Distance(Directrix(G), H) = 3", "query_expressions": "Expression(G)", "answer_expressions": "{y^2=8*x, y^2=-16*x}", "fact_spans": "[[[1, 15], [36, 39]], [[1, 15]], [[4, 15]], [[19, 26]], [[19, 26]], [[1, 34]]]", "query_spans": "[[[36, 44]]]", "process": "Solution: When m > 0, the equation of the directrix is x = -\\frac{m}{4}, hence 1 - (-\\frac{m}{4}) = 3, so m = 8. At this time, the equation of the parabola is y^{2} = 8x. When m < 0, the equation of the directrix is x = -\\frac{m}{4}, hence -\\frac{m}{4} - 1 = 3, so m = -16. At this time, the equation of the parabola is y^{2} = -16x. Therefore, the required equations of the parabola are y^{2} = 8x or y^{2} = -16x." }, { "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{3}=1$ $(m>0)$ is $y=\\frac{1}{2} x$, find the value of $m$.", "fact_expressions": "G: Hyperbola;m: Number;m>0;Expression(G) = (-y^2/3 + x^2/m = 1);Expression(OneOf(Asymptote(G))) = (y = x/2)", "query_expressions": "m", "answer_expressions": "12", "fact_spans": "[[[2, 45]], [[74, 77]], [[5, 45]], [[2, 45]], [[2, 71]]]", "query_spans": "[[[74, 81]]]", "process": "Problem Analysis: From the given condition, $\\frac{\\sqrt{3}}{\\sqrt{m}}=\\frac{1}{2}\\Rightarrow m=12$" }, { "text": "If the point $(\\sqrt{2}, 1)$ lies on the parabola $y=2 p x^{2}$, then what is the standard equation of this parabola?", "fact_expressions": "G: Parabola;p: Number;H: Point;Expression(G) = (y = 2*(p*x^2));Coordinate(H) = (sqrt(2), 1);PointOnCurve(H, G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2 = 2*y", "fact_spans": "[[[18, 34], [38, 41]], [[21, 34]], [[1, 17]], [[18, 34]], [[1, 17]], [[1, 35]]]", "query_spans": "[[[38, 48]]]", "process": "Substituting the point (\\sqrt{2},1) into the parabola equation gives 1=2p\\times(\\sqrt{2})^{2}, p=\\frac{1}{4}, hence y=\\frac{1}{2}x^{2}, x^{2}=2y[" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{12}+\\frac{y^{2}}{9}=1$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. $P$ is a point on the ellipse $C$, and $\\angle F_{1} P F_{2}=60^{\\circ}$. What is the area of $\\Delta P F_{1} F_{2}$?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/12 + y^2/9 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C) = True;AngleOf(F1, P, F2) = ApplyUnit(60, degree)", "query_expressions": "Area(TriangleOf(P, F1, F2))", "answer_expressions": "3*sqrt(3)", "fact_spans": "[[[2, 45], [74, 79]], [[2, 45]], [[54, 61]], [[62, 69]], [[2, 69]], [[2, 69]], [[70, 73]], [[70, 83]], [[85, 118]]]", "query_spans": "[[[120, 146]]]", "process": "a=2\\sqrt{3},b=3,c=\\sqrt{3}, let |PF_{1}|=m,|PF_{2}|=n, then \\begin{cases}m+n=2a=4\\sqrt{3}\\\\(2c)^{2}=m^{2}+n^{2}-2mn\\cos60\\end{cases}\\begin{cases}m+n=4\\sqrt{3}\\\\12=(m+n)^{2}-3mn\\end{cases},12=(4\\sqrt{3})^{2}-3mn,mn=12, so the area of triangle \\triangle PF_{1}F_{2} is \\frac{1}{2}mn\\sin60^{\\circ}=\\frac{1}{2}\\times12\\times\\frac{\\sqrt{3}}{2}=3\\sqrt{3}" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the right focus is $F$. A perpendicular is drawn from $F$ to one of its asymptotes, with the foot of the perpendicular being $P$. If the midpoint of the segment $PF$ lies on this hyperbola, then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(G) = F;L: Line;PointOnCurve(F,L)=True;IsPerpendicular(Asymptote(G),L)=True;FootPoint(Asymptote(G),L) = P;P: Point;PointOnCurve(MidPoint(LineSegmentOf(P,F)),G) = True", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 58], [72, 73], [99, 102], [106, 109]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[63, 66], [68, 71]], [[2, 66]], [], [[67, 79]], [[67, 79]], [[67, 86]], [[83, 86]], [[88, 103]]]", "query_spans": "[[[106, 115]]]", "process": "Test analysis: F(c,0), one asymptote of the hyperbola is given by y=\\frac{b}{a}x, then the equation of the line passing through F and perpendicular to this asymptote is y=-\\frac{a}{b}(x-c). Solving these equations simultaneously gives P(\\frac{a^{2}}{c},\\frac{ab}{c}). Therefore, the midpoint of PF is (\\frac{a^{2}+c^{2}}{2c},\\frac{ab}{2c}). Substituting this point into the hyperbola equation yields \\frac{c}{a}=\\sqrt{2}, so the eccentricity of the hyperbola is \\sqrt{2}." }, { "text": "Given the ellipse $\\Gamma$: $\\frac{x^{2}}{a^{2}}+y^{2}=1(a>1)$, the line $y=x+1$ intersects the $x$-axis at point $P$, and intersects the ellipse at points $A$ and $B$. If $\\overrightarrow{P A}+2 \\overrightarrow{P B}=\\overrightarrow{0}$, then $a=?$", "fact_expressions": "Gamma: Ellipse;H: Line;P: Point;A: Point;B: Point;a:Number;a>1;Expression(Gamma)=(x^2/a^2+y^2=1);Expression(H) = (y = x + 1);Intersection(H, xAxis) = P;Intersection(H,Gamma)={A,B};VectorOf(P, A) + 2*VectorOf(P, B) = 0", "query_expressions": "a", "answer_expressions": "sqrt(3)", "fact_spans": "[[[2, 48], [71, 73]], [[49, 58]], [[65, 69]], [[75, 78]], [[79, 82]], [[152, 155]], [[14, 48]], [[2, 48]], [[49, 58]], [[49, 69]], [[49, 84]], [[86, 150]]]", "query_spans": "[[[152, 157]]]", "process": "From \\begin{cases}y=x+1\\\\\\frac{x^{2}}{a^{2}}+y^{2}=1\\end{cases}, solving gives \\begin{cases}x=0\\\\y=1\\end{cases} or \\begin{cases}x=-\\frac{2a^{2}}{1+a^{2}}\\\\y=\\frac{1-a^{2}}{1+a^{2}}\\end{cases}. Since |\\overrightarrow{PA}|>|\\overrightarrow{PB}|, the points are A(0,1) and B(-\\frac{2a^{2}}{1+a^{2}},\\frac{1-a^{2}}{1+a^{2}}). Given P(-1,0), we have \\overrightarrow{PA}=(1,1), \\overrightarrow{PB}=(\\frac{1-a^{2}}{1+a^{2}},\\frac{1-a^{2}}{1+a^{2}}). Also, since \\overrightarrow{PA}+2\\overrightarrow{PB}=\\overrightarrow{0}, it follows that 1+2\\cdot\\frac{1-a^{2}}{1+a^{2}}=0. Solving yields a^{2}=3, so a=\\sqrt{3}." }, { "text": "Given the parabola $ C $: $ y^{2} = 4x $, a line $ l $ with slope $ \\frac{1}{3} $ intersects $ C $ at points $ A $ and $ B $. If the circle centered at point $ E(1,1) $ is the incircle of $ \\triangle OAB $, then what is the radius of circle $ E $?", "fact_expressions": "l: Line;C: Parabola;E:Circle;E1: Point;O: Origin;A: Point;B: Point;Expression(C) = (y^2 = 4*x);Coordinate(E1) = (1, 1);Slope(l)=1/3;Intersection(l,C)={A,B};Center(E)=E1;InscribedCircle(TriangleOf(O,A,B))=E", "query_expressions": "Radius(E)", "answer_expressions": "sqrt(10)/5", "fact_spans": "[[[39, 44]], [[2, 21], [45, 48]], [[100, 104], [75, 76]], [[62, 71]], [[77, 94]], [[51, 54]], [[55, 58]], [[2, 21]], [[62, 71]], [[22, 44]], [[39, 60]], [[61, 76]], [[75, 98]]]", "query_spans": "[[[100, 109]]]", "process": "Let the equation of line $ l $ be $ y = \\frac{1}{3}x + m $, that is, $ x - 3y + 3m = 0 $. Since the line is tangent to the circle, then $ r = \\frac{|1 - 3 + 3m|}{\\sqrt{10}} = \\frac{|3m - 2|}{\\sqrt{10}} $. Let the equations of lines $ OA $ and $ OB $ be $ y = k_{1}x $, $ y = k_{2}x $, respectively. Since lines $ OA $ and $ OB $ are tangent to circle $ E $, we have $ \\frac{|k_{1} - 1|}{\\sqrt{k_{1}^{2} + 1}} = r $, $ \\frac{|k_{2} - 1|}{\\sqrt{k_{2}^{2} + 1}} = r $, that is, $ k_{1} $ and $ k_{2} $ are two distinct real roots of the equation $ (1 - r)x^{2} - 2x + 1 - r^{2} = 0 $. Then $ k_{1}k_{2} = \\frac{y_{1}}{x_{1}} \\cdot \\frac{y_{2}}{x_{2}} = \\frac{y_{1}y_{2}}{x_{1}x_{2}} = 1 $, that is, $ x_{1}x_{2} = y_{1}y_{2} $. Then, by considering the intersection of line $ l $ with the parabola, solve using Vieta's formulas.\n\nLet the equation of line $ l $ be $ y = \\frac{1}{3}x + m $, that is, $ x - 3y + 3m = 0 $, and let the radius of the incircle be $ r $, then $ r = \\frac{|1 - 3 + 3m|}{\\sqrt{10}} = \\frac{|3m - 2|}{\\sqrt{10}} $. Let the equations of lines $ OA $, $ OB $ be $ y = k_{1}x $, $ y = k_{2}x $, that is, $ k_{1}x - y = 0 $, $ k_{2}x - y = 0 $. Since line $ OA $ is tangent to circle $ E $, we have $ \\frac{|k_{1} - 1|}{\\sqrt{k_{1}^{2} + 1}} = r $, which simplifies to $ (1 - r^2)k_{1}^{2} - 2k_{1} + 1 - r^{2} = 0 $. Similarly, we obtain $ (1 - r^{2})k_{2}^{2} - 2k_{2} + 1 - r^{2} = 0 $. Therefore, $ k_{1} $ and $ k_{2} $ are two distinct real roots of the equation $ (1 - r^{2})x^{2} - 2x + 1 - r^{2} = 0 $. Hence,\n\n$$\n\\begin{cases}\nr \\neq 1 \\\\\n\\Delta = 4 - 4(1 - r)^{2} \\\\\nk_{1} + k_{2} = \\frac{2}{1 - r^{2}} \\\\\nk_{1}k_{2}\n\\end{cases}\n$$\n\nLet $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ k_{1}k_{2} = \\frac{y_{1}}{x_{1}} \\cdot \\frac{y_{2}}{x_{2}} = \\frac{y_{1}y_{2}}{x_{1}x_{2}} = 1 $, that is, $ x_{1}x_{2} = y_{1}y_{2} $. From\n\n$$\n\\begin{cases}\ny = \\\\\ny =\n\\end{cases}\n\\frac{1}{3}x + m,\n$$\n\nwe get $ y^{2} - 12y + 12m = 0 $, therefore\n\n$$\n\\begin{cases}\ny_{1} + y_{2} = 12 \\\\\ny_{1}y_{2} = 12m\n\\end{cases}\n\\quad\n\\begin{cases}\ny^2 = 4x \\\\\n\\Delta = 144 - 48m > 0 \\\\\ny_{1} + y_{2} = 12\n\\end{cases},\n$$\n\nthus $ m < 3 $. $ x_{1}x_{2} = \\frac{1}{16}(y_{1}y_{2})^{2} = 9m^{2} $, therefore $ 9m^{2} = 12m $. Given $ m \\neq 0 $, we have $ m = \\frac{4}{3} $, which satisfies the condition. Therefore, $ r = \\frac{|3m - 2|}{\\sqrt{10}} = \\frac{2}{\\sqrt{10}} = \\frac{\\sqrt{10}}{5} $." }, { "text": "The point $P(1 , 1)$ bisects a chord of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$. Then the equation of the line containing this chord is?", "fact_expressions": "P: Point;Coordinate(P) = (1, 1);H: LineSegment;G: Ellipse;Expression(G) = (x^2/4 + y^2/2 = 1);IsChordOf(H, G) = True;MidPoint(H) = P", "query_expressions": "Expression(OverlappingLine(H))", "answer_expressions": "x+2*y-2=0", "fact_spans": "[[[0, 11]], [[0, 11]], [], [[13, 50]], [[13, 50]], [[13, 54]], [[0, 54]]]", "query_spans": "[[[13, 68]]]", "process": "" }, { "text": "Given the hyperbola $x^{2}-y^{2}=1$, points $F_{1}$ and $F_{2}$ are its two foci, and point $P$ is a point on the hyperbola. If $P F_{1} \\perp P F_{2}$, then what is the value of $|P F_{1}|+|P F_{2}|$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))", "query_expressions": "Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2))", "answer_expressions": "2*sqrt(3)", "fact_spans": "[[[2, 20], [38, 39], [49, 52]], [[2, 20]], [[21, 29]], [[30, 37]], [[21, 43]], [[44, 48]], [[44, 55]], [[57, 80]]]", "query_spans": "[[[82, 107]]]", "process": "Problem Analysis: From the conditions we know: c=\\sqrt{2}, while a=b=1,\\begin{cases}PF'+|PF_{2}^{2}=8\\\\|PF|-|PF_{2}=2\\end{cases}.|PF|PF_{2}=2,\\therefore|PF_{1}|+|PF_{2}|^{2}=|PF|^{2}+|PF_{2}^{2}+2|PF_{1}|PF_{2}|=8+4=12,\\therefore|PF|+|PF_{2}|=2\\sqrt{3}" }, { "text": "Given that $O$ is the coordinate origin, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, a point $P$ on the hyperbola $C$ satisfies $(\\overrightarrow{O P}+\\overrightarrow{O F_{2}}) \\cdot \\overrightarrow{F_{2} P}=0$, and $|\\overrightarrow{P F_{1}}| \\cdot|\\overrightarrow{P F_{2}}|=2 a^{2}$, then the asymptotic equations of the hyperbola $C$ are?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;O: Origin;P: Point;F2: Point;F1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P,C);DotProduct((VectorOf(O,P)+VectorOf(O,F2)),VectorOf(F2,P))=0;Abs(VectorOf(P, F1))*Abs(VectorOf(P, F2)) = 2*a^2", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "y=pm*x", "fact_spans": "[[[27, 89], [96, 102], [264, 270]], [[35, 89]], [[35, 89]], [[2, 5]], [[105, 108]], [[19, 26]], [[11, 18]], [[35, 89]], [[35, 89]], [[27, 89]], [[11, 95]], [[11, 95]], [[96, 108]], [[110, 192]], [[194, 262]]]", "query_spans": "[[[264, 278]]]", "process": "According to $(\\overrightarrow{OP}+\\overrightarrow{OF}_{2})\\cdot\\overrightarrow{F_{2}P}=0$, it is known that $|OP|=|OF_{2}|=|OF_{1}|$, i.e., $\\triangle PF_{1}F_{2}$ is a right triangle. Let $|PF_{1}|=m$, $|PF_{2}|=n$. According to the given conditions, we have\n$$\n\\begin{cases}\nmn=2a^{2} \\\\\nm-n=2a\n\\end{cases}\n$$\nSolving gives $m=(\\sqrt{3}+1)a$, $n=(\\sqrt{3}-1)a$. By the Pythagorean theorem, $m^{2}+n^{2}=4c^{2}$, solving yields $c=\\sqrt{2}a=\\sqrt{2}b$, $a=b$. Hence, the hyperbola is an equilateral hyperbola with asymptotes $y=\\pm x$. [Comment] This question mainly examines the definition of hyperbola and the solution of directrix equations. It tests geometric properties of right triangles and computational problem-solving ability. It is a medium-difficulty problem." }, { "text": "The equation $k x^{2}+y^{2}=9 k+4$ represents an ellipse with foci on the $x$-axis. What is the range of values for $k$?", "fact_expressions": "G: Ellipse;Expression(G) = (k*x^2 + y^2 = 9*k + 4);PointOnCurve(Focus(G), xAxis);k:Number", "query_expressions": "Range(k)", "answer_expressions": "(0, 1)", "fact_spans": "[[[34, 36]], [[0, 36]], [[25, 36]], [[37, 40]]]", "query_spans": "[[[37, 47]]]", "process": "Since the equation $ kx^{2} + y^{2} = 9k + 4 $ represents an ellipse with foci on the x-axis, then $ 9k + 4 \\neq 0 $. Converting the ellipse equation into standard form gives $ \\frac{x^{2}}{\\frac{9k+4}{k}} + \\frac{y^{2}}{9k+4} = 1 $. From the given conditions, we have\n\\[\n\\begin{cases}\n9k + 4 > 0 \\\\\n\\frac{9k+4}{k} > 9k+4\n\\end{cases}\n\\]\nSolving this yields $ 0 < k < 1 $. Therefore, the range of real values for $ k $ is $ (0, 1) $." }, { "text": "The line $y=k(x+1)$ intersects the parabola $C$: $y^{2}=4x$ at two points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. Let the focus of the parabola $C$ be $F$. If $|FA|+|FB|=16$, then $k=?$", "fact_expressions": "G: Line;Expression(G) = (y = k * (x + 1));C: Parabola;Expression(C) = (y^2 = 4*x);A: Point;Coordinate(A) = (x1, y1);x1: Number;y1: Number;B: Point;Coordinate(B) = (x2, y2);x2: Number;y2: Number;Intersection(G, C) = {A, B};Focus(C) = F;F: Point;Abs(LineSegmentOf(F, A)) + Abs(LineSegmentOf(F, B)) = 16;k: Number", "query_expressions": "k", "answer_expressions": "pm*(1/2)", "fact_spans": "[[[0, 12]], [[0, 12]], [[13, 32], [75, 81]], [[13, 32]], [[34, 51]], [[34, 51]], [[34, 51]], [[34, 51]], [[54, 71]], [[54, 71]], [[54, 71]], [[54, 71]], [[0, 73]], [[75, 88]], [[85, 88]], [[90, 106]], [[108, 111]]]", "query_spans": "[[[108, 113]]]", "process": "Solving the system of equations \\begin{cases}y=k(x+1)\\\\y^{2}=4x\\end{cases}, eliminating $ y $ gives $ k^{2}x^{2}+(2k^{2}-4)x+k^{2}=0 $. Then $ k\\neq0 $, and $ a=(2k^{2}-4)^{2}-4k^{4}>0 $. That is, $ -10)$, the focus is $F$. A line $l$ passing through point $F$ intersects the parabola $C$ and its directrix at points $P$ and $Q$, respectively. $\\overrightarrow{Q F}=3 \\overrightarrow{F P}$. Find the slope of line $l$.", "fact_expressions": "l: Line;C: Parabola;p: Number;Q: Point;F: Point;P: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = P;Intersection(l, Directrix(C)) = Q;VectorOf(Q, F) = 3*VectorOf(F, P)", "query_expressions": "Slope(l)", "answer_expressions": "pm*sqrt(15)", "fact_spans": "[[[43, 48], [120, 125]], [[2, 29], [49, 55], [56, 57]], [[10, 29]], [[67, 70]], [[33, 36], [38, 42]], [[63, 66]], [[10, 29]], [[2, 29]], [[2, 36]], [[37, 48]], [[43, 72]], [[43, 72]], [[73, 118]]]", "query_spans": "[[[120, 130]]]", "process": "From \\overrightarrow{QF}=3\\overrightarrow{FP},x_{Q}=-\\frac{p}{2},x_{F}=\\frac{p}{2}\\Rightarrowx_{P}=\\frac{5p}{6},y_{P}=\\pm\\sqrt{\\frac{5}{3}}p,\\thereforek=\\frac{y_{P}}{x_{P}}-x_{F}=\\pm\\sqrt{15}" }, { "text": "Point $P$ is a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and $F_{1}$, $F_{2}$ are the two foci of the ellipse. The inradius of triangle $\\Delta P F_{1} F_{2}$ is $1$. When $P$ lies in the first quadrant, what is the $y$-coordinate of point $P$?", "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2};Radius(InscribedCircle(TriangleOf(P, F1, F2))) = 1;Quadrant(P) = 1", "query_expressions": "YCoordinate(P)", "answer_expressions": "8/3", "fact_spans": "[[[5, 44], [64, 66]], [[0, 4], [107, 110], [118, 122]], [[48, 55]], [[56, 63]], [[5, 44]], [[0, 47]], [[48, 71]], [[73, 105]], [[107, 116]]]", "query_spans": "[[[118, 128]]]", "process": "Solution: According to the definition of an ellipse, we know |PF_{1}| + |PF_{2}| = 10, |F_{1}F_{2}| = 6. Let the incenter of the incircle be O. Then S_{APF_{1}F_{2}} = S_{APOF_{1}} + S_{APOF_{2}} + S_{AOF_{1}F_{2}} = \\frac{1}{2}(|PF_{1}|r + |PF_{2}|r + |F_{1}F_{2}|r) = \\frac{1}{2}(|PF_{1}| + |PF_{2}| + |F_{1}F_{2}|)\\cdot1 = 8. Also, \\because S_{APF_{1}F_{2}} = \\frac{1}{2}|F_{1}F_{2}|\\cdot y_{P} = 3y_{P}, so 3y_{p} = 8, y_{p} = \\frac{8}{3}." }, { "text": "The parabola $C$: $y^{2}=4x$ has focus $F$. The line $y=k(x-2)$ $(k>0)$ intersects the parabola $C$ at two distinct points $A$ and $B$, and $\\frac{|AF|}{|BF|}=\\frac{2}{5}$. Then $k=?$", "fact_expressions": "C: Parabola;G: Line;A: Point;F: Point;B: Point;k:Number;Expression(C) = (y^2 = 4*x);k>0;Expression(G) = (y = k*(x - 2));Focus(C) = F;Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(B, F)) = 2/5;Negation(A=B)", "query_expressions": "k", "answer_expressions": "2", "fact_spans": "[[[0, 19], [45, 51]], [[27, 44]], [[56, 59]], [[23, 26]], [[60, 63]], [[102, 105]], [[0, 19]], [[29, 44]], [[27, 44]], [[0, 26]], [[27, 65]], [[67, 100]], [53, 64]]", "query_spans": "[[[102, 107]]]", "process": "Solve the system of equations consisting of the line $ y = k(x - 2) $ ($ k > 0 $) and the parabola equation, and use the definition of the parabola to find the coordinates of point B, thereby determining the value of $ k $. According to the problem, $ F(1, 0) $, $ p = 2 $, $ \\frac{p}{2} = 1 $. Let $ A(x_1, y_1) $, $ B(x_2, y_2) $. Since $ k > 0 $, we have $ x_1 > 0 $, $ x_2 > 0 $. From\n$$\n\\begin{cases}\ny = k(x - 2), \\\\\ny^2 = 4x,\n\\end{cases}\n$$\neliminate $ y $ and simplify to obtain $ k^2x^2 - (4k^2 + 4)x + 4k^2 = 0 $, so $ x_1 \\cdot x_2 = 4 $. According to the definition of the parabola, $ \\frac{|AF|}{|BF|} = \\frac{x_1 + 1}{x_2 + 1} = \\frac{2}{5} $, which simplifies to $ x_1 = \\frac{2}{5}x_2 - \\frac{3}{5} $. Substituting into $ x_1 \\cdot x_2 = 4 $ gives $ \\left(\\frac{2}{5}x_2 - \\frac{3}{5}\\right) \\cdot x_2 = 4 $, solving yields $ x_2 = 4 $ (the negative root is discarded). Since $ \\frac{|AF|}{|BF|} = \\frac{2}{5} \\Rightarrow |BF| > |AF| $, and $ k > 0 $, point B lies in the first quadrant. Thus, $ y_2^2 = 4x_2 = 16 \\Rightarrow y_2 = 4 $, so $ B(4, 4) $. Substituting into $ y = k(x - 2) $ gives $ 4 = k(4 - 2) \\Rightarrow k = 2 $." }, { "text": "Given that $F$ is the left focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, a line passing through point $F$ intersects the two asymptotes of hyperbola $C$ at points $Q$ and $P$, respectively, and $2 \\overrightarrow{F Q}=\\overrightarrow{F P}$. A circle centered at the origin $O$ is tangent to the line $F P$, and the point of tangency is exactly $Q$. Then the eccentricity of the hyperbola is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;LeftFocus(C) = F;H: Line;PointOnCurve(F, H);Z1: Line;Z2: Line;Asymptote(C) = {Z1, Z2};Q: Point;Intersection(H, Z1) = Q;P: Point;Intersection(H, Z2) = P;2*VectorOf(F, Q) = VectorOf(F, P);O: Origin;G: Circle;Center(G) = O;IsTangent(G, LineOf(F, P));TangentPoint(G, LineOf(F, P)) = Q", "query_expressions": "Eccentricity(C)", "answer_expressions": "2", "fact_spans": "[[[6, 67], [81, 87], [184, 187]], [[6, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[73, 77], [2, 5]], [[2, 71]], [[78, 80]], [[72, 80]], [], [], [[81, 93]], [[97, 100], [179, 182]], [[78, 104]], [[101, 104]], [[78, 104]], [[106, 151]], [[153, 158]], [[162, 163]], [[152, 163]], [[162, 173]], [[162, 182]]]", "query_spans": "[[[184, 193]]]", "process": "\\because\\overrightarrow{FP}=2\\overrightarrow{FQ}.\\therefore Q is the midpoint of PF, and it is also given that OQ\\bot PF.\\therefore OQ is the perpendicular bisector of segment PF,\\therefore\\angleFOQ=\\anglePOQ=\\anglePOx=60^{\\circ}\\therefore k_{OP}=\\tan60^{\\circ}=\\sqrt{3}, i.e. \\frac{b}{a}=\\sqrt{3},\\therefore e=2," }, { "text": "The parabola $y=ax^{2}$ intersects the line $l$: $y=kx+b(k \\neq 0)$ at two points $A$ and $B$, whose x-coordinates are $x_{1}$ and $x_{2}$ respectively. Then the x-coordinate of the intersection point between line $l$ and the x-axis is equal to (expressed in terms of $x_{1}$, $x_{2}$, without involving $a$, $b$, $k$)?", "fact_expressions": "l: Line;G: Parabola;a: Number;b: Number;k: Number;Negation(k = 0);A: Point;B: Point;Expression(G) = (y = a*x^2);Expression(l) = (y = b + k*x);Intersection(G, l) = {A, B};x1: Number;x2: Number;XCoordinate(A) = x1;XCoordinate(B) = x2", "query_expressions": "XCoordinate(Intersection(l, xAxis))", "answer_expressions": "(x1*x2)/(x1 + x2)", "fact_spans": "[[[14, 39], [83, 88]], [[0, 13]], [[125, 128]], [[130, 133]], [[135, 138]], [[21, 39]], [[41, 44]], [[47, 50]], [[0, 13]], [[14, 39]], [[0, 52]], [[64, 71], [103, 110]], [[74, 81], [111, 118]], [[0, 81]], [[0, 81]]]", "query_spans": "[[[83, 140]]]", "process": "y=ax^{2}k\\neq0, rearrange to get ax^{2}-kx-b=0. From the given conditions, we know x_{1}+x_{2}=\\frac{k}{a}, x_{1}x_{2}=-\\frac{b}{a}, x_{3}=-\\frac{b}{k}. From this it follows that x_{1}x_{3}+x_{2}x_{3}=(x_{1}+x_{2})x_{3}=\\frac{k}{a}\\times(-\\frac{b}{k})=-\\frac{b}{a}=x_{1}x_{2}. Solve y=kx+b, k\\neq0 \\therefore ax^{2}=kx+b, rearrange to get ax^{2}-kx-b=0. From the given conditions, we know x_{1}+x_{2}=\\frac{k}{a}, x_{1}x_{2}=-\\frac{b}{a}. Let the x-coordinate of the intersection point between line l and the x-axis be x_{3}, then x_{3}=-\\frac{b}{k}. x_{1}x_{3}+x_{2}x_{3}=(x_{1}+x_{2})x_{3}=\\frac{k}{a}\\times(-\\frac{b}{k})=-\\frac{b}{a}=x_{1}x_{2}. \\therefore x_{3}=\\frac{x_{1}+x_{2}}{x_{1}+x_{2}}" }, { "text": "The equation of the tangent line to the curve $y=x^{2}$ at the point $(1 , 1)$ is?", "fact_expressions": "G: Curve;Expression(G) = (y = x^2);H: Point;Coordinate(H) = (1, 1)", "query_expressions": "Expression(TangentOnPoint(H, G))", "answer_expressions": "y=3*x-2", "fact_spans": "[[[0, 11]], [[0, 11]], [[12, 22]], [[12, 22]]]", "query_spans": "[[[0, 30]]]", "process": "" }, { "text": "Given that point $P$ is a moving point on the parabola $y^{2}=2 x$, the projection of point $P$ onto the $y$-axis is $M$, and the coordinates of point $A$ are $A(\\frac{7}{2}, 4)$. Then, the minimum value of $|P A|+| PM |$ is?", "fact_expressions": "G: Parabola;A: Point;P: Point;M: Point;Expression(G) = (y^2 = 2*x);Coordinate(A) = (7/2, 4);PointOnCurve(P, G);Projection(P, yAxis) = M", "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, M)))", "answer_expressions": "9/2", "fact_spans": "[[[7, 21]], [[44, 48], [52, 71]], [[2, 6], [26, 30]], [[40, 43]], [[7, 21]], [[52, 71]], [[2, 25]], [[26, 43]]]", "query_spans": "[[[73, 93]]]", "process": "" }, { "text": "It is known that the focus of the parabola $y^{2}=2 p x(p>0)$ coincides with the right focus of the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$. Find the value of $p$.", "fact_expressions": "G: Parabola;p: Number;H: Ellipse;p>0;Expression(G) = (y^2 = 2*p*x);Expression(H) = (x^2/6 + y^2/2 = 1);Focus(G) = RightFocus(H)", "query_expressions": "p", "answer_expressions": "4", "fact_spans": "[[[2, 23]], [[72, 75]], [[27, 64]], [[5, 23]], [[2, 23]], [[27, 64]], [[2, 70]]]", "query_spans": "[[[72, 79]]]", "process": "The right focus of the ellipse \\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1 is (2,0), so the focus of the parabola is also this point (\\frac{p}{2},0), \\frac{p}{2}=2, p=4." }, { "text": "Given that the standard equation of line $l$ is $x + y + 1 = 0$, and point $P$ is an arbitrary point on the curve $C$: $\\frac{x^{2}}{3} + y^{2} = 1$, then the maximum distance from point $P$ to line $l$ is?", "fact_expressions": "l: Line;C: Curve;P: Point;Expression(C) = (x^2/3 + y^2 = 1);Expression(l) = (x + y + 1 = 0);PointOnCurve(P, C)", "query_expressions": "Max(Distance(P, l))", "answer_expressions": "3*sqrt(2)/2", "fact_spans": "[[[2, 7], [73, 78]], [[28, 60]], [[23, 27], [68, 72]], [[28, 60]], [[2, 22]], [[23, 66]]]", "query_spans": "[[[68, 87]]]", "process": "Draw a line parallel to line $ l $: $ x + y + t = 0 $, such that this line is tangent to the ellipse $ C $. Solve the system \n\\begin{cases}x+y+t=0\\\\\\frac{x^{2}}{3}+y^{2}=1\\end{cases}, \neliminate $ y $ to obtain $ 4x^{2} + 6tx + 3(t^{2}-1) = 0 $. The discriminant is $ \\Delta = 36t^{2} - 4\\times4\\times3(t^{2}-1) = 48 - 12t^{2} = 0 $, solving gives $ t = \\pm 2 $. Therefore, the maximum distance from point $ P $ to line $ l $ equals the distance between the line $ x + y - 2 = 0 $ and line $ l $, which is $ d = \\frac{3}{\\sqrt{2}} = \\frac{3\\sqrt{2}}{2} $." }, { "text": "Given the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has its right vertex at point $A$, and the focus of the parabola $C$: $y^{2}=8 a x$ is point $F$. If there exists a point $P$ on an asymptote of $E$ such that $\\overrightarrow{A P} \\perp \\overrightarrow{F P}$, then the range of the eccentricity of hyperbola $E$ is?", "fact_expressions": "E: Hyperbola;a: Number;b: Number;C: Parabola;A: Point;P: Point;F: Point;a>0;b>0;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);Expression(C) = (y^2 = 8*(a*x));RightVertex(E) = A;Focus(C) = F;PointOnCurve(P, Asymptote(E));IsPerpendicular(VectorOf(A,P),VectorOf(F,P))", "query_expressions": "Range(Eccentricity(E))", "answer_expressions": "(1, 3*sqrt(2)/4]", "fact_spans": "[[[2, 63], [104, 107], [172, 178]], [[10, 63]], [[10, 63]], [[72, 93]], [[68, 71]], [[114, 118]], [[97, 100]], [[10, 63]], [[10, 63]], [[2, 63]], [[72, 93]], [[2, 71]], [[72, 100]], [[104, 118]], [[121, 170]]]", "query_spans": "[[[172, 189]]]", "process": "Hyperbola $ E: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) has its right vertex at $ A(a, 0) $. The focus of the parabola $ C: y^{2} = 8ax $ is $ F(2a, 0) $. The asymptotes of the hyperbola are given by $ y = \\pm \\frac{b}{a}x $. Let point $ P(m, \\frac{b}{a}m) $, so that $ \\overrightarrow{AP} = (m - a, \\frac{b}{a}m) $, $ \\overrightarrow{FP} = (m - 2a, \\frac{b}{a}m) $. It follows that $ \\overrightarrow{AP} \\cdot \\overrightarrow{FP} = 0 $, which gives $ (m - a)(m - 2a) + \\frac{b^{2}}{a^{2}}m^{2} = 0 $, simplifying to $ \\left(1 + \\frac{b^{2}}{a^{2}}\\right)m^{2} - 3ma + 2a^{2} = 0 $. According to the problem, the discriminant $ \\Delta = 9a^{2} - 4\\left(1 + \\frac{b^{2}}{a^{2}}\\right) \\cdot 2a^{2} \\geqslant 0 $, leading to $ a^{2} \\geqslant 8b^{2} = 8(c^{2} - a^{2}) $, i.e., $ 8c^{2} \\leqslant 9a^{2} $. Therefore, the eccentricity $ e = \\frac{c}{a} \\leqslant \\frac{3\\sqrt{2}}{4} $. Since $ e > 1 $, we obtain $ 1 < e \\leqslant \\frac{3\\sqrt{2}}{4} $." }, { "text": "The equation of the locus of a moving point whose distance to the left focus of the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$ is equal to its distance to the fixed line $x=2$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/8 + y^2/4 = 1);P: Point;Distance(P, LeftFocus(G)) = Distance(P, H);H: Line;Expression(H) = (x = 2)", "query_expressions": "LocusEquation(P)", "answer_expressions": "y^2=-8*x", "fact_spans": "[[[1, 38]], [[1, 38]], [[59, 61]], [[0, 61]], [[49, 54]], [[49, 54]]]", "query_spans": "[[[59, 67]]]", "process": "" }, { "text": "Let the left and right vertices of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ ($a>b>0$) be $A$ and $B$, respectively. Let point $P$ lie on the ellipse and be distinct from points $A$ and $B$, and let $O$ be the origin. If the product of the slopes of lines $PA$ and $PB$ is $-\\frac{1}{2}$, then what is the eccentricity of the ellipse?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;A: Point;B: Point;O: Origin;a > b;b > 0;LeftVertex(G)=A;RightVertex(G)=B;P:Point;PointOnCurve(P, G);Negation(P=A);Negation(P=B);Slope(LineOf(P, A)) *Slope(LineOf(P, B)) = -1/2", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[1, 53], [77, 79], [139, 141]], [[1, 53]], [[3, 53]], [[3, 53]], [[62, 65], [83, 86]], [[68, 71], [89, 92]], [[95, 98]], [[3, 53]], [[3, 53]], [[1, 71]], [[1, 71]], [[72, 76]], [[72, 80]], [[72, 94]], [[72, 94]], [[106, 137]]]", "query_spans": "[[[139, 147]]]", "process": "Let the coordinates of point P be (x_{0}, y_{0}). From the given conditions, we have \\frac{x_{0}^{2}}{a^{2}}+\\frac{y_{0}^{2}}{b^{2}}=1,\\textcircled{1}. Given A(-a,0), B(a,0), we obtain k_{AP}=\\frac{y_{0}}{x_{0}+a}, k_{BP}=\\frac{y_{0}}{x_{0}-a}. From k_{AP}\\cdot k_{BP}=-\\frac{1}{2}, we get x_{0}^{2}=a^{2}-2y_{0}^{2}. Substituting into \\textcircled{1} and simplifying yields (a^{2}-2b^{2})y_{0}^{2}=0. Since y_{0}\\neq0, it follows that a^{2}=2b^{2}. Thus, e^{2}=\\frac{a^{2}-b^{2}}{a^{2}}=\\frac{1}{2}. Therefore, the eccentricity of the ellipse is e=\\frac{\\sqrt{2}}{2}." }, { "text": "Given that the perimeter of $\\triangle ABC$ is $26$ and the coordinates of points $A$, $B$ are $(-6,0)$, $(6,0)$ respectively, then the trajectory equation of point $C$ is?", "fact_expressions": "A: Point;B: Point;C: Point;Perimeter(TriangleOf(A, B, C)) = 26;Coordinate(A) = (-6, 0);Coordinate(B) = (6, 0)", "query_expressions": "LocusEquation(C)", "answer_expressions": "(x^2/49+y^2/13=1)&Negation(x=0)", "fact_spans": "[[[28, 32]], [[33, 36]], [[60, 64]], [[2, 27]], [[28, 58]], [[28, 58]]]", "query_spans": "[[[60, 71]]]", "process": "Problem Analysis: Since the perimeter of \\triangle ABC is 26, and vertex B is at (-6,0), C is at (6,0), \\therefore BC=12, AB+AC=26-12=14. \\because 14>12, \\therefore the sum of distances from point A to two fixed points is constant, \\therefore the locus of point A is an ellipse. \\because a=7, c=6, b^{2}=13, \\therefore the equation of the ellipse is \\frac{x2}{49}+\\frac{y^{2}}{13}=1(x\\neq0)." }, { "text": "The minimum distance from point $M(8,0)$ to a point on the parabola $y^{2}=10 x$ is?", "fact_expressions": "G: Parabola;M: Point;P0: Point;Expression(G) = (y^2 = 10*x);Coordinate(M) = (8, 0);PointOnCurve(P0, G)", "query_expressions": "Min(Distance(M, P0))", "answer_expressions": "sqrt(55)", "fact_spans": "[[[10, 25]], [[0, 9]], [[27, 28]], [[10, 25]], [[0, 9]], [[10, 28]]]", "query_spans": "[[[0, 37]]]", "process": "Let the coordinates of point A on the parabola \\( y^{2} = 10x \\) be \\( A\\left(\\frac{n^{2}}{10}, n\\right) \\), then the minimum value exists, and the minimum value is 55." }, { "text": "The distance from the focus of the parabola $y^{2}=2 m x(m>0)$ to an asymptote of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ is $3$, then the equation of this parabola is?", "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 2*m*x);m: Number;m>0;G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1);Distance(Focus(H),OneOf(Asymptote(G))) = 3", "query_expressions": "Expression(H)", "answer_expressions": "y^2=20*x", "fact_spans": "[[[0, 21], [81, 84]], [[0, 21]], [[3, 21]], [[3, 21]], [[25, 64]], [[25, 64]], [[0, 77]]]", "query_spans": "[[[81, 89]]]", "process": "The focus of the parabola is given by $(\\frac{m}{2},0)$. The asymptotes of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ are $y=\\pm\\frac{3}{4}x$. Taking one of them as $y=\\frac{3}{4}x$, which is equivalent to $3x-4y=0$. According to the problem, $\\frac{|3\\times\\frac{m}{2}|}{\\sqrt{3^{2}+4^{2}}}=\\frac{|3m|}{10}=3$. Solving this equation gives $m=10$. Therefore, the equation of the parabola is $y^2=20x$. Answer: $y^2=20x$." }, { "text": "Given point $M(-2,2)$ and the parabola $C$: $y^{2}=8 x$, a line passing through the focus of $C$ with slope $k$ intersects $C$ at points $A$ and $B$. If $\\overrightarrow{M A} \\cdot \\overrightarrow{M B}=0$, then $k=?$", "fact_expressions": "C: Parabola;G: Line;M: Point;A: Point;B: Point;k:Number;Expression(C) = (y^2 = 8*x);Coordinate(M) = (-2, 2);PointOnCurve(Focus(C), G);Slope(G) = k;Intersection(G, C) = {A, B};DotProduct(VectorOf(M, A), VectorOf(M, B)) = 0", "query_expressions": "k", "answer_expressions": "2", "fact_spans": "[[[13, 32], [34, 37], [51, 54]], [[48, 50]], [[2, 12]], [[56, 59]], [[60, 63]], [[44, 47], [122, 125]], [[13, 32]], [[2, 12]], [[33, 50]], [[41, 50]], [[48, 65]], [[68, 119]]]", "query_spans": "[[[122, 127]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), then \\begin{cases}y_{1}^{2}=8x_{1},\\\\y_{2}^{2}=8x_{2},\\end{cases} so y_{1}^{2}-y_{2}^{2}=8x_{1}-8x_{2}. Therefore, k=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{8}{y_{1}+y_{2}}. Let M(x_{0},y_{0}) be the midpoint of AB, and draw perpendiculars from points A and B to the directrix x=-2, with feet of perpendiculars denoted as A and B'. Since \\overrightarrow{MA}\\cdot\\overrightarrow{MB}=0, \\angle AMB=90^{\\circ}, it follows that |MM|=\\frac{1}{2}|AB|=\\frac{1}{2}(|AF|+|BF|)=\\frac{1}{2}(|AA|+|BB|). Because M is the midpoint of AB, MM is parallel to the x-axis. Given M(-2,2), the answer is: 2" }, { "text": "Given that the vertex of the parabola $C$ is at the origin, the focus is on the $y$-axis, and it passes through the point $(-1 , 4)$, then the equation of the directrix of the parabola is?", "fact_expressions": "C: Parabola;G: Point;O: Origin;Coordinate(G) = (-1, 4);Vertex(C) = O;PointOnCurve(Focus(C), yAxis);PointOnCurve(G,C)", "query_expressions": "Expression(Directrix(C))", "answer_expressions": "y=-1/16", "fact_spans": "[[[2, 8], [40, 43]], [[27, 38]], [[12, 14]], [[27, 38]], [[2, 14]], [[2, 23]], [[2, 38]]]", "query_spans": "[[[40, 50]]]", "process": "" }, { "text": "Given a hyperbola $ C $: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 $ ($ a>0, b>0 $), let point $ P $ be on the right branch, and $ F_{1} $, $ F_{2} $ be its left and right foci respectively. Let circle $ M $ be the incircle of $ \\Delta P F_{1} F_{2} $, and $ P F_{1} $ is tangent to circle $ M $ at point $ A $, with $ |P A| = \\frac{c^{2}}{2 a} $ ($ c $ is the semi-focal length). If $ \\frac{| P F_{1} |}{|PF_{2}|} > 2 $, then what is the range of values for the eccentricity of the hyperbola?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;P: Point;PointOnCurve(P,RightPart(C)) = True;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;M: Circle;IsInscribedCircle(M,TriangleOf(P,F1,F2)) = True;A: Point;TangentPoint(LineSegmentOf(P, F1), M) = A;Abs(LineSegmentOf(P,A)) = c^2/2*a;c: Number;HalfFocalLength(C) = c;Abs(LineSegmentOf(P, F1))/Abs(LineSegmentOf(P, F2)) > 2", "query_expressions": "Range(Eccentricity(C))", "answer_expressions": "(1,sqrt(7)-1)", "fact_spans": "[[[2, 63], [90, 91], [220, 223]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 71]], [[2, 71]], [[72, 79]], [[80, 87]], [[72, 95]], [[72, 95]], [[96, 100], [138, 142]], [[96, 126]], [[145, 149]], [[128, 149]], [[150, 175]], [[176, 179]], [[90, 184]], [[186, 218]]]", "query_spans": "[[[220, 233]]]", "process": "First, use the definition of the hyperbola and the properties of the incircle to prove that the incircle is tangent to the x-axis at the vertex. Then express |PF_{1}| and |PF_{2}| respectively, and set up a homogeneous inequality in terms of a and c to find the range of the eccentricity of the hyperbola. Let the center of the circle be M(x,y), and let the incircle be tangent to PF_{1}, PF_{2}, and F_{1}F_{2} at points A, B, C, respectively, as shown in the figure: According to the properties of the incircle, we have |PA| = |PB|, |F_{1}A| = |F_{1}C|, |F_{2}B| = |F_{2}C|. Therefore, |PF_{1}| - |PF_{2}| = |PA| + |AF_{1}| - |PB| - |BF_{2}| = |CF_{1}| - |CF_{2}| = 2a. Hence, point C is the vertex of the hyperbola, so |F_{1}A| = |F_{1}C| = c + a, |F_{2}B| = |F_{2}C| = c - a, |PA| = |PB| = \\frac{c^{2}}{2a}. Since \\frac{|PF_{1}|}{|PF_{2}|} = \\frac{c + a + \\frac{c^{2}}{2a}}{c - a + \\frac{c^{2}}{2a}} > 2, simplifying gives: c^{2} + 2ac - 6a^{2} < 0. Dividing both sides by a^{2} yields e^{2} + 2e - 6 < 0. Solving this inequality gives: -1 - \\sqrt{7} < e < -1 + \\sqrt{7}, and since e > 1, the range of eccentricity is (1, \\sqrt{7} - 1)." }, { "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$. A line passing through $F$ intersects the parabola $C$ at points $A$ and $B$. The circle with diameter $AB$ is tangent to the directrix of the parabola $C$ at $M(-\\frac{p}{2}, 3)$, and the area of $\\triangle A O B$ is $\\sqrt{13}$. Then the equation of the parabola $C$ is?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;H: Line;PointOnCurve(F, H);A: Point;B: Point;Intersection(H, C) = {A, B};G: Circle;IsDiameter(LineSegmentOf(A, B), G);M: Point;Coordinate(M) = (-p/2, 3);TangentPoint(Directrix(C), G) = M;O: Origin;Area(TriangleOf(A, O, B)) = sqrt(13)", "query_expressions": "Expression(C)", "answer_expressions": "y^2=4*x", "fact_spans": "[[[2, 28], [44, 50], [75, 81], [143, 149]], [[2, 28]], [[10, 28]], [[10, 28]], [[32, 35], [37, 40]], [[2, 35]], [[41, 43]], [[36, 43]], [[51, 54]], [[55, 58]], [[41, 60]], [[73, 74]], [[61, 74]], [[86, 106]], [[86, 106]], [[73, 106]], [[109, 126]], [[109, 141]]]", "query_spans": "[[[143, 154]]]", "process": "Let the equation of the line passing through the focus $ F\\left(\\frac{p}{2},0\\right) $ be $ y = k\\left(x - \\frac{p}{2}\\right) $. Substituting $ x = \\frac{y^{2}}{2p} $ into it, we get $ y = k\\left(\\frac{y^{2}}{2p} - \\frac{p}{2}\\right) $, which simplifies to $ ky^{2} - 2py - kp^{2} = 0 $. Let the intersection points be $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Then $ y_{1} + y_{2} = \\frac{2p}{k} = -6 $, $ y_{1}y_{2} = -p^{2} $, $ |y_{1} - y_{2}| = \\sqrt{(y_{1} + y_{2})^{2} - 4y_{1}y_{2}} = \\sqrt{36 + 4p^{2}} $. From the given condition, $ p^{2} = 4 \\Rightarrow p = 2 $. Hence, $ y^{2} = 2px = 4x $. The answer to be filled is $ y^{2} = 4x $." }, { "text": "The line passing through the focus $F$ of the parabola $y^{2}=2 p x(p>0)$ is parallel to an asymptote of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, and intersects the parabola at points $A$ and $B$. If $|A F|>|B F|$ and $|A F|=2$, then the value of $p$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1);H: Parabola;Expression(H) = (y^2 = 2*p*x);p: Number;A: Point;F: Point;B: Point;p>0;Focus(H)=F;l:Line;PointOnCurve(F,l);IsParallel(l,OneOf(Asymptote(G)));Intersection(l,H)={A,B};Abs(LineSegmentOf(A, F)) > Abs(LineSegmentOf(B, F));Abs(LineSegmentOf(A, F)) = 2", "query_expressions": "p", "answer_expressions": "1", "fact_spans": "[[[31, 59]], [[31, 59]], [[1, 22], [70, 73]], [[1, 22]], [[111, 114]], [[74, 77]], [[24, 27]], [[78, 81]], [[4, 22]], [[1, 27]], [[28, 30]], [[0, 30]], [[28, 67]], [[28, 83]], [[85, 98]], [[100, 109]]]", "query_spans": "[[[111, 118]]]", "process": "According to the definition of a parabola and the definition of a hyperbola, assume without loss of generality that line AB is $ y = \\sqrt{3}(x - \\frac{p}{2}) $. Let $ A(x_{0}, y_{0}) $, then $ |AF| = x_{0} + \\frac{p}{2} = 2 $, thus obtaining the coordinates of point A. Finally, substitute into the equation of the parabola to find the value of $ p $.\n\n[SOLUTION] Solution: The focus F of the parabola $ y^{2} = 2px $ ($ p > 0 $) has coordinates $ (\\frac{p}{2}, 0) $, and the directrix equation is $ x = -\\frac{p}{2} $. The asymptotes of the hyperbola $ x^{2} - \\frac{y^{2}}{3} = 1 $ are $ y = \\sqrt{3}x $. A line passing through the focus F of the parabola $ y^{2} = 2px $ ($ p > 0 $) is parallel to the asymptote $ y = \\sqrt{3}x $ and intersects the parabola at points A and B. Without loss of generality, assume line AB is $ y = \\sqrt{3}(x - \\frac{p}{2}) $. Let $ A(x_{0}, y_{0}) $. Therefore, $ |AF| = x_{0} + \\frac{p}{2} $. Since $ |AF| > |BF| $ and $ |AF| = 2 $, it follows that $ x_{0} = 2 - \\frac{p}{2} $, $ x_{0} > \\frac{p}{2} $, so $ 0 < p < 2 $. Thus, $ y_{0} = \\sqrt{3}(2 - p) $. Then $ A(2 - \\frac{p}{2}, \\sqrt{3}(2 - p)) $. Since point A lies on the parabola $ y^{2} = 2px $ ($ p > 0 $), we have $ 3(2 - p)^{2} = 2p(2 - \\frac{p}{2}) $. Simplifying yields $ p^{2} - 4p + 3 = 0 $, solving gives $ p = 1 $ or $ p = 3 $ (discarded)." }, { "text": "Given that $F$ is the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, a line passing through $F$ and perpendicular to the $x$-axis intersects the hyperbola at points $A$ and $B$. If the circle with diameter $AB$ passes through the origin, then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;L: Line;A: Point;B: Point;F: Point;O:Origin;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;PointOnCurve(F, L);IsPerpendicular(L,xAxis);Intersection(L, G) = {A, B};IsDiameter(LineSegmentOf(A,B),H);PointOnCurve(O,H)", "query_expressions": "Eccentricity(G)", "answer_expressions": "(sqrt(5)+1)/2", "fact_spans": "[[[6, 62], [83, 86], [117, 120]], [[9, 62]], [[9, 62]], [[108, 109]], [[80, 82]], [[87, 90]], [[91, 94]], [[68, 71], [2, 5]], [[110, 114]], [[9, 62]], [[9, 62]], [[6, 62]], [[2, 66]], [[67, 82]], [[72, 82]], [[80, 96]], [[98, 109]], [[108, 114]]]", "query_spans": "[[[117, 126]]]", "process": "Let F(c,0), substitute x=c into \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 to get \\frac{y^{2}}{b^{2}}=\\frac{c^{2}}{a^{2}}-1=\\frac{b^{2}}{a^{2}}, y=\\pm\\frac{b^{2}}{a}, that is, point A(c,\\frac{b^{2}}{a}), B(c,-\\frac{b^{2}}{a}), AB=\\frac{2b^{2}}{a}, and the circle with AB as diameter passes through the origin, then c=\\frac{b^{2}}{a}, and b^{2}=c^{2}-a^{2} \\therefore c^{2}-ac-a^{2}=0 \\Rightarrow e^{2}-e-1=0, and e>1, solving gives e=\\frac{\\sqrt{5}+1}{2}" }, { "text": "$A B$ is a chord of the parabola $y^{2}=x$. If the distance from the midpoint of $A B$ to the $y$-axis is $1$, then the maximum length of the chord $A B$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = x);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G) = True;Distance(MidPoint(LineSegmentOf(A, B)), yAxis) = 1", "query_expressions": "Max(Length(LineSegmentOf(A, B)))", "answer_expressions": "5/2", "fact_spans": "[[[6, 18]], [[6, 18]], [[0, 5]], [[0, 5]], [[0, 22]], [[24, 44]]]", "query_spans": "[[[47, 61]]]", "process": "" }, { "text": "Given that the distance from a point $M(1, y_{0})$ on the parabola $y^{2}=2 p x(p>0)$ to its focus is $5$, and the left vertex of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{b^{2}}=1 (b>0)$ is $A$. If an asymptote of the hyperbola $C$ is perpendicular to the line $A M$, then its eccentricity is?", "fact_expressions": "C: Hyperbola;b: Number;G: Parabola;p: Number;M: Point;A: Point;b>0;Expression(C) = (x^2 - y^2/b^2 = 1);p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(M) = (1, y0);y0:Number;PointOnCurve(M, G);Distance(M, Focus(G)) = 5;LeftVertex(C) = A;IsPerpendicular(OneOf(Asymptote(C)),LineOf(A,M))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)/2", "fact_spans": "[[[51, 95], [105, 111], [129, 130]], [[59, 95]], [[2, 23], [40, 41]], [[5, 23]], [[26, 39]], [[100, 103]], [[59, 95]], [[51, 95]], [[5, 23]], [[2, 23]], [[26, 39]], [[26, 39]], [[2, 39]], [[26, 50]], [[51, 103]], [[105, 127]]]", "query_spans": "[[[129, 135]]]", "process": "\\because the distance from point M(1,m) on the parabola y^{2}=2px(p>0) to its focus is 5, \\therefore 1+\\frac{p}{2}=5, p=8, the equation of the parabola is y^{2}=16x, m=\\pm4. Take M(1,4), the left vertex of the hyperbola is A(a,0), the slope of line AM is \\frac{4}{1+a}. \\because one asymptote of the hyperbola is perpendicular to line AM, \\frac{4}{1+a}\\times b=-1, and a=1, solving gives: b=\\frac{1}{2}, then: e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{a^{2}+b^{2}}{a^{2}}=\\frac{5}{4}, e=\\frac{\\sqrt{5}}{2}" }, { "text": "The sum of the distances from the points on curve $C$ to $F_{1}(0,-1)$, $F_{2}(0,1)$ is $4$, then the equation of curve $C$ is?", "fact_expressions": "C: Curve;P: Point;PointOnCurve(P,C) = True;F1: Point;Coordinate(F1) = (0, -1);F2: Point;Coordinate(F2) = (0, 1);Distance(P,F1) + Distance(P,F2) = 4", "query_expressions": "Expression(C)", "answer_expressions": "y^2/4 + x^2/3 = 1", "fact_spans": "[[[0, 5], [47, 52]], [[7, 8]], [[0, 8]], [[9, 22]], [[9, 22]], [[24, 36]], [[24, 36]], [[7, 45]]]", "query_spans": "[[[47, 57]]]", "process": "" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the distance from the right focus $F(5,0)$ to the asymptote is $4$, then the length of the real axis is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;F: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (5, 0);RightFocus(C)=F;Distance(F,Asymptote(C))=4", "query_expressions": "Length(RealAxis(C))", "answer_expressions": "6", "fact_spans": "[[[2, 63]], [[10, 63]], [[10, 63]], [[67, 75]], [[10, 63]], [[10, 63]], [[2, 63]], [[67, 75]], [[2, 75]], [[2, 86]]]", "query_spans": "[[[2, 93]]]", "process": "Solution: The asymptotes of hyperbola C have equations y = \\pm\\frac{b}{a}x, so the distance d from focus F(5,0) to the asymptote bx \\pm ay = 0 is d = \\frac{|5b|}{\\sqrt{b^{2}+a^{2}}} = 4. Therefore, \\frac{5b}{c} = 4, which implies b = \\frac{4c}{5} = \\frac{4 \\times 5}{5} = 4. Hence, a^{2} = c^{2} - b^{2} = 25 - 16 = 9, so a = 3, and thus the length of the real axis is 2a = 6." }, { "text": "The center of circle $C$ is the focus of the parabola $x^{2}=4 y$. The line $4 x-3 y-2=0$ intersects circle $C$ at points $A$ and $B$, and $|A B|=6$. Then the standard equation of circle $C$ is?", "fact_expressions": "C: Circle;G: Parabola;Expression(G) = (x^2 = 4*y);Center(C) = Focus(G);H: Line;Expression(H) = (4*x - 3*y - 2 = 0);A: Point;B: Point;Intersection(H, C) = {A, B};Abs(LineSegmentOf(A, B)) = 6", "query_expressions": "Expression(C)", "answer_expressions": "x^2+(y-1)^2=10", "fact_spans": "[[[2, 6], [44, 48], [74, 78]], [[10, 24]], [[10, 24]], [[2, 27]], [[28, 43]], [[28, 43]], [[51, 54]], [[55, 58]], [[28, 60]], [[62, 72]]]", "query_spans": "[[[74, 85]]]", "process": "Solution: From the given conditions, the center of the circle is C(0,1). Therefore, the distance from the center C(0,1) to the line 4x-3y-2=0 is d = \\frac{|-3-2|}{\\sqrt{4^{2}+(-3)^{2}}} = 1. Since the line 4x-3y-2=0 intersects the circle C at points A and B, and |AB| = 6, the radius r of circle C is r = \\sqrt{(\\frac{1}{2}|AB|)^{2}+d^{2}} = \\sqrt{9+1} = \\sqrt{10}. The standard equation of circle C is: x^{2}+(y-1)^{2}=10." }, { "text": "Given that one asymptote of the hyperbola $C$ is the line $l$: $y=2x$, and its transverse axis length is less than $4$, then a standard equation of $C$ could be?", "fact_expressions": "C: Hyperbola;l: Line;OneOf(Asymptote(C)) = l;Expression(l) = (y = 2*x);Length(RealAxis(C)) < 4", "query_expressions": "Expression(C)", "answer_expressions": "x^2 - y^2/4 = 1", "fact_spans": "[[[2, 8], [42, 45], [31, 32]], [[17, 29]], [[2, 20]], [[17, 29]], [[31, 40]]]", "query_spans": "[[[42, 56]]]", "process": "Solution: The equation of the hyperbola can be assumed as x^{2}-\\frac{y^{2}}{4}=2(\\lambda\\neq0), that is, \\frac{x^{2}}{2}-\\frac{y^{2}}{4\\lambda}=1(\\lambda\\neq0). When \\lambda>0, the real axis of the hyperbola is 2\\sqrt{\\lambda}, then 2\\sqrt{2}<4, so 0<\\lambda<4. We can take \\lambda=1, then a standard equation of C can be x^{2}-\\frac{y^{2}}{4}=1." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$, $F_{2}$ respectively. $P$ is a point on the hyperbola $C$. The circle with diameter $F_{1} F_{2}$ intersects the asymptotes of the hyperbola at point $Q$. Both $P$ and $Q$ lie in the first quadrant, and $P$ is the midpoint of $Q F_{2}$. Then the eccentricity of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C) = True;IsDiameter(LineSegmentOf(F1, F2), G) = True;G: Circle;Intersection(G, Asymptote(C)) = Q;Q: Point;Quadrant(P) = 1;Quadrant(Q) = 1;MidPoint(LineSegmentOf(Q, F2)) = P", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)-1", "fact_spans": "[[[2, 63], [92, 98], [122, 125], [170, 176]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[72, 79]], [[80, 87]], [[2, 87]], [[2, 87]], [[88, 91], [136, 139], [152, 155]], [[88, 101]], [[102, 121]], [[120, 121]], [[120, 135]], [[131, 135], [140, 143]], [[136, 150]], [[136, 150]], [[152, 168]]]", "query_spans": "[[[170, 182]]]", "process": "" }, { "text": "The standard equation of the parabola with focus at the right focus of the hyperbola $\\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/5 - y^2/4 = 1);RightFocus(G) = Focus(H);H: Parabola", "query_expressions": "Expression(H)", "answer_expressions": "y^2 = 12*x", "fact_spans": "[[[1, 39]], [[1, 39]], [[0, 50]], [[47, 50]]]", "query_spans": "[[[47, 57]]]", "process": "In the hyperbola, $ c = \\sqrt{5+4} = 3 $, so the right focus is $ F(3,0) $. The focus of the parabola is also $ (3,0) $, so $ \\frac{p}{2} = 3 $, $ p = 6 $. The standard equation of the parabola is: $ y^{2} = 12x $." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, let $D$ be one vertex of the conjugate axis. The line $x=2a$ intersects $C$ at points $A$ and $B$. If the orthocenter of $\\triangle ABD$ lies on an asymptote of $C$, then the eccentricity of $C$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;D: Point;OneOf(Vertex(ImageinaryAxis(C))) = D;G: Line;Expression(G) = (x = 2*a);A: Point;B: Point;Intersection(G, C) = {A, B};PointOnCurve(Orthocenter(TriangleOf(A, B, D)), OneOf(Asymptote(C)))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 63], [85, 88], [122, 125], [134, 137]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[71, 74]], [[2, 74]], [[75, 84]], [[75, 84]], [[90, 93]], [[94, 97]], [[75, 99]], [[101, 132]]]", "query_spans": "[[[134, 143]]]", "process": "Solution: Let the orthocenter of triangle ABD be H, then DH ⊥ AB. Without loss of generality, let D(0, b), then H(x, b). Substituting into the asymptote equation y = \\frac{b}{a}x, we solve to get x = a, so H(a, b). Since the line x = 2a intersects the hyperbola at points A and B, the coordinates of A and B are respectively: A(2a, \\sqrt{3}b), B(2a, -\\sqrt{3}b). Because k_{AD} \\cdot k_{BH} = \\frac{(\\sqrt{3}-1)b}{2a} \\times \\frac{(\\sqrt{3}+1)b}{-a} = -1, simplifying yields a^{2} = b^{2}. Therefore, the eccentricity of the hyperbola is e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{2}." }, { "text": "Given the hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$, its left directrix is $l$, and the left and right foci are $F_{1}$ and $F_{2}$ respectively. The parabola $C_{2}$ has directrix $l$ and focus $F_{2}$. Let $M$ be the intersection point of $C_{1}$ and $C_{2}$. Then $\\frac{|F_{1} F_{2}|}{|M F_{1}|}-\\frac{|M F_{1}|}{|M F_{2}|}$=?", "fact_expressions": "C1: Hyperbola;Expression(C1) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;l: Line;LeftDirectrix(C1) = l;F1: Point;F2: Point;LeftFocus(C1) = F1;RightFocus(C1) = F2;C2: Parabola;Directrix(C2) = l;Focus(C2) = F2;M: Point;Intersection(C1, C2) = M", "query_expressions": "Abs(LineSegmentOf(F1, F2))/Abs(LineSegmentOf(M, F1)) - Abs(LineSegmentOf(M, F1))/Abs(LineSegmentOf(M, F2))", "answer_expressions": "-1", "fact_spans": "[[[2, 70], [132, 139]], [[2, 70]], [[14, 70]], [[14, 70]], [[14, 70]], [[14, 70]], [[75, 78], [117, 120]], [[2, 78]], [[87, 94]], [[95, 102], [124, 131]], [[2, 102]], [[2, 102]], [[103, 113], [140, 147]], [[103, 120]], [[103, 131]], [[151, 154]], [[132, 154]]]", "query_spans": "[[[156, 219]]]", "process": "" }, { "text": "The asymptote equations of the hyperbola $\\frac{x^{2}}{36}-\\frac{y^{2}}{45}=1$ are? The eccentricity is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/36 - y^2/45 = 1)", "query_expressions": "Expression(Asymptote(G));Eccentricity(G)", "answer_expressions": "y=pm*x*sqrt(5)/2\n3/2", "fact_spans": "[[[0, 40]], [[0, 40]]]", "query_spans": "[[[0, 48]], [[0, 53]]]", "process": "" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, a line passing through the left focus $F_{1}$ intersects the ellipse $C$ at points $A$ and $B$, such that $|A F_{1}|=3|B F_{1}|$, $|A B|=|B F_{2}|$. Then the eccentricity of the ellipse $C$ is?", "fact_expressions": "C: Ellipse;a: Number;b: Number;G: Line;A: Point;F1: Point;B: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, G);Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, F1)) = 3*Abs(LineSegmentOf(B, F1));Abs(LineSegmentOf(A, B)) = Abs(LineSegmentOf(B, F2))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(10)/5", "fact_spans": "[[[18, 75], [97, 102], [158, 163]], [[24, 75]], [[24, 75]], [[94, 96]], [[104, 107]], [[2, 9], [86, 93]], [[108, 111]], [[10, 17]], [[24, 75]], [[24, 75]], [[18, 75]], [[2, 81]], [[2, 81]], [[82, 96]], [[94, 113]], [[115, 137]], [[139, 156]]]", "query_spans": "[[[158, 169]]]", "process": "Connect $AF_{2}$, let $|BF_{1}|=k$, use the ellipse property $|BF_{1}|+|BF_{2}|=|AF_{1}|+|AF_{2}|=2a$, obtain the length of $AF_{2}$, then apply the cosine law in $\\triangle ABF_{2}$ and $\\triangle F_{1}AF_{2}$ respectively to find the length of $c$, and compute the answer using the definition of eccentricity. Let $|BF_{1}|=k$, then $|AF_{1}|=3k$, $|BF_{2}|=4k$. From $|BF_{1}|+|BF_{2}|=|AF_{1}|+|AF_{2}|=2a$, we get $2a=5k$, $|AF_{2}|=2k$. In $\\triangle ABF_{2}$, $\\cos\\angle BAF_{2}=\\frac{1}{4}$, hence the eccentricity $e=\\frac{c}{a}=\\frac{\\sqrt{10}}{5}$." }, { "text": "The equation of the line passing through the focus of the parabola $y^{2}=4x$ and intersecting the circle $x^{2}+y^{2}-4x+2y=0$ such that the chord cut off is the longest possible is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);L: Line;PointOnCurve(Focus(G), L);H: Circle;Expression(H) = (2*y - 4*x + x^2 + y^2 = 0);WhenMax(Length(InterceptChord(L, H)))", "query_expressions": "Expression(L)", "answer_expressions": "x+y-1=0", "fact_spans": "[[[1, 15]], [[1, 15]], [[51, 53]], [[0, 53]], [[21, 45]], [[21, 45]], [[20, 53]]]", "query_spans": "[[[51, 58]]]", "process": "" }, { "text": "If a point $P$ on the hyperbola $\\frac{x^{2}}{64}-\\frac{y^{2}}{36}=1$ is at a distance of $8$ from its right focus, then what is the distance from point $P$ to its right directrix?", "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2/64 - y^2/36 = 1);PointOnCurve(P, G);Distance(P,RightFocus(G)) = 8", "query_expressions": "Distance(P,RightDirectrix(G))", "answer_expressions": "32/5", "fact_spans": "[[[1, 41], [48, 49], [67, 68]], [[44, 47], [62, 66]], [[1, 41]], [[1, 47]], [[44, 60]]]", "query_spans": "[[[62, 77]]]", "process": "" }, { "text": "The focal distance of the curve $\\frac{x^{2}}{25-k}+\\frac{y^{2}}{9-k}=1 (90, it follows that t>-1. \nThen \n\\begin{cases}x_{1}+x_{2}=2t+4\\\\x_{1}\\cdot x_{2}=t^{2}\\end{cases}. \nThus, y_{1}\\cdot y_{2}=(x_{1}-t)(x_{2}-t)=x_{1}\\cdot x_{2}-t(x_{1}+x_{2})+t^{2}=-4t. \nTherefore, \\overrightarrow{OA}\\cdot\\overrightarrow{OB}=x_{1}\\cdot x_{2}+y_{1}\\cdot y_{2}=t^{2}-4t=-3, which gives t=1 or 3, both satisfying \\triangle>0." }, { "text": "Given the parabola equation $2 x^{2}=y$, what are the coordinates of its focus?", "fact_expressions": "G: Parabola;Expression(G) = (2*x^2 = y)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0,1/8)", "fact_spans": "[[[2, 5], [21, 22]], [[2, 19]]]", "query_spans": "[[[21, 28]]]", "process": "" }, { "text": "The standard equation of an ellipse that has the same eccentricity as the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ and passes through the point $(2,-\\sqrt{3})$ is?", "fact_expressions": "G1: Ellipse;Expression(G1) = (x^2/4 + y^2/3 = 1);G2: Ellipse;Eccentricity(G1) = Eccentricity(G2);H: Point;Coordinate(H) = (2, -sqrt(3));PointOnCurve(H,G2) = True", "query_expressions": "Expression(G2)", "answer_expressions": "{(x^2/8+y^2/6=1),(y^2/25/3+x^2/25/4=1)}", "fact_spans": "[[[1, 38]], [[1, 38]], [[64, 66]], [[0, 66]], [[47, 63]], [[47, 63]], [[45, 66]]]", "query_spans": "[[[64, 72]]]", "process": "Method 1: \\because e=\\frac{c}{a}=\\frac{\\sqrt{a^{2}-b^{2}}}{a}=\\sqrt{1-\\frac{b^{2}}{a^{2}}}=\\sqrt{1-\\frac{3}{4}}=\\frac{1}{2}. If the foci are on the x-axis, assume the required ellipse equation is \\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{n^{2}}=1 (m>n>0), then 1-(\\frac{n}{m})^{2}=\\frac{1}{4}, thus (\\frac{n}{m})^{2}=\\frac{3}{4}, \\frac{n}{m}=\\frac{\\sqrt{3}}{2}. Also \\frac{4}{m^{2}}+\\frac{3}{n^{2}}=1, \\therefore m^{2}=8, n^{2}=6. \\therefore The standard equation of the required ellipse is \\frac{x^{2}}{8}+\\frac{y^{2}}{6}=1. If the foci are on the y-axis, assume the ellipse equation is \\frac{y^{2}}{m^{2}}+\\frac{x^{2}}{n^{2}}=1 (m>n>0), then \\frac{3}{m^{2}}+\\frac{4}{n^{2}}=1, and \\frac{n}{m}=\\frac{\\sqrt{3}}{2}, solving gives m^{2}=\\frac{25}{3}, n^{2}=\\frac{25}{4}. Hence, the standard equation of the required ellipse is \\underline{\\frac{y^{2}}{25}}+\\frac{x^{2}}{\\frac{25}{2}}=1." }, { "text": "If the equation $x^{2}+(k-1) y^{2}=k+1$ represents a hyperbola with foci on the $x$-axis, then the range of real values for $k$ is?", "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (x^2 + y^2*(k - 1) = k + 1);PointOnCurve(Focus(G), xAxis)", "query_expressions": "Range(k)", "answer_expressions": "(-1,1)", "fact_spans": "[[[37, 40]], [[42, 47]], [[1, 40]], [[28, 40]]]", "query_spans": "[[[42, 54]]]", "process": "Problem Analysis: Transforming the equation $x^{2}+(k-1)y^{2}=k+1$, we obtain $\\frac{x^2}{k+1}+\\frac{y^{2}}{\\frac{k+1}{k-1}}=1$. From the given conditions, we have \n$$\n\\begin{cases}\nk+1>0 \\\\\n\\frac{k+1}{k-1}<0\n\\end{cases}\n\\Rightarrow\n\\begin{cases}\nk+1>0 \\\\\nk-1<0\n\\end{cases}\n\\Rightarrow\n\\begin{cases}\nk>-1 \\\\\nk<1\n\\end{cases}\n\\Rightarrow -10)$, its two foci are denoted by $F_{1}$ and $F_{2}$. Let point $P(x_{0}, y_{0})$ be a point on the first quadrant branch of the hyperbola. Draw the tangent line $l$ to hyperbola $C$ at point $P$. If the product of the distances from points $F_{1}$ and $F_{2}$ to the tangent line $l$ is $3$, then what is the eccentricity of hyperbola $C$?", "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/4 - y^2/b^2 = 1);b: Number;b>0;F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;x0: Number;y0: Number;Coordinate(P) = (x0, y0);Quadrant(P) = 1;PointOnCurve(P, C);l: Line;TangentOnPoint(P, C) = l;Distance(F1, l)*Distance(F2, l) = 3", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(7)/2", "fact_spans": "[[[2, 54], [97, 100], [115, 121], [162, 168]], [[2, 54]], [[10, 54]], [[10, 54]], [[62, 69], [129, 137]], [[70, 77], [138, 145]], [[2, 77]], [[78, 96], [109, 113]], [[79, 96]], [[79, 96]], [[78, 96]], [[78, 107]], [[78, 107]], [[124, 127], [148, 151]], [[108, 127]], [[129, 160]]]", "query_spans": "[[[162, 174]]]", "process": "Let point P(x_{0},y_{0}) (x_{0}>2, y_{0}>0), we have \\frac{x^{2}}{4}-\\frac{y_{0}^{2}}{b^{2}}=1 \\Rightarrow 4y_{0}^{2}=b^{2}x_{0}^{2}-4b^{2}. Let the tangent line at point P be y-y_{0}=k(x-x_{0}), combining with the hyperbola equation, solving by \\triangle=0 gives k=\\frac{b^{2}x_{0}}{4y_{0}}, so the tangent line equation is (b^{2}x_{0})x-(4y_{0})y-4b^{2}=0. The distance from F_{1}(-c,0) to the tangent line is d_{1}=\\frac{|-b^{2}cx_{0}-4b^{2}|}{\\sqrt{b^{4}x_{0}^{2}+16y_{0}^{2}}}=\\frac{b^{2}cx_{0}+4b^{2}}{\\sqrt{b^{4}x_{0}^{2}+16y_{0}^{2}}}, and the distance from F_{2}(c,0) to the tangent line is d_{2}=\\frac{|b^{2}c}{\\sqrt{b^{4}x}}. Therefore, d_{1}d_{2}=\\frac{b^{4}c^{2}x_{0}^{2}-16b^{4}}{b^{4}x_{0}^{2}+16y_{0}^{2}}=\\frac{b^{4}(b^{2}+4)x_{0}^{2}-16b^{4}}{b^{4}x_{0}^{2}+4b^{2}x_{0}^{2}-16b^{2}}=b^{2}=3, thus b=\\sqrt{3}, so a=2, c=\\sqrt{7}, hence e=\\frac{\\sqrt{7}}{2}." }, { "text": "The equation of the directrix of the parabola $x^{2}=12 y$ is?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 12*y)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "y=-3", "fact_spans": "[[[0, 15]], [[0, 15]]]", "query_spans": "[[[0, 22]]]", "process": "\\becausex^{2}=2py=12y,\\therefore\\frac{p}{2}=3,\\therefore the directrix equation of the parabola x^{2}=12y is y=-\\frac{p}{2}=-3," }, { "text": "The coordinates of the foci of the ellipse $\\frac{x^{2}}{2}+\\frac{y^{2}}{3}=1$ are?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2/3 = 1)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0, pm*1)", "fact_spans": "[[[0, 37]], [[0, 37]]]", "query_spans": "[[[0, 44]]]", "process": "From \\frac{x^{2}}{2}+\\frac{y^{2}}{3}=1, we obtain c^{2}=3-2=1, c=1, and the foci lie on the y-axis, so the coordinates of the foci are (0,\\pm1)" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$, respectively, if there exists a point $P$ on the left branch of the hyperbola such that $\\frac{|P F_{2}|^{2}}{|P F_{1}|}=8 a$, then the range of the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F2: Point;F1: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P,LeftPart(G));Abs(LineSegmentOf(P,F2))^2/Abs(LineSegmentOf(P,F1))=8*a", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1,3]", "fact_spans": "[[[20, 79], [87, 90], [141, 144]], [[23, 79]], [[23, 79]], [[97, 100]], [[10, 17]], [[2, 9]], [[23, 79]], [[23, 79]], [[20, 79]], [[2, 85]], [[2, 85]], [[87, 100]], [[102, 139]]]", "query_spans": "[[[141, 155]]]", "process": "" }, { "text": "Given that $M(x_{0}, y_{0})$ is a point on the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, the semi-focal distance is $c$, and if $|M O| \\leq c$ (where $O$ is the coordinate origin), then the range of values for $y_{0}^{2}$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;c:Number;M: Point;O: Origin;x0:Number;y0:Number;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(M) = (x0, y0);PointOnCurve(M, C);HalfFocalLength(C) = c;Abs(LineSegmentOf(M, O)) <= c", "query_expressions": "Range(y0^2)", "answer_expressions": "[0,b^4/c^2]", "fact_spans": "[[[20, 71]], [[28, 71]], [[28, 71]], [[80, 83]], [[2, 19]], [[102, 105]], [[2, 19]], [[2, 19]], [[20, 71]], [[2, 19]], [[2, 75]], [[20, 83]], [[85, 99]]]", "query_spans": "[[[113, 131]]]", "process": "Because |MO| ≤ c, so |MO| ≤ √(a² + b²), thus x₀² + y₀² ≤ a² + b². Also, (x₀² / a²) - (y₀² / b²) = 1, we obtain x₀² = a² + (a² y₀²) / b². Therefore, x₀² + y₀² = a² + (a² y₀²) / b² + y₀² = a² + (c² y₀²) / b² ≤ a² + b², so 0 ≤ y₀² ≤ b⁴ / c²." }, { "text": "Through the right vertex $A$ of the hyperbola $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a line with slope $-1$. The points of intersection of this line with the two asymptotes of the hyperbola are $B$ and $C$, respectively. If $\\overrightarrow{A B}=\\frac{1}{2} \\overrightarrow{B C}$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;A: Point;B: Point;C: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightVertex(G)=A;Slope(H)=-1;PointOnCurve(A,H);L1:Line;L2:Line;Asymptote(G)={L1,L2};Intersection(H,L1)=B;Intersection(H,L2)=C;VectorOf(A, B) = VectorOf(B, C)/2", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[1, 60], [82, 85], [165, 168]], [[4, 60]], [[4, 60]], [[75, 77], [79, 81]], [[64, 67]], [[97, 100]], [[101, 104]], [[4, 60]], [[4, 60]], [[1, 60]], [[1, 67]], [[68, 77]], [[0, 77]], [], [], [[82, 91]], [[79, 104]], [[79, 104]], [[107, 162]]]", "query_spans": "[[[165, 174]]]", "process": "" }, { "text": "The coordinates of the two foci of an ellipse are $(5,0)$ and $(-5, 0)$. The sum of the distances from a point $P$ on the ellipse to the two foci is $26$. Find the standard equation of the ellipse.", "fact_expressions": "G: Ellipse;F1: Point;F2: Point;P: Point;Coordinate(F1) = (5, 0);Coordinate(F2) = (-5, 0);Focus(G)={F1,F2};PointOnCurve(P, G);Distance(P, F1)+Distance(P,F2) = 26", "query_expressions": "Expression(G)", "answer_expressions": "x^2/169+y^2/144=1", "fact_spans": "[[[2, 4], [33, 35], [58, 60]], [[13, 20]], [[22, 32]], [[38, 41]], [[13, 20]], [[22, 32]], [[2, 32]], [[33, 41]], [[33, 56]]]", "query_spans": "[[[58, 67]]]", "process": "Test analysis: By the definition of an ellipse, we have 2a = 26, so a = 13. Since c = 5, it follows that b = 12. The equation of the ellipse is \\frac{x^{2}}{169}+\\frac{y^{2}}{144}=1" }, { "text": "Given that the eccentricity of hyperbola $C$ is $\\sqrt{5}$, write a standard equation of hyperbola $C$?", "fact_expressions": "C: Hyperbola;Eccentricity(C) = sqrt(5)", "query_expressions": "Expression(C)", "answer_expressions": "x^2 - y^2/4 = 1", "fact_spans": "[[[2, 8], [26, 32]], [[2, 23]]]", "query_spans": "[[[24, 40]]]", "process": "The eccentricity of the hyperbola is given as $\\sqrt{5}$, that is, $c=\\sqrt{5}a$. Let $a=1$, then $c=\\sqrt{5}$, and $b=\\sqrt{c^{2}-a^{2}}=2$. Thus, one possible answer is $x^{2}-\\frac{y^{2}}{4}=1$ (the answer is not unique)." }, { "text": "The vertex of parabola $C$ is at the coordinate origin, and the focus lies on the $x$-axis. Moreover, $C$ passes through the point $(-2,3)$. Then the equation of $C$ is?", "fact_expressions": "C: Parabola;G: Point;O: Origin;Coordinate(G) = (-2, 3);Vertex(C) = O;PointOnCurve(Focus(C), xAxis);PointOnCurve(G, C)", "query_expressions": "Expression(C)", "answer_expressions": "y^2=(-9/2)*x", "fact_spans": "[[[0, 6], [25, 28], [40, 43]], [[29, 38]], [[10, 14]], [[29, 38]], [[0, 14]], [[0, 23]], [[25, 38]]]", "query_spans": "[[[40, 48]]]", "process": "Given: The vertex of parabola C is at the coordinate origin, and the focus lies on the x-axis. It passes through the point (-2, 3). Let the equation of the parabola be y^{2} = -2px. Since it passes through the point (-2, 3), we have 3^{2} = -2p \\times (-2). Solving gives: p = \\frac{9}{4}. Therefore, the equation of the parabola is: y^{2} = -\\frac{9}{2}x" }, { "text": "Let $F$ be the focus of the parabola $y=-\\frac{1}{4} x^{2}$, and let the line $l$, tangent to the parabola at point $P(-4, -4)$, intersect the $x$-axis at point $Q$. Then the value of $\\angle P Q F$ is?", "fact_expressions": "l: Line;G: Parabola;P: Point;Q: Point;F: Point;Expression(G) = (y = -x^2/4);Coordinate(P) = (-4, -4);Focus(G) = F;TangentPoint(l, G) = P;Intersection(l, xAxis) = Q", "query_expressions": "AngleOf(P, Q, F)", "answer_expressions": "pi/2", "fact_spans": "[[[54, 59]], [[5, 30], [35, 38]], [[41, 53]], [[68, 71]], [[1, 4]], [[5, 30]], [[41, 53]], [[1, 33]], [[34, 59]], [[54, 71]]]", "query_spans": "[[[73, 91]]]", "process": "" }, { "text": "If the equation of hyperbola $C$ is $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, denote the left and right vertices of hyperbola $C$ as $A$ and $B$. Chord $P Q \\perp x$-axis, denote the intersection point of line $P A$ and line $Q B$ as $M$, whose trajectory is curve $T$, then the eccentricity of curve $T$ is?", "fact_expressions": "C: Hyperbola;A: Point;P: Point;Q: Point;B: Point;T: Curve;M: Point;Expression(C) = (x^2/4 - y^2/5 = 1);LeftVertex(C) = A;RightVertex(C) = B;IsChordOf(LineSegmentOf(P, Q), C);IsPerpendicular(LineSegmentOf(P, Q), xAxis);Intersection(LineOf(P, A), LineOf(Q, B)) = M;Locus(M)=T", "query_expressions": "Eccentricity(T)", "answer_expressions": "sqrt(5)/5", "fact_spans": "[[[1, 7], [48, 54]], [[61, 64]], [[89, 94]], [[97, 102]], [[65, 68]], [[113, 118], [120, 125]], [[105, 108], [109, 110]], [[1, 46]], [[48, 68]], [[48, 68]], [[48, 85]], [[71, 85]], [[87, 108]], [[109, 118]]]", "query_spans": "[[[120, 131]]]", "process": "Let P(x_{0},y_{0}), M(x,y). According to the point-slope form of the equation of a line, express the equations of lines PA and QB. By simplifying the equations of the two lines, we obtain \\frac{y^{2}}{x^{2}-4}=-\\frac{y_{0}^{2}}{x_{0}^{2}-4}. Combining with the fact that point P(x_{0},y_{0}) lies on the hyperbola \\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1, we get \\frac{y_{0}^{2}}{x_{0}^{2}-4}=\\frac{5}{4}, thereby deriving the equation of curve T, and thus the eccentricity can be found.\n\nSolution: Let P(x_{0},y_{0}), then Q(x_{0},-y_{0}). Let point M(x,y), and A(-2,0), B(2,0). Therefore, the equation of line PA is y=\\frac{y_{0}}{x_{0}+2}(x+2)\\textcircled{1}, and the equation of line QB is y=\\frac{-y_{0}}{x_{0}-2}(x-2)\\textcircled{2}. From \\textcircled{1}, we have \\frac{y}{x+2}=\\frac{y_{0}}{x_{0}+2}; from \\textcircled{2}, we have \\frac{y}{x-2}=-\\frac{y_{0}}{x_{0}-2}. Multiplying these two equations yields \\frac{y^{2}}{x^{2}-4}=-\\frac{y_{0}^{2}}{x_{0}^{2}-4}. Since P(x_{0},y_{0}) lies on the hyperbola \\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1, \\therefore \\frac{x_{0}^{2}}{4}-\\frac{y_{0}^{2}}{5}=1, which gives \\frac{y_{0}^{2}}{5}=\\frac{x_{0}^{2}-4}{4}, \\therefore \\frac{y_{0}^{2}}{x_{0}^{2}-4}=\\frac{5}{4}. Thus, the equation of curve T is \\frac{x^{2}}{4}+\\frac{y^{2}}{5}=1, and its eccentricity is \\frac{\\sqrt{5}}{5}." }, { "text": "The vertex of the parabola $C$ is at the origin, and its focus $F$ coincides with the right focus of the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{6}=1$. The line $l$ passing through point $P(2,0)$ with slope $1$ intersects the parabola $C$ at points $A$ and $B$. Then, the distance from the midpoint of chord $AB$ to the directrix of the parabola is?", "fact_expressions": "l: Line;G: Hyperbola;C: Parabola;A: Point;B: Point;P: Point;F:Point;O: Origin;Expression(G) = (x^2/3 - y^2/6 = 1);Coordinate(P) = (2, 0);Vertex(C) = O;Focus(C) =F;RightFocus(G)=F;PointOnCurve(P, l);Slope(l) = 1;Intersection(l, C) = {A, B};IsChordOf(LineSegmentOf(A,B),C)", "query_expressions": "Distance(MidPoint(LineSegmentOf(A,B)),Directrix(C))", "answer_expressions": "11", "fact_spans": "[[[82, 87]], [[19, 57]], [[0, 6], [88, 94], [117, 120]], [[96, 99]], [[100, 103]], [[65, 74]], [[15, 18]], [[10, 12]], [[19, 57]], [[65, 74]], [[0, 12]], [[0, 18]], [[15, 63]], [[64, 87]], [[75, 87]], [[82, 105]], [[88, 113]]]", "query_spans": "[[[108, 127]]]", "process": "Problem Analysis: Since the coordinates of the right focus of the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{6}=1$ are $(3,0)$, we have $\\frac{p}{2}=3$, therefore $p=6$. Thus, the standard equation of the parabola is $y^{2}=12x$. Let the line $l$ passing through the point $P(2,0)$ with slope $1$ be given by $y=x-2$, and let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$. By solving simultaneously with $y^{2}=12x$ and eliminating $y$, we obtain $x^{2}-16x+4=0$, so $x_{1}+x_{2}=16$. Moreover, the distance from the midpoint of segment $AB$ to the directrix of the parabola is $\\frac{x_{1}+x_{2}+p}{2}=\\frac{16+6}{2}=11$. Hence, fill in $11$." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{1}$ intersects $C$ at points $A$ and $B$, satisfying $A F_{1} \\perp A F_{2}$ and $|A F_{2}|=2|A F_{1}|$. Then $\\tan \\angle B F_{2} F_{1}$=?", "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;G: Line;PointOnCurve(F1, G);A: Point;B: Point;Intersection(G, C) = {A, B};IsPerpendicular(LineSegmentOf(A, F1), LineSegmentOf(A, F2));Abs(LineSegmentOf(A, F2)) = 2*Abs(LineSegmentOf(A, F1))", "query_expressions": "Tan(AngleOf(B, F2, F1))", "answer_expressions": "2/11", "fact_spans": "[[[2, 59], [96, 99]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[68, 75], [85, 92]], [[76, 83]], [[2, 83]], [[2, 83]], [[93, 95]], [[84, 95]], [[101, 104]], [[105, 108]], [[93, 110]], [[113, 136]], [[137, 159]]]", "query_spans": "[[[161, 190]]]", "process": "Since |AF₂| = 2|AF₁|, and |AF₁| + |AF₂| = 2a, it follows that |AF₁| = (2a)/3, |AF₂| = (4a)/3. Let |BF₁| = m > 0. According to the definition of an ellipse, we have |BF₁| + |BF₂| = 2a, so |BF₂| = 2a − m. Since AF₁ ⊥ AF₂, we have ∠BAF₂ = π/2, so |AB|² + |AF₂|² = |BF₂|², that is, ((2a)/3 + m)² + ((4a)/3)² = (2a − m)². Solving this equation gives a = 3m. Therefore, |BF₁| = a/3. Then tan∠AF₁F₂ = |AF₂| / |AF₁| = 2, tan∠ABF₂ = |AF₂| / |AB| = (4a/3) / (2a/2 + a/2) = 4/3, so tan∠BF₂F₁ = tan(∠AF₁F₂ − ∠ABF₂) = (tan∠AF₁F₂ − tan∠ABF₂) / (1 + tan∠AF₁F₂ tan∠ABF₂) = (2 − 4/3) / (1 + 2 × 4/11)" }, { "text": "Given the circle $x^{2}+y^{2}=1$ intersects the parabola $y^{2}=2 p x(p>0)$ at points $A$ and $B$, and intersects the directrix of the parabola at points $C$ and $D$. The coordinate origin $O$ is the midpoint of $A C$. Then the value of $p$ equals?", "fact_expressions": "G: Parabola;p: Number;H: Circle;A: Point;C: Point;D:Point;B: Point;O: Origin;p>0;Expression(G) = (y^2 = 2*p*x);Expression(H) = (x^2 + y^2 = 1);Intersection(H, G) = {A, B};Intersection(H,Directrix(G))={C,D};MidPoint(LineSegmentOf(A, C)) = O", "query_expressions": "p", "answer_expressions": "2*sqrt(5)/5", "fact_spans": "[[[19, 40], [53, 56]], [[90, 93]], [[2, 18]], [[42, 45]], [[61, 64]], [[65, 68]], [[46, 49]], [[72, 79]], [[22, 40]], [[19, 40]], [[2, 18]], [[2, 51]], [[2, 70]], [[72, 88]]]", "query_spans": "[[[90, 98]]]", "process": "Since the equation of the directrix of the parabola is $ x = -\\frac{p}{2} $, by symmetry, substituting point $ A\\left(\\frac{p}{2}, \\pm p\\right) $ into the equation of the circle gives $ \\left(\\frac{p}{2}\\right)^2 + (\\pm p)^2 = 1 $. Solving this equation yields $ p = \\frac{2\\sqrt{5}}{5} $." }, { "text": "Given the line $ l $: $ 4x - 3y + 6 = 0 $, and a moving point $ P(x_0, y_0) $ on the parabola $ y^2 = 8x $, the distance from $ P $ to the line $ l $ is $ d $. Then the minimum value of $ d + |x_0| $ is?", "fact_expressions": "l: Line;Expression(l) = (4*x - 3*y + 6 = 0);G: Parabola;Expression(G) = (y^2 = 8*x);P: Point;Coordinate(P) = (x0, y0);x0: Number;y0: Number;PointOnCurve(P, G);Distance(P, l) = d;d: Number", "query_expressions": "Min(d + Abs(x0))", "answer_expressions": "4/5", "fact_spans": "[[[2, 22], [59, 64]], [[2, 22]], [[23, 37]], [[23, 37]], [[41, 58]], [[41, 58]], [[41, 58]], [[41, 58]], [[23, 58]], [[41, 71]], [[68, 71]]]", "query_spans": "[[[73, 90]]]", "process": "As shown in the figure: If PC\\bot line l, PB\\bot the directrix of the parabola and intersects the y-axis at point A, then d=|PC|, |x_{0}|=|PA|. By the definition of the parabola, we know: |PF|=|PB|=|PA|+\\frac{p}{2}, so |PA|=|PF|-\\frac{p}{2}=|PF|-2. Therefore, d+|x_{0}|=|PC|+|PI|-2. To minimize the objective expression, we need to minimize |PC|+|PF|. When F, P, C are collinear, and given F(2,0), then (d+|x_{0}|)_{\\min}=\\frac{|8-0+6|}{5}-2=\\frac{4}{5}." }, { "text": "If the focus of the parabola $y^{2}=m x$ coincides with the right focus of the hyperbola $\\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1$, then what is the value of the real number $m$?", "fact_expressions": "G: Hyperbola;H: Parabola;m: Real;Expression(G) = (x^2/6 - y^2/3 = 1);Expression(H) = (y^2 = m*x);Focus(H) = RightFocus(G)", "query_expressions": "m", "answer_expressions": "12", "fact_spans": "[[[19, 57]], [[1, 15]], [[65, 70]], [[19, 57]], [[1, 15]], [[1, 63]]]", "query_spans": "[[[65, 74]]]", "process": "Hyperbola \\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1, c=\\sqrt{6+3}=3, right focus (3,0), the focus of the parabola y^{2}=mx is (3,0), so \\frac{m}{4}=3, m=12" }, { "text": "The left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ are $F_{1}$ and $F_{2}$, respectively. A line passing through $F_{2}$ with an inclination angle of $120^{\\circ}$ intersects the ellipse at a point $M$. If $MF_{1}$ is perpendicular to the $x$-axis, then what is the eccentricity of the ellipse?", "fact_expressions": "G: Ellipse;a: Number;b: Number;H: Line;M: Point;F1: Point;F2:Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(F2,H);Inclination(H)=ApplyUnit(120,degree);OneOf(Intersection(H,G))=M;IsPerpendicular(LineSegmentOf(M,F1),xAxis)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2-sqrt(3)", "fact_spans": "[[[0, 52], [110, 112], [140, 142]], [[2, 52]], [[2, 52]], [[107, 109]], [[118, 121]], [[61, 68]], [[71, 78], [81, 88]], [[2, 52]], [[2, 52]], [[0, 52]], [[0, 78]], [[0, 78]], [[80, 109]], [[89, 109]], [[107, 121]], [[123, 138]]]", "query_spans": "[[[140, 148]]]", "process": "" }, { "text": "The chord $AB$ of the parabola $y^{2}=4 x$ is perpendicular to the $x$-axis. If $|AB|=4 \\sqrt{3}$, then the distance from the focus to $AB$ is?", "fact_expressions": "G: Parabola;A: Point;B: Point;Expression(G) = (y^2 = 4*x);IsChordOf(LineSegmentOf(A,B),G);IsPerpendicular(LineSegmentOf(A,B),xAxis);Abs(LineSegmentOf(A, B)) = 4*sqrt(3)", "query_expressions": "Distance(Focus(G), LineSegmentOf(A, B))", "answer_expressions": "2", "fact_spans": "[[[0, 14]], [[16, 20]], [[16, 20]], [[0, 14]], [[0, 20]], [[16, 26]], [[28, 45]]]", "query_spans": "[[[0, 59]]]", "process": "" }, { "text": "Given that $F$ is the right focus of the hyperbola $C$: $x^{2}-y^{2}=2$, $P$ is a point on the left branch of $C$, and $A(0 , 2)$. When the perimeter of $\\triangle APF$ is minimized, what is the area of this triangle?", "fact_expressions": "C: Hyperbola;A: Point;P: Point;F: Point;Expression(C) = (x^2 - y^2 = 2);Coordinate(A) = (0, 2);RightFocus(C) = F;PointOnCurve(P, LeftPart(C));WhenMin(Perimeter(TriangleOf(A, P, F)))", "query_expressions": "Area(TriangleOf(A, P, F))", "answer_expressions": "3", "fact_spans": "[[[6, 29], [38, 41]], [[48, 58]], [[34, 37]], [[2, 5]], [[6, 29]], [[48, 58]], [[2, 33]], [[34, 47]], [[60, 81]]]", "query_spans": "[[[83, 91]]]", "process": "Problem Analysis: Using the definition of a hyperbola, determine the coordinates of point P when the perimeter of triangle APF is minimized, then find the area of this triangle when its perimeter is minimal. \nSolution: Let the left focus be F₁(-2, 0), and the right focus be F(2, 0). The perimeter of triangle APF is |AF| + |AP| + |PF| = |AF| + |AP| + (|PF₁| + 2a) = |AF| + |AP| + |PF₁| + 2a ≥ |AF| + |AF₁| + 2a. The equality holds if and only if points A, P, and F₁ are collinear, that is, when P is at P₀, the perimeter of the triangle is minimized. At this time, the equation of line AF₁ is y = x + 2. Substituting into x²·y² = 2, we can solve to obtain P₀(-3/2, 1/2). Hence, S_{△AP₀F} = S_{△AF₁F} - S_{△P₀F₁F^{-1/2}}×4×2 - 1/2×4×1/2 = 3." }, { "text": "The line $ l $ with slope $ \\frac{4}{3} $ passes through the focus $ F(1,0) $ of the parabola $ y^{2}=2 p x $ ($ p>0 $) and intersects the parabola at points $ A $ and $ B $. What is the length of segment $ AB $?", "fact_expressions": "l: Line;G: Parabola;p: Number;B: Point;A: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(F) = (1, 0);Slope(l)=4/3;Focus(G) = F;PointOnCurve(F,l);Intersection(l, G) = {A, B}", "query_expressions": "Length(LineSegmentOf(A,B))", "answer_expressions": "25/4", "fact_spans": "[[[17, 22]], [[24, 45], [58, 61]], [[27, 45]], [[67, 70]], [[63, 66]], [[48, 56]], [[27, 45]], [[24, 45]], [[48, 56]], [[0, 22]], [[24, 56]], [[17, 56]], [[17, 72]]]", "query_spans": "[[[74, 85]]]", "process": "First, derive the standard equation of the parabola from the coordinates of its focus. Let points A(x_{1},y_{1}) and B(x_{2},y_{2}). Combine the equation of line l with the equation of the parabola, and use Vieta's formulas together with the chord length formula for a parabola passing through its focus to calculate the length of segment AB. Since the parabola y^{2}=2px (p>0) has focus F(1,0), then \\frac{p}{2}=1 \\Rightarrow p=2. Therefore, the equation of the parabola is y^{2}=4x. Let points A(x_{1},y_{1}) and B(x_{2},y_{2}). The equation of line l is y=\\frac{4}{3}(x-1). Combining \\begin{cases} y=\\frac{4}{3}(x-1) \\\\ y^{2}=4x \\end{cases}, eliminating y yields 4x^{2}-17x+4=0. Therefore, x_{1}+x_{2}=\\frac{17}{4}, |AB|=x_{1}+x_{2}+2=2+\\frac{17}{4}=\\frac{25}{4}." }, { "text": "Given that the focus $F$ of the parabola $y^{2}=2 p x(p>0)$ is exactly the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, and the hyperbola passes through the point $(\\frac{3 a^{2}}{p} , \\frac{2 b^{2}}{p})$, then the asymptotes equation of this hyperbola is?", "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(H) = F;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;RightFocus(G) = F;I: Point;Coordinate(I) = ((3*a^2)/p, (2*b^2)/p);PointOnCurve(I, G)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*(sqrt(10)/4)*x", "fact_spans": "[[[2, 23]], [[2, 23]], [[5, 23]], [[5, 23]], [[25, 28]], [[2, 28]], [[31, 77], [83, 86], [132, 135]], [[31, 77]], [[34, 77]], [[34, 77]], [[25, 81]], [[87, 129]], [[87, 129]], [[83, 129]]]", "query_spans": "[[[132, 143]]]", "process": "" }, { "text": "Given that the equation $(m+1) x^{2}+m y^{2}=1$ represents a hyperbola with foci on the $x$-axis, find the range of real values for $m$.", "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (m*y^2 + x^2*(m + 1) = 1);PointOnCurve(Focus(G), xAxis)", "query_expressions": "Range(m)", "answer_expressions": "(-1,0)", "fact_spans": "[[[38, 41]], [[43, 48]], [[2, 41]], [[29, 41]]]", "query_spans": "[[[43, 55]]]", "process": "" }, { "text": "Given point $P(0 , 1)$, two points $A$, $B$ on the ellipse $\\frac{x^{2}}{4}+y^{2}=m(m>1)$ satisfy $\\overrightarrow{A P}=2\\overrightarrow{P B}$. Then when $m=$?, the absolute value of the horizontal coordinate of point $B$ is maximized", "fact_expressions": "G: Ellipse;m: Number;P: Point;A: Point;B: Point;m>1;Expression(G) = (x^2/4 + y^2 = m);Coordinate(P) = (0, 1);PointOnCurve(A,G);PointOnCurve(B,G);VectorOf(A,P)=2*VectorOf(P,B);WhenMax(Abs(XCoordinate(B)))", "query_expressions": "m", "answer_expressions": "5", "fact_spans": "[[[14, 46]], [[105, 108]], [[2, 13]], [[49, 52]], [[53, 56], [112, 116]], [[16, 46]], [[14, 46]], [[2, 13]], [[14, 56]], [[14, 56]], [[58, 102]], [[112, 125]]]", "query_spans": "[[[105, 110]]]", "process": "Analysis: First, obtain the relationship between the coordinates of A and B based on the given conditions, substitute into the ellipse equation to solve for the y-coordinate of point B, then derive the functional relationship of the x-coordinate of B with respect to m; finally, determine the method for obtaining the maximum value according to the properties of quadratic functions. \nDetailed solution: Let A(x_{1},y_{1}), B(x_{2},y_{2}). From \\overrightarrow{AP}=2\\overrightarrow{PB}, we get -x_{1}=2x_{2}, 1-y_{1}=2(y_{2}-1), so y_{1}=2y_{2}-3. Since A and B lie on the ellipse, \\frac{x_{1}^{2}}{4}+y_{1}^{2}=m, \\frac{x_{2}^{2}}{4}+y_{2}^{2}=m, (2y_{2}-3)^{2}=m, therefore \\frac{x_{2}^{2}}{4}+(y_{2}-\\frac{3}{2})^{2}=\\frac{m}{4}=m. Subtracting these equations gives y_{2}=\\frac{3+m}{4}, x_{2}^{2}=-\\frac{1}{4}(m^{2}-10m+9)\\leqslant4, and the maximum value is attained if and only if m=5." }, { "text": "The parabola $y=a x^{2}$ passes through the point $M(2,1)$, then the distance from $M$ to the focus $F$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y = a*x^2);a: Number;M: Point;Coordinate(M) = (2, 1);PointOnCurve(M, G);F: Point;Focus(G) = F", "query_expressions": "Distance(M, F)", "answer_expressions": "2", "fact_spans": "[[[0, 14]], [[0, 14]], [[3, 14]], [[16, 25], [27, 30]], [[16, 25]], [[0, 25]], [[33, 36]], [[0, 36]]]", "query_spans": "[[[27, 41]]]", "process": "The parabola y=ax^{2} passes through point M(2,1), hence 1=4a, solving gives a=\\frac{1}{4}. Since the parabola is y=\\frac{x^{2}}{4}, the standard equation is: x^{2}=4y. The equation of the directrix is: y=-1. The distance from point M(2,1) to the focus F is equal to the distance from M to the directrix: 1+1=2" }, { "text": "The curve $C$ is the locus of a moving point $P$ in the plane such that the sum of the distance from $P$ to the fixed point $A(1,0)$ and the distance from $P$ to the fixed line $x=-1$ is equal to $3$. Then, what are the coordinates of the points where the curve $C$ intersects the $y$-axis?", "fact_expressions": "C: Curve;A: Point;Coordinate(A) = (1, 0);G: Line;Expression(G) = (x = -1);P: Point;Distance(P, A) + Distance(P, G) = 3;C = Locus(P)", "query_expressions": "Coordinate(Intersection(C, yAxis))", "answer_expressions": "(0, pm*sqrt(3))", "fact_spans": "[[[0, 5], [51, 56]], [[12, 20]], [[12, 20]], [[28, 34]], [[28, 34]], [[43, 46]], [[9, 46]], [[0, 49]]]", "query_spans": "[[[51, 69]]]", "process": "Let P(x,y), from the given condition we have: \\sqrt{(x-1)^{2}+y^{2}}+|x+1|=3. When x<-4 or x>2, 3-|x+1|<0, so there is no trajectory; when -4\\leqslant x\\leqslant-1, \\sqrt{(x-1)^{2}+y^{2}}+|x+1|=3 \\Rightarrow y^{2}=10x+15, which has no intersection with the y-axis; when -1|AB|$. Therefore, the trajectory of the center M of the moving circle satisfies an ellipse with foci at A and B. Here, $a=5$, $c=3$, so $b^{2}=a^{2}-c^{2}=16$. Hence, the trajectory equation of the center M of the moving circle is: $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$." }, { "text": "The distance from the point $M(4, y_{0})$ on the parabola $y^{2}=4 x$ to its focus $F$ is?", "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = 4*x);Coordinate(M) = (4, y0);y0:Number;PointOnCurve(M, G);Focus(G)=F;F: Point", "query_expressions": "Distance(M,F)", "answer_expressions": "5", "fact_spans": "[[[0, 14], [31, 32]], [[16, 30]], [[0, 14]], [[16, 30]], [[17, 30]], [[0, 30]], [[31, 37]], [[34, 37]]]", "query_spans": "[[[16, 42]]]", "process": "" }, { "text": "It is known that point $P$ is a moving point on the parabola $y^{2}=4x$, $A(2,2)$ is a fixed point in the plane, and $F$ is the focus of the parabola. When the coordinates of point $P$ are ?, $PA + PF$ is minimized.", "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (2, 2);PointOnCurve(P, G);Focus(G) = F;WhenMin(LineSegmentOf(P,A)+LineSegmentOf(P,F))", "query_expressions": "Coordinate(P)", "answer_expressions": "(1,2)", "fact_spans": "[[[7, 21], [48, 51]], [[27, 35]], [[2, 6], [56, 60]], [[44, 47]], [[7, 21]], [[27, 35]], [[2, 26]], [[44, 54]], [[65, 76]]]", "query_spans": "[[[56, 64]]]", "process": "From point P, draw a perpendicular to the directrix of the parabola, and let the foot of the perpendicular be M. Then |PM| = |PF|, and |PA| + |PF| = |PA| + |PM|. When points P, A, and M are collinear, |PA| + |PM| is minimized, which means |PA| + |PF| is minimized. At this time, the y-coordinate of P is y = 2; substituting into the parabolic equation y^{2} = 4x gives the x-coordinate x = 1. Therefore, P(1,2)." }, { "text": "Given that the hyperbola passes through point $A(3,-2)$, and has the same foci as the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, then the equation of this hyperbola is?", "fact_expressions": "G: Hyperbola;H: Ellipse;A: Point;Expression(H) = (x^2/9 + y^2/4 = 1);Coordinate(A) = (3,-2);PointOnCurve(A, G);Focus(G)=Focus(H)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/3-y^2/2=1", "fact_spans": "[[[2, 5], [65, 68]], [[19, 56]], [[6, 16]], [[19, 56]], [[6, 16]], [[2, 16]], [[2, 62]]]", "query_spans": "[[[65, 73]]]", "process": "It is easy to find that the foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ are $(\\pm\\sqrt{5},0)$; thus, suppose the equation of the hyperbola is $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a,b>0)$. Hence, \n$$\n\\begin{cases}\n\\frac{3^{2}}{a^{2}}-\\frac{(-2)^{2}}{b^{2}}=1 \\\\\na^{2}+12-\\leqslant\n\\end{cases}\n$$\nSolving gives $a^{2}=3$, $b^{2}=2$. Therefore, the equation of the hyperbola is $\\frac{x^{2}}{3}-\\frac{y^{2}}{2}=1$." }, { "text": "Given that one focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1$ ($a>0$) coincides with the focus of the parabola $y^{2}=8x$, then $a=?$", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/3 + x^2/a^2 = 1);a: Number;a>0;H: Parabola;Expression(H) = (y^2 = 8*x);OneOf(Focus(G)) = Focus(H)", "query_expressions": "a", "answer_expressions": "1", "fact_spans": "[[[2, 49]], [[2, 49]], [[76, 79]], [[5, 49]], [[55, 69]], [[55, 69]], [[2, 74]]]", "query_spans": "[[[76, 81]]]", "process": "Given the parabola y^{2}=8x, the coordinates of the focus are (2,0). For the hyperbola, c=2. Combining a^{2}+b^{2}=c^{2} and substituting, we obtain a^{2}+3=4, yielding a=1." }, { "text": "Given that $F$ is the left focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, point $A(1,6)$, and $P$ is a moving point on the right branch of the hyperbola, then the minimum value of $|P F|+|P A|$ is?", "fact_expressions": "G: Hyperbola;A: Point;P: Point;F: Point;Expression(G) = (x^2/4 - y^2/12 = 1);Coordinate(A) = (1, 6);LeftFocus(G) = F;PointOnCurve(P, RightPart(G))", "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))", "answer_expressions": "4 + 3*sqrt(5)", "fact_spans": "[[[6, 45], [67, 70]], [[50, 60]], [[62, 65]], [[2, 5]], [[6, 45]], [[50, 59]], [[2, 49]], [[62, 78]]]", "query_spans": "[[[80, 99]]]", "process": "From the given conditions, we have: a=2, b=2\\sqrt{3}, c=4. Denote the right focus of the hyperbola as F, and draw a schematic diagram as follows. Therefore, by the definition of a hyperbola, we obtain: |PF| - |PF| = 2a = 4-\\textcircled{1}. Also, \\because in triangle PAF, \\therefore |PA| + |PF| \\geqslant |AF| = \\sqrt{(4-1)^{2}+(0-6)^{2}} = 3\\sqrt{5}-\\textcircled{2}. Adding equation \\textcircled{1} and inequality \\textcircled{2}, we get: |PF| + |PA| \\geqslant 4 + 3\\sqrt{5}. The equality holds if and only if points A, P, and F are collinear. Therefore, the minimum value of |PF| + |PA| is 4 + 3\\sqrt{5}." }, { "text": "The asymptote equations of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{16}=1$ are?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/16 = 1)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*2*x", "fact_spans": "[[[0, 39]], [[0, 39]]]", "query_spans": "[[[0, 47]]]", "process": "Hyperbola \\frac{x^{2}}{4}-\\frac{y^{2}}{16}=1, we have a=2, b=4, then the asymptotes of this hyperbola are: y=\\pm2x." }, { "text": "Given that point $P$ is on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola respectively, $I$ is the incenter of $\\Delta P F_{1} F_{2}$, and if $S_{\\Delta I P F_{1}}=\\frac{1}{2} S_{\\Delta I F_{1} F_{2}}+S_{\\Delta I P F_{2}}$ holds, then what is the eccentricity of the hyperbola?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;P: Point;PointOnCurve(P, RightPart(G));F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;I: Point;Incenter(TriangleOf(P, F1, F2)) = I;TriangleOf(I, P, F1) = (1/2)*TriangleOf(I, F1, F2) + TriangleOf(I, P, F2)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[7, 66], [90, 93], [215, 218]], [[7, 66]], [[10, 66]], [[10, 66]], [[10, 66]], [[10, 66]], [[2, 6]], [[2, 71]], [[72, 79]], [[80, 87]], [[72, 99]], [[72, 99]], [[100, 103]], [[100, 129]], [[131, 211]]]", "query_spans": "[[[215, 224]]]", "process": "Using the inradius, $S_{\\triangle IPF_{1}} = \\frac{1}{2}S_{AIF_{1}F_{2}} + S_{\\triangle APF_{2}}$ is transformed into $|PF_{1}| = \\frac{1}{2}|F_{1}F_{2}| + |PF_{2}| = c + |PF_{2}|$. According to the definition of the hyperbola, $2a = c$, so the eccentricity of the hyperbola is $e = \\frac{c}{a} = 2$." }, { "text": "The distance from point $M$ to point $F(0 , 4)$ is less by $1$ than its distance to the line $L$: $y+5=0$. Then, the equation of the trajectory of $M$ is?", "fact_expressions": "L: Line;F: Point;M: Point;Expression(L) = (y + 5 = 0);Coordinate(F) = (0, 4);Distance(M, F) = Distance(M, L) - 1", "query_expressions": "LocusEquation(M)", "answer_expressions": "x^2=16*y", "fact_spans": "[[[22, 35]], [[5, 16]], [[0, 4], [20, 21], [44, 47]], [[22, 35]], [[5, 16]], [[0, 42]]]", "query_spans": "[[[44, 54]]]", "process": "" }, { "text": "The hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{n}=1(m n \\neq 0)$ has an eccentricity of $2$, and one of its foci coincides with the focus of the parabola $y^{2}=4 x$. What is the value of $m \\cdot n$?", "fact_expressions": "G: Hyperbola;m: Number;n: Number;H: Parabola;Negation(m*n=0);Expression(G) = (-y^2/n + x^2/m = 1);Expression(H) = (y^2 = 4*x);Eccentricity(G) = 2;OneOf(Focus(G)) = Focus(H)", "query_expressions": "m*n", "answer_expressions": "3/16", "fact_spans": "[[[0, 50]], [[86, 97]], [[86, 97]], [[65, 79]], [[3, 50]], [[0, 50]], [[65, 79]], [[0, 58]], [[0, 84]]]", "query_spans": "[[[86, 101]]]", "process": "From the given problem: the focus of the parabola $ y^{2} = 4x $ has coordinates $ (1, 0) $, so for the hyperbola, $ c = 1 $. The equation $ \\frac{x^{2}}{m} - \\frac{y^{2}}{n} = 1 $ ($ mn \\neq 0 $) represents a hyperbola, so $ m $ and $ n $ have the same sign. When $ m $ and $ n $ are both positive, $ a^{2} = m $, $ b^{2} = n $, then $ e = \\frac{c}{a} = \\frac{1}{\\sqrt{m}} = 2 $, solving gives: $ m = \\frac{1}{4} $, then $ n = b^{2} = c^{2} - a^{2} = 1 - m = \\frac{3}{4} $, at this time $ m \\cdot n = \\frac{1}{4} \\times \\frac{3}{4} = \\frac{3}{16} $. Then $ -m = b^{2} = c^{2} - a^{2} = 1 + n = \\frac{3}{4} $, at this time $ m \\cdot n = (-\\frac{1}{4}) \\times (-\\frac{3}{4}) = \\frac{3}{16} $. In summary: $ m \\cdot n = \\frac{3}{16} $. This problem mainly examines the simple properties of a parabola, as well as the simple properties of a hyperbola and the idea of classification, tests the identification of $ a, b, c $ in the standard equation of a hyperbola, and examines computational ability; it is a medium-difficulty problem." }, { "text": "The distance from a point $P$ on the hyperbola $4x^{2}-y^{2}+64=0$ to one of its foci is equal to $1$. What is the distance from point $P$ to the other focus?", "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (4*x^2 - y^2 + 64 = 0);PointOnCurve(P, G);F1:Point;F2:Point;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2);Distance(P, F1) = 1", "query_expressions": "Distance(P, F2)", "answer_expressions": "17", "fact_spans": "[[[0, 23], [30, 31]], [[26, 29], [47, 51]], [[0, 23]], [[0, 29]], [], [], [[30, 36]], [[30, 57]], [[30, 57]], [[26, 44]]]", "query_spans": "[[[30, 62]]]", "process": "Convert the hyperbola $4x^{2}\\cdot y^{2}+64=0$ into standard form: $\\frac{y^{2}}{64}-\\frac{x}{16}=1 \\therefore a^{2}=64, b^{2}=16$. The distance from point $P$ to one of its foci is equal to 1. Let $PF_{1}=1$. Since $|PF_{1}\\cdot PF_{2}|=2a=16$, therefore $PF_{2}=PF_{1}\\pm16=17$ (negative value discarded)." }, { "text": "The left focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ is $F$, and the line $y=k x-1$ intersects the ellipse at points $A$ and $B$. When the perimeter of $\\Delta F A B$ is maximized, what is the area of $\\Delta F A B$?", "fact_expressions": "G: Ellipse;H: Line;k: Number;F: Point;A: Point;B: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Expression(H) = (y = k*x - 1);LeftFocus(G) = F;Intersection(H, G) = {A, B};WhenMax(Perimeter(TriangleOf(F,A,B)))", "query_expressions": "Area(TriangleOf(F, A, B))", "answer_expressions": "12*sqrt(2)/7", "fact_spans": "[[[0, 37], [58, 60]], [[46, 57]], [[48, 57]], [[42, 45]], [[63, 66]], [[67, 70]], [[0, 37]], [[46, 57]], [[0, 45]], [[46, 72]], [[73, 94]]]", "query_spans": "[[[95, 114]]]", "process": "First, using the definition of the ellipse, the perimeter of triangle FAB is |AF| + |BF| + |AB| = 2a - |AF| + 2a - |BF| + |AB| = 8 + |AB| - (|AF| + |BF|) ≤ 8 + |AB| - |AB| = 8. It follows that the perimeter is maximized when line AB passes through the right focus of the ellipse. Then, the equation of line AB is found to be: y = x - 1. By solving this equation simultaneously with the ellipse equation, the value of |y₁ - y₂| is obtained. Using S_{\\triangle FAB} = \\frac{1}{2}|FF| \\times |y_1 - y_2|, the solution can then be determined. From the standard equation of the ellipse, we have: a = 2, c = 1. As shown in the figure, let the right focus of the ellipse be F. Then, the perimeter of triangle FAB is |AF| + |BF| + |AB| = 2a - |AF| + 2a - |BF| + |AB| = 8 + |AB| - (|AF| + |BF|) ≤ 8 + |AB| - |AB| = 8, with equality if and only if line AB passes through the right focus F of the ellipse. Substituting (1,0) into y = kx - 1 gives k = 1. Therefore, the equation of line AB is: y = x - 1. Let A(x₁, y₁), B(x₂, y₂). Solving simultaneously with ()2 - \\frac{y^{2}}{3} = 1, eliminating x gives: 7y² + 6y - 9 = 0, -\\frac{6}{7}, 3. Thus, y₁ + y₂ = -\\frac{6}{7}, y₁y₂ = -\\frac{9}{7}, |y₁ - y₂| = \\sqrt{(y₁ + y₂)² - 4y₁y₂} = \\sqrt{(-\\frac{6}{7})² - 4(-\\frac{9}{7})} = \\sqrt{\\frac{36}{49} + \\frac{36}{7}} = \\sqrt{\\frac{36 + 252}{49}} = \\sqrt{\\frac{288}{49}} = \\frac{12\\sqrt{2}}{7}. Therefore, the area of triangle FAB = \\frac{1}{2}|FF| \\times |y_1 - y_2| = \\frac{1}{2} \\times 2 \\times \\frac{12\\sqrt{2}}{7} = \\frac{12\\sqrt{2}}{7}." }, { "text": "Given the hyperbola $\\frac{y^{2}}{4}-\\frac{x^{2}}{3}=1$, then the distance from the focus to the asymptote is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/3 + y^2/4 = 1)", "query_expressions": "Distance(Focus(G), Asymptote(G))", "answer_expressions": "sqrt(3)", "fact_spans": "[[[2, 40]], [[2, 40]]]", "query_spans": "[[[2, 53]]]", "process": "From the hyperbola equation, we know: the foci of the hyperbola lie on the y-axis, a=2, b=\\sqrt{3}, c=\\sqrt{7}; thus, the coordinates of the foci are (0,\\pm\\sqrt{7}), and the equations of the asymptotes are y=\\pm\\frac{2\\sqrt{3}}{3}x, i.e., 2x\\pm\\sqrt{3}y=0. Therefore, the distance from a focus to its asymptote is d=\\frac{|2\\times0\\pm\\sqrt{3}\\times\\sqrt{7}|}{\\sqrt{4+3}}=\\sqrt{3}." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $3 x^{2}-y^{2}=3 a^{2}(a>0)$, respectively, and $P$ is an intersection point of the parabola $y^{2}=8 a x$ and the hyperbola. If $|P F_{1}|+|P F_{2}|=12$, then the equation of the directrix of the parabola is?", "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;P: Point;F1: Point;F2: Point;a>0;Expression(G) = (3*x^2 - y^2 = 3*a^2);Expression(H) = (y^2 = 8*(a*x));LeftFocus(G) = F1;RightFocus(G) = F2;OneOf(Intersection(G, H)) = P;Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 12", "query_expressions": "Expression(Directrix(H))", "answer_expressions": "x = -2", "fact_spans": "[[[20, 51], [79, 82]], [[23, 51]], [[62, 78], [115, 118]], [[58, 61]], [[2, 9]], [[10, 17]], [[23, 51]], [[20, 51]], [[62, 78]], [[2, 57]], [[2, 57]], [[58, 87]], [[89, 113]]]", "query_spans": "[[[115, 125]]]", "process": "Convert the hyperbola equation into standard form to obtain \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3a^{2}}=1. The directrix of the parabola is x=-2a. Solving the system \\begin{cases}\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3a^{2}}=1\\\\y2=8ax\\end{cases}, we obtain x=3a, so the x-coordinate of point P is 3a. From \\begin{cases}|PF_{1}|+|PF_{2}|=12\\\\|PF_{1}|-|PF_{2}|=2a\\end{cases}, solving gives |PF_{2}|=6-a. Also, |PF_{2}|=3a+2a=6-a, solving yields a=1. Therefore, the directrix equation of the parabola is x=-2." }, { "text": "A moving circle is externally tangent to the circle $Q_{1}$: $(x+3)^{2}+y^{2}=1$ and internally tangent to the circle $Q_{2}$: $(x-3)^{2}+y^{2}=81$. Then, the trajectory equation of the center of this moving circle is?", "fact_expressions": "G: Circle;IsOutTangent(G,Q1) = True;Q1: Circle;Expression(Q1) = ((x+3)^2+y^2=1);Q2: Circle;Expression(Q2) = ((x-3)^2+y^2=81);IsInTangent(G,Q2) = True", "query_expressions": "LocusEquation(Center(G))", "answer_expressions": "x^2/25+y^2/16=1", "fact_spans": "[[[2, 4], [72, 74]], [[2, 35]], [[5, 33]], [[5, 33]], [[37, 66]], [[37, 66]], [[2, 68]]]", "query_spans": "[[[72, 83]]]", "process": "Let the center of the moving circle be Q(x, y), with radius R. Since the moving circle is externally tangent to the circle Q_{1}:(x+3)^{2}+y^{2}=1 and internally tangent to the circle Q_{2}:(x-3)^{2}+y^{2}=81, we have |QQ_{1}| = R + 1, |QQ_{2}| = 9 - R. Therefore, |QQ_{1}| + |QQ_{2}| = 10 > 6 = |Q_{1}Q_{2}|. Thus, the locus of the center of the moving circle is an ellipse with foci at Q_{1} and Q_{2}. Hence, 2a = 10, a = 5, c = 3, b^{2} = 16. Therefore, the equation of the locus of the center of the moving circle is \\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1." }, { "text": "Given that the hyperbola $C$ passes through the point $(3, \\sqrt{2})$ and has asymptotes $y=\\pm \\frac{\\sqrt{3}}{3} x$, then the standard equation of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;G: Point;Coordinate(G) = (3, sqrt(2));PointOnCurve(G, C);Expression(Asymptote(C)) = (y = pm*x*(sqrt(3)/3))", "query_expressions": "Expression(C)", "answer_expressions": "x^2/3-y^2=1", "fact_spans": "[[[2, 8], [60, 66]], [[9, 25]], [[9, 25]], [[2, 25]], [[2, 58]]]", "query_spans": "[[[60, 73]]]", "process": "According to the asymptote equations of the hyperbola, the hyperbola equation can be assumed as x^{2}-3y^{2}=\\lambda(\\lambda\\neq0). Substituting the point (3,\\sqrt{2}) into the equation to solve for \\lambda, the hyperbola equation can then be determined. Solution: According to the given conditions, the asymptote equations of the hyperbola are y=\\pm\\frac{\\sqrt{3}}{3}x, which can be rewritten as: x\\pm\\sqrt{3}y=0. Then, the hyperbola equation can be assumed as x^{2}-3y^{2}=\\lambda(\\lambda\\neq0). Substituting the point (3,\\sqrt{2}) into x^{2}-3y^{2}=\\lambda(\\lambda\\neq0), we get 3^{2}-3\\sqrt{2}^{2}=\\lambda(\\lambda\\neq0), thus \\lambda=3. Therefore, the hyperbola equation is: \\frac{x^{2}}{3}-y^{2}=1." }, { "text": "The focus $F$ of the parabola $C$: $y^{2}=4x$ is used to draw a line $l$ intersecting the parabola $C$ at points $A$, $B$. If $|AF|=4|BF|$, then what is the slope of line $l$?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, F)) = 4*Abs(LineSegmentOf(B, F))", "query_expressions": "Slope(l)", "answer_expressions": "pm*(4/3)", "fact_spans": "[[[1, 19], [32, 38]], [[1, 19]], [[22, 25]], [[1, 25]], [[26, 31], [62, 67]], [[0, 31]], [[39, 42]], [[43, 46]], [[26, 46]], [[48, 60]]]", "query_spans": "[[[62, 72]]]", "process": "The equation of parabola C is $ y^{2} = 4x $, and the coordinates of the focus F are $ (1,0) $. Therefore, let the equation of line $ l $ be $ y = k(x - 1) $. By eliminating from \n$$\n\\begin{cases}\ny = k(k - 1) \\\\\ny^{2} = 4x\n\\end{cases}\n$$\nwe obtain $ \\frac{k}{4}y^{2} - y - k = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then \n$$\n\\begin{cases}\ny_{1} + y_{2} = \\frac{4}{k} \\\\\ny_{1}y_{2} = -4\n\\end{cases} \\textcircled{1}\n$$\nSince $ |AF| = 4|BF| $, we have $ y_{1} + 4y_{2} = 0 $, which gives $ y_{1} = -4y_{2} $. Substituting into \\textcircled{1} yields $ -3y_{2} = \\frac{4}{k} $ and $ -4y_{2}^{2} = -4 $. Since $ y_{2}^{2} = 1 $, that is, $ y_{2} = \\pm1 $, therefore $ k = \\pm\\frac{4}{3} $." }, { "text": "Let $ P $ be a point on the ellipse $ C $: $ \\frac{x^{2}}{4} + y^{2} = 1 $, and let $ A(1,0) $. Then the minimum value of $ |PA| $ is?", "fact_expressions": "C: Ellipse;A: Point;P: Point;Expression(C) = (x^2/4 + y^2 = 1);Coordinate(A) = (1, 0);PointOnCurve(P,C)", "query_expressions": "Min(Abs(LineSegmentOf(P, A)))", "answer_expressions": "sqrt(6)/3", "fact_spans": "[[[4, 36]], [[40, 48]], [[0, 3]], [[4, 36]], [[40, 48]], [[0, 39]]]", "query_spans": "[[[50, 62]]]", "process": "" }, { "text": "Let the hyperbola with foci on the $x$-axis be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$. The right directrix intersects the two asymptotes at points $A$ and $B$, and the right focus is $F$. Given that $\\overrightarrow{F A} \\cdot \\overrightarrow{F B}=0$, find the eccentricity $e$ of the hyperbola.", "fact_expressions": "G: Hyperbola;b: Number;a: Number;F: Point;A: Point;B: Point;e: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(Focus(G),xAxis);L1: Line;L2: Line;Asymptote(G) = {L1, L2};Intersection(RightDirectrix(G), L1) = A;Intersection(RightDirectrix(G), L2) = B;RightFocus(G) = F;DotProduct(VectorOf(F, A), VectorOf(F, B)) = 0;Eccentricity(G) = e", "query_expressions": "e", "answer_expressions": "sqrt(2)", "fact_spans": "[[[10, 56], [140, 143]], [[13, 56]], [[13, 56]], [[82, 85]], [[68, 71]], [[72, 75]], [[147, 150]], [[10, 56]], [[1, 56]], [], [], [[10, 66]], [[10, 77]], [[10, 77]], [[10, 85]], [[87, 138]], [[140, 150]]]", "query_spans": "[[[147, 152]]]", "process": "" }, { "text": "Through the right focus $F$ of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, draw a straight line with positive slope intersecting the asymptotes at points $A$ and $B$. $O$ is the origin. If the area of $\\triangle OAB$ is $16\\sqrt{3}$, then what is the slope of line $AB$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1);F: Point;RightFocus(G) = F;H: Line;PointOnCurve(F, H);Slope(H)>0;A: Point;B: Point;Intersection(H, Asymptote(G)) = {A, B};O: Origin;Area(TriangleOf(O, A, B)) = 16*sqrt(3)", "query_expressions": "Slope(LineOf(A, B))", "answer_expressions": "{2, (2*sqrt(15))/5}", "fact_spans": "[[[1, 29]], [[1, 29]], [[32, 35]], [[1, 35]], [[41, 43]], [[0, 43]], [[36, 43]], [[48, 51]], [[52, 55]], [[1, 57]], [[58, 61]], [[68, 102]]]", "query_spans": "[[[104, 116]]]", "process": "The asymptotes of the hyperbola are given by y=\\sqrt{3}x and y=-\\sqrt{3}x. Let the equation of line AB be y=kx-2k, (k\\neq\\pm\\sqrt{3}). Solving the systems \\begin{cases}y=\\sqrt{3}x\\\\y=kx-2k\\end{cases} and \\begin{cases}y=-\\sqrt{3}x\\\\y=kx-2k\\end{cases}, we obtain y_{A}=\\frac{2\\sqrt{3}k}{k-\\sqrt{3}}, y_{B}=\\frac{-2\\sqrt{3}k}{k+\\sqrt{3}}. Since the area of \\triangle OAB is 16\\sqrt{3}, it follows that \\frac{1}{2}\\times2|y_{A}-y_{B}|=16\\sqrt{3}, that is, |y_{A}-y_{B}|=|\\frac{4\\sqrt{3}k^{2}}{k^{2}-3}|=16\\sqrt{3}. Solving this yields k=2 or k=\\frac{2\\sqrt{15}}{5}." }, { "text": "The ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ has its right focus at $F(c, 0)$. Points $P$ and $Q$ lie on the ellipse $C$. The distance from point $M(-\\frac{c}{2}, 0)$ to the line $FP$ is $\\frac{c}{2}$, and the incenter of $\\Delta P Q F$ is exactly the point $M$. Then the eccentricity $e$ of the ellipse $C$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;c: Number;F: Point;RightFocus(C) = F;Coordinate(F) = (c, 0);P: Point;Q: Point;PointOnCurve(P, C);PointOnCurve(Q, C);M: Point;Coordinate(M) = (-c/2, 0);Distance(M, LineOf(F, P)) = c/2;Incenter(TriangleOf(P, Q, F)) = M;e: Number;Eccentricity(C) = e", "query_expressions": "e", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[0, 57], [81, 86], [162, 167]], [[0, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[62, 71]], [[62, 71]], [[0, 71]], [[62, 71]], [[72, 76]], [[77, 80]], [[72, 87]], [[77, 87]], [[88, 109], [156, 160]], [[88, 109]], [[88, 134]], [[136, 160]], [[171, 174]], [[162, 174]]]", "query_spans": "[[[171, 176]]]", "process": "Let PQ intersect the x-axis at point F. Analysis shows that point F is the left focus of the ellipse. Then compute |PF| = \\frac{b^{2}}{a}, |PF| = a + \\frac{c^{2}}{a}. Then, according to \\frac{|ME|}{|MF|} = \\frac{|PF|}{|PF|}, the solution is obtained. As shown in the figure, the incenter of \\triangle PQF is exactly the point. By symmetry, region, so P and Q are symmetric with respect to the x-axis, hence the axis. Let PQ intersect the x-axis at point F, then \\overrightarrow{BC}, so point F is the left focus of the ellipse. Substitute into the ellipse equation to get \\otimes. Therefore, from point M draw ME \\bot PF, with foot at E, then B, so region." }, { "text": "Ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ has left and right foci $F_{1}$, $F_{2}$ respectively. If there exists a point $P$ on ellipse $C$ such that $P F_{1}=2 e P F_{2}$, ($e$ is the eccentricity of the ellipse), then the range of values for the eccentricity of ellipse $C$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C);e: Number;Eccentricity(C) = e;LineSegmentOf(P, F1) = (2*e)*LineSegmentOf(P, F2)", "query_expressions": "Range(Eccentricity(C))", "answer_expressions": "[(sqrt(17)-3)/4, 1)", "fact_spans": "[[[0, 50], [77, 82], [131, 136], [121, 123]], [[0, 50]], [[7, 50]], [[7, 50]], [[58, 65]], [[67, 74]], [[0, 74]], [[0, 74]], [[85, 89]], [[77, 89]], [[117, 120]], [[117, 127]], [[93, 114]]]", "query_spans": "[[[131, 147]]]", "process": "There exists a point P on the ellipse C such that PF_{1}=2ePF_{2}, and since PF_{1}+PF_{2}=2a, then PF_{2}=\\frac{2a}{2c+1}=\\frac{2a^{2}}{2c+a}\\cdot PF_{2}\\in[a-c,a+c]" }, { "text": "The standard equation of the hyperbola passing through two points $A(3, \\frac{\\sqrt{5}}{2})$, $B(-4, \\sqrt{3})$ is?", "fact_expressions": "A: Point;Coordinate(A) = (3, sqrt(5)/2);B: Point;Coordinate(B) = (-4, sqrt(3));G: Hyperbola;PointOnCurve(A, G);PointOnCurve(B, G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4-y^2=1", "fact_spans": "[[[4, 30]], [[4, 30]], [[33, 50]], [[33, 50]], [[51, 54]], [[0, 54]], [[0, 54]]]", "query_spans": "[[[51, 61]]]", "process": "Let the equation of the hyperbola be $mx^{2}+ny^{2}=1$ $(mn<0)$. Since the desired hyperbola passes through points $A(3,\\frac{\\sqrt{5}}{2})$, $B(-4,\\sqrt{3})$, we have\n\\[\n\\begin{cases}\n9m+\\frac{5}{4}n=1,\\\\\n16m+3n=1,\n\\end{cases}\n\\]\nsolving which yields\n\\[\n\\begin{cases}\nm=\\frac{1}{4},\\\\\nn=-1.\n\\end{cases}\n\\]\nTherefore, the equation of the required hyperbola is $\\frac{x^{2}}{4}-y^{2}=1$." }, { "text": "The line perpendicular to the $x$-axis passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. Then, the equation of the circle with center $F$ and diameter $AB$ is?", "fact_expressions": "G: Parabola;H: Circle;L: Line;A: Point;B: Point;F:Point;Expression(G) = (y^2 = 4*x);Focus(G)=F;PointOnCurve(F,L);IsPerpendicular(L,xAxis);Intersection(L,G)={A,B};Center(H)=F;IsDiameter(LineSegmentOf(A,B),H)", "query_expressions": "Expression(H)", "answer_expressions": "(x-1)^2+y^2=4", "fact_spans": "[[[1, 15], [34, 37]], [[66, 67]], [[30, 32]], [[38, 41]], [[42, 45]], [[18, 21], [50, 53]], [[1, 15]], [[1, 21]], [[0, 32]], [[22, 32]], [[30, 47]], [[49, 67]], [[57, 67]]]", "query_spans": "[[[66, 72]]]", "process": "" }, { "text": "Given the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively, with eccentricity $e$. If point $P$ on the hyperbola satisfies $\\frac{\\sin \\angle P F_{2} F_{1}}{\\sin \\angle P F_{1} F_{2}}=e$, then the value of $\\overrightarrow{F_{2} P} \\cdot \\overrightarrow{F_{2} F_{1}}$ is?", "fact_expressions": "G: Hyperbola;P: Point;F2: Point;F1: Point;Expression(G) = (x^2 - y^2/3 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);Sin(AngleOf(P, F2, F1))/Sin(AngleOf(P, F1, F2)) = e;e:Number;Eccentricity(G)=e", "query_expressions": "DotProduct(VectorOf(F2, P), VectorOf(F2, F1))", "answer_expressions": "2", "fact_spans": "[[[2, 30], [64, 67]], [[68, 72]], [[47, 54]], [[39, 46]], [[2, 30]], [[2, 54]], [[2, 54]], [[64, 72]], [[74, 137]], [[59, 62]], [[2, 62]]]", "query_spans": "[[[139, 204]]]", "process": "From the hyperbola equation $ x^{2}-\\frac{y^{2}}{3}=1 $, we obtain $ a=1 $, $ c=2 $. By the definition of the hyperbola, $ ||\\overrightarrow{PF}|-|\\overrightarrow{PF_{2}}||=2 $. Since $ \\frac{\\sin\\angle PF_{2}F_{1}}{\\sin\\angle PF_{1}F_{2}}=e=2 $, by the law of sines, $ \\frac{|\\overrightarrow{PF}|}{|\\overrightarrow{PF}|}=2 $, which gives $ |\\overrightarrow{PF_{1}}|=4 $, $ |\\overrightarrow{PF_{2}}|=2 $. Also, $ |\\overrightarrow{F_{1}F_{2}}|=4 $. Using the law of cosines, we get $ \\cos\\angle PF_{2}F_{1}=\\frac{1}{4} $. Therefore, $ \\overrightarrow{F_{2}P}\\cdot\\overrightarrow{F_{2}F_{1}}=|\\overrightarrow{F_{2}P}|\\cdot|\\overrightarrow{F_{2}F_{1}}|\\cdot\\cos\\angle PF_{2}F_{1}=2\\times4\\times\\frac{1}{4}=2 $." }, { "text": "The circle $(x-a)^{2}+y^{2}=1$ is tangent to the asymptotes of the hyperbola $x^{2}-y^{2}=1$, then the value of $a$ is?", "fact_expressions": "G: Hyperbola;H: Circle;a: Number;Expression(G) = (x^2 - y^2 = 1);Expression(H) = (y^2 + (-a + x)^2 = 1);IsTangent(H,Asymptote(G))", "query_expressions": "a", "answer_expressions": "pm*sqrt(2)", "fact_spans": "[[[21, 39]], [[0, 20]], [[47, 50]], [[21, 39]], [[0, 20]], [[0, 45]]]", "query_spans": "[[[47, 54]]]", "process": "" }, { "text": "The left and right foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ are $F_{1}$ and $F_{2}$ respectively. A line passing through focus $F_{1}$ intersects the ellipse at points $A$ and $B$. What is the perimeter of $\\Delta ABF_{2}$?", "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2/16 + y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, H);Intersection(H, G) = {A,B}", "query_expressions": "Perimeter(TriangleOf(A, B, F2))", "answer_expressions": "4*sqrt(7)/7", "fact_spans": "[[[0, 38], [77, 79]], [[74, 76]], [[80, 83]], [[84, 87]], [[55, 62]], [[47, 54], [66, 73]], [[0, 38]], [[0, 62]], [[0, 62]], [[63, 76]], [[74, 89]]]", "query_spans": "[[[91, 112]]]", "process": "" }, { "text": "The equation of the locus of points equidistant from $A(2,-3)$ and $B(4,-1)$ is?", "fact_expressions": "A: Point;B: Point;Coordinate(A) = (2, -3);Coordinate(B) = (4, -1);H:Point;Distance(H,A)=Distance(H,B)", "query_expressions": "LocusEquation(H)", "answer_expressions": "x+y-1=0", "fact_spans": "[[[1, 10]], [[11, 20]], [[1, 10]], [[11, 20]], [[26, 27]], [[0, 27]]]", "query_spans": "[[[26, 34]]]", "process": "Since point P satisfies |PA| = |PB|, the trajectory of point P is the perpendicular bisector of points A(2,-3) and B(4,-1). Given A(2,-3) and B(4,-1), using the midpoint formula, the midpoint of AB is (3,-2), k_{AB} = \\frac{1+3}{4-2} = 1, therefore the slope of its perpendicular bisector is -1. Therefore, the trajectory equation of point P is y+2 = -(x-3), that is, x+y-1=0." }, { "text": "A line with slope $1$ passes through the focus of the parabola $y^{2}=4 x$ and intersects the parabola at points $A$ and $B$. Then $|A B|$ equals?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Line;Slope(H) = 1;PointOnCurve(Focus(G), H);A: Point;B: Point;Intersection(H, G) = {A, B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "8", "fact_spans": "[[[11, 25], [31, 34]], [[11, 25]], [[7, 9]], [[0, 9]], [[7, 28]], [[37, 40]], [[41, 44]], [[7, 46]]]", "query_spans": "[[[48, 58]]]", "process": "" }, { "text": "Point $P$ is a moving point on the parabola ${y}^{2}=4 x$. Then, the minimum value of the sum of the distance from point $P$ to point $A(0,-1)$ and the distance from point $P$ to the line $x=-1$ is?", "fact_expressions": "G: Parabola;H: Line;A: Point;P: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (x = -1);Coordinate(A) = (0, -1);PointOnCurve(P, G)", "query_expressions": "Min(Distance(P, A) + Distance(P, H))", "answer_expressions": "5", "fact_spans": "[[[5, 21]], [[52, 60]], [[33, 43]], [[0, 4], [28, 32], [47, 51]], [[5, 21]], [[52, 60]], [[33, 43]], [[0, 26]]]", "query_spans": "[[[28, 71]]]", "process": "" }, { "text": "Given that the center of circle $C$ lies on the parabola $y^{2}=4 x$ and is tangent to both the $x$-axis and the directrix of the parabola, find the standard equation of circle $C$.", "fact_expressions": "G: Parabola;C: Circle;Expression(G) = (y^2 = 4*x);PointOnCurve(Center(C),G);IsTangent(C,xAxis);IsTangent(C,Directrix(G))", "query_expressions": "Expression(C)", "answer_expressions": "{(x-1)^2+(y-2)^2=4,(x-1)^2+(y+2)^2=4}", "fact_spans": "[[[10, 24], [33, 36]], [[2, 6], [2, 6]], [[10, 24]], [[2, 25]], [[2, 42]], [[2, 42]]]", "query_spans": "[[[44, 55]]]", "process": "The center of the circle can be set as $(\\frac{b^{2}}{4}, b)$, with radius $r = |b|$. The focus of the parabola is $F(1, 0)$, and the equation of the directrix is $x = -1$. Then $\\frac{b^{2}}{4} + 1 = |b|$, solving gives: $b = 2$ or $b = -2$, so the radius is $r = |b| = 2$." }, { "text": "$A B$ is a chord passing through the left focus $F_{1}$ on the left branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, $|A B|=m$, $F_{2}$ is the right focus, then the perimeter of $\\triangle A B F_{2}$ is?", "fact_expressions": "G: Hyperbola;a: Number;b: Number;A: Point;B: Point;F1:Point;F2: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F2;Focus(G)={F1,F2};Abs(LineSegmentOf(A, B)) = m;IsChordOf(LineSegmentOf(A, B),LeftPart(G));PointOnCurve(F1,LineSegmentOf(A, B));m:Number", "query_expressions": "Perimeter(TriangleOf(A, B, F2))", "answer_expressions": "4*a+2*m", "fact_spans": "[[[6, 52]], [[9, 52]], [[9, 52]], [[0, 5]], [[0, 5]], [[58, 65]], [[80, 87]], [[6, 52]], [[6, 91]], [[6, 91]], [[68, 78]], [[0, 67]], [[0, 67]], [[68, 78]]]", "query_spans": "[[[93, 120]]]", "process": "" }, { "text": "The line $x+2 y-2=0$ passes through one focus and one vertex of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$. Then the eccentricity of this ellipse is equal to?", "fact_expressions": "H: Line;Expression(H) = (x + 2*y - 2 = 0);G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;PointOnCurve(OneOf(Focus(G)), H);PointOnCurve(OneOf(Vertex(G)), H)", "query_expressions": "Eccentricity(G)", "answer_expressions": "(2*sqrt(5))/5", "fact_spans": "[[[0, 13]], [[0, 13]], [[15, 67], [80, 82]], [[15, 67]], [[17, 67]], [[17, 67]], [[17, 67]], [[17, 67]], [[0, 72]], [[0, 77]]]", "query_spans": "[[[80, 89]]]", "process": "For the line x+2y-2=0, let x=0, solve to get y=1; let y=0, solve to get x=2. Therefore, the coordinates of the right focus of the ellipse are (2,0), and the coordinates of the upper vertex are (0,1). Then c=2, b=1, so a=\\sqrt{b^{2}+c^{2}}=\\sqrt{5}. Hence, the eccentricity of the ellipse is e=\\frac{c}{a}=\\frac{2}{\\sqrt{5}}=\\frac{2\\sqrt{5}}{5}." }, { "text": "Given the equation of the hyperbola is $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1$, $F_{1}$, $F_{2}$ are the left and right foci of the hyperbola. Point $P$ lies on the hyperbola, $|PF_{1}|=8$, then $|P F_{2}|$=?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/4 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 8", "query_expressions": "Abs(LineSegmentOf(P, F2))", "answer_expressions": "{14, 2}", "fact_spans": "[[[2, 5], [63, 66], [78, 81]], [[2, 44]], [[46, 53]], [[55, 62]], [[46, 71]], [[46, 71]], [[73, 77]], [[73, 82]], [[83, 95]]]", "query_spans": "[[[97, 110]]]", "process": "" }, { "text": "The line $ l $ with slope $ 2 $ passes through the focus $ F $ of the parabola $ y^{2} = 8x $, and intersects the parabola at points $ A $ and $ B $. What is the length of segment $ AB $?", "fact_expressions": "l: Line;Slope(l) = 2;G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, G) = {A, B}", "query_expressions": "Length(LineSegmentOf(A, B))", "answer_expressions": "10", "fact_spans": "[[[7, 12]], [[0, 12]], [[14, 28], [38, 41]], [[14, 28]], [[31, 34]], [[14, 34]], [[7, 34]], [[44, 47]], [[48, 51]], [[7, 53]]]", "query_spans": "[[[55, 66]]]", "process": "】Solve the system of equations of the line and the parabola, and use the formula for calculating the focal chord of a parabola: $x_{A}+x_{B}+p$, to find the length of the chord passing through the focus. Since the focus $F(2,0)$, the line $l: y=2(x-2)$. Solving the system of the line and the parabola gives:\n\\[\n\\begin{cases}\ny^{2}=8x \\\\\ny=2x-4\n\\end{cases}\n\\]\nThus, $4x^{2}-24x+16=0$, or equivalently $x^{2}-6x+4=0$. Therefore, $x_{A}+x_{B}=6$, and so $AB = x_{A}+x_{B}+p = 6+4 = 10$." }, { "text": "The hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ has two foci $F_{1}$, $F_{2}$. Point $P$ lies on the hyperbola. If $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, then what is the distance from point $P$ to the $x$-axis?", "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0", "query_expressions": "Distance(P, xAxis)", "answer_expressions": "16/5", "fact_spans": "[[[0, 39], [67, 70]], [[62, 66], [134, 138]], [[45, 52]], [[53, 60]], [[0, 39]], [[0, 60]], [[62, 71]], [[73, 132]]]", "query_spans": "[[[134, 148]]]", "process": "" }, { "text": "The foci of the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{2}=1$ are?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2/2 = 1)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(pm*sqrt(5), 0)", "fact_spans": "[[[0, 38]], [[0, 38]]]", "query_spans": "[[[0, 43]]]", "process": "Given that $ a = \\sqrt{3} $, $ b = \\sqrt{2} $, so $ c = \\sqrt{a^{2} + b^{2}} = \\sqrt{5} $, the foci are $ (\\pm\\sqrt{5}, 0) $." }, { "text": "If the line $l$: $y=x+\\frac{3}{2}$ and the parabola $C$: $y^{2}=2 p x(p>0)$ have only one common point, then the equation of the directrix of the parabola $C$ is?", "fact_expressions": "l: Line;C: Parabola;P: Number;P>0;Expression(l)=(y = x + 3/2);Expression(C) = (y^2 = 2*(P*x));NumIntersection(l,C)=1", "query_expressions": "Expression(Directrix(C))", "answer_expressions": "x=-3/2", "fact_spans": "[[[1, 25]], [[26, 52], [61, 67]], [[34, 52]], [[34, 52]], [[1, 25]], [[26, 52]], [[1, 59]]]", "query_spans": "[[[61, 74]]]", "process": "From $ y = x + \\frac{3}{2} $, we get $ x = y - \\frac{3}{2} $. Substituting $ x = y - \\frac{3}{2} $ into $ y^2 = 2px $, eliminating $ x $ gives $ y^2 - 2py + 3p = 0 $. Since line $ l $ and parabola $ C $ have only one common point, $ A = (-2p)^2 - 12p = 0 $, that is, $ p^2 - 3p = 0 $. Solving yields $ p = 0 $ (discarded) or $ p = 3 $. Therefore, the directrix equation of parabola $ C $ is $ x = -\\frac{3}{2} $." }, { "text": "The eccentricity of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/9 = 1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)/3", "fact_spans": "[[[0, 37]], [[0, 37]]]", "query_spans": "[[[0, 43]]]", "process": "From the given conditions, according to the equation of the ellipse, we have a=3, b=2, c=\\sqrt{9-4}=\\sqrt{5}, so the eccentricity is e=\\frac{c}{a}=\\frac{\\sqrt{5}}{3}" }, { "text": "If the foci and the endpoints of the minor axis of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0) $ lie on the same circle, then the eccentricity of the ellipse is equal to?", "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Circle;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(Focus(G),H);PointOnCurve(Endpoint(MinorAxis(G)),H)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[1, 56], [72, 74]], [[3, 56]], [[3, 56]], [[68, 69]], [[3, 56]], [[3, 56]], [[1, 56]], [[1, 70]], [[1, 70]]]", "query_spans": "[[[72, 81]]]", "process": "" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the distance between its left vertex and the focus of the parabola $y^{2}=2 p x(p>0)$ is $4$, and the coordinates of the intersection point between one asymptote of the hyperbola and the directrix of the parabola are $(-2,-1)$. Find the focal length of the hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;p: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);p>0;Expression(H) = (y^2 = 2*(p*x));Distance(LeftVertex(G), Focus(H)) = 4;Coordinate(Intersection(OneOf(Asymptote(G)),Directrix(H)))=(-2,-1)", "query_expressions": "FocalLength(G)", "answer_expressions": "2*sqrt(5)", "fact_spans": "[[[2, 58], [96, 99], [128, 131]], [[5, 58]], [[5, 58]], [[63, 84], [106, 109]], [[66, 84]], [[5, 58]], [[5, 58]], [[2, 58]], [[66, 84]], [[63, 84]], [[2, 94]], [[96, 126]]]", "query_spans": "[[[128, 136]]]", "process": "" }, { "text": "The coordinates of the focus of the parabola $y=16 x^{2}$ are?", "fact_expressions": "G: Parabola;Expression(G) = (y = 16*x^2)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0, 1/64)", "fact_spans": "[[[0, 15]], [[0, 15]]]", "query_spans": "[[[0, 22]]]", "process": "From the given information, the standard equation of the parabola is $x^{2}=\\frac{1}{16}y$, opening upwards, with the focus on the y-axis and $2p=\\frac{1}{16}$, so the coordinates of the focus are $(0,\\frac{1}{64})$." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1$, the right focus is $F_{2}$. Let $P$ and $Q$ be two moving points on the ellipse $C$ such that $PQ = 2\\sqrt{2}$. Then, when $PF_{2} + QF_{2}$ attains its maximum value, what is the perimeter of $\\triangle PQF_{2}$?", "fact_expressions": "C: Ellipse;P: Point;Q: Point;F2: Point;Expression(C) = (x^2/3 + y^2/2 = 1);RightFocus(C) = F2;PointOnCurve(P, C);PointOnCurve(Q, C);LineSegmentOf(P, Q) = 2*sqrt(2);WhenMax(LineSegmentOf(P,F2)+LineSegmentOf(Q,F2))", "query_expressions": "Perimeter(TriangleOf(P,Q,F2))", "answer_expressions": "4*sqrt(3)", "fact_spans": "[[[2, 44], [66, 71]], [[58, 61]], [[62, 65]], [[49, 56]], [[2, 44]], [[2, 56]], [[58, 77]], [[58, 77]], [[79, 95]], [[97, 121]]]", "query_spans": "[[[122, 148]]]", "process": "" }, { "text": "Point $F$ is the focus of the parabola $y^{2}=4x$. A line passing through point $F$ with an inclination angle of $\\frac{\\pi}{3}$ intersects the parabola at points $A$ and $B$. Then the chord length $|AB|=$?", "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, H);Inclination(H) = pi/3;Intersection(H, G) = {A, B};IsChordOf(LineSegmentOf(A,B),G)", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "16/3", "fact_spans": "[[[5, 19], [52, 55]], [[49, 51]], [[56, 59]], [[60, 63]], [[0, 4], [24, 28]], [[5, 19]], [[0, 22]], [[23, 51]], [[29, 51]], [[49, 65]], [[52, 76]]]", "query_spans": "[[[69, 78]]]", "process": "From the given conditions, the focus of the parabola is $ F(1,0) $. Since the angle of inclination of the line is $ \\frac{\\pi}{3} $, the slope of line $ AB $ is $ \\sqrt{3} $. Therefore, the equation of line $ AB $ is $ y = \\sqrt{3}(x - 1) $. Solving the system of equations\n\\[\n\\begin{cases}\ny = \\sqrt{3}(x - 1) \\\\\ny^2 = 4x\n\\end{cases}\n\\]\nyields $ 3x^2 - 10x + 3 = 0 $. Solving this equation gives $ x_1 = 3 $ or $ x_2 = \\frac{1}{3} $. By the definition of the parabola, $ |AB| = |AF| + |BF| = x_1 + 1 + x_2 + 1 = \\frac{16}{3} $." }, { "text": "The line $ l $ passing through the point $ M(-1,0) $ intersects the parabola $ C $: $ y^{2}=4x $ at points $ A $ and $ B $ (with $ A $ between $ M $ and $ B $), $ F $ is the focus of $ C $, and point $ N $ satisfies $ \\overrightarrow{NF} = 6\\overrightarrow{AF} $. Then the minimum value of the sum of the areas of $ \\Delta ABF $ and $ \\Delta AMN $ is?", "fact_expressions": "l: Line;C: Parabola;M: Point;A: Point;B: Point;F: Point;N: Point;Expression(C) = (y^2 = 4*x);Coordinate(M) = (-1, 0);PointOnCurve(M, l);Intersection(l, C) = {A, B};Between(A, M, B);Focus(C) = F;VectorOf(N, F) = 6*VectorOf(A, F)", "query_expressions": "Min(Area(TriangleOf(A, B, F)) + Area(TriangleOf(A, M, N)))", "answer_expressions": "8", "fact_spans": "[[[12, 17]], [[18, 37], [68, 71]], [[1, 11], [53, 56]], [[39, 42], [49, 52]], [[43, 46], [57, 60]], [[64, 67]], [[75, 79]], [[18, 37]], [[1, 11]], [[0, 17]], [[12, 48]], [[49, 62]], [[64, 74]], [[81, 126]]]", "query_spans": "[[[128, 168]]]", "process": "Given that line $ l $ passes through point $ M(-1,0) $, set up the equation of line $ l $. After combining with the parabola, the vertical coordinates of points $ A $ and $ B $ can be expressed. Using $ \\overrightarrow{NF}=6\\overrightarrow{AF} $, the vertical coordinate of point $ N $ can be expressed. The sum of the areas of $ \\triangle ABF $ and $ \\triangle AMN $ can be represented by the triangle area formula. By differentiating the expression, the minimum value of the sum of the areas can be found based on the derivative.\n\nLet the equation of the line be $ x = ty - 1 $. Combine with\n$$\n\\begin{cases}\ny^2 = 4x \\\\\nx = ty - 1\n\\end{cases}\n$$\nSimplifying yields $ y^2 - 4ty + 4 = 0 $. Then $ \\Delta_A = 16t^2 - 16 > 0 $, solving gives $ t > 1 $ or $ t < -1 $. Without loss of generality, assume $ t > 1 $, then\n$$\ny_A = 2t - 2\\sqrt{t^2 - 1}, \\quad y_B = 2t + 2\\sqrt{t^2 - 1}\n$$\nSince $ \\overrightarrow{NF} = 6\\overrightarrow{AF} $, we have\n$$\ny_N = 6y_A = 12t - 12\\sqrt{t^2 - 1}\n$$\nThus,\n$$\nS_{\\triangle ABF} = S_{\\triangle MBF} - S_{\\triangle AMF} = \\frac{1}{2} \\times 2 \\times (y_B - y_A) = y_B - y_A\n$$\n$$\nS_{\\triangle AMN} = S_{\\triangle MNF} - S_{\\triangle AMF} = \\frac{1}{2} \\times 2 \\times (y_N - y_A) = y_N - y_A\n$$\nThen\n$$\nS_{\\triangle ABF} + S_{\\triangle AMN} = y_B - y_A + y_N - y_A = 2t + 2\\sqrt{t^2 - 1} + 12t - 12\\sqrt{t^2 - 1} - 2(2t - 2\\sqrt{t^2 - 1}) = 10t - 6\\sqrt{t^2 - 1} \\quad (t > 1)\n$$\nLet $ f(t) = 10t - 6\\sqrt{t^2 - 1} $, then\n$$\nf'(t) = 10 - \\frac{6t}{\\sqrt{t^2 - 1}}\n$$\nSet $ f'(t) = 0 $, solving gives $ t = \\frac{5}{4} $. When $ 1 < t < \\frac{5}{4} $, $ f'(t) < 0 $, so $ f(t) $ is decreasing on $ (1, \\frac{5}{4}) $. When $ t > \\frac{5}{4} $, $ f'(t) > 0 $, so $ f(t) $ is increasing on $ (\\frac{5}{4}, +\\infty) $. Therefore, $ f(t) $ achieves its minimum at $ t = \\frac{5}{4} $. Hence, the minimum value of $ S_{\\triangle ABF} + S_{\\triangle AMN} $ is 8.\n\nFinal answer: $ 8 $\n\nThis question examines the application of the positional relationship between a line and a parabola, the method for solving in parabola problems, and the use of derivatives to find maximum or minimum values." }, { "text": "Point $P$ is a point on the ellipse $\\frac{x^{2}}{100}+\\frac{y^{2}}{36}=1$, and $F_{1}$, $F_{2}$ are the left and right foci respectively. If $|P F_{1}|=6$, then the value of $|P F_{2}|$ is?", "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/100 + y^2/36 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G)=F2;Abs(LineSegmentOf(P, F1)) = 6", "query_expressions": "Abs(LineSegmentOf(P, F2))", "answer_expressions": "14", "fact_spans": "[[[5, 45]], [[0, 4]], [[49, 56]], [[58, 65]], [[5, 45]], [[0, 48]], [[5, 73]], [[5, 73]], [[75, 88]]]", "query_spans": "[[[90, 105]]]", "process": "" }, { "text": "If the equation $\\frac{x^{2}}{3-t}-\\frac{y^{2}}{1-t}=1$ represents an ellipse, then the range of real values for $t$ is?", "fact_expressions": "G: Ellipse;H: Curve;t: Real;Expression(H)=(x^2/(3-t)-y^2/(1-t)=1);G=H", "query_expressions": "Range(t)", "answer_expressions": "(1,2)+(2,3)", "fact_spans": "[[[49, 51]], [[46, 48]], [[53, 58]], [[1, 48]], [[46, 51]]]", "query_spans": "[[[53, 65]]]", "process": "\\frac{x2}{3-t}-\\frac{y^{2}}{3-t}=1\\Rightarrow\\frac{x^{2}}{3-t}+\\frac{y^{2}}{t-1}=1 represents an ellipse," }, { "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively. If there exists a point $P$ on the right branch of the hyperbola such that $|P F_{1}|-|P F_{2}|=\\frac{3}{5}|F_{1} F_{2}|$, then the equations of the asymptotes of this hyperbola are?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)) = (3/5)*Abs(LineSegmentOf(F1, F2))", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*(4/3)*x", "fact_spans": "[[[19, 75], [148, 151], [84, 87]], [[22, 75]], [[22, 75]], [[92, 96]], [[1, 8]], [[9, 16]], [[22, 75]], [[22, 75]], [[19, 75]], [[1, 81]], [[1, 81]], [[84, 96]], [[99, 145]]]", "query_spans": "[[[148, 159]]]", "process": "Solution: \\because|PF_{1}|-|PF_{2}|=\\frac{3}{5}|F_{1}F_{2}|,\\therefore2a=\\frac{3}{5}\\cdot2c,\\thereforea=\\frac{3}{5}c^{'}, so b=\\frac{4}{8}c\\Rightarrow\\frac{b}{a}=\\frac{4}{3}, the equations of asymptotes are y=\\pm\\frac{4}{3}x" }, { "text": "Point $P$ is a point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, and $F_{1}$, $F_{2}$ are the left and right foci of the ellipse, respectively. If $\\angle F_{1} PF_{2}=60^{\\circ}$, then $|P F_{1}| \\cdot |P F_{2} |$=?", "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/9 + y^2/4 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;AngleOf(F1, P, F2) = ApplyUnit(60, degree)", "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))", "answer_expressions": "16/3", "fact_spans": "[[[5, 42], [65, 67]], [[47, 54]], [[0, 4]], [[55, 62]], [[5, 42]], [[0, 46]], [[47, 72]], [[47, 72]], [[74, 106]]]", "query_spans": "[[[108, 138]]]", "process": "Problem Analysis: Using the given conditions, utilize the ellipse definition and the cosine law to set up a system of equations, from which |PF₁||PF₂| can be solved. \nSolution: \n∵ Point P is a point on the ellipse \\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1, F_{1}, F_{2} are the left and right foci of the ellipse respectively, \\angle F_{1}PF_{2}=60^{\\circ}, \n∴ \\begin{cases} |PF_{1}|+|PF_{2}|=6 \\\\ \\cos60^{\\circ}=\\frac{|PF_{1}|^{2}+|PF_{2}|^{2}-|F_{1}F_{2}|^{2}}{16}^{2|PF_{1}|}\\cdot|PF_{2}| \\end{cases} \nSolving gives |PF_{1}||PF_{2}|=\\frac{16}{3}." }, { "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$ and directrix $l$, let $P$ be a point on $l$, and $Q$ be an intersection point of line $PF$ and $C$. If $\\overrightarrow{FP}=4\\overrightarrow{FQ}$, then what is $|QF|$?", "fact_expressions": "C: Parabola;F: Point;P: Point;Q: Point;l: Line;Expression(C) = (y^2 = 8*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(P, l);OneOf(Intersection(LineOf(P,F),C))=Q;VectorOf(F, P) = 4*VectorOf(F, Q)", "query_expressions": "Abs(LineSegmentOf(Q, F))", "answer_expressions": "3", "fact_spans": "[[[2, 20], [59, 62]], [[24, 27]], [[36, 39]], [[47, 50]], [[31, 34], [40, 43]], [[2, 20]], [[2, 27]], [[2, 34]], [[35, 46]], [[47, 67]], [[69, 114]]]", "query_spans": "[[[116, 126]]]", "process": "Problem Analysis: Obtain the equation of line PF, and solve it together with y^{2}=8x to get x=1. Then use |QF|=d to find the value.\n\nSolution: Let the distance from Q to l be d, then |QF|=d. Since \\overrightarrow{FP}=4\\overrightarrow{FQ}, we have |PQ|=3d. Therefore, assume without loss of generality that the slope of line PF is \\cdot\\frac{2\\sqrt{2}d}{d}=2\\sqrt{2}. Since F(2,0), the equation of line PF is y=-2\\sqrt{2}(x-2). Solving this together with y^{2}=8x yields x=1. Hence, |QF|=d=1+2=3." }, { "text": "Given $M(-2,0)$, $N(2,0)$, $|P M|-|P N|=3$, then the trajectory of the moving point $P$ is?", "fact_expressions": "M: Point;N: Point;P: Point;Coordinate(M) = (-2, 0);Coordinate(N) = (2, 0);Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)) = 3", "query_expressions": "Locus(P)", "answer_expressions": "the right branch of the hyperbola", "fact_spans": "[[[2, 11]], [[14, 22]], [[43, 46]], [[2, 11]], [[14, 22]], [[24, 39]]]", "query_spans": "[[[43, 51]]]", "process": "" }, { "text": "The coordinates of the focus of the parabola $4 y^{2}=x$ are?", "fact_expressions": "G: Parabola;Expression(G) = (4*y^2 = x)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(1/16, 0)", "fact_spans": "[[[0, 14]], [[0, 14]]]", "query_spans": "[[[0, 21]]]", "process": "" }, { "text": "Given the hyperbola $\\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1$, if line $l$ intersects the hyperbola at points $P$ and $Q$, and the midpoint of segment $PQ$ is point $A(1,1)$, then the slope of line $l$ is?", "fact_expressions": "l: Line;G: Hyperbola;Q: Point;P: Point;A: Point;Expression(G) = (x^2/5 - y^2/4 = 1);Coordinate(A) = (1, 1);Intersection(l, G) = {P, Q};MidPoint(LineSegmentOf(P,Q))=A", "query_expressions": "Slope(l)", "answer_expressions": "4/5", "fact_spans": "[[[86, 91], [42, 47]], [[2, 40], [49, 52]], [[57, 60]], [[53, 56]], [[75, 84]], [[2, 40]], [[75, 84]], [[42, 62]], [[64, 84]]]", "query_spans": "[[[86, 96]]]", "process": "Let P(x_{1},y_{1}), Q(x_{2},y_{2}), then \\begin{cases}\\frac{x_{1}2}{5}-\\frac{y_{1}}{4}=1\\\\\\frac{x_{2}2}{x_{2}}-\\frac{y_{2}}{2}=1\\end{cases}\\Rightarrow\\frac{(x_{1}-x_{2})(x_{1}+x_{2})}{5}-\\frac{(}{\\frac{-y_{2})(y_{1}+y_{2})}{4}=\\Rightarrow\\frac{2(x_{1}}{}\\frac{-x_{2})}{-\\frac{2(y_{1}-y_{2})}{4}}{\\frac{2^{2}}{5}-\\frac{2}{4}}=0\\Rightarrowk=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{4}{5}" }, { "text": "Given that one of the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $y=2x$, what is the eccentricity of this hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G)))=(y=2*x)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[2, 58], [77, 80]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 74]]]", "query_spans": "[[[77, 86]]]", "process": "Since one of the asymptotes of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0, b>0) is given by y=2x, it follows that \\frac{b}{a}=2^{n}, so e=\\frac{c}{a}=\\sqrt{1+(\\frac{b}{a})^{2}}=\\sqrt{5}" }, { "text": "Given the parabola $E$: $y^{2}=12x$ has focus $F$, directrix $l$, a line $m$ passing through $F$ intersects $E$ at points $A$ and $B$. Draw $AM \\perp l$, with foot of perpendicular at $M$, let $N$ be the midpoint of $AM$. If $AM \\perp FN$, then $|AB|=$?", "fact_expressions": "m: Line;E: Parabola;A: Point;M: Point;F: Point;N: Point;B: Point;l: Line;Expression(E) = (y^2 = 12*x);Focus(E) = F;Directrix(E) = l;PointOnCurve(F,m);Intersection(m, E) = {A, B};PointOnCurve(A,LineSegmentOf(A,M));IsPerpendicular(LineSegmentOf(A,M),l);FootPoint(LineSegmentOf(A,M),l)=M;MidPoint(LineSegmentOf(A, M)) = N;IsPerpendicular(LineSegmentOf(A, M), LineSegmentOf(F, N))", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "16", "fact_spans": "[[[42, 47]], [[2, 22], [48, 51]], [[53, 56], [64, 67]], [[85, 88]], [[26, 29], [26, 29]], [[98, 101]], [[57, 60]], [[33, 36]], [[2, 22]], [[2, 29]], [[2, 36]], [[37, 47]], [[42, 62]], [[63, 81]], [[68, 81]], [[68, 88]], [[89, 101]], [[103, 118]]]", "query_spans": "[[[120, 129]]]", "process": "\\because AF=AM, N is the midpoint of AM, and FN\\bot AM, \\therefore \\angle AFN=30^{\\circ}, then the inclination angle of line AB is 60^{\\circ}, and the slope is \\sqrt{3}. From the parabola y^{2}=12x, we obtain F(3,0), then the equation of line AB is y=\\sqrt{3}(x-3). Solving simultaneously \\begin{cases}y=\\sqrt{3}(x-3)\\\\y^{2}=12x\\end{cases}, we get x^{2}-10x+9=0. Then x_{A}+x_{B}=10, \\therefore |AB|=x_{A}+x_{B}+p=16" }, { "text": "Given the circle $(x+2)^{2}+y^{2}=36$ with center $M$, let $A$ be any point on the circle, and let $N(2,0)$. The perpendicular bisector of segment $AN$ intersects $MA$ at point $P$. Then the trajectory equation of the moving point $P$ is?", "fact_expressions": "G: Circle;A: Point;N: Point;M: Point;P: Point;Expression(G) = (y^2 + (x + 2)^2 = 36);Coordinate(N) = (2, 0);Center(G)=M;PointOnCurve(A, G);Intersection(PerpendicularBisector(LineSegmentOf(A,N)), LineSegmentOf(M, A)) = P", "query_expressions": "LocusEquation(P)", "answer_expressions": "x^2/9+y^2/5=1", "fact_spans": "[[[2, 23], [36, 37]], [[32, 35]], [[43, 52]], [[27, 30]], [[81, 84], [73, 77]], [[2, 23]], [[43, 52]], [[2, 30]], [[32, 41]], [[53, 77]]]", "query_spans": "[[[81, 91]]]", "process": "First, from the equation of the circle, we obtain the center at M(-2,0) and radius r=6. Let point P(x,y). According to the given conditions and by symmetry, we have |PM| + |PN| = |PM| + |PA| = 6 > |MN| = 4. Therefore, the trajectory of point P is an ellipse with foci at M(-2,0) and N(2,0), and major axis length 2a=6. Then the result can be found. \nDetailed explanation: Since the circle (x+2)^{2}+y^{2}=36 has center M(-2,0) and radius r=6, let point P(x,y). Thus |PN| = |PA|, so |PM| + |PN| = |PM| + |PA| = |MA| = 6 > |MN| = 4. Therefore, the trajectory of point P is an ellipse with foci at M(-2,0) and N(2,0), and major axis length 2a=6. That is, b=\\sqrt{a^{2}-c^{2}}=\\sqrt{9-4}=\\sqrt{5}. Hence, the trajectory equation of moving point P is \\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1." }, { "text": "Let $F_{1}$, $F_{2}$ be the two foci of the hyperbola $x^{2}-\\frac{y^{2}}{24}=1$, and let $P$ be a point on the hyperbola such that $3 P F_{1}=4 P F_{2}$. Then the area of $\\triangle P F_{1} F_{2}$ equals?", "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2/24 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);3*LineSegmentOf(P, F1) = 4*LineSegmentOf(P, F2)", "query_expressions": "Area(TriangleOf(P, F1, F2))", "answer_expressions": "24", "fact_spans": "[[[17, 46], [56, 59]], [[52, 55]], [[1, 8]], [[9, 16]], [[17, 46]], [[1, 51]], [[52, 63]], [[65, 86]]]", "query_spans": "[[[88, 119]]]", "process": "First, find |F_{1}F_{2}| = 10 from the equation of the hyperbola; then, using 3|PF_{1}| = 4|PF_{2}|, solve to obtain |PF_{1}| = 8, |PF_{2}| = 6; from this, the area of \\triangle PF_{1}F_{2} can be determined. \nDetailed Solution: The two foci of the hyperbola x^{2} - \\frac{y^{2}}{24} = 1 are F_{1}(-5,0), F_{2}(5,0), and |F_{1}F_{2}| = 10. From 3|PF_{1}| = 4|PF_{2}|, let |PF_{2}| = x, then |PF_{1}| = \\frac{4}{3}x. By the property of the hyperbola, we have \\frac{4}{3}x - x = 2, solving gives x = 6. Therefore, |PF_{1}| = 8, |PF_{2}| = 6. Since |F_{1}F_{2}| = 10, it follows that \\angle F_{1}PF_{2} = 90^{\\circ}, thus the area of \\triangle PF_{1}F_{2} = \\frac{1}{2} \\times 8 \\times 6 = 24. The answer is 24. This problem examines the properties and applications of hyperbolas, and the calculation of triangle area, and is a basic problem." }, { "text": "Given that one of the asymptotes of hyperbola $C$ has the equation $y=2x$, write a standard equation for hyperbola $C$.", "fact_expressions": "C: Hyperbola;Expression(OneOf(Asymptote(C))) = (y = 2*x)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/2 - y^2/8 = 1", "fact_spans": "[[[2, 8], [27, 33]], [[2, 24]]]", "query_spans": "[[[27, 42]]]", "process": "From the hyperbola equation \\frac{x^{2}}{2}-\\frac{y^{2}}{8}=1, the asymptote equations are obtained as y=\\pm2x, satisfying the given condition." }, { "text": "The equation of a hyperbola that shares common foci with the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ and has two asymptotes perpendicular to each other is?", "fact_expressions": "H: Ellipse;Expression(H) = (x^2/9 + y^2/5 = 1);G: Hyperbola;Focus(H) = Focus(G);L1: Line;L2: Line;Asymptote(G) = {L1, L2};IsPerpendicular(L1, L2)", "query_expressions": "Expression(G)", "answer_expressions": "x^2-y^2=2", "fact_spans": "[[[1, 38]], [[1, 38]], [[55, 58]], [[0, 58]], [], [], [[45, 58]], [[45, 58]]]", "query_spans": "[[[55, 62]]]", "process": "" }, { "text": "The hyperbola $8 k x^{2}-k y^{2}=8$ has a focus at $(3,0)$. Find $k=?$", "fact_expressions": "G: Hyperbola;k: Number;Expression(G) = (8*(k*x^2) - k*y^2 = 8);Coordinate(OneOf(Focus(G)))=(3,0)", "query_expressions": "k", "answer_expressions": "1", "fact_spans": "[[[0, 24]], [[39, 42]], [[0, 24]], [[0, 37]]]", "query_spans": "[[[39, 44]]]", "process": "" }, { "text": "Given that $P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and $M$, $N$ are points on the circles $(x+5)^{2}+y^{2}=4$ and $(x-5)^{2}+y^{2}=1$ respectively, then the maximum value of $|P M|-|P N|$ is?", "fact_expressions": "G: Hyperbola;H: Circle;P: Point;M: Point;N: Point;Expression(G) = (x^2/9-y^2/16 = 1);Expression(C) = (y^2 + (x + 5)^2 = 4);PointOnCurve(P,RightPart(G));PointOnCurve(M,C);PointOnCurve(N,H);C:Circle;Expression(H)=(y^2 + (x-5)^2 = 1)", "query_expressions": "Max(Abs(LineSegmentOf(P,M)) - Abs(LineSegmentOf(P,N)))", "answer_expressions": "9", "fact_spans": "[[[6, 45]], [[83, 102]], [[2, 5]], [[52, 55]], [[56, 59]], [[6, 45]], [[62, 82]], [[2, 51]], [[52, 105]], [[52, 105]], [[62, 82]], [[83, 102]]]", "query_spans": "[[[107, 126]]]", "process": "\\because\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1,\\thereforea^{2}=9,b^{2}=16, then c^{2}=25, hence the two foci of the hyperbola are F_{1}(-5,0),F_{2}(5,0). F_{1}(-5,0),F_{2}(5,0) are also the centers of the two circles, with radii r_{1}=2 and r_{2}=1 respectively. Therefore, |PM|_{\\max}=|PF_{1}|+2, |PN|_{\\min}=|PF_{2}|-1, then (|PM|-|PN|)_{\\max}=|PM|_{\\max}-|PN|_{\\min}=(|PF_{1}|+2)-(|PF_{2}|-1)=|PF_{1}|-|PF_{2}|+3=2\\times3+3=9," }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are denoted by $F_{1}$ and $F_{2}$ respectively. Let $P$ be a point on the hyperbola $C$. If $|\\overrightarrow{P F_{1}}+\\overrightarrow{P F_{2}}|=\\sqrt{| \\overrightarrow{P F_{1}}|^{2}+|\\overrightarrow{P F_{2}}|^{2}}$ and $|\\overrightarrow{P F_{1}}|=2|\\overrightarrow{P F_{2}}|$, then what is the eccentricity of the hyperbola $C$?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) =F1;RightFocus(C)=F2;PointOnCurve(P, C);Abs(VectorOf(P, F1) + VectorOf(P, F2)) = sqrt(Abs(VectorOf(P, F1))^2 + Abs(VectorOf(P, F2))^2);Abs(VectorOf(P, F1)) = 2*Abs(VectorOf(P, F2))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[2, 63], [92, 98], [287, 293]], [[10, 63]], [[10, 63]], [[88, 91]], [[72, 79]], [[80, 87]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 87]], [[2, 87]], [[88, 102]], [[104, 228]], [[229, 285]]]", "query_spans": "[[[287, 299]]]", "process": "Let $|\\overrightarrow{PF_1}| + |\\overrightarrow{PF_2}| = \\sqrt{|\\overrightarrow{PF_1}|^2 + |\\overrightarrow{PF_2}|^2}$. Square both sides simultaneously, then $\\triangle PF_1F_2$ is a right triangle. By the definition of the hyperbola, we obtain $|\\overrightarrow{PF_2}| = 2a$, $|\\overrightarrow{PF_1}| = 2c$. Using the Pythagorean theorem to set up an equation yields an equation in terms of $a$ and $c$, and then the eccentricity can be found using the eccentricity formula.\n\nSolution: From $|\\overrightarrow{PF_1}| + |\\overrightarrow{PF_2}| = \\sqrt{|\\overrightarrow{PF_1}|^2 + |\\overrightarrow{PF_2}|^2}$, we get $(|\\overrightarrow{PF_1}| + |\\overrightarrow{PF_2}|)^2 = |\\overrightarrow{PF_1}|^2 + |\\overrightarrow{PF_2}|^2$, that is, \n$|\\overrightarrow{PF_1}|^2 + |\\overrightarrow{PF_2}|^2 + 2\\overrightarrow{PF_1} \\cdot \\overrightarrow{PF_2} = |\\overrightarrow{PF_1}|^2 + |\\overrightarrow{PF_2}|^2$. \nTherefore, $|\\overrightarrow{PF_1}|^2 + |\\overrightarrow{PF_2}|^2 + 2\\overrightarrow{PF_1} \\cdot \\overrightarrow{PF_2} = |\\overrightarrow{PF_1}|^2 + |\\overrightarrow{PF_2}|^2$, \nwe obtain $\\overrightarrow{PF_1} \\cdot \\overrightarrow{PF_2} = 0$, so $\\overrightarrow{PF_1} \\perp \\overrightarrow{PF_2}$, hence $\\triangle PF_1F_2$ is a right triangle. \nLet $|\\overrightarrow{PF_1}| = m$, then $|\\overrightarrow{PF_2}| = 2m$. By the definition of the hyperbola, we have $|\\overrightarrow{PF_2}| - |\\overrightarrow{PF_1}| = 2m - m = 2a$, so $m = 2a$, $|\\overrightarrow{PF_2}| = 2a$, $|\\overrightarrow{PF_1}| = 4a$, $|\\overrightarrow{F_1F_2}| = 2c$. \nIn the right triangle $\\triangle PF_1F_2$, $PF_1^2 + PF_2^2 = F_1F_2^2$, that is, \n$(4a)^2 + (2a)^2 = (2c)^2$, \nso $20a^2 = 4c^2$, thus $e^2 = \\frac{c^2}{a^2} = \\frac{20}{4} = 5$, therefore $e = \\sqrt{5}$." }, { "text": "If point $P$ lies on the curve represented by the equation $\\sqrt{(x-5)^{2}+y^{2}}-\\sqrt{(x+5)^{2}+y^{2}}=6$, and point $P$ also lies on the line $y=4$, then what is the x-coordinate of point $P$?", "fact_expressions": "G: Curve;H: Line;P: Point;Expression(G) = (sqrt((x-5)^2+y^2)-sqrt((x+5)^2+y^2)=6);Expression(H) = (y = 4);PointOnCurve(P, G);PointOnCurve(P,H)", "query_expressions": "XCoordinate(P)", "answer_expressions": "-3*sqrt(2)", "fact_spans": "[[[61, 63]], [[73, 80]], [[1, 5], [67, 71], [85, 89]], [[6, 63]], [[73, 80]], [[1, 66]], [[67, 83]]]", "query_spans": "[[[85, 95]]]", "process": "Problem Analysis: The equation $\\sqrt{(x-5)^{2}+y^{2}}-\\sqrt{(x+5)^{2}+y^{2}}=6$ simplifies to $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ $(x<0)$. Substituting $y=4$ gives $x^{2}=18$, $\\therefore x=-3\\sqrt{2}$" }, { "text": "If the chord of the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$ is bisected by the point $(2,1)$, then the equation of the line on which this chord lies is?", "fact_expressions": "G: Ellipse;H: LineSegment;Expression(G) = (x^2/8 + y^2/4 = 1);Coordinate(MidPoint(H)) = (2, 1);IsChordOf(H,G)", "query_expressions": "Expression(OverlappingLine(H))", "answer_expressions": "y=-x+3", "fact_spans": "[[[2, 39]], [], [[2, 39]], [[2, 52]], [[2, 41]]]", "query_spans": "[[[2, 66]]]", "process": "Let the coordinates of the two intersection points between the line and the ellipse be (x_{1},y_{1}), (x_{2},y_{2}). Substituting into the ellipse equation, use the point difference method to find the slope, then solve for the line equation in detail. Let the coordinates of the two intersection points between the line and the ellipse be (x_{1},y_{1}), (x_{2},y_{2}). Then \\frac{x_{1}^{2}}{8}+\\frac{y_{1}^{2}}{4}=1, \\frac{x_{2}^{2}}{8}+\\frac{y_{2}^{2}}{4}=1. Subtracting these two equations gives \\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{8}+\\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{4}=0. Since the chord is bisected by the point (2,1), we have x_{1}+x_{2}=4, y_{1}+y_{2}=2. Therefore, \\frac{4(x_{1}-x_{2})}{8}+\\frac{2(y_{1}-y_{2})}{4}=0, which simplifies to \\frac{x_{1}-x_{2}}{2}+\\frac{y_{1}-y_{2}}{2}=0, i.e., \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-1. Hence, the slope of the line is -1. Therefore, the equation of the line is: y-1=-1\\times(x-2), i.e., y=-x+3." }, { "text": "If $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, respectively, and a line passing through point $F_{1}$ intersects the ellipse at points $A$ and $B$, then what is the perimeter of $\\Delta A F_{2} B$?", "fact_expressions": "G: Ellipse;H: Line;A: Point;F2: Point;B: Point;F1: Point;Expression(G) = (x^2/16 + y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, H);Intersection(H, G) = {A, B}", "query_expressions": "Perimeter(TriangleOf(A, F2, B))", "answer_expressions": "16", "fact_spans": "[[[19, 57], [77, 79]], [[74, 76]], [[81, 84]], [[9, 16]], [[85, 88]], [[1, 8], [65, 73]], [[19, 57]], [[1, 63]], [[1, 63]], [[64, 76]], [[74, 90]]]", "query_spans": "[[[92, 115]]]", "process": "From the definition of the ellipse, we know |AF_{1}| + |AF_{2}| = 2a = |BF_{1}| + |BF_{2}|. Also, since |AB| = |AF_{1}| + |BF_{1}|, the perimeter of quadrilateral AF_{2}B is |AF_{2}| + |BF_{2}| + |AB| = |AF_{2}| + |BF_{2}| + |AF_{1}| + |BF_{1}| = 4a. For the ellipse \\frac{x^{2}}{16} + \\frac{y^{2}}{9} = 1, a = 4, so the perimeter of quadrilateral AF_{2}B is 4a = 16." }, { "text": "It is known that point $P$ is a moving point on the parabola $y^{2}=4 x$, and point $P$ lies in the first quadrant, $F$ is the focus of the parabola, the coordinates of point $A$ are $(-1,0)$. When $\\frac{|P F|}{|P A|}$ takes the minimum value, what is the equation of line $A P$?", "fact_expressions": "G: Parabola;P: Point;A: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (-1, 0);PointOnCurve(P, G);Quadrant(P)=1;Focus(G) = F;WhenMin(Abs(LineSegmentOf(P, F))/Abs(LineSegmentOf(P, A)))", "query_expressions": "Expression(LineOf(A,P))", "answer_expressions": "x-y+1=0", "fact_spans": "[[[7, 21], [40, 43]], [[2, 6], [26, 30]], [[47, 51]], [[36, 39]], [[7, 21]], [[47, 63]], [[2, 24]], [[26, 35]], [[36, 46]], [[64, 91]]]", "query_spans": "[[[92, 104]]]", "process": "" }, { "text": "If an ellipse centered at the origin with coordinate axes as symmetry axes passes through the point $P(3,0)$, and the length of the major axis is $\\sqrt{3}$ times the length of the minor axis, then its standard equation is?", "fact_expressions": "G: Ellipse;P: Point;O: Origin;Coordinate(P) = (3, 0);Center(G) = O;SymmetryAxis(G)=axis;PointOnCurve(P,G);Length(MajorAxis(G))=sqrt(3)*Length(MinorAxis(G))", "query_expressions": "Expression(G)", "answer_expressions": "{x^2/9+y^2/3=1,y^2/27+x^2/9=1}", "fact_spans": "[[[15, 17], [50, 51]], [[18, 27]], [[4, 6]], [[18, 27]], [[1, 17]], [[7, 17]], [[15, 27]], [[15, 48]]]", "query_spans": "[[[50, 57]]]", "process": "Discuss separately the two cases where the foci of the ellipse lie on the x-axis and on the y-axis, using the method of undetermined coefficients. According to $a=\\sqrt{3}b$ and the fact that the ellipse passes through $P(3,0)$, the result can be obtained.\n\n① When the foci of the ellipse lie on the x-axis, let its equation be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. Since the length of the major axis is $\\sqrt{3}$ times the length of the minor axis, we have $a=\\sqrt{3}b$. Also, since the ellipse passes through $P(3,0)$, we get $\\frac{9}{a^{2}}=1$, solving gives: $a^{2}=9$, therefore $b^{2}=\\frac{a^{2}}{3}=3$, thus the standard equation is $\\frac{x^{2}}{9}+\\frac{y^{2}}{3}=1$.\n\n② When the foci of the ellipse lie on the y-axis, let its equation be $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$ $(a>b>0)$. Since the length of the major axis is $\\sqrt{3}$ times the length of the minor axis, we have $a=\\sqrt{3}b$. Also, since the ellipse passes through $P(3,0)$, we get $\\frac{9}{b^{2}}=1$, solving gives: $b^{2}=9$, therefore $a^{2}=3b^{2}=27$, thus the standard equation is $\\frac{y^{2}}{27}+\\frac{x^{2}}{9}=1$.\n\nIn summary, the required standard equations of the ellipse are $\\frac{x^{2}}{9}+\\frac{y^{2}}{3}=1$ or $\\frac{y^{2}}{27}+\\frac{x^{2}}{9}=1$." }, { "text": "The focus of the parabola $C$: $y^{2}=4 x$ is $F$. Let the line $l$ passing through point $F$ intersect the parabola at points $A$ and $B$, and $|A F|=\\frac{4}{3}$. Then $|B F|$=?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, F)) = 4/3", "query_expressions": "Abs(LineSegmentOf(B, F))", "answer_expressions": "4", "fact_spans": "[[[0, 18], [40, 43]], [[0, 18]], [[22, 25], [29, 33]], [[0, 25]], [[34, 39]], [[28, 39]], [[44, 47]], [[48, 51]], [[34, 53]], [[55, 74]]]", "query_spans": "[[[77, 86]]]", "process": "Let the x-coordinates of points A and B be $x_{A}$, $x_{B}$. By the focal radius formula, we get $|AF| = \\frac{4}{3} = x_{A} + 1 \\Rightarrow x_{A} = \\frac{1}{3}, y_{A} = \\pm\\frac{2\\sqrt{3}}{3}$. When $y_{A} = \\frac{2\\sqrt{3}}{3}$, $k_{AB} = k_{AF} = \\frac{\\frac{2\\sqrt{3}}{3}}{1 - \\frac{1}{3}} = -\\sqrt{3}$, the equation of AB is $y = -\\sqrt{3}(x - 1)$, solving simultaneously with $y^{2} = 4x$ gives $3x^{2} - 10x + 3 = 0$, solving yields $x_{B} = 3$, so $|BF| = 3 + 1 = 4$. Similarly, when $y_{A} = -\\frac{2\\sqrt{3}}{3}$, $|BF| = 4$." }, { "text": "Write an equation of a parabola $C$ that simultaneously satisfies the following conditions: \n($1$) The vertex of $C$ is at the coordinate origin; ($2$) The axis of symmetry of $C$ is one of the coordinate axes; ($3$) The distance from the focus of $C$ to its directrix is $\\frac{3}{4}$.", "fact_expressions": "C: Parabola;O:Origin;Vertex(C)=O;SymmetryAxis(C)=axis;Distance(Focus(C),Directrix(C))=3/4", "query_expressions": "Expression(C)", "answer_expressions": "y^2=(3/2)*x", "fact_spans": "[[[13, 19], [30, 33], [47, 50], [64, 67], [71, 72]], [[37, 41]], [[30, 41]], [[47, 58]], [[64, 91]]]", "query_spans": "[[[13, 24]]]", "process": "" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{2}=1(a>0)$ has eccentricity $\\sqrt{3} a$, then the asymptotes of this hyperbola are?", "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2/2 + x^2/a^2 = 1);Eccentricity(G) = sqrt(3)*a", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*sqrt(2)*x", "fact_spans": "[[[2, 49], [69, 72]], [[5, 49]], [[5, 49]], [[2, 49]], [[2, 66]]]", "query_spans": "[[[69, 78]]]", "process": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{2}=1$ ($a>0$) has eccentricity $\\sqrt{3}a$, which gives: $\\frac{\\sqrt{a^{2}+2}}{a}=\\sqrt{3}a$. Solving for $a$ yields $a=1$. Therefore, the equation of the hyperbola is: $\\frac{x^{2}}{1}-\\frac{y^{2}}{2}=1$. Hence, the asymptotes of this hyperbola are $y=\\pm\\sqrt{2}x$." }, { "text": "The distance from the focus $F$ of the parabola $x^{2}=2 y$ to the directrix $l$ is?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 2*y);F: Point;Focus(G) = F;l: Line;Directrix(G) = l", "query_expressions": "Distance(F, l)", "answer_expressions": "1/2", "fact_spans": "[[[0, 14]], [[0, 14]], [[17, 20]], [[0, 20]], [[23, 26]], [[0, 26]]]", "query_spans": "[[[0, 31]]]", "process": "" }, { "text": "The ellipse $a x^{2}+b y^{2}=1$ intersects the line $y=1-x$ at points $A$ and $B$. The slope of the line passing through the origin and the midpoint of segment $AB$ is $\\frac{\\sqrt{3}}{2}$, \nthen the value of $\\frac{a}{b}$ is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;B: Point;H: Line;O: Origin;V: Line;Expression(G) = (a*x^2 + b*y^2 = 1);Expression(H) = (y = 1 - x);Intersection(G, H) = {A, B};PointOnCurve(O, V);PointOnCurve(MidPoint(LineSegmentOf(A, B)), V);Slope(V) = sqrt(3)/2", "query_expressions": "a/b", "answer_expressions": "sqrt(3)/2", "fact_spans": "[[[0, 21]], [[2, 21]], [[2, 21]], [[33, 36]], [[37, 40]], [[22, 31]], [[44, 46]], [[57, 59]], [[0, 21]], [[22, 31]], [[0, 42]], [[43, 59]], [[43, 59]], [[57, 83]]]", "query_spans": "[[[86, 103]]]", "process": "" }, { "text": "Given that the right focus of the hyperbola $x^{2}-y^{2}=4 a(a \\in R , a \\neq 0)$ is a vertex of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, then $a$=?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 4*a);a: Real;Negation(a=0);H: Ellipse;Expression(H) = (x^2/16 + y^2/9 = 1);RightFocus(G) = OneOf(Vertex(H))", "query_expressions": "a", "answer_expressions": "2", "fact_spans": "[[[2, 42]], [[2, 42]], [[92, 95]], [[5, 42]], [[47, 85]], [[47, 85]], [[2, 90]]]", "query_spans": "[[[92, 97]]]", "process": "" }, { "text": "If the eccentricity of the hyperbola $x^{2}-\\frac{y^{2}}{m}=1$ is $\\sqrt{3}$, then the real number $m$=?", "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (x^2 - y^2/m = 1);Eccentricity(G) = sqrt(3)", "query_expressions": "m", "answer_expressions": "2", "fact_spans": "[[[1, 29]], [[46, 51]], [[1, 29]], [[1, 44]]]", "query_spans": "[[[46, 53]]]", "process": "a^{2}=1,b^{2}=m,e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{a^{2}+b^{2}}{a^{2}}=1+m=3,m=2. The equations of the asymptotes are y=\\pm\\sqrt{m}x=\\pm\\sqrt{2}x" }, { "text": "The standard equation of a parabola with focus coordinates $(-2,0)$ is?", "fact_expressions": "G: Parabola;Coordinate(Focus(G)) = (-2, 0)", "query_expressions": "Expression(G)", "answer_expressions": "y^2=-8*x", "fact_spans": "[[[14, 17]], [[0, 17]]]", "query_spans": "[[[14, 24]]]", "process": "Problem Analysis: According to the given conditions, \\frac{p}{2}=2, p=4. Since the parabola opens to the left, the standard equation of the parabola is y^{2}=-8x" }, { "text": "Given that the circle $x^{2}+y^{2}-6 x-7=0$ is tangent to the directrix of the parabola $y^{2}=2 p x(p>0)$, then $p=?$", "fact_expressions": "G: Parabola;p: Number;H: Circle;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (-6*x + x^2 + y^2 - 7 = 0);IsTangent(H, Directrix(G))", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[25, 46]], [[53, 56]], [[2, 24]], [[28, 46]], [[25, 46]], [[2, 24]], [[2, 51]]]", "query_spans": "[[[53, 58]]]", "process": "Problem analysis: $x^{2}+y^{2}-6x-7=0 \\therefore (x-3)^{2}+y^{2}=16$, the center is $(3,0)$, radius is 4, the parabola's directrix is $x=-\\frac{p}{2}$, since the circle is tangent to the line, we have $\\frac{p}{2}=1 \\therefore p=2$" }, { "text": "Given the parabola $C$: $x^{2}=2 p y(p>0)$ with focus $F$. Point $P(4,4)$ lies on $C$, then $|P F|$=?", "fact_expressions": "C: Parabola;p: Number;P: Point;F: Point;p>0;Expression(C) = (x^2 = 2*(p*y));Coordinate(P) = (4, 4);Focus(C) = F;PointOnCurve(P, C)", "query_expressions": "Abs(LineSegmentOf(P, F))", "answer_expressions": "5", "fact_spans": "[[[2, 28], [46, 49]], [[10, 28]], [[36, 45]], [[32, 35]], [[10, 28]], [[2, 28]], [[36, 45]], [[2, 35]], [[36, 50]]]", "query_spans": "[[[52, 61]]]", "process": "Substituting the coordinates of point P(4,4) into the parabola C: x^{2}=2py, we obtain p=2, so the equation of the parabola is x^{2}=4y. Therefore, |PF|=4+\\frac{p}{2}=5." }, { "text": "It is known that the center of ellipse $G$ is at the origin, the major axis lies on the $x$-axis, the eccentricity is $\\frac{\\sqrt{3}}{2}$, and the sum of the distances from a point on $G$ to the two foci of $G$ is $12$. Then the equation of ellipse $G$ is?", "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;OverlappingLine(MajorAxis(G), xAxis);Eccentricity(G) = sqrt(3)/2;P: Point;F1: Point;F2: Point;PointOnCurve(P, G);Distance(P, F1) + Distance(P, F2) = 12;Focus(G) = {F1, F2}", "query_expressions": "Expression(G)", "answer_expressions": "x^2/36 + y^2/9 = 1", "fact_spans": "[[[2, 7], [49, 52], [56, 59], [75, 80]], [[11, 13]], [[2, 13]], [[2, 22]], [[2, 47]], [[54, 55]], [[62, 63]], [[62, 63]], [[49, 55]], [[49, 73]], [[56, 63]]]", "query_spans": "[[[75, 85]]]", "process": "" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{3}+y^{2}=1$, a chord $AB$ passes through the point $(-1,0)$. Then the trajectory equation of the midpoint of chord $AB$ is?", "fact_expressions": "C: Ellipse;G: Point;A: Point;B: Point;Expression(C) = (x^2/3 + y^2 = 1);Coordinate(G) = (-1, 0);IsChordOf(LineSegmentOf(A,B),C);PointOnCurve(G,LineSegmentOf(A,B))", "query_expressions": "LocusEquation(MidPoint(LineSegmentOf(A, B)))", "answer_expressions": "x^2+x+3*y^2=0", "fact_spans": "[[[2, 34]], [[42, 51]], [[36, 41]], [[36, 41]], [[2, 34]], [[42, 51]], [[2, 41]], [[36, 51]]]", "query_spans": "[[[54, 68]]]", "process": "Problem Analysis: Let A(x_{1},y_{1}), B(x_{2},y_{2}), and the midpoint of AB be C(x,y). If the slope of line AB exists: \\frac{x_{1}^{2}}{3}+y_{1}^{2}=1\\textcircled{1}, \\frac{x_{2}^{2}}{3}+y_{2}^{2}=1\\textcircled{2}, \\textcircled{1}-\\textcircled{2} yields \\frac{x_{1}+x_{2}}{3}+(y_{1}+y_{2})\\times\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=0, that is, \\frac{2x}{3}+2y\\times\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=0, leading to x^{2}+x+3y^{2}=0; if the slope of AB does not exist: C(-1,0) also satisfies the equation. In conclusion, the trajectory equation of the midpoint of AB is x^{2}+x+3y^{2}=0." }, { "text": "A hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$, what is the equation of one of its asymptotes?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/2 = 1)", "query_expressions": "Expression(OneOf(Asymptote(G)))", "answer_expressions": "y=pm*(sqrt(2)/2)*x", "fact_spans": "[[[0, 38]], [[0, 38]]]", "query_spans": "[[[0, 48]]]", "process": "Problem Analysis: Find the values of a=2, b=\\sqrt{2} for the hyperbola, then use the asymptote equations y=\\pm\\frac{b}{a} to obtain the result. \nSolution: For the hyperbola \\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1, we have a=2, b=\\sqrt{2}. Thus, the equations of the asymptotes are y=\\pm\\frac{\\sqrt{2}}{2}x" }, { "text": "Given that point $M$ is a moving point on the ellipse $\\frac{y^{2}}{25}+\\frac{x^{2}}{16}=1$, the coordinates of point $T$ are $(0,-3)$, and point $N$ satisfies $|NT|=1$, and $\\angle MNT=90^{\\circ}$, then the maximum value of $|MN|$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/25 = 1);M: Point;PointOnCurve(M, G) = True;T: Point;Coordinate(T) = (0, -3);N: Point;Abs(LineSegmentOf(N, T)) = 1;AngleOf(M, N, T) = ApplyUnit(90, degree)", "query_expressions": "Max(Abs(LineSegmentOf(M, N)))", "answer_expressions": "3*sqrt(7)", "fact_spans": "[[[7, 46]], [[7, 46]], [[2, 6]], [[2, 51]], [[52, 56]], [[52, 68]], [[69, 73]], [[75, 84]], [[86, 111]]]", "query_spans": "[[[113, 126]]]", "process": "Solution: As shown in the figure, M lies on the ellipse, and N lies on a circle centered at T with radius 1. From the ellipse equation \\frac{y^{2}}{25}+\\frac{x^{2}}{16}=1, we find that point T is the lower focus of the ellipse. To maximize |MN|, we need to maximize |MT|. The maximum value of |MT| is |MT|_{\\max}=5+3=8. Therefore, |MN|_{\\max}=\\sqrt{8^{2}-1^{2}}=3\\sqrt{7}." }, { "text": "Given the parabola $y^{2}=4x$ with focus $F$, $A(-1,0)$, and point $P$ being a moving point on the parabola, when the value of $|PF|$ is minimized, $|PF|=$?", "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (-1, 0);Focus(G) = F;PointOnCurve(P, G);WhenMin(Abs(LineSegmentOf(P, F)))", "query_expressions": "Abs(LineSegmentOf(P, F))", "answer_expressions": "2", "fact_spans": "[[[2, 16], [41, 44]], [[26, 35]], [[36, 40]], [[20, 23]], [[2, 16]], [[26, 35]], [[2, 23]], [[36, 48]], [[50, 63]]]", "query_spans": "[[[64, 73]]]", "process": "The equation of the directrix of the parabola is x = -1. Let the distance from P to the directrix be |PQ|, then |PQ| = |PF|, \\therefore \\frac{|PF|}{|PA|} = \\frac{|PQ|}{|PA|} = \\sin\\angle PAQ. Therefore, when PA is tangent to the parabola y^{2} = 4x, \\angle PAQ is minimized, i.e., \\frac{|PF|}{|PA|} reaches its minimum value. Suppose the line passing through point A, y = kx + k, is tangent to the parabola (k \\neq 0). Substituting into the parabola equation gives k^{2}x^{2} + (2k^{2} - 4)x + k^{2} = 0. \\therefore \\Delta = (2k^{2} - 4)^{2} - 4k^{4} = 0. Solving yields k = \\pm 1. Then we have x^{2} - 2x + 1 = 0, solving gives x = 1. Substituting x = 1 into y^{2} = 4x gives y = \\pm 2. \\therefore P(1,2) or P(1,-2), \\therefore |PF| = 2." }, { "text": "Given that $A$ and $B$ are two points on the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and the midpoint of segment $AB$ lies on the circle $x^{2}+y^{2}=1$, then the range of the $y$-intercept of line $AB$ is?", "fact_expressions": "C: Ellipse;G: Circle;A: Point;B: Point;Expression(C) = (x^2/4 + y^2/3 = 1);Expression(G) = (x^2 + y^2 = 1);PointOnCurve(A,C);PointOnCurve(B,C);PointOnCurve(MidPoint(LineSegmentOf(A,B)), G)", "query_expressions": "Range(Intercept(LineOf(A,B),yAxis))", "answer_expressions": "(-oo,-1]+[1,+oo)", "fact_spans": "[[[10, 52]], [[67, 83]], [[2, 5]], [[6, 9]], [[10, 52]], [[67, 83]], [[2, 55]], [[2, 55]], [[56, 84]]]", "query_spans": "[[[86, 108]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), the midpoint of AB be D(x_{0},y_{0}), and the line AB: y = kx + m. Substitute into the ellipse equation and eliminate y to obtain a quadratic equation in x. Using Vieta's formulas, find x_{1}+x_{2} and x_{1}x_{2}, then compute x_{0} = \\frac{x_{1}+x_{2}}{2}, y_{0} = \\frac{y_{1}+y_{2}}{2} expressed in terms of k and m. Substitute these into x^{2}+y^{2}=1, treat it as an equation in k^{2}, and use the distribution of roots of a quadratic function to determine the range of m.\n\nLet the line AB: y = kx + m, A(x_{1},y_{1}), B(x_{2},y_{2}), the midpoint of AB be D(x_{0},y_{0}). From \n\\begin{cases} y = kx + m \\\\ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 \\end{cases} \nwe get: \n3x^{2} + 4(kx + m)^{2} = 12, \ni.e., (3 + 4k^{2})x^{2} + 8mkx + 4m^{2} - 12 = 0. \n\nThus, \nx_{1} + x_{2} = \\frac{-8km}{3 + 4k^{2}}, \\quad \nx_{1}x_{2} = \\frac{4m^{2} - 12}{3 + 4k^{2}}, \\quad \ny_{1} + y_{2} = k(x_{1} + x_{2}) + 2m = \\frac{-8k^{2}m}{3 + 4k^{2}} + 2m = \\frac{6m}{3 + 4k^{2}}. \n\nTherefore, \nx_{0} = \\frac{-4km}{3 + 4k^{2}}, \\quad \ny_{0} = \\frac{3m}{3 + 4k^{2}}. \n\nSince point D(x_{0}, y_{0}) lies on the circle x^{2} + y^{2} = 1, \n\\left(\\frac{-4km}{3 + 4k^{2}}\\right)^{2} + \\left(\\frac{3m}{3 + 4k^{2}}\\right)^{2} = 1. \nSimplifying yields: \n16k^{4} + (24 - 16m^{2})k^{2} + 9 - 9m^{2} = 0. \n\nLet t = k^{2}, then \n16t^{2} + (24 - 16m^{2})t + 9 - 9m^{2} = 0 \nhas a solution for t \\geqslant 0. Therefore, \n\\Delta = (24 - 16m^{2})^{2} - 4 \\times 16 \\times (9 - 9m^{2}) \\geqslant 0. \nSimplifying gives: \nm^{2} \\geqslant \\frac{3}{4}, \nso \nm \\geqslant \\frac{\\sqrt{3}}{2} \\quad \\text{or} \\quad m \\leqslant -\\frac{\\sqrt{3}}{2}. \\quad \\textcircled{1} \n\nThe axis of symmetry is \nx = \\frac{16m^{2} - 24}{32} \\geqslant 0, \nsolving gives: \nm \\geqslant \\frac{\\sqrt{6}}{2} \\quad \\text{or} \\quad m \\leqslant -\\frac{\\sqrt{6}}{2}, \nin which case the equation certainly has a solution for t \\geqslant 0. \n\n\\textcircled{2} From \n\\begin{cases} \nx = \\frac{16m^{2} - 24}{32} < 0 \\\\ \nf(0) = 9 - 9m^{2} \\leqslant 0 \n\\end{cases}, \nwe get \n-\\frac{\\sqrt{6}}{2} < m \\leqslant -1 \\quad \\text{or} \\quad 1 \\leqslant m < \\frac{\\sqrt{6}}{2}. \n\nIn summary: \nm \\leqslant -1 \\quad \\text{or} \\quad m \\geqslant 1. \n\nTherefore, the range of the intercept of line AB on the y-axis is (-\\infty, -1] \\cup [1, +\\infty)." }, { "text": "If the hyperbola $x^{2}+m y^{2}=1$ passes through the point $(-\\sqrt{2}, 2)$, then the length of the imaginary axis of this hyperbola is?", "fact_expressions": "G: Hyperbola;m: Number;H: Point;Expression(G) = (m*y^2 + x^2 = 1);Coordinate(H) = (-sqrt(2), 2);PointOnCurve(H, G)", "query_expressions": "Length(ImageinaryAxis(G))", "answer_expressions": "4", "fact_spans": "[[[1, 21], [43, 46]], [[4, 21]], [[22, 40]], [[1, 21]], [[22, 40]], [[1, 40]]]", "query_spans": "[[[43, 52]]]", "process": "Problem Analysis: From the given condition, we have 2 + 4m = 1, so m = -\\frac{1}{4}; therefore, the length of the imaginary axis of the hyperbola is 2 \\times 2 = 4." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. Point $A$ is a point on the left branch of the hyperbola. The line $A F_{1}$ is parallel to the line $y=\\frac{b}{a} \\cdot x$, and the perimeter of $\\triangle A F_{1} F_{2}$ is $8 a$. Then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;A: Point;PointOnCurve(A, LeftPart(G));IsParallel(LineOf(A, F1), H);H: Line;Expression(H) = (y = x*(b/a));Perimeter(TriangleOf(A, F1, F2)) = 8*a", "query_expressions": "Eccentricity(G)", "answer_expressions": "2*sqrt(2)-1", "fact_spans": "[[[2, 58], [88, 91], [174, 177]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[67, 74]], [[75, 82]], [[2, 82]], [[2, 82]], [[83, 87]], [[83, 97]], [[98, 137]], [[110, 135]], [[110, 135]], [[138, 172]]]", "query_spans": "[[[174, 183]]]", "process": "According to the definition of hyperbola, analyze the relationship between AF_{1} and AF_{2}; then, since the line AF_{1} is parallel to the line y=\\frac{b}{a}\\cdot x, we can obtain the expression for \\tan\\angle AF_{1}F_{2}, and further find \\cos\\angle AF_{1}F_{2}; finally, use the cosine law in the focal triangle to simplify and solve. From the given condition, let AF_{1}=x, then AF_{2}=x+2a. Also, since the perimeter of triangle AF_{1}F_{2} is 8a, we have x+x+2a+2c=8a \\Rightarrow x=3a-c. Since the line AF_{1} is parallel to the line y=\\frac{b}{a}\\cdot x, it follows that \\tan\\angle AF_{1}F_{2}=\\frac{b}{a}, hence \\cos\\angle AF_{1}F_{2}=\\frac{a}{\\sqrt{a^{2}+b^{2}}}=\\frac{a}{c}. By the cosine law, we get (5a-c)^{2}=(3a-c)^{2}+(2c)^{2}-2(3a-c)\\cdot(2c)\\cos\\angle AF_{2}F_{1}. Simplifying yields c^{2}+2ac-7a^{2}=0. Dividing through by a^{2} gives e^{2}+2e-7=0, hence e=\\frac{-2\\pm\\sqrt{4+28}}{2}=-1\\pm2\\sqrt{2}." }, { "text": "If the focus of the parabola $y^{2}=2 p x(p>0)$ is a focus of the hyperbola $\\frac{x^{2}}{2 p}-\\frac{y^{2}}{p}=1$, then $p$=?", "fact_expressions": "G: Hyperbola;p: Number;H: Parabola;Expression(G) = (x^2/(2*p) - y^2/p = 1);p>0;Expression(H) = (y^2 = 2*(p*x));Focus(H) = OneOf(Focus(G))", "query_expressions": "p", "answer_expressions": "12", "fact_spans": "[[[26, 66]], [[73, 76]], [[1, 22]], [[26, 66]], [[4, 22]], [[1, 22]], [[1, 71]]]", "query_spans": "[[[73, 78]]]", "process": "Since the focus of the parabola $y^{2}=2px$ ($p>0$) is $\\left(\\frac{p}{2},0\\right)$, the right focus of the hyperbola is $\\left(\\frac{p}{2},0\\right)$. Also, since the coordinates of the right focus of the hyperbola $\\frac{x^{2}}{2p}-\\frac{y^{2}}{p}=1$ are $\\left(\\sqrt{3p},0\\right)$, we have $\\frac{p}{2}=\\sqrt{3p}$. Solving this gives $p=12$. The final answer is: 12" }, { "text": "Given that one of the directrices of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ ($a>0$) is $x=\\frac{3}{2}$, then what is the value of $a$? What is the eccentricity of this hyperbola?", "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(OneOf(Directrix(G))) = (x = 3/2)", "query_expressions": "a;Eccentricity(G)", "answer_expressions": "sqrt(3)\n2*sqrt(3)/3", "fact_spans": "[[[2, 39], [71, 74]], [[64, 67]], [[5, 39]], [[2, 39]], [[2, 62]]]", "query_spans": "[[[64, 70]], [[71, 80]]]", "process": "" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{2}=1(a>0)$ has eccentricity $\\sqrt{3 a}$, then the asymptotes of this hyperbola are given by?", "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2/2 + x^2/a = 1);Eccentricity(G) = sqrt(3*a)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y = pm*sqrt(2)*x", "fact_spans": "[[[2, 45], [65, 68]], [[5, 45]], [[5, 45]], [[2, 45]], [[2, 62]]]", "query_spans": "[[[65, 76]]]", "process": "According to the problem, we have \\frac{\\sqrt{a+2}}{\\sqrt{a}}=\\sqrt{3a}, which leads to 3a^{2}-a-2=0. Solving gives a=1, so the equations of the asymptotes are y=\\pm\\frac{b}{a}x=\\pm\\sqrt{2}x" }, { "text": "Given that the moving point $M$ is equidistant from $A(2,1)$ and $B(3,4)$, then the equation of the trajectory of point $M$ is?", "fact_expressions": "A: Point;B: Point;M:Point;Coordinate(A) = (2, 1);Coordinate(B) = (3, 4);Distance(M,A)=Distance(M,B)", "query_expressions": "LocusEquation(M)", "answer_expressions": "x+2*y-10=0", "fact_spans": "[[[8, 16]], [[21, 29]], [[36, 40], [4, 7]], [[8, 16]], [[21, 29]], [[4, 34]]]", "query_spans": "[[[36, 47]]]", "process": "Let the coordinates of point M be (x, y). According to the given condition, |MA| = |MB|, that is, \\sqrt{(x-2)^{2}+(y-1)^{2}}=\\sqrt{(x-3)^{2}+(y-4)^{2}}. Squaring both sides and simplifying, we obtain x + 3y - 10 = 0. Hence, the trajectory equation of point M is x + 3y - 10 = 0." }, { "text": "The focal width of the hyperbola $\\frac{x^{2}}{m^{2}+12}-\\frac{y^{2}}{4-m^{2}}=1$ is?", "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/(m^2 + 12) - y^2/(4 - m^2) = 1)", "query_expressions": "FocalLength(G)", "answer_expressions": "8", "fact_spans": "[[[0, 51]], [[3, 51]], [[0, 51]]]", "query_spans": "[[[0, 56]]]", "process": "" }, { "text": "Given that the eccentricity of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{6}=1$ is $\\frac{1}{2}$, then $m=?$", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/6 + x^2/m = 1);m: Number;Eccentricity(G) = 1/2", "query_expressions": "m", "answer_expressions": "{9/2, 8}", "fact_spans": "[[[2, 39]], [[2, 39]], [[59, 62]], [[2, 57]]]", "query_spans": "[[[59, 64]]]", "process": "Solution: \n∵ The ellipse is $\\frac{x^2}{m} + \\frac{y^2}{6} = 1$, \n∴ $\\textcircled{1}$ When the foci of the ellipse lie on the x-axis, $a^2 = m$, $b^2 = 6$, we obtain $c = \\sqrt{m - 6}$; the eccentricity is $e = \\frac{\\sqrt{m - 6}}{\\sqrt{m}} = \\frac{1}{2}$, solving gives $m = 8$; \n$\\textcircled{2}$ When the foci of the ellipse lie on the y-axis, $a^2 = 6$, $b^2 = m$, we obtain $c = \\sqrt{6 - m}$; the eccentricity is $e = \\frac{\\sqrt{6 - m}}{\\sqrt{6}} = \\frac{1}{2}$, solving gives $m = \\frac{9}{2}$. \nIn summary, $m = \\frac{9}{2}$ or $8$." }, { "text": "Let $F$ be the focus of the parabola $C$: $y^{2}=4 x$. The line passing through $F$ with an inclination angle of $60^{\\circ}$ intersects $C$ at points $A$ and $B$. Then $|A B|=$?", "fact_expressions": "C: Parabola;G: Line;A: Point;B: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F, G);Inclination(G) = ApplyUnit(60, degree);Intersection(G, C) = {A, B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "16/3", "fact_spans": "[[[5, 24], [53, 56]], [[50, 52]], [[57, 60]], [[61, 64]], [[1, 4], [29, 32]], [[5, 24]], [[1, 27]], [[28, 52]], [[33, 52]], [[50, 66]]]", "query_spans": "[[[68, 77]]]", "process": "According to the given conditions, the equation of line AB is: $ y = \\sqrt{3}x - \\sqrt{3} $, and by combining it with the parabola we obtain: $ 3x^{2} - 10x + 3 = 0 $. Use the chord length formula to find $ |AB| $. Since the focus $ F(1,0) $, $ k_{AB} = \\tan60^{\\circ} = \\sqrt{3} $, therefore the equation of line AB is: $ y = \\sqrt{3}x - \\sqrt{3} $. Substituting into $ y^{2} = 4x $, simplifying gives: $ 3x^{2} - 10x + 3 = 0 $. Thus $ x = 3 $ or $ x = \\frac{1}{3} $, so we can take $ A\\left(\\frac{1}{3}, -\\frac{2\\sqrt{3}}{3}\\right) $, $ B(3, 2\\sqrt{3}) $. Therefore, $ |AB| = \\sqrt{\\left(\\frac{1}{3} - 3\\right)^{2} + \\left(-\\frac{2\\sqrt{3}}{3} - 2\\sqrt{3}\\right)^{2}} = \\frac{16}{3} $." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{3}=1$, a line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. If $|A F_{2}|+|B F_{2}|=8$, then $|A B|=$?", "fact_expressions": "F1: Point;F2: Point;Focus(G) = {F1, F2};G: Ellipse;Expression(G) = (x^2/9 + y^2/3 = 1);H: Line;PointOnCurve(F1, H);Intersection(H, G) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(A, F2)) + Abs(LineSegmentOf(B, F2)) = 8", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "4", "fact_spans": "[[[2, 9], [62, 69]], [[10, 17]], [[2, 60]], [[18, 55], [74, 76]], [[18, 55]], [[70, 72]], [[61, 72]], [[70, 86]], [[77, 80]], [[81, 84]], [[89, 112]]]", "query_spans": "[[[114, 123]]]", "process": "Since |AF_{2}|+|BF_{2}|=8, a=3(|AF_{1}|+|AF_{2}|)+(|BF_{1}|+|BF_{2}|)=|AB|+(|AF_{2}|+|BF_{2}|)=|AB|+8=12, therefore |AB|=4." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$, the line $l$: $x=4 a$ intersects the two asymptotes of hyperbola $C$ at points $A$ and $B$, respectively. If the area of $\\Delta O A B$ (point $O$ being the origin) is $32$, and the focal distance of hyperbola $C$ is $2 \\sqrt{5}$, then the eccentricity of hyperbola $C$ is?", "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;O: Origin;A: Point;B: Point;a>0;b>0;L1:Line;L2:Line;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(l)=(x=4*a);Asymptote(C)={L1,L2};Intersection(l,L1)=A;Intersection(l,L2)=B;Area(TriangleOf(O,A,B))=32;FocalLength(C) = 2*sqrt(5)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5), sqrt(5)/2", "fact_spans": "[[[65, 79]], [[2, 64], [80, 86], [142, 148], [166, 172]], [[10, 64]], [[10, 64]], [[121, 125]], [[96, 99]], [[100, 103]], [[10, 64]], [[10, 64]], [], [], [[2, 64]], [[65, 79]], [[80, 92]], [[65, 105]], [[65, 105]], [[106, 140]], [[142, 164]]]", "query_spans": "[[[166, 178]]]", "process": "" }, { "text": "Given $\\frac{1}{m}+\\frac{2}{n}=1(m>0 , n>0)$, when $m \\cdot n$ attains its minimum value, what is the eccentricity of the ellipse $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{n^{2}}=1$?", "fact_expressions": "G: Ellipse;n: Number;m: Number;m>0;n>0;2/n + 1/m = 1;WhenMin(m*n);Expression(G) = (y^2/n^2 + x^2/m^2 = 1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)/2", "fact_spans": "[[[61, 106]], [[43, 54]], [[43, 54]], [[2, 40]], [[2, 40]], [[2, 40]], [[42, 60]], [[61, 106]]]", "query_spans": "[[[61, 112]]]", "process": "" }, { "text": "Given that the distance from the focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ to its asymptote is $\\frac{a^{2}}{b}$, then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Distance(Focus(G),Asymptote(G))= a^2/b", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 58], [89, 92]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 86]]]", "query_spans": "[[[89, 98]]]", "process": "Solution: Take the right focus F(c,0) of the hyperbola, and take the asymptote of the hyperbola y=\\frac{b}{a}x, that is, bx-ay=0. According to the given condition, \\frac{bc}{\\sqrt{a^{2}+b^{2}}}=\\frac{a^{2}}{b}, which implies b=a. Therefore, the eccentricity e of the hyperbola is e=\\frac{c}{a}=\\frac{\\sqrt{a^{2}+b^{2}}}{a}=\\sqrt{2}." }, { "text": "The standard equation of a hyperbola with foci on the $x$-axis, focal distance $6$, and passing through the point $(\\sqrt{5}, 0)$ is?", "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (sqrt(5), 0);PointOnCurve(Focus(G), xAxis);FocalLength(G) = 6;PointOnCurve(H, G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/5 - y^2/4 = 1", "fact_spans": "[[[36, 39]], [[19, 35]], [[19, 35]], [[0, 39]], [[9, 39]], [[17, 39]]]", "query_spans": "[[[36, 46]]]", "process": "The foci are on the x-axis, the focal distance is 6, so c=3, and passing through the point (\\sqrt{5},0) gives a=\\sqrt{5}; thus, b^{2}=9-5=4. The standard equation of the hyperbola is: \\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1" }, { "text": "Point $P$ lies on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$. What are the maximum and minimum distances from point $P$ to the line $3x-4y=24$?", "fact_expressions": "G: Ellipse;H: Line;P: Point;Expression(G) = (x^2/16 + y^2/9 = 1);Expression(H) = (3*x - 4*y = 24);PointOnCurve(P, G)", "query_expressions": "Max(Distance(P,H));Min(Distance(P,H))", "answer_expressions": "12*(2+sqrt(2))/5\n12*(2-sqrt(2))/5", "fact_spans": "[[[5, 43]], [[50, 64]], [[0, 4], [45, 49]], [[5, 43]], [[50, 64]], [[0, 44]]]", "query_spans": "[[[45, 76]], [[45, 76]]]", "process": "Let the coordinates of point P be (4\\cos\\theta,3\\sin\\theta). The distance d from point P to the line 3x-4y=24 reaches its maximum when \\cos(\\theta+\\frac{\\pi}{4})=-1, and reaches its minimum when \\cos(\\theta+\\frac{\\pi}{4})=1, with the minimum value being \\frac{12(2-}{" }, { "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ have no common points with the circle $x^{2}+y^{2}-6 y+5=0$, then the range of the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;H: Circle;Expression(H) = (-6*y + x^2 + y^2 + 5 = 0);NumIntersection(Asymptote(G), H) = 0", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1, 3/2)", "fact_spans": "[[[0, 56], [90, 93]], [[0, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[61, 83]], [[61, 83]], [[0, 88]]]", "query_spans": "[[[90, 103]]]", "process": "First, find the asymptotes of the hyperbola from its equation, then use the condition that the distance from the center of the circle to the asymptote is greater than the radius to obtain the relationship between $a$ and $b$. Then, using $c^{2} = a^{2} + b^{2}$, derive the relationship between $a$ and $c$, so that the eccentricity of the hyperbola can be determined. \n【Solution】 $\\because$ the asymptotes of the hyperbola are $bx \\pm ay = 0$, and they have no common points with the circle $x^{2} + (y - 3)^{2} = 4$, \n$\\therefore$ the distance from the center of the circle to the asymptote is greater than the radius, i.e., $\\frac{3a}{\\sqrt{a^{2} + b^{2}}} > 2$. \n$\\therefore$ $9a^{2} > 4c^{2}$, hence $e = \\frac{c}{a} < \\frac{3}{2}$. \n(This question mainly examines the simple properties of hyperbolas, the positional relationship between a line and a circle, and the point-to-line distance formula. It tests the student's ability to apply number-shape combination reasoning, and is a medium-difficulty problem.)" }, { "text": "Let point $O$ be the vertex of the parabola, $F$ be the focus, and $P Q$ be a chord passing through the focus. Given that $|O F|=1$, $|P Q|=5$, then the area of $\\triangle O P Q$ is?", "fact_expressions": "G: Parabola;P: Point;Q: Point;O: Origin;F: Point;Vertex(G) = O;IsChordOf(LineSegmentOf(P,Q),G);PointOnCurve(F,LineSegmentOf(P,Q));Abs(LineSegmentOf(O, F)) = 1;Abs(LineSegmentOf(P,Q))=5;Focus(G)=F", "query_expressions": "Area(TriangleOf(O, P, Q))", "answer_expressions": "sqrt(5)", "fact_spans": "[[[6, 9]], [[20, 25]], [[20, 25]], [[1, 5]], [[13, 16]], [[1, 12]], [[6, 31]], [[13, 31]], [[34, 43]], [[44, 53]], [[6, 19]]]", "query_spans": "[[[55, 77]]]", "process": "" }, { "text": "Through the focus $F$ of the parabola $y^{2}=2 p x(p>0)$, draw a line with an inclination angle of $45^{\\circ}$ intersecting the parabola at points $A$ and $B$. If the length of segment $AB$ is $8$, then $p=?$", "fact_expressions": "A: Point;B: Point;G: Parabola;p: Number;H: Line;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G)=F;PointOnCurve(F, H);Inclination(H)=ApplyUnit(45, degree);Intersection(H,G) = {B, A};Length(LineSegmentOf(A, B)) = 8;F: Point", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[53, 56]], [[59, 62]], [[1, 22], [49, 52]], [[80, 83]], [[46, 48]], [[4, 22]], [[1, 22]], [[1, 28]], [[0, 48]], [[29, 48]], [[46, 64]], [[66, 78]], [25, 27]]", "query_spans": "[[[80, 85]]]", "process": "" }, { "text": "The hyperbola $\\frac{y^{2}}{4}-\\frac{x^{2}}{b^{2}}=1(b>0)$ has a focal distance of $8$, what is the equation of its asymptotes?", "fact_expressions": "G: Hyperbola;b: Number;b>0;Expression(G) = (y^2/4 - x^2/b^2 = 1);FocalLength(G) = 8", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*(sqrt(3)/3)*x", "fact_spans": "[[[0, 47]], [[3, 47]], [[3, 47]], [[0, 47]], [[0, 54]]]", "query_spans": "[[[0, 62]]]", "process": "Solution: Given the hyperbola \\frac{y^{2}}{4}-\\frac{x^{2}}{b^{2}}=1 (b>0) has a focal distance of 8, then c=4, a=2. Since a^{2}+b^{2}=c^{2}, we have 4+b^{2}=16, solving gives b=2\\sqrt{3}. Thus, the equation of the hyperbola is \\frac{y^{2}}{4}-\\frac{x^{2}}{12}=1, that is, the equations of the asymptotes of the hyperbola are y=\\pm\\frac{\\sqrt{3}}{3}x." }, { "text": "The left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ are $F_{1}$ and $F_{2}$, respectively. A point $M$ on the ellipse satisfies $\\angle F_{1} M F_{2}=60^{\\circ}$, and $\\overrightarrow{M F_{1}} \\cdot \\overrightarrow{M F_{2}}=4$, then $b$=?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;M: Point;PointOnCurve(M, G);AngleOf(F1, M, F2) = ApplyUnit(60, degree);DotProduct(VectorOf(M, F1), VectorOf(M, F2)) = 4", "query_expressions": "b", "answer_expressions": "sqrt(6)", "fact_spans": "[[[0, 52], [77, 79]], [[0, 52]], [[183, 186]], [[2, 52]], [[2, 52]], [[2, 52]], [[61, 68]], [[69, 76]], [[0, 76]], [[0, 76]], [[81, 85]], [[77, 85]], [[87, 120]], [[122, 181]]]", "query_spans": "[[[183, 188]]]", "process": "" }, { "text": "The shortest chord length passing through the right focus $F_{2}$ of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);F2: Point;RightFocus(G) = F2;H: LineSegment;IsChordOf(H, G);PointOnCurve(F2, H)", "query_expressions": "Min(Length(H))", "answer_expressions": "32/5", "fact_spans": "[[[1, 40]], [[1, 40]], [[43, 50]], [[1, 50]], [], [[1, 54]], [[0, 54]]]", "query_spans": "[[[1, 57]]]", "process": "By the property of the ellipse, the shortest chord passing through the focus of the ellipse is the perpendicular diameter. From the ellipse equation: a=5, b=4, ∴c=3, hence the coordinates of the right focus are (3,0). Let x=3, substitute into the equation to obtain: y=\\pm\\frac{16}{6}" }, { "text": "Given that the right focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{b^{2}}=1$ is $(\\sqrt{13}, 0)$, then the asymptotes of this hyperbola are given by?", "fact_expressions": "G: Hyperbola;b: Number;H: Point;Expression(G) = (x^2/9 - y^2/b^2 = 1);Coordinate(H) = (sqrt(13), 0);RightFocus(G) = H", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*(2/3)*x", "fact_spans": "[[[2, 44], [68, 71]], [[5, 44]], [[49, 65]], [[2, 44]], [[49, 65]], [[2, 65]]]", "query_spans": "[[[68, 79]]]", "process": "" }, { "text": "The center of the hyperbola is at the origin of the coordinate system, and its eccentricity is equal to $2$. One of its vertices coincides with the focus of the parabola $y^{2}=-8 x$. Then the equation of the hyperbola is?", "fact_expressions": "O: Origin;G:Hyperbola;H: Parabola;Expression(H) = (y^2 = -8*x);Eccentricity(G)=2;Center(G)=O;OneOf(Vertex(G))=Focus(H)", "query_expressions": "Expression(G)", "answer_expressions": "x^2-y^2/3=1", "fact_spans": "[[[7, 11]], [[0, 3], [21, 22], [50, 53]], [[28, 43]], [[28, 43]], [[0, 20]], [[0, 11]], [[21, 48]]]", "query_spans": "[[[50, 58]]]", "process": "" }, { "text": "If for the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$, the distance from the point with abscissa $\\frac{3 a}{2}$ to the right focus is greater than its distance to the left directrix, then what is the range of values for the eccentricity of the hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;P:Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P,G);XCoordinate(P)=3*a/2;Distance(P,RightFocus(G))>Distance(P,LeftDirectrix(G))", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(2,+oo)", "fact_spans": "[[[1, 60], [101, 104]], [[4, 60]], [[4, 60]], [[81, 82], [91, 92]], [[4, 60]], [[4, 60]], [[1, 60]], [[1, 82]], [[61, 82]], [[1, 99]]]", "query_spans": "[[[101, 114]]]", "process": "" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ (where $a>0 , b>0$), the focal distance is $4 \\sqrt{5}$, and the slope of one of its asymptotes is $2$. Then $a=?$", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;FocalLength(C) = 4*sqrt(5);Slope(OneOf(Asymptote(C))) = 2", "query_expressions": "a", "answer_expressions": "2", "fact_spans": "[[[2, 52]], [[2, 52]], [[57, 68]], [[102, 105]], [[57, 68]], [[57, 68]], [[2, 85]], [[2, 100]]]", "query_spans": "[[[102, 107]]]", "process": "The value of b is obtained from the slope of the asymptote as b = 2a. The value of c is determined from the focal distance. Using the square relationship among a, b, and c, an equation in terms of a is obtained, and the value of a is solved. [Detailed Solution] The hyperbola C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 has asymptotes given by y = \\pm\\frac{b}{a}x. Since the slope of one asymptote is 2, \\therefore \\frac{b}{a} = 2, i.e., b = 2a. Also, \\because 2c = 4\\sqrt{5}, \\therefore c = 2\\sqrt{5}, \\therefore c^{2} = a^{2} + b^{2} = 5a^{2} = 20, \\therefore a = 2. The final answer is: 2" }, { "text": "Given points $M(-2,0)$, $N(2,0)$, a moving point $P$ satisfies the condition $|P M|-|P N|=2 \\sqrt{2}$, then what is the trajectory equation of the moving point $P$?", "fact_expressions": "M: Point;N: Point;P: Point;Coordinate(M) = (-2, 0);Coordinate(N) = (2, 0);Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)) = 2*sqrt(2)", "query_expressions": "LocusEquation(P)", "answer_expressions": "(x^2/2-y^2/2=1)&(x>0)", "fact_spans": "[[[2, 12]], [[13, 22]], [[25, 28], [60, 63]], [[2, 12]], [[13, 22]], [[32, 56]]]", "query_spans": "[[[60, 69]]]", "process": "Problem Analysis: According to the given conditions, the trajectory of point P is the right branch of a hyperbola with foci at M and N. Given that M(-2,0), N(2,0), and |PM| - |PN| = 2\\sqrt{2}, we have c = 2 and a = \\sqrt{2}. Therefore, the required equation is: \\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1 (x>0)." }, { "text": "Given that point $P$ is a moving point on the parabola $y^{2}=2 x$, the projection of point $P$ onto the $y$-axis is $M$, and $A(\\frac{7}{2}, 4)$, then the minimum value of $|P A|+|P M|$ is?", "fact_expressions": "G: Parabola;A: Point;P: Point;M: Point;Expression(G) = (y^2 = 2*x);Coordinate(A) = (7/2, 4);PointOnCurve(P, G);Projection(P, yAxis) = M", "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, M)))", "answer_expressions": "9/2", "fact_spans": "[[[7, 21]], [[46, 65]], [[2, 6], [26, 30]], [[40, 43]], [[7, 21]], [[46, 65]], [[2, 25]], [[26, 43]]]", "query_spans": "[[[67, 86]]]", "process": "Problem Analysis: From the given information, the focus of the parabola is $ F\\left(\\frac{1}{2}, 0\\right) $, and the directrix is $ x = -\\frac{1}{2} $. Extend $ PM $ to meet the directrix at point $ H $. Then $ |PF| = |PH| $, and $ |PM| = |PH| - \\frac{1}{2} = |PF| - \\frac{1}{2} $. Therefore, $ |PM| + |PA| = |PF| + |PA| - \\frac{1}{2} $. Thus, we only need to find the minimum value of $ |PF| + |PA| $. Since $ |PF| + |PA| \\geqslant |AF| = \\sqrt{\\left(\\frac{7}{2} - \\frac{1}{2}\\right)^{2} + (4 - 0)^{2}} = 5 $, the minimum value of $ |PA| + |PM| $ is $ 5 - \\frac{1}{2} = \\frac{9}{2} $." }, { "text": "Given fixed points $A(-5,0)$, $B(5,4)$, and point $P$ being any point on the right branch of the hyperbola $C$: $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, then the maximum value of $|PB|-|PA|$ is?", "fact_expressions": "C: Hyperbola;A: Point;B: Point;P: Point;Expression(C) = (x^2/16 - y^2/9 = 1);Coordinate(A) = (-5, 0);Coordinate(B) = (5, 4);PointOnCurve(P, RightPart(C))", "query_expressions": "Max(-Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B)))", "answer_expressions": "-4", "fact_spans": "[[[30, 74]], [[4, 13]], [[16, 24]], [[25, 29]], [[30, 74]], [[4, 13]], [[16, 24]], [[25, 81]]]", "query_spans": "[[[83, 102]]]", "process": "According to the given conditions, a=4, b=3, so c=\\sqrt{a^{2}+b^{2}}=5; therefore, A(-5,0) is the left focus of the hyperbola, and let the right focus be F(5,0). Thus, |PB|-|PA|=|PB|-(|PF|+2a)=|PB|-|PF|-8\\leqslant|BF|-8=4-8=-4, with equality if and only if points P, F, and B are collinear." }, { "text": "The line $l$ passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) intersects the parabola at points $A$ and $B$, with point $A$ above the $x$-axis, and $O$ being the origin. When $|\\overrightarrow{O A}| \\geq|\\overrightarrow{F A}|$, what is the range of possible values for the slope of line $l$?", "fact_expressions": "l: Line;G: Parabola;p: Number;O: Origin;A: Point;F: Point;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};PointOnCurve(A, xAxis);Abs(VectorOf(O, A)) >= Abs(VectorOf(F, A))", "query_expressions": "Range(Slope(l))", "answer_expressions": "(-oo, -2*sqrt(2)] + (0, +oo)", "fact_spans": "[[[29, 34], [124, 129]], [[1, 22], [35, 38]], [[4, 22]], [[61, 64]], [[39, 42], [49, 53]], [[25, 28]], [[43, 46]], [[4, 22]], [[1, 22]], [[1, 28]], [[0, 34]], [[29, 48]], [[49, 60]], [[71, 122]]]", "query_spans": "[[[124, 138]]]", "process": "According to |\\overrightarrow{OA}|\\geqslant|\\overrightarrow{FA}|, we obtain x_{A}\\geqslant\\frac{p}{4}, and further derive y_{A}; then, based on the slope of line l given by \\frac{y_{A}}{x_{A}-\\frac{p}{2}}, we obtain the result. From the given condition: when the horizontal coordinate of point A satisfies x_{A}\\geqslant\\frac{p}{4}, the inequality |\\overrightarrow{OA}|\\geqslant|\\overrightarrow{FA}| holds, and at this time y_{A}\\geqslant\\frac{\\sqrt{2}}{2}p; hence, the range of the slope of line l is (-\\infty,-2\\sqrt{2}]\\cup(0,+\\infty)." }, { "text": "The equation of the hyperbola that has the same asymptotes as the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ and passes through the point $A(3,-2 \\sqrt{3})$ is?", "fact_expressions": "G: Hyperbola;C:Hyperbola;A: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Coordinate(A) = (3, -2*sqrt(3));Asymptote(C) = Asymptote(G);PointOnCurve(A, C)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/(9/4)-y^2/4=1", "fact_spans": "[[[1, 40]], [[72, 75]], [[52, 71]], [[1, 40]], [[52, 71]], [[0, 75]], [[50, 75]]]", "query_spans": "[[[72, 79]]]", "process": "Solution: Let the equation of the hyperbola that shares the same asymptotes with the hyperbola \\frac{x^2}{9}-\\frac{y^{2}}{16}=1 be \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=m (m\\neq0). Substituting the point A(3,-2\\sqrt{3}), we obtain m=\\frac{1}{4}. Therefore, the equation of the desired hyperbola is \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=\\frac{1}{4}." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left and right foci are $F_{1}$, $F_{2}$ respectively, the eccentricity is $\\frac{3}{4}$, and point $P$ lies on $C$. If the area of $\\Delta F_{1} P F_{2}$ is $7$, and the inradius of $\\Delta F_{1} P F_{2}$ is $1$, then the equation of $C$ is?", "fact_expressions": "C: Ellipse;b: Number;a: Number;F1: Point;P: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C)=F1;RightFocus(C)=F2;Eccentricity(C)=3/4;PointOnCurve(P, C);Area(TriangleOf(F1, P, F2)) = 7;Radius(InscribedCircle(TriangleOf(F1,P,F2)))=1", "query_expressions": "Expression(C)", "answer_expressions": "x^2/16+y^2/7=1", "fact_spans": "[[[2, 59], [107, 110], [180, 183]], [[8, 59]], [[8, 59]], [[68, 75]], [[102, 106]], [[76, 83]], [[8, 59]], [[8, 59]], [[2, 59]], [[2, 83]], [[2, 83]], [[2, 101]], [[102, 113]], [[115, 144]], [[146, 178]]]", "query_spans": "[[[180, 188]]]", "process": "From the area of \\triangle F_{1}PF_{2} being 7, we obtain \\frac{1}{2}(2a+2c)r=7, that is, a+c=7. Also, the eccentricity is \\frac{c}{a}=\\frac{3}{4}, so a=4, c=3, thus b^{2}=a^{2}-c^{2}=7. Therefore, the equation of C is \\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1" }, { "text": "Given that $F$ is the focus of the parabola $y^{2}=4 x$, $M$ is a moving point on this parabola, and $P(3,1)$ is a fixed point, then the minimum value of $|M P|+|M F|$ is?", "fact_expressions": "F: Point;G: Parabola;Expression(G) = (y^2 = 4*x);Focus(G) = F;M: Point;PointOnCurve(M, G);P: Point;Coordinate(P) = (3, 1)", "query_expressions": "Min(Abs(LineSegmentOf(M, F)) + Abs(LineSegmentOf(M, P)))", "answer_expressions": "4", "fact_spans": "[[[2, 5]], [[6, 20], [30, 33]], [[6, 20]], [[2, 23]], [[24, 27]], [[24, 39]], [[40, 48]], [[40, 48]]]", "query_spans": "[[[55, 74]]]", "process": "Let the projection of point M on the directrix be D. Then, according to the definition of a parabola, we have: |MF| = |MD|. Therefore, the minimum value of |MP| + |MF| is equivalent to the minimum value of |PM| + |MD|. The sum |PM| + |MD| is minimized when points D, M, and P are collinear, and its value is 3 - (-1) = 4." }, { "text": "A hyperbola has an asymptote given by $y=\\sqrt{3} x$ and a focus at $(4,0)$. Find the standard equation of the hyperbola.", "fact_expressions": "G: Hyperbola;Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x);Coordinate(Focus(G))=(4,0)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4 - y^2/12 = 1", "fact_spans": "[[[34, 37]], [[0, 37]], [0, 36]]", "query_spans": "[[[34, 44]]]", "process": "Problem Analysis: From the given conditions, we have \n\\begin{cases}\\frac{b}{a}=\\sqrt{3}\\\\c=4\\\\c^{2}=a^{2}+b^{2}\\end{cases} \nIt follows that $ a^{2}=4 $, $ b^{2}=12 $. Therefore, the equation of the hyperbola is $ \\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1 $." }, { "text": "The distance from the focus to the directrix of the parabola $x^{2}=8 y$ is?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 8*y)", "query_expressions": "Distance(Focus(G), Directrix(G))", "answer_expressions": "4", "fact_spans": "[[[0, 14]], [[0, 14]]]", "query_spans": "[[[0, 25]]]", "process": "Problem Analysis: The focus of the parabola $x^{2}=8y$ is $(0,2)$, and the equation of the directrix is $y=-2$. Therefore, the distance from the focus to the directrix is 4." }, { "text": "Given that $M(x1, y1)$ is an arbitrary point on the curve $C$: $\\sqrt{\\frac{x^{2}}{16}}+\\sqrt{\\frac{y^{2}}{7}}=1$, and points $A(-3,0)$, $B(3,0)$, then the maximum value of $|M A|+|M B|$ is?", "fact_expressions": "C: Curve;M: Point;Expression(C) = (sqrt(y^2/7) + sqrt(x^2/16) = 1);Coordinate(M) = (x1, y1);PointOnCurve(M, C) = True;A: Point;B: Point;Coordinate(A) = (-3, 0);Coordinate(B) = (3, 0);x1: Number;y1: Number", "query_expressions": "Max(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, B)))", "answer_expressions": "8", "fact_spans": "[[[14, 71]], [[2, 13]], [[14, 71]], [[2, 13]], [[2, 76]], [[77, 86]], [[88, 96]], [[77, 86]], [[88, 96]], [[2, 13]], [[2, 13]]]", "query_spans": "[[[98, 117]]]", "process": "The curve $ C: \\sqrt{\\frac{x^{2}}{16}} + \\sqrt{\\frac{y^{2}}{7}} = 1 $ can be rewritten as $ \\frac{|x|}{4} + \\frac{|y|}{\\sqrt{7}} = 1 $. Its graph is a parallelogram with vertices at $ (\\pm4,0) $ and $ (0,\\pm\\sqrt{7}) $, and it is an inscribed parallelogram of the ellipse $ \\frac{x^{2}}{16} + \\frac{y^{2}}{7} = 1 $. Using the properties of ellipses, it is known that when point $ M $ lies on the ellipse, $ |MA| + |MB| = 8 $, while when $ M $ lies inside the ellipse, $ |MA| + |MB| < 8 $. In conclusion, the maximum value of $ |MA| + |MB| $ is 8." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, what is the radius of the circle centered at the right focus of $C$ and tangent to the asymptotes of $C$?", "fact_expressions": "C: Hyperbola;G: Circle;Expression(C) = (x^2/16 - y^2/9 = 1);RightFocus(C)=Center(G);IsTangent(G, Asymptote(C))", "query_expressions": "Radius(G)", "answer_expressions": "3", "fact_spans": "[[[2, 46], [48, 51], [60, 63]], [[70, 71]], [[2, 46]], [[47, 71]], [[59, 71]]]", "query_spans": "[[[70, 76]]]", "process": "" }, { "text": "Given the ellipse $C$: $ \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, with right focus $F$, the line $y=\\sqrt{3} x$ intersects the ellipse $C$ at points $A$ and $B$ (point $A$ lies in the first quadrant), and $OA=OF$ ($O$ is the origin). Then the eccentricity of the ellipse is?", "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;O: Origin;A: Point;B:Point;F: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y = sqrt(3)*x);RightFocus(C) = F;Intersection(G,C)={A,B};Quadrant(A)=1;LineSegmentOf(O, A) = LineSegmentOf(O, F)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3)-1", "fact_spans": "[[[2, 59], [84, 89], [135, 137]], [[8, 59]], [[8, 59]], [[67, 83]], [[124, 127]], [[91, 94], [102, 106]], [[95, 98]], [[63, 66]], [[8, 59]], [[8, 59]], [[2, 59]], [[67, 83]], [[2, 66]], [[67, 101]], [[102, 111]], [[114, 123]]]", "query_spans": "[[[135, 143]]]", "process": "Let the left focus of the ellipse be $ F_{1} $, connect $ F_{1}A $, $ F_{1}B $. In $ \\triangle AFB $, $ OA = OB = OF = c $, $ \\angle AOF = \\frac{\\pi}{3} $, then $ AF \\perp BF $, $ AF = c $, $ BF = \\sqrt{3}c $. Since $ F_{1}O = FO $, $ OA = OB $, it can be seen that quadrilateral $ F_{1}AFB $ is a parallelogram, then $ AF + BF = AF + AF_{1} = 2a $. Also $ AF + BF = c + \\sqrt{3}c $, so $ c + \\sqrt{3}c = 2a $. Hence, the eccentricity of the ellipse $ e = \\frac{c}{a} = \\frac{2}{1 + \\sqrt{3}} = \\sqrt{3} - 1 $." }, { "text": "The foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ are $F_{1}$ and $F_{2}$, point $P$ lies on the ellipse. If $|P F_{1}|=6$, then $|P F_{2}|$ equals?", "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/16 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 6", "query_expressions": "Abs(LineSegmentOf(P, F2))", "answer_expressions": "4", "fact_spans": "[[[0, 39], [64, 66]], [[59, 63]], [[43, 50]], [[51, 58]], [[0, 39]], [[0, 58]], [[59, 67]], [[69, 82]]]", "query_spans": "[[[84, 98]]]", "process": "In the ellipse \\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1, a=5, and by the definition of an ellipse, |PF_{2}|=2a-|PF_{1}|=4." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{6}=1$, and point $P$ moves on the ellipse. When the value of $\\frac{1}{|P F_{1}|}+\\frac{4}{|P F_{2}|}$ is minimized, what is the area of $\\Delta P F_{1} F_{2}$?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/6 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);WhenMin(1/Abs(LineSegmentOf(P, F1))+4/Abs(LineSegmentOf(P, F2)))", "query_expressions": "Area(TriangleOf(P, F1, F2))", "answer_expressions": "2*sqrt(3)", "fact_spans": "[[[18, 55], [66, 68]], [[18, 55]], [[2, 9]], [[10, 17]], [[2, 61]], [[2, 61]], [[62, 65]], [[62, 69]], [[72, 119]]]", "query_spans": "[[[120, 147]]]", "process": "From the definition of an ellipse, |PF_{1}| + |PF_{2}| = 2a = 6, so (|PF_{1}| + |PF_{2}|)(\\frac{1}{|PF_{1}|} + \\frac{4}{|PF_{2}|}) = 5 + \\frac{|PF_{2}|}{|PF_{1}|} + \\frac{4|PF_{1}|}{|PF_{2}|} \\geqslant 5 + 2\\sqrt{\\frac{|PF_{2}|}{|PF_{1}|} \\cdot \\frac{4|PF_{1}|}{|PF_{2}|}} = 9. Therefore, \\frac{1}{|PF_{1}|} + \\frac{4}{|PF_{2}|} \\geqslant \\frac{9}{6} = \\frac{3}{2}, with equality if and only if \\frac{|PF_{2}|}{|PF_{1}|} = \\frac{4|PF_{1}|}{|PF_{2}|}, that is, when |PF_{1}| = 2, |PF_{2}| = 4. Also, c^{2} = a^{2} - b^{2} = 3 \\Rightarrow c = \\sqrt{3}, so |F_{1}F_{2}| = 2\\sqrt{3}. Thus, |PF_{1}|^{2} + |F_{1}F_{2}|^{2} = |PF_{2}|^{2}. By the Pythagorean theorem, PF_{1} \\bot F_{1}F_{2}, so S_{\\triangle PF_{1}F_{2}} = \\frac{1}{2} \\times 2\\sqrt{3} \\times 2 = 2\\sqrt{3}. The answer is: 2\\sqrt{3}." }, { "text": "Given that the hyperbola $C$ passes through the point $(1, 1)$, and one of its asymptotes has the equation $y = \\sqrt{3}x$. Then the standard equation of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;C1: Point;Coordinate(C1) = (1,1);Expression(OneOf(Asymptote(C))) = (y = sqrt(3)*x)", "query_expressions": "Expression(C)", "answer_expressions": "(3*x^2/2-y^2/2 = 1)", "fact_spans": "[[[2, 9], [53, 60], [27, 28]], [[12, 25]], [[12, 25]], [[27, 50]]]", "query_spans": "[[[53, 67]]]", "process": "" }, { "text": "The equation of a hyperbola with vertices at the foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ and eccentricity $2$ is?", "fact_expressions": "H: Ellipse;Expression(H) = (x^2/25 + y^2/16 = 1);G: Hyperbola;Focus(H) = Vertex(G);Eccentricity(G) = 2", "query_expressions": "Expression(G)", "answer_expressions": "x^2/9 - y^2/27 = 1", "fact_spans": "[[[1, 40]], [[1, 40]], [[55, 58]], [[0, 58]], [[47, 58]]]", "query_spans": "[[[55, 62]]]", "process": "" }, { "text": "Given the parabola $C$: $y^{2}=4x$ has focus $F$, and $A(x_{0}, y_{0})$ is a point on $C$ such that $|AF| = \\frac{5}{4}x_{0}$, find $x_{0}$=?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);Focus(C) = F;F: Point;A: Point;Coordinate(A) = (x0, y0);x0: Number;y0: Number;PointOnCurve(A, C) = True;Abs(LineSegmentOf(A, F)) = (5/4)*x0", "query_expressions": "x0", "answer_expressions": "4", "fact_spans": "[[[2, 21], [49, 52]], [[2, 21]], [[2, 28]], [[25, 28]], [[31, 48]], [[31, 48]], [[83, 90]], [[31, 48]], [[31, 55]], [[56, 81]]]", "query_spans": "[[[83, 92]]]", "process": "According to the definition of a parabola, the distance from point A to the focus F is equal to its distance to the directrix x = -1. Hence, x_{0} + 1 = \\frac{5}{4}x_{0}, solving gives x_{0} = 4." }, { "text": "Given that the distance from a point $P$ on the parabola $y^{2}=8x$ to the focus is $6$, then the coordinates of point $P$ are?", "fact_expressions": "G: Parabola;P: Point;Expression(G) = (y^2 = 8*x);PointOnCurve(P, G);Distance(P, Focus(G)) = 6", "query_expressions": "Coordinate(P)", "answer_expressions": "(4,pm*4*sqrt(2))", "fact_spans": "[[[2, 16]], [[19, 22], [34, 38]], [[2, 16]], [[2, 22]], [[2, 32]]]", "query_spans": "[[[34, 43]]]", "process": "Using the definition of a parabola, the x-coordinate of point P can be found, and then the y-coordinate of the point can be determined; thus, the result is obtained. Let the coordinates of point P be (x_{0},y_{0}). The directrix equation of the parabola y^{2}=8x is x=-2. By the definition of a parabola, we have x_{0}+2=6\\Rightarrow x_{0}=4, \\therefore y_{0}^{2}=8x_{0}=32, solving gives y_{0}=\\pm4\\sqrt{2}. Therefore, the coordinates of point P are (4,\\pm4\\sqrt{2})." }, { "text": "Let the foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ be $F_{1}$ and $F_{2}$, and let $P$ be any point on the ellipse. If the maximum value of $\\angle F_{1} P F_{2}$ is $\\frac{2 \\pi}{3}$, then the eccentricity of this ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;P: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Focus(G)={F1,F2};PointOnCurve(P,G);Max(AngleOf(F1, P, F2)) = (2*pi)/3", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)/2", "fact_spans": "[[[1, 53], [77, 79], [132, 134]], [[3, 53]], [[3, 53]], [[57, 64]], [[73, 76]], [[65, 72]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 72]], [[73, 83]], [[85, 129]]]", "query_spans": "[[[132, 140]]]", "process": "Problem Analysis: According to the properties of an ellipse, when point P is taken at one endpoint of the minor axis of the ellipse, $\\angle F_{1}PF_{2}$ reaches its maximum value of $\\frac{2\\pi}{3}$. Thus, $\\tan\\angle OPF_{2} = \\tan\\frac{\\pi}{3} = \\sqrt{3} = \\frac{c}{b}$. Simplifying and rearranging this equation leads to the solution: From the properties of the ellipse, when point P is taken at one endpoint of the minor axis of the ellipse, $\\angle F_{1}PF_{2}$ reaches its maximum value of $\\frac{2\\pi}{3}$. Therefore, $\\tan\\angle OPF_{2} = \\tan\\frac{\\pi}{3} = \\sqrt{3} = \\frac{c}{b}$, hence $c^{2} = 3b^{2} = 3(a^{2} - c^{2})$." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a, b>0)$, the line $l$: $y=\\sqrt{3} x$ intersects the hyperbola $C$ at points $A$ and $B$, and $|A B|=|F_{1} F_{2}|$, then the eccentricity of the hyperbola $C$ is?", "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;A: Point;B: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;Expression(l)=(y = sqrt(3)*x) ;Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, B)) = Abs(LineSegmentOf(F1, F2))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3)+1", "fact_spans": "[[[83, 104]], [[18, 77], [105, 111], [147, 153]], [[26, 77]], [[26, 77]], [[113, 116]], [[117, 120]], [[2, 9]], [[10, 17]], [[26, 77]], [[26, 77]], [[18, 77]], [[2, 82]], [[2, 82]], [[83, 104]], [[83, 122]], [[124, 145]]]", "query_spans": "[[[147, 159]]]", "process": "Without loss of generality, assume points A and B lie in the first and third quadrants respectively; then ∠AOF₂ = 60°. From |AB| = |F₁F₂| = 2c, it follows that |OA| = |OF₂|, and quadrilateral AF₁BF₂ is a rectangle. Hence, triangle AOF₂ is equilateral, |AF₂| = c, |AF₁| = √3c. By the definition of a hyperbola, √3c − c = 2a. Thus, e = c/a = √3 + 1." }, { "text": "The coordinates of the foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{8}=1$ are?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/8 = 1)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(pm*2*sqrt(2), 0)", "fact_spans": "[[[0, 38]], [[0, 38]]]", "query_spans": "[[[0, 45]]]", "process": "" }, { "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{25}=1$ are?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/25 = 1)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*(5/3)*x", "fact_spans": "[[[0, 39]], [[0, 39]]]", "query_spans": "[[[0, 47]]]", "process": "In the standard equation of a hyperbola, replacing 1 with 0 yields the asymptote equation of this hyperbola. [Detailed explanation] Setting \\frac{x^{2}}{9}\\cdot\\frac{y^{2}}{2<}=0 gives \\therefore the asymptote equations of the hyperbola \\frac{x2}{9}\\cdot\\frac{y^{2}}{25}=1 are y=\\pm\\frac{5}{3}x" }, { "text": "If hyperbola $C$ shares the same asymptotes with the hyperbola $\\frac{x^{2}}{12}-\\frac{y^{2}}{8}=1$ and passes through the point $A(3, \\sqrt{2})$, then the equation of hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;H: Hyperbola;A: Point;PointOnCurve(A,C);Expression(H) = (x^2/12-y^2/8=1);Coordinate(A) = (3, sqrt(2));Asymptote(C) = Asymptote(H)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/6-y^2/4=1", "fact_spans": "[[[1, 7], [73, 79]], [[8, 47]], [[54, 71]], [[1, 71]], [[8, 47]], [[54, 71]], [[1, 51]]]", "query_spans": "[[[73, 84]]]", "process": "Since hyperbola C shares asymptotes with the hyperbola \\frac{x^{2}}{12}-\\frac{y^{2}}{8}=1, we assume the equation of hyperbola C is \\frac{x^{2}}{12}-\\frac{y^{2}}{8}=\\lambda (\\lambda\\neq0). Since hyperbola C passes through point A(3,\\sqrt{2}), we have \\frac{9}{12}-\\frac{2}{8}=\\lambda, which gives \\lambda=\\frac{1}{2}. Therefore, the equation of hyperbola C is \\frac{x^{2}}{12}-\\frac{y^{2}}{8}=\\frac{1}{2}, or equivalently \\frac{x^{2}}{6}-\\frac{y^{2}}{4}=1." }, { "text": "Given that the intersection point of the right directrix and the asymptote of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$ lies on the parabola $y^{2}=2 p x$, find the value of the real number $p$.", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/12 = 1);H: Parabola;Expression(H) = (y^2 = 2*p*x);p: Real;PointOnCurve(Intersection(RightDirectrix(G), Asymptote(G)), H)", "query_expressions": "p", "answer_expressions": "3/2", "fact_spans": "[[[2, 41]], [[2, 41]], [[53, 69]], [[53, 69]], [[72, 77]], [[2, 70]]]", "query_spans": "[[[72, 81]]]", "process": "Solution: The right directrix of the hyperbola \\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1 is x=\\frac{a^{2}}{c}=\\frac{4}{4}=1, and the asymptotes are y=\\pm\\sqrt{3}x. The intersection points of the right directrix and the asymptotes of the hyperbola \\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1, which are (1,\\pm\\sqrt{3}), lie on the parabola y^{2}=2px. Thus, we have: 3=2p, solving for p gives p=\\frac{3}{2}." }, { "text": "It is known that the ellipse $E$ has its center at the origin of coordinates, with eccentricity $\\frac{\\sqrt{3}}{2}$. The right focus of $E$ coincides with the focus of the parabola $C$: $y^{2}=12 x$, and $A$, $B$ are the two intersection points of the directrix of $C$ and $E$. Then $|A B|$=?", "fact_expressions": "C: Parabola;E: Ellipse;A: Point;B: Point;O:Origin;Center(E)=O;Eccentricity(E)=sqrt(3)/2;Expression(C) = (y^2 = 12*x);RightFocus(E) = Focus(C);Intersection(Directrix(C),E)={A,B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "sqrt(3)", "fact_spans": "[[[49, 69], [83, 86]], [[2, 7], [41, 44], [41, 44]], [[75, 78]], [[79, 82]], [[11, 15]], [[2, 15]], [[2, 40]], [[49, 69]], [[41, 74]], [[75, 98]]]", "query_spans": "[[[100, 109]]]", "process": "Test Analysis: The focus coordinates of the parabola can be used to find the focus coordinates of the ellipse (3,0), so c=3. Then, from the eccentricity, we get a=2\\sqrt{3}, and thus the equation of the ellipse is \\frac{x^{2}}{12}+\\frac{y^{2}}{3}=1. It is known that the length of chord AB is exactly the latus rectum length \\frac{2b^{2}}{a}=\\sqrt{3}." }, { "text": "There is a point $A$ on the parabola $y^{2}=2 p x(p>0)$, its horizontal coordinate is $3$, and its distance to the focus is $5$. Then the equation of the parabola is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;A: Point;PointOnCurve(A, G);XCoordinate(A) = 3;Distance(A, Focus(G)) = 5", "query_expressions": "Expression(G)", "answer_expressions": "y^2=8*x", "fact_spans": "[[[0, 21], [51, 54]], [[0, 21]], [[3, 21]], [[3, 21]], [[23, 27], [28, 29], [38, 39]], [[0, 27]], [[28, 37]], [[0, 49]]]", "query_spans": "[[[51, 59]]]", "process": "" }, { "text": "The eccentricity of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ is $\\frac{1}{2}$, then $m=?$", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/m = 1);m: Number;Eccentricity(G) = 1/2", "query_expressions": "m", "answer_expressions": "{3, 16/3}", "fact_spans": "[[[0, 37]], [[0, 37]], [[57, 60]], [[0, 55]]]", "query_spans": "[[[57, 62]]]", "process": "\\textcircled{1} When the foci of the ellipse lie on the x-axis, it follows from the given condition that \\frac{\\sqrt{4-m}}{2}=\\frac{1}{2}, solving gives m=3; \\textcircled{2} When the foci of the ellipse lie on the y-axis, it follows from the given condition that \\frac{\\sqrt{m-4}}{\\sqrt{m}}=\\frac{1}{2}, solving gives m=\\frac{16}{3}. In conclusion, m=3 or m=\\frac{16}{3}." }, { "text": "Given fixed point $P(-4,0)$ and fixed circle $Q$: $x^{2}+y^{2}=8 x$, a moving circle $M$ is externally tangent to circle $Q$ and passes through point $P$. Find the trajectory equation of the center of circle $M$.", "fact_expressions": "Q: Circle;P: Point;M: Circle;Expression(Q) = (x^2 + y^2 = 8*x);Coordinate(P) = (-4, 0);PointOnCurve(P,M);IsOutTangent(M,Q);Center(M)=M1;M1:Point", "query_expressions": "LocusEquation(M1)", "answer_expressions": "English:", "fact_spans": "[[[16, 38], [45, 49]], [[4, 13], [55, 59]], [[41, 44]], [[16, 38]], [[4, 13]], [[39, 59]], [[39, 51]], [[39, 66]], [[63, 66]]]", "query_spans": "[[[63, 72]]]", "process": "Combining the graph, we obtain |MQ|·|MP| = 4, so a = 2, c = 4, then b = 2\\sqrt{3}. The trajectory of M is the hyperbola \\frac{x^{2}}{4}-\\frac{y}{1}" }, { "text": "Given a point $P$ on the hyperbola $x^{2}-4 y^{2}=4$ such that the distance from $P$ to one focus of the hyperbola is $6$, then the distance from point $P$ to the other focus is?", "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2 - 4*y^2 = 4);PointOnCurve(P, G);F1:Point;F2:Point;OneOf(Focus(G))=F1;OneOf(Focus(G)) = F2;Negation(F1=F2);Distance(P, F1) = 6", "query_expressions": "Distance(P,F2)", "answer_expressions": "{10,2}", "fact_spans": "[[[2, 22], [29, 32]], [[25, 28], [48, 52]], [[2, 22]], [[2, 28]], [], [], [[29, 37]], [[29, 57]], [[29, 57]], [[25, 45]]]", "query_spans": "[[[29, 63]]]", "process": "" }, { "text": "Given the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has a semi-focal length $c$, and $c=\\sqrt{3} b$. If the ellipse $E$ passes through points $A$ and $B$, and $AB$ is a diameter of the circle $M$: $(x+2)^{2}+(y-1)^{2}=r^{2}$, then what is the equation of line $AB$?", "fact_expressions": "E: Ellipse;b: Number;a: Number;M: Circle;r: Number;B: Point;A: Point;c: Number;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);Expression(M) = ((x + 2)^2 + (y - 1)^2 = r^2);HalfFocalLength(E) = c;c = sqrt(3)*b;PointOnCurve(A, E);PointOnCurve(B, E);OneOf(Diameter(LineSegmentOf(A, B), M))", "query_expressions": "Expression(LineOf(A, B))", "answer_expressions": "x - 2*y + 4 = 0", "fact_spans": "[[[2, 59], [85, 90]], [[8, 59]], [[8, 59]], [[109, 141]], [[114, 141]], [[96, 99]], [[92, 95]], [[64, 67]], [[8, 59]], [[8, 59]], [[2, 59]], [[109, 141]], [[2, 67]], [[69, 83]], [[85, 101]], [[85, 101]], [[103, 146]]]", "query_spans": "[[[148, 160]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Substitute into the ellipse equation and subtract; according to the slope formula of the line and the midpoint M of AB, find the slope of the line, then the equation of the line can be obtained. Let A(x_{1},y_{1}), B(x_{2},y_{2}), substituting into the ellipse equation gives: \\frac{x_{1}^2}{a^{2}}+\\frac{y_{1}^2}{b^{2}}=1\\textcircled{1}, \\frac{x_{2}^2}{a^{2}}+\\frac{y_{2}^2}{b^{2}}=1\\textcircled{2}, \\textcircled{2}-\\textcircled{1} yields: \\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=-\\frac{b^{2}(x_{2}+x_{1})}{a^{2}(y_{2}+y_{1})}. From c=\\sqrt{3}b we get a^{2}-b^{2}=c^{2}=3b^{2}, thus \\frac{b^{2}}{a^{2}}=\\frac{1}{4}. Also, the midpoint M of AB is (-2,1), so k_{AB}=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=-\\frac{b^{2}(x_{2}+x_{1})}{a^{2}(y_{2}+y_{1})}=-\\frac{1}{4}\\times(-2)=\\frac{1}{2}. Therefore, the equation of line AB is y-1=\\frac{1}{2}(x+2), that is, x-2y+4=0." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has eccentricity $\\frac{1}{2}$, the right focus is $F_{2}$, point $M$ lies on the circle $x^{2}+y^{2}=b^{2}$ and $M$ is in the first quadrant. The tangent line to the circle $x^{2}+y^{2}=b^{2}$ at $M$ intersects the ellipse at points $P$, $Q$. If the perimeter of $\\Delta P F_{2} Q$ is $4$, then the equation of ellipse $C$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Eccentricity(C) = 1/2;F2: Point;RightFocus(C) = F2;M: Point;G: Circle;Expression(G) = (x^2 + y^2 = b^2);PointOnCurve(M, G) = True;Quadrant(M) = 1;P: Point;Q: Point;l: Line;TangentOnPoint(M,G) = l;Perimeter(TriangleOf(P, F2, Q)) = 4;Intersection(l,C) = {P,Q}", "query_expressions": "Expression(C)", "answer_expressions": "x^2/4 + y^2/3 = 1", "fact_spans": "[[[2, 59], [198, 203], [156, 158]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[2, 77]], [[82, 89]], [[2, 89]], [[90, 94], [118, 121], [128, 131]], [[95, 115], [132, 152]], [[95, 115]], [[90, 116]], [[118, 126]], [[159, 162]], [[163, 166]], [], [[127, 155]], [[171, 196]], [[127, 168]]]", "query_spans": "[[[198, 208]]]", "process": "" }, { "text": "Let the two foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ be $F_{1}$ and $F_{2}$. If there exists a point $P$ on the ellipse such that $PF_{1} \\perp PF_{2}$, then the range of values for the eccentricity of this ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "[sqrt(2)/2, 1)", "fact_spans": "[[[1, 54], [81, 83], [117, 119]], [[3, 54]], [[3, 54]], [[88, 91]], [[61, 68]], [[71, 78]], [[3, 54]], [[3, 54]], [[1, 54]], [[1, 78]], [[81, 91]], [[93, 114]]]", "query_spans": "[[[117, 130]]]", "process": "" }, { "text": "There are three points on the plane: $A(-2, y)$, $B(0, \\frac{\\gamma}{2})$, $C(x, y)$. If $\\overrightarrow{A B} \\perp \\overrightarrow{B C}$, then what is the trajectory equation of the moving point $C$?", "fact_expressions": "A: Point;B: Point;C: Point;x:Number;y:Number;gamma:Number;Coordinate(A) = (-2,y) ;Coordinate(B) =(0, gamma/2);Coordinate(C) = (x, y);IsPerpendicular(VectorOf(B, C),VectorOf(A, B))", "query_expressions": "LocusEquation(C)", "answer_expressions": "y^2=8*x", "fact_spans": "[[[7, 17]], [[19, 43]], [[45, 54], [111, 114]], [[45, 54]], [[45, 54]], [[19, 43]], [[7, 17]], [[19, 43]], [[45, 54]], [[57, 106]]]", "query_spans": "[[[111, 121]]]", "process": "" }, { "text": "$P$ is a point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, and $F_{1}$, $F_{2}$ are the two foci. Then the difference between the maximum and minimum values of $|PF_{1}| \\cdot |PF_{2}|$ is?", "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 + y^2/4 = 1);PointOnCurve(P, G);Focus(G)={F1,F2}", "query_expressions": "Max(Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)))-Min(Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)))", "answer_expressions": "5", "fact_spans": "[[[4, 41]], [[0, 3]], [[45, 52]], [[53, 60]], [[4, 41]], [[0, 44]], [[4, 65]]]", "query_spans": "[[[67, 104]]]", "process": "" }, { "text": "Given the parabola $C$: $x^{2}=2 p y(p>0)$ with focus $F$, if a point $P(4, y_{0})$ on the parabola $C$ satisfies $|F P|=4$, then what is the value of $p$?", "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 2*p*y);p: Number;p > 0;F: Point;Focus(C) = F;P: Point;Coordinate(P) = (4, y0);y0: Number;PointOnCurve(P, C) = True;Abs(LineSegmentOf(F, P)) = 4", "query_expressions": "p", "answer_expressions": "4", "fact_spans": "[[[2, 28], [37, 43]], [[2, 28]], [[72, 75]], [[10, 28]], [[32, 35]], [[2, 35]], [[46, 59]], [[46, 59]], [[46, 59]], [[37, 59]], [[61, 70]]]", "query_spans": "[[[72, 79]]]", "process": "Since point P(4,y_{0}) lies on the parabola C: x^{2}=2py (p>0), we have y_{0}=\\frac{8}{p}. By the definition of the parabola, |FP|=y_{0}+\\frac{p}{2}=\\frac{8}{p}+\\frac{p}{2}=4. Solving this equation gives p=4." }, { "text": "The coordinates of the foci of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ are?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(pm*1, 0)", "fact_spans": "[[[0, 37]], [[0, 37]]]", "query_spans": "[[[0, 44]]]", "process": "c^{2}=4-3=1, c=1, therefore the coordinates of the foci are: (-1,0), (1,0)." }, { "text": "The center of the ellipse $C$ is at the origin of the coordinate system, the major axis lies on the $y$-axis, the eccentricity is $\\frac{\\sqrt{3}}{2}$, and the sum of the distances from a point on $C$ to the two foci is $12$. What is the equation of the ellipse?", "fact_expressions": "C: Parabola;C1: Point;O: Origin;Center(C) = O;OverlappingLine(MajorAxis(C), yAxis);Eccentricity(C) = sqrt(3)/2;F1: Point;F2: Point;Focus(C) = {F1,F2};PointOnCurve(C1,C);Distance(C1, F1) + Distance(C1, F2) = 12", "query_expressions": "Expression(C)", "answer_expressions": "y^2/36 + x^2/9 = 1", "fact_spans": "[[[2, 7], [51, 54], [78, 80]], [[57, 60]], [[11, 15]], [[2, 15]], [[2, 24]], [[2, 49]], [], [], [[51, 66]], [[51, 60]], [[51, 76]]]", "query_spans": "[[[78, 85]]]", "process": "Let the equation of the ellipse be: \\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1. From the given conditions, we obtain: \n\\begin{cases}e=\\frac{c}{a}=\\frac{\\sqrt{3}}{2}\\\\2a=12\\\\a^{2}=b^{2}+c^{2}\\end{cases} \nSolving this system of equations yields: \n\\begin{cases}a^{2}=36\\\\b^{2}=9\\\\c^{2}=27\\end{cases} \nBased on this, the equation of the ellipse is \\frac{y^{2}}{36}+\\frac{x^{2}}{9}=1" }, { "text": "The minimum distance from a point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$ to the line $x-2 y-12=0$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/12 = 1);H: Line;Expression(H) = (x - 2*y - 12 = 0);D: Point;PointOnCurve(D, G) = True", "query_expressions": "Min(Distance(D, H))", "answer_expressions": "4*sqrt(5)/5", "fact_spans": "[[[0, 39]], [[0, 39]], [[43, 57]], [[43, 57]], [[41, 42]], [[0, 42]]]", "query_spans": "[[[41, 66]]]", "process": "" }, { "text": "The distance from a moving point $P$ to the point $F(-1,0)$ is greater than its distance to the $y$-axis by $1$. What is the trajectory equation of the moving point $P$?", "fact_expressions": "F: Point;P: Point;Coordinate(F) = (-1, 0);Distance(P, F) = Distance(P, yAxis) + 1", "query_expressions": "LocusEquation(P)", "answer_expressions": "English:", "fact_spans": "[[[6, 16]], [[2, 5], [21, 22], [37, 40]], [[6, 16]], [[2, 34]]]", "query_spans": "[[[37, 47]]]", "process": "Let P(x,y). According to the given condition, we have \\sqrt{(x+1)^{2}+y^{2}}=|x|+1. Square both sides and simplify, then remove the absolute value to solve. Let P(x,y). According to the given condition, we have \\sqrt{(x+1)^{2}+y^{2}}=|x|+1. Squaring both sides and simplifying yields y^2=2|x|-2x. When x>0, y^2=0; when x\\leqslant0, y^{2}=-4x. In summary: y^{2}=\\begin{cases}0,x>0\\\\-4x,x\\leqslant0\\end{cases}" }, { "text": "Let $ P $ be any point on the ellipse $ \\frac{x^{2}}{16} + \\frac{y^{2}}{15} = 1 $, and let $ EF $ be any diameter of the circle $ N $: $ (x-1)^{2} + y^{2} = 4 $. Then the maximum value of $ \\overrightarrow{P E} \\cdot \\overrightarrow{P F} $ is?", "fact_expressions": "G: Ellipse;N: Circle;E: Point;F: Point;P: Point;Expression(G) = (x^2/16 + y^2/15 = 1);Expression(N) = (y^2 + (x - 1)^2 = 4);PointOnCurve(P, G);IsDiameter(LineSegmentOf(E,F),N)", "query_expressions": "Max(DotProduct(VectorOf(P, E), VectorOf(P, F)))", "answer_expressions": "21", "fact_spans": "[[[4, 43]], [[54, 78]], [[48, 53]], [[48, 53]], [[0, 3]], [[4, 43]], [[54, 78]], [[0, 47]], [[48, 85]]]", "query_spans": "[[[87, 141]]]", "process": "Circle N: (x-1)^{2}+y^{2}=4 has center N(1,0) and radius length 2. Let point P(x,y), then y^{2}=15-\\frac{15}{16}x^{2} and -4\\leqslant x \\leqslant 4, \\overrightarrow{PE}=\\overrightarrow{PN}+\\overrightarrow{NE}, \\overrightarrow{PF}=\\overrightarrow{PN}+\\overrightarrow{NF}=\\overrightarrow{PN}-\\overrightarrow{NE}, so, \\overrightarrow{PE}\\cdot\\overrightarrow{PF}=(\\overrightarrow{PN}+\\overrightarrow{NE})\\cdot(\\overrightarrow{PN}-\\overrightarrow{NE})=\\overrightarrow{PN}^{2}-\\overrightarrow{NE}^{2}=(x-1)^{2}+y^{2}-4=x^{2}-2x+1+15-\\frac{15}{16}x^{2}-4=\\frac{1}{16}x^{2}-2x+12=\\frac{1}{16}(x-16)^{2}-4, so, when x=-4, \\overrightarrow{PE}\\cdot\\overrightarrow{PF} reaches its maximum value, i.e., (\\overrightarrow{PE}\\cdot\\overrightarrow{PF})_{\\max}=\\frac{1}{16}\\times(-4)^{2}+8+12=21" }, { "text": "Given that the line $l$ intersects the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{6}=1$ at points $A$ and $B$, and the midpoint of segment $AB$ is $(-1,1)$, what is the slope of line $l$?", "fact_expressions": "l: Line;G: Ellipse;Expression(G) = (x^2/8 + y^2/6 = 1);A: Point;B: Point;Intersection(l, G) = {A, B};Coordinate(MidPoint(LineSegmentOf(A,B)))=(-1,1)", "query_expressions": "Slope(l)", "answer_expressions": "3/4", "fact_spans": "[[[2, 7], [78, 83]], [[8, 45]], [[8, 45]], [[46, 49]], [[50, 53]], [[2, 55]], [53, 73]]", "query_spans": "[[[78, 88]]]", "process": "According to the problem, let A(x_{1},y_{1}), B(x_{2},y_{2}). Since the midpoint of AB is (-1,1), therefore (\\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{=0}\\frac{2(y_{1}-y_{2})}{6}=0\\Rightarrow\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{3}{4}, that is, the slope of the required line is \\frac{3}{4}" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$, the right focus is $F$, the right directrix is $l$, point $A\\in l$, segment $AF$ intersects $C$ at point $B$, if $\\overrightarrow{FA}=3 \\overrightarrow{FB}$, then $|\\overrightarrow{AF}|$=?", "fact_expressions": "A: Point;F: Point;C: Ellipse;B: Point;l: Line;Expression(C) = (x^2/2 + y^2 = 1);RightFocus(C) = F;RightDirectrix(C) = l;In(A, l) = True;Intersection(LineSegmentOf(A, F), C) = B;VectorOf(F, A) = 3*VectorOf(F, B)", "query_expressions": "Abs(VectorOf(A, F))", "answer_expressions": "sqrt(2)", "fact_spans": "[[[63, 67]], [[39, 42]], [[2, 34], [68, 71]], [[72, 76]], [[47, 50]], [[2, 34]], [[2, 42]], [[2, 50]], [[51, 60]], [[61, 76]], [[78, 121]]]", "query_spans": "[[[123, 148]]]", "process": "" }, { "text": "$O$ is the coordinate origin, $F$ is the focus of the parabola $C$: $y^{2}=2 x$, $P(x_{0} , y_{0})$ is a point on $C$, if $|P F|=\\frac{3}{2} x_{0}$, then the area of $\\triangle P O F$ is?", "fact_expressions": "C: Parabola;P: Point;O: Origin;F: Point;x0: Number;y0: Number;Expression(C) = (y^2 = 2*x);Coordinate(P) = (x0, y0);Focus(C) = F;PointOnCurve(P, C);Abs(LineSegmentOf(P, F)) = (3/2)*x0", "query_expressions": "Area(TriangleOf(P, O, F))", "answer_expressions": "sqrt(2)/4", "fact_spans": "[[[13, 32], [55, 58]], [[36, 54]], [[0, 3]], [[9, 12]], [[36, 54]], [[36, 54]], [[13, 32]], [[36, 54]], [[9, 35]], [[36, 61]], [[63, 88]]]", "query_spans": "[[[90, 112]]]", "process": "Given the parabola C: y^{2}=2x, its focus is F(\\frac{1}{2},0). Let P(x_{0},y_{0}) be a point on C. If |PF|=\\frac{3}{2}x_{0}, then we have: \\frac{1}{2}+x_{0}=\\frac{3}{2}x_{0}. Solving gives x_{0}=1, so P(1,\\pm\\sqrt{2}). Then the area of triangle POF is: \\frac{1}{2}\\times\\frac{1}{2}\\times\\sqrt{2}=\\frac{\\sqrt{2}}{4}." }, { "text": "The equation of the conic section sharing foci with the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$ and passing through the point $(2 , 1)$ is?", "fact_expressions": "G: Hyperbola;H: ConicSection;P:Point;Expression(G) = (x^2/4 - y^2/2 = 1);Coordinate(P) = (2, 1);Focus(G)=Focus(H);PointOnCurve(P,H)", "query_expressions": "Expression(H)", "answer_expressions": "{(x^2/3-y^2/3=1),(x^2/8+y^2/2=1)}", "fact_spans": "[[[1, 39]], [[56, 60]], [[45, 55]], [[1, 39]], [[45, 55]], [[0, 60]], [[44, 60]]]", "query_spans": "[[[56, 65]]]", "process": "The foci of the given curve are $ F(\\pm\\sqrt{6},0) $. If the curve is an ellipse, by the given conditions, assume the required ellipse equation is $ \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 $, $ (a>b>0) $. Since the ellipse passes through the point $ (2,1) $, we have\n$$\n\\begin{cases}\na^2 - b^{2} = (\\pm\\sqrt{6})^{2} \\\\\n\\frac{4}{a^{2}} + \\frac{1}{b^{2}} = 1\n\\end{cases}\n$$\nSolving gives $ a^{2}=8 $, $ b^{2}=2 $. Therefore, the required ellipse is $ \\frac{x^{2}}{8}+\\frac{y^{2}}{2}=1 $.\n\nIf the curve is a hyperbola, assume the required hyperbola equation is $ \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 $, $ (a>0,b>0) $. Since the hyperbola passes through the point $ (2,1) $,\n$$\n\\begin{cases}\na^{2} + b^{2} = (\\pm\\sqrt{6})^{2} \\\\\n\\frac{4}{a^{2}} - \\frac{1}{b^{2}} = 1\n\\end{cases}\n$$\nSolving gives $ a^{2}=3 $, $ b^{2}=3 $. Therefore, the required hyperbola is $ \\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1 $." }, { "text": "Given the parabola $y^{2}=4x$ has focus $F$ and directrix $l$. A line passing through the focus $F$ intersects the parabola at points $A$ and $B$, such that $\\overrightarrow{AF}=\\overrightarrow{FB}$. If the projections of points $A$ and $B$ onto $l$ are $M$ and $N$ respectively, then what is the inradius of $\\Delta MFN$?", "fact_expressions": "G: Parabola;H: Line;M: Point;F: Point;N: Point;A: Point;B: Point;l: Line;Expression(G) = (y^2 = 4*x);Focus(G) = F;Directrix(G) = l;PointOnCurve(F,H);Intersection(H,G) = {A,B};VectorOf(A, F) = VectorOf(F, B);Projection(A,l)=M;Projection(B,l)=N", "query_expressions": "Radius(InscribedCircle(TriangleOf(M,F,N)))", "answer_expressions": "2*(sqrt(2)-1)", "fact_spans": "[[[2, 16], [41, 44]], [[38, 40]], [[124, 127]], [[20, 23], [20, 23]], [[128, 131]], [[45, 48], [103, 107]], [[51, 54], [110, 113]], [[27, 30], [27, 30]], [[2, 16]], [[2, 23]], [[2, 30]], [[31, 40]], [[38, 56]], [[58, 101]], [[103, 131]], [[103, 131]]]", "query_spans": "[[[133, 155]]]", "process": "The focus of the parabola y^{2}=4x is F(1,0). Since \\overrightarrow{AF}=\\overrightarrow{FB}, the line l is perpendicular to the x-axis. Therefore, A(1,2), B(1,-2), so M(-1,2), N(-1,-2), \\overrightarrow{FM}=(-2,2), \\overrightarrow{FN}=(-2,-2). Since \\overrightarrow{FM}\\cdot\\overrightarrow{FN}=0, triangle MFN is a right triangle, and |\\overrightarrow{FM}|=|\\overrightarrow{FN}|=2\\sqrt{2}. Let r be the inradius. Then \\frac{1}{2}\\times2\\sqrt{2}\\times2\\sqrt{2}=\\frac{1}{2}(2\\sqrt{2}+2\\sqrt{2}+4)r, solving gives r=\\frac{2}{\\sqrt{2}+1}=2(\\sqrt{2}-1)." }, { "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1$ are?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/4 = 1)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*(2/3)*x", "fact_spans": "[[[0, 38]], [[0, 38]]]", "query_spans": "[[[0, 46]]]", "process": "" }, { "text": "Given the parabola $C$: $x^{2}=4 y$, draw tangents $PA$ and $PB$ from point $P(0,-1)$ to the parabola, where $A$ and $B$ are the points of tangency. Then $\\overrightarrow{PA} \\cdot \\overrightarrow{PB}$=?", "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 4*y);P: Point;Coordinate(P) = (0, -1);TangentOfPoint(P, C) = {LineOf(P, A), LineOf(P, B)};TangentPoint(LineOf(P, A), C) = A;TangentPoint(LineOf(P, B), C) = B;A: Point;B: Point", "query_expressions": "DotProduct(VectorOf(P, A), VectorOf(P, B))", "answer_expressions": "0", "fact_spans": "[[[2, 21], [34, 37]], [[2, 21]], [[23, 33]], [[23, 33]], [[22, 52]], [[22, 65]], [[22, 65]], [[54, 57]], [[59, 62]]]", "query_spans": "[[[67, 118]]]", "process": "\\because the tangent line passes through point P(0,-1) and intersects the parabola C: x^{2}=4y at two points A and B, \\therefore the slope of the tangent line exists. Let the slope be k, then the equation of the tangent line passing through P(0,-1) is y=kx-1. Solving simultaneously \n\\begin{cases}y=kx-1\\\\x^{2}=4y\\end{cases}, we get x^{2}-4kx+4=0. Let d=16k^{2}-16=0, \\therefore k=\\pm1. \\therefore the equations of tangents PA and PB are respectively: y=x-1 and y=-x-1. Solving simultaneously \n\\begin{cases}y=x-1\\\\x^{2}=4y\\end{cases}, we get x^{2}-4x+4=0, \\therefore x=2, \\therefore A(2,1). Similarly, B(-2,1). \\because P(0,-1), \\therefore \\overrightarrow{PA}=(2,2), \\overrightarrow{PB}=(-2,2), \\therefore \\overrightarrow{PA}\\cdot\\overrightarrow{PB}=-4+4=0xx." }, { "text": "The equation of the locus of points whose sum of distances to $(-2,1)$ and $(2 , 1)$ is $4$ is?", "fact_expressions": "A: Point;B: Point;Coordinate(A) = (-2, 1);Coordinate(B) = (2, 1);P: Point;Distance(P, A) + Distance(P, B) =4", "query_expressions": "LocusEquation(P)", "answer_expressions": "(y=1)&(-2<=x<=2)", "fact_spans": "[[[1, 10]], [[11, 20]], [[1, 10]], [[11, 20]], [[30, 31]], [[0, 31]]]", "query_spans": "[[[30, 38]]]", "process": "Denote points A(-2,1), B(2,1), and let the desired point be P. Since |PA| + |PB| = |AB|, the trajectory of point P is the line segment AB. Thus, the equation of the locus of the moving point is y = 1 (-2 \\leqslant x \\leqslant 2)." }, { "text": "Through the focus $F$ of the parabola $y^{2}=4x$, draw a line intersecting the parabola at points $A$ and $B$. If $|AB|=8$ and $|AF|<|BF|$, then $|BF|=$?", "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 8;Abs(LineSegmentOf(A, F)) < Abs(LineSegmentOf(B, F))", "query_expressions": "Abs(LineSegmentOf(B, F))", "answer_expressions": "4+2*sqrt(2)", "fact_spans": "[[[1, 15], [25, 28]], [[22, 24]], [[29, 32]], [[33, 36]], [[18, 21]], [[1, 15]], [[1, 21]], [[0, 24]], [[22, 38]], [[40, 49]], [[51, 64]]]", "query_spans": "[[[66, 75]]]", "process": "From $ y^{2} = 4x $, we obtain the focus $ F(1,0) $. Also, since $ |AB| = 8 $, the slope of $ AB $ exists (otherwise $ |AB| = 4 $). Let the equation of line $ AB $ be $ y = k(x - 1) $ ($ k \\neq 0 $), and let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Substituting $ y = k(x - 1) $ into $ y^{2} = 4x $, we get $ k^{2}x^{2} - (2k^{2} + 4)x + k^{2} = 0 $. Hence, $ x_{1} + x_{2} = 2 + \\frac{4}{k^{2}} $. From $ |AB| = |AF| + |BF| = x_{1} + x_{2} + 2 = 8 $, we have $ x_{1} + x_{2} = 2 + \\frac{4}{k^{2}} = 6 $, so $ k^{2} = 1 $. Then $ x^{2} - 6x + 1 = 0 $. Since $ |AF| < |BF| $, we have $ x_{1} = 3 - 2\\sqrt{2} $, $ x_{2} = 3 + 2\\sqrt{2} $. Therefore, $ |BF| = x_{1} + 1 = 3 + 2\\sqrt{2} + 1 = 4 + 2\\sqrt{2} $." }, { "text": "The eccentricity $e$ of the hyperbola $\\frac{x^{2}}{9}-y^{2}=1$ is? The equations of the asymptotes are?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2 = 1);e: Number;Eccentricity(G) = e", "query_expressions": "e;Expression(Asymptote(G))", "answer_expressions": "sqrt(10)/3\nx+pm*3*y=0", "fact_spans": "[[[0, 28]], [[0, 28]], [[32, 35]], [[0, 35]]]", "query_spans": "[[[32, 37]], [[0, 44]]]", "process": "" }, { "text": "The standard equation of the ellipse passing through the point $(2, -3)$ and sharing the same foci with the ellipse $9x^{2} + 2y^{2} = 36$ is?", "fact_expressions": "G: Ellipse;H: Point;C: Ellipse;Expression(G) = (9*x^2 + 2*y^2 = 36);Coordinate(H) = (2, -3);PointOnCurve(H, C);Focus(G) = Focus(C)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/7 + y^2/21 = 1", "fact_spans": "[[[14, 36]], [[2, 12]], [[42, 44]], [[14, 36]], [[2, 12]], [[0, 44]], [[13, 44]]]", "query_spans": "[[[42, 51]]]", "process": "" }, { "text": "If the focal distance of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{6}=1$ is equal to $6$, then $m=?$", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/6 + x^2/m = 1);m: Number;FocalLength(G) = 6", "query_expressions": "m", "answer_expressions": "3", "fact_spans": "[[[1, 39]], [[1, 39]], [[49, 52]], [[1, 47]]]", "query_spans": "[[[49, 54]]]", "process": "From the equation of the hyperbola, we know that m+6=(\\frac{6}{2})^{2}=9, so m=3. Therefore, fill in 3." }, { "text": "The length of the major axis of the ellipse $x^{2}+4 y^{2}=100$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 4*y^2 = 100)", "query_expressions": "Length(MajorAxis(G))", "answer_expressions": "20", "fact_spans": "[[[0, 21]], [[0, 21]]]", "query_spans": "[[[0, 27]]]", "process": "The ellipse $x^{2}+4y^{2}=100$ is converted into standard form, yielding: $\\frac{x^2}{100}+\\frac{y^{2}}{25}=1$, $\\therefore a=10$, $b=5$. The major axis length of the ellipse $x^{2}+4y^2=100$ is $2a=20$." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, its asymptotes intersect the circle $(x-2)^{2}+y^{2}=2$, and the lengths of the two chords cut by the circle on the asymptotes are both $2$. Then the eccentricity of the hyperbola is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;G: Circle;Expression(G) = (y^2 + (x - 2)^2 = 2);IsIntersect(Asymptote(C), G);L1: Line;L2: Line;Asymptote(C) = {L1, L2};Length(InterceptChord(L1, G)) = 2;Length(InterceptChord(L2, G)) = 2", "query_expressions": "Eccentricity(C)", "answer_expressions": "2*sqrt(3)/3", "fact_spans": "[[[2, 63], [64, 65], [112, 115]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[69, 89], [97, 98]], [[69, 89]], [[64, 91]], [], [], [[64, 68]], [[64, 110]], [[64, 110]]]", "query_spans": "[[[112, 121]]]", "process": "Hyperbola C: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0, b>0) has an asymptote given by bx+ay=0, which intersects the circle (x-2)^{2}+y^{2}=2, with chord length 2, meaning the distance from the center of the circle (2,0) to the asymptote bx+ay=0 is 1. Therefore, \\frac{|2b|}{\\sqrt{a^{2}+b^{2}}}=1, yielding 3b^{2}=a^{2}. In the hyperbola, b^{2}=c^{2}-a^{2}, so 3c^{2}-3a^{2}=a^{2}, i.e., 3c^{2}=4a^{2}. Hence, e=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{\\frac{4}{3}}=\\frac{2\\sqrt{3}}{3}" }, { "text": "If the focus of the parabola $y^{2}=2 p x$ coincides with the right focus of the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$, then the value of $p$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;H: Ellipse;Expression(H) = (x^2/6 + y^2/2 = 1);Focus(G) = RightFocus(H)", "query_expressions": "p", "answer_expressions": "4", "fact_spans": "[[[1, 17]], [[1, 17]], [[66, 69]], [[21, 58]], [[21, 58]], [[1, 64]]]", "query_spans": "[[[66, 73]]]", "process": "Test Analysis: According to the given ellipse equation $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$, it is known that $a^{2}=6$, $b^{2}=2$, so $c=2$, thereby determining the right focus of the ellipse as $(2,0)$. Since the focus of the parabola $y^{2}=2px$ is $(\\frac{p}{2},0)$, we have $\\frac{p}{2}=2$, thus $p=4$." }, { "text": "Point $P$ lies on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, with its left and right foci denoted by $F_{1}$ and $F_{2}$ respectively. The line $P F_{1}$ is tangent to the circle centered at the origin $O$ with radius $a$ at point $A$. The perpendicular bisector of segment $P F_{1}$ passes exactly through point $F_{2}$. What is the value of $\\frac{S_{\\Delta O F_{1} A}}{S_{\\triangle P F_{1} F_{2}}}$?", "fact_expressions": "P: Point;PointOnCurve(P,RightPart(G)) = True;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Circle;Center(H) = O;O: Origin;Radius(H) = a;TangentPoint(LineOf(P,F1),H) = A;A: Point;PointOnCurve(F2,PerpendicularBisector(LineSegmentOf(P,F1))) = True;IsTangent(LineOf(P,F1),H) = True", "query_expressions": "Area(TriangleOf(O,F1,A))/Area(TriangleOf(P,F1,F2))", "answer_expressions": "1/8", "fact_spans": "[[[0, 4]], [[0, 65]], [[5, 61], [66, 67]], [[5, 61]], [[8, 61]], [[115, 118]], [[8, 61]], [[8, 61]], [[75, 82]], [[83, 90], [151, 159]], [[66, 90]], [[66, 90]], [[122, 123]], [[103, 123]], [[104, 111]], [[115, 123]], [[91, 130]], [[126, 130]], [[131, 159]], [[91, 125]]]", "query_spans": "[[[160, 222]]]", "process": "Since |OA| = a, it follows that |BF_{2}| = 2a. Also, |F_{1}F_{2}| = 2c, so |BF_{1}| = 2b, |PF_{1}| = 4b. Moreover, |AF_{1}| = b, thus \\frac{S_{\\DeltaOFA}}{S_{\\DeltaPF_{1}F_{2}}}=\\frac{0.5bc\\cdot\\sinF_{1}}{0.5\\times4b\\times2c\\sinF_{1}}=\\frac{1}{8}" }, { "text": "If the vertex of the parabola $x^{2}=4 y$ is the point on the parabola closest to the point $A (0 , a)$, then the range of real values for $a$ is?", "fact_expressions": "G: Parabola;a: Real;Expression(G) = (x^2 = 4*y);A:Point;Coordinate(A)=(0,a);WhenMin(Distance(Vertex(G),A))", "query_expressions": "Range(a)", "answer_expressions": "(-oo,2]", "fact_spans": "[[[1, 15], [19, 22]], [[46, 51]], [[1, 15]], [[24, 37]], [[24, 37]], [[16, 44]]]", "query_spans": "[[[46, 58]]]", "process": "Let point P(x, y) be an arbitrary point on the parabola, then the square of the distance from point P to point A(0, a) is |AP|^{2} = x^{2} + (y - a)^{2} = x^{2} + y^{2} - 2ay + a^{2}. Since x^{2} = 4y, therefore |AP|^{2} = 4y + y^{2} - 2ay + a^{2} (y \\geqslant 0) = y^{2} + 2(2 - a)y + a^{2} (y \\geqslant 0). Therefore, the axis of symmetry is y = a - 2. Since the point closest to point A(0, a) is exactly the vertex, we have a - 2 \\leqslant 0, solving gives a \\leqslant 2." }, { "text": "Given the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{b^{2}}=1(b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. One of its asymptotes has the equation $y=x$. The point $P(\\sqrt{3}, y_{0})$ lies on this hyperbola. Find $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$=?", "fact_expressions": "G: Hyperbola;b: Number;P: Point;F1: Point;F2: Point;b>0;Expression(G) = (x^2/2 - y^2/b^2 = 1);y0:Number;Coordinate(P) = (sqrt(3), y0);LeftFocus(G) = F1;RightFocus(G) = F2;Expression(OneOf(Asymptote(G))) = (y = x);PointOnCurve(P, G)", "query_expressions": "DotProduct(VectorOf(P, F1), VectorOf(P, F2))", "answer_expressions": "0", "fact_spans": "[[[2, 49], [74, 75], [112, 115]], [[5, 49]], [[89, 110]], [[57, 64]], [[65, 72]], [[5, 49]], [[2, 49]], [[90, 110]], [[89, 110]], [[2, 72]], [[2, 72]], [[74, 88]], [[89, 116]]]", "query_spans": "[[[118, 177]]]", "process": "Problem Analysis: Since the asymptotes of the hyperbola are $ y = x $, we have $ \\frac{b}{\\sqrt{2}} = 1 $, so $ b = \\sqrt{2} $. The equation is $ \\frac{x^{2}}{2} - \\frac{y^{2}}{2} = 1 $. Substituting $ x = \\sqrt{3} $, we get $ \\frac{(\\sqrt{3})^{2}}{2} - \\frac{y_{0}^{2}}{2} = 1 $, hence $ y_{0} = \\pm1 $. Also, $ F_{1}(-2,0) $, $ F_{2}(2,0) $, $ |PF_{1}|^{2} = (-2-\\sqrt{3})^{2} + (0-1)^{2} = 8+4\\sqrt{3} $. Similarly, $ |PF_{2}|^{2} = 8-4\\sqrt{3} $. Therefore, $ |PF_{1}|^{2} + |PF_{2}|^{2} = 16 = |F_{1}F_{2}|^{2} $, which implies $ PF_{1} \\perp PF_{2} $, so $ \\overrightarrow{PF}_{1} \\cdot \\overrightarrow{PF_{2}} = 0 $." }, { "text": "The coordinates of the moving point $M(x, y)$ satisfy $\\sqrt{(x-2)^{2}+y^{2}}=|x+2|$. What is the trajectory equation of the moving point $M$?", "fact_expressions": "x_: Number;y_: Number;M: Point;Coordinate(M)=(x_, y_);sqrt((x_-2)^2 + y_^2) = Abs(x_+2)", "query_expressions": "LocusEquation(M)", "answer_expressions": "y^2=8*x", "fact_spans": "[[[4, 13]], [[4, 13]], [[4, 13], [52, 55]], [[4, 13]], [[18, 48]]]", "query_spans": "[[[52, 62]]]", "process": "Let F(2,0) and the line l: x = -2. Then the distance from a moving point M to the point F is \\sqrt{(x-2)^{2}+y^{2}}, and the distance from the moving point M to the line l: x = -2 is |x+2|. Since \\sqrt{(x-2)^{2}+y^{2}} = |x+2|, the trajectory of the moving point M is a parabola with focus F(2,0) and directrix x = -2, and its trajectory equation is y^{2} = 8x." }, { "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F(1,0)$, point $A$ is the intersection point of the directrix of the parabola and the coordinate axis, point $M$ lies on the parabola, when $\\frac{|M F|}{|M A|}$ is minimized, $|M A|$=?", "fact_expressions": "C: Parabola;p: Number;F: Point;M: Point;A: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(F) = (1, 0);Focus(C) = F;Intersection(Directrix(C),axis)=A;PointOnCurve(M, C);WhenMin(Abs(LineSegmentOf(M,F))/Abs(LineSegmentOf(M,A)))", "query_expressions": "Abs(LineSegmentOf(M, A))", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[2, 28], [46, 49], [65, 68]], [[10, 28]], [[32, 40]], [[60, 64]], [[41, 45]], [[10, 28]], [[2, 28]], [[32, 40]], [[2, 40]], [[41, 59]], [[60, 69]], [[70, 95]]]", "query_spans": "[[[96, 105]]]", "process": "Solution: Since the focus of the parabola $ C: y^{2} = 2px $ ($ p > 0 $) is $ F(1,0) $, we have $ y^{2} = 4x $, and the equation of the directrix is $ x = -1 $, $ A(-1,0) $. Let $ M\\left(\\frac{y^{2}}{4}, y\\right) $, then $ |MF| = \\sqrt{\\left(\\frac{y^{2}}{4} - 1\\right)^{2} + y^{2}} $, $ |MA| = $. Therefore, when $ y = 0 $, it is clear that $ \\frac{|MF|}{|MA|} = 1 $. When $ y \\neq 0 $, \n$$\n\\left(\\frac{|MF|}{|MA|}\\right)^{2} = \\frac{\\left(\\frac{y^{2}}{4} - 1\\right)^{2} + y^{2}}{\\left(\\frac{y^{2}}{4} + 1\\right)^{2} + y^{2}} = \\frac{\\frac{1}{16}y^{4} + \\frac{1}{2}y^{2} + 1}{\\frac{1}{16}y^{4} + \\frac{3}{2}y^{2} + 1} = 1 - \\frac{y^{2}}{\\frac{1}{16}y^{4} + \\frac{3}{2}y^{2} + 1}.\n$$\nThe equality holds if and only if $ \\frac{y^{2}}{16} = \\frac{1}{y^{2}} $, so when $ y^{2} = 4 $, $ \\frac{|MF|}{|MA|} $ attains its minimum value $ \\frac{\\sqrt{2}}{2} $, at which point $ M(1, \\pm 2) $, $ |MA| = 2\\sqrt{2} $." }, { "text": "$\\odot O$: $x^{2}+y^{2}=16$, $A(-2,0)$, $B(2,0)$ are two fixed points, $l$ is a tangent line to $\\odot O$. If a parabola passing through points $A$ and $B$ has the line $l$ as its directrix, then the equation of the locus of the focus of this parabola is?", "fact_expressions": "O: Circle;Expression(O) = (x^2 + y^2 = 16);A: Point;Coordinate(A) = (-2, 0);B: Point;Coordinate(B) = (2, 0);IsTangent(l, O) = True;l: Line;PointOnCurve(A, G) = True;PointOnCurve(B, G) = True;G: Parabola;Directrix(G) = l", "query_expressions": "LocusEquation(Focus(G))", "answer_expressions": "(x^2/16+y^2/12=1)&Negation(y=0)", "fact_spans": "[[[0, 28]], [[0, 28]], [[30, 39], [76, 79]], [[30, 39]], [[41, 49], [80, 83]], [[41, 49]], [[55, 73]], [[55, 58], [90, 95]], [[75, 89]], [[75, 89]], [[86, 89], [101, 104]], [[86, 98]]]", "query_spans": "[[[101, 114]]]", "process": "Draw the graph according to the problem, arbitrarily draw a tangent line, and draw AB perpendicular to the tangent line at points C and D. Let F be the focus of the parabola. According to the definition of the parabola, we obtain FA = AE, FB = BC, and thus FA + FB = AE + BC. Since O is the midpoint of AB, draw OG perpendicular to the tangent line at point G. OG is the median line of trapezoid ABCE, hence BF + AF = 2OG = 8. Therefore, according to the definition of the ellipse, the trajectory is an ellipse with AB as foci, semi-major axis length 4, and c = 2. Thus, the trajectory equation is \\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1(y\\neq0). From the given condition, the foci must not lie on the x-axis; therefore, two points must be excluded." }, { "text": "Given a point $A$ on the parabola $y^{2}=2 p x(p>0)$, the area of triangle $O A F$ formed by point $A$, the focus $F$, and the origin $O$ is $p$, and $|A F|=4$. Then $p=?$", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;A: Point;PointOnCurve(A, G);F: Point;Focus(G) = F;O: Origin;Area(TriangleOf(O, A, F)) = p;Abs(LineSegmentOf(A, F)) = 4", "query_expressions": "p", "answer_expressions": "4", "fact_spans": "[[[2, 23]], [[2, 23]], [[61, 64], [77, 80]], [[5, 23]], [[26, 29]], [[2, 29]], [[32, 35]], [[2, 35]], [[37, 44]], [[47, 64]], [[65, 74]]]", "query_spans": "[[[77, 82]]]", "process": "" }, { "text": "If the line $y = kx + 1$ ($k \\in \\mathbb{R}$) always has common points with the ellipse $\\frac{x^{2}}{5} + \\frac{y^{2}}{t} = 1$ whose foci lie on the $x$-axis, then what is the range of values for $t$?", "fact_expressions": "G: Ellipse;t: Number;H: Line;k: Real;Expression(G) = (x^2/5 + y^2/t = 1);Expression(H) = (y = k*x + 1);PointOnCurve(Focus(G),xAxis);IsIntersect(G,H)", "query_expressions": "Range(t)", "answer_expressions": "[1,5)", "fact_spans": "[[[31, 68]], [[75, 78]], [[1, 21]], [[3, 21]], [[31, 68]], [[1, 21]], [[22, 68]], [[1, 73]]]", "query_spans": "[[[75, 85]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{b^{2}}=1$ $(4>b>0)$ with left and right foci $F_{1}$, $F_{2}$, and eccentricity $\\frac{\\sqrt{3}}{2}$. If $P$ is a point on the ellipse $C$ such that $\\angle F_{1} P F_{2}=90^{\\circ}$, then the area of $\\Delta F_{1} P F_{2}$ equals?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/16 + y^2/b^2 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;Eccentricity(C) = sqrt(3)/2;PointOnCurve(P,C) = True;P: Point;AngleOf(F1, P, F2) = ApplyUnit(90, degree);b:Number;4>b;b>0", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "4", "fact_spans": "[[[2, 51], [103, 108]], [[2, 51]], [[57, 64]], [[65, 72]], [[2, 72]], [[2, 72]], [[2, 97]], [[99, 111]], [[99, 102]], [[113, 146]], [[2, 51]], [[2, 51]], [[2, 51]]]", "query_spans": "[[[148, 176]]]", "process": "Analysis: From the ellipse $ C: \\frac{x^{2}}{16}+\\frac{y^{2}}{b^{2}}=1 $, we can obtain $ a=4 $. Given the eccentricity $ \\frac{\\sqrt{3}}{2} $, we can find the value of $ c $. Since $ \\angle F_{1}PF_{2}=90^{\\circ} $, combining the definition of the ellipse and the Pythagorean theorem forms a system of equations to solve for the value of $ |PF_{1}|\\cdot|PF_{2}| $, and then the area can be calculated.\n\nDetailed solution: From the given conditions, $ a=4 $, $ e=\\frac{c}{a}=\\frac{\\sqrt{3}}{2} $, we get $ a=4 $, $ b=2 $, $ c=2\\sqrt{3} $. Since $ P $ is a point on the ellipse $ C $ and $ \\angle F_{1}PF_{2}=90^{\\circ} $, we have $ |PF_{1}|+|PF_{2}|=2a=8 $, $ |PF_{1}|^{2}+|PF_{2}|^{2}=4c^{2}=48 $. Therefore, $ |PF_{1}|+|PF_{2}|^{2}-2|PF_{1}|\\cdot|PF_{2}|=48 $, that is, $ 64-2|PF_{1}|\\cdot|PF_{2}|=48 $, so $ |PF_{1}|\\cdot|PF_{2}|=8 $. Hence, the area of $ \\triangle F_{1}PF_{2} $ is $ S=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|=\\frac{1}{2}\\times8=4 $." }, { "text": "Given that the coordinate of endpoint $B$ of segment $AB$ is $(4,0)$, and endpoint $A$ moves on the circle $x^{2}+y^{2}=1$, then the trajectory equation of the midpoint of segment $AB$ is?", "fact_expressions": "G: Circle;B: Point;A: Point;Endpoint(LineSegmentOf(A,B))={A,B};Expression(G) = (x^2 + y^2 = 1);Coordinate(B)=(4,0);PointOnCurve(A, G)", "query_expressions": "LocusEquation(MidPoint(LineSegmentOf(A,B)))", "answer_expressions": "Preserve all mathematical expressions, LaTeX formulas, symbols, and equations EXACTLY. \nDo not add explanations, do not rewrite formulas, do not omit any content. \n\nText:", "fact_spans": "[[[33, 49]], [[12, 15]], [[29, 32]], [[2, 32]], [[33, 49]], [[12, 26]], [[29, 52]]]", "query_spans": "[[[54, 71]]]", "process": "" }, { "text": "Let $F_{1}$ and $F_{2}$ be the two foci of the hyperbola $4 x^{2}-2 y^{2}=1$, point $P$ lies on the hyperbola and satisfies $\\angle F_{1} P F_{2}=60^{\\circ}$, then the area of $\\Delta F_{1} P F_{2}$ is?", "fact_expressions": "G: Hyperbola;F1: Point;P: Point;F2: Point;Expression(G) = (4*x^2 - 2*y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree)", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "sqrt(3)/2", "fact_spans": "[[[17, 39], [50, 53]], [[1, 8]], [[45, 49]], [[9, 16]], [[17, 39]], [[1, 44]], [[45, 54]], [[58, 91]]]", "query_spans": "[[[93, 120]]]", "process": "From the given, $ a^{2} = \\frac{1}{4} $, $ b^{2} = \\frac{1}{2} $, $ \\therefore c^{2} = \\frac{1}{4} + \\frac{1}{2} = \\frac{3}{4} $, $ \\therefore c = \\frac{\\sqrt{3}}{2} $. From the given, \n$$\n\\begin{cases}\n3 = |PF_{1}|^{2} + |PF_{2}|^{2} - 2|PF_{1}||PF_{2}| \\times \\frac{1}{2} \\\\\n|PF_{1}| - |PF_{2}| = 1\n\\end{cases}\n$$\nSo $ s = \\frac{1}{2} \\cdot |PF_{1}||PF_{2}| \\cdot \\sin\\frac{\\pi}{3} = \\frac{1}{2} \\cdot 2 \\cdot \\frac{\\sqrt{3}}{2} = \\frac{\\sqrt{3}}{2} $." }, { "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ have no common points with the circle $x^{2}+(y-2)^{2}=1$. Then, the range of the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Circle;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + (y - 2)^2 = 1);NumIntersection(Asymptote(G), H) = 0", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1, 2)", "fact_spans": "[[[0, 46], [78, 81]], [[3, 46]], [[3, 46]], [[51, 71]], [[0, 46]], [[51, 71]], [[0, 76]]]", "query_spans": "[[[78, 91]]]", "process": "" }, { "text": "Given that the hyperbola $C$ with center at the origin has its right focus at $F(3,0)$ and eccentricity equal to $\\frac{3}{2}$, then the equation of $C$ is?", "fact_expressions": "C: Hyperbola;O:Origin;Center(C)=O;F: Point;Coordinate(F) = (3, 0);RightFocus(C) = F;Eccentricity(C)=3/2", "query_expressions": "Expression(C)", "answer_expressions": "x^2/4 - y^2/5 = 1", "fact_spans": "[[[8, 14], [48, 51]], [[5, 7]], [[2, 14]], [[19, 27]], [[19, 27]], [[8, 27]], [[8, 46]]]", "query_spans": "[[[48, 56]]]", "process": "" }, { "text": "Draw a perpendicular line from a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ to one of its asymptotes. The foot of the perpendicular lies exactly on the curve $\\frac{x^{2}}{b^{2}}+\\frac{y^{2}}{a^{2}}=1$. Then, the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Z: Line;PointOnCurve(OneOf(Focus(G)), Z);IsPerpendicular(Z, OneOf(Asymptote(G)));H: Curve;Expression(H) = (x^2/b^2 + y^2/a^2 = 1);PointOnCurve(FootPoint(Z, OneOf(Asymptote(G))), H)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[1, 57], [126, 129]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [], [[0, 71]], [[0, 71]], [[78, 123]], [[78, 123]], [[0, 124]]]", "query_spans": "[[[126, 135]]]", "process": "" }, { "text": "Given the ellipse $ C $: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 $ $(a>b>0)$, the left and right foci are $ F_{1} $, $ F_{2} $ respectively. Point $ P $ is any point on the ellipse that is not an endpoint of the major axis. If $ M $ is a point on the segment $ P F_{1} $ such that $ \\overrightarrow{M F_{1}}=2 \\overrightarrow{P M} $, and $ \\overrightarrow{M F_{2}} \\cdot \\overrightarrow{O P}=0 $, then what is the range of possible values for the eccentricity of the ellipse?", "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C);Negation(P=Endpoint(MajorAxis(C)));M: Point;PointOnCurve(M, LineSegmentOf(P, F1));O: Origin;VectorOf(M, F1) = 2*VectorOf(P, M);DotProduct(VectorOf(M, F2), VectorOf(O, P)) = 0", "query_expressions": "Range(Eccentricity(C))", "answer_expressions": "(1/2, 1)", "fact_spans": "[[[2, 59], [89, 91], [235, 237]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[68, 75]], [[76, 83]], [[2, 83]], [[2, 83]], [[84, 88]], [[84, 103]], [[84, 103]], [[105, 108]], [[105, 123]], [[178, 233]], [[127, 176]], [[178, 233]]]", "query_spans": "[[[235, 247]]]", "process": "Problem Analysis: According to the given conditions, let P(x,y), take the midpoint N of MF_{1}. Since \\overrightarrow{MF_{1}}=2\\overrightarrow{PM}, it follows that \\overrightarrow{NF_{1}}=\\frac{1}{2}\\overrightarrow{PN}. Solving gives point N(\\frac{x-c}{3},\\frac{y}{3}). Also, since \\overrightarrow{MF_{2}}\\cdot\\overrightarrow{OP}=0, we have \\overrightarrow{MF_{2}}\\bot\\overrightarrow{OP}. From the midline theorem of triangles, it follows that \\overrightarrow{ON}\\bot\\overrightarrow{OP}, that is, (x,y)\\cdot(\\frac{x-c}{3},\\frac{y}{3})=0. Simplifying yields (x-c)^{2}+y^{2}=c^{2}. Therefore, the trajectory of point P lies on a circle centered at (c,0) with radius c. For this circle and the ellipse to have common points, we require a-c\\frac{1}{2}. Hence, the eccentricity of the ellipse is e\\in(\\frac{1}{2},1)." }, { "text": "Given the parabola $E$: $y^{2}=4x$ with focus $F$ and directrix $l$. A line $m$ passing through $F$ intersects $E$ at points $A$ and $B$. The perpendicular bisector of $AF$ intersects $l$ and the $x$-axis at points $P$ and $Q$, respectively. If $\\angle AFP = \\angle AFQ$, then $|AB|=$?", "fact_expressions": "m: Line;E: Parabola;A: Point;F: Point;P: Point;Q: Point;B: Point;l: Line;Expression(E) = (y^2 = 4*x);Focus(E) = F;Directrix(E) = l;PointOnCurve(F, m);Intersection(m, E) = {A, B};Intersection(PerpendicularBisector(LineSegmentOf(A,F)),l)=P;Intersection(PerpendicularBisector(LineSegmentOf(A,F)),xAxis)=Q;AngleOf(A,F,P)=AngleOf(A,F,Q)", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "16/3", "fact_spans": "[[[42, 47]], [[2, 21], [48, 51]], [[53, 56]], [[25, 28], [38, 41]], [[86, 89]], [[90, 93]], [[57, 60]], [[32, 35], [77, 80]], [[2, 21]], [[2, 28]], [[2, 35]], [[37, 47]], [[42, 62]], [[63, 95]], [[63, 95]], [[96, 124]]]", "query_spans": "[[[126, 135]]]", "process": "From the given conditions, it can be deduced that $\\triangle APF$ is an equilateral triangle, then the inclination angle of line $AB$ can be obtained, and the chord length formula can be used to solve the problem. Let $M$ be the midpoint of $AF$. Since $\\angle AFP = \\angle AFQ$, and the perpendicular bisector of $AF$ intersects line $l$ and the $x$-axis at points $P$ and $Q$ respectively, therefore $\\triangle PMF \\cong \\triangle QMF$, so $PA = PF = FQ$, and $M$ is the midpoint of $PQ$. Hence, quadrilateral $APFO$ is a rhombus, so $AP \\parallel FO$. Thus, by the definition of the parabola, $PA = AF$, so $\\triangle APF$ is an equilateral triangle, so $\\angle PAF = \\angle AFQ = 60^{\\circ}$. Therefore, the equation of line $AB$ is $y = \\sqrt{3}(x - 1)$. Let $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$, from \n\\[\n\\begin{cases}\ny = \\sqrt{3}(x - 1)\n\\end{cases}\n\\]\nit follows that $3x^{2} - 10x + 3 = 0$. So $x_{1} + x_{2} = \\frac{10}{3}$, $x_{1}x_{2} = 1$, so \n\\[\n|AB| = \\sqrt{(1+3)\\left(\\frac{100}{9} - 4\\right)} = \\frac{16}{3}.\n\\]\nThe answer is: $\\frac{16}{3}$." }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0)$, circle $O$: $x^{2}+y^{2}=b^{2}$, from any point $P$ on the ellipse that does not coincide with a vertex, draw two tangents to circle $O$, with points of tangency $A$ and $B$ respectively. The line $AB$ intersects the $x$-axis and $y$-axis at points $M$ and $N$ respectively. Then $\\frac{a^{2}}{|O N|^{2}}+\\frac{b^{2}}{|O M|^{2}}$=?", "fact_expressions": "G: Ellipse;b: Number;a: Number;O: Circle;B: Point;A: Point;N: Point;M: Point;P:Point;a > b;b > 0;L1:Line;L2:Line;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(O)=(x^2+y^2=b^2);PointOnCurve(P,G);Negation(P=Vertex(G));TangentOfPoint(P,O)={L1,L2};TangentPoint(L1,O)=A;TangentPoint(L2,O)=B;Intersection(LineOf(A,B), xAxis) = M;Intersection(LineOf(A,B), yAxis) = N", "query_expressions": "a^2/Abs(LineSegmentOf(O, N))^2 + b^2/Abs(LineSegmentOf(O, M))^2", "answer_expressions": "a^2/b^2", "fact_spans": "[[[2, 56], [84, 86]], [[4, 56]], [[4, 56]], [[57, 82], [101, 105]], [[120, 123]], [[116, 119]], [[150, 153]], [[145, 149]], [[96, 100]], [[4, 56]], [[4, 56]], [], [], [[2, 56]], [[57, 82]], [[83, 100]], [[84, 100]], [[83, 110]], [[83, 123]], [[83, 123]], [[124, 153]], [[124, 153]]]", "query_spans": "[[[155, 206]]]", "process": "" }, { "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=2 p x(p>0)$, $M$ is a point on $C$, and the extension of $F M$ intersects the $y$-axis at point $N$. If $M$ is the midpoint of segment $F N$ and $|F N|=6$, then $p=?$", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;M: Point;PointOnCurve(M, C);N: Point;Intersection(OverlappingLine(LineSegmentOf(F, M)), yAxis) = N;MidPoint(LineSegmentOf(F, N)) = M;Abs(LineSegmentOf(F, N)) = 6", "query_expressions": "p", "answer_expressions": "4", "fact_spans": "[[[6, 32], [40, 43]], [[6, 32]], [[95, 98]], [[14, 32]], [[2, 5]], [[2, 35]], [[36, 39], [68, 71]], [[36, 46]], [[62, 66]], [[47, 66]], [[68, 82]], [[84, 93]]]", "query_spans": "[[[95, 100]]]", "process": "" }, { "text": "Given that the parabola $C$ has the coordinate origin $O$ as its vertex and $(\\frac{p}{2}, 0)$ as its focus, the line $x - m y - 2p = 0$ intersects the parabola $C$ at two points $A$, $B$, and the point $M(1,1)$ on the line $AB$ satisfies $OM \\perp AB$. Then the equation of the parabola $C$ is?", "fact_expressions": "C: Parabola;O: Origin;Vertex(C) = O;Coordinate(Focus(C)) = (p/2,0);p: Number;G: Line;Expression(G) = (-2*p - m*y + x = 0);m: Number;Intersection(C,G) = {A,B};A: Point;B: Point;PointOnCurve(M,LineOf(A,B)) = True;Coordinate(M) = (1, 1);M: Point;IsPerpendicular(LineSegmentOf(O,M),LineSegmentOf(A,B)) = True", "query_expressions": "Expression(C)", "answer_expressions": "y^2=2*x", "fact_spans": "[[[2, 8], [59, 65], [117, 123]], [[9, 16]], [[2, 19]], [[2, 42]], [[21, 39]], [[43, 58]], [[43, 58]], [[45, 58]], [[43, 78]], [[69, 73]], [[75, 78]], [[80, 98]], [[89, 98]], [[89, 98]], [[100, 115]]]", "query_spans": "[[[117, 128]]]", "process": "From the given condition, when y=0, x=2p. That is, the line x-my-2p=0 always passes through the fixed point (2p,0). Therefore, k_{AB}=\\frac{1-0}{1-2p}=\\frac{1}{1-2p}, and k_{OM}=\\frac{1-0}{1-0}=1. Since OM\\bot AB, it follows that k_{AB}\\cdot k_{OM}=-1, so \\frac{1}{1-2p}\\times 1=-1. Solving this equation gives p=1. Hence, the equation of parabola C is: y^2=2x" }, { "text": "The standard equation of the hyperbola with vertices at the foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ and foci at the vertices of the ellipse is?", "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/25 + y^2/16 = 1);Vertex(H)=Focus(G);Focus(H)=Vertex(G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/9 - y^2/16 = 1", "fact_spans": "[[[53, 56]], [[1, 40]], [[1, 40]], [[0, 56]], [[0, 56]]]", "query_spans": "[[[53, 65]]]", "process": "Analysis: From the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, the foci are $F(\\pm3,0)$ and the vertices are $A(\\pm5,0)$. Thus, the foci and vertices of the hyperbola can be obtained, which leads to the equation of the hyperbola. \nDetailed solution: Since the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ has foci at $F(\\pm3,0)$ and vertices at $A(\\pm5,0)$, the vertices and foci of the hyperbola are $F(\\pm3,0)$ and $A(\\pm5,0)$ respectively. We obtain $c=5$, $a=3 \\Rightarrow b=4$. Therefore, the equation of the hyperbola is $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$." }, { "text": "If the eccentricity of the hyperbola $x^{2}+ky^{2}=1$ is $2$, then what is the value of the real number $k$?", "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (k*y^2 + x^2=1);Eccentricity(G) = 2", "query_expressions": "k", "answer_expressions": "-1/3", "fact_spans": "[[[1, 20]], [[30, 35]], [[1, 20]], [[1, 28]]]", "query_spans": "[[[30, 39]]]", "process": "" }, { "text": "Given that point $P(m, 4)$ lies on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, $F_{1}$, $F_{2}$ are the two foci of the ellipse, and if the radius of the incircle of triangle $\\triangle P F_{1} F_{2}$ is $\\frac{3}{2}$, then what is the eccentricity of this ellipse?", "fact_expressions": "G: Ellipse;a: Number;b: Number;m: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (m, 4);PointOnCurve(P, G);Focus(G) = {F1, F2};Radius(InscribedCircle(TriangleOf(P, F1, F2))) = 3/2", "query_expressions": "Eccentricity(G)", "answer_expressions": "3/5", "fact_spans": "[[[13, 65], [86, 88], [144, 146]], [[15, 65]], [[15, 65]], [[3, 12]], [[2, 12]], [[70, 77]], [[78, 85]], [[15, 65]], [[15, 65]], [[13, 65]], [[2, 12]], [[2, 69]], [[70, 93]], [[95, 141]]]", "query_spans": "[[[144, 152]]]", "process": "Test Analysis: Since the radius of the incircle of triangle $ PF_{1}F_{2} $ is $ \\frac{3}{2} $, the area $ S $ of triangle $ PF_{1}F_{2} $ is given by $ S = \\frac{1}{2}(|PF_{1}| + |PF_{2}| + |F_{1}F_{2}|)r $, that is, $ S = (a + c)r = \\frac{3}{2}(a + c) $. Also, since the area $ S $ of triangle $ PF_{1}F_{2} $ is $ S = \\frac{1}{2} \\times |F_{1}F_{2}| \\times 4 = 4c $, we have $ \\frac{3}{2}(a + c) = 4c $, which implies $ e = \\frac{c}{a} = \\frac{3}{5} $. Therefore, the answer to be filled in is $ \\frac{3}{5} $." }, { "text": "Given that $F_{1}$, $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, if point $P(1, \\frac{3}{2})$ lies on the ellipse and satisfies $|P F_{1}|+|P F_{2}|=4$, then what is the equation of ellipse $C$?", "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Coordinate(P) = (1, 3/2);Focus(C) = {F1, F2};PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 4", "query_expressions": "Expression(C)", "answer_expressions": "x^2/4+y^2/3=1", "fact_spans": "[[[18, 75], [102, 104], [134, 139]], [[18, 75]], [[25, 75]], [[25, 75]], [[82, 101]], [[2, 9]], [[10, 17]], [[25, 75]], [[25, 75]], [[82, 101]], [[2, 80]], [[82, 105]], [[109, 132]]]", "query_spans": "[[[134, 144]]]", "process": "From the definition of the ellipse and the point lying on the ellipse, we can establish a system of equations. Solving it, from |PF₁| + |PF₂| = 4 we obtain 2a = 4, solving gives a = 2. Also, since P(1, \\frac{3}{2}) lies on the ellipse C: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a > b > 0), substituting gives \\frac{1}{2} + \\frac{(\\frac{3}{2})^{2}}{b^{2}} = 1, solving yields b = \\sqrt{3}. Therefore, the equation of the ellipse C is \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1." }, { "text": "Given that line $l$ passes through the point $(1,0)$ and is perpendicular to the $x$-axis, if the segment of $l$ intercepted by the parabola $y^{2}=4 a x$ has length $4$, then what are the coordinates of the focus of the parabola?", "fact_expressions": "l: Line;G: Parabola;a: Number;P: Point;Expression(G) = (y^2 = 4*(a*x));Coordinate(P) = (1, 0);PointOnCurve(P,l);IsPerpendicular(l, xAxis);Length(InterceptChord(l,G))=4", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(1,0)", "fact_spans": "[[[2, 7], [26, 29]], [[30, 46], [58, 61]], [[33, 46]], [[8, 16]], [[30, 46]], [[8, 16]], [[2, 16]], [[2, 24]], [[26, 56]]]", "query_spans": "[[[58, 68]]]", "process": "Analysis: According to the description in the problem, draw the corresponding graph. It can be analyzed that the parabola passes through the point (1,2). Substituting the coordinates of point (1,2) into the equation yields the value of parameter a, and then the focus coordinates can be determined. \nDetailed: From the given conditions, point P(1,2) lies on the parabola. Substituting P(1,2) into $ y^{2}=4ax $ gives: $ a=1 $, $ \\therefore y^{2}=4x $. From the parabola equation, we obtain: $ 2p=4 $, $ p=2 $, $ \\frac{p}{2}=1 $, $ \\therefore $ the focus coordinates are $ (1,0) $." }, { "text": "Given that point $P$ lies on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ whose asymptotes are given by $4x \\pm 3y = 0$, where $F_{1}$ and $F_{2}$ are its left and right foci respectively, if the area of $\\Delta P F_{1} F_{2}$ is $16$ and $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}} = 0$, then what is the value of $a+b$?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(Asymptote(G))=(4*x+pm*3*y=0);PointOnCurve(P,G);LeftFocus(G) = F1;RightFocus(G) = F2;Area(TriangleOf(P, F1, F2)) = 16;DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0", "query_expressions": "a + b", "answer_expressions": "7", "fact_spans": "[[[29, 85], [107, 108]], [[207, 212]], [[207, 212]], [[2, 6]], [[89, 96]], [[97, 104]], [[32, 85]], [[32, 85]], [[29, 85]], [[7, 85]], [[2, 86]], [[89, 113]], [[89, 113]], [[115, 145]], [[146, 205]]]", "query_spans": "[[[207, 216]]]", "process": "Problem Analysis: From the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$), the equations of the asymptotes are $4x\\pm3y=0$, which implies $\\frac{b}{a}=\\frac{4}{3}$ ①; from $\\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=0$, we know $\\overrightarrow{PF_{1}}\\bot\\overrightarrow{PF_{2}}$, so triangle $PF_{1}F_{2}$ is a right triangle. Therefore, $|PF_{1}|^{2}+|PF_{2}|^{2}=(2c)^{2}$. By the definition of a hyperbola, $||PF_{1}|-|PF_{2}||=2a$, so $(|PF_{1}|-|PF_{2}|)^{2}=(2a)^{2}$, which gives $|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}|\\cdot|PF_{2}|=4a^{2}$. Hence, $|PF_{1}|\\cdot|PF_{2}|=2(c^{2}-a^{2})$. Then, $\\frac{1}{2}\\cdot|PF_{1}|\\cdot|PF_{2}|=\\frac{1}{2}\\times2(c^{2}-a^{2})=16$, so $c^{2}-a^{2}=16$. Since $a^{2}+b^{2}=c^{2}$, it follows that $b^{2}=16$, thus $b=4$. Combining this with ①, we get $a=3$, and therefore $a+b=7$." }, { "text": "The left focus of the ellipse $\\frac{x^{2}}{4 a^{2}}+\\frac{y^{2}}{3 a^{2}}=1 (a>0)$ is $F$, and the line $x=m$ intersects the ellipse at points $A$ and $B$. When the perimeter of $\\triangle F A B$ is maximized, what is the area of $\\triangle F A B$?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/((4*a^2)) + y^2/((3*a^2)) = 1);a: Number;a>0;F: Point;LeftFocus(G) = F;H: Line;Expression(H) = (x = m);m: Number;A: Point;B: Point;Intersection(H, G) = {A, B};WhenMax(Perimeter(TriangleOf(F,A,B))) = True", "query_expressions": "Area(TriangleOf(F, A, B))", "answer_expressions": "3*a^2", "fact_spans": "[[[0, 56], [73, 75]], [[0, 56]], [[2, 56]], [[2, 56]], [[61, 64]], [[0, 64]], [[65, 72]], [[65, 72]], [[67, 72]], [[78, 82]], [[83, 86]], [[65, 86]], [[87, 111]]]", "query_spans": "[[[112, 134]]]", "process": "Solution: Let the right focus of the ellipse be E, connect AE and BE, as shown in the figure. By the definition of the ellipse, the perimeter of quadrilateral AFAB is |AB| + |AF| + |BF| = |AB| + 4a - |AE| + 4a - |BE| = 8a + |AB| - |AE| - |BE|. Since |AE| + |BE| \\geqslant |AB|, it follows that |AB| - |AE| - |BE| \\leqslant 0, with equality if and only if AB passes through point E. Therefore, |AB| + |AF| + |BF| \\leqslant 8a, meaning the perimeter of quadrilateral AFAB is maximized when the line x = m passes through the right focus E of the ellipse. At this time, m = a. Substituting x = a into \\frac{x^{2}}{4a^{2}} + \\frac{y^{2}}{3a^{2}} = 1 gives y = \\pm\\frac{3}{2}a, so |AB| = 3a. Also, |EF| = 2a. Hence, the area of quadrilateral AFAB is S_{AFAB} = \\frac{1}{2}|AB| \\cdot |EF| = 3a^{2}." }, { "text": "If a line $l$ with slope $\\frac{\\sqrt{2}}{2}$ intersects the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ at two distinct points, and the projections of these two points onto the $x$-axis are exactly the two foci of the ellipse, then the eccentricity of the ellipse is?", "fact_expressions": "Slope(l) = sqrt(2)/2;l: Line;G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Intersection(l, G) = {A, B};A: Point;B: Point;Focus(G) = {Projection(A, xAxis), Projection(B, xAxis)}", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[1, 30]], [[25, 30]], [[31, 83], [110, 112], [120, 122]], [[31, 83]], [[33, 83]], [[33, 83]], [[33, 83]], [[33, 83]], [[25, 91]], [], [], [[25, 117]]]", "query_spans": "[[[120, 128]]]", "process": "From the given conditions, it is easy to see that the horizontal coordinates of the two intersection points are -c and c, and their vertical coordinates are -\\frac{b}{a} and \\frac{b^{2}}{a}, respectively. Therefore, from \\frac{b}{c-(\\frac{b}{a})}=\\frac{\\sqrt{2}}{2}, we obtain 2b^{2}=\\sqrt{2}ac=2(a^{2}-c^{2}), which leads to 2e^{2}+\\sqrt{2}c-2=0. Solving this equation gives c=\\frac{\\sqrt{2}}{2} or c=-\\sqrt{2} (the negative root is discarded)." }, { "text": "The standard equation of the hyperbola that shares asymptotes with $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$ and passes through the point $M(2 \\sqrt{6}, 6)$ is?", "fact_expressions": "G: Hyperbola;H: Hyperbola;M: Point;Expression(G) = (x^2/4 - y^2/9 = 1);Coordinate(M) = (2*sqrt(6), 6);PointOnCurve(M,H);Asymptote(G) = Asymptote(H)", "query_expressions": "Expression(H)", "answer_expressions": "x^2/8 - y^2/18 = 1", "fact_spans": "[[[1, 39]], [[66, 69]], [[46, 65]], [[1, 39]], [[46, 65]], [[44, 69]], [[0, 69]]]", "query_spans": "[[[66, 76]]]", "process": "Let the standard equation of the hyperbola be \\frac{x^{2}}{4}-\\frac{y^{2}}{9}=t (t\\neq0). Since it passes through the point M(2\\sqrt{6},6), we have \\frac{24}{4}-\\frac{36}{9}=t, that is, t=2. Therefore, the required standard equation of the hyperbola is \\frac{x^{2}}{8}-\\frac{y^{2}}{18}=1." }, { "text": "Given the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, $M(1,8)$. A line passing through point $M$ intersects $E$ at points $A$ and $B$, with $M$ being the midpoint of $AB$, and the line $AB$ is perpendicular to one of the asymptotes of $E$. Then, the eccentricity of $E$ is?", "fact_expressions": "E: Hyperbola;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;M: Point;Coordinate(M) = (1, 8);PointOnCurve(M, G);G: Line;Intersection(G, E) = {A, B};B: Point;A: Point;MidPoint(LineSegmentOf(A, B)) = M;IsPerpendicular(LineOf(A, B), OneOf(Asymptote(E))) = True", "query_expressions": "Eccentricity(E)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[2, 63], [84, 87], [120, 123], [133, 136]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[66, 74], [76, 80], [98, 101]], [[66, 74]], [[75, 83]], [[81, 83]], [[81, 97]], [[92, 95]], [[88, 91]], [[98, 110]], [[112, 131]]]", "query_spans": "[[[133, 142]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), substituting into the hyperbola equation gives: \\frac{x_{1}^2}{a^{2}} - \\frac{y_{1}^2}{b^{2}} = 1, \\frac{x_{2}^2}{a^{2}} - \\frac{y_{2}^2}{b^{2}} = 1. Subtracting these two equations yields \\frac{x_{1}^2 - x_{2}^2}{a^{2}} = \\frac{y_{1}^2 - y_{2}^2}{b^{2}} \\Rightarrow \\frac{b^{2}}{a^{2}} = \\frac{y_{1}^2 - y_{2}^2}{x_{1}^2 - x_{2}^2} = \\frac{(y_{1}-y_{2})(y_{1}+y_{2})}{(x_{1}-x_{2})(x_{1}+x_{2})}. Since M is the midpoint of AB, we have x_{1}+x_{2}=2, y_{1}+y_{2}=16, so k_{AB} = \\frac{b^{2}}{8a^{2}}. According to the given condition: \\frac{b^{2}}{8a^{2}} \\cdot (-\\frac{b}{a}) = -1 \\Rightarrow b^{3} = 8a^{3} \\Rightarrow b = 2a \\Rightarrow b^{2} = 4a^{2} \\Rightarrow c^{2} - a^{2} = 4a^{2}, which implies c^{2} = 5a^{2} \\Rightarrow c = \\sqrt{5}a \\Rightarrow e = \\frac{c}{a} = \\sqrt{5}." }, { "text": "Given a point $P(3 , y)$ on the parabola $y^{2}=4 x$, then the distance from point $P$ to the focus of the parabola is?", "fact_expressions": "G: Parabola;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(P) = (3, y1);PointOnCurve(P,G);y1:Number", "query_expressions": "Distance(P, Focus(G))", "answer_expressions": "4", "fact_spans": "[[[2, 16], [36, 39]], [[19, 29], [31, 35]], [[2, 16]], [[19, 29]], [[2, 29]], [[19, 29]]]", "query_spans": "[[[31, 46]]]", "process": "Solution: \\because the parabola $ y^{2} = 4x $, comparing with the standard equation of a parabola $ y^{2} = 2px $, we get $ p = 2 $. \\therefore the directrix of the parabola is $ x = -1 $. By the definition of a parabola, the distance from point P to the focus of the parabola equals the distance from P to the directrix of the parabola. Since the x-coordinate of P is 3, the distance from P to the directrix of the parabola is $ 3 + 1 = 4 $. \\therefore the distance from P to the focus of the parabola is 4." }, { "text": "Through the focus of the parabola $x^{2}=\\frac{1}{8} y$, a line is drawn intersecting the parabola at points $A$ and $B$. The ordinate (y-coordinate) of the midpoint $M$ of segment $AB$ is $2$. What is the length of segment $AB$?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = y/8);H: Line;PointOnCurve(Focus(G), H);A: Point;B: Point;Intersection(H, G) = {A, B};M: Point;MidPoint(LineSegmentOf(A, B)) = M;YCoordinate(M) = 2", "query_expressions": "Length(LineSegmentOf(A, B))", "answer_expressions": "65/16", "fact_spans": "[[[1, 25], [32, 35]], [[1, 25]], [[29, 31]], [[0, 31]], [[36, 39]], [[41, 44]], [[29, 46]], [[56, 59]], [[47, 59]], [[56, 67]]]", "query_spans": "[[[69, 78]]]", "process": "" }, { "text": "The line $l$ passes through the focus $F$ of the parabola $y^{2}=8x$, and intersects the parabola at points $A$ and $B$. If the distance from the midpoint of segment $AB$ to the $y$-axis is $2$, then $|AB|=$?", "fact_expressions": "l: Line;G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;PointOnCurve(F, l);A: Point;B: Point;Intersection(l,G) = {A,B};Distance(MidPoint(LineSegmentOf(A,B)), yAxis) = 2", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "8", "fact_spans": "[[[0, 5]], [[29, 32], [6, 20]], [[6, 20]], [[23, 26]], [[6, 26]], [[0, 26]], [[34, 37]], [[38, 41]], [[0, 43]], [[45, 67]]]", "query_spans": "[[[69, 78]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), then using the midpoint coordinate formula, we know M(\\frac{x_{1}+x_{2}}{2},\\frac{y_{1}+y_{2}}{2}). Since the distance from point M to the axis is 2, it follows that \\frac{x_{1}+x_{2}}{2}=2, that is, x_{1}+x_{2}=4. Also, 2p=8, so p=4. Therefore, using the chord length formula for a parabola passing through the focus, |AB|=x_{1}+x_{2}+p=4+4=8" }, { "text": "For any point $P$ on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, draw a line through $P$ parallel to the real axis, intersecting the two asymptotes at points $M$ and $N$. If $\\overrightarrow{P M} \\cdot \\overrightarrow{P N}=2 b^{2}$, then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;P: Point;PointOnCurve(P, G) ;L: Line;PointOnCurve(P,L);IsParallel(L,RealAxis(G)) ;L1: Line;L2: Line;Asymptote(G) = {L1,L2};Intersection(L,L1) = M;M: Point;Intersection(L,L2) = N;N: Point;DotProduct(VectorOf(P, M), VectorOf(P, N)) = 2*b^2", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(6)/2", "fact_spans": "[[[1, 57], [152, 155]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[62, 65]], [[1, 65]], [[73, 75]], [[0, 75]], [[1, 75]], [], [], [[1, 81]], [[1, 90]], [[81, 84]], [[1, 90]], [[85, 88]], [[92, 149]]]", "query_spans": "[[[152, 161]]]", "process": "" }, { "text": "Given the curve $\\frac{x^{2}}{a}-\\frac{y^{2}}{b}=1$ intersects the line $x+y-1=0$ at points $P$ and $Q$, and $\\overrightarrow {O P} \\cdot \\overrightarrow{O Q}=0$ ($O$ is the origin), then the value of $\\frac{1}{a}-\\frac{1}{b}$ is?", "fact_expressions": "G: Line;H: Curve;b: Number;a: Number;O: Origin;P: Point;Q: Point;Expression(G) = (x + y - 1 = 0);Expression(H) = (-y^2/b + x^2/a = 1);Intersection(H, G) = {P, Q};DotProduct(VectorOf(O,P),VectorOf(O,Q)) = 0", "query_expressions": "-1/b + 1/a", "answer_expressions": "2", "fact_spans": "[[[40, 51]], [[2, 39]], [[4, 39]], [[4, 39]], [[118, 121]], [[54, 57]], [[58, 61]], [[40, 51]], [[2, 39]], [[2, 63]], [[65, 117]]]", "query_spans": "[[[127, 156]]]", "process": "" }, { "text": "Given the ellipse $E$: $\\frac{x^{2}}{2}+y^{2}=1$, point $P(2 , t)$, $F$ is the left focus of the ellipse. Draw two tangent lines from point $P$ to the ellipse, denoted as $PA$ and $PB$, with points of tangency $A$ and $B$ respectively. Then, what is the range of the area of $\\triangle ABF$?", "fact_expressions": "E: Ellipse;A: Point;B: Point;F: Point;P: Point;t:Number;Expression(E) = (x^2/2 + y^2 = 1);Coordinate(P)=(2,t);LeftFocus(E) = F;TangentOfPoint(P,E)={LineOf(P,A),LineOf(P,B)};TangentPoint(LineOf(P,B),E)=A;TangentPoint(LineOf(P,A),E)=B", "query_expressions": "Range(Area(TriangleOf(A,B,F)))", "answer_expressions": "(0,sqrt(2)]", "fact_spans": "[[[2, 33], [50, 52], [50, 52]], [[87, 90]], [[91, 94]], [[46, 49]], [[33, 44], [58, 62]], [[34, 44]], [[2, 33]], [[33, 44]], [[45, 56]], [[57, 81]], [[63, 94]], [[63, 94]]]", "query_spans": "[[[96, 120]]]", "process": "Solution: Let point A(x_{1},y_{1}), B(x_{2},y_{2}). Therefore, the equation of tangent line PA is \\frac{x_{1}x}{2}+y_{1}y=1, and the equation of tangent line PB is \\frac{x_{2}x}{2}+y_{2}y=1. Since point P lies on both tangent lines PA and PB, we have \\begin{cases}x_{1}+ty_{1}=1\\\\x_{2}+ty_{2}=1\\end{cases}. Therefore, the equation of line AB is x+ty=1. Hence, line AB passes through the fixed point (1,0), which is the right focus F_{2} of ellipse E. (x+ty=1 and \\begin{cases}x+ty=1\\\\\\frac{x^{2}}{2}+y^{2}=1\\end{cases}, eliminating x gives: (t^{2}+2)y^{2}-2ty-1=0, y_{1}+y_{2}=\\frac{2t}{t^{2}+2}'. |x|-1\\begin{matrix}t-+2&t\\\\=\\frac{1}{2}\\times2\\times\\sqrt{(y_{1}+y_{2})^{2}-4y_{1}y_{2}}=\\sqrt{(\\frac{2t}{t^{2}+2})^{^{2}}-4\\times\\frac{-1}{t^{2}+2}}&=\\frac{2\\sqrt{2}\\sqrt{t^{2}+1}}{t^{2}+2}\\end{matrix}. Let \\sqrt{t^{2}+1}=m\\geqslant1, then t^{2}=m^{2}-1, m+\\frac{1}{m}\\geqslant2, then 0<\\frac{1}{m+\\frac{1}{m}}\\leqslant\\frac{1}{2}. Then S_{\\DeltaABF}=\\frac{2\\sqrt{2}\\sqrt{t^{2}+1}}{t^{2}+2}=\\frac{2\\sqrt{2}m}{m^{2}+1}=\\frac{2\\sqrt{2}}{m+\\frac{1}{m}}\\in(0,\\sqrt{2}]." }, { "text": "The distance from the focus of the hyperbola $x^{2}-y^{2}=1$ to its asymptote is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 1)", "query_expressions": "Distance(Focus(G), Asymptote(G))", "answer_expressions": "1", "fact_spans": "[[[0, 18], [22, 23]], [[0, 18]]]", "query_spans": "[[[0, 31]]]", "process": "From the given problem: The coordinates of the foci are $(-\\sqrt{2},0)$, $(\\frac{\\sqrt{2}}{2},0)$. The equations of the asymptotes are $y = \\pm x$. Therefore, the distance $d$ from a focus to its asymptote is $d = \\frac{|\\pm\\sqrt{2}|}{\\sqrt{1^{2}+(\\pm1)^{2}}} = 1$." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a , b>0)$, the left and right foci are $F_{1}$, $F_{2}$ respectively. A line $l$ passing through point $F_{1}$ with an inclination angle of $\\frac{\\pi}{6}$ intersects the left and right branches of the hyperbola at points $A$, $B$ respectively. Given that $|A F_{2}|=|B F_{2}|$, what is the eccentricity of this hyperbola?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;l: Line;PointOnCurve(F1, l);Inclination(l) = pi/6;A: Point;Intersection(l, LeftPart(G)) = A;B: Point;Intersection(l, RightPart(G)) = B;Abs(LineSegmentOf(A, F2)) = Abs(LineSegmentOf(B, F2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 57], [120, 123], [167, 170]], [[2, 57]], [[5, 57]], [[5, 57]], [[5, 57]], [[5, 57]], [[66, 73], [85, 93]], [[75, 82]], [[2, 82]], [[2, 82]], [[114, 119]], [[83, 119]], [[94, 119]], [[132, 136]], [[114, 140]], [[137, 140]], [[114, 140]], [[143, 164]]]", "query_spans": "[[[167, 176]]]", "process": "Draw F_{2}N\\botAB at point N, let |AF_{2}|=|BF_{2}|=m. Since the inclination angle of line l is \\frac{\\pi}{6}, in the right triangle F_{1}F_{2}N, |NF_{2}|=\\frac{1}{2}|F_{1}F_{2}|=c, |NF_{1}|=\\sqrt{3}c. By the definition of the hyperbola, |BF_{1}|-|BF_{2}|=2a, so |BF_{1}|=2a+m. Similarly, |AF_{1}|=m-2a. Thus, |AB|=|BF_{1}|-|AF_{1}|=4a, so |AN|=2a. Therefore, |AF_{1}|=\\sqrt{3}c-2a, and hence m=\\sqrt{3}c. In the right triangle ANF_{2}, AF_{2}^{2}=|NF_{2}|^{2}+|AN|^{2}, so (\\sqrt{3}c)^{2}=4a^{2}+c^{2}, thus c=\\sqrt{2}a, then e=\\frac{c}{a}=\\sqrt{2}. Hence the answer is: \\sqrt{2}." }, { "text": "Given two fixed points $A(-2,0)$, $B(2,0)$, point $P$ lies on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$ and satisfies $|P A|-|P B|=2$, then $\\overrightarrow{P A} \\cdot \\overrightarrow{P B}$=?", "fact_expressions": "G: Ellipse;A: Point;B: Point;P: Point;Expression(G) = (x^2/16 + y^2/12 = 1);Coordinate(A) = (-2, 0);Coordinate(B) = (2, 0);PointOnCurve(P, G);Abs(LineSegmentOf(P, A)) - Abs(LineSegmentOf(P, B)) = 2", "query_expressions": "DotProduct(VectorOf(P, A), VectorOf(P, B))", "answer_expressions": "9", "fact_spans": "[[[31, 70]], [[5, 14]], [[17, 25]], [[26, 30]], [[31, 70]], [[5, 14]], [[17, 25]], [[26, 71]], [[75, 90]]]", "query_spans": "[[[92, 143]]]", "process": "Solution: Let P(x, y). Given A(-2, 0), B(2, 0), and |PA| - |PB| = 2, the trajectory of point P is the right branch of a hyperbola with foci at points A and B, where 2a = 2, c = 2. Therefore, b = \\sqrt{3}. Thus, the trajectory equation of P is: x - \\frac{y^{2}}{3} = 1 (x \\geqslant 1). Solving the system of equations of the ellipse and hyperbola simultaneously:\n\n\\begin{cases}\n\\frac{x^2}{16} + \\frac{y^{2}}{12} = 1 \\\\\nx^2 - \\frac{y^{2}}{3} = 1\n\\end{cases}\n\nWe obtain x^{2} = 4, y^{2} = 9. Therefore, \\overrightarrow{PA} \\cdot \\overrightarrow{PB} = (x+2)(x-2) + y^{2} = x^{2} - 4 + y^{2} = . Hence, the answer is: 9. (Note: This problem examines using the definition method to find the standard equation of a hyperbola, finding coordinates of intersection points of conic sections, and applying the formula for the dot product of two vectors. It is a medium-difficulty problem.)" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has eccentricity $\\frac{\\sqrt{2}}{2}$, and the distance from the right vertex to the line $x-y+2 \\sqrt{2}=0$ is $3$, then the equation of ellipse $C$ is?", "fact_expressions": "C: Ellipse;G:Line;a: Number;b: Number;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x - y + 2*sqrt(2) = 0);Eccentricity(C) = sqrt(2)/2;Distance(RightVertex(C),G)=3", "query_expressions": "Expression(C)", "answer_expressions": "x^2/2+y^2=1", "fact_spans": "[[[2, 59], [118, 123]], [[89, 109]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[2, 59]], [[89, 109]], [[2, 84]], [[2, 116]]]", "query_spans": "[[[118, 128]]]", "process": "Since the eccentricity of the ellipse is \\frac{\\sqrt{2}}{2}, we have \\sqrt{1-\\frac{b^{2}}{a^{2}}}=\\frac{\\sqrt{2}}{2}. Also, the coordinates of the right vertex are (a,0), hence \\frac{|a-0+2\\sqrt{2}|}{\\sqrt{1+1}}=3, so a=\\sqrt{2}, therefore b=1. The equation of the ellipse is: \\frac{x^{2}}{2}+y^{2}=1." }, { "text": "The length of the major axis of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{16}=1$ is? The length of the minor axis is? The coordinates of the foci are? The coordinates of the vertices are? The eccentricity is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/16 = 1)", "query_expressions": "Length(MajorAxis(G));Length(MinorAxis(G));Coordinate(Focus(G));Coordinate(Vertex(G));Eccentricity(G)", "answer_expressions": "8;4;(0,pm*2*sqrt(3));{(pm*2,0),(0,pm*4)};sqrt(3)/2", "fact_spans": "[[[0, 38]], [[0, 38]]]", "query_spans": "[[[0, 44]], [[0, 49]], [[0, 55]], [[0, 61]], [[0, 66]]]", "process": "" }, { "text": "The line $l$ passing through the point $P(1,1)$ intersects the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$ at points $A$ and $B$, and $P$ is the midpoint of segment $AB$. Then the slope of line $l$ is?", "fact_expressions": "P: Point;Coordinate(P) = (1, 1);PointOnCurve(P, l) = True;l: Line;G: Ellipse;Expression(G) = (x^2/8 + y^2/4 = 1);Intersection(l, G) = {A, B};A: Point;B: Point;MidPoint(LineSegmentOf(A, B)) = P", "query_expressions": "Slope(l)", "answer_expressions": "-1/2", "fact_spans": "[[[1, 10], [67, 70]], [[1, 10]], [[0, 16]], [[11, 16], [83, 88]], [[17, 54]], [[17, 54]], [[11, 65]], [[56, 59]], [[60, 63]], [[67, 81]]]", "query_spans": "[[[83, 93]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), P(1,1) be the midpoint of segment AB, then x_{1}+x_{2}=2, y_{1}+y_{2}=2; substituting points A and B into the ellipse equation and taking the difference, we obtain: \\frac{1}{8}(x_{1}+x_{2})(x_{1}-x_{2})+\\frac{1}{4}(y_{1}+y_{2})(y_{1}-y_{2})=0, so \\frac{1}{8}\\times2\\times(x_{1}-x_{2})+\\frac{1}{4}\\times2\\times(y_{1}-y_{2})=0. According to the problem, the slope of line l exists, thus \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{1}{2} \\therefore k_{AB}=-\\frac{1}{2}" }, { "text": "Point $P$ is a moving point on the parabola $C$: $y^{2}=4 x$. What is the minimum value of the sum of the distance from point $P$ to the point $(6,12)$ and the distance from point $P$ to the $y$-axis?", "fact_expressions": "C: Parabola;P: Point;G: Point;Expression(C) = (y^2 = 4*x);PointOnCurve(P, C);Coordinate(G) = (6, 12)", "query_expressions": "Min(Distance(P, G) + Distance(P, yAxis))", "answer_expressions": "12", "fact_spans": "[[[5, 23]], [[0, 4], [29, 33]], [[34, 43]], [[5, 23]], [[0, 27]], [[34, 43]]]", "query_spans": "[[[29, 62]]]", "process": "" }, { "text": "In the rectangular coordinate system $x O y$, the distance from the right vertex of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{16}=1(a>0)$ to one of the asymptotes of the hyperbola is $\\frac{4 \\sqrt{5}}{3}$. Then the equation of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;a: Number;a > 0;Expression(C) = (x^2/a^2 - y^2/16 = 1);Distance(RightVertex(C), OneOf(Asymptote(C))) = 4*sqrt(5)/3", "query_expressions": "Expression(C)", "answer_expressions": "x^2/20 - y^2/16 = 1", "fact_spans": "[[[17, 70], [75, 78], [112, 118]], [[25, 70]], [[25, 70]], [[17, 70]], [[17, 110]]]", "query_spans": "[[[112, 123]]]", "process": "From the hyperbola equation, the right vertex is at (a,0), and the asymptote equations are: y=\\pm\\frac{4}{a}x, that is, \\pm4x-ay=0. Therefore, the distance d from the right vertex to the hyperbola's asymptote is d=\\frac{|\\pm4a|}{\\sqrt{16+a^{2}}}=\\frac{4\\sqrt{5}}{3}. Solving gives: a^{2}=20. Therefore, the equation of hyperbola C is: \\frac{x^{2}}{20}-\\frac{y^{2}}{16}=1. The correct result for this problem: \\frac{x^{2}}{20}-\\frac{y^{2}}{16}=1" }, { "text": "The equation of the locus of points equidistant from the point $F(0,-3)$ and the line $y-3=0$ is?", "fact_expressions": "G: Line;F: Point;Expression(G) = (y - 3 = 0);Coordinate(F) = (0, -3);P:Point;Distance(P,F)=Distance(P,G)", "query_expressions": "LocusEquation(P)", "answer_expressions": "x^2 = -12 * y", "fact_spans": "[[[12, 21]], [[1, 11]], [[12, 21]], [[1, 11]], [[27, 28]], [[0, 28]]]", "query_spans": "[[[27, 35]]]", "process": "Solution: By the definition of a parabola, the locus of points in the plane that are equidistant from the point F(0,-3) and the line y-3=0 is a parabola. Moreover, F(0,-3) is the focus and the line y=3 is the directrix. Let the equation of the parabola be x^{2}=-2py (p>0). We know that \\frac{p}{2}=3, solving gives p=6. Therefore, the equation of this parabola is x^{2}=-12y." }, { "text": "If the focus of the parabola $y=m x^{2}$ coincides with the upper focus of the ellipse $\\frac{x^{2}}{2}+\\frac{y^{2}}{6}=1$, then $m$=?", "fact_expressions": "G: Parabola;m: Number;H: Ellipse;Expression(G) = (y = m*x^2);Expression(H) = (x^2/2 + y^2/6 = 1);Focus(G) = UpperFocus(H)", "query_expressions": "m", "answer_expressions": "1/8", "fact_spans": "[[[1, 15]], [[64, 67]], [[19, 56]], [[1, 15]], [[19, 56]], [[1, 62]]]", "query_spans": "[[[64, 69]]]", "process": "In the ellipse $\\frac{x^{2}}{2}+\\frac{y^{2}}{6}=1$, $a=\\sqrt{6}$, $b=\\sqrt{2}$, $c=\\sqrt{6-2}=2$, so the foci are $(0,\\pm2)$. The standard equation of the parabola $y=mx^{2}$ is $x^{2}=\\frac{1}{m}y$, $2p=\\frac{1}{m}$, $p=\\frac{1}{2m}$. Since the focus of the parabola coincides with the focus of the ellipse, $\\frac{p}{2}=\\frac{1}{4m}=2$, $m=\\frac{1}{8}$." }, { "text": "Let $F$ be the focus of the parabola $C$: $y^{2}=3 x$. A line passing through $F$ intersects the parabola $C$ at points $A$ and $B$, and $O$ is the origin. Then the minimum area of $\\triangle A O B$ is?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 3*x);F: Point;Focus(C) = F;G: Line;PointOnCurve(F, G);A: Point;B: Point;Intersection(G, C) = {A, B};O: Origin", "query_expressions": "Min(Area(TriangleOf(A, O, B)))", "answer_expressions": "9/8", "fact_spans": "[[[5, 24], [36, 42]], [[5, 24]], [[1, 4], [29, 32]], [[1, 27]], [[33, 35]], [[28, 35]], [[43, 46]], [[47, 50]], [[33, 52]], [[53, 56]]]", "query_spans": "[[[63, 88]]]", "process": "The focus of the parabola is $(\\frac{3}{4},0)$. When the slope of the line does not exist, i.e., when it is perpendicular to the x-axis, the area is minimized. Substitute $x=\\frac{3}{4}$ into $y^{2}=3x$, solving gives $y=\\pm\\frac{3}{2}$, hence $S_{\\triangle OAB}=\\frac{1}{2}\\times\\frac{3}{4}\\times2\\times\\frac{3}{2}=\\frac{9}{8}$." }, { "text": "The moving circle $M$ is internally tangent to the circle $C_{1}$: $(x+1)^{2}+y^{2}=36$, and externally tangent to the circle $C_{2}$: $(x-1)^{2}+y^{2}=4$. Then, the trajectory equation of the center of the moving circle $M$ is?", "fact_expressions": "M:Circle;C1:Circle;Expression(C1) = ((x+1)^2 + y^2 = 36);IsInTangent(M,C1);C2: Circle;Expression(C2) = ((x-1)^2 + y^2 = 4);IsOutTangent(M,C2);Center(M) = M1;M1: Point", "query_expressions": "LocusEquation(M1)", "answer_expressions": "x^2/16 + y^2/15 = 1", "fact_spans": "[[[2, 5], [72, 74]], [[6, 35]], [[6, 35]], [[2, 37]], [[40, 68]], [[40, 68]], [[2, 70]], [[72, 79]], [[76, 79]]]", "query_spans": "[[[76, 86]]]", "process": "Let the coordinates of the center M of the moving circle be (x, y), and its radius be r. According to the given conditions, we obtain |MC₁| = 6 - r, |MC₂| = r + 2, and thus |MC₁| + |MC₂| = 8 > |C₁C₂| = 2. Combining this with the definition of an ellipse, the solution can be found. Let the coordinates of the center M of the moving circle be (x, y), and its radius be r. Since the moving circle M is internally tangent to the circle C₁: (x + 1)² + y² = 36, and externally tangent to the circle C₂: (x - 1)² + y² = 4, we get |MC₁| = 6 - r, |MC₂| = r + 2. Therefore, |MC₁| + |MC₂| = 8 > |C₁C₂| = 2. According to the definition of an ellipse, the locus of the moving point M is an ellipse with foci at C₁ and C₂, where 2a = 8, 2c = 2. Thus, a = 4, c = 1, and then b = √(a² - c²) = √15. Hence, the equation of the locus of the moving point M is x²/16 + y²/15 = 1." }, { "text": "Given point $A(8 , 2)$ and the parabola $y^{2}=8 x $, where $F$ is the focus of the parabola and $P$ lies on the parabola, find the minimum value of $|PA|+|PF|$.", "fact_expressions": "A: Point;Coordinate(A) = (8, 2);G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;P: Point;PointOnCurve(P, G) = True", "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))", "answer_expressions": "10", "fact_spans": "[[[2, 13]], [[2, 13]], [[14, 29], [35, 38], [46, 49]], [[14, 29]], [[31, 34]], [[31, 41]], [[42, 45]], [[42, 50]]]", "query_spans": "[[[52, 69]]]", "process": "" }, { "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{n}-\\frac{y^{2}}{3-n}=1$ are given by $y=\\pm 2 x$, then $n=?$", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/(3 - n) + x^2/n = 1);n: Number;Expression(Asymptote(G)) = (y = pm*(2*x))", "query_expressions": "n", "answer_expressions": "3/5", "fact_spans": "[[[0, 40]], [[0, 40]], [[60, 63]], [[0, 58]]]", "query_spans": "[[[60, 65]]]", "process": "" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, line $l$ passing through the origin $O$ with an inclination angle of $30^{\\circ}$ intersects the ellipse $C$ at a point $A$. If $A F_{1} \\perp A F_{2}$ and $S_{\\Delta F_{1} A F_{2}}=2$, then the equation of the ellipse $C$ is?", "fact_expressions": "l: Line;C: Ellipse;b: Number;a: Number;A: Point;F1: Point;F2: Point;O:Origin;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(O,l);Inclination(l)=ApplyUnit(30,degree);OneOf(Intersection(l,C))=A;IsPerpendicular(LineSegmentOf(A, F1), LineSegmentOf(A, F2));Area(TriangleOf(F1,A,F2)) = 2", "query_expressions": "Expression(C)", "answer_expressions": "x^2/6+y^2/2=1", "fact_spans": "[[[106, 111]], [[18, 75], [112, 117], [182, 187]], [[24, 75]], [[24, 75]], [[123, 126]], [[2, 9]], [[10, 17]], [[83, 88]], [[24, 75]], [[24, 75]], [[18, 75]], [[2, 81]], [[2, 81]], [[82, 111]], [[89, 111]], [[106, 126]], [[128, 151]], [[152, 180]]]", "query_spans": "[[[182, 192]]]", "process": "Using the definition of an ellipse and the Pythagorean theorem, we obtain |AF_{1}||AF_{2}|=2b^{2}; combining with the area of the triangle gives b^{2}=2. From the property of a right triangle, we know |OA|=c. Without loss of generality, assume point A lies in the first quadrant. Given \\angle AOF_{2}=30^{\\circ}, the coordinates of point A (expressed in terms of c) can be obtained. Then, using the area of the triangle, we can find c, and subsequently determine a^{2}, thus obtaining the equation of the ellipse. Since point A lies on the ellipse, |AF_{1}|+|AF_{2}|=2a. Squaring both sides yields |AF_{1}|^{2}+|AF_{2}|^{2}+2|AF_{1}||AF_{2}|=4a^{2}. Since AF_{1}\\bot AF_{2}, we have |AF_{1}|^{2}+|AF_{2}|^{2}=4c^{2}, so 2|AF_{1}||AF_{2}|=4a^{2}-4c^{2}=4b^{2}, which implies |AF_{1}||AF_{2}|=2b^{2}. Therefore, S_{\\triangle F_{1}AF_{2}}=\\frac{1}{2}|AF_{1}||AF_{2}|=b^{2}=2. Also, \\triangle AF_{1}F_{2} is a right triangle with \\angle F_{1}AF_{2}=90^{\\circ}, and O is the midpoint of F_{1}F_{2}, so |OA|=\\frac{1}{2}|F_{1}F_{2}|=c. Without loss of generality, assume A lies in the first quadrant; then \\angle AOF_{2}=30^{\\circ}, so A(\\frac{\\sqrt{3}}{2}c,\\frac{1}{2}c). Then S_{\\triangle F_{1}AF_{2}}=\\frac{1}{2}|F_{1}F_{2}|\\cdot\\frac{1}{2}c=\\frac{1}{2}c^{2}=2, hence c^{2}=4. Therefore, a^{2}=b^{2}+c^{2}=6. Thus, the equation of the ellipse is \\frac{x^{2}}{6}+" }, { "text": "Given the ellipse $C_{1}$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ and the hyperbola $C_{2}$: $ x^{2}-y^{2}=4$ have the same right focus $F_{2}$, point $P$ is a common point of ellipse $C_{1}$ and hyperbola $C_{2}$ in the first quadrant. If $|P F_{2}|=2$, then the eccentricity of ellipse $C_{1}$ equals?", "fact_expressions": "C2: Hyperbola;C1: Ellipse;a:Number;b:Number;a>b;b>0;P: Point;F2: Point;Expression(C1) = (x^2/a^2+y^2/b^2=1);Expression(C2) = (x^2-y^2=4);RightFocus(C1)=F2;RightFocus(C2)=F2;Intersection(C1, C2)=P;Quadrant(P) = 1;Abs(LineSegmentOf(P, F2)) = 2", "query_expressions": "Eccentricity(C1)", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[64, 91], [121, 131]], [[2, 63], [111, 120], [157, 166]], [[13, 63]], [[13, 63]], [[13, 63]], [[13, 63]], [[106, 110]], [[98, 105]], [[2, 63]], [[64, 91]], [[2, 105]], [[2, 105]], [[106, 140]], [[106, 140]], [[142, 155]]]", "query_spans": "[[[157, 173]]]", "process": "By the given condition, without loss of generality, assume point P lies in the first quadrant. From the equation of hyperbola $ C_{2}: x^{2} - y^{2} = 4 $, we know $ |PF_{1}| - |PF_{2}| = 4 $, $ c = \\because |PF_{2}| = 2 $, $ \\therefore |PF_{1}| = 6 $, $ \\therefore 2a = |PF_{2}| + |PF_{2}| = 8 $, $ \\therefore a = 4 $. $ \\because $ ellipse $ C_{1}: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ ($ a > b > 0 $) and hyperbola $ C_{2}: x^{2} - y^{2} = 4 $ have the same right focus $ F_{2} $, $ c = 2\\sqrt{2} $. $ \\because $ the eccentricity of ellipse $ C_{1} $ is $ e = \\frac{c}{a} = \\frac{\\sqrt{2}}{} $," }, { "text": "The asymptotes of a hyperbola with foci on the $x$-axis are given by $y=\\pm \\frac{3}{4} x$. Then, the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), xAxis);Expression(Asymptote(G)) = (y = pm*(3/4)*x)", "query_expressions": "Eccentricity(G)", "answer_expressions": "5/4", "fact_spans": "[[[9, 12], [42, 45]], [[0, 12]], [[9, 40]]]", "query_spans": "[[[42, 51]]]", "process": "Solve for \\frac{b}{a}=\\frac{3}{4}, then use the formula e=\\sqrt{1+(\\frac{b}{a})^{2}} to find the eccentricity of the hyperbola. Since the equations of the asymptotes of a hyperbola with foci on the x-axis are y=\\pm\\frac{3}{4}x, it follows that \\frac{b}{a}=\\frac{3}{4}. Therefore, the eccentricity of this hyperbola is \\frac{5}{4}." }, { "text": "Given that the point $(1,2)$ lies on the asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1(a>0)$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/4 + x^2/a^2 = 1);a: Number;a>0;H: Point;Coordinate(H) = (1, 2);PointOnCurve(H, Asymptote(G))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[11, 58], [67, 70]], [[11, 58]], [[14, 58]], [[14, 58]], [[2, 10]], [[2, 10]], [[2, 65]]]", "query_spans": "[[[67, 76]]]", "process": "" }, { "text": "Let the line $l$ passing through the focus $F$ of the parabola $y^{2}=4x$ intersect the parabola at points $A$ and $B$. If $|AB|=6$, then what is the x-coordinate of the midpoint of segment $AB$?", "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, B)) = 6", "query_expressions": "XCoordinate(MidPoint(LineSegmentOf(A, B)))", "answer_expressions": "2", "fact_spans": "[[[22, 27]], [[1, 15], [28, 31]], [[32, 35]], [[36, 39]], [[18, 21]], [[1, 15]], [[1, 21]], [[0, 27]], [[22, 41]], [[43, 52]]]", "query_spans": "[[[54, 69]]]", "process": "The parabola $ y^{2} = 4x $, $ \\therefore p = 2 $. Let a line passing through point $ F $ intersect the parabola at points $ A $ and $ B $, with their x-coordinates being $ x_{1}, x_{2} $ respectively. Using the definition of a parabola, the x-coordinate of the midpoint of $ AB $ is $ x_{0} = \\frac{1}{2}(x_{1} + x_{2}) = \\frac{1}{2}(|AB| - p) = 2 $." }, { "text": "Let $F_{1}$, $F_{2}$ be the two foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$. If there exists a point $P$ on $C$ such that $PF_{1} \\perp PF_{2}$ and $\\angle PF_{1} F_{2}=30^{\\circ}$, then the eccentricity of $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Focus(C)={F1,F2};PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P,F1),LineSegmentOf(P,F2));AngleOf(P,F1,F2)=ApplyUnit(30,degree)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3)+1", "fact_spans": "[[[18, 78], [87, 90], [157, 160]], [[25, 78]], [[25, 78]], [[95, 98]], [[1, 8]], [[10, 17]], [[25, 78]], [[25, 78]], [[18, 78]], [[1, 83]], [[87, 98]], [[101, 122]], [[123, 155]]]", "query_spans": "[[[157, 166]]]", "process": "" }, { "text": "Given that the two foci of a hyperbola are $F_{1}(-\\sqrt{10}, 0)$, $F_{2}(\\sqrt{10}, 0)$, and the asymptotes are $y=\\pm \\frac{1}{3} x$, then the standard equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;Coordinate(F1) = (-sqrt(10), 0);Coordinate(F2) = (sqrt(10), 0);Focus(G) = {F1, F2};Expression(Asymptote(G)) = (y = pm*(x/3))", "query_expressions": "Expression(G)", "answer_expressions": "x^2/9 - y^2 = 1", "fact_spans": "[[[2, 5], [85, 88]], [[11, 33]], [[36, 57]], [[11, 33]], [[36, 57]], [[2, 57]], [[2, 83]]]", "query_spans": "[[[85, 95]]]", "process": "From the coordinates of the foci of the hyperbola, we know $ c = \\sqrt{10} $. Also, since the equations of the asymptotes are $ y = \\pm\\frac{1}{3}x $, it follows that $ \\frac{b}{a} = \\frac{1}{3} $. Moreover, since $ c^{2} = a^{2} + b^{2} $, solving gives $ b^{2} = 1 $, $ a^{2} = 9 $. Therefore, the standard equation of the hyperbola is $ \\frac{x^{2}}{9} - y^{2} = 1 $. Hence, fill in $ \\frac{x^{2}}{9} - y^{2} = 1 $. \n[Analysis] This problem mainly examines the standard equation of a hyperbola and the simple geometric properties of hyperbolas, and belongs to a medium-difficulty problem." }, { "text": "The hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ has right focus $F$, point $P$ is a point on the asymptote, and $|O P|=2$, then $|P F|$=?", "fact_expressions": "G: Hyperbola;O: Origin;P: Point;F: Point;Expression(G) = (x^2 - y^2/3 = 1);RightFocus(G) = F;PointOnCurve(P,Asymptote(G));Abs(LineSegmentOf(O, P)) = 2", "query_expressions": "Abs(LineSegmentOf(P, F))", "answer_expressions": "2", "fact_spans": "[[[0, 28]], [[49, 58]], [[36, 40]], [[32, 35]], [[0, 28]], [[0, 35]], [[0, 47]], [[49, 58]]]", "query_spans": "[[[60, 69]]]", "process": "" }, { "text": "If the ellipse $\\frac{x^{2}}{a^{2}}+y^{2}=1(a>0)$ and the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$ have the same foci, then $a$=?", "fact_expressions": "G: Hyperbola;H: Ellipse;a: Number;Expression(G) = (x^2/2 - y^2 = 1);a>0;Expression(H) = (y^2 + x^2/a^2 = 1);Focus(G) = Focus(H)", "query_expressions": "a", "answer_expressions": "2", "fact_spans": "[[[38, 66]], [[1, 37]], [[74, 77]], [[38, 66]], [[3, 37]], [[1, 37]], [[1, 72]]]", "query_spans": "[[[74, 79]]]", "process": "" }, { "text": "Given point $A(1,1)$, and $F$ is the left focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, $P$ is any point on the ellipse. Find the minimum value of $|P F|+|P A|$.", "fact_expressions": "G: Ellipse;A: Point;P: Point;F: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Coordinate(A) = (1, 1);LeftFocus(G) = F;PointOnCurve(P, G)", "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))", "answer_expressions": "3", "fact_spans": "[[[17, 54], [63, 65]], [[2, 11]], [[59, 62]], [[13, 16]], [[17, 54]], [[2, 11]], [[13, 58]], [[59, 70]]]", "query_spans": "[[[72, 91]]]", "process": "From the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, we know $a=2$, $c=\\sqrt{a^{2}-b^{2}}=1$. Let the right focus of the ellipse be $F_{1}(1,0)$, then $|AF_{1}|=1$. As shown in the figure, so $|PF|+|PA|=2a-|PF_{1}|+|PA|=4+|PA|-|PF_{1}|\\geqslant4-|AF_{1}|=4-1=3$, that is, the minimum value is attained when $P$ lies on the extension line of $F_{1}A$." }, { "text": "Point $A$ is an intersection point of the ellipse $C_{1}$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ and the hyperbola $C_{2}$: $\\frac{x^{2}}{4}-y^{2}=1$, and the sum of the distances from point $A$ to the two foci of ellipse $C_{1}$ is $m$, while the absolute value of the difference of these distances is $n$. Then the value of $m+n$ is?", "fact_expressions": "C2: Hyperbola;C1: Ellipse;A: Point;Expression(C1) = (x^2/9 + y^2/4 = 1);Expression(C2) = (x^2/4 - y^2 = 1);OneOf(Intersection(C1, C2)) = A;F1: Point;F2: Point;Focus(C1) = {F1, F2};Distance(A, F1) + Distance(A, F2) = m ;m: Number;n: Number;Abs(Distance(A, F1) - Distance(A, F2)) = n", "query_expressions": "m + n", "answer_expressions": "10", "fact_spans": "[[[52, 89]], [[5, 51], [101, 110]], [[0, 4], [96, 100]], [[5, 51]], [[52, 89]], [[0, 94]], [], [], [[101, 113]], [[96, 121]], [[118, 121]], [[131, 134]], [[96, 134]]]", "query_spans": "[[[136, 145]]]", "process": "The ellipse $ C_{1}: \\frac{x^{2}}{9} + \\frac{y^{2}}{4} = 1 $ has foci at $ F_{1}(\\sqrt{5},0), F_{2}(-\\sqrt{5},0) $. The hyperbola $ C_{2}: \\frac{x^{2}}{4} - y^{2} = 1 $ has foci at $ (\\sqrt{5},0), (-\\sqrt{5},0) $. Therefore, the ellipse and the hyperbola have the same foci. Point A is an intersection point of the ellipse $ C_{1}: \\frac{x^{2}}{9} + \\frac{y^{2}}{4} = 1 $ and the hyperbola $ C_{2}: \\frac{x^{2}}{4} - y^{2} = 1 $. Hence, $ m = 6 $, $ n = 4 $, $ m + n = 10 $." }, { "text": "Given the line $ l $: $ y = k(x + 4) $ intersects the circle $ (x + 2)^2 + y^2 = 4 $ at two points $ A $ and $ B $, and $ M $ is the midpoint of segment $ AB $. Then the minimum distance from $ M $ to the line $ 3x + 4y - 6 = 0 $ is?", "fact_expressions": "l: Line;Expression(l) = (y = k*(x + 4));k:Number;G: Circle;Expression(G) = (y^2 + (x + 2)^2 = 4);A: Point;B: Point;Intersection(l, G) = {A, B};M: Point;MidPoint(LineSegmentOf(A, B)) = M;H: Line;Expression(H) = (3*x + 4*y - 6 = 0)", "query_expressions": "Min(Distance(M, H))", "answer_expressions": "2", "fact_spans": "[[[2, 19]], [[2, 19]], [[9, 19]], [[20, 40]], [[20, 40]], [[43, 46]], [[47, 50]], [[2, 52]], [[53, 56], [69, 72]], [[53, 67]], [[73, 88]], [[73, 88]]]", "query_spans": "[[[69, 97]]]", "process": "Since line $ l $ always passes through the point $(-4,0)$ and this point lies on the circle, let $ A(-4,0) $, $ M(x,y) $, $ B(x_{1},y_{1}) $. The trajectory method can be used to find the trajectory equation of $ M $ as $ (x+3)^{2}+y^{2}=1 $. The solution can then be obtained using the distance from a point to a line. [Detailed Solution] The center of the circle $ (x+2)^{2}+y^{2}=4 $ is $ C(-2,0) $, with radius $ r=2 $. The distance from the center $ C(-2,0) $ to the line $ y=k(x+4) $ is $ d = \\frac{|-2k+4k|}{\\sqrt{k^{2}+1}} = \\frac{|2k|}{\\sqrt{k^{2}+1}} < 2 $. Line $ l: y=k(x+4) $ passes through the fixed point $ A(-4,0) $. Let $ M(x,y) $, $ B(x_{1},y_{1}) $, then \n$$\n\\begin{cases}\nx_{1}=2x+4 \\\\\ny_{1}=2y\n\\end{cases}\n$$\nSubstituting into $ (x+2)^{2}+y^{2}=4 $, we obtain $ (x+3)^{2}+y^{2}=1 $. Therefore, the trajectory of $ M $ is a circle centered at $ (-3,0) $ with radius 1, excluding the point $ (-4,0) $. Then, the minimum distance from $ M $ to the line $ 3x+4y-6=0 $ is $ \\frac{|-3\\times3-6|}{5}-1=2 $. This problem examines the relationship between a line and a circle, the use of the trajectory method to find curve equations, and the ability to combine numerical and geometric reasoning; it is a medium-difficulty problem." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the right focus is $F(c, 0)$. A perpendicular is drawn from $F(c, 0)$ to one of the asymptotes, with foot of perpendicular at point $A$, and $O$ is the origin. When the area of $\\Delta O A F$ is $\\frac{c^{2}}{4}$, what is the eccentricity of this hyperbola?", "fact_expressions": "G:Hyperbola;b: Number;a: Number;F: Point;O: Origin;A: Point;c:Number;L:Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G)=F;Coordinate(F) = (c, 0);PointOnCurve(F,L);IsPerpendicular(OneOf(Asymptote(G)),L);FootPoint(OneOf(Asymptote(G)),L)=A;Area(TriangleOf(O, A, F)) = c^2/4", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 58], [149, 152]], [[5, 58]], [[5, 58]], [[63, 72], [74, 83]], [[100, 103]], [[96, 99]], [[63, 72]], [], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 72]], [[63, 72]], [[2, 92]], [[2, 92]], [[2, 99]], [[110, 145]]]", "query_spans": "[[[149, 158]]]", "process": "According to the problem, take one asymptote of the hyperbola: $ l: bx - ay = 0 $. Using the point-to-line distance formula, we find $ |FA| $. By the Pythagorean theorem, we obtain $ |OA| $. Based on the area formula, we derive a relation between $ a $ and $ c $, and solve accordingly. \nAs per the problem, take one asymptote of the hyperbola: $ l: bx - ay = 0 $, then $ |FA| = \\frac{bc}{\\sqrt{a^{2}+b^{2}}} = b $. By the Pythagorean theorem, $ |OA| = \\sqrt{OF^{2}-AF^{2}} = \\sqrt{c^{2}-b^{2}} = a $. The area $ S_{AOAF} = \\frac{1}{2}ab = \\frac{1}{4}c^{2} $. Solving gives $ 2ab = c^{2} = a^{2} + b^{2} $, so $ a = b $. Then $ c = \\sqrt{2}a $, and the eccentricity is $ e = \\frac{c}{a} = \\sqrt{2} $." }, { "text": "The coordinates of the focus of the parabola $y^{2}=-8 x$ are?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -8*x)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(-2,0)", "fact_spans": "[[[0, 15]], [[0, 15]]]", "query_spans": "[[[0, 22]]]", "process": "" }, { "text": "The line passing through the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$ and perpendicular to the $x$-axis intersects the hyperbola at two points $M$ and $N$. The circle with $MN$ as diameter passes exactly through the left vertex of the hyperbola. Then, the eccentricity of the hyperbola is equal to?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;L: Line;M: Point;N: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(RightFocus(G), L);IsPerpendicular(L,xAxis);Intersection(L,G) = {N, M};IsDiameter(LineSegmentOf(M,N),H);PointOnCurve(LeftVertex(G),H)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[1, 60], [76, 79], [106, 109], [115, 118]], [[4, 60]], [[4, 60]], [[102, 103]], [[73, 75]], [[82, 85]], [[86, 89]], [[4, 60]], [[4, 60]], [[1, 60]], [[0, 75]], [[65, 75]], [[73, 92]], [[93, 103]], [[102, 113]]]", "query_spans": "[[[115, 125]]]", "process": "" }, { "text": "Let $F_{1}$, $F_{2}$ be the two foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, point $P$ lies on the ellipse, and $F_{1} P \\perp P F_{2}$, then the area of $\\Delta F_{1}PF_{2}$ is?", "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(F1, P), LineSegmentOf(P, F2))", "query_expressions": "Area(TriangleOf(F1,P,F2))", "answer_expressions": "1", "fact_spans": "[[[17, 44], [55, 57]], [[1, 8]], [[50, 54]], [[9, 16]], [[17, 44]], [[1, 49]], [[50, 58]], [[60, 83]]]", "query_spans": "[[[85, 110]]]", "process": "Problem Analysis: From the given, |PF_{1}| + |PF_{2}| = 4, |F_{1}F_{2}| = 2\\sqrt{3}. By the Pythagorean theorem, |PF_{1}||PF_{2}| = 2. From this, the area of \\triangle F_{1}PF_{2} can be found. \nSolution: Since F_{1}, F_{2} are the two foci of the ellipse \\frac{x^{2}}{4} + y^{2} = 1, and point P lies on the ellipse such that F_{1}P \\perp PF_{2}, \n\\therefore |PF_{1}| + |PF_{2}| = 4, |F_{1}F_{2}| = 2\\sqrt{3}, \n\\therefore |PF_{1}|^{2} + |PF_{2}|^{2} + 2|PF_{1}| \\times |PF_{2}| = 16, \n\\therefore |F_{1}F_{2}|^{2} + 2|PF_{1}| \\times |PF_{2}| = 16, \n\\therefore 12 + 2|PF_{1}| \\times |PF_{2}| = 16, \n2|PF_{1}| \\times |PF_{2}| = 4, \n\\therefore |PF_{1}| \\times |PF_{2}| = 2 \n\\therefore The area S of \\triangle F_{1}PF_{2} is \\frac{1}{2}|PF_{1}| \\times |PF_{2}| = \\frac{1}{2} \\times 2 = 1." }, { "text": "Given the parabola $C$: $y^{2}=-4x$ with focus $F$, $A(-2,1)$, and $P$ a moving point on the parabola $C$, find the minimum value of $|PF| + |PA|$.", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = -4*x);F: Point;Focus(C) = F;A: Point;Coordinate(A) = (-2, 1);P: Point;PointOnCurve(P, C)", "query_expressions": "Min(Abs(LineSegmentOf(P, F)) + Abs(LineSegmentOf(P, A)))", "answer_expressions": "3", "fact_spans": "[[[2, 22], [48, 54]], [[2, 22]], [[26, 29]], [[2, 29]], [[32, 41]], [[32, 41]], [[44, 47]], [[44, 58]]]", "query_spans": "[[[60, 79]]]", "process": "" }, { "text": "If the line $a x+y+b-1=0$ ($a>0$, $b>0$) passes through the focus $F$ of the parabola $y^{2}=4 x$, then the minimum value of $\\frac{1}{a}+\\frac{1}{b}$ is?", "fact_expressions": "G: Parabola;H: Line;b: Number;a: Number;F: Point;Expression(G) = (y^2 = 4*x);a>0;b>0;Expression(H) = (b + a*x + y - 1 = 0);Focus(G) = F;PointOnCurve(F, H)", "query_expressions": "Min(1/b + 1/a)", "answer_expressions": "4", "fact_spans": "[[[28, 42]], [[1, 27]], [[50, 75]], [[50, 75]], [[45, 48]], [[28, 42]], [[3, 27]], [[3, 27]], [[1, 27]], [[28, 48]], [[1, 48]]]", "query_spans": "[[[50, 81]]]", "process": "From the parabola $ y^{2} = 4x $, we obtain the focus $ F(1,0) $. Substituting into the line equation $ ax + y + b - 1 = 0 $, we get: $ a + b = 1 $. Since $ a > 0 $, $ b > 0 $, \n$ \\therefore \\frac{1}{a} + \\frac{1}{b} = (a + b)\\left( \\frac{1}{a} + \\frac{1}{b} \\right) = 2 + \\frac{b}{a} + \\frac{a}{b} \\geqslant 2 + 2\\sqrt{\\frac{b}{a} \\cdot \\frac{a}{b}} = 4 $. Equality holds if and only if $ a = b = \\frac{1}{2} $. \n$ \\therefore $ The minimum value of $ \\frac{1}{a} + \\frac{1}{b} $ is $ 4 $." }, { "text": "Let the parabola $C$: $y^{2}=4x$ have focus $F$, and let $M$ be a point on the parabola $C$ such that the horizontal coordinate of $M$ is $2$. Then $|MF|=$?", "fact_expressions": "C: Parabola;M: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(M, C);XCoordinate(M) = 2", "query_expressions": "Abs(LineSegmentOf(M, F))", "answer_expressions": "3", "fact_spans": "[[[1, 19], [31, 37]], [[27, 30], [42, 46]], [[23, 26]], [[1, 19]], [[1, 26]], [[27, 40]], [[42, 54]]]", "query_spans": "[[[56, 65]]]", "process": "" }, { "text": "Given point $A(2, \\frac{3}{2})$, point $F$ is the focus of the parabola $C$: $y^{2}=4 x$, and point $M$ is a point on the parabola $C$. What are the coordinates of point $M$ when $|M A|+|M F|$ takes the minimum value?", "fact_expressions": "C: Parabola;A: Point;M: Point;F: Point;Expression(C) = (y^2 = 4*x);Coordinate(A) = (2, 3/2);Focus(C) = F;PointOnCurve(M, C);WhenMin(Abs(LineSegmentOf(M,A))+Abs(LineSegmentOf(M,F)))", "query_expressions": "Coordinate(M)", "answer_expressions": "{9/16, 3/2}", "fact_spans": "[[[28, 47], [56, 62]], [[2, 22]], [[51, 55], [86, 90]], [[23, 27]], [[28, 47]], [[2, 22]], [[23, 50]], [[51, 65]], [[68, 86]]]", "query_spans": "[[[86, 95]]]", "process": "" }, { "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1(b>0)$ are given by $y=\\pm \\frac{1}{2} x$, then what is the value of $b$?", "fact_expressions": "G: Hyperbola;b: Number;b>0;Expression(G) = (x^2/4 - y^2/b^2 = 1);Expression(Asymptote(G)) = (y = pm*(x/2))", "query_expressions": "b", "answer_expressions": "English:", "fact_spans": "[[[1, 48]], [[78, 81]], [[4, 48]], [[1, 48]], [[1, 76]]]", "query_spans": "[[[78, 84]]]", "process": "" }, { "text": "Given the parabola $ C $: $ y^{2} = 2px $ ($ p > 0 $), the focus is $ F $, the directrix intersects the $ x $-axis at point $ M $. A line $ l $ with slope $ k $ passing through point $ M $ intersects the parabola $ C $ at points $ A $ and $ B $. If $ |AM| = \\frac{5}{4}|AF| $, what is the value of $ k $?", "fact_expressions": "l: Line;C: Parabola;p: Number;k:Number;A: Point;B: Point;M: Point;F:Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(M, l);Intersection(Directrix(C), xAxis) = M;Slope(l)=k;Intersection(l,C)={A,B};Abs(LineSegmentOf(A, M)) = (5/4)*Abs(LineSegmentOf(A, F))", "query_expressions": "k", "answer_expressions": "pm*(3/4)", "fact_spans": "[[[60, 65]], [[2, 26], [66, 72]], [[9, 26]], [[111, 114], [56, 59]], [[74, 78]], [[79, 83]], [[43, 47], [49, 53]], [[30, 33]], [[9, 26]], [[2, 26]], [[2, 33]], [[48, 65]], [[2, 47]], [[53, 65]], [[60, 85]], [[87, 109]]]", "query_spans": "[[[111, 117]]]", "process": "" }, { "text": "Given that the line $l$ intersects the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$ at points $A$ and $B$, if the midpoint of $AB$ has coordinates $(-1,1)$, then what is the equation of line $l$?", "fact_expressions": "l: Line;G: Ellipse;A: Point;B: Point;Expression(G) = (x^2/8 + y^2/4 = 1);Intersection(l, G) = {A, B};Coordinate(MidPoint(LineSegmentOf(A, B))) = (-1,1)", "query_expressions": "Expression(l)", "answer_expressions": "x-2*y+3=0", "fact_spans": "[[[2, 7], [79, 84]], [[8, 45]], [[47, 50]], [[51, 54]], [[8, 45]], [[2, 56]], [[58, 77]]]", "query_spans": "[[[79, 89]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}) satisfy \\frac{x_{1}^2}{8}+\\frac{y_{1}^2}{4}=1, subtracting the two equations yields \\frac{x_{1}^2-x_{2}^2}{8}+\\frac{y_{1}^2-y_{2}^2}{4}=0, which can be rewritten as \\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{8}+\\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{4}=0. Since the midpoint coordinates of AB are (-1,1), we have x_{1}+x_{2}=-2, y_{1}+y_{2}=2. Thus, \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{1}{2}, meaning the slope of line l is \\frac{1}{2}. Therefore, the equation of line l is y-1=\\frac{1}{2}(x+1) \\Rightarrow x-2y+3=0." }, { "text": "It is known that the right focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ coincides with the focus of the parabola $y^{2}=12 x$. Then the asymptotes of this hyperbola are given by?", "fact_expressions": "G: Hyperbola;b: Number;H: Parabola;Expression(G) = (x^2/4 - y^2/b^2 = 1);Expression(H) = (y^2 = 12*x);RightFocus(G) = Focus(H)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y = pm*(sqrt(5)/2)*x", "fact_spans": "[[[2, 44], [72, 75]], [[5, 44]], [[49, 64]], [[2, 44]], [[49, 64]], [[2, 69]]]", "query_spans": "[[[72, 83]]]", "process": "Since the focus of the parabola $ y^{2} = 12x $ is at $ (3,0) $, the right focus of the hyperbola $ \\frac{x^{2}}{4} - \\frac{y^{2}}{b^{2}} = 1 $ is also $ (3,0) $, that is, $ c = 3 $. Since $ c^{2} = a^{2} + b^{2} \\Rightarrow 9 = 4 + b^{2} \\Rightarrow b = \\sqrt{5} $, the asymptotes of this hyperbola are given by $ y = \\pm\\frac{\\sqrt{5}}{2}x $." }, { "text": "Let the hyperbolas $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ and $-\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ have eccentricities $e_{1}$ and $e_{2}$, respectively. Then, when $a$ and $b$ vary, what is the minimum value of $e_{1}+e_{2}$?", "fact_expressions": "G1: Hyperbola;Expression(G1) = (-y^2/b^2 + x^2/a^2 = 1);G2: Hyperbola;Expression(G2) = (y^2/b^2 - x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;e1: Number;e2: Number;Eccentricity(G1) = e1;Eccentricity(G2) = e2", "query_expressions": "Min(e1 + e2)", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[1, 47]], [[1, 47]], [[48, 103]], [[48, 103]], [[135, 138]], [[129, 132]], [[48, 103]], [[48, 103]], [[109, 116]], [[118, 126]], [[1, 126]], [[1, 126]]]", "query_spans": "[[[143, 161]]]", "process": "" }, { "text": "If the equation of the parabola is $7 x+4 y^{2}=0$, then the coordinates of the focus are?", "fact_expressions": "G: Parabola;Expression(G) = (7*x + 4*y^2 = 0)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(-7/16, 0)", "fact_spans": "[[[1, 4]], [[1, 22]]]", "query_spans": "[[[1, 30]]]", "process": "The equation becomes y^{2}=-\\frac{7}{4}x, the parabola opens to the left, so 2p=\\frac{7}{4}, that is, \\frac{p}{2}=\\frac{7}{16}, therefore the focus coordinates are (-\\frac{7}{16},0)" }, { "text": "A circle centered at the vertex of parabola $C$ intersects $C$ at points $M$, $N$, and intersects the directrix of $C$ at points $A$, $B$. Given $|M N|=4 \\sqrt{2}$, $|A B|=2 \\sqrt{11}$, then the distance from the focus of $C$ to its directrix is?", "fact_expressions": "C: Parabola;G: Circle;M: Point;N: Point;A: Point;B: Point;Vertex(C)=Center(G);Intersection(G, C) = {M, N};Intersection(G, Directrix(C)) = {A, B};Abs(LineSegmentOf(M, N)) = 4*sqrt(2);Abs(LineSegmentOf(A, B)) = 2*sqrt(11)", "query_expressions": "Distance(Focus(C), Directrix(C))", "answer_expressions": "2", "fact_spans": "[[[1, 7], [16, 19], [31, 34], [88, 91]], [[14, 15]], [[20, 23]], [[24, 27]], [[38, 41]], [[42, 45]], [[0, 15]], [[14, 29]], [[14, 47]], [[47, 65]], [[67, 86]]]", "query_spans": "[[[88, 102]]]", "process": "Solution: Let the parabola be y^{2}=2px, as shown in the figure: |MN|=4\\sqrt{2}, |AB|=2\\sqrt{11}(\\frac{2\\sqrt{2}}{2})^{2}=\\frac{4}{P}, r=\\sqrt{3}(2\\sqrt{2})^{2}=2px_{M}', F=\\sqrt{(-\\frac{p}{2})^{2}+(\\sqrt{11})^{2}}=\\frac{4}{p}, \\therefore M(\\frac{4}{P}, 2\\sqrt{2}), solving gives: p=2, that is, the distance from the focus of C to the directrix is: 2" }, { "text": "The hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a, b>0)$ passes through the point $(2,2)$. The left and right foci of $C$ are $F_{1}$ and $F_{2}$, respectively. A line perpendicular to the $x$-axis is drawn through $F_{2}$, intersecting $C$ at points $A$ and $B$. The line $F_{1} B$ intersects the $y$-axis at $D$. If $A D \\perp F_{1} B$, then the equation of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;F1: Point;B: Point;A: Point;D: Point;F2: Point;L:Line;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (2, 2);PointOnCurve(P,C);LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(F2,L);IsPerpendicular(L,xAxis);Intersection(L,C)={A,B};Intersection(LineSegmentOf(F1, B), yAxis) = D;IsPerpendicular(LineSegmentOf(A, D), LineSegmentOf(F1, B))", "query_expressions": "Expression(C)", "answer_expressions": "x^2/2 - y^2/4 = 1", "fact_spans": "[[[10, 69], [108, 111], [168, 174]], [[18, 69]], [[18, 69]], [[1, 9]], [[75, 82]], [[118, 121]], [[114, 117]], [[141, 144]], [[83, 90], [92, 99]], [], [[18, 69]], [[18, 69]], [[10, 69]], [[1, 9]], [[0, 69]], [[10, 90]], [[10, 90]], [[91, 107]], [[91, 107]], [[91, 123]], [[124, 144]], [[147, 166]]]", "query_spans": "[[[168, 179]]]", "process": "According to the perpendicular relationship, we have \\overrightarrow{AD}\\cdot\\overrightarrow{F_{1}B}=0. Based on the coordinate operation of the dot product, a homogeneous equation in terms of a and c can be constructed to solve for the eccentricity e, thereby obtaining the relationship between a and c. Using the fact that the hyperbola passes through the point (2,2) and c^{2}=a^{2}+b^{2}, a system of equations can be constructed to find the result. Let x=c, substitute into the hyperbola equation: A(c,\\frac{b^{2}}{a}), B(c,-\\frac{b^{2}}{a}). Since OD//AB and O is the midpoint of F_{1}F_{2}, OD is the midline of quadrilateral AF_{1}BF_{2}. Therefore, |OD|=\\frac{1}{2}|BF_{2}|=\\frac{b^{2}}{2a}, so D(0,-\\frac{b^{2}}{2a}). Thus, \\overrightarrow{AD}=(-c,-\\frac{3b^{2}}{2a}), \\overrightarrow{F_{1}B}=(2c,-\\frac{b^{2}}{a}). Since AD\\bot F_{1}B, \\overrightarrow{AD}\\cdot\\overrightarrow{F_{1}B}=0, i.e., -2c^{2}+\\frac{3b^{4}}{2a^{2}} = -2c^{2}+\\frac{3(c^{2}-a^{2})^{2}}{2a^{2}}=0. Therefore, 3c^{4}-10a^{2}c^{2}+3a^{4}=0, so 3e^{4}-10e^{2}+3=0. Solving gives: e^{2}=3 or e^{2}=\\frac{1}{3} (discarded). Therefore, e=\\sqrt{3}, i.e., c=\\sqrt{3}a. Also, \\begin{cases}\\frac{4}{a^{2}}-\\frac{4}{b^{2}}=1\\\\c^{2}=a^{2}+b^{2}\\end{cases} \\Rightarrow a^{2}=2, b^{2}=4. Therefore, the equation of hyperbola C is: \\frac{x^{2}}{2}-\\frac{y^{2}}{4}=1." }, { "text": "The line $ l $ with slope $ -\\frac{1}{3} $ intersects the ellipse $ C $: $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 \\ (a > b > 0) $ at points $ A $ and $ B $. The coordinates of the midpoint of segment $ AB $ are $ (1, 2) $. What is the eccentricity of ellipse $ C $?", "fact_expressions": "l: Line;C: Ellipse;b: Number;a: Number;B: Point;A: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(MidPoint(LineSegmentOf(A,B))) = (1, 2);Slope(l)=-1/3;Intersection(l, C) = {A, B}", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[18, 23]], [[24, 83], [120, 125]], [[31, 83]], [[31, 83]], [[90, 93]], [[86, 89]], [[31, 83]], [[31, 83]], [[24, 83]], [[96, 118]], [[0, 23]], [[18, 95]]]", "query_spans": "[[[120, 132]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), then \\frac{x_{1}^{2}}{a^{2}}+\\frac{y_{1}^{2}}{b^{2}}=1\\textcircled{1}, \\frac{x_{2}^{2}}{a^{2}}+\\frac{y_{2}^{2}}{b^{2}}=1\\textcircled{2}. Since (1,2) is the midpoint of segment AB, \\therefore \\frac{1}{2}(x_{1}+x_{2})=1, \\frac{1}{2}(y_{1}+y_{2})=2. Since the equation of line AB is y=-\\frac{1}{3}(x-1)+2, \\therefore y_{1}-y_{2}=-\\frac{1}{3}(x_{1}-x_{2}). Subtracting equation \\textcircled{2} from \\textcircled{1} gives: \\frac{1}{a^{2}}(x_{1}^{2}-x_{2}^{2})+\\frac{1}{b^{2}}(y_{1}^{2}-y_{2}^{2})=0, \\therefore \\frac{1}{a^{2}}(x_{1}-x_{2})(x_{1}+x_{2})+\\frac{1}{b^{2}}(y_{1}-y_{2})(y_{1}+y_{2})=0, \\therefore 2\\times\\frac{1}{a^{2}}(x_{1}-x_{2})+4\\times\\frac{1}{b^{2}}(y_{1}-y_{2})=0, \\therefore \\frac{b^{2}}{a^{2}}=\\frac{2}{3}, e^{a-1}=\\frac{1}{2} \\therefore e=\\frac{\\sqrt{3}}{3}" }, { "text": "Through the right focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a perpendicular to one of its asymptotes, with foot of perpendicular at $Q$. The line $FQ$ intersects the left and right branches of the hyperbola at points $M$ and $N$, respectively. If $|MQ|=3|QN|$, $|FN|=4$, then the standard equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;a: Number;b: Number;Q: Point;F: Point;M: Point;N: Point;l:Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G)=F;PointOnCurve(F,l);IsPerpendicular(l,OneOf(Asymptote(G)));FootPoint(l,OneOf(Asymptote(G)))=Q;Intersection(LineOf(F,Q), LeftPart(G)) = M;Intersection(LineOf(F,Q), RightPart(G)) = N;Abs(LineSegmentOf(M, Q)) = 3*Abs(LineSegmentOf(Q, N));Abs(LineSegmentOf(F, N)) = 4", "query_expressions": "Expression(G)", "answer_expressions": "x^2/9 - y^2/36 = 1", "fact_spans": "[[[1, 57], [91, 94], [140, 143]], [[4, 57]], [[4, 57]], [[79, 82]], [[61, 64]], [[104, 108]], [[109, 112]], [], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 64]], [[0, 75]], [[0, 75]], [[0, 82]], [[83, 112]], [[83, 112]], [[114, 128]], [[129, 138]]]", "query_spans": "[[[140, 150]]]", "process": "" }, { "text": "Given the circle $C$: $x^{2}+y^{2}=25$, draw a line $l$ through point $M(-2,3)$ intersecting the circle $C$ at points $A$ and $B$. Tangents to the circle are drawn at points $A$ and $B$, respectively. When the two tangents intersect at point $N$, what is the trajectory equation of point $N$?", "fact_expressions": "C: Circle;Expression(C) = (x^2 + y^2 = 25);M: Point;l: Line;Coordinate(M) = (-2, 3);PointOnCurve(M, l);A: Point;B: Point;Intersection(l, C) = {A, B};L1: Line;L2: Line;TangentOnPoint(A,C)=L1;TangentOnPoint(B,C)=L2;N: Point;Intersection(L1, L2) = N", "query_expressions": "LocusEquation(N)", "answer_expressions": "2*x-3*y+25=0", "fact_spans": "[[[2, 24], [43, 47], [71, 72]], [[2, 24]], [[26, 36]], [[37, 42]], [[26, 36]], [[25, 42]], [[48, 51], [61, 64]], [[52, 55], [65, 68]], [[37, 57]], [], [], [[60, 75]], [[60, 75]], [[84, 88], [91, 95]], [[60, 88]]]", "query_spans": "[[[91, 102]]]", "process": "Consider the following problem: Given a circle $ C: x^{2} + y^{2} = r^{2} $ ($ r > 0 $) and a point $ P(a, b) $. If point $ P $ lies inside $ C $, draw a line $ l $ through $ P $ intersecting $ C $ at points $ A $ and $ B $. Tangents to $ C $ are drawn at points $ A $ and $ B $ respectively. When these two tangents intersect at point $ Q $, find the locus equation of point $ Q $. The center of circle $ C: x^{2} + y^{2} = r^{2} $ is $ (0, 0) $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, $ Q(x_{0}, y_{0}) $. Since $ AQ $ is tangent to circle $ C $, we have $ AQ \\perp CA $. Therefore, $ (x_{1} - x_{0})(x_{1} - 0) + (y_{1} - y_{0})(y_{1} - 0) = 0 $, that is, $ x^{2}_{1} - x_{0}x_{1} + y^{2}_{1} - y_{0}y_{1} = 0 $. Since $ x^{2} + y^{2} = r^{2} $, it follows that $ x_{0}x_{1} + y_{0}y_{1} = r^{2} $. Similarly, $ x_{0}x_{2} + y_{0}y_{2} = r^{2} $. Hence, the equation of the line passing through points $ A $ and $ B $ is $ xx_{0} + yy_{0} = r^{2} $. Since line $ AB $ passes through point $ (a, b) $, substituting gives $ ax_{0} + by_{0} = r^{2} $. Therefore, the locus equation of point $ Q $ is: $ ax + by = r^{2} $. According to the problem, the locus equation of point $ N $ is $ 2x - 3y + 25 = 0 $." }, { "text": "If the parabola passes through the point $(-1,3)$, then what is the standard equation of the parabola?", "fact_expressions": "G: Parabola;H: Point;Coordinate(H) = (-1, 3);PointOnCurve(H, G)", "query_expressions": "Expression(G)", "answer_expressions": "{(y^2=-9*x),(x^2=y/3)}", "fact_spans": "[[[1, 4], [16, 19]], [[5, 14]], [[5, 14]], [[1, 14]]]", "query_spans": "[[[16, 26]]]", "process": "Let the equation of the parabola be y^{2}=ax, then 3^{2}=-a, a=-9, the equation is y^{2}=-9x. Let the equation of the parabola be x^{2}=ay, then (-1)^{2}=3a, a=\\frac{1}{3}, the equation is x^{2}=\\frac{1}{3}y." }, { "text": "The focal distance of the hyperbola $2 x^{2}-y^{2}=k$ is $6$, then what is the value of $k$?", "fact_expressions": "G: Hyperbola;k: Number;Expression(G) = (2*x^2 - y^2 = k);FocalLength(G) = 6", "query_expressions": "k", "answer_expressions": "pm*6", "fact_spans": "[[[0, 20]], [[29, 32]], [[0, 20]], [[0, 27]]]", "query_spans": "[[[29, 36]]]", "process": "" }, { "text": "Given the parabola $y=\\frac{1}{4} x^{2}$ with focus $F$, and let $P$ be a moving point on the parabola, point $Q(1,1)$. When the perimeter of $\\triangle P Q F$ is minimized, what are the coordinates of point $P$?", "fact_expressions": "G: Parabola;Q: Point;P: Point;F: Point;Expression(G) = (y = x^2/4);Coordinate(Q) = (1, 1);Focus(G) = F;PointOnCurve(P, G);WhenMin(Perimeter(TriangleOf(P, Q, F)))", "query_expressions": "Coordinate(P)", "answer_expressions": "(1, 1/4)", "fact_spans": "[[[2, 26], [38, 41]], [[46, 55]], [[34, 37], [81, 85]], [[30, 33]], [[2, 26]], [[46, 55]], [[2, 33]], [[34, 45]], [[56, 80]]]", "query_spans": "[[[81, 90]]]", "process": "As shown in the figure, let l: y = -1 be the directrix of the parabola. Draw PH ⊥ l at H from point P, and draw QN ⊥ l at N from point Q. Then |PF| = |PH|, F(0,1), |FQ| = 1, |PF| + |PQ| = |PQ| + |PH|. It is clear that when points Q, P, H are collinear, |PQ| + |PH| is minimized, and the minimum value is 1 + 1 = 2. Therefore, the minimum perimeter of triangle PQF is 3. At this time, x_{P} = 1, y_{P} = \\frac{1}{4}, that is, P(1,\\frac{1}{4})." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the right branch of the hyperbola. If $|P F_{1}|-|P F_{2}|=b$, and the focal distance of the hyperbola is $2 \\sqrt{5}$, then the equation of this hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)) = b;FocalLength(G) = 2*sqrt(5)", "query_expressions": "Expression(G)", "answer_expressions": "x^2-y^2/4=1", "fact_spans": "[[[2, 59], [89, 92], [123, 126], [145, 148]], [[2, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[68, 75]], [[76, 83]], [[2, 83]], [[2, 83]], [[84, 88]], [[84, 96]], [[98, 121]], [[123, 142]]]", "query_spans": "[[[145, 153]]]", "process": "From the given condition, we have |PF₁| - |PF₂| = 2a = b, c² = a² + b². Solving gives \n\\begin{cases} a^2 = 1, \\\\ b^{2} = 4 \\end{cases} \n2c = 2\\sqrt{5}. Then the equation of this hyperbola is x^{2} - \\frac{y^{2}}{4} =" }, { "text": "The focus of the parabola $y^{2}=2 p x(p>0)$ is $F$, the directrix is $l$, $A$ and $B$ are two moving points on the parabola satisfying $\\angle A F B=\\frac{\\pi}{3}$. Let the midpoint $M$ of segment $A B$ have projection $N$ on $l$. Then the maximum value of $\\frac{|M N|}{|A B|}$ is?", "fact_expressions": "G: Parabola;p: Number;l: Line;A: Point;B: Point;F: Point;M: Point;N: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(A, G);PointOnCurve(B, G);AngleOf(A, F, B) = pi/3;Directrix(G) = l;MidPoint(LineSegmentOf(A, B)) = M;Projection(M, l) = N", "query_expressions": "Max(Abs(LineSegmentOf(M, N))/Abs(LineSegmentOf(A, B)))", "answer_expressions": "1", "fact_spans": "[[[0, 21], [47, 50]], [[3, 21]], [[32, 35]], [[37, 40]], [[43, 46]], [[25, 28]], [[99, 102]], [[111, 114]], [[3, 21]], [[0, 21]], [[0, 28]], [[37, 55]], [[37, 55]], [[59, 87]], [[0, 35]], [[89, 102]], [[99, 114]]]", "query_spans": "[[[116, 143]]]", "process": "Draw AQ perpendicular to line l at Q from point A, draw BP perpendicular to line l at P from point B. Let |AF| = a, |BF| = b. As shown in the figure, according to the definition of a parabola, we have |MN| = \\frac{1}{2}(a + b). In triangle ABF, |AB| = a^{2} + b^{2} - 2ab \\cdot \\cos\\frac{\\pi}{2} = a^{2} + b^{2} - ab = (a + b)^{2} - 3ab. Since ab \\leqslant \\left(\\frac{a + b}{2}\\right)^{2}, it follows that |AB| \\geqslant \\frac{(a + b)^{2}}{4} \\Rightarrow |AB| \\geqslant \\frac{a + b}{2}. Therefore, \\frac{|MN|}{|AB|} \\leqslant \\frac{\\frac{1}{2}(a + b)}{\\frac{a + b}{2}} = 1. Hence, the maximum value of \\frac{|MN|}{|AB|} is 1." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has an eccentricity of $2$. A line passing through the right focus $F$ with slope $k$ intersects the right branch of the hyperbola $C$ at points $A$ and $B$. If $\\overrightarrow{A F}=2 \\overrightarrow{F B}$, then $k=?$", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(C) = 2;F: Point;RightFocus(C) = F;G: Line;PointOnCurve(F, G);k: Number;Slope(G) = k;A: Point;B: Point;Intersection(G, RightPart(C)) = {A, B};VectorOf(A, F) = 2*VectorOf(F, B)", "query_expressions": "k", "answer_expressions": "pm*sqrt(35)", "fact_spans": "[[[2, 63], [90, 96]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 71]], [[76, 79]], [[2, 79]], [[87, 89]], [[72, 89]], [[83, 86], [159, 162]], [[80, 89]], [[101, 104]], [[105, 108]], [[87, 110]], [[112, 157]]]", "query_spans": "[[[159, 164]]]", "process": "Let l be the right directrix of the ellipse, draw AA_{1}, BB_{1} perpendicular to l from points A and B respectively, with A_{1}, B_{1} as the feet of the perpendiculars. Draw BE \\perp AA_{1} from point B meeting at E. According to the second definition of the hyperbola, we obtain |AA_{1}| = \\frac{|AF|}{e}, |BB_{1}| = \\frac{|BF|}{e}. Since \\overrightarrow{AF} = 2\\overrightarrow{FB}, then |AA_{1}| = 2|BB_{1}| = |BF|. Thus, \\cos\\angle BAE = \\frac{|AE|}{|AB|} = \\frac{|AA_{1}| - |BB_{1}|}{3|BF|} = \\frac{|BF|}{3|BF|} = \\frac{1}{6}. Therefore, \\sin\\angle BAE = \\frac{\\sqrt{35}}{6}, hence \\tan\\angle BAE = \\sqrt{35}, so k = \\pm\\sqrt{35}." }, { "text": "Let $F_{1}$, $F_{2}$ be the two foci of the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1$, and let $P$ be a point on the ellipse such that $|PF_{1}|-|PF_{2}|=1$. Then $\\cos \\angle F_{1} PF_{2}$=?", "fact_expressions": "F1: Point;F2: Point;G: Ellipse;Expression(G) = (x^2/3 + y^2/4 = 1);Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)) = 1", "query_expressions": "Cos(AngleOf(F1, P, F2))", "answer_expressions": "3/5", "fact_spans": "[[[1, 8]], [[9, 16]], [[17, 54], [64, 66]], [[17, 54]], [[1, 59]], [[60, 63]], [[60, 69]], [[71, 93]]]", "query_spans": "[[[96, 124]]]", "process": "" }, { "text": "Given the equation of the hyperbola is $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, points $F_{1}$ and $F_{2}$ are its left and right foci, respectively, point $A$ is a point on the circle $x^{2}+(y-6)^{2}=4$, and point $M$ lies on the right branch of the hyperbola. Then the range of values for $|M F_{1}|+|M A|$ is?", "fact_expressions": "G: Hyperbola;H: Circle;M: Point;F1: Point;F2: Point;A: Point;Expression(H) = (x^2 + (y - 6)^2 = 4);Expression(G) = (x^2/9 - y^2/16 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(A, H);PointOnCurve(M, RightPart(G))", "query_expressions": "Range(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F1)))", "answer_expressions": "[4 + sqrt(61), +oo)", "fact_spans": "[[[2, 5], [63, 64], [103, 106]], [[73, 93]], [[98, 102]], [[46, 54]], [[55, 62]], [[69, 72]], [[73, 93]], [[2, 45]], [[46, 68]], [[46, 68]], [[69, 97]], [[98, 110]]]", "query_spans": "[[[112, 136]]]", "process": "" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1(a>0)$ has its left focus at $(-2,0)$, what is the value of $a$?", "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2/3 + x^2/a^2 = 1);Coordinate(LeftFocus(G))=(-2,0)", "query_expressions": "a", "answer_expressions": "1", "fact_spans": "[[[2, 49]], [[64, 67]], [[5, 49]], [[2, 49]], [2, 59]]", "query_spans": "[[[64, 71]]]", "process": "From the given condition, c=2, that is, c^{2}=4. Since b^{2}=3, therefore c^{2}=b^{2}+a^{2}, then a^{2}=1. Also, a>0, therefore a=1" }, { "text": "The equation of the hyperbola that shares a common focus with the parabola $y^{2}=-8 \\sqrt{3} x$ and has an asymptote given by $x+\\sqrt{3} y=0$ is?", "fact_expressions": "G: Hyperbola;H: Parabola;Expression(H) = (y^2 = -8*sqrt(3)*x);Focus(G) = Focus(H);Expression(OneOf(Asymptote(G))) = (x+sqrt(3)*y=0)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/9 - y^2/3 = 1", "fact_spans": "[[[57, 60]], [[1, 25]], [[1, 25]], [[0, 60]], [[32, 60]]]", "query_spans": "[[[57, 65]]]", "process": "" }, { "text": "Let the parabola $y^{2}=4x$ have a chord $AB$ with midpoint $P\\left(\\frac{3}{2}, 1\\right)$. Then the slope of the line containing this chord is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);P: Point;Coordinate(P) = (3/2, 1);MidPoint(LineSegmentOf(A, B)) = P", "query_expressions": "Slope(OverlappingLine(LineSegmentOf(A, B)))", "answer_expressions": "2", "fact_spans": "[[[1, 15]], [[1, 15]], [[19, 23]], [[19, 23]], [[1, 23]], [[24, 44]], [[24, 44]], [[19, 47]]]", "query_spans": "[[[1, 60]]]", "process": "" }, { "text": "Given point $P(0 , 1)$ and the parabola $y=x^{2}+2$, where $Q$ is a moving point on the parabola, find the minimum value of $|PQ|$.", "fact_expressions": "P: Point;Coordinate(P) = (0, 1);G: Parabola;Expression(G) = (y = x^2 + 2);Q: Point;PointOnCurve(Q, G)", "query_expressions": "Min(Abs(LineSegmentOf(P, Q)))", "answer_expressions": "1", "fact_spans": "[[[2, 13]], [[2, 13]], [[14, 28], [34, 37]], [[14, 28]], [[30, 33]], [[30, 41]]]", "query_spans": "[[[43, 55]]]", "process": "" }, { "text": "Let a point $P$ on the parabola ${y}^{2}=4 x$ be at a distance of $5$ from the line $x+2=0$. Then, what is the distance from point $P$ to the focus $F$ of the parabola?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G) = True;H: Line;Expression(H) = (x + 2 = 0);Distance(P, H) = 5;F: Point;Focus(G) = F", "query_expressions": "Distance(P,F)", "answer_expressions": "4", "fact_spans": "[[[1, 17], [47, 50]], [[1, 17]], [[20, 23], [42, 46]], [[1, 23]], [[24, 33]], [[24, 33]], [[20, 40]], [[52, 55]], [[47, 55]]]", "query_spans": "[[[42, 60]]]", "process": "" }, { "text": "The coordinates of the focus of the parabola $x^{2}=-4 y$ are?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = -4*y)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0, -1)", "fact_spans": "[[[0, 15]], [[0, 15]]]", "query_spans": "[[[0, 22]]]", "process": "Solution: From the parabola equation $x^{2}=-4y$, we know that the focus of the parabola lies on the $y$-axis. From $2p=-4$, we obtain $\\frac{p}{2}=-1$, so the coordinates of the focus are $(0,-1)$." }, { "text": "If real numbers $a$, $b$, $c$, $d$ satisfy $\\frac{3 \\sqrt{16-a^{2}}}{4 b}=\\frac{9-c}{d}=1$, then the minimum value of $\\sqrt{(a-c)^{2}+(b-d)^{2}}$ is?", "fact_expressions": "a: Real;b: Real;c: Real;d: Real;3*sqrt(16 - a^2)/4*b = (9 - c)/d;(9 - c)/d = 1", "query_expressions": "Min(sqrt((a - c)^2 + (b - d)^2))", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[1, 7]], [[9, 12]], [[14, 18]], [[21, 24]], [[26, 73]], [[26, 73]]]", "query_spans": "[[[75, 109]]]", "process": "" }, { "text": "$A$, $B$ are two points on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, $F_{1}$, $F_{2}$ are its left and right foci, respectively, and satisfy $\\overrightarrow{A F_{1}}=2 \\overrightarrow{F_{1} B}$. When $\\angle F_{1} A F_{2}=\\frac{\\pi}{3}$, what is the eccentricity of the ellipse?", "fact_expressions": "A: Point;B: Point;G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;PointOnCurve(A, G) = True;PointOnCurve(B, G) = True;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;VectorOf(A, F1) = 2*VectorOf(F1, B);AngleOf(F1, A, F2) = pi/3", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(21)/9", "fact_spans": "[[[0, 3]], [[4, 7]], [[8, 60], [81, 82], [183, 185]], [[8, 60]], [[10, 60]], [[10, 60]], [[10, 60]], [[10, 60]], [[0, 64]], [[0, 64]], [[65, 72]], [[73, 80]], [[65, 86]], [[65, 86]], [[90, 143]], [[145, 181]]]", "query_spans": "[[[183, 191]]]", "process": "Solution: Let |\\overrightarrow{F_{1}B}|=m (m\\neq0), then |\\overrightarrow{AF}|=2m, |F_{1}F_{2}|=2c, \\angle F_{1}AF_{2}=\\frac{\\pi}{3}. Thus, by the definition of the ellipse, we obtain |AF_{2}|=2a-2m, |BF_{2}|=2a-m. In \\triangle F_{1}AF_{2}, applying the law of cosines gives |F_{1}F_{2}|^{2}=|AF_{1}|^{2}+|AF_{2}|^{2}-2|AF_{1}||AF_{2}|\\cos\\angle F_{1}AF_{2}, that is, 4c^{2}=4m^{2}+(2a-2m)^{2}-2\\cdot2m\\cdot(2a-2m)\\cos\\frac{\\pi}{3}. Simplifying yields c^{2}=a^{2}+3m^{2}-3am. In \\triangle ABF_{2}, applying the law of cosines gives |BF_{2}|^{2}=|AB|^{2}+|AF_{2}|^{2}-2|AB||AF_{2}|\\cos\\angle BAF_{2}, that is, (2a-m)^{2}=9m^{2}+(2a-2m)^{2}-2\\cdot3m\\cdot(2a-2m)\\cos\\frac{\\pi}{3}. Simplifying yields 9m^{2}-5am=0. Since m\\neq0, we have m=\\frac{5}{9}a. Therefore, c^{2}=a^{2}+3\\times\\frac{25}{81}a^{2}-3a\\cdot\\frac{5}{9}a, resulting in c^{2}=\\frac{7}{27}a^{2}. Hence, \\frac{c}{a}=\\frac{\\sqrt{21}}{9}." }, { "text": "Let $F$ be the focus of the parabola $C$: $y^{2}=3 x$, and let a line passing through $F$ with an inclination angle of $30^{\\circ}$ intersect $C$ at points $A$ and $B$. Then $|A B|$=?", "fact_expressions": "C: Parabola;G: Line;A: Point;B: Point;F: Point;Expression(C) = (y^2 = 3*x);Focus(C) = F;PointOnCurve(F, G);Inclination(G) = ApplyUnit(30, degree);Intersection(G, C) = {A, B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "12", "fact_spans": "[[[5, 24], [53, 56]], [[50, 52]], [[57, 60]], [[61, 64]], [[1, 4], [29, 32]], [[5, 24]], [[1, 27]], [[28, 52]], [[33, 52]], [[50, 66]]]", "query_spans": "[[[68, 77]]]", "process": "From y = 3x, the focus F is at (\\frac{3}{4}, 0), so assume the equation of line AB is y = \\frac{\\sqrt{3}}{3}(x - \\frac{3}{4}). Solving the parabola and line equations simultaneously and eliminating variables yields: 16x^{2} - 168x + 9 = 0. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), then x_{1} + x_{2} = \\frac{21}{2}. According to the definition of a parabola, |AB| = x_{1} + x_{2} + p = \\frac{21}{2} + \\frac{3}{2} = 12. Therefore, fill in: 12." }, { "text": "It is known that line $l$ passes through the focus of the parabola $C$: $y^{2}=4x$, and intersects the parabola $C$ at points $A$ and $B$. If $|AB|=8$, then the equation of line $l$ is?", "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;Expression(C) = (y^2 = 4*x);PointOnCurve(Focus(C),l);Intersection(l,C)={A,B};Abs(LineSegmentOf(A, B)) = 8", "query_expressions": "Expression(l)", "answer_expressions": "{x+y-1=0,x-y-1=0}", "fact_spans": "[[[2, 7], [63, 68]], [[8, 27], [33, 39]], [[41, 44]], [[45, 48]], [[8, 27]], [[2, 30]], [[2, 50]], [[52, 61]]]", "query_spans": "[[[63, 73]]]", "process": "It is easy to see that the focus coordinates of the parabola $ C: y^{2} = 4x $ are $ (1, 0) $. If $ l $ is perpendicular to the $ x $-axis, then $ |AB| = 4 $, which does not meet the requirement. Thus, we can assume the equation of the desired line $ l $ is $ y = k(x - 1) $. From \n$$\n\\begin{cases}\ny = k(x - 1) \\\\\ny^2 = 4x\n\\end{cases},\n$$\nwe obtain $ k^{2}x^{2} - (2k^{2} + 4)x + k^{2} = 0 $. By the relationship between roots and coefficients, we get $ x_{1} + x_{2} = \\frac{2k^{2} + 4}{k^{2}} $. Since $ AB $ passes through the focus, we have $ |AB| = x_{1} + x_{2} + p = \\frac{2k^{2} + 4}{k^{2}} + 2 = 8 $, so $ \\frac{2k^{2} + 4}{k^{2}} = 6 $. Solving this gives $ k = \\pm 1 $. Therefore, the equations of the required line $ l $ are $ x + y - 1 = 0 $ or $ x - y - 1 = 0 $. \n\n[Note] This problem mainly examines the definition and geometric properties of a parabola. For the parabola $ y^{2} = 2px $ ($ p > 0 $), if the chord $ AB $ (a chord passing through the focus) has endpoints $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then the following conclusions hold: \n(1) $ |AB| = x_{1} + x_{2} + p $; \n(2) $ y_{1}y_{2} = -p^{2} $, $ x_{1}x_{2} = \\frac{p^{2}}{4} $." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the hyperbola $C$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, point $A(x_{0}, 4)$ $(x_{0}>0)$ lies on hyperbola $C$, and the area of $\\Delta F_{1} A F_{2}$ is $20$, find the eccentricity $e$ of hyperbola $C$.", "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/9 - y^2/b^2 = 1);b: Number;b>0;F1: Point;F2: Point;Focus(C) = {F1, F2};A: Point;x0: Number;Coordinate(A) = (x0, 4);x0>0;PointOnCurve(A, C);Area(TriangleOf(F1, A, F2)) = 20;e: Number;Eccentricity(C) = e", "query_expressions": "e", "answer_expressions": "5/3", "fact_spans": "[[[18, 70], [100, 106], [141, 147]], [[18, 70]], [[26, 70]], [[26, 70]], [[2, 9]], [[10, 17]], [[2, 75]], [[76, 99]], [[77, 99]], [[76, 99]], [[77, 99]], [[76, 107]], [[109, 139]], [[151, 154]], [[141, 154]]]", "query_spans": "[[[151, 156]]]", "process": "Given that the area of triangle $AF_{1}AF_{2}$ is 20, we obtain $b^{2}=16$, then use the eccentricity formula $e=\\sqrt{1+\\frac{b^{2}}{a^{2}}}$ to find the eccentricity value. Since the area of triangle $4F_{1}AF_{2}$ is 20, we have $\\frac{1}{2}\\cdot|F_{1}F_{2}|\\cdot4=\\frac{1}{2}\\cdot(2\\sqrt{9+b^{2}})\\cdot4=20 \\Rightarrow b^{2}=16$." }, { "text": "Given point $P$ is a moving point on the parabola $y^{2}=4 x$. From point $P$, draw a tangent line to the circle $x^{2}+(y-3)^{2}=1$, with the point of tangency being $A$. Then, the minimum value of $P A$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G);H: Circle;Expression(H) = (x^2 + (y - 3)^2 = 1);L: Line;TangentOfPoint(P, H) = L;A: Point;TangentPoint(L, H) = A", "query_expressions": "Min(LineSegmentOf(P, A))", "answer_expressions": "1", "fact_spans": "[[[7, 21]], [[7, 21]], [[2, 6], [27, 31]], [[2, 25]], [[32, 52]], [[32, 52]], [], [[26, 55]], [[59, 62]], [[26, 62]]]", "query_spans": "[[[64, 75]]]", "process": "The circle $ x^{2}+(y-3)^{2}=1 $ has center $ C(0,3) $ and radius $ r=1 $. From the property of tangents, we have $ |PA| = \\sqrt{|PC|^{2}-r^{2}} = \\sqrt{|PC|^{2}-1} $. To minimize $ |PA| $, $ |PC| $ must be minimized. Let $ P\\left(\\frac{y_{0}}{4}, y_{0}\\right) $, then \n$$\n|PC| = \\sqrt{\\left(\\frac{y_{0}}{4}\\right)^{2} + (y_{0}-3)^{2}} = \\sqrt{\\frac{y_{0}^{4}}{16} + y_{0}^{2} - 6y_{0} + 9}.\n$$\nLet $ f(x) = \\frac{x^{4}}{16} + x^{2} - 6x + 9 $, then \n$$\nf'(x) = \\frac{x^{3}}{4} + 2x - 6.\n$$\nIt is clear that $ f'(x) $ is monotonically increasing, and $ f'(2) = 0 $. Therefore, $ f(x) $ is monotonically decreasing on $ (-\\infty, 2) $ and monotonically increasing on $ (2, +\\infty) $. Thus, $ f(x) \\geqslant f(2) = 2 $, so the minimum value of $ \\frac{y_{0}^{4}}{16} + y_{0}^{2} - 6y_{0} + 9 $ is 2, i.e., the minimum value of $ |PC| $ is $ \\sqrt{2} $. Hence, \n$$\n|PA|_{\\min} = \\sqrt{2 - 1} = 1.\n$$" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$, the right focus is $F$, and $A$ is a point on the line $x=2$. The line segment $AF$ intersects $C$ at point $B$. If $\\overrightarrow{F A}=3 \\overrightarrow{F B}$, then $|\\overrightarrow{A F}|$=?", "fact_expressions": "C: Ellipse;G: Line;A: Point;F: Point;B: Point;Expression(C) = (x^2/2 + y^2 = 1);Expression(G) = (x = 2);RightFocus(C)=F;PointOnCurve(A, G);Intersection(LineSegmentOf(A,F), C) = B;VectorOf(F, A) = 3*VectorOf(F, B)", "query_expressions": "Abs(VectorOf(A, F))", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 34], [66, 69]], [[47, 54]], [[43, 46]], [[39, 42]], [[70, 74]], [[2, 34]], [[47, 54]], [[2, 42]], [[43, 57]], [[58, 74]], [[76, 121]]]", "query_spans": "[[[123, 149]]]", "process": "" }, { "text": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has an asymptote with an inclination angle of $\\frac{\\pi}{3}$, and eccentricity $e$. Then the minimum value of $\\frac{a^{2}+e}{b}$ is?", "fact_expressions": "G: Hyperbola;a: Number;b: Number;e:Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Inclination(OneOf(Asymptote(G)))=pi/3;Eccentricity(G)=e", "query_expressions": "Min((a^2 + e)/b)", "answer_expressions": "(2/3)*sqrt(6)", "fact_spans": "[[[0, 56]], [[3, 56]], [[3, 56]], [[86, 89]], [[3, 56]], [[3, 56]], [[0, 56]], [[0, 81]], [[0, 89]]]", "query_spans": "[[[91, 116]]]", "process": "Test analysis: Since the inclination angle of one asymptote is $\\frac{\\pi}{3}$, we have $\\frac{b}{a}=\\sqrt{3}$, therefore $b=\\sqrt{3}a$, thus $c=2a$, so $e=2$. Hence, $\\frac{a_{2}+e}{b}=\\frac{\\frac{b^{2}}{3}+2}{b}=\\frac{b}{3}+\\frac{2}{b}\\geqslant2\\sqrt{\\frac{b}{3}\\cdot\\frac{2}{b}}=\\frac{2\\sqrt{6}}{3}$" }, { "text": "The equation of the ellipse passing through the point $(-3, 2)$ and having the same foci as $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ is?", "fact_expressions": "G: Ellipse;C:Curve;H: Point;Coordinate(H) = (-3, 2);PointOnCurve(H, G);Expression(C)=(x^2/9 + y^2/4 = 1);Focus(G)=Focus(C)", "query_expressions": "Expression(G)", "answer_expressions": "Preserve all mathematical expressions, LaTeX formulas, symbols, and equations EXACTLY. \nDo not add explanations, do not rewrite formulas, do not omit any content. \n\nText:", "fact_spans": "[[[55, 57]], [[14, 49]], [[1, 12]], [[1, 12]], [[0, 57]], [[14, 49]], [[13, 57]]]", "query_spans": "[[[55, 61]]]", "process": "" }, { "text": "If the line $y=kx+2$ always has two common points with the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{b^{2}}=1(b>0)$ whose focus lies on the $x$-axis, what is the range of the real number $b$?", "fact_expressions": "G: Ellipse;b: Real;H: Line;k: Number;b>0;Expression(G) = (x^2/16 + y^2/b^2 = 1);Expression(H) = (y = k*x + 2);PointOnCurve(Focus(G), xAxis);NumIntersection(H, G) = 2", "query_expressions": "Range(b)", "answer_expressions": "(2, 4)", "fact_spans": "[[[21, 68]], [[77, 82]], [[1, 12]], [[3, 12]], [[23, 68]], [[21, 68]], [[1, 12]], [[13, 68]], [[1, 75]]]", "query_spans": "[[[77, 86]]]", "process": "The line y = kx + 2 always passes through the fixed point (0,2). To ensure that the line intersects the ellipse at two points, the fixed point must lie inside the ellipse: \\frac{0}{16}+\\frac{4}{b^{2}}<1, \\therefore b>2. Also, since the foci of the ellipse lie on the x-axis, \\therefore b^{2}<16 \\Rightarrow b<4, \\therefore b\\in(2,4)" }, { "text": "Through the focus of the parabola $x^{2}=4 y$, draw a line $l$ intersecting the parabola at points $A(x_{1} , y_{1})$ and $B(x_{2} , y_{2})$. Then $x_{1} x_{2}$?", "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;Expression(G) = (x^2 = 4*y);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);PointOnCurve(Focus(G), l);Intersection(l, G) = {A,B};x1:Number;y1:Number;x2:Number;y2:Number", "query_expressions": "x1*x2", "answer_expressions": "-4", "fact_spans": "[[[19, 24]], [[1, 15], [25, 28]], [[29, 47]], [[50, 68]], [[1, 15]], [[29, 47]], [[50, 68]], [[0, 24]], [[19, 68]], [[29, 47]], [[29, 47]], [[50, 68]], [[50, 68]]]", "query_spans": "[[[70, 84]]]", "process": "" }, { "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{m}=1$ are given by $y=\\pm \\frac{\\sqrt{3}}{2} x$, then what are its foci coordinates?", "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/4 - y^2/m = 1);Expression(Asymptote(G)) = (y = pm*(x*(sqrt(3)/2)))", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(pm*sqrt(7),0)", "fact_spans": "[[[1, 39], [75, 76]], [[4, 39]], [[1, 39]], [[1, 73]]]", "query_spans": "[[[75, 82]]]", "process": "" }, { "text": "The equation of the directrix of the parabola $y=2 x^{2}$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y = 2*x^2)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "y=-1/8", "fact_spans": "[[[0, 14]], [[0, 14]]]", "query_spans": "[[[0, 21]]]", "process": "Problem Analysis: From y-2x^{2}, we get x^{2}=\\frac{1}{2}y, so the directrix equation is y=-\\frac{1}{8}" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the right branch of the hyperbola, and $|P F_{1}|=6|P F_{2}|$. What is the maximum value of the eccentricity $e$ of this hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;e: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F1)) = 6*Abs(LineSegmentOf(P, F2));Eccentricity(G) = e", "query_expressions": "Max(e)", "answer_expressions": "7/5", "fact_spans": "[[[2, 58], [88, 91], [122, 125]], [[5, 58]], [[5, 58]], [[83, 87]], [[67, 74]], [[75, 82]], [[129, 132]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 82]], [[2, 82]], [[83, 95]], [[97, 119]], [[122, 132]]]", "query_spans": "[[[129, 138]]]", "process": "Using the law of cosines, obtain the expression of 2c in terms of |PF_{2}|, express 2a also in terms of |PF_{2}|, thereby derive the expression for the eccentricity e, and further find the maximum value of the eccentricity. Since |PF_{1}| = 6|PF_{2}|, in triangle F_{1}F_{2}P, by the law of cosines we have -2|PF_{1}|\\cdot|PF_{2}|\\cdot\\cos\\angle F_{1}PF_{2}, so that the hyperbola's eccentricity is e = \\frac{2c}{2a} = \\frac{\\sqrt[2\\cos\\angle F}{37|PF_{2}}}|PF_{2}| = 5|PF_{2}|.\\frac{1PF_{2}}{\\textcircled{1}} When P is at the right vertex of the hyperbola, \\angle F_{1}PF_{2} = \\pi, and at this time \\textcircled{1} attains its maximum value of \\underline{\\sqrt{37+12}} = \\frac{7}{5}, which means the maximum eccentricity of the hyperbola is \\frac{7}{6}." }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, the right focus is $F$, and the intersection point of the right directrix with the $x$-axis is $D$. There exists a point $P$ on the ellipse such that $\\angle P F D=60^{\\circ}$, $\\sin \\angle PDF=\\frac{3}{5}$. Then the eccentricity of this ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F: Point;D: Point;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;Intersection(RightDirectrix(G), xAxis) = D;PointOnCurve(P,G);AngleOf(P,F,D)=ApplyUnit(60,degree);Sin(AngleOf(P, D, F)) = 3/5", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)/2", "fact_spans": "[[[2, 47], [74, 76], [144, 146]], [[4, 47]], [[4, 47]], [[79, 82]], [[52, 55]], [[68, 71]], [[2, 47]], [[2, 55]], [[2, 71]], [[74, 82]], [[84, 109]], [[112, 141]]]", "query_spans": "[[[144, 152]]]", "process": "" }, { "text": "Given that the perimeter of $\\triangle ABC$ is $20$, and the vertices are $B(0,-3)$, $C(0,3)$, then the trajectory equation of vertex $A$ is?", "fact_expressions": "B: Point;C: Point;A: Point;Coordinate(B) = (0, -3);Coordinate(C) = (0, 3);Perimeter(TriangleOf(A, B, C)) = 20", "query_expressions": "LocusEquation(A)", "answer_expressions": "(x^2/40+y^2/49=1)&Negation(x=0)", "fact_spans": "[[[31, 40]], [[43, 51]], [[55, 58]], [[31, 40]], [[43, 51]], [[2, 27]]]", "query_spans": "[[[55, 65]]]", "process": "Since the perimeter determines |AB| + |AC| = 14 > 6, the trajectory is an ellipse; just pay attention to the position of the foci and exclude points that do not satisfy the conditions. Given that the perimeter of $\\triangle ABC$ is 20, and vertex B(0,-3), C(0,3), $\\therefore |BC| = 6$, $|AB| + |AC| = 14 > 6$, so the sum of distances from point A to two fixed points is constant. Hence, the trajectory of point A is an ellipse with foci on the y-axis. $2a = 14 \\Rightarrow a = 7$, $c = 3$, $b^{2} = a^{2} - c^{2} = 49 - 9 = 40$. Then the equation of the trajectory of vertex A is $\\frac{x^{2}}{40} + \\frac{y^{2}}{49} = 1$ $(x \\neq 0)$." }, { "text": "It is known that the focus of the parabola lies on the $x$-axis, and the line segment intercepted by the parabola from the line $y=2x-4$ has a length of $3\\sqrt{5}$. Then, what is the standard equation of the parabola?", "fact_expressions": "G: Parabola;PointOnCurve(Focus(G), xAxis) = True;H: Line;Expression(H) = (y = 2*x - 4);Length(InterceptChord(H, G)) = 3*sqrt(5)", "query_expressions": "Expression(G)", "answer_expressions": "{(y^2=4*x),(y^2=-36*x)}", "fact_spans": "[[[2, 5], [27, 30], [51, 54]], [[2, 14]], [[15, 26]], [[15, 26]], [[15, 49]]]", "query_spans": "[[[51, 61]]]", "process": "" }, { "text": "Given point $A(1,0)$, line $l$: $x=-1$, two moving circles both pass through point $A$ and are tangent to $l$, with centers $C_{1}$, $C_{2}$. If moving point $M$ satisfies $2 \\overrightarrow{C_{2} M}=\\overrightarrow{C_{2} C_{1}}+\\overrightarrow{C_{2} A}$, then the equation of the trajectory of $M$ is?", "fact_expressions": "A: Point;Coordinate(A) = (1, 0);l: Line;Expression(l) = (x=-1);D: Circle;Z: Circle;PointOnCurve(A, D);PointOnCurve(A, Z);IsTangent(D, l);IsTangent(Z, l);C1: Point;C2: Point;Center(D) = C1;Center(Z) = C2;M: Point;2*VectorOf(C2, M) = VectorOf(C2, A) + VectorOf(C2, C1)", "query_expressions": "LocusEquation(M)", "answer_expressions": "y^2=2*x-1", "fact_spans": "[[[2, 11], [32, 36]], [[2, 11]], [[12, 25], [38, 41]], [[12, 25]], [], [], [[26, 36]], [[26, 36]], [[26, 43]], [[26, 43]], [[50, 57]], [[58, 65]], [[44, 65]], [[44, 65]], [[69, 72], [158, 161]], [[74, 156]]]", "query_spans": "[[[158, 168]]]", "process": "From the definition of a parabola, the trajectory equation of the center of the moving circle is y^{2}=4x. Let C_{1}(a,b), C_{2}(m,n), M(x,y). According to 2\\overrightarrow{C_{2}M}=\\overrightarrow{C_{2}C_{1}}+\\overrightarrow{C_{2}A}, we obtain a=2x-1, b=2y. Using b^{2}=4a, the result can be found. From the definition of a parabola, the trajectory of the center of the moving circle is a parabola with focus A(1,0) and directrix l: x=-1, whose equation is y^{2}=4x. Let C_{1}(a,b), C_{2}(m,n), M(x,y). Since the moving point M satisfies 2\\overrightarrow{C_{2}M}=\\overrightarrow{C_{2}C_{1}}+\\overrightarrow{C_{2}A}, it follows that 2(x-m,y-n)=(a-m,b-n)+(1-m,-n), which implies 2x=a+1, 2y=b. Therefore, a=2x-1, b=2y. Since b^{2}=4a, we have (2y)^{2}=4(2x-1), so y^{2}=2x-1. Thus, the trajectory equation of M is y^{2}=2x-1." }, { "text": "If the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ and the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ have common foci $F_{1}$, $F_{2}$, and $P$ is an intersection point of the two curves, then what is the value of $|P F_{1}| \\cdot|P F_{2}|$?", "fact_expressions": "E: Ellipse;Expression(E) = (x^2/25 + y^2/16 = 1);G: Hyperbola;Expression(G) = (x^2/4 - y^2/5 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};Focus(E) = {F1, F2};P: Point;OneOf(Intersection(E, G)) = P", "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))", "answer_expressions": "21", "fact_spans": "[[[1, 40]], [[1, 40]], [[41, 79]], [[41, 79]], [[85, 92]], [[93, 100]], [[1, 100]], [[1, 100]], [[101, 104]], [[101, 113]]]", "query_spans": "[[[115, 145]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{10-m}+\\frac{y^{2}}{m-2}=1$, if its major axis lies on the $y$-axis and the focal distance is $4$, then what is the value of $m$?", "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/(10 - m) + y^2/(m - 2) = 1);OverlappingLine(MajorAxis(G),yAxis) = True;FocalLength(G) = 4", "query_expressions": "m", "answer_expressions": "8", "fact_spans": "[[[2, 44], [46, 47]], [[65, 68]], [[2, 44]], [[46, 55]], [[46, 63]]]", "query_spans": "[[[65, 71]]]", "process": "" }, { "text": "$F_{1}$ is a focus of the ellipse $\\frac{y^{2}}{4}+\\frac{x^{2}}{2}=1$. A line $l$ not passing through the point $F_{1}$ intersects the ellipse at points $A$ and $B$. Then, the maximum value of the perimeter of $\\triangle F_{1} A B$ is?", "fact_expressions": "l: Line;G: Ellipse;F1: Point;A: Point;B: Point;Expression(G) = (x^2/2 + y^2/4 = 1);OneOf(Focus(G)) = F1;Negation(PointOnCurve(F1, l));Intersection(l, G) = {A, B}", "query_expressions": "Max(Perimeter(TriangleOf(F1, A, B)))", "answer_expressions": "8", "fact_spans": "[[[62, 67]], [[8, 45], [68, 70]], [[0, 7], [53, 61]], [[71, 74]], [[75, 78]], [[8, 45]], [[0, 50]], [[51, 67]], [[62, 80]]]", "query_spans": "[[[82, 112]]]", "process": "From the given, we have $a^{2}=4$, so $a=2$. Let the other focus of the ellipse be $F_{2}$. The perimeter of $\\triangle F_{1}AB$ is \n$$\nL = |F_{1}A| + |F_{1}B| + |AB| = (2a - |F_{2}A|) + (2a - |F_{2}B|) + |AB| = 4a + |AB| - (|F_{2}A| + |F_{2}B|) \\leqslant 4a + |AB| - |AB| = 4a = 8,\n$$\nwith equality if and only if points $A$, $F_{2}$, and $B$ are collinear." }, { "text": "The coordinates of the foci of the hyperbola $y^{2}-\\frac{x^{2}}{2}=1$ are? The equations of the asymptotes are?", "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/2 + y^2 = 1)", "query_expressions": "Coordinate(Focus(G));Expression(Asymptote(G))", "answer_expressions": "(0,pm*sqrt(3))\ny=pm*(sqrt(2)/2)*x", "fact_spans": "[[[0, 28]], [[0, 28]]]", "query_spans": "[[[0, 35]], [[0, 42]]]", "process": "Test analysis: From the hyperbola equation, it can be seen that its foci lie on the y-axis, and $ a^{2}=1 $, $ b^{2}=2 $, $ \\therefore c=\\sqrt{a^{2}+b^{2}}=\\sqrt{3} $, so the foci are $ (0,\\sqrt{3}) $, $ (0,-\\sqrt{3}) $. The asymptote equations are $ y=\\pm\\frac{b}{a}x $, that is, $ y=\\pm\\frac{\\sqrt{2}}{2}x $." }, { "text": "The sum of the length of the real axis, the length of the imaginary axis, and the focal distance of a hyperbola is $8$. Find the range of values for the semi-focal distance (express your answer in interval notation).", "fact_expressions": "G: Hyperbola;Length(RealAxis(G)) + Length(ImageinaryAxis(G)) + FocalLength(G) = 8", "query_expressions": "Range(HalfFocalLength(G))", "answer_expressions": "[4*sqrt(2)-4,2)", "fact_spans": "[[[0, 3]], [[0, 20]]]", "query_spans": "[[[0, 42]]]", "process": "" }, { "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are given by $y=\\pm x$ and one focus is $(4 \\sqrt{2}, 0)$, then $a=?$", "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(Asymptote(G))=(y=pm*x);Coordinate(OneOf(Focus(G))) = (4*sqrt(2), 0)", "query_expressions": "a", "answer_expressions": "4", "fact_spans": "[[[1, 57]], [[4, 57]], [[98, 101]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 73]], [[1, 96]]]", "query_spans": "[[[98, 103]]]", "process": "Solution: Since the asymptotes of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0, b>0) are given by y=\\pm x, we have \\frac{b}{a}=1, that is, a=b. Since one focus of the hyperbola is (4\\sqrt{2},0), it follows that c=4\\sqrt{2}, so a^{2}+b^{2}=32. Solving gives a=b=4." }, { "text": "If the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{n}=1(m>n>0)$ and the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ have the same left and right foci $F_{1}$, $F_{2}$, and $P$ is an intersection point of the two curves, then what is the value of $P F_{1} \\cdot P F_{2}$?", "fact_expressions": "H: Ellipse;Expression(H) = (y^2/n + x^2/m = 1);m: Number;n: Number;m > n;n > 0;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(H) = F1;RightFocus(H) = F2;LeftFocus(G) = F1;RightFocus(G)= F2;P: Point;OneOf(Intersection(H, G)) = P", "query_expressions": "LineSegmentOf(P, F1)*LineSegmentOf(P, F2)", "answer_expressions": "m - a^2", "fact_spans": "[[[1, 45]], [[1, 45]], [[3, 45]], [[3, 45]], [[3, 45]], [[3, 45]], [[46, 102]], [[46, 102]], [[49, 102]], [[49, 102]], [[49, 102]], [[49, 102]], [[111, 118]], [[119, 126]], [[1, 126]], [[1, 126]], [[1, 126]], [[1, 126]], [[127, 130]], [[127, 140]]]", "query_spans": "[[[142, 169]]]", "process": "Without loss of generality, let point P lie on the right branch of the hyperbola. |PF_{1}| + |PF_{2}| = 2\\sqrt{m}, |PF_{1}| - |PF_{2}| = 2a. Therefore, PF_{1}^{2} + PF_{2}^{2} + 2PF_{1}\\cdot PF_{2} = 4m, PF_{1}^{2} - 2PF_{1}\\cdot PF_{2} + PF_{2}^{2} = 4a^{2}. Subtracting these two equations yields: 4PF_{1}\\cdot PF_{2} = 4m - 4a^{2}, thus PF_{1}\\cdot PF_{2} = m - a^{2}. Hence, the answer is: m - a^{2}. This problem mainly examines comprehensive problems involving conic sections. The key to solving this problem lies in using the definitions to simplify, given that the ellipse and hyperbola share the same foci F_{1}, F_{2}." }, { "text": "The circle with the left focus $F(-c, 0)$ of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ as its center and radius $c$ intersects the left directrix of the ellipse at two distinct points. Then the range of values for the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;LeftFocus(G) = F;Coordinate(F) = (-c, 0);c: Number;H: Circle;Center(H) = F;Radius(H) = c;NumIntersection(H,LeftDirectrix(G)) = 2", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(sqrt(2)/2,1)", "fact_spans": "[[[1, 53], [80, 82], [97, 99]], [[1, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[57, 67]], [[1, 67]], [[57, 67]], [[71, 74]], [[78, 79]], [[57, 79]], [[71, 79]], [[78, 94]]]", "query_spans": "[[[97, 110]]]", "process": "" }, { "text": "The asymptotes of the hyperbola pass through the point $(1,2)$, and the hyperbola passes through the point $(2 \\sqrt{2}, 4)$. What is the eccentricity of the hyperbola?", "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (1, 2);PointOnCurve(H, Asymptote(G));I: Point;Coordinate(I) = (2*sqrt(2), 4);PointOnCurve(I, G)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[0, 3], [18, 21], [43, 46]], [[9, 17]], [[9, 17]], [[0, 17]], [[23, 41]], [[23, 41]], [[18, 41]]]", "query_spans": "[[[43, 52]]]", "process": "According to the problem, let the hyperbola equation be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0), then its asymptotes are given by y=\\pm\\frac{b}{a}x'\\begin{cases}2=\\frac{b}{a}\\\\\\frac{8}{a2}-\\frac{16}{12}=1\\end{cases}, solving yields a=2, b=4, then c=2\\sqrt{5}, so the eccentricity of the hyperbola is e=\\frac{2\\sqrt{5}}{2}=\\sqrt{5}" }, { "text": "Given that $F_{1}$, $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and $A$, $B$ are the left vertex and the upper vertex of this ellipse respectively, point $P$ lies on the line segment $AB$. Then the range of values of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ is?", "fact_expressions": "A: Point;B: Point;G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);Focus(G) = {F1, F2};LeftVertex(G)=A;UpperVertex(G)=B;PointOnCurve(P, LineSegmentOf(A, B))", "query_expressions": "Range(DotProduct(VectorOf(P, F1), VectorOf(P, F2)))", "answer_expressions": "[-11/5,1]", "fact_spans": "[[[51, 54]], [[55, 58]], [[18, 45], [62, 64]], [[73, 77]], [[2, 9]], [[10, 17]], [[18, 45]], [[2, 50]], [[51, 72]], [[51, 72]], [[73, 85]]]", "query_spans": "[[[87, 151]]]", "process": "Given that $F_{1}, F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and $A$, $B$ are the left vertex and upper vertex of the ellipse respectively, we can obtain $F_{1}(-\\sqrt{3},0)$, $F_{2}(\\sqrt{3},0)$, $A(-2,0)$, $B(0,1)$. Let $P(x,y)$. Since point $P$ lies on the line segment $AB$, it follows that $y=\\frac{1}{2}x+1$, $-2\\leqslant x\\leqslant 0$. Then, $\\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=(x+\\sqrt{3},y)\\cdot(x-\\sqrt{3},y)=x^{2}+y^{2}-3=\\frac{5}{4}x^{2}+x-2=\\frac{5}{4}(x+\\frac{2}{5})^{2}-\\frac{11}{5}\\in[-\\frac{11}{5},1]$." }, { "text": "If the asymptotes of a hyperbola are given by $y=\\pm \\frac{3}{2} x$, then its eccentricity is?", "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*(3/2)*x)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(13)/2, sqrt(13)/3", "fact_spans": "[[[1, 4], [34, 35]], [[1, 32]]]", "query_spans": "[[[34, 40]]]", "process": "" }, { "text": "Given that the focus of the parabola $y^{2}=16 x$ is exactly the right focus of the hyperbola $\\frac{x^{2}}{12}-\\frac{y^{2}}{b^{2}}=1$, then the asymptotes equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;C:Parabola;b: Number;Expression(C)=(y^2 = 16*x);Expression(G) = (x^2/12 - y^2/b^2 = 1);Focus(C) = RightFocus(G)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*(sqrt(3)/3*x)", "fact_spans": "[[[23, 66], [72, 75]], [[2, 17]], [[26, 66]], [[2, 17]], [[23, 66]], [[2, 70]]]", "query_spans": "[[[72, 83]]]", "process": "From the given condition, the focus of the parabola $ y^{2} = 16x $ is at $ (4, 0) $. Therefore, the coordinates of the right focus of the hyperbola $ \\frac{x^{2}}{12} - \\frac{y^{2}}{b^{2}} = 1 $ are $ (4, 0) $, so $ \\sqrt{12 + b^{2}} = 4 $, which gives $ b = 2 $. Thus, the equation of the hyperbola is $ \\frac{x^{2}}{12} - \\frac{y^{2}}{4} = 1 $, and the asymptotes of this hyperbola are $ y = \\pm\\frac{\\sqrt{3}}{3}x $." }, { "text": "Given that the directrix $l$ of the parabola $y^{2}=8x$ is tangent to the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-y^{2}=1$, then the eccentricity $e$ of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;a: Number;G: Parabola;l: Line;e: Number;Expression(C) = (-y^2 + x^2/a^2 = 1);Expression(G) = (y^2 = 8*x);Directrix(G) = l;IsTangent(l, C);Eccentricity(C) = e", "query_expressions": "e", "answer_expressions": "sqrt(5)/2", "fact_spans": "[[[23, 60], [64, 70]], [[31, 60]], [[2, 16]], [[19, 22]], [[74, 77]], [[23, 60]], [[2, 16]], [[2, 22]], [[19, 62]], [[64, 77]]]", "query_spans": "[[[74, 79]]]", "process": "The directrix of the parabola $ y^{2}=8x $ is $ l: x=-2 $. According to the problem, the line $ l: x=-2 $ passes through the left vertex of the hyperbola $ C: \\frac{x^{2}}{a^{2}}-y^{2}=1 $, so $ a=2 $. Therefore, $ c=\\sqrt{a^{2}+b^{2}}=\\sqrt{5} $, hence $ e=\\frac{c}{a}=\\frac{\\sqrt{5}}{2} $." }, { "text": "The hyperbola that shares the same asymptotes with $x^{2}-2 y^{2}=2$ and passes through the point $M(2,-2)$ has the equation?", "fact_expressions": "G: Hyperbola;M: Point;C:Hyperbola;Expression(G) = (x^2 - 2*y^2 = 2);Coordinate(M) = (2,-2);Asymptote(C) = Asymptote(G);PointOnCurve(M,C)", "query_expressions": "Expression(C)", "answer_expressions": "y^2/2 - x^2/4 = 1", "fact_spans": "[[[1, 21]], [[31, 41]], [[42, 45]], [[1, 21]], [[31, 41]], [[0, 45]], [[30, 45]]]", "query_spans": "[[[42, 49]]]", "process": "Let the hyperbola equation be x^{2}-2y^{2}=m (m\\neq0), \\therefore 2^{2}-2(-2)^{2}=m, \\therefore m=-4, so the hyperbola equation is \\frac{y^{2}}{2}-\\frac{x^{2}}{4}=1" }, { "text": "The coordinates of the focus of the parabola $y=-2 x^{2}$ are?", "fact_expressions": "G: Parabola;Expression(G) = (y = -2*x^2)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0,-1/8)", "fact_spans": "[[[0, 15]], [[0, 15]]]", "query_spans": "[[[0, 22]]]", "process": "" }, { "text": "Let the point $M$ on the hyperbola $\\frac{y^{2}}{9}-\\frac{x^{2}}{16}=-1$ be at a distance of $25$ from the point $A(5 , 0)$. Then, what is the distance from $M$ to the point $B(-5 , 0)$?", "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/16 + y^2/9 = -1);M: Point;PointOnCurve(M, G);A: Point;Coordinate(A) = (5, 0);Distance(M, A) = 25;B: Point;Coordinate(B) = (-5, 0)", "query_expressions": "Distance(M, B)", "answer_expressions": "{17, 33}", "fact_spans": "[[[1, 41]], [[1, 41]], [[43, 47], [69, 72]], [[1, 47]], [[48, 59]], [[48, 59]], [[43, 67]], [[73, 85]], [[73, 85]]]", "query_spans": "[[[69, 90]]]", "process": "" }, { "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$, a circle with its center on $C$ passes through the origin and is tangent to the directrix of the parabola $C$. If the chord length intercepted by the circle on the line $x-\\sqrt{2} y=0$ is $2 \\sqrt{6}$, then the equation of $C$ is?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;G: Circle;PointOnCurve(Center(G), C);O: Origin;PointOnCurve(O, G);IsTangent(G, Directrix(C));H: Line;Expression(H) = (x - sqrt(2)*y = 0);Length(InterceptChord(H, G)) = 2*sqrt(6)", "query_expressions": "Expression(C)", "answer_expressions": "y^2=8*x", "fact_spans": "[[[2, 28], [43, 49], [32, 35], [96, 99]], [[2, 28]], [[10, 28]], [[10, 28]], [[37, 38], [57, 58]], [[29, 38]], [[39, 41]], [[37, 41]], [[37, 54]], [[59, 77]], [[59, 77]], [[57, 94]]]", "query_spans": "[[[96, 104]]]", "process": "From the given problem, the coordinates of the circle center are P(\\frac{p}{4},\\frac{p}{\\sqrt{2}}), so the distance from P to x-\\sqrt{2}y=0 is d=\\frac{|\\frac{p}{4}-p|}{\\sqrt{3}}=\\frac{\\sqrt{3}p}{4}. Also, from l=2\\sqrt{r^2-d^{2}}, set up the equation and solve for p to obtain the answer. As shown in the figure, let the center of the circle be P. Since |PO|=|PE|=|PF|, it follows that P(\\frac{p}{4},\\frac{p}{\\sqrt{2}}) and r=\\frac{3p}{4}. Then the distance from P to x-\\sqrt{2}y=0 is d=\\frac{|\\frac{p}{4}-\\sqrt{2}\\cdot\\frac{p}{\\sqrt{2}}|}{\\sqrt{1+(\\sqrt{2})^2}}=\\frac{|\\frac{p}{4}-p|}{\\sqrt{3}}=\\frac{\\sqrt{3}p}{4}. Again, from l=2\\sqrt{r^2-d^{2}}, we have 2\\sqrt{6}=2\\sqrt{(\\frac{3p}{4})^{2}-(\\frac{\\sqrt{3}p}{4})^{2}} \\Rightarrow p=4. Therefore, the standard equation of the parabola is y^{2}=8x." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, one of its asymptotes is perpendicular to the line $l$: $x-2 y+2020=0$. Then the eccentricity $e$ of the hyperbola $C$ is?", "fact_expressions": "C:Hyperbola;l:Line;a,b:Number;e:Number;a>0;b>0;Expression(C)=(x^2/a^2-y^2/b^2=1);Expression(l)=(x-2*y+2020=0);IsPerpendicular(OneOf(Asymptote(C)),l)=True;Eccentricity(C)=e", "query_expressions": "e", "answer_expressions": "sqrt(5)", "fact_spans": "[[[2, 63], [95, 101]], [[70, 91]], [[10, 63]], [[105, 108]], [[10, 63]], [[10, 63]], [[2, 63]], [[70, 91]], [[2, 93]], [[95, 108]]]", "query_spans": "[[[95, 110]]]", "process": "" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, respectively, $P$ is an intersection point of the circle with diameter $F_{1} F_{2}$ and the ellipse, and $\\angle PF_{1} F_{2}=2 \\angle PF_{2} F_{1}$, then what is the eccentricity of this ellipse?", "fact_expressions": "F1: Point;F2: Point;G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;LeftFocus(G) = F1;RightFocus(G) = F2;OneOf(Intersection(H, G)) = P;P: Point;IsDiameter(LineSegmentOf(F1, F2), H);H: Circle;AngleOf(P, F1, F2) = 2*AngleOf(P, F2, F1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)-1", "fact_spans": "[[[2, 9]], [[10, 17]], [[20, 65], [97, 99], [153, 155]], [[20, 65]], [[22, 65]], [[22, 65]], [[2, 71]], [[2, 71]], [[72, 104]], [[72, 75]], [[76, 95]], [[94, 95]], [[106, 149]]]", "query_spans": "[[[153, 161]]]", "process": "Since P is an intersection point of the circle with diameter F₁F₂ and the ellipse, ∠F₁PF₂ = π/2. Because ∠PF₁F₂ = 2∠PF₂F₁, it follows that ∠PF₂F₁ = π/6. Therefore, |PF₁| = |F₁F₂| sin∠PF₂F₁ = 2c sin(π/6) = c, |PF₂| = |F₁F₂| cos∠PF₂F₁ = 2c × cos(π/6) = √3c. Hence, a = (|PF₁| + |PF₂|)/(e = c/a) = (c + √3c)/2 = c(1 + √3)/2" }, { "text": "Through the focus $F$ of the parabola $y=\\frac{x^{2}}{8}$, draw a tangent to the circle: $(x+3)^{2}+(y+3)^{2}=16$, with the point of tangency being $P$. Find $|F P|=?$", "fact_expressions": "G: Parabola;H: Circle;F: Point;P: Point;Expression(G) = (y = x^2/8);Expression(H) = ((x + 3)^2 + (y + 3)^2 = 16);Focus(G) = F;L: Line;TangentOfPoint(F, H) = L;TangentPoint(H, L) = P", "query_expressions": "Abs(LineSegmentOf(F, P))", "answer_expressions": "3*sqrt(2)", "fact_spans": "[[[1, 23]], [[30, 56]], [[25, 28]], [[63, 66]], [[1, 23]], [[30, 56]], [[1, 28]], [], [[0, 59]], [[0, 67]]]", "query_spans": "[[[68, 77]]]", "process": "From the parabola $ y = \\frac{x^2}{8} $, we get $ x^2 = 8y $, so the focus is $ F(0,2) $. Let the center of the circle $ (x+3)^2 + (y+3)^2 = 16 $ be $ (-3,-3) $, and the radius be $ R = 4 $. Then $ |FQ| = \\sqrt{(-3)^2 + (2+3)^2} = \\sqrt{34} $, $ \\sqrt{(\\sqrt{34})^2 - 16} = 3\\sqrt{2} $." }, { "text": "Let $P$ be a point on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and let $F_{1}$, $F_{2}$ be the two foci of the ellipse. Then what is the maximum value of $|P F_{1}| | P F_{2}|$? What is the minimum value?", "fact_expressions": "P: Point;PointOnCurve(P, G);G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2}", "query_expressions": "Max(Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)));Min(Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)))", "answer_expressions": "4\n1", "fact_spans": "[[[1, 4]], [[1, 36]], [[5, 32], [53, 55]], [[5, 32]], [[37, 44]], [[45, 52]], [[37, 60]]]", "query_spans": "[[[62, 90]], [[62, 95]]]", "process": "" }, { "text": "It is known that a moving point $P(x, y)$ lies on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$. If the coordinates of point $A$ are $(3,0)$, $|\\overrightarrow{A M}|=1$, and $\\overrightarrow{P M} \\cdot \\overrightarrow{A M}=0$, then the minimum value of $|\\overrightarrow{P M}|$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);x1: Number;y1: Number;P: Point;Coordinate(P) = (x1, y1);PointOnCurve(P, G);A: Point;Coordinate(A) = (3, 0);M: Point;Abs(VectorOf(A, M)) = 1;DotProduct(VectorOf(P, M), VectorOf(A, M)) = 0", "query_expressions": "Min(Abs(VectorOf(P, M)))", "answer_expressions": "sqrt(3)", "fact_spans": "[[[14, 53]], [[14, 53]], [[4, 13]], [[4, 13]], [[4, 13]], [[4, 13]], [[4, 54]], [[56, 60]], [[56, 70]], [[72, 98]], [[72, 98]], [[100, 151]]]", "query_spans": "[[[153, 183]]]", "process": "\\because\\overrightarrow{PM}\\cdot\\overrightarrow{AM}=0,\\therefore\\overrightarrow{PM}\\bot\\overrightarrow{AM},|\\overrightarrow{PM}|=|A|1,\\therefore|\\overrightarrow{AN}|\\frac{AM}{2}=1 then |\\overrightarrow{PM}|^{2}=|\\overrightarrow{AP}|^{2}-|\\overrightarrow{AM}|^{2}=|\\overrightarrow{AP}|^{2}-1\\because|\\overrightarrow{AM}|=1,\\therefore the trajectory of point M is a circle with point A as the center and radius 1, where point A is simultaneously the right focus of the ellipse. |\\overrightarrow{PM}|^{2}=|\\overrightarrow{AP}|^{2}-1' the smaller |\\overrightarrow{AP}|, the smaller |\\overrightarrow{PM}|. From the graph, when point P is the right vertex of the ellipse, |\\overrightarrow{AP}| takes the minimum value a-c=5-3=2,\\therefore the minimum value of |\\overrightarrow{PM}| is \\sqrt{4-1}=\\sqrt{3}" }, { "text": "The equation of the hyperbola that shares foci with the ellipse $\\frac{x^{2}}{64}+\\frac{y^{2}}{16}=1$ and has asymptotes $x \\pm \\sqrt{3} y=0$ is?", "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/64 + y^2/16 = 1);Focus(G)=Focus(H);Expression(Asymptote(G))=(x+pm*sqrt(3)*y=0)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/36 - y^2/12 = 1", "fact_spans": "[[[70, 73]], [[1, 40]], [[1, 40]], [[0, 73]], [[44, 73]]]", "query_spans": "[[[70, 77]]]", "process": "" }, { "text": "Point $P(x_{0}, 1)$ is a point on the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{4}=1 (a>2)$. The area of the triangle with vertices at point $P$ and the foci $F_{1}$, $F_{2}$ is $1$. Then, what is the length of the major axis of ellipse $C$?", "fact_expressions": "C: Ellipse;a: Number;P: Point;F1:Point;F2:Point;x0:Number;a>2;Expression(C) = (y^2/4 + x^2/a^2 = 1);Coordinate(P) = (x0, 1);PointOnCurve(P, C);Focus(C)={F1,F2};Area(TriangleOf(P,F1,F2))=1", "query_expressions": "Length(MajorAxis(C))", "answer_expressions": "2*sqrt(5)", "fact_spans": "[[[15, 68], [113, 118]], [[22, 68]], [[0, 14], [75, 78]], [[82, 89]], [[90, 97]], [[1, 14]], [[22, 68]], [[15, 68]], [[0, 14]], [[0, 72]], [[15, 97]], [[73, 111]]]", "query_spans": "[[[113, 124]]]", "process": "According to the area of \\trianglePF_{1}F_{2}, first find c, and combining with the ellipse equation, derive a, then the major axis length of the ellipse can be found. Since the area of the triangle with vertices at point P(x_{0},1) and foci F_{1}, F_{2} is 1, we have S_{\\trianglePF_{1}F_{2}}=\\frac{1}{2}|F_{1}F_{2}|\\cdot y_{p}=\\frac{1}{2}\\cdot2c\\cdot1=c=1. Also, the ellipse equation is \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{4}=1 (a>2), so a^{2}=b^{2}+c^{2}=4+1=5, then a=\\sqrt{5}. Therefore, the major axis length of ellipse C is 2a=2\\sqrt{5}." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has foci $F_{1}$, $F_{2}$. If there exists a point $P$ on the ellipse $C$ such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, and the area of $\\Delta F_{1} P F_{2}$ equals $4$, then the value of the real number $b$ is?", "fact_expressions": "C: Ellipse;a: Number;b: Real;F1: Point;P: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Focus(C)={F1,F2};PointOnCurve(P, C);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Area(TriangleOf(F1, P, F2)) = 4", "query_expressions": "b", "answer_expressions": "2", "fact_spans": "[[[2, 59], [80, 85]], [[8, 59]], [[189, 194]], [[63, 70]], [[90, 93]], [[71, 78]], [[8, 59]], [[8, 59]], [[2, 59]], [[2, 78]], [[80, 93]], [[96, 155]], [[157, 187]]]", "query_spans": "[[[189, 198]]]", "process": "From the triangle area formula, the coordinate representation of the dot product of vectors, and the equation derived from point P lying on the ellipse, we obtain c|y_{P}|=4, |y_{P}|=\\frac{b^{2}}{c}, thus parameter b can be found. Solution: According to the given condition, \\frac{1}{2}\\times2c\\times|y_{p}|=c|y_{p}|=4, and (-c-x_{P},-y_{P})\\cdot(c-x_{P},-y_{P})=0, we obtain x_{P}^{2}=c^{2}-y_{P}^{2}. Also, for ellipse C: \\frac{x_{P}^{2}}{a^{2}}+\\frac{y_{P}^{2}}{k^{2}}=\\frac{c^{2}-y_{P}^{2}}{a^{2}}+\\frac{y_{P}^{2}}{k^{2}}=1, then y_{P}=\\frac{b^{2}}{c}. In summary, b^{2}=4, and since b>0, we have b=2. Final answer is: 2" }, { "text": "Given the curve $C$: $x^{2}-y|y|=1$, if the line $l$: $y=x+t$ intersects the curve at exactly one point, then what is the range of real values for $t$?", "fact_expressions": "l: Line;C: Curve;t: Real;Expression(C) = (x^2 - y*Abs(y) = 1);Expression(l)=(y=x+t);NumIntersection(l,C)=1", "query_expressions": "Range(t)", "answer_expressions": "{-sqrt(2)}+[0,+oo)", "fact_spans": "[[[25, 39]], [[2, 23], [40, 42]], [[53, 58]], [[2, 23]], [[25, 39]], [[25, 51]]]", "query_spans": "[[[53, 65]]]", "process": "When $ y \\geqslant 0 $, $ x^{2} - y^{2} = 1 $; when $ y < 0 $, $ x^{2} + y^{2} = 1 $; thus, the graph of curve $ C $ is shown in the following figure. \n① When $ t \\geqslant 0 $, since $ y = x $ is an asymptote of the hyperbola $ x^{2} - y^{2} = 1 $, and it intersects $ x^{2} + y^{2} = 1 $ ($ y < 0 $) at a unique point, therefore when $ t \\geqslant 0 $, $ y = x + t $ intersects $ C $ at a unique point. \n② When $ t < 0 $, if $ y = x + t $ intersects $ C $ at a unique point, then $ y = x + t $ is tangent to $ x^{2} + y^{2} = 1 $ ($ y < 0 $), so $ \\frac{|t|}{\\sqrt{2}} = 1 $, solving gives: $ t = \\sqrt{2} $ (discarded) or $ t = -\\sqrt{2} $. \nIn summary, the range of real number $ t $ is $ \\{-\\sqrt{2}\\} \\cup [0, +\\infty) $." }, { "text": "The eccentricity of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "3/5", "fact_spans": "[[[0, 39]], [[0, 39]]]", "query_spans": "[[[0, 45]]]", "process": "From the ellipse equation, we have: a^{2}=25, b^{2}=16, then c^{2}=a^{2}-b^{2}=9, e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{9}{25}, e=\\frac{3}{5}" }, { "text": "If point $M$ is a point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, and $O$ is the origin, then what are the maximum and minimum values of $| O M |$?", "fact_expressions": "G: Ellipse;O: Origin;M: Point;Expression(G) = (x^2/9 + y^2/5 = 1);PointOnCurve(M, G)", "query_expressions": "Min(Abs(LineSegmentOf(O, M)));Max(Abs(LineSegmentOf(O, M)))", "answer_expressions": "sqrt(5);3", "fact_spans": "[[[6, 43]], [[48, 51]], [[1, 5]], [[6, 43]], [[1, 47]]]", "query_spans": "[[[58, 79]], [[58, 79]]]", "process": "From the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, we obtain $a=3$, $b=\\sqrt{5}$. Let point $M(x,y)$, then $|OM|=\\sqrt{x^{2}+y^{2}}=\\sqrt{\\frac{4}{9}x^{2}+5}$. Since $x\\in[-3,3]$, when point $M$ is at the endpoints of the major axis and minor axis respectively, $|OM|$ attains its maximum and minimum values. Therefore, the maximum and minimum values of $|OM|$ are $3$ and $\\sqrt{5}$. The answer is: $3,\\sqrt{5}$." }, { "text": "The perpendicular bisector of chord $AB$ of the parabola $y^2 = x$, where $AB$ is not perpendicular to the $x$-axis, intersects the $x$-axis. The range of the $x$-coordinate of this intersection point is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = x);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);Negation(IsPerpendicular(LineSegmentOf(A, B), xAxis))", "query_expressions": "Range(XCoordinate(Intersection(PerpendicularBisector(LineSegmentOf(A, B)), xAxis)))", "answer_expressions": "(1/2, +oo)", "fact_spans": "[[[0, 12]], [[0, 12]], [[14, 19]], [[14, 19]], [[0, 19]], [[14, 27]]]", "query_spans": "[[[29, 56]]]", "process": "Let A(y_{1},y_{1}), B(y_{2}2,y_{2}), y_{1}^{2}\\neq y_{2}^{2}, so the midpoint coordinates of AB are: (\\frac{y_{1}+y_{2}}{2},\\frac{y_{1}+y_{2}}{2}). The slope of line AB is: \\frac{y_{1}-y_{2}}{y_{1}^{2}-y_{2}^{2}}=\\frac{1}{y_{1}+y_{2}}, therefore the slope of the perpendicular bisector of segment AB is: -(y_{1}+y_{2}), so the equation of the perpendicular bisector of AB is: y-\\frac{y_{1}+y_{2}}{2}=-(y_{1}+y_{2})(x-\\frac{y_{1}^{2}+y_{2}2}{2}). Letting y=0, x=\\frac{1}{2}+\\frac{y_{1}^{2}+y_{2}^{2}}{2}>\\frac{1}{2}" }, { "text": "Given a line with slope $-\\frac{1}{3}$ that does not pass through the origin $O$, intersecting the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{7}=1$ at points $A$ and $B$, and let $M$ be the midpoint of segment $AB$. Then, what is the slope of line $OM$?", "fact_expressions": "G: Ellipse;H: Line;M: Point;O: Origin;B: Point;A: Point;Expression(G) = (x^2/9 + y^2/7 = 1);Slope(H)=-1/3;Negation(PointOnCurve(O,H));Intersection(H, G) = {A, B};MidPoint(LineSegmentOf(A,B))=M", "query_expressions": "Slope(LineOf(O,M))", "answer_expressions": "7/3", "fact_spans": "[[[34, 71]], [[31, 33]], [[84, 87]], [[23, 30]], [[78, 81]], [[74, 77]], [[34, 71]], [[2, 33]], [[20, 33]], [[31, 83]], [[84, 98]]]", "query_spans": "[[[100, 112]]]", "process": "Let the equation of line AB be y = -\\frac{1}{3}x + b. Solving simultaneously: \n\\begin{matrix} y = -\\frac{1}{3}x + b \\\\ \\frac{x^{2}}{9} + \\frac{y^{2}}{7} = 1 & \\frac{x^{2}}{9} + \\frac{(-\\frac{1}{3}x + b)^{2}}{7} = 1 \\end{matrix} \nThis gives 8x^{2} - 6bx + 9b^{2} - 63 = 0. From \\triangle = 36b^{2} - 32(9b^{2} - 63) > 0, we obtain -2\\sqrt{2} < b < 2\\sqrt{2}. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), M(x_{0}, y_{0}), then x_{0} = \\frac{x_{1} + x_{2}}{2} = \\frac{3b}{8}, y_{0} = -\\frac{1}{3}x_{0} + b = -\\frac{1}{3} \\times \\frac{3b}{8} + b = \\frac{7b}{8}, so M(\\frac{3b}{8}, \\frac{7b}{8}). Then the slope of line OM is k = \\frac{y_{0}}{x_{0}} = \\frac{7}{3}." }, { "text": "Let the semi-focal length of the hyperbola $\\frac {x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1,(00, b>0)$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. A perpendicular line is drawn from the right focus $F_{2}$ to one of its asymptotes, with foot of perpendicular at $M$, intersecting the right branch of hyperbola $C$ at point $P$. If $\\overrightarrow{F_{2} P}=2 \\overrightarrow{P M}$, and $\\angle F_{1} P F_{2}=120^{\\circ}$, then the eccentricity of hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;F2: Point;P: Point;M: Point;F1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;L:Line;PointOnCurve(F2,L);IsPerpendicular(Asymptote(C),L);FootPoint(Asymptote(C),L)=M;Intersection(L,RightPart(C))=P;VectorOf(F2, P) = 2*VectorOf(P, M);AngleOf(F1, P, F2) = ApplyUnit(120, degree)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(13)/2", "fact_spans": "[[[2, 62], [99, 100], [115, 121], [217, 223]], [[9, 62]], [[9, 62]], [[79, 86], [91, 98]], [[124, 128]], [[110, 113]], [[71, 78]], [[9, 62]], [[9, 62]], [[2, 62]], [[2, 86]], [[2, 86]], [], [[87, 106]], [[87, 106]], [[87, 113]], [[87, 128]], [[130, 179]], [[181, 215]]]", "query_spans": "[[[217, 229]]]", "process": "From $ F_{2}M\\bot OM $, and $ OM $ being the asymptote, find $ |F_{2}M| $, thereby obtaining $ |F_{2}P| $. From the hyperbola definition, obtain $ |F_{1}P| $, then use the cosine law to derive an equation in $ a,b,c $, thus finding the eccentricity. \n**Detailed Solution:** From the given conditions, $ F_{2}(c,0) $, the asymptote equation is $ y=\\frac{b}{a}x $, i.e., $ bx-ay=0 $, \n$ \\therefore |F_{2}M| = \\frac{bc}{\\sqrt{a^{2}+b^{2}}} = b $. \nSince $ \\overrightarrow{F_{2}P} = 2\\overrightarrow{PM} $, $ \\therefore |PF_{2}| = \\frac{2}{3}|MF_{2}| = \\frac{2}{3}b $. \nFrom the hyperbola definition, $ |PF_{1}| - |PF_{2}| = 2a $, so $ |PF_{1}| = 2a + \\frac{2}{3}b $. \nAlso, $ |F_{1}F_{2}| = 2c $, $ \\angle F_{1}PF_{2} = 120^{\\circ} $. \nIn triangle $ PF_{1}F_{2} $, by the cosine law: \n$$\n4c^{2} = \\frac{4}{9}b^{2} + \\left(\\frac{2}{3}b + 2a\\right)^{2} - 2 \\times \\frac{2}{3}b \\times \\left(\\frac{2}{3}b + 2a\\right) \\times \\left(-\\frac{1}{2}\\right),\n$$\nand $ c^{2} = a^{2} + b^{2} $, $ e = \\sqrt{1 + \\left(\\frac{b}{a}\\right)^{2}} = \\frac{\\sqrt{13}}{2} $." }, { "text": "Given that points $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. A line $l$ passing through the origin $O$ with an inclination angle of $60^{\\circ}$ intersects the ellipse $C$ at a point $M$, and $|\\overrightarrow{M F_{1}}+\\overrightarrow{M F_{2}}|=|\\overrightarrow{M F_{1}}-\\overrightarrow{M F_{2}}|$. Find the eccentricity of the ellipse $C$.", "fact_expressions": "F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;O: Origin;l: Line;PointOnCurve(O, l);Inclination(l) = ApplyUnit(60, degree);M: Point;OneOf(Intersection(l, C)) = M;Abs(VectorOf(M, F1) + VectorOf(M, F2)) = Abs(VectorOf(M, F1) - VectorOf(M, F2))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(2)-1", "fact_spans": "[[[2, 10]], [[11, 18]], [[2, 84]], [[2, 84]], [[21, 78], [115, 120], [238, 243]], [[21, 78]], [[28, 78]], [[28, 78]], [[28, 78]], [[28, 78]], [[86, 91]], [[109, 114]], [[85, 114]], [[92, 114]], [[126, 129]], [[109, 129]], [[131, 236]]]", "query_spans": "[[[238, 249]]]", "process": "From |\\overrightarrow{MF_{1}}+\\overrightarrow{MF_{2}}|=|\\overrightarrow{MF_{1}}-\\overrightarrow{MF_{2}}|, analyze that \\triangle MF_{1}F_{2} is a right triangle. Solve the triangle to obtain the side lengths, and use the ellipse definition to get c+\\sqrt{3}c=2a, then find the eccentricity. Without loss of generality, assume M lies in the first quadrant. From |\\overrightarrow{MF}_{1}+\\overrightarrow{MF_{2}}|=|\\overrightarrow{MF_{1}}-\\overrightarrow{MF_{2}}|, square both sides and simplify to obtain: \\overrightarrow{MF}\\cdot\\overrightarrow{MF_{2}}=0, therefore \\overrightarrow{MF_{1}}\\bot\\overrightarrow{MF_{2}}. In the right triangle \\triangle MF_{1}F_{2}, \\because \\angle MOF_{2}=60^{\\circ}, |OM|=c, |OF_{2}|=\\cdot \\angle MF_{2}F_{1}=60^{\\circ}, |MF_{2}|=|F_{2}F_{1}|\\cos\\angle MF_{2}F_{1}=2c\\cos60^{\\circ}=c, |MF_{1}|=|F_{2}F_{1}|\\sin\\angle MF_{2}F_{1}=2c\\sin60^{\\circ}=\\sqrt{3}c. From the definition of the ellipse: |MF_{2}|+|MF_{1}|=c+\\sqrt{3}c=2a, so the eccentricity e=\\frac{c}{a}=\\frac{2}{1+\\sqrt{3}}=\\sqrt{3}-" }, { "text": "Given the hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$ has eccentricity $2$. If the distance from the focus of the parabola $C_{2}$: $ x^{2}=2 p y (p>0)$ to the asymptotes of the hyperbola $C_{1}$ is $2$, then what is $p$=?", "fact_expressions": "C1: Hyperbola;C2: Parabola;a:Number;b:Number;p:Number;p>0;a>0;b>0;Expression(C1)=(x^2/a^2-y^2/b^2=1);Eccentricity(C1)=2;Expression(C2)=(x^2=2*(p*y));Distance(Focus(C2),Asymptote(C1))=2", "query_expressions": "p", "answer_expressions": "8", "fact_spans": "[[[2, 70], [116, 126]], [[80, 112]], [[14, 70]], [[14, 70]], [[139, 142]], [[91, 112]], [[14, 70]], [[14, 70]], [[2, 70]], [[2, 78]], [[80, 112]], [[80, 137]]]", "query_spans": "[[[139, 144]]]", "process": "Test analysis: $ e = \\frac{c}{a} = 2, \\therefore c = 2a, b = \\sqrt{c^{2} - a^{2}} = \\sqrt{3}a $, so the asymptotes of the hyperbola are $ y = \\pm\\sqrt{3}x $. The focus coordinates of the parabola are $ (0, \\frac{p}{2}) $. Using the point-to-line distance formula, $ \\frac{p}{2} = 2, \\therefore p = 8 $." }, { "text": "The ellipse $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{3-m}=1$ has a focus at $(0,1)$, then its major axis length =?", "fact_expressions": "G: Ellipse;m: Number;H: Point;Expression(G) = (y^2/(3 - m) + x^2/m^2 = 1);Coordinate(H) = (0, 1);OneOf(Focus(G)) = H", "query_expressions": "Length(MajorAxis(G))", "answer_expressions": "{2*sqrt(2),2*sqrt(5)}", "fact_spans": "[[[0, 43], [58, 59]], [[2, 43]], [[49, 56]], [[0, 43]], [[49, 56]], [[0, 56]]]", "query_spans": "[[[58, 64]]]", "process": "" }, { "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ is $\\frac{1}{2}$, then what is the value of $m$?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/m = 1);m: Number;Eccentricity(G) = 1/2", "query_expressions": "m", "answer_expressions": "{3,16/3}", "fact_spans": "[[[1, 38]], [[1, 38]], [[58, 61]], [[1, 56]]]", "query_spans": "[[[58, 65]]]", "process": "\\textcircled{1} When the foci of the ellipse lie on the x-axis, it follows from the given condition that \\underline{\\sqrt{4-m}}=\\frac{1}{2}, solving gives m=3; \\textcircled{2} When the foci of the ellipse lie on the y-axis, it follows from the given condition that \\frac{\\sqrt{m-4}}{\\sqrt{m}}=\\frac{1}{2}, solving gives m=\\frac{16}{3}. In conclusion, m=3 or \\frac{16}{3}." }, { "text": "If the point $(a, \\frac{a}{2})$ lies on the curve represented by the equation $y=3 x^{2}+x$, then $a=?$", "fact_expressions": "G: Curve;H: Point;Coordinate(H) = (a, a/2);Expression(G)=(y = 3*x^2 + x);PointOnCurve(H, G);a:Number", "query_expressions": "a", "answer_expressions": "{0,-1/6}", "fact_spans": "[[[40, 42]], [[1, 20]], [[1, 20]], [[21, 42]], [[1, 43]], [[45, 48]]]", "query_spans": "[[[45, 50]]]", "process": "The point (a, \\frac{a}{2}) lies on the curve represented by the equation y = 3x^{2} + x, then \\frac{a}{2} = 3 \\cdot a^{2} + a, solving gives a = 0 or a = -\\frac{1}{6}" }, { "text": "Given points $A(1, y_{1})$, $B(9, y_{2})$ are two points on the parabola $y^{2}=2 p x(p>0)$, $y_{2}>y_{1}>0$, point $F$ is the focus of the parabola. If $|B F|=5|A F|$, then the value of $y_{1}^{2}+y_{2}$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;A: Point;B: Point;y1: Number;y2: Number;Coordinate(A) = (1, y1);Coordinate(B) = (9, y2);PointOnCurve(A, G);PointOnCurve(B, G);y2>y1;y1>0;F: Point;Focus(G) = F;Abs(LineSegmentOf(B, F)) = 5*Abs(LineSegmentOf(A, F))", "query_expressions": "y1^2 + y2", "answer_expressions": "10", "fact_spans": "[[[31, 52], [78, 81]], [[31, 52]], [[34, 52]], [[34, 52]], [[2, 16]], [[17, 30]], [[3, 16]], [[17, 30]], [[2, 16]], [[17, 30]], [[2, 56]], [[2, 56]], [[57, 72]], [[57, 72]], [[73, 77]], [[73, 84]], [[86, 100]]]", "query_spans": "[[[102, 123]]]", "process": "From the definition of the parabola, we have |AF| = 1 + \\frac{p}{2}, |BF| = 9 + \\frac{p}{2}. According to the given condition, 9 + \\frac{p}{2} = 5 + \\frac{5p}{2} \\Rightarrow p = 2. Then y_{1}^{2} = 4 \\times 1 = 4, y_{2}^{2} = 4 \\times 9 = 36 \\Rightarrow y_{2} = 6 (negative value discarded), hence y_{1}^{2} + y_{2} = 10. Therefore, fill in 10." }, { "text": "For any point $Q$ on the parabola $y^{2}=4x$, the point $P(a, 0)$ satisfies $|PQ| \\geq |a|$. Then the range of values for $a$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);Q: Point;PointOnCurve(Q, G);P: Point;Coordinate(P) = (a, 0);a: Number;Abs(LineSegmentOf(P, Q)) >= Abs(a)", "query_expressions": "Range(a)", "answer_expressions": "(-oo,2]", "fact_spans": "[[[2, 16]], [[2, 16]], [[21, 24]], [[2, 24]], [[25, 36]], [[25, 36]], [[56, 59]], [[39, 54]]]", "query_spans": "[[[56, 66]]]", "process": "" }, { "text": "If points $O$ and $F$ are the center and the left focus of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$, respectively, and point $P$ is an arbitrary point on the ellipse, then the minimum value of $|O P|^{2}+|P F|^{2}$ is?", "fact_expressions": "G: Ellipse;O: Point;P: Point;F: Point;Expression(G) = (x^2/2 + y^2 = 1);Center(G) = O;LeftFocus(G) = F;PointOnCurve(P, G)", "query_expressions": "Min(Abs(LineSegmentOf(O, P))^2 + Abs(LineSegmentOf(P, F))^2)", "answer_expressions": "2", "fact_spans": "[[[13, 40], [53, 55]], [[1, 5]], [[48, 52]], [[6, 10]], [[13, 40]], [[1, 47]], [[1, 47]], [[48, 61]]]", "query_spans": "[[[63, 90]]]", "process": "This problem examines the standard equation of an ellipse, its geometric properties, and the application of functional thinking. The ellipse $\\frac{x^{2}}{2}+y^{2}=1$ has center $O(0,0)$ and left focus $F(-1,0)$. Let $P(x,y)$, then $y=1-\\frac{x^{2}}{2}$ $(-\\sqrt{2}\\leqslant x \\leqslant \\sqrt{2})$. Thus, $|OP|^{2}+|PF|^{2}=x^{2}+y^{2}+(x+1)^{2}+y^{2}=x^{2}+2(1-\\frac{x^{2}}{2})+(x+1)^{2}=(x+1)^{2}+2$ $(-\\sqrt{2}\\leqslant x \\leqslant \\sqrt{2})$. When $x=-1$, $|OP|^{2}+|PF|^{2}$ attains its minimum value, which is 2." }, { "text": "If the focus of the parabola $C$: $y^{2}=2 p x$ lies on the line $x+2 y-4=0$, then what is $p$? What is the equation of the directrix of $C$?", "fact_expressions": "C: Parabola;p: Number;G: Line;Expression(C) = (y^2 = 2*(p*x));Expression(G) = (x + 2*y - 4 = 0);PointOnCurve(Focus(C), G)", "query_expressions": "p;Expression(Directrix(C))", "answer_expressions": "8;x = -4", "fact_spans": "[[[1, 22], [48, 51]], [[42, 45]], [[26, 39]], [[1, 22]], [[26, 39]], [[1, 40]]]", "query_spans": "[[[42, 47]], [[48, 58]]]", "process": "" }, { "text": "The ellipse $m x^{2}+n y^{2}=1$ intersects the line $y=1-x$ at points $M$ and $N$. The slope of the line connecting the origin to the midpoint of segment $MN$ is $\\frac{\\sqrt{2}}{2}$. What is the value of $\\frac{m}{n}$?", "fact_expressions": "M: Point;N: Point;G: Ellipse;n: Number;m: Number;H: Line;Expression(G) = (m*x^2 + n*y^2 = 1);Expression(H) = (y = 1 - x);Intersection(G, H) = {M, N};O:Origin;Slope(LineSegmentOf(O,MidPoint(LineSegmentOf(M,N))))=sqrt(2)/2", "query_expressions": "m/n", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[33, 36]], [[37, 40]], [[0, 21]], [[84, 97]], [[84, 97]], [[22, 31]], [[0, 21]], [[22, 31]], [[0, 42]], [[43, 45]], [[43, 81]]]", "query_spans": "[[[84, 101]]]", "process": "" }, { "text": "Given that $F$ is the focus of the parabola $y^{2}=8 x$, $M$ is a point on the parabola such that $|M F|=3$, and $N(-2,0)$, then what is the slope of the line $M N$?", "fact_expressions": "G: Parabola;M: Point;N: Point;F: Point;Expression(G) = (y^2 = 8*x);Focus(G) = F;PointOnCurve(M, G);Abs(LineSegmentOf(M, F)) = 3;Coordinate(N) = (-2, 0)", "query_expressions": "Slope(LineOf(M,N))", "answer_expressions": "pm*3*sqrt(2)/3", "fact_spans": "[[[6, 20], [28, 31]], [[24, 27]], [[47, 56]], [[2, 5]], [[6, 20]], [[2, 23]], [[24, 34]], [[35, 45]], [[47, 56]]]", "query_spans": "[[[58, 70]]]", "process": "Let M(x,y), the directrix equation of the parabola y^{2}=8x is x=-2. Given |MF|=3, then x+2=3, we get x=1, y=\\pm2\\sqrt{2}, hence the slope of MN is \\frac{\\pm2\\sqrt{2}}{1-(-2)}=\\pm\\frac{2\\sqrt{2}}{3}" }, { "text": "The length of the real axis is $12$, the eccentricity is $2$, and the standard equation of the hyperbola with foci on the $x$-axis is?", "fact_expressions": "G: Hyperbola;Length(RealAxis(G))=12;Eccentricity(G)=2;PointOnCurve(Focus(G),xAxis)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/36 - y^2/108 = 1", "fact_spans": "[[[26, 29]], [[0, 29]], [[9, 29]], [[17, 29]]]", "query_spans": "[[[26, 36]]]", "process": "According to the problem, the length of the real axis of the hyperbola is 12, so 2a = 12, that is, a = 6. Also, given its eccentricity e = 2, which means e = \\frac{c}{a} = 2, we have c = 12. Then b = \\sqrt{c^{2}-a^{2}} = \\sqrt{108}. Since the foci of the hyperbola lie on the x-axis, its standard equation is: \\frac{x^{2}}{36}-\\frac{y^{2}}{108}=1;" }, { "text": "The equation of a parabola that has the same focus as the right focus of the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$ is?", "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (x^2 - y^2/4 = 1);Focus(H) = RightFocus(G)", "query_expressions": "Expression(H)", "answer_expressions": "y^2 = 4*sqrt(5)*x", "fact_spans": "[[[3, 31]], [[37, 40]], [[3, 31]], [[0, 40]]]", "query_spans": "[[[37, 44]]]", "process": "Find the right focus of the hyperbola x^{2}-\\frac{y^{2}}{4}=1 as (\\sqrt{5},0). Let the equation of the parabola be y^{2}=2px (p>0). By solving \\frac{p}{2}=\\sqrt{5}, we obtain p=2\\sqrt{5}, which allows us to solve the problem. Detailed solution: From x^{2}-\\frac{y^{2}}{4}=1, we get: a^{2}=1, b^{2}=4, so c=\\sqrt{a^{2}+b^{2}}=\\sqrt{1+5}=\\sqrt{5}. Therefore, the right focus of the hyperbola x^{2}-\\frac{y^{2}}{4}=1 is (\\sqrt{5},0). Let the equation of the parabola be y^{2}=2px (p>0). According to the given condition, \\frac{p}{2}=\\sqrt{5}, so p=2\\sqrt{5}. Thus, the equation of the parabola is y^{2}=4\\sqrt{5}x." }, { "text": "Given the hyperbola $C$: $ \\frac{x^{2}}{a^{2}}-y^{2}=1(a>0)$ intersects with line $l$: $x+y=1$ at two distinct points $A$, $B$, and line $l$ intersects the $y$-axis at point $P$. If $\\overrightarrow{P A}=\\frac{5}{12} \\overrightarrow{P B}$, then $a=?$", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2 + x^2/a^2 = 1);a: Number;P: Point;A: Point;B: Point;a>0;l:Line;Intersection(C,l)={A,B};Negation(A=B);Expression(l)=(x+y=1);Intersection(l, yAxis) = P;VectorOf(P, A) = (5/12)*VectorOf(P, B)", "query_expressions": "a", "answer_expressions": "17/13", "fact_spans": "[[[2, 44]], [[2, 44]], [[150, 153]], [[86, 90]], [[65, 69]], [[71, 74]], [[9, 44]], [[45, 57], [76, 79]], [[2, 74]], [[60, 74]], [[45, 57]], [[76, 90]], [[92, 148]]]", "query_spans": "[[[150, 155]]]", "process": "Since the hyperbola C and the line l have two distinct intersection points, the system of equations: \n\\begin{cases}\\frac{x^{2}}{a^{2}}-y^{2}=1\\\\x+y=1\\end{cases} \nhas two distinct real solutions. Eliminating $ y $ and simplifying yields: \n$ (1-a^{2})x^{2}+2a^{2}x-2a^{2}=0 $. \nTherefore, the real number $ a $ must satisfy: \n\\begin{cases}1-a^{2}\\neq0\\\\4a^{4}+8a^{2}(1-a^{2})>0\\end{cases}. \nSolving gives: $ 0b>0)$, respectively. If there exists a point $P$ on the line $x=\\frac{2 a^{2}}{c}$ such that $|P F_{2}|=2 c$, then the range of the eccentricity of the ellipse is?", "fact_expressions": "F1: Point;F2: Point;b: Number;a: Number;a > b;b > 0;H: Line;P: Point;c: Number;G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);Expression(H) = (x = (2*a^2)/c);PointOnCurve(P, H);Abs(LineSegmentOf(P, F2)) = 2*c;LeftFocus(G) = F1;RightFocus(G) = F2", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(0, sqrt(6)/3]", "fact_spans": "[[[1, 16]], [[17, 31]], [[36, 86]], [[36, 86]], [[36, 86]], [[36, 86]], [[94, 117]], [[120, 124]], [[126, 141]], [[34, 86], [143, 145]], [[34, 86]], [[1, 16]], [[17, 31]], [[94, 117]], [[94, 124]], [[126, 141]], [[1, 92]], [[1, 92]]]", "query_spans": "[[[143, 155]]]", "process": "By the given condition, |PF_{2}| = 2c \\geqslant \\frac{2a^{2}}{c} - c, then e^{2} = \\frac{c^{2}}{a^{2}} \\leqslant \\frac{2}{3}, while 0 < e < 1. Therefore, 0 < e \\leqslant \\frac{\\sqrt{6}}{3}." }, { "text": "Given that $F$ is the focus of the parabola $x^{2}=4 y$, $B(0,-1)$, and $A$ is an arbitrary point on the parabola. When $\\frac{|A F|}{|A B|}$ takes its minimum value, $|A B|=$?", "fact_expressions": "F:Point;G:Parabola;Expression(G) = (x^2=4*y);B: Point;Coordinate(B) = (0,-1);A:Point;PointOnCurve(A,G) = True;WhenMin(Abs(LineSegmentOf(A,F)) / Abs(LineSegmentOf(A,B))) = True;Focus(G) = F", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[2, 5]], [[6, 20], [40, 43]], [[6, 20]], [[24, 33]], [[24, 33]], [[36, 39]], [[35, 48]], [[49, 76]], [[2, 23]]]", "query_spans": "[[[77, 86]]]", "process": "From the given conditions, we have F(0,1). The directrix of the parabola $x^{2}=4y$ is $y=-1$, and point B(0,-1) lies on the directrix. As shown in the figure, by the definition of a parabola, $|AF|=|AA_{1}|$. Therefore, $\\frac{|AF|}{|AB|}=\\frac{|AA_{1}|}{|AB|}=\\sin\\angle ABA_{1}'$, meaning the problem transforms into finding the value of $|AB|$ when the sine of the inclination angle of line AB is minimized. Let $l_{AB}: y=kx-1$. The sine of the inclination angle is minimized when the line is tangent to the parabola. Solving the system: \n$$\n\\begin{cases}\nx^{2}=4y \\\\\ny=kx-1\n\\end{cases}\n\\Rightarrow x^{2}-4kx+4=0,\n$$\nthe discriminant $\\Delta=(-4k)^{2}-4\\times4=0$ gives $k=\\pm1$." }, { "text": "If the coordinates of the focus of the parabola $y^{2}=m x$ are $(\\frac{1}{2}, 0)$, then what is the value of the real number $m$?", "fact_expressions": "G: Parabola;m: Real;Expression(G) = (y^2 = m*x);Coordinate(Focus(G))=(1/2,0)", "query_expressions": "m", "answer_expressions": "2", "fact_spans": "[[[1, 15]], [[41, 46]], [[1, 15]], [[1, 39]]]", "query_spans": "[[[41, 50]]]", "process": "Directly write the coordinates of the focus from the parabolic equation, and solve for the value of $ m $ according to the given conditions. \nSolution: From the parabolic equation, the focus coordinates are $ \\left( \\frac{m}{4}, 0 \\right) $. \n$ \\therefore \\frac{m}{4} = \\frac{1}{2} $, $ \\therefore m = 2 $." }, { "text": "Given the equation of the ellipse is $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{9}=1$, its two foci are $F_{1}$, $F_{2}$. If $|F_{1} F_{2}|=8$, and chord $AB$ passes through $F_{1}$, then the perimeter of $\\triangle ABF_{2}$ is?", "fact_expressions": "G: Ellipse;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (y^2/9 + x^2/a^2 = 1);Focus(G) = {F1, F2};Abs(LineSegmentOf(F1, F2)) = 8;PointOnCurve(F1, LineSegmentOf(A, B));IsChordOf(LineSegmentOf(A, B), G);a:Number", "query_expressions": "Perimeter(TriangleOf(A, B, F2))", "answer_expressions": "20", "fact_spans": "[[[2, 4], [48, 49]], [[93, 97]], [[93, 97]], [[65, 72]], [[55, 62], [98, 105]], [[2, 47]], [[48, 72]], [[74, 91]], [[93, 105]], [[48, 97]], [[2, 47]]]", "query_spans": "[[[107, 131]]]", "process": "" }, { "text": "Given that $(4,2)$ is the midpoint of the line segment cut by line $l$ on the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$, then the equation of $l$ is?", "fact_expressions": "l: Line;G: Ellipse;Expression(G) = (x^2/36 + y^2/9 = 1);Coordinate(MidPoint(InterceptChord(l,G)))=(4,2)", "query_expressions": "Expression(l)", "answer_expressions": "x+2*y-8=0", "fact_spans": "[[[10, 15], [65, 68]], [[16, 54]], [[16, 54]], [[2, 63]]]", "query_spans": "[[[65, 73]]]", "process": "Let the line l intersect the ellipse at points A(x_{1},y_{1}), B(x_{2},y_{2}); substitute into the ellipse equation and subtract to obtain \\frac{y_{1}-y_{2}}{x_{1}}-x_{2}=k=-\\frac{1}{2}, then compute the line equation to get the answer. Let the line l intersect the ellipse at points A(x_{1},y_{1}), B(x_{2},y_{2}); then \\frac{x_{1}^{2}}{36}+\\frac{y_{1}^{2}}{9}=1 and \\frac{x_{2}^{2}}{36}+\\frac{y_{2}^{2}}{9}=1. Subtracting these two equations gives \\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{36}+\\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{9}=0. Since (x_{1}+x_{2}=8, y_{1}+y_{2}=4), it follows that \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=k=-\\frac{1}{2}. Hence, the equation of line l is y-2=-\\frac{1}{2}(x-4), or equivalently, x+2y-8=0." }, { "text": "Given that the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1$ coincides with the focus of the parabola $y^{2}=12 x$, then the distance from the focus of this hyperbola to its asymptote is equal to?", "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;Expression(G) = (-y^2/4 + x^2/a^2 = 1);Expression(H) = (y^2 = 12*x);RightFocus(G) = Focus(H)", "query_expressions": "Distance(Focus(G), Asymptote(G))", "answer_expressions": "2", "fact_spans": "[[[2, 44], [72, 75], [79, 80]], [[5, 44]], [[49, 64]], [[2, 44]], [[49, 64]], [[2, 69]]]", "query_spans": "[[[72, 89]]]", "process": "" }, { "text": "If $F$ is the left focus of the hyperbola $M$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and a line $l$ passing through the origin intersects the left and right branches of the hyperbola $M$ at points $A$ and $B$ respectively, then the range of values of $\\frac{1}{|F A|}-\\frac{9}{|F B|}$ is?", "fact_expressions": "F: Point;M: Hyperbola;Expression(M) = (x^2/9 - y^2/16 = 1);LeftFocus(M) = F;O: Origin;PointOnCurve(O, l) = True;l: Line;Intersection(l,LeftPart(M)) = A;Intersection(l, RightPart(M)) = B;A: Point;B: Point", "query_expressions": "Range(-9/Abs(LineSegmentOf(F, B)) + 1/Abs(LineSegmentOf(F, A)))", "answer_expressions": "[-2/3,0)", "fact_spans": "[[[1, 4]], [[5, 49], [64, 70]], [[5, 49]], [[1, 53]], [[55, 57]], [[54, 63]], [[58, 63]], [[58, 88]], [[58, 88]], [[79, 82]], [[83, 86]]]", "query_spans": "[[[90, 130]]]", "process": "As shown in the figure: Let |AF| = m, |BF| = n, F be the right focus of the hyperbola, connect AF, BF, then AFBF is a parallelogram, so |AF| = |BF| = m. By the definition of hyperbola, n - m = 2a = 6, thus n = m + 6, and m \\geqslant c - a = 2, so |\\frac{1}{|F}| - \\frac{9}{\\frac{9}{1}} = \\frac{1}{m} - \\frac{9}{n} = \\frac{1}{m} - \\frac{9}{m+6}, = \\frac{|FB|}{m} - \\frac{9}{m+6}\\frac{1}{2}(m \\geqslant 2), then f'(m) = -\\frac{1}{m^{2}} + \\frac{9}{(m+6)}\\frac{8m2-12m-1}{m^{2}(m+6)}\\frac{2m-36}{+6}^{2} = \\frac{8(2m+3)(m}{m2(m+6)}\\_. When 2 \\leqslant m < 3, f'(m) < 0, when m > 3, f'(m) > 0, so when m = 3, f(m)_{\\min} = -\\frac{2}{3}. When m \\rightarrow +\\infty, f(m) \\rightarrow 0, when m = 2, f(2) = \\frac{1}{2} - \\frac{9}{2+6} = -\\frac{5}{8}, so the range of \\frac{1}{FA} - \\frac{9}{FB} is (-\\frac{2}{3}, 0)." }, { "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=2 p x (p>0)$, the directrix of $C$ intersects the $x$-axis at point $A$. A tangent line $AB$ to the curve $C$ is drawn from point $A$. If the point of tangency $B$ lies in the first quadrant, and $D$ is a point on $C$ in the fourth quadrant such that $DF \\parallel AB$, then $\\frac{|AB|}{|DF|}$=?", "fact_expressions": "C: Parabola;p: Number;A: Point;B: Point;D: Point;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;Intersection(Directrix(C), xAxis) = A;TangentOfPoint(A,C)=LineOf(A,B);TangentPoint(LineOf(A,B),C)=B;Quadrant(B)=1;PointOnCurve(D,C);Quadrant(D)=4;IsParallel(LineSegmentOf(D,F),LineSegmentOf(A,B))", "query_expressions": "Abs(LineSegmentOf(A, B))/Abs(LineSegmentOf(D, F))", "answer_expressions": "sqrt(2)+1", "fact_spans": "[[[6, 34], [38, 41], [62, 67], [95, 98]], [[14, 34]], [[51, 55], [57, 61]], [[81, 84]], [[91, 94]], [[2, 5]], [[14, 34]], [[6, 34]], [[2, 37]], [[38, 55]], [[56, 77]], [[62, 84]], [[81, 90]], [[91, 107]], [[91, 107]], [[109, 122]]]", "query_spans": "[[[124, 147]]]", "process": "" }, { "text": "Given that one of the asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1 (b>0)$ has the equation $y=2 x$, then $b$=?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/b^2 = 1);b: Number;b>0;Expression(OneOf(Asymptote(G))) = (y = 2*x)", "query_expressions": "b", "answer_expressions": "2", "fact_spans": "[[[2, 41]], [[2, 41]], [[60, 63]], [[5, 41]], [[2, 58]]]", "query_spans": "[[[60, 65]]]", "process": "" }, { "text": "Let $F_{1}$ and $F_{2}$ be the two foci of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and let point $P$ lie on the ellipse such that $\\angle F_{1} P F_{2}=60^{\\circ}$. Then the area of $\\Delta F_{1} P F_{2}$ is?", "fact_expressions": "F1: Point;F2: Point;G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G) = True;AngleOf(F1, P, F2) = ApplyUnit(60, degree)", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "sqrt(3)", "fact_spans": "[[[1, 8]], [[9, 16]], [[17, 54], [65, 67]], [[17, 54]], [[1, 59]], [[60, 64]], [[60, 68]], [[72, 105]]]", "query_spans": "[[[107, 134]]]", "process": "Solution: By the definition of the ellipse, we have |PF₁| + |PF₂| = 4. In triangle PF₁F₂, by the cosine law: \n|F₁F₂|² = |PF₁|² + |PF₂|² - 2|PF₁||PF₂|cos∠F₁PF₂. \nGiven |F₁F₂| = 2√(4−3) = 2, ∠F₁PF₂ = 60°, we obtain \n|PF₁|² + |PF₂|² − |PF₁||PF₂| = 4, \nwhich implies (|PF₁| + |PF₂|)² = 3|PF₁||PF₂| + 4, \nso 3|PF₁||PF₂| + 4 = 4² = 16, \nthus |PF₁||PF₂| = 4. \nBy the triangle area formula, we get \nS_{ΔF₁PF₂} = (1/2)|PF₁||PF₂|sin∠F₁PF₂ = (1/2) × 4 × (√3/2) = √3." }, { "text": "The equation of the hyperbola passing through the points $(-\\sqrt{2} , \\sqrt{3})$ and $(\\frac{\\sqrt{15}}{3} , \\sqrt{2})$ is?", "fact_expressions": "H: Point;Coordinate(H) = (-sqrt(2), sqrt(3));I: Point;Coordinate(I) = (sqrt(15)/3, sqrt(2));G: Hyperbola;PointOnCurve(H, G);PointOnCurve(I, G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2-y^2/3=1", "fact_spans": "[[[2, 27]], [[2, 27]], [[29, 63]], [[29, 63]], [[64, 67]], [[1, 67]], [[1, 67]]]", "query_spans": "[[[64, 71]]]", "process": "" }, { "text": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has one of its asymptotes given by $y=2x$, and the distance from a focus to the asymptote is $2$. Then the equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;Expression(OneOf(Asymptote(G))) = (y = 2*x);Distance(Focus(G), Asymptote(G)) = 2", "query_expressions": "Expression(G)", "answer_expressions": "x^2-y^2/4=1", "fact_spans": "[[[0, 46], [81, 84]], [[0, 46]], [[3, 46]], [[3, 46]], [[0, 64]], [[0, 79]]]", "query_spans": "[[[81, 89]]]", "process": "From the given conditions: \n\\begin{cases}\\frac{b}{a}=2,\\\\b=2,\\end{cases}\\Rightarrow a=1,\\therefore the equation of the hyperbola is x^{2}-\\frac{y^{2}}{4}=1," }, { "text": "Let $F$ be the right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$, and let line $l$ passing through $F$ with slope $\\frac{a}{b}$ intersect the two asymptotes of hyperbola $C$ at points $A$ and $B$, respectively, such that $|\\overrightarrow{A F}|=2|\\overrightarrow{B F}|$. Then the eccentricity of hyperbola $C$ is?", "fact_expressions": "F: Point;C: Hyperbola;a: Number;b: Number;a>0;b>0;l: Line;A: Point;B: Point;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;PointOnCurve(F, l);Slope(l) = a/b;l1: Line;l2: Line;Asymptote(C) = {l1, l2};Intersection(l, l1) = A;Intersection(l,l2) = B;Abs(VectorOf(A, F)) = 2*Abs(VectorOf(B, F))", "query_expressions": "Eccentricity(C)", "answer_expressions": "{2,(2*sqrt(3)/3)}", "fact_spans": "[[[1, 4], [75, 78]], [[5, 69], [102, 108], [179, 185]], [[13, 69]], [[13, 69]], [[13, 69]], [[13, 69]], [[96, 101]], [[118, 121]], [[122, 125]], [[5, 69]], [[1, 73]], [[74, 98]], [[79, 101]], [], [], [[102, 114]], [[96, 127]], [[96, 127]], [[129, 177]]]", "query_spans": "[[[179, 191]]]", "process": "If $\\overrightarrow{AF}=-2\\overrightarrow{BF}$, then from Figure 1 it can be seen: the slope of asymptote $OB$ is $-\\frac{b}{a}$, $l \\perp OB$. In right triangle $AOBA$, by the angle bisector theorem we obtain $\\frac{|OA|}{|OB|}=\\frac{|FA|}{|FB|}=2$, so $\\angle AOB=60^{\\circ}$, $\\angle xOA=30^{\\circ}$, thus $\\frac{b}{a}=\\frac{\\sqrt{3}}{3}$, $e=\\frac{c}{a}=\\sqrt{1+(\\frac{b}{a})^{2}}=\\frac{2\\sqrt{3}}{3}$. If $\\overrightarrow{AF}=2\\overrightarrow{BF}$, then from Figure 2 it can be seen: asymptote $OB$ is the perpendicular bisector of side $AF$ of $AAOF$, hence $\\triangle AOF$ is an isosceles triangle, therefore $\\angle AOB=\\angle BOF=60^{\\circ}$, $\\frac{b}{a}=\\sqrt{3}$, $e=\\frac{c}{a}=\\sqrt{1+(\\frac{b}{a})^{2}}=2$. Thus, the eccentricity of this hyperbola is $2$ or $\\frac{2\\sqrt{3}}{}$." }, { "text": "Given that one of the asymptotes of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{3}=1$ $(m>0)$ is $y=\\frac{\\sqrt{3}}{2} x$, find the value of $m$.", "fact_expressions": "G: Hyperbola;m: Number;m>0;Expression(G) = (-y^2/3 + x^2/m = 1);Expression(OneOf(Asymptote(G))) = (y = x*(sqrt(3)/2))", "query_expressions": "m", "answer_expressions": "4", "fact_spans": "[[[2, 45]], [[80, 83]], [[5, 45]], [[2, 45]], [[2, 78]]]", "query_spans": "[[[80, 87]]]", "process": "" }, { "text": "Given that one focus of the hyperbola $\\frac{y^{2}}{t^{2}}-\\frac{x^{2}}{3}=1$ ($t>0$) coincides with the focus of the parabola $y=\\frac{1}{8} x^{2}$, find the real number $t$.", "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/3 + y^2/t^2 = 1);t: Real;t>0;H: Parabola;Expression(H) = (y = x^2/8);OneOf(Focus(G)) = Focus(H)", "query_expressions": "t", "answer_expressions": "1", "fact_spans": "[[[2, 49]], [[2, 49]], [[86, 91]], [[5, 49]], [[55, 79]], [[55, 79]], [[2, 84]]]", "query_spans": "[[[86, 94]]]", "process": "" }, { "text": "Given that the hyperbola $E$ has its center at the origin, $P(3 , 0)$ is a focus of $E$, a line $l$ passing through $F$ intersects $E$ at points $A$ and $B$, and the midpoint of $AB$ is $N(-12 ,-15)$, then the equation of $E$ is?", "fact_expressions": "E: Hyperbola;Center(E) = O;O: Origin;P: Point;Coordinate(P) = (3, 0);Focus(E) = P;l: Line;F: Point;PointOnCurve(F, l);Intersection(l, E) = {A, B};A: Point;B: Point;MidPoint(LineSegmentOf(A, B)) = N;N: Point;Coordinate(N) = (-12, -15)", "query_expressions": "Expression(E)", "answer_expressions": "x^2/4 - y^2/5 = 1", "fact_spans": "[[[2, 8], [26, 29], [44, 47], [85, 88]], [[2, 14]], [[12, 14]], [[15, 25]], [[15, 25]], [[15, 32]], [[38, 43]], [[34, 37]], [[33, 43]], [[38, 59]], [[50, 53]], [[54, 57]], [[61, 83]], [[70, 83]], [[70, 83]]]", "query_spans": "[[[85, 94]]]", "process": "" }, { "text": "Let the circle pass through the right vertex and the right focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, with its center on the hyperbola. What is the distance from the center of the circle to the center of the hyperbola?", "fact_expressions": "G: Hyperbola;H: Circle;Expression(G) = (x^2/9 - y^2/16 = 1);PointOnCurve(Center(H), G);PointOnCurve(RightVertex(G),H);PointOnCurve(RightFocus(G),H)", "query_expressions": "Distance(Center(H),Center(G))", "answer_expressions": "Preserve all mathematical expressions, LaTeX formulas, symbols, and equations EXACTLY. \nDo not add explanations, do not rewrite formulas, do not omit any content. \n\nText:", "fact_spans": "[[[3, 42], [54, 57], [63, 66]], [[1, 2]], [[3, 42]], [[1, 58]], [[1, 46]], [[1, 50]]]", "query_spans": "[[[1, 72]]]", "process": "" }, { "text": "Given the line $y=\\frac{1}{2} x+1$ intersects the parabola $C$: $x^{2}=4 y$ at points $A$ and $B$, $O$ is the origin. What is the area of triangle $O A B$?", "fact_expressions": "G: Line;Expression(G) = (y = (1/2)*x + 1);C: Parabola;Expression(C) = (x^2 = 4*y);A: Point;B: Point;Intersection(G, C) = {A, B};O: Origin", "query_expressions": "Area(TriangleOf(O, A, B))", "answer_expressions": "sqrt(5)", "fact_spans": "[[[2, 23]], [[2, 23]], [[24, 43]], [[24, 43]], [[46, 49]], [[50, 53]], [[2, 55]], [[56, 59]]]", "query_spans": "[[[66, 81]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). From \\begin{cases}y=\\frac{1}{2}x+1\\\\x^{2}=4y\\end{cases}, rearranging gives x^{2}-2x-4=0. By Vieta's formulas, we know x_{1}+x_{2}=2, x_{1}x_{2}=-4. \\begin{matrix}--&^{\\\\}AB=\\sqrt{1+k^{2}}\\times|x_{1}-x_{2}|=\\sqrt{1+\\frac{1}{4}}\\times\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\frac{\\sqrt{5}}{2}\\times2\\sqrt{5}=5. The distance from point O to the line y=\\frac{1}{2}x+1 is d=\\frac{2}{\\sqrt{5}}, \\therefore the area S of triangle AOB is S=\\frac{1}{2}.|AB|\\cdot d=\\sqrt{5}" }, { "text": "The asymptotes of a hyperbola with foci on the $x$-axis are given by $2x \\pm 3y = 0$, and the focal distance is $2\\sqrt{13}$. Find the equation of the hyperbola.", "fact_expressions": "PointOnCurve(Focus(G), xAxis);G: Hyperbola;Expression(Asymptote(G)) = (2*x + pm*3*y= 0);FocalLength(G) = 2*sqrt(13)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/9 - y^2/4 = 1", "fact_spans": "[[[2, 14]], [[11, 14], [55, 58]], [[11, 35]], [[11, 53]]]", "query_spans": "[[[55, 63]]]", "process": "Solution: Since the foci of the hyperbola lie on the x-axis, we assume the equation of the hyperbola is \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0, b>0). Because the asymptotes of the hyperbola are given by 2x\\pm3y=0, it follows that \\frac{b}{a}=\\frac{2}{3}. Since the focal distance is 2\\sqrt{13}, we have 2c=2\\sqrt{13}, so c=\\sqrt{13}. Therefore, a^{2}+b^{2}=13. Solving gives a=3, b=2. Hence, the equation of the hyperbola is \\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1." }, { "text": "Let $P_{1} P_{2}$ be a chord of the parabola $x^{2}=y$, and let the perpendicular bisector $l$ of $P_{1} P_{2}$ have the equation $y=x+3$. Then what is the equation of the line on which $P_{1} P_{2}$ lies?", "fact_expressions": "G: Parabola;l: Line;P1: Point;P2: Point;Expression(G) = (x^2 = y);IsChordOf(LineSegmentOf(P1, P2), G);PerpendicularBisector(LineSegmentOf(P1, P2)) = l;Expression(l) = (y = x + 3)", "query_expressions": "Expression(OverlappingLine(LineSegmentOf(P1, P2)))", "answer_expressions": "x - y + 2 = 0", "fact_spans": "[[[15, 27]], [[47, 50]], [[1, 14]], [[1, 14]], [[15, 27]], [[1, 29]], [[30, 50]], [[47, 61]]]", "query_spans": "[[[64, 85]]]", "process": "" }, { "text": "Given that the parabola $y^{2}=2 a x$ passes through the point $(-1,4)$, then the coordinates of the focus of the parabola are?", "fact_expressions": "G: Parabola;a: Number;H: Point;Expression(G) = (y^2 = 2*(a*x));Coordinate(H) = (-1, 4);PointOnCurve(H, G)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(-4,0)", "fact_spans": "[[[2, 18], [30, 33]], [[5, 18]], [[19, 28]], [[2, 18]], [[19, 28]], [[2, 28]]]", "query_spans": "[[[30, 40]]]", "process": "Solution: Since the parabola l passes through point B, solving yields: a = -8, so the equation of the parabola is y² = -16x, therefore the focus coordinates of the parabola are region." }, { "text": "Given the hyperbola $x^{2}-y^{2}=1$, points $F_{1}$, $F_{2}$ are its two foci, and point $P$ is a point on the hyperbola. If $P F_{1} \\perp P F_{2}$, then the value of $|P F_{1}|+|P F_{2}|$ is?", "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))", "query_expressions": "Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2))", "answer_expressions": "2*sqrt(3)", "fact_spans": "[[[2, 20], [39, 40], [50, 53]], [[45, 49]], [[21, 29]], [[31, 38]], [[2, 20]], [[21, 44]], [[45, 56]], [[58, 81]]]", "query_spans": "[[[84, 109]]]", "process": "" }, { "text": "Given $A_{1}$, $A_{2}$ are the left and right vertices of the ellipse $\\frac{x^{2}}{a^{2}}+y^{2}=1(a>1)$, point $P$ is a point on the ellipse, the slope of line $PA_{1}$ is $k_{1}$, the slope of line $P A_{2}$ is $k_{2}$. If $k_{1} k_{2}=-\\frac{1}{4}$, then the real number $a$=?", "fact_expressions": "A1: Point;A2: Point;G: Ellipse;Expression(G) = (y^2 + x^2/a^2 = 1);a>1;LeftVertex(G) = A1;RightVertex(G) = A2;P: Point;PointOnCurve(P, G) = True;k1: Number;Slope(LineOf(P,A1)) = k1;k2: Number;Slope(LineOf(P,A2)) = k2;k1 * k2 = -1/4;a: Real", "query_expressions": "a", "answer_expressions": "2", "fact_spans": "[[[2, 9]], [[12, 19]], [[20, 56], [68, 70]], [[20, 56]], [[22, 56]], [[2, 62]], [[2, 62]], [[63, 67]], [[63, 73]], [[88, 95]], [[74, 95]], [[111, 118]], [[96, 118]], [[121, 147]], [[149, 154]]]", "query_spans": "[[[151, 156]]]", "process": "\\because the ellipse \\frac{x^2}{a^{2}}+y^{2}=1 (a>1) \\therefore A(-a,0), B(a,0), let P(x,y), \\because the slope of line PA_{1} is k_{1}, the slope of line PA_{2} is k_{2}, if k_{1}k_{2}=-\\frac{1}{4}. \\therefore \\frac{y-0}{x+a} \\cdot \\frac{y-0}{x-a} = -\\frac{1}{4}, simplifying gives: x^{2}+4y^{2}-a^{2}=0, i.e. \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{\\frac{a^{2}}{4}}=1 \\therefore a^{2}=4, yielding a=2." }, { "text": "The ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ has left and right foci $F_{1}$, $F_{2}$. The line $x=m$ passes through $F_{2}$ and intersects the ellipse at points $A$, $B$. Then the area of $\\triangle F_{1} A B$ is equal to?", "fact_expressions": "G: Ellipse;H: Line;m: Number;F1: Point;A: Point;B: Point;F2: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Expression(H) = (x = m);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, H);Intersection(H, G) = {A, B}", "query_expressions": "Area(TriangleOf(F1, A, B))", "answer_expressions": "3", "fact_spans": "[[[0, 37], [77, 79]], [[60, 67]], [[62, 67]], [[44, 51]], [[82, 85]], [[86, 89]], [[52, 59], [68, 75]], [[0, 37]], [[60, 67]], [[0, 59]], [[0, 59]], [[60, 75]], [[60, 91]]]", "query_spans": "[[[93, 120]]]", "process": "" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a, b>0)$, a line passing through $F_{2}$ intersects the right branch of the hyperbola at points $A$ and $B$, and $|A F_{1}|=2|A F_{2}|$, $\\angle A F_{1} F_{2}=\\angle F_{1} B F_{2}$. Then the eccentricity of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;a: Number;b: Number;G: Line;A: Point;F1: Point;F2: Point;B: Point;a>b;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F2, G);Intersection(G,RightPart(C)) = {A, B};Abs(LineSegmentOf(A, F1)) = 2*Abs(LineSegmentOf(A, F2));AngleOf(A, F1, F2) = AngleOf(F1, B, F2)", "query_expressions": "Eccentricity(C)", "answer_expressions": "2", "fact_spans": "[[[18, 77], [96, 99], [182, 188]], [[26, 77]], [[26, 77]], [[93, 95]], [[103, 106]], [[2, 9]], [[10, 17], [85, 92]], [[107, 110]], [[26, 77]], [[26, 77]], [[18, 77]], [[2, 83]], [[2, 83]], [[84, 95]], [[93, 112]], [[114, 136]], [[137, 180]]]", "query_spans": "[[[182, 194]]]", "process": "By the property of hyperbola, we know |AF₁| - |AF₂| = 2a, |F₁F₂| = 2c. Since |AF₁| = 2|AF₂|, it follows that |AF₁| = 4a, |AF₂| = 2a. Also, ∠AF₁F₂ = ∠F₁BF₂, and ∠F₁AF₂ = ∠F₁AB, so △F₁AF₂ ∼ △BAF₁, hence |AF₁|/|AB| = |AF₂|/|AF₁| = |F₁F₂|/|BF₁|, so |AB| = |AF₁|² / |AF₂| = (16a²)/(2a) = 8a, |BF₁| = (|AB||F₁F₂|)/|AF₁| = (8a·2c)/(4a) = 4c. Since |AB| = |AF₂| + |BF₂|, it follows that |BF₂| = 6a. Also, |BF₁| - |BF₂| = 2a, so |BF₁| = 8a, thus 8a = 4c, so e = c/a = 2." }, { "text": "Point $P(a, b)$ lies on the right branch of the hyperbola $x^{2}-y^{2}=1$, and the distance from $P$ to the asymptotes is $\\sqrt{2}$, then $a+b=$?", "fact_expressions": "P: Point;a: Number;b: Number;Coordinate(P) = (a, b);G: Hyperbola;Expression(G) = (x^2 - y^2 = 1);PointOnCurve(P, RightPart(G));Distance(P,Asymptote(G)) = (sqrt(2))", "query_expressions": "a + b", "answer_expressions": "1/2", "fact_spans": "[[[0, 10], [36, 39]], [[1, 10]], [[1, 10]], [[0, 10]], [[11, 29]], [[11, 29]], [[0, 34]], [[11, 56]]]", "query_spans": "[[[58, 65]]]", "process": "" }, { "text": "The equation of the hyperbola that shares common foci with the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{5}=1$ and passes through the point $(3 \\sqrt{2}, 2)$ is?", "fact_expressions": "H: Ellipse;Expression(H) = (x^2/25 + y^2/5 = 1);Focus(H) = Focus(G);I: Point;Coordinate(I) = (3*sqrt(2), 2);PointOnCurve(I, G);G: Hyperbola", "query_expressions": "Expression(G)", "answer_expressions": "x^2/12-y^2/8=1", "fact_spans": "[[[1, 39]], [[1, 39]], [[0, 69]], [[47, 65]], [[47, 65]], [[46, 69]], [[66, 69]]]", "query_spans": "[[[66, 73]]]", "process": "\\because the hyperbola and the ellipse \\frac{x2}{25}+\\frac{y^{2}}{5}=1 have common foci. \\therefore the foci coordinates of the hyperbola are F_{1}(-2\\sqrt{5},0),F_{2}(2\\sqrt{5},0). Let the equation of the hyperbola be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{20-a^{2}}=1(00)$ have focus $F$, and let line $l$ passing through point $F$ with inclination angle $\\frac{\\pi}{4}$ intersect the parabola at points $A$ and $B$, where $|A B|=4$. Then the equation of the parabola is?", "fact_expressions": "l: Line;G: Parabola;p: Number;A: Point;B: Point;F: Point;p>0;Expression(G) = (y^2 = 2*p*x);Focus(G) = F;PointOnCurve(F, l);Inclination(l) = pi/4;Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, B)) = 4", "query_expressions": "Expression(G)", "answer_expressions": "y^2=2*x", "fact_spans": "[[[56, 61]], [[1, 22], [62, 65], [90, 93]], [[4, 22]], [[68, 71]], [[72, 75]], [[31, 35], [26, 29]], [[4, 22]], [[1, 22]], [[1, 29]], [[30, 61]], [[36, 61]], [[56, 77]], [[78, 87]]]", "query_spans": "[[[90, 98]]]", "process": "Analysis: From the coordinates of the focus, write the equation of line AB as $ y = x - \\frac{p}{2} $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Substitute the line equation into the parabola equation, rearrange, and use Vieta's formulas to obtain $ x_{1} + x_{2} $. Then, by the definition of a parabola, express the focal chord length as $ x_{1} + x_{2} + p $, thereby solving for $ p $. Detailed solution: The equation of line AB is $ y = x - \\frac{p}{2} $. Substituting into the parabola equation and simplifying yields $ x^{2} - 3px + \\frac{p^{2}}{4} = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = 3p $. Also, $ |AB| = x_{1} + x_{2} + p $, so $ 3p + p = 4 $, $ p = 1 $. Therefore, the equation of the parabola is $ y^2 = 2x $." }, { "text": "Given the parabola $y^{2}=4x$, $P$ is a point on the parabola, $F$ is the focus, and $|PF|=5$. Then the coordinates of point $P$ are?", "fact_expressions": "G: Parabola;P: Point;F: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(P, G);Focus(G)=F;Abs(LineSegmentOf(P, F)) = 5", "query_expressions": "Coordinate(P)", "answer_expressions": "{(4,4),(4,-4)}", "fact_spans": "[[[2, 16], [23, 26]], [[18, 22], [49, 53]], [[30, 33]], [[2, 16]], [[18, 29]], [[23, 36]], [[38, 47]]]", "query_spans": "[[[49, 58]]]", "process": "" }, { "text": "The line $l$ intersects the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ at points $A$ and $B$. If $M(1,-1)$ is the midpoint of segment $AB$, then the slope of line $l$ is?", "fact_expressions": "l: Line;G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);Intersection(l, G) = {A, B};B: Point;A: Point;M: Point;Coordinate(M)=(1,-1);MidPoint(LineSegmentOf(A,B)) = M", "query_expressions": "Slope(l)", "answer_expressions": "3/4", "fact_spans": "[[[0, 5], [77, 82]], [[6, 43]], [[6, 43]], [[0, 53]], [[48, 51]], [[44, 47]], [[55, 64]], [[55, 64]], [[55, 75]]]", "query_spans": "[[[77, 87]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), so \\begin{cases}3x_{1}^2+4y_{1}^2=12\\\\3x_{2}^2+4y_{2}^2=12\\end{cases}, \\therefore 3(x_{1}+x_{2})(x_{1}-x_{2})+4(y_{1}+y_{2})(y_{1}-y_{2})=0, so 3\\times2\\times1\\times(x_{1}-x_{2})+4\\times2\\times(-1)\\times(y_{1}-y_{2})=0, therefore 6(x_{1}-x_{2})-8(y_{1}-y_{2})=0, \\therefore \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=k=\\frac{3}{4}, so the slope of line l is \\frac{3}{4}." }, { "text": "If point $H(-2,4)$ lies on the directrix of the parabola $y^{2}=2 p x$, then the value of the real number $p$ is?", "fact_expressions": "G: Parabola;p: Real;H: Point;Expression(G) = (y^2 = 2*(p*x));Coordinate(H) = (-2, 4);PointOnCurve(H, Directrix(G))", "query_expressions": "p", "answer_expressions": "4", "fact_spans": "[[[12, 28]], [[34, 39]], [[1, 11]], [[12, 28]], [[1, 11]], [[1, 32]]]", "query_spans": "[[[34, 43]]]", "process": "The directrix equation of the parabola y^{2}=2px is x=-\\frac{p}{2}. According to the problem, we have -\\frac{p}{2}=-2, solving gives p=4." }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, $c=\\sqrt{a^{2}-b^{2}}$, the circle $(x-c)^{2}+y^{2}=c^{2}$ intersects the ellipse at exactly two common points. Then the range of values for the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Circle;c: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 + (-c + x)^2 = c^2);c=sqrt(a^2-b^2);NumIntersection(H,G)=2", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1/2,1)", "fact_spans": "[[[2, 54], [104, 106], [116, 118]], [[4, 54]], [[4, 54]], [[79, 103]], [[56, 78]], [[4, 54]], [[4, 54]], [[2, 54]], [[79, 103]], [[56, 78]], [[79, 114]]]", "query_spans": "[[[116, 129]]]", "process": "" }, { "text": "Given that $O$ is the coordinate origin, $F$ is the focus of the parabola $y^{2}=4x$, the line $l$: $y=m(x-1)$ intersects the parabola at points $A$ and $B$, point $A$ lies in the first quadrant, and $|FA|=3|FB|$, then what is the value of $m$?", "fact_expressions": "O: Origin;F: Point;Focus(G) = F;G: Parabola;Expression(G) = (y^2 = 4*x);l: Line;Expression(l) = (y = m*(x - 1));m: Number;Intersection(l, G) = {A, B};A: Point;B: Point;Quadrant(A) = 1;Abs(LineSegmentOf(F, A)) = 3*Abs(LineSegmentOf(F, B))", "query_expressions": "m", "answer_expressions": "sqrt(3)", "fact_spans": "[[[2, 5]], [[11, 14]], [[11, 32]], [[15, 29], [51, 54]], [[15, 29]], [[33, 50]], [[33, 50]], [[93, 96]], [[33, 65]], [[56, 59], [66, 70]], [[60, 63]], [[66, 75]], [[77, 91]]]", "query_spans": "[[[93, 100]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). By solving the system of equations consisting of the line equation and the ellipse equation, eliminating one variable, and applying Vieta's formulas, we obtain y_{1}+y_{2} and y_{1}y_{2}. Using the given condition y_{1}=-3y_{2}, the coordinates of intersection points can be found, thus determining m. Solution: Let A(x_{1},y_{1}), B(x_{2},y_{2}). Solve the system \\begin{cases}y=m(x-1)\\\\y2=4x\\end{cases}. Eliminating x gives my^{2}-4y-4m=0. Then y_{1}+y_{2}=\\frac{4}{m}, y_{1}y_{2}=-4. From |AF|=3|BF|, we get y_{1}=-3y_{2}. Therefore, -2y_{2}=\\frac{4}{m}, 3y^{2}. Solving yields m=\\sqrt{3} (discarding m=-\\sqrt{3})." }, { "text": "The distance from the focus of the parabola $x^{2}=8 y$ to the asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ is?", "fact_expressions": "H: Parabola;Expression(H) = (x^2 = 8*y);G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1)", "query_expressions": "Distance(Focus(H),Asymptote(G))", "answer_expressions": "1", "fact_spans": "[[[0, 14]], [[0, 14]], [[18, 46]], [[18, 46]]]", "query_spans": "[[[0, 55]]]", "process": "The focus of the parabola $x^{2}=8y$ is $(0,2)$, the asymptote of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ is $y=\\sqrt{3}x$, the required distance is $d=\\frac{|2|}{\\sqrt{3+1}}=1$." }, { "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, let point $P$ be any point on the parabola $C$. If point $A(4,2)$, then the minimum value of $|PF| + |PA|$ is?", "fact_expressions": "C: Parabola;A: Point;P: Point;F: Point;Expression(C) = (y^2 = 4*x);Coordinate(A) = (4, 2);Focus(C) = F;PointOnCurve(P, C)", "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))", "answer_expressions": "5", "fact_spans": "[[[2, 20], [33, 39]], [[46, 55]], [[28, 32]], [[24, 27]], [[2, 20]], [[46, 55]], [[2, 27]], [[28, 44]]]", "query_spans": "[[[57, 76]]]", "process": "The directrix of the parabola $ C: y^{2} = 4x $ is $ x = -1 $. Let point $ P $ have its projection on the directrix at point $ D $. Then, according to the definition of a parabola, we know $ |PF| = |PD| $. To minimize $ |PA| + |PF| $, it is equivalent to minimizing $ |PA| + |PD| $. The sum $ |PA| + |PD| $ reaches its minimum when points $ D $, $ P $, and $ A $ are collinear, and the minimum value is $ 4 - (-1) = 5 $." }, { "text": "Given that $F$ is the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, points $A$ and $B$ lie on its two asymptotes respectively, satisfying $\\overrightarrow{B F}=2 \\overrightarrow{F A}$, $\\overrightarrow{O A} \\cdot \\overrightarrow{A B}=0$ ($O$ being the origin), then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;B: Point;F: Point;A: Point;O: Origin;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;L1:Line;L2:Line;Asymptote(G)={L1,L2};PointOnCurve(A,L1);PointOnCurve(B,L2);VectorOf(B,F)=2*VectorOf(F,A);DotProduct(VectorOf(O, A), VectorOf(A, B)) = 0", "query_expressions": "Eccentricity(G)", "answer_expressions": "2*sqrt(3)/3", "fact_spans": "[[[6, 52], [68, 69], [193, 196]], [[9, 52]], [[9, 52]], [[62, 65]], [[2, 5]], [[57, 61]], [[180, 184]], [[6, 52]], [[2, 56]], [], [], [[68, 74]], [[57, 75]], [[57, 75]], [[79, 124]], [[126, 179]]]", "query_spans": "[[[193, 202]]]", "process": "" }, { "text": "The line passing through the left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ and perpendicular to the $x$-axis intersects the hyperbola at points $M$ and $N$. The circle with diameter $MN$ passes exactly through the right vertex of the hyperbola. Then the eccentricity of the hyperbola is equal to?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;M: Point;N: Point;C:Circle;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(LeftFocus(G), H);IsPerpendicular(H,xAxis);Intersection(H, G) = {M, N};IsDiameter(LineSegmentOf(M,N),C);PointOnCurve(RightVertex(G),C)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[1, 57], [73, 76], [103, 106], [112, 115]], [[4, 57]], [[4, 57]], [[70, 72]], [[79, 82]], [[83, 86]], [[99, 100]], [[4, 57]], [[4, 57]], [[1, 57]], [[0, 72]], [[62, 72]], [[70, 88]], [[89, 100]], [[99, 110]]]", "query_spans": "[[[112, 122]]]", "process": "" }, { "text": "Given an ellipse with foci on the $y$-axis, $\\frac{x^{2}}{9}+\\frac{y^{2}}{4+k}=1$, whose eccentricity is $\\frac{4}{5}$, find the value of $k$.", "fact_expressions": "G: Ellipse;k: Number;Expression(G) = (x^2/9 + y^2/(k + 4) = 1);PointOnCurve(Focus(G), yAxis);Eccentricity(G)= (4/5)", "query_expressions": "k", "answer_expressions": "21", "fact_spans": "[[[10, 49]], [[69, 72]], [[10, 49]], [[2, 49]], [[10, 67]]]", "query_spans": "[[[69, 76]]]", "process": "From the problem, we know that $a^{2}=4+k$, $b^{2}=9$, and $c^{2}=a^{2}-b^{2}$, so $c^{2}=k-5$. Also, since $e=\\frac{c}{a}=\\frac{4}{5}$, it follows that $e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{16}{25}$, that is: $\\frac{k-5}{4+k}=\\frac{16}{25}$. Solving gives $k=2$." }, { "text": "If the length of the major axis of an ellipse is 2 times the length of its minor axis, and one of its foci is $F_{1}(0,-3)$, then the standard equation of the ellipse is?", "fact_expressions": "G: Ellipse;F1: Point;Coordinate(F1) = (0, -3);OneOf(Focus(G))=F1;Length(MajorAxis(G)) = 2 * Length(MinorAxis(G))", "query_expressions": "Expression(G)", "answer_expressions": "y^2/12 + x^2/3 = 1", "fact_spans": "[[[1, 3], [40, 42], [18, 19]], [[25, 38]], [[25, 38]], [[18, 38]], [[1, 17]]]", "query_spans": "[[[40, 49]]]", "process": "According to the problem, assume the equation of the ellipse is \\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1 (a>b>0), then \\begin{cases}2a=4b\\\\c=3\\\\a=b^{2}+c^{2}\\end{cases}. Solving gives \\begin{cases}a=2\\sqrt{3}\\\\b=\\sqrt{3}\\end{cases}, so the equation of the ellipse is \\frac{y^{2}}{12}+\\frac{x^{2}}{3}=1" }, { "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ passes through the point $(2, \\sqrt{3})$, what is the eccentricity of the hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(H) = (2, sqrt(3));PointOnCurve(H, OneOf(Asymptote(G)))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(7)/2", "fact_spans": "[[[2, 58], [83, 86]], [[5, 58]], [[5, 58]], [[65, 81]], [[5, 58]], [[5, 58]], [[2, 58]], [[65, 81]], [[2, 81]]]", "query_spans": "[[[83, 92]]]", "process": "From the condition, it follows that the line y = \\frac{b}{a}x passes through the point (2,\\sqrt{3}), then we obtain \\frac{b}{a} = \\frac{\\sqrt{3}}{2}, and thus get the answer. [Detailed solution] The asymptotes of the hyperbola \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 are given by y = \\pm\\frac{b}{a}x. Therefore, the line y = \\frac{b}{a}x passes through the point (2,\\sqrt{3}); substituting gives \\frac{b}{a} = \\frac{\\sqrt{3}}{2}. Hence, e = \\sqrt{1 + (\\frac{b}{a})^{2}} = \\frac{\\sqrt{7}}{2}." }, { "text": "Let the two foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ be $F_{1}$ and $F_{2}$, respectively. Point $P$ lies on the ellipse such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$ and $\\tan \\angle P F_{1} F_{2}=2$. Then the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G) = True;DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Tan(AngleOf(P, F1, F2)) = 2", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)/3", "fact_spans": "[[[1, 53], [82, 84], [178, 180]], [[1, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[61, 68]], [[69, 76]], [[1, 76]], [[77, 81]], [[77, 85]], [[87, 146]], [[147, 176]]]", "query_spans": "[[[178, 187]]]", "process": "" }, { "text": "Given the hyperbola $C: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{1}$ intersects the two asymptotes of $C$ at points $A$ and $B$ respectively. If $3 \\overrightarrow{F_{1} A}=2 \\overrightarrow{A B}$ and $\\overrightarrow{F_{1} B} \\cdot \\overrightarrow{F_{2} B}=0$, then the eccentricity of $C$ is?", "fact_expressions": "C: Hyperbola;a: Number;b: Number;a > 0;b > 0;Expression(C) = (x^2/a^2 - y^2/b^2 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;l: Line;PointOnCurve(F1, l);l1, l2: Line;Asymptote(C) = {l1, l2};A: Point;B: Point;Intersection(l, l1) = A;Intersection(l, l2) = B;3*VectorOf(F1, A) = 2*VectorOf(A, B);DotProduct(VectorOf(F1, B), VectorOf(F2, B)) = 0", "query_expressions": "Eccentricity(C)", "answer_expressions": "4/3", "fact_spans": "[[[2, 61], [97, 100], [238, 241]], [[5, 61]], [[5, 61]], [[5, 61]], [[5, 61]], [[2, 61]], [[69, 76], [86, 93]], [[77, 84]], [[2, 84]], [[2, 84]], [[94, 96]], [[85, 96]], [[101, 106]], [[97, 106]], [[110, 113]], [[116, 119]], [[94, 121]], [[94, 121]], [[123, 174]], [[177, 236]]]", "query_spans": "[[[238, 247]]]", "process": "" }, { "text": "The focus of the parabola $y^{2}=2 p x(p>0)$ is $F$, points $A$ and $B$ lie on the parabola such that $\\angle A F B=\\frac{\\pi}{2}$. The projection of the midpoint $M$ of chord $AB$ onto the directrix is $N$. Then the maximum value of $\\frac{|M N|}{|A B|}$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;A: Point;B: Point;PointOnCurve(A,G) = True;PointOnCurve(B, G) = True;AngleOf(A, F, B) = pi/2;IsChordOf(LineSegmentOf(A, B), G) = True;MidPoint(LineSegmentOf(A,B)) = M;M: Point;Projection(M,Directrix(G)) = N;N: Point", "query_expressions": "Max(Abs(LineSegmentOf(M, N))/Abs(LineSegmentOf(A, B)))", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[0, 21], [40, 43], [87, 88]], [[0, 21]], [[3, 21]], [[3, 21]], [[25, 28]], [[0, 28]], [[30, 34]], [[36, 39]], [[30, 44]], [[30, 44]], [[46, 74]], [[40, 80]], [[76, 86]], [[83, 86]], [[83, 99]], [[96, 99]]]", "query_spans": "[[[101, 128]]]", "process": "" }, { "text": "Given that $F$ is a focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, if the line $y=k x$ intersects the ellipse at points $A$ and $B$, and $\\angle A F B=135^{\\circ}$, denote the eccentricity of the ellipse as $e$, then the range of values for $e^{2}$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;OneOf(Focus(G)) = F;H: Line;Expression(H) = (y = k*x);k: Number;A: Point;B: Point;Intersection(H, G) = {A, B};AngleOf(A, F, B) = ApplyUnit(135, degree);e: Number;Eccentricity(G) = e", "query_expressions": "Range(e^2)", "answer_expressions": "((2-sqrt(2))/4, 1)", "fact_spans": "[[[6, 58], [75, 77], [119, 121]], [[6, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[2, 5]], [[2, 63]], [[65, 74]], [[65, 74]], [[67, 74]], [[80, 83]], [[84, 87]], [[65, 89]], [[91, 117]], [[126, 129]], [[119, 129]]]", "query_spans": "[[[131, 145]]]", "process": "Let F be the other focus of the ellipse. As shown in the figure, connect AF, BF, BF, AF. According to the symmetry of the ellipse and the line, it can be seen that quadrilateral AFBF is a parallelogram. Moreover, since \\angle AFB = 135^{\\circ}, it follows that \\angle FAF = 45^{\\circ}. In \\triangle AFF, |FF|^{2} = |AF| + |AF|^{2} - 2|AF|\\cdot|AF|\\cos\\angle FAF = (|AF| + |AF|)^{2} - (2+\\sqrt{2})\\times|AF|\\cdot|AF|. Therefore, (|AF| + |AF|)^{2} - (2+\\sqrt{2})\\times\\left(\\frac{|AF| + |AF|}{2}\\right)^{2} \\leqslant (|FF|)^{2}, with equality holding if and only if |AF| = |AF|, i.e., \\frac{2-\\sqrt{2}}{4} \\leqslant \\left(\\frac{|FF|}{|AF| + |AF|}\\right)^{2}. Furthermore, since |FF| = 2c and |AF| + |AF| = 2a, it follows that e^{2} \\geqslant \\frac{2-\\sqrt{2}}{4}. Also, because e^{2} < 1, we have \\frac{2-\\sqrt{2}}{4} \\leqslant e^{2} < 1: \\frac{2-\\sqrt{2}}{4} < e^{2} < 1." }, { "text": "The moving point $P$ travels along the parabola $y=2 x^{2}+1$. Then, the equation of the trajectory of the midpoint $M$ of the segment joining point $P$ and point $A(0,-1)$ is?", "fact_expressions": "G: Parabola;A: Point;P: Point;M:Point;Expression(G) = (y = 2*x^2 + 1);Coordinate(A) = (0, -1);PointOnCurve(P, G);MidPoint(LineSegmentOf(P,A))=M", "query_expressions": "LocusEquation(M)", "answer_expressions": "Preserve all mathematical expressions, LaTeX formulas, symbols, and equations EXACTLY. \nDo not add explanations, do not rewrite formulas, do not omit any content. \n\nText:", "fact_spans": "[[[6, 22]], [[32, 42]], [[2, 5], [27, 31]], [[48, 51]], [[6, 22]], [[32, 42]], [[2, 25]], [[27, 51]]]", "query_spans": "[[[48, 58]]]", "process": "" }, { "text": "Through the focus $F$ of the parabola $C$: $y^{2}=4x$, draw a line with slope $\\frac{\\sqrt{3}}{3}$ intersecting the parabola $C$ at points $A$ and $B$. Then $|AB|=$?", "fact_expressions": "C: Parabola;G: Line;A: Point;B: Point;F:Point;Expression(C) = (y^2 = 4*x);Focus(C)=F;PointOnCurve(F,G);Slope(G) = sqrt(3)/3;Intersection(G,C)={A,B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "16", "fact_spans": "[[[1, 20], [55, 61]], [[52, 54]], [[63, 66]], [[68, 71]], [[23, 26]], [[1, 20]], [[1, 26]], [[0, 54]], [[27, 54]], [[52, 73]]]", "query_spans": "[[[75, 84]]]", "process": "Substitute the equation of the line into the equation of the parabola, use the focal chord length formula of the parabola, and combine Vieta's formulas to obtain the result. From the equation of the parabola, we know: F(1,0), then the equation of line AB is: y=\\frac{\\sqrt{3}}{3}(x-1). Substituting into the equation of the parabola gives: \\frac{1}{3}(x^{2}-2x+1)=4x, simplifying yields: x^{2}-14x+1=0, \\therefore|AB|=x_{1}+x_{2}+2=14+2=16." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, point $M$ lies on $E$, $M F_{1}$ is perpendicular to the $x$-axis, $\\sin \\angle M F_{2} F_{1}=\\frac{1}{3}$, then the eccentricity of $E$ is?", "fact_expressions": "E: Hyperbola;b: Number;a: Number;M: Point;F1: Point;F2: Point;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(E)=F1;RightFocus(E)=F2;PointOnCurve(M, E);IsPerpendicular(LineSegmentOf(M,F1),xAxis);Sin(AngleOf(M, F2, F1)) = 1/3", "query_expressions": "Eccentricity(E)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[18, 68], [80, 83], [143, 146]], [[25, 68]], [[25, 68]], [[75, 79]], [[2, 9]], [[10, 17]], [[18, 68]], [[2, 74]], [[2, 74]], [[75, 84]], [[85, 101]], [[102, 141]]]", "query_spans": "[[[143, 152]]]", "process": "" }, { "text": "The shortest distance from a point $P$ on the parabola $y^{2}=4 x$ to its focus $F$ is?", "fact_expressions": "G: Parabola;P: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(P, G) ;Focus(G) = F;F:Point", "query_expressions": "Min(Distance(P, F))", "answer_expressions": "1", "fact_spans": "[[[0, 14], [21, 22]], [[16, 20]], [[0, 14]], [[0, 20]], [[21, 28]], [[25, 28]]]", "query_spans": "[[[16, 35]]]", "process": "Problem Analysis: P(x_{0},y_{0}), (x_{0}\\geqslant0), according to the focal radius formula |PF|=x_{0}+1\\geqslant1." }, { "text": "Hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has left and right foci $F_{1}$, $F_{2}$ respectively. Points $M$, $N$ lie on the left and right branches of the hyperbola respectively, and $M N \\parallel F_{1} F_{2}$, $|M N|=\\frac{1}{2}|F_{1} F_{2}|$. The line segment $F_{1} N$ intersects hyperbola $C$ at point $Q$, and $|F_{1} Q|=\\frac{2}{5}|F_{1} N|$. Then the eccentricity of this hyperbola is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;M: Point;N: Point;PointOnCurve(M, LeftPart(C)) = True;PointOnCurve(N, RightPart(C)) = True;IsParallel(LineSegmentOf(M, N), LineSegmentOf(F1, F2)) = True;Abs(LineSegmentOf(M, N)) = Abs(LineSegmentOf(F1, F2))/2;Intersection(LineSegmentOf(F1, N), C) = Q;Q: Point;Abs(LineSegmentOf(F1, Q)) = (2/5)*Abs(LineSegmentOf(F1, N))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(7)", "fact_spans": "[[[0, 61], [97, 100], [177, 183], [224, 227]], [[0, 61]], [[7, 61]], [[7, 61]], [[7, 61]], [[7, 61]], [[70, 77]], [[78, 85]], [[0, 85]], [[0, 85]], [[86, 90]], [[91, 94]], [[86, 106]], [[86, 106]], [[108, 129]], [[131, 163]], [[165, 188]], [[184, 188]], [[189, 221]]]", "query_spans": "[[[224, 233]]]", "process": "Analysis: Using the symmetry of the hyperbola, the coordinates of point N can be set according to the conditions. The coordinates of point Q can be obtained by the midpoint coordinate formula. Then, since points N and Q lie on the hyperbola, a relationship among a, b, c is obtained, thus giving the eccentricity of the hyperbola. ∴ |MN| = c. Since MN // F₁F₂, we can set N(c/2, t). ∵ |F₁Q| = (2/5)|F₁N|, ∴ the coordinates of point Q are (−2c/5, 2t/5). ∵ points N and Q lie on the hyperbola, c²/(4a²) − t²/b² = 1. Eliminating t and simplifying gives c²/a² = 7, ∴ the eccentricity e = √7." }, { "text": "Let the left and right foci of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{6}=1$ be $F_{1}$ and $F_{2}$, respectively. The line $l$ passing through $F_{1}$ intersects the left branch of the hyperbola at points $A$ and $B$. Then the minimum value of $|AF_{2}|+|BF_{2}|$ is equal to?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/6 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, l) = True;l: Line;Intersection(l, LeftPart(G)) = {A, B};A: Point;B: Point", "query_expressions": "Min(Abs(LineSegmentOf(A, F2)) + Abs(LineSegmentOf(B, F2)))", "answer_expressions": "16", "fact_spans": "[[[1, 39], [78, 81]], [[1, 39]], [[47, 54], [64, 71]], [[55, 62]], [[1, 62]], [[1, 62]], [[63, 77]], [[72, 77]], [[72, 93]], [[84, 87]], [[88, 91]]]", "query_spans": "[[[95, 121]]]", "process": "Test analysis: |AF_{2}|+|BF_{2}|=2a+|AF_{1}|+2a+|BF_{1}|=4a+|AB|\\geqslant4a+\\frac{2b^{2}}{a}=4\\times3+\\frac{2\\times6}{3}=16" }, { "text": "Given the ellipse $C_{1}$: $\\frac{x^{2}}{36}+\\frac{y^{2}}{b^{2}}=1$ has foci $F_{1}$, $F_{2}$, and $F_{2}$ is the focus of the parabola $C_{2}$: $y^{2}=2 p x(p>0)$. If $P$ is an intersection point of $C_{1}$ and $C_{2}$, and $|P F_{1}|=7$, then the value of $\\cos \\angle P F_{1} F_{2}$ is?", "fact_expressions": "C1: Ellipse;Expression(C1) = (x^2/36 + y^2/b^2 = 1);b: Number;F1: Point;F2: Point;Focus(C1) = {F1, F2};C2: Parabola;Expression(C2) = (y^2 = 2*p*x);p: Number;p > 0;Focus(C2) = F2;P: Point;Intersection(C1, C2) = P;Abs(LineSegmentOf(P, F1)) = 7", "query_expressions": "Cos(AngleOf(P, F1, F2))", "answer_expressions": "5/7", "fact_spans": "[[[2, 53], [122, 129]], [[2, 53]], [[13, 53]], [[59, 66]], [[67, 74], [76, 83]], [[2, 74]], [[84, 114], [130, 137]], [[84, 114]], [[96, 114]], [[96, 114]], [[76, 116]], [[118, 121]], [[118, 140]], [[142, 155]]]", "query_spans": "[[[158, 189]]]", "process": "According to the problem, by the definition of an ellipse, we have |PF₁| + |PF₂| = 12. Given |PF₁| = 7, then |PF₂| = 5. Since point F₂ is the focus of the parabola C₂: y² = 2px (p > 0), the directrix l of this parabola passes through point F₁. As shown in the figure, draw PQ ⊥ l from point P, meeting l at Q. By the definition of a parabola, we know |PQ| = |PF₂| = 5. Since F₁F₂ // PQ, it follows that ∠PF₁F₂ = ∠F₁PQ. Therefore, cos∠PF₁F₂ = cos∠F₁PQ = |PQ| / |PF| = 5 / 7." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has eccentricity $\\sqrt{3}$, then the asymptote equations of hyperbola $C$ are?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(C) = sqrt(3)", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "y=pm*sqrt(2)*x", "fact_spans": "[[[2, 63], [80, 86]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 78]]]", "query_spans": "[[[80, 94]]]", "process": "Since the eccentricity of the hyperbola is $\\sqrt{3}$, from $e = \\frac{c}{a} = \\sqrt{1 + \\left(\\frac{b}{a}\\right)^2} = \\sqrt{3}$, solve as follows. Because the hyperbola $C: \\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1$ ($a > 0$, $b > 0$) has eccentricity $\\sqrt{3}$, so $e = \\frac{c}{a} = \\sqrt{1 + \\left(\\frac{b}{a}\\right)^2} = \\sqrt{3}$. Solving gives $\\frac{b}{a} = \\pm\\sqrt{2}$. Therefore, the asymptotes of hyperbola $C$ are $y = \\pm\\sqrt{2}x$." }, { "text": "If the asymptotes of a hyperbola with foci on the $x$-axis are given by $y=\\pm \\frac{\\sqrt{3}}{2} x$, then what is the eccentricity of this hyperbola?", "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), xAxis);Expression(Asymptote(G)) = (y = pm*x*(sqrt(3)/2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(7)/2", "fact_spans": "[[[10, 13], [51, 54]], [[1, 13]], [[10, 48]]]", "query_spans": "[[[51, 60]]]", "process": "\\because the asymptotes of a hyperbola with foci on the x-axis are given by y=\\pm\\frac{\\sqrt{3}}{2}x, \\therefore\\frac{b}{a}=\\frac{\\sqrt{3}}{2}; then the eccentricity is e=\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{1+\\frac{3}{4}}=\\frac{\\sqrt{7}}{2}" }, { "text": "$F_{1}$, $F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and $P$ is a point on the hyperbola such that $|P F_{1}| \\cdot |P F_{2}|=32$. Then what is the measure of $\\angle F_{1} P F_{2}$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 32", "query_expressions": "AngleOf(F1, P, F2)", "answer_expressions": "ApplyUnit(90, degree)", "fact_spans": "[[[16, 55], [65, 68]], [[16, 55]], [[0, 7]], [[8, 15]], [[0, 60]], [[61, 64]], [[61, 69]], [[72, 101]]]", "query_spans": "[[[103, 130]]]", "process": "From \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1 we get: c^{2}=9+16=25, so 4c^{2}=100. Let |PF_{1}|=m, |PF_{2}|=n, then |m-n|=6, mn=32, solving gives: m^{2}+n^{2}=100. In \\triangle F_{1}PF_{2}, since m^{2}+n^{2}=(2c)^{2}, that is, |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}, therefore \\triangle F_{1}PF_{2} is a right triangle with \\angle F_{1}PF_{2}=90^{\\circ}, i.e., \\angle F_{1}PF_{2}=90^{\\circ}. The answer is \\cdot OO." }, { "text": "If the distance from a focus of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1$ to one of its asymptotes is $2 \\sqrt{2}$, then the focal length of this hyperbola is equal to?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/b^2 = 1);b: Number;Distance(OneOf(Focus(G)), Asymptote(G)) = 2*sqrt(2)", "query_expressions": "FocalLength(G)", "answer_expressions": "6", "fact_spans": "[[[1, 33], [39, 40], [62, 65]], [[1, 33]], [[4, 33]], [[1, 59]]]", "query_spans": "[[[62, 71]]]", "process": "Solution: Given the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1$, the asymptotes of the hyperbola are $y=\\pm bx$. Without loss of generality, assume $y=-bx$, that is, $bx+y=0$. The coordinates of the focus are $F(c,0)$. Then the distance from the focus to its asymptote is $d=\\frac{bc}{\\sqrt{1+b^{2}}}=\\frac{bc}{c}=b=2\\sqrt{2}$. Therefore, the focal length of the hyperbola equals $2c=6$." }, { "text": "Let the focus of the parabola $x^{2}=12 y$ be $F$. A line $l$ passing through point $P(2,1)$ intersects the parabola at points $A$ and $B$, and point $P$ is exactly the midpoint of $AB$. Then $|A F|+|B F|$=?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 12*y);F: Point;Focus(G) = F;P: Point;Coordinate(P) = (2, 1);A: Point;B: Point;l: Line;PointOnCurve(P, l);Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = P", "query_expressions": "Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F))", "answer_expressions": "8", "fact_spans": "[[[1, 16], [42, 45]], [[1, 16]], [[20, 23]], [[1, 23]], [[26, 35], [58, 62]], [[26, 35]], [[48, 51]], [[52, 55]], [[36, 41]], [[25, 41]], [[36, 55]], [[58, 72]]]", "query_spans": "[[[74, 89]]]", "process": "" }, { "text": "Given the ellipse $ C $: $ \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 $ ($ a>b>0 $) has eccentricity $ \\frac{\\sqrt{2}}{2} $. If the maximum radius of a circle centered at $ N(0,2) $ that has common points with the ellipse $ C $ is $ \\sqrt{26} $, then what is the equation of the ellipse $ C $?", "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Circle;N: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(N) = (0, 2);Eccentricity(C) = sqrt(2)/2;IsIntersect(G,C) = True;Center(G) = N;Max(Radius(G)) = sqrt(26)", "query_expressions": "Expression(C)", "answer_expressions": "Preserve all mathematical expressions, LaTeX formulas, symbols, and equations EXACTLY. \nDo not add explanations, do not rewrite formulas, do not omit any content. \n\nText:", "fact_spans": "[[[2, 59], [100, 105], [131, 136]], [[8, 59]], [[8, 59]], [[110, 111]], [[87, 95]], [[8, 59]], [[8, 59]], [[2, 59]], [[87, 95]], [[2, 84]], [[99, 111]], [[86, 111]], [[110, 128]]]", "query_spans": "[[[131, 141]]]", "process": "" }, { "text": "Given that point $P$ lies on the parabola $x^{2}=4 y$, then when the sum of the distance from point $P$ to point $Q(1,2)$ and the distance from point $P$ to the focus of the parabola is minimized, what are the coordinates of point $P$?", "fact_expressions": "G: Parabola;Q: Point;P: Point;Expression(G) = (x^2 = 4*y);Coordinate(Q) = (1, 2);PointOnCurve(P, G);WhenMin(Distance(P, Q) + Distance(P, Focus(G)))", "query_expressions": "Coordinate(P)", "answer_expressions": "(1, 1/4)", "fact_spans": "[[[7, 21], [48, 51]], [[30, 39]], [[2, 6], [25, 29], [43, 47], [64, 68]], [[7, 21]], [[30, 39]], [[2, 22]], [[24, 63]]]", "query_spans": "[[[64, 73]]]", "process": "According to the definition of a parabola, the distance from point P to the focus of the parabola is equal to the distance from P to the directrix. Combining this with the graph, the position of P when PQ + PM reaches its minimum value can be determined, thus obtaining the answer. As shown in the figure: based on the definition of a parabola, the distance from point P to the focus of the parabola equals the distance from P to the directrix, that is, PF = PM. Therefore, the required expression becomes PQ + PF = PQ + PM. From the graph, it can be seen that when P moves to the point P' directly below Q, QP' + P'M is minimized, which equals QM'. Hence, $x_{p} = 1$. Substituting into the parabolic equation gives $y_{p} = \\frac{x_{p}^{2}}{4} = \\frac{1}{4}$. Therefore, the coordinates of point P are $(1, \\frac{1}{4})$." }, { "text": "Given that the vertex of the parabola $C$ is at the origin and its focus lies on the $x$-axis, the equation of the parabola $C$ has its vertex at the origin. The line $y = x$ intersects the parabola $C$ at points $A$ and $B$. If $P(2 , 2)$ is the midpoint of $AB$, then what is the equation of the parabola $C$?", "fact_expressions": "C: Parabola;G: Line;A: Point;B: Point;P: Point;O:Origin;Expression(G) = (y = x);Coordinate(P) = (2, 2);Vertex(C)=O;PointOnCurve(Focus(C), xAxis);Intersection(G, C) = {A, B};MidPoint(LineSegmentOf(A, B)) = P", "query_expressions": "Expression(C)", "answer_expressions": "y^2=4*x", "fact_spans": "[[[8, 14], [40, 46], [79, 85]], [[32, 39]], [[48, 51]], [[52, 55]], [[59, 69]], [[5, 7], [20, 22]], [[32, 39]], [[59, 69]], [[2, 14]], [[8, 31]], [[32, 57]], [[59, 77]]]", "query_spans": "[[[79, 90]]]", "process": "" }, { "text": "If the eccentricity of the hyperbola $\\frac{y^{2}}{16}-\\frac{x^{2}}{m}=1$ is $3$, then the equations of the asymptotes of this hyperbola are?", "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (y^2/16 - x^2/m = 1);Eccentricity(G) = 3", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y = pm*(sqrt(2)/4)*x", "fact_spans": "[[[1, 40], [51, 54]], [[4, 40]], [[1, 40]], [[1, 48]]]", "query_spans": "[[[51, 62]]]", "process": "Analysis: First, find the value of m, then solve for the asymptote equations. \nDetailed solution: From the hyperbola equation, we know: m > 0, and a^{2} = 16, b^{2} = m^{1}, then c^{2} = 16 + m. The eccentricity of the hyperbola is: e = \\sqrt{\\frac{c^{2}}{a^{2}}} = \\sqrt{\\frac{16 + m}{16}} = 3. Solving gives: m = 128. Then the asymptotes of the hyperbola satisfy: \\frac{y^{2}}{16} - \\frac{x^{2}}{128} = 0. Rearranging yields the asymptote equations: y = \\pm\\frac{\\sqrt{2}}{4}x" }, { "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, point $P(x_{0}, y_{0})$ lies on $C$, and $|PF|=5$, then $x_{0}=?$", "fact_expressions": "C: Parabola;P: Point;F: Point;x0:Number;y0:Number;Expression(C) = (y^2 = 4*x);Coordinate(P) = (x0, y0);Focus(C) = F;PointOnCurve(P, C);Abs(LineSegmentOf(P, F)) = 5", "query_expressions": "x0", "answer_expressions": "4", "fact_spans": "[[[2, 21], [48, 51]], [[29, 47]], [[25, 28]], [[65, 72]], [[30, 47]], [[2, 21]], [[29, 47]], [[2, 28]], [[29, 52]], [[54, 63]]]", "query_spans": "[[[65, 74]]]", "process": "The directrix equation of parabola C is x = -1. By the focal radius formula of a parabola, we have |PF| = x_{0} + 1 = 5, solving which yields x_{0} = 4." }, { "text": "The line $l$: $x-y+1=0$ and the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ intersect at points $A$ and $B$. Find the length of chord $|A B|=?$", "fact_expressions": "l: Line;G: Ellipse;A: Point;B: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Expression(l) = (x - y + 1 = 0);Intersection(l, G) = {A, B};IsChordOf(LineSegmentOf(A,B),G)", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "24/7", "fact_spans": "[[[0, 16]], [[17, 54]], [[57, 60]], [[61, 64]], [[17, 54]], [[0, 16]], [[0, 66]], [[17, 76]]]", "query_spans": "[[[69, 78]]]", "process": "Analysis: By solving the system of equations consisting of the line equation and the ellipse equation, and using the relationship between roots and coefficients along with the chord length formula, the answer can be obtained. \nDetailed solution: Let A(x₁, y₁), B(x₂, y₂). Solve the system \n\\begin{cases}x - y + 1 = 0 \\\\ \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1\\end{cases}, \nwhich simplifies to 7x² + 8x - 8 = 0, \n\\frac{x_{2} = -\\frac{8}{7}}{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}} = -\\frac{2}{1} = \\sqrt{2} \\cdot \\sqrt{\\left(-\\frac{8}{7}\\right)^{2} + 4 \\times \\frac{8}{7}} = \\frac{24}{7}" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the right focus is $F$. Point $P$ lies on the hyperbola $C$. If $|P F|=5 a$, $\\angle P F O=120^{\\circ}$, where $O$ is the coordinate origin, then the eccentricity of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(C) = F;P: Point;PointOnCurve(P, C);Abs(LineSegmentOf(P, F)) = 5*a;AngleOf(P, F, O) = ApplyUnit(120, degree);O: Origin", "query_expressions": "Eccentricity(C)", "answer_expressions": "3/2", "fact_spans": "[[[2, 63], [77, 83], [138, 144]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 71]], [[2, 71]], [[72, 76]], [[72, 84]], [[86, 97]], [[99, 125]], [[128, 131]]]", "query_spans": "[[[138, 150]]]", "process": "Let the left focus be F_{1}. By the definition of the hyperbola, we have |PF_{1}| = |PF| + 2a = 7a, |F_{1}F| = 2c, \\angle PFF_{1} = 120^{\\circ}. By the law of cosines, \\cos\\angle PFF_{1} = \\frac{|PF|^{2} + |FF_{1}|^{2} - |PF_{1}|^{2}}{2|PF||FF_{1}|}. Substituting the values gives -\\frac{1}{2} = \\frac{25a^{2} + 4c^{2} - 49a^{2}}{2 \\times 2c \\times 5a}. Simplifying yields 2c^{2} + 5ac - 24a^{2} = 0. Dividing through by a^{2} gives 2e^{2} + 5e - 24 = 0, which factors as (2e - 3)(e + 4) = 0. Therefore, e = \\frac{3}{2}. This problem examines the definition of a hyperbola and the method for finding eccentricity, as well as a simple application of the law of cosines; it is a medium-difficulty problem." }, { "text": "Let the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 ,(a>0, b>0)$ have a length of the imaginary axis equal to $2$ and a focal distance equal to $2 \\sqrt{3}$. Then the equation of this hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Length(ImageinaryAxis(G)) = 2;FocalLength(G) = 2*sqrt(3)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/2-y^2=1", "fact_spans": "[[[1, 59], [86, 89]], [[1, 59]], [[4, 59]], [[4, 59]], [[4, 59]], [[4, 59]], [[1, 67]], [[1, 83]]]", "query_spans": "[[[86, 94]]]", "process": "From the problem, we have 2b=2\\Rightarrow b=1, 2c=2\\sqrt{3}\\Rightarrow c=\\sqrt{3}. From c^{2}=a^{2}+b^{2}, we obtain a=\\sqrt{2}. Therefore, the equation of the hyperbola is: \\frac{x^{2}}{2}-y^{2}=1" }, { "text": "Given the parabola $y^{2}=4 x$ and point $A(0,1)$, let point $P$ be a moving point on the parabola, and let $d$ be the distance from $P$ to the directrix of the parabola. Then the minimum value of $|P A|+d$ is?", "fact_expressions": "G: Parabola;A: Point;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (0, 1);PointOnCurve(P, G);Distance(P,Directrix(G))=d;d:Number", "query_expressions": "Min(d + Abs(LineSegmentOf(P, A)))", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 16], [31, 34], [44, 47]], [[17, 25]], [[26, 30], [39, 42]], [[2, 16]], [[17, 25]], [[26, 38]], [[39, 56]], [[53, 56]]]", "query_spans": "[[[58, 73]]]", "process": "According to the definition of a parabola, d = |PF|, then (|PA| + d)_{\\min} = (|PA| + |PF|)_{\\min} = |AF| = \\sqrt{2}" }, { "text": "If the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1$ passes through the point $(-3 \\sqrt{2}, 2)$, then the focal length of this hyperbola is?", "fact_expressions": "G: Hyperbola;a: Number;H: Point;Expression(G) = (-y^2/4 + x^2/a^2 = 1);Coordinate(H) = (-3*sqrt(2), 2);PointOnCurve(H, G)", "query_expressions": "FocalLength(G)", "answer_expressions": "2*sqrt(13)", "fact_spans": "[[[1, 43], [67, 70]], [[4, 43]], [[44, 63]], [[1, 43]], [[44, 63]], [[1, 63]]]", "query_spans": "[[[67, 75]]]", "process": "" }, { "text": "Given that one of the asymptotes of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{4}=1$ is $y=x$, then the real number $m$ equals?", "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (-y^2/4 + x^2/m = 1);Expression(OneOf(Asymptote(G))) = (y = x)", "query_expressions": "m", "answer_expressions": "4", "fact_spans": "[[[2, 40]], [[56, 61]], [[2, 40]], [[2, 54]]]", "query_spans": "[[[56, 64]]]", "process": "Problem Analysis: From the asymptote equation, it can be seen that a = b, therefore m = 4" }, { "text": "The line passing through the focus of the hyperbola $x^{2}-y^{2}=1$ and perpendicular to the $x$-axis intersects the hyperbola at points $A$ and $B$. Then $|AB|=$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 1);H: Line;PointOnCurve(Focus(G), H);IsPerpendicular(H, xAxis);A: Point;B: Point;Intersection(H, G) = {A, B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "2", "fact_spans": "[[[1, 19], [36, 39]], [[1, 19]], [[32, 34]], [[0, 34]], [[24, 34]], [[40, 43]], [[44, 47]], [[32, 49]]]", "query_spans": "[[[51, 60]]]", "process": "The analytical process is omitted" }, { "text": "The equation of curve $C$ is $x^{2}+\\frac{y^{2}}{3}=1$, and for a point $P(x , y)$ on it, the maximum value of $3 x+y$ is?", "fact_expressions": "C: Curve;P: Point;Coordinate(P) = (x1, y1);Expression(C) = (x^2 + y^2/3 = 1);x1:Number;y1:Number;PointOnCurve(P, C)", "query_expressions": "Max(3*x1 + y1)", "answer_expressions": "2*sqrt(3)", "fact_spans": "[[[0, 5], [35, 36]], [[39, 49]], [[39, 49]], [[0, 34]], [[39, 49]], [[39, 49]], [[35, 49]]]", "query_spans": "[[[51, 64]]]", "process": "Let 3x+y=n, substitute into x^{2}+\\frac{y^{2}}{3}=1, eliminate y and simplify, we obtain 12x^{2}-6nx+n^{2}-3=0, A=36n^{2}-4\\times12(n^{2}-3)\\geqslant0, solving gives -2\\sqrt{3}\\leqslant n \\leqslant 2\\sqrt{3}. Answer: 2\\sqrt{3}" }, { "text": "Let the endpoints of the major axis of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ be $M$ and $N$, and let point $P$ on the ellipse be different from $M$ and $N$. Then the product of the slopes of $PM$ and $PN$ is?", "fact_expressions": "G: Ellipse;P: Point;M: Point;N: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Endpoint(MajorAxis(G))={M,N};PointOnCurve(P,G);Negation(M=P);Negation(N=P)", "query_expressions": "Slope(LineSegmentOf(P, M))*Slope(LineSegmentOf(P, N))", "answer_expressions": "Preserve all mathematical expressions, LaTeX formulas, symbols, and equations EXACTLY. \nDo not add explanations, do not rewrite formulas, do not omit any content. \n\nText:", "fact_spans": "[[[1, 38], [68, 70]], [[63, 67]], [[45, 48], [55, 58]], [[49, 52], [59, 62]], [[1, 38]], [[1, 52]], [[63, 71]], [[53, 67]], [[53, 67]]]", "query_spans": "[[[73, 92]]]", "process": "" }, { "text": "Let the focus of the parabola $y^{2}=2 p x$ ($p>0$) be $F$. Given that $A$ and $B$ are two moving points on the parabola such that $\\angle A F B=60^{\\circ}$, draw a perpendicular line $M N$ from the midpoint $M$ of chord $A B$ to the directrix of the parabola, with foot of perpendicular at $N$. Then the maximum value of $\\frac{|M N|}{|A B|}$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;A: Point;B: Point;M: Point;N: Point;F: Point;p>0;Focus(G)=F;PointOnCurve(A,G);PointOnCurve(B,G);AngleOf(A, F, B) = ApplyUnit(60, degree);IsChordOf(LineSegmentOf(A,B),G);MidPoint(LineSegmentOf(A,B))=M;IsPerpendicular(LineSegmentOf(M,N),Directrix(G));FootPoint(LineSegmentOf(M,N),Directrix(G))=N;PointOnCurve(M, LineSegmentOf(M, N))", "query_expressions": "Max(Abs(LineSegmentOf(M, N))/Abs(LineSegmentOf(A, B)))", "answer_expressions": "1", "fact_spans": "[[[1, 22], [40, 43], [94, 97]], [[1, 22]], [[4, 22]], [[32, 35]], [[36, 39]], [[90, 93]], [[112, 115]], [[26, 29]], [[4, 22]], [[1, 29]], [[32, 49]], [[32, 49]], [[53, 78]], [[40, 87]], [[81, 93]], [[94, 107]], [[94, 115]], [[80, 107]]]", "query_spans": "[[[117, 144]]]", "process": "Problem Analysis: Draw perpendiculars from A and B to the directrix, with feet of perpendiculars P and Q, respectively. From the figure, it follows that $ MN = \\frac{1}{2}(AP + BQ) $. According to the definition of a parabola, $ AP = AF $, $ BQ = BF $, so $ MN = \\frac{1}{2}(AF + BF) $. In triangle $ \\triangle ABF $, by the law of cosines, we have $ \\cos 60^{\\circ} = \\frac{AF^{2} + BF^{2} - AB^{2}}{2 \\cdot AF \\cdot BF} = \\frac{1}{2} $, hence $ (AF + BF)^{2} - 3 \\cdot AF \\cdot BF = AB^{2} $. Using the property of the AM-GM inequality, $ AF \\cdot BF \\leqslant \\left( \\frac{AF + BF}{2} \\right)^{2} $, thus the above expression can be transformed into $ \\frac{1}{4}(AF + BF)^{2} \\leqslant AB^{2} $, i.e., $ MN^{2} \\leqslant AB^{2} $, therefore $ \\frac{MN}{AB} \\leqslant 1 $." }, { "text": "Given circle $C_{1}$: $(x+2)^{2}+y^{2}=4$ and circle $C_{2}$: $(x-2)^{2}+y^{2}=36$, a moving circle $M$ is externally tangent to circle $C_{1}$ and internally tangent to circle $C_{2}$. Then the trajectory equation of the center $M$ of the moving circle is?", "fact_expressions": "C1: Circle;Expression(C1) = ((x + 2)^2 + y^2 = 4);C2: Circle;Expression(C2) = ((x - 2)^2 + y^2 = 36);M: Circle;IsOutTangent(M, C1);IsInTangent(M, C2);M1: Point;Center(M) = M1", "query_expressions": "LocusEquation(M1)", "answer_expressions": "x^2/16+y^2/12=1", "fact_spans": "[[[2, 30], [67, 75]], [[2, 30]], [[31, 60], [79, 87]], [[31, 60]], [[63, 66], [91, 93]], [[63, 77]], [[63, 89]], [[95, 98]], [[91, 98]]]", "query_spans": "[[[95, 105]]]", "process": "Let the center of the moving circle M be M(x, y), with radius r₁. According to the problem, MC₁ = r + 2, MC₂ = 6 - r, so MC₁ + MC₂ = 8 > 4. Therefore, the trajectory of M is an ellipse with foci at C₁ and C₂, where a = 4, c = 2, so b² = a² - c² = 12. Hence, the equation of the trajectory of the center M of the moving circle is \\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1" }, { "text": "Given the equation of ellipse $ C $ is $ \\frac{x^{2}}{9-k} + \\frac{y^{2}}{k-1} = 1 $. If the eccentricity of ellipse $ C $ is $ e = \\sqrt{\\frac{6}{7}} $, then what is the set of all possible values of $ k $?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/(9 - k) + y^2/(k - 1) = 1);Eccentricity(C) = e ;e= sqrt(6/7);e:Number;k:Number", "query_expressions": "Range(k)", "answer_expressions": "{2,8}", "fact_spans": "[[[2, 7], [52, 57]], [[2, 50]], [[52, 83]], [[61, 83]], [[61, 83]], [[85, 88]]]", "query_spans": "[[[85, 100]]]", "process": "Test analysis: When $9 - k > k - 1 > 0$, $\\frac{\\sqrt{9 - k - (k - 1)}}{\\sqrt{9 - k}} = \\sqrt{\\frac{6}{7}}$, solving gives $k = 2$. When $k - 1 > 9 - k > 0$, $\\frac{(k - 1) - (9 - k)}{\\sqrt{k - 1}} = \\sqrt{6}$, solving gives $k = 8$. Therefore, the values of $k$ are $\\{2, 8\\}$." }, { "text": "Given the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, $A$, $B$ are its right vertex and upper vertex respectively, $M$ is a point on the ellipse $E$. A perpendicular line from $M$ to the $x$-axis passes exactly through the left focus $F_{1}$ of the ellipse. If $AB \\parallel OM$, then what is the eccentricity of the ellipse $E$?", "fact_expressions": "E: Ellipse;b: Number;a: Number;A: Point;B: Point;O: Origin;M: Point;F1:Point;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);RightVertex(E)=A;UpperVertex(E)=B;LeftFocus(E)=F1;PointOnCurve(M, E);L:Line;PointOnCurve(M,L);IsPerpendicular(L,xAxis);PointOnCurve(F1,L);IsParallel(LineSegmentOf(A, B),LineSegmentOf(O,M))", "query_expressions": "Eccentricity(E)", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[2, 59], [74, 75], [88, 93], [114, 116], [144, 149]], [[9, 59]], [[9, 59]], [[61, 66]], [[68, 71]], [[129, 142]], [[99, 102], [84, 87]], [[120, 127]], [[9, 59]], [[9, 59]], [[2, 59]], [[61, 83]], [[61, 83]], [[114, 127]], [[84, 97]], [], [[98, 110]], [[98, 110]], [[98, 127]], [[129, 142]]]", "query_spans": "[[[144, 155]]]", "process": "According to the given conditions: A(a,0), B(0,b), M(-c,\\frac{b^{2}}{a}), then k_{AB}=-\\frac{b}{a}, k_{MO}=-\\frac{b^{2}}{ac}. Since AB//OM, then -\\frac{b}{a}=-\\frac{b^{2}}{ac}, hence b=c, therefore the eccentricity is \\frac{\\sqrt{2}}{2}." }, { "text": "Given the parabola equation $x^{2}=-2 y$, then the directrix equation of this parabola is?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = -2*y)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "2*y - 1 = 0", "fact_spans": "[[[2, 5], [23, 26]], [[2, 20]]]", "query_spans": "[[[23, 33]]]", "process": "" }, { "text": "If point $P(2021, t)$ lies on the parabola $y^{2}=4 x$, and point $F$ is the focus of this parabola, then the value of $|P F|$ is?", "fact_expressions": "G: Parabola;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(P) = (2021, t);PointOnCurve(P, G);Focus(G) = F;t:Number", "query_expressions": "Abs(LineSegmentOf(P, F))", "answer_expressions": "2022", "fact_spans": "[[[15, 29], [37, 40]], [[1, 14]], [[31, 35]], [[15, 29]], [[1, 14]], [[1, 30]], [[31, 43]], [[2, 14]]]", "query_spans": "[[[45, 56]]]", "process": "" }, { "text": "Given that ellipse $C_{1}$ and hyperbola $C_{2}$ share common foci $F_{1}$, $F_{2}$, and $M$ is an intersection point of $C_{1}$ and $C_{2}$, with $M F_{1} \\perp M F_{2}$. The eccentricity of ellipse $C_{1}$ is $e_{1}$, the eccentricity of hyperbola $C_{2}$ is $e_{2}$. If $e_{2}=2 e_{1}$, then $e_{1}$=?", "fact_expressions": "C1: Ellipse;C2: Hyperbola;F1: Point;F2: Point;Focus(C1) = {F1, F2};Focus(C2) = {F1, F2};M: Point;OneOf(Intersection(C1, C2)) = M;IsPerpendicular(LineSegmentOf(M, F1), LineSegmentOf(M, F2));e1: Number;e2: Number;Eccentricity(C1) = e1;Eccentricity(C2) = e2;e2 = 2*e1", "query_expressions": "e1", "answer_expressions": "sqrt(10)/4", "fact_spans": "[[[2, 11], [92, 101], [47, 54]], [[12, 22], [114, 124], [55, 62]], [[27, 34]], [[35, 42]], [[2, 42]], [[2, 42]], [[43, 46]], [[43, 67]], [[68, 91]], [[106, 113], [155, 162]], [[129, 136]], [[92, 113]], [[114, 136]], [[138, 153]]]", "query_spans": "[[[155, 164]]]", "process": "As shown in the figure, by the definition of ellipse and the Pythagorean theorem, we obtain \n\\begin{cases}|PF_{1}|+|PF_{2}|=2a_{1}\\\\|PF_{1}|^{2}+|PF_{2}|^{2}=4c^{2}\\end{cases}, \nit follows that $S_{\\triangle PF_{1}F_{2}}=b_{1}^{2}$. \nSince $e_{1}=\\frac{c}{a_{1}}$, $\\therefore a_{1}=\\frac{c}{e_{1}}$, $\\therefore b_{1}^{2}=a_{1}^{2}-c^{2}=c^{2}\\left(\\frac{1}{e_{1}^{2}}-1\\right)$. \nSimilarly, we obtain $S_{\\triangle PF_{1}F_{2}}=b_{2}^{2}$, since $e_{2}=\\frac{c}{a_{2}}$, $\\therefore a_{2}=\\frac{c}{e_{2}}$, $\\therefore b_{2}^{2}=a_{2}^{2}-c^{2}=c^{2}\\left(\\frac{1}{e_{2}^{2}}-1\\right)$. \nThen $c^{2}\\left(\\frac{1}{e_{1}^{2}}-1\\right)=c^{2}\\left(1-\\frac{1}{e_{2}^{2}}\\right)$, i.e., $\\frac{1}{e_{1}^{2}}+\\frac{1}{e_{2}^{2}}=2$. Given $e_{2}=2e_{1}$, $\\therefore e_{1}=\\frac{\\sqrt{10}}{4}$. \nThis problem examines simple properties of ellipses and hyperbolas; using the equality of triangle areas is the key to solving this problem, and it is a medium-difficulty question." }, { "text": "If point $O$ and point $F(-2,0)$ are respectively the center and the left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1(a>0)$, and point $P$ is an arbitrary point on the right branch of the hyperbola, then the range of values of $\\overrightarrow{O P} \\cdot \\overrightarrow{F P}$ is?", "fact_expressions": "G: Hyperbola;a: Number;F: Point;P: Point;O: Origin;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Coordinate(F) = (-2, 0);Center(G) = O;LeftFocus(G) = F;PointOnCurve(P, RightPart(G))", "query_expressions": "Range(DotProduct(VectorOf(O, P), VectorOf(F, P)))", "answer_expressions": "[3+2*sqrt(3),+oo)", "fact_spans": "[[[19, 56], [69, 72]], [[22, 56]], [[6, 16]], [[64, 68]], [[1, 5]], [[22, 56]], [[19, 56]], [[6, 16]], [[1, 63]], [[1, 63]], [[64, 80]]]", "query_spans": "[[[82, 138]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, the left and right foci are $F_{1}(-c, 0)$, $F_{2}(c, 0)$ respectively. If there exists a point $P$ on the ellipse such that $\\frac{a}{\\sin \\angle P F_{1} F_{2}}=\\frac{c}{\\sin \\angle P F_{2} F_{1}}$, then the range of values for the eccentricity of this ellipse is?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;c: Number;Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);a/Sin(AngleOf(P, F1, F2)) = c/Sin(AngleOf(P, F2, F1))", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(sqrt(2)-1, 1)", "fact_spans": "[[[2, 54], [94, 96], [181, 183]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[63, 77]], [[79, 92]], [[63, 77]], [[63, 77]], [[79, 92]], [[2, 92]], [[2, 92]], [[99, 103]], [[94, 103]], [[104, 177]]]", "query_spans": "[[[181, 194]]]", "process": "" }, { "text": "The coordinates of the foci of the hyperbola $x^{2}-y^{2}=1$ are? The equations of the asymptotes are?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 1)", "query_expressions": "Coordinate(Focus(G));Expression(Asymptote(G))", "answer_expressions": "(pm*sqrt(2), 0)\ny = pm*x", "fact_spans": "[[[0, 18]], [[0, 18]]]", "query_spans": "[[[0, 25]], [[0, 32]]]", "process": "From c^{2}=a^{2}+b^{2}, we obtain c=\\sqrt{2}; the coordinates of the foci of the hyperbola x^{2}-y^{2}=1 are (\\pm\\sqrt{2},0). Letting x^{2}-y^{2}=0, we obtain the asymptotes y=\\pmx" }, { "text": "Given an ellipse $\\frac{x^{2}}{a^{2}}+y^{2}=1$ ($a>1$) with $F_{1}$ and $F_{2}$ as the left and right foci, respectively, the left vertex is denoted by $A$, and the upper vertex by $B$. Points $M$ and $N$ are arbitrary points on the ellipse. If the maximum area of $\\Delta M A B$ is $4(\\sqrt{2}+1)$, then the minimum value of $\\frac{|N F_{1}|+9|N F_{2}|}{4|N F_{1}||N F_{2}|}$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2 + x^2/a^2 = 1);F1:Point;F2:Point;LeftFocus(G)=F1;RightFocus(G)=F2;a: Number;a>1;A:Point;B:Point;M:Point;N:Point;PointOnCurve(M, G);PointOnCurve(N, G);Max(Area(TriangleOf(M, A, B))) = 4*(1 + sqrt(2)) ;LeftVertex(G)=A;UpperVertex(G)=B", "query_expressions": "Min((Abs(LineSegmentOf(N, F1)) + 9*Abs(LineSegmentOf(N, F2)))/(4*(Abs(LineSegmentOf(N, F1))*Abs(LineSegmentOf(N, F2)))))", "answer_expressions": "1/4", "fact_spans": "[[[25, 61], [87, 89]], [[25, 61]], [[3, 10]], [[11, 18]], [[2, 61]], [[2, 61]], [[27, 61]], [[27, 61]], [[66, 69]], [[74, 77]], [[78, 82]], [[83, 86]], [[78, 94]], [[78, 94]], [[96, 132]], [[25, 69]], [[25, 77]]]", "query_spans": "[[[134, 190]]]", "process": "Given the conditions, A(-a,0), B(0,1), then |AB| = \\sqrt{1+a^{2}}, k_{AB} = \\frac{1}{a}. Let line l be a tangent line to the ellipse parallel to AB, with equation: y = \\frac{1}{a}x + m. From \\begin{cases} y = \\frac{1}{a}x + m \\\\ x^{2} + a^{2}y^{2} = a^{2} \\end{cases}, we obtain 2x^{2} + 2amx + a^{2}m^{2} - a^{2} = 0. Therefore, \\Delta = (2am)^{2} - 4 \\times 2 \\times (a^{2}m^{2} - a^{2}) = 0, solving gives m = \\pm\\sqrt{2}. According to the problem, m = -\\sqrt{2}. The distance from tangent line l to line AB is: d = \\frac{\\sqrt{2}+1}{\\sqrt{1+\\frac{1}{a^{2}}}}. Therefore, the maximum area is (S_{\\triangle ABM})_{\\max} = \\frac{1+\\frac{1}{a^{2}}}{2\\sqrt{a^{2}+1}\\frac{\\sqrt{2}+1}{\\sqrt{1}+1}} = 4(\\sqrt{2}+1), solving gives a = 8. Let |NF| = m, |NF_{2}| = n, then m + n = 16, so \\frac{|NF| + 9|NF_{2}|}{4|NF|} = \\frac{m + 9n}{4mn} = \\frac{1}{4}\\left(\\frac{1}{n} + \\frac{9}{m}\\right) = \\frac{1}{64}\\left(\\frac{1}{n} + \\frac{9}{m}\\right)(m + n) = \\frac{1}{6}\\left(1 + 9 + \\frac{m}{n} + \\frac{9n}{m}\\right) \\geqslant \\frac{1}{74}\\left(10 + 2\\sqrt{\\frac{m}{n} \\cdot \\frac{9n}{m}}\\right) = \\frac{1}{4}, equality holds if and only if \\frac{m}{n} = \\frac{9n}{m}, i.e., m = 3n, both answer letters are 1." }, { "text": "The equation of line $l$ is $y=x+3$. Take an arbitrary point $P$ on $l$. If an ellipse is constructed passing through point $P$ with the foci of the hyperbola $12 x^{2}-4 y^{2}=3$ as the foci of the ellipse, then what is the equation of the ellipse with the shortest major axis?", "fact_expressions": "l: Line;G: Hyperbola;H: Ellipse;P: Point;Expression(G) = (12*x^2 - 4*y^2 = 3);Expression(l) = (y = x + 3);PointOnCurve(P, l);PointOnCurve(P, H);Focus(G) = Focus(H);WhenMin(MajorAxis(H))", "query_expressions": "Expression(H)", "answer_expressions": "x^2/5 + y^2/4 = 1", "fact_spans": "[[[0, 5], [18, 21]], [[39, 62]], [[66, 68], [72, 74], [84, 86]], [[27, 30], [33, 37]], [[39, 62]], [[0, 16]], [[17, 30]], [[32, 74]], [[39, 71]], [[79, 86]]]", "query_spans": "[[[84, 90]]]", "process": "" }, { "text": "Let $F$ be the focus of the parabola $y^{2}=x$, and let a line passing through $F$ intersect the parabola at points $A$ and $B$. Then the minimum value of $|A F|+3| B F |$ is?", "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;Expression(G) = (y^2 = x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B}", "query_expressions": "Min(Abs(LineSegmentOf(A, F)) + 3*Abs(LineSegmentOf(B, F)))", "answer_expressions": "sqrt(3)/2 + 1", "fact_spans": "[[[5, 17], [29, 32]], [[26, 28]], [[34, 37]], [[1, 4], [22, 25]], [[38, 41]], [[5, 17]], [[1, 20]], [[21, 28]], [[26, 43]]]", "query_spans": "[[[45, 67]]]", "process": "" }, { "text": "Through the focus $F$ of the parabola $y^{2}=2 p x(p>0)$, draw a line $l$ with slope $2$, intersecting the parabola at points $A$ and $B$. If the area of $\\Delta O A B$ is equal to $\\sqrt{5}$ ($O$ being the origin), then $p=?$", "fact_expressions": "l: Line;G: Parabola;p: Number;O: Origin;A: Point;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));F: Point;Focus(G) = F;PointOnCurve(F, l);Slope(l) = 2;Intersection(l, G) = {A, B};Area(TriangleOf(O, A, B)) = sqrt(5)", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[36, 41]], [[1, 22], [44, 47]], [[101, 104]], [[90, 93]], [[49, 52]], [[53, 56]], [[4, 22]], [[1, 22]], [[25, 28]], [[1, 28]], [[0, 41]], [[29, 41]], [[36, 58]], [[60, 89]]]", "query_spans": "[[[101, 106]]]", "process": "First, write the coordinates of the focus F from the parabola equation, then obtain the equation of line l. Substitute the equation of the line into the parabola equation to get a relation for the y-coordinates of points A and B. Combine this with the given information and use the triangle area formula to set up an equation, which allows solving for the value of p. \n[Solution] From the given conditions, the focus of the parabola is F(\\frac{p}{2},0), thus the equation of line l is x=\\frac{1}{2}y+\\frac{p}{2}. Substituting into the parabola equation yields y^{2}-py-p^{2}=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), then y_{1}+y_{2}=p, y_{1}y_{2}=-p^{2}. The area of \\Delta OAB is \\frac{1}{2}\\times\\frac{p}{2}|y_{1}-y_{2}|=\\frac{p}{4}\\sqrt{(y_{1}+y_{2})^{2}-4y_{1}y_{2}}=\\frac{\\sqrt{5}}{4}p^{2}=\\sqrt{5}, solving gives p=2." }, { "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $C$: $\\frac{x^{2}}{25} + \\frac{y^{2}}{16}=1$, respectively. Point $P$ lies on the ellipse $C$ and satisfies $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$. Then $|\\overrightarrow{P F_{1}}| \\cdot |\\overrightarrow{P F_{2}}|$=?", "fact_expressions": "C: Ellipse;P: Point;F2: Point;F1: Point;Expression(C) = (x^2/25 + y^2/16 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);DotProduct(VectorOf(P, F1),VectorOf(P, F2))=0", "query_expressions": "Abs(VectorOf(P, F1))*Abs(VectorOf(P, F2))", "answer_expressions": "32", "fact_spans": "[[[19, 65], [77, 82]], [[72, 76]], [[9, 16]], [[1, 8]], [[19, 65]], [[2, 71]], [[2, 71]], [[72, 83]], [[87, 146]]]", "query_spans": "[[[148, 211]]]", "process": "" }, { "text": "Let real numbers $x$, $y$ satisfy $x^{2}=4 y$, then the minimum value of $\\sqrt{(x-3)^{2}+(y-1)^{2}}+y$ is?", "fact_expressions": "x_: Real;y_:Real;x_^2 = 4*y_", "query_expressions": "Min(sqrt((x_ - 3)^2 + (y_ - 1)^2)+y_)", "answer_expressions": "2", "fact_spans": "[[[1, 7]], [[9, 12]], [[14, 25]]]", "query_spans": "[[[27, 63]]]", "process": "" }, { "text": "Given two points $A(-3 , 0)$ and $B(3 , 0)$, if $|PA|-|PB|=2$, then what is the trajectory equation of point $P$?", "fact_expressions": "A: Point;Coordinate(A) = (-3, 0);B: Point;Coordinate(B) = (3, 0);P: Point;Abs(LineSegmentOf(P, A)) - Abs(LineSegmentOf(P, B)) = 2", "query_expressions": "LocusEquation(P)", "answer_expressions": "(x^2-y^2/8=1) & (x>0)", "fact_spans": "[[[4, 15]], [[4, 15]], [[16, 26]], [[16, 26]], [[44, 48]], [[28, 41]]]", "query_spans": "[[[44, 55]]]", "process": "" }, { "text": "If a point on the parabola $y^{2}=-2 p x(p>0)$ has distances of $10$ and $6$ to the focus and the axis of symmetry of the parabola, respectively, then what is the value of $p$?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -2*p*x);p: Number;p>0;D: Point;PointOnCurve(D, G) = True;Distance(D, Focus(G)) = 10;Distance(D, SymmetryAxis(G)) = 6", "query_expressions": "p", "answer_expressions": "2, 18", "fact_spans": "[[[1, 23], [30, 33]], [[1, 23]], [[53, 56]], [[4, 23]], [], [[1, 26]], [[1, 51]], [[1, 51]]]", "query_spans": "[[[53, 60]]]", "process": "The focus of the parabola is $(-\\frac{p}{2},0)$, the axis of symmetry is the x-axis, $\\sqrt{10^{2}-6^{2}}=8$. Therefore, the coordinates of the point satisfying the condition can be set as $(-\\frac{p}{2}\\pm8,\\pm6)$. Substituting into the parabolic equation gives $36=-2p(-\\frac{p}{2}+8)$, solving yields $p=2$ or $p=18$, the negative root is discarded. [Note] This question mainly examines the method of finding the parabolic equation, the geometric properties of a parabola, and the application of equations, belonging to basic problems." }, { "text": "The length of the chord intercepted by the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$ from the line $y=x+1$ is?", "fact_expressions": "G: Hyperbola;H: Line;Expression(G) = (x^2 - y^2/4 = 1);Expression(H) = (y = x + 1)", "query_expressions": "Length(InterceptChord(H,G))", "answer_expressions": "8*sqrt(2)/3", "fact_spans": "[[[0, 28]], [[29, 38]], [[0, 28]], [[29, 38]]]", "query_spans": "[[[0, 44]]]", "process": "" }, { "text": "The eccentricity of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$ is $e=\\frac{\\sqrt{15}}{5}$, then $m=?$", "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/5 + y^2/m = 1);Eccentricity(G) = e ;e:Number;e= sqrt(15)/5", "query_expressions": "m", "answer_expressions": "{2,25/2}", "fact_spans": "[[[0, 37]], [[66, 69]], [[0, 37]], [[0, 64]], [[41, 64]], [[41, 64]]]", "query_spans": "[[[66, 71]]]", "process": "If 05, then e^{2}=\\frac{m-5}{m}=\\frac{3}{5}\\therefore m=\\frac{25}{2}\\therefore the value of m is: 2 or \\frac{25}{2}" }, { "text": "Given that the moving point $P(x, y)$ lies on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$. A tangent is drawn from point $P$ to the circle $(x-3)^{2}+y^{2}=1$, with the point of tangency being $M$. Then, the minimum value of $PM$ is?", "fact_expressions": "G: Ellipse;H: Circle;P: Point;M: Point;x1:Number;y1:Number;l1:Line;Expression(G) = (x^2/25 + y^2/16 = 1);Expression(H) = (y^2 + (x - 3)^2 = 1);Coordinate(P) = (x1, y1);PointOnCurve(P, G);TangentOfPoint(P,H)=l1;TangentPoint(l1,H)=M", "query_expressions": "Min(LineSegmentOf(P, M))", "answer_expressions": "sqrt(3)", "fact_spans": "[[[14, 53]], [[61, 81]], [[4, 13], [56, 60]], [[88, 91]], [[4, 13]], [[4, 13]], [], [[14, 53]], [[61, 81]], [[4, 13]], [[4, 54]], [[55, 84]], [[55, 91]]]", "query_spans": "[[[93, 104]]]", "process": "Let the center of the circle $(x-3)^{2}+y^{2}=1$ be $A(3,0)$. According to the problem, $PM \\perp AM$, therefore $|PM^{2}| = |AP|^{2} - |AM^{2}|$. Since $|AM|^{2} = 1$, the smaller $|AP|$ is, the smaller $|PM|$ is. Moreover, $A$ is the right focus of the ellipse, so the minimum value of $|AP|$ is $5-3=2$. Therefore, the minimum value of $|PM|$ is $\\sqrt{4-1} = \\sqrt{3}$." }, { "text": "Given that $F_{1}(-3,0)$, $F_{2}(3,0)$ are the two foci of hyperbola $C$, and the line $y=\\frac{\\sqrt{5}}{2} x$ is an asymptote of this hyperbola, then the standard equation of this hyperbola is?", "fact_expressions": "C: Hyperbola;F1: Point;F2: Point;Coordinate(F1) = (-3, 0);Coordinate(F2) = (3, 0);Focus(C) = {F1,F2};G: Line;Expression(G) = (y = x*(sqrt(5)/2));OneOf(Asymptote(C)) = G", "query_expressions": "Expression(C)", "answer_expressions": "x^2/4 - y^2/5 = 1", "fact_spans": "[[[31, 37], [72, 75], [84, 87]], [[2, 15]], [[18, 30]], [[2, 15]], [[18, 30]], [[2, 42]], [[44, 70]], [[44, 70]], [[44, 81]]]", "query_spans": "[[[84, 94]]]", "process": "From the asymptotes, we get $\\frac{b}{a}=\\frac{\\sqrt{5}}{2}$; from the focus, we get $c=3$; combining with $c^{2}=a^{2}+b^{2}$, we can solve for $a$ and $b$, thus obtaining the standard equation. According to the given conditions:\n\\[\n\\begin{cases}\n\\frac{b}{a}=\\frac{\\sqrt{5}}{2}\\\\\nc=3\\\\\nc^2=a^2+b^{2}\n\\end{cases}\n\\]\nSolving this system yields\n\\[\n\\begin{cases}\na=2\\\\\nb=\\sqrt{5}\n\\end{cases}\n\\]\nTherefore, the standard equation of the hyperbola is $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$." }, { "text": "The focus of the parabola $y^{2}=2 p x(p>0)$ is $F$, the intersection point of the line $y=2$ and the $y$-axis is $M$, and the intersection point with the parabola is $N$. Given that $4|N F|=5|M N|$, find the value of $p$.", "fact_expressions": "G: Parabola;p: Number;H: Line;N: Point;F: Point;M: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (y = 2);Focus(G) = F;Intersection(H, yAxis) = M;Intersection(H,G)=N;4*Abs(LineSegmentOf(N, F)) = 5*Abs(LineSegmentOf(M, N))", "query_expressions": "p", "answer_expressions": "1", "fact_spans": "[[[0, 21], [50, 53]], [[79, 82]], [[29, 36]], [[57, 60]], [[25, 28]], [[45, 48]], [[3, 21]], [[0, 21]], [[29, 36]], [[0, 28]], [[29, 48]], [[29, 60]], [[62, 77]]]", "query_spans": "[[[79, 86]]]", "process": "Analysis: First, substitute y=2 into the equation of the parabola to find the coordinates of the corresponding point N, thus obtaining |MN|. Using the definition of the parabola, express |NF| as the distance from point N to the directrix of the parabola, and compute |NF|. Then apply the given equality in the problem to obtain an equation in terms of p, and solve for the result. \nDetailed solution: Substitute y=2 into the parabolic equation to obtain x=\\frac{2}{p}. Using the given condition and combining with the definition of the parabola, we get 4(\\frac{2}{p}+\\frac{p}{2})=5\\times\\frac{2}{p}, solving which yields p=1." }, { "text": "The distance from point $M(5,3)$ to the directrix of the parabola $y^{2}=a x(a<0)$ is $6$, then the standard equation of the parabola is?", "fact_expressions": "G: Parabola;a: Number;M: Point;a<0;Expression(G) = (y^2 = a*x);Coordinate(M) = (5, 3);Distance(M, Directrix(G)) = 6", "query_expressions": "Expression(G)", "answer_expressions": "y^2 = -44*x", "fact_spans": "[[[10, 29], [42, 45]], [[13, 29]], [[0, 9]], [[13, 29]], [[10, 29]], [[0, 9]], [[0, 39]]]", "query_spans": "[[[42, 52]]]", "process": "From the given condition: the directrix of the parabola is $ x = -\\frac{a}{4} $ ($ a < 0 $), and the distance from point $ M(5,3) $ to the directrix is 6, therefore $ \\left|5 + \\frac{a}{4}\\right| = 6 $. Solving this gives $ a = 4 $ (discarded) or $ a = -44 $. Therefore, the standard equation of the parabola is $ y^2 = -44x $. Answer: $ y^2 = -44x $." }, { "text": "Given that the focus of the parabola $x^{2}=a y$ coincides exactly with the upper focus of the hyperbola $y^{2}-x^{2}=2$, then $a$=?", "fact_expressions": "H: Parabola;Expression(H) = (x^2 = a*y);a: Number;G: Hyperbola;Expression(G) = (-x^2 + y^2 = 2);Focus(H) = UpperFocus(G)", "query_expressions": "a", "answer_expressions": "8", "fact_spans": "[[[2, 16]], [[2, 16]], [[46, 49]], [[22, 40]], [[22, 40]], [[2, 44]]]", "query_spans": "[[[46, 51]]]", "process": "The focus of the parabola \\(x^{2}=ay\\) (\\(a>0\\)) is \\((0,\\frac{a}{4})\\); the foci of the hyperbola \\(y^{2}-x^{2}=2\\) are \\((0,\\pm2)\\). Since \\(a>0\\), \\(\\frac{a}{4}=2\\), therefore \\(a=8\\)." }, { "text": "On the parabola $C$: $y^{2}=4x$, a point $Q$ is such that the sum of the distance from $Q$ to the point $B(4, 1)$ and the distance from $Q$ to the focus $F$ is minimized. Then the coordinates of point $Q$ are?", "fact_expressions": "C: Parabola;B: Point;Q: Point;F: Point;Expression(C) = (y^2 = 4*x);Coordinate(B) = (4, 1);PointOnCurve(Q, C);Focus(C) = F;WhenMin(Distance(Q, B) + Distance(Q, F))", "query_expressions": "Coordinate(Q)", "answer_expressions": "(1/4, 1)", "fact_spans": "[[[0, 19]], [[26, 37]], [[22, 25], [52, 56]], [[41, 44]], [[0, 19]], [[26, 37]], [[0, 25]], [[0, 44]], [[0, 50]]]", "query_spans": "[[[52, 61]]]", "process": "" }, { "text": "If the eccentricity of a hyperbola is $2$, and the coordinates of the two foci are $(-2 , 0)$, $(2 , 0)$, then what is the equation of this hyperbola?", "fact_expressions": "E: Hyperbola;Eccentricity(E) = 2;F1: Point;F2: Point;Focus(E) = {F1, F2};Coordinate(F1) = (-2, 0);Coordinate(F2) = (2, 0)", "query_expressions": "Expression(E)", "answer_expressions": "x^2 - y^2/3 = 1", "fact_spans": "[[[1, 4], [42, 45]], [[1, 12]], [], [], [[1, 39]], [[1, 39]], [[1, 39]]]", "query_spans": "[[[42, 50]]]", "process": "" }, { "text": "A line $ l $ with an inclination angle of $\\frac{\\pi}{6}$ is drawn through the focus of the parabola $ C $: $ y^{2} = 4x $. The line $ l $ intersects the parabola $ C $ at points $ A $ and $ B $. Then $ |AB| $ = ?", "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;Expression(C) = (y^2 = 4*x);Intersection(l, C) = {A, B};PointOnCurve(Focus(C),l);Inclination(l)=pi/6", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "16", "fact_spans": "[[[46, 51], [52, 57]], [[1, 20], [58, 64]], [[66, 69]], [[70, 73]], [[1, 20]], [[52, 75]], [[0, 51]], [[26, 51]]]", "query_spans": "[[[77, 86]]]", "process": "The coordinates of the focus of the parabola are F(1,0), p=2. The slope of the line passing through the focus is k=\\tan\\frac{\\pi}{6}=\\frac{\\sqrt{3}}{3}. Then the equation of the line is y=\\frac{\\sqrt{3}}{3}(x-1). Substituting into y^{2}=4x gives \\frac{1}{3}(x-1)^{2}=4x. Rearranging yields x^{2}-14x+1=0. Let the coordinates of points A and B be (x_{1},y_{1}) and (x_{2},y_{2}), respectively. Then x_{1}+x_{2}=14. Thus, |AB|=x_{1}+x_{2}+p=14+2=16." }, { "text": "If the focus of the parabola $y^{2}=2 p x(p>0)$ is exactly the right focus of the hyperbola $\\frac{x^{2}}{16-m}-\\frac{y^{2}}{m+20}=1$, then the value of the real number $p$ is?", "fact_expressions": "G: Hyperbola;m: Number;H: Parabola;p: Real;Expression(G) = (x^2/(16 - m) - y^2/(m + 20) = 1);p>0;Expression(H) = (y^2 = 2*(p*x));Focus(H) = RightFocus(G)", "query_expressions": "p", "answer_expressions": "12", "fact_spans": "[[[28, 72]], [[31, 72]], [[1, 22]], [[78, 83]], [[28, 72]], [[4, 22]], [[1, 22]], [[1, 76]]]", "query_spans": "[[[78, 87]]]", "process": "c^{2}=(\\frac{p}{2})^{2}=16-m+m+20=36,p=12" }, { "text": "It is known that the hyperbola $C$ is centered at the origin and its foci lie on the $y$-axis. If one asymptote of the hyperbola $C$ is parallel to the line $\\sqrt{2} x - y - 1 = 0$, then what is the eccentricity of the hyperbola $C$?", "fact_expressions": "C: Hyperbola;O: Origin;Center(C) = O;PointOnCurve(Focus(C), yAxis);G: Line;Expression(G) = (sqrt(2)*x - y - 1 = 0);IsParallel(OneOf(Asymptote(C)), G)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(6)/2", "fact_spans": "[[[2, 8], [25, 31], [62, 68]], [[12, 14]], [[2, 14]], [[2, 23]], [[38, 58]], [[38, 58]], [[25, 60]]]", "query_spans": "[[[62, 74]]]", "process": "Problem Analysis: Use the given conditions to set up an equation relating $a$ and $b$, then solve for the eccentricity of the hyperbola. \nDetailed Solution: The hyperbola $C$ is centered at the origin, with foci on the $y$-axis. If one asymptote of the hyperbola $C$ and the line $\\sqrt{2}x - y - 1 = 0$ yield $\\frac{a}{b} = \\sqrt{2}$, then $a^{2} = 2b^{2} = 2c^{2} - 2a^{2}$, from which the eccentricity is obtained as $e = \\frac{\\sqrt{6}}{2}$." }, { "text": "Given that $A$, $B$ are two points on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{25 y^{2}}{9 a^{2}}=1$, $F_{2}$ is the right focus of the ellipse, if $|A F_{2}|+|B F_{2}|=\\frac{8}{5} a$, and the distance from the midpoint of $AB$ to the left directrix of the ellipse is $\\frac{3}{2}$, then what is the equation of the ellipse?", "fact_expressions": "G: Ellipse;a: Number;A: Point;B: Point;F2: Point;Expression(G) = ((25*y^2)/(9*a^2) + x^2/a^2 = 1);PointOnCurve(A, G);PointOnCurve(B, G);RightFocus(G) = F2;Abs(LineSegmentOf(A,F2))+Abs(LineSegmentOf(B,F2))=(8/5)*a;Distance(MidPoint(LineSegmentOf(A, B)), LeftDirectrix(G)) = 3/2", "query_expressions": "Expression(G)", "answer_expressions": "x^2+(25*y^2)/9=1", "fact_spans": "[[[10, 60], [73, 75], [126, 128], [149, 151]], [[12, 60]], [[2, 5]], [[6, 9]], [[65, 72]], [[10, 60]], [[2, 64]], [[2, 64]], [[65, 79]], [[82, 117]], [[118, 147]]]", "query_spans": "[[[149, 155]]]", "process": "" }, { "text": "The line $l$ passes through a vertex and a focus of an ellipse. If the distance from the center of the ellipse to $l$ is $\\frac{1}{4}$ of the length of its minor axis, then the eccentricity of the ellipse is?", "fact_expressions": "l: Line;G: Ellipse;PointOnCurve(OneOf(Vertex(G)), l);PointOnCurve(OneOf(Focus(G)), l);Distance(Center(G), l) = (1/4)*Length(MinorAxis(G))", "query_expressions": "Eccentricity(G)", "answer_expressions": "1/2", "fact_spans": "[[[0, 5], [26, 29]], [[7, 9], [21, 23], [33, 34], [54, 56]], [[0, 14]], [[0, 19]], [[21, 51]]]", "query_spans": "[[[54, 62]]]", "process": "Method 1: Without loss of generality, assume the line passes through the upper vertex (0,b) and the left focus (-c,0) of the ellipse, where b>0, c>0. Then the equation of line l is bx - cy + bc = 0. From the given condition, we have \\frac{bc}{\\sqrt{b^{2}+c^{2}}} = \\frac{1}{4} \\times 2b. Solving this gives b^{2} = 3c^{2}. Since b^{2} = a^{2} - c^{2}, it follows that e^{2} = \\frac{c^{2}}{a^{2}} = \\frac{1}{4}, so e = \\frac{1}{2}. \nMethod 2: Without loss of generality, assume the line l passes through the upper vertex (0,b) and the left focus (-c,0), where b>0, c>0. Then the equation of line l is bx - cy + bc = 0. From the given condition, we have \\frac{bc}{\\sqrt{b^{2}+c^{2}}} = \\frac{1}{4} \\times 2b, so \\frac{bc}{a} = \\frac{1}{4} \\times 2b, hence e = \\frac{c}{a} = \\frac{1}{2}." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has an asymptote line $l$: $y=2x$, and its transverse axis length is less than $6$, then a standard equation of $C$ can be?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;l:Line;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);OneOf(Asymptote(C))=l;Expression(l)=(y=2*x);Length(RealAxis(C))<6", "query_expressions": "Expression(C)", "answer_expressions": "x^2/4 - y^2/16 = 1", "fact_spans": "[[[2, 63], [86, 87], [97, 100]], [[10, 63]], [[10, 63]], [[72, 84]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 84]], [[72, 84]], [[86, 95]]]", "query_spans": "[[[97, 111]]]", "process": "According to the given condition, \\frac{b}{a}=2; also, the length of its real axis 2a<6, hence a<3. Let a=2, then b=4. Thus, a standard equation of C is \\frac{x^{2}}{4}-\\frac{y^{2}}{16}=1" }, { "text": "Given that a focus of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{3 m}=1$ is $(2 \\sqrt{3}, 0)$, then $m=?$", "fact_expressions": "G: Hyperbola;Expression(G) = (-1/(3*m)*y^2 + x^2/m = 1);m: Number;Coordinate(OneOf(Focus(G))) = (2*sqrt(3),0)", "query_expressions": "m", "answer_expressions": "3", "fact_spans": "[[[2, 42]], [[2, 42]], [[67, 70]], [[2, 65]]]", "query_spans": "[[[67, 72]]]", "process": "The hyperbola \\frac{x^{2}}{m}-\\frac{y^{2}}{3m}=1 has a focus at (2\\sqrt{3},0), that is, c=\\sqrt{m+3m}=2\\sqrt{3}, solving gives m=3," }, { "text": "It is known that the directrix of the parabola $y^{2}=2 p x(p>0)$ is tangent to the circle $(x-3)^{2}+y^{2}=16$, then the value of $p$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;H: Circle;Expression(H) = (y^2 + (x - 3)^2 = 16);IsTangent(Directrix(G), H)", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[2, 23]], [[2, 23]], [[52, 55]], [[5, 23]], [[27, 48]], [[27, 48]], [[2, 50]]]", "query_spans": "[[[52, 59]]]", "process": "The directrix of the parabola is x=-\\frac{p}{2}, and it is tangent to the circle, so 3+\\frac{p}{2}=4, p=2." }, { "text": "It is known that one focus of a hyperbola coincides with the focus of the parabola $x^{2}=24 y$, and the angle of inclination of one asymptote is $30^{\\circ}$. Find the standard equation of this hyperbola?", "fact_expressions": "G: Hyperbola;H: Parabola;Expression(H) = (x^2 = 24*y);OneOf(Focus(G)) = Focus(H);Inclination(OneOf(Asymptote(G)))=ApplyUnit(30,degree)", "query_expressions": "Expression(G)", "answer_expressions": "y^2/9 - x^2/27 = 1", "fact_spans": "[[[2, 5], [57, 60]], [[11, 26]], [[11, 26]], [[2, 31]], [[2, 54]]]", "query_spans": "[[[57, 67]]]", "process": "Problem Analysis: Find the focus of the parabola, assume the equation of the hyperbola is \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1(a,b>0), find the equations of the asymptotes and the relationship among a, b, c, then solve the equations to obtain the required result. \nSolution: The focus of the parabola x^{2}=24y is (0,6). Assume the equation of the hyperbola is \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1(a,b>0). Thus, c=6, that is, a^{2}+b^{2}=36. The equations of the asymptotes are y=\\pm\\frac{a}{b}x. From the given condition, we have \\tan30^{\\circ}=\\frac{a}{b}, which implies b=\\sqrt{3}a. Solving gives a=3, b=3\\sqrt{3}. Therefore, the standard equation of the hyperbola is: \\frac{y^{2}}{9}-\\frac{x^{2}}{27}=1" }, { "text": "Given that points $A$ and $B$ are two distinct points on the parabola $C$: $y^{2}=2 p x(p>0)$, and the sum of the distances from points $A$ and $B$ to the focus $F$ of the parabola $C$ is $6$, the midpoint of segment $AB$ is $M(2,-1)$. Then, what is the distance from the focus $F$ to the line $AB$?", "fact_expressions": "C: Parabola;p: Number;B: Point;A: Point;M: Point;F:Point;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(M) = (2, -1);PointOnCurve(A, C);PointOnCurve(B, C);Focus(C)=F;Distance(A,F)+Distance(B,F)=6;MidPoint(LineSegmentOf(A,B)) = M;Negation(A=B)", "query_expressions": "Distance(F, LineOf(A, B))", "answer_expressions": "sqrt(5)/5", "fact_spans": "[[[11, 37], [55, 61]], [[19, 37]], [[7, 10], [49, 52]], [[2, 6], [45, 48]], [[88, 97]], [[64, 67], [101, 104]], [[19, 37]], [[11, 37]], [[88, 97]], [[2, 43]], [[2, 43]], [[55, 67]], [[45, 76]], [[77, 97]], [0, 42]]", "query_spans": "[[[101, 117]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). From the definition of the parabola, we know: x_{1}+x_{2}+p=6, then x_{1}+x_{2}=6-p. Since M(2,-1) is the midpoint of AB, then \\frac{6-p}{2}=2 \\Rightarrow p=2 \\Rightarrow F(1,0). Therefore, the equation of the parabola is y^{2}=4x. Then: \n\\begin{cases}y_{1}^{2}=4x_{1}\\\\y_{2}^{2}=4x_{2}\\end{cases}, subtracting these two equations gives: (y_{1}+y_{2})(y_{1}-y_{2})=4(x_{1}-x_{2}). \nThen k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{4}{y_{1}+y_{2}}=\\frac{4}{-2}=-2. \nTherefore, the equation of line AB is: y+1=-2(x-2), i.e., 2x+y-3=0. \nTherefore, the distance from point F to line AB is d=\\frac{|2+0-3|}{\\sqrt{5}}=\\frac{\\sqrt{5}}{5}. \nThe correct answer to this problem: \\frac{\\sqrt{5}}{5}" }, { "text": "The center is at the origin, and for a hyperbola with foci on the $x$-axis, one of its asymptotes passes through the point $(4,-2)$. What is its eccentricity?", "fact_expressions": "G: Hyperbola;H: Point;O: Origin;Coordinate(H) = (4, -2);PointOnCurve(Focus(G), xAxis);PointOnCurve(H,OneOf(Asymptote(G)));Center(G)=O", "query_expressions": "Eccentricity(G)", "answer_expressions": "3*sqrt(2)/4", "fact_spans": "[[[15, 18], [37, 38]], [[26, 35]], [[3, 5]], [[26, 35]], [[6, 18]], [[15, 35]], [[0, 18]]]", "query_spans": "[[[37, 44]]]", "process": "Let the hyperbola equation be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0, b>0); then its asymptotes are given by y=\\pm\\frac{b}{a}x. Substituting the coordinates of the point (4,-\\sqrt{2}) into the above expression, we get \\frac{b}{a}=\\frac{\\sqrt{2}}{4}. Therefore, e=\\frac{c}{a}=\\frac{\\sqrt{a^{2}+b^{2}}}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{1+(\\frac{\\sqrt{2}}{4})^{2}}=\\frac{3\\sqrt{2}}{4}. Answer: \\frac{3\\sqrt{2}}{4}" }, { "text": "The hyperbola $C$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1(a>0, b>0)$ has an asymptote whose angle of inclination is $60^{\\circ}$; then the eccentricity of $C$ is?", "fact_expressions": "C: Hyperbola;a: Number;b: Number;a>0;b>0;Expression(C) = (-x^2/b^2 + y^2/a^2 = 1);Inclination(OneOf(Asymptote(C)))=ApplyUnit(60,degree)", "query_expressions": "Eccentricity(C)", "answer_expressions": "2*sqrt(3)/3", "fact_spans": "[[[0, 61], [86, 89]], [[8, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[0, 61]], [[0, 84]]]", "query_spans": "[[[86, 95]]]", "process": "From the hyperbola equation \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1, it follows that its foci lie on the y-axis. Since the inclination angle of one asymptote is 60^{\\circ}, its equation is y=\\sqrt{3}x, so \\frac{a}{b}=\\sqrt{3}, \\frac{b}{a}=\\frac{\\sqrt{3}}{3}. Let the semi-focal length be c, then e=\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\frac{2\\sqrt{3}}{3}" }, { "text": "Given the hyperbola $\\frac{x^{2}}{2 m}-\\frac{y^{2}}{m+4}=1$ has an asymptote with equation $y=\\sqrt{3} x$, find the value of the real number $m$.", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/(m + 4) + x^2/(2*m) = 1);m: Real;Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x)", "query_expressions": "m", "answer_expressions": "4/5", "fact_spans": "[[[2, 44]], [[2, 44]], [[69, 74]], [[2, 67]]]", "query_spans": "[[[69, 78]]]", "process": "Problem Analysis: Since the two asymptotes of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 are y=\\pm\\frac{b}{a}x, the asymptotes of \\frac{x^{2}}{2m}-\\frac{y^{2}}{m+4}=1 are y=\\pm\\sqrt{\\frac{m+4}{2m}}x. Then, \\sqrt{\\frac{m+4}{2m}}=\\sqrt{3} \\Rightarrow m=\\frac{4}{5}." }, { "text": "Let $F_{1}$, $F_{2}$ be the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{16}=1$ ($a>0$), and let point $P$ be a point on the hyperbola $C$. If $|P F_{1}|-|P F_{2}|=4$, then what is the equation of the hyperbola $C$? What is its eccentricity?", "fact_expressions": "C: Hyperbola;a: Number;P: Point;F1: Point;F2: Point;a>0;Expression(C) = (-y^2/16 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)) = 4", "query_expressions": "Expression(C);Eccentricity(C)", "answer_expressions": "x^2/4-y^2/16=1\nsqrt(5)", "fact_spans": "[[[120, 126], [82, 88], [17, 70]], [[25, 70]], [[77, 81]], [[1, 8]], [[9, 16]], [[25, 70]], [[17, 70]], [[1, 76]], [[1, 76]], [[77, 91]], [[94, 117]]]", "query_spans": "[[[120, 131]], [[120, 136]]]", "process": "Problem Analysis: Since |PF_{1}|-|PF_{2}|=4, we have 2a=4, thus a=2, so the equation of hyperbola C is \\frac{x^{2}}{4}-\\frac{y^{2}}{16}=1; therefore, the eccentricity is e=\\frac{c}{a}=\\frac{\\sqrt{4+16}}{2}=\\sqrt{5}" }, { "text": "It is known that the focus of the parabola $y=\\frac{1}{2} x^{2}$ coincides with one focus of the ellipse $\\frac{y^{2}}{m}+\\frac{x^{2}}{2}=1$. Find $m=?$", "fact_expressions": "G: Parabola;Expression(G) = (y = x^2/2);H: Ellipse;Expression(H) = (x^2/2 + y^2/m = 1);m: Number;Focus(G) = OneOf(Focus(H))", "query_expressions": "m", "answer_expressions": "9/4", "fact_spans": "[[[2, 26]], [[2, 26]], [[30, 67]], [[30, 67]], [[76, 79]], [[2, 74]]]", "query_spans": "[[[76, 81]]]", "process": "" }, { "text": "Point $P$ is a moving point on the parabola $y^{2}=4x$. Then, the minimum value of the sum of the distance from point $P$ to point $A(0,-1)$ and the distance to the line $x=-1$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G);A: Point;Coordinate(A) = (0, -1);H: Line;Expression(H) = (x = -1)", "query_expressions": "Min(Distance(P, A)+Distance(P, H))", "answer_expressions": "sqrt(2)", "fact_spans": "[[[5, 19]], [[5, 19]], [[0, 4], [25, 29]], [[0, 23]], [[30, 40]], [[30, 40]], [[45, 53]], [[45, 53]]]", "query_spans": "[[[25, 64]]]", "process": "According to the definition of a parabola, when points A, P, and F are collinear, the sum of the distance from point P to point A(0,-1) and the distance to the line x = -1 is minimized. [Detailed Solution] Since the equation of the parabola is $ y^{2} = 4x $, the focus of the parabola has coordinates $ F(1,0) $, and the directrix equation is: $ x = -1 $. As shown in the figure: By the definition of the parabola, the distance from point P to the focus $ F(1,0) $ is equal to the distance from point P to the directrix $ x = -1 $. Therefore, when points A, P, and F are collinear, the sum of the distance from point P to point A(0,-1) and the distance to the line $ x = -1 $ is minimized. The minimum value is $ |AF| = \\sqrt{2} $." }, { "text": "Given that a chord $AB$ of the parabola $y^{2}=4x$ has $P(1,1)$ as its midpoint, what is the equation of the line containing the chord $AB$?", "fact_expressions": "G: Parabola;A: Point;B: Point;P: Point;Expression(G) = (y^2 = 4*x);IsChordOf(LineSegmentOf(A,B),G);Coordinate(P) = (1, 1);MidPoint(LineSegmentOf(A,B))=P", "query_expressions": "Expression(OverlappingLine(LineSegmentOf(A,B)))", "answer_expressions": "2*x-y-1=0", "fact_spans": "[[[2, 16]], [[20, 25]], [[20, 25]], [[28, 36]], [[2, 16]], [[2, 25]], [[28, 36]], [[20, 39]]]", "query_spans": "[[[42, 55]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), the equation of the line containing chord AB is y = k(x - 1) + 1, then x_{1} + x_{2} = 2, y_{1} + y_{2} = 2. Since A, B lie on the parabola y^{2} = 4x, (y_{1}^{2} = 4x_{1}, y_{2}^{2} = 4x_{2}), (y_{1} - y_{2}) = 4(x_{1} - x_{2}) = 2, so k = 2. Therefore, the equation of the line containing chord AB is 2x - y - 1 = 0." }, { "text": "The equation of the directrix of the parabola $y^{2}=4 x$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "x=-1", "fact_spans": "[[[0, 14]], [[0, 14]]]", "query_spans": "[[[0, 21]]]", "process": "For the parabola y^{2}=4x, since 2p=4, therefore p=2; and because the parabola opens to the right, the directrix equation is x=-1" }, { "text": "A point on the parabola $y^{2}=4 x$ is at a distance of $3$ from its focus; then the coordinates of this point are?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G);Distance(P, Focus(G)) = 3", "query_expressions": "Coordinate(P)", "answer_expressions": "(2, pm*2*sqrt(2))", "fact_spans": "[[[0, 14], [19, 20]], [[0, 14]], [[31, 32]], [[0, 18]], [[0, 28]]]", "query_spans": "[[[31, 36]]]", "process": "Problem Analysis: From the given conditions, the focus of the parabola is (1,0) and the directrix is x = -1. According to the definition of a parabola—the distance from any point on the parabola to the focus equals the distance from that point to the directrix—the x-coordinate of the point is 2. Substituting into the parabolic equation yields the coordinates of the point as (2,\\pm2\\sqrt{2})." }, { "text": "The equation of the directrix of the parabola $16 y^{2}=x$ is?", "fact_expressions": "G: Parabola;Expression(G) = (16*y^2 = x)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "x=-1/64", "fact_spans": "[[[0, 15]], [[0, 15]]]", "query_spans": "[[[0, 22]]]", "process": "According to the problem, y^{2}=\\frac{1}{16}x, hence 2p=\\frac{1}{16}, \\frac{p}{2}=\\frac{1}{64}, the equation of the directrix is x=-\\frac{1}{64}." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, the line $a x+b y-\\sqrt{a b}=0$ is tangent to the circle $x^{2}+y^{2}=\\frac{2}{5}$. Then, the eccentricity of the ellipse $C$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;H: Line;Expression(H) = (-sqrt(a*b) + a*x + b*y = 0);G: Circle;Expression(G) = (x^2 + y^2 = 2/5);IsTangent(H, G)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3)/2", "fact_spans": "[[[2, 59], [115, 120]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[60, 84]], [[60, 84]], [[85, 111]], [[85, 111]], [[60, 113]]]", "query_spans": "[[[115, 126]]]", "process": "Since the line $ ax + by - \\sqrt{ab} = 0 $ is tangent to the circle $ x^{2} + y^{2} = \\frac{2}{5} $, the distance from the center of the circle $ (0,0) $ to the line $ ax + by - \\sqrt{ab} = 0 $ is $ \\frac{|-\\sqrt{ab}|}{\\sqrt{a^{2}+b^{2}}} = \\sqrt{\\frac{2}{5}} $. Therefore, $ a = 2b $, so $ c = \\sqrt{a^{2} - b^{2}} = \\sqrt{a^{2} - \\frac{1}{4}a^{2}} = \\frac{\\sqrt{3}}{2}a $. Hence, the eccentricity of the ellipse $ C $ is $ e = \\frac{c}{a} = \\frac{\\sqrt{3}}{2} $." }, { "text": "Given that the hyperbola $C$ passes through the point $M(2, 2)$ and has the same asymptotes as $x^{2}-4 y^{2}=4$, find the equation of the hyperbola $C$.", "fact_expressions": "C: Hyperbola;M: Point;Coordinate(M) = (2, 2);PointOnCurve(M, C);Z: Curve;Expression(Z) = (x^2 - 4*y^2 = 4);Asymptote(Z) = Asymptote(C)", "query_expressions": "Expression(C)", "answer_expressions": "y^2/3 - x^2/12 = 1", "fact_spans": "[[[2, 8], [49, 55]], [[9, 20]], [[9, 20]], [[2, 20]], [[23, 40]], [[23, 40]], [[2, 47]]]", "query_spans": "[[[49, 60]]]", "process": "Hyperbola C has the same asymptotes as $x^{2}-4y^{2}=4$. Let the equation of hyperbola C be $x^{2}-4y^{2}=4\\lambda$ ($\\lambda\\neq0$). Hyperbola C passes through point $M(2,2)$, $\\therefore 2^{2}-4\\cdot2^{2}=-12=4\\lambda$, $\\therefore \\lambda=-3$. Therefore, the equation of hyperbola C is $\\frac{y^{2}}{3}-\\frac{x^{2}}{12}=1$." }, { "text": "Let the focus of the parabola $x^{2}=4 y$ be $F$, points $A$, $B$ lie on the parabola, and satisfy $\\overrightarrow{A F}=\\lambda \\overrightarrow{F B}$. If $|\\overrightarrow{A F}|=\\frac{3}{2}$, then the value of $\\lambda$ is?", "fact_expressions": "G: Parabola;A: Point;F: Point;B: Point;lambda:Number;Expression(G) = (x^2=4*y);Focus(G) = F;PointOnCurve(A,G);PointOnCurve(B,G);VectorOf(A, F) = lambda*VectorOf(F, B);Abs(VectorOf(A, F)) = 3/2", "query_expressions": "lambda", "answer_expressions": "1/2", "fact_spans": "[[[1, 15], [32, 35]], [[23, 27]], [[19, 22]], [[28, 31]], [[131, 140]], [[1, 15]], [[1, 22]], [[23, 36]], [[23, 36]], [[40, 91]], [[93, 129]]]", "query_spans": "[[[131, 144]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since the focus of the parabola x^{2}=4y is F(0,1) and the directrix is y=-1, so from |\\overrightarrow{AF}|=\\frac{3}{2}, we obtain y_{1}+1=\\frac{3}{2}, hence y_{1}=\\frac{1}{2}, x_{1}^{2}=4y_{1}=2. From \\overrightarrow{AF}=\\lambda\\overrightarrow{FB}, we get \\begin{cases}-x_{1}=\\lambda x_{2},\\\\1-y_{1}=\\lambda(y_{2}-1)\\end{cases}, that is, \\begin{cases}x_{2}=-\\frac{1}{\\lambda}x_{1},\\\\y_{2}=\\frac{1-y_{1}}{\\lambda}+1=\\frac{1}{2\\lambda}+1\\end{cases}. Since x_{2}^{2}=4y_{2}, we have (-\\frac{1}{\\lambda}x_{1})^{2}=4(\\frac{1}{2\\lambda}+1). Solving yields \\lambda=\\frac{1}{2} or \\lambda=-1 (discarded)." }, { "text": "Given that one focus of the hyperbola $E$: $b x^{2}+y^{2}=-2 b$ coincides with the focus of the parabola $C$: $x^{2}=4 \\sqrt{6} y$, find the equations of the asymptotes of the hyperbola $E$.", "fact_expressions": "E: Hyperbola;Expression(E) = (b*x^2 + y^2 = -2*b);b: Number;C: Parabola;Expression(C) = (x^2 = 4*(sqrt(6)*y));OneOf(Focus(E)) = Focus(C)", "query_expressions": "Expression(Asymptote(E))", "answer_expressions": "y=pm*sqrt(2)*x", "fact_spans": "[[[2, 30], [71, 77]], [[2, 30]], [[10, 30]], [[36, 64]], [[36, 64]], [[2, 69]]]", "query_spans": "[[[71, 85]]]", "process": "The focus of the parabola C: x^{2}=4\\sqrt{6}y is (0,\\sqrt{6}), so one focus of the hyperbola E: \\frac{x^{2}}{-2}+\\frac{y^{2}}{-2b}=1 is (0,\\sqrt{6}), hence b<0, 2-2b=6, solving gives b=-2, that is, \\frac{x^{2}}{-2}+\\frac{y^{2}}{4}=1. Then the asymptotes of the hyperbola E are given by \\frac{x^{2}}{-2}+\\frac{y^{2}}{4}=0, i.e., y=\\pm\\sqrt{2}x." }, { "text": "The directrix equation of the parabola $y=-\\frac{1}{6} x^{2}$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y = -x^2/6)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "y= 3/2", "fact_spans": "[[[0, 25]], [[0, 25]]]", "query_spans": "[[[0, 32]]]", "process": "" }, { "text": "The length of the line segment cut from an asymptote of the hyperbola $x^{2}-y^{2}=1$ by the circle $(x-2)^{2}+y^{2}=4$ is?", "fact_expressions": "G: Hyperbola;H: Circle;Expression(G) = (x^2 - y^2 = 1);Expression(H) = (y^2 + (x - 2)^2 = 4)", "query_expressions": "Length(InterceptChord(OneOf(Asymptote(G)), H))", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[0, 18]], [[25, 45]], [[0, 18]], [[25, 45]]]", "query_spans": "[[[0, 52]]]", "process": "A asymptote of the hyperbola $ x^{2} \\cdot y^{2} = 1 $ can be taken as: $ x + y = 0 $. The circle $ C: (x - 2)^{2} + y^{2} = 4 $ has center $ (2, 0) $ and radius $ 2 $. The distance from the center to the line is: $ d = \\frac{|2 - 0|}{\\sqrt{2}} = \\sqrt{2} $. Therefore, the length of the segment intercepted by the circle $ C $ is $ 2\\sqrt{2 - d^{2}} = 2\\sqrt{4 - (\\sqrt{2})^{2}} = 2\\sqrt{2} $." }, { "text": "The sum of the distances from two points $A$ and $B$ on the parabola $y^{2}=5 x$ to the focus is $10$. Then, what is the distance from the midpoint of segment $AB$ to the $y$-axis?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 5*x);A: Point;B: Point;PointOnCurve(A, G) = True;PointOnCurve(B, G) = True;Distance(A, Focus(G)) + Distance(B, Focus(G)) = 10", "query_expressions": "Distance(MidPoint(LineSegmentOf(A, B)), yAxis)", "answer_expressions": "15/4", "fact_spans": "[[[0, 14]], [[0, 14]], [[18, 21]], [[22, 25]], [[0, 25]], [[0, 25]], [[0, 38]]]", "query_spans": "[[[40, 60]]]", "process": "Problem Analysis: According to the given conditions, the sum of the distances from points A and B to the directrix of the parabola $ x = -\\frac{5}{4} $ is 10. Therefore, the distance from the midpoint of segment AB to the directrix of the parabola $ x = -\\frac{5}{4} $ is 5. The distance from the midpoint of segment AB to the y-axis is $ 5 - \\frac{5}{4} = \\frac{15}{4} $." }, { "text": "The eccentricity of the hyperbola $y^{2}-x^{2}=m(m \\neq 0)$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2 + y^2 = m);m: Number;Negation(m=0)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[0, 28]], [[0, 28]], [[3, 28]], [[3, 28]]]", "query_spans": "[[[0, 34]]]", "process": "When $ m > 0 $, then $ y^{2} - x^{2} = m $, we get $ \\frac{y^{2}}{m} - \\frac{x^{2}}{m} = 1 $. Therefore, $ a^{2} = b^{2} = m $, so $ a = b = \\sqrt{m} $, $ c = \\sqrt{a^{2} + b^{2}} = \\sqrt{2m} $, so the eccentricity $ e = \\frac{c}{a} = \\sqrt{2} $. When $ m < 0 $, $ \\frac{x^{2}}{m} - \\frac{y^{2}}{m} = 1 $, so $ a^{2} = b^{2} = -m $, thus the eccentricity $ e = \\frac{c}{a} = \\sqrt{2} $. In summary, the eccentricity of the hyperbola is $ \\sqrt{2} $. The answer is: $ \\sqrt{2} $." }, { "text": "If the line $l$ passes through the focus $F$ of the parabola $y^{2}=2 p x$, and intersects the parabola at two distinct points $A$ and $B$, where point $A(2, y_{0})$, and $|A F|=4$, then $p=?$", "fact_expressions": "l: Line;G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;F: Point;Focus(G) = F;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, G) = {A, B};Negation(A=B);y0: Number;Coordinate(A) = (2, y0);Abs(LineSegmentOf(A, F)) = 4", "query_expressions": "p", "answer_expressions": "4", "fact_spans": "[[[1, 6]], [[7, 23], [32, 35]], [[7, 23]], [[79, 82]], [[26, 29]], [[7, 29]], [[1, 29]], [[42, 45], [52, 66]], [[46, 49]], [[1, 49]], [[37, 49]], [[53, 66]], [[52, 66]], [[68, 77]]]", "query_spans": "[[[79, 84]]]", "process": "According to the definition of a parabola, the result can be obtained. Based on the given conditions: the directrix equation of the parabola $ y^{2} = 2px $ is $ x = -\\frac{p}{2} $. Since point $ A(2, y_{0}) $ lies on the parabola, we have $ |AF| = 2 - (-\\frac{p}{2}) = 2 + \\frac{p}{2} $. Also, $ |AF| = 4 $. Therefore, $ 2 + \\frac{p}{2} = 4 $, then $ p = 4 $." }, { "text": "The equation of the hyperbola that has the same asymptotes as $\\frac{x^2}{2}-y^{2}=1$ and passes through $(2,0)$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/2 - y^2 = 1);Z: Hyperbola;Asymptote(Z) = Asymptote(G);H: Point;Coordinate(H) = (2, 0);PointOnCurve(H, Z)", "query_expressions": "Expression(Z)", "answer_expressions": "x^2/4 - y^2/2 = 1", "fact_spans": "[[[1, 27]], [[1, 27]], [[44, 47]], [[0, 47]], [[36, 43]], [[36, 43]], [[35, 47]]]", "query_spans": "[[[44, 51]]]", "process": "Let the equation of the hyperbola having the same asymptotes as $\\frac{x^2}{2}-y^2=1$ be: $\\frac{x^2}{2}-y^2=m$ $(m\\neq0)$. Since the hyperbola passes through $(2,0)$, $\\therefore m=2$, $\\frac{x^2}{4}-\\frac{y^2}{2}=1$." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0, b>0)$, the left and right foci are $F_{1}$, $F_{2}$ respectively. $P$ is a point on the hyperbola such that $P F_{1} \\perp P F_{2}$, and the area of $\\Delta P F_{1} F_{2}$ is $2 a b$. Find the eccentricity $e$ of the hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;e: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P,G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));Area(TriangleOf(P, F1, F2)) = 2*(a*b);Eccentricity(G) = e", "query_expressions": "e", "answer_expressions": "sqrt(5)", "fact_spans": "[[[2, 60], [88, 91], [158, 161]], [[5, 60]], [[5, 60]], [[84, 87]], [[68, 75]], [[76, 83]], [[165, 168]], [[5, 60]], [[5, 60]], [[2, 60]], [[2, 83]], [[2, 83]], [[84, 95]], [[97, 120]], [[123, 156]], [[158, 168]]]", "query_spans": "[[[165, 170]]]", "process": "Since PF_{1} \\perp PF_{2}, the area of right triangle PF_{1}F_{2} is \\frac{1}{2}PF_{1}\\cdot PF_{2}=2ab, so PF_{1}\\cdot PF_{2}=4ab and PF_{1}^{2}+PF_{2}^{2}=F_{1}F_{2}^{2}=4c^{2}. Since P is a point on the hyperbola, PF_{1}-PF_{2}=2a. Squaring both sides gives PF_{1}^{2}+PF_{2}^{2}-2PF_{1}\\cdot PF_{2}=4a^{2}, so 4c^{2}-8ab=4a^{2}. Since c^{2}=a^{2}+b^{2}, we have 4a^{2}+4b^{2}-8ab=4a^{2}. Simplifying yields b=2a. Therefore, c^{2}=a^{2}+b^{2}=5a^{2}. Thus, the eccentricity e=\\frac{c}{a}=\\sqrt{5}." }, { "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{m}=1$ are given by $y=\\pm \\frac{\\sqrt{3}}{2} x$, then what are the coordinates of the foci of the hyperbola?", "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/4 - y^2/m = 1);Expression(Asymptote(G)) = (y = pm*(x*(sqrt(3)/2)))", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(pm*sqrt(7),0)", "fact_spans": "[[[1, 39], [76, 79]], [[4, 39]], [[1, 39]], [[1, 74]]]", "query_spans": "[[[76, 86]]]", "process": "Test Analysis: According to the given conditions, since the asymptotes of the hyperbola \\frac{x^2}{4}-\\frac{y^{2}}{m}=1 are y=\\pm\\frac{\\sqrt{3}}{2}x, it follows that \\frac{\\sqrt{3}}{2}=\\frac{\\sqrt{m}}{2}. Solving this gives m=3. Therefore, c=\\sqrt{4+m}=\\sqrt{7}, so the coordinates of the foci of the hyperbola are (\\pm\\sqrt{7},0)." }, { "text": "If points $O$ and $F$ are the center and left focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, respectively, and point $P$ is an arbitrary point on the ellipse, then the range of values of $\\overrightarrow{O P} \\cdot \\overrightarrow{F P}$ (expressed in interval form) is?", "fact_expressions": "G: Ellipse;O: Point;P: Point;F: Point;Center(G)=O;Expression(G) = (x^2/4 + y^2/3 = 1);LeftFocus(G)=F;PointOnCurve(P, G)", "query_expressions": "Range(DotProduct(VectorOf(O,P),VectorOf(F,P)))", "answer_expressions": "[2,6]", "fact_spans": "[[[13, 50], [63, 65]], [[1, 5]], [[58, 62]], [[6, 10]], [[1, 57]], [[13, 50]], [[1, 57]], [[58, 70]]]", "query_spans": "[[[72, 139]]]", "process": "Let P(x,y), we can obtain the coordinate formula of the dot product of vector $\\overrightarrow{OP}$, and combining it with the equation of the ellipse, we can find its solution: \n$\\because$ point P is an arbitrary point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, let $P(x,y)$ $(-2\\leqslant x \\leqslant 2, -\\sqrt{3}\\leqslant y \\leqslant \\sqrt{3})$. \nAccording to the problem, the left focus is $F(-1,0)$, \n$\\therefore \\overrightarrow{OP} = (x,y)$, $\\overrightarrow{FP} = (x+1,y)$, \n$\\therefore \\overrightarrow{OP} \\cdot \\overrightarrow{FP} = x(x+1) + y^{2} = x^{2} + x + \\frac{12--}{4} = \\frac{1}{4}x^{2} + x + 3 = (\\frac{1}{2}x + 1)^{2} + 2$. \nSince $-2 \\leqslant x \\leqslant 2$, \n$\\therefore 0 \\leqslant \\frac{1}{2}x + 1 \\leqslant 2$, \n$\\therefore 0 \\leqslant (\\frac{1}{2}x + 1)^{2} \\leqslant 4$, \n$\\therefore 2 \\leqslant (\\frac{1}{2}x + 1)^{2} + 2 \\leqslant 6$, \nthat is, $2 \\leqslant \\overrightarrow{OP} \\cdot \\overrightarrow{FP} \\leqslant 6$." }, { "text": "The standard equation of the parabola with focus at the left focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ is?", "fact_expressions": "E: Hyperbola;F: Parabola;Expression(E) = (x**2/4-y**2/5=1);Focus(F) = LeftFocus(E)", "query_expressions": "Expression(F)", "answer_expressions": "y**2=-12*x", "fact_spans": "[[[1, 4]], [[49, 52]], [[1, 40]], [[0, 52]]]", "query_spans": "[[[49, 58]]]", "process": "" }, { "text": "If the left focus of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{p^{2}}=1$ lies on the directrix of the parabola $y^{2}=2 px$, then the value of $p$ is?", "fact_expressions": "G: Parabola;p: Number;H: Ellipse;Expression(G) = (y^2 = 2*p*x);Expression(H) = (x^2/5 + y^2/p^2 = 1);PointOnCurve(LeftFocus(H),Directrix(G))", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[47, 62]], [[68, 71]], [[1, 42]], [[47, 62]], [[1, 42]], [[1, 66]]]", "query_spans": "[[[68, 75]]]", "process": "" }, { "text": "Given that the distance from a vertex of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{m^{2}}=1$ $(m>0)$ to one of its asymptotes is $1$, find the value of $m$.", "fact_expressions": "G: Hyperbola;m: Number;m>0;Expression(G) = (x^2/9 - y^2/m^2 = 1);Distance(OneOf(Vertex(G)), OneOf(Asymptote(G))) = 1", "query_expressions": "m", "answer_expressions": "3*sqrt(2)/4", "fact_spans": "[[[2, 49], [55, 56]], [[71, 74]], [[5, 49]], [[2, 49]], [[2, 69]]]", "query_spans": "[[[71, 76]]]", "process": "First, find the equations of the asymptotes and the coordinates of the vertices. Use the point-to-line distance formula to derive the expression for the distance from each vertex to one of its asymptotes. By setting this expression equal to 1, solve for m. [Detailed solution] For $\\frac{x^{2}}{9}-\\frac{y^{2}}{m^{2}}=1$ ($m>0$), an asymptote equation is $y=\\frac{m}{3}x$, i.e., $3x-my=0$, and a vertex coordinate is $(0,3)$. Solving gives $m=\\frac{3\\sqrt{2}}{4}$." }, { "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, $F_{1}$ and $F_{2}$ are the two foci of the ellipse, and $\\angle F_{1} P F_{2}=60^{\\circ}$, find the area of $\\Delta F_{1} P F_{2}$.", "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/16 + y^2/9 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2};AngleOf(F1, P, F2) = ApplyUnit(60, degree)", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "3*sqrt(3)", "fact_spans": "[[[6, 44], [64, 66]], [[48, 55]], [[2, 5]], [[56, 63]], [[6, 44]], [[2, 47]], [[48, 71]], [[72, 105]]]", "query_spans": "[[[107, 134]]]", "process": "Using the definition of the ellipse, we obtain PF_{1}+PF_{2}=2a=8, F_{1}F_{2}=2c=2\\sqrt{7}. The value of PF_{1}\\cdot PF_{2} can be calculated using the law of cosines, and then the area of the triangle can be found using the triangle area formula. From the equation of the ellipse, we get: a=4, b=3, c=\\sqrt{a^{2}-b^{2}}=\\sqrt{7}. In \\Delta F_{1}PF_{2}, PF_{1}+PF_{2}=2a=8, F_{1}F_{2}=2c=2\\sqrt{7}. In \\triangle F_{1}PF_{2}, by the law of cosines: F_{1}F_{2}^{2}=PF_{1}^{2}+PF_{2}^{2}-2PF_{1}\\times PF_{2}\\times \\cos60^{\\circ}, which gives 4c^{2}=(PF_{1}+PF_{2})^{2}-3PF_{1}\\times PF_{2}, leading to 28=64-3PF_{1}\\times PF_{2}. Solving yields: PF_{1}\\times PF_{2}=12. Using the triangle area formula, the area of \\triangle F_{1}PF_{2} is \\frac{1}{2}\\times PF_{1}\\times PF_{2}\\sin60^{\\circ}=\\frac{1}{2}\\times12\\times\\frac{\\sqrt{3}}{2}=3\\sqrt{3}." }, { "text": "In $\\triangle A B C$, $B(-3,0)$, $C(3,0)$, if the perimeter is $16$, then the trajectory equation of vertex $A$ is?", "fact_expressions": "A: Point;B: Point;C: Point;Coordinate(B) = (-3, 0);Coordinate(C) = (3, 0);Perimeter(TriangleOf(A, B, C)) = 16", "query_expressions": "LocusEquation(A)", "answer_expressions": "(x^2/25+y^2/16=1)&Negation(y=0)", "fact_spans": "[[[52, 55]], [[20, 29]], [[31, 39]], [[20, 29]], [[31, 39]], [[1, 48]]]", "query_spans": "[[[52, 62]]]", "process": "" }, { "text": "Given that the curve represented by $m(x^{2}+y^{2}+2 y+1)=(x-2 y+3)^2$ is an ellipse, find the range of values for $m$.", "fact_expressions": "G: Ellipse;H: Curve;Expression(H)=(m*(x^2+y^2+2*y+1)=(x-2*y+3)^2);G=H;m:Number", "query_expressions": "Range(m)", "answer_expressions": "(5,+oo)", "fact_spans": "[[[44, 46]], [[39, 41]], [[2, 41]], [[39, 46]], [[48, 51]]]", "query_spans": "[[[48, 58]]]", "process": "" }, { "text": "Let $F_{1}$, $F_{2}$ be the left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively, $B$ be the bottom vertex of the ellipse, and $P$ be an intersection point of the circle passing through points $F_{1}$, $F_{2}$, $B$ and the ellipse $C$, such that $P F_{1} \\perp F_{1} F_{2}$. Then the value of $\\frac{b}{a}$ is?", "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Circle;P: Point;F1: Point;F2: Point;B: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;LowerVertex(C) = B;PointOnCurve(F1, G);PointOnCurve(F2, G);PointOnCurve(B, G);Intersection(G, C) = P;IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(F1, F2))", "query_expressions": "b/a", "answer_expressions": "(sqrt(5) - 1)/2", "fact_spans": "[[[19, 76], [121, 126], [86, 88]], [[26, 76]], [[26, 76]], [[119, 120]], [[93, 96]], [[1, 8], [98, 106]], [[9, 16], [107, 114]], [[82, 85], [115, 118]], [[26, 76]], [[26, 76]], [[19, 76]], [[1, 81]], [[1, 81]], [[82, 92]], [[97, 120]], [[97, 120]], [[97, 120]], [[93, 131]], [[133, 160]]]", "query_spans": "[[[162, 179]]]", "process": "Let the center of the circle passing through points $F_{1}, F_{2}, B$ be $M$. Since $PF_{1} \\perp F_{1}F_{2}$, $PF_{1}$ is half of the latus rectum, so $PF_{1} = \\frac{b^{2}}{a}$. Since $PF_{1}$ is a chord of circle $M$, by symmetry of the circle, the coordinates of $M$ are $M(0, \\frac{b^{2}}{2a})$. Since $MB^{2} = MF_{1}^{2} = R^{2}$, we have $(\\frac{b^{2}}{2a})^{2} + c^{2} = (\\frac{b^{2}}{2a} + b)^{2}$. Simplifying gives $ac^{2} = b^{3} + ab^{2}$. Since $c^{2} = a^{2} - b^{2}$, substituting yields $a(a^{2} - b^{2}) = b^{3} + ab^{2}$. Simplifying gives $b^{2} + ab - a^{2} = 0$. Therefore, $(\\frac{b}{a})^{2} + (\\frac{b}{a}) - 1 = 0$. Solving gives $\\frac{b}{a} = \\frac{\\sqrt{5}-1}{2}$, discarding the negative root." }, { "text": "Given that $P$ is an arbitrary point on the right branch of hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-y^{2}=1(a>0)$, and $Q$ is an arbitrary point on the circle $C_{2}: x^{2}+y^{2}-6 y+8=0$. If $|P Q|$ is minimized when the ordinate (y-coordinate) of point $P$ is $1$, then what is the eccentricity of the hyperbola?", "fact_expressions": "P: Point;C1: Hyperbola;Expression(C1) = (x^2/a^2-y^2=1);a: Number;a>0;PointOnCurve(P,RightPart(C1)) = True;Q: Point;C2: Circle;Expression(C2) = (x^2+y^2-6*y+8=0);PointOnCurve(Q,C2) = True;WhenMin( Abs( LineSegmentOf(P,Q) )) = True;YCoordinate(P) = 1", "query_expressions": "Eccentricity(C1)", "answer_expressions": "sqrt(6)/2", "fact_spans": "[[[2, 5], [111, 115]], [[6, 51], [125, 128]], [[6, 51]], [[9, 51]], [[9, 51]], [[2, 59]], [[60, 63]], [[64, 93]], [[64, 93]], [[60, 98]], [[100, 110]], [[111, 123]]]", "query_spans": "[[[125, 134]]]", "process": "The center coordinates of circle $ C_{2}: x^{2} + y^{2} - 6y + 8 = 0 $ are $ C_{2}(0,3) $. When $ |PQ| $ is shortest, $ |PC_{2}| $ is minimized. Let the coordinates of point $ P $ be $ (x,y) $, then $ |PC_{2}|^{2} = x^{2} + (y-3)^{2} = (a^{2}+1)y^{2} - 6y + 9 + a^{2} $. When $ y = \\frac{3}{a^{2}+1} $, $ |PC_{2}| $ is minimized. Therefore, $ \\frac{3}{a^{2}+1} = 1 $, solving gives $ a = \\sqrt{2} $. Since $ b = 1 $, we have $ c = \\sqrt{a^{2} + b^{2}} = \\sqrt{3} $, and $ e = \\frac{\\sqrt{3}}{\\sqrt{2}} = \\frac{\\sqrt{6}}{2} $." }, { "text": "For the equation $\\frac{x^{2}}{k+2}+\\frac{y^{2}}{5-k}=-1$ to represent a hyperbola, what is the range of values for $k$?", "fact_expressions": "G: Hyperbola;k:Number;Expression(G)=(x^2/(k+2)+y^2/(5-k)=-1)", "query_expressions": "Range(k)", "answer_expressions": "(-oo,-2)+(5,+oo)", "fact_spans": "[[[45, 48]], [[50, 53]], [[1, 48]]]", "query_spans": "[[[50, 60]]]", "process": "" }, { "text": "The asymptote equations of the hyperbola $\\frac{y^{2}}{4}-\\frac{x^{2}}{2}=1$ are?", "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/2 + y^2/4 = 1)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*sqrt(2)*x", "fact_spans": "[[[0, 38]], [[0, 38]]]", "query_spans": "[[[0, 46]]]", "process": "From the hyperbola equation, determine a and b, and simultaneously identify the axis on which the foci lie. The asymptote equations can be derived. Given a=2, b=\\sqrt{2}, and the foci of the hyperbola are on the y-axis, \\therefore the asymptote equations are y=\\pm\\frac{a}{b}x=\\pm\\sqrt{2}x." }, { "text": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has two foci $F_{1}$ and $F_{2}$. If there exists a point $P$ on the hyperbola, which is not a vertex, such that $\\angle P F_{2} F_{1}=3 \\angle P F_{1} F_{2}$, then what is the range of values for the eccentricity of the hyperbola?", "fact_expressions": "G: Hyperbola;a: Number;b: Number;P: Point;F1: Point;F2: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);Negation(Vertex(G) = P);AngleOf(P, F2, F1) = 3*AngleOf(P, F1, F2)", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1, 2)", "fact_spans": "[[[0, 46], [70, 73], [135, 138]], [[3, 46]], [[3, 46]], [[81, 85]], [[53, 60]], [[61, 68]], [[0, 46]], [[0, 68]], [[70, 85]], [[70, 85]], [[88, 133]]]", "query_spans": "[[[135, 148]]]", "process": "" }, { "text": "Given that an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $y=\\frac{\\sqrt{3}}{3} x$, and one focus lies on the directrix of the parabola $y^{2}=8 x$, then the equation of this hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 8*x);Expression(OneOf(Asymptote(G))) = (y = x*(sqrt(3)/3));PointOnCurve(OneOf(Focus(G)), Directrix(H))", "query_expressions": "Expression(G)", "answer_expressions": "x^2/3-y^2=1", "fact_spans": "[[[2, 58], [119, 122]], [[5, 58]], [[5, 58]], [[98, 112]], [[5, 58]], [[5, 58]], [[2, 58]], [[98, 112]], [[2, 91]], [[2, 116]]]", "query_spans": "[[[119, 127]]]", "process": "Using the asymptote equation of the hyperbola, we obtain \\frac{b}{a}=\\frac{\\sqrt{3}}{3}, then find the coordinates of the foci of the hyperbola; solving for a and b gives the equation of the hyperbola. Solution: \\because one asymptote equation of the hyperbola is y=\\frac{\\sqrt{3}}{3}x, \\therefore \\frac{b}{a}=\\frac{\\sqrt{3}}{3}, \\textcircled{1} \\because the directrix equation of the parabola y^{2}=8x is x=-2, and a focus of the hyperbola lies on the directrix of the parabola y^{2}=8x, \\therefore c=2, and since c=\\sqrt{a^{2}+b^{2}}, \\therefore a^{2}+b^{2}=4, \\textcircled{2} from \\textcircled{1} and \\textcircled{2}, we get a^{2}=3, b^{2}=1, \\therefore the equation of the hyperbola is \\frac{x^{2}}{3}-y^{2}=1." }, { "text": "Given that the center of the ellipse is at the origin, one focus is at $(\\sqrt{3}, 0)$, and the length of the major axis is twice the length of the minor axis, then the standard equation of the ellipse is?", "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;OneOf(Focus(G)) = F ;F: Point;Coordinate(F) = (sqrt(3), 0);Length(MajorAxis(G)) = Length(MinorAxis(G))*2", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4 + y^2 = 1", "fact_spans": "[[[2, 4], [47, 49]], [[7, 9]], [[2, 9]], [[2, 30]], [[15, 30]], [[15, 30]], [[2, 44]]]", "query_spans": "[[[47, 56]]]", "process": "" }, { "text": "The equation of the ellipse, whose vertices are the foci of the hyperbola $-3 x^{2}+y^{2}=12$ and whose foci are the vertices of this hyperbola, is?", "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(G) = (-3*x^2 + y^2 = 12);Focus(G) = Vertex(H);Vertex(G) = Focus(H)", "query_expressions": "Expression(H)", "answer_expressions": "x^2/4 + y^2/16 = 1", "fact_spans": "[[[1, 23]], [[36, 38]], [[1, 23]], [[0, 38]], [[0, 38]]]", "query_spans": "[[[36, 43]]]", "process": "" }, { "text": "The distance from a point $M(x_{1}, y_{1})$ on the parabola $y^{2}=8x$ to its focus is $3$. Then, what is the distance from point $M$ to the coordinate origin?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);x1: Number;y1: Number;M: Point;Coordinate(M) = (x1, y1);PointOnCurve(M, G);Distance(M, Focus(G)) = 3;O: Origin", "query_expressions": "Distance(M, O)", "answer_expressions": "3", "fact_spans": "[[[0, 14], [35, 36]], [[0, 14]], [[17, 34]], [[17, 34]], [[17, 34], [47, 51]], [[17, 34]], [[0, 34]], [[17, 45]], [[52, 56]]]", "query_spans": "[[[47, 61]]]", "process": "Since the distance from a point M(x_{1},y_{1}) on the parabola y^{2}=8x to its focus F is 3, we have MF = x_{1} + \\frac{p}{2} = x_{1} + 2 = 3. Solving gives x_{1}=1, y_{1}=8. Therefore, OM = \\sqrt{x_{1}^{2}+y_{1}^{2}} = \\sqrt{1+8} = 3." }, { "text": "If one of the asymptotes of a hyperbola is $x+2y=0$, and the hyperbola intersects the directrix of the parabola $y=x^{2}$ at exactly one common point, then the standard equation of this hyperbola is?", "fact_expressions": "G: Hyperbola;H: Parabola;Expression(H) = (y = x^2);Expression(OneOf(Asymptote(G)))=(x+2*y=0);NumIntersection(G,Directrix(H)) = 1", "query_expressions": "Expression(G)", "answer_expressions": "x^2/(1/16)-y^2/(1/4)=1", "fact_spans": "[[[1, 4], [22, 25], [51, 54]], [[26, 38]], [[26, 38]], [[1, 20]], [[22, 48]]]", "query_spans": "[[[51, 61]]]", "process": "Since the directrix of the parabola $ y = x^{2} $ is given by the equation: $ y = -\\frac{1}{4} $, and the hyperbola shares only one common point with the directrix of the parabola $ y = x^{2} $, the real semi-axis length of the hyperbola is $ a = \\frac{1}{4} $, and the foci lie on the $ y $-axis. Since one asymptote of the hyperbola is $ x + 2y = 0 $, it follows that $ \\frac{a}{b} = \\frac{1}{2} $, so $ b = \\frac{1}{2} $. Therefore, the standard equation of the hyperbola is $ \\frac{y^{2}}{16} - \\frac{x^{2}}{\\frac{1}{4}} = $" }, { "text": "The equation $\\frac{x^{2}}{m-3}+\\frac{y^{2}}{m-4}=1$ represents an ellipse; then the range of real values for $m$ is?", "fact_expressions": "H: Curve;Expression(H) = (x^2/(m - 3) + y^2/(m - 4) = 1);m: Real;G: Ellipse;H = G", "query_expressions": "Range(m)", "answer_expressions": "(4, +oo)", "fact_spans": "[[[44, 46]], [[0, 46]], [[51, 56]], [[47, 49]], [[44, 49]]]", "query_spans": "[[[51, 63]]]", "process": "According to the problem, \\begin{cases}m-4>0\\\\m-3>0\\end{cases} and m-3\\neq m-4, solving gives m>4" }, { "text": "P is a point on the hyperbola $\\frac{x^{2}}{64}-\\frac{y^{2}}{36}=1$, $F_{1}$, $F_{2}$ are the two foci of the hyperbola, and $|P F_{1}|=15$, then the value of $|P F_{2}|$ is?", "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/64 - y^2/36 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2};Abs(LineSegmentOf(P, F1)) = 15", "query_expressions": "Abs(LineSegmentOf(P, F2))", "answer_expressions": "31", "fact_spans": "[[[4, 44], [64, 67]], [[0, 3]], [[48, 55]], [[56, 63]], [[4, 44]], [[0, 47]], [[48, 72]], [[74, 88]]]", "query_spans": "[[[90, 105]]]", "process": "By the definition of a hyperbola, we know |PF_{1}|-|PF_{2}|=\\pm2a, and 2a=16, so |PF_{2}|=31. Therefore, fill in 31." }, { "text": "Let $ m>0 $, point $ A(4, m) $ be a point on the parabola $ y^{2}=2 p x $ ($ p>0 $), $ F $ be the focus. The circle $ C $ centered at $ A $ with radius $ |A F| $ is intersected by the $ y $-axis, and the length of the chord is $ 6 $. Then the standard equation of circle $ C $ is?", "fact_expressions": "G: Parabola;p: Number;C: Circle;A: Point;F: Point;m: Number;p > 0;m > 0;Expression(G) = (y^2 = 2*(p*x));Coordinate(A) = (4, m);PointOnCurve(A, G);Focus(G) = F;Center(C) = A;Radius(C) = Abs(LineSegmentOf(A, F));Length(InterceptChord(C, yAxis)) = 6", "query_expressions": "Expression(C)", "answer_expressions": "(x - 4)^2 + (y - 4)^2 = 25", "fact_spans": "[[[18, 39]], [[21, 39]], [[68, 72], [88, 92]], [[7, 17], [51, 54]], [[43, 46]], [[1, 6]], [[21, 39]], [[1, 6]], [[18, 39]], [[7, 17]], [[7, 42]], [[18, 49]], [[50, 72]], [[57, 72]], [[68, 86]]]", "query_spans": "[[[88, 99]]]", "process": "From the given condition, the distance from point A(4, m) to the y-axis is 4. It is also known that the chord length intercepted by the y-axis on circle C is 6, so |AF| = \\sqrt{4^{2} + \\left(\\frac{6}{2}\\right)^{2}} = 5. Then 4 + \\frac{p}{2} = 5, thus p = 2. Since point A(4, m) lies on the parabola y^{2} = 2px (p > 0), and m > (\\times 2 \\times 4 = 4, the standard equation of circle C is: (x - 4)^{2} + (y - 4)^{2} = 25" }, { "text": "If the hyperbola $x^{2}-y^{2}=1$ and the circle $(x-1)^{2}+y^{2}=a^{2}$ ($a>0$) have exactly three distinct common points, then $a=$?", "fact_expressions": "G: Hyperbola;H: Circle;a: Number;Expression(G) = (x^2 - y^2 = 1);a > 0;Expression(H) = (y^2 + (x - 1)^2 = a^2);NumIntersection(G, H) = 3", "query_expressions": "a", "answer_expressions": "2", "fact_spans": "[[[1, 19]], [[20, 49]], [[61, 64]], [[1, 19]], [[21, 49]], [[20, 49]], [[1, 59]]]", "query_spans": "[[[61, 66]]]", "process": "" }, { "text": "Given that both asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are tangent to the circle $C$: $x^{2}+y^{2}-8 x+12=0$, and the right focus of the hyperbola is the center of circle $C$, then the equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;C: Circle;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(C) = (-8*x + x^2 + y^2 + 12 = 0);IsTangent(Asymptote(G), C);RightFocus(G) = Center(C)", "query_expressions": "Expression(G)", "answer_expressions": "Preserve all mathematical expressions, LaTeX formulas, symbols, and equations EXACTLY. \nDo not add explanations, do not rewrite formulas, do not omit any content. \n\nText:", "fact_spans": "[[[2, 58], [98, 101], [115, 118]], [[5, 58]], [[5, 58]], [[66, 94], [106, 110]], [[5, 58]], [[5, 58]], [[2, 58]], [[66, 94]], [[2, 96]], [[98, 113]]]", "query_spans": "[[[115, 123]]]", "process": "Using the condition that the asymptotes are tangent to the circle, find the value of \\frac{b}{a}. The center coordinates of circle C can be used to determine the value of \\sqrt{a^{2}+b^{2}}. Thus, a system of equations in terms of a and b can be established. Solving this system gives the values of a and b, allowing us to obtain the equation of the hyperbola. The standard equation of circle C is (x-4)^{2}+y^{2}=4, with center C(4,0) and radius 2. The asymptotes of the hyperbola \\frac{x^2}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0) are given by y=\\pm\\frac{b}{a}x. Since both asymptotes of the hyperbola are tangent to circle C, we have \\frac{4b}{\\sqrt{a^2+b^2}}=2, solving which gives \\frac{b}{a}=\\frac{\\sqrt{3}}{3}. Also, since the right focus of the hyperbola coincides with the center of circle C, we have \\sqrt{a^{2}+b^{2}}=4. Therefore, we have the system:\n\\begin{cases}\nb=\\frac{\\sqrt{3}}{3}a \\\\\n\\sqrt{a^2+b^{2}}=4\n\\end{cases}\nSolving yields\n\\begin{cases}\na=2\\sqrt{3} \\\\\nb=2\n\\end{cases}\nHence, the equation of the hyperbola is \\frac{x^{2}}{12}-\\frac{y^{2}}{4}=1." }, { "text": "It is known that the right focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ coincides with the focus of the parabola $y^{2}=12 x$. Then, the distance from the focus of this hyperbola to its asymptote is equal to?", "fact_expressions": "G: Hyperbola;b: Number;H: Parabola;Expression(G) = (x^2/4 - y^2/b^2 = 1);Expression(H) = (y^2 = 12*x);RightFocus(G) = Focus(H)", "query_expressions": "Distance(Focus(G), Asymptote(G))", "answer_expressions": "sqrt(5)", "fact_spans": "[[[2, 44], [72, 75], [79, 80]], [[5, 44]], [[49, 64]], [[2, 44]], [[49, 64]], [[2, 69]]]", "query_spans": "[[[72, 89]]]", "process": "" }, { "text": "Given that $F$ is the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$, the line $l$ passing through the origin intersects the hyperbola at points $M$ and $N$, and $\\overrightarrow{M F} \\cdot \\overrightarrow{N F}=0$, the area of $\\Delta M N F$ is $a b$. Then the eccentricity of this hyperbola is?", "fact_expressions": "F: Point;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;RightFocus(G) = F;O: Origin;PointOnCurve(O, l) = True;l: Line;Intersection(l, G) = {M, N};M: Point;N: Point;DotProduct(VectorOf(M, F), VectorOf(N, F)) = 0;Area(TriangleOf(M, N, F)) = a*b", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 5]], [[6, 63], [78, 81], [174, 177]], [[6, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[2, 67]], [[69, 71]], [[68, 77]], [[72, 77]], [[72, 92]], [[83, 86]], [[87, 90]], [[94, 145]], [[148, 171]]]", "query_spans": "[[[174, 183]]]", "process": "\\because\\overrightarrow{MF}\\cdot\\overrightarrow{NF}=0,\\therefore\\overrightarrow{MF}\\bot\\overrightarrow{NF}. Let the left focus of the hyperbola be F', connect MF', NF, then quadrilateral F'MFN is a rectangle, \\therefore|MF|=|NF'|, |MN|=|FF'|=2c. Assume point N lies on the right branch of the hyperbola, by the definition of hyperbola we have |NF'|-|NF|=2a, \\therefore|MF|-|NF|=2a. \\because S_{\\triangle MNF}=\\frac{1}{2}|MF||NF|=ab, \\therefore|MF||NF|=2ab. Also in right triangle \\triangle MNF, |MF|^{2}+|NF|^{2}=|MN|^{2}, i.e., (|MF|-|NF|)^{2}+2|MF||NF|=|MN|^{2}, \\therefore(2a)^{2}+2\\cdot2ab=(2c)^{2}, i.e., 4a^{2}+4ab=4c^{2}=4(a^{2}+b^{2}), simplifying yields \\frac{b}{a}=1, \\therefore e=\\frac{c}{a}=\\sqrt{1+(\\frac{b}{a})^{2}}=\\sqrt{2}. Answer: \\sqrt{2}" }, { "text": "If point $P$ lies on the curve $C_{1}$: $y^{2}=8 x$, point $Q$ lies on the curve $C_{2}$: $(x-2)^{2}+y^{2}=1$, and point $O$ is the coordinate origin, then the maximum value of $\\frac{|P O|}{|P Q|}$ is?", "fact_expressions": "P: Point;PointOnCurve(P,C1) = True;C1: Curve;Expression(C1) = (y^2 = 8*x);Q: Point;PointOnCurve(Q,C2) = True;C2: Curve;Expression(C2) = (y^2 + (x - 2)^2 = 1);O: Origin", "query_expressions": "Max(Abs(LineSegmentOf(P, O))/Abs(LineSegmentOf(P, Q)))", "answer_expressions": "4*sqrt(7)/7", "fact_spans": "[[[1, 5]], [[1, 29]], [[6, 28]], [[6, 28]], [[30, 34]], [[30, 65]], [[35, 64]], [[35, 64]], [[66, 70]]]", "query_spans": "[[[77, 104]]]", "process": "" }, { "text": "Let the parabola $K$: $x^{2}=4 y$ have focus $F$, and let $A$, $B$ be two points on $K$ with unequal $y$-coordinates, satisfying $|A F|+|B F|=4$. Then the intercept on the $y$-axis made by the perpendicular bisector of segment $A B$ is?", "fact_expressions": "K: Parabola;A: Point;B: Point;F: Point;Expression(K) = (x^2 = 4*y);PointOnCurve(A,K);PointOnCurve(B,K);Negation(YCoordinate(A)=YCoordinate(B));Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F)) = 4;Focus(K)=F", "query_expressions": "Length(Intercept(PerpendicularBisector(LineSegmentOf(A,B)),yAxis))", "answer_expressions": "3", "fact_spans": "[[[1, 20], [36, 39]], [[28, 31]], [[32, 35]], [[24, 27]], [[1, 20]], [[28, 49]], [[28, 49]], [[28, 49]], [[52, 67]], [[1, 27]]]", "query_spans": "[[[69, 92]]]", "process": "" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, and a point $P$ on the hyperbola satisfies $|P F_{1}|+|P F_{2}|=6$, then the area of $\\triangle P F_{1} F_{2}$ is?", "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 - y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P,G);Abs(LineSegmentOf(P,F1))+Abs(LineSegmentOf(P,F2))=6", "query_expressions": "Area(TriangleOf(P, F1, F2))", "answer_expressions": "2", "fact_spans": "[[[18, 46], [53, 56]], [[59, 62]], [[2, 9]], [[10, 17]], [[18, 46]], [[2, 52]], [[2, 52]], [[53, 62]], [[64, 87]]]", "query_spans": "[[[89, 119]]]", "process": "Without loss of generality, assume that point P lies on the left branch; then |PF₂| - |PF₁| = 2a = 4. Given |PF₁| + |PF₂| = 6, it follows that |PF₂| = 5, |PF₁| = 1. Also, |F₂F₁| = 2c = 2√5. Then, cos∠F₁PF₂ = (25 + 1 - 20)/10 = 3/5. Therefore, ∠F₁PF₂ ∈ (0, π/2), so sin∠F₁PF₂ = 4/5. In conclusion, the area of triangle PF₁F₂ is (1/2) × |PF₂| × |PF₁| × sin∠F₁PF₂ = 2." }, { "text": "Given that point $F$ is the right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a, b>0)$, the line $y=k x$, $k \\in[\\frac{\\sqrt{3}}{3}, \\sqrt{3}]$ intersects the hyperbola $C$ at points $A$ and $B$. If $A F \\perp B F$, then the range of values for the eccentricity of the hyperbola is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(C) = F;G: Line;Expression(G) = (y = k*x);k: Number;In(k, [sqrt(3)/3, sqrt(3)]);A: Point;B: Point;Intersection(G, C) = {A, B};IsPerpendicular(LineSegmentOf(A, F), LineSegmentOf(B, F))", "query_expressions": "Range(Eccentricity(C))", "answer_expressions": "[sqrt(2), sqrt(3)+1]", "fact_spans": "[[[7, 66], [120, 126], [158, 161]], [[7, 66]], [[15, 66]], [[15, 66]], [[15, 66]], [[15, 66]], [[2, 6]], [[2, 70]], [[71, 80]], [[71, 80]], [[82, 119]], [[82, 119]], [[129, 132]], [[133, 136]], [[71, 138]], [[140, 155]]]", "query_spans": "[[[158, 172]]]", "process": "Let the angle of inclination of the line y = kx be \\alpha, then k = \\tan\\alpha \\in [\\frac{\\sqrt{3}}{3}, \\sqrt{3}], so \\alpha = \\angle AOF. Let F be the left focus of the hyperbola, and let point A lie in the first quadrant. Connect FB and FA. By the symmetry of the hyperbola, quadrilateral FBFA is a rectangle. Therefore, FF' = BA = 2c, so OA = c, then A(c \\cdot \\cos\\alpha, c \\cdot \\sin\\alpha). Substituting into the hyperbola equation gives \\frac{c^{2} \\cdot \\cos^{2}\\alpha}{a^{2}} - \\frac{c^{2} \\cdot \\sin^{2}\\alpha}{c^{2}-a^{2}} = 1, so e^{2}\\cos2\\alpha - \\frac{e^{2}\\sin^{2}\\alpha}{e^{2}-1} = 1, thus e^{4}\\cos2\\alpha - 2e^{2} + 1 = 0. Solving yields e = \\frac{1}{1-\\cos\\alpha}. From \\tan\\alpha \\in [\\frac{\\sqrt{3}}{3}, \\sqrt{3}], it follows that" }, { "text": "From a point $P$ on the right branch of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, tangents are drawn to circle $C_{1}$: $(x+5)^{2}+y^{2}=4$ and circle $C_{2}$: $(x-5)^{2}+y^{2}=r^{2}(r>0)$, with points of tangency $M$ and $N$ respectively. If the minimum value of $|P M|^{2}-|P N|^{2}$ is $58$, then what is $r$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);P: Point;PointOnCurve(P, RightPart(G));C1: Circle;Expression(C1) = ((x + 5)^2 + y^2 = 4);C2: Circle;Expression(C2) = ((x - 5)^2 + y^2 = r^2);r: Number;r>0;Z1: Line;Z2: Line;TangentOfPoint(P, C1) = Z1;TangentOfPoint(P, C2) = Z2;M: Point;TangentPoint(Z1, C1) = M;N: Point;TangentPoint(Z2, C2) = N;Min(Abs(LineSegmentOf(P, M))^2 - Abs(LineSegmentOf(P, N))^2) = 58", "query_expressions": "r", "answer_expressions": "sqrt(2)", "fact_spans": "[[[1, 40]], [[1, 40]], [[46, 49]], [[1, 49]], [[53, 81]], [[53, 81]], [[82, 119]], [[82, 119]], [[169, 172]], [[91, 119]], [], [], [[0, 122]], [[0, 122]], [[128, 131]], [[0, 135]], [[132, 135]], [[0, 135]], [[137, 167]]]", "query_spans": "[[[169, 174]]]", "process": "From the given conditions, it follows that $ C_{1}C_{2} $ are the left and right foci of the hyperbola. By the property of circle tangents, we have $ |PM|^{2}-|PN|^{2}=6(|PC_{1}|+|PC_{2}|)+r^{2}-4 $. Using the geometric properties of the hyperbola, the minimum value can be found. Thus, solving: from $ c^{2}=9+16=25 $, we get $ c=5 $, so $ C_{1}C_{2} $ are the left and right foci of the hyperbola, and also the centers of the circles in the problem. Therefore, $ |PM|^{2}-|PN|^{2}=|PC_{1}|^{2}-4-(|PC_{2}|^{2}-r^{2})=(|PC_{1}|-|PC_{2}|)(|PC_{1}|+|PC_{2}|)+r^{2}-4=6(|PC_{1}|+|PC_{2}|)+r^{2}-4 $. When point $ P $ lies on the x-axis, $ |PC_{1}|+|PC_{2}| $ is minimized to $ |C_{1}C_{2}|=10 $. Then the minimum value of $ |PM|^{2}-|PN|^{2} $ is $ 6\\times10+r^{2}-4=58 $, solving gives $ r=\\sqrt{2} $." }, { "text": "Let a line be drawn through the focus $F$ of the parabola $C$: $y^{2}=8x$, intersecting $C$ at points $A$ and $B$. The perpendicular bisector of segment $AB$ intersects the $x$-axis at point $P$. Then $\\frac{|AB|}{|PF|}$=?", "fact_expressions": "C: Parabola;G: Line;B: Point;A: Point;P: Point;F: Point;Expression(C) = (y^2 = 8*x);Focus(C)=F;PointOnCurve(F, G);Intersection(G, C) = {A, B};Intersection(PerpendicularBisector(LineSegmentOf(A,B)), xAxis) = P", "query_expressions": "Abs(LineSegmentOf(A, B))/Abs(LineSegmentOf(P, F))", "answer_expressions": "2", "fact_spans": "[[[1, 20], [30, 33]], [[27, 29]], [[39, 42]], [[35, 38]], [[64, 68]], [[23, 26]], [[1, 20]], [[1, 26]], [[0, 29]], [[27, 44]], [[45, 68]]]", "query_spans": "[[[70, 93]]]", "process": "It is given that the focus of the parabola is (2,0), and the line drawn has a slope. Let the equation be y-0=k(x-2), i.e., kx-y-2k=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}). Solving the line and parabola equations simultaneously and eliminating y yields k^{2}x^{2}-(4k^{2}+8)x+4k^{2}=0. Then x_{1}+x_{2}=\\frac{4k^{2}+8}{k^{2}}, x_{1}x_{2}=4. Further, we obtain y_{1}+y_{2}=\\frac{8}{k}, so the midpoint of AB has coordinates (\\frac{2k^{2}+4}{k^{2}},\\frac{4}{k}). The perpendicular bisector of segment AB has equation y-\\frac{4}{k}=-\\frac{1}{k}(x-\\frac{2k^{2}+4}{k^{2}}). Setting y=0, the x-coordinate of point P is x_{p}=4+\\frac{2k^{2}+4}{k^{2}}, so PF=x_{p}-2=\\frac{4k^{2}+4}{k^{2}}. Using the focal chord formula, we get AB=\\frac{8(1+k^{2})}{k^{2}}. Hence \\frac{|AB|}{|PF|}=2. Therefore, fill in 2 for this problem." }, { "text": "It is known that on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, there exists a point $P$ such that the distance from $P$ to the left focus is equal to 6 times the distance from $P$ to the right directrix. Then, the range of values for the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);P: Point;PointOnCurve(P, RightPart(G));Distance(P, LeftFocus(G)) = 6*Distance(P, RightDirectrix(G))", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1, 2] + [3, 6)", "fact_spans": "[[[2, 58], [98, 101]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 58]], [[63, 67], [69, 72], [81, 84]], [[2, 67]], [[2, 96]]]", "query_spans": "[[[98, 112]]]", "process": "" }, { "text": "If the center of a circle is the focus of the parabola $x^{2}=4 y$, and the chord length intercepted by the line $y=x+3$ is $2$, then what is the standard equation of the circle?", "fact_expressions": "G: Parabola;H: Circle;I: Line;Expression(G) = (x^2 = 4*y);Expression(I) = (y = x + 3);Center(H)=Focus(G);Length(InterceptChord(I,H))=2", "query_expressions": "Expression(H)", "answer_expressions": "x^2+(y-1)^2=3", "fact_spans": "[[[8, 22]], [[3, 4], [3, 4]], [[28, 37]], [[8, 22]], [[28, 37]], [[3, 25]], [[3, 46]]]", "query_spans": "[[[49, 57]]]", "process": "According to the focus of the parabola, the coordinates of the center of the circle can be found. Given that the chord length is 2 and using the chord length formula, $ r_{2} $ can be determined. Substituting into the equation yields the answer. Since the focus of $ x^{2}=4y $ is $ (0,1) $, the center of the required circle is $ (0,1) $. Let the radius of this circle be $ r $. Then the distance from the center $ (0,1) $ to the line $ x-y+3=0 $ is $ d=\\frac{|0-1+3|}{\\sqrt{1^{2}+1^{2}}}=\\sqrt{2} $. Therefore, the chord length $ 2=2\\sqrt{r^{2}-d^{2}}=2\\sqrt{r^{2}-2} $, solving gives $ r^{2}=3 $. Hence, the standard equation of the circle is: $ x^{2}+(y-1)^{2}=3 $." }, { "text": "The area of the triangle formed by one focus of the ellipse $\\frac{x^{2}}{t}+y^{2}=1(t>1)$ and the two endpoints of the minor axis is $1$, then $t=?$", "fact_expressions": "G: Ellipse;t: Number;t>1;Expression(G) = (y^2 + x^2/t = 1);F:Point;A:Point;B:Point;OneOf(Focus(G))=F;Endpoint(MinorAxis(G))={A,B};Area(TriangleOf(F,A,B))=1", "query_expressions": "t", "answer_expressions": "2", "fact_spans": "[[[0, 32]], [[57, 60]], [[2, 32]], [[0, 32]], [], [], [], [[0, 36]], [[0, 42]], [[0, 55]]]", "query_spans": "[[[57, 62]]]", "process": "\\because the ellipse equation \\frac{x^{2}}{t}+y^{2}=1(t>1), \\therefore its foci lie on the x-axis, its semi-minor axis b=1, let the semi-focal distance be c, the area of the triangle formed by one focus and the two endpoints of the minor axis is 1, and \\frac{1}{2}\\times2b\\times c=1, while b=1, \\therefore c=1, \\therefore t=b^{2}+c^{2}=1+1=2," }, { "text": "The coordinates of the midpoint of the chord intercepted by the line $y=x+1$ on the ellipse $x^{2}+2 y^{2}=4$ are?", "fact_expressions": "G: Ellipse;H: Line;Expression(G) = (x^2 + 2*y^2 = 4);Expression(H) = (y = x + 1)", "query_expressions": "Coordinate(MidPoint(InterceptChord(H, G)))", "answer_expressions": "(-2/3, 1/3)", "fact_spans": "[[[10, 29]], [[0, 9]], [[10, 29]], [[0, 9]]]", "query_spans": "[[[0, 41]]]", "process": "Problem Analysis: Problem Analysis: Solving the system \\begin{cases}y=x+1\\\\x^{2}+2y^{2}=4\\end{cases}, we get $x^{2}+2(x+1)^{2}-4=0$, which simplifies to $3x^{2}+4x-2=0$. The x-coordinate of the midpoint of the chord is $-\\frac{2}{3}$, and the y-coordinate is $-\\frac{2}{3}+1=\\frac{1}{3}$, so the midpoint is $\\left(-\\frac{2}{3},\\frac{1}{3}\\right)$; thus, fill in $\\left(-\\frac{2}{3},\\frac{1}{3}\\right)$." }, { "text": "Given that line $l$ passes through the focus $F$ of the parabola $C$: $x^{2}=2 p y(p>0)$, intersects $C$ at points $A$, $B$, and intersects the directrix of $C$ at point $M$. If $\\overrightarrow{A F}=\\overrightarrow{F M}$, then $\\frac {p}{|B F|}$=?", "fact_expressions": "l: Line;C: Parabola;p: Number;A: Point;F: Point;M: Point;B: Point;p>0;Expression(C) = (x^2 = 2*(p*y));Focus(C) = F;PointOnCurve(F, l);Intersection(l,C)={A,B};Intersection(l,Directrix(C))= M;VectorOf(A, F) = VectorOf(F, M)", "query_expressions": "p/Abs(LineSegmentOf(B, F))", "answer_expressions": "3/2", "fact_spans": "[[[2, 7]], [[8, 34], [42, 45], [57, 60]], [[16, 34]], [[46, 49]], [[37, 40]], [[64, 68]], [[50, 53]], [[16, 34]], [[8, 34]], [[8, 40]], [[2, 40]], [[2, 55]], [[2, 68]], [[71, 114]]]", "query_spans": "[[[116, 136]]]", "process": "As shown in the figure, draw a perpendicular line AD from point A to the directrix of the parabola. Since \\overrightarrow{AF}=\\overrightarrow{FM}, we have |AD|=|AF|=|FM|. Thus, the inclination angle of line l is \\frac{5\\pi}{6}. Let the equation of line l be y=-\\frac{\\sqrt{3}}{3}x+\\frac{p}{2}. Let A(x_{1},y_{1}), B(x_{2},y_{2}). Solving the system of equations: \n\\begin{cases}y=-\\frac{\\sqrt{3}}{3}x+\\frac{p}{2},\\\\x^2=2py\\end{cases} \nwe obtain 12y^{2}-20py+3p^{2}=0\\textcircled{1}. Solving gives y_{1}=\\frac{3}{2}p. Therefore, |BF|+\\frac{p}{2}=\\frac{p}{6}+\\frac{p}{2}=\\frac{2}{3}p, hence \\frac{p}{|BF|}=\\frac{3}{2}." }, { "text": "Given that a line with an inclination angle of $60^{\\circ}$ passes through the focus $F$ of the curve $C$: $y^{2}=2 x$, and intersects $C$ at two distinct points $A$, $B$ ($A$ is in the first quadrant), then $|A F|$=?", "fact_expressions": "G: Line;C: Curve;A: Point;B: Point;F: Point;Expression(C) = (y^2 = 2*x);Inclination(G) = ApplyUnit(60, degree);PointOnCurve(F, G);Intersection(G, C) = {A, B};Quadrant(A) = 1;Negation(A = B);Focus(C) = F", "query_expressions": "Abs(LineSegmentOf(A, F))", "answer_expressions": "2", "fact_spans": "[[[19, 21]], [[22, 40], [49, 52]], [[60, 63], [68, 71]], [[64, 67]], [[43, 46]], [[22, 40]], [[2, 21]], [[19, 46]], [[19, 67]], [[68, 76]], [[55, 67]], [[22, 46]]]", "query_spans": "[[[79, 88]]]", "process": "" }, { "text": "Given that $O$ is the coordinate origin, the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $F$, and the circle with diameter $OF$ intersects an asymptote of the hyperbola at a point $A$ distinct from the origin. If the line connecting point $A$ and the midpoint of $OF$ is perpendicular to $OF$, then the eccentricity $e$ of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;O: Origin;F: Point;A:Point;e: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;IsDiameter(LineSegmentOf(O,F),H);Intersection(OneOf(Asymptote(G)),H)=A;Negation(A=O);IsPerpendicular(LineSegmentOf(A,MidPoint(LineSegmentOf(O,F))),LineSegmentOf(O,F));Eccentricity(G) = e", "query_expressions": "e", "answer_expressions": "sqrt(2)", "fact_spans": "[[[11, 67], [88, 91], [133, 136]], [[14, 67]], [[14, 67]], [[86, 87]], [[2, 5]], [[72, 75]], [[103, 106]], [[140, 143]], [[14, 67]], [[14, 67]], [[11, 67]], [[11, 75]], [[76, 87]], [[86, 106]], [[98, 106]], [[108, 131]], [[133, 143]]]", "query_spans": "[[[140, 145]]]", "process": "Since the line connecting point A and the midpoint of OF is perpendicular to OF, triangle OAF is an isosceles right triangle, so the base angle AOF is 45 degrees. Hence, a = b, and the eccentricity is \\sqrt{2}." }, { "text": "The focus of the parabola $y^{2}=2 p x(p>0)$ is $F$. Points $A$ and $B$ are two moving points on the parabola such that $\\angle A F B=60^{\\circ}$. From the midpoint $M$ of chord $AB$, a perpendicular $MN$ is drawn to the directrix of the parabola, with $N$ being the foot of the perpendicular. Then the maximum value of $\\frac{|M N|}{|A B|}$ is?", "fact_expressions": "G: Parabola;p: Number;A: Point;B: Point;M: Point;N: Point;F: Point;p>0;Expression(G) = (y^2 = 2*p*x);Focus(G) = F;PointOnCurve(A,G);PointOnCurve(B,G);AngleOf(A, F, B) = ApplyUnit(60, degree);IsChordOf(LineSegmentOf(A,B),G);MidPoint(LineSegmentOf(A,B))=M;PointOnCurve(M,LineSegmentOf(M,N));IsPerpendicular(LineSegmentOf(M,N),Directrix(G));FootPoint(LineSegmentOf(M,N),Directrix(G))=N", "query_expressions": "Max(Abs(LineSegmentOf(M, N))/Abs(LineSegmentOf(A, B)))", "answer_expressions": "1", "fact_spans": "[[[0, 21], [42, 45], [42, 45]], [[3, 21]], [[31, 35]], [[38, 41]], [[91, 94]], [[111, 114]], [[25, 28]], [[3, 21]], [[0, 21]], [[0, 28]], [[31, 51]], [[31, 51]], [[55, 80]], [[42, 88]], [[84, 94]], [[82, 107]], [[95, 107]], [[95, 114]]]", "query_spans": "[[[116, 143]]]", "process": "Let |AF| = a, |BF| = b. By the definition of a parabola, we have |AF| = |AQ|, |BF| = |BP|. In trapezoid ABPQ, ∴ 2|MN| = |AQ| + |BP| = a + b. By the cosine law, |AB|² = a² + b² − 2ab cos60° = a² + b² − ab. Completing the square gives |AB|² = (a + b)² − 3ab. Since ab ≤ ((a + b)/2)², ∴ (a + b)² − 3ab ≥ (a + b)² − (3/4)(a + b)² = (1/4)(a + b)², yielding |AB| ≥ (1/2)(a + b). ∴ |MN| / |AB| ≤ 1, i.e., the maximum value of |MN| / |AB| is 1." }, { "text": "The hyperbola $C$ with foci on the $x$-axis shares the same asymptotes as the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$, and the distance from a focus of $C$ to one of the asymptotes is $3 \\sqrt{2}$. Find the equation of the hyperbola $C$?", "fact_expressions": "C: Hyperbola;PointOnCurve(Focus(C), xAxis);Z: Hyperbola;Expression(Z) = (x^2/4-y^2/9=1);Asymptote(C) = Asymptote(Z);Distance(Focus(C), OneOf(Asymptote(C))) = 3*sqrt(2)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/8 - y^2/18 = 1", "fact_spans": "[[[9, 15], [63, 66], [93, 99]], [[0, 15]], [[16, 54]], [[16, 54]], [[9, 61]], [[63, 91]]]", "query_spans": "[[[93, 104]]]", "process": "Analysis 1: From the family of hyperbolas sharing asymptotes, we can assume the equation of C is: \\frac{x^{2}}{4}-\\frac{y^{2}}{9}=\\lambda (\\lambda>0). According to the condition that the distance from the focus to the asymptote is b, an equation can be constructed to solve for \\lambda, thereby obtaining the hyperbola's equation. As per the given conditions, assume the equation of hyperbola C is: \\frac{x^{2}}{4}-\\frac{y^{2}}{9}=\\lambda (\\lambda>0), which can be rewritten as \\frac{x^{2}}{4\\lambda}-\\frac{y^{2}}{9\\lambda}=1; then a^{2}=4\\lambda, b^{2}=9\\lambda. \\because the distance from the focus of the hyperbola to its asymptote is b, \\therefore 3\\sqrt{2}=\\sqrt{9\\lambda}, solving gives: \\lambda=2. \\therefore The equation of hyperbola C is: \\frac{x^{2}}{8}-\\frac{y^{2}}{18}=1" }, { "text": "If the point $(3,1)$ is the midpoint of a chord of the parabola $y^{2}=2 p x$, and the slope of the line containing this chord is $2$, then $p$=?", "fact_expressions": "G: Parabola;p: Number;H: LineSegment;Expression(G) = (y^2 = 2*p*x);Coordinate(Z) = (3, 1);Z: Point;MidPoint(H) = Z;IsChordOf(H,G);Slope(OverlappingLine(H))=2", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[10, 26]], [[51, 54]], [], [[10, 26]], [[1, 9]], [[1, 9]], [[1, 33]], [[10, 30]], [[10, 49]]]", "query_spans": "[[[51, 56]]]", "process": "Problem Analysis: The equation of the line passing through the point (3,1) with slope 2 is y=2x-5. Substituting into the parabola y^{2}=2px, we obtain (2x-5)^{2}=2px^{2}, which simplifies to 4x^{2}-(20+2p)x+25=0\\cdot\\frac{20+2p}{4}=6,\\therefore p=2." }, { "text": "Given the hyperbola $C_{1}$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{8}=1$, the hyperbola $C_{2}$ that shares the same asymptotes as $C_{1}$ passes through $(2,4)$. Then the equation of $C_{2}$ is?", "fact_expressions": "C1: Hyperbola;C2: Hyperbola;H: Point;Expression(C1) = (x^2/4 - y^2/8 = 1);Coordinate(H) = (2, 4);Asymptote(C1)=Asymptote(C2);PointOnCurve(H,C2)", "query_expressions": "Expression(C2)", "answer_expressions": "y^2/8 - x^2/4 = 1", "fact_spans": "[[[2, 48], [50, 57]], [[62, 72], [82, 89]], [[73, 80]], [[2, 48]], [[73, 80]], [[49, 72]], [[62, 80]]]", "query_spans": "[[[82, 94]]]", "process": "Let the equation of hyperbola $ C_{2} $ be: $\\frac{x^{2}}{4}-\\frac{y^{2}}{8}=\\lambda $. From the given condition, we have $\\frac{2}{4}-\\frac{4^{2}}{8}=\\lambda$, $\\therefore \\lambda=1-2=-1$. Therefore, the equation of hyperbola $ C_{2} $ is: $\\frac{x^{2}}{4}-\\frac{y^{2}}{8}=-1$, that is: $\\frac{y^{2}}{8}-\\frac{x^{2}}{4}=1$." }, { "text": "Point $Q$ lies on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{7}=1$. Then, the maximum distance from point $Q$ to the line $3x-2y-16=0$ is?", "fact_expressions": "G: Ellipse;H: Line;Q: Point;Expression(G) = (x^2/4 + y^2/7 = 1);Expression(H) = (3*x - 2*y - 16 = 0);PointOnCurve(Q, G)", "query_expressions": "Max(Distance(Q,H))", "answer_expressions": "24*sqrt(13)/13", "fact_spans": "[[[5, 42]], [[50, 66]], [[0, 4], [0, 4]], [[5, 42]], [[50, 66]], [[0, 43]]]", "query_spans": "[[[45, 75]]]", "process": "Let the parametric equations of the ellipse be \\begin{cases}x=2\\\\v=\\end{cases}\\frac{6\\cos\\theta-2\\sqrt{7}\\sin\\theta-1}{\\sqrt{13}}=\\frac{\\sqrt[0]{si}(\\theta \\text{ as parameter})}{|2\\sqrt{7}\\sin\\theta-6\\cos\\theta+16|}=\\frac{|8si(\\theta-\\varphi)+16|}{\\sqrt{13}} \\left(\\text{where } \\tan\\varphi=\\frac{3}{\\sqrt{7}}\\right). When \\sin(\\theta-\\varphi)=1, d_{\\max}=\\frac{24}{13}=\\frac{24\\sqrt{13}}{13}." }, { "text": "Given the circle $(x-2)^{2}+y^{2}=4$, then the shortest chord length of the intersection between a line passing through the focus of the parabola $y^{2}=4 x$ and the given circle is equal to?", "fact_expressions": "G: Parabola;H: Circle;Expression(G) = (y^2 = 4*x);Expression(H) = (y^2 + (x - 2)^2 = 4);C:Line;PointOnCurve(Focus(G),C)", "query_expressions": "Min(Length(InterceptChord(C,H)))", "answer_expressions": "2*sqrt(3)", "fact_spans": "[[[26, 40]], [[2, 22], [48, 49]], [[26, 40]], [[2, 22]], [[43, 45]], [[25, 45]]]", "query_spans": "[[[43, 59]]]", "process": "The focus of the parabola $ y^{2}=4x $ is $ (1,0) $, and the equation of the directrix is $ x=1 $. Since $ (1-2)^{2}+0^{2}=1<4 $, the focus lies inside the circle. A line passing through the focus of the parabola $ y^{2}=4x $ intersects the given circle, and when the chord length is shortest, the distance from the center of the circle to the line is maximized. At this time, the equation of the line is $ x=1 $, and the distance from the center of the circle to the line is 1. Therefore, the shortest chord length equals $ 2\\sqrt{4-1}=2\\sqrt{3} $." }, { "text": "Let the angle of inclination of one asymptote of the hyperbola $\\frac{y^{2}}{a^{2}}-x^{2}=1(a>0)$ be $30^{\\circ}$. Then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2 + y^2/a^2 = 1);a: Number;a>0;Inclination(OneOf(Asymptote(G))) = ApplyUnit(30, degree)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[1, 38], [64, 67]], [[1, 38]], [[4, 38]], [[4, 38]], [[1, 61]]]", "query_spans": "[[[64, 73]]]", "process": "The hyperbola \\frac{y^2}{a^{2}}-x^{2}=1 (a>0) has an asymptote with an inclination angle of 30^{\\circ}, so a=\\tan30^{\\circ}=\\frac{\\sqrt{3}}{3}, e=\\frac{\\sqrt[2]{3}}{\\frac{\\sqrt{3}}{3}}=2" }, { "text": "Given fixed points $M(3,1)$, $N(-5,0)$, if a moving point $P(x, y)$ satisfies the equation $\\sqrt{(x+5)^{2}+y^{2}}+\\sqrt{(x-5)^{2}+y^{2}}=20$, then the minimum value of $|P M|+|P N|$ is?", "fact_expressions": "M: Point;N: Point;P: Point;Coordinate(M) = (3, 1);Coordinate(N) = (-5, 0);Coordinate(P) = (x, y);x: Number;y: Number;sqrt(y^2 + (x - 5)^2) + sqrt(y^2 + (x + 5)^2) = 20", "query_expressions": "Min(Abs(LineSegmentOf(P, M)) + Abs(LineSegmentOf(P, N)))", "answer_expressions": "20 - sqrt(5)", "fact_spans": "[[[4, 12]], [[14, 23]], [[27, 36]], [[4, 12]], [[14, 23]], [[27, 36]], [[27, 36]], [[27, 36]], [[40, 90]]]", "query_spans": "[[[92, 111]]]", "process": "" }, { "text": "Given the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, the left and right foci are denoted as $F_{1}$, $F_{2}$ respectively. Let $P$ be a point on the right branch of the hyperbola, and the coordinates of point $Q$ are $(-2,3)$. Then the minimum value of $|P Q|+|P F_{1}|$ is?", "fact_expressions": "G: Hyperbola;P: Point;Q: Point;F1: Point;F2:Point;Expression(G) = (x^2 - y^2/3 = 1);Coordinate(Q) = (-2, 3);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P,RightPart(G))", "query_expressions": "Min(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, Q)))", "answer_expressions": "7", "fact_spans": "[[[2, 30], [58, 61]], [[54, 57]], [[67, 71]], [[38, 45]], [[46, 53]], [[2, 30]], [[67, 83]], [[2, 53]], [[2, 53]], [[54, 66]]]", "query_spans": "[[[85, 108]]]", "process": "Problem Analysis: By the definition of a hyperbola, we know that |PF_{1}| = |PF_{2}| + 2; hence, |PQ| + |PF_{1}| = |PQ| + |PF_{2}| + 2. It can be seen that when points Q, P, and F_{2} are collinear, |PQ| + |PF_{1}| is minimized, and the minimum value is |QF_{2}| + 2 = 5 + 2 = 7." }, { "text": "The line segment $AB$ of fixed length $2$ has its two endpoints moving on the parabola $C$: $y = x^2$, and point $M$ is the midpoint of segment $AB$. Then the trajectory equation of point $M$ is?", "fact_expressions": "C: Parabola;B: Point;A: Point;P1: Point;P2: Point;M: Point;Expression(C) = (y = x^2);Length(LineSegmentOf(A, B)) = 2;Endpoint(LineSegmentOf(A, B)) = {P1, P2};PointOnCurve(P1, C);PointOnCurve(P2, C);MidPoint(LineSegmentOf(A, B)) = M", "query_expressions": "LocusEquation(M)", "answer_expressions": "(y - x^2)*(1 + 4*x^2) = 1", "fact_spans": "[[[20, 37]], [[9, 14]], [[9, 14]], [], [], [[41, 45], [58, 62]], [[20, 37]], [[0, 14]], [[7, 19]], [[7, 38]], [[7, 38]], [[41, 56]]]", "query_spans": "[[[58, 69]]]", "process": "" }, { "text": "If the asymptotes of a hyperbola are given by $y=\\pm 3 x$, and one of its foci is $(\\sqrt{10}, 0)$, then the equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*(3*x));Coordinate(OneOf(Focus(G))) = (sqrt(10), 0)", "query_expressions": "Expression(G)", "answer_expressions": "x^2 - y^2/9 = 1", "fact_spans": "[[[1, 4], [23, 24], [48, 51]], [[1, 22]], [[23, 46]]]", "query_spans": "[[[48, 56]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{2}=1(a>\\sqrt{2})$, the left and right foci are $F_{1}$, $F_{2}$ respectively. A line with slope $-2$ passing through the left focus $F_{1}$ intersects the ellipse at points $A$, $B$. $P$ is the midpoint of $AB$, $O$ is the coordinate origin. If the slope of line $OP$ is $\\frac{1}{4}$, then the value of $a$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/2 + x^2/a^2 = 1);a: Number;a>sqrt(2);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Line;PointOnCurve(F1, H) = True;Slope(H) = -2;Intersection(H, G) = {A, B};A: Point;B: Point;MidPoint(LineSegmentOf(A, B)) = P;P: Point;O: Origin;Slope(LineOf(O, P)) = 1/4", "query_expressions": "a", "answer_expressions": "2", "fact_spans": "[[[2, 55], [103, 105]], [[2, 55]], [[166, 169]], [[4, 55]], [[64, 71], [84, 91]], [[72, 79]], [[2, 79]], [[2, 79]], [[100, 102]], [[80, 102]], [[92, 102]], [[100, 118]], [[107, 110]], [[113, 116]], [[119, 130]], [[119, 122]], [[131, 134]], [[141, 164]]]", "query_spans": "[[[166, 173]]]", "process": "The ellipse $\\frac{x^{2}}{a^2}+\\frac{y^{2}}{2}=1$ ($a>\\sqrt{2}$), so the foci lie on the x-axis. Since a line passing through the left focus $F_{1}$ has slope $-2$, and $P$ is the midpoint of $AB$, let $P(x_{0},y_{0})$, $A(x_{1},y_{1})$, $B(x_{2},y_{2})$. Substituting the coordinates of $A$ and $B$ into the ellipse equation, we obtain\n\\[\n\\begin{cases}\n\\frac{x_{1}^2}{a^2}+\\frac{y_{1}^2}{2}=1 \\\\\n\\frac{x_{2}^2}{a^2}+\\frac{y_{2}^2}{2}=1\n\\end{cases}\n\\]\nSubtracting these two equations and simplifying yields\n\\[\n-\\frac{2(x_{1}+x_{2})}{a^{2}(y_{1}+y_{2})}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}},\n\\]\nthat is,\n\\[\n-\\frac{2x_{0}}{a^{2}y_{0}}=k.\n\\]\nFurther simplification gives\n\\[\n-\\frac{2}{a^{2}}=k\\times\\frac{y_{0}}{x_{0}}.\n\\]\nSubstituting $-\\frac{2}{a^{2}}=-2\\times\\frac{1}{4}$, we solve to get $a=2$. \n【Analysis】This problem examines the positional relationship between a line and an ellipse, and the application of the point-difference method in solving chord midpoint problems, belonging to a medium-difficulty question." }, { "text": "Given that the equation $\\frac{x^{2}}{m+2}-\\frac{y^{2}}{m+1}=1$ represents a hyperbola, what is the range of values for $m$?", "fact_expressions": "G: Hyperbola;m:Number;Expression(G)=(x^2/(m + 2) - y^2/(m + 1) = 1)", "query_expressions": "Range(m)", "answer_expressions": "(-oo,-2)+(-1,oo)", "fact_spans": "[[[45, 48]], [[50, 53]], [[4, 48]]]", "query_spans": "[[[50, 60]]]", "process": "" }, { "text": "Given the parabola $C$: $y^{2}=4x$, point $P$ is an arbitrary point on the directrix of parabola $C$. Two tangent lines are drawn from point $P$ to parabola $C$, with points of tangency $A$ and $B$, respectively. Then, the maximum distance from point $M(0,1)$ to line $AB$ is?", "fact_expressions": "C: Parabola;B: Point;A: Point;M: Point;P: Point;Expression(C) = (y^2 = 4*x);Coordinate(M) = (0, 1);PointOnCurve(P, Directrix(C));L1:Line;L2:Line;TangentOfPoint(P, C) = {L1,L2};TangentPoint(L1,C)=A;TangentPoint(L2,C)=B", "query_expressions": "Max(Distance(M, LineOf(A,B)))", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 21], [27, 33], [49, 55]], [[70, 73]], [[66, 69]], [[75, 84]], [[22, 26], [44, 48]], [[2, 21]], [[75, 84]], [[22, 42]], [], [], [[43, 60]], [[43, 73]], [[43, 73]]]", "query_spans": "[[[75, 101]]]", "process": "Let P(-1, m), A(x_{1}, y_{1}), B(x_{2}, y_{2}). According to the problem, the slope of the tangent line at point A is defined and not zero. Let the slope of the tangent line at point A be k; then the equation of the tangent line is y - y_{1} = k(x - x_{1}). Therefore,\n\n\\[\n\\begin{cases}\ny - y_{1} = k(x - x_{1}) \\\\\ny^{2} = 4x\n\\end{cases}\n\\]\n\nRearranging gives:\n\n\\[\ny^{2} - \\frac{4}{k}y + \\frac{4y_{1}}{k} - 4x_{1} = 0\n\\]\n\nFrom Δ = 0, solving yields k = \\frac{2}{y_{1}}. Thus, the equation of the tangent line at point A is:\n\n\\[\n2x - y_{1}y + 2x_{1} = 0\n\\]\n\nSimilarly, the equation of the tangent line at point B is:\n\n\\[\n2x - y_{2}y + 2x_{2} = 0\n\\]\n\nSince both pass through point P, we have:\n\n\\[\n-2 - m y_{1} + 2x_{1} = 0, \\quad -2 - m y_{2} + 2x_{2} = 0\n\\]\n\nThus, the equation of line AB is:\n\n\\[\n-2 - m y + 2x = 0, \\quad \\text{i.e.,} \\quad 2x - m y - 2 = 0\n\\]\n\nLine AB always passes through the fixed point (1, 0). Therefore, the maximum distance from point M(0, 1) to line AB is the distance from point M(0, 1) to the fixed point (1, 0), which is \\sqrt{2}." }, { "text": "The focus of the parabola $x^{2}=4 y$ is $F$, and a line passing through the point $Q$, which is the intersection of the directrix and the $y$-axis, is tangent to the parabola at point $P$. Then the standard equation of the circumcircle of $\\triangle F P Q$ is?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);F: Point;Focus(G) = F;Intersection(Directrix(G), yAxis) = Q;Q: Point;PointOnCurve(Q, L) = True;L: Line;TangentPoint(L, G) = P;P: Point", "query_expressions": "Expression(CircumCircle(TriangleOf(F, P, Q)))", "answer_expressions": "{((x-1)^2+y^2=2),((x+1)^2+y^2=2)}", "fact_spans": "[[[0, 14], [24, 25], [42, 45]], [[0, 14]], [[18, 21]], [[0, 21]], [[24, 38]], [[35, 38]], [[22, 41]], [[39, 41]], [[39, 51]], [[47, 51]]]", "query_spans": "[[[53, 80]]]", "process": "Problem Analysis: From the given conditions, we know F(0,1). Let P(x_{0},\\frac{1}{4}x_{0}). Since y'=\\frac{1}{2}x, the equation of the tangent line is y-\\frac{1}{4}x_{0}^{2}=\\frac{1}{2}x_{0}(x-x_{0}). Substituting (0,-1) yields x_{0}=\\pm2. Then P(2,1), (-2,1). It follows that PF\\perp FQ. Therefore, the circumcircle of \\triangle FPQ has PQ as diameter, giving (x-1)^{2}+y^{2}=2 or (x+1)^{2}+y^{2}=2. Hence, the answer to this problem is (x-1)^{2}+y^{2}=2 or (x+1)^{2}+y^{2}=2." }, { "text": "Let the line $l$ pass through a focus of the hyperbola $C$, and be perpendicular to one of the symmetry axes of $C$. The line $l$ intersects $C$ at points $A$ and $B$, and $|AB|$ is twice the length of the real axis of $C$. Then the eccentricity of the hyperbola $C$ is?", "fact_expressions": "l: Line;C: Hyperbola;A: Point;B: Point;PointOnCurve(OneOf(Focus(C)),l);IsPerpendicular(l,OneOf(SymmetryAxis(C)));Intersection(l, C) = {A, B};Abs(LineSegmentOf(A,B))=2*Length(RealAxis(C))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3)", "fact_spans": "[[[1, 6], [33, 36]], [[7, 13], [21, 24], [37, 40], [60, 63], [74, 80]], [[42, 45]], [[46, 49]], [[1, 18]], [[1, 32]], [[33, 51]], [[52, 72]]]", "query_spans": "[[[74, 86]]]", "process": "Without loss of generality, let the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $, focus $ F(-c,0) $, axis of symmetry $ y = 0 $. From the given condition, we know that $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $, $ x = c \\Rightarrow y = \\pm \\frac{b^{2}}{a} $. Since the length of $ |AB| $ is twice the length of the real axis, $ \\therefore \\frac{2b^{2}}{a} = 4a $, $ b^{2} = 2a^{2} $, $ c^{2} - a^{2} = 2a^{2} $, $ c^{2} = 3a^{2} $, $ \\therefore e = \\frac{c}{a} = \\sqrt{3} $," }, { "text": "Let the hyperbola $C$: $\\frac{x^{2}}{4}-y^{2}=1$ have its right focus at $F_{2}$, then the distance from $F_{2}$ to the asymptote is?", "fact_expressions": "C: Hyperbola;F2: Point;Expression(C) = (x^2/4 - y^2 = 1);RightFocus(C) = F2", "query_expressions": "Distance(F2, Asymptote(C))", "answer_expressions": "1", "fact_spans": "[[[1, 34]], [[39, 46], [48, 55]], [[1, 34]], [[1, 46]]]", "query_spans": "[[[1, 64]]]", "process": "" }, { "text": "The standard equation of the hyperbola passing through points $M(-4,3)$, $N(\\frac{4 \\sqrt{3}}{3},-1)$ is?", "fact_expressions": "M: Point;N: Point;Coordinate(M) = (-4, 3);Coordinate(N) = ((4*sqrt(3))/3, -1);G: Hyperbola;PointOnCurve(M,G) = True;PointOnCurve(N,G) = True", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4 - y^2/3 = 1", "fact_spans": "[[[2, 11]], [[13, 41]], [[2, 11]], [[13, 41]], [[44, 47]], [[0, 47]], [[0, 47]]]", "query_spans": "[[[44, 54]]]", "process": "Let the standard equation of the hyperbola be obtained by substituting point coordinates to solve for parameters, thus determining the standard equation (detailed explanation). Let \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1, then \\begin{cases}\\frac{16}{a^{2}}-\\frac{9}{b^{2}}=1\\\\\\frac{16}{3a^{2}}-\\frac{1}{b^{2}}=1\\end{cases}, which yields \\begin{cases}a^{2}=4\\\\b^{2}=3\\end{cases}. Let \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1, then \\begin{cases}\\frac{9}{a^{2}}-\\frac{16}{b^{2}}=1\\\\\\frac{1}{a^{2}}-\\frac{16}{212}=1\\end{cases}, which has no solution. Therefore, the standard equation of the hyperbola is \\frac{x^{3}}{4}-\\frac{y^{2}}{3}=1" }, { "text": "Given the ellipse $\\frac{x^{2}}{81}+\\frac{y^{2}}{25}=1$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. If there exists a point $P$ on the ellipse such that $|P F_{1}|=2|P F_{2}|$, then $|P F_{1}|=$?", "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/81 + y^2/25 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))", "query_expressions": "Abs(LineSegmentOf(P, F1))", "answer_expressions": "12", "fact_spans": "[[[2, 41], [67, 69]], [[72, 76]], [[50, 57]], [[58, 65]], [[2, 41]], [[2, 65]], [[2, 65]], [[67, 76]], [[78, 100]]]", "query_spans": "[[[102, 115]]]", "process": "From the standard equation of the ellipse, we know: a=9. From the definition of the ellipse, we have: |PF_{1}|+|PF_{2}|=2a=18, and |PF_{1}|=2|PF_{2}|. Solving gives |PF_{2}|=6, therefore |PF|=2\\times6=12." }, { "text": "The distance from the focus to the directrix of the parabola $y^{2}=20 x$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 20*x)", "query_expressions": "Distance(Focus(G), Directrix(G))", "answer_expressions": "10", "fact_spans": "[[[0, 15], [19, 20]], [[0, 15]]]", "query_spans": "[[[0, 27]]]", "process": "\\because the parabola y^{2}=20x,\\therefore p=10, then the distance from the focus to the directrix is 10." }, { "text": "Given that a vertex of the hyperbola $2 m x^{2}-m y^{2}=2$ is $(0,1)$, then the value of $m$ is?", "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (2*m*x^2 - m*y^2 = 2);Coordinate(OneOf(Vertex(G))) = (0, 1)", "query_expressions": "m", "answer_expressions": "-2", "fact_spans": "[[[2, 26]], [[41, 44]], [[2, 26]], [[2, 39]]]", "query_spans": "[[[41, 48]]]", "process": "From the given information, it can be determined that the focus lies on the y-axis, and $-\\frac{2}{m}=1$, which allows us to solve for $m$. [Detailed Solution] The hyperbola $2mx^{2}-my^{2}=2$ is rewritten as $\\frac{x^{2}}{m}-\\frac{y^{2}}{m}=1$. Since one vertex is $(0,1)$, the focus lies on the y-axis and $a=1$. Therefore, $m<0$, and $-\\frac{2}{m}=1$. Solving gives $m=-2$." }, { "text": "The equation of the parabola is $y^{2}=8x$, the right focus of the hyperbola coincides with the focus of the parabola, and the eccentricity is $2$. Then the standard equation of the hyperbola is?", "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 8*x);G: Hyperbola;RightFocus(G) = Focus(H);Eccentricity(G) = 2", "query_expressions": "Expression(G)", "answer_expressions": "x^2-y^2/3=1", "fact_spans": "[[[2, 5], [29, 32]], [[2, 20]], [[21, 24], [45, 48]], [[21, 35]], [[21, 43]]]", "query_spans": "[[[45, 55]]]", "process": "" }, { "text": "$A(-2 , 0)$, $B(2 , 0)$, $P$ is a moving point on the line $x=-1$, then the range of the eccentricity $e$ of the hyperbola with foci at $A$, $B$ and passing through point $P$ is?", "fact_expressions": "G: Hyperbola;H: Line;P: Point;A:Point;B:Point;Focus(G)={A,B};Eccentricity(G)=e;Coordinate(A)=(-2,0);Coordinate(B)=(2,0);Expression(H) = (x = -1);PointOnCurve(P,H);PointOnCurve(P,G);e:Number", "query_expressions": "Range(e)", "answer_expressions": "[2,+oo)", "fact_spans": "[[[63, 66]], [[31, 39]], [[58, 62], [27, 30]], [[0, 11], [46, 49]], [[50, 53], [13, 25]], [[45, 66]], [[63, 71]], [[0, 11]], [[13, 25]], [[31, 39]], [[27, 43]], [[57, 66]], [[70, 73]]]", "query_spans": "[[[70, 80]]]", "process": "" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, the left focus is $F$, and $P$ is a moving point on the right branch of hyperbola $C$. $A(0,4)$. Then the minimum perimeter of $\\triangle P A F$ is?", "fact_expressions": "C: Hyperbola;A: Point;P: Point;F: Point;Expression(C) = (x^2/4 - y^2/5 = 1);Coordinate(A) = (0, 4);LeftFocus(C)=F;PointOnCurve(P, RightPart(C))", "query_expressions": "Min(Perimeter(TriangleOf(P, A, F)))", "answer_expressions": "14", "fact_spans": "[[[58, 64], [2, 45]], [[71, 79]], [[54, 57]], [[50, 53]], [[2, 45]], [[71, 79]], [[2, 53]], [[54, 70]]]", "query_spans": "[[[81, 106]]]", "process": "Problem Analysis: Find the coordinates of the right focus H. According to the definition of a hyperbola, we have |PF| + |PA| = 2a + |PH| + |PA| \\geqslant 2a + |AH|. Calculate the value of 2a + |AH|, and then the minimum perimeter of triangle PAF can be determined. Since F is the left focus of the hyperbola C: \\frac{x^{2}}{4} - \\frac{y^{2}}{5} = 1, we have a = 2, b = \\sqrt{5}, c = 3, F(-3, 0), and the right focus is H(3, 0). By the definition of the hyperbola, |PF| + |PA| = 2a + |PH| + |PA| \\geqslant 2a + |AH| = 4 + \\sqrt{(3-0)^{2} + (0-4)^{2}} = 4 + 5 = 9. Since |AF| = \\sqrt{(-3-0)^{2} + (0-4)^{2}} = 5, the minimum perimeter of triangle PAF is 14." }, { "text": "If the curve represented by the equation $\\frac{x^{2}}{1+k}-\\frac{y^{2}}{k-1}=1$ with respect to $x$ and $y$ is an ellipse with foci on the $x$-axis, then what is the range of values for $k$?", "fact_expressions": "G: Curve;Expression(G) = (x^2/(1 + k) - y^2/(k - 1) = 1);k: Number;H: Ellipse;PointOnCurve(Focus(H), xAxis);G = H", "query_expressions": "Range(k)", "answer_expressions": "(0, 1)", "fact_spans": "[[[57, 59]], [[13, 59]], [[73, 76]], [[69, 71]], [[60, 71]], [[57, 71]]]", "query_spans": "[[[73, 83]]]", "process": "" }, { "text": "Given that point $P$ is any point on the right branch of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{3}=1$, and from point $P$ perpendiculars are drawn to the two asymptotes, with the feet of the perpendiculars being $M$ and $N$, then what is the area of $\\triangle PMN$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/3 = 1);P: Point;PointOnCurve(P, RightPart(G));Z1: Line;Z2: Line;K1: Line;K2: Line;Asymptote(G) = {K1, K2};PointOnCurve(P, Z1);PointOnCurve(P, Z2);IsPerpendicular(K1, Z1);IsPerpendicular(K2, Z2);M: Point;N: Point;FootPoint(K1, Z1) = M;FootPoint(K2, Z2) = N", "query_expressions": "Area(TriangleOf(P, M, N))", "answer_expressions": "9*sqrt(3)/16", "fact_spans": "[[[7, 45]], [[7, 45]], [[2, 6], [54, 58]], [[2, 52]], [], [], [], [], [[7, 64]], [[7, 67]], [[7, 67]], [[7, 67]], [[7, 67]], [[73, 76]], [[79, 82]], [[7, 82]], [[7, 82]]]", "query_spans": "[[[84, 104]]]", "process": "" }, { "text": "The hyperbola $8 k x^{2}-ky^{2}=8$ has a focus at $(0 , 3)$. Find the value of $k$.", "fact_expressions": "G: Hyperbola;k: Number;Expression(G) = (8*k*x^2 - k*y^2 = 8);Coordinate(OneOf(Focus(G))) = (0, 3)", "query_expressions": "k", "answer_expressions": "-1", "fact_spans": "[[[0, 23]], [[40, 43]], [[0, 23]], [[0, 38]]]", "query_spans": "[[[40, 47]]]", "process": "Problem Analysis: First, convert the equation of the hyperbola $8kx^{2}-ky^{2}=8$ into standard form. From the coordinates of the foci, obtain $c^{2}=9$, and then use the relationship among $a$, $b$, and $c$ in the standard equation of a hyperbola to find the value of $k$ in the hyperbola's equation. \nSolution: According to the given conditions, the hyperbola $8kx^{2}-ky^{2}=8$ lies on the y-axis, that is, $\\frac{y^{2}}{\\frac{8}{k}}-\\frac{x^{2}}{-\\frac{1}{k}}$. Therefore, the coordinates of the foci are $(0,3)$, so $c^{2}=9$. Then $\\frac{8}{k}-\\frac{1}{k}=9$, therefore $k=-1$." }, { "text": "If the focus of the parabola $y^{2}=a x$ coincides with the right focus of the hyperbola, then what is the value of $a$?", "fact_expressions": "G: Hyperbola;H: Parabola;a: Number;Expression(H) = (y^2 = a*x);Focus(H) = RightFocus(G)", "query_expressions": "a", "answer_expressions": "8", "fact_spans": "[[[19, 22]], [[1, 15]], [[30, 33]], [[1, 15]], [[1, 28]]]", "query_spans": "[[[30, 37]]]", "process": "Problem Analysis: Since the focus of the parabola y^{2}=ax is (\\frac{a}{4},0), it follows that \\frac{a}{4}=2, a=8" }, { "text": "Given point $F(c , 0)$ is the right focus of hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$, point $B$ is an endpoint of the imaginary axis of the hyperbola, and line $BF$ is perpendicular to one asymptote of the hyperbola. Then the eccentricity of hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;c: Number;B: Point;F: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (c, 0);RightFocus(C) = F;OneOf(Endpoint(ImageinaryAxis(C))) = B;IsPerpendicular(LineOf(B, F), OneOf(Asymptote(C)))", "query_expressions": "Eccentricity(C)", "answer_expressions": "(1 + sqrt(5))/2", "fact_spans": "[[[14, 78], [88, 91], [107, 110], [120, 126]], [[22, 78]], [[22, 78]], [[3, 13]], [[83, 87]], [[2, 13]], [[22, 78]], [[22, 78]], [[14, 78]], [[2, 13]], [[2, 82]], [[83, 98]], [[99, 118]]]", "query_spans": "[[[120, 132]]]", "process": "By symmetry, select one asymptote of hyperbola C with equation $ y = \\frac{b}{a}x $. The corresponding line BF has equation $ \\frac{x}{c} + \\frac{y}{b} = 1 $, so $ k_{BF} = -\\frac{b}{c} $. Thus, $ \\frac{b}{a} \\cdot \\left(-\\frac{b}{c}\\right) = -1 $, which implies $ ac = b^{2} $. Then $ ac = c^{2} - a^{2} $. Dividing both sides by $ a^{2} $, we obtain $ e^{2} - e - 1 = 0 $. Since $ e > 1 $, the eccentricity of the hyperbola is $ e = \\frac{1+\\sqrt{5}}{2} $." }, { "text": "The distance from the focus of the parabola $y^{2}=4 x$ to the asymptote of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ is?", "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 4*x);G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1)", "query_expressions": "Distance(Focus(H), Asymptote(G))", "answer_expressions": "sqrt(5)/5", "fact_spans": "[[[0, 14]], [[0, 14]], [[18, 46]], [[18, 46]]]", "query_spans": "[[[0, 55]]]", "process": "Problem Analysis: Given that the focus of the parabola is F(1,0), and the asymptotes of the hyperbola have equations x\\pm2y=0, the required distance is \\frac{|1\\pm0|}{\\sqrt{1^{2}+2^{2}}}=\\frac{\\sqrt{5}}{5}" }, { "text": "The center of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is at the origin, $F_{1}$ and $F_{2}$ are the left and right foci respectively, $A$ and $B$ are the upper vertex and the right vertex of the ellipse respectively, $P$ is a point on the ellipse such that $P F_{1} \\perp x$-axis and $P F_{2} / / A B$, then the eccentricity of this ellipse is?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;O: Origin;Center(G) = O;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;A: Point;B: Point;UpperVertex(G) = A;RightVertex(G) = B;P: Point;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), xAxis);IsParallel(LineSegmentOf(P, F2), LineSegmentOf(A, B))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)/5", "fact_spans": "[[[0, 52], [93, 95], [108, 110], [154, 156]], [[0, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[56, 58]], [[0, 58]], [[59, 66]], [[67, 74]], [[0, 82]], [[0, 82]], [[83, 86]], [[87, 90]], [[83, 103]], [[83, 103]], [[104, 107]], [[104, 113]], [[115, 133]], [[134, 151]]]", "query_spans": "[[[154, 162]]]", "process": "As shown in the figure, substitute x = -c into the ellipse equation \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0), we obtain P(-c,\\frac{b^{2}}{a}). Also, A(0,b), B(a,0), F_{2}(c,0), \\therefore k_{AB}=-\\frac{b}{a}. Since PF_{2} \\parallel AB, \\therefore -\\frac{b}{a}=-\\frac{b^{2}}{2ac}, simplifying yields b=2c. Then 4c^{2}=b^{2}=a^{2}-c^{2}, that is, a^{2}=5c^{2}, \\therefore e=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\frac{\\sqrt{5}}{5}" }, { "text": "The equation $\\frac{x^{2}}{k+2}-\\frac{y^{2}}{k+1}=1$ represents a hyperbola with foci on the $y$-axis. Then the range of real values for $k$ is?", "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (x^2/(k + 2) - y^2/(k + 1) = 1);PointOnCurve(Focus(G), yAxis)", "query_expressions": "Range(k)", "answer_expressions": "(-oo, -2)", "fact_spans": "[[[52, 55]], [[57, 62]], [[0, 55]], [[43, 55]]]", "query_spans": "[[[57, 69]]]", "process": "" }, { "text": "The standard equation of a circle with center at point $A(0,5)$ and tangent to the asymptotes of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ is?", "fact_expressions": "G: Hyperbola;H: Circle;A: Point;Expression(G) = (x^2/16 - y^2/9 = 1);Coordinate(A) = (0, 5);Center(H)=A;IsTangent(Asymptote(G),H)", "query_expressions": "Expression(H)", "answer_expressions": "x^2+(y-5)^2=16", "fact_spans": "[[[14, 53]], [[61, 62]], [[1, 10]], [[14, 53]], [[1, 10]], [[0, 62]], [[14, 62]]]", "query_spans": "[[[61, 69]]]", "process": "The equations of the asymptotes of the hyperbola are: y=\\pm\\frac{3}{4}x. The distance d from point A to the asymptote of the hyperbola is d=\\frac{|0-5|}{\\sqrt{1+(\\frac{3}{4})^{2}}}=4, so the radius of the circle is 4. Therefore, the standard equation of the required circle is x^{2}+(y-5)^{2}=16" }, { "text": "Given that the distance from a point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ to one focus is $3$, then the distance from $P$ to the other focus is?", "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(P, G);F1: Point;F2: Point;OneOf(Focus(G)) = F1;Distance(P, F1) = 3;OneOf(Focus(G)) = F2;Negation(F1 = F2)", "query_expressions": "Distance(P, F2)", "answer_expressions": "7", "fact_spans": "[[[2, 41]], [[43, 47], [61, 64]], [[2, 41]], [[2, 47]], [], [], [[2, 52]], [[2, 59]], [[2, 70]], [[2, 70]]]", "query_spans": "[[[2, 75]]]", "process": "" }, { "text": "Given the parabola $C$: $y^{2}=12x$ and the point $M(-3, 4)$, a line passing through the focus of $C$ with slope $k$ intersects $C$ at points $A$ and $B$. If $\\overrightarrow{MA} \\cdot \\overrightarrow{MB} = 0$, then what is the value of $k$?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 12*x);M: Point;Coordinate(M) = (-3, 4);PointOnCurve(Focus(C), L);Slope(L) = k;k: Number;L: Line;Intersection(L, C) = {A, B};A: Point;B: Point;DotProduct(VectorOf(M, A), VectorOf(M, B)) = 0", "query_expressions": "k", "answer_expressions": "3/2", "fact_spans": "[[[2, 22], [37, 40], [54, 57]], [[2, 22]], [[23, 35]], [[23, 35]], [[36, 53]], [[44, 53]], [[47, 50], [123, 126]], [[51, 53]], [[51, 70]], [[59, 62]], [[64, 68]], [[72, 121]]]", "query_spans": "[[[123, 130]]]", "process": "Problem Analysis: From the given information, the equation of the line passing through points A and B can be found as y = k(x - 3). Then, by solving the system of equations:\n\\[\n\\begin{cases}\ny^{2} = 12x \\\\\ny = k(x - 3)\n\\end{cases}\n\\]\nwe obtain:\n\\[\nk^{2}x^{2} - 2(6 + 3k^{2})x + 9k^{2} = 0\n\\]\nFrom this, we can express \\( x_{1} + x_{2} \\), \\( x_{1}x_{2} \\), \\( y_{1} + y_{2} \\), and \\( y_{1}y_{2} \\). Using the condition \\( \\overrightarrow{MA} \\cdot \\overrightarrow{MB} = 0 \\), substituting and simplifying allows us to solve for \\( k \\).\n\nSolution:\nSince the focus of the parabola \\( C: y^{2} = 12x \\) is \\( F(3, 0) \\), the equation of the line passing through points A and B is \\( y = k(x - 3) \\). \nSolving the system:\n\\[\n\\begin{cases}\ny^{2} = 12x \\\\\ny = k(x - 3)\n\\end{cases}\n\\]\nwe get:\n\\[\nk^{2}x^{2} - 2(6 + 3k^{2})x + 9k^{2} = 0\n\\]\nLet \\( A(x_{1}, y_{1}) \\), \\( B(x_{2}, y_{2}) \\). Then:\n\\[\nx_{1} + x_{2} = \\frac{12 + 6k^{2}}{k^{2}}, \\quad x_{1}x_{2} = 9\n\\]\n\\[\ny_{1} + y_{2} = k(x_{1} + x_{2} - 6) = \\frac{12}{k}, \\quad y_{1}y_{2} = k^{2}(x_{1} - 3)(x_{2} - 3) = k^{2}[x_{1}x_{2} - 3(x_{1} + x_{2}) + 9] = -36\n\\]\nGiven \\( M(3, 4) \\), we have:\n\\[\n\\overrightarrow{MA} = (x_{1} - 3, y_{1} - 4), \\quad \\overrightarrow{MB} = (x_{2} - 3, y_{2} - 4)\n\\]\nSince \\( \\overrightarrow{MA} \\cdot \\overrightarrow{MB} = 0 \\),\n\\[\n(x_{1} - 3)(x_{2} - 3) + (y_{1} - 4)(y_{2} - 4) = 0\n\\]\nSimplifying gives:\n\\[\nx_{1}x_{2} - 3(x_{1} + x_{2}) + 9 + y_{1}y_{2} - 4(y_{1} + y_{2}) + 16 = 0\n\\]\n\\[\nx_{1}x_{2} - 3(x_{1} + x_{2}) + y_{1}y_{2} - 4(y_{1} + y_{2}) + 25 = 0\n\\]\nSubstituting values:\n\\[\n9 - 3 \\times \\frac{12 + 6k^{2}}{k^{2}} - 36 - 4 \\times \\frac{12}{k} + 25 = 0\n\\]\nSolving yields:\n\\[\nk = \\frac{3}{2}\n\\]" }, { "text": "If the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{n}=1$ $(m>n>0)$ and the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ have the same foci $F_{1}$, $F_{2}$, and $P$ is an intersection point of the two curves, then what is the value of $PF_{1} \\cdot PF_{2}$?", "fact_expressions": "H: Ellipse;Expression(H) = (y^2/n + x^2/m = 1);m: Number;n: Number;m > n;n > 0;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Focus(H) = {F1, F2};Focus(G) = {F1, F2};F1: Point;F2: Point;P: Point;OneOf(Intersection(H, G)) = P", "query_expressions": "LineSegmentOf(P, F1)*LineSegmentOf(P, F2)", "answer_expressions": "m - a^2", "fact_spans": "[[[1, 45]], [[1, 45]], [[3, 45]], [[3, 45]], [[3, 45]], [[3, 45]], [[46, 103]], [[46, 103]], [[49, 103]], [[49, 103]], [[49, 103]], [[49, 103]], [[1, 124]], [[1, 124]], [[109, 116]], [[117, 124]], [[126, 129]], [[125, 139]]]", "query_spans": "[[[141, 166]]]", "process": "" }, { "text": "Let $F_{1}$, $F_{2}$ be the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$), and let $P$ be a point on the right branch of the hyperbola satisfying $(\\overrightarrow{O P}+\\overrightarrow{O F_{2}}) \\cdot \\overrightarrow{P F_{2}}=0$ (where $O$ is the origin), and $3|\\overrightarrow{P F_{1}}|=4|\\overrightarrow{P F_{2}}|$. Then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, RightPart(G));O: Origin;DotProduct((VectorOf(O, F2) + VectorOf(O, P)),VectorOf(P, F2)) = 0;3*Abs(VectorOf(P, F1)) = 4*Abs(VectorOf(P, F2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "5", "fact_spans": "[[[17, 73], [84, 87], [250, 253]], [[17, 73]], [[20, 73]], [[20, 73]], [[20, 73]], [[20, 73]], [[1, 8]], [[9, 16]], [[1, 79]], [[1, 79]], [[80, 83]], [[80, 92]], [[180, 183]], [[95, 177]], [[191, 248]]]", "query_spans": "[[[250, 259]]]", "process": "Problem Analysis: Since point P lies on the right branch of the hyperbola, by the definition of a hyperbola, we have |PF₁| - |PF₂| = 2a and |PF₁| = (4/3)|PF₂|. Therefore, |PF₁| = 8a, |PF₂| = 6a. Since (OP + OF₂) · PF₂ = 0, it follows that (OP + OF₂) · (OF₂ - OP) = 0, so OP² = OF₂². Then in triangle PF₁F₂, |OP| = |OF₂| = |OF|, thus ∠F₁PF₂ = 90°. By the Pythagorean theorem, |PF₁|² + |PF₂|² = |F₁F₂|², which gives 64a² + 36a² = 4c², therefore c = 5a, hence e = 5." }, { "text": "Given the parabola $C$: $y^{2}=2px$ ($p>0$) with focus $F$, the point of intersection of the line $y=4$ and the $y$-axis is $P$, and the point of intersection of the line $y=4$ and the parabola $C$ is $Q$. It is given that $|QF|=2|PQ|$. Then, the equation of the parabola $C$ is?", "fact_expressions": "C: Parabola;p: Number;G: Line;Q: Point;F: Point;P: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Expression(G) = (y = 4);Focus(C) = F;Intersection(G, yAxis) = P;Intersection(G,C)=Q;Abs(LineSegmentOf(Q, F)) = 2*Abs(LineSegmentOf(P, Q))", "query_expressions": "Expression(C)", "answer_expressions": "y^2=8*x", "fact_spans": "[[[2, 28], [57, 63], [88, 94]], [[9, 28]], [[36, 43]], [[67, 70]], [[32, 35]], [[52, 55]], [[9, 28]], [[2, 28]], [[36, 43]], [[2, 35]], [[36, 55]], [[36, 70]], [[72, 86]]]", "query_spans": "[[[88, 99]]]", "process": "It is easy to see that the focus of the parabola $ C $ is $ F\\left(\\frac{p}{2},0\\right) $. Solving the system \n$$\n\\begin{cases}\ny=4 \\\\\ny^2=2px\n\\end{cases}\n$$\nyields \n$$\n\\begin{cases}\nx=\\frac{8}{p} \\\\\ny=4\n\\end{cases}\n$$\nwhich means point $ Q\\left(\\frac{8}{p},4\\right) $. From $ |QF|=2|PQ| $, we obtain \n$$\n\\frac{8}{p} + \\frac{p}{2} = 2 \\times \\frac{8}{p}.\n$$\nSince $ p>0 $, solving gives $ p=4 $. Therefore, the equation of the parabola $ C $ is $ y^2 = 8x $." }, { "text": "Through the focus $F$ of the parabola $y^{2}=m x$ ($m>0$), draw a line with slope $2 \\sqrt{2}$ intersecting the parabola at points $A$ and $B$. The circle with diameter $AB$ intersects the directrix $l$ at a common point $M$. If $|M F|=\\sqrt{2}$, then $|A B|=$?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = m*x);m: Number;m>0;F: Point;Focus(G) = F;Z: Line;PointOnCurve(F, Z);Slope(Z) = 2*sqrt(2);A: Point;B: Point;Intersection(Z, G) = {A, B};H: Circle;IsDiameter(LineSegmentOf(A, B), H);l: Line;Directrix(G) = l;M: Point;Intersection(H, l) = M;Abs(LineSegmentOf(M, F)) = sqrt(2)", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "3", "fact_spans": "[[[1, 20], [46, 49]], [[1, 20]], [[4, 20]], [[4, 20]], [[23, 26]], [[1, 26]], [[43, 45]], [[0, 45]], [[27, 45]], [[50, 53]], [[54, 57]], [[43, 59]], [[70, 71]], [[60, 71]], [[74, 77]], [[46, 77]], [[81, 84]], [[70, 84]], [[86, 102]]]", "query_spans": "[[[104, 113]]]", "process": "Solution: Without loss of generality, assume point A is above the x-axis. According to the properties of the parabola, the circle with AB as diameter intersects the directrix l at a common point M; therefore, MA ⊥ MB. Let C be the midpoint of AB, and connect MC. By the properties of the parabola, MC is parallel to the x-axis, and MF ⊥ AB, so |MF|² = |AF| ⋅ |BF|. Since line AB passes through the focus F of the parabola y² = mx (m > 0) and has slope 2√2, by the definition of the parabola and the properties of a right trapezoid, we obtain |AF| = 2|BF|. Given |MF| = √2, it follows that (√2)² = 2|BF|², so |BF| = 1, |AF| = 2, hence |AB| = 3." }, { "text": "Given $A(2,0)$, $B(0,1)$ are two vertices of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, the line $y=k x(k>0)$ intersects the line $AB$ at point $D$, and intersects the ellipse at points $E$, $F$. If $\\overrightarrow{E D}=6 \\overrightarrow{D F}$, then the value of the slope $k$ is?", "fact_expressions": "A: Point;Coordinate(A) = (2, 0);B: Point;Coordinate(B) = (0, 1);G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;In(A, Vertex(G));H: Line;Expression(H) = (y = k*x);k: Number;k>0;Intersection(H, LineOf(A, B)) = D;D: Point;Intersection(H, G) = {E, F};E: Point;F: Point;VectorOf(E, D) = 6*VectorOf(D, F);Slope(H) = k;In(B, Vertex(G))", "query_expressions": "k", "answer_expressions": "3/8, 2/3", "fact_spans": "[[[2, 10]], [[2, 10]], [[12, 20]], [[12, 20]], [[21, 66], [103, 105]], [[21, 66]], [[23, 66]], [[23, 66]], [[2, 71]], [[72, 86]], [[72, 86]], [[168, 171]], [[74, 86]], [[72, 101]], [[97, 101]], [[72, 117]], [[108, 111]], [[112, 115]], [[119, 164]], [[72, 171]], [[2, 71]]]", "query_spans": "[[[168, 175]]]", "process": "From the problem, we know that the equation of the ellipse is $\\frac{x^{2}}{4}+y^{2}=1$, and the equations of lines $AB$ and $EF$ are $x+2y=2$ and $y=kx$, respectively. Let $D(x_{0},y_{0})$, $E(x_{1},kx_{1})$, $F(x_{2},kx_{2})$, where $x_{1} 0 is possible. This problem examines finding the parameter range based on the positional relationship between a line and a hyperbola, and belongs to basic problems." }, { "text": "Given point $A(0,-1)$, when point $B$ lies on the curve $y=2 x^{2}+1$, the equation of the trajectory of the midpoint $M$ of segment $AB$ is?", "fact_expressions": "G: Curve;A: Point;B: Point;M: Point;Expression(G) = (y = 2*x^2 + 1);Coordinate(A) = (0, -1);PointOnCurve(B, G);MidPoint(LineSegmentOf(A,B)) = M", "query_expressions": "LocusEquation(M)", "answer_expressions": "y=4*x^2", "fact_spans": "[[[18, 33]], [[3, 12]], [[14, 17]], [[45, 48]], [[18, 33]], [[3, 12]], [[14, 34]], [[35, 48]]]", "query_spans": "[[[45, 55]]]", "process": "" }, { "text": "Given a point on the parabola $y^{2}=6x$ such that its distance to the focus is twice its distance to the $y$-axis, find the $x$-coordinate of this point?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 6*x);P:Point;PointOnCurve(P,G);Distance(P,Focus(G))=2*Distance(P,yAxis)", "query_expressions": "XCoordinate(P)", "answer_expressions": "3/2", "fact_spans": "[[[2, 16]], [[2, 16]], [[42, 43]], [[2, 20]], [[2, 39]]]", "query_spans": "[[[42, 49]]]", "process": "Let the x-coordinate of the point be $x_{0}$. From the problem and the definition of the parabola, we have $x_{0}+\\frac{p}{2}=x_{0}+\\frac{3}{2}=2x_{0}\\Rightarrow x_{0}=\\frac{3}{2}$." }, { "text": "The asymptote equations of the hyperbola $\\frac{x^{2}}{8}-\\frac{y^{2}}{6}=1$ are?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/8 - y^2/6 = 1)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y= pm*(sqrt(3)/2)*x", "fact_spans": "[[[0, 38]], [[0, 38]]]", "query_spans": "[[[0, 46]]]", "process": "\\because the standard equation of the hyperbola is \\frac{x^{2}}{8}-\\frac{y^{2}}{6}=1, \\therefore its asymptotes are given by \\frac{x^{2}}{8}-\\frac{y^{2}}{6}=0, which simplifies to y=\\pm\\frac{\\sqrt{3}}{2}x" }, { "text": "Given $a>b>0$, the equation of ellipse $C_{1}$ is $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, the equation of hyperbola $C_{2}$ is $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, and the product of the eccentricities of $C_{1}$ and $C_{2}$ is $\\frac{\\sqrt{3}}{2}$. Then what is the equation of the asymptotes of $C_{2}$?", "fact_expressions": "C2: Hyperbola;C1: Ellipse;a: Number;b: Number;a > b;b > 0;Expression(C1) = (y^2/b^2 + x^2/a^2 = 1);Expression(C2) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(C1)*Eccentricity(C2) = sqrt(3)/2", "query_expressions": "Expression(Asymptote(C2))", "answer_expressions": "x + pm*sqrt(2)*y = 0", "fact_spans": "[[[67, 77], [170, 177], [134, 141]], [[10, 19], [126, 133]], [[2, 9]], [[2, 9]], [[2, 9]], [[2, 9]], [[10, 66]], [[67, 123]], [[126, 168]]]", "query_spans": "[[[170, 185]]]", "process": "" }, { "text": "Given the hyperbola $\\frac{x^{2}}{64}-\\frac{y^{2}}{36}=1$ with foci $F_{1}$, $F_{2}$, point $P$ lies on the hyperbola such that $P F_{1} \\perp P F_{2}$. Then the area of $\\triangle P F_{1} F_{2}$ is?", "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/64 - y^2/36 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))", "query_expressions": "Area(TriangleOf(P, F1, F2))", "answer_expressions": "36", "fact_spans": "[[[2, 42], [66, 69]], [[61, 65]], [[45, 52]], [[53, 60]], [[2, 42]], [[2, 60]], [[61, 70]], [[72, 95]]]", "query_spans": "[[[97, 127]]]", "process": "From the standard equation of the hyperbola, we obtain: a=8, b=6, c=10. Let PF_{1}=m, PF_{2}=n. By the definition of the hyperbola, we have: m-n=2a=16, (1) By the law of cosines, we have: m^{2}+n^{2}=4c^{2}=400, (2) (2)^{2}-(1)^{2} yields: mn=72. Then the area of \\triangle PF_{1}F_{2} is \\frac{1}{2}mn=\\frac{1}{2}\\times72=36" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left vertex is $M$, the right focus is $F$. The line $l$ passing through the left vertex with slope $1$ intersects the right branch of the hyperbola $C$ at point $N$. If the area of $\\Delta M N F$ is $\\frac{3}{2} b^{2}$, then the eccentricity of the hyperbola $C$ is?", "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;M: Point;N: Point;F: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftVertex(C) = M;RightFocus(C) = F;PointOnCurve(LeftVertex(C),l);Slope(l)=1;Intersection(l, RightPart(C)) = N;Area(TriangleOf(M, N, F)) = (3/2)*b^2", "query_expressions": "Eccentricity(C)", "answer_expressions": "2", "fact_spans": "[[[92, 97]], [[2, 63], [98, 104], [156, 162]], [[10, 63]], [[10, 63]], [[68, 71]], [[109, 113]], [[76, 79]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 71]], [[2, 79]], [[2, 97]], [[85, 97]], [[92, 113]], [[116, 153]]]", "query_spans": "[[[156, 168]]]", "process": "Given the problem, the left vertex of the hyperbola is M(-a,0); then the line l: y = x + a. Solving simultaneously with the hyperbola equation:\n\\begin{cases}\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1\\\\y=x+a\\end{cases},\nwe obtain N\\left(\\frac{a^{2}+ab^{2}}{b^{2}-a^{2}},\\frac{2ab^{2}}{b^{2}-a^{2}}\\right). Then the area of quadrilateral AMNF is \\frac{1}{2}\\cdot(a+c)\\cdot\\frac{b^{2}-a^{2}}{b^{2}-a^{2}}=\\frac{3}{2}b^{2}, \\therefore 2(a^{2}+ac)=3(c^{2}-2a^{2}), \\therefore 2(1+e)=3(e^{2}-2), i.e., 3e^{2}-2e-8=0, \\therefore e=2, so the answer is 2." }, { "text": "If the equation $\\frac{x^{2}}{k-5}+\\frac{y^{2}}{10-k}=1$ represents a hyperbola with foci on the $y$-axis, then the range of real values for $k$ is?", "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (x^2/(k - 5) + y^2/(10 - k) = 1);PointOnCurve(Focus(G), yAxis)", "query_expressions": "Range(k)", "answer_expressions": "(-oo,5)", "fact_spans": "[[[54, 57]], [[59, 64]], [[1, 57]], [[45, 57]]]", "query_spans": "[[[59, 71]]]", "process": "Because the equation \\frac{x^{2}}{k-} from the problem we get \\begin{cases}10-k>0\\\\b is the solution of the equation x-5<0\\\\10-k>0\\\\k-5<0\\end{cases}, solving gives k<5." }, { "text": "It is known that the perimeter of triangle $A B C$ is $8$, and the coordinates of vertices $B$ and $C$ are $(-1,0)$ and $(1 , 0)$, respectively. Then the trajectory equation of vertex $A$ is?", "fact_expressions": "A: Point;B: Point;C: Point;Coordinate(B) = (-1, 0);Coordinate(C) = (1, 0);Perimeter(TriangleOf(A, B, C)) = 8", "query_expressions": "LocusEquation(A)", "answer_expressions": "(x^2/9+y^2/8=1)&Negation(y=0)", "fact_spans": "[[[58, 61]], [[22, 25]], [[26, 29]], [[22, 54]], [[22, 54]], [[2, 19]]]", "query_spans": "[[[58, 68]]]", "process": "Using the perimeter of the triangle, we obtain |AB| + |AC| = 8 - 2 = 6 > 2. Combining this with the definition of an ellipse, we can find the trajectory equation of vertex A. [Detailed solution] Let A(x, y), |BC| = 2, so |AB| + |AC| = 8 - 2 = 6 > 2, which means point A lies on an ellipse with foci at points B and C, where 2a = 6, 2c = 2. Then b^{2} = a^{2} - c^{2} = 8, so the ellipse equation is \\frac{x^{2}}{9} + \\frac{y^{2}}{8} = 1. Since points A, B, and C cannot be collinear, y \\neq 0. Therefore, the trajectory equation of vertex A is \\frac{x^{2}}{9} + \\frac{y^{2}}{8} = 1 (y \\neq 0)." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively, a line passing through $F_{2}$ and parallel to one asymptote of the hyperbola intersects the hyperbola at point $P$. If $|P F_{1}|=3|P F_{2}|$, then the eccentricity of the hyperbola is?", "fact_expressions": "F1: Point;F2: Point;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, H);IsParallel(H, OneOf(Asymptote(G)));H: Line;Intersection(H, G) = P;P: Point;Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, F2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)", "fact_spans": "[[[2, 9]], [[10, 17], [84, 91]], [[20, 76], [92, 95], [107, 110], [141, 144]], [[20, 76]], [[23, 76]], [[23, 76]], [[23, 76]], [[23, 76]], [[2, 82]], [[2, 82]], [[83, 106]], [[91, 106]], [[104, 106]], [[104, 115]], [[111, 115]], [[117, 139]]]", "query_spans": "[[[141, 150]]]", "process": "Solution: Let the line passing through $ F_{2} $ and parallel to one of the asymptotes of the hyperbola, $ y = \\frac{b}{a}x $, intersect the hyperbola at point $ P $. From the definition of the hyperbola, we have $ |PF_{1}| - |PF_{2}| = 2a $. Given $ |PF_{1}| = 3|PF_{2}| $, it follows that $ |PF_{1}| = 3a $, $ |PF_{2}| = a $, and $ |F_{1}F_{2}| = 2c $. From $ \\tan\\angle F_{1}F_{2}P = \\frac{b}{a} $, we obtain $ \\cos\\angle F_{1}F_{2}P = \\frac{1}{\\sqrt{1 + \\frac{b^{2}}{a^{2}}}} = \\frac{a}{c} $. In triangle $ PF_{1}F_{2} $, applying the law of cosines gives: \n$ |PF_{1}|^{2} = |PF_{2}|^{2} + |F_{1}F_{2}|^{2} - 2|PF_{2}| \\cdot |F_{1}F_{2}| \\cos\\angle F_{1}F_{2}P $. \nThus, \n$ 9a^{2} = a^{2} + 4c^{2} - 2a \\cdot 2c \\cdot \\frac{a}{c} $. \nSimplifying yields $ c^{2} = 3a^{2} $. Therefore, the eccentricity of the hyperbola is $ e = \\frac{c}{a} = \\sqrt{3} $." }, { "text": "Given that the center of the hyperbola is at the origin, one focus is $F_{1}(-\\sqrt{5}, 0)$, point $P$ lies on the hyperbola, and the midpoint coordinates of segment $P F_{1}$ are $(0,2)$, then the equation of this hyperbola is?", "fact_expressions": "G: Hyperbola;F1: Point;P: Point;O:Origin;Center(G)=O;Coordinate(F1) = (-sqrt(5), 0);OneOf(Focus(G))=F1;PointOnCurve(P, G);Coordinate(MidPoint(LineSegmentOf(P,F1))) = (0,2)", "query_expressions": "Expression(G)", "answer_expressions": "x^2-y^2/4=1", "fact_spans": "[[[2, 5], [43, 46], [76, 79]], [[16, 37]], [[38, 42]], [[8, 10]], [[2, 10]], [[16, 37]], [[2, 37]], [[38, 47]], [[49, 73]]]", "query_spans": "[[[76, 84]]]", "process": "" }, { "text": "The focal distance of the hyperbola $\\frac{y^{2}}{2}-\\frac{x^{2}}{14}=1$ is? The equations of the asymptotes are?", "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/14 + y^2/2 = 1)", "query_expressions": "Expression(Asymptote(G));FocalLength(G)", "answer_expressions": "y=pm*(sqrt(7)/7)*x\n8", "fact_spans": "[[[0, 39]], [[0, 39]]]", "query_spans": "[[[0, 51]], [[0, 44]]]", "process": "(1) It is easy to obtain that for the hyperbola, c^{2}=2+14=16, hence the focal distance is 2c=8; (2) It is easy to obtain that for the hyperbola, a^{2}=2, b^{2}=14, and since the foci lie on the y-axis, the equations of the asymptotes are y=\\pm\\frac{a}{b}x=\\pm\\frac{\\sqrt{7}}{7}x" }, { "text": "Write the equation of a hyperbola with focus on the $y$-axis and eccentricity $\\sqrt{3}$?", "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G),yAxis);Eccentricity(G)=sqrt(3)", "query_expressions": "Expression(G)", "answer_expressions": "y^2 - x^2 / 2 = 1", "fact_spans": "[[[27, 30]], [[3, 30]], [[12, 30]]]", "query_spans": "[[[27, 33]]]", "process": "Take $ c = \\sqrt{3} $, then $ e = \\frac{c}{a} = \\sqrt{3} $, we obtain $ a = 1 $, $ \\therefore b = \\sqrt{c^{2} - a^{2}} = \\sqrt{2} $. Therefore, the hyperbola equation satisfying the condition is $ y^{2} - \\frac{x^{2}}{2} = 1 $." }, { "text": "If the sum of the distances from the points on the ellipse $\\frac{x^{2}}{m+2}-\\frac{y^{2}}{m}=1$ $(-1 -1;m < 0;P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};Distance(P, F1) + Distance(P, F2) = 8/3", "query_expressions": "Length(MinorAxis(G))", "answer_expressions": "2*sqrt(2)/3", "fact_spans": "[[[1, 48], [76, 78]], [[1, 48]], [[3, 48]], [[3, 48]], [[3, 48]], [[50, 51]], [[1, 51]], [], [], [[1, 55]], [[1, 73]]]", "query_spans": "[[[76, 84]]]", "process": "According to the definition of an ellipse, the sum of the distances from any point on the ellipse to the two foci is 2a, thus we have 2a = \\frac{8}{3}, that is, 2\\sqrt{m+2} = \\frac{8}{3}. Solve for the value of m, then use b^{2} = -m to find the length of the minor axis of the ellipse. [Solution] \\because -1 < m < 0, \\therefore m+2 > -m, \\therefore 2a = 2\\sqrt{m+2} = \\frac{8}{3}, \\therefore m = -\\frac{2}{9}, \\therefore b^{2} = -m = \\frac{2}{9}, 2b = \\frac{2\\sqrt{2}}{3}." }, { "text": "Let $F_{1}$, $F_{2}$ be the foci of the curve $C_{1}$: $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$, and let $P$ be an intersection point of the curves $C_{2}$: $\\frac{x^{2}}{3}-y^{2}=1$ and $C_{1}$. Then the area of $\\triangle P F_{1} F_{2}$ is?", "fact_expressions": "C1: Curve;C2: Curve;P: Point;F1: Point;F2: Point;Expression(C1) = (x^2/6+y^2/2=1);Expression(C2) = (x^2/3-y^2=1);Focus(C1) = {F2,F1};OneOf(Intersection(C1,C2))=P", "query_expressions": "Area(TriangleOf(P, F1, F2))", "answer_expressions": "sqrt(2)", "fact_spans": "[[[19, 65], [110, 117]], [[73, 109]], [[69, 72]], [[1, 8]], [[11, 18]], [[19, 65]], [[73, 109]], [[1, 68]], [[69, 122]]]", "query_spans": "[[[124, 154]]]", "process": "" }, { "text": "Let $F_{1}$, $F_{2}$ be the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, point $M(a, b)$, $\\angle M F_{1} F_{2}=30^{\\circ}$, then the eccentricity of the hyperbola is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;M: Point;F1: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;Coordinate(M)=(a,b);AngleOf(M,F1,F2)=ApplyUnit(30,degree)", "query_expressions": "Eccentricity(C)", "answer_expressions": "2", "fact_spans": "[[[19, 80], [132, 135]], [[26, 80]], [[26, 80]], [[86, 96]], [[1, 8]], [[9, 16]], [[26, 80]], [[26, 80]], [[19, 80]], [[1, 85]], [[1, 85]], [[86, 96]], [[97, 130]]]", "query_spans": "[[[132, 141]]]", "process": "" }, { "text": "The equation of the parabola with the $x$-axis as the directrix and focus $F(-1,-4)$?", "fact_expressions": "G: Parabola;F: Point;Coordinate(F) = (-1, -4);Focus(G) = F;Directrix(G)=xAxis", "query_expressions": "Expression(G)", "answer_expressions": "(x+1)^2=-8*(y+2)", "fact_spans": "[[[23, 26]], [[9, 19]], [[9, 19]], [[9, 26]], [[0, 26]]]", "query_spans": "[[[23, 29]]]", "process": "" }, { "text": "Given that the two foci of a hyperbola are $F_{1}(-\\sqrt{10}, 0)$ and $F_{2}(\\sqrt{10}, 0)$, and $M$ is a point on this hyperbola. If $\\overrightarrow{M F_{1}} \\cdot \\overrightarrow{M F_{2}}=0$ and $|\\overrightarrow{M F_{1}}| \\cdot |\\overrightarrow{M F_{2}}|=2$, then what is the equation of this hyperbola?", "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;M: Point;Coordinate(F1) = (-sqrt(10), 0);Coordinate(F2) = (sqrt(10), 0);Focus(G) = {F1,F2};PointOnCurve(M, G);DotProduct(VectorOf(M, F1), VectorOf(M, F2)) = 0;Abs(VectorOf(M, F1))*Abs(VectorOf(M, F2)) = 2", "query_expressions": "Expression(G)", "answer_expressions": "x^2/9 - y^2 = 1", "fact_spans": "[[[2, 5], [65, 68], [199, 202]], [[11, 33]], [[36, 57]], [[60, 63]], [[11, 33]], [[36, 57]], [[2, 58]], [[60, 71]], [[73, 132]], [[134, 196]]]", "query_spans": "[[[199, 207]]]", "process": "" }, { "text": "Given the parabola $C$: $y=x^{2}$ with focus $F$, point $M$ lies on $C$, and $|M F|=\\frac{5}{4}$. Then the coordinates of $M$ are?", "fact_expressions": "C: Parabola;M: Point;F: Point;Expression(C) = (y = x^2);Focus(C) = F;PointOnCurve(M, C);Abs(LineSegmentOf(M, F)) = 5/4", "query_expressions": "Coordinate(M)", "answer_expressions": "{(1,1),(-1,1)}", "fact_spans": "[[[2, 19], [32, 35]], [[27, 31], [59, 62]], [[23, 26]], [[2, 19]], [[2, 26]], [[27, 36]], [[38, 57]]]", "query_spans": "[[[59, 67]]]", "process": "Let M(x_{0},y_{0}). By the definition of the parabola, y_{0}+\\frac{p}{2}=\\frac{5}{4}. Since p=\\frac{1}{2}, it follows that y_{0}=1. Substituting into C gives: x_{0}=\\pm1. Hence, M(1,1) or (-1,1)." }, { "text": "Given the ellipse $x^{2}+3 y^{2}=9$, the left focus is $F_{1}$, point $P$ is any point on the ellipse distinct from the vertices, and $O$ is the origin. If point $D$ is the midpoint of segment $P F_{1}$, then the perimeter of $\\Delta F_{1} O D$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 3*y^2 = 9);F1: Point;LeftFocus(G) = F1;P: Point;PointOnCurve(P, G);Negation(In(P,Vertex(G)));O: Origin;D: Point;MidPoint(LineSegmentOf(P, F1)) = D", "query_expressions": "Perimeter(TriangleOf(F1, O, D))", "answer_expressions": "3+sqrt(6)", "fact_spans": "[[[2, 21], [39, 41]], [[2, 21]], [[26, 33]], [[2, 33]], [[34, 38]], [[34, 51]], [[34, 51]], [[52, 55]], [[62, 66]], [[62, 81]]]", "query_spans": "[[[83, 106]]]", "process": "Problem Analysis: Transform $x^{2}+3y^{2}=9$ into standard form, yielding $\\frac{x^{2}}{9}+\\frac{y^{2}}{3}=1$, so $OF_{1}=\\sqrt{6}$. Let its right focus be $F_{2}$, then $PF_{1}+PF_{2}=6$. Since point $D$ is the midpoint of segment $PF_{1}$, by the midline theorem, the perimeter of quadrilateral $AF_{1}OD$ is $DF_{1}+DO+OF_{1}=\\frac{1}{2}(PF_{1}+PF_{2})+OF_{1}=3+\\sqrt{6}$." }, { "text": "In $\\triangle A B C$, $C A=3$ , $C B=5$ , $\\angle A C B=\\frac{2 \\pi}{3}$, an ellipse $C_{1}$ and a hyperbola $C_{2}$ with common foci at $A$ and $B$ both pass through point $C$. What is the sum of the eccentricities of $C_{1}$ and $C_{2}$?", "fact_expressions": "C1:Ellipse;C2:Hyperbola;A: Point;B: Point;C: Point;LineSegmentOf(C,B) = 5;LineSegmentOf(C, A) = 3;Focus(C1)={A,B};Focus(C2)={A,B};AngleOf(A,C,B)=2*pi/3;PointOnCurve(C,C1);PointOnCurve(C,C2)", "query_expressions": "Eccentricity(C1)+Eccentricity(C2)", "answer_expressions": "35/8", "fact_spans": "[[[86, 95], [115, 122]], [[96, 106], [123, 130]], [[73, 76]], [[77, 80]], [[109, 113]], [[30, 37]], [[20, 27]], [[72, 106]], [[72, 106]], [[40, 70]], [[86, 113]], [[86, 113]]]", "query_spans": "[[[115, 138]]]", "process": "In triangle ABC, CA=3, CB=5, ∠ACB=2π/3. By the law of cosines we obtain: AB² = CA² + CB² - 2·CA·CB·cos(2π/3). Solving gives AB=7, so 2c=7, c=7/2. The ellipse C₁ and hyperbola C₂ both pass through point C. According to the definition of an ellipse, CB+CA=2a₁=8, so a₁=4. According to the definition of a hyperbola, CB−CA=2a₂=2, so a₂=1. Therefore, the sum of the eccentricities of C₁ and C₂ is c/a₁ + c/a₂ = 7/4 + 7/1 = 7/8 + 7/2 = 35/8." }, { "text": "The focal distance of the hyperbola $x^{2}-4 y^{2}=1$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - 4*y^2 = 1)", "query_expressions": "FocalLength(G)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[0, 20]], [[0, 20]]]", "query_spans": "[[[0, 25]]]", "process": "Since $x^{2}-4y^{2}=1$, we have $a=1$, $b=\\frac{1}{2}$, so $c^{2}=a^{2}+b^{2}=\\frac{5}{4}$, thus $c=\\frac{\\sqrt{5}}{2}$, so the focal distance is $2c=\\sqrt{5}$." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has eccentricity $\\frac{\\sqrt{5}}{2}$, then the asymptotes of the hyperbola $C$ have the equation?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(C) = sqrt(5)/2", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "x+pm*2*y=0", "fact_spans": "[[[2, 63], [90, 96]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 88]]]", "query_spans": "[[[90, 104]]]", "process": "Since the eccentricity of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) is $ \\frac{\\sqrt{5}}{2} $, we have $ e = \\frac{c}{a} = \\sqrt{1 + \\left( \\frac{b}{a} \\right)^{2}} = \\frac{\\sqrt{5}}{2} $. Solving gives $ \\frac{b}{a} = \\frac{1}{2} $. Moreover, since the foci of the hyperbola lie on the x-axis, the asymptotes of hyperbola $ C $ are given by $ y = \\pm \\frac{1}{2}x $." }, { "text": "If the length of the minor axis of an ellipse is $6$, and the distance from the right focus to the left vertex is $9$, then what is the eccentricity of the ellipse?", "fact_expressions": "G: Ellipse;Length(MinorAxis(G))=6;Distance(RightFocus(G),LeftVertex(G))=9", "query_expressions": "Eccentricity(G)", "answer_expressions": "4/5", "fact_spans": "[[[1, 3], [1, 3]], [[1, 11]], [[1, 26]]]", "query_spans": "[[[28, 36]]]", "process": "Solution: According to the problem, 2b=6, a+c=9, and c^{2}=a^{2}-b^{2}, so a=5, c=4. Therefore, the eccentricity of the ellipse is e=\\frac{c}{a}=\\frac{4}{5}." }, { "text": "Given that $F$ is the left focus of the hyperbola $C$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and $P$, $Q$ are points on $C$. If the length of $PQ$ is equal to twice the length of the imaginary axis, and point $A(5 , 0)$ lies on the segment $PQ$, then what is the perimeter of $\\triangle PQF$?", "fact_expressions": "F: Point;C: Hyperbola;Expression(C) = (x^2/9 - y^2/16 = 1);LeftFocus(C) = F;Q: Point;P: Point;PointOnCurve(P, C);PointOnCurve(Q, C);Length(LineSegmentOf(P, Q)) = Length(ImageinaryAxis(C)) * 2;A: Point;Coordinate(A) = (5, 0);PointOnCurve(A, LineSegmentOf(P, Q))", "query_expressions": "Perimeter(TriangleOf(P, Q, F))", "answer_expressions": "44", "fact_spans": "[[[2, 5]], [[6, 50], [63, 66]], [[6, 50]], [[2, 54]], [[59, 62]], [[55, 58]], [[55, 69]], [[55, 69]], [[63, 88]], [[90, 101]], [[90, 101]], [[90, 110]]]", "query_spans": "[[[112, 134]]]", "process": "" }, { "text": "The standard equation of the hyperbola that has the same asymptotes as the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$ and passes through the point $(2,3)$ is?", "fact_expressions": "G: Hyperbola;H: Point;Q: Hyperbola;Expression(G) = (x^2/2 - y^2 = 1);Coordinate(H) = (2, 3);PointOnCurve(H, Q);Asymptote(G) = Asymptote(Q)", "query_expressions": "Expression(Q)", "answer_expressions": "y^2/7 - x^2/14 = 1", "fact_spans": "[[[1, 29]], [[40, 48]], [[49, 52]], [[1, 29]], [[40, 48]], [[39, 52]], [[0, 52]]]", "query_spans": "[[[49, 59]]]", "process": "" }, { "text": "Given a hyperbola with foci on the $x$-axis: $\\frac{x^{2}}{8-m}+\\frac{y^{2}}{4-m}=1$, what is the range of values for the distance from its foci to the asymptotes?", "fact_expressions": "PointOnCurve(Focus(G), xAxis);G: Hyperbola;Expression(G) = (x^2/(8 - m) + y^2/(4 - m) = 1);m: Number", "query_expressions": "Range(Distance(Focus(G), Asymptote(G)))", "answer_expressions": "(0, 2)", "fact_spans": "[[[2, 53]], [[11, 53], [54, 55]], [[11, 53]], [[14, 53]]]", "query_spans": "[[[54, 72]]]", "process": "For the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) with foci on the x-axis, the distance from one of its foci $(c,0)$ to the asymptote $bx-ay=0$ is $\\frac{|bc|}{\\sqrt{b^{2}+a^{2}}}=b$. In this problem, the hyperbola $\\frac{x^{2}}{8-m}+\\frac{y^{2}}{4-m}=1$, i.e., $\\frac{x^{2}}{8-m}-\\frac{y^{2}}{m-4}=1$, has its foci on the x-axis, so $\\begin{cases}8-m\\\\m-4\\end{cases}>0$, solving gives $40)$ and intersects the parabola at points $A$ and $B$, then $|A F|=4$, and a point $C$ on the directrix of the parabola satisfies $\\overrightarrow{C B}=2 \\overrightarrow{B F}$, find $p=?$", "fact_expressions": "G: Parabola;p: Number;H: Line;A: Point;F: Point;C: Point;B: Point;p>0;Expression(G) = (y^2 = 2*p*x);Focus(G) = F;PointOnCurve(F,H);Intersection(H,G) = {A, B};Abs(LineSegmentOf(A,F))=4;PointOnCurve(C,Directrix(G));VectorOf(C,B)=2*VectorOf(B,F)", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[6, 27], [35, 38], [62, 65]], [[123, 126]], [[2, 4]], [[39, 42]], [[30, 33]], [[72, 75]], [[43, 46]], [[9, 27]], [[6, 27]], [[6, 33]], [[2, 33]], [[2, 48]], [[50, 59]], [[62, 75]], [[77, 122]]]", "query_spans": "[[[123, 128]]]", "process": "Draw perpendiculars from points A and B to the directrix $ x = -\\frac{p}{2} $ of the parabola $ y^2 = 2px $ ($ p > 0 $), with feet of perpendiculars being N and M, respectively. Let the directrix intersect the x-axis at point K, as shown in the figure. Then we have $ |AN| = |AF| $, $ |BM| = |BF| $. Since point C lies on the directrix and satisfies $ \\overrightarrow{CB} = 2\\overrightarrow{BF} $, i.e., point C is the intersection point of line AB and the directrix, it follows that $ |CB| = 2|BM| $, yielding $ \\angle ACN = 30^{\\circ} $. Thus, $ |AC| = 2|AN| = 2|AF| $, meaning point F is the midpoint of segment AC. Since $ FK \\parallel AN $, we have $ |FK| = \\frac{1}{2}|AN| = \\frac{1}{2}|AF| = 2 $. Also, $ |FK| = p $, so $ p = 2 $." }, { "text": "Let $F_{1}$, $F_{2}$ be the two foci of the ellipse $C$: $\\frac{x^{2}}{12}+\\frac{y^{2}}{3}=1$, $O$ be the coordinate origin, and point $P$ lies on $C$ such that $|O P|=3$. Then the area of $\\Delta P F_{1} F_{2}$ is?", "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;O: Origin;Expression(C) = (x^2/12 + y^2/3 = 1);Focus(C) = {F1, F2};PointOnCurve(P, C);Abs(LineSegmentOf(O, P)) = 3", "query_expressions": "Area(TriangleOf(P, F1, F2))", "answer_expressions": "3", "fact_spans": "[[[17, 60], [80, 83]], [[75, 79]], [[1, 8]], [[9, 16]], [[66, 69]], [[17, 60]], [[1, 65]], [[75, 83]], [[85, 94]]]", "query_spans": "[[[96, 123]]]", "process": "Given $ F_{1}(-3,0) $, $ F_{2}(3,0) $, let $ P(x_{0},y_{0}) $. Since point $ P $ lies on $ C $, we have $ \\frac{x_{0}^{2}}{12} + \\frac{y_{0}^{2}}{3} = 1 \\cdots\\cdots (1) $. Given $ |OP| = 3 $, we have $ x_{0}^{2} + y_{0}^{2} = 9 \\cdots\\cdots (2) $. Solving these two equations simultaneously gives $ \\frac{9 - y_{0}^{2}}{12} + \\frac{y_{0}^{2}}{3} = 1 $, solving which yields $ y_{0}^{2} = 1 $, that is, $ |y_{0}| = 1 $. Therefore, $ S_{\\Delta PF_{1}F_{2}} = \\frac{1}{2} \\times |F_{1}F_{2}| \\times |y_{0}| = \\frac{1}{2} \\times 6 \\times 1 = 3 $." }, { "text": "The hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has two foci $F_{1}$ and $F_{2}$. Point $P$ lies on the hyperbola $C$, and $\\angle P F_{1} F_{2}=15^{\\circ}$. If the area of $\\triangle P F_{1} F_{2}$ is $b^{2}$, then the eccentricity of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C) = True;AngleOf(P, F1, F2) = ApplyUnit(15, degree);Area(TriangleOf(P, F1, F2)) = b^2", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[0, 61], [90, 96], [172, 178]], [[0, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[69, 76]], [[77, 84]], [[0, 84]], [[85, 89]], [[85, 97]], [[98, 131]], [[134, 170]]]", "query_spans": "[[[172, 184]]]", "process": "Solution: Let \\angle F_{1}PF_{2}=\\theta, |PF_{1}|=n, |PF_{2}|=m. From the given conditions, we have \\frac{1}{2}mn\\sin\\theta=b^{2}. From the properties of the hyperbola, we know that n-m=2a, so m^{2}-2mn+n^{2}=4a^{2}. By the law of cosines, we have m^{2}+n^{2}-2mn\\cos\\theta=4c^{2}, thus 2mn-2mn\\cos\\theta=4(c^{2}-a^{2})=4b^{2}. Hence, 2mn-2mn\\cos\\theta=2mn\\sin\\theta. That is, \\sin\\theta+\\cos\\theta=\\sqrt{2}\\sin(\\theta+\\frac{\\pi}{4})=1. Solving gives \\theta=\\frac{\\pi}{2}, i.e., \\angle F_{1}PF_{2}=90^{\\circ}. In right triangle \\triangle PF_{1}F_{2}, n=|PF_{1}|=2c\\cdot\\cos15^{\\circ}, m=|PF_{2}|=2c\\cdot\\sin15^{\\circ}. Then n-m=2c\\cdot\\cos15^{\\circ}-2c\\cdot\\sin15^{\\circ}=2a. Therefore, e=\\frac{c}{a}=\\frac{1}{\\cos15^{\\circ}-\\sin15^{\\circ}}=\\sqrt{2}." }, { "text": "The equation of the directrix of the parabola $y=\\frac{1}{2} x^{2}$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y = x^2/2)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "y = -1/2", "fact_spans": "[[[0, 24]], [[0, 24]]]", "query_spans": "[[[0, 31]]]", "process": "From $ y = \\frac{1}{2}x^{2} \\Rightarrow x^{2} = 2y $, the parabola's focus lies on the positive $ y $-axis; therefore, the directrix of this parabola is $ y = $" }, { "text": "The line $y = m x + 1$ intersects the hyperbola $x^{2} - y^{2} = 1$ at two distinct points; then the range of real values for $m$ is?", "fact_expressions": "G: Hyperbola;H: Line;m: Real;Expression(G) = (x^2 - y^2 = 1);Expression(H) = (y = m*x + 1);NumIntersection(H, G) = 2", "query_expressions": "Range(m)", "answer_expressions": "(-sqrt(2), sqrt(2))&Negation(m=pm*1)", "fact_spans": "[[[12, 30]], [[0, 11]], [[41, 46]], [[12, 30]], [[0, 11]], [[0, 39]]]", "query_spans": "[[[41, 53]]]", "process": "" }, { "text": "Given the ellipse $C$: $ \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has an eccentricity of $\\frac{\\sqrt{3}}{2}$. The line $l$: $y=x+2$ intersects the ellipse $C$ at points $A$ and $B$. If the sum of the slopes of lines $OA$ and $OB$ is $4$, where $O$ is the origin, then what is the equation of the ellipse $C$?", "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Eccentricity(C) = sqrt(3)/2;l: Line;Expression(l) = (y = x + 2);A: Point;B: Point;Intersection(l, C) = {A, B};O: Origin;Slope(LineOf(O, A)) + Slope(LineOf(O, B)) = 4", "query_expressions": "Expression(C)", "answer_expressions": "x^2/32+y^2/8=1", "fact_spans": "[[[2, 59], [100, 105], [155, 160]], [[2, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[2, 84]], [[85, 99]], [[85, 99]], [[107, 110]], [[111, 114]], [[85, 116]], [[145, 148]], [[118, 142]]]", "query_spans": "[[[155, 165]]]", "process": "Solution: Let A(x_{1},y_{1}), B(x_{2},y_{2}), then y_{1}=x_{1}+2, y_{2}=x_{2}+2. Since k_{OA}=\\frac{y_{1}}{x_{1}}, k_{OB}=\\frac{y_{2}}{x_{2}}, and k_{OA}+k_{OB}=\\frac{y_{2}}{x_{2}}+\\frac{y_{1}}{x_{1}}=4, that is, \\frac{x_{2}+2}{x_{2}}+\\frac{x_{1}+2}{x_{1}}=4, so \\frac{1}{x_{1}}+\\frac{1}{x_{2}}=\\frac{x_{1}+x_{2}}{x_{1}x_{2}}=1. \nSolving the system of equations of the line and the ellipse: \n\\begin{cases}y=x+2\\\\\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1\\end{cases} \nEliminating y gives (a^{2}+b^{2})x^{2}+4a^{2}x+4a^{2}-a^{2}b^{2}=0, so x_{1}+x_{2}=\\frac{-4a^{2}}{a^{2}+b^{2}}, x_{1}x_{2}=\\frac{4a^{2}-a^{2}b^{2}}{a^{2}+b^{2}}. Therefore, \\frac{4a^{2}-a^{2}b^{2}}{\\frac{a^{2}+b^{2}}{-4a^{2}}}=1, solving gives b^{2}=8. \nSince the eccentricity e=\\frac{c}{a}=\\frac{\\sqrt{3}}{2} and c^{2}=a^{2}-b^{2}, i.e., \\frac{c^{2}}{a^{2}}=\\frac{a^{2}-b^{2}}{a^{2}}=\\frac{3}{4}, solving gives a^{2}=32. Therefore, the equation of the ellipse is \\frac{x^{2}}{32}+\\frac{y^{2}}{8}=1." }, { "text": "A moving point's distance to the $y$-axis is less than its distance to the point $(2,0)$ by $2$. What is the trajectory equation of this moving point?", "fact_expressions": "P: Point;G:Point;Coordinate(G) = (2, 0);Distance(P,G)=Distance(P,yAxis)+2", "query_expressions": "LocusEquation(P)", "answer_expressions": "{(y^2=8*x)&(x>=0),(y=0)&(x<0)}", "fact_spans": "[[[1, 3], [31, 33]], [[12, 20]], [[12, 20]], [[1, 27]]]", "query_spans": "[[[31, 40]]]", "process": "Let the moving point be M(x, y), then from the condition we have \\sqrt{(x-2)^{2}+y^{2}}=|x|+2. Squaring both sides gives y^{2}=4x+4|x|. When x\\geqslant0, y^{2}=8x; when x<0, y=0. Therefore, the trajectory equation of the moving point is y=0 (x<0)." }, { "text": "On the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, a point $P$ is such that the lines connecting $P$ to the two foci $F_{1}$, $F_{2}$ of the ellipse are perpendicular to each other. Then the area of $\\triangle P F_{1} F_{2}$ is?", "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/9 = 1);PointOnCurve(P,G);Focus(G)={F1,F2};IsPerpendicular(LineSegmentOf(P,F1),LineSegmentOf(P,F2))", "query_expressions": "Area(TriangleOf(P, F1, F2))", "answer_expressions": "9", "fact_spans": "[[[0, 38], [45, 47]], [[41, 44]], [[52, 59]], [[60, 67]], [[0, 38]], [[0, 44]], [[45, 67]], [[41, 74]]]", "query_spans": "[[[76, 106]]]", "process": "Problem Analysis: For the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, we have $a=5$, $b=3$, $c=\\sqrt{a^{2}-b^{2}}$. Let $|PF_{1}|=m$, $|PF_{2}|=n$, then $m+n=2a=10$, $m^{2}+n^{2}=(2c)^{2}$. Solving these equations simultaneously yields the result. \nSolution: $\\because$ the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, $\\therefore a=5$, $b=3$, $c=\\sqrt{a^{2}-b^{2}}=4$. Let $|PF_{1}|=m$, $|PF_{2}|=n$, then $m+n=2a=10$, $m^{2}+n^{2}=(2c)^{2}=64$, $\\therefore mn=18$. $\\therefore$ the area of $\\triangle PF_{1}F_{2} = \\frac{1}{2}mn=$" }, { "text": "The equation of the directrix of the parabola $y=8 x^{2}$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y = 8*x^2)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "y=-1/32", "fact_spans": "[[[0, 14]], [[0, 14]]]", "query_spans": "[[[0, 21]]]", "process": "" }, { "text": "Let $AB$ be the major axis of ellipse $\\Gamma$, point $C$ lies on $\\Gamma$, and $\\angle C B A=\\frac{\\pi}{4}$. If $AB=4$, $B C=\\sqrt{2}$, then the distance between the two foci of $\\Gamma$ is?", "fact_expressions": "Gamma: Ellipse;A: Point;B: Point;C: Point;F1: Point;F2: Point;MajorAxis(Gamma) = LineSegmentOf(A, B);PointOnCurve(C, Gamma);AngleOf(C, B, A) = pi/4;LineSegmentOf(A, B) = 4;LineSegmentOf(B, C) = sqrt(2);Focus(Gamma) = {F1, F2}", "query_expressions": "Distance(F1, F2)", "answer_expressions": "4*sqrt(6)/3", "fact_spans": "[[[6, 16], [25, 33], [91, 99]], [[1, 5]], [[1, 5]], [[20, 24]], [], [], [[1, 19]], [[20, 34]], [[36, 64]], [[66, 72]], [[75, 89]], [[91, 104]]]", "query_spans": "[[[91, 111]]]", "process": "Without loss of generality, assume the standard equation of the ellipse is \\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1. Then we can calculate C(1,1), obtaining b^{2}=\\frac{4}{3}, 2c=\\frac{4\\sqrt{6}}{3}.【" }, { "text": "Given the hyperbola equation $x^{2}-\\frac{y^{2}}{4}=1$, the line $y=x+b$ ($b>0$) intersects the left and right branches of the hyperbola at points $A$ and $B$ respectively, and intersects the $y$-axis at point $C$. Then the range of $\\frac{|A C|}{|B C|}$ is?", "fact_expressions": "G: Hyperbola;H: Line;b: Number;A: Point;C: Point;B: Point;b>0;Expression(H) = (y = b + x);Expression(G) = (x^2 - y^2/4 = 1);Intersection(H,LeftPart(G))=A;Intersection(H,RightPart(G))=B;Intersection(H,yAxis)=C", "query_expressions": "Range(Abs(LineSegmentOf(A, C))/Abs(LineSegmentOf(B, C)))", "answer_expressions": "(1/3, 1)", "fact_spans": "[[[2, 5], [51, 54]], [[34, 48]], [[36, 48]], [[59, 62]], [[76, 80]], [[63, 66]], [[36, 48]], [[34, 48]], [[2, 33]], [[34, 68]], [[34, 68]], [[34, 80]]]", "query_spans": "[[[82, 108]]]", "process": "Solve the system \n\\[\n\\begin{cases}\nx_{1}x_{1}x_{2} \\\\\ny = x + b\n\\end{cases}\n\\]\neliminate $ y $ and simplify to obtain \n\\[\n3x^{2} - 2bx - b^{2} - 4 = 0.\n\\]\nThe discriminant is \n\\[\n\\Delta = 4b^{2} + 12(b^{2} + 4) = 16b^{2} + 48 > 0,\n\\]\nwhich always holds. Let $ A(x_{1}, y_{1}), B(x_{2}, y_{2}) $, with $ x_{1} < 0, x_{2} > 0 $. Then \n\\[\nx_{1} + x_{2} = \\frac{2}{3}b, \\quad x_{1}x_{2} = -\\frac{b^{2} + 4}{3}.\n\\]\nThus, \n\\[\n\\frac{(x_{1} + x_{2})^{2}}{x_{1}x_{2}} = \\frac{\\frac{4}{9}b^{2}}{-\\frac{b^{2} + 4}{3}} = -\\frac{4b^{2}}{3(b^{2} + 4)} = -\\frac{4(b^{2} + 4) - 16}{3(b^{2} + 4)} = -\\frac{4}{3} + \\frac{16}{3(b^{2} + 4)} \\in \\left(-\\frac{4}{3}, 0\\right].\n\\]\nTherefore, \n\\[\n-\\frac{4}{3} < \\frac{x_{1}^{2} + 2x_{1}x_{2} + x_{2}^{2}}{x_{1}x_{2}} = \\frac{x_{1}}{x_{2}} + 2 + \\frac{x_{2}}{x_{1}} \\leq 0.\n\\]\nLet \n\\[\n\\frac{|AC|}{|BC|} = \\frac{-x_{1}}{x_{2}} = t, \\quad t > 0.\n\\]\nThen \n\\[\n-\\frac{4}{3} < -t + 2 - \\frac{1}{t} \\leq 0,\n\\]\nso we have \n\\[\n\\begin{cases}\nt + \\frac{1}{t} - 2 \\geq 0 \\\\\nt + \\frac{1}{t} - \\frac{10}{3} < 0\n\\end{cases},\n\\]\nwhich implies \n\\[\n\\begin{cases}\nt^{2} - 2t + 1 \\geq 0 \\\\\nt^{2} - \\frac{10}{3}t + 1 < 0\n\\end{cases}.\n\\]\nHence, \n\\[\n\\frac{1}{3} < t < 3, \\quad \\text{i.e.,} \\quad \\frac{1}{3} < \\frac{-x_{1}}{x_{2}} < 3.\n\\]\nSince $ b > 0 $, we have $ x_{1} + x_{2} > 0 $, so $ x_{2} > -x_{1} > 0 $, which gives \n\\[\n0 < \\frac{-x_{1}}{x_{2}} < 1.\n\\]\nTherefore, \n\\[\n\\frac{1}{3} < \\frac{-x_{1}}{x_{2}} < 1,\n\\]\ni.e., \n\\[\n\\frac{1}{3} < \\frac{|AC|}{|BC|} < 1.\n\\]" }, { "text": "A line segment has length $4$, with endpoints $A$ and $B$ moving on the $x$-axis and $y$-axis respectively. If a moving point $P$ satisfies $\\overrightarrow{A P}=3 \\overrightarrow{P B}$, then the trajectory equation of point $P$ is?", "fact_expressions": "A: Point;B: Point;Length(LineSegmentOf(A, B)) = 4;Endpoint(LineSegmentOf(A, B)) = {A, B};PointOnCurve(A, xAxis);PointOnCurve(B, yAxis);P: Point;VectorOf(A, P) = 3*VectorOf(P, B)", "query_expressions": "LocusEquation(P)", "answer_expressions": "x^2+y^2/9=1", "fact_spans": "[[[15, 18]], [[19, 22]], [[2, 22]], [[2, 22]], [[15, 35]], [[15, 35]], [[41, 44], [95, 98]], [[46, 91]]]", "query_spans": "[[[95, 105]]]", "process": "Solution: Let P(x, y), A(a, 0), B(0, b). Then a^{2}+b^{2}=16. Then \\overrightarrow{AP}=(x-a, y), \\overrightarrow{PB}=(-x, b-y). Since \\overrightarrow{AP}=3\\overrightarrow{PB}, that is (x-a, y)=(-3x, 3b-3y), we have \\begin{cases}x-a=-3x\\\\y=3b-3y\\end{cases}. Solving gives \\begin{cases}a=4x\\\\b=\\frac{4}{3}y\\end{cases}. Therefore, a^{2}+b^{2}=16x^{2}+\\frac{16}{9}y^{2}=16, so the trajectory equation of the moving point P is x^{2}+\\frac{y^{2}}{9}=1" }, { "text": "Points $A$ and $B$ are two points on the parabola $C$: $y^{2}=2 p x(p>0)$, $F$ is the focus of the parabola $C$. If $\\angle A F B=120^{\\circ}$, and the distance from the midpoint $D$ of $AB$ to the directrix of the parabola $C$ is $d$, then the maximum value of $\\frac{d^{2}}{|A B|^{2}}$ is?", "fact_expressions": "C: Parabola;p: Number;A: Point;B: Point;F: Point;D: Point;p>0;Expression(C) = (y^2 = 2*(p*x));PointOnCurve(A, C);PointOnCurve(B, C);Focus(C) = F;AngleOf(A, F, B) = ApplyUnit(120, degree);MidPoint(LineSegmentOf(A, B))=D;Distance(D, Directrix(C)) = d;d:Number", "query_expressions": "Max(d^2/Abs(LineSegmentOf(A, B))^2)", "answer_expressions": "1/3", "fact_spans": "[[[9, 35], [44, 50], [94, 100]], [[17, 35]], [[0, 4]], [[5, 8]], [[40, 43]], [[90, 93]], [[17, 35]], [[9, 35]], [[0, 39]], [[0, 39]], [[40, 53]], [[55, 81]], [[83, 93]], [[90, 110]], [[107, 110]]]", "query_spans": "[[[112, 143]]]", "process": "Let the lengths of AF and BF be a and b respectively. From the given conditions, we obtain the relationship between d and a, b. In triangle ABF, using the law of cosines, we derive the relationship between AB and a, b, compute the ratio, and find the maximum value using the basic inequality. Let |AF| = a, |BF| = b, then \\frac{|AC|+|BE|}{2}=\\frac{|AF|+|BF|}{2}=\\frac{a+b}{2}. In triangle ABF, \\therefore\\frac{d}{|AB|}=\\sqrt{\\frac{AB^{2}}{(a+b)^{2}}}+b^{2}-2ab\\cos\\angle AFB=a^{2}+b^{2}+ab'}, with equality if and only if a = b. Therefore, the maximum value of \\frac{d^{2}}{|AB|^{2}} is \\frac{1}{3}." }, { "text": "The line $ l $ passing through point $ M(-1,0) $ intersects the parabola $ C $: $ y^{2}=4x $ at points $ A $ and $ B $ ($ A $ lies between $ M $ and $ B $), $ F $ is the focus of the parabola $ C $. If $ S_{\\triangle MBF} = 4 S_{\\triangle MAF} $, then the area of $ \\triangle ABF $ is?", "fact_expressions": "M: Point;Coordinate(M) = (-1, 0);PointOnCurve(M, l) = True;l: Line;C: Parabola;Expression(C) = (y^2 = 4*x);Intersection(l, C) = {A, B};A: Point;B: Point;Between(A, M, B) = True;F: Point;Focus(C) = F;Area(TriangleOf(M, B, F)) = 4*Area(TriangleOf(M, A, F))", "query_expressions": "Area(TriangleOf(A, B, F))", "answer_expressions": "3", "fact_spans": "[[[1, 11], [54, 57]], [[1, 11]], [[0, 17]], [[12, 17]], [[18, 37], [71, 77]], [[18, 37]], [[12, 48]], [[50, 53], [39, 42]], [[43, 46], [58, 61]], [[50, 63]], [[67, 70]], [[67, 80]], [[82, 125]]]", "query_spans": "[[[127, 149]]]", "process": "Without loss of generality, assume A and B are in the first quadrant, and let A(x_{1},y_{1}), B(x_{2},y_{2}). From S_{AMBF}=4S_{\\triangle MAF}, we obtain \\frac{1}{2}|MF|y_{2}=4\\times\\frac{1}{2}|MF|y_{1}, thus y_{2}=4y_{1}. Since A, B, M are collinear and A, B lie on the parabola, we can solve for y_{1}, y_{2}. Solution: Without loss of generality, assume A and B are in the first quadrant, as shown in the figure. Let A(x_{1},y_{1}), B(x_{2},y_{2}), from the given condition F(1,0), \\because S_{AMBF}=4S_{\\Delta MAF}, \\therefore \\frac{1}{2}|MF|y_{2}=4\\times\\frac{1}{2}|MF|y_{1}, \\therefore y_{2}=4y_{1}. Also, M, A, B are collinear, \\therefore \\frac{y_{1}}{x_{1}+1}=\\frac{y_{2}}{x_{2}+1}, i.e., \\frac{y_{1}}{\\frac{1}{4}y_{1}^{2}+1}=\\frac{y_{2}}{\\frac{1}{4}y_{2}^{2}+1}, substituting y_{2}=4y_{1} yields: \\frac{y_{1}}{\\frac{1}{4}y_{1}^{2}+1}=\\frac{4y_{1}}{4y_{1}^{2}+1}, clearly y_{1}\\neq0, solving gives y_{1}=1, \\therefore y_{2}=4, \\therefore S_{\\Delta MAF}=\\frac{1}{2}\\times2\\times1=1, S_{\\Delta MBF}=4, \\therefore S_{\\Delta FAB}=S_{\\Delta MBF}-S_{\\Delta MAF}=4-1" }, { "text": "Given the curve $\\Gamma$: $x^{2}=8 y$ with focus $F$, point $P$ moves along the curve $\\Gamma$, and fixed point $A(0,-2)$. Then the minimum value of $\\frac{|P F|}{|P A|}$ is?", "fact_expressions": "Gamma: Curve;Expression(Gamma) = (x^2 = 8*y);F: Point;Focus(Gamma) = F;P: Point;PointOnCurve(P,Gamma) = True;A: Point;Coordinate(A) = (0, -2)", "query_expressions": "Min(Abs(LineSegmentOf(P, F))/Abs(LineSegmentOf(P, A)))", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[2, 25], [38, 48]], [[2, 25]], [[29, 32]], [[2, 32]], [[33, 37]], [[33, 49]], [[54, 63]], [[54, 63]]]", "query_spans": "[[[65, 92]]]", "process": "Let P(x,y). When y=0, |\\frac{PF}{PA}|=\\sqrt{\\frac{x^2+(y-2)^{2}}{x^{2}+(y+2)^{2}}}|\\frac{PF|}{|PA|}=1. When y>0, \\begin{cases} there \\\\ \\sqrt{y^{2+4y+4}}=\\sqrt{1-\\frac{8y}{y^{2}+12y+}}{y^{2}+12y+4}\\end{cases}+\\frac{4}{y}+12^{\\geqslant}\\frac{-}{2}, equality holds if and only if y=2. \\therefore the minimum value of \\frac{|PF|}{|PA|} is \\frac{\\sqrt{2}}{2}." }, { "text": "Given that $O$ is the coordinate origin, $F$ is the left focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$. Draw a perpendicular line from $F$ to the $x$-axis; this line intersects one asymptote of the hyperbola at point $A$. Take a point $B$ on the other asymptote such that $\\overrightarrow{O A}=\\lambda \\overrightarrow{B F}(\\lambda>0)$. If $\\overrightarrow{A B} \\cdot \\overrightarrow{O B}=0$, then the eccentricity of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;O: Origin;A: Point;B: Point;F: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;L:Line;PointOnCurve(F, L);IsPerpendicular(L, xAxis);L1:Line;L2:Line;Intersection(L,L1)=A;PointOnCurve(B, L2);lambda:Number;lambda>0;VectorOf(O, A) = lambda*VectorOf(B, F);DotProduct(VectorOf(A, B), VectorOf(O, B)) = 0;OneOf(Asymptote(C))=L1;OneOf(Asymptote(C))=L2;Negation(L1=L2)", "query_expressions": "Eccentricity(C)", "answer_expressions": "2*sqrt(3)/3", "fact_spans": "[[[15, 76], [98, 101], [247, 253]], [[22, 76]], [[22, 76]], [[2, 5]], [[108, 112]], [[124, 127]], [[82, 85], [11, 14]], [[22, 76]], [[22, 76]], [[15, 76]], [[11, 80]], [], [[81, 93]], [[81, 93]], [], [], [[81, 112]], [[98, 127]], [[130, 192]], [[130, 192]], [[130, 192]], [[194, 245]], [95, 104], [95, 117], [95, 117]]", "query_spans": "[[[247, 259]]]", "process": "Let the semi-focal length of hyperbola C be c, and without loss of generality, let A(-c, \\frac{bc}{a}). From \\overrightarrow{OA} = \\lambda \\overrightarrow{BF} (\\lambda > 0), it follows that OA // FB, so the equation of line FB is y = -\\frac{b}{a}(x + c). Solving the system\n\\begin{cases}\ny = -\\frac{b}{a}(x + c) \\\\\ny = \\frac{b}{a}x\n\\end{cases},\nwe obtain B(-\\frac{c}{2}, -\\frac{bc}{2a}). Also, since \\overrightarrow{AB} \\cdot \\overrightarrow{OB} = 0, we have (-\\frac{c}{2} - (-c), -\\frac{bc}{2a} - \\frac{bc}{a}) \\cdot (-\\frac{c}{2}, -\\frac{bc}{2a}) = 0. Solving this yields a^{2} = 3b^{2}. Therefore, the eccentricity of hyperbola C is e = \\sqrt{\\frac{a^{2} + b^{2}}{a^{2}}} = \\frac{2\\sqrt{3}}{3}." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ has two asymptotes with equations $y=\\pm \\frac{\\sqrt{3}}{3} x$. If the distance from a vertex to an asymptote is $1$, then the equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;Expression(Asymptote(G)) = (y = pm*((sqrt(3)/3)*x));Distance(Vertex(G), Asymptote(G)) = 1", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4-(3*y^2)/4=1", "fact_spans": "[[[2, 59], [113, 116]], [[2, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[2, 96]], [[2, 111]]]", "query_spans": "[[[113, 120]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ has its right vertex at point $A$, upper vertex at point $B$, and point $C$ is $(2 , 5)$, then the standard equation of the circle passing through points $A$, $B$, $C$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1);A: Point;B: Point;C: Point;RightVertex(G) = A;UpperVertex(G) = B;Coordinate(C) = (2, 5);H: Circle;PointOnCurve(A, H);PointOnCurve(B, H);PointOnCurve(C, H)", "query_expressions": "Expression(H)", "answer_expressions": "(x-5/2)^2+(y-5/2)^2=13/2", "fact_spans": "[[[2, 39]], [[2, 39]], [[44, 47], [73, 77]], [[52, 55], [79, 82]], [[56, 60], [84, 87]], [[2, 47]], [[2, 55]], [[56, 70]], [[88, 89]], [[72, 89]], [[72, 89]], [[72, 89]]]", "query_spans": "[[[88, 96]]]", "process": "The right vertex of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ is $A(3,0)$, and the upper vertex is $B(0,2)$. Therefore, the equation of the circle $M$ passing through points $A$, $B$, and $C$ can be written as $x^{2}+y^{2}+Dx+Ey+F=0$ ($D^{2}+E^{2}-4F>0$). Thus,\n$$\n\\begin{cases}\n9+3D+F=0, \\\\\n4+2E+F=0, \\\\\n29+2D+5E+F=0\n\\end{cases}\n\\Rightarrow\n\\begin{cases}\n5+3D-2E=0, \\\\\n20-D+5E=0\n\\end{cases}\n\\Rightarrow\nD=-5, E=-5, F=6\n$$\nTherefore, the equation of circle $M$ is $x^{2}+y^{2}-5x-5y+6=0$. The standard form is $(x-\\frac{5}{2})^{2}+(y-\\frac{5}{2})^{2}=\\frac{13}{2}$." }, { "text": "The center of the circle lies on the parabola $y=\\frac{1}{2} x^{2}(x<0)$, and the circle is tangent to both the directrix of this parabola and the $y$-axis. What is the standard equation of the circle?", "fact_expressions": "G: Parabola;H: Circle;Expression(G) = (y = x^2/2)&(x<0);PointOnCurve(Center(H), G);IsTangent(Directrix(G),H);IsTangent(yAxis,H)", "query_expressions": "Expression(H)", "answer_expressions": "(x+1)^2+(y-1/2)^2=1", "fact_spans": "[[[3, 32], [38, 41]], [[53, 54]], [[3, 32]], [[0, 54]], [[36, 54]], [[36, 54]]]", "query_spans": "[[[53, 61]]]", "process": "According to the problem, let P(t, \\frac{1}{2}t^{2}) be the center of the circle with t < 0, and the equation of the directrix is y = \\frac{1}{2}. Since the circle is tangent to the directrix of the parabola and the y-axis, therefore -t = \\frac{1}{2}t^{2} + \\frac{1}{2} \\Rightarrow t = -1. Thus, the center is (-1, \\frac{1}{2}), and the radius is 1. The answer is: (x+1)^{2}+(y-\\frac{1}{2})^{2}=1." }, { "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$, point $A$ lies on $C$, and the midpoint coordinates of $AF$ are $(2,2)$, then the equation of $C$ is?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(C) = F;A: Point;PointOnCurve(A, C);Coordinate(MidPoint(LineSegmentOf(A, F))) = (2, 2)", "query_expressions": "Expression(C)", "answer_expressions": "y^2 = 8*x", "fact_spans": "[[[2, 28], [41, 44], [66, 69]], [[2, 28]], [[10, 28]], [[10, 28]], [[32, 35]], [[2, 35]], [[36, 40]], [[36, 45]], [[46, 64]]]", "query_spans": "[[[66, 74]]]", "process": "Since the focus of the parabola $ C: y^{2} = 2px $ ($ p > 0 $) is $ F $, we have $ F\\left(\\frac{p}{2}, 0\\right) $. Given that the midpoint of $ AF $ has coordinates $ (2, 2) $, it follows that $ A\\left(4 - \\frac{p}{2}, 4\\right) $. Since point $ A $ lies on $ C $, we have $ 4^{2} = 2p\\left(4 - \\frac{p}{2}\\right) $, which simplifies to $ p^{2} - 8p + 16 = 0 $. Therefore, $ p = 4 $, and hence the equation of the parabola is $ y^{2} = 8x $." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $x^{2}-y^{2}=1$, respectively, $P$ is a point on one of its asymptotes, and the circle with diameter $F_{1} F_{2}$ passes through point $P$. Then the area of $\\Delta P F_{1} F_{2}$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, OneOf(Asymptote(C)));G: Circle;IsDiameter(LineSegmentOf(F1, F2), G);PointOnCurve(P, G)", "query_expressions": "Area(TriangleOf(P, F1, F2))", "answer_expressions": "sqrt(2)", "fact_spans": "[[[20, 43], [54, 55]], [[20, 43]], [[2, 9]], [[10, 17]], [[2, 49]], [[2, 49]], [[50, 53], [87, 91]], [[50, 64]], [[84, 85]], [[66, 85]], [[84, 91]]]", "query_spans": "[[[93, 120]]]", "process": "According to the problem, the asymptotes are given by: $ y = \\pm x $, $ c = \\sqrt{2} $, and the equation of the circle is: $ x^{2} + y^{2} = 2 $. Solving the system \n\\[\n\\begin{cases}\nx^{2} + y^{2} = 2 \\\\\ny = \\pm x\n\\end{cases}\n\\Rightarrow y_{P} = \\pm 1\n\\]\nThus, $ S_{\\Delta PF_{1}F_{2}} = \\frac{1}{2} \\times 2\\sqrt{2} \\times |\\pm 1| = \\sqrt{2} $." }, { "text": "Given that the axis of symmetry of the parabola is $x=-1$, the distance between its intersection points with the $x$-axis is equal to $4$, and its intercept on the $y$-axis is $-6$, then its analytical expression is?", "fact_expressions": "G: Parabola;Expression(SymmetryAxis(G)) = (x = -1);D1: Point;D2: Point;Intersection(G, xAxis) = {D1, D2};Distance(D1, D2) = 4;Intercept(G, yAxis) = -6", "query_expressions": "Expression(G)", "answer_expressions": "y=2*x^2+4*x-6", "fact_spans": "[[[2, 5], [17, 18], [36, 37], [53, 54]], [[2, 16]], [], [], [[17, 26]], [[17, 35]], [[36, 51]]]", "query_spans": "[[[53, 60]]]", "process": "" }, { "text": "Given the ellipse equation $\\frac{x^{2}}{16}+\\frac{y^{2}}{m^{2}}=1(m>0)$, the line $y=\\frac{\\sqrt{2}}{2} x$ intersects the ellipse at a point $M$, and the projection of $M$ onto the $x$-axis is exactly the right focus of the ellipse. Find the value of $m$.", "fact_expressions": "G: Ellipse;H: Line;m:Number;M:Point;m>0;Expression(H) = (y = x*(sqrt(2)/2));Expression(G) = (x^2/16 + y^2/m^2 = 1);OneOf(Intersection(G,H))=M;Projection(M,xAxis)=RightFocus(G)", "query_expressions": "m", "answer_expressions": "Preserve all mathematical expressions, LaTeX formulas, symbols, and equations EXACTLY. \nDo not add explanations, do not rewrite formulas, do not omit any content. \n\nText:", "fact_spans": "[[[2, 4], [81, 83], [104, 106]], [[53, 79]], [[112, 115]], [[88, 91]], [[7, 52]], [[53, 79]], [[2, 52]], [[53, 91]], [[88, 110]]]", "query_spans": "[[[112, 119]]]", "process": "" }, { "text": "Draw a line $l$ through the focus of the parabola $y^{2}=4 x$, intersecting the parabola at points $A$ and $B$. If the horizontal coordinate of the midpoint of segment $AB$ is $3$, then what is $|AB|$ equal to?", "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), l);Intersection(l, G) = {A, B};XCoordinate(MidPoint(LineSegmentOf(A, B))) = 3", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "8", "fact_spans": "[[[19, 24]], [[1, 15], [25, 28]], [[29, 32]], [[33, 36]], [[1, 15]], [[0, 24]], [[19, 38]], [[40, 57]]]", "query_spans": "[[[59, 69]]]", "process": "Problem Analysis: The focus of the parabola $ y^{2} = 4x $ is $ (1,0) $. Let the line be $ y = k(x - 1) $. Solving the system and simplifying yields $ k^{2}x^{2} - (2k^{2} + 4)x + k^{2} = 0 $. Therefore, $ x_{1} + x_{2} = \\frac{2k^{2} + 4}{2k^{2}} = 6 $, so $ k^{2} = \\frac{2}{5} $. The equation becomes $ x^{2} - 12x + 1 = 0 $. Hence, $ |AB| = \\sqrt{1 + k^{2}} |x_{1} - x_{2}| = 8 $." }, { "text": "The line $l$ passes through the right focus of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{4}=1$ and intersects the right branch of the hyperbola at points $A$ and $B$, with $|A B|=4$. Then, what is the perimeter of the triangle formed by $A$, $B$, and the left focus of the hyperbola?", "fact_expressions": "l: Line;G: Hyperbola;A: Point;B: Point;Expression(G) = (x^2/16 - y^2/4 = 1);PointOnCurve(RightFocus(G),l);Intersection(l,RightPart(G))={A,B};Abs(LineSegmentOf(A, B)) = 4;F:Point;LeftFocus(G)=F", "query_expressions": "Perimeter(TriangleOf(A,B,F))", "answer_expressions": "24", "fact_spans": "[[[0, 5]], [[6, 45], [51, 54], [90, 93]], [[59, 62], [80, 83]], [[63, 66], [86, 89]], [[6, 45]], [[0, 49]], [[0, 68]], [[69, 78]], [], [[90, 97]]]", "query_spans": "[[[80, 107]]]", "process": "" }, { "text": "Given that the focus of the parabola $y^{2}=4 x$ is $F$, and $P$ is a point on the parabola such that $|P F|=3$, what is the horizontal coordinate of point $P$?", "fact_expressions": "G: Parabola;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G)=F;PointOnCurve(P, G);Abs(LineSegmentOf(P, F)) = 3", "query_expressions": "XCoordinate(P)", "answer_expressions": "2", "fact_spans": "[[[2, 16], [28, 31]], [[24, 27], [47, 51]], [[20, 23]], [[2, 16]], [[2, 23]], [[24, 34]], [[36, 45]]]", "query_spans": "[[[47, 57]]]", "process": "From the parabola equation, we obtain 2p = 4, that is, p = 2, then |PF| = x_{P} + \\frac{p}{2} = x_{P} + 1 = 3, solving gives x_{P} = 2." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and point $P$ is a moving point on the ellipse, then the minimum value of $|PF_{1}| \\cdot|P F_{2}|$ is?", "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G)", "query_expressions": "Min(Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)))", "answer_expressions": "9", "fact_spans": "[[[18, 56], [67, 69]], [[62, 66]], [[2, 9]], [[10, 17]], [[18, 56]], [[2, 61]], [[62, 75]]]", "query_spans": "[[[77, 108]]]", "process": "" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ shares foci with the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$, and their eccentricities are reciprocals of each other. Let $F_{1}$, $F_{2}$ denote the left and right foci of the hyperbola $C$, respectively, and let $P$ be an arbitrary point on the right branch. Then the minimum value of $\\frac{P F_{1}^{2}}{P F_{2}}$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;G: Ellipse;Expression(G) = (x^2/16 + y^2/12 = 1);Focus(C) = Focus(G);InterReciprocal(Eccentricity(C), Eccentricity(G));F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P,RightPart(C))", "query_expressions": "Min(LineSegmentOf(P, F1)^2/LineSegmentOf(P, F2))", "answer_expressions": "8", "fact_spans": "[[[2, 64], [137, 143]], [[2, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[10, 64]], [[65, 104]], [[65, 104]], [[2, 109]], [[2, 117]], [[119, 126]], [[127, 134]], [[119, 149]], [[119, 149]], [[150, 153]], [[137, 161]]]", "query_spans": "[[[163, 198]]]", "process": "Find the eccentricity and foci of the ellipse, thereby obtaining the eccentricity of the hyperbola, and the real semi-axis length $ a $ of the hyperbola, which yields $ PF_{2} \\geqslant 1 $. By the definition of the hyperbola, $ PF_{1} = PF_{2} + 2 $. Thus, $ \\frac{PF_{1}^{2}}{PF_{2}} $ can be expressed as a function of $ PF_{2} $. Then, the basic inequality can be used to find the minimum value. Let the semi-major axis of the ellipse be $ a_{1} $, the semi-minor axis be $ b_{1} $, and the semi-focal distance be $ c $. Then, $ c = \\sqrt{a_{1}^{2} - b_{1}^{2}} = \\sqrt{16 - 12} = 2 $. Hence, the eccentricity of the ellipse is $ e_{1} = \\frac{c}{a_{1}} = \\frac{1}{2} $. Consequently, the eccentricity of the hyperbola is $ e = \\frac{c}{a} = \\frac{2}{a} = 2 $, yielding $ a = 1 $. According to the definition of the hyperbola, $ PF_{1} - PF_{2} = 2a $, that is, $ PF_{1} = PF_{2} + 2 $. Therefore, \n$$\n\\frac{PF_{1}^{2}}{PF_{2}} = \\frac{(PF_{2} + 2)^{2}}{PF_{2}} = \\frac{PF_{2}^{2} + 4PF_{2} + 4}{PF_{2}} = PF_{2} + \\frac{4}{PF_{2}} + 4.\n$$\nFrom the range of the hyperbola, we have $ PF_{2} \\geqslant c - a = 1 $. By the basic inequality, \n$$\nPF_{2} + \\frac{4}{PF_{2}} + 4 \\geqslant 2\\sqrt{PF_{2} \\times \\frac{4}{PF_{2}}} + 4 = 8,\n$$\nwith equality if and only if $ PF_{2} = \\frac{4}{PF_{2}} $, i.e., when $ PF_{2} = 2 $. Therefore, the minimum value of $ \\frac{PF_{1}^{2}}{PF_{2}} $ is 8." }, { "text": "The eccentricity of the ellipse $x^{2}+4 y^{2}=1$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 4*y^2 = 1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)/2", "fact_spans": "[[[0, 19]], [[0, 19]]]", "query_spans": "[[[0, 25]]]", "process": "" }, { "text": "Given the parabola $C_{1}$: $y^{2}=2 p x(p>0)$, the focus $F$ of $C_{1}$ coincides with one focus of the hyperbola $C_{2}$: $\\frac{x^{2}}{3}-\\frac{y^{2}}{b^{2}}=1$. If the distance from point $F$ to an asymptote of the hyperbola $C_{2}$ is $1$, then what is the distance from the focus $F$ of $C_{1}$ to its directrix?", "fact_expressions": "C1:Parabola;C2:Hyperbola;Expression(C1) = (y^2=2*p*x);p:Number;p>0;Expression(C2) = (x^2/3-y^2/b^2=1);F:Point;Focus(C1)=F;OneOf(Focus(C2))=F;Distance(F,OneOf(Asymptote(C2)))=1", "query_expressions": "Distance(F,Directrix(C1))", "answer_expressions": "4", "fact_spans": "[[[2, 32], [129, 136], [143, 144]], [[39, 90], [104, 114]], [[2, 32]], [[14, 32]], [[14, 32]], [[39, 90]], [[35, 38], [99, 103], [139, 142]], [[2, 38]], [[35, 97]], [[99, 127]]]", "query_spans": "[[[129, 151]]]", "process": "The focus of the parabola $ C_{1}: y^{2} = 2px $ ($ p > 0 $) coincides with one focus of the hyperbola $ C_{2}: \\frac{x^{2}}{3} - \\frac{y^{2}}{b^{2}} = 1 $, then $ \\frac{p}{2} = c $, and $ c^{2} = 3 + b^{2} = \\frac{p^{2}}{4} $. The distance from the point $ F(c, 0) $ to the asymptote line $ bx - \\sqrt{3}y = 0 $ of the hyperbola is 1, that is, $ \\frac{|bc|}{\\sqrt{b^{2} + 3}} = 1 $. Since $ c^{2} = 3 + b^{2} $, solving gives $ b = 1 $, thus $ c = 2 $, so $ p = 2c = 4 $. Therefore, the distance from the focus of the parabola to its directrix is $ p = 4 $." }, { "text": "Hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$ has right focus $F$, point $P$ lies on one of the asymptotes of $C$, $O$ is the origin, and if $|PO|=|PF|$, then the area of $\\triangle PFO$ is?", "fact_expressions": "C: Hyperbola;P: Point;F: Point;O: Origin;Expression(C) = (x^2/4 - y^2/2 = 1);RightFocus(C) = F;PointOnCurve(P, OneOf(Asymptote(C)));Abs(LineSegmentOf(P, O)) = Abs(LineSegmentOf(P, F))", "query_expressions": "Area(TriangleOf(P, F, O))", "answer_expressions": "3*sqrt(2)/4", "fact_spans": "[[[0, 43], [57, 60]], [[52, 56]], [[48, 51]], [[68, 71]], [[0, 43]], [[0, 51]], [[52, 67]], [[78, 91]]]", "query_spans": "[[[93, 115]]]", "process": "Hyperbola C: \\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1, one asymptote equation is: y=\\frac{\\sqrt{2}}{2}x OF=\\sqrt{6}. Draw a perpendicular PM from point P to the x-axis, |PO|=|PF|, OM=\\frac{1}{2}OF=\\frac{1}{2}\\sqrt{6}, PM=\\frac{\\sqrt{2}}{2}\\times\\frac{\\sqrt{6}}{2}=\\frac{\\sqrt{3}}{2}. The area of \\triangle PFO is \\frac{1}{2}\\times\\sqrt{6}\\times\\frac{\\sqrt{3}}{2}=\\frac{3\\sqrt{2}}{4}" }, { "text": "Through the focus $F$ of the parabola $y^{2}=2 p x(p>0)$, draw a line with an inclination angle of $45^{\\circ}$ intersecting the parabola at points $A$ and $B$. If the length of segment $AB$ is $8$, then $p=?$", "fact_expressions": "A: Point;B: Point;G: Parabola;p: Number;H: Line;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(F, H);Inclination(H) = ApplyUnit(45, degree);Intersection(H, G) = {A, B};Length(LineSegmentOf(A, B)) = 8", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[53, 56]], [[57, 60]], [[1, 22], [49, 52]], [[78, 81]], [[46, 48]], [[25, 28]], [[4, 22]], [[1, 22]], [[1, 28]], [[0, 48]], [[29, 48]], [[46, 62]], [[64, 76]]]", "query_spans": "[[[78, 83]]]", "process": "Let the coordinates of points A and B be (x_{1},y_{1}) and (x_{2},y_{2}), respectively. The equation of the line passing through the focus F of the parabola y^{2}=2px (p>0) with an inclination angle of 45^{\\circ} is y=x-\\frac{p}{2}. Substituting y=x-\\frac{p}{2} into y^{2}=2px gives x^{2}-3px+\\frac{1}{4}p^{2}=0, \\therefore x_{1}+x_{2}=3p. Since |AB|=x_{1}+x_{2}+p=4p=8, \\therefore p=2." }, { "text": "Given the ellipse $ E $: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the right focus is $ F(3,0) $. A line passing through point $ F $ intersects the ellipse at points $ A $ and $ B $. If the midpoint of $ AB $ has coordinates $ (1, -1) $, then the equation of $ E $ is?", "fact_expressions": "E: Ellipse;b: Number;a: Number;G: Line;A: Point;B: Point;F: Point;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(MidPoint(LineSegmentOf(A,B))) = (1, -1);Coordinate(F) = (3, 0);RightFocus(E) = F;PointOnCurve(F, G);Intersection(G, E) = {A, B}", "query_expressions": "Expression(E)", "answer_expressions": "x^2/18+y^2/9=1", "fact_spans": "[[[2, 59], [82, 84], [119, 122]], [[9, 59]], [[9, 59]], [[79, 81]], [[85, 88]], [[89, 92]], [[64, 72], [74, 78]], [[9, 59]], [[9, 59]], [[2, 59]], [[97, 117]], [[64, 72]], [[2, 72]], [[73, 81]], [[79, 94]]]", "query_spans": "[[[119, 127]]]", "process": "" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right foci are $F_{1}$, $F_{2}$ respectively. A line $l$ passing through $F_{1}$ with slope $\\frac{a}{b}$ intersects the right branch of the hyperbola at point $P$ and one of its asymptotes at point $M$, and satisfies $\\overrightarrow{P M}=3 \\overrightarrow{M F_{1}}$. Then the equations of the asymptotes of the hyperbola are?", "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;P: Point;M: Point;F1: Point;F2:Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(F1,l);Slope(l)=a/b;Intersection(l,RightPart(C))=P;Intersection(l,OneOf(Asymptote(C)))=M;VectorOf(P, M) = 3*VectorOf(M, F1)", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "y=pm*(4/3)*x", "fact_spans": "[[[116, 121]], [[2, 63], [122, 125], [203, 206]], [[9, 63]], [[9, 63]], [[130, 134]], [[145, 149]], [[72, 79], [91, 98]], [[81, 88]], [[9, 63]], [[9, 63]], [[2, 63]], [[2, 88]], [[2, 88]], [[90, 121]], [[99, 121]], [[116, 134]], [[116, 149]], [[152, 201]]]", "query_spans": "[[[203, 214]]]", "process": "Let the left focus of the hyperbola be (-c,0), so the equation of line l is: y=\\frac{a}{b}(x+c). Since line l intersects one of the asymptotes at point M, and \\overrightarrow{PM}=3\\overrightarrow{MF_{1}}, we have \\begin{cases}y=\\frac{a}{b}(x+c)\\\\y=-\\frac{b}{a}x\\end{cases}. Solving this system of equations gives \\begin{cases}x=-\\frac{a^{2}}{c}\\\\y=\\frac{ab}{c}\\end{cases}, that is, M(-\\frac{a^2}{c},\\frac{ab}{c}). Draw perpendiculars from points M and P to the x-axis, with feet at A and B respectively. Since \\overrightarrow{PM}=3\\overrightarrow{MF_{1}}, it follows that |AF_{1}|=\\frac{1}{4}|BF_{1}|, |AM|=\\frac{1}{4}|BP|. Without loss of generality, assume P(x_{0},\\frac{4ab}{c}), then c-\\frac{a^{2}}{c}=\\frac{c+x_{0}}{4}, solving gives x_{0}=3c-\\frac{4a^{2}}{c}. Thus, P(3c-\\frac{4a^{2}}{c},\\frac{4ab}{c}). Substituting point P(3c-\\frac{4a^{2}}{c},\\frac{4ab}{c}) into \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0), we get \\underline{(3c-\\frac{4a^{2}}{a^{2}})^{2}}-\\frac{(4ab}{b^{2}})^{2}=1 (a>0,b>0). Simplifying yields 9c^{2}=25a^{2}, that is, 9(a^{2}+b^{2})=25a^{2}, solving gives \\frac{b^{2}}{a^2}=\\frac{16}{9}, \\therefore \\frac{b}{a}=\\frac{4}{3}. Therefore, the asymptotic equations of the hyperbola are y=\\pm\\frac{4}{3}x." }, { "text": "If the line $y=kx+1$ ($k \\in R$) and the hyperbola $x^{2}-y^{2}=2$ have exactly one common point, then $k=$?", "fact_expressions": "G: Hyperbola;H: Line;k: Real;Expression(G) = (x^2 - y^2 = 2);Expression(H) = (y = k*x + 1);NumIntersection(H, G)=1", "query_expressions": "k", "answer_expressions": "{pm*1,pm*sqrt(6)/2}", "fact_spans": "[[[22, 40]], [[1, 21]], [[51, 54]], [[22, 40]], [[1, 21]], [[1, 49]]]", "query_spans": "[[[51, 56]]]", "process": "Solving the system \n\\begin{cases}y=kx+1\\\\x^2-y^2=2\\end{cases}, \nwe obtain \n(1-k^{2})x^{2}-2kx-3=0. \n① When $1-k^{2}=0$, we get $k=\\pm1$. In this case, the line is parallel to the asymptotes of the hyperbola, and the line intersects the hyperbola at exactly one point, which satisfies the condition. \n② When $1-k^{2}\\neq0$, since the line and the hyperbola have exactly one common point, we have \n$\\triangle=4k^{2}+12(1-k^{2})=0$, \nsolving gives $k=\\pm\\frac{\\sqrt{6}}{2}$, which satisfies the condition. \nIn conclusion, $k=\\pm1,\\pm\\frac{\\sqrt{6}}{2}$." }, { "text": "If the equation $\\frac{x^{2}}{k-1}+\\frac{y^{2}}{k^{2}-3}=-1$ represents an ellipse, then what is the range of values for $k$?", "fact_expressions": "G: Ellipse;Expression(G)=(x^2/(k - 1) + y^2/(k^2 - 3) = -1);k:Number", "query_expressions": "Range(k)", "answer_expressions": "(-sqrt(3),-1)+(-1,1)", "fact_spans": "[[[49, 51]], [[1, 51]], [[53, 56]]]", "query_spans": "[[[53, 63]]]", "process": "Solution: Since the equation $\\frac{x^2}{k-1} + \\frac{y^2}{k^2 - 3} = -1$ represents an ellipse, we have $\\begin{cases} 3 - k^2 > 0 \\\\ 1 - k > 0 \\\\ 1 - k \\neq 3 - k^2 \\end{cases}$. Therefore, $-\\sqrt{3} < k < 1$ and $k \\neq -$" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has a focus at $F(2 \\sqrt{5} , 0)$, and eccentricity $e=\\frac{\\sqrt{5}}{2}$, then the standard equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;F: Point;e:Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F)=(2*sqrt(5),0);OneOf(Focus(G))=F;Eccentricity(G)=e;e=sqrt(5)/2", "query_expressions": "Expression(G)", "answer_expressions": "x^2/16 - y^2/4 = 1", "fact_spans": "[[[2, 58], [113, 116]], [[5, 58]], [[5, 58]], [[64, 83]], [[89, 111]], [[5, 58]], [[5, 58]], [[2, 58]], [[64, 83]], [[2, 83]], [[2, 111]], [[89, 111]]]", "query_spans": "[[[113, 123]]]", "process": "The hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0,b>0) has a focus at F(2\\sqrt{5},0), so c=2\\sqrt{5}. Moreover, since the eccentricity of the hyperbola C:\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0,b>0) is \\frac{c}{a}=\\frac{\\sqrt{5}}{2}=\\frac{2\\sqrt{5}}{a}, it follows that a=4, b=2; the equation of the hyperbola is \\frac{x^{2}}{16}-\\frac{y^{2}}{4}=1." }, { "text": "The line $l$ passes through the focus $F$ of the parabola $y=\\frac{1}{4} x^{2}$ and intersects the parabola at points $A$ and $B$. If $|A B|=16$, then what is the distance from the midpoint $D$ of $A B$ to the $x$-axis?", "fact_expressions": "l: Line;G: Parabola;Expression(G) = (y = (1/4)*x^2);F: Point;Focus(G) = F;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, B)) = 16;D: Point;MidPoint(LineSegmentOf(A, B)) = D", "query_expressions": "Distance(D, xAxis)", "answer_expressions": "7", "fact_spans": "[[[0, 5]], [[6, 30], [38, 41]], [[6, 30]], [[33, 36]], [[6, 36]], [[0, 36]], [[43, 46]], [[47, 50]], [[0, 52]], [[54, 64]], [[74, 77]], [[66, 77]]]", "query_spans": "[[[74, 87]]]", "process": "The focus of the parabola y=\\frac{1}{4}x^2 is F(0,1). Let A(x_{1},y_{1}), B(x_{2},y_{2}), then |AB|=y_{1}+y_{2}+2=16, that is, y_{1}+y_{2}=14. Therefore, the distance from the midpoint D(\\frac{x_{1}+x_{2}}{2},\\frac{y_{1}+y_{2}}{2}) of AB to the point is \\frac{y_{1}+y_{2}}{2}=7." }, { "text": "Given the ellipse $m x^{2}+n y^{2}=1$ $(m>0 , n>0)$ intersects the line $x+y-1=0$ at points $A$ and $B$. If $\\frac{n}{m}=\\sqrt{2}$, then what is the slope of the line connecting the origin and the midpoint $M$ of segment $AB$?", "fact_expressions": "G: Ellipse;n: Number;m: Number;H: Line;B: Point;A: Point;m>0;n>0;Expression(G) = (m*x^2 + n*y^2 = 1);Expression(H) = (x + y - 1 = 0);Intersection(G, H) = {A, B};n/m = sqrt(2);O: Origin;M: Point;MidPoint(LineSegmentOf(A, B)) = M", "query_expressions": "Slope(LineSegmentOf(O, M))", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[2, 34]], [[59, 81]], [[59, 81]], [[35, 46]], [[52, 55]], [[48, 51]], [[4, 34]], [[4, 34]], [[2, 34]], [[35, 46]], [[2, 57]], [[59, 81]], [[84, 86]], [[97, 100]], [[87, 100]]]", "query_spans": "[[[83, 108]]]", "process": "From the line equation x + y - 1 = 0, we obtain y = -x + 1. Substituting into mx^{2} + ny^{2} = 1 gives: (m + n)x^{2} - 2nx + n - 1 = 0. Using Vieta's formulas, determine the coordinates of M, then find the slope of the line passing through the origin and the midpoint of segment AB. [Detailed solution] From the line equation x + y - 1 = 0, we obtain y = -x + 1. Substituting into mx^{2} + ny^{2} = 1 gives: (m + n)x^{2} - 2nx + n - 1 = 0. Let the coordinates of A and B be (x_{1}, y_{1}), (x_{2}, y_{2}). Then we have x_{1} + x_{2} = \\frac{2n}{m+n}, y_{1} + y_{2} = 1 - x_{1} + 1 - x_{2} = 2 - (x_{1} + x_{2}) = \\frac{2m}{m+n}. Therefore, the coordinates of M are (\\frac{n}{m+n}, \\frac{m}{m+n}). Hence, the slope of the line passing through the origin and the midpoint M of segment AB is |\\frac{m}{n} = \\frac{\\sqrt{2}}{2}." }, { "text": "From a point $P$ on the parabola $y^2=4x$, draw a perpendicular to its directrix, with foot of perpendicular at $M$. Let the focus of the parabola be $F$, and $|PF|=5$. Then the area of $\\triangle MPF$ is?", "fact_expressions": "C: Parabola;P: Point;PointOnCurve(P, C);H: Line;PointOnCurve(P, H);IsPerpendicular(H, Directrix(C)) = True;M: Point;FootPoint(H, Directrix(C)) = M;F: Point;Focus(C) = F;Abs(LineSegmentOf(P, F)) = 5;Expression(C)=(y^2=4*x)", "query_expressions": "Area(TriangleOf(M, P, F))", "answer_expressions": "10", "fact_spans": "[[[1, 13], [36, 39]], [[16, 19]], [[1, 20]], [[25, 27]], [[0, 27]], [[0, 27]], [[31, 34]], [[1, 34]], [[44, 47]], [[36, 47]], [[49, 58]], [[1, 13], [36, 39]]]", "query_spans": "[[[60, 82]]]", "process": "" }, { "text": "Given that $P$ is a point on the parabola $y^{2}=16 x$, if the distance from point $P$ to the $x$-axis is $12$, then the distance from point $P$ to the focus is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 16*x);P: Point;PointOnCurve(P, G);Distance(P, xAxis) = 12", "query_expressions": "Distance(P, Focus(G))", "answer_expressions": "13", "fact_spans": "[[[6, 21]], [[6, 21]], [[2, 5], [27, 31], [46, 50]], [[2, 25]], [[27, 44]]]", "query_spans": "[[[6, 58]]]", "process": "Since point P lies on the parabola $ y^{2} = 16x $ and its distance to the x-axis is 12, the y-coordinate of point P is $ \\pm12 $. Substituting into $ y^2 = 16x $, we get $ x = 9 $. The distance from point P to the focus equals its distance to the directrix. The directrix of $ y^{2} = 16x $ is $ x = -4 $, so the distance from point P to the focus is $ 9 + 4 = 13 $. \nNote: This problem mainly examines the standard equation of a parabola and the definition of a parabola, and belongs to a medium-difficulty question." }, { "text": "Let the left and right foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ be $F_{1}$, $F_{2}$ respectively. Point $P$ moves on the ellipse, the maximum value of $|\\overrightarrow{P F_{1}}|| \\overrightarrow{P F_{2} } |$ is $m$, the minimum value of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ is $n$, and $m \\geq 2 n$. Then the range of the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;m:Number;n:Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P, G);Max(Abs(VectorOf(P,F1))*Abs(VectorOf(P,F2)))=m;Min(DotProduct(VectorOf(P, F1), VectorOf(P, F2))) = n;m >= 2*n", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "[1/2,1)", "fact_spans": "[[[1, 53], [83, 85], [238, 240]], [[3, 53]], [[3, 53]], [[78, 82]], [[61, 68]], [[69, 77]], [[151, 154]], [[218, 221]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 77]], [[1, 77]], [[78, 88]], [[89, 154]], [[156, 221]], [[223, 235]]]", "query_spans": "[[[238, 251]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{20}+\\frac{y^{2}}{8}=1$ with two foci $F_{1}$, $F_{2}$, and a point $P$ on this ellipse such that $P F_{1} \\perp P F_{2}$, find the area of $\\Delta P F_{1} F_{2}$.", "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/20 + y^2/8 = 1);Focus(G)={F1,F2};PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))", "query_expressions": "Area(TriangleOf(P, F1, F2))", "answer_expressions": "8", "fact_spans": "[[[2, 40], [68, 70]], [[62, 66]], [[46, 53]], [[54, 61]], [[2, 40]], [[2, 61]], [[62, 71]], [[73, 96]]]", "query_spans": "[[[98, 125]]]", "process": "From the given conditions, we have: $ a=2\\sqrt{5}, b=2\\sqrt{2}, c=2\\sqrt{3} $. Since $ |PF_{1}| + |PF_{2}| = 2a = 4\\sqrt{5} $, it follows that $ |PF_{1}|^{2} + |PF_{2}|^{2} + 2|PF_{1}||PF_{2}| = 80 $—$\\textcircled{1}$. Also, since $ PF_{1} \\perp PF_{2} $, we have $ |PF_{1}|^{2} + |PF_{2}|^{2} = |F_{1}F_{2}|^{2} = (2c)^{2} = (4\\sqrt{3})^{2} = 48 $—$\\textcircled{2}$. Solving equations $\\textcircled{1}$ and $\\textcircled{2}$ simultaneously, we obtain $ |PF_{1}||PF_{2}| = 16 $. Therefore, $ S_{\\Delta PF_{1}F_{2}} = \\frac{1}{2}|PF_{1}||PF_{2}| = \\frac{1}{2} \\times 16 = 8 $." }, { "text": "The eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0)$ is $\\frac{\\sqrt{2}}{2}$. If the x-coordinate of an intersection point between the line $y=k x$ and the ellipse is $b$, then what is the value of $k$?", "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;k: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = k*x);Eccentricity(G) = sqrt(2)/2;XCoordinate(OneOf(Intersection(H,G))) = b", "query_expressions": "k", "answer_expressions": "pm*sqrt(2)/2", "fact_spans": "[[[0, 54], [91, 92]], [[101, 104]], [[2, 54]], [[81, 90]], [[106, 109]], [[2, 54]], [[2, 54]], [[0, 54]], [[81, 90]], [[0, 79]], [[81, 104]]]", "query_spans": "[[[106, 113]]]", "process": "" }, { "text": "The line passing through the focus $F$ of the parabola $y=2 x^{2}$ intersects the parabola at points $A$ and $B$. If $\\overrightarrow{A F}=2 \\overrightarrow{F B}$, then $|\\overrightarrow{A F}|$=?", "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;Expression(G) = (y = 2*x^2);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};VectorOf(A, F) = 2*VectorOf(F, B)", "query_expressions": "Abs(VectorOf(A, F))", "answer_expressions": "3/8", "fact_spans": "[[[1, 15], [25, 28]], [[21, 23]], [[29, 32]], [[17, 20]], [[33, 36]], [[1, 15]], [[1, 20]], [[0, 23]], [[21, 38]], [[40, 85]]]", "query_spans": "[[[87, 113]]]", "process": "Analysis: First, from the parabola equation, we obtain p = \\frac{1}{4}; then, using the property of the parabola \\frac{1}{|AF|} + \\frac{1}{|BF|} = \\frac{2}{p}, and combining with the given conditions, the solution can be obtained. \nDetailed Solution: The focus coordinates of the parabola y = 2x^{2} are: F(0, \\frac{1}{8}). \nFrom the property of the parabola, we have: \\frac{1}{|AF|} + \\frac{1}{|BF|} = \\frac{2}{p} = 8. \nAlso, \\overrightarrow{AF} = 2\\overrightarrow{FB}, so |AF| = 2|FB|. Substituting |FB| = \\frac{1}{2}|AF| into the above equation, we obtain |\\overrightarrow{AF}| = \\frac{3}{8}." }, { "text": "Let $F_{1}$, $F_{2}$ be the two foci of the curve $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{n^{2}}=1$ $(m>0, n>0)$, the perimeter of the triangle formed by a point on the curve and $F_{1}$, $F_{2}$ is $16$, and the minimum distance from a point on the curve to $F_{1}$ is $2$, then $n=?$", "fact_expressions": "G: Curve;n: Number;m: Number;F1: Point;F2: Point;P: Point;m>0;n>0;Expression(G) = (y^2/n^2 + x^2/m^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P,G);Perimeter(TriangleOf(P,F1,F2))=16;Min(Distance(P,F1))=2", "query_expressions": "n", "answer_expressions": "{4,5}", "fact_spans": "[[[17, 72], [78, 80], [114, 116]], [[138, 141]], [[19, 72]], [[1, 8], [84, 91]], [[9, 16], [92, 99]], [[118, 119]], [[19, 72]], [[19, 72]], [[17, 72]], [[1, 77]], [[78, 83]], [[78, 113]], [[114, 136]]]", "query_spans": "[[[138, 143]]]", "process": "Given that the curve $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{n^{2}}=1$ ($m>0$, $n>0$) represents an ellipse, and the perimeter of the triangle formed by a point on the curve and $F_{1}$, $F_{2}$ is 16, therefore $2a+2c=16 \\Rightarrow a+c=8$. Since the minimum distance from a point on the curve to $F_{1}$ is 2, we have $a-c=2$. Thus,\n\\[\n\\begin{cases}\na+c=8 \\\\\na-c=2\n\\end{cases}\n\\Rightarrow a=5, c\n\\]\nBecause $a^{2}=b^{2}+c^{2} \\Rightarrow b=4$, and since the axis containing the foci of the ellipse cannot be determined, $n=4$ or $n=5$." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{3}-y^{2}=\\lambda(\\lambda>0)$, let $P$ be a point on the right branch. A line passing through point $P$ intersects the two asymptotes of hyperbola $C$ at points $A$ and $B$, respectively. Suppose point $A$ lies in the first quadrant, point $B$ lies in the fourth quadrant, and $O$ is the origin. When $\\overrightarrow{A P}=\\frac{1}{2} \\overrightarrow{P B}$, the area of $\\triangle A O B$ is $\\sqrt{3}$. Then $\\lambda=?$", "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/3 - y^2 = lambda);lambda: Number;lambda>0;P: Point;PointOnCurve(P, RightPart(C));G: Line;PointOnCurve(P, G);L1: Line;L2: Line;Asymptote(C) = {L1, L2};A: Point;B: Point;Intersection(G, L1) = A;Intersection(G, L2) = B;Quadrant(A) = 1;Quadrant(B) = 4;O: Origin;VectorOf(A, P) = (1/2)*VectorOf(P, B);Area(TriangleOf(A, O, B)) = sqrt(3)", "query_expressions": "lambda", "answer_expressions": "8/9", "fact_spans": "[[[2, 52], [72, 78]], [[2, 52]], [[220, 229]], [[10, 52]], [[58, 61], [64, 68]], [[2, 61]], [[69, 71]], [[62, 71]], [], [], [[72, 84]], [[89, 92], [100, 104]], [[93, 96], [105, 108]], [[69, 98]], [[69, 98]], [[100, 119]], [[100, 119]], [[120, 123]], [[130, 185]], [[187, 218]]]", "query_spans": "[[[220, 231]]]", "process": "" }, { "text": "Given point $P(1,-1)$ and parabola $C$: $y=\\frac{1}{4} x^{2}$, a line passing through the focus of parabola $C$ with slope $k$ intersects parabola $C$ at points $A$ and $B$. If $\\overrightarrow{P A} \\cdot \\overrightarrow{P B}=0$, then $k=?$", "fact_expressions": "P: Point;Coordinate(P) = (1, -1);C: Parabola;Expression(C) = (y = x^2/4);G: Line;PointOnCurve(Focus(C), G);k: Number;Slope(G) = k;A: Point;B: Point;Intersection(G, C) = {A, B};DotProduct(VectorOf(P, A), VectorOf(P, B)) = 0", "query_expressions": "k", "answer_expressions": "1/2", "fact_spans": "[[[2, 12]], [[2, 12]], [[13, 42], [44, 50], [64, 70]], [[13, 42]], [[61, 63]], [[43, 63]], [[57, 60], [138, 141]], [[54, 63]], [[74, 77]], [[78, 81]], [[61, 83]], [[85, 136]]]", "query_spans": "[[[138, 143]]]", "process": "Let the parabola C: y = \\frac{1}{4}x^{2} have focus F; then the coordinates of F are (0,1). Thus, the equation of line AB is y = kx + 1. Let points A(x_{1},y_{1}) and B(x_{2},y_{2}). Substituting the equation of line AB into the equation of parabola C and simplifying yields x^{2} - 4kx - 4 = 0. Then x_{1} + x_{2} = 4k, x_{1}x_{2} = -4. From \\overrightarrow{PA} \\cdot \\overrightarrow{PB} = 0, we obtain (x_{1}-1)(x_{2}-1) + (y_{1}+1)(y_{2}+1) = 0, that is, (x_{1}-1)(x_{2}-1) + (kx_{1}+2)(kx_{2}+2) = 0, which simplifies to (k^{2}+1)x_{1}x_{2} + (2k-1)(x_{1}+x_{2}) + 5 = 0. Therefore, -4(k^{2}+1) + 4k(2k-1) + 5 = (2k-1)^{2} = 0. Solving gives k = \\frac{1}{2}." }, { "text": "Given that the line $y=m$ intersects the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1$ at two distinct points, find the range of real values for $m$.", "fact_expressions": "G: Ellipse;H: Line;m: Real;Expression(G) = (x^2/3 + y^2/4 = 1);Expression(H) = (y = m);NumIntersection(H, G) = 2", "query_expressions": "Range(m)", "answer_expressions": "(-2,2)", "fact_spans": "[[[10, 47]], [[2, 9]], [[57, 62]], [[10, 47]], [[2, 9]], [[2, 55]]]", "query_spans": "[[[57, 69]]]", "process": "" }, { "text": "The focal distance of the ellipse is $2 \\sqrt{5}$, and the distance between the directrices is $\\frac{18 \\sqrt{5}}{5}$. Then the standard equation of the ellipse is?", "fact_expressions": "G: Ellipse;FocalLength(G) = 2*sqrt(5);L1:Line;L2:Line;Directrix(G)={L1,L2};Distance(L1,L2)=18*sqrt(5)/5", "query_expressions": "Expression(G)", "answer_expressions": "x^2/9+y^2/4=1", "fact_spans": "[[[0, 2], [52, 54]], [[0, 18]], [], [], [[0, 21]], [[0, 50]]]", "query_spans": "[[[52, 61]]]", "process": "" }, { "text": "Given that line $l$ passes through point $P(2, 1)$ and intersects the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$ at points $A$ and $B$, if $P$ is the midpoint of $AB$, then what is the equation of line $l$?", "fact_expressions": "l: Line;P: Point;Coordinate(P) = (2, 1);PointOnCurve(P, l);G: Hyperbola;Expression(G) = (x^2 - y^2/4 = 1);Intersection(l, G) = {A, B};A: Point;B: Point;MidPoint(LineSegmentOf(A, B)) = P", "query_expressions": "Expression(l)", "answer_expressions": "8*x-y-15=0", "fact_spans": "[[[2, 7], [76, 81]], [[8, 19], [62, 65]], [[8, 19]], [[2, 19]], [[20, 48]], [[20, 48]], [[2, 60]], [[51, 54]], [[55, 58]], [[62, 74]]]", "query_spans": "[[[76, 86]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{1}$ and perpendicular to the major axis intersects the ellipse at points $A$ and $B$. Find the radius of the incircle of triangle $ABF_{2}$.", "fact_expressions": "G: Ellipse;H:Line;A: Point;B: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/9 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(F1,H);IsPerpendicular(H,MajorAxis(G));Intersection(H,G) = {A, B}", "query_expressions": "Radius(InscribedCircle(TriangleOf(A,B,F2)))", "answer_expressions": "36/25", "fact_spans": "[[[2, 40], [85, 87]], [[82, 84]], [[88, 91]], [[92, 95]], [[49, 56], [68, 75]], [[59, 66]], [[2, 40]], [[2, 66]], [[2, 66]], [[67, 84]], [[2, 84]], [[82, 97]]]", "query_spans": "[[[99, 118]]]", "process": "According to the problem, let the inradius of $\\triangle ABF_{2}$ be $r$, and the perimeter of the triangle be $4a$. Then, derive the expression for the area of the triangle, use $AF_{1}, BF_{1}$ to find the area of $\\triangle ABF_{2}$, and then determine the inradius. [Detailed Solution] According to the problem, let the inradius of $\\triangle ABF_{2}$ be $r$; the equation of the ellipse is $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and the perimeter of the triangle is $20$. Hence, $S=\\frac{1}{2}\\times20\\times r=10r$. The line passing through $F_1$ and perpendicular to the major axis intersects the ellipse at points $A$ and $B$. Then, $AB=2\\sqrt{9(1-\\frac{16}{25})}=\\frac{18}{5}$. Therefore, $S=\\frac{1}{2}\\times8\\times\\frac{18}{5}=\\frac{72}{5}$. Thus, $10r=\\frac{72}{5}$, solving gives $r=\\frac{36}{25}$. Therefore, the inradius is $\\frac{36}{25}$." } ]